This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Physics Editor: Chris Hall Development Editor: Ed Dodd Assistant Editor: Brandi Kirksey Editorial Assistant: Stefanie Beeck
ALL RIGHTS RESERVED. No part of this work covered by the copyright hereon may be reproduced or used in any form or by any means-graphic, electronic, or mechanical, including but not limited to photocopying, recording, taping, Web distribution, information networks, or information storage and retrieval systems-without the written permission of the publisher.
Technology Project Manager: Sam Subity Marketing Manager: Mark Santee Marketing Assistant: Melissa Wong Marketing Communications Manager: Darlene Amidon Brent Project Manager, Editorial Production:
Teri Hyde Creative Director: Rob Hugel Art Director: John Walker Print/Media Buyer: Rebecca Cross Permissions Editor: Roberta Brayer Production Service: Lachina Publishing
Services
For product information and technology assistance, contact us at Cengage learning Academic Resource Center 1-800-423-0563 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions. Further permissions questions can be e·mailed to [email protected].
Library of Congress Control Number: 2007937232 ISBN-13: 978'0-495-38693-3
Text Designer: Patrick Devine Design
Photo Researcher: Jane Sanders Miller Copy Editor: Kathleen Lafferty Illustrator: Lachina Publishing Services,
Brooks/Cole 10 Davis Drive Belmont, CA 94002-3098 USA
(engage Learning is a leading prOVider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil and Japan. Locate your local office at international. cengage.com/region
fxamView~ and ExamView Pro@are trademarks of FSCreations, Inc. Windows is
a registered trademark of the Microsoft Corporation used herein under Iicense_ Macintosh and Power Macintosh are registered trademarks of Apple Computer, Inc. Used herein under license.
Cengage Learning products are represented in Canada by Nelson Education, ltd.
For your course and learning solutions, visit academic.cengage.com
Purchase any of our products at your local college store or at our preferred online store www.ichapters.com.
Printed in Canada 123-15671211100908
We dedicate this book to our colleagu e Jerry S. Faughn, whose dedicati on to all aspects of the project and tireless eHorts through the years are deeply appreci ated.
CONTENTS OVERVIEW
~1echanics
PART 1 CHAPTER I CHAPTER 2 CHAPTER 3 CHAPTER 4 CHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8 CHAPTER 9
Introduction 1 Motion in One Dimension 24
Vectors and Two~Dimen5ionQIMotion 54 The Laws of Motion 83 Energy 119 Momentum and Collisions 161
Rotational Motion and the Low of Gravity 190 Rotational Equilibrium and Rotational Dynamics 228
Solids and Fluids 268
PART 2
The -modynamics
CHAPTER 10 CHAPTER 11 CHAPTER 12
Thermol Physics 322 The Lows of Thermodynamics 385
PART 3
Vibrations and Waves
CHAPTER 13 CHAPTER 14
Vibrations and Waves 425
Sound 459
Electricity and Magnetism
PART 4 CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER
Energy in Thermol Processes 352
15 16 17 18 19
20 21
Electric Forces and Electric Fields
497
Electrical Energy and Capacitance
531
Current and Resistance
570
Direct-Current Circuits
594
Magnetism 626 Induced Voltages and Inductance 663 Alternating-Current Circuits and Electromagnetic Waves
PARTS
Light and Optics
CHAPTER 22 CHAPTER 23 CHAPTER 24 CHAPTER 25
Reftectian and Refraction of Light Mirrars and Lenses 759 Wave Optics 790 Optical Instruments 823
PART 6
tv odem Physics
CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER
26 27 28 29 30
APPENDIX A APPENDIX B APPENDIXC APPENDIX D APPENDIX E
Relativity 847 Quantum Physics 870 Atomic Physics 891 Nuclear Physics 913 Nuclear Energy and Elementary Particles Mathematics Review A.l An Abbreviated Table af Isotopes A.14 Some Useful Tables A.19 SI Units A.21
MeAT Skill Builder Study Guide A.22
Answers to Quick Quizzes, Example Questions, Odd~Numbered Multiple-Choice Questions, Conceptual Questions, and Problems A.52 Index 1.1
Standards of Length, Mass, and Time 1.2 The Building Blocks of Matter 4 1.3 Dimensional Analysis 5 1.4 Uncertainty in Measurement and Significant Figures 1.5 Conversion of Units 9 1.6 Estimates and Order-of-Magnitude Calculations 11 1.7 Coordinate Systems 13 1.8 Trigonometry 14 1.9 Problem-Solving Strategy 16 Summary 18
7
Rotational Equilibrium and Rotational Dynamics 228 8.1
Torque 228 Torque and the Two Conditions for Equilibrium 8.3 The Center of Gravity 234 8.4 Examples of Objects in Equilibrium 236 8.5 Relationship Between Torque and Angular Acceleration 239 8.6 Rotational Kinetic Energy 246 8.7 Angular Momentum 249 Summary 254
Vectors and Their Properties 54 3.2 Components of a Vector 57 3.3 Displacement, Velocity, and Acceleration in Two Dimensions 60 3.4 Motion in Two Dimensions 62 3.5 Relative Velocity 70 Summary 74
CHAPTER 4
The Laws of Motion
232
CHAPTER 9
Solids and Fluids
CHAPTER 3
196
CHAPTER 8
CHAPTER 2
83 Forces 83 4.2 Newton's First Law 85 4.3 Newton's Second Law 86 4.4 Newton's Third Law 92 4.5 Applications of Newton's Lows 4.6 Forces of Friction 101 Summary 108
Angular Speed and Angular Acceleration \90 7.2 Rotational Motion Under Constant Angular Acceleration 194 7.3 Relations Between Angular and linear Quantities 7.4 Centripetal Acceleration 199 7.5 Newtonian Gravitation 207 7.6 Kepler's Laws 215 Summary 218
190
7.1
268
9.1 9.2 9.3
States of Matter 268 The Deformation af Solids 270 Density and Pressure 276 9.4 Variation of Pressure with Depth 279 9.5 Pressure Measurements 283 9.6 Buoyant Forces and Archimedes' Principle 284 9.7 Fluids in Motion 290 9.8 Other Applications of Fluid Dynamics 296 9.9 Surface Tension, Capillary Action, and Viscous Fluid Flow 299 9.10 Transport Phenomena 307 Summary 311
4.1
Part 2: Thermodynamics CHAPTER 10
94
Thermal Physics
CHAPTER 5
Energy
119
5.1 Work 119 5.2 Kinetic Energy and the Work-Energy Theorem 5.3 Gravitational Potential Energy 127 5.4 Spring Potential Energy 135 5.5 Systems and Energy Conservation 141 5.6 Power 143 5.7 Work Done by Q Varying Force 147 Summary 150
CHAPTER 6
Momentum and Collisions 6.1 6.2
Momentum and Impulse 161 Conservation of Momentum 166
161
124
322 ) 0.1 Temperature and the Zeroth Law of Thermodynamics 10.2 Thermometers and Temperature Scoles 324 10.3 Thermal Expansion of Solids and Liquids 328 10.4 Macroscopic Description of an Ideal Gas 335 10.5 The Kinetic Theory of Gases 340 Summary 345
322
CHAPTER 11
Energy in Thermal Processes
352 Heat and Internal Energy 352 11.2 Specific Heat 355 11.3 Calorimetry 357 11.4 Latent Heat and Phose Change 359 11.5 Energy Transfer 366 11.6 Global Warming and Greenhouse Gases 375 Summary 377
11.1
y
vi
Contents
CHAPTER 12
The Laws of Thermodynamics
16.7 The Parallel-Plate Capacitor 547 16.8 Combinations of Capacitors 549 16.9 Energy Stored in a Charged Capacitor 16.IOCapacitors with Dielectrics 557 Summary 562
385
12.1 Work in Thermodynamic Processes 385 12.2 The First Low of Thermodynamics 388 12.3 Thermal Processes 390 12.4 Heat Engines and the Second Low of Thermodynamics 12.5 Entropy 408 12.6 Human Metabolism 413 Summary 416
399
Part 3: Vibrations and Waves CHAPTER 13
Vibrations and Waves
425
13.1 Hooke's Law 425 13.2 Elastic Potential Energy 428 13.3 Comparing Simple Harmonic Motion with Uniform Circular Motion 432 13.4 Position, Velocity, ond Acceleration as a Function of Time 436 13.5 Motion of a Pendulum 439 13.6 Damped Oscillations 442 13.7 Waves 443 13.8 Frequency, Amplitude, and Wavelength 445 J3.9 The Speed of Waves on Strings 447 13.10 Interference of Waves 448 13.11 Reflection of Waves 449 Summary 450
17.1 Electric Current 570 17.2 A Microscopic View: Current and Drift Speed 572 17.3 Current and Voltage Measurements in Circuits 574 17.4 Resistance, Resistivity, and Ohm's Low 575 17.5 Temperature Variation of Resistance 579 17.6 Electrical Energy and Power 580 17.7 Superconductors 584 17.8 Electrical Activity in the Heart 585 Summary 588
CHAPTER 18
Direct-Current Circuits
594
18.1 Sources of emf 594 18.2 Resistors in Series 595 18.3 Resistors in Parallel 598 18.4 Kirchhoff's Rules and Complex DC Circuits 603 18.5 RC Circuits 607 18.6 Household Circuits 611 18.7 Electrical Safety 612 18.8 Conduction of Electrical Signals by Neurons 613 Summary 615
626
19.1 19.2 19.3 19.4 19.5 19.6 19.7
Magnets 626 Earth's Magnetic Field 628 Magnetic Fields 630 Magnetic Force on a Current-Carrying Conductor 633 Torque on a Current Loop and Electric Motors 636 Motion of a Charged Particle in a Magnetic Field 639 Magnetic Field of a Long, Straight Wire and Ampere's Low 642 19.8 Magnetic Force Between Two Parallel Conductors 645 19.9 Magnetic Fields of Current Loops and Solenoids 646 19.10 Magnetic Domains 650 Summary 652
14.1 Producing a Sound Wave 459 14.2 Characteristics of Sound Waves 460 14.3 The Speed of Sound 461 14.4 Energy and Intensity of Sound Waves 463 14.5 Spherical and Plane Waves 466 14.6 The Doppler Effect 468 14.7 Interference of Sound Waves 473 14.8 Standing Waves 475 14.9 Forced Vibrations and Resonance 479 14.lOStanding Waves in Air Columns 480 14.11Beats 484 14.12 Quality of Sound 486 14.13The Ear 487 Summary 489
CHAPTER 20
Induced Voltages and Inductance
Part 4: Electricity and Magnetism CHAPTER 15
497
15.' Properties of Electric Charges 497 15.2 Insulators and Conductors 499 15.3 Coulomb's Law 500 15.4 The Electric Field 505 15.5 Electric Field Lines 510 15.6 Conductors in Electrostatic Equilibrium 513 15.7 The Millikan Oil-Drop Experiment 515 15.8 The Van de Graaff Generator 516 15.9 Electric Flux and Gauss's Law 517 Summary 523
663
20.1 20.2 20.3 20.4
Induced emf and Magnetic Flux 663 Faraday's Law of Induction 666 Motional emf 670 Lenz's Low Revisited (The Minus Sign in Faraday's Law) 674 20.5 Generators 676 20.6 Self-Inductance 680 20.7 RL Circuits 683 20.8 Energy Stored in a Magnetic Field 686 Summary 687
CHAPTER 21
Alternating-Current Circuits and Electromagnetic Waves 696
CHAPTER 16
Electrical Energy and Capacitance
Current and Resistance 570
Magnetism
459
Electric Forces and Electric Fields
CHAPTER 17
CHAPTER 19
CHAPTER 14
Sound
555
531
16.1 Potentiol Difference and Electric Potential 531 16.2 Electric Potential and Potential Energy Due to Point Charges 538 16.3 Potentiols and Charged Conductors 542 16.4 Equipotential Surfaces 543 16.5 Applications 544 16.6 Capocitance 546
21,1 Resistors in an AC Circuit 696 21.2 Capacitors in an AC Circuit 699 21.3 Inductors in on AC Circuit 701 21.4 The RLC Series Circuit 702 21.5 Power in on AC Circuit 707 21.6 Resonance in a Series RLC Circuit 708 21.7 The Transformer 710 21.8 Maxwell's Predictions 712 21.9 Hertz's Confirmation of Maxwell's Predictions 21.10 Production of Electromagnetic Waves by on Antenna 714
713
Contents
21.11 Properties of Electromagnetic Waves 715 21.12 The Spectrum of Electromagnetic Waves 720 21.13 The Doppler Effect for Electromagnetic Waves 722 Summary 723
CHAPTER 22
732
22.1 The Nature of Light 732 22.2 Reflection and Refraction 733 22.3 The Low of Refraction 737 22.4 Dispersion ond Prisms 742 22.5 The Rainbow 745 22.6 Huygens's Principle 746 22.7 Total Internal Reflection 748 Summary 751
Nuclear Physics
790
913
847
Galilean Relativity 847 The Speed of Light 848 Einstein's Principle of Relativity 850 Consequences of Special Relativity 851 Relativistic Momentum 858 Relativistic Energy and the Equivalence of Mass and Energy 859 26.7 General Relativity 863 Summary 865
Nuclear Fission 937 Nuclear Fusion 941 Elementary Particles and the Fundamental Forces Positrons and Other Antiparticles 944 Classification of Particles 945 Conservation Laws 947
30.7 The Eightfold Way 949 30.8 Quarks and Color 950 30.9 Eleetroweak Theory and the Standard Model 30.lOThe Cosmic Connection 954 30.11 Problems and Perspectives 955 Summary 956
823
CHAPTER 26
929
Nuclear Energy and Elementary Particles 30.1 30.2 30.3 30.4 30.5 30.6
823
Part 6: Modern Physics 26.1 26.2 26.3 26.4 26.5 26.6
899 902
CHAPTER 30
800
25.2 The Eye 824 25.3 The Simple Magnifier 829 25.4 The Compound Microscope 830 25.5 The Telescope 832 25.6 Resolution of Single-Slit and Circular Apertures 25.7 The Michelson Interferometer 840 Summary 841
Relativity
891
29.1 Some Properties of Nuclei 913 29.2 Binding Energy 916 29.3 Radioactivity 918 29.4 The Decay Processes 921 29.5 Natural Radioactivity 926 29.6 Nuclear Reactions 927 29.7 Medical Applications of Radiation Summary 931
CHAPTER 25 25.1 The Camera
891
CHAPTER 29
CHAPTER 24
Optical Instruments
27.4 Diffroction of X-Rays by Crystals 876 27.5 The Compton Effect 879 27.6 The Dual Nature of Light and Motter 880 27.7 The Wove Function 883 27.8 The Uncertainty Principle 884 Summary 886
28.3 The Bohr Madel 894 28.4 Quantum Mechanics and the Hydrogen Atom 28.5 The Exclusion Principle and the Periodic Table 28.6 Characteristic X-Rays 905 28.7 Atomic Transitions and Lasers 906 Summary 908
759
24.1 Conditions for Interference 790 24.2 Young's Double-Slit Experiment 791 24.3 Change of Phase Due to Reflection 795 24.4 Interference in Thin Films 796 24.5 Using Interference to Read CDs and DVDs 24.6 Diffraction 802 24.7 Single-Slit Diffraction 803 24.8 The Diffraction Grating 805 24.9 Polarization of Light Waves 808 Summary 815
870
27.1 Blackbody Radiation and Planck's Hypothesis 870 27.2 The Photoelectric Effect and the Particle Theory
28.1 Early Models of the Atom 28.2 Atomic Spectra 892
23.1 Flat Mirrors 759 23.2 Images Farmed by Concave Mirrors 762 23.3 Convex Mirrors and Sign Conventions 764 23.4 Images Formed by Refraction 769 23.5 Atmospheric Refraction 772 23.6 Thin Lenses 773 23.7 Lens and Mirror Aberrations 781 Summary 782
Wave Optics
Quantum Physics
CHAPTER 28 Atomic Physics
CHAPTER 23
Mirrors and Lenses
CHAPTER 27
of Ught 872 27.3 X-Rays 875
Part 5: Light and Optics Reflection and Refraction af Light
vii
952
835 Appendix Appendix Appendix Appendix Appendix
A: B: C: D: E:
Mathematics Review A.l An Abbreviated Table af Isotopes Some Useful Tables A.19 SI Units A.21
MeAT Skill Builder Study Guide
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems A.52 Index 1.1
A.14
A.22
937
943
_~AC'BOUT THE AUTHORS
Raymond A. Serway received his doctorate at Illinois Instimte of Technology and is Professor Emeritus aLJames rvladison University. In 1990 he received the Madison Scholar Award atJames Madison University, ,..· here he taught for 17 years. Dr. Serway began his teaching career at Clarkson University, where he conducted research and taught from 1967 to 1980. He was the recipient. of the Distinguished Teaching Award at Clarkson University in 1977 and of the Alumni Achievement Award from Utica College in 1985. As Guest Scientist at the IBM Research Laboratory in Zurich, S",!itzeriand, he worked ,·,'ith K. Alex t\'lliller, 1987 Nobel Prize recipient. Dr. Serway also was a visiting scientist at Argonne National Laboratory, where he collaborated with his mentor and friend, Sam Marshall. In addition to earlier editions of this textbook, Dr. Sen,,'a)' is the coauthor of Principles of Ph)lSics, fourth edition; Physics for Scient.ists and Engineers, seventh edition; Essentials o/College Physics; and Alodenz Ph)'sics, third edition. He also is the coauthor of the high school t.extbook Physics, published by Holt, Rinehart and 'Vinston. In addition, Dr. Senvay has published more than 40 research papers in the field of condensed matter physics and has given more than 70 presentations at professional meetings. Dr. Serway and his wife, Elizabeth, enjoy traveling, golf, gardening, singing in a church choir, and spending time with their four children and eight grandchildren. Chris Yuille is an associate professor of physics at Embry-Riddle Aeronautical University (ERAU), Daytona Beach, Florida, the world's premier institution for aviation higher education. T-Te received his doctorate in physics from the University of Florida in 1989 and moved to Daytona after a year at ERAU's Frescau, Arizona, campus..t\1though he has taught courses at all levels, including postgraduate, his primary interest has been the delivery of introductory physics. He has received several awards for teaching excellence, including the Senior Class Appreciation Award (three times). He conducts research in general relativity and quantum theory, and was a participant in the JOVE program, a special three-year NASA grant program during \vhich he studied neutron stars. His work has appeared in a number of scientific journals, and he has been a featured science writer in Analog Science Fiction/Science Fact magazi ne. 1n add ition to th is textbook, he is coauthor of Esse'ntials of College Physics. Dr. Vuille el~oys tennis, swimming, and playing classical piano, and he is a fanner chess champion of SL Petersburg and Atlanta. In his spare time he writ.es fiction and goes to the beach. His '\life, Dianne Kowing, is an optometrisl for a local Veterans' Administration clinic. His daughter, Kira VllilleKml,:ing, is a meteorology/communications double m~jor at ERAU and a graduate of her father's first-year physics course. He has two sons, Christopher, a cellist and fisherman, and James, avid reader of Disney comics. Jerry S. Faughn earned his doctorat.e at the University of\rfississippi. I-Ie is Profes-
sor Emeritus and fonner chair of the Department of Physics and Astronomy at Eastern Kenmcky University. Dr. Faughn has also ,...' ritten a microprocessor interfacing text [or upper-division physics students. He is coauthor ofa nonmathematical physics text and a physical science text for general education students, and (with Or. Scrway) the high-school textbook Physics, published by Holt, Reinhart and Winston. He has taught courses ranging from the lower division to the graduate level, but his primary interest is in studentsjusr beginning 1.0 learn physics. Dr. Faughn has a wide variety of hobbies, among \vhich are reading, travel, genealogy, and old-time radio. His 'tvife, Mary Ann, is an avid gardener, and he contributes to her efforts by staying out of the way. His daughter, Laura, is in family practice, and his son, David, is an attorney.
viii
PREFACE
Coltege PhJsi.cs is written for a one-year course in introductory physics usually taken by students majoring in biology, the health professions, and other disciplines including environmental, earth, and social sciences, and technical fields such as architecture. The mathematical techniques used in this book include algebra, geometry, and trigonometry, but not calculus. This textbook, which covers the standard topics in classical physics and 20thcentury physics, is divided into six pans. Part I (Chapters 1-9) deals with Ne,..,,~ tonian mechanics and the physics of fluids; Part 2 (Chapters ]0-L2) is concerned 'with heat and thermodynamics; Pan 3 (Chapters 13 and 14) covers wave motion and sound; Pan 4 (Chapters 15-21) develops the concepts of electricity and magnetism; Part 5 (Chapters 22-25) treats the propen.ies or light and the field of geometric and \vave optics; and Part 6 (Chapters 26-30) provides an introduction to special relativity, quantum physics, atomic physics, and nuclear physics.
OBJECTIVES The main o~jectives of this introductory textbook are twofold: to provide the student with a clear and logical presentation of the basic concepts and principles of physics, and to strengthen an understanding of the concepts and principles through a broad range of interesting applications to the real world. To meet those objectives, we have emphasized sound physical arguments and problem-solving methodology. At the same time, \-\le have attempted to motivate the student through practical examples that demonstrate the role of physics in other disciplines.
CHANGES TO THE EIGHTH EDITION A number of changes and improvements have been made to this edition. Based on comments [rom users of the sevel1lh edition and reviewers' suggestions, a m~~jor c1Tort \vas made to increase the emphasis on conceptual understanding, to add new end-of-chapter questions and problems that are informed by research, and to improve the clarity of the presentation. The new pedagogical features added to this edition are based on current trends in science education. The following represent the major changes in the eighth edition.
Questions and Problems We have substantially revised the end-of-chapter questions and problems [or this edition. Three new types of quest.ions and problems have been added: •
Multiple-Choice Questions have been introduced with several purposes III mind. Some require calculations designed 1.0 facilitate students' familiarity with the equations, the variables used, the concepts the variables represent, and the relationships benveen the concepts. The rest are conceptual and are designed to encourage conceptual thinking. Finally, many student.s are required to take multiple-choice tests, so some practice with that form of question is desirable. Here is an example ofa multiple-choice question: 12. A truck loaded with sand accelerates along a highway. The driving force on the truck remains constant. "Vhat happens to the acceleration or the trLlck as its trailer leaks sand al a constant rate through a hole in its bot.tom? (a) It decreases at a steady ratc. (h) It increases at a steady rate. (c) It increases and then decreases. (d) It decreases and then increases. (e) 11. remains constant.
ix
x
Preface
The illstrtlCLOf may select multiple-choice questions to assign as homework or usc them in the classroom, possibly with -peer instruction" methods or in conjunction with "'clicker" s}'stcms. More than 350 multiple-choice questions are included in this edition. Answers LO odd-numbered Illultiple-choice questions dl-C included in the Answers ~eLliulI at tile end uf the book, and answers to all questions are found in the l1JslnLCtor~'i Solutions Afmlllal and on the instrucLOr's POllJPrl-Prlurp CD-ROlvI.
•
Enhanced Content problems require symbolic or conceplllal responses from the slllden t. A s)'l1Ibo(;( Enhanced Conlent problem requires the sludent to obtain an answer in terms of symbols_ In general, some g·uidance is built into the problem stateIllent. The Roal is to belter train the studenlto deal with mathematics at a level appropriate to this course. Most studcnts at this Icvel are uncomfortable \,..-ith symbolic equations, which is unfortunate because symbolic equations are the most efficient vehicle (or presenting relationships between physics concepts. Once students understand the physical concepts, their ability [Q solve problems is greatly en hallCed. As soon as the numbers ,lre substilUted into an equation, however, allthe concepts and their relationships to one another are lost. melded together in the student's calculator. The symbolic Enhanced Content problems train students LO postpone substitution of values, facilitating their ability to think conceptually lIsing the equations. An example of a symbolic Enhanced Content prOblem is provided here: 14'- An objeCl of mass
In is dropper! from the roof of a building of heighl 11_ \·Vhile the objecl is falling, a wind hlowing parallel to the face of the building cxcns a conslane horizollLal force Fon the ol~jccl. (a) Ilow long does it take lhe objcclto slrike the grolllld? Express lhe time I in terms of g and h. (b) Find an cxprcssiUll in lcrms of "/Ii and Ffar the accc1cralion fl) of tllC ohjcCl in the horizontal direction (taken as lhe positive x-direclion). (c) How far is the object displaced horil.olltall}' before hilling lhc ground? Answer in terms of HI, g, I': and II. (d) Find lhe magnilude of the object's accclcralioll while il is falling, using [he variables F, 111, and g.
A cOllcrlJlUal Enhanced Content ,noblem encourages the student to think \'erballr and conceptually about a gi\·en physics problem rather than rely solely on compUl
Preface
•
Guided Problems help students break problems into steps. A physics problem typically asks for one physical quantity in a given context. Often, hmvever, several concepts must be used and a number of calculations are required to get that final answer. Many swdents are not accustomed to this level of complexity and often don't know ,...· here to star\.. A Guided Problem br'Caks a standard pl-oblem into smaller steps, enabling students to grasp all the concepts and strategies required to arrive at a correct solution. Unlike standard physics problems, guidance is often built into the problem statement. For example, the problem might say "Find the speed using conservation of energy" rather than only aski ng for the speed. Tn any given chapter there are usually two or three problem types that are particularly suited to this problem {-arm. The problem must have a certain level of complexity, v·,lith a similar problem-solving strategy involved each time it appears. Guided Problems are reminiscent of how a student might interact with a professor in an office visit. These problems help train students to break down complex problems into a series of simpler problems, an essential problem-solving skill. An example of a Guided Problem is provided here: 32.
E!I
Two blocks of masses 11/,1 and 111'2 (111 1 > 1n.) are placed on a fJ"ictionless table in contace \"jth each other. A horizontal force of magnitude Fis applied to lhe block ofm
11~
FIGURE P4.32
In addition to these three new question and problem types, \ve carefully reviewed all other questions and problems for this revision to improve their variety, interest, and pedagogical value \'\-'hile maintaining their clarity and quality. Approximately 30% of the questions and problems in this edition are ne\'·...
Examples In the last edition all in-text worked examples were reconstituted in a two-column format to better aid student learning and help reinforce physical concepts. For this eighth edition we have review'ed all the worked examples, made improvements, and added a nc'v Question at the end of each \\"orked example. The Questions usually require a conceptual response or determination, or estimates requiring knowledge of the relationships between concepts. The answers for the new Questions can be found at the back of the book. A sample of an in-text worked example follows on the next page, with an explanation of each of the example's main parts:
The Solution section uses a twocolumn format that gives the explanation lor each step of the
Gool
solution in the left-hand column, while giving each accompanying
Problem Using a ;\11 pendululIl of length O.! 71 \\"ha! is the \", . of gin this local ion?
mothematical step in the right-hand column. This layout facilitates matching the idea with its execution and helps students learn how to organize their work. Another benefit: "tudents can easily use this format as a troining tool, covering up the solution on the right and salving the problem using the comments on the left as a guide
Strategy ;irsl calculate the period of the pendulum hy dividing the total Lime by [he llumber of complete s\\"ing_. Solve Equation 13.15 lor g;lnd Mlbslitutc values.
EXAMP~tl3.7
Measud~Val~f.g
~~:rlTline ff from
....
ndulum mOlion. Ill.
a geophysicist
COUIlLS
72.0 complete swings in a time of 60.0 s.
,.. ..... Solution Calculate the peri(,d by dividing the total elapsed time by the number of complete oscillalion;:,;
T =
time
# of o;:,cillat.ion;:,
= 60.0 s = 0.833;:,
72.0
Solve Equation 13.15 li,r gand suhstitmc values;
(39.5)(O,17Im) = 9.73 mist (0.833 s)t
, .. ~
Remork
.,
\1easurillg" sudl a vibration is a good \"ar of r1etermining the local value of the "ccderation of gravity.
Remarks follow each Solution
QUESTION 13.7 True or False: A simple penduluill of length 0.50 rn has a larger frequenq' of vibration llian a simple pendulum of length l.0 Ill.
ond highlight "orne of the underlying concept" ond methodology used in arriving at a
EXERCISE 13.7t~ What would he the I . 'iot! of the O.I71-m pendulum 011 the :\100n, where the acceleration ofgra\'llY is 1.62 rnls 2?
correct solution. In oddition, the remorks ore often used to put the
Answer
problem into a larger, real-world context.
2.0·1 s
I
Exercise/Answer Every worked example is followed immediately by on exercise with an answer. These exercises
Question New to this edition,
allow students to reinforce their understonding by working
each worked example will feature
a similar or related problem, with the onswers giving them instant feedback. At the option of the instrudor, the
a conceptual question that promotes student understonding
exercises con also be assigned as homework. Students who work through these exercises on a regular basis will find the
of the underlying concepts contained in the example
end-of-chapter problems less intimidating.
-
WebAssign
Many Worked Examples are olso available to be assigned as Active Examples in the Enhanced WebAssign homework management system (visit
wwwoserwoyphysics.com for more details),
Preface
Online Homework It is now easier to assign online homework with Serway and Vuille using the widely acclaimed program Enhanced vVebAssign. All end-of-chapter problems, active figures, quick quizzes, and most questions and worked examples in this book are available in vVebAssign. Most problems include hints and feedback to provide instantaneous reinforcement or direction for that problem. We have also added math remediation tools to help students get up to speed in algebra and trigonometry, animated Active Figure simulations to strengthen students' visualization skills, and video to help students better understand the concepts. Visit www.serwayphysics. com to vic"..' an interactive demo of this innovative online homework solution.
Content Changes The text has been carefully edited to improve clarity of presentation and precision of language. vVe hope that the result is a book both accurate and er~joyable to read. Although the overall content and organization of the textbook are similar to the seventh edition, a few changes were implemented. • Chapter 1, Introduction, has a ne,.., biological example involving an estimate. • Chapter 2, Motion in One Dimension, has an improved first example. Quick Quiz 2.) was given another part so that students ,·\!ould understand the distinction between average speed and average velocity. Quick Quiz 2.2 V'las completely rewritten to improve its effectiveness. An extra pan was added to Example 2.4, and an example from the last edition was eliminated because it was not suf~ ficiently illustrative and somewhat redundant. It was replaced with a new symbolic example. • Chapter 3, Vectors and T,vo-Dimensional fi10tion, features a new symbolic example on the range equation. • Chapter 4, The Laws of Motion, contains several improved Quick Quizzes and a revised and improved example. The first three quick quizzes were combined into one master quick quiz, requiring the student to answer five related truefalse questions on the concept of a force. Quick Quizzes 4.4 and 4.5 were rewritten, and Example 4.6 was improved. • In Chapter 5, Energy, two definitions of \'wrk and the definitions of average power and instantaneous power were clarified. The Problem-Solving Strategy on conservation of energy was improved, resulting in positive changes to Example 5.5. A new part was added to Example 5.14 to enhance student comprehension of instantaneous as opposed to average power. • Tn Chapter 6, Momentum and Collisions, the connection between kinetic energy and momentum was made explicit early in the chapter and then lIsed in a Quick Quiz and elsewhere in the problem set. • In Chapter 7, Rotational Motion and the La'''' of Gravity, the definitions of the radian and radian measure were clarified. A new pan was added to Example 7.1, dealing with arc length. • Chapter 9, Solids and Fluids, features a new discussion of dark matter and dark energy in Sectlon 9.1, States of Matter. Example 9.2 is a new biological example about sports injuries. • Chapter 12, The Laws ofThennodynamics, has been reorganized slightly, and a new section (Section 12.3, Thermal Processes) has been added. Another equivalent statement of the second law of thermodynamics was included along with further explanation. • Chapter 14, Sound, has a new, more instructive Example 14.1, replacing the previous example. • Chapter 15, Electric Forces and Electric Fields, has two worked examples that were upgraded with new parts. • Chapter 16, Electrical Energy and Capacitance, has a new worked example that illustrates particle dynamics and electric potential. Three other . . " orked examples were upgraded with new pans, and two new quick quizzes were added.
xiii
xiv
Preface
• Chapter 17, Current and Resistance, was reorganized slightly, putting the subsection on power ahead of superconductivity. It also has two new quick quizzes. • Chapter 18, Direct-Current Circuits, has both a new and a reorganized quick quiz. • Chapter 19, JvIagnctism, has a new section on types uf 11Iag-llelic materials as well as a ne,,,"' quick quiz. • Chapter 20, Induced Voltages and Inductance, has new material on RL circuits, along "vith a new example and quick quiz. • Chapter 21, Alternating-Current Circuits and Electromagnetic ,,,raves, has a new series of four quick quizzes that 'vere added to drill the fundamentals of AC circuits. The problem-solving strategy for RLCcircuits was completely revised, and a new physics application on using alternating electric fields in cancer treatment was added. • Chapter 24, "Vave Optics, has an improved example and two new quick quizzes. • Chapter 26, Relativity, no longer covers relativistic addition of velocities. Three new quick quizzes were added to the chapter. • Chapter 27, Quantum Physics, was rewritten and streamlined. Two superfluous w·orked examples were eliminated (old Examples 27.1 and 27.2) because both are discussed adequately in the text. One of two worked examples on the Heisenberg uncertainty principle was deleted and a new quick quiz was added. The scanning tunneling microscope application was deleted. • Chapter 28, Atomic Physics, was rewritten and streamlined, and the subsection on spin was transferred to the section on quantum mechanics. The section 011 electron clouds was shortened and made into a subsection. The sections on al.Omic transitions and lasers were combined into a single, shorter seerion. • Chapter 29, Nuclear Physics, was reduced in size by deleting less essential worked examples. Old worked examples 29.1 (Sizing a Neutron Star), 29.4 (Radon Gas), 29.5 (The Bela Decay of Carbon-14), and 29.9 (Synthetic Elements) were eliminated because they were similar to other examples already in the text. The medical application of radiation '...·as shortened, and a new quick quiz ,vas developed. • Chapter 30, Nuclear Energy and Elementary Particles, 'vas rewritten and streamlined. The section on nuclear reactors was combined "\lith the one on nuclear fission. The historlcal section and old Section 30.7 on the meson were e1iminaled, and the beginning of the section on panicle physics was eliminated. The section on strange panicles and strangeness was combined with the section on conservation laws. The sections on quarks and colored quarks were also combined into Section 30.8, Quarks and Color.
TEXTBOOK FEATURES Most instructors would agree that the textbook assigned in a course should be the student's primary guide far understanding and learning the sul~ject matter. Further, the textbook should be easily accessible and written in a style that facilitates instruction and learning. vVith that in mind, we have included many pedagogical features that are intended to enhance the textbook's usefulness to both students and instructors. The follmv'ing features are included. QUICK QUIZZES All the Quick Quizzes (see example below) are cast in an objective format, including multiple-choice, true-false, matching, and ranking questions. Quick Quizzes provide students with opportunities to test their understanding of the physical concepts presented. The questions require students to make decisions on t.he basis of sound reasoning, and some have been writlen to help students overcome common misconceptions. Answers to all Quick Quiz questions are taund at the end of the textbook, and ans\vers with detailed explanations are provided in the fnslruclo-r's S'oll.ll"ions 1'vlanual. ~1any instructors choose LO use Quick Quiz questions in a "peer instruction" teaching style.
Preface
QUICK QUIZ 4.3
A small sports car collides head-oil with a massive truck. The greater impact force (in magnitude) acts on (a) the car, (b) the truck, (c) neither, the force is the same on both. v\lhich vehicle undergoes the greater magnitude acceleration? (d) the car, (e) tIle truck, (f) the accelerations are the same.
PROBLEM-SOLVING STRATEGIES A general problem-solving strategy to be followed by the student is outlined atlhe end of Chapter 1. This strategy provides students with a structured process for solving problems. Tn Illost chapters more specific strategies and suggestions (see example below) are included for solving the types of problems featured in both the ,"vorked examples and the end-of-chapter problems. This feature helps students identify the essential steps in solving problems and increases their skills as problem solvers. R BLEM-SOLVING STRATEGY NEWTON'S SECOND LAW Problems involving Newton's second 1<1\'\1 can be very complex. The following protocol breaks the solution process dm'\"n into smaller, intermediate goals: 1. Read the problem carefully at least once.
2. Draw a picture of the system, identify the object of primary interest, and indicate forces with arrows. 3. Label each force in the picture in a ,vay thal will bring to mind what physical quantity the label stands for (e.g., Tfor tension). 4, Draw a Cree-body diagram of the ol~iect of interest, based on the labeled picture. If additional objects are involved, drm.. . separate free-body diagrams for them, Choose convenient coordinates for each object. 5. Apply Newton's second law. The x- and )i-components of Newton's second law should be taken from the vector equation and ,..'ritten individually. This usually results in two equations and two unknowns. 6. Solve for the desired unknown quantity, and substitute the numbers.
II icons point the way to various practical and interesting applications of physical principles to biology and medicine, vVhenever possible, more problems that are relevant to these disciplines are included. BIOMEDICAL APPLICATIONS For biology and pre-med students,
MCAT SKILL BUILDER STUDY GUIDE The eighth edition of College Physics contains a special skill-building appendix (Appendix E) to help premed students prepare for the MeAT exam. The appendix contains examples written by the text authors that help students build conceptual and quantitative skills, These skillbuilding examples are followed by MCAT-style questions written by test prep experts to make sure students are ready to ace the exam.
MCAT TEST PREPARATION GUIDE
Located after the "To the Student" section in the front of the book, this guide outlines 12 concept-based sLUdy courses for the physics part of the MeAT exam. Students can use the guide to prepare for the MeAT exam, class tests, or homework assignlllellts.
APPLYING PHYSICS The Applying Physics features provide students with an additional means of reviewing concepLS presented in that section. Some Applying Physics examples demonstrate the connection between the concepts presented in that chapter and other scientific disciplines. These examples also serve as models for students when assigned the task of responding to the Conceptual Questions
xv
xvi
Preface
presented at the end of each chapter. For examples of Applying Physics boxes, see Applying Physics 9.5 (Home Plumbing) on page 299 and Applying Physics 13.1 (BungeeJumping) on page 435.
TIP 4.3 Newton's Second Low Is a Vector Equation In applying Newlon's second law, add all of the forces all the object as vectors and then lind the resultanl vector acceleration by dividing by m. Don'[ find llw individual magnitudes of the forces and add them like scalars.
Newton's third law
APPLICATION
II
Diet Versus Exercise in Weight-loss Programs
TIPS PI;:tf"f'r1 in the margins of the text, Tips address common student misconceptions and situations in \'vhich students often follow unproductive paths (see example at the left). More than ninety-five Tips are provided in this edition to help students avoid common mistakes and misunderstandings.
~
MARGINAL NOTES Comments and notes appearing in the margin (see example at the left) can be used to locate imponant statements, equations, and concepts in the text.
Although physics is relevant to so much in our modern lives, it may not be obvious to students in an introductory course. Application margin notes (see example at the left) make the relevance of physics to ever),day life more obvious by pointing out specific applications in the text. Some of these applications pertain to the life sciences and are marked with a II icon. APPLICATIONS
MULTIPLE·CHOICE QUESTIONS New to this edition are end-of-chapter lTIultiplechoice questions. The instructor may select items to assign as homework or use them in the classroom, possibly with ';peer instruction" methods or with "clicker" systems. More than 350 multiple-choice questions are included in this edition. Answers to odd-numbered multiple-choice questions are included in the ans\.\'er section at the end of the book, and ans\vers to all questions are found in the InstTuct.ors Solutions lWanuaL
.
CONCEPTUAL QUESTIONS At the end of each chapter there are 10-15 conceptual questions. The Applying Physics examples presented in the text serve as models for students when conceptual questions are assigned and show how the concepts can be applied to understanding the physical world. The conceptual questions provide the student \,vith a means of self-testing the concepts presented in the chapter. Some conceptual questions are appropriate for initiating classroom discussions. Answers to odd-numbered conceptual questions are included in the AnS\\lerS section at the end of the book, and anS\'lerS to all questions are found in the Instru..ct01·S Solulion.s J\1anuaL An extensive set oCproblems is included at the end of each chapter (in all, almost 2 000 problems are provided in this edition). Ans\vers to odd-numbered problems are given at the end of the book. For the convenience of both the student and instructor, about two-thirds of the problems are keyed to specific sections of the chapter. The remaining problems, labeled ''Additional Problems," are not keyed to specific sections. The three levels of problems are graded according to t.heir difficulty. Straightfon,vard problems are numbered in black, intermediate.· level problems are numbered in blue, and the most challenging problems are numbered in magenta. The II icon identifies problems dealing with applications to the life sciences and medicine. Solutions to approximately 12 problems in each chapter are in the Student Sol'l.l.tions iWa.nual/Slurry Guide.
PROBLEMS
STYLE To facilitate rapid comprehension, we have attempted to write the book in a style that is clear, logical, relaxed, and engaging. The somewhat informal and relaxed \"TiLing st.yle is designed to connect better ,vith students and enhance their reading enjoyment. New terms are carefully defined, and we have tried to avoid the use ofjargon. INTRODUCTIONS All chapters begin with a brief previc\v that includes a discllssion of the chaptel-'s o~jectjves and content.
Preface
UNITS The international system of units (Sf) is used throughout the text. The U.S. customary system of units is used only to a limited extent in the chapters on mechanics and thermodynamics. PEDAGOGICAL USE OF COLOR Readers should consult the pedagogical color chart (inside the front cover) for a lisLing of the color-coded symbols uscd in the text diagrams. This SYSICITl is follm,,·cd consistently throughout thc text. IMPORTANT STATEMENTS AND EQUATIONS Most important statements and definitions are set in boldface type or are highlighted with a background screen [or added cmphasis and ease of review. Similarly, important equations are highlighted ,<\'ith a ran background screen to facilitate location. ILLUSTRATIONS AND TABLES The readability and effectiveness of the text material, '\.Iorked exarnples, and cnd-ol:'chapter concepLUal questions and problems are enhanced by the large number of figures, diagrams, photographs, and tables. Full color adds clarity to the artwork and makes illustrations as realistic as possible. Three-dimensional effects (Ire rendered with the use of shaded and lightened areas where appropriate. Vectors are color coded, and curves in graphs are drawn in color. Color photographs have been carefully selcCled, and their accompanying captions have been written to serve as an added instructional tool. A complete description of the pedagogical use of color appears on the inside front cover.
SUMMARY The end-oJ:'chapter Summary is organized by individual section headings for ease of reference. SIGNIFICANT FIGURES Signiftcant figures in both worked examples anel enelof-chapter problems have been handled "vith care. Most numerical examples and problems are worked out to either two or three signil1cant figures, depending on the accuracy of the data provided. rntermediate results presented in the examples are rounded to the proper number of significant figures, and on Iy those digits are carried fonvard. APPENDICES AND ENDPAPERS Several appendices are provieled al the enel or the textbook. Most of the appendix material represents a review of mathematical concepts and techniques used in the text, including scientific notation, algebra, geometry, trigonometry, differentia] calculus, and integral calculus. Reference to these appendices is made as needed throughout the text. Most of the math~ ernatical review senions include worked examples and exercises with ans'...·ers. In addition to the mathematical reviev\o', some appendices contain useful tables that supplement textual information. For easy reference, the front endpapers contain a chart explaining the use of color throughout the book and a list of frequently used conversion factors.
ACTIVE FIGURES Many diagrams from the text have been animated to become Active Figures (identified in the figure legend), part of the Enhmu;ed lVebAssign online homework system. By vie\ving animations of phenomena and processes that cannot be fully represented on a st.atic page. students greatly increase their conceptual understanding. In addition to viewing animations of the figures, students can see the outcome of changing variables to see the effects, conduct suggested explorations of the principles involved in the figure, and take and receive feedback on quizzes related to the figure. All Active Figures are included on the instructor's PowerLeclllTe CD-ROkE for in-class lecture presentation.
TEACHING OPTIONS This book contains rnorc than enough material for a one-year course in introducLOry physics, which serves two purposes. First, il gives {he instructor more flexibility
xvii
xviii
Preface
in choosing topics for a specific course. Second, the book becomes more useful as a resource for sLUdcnts. On average, it should be possible to cover about one chapter each ,,\!eek for ..1 class that meets thrce hours per week. Those sections, examples, and end-of-chapter problems dealing with applications of physics to life sciences aloe identified 'with the DNA icon ft. We ofTel' the follmving suggestions for shaner courses for those instructors ,vho choose to move at a slower pace through the year.
Option A: If yOIl choose to place more emphasis on contemporary topics in physics, you could omit all or parts of Chapter 8 (Rotational Equilibrium and Rotational Dynamics), Chapt.er 21 (Alternating-Current Circuits and Electromagnetic "Vaves), ancl Chapter 25 (Optical Instruments). Opt-ionB: If you choose to place more emphasis on classical physics, you could omit all or parts of Pan 6 of the textbook, which deals with speciaJ relati\'ity and other topics in 20th-century physics_ The JnslruclO'r'.s Solulions Manual offers additional suggeslions for specific sections and topics that may be omitted \vitholltloss orcontinuit), if time presses.
COURSE SOLUTIONS THAT FIT YOUR TEACHING GOALS AND YOUR STUDENTS' LEARNING NEEDS Recent advances in educational technology have made hOmC\\lOrk 111anagemclll systems and audiencc response systems powelfu] and affordable tools to enhance the way you teach your course. Whether you offer a more traditional text-based course, are int.el"ested in using or are currently using an online homework management system such as VlebAssign, or are ready to turn your lecture illlo an interactive learning environment with an audience response system, yOll can be confident. that the text's proven content provides the foundation for each and every component 01" our technology and ancillary package. VISUALIZE WHERE YOU WANT TO TAKE YOUR COURSE
WE PROVIDE YOU WITH THE FOUNDATION TO GET THERE SerwoylVuille, College Physics, 8e
Preface
xix
Homework Management Systems ENHANCED WEBASSIGN Enhanced WebAssign is the perfect solution to your homework management needs. Designed by physicists for physicists, this system is a reliable and user-friendly teaching companion. Enhanced WebAssign is available [or College Physics, giving you the freedom to assign
• every end-or-chapter Problem, Multiple-Choice Question, and Conceptual Question, enhanced "lIith hints and feedback • most worked examples, enhanced with hints and reedback, to help strengthen students' problem-solving skills • every Quick Quiz, giving your students ample opportunity to tesllheir conceptual understanding • animated Active Figures, enhanced ,,,,ith hints and feedback, La help students develop their visualization skills • a math revie\,.: to help sllldents brush up on key quantitative concepts Please visit www.serwayphysies.eom to view an interactive demonstration of Enhanced \VebAssign. The text is also supported by the following HOlllc,·\,ork Management Systems. Contact your local sales representative for more information. CAPA: A Computer-Assisted Personalized Approach and LON-CAPA, http;//www.lon-capa.org/ The University of Texas Homework Service
Audience Response Systems AUDIENCE RESPONSE SYSTEM CONTENT
Regardless of the response system you are using, we provide the tested content to support it. Our ready-lo-go content includes all the questions from the Quick Quizzes, all the end-oE-chapter MultipleChoice Questions, test questions, and a selection of end-of-chapter questions LO provide helpful concepmal checkpoints to drop into your leClure. Our Active Figure animations have also been enhanced with multiple-choice questions to help test students' observational skills. vVe also feature the Assessing to Learn in the Classroom content from the University of Massachusetts. This collection of250 advanced conceptual questions has been tested in the classroom for more than ten years and takes peer learning to a ne,,·,' level. Contact your local sales representative to learn more about our audience response software and hardware. Visit wwwoserwayphysicsocom to download samples of our audience response system content.
Lecture Presentation Resources The following resources provide support !-or your presentations in lecture. POWERLECTURE CD-ROM An easy-t.o-use multimedia lecture tool, the PowerLecture CD-ROl\!l allows you t.o quickly assemble art, animations, digit.al video, and
database files \vith notes 1.0 create fluid lectures. The two-volume seL (Volume 1: Chapters 1-14; Volume 2: Chapters 15-30) includes prebuilt PowerPoint® lectures, a database of animations, video clips, and digital an from the text as well as editable eleClronic files of the fnstnu;l-or~s Solutions iHrwual. Also included is the easy-LOuse test generalor ExamView, which features all the questions from the printed Test Bank in an editable formal.
TRANSPARENCY ACETATES Each volume contains approximately 100 transparency acetates featuring art from the text. Volume I contains Chapters 1 through 14, and Volume 2 contains Chapters 15 through 30.
1MUOd3$...
WebAssign
xx
Preface
Assessment and Course Preparation Resources: A number of the resources listed belmv will help assist with your assessment and preparation processes, and are available to qualified adopters. Please contact your local Cengage • Brooks/Cole sales representative {-or details. Ancillaries offered in nl.'o volumes are split as follows: Volume I contains Chapters '1 through 14, and Volume 2 contains Chapters 15 through 30. INSTRUCTOR'S SOLUTIONS MANUAL by Charles Teague and Jerry S. Faughn. Available in t\vo volumes, the Tnstnl-cLor's Soltttions Nlanual consists of complete solu-
tions to all th~ problems, multiple-choice questions, and conceptual questions in the text, and full answers with explanations to the Quick Quizzes. An editable version of the complete instructor's solutions is also available electronically on the PownLeclure CD-R01\1.
PRINTED TEST BANK by Ed Oberhofer. This test bank contains approximately 1 750 multiple-choice problems and questions. Answers are provided in a separate key. The tcst bank is provided in print form (in two volumes) for the instructor v~'ho does not have access to a computer, and instructors may duplicate pages for distribution to students. These questions are also available on the PowerLecture CD-ROJ."v1 as either editable \Vord@ files (with complete answers and solutions) or via the ExmnView test software. WEBCT AND BLACKBOARD CONTENT For users of either course management system, we provide our test bank questions in proper \VebCT and Blackboard content format for easy upload into your online course. INSTRUCTOR'S COMPANION WEB SITE Consult the instructor's Web site at www, serwayphysicsocom for additional Quick Quiz questions, a problem correlation guide, images from the text, and sample PowerPoint® lectures. Instructors adopting the eighth edition of College Physics may download these materials after securing the appropriate pass\vord from thei r local Brooks/Cole sales representative.
Student Resources Brooks/Cole offers several items to supplement and enhance the classroom experience. These ancillaries allo\'..' instructors to customize the textbook to their student.s' needs and to their own style of instruction. One or more of the follm.. · ing ancillaries may be shrink-wrapped with the text at a reduced price: STUDENT SOLUTIONS MANUAL/STUDY GUIDE by John R. Gordon, Charles Teague, and Raymond A. Ser\vay. Now offered in two volumes, the Student Solutions iWanual/SLudy Guide features detailed solutions to approximatel)"' 12 problems per chapter. Boxed numbers identify those problems in the textbook for \·~. hich complete solutions arc found in the manual. The manual also features a skills section, important notes from key sections of the text, and a list of important equations and concepts. Volume j contains Chapters 1 through 14, and Volume 2 contains Chapters 15 through 30. PHYSICS LABORATORY MANUAL, 3rd edition, by David Loyd. The Physics Labo-ralory Manual supplements the learning of basic physical principles while imroduc-
ing laboratory procedures and equipment. Each chapter of the manual includes a prelaboratory assignment, objectives, an equipment list, the theory behind the experiment, experimental procedures, graphs, and questions. A laboratory report is provided for each experiment so that the student can record data, calculations, and experimental results. To develop their ability to judge the validity of their results, students are encouraged to apply statistical analysis to t.heir data. A complete instructor's manual is also available to facilitate lise of this manual.
Preface
ACKNOWLEDGMENTS In preparing the eighlh edition of Lhis textbook, \\,'C have been guided by the expertise or many people \\'ho have reviewed manuscript or provided prerevisi?1l suggestions, \fl"Te wish to acknowledge the following reviewers and express our SIncere appreciation for their helpful suggestions, criticism, and encouragement.
Eighth edition reviewers: Gary Blanpied, University of South Carolina
Gardner Friedlander, University School (!! Alilwalt/we Dolores Gende, Parish l~j}iscol)al School Grant v"'/' Hart, Brigham Young UniverJity
Joey Huston, lvlichigan Stale Universit)' MarkJames. Northern AT/>..ona UniveTsil.y TCl"uki Kamon, Texas A & M Universit)'
Mark Lucas, Ohio Ul1iversily Mark E. Mauson,Jmnes lV1ad'ison Universit)l
J.
Patrick Policy, Belo;t College Eugene Surdutovich, l¥a)me State Univprsily
Marshall Thomsen, Eastern Michigan University
David P. Young, Louisiana ,)'Iale Universit)'
Collt'gp Physics, eighth edition, was carefully checked for accuracy by Philip '1"Adams, Louisiana State University; Grant ''''. Hart, Brigham Young Univenity; Thomas K. Hemmick, Stony Brook University; Ed Obcrhorer, [~ake Sumter Comm:u.nilJ College; M. Anthony Reynolds, Em.b~)'-Riddle Aeronautical University; Eugene SurduLOvich, Woyne State Univenity; and David P. Young, Louisiana Slale University. Although responsibility for any remaining errors rests with us, we thank them for their dedication and vigilance. Prior to our \\lork on this revision, we conducted a survey of' professors to gauge hO\\l they used student assessment in [heir classroom. We were overwhelmed not only by the number of professors who wantcd to take part in Lhe survey, but also by their insightful comments. Their feedback and suggestions helped shape the revision of the end-o(:'chapter questions and problems in this edition, and so we \.,rould like to thank the survey participants: Eli<;e Adamson, IVft)'tmul na/Jlist U1Iiversit)'; Rhett Allain, SOlllhwstcl'II LouisiftJla Univt'rsity; ~'lichael Andel"SOn, Univenil)' of California, SOl/ Diego;.Iames Andrcws, YoungslowJI Stall' Uuiversil)'; Bradlcy Antanaitis, Lafa)"elte Colll'ge:. Robcn Astalos, Adams Slate College; Charles Atchley, SrU/1! VallI!)' Cmmllll1lity College; Kandiah Balachandran, Kalrwlfl1.o0 VaLft, COl/lI/nwily Collrge; Colley Baldwin, Sl. Jolin's Ulliversit)'; f\hhmoud Bashara1, Houston COJflmunit)' ColIl'ge ,Vorlhmst; Celso Batalha, Evrrgrten Valliy College; Natatie Bilfalha, SanJosrStale VI/ivan/y; Charles Benesh. Wesll)'fl1i College; Raymond Benge, Taruml County College i\'orlheast; Lce Benjamin. fHarywoorf UlIiVer,11tJ; Edgar Bering, U1/ivenil)' of 1-IOIuton; Ron Bingaman, Indiana UllillCrsity !':asl;jennifer Birriel, /'vlo,1,head Slate Unillersit)l; Earl Blodg<.;lt, Univrnit)' oj "':isconsin-Riller Falls; Ant hony Blose, UniwI',I;tJ of Norlll Alabmun; Jeff Bodan, Chi/Jola College; Ken Bolland, Tile Ohio Slate Universit),; Roscoe Bowen, Citm/Jbdlsville U1/ivrrsity; Shane Brower, Grove City College; Ch,lrlCS Burkhardt, St. Louis Connmmily Colli'ge; Richard Cardenas, ."''1. MaI"'Y\ Uni<wrsity; Kelly Casey, rahill/it l'alley COlJwlll1lity Collegt;, ClifT Castlc, jefferson Coflegl': ~'1arco Cavaglia, University of ll1ississippi; Eugene Chaffin, Bob jones lIniversit)'; Chan~ Chang, Drexel Ul/iversifJ; .ling Chang, Culver-Stockton Colll'l,re; Hirendra Chaucljee, Camdell Cou7I/y Cuff.ege; SOlllllitra Chauopadh>'ay, Ccrngia I-Jighlal/(Is Collegf; Anastasia Chopclas, UniversilJ oj Washhlgtrm; Krishna Chowdar)', Bm-hull UllivasilJ; Kelvin Chu, University (1 Vermonl; Alice D, Churukian, Conrrmlia ColIl'ge; David Cinahro, l\'aY/II' State University; Gar)' Copeland, Old DOJllhlioTl University; Sean Cordry, Norillwestl'rlt Colfl'ge oj Iowa; Viclur Coronel, Sl1;...ry Rorkland Communit)' College; Douglas COrleville, Iowa HleS/I'17l Communil)' College; Randy Criss, Saint Leo UnilJl'rsily;.John Cr1ltchfield. l?ocliili/{IUllTl COl/llIlunily College; Oanielle [)alafave, College 0J:'VelJljene')'; Lawrcncc D<'t}', Ulica Collel"re;Joe DeLeonl.:. Corning CO/Jlllllmil)1 Collegf'; Ton)' DeLia, ,\lor/Ii Florida CoUmlllnil)' College; Du>gu Delilidioglu. !lolJ Names Universily; Sandra Desmarais. !Jaytona Beach (;01ll11l1l1lity College; Gregory Dolise, J-!arnsl!uIg Area Community Coffe{!;e; Duane Doyle, Arkonms State Ulliversity-l'I/ewfml't; J,jacksonville U'lliversitJ; Ted Eltzroth, Elgin Commul/it)' Coffege; i\·lanin Epstci II, California Slate U1!iversit)', Los Angelr's: Florence [top, Virginia Slate UnivI'T.I'il)'; ~'I ike [ydenbeq~, /liew iVII'xi,/) Stall' Universit), at Alamogordo; Davene Eyrcs, North Sealtle Community College; BreI! Fadem, MllhLenbF.rg Collegl'". Greg: Falabella, Wagner College; 1\'lichaeJ Falcski, Delta Co/ll'ge; .Jacqueline Faridani. Shiptll'1lSburg UniT'ersily; Abu Fasihuddin, Vnivl"f.\ity oleO/mediCl/t; Scott Fedorchak. CfllJI/Jb(,1t Uniwniry; Frank Ferronc, Drexl'/
xxi
xxii
Preface
UnivN.I'II)'; l.~arland Fish, KaLall!azoo Valley COl/wnwily Collf'gt:; KCIll Fisher, CollIlI/bus S/nfe Communily Col~ lege; Allen I-lora, !/ood Collegr.Jlllles Friedrichsen. Austill Community Collegf'", Cylllhi,l Galo\'ich. UniT'ersify U/Ai01 -lhern Cohmu/o: Ticu G,lmalie, ArJwllSu.\· Sif/le Ulliversil)'-LRAF!~: And)' Gavrin, Indiana UniverSil)" Purdue Univn\'ity hl([irtnapoli~; Michael Giangnlildc, OaJdulld Com1llullily Collegl': Wdls Gordon, Ohio Fa/lfy UlIiVl'r.litJ; Charlt:S Grabowski, CaITol! CommulIit)' College; Roben Cramer, [.aile (;il} Commullity ColllKI'", Janusl Grcbowicz, Ulltun,lily oj Ilou~lulI-J)uwnluwn;~:loITis Greellll'ood, Sa 1/ jarinlo College Cen/ml; David Grah, Cal/lloll. UTlitlersily; Fred Crossc, SuSqlll'lWlI11fl Univnsil)': 1-Iil)"; Charles Kao, Colllllllnu Siall' UlIiTJenit}·; David IZardel is, Collegl' of Eastf'17I Utah: Edw;lrd Kearlls, 1J()~·tOll UniIler~'i/y: Roben Keeler, I.alle SUIIl/e,: Communil)' College:. \lamadoll Kcita, SllI'ridal/ College, Cillellr' Calli/JUs; Luke Keller, Jllwm Co{{rge; Andrew Kerr. Univl'I:~il)' ofFi1ldltty; Kinne)' Kim, ;Vorlh Caroli'na (,"1''/Ilral Ulli7.lel".l'il)'; Kevill Kimberlin, Bradl!')' Ulliva~ily; Ceorge KnOll, COSIlJ1/1Ie~ Hiuer Collegr; Corinne Krauss, VickillsrJ1} Slate Ullivcr~it)~ Chrislopher Kulp, East.em Kel/llIcky (.Jlliversil): A. Anil KUlnar, Pmine VincA & NT UniversilY;Jo~cphinc Lame!a, :'rliddll'sex COUIII)' Cotll'ge; Eric Lane. Vllivenify ofTmnessee; Grt:gory Lapicki. Easl Camlina UrlilJt'rsiIJ~ Byron Ldes. Snead Stale Communi/)' College; David Lieberman, Quee1/SIJOl"Ollgh Com1ll((1Ii(., Collrge; Marilyll List\'an, NornuJlulalt, (;ommU"flil)' Cullege; Rafael Lopcz-I\:lobilia, UI/hm".';it)" oj'Te.wH al Still Anlollio; Jose Lo:rano, Brtu//ry Uniunsit),: ;vlark Luca1>, Ohio Univn:~il)~ Ntungwa \'1aaslia, Coastal (;I'(Jlgill CO/l/mlmit) College; Keil h l'vlacAdam, (.illivNSit), of KlnilllCky; Kevin Mack;ly, G-,-{)ve Cil)' Coll('ge; Sieve Maier, :Vorlhw/'stenl Vldah01lla Slall! UnitlersifJ; Helen Major, LiT/mln UnivltrsilT Igor Makas)"uk, Srm FruJlciscoSlafe Universil); Cary Malek,.Io!tns()!1 COIlIII)' COlJlmulIity Colligr; Frallk \'hlllll, 1~/IIlIwnllrl Colleg"e; Ronald r-hrks, Norlh Gr~eJl ville Uuiversit),; Perr}' Mason, I-libbocli Chri~tian Ullivenity; Mark J\·laLlson,.Inme~ A'Tadiwn Unit'l'I"sit)~.Jllhn McClain, PaJlola College;.Jamcs McDonald, Universily oj'Hm·!fonl; Linda MeDon,lld, Xorlli Park Universil)'; Ralph V. \'lcCrcw, IJrooll/e Community College; Janet McLany-Schroeder, Cerritos College; Rahul t\ktHa. Univet:~il)' of r:enlml Arkamas; Mike Mikhaiel, /'assoir Coul/l)' Commu.nity Collegr, Lanc)" Mills, Coflt'gl' of Chmh:.I·lon:.John Millon, J)ePaul UnivN.I'il)'; Stephen Minnick, Knit Slalr Uniul'T,I'ily, 'ttlScamwas Call/pus: Dominick 1\'lisciascio, ,\.ll'rcerCoulIl.l Cmmntlllit)' College; Anhur Millin, Univnsit), I!fMassachu.leIIs l..iJwell; Glenn Modrak, Broome Commu1Iil)' Collf'gp; Toby Moleski, :\Il1slwgoll Com!/ll/lIily College: C. Da\'id ~'Iollre, Reill!lardt College:; Hassan Moore,'/ollll.\"IJ1) C. Smilh UuiveTsit_~; David \'Ioran, Brf')'erSlale Univl'YsitJ~ Laurie Morgus, Drew UniTlasily; David Murdock, Ten1l('ssee T('dmological Universily; Dennis Nemcschansky, Urlivasity of SOI/.t!lI'm California; Bob t\erhull, Univenity r1 South Carolina SU/l/II'r. Lorin ~euldd. Frt'S11O I'acijir Ul!iun:~i/)~ K. \''11. :"Jicholsotl, Gel/lml Alaumlla Commullil)' Collrge; Ch;lrles .\Iiekles, Un;vrNily oJ.'Uas~(ICJIll.\·l'!ts Darll/wtil!l; Paul Nienabt:r, SOl11l MalJ's UnivnsilJ o/IHi1llie.\"ola: Ralph Oberly. i\-lan!lall Univer· sity; Tt:tTy F. O'Ow)'cr, /I;"l-SS(lJ( COIIIII!/Hiil)' CoUegf'", Don Olivc, Ganlner-1Vebh UlIi"rrsil)~ Jacqueline Ornland, Nor(hf'r!I Siale Uniuen'il)'; Paige OU/.tS, Lfi/lder (}nivenily; Vaheribh:li Patel. Tomball Colll'ge; I\Uoy Palnaik, H'lqlV~ COII/muni/)' Collr'w; Philip PaLlerson, SOll/lIeni Pol)'ledmir Slale U1livl'TsilJ; James l\lZlltl, Pfeij!r,. UnivNsit)'; Chuck Pearson, Shorler Collr'gf'", Todd Pedlar, Lllllin College; Amhon)" Peel", !Jel· aware "l'ecJmiml & Communily College; Frtederick Phelps, Cell/ral AlichigrlJl Uniuersil.V; Roben Philbin, Trinidad Sialejll"llior Colhge;.Joshlia Phiri, Floreno'-Dar/iliKtoll Tee/milllf Co/f.f'gr, ell Phung. Me/hodisl Coltl'ge; Alberto Pi n kas. IV'l'TojeTSeJ City Uuir!ersily; AI i Piran, Slephr/l F Austill Statr Universily; Marie Pllllllb,.Ial/les~ lown Communit)" Coflrgr; Dwight Portman, Miami Uniuer.l'il)' Mir/dletown; Rose Rakers, 'fri/liIJ Chris/iall Col!rgr; Periasamy Ramalingam, AlbfWY Slate Universil)': i\hrilyn Rands, 1.1II111"1'IICe 'Irdmologiml Ullivrr5ily; Tom Richardson, Marian College; Herbert Ringel, Borough of Manhalfall COll/munily COl/fIJI'", ,sal\'alore Rodano, Ilar/oni COII/llmni/y Co/lrw; John Rollino, Rutgers Ulliversil)'~IVewf{rk; Fernando ROIllt:roI~orja, I-Iolll/oll CO/l!1nunily ColIl'ge-Centm/; Michael Rulison, OgletllOr/Je Universil)'; Marylyn Russ, A'lm-)'gruue College; Craig Rutan, Sanliago CmlJoll Colll'ge; J)'otsna Sau, Delmvarr Tl'tJmiml & COrlunll1lily Collegl", Charlcs ,sawicki, Norlh /Jaliola Slate UnilJe1'Sity; Daniel SchoL\n, Kelll'1'ing Gotlrge oj Mediad IIrL5: Ancll-ia Schwanz, Qlli"llSigamonrl COlnllll/ni/) Collegr.. Da\'id Seely, Albio/l Colhgr; Ross St:lze, Pearl Hiul'Y Community Colle!:,re; Ban Sht:inberg; Pelcr Sheldon, IftmdollJh-Maron WOII/flIIS Co!ll'gr'; \VCIl ShCIl, COrl/muIlily Collegl' OJ'SOl/tllI'I'n ,,\icuada; :-\m,'ar Shiekli, Dine Colll'ge; Marilin Simon. AubllnJ UlIivl'fsily; Don Sparks. Pil'l"(f' ColIl'l!/': Philip Spickler, I1ridgrwalt:r (.'(Jllegr; Flelcha Sryglt::y, /.i/J~com/J Uuiwnity: SCOll Sleckenridter, IIlil/ois Collegf'; Donna Slokes, University oj HUllS/on; L<mrence StOlle, Dakota COllllfy Trdwjral Colll-ge; Yang Sun, Un;lIf'rsily of Notrl' Df/lllr; Gregor)' SllI·an, Rarilan \frtlll') C(!I1wlIlIlily College; Vahe Taloiall, MI. San Anlo1lio Collegt'; Alem 'Tcklu, Collew' oj' Charles/on: Paul TeSla, Tom/JkilB Corlland Commlln;f) (;ollf'ge:. t\·lichacl ThacksLOn, S(lIIllIernl'ol)'teclmi(' Slate UlIivenitJ; ;vlc1odyThomas, ,\'orlhll'est Arkall.1"fI~ Cmlll/nmify Collfgr. Chcng Ting, /-fouston Conmmni/.'r Co/teb'11~S(JIlIIu:tI\/; Donn To\'msend, Penn S/ille Shell/wgo; llerman Trivilino; Cajendra Tulsiall, Da)'lolla Bl'flch Community ColIl'ge; Reill Urilaltl, Rosltm Colll:gr; Daniel Van \Vingerdell, };·{/.slem Michig(/n UllilJersitJ; Asliok Vaseashla. !lIars/wI! Um'lJI,,..lil)'; Rohcrl Vaughn, Grace/and Univn..il}; Robt:rt Wanl1>ila, Sl/jJolk COllnt)" Commllnity College; Rohcn ",",ebb, TI'X(l-S A &' AI UlIiTwnily; Zudiac Webster, Colltlllfms Slafe UniTICTSity; Rrian Weiner, PI'1JIl Siale 011 Bois;Jack Wells, Thoma.l· Mon' Collef.{l'; Ronnic \\'hilener, Tn'-Coulll)' r:oJn1nllllil)' College; Tum \Vilhur, Aline A1'Iwdel ComII/linily Colll'gf'", Sam \Vilcy, California Stall' U11iT'l'rsily, /}omi1lgUf7. J-tiUs;.Iudith \.filliams, H'illium Pell1I ffni111''"-111)": Mark Willianls; Don \\iilliamsoll, Char/mn Sfnll' Cot!egr. Neal \\lilst:}'. ColIl'{JI' oJSoulhem Mm)'land:
Preface
Lowell Wood, Universily of Housloll;.Jainshi \,Vu; Pei Xiong-Skiba, Anslin Peay State L/lIiva.lily; r.,'ling Yin, Hennlicl Collegp; David Young, Louisiana Stale University; Douglas Young, A1erca University; T. \Valdck Zenia, Texas Christian UlI'iversit)'; Peizhen Zhao, Fdi.mn Conltll1lnity College; Sleven Zides, Wofford College; and U [rich Zurcher, Cleveland Stale Ulliversil)l.
Finally, we would like 1.0 thank the following people for their suggeslions and assist.ance during the preparation of earlier editions of this textbook: Gary B. Adams, Arizona Siale University; iVfarilYll Akins, Bmome Communil)1 College; Ricardo Alarcon. At'iz~ma Stale Univnsity; Albert:\ 1Iman, UJliV(~rsil)' ofLowrll;Johll Anderson, Univer.l-ity of PillsbUlgh; Lawrence Anderson-Huang, [j'Niversily of Toledo; Sllblnsh Antani, Edgewood Collep;e; Neil ,,,,r, Ashcrort, Comell Univasil)'; Charles R. Bacon, Fen'is Slale Universil); Dilip Balamorc, Nfl.I·.IWi Communily College; Ralph Barnett, Florissant \'allf'J Cmmnu.nity College; Lois Barrelt, Weslern Washington Uuiversit)'; Natalie llatalha, ,',an Jose Stale UniT!enity; Paul D. Beale, University of Col(Jmdo (jt Bo'ulder, Paul Bender, Washington Slale UnivfHi/J; David H. Bennull1, University o(Nevada rtl Rnw; Ken Bolland, The Ohio Slate Univel'sii),;JefTery Braun, Univenil) of Evmlsville; John Brennan, University of Central Norida; Michael Bretz, University of Michigan, Ann ATb01~ Michael E. Browne, Urliversity 0/ fdaho;.Joscph Cantazarile, Cypress College; Ronald V./. Canterna, Universily of H':)'Oming; Clinton M. Case, Weslern Nevada COlllumnily College; Neal M. Cason, Universit), (if l'/olre Oame, Kapila Clara Caswldi, Oakland University; Roger \V, Clapp, University of Soulh Florida; Giuseppe Colaccico, University t:!ISou{h Florida; Lall.ie F. Collins, Easl 'li:nllessee Slate University; Lawt-cncc B. Colrnau, University of Cfllijomia, Davis; Andrew Cornelius, Univel'sity (if!Vevada, Las Vegas; .Jorge Cossio, Miami Dade Community College; Terry T. Crow, klis.\-i,,·si/JI)i Statr: ColLege; Yesim Darici, Floridal7lterllatiolwl Universily; Stephen D. Davis, Unillersit)' ofArkansas al Little Rock;john DeFord, Univelsit) of Utah; Chris.J. De.r...1arco, jachson Communily College; Michael Dennin, Uni1Jersit), of Califomia, Imine; [\I. John DiNardo, Drexel University; Sleve Ellis, Universily of KP!lll1c1i~'1; Roberl.J. Endorf, University of Cincinnali; Stevc Ellis, Univnrit)' (if Kmturk)'; Hasan Fakhruddin, Bali Slale Uni-versit)'/Indimw .r1cadem)'; Paul Feldker, FlorlsSftnl \lalle)' Communily Co!Ll:ge; Leonard X. Finegold, Dn:xel Uniuersily; Emily Flynn; Lewis ford, Te:ms il &' M University; Tom French, iHontgomery COllnty Comuwnity Coll-ege; Albert Thomas Frommhold, jr., Aumt17l Uniupr.lily; Lothar Frommhnld, Univenity of Texas al Austin; Eric Ganz, Univenit)' of Minnesota; Teymoor Gedayloo, Calijomia Polytechnic Siale Unive-nil)'; Simon George, California Siale University, L(mg Beach; James R. Gorr, JJf/lla COlllJnun,itj' Co!l('ge; Yadin Y. Goldschmidl, Univen-it)' 0fPillsbU1gh;.John R. Gordoll,james Madison Uilivenit)'; George \Ill Greenlees, UniversitJ ol"Minnesola; vVlodzimierz Gurvn, Jj1{Jolduwe1/. 'vationnl Lnboratory; SIeve Hagen, Univenity ofFlmida; Raymond Hall, Calijornia Siate UllillC1~ril)', Frpsl/o; Patrick Hami 11. San jm-e Stall' UlIimrsil)';.Ioel Handley;James Harmon, Oklahoma Slate Universily; Grant '\-V. Hart, Brigham )'Ollllg Un/ver.. . ity; Jamcs L Heath, Austin Communil.., College; Grady Hendricks, Blinn Coll.ege; CbrislOpber Herbert, "'ew jPHey Cil)' University; Rhett Herman, Radford UnivPI~'1'it),; john Bo, S/.ate University of New Yorl! al Buffalo; Alehcy Holloway, University of Nebroslw at Omaha; Murshed Hossain, HOll/(111 Universil_y; Roben C. Hudson, ROfl7wke College; Joey Huston, Michigan Siale Universit); Fred Inman, A1anlwto Slate Univwr.l-ity; i\.'1ark .lames, Northern Aliwna Univenity; Ronald E. jodoin, Rochester 'nstilule of T'edmology; Randall Jones, Loyola College in Mm)'fand; Drasko Jm'anovic, Fermilalr, George V'... Kallawar, Texas A & M Univ('rsity; Joseph Keane, St. Thomas Aquinas College; Frank Kolp, Trenlon Stale Univasit); DOl'ina KoszLin, Univen-il_y of Missouri-Colu./l/bia; .loan P. S. Kowalski, G('()rge Mason Univenily; Ivan Kramer, Univel:'1'il)' 0/ Maryfand, Ballimore COUnl)'; Sol KJ"a,~ner, Universily of Chicago; Karl F. Kuhn, '~'astem Kentlicky Universit)'; David Lamp, Texas Tedt University; Harvey S. Leff, Califomia Slafe Polj"lecJmir University; Joel Levine, Omllge Coast College; Michael Lieber, University ofArliansas; Martha Lietz, ,Viles I'Vest /-ligh School;james Li nbald, Saddlebncl! COmtnum:ly College; Edwin 1..0; Uill Lochslet, Pennsylvania Slal/1 U71iVN.lilJ; Rafael Lopez-Mobilia, Universil)' of Texm al San Antouio; Michael LoPresto, flenryFonl Community Coll('gr; Bo L.Oll, Ferris Stale UniIJeI:~ity;.reffreyV. Mallow, Loyola Universil)' of Chicago; David j'vfarkowitz, Unhlcrsil)' 0fCml1leclicnl;.Ioe McCau1cy,jr., Universily of Houston; Sleven McCauley, California Slale Po(ylechnic Universil)', PO/l/ona; Ralph V. McGrew, ~roo1Jle Cmnmunil)' College: Bill F. Mellon, Universily ofNorlh Carolina al Charlotle;John A. Milsom, Universil)'/if :1rizo'lla; Monty Mob, lIumboldl Stale Universil); H. Kent Moore,jame.1 kladi,<;011 Ulliversil)';John Morack, UnivrTSily t:!( AlasJIfI, Fairbank... ; Steven Morris, Los Angeles Harbor Coflege; Charles W, Myles, Texas Tech Universily; Carl R. Na\'C, Georgia Siale UniIJersil)'; !\'Ianin Nikolo, SainI Louis UnivI'nit); Blaine Norum, Universily of \litgiuia; 1\:1. Eo. Oakes, Univer.l-ity of Trxas at /l-l/.slill; LewisJ. Oakland, University 0IMinnesol(/.; Ed Ohc.-ll0fer, Lake Sumler Community Collegl;, Lewis O'Kclly, Memj)his Stale University; David G. Onn, Utrivenit)' oj Delaware;j. Scolt Payson, Wa)lne Slate University; Chris Pearson, Univenit)" 0/ Mhhiga.n-Flinl; Alexey A. Peu"ov. i!-'ttYlle Slate Univenit)"; T. A. K. Pillai, Universil)' 0/ Wisl"Ousin, La Cros,"'r>; Lawrence S. Pinsky, U11ivenity oj [JOIlSton; \Villiam D. Ploughe, TIU' Ohio Slntp Universily; Patrick Polley, Beloit Collegr>; Brooke M. Pridmore, Clayloll State Universily; joseph Pries1, Mia'flli University; james Purcell, GeOl"{{ia Siale [Jniversity; "V. Sieve Quon, Feulura Coffege; Michael Ram, Stall! Uninasity OliVeT/} YOI'll at BuJ/alo; Kl-'J"l R_cihd, The Ohio Slnle UlIillersil)'; }.'I. AnLhony Reynolds, Embry-Riddle Aeronautical University; Barry Robertson, Queen's University; Virginia Roundy, Califinnia ,\'tate University, Flllh'rlon; Larry Rowan, UnivrrsiiJ ofNo'rtlt CaroliHa, CllllfJd /-lill; Dubravka Rupnik, I.ollisiana Slale UniV/:rsit)'; William R. Savage:, Universil)' of Iowa; Reinhard A, Schumacher, Carneg'; Donald D. Snyder, Indimw Univenit)" at SolttMend; George Strobel, Univenit)' of Cnll"/:;rin; Carey E. Strotl<Jeh, Virginia Stnle Universit)'; Thomas \V. Taylor, Clevdalld Stale UlIiversit~'1; Perry A. Tompkins, Samford
xxiii
xxiv
Preface
l..hriversit): L. L. Van Zandt. P/t1dllt' University; Howard G. Voss, Al'izol/a Slate UTliTlf.rsil),;James Wanliss, Embry-niddlR Aeronll/ltical University; Larry Weaver, Kansas Slale Ulliucrsit),; Donald II. \"'hite. Western Oregon State Collegt"; Bernard Whiting, Ulliversi(v ojFlon-da; George A. Williams, VlIivm·ity oj l./tllh;Jerry H. Wilson. Metropolitan Statp C.ollege; Roben M. Wood, UnivprsitJ' ojGporgin: and Clyde A. Zaidins, UlliversitJ ojColnmrlo af Df1Iver.
Gerd Konemeyer and Randall Jones contributed several end-oF-chapter problems. especially those or interest to the life sciences. Ed\'.:ard F. Redish of the Cniversity of Maryland graciously allowed us to list some of his problems rrom the Activity Based Physics Project. ',Ve are extremely grateful to the publishing team at the Brooks/Cole Publishing Company for their expertise and outstanding work in all aspects orthis project. In particular, we thank Ed Dodd, who tirelessly coordinated and directed our efforts in preparing the manuscript in its \'arious stages, and Sylvia Krick, who transmitted all the print ancillaries. Jane Sanders Miller, the photo researcher, did a great job finding phoLOs or physical phenomena, Sam Subity coordinated the media program for the text, and Rob Hugel helped translate our rough sketches into accurate, compelling art. Katherine '''Tilson of Lachina Publishing Services managed the difficult task of keeping produClion moving and on schedule. :\tfark Santee, Teri Hyde, and Chris Hall also made numerous valuable comribmions. Mark, the book's marketing manager, was a tireless advocate for the text. Tcri coordinated the entire production and manufacturing of the text, in all its various incarnations, from start to finish. Chris provided justtbe right amoum of guidance and vision throughout the project. \·Ve also thank David Harris, a great team builder and motivator with loads of enthusiasm and an infectious sense of humor. Finally, we are deeply indebted to our wives and children [or their love, support, and langterm sacrifices.
Raymond A. Serway Sl. Petersburg, Florida Chris Yuille Daytona Beach, Florida
ENGAGINii.ApPLICATIONS
:\ll1lOugh pll\sics is relc\-ant to so much in Ollr modem lives. it Ill.n not be obvious to sluclclllS in an imroducLOry course. In this eighth t'diLion of Colhgl' Physirs, h(' continue a d{'~ilin feature beg-ull in the ~t'vt'nlh edition. This feature makes the relevance of physics LO everrday life more obvious by pointing oul specific applications in tILe form of a mal'ginal notc. Some of these .-.pplicalions pertain to lhe life sciencc:-. and are marked with Ihe DNA icon The lisl below is not illlended to be a complete listing of all the applications of the principles of physics found in this textbook. Man~' other applic:niolls are LO be found within the text and especially in the worked l'xamples, conn:pwal qlle~lions, and cnd-ol~chaplerproblems .
II.
Chapter 3 The longjump. p. 66 Chapter 4 Se.-'al belts. p. HI. Iidicopler Hight. p. 93 Colliding \~hiclt:s. p. 94 Sl..ydi\ing, p. lOR
II
a III
II II
• "
II
Chapter 5 Fl:lgellar llloWllIelll: bioluminescence, p. 142 I"-I('roid impac!' p. 142 Dict versus ("Xl'ITi~(' in weighl-Ios~ programs, p.147
Chapter 6 Boxing and brain ir~jury, p. 11';3 11~jllry to passcng:c..:rs in car collisions. p. Ili5 (;Iaucoma testillg:, p. 169 Professor Goclcl:rrd \'Jas righl all along: Rockeb \\'clrk in spacer p. 178 IVlullislage rod..l'I:.. p. 179
pp.330-331
II III
II III •
198
ArtifIcial gr;wil)'. p. 203 I\anked madw:ns. p. 205 Wh), is the Sun hot? p. 213 Geosynchronolls orbit and telecommunications salellitcs. p. 21 i
111 III
\\'ol'king oITbn;akra~t. p. 354 Physiolllg-y of cxercise, p. ~-\54 Sea breaesand Ihcl'lllOlb. p. 355 Home il1SlIlatioll, pp. :-\liH-369 Staying warlll in the arnie, p. 370 Cooling automobile clIgilll·S. p. 37\ Algal hlooms in ponds ancllakes. p. 371 Hod)' tClltperalllrc. p. :n2 Light-culored Sllllllllt:l dothing, p. :{7:l Thennog:rapln·. p. 3i~ Radi;lIioll IhermOllletl'rs for mcasuring boch lemperatllre. p. 3;3 Thermal radiation :tnd Ili~ht vision, p. 374 Thermo:-. huttles. p. 37:, Global w,lIllling and g-n:t:nhollse gases. p. 375
"
Refriger;uurs and he.-·;u I}umps, PI). 102-·103 ~Peqwlllal lIlolion~ machines, p. 409 The di I"('(tion of tillle. p. ·112 Humall metabolism. pp. 113-416 Fighting- fat, p. 41f}
Chapter 13
Archery. p. '129 Pi.<;toll~ and drive whcl'h, p. '133 Arch structUl"l'S in huildings, p. 275 BUllgee jumping. p. 4~\5 Sno\'Jshoes, I). '277 Pendulum clocks. p. ·110 Bt:d ofnaib Irick p. 278 Use orpt:ndululll in pro~pcctlng, p. 'HO Ilydraulic lifts. p. '281 Shock ahsorhers, p. H~ Building lhe p)'ramids. pp. ~H~-283 Bas.'>g-llilar string.... p. +17 (I t\'lcasuring blo(ld prcssurt:, p. 28~-284 Ballpoilll pt:nl>. p. 284 Chapter 14 III Swim bladder~ in li~h. p. 2flfi (I ~ledical u'i('s of ulli"a.'>ouml. p. 460 II Ccrcbrospin.d fluid. p. 286 (I Cavilroll ultr;lsnnic "urg-ical aspir..lIor, Testing ~our car'~ antirrcel.e. p. 286 p. 'IIi I Checking lhe l),ltlt:r~, chaq{c. p. 287 Uhrasonic ranging unil fOl' cameras, p. 461 Flighl of a gellf ball. p. 296 The sounds heard dllrill~ a SLorm, ~Atomizers~ in perfume boules and paim pp. ,!l,2-i6;'\ sprayers. p. 29i III OSHA noise Ic\d re~ul,lIil)ns, p. 466 III Vascular nmtcl' ;tllel ancury"rns, p. 297 Sonic hoon}'•. p. 473 Lift on aircran willl{s. p. ~!l7 C:onnc<:ling: youI' stl'l'l'Il,"'peakers, p. 474 Sailing up\\'ind, p. 298 TlInirlp;;1 mllskal ill~lnlllJClIl, p. ·177 I-lome plumbing, p. 299 Gllilal' fllmbmcl1lah, p. 'li7 Rockel engines. p. 299 Shatlt:ring goblet s \\'11 h 1he voice, p. 480 fIJI Air sac sUl'face lension. p. 101 SU'lIClur;t1 rcsonancl' in hridges and DClcrgents and \\',Herproorill~ag-cnts. p. :\0:-\ buildings, p. 480
Chapter 9
• "
Oscillalions in a h"rbor, p. 482 Why arc instrullwnts \\'armed IIp? p. 482 I low do bugles work? p. 482 Using beats 10 tUlle a musical instrument. p.485 'o\'hy does the I)rofcssor sound like Donald Duck? p ... Hi The ear, pp. 487-48~1 eochlt:ar implaIlL'i. r). 489
Chapter 15 Mcasuring atmosphcric e1cctric fields. p. ;')12 Liglllning rods. p. f)14 Drin.'r safety during electrical Simms, p. 515
Chapter 11
Chapter 12
Chapter 8 1\ weighted (CIrcaI'm. p. 237 nicrde gears. p. 241 Wanning up, p. 2·'4 Figlu'c skating:, p. 250 Acrial somersaults. p. 250 Rol
Chapter 10 Skin tempt:ralllre, p. :~~n Thermal expansion joints, p. 328 Pyrex glass, p. 329 Bimelallic strips and Ihl·rmost;lIs. Ril>ing sea len"ls. p. 3:':\ Bursting water pipe:. in \\'inlel', p. 334 Expansion and tcmpel'lIure. p. 344
Chapter 7 ESA laullch ~itl·~. p. 197 Phonograph records and compacl discs, p.
Turhuknt 111m' of blood. p. ;l06 EffcCt 01 o~lIlosison living' cells, p. 308 Kidnt:\ function and di;i1ysis. p. 309
Chapter 16 Autumobile baut:rics, p. 537 The clectrostalic pn.;cipit
Chapter 17
III III " II
Dimming of82 Electrical acti\'in in lhe hear!. pp. 585-587 Electrocardiograms. p. 5 5 Cardiac pacemal..t:f~. p. 586 Implanted cardio\'crter defibrillaLOI's. p. 586
Chapter 18
II II
Christmas lights ill series, p. 596 Circuit breakers. p. 600 Tlll-ee-way liglubllibs. p. 601 Tillied windshield wipers, p. 608 Ibclcrial growlh. p. 608 Road\\'ay nashers, p. 608 Fuses and circuit bl"e:lkers. p. 612 Third wire on consumer appliances, p. 612 ConduClion of electrical signal:. hy neurons. pp.613-6E>
Chapter 19 Du~ting for Ilngnprims. p. 62H MagneLic bacleri.l. p. 629 I.abeling airport rUIlWd)·S. p. 629 Compasses down III1c1er, p. 6~O I.Otld...pcakcr op<.Tation, p. 6::\4 Electromagnetic pumps for artificial hcarL;" and kidneH. p. 635 I.ighlning SLri1..c~. p. 635 Electric moton.I). 638 r-.I;ISS SPCClHlIllCll'l'S, p. 641
Chapter 20 Ground fault interrupters, p. 6GS Electric guitar pickllp:', p. G69
xxv
xxvi
Apnea
List of Active Figures motli")f.~, p,
61i9
Concave vs. comcx, p. 7fi6 Rl;versible wa....es, p. ;lili Unde,"water vision. p. 770 Vision and di.. . ing Ill:ll>ks. p. 776
Space catapult, p. 671 Magnetic lapc recorders, p. 67.::; Altcrnalillg-currcnt g(:ncralOrs, p. 676 DircCI-currCItl generators, p. 67; Motol"S, p. (i7~J
Chapter 24 A smoky Young's experiment.
Chapter 21
Chapter 22 Secing the road un a raill)' night. p. 734 Red C\CS in l1ash phoLOgraphs. p. 735 The color.. of waleI' ripples al sunset, p. i35 Double imagt·~. p. 73S Refr.lClion 01 laser light in a digital video di"k (DVD). p. 741 Idelllifving g~\1oCS with" spectrometer, p. 7'13 Suhm<\rinc periscopes, !l. 749 Fiber oplic~ ill medical diagnosis and ...urgery. p. 750 Fiber optics ill lelecollillmnici-ltiollS. p. 750 Oe..ign orall uptical fiher, p. 751
III
The camera, pp. 823-W,N Thl: e~'e, pp. 82'1-829 USillg optical lenses to correCl fiJr defects, p.826 Pre~cribiJlg a correcl iH: lens for;t farsighted patienl" pp. 827-R2H A cOITecti\,(' It:ns for' IIcarsighlcdncss, p. 828 Vision oflht: ill\'isiblt' man, p. 828 Cll's eyes, p. H36
Chapter 26 Fa~tel" clock-.
I{,r rearvicw mirrors,
in a "mile-high city." p. 86::>
Chapter 27
p. i61
SIal'
Ilithionisl's Lric.:k, p. 762
Chapter 28 Discmcl-y of heliulll, p. 893 The,'mill or speclral, II. fl93 Aurora~, p. 894 La:oocr technolog~. p. !I08
794
Chapter 25
Chapter 23 sl;lIing~
p.
Television ..ignal inlel'ference, p. 794 Checking lor irnperlcctiol1s ill optical lenses, p. 798 The phy10ics of CDs and DVDs. p. 800 Diffraction 01 sound \\a\·es. p. RO-l Prism \'5. grating, p. R()6 Rainhows fmm a CD. p" H07 Tracking infonnalion 01l.1 CO, p. H07 Pobrizing microwavcs. p. 810 I'ol:lroid slinglasses. p. 812 Finding lhe cOllcentr.ltions of solutions by means oflhcir oplical aCli\·it}'. p. 813 Liquid crpOlal displap (LCOs). p. 813
EI«:Clric fields and c;lllu:r freMJIINU. p. 699 Shilling phase til deli\cr more power. p. 707 Tuning your radio, p. i08 i'lt-tal deteclors althe C'tlUnhousc, p. 709 Lung-dista'ICt· c1(:t:tric I}()\\cr I rallMllission, p.7l1 Radio-\\-a\{" Iransmissiun. p. 714 Solar system dust, p. ; 17 A, hal tin mnf (sohll"powered hOllies), p. 718 Tht: sun alld thc c\'olUlion uf the c~·c, p. 72"2
Day and night
I'hOLOCdls. p. H75 Using x-rap to stll(ly lhe wurk oflllastel" pailllcr~. p. 876 Electron microscope... p. SA2 X-nl)' microscope~. p. H83
Chapter 29 Hinding lI11clcons anrl electrullS. p. 917 Energy and half-life. p. 921 Carbon dating, p. 924 Smuke dell:ClOl"s. p. 925 Radon pollution, p. 92S :l.ledical applicalion .. llf mdiatioll.
pp.929-931
11.1 I/J
Occupational r.ldiation exposure limiu. p.
I
R:ldioaClivc Il"aCCI'1o in Illedicillt'. p. 930 :l.lagnclic resonance imaging (MRI), p. 931
930 Irradiation 01 food 'll1d medical cquipment,
p.930
Chapter 30 UlIStabk products, p. 938 Nuclear rcaClor de~igll, p. 940 Fusion reactors. p. 911 PClsitron emission lClllU)graphy (1'£'1' scanlling). p. 94:) Ikeaking l:ol1serV(ltioll laws. p. 949 Conser\'ation ol"nlesullllllmber. p. 95J
colors, p. 871
LIST OF ACTIVE FIGURES
Chapter 17 AClive Figures 17,4 and 17.9
Chapter 1
AClhc Figtlres 1.6and 1.7
Chapter 2
Acti\'c Figures 2.2, 2.12,
Chapter 3
AClive Figures 3.3. 3.14, and 3.15
Chapter 19 AClh'c Figures 19.2, 19.17, 19.19, 19.20, and 19.23
Chapter 4
AClive FiKures 4.6. 4.18. and 4.19
Chapter 5
AClin' Figures 5.5.5.'5,5.20, and 5.29
Chapter 20 Active Figun::s 20.4, 20.13, 20.20, 20.22, 20.27. and 20.28
Chapter 13 Active Figurcs 13.1, 13.8, 13.12,13.13. 1~.15, I~U6, 13.19,13.24.13.26,13.32,13.33.13.34, and 13.3.'1
Chapter 27 Active figures 27.2, 27.3, alld 27.4
Chapter 14 t\cli\'c Figures 1'1.8, 14.J0. 14.18, and 14.25 Chapter 15 I\ctivc Figures 15.6.15.11, IFJ.16. 1!:i.2 I. and 15.28 Chapter 16 ACtive Figures 16.7, 16.18. and 16.20
2!U~.
and 2!U5
Chapter 28 Active Figul"cs 28.7. 28.8. and 28.17 Chapter 29 Active Figurcs 29.1, 20.6, :ancl29.7 Chapter 30 Active Figmcs 30.2 am130.S
2~.25
TO THE STUDENT
As a student, it's important that you understand haw to use this book most effectively and hm'\-' best to go about learning physics. Scanning through the pref· ace will acquaint you with the various features available, both in the book and online. Awareness of your educational resources and how to use them is essential. Although physics is challenging, it can be mastered ,... ith the correct approach.
HOW TO STUDY Students often ask how best to study physics and prepare for examinations. There is no simple answer to this question, but we'd like to offer some suggestions based on our own experiences in learning and teaching over the years. First and foremost, maintain a positive at.titude tmvard t.he subject matter. Like learning a language, physics takes time. Those ".. ho keep applying themselves on a daily basis can expect to reach underst.anding and succeed in the course. Keep in mind that physics is the most fundamental of all natural sciences. Other science courses that follow ,.. . ilI use the same physical principles, so it is important that you understand and are able to apply the various concepts and theories discussed in the text. They're relevant!
CONCEPTS AND PRINCIPLES Students often try to do their home\vork without Erst studying t.he basic concepts. It is essential that you understand the basic concepts and principles bif01-e attempting to solve assigned problems. YOll can best accomplish this goal by carefully reading the textbook before you attend your lecture on the covered material. When reading the t.ext, you should jot down those points that are not clear to you. Also be sure to make a diligent attempt at ans\vering the questions in the Quick Quizzes as you come to them in your reading. \'\1e have worked hard to prepare questions that help you judge for yourself how well you understand the mat.erial. Pay careful attention to the many Tips throughoullhe text. They will help you avoid misconceptions, mistakes, and misunderstandings as well as maximize the efficiency of your time by minimizing adventures along fruitless paths. During class, take careful notes and ask questions about those ideas that are unclear to you. Keep in mind thal few people are able to absorb the full meaning of scientific material after only one reading. Your lectures and laboratory "\lark supplement your textbook and should clarify some of the more difficullmaterial. You should minimize rate memorization of material. Successful memorization of passages from the text, equations, and derivations does not necessarily indicate that you understand the fundamental principles. Vour understanding will be enhanced through a combinarion of efficient study habits, discussions 'with other students and with instructors, and your ability 1.0 solve the problems presented in the textbook. Ask questions whenever you think clarification of a concept is necessary.
STUDY SCHEDULE It is important for you to set up a regular study schedule, preferably a daily one. Make sure you read the syllabus for the course and adhere to the schedule set by your inst.ructor. As a general rule, you should devote about lWO hours of study time for everyone hour you are in class. Tfyou are having trouble with the course, seek the advice of the instructor or other students who have taken the course. YOll
xxvii
xxviii
Preface
may find it necessary to seek further instruction from experienced students. Verv often, instrucLOrs offer review sessions in addition to regular class periods. It i~ imponantthat you avoid the practice of delaying study until a day or two before an exam. One hour of study a day for 14 days is far marc effective than 14 hours the day before the exam. HCramming" usually produces disastrous results, especially in science. Rather than undertake an all-night study session immediately before al~ exam, briefly revic\'.,· the basjc concepts and equations and get a good night's resl. If you think you need additional help in understanding the concepts, in preparing for exams, or in problem solving, we suggest yOll acquire a copy of the Student Solutions !\;!rwual/Sludy Guide that accompanies t.his textbook; this Illanual should be available at your college bookstore.
USE THE FEATURES You should make full use of the various features of the text discussed in the preface. For example, marginal notes are useful for locating and describing important equations and concepts, and boldfaced type indicates important statements and defmitions. Many useful tables arc contained in the appendices, bUl most tables arc incorporated in the text where they are most often referenced. Appendix A is a con\'enient revie\'.' of mathematical techn iqucs. Answers to all Quick Quizzes and Example Questions, as well as odd-numbered multiple-choice questions, conceptual questions, and problems, are gi.. .en at the end of the textbook. Answers to selected end~of-chapter problems are provided in the Student Solutions NJanual/Slud)' Guide. Problem-Solving Strategies included in selected chapters throughout the text give yOll additional information about how you should solve problems. The contents provides an o\'ervie\v of the entire text, and the index enables you to locate specific material quickly. Footnotes sometimes are used to supplement the text or to cite other references on the subject discllssed. After ,-eading a chapter, you should be able to def-Ine ~lllY new quantities introduced in I.hat. chapter and to discuss the principles and assumptions Llsed to arrive at certain key relations. The chapter summaries and the revic\',,' sections of the Student Solutions Nlan'Ual/Stud)' Guide should help you in this regard. In some cases, it. may be necessary for you to refer to the index of the text. t.o locate certain topics. You should be able to correctly associate \vith each physical quantit.y the symbol used to represent that quantity and the unit in . vhich the quantity is specified. Further, you should be able to express each imponant relation in a concise and accurate prose st.atement.
PROBLEM SOLVING R. P. Feynman, Nobel laureate in physics, once said, "You do not know anything until you ha\'e practiced." In keeping with this Sl.aternent, \ve strongly lldvise that you develop the skills necessary to solve a wide range of problems. Your ability to solve problems will be one of the main tests of your knm.. . ledge of physics, so you should try to solve as many problems as possible. It is essential that. you understand basic concept.s and principles before attempting to solve problems. It is good practice to try to find aJternate solutions to the same problem. For example, you can solve problems in mechanics using Newton's laws, but very often an alternate method t.hat. drmvs on energy considerations is more direct. YOLl should not decei\'e yourself into thinking you understand a problem merely because you have seen it solved in class. YOli must be able to solve t.he problem and similar problems on your own. \,Vc have cast the examples in this book in a special, two-column format. t.o help you in this regard. After studying an example, see ifyOlI can cover up the right-hand side and do it yourself, using only the wriucn descriptions on the left as hint.s. Oncc you succeed at that, try solving the example cornplet.e1r on rour m·m. Finally, answer t.he question and solve the exercise. Once you ha\·e accomplished
Preface
all these steps, you will have a good mastery or the problem, its concepts, and mathematical technique. After studying all the Example Problems in this \·. . ay, you are ready to tackle the problems at the end of the chapter. or these, the Guided Problems provide another aid to learning how to solve some of the morc complex p,ooblems. The approach to solving problems should be carefully planned. A systematic plan is especially important when a problem involves several concepts. First, read the problem sc\'eral times until YOll arc confident you understand what is being asked. Look for an)' key words that will help yOll interpret the problem and perhaps allow you LO make certain assumptions. Your ability LO interpret a question properly is an integral pan of problem solving. Second, yOll should acquire the habit of writing down the information given in a problem and those quantities that need to be found; for example, yOll rnight construct a table listing both the quantities given and the quantities LO be found. This procedure is sometimes used in the worked examples of the textbook. After you have decided on thc method you think is appropriate for a given problem, proceed with your solution. Finally, check your results to see if they are reasonable and consistent with your initial \lnderstanding of' the problem. General prOblem-solving strategies of this type are included in the text and are highlighted with a surrounding boxo rfyou follow the steps of this procedure, you will find it easier to corne up with a solution and will also gain rnore from your efforts. Often, students fail to recognize the limitations of certain equations or physical laws in a particular situation. It is very important that you understand and remember the assumptions underlying' a particular theory or formalism. For example, certain equations in kinematics apply only to a pttniclc moving with constant acceleration. These equations £lrc not valid for describing motion whose acceleration is not constant, such as the motion of' an object connected to a spring or the motion of an object through a fluid.
EXPERIMENTS Because physics is a science based all experimental observations, we recommend that you supplement the text by performing various types of "hands-on" experiments, either at home or in the laboratory. For example, the common SlinkyTM lO)' is excellent for studyi ng travel i ng waves, a ba II swi ngi ng on the end of a long stri ng can be used to investigate pendulum motion, various masses attached to the end of a vertical spring or rubber band can be L1sed to determine their elastic nature, an old pair of Polaroid sunglasses and some discarded lenses and a magnifying glass are the componelus of various experiments in optics, and the approximate measure of the free-fall acceleration can be determined simply by measuring with a stopwatch the time it takcs for a ball to drop from a known height. The list of' such experiments is endless. \Vhen physical models are not available, be imaginative and try to develop models of rOll I' own.
An Invitatian to Physics It is our hope that you too will find physics an exciting and enjoyable experience and that you will profit from this experience, regardless of your chosen profession. V\'elcome to the exciting world of physics! 'fo SI'f:' the ~Vo)"Ul in a Gmin ofSmu/ And a /-leaven in (L Wild Flower, l/fJ/d infinity in the palm. ojyour luwd And I<.-'Ienzity in an how:
-William Blake, "Auguries of Innocence"
xxix
Welcome to your MeAT Test Preparation Guide The MeAT Test Preparation Guide makes your copy of Collegp PhJsics, eighth edition, the
1110st
comprehensive
~\'ICAT slUdy LOO! and classroom resource in introducLOry ph)'sics. The grid, which begins below and continues
on the nextlwa pages, ollllines ]2 concept-based study courses for the physics part ofyollr
~lCAT cxarn.
Csc it
to prepare for the MeAT, class tf'sIS, and YOllr hOIllf'\\'ork
Vectors
Force
Skill Objectives: To calculate distance, angles between "eetors, and magnillides.
Skill Objectives: To know and understand Xewton's laws and to calculate resultant forces and weight.
Review Plan: Distance and Angles: Chapter I, Sections 1.7, 1.8 Active Figure 1.6 Chapter Problems 35, 41, 44 Using Vectors: Chapter 3, Sections 3.1, 3.2 Quick Quizzes 3.1, 3.2 Examples 3.1-3.3 Active Figure 3.3 Chapler Problems 8, 13
Stonehenge, in southern England, was built thousands of years ago ta help keep track of the seasons. At dawn on the summer solstice the sun can be seen through these giant stone slobs.
1.1
Standards of Length, Mass, and Time
1.2
The Building Blocks of Motter
1.3
Dimensional Analysis
1.4
Uncertainty in Measurement and Significant Figures
1.5
Conversion of Units
1.6
Estimates and Order-ofMagnitude Calculations
1.7
Coordinate Systems
The goal of physics is to provide on understanding of the physical world by developing theo-
1.8
Trigonometry
ries based on experiments. A physical theory is essentially a guess, usually expressed math-
1.9
Problem~Solving Strategy
INTRODUCTION ematically, about how
given physical system works. The theory makes certain predictions about the physical system which can then be checked by observations and experiments. If the predictions turn out to correspond closely to what is actually observed, then the theory 0
stands, although it remains provisional. No theory to date has given a complete description of all physical phenomena, even within a given subdiscipline of physics. Every theory is a work in progress. The basic laws of physics involve such physical quantities as force, velocity, volume, and acceleration, all of which can be described in terms of more fundamental quantities. In mechanics, the three most fundamental quantities are length (L), mass (M), and time (T); all other physical quantities can be constructed from these three.
1.1
STANDARDS OF LENGTH, MASS, AND TIME
To communicate the result ora rneasuremCl1r ora certain physical quantity, a unit for the quantity must be defined. If our fundamental unit of length is defined to be 1.0 meter, for example, and someone familiar with our system or measurement reports that a wall is 2.0 meters high, we know thalthe height or the wall is l'..vice the fundamental unit of' length. Likcwise, if our fundamental unit of mass is defined as 1.0 kilogram and we are told that a person has a mass of 75 kilograms, then that person has a mass 75 times as grcat as the fund,HTIental unil o[mass. III 1960 an illlernational committee agreed on a standard system of units for the fundamcntal quantities of science, called SI (Systemc Intcrnational). Its units of length, mass, and time are the meter, kilogram, and second, respectively.
Length In 1799 the legal stflndard of length in Fr;'lnce became the meter, defined as one len-milliont.h of lite distance from the equator 10 rhe North Pole. Ulllil 1960,
2
Chapter 1
Introduction
Definition of the meter ~
the orficial length of the meter was the distance between two lines on a specific bar of platinum-iridium alloy stored under controlled condit.ions. This standard was abandoned for several reasons, the principal one being that measurements of the separ"'1tion between the lines are not precise enough. In 1960 the meter was ddllled as I 650 7G3.73 wavelengths of OI-ange-red liglu cmiucd rrol11 a krypton-86 lamp. In October 1983 this definition ,\'as abandoned also, and the meter was redefined as the distance traveled by light in vacuum during a time interval of 1/299 792458 second. This latest definition establishes the speed of light at 299 792 -158 meters per second.
Moss Definition of the kilogram
-+
TI P 1.1 No Commos in Numbers with Mony Di9its
The SI unit of mass, the kilogram, is defined as the mass of a specific platinumiridium alloy cylinder kept at the International Bureau of Weights and Measures at Sevres, France (similar to that shown in Fig. l.1a). As we'll see in Chapter 4, mass is a quantity used to measure the resistance to a change in the motion of an object. It's more difficult to cause a change in the mOlion oran object with a large mass than an objcct with a small mass.
In :o.eicncc. numbers wilh more
than three digits arc wriucn in of three digiLS separated by "paces rather than commas; so Ihal 10 000 is the ~alllC as the grnllp~
Time
comnlon American llotation 10.000. Similarly. 7T = :~.14159265
is wriucn as 3.141 59265.
Definition of the second
-+
Bcfore 1960, the time standard was defined in terms of the average lengrh of a solar day in thc year 1900. (A solar day is the time between sllccessi\'c appearances of the Sun at the highest point it reaches in the sky each day.) The basic unit of time, the second, was de filled to be (1/60)(1/60)(1/24) = 1/86400 of the average solar day. In 1967 the second was redefined to take advantage of the high precision auainablc with an atomic clock, which uses the charactcristic frequency or the light emittcd fran) the cesium-133 atom as its "reference clock." The second is now defined as 9 192631 700 times the period of oscillation of radiation from the cesium atom. The newest type of cesium atomic clock is shown in Figure 1.1 b.
Approximate Values for Length, Moss, and Time Intervals Approximate values of some lengths, masses, and timc intervals are presented in Tables 1.1, 1.2. and 1.3, respectively. Note the wide ranges or values. Study these tables to gel a feel for a kilogram of mass (this book has a mass of about 2 kilograms), a time interval of 10 10 scconds (one century is about 3 X 109 seconds), or two meters of length (the approximate height of a [orw(lrd on a basketball
FIGURE 1.1 (a) The ~alioll;J1 Sland••rd Kilogr;ll11 l'o. 20. an .Iccuralt: cup.. of , he 11Ilcrn:Hional S';lIlcl.lrd l\.il0J{r:un kcpt ,If SC\Te~. FnHln~. is hou,ed under a double bdljar in ;l ,-alllt;1t the Nalionallll~liWI('of Sl.llldarris and Technologv. (b) The 1I,lIion'" primary time standard j" a
~
~
~
"• ~
.;;
Cl,..iulll fnumain 3lOmic clod.. den:lopt. d.1I the N;lliolli.llnstilllle of
"
o
Staudards :Illd Technolog} labor;:lIories ill Boulder. Color-Ido. rhi\('lock will lI~ilher gaill lIor luse;1 sl'('(llld 11\ 20 million }C'lrs.
~ ~
~ ~ ~~
Ii -~
~;
(a)
(b)
1.1
TABLE 1.1
Standards of Length, Mass, and Time
_
Approximate Values of Some Measured Lengths Length (m) Distance from E.anh to most rcmoll' known
Distance from Eanh
\
to
LX IO:>t;
qU
4 X IO:.!·-'
most remote known normal galaxies
Distance from Earth to nearest large galaxy (M31, the Andromeda Distance from Earth
to
~al(lxy)
nearest star (Proxima CCIlI
2 X IO:!:.! 4X IOH; 9x 10'"
One light rcar Mean orbit r"dills of E.arth abollt Sun
2
Mean distance from Eanh to Moon
4 X 10"
TABLE 1.2 Approximate Values of Some Masses Mass (kg) Observable Universe \!1ilky W;lY Ralaxy
I X 10" 7X lO"n
Sun
2 X
£anh
6X 10:!' 7 X 102:!
.Moon
X 1011
3
lo~t)
Shark
I X 10:!
Human
7X 10'
Frog
I X 10
rVlosquito
I X 10- 5
,
Mean radius of Earth
fix lOll
'l\'pical a Itiluclc of satellite orbiting Earth
2
Length of football field
9X 10'
[.ength of housefly
5X 10 :I 1 X 10 .,
B'lcterium
I X
I fydrogcn atom
2X 10-27
Electron
9X 10-31
Diameter of hydnJg'cn atom
1 X 10 :, 1 X 10
Diameter of atomic nucleus
I X 10
Diameter of proton
I X 10
Size of smallest dust particles Size of cells in Illoslliving organisms
X
I (I"
1O-\.';
"' "15
tc"lrn). Appendix A reviews the notation for powers oflO, such as the expression of thc number 50 000 in the form 5 X 10-1. Systems of units commonl}' used in physics arc the Systcmc International, in which the units of length, mass, and time arc the meter (m), kilogram (kg), and second (s); the cgs, or Gaussian, system, in ",'hich the units of length, mass, and time are the ccntimeter (cm), gram (g), and sccond; and the U.S. customary systcm, in which thc units of Iellgth, mass, and time arc the foot (ft) , slug, and second. 51 units are almost universally accepted in science and industry, and ""ill be lIsed throughoutlhc book. Limitcd lise will be made of Gaussian and U.S. customary units. Some of the most frequently lised "metric" (S1 and cgs) prcfixes representing powers of JO and their abbreviations are listed in Table 1.4. por example, 10-:1 III is
TABLE 1.4 Some Prefixes for Powers of Ten Used with "Metric" (51 and cgs) Units Power
TABLE 1.3
Approximate Values of Some Time Intervals Time Interv.d (s)
Period'" of t}'pical radio waves Period" ofvibratioll of atom in solid
10-1)
micro-
IJ.
10-:\
milli-
m
10- 2
cenl.i-
c
X 10 I
](]
dcci-
d
10
:1
](]'
deka-
da
1 X 10
11
10 3
kilo-
k
1:1
10"
Il1cga-
M
10 9
G
101~
Rig'llera-
1015
pCLa-
T P
1018
cxa-
E
1 X 10li
3 X IOJ
9 8
X 10"
X
1 X 10
Period" 01" visible light W;'lves
2 X 10 "
Dural ion of nucle'lr collision
] X 10 "0'
Time required for light to [ravel :1cross a prawn
:1 X 10
"'A jH'riod is defined ,I' the lime reqllin;fllor 0111" complete vibr-;II;"Il.
r
12
P
6 X 10"
Period" of audible :'\ol'ncl waves
a
fcml.O-
n
Average age or collcRe student nile day
auo-
10-1:,
nana-
5 X 10"
Time hetween normal heartbeats
Abbreviation
pico-
Age ofCni\'ersc Age of Eanh One year
Prefix
lO-]fl
~l
10
IO-~l
,
4
Chapter 1
Introduction
equivalent to 1 millimeter (mm), and 10 3 m is 1 kilometer (km). Likewise, I kg is equal [Q 10 3 g, and I megavoll (MV) is 10 6 volts (V).
I
1.2 THE BUILDING BLOCKS OF MATTER
Gold cube
NucleliS
. .. '
(.. (."r. '/ ( (
(
( ... •
J
.{ ~
- (. /
-';'. /
(.
, ,
a
/
I I' 'f.' Quark composition of a proton FIGURE 1.2 Lc\'c1soforganizaLioll in malleI'. Ordinary matter consists of atoms. and
A I-kg (= 2-lb) clIbe or solid gold has a lenglh of abolll 3.73 cm (= 1.5 in.) on a side. If the cube is cut in half, the two resulting pieces retain their chemical identit)' as solid gold. But what happens if the pieces of the cube arc cut again and again, indefinitely? The Greek philosophers Leucippus and Democritus couldn't accept the idea that such cutting could go on forever. They speculated that the process ultimately ""ould end w'hen it produced a panicle that could no longer be cuL In Greek, alomos means "not sliceable." From this term comes our English word alom, once believed to be the smallest particle of matter but since found to be a composite of more elementary particles. The atom can be naively visualized as a miniature Solar System, with a dense, positively charged nucleus occupying the position of the Sun and negatively charged electrons orbiting like planets. This model of the atom, first developed by the great Danish physicist Niels Bohr nearly a century ago, led to the understanding of certain properties of the simpler atoms such as hydrogen but failed to explain many fine details of atomic structure. Notice the size or a hydrogen atom, listed in Table 1.1, and the size of a proton-the nucleus of a hydrogen atom-one hundred thollsand times smaller. If the proton were the size of a Ping Pong ball, the electron would be a tiny speck about the size of a bacterium, orbiting the prot.on a kilometer away! Other atoms are similarly constructed. So there is a surprising amount of empt)' space in ordinary matter. After the discovery of the nucleus in the early 1900s, quest.ions arose concerning its structure. The exact composition of the nucleus hasn't been defined completely e"en today, but by the early 19305 scientists determined that two basic entitiesprotons and neutrons-occupy the nucleus. The pmlon is nature's fundamental carrier of positive charge, equal in magnitude but opposite in sign to the charge on the electron. The T1urnber of protons in a nucleus determines ,·.. hat the element is. For instance, a nucleus containing only one proton is the nucleus of an atom of hydrogen, regardless of how many neutrons may be present. Extra neuU·ons correspond to different isotopes of hydrogen-deuterium and tritium-which react chemically in exactly the same way as hydrogen, but are more massive. An atom having t"".-o protons in its nucleus, similarly, is always helium, ahhough again, differing numbers of neutrons are possible. The existence of neu.lrollS was verified conclusively in 1932. A neutron has no charge and has a mass about equal to that of a proton. One orits primary purposes is to act as a "glue" to hold the nucleus together. If neutrons were not present, the repulsive electrical force between the positively charged proLOns would cause the nucleus to fly apart. The division doesn't stop here; it turns out that protons, neutrons, and a zoo of other exotic panicles are now thought to be composed of six panicles called quarks (rhymes with "'forks," though some rhyme it with "sharks"). These particles have been given the names up, down, strange, charm, bottom., and (0/). The up, charm, and top quarks each carr)' a charge equal to +~ that of the proton, whereas the down, strange, and bottom quarks each carry a charge equal to the proton charge. The proton consists of two up quarks and one down quark (see Fig. 1.2), giving the correct charge for the proton, +]. The neutron is composed of two down quarks and one up quark and has a net charge of zero. The up and down quarks are sufficient to describe all normal matter, so the existence of the other four quarks, indirectly observed in high-energy experimems, is something of a mystery. It's also possible that quarks themselves have imernal
-*
I
1.3
structure. Many physicists believe that the most fundamental particles may be tiny loops of vibrating string.
1.3
DIMENSIONAL ANALYSIS
In physics the word dimension denotes the physical nature of a quantity. The distance betwecn two points, for cxample, can be measured in fect, meters, or furlongs, which are diffcrent ways of expressing the dimcnsion of length. The symbols used in this section to specify thc dirncllsions of length, mass, and time are L, M, and T, respectively. Brackets [ ] will often be used to denote the dimensions ofa physical quantity. In this not.ation, for exam pie, the dimensions of velocity v are '''Titten [v] = LIT, and the dimensions of area A are [A] = U~. The dimensions of area, volume, velocity, and acceleration are list.ed in Table 1.5, along with their units in the t.hree com mOIl systems. The di mensions or other quantit.ies, such as force and energy, will be described later as they are introduced. In physics it's often necessary either to derive a mathematical expression or equation or to check its correctness. A useful procedure for doing this is called dimensional analysis, which makes use of the fact that dimensions can be treated as algebraic quantities. Such quantit.ies can be added or subtracted only if they have the same dimcnsions. It follov.,'5 that the terms on the opposite sides of an equation must have the same dimensions. If t.hey don't, the equation is wrong. If they do, the equation is probably con-eel, except. for a possible constant facLDr. To illustrate this procedure, suppose we wish to derive a formula for the distance x traveled by a car in a t.ime l if l.he car starts [rom rest. and moves with constant. acceleration a. The quantity."C has the dimension length: [x] = L. Time l, of course, has dimension [I] = T. Acceleration is the change in velocity vwith lime. Because v has dimensions of length per unit timc, or [v] = LIT, acceleration must have dimensions [a] = L/T2. We organize this informat.ion in the form oran equation:
[v]
[aJ
=
LIT
111 = T
=
L [x] T' ~ [I]'
Looking at the left- and right.-hand sides of this equat.ion, we might now guess that
This expression is not quite correct, however, because there's a constant of" proportionality-a simple numerical faclor-that. can't be dctermined solely through dimensional analysis. As will be seen in Chapter 2, it turns out that the correct .. I 2 expression IS x = 'la-I. When we work algcbraically with physical quantitics, dimensional analysis allows us La check for errors in calculation, \-\'hich often shm.,: lip as discrepancies in units. If, for example, the left-hand side of an equation is in meters and the right-hand side is in meters per second, we know immediately that we've made an error.
T BlE 1.5 Dimensions and Some Units of Area, Volume, Velocity, and Acceleration System
Area (L2 )
Volume (L3 )
Velocity (L/T)
51
m2 cm 2 rt 2
m Cl em:;
em/...
f(l
fl/s
cgs
U.S. cuslomary
m/s
Acceleration (L/T 2 ) m/5 2 em/5 2 fl/S 2
Dimensional Analysis
5
Chapter 1
6
EXAMPLE 1.1 Goal
Introduction
Anal sis af an
Eq!.':u~a~ti~a::.:n~
Check an equation using dimensional analysis.
Problem Shmv that the expression v = Vn acceleration, and l is a time interval.
Analyze each term, finding its dimensions, and then check to see if all the terms agree with each other.
Strategy
.....
",
~
L
=
T
...
.
...
All the terms agree, so the equation is din1cnsionally correct.
QUESTION 1.1 True or False. An equation that is dimensionally correct is always physically correct, up to a constanl of proportionality. EXERCISE 1.1 Determine whether the equation all constant of proportionality. Answer
is dimensionally correct. If not, provide a correct expression, up to an over-
Incorrect. The cxpression x = vI is dimensionally correct.
EXAMPLE 1.2 Goal
x = vI'.!.
_
Fin-"d'"'a"'n"-"E"'-"ua:tot:.:;io~nc:...
Derive an equation by llsing dimensional analysis.
Problem Find a relationship between a constant acceleration a, speed v, and distance r from the origin ticIe traveling in a circle.
1'01'
a
par~
Strategy St
.
,..-' .. Solution "\'rite down the dimensions of a:
[a]
=
T'
Solve the dimensions of speed [or T:
[v]
=
T
I. T=-
SubSlilUte the expression for T into the equation for [a]:
[]
L
=
L 1"
Substitute L = ld, and guess at the equation:
[a]
~
...
a
.,
,
L
L
[v]
= (L/[v])'
[v]' L
[v]' [ r] ..
,~
Remarks This is the correct equation for centripetal acceleration-acceleration to\vtlrds the center of motion-to be discussed in Chapter 7. In this case it isn't necessary to introdllce a numerical factor. Such a factor is oft.en displayed explicitly as a cOllstantll in front of the right-hand side-for example, (L = !al/I: As it turns out, k = I gives the correct expression.
1.4
Uncertainty in Measurement and Significant Figures
7
QUESTION 1.2 . True or False: Replacing v by 1!t in the fin~l answer also gives a dimensionally correct equation. EXERCISE 1.2 In physics, energy E carries dimensions of m~ss limes length squared divided by time squar~d. Use ~imeI~sional analysis to derive a relationship for energy in terms of mass 111 and speed v, up to a constant ot proporuonahty. Set the speed equal to c, the speed of light, and the constant of proportionality equal to 1 to get the most famous equa-
tion in physics. Answer
1.4
E = kmv'J.
---*
E = mc'J. when h
=
1 and v
= c.
UNCERTAINTY IN MEASUREMENT AND SIGNIFICANT FIGURES
Physics is a science in which mal.hematicallaws are tested by experiment. No physical quantity can be determined with complete accuracy because our senses are physically limited, even when extended with microscopes, cyclotrons, and other gadgets. Knowing the experimental uncertainties in all)' measurement is very irnportant. vVithout this information, little can be said about the final measurement. Using a crude scale, for example, we might find that a gold nugget has a mass of 3 kilograms. A prospective client interested in purchasing the nugget vmuld naturally want to know about the accuracy of the measurement, to ensure paying a fair price. He wouldn't be happy to find that the measurement was good only to within a kilogram, because he might pay for three kilograms and get only two. Of course, he might get four kilograms for the price of three, but most people would be hesitant to gamble that an error would LUrn out in their favor. Accuracy of measurement depends on the sensitivity of the apparatus, the skill of the person carrying out the measurement, and the number of times the measurement is repeated. There are many ways of handling uncertainties, and here we'll develop a basic and reliable method of keeping track of them in the measuremellt itself and in subsequent calculations. Suppose that in a laboratory experiment we measure the area of a rectangular plate with a meter stick. Let's assume that the accuracy to ,'vhich we can measure a particular dimension of the plate is ±O.I em. I r the length of the plate is measured to be 16.3 cm, we can claim only that it lies somewhere between 16.2 cm and 16.4 cm. In this case, 'loVe say that the measured value has three significant. figures. Likewise, if the plate's width is measured to be 4.5 cm, the actual value lies betv.!cen 4.4 cm and 4.6 cm. This measured value has only two significant figures. We could write the measured values as 16.3 :!:: 0.1 cm and 4.5 ± 0.1 em. In general, a significant figure is a reliably known digit (other than a zero used to locate a decimal point). Suppose we would like to find the area of the plate by multiplying the two measured values together. The final value can range between (16.3 - 0.1 cm)(4.5 0.1 em) = (16.2 em)(4.4 em) = 71.28 em' and (16.3 + 0.1 em)(4.5 + 0.1 em) ~ (16.4 cm)(4.6 em) = 75.44 cm 2 . Claiming to know anything about the hundredths place, or even the lent.hs place, doesn't make any sense, because it's clear we can't even be certain of the units place, whether it's the] in 7], the 5 in 75, or somewhere in between. The tenths and the hundredths places are clearly not significant. \"'e have some information about the units place, so that number is significant.. rvlultiplying the numbers at t.he middle of the uncertainty ranges gives t16.3 cm)(4.5 cm) = 73.35 cm 2 , which is also in the middle of the area's uncertainty range. Because the hundredths and t.enl hs are not significant, we drop them and take the answer to be 73 crn 2 , ''1lith an uncertaitlly of ±2 cm 2 . Note that the answer has lWo signiflcant figures, the same number of fignres as the least accurately known quantit), being multiplied, the 4.5-cm \.vidth.
8
Chapter 1
Introduction
There arc two useful rules of I humb for determining t he number of significant figures. The first, concerning IIlUlliplication and division, is as follows: In multiply~ iug (dividing) two or more quantities, the number of significant figures in tbe final product (quotient) is tbe same as the number of significant figures in the least accurate of the factors being combined, where least accurate means having the lowest number of ~ignificalltfigures. To get the final number of significant figures, it's usually necessary lO do some rounding_ If the last digit dropped is less than 5, simply drop the digil. If the last digit dropped is grealer than or equal to 5. raise the last retained digit by one.
EXAMPLE 1.3 Goal
Installing a Carpet
.\ppl~ the 11111ltiplication rule for significant figures.
Problem A carpel is to be installed in a room oflenglh 12.7] m and \\'idth 3A-6m. Find the area orthe room, retaining the proper number of significant figures. Strategy answer.
.
(OUIll
the significant figures in each number. The smaller result is the number ofsignificam ligures in the
.
• • • • • • "'II
Solution 12.71
Count significant figures:
3.46
tvfulliply the numbers, keeping ollly three digit.s:
----+
In
IT'l
~
4 significant figures 3 significant figures
12.71 m X 3.46 m = 43.9766 m'
.....
-->
44.0 Ill'
............ - ... ~
Remarks In reducing 43.9766 to three signiflcant ligures, we used our rounding rule, adding I to the 9, which made 10 and resulted in carrying I to the unit's place. QUESTION 1.3 ,",'hat would the answer ha\-e been if the width were given as 3.460 m? EXERCISE 1.3 Repeat this problem, but with a room measuring 9.72111 long by 5.3111 wide.
Answer
52 J1l~
TIP 1.2 Using Calculators CalculalOI .. were dcsi~nl'd by engineers 10 \'ield a'l many digits as lhe memol"\' of Ihe calculator chip permitted, so be mre (0 round 11ll' final 31l')\\'cr down 10 lhe COITcn Illllllbcro/1iiv;nificanl figun:s_
Zeros mayor mar not be signit-icant figures. Zeros used to position the decimal point in such numbers as 0.03 and 0.0075 are not significant (but are useful in avoiding errors). Hence, 0_03 has one significant figure, and 0.0075 has two. \\'hen leras are placed after other digits in a whole number, there is a possibil· ity of misinterpretation_ For example. suppose the mass of an object is gi\-en as I 500 g. This value is ambiguous, because we don't know whether the lastt\\'o zeros are being used to locate the decimal point or whether they represelll significant figures in the measurcmenl. Csing scientific notation to indicate the numbcl' of significam figures rel11O\-cs this ambiguity. In this case, we express lhe mass as 1.5 X 1O:l g if there are two significant figures in the measured "alue, 1.50 X 10 3 g if there are three significanl figures, and 1.300 X 10:\g if there are four. Likewise, 0.000 15 is expressed in scien~ lific notation as 1.5 X 10 -.j ifit has two significant figures or as 1.50 X 10--1 ifit has three significant figurcs. The three zeros between the decimal point and the digit I in the number 0.000 15 arc nOl coullted as significant figures because they only locate the decimal poinl. Tn this book, most of the numerical exmnplcs and endof-chapter problems will yield answers having two or three significant figures.
1.5
For places When result
Conversion of Units
9
addition and subtraction, it's best to foclis on the number of decimal in the quantities involved rat.her than all the number of significant figures. numbers are added (subtracted), the number of decimal places in the should equal the smallest number of decimal places of any tcrm in the
sum (difference). For cxarnple, if \ve ,.. . ish to compute 123 (zero decimal places) + .1).35 (two decimal places), the <1ns\vcr is ]28 (zero decimal places) and not 128.35. If we compute the sum 1.000 I (four decimal places) + 0.0003 (four deci-
\ \
ma\ places) = L.OOO 4, the resull has the correci number of decimal places, namely four. Observe tl1m the rules for multiplying significam figures don't work here because the answer has five significant figures even though onc or the lerms in the sum, 0.0003, has only one significant figure. Likewise, if we perform the subtrac~ tion 1.002 - 0.998 = 0.004, the result h
+
:J.35x
I 28. xxx
The answer of l28.xxx means that we are justified only in kceping the number 128 because cverything after the dccimal point in the sum is aClUally unknown. The eX;'lmple shows that the controlling unccrtainly is introduced into an addition or subtraction by the term with the smallest number of decimal places. I n performing any calculation, especially one involving a number of stcps, there will always be slight discrepancies introchlced by both the rounding process and the algebraic order in which steps are carried out. For exan1ple, consider 2.35 X 5.89/1.57. This computation can be performed in three difTerent orders. First, we have 2.35 X 5.89 = 13.842, which rounds 1.0 13.8, followed by 13.8/157 = 8.7898, rounding to 8.79. Second, 5.89/1.57 = 3.751 6, which rounds to 3.75, resulting in 2.35 X 3.75 ~ 8.8125, rounding 10 8.81. Finally, 235/1.57 = 1.496 8 rounds 10 1.50, and 1.50 X 5.89 = 8.835 rounds to 8.84. So three different algebraic orders, following the rules of rounding, lead [() answers of8.7~), 8.81. and 8.84, rcspeClively. Such minor discrepancies arc to be expected, because the last significant digit is unly one represel1laLive from a range of possible values, depending on experimcn~ lal uncertainty. The discrepancies can be reduced by carrying one or more extra digits during the calculation. Tn our examples, however, intermediate results will be rounded off to the proper number of significant flgures, and only those digits will be carried forward. In experimental work, more sophisticated techniques are llsed to determine the accuracy of an cxperimental resull.
1.5
CONVERSION OF UNITS
SOlTletimcs it's necessary to convert units from one system to anOlher. Conversion factors bCl\veen the SI and U.S. customary systems for units of length are as follows: mile =1 609 m = 1.609 kill 1 m = 39.37 in.
~
3.281 ft
I ft
~
0.304 8
III
= 30.48 cm
I in. = 0.025 4m = 2.54
CIll
A more extensive list of cOIl\'ersion factors can be found on the inside front cover
of th is book. Unjl.s can be treated as algebraic quantities that can "cancel" each other. Vv'e can make a fraction with the conversion that will cancel the units we don't ',:ant,
This rD,ld sign 111:<11' Raleigh, :,\lorLh Carolina. shows dislances in miles ,llld kilomel<:rs. How accur,lle an; lhe cOlwcrsiollS?
10
Chapter 1
Introduction
and multiply that fraction by the quantity in question. For example, suppose we want to convert 15.0 in. to centimeters. Because 1 in. = 2.54 em, we find that 15.0 in.
~
2.54 cm) 15.0 iff: X ( 1.00i".
~
38.1 cm
The next two examples show how to deal with problems involyillg morc than one cOIwersion a nel with powers.
EXAMPLE 1.4 Goal
Pull Over, Budd !
Comert units lIsing several conversion [aClors.
Problem
If a car is traveling at a speed of28.0
m/s. is the driver exceeding the speed limit 01'55.0 milh?
Strategy Meiers must be convened to miles and seconds to hours, lIsing the conversion facLOrs listed on the inside from cover of the book. Here, three factors ,.. . ill be lIsed. r. ... 0
••
Solution COIwcrt meters to miles:
Connon seconds
10
28.0 m/s
hours:
1.74 X 10
:!
~
mils ~
1.00 _ mi) (28.0 -ni)( s 1609m
(
.) (
1.74 X 10-' '."'. g
~
1.74 X 10 "- mils
)(
60.0 -.:'...... mrrf
60.0
.
1111n
h
)
62.6 mi/h
.......... Remarks
.........................~ The driycr should slow down because he's exceeding lhe speed limit.
QUESTION 1.4
Repeat the conversion, lIsing lhe relationship l.00 m/s = 2.2l1 mi/h. "Vhy is the answer slightly c1irrercm? EXERCISE 1.4
Comcn 152 mi/h to m/s.
Answer 68.0 m/s
EXAMPLE 1.5 Gool
Press the Pedal to the Metal
Com'en a qWHlLit) fC
Problem The trame light LUrns green, and the dri\'cr of a high-performance car slams lhe accelerator lO the floor. The accelerometcr registers 22.0 m/s2 , COllver! this reading to kill/min:!. Strategy Here we need one factor squared to minUlcs squared.
Lo
convert meters to kilometers and another two
f£lcLOrs to COlwerl
.
r························
Solution Multiply by the three facLOrs: Irrr... • • • • • • •
ni(
km )(
22.0 1.00 1.00,' 1.00 X 10' ni
60.0" )' 1.00 min
seconds
...
_
km
79.2-.-" mlO~
..........
Remarks Notice that in each conversion faclOr the numerator equals the denominator when units are taken into account. A cOlllman error in dealing with squares is to sqllare the units inside the parenlheses ,,'hile rorgetting to square the nurnbcrs~
I
1.6
Estimates and Order-af-Magnitude Calculations
11
QUESTION 1.5 ''\I"hat lime com'ersion factor would he used to further COl l\'ert the ans\.. .er to km/h~? EXERCISE 1.5 Convert 4.:>0 X 103 kg/m:\ to g/cm 1 .
4.50 g/cm i
Answer
1.6
ESTIMATES AND ORDER·Of·MAGNITUDE CALCULATIONS
Celling an exact answer to a calculation may One!1 be difficult or impossible, either for mathematical reasons or because limited information is .n-ailable. In these GISeS, estimates can yield lIseful approximate answers thaI can determine whether a mor(> precise calculation is nccessalT Estimmes also sen'e as a partial check if the exact calculations are actuall), carried oul. If a large answer is expected bllt a small exact answer is obtained, there's an error ~omc\'iherc. For many problems, knowing the approximme value of a quantity-within a factor of 10 or so-is sufficient. This approximate valuc is called an order-ofmagnitude estimate. and requires finding' the power of to that is closest to the actual value of the quantity. For example, 75 kp; - 10 2 kg, where the symbol means "is on the order of" or "is approximately." Increasing a quan[itr by three orders ofmagniwde means that its value increases by £1 [actor of 10~ = 1 000. Occasionally the process or making such estimates results in fairly crudc answers, bill answers tcn times or morc too large or small are still useful. For example, suppose you're interested ill how mallY people have contracled a certain disease. Any estimates under ten thousand arc small compared with Eanh's tOlal population, bUl a million or more would be alarming. So even relatively imprecise inrorl1laLion can provide valuable guidance. In developing these estimates, you can take considerable liberties WiLh the numhers. For example, '1T - 1, 27 ~ 10, alld 6.:) -- 100. To gel. it less crude estima[c, it's permissiblc [Q use slightly more accurate numbers (e.g., '1T - 3, 27 - 30, 65 -- 70). Hettel' accuracy can also be obtained by systematically underestimating as man>' numbers as YOli o\,crcstim;He. Some quantities may be completely unknown, but ii's slandard to make reasonable guesses, as the examples show.
EXAMPLE 1.6 Goal
Brain Cells Estimate:--
_
Develop a simple estimate.
Problem
Estirnate the nurllber of cells in the human
brain. Strategy Estima[e the volume of a human brain and divide by the estimated volume of one cell. The brain is located in the tipper portion of the head, with a \'olumc lhaLcould be approximated bya cube = 20 cm 011 a side_
..
Brain cells, consisting of about 10% neurons and 90% glia, \'ary greatly in size. with dimensions ranging from a few microns to a meter or so. As a guess, take d = 10 miCl-ons as a typical dimension and consider a cell to be a cubc with each side having [hat length.
e
.........,
'
Solution Estimate of the volume ora human brain:
~"'"'' = f'
ESlimaLe the volume ora cell:
\~."" =
Divide Lhe volume ora brain by the volume ora cell:
number of cells
~,
= (0,2 Ill)" =
,l" =
(10 X 10
6
8 X 10-3 111"
m)"
\~)I'
=
= I X 10
0.0 I J X 10
"m'
1 111: l.'i
I X 10 'Ill'
m1
=
I X
101:~ccJls ,
,.-
12
Chapter 1
Remarks
Introduction
NOlice how little attention was paid to obtaining precise values. That's the nature oran estimate.
QUESTION 1.6
\,\'ould 10 12 cells also be a reasonable estimate? "Vhat aboul. 10'J cells? Explain. EXERCISE 1.6
Estimate the number of stacked objects required to reach a given heighL
Problem
Ilow many
onc~dollar
bills, Slacked one on LOp of the olher, would reach the Moon?
Strategy The distance (0 the Moon is about 400 000 km. Guess at the number of dollar bills in a millimeter, and multiply dIe djslan«~ by this number, after convening to consistent units. ............................
r··························· .
Solution \"'e estin1(l1.C that ten stacked bills form a layer of 1 mm. COllven mm to km:
10 bills (lOS J>l1fi)(lOS HI) = 10' bills
t\lultiply this value by the approximate lunar distance:
# of" dollar bills - (4 X 10"_ kill) (10' billS) lkm
That's the same order of magniwde as the U.S. national debt!
QUESTION 1.7
Based on the answer, about how many stacked pennies would reach the Moon? EXERCISE 1.7
lTow man)' pieces of cardboard, typically found at the back of a bound pad of paper, would match thc height or the \Vashington monument, about 170 m laB? Answer
havc
[Q
stack up
LO
....... 10:· (Answers may vary.)
EXAMPLE 1.8 Goal
yOll
Estimale a
Number of Galaxies in the Universe \"oIUIllC
and a number density, and combine.
Problem GiH'n that astronomers can sec about 10 billion light years into space and that there are 14 galaxies in our local group, 2 million light years from the next local group, estimate the number of galaxies in thc observable universe. (NOIc: One light )'ear is the distance traveled by light in one ~·ear. ~lbollt9.5 X 10 15 m.) (See Fig. 1.3.)
Strategy From thc known information, we can estimate the number of galaxies per unit volume. The local group of J4 galaxies is contained in a sphere a million light years in radius, with the Andromeda group in a similar sphere. so there are about JO g£llaxics within a volume of radius Imillion light years. Multiply that number density by the volume of the observable universe.
~
....
"
... , ..
,
Solution Compute lhe approximale volume of galaxies:
FIGURE 1.3 In tllisdcep-spacc photugraph. theH' are few stars-jml galaxies w;thout ('Itt!.
l'fg
of the local group
v" = ~1TTS -
,
(10" ly):I
~
=
10'" Iy'
1.7
E.stimate the density of galaxies:
Coordinate Systems
13
# of' galaxies density of galaxies = -----''-:--V", 10 galaxies
Iy'
I OIB
Compute t.he approxi mate volume or the observable universe: Multiply the density or galaxies by \I,:
# of galaxies
~
(density of galaxies)V;,
(
_ aalaXieS) . (lo,oly')
IO-I'~
10
-
~
I)~
galaxies
.
Notice the approximate nature of Iht: computal ion, which uses 41T/3 -- 1 on two occasions and l'l 10 for the number of galaxies in the local group. This is completely justified: Using the actual numbers would be pointless, because the other assumptions in the problem-the size of the observable universe and the idea that the local galaxy density is representative of the density everyv,,'here-are also vcry rough approximations. Further, there was nothing in the problem that refJuired using volumes of spheres rather than volumes of cubes. Despite all these arbitrary choices, the ans\ver still gives useful informatlon, becallse it rules out a lot of reasonable possible answers. Before doing the calculation, a guess ofa billion galaxies might have seemed plausible. Remarks
1.7
13
............. ~
QUESTION 1.8 Of the fourteen galaxies in the local group, only one, the Milky Way, is not a dwarf gablxy. Estimate the number of galaxies in the universe that are not dwarfs. EXERCISE 1.8 Given that the nearest star is about 4 light years "I\o\'ay and thatlhe galaxy is roughly a disk 100 000 light years across and a thousand tight years thick, estimate the number of stars in the Milky \·Vay galaxy. ~ 10 12 stars (Estimates will vary_ The actual answer is prob:lbly close 1.0 4 X lOll stars.)
Answer
COORDINATE SYSTEMS
Many aspects of physics deal \\lith locations in space, \vhich require the dcllnition of a coordinate system. A poilU on a line can be located with one coordinate, a point in a plane with two coordinates, and a point in space \vith three. A coordinate system used to specify locations in space consists of the following: • A fl xed reference point 0, called the origi1J • A sct of specified axes, or directions, with an appropriate scale and labels all the axes • rnsl:l'uctions on labeling a point in space relative to the origin and axes
One convenient and commonly used coordinat.e system is the Cartesian coordi· nate system, sometimes called the rectangular coordinate system. Such a system in two dimensions is illust.rated in Figure 104. An arbitrary point in this system is labeled with the coordinates (.~, y). For example, the point P in the figure has coordinates (5, 3). If we start at the origin 0, we can reach P by moving 5 meters horizontally 1.0 the right and then 3mel.ers vertically upv..' ards. In t.he same vvay, the point Qhas coordinates (-3, 4), which corresponds to going 3 meters horizontally to t he left of the origin and 4 meters vertically upwards frolll there. Positive x is usually selected as right. of the origin and positive y upward frolll the origin, but in two dimensions this choice is largely a Illallcr of tasle. (In t.hree dimensions, however, there are "right-handed" and "left-handed" coordinates, which lead 10 minus sign differenccs in certain operalions. These will be addressed as necded.)
>'
(m)
10
Q
• (-3.4)
3 p
.(5,3) :r(m)
()
5
10
FIGURE 1.4 Dcsigll;:J1jon of poi illS in a Iwo-dimcllsional Cartesian coordinale syslem. Ever}' poim is Iabc/cd witli roordinalcs (x. y).
14
Chapter 1
Introduction
Sometimes it's more convenient to locate a point in space by its plane polar coordinates (1; 0), as in Figure ].5. In this coordinate system, an origin Ganci a reference line are scleClcd as shown. A poin1 is then specified by the distance r from the origin to the point and by the angle fJ between the reference line and a line dnlwn from the origin to the point. The standard rcference line is usually selected to be the positive x-axis of a Cartesian coordinate system. The angle 8 is considered positive when measured counterclockwise from the reference line and negative when mcasured clockwise. For example, if a point is specified by the polar coordinates 3 m and 60 we locate this point by moving alit 3 m from the origin 0 at an angle of 60 above (coullterclocbvise from) the reference line. A point specined by polar coordinates 3 m and -60 is located 3 m out ['rom the origin and 60 belm\' (c1ocbvise [rom) the reference line. 0
Refi.:rcllcc line
,
0
FIGURE 1.S
A poJarcllordillate
S"~[f'm.
0
1.8 TRIGONOMETRY Consider the riglll triangle shown in Active Figure 1.6, where side)' is opposite the angle 8, side;r is adjacent to the angle 8, and side ris the hypotenuse orthe triangle. The basic trigonometric functions de filled by such a triangle are the ratios of the lengths of the sides or the triangle. These relationships are called the sine (sin), cosine (cos), and tangent (tan) funClions. fn terms 01'8, the basic trigonometric functions arc as follows: 1
)'
. Sill () '=
,
"'r
,
co<;() =.!
Utne
,
y
sin ()
=..:..
-'
...I_,
1L...I. ACTIVE FIGURE 1.6
Certain Irig(1I1011H'lric fUllClioll:; 01';\
,"il{ht rri
cos 8 tan
f) =
side opposite
e
)'
.
hypotenuse side adjacentLO 8
x
[1.1]
~
r
hypotenuse side opposite
f)
side acUacent to
)'
x
f)
For example, if the ang·le 0 is equal to 30 0 , then the ratio of)' to r is always 0.50; then is, sin :W o = 0..50. Note that I he sine, cosi ne, and tangent Functions are quantities \-vithout units because each represents the ratio of tvlO Icngths. Another important relationship, called the Pythagorean theorem, exists betwecn the lengths of the sides of a right triangle: [1.2]
Finally, it. will often be nccessary to find thc values of inverse relationships. For example, suppose you kllO\'i thal the sine of an angle is 0.866, but you need to know the value of the angle itself. The inverse sine Function may be expressed as sin-I (0.866), which is a shonhand way of asking the question "'What angle has a sine of 0.866?" Punching a couple of but.tons on your calcubtor reveals that this angle is 60.0°. Try it for yourself and show then tan- 1 (0.400) = 21.8 0 • Be sure that your calculator is set for degrees and not. radians. In addition, the inverse tangent function Gl1l return only values bet\veen -90 and +90 so when an angle is in the second or third quadrant, it·s necess31-y l.o add 180 to the answer in the calculator window. The derinitions uf the trigonomctric functions and the inverse trigonometric functions, as \'\'ell as the Pythagorean theorem, can be applied to any right triangle, regardless of whether its sjdcs correspond to x- and y-coordinatcs. These results from trigonometry are useful in converting from rectangular coordinates to polar coordinates, or vice versa, as the next example shows. 0
0
,
TIP 1.3 Degrees vs. Radians
0
When calc\llaLillg trigonometric functions, make sure your calculator sctling-dc~reesor radians-is con.<;istent \,'1,11 the
dcgn:c measure you're a givC'll problem.
usin~
in
1;"·laIIY pcclplc liSt.: lht.: mne1110niL SOf-((:/\/-fT{)A to rrnlcn,l,cr lilt.' basic: Ll"ig,moIHt'tric fOI ml11a~: Sinc = Oppositc/ lh'IHJtcnllse, r:c'.,inc = /ldj;ICCIl1/ 11~'POlt.:llll~\" and "lilllgem = Oppo"itc/Jk!jaccnl. (Th;'l1l-s go 10 I'n>rc.~s"r Dun Cliodro\\ for pojlltiug-lhis nuL.)
J.8
EXAMPLE 1.9
Trigonometry
15
Cartesian and Polar Coordinates
Goal Understand hO\.. . to convert from plane rectangular coordinates to plane polar coordinates and vice versa.
-" (m)
Problem (a) The Cartesian coordinales of a point in the ;\)'-plane are (x, )') = (-3.50 In, -2.50 m), as shown in Active figure 1.7. Find the polar coordinates of this point. (b) Convert (/; 8) = (5.00 Ill, 37.0°) to rectangular coordinates.
-----+--"f----'---- :..-( m)
e ACTIVE FIGURE 1.7 (Example 1.9) COllverting from Cancsian coordillales to po!;\] nloH\inalt·s.
Strategy Apply the trigonometric functions and their inverses, together with lilt: Pythagorean theorem.
,.. ...
(-3.50,
-2.~O)
.
"""
Solution (a) Cartesian to Polar
v' x:! + )1:'!
'V( -350
m)~
Take the square root or both sides or Equation 1.2 to fInd the radial coordinate:
,. =
Use Equation 1.1 for the tangenl function to lind the angle with the im'erse tangent, adding 1800 because the angle is aClllally in third quadrant:
tan IJ = - = -,--,--- = 0.714 x -350 III
)'
e=
=
+ (-2.50 mf
4.30
III
-2.50 m
Ian '(0.714) = 35.5
0
+
180 ~ 0
(b) Polar to Cartesian
Use the trigonometric deflnitions, Equation L.1.
x
=
)'
~
rcos 8
=
rsin 8
~
(5.00 m) cos 37.0
0
0
~
(5.00 m) sin 37.0 =
3.99 111
3.01 m
.
............
~
Remarks \,Vhen '',Ie take up vectors in t,,",o dimensions in Chapter 3, '',Ie will rOUlinely use a sinlilar process to flnd the direction and magnitude ora given vector from its components, or, conversely, to find the components froln the vector's magnilllde and direction. QUESTION 1.9 Starting ,\-'ith the answers to part (b), work backwards to recover the given radius and angle. ·Why are there slight differences from the orig'inal quantities? EXERCISE 1.9 (a) hna llle polar coordinates corrcsponding- to (.x:. y) = (-3.25111,1.50 m). (b) Find the Cartesian coordinates cor· responding to (,; 8) = (4.00 Ill, 53.00)
Answers
(a) (r. 8) = (3.58
Ill,
155°) (b) ('., y)
~
(2.41 111,3.19111)
EXAMPLE 1.10 How Hi h Is the Buildin ? Goal
Apply basic results of trigonometry.
Problem A person mcasures the height of a building by walking out a dist,lJ1cC of 40.0 m from its base and shining a flashlight beam to\·\,ard the top. \,Vhen the beam is elevated at an angle of 39.0° with re::;pect. to the horizontal, as shown in Figure 1.8, the beam just strikes the top of the building. find the height of the building and the distance the flashlight beam has to travel before it slrikes the top of Ihe building. Strategy Refer to thc right triangle shown in the fig·urc. \Ve knm·\' the angle, 39.0°, and the length of the side a(!jacent to it.. Because thc height of the building is the side opposite the angle, we can use the tangent funclion. 'vVith the adjacent and opposite sides known, we can then find the hypotenuse with the PYlhag'orean theorem.
1{-J..::3:::9:::.0_0__;;:-;;-:: rt FIGURE
4(iO
1.8
III
(E.>:ampk l.10)
....:3,;l~~J]~'oI'-
16 ~.
Chapter 1
Introduction
.
_
.
Solution Use the tangent of the given angle:
Ian
Solve for the height:
Height = (tan 39.0')(46.0 111)
39.0 0
...,
-
hciglll 40.0 111 (0.810)(46.0 m)
37.3 111
Find the hypotenuse of the triangle:
....
)'=
\/x2 +
l
= V(37.3111)'
+
(46.0111)' = .
Remarks
Jn
59.2m
...
a later chapter, righl-lrianglc trigonometry is oftell used ,vhen '.. .orking with vectors.
QUESTION 1.1 0 Could the distance traveled by the light beam be found \V'it hOllt Ilsing the Pythagorean Theorem? Hmv?
EXERCISE 1.10 VVhilc standing atop a building 50.0 III tall, you spot a friend standing on a street corner. Using a protractor and dangling a plumb bob, you find that the angle between the horizontal and the direction 1.0 the spot on the sidewalk where YOllr friend is standing is 25.0°. Your eyes are located 1.75 m above the LOp of the building. I-IO\\' far away from the foot or the building is your friend? Answer
III
III
1.9
PROBLEM-SOLVING STRATEGY
~:lost courses in general physics require the S!UdeI1lLO learn the skills used in sol\'ing problems, and examinations usually include problems that test such skills. This brief section presents some useful suggestions that \ViII help increase your success in solving prOblems. An org(lnized approach to problem solving will also enhance your underslanding of physical concepts and reduce exam stress. Throughout the book, there will be a number of sections labeled "Problem-Solving Strategy," Illany of lhemjust a specializing of the list given below (and illustrated in Fig. 1.9).
Read Problem
General Problem-Solving Strategy Ora\\' Diagram
Label physical qU
Idcllliry principle(s); list data
Chuose Equation (s)
Solvc Equadon(s)
$ubSliluLe known values
Chcck Answer FIGURE 1.9 sohing:.
A gllide
\0
prnhktll
1. Read the problem carefully at least twice. Be sure you understand the nature of' the problem before proceeding further. 2. Draw a di ..l gram while rereading the problem. 3. Label all physical quantities in the diagram, using letters that remind yOll what the quantity is (e.g., m for mass). Choose a coordinate system and label it. 4. Identify physical principles, the knm.. . ns and unknowns, and list them. Put circles around the unknowns. 5. Equations, the relationships bCl\'lecn the labeled physical quamities, should be v,:rittcll clown next. Naturally, the selected equations should be consistelll with the physical principles identified in the previous step. 6. Solve the sctol"equations for the unknown quantities inlcrmsofthc known. Do this algebraically, \\'ithoUl substituting values until the next step, except '\'here terms are lera. 7. Substitute the kno\\'11 values, together with their units. Obtain a numerical value \\lith units for each unknmvl1. 8. Check your answer. Do the units match? Is the answer reasollable~ Docs rhe pIllS or Illinus sign make sense? Is your ans\ver consistent \\lith <111 order or magn itude esti matc?
This same procedure, \'lith minor variations, should be followed throughout the course. The rirst three steps are extremely important, because tlley get yOll men-
\.9
\
Problem-Solving Strategy
17
tally oriented. Identifying the proper concepts and physical principles assists you in choosing the correct equations. The equations themselves are essential. because when you understand them, yOll also understand the relationships between the physical quantities. This understanding comes through a lot 01" daily practice. Equalion:) arc the tools or physics: To solve problems, you have to have them at hand, like a plumber and his wrenches. Know the; equatiuns, and understand what they rne;-Hl and hmv 10 use them. JUS! as you can't have a conversation without k 1l00vi ng the loca I lang\ lage, you ca n't solve physics problems without knowing a nel understanding the equations. This understanding grm,,'s as ~'ou sllIdy and apply the concepts and the equat.ions relining them. Carrying through the algebra for as long as possible, ~ubstitllting numbers only <-it the end, is also important, because it helps you think in terms or the physical quantities involved, nol mcrely the numbers that represelltthelll. Many beginning physics slUdcnts are eager to substitute, but once numbers are substituted it.'s harder to understand relationships and easier to make mistakes. The physical layout and organization uf your work will make the final producl 1110rc understandable and casier to follmv. Althuugh physics is a challenging discipline, your chances of success an: excellent if you maintain a positi\·c altitude and keep Iryi ng.
EXAMPLE 1.11 Goal
A Round Trip, b Air
lliustrate the Problem-Solving Strategy.
Problem .-\n airplane travels 4.50 X 10~ km due east and then I r::wels an unknown distance due north. Finally, it returns to It.S starting poi111 by traveling a distance or 525 kill. How far did the airplane travel in the northerly direction? Strategy vVe've fi.nished reading the problem (step ]), and have drawn a diagram (st.ep 2) in Figure 1.10 and labeled il (step 3). From the diagram, we recognize {\ right triangle and identify (step 4) the principle involved: the Pythagorean theorem. Side y is the uliknov,ill quantity, and the other sides are known.
FIGURE 1.10
x=450km r= 525 km
(Ex
......... ...,
r"'
Solution '-Vrite Ihe Pylhagorean theorem (step 5):
r~ = x~
+ )'2
Solve s),mbolically for J (step 6): Substitute the numbers, with units (step 7):
) = V(525 km)' - (4.50 X 10' kill)' =
270 km
.... Remarks Note that the negative solution has been disreJ:{arded, because it.'s not physically meaningful. Ln checking (step 8), note that the units are correct and that an approximate answer can be obtained by using the easier quantities, .~OO km and 400 km. Doing so gives all answer of 300 km, which is approximately the same as our calculat.ed ans\vel' of :270 kill. QUESTION 1.11 ''''hat is the answer ifbolh the distance Ira\'ckd due east and the direct. relllrn distance arc both doubled? EXERCISE 1.11 A plane flies 345 kill due south, then turns and flies 615 kill at a heading 45.0° north of east, until it's due cast ol"its slarting point. If"the plane now turns and heads for home, hm\' rar \-vill it. havc to go?
Answer
SOg kill
Chapter 1
18
._-----'_ SUMMARY::
Introduction
.... ~
--.
~--.~
.
1.1 Standards 01 Length, Mass, and Time The physical quantities in the study of mechanics can be expressed in terms of three fundamental quantities: length. mass, and time, which hm"c the 51 units meters (m), kilogram!; (kg), and seconds (s), rcspcClively.
1.2
The Building Blocks 01 Matter
:'\1atter is made of aLOIllS, which ill l.Urn arc made up of a relatively small nucleus of proLons and neutrons within a cloud of electrons. Protons and neutrons arc composed of still smaller panicles. called quarks. 1.3 Dimensional Analysis Dimensional analysis can be lIsed LO check equations and to assist in deriving them. When the dimensions 011 both sides of lhe equation agree, the equation is often correct up lO a numerical factor. When the dimensions don't agree, the equation must be wrong.
1.4 Uncertainty in Measurement and Significant Figures No physical quantity C;ln be determined with complete accuracy. The concept of sign irlcan t I-Igu res affords a basic method of handling these unccnainties. A significalll figUfe is a reliably known digit, other than a zero, used to locate the decimal poi Ill. The l\\;0 ru les of signi fkalll figures arc as follows: 1.
2.
When Illll1tiplyingordividing using two or more quantities, the result should have the same Ilumber of signif-icant figures as the quantity having the fewest signiflcant figures. \-"hen quamitics are added or subtracted, the number of decimal places in the result should be tile same as in the quantit), with the fewest decimal places.
Use of sciel'llific Ilot,uion can avoid ambiguity in sig~ nificant figures. In rounding. if the last digit dropped is less than 5. simply drop the digit, otherwise raise the last retained digit by one.
1.5 Conversion of Units Units in physics equations must always be consistent. III solving a physics problem, it's best to start with consistem units, IIsing the table of conversion factors on the inside front cover as necessary. Converting units is a matter of ITlultiplying the given quantity by a fraction, with onc unit in the numerator
and its equivalent in the other units in the denominator. arranged so the unwanted units in the given quantity are cancelled ou[ in hn'or orthe desired units.
1.6
Estimates and Order·ol-Magnitude Calculations
Sometimes it's uscful to find an approximate answer to a question, either becallse the math is diffICult or because infunnation is incomplete. A quick estimate call also be used to check a more detailed calculation. in an ordcrof-lIlagnillide calculat.ion, each value is replaced by the closest power of ten, '.vhich sometimes must be guessed or estimated when the vahle is unkno\.. . n. The complll
1.7
Coordinate Systems
Thc Cartesian coordinate system consists of two perpendicular axes, usually called the x-axis and y-axis, with each axis labeled with all numbers from negative infinity to positive infinity. Point.s are located by specifying the x- and )'-vlllucs. Polar coordinates consist of a radial coordinate r which is the distance ['rom the origin, and an angular coordinate 8 ,·.. hich is the angular displacement from the positive x-axis.
1.8 Trigonometry Tile threc most basic trigonomet.ric functions of a right tri~ angle are the sine, cosine, and tangent, defined as follows: side opposite 8 sin 8
hypotenuse side acljacent to 0
cos 8
tan 0
y
,. x
hypotenuse
,.
side opposite ()
y
side adjacent to 0
x
[l.l]
The Pythagorean theorem is an important relationship bCl\yeen I he lengths of the sides of a righl triangle: [I. 2]
where 'I' is the hypotenuse of the triangle and x and J are the other two sides.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
,; ~':!~:rIPLE-CHOICE QUESTIONS 1. Newton's second law of Illation (Chapler ~I) says that tile mass of an objecl times its acceleration is equal to the net force 011 [he object. \Vhich or the following gives the correcl units for force? (a) kg' m/5 2 (b) kg' Ill:! Is:! (c) kg/Ill' s:! (d) kg' m:!/s (e) none of these
2. Suppose IWO quantities, A and 13, have different dimensions. Determine ,.".hich of" the foJlmving arithmetic operations ('(wid be physically meaninl4'ful. (a) A + B (b) B - 1\ (c) A - B (eI) AIIl (eJ lIB
Problems
Ill, with accurately than the length. acc-oun1 significanl figures. 6.783 X 10' 111' (c) 6.78 X 7 X 103 m 2
8. Find the polar coordinates corresponding to a point located al (-5.00, 12.00) in Cartesian coordinates. (a) (130, -67.4°) (b) (13.0, 113°) (c) (14.2. -67.4°) (e1) (1'1.2. 113°) (0) (19, -72.5")
Use the rules lor significant figllres lO lind the ,1Ils\ver + l!'l + 17.17 + 4.003. (a) 57.573 (b) 57.57 (c) 576 (d) 58 (c) 60
9. At a horizontal distance or 45 m rrom a tree, the angle of' elevation to the top of the tree is 26°. Hm,\' tall is the tree' (a) 22 In (b) 31 111 (c) 45 m (e1) 16 m (e) 11 In
3. A rectangular airstrip measures 32.30 m by 21O
the width measured more find the area, taking into ('II 6.783 0 X 10' 111' (b) 1O:i m 2 (d) 6.8 X H':~ m:.! (e) 4.
19
to the addition problem 21.4
5. The Roman cubitus is an ancientlillit or measure equivalt:ntlo about 445 mm. COllvert the 2.00-m-height oj' a basketball I'orw<'lrd 1.0 cubiti. (a) 2.52 cubiti (b) 3.12 cubiti (c) 4.4Y cubiti (d) 5.33 cubiti (e) none or these 6. A !louse is advertised as having 1420 square reetllnder ,·oof. \'Vhal is the area of this house in square meters? (a) ll:lm~ (b) 132 Ill:! (c) 176 m\? (d) 22~ m~ (e) none of these 7. Which of the following is the best estimate lor the mass of all the pcopk living on Earth? (a) 2 X 10 8 kg (b) I X 10" kg' (c) 2 X 101ll kg (e1) 3 X lO" kg (e) 4 X LOI2 kg
10. What is the approximate number of' breaths a person takes over a period or 70 years? (a) 3 X 10 1' breaths (b) 3 X lO' breaths (c) 3 X 10' breaths (e1) 3 X 10" breaths (c) 3 X 10 lO breaths 11. Which of the follO\..· ing relalionships is dimensionally consistent. with an expression yielding a value for acceleration? Acceleration has the units of' distance divided by time squared. In these equations, x is a diS1.
~. QUESTIONS 1. Est ima te the order or magn ilude oft he length, in meters, of each of' the rollO\.. . ing: (a) a mouse. (b) a pool cue, (c) a basketball court, (<1) an elephant, (e) a city block.
7. The height of" a horse is sometimes given in units of'
2. What types of natul'al phenomena could serve standards?
8. How man)' or thc lengths or time intervals given in Tables 1.2 and 1.3 could you verify, llsing only equipIllentl()und in a typical dormitory room?
;IS
time
'·hands." Why is this a poor standard of length?
3. Find the order oflllaglliludc of your age in seconds. 4. An ol~ject with a mass or 1 k~weighs approximalely 21b. Use this inlornliltion to estimale the mass or the rollo\V'ing o~jects: (a) a baseball; (b) your physics teXlbook; (c) a pickup truck.
9. Ifan equation is dimensionally con'eel, does this mean that the equation mllst be true? If an equation is not dimensionally correct, docs this mean that the equation can't be true?
5. (a) Estimate the number 01' limes your heart beats ill a month. (b) Estimate the number of human heartbeats in an average lifetime.
10. \o\lhy is the metric system of units considered superior to mosl other systems or units?
6. E.stimate the llumber ol'atolJ1s in 1 cmol ora solid. (~ole th(llthc diamet.er of"an atom is
11. How can an estimate be or value even when it is ofT by an order of magnitudc? Explain and give an example.
PROBLEMS ~
WebAssign
I.~,
mil _
(II
Prublems ror thi:-. chapter assignee! online al vVebAs1'>ig'n.
= dcnolcs guided problem = dt:lloles CllhanC!;d contcilt probkm
= biomedical application
0= dCllotcs rul1 solution available in
SllIdI'II15io/llli()1I1 ,\lallllalj
Stud)' (;lIidl'
SECTION 1.3 DIMENSIONAL ANALYSIS I. The period of a simple pendulum, defined
;lS the lime necessary for one complete oscillatiun. is measured in time units and is given by
where e is the length or the pendulum and g is the ac.:celcration due to gravity. in units or length di\'ided by time squared. Show th;:}l this equation is dimensionally consistenL. (YOIl might want 1O check the [ormula using your keys ;11 the end of a strillg alld a slOpwaLch.) 2. (a) Suppose Lhal the displacemcnt of an object is re1
20
Chapter 1
Introduction
3. A shape thaL covers an area A and has a uniform heiglll II has a volume V = All. (a) Shm.. . thal V = Ali is dimensionally COITeel. (b) Show that the volumes of a cylinder and of a rectangular box can be wriLLcn in the form V = All, idclUirying A in each case. (Note that A. sometimes called the "fooLpriIIL" orthe object, can Ilave allY shape ~llid 1Iiat the height can, ill general, be replaced by the average thickness of the object.) UJE
')
<j11ltr
I
,>
= :i1n1Jo~ +
r----., V mgn
~
v =
' Vo + al-
Do a dimensional analysis or each equation and explain why liH: equation can't be correcl. 5. Newlon's law of universal gravitation is represented by
Mm F=G,.2 where F is the gravitational force, tv! and m are masses, and 1';S::l k:llgth. Force has the Slunits kg· lll/s2 . What are thc 51 units o[ the proponionality constant C?
6.
am Kinetic
energy KE (Chapter 5) has dimensions kg·m 2/s:!. It can be written in terms of the momentum f) (Chapter 6) and Illass III as
1/
KE~
2m
(a) Determine the proper units [or momentum using dimensional analysis. (b) Refer to Problem 5. Given the units of' force, write a simple equation relating a constant force Fexerted on an object. an interval oftimc {during which the force is applied, and the resulting momentum of the object, p.
the volume by llluitiplying by the Icngth. (d) Explain why the answers don't agree in the Ihird signillcant figurc. [I[]Thc radius of a circle is measured to be nO.5 ± 0.2) rll. Calculate (a) the arca and (b) the circumference of the circle, and givc the llncertainty in each value. 13. Carry Ollt the following arithmctic operations: (a) the sum of the measured values 756, 37.2, 0.83, and 2.5; (b) the product 0.003 2 X 356.3; (c) the product 5.620 X 1T. 14. (a) Using your calculator, find, in scienlific notation \.. . ith approprialc rounding, (a) the value of (2.437 X lO-t) (6.521 1 X 109)/(5.37 X 10") and (b) Ihe value of (3.14J 59 X 10')(27.01 X 10·')/(1234 X 10'l
SECTION 1.5 CONVERSION OF UNITS
115.IA fathom
is a unit of length, usually reserved 1'01' measuring the depth of water. A fathom is approximately 6 ft in length. Take the distance from Earth to the r"loon to be 250000 miles, and use the given
16. A furlong is an old British unit of ICllgth equal 10 0.J25 mi, derived from the length ofa furrow in an acre of ploughed land. A fortnight is a unit of time corresponding to two weeks, or 14 days and nights. Find the speed of light in megafurlollgs per fortnight. (One megafurlong equals a million furlongs.)
[J1JA {-irkin is an old British unit of"volume equal to 9 gallons. How many cubic meters are there in 6.00 firkins?
18. Find thc height or length or these nawra] wonders in kilometers, meters, and centimeters: (a) The longest cave system in the world is the .\lammoth Cave system in Central Kentucky, \\rith a mapped length of 348 miles. (b) In t.he United States, the waterf~dl with the greatest single drop is Ribbon Falls in California, which drops] 612 1'1. (c) At 20 320 feet, I'vlount McKinley in Alaska is America's highest mountain. (d) The dcepest canyon in the United States is King's Canyon in California, with a depth of 8200 fL
SECTION 1.4 UNCERTAINTY IN MEASUREMENT AND SIGNIFICANT FIGURES 7. A fishcrm
20. Using the data in Table 1.3 and the appropriate com'cr-
8. A rectangular platc has a length of (21.3 ::t 0.2) em and a width of (9.8 ::t 0.1) em. Calculate the area of the plate, including its uncertainLy.
21. Using the data in Table 1.1 and the appropriate conversion factors, find the distance to the nearest star ill leet.
[]JI-Iow many significant figures are thcre in (a) 78.9 ::t 0.2, (b) 3.788 X 10', (e) 2A(j X 10-'\ (d) 0.0032 10. The speed of light is now defined to be 2.997924 58 X 10 8 m/s. Express lile speed of light to (a) three significant figures, (b) Fl\'e significant figures, and (c) sevcn significant l-lgures.
11.
am A block of gold has length 5.62 em, ' . . idth
6.35 cm, and height 2.78 em. (a) Calculate the length times the width and roulld the answer 10 the appropriate llumber of significant figures. (b) Now multiply the rounded result of part (a) by llle height and again round, obtaining the volume. (c) Repeal the process, first finding the width times the height, rounding it, and then obtaining
EJ. A rectangubr building lot measures 1.00 X 10 2 ft by 1.50 X 10~ ft. Detcrmine the area of this lot in square meters (m:'}
sion factot"s, find the agc of Eanh in years.
22.
II Suppose your hair grows at the rate of 1/32 inch
per
day. Find the rate;Ol! which it grows in nanometers per sec-
ond. Because the distance between atoms in a molecule is on the order of 0.1 nm, rour answer suggests how rapidl)' atoms arc assembled in this protein synthesis.
23. Tile spced of light. is about 3.00 X figure
to
lO~
m/s. Convert this
miles per hour.
24, A house is 50.0 f1. long and ~6 ft wide and has 8.0-ft-higil ceilings. V\.'hat is the volume of the interior of the house in cubic meters and in cubic centimetcrs? 125.IThe amount of water in reservoirs is orlen measured in acre-ft. One acre·ft is a volu me that covers an area of one acre to a depth or one f"ooL An acre is 43 560 rt 2 . Find the
Problems volume in 51 units of a reservoi r contai ni ng" 25.0 acre-l"t of water.
26. The base of a pyralllid cover'S an area of 13.0 acres (I acre = 43560 f( 2) and has a height of 48] ft (Fig. P 1.26). II' the volume of a pyramid is givcn by the expression V = b1l/3, where b is the arca of the base and h is the height, find the volume of this pyramid in cubic melers.
34.
21
II Bacteria and otller prokaryOles arc found deep underground, in water, and in the air. One micron (10-6 m) is a typical length scale associated with thesc microbes. (a) Estimate the LOtal number of bacteria and Olher prokaryotcs in the biosphere or the EarLh. (b) Estimat.e t.he total mass of all such microbes. (c) Discuss the relative importance or humam and microbes to the ecology of planet Earth. Can Homo salJiew· survive without I hem?
m:J
SECTION 1.7 COORDINATE SYSTEMS 35. A point is located in a polar coordinate system by thc coordinates r = 2.5 m and () = 35°. Find the x- and )'coordinates 01' Ihis point, assuming that the tv..-o coordinate systems h,lve the same origin.
FIGURE P1.26
36. A certain corner of a room is selected as the origin of a rectangular coordinate system. ]f a fly is navv'ling on an adjacent wall at a point having coordinates (2.0, 1.0), where the units are meters, what is the distance of, he fly from the corner of the room? 37. Express the location of the rty in Problem 36 in polar coordillales.
27. A quan container orice cream is to be made in the form of a cube. Wh
SECTION 1.6 ESTIMATES AND ORDEROF-MAGNITUDE CALCULATIONS Note: In dcvclolJing answers to the problcrns in this section, yOll should state )'01.11' important assumptions, including the numerical values assigned to parameters uscd in the solution.
28. A hamburger chain advertises that it has sold more than 50 billion hamburgers. Estimate how many pounds of hamburger meat musl have been uscd by the chain and how many head of cattle wcre reqllired to furnish t.he meal.
29. Estimatc the number of Ping-Pong balls Ihat vv'otdd fit into a typical-sizc room (without being crushed). In your solution, state the quantities you mea.'Hlre or estimatc and lhe valucs you take for them. 30.
139.ITwo points are given in polar coordinates by (1; 0) = (2.00 m, 50.0°) and (1; 0 = (5.00 m, -50.0°), l·especlively. \Vhat is the distance between them? 40. _ Given points (r i , 0)) and (1'2' ()2) in polar coordinates, obtain a general for'mula for the distance between them. Simplify it as much as possible using t.he identit.y cos2 () + sin 2 () = l. Hint: Write the expressions for the two points in Cartesian coordinates and substitute into the usual distance f(Jrmula.
SECTION 1.8 TRIGONOMETRY 41. For the triangle shown in Figure P1.41, what are (a) the length or the unknown side, (b) the t.angent of 0, and (c) the sine of '!J?
II
Estimate the number of people in thc world who are suffering from the common cold on any given day. (Answers rna}' vary. Remcmber that a person suffers from it cold for abom a week.)
III (a) Ahout how mall)' microorganisms arc lound 31. _ in the human intestinal tract? (A rypical bacteriallcngt.h scale is 10--6 m . Estimat.c the intestinal volume and assume one hundredth of it is occupied by bacteria.) (b) Discuss your ans'.. .el· to pan (a). Are these bacteria beneficial, dangerous, or lIeutral? \'\'hat functions could they serve? 32.
38. Two points in a rectangular coordinate systcm have the coordinates (5.0, 3.0) and (-3.0, 4.0), ,...· here the units are centimet.ers. Determine t.he dist.ance between these points.
aim
Grass grows densely everywhere on a quarter-acre plot of land. What is the order of magllitlldc of the number of blades of'grass? Explain your reasoning. Note that I acre = . . 3 560 ft 2 .
133.IAn alllomobile tire is rated to last for 50 000 miles. Estimale the number of revolutions the tire will make in its lifetime.
e 6.00111 9.00111
FIGURE PI.41
42. A ladder 9.001ll long leans against the side or a building. If the ladder is inclined at an angle of 75.0° to the horizontal, what is the horizontal distance from the bottom of the ladder to the building? 143.\A high fountain ohvater is located at Ihe center ora circu· lar pool as shown in Figure PI.43. Not wishing to gCt his reel. wet, a s(Uc!em walks around the pool and measures il.s circllmference to be 15.0 m. Next, the student stands
22
Chapter 1
Introduction
at the edge of" the pool and uses a prmracwr La gallge the angle ofclc\',ltion at the bOlLom of the fountain to be 55.0°. How high is the fountain?
50.
mJ
Refer 10 Problem 48. Suppose the mOllntain height is y, the woman's origin~l diswncl: from the mountain is x, and the angle of c1e"
ADDITIONAL PROBLEMS
FIGURE P1.43
44. .-Ii riglll trianglt.: has a hypotenuse or length 3.00 In, and one of its angles is 30.0". What arc the lengths of (a) the side opposite 1he 30.0°
51. (a) One of thc fundamelllal laws of mOLion states thal the acceleration of an object is directly proportional to the resultanl force on it and inversely proportional to its mass. I r the proportionality constant is defined to have no dimensions, determine the dimensions of force. (b) The newton is the SI unit of" force. According La the results li)1' (a), bow can you express a force ha"ing lInits of newLons by usi ng the fumlamental units of mass, leng'L!l, ami time?
52.
45. In Figure Pl.45, find (a) the side opposite e. (b) the side (lcUaccnt to ¢, (c) (Os (d) sin cb, and (c) tan 4),
e,
(;1) Fincl a conversion faClOl- (a converL from miles per hour to kilometers per hour. (b) For a while. federal law mandated that the maximum highway speed would be 55 Illi/h. Use the conversion factor from part (a) to find the speed in kilOlneters pel' hour. (c) The maxilllum highway speed has been raised 10 65 mi/h in some places. (11 kilometers per hou 1', how much of an inCl'ease is this O\'er the 55~ll1ijh
53.
3.00
4.00 FIGURE PUS
46. In a certain right triangle, lhe two sides that are perpendicular to each other arc 5.00 m and 7.00 In long. \·Vhat is the length of tile third side..: orlhe triangle?
47. In Problem 46, whal is the Langenl of the ang'le lor which 5.00 m is the opposite side? 48.
EI
A woman measures the angle of eleV
149.IA sllrn~ror measures the distance across a stnlight river by thc f(lllowing mcthod: Starting directly across from a trce on the opposite bank, he walks 100 III along the 1'iverbank to establish a baseline. Theil he sights
limil?
II
One cubic centimeter (1.0 CIll~) ofwatc.:r has a mass of 1.0 X JO ~l kg. (a) Detnmine lhe lllass of 1.0 m~ of water. (h) Assuming that biological substances are 98% water, estimale the masscs of a cell with a diamc.:ler of 1.0 ,um, a human kidney, and a fly. Take a kidney to be roughly a sphere with a radius of 4.0 cm and a fly to be roughly a q,linder '1.0 mill long and 2.0 mm in diameter.
54. Soft drinks are cOlllmonly sold in aluminum containcr'i. To an ordcr of magnitude. how mall}' such containers arc I hl"Own away or recycled each ycar by U.S. consumcrs? How mallY tOllS ofalulliilllll11 docs this represent? In your solmion. stalC the qUi1l1lilies you measure or estimate anel I he valucs you takc for Lhcm. 155.IThe displflccillent of an object moving under uniform acceleratioll is somc fUllction 01" time and the acceleration. Suppose we writc this displacement as s = /1(f"'1", where k is a dimensionless constant. Show by dimensional analysis that this expression is slltisfierl if' 'Ill = I aile! 1/ = 2. Can the analysis give the ",tluc of Ii? 56. Compute I he order of magnilude of the mass of (a) a bathtub filled with water and (b) a ba1.hlllb filled with pennies. In )'our solution. list the quantities you estimate
57. You can obtain a rough estimatc of the size of a l11ole
4
cllle by the following simple expcrimenl: LeI a droplcl of oil spread OUl on a smooth surface of water. The l"esu!ling oil slick will be approximately one molecule lhick. Given an oil dmplct of mass 9.00 X 1O~7 k~ and density 91S kg/J11:~ thm spreads out into
Problems
58. _ Sphere 1 has surfacc area Al and '"OIUlllC \"1' and sphere 2 has surface area A 2 and volume \~. If the radius of sphere 2 is double the radius of spherc I, what is lhe rmio of (3) the areas, ,4:/A. and (b) the volumcs, V2/VI ? 59, E.Slimale the number of piano tuners living in New York City. This queslion was raised by the physicist Enrico Fermi, who was well known for making order·ofmagnitude calculations.
60. 111 2007, thc C.S. national debt was about $9 trillion. (a) If payments were made at the rate of 1000 per second, how man) ycars would il take to payoff the debt, assuming that no interest were ch::u'gcd? (b) A dollar bill is about 15.5 em long. lfnine trillion dollar bills ,..' ere laid end (Q end ilround the Earth"s equator, how many times would they encircle the planet? Take the radius of t.he Earth al the equaLOf 10 be 6378 km. (!Vole: Before doing
23
anv of these calculations, try to guess at the answers. You may be very surprised_) ffiIJ{a) How many seconds arc there in a year? (b) II' one micrometeorite (a sphere with a diamcter on the order of 10 -(0111) st.-uck each sqlla,,: meter of the :\Ioon each second. estimate the number of years it would take to CO\-<:,' the Y1001l with micrometeoritcs to a depth of one metc!"_ (flint: Consider a cubic box, 1 III 011 a side, on the l\.loon. and find how long it would lake 10 fillihe box.)
62. Imagine that you are the <,C]uipmcllllllanager of a professional baseball team. One of your jobs is to keep baseballs on hand for games. Balls are sometimes lost ..... hen pl<\\-l'rs hit them into the slallds as either home runs or foul halls. Estimate how man}' baseballs rou ha\-c to buy pcr season in order 10 make up for such losses. Assume that ),our team plays an 8]-gaml.' home schedule in a seasoll.
Croig Breedlove, five times world land speed record holder, accelerates across the Block Rock Desert in Gerlach, Nevada, in his jetpowered car, Spirit of America, on its first test run on September 6, 1997. Subsequent jet-powered cars hove broken the sound borrier on
land
2.1
Displacement
2.2
Velocity
2.3
Acceleration
2.4
Motion Diagrams
2.5
One-Dimensional Motion with Constant Acceleration
2.6
Freely Falling Objects
MOTION IN ONE DIMENSION Ufe is motion. Our muscles coordinate motion microscopically to enable us to walk and jog. Our hearts pump tirelessly for decades, moving blood through our bodies. Cell wall mechanisms move select atoms and molecules in and out of cells. From the prehistoric chose of antelopes across the savanna to the pursuit of satellites in space, mastery of motion has been critical to our survival and success as a species. The study of motion and of physical concepts such as force and moss is coiled dynamics. The part of dynamics that describes motion without regard to its causes is called kinematics. In this chapter the focus is on kinematics in one dimension: motion along a straight line. This kind of motion-and, indeed, any motion-involves the concepts of displacement, velocity, and acceleration. Here, we use these concepts to study the motion of objects undergoing constant acceleration. In Chapter 3 we will repeat this discussion for objects moving in two dimensions. The first recorded evidence of the study of mechanics can be traced to the people of ancient Sumeria and Egypt, who were interested primarily in understanding the motions of heavenly bodies. The most systematic and detailed early studies of the heavens were conducted by the Greeks from about 300 S.c. to A.D. 300. Ancient scientists and laypeople regarded the Earth as the center of the Universe. This geocentric model was accepted by such notables as Aristotle (384-322 s.c.) and Claudius Ptolemy (about A.D. 140). Largely because of the authority of Aristotle, the geocentric model become the accepted theory of the Universe until the 17th century. About 250 B.C., the Greek philosopher Aristorchus worked out the details of a model of the Solar System based on a spherical Earth that rotated on its axis and revolved around the Sun. He proposed that the sky appeared to turn westward because the Earth was turning eastward. This model wasn't given much consideration because it was believed that a turning Earth would generate powerful winds as it moved through the air. We now know that the Earth carries the air and everything else with it as it rotates. The Polish astronomer Nicolaus Copernicus (l473-1543) is credited with initiating the revolution that finally replaced the geocentric model. In his system, called the heliocentric
model, Earth and the other planets revolve in circular orbits around the Sun.
24
1.\
This early knowledge formed the foundation for the work of Galileo Galilei (1564-1642),
Displacement
2S
SCHWADRON
USA
who stands out as the dominant facilitator of the entrance of physics into the modern era. In 1609 he become one of the first to make astronomical observations with a telescope. He observed mountains on the Moon, the larger satellites of Jupiter, spots on the Sun, and the phases of Venus. Galileo's observations convinced him of the correctness of the Copernican theory. His quantitative study of motion formed the foundation of Newton's revolutionary work in the next century. ;; ~
2.1
~
DISPLACEMENT
i
Motion involves the displacement of an object from one place in space and time ~ to another. Describing motion requires some convenient coordinate system and a -"'iss Oempswell, send in • frMne of reIerenee.specified origin. A frame of reference is a choice of coordinate axes that defines FIGURE 2.1 the starting point for measuring any quantity, an essemial first step in solying vir· (ually any problem in mechanics (Fig. 2.1). In Active Figure 2.2a, for example. a car moves along the x-axis. The coordinates of the car at any time describe its position in space and, more importantly. its disjJlarP1npnt at some given time of interest.
""
~
The displacement Ilx of an object is deAned as its change in position, and is given by t>x
= .'1- .<,
Definition of displacement
[2.1]
where the initial position of the car is labeled ;"'j and the final position is xf' (The indices iandJstand [or initial and final, respectively.)
Tip 2.1 A Displacement Isnlt a Distance! The displacclllclll of an object is not the same as the distallce it travels. To.,>s a tennis ball up and catch it. The ball travels a tlif.lo.nce equal to twice the maximum height reached, but its dis/J!aumell{ is tcro.
(a) t\ car lUO\-es back and forth along a ~traighlline takeuto be the x-axis. Becau~e W~ arc interested ani}' ill the car's lranslalional mOlion, we can model it as a particle. (I» Graph or position \"s. time for the mol ion of lhe -particle .
@
0
-:,W
..
®
-40
® I (5)
-60 (a)
()O :I:(m)
0
\0
30
20 (bl
40
50
26
Chapter 2
Motion in One Dimension
Tip 2.2 Vectors Hove Both 0 Mognitude ond Direction.
0
Scalal-S have size. VCUOfS, too, ha\'c size, but they also indicate a direclion.
has magnitude, but no direction. Scalar quantities such as mass and temperature are completely specified by a numerjc value ''',lith appropriate units; no direction is involved. Vect.or quantities will be usually denoted in boldface lype with an arrow over the top of the letter. For example, v represents velocit)' and it denotes an acceleration, bOlh vector quantities. Tn this chapter, however, it won't be necessary to use that notation because in one-dimensional motion an object can only move in one of two directions, and these direct.ions arc easily specified by plus and minus signs.
2.2
VELOCITY
Tn everyday usage the terms sj)ccrl and velocity are interchangeable. In physics, however, there's a clear distinction between them: Speed is a scalar quantity, having on ly magnitude, whereas velocity is a vector, having both magnitude and direction. vVhy must velocity be a vector? If you want to get to a town 70 km away in an hour's Lime, it's not enough to drive at a speed of 70 km/h; you must travel in the correct direction as well. This is obvious, but shows that velocity gives considerably more information than speed, as 'wi 11 be made more precise in the formal definitions. Definition of overage speed -i>
The average speed of an object over a given time interval is lhe LOlal dislance traveled divided by the total time elapsed: Average speed
total distance total time
81 unit: meter per second (m/s) symbols, this equation might be written 11 = dll, ",,,ith the letter rJ understood in context to be the average speed, not a velocity. Because total distance and total time arc ahlo,'ays positive, the average speed will be positive, also. The definition of average speed completely ignores 'what may happen between the beginning and the end of the motion. For example, you might drive from Atlanta, Georgia, to St. Petersburg, Florida, a distance of about 500 miles, in 10 hours. Your average speed is 500 mi/l0 h = 50 mi/h. It doesn't matter if you spent two hours in a traffic jam traveling only 5 mi/h and another hour at a rest stop. For average speed, only the total distance traveled and total elapsed time are important. ]11
EXAMPLE 2.1 Goal
The Tortoise and the Hare
Apply the concept of average speed.
Problem A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops [or a gO.O-min nap. Upon awakening, he remembers the race and runs twice as fast. Finishing the course in a lotal time of 1.75 h, the rabbit wins the race. (a) Calculate the average speed of the rabbit. (b) "Vhat was his average speed before he stopped for a nap? Strategy Finding the overall average speed in part (a) isjust a matter of dividing the total distance by the total time. Part (b) requires lWO equations and two unknowns, the latter turning out to be the two different average speeds: VI before the nap and v2 after the nap. One equation is given in the statement of the problem (v2 = 2v l ), whereas the other comes from the factlhe rabbit ran for only 15 minutes because he napped for gO minutes. ".
~
Solution (a) Find the rabbit's overall average speed.
Apply the equation for average speed:
Average speed
total distance total time 2.29 km/h
4.00 km 1.75 h
2.2
Velocity
27
(b) Find the rabbit's average speed before his nap. Surn the running times, and seL the Slim equal to 0.25 h:
II
+
= 0.250 h
12
(I)
Substitute V2 = 2v, and the values of til and d';!. inLO Equation (1):
Remark As seen in this example, average speed can be calculated regardless of any variation in speed over the given time interval. QUESTION 2.1 Does a doubling of an object's average speed always double the magnitude of its displacement in a given amount of time? Explain. EXERCISE 2.1 Estimate the average speed of the Apollo spacecraft in meters per second, given thal the craft LOok five days to reach the Moon from Earth. (The Moon is 3.8 x 1Qli rn from Earth.)
Answer
- 900 m/s
Unlike average speed, average velocity is a veeLOr quantity. having both a magnitude and a direction. Consider again the car of Figure 2.2, moving along the road (the x-axis). Let the car's position be xj at some time and x/at a later time 'I' In the Lime interval at = t.r - ti , the displacerncl1l of the car is.6.x = -'i - Xi"
'j
The average velocity divided by tJ./:
v during
a tirne interval
at
is the displacement
ax
TABLE 2.1
51 unit: meter per second (m/s)
Unlike the average speed, which is always positive. the average velocity of an object in one dimension can be either positive or negative. depending on the sign of the displacement. (The lime interval ~I is always positive.) In Figure 2.241, for example, the average ,'clocil)' of the car is positive in the upper illustration, a positive sign indicating motion to the right along the x-axis. Similarly, a negative average velocity for the car in the lower illustration of rhe figure indicates that it moves to the left along the x-axis. As an example, we can use the data in Table 2.1 to find the average velocity in the time interval from point ® to point@ (assume two digits are significant):
ax
52
m -
Definition of average velocity
[2.2]
v==:tJ.I
V ~ -tJ.-I
+-
30
Position of the Car at Various Times Position I (s) @ @
Aside froll) meters per second, other common units for average velocity are feet per second (ft/s) in the U.S. customary system and centimeters per second (cm/s) in the cgs system. To further illustrate the distinction between speed and velocity, suppose we're watching a drag race from the Goodyear blimp. In one run we see a car follow the straight-line path from ® to @ shown in Figure 2.3 during the time interval 1J.t,
®c:>Xi
@
~l
x
FIGURE 2.3 A drag race viewed from a blimp. One car follows the red slraighl-line palh from ® [email protected] :l second car follows lhe h[lJe curved p:llh.
28
Chapter 2
Motion in One Dimension
and in a second run a car follm.. . s the curved path during the same interval. From the definition in Equation 2.2, the two cars had the same average velocity because they had the same displacement = XI ~ Xi during the same time interval dt. The car taking the curved route, however, traveled a greater distance and had the higher average speed.
ax
( 100 yd
FIGURE 2.4 (Quick Quiz 2.1) The path followed by a confused football player.
(TIP 2.3 Slopes of Graphs '\ The word slope is often lIsed in reference to the graphs of physical data. Regardless of the type of data, the stolle is given by
change in vertical axis
Slope
=
change
.. ttl
honzontal
. aXIS
Slope carries units.
TIP 2.4 Average Velocity vs. Average Speed Average velocity is nollhe same as average speed. If you nm from x= 0 III 10 x= 25 malld back to your starting point in a Lime
int.erva\ of 5 s, the average velocity is zero, whereas the average
speed is 10 mho
QUICK QUIZ 2.1 Figure 2.4 shows the unusual path of a confused football player. AFter receiving a kickoff at his own goal, he runs down field to within inches of a touchdown, then reverses direction and races back until he's tackled at the exacllocation where he first caught the ball. During this run, which took 25 s, what is (a) the total distance he travels, (b) his displacement, and (c) his average velocity in the x-direction? (d) Whal is his average speed?
Graphical Interpretatian of Velocity If a car moves along the x-axis from ® to @
Instantaneous Velocity Average velocity doesn't take into account the details of what happens during an interval of time. On a car trip, for example, you may speed up or slmv down a number of times in response to the traffic and the condition of the road, and on rare occasions even pull over to chat with a police officer about your speed. '<\That is most important to the police (and to your own safety) is the speed of your Cal" and the direction it was going at a particular instant in time, which together determine the car's instantaneous velocity.
--liO l-_---'-,-_--'-:-_ _L-_----',-_-' I (s) o 10 ~O :~() 40 50
'10
(b)
(a)
So in driving a car bctwcen two points, the average velocity must be computcd over an intcrval of timc, but the magniLUde of instantaneous velocity can be read on the car's spcedomctcr. The instantancous velocity v is the limit of the a\'erage velocity as the time interval !:J.t becomes infinitesimally small:
v
~>.
= lim -!:it
~
Definition of instantaneous velocity
[2.3]
~I-.O
51 unit: meter per second (m/s) The notation lim mcans that thc ratio a"jD.l is repeatedly evaluated for sl11
v
TABLE 2.2 Positions of a Car at Specific Instants of Time
TABLE 2.3 Calculated Values of the Time Intervals, Displacements, and Average Velocities for the Car of Table 2.2
I (s)
x (01)
1.00 1.01
5.00 5.-17
1.00
3.00
2.00
'17.3
1.10
9.67
1.00 102.00
1.00 0.50 0.20
29.7
1.20 1.50 2.00 3.00
10
1'1.3
1.00 to 1.50
26.3 3'1.7
1.00 to 1.20 1.00 to 1.10
5~.5
1.00 to 1.01
29
FIGURE 2.5 (a) Position "S. time graph for the motion ofa car moving along the x-axis at constant \'elodty. (b) Position \'S. time graph for the mOl ion of a car with changing \·e1odty. using the data in Table 2.1. The a\'crage \'c1ocity in the time interval from t = 0 s to t = 30 s is the slope of the blue straight linc connecting ® and @.
FIGURE 2.6 Graph represeming the motion oflile car from the data in Table 2.2. The slope of the blne line represents the average \'e1ocitr for smaller
x (m)
As can be seen 111 Figure 2.6, the chords formed by the blue lines gradually approach a tangent line as the time interval becomes smaller. The slope of the line tangent to the position vs. time curve at "a given time" is defined to be the instantaneous velocity at that time. The instantaneous speed of an object, which is a scalar quantity, is defined as the magnitude of the instantaneous velocity. Like average speed, instantaneous speed (which we will usually call, simply, "speed") has no direnion associated with it and hence carries no algebraic sign. For example, if one o~iect has an instantaneous velocity of + 15 mls along a given line and another object has an instamaneous velocity of -15 mls along the same line, both have an instantaneous speed of 15 m/s.
EXAMPLE 2.2
Slowly Moving Train
Goal Obtain average and instantaneous velocities from a graph.
Problem A train moves slowly along a straight portion of track according to the graph ofposiLion versus time in Figure 2.7a. Find (a) the average velocity for the total trip, (b) the average velocity during the first 4.00 s of motion, (c) the average velocity during the next 4.00 s of motion, (d) the instantaneous velocity at t = 2.00 s, ancl (e) the instantaneous velocity at / = 9.00,.
10 x (Ill)
8
8
6
6
4
4
2
2 2
4
(;
8
[0 12
t (s)
@
(a) FIGURE 2.7
2
4
6
8
(hi
(a) (Example 2.2) (b) (Exercise 2.2)
Strategy The average velocities can be obtained by substituting the data into the definition. The instantaneous velocity at t = 2.00 s is the same as the average velocity at that point becallse the position \'s. time graph is a straight line, indicating constant velocity. Finding t.he instantaneous velocity when {= 9.00 s requires sketching a line tangent to the curve at that point and finding its slope. ~
(b) Find the average velocity during the first 4 seconds of
the t.rain's motion. Again, find the slope:
~
II.x
v~-~---=
11./
(c) Find the average velocity during the next 4 seconds.
Llx
Here, there is no change in position, so the displacement ~x is zero:
V ~ -
11./
0 In 4.00 s
= -- =
0 ml'
-.~
2.3
Acceleration
31
(d) Find the instantaneous velocity at t = 2.00 s. This is the same as the average velocity found in (b), because the graph is a straight line: (e) Find the instantaneous velocit.y at
(=
v=
9.00 s.
The tangent line appears to intercept the ;\'-axis at (3.0 s. o Ill) and graze the curve at (9.0 s, 4.5 m). The instantaneous velocity at ( = 9.00 s equals the slope of the tangent line through these points:
(a) 0 m/s (b) both are +0.5 m/s (c) 2 m/s (d) balh are -2.5 m/s
ACCELERATION
Going from place to place in your car, yOll rarely travel long distances al constant velocity. The velocity of the car increases when you step harder on the gas pedal and decreases when yOll apply the brakes. The velocity also changes when you round a curve, altering your direction of motion. The changing of an object'S velocity with time is called acceleration.
Average Acceleration A car mo\'es along a straight highway as in Figure 2.8. At time tj it has a velocity of v" and at time t.rits velocity is v/' with 6.v = VI - VI and a, = i.r - Ii' The average acceleration ity Il.vdivided by 1l.1:
a during the time intervalllI is the change in veloc-
t- Definition of averoge acceleration
_
llv
VI-Vj
Il.l
If - I,
a;::::-=---
[2.4)
51 unit: meter per second per second (m/s~) For example, suppose the car shown in Figure 2.8 accelenHes from an initial velocit)' of vi = +10 m/s to a final velocity of VI = +20 m/s in a time interval 0[2 s. (Both velocities are toward the right, selected as the positive direction.) These values can be inserted into Equation 2.4 to find the average acceleration:
_
a
~
Il.v
Il.!
=
20 m/s - lO mi. 2.
=
" --"lI,
<.
+5 m/s'
Acceleration is a vector quantity having dimensions of length divided by the lime squared. Common units of acceleration are meters per second per second «m/s)/s, which is usually written m/s2 ) and [eet per second per second (ft/s2 ). An
FIGURE 2.8 A car moving to the ri~hl accelerates from a velocity or Tli 10 a ....elocily or v} in the lime interval !:J.t = 1/- I,.
32
Chapter 2
Motion in One Dimension
average acceleralion oE +5 m/s~ means that, on average, the car increases its velocity by.5 m/s every second in the positive x-direction. For the case oEmotion in a straight line, the direction of the velocity oEan object and the direction of its acceleral.ion are related as foIIO\,,'s; When the object's velocity and acceleration are in the same direction, the speed of the object increases with time. ''''hen the object's velocity and acceleration are in opposite directions, the speed of the object decreases with time. To clarify this point, suppose the velocity of a car changes from -]0 m/s to -20 m/s in a time interval of 2 s. The minus signs indicate that the velocities of the car are in the negative x-direction; they do not mean that the car is slmving dov..'n! The average acceleration of the car in this time interval is
TIP 2.5 Negative Acceleration Negau\'c acceleral ion doesn't necmean an ()1~jccl is slowing down. I r the acceleration is negath'c and the velocity is also negat.ive, the ol~jcct is speeding up! essaril~'
_ Liv -20 a=-= Lit
'I
TIP 2.6 Deceleration The word deceleration means (l reduction in speed. a slowing down. Some confuse it with a negative acceleration, which can speed something up. (See Tip
,
2.5.)
)
m/s - (-10 m/5) 2s
,
=-.?m/s-
The minus sign indicates that the acceleration vector is also in the negative x-direction. Because the velocity and acceleration vectors are in the same direction, the speed of the car must increase as the car moves to the left. Positive and negative accelerations specify directions relative to chosen axes, not "speeding up" or "slm'\'ing down." The terms "speeding up" or "slm'\'ing down" refer to an increase and a decrease in speed, respectively.
QUICK QUIZ 2.2 True or False? (a) A car must always have an acceleration in the same direction as its velocity. (b) It's possible for a slowing car to have a positive acceleration. (c) An o~ject 'with constant nonzero acceleration can never stop and remain at rest.
Instantaneous Acceleration The value of the average acceleration often diEters in diflerent time intervals, so it's useful to define the instantaneous acceleration, which is analogous to the instantaneous velocity discussed in Section 2.2. Definition of instantaneous
acceleration
-+
The instantaneous acceleration a is the limit of the average acceleration as the time intervalul goes to zero:
a
=
Liv
lim -
~(---+()
D.i
[2.5]
Sf unit: meter per second per second (m/s2 ) Here again, the nOlation lim means that the ratio Li.v/flt is evaluated for smaller ()
~v
Ti
Slope
=
a = t:i
, Vi
® ' -~- ------------~
,I • ,, " FIGURE 2.9
,
~ I ---------»j
,
,I 'J Velocity\'s. time graph
for an object moving ill a straight line.
The slope oflhe blue line connecting points
®
and @isdefinedasthe<\\"cr-
age acceleration in the time inLen'al ~/ = 11 = ',"
and smaller values of nt. The closer C1t gets to zero, the closer the ratio gets to a fixed number, which is the instantaneous acceleration. Figure 2.9, a velocityvs. time graph, plots the velocity' oran o~jectagainst time. The graph could represent, [or example, the motion of a car along a busy street. The average acceleration of the car between times t i and trcan be found by determining the slope of the line joining points ® and @. Ifwe imagine t.hal. point@ is brought closer and closer to point ®, the line comes closer and closer to becoming tangent at ®. The instantaneous acceleration of an object at a given time equals the slope of the tangent to the velocity vs. time graph at that time. From nm.,.. on, we will use the term accelemt.ion to mean "instantaneous acceleration." In the special case where the velocity vs. time graph of an ol~ieet's motion is a straight jine, the instantaneous <-lcceJeration of the object at any point is equal to its average acceleration. This also means thaI. I.he tangent line to the graph overlaps the graph itself. Tn that case, the object's acceleration is said to be lm~forJn, which means that it has a constant value. Constant acceleration problems are important in kinematics and ,"viII be studied extensively in this and the next chapter.
2.3 FIGURE 2.10 (Quick Quiz 2.3) Malch each veiocilY \'s. lime graph 10 ils corresponding acccler,llion \'s. lime graph.
v
(a)
(bl
(d)
Ie)
33
Acceleration
lei (c)
(f)
QUICK QUIZ 2.3 Parts (a), (b), and (c) of Figure 2.10 represent three graphs of the velocities of ditferent objects moving in straight-line paths as functions of time. The possible accelerations of each object as functions of tilTIC an.~ shown in parts (d), (e), and (f). Match each velocity vs. time graph "','lth the acceleration vs. time graph that best describes the motion.
Catching a Fly Ball
EXAMPLE 2.3 Goal
Apply the definition of instantaneous acceleration.
Solution Acceleration at ®. The acceleration at ® equals the slope of the line connecting the points (0 s, 0 m/s) and (2.0 s, 4.0 m/s): Acceleration at
Assume the player is initially moving in the positive x-direction. For the first 2.0 s, the ballplayer moves III the positive x-direction (the velocity is positive) and steadily accelerates (the curve is steadily rising) to a maximum speed of" 4.0 m/s_ He moves for 1.0 s at a steady speed of 4.0 m/s and then slmvs down in the last second (the v VS. t curve is [aIling), still moving in t.he positive x-direction (v is always positive). Remarks
QUESTION 2.3 Can the tangent line to a velocity vs. time graph evel be vertical? Explain. EXERCISE 2.3 Repeat the problem, llsing Figure 2.11b. Answer
ACTIVE FIGURE 2.12 (a) :\1otion diagram for a car moving at conSlant velocity (zero acceleration). (b) r-dotion diagram fur a cal undergoing constant acceleration ill the direction or ils vclocity. The rdocit)' "enol' at each instant is indicated by a red arrow and the constant acceleration vector by a viulet arrow. (e) Motion diagram for a car undergoing constant acceleration in the direction o!)!){)sile Llle velocity al each instanL
2.4 MOTION DIAGRAMS Velocity and acceleration arc sometimes confused with each other, but they're very different concepts, as can be illustrated with the help of motion diagrams. A motion diagram is a representation of a moving object at successive time intervals, with velocity and acceleration vectors sketched at each position, red for velocity vectors and violet for acceleration vectors, as in Active Figure 2.12. The time intervals between adjacent positions in the motion diagram are assumed equal. A motion diagram is analogous to images resulting from a stroboscopic photograph of a moving object. Each image is made as the strobe light flashes. Active Figure 2.12 represents three sets of strobe photographs of cars moving along a straight roadway from left to right. The time intervals between flashes of the stroboscope are equal in each diagram. In Active Figure 2.12a, the images of the car are equally spaced: The car moves the same distance in each time interval. This means Lhat Lhe car moves ,,·:ith conslant jJosilive veloell)l and has zero acceleration. The red arrows are all the same leng·th (constant velocity) and there are no violet arrows (zero acceleration). In Active Figure 2.1 2b, the images of the car become farther apart as time progresses and the velocity vecLor increases with time, because the car's displacement between adjacent positions increases as time progresses. The car is moving with a jJositive velocity and a constant positive accelemtion. The red arruws are successively longer in each image, and the violet arrows point to the right. In Active Figure 2.12c, the car slo\1ls as it moves to the right because its displacement betv.'een a~jacenL positions decreases with time. In this case, the car moves initially Lo the right with a constant negative acceleration. The velocity vector decreases in time (lhe red arrows gel shaner) and eventually reaches zero, as \-vould happen when the brakes are applied. Note lhat the acceleration and velocity' vectors are not in the same direction. The car is moving with a !Jositive ve[ocil)', but with a negative acceleration. Try constructing your own diagrams for various problems involving kinematics.
QUICK QUIZ 2.4 The three graphs in Active Figure 2.13 represent the position vs. time for objects moving along the x-axis. vVhich, if any, of these graphs is not physically possible? ACTIVE FIGURE 2.13 (Quick Quiz 2.4) \\'hich position lime curve is impossible?
x
x
x
\"5.
(a)
(b)
(e)
2.5
One-Dimensional Motion with Constant Acceleration
QUICK QUIZ 2.5 Figure 2.14a is a diagram of· a multi nash image or an air puck moving to the right on a horizuntal surface. The images sketched are separated by equal time intervals, and the first. and last images show the puck at rest. (a) Tn Figure 2.14b, 'which color graph best shows the puck's posiLion as a function of time? (b) In Figure ~.14c, which color graph best shows the puck's velocity as a function of time? (c) In Figure 2.14d, "lhich color graph best sho"..'s the puck's acceleration as a function of time?
(a)...
•
•
35
.
•
-
(b)~~ v
+
2.5 ONE.DIMENSIONAL MOTION
(e)
WITH CONSTANT ACCELERATION Many applications of mechanics involve objects moving "\lith constant acceleration. This type of motion is irnponam because it applies to numerous objects in nature, such as an o~jcct in free fall near Earth's surface (assuming air resistance can be neglected). A graph of acceleration versus time for motion with constant acceleration is sho\',ln in Active Figure 2.15a. When an object moves with constant acceleration, the instantaneous acceleration at any point in a time interval is equal to the value of the average acceleration over the entire time interval. Consequently, the velocity increases or decreases at the same rate throughout the motion, and a plot of v versus l gives a straight line Wilh either positive, zero, or negative slope. Because the average acceleration equals the instantaneous acceleration when a is constant, we can eliminate the bar used to denote average values from our defining equation for acceleration, \..-riting = a, so that Equation 2.4 becomes
..,.i'~~----l-~ o a
+ f-..,..,,-==:==~
FIGURE 2.14 (Quick Quiz 2.5) Choose the correCl graphs.
a
Slope
V j - VI
'=
0
a=--If - I,
The observer timing the motion is always at liberty to choose the initial time, so for convenience, let ti = 0 and '.r be any arbitrary time t. Also, let Vi = Vo (the initial velocity alt = 0) and vJ = V (the velocity at any arbitrary time t). \tVith this notation, we can express the acceleration as V -
Vo
a=---
a
o (a)
v
I
or V=Vo+a.t
[2.6]
(Jor constant a)
Equation 2.6 stales that the acceleration a steadily changes the initial velocity V o by an amount at. For example, if a car starts with a velocity of +2.0 mls to the right and accelerates to the right 'with ({. = +6.0 m/s~, it ,..,ill have a velocity of +14 mls after 2.0 s have elapsed: v
=
V
o + at ~ + 2.0 m/s + (6.0 m/s')(2.0 s)
= +14 m/s
The graphical interpretation of v is shown in Active Figure 2.15b. The velocity varies linearly with time according to Equation 2.6, as it should for constant ~1Ccelera tion. Because the velocity is increasing or decreasing uniformly with time, we can express the average velocity in any time interval as the arithmetic average of the initial velocity V o and the final velocity v: Vo + v v=---
2
(b)
x
XO>l-_"""~
----- Slope,= Vo
o (for constant a)
[2.7]
Remember that this expression is valid only ,'vhen the acceleration is constant, in which case the velocity increases uniformly. \'Ve can now use this result along with the defining equation for average velocit.y, Equation 2.2, to obtain an expression for the displacement of an object as a
(e) ACTIVE FIGURE 2.15 A particle moving along the x-axis wir.h constallt acedcraliOIl a. (a) the acceleration \'s. lillie graph, (b) the \'e!ocity \'s. time graph, and (e) the position vs. tillle graph.
36
Chapter 2
Motion in One Dimension
ABLE 2.4 Equations for Motion in a Slraight Line Under Constant Acceleration Information Given by Equation
Equation
v - Uo ~x
if
= =
+
vol
v o::!
at
+ ~al::! + 2a8.x
Velocity as a function of time Displacement as a function of lime VelociLr as a [unction of displacement
NOle: \ilolilltl is along the x-axis. All- U, the velocityofthc panicle is
Voo
function of time. Again, we choose Ii = 0 and t = t, and [or convenience, we write f ax = Xf - Xi = X - xo- This results in
Ilx =
v{ =
(
vo 2+
V)I
(for constant a)
[2.8]
We can obtain another useful expression for displacement by substituting the equation for v (Eq. 2.6) into Equation 2.8:
Ilx ~
Hvo +
110
+
al)1
(for constant a)
[2.9]
This equation can also be written in terms of the position x, since ax = x - xo' Active Figure 2.15c shows a plot of xversus I [or Equation 2.9, which is related to the graph o[velocity vs. time: The area under the curve in Active Figure 2.15b is equal to vol + ~al2, which is equal to the displacement .:lx. In fact, the area under the graph of v versus t for any object is equal to the displacement .6.x of the object. Finally, we can obtain an expression that doesn't contain time by solving Equation 2.6 for I and substituting into Equation 2.8, resulting in
n.x
= '2I( 11
+
Vo
)(v--a- vo) =
20.
(for constant a.)
[2.10]
Equations 2.6 and 2.9 rogether can solve any problem in one-dimensional motion with constant acceJermion, but Equations 2.7, 2.8, and, especially, 2.10 are sometimes convenient. The three most useful equations-Equations 2.6, 2.9, and 2.10-are listed in Table 2.4. The best \vay to gain confidence in the use of these equations is to work a number of problems. There is usually more than one way to solve a given problem, depending on which equations are selected and what quantities are given. The difference lies mainly in the algebra.
GTA:EY ACCELERATED MOTION The following procedure is recommended for solving problems involving accelerated motion. I. Read the problem.
2. Draw a diagram, choosing a coordinate system, labeling initial and final points, and indicating direClions of velocities and accelerations with arrows. 3. Label all quantities, circling the unknowns. Convert units as needed.
2.5
One-Dimensional Motion with Constant Acceleration
4. Equations from Table 2.4 should be selected next. All kinernatics problems in this chapter can be solved with the first t\'1'0 equations, and the third is often convenient.
/
37
\
TIP 2.7 Pigs Don't Fly After solving a problem, }'ou should think about yOllr answer and decide whether it seems reasonable.If it isn·t. look for yOLlr
5. Solve for the unknm\'ns. Doing so often involves solving two equations for two unknowns. It's usually more convenient to substitute all kno'\1Jl values before solving.
mistake! ,/
6. Check your ans'.. .er, using common sense and estimates. Most of these problems reduce to writing the kinematic equations from T<:lble 2.4 and then substituting the correct values into the constants a, v o, and xl) from the given information. Doing this produces two equations-one linear and one quadratic-for two unknown quantities.
EXAMPLE 2.4 Goal
The Do
tono'..;5~O::.:O~
,."
~
Apply the basic kinematic equations.
v=?
Problem (a) A race car start.ing from rest accelerates at a constant rate of 5.00 m/s:.!. What is the velocity orthe car after it has traveled 1.00 X IO:.! ft? (b) How rnuch time has elapsed? .>:=
x=30.5m
0
Strategy (a) \'Ve've read the problem, drawn [}1e diagram in Figure 2.16, and -------~. X chosen a coordinate system (steps 1 and 2). W'e'd like to find the velocity v arter a FIGURE 2.16 (Example 2.4) certain known displacement dx. The acceleration a is also known, as is t.he initial velocity V o (step 3, labeling, is cornplete), so the third equation in Table 2.4 looks most useful for solving parI. (<-1). Given the velocity, the first equation in Table 2.4 can then be used to find the time in pan (b). T
,..
.
• • • '""'111
Solution (a) Conven units or iix to SI, using the information in the inside front cover.
1.00 X j 0' I't = (1.00 X 10' fL)(~) 3.28 f,
vVrite the kinematics equal.ion lor v~ (step 4):
v:.! = va'.!
Solve for v, taking the positive square root because the car moves to the right (step 5):
v =
V VI)'2
Substitute V u = O. a = 5.00 Ill/S'. and ~.< = 30.5 Ill:
v=
V v/ + 2a ~x = V( 0)' + 2(5.00 m/5')(30.5 m)
~
30.5 m
+ 2a UX +
2a ax
1i.5 Ill/S (b) How much time has elapsed? Apply the f-jrst equation of Table 2.4:
v=al+vo
Substitute values and solve r'or time l:
17.5 111/5
~
(5.00 m/,'),
L7.5 m/s 5.0 mis'
3.505
. .... ~
Ilo. • • • • • • •
Remarks Tbe answers (lrc easy to check. An alternate techniquc is to use equation u = Vo + at to rilld v. QUESTION 2.4 \J\lhat is the final speed if the displacement is increased by a [aclar of ~l?
ax =
vel1
+
bat~
[Q
find t and then use the
38
Chapter 2
Motion in One Dimension
EXERCISE 2.4 Suppose the driver in this example now slams on the brakes, stOpping the car in 4-.00 s. Find (a) the acceleration and (b) the distance the car travels while braking, assuming the acceleration is constanL.
Answers
(a) {/ =
EXAMPLE
-~.38
mis'
(b) d = 35.0 m
2.5 Car Chase
Gool Sohre a problem im'olving two ObjcCLS. olle moving at constant acceleration and the other at constant velocity.
V,.ll =2'1.0 lIl/s
l' ",0
(f....
fll"",.....r
Problem A car (ra\'cling at a constanL speed of 24.0 m/s passes a trooper hidden behind a billboard. as in Figure 2.17. One second after the speeding car passes the billboard. the trooper sets off in chase with a constalll
= 3.00 ill/S 2 t~"'?
FIGURE 2.17 (Exalllple 25).-\ spc.::t:dillg" cal' pilSSCS a hidden trooper. \Vhell dOl:s the IrooperC;II(111Ip t(l tIle rar?
...
,.. Solution (a) How long does il take [he trooper to overtake the cal-? \"Trite the equation for Lhe car's displacement: Take Xo = 24.0
111, P u
= 24.0 mls and (ll:ar
= O. Solve for
.<'" =
'<0
+ vi
= 24.0111
+
(24.0 m/s)/
Vhite the equation for the trooper's position. taking Xu = O. Vo = 0, and a,rooper = 3.00 m/s:.!: Sel Xtruoper = '\'1::11"' and soh'c Lhe quadratic equation. (The quadratic formula appears in Appendix A, E.quation A.S.) Only the positive root is meaningful.
(1.50111/5')1' = 24.0 m
+ (24.0 m/s)/
(1.50 m/s')I' - (24.0 m/s)1 - 24.0111 = 0 /=
16.95
(b) Find the (rooper's speed at this lime. Substitute lhe time into the trooper's velocity equation:
TJ1l"Oopcr = V o
= ~
+
aln....
p~1 l
= 0 + (3.00111/5 2)(16.9 s)
50.7 111/5 .
......................~
Remarks The trooper, Il"rl\'eling about Iwice as fast as the car, must swerve or apply his brakes strongly to m'oid a collision! This problem can also be solved graphically by plotting position versus time for each vehicle on the same graph. The imerseclion of the IWO graphs corresponds to the time and posilion at which the trooper overtakes the car.
2.5
One-Dimensional Motion with Constant Acceleration
39
QUESTION 2.5 The graphical solution corresponds to finding the interscnion of what two types of curves in the xl-plane? EXERCISE 2.5 A motorist with an expired license tag is traveling at 10.0 mls down a street, and a policeman on a motorcycle. taking another 5.00 s to finish his donut, gives chase at an acceleration of 2.00 m/s2 . Find (a) the time required to catch the
car and (b) the distance the troopet' travels while overtaking the motorist. Answers
(a) 13.7 s (b) 188 m
EXAMPLE 2.6
Runwo length
Goal
Apply kinematics to horizontal motion with two phases.
Origln
i/
Problem A typical jetliner lands at a speed of 160 milh and decelerates at the rate of (10 mi/h)/s. If the plane travels at a constant speed of ]60 mi/h for 1.0 s after landing before applying the brakes, what is the total displacement of the aircraft betwecn touchdown on the runway and coming to rest?
(Example 2.6) Coasling alld braking distances for a
Strategy See Figure 2.18. First, convert all quantities to SI landingjcllilH:r. units. The problem must be solved in two parts, or phases, corresponding LO the initial coast after touchdown, follm\!ed by braking. Using the kincmatic equations, find the displacemcnt during each part and add the two displacemcnts.
.
.
~
Solution Convcrt unit.s of specd and acceleration to SI:
O~~7 m/,) = I. mi h
v"
~
(160 mt/h) (
a
=
( -10.0 ( ,ni/lf) Is )( 0.447 m/s) . 1.00 nl1/11
...,
71.5 mls
=
-4.47 mls-?
Taking a. = 0, Vo = 71.5 mis, and l = 1.00 s, fmd the displacement while the plane is coasting: Use the time-independent kinematic equation to find the displacement while the plane is braking. Take a = -4.47 m/52 and Vo = 71.5 m/s. The negative sign on a means that the plane is slowing dO\vll.
Remarks To find the displacement while braking, we could have lIsed the two kinematics equations involving time, namely, .6.x = vol + ~al2 and v = va + al, but because we weren't interested in time, the time-independent equation was easier to use. QUESTION 2.6 How would t.he answer change if the plane coasted for 2.0 s beforc the pilot applied the brakes? EXERCISE 2.6 Ajellands at80.D mis, the pilot applying the brakes 2.00 s after landing. Find the acceleration needed within 5.00 X 10 2 m.
Answer
a
= -9.41
m/s 2
[0
stop the jet
40
Chapter 2
Motion in One Dimension
EXAMPLE 2.7 The Acela: The Porsche of American Trains Goal
Find accelerations and displacements from a velocity vs. time graph.
Problem The sleek high-speed electric train known as the Acela (pronounced ahh-sell-ah) is currently in service on the vVashlngton-New York-Boston run. The Acela consists of tVI"O power cars and six coaches and can carry 304 passengers at speeds lip to 170 mi/h. In order to negotiate curves comfortably at high speeds, the train carriages LilL as much as 6° from the vertical, preventing passengers from being pushed Lo the side. A velocity vs. lime graph for the Acela is shown in Figure 2.19a. (a) Describe the motion of the Acela. (b) Find the peak acceleration of the Acela in miles per hour per second ((mi/h)/s) as the train speeds up from 45 mi/h to 170 mi/h. (c) Find the train's displacement in miles between l = 0 and t = 200 s. (d) Find the average acceleration of the Acela and its displacement in miles in the interval from 200 s to 300 s. (The train has regeneraLive ~
braking, which means that it feeds energ}' back into the utility lines each time it stops!) (e) Find the total displacement in the interval from 0 to 400 s. Strategy For part (a), remember that the slope of the tangent line at any point of the velocity vs. time graph gi\:es the acceleration at that time. To nnd the peak acceleration in part (b), study the graph and locate the point at 'which the slope is steepest. In parts (c) through (e), estimating the area under the curve gives the displacernent during a given period, with areas below the time axis, as in part (e), subtracted from the total. The average acceleration in part (d) can be obtained by substituting numbers taken from the graph into the definition of average acceleration, = D.v/ Uf..
a
-
_
~
Solution (a) Describe the motion. From abollt -50 s to 50 s, the Acela cruises at a constant velocity in the +x-direction. Then the train accelerates in the +x-direcrioll from 50 s to 200 s, reaching a top speed of about ]70 mi/h, whereupon it brakes Lo rest at 350 sand reverses, steadily gaining speed in the -x-direction. (b) Find the peak acceleration.
Calculate the slope of the steepest tangent line, which connects Lhe points (50 S, 50 mi/h) and (100 s, 150 mi/h) (the light blue line in Figure 2.19b):
a ~ ,lope =
=
(1.5 X 10' - 5.0 X 10') mi/h L'>v L'>t = (1.0 X 10' - 5.0 X lO'ls
2.0 (mi/h)js
(c) Find the displacement between 0 sand 200 s.
Using triangles and rectangles, approximate the area in Figure 2.19c:
ii.xCj ~ 200 ~ =
~
Convert units to seconds:
1.0
miles by converting hours
areal + area~ + area3 + area'l + areas = (5.0 X 10' llli/h)(5.0 X 10' ,) + (5.0 X 10' llli/h) (5.0 X la' ,) + (1.6 X 10' mi/h)(1.0 X la's) + H5.0 X la' s)(1.0 X la' mi/h) + H1.0 X 10' s)(1.7 X la' mi/h - 1.6 X 10'mi/h) 2.4 X 10" (mi/h)s
ii.XO---+20D~ = 2.4
X
L'>v slope = L'>t
~
Ih ) 4 mi.S( 10 -h- 3600 s
=
6.7 mi
(d) Find the average acceleration from 200 s to 300 s, and fll1d the displacement..
The slope of the green line is the average acceleration from 200 ,to 300 s (Fig. 2.19b):
a= ~
(1.0 X 10' - 1.7 X 10') llli/h LOX 10',
-1.6 (mi/h)/s
2.5
One-Dimensional Motion with Constant Acceleration
41
200
ISO _ 100
~
5,
50 0 -50 -50
t"
-100 (h)
(a) !'")
200
~tJu
150
ElO
_ 100 ~
::, 50
S,
0
o
t
-50 -50
50
100
0 -100
-100
(d)
(e)
FIGURE 2.19
(Example 2.7) (a) Velocil}'\". time graph Ii)! the Aceb. (b) Thl.:
~lopl.: ofthe"lt::epe~l
1:111genl blue litH: g-ivcs the peak a,cdt.:r:ttioll. and lhe slope 01 lhe greell lim: i~ Ill(' average acn.·lcl",uitlll between 200 sand :H10s. (1.:) The area llllder the vclocityv.,. lillle gr<Jph ill some lilllC illlen'
displacemellL or Ilie :Vela in th;1\ time illterval. (d) (Exercise 2.7).
The displacement. from 200 s to 300 s is equal LO area 6, ~vhich is the area of a triangle plus the area or a very n;.1rTO\v rectangle beneath the triangle;
ilx,o"~"",,, = &(1.0 X 10's)(1.7 X 10' - 1.0 X 10') mi/h
+ (1.0 X ~
10' mi/h)(I.O X 10' s)
9.0 X IO'J(mi/h)(s) =
2.5 mi
(e) Find the total displacement from 0 s to 400 s. The total displacement is the SUlll of all the individual displacements. \Ve still need LO calculate the c1isplace~ ments for the time intervals from 300 s to 350 s and from 350 s to 4-00 s. The latter is negmive because it's below the time axis.
ilx""J_"", = ~(5.0 X 10' s)(1.0 X 10' mi/h)
Find the LOtal displacemcnt by summing the parts:
ilx"_'1<""
~
25 X 1O'(mi/h)(s)
ilx3f>"~ 1("', = :(5.0 X 10' s)( -5.0 X 10' mi/h) ~
-1.3 X 103(mi/h)(s)
= (2.4 X 10"
+ 9.0 X JO" + 2.5 X 10'
-1.3 X 10')(mi/h)(s) = lrrrr..- • • • • •
8.9 mi •
• • • • • • • ...011I
Remarks There are a number of ways to find the approximate area uncleI' a graph. Choice of technique is a sonal pre[ercnce.
per~
QUESTION 2.7 According tu the graph in Figure 2.19a, at whal different times is the acceleration zero? EXERCISE 2.7 Suppose the velocity vs. time graph of another train is given in Figure 2.l9d. Find (a) the maximum instantaneous acceleration and (b) the total displacement in the imer\'al ['1"0111 0 S [Q 4.00 X IO't s.
Answers
(a) 1.0 (mi/h)/s
(b) 4.7 mi
42
Chapter 2
Motion in One Dimension
2.6 FREELY FALLING OBJECTS
GALILEO GAl/LEI ftolian Physicist and Astronomer (1564-1642) Galileo formulated the lows that
govern the motion of objects in free fall. He olso investigated the motion
of on object on on inclined plane, established the concept of relative motion, invented the thermometer, and discovered that the motion of a
swinging pendulum could be used to measure time intervals. After designing and constructing his own
telescope, he discovered four of
Jupiter's moons, found that our own Moon's surface is rough, discovered sunspots and the phases of Venus, and showed that the Milky Way consists of an enormous number of stars. Galilee publicly defended Nicolaus
Copernicus's assertion that the Sun is at the center of the Universe (the heliocentric system). He published Dialogue Concerning Two New World Systems to support the Copernican model, a view the Church declared to be heretical. After being taken to Rome in 1633 on a charge of heresy, he was sentenced ta life imprisonment and loter was confined to his villa ot Arcetri, near Florence, where he died in 1642.
'Vhen air resistance is negligible, all objects dropped under the influence of gravity near Earth's surface fall toward Earth with the same constant acceleration. This idea may seem obvious mclay, but it wasn't until about 1600 that it \'vas accepted. Prior to that time, the teachings of the great philosopher Aristotle (384-322 H.C.) had held that heavier objects fell faster than lighter ones. According to legend, Galileo discovered the law of falling objects by observing that two different weights dropped simultaneously from the Leaning Tower of Pisa hit the ground at approximately the same time. Although it's unlikely that this particular experiment was carried out, we know that Galileo performed Illany systematic experiments with objects moving on inclined planes. In his experimenLS he rolled balls dO\;,.'n a slight incline and measured the distances they covered in successive time intervals. The purpose of the incline ,\o'as to reduce the acceleration and enable Galileo to make accurate measurements of the intervals. (Some people refer to this experiment as "diluting gravity.") By gradually increasing the slope of the incline he was finally able to draw mathematical conclusions about freely falling objects, because a falling ball is equivalent to a ball going down a vertical incline. Galileo's achievements in the science of mechanics paved the \-va)' for Newton in his development of the Im.. . s of motion, \vhich 've will study in Chapter 4. Try the following experiment: Drop a hammer and a feather simultaneously from the same height. The hammer hits the floor first because air drag has a greater effect. on the much lighter feather. On August 2, 1971, this same experiment was conducted on the Moon by astronaut David Scott, and the hammer and feather fell with exactly the same acceleration, as expected, hitting the lunar surface at the same time. Tn the idealized case ,vhere air resistance is negligible, such marion is called free fall. The expression freely faMing objer' doesn't necessarily refer to an object dropped from rest. A freely falling o~ject is any object moving freely under the influence of gravity alone, regardless of its initial motion. Objects thrown upward or downward and those released from rest are all considered freely falling. ',Ve denote the magnitude of tile free-fall acceleration by the symbol g. The value of g decreases with increasing altitude, and varies slightly 'with latitude as well. At Earth's surface, the value of gis approximately 9_80 l11/s2 . Unless stated otherwise, we willllse this value for gin doing calculations. For quick estimates, use g= 10 111/52 . If \-\fe neglect air resistance and assume that the free-fall acceleration doesn't vary \.,..ith altitude over short vertical dislances, then the motion of a [reely falling object is the same as motjol1 in one dimension under constam acceleration. This means that the kinematics equations developed in Section 2.6 can be applied. It's conventional to define "up" as the + )I-direction and to use )' as the position variable. In that case the acceleration is a = -g = -9.80 m/s2 . In Chapter 7, we study the variation in gwith altitude. QUICK QUIZ 2.6 A tennis player on serve tosses a ball straighlup. While the ball is in free fall, does its acceleration (a) increase, (b) decrease, (c) increase and then decrease, (d) decrease and then increase, or (e) remain constant? QUICK QUIZ 2.7 As the tennis ball of Quick Quiz 2.6 travels through the air, does its speed (a) increase, (b) decrease, (c) decrease and then increase, (d) increase and then decrease, or (e) remain the same? QUICK QUIZ 2.8 A skydiver jumps out of a hovering helicopter. A few seconds later, anolher skydiverjumps out, so they both fall along the same vertical Jine relative to the helicopter. Both skydivers fall vlo'ith the same acceleration. Does the vertical distance bet,.,..een them (a) increase, (b) decrease, or (c) stay the same? Does the difference in their velocities (d) increase, (e) decrease, or (f) stay the same? (Assume g is constant.)
2.6
EXAMPLE 2.8
Not
0
43
Bad Throw for a Rookie!
Goal Apply the kinematic equ<:ltiom,lO a freely falling object with a nonzero initial ,"e1ocil)'.
1=
2.0-1 ~ - 20.4
)"'.,.~
III
1'= (}
above the ground. The stone JUSt misses the edge of the roof on its W3)' down, as shown in Figure 2.20. Determine (a) the time needed lor
FIGURE 2.20 (Example 2.8) A Ireel)' tailing object is thrown upward with ,Ill initial "c1ocit)' ofvo = -20 U 1llS POSilioIH' and velocitie..
Problem A stone is thrown from the LOp 01 a building with an initial velOCil) of 20.0 Ill/s straight upward, at an initi<:ll height of 50.0 III
;II·C gin:n for se\eral times.
I=O..\U=() Vo = :!O.lJ IIIls
1:-: ·1.0H .. )-= 0 11 -:W.lJ IIl/s
the stone La reach its maximulll heigl-H. (b) the rnaximulll height (c) the time needed for the SLone La return to the height from which il was thrown and the velocity or the Slone at lhaL instant, (d) the Lillle needed fur the sLone La reach the ground, and (e) the velocity and posi-
tion of the SLOne all
Freely Falling Objects
= 5.00 s.
Strategy The diagram in Figure 2.20 establishes a coordinate system with Yo = 0 at (he level at which the stone is released rrom lhe thrm...· er's hand, with J positi,'c up''¥'ard, \\'rile the Yeloci,y and position kinematic equations ror the stone, and substitute the given information, All the answers come from (hese (w'o equations by using simple algcbra or by just SllbSlilUting the time. In pan (a), ror exarnplc, the stone comes La rcst rar an inst.an!. at its ITlllXimum height, so set v = 0 at. this poi nt and solve ror time. Then substitute thc Lilllc into the displacement equation, obtaining the maximum height.
(= ~.O()
s
)" = -22.5 111 tJ= -2!I.O IIl/~
1
=5,H:-\:..
y =-50.0 III v= -:-\7.1 ~~ ".
III
!l
.
...
....
Solution (a) Find the time when the stone reaches its maxirl1l1J11
height. \Vrite lhe velocity and position kinematic equations:
Substitute a = -9.80 Ill/52 , iJo = 20.0 mis, and Ytl = 0 into the preceding two equations: Substilll(C v = 0, the vclociL}' at maximulII height, into EquaLion (I) and soh'c for time:
Apply the quadratic formula and take the positive root:
t=
5.835
(e) Find the velocity and position of the stone at t = 5.00 s. Substitute values into Equations (I) and (2):
v = (-9.80 m/5') (5.005)
+
20.0 m/5 =
-29.0 m/5
y ~ (20.0 m/5)(5.00 5) - (4.90 m/5')(5.00 5)' ~
=
-22.5
III ~
Remarks Notice how everything follows from the two kinematic equations. Once they are written down and the constants correctly identil1ed as in Equations (I) and (2), the rest is relatively easy. If the SLOne were thrown downward, the initial velocity would have been negative. QUESTION 2.8 I-low would the answer to part (b), the maximum height, change if the person thrO\ving the balljllmped upward at the instant he released the ball? EXERCISE 2.8 A projectile is launched straighlllp at 60.0 m/5 from a height of 80.0 m, at the edge or a sheer cliff. The projectile falls, just missing the cliff and hilting the ground belm\'. Find (a) the maximum height of the projectile above the point of fi ri ng, (b) the ti me it ta kes to h it the groll nd at the base of the cI iff, and (c) its velocity at impact. Answers
(a) 184 m
EXAMPLE 2.9 Goal
(b) 13.55
(c) -72.3 m/5
Maximum Height Derived
Find the maximum height of a thrO\vn projectile using symbols.
Problem Refer to Example 2.8. Use symbolic manipulation to I1nd (a) the time Lilla" it takes the ball to reach its maximum height and (b) an expression for the maximum height that doesn't depend on time. Answers should be expressed in terms of the quantities 110' g, and Yo only. Strategy ""hen the ball reaches ils maximum height, its velocity is zero, so for part (a) solve the kinematics velocity equation for time t and set v = O. For part (b), substitute the expression for time found ill pan (a) into the displace· ment equation, solving it for the maximurn height.
r····················································· Solution (a) Find the time it takes the ball to reach its maximum height.
\"'rite the velocity kinematics equation:
v=at+vo
Move V o to the left side of lhe equation:
v -
Divide both sides by a:
Vo
= aL
v-vo tit --=-=1 a
n
~
2.6
Turn the equation around so that t is on the left and substitute u = 0, corresponding to the velocity at maximum height: Replace I by
lm'lx
and substitute a
=
-g:
(1)
Freely Falling Objects
45
-Uti
1= fl
(2)
'lI,a" -
g
(b) Find the maximum height. \,Vrite the equation for the position y at any time: Substitute I = -vola" which corresponds to thc time it takes to reach Ym;ox' the maximum height;
Combine the last
1.\\'0
terms and substitutc a = -g:
)'",,, = )'0
+
;u) +,a, (- -) I
vo( -
- VtI
"-
(/
., )'0 + vu-
(3)
2g
...
lo.. •• _
....
,
Remarks Notice thatg= +9.8 m/s 2 , so the second term is positive overall. Equations (1)-(3) are much more useful than a numerical ,1I1S\Ver because the dTect or changing one value can be seen immediately. for example. doubling the initial velocity V o quadruples the displacernent above the point of release. Notice also that )II1I<1X could be obtained more readily from the time-independent equation, u 2 ~ uo:! = 2a.ti),.
QUESTION 2.9 By what factor would the maximum displacement above the rooftop be increased if the building were transported to the Moon, where a = -tg? EXERCISE 2.9 (a) Using symbols, fi nd the time I.E it takes for a ball to reach the ground on Earth if released [rom rest at height. Yo' (b) 1n terms of' /./~" how much time lM would be required if the building \vere on Mars, where a = -O.385g?
Answers
I,. J¥9)'H
(a) . =
EXAMPLE 2,10
~
b
(b) ttl
~ 1.6J I,:
A Rocket Goes Ballistic ~_--::=~<~ Maximum
Goal Solve a problem involving a powered ascent followed by free-fall motioll.
,' ,,, \\I , , \ ,, ,, ,,
Problem A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s 2 . 1t rUlIS out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximulll height before falling back to Earth. (a) Find the rocket's "elocit)' and position at the end of 4.00 s. (b) Find the maximum height the rocket reaches. (c) Find the velocit), the instant before the rocket crashes on the ground. Strategy Take y = 0 at the launch point and)' positive up''1'arcl, as in Figure 2.21. Tile problem conshts of" two phases. In phase I the rocket has a net upward accelen.ttion of' 29.4 m/s~, and ' ' C can use the kinematic equations with constant acceleration a to find the height and velocit)' of the rocket al the end of phase 1, when the fuel is burned up. In phase 2 the rockel is in free fall and has an acceleration of -9.80 m/s 2 , with initial velocity and position given by the results of phase 1. Apply the kinematic equations for I'ree l'all. FIGURE 2.21 (Examplt: ~.1O) Two linked phasesuflllolioll for a rockclll';ll is launched. uses lip its filL:!,
heighl )'no;,x
Substitute t = 4.005 imo Equations (3) and (4) to fInd the rocket's velocity v and position )l ~ll the time of burnollt. These will be called vb and y". respectively.
Vb
= 1]8
m/s and
'Yb
= 235 m
(b) Phase 2: Find the maximum height the rocket attains. v = (-9.8 Ill/S') I + 118 Ill/S
Adapt Equations (I) and (2) [0 phase 2, substituting a = -9.801/52, V o = Vb = US mis, and )'u = ')'11 = 235 Ill:
(5) (6)
Substillltc v = 0 (the rocket's velocit)' at maximum height) in Equation (5) to gel lhe time it takes the rucket to reach its maximum height:
o=
Substitute l = 12.0 S illlo Equation (6) to f-ind the rocket's maximum height:
Y,,,,,x = 235
Y = 235
III
+ (118 Ill/S)I - (4.90 m/s'),'
(-9.8 Ill/S') ,
=
III
+
+ ] 18
m/s
(118 Ill/s)(12.0
-->
5) -
,=
118 m/s
9.80 m/s
,=12.0s
(4.90 m/s')(L2.0 5)'
945111
(c) Phase 2: Find the velocity of the rocket just prior to impact.
+ (118 Ill/S) , - (4.90 m/,)I'
Find the tin1e to impacl by selling)' = 0 in Equation (6) and using the quadratic formula:
0= 235
Substitute this value of t into Equat.ion (5):
v = (-9.80 m/s')(25.9 s)
t =
III
25.9 s
+ 118 Ill/S = -136m/s
~
~
Remarks You may think thal it is more natural to break this problem into three phases, with the second phase ending at the maximum height and the third phase a free fall from maximum heig·hlLO the ground. Although t.his approach gives the correct answer, it's all unnecessary complication. Two phases arc sufficient, one lor each different acceleration. QUESTION 2.10 [f, instead, some fuel remains, at what height should the engines be fired again to bri'lkc the rocket's fall and allO\v a
pcrfcClly soft landing? (Assume the samc acceleration as during t.he initial ascent.) EXERCISE 2.10
An experimental rocket designed to land upright falls freely from a height of 2.00 X 10~ 111, starting at rest. At a height of 80.0 m, the rocket's engines start and pro\'ide constant upward accelcration until the rockellands. Vlhat acceleralion is reguired if the speed on touchdowil is to be zero? (Neglect air resislance.) Answer
14.7 m/s~
Multiple-Choice Questions
2.1
_ a
Displacement
The displacement of all object moving along the defined as the change in position or the object, "'" ~ xJ - xi
x-aX1S IS
[2.1]
where Xi is the initial position of the object and Xi is its final position. A vector quantity is characterized by both a magnitude and a direction. Ascalar quantity has a magniLUde only.
2.2
o~icct
Average speed ==
[2.4]
Ii
2.5 One-Dimensional Motion with Constant Acceleration
is given by total distance
u=vo+a./
total time
v
"'x
[2.2]
v==:-~
tJ - l,
The average velocity is equal to the slope of the straight line .joining the initial and final points on a graph of the position of the object versus time. The slope of the line tangent to the position \IS. time curve at some point is equal to the instantaneous velocity at that time. The instantaneous speed of an objecl is defined as the magnitude of the instantaneous velocity.
2.3
IJ -
The instantaneous acceleration of an object at a certain time equals the slope of a velocity vs. time graph at that instant.
The average velocity during a time int.erva\ 6.l is the displacement 6.x divided by D.t.
"'t
"'t
l'j- v, --
The most useful equations that describe the motion of an object moving \\"ith constant acceleration along the x-aXIS are as follows:
Velocity
The average speed or an
=-tiu =
47
Acceleration
The average acceleration a. of an object undergoing a change in \'elocity Av during a time interval tit is
[2.6]
ti.x = vol.
+ ~aL2
[2.9]
v2 =
+ 2atix
[2.10]
uo'.!
All problems can be solved with the first two equations alone, the last being convenient when time doesn't explic· itly enter the problem. After the constants are properly identified, most problems reduce [0 one or two equations in as many unknov,,·ns.
2.6
Freely Falling Objects
An object falling in the presence of Earth's gravity exhibits a free-fall acceleration directed toward Earth's center. If air friction is neglected and if the altitude of the falling o~iect is small compared with Earth's radius, then we can assume that the free-fall acceleration g = 9.8 111/s2 is constant over the range of Illation. Equations 2.6, 2.9, and 2.10 apply, with a. = -g.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
•
•
1. An arrow is shot straight up in the air at an init.ial spced of 15.0 m/s. After how much t.ime is the arrow heading clowl1\vard at a speed of 8.00 m/s? (a) 0.714 s (b) 1.24 s (c) 1.87 s (d) 2.355 (e) 3.22 s 2. Acannon shell is fired straight up in the air at an initial speed of 225 m/s. After how much timc is the shell at a height of 6.20 X ]02 III and heading down? (a) 2.96 s (b) 17.3 s (c) 25.4 5 (d) 33.6 s (e) 43.05 3. ''''Then applying the equations of kinematics for an object moving in one dimension, which of the following statements must be true? (a) The velocity of the object must remain constant. (b) The acceleration of the object must remain constant. (c) The velocity of the object must increase with time. (d) The position of the o~ject must increase with time. (e) The velocity of the object must ahvays be in the same direction as its acceleration. 4.Ajuggler throws a bmvling pin straight up in the air. After the pin leaves his hand and while it is in the air, \vhich statcmCIll is tfue? (a) The velocity of the pin is always in the samc direction as its acceleration. (b) The velocity of the pin is ncver in the same direct.ion as its
acceleration. (c) The acceleration of the pin is zero. (d) The velocity of the pin is opposite its acceleration on the way up. (e) The velocity oftlle pin is in the same direction as its acceleration on the way up. 5. A racing car starts from rest and reaches a final speed v in a time t. If the acceleration of the car is constant during this rime, which of the roJlowing statements must be true? (a) The car travels a distance vt. (b) The average speed of the car is v12. (c) The acceleration of the car is viI. (d) The velocity of the car remains constant. (e) None of these 6. A pebble is dropped from rest from lhe top of a tall cliff and falls 4.9 m after 1.0 s has elapsed. How much farther does it drop in the next 2.0 seconds? (a) 9.8 In (b) 19.6 m (c) 39 m (d) 44 m (ej 27 m 7. An object moves along the x-axis, its position measured at each instant of time. The data are organized intO an accurate graph of x vs. t. ·Which of the following quantities cannol be obtained from this graph? (a) the velocity at any instant (b) the acceleration at any instant (c) the displacement during some time interval (d) the average
48
Chapter 2
Motion in One Dimension
\"clocity during some time interval (e) the speed panicle at any instant
or the
8. People become ul1collllOl"Lable in an elevator if" it accelerates fmm rest al a rate such that it attains a speed of abOlU 6 rn/s after descending ten stories (abo11130 111).
\\'hat is the approximate magnitude of its acceleration? (Choose the closest answer.) (a) 10 111/52 (b) 0.3 m/s:! (e) 0.6 mis' (el) I m/s' (e) 0.8 mis'
neglect air rl·iction.) (a) The speed of the red ball is less than thai of Ihe blue ball. (b) The speed of the red ball is greatcr than that of the blue ball. (c) Their velocities are equal. (d) The speed of each ball is greater than 1/0. (e) The acceleration of the blue Inll is greater than that or the red b<111. 11. A rock is thnm'n downward from the tOP of a 40.0 m
9. Races are timed lO an accuracy of 1/1 000 or a second. \Vhat distance could a person rollerblading at a speed of 8.5 m/s travel in lha! period of time? (a) 8F> 111m (b) 85 em (e) 8.5 m (el) 8.5 mm (e) 8.5 km
tower with an initial speed of 12 111/5. Assuming negligible air resistance, what is the speed of the rock just before hilling the ground? (a) 28 m/s (b) 30 m/s (c) 56 m/s (e1) 784 m/s (e) ylore information is needed.
10. A student at the top of a building throws a red ball upward with speed 1Iu and then throws a blue ball downward with the same initial speed v o' Immediately before the (\vo balls reach the ground, \vhich of the following statements are true? (Choose all correct statements;
12. A ball is thl"O\-\'n straight up in the air. For \'¥'hich situation ~lre both the instantaneous velocity and the acceleration zero? (a) on the \',.:ay up (b) at the top of the flight path (c) on the way down (d) halfway up and halfway down (e) none of these
CON~EPTUAL QUESTIQ~_~ 1. If the velocity ofa particle is nonzero, can the particle's acceleration be zero? Explain.
(a) the accck:ration positive? (b) the acceleration negative? (c) Ihe velocity constant?
2. If the velocity of a panicle is zero, can Ihe particle's acceleration be zero? Explain.
7. Can the instantaneous velocity of an ol~jeC[ at an instant of time C\'cr be greater in magnilllde than the average velocit}' over a time interval containing that instant? Can it C\'cr be less?
3. If a car is traveling eastward, can its acceleration be wes[ward? Explain. 4. Can the equations or kincmatics be used in a situation where the acceleration varies \.. . ith time? Can they be uscd when the acceler;Hion is zero? 5.lfthe average velocity oran object is zero in some time interval, what can you say about the displacement of the object during that inter\'al? 6. Figure CQ2.6 shows strobe photographs taken of a disk moving from left to righl under different cOllditions. The rime imen'al between images is (On5t(\n1. Taking the direction \0 the right to be positive, describe the motion of lhe disk in each case. For which case is
8. A ball is thrown vertically upward. (a) \Vhat are its velocity and acceleration when it reaches its maximum altitucle? (b) What is the acceleration 01" the ball just before it hits the ground? 9. Consider the tollowing combinations of signs and values for the velocity and acceleration of panicle with respect to a one-dimensional x-axis:
,I
a.
b. c.
d. e.
r. g. h.
"
,
(h)
(e)
FIGURE CQ2.6
.'
~- r s~ I
Velocity
Acceleration
Posit ive Positivc Posit.ive
Positive I\cgati\'c Zero Positive Ncgativc Zero Positive Negalive
Ncgative Ncg
Describe what the particle is doing in each case and give a real-lif'c example for an automobile on an eastwest one-dimensional axis, with east considered the positive direction. 10. A ball rolls in a straight linc along the horizontal direction. Csing motion diagrams (or llluJ(iflash photographs), describe the velocity
Problems
49
fP.RoBLEM'5"""""~ ~>_
~
L
~
~<.
The Problems for this chapter may be assigned online at \VebAssign. straightforward, intermediate, clultcllging
WebAssign
1,
~,
m _
3 = =
=
denotes guided problem denotes enhanced conlent prohlem
III =
biomedical application 0= denotes full solution available in SlUdclil Soil/lions Malll/((1/ Slud)' Guidi'
SECTION 2.1 DISPLACEMENT SECTION 2.2 VELOCITY
average velocity in the time intervals from (a) 0 (Q 2.00 oS, (b) 0 to 4.00 s, (e) 2.00 s to 4.00 s, (d) 4.00 s to 7.00 s. anel (ei 0 to 8.00 s.
rnA motorist drives north
for 35.0 minmes at 85.0 km/h and then stops for 15.0 minutes. He then continues nonh, traveling 130 km in 2.00 h. (a) vVhaL is his total displacement? (b) What is his average velocity?
8. A tennis player moves in a straight-line path as shown in Figure P2.8. Find her average velocity in the time intervals from (a) 0 1O 1.0 s, (b) 0 to 4.0 s, (c) 1.0 s to .5.0 s, and (d) 0 to 5.0 s.
1. " The speed of a nerve impulse in the human hody is abollt lOO m/s. If you accidenl
X
2. Light travels at a speed of abollt 3 X 101> m/s. Ho\\' rn
Two boats start together and race across a 60-km-wide lake and back. Boat A goes across al 60 km/h and returns at 60 km/h. Boat B goes auoss at 30 km/h, and its crew. realizing how far behind it is getting. returns at 90 km/h. Turnaround times are negligible, and the boat Ihat completes the round trip first wins. (a) vVhich boa 1. wins and by ho\~' much? (Or is il a tie?) (b) What is the average velocity of the winning boat?
6. A graph of position versus time for a certain particle moving along the x-axis is shown in Figure P2.6. Find the
(Ill)
FIGURE P2.8
(Prohlems Rand 9)
9. Find the instantaneous velocities of the tennis player of Figure 1'2.8 at (a) 0.50 s. (b) 2.0 S, (e) 3.0 S, and (d) 4.5 s. 10. Two cars travel in the same direction along a straight highway. one at a constant speed of 55 mi/h and the other at 70 mi/h. (a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 10 mi m.. . ay? (b) How l~lr must the faster car travel before it has a 15-min lead on the slower car? II. If the average speed of an orbiting space shuttle is
19800 mi/h, determine the lime required for it 1.0 circle Earth. Make sure you consider that the shuttle is orbiting abollt 2.00 X 10:! mi above Earth's surface and assume th(lt Earth's radius is 3 963 miles. An athlete swims the length L of a pool in a time and makes the return trip to the starting position in a time t']. If she is swimming initially in the positive xdirection, determine her average velocities symbolically in (a) (he first half of the swim, (b) the second half or the swim, and (c) the round trip. (d) "\That is her average speed for the round trip?
12. _
II
[IIJA person takes a trip, driving with a constam speed of 89.5 km/h, except for a 22.0-min rest stop. l[the person's average speed is 77.8 km/h, ho\\' much time is spent on the trip and how far docs the person travel? 14. A tortoise can rUll with a speed of 0.10 mis, and a hare can rUll 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2.0 minutes. The tortoise wins by a shell (20 cm). (a) How long does the race take? (b) \Vhat is the length of the race?
FIGURE P2.6
(Prohlems 6 and 17)
15. To qualify for the finals in a racing event, a racc car must achieve an average speed of 250 km/h on a track with a total length of 1600 m. If a panicular car covers the first half of the track at an average speed of 230 km/h, what minimum average speed must it have in the .~ecolld half 0(" the event in order to qualify?
50
Chapter 2
Motion in One Dimension
16. _ One athlete in a race running on a long, straight track with a constant speed VI is a distance d behind a second athlete running with a constant speed v 2 ' (a) Under what circumstances is the first athlete able to overtake the second athlete? (b) Find the time t it takes the first athlete to overtake the second athlete, in terms of d, vi' and v2 " (c) At what minimum distance d'l [rom the leading athlete must the finish line be localecl so. that the trailing athlete can at least tie lor first place? Express d 2 in terms of d, v j ) and v 2 by using the result orpart (b). 17. A graph of position versus time for a cerrain particle moving along the x-axis is shown in Figure P2.6. Find the instantaneous "dacil}' at. the instants (a) I = 1.00 s, (b) I ~ 3.00 S, (e) I ~ 4.50 S, and (d) I ~ 7.50 s. 18. A race car moves such that its position fits the relationship
x
=
(5.0 l11/s)/
+
orpan (b). [l[]Runner A is initially 4.0 mi west of a Oagpole and is rUllning with a constant velocity of 6.0 mi/h due east. Runner B is initially 3.0 rni east of the rlagpole and is running with a cOllStantvelocity of5.0 mi/h due west. I-Iow far are the runners from the Oagpole when they meet?
SECTION 2.3 ACCELERATION
mil
Assume a canister in a straight l.ube moves with a constant acceleration of -4.00 m/s2 and has a velocity of ] 3.0 m/s at I = O. (a) What is its velocity aL t = 1.00 s? (bl AI I ~ 2.00 s' (e) At I ~ 2.50 5? (d) At I ~ 4.00 s' (e) Describe the shape of the canister's velocity versus time graph. (f) What lwo things must be known at a given time to predict the canister's velocity at any later time?
21. Secreta rim ran the Kentucky Derby with times of 25.2 s, 24.0 s, 23.8 s, 23.2 s, and 23.0 s for the quarLer mile. (a) Find his average speed during each quaner-mile scgmem in ft/s. (b) Assuming that Secretariat's instantaneous speed at the finish line was the same as his average speed during the fmal quarter mile, find his average acceleration for the entire race in fl/s 2 . (Hint: Recall that horses in the Derby start [rom rest.)
22.
8 6 4 2
-2
II
The average person passes out at an acceleration of 7 g (that is, seven times the graviLational acceleration on Eanh). Suppose a car is designed to accelerate at this rate. How much time would be required for the car 1.0 accelerate from rest to 60.0 miles per hour? (The car would need rocket boosters!)
23. A certain car is capablc of accelerating at a rate of 10.60 m/s? HD\\llong does it take for this car to go from a speed of 55 mi/h to a speed of 60 mi/h? 24. The velocity vs. time graph for an object moving along a straight path is shown in Figure P2.24. (a) Find the average acceleration of the oqjeCl during the time intervals 0 to 5.0 s, 5.0 s 1015 s, and 0 to 20 s. (b) Find the instantaneous acceler,nion at 2.0 s, 10 S, and 18 s.
15
5
-4
-6
-81---' FIGURE P2,24
125.1:\ steam
catapult launches a jet aircraft from the aircraft carrier John C. Stennis, giving it a speed of 175 mi/h in 2.50 s. (a) Find the average acceleration of the plane. (b) Assuming the acceleration is constant, find the distance the plane moves.
(0.75 m/s 3)(1
where x is measured in meters and I in seconds. (a) Pial a graph of the car's position versus time. (b) Determine the instantaneous velocity of the car at I = 4.0 s, using time intervals of 0040 s, 0.20 s, and 0.10 s. (c) Compare the average velocity during the first 4.0 s \,.. . ith the results
20.
v (m/5)
SECTION 2.5 ONE-DIMENSIONAL MOTION WITH CONSTANT ACCELERATION 26. A car is traveling clue east at 25.0 m/s at some instant. (a) If its constant accelcrmion is 0.750 Ill/S2 due east, find its velocity after 8.50 s have elapsed. (b) If its constant acceleration is 0.750 Ill/S2 due west, find it.s velocity after 8.50 s have elapsed. 27. A car traveling east a1.40.0 m/s passes a trooper hiding at Lhe roadside. The driver uniformly reduces his speed to 25.0 m/s in 3.50 s. (a) is the magnitude and direction of the car's acceleration as it slows down? (b) How far docs the car travel in the 3.5-s time period?
''''h,H
28. In 1865 Jules Verne proposed sending men to the ~'loon by firing a space capsule from a 220-m-Iong cannon with final speed or 10.97 km/s. \'Vhat would have been the unrealistically large acceleration experienced by the space travelers during their launch? (A human can stand an acceleration of 15g for a short time.) Compare your answer with the free-bll accelerat.ion, 9.80 01/52. 29. A truck covers 40.0111 in 8.50 s while smoothly slowing down to a final velocity of 2.80 m/s. (a) Find the truck's original speed. (b) Find its acceleration. 30.
m
A speedboat increases its speed uniformly from Vi = 20.0 m/s to vJ = 30.0 m/s in a distance of 2.00 X 10 2 m. (a) Draw a coordinate system for this situation and label the relevam quantities, including vectors. (b) For the given infonnation, what single equation is most appropl'iate for finding the acceleration? (c) Solve the equation selected in part (b) symbolically for the boat's acceleration in terms of vJ' Vii' and ~x. (d) Substitute given values, obtaining that acceleration. (e) Find the time it [akes the boat to travel the given distance.
S!. A Cessna aircraft has a liftoff speed of 120 km/h. (a) ""hat minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of 240 m? (b) Hmv long does it take the aircraft to become airborne? 32. A truck on a straight road starts from rest and accelerates at 2.0 Ill/S 2 until it reaches a speed of 20 m/s. Then the
Problems truck travels for 20 s at constalll !'>peed until are applied, stopping the truck in a uniform an additional 5.0 s. (a) How long is the truck (b) \'\'hat is the average velocity of the truck motion desnibed?
the brakes manner in in mOl ion? during- the
braking, whatlotitl distance does the car travel? Answers are in terms of the variables a l. (12' II' and f'1. 41. In the Daytona 500 auto race. a Ford Thunderbird and a
Mercedes Benz are moving side by side down a straight::lway at 71.5 m/s. The dl'i\'cr of the Thundcl'bird I·calizcs
that she Illust make a pit stop. and she smoothly slows to a SLOp o\'er a distance of 2.50 m. She spends 5.00 s in the pil and then accelerates Ollt. reaching her previous speed of 71.5 m/s after a dislance of 350 m. AI this point. how· fal· has the Thunderbird fallen behind lhe ~lercedes Benz. which has continucd at a constant spced?
33, _ In a lest run, a certain car accelerates uniformly from zero lO 24.0 m/s in 2.95 s. (a) Wh;:u is the magnitude of the car's acceleration? (b) How long does iltake the car to change its speed fmm 10.0 m/s 10 20.0 m/s? (c) Will douhling the lime always double the change in speed? Why? 34.
E
A jel plane lands with a speed of 100 m/s and can accelerate ata maximum rate of -5.00 1'n/s2 as it comes to rest. (a) From the instant the plane touches the f\lllwa)r. what is the minimum time needed before it can come to rest? (b) Can this plane land on a small tropical island airpon where the runway is 0.800 km longo?
135.\_ Speedy Sue, driving at 30.0 mis, enters a one-lane tUllnel. She then obsenles a slow-moving" van 155 m ahe'ld traveling at 5.00 m/s. Sue applies her hrakes but can accelerate onl}' at -2.00 m/s 2 bec
36. A record of travel along a straight path is as follows: I. Start from rest with a constant accelerilt ion 01"2.77
ll1/s~
42. A certain cable car in San Francisco can stop in 10 s when tr;n-e1ing at maximum speed. On one occasion. the drivcr sees a dog a distance d m in front of lhe car and slams on the brakes instantly. The car reaches the dog 8.0 slater. and the dog jumps off the track just in time. If the car t,'a\·e1s 4.0 m beyond the position of the dog befol-e coming to a stop, how far was the car from the dog? (Him: You will need thrce equations.) 143.IA hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 12 m/s, skates b)'with the puck. Aftel' 3.0 s, the first player makes up his mind t.o chase his opponent. If he acccler* ates lIniformly at4.0 m/s 2 , (a) how long does it take him to catch his opponent, and (b) how far has he traveled in that t.ime? (Assume the player with the puck remains in motion at consl.allL speed.) 44. A train 400111 long is moving all ,I straight track with a
for 15.0 s.
speed of 82.4 kill/h. The engineer applies the brakes at a crossing, and laLcr lhe last car passes the crossing with a speed of 16.4 kill/h. f\ssutning constant acceleration. determine how long the train blocked the crossing. Disregard the width orlhe crossing.
2. Maintain a constant velocity for the ncxt2.05 min. 3. Apply a constant negative acceleration of -9.'17 l"or4.39s.
m/s~
(a) ,"Vhat. was the total displacement for the t.rip? (b) What were the average speeds for legs 1, 2, and 3 of the Irip, as well as for the complete trip?
37. A train is traveling down a straight track at 20 m/s when the engineer applies Ihe brakes, resulting in an accelenltion of -1.0 m/5 2 as long as the u'ain is in motion. How far docs the train move during- a 40-s lime interval st.tt·ting at the instant lhe brakes are applied? 38. A car accelerates uniformly from rest to a speed of 40.0 mi/h in 12.0 s. Find (a) the distance Ihe car tra\'e1s during this lime and (b) the constal1l acceleration of lhe car. 139.IA car starts from rest and tra\'e!s for 5.0 s with a uniform acceleration of +1.5 m/s~. The driver then applies lhc brakes, causing a uniform acceleration of -2.0 m/s2 . If lhe brakes are applied for 3.0 s, (a) how fast is the (;lr going at the end of the braking pel'iod, and (h) how f;ll" has the car gone? 40.
Sl
mil A car st.lrts from ,'est and Inwcls for /, seconds with a uniform acceleration at. The dri . . er thell applies the brakes, causing a uniform acceleration (1'2' If the brakes arc applied for /2 seconds, (a) how fast is lhe car going just before the beginning of the braking period? (h) How rar docs the car go before the ciriver begins to brtlke? (c) Using the answers to pans (a) and (b) as the inilial velocity and [losition [or the rnotion of the C(lr during
SECTION 2.6 FREELY FALLING OBJECTS 45. A ball is thrown ....ertically upward with a speed 0[25.0 m/s. (a) How high does it rise? (b) How long does it take to reach ilS highest poilH? (c) How long does the ball take to hit the gmund after it reaches its highest point? (eI) \on,at is its velocity when it returns LO the le\'el from which it started? 46. It is possihle to shoot an arrow at a speed as high as JOO m/s. (a) If friction is negleCled, how high would an arrow launched at this speed rise if shot straight up? (h) How long \..· ould the alTow be in the air? 147.IA cCl"tain freely falling object requires 1.50 s to travel the last 30.0 m before it hiLl; the grollnd. Fl"Om what height above the ground did it fall? 48.
mm An attacker al the base of a castle wall 3.65 m high throws a mck straight up with speed 7.40 m/s at a height of 1.55 m above the ground. (a) Will the rock reach the top of the wall? (b) If so. what is the mck's speed at the top? Ifllot. what initial speed muslthe rock have to reach the top? (c) Find the change in the speed ofa mck thmwn straight down fmm the top of the wall at an initial speed o[ 7.40 m/s and moving hetween the same two points. (d) Does the change in speed of the ciownward-moving rock agree with the magniwde of the speed change of the rock moving upward between the same elevations? Explain physically why or why not.
52 49.
Chapter 2
Motion in One Dimension
II
Traumatic brain il"tiury sHch as COIlCLlssion results when the head undergoes a very brge acceleration. Generally, an acceleration less than SOO rn/s 2 lasting for any length of Lime will not cause injury, whereas an acceleration greater than] 000 m/5 2 last i ng for at least I inS wi 11 cause injury. Suppose a small child rolls ofr a bed that is DAD m above the noor. If the noor is hardwood, the child's head is brougl1l10 rest in approximately 2.0 mm. If the floor is carpeted, this slopping distance is increased to about 1.0 em. Calculate the magnitude and duration OrtJ1C deceleration in bOlh cases, LO determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicucd r~i11 could I-esuh in a head velocity greater or less th:ln lite speed you calculate.
50. A small mailbag is released from a helicopt.er thaI is descending steadily at 150 m/s. Aftcr 2.00 s, (a) what is the speed of the mailbag, and (b) how far is it below the helicopter? (c) \Vhat are )'our answers 10 parts (a) and (b) if the helicopter is rising steadily at 1.50 m/s? 51. A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height. as Ihe point of release. (a) \Vhat is the acceleration of t.he ball w'hile it is in flight? (b) ,,,,'hat is the velocit), of the ball when it reaches its maximum heigh!? Find (c) tlte initial velocity OfU1C ball and (d) the maximum height it reaches. 52.
mm A
package is dropped from a helicopter th,u is descending steadily at a speed v o' After I seconds have elapsed, (a) what. is the speed of the package in terms of vo, g, and t? (b) What distance dis il frorn the helicopter in terms of gand I? (c) \\ihat are the ans'.. .ers to parts (a) and (b) if the helicopter is rising stcadil)' al the same speed?
~ _ A model rocket is launched straighl upward with an
initial speed of 50.0 m/s. It
54. A parachutist with a camera dC:icend:i in free fall at a speed of 10 m/s. Thc parachutist releascs the camcra at an alumde of 50 m. (a) How long docs it take lhe camera to reach the groundr (b) \Vhat is the velocity of the camcrajust before it hits the ground?
ADDITIONAL PROBLEMS 55. A truck tractor pulls t\\'o trailers, onc behind Ihe Olher, at a constant speed 01" 100 km/h. It. l.akes 0.600 s for the big rig to completely pass OnLO a bridge 400 tTl long. For what dUl-ation oft.ime is all or pan of !he I.rtu..:k-trailcr combination on the bridge?
56. A speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot. slows the boat with a constant acceleration of -3.50 1ll/s2 hy reducing Ihe throttle. (a) How long does it take the boat 10 reach t.he buoy? (b) What is the velocity of the bom whell it reaches the buoy?
57. A bullet is fired through a board 10.0 em thick in such a way that the bullet's line of motion is perpendicular 10 the face of the board. If' the initial speed of the bullet. is 400 m/s ancl it. emerges from t.he othcr side of the board wit.h a speed of 300 mis, lind (a) the accelerat.ion of the bullet as it passes throllgh the board and (b) the total t.ime the bullet is in conlact with the board.
58. An indestructihle bullet 2.00 em long is [ired straight through a board that is 10.0 cm thick. Thc bullet strikes the board with a speed of 420 m/s and emerges with a speed or 280 m/s. (a) \I\,'h'l1. is the avcrage acceleration of the bullet through Lhe board? (b) \"vhat is the t.olal time that the bullet is in conl;'lCl with the board? (c) ''''hat thickncss of board (calculated to O.J cm) would it take to stop t.he bullet.,
159.IA student throws a set of ke),s vertically upward to his haternit)' brother, who is in a window 4.00 m abO\'e. The brother's OLIL<;Lretched hand cat.ches [he keys 1.50 s later. (a) \Nith what initial velocity wcre I.he keys thrown? (b) What was the velocity of the keys just before they were caught? 60.
mil
A sludent.throws a set. of keys vcrtically upward to his rraternir)' brother, who is in a window a distance II above. The brot.her's outstretched hand catches the keys on their way up a time flaler. (a) With what initial velocity werc the keys thrown? (h) \1Ir'h
61.111 It. has been claimed th;:ll an insect called the froghopper (Pltilnen1l.s sjJlImadlls) is the bestjurnpcr in the animal kingdom. This insect can accelerate at4 000 m/s2 over a distance of 2.0 mm as it straightens it.s specially designed "jumping le~s." (
163.IA ball is t.hrown upward from the ground with an initial speed 01" 25 m/s; at Ihe same instant, another ball is dropped from a building 15 m high. After how long will the balls be at the same height? 64. To pass a physical education cl
65. Two students are on a balcony J9.6 m above ule street. One student throws a ball vertically dowmvard at 1'1.7 m/s; at the same inSlant, the other student throws a ball ver-
Problems
lically upward at the same speed. The second ball just misses the balcony on the way down. (a) ,.vhal is the difference in the n't'o balls' time in the air? (b) \Vhat is the velocity of each ball as it strikes the ground? (c) Hmv far apart are the balls 0.800 s after they arc thrown? 66. _ Two students arc on a balcony a distancc h above Lhe streel. One swdent throws a ball vertically downward at a speed Vo; at lhe same time, the other student throws a ball venically upward at the same speed. Answer the following symbolically in terms of va, g, It, and I. (a) \·Vrite the kinematic equation for the )'-cool"dinatc of each ball. (b) Sctthe equations found in part (a) equal to height 0 and solve e
67. You drop a ball from a window on all upper noor of a building and it is callght by a friend on the ground when the ball is moving with speed vI" You now repeal. the drop, but you have a friend on the street bc10w throw another ball upward at speed vfcxactl), al the same time thai you drop your ball from the window. The two balls arc initially separated by 28.7 m. (a) At whal time do they pass c.lCh other? (b) At what location do Lhe}' p
mil
Emily challenges her husband, David. to catch a $1 bill as follows. She holds t.he hill venically as in Figure 1'2.69, with Ille center 01" the bill between David's index finger and thumb. David must catch Ihe bill afLer Emily releases it wilhollt moving his hand downward. Ifhis reaction time is 0.2 s, will he succeed? Explain your reasoll-
ing. (This chal1enge is a gnod trick yOll might walllLO try with you r friends.) 170.IA mountain climber stancls at the top ofa 50.0-m clifT that overhangs a calm pool or \vater. She throws two stones venically downward 1.00 s apan and obscn'es that the)' cause a single splash. The first stone had an initial \'elocit]' of" -2.00 m/s. (a) How long after release of the firsl stone did the two SLOnes hit the water? (b) \-Vhat initial velocity must the second stone have had, given that they hit the water simultaneously? (c) What was the velocit), of each stone aL the instant it hit lhe water? 71. An icc sled powercd by a rocket eng"ine stans from resl on a large frozen lake and accelerates at +~IO ft/s:':. After some time 'I' the rocket engine is sinH clown and the sled moves wiLh const,lIlt velocity lJ for a time '2. If the total distance traveled by the sled is 17500 rr and t.he LOt.a1 time is 90 s, find (a) Ihe times t l ,md l2 and (b) the vclocity iJ. At the 17500-("1. mark, the sled begins to accelerate at -20 ft/s 2 . (c) What is the final position of the sled when il comes La rest? (d) I-!m...· long does iL take to come to rest.? 72. In Bosnia, the ultimale lesLofa young man'scour
FIGURE P2.69
53
mil
A glider on an air Lrack carries a nag of length ( throug-h a stationary photogare, which measures the time illlerval CJ.t" during which the flag blocks a beam of infrared lig·hl. passing across the photogatc. The ral.io v d = f/CJ.fr/ is the average velocity of the glider over this part of its mOl.ion. Suppose the ~lider moves with cunstant acceleratioll. (a) Is v d necessarily equal La the instantaneous velocity of Ihe glider when it is halfw;-,y through the phoLOgale in space? Explain. (b) Is v d equal to the insLanLaneolls velocity of the glider when it is halfway thmugh the pho!.agate in lime? Explain.
75. A Sllllllman siLLing on a tree limb wishes to drop vertically OlltO
Legendary motorcycle stuntman Evel Knievel blosts off in his custom rocket-powered Harley~Dovidson Skycycle in on attempt to jump
the Snake River Canyon in 1974 A parachute prematurely deployed and caused the croft to fall into the canyon, just short of the other side. Knievel survived.
3.1
Vectors and Their Properties
3.2
Components of a Vector
3.3
Displacement, Velocity, and Acceleration in Two Dimensions
3.4
Motion in Two Dimensions
3.5
Relative Velocity
VECTORS AND TWO-DIMENSIONAL MOTION In our discussion of one-dimensional motion in Chapter 2, we used the concept of vectors only to a limited extent. In our further study of motion, manipulating vector quantities will become increasingly important, so much of this chapter is devoted to vector techniques. We'll then apply these mathematical tools to two-dimensional motion, especially that of projectiles, and to the understanding of relative motion.
3.1
VECTORS AND THEIR PROPERTIES
Each of the physical quantities we will encounter in this book can be categorized as either a vector quantity or a scalar quantit),. As noted in Chapter 2, a \-ector has both direction and magnitude (size). A scalar can be completel}' specified by its magnitude \\Iith appropriate units; it has no direction. An example of each kind of quantity is shown in Figure 3.J. As descl-ibed in Chapter 2, displacement, velocity, and acceleration are vector quantities. TernperalUre is an example of a scalar quantit)'. J f the tempel-attire of an object is -5°C, that information completely specifies the temperallire of the object; no direction is required. Masses, time inter\'als. and volumes are scalars as well. Scalar qtlantities can be manipulated with the rules of ordinary arithmetic. Vectors can also be added and subtracted from each other, and multiplied, but there are a number of imponant differences, as will be seen in the following sections. \Vhen a vector quantity is handwritten, it is often represented with an arrow over the letter (A). As mentioned in Scnion 2.1, a vector quantity in this book will be represented by boldface type with an arrow on top (for example, A). The magnitude of the vector A will be represented by italic type, as A. It.alic type will also be L1sed 1.0 represent scalars. ~
~
54
3,1
Vectors and Their Properties
S5
Equality of Two Vectors. Two veCLOrs Aand B arc equal if they have thc same magnitude and the same din:ctioll. This property allows us to translate a vector parallel to itself in a diagram withoLlt affecting the vector. ln fact, for most purposes, any vcctor can be moved paral1<.:ILO itselfwitholiL being affectcd. (See Fig. 3.2.) Adding Vectors. \-Vhen t,.. . o or more vectors arc added. the)' must all have the same units. For example, it doesn't make scnse to add a velocity vector, carrying un its of mc[crs per second. to a displacement vector, carrying units of meters. Scalars obey thc same rule: II would be similarly meaninglcss LO add temperatures to volumes or masses to time illlervais. Vectors can be added geomctrically or algebraically. _{The lalter is discusscd ;'It lhe end of the next section.) To add vector B to vector A geometrically, first draw Aon a piece of graph paper to some scale, such as 1 cm = I Ill, so that its direction is specificd relative a coordinate svstem. Then dra,'.. vector B to the samc scale with the tail of Ii starting at the tip of A, as in Active Figure 3.3a. Vector Ii must be dra'.. ·. n along the direction that makes the proper angle relative vector A. The resultant vector Ii = A + Ii is the vector drawn from the tail of A to the tip orB. This procedure is known as the triangle method of addition. \Nhen two vectors arc added, their sum is independent of I he order of the addition: A + Ii = Ii + A. This relationship can be seen from the geometric construction in Active Figure 3.3b. and is called the commutative law of addition. This same general approach can also be used to add more than two vectors, - l o in _ _ asl ois- + done Figure 3.4- (page 56) [or 1'0111' vectors. The resultant vector sum R = A + B + C + D is the vector dra\vn fr01l1 I he tail of the first vector to the tip of the last. Age:lin, the order in which the vectors are added is unimportant. Negative of a Vector. The negative of the \'ector A is den ned as the vector that gives zero when added to A. This means that A and -A have the same magnitude but opposite directions. ~
~
~
Subtracting Vectors. Vector subtraction makes use of (he definition of the negative of a vector. \Ve define the operation A - B as the veCLOr - B added to the vector A: ~
--7
--7
--4
A - B ~ A
~
+
~
(a)
(h)
FIGURE 3.1 (a) The numbl:1" of grapes ill this bUllch ripe for picking is olle e:o:ample of a scal'lr quanl il\'. Can you think of other eX
--4
(-B)
[3.l]
Vcctor subtraction is really
TI P 3.1 Vector Addition vs. Scolor Addition -~
->
A
- .....
= C is very different from + B = C. The first is a vector
A + B
sum, which must be handkd graphically or with components. whereas the second is'l simple ar;th melie sum of numbers.
y
o
---'+,~------,,£--.,
FIGURE 3.2
fOllr
\·CCIO/"S
are l:qual
because they have eqllallcllgths and
poillt in the sallie
direclion.
(a)
ACTIVE FIGURE 3.3 (a) \'1'lw11 \L"ClOr B i~ added to \l:(;[or
(hi
A. the H'Clllr ~llm R is the \"l'UU!" liial runs frolllllll' tail of A to 111l' lipofn. (b) Here the n.'sllltalll runs lrulll the tail orB to Ill(' lipllfA, These COllstfllctiOIlS prove lhal A + Ii "" Ii + A.
S6
Chapter 3
Vectors and Two-Dimensional Motion
FIGURE 3.4
A geometric
FIGURE 3.5 Thi~COllstruclj..on how lO subt ract \'(;clOr B from veclor A. The vennr II;\s-!he same magnitlloe as Ille "cClO!" n, but points in the opposite direnion.
COIl-
~ho\\'s
~I l"ll("lion for sum III i.!]g fOllr veClOr~.
"l'lle resilltant V{~Cl()r R is lhe vector
lhat completes [he polyg-on.
~
is -~A, a vector \\.. ith a magnitude three times that site eli recl'ion (because or the negative sign).
or A and point.ing in the oppo-
QUICK QUIZ 3.1 The magnitudes of[wo vectors Aand Ii arc 12 units and 8 units, respcClivcly. \,Vhat arc the 1~.rges!:..and.Jn1allestpossible values for the magnitude of the resultant vector R ~ A + B' (a) 14.4 and 4 (1)) 12 and 8 (c) 20 and 4 (d) none of these.
EXAMPLE 3.1 Goal
Taking a Trip
Find the slim ofl\\'o vectors by lIsing a graph.
Problem A car travels 20.0 km due north and thcn ?5.0 km in a direction 60 0 west of north, as in Figure 3.6. Using a graph, find the magnitude and direction of a single vector that gives the net effect of the car's trip. This veClor is called the car's resulta11t di~l)I{f(:emenL Strategy Draw a graph and represent the displacement vectors as arrows. Graphically locate the vector resulting from the sum of lhe two displacement vectors. Measure its lengLl1 and angle with respect to the vertical. ~
y(km)
FIGURE 3.6 (Example ~.l) 1\ gr
---'_...L_'--'l-_-'-x (km)
-20
0
................
.
-
~
Solution
Let Arepresent the first displacement. vector, 20.0 kill nort.h, and B t.hc second displacement vector, extending west of north. Carefully graph the two vectors, drawing a resultant vector Ii wit.h its base touching the base of Aand extending to t.he tip or"i~. Measure the length or this vector, which LUrns Ollt. t.o be about 48 km. The angle {3. measured with a protractor, is about ~9° west of l1onh. ~...
.
.
.
Remarks Not.ice that ordinary arithmetic doesn't work here: the correct answer or 48 km is not equal to 20.0 km 35.0 kill = 55.0 km! QUESTION 3.1 Suppose two vectors are added. Cndcr \\I'hat conditions would the sum magnitude of the resultant vector?
or the
....
+
magnitudes of the vectors equal the
EXERCISE 3.1 Graphical!)' determine lhe rnagniwdc and direction of' the displacement if a man walks 30.0 kill 45 0 north of east and then walks clue east 20.0 km. Answer
46 km, 27 north or cast 0
3.2
57
Components of a Vector
3.2 COMPONENTS OF A VECTOR One method of adding vectors makes use of the projections of a vector along the axes ofa rectangular coordinate system. These projections arc calJed components. Any vector can be completely described by its compollcills. Consider a vector Ain a rectangular coordinate system, as shown in Figure 3.7. --> A can be expressed as the SlIlll of (\'0'0 vectors: A.\" parallel 10 the x~axis; and A v• parallel to the )I-axis. \tla1.hcmal ically, ~'---j.
~
~
A = Ax
+
iJ!
~
A 1•
\'ihelc Ax and Ay ;,He the component \CC[OIS of A The projection of A along the x-axis, Ax. is called the x-componen~, of A, and the projection of along the )'-axis, A)., is called the y-component of A. These components can be either positive or negative numbers \I.lith units. From the dellnit.ions of~sine and cosme, "Ie see that cos 8 = AJA and sin () = A/A, so thc components of A are
[3,3]
rcctanglliar components
= ---:.. Ax
[3.4]
To solve for the angle 8, which is measured from the positi,'e x-axis by convention, we ca n writ.e Equation 3.4 in the form
I
y-components Equat ion 3.2 It1l' the .\'- and Jcompollellls 01 a \'coor a$socimes cosine with the x-component and sine with the J-compolll..' llt, Fi~ure
3.8a. This assotia-
(II,) A,
is due so/('I)' LO the fact ihm we chus" to measure the angle 0 with respeolo the positive x-axis. If the angle were measured with rcspecltO the J-axis, a~ in Figure 3.8b.thc components would be given b)' A, A sin () and.l1,
=
=
A cosO.
tan- l ~
This formula giYes the right anS'\ler only half the time! The inverse langelll function rewrns values only from ~90° to +90". so the answer in your calculator ,.. . ine10w will only be correct if the veclQr happens to lie ill the first or fourth quadrant. Ifitlies in the seconcl or third quadrant. addillg 180 0 to the number in the calculaLOr ,<.. indow will always give the right answer. The angle in Equations 3.2 and 3.4 rnust be measured from the positive x-axis. Other choices of" reference line are possible, but certain adjustments must then be made. (Sec Tip 3.2 and Fig. 3.8.) Ira coordinate system other than the one shown in Figure 3.7 is chosen, the components of the yector must be modified accordingly. In Illany applications it's more convenient to express the component.s ora vector in a coordinate system having axes that are nOl hg,rizontal and "cnical, but are still perpendicular to each other. Suppose a vector B makes an angle fJ' with the x'-axis defined in Figure 3,9 (page 58).
.'.
J
I",
----x A.~= A cos f) (
and AT'
I ion
II,
o
A~
(TIP 3.2 x- and
as in
o=
Alying in
II sin (J
These components form tw~ sides of a right triangle having a hypotenuse ,\'ith magnitude A. It fo\1m\l5 that A's magnitude and direction are related to its component.s through the Pythagorean theorcm and the definition of the tangent:
tan ()
AllY \,ector
[3.2]
A.-.:::::: A cos 0
,1, =
FIGURE 3.7
111<' .,)'-plane call be representcd by il~
= A
~ill (J
----;c-/'---------x o (h)
TIP 3.3 Inverse Tangents an Calculators: Right Half the Time The inwrse tangent function on calculators returns an angle hetween -900 and +~lO°. 1r Lile \,ector lie~ in lhe second or third quadrant, the angle, a, measured from the positive x-axis, \\-ill be the angle returned b~ )'our calcllIcttor plus 1800 •
FIGURE 3,8 The angh: 0 need nOI alw;lys bl: dCrlllCd from IIH: positive
x-axis.
58
Chapter 3
Vectors and Two-Dimensional Motion FIGURE 3.9 The components ofvcclor B in a lilted coordinat,e syslem,
9'
The rectangular components of Ii along the axes of the figure are given by Bx ' : : B cos 8' and B)', = B sin 0', as in Equations 3.2. The magnitude and direction of B arc then obtained from expressions equivalent to Equations 3.3 and 3.4.
y
QUICK QUIZ 3,2 Figure 3.10 shows two vectors lying in the .
FIGURE 3.10
(Quick Quiz
EXAMPLE 3.2
x-component
~t~)
Help' Is on the Wa !
Goal Find vector components, given a magnitude and direction, and vice versa.
J
(a) Find the horizontal and venical components of the 1.00 X .102 m displacement or a superhero who flies from the top of a tall building along the path shown in Figure 3.lla. (h) Suppose instead the superhero leaps in the other direction along a displaceluent veeror Ii [Q the top of a nagpole \vhere the displacement components arc given by E." = -25.0 In and By = 10.0 m. Find the magnitude and direction of the displrlcement vector. Problem
(a)
y
Strategy (a) The triangle formcd by the displacement and its components is shown in Figure 3.1 tb. Simple trigonometry gives the components relative to the standard x-y coordinate system: Ax = A cos 0 and A, = A sin 0 (Eqs. 3.2). Note that 0 = -30.00, negative because it's measured clockwise from the positive x-axis. (b) Apply Equations 3.3 and 3.4 to find the m
,..
(b)
.
. '--'
Solution (a) Find the veCLOr components of A from its magnitude and direction. Use Equations 3.2 to find the components of the displacement vector A:
II, = A cos 0 =(1.00 X
]0'
m) cos (-30.00) =
Ay = A sin 0 ~ (l.OO X 10 2 m) sill (-30.0°) ~
+86.6 m
-50.0 m
3.2
Components of a Vector
59
(b) Find the magnitude and direction of the displacement vector rrom its component.s.
B
13
Compute the magnitude or B from the Pythagorean
='11 B/ + IJ,' = V( -25.0 m)' + (10.0111)' =
26.9111
theutcm:
Calculate the direction of B using the inverse tangelll. rcmembering to add 1800 to the answer in your calculator windmv, because thc vector lies in the second quadrant:
8 = t.n-
8=
.
~
I
(~)
tan _1(10.0) ---25.0
J58°
.
~
Remarks Tn part (a), note that cos (-0) = cos 0; however, sin (-8) = -sin O. The negative sign of 1\" renects the fact thal displacemcnt in the )'-dircction is downwa.rd. QUESTION 3.2 \-\lhat othcr functions, if any, can be llsed to nnd the angle in part (b)? EXERCISE 3.2 (a) Suppose the superhero had flown 150 m at a 120 angle with respect to the positive x-axis. Find the components of the displacemcJlt vector. (b) Suppose instead the superhero had leaped with .a displacernenl having an x-componcnt of 32.5 m and a J-component of 24.3 Ill. Find the maRnitucLe and di rcction of the displacement vector. 0
Answers
<.l A.
=
-75111,
'".I' =
I~O
m (b) 40.6111.36.8
0
Adding Vectors Algebraically The graphiGll method of adding vectors is valuable in understanding how vcctors can bc manipulated, but most of the tin1c vectors arc added ~llgebraically in terms or theiS components. Suppose Ii : : : A + B. Then the componcnts of the resultant vector R are given by R = A."<: + Bx
[3.5.]
A, + 13,
[3.5b]
J;
R,
=
So x-components are added only to x-componsnts, and J-componcnls only to)'components. Thc magnitude and direction of R can subsequcntly be found . .vith Equations 3.3 and ~.4. Subtl'acting two vectors works the same \,,'a)' because it's a matter of adding the negative of one \'ector to another vector. You should make a rOllMh skctch whcn adding or subtracting vectors, in order to get an approximate geometric solution as a chcck.
EXAMPLE 3.3
Take a Hike
Goal Add vectors algebraically and Ii nd t he resultant vector.
)'(klll>
Problem A hiker hegins. trip by first walking 25.0 km 45.0° south of cast rrom her base camp. On the second day she w(llks 'JO.O km in a direction 60.0° north of east, at which point she discovers a forest rang-er's lower. (a) Determine thc componcnts of" the hiker's displacements in the firsl and second days. (b) Determinc I he components or the hikcr's total displacemcnt for the trip. (c) rind the
20
K
11'+"
Tower
)"(klll)
30
S
10
20 '0
7.!"'=;""'::--7!'::--:!"'.!.:';;\"{ kill) o 10 20 ~O 40
-10
H,,:= '!o7.7 km
-20 (a)
(hi
FIGURE 3.12 (J::xarnplc '!o.:~) (a) II iker'~ pal h ami 1 h~ re"ultant H"Clor. (b) Component" of the hik~r's lotal disp!:lcemellt ["rom Gllllp.
60
Chapter 3
Vectors and Two-Dimensional Motion
magnitude ;mcl direction of the displacement from base camp.
of the coordinates, we get the veClors shown in Figure 3.l2a. After finding x- and y-components for each vector, we acid them "componclllwise." Finally. we determi~e the magnilUde and direction of the resultant vector R,
Strategy This problem is just an application o[ vector addition IJsing- components, Equations 3.5. \,Vc denote the displacement vect.ors on the first and second days ~ by A and n, respectively. Using the ctlmp as the origin .,,-
using the Pythagorean theorem and the inverse tangent
-
fUllction. ,
~
Solution (a) Find the componcnLs of A. Use Equations 3.2 to find the components of A:
Find the cornponems orB:
A,
~
A cos (-45.0')
A,
~
A sin (-45.0')
= ~
(25.0 km)(0.707)
=
-(250 km)(0.707)
~
17.7 km -17.7 km
13., = B cos 50.00 = (40.0 km)(0.500) =
20.0 km
~
34.5 km
13, = Bsin 50.0' = (40.0 km)(0.855) (b) Find t.he components of the resuilant vector,
Ii = A+ B. To fi nd R.t , add the .\"~componel1ls of Aand
B:
R, = Ax + Bx = 17.7 km + 20.0 km =
add the )I-componenl.s of Aand
B:
R, = A, + B,. = -17.7 km + 34.5 km = 15.9 kill
To find
!S"
37.7 kill
(c) Find the magnitude and direction oER.
v'Rx' +
Use the Pythagorean theorem to get the magnitude:
R=
Calculate the direction ofR using the inverse tangent [unClion:
km) = o = lan _I (l5.9 37.7 km
R,' = V(37.7 km)'
~
+ (15.9 km)'
= 41.3 km
24.] , ~
->
Remarks Figure 3.12b shows a sketch of the components of R and their directions in space. The magnitude and direction of the resultant can also be determined from such a sketch. QUESTION 3.3 A second hiker follows the same path the first day, but then \valks 15.0 km east on the second day before turning and reaching the ranger's tower. Is the second hiker's resultant displacement vector the same as the first hiker's, or different? EXERCISE 3.3
A cruise ship leaving port travels 50.0 km 4·5.0° north oEwest and then 70.0 km at a heading 30.0° north of east. Find (a) the ship's displacement vector and (b) the displacement vector's magnitude and direclion. Answer
(<1) Rx = 25.3 km.
R, =
70.4 kill (b) 74.8 km, 70.2' nonh oreasl
3.3
DISPLACEMENT, VELOCITY, AND ACCELERATION IN TWO DIMENSIONS
In one~dilllensional motion, as discussed in Chapter 2, the direction of a vector quantity such as a velocity or acceleration can be taken into account b)1 specifying whether the quantity is positive or negative. The velocity of a rocket, for example, is positive ifl.he rocket is going lip and negative if' it's going down. This simple solu-
3.3
Displacement, Velocity, and Acceleration in Two Dimensions
lion is no longer available in two or three dimensions. InsLCad, we mllst make filII usc of the vcctor concept. Consider an object moving through space as shown in Figure 3.13. \Vhen the o~ject is at somc point ® at timc 11' its position is described by the position vcctor rj, dt-awn from the origin to ®. \-Vhen the ol~jcct has moved LO some other point @ 3ttime If' its position yeetor is rIo From the vcctor diagram in Figure 3.13, the final position yector is the sum of the initial position vector and the displacement tJ.T: 1/ == j + 6.1. From this relationship, we obtain the follO\\'ing one:
r
An Object'S displacement is defined as the change in its position vector. or
[3.6J SI unit: meter (m)
)
Palh of an objt:Cl -"j"-----------x FIGURE 3.13 An object moving along: some CUf\Cc! p.uh between puint.. ® and @. The displacement \'ector dr is lhe difference in the position \l·CWrS:.lr == 1",- r·,.
We now present several generalizations of the definitions of velocity and acceleration given in Chapter 2. An object's average velocity during a time interval 6.1 is its displacement divided by Ilt:
~ Average velocity
[3.7]
51 unit: meter per second (m/s) Hecause the displaccment is a vector quantity anclthe timc interval is a scalar quantity, we conclude that the average velocity is a vf'tlorqualllity directed along ~ T. An object'S instantaneous velocity ~ is the lirnit or its average velocity as 6.t gocs to zero:
1= lim
111
~
Instantaneous velocity
[3.8]
3/-+06.t
SI unit: meIer per second (m/s) Thc dircction of the instantaneous \'c1ocit}' vector is along a line that is tangent to the Object'S path and in the dircction orits motion. An object's average acceleration during a time intervill 6.1 is the change in its velocity Il divided by Ill, or
v
+- Average acceleration
[3.9] SI unit: meIer per second squared (m/s')
An object'S instantaneous acceleration vector acceleration vector as 6.1 goes to zero:
A-;
-+
a= lim6/-+0
AI
a is the limit of its average [3.10]
SI unit: meter per second squared (m/s 2 ) II.'s important to ,"ccognize that all object can accelerate in several ways. First, lhe magnitude of the velocity vector (the speed) may change 'with time. Second, the direct ion of the velocity vector may change with time, even though the speed is
61
+- Instantaneous acceleration
62
Chapter 3
Vectors and Two-Dimensional Motion
cOnSlanl.. as can happen (Ilong a curved path. Third, bOlh t.he magnitude and the direction of the velocity vector may change aL the same lime.
QUICK QUIZ 3.3
Which of the following objects can't be accelerating? moving with a consranl speed; (b) an object moving with a constant velocity; (c) an ol~jcct moving along a curve. (a) An
o~jcct
QUICK QUIZ 3.4 Consider the following conlrols in an aULOlnobile: gas pedal, brake, steering \vhecl. The controls in this listlhat cause an acceleration of the car are (a) all three controls, (b) the gas pedal and the brake, (c) only the brake, or (d) only the gas pedal.
3.4 MOTION IN TWO DIMENSIONS
Projectile motion
-+
TIP 3.4 Acceleration at the Highest Paint The acceleration in the ydirection is nofzero at the tOP ora prC!.jeclilc's tr.:Uecwry. Only the y-compollenl of tIle vclocit~· j .. zero there. If tbe acceleration were lero, lOa, the projectik
would '-
ne\(~r cOll)e
1n Chapter 2 \\Ie studied objects moving along straight·line paths, such as the x-axis. In this chelprer, we look at objects that move in both the x- and )'-dircctions simultaneously under constant acceleration. An important special case of this t\\'odimensional Illotion is called projectile motion. Anyone who has tossed any kind of object into the air has observed projectile motion. If the effects of air resistancc and the rotation of Earth are neglected, the path ofa projectile in Earth's gravity field is curved in the shape ofa parabola, as shown in Active Figure 3.14-. The positivc x-direction is horizontal and to thc right. and the )'-direnioll is vcnical and positive upward. The most important expcrimental fact about projectile Illotion in two dimensions is that the horizontal and vertical motions are completely independent of each other. This mcans that Illotion in one direction has no effect on motion in the other direction. If a baseball is tossed in a parabolic path, as in Anive Figure 3.14, the motion in the )'-dircction \-.. . ill look just like a ball tossed straight up uncleI' the influence of gravity. Active Figure 3.15 shows the cffect or various initial angles; note that complcl1lentar}' angles give the same horizontal nlngc. J n general, the equations of constant acceleration developed in Chapter 2 follow separately for both the ;\;-direction and the J-direction. An important differencc is that the initial velocity now has twO components, not just one as in that chapter. \,Ve assume t.hat at I = 0 the projectile leavcs the origin with an initial velocity Yo. If the velocity vector makes an angle 80 with the horizontal, where 00 is called the Jnvjn:tion a:ngl.t:, then from the definitions or the cosine and sinc funnions and Active Figure 3.14 we have
down! Voy
--------"
= Vo sin 80
where vox is the initial velocity (at t = 0) in the x-direction anel vO)' is the initial velocity in the y-direction.
ACTIVE FIGURE 3.14 The parabolic lr,u~ctory of ,1 pari icle thalleavcs the orig-ill with a vcloch)' ofv u. Note llial V changes with lilLle. Tlowcvcr. the x-component of the velocity. I'.. , remains conS!
y
8
ans ,-eflieall} dOWII\\"
II-:..,;.:..
-\_-,;;~x
3.4
63
ACTIVE FIGURE 3.15 r\ proj<'Clile !alltH lied from lite urigin wilh all illiliill .;peed 01:;0 III!S at \'ariOlI~ angks of projt'l,tillli. NCIIl" tllal Lomplclllclltarv \·alue.; of tile illili.11 ~1II~le 0 rnull in the '1.lIllt' \-alut" lIf R (lhe r:IIII-;I' uldu: plOjL·etile).
" (111)
t·,
Motion in Two Dimensions
= 50 1lI/~
If)O
100
50
100
1.:;0
200
2:;0
ow. Equations 2.6, 2.9, and 2.10 developed in Chapter 2 for motion with constant acceleration in one dimension carr;' m·er to the n\"o-dimensional case; there is olle set of three equations for each direction, with the initial velocities modified as just discussed. In the x-direction, with fix COII'\1.;\l1t, we have
I he indi\ic!u well as il~ initial Spl'l'<1.
[3.11.] [3. lib] [3.lIc]
where vn\, =
110
cos O!l'
III
the y-directiol1, we
h~vc
[3.12.] ,\
L1Y
vni
=
I
.,
+ 2a\'-
[3.12b] [3.12c]
where v{)¥ = v() sin 00 and ay is constant. The object's speed v can be calculated from the components of the velociLy Llsing the Pythagorean theorem: V
• I'J v~ -
= 'v
+
'j
v~-
The angle I haL the velocity vector mak('~ with the x-axis is given by
e=
'all 1 (:::)
This formula for 0, as previollsly stated. must be used with care. because the inverse tangent funnion returns values only bCl\\'een -90 and +90". Adding 180" is necessary for vectors lying in the second or third quadrant. The kinelllatic equations arc easilv adapted and simplillcd for projectiles close to lhe surface of the Eartho In that case, assuming air friction is negligible, [he accelef
=
v,l(
= vox =
Vo
cos 80
= constant
[3.13.]
whereas the horizontal displacement is !loimply Lh.· = vlh·t = (vo cos 0ll)l
For the l11otion ill the J,-direction, we make the substitution vo) V o sin 00 in Equations 3.12. giving vf =
VI)
si n 00
Do)' = (VI)
-
gf
sin ( 0 )1 - ~gl:!
[3.13b] tt'J
-g and
[3.14.] [3. 14b] 13.14c]
~
64
Chapter 3
Vectors and Two-Dimensional Motion
The important ["(lets of projectile motion can be summarized as follows: 1. Provided air resistance is negligible, the horizolllal componenl. of the velocit)' v.. remains conSlanl because there is no horizontal componenlofacceleralion. 2. The vertical component of the acceleration is equal LO the free-fall acceleration -g. 3. The vertical component of the velocity v y and the clisplaccrncnL in the y-direction are identical to those ora freely Calling body. 4. Prqjcclilc mOlion call be described as a superposition of two independent mOlions in the x-and J-directions.
EXAMPLE 3.4 Goal
Projectile Motion with Dia rams
Approximate answers in projectile motion using a Illotion diagram.
Problem A ball is thrown so that its initial vertical and horizontal components of velocity are 40 m/s and 20 mis, respectively. Cse a motion diagram to estimate the ball's total time of flight and the distance it traverses before hiuing the ground. Strategy Use the diagram, estimating the acceleration of gravity as -10 m/s 2 . By symmetry, the ball goes up and comes back down to the ground at the same )'-velocity as when it left. except with opposite sign. \'Vith this faCl and the fact that the accelera[ ion of gravity decreases the velocity in the y-direction by 10 m/s every second, we can find thc tot.al time of flight and then the horizontal range.
FIGURE 3.16 (Exampll' 3.1) Motion dia· gram for a projectile.
Solution -10 mls every second; hence, approximately 8 seconds Tn the Illmion diagram shO\\in in Figure 3.16, the accel- elapse during the motion. eration vectors are all the same, pointing dmvl1\vard with The \'elocit}' vector constantly changes direction, but magnitude or nearly 10 tn/s 2 . By symmctry, we kno\\llhat the horizontal velocity never changes because the accelthe ball will hit the grollnd at thc saine speed in the y- eration in the horizontal direction is zero. Therefore, direction as when it was thrown, so the \'elocity in [he the displacement of the ball in the x-direction is given by ),-direclion goes from 40 m/s to -40 mls in steps or Equation 3.13b, t.x = 1)0.,' = (20 mjs) (8 s) = 150 m . ~ ...... . ........................................... _ Remarks
This example emphasizes the i ndepenc1encc of the x- and )'-components in
pr~jeetile
motion problems.
QUESTION 3.4
Is the magnitude of the velocit)' vector at impact grcaLcr than, less than, or equal
to the magnitude of the initial
velocity vector? 'Why? EXERCISE 3.4 "Estimate the maximluTI height in this same problem.
Answer
80 m
QUICK QUIZ 3.5 Suppose you arc carrying a ball and running at constant speed, and wish to throw the ball and catch it as it comes back down. Should yOll (a) throw the ball al an angle of abollt 45° above the horizol1l.al and maintain the same speed, (b) throw the ball straight up in the
3.4
PROBLEM-
LVIN
Motion in Two Dimensions
65
TRATE Y
PROJECTILE MOTION I. Select a coordinate system and sketch the path of the projectile, including
initial and final positions, velocities, and accelerations. 2. Resolve the initial velocity vector into x- and y-componems. 3. Treat the horizontal motion and the vertical motion independently. 4,. Follow the techniques for solving problems with constant velocity the horizontal motion or [J1e projectile.
to
analyze
5. Follm.. . the techniques Cor solving problems with constant acceleration to analyze the venical motion of the projectile.
EXAMPLE 3.5
Stranded Ex lorers
Goal Solve a two-dimensional projectile motion problem in which an object has an initial horizontal velocity.
J
1---" 40.0 m/s
-1---<' "
Problem An Alaskan rescue plane drops a package of emergency rations to a stranded hiker, as shm.,.'n in Figure 3.17. The plane is traveling horizontally at 40.0 m/s at a height of l.00 X 10 2 m above the ground. (a) "Vhere does the package strike the ground relative to the point at which it was released? (b) V\'hat are the horizontal and vertical components orthe velocity of the package just before it hits the ground? Strategy Here, \\'c'rejust taking Equations 3.13 and 3.14, fi1ling in kno""n quantities, and solving for the remaining unknown quantities. Skctch the problem using a coordinate system as in Figure 3.17. In pan (a), set the y-component of the displacement equations equal to -1.00 X 10~ m-the ground le\'cl \vhere the package lands-and solve for the time it takes the package to reach the ground. Substitute this time into the displacernent equation for the x-component to fll1d the range. In pan (b), substitute lhe time found in part (a) into the velocity components. Notice that the initial velocity has only an .x:-component, which simplifies the math. ~
100m
"
,
FIGURE 3.17 (Example 3.5) From the POilil of"icw oran obsen'er on the ground, a package n..:leased from Ih~ rescue plane travels along the path shown.
.....
.
....
Solution (a) Find the range orlhc package. Use Equation 3.14b
to
find the y-displacel1lent:
Substitute Yo = 0 and uo) = 0, set)' = -LOa X 10 2 mthe final vertical posit.ion of t.he package relative the airplane-and solve [or time: Use Equation Substitute
.\'0
~.13b to
y = -(4.90 m/5')1'
= -1.00 X 10' III
{~4.52s
find lhe x-displacement:
= 0, vox = 40.0 m/s, and the time:
x = (40.0 Ill/s)(4.52 s)=
181
III
(b) Find lhe components of' the package's velocity at impact:
Find the x-componcnt of the velocity at the time of
v.'t =
1)0
'\ =
V
cos 8 = (40.0 m/s) cos 00 =
40.0 m/s
i mpacc
rind the y·component of thc velocity at the time of' impact: >.
n sin 8 - gt = 0 - (9.80 m/s:!) (4.52 s)
- ..4.3 m/s • • • • • • • • ..tI.
66
Chapter 3
Remarks
Vectors and Two-Dimensional Motion
NOLicc how motion in the x-direction and motion in the J-direcl.ion are handled
sep~lra[el)'_
QUESTION 3.5 ~eglecting air
h-iClion effects, what path does the package travel as observed by the pilot? \..vh)'?
EXERCISE 3.5 A bartender slides a beer Illug at 1.50 111/S toward a customer at the end of a frictionless bar that is 1.20 III tall. The customer makes a grab for the Illug and misses, and the mug sails olT the end of the bar. (a) HQ\\l far away from the end of the bar does the mug hit the noor? (b) \'\That are the speed and direction of the mug at impact?
Answers
(aJ 0.742 m (b) 5.08 m/s, 8 = -72.8'
EXAMPLE 3,6 The Long Jump' Goal Solve a two-dimensional projectile motion problem involving an object starLing and ending at the same height. A long jumper (Fig. 3.18) leaves the ground at an angle of 20.0 0 10 the horizontal and at a speed or ILO m/s. (a) How long does it take for him to reach maximum height? (b) \'\7hat is the maximum height? (c) How far docs hejump? (Assume his motion is equivalent to that or a particle, disn::garding the Illotion of his arms and legs.) (d) Use Equation 3.l4c to flnd the maximum height he reaches.
Problem
Strategy Again, we take the projectile equations, fill in the known quantities, and solve [or the unknowns. At the maximum height, the velocity in the )'-direction is zero, so setLing Equation 3.14a equal to zero and solYing gives the time it takes him to reach his maximum height. By symmetry, given that his trajectory starts and ends at the same height, doubling this time gives the total time of the jump.
Solution (a) Find the time Set
.
,
~
VI'
'max
FIGURE 3.18 (E:\.arnple 3.6) ~Ijke Powell, current holder Ilflhe world IOllgjump record OrS.9:l rn.
.
-
,
taken to reach maximum height.
= 0 in Equalion 3.14b and solve for
'max:
Vy
=
~lax
=
Un
sin On -
110
sin 8" g
g'lIlax
=0
(11.0 m/s)(sin 20.0')
9.80 rn/s~
0.384 s (b) Find the maximulll height he reaches.
Substitute the time y-displacclll en t:
'rll:lx
into the equation for Lhe
y"""
= (11.0 m/s)(sin 20.0')(0.384 s)
- ~(9.80 m/s')(0.384 s)' Jl rnax =
0.722
III
(c) Find the horizomal distance hejulllps. First flnd the time for l.hcjump, \"hich is twice
'11I<1X:
Substitute this result into the equation for Lhe x-displacement: (d) Use an alternate mClhod to find the maximum height. Use Equation 3.14c, solving for .o.y:
Substitute v). = 0 at maximum height, and the fact that v"J = (l1.0 ,;,/s) sin 20.0°:
67
Motion in Two Dimensions
0.722 m
-2(9.80 m/s')
~
_
Remarks Although modeling the long jumper's motion as that of a projectile is an oversimplification, the values obtained are reasonable.
QUESTION 3.6 True or False: Because the x-component of the displacement doesn't depend explicitly on g, the horizontal distance traveled doesn't depend on the acceleration of gravity. EXERCISE 3.6 A grasshopper jumps a horizontal dist.ance of ] .00 III from rest, ",lith an initial velocity at a 45.0" angle with respect to the horizontal. Find (a) the initial speed of the grasshoppcr and (b) the maximum height reached.
Answers
(a) 3.13 m/s (b) 0.250 m
EXAMPLE 3.7 Goal
The Range Equation
Find an equation for the maximum horizolllal displacemcm ofa projectile nred from ground level.
Problem An athletc participates in a long:jump compctition, leaping into the air with a velocity V o at an angle with t.he horizonLal. Obtain an exprcssion for the length of 1he jump in tcrms of vo, 00 , and g. Strategy
eo
Usc the results of Example 3.6, eliminating the time l from Equations (I) and (2).
.
.
,
~
~
Solution Use Equation C1) of Exarnple 3.6 to nnd the time of night, l:
Substitute t.his expression for t into Equation (2) of
~x =
() en
ax =
9 1/- V"
Vo
cos
l
=
(110 cos 0 0
Example 3.6:
Simplify:
Remarks
sin 00) g
0I) cos O' j) SIJ)
g
Substitute the identity 2 cos On sin 00 = sin 20 0 to reduce t he foregoing expression to a single t.rigonomctric function: ~
l( 2Vo
,
,
-,
,
-
(I)
.
.
,
~
The usc of a trigonometric identity in t.he nnal step isn't necessary, but it makcs Question 3.7 easier to
ans,·\'cr. QUESTION 3.7 VVhal. angle On produces the longesljurnp? EXERCISE 3.7 Obtain an expression for the athlete's maximum displacement in the vertical direction, and g.
Answer
6.)'111.11\ =
vo2 sin 2 0 0
2g
6.Ym;n;
in terms of
Vn,
00 ,
68
Chapter 3
Vectors and Two-Dimensional Motion
EXAMPLE 3.8 That's Quite an Arm Goal Solve a two-dimensional kinematics problem with a non horizontal initial \'clocity, starting and ending at different heights.
110'::
20.0 m/s ......
Problem A stone is thrm..·!1 upward from the LOp of a building at an angle of30.0° to the horizontal and with <'In initial speed of20.0 m/s, as in f'igurc 3.19. The point of release is 45.0 In above the ground. (a) How long does it take [or the stone to hit the grollnd? (b) Find the SLone's speed at impact. (c) Find the horizontal range of the stonc. Neglect air resistance.
,,
>:
,,
,,
,, ,
,
",,,,
Strategy Choose coordinates as in the figure. with the origin at the point of release. (a) Fill in the constants of Equ3tion 3.14b [or the y-displaccmcnl and set the displacement equal to -4!'>.0 m, the )'-displacement ""hen the stone hits the ground. Using the quadratic formula, solve for the time. To solve pan (b), subsl.iwte the time [rom pan (a) into the componems of the \'e1ocity, and substitute the same time into the equation for the x-displacement to solve pan (c).
,
\
\ I \ \ \
\ I
(>:.-45.0 Ill) I
-, FIGURE 3.19
(Example .3.8)
.,
".
Solution (a) Find the time of flight. Find the initial x- and J-components of'the velocity:
Find the y-displacemcnt, taking Yo vo)' = 10.0 m/s:
=
= vo,. =
V o•
0,)' = -45.0111, and
Reorganize the equation illLo standard form and use the quadratic formula (see Appendix A) 1.0 find the posit.ive
vosin
= (20.0 m/s) (cos 30.0°) = +17.3 m/s eo = (20.0 m/s)(sin 30.0°) ~ +10.0 m/s
6} =
J -
eo
Vo cos
)'0
= vo./ - ~gt'2
-45.0 m = (10.0 m/s)/- (4.90 m/s')/' 4.22s
/=
l'OOt:
(b) Find the speed
at.
impact.
=
Substitute the value of I found in part (a) into Equation 3.14a to find the y~col1lponentof the velocity at impact:
v,
Use this value of vJ ' the Pythagorean theorem, and the fact that V x = vO.t = 17.3 m/s to find t.he speed of t.he stone at impact.:
v = V V, 2
v nJ -
g/
= -31.4
~
=
10.0 m/s - (9.80 m/s') (4.22 5)
m/s
v,' = V( 17.3 m/s)' +
+
(-31.4 m/s)'
35.9 m/s
(c) Find the horizontal range of the stone. Substitute the time of" night. into the range equation:
.:'»- = x =
Xo
=
(v o cos
ell = (20.0 m/s)(eos 30.0°)(4.22 s)
73.1 m
~
~
Remarks The angle at which the b~lll is lhrown affects the vclocity vect.or throughout it.s subsequent. motion. but doesn't affect the speed at a given height. This is a consequence of the conservat.ion of energy, described in Chapter 5. QUESTION 3.8 True or False: All other things being equal, if the ball is thrmvn at hal r the given speed it will t.ravel half as far.
3.4
Motion in Two Dimensions
69
EXERCISE 3.8 Suppose the stone is thrown from the same heighl as in the example at an angle of 30.0 degrees below the horizontal. If it strikes the ground 57.0 m away, find (a) the time of flight, (b) the initial speed, and (c) the speed and the angle of the velocity vector with respcct to the horizontal at impact. (Hinl: For part (a), use the equation for the xdisplacement to eliminate vol [,"om the equation [or the J-displacemcnt.) 0
Answers
(a) 1.57 s (b) 41.9 m/s (c) 51.3 mis, -45.0"
So far we have studied only problems in which an object with an initial velocity (ollmvs a trajectory del.errnilled by the acceleration or gravity alone. Tn the more general case, other agents, such as air drag, surface friction. or engines, can cause accelerations. These accelerations. taken together, form a vector quantity with components ax and fl),. \·Vhen both components are constant, we can lise Equations ~.ll and 3.12 to study the motion, as in the next example.
EXAMPLE 3,9 Gool
The Rocket
Solve a problem involving accelerations in two directions.
Problem Ajet plane traveling horizontally at 1.00 X 102 m/s drops a rocket from a considerable height. (See Fig. 3.20.) The rocket immediately fires its engines, accelenlting at 20.0 m/s:! in the x-direction while falling under the influence of gravity in the y-direct:ion. When the rockct has fallen 1.00 kill, lind (a) its velocity in the y-direcl.ion, (b) it.s velocity in thc x-direction, and (c) t.he magnitude and direction of its velocity. Neglect air drag and aerodynamic lin.
,,
,,
,,
!1y= -1.OOX
,, \, 103m\
,,
Strategy Because the rocket maintains a horizontal orientation (say, through ~ gyroscopes), the x- and )'-components of acceleration are independent of" each FIGURE 3.20 (Example 3.9) other. Usc the time-independent equation for [he velocily in the y-d i rection LO find the y-component. of the velocity after the rocket falls 1.00 kill. Then calculate t.he time of the rail and use that time to find the velocity in the x-direction .
,..
.
. . . .... . .
.....
....
"
Solution (a) Find the velOCity in the )I-direction. Use Equation 3.14c:
vv' -
Substilllt:e VOl' = 0, g = -9.80 m/s2 , Clnc! ily = ~ 1.00 X 1O.'l m, and solve for V.I.:
0 = 2(-9.8 m/s')(-1.00
v,. =
X
10' m)
-lAO X 10' m/s
(b) Find the velocity in the x-direCLion. Find thc time il takes the rocket 1.0 drop LOO X -10 3 m, using the y-componcnl of the velocity:
SllbslitUl.e I, vox, and a.~ inl.O Equation 3.11a velocity in the x-direction:
10
find the
-lAO X lQ-,> m/s = 0 Vx
= v o, ~
+
(
9.80 m/s-' ) I
1/) ~ 1.00 X 10' m/s
I =
14.3 s
+ (20.0 m/s')(14.3 s)
386 m/s
(c) Find the magnitude and direction or the velocity. Find the magnitude lIsing the Pythagorean theorem and the results of paris (a) and (b):
v ~ Vv,' + v,' ~
411 m/s
=
V( -lAO
X 10'm/s)'
+ (386 m/,)'
70
Chapter 3
Vectors and Two-Dimensional Motion
e=
Use the inverse tangent function to find the angle:
tan-
l
( lJ,) = V
tan
x
.
,
~
Remarks
_,(-1.40X10'm/s) 386 m/s
=
-19. go
.............................~
Notice the symmetry: The kinematic equations for the x- and y-c1il-eclions are handled in exacLiy the same
way. Having a nonzero acceleration in the x-eli rcctioll doesn't greatly increase the difficulty of the problem. QUESTION 3.9
True or False: Neglecting air friction, a projectile ,,\·jth a horizontal acceleration slays in the air longer than a projectile thaL is freely railing. EXERCISE 3.9
Suppose a rockel-propelled motorcycle is fired from rest horizontally across a canyon l.00 km wide. (a) ',"hat minimum constam acceleration in the x-direction must be provided by the engines so the cycle crosses safely if the opposite side is 0.750 kill lower than the starting point? (b) At ""hat speed docs the motorcycle land if it maintains this constant horizontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction. Answers
(a) 13.1 mis' (b) 202 m/s
In a stunt similar to that described in Exercise 3.9, motorcycle daredevil Evel Knievel tried to vault across Hells Canyon, part of the Snake River system in Idaho, on his rocket-powered Harley-Davidson X-2 "Skycycle." (See the chapter-opening photo on page 54). He lost consciousness at takeoff and released a lever, premaLUrel)' deploying his parachute and falling short of the other side. He landed safely in the canyon.
3.5
)'
RELATIVE VELOCITY
Relative velocit)' is all about relating the measurements of two different observers, one moving '.vith respect to the other. The measured velocity of an object depends on t.he vclocity of the observer with respect to the object. On higlnvays, for example, cars moving in the saine direct.ion are often rnoving at high speed relative to Earth, but. relative to each other they hardly move at all. To an observer at rest at the side of the road, a car might be traveling at 60 mi/h, but. to an observer in a truck travcling in the same direction at 50 mi/h, the car would appear to be traveling only]O mi/h. So measurements of velocity depend on the reference frame of' the observer. Reference frames are just coordinate systems. Most of lhe time, we use a stationary frame of reference relative to Earth, but occasionally 1ve usc a moving frame of reference associated with a bus, car, or plane moving with constant velocity relative to Earth. In t,.".o dimensions relative velocity calculations can be confusing, so a systematic approach is important and useful. Let E be an observer, assumed stationary with respect to Earth. Let l.\VO cars be labeled A and B, and int.roduce the following notation (see Fig. 3.21): rAI~ = the position of Car A as measured by E (in a coordinate system fixed with
respect to Earth).
-::k:-:-----jf-----x E
FIGURE 3.21 The position orCar A rehtliw: to Car B can be round by vector subl raction. The rate or change or the rCSll1tant \'ector with respen to lime is the rdative vdncilr equation.
rBE
= the position orCaI' B as measured by E.
fAll
= t.he position orCaI' A as measured by an observer in Car B.
According to the preceding notation, the nrst letter tells us "..· hat the veclor IS pointing at and the second lett.er tells us where the position vector starts. The posiand rBE, are given in the Agure. tion vectors of Car A and Car B relati\'e to E, How do \ove find 1,,1\' the position of Car A as measured by an observer in Car B? 'Ve simply draw an arrO\'v pointing from Car B to Car A, which can be obtained b)' subtracting THE from r.'\E:
rr\E
3.5 ~
I"AI\
=
~
rAE -
Relative Velocity
71
~
rBE
[3.15]
Now, the rale of change of these quantities ,.. . ith lime gives us the rel
The coordinate system of observer E need not be fixed 10 Eanh, although it oflen is. Take careful note of the pattern of subscripls; rather than memorize Equation 3.16, it's better to study the shon derivation based on Figure 3.2L Notc also thal t.he equation doesn't. 'work for observers traveling a si/ablc fraction of the speed of light, \vhcn Einstein's theory of special relmi\'ily comes into play.
PROBLEM-SOLVING STRATEGY RELATIVE VELOCITY
1. Label each oqjecl involved (usually three) with a letter that reminds you of what it is (for example, E for Earth). 2. Look through the problem lor phrases such as "The velocity of A relative to B" and write the velocities as VAB . vVhen a velocit}, is mentioned but it isn't explicitly stated as relative to something, it's almost always relative to Earth. 3. Take the three velocities you've found and assemble them int.o an equation just like Equation 3.16, with subscripts in an analogous order. 4. There will be two unknown components. Solvc for them with lhc x- and y-components of the equation developed in step 3.
EXAMPLE 3.10 Gool
Pitching Practice an the Train'--
~_~
~
__
Solve a one-dimensional relative velocit)' problem.
Problem A train is traveling with a speed of 15.0 ll1/s rcla, ive LO Earth. A passenger standing a\. the rear of the tr"in pitches a baseball 'with a speed of 15.0 m/s rclmiye La the train off the back end, in the direction opposite the motion of the train. \'Vhat is the velocity of the baseba 11 rel ali ve to Ea nh?
Strategy Solving' these problems involves putting the proper subscripts on the velocities and arranging them as in Equation 3.16. In the first sentence of the problem
,.
~
stat.ement, \\!e are informed that the train travels at "15.0 m/s relative 1.0 Earth." This quantity is VrE, with T [or train and E for Earth. The passenger throws the baseball al "15 m/s relative to the train," so this quantity is VUT , where B stands for baseball. The second sentence asks for the velocity of the baseball relative to Earth, VeE' The rest of the problem can be solved by identifying the correct components of the known quantities and solving [or the unknowns, llsing an analog of Equation 3.16.
.
.•• ''''''l
Solution ","'rite the x-components of I. he knmvn quantities:
("\In:L = +]5 m/s (Vir,,), = -J5 m/s
Follmv Equation 3.16: 1nsert the given values a nd solve:
-15 m/s = (V BI.·),
(v,,,),
=
15 m/s
0
............ -
Irrr... • • • • • • • • • • • • • • •
QUESTION 3.10 Describe the motion
-
or the ball as related by an observer on the ground.
~
Chapter 3
72
Vectors and Two-Dimensional Motion
EXERCISE 3.10 A train is traveling at 27 mls relative to Earth, and a passenger standing in the train thrmvs a ball at 15m/s relatiyc to the train in the same direction as the train's motion. Find the speed of the ball relative to Earth. Answer
42 m/s
EXAMPLE 3.11 Goal
Crossin
0
River
Solve a simple two-dimensional relative motion problem.
Problem The boat in Figure 3.22 is heading due Ilonh as it crosses a wide river '"\lith a velocity of 10.0 km/h relative to the water. The river has a uniform velocity of 5.00 km/h due east. Determine the velocity orthe boat ".. jth respect to an observer on the riverbank. Strategy Again, we look for key phrases. "The boat (has) ... a velocity of lO.O km/h relati\'c lO the water" gives VBR. "The river has a uniform velocit), of 5.00 km/h due east" gives VRE' because this implies velocity with respect LO Earth. The observer all the riverbank is in a reference frame at rest \'iith respect to Earth. Because we're looking for the velocity of the boat with respect to that observer, rhis last velocity is designated VBI'. Take east to be the +'\·-direction, north the +y-di rectiol1.
=,-
~.~.
FIGURE 3.22
,..
(Example 1.10)
.
.~
Solution Arrange the three qualllities into rhe proper relaLi\·c velocityequat.ion: '''·rite the velocity vect.ors in terms of their components. For convenience, these are organized in the following
Vector
x·Component (km/h)
y-ComponenL (km/h)
o
10.0
~
VB'
table:
~
v, 5.00
V BE ~
VIU:
Find the x*componenl of\'e1ocity:
o=
5.00 km/h
-:,.
Vx
Find the y-componellt of velocity:
10.0 km/h = v, - 0
~
v, = 10.0 km/h
1Jx -
=
5.00 klll/h
Find the lllagninlde orvlu::
= Find the direction o[V UE :
V(5.00 km/h)'
8 = tan- J
(v,)
..... Remark
v)
+ (l0.0 km/h)' = 11.2 km/h
= tan-J
I11/S) =
(5.00 10.0 mh
26.60 • • • .,J.
The boat travels at a speed of 11.2 km/h in the direction 26.6 east of north with respecllo Earth. 0
QUESTION 3.11 Uthe speed of the boat relative to the waler is increased, what. happens
1.0
lhe angle?
EXERCISE 3.11 Suppose the rh'er is flO\ving eaSl at 3.00 111/S and the bOaL is lraveling south at 4.00 m/s with respect to the river. Find rhe speed and direct.ion of the boal relative lo Earth. Answer
5.00 mis, 53.1 south of eaSl C
3.5
EXAMPLE 3.12 Goal
Relative Velocity
73
Bucking the Current
Solve a complex t\\:o-dimensional relative mOlion problem.
Problem
I r the skipper of the bam of Example
3.11 moves with the same speed of 10.0 km/h relative to the water but now wants to travel due
north, as in Figure 3.23, in what direction should he head? \·Vhat is the speed of the boat, accordin~ LO an observer on t he shore? The river is
nov,,'ing east at 5.00 km/h. Strategy Proceed as in the previous example. In this siwation, we must find the heading of the boat and its \'elocit)' with respect to the \\-'ater, using the fact that the boat tra\"cls due north.
" W+E
, \\4'+ ~
S
s
Q; (a)
FIGURE 3.23
(b)
(
.
~
.
.. - - ....
~
Solution ArrClnge the three f]uantities, as before: Organize a table or velocity components:
Vector
x-Component (km/h)
y-Component (km/h)
-(IO.Okm/h) sinO
(10.0 km/h) cos 0 v
o
o
5.00 km/h The x-component of the relative velocity equation can be used to find 0:
-(10.0 m/s) sin 0 = 0 - 5.00 km/h sin (} =
Apply the inverse sine function and find (}, which is the boat's heading, east of north:
0= sin- I (1.00) --
The )l-COmpOncl1l used LO fi nd v:
(10.0 km/h) cos () =
of the
relative vclocily equation can be
2.00
.......... Remarks From Figure 3.23, we sec that this problem call be solved with 1he Pythagorean ! hcorem, because the problem involves a riglll triangle: the boat's x-component ofvelocit)' exactly cancels the r;\'cr's velocity. \'Vhen this is not the case, a morc generallechniquc is necessary, asshm'm in the foIl m . .·i ng exercise. Notice that. in the x-component of the relative velocity equation a minus sign had to be included in the term -(10.0 km/h)sin () because the x~colllponellt of the boat's velocity with rcspeclto the river is negative.
QUESTION 3.12 The speeds in this example <'Ire the sarne as in Example 3.11. vVhy isn't the angle the same as before?
=
5.00 klll/h 10.0 km/h
1.00 2.00
30.0'
V
-7
V
= 8.66 km/h .
.J
EXERCISE 3.12 Suppose the river is lTIm'ing east at 5.00 km/h and the boat is traveling 45.0° south of east with respect to Earth. find (a) the speed of the bOlH with respect to Earth and (b) the speed of the boat with respect LO the river if the boat's heading in the water is 60.0° south or cast. (See Fig. 3.23b.) You will have Lo solve two equaLions with 1.\\'0 unknowns. Answers
(a) J6.7 km/h (b) 13.7 km/h
74
Chapter 3
Vectors and Two-Dimensional Motion
.
---~
SUMMARY 3.1
Vectors ond Their Properties
Two vectors Aand Bcan be added geometrically with the triangle method. The 1wo vectors arc drawn to scale on graph paper, with the tail of the second wetor located at the tip of the first. The resultant vector is the vector dnn\'11 from the tail orthe first VCCLOI~to the lip orthe second. The negative of a vceLOr A is a veCLOr with the same magnilllde as A, bUl pointing in the opposite direction. A vector can be multiplied by a scalar, changing its magnitude, and its direction if the scalar is negative.
3.2
Components of
Taking the limit of this expression as ill gelS arbitrarily small gives the instantaneous acceleration vector
a:
---->
t,·..
:=
3.4
Motion in Two Dimensions
v~: = v rjx
vx '!. ''''!Jere vox =
TJn
[3.2]
-
The magnitude and direction or A arc related to its components through the Pythagorean theorem and the definition orthe tangcnt:
-
If Ii = R arc
~
fly
[3.4]
-
A.~
1
[3.5b]
[3. lib]
[3.l1c]
VOl
+
[3.12aJ
(l.J!
vo"l + !a/2
13.12b]
va_/" +
[3.1 2c]
211 1, Ll)l
\.. . here vo~ = v n sin 00 , The speed vo[thc object atany instant call be calculated from thc components of vcloci[y at lhat instant using the Pythagorean theorcm: v=
Vv'.!.x + v:!y
The angle that the velocity ,"ector makes with the x-axis is given by
o = tan
A + Ii, then the componcllls of the resultant vector [3.5aJ
[3.l1a]
+ !o.i va/" + 2a x .6.x
:=
vJ '.!. =
[3.3] 0
=
v) :=
,\.=AsinO
loll
+ 0.:/
cos Bt)' and, for the y-direcrion,
6.)'
A cos 0
[3.10]
The general ki nematic equations in t\-\'o dimensions [or objects wit.h constant acceleration are, for the x-direction,
!J"x = vox!
A vector A can be split into two components, one pointing in the x-direction and the Olhe.. in the )'-direcLion. These components form o sides of a right trianglc having a hypOI.cnllse with magnitude A and arc given by Il~.
D..v
•
11111 61
6,.1- ()
Vector
0
=
a
'('-'-'-) Ux
The kinematic equations arc easily adapted and simplified for projectiles close LO the surface of lhe Earth. The equations for lhc motion in the horizonral or x-direction are [3.13a]
3.3 Displacement, Velocity, and Acceleration in Two Dimensions
[3.13b]
The d isplaccmem of an object in two eli l11cnsions is dcn ned as the change in the object's posit ion vcernr:
while tht equations ror the motion in [he vertical or y-direction are
[3.6]
[3.14a]
The averag-e velocit)' of an object during the time interval n/ is
[3.14b]
[3.7]
Problems are solved by algebraically manipulating one or more ofl.hese equations, which often reduces the system LO t,-,;o equations and 1WO unknm.. ·ns.
-
" al' -
iii
Taking the limit or this expression as D.l gcts arbitrarily small gives the instantaneous velocity
v:
-jo
v
6.1 = ~i-O lim D.l
[3.8]
The direction of thc instamaneous velocity vector is along a line that is tangcllt to the path of the object and in the direction of its mOlion. The averagc acccleration of an object with a velocity changing by!1-V in the time interval!J", is
_
a",
D.v
=r;;
13.9]
[3.14c]
3.5
Relotive Velocity
Lei E be an observer, and B a second observer traveling with vclocilV rw as mcasured bv E. IfE measures the velocity or an o~iect.~ as V,lE' then B' will measure A's vclocity as
v
[3.16] Solving relative velocity problems ilwolvcs identifying the velocities properly and labeling lhem correctly, subsrilu(ing into Equation 3.16, and (hen solving for unknown qU
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
75
Conceptual Questions
MULTIPLE-CHOICE QUESTIONS l. A catapult launches
(l large stone at a speed of 45.0 Ill/s at an angle of 55.0° with the horizontal. What l1l<1xi-
11ll111l
height does the stone reach? (t\eg1ect ,1 i r rriction.) III (b) ~2.7 III (e) r,9.~ III (d) 83.2 III (e) ]02 III
(a) 45.7
2. A skier Icaves the end of a hurilolltal ski jUlllp at 22.0 m/s and falls :t20 111 before landing. NeglcCling friction, how rar hori7omally does the skier [ravel in the air before landingr (a) 9.8 J11 (b) 12.2 In (c) 14.3 m (d) 17.8 III (e) 21.6 III
3. A cruise ship sails due north at 4.50 m/s while a coast guard parral boat heads 45.0° north of weSI at 5.20 mls. What is Lhe velocity of the cruise ship relative to the palrol boat? (a) v~ = 3.68 m/s; 11,_ = 0.823 m/s (b) V x = -3.68 111/5: VI" = 8.18 m/s (c) V x = 3.68 111/5; ", = 8.18 Ill/s (d) ", '= -3.68 m/s: v"~ = -0,823 m/s (e) V x = 3.68 m/s; v). = 1.82 m/s 4. A \'ccrar lying in the xJ-plane has components of opposite sign. The venol' must lie in which quadrant? (a) the !'irsr quadrant (b) the second quadrant (c) the third quadrant (d) rhe fourth qnadram (e) either thc sccond or the fourth quadrant 5. An alhlete runs three-fourths of the way around a circilIaI' track. Which of the following statements is true? (a) His average speed is greater than the magnitude or his average velocit}'. (b) The magniulde of his average velocity is greater than his average speed. (c) His average speed is equal to the magnitude of his
eration of gravity is one-sixth that of Earth. :'Jeglecting air resistance (on Earth) and \'ariations in the acceleration of gravity with height, how long does it take the projectile on the .\1oon to reach its maximum height? (a) 'max (b) 'm:1x /6 (c) Y/6t m
/ (a)
(b)
(e)
(d)
(e)
FIGURE MCQ3,12
13. \Vilich of the following quantities are "ectors? (a) the velocity of a sports car (b) temperature (c) the volume of water in a can (d) the displacement ofa tennis pla)'er from the backline of the court LO the net (e) the height ofa l)[1iIding
~'S:0NCEPTUAL QUESTIONS
-
-
1. Il"B is added to A, under wh<1t conditions does the resultant vector have a magnitude equal 10 A + B? Under what conditions is the reslillant vcctor equal 10 zero?
3. As a pr~jectile moves in its path. is there an) point along the paTh where the "clocil:y and acceleration \'eClors are (a) perpendicular to each other? (b) Parallel to each other?
2. Uncle]' what circumstances would a vect.or have nents that are equal in magnitude?
4. Two vectors have unequal magnitudes. Can their surn be zcro? Explain.
COIllPO-
76
Chapter 3
Vectors and Two-Dimensional Motion
5. Explain whether t.he following panicles do or do not. have an acceleration: (a) a particle moving in a straight line with consram speed and (b) a particle moving around a curve with conSlant speed. 6 . .-\ ball is projected horizomally from the lOp ora building. One second later, another ball is projected horizontally from the same point. with the same velocity. At. what point in the mol ion will the balls be closest to each other? Will the first ball alw
7. A spacecrart drifts through space at a constant. velocity. Suddenly. a gas leak in rhe side of the spacecraft causes it to constantly accelerate in a direction perpendicular to the initial velocity. The oricmation of the spacecraf't does not change, so the acceleration remains perpendicular to the original direction of the velocity. \""hat is the shape of the path followed by the spacecraft?
8. Determine which or the following moving objects obey the equations of prqjec1.ile motion developed in this chapter. (a) A ball is thrown in an arbitrary direction. (b) A jet airplane crosses the skY' with its engines thrusting the plane forward. (c) A rocket lea\"es the launch pad. (d) A rocketllloves through the sky after its engines have failed. (e) A stone is thrown under water. 9.1\\10 projectiles are thrown with thc same initial speed, one at all angle 0 with respect to the level ground and the other at angle 90" - O. Both projectiles strike the ground ar the same distance from the projection point. Are both projectiles in the air for the same length of time? 10. A bat! is thrown upward in the air by a passenger on a train that is moving wir.h constant velocity. (a) Describe the path or the ball as seen by the passenger. Describe the path as seen by a stationary observcr OLilside the train. (b) I-Iow would these obscr"vations change if" the train were accelerating along the track?
-"~~~.,..-
PROBLEMS . ~
-~~
~
WebAssign
The Problems for this chaptel' mar be assigned onlinc al WebAssign.
denotes guided pr(,blcm denotes enhanct:d content problem
III =
biomedical application 0= denotes full solLHion available in ....· ,udenl Solllliun.\ /H(jIl1UlI/ S/lId)' (;uide
SECTION 3.1 VECTORS AND THEIR PROPERTIES 1. \'ector A has" magnitudc of 29 units and points in the positivc J~dircc[ion. \A,'hen yector B is added ro A. the resultalll \"ector A+ B poinrs in the negative y-direction with a magnitude of 14 units. Find the magnitude and di recti 011 of B. 2. Vector A has a rnagniLUde of 8.00 units and makes an angle of 45.W Wilh the positive x-axis. Vector Ii also has a magnitude of 8.00 units and is directed along the negative x-axis. L'sing graphical methods, !-ind (a) the vector sum A + B and (b) the vector difference A - 8. [TI\'ector A is 3.00 units in length and points along the positive x-axis. Vector Ii is 4.00 units in length and points along the negative J-axis. Usc graphical methods to find the magniLUde and direction of" the vcctors (a) A + Band (b)
A - il
4. Each of the displaccment vectors A and Ii shown in Fig~!"e Pl.4 ha~.a n~'l.gnit~dc~·3.00 m. ~raph~:al1}' find (a) A
+ B,
(b) A-B. (c) B - A. and (d) A - 2B. )'
3.00 In
5. A roller coaster moves 200 [t horizont
6.
am An airplane nies 200 kill clue west from city A to city B and then 300 km in the direction 0[30.0" north of west from cit)' B 1.0 city C. (a) In straight-line distance, how far is city C from city A? (b) Relat.ivc ro city A, in what direc· tioll is city C? (c) Why is the ans\.. .cr only approximately correct?
rnA plane flies from
base calllp to lake A, a distance of 280 km 'Ita direction of20.0" north of cast. After dropping off supplies, the plane nics to lake B. which is 190 km and 30.0" west or north from lake A. Graphically determine the distance and direcljon from lake B to lhe base camp.
8. Ajogger runs 100 In due west, then changes direction for the second leg of the ntn. Atlhe end of thc run, she is 175 m away from the starling point at an angle of 15.0 a north oh....est. What wcre the direction and lengLh of her second displacement? Use graphical techniques. 9. A man lost in a malC makes three consecutivc displacements so that at the end ofh is tra\'cl he is right back where he started. The first displacement is 8.00 m westward. and the second is 13.0 III northward. Usc the graphical method to find the magnitude and direction orthe third displaccment.
SECTION 3,2 COMPONENTS OF A VECTOR
A
10. Thc magnitude of vector A is 35.0 units and points in the direction 3250 counterdock\visc from lhe positivc x-axis. Calculate the x- and )'-componcnts of this vector. 11. Agolt'cr takes two putlstogethis ball into the hole once he
-0;;-1'''----'-------- x FIGURE P3.4
is on the green. The Ill'Sl putt displaces the ball 6.00 m east. tJ1C second 5A-O III south. \'\!J13l displacement would ha\"e been needed to g-et thc ball illto the hole on the first putt?
Problems 12. A figure skaler glides along a circular pmh of radius 5.00111. [f she coasls around one h,ilf 01" the circle, lind (a) the magnitude of" the displacement veCLQr and (b) what distance she skated. (c) What is the magn itude of 1he displacement if she skates all the way around the circle?
77
is equivalent 10 the two forces shown and (b) the force a third person would have (0 exert on the mule (Q make the net force equal to zero. The forces are measured in units of newlQlls (N).
13. A girl delivering newspapers covers her rome by traveling 3.00 blocks \.. ; (:'SI, 4.00 blocks norlh. and then 6.00 blocks east. (a) 'What is her resulranl displacclllem? (b) Wh"H is the tOlal distance she travels? ]4.
mil
A hiker starts at his camp and moves the following distances while exploring- his su rroundings: 7!'i.0 m non h. 2..::i0 X 10 2 m east, 125 m at an angle 30.00 north orcas!.. and 1.50 X 10 2 m 50mh. (a) Find his resultant displacement from camp. (Take east. as lhe positive x-direction and north as the positivc y~direction.) (b) \'\-'auld changes in the order in which the hiker makes the given dispbcemel1ls alter his final position? Explain.
15. A vector has an x-component of -25.0 units and a )'component of 40.0 units. Find thc magnitude and direction of the veClOL 16. A quanerhack takes the hall from the line of . . crinl1nage, runs backwards for 10.0 yards, then rllns sideways parallel to the line of scrimmage for 15.0 )'ards. At this point. he t.hrows a 50.0-yard forward pass straight downflcld, perpendicular to the line of scrimmage. \-\'hat is the mag-Ilitilde of the football's resultant displacement? lJ1JThc eye of a hurricallc passes over Grand l3ahama Island in a direction 60.0" nonh of west wilh a sl)eed of 41.0 km/h. Three hours lalcr the course of lhe hurric-lIle suddenly shifts clue nonh. and its speed slows to 25.0 km/h. How far from Grand l3ahama is the hurrinll1e 4.50 h after it passes over the island~ 18. A small map shows Atla1ll.a to be 730 miles in
) (km)
,._-'9:.1':11°+__... 1 C• c 'J 200
'
W -----Ef:t---.- E
20.0" ~
T s
_ _ _ _ _~::.JJ::::.:l-.L-Lx (kill)
o
50 100 150 200 FIGURE P3.19
20. The helicopter view in Figure P3.20 shows IwO people pulling on a stubborn mule. Find (a) t.he single force that
FIGURE P3.20
21. A 11m'ice golfer on the green takes three strokes to sink the ball. The sllccessive displacements of the ball are 4.00 III to the north. 2.00 m L15.oo north of east. and 1.00 m at .~O.oo west of south. Starling at the same initial point, an expert golfer could make the hole in what single displacement?
SECTION 3.3 DISPLACEMENT, VELOCITY, AND ACCELERATION IN TWO DIMENSIONS SECTION 3.4 MOTION IN TWO DIMENSIONS 22. One of lhe fastest recorded pitches in major-league baseball, thrown by Joel ZUlllaY£l in 2006, was docked at ]01..0 mi/h (Fig. P3.22). [f a pitch were thrown horizontally with this velocity, how far would the ball fall vertic
~
~
&
i
g §
"
~"",_-.
_ _.J
FIGURE P3.22 Joel Zumaya
lhrow~ a
baseball.
~m A stLidem stands at lhe edge of a cliff and throws a stone horil.ontally O\el' the edge with a speed of 18.0 m/s. Tile c1iffis 50.0 m above a nat. horizontal beach as shown in Figme P3.23 (page is). (
78
Chapter 3
Vectors and Two-Dimensional Motion allhe apex of its tn:~jecLor}', how fasl was Ihe ball movingwhen illeft the racket?
below the clifP (n With what speed and angle of impact does the stone land? 28.
FIGURE P3.23
24.
II
A peregrine falcon (Fig. 1'3.24) is the fastest bird, fly~aLUre has adapted the bird 10 reach such a speed by placing barnes in its nose to pre~ ,"elH air from rushing in and slowing it dO\.\"I1. Also. the bird's eyes adjust their focus faster than lhe eyes of allY 01 her creaUlre, so the falcon can focus quickly on its prey. Assullle a peregrine falcon is moving horizOlllall} at its lOp speed at a heigh I of 100 Tn above the grollnd when it brings its wings into its sides and begins 1.0 drop in free fall. How far will the bird fall vertically while traveling horizontally a distance of 100 m? ing al a speed of 200 mi/h.
m
From the window of a building. .:- ,Ind }'~colllpOneI1(S of the initial velocity. (c) Find lhe equations lor the x~and .'\.cOlnponents or the position as functions oftimc. (d) How far horizontally from the base of the building does the ball slrike the grollnd? (e) Find [he height from which the ball was thrown. (f) How long does it lake the ba1110 reach a point 10.0 m belm.. . the level oflallllching?
29. A brick is thrown upward from the top of a building at <111 angle of 25° to the horizontal and with an initial speed of 15 m/s. [f the brick is in fl ight for 3.0 S, how tall is the building? 30. All artillery shell is fired with all initial "cIocity of 300 m/s at 55.00 above the horizolltal. To clear an a\'alanche, it explodes on a moumainside 42.0 s after firing. What are Ihe x- and )'-coordinates of the shell where it explodes, "elative to its firing point? 31. A car is parked on a cliff overlooking Lhe ocean all an incline thm makes an angle of 24.0° below the horizontal. The negligcnt driver lea\'es the car in neutral, and the emergency brakes arc defcctive. The car rolls from rest down the incline ,.. . ith a constantacceleriJtion of4.00 Ill/5 2 for a distance of .10.0 III 10 the edge of the cliff, which is 30.0 Tn abo\'e the ocean. Find (a) the car's position l'clati,'c to the base of lhe c!iff,...· hen the car lands in the ocean and (b) the length or time the car is in the air. 32, A fircman 50.0 m away from a burning building directs a 51 rea III of" ""ater from a ground-Icvel fire hose at an angle of 30.0° above the horizontal. If the speed of the stream as it leaves the hose is 40.0 m/s. at whal. height will Ihe stream of water strike the building?
FIGURE P3.24 llo'lrils.
Noticl.-,the SlrllCLUre within the peregrine falcon's
25, The best leaper in the animal kingdom is the puma, which Gllljump to a hciglll of l2 rt when le;n.·ing the ground al an angle of 45°. With what speed, in 51 units, IllUSt the <-lllimallea\'e the ground to reach thal height? 26. • The record distance in the SpOrL of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United SlalCS in 19S1. Assuming the initial laullch angle was 45~ and neglecting air resistance. determine (a) the initial speed ohhe projectile and (h) the tolaltime the projectile was in flight. (c) QU
133.IA pr~jeClile is launched wilh
SECTION 3.5 RELATIVE VELOCITY 35.
mil
A jet airliner moving initially at 3.00 X 102 mi/h due cast enlcrs a region where the wind is blowing 1.00 X 10~ ll1i/h in a direclion 30.W north of east. (a) Find the components of the velocity of Ihe jet airliner relative to the air, VIA' (b) Find the componcllls of the velocity of the air relative to Earth, v,n.. (c) \Vrilc an equation analo-
Problems gous LO Equation :3.16 for the velocities v.I-\' V:\l-' and vlF . (d) What is the speed and dir-ectioll of" the aircraft rc1ath'c to th~ ground? 36. A boat moves through the \·...ater of a river at I () m/s reI ati,'c to the watel-, l-egal-dle~~ uf Llll.~ uu,lt's dilelliull. II' tile water in the river is flowing at 1.5 mis, how long cloes it take the boa I LO make a round trip consisting of
II
A chinook (king) salmon (gcnus 01lrorh)1I/chll\) can jump out ofw<-Iter with tI speed of 6.26 m/s. (Sec Problem -1-.9, page 111 for an investigation of how the fish can lean' the waleI' al a higher speed than it can swim un(krwater.) If the salmon is in a stream \vith water speed equal to 1.50 mis, how high in the air can the fish jump if it leaves thc w;lter Iraveling vertically upwards relative to the Earth?
38. A river flows clue cast at 1.50 Ill/s. A boat crosses Ihe river f"rom the south shore to the norlh shore by maintailling a constanl velocity of 10.0 m/s due nonh r-e1al.i\'e 10 the
\valer. (a) vVhat is 1he vdocit~· or the boal relative tu the shorc? (b) If the river is 300 III wide, how far downstream has the boat moved by Ihe time it reaches the north shore? 39. A rmvbo
42.
mil
A river has a slead}· speed of 0.!'i00 m/s. A student swims upstream a distance of 1.00 km and sv.. ims back to Ihe starting point. (a) If the student can sw·im al a ~peed of L.20 mls in still water, how long does the trip take? (b) How much time is required in still water for the same length swim? (c) Inluitively, why does the swim takc lon~ g-er when t here is a currcnt?
mil
A river has a steady speed of 11,. A swdem s\'v·ims upstream a distance d and back to the starling point. (a) If the studcnt can swim at a speed of v in still w,Her, how much time I"p does it li:lke the student to swim upstream a distance d? Express the answer in lcrm~ of" d, l), and v,. (b) Using the same yariabks, how milch time '.I""",, does it t,ike to swim b,lCk downstream w the starting point? (c) Slim the "lnswers found in pans (a) and (b) and show that the time I" required for the whole trip can he wrill.en as
I"~
2dl1l =
I -
lJ,:!/1J:!
(d) How much time 1" does Ihe trip lake in still wmer? (c) \Vhich is lar~er, til or lh? Is it always larger?
79
143.1:\ bomber is flying horizontally oycr level terrain at a speed 01'275 mls relative to lhe gmund and at an ahiwde of 3.00 kill. (a) The bombardier releases one bomb. How
far does the bomb tr(\\"el horizontally betwcen its relea~e and its impacl all the ground? Ignore the effects of air resislance. (b) Firing from the people on the ground suddenly incapacitates the bombardicr before he call call, ~Botllbs (tway!" Consequenlly, lhe pilol maintains the plane's original course, altitude, and speed through a storm of" flak. \Vhere is the plane relativl' to the bomb's point of impact when tlie bomb hits the g-wund? (c) The plalle has a telescopic hombsight set so that the bomb hilS the targct seen in the sighl 3t the moment of release. At whal angle from the vertical was lhe bombsight sel?
ADDITIONAL PROBLEMS 44. _ A moving walkway at an airpon has a speed VI and a kngLh L. A woman stands on the walkway as it mO\'es from one end to the other, while a man in a hurry 10 reach his flight walks on I he walkway wilh a speed of v2 relal ive to the moving walkway. (a) How long does it take the wom,lll to trawl the distance L? (b) Ho\\' long does it lake Ihe man to lravel this distance? 45. How long does it take an automobile Iraveling in the left lane of a highway at 60.0 km/h to overtake (become even \-.,..ilh) another car lhat is traveling in the right lane at -10.0 km/h when the cars' from bumpers are initially 100 III ap
46. You call llse any coordinalc s}'stem you like to solve a projectile motion problem. To demonstrate the trmh of this statement, consider a ball thrown off lhe top of ,I building Wilh a velocit), v at an angle e with respect to the horizontal. Lel the building be 50.0 m tall, the initial horilOnlal velocity be 9.00 mis, and thc initial vel'· tical velocity be 12.0 mls. Choose your cuordinates such that the positive y-;-1 xis is upward, the x-axis is to the right, and the origin is at the point. where the ball is released. (a) '<\'ith t.hese choices, flllel the ball's maximum height "bO\·c the ground (lnd the time it takes to reach the maximum height. (b) Repeal your calculations choosing the origin althe base 01" the building. 47. _
A .\Iordic jUlllper goes off a ski jump al an angle below the horizonlal, traveling 108 m horizontally and .";5.0 In vertically before landing. (a) Ignoring friction and aerodynamic effects, calculate the speed needed by lhe skier on leaving the ramp. (b) Olympic .\Iordic jUll1pers Gill make such .iump~ wit h a ju mp speed of 23.0 mis, which is considerably less than the answer found in part (a). Explain how that is possible.
or 10.0
0
-l8. _ In
80
Chapter 3
Vectors ond Two-Dimensional Motion
49. TOWliS A and B in Figure P3.49 (Ire 80.0 kill aparl. A couple arranges 1.0 drive from lawn A and meet a couple dri\"ing from lawn B at the lake. L. The two couples leave simultaneously and drive for 2.50 h in the directions shown. Car J has a speed of 90.0 kill/h. Irthe cars arri\'c simultaneously at the lake, whaL is the speed of car 2?
L
54. A daredevil decides to jUlllP a call)'on. Its walls (Ire equally high and 10 m apart. He takes orr b}' driving a motorcycle up a short ramp sloped at an angle of 15°. \lVhat minimum speed must he h
IS5.IA home run is hit in such a way thaI the baseball just clears a wall 21 111 hig'h, located 130 m from home plate, The ball is hit at an angle of 35° to the horizontal, and air resistance is negligible. Find (a) the initial speed or the ball. (b) the time it takes tile ball to reach the wall,
56. A ball is lhrown straight upw'lrd and returns to the thrower's hand after 3.00 s in the air. A second ball is thrown at ,Ill angle or 30.0° \'.;ith the horizontal. At what speed must the second ball be thrown so lhat it reaches Lhe same height as the one thrown vertically?
FIGURE P3.49
50.
II
A chinook salmon has a maximum lInderwater speed of 3.58 mis, but it can jump om of water with a speed of 6.26 m/s. To move upstream past a waterfall, the salmon does not Ilccd to jump 10 the top of (he fall, but only to a point in the fall where (he watcr speed is less than 3.58 l11/s; it Olil then swim up the fall for the remaining dislance. Because the s
57. A quarterback throws a football toward a receiver with ,Ill inilial speed of 20 m/s at an angle of" 30" above the horizontal. At that instant the receiver is 20 m from the quancrback. In what direClion and with what constant speed should the receiver run in order to catch the football al the level at wh ich il was thrown?
58. A 2.00-m-l,dl basket.ball player is standing on the floor 10.0 In from the basket, as in Figure P3.58. if he shooLS the ball at a 40.0" angle with the horizonlal, at ,,'hat initial speed lllUSt he throw the hasketball ~o liUl it goes through the hoop '''''ithout striking the backboard? The height of lhe basket is 3.05 In.
3.0:lffi 2.00
151.IA rockct is launched
1
40.0°
Jli
L
l,-----'O
1
."...-=--. . . . . 0 m--------.1
FIGURE P3.58
59.
miD In (I very popular lecture dcmollSlralion, a projectile is fired at a blling target as in figure P3.59. The projectile leaves the gun (It the same instant the target is
,
/::r ,
--~ ......
:
/'
Point 01
collision
FIGURE P3.59
81
Problems
dropped from rest. Assuming the gun is initially aimed at the target, show thai the projectile will hit the target. (One restriction of lhis experimellt is that lhe prqjertile must reach the tilrgel before the largct strikcs the Ooor.) 60.
(b) Dctermine the ratio of the times for the one-bounce and no-bounce throws.
III
Figure P3.60 illustrales the difference in propor· between tile male (Ill) and female (11 anaLOmies. The displacements dim and d lr from the bOl.lol11 of the.: feet to 1he navel have magnitudes of 10·1 cm and R4.0 cm, respectively. The displacemenls d:!Il' ;md d:!, h;n-e magniludes or 50.0 em and -J3.0 em, respcCli\'clv. (
I'
IJ
-I
FIGURE P3.64
65. A daredevil is shot out of a cannon at 45.00 to the horizontal with an initial speed of25.0 m/s. A net is positioned a horizontal distance of 50.0 III from the callnon. At what height above the canllon should the net be placed in order 10 catch the daredevil? 66.
fIJI
Chinook salmon are abJc to \llO\'e upstrealll faster by jumping OUI of the water periodically: this behavior ;s called /J01jmisillg. Suppose a salmon swimming- in still w,Her jumps out of the water with a speed of 6.26 mls al an angle of 45 c , sails thmllgh lhe air a distance L hefore rellJl"nillg to the water, and thell swims a distance I. underwater at a speed of 3.58 nl/s before beginning another porpoi~ing maneuver. Delerminc the average :o;peed of the fish.
By throwing a ball at an angle of' 'Jf>0, ;1 girl call throw the balt a maximullI hori/omal distance H. on a level field. How far can she throw lhe same ball vertically upward? Assume her musclcs give the ball the same speed in each case. (Is lhis (Issumplion valid?)
67. _ A student decides to measure the muzzle vclocity ora pellet shot from his gun. He poilll~ the gun horizontally. He places a target 011 a n~rtical wall a distance Xa\e1)' from the gun. The pellet hits the target ,1 venical distance )' below the glill. (a) Show thalthe position of the pellet whcn tl'aveling thmllgh the air is given by)' = Ax:!, where A is a UHlstaJl(. (b) Express the constanL A in terms of the inilial (muzzle) velocityand the frce-f;lll acceleration. (c) If x = 3.00 III ami )' = 0.21O m, \."hat is the initial speed of the pellet?
62. _ The equation ora parabob is y = ax'! + bx + r, \dlere a, U. and r are constants. The x- and )'-coordin
68. A sailhoat is heading directly north at a speed of 20 kllots (I knot = 0.51'1 m/s). The wind is blowillg towards the east with a speed of 17 knots. Determine the magnil udc and di rection of the wi nd velocit), as measured on the boat. ,"fhat is the component of the wind velocity in th~ direction parallel to the motion of tbe boal? (See Problem LI.5R for all expl;lIlillion of how a sailboat can t1lm'C "into the wind.~)
63. A hlllller wishes to cross a river that is 1.5 km wide and flO\vs with a speed of 5.0 km/h parallclto it~ banb. The hUlller uses a srnall powerboat tlul moves at a maximuIll speed of 12 km/h wit h respect 10 the water. What is the minimum time necessary for crossing?
69. A gall" ball with all initial speed or 50.0 m/s lands exactly 240 m downrange on a le\'e! course. (a) NcgJcct~ ing air fricl ion, whal two projection angles would achieve this result? (b) What is the maximulll height I-cached by llie ball, using the two angles determined in part (a)?
FIGURE P3.60
OITJE
64.
E
When baseball outfielders throw the ball, Ihe)' usually allow it to take one bounce, on the theory lhal the ball arrives;I1 its targel sooner thaL way. Suppose thal. af"tcr the bounce, the ball rebounds at the same ,1Ilglc 0 that it had \.. . hen it was released (as in Fig. P3.fi-l). blltloses half its speed. (a) Assuming that the ball is always IhrO\vn wit h the same inilial speed"ll \V'llaL angle e should the ball be thrown in GI'der to go thc same distance [) with one bouncc as a hall thrown upward at 45.0 c with no bOlll1ce?
70.
mil
A landscape architect is planning an artificial \vaterfall in a cit)' park. \Vater flowing al 0.750 111/S leaves the end of a hOI-i/.olllal channel al the top of a \crtical wall 2.35 III high and falls into a pool. (a) Ho\\' far from the wall will the \-\'aler land? \Vill the space behind the w;iterbll be wide enough for a pedestrian walkway? (b) To sell her plan to the cil)' council, thl' archiLect wants to build a moc!clto standard scale, one-!",,'elfth aCLual size. How fast should the waleI' flo\\' in tht' channel in the model?
82
Chapter 3
Vectors and Two-Dimensional Motion
71. One stralt'g) in a snowball fight is to throw a snowhall at a high angle over le.. .el ground. Then, while your opponellt is \\alching thaI snowball, yOll throw a second one at a low angle timed to arri\'c before or at (he same time as the first one. Assume bOlh snowballs afe L1uown with a "peed of25.0 m/s. The flrsl is thrown at an angle of70.W with respect (0 the horizontal. (al At whaL angle should the second snowball be thrown to arrive althe same POilll as the firn? (b) How man) seconds later should the second snowball be thrown aflcr the first in order for both LO arrin' althe same time? 172.1.-\ dart gun is fired while being held hori/ont.,II} al a height of 1.00 m above gmund level and while it is al rest rClali\c ((l the ground. The dan from the gun I.ra\eb a horimnlal distance of 5.00 Ill. A college SWdClll holds the '>amc !{un in a hori/ontal position while ~Iiding do\, n a ·15.W incline ~ll a constant speed of 2.00 m/s. How far will the clarl lr<1\'eI if the swdclll fires the gun when it is 1.00 III aban' the ground?
i3. The dctermined \rilc E. Coyote is out once to
mOI'C
capture the elusive roadrunner. The co)ote
new pair of Acme power roller skates, which provide a Calista III horizontal acceleration of 1:'"1 m/s:!, as shown in Figure P3.73. The corOte sLart~ orf at rest 70 III from the edge of a cliff at the instant the roadrunner ;tips by in the direction of lhe clifT. (a) If the roadrunner mO\'es with constant speed, lind thc minimulIl speed the roadnlllilcr must havc to reach the cliff before the coyote, (b) If the cliff is 100 III above the base of a canyon, find where the coyote lands in the camon. (.-\ssume hi::. skates are still in operation when he is in "lIight~ and that his horizontal component of acceleration remains constalll at 15 m/s2 .) Co)'QII'
5llllJidll5
-,,-
/(Q(uJnl'JIll'T
(uligltl1l5 ~~~~
-
~ 8/:1,/>
to try
wear~
a
FIGURE P3.73
Forces exerted by Earth, wind, and water, properly channeled by the strength and skill of these windsurfers, combine to create a non-zero net force on their surfboards, driving them forward through the waves.
4.1
Forces
4.2 Newton's First Low
" ~ J'
4.3
Newton's Second Low
4.4
Newton's Third Low
j
4.5 Applications of Newton's
"
4.6 Forces of Friction
Lows
THE LAWS OF MOTION Classical mechanics describes the relationship between the motion of objects found in our everyday world and the forces acting on them, As long as the system under study doesn't involve objects comparable in size to an atom or traveling close to the speed of light, classical mechanics provides an excellent description of nature. This chapter introduces Newton's three laws of motion and his law of gravity. The three lows are simple and sensible, The first low states that a force must be applied to on object in order to change its velocity. Changing on object's velocity means accelerating it, which implies a relationship between force and acceleration. This relationship, the second low, states thot the net force on on object equals the object's mass times its acceleration. Finally, the third low says that whenever we push on something, it pushes bock with equal force in the opposite direction, These are the three laws in a nutshell. Newton's three laws, together with his invention of calculus, opened avenues of inquiry and discovery that are used routinely today in virtually all areas of mathematics, science, engineering, and technology. Newton's theory of universal gravitation had 0 similar impact, starting a revolution in celestial mechanics and astronomy that continues to this day. With the advent of this theory, the orbits of all the planets could be calculated to high precision ond the tides understood. The theory even led to the prediction of "dork stars," now called block holes, more than two centuries before any evidence for their existence was observed. 1 Newton's three lows of motion, together with his law of gravitation, ore considered among the greatest achievements of the human mind.
4.1
FORCES
A force is commonly imagined as a push or a putl on some ol~jecl, perhaps rapidly. as when we hit a tennis ball \vith a racket. (Sec Fig, 4,1.) \Ve can hit thc ball at djf~ fcrent speeds
FIGURE 4.1
Tennis champion ball WiLh his racket, applviJig a fOl-ce and directing the ball into tILe opcn P'll"lOfthe RaE.H~l \!a(LtI.~lrikes the
COllI'\.
IIIl 178:1,Johll .\lidll'lI combined Nl'" l"n's 'hc(>l"~ "I light
83
84
Chapter 4
FIGURE 4.2
Examples
or forces applied to various objects. In each case. a furce acts on t he object surrounded b~
The Lows of Motion
C01llact ror(;c~
the dashed Jines.
Something in the clwirOllment external LO the boxed area exerts the force.
(ej
(al
Field forces
M
i
Iron ~-----_
(eI)
(ej
s'
. (I)
Iryou pull on a spring (Fig. 4.2a), the spring stretches. !fyou pull hard enough on a wagon (Fig. 4.2b), the \...agon moves. \'Vhen you kick a football (Fig. 4.2c), it deforms brierJy and is set in motion. These are all examples of contact forces, so named because they resuh from physical contact bel,veen two objects. Another class of forces doesn't involve any direct physical contact. Early scient.ists, including Newton, \V"ere uneasy ,.. . ith the concept of forces lhm act between two disconnected objects. Nonetheless, Newton used this "3crion-at-a-dis(allce" concept in his law of gravity, whereby a mass at one location, such as the Sun, ancets the motion of a distant object such as Earth despite no evident physical conncction between the t",,'o objects. To overcome the conceptual difficulty associated ,\lith action at a distance, Michael Faraday (1791-1867) ii1ll'oducerl the concept of a .field. The corresponding forces are called field forces. According 1O this approach,
4.2
Newton's First Low
85
4.2 NEWTON'S FIRST LAW Consider a book lying on a table. Obviollsly. the book remains at rest if left alone. Now imagine pushing the book with a horizontal force greal enough to overcome the [OI-ce offriclion between the book and 1Ilf' lahle, Sp.lling Ihe honk in motion.
Because the magnitude of the applied force exceeds the magnitude of the friction force, the book accelerates. V\'hcn the applied force is withdrawn, friction soon slows the book to a stop. No,.. · imagine pushing the book across a smooth, waxed floor. The book again comes to rest once the force is no longer applied, but not as quickly as before. Finally, if' the book is moving on a horizontal frictionless surface, it continues to move in a straight line with constant velocity until it hits a \vall or some other obstruction. Before about 1600, scientists felt that the natural state ol'mauer was the Slate of resl. Galileo, however, devised thought experiments-such as an object moving on a frictionless surface, as just described-and concluded that it's not tbe nature of an object to stop, once set in motion, but rather to continue in its original state of motion. This approach \vas later formalized as Newton's first law of motion: An object moves with a velocity that is constant in magnitude and direction, unless acted on by a nonzero net force.
~
Newton's first law
The net force on an object is defined as the vector Slim of all external forces exerted on the object. External forces come from the object's environ menl. I f an object's velocity isn't changing in either magnitude or direction, then its acceleration and the net force acting on it must both be zero. Internal forces originate within the object itself and can't change the o~ject's velocity (although they can change the object's rate of rotation, as described in Chapter 8). As a result, internal forces aren't included in Newton's second law. It's not really possible to "pull yourself up by YOllr own bOOlstraps:' A consequence of I he first law is the feasibility of space travel. After just a few moments of powerful thrust, the spacecraft coasts for months or years, its velocity only slowly changing with time under the relatively faint influence of the distant sun and planets.
Mass and Inertia Imagine hitting a golf ball off a tee with a driver. Jf you're a good golfer, the ball will sail over two hundred yards dmvn the f~lirway. Now imagine teeing up a bm..,r}ing ball and striking it 1,-.... ith the same club (an experiment we don't recommend). You r club \vallld probably break, yOLi might sprain your wrist, and the bowling ball, at best, would falloff the tee, take half a roll, and come to rest. From this thought experiment, we conclude that although both balls resist changes in their state of motion, the bO\vling ball oners much more effective resistance. The tendency of an object to continue in its original state of mOlion is called inertia. Although inertia is the tendency of an object to continue its motion in the absence of a force, mass is a measure of the object's resistance to changes in its motion due to a force. The greater the mass ora body, the less it accelerates under the action of a given applied force. The SI unit of mass is the kilogram. Mass is a scalar quanrity that obeys the rules of ordinary arithmetic. Inertia can be used to explain the operation orone type of seat bdt.mechanism. In the event of an accident, the purpose of the seat belt is to hold the passenger firmly in place relative to the car, to prevent serious injury. Figure 4-.3 (page 86) illustrates how one type of shoulder harness operates. Under normal conditions, the ratchet turns freely to allow the harness to wind on or unwind from the pulley a.'i the passenger moves. In an accident, the car undergoes a large acceleration and
Unless acted on by all external force, all object at rest will n:main at rcst and an objt:c1 in mOlion will cOlltinue in mOl ion wilb constalll \'eJod/y. In this case, the wall of the building did not c)a:n a large enough external force UII the muving lrain to SLOp it.
86
Chapter 4
The Lows of Motion FIGURE 4.3
APPLICATION Seat Belts
A IHl:chanical
arrangelllent for :111 allLOlTlohik sell bell.
SC:II-~'"''
belt
Pulley
Rod
Il<--- PiVOl Tracks Pin cOllnenion ~"...,;...~...
ISAAC NEWTON English Physicist and Mathematician (1642-1727) Newton was one of the most brilliont
scientists in history. Before he was 3D, he formulated the basic concepts and lows of mechanics, discovered the law of universal gravitation, and invented the mathematical methods of the calculus. As 0 consequence of his theories, Newton was able to exploin the motions of the planets, the ebb and flow of the tides, and many special features of the motions of the Moon and Earth. He olso interpreted many fundamental observations concerning the nature of light. His contributions to physical theories dominated scientific thought for two centuries and remain important
today.
TIP 4.1 Force Couses Changes in Motion -:-,-Motion can occur c\"en in the
absence of forces. Force causes dumge!. ill rnotlon.
" Newton's second low
~
rapidly comes Lo rest. Because of its inertia, the large block under the seat continues lO slide forward along the tracks. The pin connection between the block and the rod causes the rod to pivot abollt its center and engage the ratchet wheel. At. this point, the ratchet wheel locks in place and the harness no longer unwinds.
4.3 NEWTON'S SECOND LAW NeWlon's firsllaw explains what happens to an object that has no net force acting on it: The object either remains at rest or continues mm-ing in a straight line with constant. speed. Nev·.'lon's second law ans\-vers the question or \'lhat happens [Q an object that dor's have a net lorce acting on it. Imagine pushing a block of ice across a frictionless horizontal surface. \Vhen you exert some horizontal force on the block, it moves with an acceleration 0[, say. 2 m/s 2. Iryou apply a force twice as large, the acceleration doubles to 4 m/s 2. Pushing three times as hard triples the acceleration. and so on. From such observations, we conclude that the acceleration of an object is directly proportional to the net force acting on it. \-tass also affects acceleration. Suppose yOll stack identical blocks of" ice on LOp of each other while pushing the stack with conSlant force. If the force applied to one block produces an acceleration of 2 m/s 2 , then the acceleration drops to half that value, 1 m/s~, when n,,'O blocks arc pushed, to one-third the initial value when three blocks are pushed, and so on. \\le conclude that the acceleration of an object is inversely proportional to its mass. These observations are summarized in Newton's second law:
Thc acceleration "'it of an object is dircctly proponional ing on it and invcrsel)' proponional LO its mass.
LO
the net force act-
The constant of proportionality is equalLO one, so in mathcmatical terms the preceding statcment can bc written ~
2;F
a=~
where "'it is the accelenllion of the objcct, 'III is its mass, and all forces acting- all it. l'vlultiplying- through by In, we have
:LF is the vccLor sum of [4.1]
4.3
Physicists commonly refer 10 this equation as 'F = ma.' The second law is;1 vector equation, equivalent to the following three component equations: L,I\ = max
2>~ =
1JUl
y L.I'~ = ma:
[4.2]
\'\Then there is no net force on an object, its acceleration is zero, which means the velocity is constant.
Units of Force and Mass The SI unit of force is the newton. When I newton of force £lClS on an ol~jeClthat has a mass of" 1 kg, it. produces an acccleralion of I Ill/s~ in the object. From this definition and Newton's second law, we see th(ll the newton can be expressed in terms of the fundamental units of mass, length, and time as
IN'" I kg· mis'
[4.3J
Ne\'.rton's Second Low
87
TIP 4.2 ma' Is Not a Force Equation 4.1 does 1101 say that the product 1111' is" force. All forces exerted on an object are sLllnmed as veclors to generate lhe net force all the ldl side of (he equation. This nel force is then equated to the product or the mass and resulting acceleration of the ohject. 00 1/01 include an '·ma rorce" in yOUl" an,liy~is.
~
Definition of newton
In the L.S. customary system, the unit of" force is the pound. The cOIwcrsion from newtons to pounds is given by I N = 0.225 110
[4.4]
The units of mass, acceleration, and forcc in the SI and U.S. customary systcms arc summarized in Table 4.1.
QUICK QUIZ 4.1 Which of the following statemel1ls are true? (a) An object. can move even when no force acts on it. (b) H an object isn't 1110vinf.{, no external forces act on it. (c) If a single force acts on an object. the object accelerates. (d) Ifan object accelerates, a force is acting on it. (e) JJan object. isn't accelerating, no external Corce is acting on it. ([) If the net force acting on an object is in the positive x-direction, the objectlllovcS only in the positive x-dircclion.
TABLE 4.1 Units of Mass, Acceleration, and Force System
Mass
Acceleration
Force
51
kg
=:
slug
m/s:! IVs:!
N
U.S. customary
lb
= slug·[t/s2
EXAMPLE 4.1
kg'm/s~
Airboot
Goal Apply Newton's law in one dimension, togcther with the equations of kinematics. Problem An airboat with mass 3.50 X 10 2 kg, including pas· scngcrs, has an engine that produces a net horizol1lal force of 7.70 X 10 2 1\', after accounting for forces ofrcsiS1.ancc (see Fig. 4.4). (a) Find the acceleration of the airboat. (b) Starting rrom rest., how long does iltake the airboat to reach a speed of 12.0 m/s? (c) After reaching this speed, the pilot I.Urns off the engine and drifts to a stop over a distance or 50.0 m. Find the resistance force, assuming iI's constant.
FIGURE 4.4
(Example '1.1)
Strategy In pan (a), apply Nev.lton's second la,'\' to nnd the accelerat.ion, and in part (b) usc this acceleration in the one-dimensional kinematics equation for the velocity. ',Vhen the engine is lUfned ofT in part (c), only the resistance rorces act on the bOaL, so their net acceleration can be found from 1/'1. ~ 1/0 2 = 2a ~x. Then Newton's second law gives the resistance force.
88
Chapter 4
The Lows of Motion
,...
"
.
Solution (a) Find the acceleration of the airboaL
Apply. ewton's second law and solve [or the accclermion:
-->
7.70 X 10' N 1':Il~1 {/=--= 3.50 X 10' kg
11/
=
2.20 m/s2
m!s')t
= 12.0 m!s
(b) Find the time necessary to reach a speed of 12.0 m/s. Apply the kinematics "dacit)' equation:
+ v"
v=
(I{
o-
(12.0 m!s)'
= (2.20
-->
1=
5.45 s
(c) Find the resistance force after the engine is turned off. Using kincmatic!i. find the net acceleration due to rC"lislance forces: Substitmc the acceleration into NeWLon's second law, finding the resistance force:
F«"" =
11/(1
= 20(50.0 m)
-->
(I
=
-1.44 mis'
= (3.50 X 10' kg)( - 1.44 m!s') =
-5041\
....
....
Remarks The propeller excns a force 011 the air, pushing it b~ICk\''''
QUESTION 4.1 \,\That other forces act 011 the airboat? Describe them.
TIP 4.3 Newton's Second Low Is a Vector Equation
EXERCISE 4.1 Suppose the pilot, staning again from rest, opens the throtLie part,\·ay. At a constam acceleration, the airboat then covers a distance of 60.0 m in 10.0 s. Find the net force acting on the boat.
I n applying :-Jewton's second Jaw, add all of the forct:son the object as vector.. and then find t he resultant ,'ector acceleration h~ dividing hy m. Don', find the individual m:lgniLUdes of the forces and add them like scalars.
Answer 4.20 X 10' '.J
EXAMPLE 4.2
Horses Pulling a Borge
Goal Appl), lewton's second problem.
la\'~"
in a two-dimensional
Problem Two horses are pulling a barge with mass 2.00 X 10J kg along a canal, as shown in Figure 4.5. The cable conneCled LO the first horse makes an angle of 30.0° with respectLO the direction of the canal, while the cable connected to the second horse makes an angle of 45.0°. Find the initial acceleration of the barge, starting at rest, if each horse exerts a force of' magnitude 6.00 X IO~:"J on the barge. Ignore forces of resistance 011 the barge.
J l I I
I
Strategy Using trigollometry, find the vector force FIGURE 4.5 {Example '1.2) exerted by each horse on the barge. Add the x-components LOgcther to get the x-componellt of' the resultant force, and then do the same ,~ith the y-componel1ls. Divide by the mass of' thc barge LO get the accelerations in the ;\:- and )'-direclions.
4.3
Newton's Second Low
89
......... ...... Solution Find the x-components of the forces exerted by the horses.
Fix
Find the total force in the x-direction by adding
1', ~ Fix
~
1'1 cos III
+ I"x
~
1'1,
1'1 sin III
(-45.0°) ~ 4.24 X 10 2 N X 10'
~
9.4'1 X \ 0' 1'\
~
(6.00 X 10':-.1) sin 30.0° ~ 3.00 X \0' ,
1'" ~ I" sin 112 ~ (6.00 X 10' l\) sin (- ..5.0 )
Find the lOla I force in the y·dircClion by adding the y-cornponcnts:
I'~ ~ F h
Find the componcnLs of the acceleration by didd-
(I
~
2.00 X 10J kg
I'; -
~
v"
(I,'
-
4.24 X 10'.
~
_,
0.472 m/s'
~
" -0.062 0 mls'
+ iI;"~V ( 0.472 m/s'")". + ( -0.0620 m/s'")'. ~
tan
~
-1.24 X 10' N 2.00 X 10' kg
~
11/
I
-1.24 X 10'
11/
,
(I
X 10'
9.-14 X 1O~
x
\0 2 N
X
F\
~ ~ ~
(I
-4.24
1'". ~ 3.00
T
~
ing the force components by the mass:
Find the dircCLion of the acceleration:
COS
+ 4.24
~ 5.20 X 10' N
~
Find the rnagn itudc of the acceleration:
(6.00 X \0' N) cos (30.0°) ~ 5.20 X 102 N
1'" ~ I'" cos 112 ~ (6.00 X 10 2 N)
the x-components:
Find the y-componems of the forces exerted by the horses:
~
ii,
0.476 mis'
-0.062 0 7 0.4 2
II
~ ~ ~
II
~ lan- I (-0.131)
ax
~
-0.131 -7.46
0
... Remarks The horses exert a force on the barge through the tension in the cables, while the barge exerts an equaJ and opposite force on the horses, again through the cables. If that were not true. the horses would easily move for\'lard, as if unburdened. This example is another illustration of forces acting in pairs. QUESTION 4.2 True or False: In general, the net acceleration is always directed toward the horse that exerts the greater magn.itude force. EXERCISE 4.2 Repeat Example 4.2, bUl assume the upper horse pulls at a 40.0 angle, the lower horse at 20.0:::. 0
Answer
0.520 m/s". 10.0
0
The Gravitational Force The gravitational force is the mUllial force of attraction bel ween any two objects in the L"ni\·erse. Although the graYitational force can be \'ery strong between \'cr)' large objects, it's the \'1'eakest of the fundamental forces. A good demonstration of how weak it is can be carried out with a slllall balloon. Rubbing the balloon in YOUl" hair gives the balloon a tiny electric charge. Through electric forces, the balloon then adheres to a wall, resisting the gra\'il~\liollal pull of the entire Earth!
90
Chapter 4
The Laws of Motion
Law of universal gravitation
~
In addition Lo contributing to the understanding of mOlion, Newwn studied gravity eXlensively. Newton's law of universal gravitation states that every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. If the particles have masses mj and JII'j, and <:Ire separated by a distance T, as in Active rigure 4.6, the magnitude of the gravitational force, FK, is [4.5]
ACTIVE FIGURE 4.6 The gravitational force particle.. is allraclive.
where C = 6.67 X 10- 11 N- rn 2/kg 2 is the universal gravitation constant. \'Ve examine the gravitational force in more detail in Chapter 7. bell'"Cell lWO
Weight The magnitude of the gravitational force acting on an object of mass Earth's surface is called the weighl, w, or the ol~jcCl, given by 71)=
mg
1n
near [4.6]
,,,,,here g is the acceleration of gravity. 51 unit: newton (N) From Equation 4.5, an alternate definition of the weight oran m can be wriucn as
ol~ject "...it.h
mass
[4.7] v.'here ME is the mass of Earth and r is the disLance from the object to Earth's center. Tf the o~ject is at rest on Earth's surface, then T is equal to Earth's radius R ESi nce 'f is in [he denominator of Equation 4.7, the ""eight decreases \\;ith increasing J: SO dlC weight of an object on a mountaintop is less than the weight of the same ol~ject at sea level. Comparing Equ
,.-
[4.8]
Unlike mass, weight is not an inherent properly of an object because it can take different values, depending on the value of g in a given location. If an object has a mass of70.0 kg, for eX<'llllple, then its weight at. a location where g= 9.80 m/s~ is mg = 686 N. In a high-altitude balloon, where K might be 9.76 Ill/5 2, the object's weight would be 683 N. The value 01" galso varies slightly due to the density of mattel' in a given locality. Equation 4.8 is a general result. that can be lIsed to calculate the acceleration of an objen falling near the surface of any massive object if the morc massive object's rrldius and Illass are known. Using the values in Table 7.3 (p. 2]7), you should be able to 5hm.. . t.hat gSllll = 274111/52 and gr.lo,," = 1.62 m/s:!. An imponant f~IC( is thal for spherical bodies, distances are calculated from the centers of the objecls, a consequence of Gauss's law (explained in Chapter 15), which holds for both gravitational and electric forces. Astronaut Edwin E. "Buu· Aldrin,Jr.. walking 011 lilt: Mooll ,trier the Apollo 11 lunar landing:. Aklrin's "-clg'hl on lhe 1loon is less thall it is Oil E.anh. but
hi~
places.
mass is the sallie in bolh
QUICK QUIZ 4.2 vVhich has greater value, a newton of gold won 011 Earth or a neWlon of gold won on the Moon? (a) The newton or gold 011 the Earth. (b) The newwn of gold on t.he Yloon. (c) The value is the same, regardless.
4.3
EXAMPLE 4.3 Gool
Newton's Second Low
91
Forces of Distant Worlds
Calculate the magnitude of a gravitational force using NewLOn's law of gravitation.
Problem Find the gravitational force exened by the Sun on a 70.0-kg man located on Earth. The distance from the Sun to the Earth is about 1.50 X 1011 m. and the Sun's mass is 1.99 X ]030 kg. Strategy
SubstiLUte numbers into
~cwton's 13\.\'
of gravitation. Equation 4.5. making sure to use the correct units.
Remarks The gra\.'it3tional attraction between the Sun and objects on Earth is easily measurable and has been exploited in experiments to determine whether gravitalional auraClion depends on the composition of the objeci. As the exercise shows, the gravitat ional forct:' on Ertrth due to the Moon is much weaker than the gravitational force on Earth due to the Sun. Paradoxically. the MoolI's effect on the tides is over twice that of the Sun because the tides depend on diJftrl!urps in the gravitatiollal force across the Earth, anclthose differences arc greater for the Moon's gravitational force beca\lse the Moon is much closer to Earth than the Sun.
EXAMPLE 4.4 Goal
(1.50 X 10" m)'
QUESTION 4.3 ~'Iars is about one and a half times as far from the Sun as Earth. To one significant digit, what is the gravitational force of the Sun on a 70.0 kg man standing on Mars? (Hint: Use the result of parI (a) and the inverse square nature of the force.) EXERCISE 4.3 To one significant digit, find the force exerted by the M.ool1 on a 70-kg man on Earth. The Moon has a mass of 7.36 X 10" kg and is 3.84 X 10' m from Eanh.
Answer FMoun = 0.002 N
Wei ht on Planet X
Cnderstand the effect of a planet'S mass and radius on the weight of an objeCl on the planet's surface.
Problem An astronaut on a space mission lands on a planet with lhree times the mass and twice the radius of Earth. What is her weight wT on this planet as a multiple of her Earth weight wf ) Strategy \\'rite J\l x and r..... the mass and radius of the planet, in terms of .\11 and Hr.• the mass and radius of Earth, respectively, and substitute into the law of gra\itat ion.
,.
.
.
Solution From the statement of the problem, we have the follow· ing relationships:
Substitute the preceding expressions into Equation ..... 5 and simplify. algebraically associating- the terms giving the weight on Earth: trrr... • • • . • • •
~\ =
x w\= G Al .:" = rx-
~
2RE
3
'4
W1-.
.......... , ..
~
Remarks This problem shows the interplay bet\vcen a planet'S mass and radius in determining the weight of o~jects its surl~lce. Because of Earth's much smaller radius, the weight of an object on .Jupiter is only 2.64 times its weight on Earth, despite the fact that.lupiter has over 300 times as Illllch mass.
all
92
Chapter 4
The Laws of Motion
QUESTION 4.4 Suppose one world is made of ice whereas another \.. . orld with the same radius is made of rock. If g is the acceleration of gravity on the surface of the ice world, what is the approximate acceleration of gravity on the rock world? (Hint: Estimate the mass of a rock in terms of the mass of an ice cube having the same size.)
EXERCISE 4.4 An astronaut lands on Ganymede, a goiant moon ofJllpiter that is larger than the planet Mercury. Ganymede has oncfonieth the mass of Earth and lwo-flflhs the radius. Find the weight of the astronaut standing on Ganymede in terms of his Eanh weiglll WE-
Answer
WG
=
(5/32) wI-:
4.4 NEWTON'S THIRD LAW In Section 4.1 we found that a force is exerted on an object when it comes into contact with some other object. Consider the task or driving a nail into a block of ,vood, for example, as illustrated in Figure 4.7a. To accelerate the nail and drive it into the block, the hammer must exerl a net force on the nail. Newton recognized, however, thal a single isolated force (such as the force exerted by the hammer all the nail) couldn't exist. Instead, forces in nature always exist in pairs. According to Ne\vton, as the nail is driven into the block by the force exerted by the hammer, the hammer is slowed down and stopped by the force exerted by the nail. Newton described such paired forces with his third la,,,,': Newton's third low
~
TIP 4.4 Action-Reaction Pairs 1n applying Newlon's lhird law, remember thaI an action and it.~ rcaction force always ilCl on dif-
jprent objects. Two external forces acting On the same object, c\'en if they arc equal in magniHlde and opposite in direction. cou', be an actiOll-reaction pair,
FIGURE 4.7 "ell'loll'S lhird law. (a) The force exerled by the hallllller on the nail is equal in rllagnilllde and opposift' in direClioll to the force l"xened bv tlie nail on the hammer. (b) The r~r('c F,~ cxerlcd by objeci 1 on object 2 is equal in magniltlde ~nd oppositt· in direction In the [uro:' F~l exerted by objcCl 2 on ohject I.
If object "I and object 2 interact, the force j\! exerted by objest ] on object 2 is equal in magnitude but opposite in direction to the force F 21 exerted by object 2 on object 1. This law, 'which is illi.lstrated in Figure 4.7b, states that a single isolated force can't exist. The force FI~ e~erted by o~ject 1 all object 2 is sometimes called the action fone, and the force F~I exerted by object 2 on object 1 is called the reaclion lorce. In reality, either force can be labeled the action or reaction force. The action force is equal in magnitude to the reaction force and opposite in direction. In all cases, the action and reaction forces act on different objects. For example, the force acting on a freely falling projectile is the force exerted by Earth on the projectils. g , and the magnitude of this force is its weight 'mg. Ife re~tion to force Fg is-tthe force exerted by the projectile on Earth, F/;" = - F 15' The reactio~ [orce F K' must accelerate the Earth towards the pr~jectile, just as the aCLion force F K accelerates the projectile towards the Eanh. Because the
F
2
(a)
(b)
4.4
Earth has such a large mass, however, its acceleration clue to this reaction force is negligibly small. Newton's third law constantly affects our activities in everyday life. vVithout it, no locomotion of any kind would be possible, whether on foot, on a bicycle, or in a motorizcd vehicle. \Vhcn walking, [01" example, "..c cxert a frictional force against the ground. The reaction force of the ground against our fOOl propels us forward. 1n the same way, the tires on a bicycle exert a frictional force against the ground, and the reaction of the ground pushes the bicycle fon.. . ard. As we'll see shorLl)', friction plays a large role in such reaction forces. For another example of NcI,vton's third law, consider the helicopter. Most helicopters have a large set of blades rotating in a horizontal plane above the body of the vehicle and another, smaller set rotating in a venical plane at the back. Other helicopters have t\VO large sets or blades above the body fOLating in opposite directions. \'Vhy do helicopters always have 1.\'.'0 sets of blades? In the flrstt)'pe of helicopter, the engine applies a force to the blades, causing them to change their rotational motion. According to Ne\,\'ton's third law, however, the blades lllust cxen a force on the engine of equallTlagnitude and in the opposite direction. This force would cause the body of the helicopter to rotate in the direction opposite the blades. A rotating helicopter would be irnpossible to control, so a second set of blades is used. The small blades in the back provide a force opposite to that tending to rotate the body of the helicopter, keeping the body oriented in a stable position. In helicopters \\lith two sets of large countelTotating blades, engines apply forces in opposite directions, so there is no net [OI~e roLating the helicopter. As mentioned earlier, the Earth exerts a force F}[ all an)' obj<:£l. If the object is a TV at rest on a table, ~ in Figure 4.8a, the reaction force to F g is the force the TV exerts on the Earth, F g' . The TV doesn't accelerate dmvnward because it's held IIp by the table. The [able therefore exerts an upward force Ii, called the normal force, on the TV. (Normal, a technicallerm from m,lIhematics, means "perpendic~ ular" in this context.) The normal force is an elastic force arising from the cohesion of matter and is electromagnetic in origin. It balances the gravitational force acting on the TV, preventing the TV from falling through the table, and can have any value needed, up to the point of breaking the table. The reaction to n is the force exerted by the TV on the table, n' . Therefore, ~
~
FJ; = -F/,r'
and
n=
Newton's Third Low
93
APPLICATION Helicopter Flight
-It' ~
The forces nand n' bmh have the same magnitude as F,. Note that the forces acting on the TV are F g and Ii, as shown in Figure 4.8b. The two reaction forces, F g ' and Iii, are exerted by the TV on objects other than the TV. Remember that the two forces in an action-reaction pair ahvays act on two different ol~jecls. ~
~
(b)
FIGURE 4.8 When a TV set is sitting on :ltable, the forces acting on Ihe Sci are the normal force ii exene~)by Ihe table and Ihl.: force of gravily, F 1:' as illustrated in (b). The reaction LO n is tlte force exerted by lhe TV-.:!.el on the tabh:. ii'. The reaction La F Kis Ihe foro.: e~·rted by the TV sct on Eal"lh. F"':.
94
Chapter 4
The Laws of Motion
Because the TV is not accelcra.!Jng in any direction Cit = 0), il follows frolll NewLOn's second law Lhal = 0 = Fg + However, Fg = -mg, so 11 = mg, a lIseful result.
ilia
QUICK QUIZ 4.3 A small sporrs car collides head-on wilh a massive lruck. The greater impact force (in magnitude) acts on (a) the car, (b) the truck, (c) neither, the force is the same on both. vVhich "ehicle undergoes the greater magnitude acceleration? (d) the car, (e) the truck, (r) the accelerations are the same.
APPLICATION Colliding Vehicles
FIGURE 4.9 Newton's second law applied to a rope gi\'cs T - T' = ma. Howc\"cr, if m:= 0, then T= T'. Thus, the tension in a rnas~lcss rope is the same at all palms in the rope.
(a)
n J
L,
III ; I
n.
4.5 APPLICATIONS OF NEWTON'S LAWS This section applies Newton's laws to objects rnoving under the influence of constant external forces. 'Ne assume that objects behave as particles, 50 we need not consider the pos5ibilil}' of rotational motion. 'VVe also neglect any friction effects and the masses of any ropes or strings involved. ''''ith these approximations, the rnagnitude of the force exerted along a rope, called the tension, is the same at all points in the rope. This is illustrated by the rope in Figure 4.9, showing the forces T and T' acting on it. Jf the rope has mass m, then Newton's second law applied to the rope gives T - T' = ma. If the mass /Jt. is taken to be negligible, hmvever, as in the upcoming examples, then T= T'. When we apply Newton's 1£1"" to an object, we are interested only in those forces ""hich aet on the object:_For example, in Figure 4.8b, the only e~ernal forces acting on the TVare Ii and Fg . The reactions to these forces, Ii' and Fg', act on the table and on Earth, respectively, and don't appear in Nevvton's second law applied to lhe TV. Consider a crate being pulled to the right on a frictionless, horizontal surface, as in Figure 4.1Oa. Suppose you wish to nnd the acceleration of the crace and the force the surface exerts on it. The horizontal force exerted on the crate acts through the rope. The force that the rope exerts on the crate is denoted by T (because it's a tension force). The magnitude of T is equal to the tension in the rope. What we mean by the words "tension in the rope" is just the force read by ,1 spring scale when the rope in question has been cut and the scale inserted between the cuI. ends. A dashed circle is drawn around the crate in Figure 4.IOa (Q emphasize the importance of isolating the crate from its surroundings. Because we are interested only in the motion of the crate, we must be able to identify aU forces acting on it~These forces are illustrated in Figure 4.lOb. In addition to clisp..!:lying the force T, tile !orce diagram for the crate includes the force of gravity F g exerted by Earth and the normal force -0 exerted by the rIoor. Such a force diagram is called a free-body diagram because the environment is replaced by a series 01" forces on an otherwise free boely. The construction of a correct [ree·body diagram is an essential step in applying Newton's laws. An incorrect diagram will most likely lead to incorrect ansv.'ers! The reactions to the forces we have listed-namely, the force exerted by the rope on the hand doing the pulling, the force exerted by the crate on Earth, and the force exerted by the craLe on the floor-aren't included in the free-body diagram because they acL un other ol~jects and not. on the crate. Consequently, they don't directly influence the crate's mOlion. Only forces acting directl), on the crate are included. Now let's apply NeWl.On's second 1<1\-\' to the crate, First we choose an appropriate coordinate system. Tn this case it's convenient to use the one shown in Figure 4.10b, \vith the x-axis horizontal and the y-axis vertical. '0le can apply Newton's second law in the x-direction, y-c1ireclion, or both, depending on what we're asked
~
T
(b)
FIGURE 4.10 (a) A crate beillg pulled 10 the right 011 a frictionless surface. (b) Tile free-bod}" diagram Ul.\( represents the forces exerted 011 the crat('.
TIP 4.5 Free-Body Diagrams The most importalll slep ill solving a problt.:m by means ofNewlon's second law is to draw the correCl free-body diagram. Include only I hose forces that act directly on Ihe objecl or interest.
~
4.5
Applications of Newton's Laws
LO find in a problem. Ne\'iLOn'S second law applied to the crate in the x- and ydirections yields the following two equations: ma,,= T
"'Ul~~=
n- mg= 0
From these equations, \.."e find that the ;-lccc!el'ation in the x-direction is constant,
given by ax = T/m, and that the normal force is gi\'cn by n = mg. Because the acceleration is constant, the equations of kinemmics can be applied 10 obtain further in formation about the velocity a nel d isplaccmcllt of tile object.
95
TIP 4.6 A Particle in Equilibrium A zero Ilellorce on a panicle docs not mean that the particle isn't mo\ing. It means I hat 1he particle iSll't ouelemlillg. I f the panicle has a nOIlJ'(::ro initial velocit~ and is acted upon bya zero net force, it continues to move wilh the same velocily.
NEWTON'S SECOND LAW
Problems involving );'ewt.on's second la\'\' can be very cornplex. The following protocol breaks the solution process down into smaller, intermediate goals: I. Read the problem carefully at Icast once. 2. Draw a picture uf the system, idelllif}' the indicatc forces with arrows.
ol~jecl
of primary interest, and
3. Label each force in the picture in a \vay that will bring to mind what physical quant.ity the Iabcl stands for (e.g., Tfor tension). 4. Draw a free-boely diagram of the object of interest, based on the labeled picture. If additional objects are involved, draw separatc free-body diagrams for them. Choose convenient. coordinates for each objcct. 5. Apply Newton's second law. The x- and )'-componcnt.s of NC\'·.'I.on's second law should be taken from the vector equation and vvritten individu
Objects in Equilibrium Objects that are either at rest or moving with constant velocity are said to be in equilibrium. Because l' = 0, NeWLOn's second law applied to an object in equilibrium givcs
2:F=O
[4.9]
(il
This statement signifies thaL the valor sum of all the forces (the net force) acting on an object in equilibrium is zero. Equation 4.9 is equivalent to the set of component equations given by and
[4.10]
""Ie
won't consider three-dimensional problems in this book, but the extension of Equation 4.10 to a three-dimensional problem can be made by adding a third equation: Ll'~ = O.
QUICK QUIZ 4.4 Consider the two situations sho""11 in Figure 4.11, in which there is no acceleration. In both cases the mc.:n pull \vil.h a {-arc'.: or magnitude F ls the reading on lhe scale in pan (i) or the figure (a) grei::1ter than, (b) less than, or (c) cqualLO the reading in part (ii)?
(ii) FIGURE 4.11 (Quick Quil 4.'1) (i) A persoll pulls with a force of mag~ nitude FOll a sprillg scale allachcd to a wall. (ii) Two people pull wilh forces ofrnagnitude Fin opposite directions on a spring scale
96
Chapter 4
EXAMPLE 4.5
The Laws of Motion
A T,o",f",fi",c""L",igt;.h:,:.t""o:,:.t.:.:R:,:;e;:;st'--
_
Goal Use the second law in an equilibrium problem requiring two free-body diagrams. Problem
A traffic light "'eighing LOO x 102 0J hangs from
a vertical cable tied to two other cables that are fastened to a support, as in Figure 4.12a. The Lipper cables make angles of 37.0° and 53.0° with the horizontal. Find t.he tension in each of the three cables. Strategy There are three unknmvns, so V·le lleed to generate three equations relating them, \""Ilich can then be solved. One equation can be obtained by applying NeWLOn's second law to the traffic light., \vhich has forces in the )I-direction only. Two more equations can be obtained by applying the second law to the knotjoinillg the cables-one equa· tion from Lhe x-component and one equation from the ycomponent.
T, (aj
(b)
FIGURE 4.12 (Example 4.5) (a) A traffic lightsu~p(;ndcd by c"bles. (b) A fn:c-bod}" diagram rar the traffic lighl. (e) A free-bod) diagram for the knot joining: the cables.
.....
-
.
Solution Find 1; froll"! Figure 4.12b, using the condition of equilibrium:
(e)
2>~=0
'I; =
Using Figure 4.12c, resolve all three tension forces into componems and construcL a table for convenience:
--->
J~ =
~
T,-Fg=O
1.00 X 10' :-I
Force
x-Component
J.Component
T,
-'/; cos 37.0°
'/; sin 37.0
T
'1"; cos 53.0°
T2 sin 53.0"
°
-1.00 X 102:-.!
~
2
~
T,
L F.~ = 2: F, =
Apply the conditions for equilibrium to the knot, using (1) the components in the table:
(2)
-7'1 COS
37.0°
7; sin 37.0"
+ J; cos 53.0" =
c
0
+ 7; sin 53.0" - 1.00 X
10':-1 = 0
0
37.0 There are two equations and n,,'o remaining unknowns. T, = T J (cos cos 53.00 Solve Equalion (I) (0,7;: Substitute the result for 1; into Equation (2):
'Ii
sin 37.0°
7;
=
7; =
)
= T (0.799) = 133T '0.602 .,
+ (l.33T,)(sin 53.0°) - 1.00 X 10' 1\ =
60.1 N 1.337;
1.33(60.0:-1) =
...... -
....
°
79.9 N ~
Remarks It's very easy to make sign errors in this kind of' problem. One way to avoid them is to ahvays measure the angle of a "ector from the positive x-direction. The ~igonomctric functions of the angle will then automatically give the correct signs for the components. For example, T 1 makes an angle of L80° - 37" = 143 0 with respect to the positi"e )"-axis, and its x-component, 7~ cos 143°, is negati,'c, as it should be.
QUESTION 4.5 How would the answers change if a second traffic light were attached beneath the first? EXERCISE 4.5 Suppose the traffic light is hung so that the tensions 71 and 7; are both equal La 80.0 ~. Find the new angles they make with respect to t.he x-axis. (By symmetry, these angles will be the same.) Answer
Both angles arc 38.7°.
EXAMPLE 4.6
4.5
Applications of Newton's Lows
(a)
(b)
97
Sled on a Frictionless Hill
Goal Use the second law and the normal force in an equilibrium problem. Problem A sled is tied to a tree on a [riclionless, sno\\,covered hill, as shmvl1 in Figure 4.l3a. If the sled weighs 77.0 N, find the force exerted by the rope on the sled and the magniwde of the force Ii exerted by the hill on the sled. Strategy \·Vhen an object is on a slope, it's conven iell! to use tilted coordinates, as in Figure 4.13b, so that the nannal force n is in the y-direction and the tension force T is in the x-direction. In thc absence of friction, the hill exerts no force on the slcd in the x-direction. Because the sled is at rest, the conditions for equilibrium, 2.J~V; = 0 and "iF), = 0, apply, giving two equations for the two unknm.. . ns-the tension and the normal force. ~
r···························
FIGURE 4.13 (£xamplo.: 4.6) (a) A sled tied to a tree on a frictionless hill. (h) A free-hody diagram ror the ~led.
.
.
~
Solution Apply Newton's second law to the sled, with -; = 0: Extract the x-component from this equation to find T The x-component of the normal force is zero, and the sled's \veight is given by mg = 77.0 :"!. Writc the y-component of )..Icwton's second law. The y-component of the tension is zero, so this equation wil1 give the normal force.
2: I~=
38.5 N
T=
2: I~ = 11,
=
T - (77.0 N) sin 30.0° = 0
T+ 0 - mgsinfJ
0
+
11, -
mgcos 0 =
11, -
(77.0 N)(cos 30.0°) = 0
6fi.7 N
.
lrrr.... • • • • • • • • • •
....
Remarks Unlike its value on a horizontal surface, n is less than the weight 0(' the sled when the sled is on the slope. This is because only pan of the force of gravity (the X~cOll1pOnellt) is acting to pull the sled down the slope. The y-component of the force of gravity balances the normal force. QUESTION 4.6
Consider the same scenario on a hill with a steeper slope. ''''ould the magnitude of the tension in the rope get larger, smaller, or rcmain the same as before? How would thc normal force be affected? EXERCISE 4.6
Sllppase a child of weight w climbs onto Lhe sled. If Lhe tension force is measured child and the magnitude of the normal force acting on the sled. Answers
ill
= 43.0 ~,
II
= 104
1O
be 60.0 N, find the weight of the
~
QUICK QUIZ 4.5 For the woman being pulled forwarcl on the toboggan in Figure -:1-.14, is the magnitude of the normal force exerted by the ground on the toboggan (a) equallo the LOtal weight of the woman plus the toboggan, (b) greater than the total \-"eight, (c) less than the total weight, or (d) possibly greater than or less than the Lotal weight., depending on the size of the '.. .· eight relative to the tension in the rope?
Accelerating Objects and Newton's Second Law "Vhen a net force acts on an object, Lhe object accelerates, and we lise NeWLOn's second law to analyze the motion.
FIGURE 4.14
(Quick Quit ·\.5)
98
Chapter 4
EXAMPLE 4.7
The
Lows 01 Motion
Moving
0
Crote
Goal Apply the second law of motIon for a system nOt in equilibrium, LOgether with a kinerllalics equation. The combined weight of" the cr"'lte and dolly in Figure 4.15 is 3.00 X 10 9 N. If" the man pulls on the rope with a constant force or 20.0 N, \vhat is thc acccleration of thc
Problem
system (crate plus dolly), and how far will it move in 2.00 s? Assume lhe system starts from rest and that there are no friclion forccs opposing the motion. Strategy \Ve can find the acceleration of the system from NCWlon's second law. Because the force exerted on the system is constant, its accelcration is constant. Therefore, we can apply a kinemal ics equation LO find the distance travcled in 2.00 s.
FIGURE 4.15
(Example 4.7)
,...-
~
Solution Find the mass of'lhe system frol11 the definition of\...' eight, tu =
W
11/=-= g
mg:
Find [he acceleration of the system from the second law: Use kinemalics to find the distance moved in 2.00 s, v.lith V o = 0:
{l
x
=
1':
3.00 X 10' N = 30.6 kg 9.80 Ill/s2 20.0 N = 30.6 kg
m
, 0.654 m/s'
~x = ~ a,I' = & (0.654 m/s')(2.00 sf =
1.31
In
.....
....................
~
Remarks Note thal the constant app! ied force of' 20.0 N is assumed to act on the system at all times during its motion. If the force were removed at some instant, the system would continue to move with constant velocity and hence zero acceleration. The rollers have an effeClthat was neglected, here.
QUESTION 4.7 What effect docs doubling the weight have on the acceleration and thc displacement?
EXERCISE 4.7 A man pulls a 50.0-kg box horizont.ally from rest \,,'hile exerting a constant horizontal force, displacing the box 3.00 min 2.00 s. Find the force the man exerts on the box. (Ignore friClion.)
Answer
75.0 N
EXAMPLE 4.8 The Runaway Car Goal
Apply the secondlav.. and kinemat.ic eqUal ions to a problem involving an object moving 011 an incline.
Problem (a) A car or mass '!II is on an icy driveway inclined at an angle tJ = 20.0°, as in Figure 4.16£1. Determine the acceleration of the car, assuming the incline is frictionless. (b) If the length of the drireway is 25.0 In and the car starts from rest at the top, how long does it take to travel to the bottom? (c) What is the car's speed at the botLOm? Strategy Choose tilted coordinates as in Figure 4.16b so thal the normal force Ii is in the positive y-dircction, perpendicular to the drive\vay, and the positive ~'-axis is down the slope. The force of gravity Fg then has all x-component, mgsin e, and ay-component, -mKcos e. The components of Newton's second law form a system of two equations and two unknowns for the accel- FIGURE 4.16 eration down thc slope, {Ix, and the normal force. Parts (b) and (c) can be solved with the kinematics equations.
(a)
(Ex;unplc 4.8)
4.5
Applications of Newton's Laws
.
.
~
99
...,
Solution (a) Find the acceleration of the car. -'>
,,~
Appl)' Newton's second law:
ma
L..;F
Extract the x- and y-components from the second law:
(1)
max =
(2)
0 =
Divide Equation (J) by
1n
and substitute the given values:
a.., = gsin
---+
=
Fg
+
L F.l: = L 1~ =
e=
-'>
n
tllgsin
(J
-1Ilgcos
e + II
(9.80 m/s') sin 20.0° = 3.35 mis'
(b) Find the time taken for the car to reach the boltom.
Use Equation 3.L1b (or displacement, with
v(h
= 0: 1=
3.86 s
(c) Find the speed of the car at the bottom of the driveway. Use Equation 3.11a for velocity, again with vox = 0:
'\ = a) =
(3.35 m/s')(3.86 s) =
12.9 m/s
.
~
•••• ...A
)Jotice that the final answer for the acceleration depends only on gand the angle (J, not the mass. Equation (2), which gives the normal force, isn't useful here, but is essential \.. . hen friction plays a role.
Remarks
QUESTION 4.8 H the car is parked on a more gentle slope, how will the time required for it to slide to the bottom of the hill be affected? Explain. EXERCISE 4.8 (a) Suppose a hockey puck slides clown a frictionless ramp with an acceleration of 5.00 mjs2. What angle docs the ramp make \vith respect to the horizontal? (b) If the ramp has a length of 6.00 m, how long does it take the puck to reach the bottom? (c) Now suppose the mass of the puck is doubled. ',\'hat's the puck's new acceleration down the ramp? Answer
(a) 30.7° (b) J .55 s (c) unchanged, 5.00 mis'
EXAMPLE 4,9
Weighing a Fish in an Elevator
Goal Explore the eHect of acceleration on the apparent weight oran object.
A man \veighs a fish with a spring scale auached to the ceiling oran elevator, as shmvn in Figure 4.17<1. "Vhile the elevator is at rest, he measures a ,,,,'eight of 40.0 N. (a) What weight does the scale read if the elevator accelerates upward at 2.00 m/s2 ? (b) What does the scale read if the elevator accelerates downward at 2.00 rn/s 2 , as in Pigure 4.17b? (c) If the elevator cable breaks, what docs d,c scale read?
Problem
\'Vrite dO\~l Newton's second law for the fish, including the force T exened by the spring scale and the force of gravity, mg. The scale doesn't measure the true weight, itmcasures the force Tthat it exerts on the fish, so in each case solve for this force, which is the apparent weight as measured by the scale. Strategy
Solution (a) Find the scale reading as the elevator accelerates upward, as in Figure 4.17a.
LF =
Apply NeWLOn's second la\\,' to the fish, laking upward as the positive direction:
ma =
Solve for T
T= lila
Find the mass of the fish from its \veighl of 40.0 N:
UI.= -
Compute the value of 7; substituting a
= +2.00 m/s~:
T - I/Ig
+ mg= m(a + g) 40.0 N 9.80 !TI/s'
w
g
T = m(a + g)
= 4.08 kg
+ 9.80 !TI/s')
(4.08 kg) (2.00 mis' =
4S.1 N
(b) Find the scale reading as the elevator accelerates dowm,,-ard, (IS in Figure 4.17b.
+ g)
The analysis is the same, the only change being the accelerarioll, which is now negat.ive: a = -2.00 mig::!.
T = m(a
(c) Find the scale reading after the elevator cable breaks. Now n = -9.80 m/g 2 , the acceleration due [Q gravity:
F=m(a+g)
(4.0S kg)(-2.00 mis'
+ 9.S0 m/s')
31.S N (4.08 kg)(-9.S0
m/s' + 9.S0 I11/S')
ON
~
~
_
l\'otice how important it is to have correct signs in this problem! Accelerations can increase or decrease the apparent weight 01" an object. Asu'onauts experience very large changes in apparent weight, from several times normal weight during ascent to weightlessness in free fall. Remarks
QUESTION 4.9 Starting from rest, an elevator accelerates upward, reaching and maintaining a constant velocity thereafter until it reaches the desired floor, when it begins to slow down. Describe the scale reading during this time. EXERCISE 4.9 Find the initial acceleration ora rocket if the astronauts on board experience eight limes their normal weight during an initial vertical ascent. (Hint: in ("his exercise, the scale force is replaced by the normal force.) Answer
68.6
Ill/5 2
EXAMPLE 4.10 Atwood's Machine Goal
Use the second hn.. . to solve a simple two-body problem symbolically.
Problem Two oqjccts of mass In] and 1n2 , \-"ith 1112 > lilt, arc connected by a light, lnextensible cord and hung over a frictionless pulley, as in Active Figure 4-.18a. Both cord and pulley have negligible mass. Find the magnitude of the accelerat.ion of the system and the tension in the cord.
The heavier mass, 1fI'1.' acceleratcs dm"lwiard, in the negative )'-dircction. Because the cord can't. be stretched, the accelerat.ions of the t\'lD masses are equal in rllagnit.ude, bm o/Jposile in direction, so that (l] is p,8.sil.ivc and (l2 is negative, and 0.2 = -a,. Each mass is acted on by a force of tension T in the upward direction and a force of gr~lVity in the downward direction. Active Figure 4.18b shows free-body diagrams for the two masses. NeWLOn's second 1m,,· for each mass, together with the equation relating the accelerations, constitutes a sel or [hrce equations for the three unknowns-at, a'l' and T. Strategy
(al
(b)
ACTIVE FIGURE 4.18 (Example 4.l0) Alwoud's machine. ~a) Two hanging ol~ject.s COllnCI;ted by
4.6
,...
.
Forces of Friction
101 . • """'l
Solution Apply the second law to each of the two masses individually: Substitute fly = -a l into Equation (2) and multiply both sides by -1:
(3)
lit,",
-T+ 'IIlzK
Add Equations (1) and (3), and solve ror ",:
Substitut.e this result into Equation (1)
[0
find T -
~
.. -
~
Remarks The acceleration of the second block is I he same as thaL of' the fi rst, but negative. \,Vhen 1112 gets very large compared with ml , the acceleration of I he system approaches g, as expected. because 1Il 2 is falling nearly freely under the influence of gravity. Indeed, m,..!. is only slightly restrained by the much lighter mI' QUESTION 4.10 I-low could this simple machine be used to raise objects too heav)' [or a person to lift? EXERCISE 4.10 Suppose in the same Atwood setup another string is attached [Q the bottom of" 111 1 and a conS[~lIl[ force/is applied, rctarding the upward Illotion of nil' If nil = 5.00 kg and 1111. = 10.00 kg, what value of/will reduce the acceleration of the system by 50%?
Answer
24.5 i\'
4.6 FORCES OF FRICTION An object moving on a surface or through a viscolls medium such as air or \-vater encounters resistance as it interacts with it.s surroundings. This resistance is called friction. Forces of friction arc essential in our cveryday lives. friction makes it. possjble to grip and hold things, drive a car, ,.. . alk, and run. Even standing in one spot would be impossible without friction, as the slightest shift ,.. . ould instantly cause yOLl to slip and fall. Imagine that you've filled a plastic trash can with yard clippings and want to drag the can acr~ss the surface of your concrete patio. If you apply an external horizontal force F to the can, acting to the right as shm.. . n in Active Figure 4.J9a (page 102). the can remains stationary if F is small. The force that cOllnteracts F and keeps the can l"rolll moving acts to the left, opposite the direction of F, and is called the force of static friction, f" As long' as the can iSIl't movinz, f~ = - F. If ----) ---+ ---+ ...., F is increased, f$ also increases. Likewise, if" F decreases, f, decreases. Experimen[s show Lhat the friction force arises from the naLUre or the two surfaces: Because of their roughness. contact is made at only a few poims, as shown ill the magnified vic\'\-' o[ the surfaces in Active Figure 4..I9a. Ifwe increase the magnitude of F, as in Active Figure 4.19b, the trash can eventually slips. 'When the can is on the verge of slipping, 1: is a maximum, as shown in Figure 4.19c. ''''hen Fcxccedsj~.Jllax.' the can accelerates to the righl. \""hen the can is in motion, the ['fiction force is less than ;;.max (fig. 4.19c)~We call the friction force fOI' an object in motion the force of kinetic friction, f k' The net force F - fk in t.he x-direction produces an acceleration LO [he right, according to Newton's ~
~
~
~
~
~
~
102
Chapter 4
The Laws of Motion
ACTIVE FIGURE 4.19 (<:I) Tht: furce of friction f, exencd bva concrete surface on a trash can direcu::d oppo~ite tht force F that YOU exen on the can. As long as the
...
i;
-n
l'vlolioll
call i~ TlOllllOvillK. lhe maj.{nitudeof t he force of static friSlion equals that of the appliedJ,orct: F. (b) Whenlht:: lllagllit~1t: of F exceeds the llI fit-
1
.f.
mg (b) IfI
15.."'.,,, ------------
01£-------"------'"---1-"
f.--
Sl
(e)
second law. If F' = //.:1 the acceleration is zero, and the can moves to the right with constant speed. If the applied force is removed, the friction force acting to the left provides an acceleration or the can in the -x-direction and eventually brings it to rest, again consistent with Newton's second law. Experimentally, to a good approximation, both Is,max and Jk for an object on a surface are proportional to the normal force exerted by lhe surface on the objecl. The experimental observations can be summarized as follows: • The magnitude of the force or static friction between any two surfaces in contact can have the values
TIP 4.7 Use the Equals Sign in Limited Situations In Equation 'I.JJ, Iheequabsign is llsed mil)' when Ihe stlrf
[4.1I]
where the dimensionless constant f.Ls is called the coefficient of static friction and n is the magnitude of the normal force exerted by one surface on the other. Equation ~1.11 also holds for.£ = J~.llIax === fL5n when an object is on the verge of slipping. This situation is called imjJending molion. The strict inequality holds \.;hen the componelll of the applied force parallel to the surfaces is less than I-L.• n. • The magniwde of the lorcc of kinetic friction acting between twO surfaces is [4.12] \V'here J.Lk is the coefficienl of kinetic friction. • The values of I-Lk and fL" depend on the naLUre of the surfaces, but f.Lk is generally less than f.L.\' Table 4.2 lists some reported values. • The direction of the friction force exerted by a surface on an object is opposite the actual Illotion (kinetic friction) or the impending Illotion (static friction) or the object relative to the surface. • The coefficients of friction are nearly independent of the an..:a of contact between the surfaces. Although the coefficient of kinetic friction varies \,vith the speed of the object, \'\'e will neglect any such variations. The approximate naLUre of Equations 4.11 and 4.12 is easily demonstrated by trying [0 get an object to slide down an incline at constant acceleration. Especially at low speeds, the motion is likel)' to be characterized by alternate stick and slip episodes.
4.6
Forces of Friction
103
TABLE 4.2 Coefficients of Friction;! iL, Aluminum on steel
Steel un steel
0.74 0.6 I
Copper on sleel
0..:=)3
1.0 0.20-0.5
Rllbber 011 concrete \Vood on wood
Class on glass \~/axed ,.. . . ood on wet snow
O.~14
0.14
0.04
0.111
006 0.03
Icc on ice
0.1
Tdlon on Teflon
0.04 0.01
Synovial joints in humans "All
vaIIlC.~ ,liT
0.4
0.1
\·Vaxecl wood on dry snow ylclal on metal (lubricated)
iL, 0.07 0.47 0.36 0.8 0.2
0.04 0.003
approxllll:ltc.
QUICK QUIZ 4.6 If you press a book lIat against a \"erlical wall with your hand, in what direction is the friction force exerted by the wall on the book? (a) downward (b) up","ard (c) Ollt frolll the \Vall (d) into the wall
QUICK QUIZ 4.7
A crate is sitting in the center of a flatbed t.ruck. As the truck accelerates to the cast, the crate moves with it, not sliding on the bed of the truck. In whal direction is the friClion force exerted by the bed of the truck on the crate? (a) To the \\,·csl. (b) To the east. (c) There is no friction lorce, because the crate isn't sliding.
(aJ
QUICK QUIZ 4.8 Suppose YOllr friend is silting on a sled and asks you LO move her across a flat, horizontal field. You have a choice of (a) pushing her from behind by applying a force downward on her shoulders at 30° belm\' the horizontal (Fig. 4-.20a) or (b) allaching a rope to the front of the sled and pulling with a force at 30° above the horizontal (Fig 4.20b). \Vhich option would be easier and why?
(h) FIGURE 4.20
EXAMPLE 4.11 Goal
A Block on
0
(Quick Quil 4.8)
Rom
Apply the concept of static friction to an
o~ject
J
resting on an incline.
Problem Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coef~ ficient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block slarlS to slip down? Strategy This is an application of NewLOn·s second law involving an object in cqlli~ librium. Choose tilted coordinates, as in Figure 4.21. Use the fact that the block is just abolitLO slip when the force of static friction takes its maximum vallie,.!: = !-L.ln. FIGURE 4.21
r··············· .
.~
Solution Writc Newton's ]a\,,'s for a static systcm in cOmpOIlCl1l form. The gravilY force has two components, just as in Examples 4.6 and 4.8.
Substitute lhe expression for n into Equation (1) and solve [or tan e: Apply the inverse tangent fUIlClion LO get the answer:
....
n
= mgcos f)
2: F.>: =
mgsin
e=
0.350
[an
e-
J1.smgcos
e=
0
Ij = lan- l
~
Lan
B=
f.J.,~
(0.350) = 19.3°
.......................... ~
It's il1leresting that the final result depends on lyon the coefficient of static friction. I\mice also how similar Equations (1) and (2) are La the equations developed in Examples 4.6 and 4.8. Recognizing such paucrns is key to solving problems successfully. Remark
QUESTION 4.11 Hm\' would a larger stalic friction coefficient affect the maximum angle? EXERCISE 4.11 The ramp in Example 4.11 is roughed up and the experimem repeated. (a) \-Vhat is the new coefficient of static fric~ lion if the maximum angle turns out to be 30.0 0 ? (b) Find the maximum static friClion force that acts on the block. Answer
(a) 0.577 (b) 12.2 :-<
EXAMPLE 4.12 Goal
The Sliding Hockey Puck
Apply the concept of kinetic friction.
n
The hockey puck in Figure 4.22, struck by a hockey stick, is givcn an initial speed of 20.0 m/s on a frozen pond. The puck rernains on the ice and slides 1.20 X ]O~ 111, slowing down steadily until it comes to rest. Determinc the coefficient of kinctic friction bctween the puck and the icc. Problem
Strategy The puck slows "steadily," which means that the acceleration is constant. Consequently, ''lie can lise the kinematic equation v 2 = v0 2 + 2a to find G the acceleration in the x-direction. The x- and )'~COl11pOllellts of Newton's second law then give two equations and nvo unknowns for the coefficient of kinetic friction, f.Lk' and the normal force n.
ux
~
~"lIIg FIGURE 4.22 (Example 4.12) Ajlenhe puck is given all initial \'elocilY to till" right, lhe eXlernal forces aeting211 il arc the forccofgravitr F~,lhenormal force Ii. '!.!,ld tile force of kinelie friction, f}.
J
.
...................
Solution
Solve the time-independent kinematic equation for the acceleration a: "
v~
a=
Substitute v = 0, vo = 20.0 mis, and ~x = 1.20 X JO~ m. Notc the negative sign in the answer: a is opposite ~: Find the normal force from the y-component of [he second law: Obtain an expression for the forcc of kinetic friction, and substitute it into the x-component of the second law:
...
~lotion
a=
- Vu-
<)
2 Llx
0- (20.0111/5)' 2(1.20 X 10'111)
:L F) = 11
=
n - Fg = n -
mg
.Ii, = 1-',," = I-'"mg ma = LJ ~ F.\' = -{ = Jk
-1.67 mis'
J/W= b
0
,
4.6
Solve for
JJ-k
and substitute values:
JJ-/:
=
Ii
1.67 mis' 9.80 mis'
........... Remarks
Forces of Friction
lOS
0.170 .......................~
Nmice hO\.. . the problem breaks dO\vn into three parts: kinematics, Newton's second law in the y-dircction,
and then Newton's law in the .,,·-direction.
QUESTION 4.12 I-Iow would the answer be affectl:c! if t.he puck were struck by an astronaut on a patch of ice on Mars, where the acceleration of gravity is O.35g, with a II other given quantities remain i ng the same? EXERCISE 4.12 An experimental rocket plane lands on skids all a dry lake bed. Hit's traveling £1180.0 m/s when it touches dO\v"n, how far does it slide before coming to rest? Assume the coefficient or kinetic friction between the skids and the lake bed is 0.600. Answer
544
In
Two-body problems can often be treated as single oqjccts and solved with a system approach. 'When the objects are rigidly connected-say, by a string of negligible mass that doesn't stretch-this ~Ipproach can greatly simplify the analysis. When the two bodies arc considered together, one or more of the forces end up becollling forces that are internal 1.0 the system, rather than external forces affecting each of the individual bodies. Both approaches will be used in Example 4.13.
EXAMPLE 4.13
Connected Objects
Goal Use both the general method and the system app,"oach to solve a connccted two-body problem involving gravity and friction. Problem (a) A block \'lith mass JIll = 4.00 kg and a ball with mass 111':.!. = 7.00 kg are connectcd by a light string that passes over a frictionless pulley, as shO\vn in figure 4.23£1. The coefficient of kinetic friction bel\veen the block and the surface is 0.300. Find the acceleration of the two objects and the tension in the string. (b) Check the answer for the acceleration by using the system approach. Strategy Connccted objects are handled by applying Ne\vton's second law separately to each objecl. The free~body diagrams for lhe block and the ball are shown in Figure 4.23b, with the +x-direction to the right and the +)1direction upwards. The magnitude of the acceleration for both o~iects has the same value, la,1 = )(121 = a. The block with mass 1n] moves in the positive x-direction, and the ball \\:ith mass 111"2 moves in the negative )'-direction, so (fl = -o":.!.. Using Newton's second la",,,. we can de\"elop two equations involving the unknowns Tand a that can be solved simull
,
.
Solution (a) Fi lid the acceleration ohhe objects and the tension in the string. vVritc the components of Newton's second law [or the block of mass "Ill]: The equation {-or the J-componem gives 1/ = JIIIg. Sub· StilUte this value for 11 and!k = JJ-k"l"l into the equation lor the x-componcnt:
1n2
(a)
+, y
11I1g'
r
III:!
L,g (b)
FIGURE 4.23 (Example 4.13) (a) Two ubjects cUllnected b}' a lighl.'Jlring lhal passes over a IriClionle.~s pullcy. (b) Frc~~ body diagrams far the abjecls.
• • , • • • • • • • • '"'IIl
106
Chapter 4
The Laws of Motion
Apply Newton's second Ill\-\' to the hall, recalling
that a2 = -a 1:
Subtract Equation (2) from Equation (I), eliminating Tand leaving an equation that can be
In.!.g - J.Lk11llg
solved for a l :
'ml
lll.-z
(7.00 kg)(9.80 111!S') - (0300)(4.00 kg)(9.80 m!s') (4.00 kg + 7.00 kg)
Substitute the given values to obtain the acceleration:
SubSliLute the value rol' a l inw Equation ("I) to find the tension T:
+
5.17 m/52
1'=
32.4 N
(b) Find the acceleration llsing the system approach, where the system consists of the two blocks.
Apply Newwn's second law to the system and solve for a: a=
11I2g 1n1
Irrrr..- • • • • • • • • • • ,
/-L/:1nlg
+
m'!
.....
• , ••••••••••••••••••••••••••••
....
,
Remarks Although the system approach appears quick and eas)" it can be applied only in special cases and can't give any information about the intern<:tl forces, such as the tension. To rlnclthe tension, you must consider the freebody diagram or one of the blocks separatcly. QUESTION 4.13
U mass J!1 2 is increased, does the acceleration of the system increase, decrease, or remain the same? Does the tension increase, decrease, or remain the same? EXERCISE 4.13
"\'hat ifan additional mass is altached to the ball in Example 4.13? How large must this mass be to increase the downward acceleration by 50%? \Vhy isn't it possible La add enough mass LO double the acceleration? Answer ]4.0 kg. Doubling the acceleration to ]0.3 m/52 isn't possible simply by suspending more mass because all objects, regardless of their mass, ["all freely at 9.8 rn/5 2 near the Eal'l.h's surface.
EXAMPLE 4.14 Two Blocks and a Cord Goal Apply NeWlon's second hlW and static friction in body system.
two-
Problem A block of mass 5.00 kg rides on LOp of a second block of mass ]0.0 kg. A person attaches a string to the botlom block and puns the system horizontally across a friction less surface, as in Figure 4.24a. Friction between the twO blocks keeps the S.OO-kg block from slipping off. If the coerflcient of static friction is 0.350, what maximum force can be exerted by the string on the lO.O-kg block without causing the 5.00-kg block to slip?
'" M
x
'11/\====
Strategy Draw a free~body diagram fOI' each block. The static (b) (a) friction force causes the LOp block to lTIove horizontally, and the FIGURE4.24 (:it) (Example 4.14) (b) (Exercise 4.14) maximum such force corresponds LO /, = f.1.;n. This same static friction retards the motion of thc bottom block. As long ,IS the top block isn't slipping, the acceleration of both blocks is the same. 'Write Newton's second law for each block, and eliminate the acceleration a by substitution, solving for the tension T
4.6
,...-' .....
Forces of Friction
107
• • • • • • • • • • '''11
Solution \·Vrite the l\VO components of NC\VlOn'S second law for the LOp block: Solve the )'-component for 11., substitute the result into the x-component, and then solve for a: \'Vrile the x-component of bottom block:
~ewton's
Substitute the expression for (( and solve for the tension ~I':
Remarks Notice that the )'-component [or the lO.O-kg block \'I'asn't needed because there was no friction between that block and the underlying surface. It's also interesting to note that the lOp block was accelerated by the force of slatic friction.
QUESTION 4.14 \,Vhat would happen if the tension rorce exceeded 51.5
N~
EXERCISE 4.14 Suppose instead the string is attached to the top block in Example 14.4 (see Fig. 4.24b). Find the maximum force that can be exerted by lhe string on the block without causing lhe lOp block to slip. Answer
25.7 i'\
APPLYING PHYSICS 4.1
CARS AND FRICTION
Forces offriction are important in the anaJrsis of the lllotion of cars and other wheeled vehicles. How do such forces both help and hinder the motion of a car? Explanation There are several types of friction forces to consider, the main ones being the force of friction between the tires and the road surface and the rewrding force produced by air resistancf-:. Assuming the car is a four-\vheel-drive vchicle o[ rnass In, as each wheclwrns to propel the car forward, the tire exerts a reanvarcl force on the road. ..., The reaClion to this rearward force is a lon\'ard force f exerted by the road on the tire (Fig. 4.25). Ifwe assume the same Jonvard force f is exerted on each tire, the net forward force on the GH ..., is 4f, and t.he car's acceleration is therefore l' = 4 f 1m. The friction between t.he rnoving car's wheels and the road is normally static friction, unless the Glr is skidding. \,Vhen the car is in Illation, ..., we must also consider the force of air resistance, R, which acts in the dircc-
FIGURE 4.25 (Appl}'ing Physics 4.1) The hod/ontal forces aClillg 011 tl~ car an: lheJOl1l1Grd forces f exerted hr rhe road 011 each lirt.: :l1.!.f1 the rorcc or'lir rCSiSl;\llCe It, which 'lcu; oPjJosill'thc car's velocity. (The car's lin:s eXCrl ,I rearward force on the road. 1l0l shown in lht: diagram.)
tion 01)IJosite the velocitY. of the ...,..., car. The net force exerted on the car is therefore 4fR, so the car's ..., acceleration is = (4 f - -!!.)/m. Atllormal driving speeds, the magnitude orR is proportional to the first pmver of the speed, R = bv, where b is a constant, so the force or air resistance increases with incrcasing speed. \,Vhen R is equal to 4f, the acceleration is zero and the car moves at a constant speed. To minimiLe this resistivc force, racing cars often have very 1m.. . profiles and streamlined contours.
a
~
108
Chapter 4
The Lows of Motion
APPLYING PHYSICS 4,2
AIR DRAG
Air resistance isn't Z11ways 1.I nclesi rable. \rVhat a re some applications that depend on it?
Consider a skydiver plunging through the air, as in Figure 4.26. Despite railing frolll a
Explanation
height of several thoLlsand meters, she never exceeds
a speed of around 120 miles per hour. This is
~
i ...
£c
because, aside from the dOwll\.. . ~lrd force orgravilY ~ l! ~ mg, there is also an upward force of air resistance, R. Before sh~. reaches a final constant speed, the mag§ nitude ofR is less than her weighL As her dov·mward FIGURE 4.26 (Applying Physics '1.2) speed increases, the force arail' resistance increases. Parachutes and paragliders create a much larger The vcctor slim of the force of gravity and t.he force orair resistance gives a tOtal force thaL decreases with drag force due to thei r large area and can reduce time, so her acceleration decreases. Once the tv·,'o the terminal speed [Q a few meters per second. Some forces balance each other, the net force is zero, so spons enthusiasts have even developed special suits the acceleration is zero, and she reaches a terminal wit.h wings, allowing a long glide to the ground. In speed. each case, a larger cross-sectional area intercepts Terminal speed is generally sl ill high enough to be more air, creating greater air drag, so the terminal fatal on impact, although there havc been amazing speed is lower. stories of survival. In one case, a man fell flat on his Air drag is also important in space lravel. 'Without back in a freshly plowed ficld and survived. (He did, it, returning to Eanh would require a considerable however, break virtually every bone in his body.) In amount of fuel. Air drag helps slow capsules and another case, a flight at1.cndalll survived a fall [rom spaceships. and aerocapLUre techniques have been thirty thousand fecI. into a snowbank. In neither case proposed for trips to other planets. These techniques would the person have had any chance of surviving significantly reduce fuel requirements b}' using air ,,,·ithout. the effects orair drag. drag to slow the spacccraft dm\il1.
I
.
-~.,......
~
, SUMMARY . -_.~
4,1
Forces
There arc four known rundamental forces or nalUre: (1) the strong nuclear force between subatomic panicles; (2) the electromagnetic forces between electric charges; (3) the weak nuclear forces, which arise in cel'lain radioactive decay processes; and (4) the gravitational force between objects. These are collectivel}' called field forces. Classical physics deals only with the gravilational and electromagnetic forces. Forces such as friction or that charactcrizi ng a bat h iuing a ball are called contact torces. On a marc fundamental level, conlan forces have an e!c((rOlnaglletic nal.ure.
4,2
Newton's First Low
Newton's first law states that an o~iect moves at constant velocity unless aet.ed on by a furce. The tendency for an object to maimain its original state of motion is called inertia. Mass is the physical qU
4.3
Newton's Second Law
Newton's second law states that the acceleration of" an object is directly proportional LO the net force acting on it and inversely proportional to its mass. The net force acting on an objcct equals the product of its mass and acceleration:
L: F = m.
[4,1]
:'\Jewton's universal I a"..' of gravitation is [4.5]
The weight w or an ol~jeet is the magnitude of the force of gravity exerted on thal ol~jcct and is given by lU
=
mg
[4,6]
where g = }~/m is the acceleration of gravity ncar Earth's surface. Solving problems '\iith Newton's second law involves finding all the forces acting on a system and writing Equa-
Summary
t ion 4.1 lor the x-component and J-component scpanl1ely. These two equations are then solved algebraically for the unknown quamities.
4.4
Newton's Third Law
Newt~'s
third law states that if t,.. .o objects int('ract, thc forcc F 12 exerted by object I on object 2 is cq~d in magnitude and oppositc in direction to the force F~l exerted by object 2 on object];
F\~ =
-
F\~I
An isolated force can never occur in naturc.
4.5
Applications of Newton's Laws
An object in equilibrium has no net external force acting on iI., and the second la".', in component for III , implies that "5.I~. = 0 and "i.}~ = 0 lor such an ol~ject. Thcse IWO eqllalions are IlscCl.l1 for solving problems in statics. in which the object is at resl or moving at constant velocity. An objectllnder accclenuion requires the same tvm equalions, but ,vith the acceleration terms included: 'i.F.~ = max and "iF.v = mil). Whcn the acceler;uion is constant, the cClua+ 1ions of kinematics can supplement Newton's second law.
109
4.6 Forces of Friction The magnitude of the maximum force of static friction, l.max' bt.:L'.. .ecn an object and a surface is proportional to lhe magnillldc of the normal force acting on the o~ject. This maximum force OCClIrf.i when the object is on the verge of slipping. Tn general. [4.11]
where
jJ." is the coefficient of static friction. When an slides ovc-.!.: a surfacc, the direction of the force of kinetic friction, rho on the objeclls opposite the direction of the I1lmion of the o~ject relath'e LO thc surface and proportional..!;o Lhe magnitude of thc normal force. The magnitude 01' fJ; is
o~ject
[4.12] where J.lk is the coefficient of kinetic friction. In general, JJ./;
<
jJ.~.
Solving problems that involve friction is a matter of using these two friction forces in Newwn's second law. The static friction force must be handled carefully bccause it refers to a maximum force, which is not always called upon in a givcn problem.
FOR ADDITIONAL STUDENT RESOURCES. GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS 1. A horizont.al force of95.0 I\ is applled (Q a 50.0-kg crate on a rough. level surface. I f the crate accelerates al 1.20 m/s 2 , whal is the magnitude of the force of kinetic friction acting on the crate? (a) 23.0 0J (b) -1.1).0 N (c) 16.0 N (d) 33.0 N (c) 880 '.J 2. A ?O.O-kg man stands on a pedef.iLal of mass 27.0 kg, which rests on a level surface. \"'hat is the normal force exerted by {he ground on the pedestal? (a) 265:\l (b) 368 N (c) 478 N (d) 624 N (e) 951 N 3. Two monkeys of cqual mass arc holding onto a single vine of negligible mass that hangs verticall}' from a tree. with one monkey
!
4. A lorce of 70.0 N is exerted at an angle of 30.0° below tile horizontal on a block of rllass 8.00 kg that is resting on a table. \'\'hat is the magnitude of the normal force acring an rhe block? (a) 4~.4 N (b) 78.'1 '.J (e) 11~ '.J (d) 126:-J (e) 92.4 N
of the follO\..: ing statements mllst be true about the magnitude of I he frictional force that acts on the crate? (a) It is larg'er than the weight of the crale. (b) 1t is at least equal to lhe wtight of the crate. (c) It is equal to j.L;n. (d) 11 is greater thall the component of the gravitational Corec acting down the ramp. (e) 1t is equal to the component or the gravitational force acting down the ramp. 7. A thrown rock hits a windO'"", breaking the glass, and ends up on the floor inside the 1'00111. \Vhich of the following statements are true? (a) The force of Ihe rock on the glass was bigger than the force of the glass on the rock. (b) The force of the rock on the glass had the same magnitude as the force of thc glass on the rock. (c) The forcc of the rock on the glass ,...as less than the forcc of the glass on the rock. (d) The rock didn't slow dO""n as it broke the glass. (e) None of these statements is true.
5. If Earth's mass and radius both suddenly doubled, what \vould the new value of thc acceleration of gravity near Earth's surface? (a) 9.80 l11/s 2 (b) 4.90 m/s 2 (c) 2.45 mis' (d) 19.6 mis' (e) 12.6111/5'
8. A manager of a restaurant pushes horizontally with a force o[ mllgnilude 150 N on a box of mclons. Thc box moves across the (-\001' with a constant acceleration in {he same direclion as the applied force. '·Vhich SLatement is most accurate concerning lhe magnitude of the force of kinctic friction acting on the box? (a) 1L is greater than 150 N. (b) It is less than 150 N. (c) It is c(jllalto 100 N. (d) The kinetic friction force is steadily decreasing. (e) The kinetic friction force must be zero.
6. A crate remains stationary after it has been placed on a ramp inclined at an angle with the horizontal. Which
9. sour forces act ol~an object, given by_A = 40 N east. B = 50 N north, C = 70 N west, and D = 90 N south.
~
Chapter 4
110
The Lows of Motion
\,Vhat is (he magnitude or the net force on the object? (a) 50 N (b) 70:\ (c) 131 N (d) 1701\ (e) 250 N
(c) rhe force of friction between the crate and lhe floor orthe truck (d) the "ma" force (e) none of these
10. If an object of mass 111 moves with constant velocity v. the net force on the object. is Ca) mg (b) lilT) (e) '11/.(1, (e1) 0 (e) None of these answers is corn::cl.
14. Which of the follo\'ving statements are true? (a) An astronaut's \veight is the same 011 the :VIoon as 011 Earth. (b) An astronaut's mass is the same on the [nternational Space Station as it is on Eanh. (c) Eanh's gravity has no effect on astl"Onauts inside the International Space Station. (d) An astronaut's mass is grearer on Earth than 011 the f\.foon. (e) None of these statements are lrue.
11. If an o~jeCl is in equilibrium, which of lhe following statemtl1ls is not true? (a) The speed of the object remains constant. (b) The acceleration ur the object is l.ero. (c) The net force
12.A truck loaded with sand accelerates along a highway. The driving torce on the truck remains constant. W'hm happens to lhe accelenuioll of the truck as its u-ailer leaks sand at. a constant rate through a hole in its botrom? (a) It decreases at a stead)' rate. (b) II. increases at a stead)' race. (c) It increases and then decreases. (d) It decreases anclthen increases. (e) II remains constant. 13. A large crate or mass m is placed on the back of" a tHick but not tied down. As the truck accelerates rOI-ward with an acceleration ft, the craIe remains at rest relative to the truck. \Vhar force causes the crate (Q accelerate forward? (a) the normal force (b) the lorce or gravity
15, Two objecr.5 are cunnected by a string that passes over a frictionless pulley as in Active Figure 4.18, where IfI j < 1/1'2. and tll and at are the respective magnitudes
of" the accelerations. ·Which mathematical statement is true concerning the magnitude of the acceleration a':!. 01" lllass m,} (a) fl'2 < g (b) fl'";l > g (c) tt.1 = g (d) a? < a l (c)
a'l
>
til
16. An object or mass m undergoes an acceleration a down a rough incline. \-Vhich of the following forces should ltol appear in t.he free-bod)' diagram for the object? Choose all correct answers. (a) the force of gravity (b) (c) rhe normal force of the incline on the objcct (d) the force of friction down the incline (c) the force of friction up the incline (f) lhe force of the object on the incline
ma
• 1. A ball is held in a person's hand. (a) Identify all the external forces acting on the ball and rhe reaclion to each. (b) lfthe ball is dropped, ,,"·hat force is exerted on it while it is l~llling? Identify the reaction forcl: in this case. (:-.Jeglect air resistance.) 2.lf gold were suld by weight, would ),ou rather buy it in Denver or in Death Valle)'? If it were sold by mass, in which of the I\VO locations would you prefer to buy it? '",'hy?
8. Analyi'.e the motion of a rock dropped in water in terms of its speed and acceleration as it falls. Assume a resistive force is acting on the rock that increases as the vclocil}' of the rock increases. 9. In the moL ion picture It I-JajJjJenn! Unp Night (Columbia Pictures, ]934), Clark Gable is st
ary bus in front or Claudette Colbert, who is seated. The bus suddenly starts moving forward and Clark falls into Claudette's lap. Why did this happen?
3. [[you push on a heavy box that is at rest, you Il1llstexcrt some lorce to start its motion. Howcver, once the box is sliding, yOIl can apply a smaller force to maintain its nlmion. \o\lhy?
10. As a rocket is fired from a launching pad, its speed and acceleration increase with time as its engines continue to operate. Explclin why this occurs even though the til rust of the engines remains consral1l.
4. A space explorer is moving through space br from (Iny planet or star. He notices a large rock, taken as a specimen from an alien planel, floating around the cabin of the ship. Should he push it gently or should he kick it LOwardthe storage compartment? ""·hy?
11. In a LUg-of-war between [""'0 athletes, each pulls on the rope with a force of 200 t\. Wh
5. A passcnger sitting in tht; rear of a bus c1airns that she was il~jured as the driver slammed on the brakes, GltlSing a suitcase to cOllle nying LO\varcl he!" from the fronL of the bus. fr you were the judge in this case, what disposition would you make? \-\'l"ly? 6. A weighl lifter stands on a bathroom scale. She PUlllPS a barbell up and down. VVhat happens to the reading on the scale? Suppose she is strong enough to actually throw the barbell upward. How does the reading on the scale vary now? 7. What force C
1.0
move? A propeller-
12. Draw a frce-body diagram for each of the follO\.,..ing objects: (a) a projectile in motion in the presence of air resistance, (b) a rocket leaving the launch pad with its engines operating, (c) an athlete running along a horizontal track. 13. Identify the iKlioll-reaction pairs in the following situations: (a) a man lakes <.l step, (b) a snowball hits a girl in the back, (c) a baseball player catches a ball, (d) a gUSt of wind strikes a window. 14. Suppose you are driving a car at a high speed. Why should yOIl avoid slamming 011 your brakes ,..· hen you want [u StOP in lhe shortesl possible distance? (Newer cars have <.llllilock brakes that avoid this problem.)
Conceptual Questions
111
PROBLEMS ~
~2
~
a speed of" 6.0 m/s. A record salmon has a length of 1.5 m and a Illas!'; 0[61 kg. \""hen swimming upward at cotlstant speed, and neglecting buoyancy, the fish experiences th ree forces: an upward force F exerted by the tail fin, the downward drag force of the water, and the downward force of'grm."ity. As the fish leaves the surface of the water, however. it experiences a net upward force causing it to accelerate from 3.0 m/s 10 6.0 m/s. Assuming the drag force disappears as soon as the head of the fish breaks lhe surface and F is exerted until t\vo-thirds o[ the fish's length has left tile water, determine the magnitude of F.
The Problems for Ihis chaplcr Illay be assigned online al VVebAssigli. 1,:!,:1, = slraighllon,>ard. il1lCrllH'rli,lh", Ch,II1{,lIgill~
WebAssign
m _
II =
= denotes guided problem = denotes enhallced content problem
SECTION 4.1 FORCES SECTION 4.2 NEWTON'S FIRST LAW SECTION 4.3 NEWTON'S SECOND LAW SECTION 4.4 NEWTON'S THIRD LAW 1. The heaviest invertebrate is the giant squid. \."hich is estimated to have a weight of aboul 2 lons spread oul over its length of70 fecI. What is its weight in newtons? 2. A foot hall pUlller accelerales a football from rest to a speed of ]() m/s during the lime in which his toe is in contact with the ball (abOllt 0.20 s). If the football has a mass of 0.50 kg, whal average force does the punter exert on the ball? 3. A 6.0-kg object undergoes an acceleration of 2.0 l11/s:1. (a) What is the magnitude of the resilltant force acting on it? (b) If th is same force is applied to a ~1.0-kg object, whal acceleration is produced? 4.
10. A 5.0-g bullet IC<1\'es the muzzle of a rille with a speed of 320 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.82·m-Iong barrel of the rifle? II. A boat moves through the water with two forces acting on it. One is a 2 OOO-~ forward push by the water on the propeller, and t he other is a I 800-N resistive force due to the water around the bO\..·. (a) \rVhat is the acceleration of the I OOO-kg boat? (b) If it starts from rest, how far will the boatlllo\'c in 10.0 s? (c) \-Vhat will its velocity be at the end or that time? 12. Two forces are applicd t.o a car in an effort to move it, as shown in Figure P4.12. (<1) What is the rcsultant of thcsc two forces? (b) If the car has a mass of 3 000 kg, what acceleration docs it h:wc? Ignore friction.
m
One or marc external forces arc exerted 011 each enclosed in C1 dashed hox shown in Figure £1.2. Jclelllify the reaction LO each of these forces. ol~ject
5. A bag of sugar weighs 5.00 lb on EarTh. What would it \vcigh in newtons on the Moon, where the free-fall acn.'lenollion is one-sixth that on Earth? Repeat for Jupiter, where g is 2.64 times that on Earth. Find the mass of the bag of sugar in ki log rams at each of the Ihrec 10GIl ions. 6. A freight train has a mass of 1.5 X 10 7 kg. H Ihe locomotive can exert a constant pull of 7.5 X 10" 1\', how long docs it take to im:rease the spced of tile train froIll rest to FIGURE P4.12
80 km/h' 7. The air excns a forward force of 10 N on the propeller of a O.20-kg model airplane. If the plane accelerates I"orward at 2.0 Ill/S 2 , what is the magniludc of Ihe rc.<;istivc force exerted by the air on the airplane? 8. _
Consider a solid metal sphere (5) a few cL'ntimein diameter and a fe;Hher (F). For each quantity in the list that follo\'..·s, indicate \\·helher the quantity is the same, greater, or lesser in the case of S or in thaI of F. Explain in each case why yOlI gave the answer you did. Here is the list: (a) the gravitational force, (b) the time it \villtake to fall a given di.<;l.ance in air, (c) the time it \·... ill take to fall a given dist.ance in vacuum, (eI) the tOlal force on the object when falling in vacuum. tel'S
mil A chinook salmon has a maximum underwat.er specd of 3.0
ITl/S,
and can jump out of the water vertically with
13. A 65.0-kg skydiver reaches a tcrminal speed of 55.0 m/s with her parachute undeployed. Suppose the drag force acting on her is proportional to the speed squared, or J'~lt.
m:J All object of mass m is dropped from the roof of a building' of height h. While the object is falling, a wind blowing parallel 10 the face of the building exerts a conSlant horizontal force F on the object. (a) How long does it take the ol~jecl. 1.0 strike the ground? E.xpress the time t in terms of 1{ and h. (b) Find an expression in terms of 11L and F for the acceleration ax of the object in the horizon[al direction (taken as the posit.ive x-dircction). (c) How
112
Chapter 4
The Lows of Motion
far is the object displaced horizolllally before hilling the ground? Answer in terms or In, g, I'~ and h. (d) Find lhe magnitude of the objcCl's acceleration while it is falling, using the variables /~ '1/1, and g. 15. Aftt:'!" [,dling" [rom rest from a height of 30 111, il O.50-kg ball rebounds upward, reaching a height of 20 m. If the contact between hal1 and grouncllasted 2.0 ms, what a"cr~ age force was exerLed on the ball?
[I§JThe force exened bv the wind on the sails ora sailboat is 390 N nonh. The waleI' exerts a force of 180 N east. If the boat (including it.s crew) Jus a lllass of' 270 kg, what arc the magnil.llde and dircnion of its acceleration?
FIGURE P4.20
21. Two
mil
(a) Find the tension in cilch cable supporting the 600-N cae burglar in Figurc P4.17. (b) Suppose the horiI.ontal cable were rcaltachccl higher up on the wall. Would the tension in the otllcr cable increase, decrease, or st.ay tile same? Why?
accelerates upward at 1.60 Ill/52 , find the tensions 7; and .,; in the upper and lower strings. (b) 1l'lhe strings can \\'i(hstand a maximum tension 01" 85.0 N, what ll1axi~ mum acceleration can the elevator have before the first string breaks?
37.00
600 N
FIGURE P4.17
22. 18.
IIJI A
certain orthodontist uses a wire brace to align a patient's crooked LOath as in Figure P4.18. The tension in the wirc is arUustcd to have a magnitude or l8.0 N. Find the magniLUde of the net force excrted by the \virc on the crooked 1001h.
\\ "C"t.\ , \
", FIGURE P4.18
19. A lSO-.'J bird feeder is supported by three c<1bles as shown
of mass
F'F'=:;:=' i='=:;:l'i .4 I
1
u
of an elevator as in Figure P4.21. (a) I f the elevator
SECTION 4.5 APPLICATIONS OF NEWTON'S LAWS 17.
blocks each
3.50 kg arc fastened to the top
I ,. ~
~ 50 kg
D~
1 60 m/s 2
1 3.50 kg FIGURE P4.21
(Problems
21 and 22)
mr:t Tv..·o blocks each of' lIlass If/. arc fastened to the top of an eleva lor as in Figure P4.21. The c1e,'ator has an upward acceleration a. The strings have negligible mass. (a) Find 1 he tensions "I; and I; ill the upper and lower strings in terms of III, fl, and g. (b) Compare the two tensions and determine which string would break first if (/ is made sur:· f1cicmly large. (c) \'\"hat are the tensions if (he e1c\'awr cable brca ks?
:23.IThe distance between IwO telephone poles is 50.0 m. 'When a I.OO-kg bird lands on (he telephone wire midway between the poles, the wire sags 0.200 m. Draw a frecbod)' diagram o[ (he bird. How much tension does the bird produce in the wire? IgnOI"e the wcight ofthc wire. 24. Two people are pulling a boallhrollgh the water as in Figurc P4.24. Each exens a force 0[600:'IJ direc(cd at a 30.0° angle relative to t.he forward mOlion of (he boaL ([ the boat moves with COnS(allt velocity, find the resistivc force exerted by 1he \vatcr on (he bo;,l.
F
in Figure P4.19. Find the tension in each cable. GOO"
li(J(J "
FIGURE P4.24
FIGURE P4.19
20.
1'1
The leg and cast in Figure 1'4.20 weig·h 220 :'IJ (WI). Determine the weiglu w~ and the angle a needed so thal no fl)l'ce is exerted on the hip joim b}' the leg plus the cast.
25. A 5.0-kg bucket of' water is raised from a well by a rope. If the upward acceleration of the bucket is 3.0 Ill/S 2 , find the force exerted by the rope on the bucket.
26. A shopper in a supermarkct pushes a loaded cart with a horizontal force or 10 1\'. The carl has a mass of 30 kg. (a) How far \...·ill it mo\'e in 3.0 s, starting from rest? (lg-nore friction.) (b) How far will it move in 3.05 if (he
Problems
on ml~ (d) What is the net force acting on JfI"1? (e) \\'rite the x-component of Newton's seeolld law for each block. ([) Solve the resulting system of two equations and t\',,·o unknowns, expressing the acceleration (l and contact force P in terms of the m(lsses and force. (g) How would the answers change if the force had been applied to m
shopper places his 30-:\1 child ill the cart bdore he hegins to push il.? 27. A 2 ODD-kg car is slowed down uniformly from 20.0 m/s to 5.00 m/s in
F ---...
FIGURE P4.28
129.IAssullle the three blocks portrayed in Fig'ul'e P4.29 Illove on a friClionless surface and a ·J2-N force aets a.') shO\m on the 3.0-kg block. Determine (a) the accelermion g-iven lhis system, (b) the tensioll in the cord connecting the 3.0-kg and the 1.0-kg blocks. and (c) the force exened b}' the 1.0-kg block Oil the 2.0-kg block.
42 N
3.0 kg
1.0 kg
111.1
FIGURE P4.32
33. An RO-kg stunllmln jumps from a window of' a building situated 30 m above a catching net. Assuming air resistance exerts a IOO-N force on the sltll1tman as he ralls, determine his vclocilyjllst before he hits the net. 34.
•
E In Figure P4.34, the lighl, taut. unstretchable cord B joins hlock 1 and the larger-mass block 2. Cord A exerts a force on block I to make it accelerate forward. (a) How does Lhc magnitude of the force exerted by cord A all block I compare with the magnitude of the force exerted by corel B on block 2? (b) How does the acceleration of block I compare \',;ith the acceleration of block 2? (c) Docs cord B cxen a fi)rct:' on block l? Explain youI' answer. A
FIGURE P4.29
30. An ohjeCl of mass 2.0 kg starts from re<;t and slides down an inclined plane 80 CIll long in 0.50 s. What nt't force is aCling on Ihe oqjecl along the incline? 31.
II
Aselllp similar to the one shown in Figurc P4.?!1 isof'ten used in hospitals to support and apply a traction force 10 an il~jured leg. (a) Dctermine the forcc of tcnsion in the rope supporting lhe leg. (b) \,Vh,,11 is the traction force exerted on the leg? Assume the traction force is horizontal.
I I
Ie .c 8.00 kg
Two blocks of masses IIt I and m~ (111 1 > 111"2) arc placed on a frictionless tablc in cont.act with each other. A horiI.ontal force ofmagllillide Fis applied to the block of mass till in figure P4.32. (a) If Pis thc magnitude of the contact force between the blocks. draw the free-body diagrams {or each block. (h) \,'hat is the net force on the system consisting of both blocks? (c) \-Vhat is the nel [orep auing
Ell
•
FIGURE P4.34
135.1_ (a) An elevator of' mass m moving upward has two forces acting on it: the upward force of tension in the cable and the down\,,'ard force due 10 gravity. When the elevator is accelerating upward, which is greater, Tor w? (b) \-\rhen lhe elevator is moving at a conStant velocity upward. \",l1ich is greater, Tor w? (c) When the elevator is moving upward, but the acceleration is downward, wl1ich is grealCl', Tor 1o? (d) Let the elevator have a mass of I !)OO kg and an upward acceleration 0[2.5 Ill/S2 . Find T Is your answer consistcnt with the ansv.:er to pan (a)? (e) The elevator of pan (d) no\\' moves wilh a constant upward velocit}, of 10 m/s. Find T Is your answer consistent with rour answer to pan (b)? (f) Having initially moved upward with a constant velocity. the elevator begins to accdc::ratc dO\...· nward at 1.50 m/s 2 . Find T Is your answer consistent with your answer to part (c)? 36.
FIGURE P4.31
32.
113
object with mass 1111 = 5.00 kg rests on a frictionless hQl-i/.olltal table and is connected to a cable that passes O\'er a pulley and is then fastened to a hanging object with mass Ill',!, = 10.0 kg, as showll in Figure P4.36. Find the acceleration of each ol~ject and the t.ension in the Glblc. All
FIGURE P4.36 36, <10. and 45)
(Problem ..
114
Chapter 4
The Lows of Motion
37. A [ OOO-kg car is pulling a 300-kg trailer. Together, the car anc! trailer have an acceleration of 2.15 11'l/s2 in the forward direction. Neglecting frictional forces on the trailer, dClCrmine (a) Ihe net force on the car, (b) the net force on the trailer, (e) the force exened br I he trailer 011 the.; car, and (d) the re.sultam force exerted by the car on lhe road.
38. Two objects \·... ith masses of 3.00 kg and 5.00 kg are connected by a light string that passes over tl frictionless pulley, as in Figure P<1.38. Determine (a) the tension in the string, (b) the acceleration of each OQjccl, and (c) the distance each ol~jec.:l will Illove in the (Jrsr second ormotion ifbolh objcClssLan from rest.
3.00 kg
5.00 kg FIGURE P4.38
SECTION 4.6 FORCES OF FRICTION
l.he load nM}' slide forward, crushing the cab, if the tfuck stops suddenly in an accidcm or even in braking. Assume. for example, a 10 ODD-kg load sits on the flatbed or a 20 ODD-kg truck moving at 12.0 m/s. Assume the load is not tied down w the truck and has a coefficient of st.atic f'riction of' O.~OO with tile trllck bed. (a) Calculate the minimum stopping distance for which thc load will not sl ide forward relative to the trtlck. (b) Is an)' piece of d;Ha unnecessary for the solution? 44. A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck's Oatbed is 0.350, and the coefficient. of kinetic friction is 0.320. (a) The truck accelerates forw,lrd on Ievcl ground. What is the maximum acceleration the truck can have so that the crate docs not slide relative to the truck's natbed? (b) The truck barely exceeds this accc1eralion and then moves with constant acceleration, with the crate sliding along its bed. 'What is the acceleration of the crate relative to the ground? 45.
39. A dockworker loading crate.s on a ship finds tlim a 20-kg crate, initially at reSI 011 a hor'izolllal surface, requires a 7!1-N hOI"izontal force to set it in motion. However, after the crate is in motion, a horizolllal force of 60 N is required to keep it moving with a constant speed. Find the cocfflcielll~ of static and kinetic friction between crate and floor.
40. In Figure P..I.~6, 111 1 = JO kg and 1112 = 4.0 kg. The coefficiclll of stalic friction between JIll and [he horizontal surface is 0.50. and Ihe coefficient of kinetic I"riction is 0.30. (a) If the system is released from rest, what will its accelcnnion be? (b) If the system is set inmol.iot1 with m'! movill~ dowllward, what will be the acceleratioll of the system? pll~1ed across a level floor at a constant speed by a force F of 300 ~ at an angle of 20.0 0 below the hol"izontal, as shown in Figun:: P4.41a. (a) What is the cocfficiel1l of kinetic friction bclween the crate and the floor? (b) If the 300-N force is instead pulling the block ;u an angle of 20.0 0 above the horizontal, as shown in Figure P4.4Ib, what will be the acceleration of the crate: Assume that! he coellicient of frict ion is the same as that found in pan (a).
41. A I OOO-N craIe is being
46.
Ol~jects with masses 11/ 1 = 10.0 kg and m~ = 5.00 kg are connectcd by a light slring thal passes over a frictionless pulley as in Figure P4.36. If, when the system starts from rest, Ill.."! falls 1.00 m in 1.20 s, determine the coefficient of kinetic Friclion bet'.. . ecn '!!II and the table.
mm
A hockey pllck struck by a hocke), stick is giyen an initial speed 1)0 in t.he positive x-direction. The coefficient of kinetic friction between the icc and the puck is JL,<" (a) Obtain an expression fOf the
47. The coefficiellt of static friction between the 3.0a-kg crate and the 3.:J.'! incline of Figure P4.47 is 0.300. What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
FIGURE P4.47
(h)
FIGURE P4.41
42. A hockey puck is hit on ,\ fmzen lake and starts movinji with a speed of 12.0 m/s. Five seconds later, its speed is 6.00 m/s. (a) What is its average acceleration? (b) What is the average '·alue of thc coefficient of kinetic friction between puck and ice? (c) How far does thc pllck t.ravel during the 5.00-s interval? 143.1Consider a large truck Glrrying a heew)' load, such as steel beams. A signilicanl hazard for the driver is that
48. A student decides to move a box of books into her dormitory room by pulling on a mpe attached to the box. SIlt: pulls wilh a force of 80.0 N at an angle of 25.0° above the horizontal. The box has a mass of 25.0 kg. and the coefficient of kinetic friction between box and floor is 0300. (a) Find the acceleration of the box. (b) The student now Slans moving the box up a 10.0 0 incline, keeping her 80.0 N force directed at 25.0:> abO\'e the line of the incline. If the coefficient of friction is unchanged, what is ,he new acceleration 01" t.he box?
An ol:~ject falling under the pull of gravity is acted upon by a frictional force of air resistance. The magnitude of this force is approximalely proportional to the
149.IE
Problems speed or the ol~icct, which can be wriuen as f:::: bv. Assume b:::: 15 kg/s and til:::: !lO kg. (a) \~/hat is the terminal speed the object reaches while rllllitlg? (b) Docs your answer LO part (a) depend all the initial speed of the object? Explain.
115
the front). J-1alfofthe person's weight is supponed by the crmches, the other half by the venical forces exerteel by the g-round on his feet. Assuming he is at rest ami the force exerted by the ground on the crutches acts along the crutches, determine (a) the smallest possible coeffi~ ciellt offrietion between crutches
50. A car is I.ravelingaL50.0 klll/h on a flat highway. (a) Irthe coefficient of friction between road and tires Oil a rainy day is 0.100. whal is the minimum distance in which the car will SlOp? (b) \<\>'haL is the stopping distance wlwn the surface is dry and I.he coefficient 01" friction is O.600? 51. A 3.00-k~ block starts from rest at the top of a :10.0° incline and slides 2.00 In down the incline in 1.!lO s. Find (a) the acceleration of the block, (ll) l.he codlicicnt of kinetic friction between the block and the incline. (c) the frictional force acting on the block, and (d) the speed of the block after it has slid 2.00 m.
52.
A 2.00-kg block is held in equilibrium ilD an incline of angle 0 = 60.0° by a horizontal force F applied in dlC direction shown in Figure P4.52. If the coefficient of static rriction between block and inclinc is J.t, = 0.300, determine (a) the minimum value of and (h) the normal force exerted by the incline on the block.
FIGURE P4.55
F
ADDITIONAL PROBLEMS
56. As a protest against the umpire's calls. a baseball pitcher throws a ball slraight lip into the airtn3 speed of20.0 m/s. In (he process, he moves his hand through a distance of 1.50 11l. If the ball has a Jllass of 0.150 kg, fi nd the force he exerts on the hallLO give it this upward speed.
\e FIGURE P4.52
5?
53. Find the acceleration reached by each of the two objects shO\\"n in Figure P4.53 if the coefficient ofkinelir friction between the j.OO-k~ o~jecl and the plane is 0.250.
7.00 kg
Three ol~jects are connected on a table as shown in Figure P4.57. The rough table has a coefficient or kinetic fl"iction of 0.350. The objccts ha\'c masses or 4.00 kg. 1.00 kg, and 2.00 kg (IS shown, and the pullers are rric· tionless. (a) Draw a free-body diagram for cach objcct. (b) Determine the acceleration or each objec! and each ol~jcct's directions. (c) Del.ermine the tensions in the two cords. (d) If the tabletop were smooth, would the tensions increase, decrease, or remain tJle same? Explain.
E
l.on kg \
:17.0'
12.0 kg
FIGURE P4.53
54. Objects or masses
lli l = 4.00 kgand 111"2 :::: 9.00 kg are COllnected by a light string lhat passes over a frictionless pulley as in Figme P4.54. The ol~jcct mj is held at rest on the noor, and 111! rests on a fixed incline FIGURE P4.54 of (J = <10.0°. The objects arc I'e!eased from reSl, and 11/2 slides 1.00 m down the incline in 4.00 s. Determine (a) the acceleratioll of" e
155.ll1
The person in Figure P4.55 weighs 170 lb. Each crutch makes an angle of 22.0° with the veniG-ll (as seen from
~"---=-:=:~II------,-r;---"'~ 4.00
k~
2.00 kg
FIGURE P4.57
58. The ('(lITe exerted by the wind 011 a sailboat is approxi~ matel)' perpelldiculal' to the sail and proponional LO the component or the wind velocity perpendicular 10 Lhe sail. For the SOO-kg sailboat shown in Figure P4 ...,R, the proponionality constant is
1~.,j1 = (5.~iO~) m/s
TJwi"d j
W..l lcr exerts a force along the keel (hoLLom) of the boal that prevents it from moving sideways, as shown in the
116
Chapter 4
The Laws of Motion
figure. Once the boat starts mO\'ing forward. water also exerts a drag force backwards on the boat, opposing [he [anvard motion. Ifa17-knol wind (1 knot = O.5l4 m/s) is blowing La the east, what is the initial acceleration of the sailbO
8501'1 /J~~.OO
"t
FIGURE P4.62
I
"F"'il ~
F~d""-""'A' 30"
FIGURE P4.S8
59. (a) What is the rc:mltam force cxcned by 1 he two cables supporting the Irarlie light in Figure P4.S9? (b) \Vhm is the weight of the light?
FIGURE P4.S9
60. (a) What is the minimum force of friction required to hold the system of Figure P4.60 in equilibrium? (b) \'Vhal coefficient of SUllie friction IWI ween the 100-N block and the t
• lOON
63.
E
A box rests on the back of a truck. The coefficient of static friction belween the box and the bed of the rruck is O.~OO. (a) When the truck accelcrates forward, what force accelerates the box? (b) Find the maximum acceleration the truck can have before the box slides.
64. Three objects are connected by light strings as shown in figure P4.64. The string connecting the 4.00-kg ol~ject and the S.OO-kg object passes ovcr a light frictionless pulley. Determine (a) the acceleration of each object and (b) the tension in the t.\\l0 stri ngs.
3.00 kg FIGURE P4.64
(§[JA frictionless plane is 10.0 III long and inclined at 35.0°, A sleel stans at lhe bOl.tOlll with an initial speed of 5.00 m/s up thc incline. When the sled reaches U1t.: point. at which it llloment.arily stops, a second sled is released from the tOP of the incline with an init.ial speed tlr • Both sleds reach the boltom of the incline at the same momcnl. (a) Dctermine thc distance lhat the first sled traveled up the incline, (b) Determine the initial speed of the seconc! sled.
66. A high di . . er of mass 70.0 kg jumps off a board 10.0 III above t.he water. I r her downward Illotion is stopped 2.00 s after she ent.ers the water, what in;erage upward force did the water exert on her? 67. A 2.00-kg aluminull1 block and a G.OO-kg copper block ;-He connected by a light string over a frictionless pulle)'. The two blocks are a Hawed to mo.. .e on a fi xed steel block wedgc (of angle 0 = 30.0°) as shown in Figure P4.67. Making lise of Table '1.2, determine (a) the acceleration orthe 1\\!0 blocks and (b) the tension in the string.
Aluminum
50.0 :"I
,"I_I_~_C_: " ' ~ : " L C e " ~ ~ [_
FIGURE P4.60
ffi[]A boy coasts down it hill on a sled, reaching a level surfacc at the bottom with a speed 01"7.0 m/s. If the coefficient of friction between the sled's runners and the snow is 0.050 and the boy
4.00 kg 5.00 kg
_
Sleel
_
FIGURE P4.67
68. A 3.0-kg object hangs
at one end ora rope t.hat is auached
to a support on a railroad car. \'\-'hen the car accelerales
the right, the rope makes an angle of 4.0 0 with the vel'· tical, as shown in Figure P4,68. Find the acceleration of the car.
In
Problems
I
4.0°, : 3.0 kg
I I I
-----------_..-
~
v.
J"U
FIGURE P4.68
69. Tv,rQ boxes of fruit on a frictionless horizontal surface arc connected by a light string as in Figure P4.69. ,.. . hcre 11/,1 = 10 kg and 111~ = 20 kg. A force of 50 N is "pplied LO the 20-kg box:. (a) Determine the acceleration of each box and the tension in the string. (b) Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.10.
FIGURE P4.69
70. _ Measuring coefficients o.ffrictioll A coi 11 is placed ncar one edge of a book lying on a table, and that edge of the book is lifted umil the coinjusL slips down the incline as shown in Figure P4.70. The angle of the incline, 0", called the critical angle, is measured. (a) Draw a free-body diagram for the coin when it is on the verge of slipping and idemify all forces acting on iL Your free-body diagram should include a force of static friction acting up the incline. (b) Is the magnitude of the friction force equal to }L,n (or angles less than 0c? Explain. \\lhat can you definitely say
117
keel. At one moment, the pole makes an angle of 35.0° with the vertical and the water exerts a horizontal drag force o[ 47.5 N on Ihe boat, opposite to its forward velocity of magnitude 0.857 m/s. The mass of the boat including its cargo and the worker is 370 kg. (a) The water exel"ts a buoyant force vertically upward on the boat. Find the magnit.llde of this force. (b) Assume the forces are constant over a short interval of time. Find the vclocity of the boat 0.450 s after the moment. described. (c) rf the angle of the pole with respect to the vertical increased but the excrted force against the boLt:om remained the same, what wOldd happen to buoyant force and the acceleration of the boat? 72. _
A rope with mass 'm, is attached to a block with mass
mil as in Figurc P4.72. Both the rope and the block rest
on a horizontal, friction less surface. Thc rope does not stretch. The free end of the rope is pulled to the right with a horizontal force F (a) Draw free-body diagrams [or the rope and the block, noting that the tension in the rope is not uniform. (b) Find the acceleration of the s)'stelll in terms of 'tn/!, tn., and F. (c) Find the magnitude of the force the rope exerts on the block. (d) \.yhat happens to the force on the block as the rope's mass approaches zero? ·WheH can you state about the tension in a light cord joining a pair of moving objects?
___--'G::::m:,::=x2FIGURE P4.72
173.IA
"an accelerates down a hill (Fig. P4.73), going from rest to 30.0 m/s in 6.00 s. During the acceleration, a toy (m = 0.100 kg) hangs by a string from the van's ceiling. The acceleration is such that the string remains perpendicular to the ceiling. Determine (a) the angle (J and (b) the tension in the string.
C"in
FIGURE P4.73
FIGURE P4.70
71. _ A fisherman poles a boat as he searches for his next catch. He pushes parallel to the length of the light pole, excrting a force of 240 N on the bottom of a shallow lake. The pole lies in the venical plane containing the boat's
74. An inquisitive physics swdent, wishing to combine pleasure with scientific inquiry, rides on a roller coaster sitting on a bathroom scale. (Do not try this yourself on a roller coaster that forbids loose, heavy packages.) The boltom of the seal. in the roller-coaster car is in a plane parallel to the track. The seat has a perpendicular back and a seat belt that fits around the student's chest in a plane parallel to the bottom of the seat. The student lifts his feet from the l-loor so that the scale reads his weight, 200 lb, when the car is horizontal. At one point during the ride, the car zooms with negligible friction down a straight slope inclined at 30.0° below the horizontal. What does the scale read at that point?
118
Chapter 4
The Laws of Motion
75. The parrlc!1utc on ;1 race car of weight 8 820 N opens
around the engines and wings of Clll ilirplane exerts an 8 DOO·N force on the plane, directed upward at an angle of 65.0° above the horizontal. The plane rises with constant "clocil)' in the \'crtical direction while cOlllinuing to accelerate in the horizollt<11 direction. (a) \,Vhat is the weiglll of the plane? (b) Whell is its horizontal acceleration?
[illThc board sandwiched between two other hoards in Figure P4.i7 weighs 95.5 N. I r the cocfficielll or friction between the boards is 0.663, what must be the magnitude of lhe compression forces (assumed to be horizolltal)
(b) During rhe first 0.805 of the elevator's ascent? (c) \""hile the elevator is traveling al constalll speed? (d) During the elevator's negmive acceleralion?
1l10Vt'?
80. A magician pulls a tablecloth from under a 200-g mug located 30.0 em [rom the edge of the clOlh. The cloth exerts a friction force 01'0.100 N 011 the lllllg and is pulled with a constam accelenu ion of 3.00 Ill/S2. How far docs the mug move relative to Ihe horizof"ll
FIGURE P4.77
78. A sled wcighing 60.0 ,\I is pulled horizolltally across snow so thaL the coclTiciclll of kinetic friction between F sled and snow is 0.100. A penguin weighing 70.0 N rides on the sled. as in Figure P·1.78. If Ihe coeflicielll FIGURE P4.78 of suuic friction between penguin and sled is 0.700, find the maximulll horizontal force tha( can be exened on the sled before the penguin begins 10 slide off.
FIGURE P4.81
79. A 72-kg man stands on a spring scale in ;111 elevator. Staft~ ing from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.80 s. The elevator travels wilh this constant speed for 5.0 s, undergoes a II ni form negative acceleration for 15 s. and then comes to resL ""'hat docs the spring scale register (a) before the elevator start.. to
82. A flre helicopter carries a 620-kg bucket or water at the end of a 20.0-m-long cable. Flying back from a fire at a cOnstanl speed of 40.0 m/s, the cable makes an angle of 40.0° with respect to the venical. Determine the force exerted by air resist
-
An asteroid plunges through Earth's otmosphere while pterodactyls watch. This artist's conception is of a catastrophic event thought to have led to the extinction of dinosaurs. During on impact, an asteroid only a kilometer across releases its awesome energy of motion as thermal energy and light, delivering the explosive equivalent of one hundred million atomic bombs.
ENERGY
5.1
Work
5.2
Kinetic Energy and the Work-Energy Theorem
5.3
Gravitational Potential Energy
5.4
Spring Potential Energy
5.5
Systems and Energy Conservation
5.6
Power
5.7
Work Done by a Varying Force
Energy is one of the most important concepts in the world of science. In everyday use energy
is associated with the fuel needed for transportation and heating, with electricity for lights and appliances, and with the foods we consume. These associations, however, don't tell us what energy is, only what it does, and that producing it requires fuel. OUf goal in this chapter, therefore, is to develop a better understanding of energy and how to Quantify it. Energy is present in the Universe in a variety of forms, including mechanical, chemical, electromagnetic, and nuclear energy. Even the inert moss of everyday matter contains a very large amount of energy. Although energy can be transformed from one kind to another, all observations and experiments to date suggest that the total amount of energy in the Universe never changes. This is also true for an isolated system-a collection of objects that can exchange energy with each other, but not with the rest of the Universe. If one form of energy in on isolated system decreases, then another form of energy in the system must increase. For example, if the system consists of a motor connected to a battery, the battery converts chemical energy to electrical energy and the motor converts electrical energy to mechanical energy. Understanding how energy changes from one form to another is essential in all the sciences. In this chapter the focus is mainly on mechanical energy, which is the sum of kinetic energy-the energy associated with motion-and potential energy-the energy associated with relative position. Using on energy approach to solve certain problems is often much easier than using forces and Newton's three lows. These two very different approaches are linked through the concept of work.
5.1
WORK
\,Vark has a different meaning in physics than it does in everyday llsage. In the physics definition, a programmer does vcry lillie \\Iork typing away at a computer. A mason, by contr~lSl, may do a lot o["l,.vork I(.lying concrete blocks. In physics, work is done only ifan OI~jCCL is moved through some displacement while a force is applied
119
120
Chapter 5
Energy
to it. If either the force or displacement is doubled, the \,lork is doubled. Double thern both, and the work is quadrupled. Doing work involves applying a force to an object """hile moving it a given distance. Figure 5.1 shows a block underg~ng a displacement !1i along a straight line ,,,,,hile acted
011
by a constant force F in the same di,·cction. 'Wc havc the foIlO'\/-
ing definition: Work by
0
constant force along the displacement
~
The "mrk It\ldone on an object by a constant force ment is given by
F during a linear displace-
W= F6.x
[5.1]
where F is the magnitude of the force, D.X is the magnitude of the displacement, and and D.X point in the same direction.
F
SI unit: joule (J) A constant force F in the same direction as the displacemellL. .lx. rlocs work FtJ.x.
FIGURE 5.1
TIP 5.1 Work Is a Scalar Quantity \'V01"k is a simple number-a scalar, nOl a vcctor-so there is no direction associated ,,'jIh iL. Energy and energy transfer are also scalars.
F
FIGURE 5.2 A CQllstallL force exerted at an angle H with respect to the displacement, di, docs work (F cos 8).1x.
Work by a constant force ot an angle to the displacement 4
= newton· meter = kg· m'js'
It's easy to see the difference benveen the physics defin ition and the everyday definition of work. The programmer exerts very little force 011 the keys of a keyboard, creating only small displacements, so relatively little physics work is done. The mason must exert much larger forces on the concrete blocks and move them significant distances, and so performs a much greater amount of work. Even very tiring tasks, hm.. .ever, may not constitute \vork according to the physics definition. A truck driver, for example, may drive for several hours, but if he doesn't exert a force, then F = 0 in Equation 5.1 and he doesn't do any work. Similarly, a student pressing against a wall for hours in an isometric exercise also does no \\'Ork, because the displacement in Equation 5.1, .:lx, is zero. 1 Atlas, of Greek mythology, bore the world on his shoulders, but that, too, wouldn't qualify as work in the physics definition. "Vork is a scalar quantity-a number rather than a vector-and consequently is easier to handle. No direction is associated with it. Further, work doesn't depend explicitly on time, ,·vhich can be an advantage in problems involving only velocities and positions. Because the units of work are those of force and distance, the SJ unit is the newton-meter (N' m). Another name for the ne\vlon-mel.er is the joule (]) (rhymes with "pool"). The U.S. customary unit of work is the foot-pound, because distances are measured in feet and forces in pounds in that system. Complications in the definition of work occur ".:hen the force exerted on an object is not in the same direction as the displacement (Fig. 5.2.) The force, however, can aJways be split into two components-one parallel and the mher perpendicular to the direction of displacement. Only the component parallel t.o the direction of displacement does work on the object. This fact can be expressed in the following more general definition: The work Hldone on an o~ject by a constant force ment is given by
w=
F during a linear displace[5.2]
(Feas e)6.x
where F is the magnitude of the force, D.X is the magnitu1e of the placement, allde is the angle between the directions ofF and
o~ject's dis-
ilx.
SI unit: joule (J)
Fg::mg FIGURE 5.3 No work is done 011 a buckel ,d1cn it is moved horizoillalir because the applied force F is perpendicular to the displacement.
In Figure 5.3 a man carries a bucket of water horizontally at constant velocity"The upward force exerted by the man's hand on the bucket is perpendicular to the direction of motion, so it does no work on the bucket. This can also be seen from Equation 5.2 because the angle bet\<\reen the force exerted by the hand and IAclually, yOll do expend energy while doing isomClric exercises because your lIluscles alT COlllillllOLlSlr cOlltfaninK ami n:bxing in lhe process. This inlernilllllHSClililr 1lI01"emenl. CJuallfies as "ark according (0 Ihe physics dclinirion
5.1
the direction of motion is 90°, giving cos 90° = 0 and W= O. Similarly, the force of gravity does no work on the bucket. \\lark always requires a s}"sl.em of more thanjusl one object. A nail, for example, can't do ,...ork on itself, but a hammer can do work on the nail by driving it into a bO<:lrcl. Tn general, an object may be moving under the innuence of several exterllal forces. In that case, the total '\lark clone all the object. as it undergoes some displacement isjust the sum of the amount of work done by each force. \,Vork can be either positive or negative. In the definition of work in Equation 5.2, F and llx are magnitudes, which arc never negative. \·Vork is therefore positive or negative depending on whether cos 0 is positive or negative. This, in turn, dcpends on the direction of" relative the direction of ~x. \,Vhcn these vcctors arc pointing in the same direction the angle between them is 0°, so cos 0° = +1 and the work is positive. For example, when a student lifts a box as in Figure 5.4, the work he does on the box is positive because the force he exerts on the box is upward, in the same direction as the displacement. In lo",:ering the box slowly back clmvn, however, the slUdcnt still exens an upw~rd force on the box, but the motion of the box is downwards. Because the vectors F and are 11m.. . in opposite directions, the angle between thern is 180°, and cos 180°.-..)= -I and the \\lork done by the student is negative. In general, when the pan ofF parallel to ~x poinls in the samc direction as ~x, the work is positivc; otherwise, it's negative. Hecause Equations 5.l and 5.2 assullle a forcc constant in bOl.h direction and size, they are only special cases ora more general definition of work-that done by a varying force-trc,Hcd bricfly in Section 5.7.
Work
121
TIP 5.2 Work Is Done by Something on Something Else l
\Vork doesn't happen hy itself. Work j;, done lrysomclhilll; ill lin: environment, 011 the object of intercst.
F
f),x
FIGURE 5.4 The studcl1t does POSilivl.' \\-ork when he liflsJhe box. because the applied force F is ill lhe sanw direction as the displacemellt. When he lowers the box to the floor, he doc... negative ,\·ork.
QUICK QUIZ 5.1 In AClive Figure 5.5 (a)-(clJ. a block moves LO the right in the positive x-direction through the displacement ~x \vhile uncleI' the innuence oCa Coree with the sallle magnitude F. \tVhich of the following is the correct order of the amount of work clone by the force P, from most positive to most negati,-e? (A) d, c, a, b (B) c, a, b, cl (C) C, a, d, b
F
•
w
• (~~)
(d)
ACTIVE FIGURE 5.5 (Quirk Qui!. ~.l) A foret.: F iscxcned on <111 object thatllnderg-on a displacL"l11ent to the right. I\oth the magll iLUde of the force and the displacemenl are the >;ame in all four case~. -
EXAMPLE 5.1 Goal
Sledding Through the Yukon
Apply (he basic definitions of work done by a constant force.
Problem An Eskimo returning [rom a successful nshing trip pulls a slcd loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of' 1.20 X I02N on the sled by pulling on the rope. (a) How much ,...' ork does he clo on the sled if the rope is horizontal 1.0 the ground (8 = 0° in Fig. 5.6) and he pulls the sled 5.00 Ill? (b) Ho\\' much work docs he do on the sled if 0 = 30.0° and he pulls the sled (he same distance? (Treat the sled as a point particle, so details such as the point of all.achmenr of the rope make no difference.) Strategy Substitute the g'iven values of Fand dx into the basic equations for work, Equations 5.\ and 5.2.
FIGURE 5.6 (Examples 5.1 anel 5.2) An Eskimo pulling a sled with a rope 011 an angle 0 to the horizolll;J1,
122
Chapter 5
Energy
,.. ..
,
~
Solution (a) Find the work clone \","hen the force is horizolllal.
W = F!lx = (1.20 X 10 2 N)(5.00
Use Equation 5.1, substituting the given values: (b) Find the work done \-vhen the force is excncd 30 0 angle.
at
Ill) =
6.00 X lO'J
a
Use Equalion 5.2, again substituting the given values:
Remarks The normal force fl, the gravitational force mg, and the upward cOl1lponcm of the applied force do no work on the sled because they're perpendicular La the clisplaccmcm. The mass of the sled didn't come into play here, but it is imporlant when the effects of friClion must be calculated and in the next section, where we introduce [he work-energy theorem.
EXERCISE 5.1 Suppose the Eskimo is pushing the same 50.0-kg sled across level tcrrain with a force of 50.0 N. (a) If he does 4.00 X 10 2 J of work on the sled while exerting the force horizontally, through what distance must he ha\'e pushed it? (b) If he exerts the same force at an angle 0("45.0° with respect to the horizontal and moves the sled through the same distancc, hO\\' much \\lork does he do on the sled?
QUESTION 5.1 How does the answer [or the \v'ork done by the applied force change if the load is doubled? Explain.
Answers
(a) 8.00
III
(b) 283 J
Work and Dissipative Forces :.;. .~
:5
(h)
The edge of
EXAMPLE 5.2 Goal
Fn.I:ctiO'fk1.(~1 w, odr~ is ekxt~'C~llel}' il~IP,orl~nlt. in ~veTrYlda)E'likf~ be~ausle d,oin g almas, t arn y at ler 'I11C 0 wor' IS ImpOSSJ) e 'willout It. le S '11T10 111 11C ast examp e, or instance, depends on surface friction to pull his sled. Othen,lise, the rope would slip in his hands and exert no force on the sled, while his feet slid out from underneath him and he fell flat on his face. Cars \vouldn't work without friction, nor could conveyor belts, nor even our muscle tissue. The \"iOrk done by pushing or pulling an object is the application of a single force. Friction, on the Olher hand, is a complex process cllused by numerous microscopic interactions over the entire area of" the surfaces in contact. Consider a metal block sliding over a metal surface. -rvricroscopic "teeth" in the block encounter equally microscopic irregularities in the underlying surface. Pressing against each other, the teeth deform, get hot, and we1dto the opposite surface. 'Nork must be done breaking these temporary bonds, and this comes at the expense of the energy of motion or the block, to be discussed in the next section. The energy lost by the block goes into heating both the block and its en\'ironmelll, with some energy converted to sou nei. The friction force of two objects in contact and in relative motion to each other always dissipates energy in these relati\'ely complex ways. For our purposes, the phrase "work done by friction" \-vill denote the effect of these processes on mechanical energy alone,
More Sledding
Calculate the work done by friction when an
ol~jeet is
Problem Suppose that in Example 5.1 the coefficient of kinetic friction bet\.. .een the loaded 50.0-kg sled and snow is 0.200. (a) The Eskimo again pulls the sled 5.00 In, exerting a force of 1.20 X ]0 2 N at an angle of 0°. Find
acted on by an applied force.
the work done on Lhc sled by friction, and the net '\Iork. (b) Repeat the calculation if the applied force is exerted
at an angle of 30.0° with the horizontal.
51
See Figure 5.h, The friClional work depends on the magnitude of the kinetic friction coefficient, the normal force, and the displacement. Use the y-component of Newton's second law to find the normal force calculate the 'work done by friction using the definitions, and
Strategy
n.
.........
r-' . . . .
"
Wo,k
123
sum with the result of Example 5.1 (a) to obtain the nct \'..-ork on thc sled. Part (b) is solved similarly, but the normal force is smallcr because it has the help of the applied force F;lPP in supporting the load.
,
~
Solution
(a) Find the \.. 'ork done bY' friction on the sled and the net work, if the applied force is horizonta1. First, find the normal force from the y-component of Newton's second law, which involves only the normal force and the force ofgnwily Use the normal force to compllte the work done by frict.ion:
"2:Fy =
"'fr.-
ic
= -Ik a.x = -Miln
----*
11
=
Ill/(
UX = -J-Lkmg Lix
= -(0.200)(50.0 kg)(9.80 m/s')(5.00 m) =
Sum the frictional work with the work done by the applied force from Example ;'.1 to get the nCI \\lork (the normal and gravity forces arc perpendicular to the displacement, so they don't contribute):
mg = 0
11, -
H~w, = =
-4.90
X
IQ'J
+ W" + Hi 6.00 X 10'.1 + (-4.90 X 10' J) + 0 + 0 \t"':,pp
H~ ric
+
1.l0
X
10']
(b) Recalculate the frictional work and net work if the appl ied force is exerted £It a .30.0° angle. Fi nd the normallorce from the y-component. of Newwn's second law:
>1\,;, Sum this answcr wit.h thc rcsult of Example 5.1(b) to gct t.he net work (again, the normal and gravity forces don't. contributc):
~
X
-4.30 X 10']
Y1~'Cl = \t\~1PP
= 520
+ nTrrk + WIl + WK X 10'.1 - 4.30 X [0'.1 + 0 + 0
.
~
10' N sin 30.0 0 )(5.00m)
=
90.0j .
... ~
Remark The most important thing to notice here is that exerting the applied force at different angles can dramatically affect the work done on the sled. Pulling at. the optimal angle (11.3° in t.his case) will result in the most nc[ work for the same applied force.
QUESTION 5.2 How does the net work change in each case if the displacement. is doubled? EXERCISE 5.2 (a) The Eskimo pushes the same 50.0-kg sled over level ground \-vith a force of" 1.75 X ]0 2 N exerted horizontally, moving it a distance of 6.00 III over new terrain. If the net work done on t.he sled is 1.50 X ]02 .1, find the coefficient of kinetic friction. (b) Repeat the exercise if the applied force is upwards at a 45.0<J anRle wilh the horizontal.
Answer
(a) 0.306
(b) 0.270
124
Chapter 5
Energy
5.2
• • FIGURE 5.7 All object undergoes:1 displacement and a change in velocity unger the aClion ora constallL net forct' F nrl'
KINETIC ENERGY AND THE WORK-ENERGY THEOREM
Solving problems lIsing Nc\vton's second law can be difficult if the forces involved are complicated. An alternative is to relate the speed of an object [Q the nct work done on it by external forces. 11' the nct ,.,.,ark can be calculated for ,I given displacement, the change in the ol:~jecL's speed is easy to evaluate. Figure .5.7 show~an ol~jecl of mass m moving to the right under the aeLlon or a constant net force F net , also directed to the right. Because the force is constant, we know from Nc,.,'ton·s second law that the object moves 'with constant acceleration . 1. If" the object is displaced by fix, the work done by F nc1 on the object is ~
[5.3]
In Chapter 2, we round that the rollO\\'·ing relationship holds ,,,,hen an object undergoes constant. acceleration: or
a.:lx
=
\Ve can substitute this expression into Equation 5.3 to get
or [5.4]
So the net work done on an object equals a change in a quantity or the [arm ~ mv 2 . Becallse this term carries units o[ energy and involves the object's speed, it
can be interpreted as energy associated with the oqject's motion, leading to the following deFlIlition: Kinetic energy
~
The kinetic energy KE of an object of mass m moving with a speed v is
KE Sf unit: joule
~ ~nw2
[5.5]
0) = kg' m'/s'
Like work, kinetic energy is a scalar quantity. Using this definition in Equation 5.4, we arrive at an important result known as the work-energy theorem: Work-energy theorem
--+
The net work done on an object is equal to the change in the object's kinetic energy: 11~", ~ KEf - Kt.', = toKE
[5.6]
,..· here the change in the kinetic energy is due entirely to the object's change in speed.
FIGURE 5.8 The moving hammer has kinetic cllergy and call du work 011 the nail, driving i1 inlo the wall.
The proviso on the specd is necessary because work that deforms or causes the object to warm up invalidates Equation 5.6, although under mOSl circumstances it remains approximately correcl.. From that. equalion, a positive net '.. .ark W;lt:t means thatlhe final kinelic energy KIT; is grealer than the initial kinetic energy KE i. This, in turn. mcans that the object's final speed is greater than its initial speed. So positive net work increascs an object'S speed, and negative net vwrk decreases its speed. \Vc can also turn the equation around and think of kinetic energy as the work a moving ol~iect can do in coming to rest. For example, suppose a hammer is on the vergc of 51 riking a nail, as in Figure 5.8. The moving hammer has kinetic energy
5.2
Kinetic Energy and the Work-Energy Theorem
125
and can therefore do work on the nail. The work done on the nail is Ffix, v...here F is the average net force exerted on the nail and L\x is the distance the nail is driven inm the wall. This work, plus small amounts of energy carried ~l\vay by heat and sound, is equal m the change in kinetic energy of the hammer, fiKE. For convenience, the work-energy theorem was derived undel' t.he assulllption t.hat t.he net force acting on t.he ol~ject. was const.ant. A morc general derivation, using calculus, would show that Equation 5.6 is valid under all circulllstances, includi ng t.he application of a variable force.
APPLYING PHYSICS 5.1
LEAVING SKID MARKS
Suppose a car traveling at. a speed v skids a distance d after its brakes lock. Estimate how far it ,,-:auld skid if it were traveling at speed 2v when its brakes locked. Explanation Assume {-or simplicity thatlhe lorce of kinetic friction betwecn t.he car and t.he road surface is constant and the same at both speeds. From the work-energy theorem, the nct force excrted on the
EXAMPLE 5.3 Goal
car times the displacement of the car, 1\11.( dX, is equal in magnitude to its initial kinetic energy, ~11/.V2. When the speed is doubled, the kinctic energy of the car is quadrupled. So for a given applied friction force, the distance tr
-
Collision Analysis
v;
Apply t.he work-energy theorem with a known force.
Problem The driver or a 1.00 X 103 kg car t.raveling on the interstatc at 35.0 m/s (nearly 80.0 mph) slarns on his brakes to avoid hit.ling a second vehiclc in front. of" him, which had come to rest because of congestion ahead (Fig. 5.9). After the brakes are applied, a constant friction force of 8.00 X 103 N acts on the car. Ignore air resistance. (a) At what minimum distance should the brakes be applied to avoid a collision \.\·ilh the other vehicle? (b) If the distance between the vehicles is init.ially olt\y 30.0 m, at what speed ",iould the collision occur?
FIGURE 5.9 (Example.' 5.3) A brak· ing vchidc jus.t prior 10 an accident.
Strategy Compute the net work, which invoh'es just t.he kinetic friction, because the normal and gravity forces are perpendicular to the motion. Then sctthe net work equal to the change in kinetic energy. To gct the minimum distance in parr (a), we take the final specd Vito be zero just as the braking vehicle reaches the rca I' of the vehicle at resl. Solve for the unkllQ\vl1, 6:<:. For pan (b) proceed similarly, except that thc unknown is the final velocity v t , .... . . . . . . . . . . . . . . . . " .. , , ,.,
,...
~
Solution (a) Find thc minimum necessary stopping distancc. Apply t.he work-energy theorem to the car: Substitute an expression for the frictional \'\-'ork and set vf = 0: Substit.ute vj = 35.0 I11/S,lk = 8.00 X 10 3 N, and III ~ 1.00 X 10" kg. Solve for
(b) Atthe given distance of 30.0111, the car is too close to the other vehicle. Find t.he speed at impacL Write clown the work-energy theorem:
-(8.00 X 10" N) ""x = -!(1.00 X 10' kg)(35.0 m/s)'
76.6 m
126
Chapter 5
Energy
Multiply by 2/mand rearrange terms, solving for the final velocity vi
?
.J
2
vr- = v- -m.Jk r Ilx I
V/,>
=
V
=
J
(3~.0 _
mls )"- -
(
3
(8.00 X 10 3 N)(30.0 m)
27.3 mls
.
llt.. • • •
2)
1.00 X 10 kg
- .. -.-
~
Remarks This calculation illustrates how important it is La remain alert on the highway, allowing for an adequate stopping distance at all Limes. ft takes about a second LO react to the brake lights of the car in rront of YOLl. On a highspeed highway, your car may travel more than 30 meters befOl'c you can engage the brakes. Bumper-to-bumper traf~ fie at high speed, as often exists on the higln"'ays near big cities, is extremely ullsafe. QUESTION 5.3 Qualitatively, how would the ans,..'cr for the final velocity change in pan (b) if it's raining during the incident? Explain. EXERCISE 5.3 A police investigator measures straight skid marks 27 m long in an accident investigation. Assuming a friction force and car mass the same as in the previous problem, what was the minimum speed of the car when the brakes locked?
FIGURE 5.10 gecall~e 1111' gn"'ity fil:ld i~COll st'l"v:llivc, lhe di\'er rcgaill~ as kinetic ent.:rgy thl: \\'ork ~hc did ag~lillst gT
A force is conservative if the work it does moving an object bet,..'een tv·/O points is the same no Illa(ter Whilt path is taken.
~ Conservative force
Nonconscrvative I""orc(.;s, as ' . . . e·ve seen, don't have this property. The ,vork-energy theorem. Equation 5.6, call be rewriuen in terms of the work done by conservative forces V\~ and the ,,"'ork done by non conservative forces H~lf because the net work is just the sum of these two:
w
'1,-
+ W(
= I1K/c'
[5.7]
It turns out that conservative forces have anot.hcr llseful property: The work they do can be rccast as somcthing called potential energy, a quantity that depends only on the begin ning and end points 01" a curve, not the path taken.
5.3 GRAVITATIONAL POTENTIAL ENERGY An o~ject with kinetic energy (cnergy of motion) can do work 011 another object, just like a moving hammer can drive a nail into a wall. A brick on a high shelf can also do work: it can falloff the shelf, accelcrate downwards, and hit a nail squarely, driving it inlo the floorboards. Thc brick is said to have pot.ential energy associated with it, because from its location on the shclf it can potentially do work. Potential energy is a property of a system, rather than of a single o~ject, because it's due to a physical position in space relati\'e a center of force, likc the falling diver and the Eanh of Figure 5.10. In this chapter wc define a systcln as a collection o[objects interacting via forces or other processcs that arc int.ernal 1.0 thc system. It turns Ollt that potential energy is anothcr way of looking at the work done by conservative forces.
Gravitational Work ond Potential Energy Using the work-cnergy theorem in problems involving gravitation reqUlrcs computing thc \,"ork done by gravity. For most lr~jectories-say,1'01' a ball traversing a parabolic arc-finding the gravitalional work done on the ball requires sophisticated techniques from calculus. Fortunately, for conservative fields there's <:1 simple alternative: potential energy. Gravity is a conservative force, and tor every conservati\·c force (l special expression called a potential energy function can be found. Evaluating that function at any t,vo points in an ol~jCCl'S path of motion and fi nding the dilference will give the negalive of thc work done by that force between those two points. I['s also advantageous that potential energy, like work and kinetic energy. is a scalar qllanlity. Our first step is to find the work done by gravity on an object when it moves from one position 1.0 anothcr. The ncgat.ive of that work is the change in the gravitational potent.ial energy of the system, and from that expression, we'll be able to identify the potential energy function.
Tn Figure 5.12, a book or mass In falls from a height )'i LO a height.}!' where the positive )'-coordinate represenLS position above the ground. We neglect the force of air friction, so the only force acting on the book is gravitation. How much work is done? The magnitude of the force is mg and that of the displaccmclll is
I I
.'"If
~
~)' = )'1 - )j (a positive nUlllbcl-), while both F and ~yare pointing downwards, so
the angle betv,!een them is zero. \Ve apply the definition of work in Equation 5.2: )'
.'
H~ =
FIGURE 5.12 The work done by till: gl"aYit,lIional force as the book falls from)'; to}/ equals 11I[..'), - mgYr
F !'iy cos 0 = IIIg()', - )'t} cos 0° = -lIIg(YJ -
)'i)
[5.8]
Factoring out the minus sign was deliberate, to clari fy the coming connection to potential energy. Equation 5.8 for gravitational v.'ark holds for any objeCl, regardless of its trajectory in space, because the gravitational force is conservative. No\\', W~ will appear as the work done by gravity in the "lork-energy theorem. For the rest of this section, assume for simplicity that we are dealing only with systems involving gravity and nonconscrvative forces. Then Equ<'ttion 5.7 can be written as Wnet
=
W'I(
+
I'llI{
= !'iKE
where M~II; is the \,"ark done by the nonconserv<:ltive forces. Substituting the expression for \~from Equation 5.8, we obtain H~"
"ext, we add IIIg()'r- y,)
La
- IIIg()'r- )',} = !'i KE
[5.9a]
both sides: W", = !J,KE
+
IIIg()'r- Y,}
[5.9b]
Now, by definition, we'll make the connection between gravitational work and gravitational potential energy. Gravitational potential energy
-+
The gravitational potential energy of a system consisting of Earth and an object of mass 111 near Earth's surface is given by
PE=mgy
[5.10]
\\!here gis the acceleration of gravity and)' is the vertical position of the mass relative the surface of Earth (or some other reference point). 51 unit: joule
UJ
In this deHnition, J = 0 is usually taken to correspond to Earth's surface, but this is not strictly necessary, as discussed in the next subsection. it (Urns out that only differences in potential energy really matter. So the gravitational potential energy associated with an object located near the surface of Earth is the Object'S weight mg times its vertical position yabove Earth. From this definition, we have the relationship between gravitational work and gravitational potential energy: [5.11]
TIP 5.3 Potential Energy Tokes Two POLcntial energy always takes a system 01" at least two imcracting objects-for example, the Earth and ~I baseball interacting \'ia 1he gravitation,ll force.
The work done by gravity is one and the same as the negative of the change in gravitational potential energy. Finally, using the relationship in Equation 5.1"1 in Equation 5.9b, we obtain an extension of the work-energy theorem: [5.12] This equation says thaL the work done by nonconservative forces, 1t:,r' is equal to the change in the kinetic energy plus the change in the gravitational potential energy. Equation 5.12 will turn Ollt to be true in general, even when other conservative forces besides gravity are present. The '''''ork clone by these additional conservative forces will again be recast as changes in potential energy and will appear on the righi-hand side along with the expression for gravitational potential energy.
5.3
129
Gravitational Potential Energy
Reference Levels for Gravitational Potential Energy In solving problems involving gravitaLional potential energy, it's important to choose a location at which to set that energy equal to zero. Given the form of' Equation 5.10, this is the same as choosing the place ,-vhere y = O. The choice is completely arbitrary because the important quantity is the tlijJ"'reTlCf in potential cnergy, and this difference will be the same regardless of the choice of zero level. However. once this position is chosen, it must remain fixcd for a given problem. While it's always possible to choose the surface of Earth as the reference position for zero potcll(ial encrgy, the staLement of a problem will usually suggest
FIGURE 5.13 An\, rcrcn::nl;c lc\'cllhe dcsktop, the floororlilc roum, ur til(' grolllld oUlside the building-can he used lO rcpresclll zero g-ra,·itatiollal potential cJ1t-'rg-y ill the hook-l~anh system.
4
EXAMPLE 5.4
Wax Your Skis
Calculate the change in gravitat.ional pOlential energy for diffcrenl choices of reference level.
Goal
Problem A 60.0-kg skier is at the top of a slope, as shown ill Figure 5.14. At lhe initial point ®, she is 10.0 m vertically above point ®. (a) Setting the zero level (or gravit;Hional potential energy at ®, (lnel the gravitational potential energy of this system when the skier is at ® and then at @. Finally, find the change in potential energy of the skier-Earth systern as the skier goes from point ® to point ®. (b) Repeat this problem \\iith the zero level at point @. (c) Repeat again, ,.. . ith the zero level 2.00 In higher than point@.
®
10.0
III
Follow the definition and be careful with signs. ® is the initial point, with gravitational potential energy PEl' and ® is the final point, wilh gravitational potential energy PJ~. The location chosen for J = 0 is also the zero point for the potential energy, because Strategy
P/~
,..
=
mgy.
® FIGURE 5.14
.
(Example: 5.4)
..,
.
Solution (a) Let)' = 0 at ®. Calcubte the potential energy at ® and at@, and calculate the change in potential energy. Finel PE i, the potential energy at ®, from Equation .rJ.lO:
PI\,
PI"J = 0 al ® by choice. Find the diffe"cnce in potential cnerg)' between ® and @:
PI'i - PE,
(b) Repeat the problem if J' point, so that PE = 0 at @.
=
= "'15)', = (60.0 kg)(9.80 mN)(lO.O m) = 588 X 1O'J = 0 - 5.88
X
10'.1
= -5.88
X
10".1
0 at ®, the new reference
Fi nd PE!, noting that point @ is now at y = -10.0 Ill:
P"f= mgyl = (60.0 kg)(9.80 m/s')(-IO.O m)
JO" J X JO".1 -
= -5.88 X
PI'r- PT>,
= -5.88
0
= -5.88
X
10"J
Chapter 5
130
Energy
(c) Repeal the problem, if y = 0 two meters above @. Find PF.i, the potential energy at ®: Find PEl' the potential energy at
PE;
®:
= mgy, =
(60.0 kg)(9.80 111/5')(8.00
Ill) ~
4.70 X 10 3 J
PEf = "'{Di = (60.0 kg)(9.8 l11/s')(-2.00 m) = -1.18 X 10 3 J
Compute the change in potential energy:
PEf - PEi = -1.18 X J03J - 4.70 X 10'J
-5.88 X 10'J .
~
......... -
~
Remarks These calculations show that the change in the gravitational potential energy when the skier goes from the top of the slope LO the bottom is -5.88 X 10 3 .1, regardless oj 111,,, zeTO Itrvel sdecled. QUESTION 5.4 If the angle of the slope is increased, docs the change of gravitational potential energy between L\...·o heights (a) increase, (b) decrease, (c) remain lhe same? EXERCISE 5.4 1f the zero level for gravitational potelltial energy is selected to be midway dO\vn the slope, 5.00 m above point ®, find the initial potential energy. the final potential energy, and the change in potential energy as the skier goes rrom point ® to ® ill Figure 5.]4·.
Answer
2.94 kj, -2.94 kJ, -5.88 kJ
Gravity and the Conservation of Mechanical Energy
TIP 5.4 Conservotion Principles There arc many conservation laws like the conservation of mechanical energy ill isohtl.ed systems, as in Equation 5.13. For example, tllOmt'lllllm, angular momentum, and ckClric charge arc all conser\'ed qU
Conservation of mechanical energy-+
Conservation principle~ playa very important role in physics. When a physi· cal quantity is conserved the numeric value of the quantity remains the same throughout the physical process. Although the form of the quantity m(ly change in some way, its final value is the same as its initial value. The kinetic energy K/!.'ofan o~ject falling only under the innuellce of gravity is constantly changing, as is the gravitational potential energy PE. Obviously, then, these quantities aren't conserved. Because all nonconscrvative forces arc assumed absent, hmvcver, we can set Wnc = 0 in Equation 5.12. Rearranging the equation, 'we arrive at the follmving very interesting result; KEI
+
PI'. = KF.'f + PI'~J -'I
[5.13]
According to this equation, the sum of the kinetic energy and the gravitational potential energy remains constant at all times and hence is a conserved quantity. We denote the total mechanical energ'y by f. = Kt; + PE, and say that the total mechanical energy is conserved. To shO\v how this concept ""orks, think or tossing a rock offa dire ignoring the drag forces. As the rock falls. its speed increases, so its kinetic energ)' increases. As the rock approaches the ground, the potential ellergy of the rock-Earth system decreases. \Vhatever potcntial energy is lost as the rock moves downward appears as kinctic energy. and Equation 5.13 says that in the absence of nonconservative forces like air drag. the trading of energy is eX
5.3
Gravitational Potentiol Energy
131
ff the force of gravity is the only force doing work within a s)'stem, then the principle ofconseryat.ion of mechanical energy takes the form I?
2" mv, -
+
I:.!
mp;y, = '2 1111'1
+
[5.14)
mf!J'.r
This form orthe equation is particularly L1seful (01- solving p,-oblcms involving only gravity. Further terms have to be added when other conservative forces are presCllt, ;'-IS we'll soon see. QUICK QUIZ 5.2 Three identical balls are thrown from the LOp ofa building, all with the sarne initial speed. The tirst hall is thrown horizontally. the second at some angle above the hOI-izontal, and the third at some angle below the horizontal. as in AClivc Figure 5.15. ::'\'eglccting air rcsistance, rank the speeds of the balls as they reach the ground, from fastest to slowest. (a) 1,2,3 (b) 2. 1.3 (c) 3, 1,2 (d) All three balls strike the ground at the same speed.
2 <-
3
,
QUICK QUIZ 5.3
Bob, of mass 111, drops frolll a tree limb at the same time thai Esther, a)so of Illass 1ft, begins her descelll down a frictionless slide. If they both start at the same height above the ground, which or the following is true about their kinetic energies as they reach the ground? (a) Bob's kinetic energy is greater than ESlher's. (b) Esther's kinetic energy is greater than nob's. (c) The) have the same kinetic energy. (d) The answer depends on the shape of the slide.
PROBLEM·SOLVI~N@GDS~T[fR~A~T!E~GrY-==========:::::§~§~~~~ APPLYING CONSERVATION OF MECHANICAL ENERGY Take the following steps when applying cOllserVmioll problems invol\'ing gravity:
or mechanical energy to
I. Define the system, including all interacting' bodies. Verify the absence of
nOllconservative forces. 2. Choose a location for y energy. ~.
= 0,
the tero point lor gravitational potential
Select the body of interest and identify two points-one point where yOll ha\'e gi"en information and the other poilll when.. you want to find out somel hing aboUl the body of interest.
4. Write down tile conservation of energy equation, Equation 5.14, for the tem. Identify the unknown quantity of interest.
S~IS
fl. Solve for the unknown quantity, which is usually either a speed or a position, and substitute known ,·alues.
As prcviouto;ly Slated, it's usually best to do the algebra with symbols rather than substitttting known numbers fi,"st, bccausc it's easicr 10 check the symbols fOI+ pas· sible errors. The exception is when a quantity is clearly l.cro, in which case immcdiatc substillltion greatly simplifies the ensuing algebra.
ACTIVE FIGURE 5.15 (Quid.. QuiL :l.2) Thn..e idelllical ball~ arc thrO\\'n willl the same inilial spl'c-'d frum LlH~ lOp ofa building.
132
Chapter 5
Energy
EXAMPLE 5.5 Platform Diver Goal
Use conservation of energy to calculate lhe speed of a body railing straight down in the presence of gravity.
Problem A diver of mass III drops fr0111 a board 10.0 m above the water's surface, as in Figure 5.16. Neglect air resistance. (a) Use conservation of mechanical energy [0 find his speed 5.00 m above t.he water's surface. (b) Find his speed as he hits the waler. Strategy Refer to the problem-solving strategy. Step 1: The system consists of the diver and the Earth. As the diver falls, only the force of gravity acts on him (neglecting air drag), so the rnechall ical energy of the system is conserved and we can use conservation of energy lor both parts (a) and (b). Step 2: Choose y = 0 lor the water's surface. Step 3: In part (a),)' = JO.O ITl and y = 5.00 III are the points of interest. while in part (b), y = 10.0 III and y = 0 m are of intcresl.
10.0 m
Iln:=ol---
5.00 m FIGURE 5.16 (l':xampic 5.5) The zero or gravitational potential energy is taken to be at the wafcr's slll"face.
Solution (3) Find lhe diver's speed halfway down, at y = 5.00 m. Step 4: V'le write the energy conservation equatioll and supply the proper terms:
KE, I
<j1IlVj
Step 5: Substitute
Vi
2
+ PE, =
KI'j + PEJ
+ 'mgYi -_ 2I 11H)r 2 + mgYJ
= 0, cancellhe mass m and solve [or
v/
(b) Find the diver's speed at. the \.. . al.er's surface,)1 = O.
o + mgYi = ~rnv/l + 0 vJ= V2g)\ =V'2.,..(9-.8-0-m-/s~·2c-c)(.,..10-.-0-mc-c)
Use the same procedure as in part (a), laking ')'.J= 0:
=14.0 m/s
.....
.
~
Remark
NOlice that the speed halfway down is not half the final speed.
QUESTION 5.5 Qualitatively, how will the answers change if the diver takes a running dive ofT the end of the board? EXERCISE 5.5 Suppose the diver vaults ofr the springboard, leaving it with an initial speed of 3.50 m/s upward. L'se energy conservation to find his speed when he strikes the waler.
Answer
14.4 m/s
EXAMPLE 5.6 The Jumpin Bu Goal Use conservation or mechanical energy and concepts from ballistics in two dimensions (Q calculate a speed. Problem A powerful grasshopper launches it.self at an angle of 45° above the horizontal and rises to a maximum height of LOO m during the leap_ (See Fig. 5.17.) \'Vith what speed VI did it leave the ground? Neglect air resistance. Strategy This problem can be soh;ed with conservation of energ)' and the relation between the initial velocity ;:lIld its x-component.. Aside from the origin, the other point ofimerest is the maximum height)' = 1.00 m, . .v hcre FIGURE 5.17 (Example 5.6) the grasshopper has a velocity V x in the x-direction only. Energy conservation then gives one equation with two llnknmvns: the initial speed Vi and speed at maximum height, there arc no forces in the x-direction, however, V x is the same as the x-component of the initial velocity.
Ym""
lJx •
==
II
Because
5.3
Gravitational Potential Energy
133
.
,..
.
...,
Solution Use energy conservation: Substitute Yi = 0, vr= v x' and)'j = II: Multiply each side by 21m. obtaining one cqu<:11ion and two unknowns: Eliminate V x by substituting V x = VI cos 45° into Equation (I), soh'iug for Vi' and substituting known values:
V,2 Y,
= (Vj cos 45°)2
+
Remarks
+
2gh
= 2vgh = 2V(9.80 m/s')(1.00 m) =
.
~
2gh = ~v12
6.26 m/s
• • • • • • • • • • • • • • ....lI
Thc final answer is a surprisingly high value and illustrates how strong insects arc rc)ati\'e to their size.
QUESTION S.6 All other gi\'cn quantities remaining the samc, hm,' would the answer change if the initial angle were smaller? Why? EXERCISE S.6 A catapUlt launchcs a rock at a 30.0° angle with respcctLO the horizontal. Find the maximum height auained if the
speed or the rock al ils highesl poinl is 30.0 m/s. Answer
15.3 m
---~,
TIP 5.5 Don't Use Work Done by the Force of
Gravity and Nonconservative Forces Whcn nonconservative forces are involved along v·.. ith gravitation, the full workenergy thcorcm must be used, often with techniques from Chapter 4. Solving problems requires the basic procedure of the problem-solving strategy for conservation-of-energy problems in the previous section. The only difference lies in substituting Equation 5.12, the work-energy equation with potential energy, for Equation 5.14.
Gravity and Gravitational
Potential Energy! Gravitational potential energy is just another wa)' of including the work clone by the force of grav~ ity in I he work-energy theorem. Don'IU'le both of them in the equation oIL the "arne time or you'll coum illwice!
_____
EXAMPLE 5.7
~.1
Der Stuka!
Goal Use the work-energy theorem with gravitational potential energy to calculate the work done by a nonconservati""e force. Problem \·\·aterslides are nearly fricLionless. hence can provide bored sludenLS wilh high-speed lhrills (Fig. 5.18). One such slide, Der Sruka, named for the terrifying German dive bombers of 'World \Nar II, is 72.0 fCel high (21.9 m), found al Six Flags in Dallas, Texas and al Wel'n Wild in Orlando, Florida. (a) Delermine the speed or a 60.0-kg woman at the boLtom of such a slide. assuming no friction i~ present. (b) If the woman is clocked at 18.0 m/s at the bottom of the slide. rind the work done 011 the woman by friction. Strategy The systcm consists of the woman, the Earth, and the slide. The normal force, always perpendicular to the displacement, does no work. Let)' = 0 m represent the hOllom of the slide. The two points of interest are y = 0 m and y = 2].9 m. Without rriction, Wile = 0, and we can apply conservation of mechanical energy, Equation 5.1~1. For pan (b), use Eflu
FIGURE 5.18 (Example 5.7) Irille slide is frictionless, the woman'!> spct.'c1 at the bottom depends Dill}' on til<.: height of the slide, 110t 011 Ihe path ill<1kes.
134
"-"
Chapter 5
Energy
..,
.
.
Solution (a) Find t.he woman's speed at the bottom of the slide, assuming no friction. \I\o'ritc down Equation 5.14, for conservation of energy: I nsen the values
Vi
= 0 and vl
I 2 ,pllV, + mgYi
I 9 "2"1.1)/ + mg)'J
=
o + mg)'t =
= 0:
''i ~
Soh'c for viand substitute values for gand )',:
v'2g); ~
I
"
"2'1JlV/
+0
V2{9.80 m/s2 ){21.9 m)
20.7 mls
(b) Find the \vark done on the ,,"'oman by friction if
vJ = 18.0 mls < 20.7 m/s. Write Equation 5.12, substituting expressions for the kinetic and potential energies:
vv,,, =
Substitute In = 60.0 kg, vJ = 18.0 mis, and and solve for V\~I(:
w'" =
Vi
= 0,
(I(£,j- KEi)
= (~1IIV/
!mv/)
+ (mg)y- mgyJ
[~·60.0 kg·{18.0 m/s)' - 0]
+ [0 tv"r =
-
+ (I'EJ - PE,)
60.0 kg· (9.80 m/s')· 21.9 m]
-3.16 X 103 J
............
.
,
....
,
Remarks The speed found in pan (a) is the same as if the woman fell venically through a distance of 21.9 Ill, consistent with our intuition in Quick Quiz 5.3. The result of part (b) is negative because the system loses mechanical energy. Friction transforms part of the mechanical energy into thermal energy and mechanical waves, absorbed partly by the system and panly by the elwlronment.
QUESTION 5.7 If the slide were not frictionless, 't\'ould the shape of the slide af'fectl.he final answer? Explain. EXERCISE 5.7 Suppose a slide similar to Del' SLUka is 35.0 meLCrs high, but is a straight slope, inclined at 45.0° with respeClto the horizontal. (a) Find the speed ofa 50.0-kg woman at the bottom of the slide, assLiming no friction. (b) If the woman has a speed of20.0 m/s at the bottom, find the change in mechanical energy due to friction and (c) the magnitude of the force or friction, assumed constant. Answers
(a) 26.2 mls
EXAMPLE 5,8
(b) -8.58 X 10"J
(c) 173 N
Hit the Ski Slop.es
Goal Combine conservation of mechanical energy with the work-energy theorem involving friction on a horizontal surface. Problem A skier st.arts from rest at the top of a frictionless incline of height 20.0 m, as in Figure .t5.19. At the bOl.Lorn or the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.210. (a) Find the skier's speed at the bottom. (b) How far does the skier travel on the horizontal surface before corning to rcst? Neglect air resistance.
~/
~r JLJ_ Ii = 20.0
III
e =_2~~~L_ -""-';®'---=o=-'~ I+--d_ I
,
Strategy Going down the frictionless incline is physically no different than going down the slide of the previous FIGURE 5.19 (Example 5.8) The skier slides down the slope and example and is handled the same way, using conservation 01110 a level slirface, stopping after Ir,lveliJlg a disl"lllCC rI from the haltom oftlH: hill. of mechanical energy to find the speed v@ at the bottom. On the flat, rough surface, use the work-energy theorem, Equation 5.l2, withl~~l( = W fric = -fkrl, where II. is the magnitude of the force of friction and rl is t.he distance traveled on the horizontal surface before coming to rest.
5.4
Spring Potential Energy
135
r"·····"· .
.. ...,
"
Solution (a) Find the skier's speed at the bottom. Follm-\' the procedure lIsed in parI (a) or the previous example as the skier moves from the top, point @, to the bottom, point @:
v® =
v2ih =
V2(9.80 m/s2)(20.0 Ill)
19.8 ml'
(b) Find the distance traveled on the horizontal, rough surface. Apply the work-energj' theorem as the skier rnoves from
Remarks Substituting the symbolic expression v® = illlo the equation for the distance d shows that d is linearly proponional to Ii: Doubling- the height doubles the distance traveled. QUESTION 5.8 Give two reasons \,\'hy skiers typically assume a crouching position down when going down a slope. EXERCISE 5.8 Filld the horizontal distance the skiel' travels before coming to rest if t he incline also has a coefficient of kinetic friction equal to 0.210.
Answer
5.4
40.3 m
SPRING POTENTIAL ENERGY
Springs are important e1cll1cnts in modern tcchnology. They arc found in machines oLII] kinds, in watches, to),s, cars, and trains. Springs will be introduced hcre, then studied in morc detail in Chaptcr ]3. Work done by an applied force in stretching or compressing a spring can be recovered by removing the applied force, so like gravity, the spring force is conservative. This means a potential energy function can be found and used in the \\lurk-energy theorem. Active Figufe 5.20a shows a spring in its equilibrium position, \\lherc the spring is neither compressed nor stretched. Pushing a block against I he spring as ill Activc Figure 5.20b compresses it a distance.'(. Although x appears to he merely a coorclinate, for springs it also represents a displacement from the equilibrium position, which for our purposes will always be taken to be at x = O. Experimentally. it turns nut that doubling a given displacement requires double the force, \-vhile tripling it takes triple the force. This means the force exerted by thc spring, I'~, must be proportional to the displacement ;lC, or [5.15]
where k is a constant of proportionalily, the !JjJring COl/slant, carrying units of newtons per meter. Equation 5.15 is called Hooke's law, after Sir Robert Hooke, who discovered the relationship. The force F:. is often called ZI Teslori:lIJ; [oJ're because the spring always exerts a force in a direction opposite Ihe displacement of its encl, tending 1.0 restore whatever is attached 1O the spring to its original position. For positive values of x, thc force is negative, pointing back towards eqllilibrium at x = 0, and for negative x, the force is positive, again pointing towards x = O. For a
(a) x= 0 !--x I I I
~\ -;r'
PE, =
'--_-.:--
(h)
-
v
'-._~_....::..____
..
t kx':!
KI'> 0
Pt.·~
=0
" Kl.j= 2II1IV-
(c)
ACTIVE FIGURE 5.20 (a):\ sprillgal l'quitibrillm, neither compressed nor sln'lehed. (h) f\ block 01 ma",s //Inn a frictionless surface i... pushed
136
Chapter 5
Energy
flexible spring, If. is a small number (about 100 N/m), ,....,hcreas for <-l sl.iffspring k is large (about 10 000 N/m). The value of the sp"ing constant k is determined by how the spring was formed, its material composition, and the thickness of the wire. The minus sign ensures that the spring force is ahvays directed back towards the equilibrium point. As in the case of gravitation, a potential energy. called the elastic potential energy, can be associated \vith the spring force. Elastic pOlcnlial energy is another way of" looking aL the \'lark clone by a spring during Illotion because it is equal [0 the negative of the work done by the spring. [L can also be considered slored energy arising from the work done to compress or strerch the spring. COllsider a horizontal spring and mass at the equilibrium position. 'y,/e determine the \'lark done by rhe spring ,...· hen compressed by an applied force from equilibrium to a displacement X, as in Active Figure 5.20b. The spring force poims in the direction opposite the motion, so ,..'e expect the work to be negat.ive. \Vhcn we st.udied the constant force of gravity near Earth's surface, we found t.he ,·.:ark done on an ol~jecr by multiplying the gravitational force b)' rhe vertical displacement of the object. However, this procedure can't be us~d with a varying force such as rhe spring force. Instead, we use the average force, F:
0- kx
2
kx 2
Therefore, the '\lark done b)' the s/)ringJoro: is H~ =
Fx =
_~hx2
1n general, when the spring is stretched or compressed from b)' the spring is
Xi
to Xi' t.he work done
WI = - (~kXJ 2 - ~hXj 2)
The work done by a spring can be included in the work-energy theorem. Assullle Equation 5.1~ now includes the work done by springs on the left-hand side. It thell reads ,"Vur
-
Ghx/ -
~ /(X,2) = D.KH
+ ilPt.~
""here P/~~ is the gravit.ational potential energy. \Ve now define the clastic potential cnergy associated with thc spring force, PE~, by [5.16]
Inserting this expression into the previous equation and rearranging gives the new form of" the work-cnergy theorem, including both gravitational and elastic potential energy: [5.17]
wherc V\~lr is the \\lark done by nonconscrvative forces, KE is kinetic energy, PEg is gravitational potcmial energy, and PE, is the clastic potential energy. PE, formerly lIsed to denote gravitational potentiaJ energy alone, will henceforth denote the toral potential energy ora system, including potential energies due to all conservative !orccs acting on the system. 11'S imponanl to remember that the work done by gravity and springs in any given physiG-d system is already included on the right-hand side of E.quation 5.L7 as potential energy and should not also be included on the left as work. Active Figure 5.20c shmvs hO\\l the stored elastic potemial energy can be recovered. W'hen the block is released, the spring snaps back to its original length and the stored elastic potential energy is converted to kineric encrg)' of the block. The elastic potential energy stored in the spring is zero 'when the spring is in the equilibrium position (x = 0). As given by Equation 5.16, poremia} energy is also stored in the spring when it's stretched. Further, the clastic potential energy is a maximum when the spring has reached irs maximum compression or extension.
5.4
Spring Potential Energy
137
Finally, because PI:;, is proponional to r, the potential energy is always positive when the spring is not in the equilibrium position. In the absence ofnonconservative forces, l:t~!r = 0, so the left-hand side ofEquation 5.l7 is lero, and an extended form [or conservation of mechanical energy results: [5.18] Problems involving springs, gravity, and other forces are handled in exactly the same way as described in the problem-solving strategy for conservation of mechanical energy, except that the equilibrium point of any spring in the problem must be defined, in addition to the zero point for gravitational potential energy
EXAMPLE 5.9
A Horizontal SPoring
Goal Use conservation of energy to calculate the speed of" a block on a horizontal spring \vith and 'without friction. Problem A block with mass of 5.00 kg is attached to a horizontal spring with spring const.ant h = 4.00 X 10 2 N/m, as in Figure 5.21. The surface the block rests upon is frictionless. Tr the block is pulled Ollt to Xi = 0.0500 III and released, (a) find the speed of the block when it first reaches the equilibrium point, (b) find the speed when x = 0.0250 m, and (c) repeat part (a) if friction acts on the block, with coefficient Jill = 0.150. Strategy In parts (a) and (b) ulerc arc no nonconservmivc forces, so conservation or-energy, Equation 5.18, can be applied. In part (c) the definition oEwork and t.he \,,:ork-energy theorem are needed to deal with the loss of mechanical encrgy due LO frictioll.
•
In
F,
m ---"'"4 -'
0
-',
!mg FIGURE 5.21 attached
lO
([xallll)]e 5.9):-\ mass
a spring.
~
~
Solution (a) Find Lhe speed oCthe block at equilibrium point. Start with Equation 5.18:
( KF•
Substitute expressions for the block's kinctic energy and the potential energy, and set thc gravity terms to zero:
( 1)
Substitute
VI
= 0,
.'e.t=
0, and multiply by 2/m:
Solve for VIand substitUle the given values:
+ PEg + PF.J ~~
I
= (KE oJ
+ p"1~g + PElIf
It '! - x- = v-
<)
111
VI.
I
f
=
!k \/-;;;
Xi
4.00 X 10' N/m ( ) 0.0500 m 5.00 kg
=
0.447 Ill/s (b) Find the speed orthe block at the halrway point. hXf~
+--
Sel Vi = 0 in Equation (I) and multiply by 2./111:
Solve lor vJ and substitut.e the given values:
UI.
V··=
J
«.
Ii x: - -1// '!
:\"J<') ~
•
4.00X10'N/m <. o [(0.050111)- - (0.025 m)'] 5.0 kg 0.387 m/s
Chapter 5
138
Energy
(c) Repeat pan (<:I), this time \..· ith friction.
Apply the work-energy theorem. The ,,'ork done by the force of gravity and the normal force is zero because these forces are perpendicular to the mOlion. Substitute
Vi
=
0,
'1"
=
0, and
l~nr = -J.lk l1X i:
Set
11
= mgand solve for 11/
v,i =
vI = \
'1 =
k
'I
, - - 9"1
'LOa x 10'N/1ll ( 500 kg
)" ()( ")( ) 0.0501 - - 2 0.150 9.8001/5- 0.050 m
0.230 m/s
...... Remarks
. Friction or drag from immersion in a fillicl clamps the motion of
ol~jeCl
~
attached to a spring, c\'cntually
bringing the objecllO rest.
QUESTION 5.9 In the case o[ friction, what percent or the mechanical energy \vas lost by the lime [he mass first reached the equilibrium point? (Hilll: use the answers to parts (a) and (c).)
EXERCISE 5.9 Suppose the spring system in the last example starts at x = a and the attached object is given ::1 kick (0 the right, so it has an initial speed of 0.600 m/s. (a) "Vhat distance ("rom the origin does the object travel before coming to rest, assuming the surface is frictionless? (b) 1-I00v does the allS'wer change if the coeffi,cielll of kinetic friction is i-Lk = 0.1.50? (Use the quadratic formula.) Answer
(a) 0.067 I In
(b) 0.051 2
In
EXAMPLE 5.10 Circus Acrobat Goal Use conservation or mechanical energy to solve a one-dimensional problem involving gravitalional potential energy and spring potential energy.
Problem A 50.0-kg circus acrobat drops from a height of 2.00 rncters straight down onto a springboard with a force constant ofS.OO X 10 3 N/m, as in Figure 5.22. By what maximum distance c10es she compress the spring? Strategy Nonconservative forces arc C1bsent, so conservation of mechanical energy can be applied. At the two points of interest, the acrobat's initial position and the point of maximum spring compression, her velocity is zero, so the kinetic energy terms will be zero. Choose)' = 0 as the point of maximum compression, so the Rnal gTavilational potential energy is zero. This choice also means that the initial position of the acrobat is )Ii = Ii. + d, 'where h is the acrobat's initial heigl11 above the platform and d is the spring's maxiIllum compression.
T
---+ "
d
--~
(a)
(h)
FIGURE 5.22 (Example 5.10) An acroh:ll dn>ps oillu a springboard, causil1g II 10 compress.
5.4
Spring Potential Energy
139
...... Solution Use
conscrv~1 ion
of mechanical energy:
The 0111\' nOlllcro terms are the initial gravitational potential energy and the final spring pOlcmial
+ 1'/;. +
(I)
(KJ...
o+
lIIg(h
+
d) ~ 0 = 0
IIIg ( h
energ}.
PE,l, = (KE
+
+0
+
P/:~
+ I'E)/
- : kd'
=, kr/-
I'
d)
SubstillllC the gi\'cn quantities and rearrange the eqwuion into standard quadratic form:
(50.0 kg)(9.80 m/s')(2.00 m + d)
Soh"c \,"ith the quadratic formula (Equation A.8):
d =
= ~ (8.00
X 10' N/m)d'
t/' - (0.123 m)d - 0.245 m' = 0
0.560 m
........... Remarks The other solution, d = -0.43i 111, can be rejected because r1was chosen to be a positi,"c number at the outsel. A change in the acrobat's center of l1)aS5, say. hy crollching as she makes contact with the springboard, also affects the spring's compression, bUllhal effeCI was neglected. Shock absorbers often involve springs. and this example illustrates hov~' the)' work. The spring action ora shock absorber turns a dangerousjoll into (l smo01.h deceleraLion, as excess kinetic energy is convened to spring potential energy
EXAMPLE 5.11
QUESTION 5.10
Is it possible for the acrobat 1.0 rebound to a height greater than her initial height? Ifso, how~ EXERCISE 5.10
An 8.00-kg block drops straight down f,"om a height of 1.00 m. striking a platform spring having force constant 1.00 X 10:t I\/m. Find Ihe maximum compression of the spring.
Answer
d = 0.482 m
A Block Pro·ected u a Frictionless Incline
Gool Use conservation or mechanical energy to solve a problem involving gravitational potential energ-y, spring potential energy, and a ramp.
Problem A 0.500-kg block rests on a horizomal, rriclionless slIrface as in Figure 5.23. The block is pressed back ag-ainst a spring hewing a constalll of k = 625 N/m, compressing the spring by 10.0 em to point@. Then Ihe block is released. (a) Find the maximum distance d the block tra\'els up the frictionless incline if 8 = 30.0°. (b) I low fast is the block going when halfway 1.0 its maximum height?
Apply conservation of mechanical energy: Substil ute x/= 0:
TIl
=
1] =
0,)'/ =
0, Yr = h = d sin 8, and
~hx,'2 = mgh
=
mgd sin 0
:(625 N/m)( -0.100 ml'
Solve lor the distance d and insert the known \allles:
(0.500 kg)(9.80 m/s') sin (30.0°) 1.28 m
Chopter 5
140
Energy
(b) Find the velocity at half the height, 1J./2. Not.e lhal Ii = dsill = (1.28111) sin 30.0' = 0.640 m.
a
Usc energy conservation again: Take
11;
= 0, J'i
= 0, J; = t h, and -'1=
Multiply by 21m and solve for
v/
0, yielding II -x·= 111, I 'J
')
+f!h '-'
1J'f~
If
.)
-xj--gh III
( V;
....
=
625 •
N/m)
O.~OO kg
,
,
(-0.100111)- - (9.80 m/,-)(0640 m)
2.50 rn/ s . •..•.• • A
Remark Nocice that it wasn't nccessar)' to compute the velocit}' gained upon leaving the spring: only the mechanical energy at each of the two points ofintercSl was required, where the block was at rest. QUESTION 511 A real spring \vi]] continue to \'ibr<-He slightly after the mass has left it. Huw would this affeCllhe answer 10 part (a), and why? EXERCISE 5.11 A ] .OO-kg block is shm horizol1lally frorn a spring, as in the previous example, and travels 0.500 m lip along a frictionless ramp before coming t.o rest anel sliding back e1m.. ·n. If the ramp makes an angle of 45.0 0 with respect to the horizont.al, and the spring was originally compressed by 0.120 m, find the spring constant.
Answer
481 N/m
APPLYING PHYSICS 5.2
ACCIDENT RECONSTRUCTION
Sometirnes people involved in automobile accidents make exaggerated claims of chronic pain due co subtle injuries to the neck or spinal column. The likelihuod of injury can be determined by finding the change in velocity of a car during the accident. The larger the change in velocity, t.he more likely it is that the person suffered spinal il~ur)' resulting in chronic paill. How can reliable estimates for (his change in \'elocity be found after the fact? Explanation The metal and plastic of an
Imv speeds. At greater relative speeds there will be permanent damagc to the vehicle. Despite the fact the struclllre of' the car may not return to its original shape, a certain force per meter is still required to deform il,just e:ts il takes a certain lorce per meter to compress a spring. The greater the original kinetic encrgy, the more the car is compressed during a collision, and the greater the damage. By using data obtained throllgh crash tests, it's possible to obtain eff'ective spri ng consta n ts lor a II I he different models of cars and detcrmine reliable estimates of the change in \'elocity of a given vehicle during an accidcnt. Medical research has established the likelihood of spinal injury for a gi\'cn change in \'clocity, and lhe estimat.ed velocity change can be used to help reduce insurancc fraud.
5.5
5.5
Systems and Energy Conservation
SYSTEMS AND ENERGY CONSERVATION
Recall that the work-energy theorem can be wriucn HllIC
+
,IS
W, = 6.KE
where l'V"c represents the \'lork done by nonconservat.ivc forces and lVc is the work done by conservati\'e forces in a given physical context. As \VC have..: seen, any work done by conservative forces, such as gravity and springs, can be accounted for by changes in potent.ial energy. The work-energy theorem can therefore be wrinen in the followi ng way:
w;" = tJ.KF + tJ.PE =
(KEf - KE,)
+ (1'1; - FE,)
[5.19]
where now, as previously stated, PI:' includes all potent.ial energies. This equation is easily rearranged to: H~" =
(KI'f + PI9 - (KE, + PE,)
[5.20]
Recall, however, that the Lotal mechanical energy is given by E = KH + PI:', rVlaking this substinJtion into Equation 5.20, we llnd I hat the work done on a system by all nonconservative forces is equal to the change in mechanical energy of that system:
[5.21] If lhe mechanical energy is changing. it has to be going somc\",'here, The energy either leaves the systcm and goes into the surrounding environment, or it stays in the s}"'stem and is convened into a non mechanical form such as thermal energy. A simple example is a block sliding along a rough surface. Friction creates thermal energy, absorbed partly by t.he block and partly by lhe surrounding cnviron~ mcnt. \"'hen the block warms up, something called inlrmal energy increases. The internal energy of a system is relat.ed 1.0 its temperature, which in turn is a consequcnce of the activity of its parts, such as the moving atoms of a gas or the vibral ion of atoms in a solid. (Internal energ)' will be studied in more detail in Chapter J2.) Energy can be transferred between a nonisolated system and its clwironmenL If positive ,,,"ork is done on the system, energy is transferred from the cn\"ironmentLO 1 he system. If negative work is done on the s}'stem, encrl=D' is transferred from the SY'stCIll to the environment. So f~lr. we have encount.ered three methods of storing energy in a system: kinetic energy, potential energy, and internal energy. On the mher hand, 'we've seen only one way of tranSferring energy into or out. of a system: through ,.. ' ork. Olher methods will be studied in later chapters. but are summarized here: • Work, in the mechanical sense of this chaptel', transfers energy to a system by displacing it with all applied force. • Heat is the process of transferring energy through microscopic collisions bel.\,,·een at.oms or molecules. For eX<1ll1ple, <1 rnctal spoon resting in a Clip of cofTee becomes hot hecause some of the kinetic energy of the molecules in thc liquid conce is transl'erredto the spoon as il1lernal cnergy. • Mechanical waves transfer energy by creating a disturbance lhat propagates through air or another medium. POI' example. energy in the form of sound leaves your stereo system through the loudspeakers and enters your cars to stimulate the hearing process, Other examples of mechanic",j waves are seismic v.'aves and ocean waves. • Electrical transmission transfers energy through eleclric currents. This is how energy enters your stereo system or any 01 her electrical de\·ice. • Electromagnetic radiation transfers energy in the form of electromagnetic waves stich as light, microwaves, and radio 'Naves. Examples of this method of transfer include cooking a pot.ato ill a microwave oven and light energy travel\ng hon\ t\1e Sun to lhe Earth t.hrough space.
141
142
Chapter 5
Energy
APPLICATION
Conservation of Energy in General
• Flagellar Movement; Bioluminescence
The most important feature of the energy approach is the idea that energy is conserved; it can't be created or destroyed, only transferred from one form into another. This is the principle of conservation of energy. The principle of conservation of energy is not confined to physics. In biology, energy transformations take place in myriad ways inside all living organisms. One example is the transformation of chern ical energy to mechanical energy that causes nagclla La move and propel an organism. Some bacteria use chemical energy to produce light. (See Fig. 5.24.) Although the mechanisms that produce these light emissions are not well undersLOod, livi ng creatures often rely all this light for their existence. for example, certain fish have sacs beneath their eyes filled \.. . ith Iightemitting bactel"ia. The ernilled light attracts creatures that become food [or the fish.
FIGURE 5.24 Thl." small plillll, round in wanll ~ollLherll waLcrs. exhibits hiolullIim'scellCe,:l process in which chemiC:l1 cnergr i~ convened La fight. The red llre:lS lIre cl110roph)l!, which glows when cxcilcd hy blue lighl.
APPLYING PHYSICS 5.3
QUICK QUIZ 5.4 A book of mass m is projecLed with a speed v across a horizontal surface. The book slides until it stops due to the friction force between the book and the surface. The surface is now tilted 30°, and the book is pr~jected lip thc surface ,vith the same initial speed v. \\-'hen the book has come to rest, how does the decrease in mechanical energy- of the book-Earth system compare with that \vhen the book slid over the horizont.al surface? (a) [t.'s the same. (b) It's larger on the t.ilted surface. (c) ft's smaller on the tiltcd surface. (d) More information is nceded.
ASTEROID IMPACT!
An asteroid about a kilometer in radius has been blamed for the extinction of the dinosaurs 63 million rears ago. I-Iow can a relatively slllall object, which could fit inside a college campus, inniet such injury all the vast biosphere or the Earth? Explanation While such all asteroid is compara~ tively small, it travels £II. a very high speed relative to the Earth. Lypically all Lhe order or 40000 m!s. A roughly spherical asteroid olle kilometer in radius and made mainly of rock has a mass ofapproximatcly 10 trillion kilograms-a small mountain of mattcr. The kinetic energy of such an asteroid would be about 1022.J, or 10 billion trillion joules. By contrast, the atomic bomb I.haldevaslated Hiroshima was equivalent to 15 kilotons of TNT, approximately 6 X 10]:3.1 of energy. On striking the Earth, the asteroid's enormous kinetic energy changes into other forms, such as thcrmal energ)', sound, and light, with a total energy release greater than 100 million Hiroshima explosions~ Aside rrom the devastation in the immediatc blast area and fires across a continent, gargantuan tidal wavcs ,,,,,ould scour low-lying regions around the world and dust would block the sun for decades. For this reason, asteroid impacts represent a thrcat to life on Earth. Asteroids large enough to cause
FIGURE 5.25
ASLt;J"oid maporlhe innersol
S}stClil.
widespread extinction hit Earth only every 60 million years or so. Smaller asteroids, of" sufflcicnt size to cause serious damage to civilization on a global scale, are thoughtlo st.rike every five to ten thousand years. There have been several ncar misses by such asteroids in the lasl ccntury and even in the last decade. In 1907, a small asteroid or cornel fragmenl struck Tunguska, Siberia, annihilating a region 60 kilometers across. Had it hit northern Europe, millions of people might have perishcd. Figure 5.25 is an asteroid map of the inner solar system. More asteroids are being discovered e\"ery ycar.
5.6
5.6
Power
143
POWER
The rate at which energy is transferred is imporlant in the design and lise of practical dC\'iccs, such ;,15 electrical appliances and engines of all kinds. The issue is particularly interesting for living- creatures bCCCHlSC the 1l1 ..lximUlll \-"ark per second, or pm'\'cr output, of an animal varies greatly with output duration. Power is defined as the rate of energy transfer with time: Lf an cXLCrnal force does ,,,·ork \tV on an object in the time interval uf, then the average power delivered to the object is the work done divided by the time interval, or
-
w
'!I'=-
f- Average power
[5.22]
D.l
81 unit: watt (W = ]/s)
It's sometimes useful ro rewrite Equation 5.22 by substitllting ,'V = F.1x and noticing 1haL ux//J.t is the average speed of the object during lhe Limc D./:
-
W
FD.x
D./
D./
~=-=--=Fv
[5.23]
According to Equation 5.23, average pmvcr is a constam fUl-cc times the average speed. The force Fis the component of force in the direction of' the average vclocilY. A more general def-inition, called the instantaneous power, call be written dmv'n with a liltle calculus and has the same form as Equation 5.23: ~f'
= Fv
[5.24]
f- Instantaneous power
Til Equatioll 5.24 both thc force F and the velocity v must be parallel, but can change with time. The 51 un it of power is the joule/sec, also called the waH, named artcrJames \lVall:
IW Th.:
.~' ~ower
= IJIs = 1 kg-m'/s"
[5.25a]
in the U.S. customary system is the horsepower (hp), where
hp
ft -Ib = 550-= s
746 W
[5.25b1
The horsepower was first defined by V\'all, who nceded a large power unit to rate the power Outpul of his new invent ion, the steam engine. The walt is cOllllllonly used in electrical applicalions, but it Gill he lIsed in other scientific areas as well. For example, European sports car engines are nl1ed in kilowal.ts. In electric power generation, it's customary to lise the kilm"att-hour as a IT1eaSllre of energy. Olle kilmv
5)
= 360 X 10" J
It's important to realize that a kilowatt-hour is a unit or energy, Tlol pmver.\Vhen yOiI pay you r electric bi II, yOll 're buyi ng energy, and that's why you r bi II Iists a charge 1'01' electricity orabout 10 ccnts/k\Vh. Thc alllount ofelcC1.ricit)' IIsed by an appliance can be calculated by rnultiplying its power rating (usually expressed in \.. . atls and valid only for llOrmal household electriGll circuits) by the length of'timc the appliance is operated. For example. an electric bllib rated al 100 \IV (= 0.100 k\V) "consumes" 3.6 X 10:'J of energy in I h.
TIP 5.6 Watts the Difference? Don't confuse the llollitalic symbol for watts, W, with lhe ilalic symbol Wfnr work. A Walt is ,1 un it, the same
144
Chapter 5
Energy
EXAMPLE 5.12 Goal
Power Delivered b an Elevator Motor
Apply the force-Limes-velocity definition ofpm.. . er.
Problem A 1.00 X 1O"-kg elevator carries a maximum load 01" 8.00 X 10 2 kg. A constant frictional force of -4.00 X 10 3 N retards its motion upward, as in Figure 5.26. \,"hat minimum power, in kilowatts and in horscpO\,,'cr. must the motor deliver to lift the fully loaded elcvaLor at a constant speed of 3.00 m/s? Strategy To solve this problem, \VC need La determine the force the elevator's Illotor must deliver th.sough the force of tension in the cable, T. Substituting this force together with the given speed 11 into 9J = Fv gives the desired power. The tension in the cable, 7~ can be [ound with Newton's second law.
,..
+
! FIGURE 5.26 (Example 5.12) The mOlor exerts an upward fonT T on the clcvaLOr. A frictional force 7 and the r"rce of gravil~' Alg acL dO\\'llI,'ard.
T
.
Solution Apply NeWLOn's second law
. to
~
"'~ ~ L.JF = ma
the elevator:
~
~
~
The velocity is constant, so the acceleration is zero. The forces acting.. . .on the clevat~r are the force of tension in the cable, T, the friction f, and gravity \vhere M is the mass of the elevator.
T+f+Mg=O
'i\Trite the equation in terms of its components:
T-.r- Mg= 0
Solve this equation for the tension Tand evaluate it:
T~.r+
Jvr-g,
Mg
= 4.00 X 10" N + (1.80 X 10' kg)(9.80 m/s')
T= 2.16 X lO":-J Substitute this value of Tfor Fin the pmver equation:
lrrrr.... • • • • • • • • • • • • • • • • • •
?i'
= Fv =
?i'
= 64.8 kW =
(2.16 X 10' N)(3.00 I11/S) ~ 6.48 X ]0' IV 86.9 hp
.
,
...
Remarks The friction force acts 10 retard the motion, requiring more power. For (\ descending elevator, the friction force can actually reduce the po\'.'er requirement.
QUESTION 5.12 In general, are the minimum power requirement.s of an elevator ascending at COnSlalll \'elocit)' (a) greater than, (b) less than, or (c) cquallo the minimum pav,lcI' requirernents of an elevator descending at. constant velocity? EXERCISE 5.12 Suppose the S
Answer
4.09 X lO'W = 54.9 hp
5.6
EXAMPLE 5.13 Goal
Power
145
Shomu 5 rint
Calcliiale the avcrage powcr needed
La
increasc an object's kineLic em:rgy.
Problem Killer whales arc kllo\VIl to reach 32 ft in length and have a mass of over 8 000 kg. They are also very quick, able to accelerate tip LO 30 mi/h in a malleI' of seconds. Disregarding the considerable drag force of water, calculate the average power a killer whale named 5hamu wi111 mass 8.00 X 103 kg would need to generate to reach a speed of ]2.0 m!s in 6.00 s.
Strategy Find the change in kinetic energy of Shalllll and use the \\:ork-encrgy theorem to obtain the minimum work Shamu has LO do to eHect this change. (fnternal and external frinion forces increase the necessary amount of energy.) Divide by the c1apsedtime to get. t.he average power. r' .
. . . . . . .
.
Solution Calculatc the change in Sha1l1u's kinetic energy. By the ~vork-energy theorem, this equals the minimulll \\lork 5hamu must do:
uK/', =
A
•
I ') 2I11V/- -
~
t· 8.00
~ 5.76
Divide by the elapsed time (Eq. 5.22), noting that W= tlKL: ...... . . .
I ') 'j1IlV,-
X J03 kg (12.0 m!s)' - 0
X 10".1
W
5.76 X 10"1
tl/
G.OO s
'f'=-~
~
.
9.60 X 10"' W
.. .
...............
Remarks This is enough power LO run a moderate-sized office building! The actual requirements are larger because of friClion in the \.. . ater and muscular tissues. Something similar can be done with gravitational potential energy, as t he exercise illustrates. QUESTION 5.13 H Shamu COl tid double his velocity in double the time, by what factor ".. . ould the average power requirement change? EXERCISE 5.] 3 "Vhatminimum average power must ([ 35-kg human ho)' generate climbing up the stairs to the top of the Wash ingLOn monument? The trip up the nearly 170-m-lall building takes him 10 minutes. lnclude only \\lork done againstgravilY, ignoring biological inefficiency.
Answer 97 IN EXAMPLE 5.14 Goal
SEeedboot Power
Combine power, the work-cnergy theorem and nonconservative forces with one-dimensional kinematics.
Problem
(a) What average power would a 1.00 X IO'-kg
speedboat nced to go from rest t.o 20.0 m/s in 5.00 s, assuming the \\later exerts a constant drag force o[ magnitudetl = 5.00 X ]0 2 N and the acceleration is constant. (b) Find an expression [or the instantaneous power in terms orthe drag [orceh/> t.he mass 11/, accclcriltion a, and time I. Strategy The power is provided by t.he engine. which creates a nonconscrvative force. L:se the \~'ork-energ-)' ,... .
theorem together wilh Ihe work done by the engine, WFIlj.;illC' and the work done by the drag force, \r\~lr
.,
.
Solution (a) \,Vrite [he work-energy theorem: Fill in the lWO work tern1S and take
Vi
= 0:
To gel the displacement llx, first find the acceleration using the velocity equation of kinematics:
(I)
'1 \"('Ilhrinc
+
Vi = at
I" 'l'dr;l~ -
+
v,
20.0 m!s = a(5.00 s)
I
'!.mvj
-->
,
Vi = at
-->
({. = 4.00 l11/s 2
146
Chapter 5
Energy
Substitute a into the time-independent kinematics equation and solve for Ilx:
that. we know' .1.x. we can find the mechanical energy lost due to the drag force: Solve equation (I) for
-Jd "',. =
H~"", =
50.0 m
- (5.00 X ]0' N)(50.0 m) = -2.50 X 10 4 J
H~ngill";
= ;(1.00 X 10' kg)(20.0
m/s)' - (-2.50 X 1O"J)
H~'llgill(" = 2.25 X -10'').1
H~'ngil1"
_
'l!'=--=
Compute the average power:
~f
2.25 X 10' 5. 00 s
I
. = 4.50 X
[Q"
W
=
60.3 hp
(b) Find a s)'Jl1bolic expression [or the instantaneous power.
Use
~-e\\'toll's
second law:
Solve [or the force exerted by the enginc,
,,~:.:
= J';..l I = (ilia + Ii) (a/) 'l!' = (ma' + aJdJ f
Substitute the expression for FI:, and v = at into Equation 5.23 to obtain the instantaneous power:
'l!'
... . Remarks
...................... In fact, drag forces general I)' gel larger with increasing speed.
QUESTION S.14
How does the instantaneous pmver at the end of 5.00 s comparc to the avcrage powcr? EXERCISE 5.14
What avcr;'lgc power must. be supplied 10 push (\ 5.DD-kg block from restLO 10.0 m/s in 5.00 s ,.. . hen the coefficient of kinetic friction between lhe block and surface is O.2!JO? Assume the accelerat.ion is uniform. Answer
111 W
Energy and Power in a Vertical Jump The stationary jump consists of two parts: extension and free night.~ In the extension phase the person jumps up from a crouch, straight.ening the legs and throwing up the arms; the free-flight phase occurs when thejuillper leaves t.he ground. Because the body is an extended ol~iecl and different parts move with different speeds, we describe the motion of the jumper in terms of' the position and velocity or the center of mass (eM), which is the point in the body ,H which all the mass may be considered to be concentrated. Figure 5.27 5hO\..·s the position and velocit)' of the eM at different stages of the jump. Using the principle of" the conservation of mechanical ellerg)'. we can find II, the maximum increase in height. of the eM, in terms of the velocily vCr..! of" the eM at liftoff'. Taking VRi, the gravitational potential energy of the jumper-Earth system just as the jumper lift.s off from the ground to be zero, and nming that the kinetic energy KI'.fof the jumper at the peak is zero, we han~ KJo.'
TABLE 5.1 Maximum Power Output from Humans over Various Periods
C~!I'
Power
Time
2 hp, or I 500 W 1 hp, or 750 W 0.2 hp, or 150 \·V
6s 60 s 35 min 5h
0.1 hp, or 75 \'\'
8h
0.35 hp, or 260 \\'
(safe daily lev!::! )
Liftoff E.xlension
Fn.:e nighl
\Ve can cs[(mate vOl by assuming that [he acceleration of thc Cl'vl is constant during the extension phase. I f the depth of the crouch is It and the time for exrension is 6.l, we find that Vo.[ = 2v = 2h/ 6.t. Mcasurements on a group of male college students shmv typical values of Ii = 0.40 m and 13.1 = 0.25 s, the lancr value being set by the fixed speed with \... hich muscle can contract. Substituting, we obtain Uc"
~
2(0.40 m)/(0.25 5) = 3.2 m/s
and fI
=
v
., L:\·I
-
2g
=
(3.2 m/s)' ") = 0.52 2 ( 9.80 m/s-
In
Measurements on this same group of students found that I-I was bct\vcen 0.45 m and 0.61 III in all cases, confirming thc basic validity of our simple calculation. To relate the abstract concepts of energy. power, and efficiency to humans, it's interesting to calculatc these values for the vcnicaljul1lp. The kinetic energy given to the body in ajul1lp is KE = ~mvo/, and [or a person of mass 68 kg, the kinetic energy is
KE
= H68 kg)(3.2 m/s)' =
3.5 X 10 2.1
Allhough this may seem like a large expenditure of energy, we can make a simple calculation LO show that jumping and exercise in general are not good ways to lose weight, in spite of their many health benefits. Because the muscles are at most 23% efficient at producing kinetic energy from chemical encrgr (muscles always produce a lot of internal energy and kinetic energy as well as work-that's why you perspire when you work out), they usc up four times the 350.I (about 1 400 J) or chcmical energy in onejuI11p. This chemical energy ultimately comes frol11 the food we eat, with energy content given in units of food calories and one food caloric equal to 4200J. So the LOtal energy supplied by the body as internal energy and kinetic energy in a vertical jump is ani)' about one-third ofa fooel calorie! Finally, it's interesting to calculate the mechanical power thal can be generated by the body in strenuous aClivit}' for brief periods. Here we find that "" = KI,' = 3.5 X 10' J
J=
----"-.
11/
0.25 s
1.4 X 10" W
or (1 '100 W)(l hp/7'l6 W) = 1.9 hI" So hUlllans can produce aboul 2 hI' or mechanical power for periods on the order of seconds. Table 5.1 shows the maxiIllUf'n power outputs from humans for various periods while bicycling anel rowing, activities in which it is possible to measure power output accurately.
5.7
147
WORK DONE BY A VARYING FORCE
Suppose an object is displaced along the :I:-axis under the action ofa force F;r; that acts in the .~-clirection and varies with position, as shown in Figure 5.28. The object
APPLICATION
II
Diet Versus Exercise in Weight-loss Programs
148
Chapter 5
Energy
FIGURE 5.28 (a) The work done by lite force COlllpont'1I1 FJor the sIllal1 displaccmellt.::l..:r is 1'~!J.x. which equals lilt" area of the- .~lla(lt:d rectallgle. The tOla] work doni:: for lhe di~placcment
from x, to "'jis approximately t:qual to the sum of I he an.:as of all tllc rectangles. (b) The "'ork dune hy the CompOJll:lll I'~ ell' the \'af) ing force as the panicle IllO\CS from x, to xtis exactlyequ,d to the area under the curve shown.
Area
r,
.6.A =
I'~ fix
\
r,
(b)
is displaced in the direction of increasing x [rom x = :\'i to x = :\"1" In such a siwation, we can't use Equation 5.1 !£ calculate the work done by the force because this relationship applies only wben F is constant in magnitude and direction. l-Tmvcver, if we imagine that the ol~ject undergoes the small displacement L\.xshown in Figure 5.28a, then lhe x-component Fx of the force is nearly constant oyer this illlerval and we can app,"oximale the work donc by the force [or this small displacement as [5.26]
This quantity isjust the arca of the shaded rectangle in Figure 5.28a. Ifwe imagine that the curve of I'~. versus x is divided into a large number of" such intervals, then the total "lOrk done for the displacement from :\:I to x/is approximately equal to the sum of the areas of a large number of small rectangles: [5.27]
-F,
F,,,,,
1'\ow imagine going through the same process with twice as many int.ervals, each half the size of the original .:lx" The rectangles then have smaller widths and will better approximate the area under the curvc. Continuing the process of increas· ing the number of intervals 'while allowing their size to approach zero, the number or terms in the SUIll increases without limiL, but the value of the sum approaches a dellnite value equal La the area under the curve bounded by I'~ and the x-axis in Figure 5.28b. In other words, the work done by a variable force acting on an objecl thal undergoes a displacement is equal to lhe area under the graph of Fx versus x. A common physical system in which force varies with position consists ofa block on a horizontal, friCtionless surface conneCled to a spring, as disclIssed in Section 5.4. "V hen the spring is stretched or compressed a small distance x from its equilibriulll position x = 0, it exerts a force on the block given by Ex = -/lx, \A/here k is the force constant of the spring. Now let's determine the work done by an external agclll on the block as the spring is stretchcd Vel)' slowly from XI = to Xf = xIlWX ' as in Active Figure 5.29a. This work can be easily calculated by noting th:lt ~t any vtllue of the displacement. r\ewLOn's thiJ] law tells us that the applied force F~,I'I' is equal in magniLUde to the spring force F, and acts in the opposite direction, so thatJ~IPP = -(-/~) = kx. A plot of I\,pp versus x is a straight line, as shown in Active Figure 5.29b. Therefore, the ,vork done by this applied force in stretching the spring from x = 0 to x = X milX is the area under Lhe straight line ill that ngure, which ill this case is the area of the shaded lri~lIlgle:
°
IL-------'-x (b) ACTIVE FIGURE 5.29 (a) .\ block being pulled from x, = 0 to."'1 = xrIn~ 011 a frictionless surface by a r01Tl' F"Pi>" IfLiu: process is carried Ollt vcry slowl)". the applied force is equal in lllagnill1dc ;lIld opposite in direction 1,0 the spring furcf" at all times. (b) A graph of '-:'1'1' \·er.~llS x.
During this same time the spring has done exactly Lhe same amount of work. but lhat ';"'ark is negative, because the spring' force painls in l.he direction opposite the Illation. The potential energy of the system is exactly equal to the work done by the applied force and is the same sign, which is why potential energy is thought of as stored ""ark.
Compute the area of [he large triangle and 'iUbtracl the area of the smaller triangle:
11' ~ ~ ~
kx®"
Hso.o N/m)(0.070 0 m)' -
0.06·) () J
O.l96J - 0.064 OJ 0.132J
..........................
trrr... • • • • •
Remarks Only simple geometries-rectangles and triangles-call be solved exactly with this method. More complex shapes require calculus or the square-coullting- technique in the next worked example. QUESTION 5.15
True or False: \'''hen stretching sprillgs, hal f the displacement requires half as llluch work. EXERCISE 5.15
Ilow much work is required to stretch this same spring from x,
Answer
= 5.00 em
to
x, = 9.00 cm?
0.22'IJ
EXAMPLE 5,16
Estimating Work by Countin
Boxes'--
_
Goal ese the graphical method and counting boxes to estimate the work done by a forcC'. 10.0
Problem Suppose the force applied to slretch a thick piece of clastic changes with position as indicucd in Figure 5.31a. Estimate the work done by the applied force. Strategy To find the work, simply count the number of boxes underneath the cllrve and multiply that number br the area of each box. The curve will pass through the middle of some boxes, in which case only an estimated fractional pan should be coullled.
Solution There are 62 complete or nearly complete boxes under the curve, 6 boxes that are about half' under the curve, and a triangular area from x = 0 m La x = 0.10 In that is equivalent to] box, for a tOlal ofabom 66 boxes. Because the area of each box is O.l0.1, (he lotal work done is approximately 66 X 0.10 J
=
6.6 J.
....
Irl.. • • • • • • •
Remarks Mathematically, there are a number of other methods for creating such esti mates, all involving
QUESTION 5.16 In developing such an estimate, is it necessary for all boxes La have the same length and width? EXERCISE 5.16 Suppose the applied force necessary to pull the drmvslring work done by counting boxes. Answer
5.1
a bow is given by Figure 5.31b. Find the approximate
Abolll50j. (Individual anS\',;ers may var),.)
5.3
Work
The work done on all object by a constalll force is lV~
[5.2]
(FeasOl!;x
where F is the magnitude of the force, .:lx is the o~jecl's displacen~nt, and () is the angle bCl\,,"cen the direction of the force F and the displacement. Si. Solving simple problems requires substituting values imo this equation. More complex problems, such as those involving rr~tion, often require llsing NeWlOn's second law, 1113 = :£F, to determine forces.
5.2
011
Grovitotionol Potentiol Energy
The gravitational force is a conscrvcnive field. Gravitational potenLial energy is another way of accounting for grad lationa I work WIi : lV, = - (Pt.[ - I't.,) = - (mgYl - mg)',)
To find the changc in gravitational potential energy as an ol~jectofmassm moves bet""een two points in a gravitational field, substiwte the values of the o~jeCl's y-coordinates. The work-energy theorem can be generalized to include gravitational potential energy: [5.12]
Kinetic Energy ond the Work-Energy Theorem
The kinetic energy of a body ",,-jlh mass given by
KE
='
~Jl/V~
'In
and speed v
[5.11]
IS
[5.5]
The work-cnerg)' theorcm states that the net ' . . ark done all an object of Illass m is equal 10 the changc in its kinctic cnergy, or [5.6] ,""ork and encrgy oran), kind carry units of joules. Sol .... ing problems involves finding the work done by cach force acting on the object and summing thcm up, ,.. . hich is H~l(:l' followed by substituting known quantitics imo Equation 5.6, soking for the unknown quantity. Conservative forces arc special: \Vork done againstlhem can be rccO\'Cred-i('s conserved. An example is gravity: The work done in lifting an objecllhrough a height is cffectively stored ill the gravity fIeld and can bc recovered in thc kinetic energy of the object simply by lelling it rail. )Ilonconser....ative forces, stich as surfacc friction and drag, dissipate energy in a form that can't bc readily recovered. To account for such forces, the work-energy Iheorem can be rewrillen as
[5.7] where H~l' is the work done by non conservative forces and H~ is the work done by conservat ive forces.
Gravitational work and gravitational potential energy should not both appear in the work-energy theorem at the same time, only one or the other, because thcy're equivalent. Setting the work due 1O nonconservative forces to zero and substiLllting the expressions for KE and PE, a form of the conservation of mechanical energ}' with gravitation can be obt(lincd: 1" -ZI1lV,-
+
mg)'l
I " + mgy/ = 'i//LV/
[5 . 14]
To solve problems \'lith this equation, identify two points in the system-one where information is known and the other ,vhel'c information is desired. Substiwte and sake for Lhe unknown quantity. The work done by othcr forces, as when frictional forces are present, isn't aiways zero. In that case, identify two points as before, calculate the work due to all other forces, and solve for the unknown in Equation 5.12.
5.4
Spring Potentiol Energy
The spring force is conservative, and its potential energy is given by [5.16] Spring potential energy can be put inlo the work-energy lheorem, which then reads
Multiple-Choice Questions When nonconscrvmi,·c forccs arc absent, mechanical energy is conserved.
5.5
o and
\"~'f
5.6
151
Power
Averagc powcr is the amount of cnel"gy transferred di,·ided by the timc taken for the transfer:
-
Systems and Energy Conservation
Thc principle of thc conservation of energy states lhal energy can't be created or dest myed. II. can be lransforIllCc1. bllt the LOLa I energy COlllelll ofcHly isolated system is always constalll. The samc is Lrue [or the universe at large. The work done by all Ilonconserval ive forces acting on a system eqnals the change in the lot~1l mechanical energy of the system: [5.20-21]
where PF:represellts all potential energies present..
W
'!P = {:,/
[5.22]
This expression can also be written
?fi = FiJ
[5.23]
whcre v is thc object's average speed. The unit of power is the watt (W = J/s). To solvc simple problems, substitute gi,·en quantities into one 01" these equations. More difficult problems usually require Gnding the work done on the object using the work-energy theorem or the definition of work.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE·CHOICE QUESTIONS 1. A ,varkel' pushes a wheelbarrow 5.0 m along a level surface, exerting a constanl horizont.al force of 50.0 N. If a frictional force of 43 N acts on the \\'hcclbarrow in a direction opposite to that of the worker, what net work is done on the wheelbarrow? (a) 250.J (b) 215 J (c) 35 J (d) 15J (c) 45.1
2. A skier \caves (\ ski jump at. 15.0 m/s at some angle 8. At what speed is he traveling at his maximum height 0[" 4.50 III above I he level of the end 01" the ski jump? (Neglect air friction.) (a) 11.7 m/s (b) 16.3 m/s (c) 12.2 Ill/S (d) 8.55 Ill/S (c) 17.4 Ill/S
3. A 40.0-N crate sLarting at rest slides down a rough 6.00-m-long ramp, inclined aL 30.0° with the horizontal. The magnitude of the force 01" friction between Lhe craLe and the ramp is 6.0 N. What is the speed 01" the crate al the bouom of the incline? (a) 1.60 m/s (b) 3.32 Ill/s (c) 4.5 Ill/S (d) 6.'12 Ill/S (e) 7.75 m/s 4. \-Vhat average mechanical powcr Illust be delivered by the muscles of a 70.0-kg mountain climber who climbs a summit 01" height 325 111 in 95.0 min? /'v'ol.p: Due LO inefficiencies in cOllverting chcmical energy to mechanical energy. the amount calculated here is only a fraction or lhe pm..· er that Illust be produced by the climbct·'s body. Sec Chapler 12. (a) 39.1 W (b) 5'1.6 W (c) 25.5 W (d) 67.0 W (e) 88.4 W 5. The 't\'ork required to acceleraLe an ol~jecl on a frictionless surface from a speed v to a speed 2v is (a) equal to the work required to accelerate lhc ol~ject from v = 0 La lJ, (b) twice the work required 1.0 accelcrau.: the object from lJ = 0 to v, (c) three times the work required to acceleratc the ol~jcct [rom v = 0 to v, (d) four timcs Lhe work required to accelerate the object from 2v to 3v. or (e) nOl known without knowledge of the accelerat ion. 6. Alex and .John are loading identical cabinets onto a I rllck. Alex lifts his cabinet straight up from the ground LO the bed of" the truck, whereasJohn slides his cabinet
lip a rough ramp
1.0 the truck. \o\'hich statement is correct? (a) Alex and John do the same amount of ,,,ork. (b) Alex does more \'mrk than John. (c) John does more work than Alex. (d) None of these statements is necessarily true because the force of friction is unknown. (e) None of these statements is necessarily true because t.he angle of the incline is unknown.
7. Mark and David arc loading identical cement blocks onto David's pickup truck. Mark lifts his block straight lip from the ground to the truck, whereas David slides his block tip a ramp on massless. friClionless rollers. \Vhich sLatement is true? (a) Mark does more work lhan David. (b) Mark and David do the same amount of work. (c) David does more work than Mark. (d) None of these statements is necessarily true because the angle or the incline is unknown. (e) None of these stalements is necessarily true because the mass of one block is not given. 8. An <\thlete jumping vertically on a trampoline leaves the surface with a ,·e\ocity of 8.S m/s upward. \>\That maxilllum heighL does she reach? (a) 13 III (b) 2.3111 (c) 3.7 m (d) 0.27 rn (e) The allS'Wer can't be determined because the Illass of the athlete isn\ given. 9. A certain Lruck has twice the mass of a car. BOth are moving at the same speed. If the kinetic energy of lhe truck is K, what is the kinetic energy of the car? (a) KI4 (b) K/2 (c) 0.71 K (d) K (e) 2K 10. If the speed ofa particle is doubled, what happens La its kinetic energy? (a) It becomes fOllr times large.,!:. (b) II. becomes two Limes larger. (c) It becomes Y2 times larger. (d) It is unchanged. (e) [t becomes halfas large. 11. rf the net work done on a panicle is zero, which of the following statemcllls must be tfue? (a) The velocity is Lero. (b) The vclocity is decreased. (c) The velocity is unchanged. (d) The speed is unchanged. (e) More informat.ion is needed.
152
Chapter 5
Energy
12. A block of mass 1Il is dropped from the f'ounh floor of an oflice building, subsequently hilLing the sidewalk at speed v. From what noor should the mass be dropped to double that impact. speed? (a) the sixth floor (b) the eighth floor (c) the [emh floor (d) the (\vell'th floor (e) the sixteenth 1I00r
13, A car accelerates uniformly from rest. When does the car require the greatest power? (a) v..·hen tbe car first accelerates from rest (b) just as the car reaches its Jllaxi~ lI1urn speed (c) when the car reaches half it.s maximum speed (d) The question is misleading because tbe power required is constant. (e) ~,lore informat.ion is needed.
CONCEPTUAL QUESTIONS 1. Consider a lug-of-war as in Figure Q5.1, in which t\\'o tcams pulling on a rope arc evenly matched so that no motion lakes place. Is work clone on the rope? On the pullers? On the ground? Is work clone on an)'lhing?
demonstrator be safe i I' the ball were given a push from its st.arting position at her nose?
FIGURE Q5.6
FIGURE OS.1
2. Discuss whether any work is being done by each or the following agents and, if so, whether the \vOl-k is positive or negative: (a) a chicken scratching the ground, (b) a person studying, (e) a crane Ii hi ng a bucket or concrete, (d) the force of gravity on the bucket in part (c), (e) the leg llluscies of a person in the act of siuing down. 3. I r the height of a playground slide is kept constant, will the length of the slide or whether it has bumps make any difference in the final speed of children playing on it? Assume thaI the slide is slick enough to be con sidereel frictionless. Repeat rhis question, assuming that the slide is nor frictionless.
7. As a simple pendulum s\\'ings back and fonh, the forces acting on the suspended object are the force of gravity, the tension in the supporling corel, and air resistallce. (a) \t\'hich of t.hese forces, if any, does no work on the pendulum? (b) \Vhich of these forces does negative work at ClII times during the pendulum's motion? (c) Describe rhe ,vork done by the force of gravit)' \'i-hile the pendulum is swinging.
8.111
During- a stress t.est of the cardiovascular system, a patient "..· alks and runs on a treadmill. (a) Is the energy expended by the paticlll equivalent. to the energy of '''l~dking and running on t.he ground? Explain. (b) What effect, if any, does tilting the treadmill llpward have? Discuss.
9. When a punter kicks a football, is he doing any work on the ball while the toe of his foot is in COlllact with it? [s
he doing any work on the ball after it loses comaCL with his toe? Are any forces doing work on the ball while it is in flight?
4. (a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.
10. The driver of a car slams on her brakes to avoid collid-
5. Roads going up mountains are funned into switchbacks, with the road weaving back and forth along the face of the slope such thm there is only a genLle rise on any ponion of the roadway. Does this con figuration require all\' less work to be clone bv an autolllobile climbing' th~ mountain, compared wi{h one traveling on a roadway thm is straight up the slope? 'Why are switchbacks used?
11. A .,.. . eight is connected to a spring that is slispended ver-
6. A bowling ball is sllspended from the ceiling of a lecture hall by a strong cord. The ball is drawn ,nvay fmm its equilibrium position and released from rest at the tip of the demonstrator's nose, as shown in Figure Q5.6. If the demonstratur remains stat.ionary. explain why the ball docs not strike her on its return swing. \Vould this
ing with a deer crossing the highway. What happens to the car's kinetic energy as it comes to rest? tically from the ceiling. Irthe weight is displaced downwa,-d fmm its equilibrium position and released, it ,..·ill oscillate up and down. Ifair resistance is neglened, will the total rnechanical energy of the system (weight plus Earth plus spring) be conserved? Ho\\! many forms of potential energy are there for this si!llation? 12. The feet of a standing person of mass m. exert a force equal to '/IIg 011 the noor, and the floor exerts an equal and opposite force upwardS on the feet, which we call the normal force. During the extension phase of a vertical jUlllP (see page 147). the feel exert a force on the 0001' that is greater than mg, so the normal force is greater I han mg. As rou learned in Chapter 4, we can
Problems LIse this result ancll\'ewton's second 1m... La calculate the acceleration of the jumper: a = F;lt:/m = (11- mg)/m. Using energy ideas, we knm...· that work is performed on the jumper to give him or her kinetic energy. But the normal force can't perform any \\'ork here because the Jeet don't undergo any displacemcnt. How is encrgy transferred to the jumper?
13. Sllppose you are reshelving books in a library. You lift a book from the floor to the top shelf. The kinetic energy of' the book on the floor was lero, and tht: ki netic energy the book on the top shelf is lero, so there is no change in kinetic energy. Yet you did some work in lifting the book. Is the work-energy theorem violat.ed?
or
153
14. In Tllost situations we have encottntcl-ed in this chap-
ter, frictional forces lend
[Q
reduce the kinetic energy
of an object. However, frictional forces Gln sometimes increase an object's kinetic cnergy Describe a fe-w situations in which rriclion causes an increase in kinctic cnergy. 15. An Eanh satellite is in a circular orbit. at an allitude of ?lOa km. Explain why the work done by the gravitational force acting on the satellitc is zero. Using the workenergy theorem. what can you sa}' about the speed of 1.he satellite?
PROBLEMS ~
1"01' this chapter ma~' be as~iglll.:d ol\lilll: at \'\'ebA.~sig-ll. I. 2. :~ = straightforward, inh"l"llWciiall", clnlkll~ill~ = dCl10tes guided problt:m ~ = denotes enhanced conlenl problem
•
The Problems
WebAsstgn
m
II =
biomedical applicalion
0= denotes lull solution aV:lilab1c in
StUr/I'111 So!utions Monuolj
.\'ludJ Cui/If'
SECTION 5.1 WORK 1. A weight lifter lifts a 350-N sci of weights from ground level to a position over his head, a vertical distance of 2.00 m. How lnuch work does the weightliftc!" do, aSSUllling he moves the '...· eights <11 constant speed?
2. In 1990 Waller Arfcllille of Belgium lifted a 281.5-kg objecllhrough a distance of 17.1 em using ollly his teeth. (a) J low much work did Arfcuillc do on the...: object? (b) ',Vhat magnit.ude force did he exert on the object during the lift, assuming the force was cOnSlaJll? 3. 'rhe record number or boal lifts, including the boat and its Len crew members, was achieved by Sami Heinonen andJuha R,isanen of Sweden in 2000. They lifted a lotal mass or 653.2 kg approximatc...:ly 4 in. oil lhe ground a total of 24 times. Estimate 1I1c total mechanical work done by the two men ill Ii ft; ng (he boat 24 1.i meso assu m· ing they applied the same force to the hoat during each lift. (~'eglecl any \V'ork they may have done allowing the boat to drop back to the ground.) 4.
5.
e
supermarket pushes a carl witll a force of 351\ directcd <11 an angle or 21)° helmv the horizontal. The force is just sufficient to overcome \'arious frictiollal forces, so the carl moves at constant speed. (a) Find the work done by the shopper
mm
A shopper in
Starting from rest, a 5.00-kg block slides 2.50111 down a rough 30.0° incline. The coefficient of kinetic ["riClion between the block and the incline is JLk = 0.436. Deter-
mine (a) the \vol·k done by the force of gravity. (b) the \vork done by the friction force between block and incline, and (c) the work clone by the normal force. (d) Qualitatively, how would 1he answers change ira shorter ramp at a Sleeper angle were used to span the same w:rtical height? 6. A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 111 on a rough horizontal surface. If the cral.e moves at constant speed, find (a) the work done by the 150-N force and (b) the coefficienl of kinetic friction between the crate and surface.
rnA sledge loaded with bricks has a LOtal mass of" 18.0 kg and is pulled at constalltspeecl by a rope inclined at. 20.0° abo\'e the hori7omal. The sledge moves a distance of ~O.O m on a horizontal surface. Thc coefficient of kinetic friction bctween the slcdg'e and surface is 0.500. (a) \Vhat is the tension in the rope? (b) How much work is done by the rope on the sledge? (c) \·Vhat is the ITlechanical energy lost due La friction? 8. A block of mass 2.ESO kg is pushed 2.20 m along a friction-
less horizontal table br
SECTION 5.2 KINETIC ENERGY AND THE WORK-ENERGY THEOREM 9. A mechanic pushes a 2.50 X I03-kg car from rest to a speed of' v, c10i ng 5 000 .J of work in the process. Du r-
ing this time. the car moves 25.0 m. Neglecting friclion between car and road, find (a) vand (b) the hori:tolllal force exerted on lhe car.
10. A 7.0D-kg bowling ball moves at 3.00 m/s. How fast must a 2,45-g Ping-Pong- ball movl' so thaI the twO balls havc the same kinetic enerb'Y'? 11. A 5.75-kg- ol~jeet is initially moving so that ilS x-component
of vclocitr is 6.00 mh, and its y-component of velocity is -2.00 m/s. (a) What is tile kinetic energy of the object at this time? (b) Find the t.:lwlIKe in kinetic energy of the ohject if its velocity changes so that its new x-component is 8.GO m/s and its new y-componellt is 5.00 m/s.
154 12.
Chapter 5
mil
Energy
A worker pushin~ a 3.=J,O-kg woodell crate at a con-
docs 350.J of work by applying a constant horizontal force or maglliSlant
speed for 12.0 rn along
wooel rIoor
wde 1';1 on the crate. (a) Determine the value of I';). (b) If
the worker now applies a force greater than J~I' describe the subsequent motion of the cralC. (c) Describe what would happen LO the crate irthe applied rOITC is less than F(]. [}I]A 70-kg base runner begins his slide illlO second base ,.. . hen he is moving at a speed of 4.0 m/s. The cocrrtcienl of friction between his clothes ;\nel Earth is 0.70. He slides so that his speed is zerojusl as he reaches the base. (a) I-1m.. . much mechanical energy is lost dllc to friction '!CLing on the nlllner? (h) How f~tr docs he slide? 14. An olltfielder throws a O.150-kg bascball at a specd of 40.0 m/s and an initial angle of~O.O°. What is the kinctic cnergy or Lhe ball at the hi~hest point or its mOlion?
15. A 7.80-g bullet moving at 575 m/s penetrates a 1rce trunk to a depth of 5.50 Cln. (a) Use work and cnergy considerations to find the avc.:rage frictional force Ihat SLOpS the bullet. (b) Assuming the frictional force is constant, determine how mtlLh time elapses between the momCllL the bullet enters the tree and the moment it stops moving. 16. A O.60-kg particle h,ls a speed of' 2.0 m/s at point A and a kinetic energy of 7.5 J at poim B. \-Vhm is (a) its kinetic energy at A? (b) Its speed <11 poim B? (c) The total ''''ork done on the particle;\s it mm"Cs from A to B?
17. A 2 OOO-kg car moves down a level highway tinder the actions of two forces: a J 000-:\ forward force exerted on the drive wheels b>' the road and ,1 950-N resistive force. Cse the work-energy theorem to find the speed of the car after it has moved ,1 distance of 20 m, assuming that it starts from res!.
18. On a rrozen pond, a 10-kg sled is given a kick Lh;-lt imparts 1.0 it all initial speed of V u = 2.0 m/s. The cod"flcient of kinetic friction bet.ween sled and icc is I-Lk = 0.10. Use t.he work-energy theorem to find the distancc the sled moves before coming to rest.
object. When subject to a st.eady acceleration o[ a.SOOg, the object must be located 0.350 cm from its equilibrium position. Find the force constant required [or the spring.
22. _
A 60.0-kg athlete leaps straight up into the air from a ll"ampolinc with an initial speed orD.O m/s. The goal of [his problem is 1.0 lind the maximum height she atl
123.1/\ 2 300-kg pile driver is llsed to drive a steel beam into the ground. The pile driver falls 7.50 III beforc coming illto COIHaCL with the top of lhe bcam, and it drives the beam 18.0 cm fanher imo the ground as it comes to rest. Using energy considerations, calculate the average force the beam exens on the pile driver while the pilc driver is hrotl~ht to res!. 24. A 3.50 X JO~-N child is in a swing thaL is attached to ropes 1.75 m long. Find t.he gravitatiolltll pOlential energy associated with thc child relative to her lowcst posilion when 0 (a) the ropes arc horizontal, (b) the ropes make a 30.0 angle With the vertical, and (c) [he child is at. the bol.lom ol'the circtllar arc. 25. A daredevil on a ll'lotorcycJe leavcs lhe end of a ramp \,..-ith a speed o[ 35.0 rnls as in Figure P5.25. II' his speed is 33.0 mls when he reaches the peak 01' the path, what is the maximulll height that he reaches? Ignore friction and air resistance.
-'-~3.011l/s
SECTION 5.3 GRAVITATIONAL POTENTIAL ENERGY
35.0
':>//--T------
SECTION 5.4 SPRING POTENTIAL ENERGY 19. Find the height from which you would have
a ball so that it would have a speed of9.0 m/sjust bcfore it hits the ground.
h
l
to drop
20. \-\,'hen a 2.50-kg oqjcct is hllng vCl'lically all a ccrlain light spring described by Hooke's law, lhc spring st.retches 2.76 cnl. (a) What is the force constant of the spring? (b) If the 2.50-kg object is removed, how far will the spring streIch if a 1.25-kg block is hung on it? (c) I-low much work must an external agellt do LO stretch the same spring 8.00 crn from its tlllslrelchcd position? 21. An accelerometer ill a control system consists of a 3.65-g objeCl sliding on a horizontal rail. A low-mass spring is connected between the o~jecl and a llange at one end of the rail. Grease on the rail makes SLllic rriclion ncgligible, bm rapidl}' d'imps out vibrations of the sliding
26. Truck suspcnsions oft.en have "helper springs" that engage at high loads. One such arrangement is a leafspring with a helper coil spring mounled on the axle, as shown in Figure P5.26. When the main leaf spring is compresscd by distance )'0' the helper spring engages and then helps 10 support any additional load. Suppose the leaf spring constant is 5.25 X 10" N/m, the helper spring constant is 3.60 X 105 N/m, and Yo = 0.500 m. (a) \\'hat is the compression of the leaf' spring ror a load of !).OO X ]O~, N? (b) How much work is done in comprc~sing the springs?
Problems
155
the spring and oscillaLes freely on a horizontal, frictionless surface as in Active Figure 5.20. The initial goal of this problem is to find the \'clocit), at the equilibrium poinl after the block is released. (a) What objects constitute the system, and through what forces do they interaCL? (b) What are the t\.. .o points of interest? (c) Find the energy stored in the spring when the mass is stretched 6.00 cm from equilibriulll and again when the mass passes through equilibrium after being released from rest. (d) \Vritc the conservation of energy equation for this situation and solve il for the speed of the mass as it passes equilibrium. Substitme to obtain a numerical value. (e)\,\fhat is the speed at the halfway point? \'\'hy isn't it half the speed at equilibrium? 1.0
FIGURE PS.26
27.11 The chin-up is one exercise tbat can be Llsed Lo strengthen the biceps muscle. This muscle can exert a force of approximately 800 N as it comraClS a distance of 7.5 em in a 75-kg malc. 3 I-Jow much work can the biceps muscles (one in each arm) perform in a single contraclion? Compare this amount or work '\lith the cncl"!D' required to lift a 75-kg person 40cm in performing a chin-up. Do you think the biceps muscle is the only muscle involved in pedorming a chin-up? 28. " A nea is able to jurnp about 0.5 m. It has been said th;:H if a nea ,vere as big as a human, it would be able Lo jump over a lOO-sLOry building! When an animal jumps, it convens lVork done in contracting llluscles into gravitational potential energy (with sotTle steps in beLween). The maximum force exerted by a muscle is proportional LO its cross~secLional area, and the work clone by the Illuscle is Ihis force times the length ofcomraction.lrwc m:lgnificd a nea by a factor of I 000, the cross section of its muscle would increase by I 000 2 and the length 01" contraClion would increase by I 000. How high would this ~supernea be able to jump? (Don't forget thm the 111,-155 of the "superflea" increases as well.) 129. )A 50.0-kg projectile is fired al an angle of 30.0° above the horizontal with an initial speed of 1.20 X 10 2 m/s fwm the LOp of a cliff 142 III above level ground, where Lhe Rround is taken to be)' = O. (a) What is the initial LOlal mechanical energy of the projectile? (b) Suppose the projectile is traveling 85.0 m/s at its maximum height of ')' = 427 m. How much work has been dOllc on I he projec\ ile by air friClion? (c) What is the speed of t he projectilc immediately before it hits the ground if air friction does one and a half times as much work on lhc projectile when it is going down as it did when it was going up?
30.
31.
~G.
mil
A prOjectile of mass '11/ is fired horizontally with an initial speed of lJ u from a height of 11 ;Lbove a l1at, desert surface. Neglecting air friction, at the installt befc)I"c the prqjeclile hits the ground, find the following in terms of 111, vo, 11., and g: (a) the work clone by the force of gravily on lhe prqjeClile, (b) the change in kinetic energy of the projectile since it. was fired, and (c) the nnal kinetic cllergy of the projectile. (d) Are any of the answers changed if Lhe iniLial angle is changed?
m
A horizontal spring attached to a wall has (\ force constant of 850 N/m. A block of mass 1.00 kg is attached
P. I'app;\s cl. aI., "Nolluniform shortcnillg: in lht: biccp~ brachii duringdb,)\\' llexi{ll1," .l0llnwl uf A1J/Jfird J'ily~-101{)gJ 92, 2:l81. :!OO;l.
SECTION 5.5 SYSTEMS AND ENERGY CONSERVATION 32. A50-kg pole val/her running at 10 mf,s vaults over the bar. Her speed when she is abovc the bar is 1.0 ntis. NeglecL air resistance, as well as any energy absorbed by the pole, and determine her ,dtilUdc as she crosses the bar. 33. A child and a sled with a combined mass of 50.0 kg slide down a l'riClionless slope. If the sled starts from reSL and has a speed of 3.00 m/s at the boltom, what is the height of the hill?
34. Hooke's law describes a certain lighL spring of un Siretched 1cnRth 35.0 em. When one end is attached to the top ora door frame and a 7.50-kg object is hung from the other end, the length of the spril1g is 41.5 em. (a) Find its spring constant. (b) The load and the spring are takcn down. Two people pull in opposite dircctions on the ends of the spring, each with a force of 190 N. Find the length of the spring in this situation.
35.
am A
0.250-kg block along a horil.ontal lrack has a speed of 1.50 m/5 immediately before colliding with a light spring of force const.ant 4.60 N/m located at the end of' the track. (a) What is the spring's maximum COJll~ pression if the track is frictionless? (b) If the Lrack is 1101 frictionless, ,,'auld the spring's maximum compression be greater than, less tholl, or equal 10 the value obtained in pan (a)?
137.ITarzan swings on a 30.0-m-long vine initially inclined at
v·
39. The launching mechanism of a lOy gun consists of a spring of unknown spring constant,
as shown in Figure P5.39a. If the spring is compressed a
dist.ance of 0.120 m and the gun fin.:d vertically as shown,
x= 0
x
the gun can launch a 20.0-g projectile from rest to a maximum height of20.0 III above the starting poim of the projectile. Neglecting all resistive forces, (
40.
(a) A block with a mass 111 is pulled along a I~rizon tal surface for a distance x b}' a constant force F at an angle 0 with respect 1.0 the horiwlllal. The coefficient of kinetic friction between block and table is ILl:' Is the force exerted by friction equal to J.l-I:mg? Jrnot, what is the force exerted by friction? QJ) How much work is donc by the friction force and by F? (Don't forget the signs,) (c) Ident..if)' all the forces lhat do no work on the block. Cd) Let In::: 2.00 kg, x == ,;1.00 In, 0 = 37.0°, F = 15.0 N, and ILl: = 0.'100, and find the answers to parts (a) and (b).
41.
ma:J
BI
(a) A child slides down a water slide at an amusement park from an initial height h. The slide can be considered frictionless because of the \...· atcr flowing down it. Can the equation for conservation of mechanical energy be lIsed on the child? (b) Is the lTIass of the child a factor in delcrmining his speed at the bOllom of the slide? Cc) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-of-energy equation be modified? (e) Find the maximulll speed of the child when the slide is frictionless if the inilial height of the slide is 12.0 ITI.
42, _ An airplane of mass 1.50 X 10" kg is moving at 60.0 m/s. The pilot then increases the engine's thrust t.o 7.50 X 10' N. The resistive forcc exerted by air on the airplane has a magllitude of 4.00 X 10" N, (3) Is the \,,,'ork done by the engine on the airplanc equal to the change in the airplane's killetic energy aftel it travels thrOl.l~h some distance through the air? Is mechanical energy COI1sen'cd? Explain. (b) Find the speed of the airplane after
it has t.raveled 5.00 X 10 2 m. Assume the airplane is in level flight throughout the motion.
43. A 70-kg diver steps off a 10-m tower and drops from rest slraight down into the water. If he comes to rest S.O m beneath the sUI'face, dctcnnine the average resisth'c torce excrted on him by the water. 44. A 25.0-kg child on a 2.00-m-long swing is released from rcst when the ropes of the swing make all
E!I
A child of mass III starts from rcst and slides without friction from a height 11. along a curved waterslidc (Fig. P5.46). She is launched from a height h/5 into the pool. (a) Is mechanical energy conserved? \Vhy? (b) Gi\'e the gravitational potential energy associated with the child
FIGURE PS.46
14i.IA skier starts from
rest at t.he lOp of a hilllhat is inclined 10.5° with respcctto the horizontal. The hillside is 200 m long, and the coefficient of friction between snow and skis is 0.075 O. At the bottolll of the hill, the snow is level and the coefllciem of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?
48, In a circus performance, a monkey is strapped to a sled and both arc given an initial speed of 4-.0 m/s up a 20°
inclined track. The combined lllass of III on key and sled is 20 kg, and the coefficient of killel ic friction between sled and incline is 0.20. How far up the indine do the monke) and sled move? 49, An SO.O-kg skydiver jumps out of a balloon al an altitude of I 000 III and opens the parachute at an altitude of
Problems 200.0111. (a) A...stlming that the total relardillg force on lhe diver is constant at 50.0 ~ with the parachlile closed anrl constalll at 3600 N with the parachutc opcn, ",hal is Ihe speed of' the diver when he lands on the ground? (b) Do you think the skydiver will gel hun? Explain. (c) AI what height should the parachutc be opent:d so that the final speed of the sk}'diver when he hilS the ground is 5.00 m/s? (d) How realistic is the assumption thm lhe LOtal retal'ding force is COnSL.'1.l1l? Explain.
SECTION 5.6 POWER
157
cruising speed, 1.75 m/s. (a) \Vhat is the average power of the elevator maLO I' during this period? (b) How does this amount of power compare with ils power during an upward trip with constant speed?
SECTION 5.7 WORK DONE BY A VARYING FORCE 159.IThe force aCling all a particle \'aries as in Figure P5,59. Find the work done by the force as the particle mm'cs (a) from x = 0 to x = 8.00111, (h) from x = 8.00 m to x = 10.0 m, and (c) from x = 0 lO x = 10.0 m.
50. A skier of mass 70 kg is pulled III' a slope h)' a motor· dri\en cable. (a) I low much work is "equircd to 1'1111 him 60 III lip a 30° slope (assumed frictionless) at a constanl speed of 2.0 m/s? (b) What power OlU<;t a moLDr have 10 perform this task?
[[h)A 3.50-ki\' piano is lifted by three workers ill constant speed to all apanmelll 25.0 m above the strcel using a pulley s}stcm fastencd to the roof or the building. Each worker is able to deliver 165 W of pm"'er, and the pllllq' s}stem is 75.0% emcient (so that 25.0% of Ihe mechanic•• 1 energy is lost due 10 friction in the pulley). Neglcnillg' the ma\s of the pilI ley, lind the time reqllin:d to lifllhc piano from the SI reet to the apartment. 52.8 While running, :l person dissipates ;-lbout O.liO J or mechanical energy per step per kilogram 01' hody mass. If a oO-kg person develops a power 01"70 W during a race, huw fast is the person running? (Assllme a running- SICP is 1.5 III long.)
-4 FIGURE PS.S9
60. An object of mass 3.00 k1{ is subject 10 a force J'x that val'ics with position as in Figure P5.60. Find the work done by the force on the ObjCCI as it moves (a) from x = 0 to .\. = 5.00 m, (b) from x = 5.00 m to .\" = 10.0 In, and (c) from x = 10.0 m lo x = 15.0 m. (d) If the object has a speed of 0.500 m/s at x = 0, find its speed at x = 5.00 m and its speed al x = 15.0 Ill.
53, The eleclric motor ora modellrain accelerates tile train from rest to 0.620 m/s in 21.0 ms. The tolal mass of the train is 875 g. Find the average power delivered lO lhe train during its acceleration. 3
54. When all automobile moves with constanl speed dm.. . n a highwa),. most of the power developed b) the engine is used to compensate for the mechanical energy loss due lo frictional forces exerted on the car by the air and Ihe road. If the power developed by an engine is 17.1) hp, estimate the LOtal frictional force acting on the car when il is moving al a speed of 29 m/s. One hor'iepower equal.'> 746 \V.
55. An older-model car accelerates frolll 0 to speed tI in 10 s. A newer, more powerful spons car urthe saint· mass accelerates from 0 10 2v in the same timc pel'iod. Assuming th(' energ} coming from the engine appears only as kinetic energy of the cars, compare the power orlhe two car.... 56. A certain min cloud at an altitude or 1.75 km contains :t20 X 107 kg of water \·apor. How long would il lake for ;:1 2.70-kW PUIllP 10 raise the same amount of water from Earth's :-.urface to the cloud's position?
2
--:-1'--::--:--:--:-:-:-:-:--,--'-.,. x (Ill)
o
2
'I
6
8
10 12
14
16
FIGURE PS.60
61.
TheforceaClingonano~ieclisgi\'enbyJ'x=
(8x-16) i\,
where x is in meter~. (a) Make a 1'101 of this force versus x from x = 0 to x = 3.00 m. (h) From your graph, find the net WOl'k done by the force as the object mo\'Cs from x = 0 to x = 3.00 m. ADDITIONAL PROBLEMS
57. A 1.50 X IO·"-kg car slarLs from rest and accelerates uni· IOl"mly to 18.0 m/s in 12.0 s. Assume that air rcsislance rcmains constant at400 ~ during Ihis timc. Find (a) the average power developed by the cngi IlC a nel (b) Ihe inSLalllaneOlis power output of Lhe cngine at t = 12.0 s, just before the car stops acceleraling.
62. A raw egg can be dropped from a third-floor window and land on a foam-rubber pad on the ground withoul breaking. If a 75.0-g egg is dropped from a window located 32.0 m above the gro1lnd and a foam-rubber pad thal is 15.0 em thick stops the egg in 9.20 m~, (a) b) how much is the pad compressed? (h) \'\'hat is the average 10l'ce exened 011 the eg-g after it strikes the pad? .".'ote: Assume constant upward acceleraliol1 as the egg compresses the foam~rubber pad.
58. A 650-kg elevator starts from rest and ])love~ upward 1"01' 3.00 s Wilh constant acceleralion unl il it rcachl'<; its
63. A person doing
Chapter 5
158
Energy
an upward force of 355 N on the lorso. If the upward movement starts from rest, what is the persoll's velocity at that point? 64.
am A boy startS at rest and slides down a frictiunless slide as in Figul'e P5.64. The bottom of the IJ'~ICk is a llf'iglll h above the ground. The boy then leaves the track hori'lomally, striking the ground a distance d as shown. Using energy methods, determine the initial height J I ortllc boy in terms of II and d.
object is released from reSl at a poi11l 4.00 III above the noo ... (a) Determine the speed of each object when the two pass each olher. (b) Uetcrmine I.he spce(! or cach object al. the moment the 5.00kg objecL hit.s lhe floor. (c) llow much higher docs FIGURE PS.71 the 3.00-kg ol~jecl t.ravel after the S.OO-kg ol~jecl hits the floor?
72. Two blocks, A and 13 (with mass 50 kg and 100 kg, respec-
(a) A 75-kg man steps out a window and falls (from rest) 1.0 m to a side",·alk. What is his speed just before his feet strike the pavement? (b) Jr the mall falls with his knees and ankles locked, the only cushion for his fall is an ;Ipproximately O.50-cm give in the pads of" his feet. Calculate the average force exerted on him by t.he ground in this situation. This average force is sufficient La C
;0. A toy gun uses a spring to project a 5.3-g soft rubber sphere horizonlally. The spring constanl is 8.0 N/m, lhe barrel of the gun is 15 cm long, and a constant frictional force of 0.032 N exisls between barrel and pr~iectile. With what speed does the projectile leave tile barrel if the spring was compressed !'l.U cm for this laullch? 71. Two ol~jecLs are connected by ,l lighl string- passing over a lig-ht, frietionless pulley as in Figure P5.71. The 5.0n-kg
100 kg
37° FIGURE P5.72
173.IA 2.00
X 102_g particle is released from res1 at point A on [he inside of a smooth hemispherical bowl of radius R = 30.0 cm (Fig. P5.i3). Calculale (a) its gravilational potential energy at A rchnivc to 11, (b) its kinet.ic energy al B. (c) it.s speed at B, (d) ils pOlcmial energy at C relative to 13, and (e) its kinetic energy at C.
c IJ I
-~
/
FIGURE P5.73
74. _ The particle described in Problem 73 (Fig. P5.73) is released from poinl A at resL Its speed at B is 1.50 m/s. (a) \Nhat. is its kinetic energy at B? (b) How much mechanical energy is lost as a resultoffriclion as the particle goes from .A [0 H? (c) Is it. possible to determine }J. from these resulls in a simple manner? Explain. 75. A lighl spring with spring conS("llll 1.20 X 10 3 :\11m hangs from an elc,,:ued supporL From its lower end hangs a second light spring, which has spring conStant 1.80 X 10 3 N/rn. A l.5D-kg object hangs at rest from the 10\'i"er end of the second spring. (a) Find the lOlal eXlensian distance 01" the pair of springs. (h) Find the effec· Live spring constant. of the pair of sprillgs as a system. We describe these springs as being in series. HIIII: Consider the forces on each spring separately.
Problems 76. _ Symbvlir \h:sion oj Problem 75 A light spring Wilh spring constalll h] hangs from an c!c\"ated SUPP0rl. From its lower end hangs a second light spring-, which has spring constant h'!.. An ol~jec( of mass III hangs 3\ rest from 1he lower end of the second spring. (a) Find the tolal exten· sion distance x of the p
displaccmcllls XI and x'2' (b) Find the cirCCli\'(~ spring constant of the pair of spring-s as a system. \Vc describe these springs as heing in I"trips. 77.llJIln (erms of saving energy. hicycling and walking ,Ire br morc <:,fficicllllllcans of lransport
159
Energy is conventionally measllr('d in Calories as well as ill joules. One Calorie in lILllrition is I kilocalorie, which we define in Chapler 11 as I kcal = ,~ 186.]. Melabolizing I gram or fat can release 9.00 kc-t1. A stllclent decides to Iry [Q lose \\ieight by exercising. She plans to run up
79. A ski jUlllper starts from rest 50.0 m above Ihe ground on a frictionless track and flies olT the track at ,Ill angle 0 of' 45.0 above the horizont,d and al a height or lO.O In above the level ground. Neglect air rcsistc\ncc. (a) Whal is her speed when she leaves the track? (b) What is Ihe maximum altitude she altains arIel' Ie;-wing lhe track? (c) \-"here docs she land relative La the end of the track?
FIGURE P5.81
82.1'1 A hummingbird is able to ho\'er because. as the wings move downwards. they exen a clowJl\varci force all the air. Newton's third law tells us thalthe air exerts an equal and opposite force (upwards) on the wings. The a\'crage of this force lIIust be cqual 10 the weight of the bird when it ho\'ers. If the ,,\'ings move thmugh a distance 01' 3.5 em with each stroke. and the wings beal80 times per second, determine the work performed by the wings on the air in I minUlc if the mass orthc hummingbird is 3.0 grams. the dangerous "sport" of bUllgce jUlllping, a daring studcnLjull1ps ['..om a hOI-air bal10011 with a specially designed clastic cord attached to his waist, as shown in Figure P5.83. The unstretchecl length of the cord is 25.0 m, Ihe student weighs 700 1\\ and t he balloon is 36.0 m abo\'e the surface of a river belm.... Cakulate Ihe required force constant of the carr! i r Ihe stlldcnl is to SLOp safely 4.00 III ahove lhc river.
8:L 111
SO. A S.O-kg block is pllshed3.0 m up a \-ertical wall with constallt speed by a constalltlorce of magnitude Fapplied at an angle of 0 = 30 0 with the hori~ zOlltal, as shown in Figure P3.S0. I r the coeflicielH of kinet ic rric· tion between block and wall is 0.30. delcrmine Ihe \vork done hy (a) F, (b) the force of gravity, and (c) the normal force between block alld wall. (d) By how much does the g-ra\'itational potential energy increase during the block's mOl ion?
tF~_
FIGURE PS.80
84.
mil
FIGURE PS.83 Hungc(' jumping. (Problems 83 and 86)
The masses of the javelin, disclis. and shot are 0.80 kg. 2.0 kg, and 7.2 kg, respeeti\·ely. and record lhrows in the corrcsponding track events arc about 98 Ill, 74 m, and 23 m, respectively. Ncglecting air resistance. (a) c;d-
160
Chapter 5
Energy
culate the minimum initial kinetic energies lhal would produce these throws, and (b) estimate the average force exerLcd on each object during the throw, assuming- the force aCl.~ over a distance 0[2.0 Ill. (c) Do your results suggest that air resistance is an imponam factor? 85. A tHIck travels uphill \vith COllstaJ1l velocity on a highway with a 7.0 0 slope. A 50-kg package sits all the flool" of the back of the truck and docs not slide, due La a slatic frictional force. During an interval in which the truck travels 340 m, what is the net work done on the package? 'What is the work done on the package b}' the force of gr:.·Wily, the normal force, and the friction force?
86. A daredevil wishes to bllngec~jump from a hot-air b
(8i.IA loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 30.0° above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 12.0 s and then continues at constant speed. (3) \'\o'hat power nlllst the winch motor providt..: when the car is moving at const.ant speed? (b) Wllat maximum power must the motor provide? (c) \'Vhat total energy transfers alit of the motor by ",;ork by the time the car moves off the end of the track, which is of length I 250 m? 88.
miJ An object of mass
111 is suspended from the top of a cart by a string of length L as in Figure P5.88a. The cart and object arc initially moving 1.0 the right at a constant speed VII. The can comes 1.0 rest after colliding and Slicking to a bumper, as in Figure P5.88b, and the suspended object swings through an angle fJ. (a) Show that the initial speedislJo= Y2gL(1 cos8). (b) IfL= 1.20mandO= 35.0°, find the initial speed of the cart. (I-linl: The force
(b)
(al FIGURE PS.88
exerted by the string on the object does no \\"ork on the object.) 89. Three objects with masses ml = 5.0 kg, 1/I'l = 10 kg. and U/.~\ = lEi kg, respectively, are attached by strings over frictionless pulleys as indicated in Figlll·e P5.81J. The horizontal surface exerts a force of friction of 30 N all ,,~. if the system is released !'mm rest, usc energy concept'S to find the speed of 1111, after it moves down 4.0 m.
"s
FIGURE PS.89
90. A cafeteria tray dispenser suppons a stack of trays on a shelf t.hat hangs [rom four identical spiral springs under tension, one near each corner of the shelf. Each tray has a mass of 580 g and is rectangular. 45.3 em by 35.6 em, and 0.450 em thick. (a) Show that the top tray in [he stack can always be at the same height above the rIoor, however man}' t.ra}'g are in the dispenser. (b) Find the spring constant each spring should have in order for the dispenser 10 function in this convcnienl way. [s any piece of data unnecessary foJ' th is c1eterm ination?
91.
1'1
In bicycling for aerobic txcrcise, (I woman wants her heart rate to be bel\veen 136 and 166 heats per minute. Assume that her hean r
92. " In a needle biops)', a narrow strip of (issue is extracted from
A smoll
buc~
from the massive
bull transfers a large amount of momentum to the cowboy, resulting in on involuntary dismount.
6.1
Momentum and Impulse
6.2
Conservation of Momentum
6.3
Collisions
6.4
Glancing Collisions
6.5
Rocket Propulsion
MOMENTUM AND COLLISIONS What happens when two automobiles collide? How does the impact affect the motion of each vehicle, and what basic physical principles determine the likelihood of serious injury? How
do rockets work, and what mechanisms con be used to overcome the limitations imposed by exhaust speed? Why do we hove to brace ourselves when firing small projectiles at high velocity? Finally, how can we use physics to improve our golf game? To begin answering such questions, we introduce momentum. Intuitively, anyone or anything that has Q lot of momentum is going to be hard to stop. In politics, the term is metaphorical. Physically, the more momentum on object has, the more force has to be applied to stop it in a given time. This concept leads to one of the most powerful principles in physics: conservation of momentum. Using this law, complex collision problems can be salved without
knowing much about the forces involved during contact. We'll also be able to derive information about the overage force delivered in on impact. With conservation of momentum, we'll have a better understanding of what choices to make when designing an automobile or a moon rocket, or when addressing
6.1
0
golf ball on a tee.
MOMENTUM AND IMPULSE
In physics, momenlUm has a precise definit.ioll. A slowly moving brontosaurus has a lot of momentum, but so does a liltle hot lead shot from the muzzle of a gUll. Vie theret-ore expect that momelltum will depend on an objcCl's mass and velocity. The linear momenlUm p or an object of Illass m moving \-",ith velocity product or its mass and velocity:
p= mv
v is the
~
Linear momentum
[6.1]
51 unit: kilogram-meter per second (kg' m/s) Doubling either the mass or the velocity of an o~jcct doubles its Illomentum; doubling both quantilies quadruples its momenturn. Momentum is a vector quantity
161
162
Chapter 6
Momentum and Collisions
"..lid1 the same direction as the object's velOCily. Irs components arc given in two dimensions by
w}lf'r~
/).," is I he moment II In of the ohjecl in r he x-d i r~c.I-ion the y-direcLion.
The magnitude of the momentum kinetic energy KE,:
;'1 nrl
p of an object of mass
!J
f
irs mOllwntulll in
. 1n
can be related to its
.,
fT
[6.2]
KE=-
2m
This relationship is easy to prove using the definitions of kinClic energy and momentum (see Problem 6.6) and is valid for objects traveling at speeds much less than the speed of lighl. Equation 6.2 is useful in grasping the interplay between the two concepts, as illustrated in Quick Quiz 6.1. Two masses 111 1 and 'IJl'1.' with 1n 1 < III?, have equal kinetic energy. How do the magnitudes of their momenta compare? (a) Not enough information is given. (b) < (c) 1', = 1'2 (d) > 1'2
QUICK QUIZ 6.1
I', I',
I',
Changing the morne[1[UITI of an object requires the application ora force. This is, in fact, hO\\l l\'ewton originally stated his second law of motion. Starting from the more common version of I he second Ia,.. .'. '\Ie have -7
F
-7
IIel
= ma =
!J.v ~ !J.(n£\t)
III -
---
!:it
!:if
\\'here the mass 'IJl and the forces are assumcd constant. The quantity ill paremhe· ses isjust the momcntum, so \,,'C have the follmving result: The change in an object's momentum divided by the elapsed time tit equals the constant net force Fm" acting on the object:
!J.p
change in momentum Newton's second law and momentum -7
=
time interval
.... F!WI
[6.3]
This equation is also valid when the forces arc not constant, provided the limit is taken as t:J.t becomes infinitesimally small. Equation 6.3 says thaI if the net force on an objcct is zero, lhe object'S momentum doesn't change. In mhcr words, the linear momentum of an object is conserved when llPt = O. Equation 6.3 also telts us that changing an ol~ject's momentum requires the cOl1linuous application or (\ force ovcr a period of time at, leading to thc derinition or impulse:
F
....
....
ff a constant force F acts on an object, the impulse [ delivered to the object
over a tirne interval dt. is given by
1 == F!:if
[6.4]
51 unit: kilogram mete!' pe!' second (kg' m/s) Impulse is <-1 vector quantity "vilh the same c.!lrection as the constant force actingon the ol.:~ject. ''''hen a single constant force F acts on an object, Equation 6.3 can be wriucn as Impulse-momentum theorem
-+
[6.5] This is a special casc or the impulse-momentum theorem. Equalion 6.5 shows that the impulse of the force acting on an object equals the change in momentum of that object. This equality is trlle even if the force is IlO\' call stant, as long as the time intcrval !J.I is taken to be arbilrarily small. (The proof of tlie general caoSe requires concepts from calculus.)
6.1
Momentum and Impulse
163
FIGURE 6.1 (3) A force aCling on an vary in lime. The irnpuhc is Ihe area under the force \'s. time ("Ilr\'('. (b) 'I he average force (horilOll I a] c1a!>hed line) gi\Oes the same impulse to the object in !.he time inlenal.11 as the real limev:I.",ing force described in (a).
F
{)1~('C1 ma~
(aj
(bl
In real-life situations, the force on an object is only I-are)y constant. Fur cxam-! pic, when a bat hiLS a baseball, the force increases sharply, reaches some maximulll ~ value, and then decreases just as rnpiclty. Figure 6.1(a) shows a lypical graph of ~ "5 force versus Lime for such incidents. The force stans out small as the bat comes in ~~ contaCl with the ball, rises to a maximum value when they arc firmly in cOIllacL,l ~ and then drops off as thc balllca\'cs the bat. In order LO....,.analyze this rather COIl1- ~ plex interaction, it's useful to define an average force FOIl' shown as the dashed ~ 8 line in Figure 6.1b. This average force is the constant force delivering the same: ~ irnpulse to the object in thc time interval !:J.t as the aCtual timc-varying force. \Ve FIGURE 6.2 An apple being pierced by a 30-caliber bullet tra\'e1illg at a can then write the impulse-momentum theorem as
J i
[6.6]
The magnitude of the impulse delivered by a force during the time intervalli.t is equal to the area under the force vs. time graph as in Figure 6.1a or, equivalently, to F;IV At. as shown in Figure 6.1 b. The hrief coil ision between a bullet and an apple is illustrated in Figure 6.2.
APPLYING PHYSICS 6.1
BOXING AND BRAIN INJURY
In boxing rnatchcs of the 19th century, bare fists \.. ' ere used. In modern boxing, fightcrs wear padded gloves. How do glm'es protect the brain of the boxer from injury? Also, why do boxers oftcn "roll with the punch"~
Explanation The brain is immel-sed ill a cushioning fluid inside the skull. If the hcad is struck suddenly by a bare fist, the skull accelerates rapidly_ The brain matches this acceleration on Iy because of the large impulsive force exerted by the skull all the brain. This large and sudden force (large J~IY and small ~/) can cause severe brain il~jury. Padded gloves extcnd the time!:J.t over which the force is applied 10 the
EXAMPLE 6.1 Goal
~Ilpcrsonic speed of900 m/s. Tlli:. cullision \"'01" photographed wilh
II
head. For a givcn impulse 1~\!'J.t, a glove results in a longer time interval than a bare fist, decreasing the average force. Because the average force is decreased. the acceleration of the skull is decreased, reducing (but not eliminating) the chance of brain injury. The same argument can be made for "rolling with the punch": lfthe head is held stcadywhile being struck, the time inter\"al O\'cr which the force is applied is relatively short and the average force is la'-ge. ff the head is allowcd to move in the same direction as the punch, Lhe time interval is lengthened and the average force reduced.
Teeing Off
Use the impulsc-momentum theorem to estimate thc ~l\'crage force excrted during an impact.
Problem A golf ball with Illass .::'.0 X 10 ~ kg is struck with a club as in Figure 6.3. The force on the ball varies from zero when contact. is made up to some maximum value (when the ball is maximally deformed) and then back to zcro
164
Chapter 6
Momentum and Collisions
when the ball leaves the club, as in the graph of force \IS. time in Figure 6.1. Assume that the baJlleaves the club face with a velocity of +44 m/s. (a) Find the magnitude of the impulse due to the collision. (b) Estimate the duration of the collision and the average force acting on the ball. Strategy In part (a), lise the fact that the impulse is equal to the change in momentum. The mass and the initial and final speeds are known, so this change can be computed. Tn pan (b), the average force is just the change in momentum computed in part (a) divided by an estimate of the duraLion of the collision. Guess at the distance the ball travels on the face of the club (abollt 2 em. roughly the same as the radius of the ball). Divide this distance by the average ,'elocicy (half the final velocity) to get an estimate of the time of contacl. r-'
,
FIGURE 6.3
(Example 6.1) A golf
hall being sHuck b)' a club.
, .. , .. -
~
Solution (a) Find the impulse delivered to the ball. The problem is essentially one dimensional. Note that Vi = 0, and calculate the change in momentum, which equals the impulse:
J=
ap
=
(5.0 X 10- 2 kg)(44m/s) - 0
1'/-1'; =
+2.2 kg'm/s
=
(b) Estimate the duration of the collision and the average force acting on the ball. I'.x
=-
2.0 X 10- 2 m
= -'--'---'-'---
Estimate the time interval of the collision, ~t, lIsing the approximate displacement (radius of the ball) and its average speed (half the maximum speed):
I'. t
Estimate the average force from Equation 6.6:
]I = ~ = a.~l 9.1 X lOs
22
Val'
aI'
2.2 kg·m/s
c:-:-----=c..,-,-;"-
+ 2.4
X 10' N
.
.............................. - ..... Remarks This estimate shows just how large such contact forces can be. A good golfer achieves maximum rnomenturn transfer by shifting weight from the back foot to the from foot. transmitting the body's momenlum through the sha[tand head of the club. This timing, involving a short movement of the hips, is marc effective than a shot powered exclusively by the arms and shoulders. Following through with the swing ensures that the mOlion isn't slowed at the critical instant of impact.
9.1 X 10- 4 s
111/ S
~
EXERCISE 6.1 A 0.J50-kg baseball, lhrown with a speed of 40.0 mis, is
hit straight back at the pitcher with a speed of 50.0 m/s. (a) What is the impulse delivered by the bat to the baseball? (b) Find the magnitude of the average force exerted by the bat on the ball if the two are in COntact [or 2.00 X 10-' s. (a) 13.5 kg· mls
Answer
(b) 6.75 kN
QUESTION 6.1 What average club speed would double the average force?
EXAMPLE 6.2
How Good Are the Bump.ers?
Goal Find an impulse and estimate a force in a collision of a movi ng object with a stationary object. Problem In a crash test, a car of mass 1.50 X 10 3 kg collides with a wall and rebounds as in Figure 6.4£1. The initial and final velocities of the car arc Vi = -15.0 mls and vJ = 2.60 mis, respectiVely. I r the call isian lasts for 0.150 s, find (a) the impulse delivered to the car due to the collision and (b) thc size and direclion of the ;:lVcragc force exerted on the car. Strategy This problem is similar to the previous example, except that the initial and f-inal momenta
.ASter
--
+2.60 m/s
(a)
(b)
FIGURE 6.4
(Example 6.2) (a) This GII"S momentullI changes as;l resull ofiLS collision wilh the h'all. (b) [n a crash test (an inelastic collision), much of\hc car's inilial ki netk energy is transforrned into lhe energy it. look [0 d'tl11age [he vchicle.
6.1
Momentum and Impulse
165
arc both nonzero. Find the momenta and substitute into the impulse-momentum theorem, Equation 6.6, solving for I~y.
,. .....
'""
.
Solution (a) Find the impulse delivered to the car. Calculate the initial and final momenta of the car:
jJ, ~ mVi ~ (1.50 X 10 3 kg)(-15.0 m/s)
-2.25 X 10 4 kg' m/s 11f = mV = (1.50 X lO' kg)(+2.60 m/s) f = +0.390 X
The impulse isjust the difference between the final and initial momenta:
I = I)t -
10' kg' m/s
p,
+0.390 X 10" kg·m/s - (-2.25 X 104 kg-m/s) 1 ~ 2.64 X 10' kg' m/s
(b) Find [he average force exerted on the car. Apply Equation 6.6, the impulse-momenLUm theorem:
F
"
2.64 X 104 kg·m/s 0.150 s
.....
+1.76 X 105 N
.....................~
Remarks vYhen the car docsn'l rebound off the wall, the average force exerted on the car is smaller than the value just calculated. With a final momentum of zero, the car undergoes a smaller change in momentum. QUESTION 6.2 \'Yhen a person is involved in a car accident, why is the likelihood of injury greater in a head-on collision as opposed to being hit from behind? Ans\,.:er using the concepts of relative velocity. momentum, and average force. EXERCISE 6.2 Suppose the car doesn't rebound off the wall, but the time interval of the collision remains at 0.150 s. Tn this case, the final velocity of the car is zero. Find the average force excned on the GIl". Answer
+1..50 X lO5 N
Injury in Automobile Collisions The main injuries that occur to a person hilling the interior uf a car in a crash are brain damage, bone fracture, and trauma to the skin, blood vessels, and internal organs. Here, \ve compare the rather imprecisely known thresholds for human injury with typical forces and acceleraLions experienced in a car crash. A force of about 90 kN (20000 Ib) compressing the tibia can cause fraclUre. Although the breaking force varies with the bonc considered, we may take this value as the threshold force for fracture. It's "veil known that rapid acceleration of the head, even withoLlt skull fracture, can be fatal. Estimates sho,,·,' that head accelerations of J50gexperienced for about 4 ms or 50g for 60 ms are 1~ILal 50% of the lime. Such injuries from rapid acceleration often resuIL in nerve damage to the spinal cord where the nerves enter the base of the brain. The threshold for damage 1.0 skin, blood vessels, and internal organs may be estimated from ,,·,rhole-body impact data, where the force is uniformly distributed over the entire front surface are~l of 0.7 m 2 to 0.9 m 2 . These data sho".' that if the collision lasts for less than about 70 ms, a person will survive if the whole-body impact pressure (force per unit area) is less t.han ].9 X 10:> N/m 2 (28 Ib/in. 2 ). Death results in 50% of cases in which the \"hole-body impact pressure reaches 3.4 X 105 N/m 2 (50 Ib/in. 2 ). Armed with the data above, we can estimate the forces and accelerations in a typical car crash ~lI1d see how scat belts, air bags, and padded intcriors can reduce thc chance of death or seriolls injury in a collision. Consider a typical collision
APPLICATION "
Injury to Passengers in Car Collisions
166
Chapter 6
Momentum and Collisions
involving a 75-kg passenger not wearing a seal belt, lraveling at 27 m/s (60 mi/h) who comes La rest in about O.OJO s after striking an unpadded dashboard. Using r~,\.llt = 11/.Vj - 'fII,V j , we find that F
_ mv; -
o-
!/lUI
(75 kg)(27 m/,)
ill'
0.010 s
-2.0 X 10' N
and
nvl
a= -
Int
=
27
m/s .
0.010 s
_
.>
= 2 100 m/s- =
2700 mis' 2 g = 280r! 9.8 m/s b
Ifwe assume the passenger crashes into the dashboard and windshield so [hat the head and chest, with a combined surface area of 0.5 m 2 , experience the force, we find a whole-body pressure of
2.0 X 10' N _
.>
0.::' mF (in units of
IO~
N)
10
FIGURE 6.5
Force 011 a car versus time for a typical collision.
(a)
:;
4
X
,.>
10'> N/m-
'V'e see that the force, the acceleration, and the ,""hole-body pressure all exceed the threshold for hltality or broken bones and that an unprotectcd collision at60 mi/h is almost certainly fatal. ''''hat can be done to reduce or eliminate the chance of dying in a car crash? The most important factor is the collision time, or the time it takes the person to come to rest. If this time can be increased by 10 to 100 times the value of 0.01 s for a hard collision, the chances of survival in a car crash are much higher because the increase in 8.l makes the contact force 10 to 100 times smaller. Seat belts restrain people so that they come LO rest in about the same amOUlll of time it takes to SLOp tIle car, typically about 0.15 s. This increases the effective collision time by an order of magnitude. Figure 6 ..1 shows the measured force on a car versus time for a car crash. Air bags also increase the collision time, absorb energy from the body as they rapidly deflate, and spread the contact force over an area of the body of about 0.5 m 2 , preventing penetration \vounds and fractures. Air bags must deploy \'ery rapidly (in less than 10 ms) in order to SLOp a human traveling at 27 m/s before he or she comes LO rest against the steering column about 0.3 III away. To achi.eve lhis rapid deployrncnl, accelerometers send a signal LO discharge a bank of capacitors (devices that store electric charge), which then ignites an explosive, thereby filling the air bag with gas very quickly. The electrical charge for ignition is stored in capacitors to ensure that the air bag continues LO operate in the event of damage to the battery or the car's electrical system in a severe collision. The important reduction in potentially fatal forces, accelerations, and pressures to tolerable levels by the simultaneous use of seat belts and air bags is summarized as follows: H a 75-kg person traveling at 27 m/s is SLOpped by a seat belt in 0.15 s, the person experiences an average force org.s kN, an average acceleration of 18g, and a whole-budy pressure of 2.8 X 10" N/m 2 [or a contact. area of 0.5 m2 . These values are about one order of magnitude less than the values estimated earlier for an unprotected person and well below the thresholds for life-threatening injuries.
p
6.2 ++
(b)
FIGURE 6.6
(a) A coilisiull between two objects rt:sulting from direct
contacl. (b) A collision bt:twcCIl two
charged objects (in this case, a protoll and a hdiulil lluclcLI1;).
CONSERVATION OF MOMENTUM
"Vhen a collision occurs in an isolated system, the total momentum of the system docsn't change 'with the passage of time. Instead, it remains constant both in magniwde and in direction. The momenta or the individual objects in the system may change, but the \'eCLOr sum of all the momenta will not change. The tOtal momenWill is therefore said to be conserved. In this sectioll, Vle will see how the laws of motion lead LIS to this important conservation law. A collision may be the result of physical contact between two objects, as illustrated in figure 6.6a. This is a common macroscopic event, as when a pair of bil-
6.2
Conservation of Momentum
liard balls or a baseball clJ1d a bal strike each other. By contrast, because contact on a submicroscopic scale is hard 10 define accurately, the 1l00ioll of rollision must be generalized to that scale. Forccs betv..'ecll t\'..' o ol~jects arise from the electrostatic interaction of the electrons in the surface atoms of" the objects. As will be
Before collision
(al
disclissed in Chapter 15, electric charges arc eirhcT posilive or negalive. Charges
with the same sign repel each oLher, \v'hilc charges \vith opposite sign attract each other. To understand the distinction bctween macroscopic and microscopic collisions, consider the collision between two positive charges, as shown in Figure 6.6b. Because the two particles in the figure arc both positiyely chargcd, they repel each Olher. During such a microscopic collision, particles need not wllch in the normal sense in order to intcract and transfer 1TI01llCIllUIll. Activc figure 6.7 shm.. . s an isolated s)"stem of t\\-·o panicles before and a ["tel' they collide. B)' aisolaLecl," wc mean that no external forces, such as Ihe gravitational force or friction, act on the system. Before the collision. the n:locities of the two panicles are Vii and V~j; after the collision, the velocities are Vlj and v~f' The impulse-momenulm thcorem applied LO 1lI 1 bccomes ~
F~1ti{=
Like\.. . ise, for
1fI'!.,
we
167
After colli.'\ion
(hi ACTIVE FIGURE 6.7 Ikfore and af'll:r it head-Oil colli:.ion belween two objccts. Tht: momentum or each 1)1~{~n changes as a resull of lht: collision, bill the lOtal momentum of the system I"t'mains conSlant.
~
1lI l V lj -
1JI1V li
h~lve
Fj~til. = 1fI~V2f-
nl':!.V'l.I
F
where j\n is the <'lverage force exerted by m'2 on 111 1 during the collision and 12 is the average force exerted by fIl"l,.on 171 2 (~Iring the collision, as in Figun.: 6.6<1. \'Ve use a"erage values for F:n and F 1'1. even though the actual forces may vary in time in a complicated way, as is the case in Figure 6.8. Ne\vton's third law states glat at a!i times these 1.\\"-0 forces are equal in magnil.ude and opposite in direction: F 21 = - F 1'1.' In acldit.ion, the t.wo forces act over the same time interval. As a result, we have
F
or ~
lll j V II -
~
1lI1Vl,
= -
(~
m.,!v~j~
~)
m'2"2,
~
after substituting the expressions obtained for F 21 and F I '1.' This equation can be rearranged to give the following important result:
FIGURE 6.8 Force as a funcLiol1 of Lime for the two colliding panicles in rigun's 6.6(<1) and 6.7. !':otl: thai
F1J
=
-Fl~'
[6.7] This result is a special casc of" the law of conservation of momentum and is true or isolated systems containing any nllmbcr of interact ing ()I~iecl.s. \Nhen no net external forcc act.s on a systcm, the total momentum of the system remains constant in time. Defining the isolated system is an important feature of applying this conservation law. A cheerleader jumping upwards from rest might appear to violatc conservation of momentum, because initially her 1TI0menLuITI is zcro and suddenly she's leaving the ground \vilh velocity V. The flav.,' in this reasoning lies in the fact lhat. the cheerleader isn't an isolated syst.em. In jurnping, she exerts a do\'vl1\vard force on thc Earth, changing its momentum. This change in the Earth's momentum isn't noticeable, however, because of the Earth's gargantuan mass compared LO the cheerleader's. When we define the system to be thi' theerlt'rtder and (hf Il.·arllt, momentum is conserved. Action and reaction, together ...vith the accompanying exchange of momentum between I\VO objects, is responsible for I he phenomenon knO\\'1l as recoil. Evcr)'~ one knows thalthrowing a baseball while standing sLraight lip, without bracing )'our feet against thc Earth, is a good way 10 fall over backwards. This reaction, an
~
Conservation of momentum
Tip 6.1 Momentum Conservation Applies to a System! The rnOmenllllll of an i.solalt;d system is consl;l"\·ed. but not lll;ccssarily the momentum of olle pal"licle within thal systell'l, hecau~c olh<:':1" panicles in the s)'stcm ma)' be interacting wil h il. Apply conscrvaLiOIl of mOltlcntum to an isolated system only.
168
Chapter 6
Momentum and Collisions
Conservation of ttlomCIllUllJ is the principle hehind a squid's propulsioll S)'SICIll. It propels ilselfb)' cxpcllilll{ waler at a high velnLity.
example of' recoil, also happens when you fire a gun or shoot an arrow. Conservation of momentum provides a straight(on,;ard ,,,,,ay La calculate such effects, as the next example shows.
EXAMPLE 6.3 Goal
The Archer FIGURE 6.9
Calculate recoil velocit), lIsing conservation of
(Example 6.3)
A II archer fires all arrow horilUmally 1.0 the right. Because he is standing on frici iOllless icc, Itt: willlwg"in La slide to the left across lIle ice.
momentum.
Problem An archer stands at rest on frictionless ice and fires a O.500-kg arrow horizolllally at 50.0 111/5. (See Fig. 6.9.) The combined mass of the archer and bow is 60.0 kg. With wh
,.. ..
................
Solution ''''rite the conservation or mOlllenturn equation. Let v lf be the archer's velocity and v2f the arrow's velocity.
,
~
Pi = 1)/ 0=
1Jl I V lr
+
"k.!.V':!.!
'In"
- - - v2f = -
Substitute 'In\ ~ 60.0 kg, 'In, = 0.500 kg. and v'lJ = 50.0 mis, and solve for vl/
1/1,
(0.500 kg) (500 m/s) 60.0 kg
-0.417 m/s ~
....................
.
Remarks The negative sign on VII indicates that the archer is moving opposite the direction of motion of the arrow, in accordance with Newton's third 1m..... Because the archer is much more massive than the arrow, his acceleration and consequent velocity are much smaller than the acceleration and velocity of" the arrow. Newton's second law, '2.F = ma, can't be used in this problem because we have no inrormaLion abollt the force on the arrO\v or its acceleration. An energy approach can't be lIsed either, because \,,'e don't know hmv much work is done in pulling the string back or how much potential energy is stored in the bow. Conservation of" momentum, however, readily solves the problem.
~
QUESTION 6.3 \-Vollld firing a heavier arrow necessarily increase the recoil velocit}'? Explain, using the result of Quick Quiz 6.1. EXERCISE 6.3 A 70.0-kg man and a 55.0-kg woman on ice skates stand facing each other. If the woman pushes the man backwards so that his final speed is 1.50 mis, at whar speed does she recoil?
Answer
1.91 m/s
6.3
Collisions
169
QUICK QUIZ 6.2 A boy standing' at one end or a floating raft that is stationary relative to the shore \valks to the opposite end of the raft, away from the shore. As a consequence, the raft. (a) remains stationary, (b) moves 3\\I<1Y from the shore, or (c) moves toward the shore. (Hint: Use conservation of momentum.)
6.3
COLLISIONS
\·Ve have seell th,ll for any type of collision, the LOtal momentum of the system just before the collision equals the LOtal momcnLUm just after the collision as long as the system may be considered isolated. The total kinetic energy, 011 the other hand, is generally not conserved in a collision because some of the kineLic energy is converted LO inlcrnal energy, sound energy, and the work needed to permanently deform the objects involved, such as cars in a car crash. We define an inelastic collision as a collision in which momentum is conserved, but kinetic energy is not. The collision of a rubber ball ,\.'ith a hard surface is inelastic, because some of the kinetic energy is lost when the ball is deformed during contact with the surface. When two objects collide and stick together, the collision is called pe,fectly inelastic. For example, if two pieces of putty collide, they stick together and move with some common velocity after the collision. lfa meteorite collides head un with the Earth, it becomes buried in the Earth and the collision is considered perfectly inelastic. Only in very special circumstances is all the initial kinetic energ}' lost in a perfectly inelastic collision. An elastic collision is defined as one in which both momentum and kinetic energy are conserved. Billiard ball collisions and the collisions of air molecules with the walls of a container at ordinary t.emperatures are highly elastic. Macroscopic collisions such as those between billiard balls are only approximately elastic, because some loss of kinetic energy takes place-for example, in the clicking sound when two balls strike each other. Perfectly elastic collisions do occur, however, between atOmic and subatomic particles. Elastic and perfectly inelastic collisions are limit.illgcases; Illost actual collisions fall into a range in between them. As a practical application, an inelastic collision is used to det.ect glaucoma, a disease in which the pressure inside the eye builds up and leacls to blindness by damaging t.he cells of the retina. In th is application, medical professionals use a clevice called a tonometer to measure the pressure inside the eyc. This device releases a puff of ai I' against the outer surface of the eye and measures the speed of the air after reflection from the eye. At normal pressure, the eye is slightly spongy, and the pulse is reflected at low speed. As the pressure inside the eye increases, the outer surface becomes more rigid, and the speed of the reflected pulse increases. In this way, the specd of the reflected puff of air can measure the internal pressure 0[-' the eye. Vtle can summarize the types of collisions as follows:
Tip 6.2 Momentum ond Kinetic Energy in Collisions The
mOIOt:nlllm
of an isolatcr\
system is conserved in all col· lisions. Howcver, the kinel Ie l:nerg)' of an isol
,
/
Tip 6.3 Inelostic vs. Perfectly Inelastic Collisions If Lhe colliding panicles 51 iek together, the c;ollision is pcrfectl)' inelastic. If they bounce orr each 01 hel- (,lnd kind ie energy is nm conserved), Lbe collision is inelaslic.
APPLICATION
II
Glaucoma Testing
•
In
and kinetic energy are conserved.
~
Elastic collision
•
In an inelastic collision, 1110mentulll is conserved but kinetic energy is
+-
Inelastic collision
•
an
clastic collision, both
mOl11el1llll11
not. In a 111f1/etl/y inela:.tic collision, mOfncnlUm is conserved, kinetic energy is not, and the two objects stick together after the collision, so their flnal velocities are the same.
In the remainder or this section, \\ie will treat perfectly inelastic collisions and elastic collisions in one dimension.
170
Chapter 6
Momentum and Collisions
QUICK QUIZ 6.3
A car and a large truck tr.weling at the same speed collide head-on and stick together. Which vehicle undergoes the larger change in the magnitude orits momentum? (a) the car (b) the truck (c) t.he change in the magnilUde of momentum is the same for both (d) impossible 10 deter-
BdiJre collision
III'!.
-.-+.\"
mine withoullllorc information.
(al
After collision
JIll
Perfectly Inelastic Collisions Consider 1'\10 ol~jccls having masses ml and 11I'2 moving with kllovm initial velocity components Vii and V'l.l along a straight line, as in Active Figure 6.10. If the two ol~jects collide head-oil, stick LOgcther, and move with a common velocity component vJ after the collision, then the collision is perfectly inelastic. Because the total 1ll0lllemUIll of the two-object isolated systelll before the collision equals the total mornentul1l of the cOlllbined·o~jectsystem after the collision, we can solve for the IInal velocit)' using conservation of momentum alone:
+ 1I~1 -,.. +x
(b) ACTIVE FIGURE 6.10 (a) Before alld (bl aftera perfectly inclaSlic head-Oil collision bClween 1\\0 objects.
[6.8] Final velocity of two objects in a one-dimensional perfectly inelastic collision
!l/.IVU
-+
nlj
+ +
U/2').b
[6.9J
ulz
It's imporlantto notice that VII' v'2I' and v/ represent the x-components of the velocity vectors, so care is needed in entering their known values, particularly with regard to signs. For exarnple, in Active Figure 6.10, vii would have a positive value (11I 1 moving to the right), ,,\-·here;·ls V'll would have a negative valuc (111~ moving to the left). Once these \'alues arc entered, Equation 6.9 can be used to find the correct fi.nal velocity, as shov.'I1 in Examples 6.4 and 6.5.
EXAMPLE 6.4 Goal
An SUV Versus a Compact
Apply conservation of momentum to a one-dimensional inelastic collision.
Problem An SUV with mass 1.80 X L03 kg is lraveling eastbound at +15.0 mis, \,:hile a compact car with mass 9.00 X 10 2 kg is traveling westbound at --15.0 m/s. (See Fig. 6.11.) The cars collide head-on, becoming entangled. (a) Find the speed of the entangled cars after t.he collision. (b) Find the change in t.he velocity of each car. (c) Find the change in the kinetic energy of the system consisting of bmh cars. Strategy The tolal momentum of the cars before the collision, Il" equals the total momentulll oflhe cars after lhe collision, Ill' if we ignore friclion and assume the two cars form an isolated system. (This is called the "impulse approximation.") Soh'c the momentum conservation equation for the final velocity of the entangled cars. Once the velocities are in hand, the other pans can be solvecl by substitution .
... .
(a)
(b)
FIGURE 6.11
(Example 6..t)
.
. . . .. . . .. . . ... . .. . . . . . . ,
Solution (a) Find the final speed after collision. Let IlIj and Vii represcnt the mass and initial velocity or the SUV, while 111'2 and v2i pertain to lhc compact. Apply conservation of momentum: Substitute the values and solvc for lhe final velocity,
v/
(l.80 X 10" kg)(15.0 m/s) + (9.00 X 10' kg)(-15.0 m/5) (1.80 X 10' kg + 9.00 X 10' kg)l] vr=
+5.00 m/s
...
6.3
171
Collisions
(b) Find the change in velocity for each car.
=
Change in velocity orthe SUY:
~V,
Change in velocity of [he compan car:
LiV2 -
vJ - Vii VI -
= 5.00 1ll/5 - 15.0 m/5
~
5.00 IJl!s - (-15.0 lIl/s)
V2i -
- 10.0 m/5 =
20.0 m/s
(c) Find the change in kinetic energy or the system.
Calculate the initial kinetic energy of the syst(;m:
'( +,9.00
X
10-., kg)(-15.0 m/,]'
.
= 3.04 X 10'J
Calculate the final kinCl.ic energy of the system and the change in kine[ic energy, 6.KE.
= ~(l.80 X \0' kg + 9.00 X lO' kg)(5.00
= 3.38
X lO'J
uK!'; = KE - Kloi = J
.
~
Remarks During the coJ1isi01l, the system lost almost 90% of its kinetic energy. The change in velocity of the SUV was only JO.O mis, compared to t,.,..ice that for the compact car. This example underscores perhaps lhe most important sarety fealUre of any car: its mass. Injury is callsed by a change in velocity, and lhe more massive vehicle undergoes a smaller velocity change in a typical accident.
QUESTION 6.4 J f the mass of both vehicles were doubled, how ,,,'ottld the final velocity be affected? The change in kinetic energy?
m/s)'
-2.70 X l05 J
.........
,
--
..
tile compacl car slams on the brakes suddenly, sIO\','ing the \'ehicle to 6.00 m/s. If the SUV traveling at 18.0 m/s crashes into the compact car, find (a) the speed of the s)'stem right aftcr the collision, assuming the two vehicles become entangled, (b) the change in velocity for both vehicles, and (c) the change in kinetic energy of lhe system, [rom the inst.ant before impact (when the compact car is traveling i:ll6.00 m/s) to the instant right after the collision.
Answers (a) 14.0 m/s (11) SUV: Ilv, = -4.0 m/5 pact car: L'lv, = 8.0 m/s (c) -4.32 X 10'J
Com-
EXERCISE 6.4 Suppose the same two "ehicles are both traveling eastward, the comp:ilct car leading [he SUv. The dri\'er of
EXAMPLE 6.5 The Ballistic Pendulum Goal
Combine the concepts of conservation of cllergy and conservation or momentum in inelastic collisions.
Problem The ballistic pendulum (Fig. 6.12a, page 172) is a device used to rneasure the speed of a fast-moving projectile such as a bullet. The bulle[ is fired imo a large block of wood suspended [rom some light wires. The bullet is Slopped by the block, and the en! il-c system swings up to a height h.. It is possible to obtain the initial speed of the bullet by measuring 11 and the two masses. As an example of the technique, assume that the mass of the bullet, ml , is 5.00 g, the mass of the pendulum, Ill":!.. is 1.000 kg, and his 5.00 Clll. Find the ini[ial speed of the bullet, vIi"
Strategy Fi rst. usc conservation of momenlum and (he properties of perfectly inelastic collisions to find the initial speed or the bullet, VII' in terms of the final \·elocity of the block-bullCl system, vp Second, use conservation or energy and the height reached by the pendulum to find vf . Finally, substitute this value of vf il1lo the previous result to obtain the initial speed of [he bullet.
172
Chapter 6
Momentum and Collisions
FIGURE 6.12 (Example 6.5) (al Diagram ofa halli~tic pendulum. Note 111,11 j, the \-e!ocity ur I he S\,SLelll JUSL (~ftnlhe perfectly illelastic collision. (h) :-"llll1iJ-l"sh phmograph of a lah(,rawry ballistic
v)
PClldullJlll.
vi
(a)
(b)
..................
r·································· Solution esc conservation of momentum, and substiwlC rhe knQ\o,Jl1 masses. Note [h;.lt V'll = 0 ancl1'I is the velocity of l1C I system (I) Jac'k + IJU 11 et) just after the collision.
p,- p) 1II1Vli
(I)
Apply conservation of energy
~
to I he block-bullet S)'SlCm
(K/~
(500 X
+
10-3
+
m2 v 'll = (111- 1
kg)v"
+ 0 ~ (l.005 kg)vr
(KE
PELflercnllisiun =
+ 1Jl'l)v}r
+ PE)top
after the collision:
! (ml
Both the potential energy al.lhe bOlLom and the kinetic energy at the top arc zero. Solve for the final velocity of the block-bullet system, Vi
2=
vJ
vJ = V
Finally. substitute Ijimo Equation (1) to find tial speed of the bullet:
Vii'
r=
-/-
mJv/ + 0
= 0
+ (11/.1 + 1II,)gh
2gh
V2ih = \/2(9.80 m/5')(5.00 X 10
m)
0.990 m/s (l.005 kg)(0.990 m/s)
the ini-
2
5.00 X 10
3
kg
L99 m/s
............. ~
..........
Because the impact is inelastic, it would be incorrect to equate the initial kineLic energy of the incollling bulletto the final gravitational potential energy associated with the bullet-block combination. The energy isn't conserved! Remarks
QUESTLON 6.5 List three ways mechanical energy can be lost from the system in this cxperimcnt.
EXERCISE 6.5 A bullet with mass 5.00 g is fired horizontally into a 2.000-kg block attached to a horizontal spring. The spring has a constant 6.00 X 1O~ N/m and reaches a maximum comprcssion of 6.00 cm. (a) Find the initial speed or the bullelblock system. (h) Fi nd the speecl of t.he bllllet. Answer
(a) 1.04 Ill/S (b) 417 m/s
QUICK QUIZ 6.4 An ol~jecl of mass '111 lTIoves to the right ,.. . ith a speed v. It collides head-on with an object of lTIass 3m moving \,\'ith speed v/3 in the oppositc directioll. If the t"w objects stick togcther, what is the specd of the combined object, of rnass 4m. after the collision? (a) 0
(b) v/2
(c)
QUICK QUIZ 6.5 thrmvs a Frisbee®
TJ
(d) 2v
A skater is lISil1g very low friction rollerbJacies. A friend al
hel', on the straight line along which she is coasting.
6.3
Collisions
173
Describe each of the following events as an elastic, an inelastic, or a perlectly inelastic collision between the skater and the Frisbee. (a) She catches the Frisbee and holds it. (b) She tries to calch the Frisbee, but it bounces off her hands and falls to the ground in front of her. (c) She catches the Frisbee and immediately throws it back \·.. . ith the same speed (relative to the ground) LO her friend.
QUICK QUIZ 6.6 In a pcrfecLly inelastic one-dimensiollal collision benveen two objects. what condition alone is necessary so Lhat ((II of the original kinetic energy of the system is gone after the collision? (a) The objects must have momenta with the same magnitude bill opposite directions. (b) The objects must have the same mass. (c) The ol~iects must have the same vcloci1 }'. (d) The objects ml/st have the same speed, \·vith velocity vectors in opposite directions.
Elastic Collisions NO\v consider 1.\'/0 objects that undergo an elastic head-on collision (Active Fig. 6.13). Tn this siLUation, both the momentum and the kinetic energy of the system of two objects are conserved. \,Ve can write tl)(;se conditions as
Bt:lixe collision
[6.10] (a)
and [6.11] After
where iJ is positive if an object moves to the right and negative if it moves to the lel'l. In a typical problem involving elastic collisions, there are 1.\...·0 unknmvn quantiLies, and Equalions 6.10 and 6.11 can be solved simultaneously 10 find them. These t\-vo equations are linear and quadratic, respcctively. An alternate approach simplir1es the quadratic equation to another linear equation, facililruing solution. Canceling the faclor in Equation 6.11, we rewrite I he equation as
t
?)
IlII(vll~ - v i / ) = m""!.(v~l- v 2
Here \ve have moved thc tcrms containing 11/\ to one sidc of the equation and thosc containing UI':!. to the other. Next, wc factor hath sides orlhe eqllation: fill (VIi -
VI!)
(v II +
VI)
= m~(v2) -
Now we separate the terms containing vation of momcntum (Eq. 6.10) 10 get fIl.l(v ll
-
VII)
III!
=
and
v 2;) (v'J.f
v'1./)
1.6.12]
in the equation for Lhe conser-
111'1
1I1':!.(v':!.i -
+
v 2)
[6.13]
To obtain our final result, wc divide Equation 6.12 by Equation 6.13, producing VII
+
VII
=
"(/'2j
+
v 21
Gathering initial
V'2I=
-(v lj -
v':!.{)
[6.14]
This equation, in combination with Equation 6.10, will be Llsed to solve problems dealing with perfecl1y elastic head-all collisions. According to Equation 6.14, the relative \'c1ocity or the two objects before thc collision, VIi - v 2 ;. equals the negative or the relative "e1ocity of the IWO objects after the collision, -(v I1 - 112/)' To better understand the equation, imagine that you arc riding along on one of the oqjCcts. As you measure the velocity of the other object. from your vant.age point, you \"iill be measuring the relative \"clucity of the two ol~jeets. In your "jew of the collision, the other object comes toward you and bounces off, leaving the collision with the same speed, but in the opposite direction. This isjust what Equation 6.14 states.
colli~ion
....I--_Vl"-f~. _+x
(b)
ACTIVE FIGURE 6.13 (a) Bd"on: and (b) after an da~lic head-on collision bel ween [\,'0 hard spheres.
The following procedure is recommended for solving one-dimensional problems involving collisions bcrv..'cen
(V·,'O
objcClS:
1. Coordinates. Choose a coordinate axis thaL lies along the direction of mOlion.
2. Diagram. Sketch the problem, representing the two objects as blocks and labeling velocity vectors and masses. 3. Conservation of Momentum. Write a general expression {-or the lolalmomentum of the system of (V',IO objects before and ajfenhe call ision, and equate the l\'\Iu, as in Equation 6.10. On the nexlline, fill in the known values. 4. Conservation of Energy. If the collision is elastic, write a general expression for the [Ola) energy before and after the coIl ision, and equate the two quantities, as in Equation 6.11 or (preferably) Equation 6.14. Fill in the knov.'n values. (Ship this step if the collision is 1I0t perfectly elastic.)
5. Solve the equations simultaneously. Equations 6.10 and 6.14 form a system of twO linear equations and £\.... 0 unknowns. If you have forgotten EquatlOl] 6.14, use Equation 6.11 instead.
Steps I alld 2 of the problem-solving strategy arc gencrall)' carried out in the process of sketching and labeling a diagram of the problem. This is clearly the case in 0\11" next example, which makes use of Figure 6.13. Other steps are pointed out as they are applied.
EXAMPLE 6.6 Goal
Let's Play Pool
Solve an elastic collision in one dimension.
Problem Two billiard balls of identical mass move lO\\;ard each other as in Active Figure 6.13. Assume that the collision between them is perfectl}" elastic. If" the initial velocities of the balls are +30.0 cm/s and -20.0 em/s, what is the velocity of each ball after the collision? Assume friction and rotation are unimportant. Strategy Solution of this problem is a matter or solving two equations, the conservation of momentum and conservation of energy equations, for two unknov,,'ns, the final velocities of the two balls. J Ilstead or using Equation 6.11 for conservation of energy, use Equation 6.14, which is linear, hence easier to handle. . ,... ....
"
Solution ""rite the conservation of mOmenLllIl1 equation. Because 1n 1 = 111 2, we can cancel the masses, then substitute "Ii = +30.0 m/s and v 2 ; = ~20.0 cm/s (Slep 3).
30.0 cm/s + (-20.0 cm/s) (1)
Next, apply conservation of" energy in the I'orm of Equalion 6.14 (Slep 4):
trr.... • • . • • •
Vii -
V'll
= -(v IJ -
=
V'lj)
"'1- v,!
30.0 cm/s - (-20.0 cm/s) =
VI!
+ "2!
10.0 cm/s = "I! + "2!
(2)
(3)
Now solve Eq uatiollS (1) and (3) si 111 ulta neollsly (Step'::;):
~ "I!
50.0 crn/s = "'2/- "I!
-20.0 cm/s
V 2!
=
+30.0 cm/s
.....................~
Remarks Notice the balls exchanged velocities-almost as if they'd passed through each other. This is ah,;ays the case ,,-hen two objects of equal mass undergo all elastic head-on collision.
6.3
Collisions
175
QUESTION 6.6 In Ihis example, is it possible to adjust the initial velocities of the balls so that both are at rest after the collision? Explain. EXERCISE 6.6 Find the final velocity of the two balls if the ball with initial velocity v ti = -20.0 em/s has a mass equal to one-half that of the ball with initial velocit)' VI, = +30.0 cm/s.
Answer
v lj = -3.33 cm/s; v2J= +4fl.7 em/s
EXAMPLE 6.7 Goal
Two Blocks ond
0
Spring
Solve an elastic collision involving spring pot.ential cncrgy.
-
v;, = -2.!lO m/s
vli=+
Problem A block of mass 111 1 = ] .60 kg, initially moving to the right with a velocity of +4.00 m/s on a frictionless horizontal track, collides with a massless spring attached to a sccond block of mass Ill'!. = 2.10 kg rnovi ng to the left witll a velocity of -2.50 mis, as in Figure 6.14a. The spring has a spring constant of 6.00 X 10 2 N/m. (a) Determine the ,·clocity of block 2 at the instant when block 1 is moving to the right \-vith a velocity of +~tOO mis, as in Figure 6.14b. (b) Find the compression of lhe spring.
(a)
- -
V 1f =+3.00 m/s
V2J
h
mlYQl~ l/I'l
Strategy We identify the system as the twO blocks and the spring. \'Vritc down the conservation of momentum equations, and solve for the fInal velocity of block 2. v'lf' Then usc conservation of energy to find the compression of the spring.
(b)
~
FIGURE 6.14
..
Solution (a) rind the ,·c1ocity vt{whcn block
(Example G.?)
...
~
has velocit)' +3.00 m/s.
\o\lritc the conservation of momentum equation for thc system and solve for v'l/
(1.60 kg)( 4.00 m/s) + (2.10 kg)( ~2.50 111/5) - (lo60 kg)(3.00 m/s) 2.10 kg v 2{ =
-1.74 m/s
(b) Find the compression orthe spring. Usc energy conservation for the system, noticing thai. potential energy is stored in the spring when it is compressed a distance x: Substitute the given ",dues and the result of part (a) into Ihc preceding expression, solving- for x: Il... • • • • • •
x=
0.173
III
. .... , ..
~
Remarks The initial velocity cornponelll of block 2 is -2.50 l11/s because the block is moving to the left. The negative value for v'!.{means thaI. block 2 is still moving to the left at the installl under consideration. QUESTION 6.7 Is it possible for both blocks to corne to rest \-vhile the spring is being compresscd? Explain. /-lint: Look at the momcntum ill Equation (I).
176
Chapter 6
Momentum and Collisions
EXERCISE 6.7 Find (a) the velocity or block 1 and (b) the compression of the spring at t he instant I hal block 2 is at resL Answer
(a) 0.7]9 mls to the right
(b) 0.25] m
6.4
GLANCING COLLISIONS
Tn Section 6.2 woe showed thallhe lotallinear rnOIT'lcnluITI DCa system is conserved when the SYStCIl1 is isolated (that is, when no external forces act all the system). For (1 general collision OrlWG o~jcets in three-dimensional space, (he conservation of momentum principle implies thal the LOlal momentum of the system in each direction is conserved. However, an irnponant subset of collisions Lakes place in a plane. The game of billiards is a familiar example involving multiple collisions of objects moving- on a t\\lo-dimensional surface, \'Ve restrict our attention to a single t,...a-dimensional collision between two objects that takes place in a plane, and ignore any possible rotation. Fat' such collisions, '\Ie obtain two component equations for the conservation of momentum: mlv li .'\"
+
}//'!.V'!.i:.: =
mlv l}:.:
+
1ll'.!Y'!.fx
"lnIV 1iJ
+
fl/'!.V'2I)'
=
mlll l/)'
+
fl/1Y2b
"Ve rnust use three subscripts in this general equation, to represcnt, respectively, (1) the object in question, and (2) the initial and final values of the components of velocity, ~ow, consider a two-dimensional problem in which an objcct of mass lUI co1licles \..Iith an o~ject or mass 'm'1. that is initially at rest, as in Active Figure 6.15. After the collision, object I moves at an angle 8 ,,,'ith respect to lhe horizontal, and ol~icct 2 moves at an angle (!J with respect to the horizontal. This is called a glancing collision. Applyi ng the Lnv of consen:alion of momentum in component form, and noting that the initial y-component of momentum is zero. we have
+0
x-component:
JJJ. 1V 1t
y-component:
0+ 0 =
= mlll l! cos 8
"1f/'lvljsin
0-
+ ·'lk.?v2f cos ¢
[6.15] [6.]6]
1J~v2fsin c~
If the collision is clastic, we can write a third equation, [or conservation of energy', in the form [6.17] (I' we know' the initial velocity
Vii and the masses, we are left with four unknowns (v lf' v'.!f' and c/». Because we have only three equations, one of the IOllr remaining quantities must be given in order to determine the mOlion after the collision from conservation principles alone. I r the collision is inelastic, dle kinetic energy of the system is not conserved, and Equation 6.17 does not apply.
e,
ACTIVE FIGURE 6.15 (a) Before ami (b) after:1 glalltlng colli"ion bctwCelllKO balk
ef--'.;.'I:..,- ••If/',!
-'-'2/ sin dl
(a) Bef'ol-c the collision
(b) Af"rc." the collision
VzJ
6.4
o
Glancing Collisions
177
ST ATEGY
TWO-DIMENSIONAL COLLISIONS To soh'e t\Vo-dimensional collisions, folio\\' this procedure: 1. Coordinate Axes. Use both x- and y~coordinales. It's convenient to have either the .\>axis or the y-axis coincide with the dircClion of one of the initial ,·elocil ies. 2. Diagram. Sketch 1he problem, la bel i ng "elacil)' vectors tlild masses. ~.
Conservation of Momentum. "Vrite a separate conservation of momen1ll1l1 equation for each of dlC x- and )'-direclions. In each case, the total initial momentulll in a given direction equals the LOlal flllal momentum in that direction.
4. Conservation of Energy. If the collision is elastic. 'write a general expression for the total energy before and after the collision, and equate the (.\\,'0 expressions, as in Equation 6.1 L Fill in the known values. (ShijJ this step if the collision is nol perfectly elastic.) The energy equation can't be simplified as in the one~dimcnsional case, so a quadratic expression such as Eqllation 6.11 or 6.17 In list be lIsed ,,,hen the coIl ision is clast ic. 5. Solve the equations simultaneously. There ,He two equations for inelastic collisions and Ihree for elastic collisions.
EXAMPLE 6.8 Goal
Collision at an Intersection
Analyze a t\"'o-dimcnsional inelastic collision.
Problem A car with mass 1.50 X 10 3 kg t.raveling cast. at a speed of 25.0 m/s collides at an intersection with a 2.50 X I O:~-kg van t.ravcling north at a speed of 20.0 mis, as shown in Figure 6.16. Find the magnitude and direction of the \'elocity of the \vreckage after lhe collision, assllming thaI the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected. Strategy Use conservation of momentum in t",·o dimensions. (Kinetic energy is 1Iot conserved.) Choose coordinates as in Figure 6.16. Before the collision, the only object haYing momeJ1l.um in the x-direction is the car, while the van carries all the momen!.um in the J-direetion. After the totally inelaslic collision, both vehicles move together al sOJ1le common speed vfand angle (). Solve for these twO lInkIlO\~'ns, using the t\\·o components of the conservation of" momentum equation.
,. ..... - .
+2fi.O m/'l
•
••.
~ t
"J.
20.0
.,
m,
FIGURE 6.16 (Example 6.H) A top \,jew of a perfect I}' inelastic collision bCl\\'cen a GIl" and a van. .~
Solution Find the x-components of the initial and final toral momenta:
L, P."
=
"'''H,U''H'
=
(1.50 X 10' kg)(25.0 m/s)
= 3.75 X 10" kg·m/s
L/J\"f= (mGt) + 1Il,,'all)V/cos()= (4.00 Set the initial x-mornenlUm equal
to
the final x-rnomentum:
Find the j-componcl1l.s of t.he initial and fInal lOtal momenta:
(1)
L.
X 10 3 kg)r'i cos ()
3.75 X 10 1 kg'm/s = (4.00 X IO.'! kg)"f cos ()
!ll)'
=
1lI\-anv';1Il
= (2.50 X
10'~ kg)(20.0 m/s)
= 5.00 X ]0' kg' m/s
L-/JI1. = (mear +
1II"
178
Chapter 6
Momentum and Collisions
5.00 X 10 4 kg·tn/s = (4.00 X 10:\ kg)vfsin ()
Set the initial J-mol11cntum equalLO the final Y-lllomcnlUm:
(2)
Divide Equation (2) by Equat ion (1) and solve for 8:
Lan () =
8 =
Substillll.e this angle back into Equation (2) to find
ill
5.00 X 10' kg·m!s 3.75 X l04 kg·Jl1
1.33
53.1°
5.00 X 10" kg' m!s vJ = -,-.,-:-:---:c:-;-:--o"---.,---,. (4.00 X JO'kg)sin53.1"
Remark IL's also possible La first find the x- and y-componcms vf...; and vfy or the resultant velocity. The lnagniwdc and dircClion or tile resultant velocity can then be found \vith the Pythagorean theorem, vf= Vv)/- + vfr2, and the inverse Langcm fUllction () = tan-I (v/y/v/;.:), Selting IIp this alternate approach is a simple matter of' substituting ii/x = Vi cos () and vrv = VI sin () in Equations (1) and (2). QUESTION 6.8
Tfthe car and van had idemical mass and speed, ,.., 'hat would the resultant angle have been? EXERCISE 6.8
A 3.00-kg ol~jeC( initially moving in the positive x-direction ''1'ith a vclocity or +5.00 m/s collides with and sticks to a 2.00-kg object initially moving in the negative y-direction ,'vith a velocity of -3.00 m/s. Find the final components of velocity of the composite o~jecL. Answer
vft
= 3.00 m/s; vJ), =
-1.20 m/s
6.5
(b) FIGURE 6.17 (a) A rocke1 reaniun chamber l\'ithoUl a noah.: has rC
ROCKET PROPULSION
vVhen ordinary vchicles such as cars and locomotives move, the driving force of the motion is friction. In the case of the car, this driving force is exerted by the road on the car, a reaction to the force exerted by the \vheels against the road. Similarly, a locomotive "pushes" against the tracks; hence, the driving force is the reaction force exerted by the tracks on the locomotive. However, a rocket moving in space has no road or tracks to push againsl. How can it move forward? In f~lCt, reaction forces also propel a rockct. (YOll should review NeWLOn's third law, discllsscd in Chapter 4_) To illustrate this point, we model our rocket with a spherical chamber containing a combustible gas, as in figure 6.17a. 'Nhen all explosion occurs in the chamber, the hot gas expands and presses against all sides of the chamber, as indicated by the arrO\vs. Because the sum of the forces exerted on the rocket is zero, it doesn't move. Now suppose a hole is drilled in the botl.Ol1l or the chamber, as in Figure 6.17b. vVhen the explosion occurs, the gas presses against the chamber in all directions, but G1l1't press against anything at the hole, where it simply escapes into space. Adding the forces on the spherical charnber now results in a nCt force upwards. Just as in the casc of cars and locomotives, this is a reaction force. A car's \'.:hccJs press against the ground, and the reaction force of the ground on the car pushes it Fonvarcl. The ""all or the rocket's combustion chamber exerts a force on the gas expanding against it. The reaction force or the gas on the wall then pushes the rocket upward. Tn a no\\' infamous article in The ,",few Yorh Times, rocket pioneer Roben Goddard was ridiculed lor thinking that rockets would work in space, \vherc, according to the Tillll'S, therc was nothing to push against. The Tiuu's retracted, rather belatedly, during the first Apollo moon landing mission in HJ69. The hot gases are not pushing against anything external, but against the rocket itself-and ironicall)', rockets actually work heuer in a vacuum. In an atmosphere, the gases have to do \-"'ork against the outside air pressure to escape the combustion chamber, slowing the exhaust velocity and reducing the reaction force.
6.5
+ Am)v =
1\11(11
+
till)
179
v
At thc microscopic levcl, this process is complicatcd, but it can be simplificd by applying conservation of momentum La the rockct and its t::jccted fuel. In principle, the solution is similar La that in Example 6.3. with the archer representing the rocket and the arro\\'s the exhaust gascs. SUppU~l: that at ~omt:: timt:: t, the momentllm of the rockct plus the rut:! b (J\J + .lm)v, \\'here.lm is an amount of fuel about 10 be burncd (Fig. 6.18a). This fuel is travcling at a speed vrelativc to, sa)', the Earth,just like the rest orthc rocket. During a short time inter\"al til, the rocket ejects fucl of mass ~m. and the rocket's speed increases to II + ..lv (Fig. 6.18b). If the fuel is ejectcd \\-'ith exhaust speed v, Trwliv" In Ihe mdwl, the speed of the fuel relative LO the Earth is 11 - vI" Equating the total initial momentum of the svstem with the total final momentum, \·...e ha\'e (M
Rocket Propulsion
M+6m
p, =
(,\1 ... .6.m)V'
(a)
.\1
•
+ !lm(n - tI,)
'V ... .6.V'
Simplifying this exprcssion gi\'es
(bl
.\I.1v =
11,
J.m
The increase D.m in the mass of thc exhaust corresponds LO an equal decrease in the mass of the rocket, so that ~m = -.l.J\1. Csing-thi~ fact, we have [6.18] This result, together with thc methods of" calculus, can be Llscd to obtain the folIm...· illg equation: vf -
Vi
=
TJfO
In
,1/,) (-M f
FIGURE 6.18 Rockel proplll"ion. (a) The illilial mass of.he rockel and fuel is.\1..,.. ~m al a lime t. and til(" rocke''" spc£'d is lJ. (b) ..... 1 a time t + .3.1. !Ill' rockc:.'I·s mass has been reduccrito .", and an amOll1H offuel.1m h.. " heen t:it"etcd. The rockel's speed inc!'l';Ises I" an ,lI110UllI .3.1'.
[6.19]
where kl, is the initial mass of the rocket plus ruel and i\1f is the final mass of lhe rocket plus its remaining fuel. This is the basic expression for rocket propulsion; it [ells us that the incrcase in velocity is proponiomll to the exhaust speed vfand t.o the natural logarithm of M,IN1,. Because the maximulll ratio of Nli to kIf for a single-stage rocket is abollt 10: I, the increase in speed can reach v, In 10 = 2.3v, or about twice the exhaust speed! For best results, therefore, the exhaust spced should be as high as possible. Currently, typical rockct exhaust speeds arc scveral kilometers per second. The thrust on the rocket is defincd as the force exerted on the rocket by the ejected exhausl gases. \Ye can obtain an expression for the instantaneous thrust by dividing Equmiol1 6.18 br 6.1: Instantaneous thrust
=
Ma = 1\1 .'w
.1/
t!.M = ITI... -I t!./
[6.20]
~
Rocket thrust
The absolute \'alue signs are used for clarity: 111 Eqll.uion fi.lR, -U1\1 is a positive quantity (as is v,., a speed). Here we see lhat the thrust increases as the exhaust velocity increases and as the rate of change of mass .1J\IIJ.I (lhe burn rate) increases.
APPLYING PHYSICS 6.2
MULTISTAGE ROCKETS
The current maximum exhaust speed of v, = 4500 mls can be realized with rocket engines fueled with liquid hydrogen and liquid oxygen. Bwthis means that the maximum speed auainable for a g-iven rocket with a Illass ratio of]O is v,. In 10 = 10000 m/s. To reach the Moon. hmve\'cr. requires <1 change in velocity of over) I 000 m/s. Further, this change must occur while working against grayit)' and atmospheric friction. Ho\"l can that be managed without developing better engines?
Explanation The answer is the multistage rocket. By dropping slages, the spacecraft becomes lighter, so that fuel burned later in the mission doesn't have to accelerate mass that no longer serves any purpose. Strap-on boosters, ~IS lIsed by 1 he Space Shuttle and a number of other rockets, such as the Titan 4 or Russian Prolon. is a similar concept. The boosters are jettisoned after their fuel is exhausted, so the rocket is no longer burdened by their \-veight.
180
Chapter 6
EXAMPLE 6.9 Goal
Momentum and Collisions
Single Stage to Orbit (SSTO)
Apply the ,'clacity and thrust equations
or a rocker.
Problem A rocket has a lata I mass of" 1.00 X 10'~ kg and a burnout mass or 1.00 X 101 kg, including engines, shell, and payload. The rocket blasts otT from Earth and exhausts all its fuel in 4.00 min, burning the fuel at a steady rate with an exhaust velocity of v" = 4.50 X 10 3 m/s. (a) H air friction and gravity arc neglected, what is the speed of the rocket at burnollt? (b) ',Vhal Lhrust does the engine develop at lifwff? (c) What is the initial acceleration of the rocket jfgravity is not neglected? (d) Estimate the speed at burnout j r gravity isn't negleCted.
,.....
Strategy Although it sounds sophisticated, this problem is l11ainly a maLler or substilUting values inLO the appropriate equations. Pan (a) requires substituting values into Equation 6.19 for the velocity. For part (b), divide the change in the rocket's mass by the LOtal time, getting ~_M/~l, then substiLUte inLO Equation 6.20 to find the thrust. (c) Using Ne\vlOn'S second la\'\o', the force of gravity, and the result of (b), \\.'e can find the initial acceleration. For p,,-lrt (d), the acceleration of gravity is approximately constant over the fev·: kilometers involved, so the velocity found in part (b) \-vilt he reduced by roughly !lv. = -gl. Add this loss to the result of part (a).
.,
.
Solution (a) Calculate the velocity at burnout. SubstiLUte v, = 0, v, = 4.50 X 10 3 m/s, AI, = 1.00 X 10 5 kg, and '\'~ = 1.00 X 10 4 kg into Equation 6.19:
5
= 0 Vf
+
100 X 10. kO") (4.5 X 10' m/s) I n ' ., ~ ( 1.00 X 10 kg
1.04 X 104 m/s
=
(b) Find the [hrusta[ liftoff.
Compute the change in the rocket's mass:
!lM
= Mf
- M,
=
1.00 X W" kg - 1.00 X J05 kg
= -9.00 X 10 4
-900 X 10"' kg
Calculate the rate at which rocket mass changes by dividing the change in mass by the time (where the time interval c(luals 4.00 min = 2.40 X 10 2 5): Substitute this rate into Equation 6.20, obtaining the th rust:
kg
2.40 X 10' s
Thrust = =
-3.75
X 10'
kg/s
t!.MI = (4.50 X 10' m/s)(375 X 10' kg/s) !ll
lJ,-
I
1.69 X 10" N
(c) Find the initial acceleration. ,,\'rite NC\'iton's second law, where Tstands for thrusl, and solve for the acceleration a:
M" = L,F= T- Mg
T
,,= =
M- g
=
1.69 X 10" N - 9.80 mis' l.00 X 10' kg
7.JO m/s'
(d) Estimate the speed at burnouL when gravity is not neglected. Find the approximale loss or-speed due to gravity:
" t!.lJ, = -gt!.1 = -(9.8 0 m/s-)(2.40 Xl 0'- s) = -2.35 X 10 3 m/s
Summory
Add this loss to the result 01" pan (b):
V
r = 1.04 X 8.05
181
10" mls - 2.35 X 10 3 m/s X
10' m/s . • • . • . •..J
lo.. • • • • • •
Even taking gravity into account. the speed is sufficient to auain orbit. Some additional boost may be I'eqlliredto overcome air ell"ag.
Remarks
QUESTION 6.9 \Vhal initial normal force \.,,·ould be excrtt:d on an astronaut or mass min ,\ rocket traveling vertically upward with an acceleration a? Answer wmbolically in tcrrns or t.he positi\'e quantities m, g, and a. EXERCISE 6.9 A spaceship with a mass of 5.00 X 10'1 kg is traveling at 6.00 X 10:\ m/s relative a space station. \Vhat mass \\fill the ship have after it fires its engines in order to reach a relati\'e speed ofS.OO X 10 3 mis, traycling the same direction? Assume an exhaust velocity of4.50 X lO:l m/s. Answer
3.21 X 10' kg
SUMMARY 6.1 Momentum and Impulse The linear momentum of an object or Illass with velocity is defined as
p
v
p ==
1/1
moving
w
[6.1J
mv
Momentum Gl-S,ries units of' kg·m/s. The impulse J' ora constalll force F deli\'ered 10 an object is equal to I he prodllct of [hc lorcc and the time illlen"al during which the force acts:
-I '" Fi\./ ~
[6.4J
Thcse two concepts are unifed in the impulse-momentum theorem, "'hich states thatlhe impulse ofa conslant force delivered Lo an object is equallO the change in momentum of the ol~jecL:
1=
F~f
6.3 Collisions Tn an inelastic collision, the momentum of the system is conserved, bllt kinetic energy is noL In a perfectly inelastic collision, the colliding objects stick together. In an elas tic collision. both thc lllomenllim and the kinetic energy of the system are conserved. A one-dimensional elastic collision bel ween tWO objects can be solved by llsing the conservalion of momentum and conservation of energy equations:
= ~p ==
mv,
+
m'!.v'!.i
=
I1I f V/J+ lJI,!v'2j
[6.10] [6.11]
The following equation, derived from Equations 6.10 and (i.ll. is u~ually more convenient to lise than the original conserv;\l ion of" encrg-y equation:
f6.5]
[6.14]
Solving problems with lhis lheorem often involves estimating speeds or contact times (or both), leading" to an a\"{~l'age f"orcc.
These equations can be solved simult.aneously for the unknO\vll velocities. Energy is not const:rved in inelastic collisions, so such problems must be solved with Equ
6.2
!/IV! -
mlv l ,
Conservation of Momentum
When no net external force acts on an isolaled system, the 10tal momentum of the system is conSlanl. This principle i, calkd conservation of momentum. In panicular, if the isolated system consists of two objects underg'oing a collision, the lOtal momentum of the systcm is lhe same bcl()re and af"ter the collision. Conservation of mOlllentum can be written mathematically for this case as [6.7J Collision and recoil problems lypically n:quire f-inding unknown \'c\ocitics in one or two dimensions. Each \"eCLor componenl gives an equation, and the resulting equations are solved simultaneotlsly.
6.4 G[ancing Collisions In glancing collisions, consen"alion of momentum can be applied "long twO perpendicular directions: an x-axis and a )'~axis. Problems can be solved by using thc x- and )'-componenrs of Eql.l
fOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
w
182
Chapter 6
Momentum and Collisions
MULTIPLE·CHOICE QUESTIONS 1. A soccer player fUllS up behind a 0.450-kg soccer ball traveling at 3.20 m/s and kicks it in the same direction as i( is moving, increasing its speed to l~.~ m/s. What magnilUde impulse did the soccer player deliver to the ball? (al 2.45 kg·m/s (b) 4.32 kg·m/s (c) 5.61 kg·m/s (d) 7.08 kg-m/s (c) 9.79 kg·m/s
2. A 57.0-g tennis ball is traveling straiglu at a player at 21.0 m/s. The player volleys the ball straight back at 25.0 m/s. If the ball remains in contact with the I-aekel for 0.060 5, Ivllal average force acts on the ball? (a) 22.6 kg·mls' (b) 32.5 kg·m/s' (c) 43.7 kg·m/s' (d) 72.1 kg·m/s' (c) 102 kg·m/s' 3. A car of mass m traveling at speed TJ crashes into the rear of a truck of lTlass 2m that is at rest and in neutral at an intersection. II" the collision is perfectly inelastic, what is the speed of the combined car and truck after the collision? (a) l! (b) v/2 (c) !J/3 (d) 2v (c) None of these
4. A small china bowl having kinetic energy I~' is sliding along a frictionless counlcnop ,.. . hen a seneI', with perfect timing, places ,I ricc hall imo the howl as it passes him. If lhe bowl and rice ball have the same mass, what is the kinetic energy of the system t.hereafter? (a) 2£ (b) E (c) E/2 (d) E/4 (e) 1,/8 5_ In a gamc of billiards, a red billiard ball is traveling in
the positive x-direction "..·il.h speed v and the cue ball is traveling in the negative .\·-direction with speed 3vw-hen the ['''''0 balls collide head on. \Vhich statement is true concerning their velocities subsequent to the collision? Neglect any effects of spin. (a) red ball: -v; Clle ball: 31/ (b) red ball: 1/; cue ball: 2v (c) red ball: -3v; cue ball: v (el) red ball: lJ; clle ball: 3v (e) The velocities can't be determined without knowing lhe mas!'; of each ball.
6. A 5-kg can moving to the right with a speed of 6 m/s collide!'; with a concrete wall and rebou nds wit.h a speed of 2 m/s. Is Ihe change in Illornentum of the cart (a) 0, (b) 40 kg·m/s, (c) -40 kg·m/s, (d) -30 kg·m/s, or (e) -10 kg·m/s'
7. A D.IO-kg object 1l10vinf{ inilially with a velocity of 0.20 m/s east'.. . a'·d makes an elasllc head-on collision ,.. . ith a 0.15-kg ol~jeet initially at rest. What is Ihe final velocity or the O.lO-kg object aft.er the collision? (a) 0.16m/s east\.. . ard (b) 0.16 mls westward (c) 0.040 m/s eastward (d) 0.0'10 Ill/S wcst.'.. . arcl (e) None of these 8. A 0.004-kg bullet is fired into a 0.200-kg block oh·,rood at rest on a horizontal surface. After impact, the block with the embedded bullet slides 8.00 In before comi ng to rest. I f the coefficient of friction is 00400, what is the speed of t.he bu llet before impact? (a) 96 m/s (b) 112 m/s (c) 286m/s (d) 404 m/s (c) 812 m/,
9. The kinetic energy of a rocket is increased by a factor of eightaftcr its engines are fircd, whereas its LOtal mass is reduced by halfl.hrollgh the burning orrllel. By what factor is the magnitude of its momentum changed? I-linl: Usc KE ~ /N2m. (a) 2 (b) 8 (c) 'I (d) 16 (e) 1 10. If t.wo panicles have equal momenta, are their kinetic energies eqnal? (a) yes, always (b) no, never (c) no, except when their masses are equal (e1) no, except when their speeds arc the same (e) )'CS, as long as they move along parallel lines 11. If twO particles have equal kinctic energies, are their momenra equal? (a) yes, ah.. . ays (b) no, never (c) yes, as long as their masses are equal (e1) yes, if both their masses and directions of motion are the same (e) no. unless they (Ire moving perpendicular to each other l2. A rocket with total mass 3.00 X 10 5 kg leaves a launch pad at Cape Kennedy, mO\'ing vertically with an acceleration of 36.0 mjs2. If the speed of the exhausted gases is 4.50 x IO:~ mis, at what. rate is the rocket initiall)' burning fuel' (a) 3.05 X 10' kg/s (b) 2.40 X 10 3 kg/s (c) 7.50 X 10' kg/s (d) 1.50 X 10' kg/s (e) None of these
C9.NCEPTUAL QUESTIONS 1. A balter bunts a pitched baseball, blocking the ball withollt swinging. (a) Can the baseball deliver more kinetic energy to the bal and bauer I him Ihe ball carries initially? (b) Can the baseball deliver more momentum to the bat and batter than the ball carries initially? Explain each of your answers.
2. Americans will never forget the terrori!:;l a1'l.ack
September 11, 200l. One commentator remarked that the force of the explosion at t.he T\vin Towers of the ,"\lorld Tracie Center was strong enough to blo",' glass and parts of the steel struc(Ure to small fragments. Yet. the television coverage showed thollsands of sh~ets of paper floating dO\.. . n, many stili intact. Explain how that could be. all
3. In perfectly inelastic collisions between two objects, there are events in which all of the original kinetic energy is lransformedto forms other than kinetic. Give an example of such an event. 4.11' two objects collide and one is initially at rest, is it
possible for both to be at rest aFter lhe collision? Is it possible for only one to be at rest afler the collision? Explilin.
5. A ball of clay of Illass '111 is thrown \vilh a speed vagainst a brick wall. The clay sticks to the wall and stops. Is the pri llciple or cotlsen:ation of momentum violated in th is example?
Problems
6. A skater is standing gti]] 011 a frictionless icc rink. Iler l"riend throws a Frisbee straight at her. In which of the following C
183
why a head-on collision is likely to be more dangerolls than other rypes of collisions.
II. A sharpshooter fires a rifle while standing with the bUll of the gun against his shoulder. If the for"'ard mOlllenIlIm ofa bullet is the same as the backward 1l10lllCIllUIll of the gun, why isn't it as dangerous to be hit by the gun as by the bllllet? 12, An air bag inflates when a collision occurs, protect.ing a passenger (the dUlllmy in Figure CQ6.L2) from serious injury. 'Why docs the air bag sofren the blow? Discuss the physics involved in this dramatic phOlograph.
8. A large bedsheet is held vertically by two students. A third student, \.. . ho happens to be the star piLcher 011 the baseball team, throws a raw egg at the sheet. Explain why t.he egg doesn·l. break when it hits the shect, regardless 01" its initial speed. (Ifyoll try (his, make sure the pitcher hits the sheet ncar its center. and don't allow the egg to f~1I1 on the floor after being callght.)
9. Your physiGl1 edtlcatioll teadler throws you a tennis baIt at a certain velocity, and you catch it. You are now given the following' choice: The teacher can throw YOll a llledicine ball (which is much more m,lssi\"(' than the tcnnis ball) with the same \·elocity, the same Illomentum, or [he same kinetic energy as the u.'l1nis ball. Which option would YOll choose in order to make the casiesl catch, and why?
10. If two automobiles collide, they uSllally do IlOt stick together. Docs [his mean the collision is elastic? Explain
FIGURE CQ6.12
13. In golf, novice players arc often advised
1.0 be sure to '·follow through" with thcir sw·ing. Why does this make the ball travel
14. An open box slides across a frictionless, ic}' surf'ace of "frozen lake. What happens to lhe speed of the box as waler from a rain shm'v·er falls venically downward into [he box? Explain.
PROBLEMS ~
WebAssign
1.
~.
m _
:', =
Tht: Prohlt:l11s rill" lhis chapter mar be a~siglH.~11 onlillt: at \'Vl"bAssil--:"n.
rnA pilcher c1aim.~ he can throw a O.145-kR basehall with a:-. much momentum as a 3.00-g bullet moving with a speed of 150 X lOS m/s. (a) \'\'h3t musl.lhe baseball's speed be if Lhe pit.cher's claim is valid? (b) \·Vhich has greaLer kinetic energy, the ball or the bullet?
= denotes guided problelll = denotes enhanced conte lit problem
II = hiornCflical application o = dl·llofe~ full solution available in Slut/rill Sollilion\ .'\Ifill/wI! Stluly Cuide
SECTION 6.1 MOMENTUM AND IMPULSE 1. Calculate the magnitude of the linear mOlllelllUIll for the follo","illg ca.'K~S: (a) a proton wirh lllass 1.67 X 10 ~i kg, llloving WiLh a speed of 5.00 X 101; tlI/s: (b) a 1!1.0-g: bullet moving with a speed of300 m/s: (c) a 7fl.O-kg sprinter running- with a speed of 10.0 l11/s; (d) lhe Llnh (lTlas~ = 5.98 X 10 21 kg) moving willi an orbital.~pccd equal Lo 2.98 X 10 4 m/s. 2. A stroboscopic photo Of;-l club hilling- a golf ball, SllCh as the photo shown in Figurc 6.3, was made by J larold EdKl'I"ton in 1~l33. The b(ll\ was initially ,1t n:Sl, and rhe club W(lS shown 10 be in conlact wilh Lht' club for about 0.0020 s. Also, lhe ball was fl.wnd 10 end up wilh a speed 01"2.0 X 10 2 [tis. Assuming that the golf ball had " llla~<; of 5r) K, f-itlclthc average force exerted by the club on the ball.
4. A (UO-kg ball is thrown straight lip into the air widl all illiti,1! speed of 15 Ill/S. Find the momentum of t.he ball (a) at its maximum height and (b) halfway to its maximum heigllL. 5, A baseball player of mass 84.0 kg running at 6.70 m/s slicks into home pt-lle. (
mJ
Show lhatlhe kinetic energy ora p.uticlc of mass In is related 10 [he maglliwde of the momentum p of [hat particle by KE = /)2/'2m. Nole: This expression is invalid for panicles tr;weling at speeds near that of light.
7. All object has a kinetic energ-y of 275 J and a momentum or magnitude 25.0 kg·m/s. Find the speed and Illass of the ol~ieCl.
8. An estimated force vs. lime CIII"\"C for a baseball struck by a bat is shown in Figtll'e P6.H. Fmm this CUITe, determine
184
Chapter 6
Momentum and Collisions
(a) the impulse delivered 1.0 the ball and (b) the average force exerted on the bal1. F(N)
"'~180001'
force, (b) the fi nal velocity of the object if it is initially at rest, and (c) the final velocity of the o~jecl. ifit is initially moving along the x-axis with a velocity of -2.0 m/s.
1\ acting in the .'(-direction on a panicle varie... in time (IS shown in Figure P6.16. Find (a) the impulse of the force, (b) the rinal velocity of the panicle if it is initially at rest, and (c) the final velocit), orthe panicle ifit is initially moving along the .'I:-axis with a velocity of -2.00 m/s.
16. A force of magnitude
20000
2.00~kg
15000
10000 5000
o'-----'-----'-'----'---'--'- t (illS) o
2
~
FIGURE P6.8
F (1')
[[]A O.280-kg \'olleybalJ approaches a player horizontally wilh a speed of 15.0 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direc-
tion with a speed of 22.0 m/s. (a) What impulse is delivered LO the b
_-f----L-'_l.--l-...JL I
o
E
A man claims he can safely hold on Lo a 12.0-kg child in a head-on collision with a relative speed of' 120-mi/h last i ng for 0.10 s as long as he has his scal belt on. (a) Find the magnitude of the average force needed to hold onto the child. (b) Based on the result to pan (a), is the man's claim valid? (c) "Vhat cloes the answer to this problem say abolltlaws requiring the use of proper safelY devices such as scat belts and speciall.Oddler scats?
12. A tennis pla>'cr receives a shot wilh the ball (0.0600 kg)
2
14. A O.SOO-kg football is thrown toward the cast with a speed of 15.0 m/s. A stalionary receiver catches the ball and brings it 10 rest in 0.0200 s. (a) What is thc impulse delivcred to the ball as it's caught? (b) What is the average force exerted 011 the receiver?
4
(,j
5
17. The forces shown in the force \IS. time diagram in Figure P6.17 act on a 1.5-kg particle. Find (a) the impulse for the interval from l = 01.0 l = 3.0 s
4
13. A GIl" is stopped for a traflic signal. When the light [Urns green, the car accelerates, increasing its speed from 0 to 5.20 m/s in 0.832 s. '''Il hat are the magnitudes of the linear impulse and the average total force experienced by a 70.0-kg passenger in the car during the time the «H accelerates?
3
FIGURE P6.16
11. A ball of mass 0.11)0 kg is dropped from rest [rom a height of 1.25 m. It rebounds from the floor to reach a height 01" 0.960111. ·What impulse was given to the ball by the floor? traveling horizontally al 50.0 m/s and l"C[UrJ1S the shot with the ball tnweling horizomally at 40.0 m/s in the opposite direclion. (a) What is the impulse delivered to the hall by the racket? (b) What work does I he racket do on the ball?
2
0
, 2 3
4
-2
51
t (s)
FIGURE P6.17
18. A 3.00-kg sleel ball slrikes a massive wall ,H 10.0 m/s at an angle of 60.0° \.... ith the plane of the wall. It bounces olT the wall with the same speed and angle (Fig. r6.18).lfthe ball is in contact with the wall for 0.200 s, what is the average force exerted by the wall on the ball? )'
60.0' ~--------....- )
------x
15. The force shown in the force vs. timc diagralll in Figure 1'6.15 acts on a 1.5-kg object. Find (a) the impillse of the I~
(N)
FIGURE P6.18
2f----""'
FIGURE P6.15
[illThe front 1.20 m of a I 400-kg car is designed as a "'crumple zone" that collapses LO absorb the shock of a collision. If' a car traveling 25.0 m/s stops uniformly in 1.20 !TI, (
Problems 20. A pilcher th rows a 0.15-kg bascl)all ~o that it crosses hnlllc plate horizolllally with (l speed of 20 m/s. Tht, ball is hit stl·aighl. back at the pitcher with a final speed of 22 1Il/~. (a) \'\'hat is the impulse delivered to Ihe ball? (h) Find the avcrag-e force exened by Ihe hat on Ihe hall if the {\\O are in contact for 2.0 X 10 :1 s.
SECTION 6.2 CONSERVATION OF MOMENTUM 21. Iligh-spced ~troboscopic photographs show that the head of a 200-g golf c1uh is lra\'cling at 55 m/s just before it strikes a 46-g golf ball at rest on a tce. Aftn th(' (olli~ion, the club head tra,els (in the same direction) at 40 m/s. Find th(' speed of the golfballjust af"ter impact. 22. A rille with a weight of 30 I fires a 5.0-g bullet \\ilh a \peed of 300 m/s. (.t) Find the recoil speed of the I illc. (h) If a 700-N m.lIl holds the rifle rirmlyagain'it hi!) slwuldef, find the recoil 'ipeed of the mall and rifle. 123.IA 45.0-kg girl is standing on a 150-kg plank. The plank, originally al rest, b free to slide on a frozen lake, which is a flm, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.50 m/s to the right relmi\e 10 the plank. (a) \,\'hat is her velocit)' rl'latiH' to tIle surface of the ice? (b) \Vhat is the velocity or I he plank rclatiH' to the Sll rface of the icc? 24. A 730-:-.J man Slands in the middle oj" a fro/ell pond of r;ldills 5.0 HI. He is ullabl<- Lv g-et La the otlwr side because oj a lack of friction between his shoes and 1he icc. To O\'e1"(lIllle Illis difTiclllty, he throws his l.~-kg ph)'sics lexthook horizolltally toward the north ~hore at a speed of 5.0 Ill/~. I [m"'long docs it take him to reach thL' soulh shore?
25. _
1.00 km/s. (a) If lhe flatcar and cannon togcther have a mass of 36.0 metric tons (not including thc projectile), wh
28. _
Two ice skaters arc holding hands at the center ofa fro7en pond when an argument ensues. Skater A shoves skater B along a horizontal direclion, Identify (a) the hori/olllal forces acting on A and (b) those aCling on B. (c) Which force is greatcr. the force on A or the force on B? (d) Can conservation 01 momentum be used for the srstem of A and B? Oefend your answer. (e) If A has a mass of 0.900 limes that of B. alld B begins to move away with a speed 01'2.00 mis, find the speed of A.
An astronaut in her space sui1 has a 10lal ma~s of
R7.0 kg. including suil and oxygen tank. ller telhcr line loses its allachmenl to her spacecraft while she·s 011 a sp:lccwalk. Iniliall)' at re~t wilh rcspeci 10 her spaccn an, she throws her 12.0-kg oxygen tank away from her space· craft with a speed of 8.00 m/s to propel herself bad. toward it (Fig. P6.25). (a) Determine the maximum (Ii~ tance she can be from the craft ami still return \\ilhin 2.00 min (the amOllnt of time the air in her hdmet remains hreathable). (b) E.xplain in terms of Ne\\ton's laws of Illation why this st rategy works.
185
A man of" mass In l = 70.0 kg is skating al til = 8.00 m/s hehind his wife or mass lllo2 = 50.0 kg, who is skating at lJ';!. = 4.00 mho In~tead of passing her, he inadvencllll)' collides with her. lie grahs her around the waist, and they maintain their balance. (a) Sketch the problem witll before-and-after diagrams, representing the skaters
BI
v,.
30.. \n archer shool.s an arrow toward a 300-g target thai is sliding in her direction at a 'Speed of 2.50 mls on a <;mooth, slippery surface. The 22.5-g arrow is shot with a <;peed of 35.0 IIlls and pa:s~cs through the larget. which is SLOpped h} the impact. What is the speed of the alTo\\" afler passing through the t.lrget? [l[]Cayle I'UIlS at a ~peed of 4.00 m/s and dives on a sled, initially at rest on the top ofa frictionless. snow-cO\'cred hill. After ~he has de~ccndcd a venical distance of 5.00 m, her brOlher, who is initiall)' at rcst, hops on her back, and they continue down lhe hill togethec What is their speed at the bottom or the hill if thc lOtal \'cnic;tl drop is 15.0 m? Garlc\ rna\s is 50.0 kg, the sled ha!\ a mass of 5.00 kg-, and her bl"Other has a ma'iS of30.0 kg.
FIGURE P6.25
26.
mJ
A cannon is mOllllled on a railroad !Ialcar, the nUll:de elevated to 30.00 and pointed in the direction of tile track. The C.1I1I10n fires ;1 l.OO-melric-I<Jll pr<~jectik ,ll
32.
III
A 75.D-kg: ice skater moving at 10.0 m/s crashes into a stationary sk~llcr of eqllal mass. After the collision, the two skaten move as a unit at 5.00 m/s. Suppose the avcrag-e force a skaler ell I experience without breaking a hone is 4 50() N. If the iJnpilCl lime is 0.100 s. does a bone break?
Momentum and Collisions
Chapter 6
186
33. A railroad GIl' of mass 2.00 X 10 1 kg moving- aL 3.00 m/s collides and couples with twO coupled railro,ld cars, each of the same mass as the single car and mm'ing in the same direction at 1.2010/5. (a) \'\'h31 is the speed of the three coupled cars after the collision? (b) How much kinetic energy is loSl in the collision? 34.
35.
mil
A railroad car of mass M moving at a speed v\ collides and couples with IwO coupled railroad cars, each of the same Ill<1SS M and moving in the same direction at a speed 111_ (a) What is the speed 11 of the three coupled cars after llle collision in terms of VI and v~) (b) J 10\';' much kinetic energy is lost in the collision? Ans\...· cr in terms of A'I, VI' and V2-
mil
Consider the ballistic pendulum device disclissed in Example 6.5 and illuSlraled in Figure 6.12. (a) Dctnmine the r.atio of lhe mOlllentum immediatcl)' after the collision to the momenlllm immediately before the collision. (b) Show thai the ratio of the kinetic energy immediate!) after lhe collision 10 the kinetic energy immediately before the collision is m.!(ml + m,!.).
41. A 12.0-g bullet is fired horizolUally into a IOO-g wooden block that is initially at rest on a friclionle!)s horil.Ontal surf~lCe and connecLed Lo a spring having spring constam 150 N/m. The bullet becomes embedded in the block. If lhe bullet-hlock s,'sLcm compresses the spring b~ ~1 maximulll of 80.0 cm, what ' ' 'lS the speed of the bullet at impact with the block? 42, r\ I 200-kg car Lraveling initially with a speed of 25.0 m/s in an easterly direction crashes into Ihe rear end of a 9 ODD-kg truck moving in the same direction at 20.0 m/s (Fig. P6A2). The ,-elocity of Ihe car righl after the collision is 18.0 rn/s LO the east. (a) \~/hat is the velocity of the truck rig'hl after the collision? (b) llow much mechanical energy is losl. in Ihe collision? Accollnt for Lhi .. loss in energy.
- -
t25.0 m/s
--
+20.0 TIll ..
tIB.O m/s
v
Mtc,.
36. A 7.0-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges from Ihe hlock with a speed of 200 mIs, and the block rise~ 1.0 a maximum height or 12 em. Find the initial speed of the bullet.
\mIn a Broadway perform
38. Two shuffleboard dbks of equal mass, one orange and Lhe olher rellow, arc im'olved in a perfecLl) clastic glancing collision. The yellow disk is initially aL I-c~t and is struck by the orange disk Illm,ing initially Lo the "iglll at 5.00 m/s. AI' LeI' the collision, the orallge disk moves in a direction lhat mal..es an angle of37.0° wilh it~ initial direction. Meanwhile, the ,clocity "ector of the yellow disk is perpendicular LO lhe posLcollision velocity vector of the orange disk. Determine the speed of each disk after the collision. 39. A 0.030-kg bullel is fired ....enically at 200 m/s imo a 0.15~ kg baseball Ihat is illiti..llly at rcsl. Ilow high docs the combined bullet
t-~250g
.....~rr
,,
I
'----- 9_. (10 1ll
FIGURE P6.40
10 the right at 20.0 cm/s makes an head-Oil collision with a 1O.O-g object that is initially at rest. Find (a) the velocity of each object afler the collision and (b) the fraction of the initial kinetic energy transferrcd to lhc IO.O-g object.
43. A 5.00-g object moving e1i1~lic
44.
m
A space probe, initiall)1 aL reSL, undergoes an internal mechanical maifullction and breaks into three pieces. One piece of mass 111 1 = 48.0 kg travels in Lhe positive x-direction at 12.0 mis, and a second piece of mass 0 1112 = 62.0 kg travels in lhe xy-plane at all angle of 105 at 15.0 m/s. The lhird piece has mass 1~ = 112 kg. (a) Sketch a diagram of the situation, labeling the difTerentmasses and their' velocities. (b) \"'rite the general expression for conservation of momenLum in Lhc x- and )'.directions in terms of 1111' Ill,!., u~, VI' l)2' and ", and [he sines and cosines of the angles, taking (J Lo be the unknown angle. (c) Calculatt.· the final x-components of the momenta of ml and 1111 , (d) Calculate the f-inal J-compnnenL'i of tile 1110menta of 111 1 and 17/"2' (e) Substitute lhe known momentum components into the general equations or momenlum for the x- and y-di,·ecLions, along with Ihe known mass 111,. (f) Solve the twO momentum equations for Us cos 0 and v:i sin 0, respectively, and use the idenlil.)' cos2 () + sin 2 0 = I to obtain V3' (g) Divide the e(lu
45. A 25.0·g object moving to dlC righl at 20.0 cmls ovcnakes anet collides e1aslically with a 10.0-g object moving in the samc direction at 15.0 em/s. Find the velocil)' of each object after lhe collision.
8.00~
_.--.
FIGURE P6.42
,
, ,
'-.1
46. A billiard ball wiling across a table at 1.50 mls makes a head-on elastic collision with an identical ball. Find the speed or each ball after the collision (a) when the second ball is initially at reSL, (h) when the second ball is moving LOward the first at a speed of' 1.00 mis, and (c) when the second ball is moving away from the first at a speed of 1.00 m/s.
Problems A gO.n-kg fullback running ea~t with a :-.peed of 5.00 Ill/S is lackled by a 95.0-kg opponelll running north with a speed of 3.00 m/s. (a) 'Why docs t he tackle con~llittlte a perfectly inelastic collision? (b) Caklllate the "docilv of the players immediately afler the tackle and (c) determine the mechanical cnelXv that is lost as a result oflhe collision. Where din the lost eller~y go?
147.lm
48. Identicallwins, each with mass 55.0 kg, ;.lre on icc 'jkate~ and al rest on a frozen lake, which lllay he' taken as rriclionless. Twin A is carq'ing a backpack of mas<; 12.0 kg. She throws il horiLontally al 3.00 Illis 10 Twin B. Neglecling an~ gravity effects, what are the subsequent 'Ipeed~ of T\\7in A and T\vin B? &9. A 2 DOD-kg car moving east at 10.0 m/s collides with a 3 ODD-kg car moving north. The cars stick togethel· and mo\'c as a unit after the collision, at an angk of 10.00 north of east and a speed of 5.22 m/s. Find the spe{'r1 of the 3 ODD-kg car before the collision. 50. Two automobiles of equal mass approach an inH...-seelion. Olle vchicle is traveling with vclocit)' 13.0 m/s toward the cast, and the other is traveling north with speed lI2,' l\'eitiler driver sees Ihe other. Theyehicles collide in the inlersection and stick togelher, leaying parallel skid marks ,II all angle or 55.0 0 north of easL. The speed limit for both roads is 35 mi/ll, and the driver of the nort hward-moving vehicle claims he was within tJle limit when the collision occurred. Is he telling the trLnh?
51. A billiard halll110ving at5.00 m/s slrikcs a sl
ADDITIONAL PROBLEMS 52. " In research in cardiology and exercise physiology. it b often important to know lhe mass of blood pumped by a person's heart in one stroke. This information can be obtained b}' means of a ballislornrdiogmjJh. The inslrument work~ .IS follows: The subject li{'s all a hnri70nlal pallet f10aling on a film of air. Friction on the pallel is negligihle. Initiall}', the momentum of the system is l{'fO. When the heart beats, it expels a mass 111 of blood illlo the aona with speed fl, and the hody and platform move in the opposite direction with speed E The speed of the blood can be determined independenlly (for example. by ohserving
187
lhe truck. To 'Iubstanliate this view. lhey point OUI that the GIl" is crushed, whereas the trllck is onlvdenled. This idea or unequal forces, of course, is false; NeWLOn's third law tells us lhat both ohjects are acted upon by forces of the same magnillide. The lruck suffers less damtlge because it is 111~1(le of stronger melal. "Ul what about the two drivers? Do they experience the same forces? To answer this question. suppose thai each vehicle is initially moving at 8.00 mls and that they underg-o a perfectl) inelastic headon collbion. Each dri\'cr has mass 80.0 kg. Including the masses of the drivers. the total masses of the \·ehicles are 800 kg for the car and 4 000 kg for the truck. If the collision lime is 0.120 s, what force docs the seat belt exerl on each driver? 54. Consider a frictionless track as shown in Figure P6.54. A block of mass 111 1 = 5.00 kg is released from @. It makes a head-on clastic collision at @ with a block of mass m'L = 10.0 kg Ihat is inilially at fest. Calculale the maximum height to which 111 1 rises after lhe collision.
155.IA 2.0-g particle moving at 8.0 m/s makes a perfeclly clastic head-on collision with a resting 1.0~g- object. (a) Find the speed of each particle after- the collision. (b) Find lhe speed of each panicle after lhe collision if the stationary parlicle has a HltlSS of 10 g'. (c) Find the final kinctic energy of the incident 2.0·g particle in the situations described in parLs (a) and (b). In which case docs the incident panicle lose morc kinetic eneqn'?
56. A hullel ofm
e
....
--- ....... ,
,,
,,
i:.--l~:~,~+ v
M
v/2
FIGURE P6.56
57. An 80-kg mall standing ereCI steps off a 3.0-m-high e1i\'ing plalrorm ,lIld begins to fall frol1l I·CSt. The man again (omes LO a rest 2.0 s afler reaching the waler. H'lwr a,,'crage force did the water exert on him? 58. A OAOO-kg bllle bead slides on tI clIl've(1 f"ictionJess wire, starting from rest
188
Chapter 6
Momentum and Collisions
At point @, the bead collides elastically with a D.fiOO-kg blue ball at rest Find the maximum height the blue ball rises as it moves up the wire.
IJII
= 0.500 kg
'1'Io.l = 1.00 kg
It, = 2.50 m ""1=
, , -==---x~
@ FIGURE P6.5B
FIGURE P6.63
59. A ball ofmassO.500 kg is dropped from a heightof2.00 m. IL bounces against the ground and rises to a height of 1.40 m. If the ball was in COnlaCl with the ground for 0.0800 . . , what average force did the ground exert on the ball? 60. An unstable nucleus of mass 1.7 X 10-26 kg. initiall}" at rest at the orig-in of a coordinale system, disintegrates imo three partic1c~. One panicle, having a mass of ffl 1 = 5.0 X 10-2 , kg, moves in the positive )'-direction with 'ipccd 1)1 = 6.0 X 10" m/s. Another panicle, of mass ''''2 = 8.4 X 10-27 kg, moves in the positive x-direction with speed v'.! = 4.0 x lOti m/s. Find the magnitude and direction of the ",·clocitr of the third panicle.
61. m:J Two blocks of masses
1ll] and 1/'l.Z approacll cach O( her on a horizontal [able with the same constant speed, fJO' as measurcd by a laboratory observer. The blocks undergo a perfcctl)' ela~tic collision, and it is observed that 1n 1 SLOpS but1Jl'.! moves opposite its original motion with some constant speed, v. (a) Determine the ratio of the two masses, m/ m'!. (b) \Vhat is the ralio oftheir speeds, vivo?
62. Two blocks of masses In\ = 2.00 kg and 1n'2 = 4.00 kg are each released from rest at a height of 5.00 m on a friClionless track, as shown in Figure P6.62, and undcrgo an elastic head-on collision. (a) Determine the velocity of each block just before the collision. (b) Detcrmine the velocity of each block immcdiately after the collision. (c) Determine lhe maximum heights to which 111 1 and 1110Z rise after the collision.
• 5.00 m
-L
III
,,
1.50 m
m\
2.00
= 2.00 kg
mz = 4.00 kS{
-
~
-
r-
5.0001
-L
FIGURE P6.62
163.IA 0.500-kg block is released from rest at the lOp of a f,-ictionlcss track 2.50 m above the top of a table_ It then collides elasticall), with a 1.OO-kg object that is inil ially at rest on the lable. as shown in Figure P6.63. (a) Determine the velocities of the two objects just after the collision. (b) I-Iow high up the track does the 0.500-kg ohjecttravcl hack after the collision? (c) How far away from the bottom of the table does the 1.00-kg ol~ject land, given that the table is 2.00 m high? (d) How far away from the bottom of thc t;'lhle does the O.50D-kg object evenlually land?
64. Two objects of masses m and 3m are moving IOwaI'd each other along the x-axis with the same initial speed Vo' The object with mass m is traveling to the left, and the object with mass 3m is traveling to the right. They undergo an clastic glancing collision such that fit is moving downward after the collision at right angles from its initial directioll. (a) Find the final speeds of the two objects. (b) \.vhat is the angle 0 at which the object with mass 3m is scattered? 65. A small block of mass "UtI = 0.500 kg is released from rest at the top of a curved wedge of mass 1"'l = 3.00 kg, wh ich sits on a friction less horizontal surface as in Figure P6.65a. When the block leaves the \...·cdge, its velocity is measured to be 4.00 m/s to the right. as in Figure P6.65b. (a) What is the velocity of the wedge aFter the block reaches the horizontal surface? (b) What is the height h of the wedge?
1 "
U (b)
(a) FIGURE P6.65
66. A cue ball u-aveling at 4.00 mls makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of 30_00 with its original direction of travel. Find (a) the angle between the velocity vectors of the two balls after the collision and (b) the speed of each ball after the collision.
167.IA
cannon is rigidly attached to a carriage, which can move along horizontal rails, but is connected to a post by a large spring, initially unstretched and with force constant k = 2.00 X "10 4 N/m, as in Figure P6.6i. The cannon
FIGURE P6.67
Problems
fires a 200-kg projcClile at a vclocity of l25 m/s directed 45.0° above ule horizontal. (a) If the mass of the cannon and its carriage is 5000 kg, find the recoil speed of the can nOll. (b) Determine the maximum extension o[ the spring. (c) Find the maximum force the spring excrts on the carriage. (d) Consider the system consisting of the cannon, the carriage, and the shell. Is the momentum of this system conserved during the firing? 'tYhy or why not?
68. The "force platform" is a tool
o[ 25 m/s. If the player was moving upward v,,'ith a speed of4.0 m/sjust before impact, what will be the speed of the ball immediately after the collision if the ball rebounds verticall)l upwards and the collision is elastic? If the ball is in contact with the player's head ror 20 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)
[lDA
tennis ball of mass 57.0 g is held just above a basketball or mass 590 g. \'Vith their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.20 m, as shown in Figure P6.73. (a) Find the magnitude of the FIGURE P6.73 downward velocity with which the basketball reaches the ground. (b) Assume that an elastic collision with the ground instantaneously reverscs the velocity of the basketball while t.he tennis ball is still moving down. Next, the two balls meet in an elastic collision. To what height. does t.he tennis ball rebound?
I.°L F (kN)
that is used to analyze the performance of athlctes by measuring the vertical force 0.8 as a function o[ time that the 0.6 athlete exerts on the ground in performing various activiI (,) -il.5 0.0 0.5 1.0 ties. A simplified force vs. t.ime graph for an athlete perFIGURE P6.68 formi ng a standing high jump is shown in Figure P6.68. The at.hlete started the jump at t = 0.0 s. How high did this ath lete jump? 69. A neutron in a reactor makes an elastic head-on collision with a cal"bon atom that is initially at rest. (The mass of the carbon nucleus is about 12 times that of the neutron.) (a) What fraction of the neutron's kjnctic energy is transfcrred to the carbon nucleus? (b) If the neutron's initial kinetic energy is 1.6 X lO-I.~J, find it.s final kinetic energy and the kinetic energy of the carbon nucleus after the collision.
70. _ T,.. .o blocks collide on a friction less surface. After the collision, the blocks stick LOget.heL Block A has a mass M and is initially moving to the right at speed v. Block B has a mass 2M and is initially at rest. System C is composed of both blocks. (a) Draw a free-body diagram [or each block at an instant. during the collision. (b) Rank the magnitudes of the horizontal forces in your diagram. Explain YOllr reasoning. (c) Calculate the change in mOll1enlllm of block A, block B, and system C. (d) Js kinetic energy conserved in this collision? Explain your answer. (This problem is courtesy of Edward F. Redish. For more such problems, visit http://www.physics.ull1d.edu/perg.) 71.
mil
(a) A car traveling due cast strikes a car traveling due north at an intersection, and the twO move together as a unit. A property owner on the southeast corner of the intersection claims that his fence was torn down in the collision. Should he be awarded damages by the insurance company? Defend your answer. (b) Let the eastward-moving car have a mass of 1300 kg and a speed of 30.0 km/h and the northward-moving car a mass or I 100 kg and a speed of20.0 km/h. Find t.he velocity after the collision. Are the results consistent with your answer to pan (a)?
72. A 60-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed
189
1
74. A 20.0-kg toboggan with 70.0-kg drivCl" is sLiding down a frictionless chute directed 30.0° below the horizontal at 8.00 mls when a 5.1).0-kg woman drops from a tree limb straight clown behind the driver. 11' she drops through a vertical displacement of 2.00 m, what is the subscgucm velocity of the t.oboggan immediately after impact? 75.
mil
!'vleasuring the speed oj a oullet. A bullet or mass -m is fired horizontally into a wooden block of mass M lying on a table. The bullet remains in the block after the collision. The coefficient or friction between the block and table is J-L, and []lC block slides a distance d before stopping. Find the initial speed V o of the bullet in t.erms of M, m, j..t, g, and d.
76. A t-lying squid (family Ommastrephidae) is able to 'JUlllP" off the surface of the sea by taking water into its body cavity and then ejecting the \ilater vertically downward. A 0.85-kg squid is able 1.0 eject 0.30 kg of water with a speed of 20 m/s. (a) Whal will be the speed of the squid immediately after ejecting the water? (h) 1-[0\.... high in the air will the squid rise? 77. A O.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 2.0 m/s. After the collision, t.he O.20-kg puck has a speed o[ 1.0 111/s at an angle of 0 = 53° to the positive x-axis. (a) Determine the velocity of the O.30-kg puck after the collision. (b) Find the fraction of kinetic energy lost in the collision. 78. A l2.0-g bullet is fired horizontally into a 100-g wooden block initially at rest on a horizontal surface. After impact., the block slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and sur[ace is 0.650, what was the speed of the bullet immediately before impact?
Astronauts fall around the Earth at thousands of meters per second, held by the centripetal force provided by gravity.
7.1
Angular Speed and Angular Acceleration
7.2
Rotational Motion Under Constant Angular Acceleration
7.3
Relations Between Angular and Linear Quantities
7.4
Centripetal Acceleration
7.5
Newtonian Gravitation
;\ ~
7.6
z
Kepler's Laws
ROTATIONAL MOTION AND THE LAW OF GRAVITY Rotational motion is an important part of everyday life. The rotation of the Earth creotes the cycle of doy and night, the rotation of wheels enables easy vehicular motion, and modern technology depends on circular motion in a variety of contexts, from the tiny gears in a Swiss watch to the operation of lathes and other machinery. The concepts of angular speed, angular acceleration, and centripetal acceleration are central to understanding the motions of a diverse range of phenomena, from a cor moving around a circular racetrack to clusters of galaxies orbiting a common center. Rotational motion, when combined with Newton's law of universal gravitation and his laws of motion, can also explain certain facts about space travel and satellite motion, such as where to place a satellite so it will remain fixed in position over the same spot on the Earth. The generalization of gravitational potential energy and energy conservation offers an easy route to such results as planetary escape speed. Finally, we present Kepler's three lows of planetary motion, which formed the foundation of Newton's approach to gravity.
7.1
ANGULAR SPEED AND ANGULAR ACCELERATION
I n the study ofli Ileal' mol ion, t.he important concepts are rh\jJ[(JrP1nenl ~x, 7)l'lor;ly v, and arcdemtiun fl. Each of these: concepts has it.s analog in rot.ational motion: anglltal' rlisjJlnremJ'J11 .1. 8, an.e:ula r velority w, and angnlaT a(;celemtion Q'. The mdirl1l, a lInit of angular measure, is essential to the understanding of these concept.s. Recall Ihat the disLance s around a circle is given by.\ = 27Tr, where r is the !'adills of" the circle. Dividing both sides by ·r results in sir = 217. This quant.ity is d imcnsionless because both sand T h.:lve dimensions of length, but the value 217 corresponds 1.0 a displacement. around a circle. A half circle would give an answer of 17. a quarter circle an answer of' n/2. The numbers 27i", 17, and 17/2 correspond
190
7.1
Angular Speed and Angular Acceleration
to angles of ~60°, 180°, and 90°, respectively, so a new unil. or angular measure, the radian, can be illlroduced, \vith 180° = 7T rad relating dcgn:es to radians. The angle 0 sll!)lcndcd by an arc length s along a circle or radius !; measured in radians coulllcrclock\\,ise from the positive x-axis, is
191
/ .}=
s
0=-
[7. J]
T
r
e I
The anglc 0 in Equation 7.1 is actually an angldar displacemcnt [rom the positive x-axis, and s the corresponding displacement along the circular arc, again mea~ sllred from the positive x-axis. Figure 7.1 illustrates the size of 1 radian, which is approximately 57°. Convening from degrees to radialiS requires multiplying by the .-aLio (7T .-aell 180°). For example, 45° (7T rael/1800) = (7T/4) rad. Generally, angular quantities in physics must be expressed in radians. Be sure 10 set YOllr calculator to radian mode; neglecling to do so is a common error. Armed with the concept of radian measure, we can now discuss angular concepts in physics. Consider Vig-ure 7.2a, a top vie,"v ofa rotating compact. disc. Such a disk is an example of a "rigid body," WiLh each parl of [he body fixed in posil.ioll relative ,1I1 other parts of the body. \,Vhen a rigid body rotates through a given angle, all part.s of" the body rotale through the same angle at the same time. For the compact disc, the axis of rOlalion is al the cenLer of the disc, 0. A point Pall the disc is at a distance rfrom the origin and moves abollt () in {\ circle of radius 1: We set up ajixed reference line, as shown in Figure 7.2<1, and assume that at time 1= 0 the point }J is on lhat reference line. After a time interval dl has elapsed, IJ has adv~lnced Lo a Ile\v pusition (Fig. 7.2b). In this interval. the line OP has moved through the angle (J \,,'ilh respect tn the reference line. The £lng-Ie 0, rneasured in radians, is called the angular position and is analogous to the linear position variable x. Likewise, }J has moved an arc length .\ measured along the circumference of the circle. In Figure 7.3, as a point 011 the rotating disc moves from ® to @ ill a time D.l, it starts at an angle 8; and ends at an angle 0,. The difference 0, - fJ , is called the angular displacement. An object's angular displacement, :10, is lhe difference in its final and initial angles:
[7.2]
x
/
\
o=! n-l
o
p
Reference linc
(,,)
p f'
e
',s
o
Refel'CllCC !illl'
(hl FIGURE 7.2 (a) Tht' poilll POll a rutal illg compact d i1>C at I = O. (I» As the disc rotau.:s. Pmovcs thruugh all arc lcnglll s.
SI unit: Tadian (rad) For example, if a point on a disc is at 0; = 4 rad and rolates to angular posillon Of= 7 rad, the angular displacement is AO = Of- 0; = 7 rad - 4 rad =.:1 rad. Note that we use angular variables to describe 1he rotating disc because each point on the disc undergoes lhe same angular displacement in any given time interval. Using lhe definition in Equation 7.2, Equation 7.1 can be ,,,,rilten more generally as 6.0 = 6.s/I; where Lis is a displacement along the circular arc subtended by lhc angular displacement. I laving dclined angular displacemellts, it's nawral to de/ine an angular speed: The average angular speed W;\I' of a rotating rigid ol~jeCl during lhe time interv,ll Lil is the angular displacemelllLifJ divided by 6.(:
w"
er -(), '" - - = l; -
I,
u/'
,.
'. ",
- - - t ' - - ' - - - ' - - - - - - ., ()
M ~
J
[7.3]
FIGURE 7.3
I\>. a point all the
(ompan disc mon·.~ frOIJl @ 10 @.
Sf unit: radian per second (rad/s)
lIlt:: diK rotates through the angk fiO = OJ- 0...
192
Chapter 7
Rotational Motion and the Law of Gravity
Tip 7.1 Remember the Radian Equation 7.1 uses angles measured in radians. Angles expressed in terms of degrees must first be cQlwerLed to
radians. Also, be sure to check whether your calcillator is in degree or radian mode when soh"ing prohlems in\'oh'ing rotation.
For very short time intervals, the average angular speed approaches the instantaneous angular speed,just as in the linear casco
The instantaneous angular speed w of a rotating rigid object is the limit of the average speed 6.8/lJ.l as the time interval fj.f, approaches zero: w
= lim .:)./ -Jo
t;e
0
[7.4]
I:1l
SI unit: radian per second (rad/s) \·Ve take w LO be positive \ovhen 8 is increasing (couTlLCrclock\.,rise motion) and negative ,..,hen e is decreasing (clockwise Illotion). \'Vhcn the angular speed is constant, the instantaneous angular speed is equal to the average angular speed.
EXAMPLE 7.1 Goal
Whirlybirds
Convert an angular speed in revolutions per minute to radians per second.
Problem The rotor on a helicopter llIrns at an angular speed of3.20 X 10 2 revolutions per minute. (In this book, we sometimes use the abbreviation rprn, but in most cases we use rev/min.) (a) Express this angular speed in radians per second. (b) If the rotor has a radius of 2.00 Ill, what arclength does the tip or the blade trace out in 3.00 X 102 s?
Strategy During one revolution, the rotor turns through an angle or 21T radians. Use this relationship as a conversion factor.
... ...,
'-' ...
Solution (a) Express this angular speed in radians per second.
Apply the conversion factors] rev = 21T rae! and 60 s = I min:
rev w = 3.20 X 10'-min = 3.20 X 10' loe<' (2'" rad)(l.oO
min
=
(b) Multiply the angular speed by the time to obtain
,'tOv
min)
60.0 s
33.3 rad/s
M
= wI = (33.5 rad/s)(3.00 X
6.s
= rt;e =
10' s)
= 1.0 I X
10" rad
the angular displacement; Multiply the angular displacement by Lhe radius to gel the arc length:
(2.00 m)(LOl X 10" rad)
2.02 X 10" m
~
~
Ln general, it's best to express angular speeds in radians per second. Consistent use of r~diall measure minimizes errors.
Remarks
QUESTION 7.1 Is it possible to express angular speed in degrees per second? H so, what's the conversion factor from radians per second? EXERCISE 7.1 A ,vat.erwheelturns at I 500 revolutions per hour. Express this rate of rotation in units of radians per second.
Answer
2.6 rad/s
QUICK QUIZ 7.1 A rigid body is rotating counterclockwise about a fixed axis. Each of" the following pairs of quantities represents an inilial angular position and a Anal angular position of the rigid body. \'Vhich of the sets can occur on/:} if the rigid body rotates through rnore than 180°? (a) 3 rad, 6 rad; (b) -I rad, ] rae!; (c) 1 rad, 5 rae!.
7.1
Angular Speed and Angular Acceleration
193
FIGURE 7.4 An accelerating biqclt· wlwel rUI'llcs \\'ith (a) angular speed w, al lime I, ami (hl allgular speed wI allimc lr
(h)
la)
QUICK QUIZ 7.2 Suppose the change in angular position for each or the pairs of values in Quick Quiz 7.1 occurred in I s. \'fhich choice represents the lowest average angular speed? Figure 7A shows a bicycle turned upside down so that a repair technician can work on the rear wheel. The bicycle pedals are turned so that at time the wheel has angular speed WI (Fig. 7.4a) and at a later time if it has angular speed wI (Fig. 7.4b).Just as a changing speed leads to the concept oran acceleration, a changing angular speed leads to the concept of an angular acceleration.
'I
An object's average angular acceleration a.l \, during the time interval ill is the change in its angular speed a.w divided by ~f: a.,I\'
WI-WI
os ,
_
,
J
~w
=~t -
+- Average angular acceleration
[7.5]
'
SI unit: radian per second squared (rad!s') As with angular velocity, positive anguLH accelerations are in the countcrclock,vise direction, negative angular accelerations in the clockwise direction. If the angular speed goes from 15 rad/s to 9.0 rad/s in 3.0 s, the average angular acceleration during that time interval is
!1w 9.0 rad!s - 15 rad!s a". = --;;:; = 3.0 s
-2.0 rad!s'
The negative sign indicates that the angular acceleration is clockwise (although the angular speed, still positive but slowing down, is in the counterclockwise direction). There is also an instantaneous version of angular acceleration: The instantaneous angular acceleration a is the limit of the average angular acceleration l:i.w/li.i as the time interval ill approaches zero: !1w a= limA/ .....
jJ
a.l
[7.6]
51 unit: radian per second squared (rad/s 2 )
When a rigid object rotates about a fixed axis, as does the bicycle wheel, every portion of the object has the same angular speed and the same angular acceleration. This fact is what makes these variables so lIseful for describing rotational motion. In contrast, the langential (linear) speed and acceleration of the ol~jecl take dinerenl values lhal depend on the distance from a gi\'en point to the axis of rOlaLion.
+-
Instantaneous angular acceleration
194
Chapter 7
Rotational Motion and the Low of Gravity
7.2
ROTATIONAL MOTION UNDER CONSTANT ANGULAR ACCELERATION
A nurnbcr of parallels exist between the equations for rotational mOlion and those for linear motion. For example, compare the defining equation for the average angular speed, w
.w
,-lith that of" the average
8J - 8, M =:---=If - i, 6.t
line~lJ- speed, XI Vall
==
'.r -
Xi
Li:f
Ii =
nt
In these equations, w takes the place of v and f) takes the place of x, so lhe equations differ only in the names of the variables. In the same way, every linear quantity we have encountered so rar has a corresponding "twin" in rotational Illotion. The procedure used in Section 2.5 LO develop the kinematic equations for lin~ ear motion under constant acceleration can be used to derive a similar set of equations for rotational motion under constant angular accelennion_ The resulting equations of rotational kinematics, along with the corresponding equations ror linear motion, are as follows: Linear Motion with a Constant (Variables: x and v)
Rotational Motion About a Fixed Axis with a Constant (Variables: 0 and w)
w = w, + al d8 = w;t + bat'.! w'l = w,'2 + 2a Li8
v = v, + at Li.x = v,I + ~at'.! v'2 = v,2
+ 2a Lix
[7.7] [7.8]
[7.9]
Notice that. every term in a given linear equation has a corresponding term in the analogous rotational equation. QUICK QUIZ 7.3 Consider again the pairs of angular positions for the rigid object in Quick Quiz 7.1.1rthe o~ject starts from rest at the initial angular position, moves counterclockwise with consram angular acceleration, and arrives at the final angular position with the same angular speed in at! three cases, for ,,,'hich choice is the angular acceleration the highest?
EXAMPLE 7.2 A Rototing Wh:::;e.:.el:.... Goal
~~_
Apply the rotational kinematic equations.
Problem A wheel roLalCS with a constant angular acceleration of 3.50 rad/s~. If thc angular speed of the wheel is 2.00 rad/s at l = 0, (a) through ,,,,hat angle does thc wheel rotate between 1= 0 and l = 2.00 s? Give your ans,...er in radians and in revolutions. (b) What is the angular speed oft-he wheel at { = 2.00 s? Strategy The angular acceleration is constant, so this problcmjusl,-equircs subslituting given values into Equations 7.7 and 7.8. pr' . . • . • • , . . . . . . . . • • • • • • • , . , • . • , , . • • • , • • • • . • , • • • • . • . . • . • • , • • . • • • • . . • -, • . . . • • . . • • • , . . '""'III
Solution (a) Find the angular displacement after 2.00 s, in both radians and revolutions. Use Equation 7.8, setting Wi = 2.00 rad/s, and t = 2.00 s:
Q'
= 3.5 rad/5 2,
~ (2.00 rad/s)(2.00 s) ~
11.0 rad
+ H3.50 rad/s')(2.00 s)'
7.2
Rotational Motion under Constant Angular Acceleration
COllvert radians to revolULiolls.
=
!18
(11.0 rad)( l.00 rev/21T rad)
=
195
1.75 rev
(b) WhaL is the angular speed of Lhe wheel at t = 2.00 s' Substhute the same values inLO Equation 7.7:
W
~
w, + at
9.00 radls
= .
~
Remarks
.
.
.
• .
.
.
2.00 radls + (3.50 rad/s')(2.00 s)
~
.. .
.
.
.
.
.
.
.
.
•
.
.
.
.
.
.
• • .
• • .
• • .
• .• • • ••
• • • ..0lIl
The result of part (b) could also be obtaint:d from Equation 7.9 and the results of part (a).
QUESTION 7.2 Suppose thc radius of the whcel is doubled. Arc the ans'.. . e rs affectcd? Ifso, in what way? EXERCISE 7.2 (a) Find the angle lhrough v,:hich the wheel rotates between l = 2.00 sand l \vhen ( = 3.00 s.
Answer
3.00 s. (b) Find the angular speed
(a) 10.8 rad (b) 12.5 radls
EXAMPLE 7.3 Goal
=
Slowing
_
Prop""e~lIe;;.;r""s
Apply the time-independent rotational kinematic equation.
An airplane propeller slows fran') an initial angular speed of" 12.5 rev/s to a final angular speed of 5.00 rev/so DIlr;ng this process, the propellcr rotates through 21.0 rcvolutions. Find the angular acceleration orthe propeller in radians per second squared, assuming it's constant.
Problem
The given quantities arc the angular speeds and the displacement, which suggests applying Equation 7.9, the time-independent rotational kinematic equation, to find CI' •
Strategy
.
,..-'
........................................
~
Solution First, convert the angular displacemellt LO radians and t.he angular speeds to rael/s:
\·Vaiting until the end to convert revolutions to radians is also possible and requires only one conversion instcad of three.
Remark
QUESTION 7.3 I r the propeller had rotated through twice as many revolutions during the process, by what [aCLOr would the angular acceleration have changed? EXERCISE 7.3 Suppose, instead, the enginc specds up so that the propeller gocs through 28.0 revolutions \\-'hile the angular speed increases uniformly from 5.00 rev/s to 15.0 rev/so Find the angular acceleration. Answer
7.1477 rad/s 2
196
Chapter 7
Rotational Motion and the Low of Gravity
7.3
y
RELATIONS BETWEEN ANGULAR AND LINEAR QUANTITIES
Angular variables a,-e closely related LO linear variables. Consider the arbitrarily shaped object in Active Figure 7.5 rotating about the z-axis through the point 0. Assume the object rotates th rough the angle ~f), and hence point Pmoves through the arc length .6.s, in the interval 1i.t. \'\lc knmv from the defining equation for radian measure that CIs
M=,. ACTIVE FIGURE 7.5 Rotarion of an o~iecL aboUl all axis through U (the :-axis) that is perpendicular 10 the plallcofthc fij:{ure. Note that ,1 POillll'OIl the object roLates in a circle of radim n:cl1lcred
Dividing both sides of this equation by .6./, the lime interval during which the rotation occurs, yields M !J.l
j
CIs
7·
!J.t
When 6.t is very small, the angle !:i.O through which the object rotates is also small and the ratio !:i.O/D.t is close to the instantaneous angular speed lV. On the other side o[ the equat.ion, similarly, the ratio !:i.s/!J.t approaches the instantaneous linear speed v for small values of !:i.t. Hence, ".. . hen !J.t gets arbitrarily small, the preceding equation is equivalent to
at O.
u
,.
w=-
as
[n Active Figure 7.5, the point Ptraverses a distance along a circular arc during the time interval at a linear speed of v. The direction of p's velocit)'vector is tangent to lhe {'irealm' j)(l.th. The magniwde of is the linear speed TJ = v" called the tangential speed ofa panicle moving in a circular path, written
at
Tangential speed
~
v
Vf
=
TllJ
v
[7.10]
The tangential speed of a point on a rotating object equals the distance of that point from the axis of rotation multiplied by the angular speed. Equation 7.1 0 shows that the linear speed ofa point on a rotating object increases as that point. is moved outv,rard from the cenler of rotation toward the rim, as expect.ed; however. every point on the rotating object has the same angular speed. Equation 7.l0, derived using the defining equation for radian measure, is valid only when w is measured in radians per unit time. Other measures of angular speed, such as degrees per second and revolutions per second, shouldn't be used. To find a second equation relating' linear and angular quantities, refer again to Active Figllre 7.5 and suppose the rotating object changes its angular speed by nw in the time interval !J.t. At. the end of this interval, the speed oCa point on the object, such as P, has changed by t.he amount From Equation 7.10 we have fiv{ =
Dividing by
av,. r n.w
ut gives Clu,
Cli As the lime interval nt is taken to be a rbitrarily small, D.w/!J.tapproaches the instantaneous angular acceleration. On the left-hand side o[ the equation, note that the ratio ~v/~t tends to the instantaneous linear acceleration, called t.he tangential acceleration of that point, given by Tangential acceleration
--+
([/ = ra
[7.11]
The tangential acceleration of a point on a rotating object equals the distance of that point from the axis of rotation multiplied by the angular acceleration. Again, radian measure must be used for the angular acceleration term in t.his equalion.
7.3
Relations between Angular and Linear Quantities
197
One last equation that relates linear quantities to angular quantities will be derived in the next section. QUICK QUIZ 7.4 Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circubu plal[onn, twice as far from the center of the circular plat[onn as Chuck, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, Andrea's angular speed is (a) twice Chuck's (b) the same as Chuck's (c) half of Chuck's (cl) impossible to determine. QUICK QUIZ 7.5 When the merry-go-round of Quick Quiz 7.4 is rotating at a constant angular speed, Andrea's tangential speed is (a) twice Chuck's (b) the same as Chuck's (c) half of Chuck's (d) impossible 1.0 determine.
APPLYING PHYSICS 7.1
ESA LAUNCH SITE
"Vhy is the launch area f·or the European Space Agency in South America and not in Europe? Explanation Satellites are boosted into orbit on top of rockets, which provide the large tangential speed necessary to achieve orbit. Due to its rotation, the surface of Earth is already traveling tmvard the east at a tangential speed of nearly I 700 mls at the equa~
EXAMPLE 7,4 Goal
LOr. This tangential speed is steadily reduced farther north because the distance to the axis of rotation is decreasing. It finally goes to zero at the North Pole. Launching eastward from the equator gives the satellite a starting initial tangential speed of I 700 mis, ,.. . hereas a European launch provides roughly half that speed (depending on the exact latitude).
Comp.0ct Discs
Apply the rotational kinematics equations in tandem with tangential acceleration and speed.
Problem A compact disc rotates from rest tip to an angular speed of3L.4 rad/s in a time of 0.892 s. (a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform? (b) Through what angle does the disc turn while coming up to speed? (c) If the radius of the disc is 4.45 em, fmd I he tangential speed of a microbe riding on the rim of the disc when t = 0.892 s. (d) \Nhat is the rnagnitude of the tangential acceleration of the III icrobe at the given ti me? . ~
Strategy We can solve pans (a) and (b) by applying the kinematic equations for angular speed and angular displacement (Eqs. 7.7 and 7.8). Multiplying the radius by the angular acceleration yields the tangential acceleration atthe rim, whereas multiplying the radius by the angular speed gives the tangential speed at that point.
. ................••......................
Solution (a) Find the angular acceleration or the elise.
a = - =
Usc Equation 7.8 for angular displacement, with ( = 0.R92 sand Wi = 0:
M = w,l
Wi
31.4 rad/s 0.892 s
W
+ at,
Apply the angular velocity equation w = laking w, = 0 at t = 0:
I
35.2 rad/s'
(b) Through what angle does the disc turn?
(e) Find r = 4.45
+
~"l' = H35.2 rad/s')(0.892
sf
=
the final tangential speed ofa microbe at cm.
Substitute into Equation 7.JO:
v, ~ rw
=
(0.044 5 m) (31.4 radls)
=
].40 m/s
14.0 rad
~
Chapter 7
198
Rotational Motion and the Low of Gravity
(d) Find the tangential acceleration of the microbe at
,.= 4.45 em.
Substitute into Equation 7.11:
0,
-
~
.
= m = (0.0445111)(35.2 rad/s') =
1.57 mis'
. .•........•.•.....•...••••.........A
Remarks Because 217 rad = 1 rev, the angular displacement in pan (b) corresponds to 2.23 rev. In general, dividing the number of radians by 6 gives a good approximation to the Ilumber or revolutions, because 2rr --- 6. QUESTION 7.4 1f the angular acceleration were doubled for the same duration, by what factor would the angular displacement change? "Vhy is the an$\·.. er true in this case but not in general? EXERCISE 7.4 (a) 'Vhat are the angular speed and angular displacement or the disc 0.300 s afLer it begins to rotate? (b) Find the tangential speed at the rim at this time. Answers
(a) 10.6 rad/s; 1.58 rae! (b) 0.472 m/s
APPLICATION
Before compact discs became the medium of choice for recorded music, phonographs were popular. There arc similarities and differences between the rotational motion of phonograph records and that of compact discs. A phonograph record rotales at a constant angular speed. Popular angular speeds were 331 rev/min for long-playing albums (hence the nickname "LP"), 45 rev/min for "singles," and 78 rev/min used in vcry early recordings. At the outer edge of the record, the pickup needle (stylus) moves over the vinyllnatcrial at a fasler tangential speed than ,,,,hen the needle is close to the center of the record. As a result, the sound information is compressed into a smaller length of" track near the center of the record than near the outer edge. CDs, on the other hand, are designed so that the disc moves under the laser pickup at a constant. tangential speed. Because the pickup moves radially as it folImvs the tracks of information, the angular speed of the compact disc must vary according to the radial posiLion of the laser. Because the tangeillial speed is fixed, the information density (per length of track) anywhere on the disc is the same. Example 7.5 demonstrates numerical calculations [or both compact discs and phonograph records.
Phonograph Records and Compact Discs
EXAMPLE 7.5 Goal
Track Len th of a Com oct Disc
Relate angular
1.0
linear variables.
Problem In a compact disc player, as the read head moves out from the center of the disc, the angular speed of the disc changes so that the linear speed at the posilion of Lhe head remains at a const.ant value of aboul. 1.3 m/s. (a) Find the £lng'ular speed of the compaCL disc ",hen the read head is at r = 2.0 em and again at r = 5.6 em. (b) An old-fashioned record player rotates at a const.ant angular speed, so the linear speed or the record groove Illoving under the detector (the styills) changes. Find the linear speed or a 45.0-rpm record at points 2.0 and 5.6 Clll from the center. (c) In both the CD and phonograph record, informat.ion is recorded in a continuous
,..
.
Solution (a) Find the angular speed of the disc when the read head is at r = 2.0 CIll and r = 5.6 Clll.
spiral track. Calculate the toral length of the track for a CD designed Lo play for 1.0 h. Strategy This problem is just a matter of substituling numbers into the appropriate equations. Part (a) requires relating angular and linear speed with Equation 7.10, VI = TW, solving for wand substituting given values. 1n pan (b), convert from rev/min 1.0 rad/s and substitute straight into Equation 7.10 to obtain the linear speeds. I n part (c), linear speed multiplied by time gives lhe total distance. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • '""'IIl
7.4
Solve VI = HrJ for wand calculate the angular speed al 1" = 2.0 cm: Likewise, find the angular speed at
f'
= 5.6 cm:
w
w
v, I"
v, I"
1.3 mls 2.0 X I 0 ~
III
1.3 mls 5.6 X 10 2
III
Centripetal Acceleration
199
65rad/s
(b) Find the linear speed in m/s ora 45.0-rpm record at points 2.0 em and 5.6 cm from the center.
rev
rad)(1.00 = 450-- (27T ~--
min) = 4.71rad
Convert rev/min Lo rad/s:
rev 45.0-min
Calculate the linear speed at r = 2.0 cm:
v, = rw = (2.0 X 10- 2 m)('1.71 rad/s)
0.094 m/s
Calculate the linear speed at 'r = 5.6 cm:
v, = rw = (5.6 X 10- 2 m)(4.71 rad/s)
0.26 m/s
In,in
l;ev
60.0 s
s
(c) Calcu late the total lengl h of Lhe track lor a CD designed to play for 1.0 h. Multiply the linear speed or the read head by the time in seconds:
d
= v,1 =
(1.3 m/s)(3 600 s)
4700 m
....
.
~
~
Remark Notice that for the record player in pan (b), even though the angular speed is constant at all points along a radial line, the tangential speed steadily increases with increasing T. The calculation ror a CD in part (c) is easy only because the linear (tangential) speed is constant. It \\lnuld be considerably more difficult for a record player, where the tangential speed ckpends on the distance from the center. QUESTION 7.5 \rVhat is the angular acceleration ora record pla}/er while it's playing a song? Can a CD player have the same angular acceleration as a record player? Explain. EXERCISE 7.5 Compute the linear speed ora record playing at 331 revolutions per minute (a) at Answers
7.4
T
= 2.00 crn and
(b) at r = 5.60 cm.
(a) 0.0698 m/s (b) 0.195 m/s
CENTRIPETAL ACCELERATION
Figure 7.6a shows a car moving in a circular path with cons/ant lil1l!llr !>J;eed v. Even though the car moves at a constant speed, it still has an acceleration. To understand this, consider the defining equation for ;wcrage acceleration: ~
---t
aav
=
vr
-
if
~
(a)
~
Vi
[7.12] /i
v,
vI
The numerator represents the difference bCl\\'een the velocity vectors and Vi' These vectors may have the same magnilude, corresponding to the same speed, but if t.hey have different direrlimu, their dirference can't equal zero. The direc[joll or the car's velocity as il moves in I he cil-culal' path is continually changing, as shown in Figure 7.6b. For circular motion at constant speed, the acceleration vector always points toward the center of the circle. Such an acceleration is called a centripetal (centcr-seeking) acceleration. Its magnil.llde is givcn by
., ,r .,.
(f=(
[7.13]
To derive Equmion 7.\3, consider Figure 7.7<1. An object is Ilrst at point ® \vith velocity Vi at time ti and thcn at point ® with velocity VI at a later time It We
@
vi
(b)
FIGURE 7.6 (a) Circular motion of;j C;jr rllllvillg wil h (·onstalll ~pccd. (h) As I he car moves ,dong the ckcubr p:l1l1 frOll1 ® 10 ®. th(" din:ction of its vclocity \-eClor change;" so the car undergoes a centripetal accelcration.
200
Chapter 7
Rotational Motion and the Law of Gravity
vI
assume Vi and differ only in direction; their magnitudes arc the same = v). To calculate the acceleration, we begin with Equation 7.12,
(Vi
= vJ
[7.14]
o
where !:J.v = vi - Vi is the change in velocity. V\'hcn ul is very small, 6.5 and lie are also very small. In Figure 7.7b is almost paraliellO Vi, and the vector is approxirnately perpendicular to them, pointing toward the cenLcr of the circle. In the limiting case when 6.1. becomes vanishingly small,!:1~ points exactly toward the center of the circle, and the average acceleration 3 av becomes the instant.:lneolls acceleration From Equation 7.14, and point in the same direction (in this limit), so the instantaneous acceleration points to the center of the circle. The triangle in Figure 7.7a, which has sides 6.sand r, is similar to the one [armed by the vectors in Figure 7.7b, so the ratios of their sides are equal:
vi
(a)
a.
FIGURE 7.7 (
or
D.v
a
t1v
II. v v
D.s ,.
or [7.15] SubstilUting the result of Equation 7.15 into a
0.,1'1
v
~s
1"
111
= I1v/D.t gives
=--
. ,IV
[7.16]
BlIl !:1s is the distance traveled along the arc of the circle in time Iii, and in the limiting case \vhen 6.t becomes very srnall, lis/at approaches the instantaneous value of the tangential speed, v. Althe same time, the average acceleration aa\ approaches (Jr' the instantaneous centripetal acceleration, so Equation 7. L6 reduces to Equation 7.13:
v'
a. r = ---;Because the tangential speed is related to the angular speed through the relation = rw (Eq. 7.10), an alternate form of Equation 7.13 is
VI
[7.17] Dimension
-+
a = Va? + a}
[7.18]
QUICK QUIZ 7.6 A racetrack is constructed such that two arcs of radius 80 m at ® and 40 m at ® are joined by two stretches of stndght track as in
7.4
Centripetal Acceleration
201
Figure 7.8. In a particular trial run, a driver travels at a constant speed of" 50 m/s [or one complete lap. 1. The ratio of the tangential acceleration at ® to that at ® is (a) ~ (b) (c) 2 (d) 4 (e) The tangential acceleralion is zero at both points. 2. The ratio of the centripetal acceleration at ® to that at@ is (a) ~ (b) (c) 2 (d) 4 (e) The centripetal acceleration is zero at both
t
t
points. 3. The angular speed is gTcatcst at (a) ® (b) ® (c) IL is equal at both
®
®
FIGURE 7.8
(Quick Quiz 7.6)
® and ®.
QUICK QUIZ 7.7 An object moves in a circular path with constant speed v. \Vhich of the foHowing statements is true concerning the object? (a) Its velocity is constant, but its accelcration is changing. (b) Its acceleration is constant, but its velocity is changing. (c) Both its velocity and acceleration arc changing. (d) lts velocity and accekration rcmain constant.
EXAMPLE 7.6 Goal
At the Racetrack
Apply the concept.s o[ccnt.ripetal acceleration and tangential speed.
A race car accelerates uniformly from a speed of40.0 mls to a speed of60.0 m/s in 5.00 s \vh ile traveling countcrclockwise around a circular track of radius 4 .00 X 102 m. \-Vhen thc car reaches a speed of 50.0 mis, fi nd (a) the mag11 itude of the car's centripet.al acceleration, (b) the angular speed, (c) the magnitude of the tangential acceleration, and (d) the magnitude of the total acceleration. Problem
Strategy Substiwte values into the definitions of centripetal acceleration (Eq. 7.13), tangential speed (Eq. 7.10), and total acceleration (Eq. i.IS). Dividing the change in linear specd by the time yields the tangential acceleration . ............................................
.
~
~
Solution (a) Find the magnitude of the centripetal accelcration when v = 50.0 m/s.
)" v'" ( 50.0 mis'
a=-=
Substitutc into Equat.ion 7.13:
,
,.
4.00 X 10'm
6.25 m/s 2
(b) Find the angular speed.
v ,.
w=-=
Solvc Equation 7.10 for wand substitute:
50.0 m/s 4.00 X 10' m
0.125 rad/s
(c) Find the magnitude of the tangential acceleration.
vr - v,
a =---= I ill
Divide the change in linear speed by the time:
60.0 m/s - 40.0 m/s 5.00 s
4.00 mis'
(d) Find the magnitude of" the total acceleration.
a = -V /" {/,~ +
Substitute into Equation 7.18:
a
.................... Remorks
.
=
7.42
{/r~.,
= -V I( 4.00 m/s-")"-
, m/s-")"+( 6.2~
Ill/5 2
.................... -
~
We can also find the centripetal accelerat.ion by substituting the derived value of w into Equation 7.17.
QUESTION 7.6 If the lorcc causing the celltripetal acceleration suddenly vanished, would the car (a) slide away along a radius, (b) proceed along a line tangent to the circular motion, or (c) proceed at an angle intermediate between the tangent and radius?
202
Rotational Motion and the Law of Gravity
Chapter 7
EXERCISE 7.6 Suppose the race car now slO\""5 down uniformly from 60.0 m/s to 30.0 m/s in 4.50 s to avoid an accident, while still traversing a circular path 4.00 X L0 2 In in radius. Find the car's (a) centripetal acceleration, (b) angular speed, (c) tangential acceleration, and (d) total acceleration when the speed is 40.0 m/s. Answers
FIGURE 7.9 (a) The righi-hand rule for delcrillillillg the direction of the angular n.:!ocilr vector (b) The
w.
direction ofw is in tht: direction of ad\'
7 Angular Quantities Are Vectors
(a)
\·Vhen we discllssed linear motion in Chapter 2, we emphasized that displacement, velocity, and acceleration are all vector quantities. In describing rotational motion, angular displacement, angular velocity, and angular acceleration arc also vector quantities. The direction orthe angular velocit)' veCLOr Wcan be found with the Tight·hand rule, as illustrated in Figure 7.9a. Grasp the axis of rotation with your right hand so that your fingers wrap in the direction of rotation. YOllr extended thumb then points in the direction of Figure 7.9b shows that is also in the direction of advance of a rotating right-handed screw. \Ve can apply this rulc to a rotating disk viewed along the axis of rotation, as in Figure 7. LO. \·Vhen the disk rotCltes clockwise (Fig. 7.10a), thc right-hand rule shows th~\t the direction of is into the page. vVhen the disk rotates counterclockwise (Fig. 7.10h), the direction ofw is out o[the page. Finally, the directions of the angular acceleration "it and the angular velocity are the same if the angular speed w (the magnitude ofw) is increasing with time, and are opposite each other if the angular speed is decreasing with time.
w.
w
w
(b) FIGURE7.10 ALOp\'ic\~ofadisk rotaling ,lbolll all axis thrtlUgli its centcr perpendicular to tbe page. (a) ·Whel) the disk rolateS clockwise, poinLS inTO the pagc. (b) VV!lcll the disk rotates cOlilltcrdock\\ise, W points Ollt of the page.
w
w
Forces Causing Centripetal Acceleration
-"
m
,, , ,, FIGURE 7.11 A ball attached (0 ,1 string of length r, rotatillg in a circular path a\ cnllsl,"l1 speed.
An object can have a centripetal acceleration only if some external force acts on it. For a ball whirling in a circle at the end ora string, that. force is the tension in the string. ll1 the case 0[£1 car moving on a flat circular track, the force is friction between the car and track. A satellite in circular orbit around Earth has a centripetal acceleration due to the gravitational force between the satellite and Earth. Some books lise the term "centripetal force," which can give the mistaken impression that il is a ne\,\' force or nature. This is nor the case: The adjective centripetal" in "centripctal force" simply means that the force in question acts lOwa,-d a center. The gravitational force and the force of tension in the string of a ),o-yo \vhirling in a circle are examples of centripetal forces, as is the force of gravity on a satellite circling the Earth. Consider a ball of mass m that is tied to a string of length,. and is being whirled at constant speed in a horizontal circular path, as illustrated in Figure 7.1l. Its \.. . e ight is supported by a frictionless table. \.,Thy does the ball move in l\ circle? Because or its inertia, the tendency of the ball is to move in a straight line; hO\...-ever, the string prevents motion along a straight line by exerting a radial force on thc ball-a tension force-that makes it follow the circular palh. The tension IS directed along the string 100vard thl' rel1(erof(he (-irclt>, as shown in the figure. H
7.4
In general, applying Newton's second law along the radial direction yields the equation relating the net centripetal force F(-the slIln of the radial components orall forces acting on a gi\'en object-with the centripetal acceleration: i"(
=
ma(
=
v-"
mT
[7.19]
A net force causing a centripetal acceleration acts toward the center of the circular path and effects a change in the direction of the velocity veClar. If that force should vanish, the object would immediately leave its circular path and move along a straight line tangent 1.0 the circle at the point where the force vanished.
APPLYING PHYSICS 7.2
Centripetal Acceleration
203
Tip 7.2 Centripetal Force Is a Type of Force, Nat a Force in Itself! "CcmripCl<'l1 force" is a classification that includes forces acting l,o\\'ard a central point, like string lension on a l.clhcrball 0)" gravity on a saH:llile. A ccnlripclallorce must be !)IIP/Jlifd b}' some aClUal, physical force.
ARTIFICIAL GRAVITY
Astronauts spending lengthy periods of Lime in space experience a number of negative effects due to weigllliessness. such as weakening of muscle tis~ sue and loss of calcium in bones. These effcet.s Illay make it very difficult for them La return to their usual environment on Earth. How could artificial gravity be generated in space to overcome such complications? Solution A rotating cylindrical space station cre~ at:cs an environment of artificial gravity. The normal force of the rigid walls provides the centripetal force, \vhich keeps the astronauts moving in a circle (Fig. 7.12). To an astronaut, the normal force can't bc easily distinguished from a gravitational force as long as the radius of the station is large compared with the astronaut's height. (Otherwise there are unpleasant inner ear effects.) This samc principle is used in certain amusement park rides in which passengers are pressed against the inside of a rotat.ing cylinder as iltilts in various directions. The visionary physicist Gerard O'Neill proposed creating a giant space colony a kilometer in radius that rot.Hes slowly, creating Earth-normal artificial gravity for the inhabitants
in its interior. These inside-out artificial worlds could enable safe transport on a several-thousand-year journey to anot.her star system.
FIGURE 7.12 A rtiHcial gr(lvity insidt: a spinning cyliJ1(ler is provided br t he normal force.
PROBLEM-SOLVING STRATEGY FORCES THAT CAUSE CENTRIPETAL ACCELERATION
Use the following steps in dealing ".,Iith centripcwl accelerations and the forces that produce them: l. Draw a free-body diagram of the object under consideration. labeling all forces that act on it.
2. Choose a coordinate system thaL has one axis perpendicular LO the circular path followed by t.he object (the radial direction) and one axis tangent 1.0 the circular path (the tangential, or angular, direction). The norlllal direction, perpendicular to the plane of motion, is also often needed. 3. Find the net force Fe toward the center of the circular path, I~ = 'il'~, ""here 'iJ~ is the sum of the radial components of" the forces. This net radial force causes the centripetal acceleration. 4. Use Newton's second law for the radial, tangential, and normal directions, as required. writing 'iFr = llW"(. "'iFt = mal' and IF,, = man. Remember that
204
Chapter 7
Rotational Motion and the Law of Gravity
the magnilUde of the centripetal acceleration Cor uniform circular motion can always be \vriuen ul' = v;2/1: 5. Solve for the unknown quantities.
EXAMPLE 7.7 Goal Calculate the frictional force that causes an o!:?jeet to have a centripetal acceleration. Problem A car travels at a constant speed of 30.0 l11i/h (13.4 m/s) on a level circular turn of radius 50.0 Ill, as shown in the bird's-eye view in Figure 7.13a. vVhal. minimum coefficient of static friction. jJ.~. bctv..'cen the tires and roadway will allmv the car to make the circular turn without sliding? (a)
Strategy In the car's free·body diagram (Fig. 7.13b) the normal direction is vertical and the tangential direction is inLO the page (Step 2). Use Newton's second 1£1".... The net force aCling on the car in t.he radial direnion is the force of static friction tmvard the center of the circular path, which causes the car to have a centripetal acceleration. Calculating the maximum static friction force requires the normal force, obtained frorn the normal component of the second law.
n
t FIGURE 7.13 (Example 7.7) (a) Top vicw oLI GIl' on a cUJ'ved palh. (b):\ free-body diagram of Lhe car, showing an end view.
Remarks The valuc of}ls for rubber on dry concrete is vcry close to I, so the car can negotiate the curve with ease. the road were \·...e t or icy, hov,:ever, the value for J.L~ could be 0.2 or lower. Under such conditions, the radial force provided by static friction \youldn't be great enough to keep the car on the circuhlr path, and it \'l"Ould slide off on a tangent, leaving the roadway.
l[
QUESTION 7.7 If the static friction coefflciem 'were increased, would the maximum sa re speed be reduced, be increased, or remain the same? EXERCISE 7.7 At what maximum speed call a car negotiate a turn on a wet road \vith coefficient of static friction 0.230 ""ithout sliding out of control? The radius of the turn is 25.0 m. Answer
7.51 mls
7.4
Centripetal Acceleration
205
EXAMPLE 7.8 Goal
Solve a centripetal force problem involving two dimensions. !leos8
Problem The Daytona International Speedway ill Daytona Beach, Florida, is r~lInous for its races, especially the Daywlla 500, held every spring. Bmh of its courses feature four-sLory, 31.0° banked curves, v.. ith maximum radius of 316 m. If a car negotiates the curve too slowly, it tends to slip down the incline of the turn, whereas if it's going loa fast, it Illay begin to slide up the incline. (a) Find the necessary centripetal acceleration on this banked curve so the car won't slip down or slide lip the incline. (Neglect friction.) (b) Calculalc the speed of the race car.
i
nsin 8
-mg FIGURE 7.14 (Example 7.8) Frollt view ofa car rounding a banked roadway. Vector components are sho\,'n to the righl.
Strategy Two forces act on the race car: the force of gravity and the normal force (See Fig. 7.14.) Use Nev:LOn's second law in the upward and radial directions to nnd the centripetal acceleration (le' Solving a,. = v'l./r for v then gives the race car's speed. ,..-' .
n.
.
...,
Solution (a) Find the centripetal acceleration. \J\lrite Newton's second law for the car: Use the )'-componcnt of Ne'.. . ton's second law LO solve for the normal force 'I:
ilia = L,F = 11
cos
e-
Ii +
mg
mg = 0 n =
mg
cos ()
Obtain an expression for the horizontal component ofn, which is the centripetal force F( in Ihis example: Substitute this expression [or Fr inLO the radial component of Newton's second la\v and divide by IlL to get the cenl ripetal acceleration:
/'~. = 11
a =
Apply Equation 7.13:
F
mg si n
e
()
cos ()
'JItgtanO
--!.. = --''---- =
(1ft
a, (b) Find the speed of the race car.
sin
m
mgtan 8
glan
e
= (9.80 m/s')(tan 31.00) =
5.89 mis'
,,-.,
-= a
r
v =
'
V;; =
V(316 m)(5.89 m/5 2)
43.1 m/s
......... ~
r.... • • • • • • • • • • • • •
Remarks
Both banking and friction assist in keeping the race car on the track.
QUESTION 7.8 \"'hal three physical quamilies determine a n1inimum safe speed on a banked racetrack?
APPLICATION Banked Roadways
EXERCISE 7.8 A racetrack is LO have a banked curve \vi1.h radius of245 1TI. V\'hal should be the angle orLhe bank if the normal force alone is 1.0 allow safe travel around the Cllrve aL58.0 m/s?
Answer
54.5°
Chapter 7
206
EXAMPLE 7.9 Gool
Rotational Motion and the Law of Gravity
Ridin the Tracks
Combine centripetal force with conservation of energy.
v
Figure 7.15a shows a roller~coaster car moving around a circular loop of radius R. (a) \,Vhat speed must the car have so that it will just make it over the LOp wi tho lit any assistance [rom the track? (b) What speed will the car subsequently have at the bottom of the loop? (c) What will be the normal force on a passenger at the bottom of the loop if the loop has a radius of ]0.0 m? Problem
, ,,
/
,
/~":,, ,, ,
/
R ----\
I
.---------
I
I \ \
,, ,,
, ,,
I I
I
,
'Iv ""'-:'<8-""/
Strategy This problem requires NeWlon's second law and centripetal acceleration to find an expression for the car's speed at (b) (a) the lOp of the loop, (ollowed by conservation of energy to find FIGURE 7.15 (a) (Example 7.9) A roller coaster lra\"eling its speed at the bottom. If the car just makes it over the lap, the around a nearly circular track. (b) (Exercise 7.9) Ajclexccutforce Ii must become zero there, so lhe only force exerted on ing a \"cnicalloop. the car at that point is the force of gravity, At the bottom of the loop, the normal force acts up LOward the center and the gravity force acts elm-vn, away from the cenler. The difference of these l\.,,-O is the centripetal force. The normal (oree can then be calculated from NeWlon's second lav·....
mg.
~
~
Solution (a) Find the speed at the top of the loop. V\7r ite Ne'.. . ton's second la\\' ror the car: 2
At the top of the loop, set n = O. The force of gra\'it)' acts tmvard the center and provides the centripetal acceleration Cl r = J!. /R:
V 11/. _WJl _
=
R
mg
Solve the foregoing equation for vlOp : (b) Find the speed at the bottom of the loop. Apply conservation of mechanical energy to find the tDtalmcchanical energy at the top of the loop:
l~op = kmv~,p
+ mgh
=
!mgR + mg(2R)
=
2.5mgR
Find the total mechanical energy at the bottom of the loop: Energy is conserved, so these two energies may be equated lind solved for vhOI :
(c) Find the normal force on a passengel- at the bottom. (This is the passenger's perceived weight.) " vbm
Use Equation (1). The net centripetal force is n - mg:
m--=
Solve ror n:
n = mg+
R
n- mg
.,
...........
,
vix)l 1/1 - -
R
=
mg +
5gR 1/1--
R
=
6mg
......................... ~
The final answer for n shows that the rider experiences a force six times normal weight al the botwm of the loop! Astronauts experience a similar force during space launches.
Remarks
7.5
Newtonian Gravitation
207
QUESTION 7.9 Suppose the car subsequently goes over a rise \-vith the same radius of curvature and at the same speed as part (a). \,Vhat is the normal force in this case? EXERCISE 7.9 Ajettraveling at a speed or 1.20 x ]0 2 mls executes a vertical loop with a radius 01'5.00 X 10'2 m. (See Fig. 7.15b.) Find the magniwcle of the force of the seat on a 70.0-kg pilot at (a) the top and (b) the bottom of the loop. Answer
(a) 1.33 X 10" N (b) 2.70 X 10' N
Fictitious Forces Anyone who has ridden a merry-go-round as a child (or as a run-loving grown-up) has experienced what reds like a "cemer-rIeeing" force. Holding onto the railing and moving toward the center feels like a walk up a steep hill. Actually, this so~cal1ed centrifugal force is JirtiliollS. in reality, the rider is exerting a centripetal force on his body with his hand and arm Illuscles. In addition, a smaller centripetal force is exerted by the static friction between his reel. and the platform. If the rider's grip slippcd, he 'wouldn't be nung radially awa),'; rather, he would go off on a straight line, tangent to the point in space where he leL go of the railing. The rider lands at a point that is hll·ther a\\:ay frorn the center, but not by "fleeing the center" along a racliallinc. Instead, he travels perpendicular to a radial line, traversing an angular displacemcnt while increasing his radial displacement. (See Fig. 7.16.)
7.5
Tip 7.3 Centrifugal Farce
so-called centrifugal force r Amost uften just the nbsf'Tu:e of all i~
adequate cpnlripelal jora, arising
from measuring phenomena from a nonincrtial (accelerating) frame ofl'derence such
;IS
a
llll::rr~·-go-I"ound.
NEWTONIAN GRAVITATION
Prior to 1686, a great deal or data had been colleCled on the motions of the Moon and planets, but no one had a clear understanding of the forces aff~cting them. In that year, isaac Newton provided the key that un locked the secrets of the heavens. He knew from the first law that a net force had to be acting on the Moon. If it were 110t, the Moon would move in a straight-line path rather than in its almos!. circular orbit around Earth. Newton reasoned that this force arose as a result of an attractive 1'orn:- between the Moon and the Earth, called the force of gravity, and t.hat it wa~ LIlt:: same kind of force that attracted objects-such as apples-close 1.0 the surface of the Earth. fn 1687 Newton published h is work on the 13\.. · of universal gravitation: I f two parLicles with masses In! and 111.2 arc separated by a dist.ance I; a gravitational force Facts along a line joining them, with magnitude given by F= G
111 "'1 2 ,.2
f- Law of universal gravitation
[7.20J
where C = 6.673 X 10- 11 kg- J • m.3· S-2 is a constant of proponion alit)' called the constant of universal gravitation. The gravitalional force is always attractive. FIGURE 7.16 A fun-lovinJ.: student loses his grip and falb along a line tangent 10 the rim ol"lhc merry-goround.
-=~=""--~-- -
--
208
Chapter 7
Rotational Motion and the Low of Gravity
Th is force law is an example of an inverse~square law, in that it varies as one over the square of the separation of the particles. From Newton's third law, we know that the force exerted by 111 1 on m'.!, designated F l '.! ill Active Figure 7.17, is equal in magniLUcie but oppositc in direClion to the force F 21 exerted by '1112 on 11/.1' forming an aCLion-l-eaction pair. Another important fact is t.hat the gravitational force exerted by a uniform sphere on a particle outside the sphere is the same as the force exerted if the entire mass of the sphere were concentrated at its center. This is called Gauss's law, after the German mathematician and astronomer Karl Friedrich Gauss, and is also true of electric fields, ,,,,hich we will encounter in Chapter 15. Gauss's law is a mathematical result, true because the force falls off as an inverse square of the separation bet\.. . cen the particles. Ne<:lr the surface of the Earth, the expression F = llIgis valid. As shown in Table 7.1. hm"'ever, the free-fall acceleration gvaries considerably ,.. . ith altitude above the Earth. ~
ACTIVE FIGURE 7.17 The graVil
TABLE 7.1
Free-Fall Acceleration g at Various Altitudes AJtitude (km)a
QUICK QUIZ 7.9 A planet has t,.. ,o Illoons with identical mass. Moon I is in a circular orbit of radius ,: Moon 2 is in a circular orbit of radius 2r. The magn itude of the gravitational force exerted by the planet on 1\'loon 2 is (a) four times as large (b) twice as large (c) the same (d) half as large (e) one-fourth as large as the gravitational force exerted by the planet on Moon 1.
QUICK QUIZ 7.8 A ball falls to the ground. Which of the following statements are raise? (a) The force that the ball exerts on Earth is equal in magnitude LO the force that Earth exerts on the ball. (b) The ball undergoes the same acceleration as Earth. (c) Earth pulls much harder on the ball than the ball pulls on Earth, so the ball falls while Earth remains stationary.
Measurement of the Gravitational Constant The gravitational constant G in Equation 7.20 was first measured in an important experiment by Henry Cavendish in 1798. His apparatus consisted of two small spheres, each of mass 1n, fixed LO the ends of a light horizontal rod suspended by a thin metal wire, as in Figure 7.18a. Two large spheres, each of mass 1\1, were placed near the smaller spheres. The attractive force between the smaller and larger sphcres caused the rod to rotate in a horizontal plane and the wire to twist. The angle through which the suspended rod rotated ,vas measured with a light beam reflected [rom a mirror attached to the vertical suspension. (Such a moving spot of light is an effective technique for amplifying the motion.) The experiment ,,,,'as carefully repeated with different masses at various separations. In addition to providing a value for G, the results showed that the force is attractive, proportional
(a) A schematic dia-
gram of the Ca\"cndish apparatLls for mt;:asuring G. The smaller spheres of mass 111 are attracted La 111e brgc: spheres of mass M, and the rod rotates through a small angle. A light beam reflected from a mirror all the rOlming apparatus measures the angle of fUl:uioll. (b) A srudent Cavendish apparatlls.
(b)
7.5
209
Newtonian Gravitation
to the product 111M, and inversely proportional to the square of the distance I: Modern forms of such experiments are carried alit regularly today in an effort to determine C with greater precision.
EXAMPLE 7.10
Billiards, An one?
Use vecLars to find the net gravitational force on an
Goal
ol~jecl.
)'
L,
~,
Problem (a) Three O.300-kg billiard balls are placed on a table at the corners of a righttrianglc, as shown from overhead in Figure 7.L9, Find the net gravitational force on the cue ball (designated as 1}/1) resulting from the forces exerted by the other t\Vo balls. (b) Find the components of the gravitational force of "Ill'}. on m"';i'
1 , 1 , 1
0.400
'
1 1 III :
,
"
"0.500 III
1
Strategy (a) To (i~d the net gravitational force..: on the clle ball of mass till' we first calculate the force F 21 exerted by 11I 2 011 mi' !his force is the )'-component of the net force acting on 111.1' Then we find the force..: F:\I exerted by J!I?o on mi' which is the .,,component of the net force acting on mI' vVith these cwo components, \\le can find t.he magnitude and direction of the net force on the cue ball. (b) In this case, we must use trigonometry La find the components of the force 2 :\.
Solution (a) Find the net gra\'itatjonal force on the cue ball. Find the rnagnitude ofl.he force F~I exerted by 1112 on '11/'1 using the law of gravitation, Equation 7.20: ~
~ 1'21
.,
2
,)(0.300 kg)(0.300 kg) ( )9 0.300 m -
= 3.75 X 10- 11 N
~
Find the magnitude of the force F~l exerted by on mil again using Newton's law of gravity:
(0.300 kg)( 0.300 kg) ( )' 0.400 m
<,
(6.67 X 10- 11 N m-/kg-)
fIlJIII[
G--.,-
n1.!)
r -
"
_
_
_II
- (6.61 X 10
<
1\'m /kg
F" = 6.67 X 10- 11 N
The net force has components Fx = F?ol and Compute the magnitude of this net force:
I~ = F'21'
Use the inverse tangent to obtain the direction of
(b) Find the components of the force of
Fil-st, cam pUle the magnitude of" F:!~:
I'
= 'Ie + J-} = '1(6.67)' + 7.65 X
=
F:
jQ-1I
X
10- 11 N
N
ran-I -I~) = tan- I (3.75
(I';
e=
(3.75)'
X10- N) 11
6.67 X 10
JI
=
N
29.3°
111" 011 Illy
-
_
._
_II
- (6.61 X 10
_I
kg
3 _, 111'5 -)
(0.300 kg)(0.300 kg)
= 2.40 X 10- 11 N
To obtain the :0;- and y-cornponcnts of 'Sol' v•.'c need cos 'P and sin <po Use the sides of the large triangle ill Figure 7.19:
adj cos
. Sill
opp hyp
=
'I' = - - =
0.300 m = 0.600 0.500 m 0.'100 m
0..')00 m
= 0.800
(-
O.~OO
r
m -
Chapter 7
210
Rotational Motion and the Low of Gravity
Compute the components of "F\n. A minus sign l1lust be supplied for the x-component because it's in the negative x-direction.
"'13.,
=
=-
-F" cos 'P
(2.40 X 10- 11 N)(0.600)
-1.44 X 10- 11 N "2")
1.92 X 10-11 N
= .
~
(2.40 X 10- 11 N)(0.800)
= F':l sin 'P =
................ -
~
Remarks Notice how small the gravity forces are bet\.. .ecn everyday objects. i\onethcless, such forces can be measured directly with torsion balances. QUESTION 7.10 15 the gravity force a significant factor in a game of billiards? Explain. EXERCISE 7.10 Find magnitude and direction of the force exerted by Answers
5.85 X 10- 11 N, -75.8
EXAMPLE 7.11 Goal
Relate
Jn l
and
'1Il~-\ all "111':!.,
0
Ceres
:NC\VLOn'S
universal law of' gravity to mgand show hov·,' gchanges with position.
Problem An astronaur standing on the surface of Ceres, the largest asteroid, drops a rock frolll a height of 10.0 m. It takes 8.06 S 1.0 hit the ground. (a) Calculate the acceleration of gravity on Ceres. (b) Find the mass of Ceres, given that the radius of Ceres is R r; = 5.l0 X 10 2 km. (c) Calculate t.he gravitational acceleraLioll 50.0 km from the surface of Ceres. Strategy Pan (a) is a rcvie\v of one-dimensional kinematics. In pan (b) the weight of an object, w = mg, is the same as the magnitude of tile force given by t.he universal law oCgravity. Solve for the unknown mass of Ceres, after which the answer for (c) can be found by substiwt.ion into the universalla\'·.. of gravity, Equation 7.20.
.......................... .
r······················
~
Solution (a) Calculate the acceleration of gravity, gr:, on Ceres. Apply the kinematics displacement equation to the raIling rock:
(1)
Substitute ilx = -lO.O In, V o = 0, a = -go and t = 8.06 s, and solve for the gravitational acceleration on Ceres, gc:
-10.0
fix
= ~a12 +
III
Vol
-igc (8.06 sF --> gc = 0.308 mis'
(b) Find the mass of Ceres. Equate the weight 01" the rock on Ceres to the gravitational force acting on the rock:
1.20 X 10" kg
Solve for the mass of' Ceres, ;\lIe;:
(c) Calculate the acceleration of gravity at a height of 50.0 km above the surface of Ceres. mNfr;
Equate the weight at 50.0 kill to t.he gravitational force:
G--.,-
,.-
7.5
C<]llcel nI, then substitute r = 5.60 X 10 5 m and the mass of Ceres:
C
Newtonian Gravitation
211
!vIc
,.-"
{6.67 X 10- 11 0.255 m/s 2
......
.
~
Remarks This is the standard method of finding the mass of a planetary body: study the motion of a falling (or orbiting) object. QUESTION 711 Give two reasons Equation (1) could not be used for every asteroid as it is used in part (a). EXERCISE 7.11 An object takes 2.40 s to fall 5.00 m on a ccnain plane!. (a) Find the acceleration due to gravity on the planel. (b) Find the planet's mass if its radius is 5 250 km.
Answers
(a) 1.7'1
mis' (b) 7.19 X ]0 23 kg
Gravitational Potential Energy Revisited Jn
Chapter 5 we introduced the concept of gravitational pOlential energy and found thal the pmential energy associated \\"ith an object could be calculated from the equation PE = mgh, where II, is the height of the object above or below some reference level. This equation, however, is valid only \vhen the object is near Earth's surface. For objects high above Earth's surface, such as a satellite, an alternative must be used because gvaries with distance from the surface, as shown in Table 7.1. The gravitational potenLial energy associated with an object of mass m at a distance rfrom the center or Earth is PE~
MFIn
-G-'j'
potential energy
[7.21]
where /VIE and REare the mass and radius of Earth, respectively, with r> 51 units: Joules
~ General form of gravitational
R/~"
U)
As before, gravitational potential energy is a property of a system, in this case the o~ject of mass 11/. and Earth. Equation 7.21 is valid for the special case ,..,here the zero level [or potential energy is at an inrinite distance from the center of Earth. Recall that the gravitational poremial energy associaled with an object is nothing more than the negative of the work done by the force of gravity in moving the object. Ifan object falls under the force ofgravitJ' from a great distance (effectively infinity), the change in gravitational potential energy is negative, which corresponds to a positive arnOllnL of gravitational v.'Ork done on the system. This posit.ive work is equal to the (also positive) change in kinetic energy, as the next example shows.
EXAMPLE 7.12 Goal
A Near-Earth Asteroid
Use gravitational potential energy to calculate the work done by gravity on a falling object.
Problem An asteroid with Illass 111 = 1.00 X 10 9 kg comes from deep space, effectively from jnfinily, and falls toward Earth. (a) Find the change in potential energy when it reaches a point 4.00 X lOfl n1 from Earth (just beyond the Moon). In addition, find the work done by the force of gravity. (b) Calculate the asteroid's speed at
that point, assuming it was initially at rest when it was arbitrarily far away. (c) How much work would have to be done on the asteroid by some other agent so the asteroid would be traveling at only half the speed found in (b) at the same point?
212
Chapter 7
Rotational Motion and the Law of Gravity
Strategy Part (a) requires simple substitution into the definition of gravitational potential energy. To find the work done by the force of gravity, recall that the \\Iork done on an objecl by a conservative force is just the nega-
,..
live of the change in potential energy. Part (b) can be solved with conservation of energy, and pan (c) is an application of the \·,;ark-energy theorem.
..,
,
Solution
(a) Find the change in potential energy and the \.. . ork done by the force of gravity.
Apply Equation 7.21:
Substitute knO\'m quantities. The asteroid's initial position is effectively infinity, so l/r, is zero:
I1PE = (6.67 X 10- 11 kg- l m'js')(5.98 X 10" kg) X (1.00 X 10" kg)(
bYE =
Compute the work done by the force of gravity:
Wgm
I
8
4.00 X 10 m
+ 0)
-9.97 X 10 14)
= -I1PE =
9.97 X 10 1.1J
(b) Find the speed of the asteroid when it reaches rf = 4.00 X 10" m. 6.KE+ I1PE= 0
Use conservation of energy:
(~1II.V' - 0) - 9.97 X 10 H
v = 1.41
X
J=0
10' ml'
(c) Find the work needed to reduce the speed to 7.05 X 10 2 m/s (half the value just found) at this point. Apply the work-energy theorem:
W= 6.KE+ I1PE
The change in potential energy remains the same as in part (a), but substitute only half the speed in the kinetic-energy term:
Remark The amount of work calculated in part (c) is negative because an external agent must exert a force against the direction of motion of the asteroid. It would take a thruster with a megawatt of Olltput about 24 years to slow dm'ln the asteroid Lo hall' its original speed. An asteroid endangering Earth need not be slowed that much: A small change in its speed, ifapplied early enough, will cause it 1.0 miss Earth. Timeliness of the applied thrust, hm..·cver, is important. By the time you can look over your shoulder and see the Earth, it's already far too late, despite how these scenarios pIa}' OUl in Hollywood. Las[-minute rescues won't vmrkl QUESTION 7.12 As the asteroid approaches Earth, does the gravitational potential energy associated with the asteroid-Earth system (a) increase, (b) decrease, (c) remain the same?
7.5
Newtonian Gravitation
213
EXERCISE 7.12 Suppose the asteroid starts from rest at a great distance (effectively infinity), falling toward Earth. How much work would have to be done on the asteroid to slow it to 425 m/s by the time it reached a distance of 2.00 X 101i m from Earth?
Answer
-1.90 X
]01.'.1
WHY IS THE SUN HOP
APPLYING PHYSICS 7.3
Explanation The Sun formed when part icles in a cloud of gas coalesced, due to gravitational attraction, into a massive astronomical objec!.. Before this occurred, the particles in the cloud ,.. 'ere widely scattered, representing a large amount of gravitational potential energy. As the particles fell closer together, their kinetic energy increased, but the gravitational potential energy of the system decreased, as required by the conservation of energy. \,Vith further slow collapse, the cloud became more dl:nse and the average
kinetic energy of the particles increased. This kinetic energ}/ is the internal energy of the cloud, which is proportional LO the temperature. If enough particles come together, the temperaturc can risc to a point at which nuclcar fusion occurs and the ball of gas bccomes a star. Otherwise, the temperature may rise, bllt not enough to ignite fusion reactions, and the object becomes a brown dwarf (a failed star) or a planet.
On inspecting Equation 7.21, some may wonder ",hat happened to mglt, the gravitational potential energy expression introduced in Chapter 5. That expression is still valid when h is small compared with Earth's raclius. To see this, we write the change in potential energy as an object is raised from the ground to height h, lIsing the general form for gravitational potential energy (see Fig. 7.20):
PI'2 - PE, =
_C
(
1VIl-fIl
RL
+
= -CM/..",[
.
h
)
(-
C i\1f
h
nl)
Rf
1 (R,.+ h)
_~] R"
After finding a common denominator and applying some algebra, ,ve obtain ~
PE, - PE,
G.Nl,-:Ulh
= -c:-=~--c, R"(R,, + h)
v\Then the height Ii is very small compared with ond factor in the denominator, yielding
R/~·'
Ii can be dropped from the secFIGURE 7.20
Rdating the general rorm ofgravitatiollClI pOlemial energy 10 mgh.
SubstitLlting this into the previoLls expression, we havc PE, - PE,
Nmv recall from Chaptc:,- 4 thal the free-faJi acceleration given by g = CIVIE/Rl, giving
~ltthe
surface of Earth is
Escape Speed If an object. is projected upward from Eanh's surf~lce ",ith a large enough speed, it can soar offinlO space and never rellirn. This speed is called Earth's escape speed. (It is also commonly called the esca/Je velocil)', but in fact is more properly a speed.)
Chapter 7
214
Rotational Motion ond the Low of Gravity
TABLE 7.2
Escape Speeds for the Planets and the Moon Plaocl
Vet
(km/s)
Merrury Venus
4.3 10.3
Earth
11.2
Moon
2.3
Mars
5.0
Jupiter
60.0
Saturn
36.0
Uranus
22.0 24.0
Neptune Pluto:!
Earth's escape speed can be found by applying conservation of energy. Suppose an object of mass m is prqjeclcd vertically up\,,'ard from Earth's surface with an initial speed Vi' The initial mechanical energy (kinetic plus potential energy) of the object-Earth system is given by .
KJ!..i
August 2U06, Ihe International ASlronomical UniOJl 'ldopred" ddillition ora planellhal sepOlrmcs Pluto frulll Ihe Mher cighl planets, Pluto i~ IlOW ddined a" a ~(h~-arr planel ~ (I ike rhe asteroid Ceres).
Gi\1£m
<.
I
= ~mvi- - - - -
RE
\Ve neglect air resistance and assume the initial speed is just large enough to allow the object to reach infinity with a speed of zero. This value of Vi is the escape speed vesco ''''hen the object is at an infinite distance from Earth, its kinetic energy is zero becausc VI = 0, and the gravitational potential energy is also zero because 1/1" goes to zero as r goes to infinity. Hence the total mechanical energy is zero, and the lav.' of conservation of energy gives I
')
'imv~~(
1.1
~In
.
+ PE;
Gl\1r m
- --"- =
Rt
so thal vesc -_
0
~2CM" ---
[7.22]
R"
The escape speed [or Earth is about 11.2 km/s, which corresponds to about 25 000 mi/h. (See Example 7.13.) Note that the expression for V CSl" doesn't depend on the mass of the object projected from Earth, so a spacecraft has the same escape speed as a molecule. Escape spceds [or the plancts, the Moon, and the Sun are listed in Table 7.2. Escape speed and temperature determine to a large extent \,"'hether a world has an atmosphere and, if so, \.. . hat the constituents of the atmosphere are. Planets \·.. ith low escape speeds, such as \rlercury. generally don't have atmospheres because the average speed of gas molecules is close to lhe escape speed. Venus has a very thick atrnosphere, but it's almost entirely carbon dioxide, a heavy gas. The atlllosphere of Earth has very little hydrogen or helium, but has retained the much heavier nit.rogen and oxygen molecules.
EXAMPLE 7.13 Goal
From the Earth to the Moon
Apply conservation of energy \·...ith the general form of Newton's universal law of gravity.
Problem rnJules Verne's classic novel From lhe Earth 10 the Iv/oon, a giant cannon dug into the Earth in Florida fired a spacecrafL all the way to the Moon. (a) If the spacecraft leaves the cannon at escape speed, at what speed is it moving when 1.50 X lOr, km from the cenler of Earth? Neglect any friction effects. (b) Approximately what constant acceleration is needed to propel the spacecraft to escape speed through a cannon bore l.00 km long? Strategy For part (a), usc conservation of energy and solve for the final speed time-independent kinematic equation: solve for the acceleration a. ~.,.,
,
..
.
,
.
vf
Part (b) is an application of the
,
..,
,
Solution (a) Find the speed at r = 1.50 X 10' kill. Apply conservation of energy:
Multiply by 21m and rcarn1l1ge, solving for v/. Then substitute known values and take the square root.
v/ = .,
, vi -
2CM 2CM +- -E - - -E 'Ii Rt
4 "U( = (1.12 X 10 1ll!5)'
=
.,
Vi~
(1 1)
+ 2 Glvh· - - 'i RE
+ 2(6.67 X 10- 11 kg- 1m's-2)
X (5.98 X 1024 kg )(
I, 1.50 X 10 m
1) 6.37 X 10' In
7.6
Kepler's Laws
215
(b) Find the acceleration through the cannon bore, assuming it's constant. Use the time-indepcndcnt kinern;:1lics equation: (1.12 x 10' m/s)' - 0 = 2n(1.00 x 10' m)
a=
6.27 X 10-' mis'
...........
ll.... •••
Remark This result corresponds to an acceleration of over 6000 times the free-fall acceleration on Eanh. Such a huge acceleration is far beyond what the hum:<\11 body can tolerate, QUESTION 7.13
Suppose the spacecraft managed to go inLO an elliptical orbit around Earth, \'lith a nearest point (perigee) and farthest point (apogee). At which point is the kinetic energy of the spacecraft higher, and ,"vhy? EXERCISE 7.13
Using the data in Table 7.3 (see page 217), find (a) the escape speed from Lhe surface of Mars and (b) the speed ofa space vehicle when it is 1.25 X 10 7 III f"rom the center of Mars if it leaves the surface at the escape speed. Answer
7.6
(a) 5.04 X 10' m/s (b) 2.62 X 10' m/s
KEPLER'S LAWS
The movements of the planets, stars, and other celestial bodies have been observed for thousands of years. In eady history scientists regarded Earth as the center of the Universe. This geocentric model was dC\'eloped extensi\'e1y by the Greek astronomer Claudius Ptolemy in the second century A.D. and was accept.ed for the next I 400 years. In 1543 Polish astronomer :\"icolaus CoperniclIs (1473-1543) showed that Earth and the other planet.s revolve in circulal- orbits around the Sun (Lhe heliocentric model). Danish astronomer Tycho Brahe (pronounced Brah or BRAH-huh; 1546-1601) made accurate astronomical measurements over a period of20 years, providing the data {or the currently accepted model of the solar system. Brahe's precise observations of the planets and 777 stars \.. .ere carried out with nothing morc elaborate than a large sextant and compass; the telescope had not yet been invented. German astronomer Johannes Kepler, who was Brahe's assistant, acquired Brahe's astronomical data and spent about 16 years trying to deduce a mathematical model for the motions of t.he planets. Aft.er many laborious calculations, he found thal Brahe's precise data on the motion or Mars about the Sun provided the answer. Kepler's analysis first shmved that the concept of" circular orbits abollt the Sun had to be abandoned. He eventually discovered that the orbit of Mars could be accurately described by an ellipse with the Sun at one foclls. He then generalized this analysis to include the motions of' all planets. The complete analysis is summarized in three statements known as Kepler's laws: 1. All planets move in elliptical orbits with the Sun at one of the focal pojnt.s. 2. A line drawn from t.he Sun to any planet s\-veeps out equal areas in equal t.ime intervals. 3. The square of the orbital period orany planet. is proportional to the cube of the average distance from the planet to the SUIl. Newton lat.er demonstrated that these laws are consequences of the gravitational force that exists bel\veen any t\VO ol~jects. Newton's law of universal gravitation, together with his laws of motion, provides the basis lor a full mathematical description of the motions of planets and satellites.
216
Chapter 7
Rotational Motion and the Law of Gravity
~ Focus
Focus
/
(a)
Sun
.-/
Planet
Kepler's First Law The first law arises as a natural consequence of the inverse-square nature of Newlon's Imv of gravitation. Any object bound Lo another b)' a rOITC lint varies as 1/1,2 will move in an elliptical orbiL. As shown in Active Figure 7.2] a, an ellipse is a curve drawn so that the sum of" the distances from any point on the curve to two internal points called focal/Joints or Jod (singular, focus) is always thc samc. The semimajor axis a is half the length of the line that goes across the ellipse and cOlllains both foci. FOI" the Sun-planet configuration (Active Fig. 7.2Jb), the Sun is at one focus and the othcr focus is empty. l3ecausc the orbit is an ellipse, the distance from the Stln to the planet continuously changes.
(bl
Kepler's Second Law
ACTIVE FIGURE 7.21 (a) The sum I) + q is rhe same for en'r)' poim all rhe ellipse. (Ill In the
Solar S'"Slem, the Sun is at OIIC focus of the elliptical orbit uf each planet and the ollwl" [OCllS is empty.
Kepler's Third Law The derivation or Kepler's third law is simple enough to carry out for the special case of a circular orbit. Consider a planet of mass Mp moving around the Sun, \vhich has a mass of Nl s, in a circular orbil. Because the orbit is circular, the planet moves at a constant speed v. Newton's second law, his law of gra\'itation, and centripetal acceleration t.hen give the follm"'ing equation: MI,v" MI,(lr= - r
GMsM" <'
r-
The speed v of the planet in its orbit is equal to the circumference of the orbir divided by the timc required 1'01" one revolution, 1~ called the period of the planer, so v = 2';n/T Substituting, the preceding expression becomes GM.,
(27T1/ T)'
r~
T
[7.23]
Kepler's third low ~
where K s is a constant given by 47T~
Ks = - - = 2.97 X 10- 19 s~ /m 3 GMs
Equation 7.23 is Kepler's third la.." for a circular orbit. The orbits of most of the planets are very nearly circular. Comets and asteroids, however, usually have elliptical orbits. For these orbits, the radius r must be replaced \vith ft, the semi major axis-half the longest distance across the elliptical orbit. (This is also the average distance of the comet or asreroid from the Sun.) A more detailed calculation shows that K s actually depends on the sum of both the mass of a given planet and the Sun's rnass. The masses of the planets, however, are negligible compared with the Sun's mass; hence can be neglected, me<1ning Equat.ion 7.23 is valid for any planet in the Sun's family. rr we consider the orbit of a satellite such as the Moon around Earth, then thc constant has a differcnt value, with the mass of the Sun replaced by the mass of Earth. In that case, KEequals 4rr2 /GM E .
7.6
Kepler's Laws
217
TABLE 7.3 Useful Planetary Data
Mass
Body
Mean Radius (10)
(kg)
Mercury
3.18 X 10'"
Venus
4.88 X 10 2.1
Earth
5.98 X 102.J 6.42 X 10 2 :'\
Mars Jupiter
Sawrn Uranus Neptune Pluto" Moon
Sun ~III
the
1.90 5.68 8.68 1.03
X 10 27 X 102r. X 10 2[, X 10"
2.4~ X 10" 6.06 X lOG 6.38 X 10"
3.37 X 106
6.99 X
la'
5.85 X 10 7
2.33 X 2.21 X
la' la'
1023
1.1'l X 10(;
7.36 X 1022 1.991 X 1030
1.74 X 10" 6.96 X LQI'
1.27 x
Period (5)
X 10" X la' X la' X 10' 3.74 X 10"
7.60 1.94 3.156 5.9'1
9.35 X IO H 2.64 X 10\1 5.22 X 10'1 7.82 X 10'1
to.'fean Distance from Sun (m) 5.79 X 1011)
1.08 X 1.496 X 2.28 X 7.7R X 1,43X
JO" 10" 10" 10" 10"
2.87 X ]0 12
10 12 5.91 X lOI'.!
4.50
X
Augllst 2006. the [Illcrnaliollal Astronomical Unitln adopu·d.1 definilion of;1 plane-tlh;H 01 her eight planets. PllUO is now defilled as a ~(h";lrrplalll·t~ like tilC asteroid (:crcs.
-1O~19
T'
r: 1
("') In 3
2.97 2.99 2.97 2.98 2.97
2.99 2.95 2.99 2.90
~ep;lr;I\"" Pluto
rwm
The mass of the Sun can be determined from Kepler's third law because the constant Ks in Equation 7.23 includes the mass of the Sun and the other variables and constants can be easily measured. The value of this constant can be found by substituting the values of a planet's period and orbital radius and solving for K s . The mass of the Sun is then 47T~
M -S CK,
This same process can be used to calculate the mass of Earth (by considering the period and orbital radius of the Moon) and the mass of other planets in the solar system that have satellites. The last column in Table 7.3 confirms that T'l./r: 1 is very nearly constant. \rVhen time is measured in Earth years and the semim~or axis in astronomical units (l AU = the distance from Eanh to the Sun), Kepler's law takes the following simple rorm: T"2 = a~\ This equation can be easily checked: Earth has a semimajor axis orone astronomical unit (by definition), and it takes one year to circle the sun. This equation, of course, is valid only for the sun and its planets, asteroids, and comets. QUICK QUIZ 7.10 Suppose an asteroid has a semimajor axis or 4- AU. How long does it take the asteroid to go around the sun? (a) 2 years (b) 4 years (c) 6 years (d) 8 years
EXAMPLE 7.14 Goal
Geosynchronous Orbit ond Telecommunicotions Sate"i;:.te;:;s::.-
Apply Kepler's third law
LO
~_ _./
an Earth satellite.
Problem From a t.elecommunications point of view, it's advantageous for satellites to remain at the same location relative to a location on Earth. This can occur only if the satellite's orbital period is the same as the Earth's period of rotation, 24.0 h. (a) At what djstance from the center of the Earth can th is geosynchronous orbit be found? (b) What's the orbital speed of the sat.ellite?
This problem can be solved ,....ith the same method that was Llsed to derive a special case of Kepler's third law, 'with Earth's mass replacing the Sun's mass. There's no need to repeat the analysis; just replace the Sun's mass with Earth's mass in Kepler's third law, substiwte the period T (com'ened to seconds), and solve for 1: For part (b), find the circumference or the circular orbit and divide by the elapsed time. Strategy
Solution (a) Find the distance no geosynchronous orbit.
Apply Kepler's third law:
T' = ( 47T' ),.,
Substitute the period in seconds, T= 86400 s, the gravity constant C = 6.67 X 10- 11 kg-11l13/S~, and the mass of the Earth, Mt: = 5.98 X 10 2.1 kg. Solve for '1':
r=
GM"
(b) Find the orbital speed:
4.23 X 10 7 m
rl
2'77"'"
27T( 4.23 X J 0 7 m)
T
T
8.64 X 10" s
1J=-=--~
.
~
. ••••• ,
3.08
X ]0'
m/s
• • • • • • • • • . • • • • • ..<11
Remarks Both these results arc independent or the mass of the smcllilc. Notjce thar Earth's mass could be found substi[Uting the .\Iloon's distance and period into this form of Kepler's third law.
b~'
QUESTION 7.14 ]f the satellite was placed in an orbit three times farther a\vay, about 11m·.... long \·...ould it take to orbit the Eanh once? Answer in days, rounding to one digit. EXERCISE 7.14 Mars rotates on its axis once every 1.02 days (almost the same as Earth does). (a) Find the distance from Mars at \"'hich a satellite would remain in one spot over the Manian surface. (b) Find the speed of the satellite.
Answer
(a) 2.03 X 10 7 m (b) 1.45 X lO"l11/s
SUMMARY 7.1
Angular Speed and Angular Acceleration
The average angular speed w;J.,. of" a rigid object is defined as the ratio of the angular dispbcemcnt tifJ to the time interval tif, or
w a,'
~
01- 0, -If -
tI
t:.o =ill
[7.3]
where w,w is in radians per second (r(l(1/s). The average angular acceleration O'av of a rotati ng ol~ject is defined as the ratio orthe chang-c in angular speed L\w 10 the time interval Ill, or W
~
IX
a'
where
O'av
J -
w,
tiw
If -
I;
fj.t
7.3 Relations Between Angular and Linear Quantities Whcn an objcct rolatcs about a fixed axis, the angular spced and angular acceleration are related to the tangential spced and tangcntial acceleration throug·h the relationships
[7.5]
is in radians per second pcr second (rad/s1 ).
[i.10]
and [7.11]
7.4 Centripetal Acceleration Any object moving in a circular path has an acceleration directed to\vard the center 01" the circular path, called a centripetal acceleration. Its magnitude is given by
., = -,. = 1)-
7.2 Rotational Motion Under Constant Angular Acceleration Ifan ol~jcct undergoes rotal.ionallllotion about a fixed axis under a constant angular acceleration 0', its mOlion can be described with the following set of equations: W = Wi
fj.8 =
w2 =
+ a.l
w,' + ~ c(/':!. wp + 2aL\fJ
[7.7] [7.8] [7.9]
Problems are sol\"cd as in one-dimensional kinematics.
fl. r
"
1"W~
[7.13,7.17]
An)' object moving in a circular path must have a net force exerted on itl-hat is directed toward the ccnter of the path. Some examples of forces that cause centripetal acceleration are the force of gravity (as in the motion of a satellite) and the force of tension in a string.
7.5 Newtonian Gravitation Newton's law of universal gravitation states thal every parlicle in the Univcrse all.racts every other particle with a force that is directl)' proportional to the product of their
Multiple-Choke Questions
masses and inversely proponional to the square of the di,,tance 1· between them: ,..~
In,m.,
G -o- -
,.-
[7.20]
where G = 6.673 X 10- 11 _ . m 2 /kg 2 is the constant of universal gravitation. A general expression for gravitational potential energ) is PI!
~
l\I/-,-m
-G-·r
1. All planets move in elliptical orbits with the Sun at one of the focal points. 2. A line drawn from the Sun to any planet sweeps out equal areas in equal time inten·als. 3. The square of the orbital period of a planet is proportional to the cube of the aver-age distance from the planet to the Sun:
" (47T' GM ) ,
T- =
[7.21]
Th is expression reduces LO PE = IIIgh clost, to t he surface of Earlh and holds for other worlds through rcpl~cement of the mass AlE. Problems such as finding the escape "cloeily from Eanh can be solved by llsing Equation 7.21 in the conservation of energy equation.
219
s
r·
[7_23]
The th ird law can be applied to any large body and its system of satellites by replacing the Sun's mass with the body's mass. In particular, it can be used to determine the mass of the central body once the avcrage distance to a satellite ,and its period are known.
7.6 Kepler's Laws Kepler derived the following three la"·5 of planetary motion: FOR ADDITIONAL STUDENT RESOURCES. GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE·CHOICE QUESTIONS 1. Find the angular speed of Earth around the Sun in radians per second. (a) 2.22 XIO 6 rad/s (b) 1.16 X 10 7 rad/s (e) 3.17 X 10-' rad/s (d) 4.5~ X 10 7 radls (e) 1.99 X 10- 7 rad/s
2. A grindstone increases in angular speed from 4.00 ,-ad/s to 12.00 rad/s in 4.00 s. Through what angle does it turn during that time if the angular acceleration is caoSlant? (a) 8.00 rad (b) 12.0 rad (e) 16.0 n,d (d) 32.0 rad (e) 64.0 rad 3. A C)'c1ist rides a bicycle with a ,,,,·heel radius of 0.500 m across campus. A piece of plastic on the front rim makes a clicking sound ever-}" time it passes through the fork. If the cyclist counts ~WO clicks between her apartment
ing this object? (ll) If the angular speed is constant, the oqjcct must have constant tangemial speed. (b) If the angular speed is constant, the object is not accelerated. (c) The object has a t<"l_ngential acceleration only if the disk has an angular acceleration. (d) If the disk has an angular acceleration, the object has both a centripetal and a tangential acceleration. (e) The object always has a centripetal acceleration except when the angular speed is zero.
7. The gnl\"itational force exerted on an astronaut on Earth's surface is 650 N down. \·\Then she is in the International Space Station, is the gra\"itational force on her (a) larger, (b) exactl)' the same, (c) smaller, (d) nearly but not exactly zero, or (e) exactly 7ero? 8. An object is located on the surface of a spherical planet of mass 1\1 and radius R. The escape speed from the planet does no! depend on ,.. . hich of the following? (a) M (b) the density of the planet (c) R (d) t.he acceleration due to gravity on that planet (e) the mass of the object 9. A satellite moves in a circulal" orbit at a constant speed around Earth. Which of the following statements is true? (a) 1'\0 force acts on the satellite. (b) The satellite mO\'es at constant speed and hence doesn't accelerate. (c) The satellite has an acceleration directed a'..·ay from Earth. (d) The satellite has an acceleration directed toward Earth. (e) \\'ork is done on the satellite by the force of gravity. 10. \,rhich of the following statements arc tl"ue of an object in orbit around Earth? (a) If the orbit is circular, the gravity f()rce is perpendicular to the object's velocity. (b) If the orbit is elliptical, the gravity force is perpendicular LO the velocity \'ector only at the nearest
220
Rotational Motion and the Law of Gravity
Chapter 7
and fttnhcSl points. (c) If the orbit is not circular, the speed is greatest when the object is fanhesL away from Earth. (d) The gravity force on the object always has components both parallel and perpendicular LO the obiecl's velocity. (e) All these statements are true.
12. A syslem consists of four particles. How many lerms appear in the expression for (he total gravitational potential energy ot" the system? (a) 4 (b) " (e) 10 (d) 12 (e) None of these
11. \"hat is the gravitational acceleration close to the surface ofa planet with twice the mass and twice the radius of Eanh? Answer as a multiple of g, the gravitational acceleration near Earth's surface. (a) O.25g (b) O.5g (c) g (d) 2g (e) 4g
and moves in an elliptical orbit. in which its distance from the Sun at closest approach is a small fraction of its maximum distance. Est.imate the comet.'s maximum distance from the Sun in astronomical units AU (the distance from Earth to the Sun). (a) 3 AU (b) 6 AU (c) 10 AU (d) 18 AI.: (e) 36AU
~.,-~-
-
13. Halley's comet has a period of approximately 76 years
---:::V---~~--"'.
CONCEPTUAL
QUE~TIONS
I. In a race like the Indianapolis 500, a drivel" circles the track counterclockwise and feels his head pulled toward one shoulder. To relieve his neck muscles from having to hold his head erect, the driver fastens a strap to one wall of the car and the other to his helmet. The length of the strap is aqjusted to keep his head vertical. (a) Which shoulder does his head tend to lean toward? (b) What force or forces produce the centripetal acceleration when there is no strap? (c) \'\"hat force or forccs do so when there is a strap?
2. Two schoolmates, Romeo andJuliet, catch each other's eye across a crowded dance floor at a school dance. Find the order of magnitude of the gravitational attraction thatJuliet exerLs on Romeo and that. Romeo exerts on Juliet. Stale the quantities you take as data and the values you measure or estimate for them.
3. If a car's wheels are replaced with wheels of greater diameter, will the reading of the speedometer change? Explain. 4. At night, yOli are Llrther away from the Sun than during the day. What's more, the force exerted by the Sun on you is downw·ard into Earth at night and upward int.o the sky during the day. If YOll had a sensitive enough bathroom scale, would you appear to weigh more al night than during lhe day?
6. Becausc of Earth's rotation aboul its axis, yOll weigh slightly less 3tthe equator than at. the poles. Explain.
7. [t has been suggested that rotating cylinders about 10 miles long and 5 miles in diameter be placed in space for colonies. The purpose of their rotation is to simulate gravit), for t.he inhabitants. Explain the concept behind this proposal.
8. Describe the path of a moving object. in the evcnlthat the object's acceleration is constant in magnitude at all times and (a) perpendicular to its velocity; (b) parallel to it.s velocity. 9. A pail of ,,,,'ater can be whirled in a vertical circular path such t.hat no waLer is spilled. Why does the water remain in the pan, even when the pail is upside down above you I" head?
10. Use Kepler's second law to convince yourself that Earth must move faster in its orbit during the nonhernhemisphere winter, when it is closest to the Sun, than during the summer, when it is farthest from the Sun. II. Is it possible for a car to move in a circular path in such a way that it has a tangential acceleration but no centripetal
12.A satellit.e in orbit is not truly traveling through a
5. Correctlhc following stat.emcnt: ;'The race car rounds
vacuum-it's moving through ver)' thin air. Does the resulting ail- friction cause the satellite 1.0 slow down?
the turn at a constant velocit.y of90 miles per hOllr." __,
-.
-
f
PROBLEMS ~
The Problems rOI-this chapler IllOl}' be
WebAssignassigned onlillC at WcbAssign. 1,2.3 = straightforward. intermediate, = d('notes guided problem
m
m:I = denotes enhanced content
II =
(h,t1leng-ill~
problem
biomedical 'lpplicOlI ion
0= denotes full solmion :wailablc in
Sfwf/'lll Solulions Manual!
Stud)" Guide
SECTION 7.1 ANGULAR SPEED AND ANGULAR ACCELERATION
[TIE (a) Find the Clngular speed of I~anh's rotation aboul its axis. (b) How does (his rotation affect the shape or
Earth?
2. A wheel has a radius of 4.1 Ill. How far (path lenglh) does a point on the circumference travel iflhe wheel is rotated thl'ough angles of 30°, 30 rad, and 30 rev, respectively?
3. The lires on a new compact car have a diameler 01'2.0 [t and are warranted for 60 000 miles. (a) Determine the angle (in radians) through which one 01" these tires will rotale during the warrant)' period. (b) How man)' revolutions of the lire are equivalent to your answer in pan (a)? 4.
E
A pott.er's wheel moves uniformly from rest 1.0 an angular speed of 1.00 rev/s in 30.0 s. (a) Find its angular acceleration in radians per second per second. (b) 'Would doubling the angular acceleration during the given period have doubled fi nal angular speed?
Problems
SECTION 7.2 ROTATIONAL MOTION UNDER CONSTANT ANGULAR ACCELERATION SECTION 7.3 RELATIONS BETWEEN ANGULAR AND LINEAR QUANTITIES
221
does the tub turn during the entire 20-s interval? Assume constanl angular acceleration while il is starting and stopping. II. A car illitially traveling al 29.0 lIl/S undergoes a constant ncgativc ~lccclcl'alioll of magnitude 1.75 m/s9 after its
5. A demisI's (II-ill SlafL'i from rest. AfLcr 3.20 S 01 constant angular acceleration, it turns at a rate of2.51 X lO~ rev/min. (a) Find the ell"ill's angular acceleration. (b) Determine the angle (in radians) through which the drill rOl.lIes during this period.
brakes arc applied. (a) How mall\ revolulions does each tire makc before the car comes to a stop. assuming the cal' docs not skid and the tires have radii of 0.330 m? (b) What is the angular speed of the wheels when lhe car has traveled half the total distance?
6. A cenu"ifugc in a medicallaborator)' rotates at an angular speed of 3 600 rev/min. When switched off, it 100ates through 50.0 rC\'OIUlions befo'"c coming to rc'it. Find the constant angular acceleration of the centrifuge.
12. A 45.0~cm diametCl" disk rotate!) with a constam angular acceleration of2.50 rad/s 2 . It starts rwm rest at t = 0, and a line dnl\'m from the center of the disk to a poim Pan the rim of the disk makes an angle of 57.30 with lhe positi\'e x-axis at this time. At t = 2.30 s, lind (a) the angular speed of the wheel, (b) the linear velocity and tangential accelennion of P, and (c) the position of P (in degrees, with respect to the positive x~axis).
7.
mil
A machine pan rotates £II an angular speed of 0.06 rad/s; its speed is then increased £0 2.2 nulls at an angular acceleration of 0.70 rad/s~. (a) Find the angle through which the pari mImes befo'"c reaching this final ~peed. (b) In general. ifbmh the initial and final angular speed arc doubled
8. _ t\ bicycle is turned upside do\\ n while it~ owner repairs a flat tire. A friend spins the other wheel and obsen'es Ihat drops of water fly off tangentially. She: measures the heights reached by drops lIloving vertically (Fig. P7.8). A drop that breaks loose from the tire 011 Olle turn rises veniealJy 54.0 em above the tangent point. A. drop lhat breaks loose on the next lUrn rises 51.0 Clll ;t1)O\'l: Ihe tangent point. The radius of the wheel is 0.381 Ill. (a) \oVhy docs the first drop rise higher thall the second drop? (b) Neglecting air friction and using only the ob:-.erved heights alld the radius oflhe wheel,lInd the wheel's al1gu~ lar accelenllion (assuming ilLO be constant).
o:IJ.-\ rotating wheel requires 3.00 s to rotate 37.0 revolutions. Its angular velocit) at the end of the 3.00-s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel? 14. An electric motor rotating a workshop grinding wheel at a rate or 1.00 X 102 rev/min is switched off. Assume the wheel has a constanl.llcgative angular acceleration of magnitude 2.00 rad/s 2 . (a) Ilow long does it take for the g-rind~ ing whcclLO stop? (b) Through how many radians has the wheellllrncd during the imenal/ound in part (a)?
SECTION 7.4 CENTRIPETAL ACCELERATION 15. _ find the centripetal accelerations due to Earth's rotation about it:) axis of a man standing (a) at the equa~ tor and (b) at the North Pole. (c) "Vhat (wo forces combine to create these centripetal accelerations? 16. It has been suggested that rotating cyJindel-s about 10 mi long and 5.0 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal accelcration at its surface equals the free·fall accelenuion on Earth?
[]I] (a)
FIGURE P7.8
Problems 8 alld 69.
[]JThe diameters orthe main rotor and I..lil rotor ora sillgleengine helicopter arc i.60 m and 1.02 m, respccti,·c1y. The respeCli,·c rotational speeds an; 450 rn min and 4 138 rev/min. Calculate the speeds of the lips of both rotors. Compare these speeds with the speed of sound, 343 m/s. 10. The tub of a washer goes into its spin-dr), cycle, startillg from rest
\'\'hat is the tangential acceleration of a bug on the rim of a IO-in.-diameter disk if the disk moves from rest to an angular speed of78,-cv/min in 3.0 s? (b) When the disk is at its final speed, what is the tangential velocity or the bug? (c) One second after the bug starts from rest. what aloe its tangential acceleration, centripetal acceleration, and total acceleration?
18. The 20-gccmrifugc at NASA's Ames Research Center in ~rotlntain View, California, is a cylind"ical tube 58 [tlong \\'ilh a radius or29 ft (Fig. Pi. IS). Ira rider sits in a chair
1 + - - - - 29 It ------1'1
FIGURE P7.18
222
Chapter 7
Rotational Motion and the Law of Gravity
at the end of one arm facing: the center, how many rc\'olu~ [ions per minute would be required to create a horizontal normal fOITC equal in magnitude to 20.0 limes the ,-icier's weight?
of one of thc habitats, relating the astronaut's mass JII, his velocity 11, his radial distance from the hub r, and the normal force /1. (c) What would n have to equal if the 50.a-kg astronaut is to experience half his normal E"rth weight? (d) Calcu[at.e the necessary tangential speed of the habitat from Newton's second law. (e) Calculate the angular speed from the tan~ential speed. (f) Calculate the period of rotat.ion I"rom the angular speed. (g) If the astronatll stands up, will his head he moving faster, slower, or at the same speed as his feet? Why? Calculate the tangential speed at lhe lOp or his head if he is [.80 m ta[l.
19. Pan of a roIlCI"-coaster ride involves coasting down an
incline and entering a loop 8.00 m in diamcLCr. For safely considerations, the roller coaster's speed al lhe top of the loop must be such that the force of the seal on a rider is equal in magnitude LO the rider's wcighl. From what height
mil A coin
rests 15.0 elll from the center of a turmable. The coefficient of static friction between the coin and turntable surface is 0.350. The tUrluable stans rrom rest at 0 and rotates with a const,lnt angular acceleration of 0.730 rad/s 2 . (a) Once the turntable stans to rotate, what rOITe causes the centripetal acceleration when dlC coin is stationary relative to the turnt,lblc? Under what condition docs the coin begin LO move relative to the turlllable? (b) After what period of time w111 the coin stan to slip on the turntable?
,=
21. A 5S.0-kg icc sk
,I
22. A race car starts frolll rest on circular track or radius 400 m. The car's speed increases at the constant rate of 0.500 Ill/S 2. Atthe pain! where the magnitudes orthe ccntripet.tl and tangemial accelerations arc equal, determine (a) the speed of the race car, (b) the distance traveled, and (c) the elapsed t.ime.
1<1.----10.0 III ------>I FIGURE P7.26
127.IAn air puck of mass 0.25 kg is tied to a string and allowcd LO revolve in a circle of radius 1.0 m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the Lable, and a mass of].a kg is tied to it (Fig. P7.27). The suspended mass rem
23. A certain liglll truck call go around a flat curve having a radius of 150 III with a maximum speed of 32.0 m/s. \'\'ith what maximum speed can it go around a curve having a radius of 75.0 m? 24.
II A sample of blood is placed
in a centrifuge of radius IS.Oem. The mass ora red blood cell is 3.0 X [O-lfikg, and the magnitude of the force acting on it as it settles out of the plasma is 4.0 X 10- 11 N. At how man)' revolutions per second should the ccmrifuge be operfncd?
FIGURE P7.27
An air puck of mass 111 J is tied to a string and allow·ed to revolve in a circle of radius R on a horizontal, frictionless table. The other end of the sU·ing passes through a small hole in the center of the table, and an object of mass m,! is tied to it (Fig. P7.27). The smpended object remains in equilibrium while the puck on the tablctop re\'olves. (a) Find a symbolic expression for lhc tension in the string in terms of UI"!. and g. (b) Write Newton's second law fOl· the air puck, using the variables ml , v, R, and T (c) Eliminate the tension T from the expressions found in parts (a) and (b) and find an expression for the speed of the puck in terms of 111 1, 111 2 , g, and R. (d) Check your answers by substittlling the values of Problem 7.27 and comparing the results with the answers [or that problem.
29.
mil
25. A 50.0-kg child stands at the rim of a merry-go-round of radius 2.0a m, rotating with an angular speed of 3.00 rad/s. (a) \,Vhat is the child's cemripctal acceleration? (b) ',I,.'hat is the minimum force between her feel and the noO!" orthe carousel that is requil"cd to keep her in the circular path? (c) Whal minimum coefficient or static friction is required? Is the answer you found reasonable? In other words, is she likely to stay on the mcrry-goround? 26.
EI
A space habitat for a long space voyage consists of two cabins each connected by a cable to a central hub as shown in Figure Pi.26. The cabins are set spinning around the hub axis, which is connected to the rest of the spacecraft to generate artificial gravity. (a) What forces are acting on an astronaut in one of the cabins? (b) W·rite Newton's second law [or an astronaml),ing on thc ~floor"
m:i1
28.
A woman places her briefcase on the backseat of her car. As she drives 1.0 work, lhe car negotiates an unbanked curve in the road that can be regarded as an arc of a circle of radius 62.0 m. While on the curve, the car's speedometer registers 15.0 m/s at the instam the briefcase slarts to slide ,IUOSS the backseat lOward the side of the car.
Problems (a) \Vhat force causes the centripetal acceleral ion of the briefCase when it is stationary relalive LO the car? Under what condition does the briefcase begin to move rclati\'c to the cae? (b) \Vhat is the coefficient of stalic friclion bel.\\'cen the briefcase and scat surface? 30. _ A pail of water is rotated in a vertical circle of r
(a) What is the cClHripclcll acceleration of the child? (b) What force (magnitude and direction) docs the seat exenon the child at the lowest point of the ride? (c) 'Vhat force docs the seat exert on the child
223
136.IAfter the Sun exhausts its nuclear fuel, its ultimate [ate may be to collapse to a while dwmfstate. In this state, it would have approximately the same mass as it has now, hut its radius wOLtld be crjl.lalto the radius of Earth. Calculate (a) the aver,lge density of the white dwarf, (b) the surface free-fall acceleration, and (c) the gravitational potential energy associated with a l.OO-kg object at the SUI"race of the white dwarf. 37. Objects with masses of 200 kg' and 500 kg are sep,lrated
by 0.100 m. (a) Find the net gravit
am
A rocket is fired straight up through the atmosphere from the South Pole, burning out at an alLitude of250 km when traveling at 6.0 km/s. (a) What maximum distance from Earth's sU1jarl' does it travel before falling back to E.arth? (b) \-\'ould its maximum distance increase if it were fired ('rom a launch sitC' on lhe equator? \t\l hy?
40. Two ol~jeCls attract each other with a gmvitalional force or magnitude 1.00 X 1O-1l N when separated by 20.0 em. If the lotal mass of the o~jects is 5.00 kg, what is the lllass of each?
SECTION 7,6 KEPLER'S LAWS
®
(
//-~--'
I
{
. 15
\ , ,
,
\ 11I \ I
I
I
'.....
41. A satellite moves in a circular orbit around Earlh at a speed of 5000 m/s. Determine (a) the satellite's altiLUde above the surface of Earth and (b) the period of the satellite's orbit.
/
...."
FIGURE P7,32
SECTION 7,5 NEWTONIAN GRAVITATION 33. The average distance sepanlting Earth and the 1\,Ioon is 384000 km. Use the data in Table 7.310 find the net gravitational force exerted by Earth and the Moon on a ~.OO X 10 4 _kg spaceship located halfway between them. 34. A satellite has a mass of 100 kg and is located at 2.00 X lOG III above the surface of Earth. (a) \,\'l1al is the potential energy associated with lhe satellite at this location? (b) 'What is the magnitude of the gravitational forct: on the satellite? 35. A coordinate system (in meters) is constructed 011 the surface of a pool table, and three o1::~jects are placed on the table as follows: a 2.0-kg o~ject at the origin orthe coordinate system, a 3.0-kg object at (0, 2.0), and a .t.O-kg object at (4.0, 0). Find the resultant gravitational force exerted by the other two objects on the object at the origin.
42. Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from its nearest point, 6 670 km from Earth's center, to its farthest point, the Moon, 385000 km from Earth's center. Nole: The average radius or "semimajor axis" is the average of the distance from Earth's center to the nearest and ral"thest points on the elliptical orbit. 43. 10, a sat.ellite ofJupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 X 10" km. From these data, determine the mass ofJupiter. 44. A 600¥kg satellite is in a circular orbit about Earth at a height
224
Rotational Motion and the Law of Gravity
Chapter 7
around Jupiter
La
study that planet's famous red spot.
Jupiter rotales once every 9.84 h. Usc the data of Table 7.3 to find the altitude orlhe satellite. 47. An artificial satellite circles Eanh in a circuhll" orbit at a location when~ the accclcnllion due La gl",wity is 9.00 Ill/52 , Determine the orbital period of the satellite. 48. Neutron stars afC extremely dense ol~jccts that arc formed from the remnants of supernova explosions. Man}' rotate
54. A OAOO-kg pendulum bob passes through the lowest paft of its path at a speed of 3.00 m/s. (a) \'''hat is the tension in the pendulum cable at this point if the pendulum is 80.0 em long? (b) When the pendulum reaches iL~ highest point, ""hat angle docs the cable make with the "'ertical? (c) What is the tension in the pendululll cahle when the pendulum reaches its highest point? 55.
m:J
56.
mil
\'ery rapidly. Suppose the mass of a certain spherical neulron star is twice the mass of the Sun and its radius is 10.0 km. Determine the greatest possible angular speed the neutron sLar can bave so thai the matter at its surface all the equator isjusl held in orhit by !he gravitational force.
149.1
One method of pitching a softball is called the "windmill" delivery melilOd. in which the pitchn's arm rotates through approximately 3600 in a vertical plane before the Ig8-gram ball is released at the lowest point of the circular motion. An experienced pilcher can throw a ball with a speed of98.0 mi/h. Assume the angular acceleralion is uniform throughout the pitching motion and take the distance between the softball and the shoulder joil1l to be 74.2 em. (a) Determine the "lng-ular· speed of the arm in rev/s
50. A digital audio compact disc carries data along a continuous spiral track from [he inner circumference of the disc to the mnside edge. E
53. The Solar Maximum r.,'fission Satellite was placed in a circular orbit aboUl 150 mi above Earth. Determine (a) the orbital speed of the satellite and (b) the time required for one complete revolution.
Show that the escape speed from the surface of a planet of uniform density is directly proportional to the radius of the planet.
ffiLJm:J
Because of Earth's rotation abollt its axis, a point on the equator has a centripetal acceleration of 0.034 0 m/s:!, whereas a point at the poles has no centripctal acceleration. (a) Show that, at the equator, the gravitational force on an object (the object's true ,,,eight) must exceed the object's apparent ,,;'eight. (b) \\That are lhe apparent weighls of a 75.0-kg person at the equator and at the poles? (Assume Earth is a uniform sphcre and take g = 9.800 Ill/s'.)
51. An athlete swings a 5.0G-kg bali horizontally on the end of a rope. The ball moves in a circle of radius 0.800 m at an angular speed of 0.500 rev/so What are (a) the tangential speed of the ball and (b) its centripetal aecelera~ tion? (c) If the maximum tension the rope can ..vilhstand before breaking is 100 ~', what is the maximum tangential speed the ball can have? 52. A cal' rounds a banked curve where the radius or curvature of the road is R, the banking angle is 8, and the coefficienl 01" static friction is J.L. (a) Octerm ine the range of speeds the car can have without slipping lip or down the road. (b) What is the range of speeds possible if R = 100 rn, () = 10 0 , and J.L = O.lO (slippery conditions)?
A car moves at speed v across a bridge made in lhe shape of a circular arc of radius I: (a) Find an expression for the normal force acting on the car when it is at the lOp of the arc. (b) At what minimum speed will the normal force become zero (causing the occupants of the car to seem weightless) if r = 30.0 m?
;-- L_: 1',
@
FIGURE P7.SS
59. In Raben Heinlein's The i\iloo'fllsa I-Iarsh ,~,tJislress, the colonial inhabitants of the Moon threaten to launch rocks down onto Earth if they are not given independence (or atleasl representation). Assuming a gun could launch a rock of mass 111 at [wice the lunar escape speed, calculate the speed of the rock as it enters Earth's atmosphere. (-iG.
mil A roller coaster travels in a circular path. (a) Identify the forces on a passenger at the top of the circular loop lhat cause centripetal acceleralion. Show the direc-
Problems tion of all forces in a sketch. (b) Identify the force~ on the passenger at the bollom of the loop that produce centripetal acceleration. Show these in {l skeL<:h. (c) Based on your alls\.. .ers (Q parts (a) and (b), at what poilll, top or bOllom, should the seat exert thc greall:st force on the passenger? (d) Assume the speed of the roller coaster is 4.00 mls at the top of the loop of radius 8.00 m. Find the force exerted by the seat on a iO.O-kg passenger at the top of the loop. Then. assuille the speed remains the same at l,he bOllom of the loop and (-ind the force exerted by the scat all the passenger Cltthis point. Are your answers consistent with your choice ol"answcrs l"or parts (a) and (b)~ you are
[§IJASSUlllC
62.
225
64. Casling 01" molLen melal is impollanl in many industrial processes. Centrifugal casting is used for manufaCturing pipcs, bearings, and many other strtlClllres. A cylindrical enclosure is rotated rapidly and steadily aboul a horizont.al axis, as in Figufe P7.64. Molten metal is poured into the rotating cylinder and then cooled, forming the finished product. Turning the cylinder at a high rotation ratc forces the solidifying metal strongly to the out· side. Any bubbles arc displaced toward the axis so that unwanted voids will not be presclll in the Glsting. Suppose a copper sleeve of inner radius 2.10 cm and outer radius 2.20 cm is to be cast. To eliminate bubbles and give high structural integrity, the centripetal acceleration of each bit of metal should be 100g. -What rate of rotation is required? State the answer in re\olutions per minute.
mJ
Figure P7.62 shows the elliptical orbil ora spacecraft around E.arth. Take the origin of your coordinate system to be at the center ofEanh.
®
FIGURE P7.64
®
FIGURE P7.62
(.1) On a copy of the figure (enlarged if necessary), draw vectors reprcsenti ng (i) the position of Ihe spacecraft when it is at@ and @; (ii) the velocity of thc spacecraft \...· heo it is at ® and @; (iii) the acceleration of the spacecraft when it is at@and
65. Suppose
® Make sure that each type of vector Gin be distinguished. Provide a legend that shows how each type is represellted. (b) Have you drawn the velocity vector at ® longer than, shoner than, or the same length as the one at@? Expl
mil
A skier starts at rcst at the top of a large hemispherical hill (Fig. P7.63). Neglecting f"ietion, show thaI the skier will leave the hill and become ail'borne at a distance It = R/3 below the top of the hill. (Hint: At this point, the normal force goes to zero.)
FIGURE P7.63
FIGURE P7.65
66. A Sllllllman whose mass is 70 kg swings I"rom lhe end of a 4.0-m-long rope along the arc ofa venical circle. AsSLtm· ing he starts from rest when the rope is horiLontal, find the tensions in the rope that arc required to make him follow his circular path (a) at the beginllingofhis motion, (b) at a hcight of 1.5 rn above the bottom of the circular arc, and (c) at the bottom ot"t.he arc. 67.
mJ
68.
mJ
A minimum-energy orbit to an outer planet consists of putting a spacecraft on an elliplicaltrajectory with the departure planet corresponding to the perihelion of the ellipse, or closest point to the Sun. and the arrival planet corresponding to the aphelion of the ellipse, or farthest point from the Sun. (a) Usc Kepler's third law to calculate hO\,,' long it ......ould take to go from Earth to r-,'lars all such an orbit. (Answer in ye'lrs.) (b) Can such an orbit be undcrtaken at any time? Explain. Thc pilot of an airplane executes a constant-speed loop-the-loop maneuver in a vertical circle as in Figure 7.15b. The speed of the airplane is 2.00 X 1O~ m/s, and
226
Chapter 7
Rotational Motion and the Low of Gravity
the radius of the circle is 3.20 X 10 3 rn. (a) What is the pilot's apparent weight
seat on his body. Under whal conditions docs this occur? (d) What speed would have resulted in the pilot experiencing weigllliessncss at the top of the loop? 69.
force must equal the bat's weiglll. The horizontal compolIent of the force provides the centripetal acceleration. (a) What is the minimum speed that the bat can have if its mass is 0,031 kg? (b) If the maximum speed of the bat is 10 m/s. what is the maximum banking angle that allo\\'5 the bat to star 111 a horizontal plane:- (c) \-"hat tS the radius of [he circle of its flight when Ihe bat flies at its ma.xilllulll speed? (d) Can the bat mrn with a smaller radius b} flying more slowl),?
am
A piece of mud is initiall}' at pain! it on the:: rim ofa bicycle wheel of radius R rotating counterclockwise abom a horizontal axis at a COnsl
170.lam
A O.275-kg object is swung in a vertical circlllar path on a string 0.850 m long as in Figure P7.iO. (a) \\'haL are the forces acting on the ball , at allY point along Ihis path? (b) Draw frce-body dia~rams for the ball when it is at the boltom \ of the circle and whl.:ll il is at the \ , top. (c) Ifits speed is 5.20 mls at ____ - ' III the tOP of the circle. what is the tension in the string there? (d) If FIGURE P7.70 lhe string breaks when ils tension exceeds 22.5 :\. what is the maximum speed the object can ha,'c al the bottom before the string breaks?
,
, ,,
,,
71. A -l.OO-kg object is attached to a vertical rod by two sU'ings as shown in Figure Pi.il. The object rotates in a horizontal circle at constant speed 6.00 m/s. Find the tension in (a) the upper string and (b) the lower string.
2.00
III
.\lg FIGURE P7.72
73. (a) A luggage carousel at all airport has lhe form of a section or a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward IOwaI'd the olltside, making an ang-Ie of 20.0° with the horizontal. A 30.0-kg piece of luggage is placed all the carousel, 7.46 III from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of static friction bet ween the bag a nd the carousel. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rota lion, and the piece of luggage is bumped to a position 7.94 m from the axis of I"Ot
L
,
-----_
2.00m
fiGURE P7.71
175.11 n a popular- amusement park ride. a rotating qlindel· 72.
The maximum lift force on a bat is proportional to the square of its flying speed v, For the hoary bat (LasiU11lS cinereus), the magnitude of the lift fOI'ce is given by FI. S (0.018 I': . S2/m :!) v 1
The bat can fir in a horiz.ol1lal circle b) "banking" it::. wings at an angle 0, as showll ill Figure P7.i2. In Ihis situation, the magnitude of the verlical component of the lift
of radius 3.00 III is set in rotation at an angular speed of 5.00 rad/s, as in Figure P7.75. The floor then drops away. leaving the riders suspended against the wall in a vertical position. \'\-'hat minimum coefficicntoffriction between a rider's clothing and the wall is needed to keep the rider from slipping? (/lint: Recall that the magnitude of the maximum force of static friction is equal to /L'I, where 11 is the normal force-in this case, the force causing the centripetal acceleration.)
Problems
227
are rrinionlcss. except for the section of track between points rl and B. Given that the coefl1cicl1t of kinetic friction between the block and the track along AB is J.J., = 0.30. and thai the length of AB is 2.5 m, determine the minimum compres:-.ion d of the spring thal enables the block to just make it thmugh tht: loop-the-loop at point C. (Hill I: The force exened b\ the u"ack on the block will be 7ero irthe block barcl)' makes it through the loop-the-Ioop.)
.c
FIGURE P7.7S
76. A massless ~prinl;" of constant Ii = 7H.4 N/m is fixed on the left side of a level track. A block of mass III = 0..:')0 l..g is pressed against the srl"ing and cOlllpres~es it a dislancc cl, as in Figure P7.76. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R = 1.5 m. The entire track and the loop-the-loop
I<
-WmJJ'r'
~""-,..-------;;"'-B 1',
FIGURE P7.76
.....
Rotational motion is key in the harnessing of energy for power and propulsion, as illustrated by Q steamboat. The rotating paddlewheel, driven by a steam engine, pushes water backwards, and the reoction force of the water thrusts the boot forward.
8.1
Torque
8.2
Torque and the Two Conditions for Equilibrium
8.3 8.4
The Center of Gravity
>
~
= ....._"e
Examples of Objects in Equilibrium
8.5
Relationship Between Torque and Angular Acceleration
8.6 8.7
Rotational Kinetic Energy
ROTATIONAL EQUILIBRIUM AND ROTATIONAL DYNAMICS
Angular Momentum In the study of linear motion, objects were treated as point particles without structure. It didn't matter where a farce was applied, only whether it was applied or not. The reality is that the point of application of
0
force does matter. In football, for example,
if the ball carrier is tackled near his midriH, he might carry the tackler several yards before foiling. If tackled well below the waistline, however, his center of mass rotates toward the ground, and he can be brought down immediately. Tennis provides another good example. If a tennis ball is struck with a strong horizontal force acting through its center of mass, it may travel a long distance before hitting the ground, far out of bounds. Instead, the same force applied in on upward, glancing stroke will impart topspin to the ball, which con cause it to land in the opponent's court. The concepts of rotational equilibrium and rotational dynamics are also important in other disciplines. For example, students of architecture benefit from understanding the forces that oct on buildings and biology students should understand the forces at work in muscles and on bones and joints. These forces create torques, which tell us how the forces affect an object's equilibrium and rate of rotation. We will find that an object remains in a state of uniform rotational motion unless acted on by a net torque. This principle is the equivalent of Newton's first law. Further, the angular acceleration of an object is proportional to the net torque acting on it, which is the analog of Newton's second law. A net torque acting on on object causes a change in its rotational energy. Finally, torques applied to an object through a given time interval can change the object's angular momentum. In the absence of external torques, angular momentum is conserved, a property that explains some of the mysterious and formidable properties of pulsars-remnants of supernova explosions that rotate at equatorial speeds approaching that of light.
8.1
TORQUE
Forces cause accelerations; torques cause angular accelerations. There is a dennite relationship, however, between the two concepts.
228
81
Figure 8.-1 depicts an overhead view of a door hinged at point 0. From this viewpoint, the door is free to rot~te around an axis perpendicular to the page and passing through 0. Tfa force F is applied to the door, there are three f~lCtors that determine the effectiveness of the force in opening the door: the magnilude of the (oITe, the IJOsition ofapplicatioll of the force, and the ang//' at which it is applied. For simplicity, we restrict ourjiscussion lO position and force vectors lying in a plane. ''''hen the applied force F is perpendicular to the outer edge of the door. as ill Figure 8.1, the door rotates counterclockwise with const.ant angular acceleration. The samc pcrpendicular force applied at a poil1lllcarer the hinge results in a smaller angular acceleration. In general, a larger radial distance ·r he tween the applied force and the axis ofrotation results in a larger angular acceleration. Similarly. a larger applicd I()l"ce will also result in (\ larger angular acceleration. These considerations motivate the basic dcfi nition of" torque for the special case of forces perpendicular to the position vector:
F
Let be a force acting 011 an object, and Ict -; be a positi~l vector from a chosen point OLD the point or application oCthe force. \vi~ F perpendicular to The magnitude of the torque;; exerted by the force F is given by
Hill"c
/" 0
r
=-rF
t- Basic definition of torque
[8.1]
where r is the length 01" the position vector and F is the magnitude of the force. 51 unit: Newton-meter (N' 01) ~
T'he vecLQrs rand F lie in a plane. As discussed in cletail shortly in conjunction with Figure 8.4, the LOrque --; is then perpendicular to this plane. The point 0 is usually chosen to coincide ,..· ith the axis the ol~ject is rotating around, such as the hinge of a door or hub of a merry-go-round. (Other choices are possible as well.) In addition, we consider only forces acting in the plane perpendicular to the axis or rotation. This crit.erion excludes. for example, a force with up'.. . ard component on a merry-go-round railing, which cannot affect the merry-go-round's rotation. LInder these conditions, an o1:~jcct can rotate around the chosen axis in one of two directions. By convention, counterclockwise is taken to be the positive direc· tion, clockwise the negative direction. \lVhen an applied force causes an object to rotate counterclocb.. . ise. the torque on the object is positive. \"'hen the force causes the ol~ject LO rotale clockwise, the LOrque on the object is negative. "Vhen ('1m or more torqlles act on an object at rest, the torques are ~Idded. If the net torque isn't zero, the object starts rotating at an ever-increasing rat.e. If the net torque is zero, the object's rate of rotation doesn't change. These considerations lead to the rotational analog of the first law: the rate of rotation of an object doesn't change, unless the object is acted on by a net torque.
EXAMPLE 8.1 Goal
Battle of the Revolving Door
Apply the basic deAnition or torque.
Problem Two disgruntled businesspeople arc trying to usc a revolving door. as in Figure 8.2. The woman on the left exerts a forcc of 625 N perpendicular to the door and 1.20111 from the hub's center, ,vhile the man on the right exerts a force of 8.50 X 10 2 N perpendicular to the c100r and 0.800 m [rom the hub's center. Find the net torquc on the revolving door. Strategy Calculate the indi\"idual torques Oil the door lIsing the definition 01" torque, Equation 8.1, and then slim to find the net torque on the door. The woman excrts a negative torque, the man a positive tnrque. Their positions of application also differ.
~
FIGURE 8.1 A bird's-c\"(: \'ic\\' ora door hinged at 0, wilh a !tlrc<: applied perpendicular to the door.
r.
T
229
Torque
FIGURE
8.2
(Example 8.1)
230
Chapter 8
Rotational Equilibrium and Rotational Dynamics
,... ...
.
Solution Calculate the torque exerted bLthe woman. A negative sign mUSt be supplied because F I> if unopposed, would
~
-(1.20 m)(625 N) = -7.50 X 10' N· m
cause a clockwise rotation:
7, ~ r,F, =
Calculate the torgue exerted by the man. The torque is positive because F~, if unopposed, would cause a counterclockwise rOlation: Sum the torques to find the net torque
011
(0.800 m)(8.50 X 10' N) = 6.80 XlO' N· m
the door:
.
................. Remark
The negative result here means that the
I1ct
~
torque will produce a clockwise rotation.
QUESTION 8.1
What happens if the \\loman suddenly slides closer
La
the hub by 0.400 m?
EXERCISE 8.1
A businessman cmcrs the same revolving door on the right, pushing with 576 N of force directed perpendicular to the door and 0.700 In from the hub, while a boy exerts a force of365 N perpendicular to the door, 1.25 m to the left of lhe hub. Find (a) lhe LOrques exerted by each person and (b) the net torque on the door. Answers
(a)
T boy
= -456 N·
= 403 N· m
01, T man
(aj
F:
~ " ----~
(b)
; Fsin 8
rt
T llel
= -53 N· m
The applied force isn't always perpendicular to lhe position vector 1. Sup· pose the force exerted all a door is directed away from the axis, as in Figure 8.3a, say, by someone's grasping lhe doorknob and pushing to the righL Exerting the force in this direC[ion couldn't possibl)' open the door. However, if the applied force acts at an angle to lhe door as in Figure B.3b, the component of the force j)(!I1Jendicular to the door \\,'ill cause il to rotate. This figure shows t.hat the component of the force perpendicular to the door is F sin 0, where (J is the angle between the position vector -; and the force F. ''''hen the force is directed away from the axis, fJ = 0°, sin (0°) = 0, and Fsin (0°) = O. When the force is directed toward the axis, () = 180° and F sin (180°) = O. The maximum absolute value of F sin fJ is attained ani)' when F is perpendicular to r-that is, when = 90° or = 270°. These considerations mot.ivate a more general definition of torque:
F
o ---,---l.~8--::"F-...,.~
o
(b)
~
e
~
e
~
(e)
FIGURE 8.3 (a) A forct: F aCI illl:{ al ,III angle.: 0 = 180 0 exert.s ,,"ern lc.Jrque abuut the pivot 0. (b) The p,trt of lhe force perpendicular til till; duor, Fsin fJ, l:xcrts torque rFsin 0 abom 0. (c) All alternate imcrprctali(lll of lOrqllC ill terms ora It'ver arlll d = Tsin 0.
Let F be a force acting on an object, and let -; be a position vect.or from a chosen point 0 to the point of application of the force. The magnitude of the lOrque T exened by the force F is T
= rFsin 8
[8.2]
where l' is the length of the position vector, F the magnitude of' the force, ~ ~ and the angle between rand F.
e
SI unit: Newton-meter (N' 00) Again the vectors 1 and F lie in a plane, and for our purposes the chosen point 0 will usually correspond to an axis or rotation perpendicular to the plane. A second \.. . a1' of' understanding the sin (J facmr is to associate it. with the magnill1c1e r of the position vector r. The quantity d = r sin e is called the lever arm, which is the perpendicular distance frolll the axis of rOlation to a line drawn along the direction of the force. This alternate interpretation is illustrated in Figure B.3e.
8.1
Torque
231
It's important LO remember that the value of T depends on the chosen axis of rotation. Torques can be cornputed around any axis, regardless of lvhclher there is some actual, physical rotation axis prcsenl. Once the point is chosen, however, it must be used consistently throughollt a given problcm. Torquc is a vcctor pcrpendicular to the plane determined by the position and force vectors, as illustrated in Figure 8.4. The direction can be determined by the n'ghl-hand I'lde: 1. Point the flngers ofy-our right hand in the direc~on ofr. 2. Curl your flngcrs tmvard the direction of'vcctor F. 3. Your thumb then points approximately in the direction of the torque, in this case out of the page.
FIGURE 8.4 The right-hand rule: Point the finger!> of yOIl l" right hand along and curl Lhem in the direction ofr.. Your thumb then poi illS in the direction of the torque (OUl ofLhe page. ill this case).
r
Problerns used in this book will be confineE to objects rotating around an axis perpcndicu lar to the plane containing -; and F, so if these vectors arc in the plane of the page, the torque will always point either into or out of the page, parallel to the axis of" rotation. If your right thumb is pointed in the direction of a torque, your fingers curl naturally in the direction of rotation that the torque \'lOuld produce on an object at resl.
EXAMPLE 8.2 Goal
The Swinging Door
Apply the more general definition oJ'torque.
Problem
(3) A lTIan applies a lorce
or F~
3.00 X 10' N at an angle 01" 60.0" La [he
c100r of Figure 8.5a. 2.00 m from the hinges. Find the torque on the door, choos-
ing the position of the hinges as the axis of rotation. (b) Suppose a ,,,,edge is placed 1..50 m frolll the hinges on the other side of the door.vVhat minimum force Illust the \vedge exert so that the force applied in part (a) ,von't open the door? Strategy Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn't open, so the sum or these torques must be zero, a condition that can be used to find the \vedge force.
,..-'
300;fO"
Hinge
/ 0[,
.
~I
2.00111 (a)
l
ZGON
Hinge
/
2.00
0
(bl
III
'8
150 N
FIGURE 8.5 (Example 8.2a) (a) Top view of ,,1 door being pushed b>' a 300-N force. (b) The components of lhe 300-r\ force.
Solution (a) Compute the torque due to the applied force exerted at. 60.0°. Substitute into the general torque equation:
'F ~
,-Fsin
e=
(2.00 m)(3.00 X LO' N) sin 60.0"
(2.00 m)(2.60 X 10' N)= 5.20 X 10' N· m (b) Calculate the force exerted by the wedge on the other side of the door. Set the sum of the l.Orqucs equal to zero: The hinge force provides no torque because it acts at the axis (r = 0). The wedge force acts at an angle of -90.0°, opposite J\,. Il... • • • • • • • • • • • •
0+ /';",,"«1.30 m) sin (-90.0°) + 5.20 X 10' N· m = 0 Fwcdge
=
347 N
.
~
Remark Notice that t.he angle from t.he position vector to the \.. . edge force is -90°. This is because, st.arting at the position vector, it's necessary to go 90° clochvise (the negative angular direction) [0 gctl.O the forcc vector. Measuring
232
Chapter 8
Rotational Equilibrium and Rotational Dynamics
the angle in this way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternate]>', the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure 8.5b illustrates the fact that the component of the force perpendicular LO lhe lever arm causes the torque. QUESTION 8.2 To make the wedge more cf'fective in keeping the door closed, should it be placed closer to the hinge or to the doorknob? EXERCISE 8.2 A man lies one end ora strong rope 8.00 m long to the bumper of his truck, 0.500 m from the ground, and the other end to a vertical tree trunk at a height of 3.00 m. He lIses the truck to create a tension of 8.00 X 101 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point.
Answer
2.28 X 10 3 N· In
8.2
TORQUE AND THE TWO CONDITIONS FOR EQUILIBRIUM
An object in mechanical equilibrium must satisfy the follmving t\-vo conditions: 1. The net external force must be zero:
2. The net external torque must be zero:
2: F =
0
L -:; =
°
The fi rst condition is a statement of lranslaLional equilibrium: The sum of all forces acting on the object rnust be zero, so the object has no translational acceleration, ~ 1" = O. The second condition is a statement of rotational equilibrium: The sum of --'W'tl ~ all torques on the object must be zero, so the object has no angular acceleration, ~._..::~, ~ = O. For an object to be in equilibriulll, it must both translate and rotate at a constant rate. This large balanced rock :lllhe Because we can choose any location for calculating tQHlues, it's usually best to Garden of the Gods in Colorado Springs. Colorado, is in mechanical select an axis thal will make at least one torque equal to zero,justLO simplify the equilibriulll. net torque equation.
a
EXAMPLE 8.3
Balancing Act
Goal Apply the conditions of equilibrium and illustrate the use of different axes for calculating the netLOrque on an object. Problem A woman of mass m. = 55.0 kg sits on the left end of a seesaw-a plank of length L = 4.00 Ill, pivoted in the middle as in Figure 8.6. (a) first compute the torques on the seesmv abom an axis that passes through the pivol. point. ''''here should a man of mass 1\1 = 75.0 kg sit if the system (seesaw pillS man and woman) is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of 1n = 12,0 kg. (c) Repeat part (b), but1.his tirnc compute the p1 torques ahout an axis through the left end of the plank. Strategy Refer La Figure 8.6b. In parI. (a), apply the second condition of equilibrium, :£7 = 0, computing tOrques around the pivot point. The mass of the plank forming the seesav·,' is distribUled evenly on either side of the pivot point, so t.he torque exerted by gravity on the plan k, 7Kravil}" can be computed as if all the plank's mass is concentrated at. the piVOl point. Then Tgr,wily is zero, as is the torque exerted by the pivot. because their lever arms are zero. In part (b) the first
(a)
I.
-x
2.00 rn
(b)
FIGURE 8.6 (a) (Example R,3) T\,·o people on a seesaw. (b) Frcc body diagram for the plank.
8.2
Torque and the Two Conditions for Equilibrium
233
condition of equilibrium, IF = 0, must be applied. Part (c) is a repeat of part (a) showing that choice of a different axis yields the same answer. .
~
Solution (a) Where should the man sit
[Q
• • • • • • • • • • • • • • • • • • • • '''11
balance the seesaw?
Apply the second condition ofequilibriulll to the plank by sClling the sum of the torques equalLO zero: Thc [-Irst two torques are zero. Let x represent. the man's distance from the pivol. The woman is at a distance L/2 from the pivot.
o+ 0 -
Solve this equation for x and evaluate it:
(b) Find thc normal force seeS
11
+
Mgx
mg(L/2) = 0
1II(L/2)
(55.0 kg)(2.00 m)
M
75.0 kg
1.47 m
exerted by the pivot on the
Apply for first condition of equilibrium to the plan k, solving the resulting equation for the unknown normal force, '17.:
mp[g + n = 0
- ..\lg - lllg -
n = (M + m
+
+ 55.0 kg + 12.0 kg)(9.80 mN)
= (75.0 kg II
=
"IIlrl)g
1.39 x lO"N
(c) Repeat part (a), choosing a new axis through the left end ofthe plan k.
Compute the torques using this axis, and set their Slim equal to zero. Nmv the pivot and gravity forces on the plank result in nonzero torques. Substitute all known quantities:
-Mg(L/2
+
x)
+
IIIg(O) - 1II",g(L/2)
+
n(L/2) = 0
-(7:>.0 kg)(9.80 m/s')(2.00 m + x) + 0 - (12.0 kg)(9.80 m/s')(2.00 Ill) -(1.47 X 10 3
~-m)
- (735
~)x-
+
n(2.00 m) = 0
(235 N-m)
+ (2.00 m)n = 0 Solve for x, substituting the normal force fOLind in pan (b):
x=
1.46 m -
~
~
Remarks The answers for x in parts (a) and (c) agree except for a small rounding discrepancy. This illustrates how choosing a different axis leads to the samc solution.
QUESTION 8.3 \·Vhat happens if the woman now leans backwards? EXERCISE 8.3 Suppose a 30.0-kg child sits 1.50 m to the left of cent.er on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find I he mass of the second child. (b) Find the normal force acting at the pivot point.
Answers
(a) 22.5 kg
(b) 632 N
234
Chapter 8
Rotational Equilibrium and Rotational Dynamics
Tip 801 Specify Your Axis
I
Choose the axis of rotation and lise that axis excJusi\"e1y throughout a given problem. The axis
need not correspond
La
a physical
axle or pivoL paim. Any Corll'CIl-
ient paim will do.
8.3
THE CENTER OF GRAVITY
[n the example of the seesQ\.\' in the previous section, we guessed thatlhe torque due to the force of gravity all the plan k ,,,'as I he same as if all the plank's weight ,vere concentrated at its center. This is a general procedure: To compute the torque on a rigid body due to the force of gravity, the body's entire weight can be thought of as concentrated at. a single point. The problem then reduces to finding the location of that point. If the body is homogeneous (its mass is distributed evenly) and symmetric, it's usually possible to guess the locmion o[thm poim, as in Example 8.3. Other· '-\-'ise, it's necessary to calculate the point's location, as explained in this section. Consider an object ol"arbitrary shape lying in the xy~plane, as in Figure 8.7. The object is divided into a large number of very small particles of weight mJg, lll'lgJ 'Jn3g, . . having coordinates (x]' )'1), (x 2 , h), (x 3 , )';I), . . if the object is free to rotate around the origin. each panicle contributes a torque about the origin that is equal to its weight multiplied by its lever arm. For example, the LOrque due LO the weight Ullgis '11I'lgxl' and so forth. vVe wish to locate the point of application of the single force of magnitude w = I~= lWg (the total weight of the object), \V'here the effect on the rotation of the object is the same as that of the individual panicles. This point is called the object.'s center of gravity. Equ.:1ling the torque exerted by IV at the center of gravity to the sum of the torques acting on the individual particles gives (mIg + nl<2g
+
}l/.3g
+ ..
·)x(.~ =
IIIlgxI
+
m'2bJX2
+ 1n315x3 + ...
\,Ve assume that g is the same everywhere in the object (v,,·hich is true for all oqjeets w'e ,\:ill encounter). Then the gfactors in the preceding equation cancel, resulting in
/IIg 1Il 1X]
FIGURE 8.7
The IWI gra\'it:ltiollal torque on all object is ~cro if COIIIpUled around the cemcr of A"ra\'il~. The o~jecl will balam:c ifsupponed 31 that point (or at allY point. along
+
'm. 1
111 2 X':!
+
Ill';!
+
+
+ + ...
111:\-"3
m.,!
LmiXi
2.1ll i
[803a]
\vhere xcg is the x-coordinate of the center of gravity. Similarly, the y-coordinate and z-coord inate of the center of gravity of the system can be found from
a \'cnicallillc above or below thaI point).
LIlI,)'; yC!{=~
[803b]
_Uli
and [803c]
c
a ,,
oj
,'oil
Center of gravity FIGURE 8.8
D
An cxperimenlallech.
niqllc for determining the Cl:l11cr of gnl\ity aLl wrench. The wrench is
hung frech' from two different pivots. A :and C. Till: intersection of tILe two \'crticallinc~. tlBand
CCilier ofgraviLy.
CD, locales the
These three equations are identical to the equations for a similar concept called center of mass, The center of mass and cenler of gravity of an object are exacLiy the same ,..,hen gdoesn't vary significantly over the object. It's often possible to guess the location of the center of gravity. The center of gravity of a homogeneous, symmetric body must lie on the axis of symmetry. For example, the center of gravity of" a homogeneous rod lies mid,vay between the ends of the rod, and the center of gravity of a homogeneous sphere or a homogeneous cube lies at the geornetric cenler of the object. The cenl.er of gravity of an irregularly shaped object, such as a wrench, can be determined experimentally by suspending the wrench from two different arbitrary points (Fig, 8.8). The wrench is first hung from point A, and a \'erticalline A.B (\-vhich can be established "lith a plumb bob) is dnnvn when the wrench is in equilibrium. The '''Tench is then hung from point C, and a second vertical line CD is drawn. The cemcr of gravity coincides with the intersection or these t,vo lines. In fact, if the wrcnch is hung frcely from any point, the center of gravity always lies straight below the point of support, so thc venical line through that point must pass through the center of gravity. Several examples in Section 8.4 involve homogeneous, symmetric objects where the centers of gravity coincidc with their geometric centers. A rigid object in a uni[orm gravitational field can be balanced by a single force equal in magnitude to the weight of the object, as long as the force is directed upward through the object's center of gravity.
8.3 The Center of Gravity
EXAMPLE 8.4 Goal
235
Where Is the Center of Grovity?
Find the center of gravity of a system of particles.
J'
Problem Three panicles are located in a coordinate system as shown in Figu"e 8.9. Find the ccnter or gravity.
S.OO k~
2.00 kg
4.00 kg
--x
Strategy The )'-coordinate and z.~coordinate of thc center of gravity are both zero because all the particles are on the x-axis. "\Ie can find the x-coordinate of the center of gravity usi ng Equation 8.3a.
,.
I.DOm
O.500m
FIGURE 8.9 (Example 8.4) Localillg the cellLcrofgravity ora sysLClll of three particles .
. . . . . . . . . . . . . . . . . ...,
.
Solution Apply Equation 8.3a to the system of three particles:
(I)
Compute the numerat.or of Equation (I): (5.00 kg)(-0.500 m)
+ (2.00 kg)(O
m)
+ (4.00 kg)(l.OO
m)
l.50 kg' m Substitute the denominator, '2tnj = 11.0 kg, and the numerator into Equation (I).
x
1.50 kg m
"
0.136 m
I 1.0 kg
.
11..- • • • • • • • • • • • • • • • _ •
~
QUESTION 8.4 If] .00 kg is added to the masses on thc left and right, does the center of mass (a) move to the left, (b) move right, or (c) remain in the same position. EXERCISE 8.4 Jra fourth particle of mass 2.00 kg is placed at x ity for this system of four panicles.
Answer
x cg =
0,)'
=
0.250
Ill,
find the x- and
y~coordinatcs of
the
the center of grav-
O. LI5 111; )'cg = 0.0385 m
EXAMPLE 8.5 Goal
=
LO
Locoting Your Lob Portner's Center of Gravity
Use torque to find a center of gravity.
L --------_.:
Problem In this example we shO\v hmv to find the location of a person's center of gravity. Suppose your lab partner has a height L of 173 em (5 ft, 8 in) and a weight W of 715 N (l60 lb). You can determine the position of his center of gravity by having him stretch out on a uniform board supported at one end by a scale, as shown in Figure 8.10. If t.he board's weight wbis 49 N and the scale reading Fis 3.50 X ]O~ N, find the distance of your lab partner's cent.er ofgravit)' from the left cnd of the board.
FIGURE 8.10
(Example 8.5) Determining yOUl" lab partner's
(t'nler or~ravil)'.
Strategy To find the position x(~ of the center of gravity, compute the torques using an axis through O. Set the sum of the torques cqual to zero and solve for xcg ' ~
-
.
Solution Apply the second condition of equilibrium. There is no torque due to the normal force n because its momcnt arm is zero about an axis through O.
........................................
- lOX"
-
lO,(L/2) + FL
= 0
~
236
Chapter 8
Solve for
xrg-
Rotational Equilibrium and Rotational Dynamics
and substitute known values:
FL - wJL/2) Xq~
=
w
(350N)(173cm) - (49N)(86.5cm) 715 N .
~
=
79
(Ill
• • • • • • • • . • • • • • • • • • .A
Remarks The given information is sufficient only to dctcrlll inc the x-coordinate of the center of gravity. The other t"'/O coordinates C
\"'hat would happen if a support is placed exactly at x = 79 em followed by the removal or the supports at the subject's head and feet? EXERCISE 8.5 Suppose a 416-kg alligaLOr of length 3.5 III is stretched out on a board of the same length weighing 65 :'1. 1r the board is supported on the ends as in Figllrc 8.10, and the scale reads] 880 :\I, nnd the x-component of the alligator's center of gravity.
Answer
1.59 m
Tip 8.2 Rotory Motion under Zero Torque I r a net torque of tero is exerted on an o~ject. itwill continue to rOlate at a constant angular speed-which need Elm be zero. However, ?era tOl"qllc does impl}· thallhe angular acceleration is zero.
8.4
EXAMPLES OF OBJECTS IN EQUILIBRIUM
Recall from Chapter 4 that when an object is treated as a geometric point, equilibrium requires only thatlhe net force on the ol~jeet is zero. In this chapter \·..-e have shmvn that for extended o~jcClS a second condition for equilibrium must also be satisfied: The net LOrque on the object must be zero. The [ollmv'ing general procedure is recommended for solving problems that involve ol~jects in equilibrium.
PR08LEM-SOLVING STRATEGY OBJECTS IN EQUILIBRIUM 1. Diagram the system. Include coordinates and choose a convenient rotation axis for computing the net torque on the object.
2. Draw a free-body diagram of the object of interest, showing all external forces acting on it. For systems \,l,'ith more than one object, draw a sejJrrralediagram for each object. C[\llost problems \·...ill have a single object of interest.) 3. Apply "I'Tj = 0, the second condition of equilibrium. This condition yields a single equation for each object of interest. If the axis of rotation has been carefully chosen, the equation often has only one unknown and can be solved immediately. 4. Apply "IFx = 0 and "I.Fy = 0, the first condition of equilibrium. This yields two more equations per object of interest. 5. Solve the system of equations. For each object, the twO conditions of equilibrium yield three equations, usually with three unknowns. Solve by substitution.
8.4
EXAMPLE 8.6
A Weighted Forearm
237
Examples of Obiects in Equilibrium
Ill!
Goal Apply the equilibrium conditions to the human body.
Problem A 50.0-:-J (II-Ib) bowling ball is held in a person's hand with the forearm horilontal, as in Figure 8.1 la. The biceps rnuscle is auachecl 0.0300 m from Lhe joilll, and Lhe ball ;,;, 0.350 m from the joint. rind the upward [OITC F exerted by the biceps on the forearm (the ulna) and the dm,vnward force Ii exerted by the humerus on the forearm, acting at the joi nl. Neglect the weight or the lorearm and slight devimion from the vertical of the biceps.
°1
-J"]
1
;~ _ ~0.0300
--l
I-c-O.03UO m
I
I.-O.?>SUrn-
m
50.0"
I.-O.350m-(h)
(
FIGURE 8,11 (Example H.6) (a) A wcig-hl held with till: {urc
Strategy The forces acting on the forearm are eCluivalent to those acting on a bar of length 0.350 In, as shown in Figure 8.11b. Choose the uSllal x- and y-coordinates as shown and the axis at 0 on the left end. (This completes Steps 1 and 2.) Use the conditions of equilibrium to generate equations for the unknowns, and solve, .
Apply the second condition for equilibrium (Step 3) and solve for the upward force F:
Apply the first condition for equilibrium (Step 4) and solve for the downward force I~: ,
~
-, ,
,. ,
.
R(O)
+ 1'(0.0300 m) - (50.0 N)(0.350 m)
1'~
583N(131Ib)
~
0
"2;1') = F- R - 500 N = 0 R = F - 50.0 N = 583 N - 50 N =
533 N (120 Ib)
.......................................
,
...
Remarks The magnitude of the lorce supplied by the biceps must be about ten times as large as the bowling ball it is supporting!
QUESTION 8.6 Suppose the biceps wefe surgically reattached three ccntirneters blrlher to\,·:ard the person's hand. If the same bowling ball were again held in the person's hand, how would the force required of the biceps be affected? Explain. EXERCISE 8.6 Suppose you ,vanted to limit the force acting on your joint to a Illi:lximum value or 8.00 x 101 N. (a) Under these circumstances, what maximum weight would you attempt to lift? (b) \o\.'hat force ,...ould your biceps apply while lifting this ~v'eight?
Answers
(a) 75.0 N
EXAMPLE 8.7 Goal
(b) 875 N
Walking a Horizontal Beam
Solve an equilibriurn problem wlth nonperpendicular torques.
A uniforrn horizontal beam 5.00 m long and weighing 3.00 X 102"!'oJ is attached to a wall by a pin connection t.hat allows the beam to rotale. Its far end is supported by a cable that makes an angle of 53_0 with the horizontal (Fig.jU2a, page 238). Ira persor!....weighing 6.00 X l02N stands 1.50 m from the wall, hnd the magnitude of the tension T in the cable and the force R exerted by the wall on the beam. Problem
0
See Figures 8.12b and 8.12c. The second condition of equilibrium, LT; = 0, ,vith-i0rques computed around the pin, can be solved for the tension Tin the cable. The I1rst condition of equilibrium, LF; = 0, gives two equations and two unknowns for the two components of the force exerted by lhe wall, H:\"" and I~. Strategy
238
Chapter 8
Rotational Equilibrium and Rotational Dynamics
o
t==;:==::::=== ~300N 6uO" (b)
Tsin 53.0~
J?y
o 53.0°
l!:/o:!l'~=;:==;:=:::'T=e=m~5~3J~oo 1.....1.50 III
~2.50
5.00 III
~100
,. 1~~~~I---->I
N
2.00 111
600 N
(e)
FIGURE 8.12 (Example ~.7) (a) A uniform beam
10
-------------
d---- 6.00 m --->I'I
III
(,)
30°
(d) wall and supporu.:d by a cable. (IJ) A free-body diagr
.................................
r·········································
~
Solution From !igure 8.12, the forces causing tOrques are the wall force R, the gravity forces on the beam and the man, WtJ and Wf\1, and the tension force T. Apply the condition of rotational equilibrium: Compute torques around the pin at 0, so T U = 0 (zero moment arm). The torque due to the beam's weight aets at the beam's center of gravity.
.2>, =
Substitute L = 5.00
- (3.00 X 10' N)(2.50 m) - (6.00 X 10' N)( 1.50 m)
In
and the weights, solving for T.
0 - wlJ(L/2) - 10",(1.50 m)
+ (Tsin 53.0°)(5.00 m) T=
'ow apply the first condition of equilibrium to the beam:
Substituting the value of Tfoune! in the previous step and the weights, obtain the components ofR: ~
(1)
(2)
+ TLsin
= 0
413 N
LF, = R, LFy = Ny -
T cos 53.0° ~ 0 1Oll -
II, =
10",
+
Tsin 53.0 0 ~ 0
5.70 X 10' N
.
~
(53°) ~ 0
...
Remarks Even j[we selected some other axis for the torque equation, the solution would be the same. For example, if the axis ",,'ere to pass through the center of gravity of the beam, the torque equation would involve both Tand ~" Together with Equations (1) and (2), however, the unknowns could still be found-a good exercise.
QUESTION 8.7 What happens to the tension in the cable if the man in Figure 8.12a moves farther away from the wall? EXERCISE 8.7 A person with mass 55.0 kg stands 2.00 J11 away from the wall on a 6.00-m beam, as shmvl1 in Figure 8.12d. The mass of the beam is 40.0 kg. Find the hinge force components and the tension in the wire.
Answers
r = 751
N, R.
= -6.50 X 10' N, II, = 556 N
8.5
EXAMPLE 8.8 Goal
Relationship between Torque and Angular Acceleration
Don't Climb the Ladder
-
---
Apply the two conditions of equilibrium.
p
Problem A uniform ladder 10.0 m long and weighing 50.0 N rests against a smooth vertical \Vall as in Figure 8.13a. If the ladder is just on 1he verge of slipping when it makes a 50.0° angle with the ground. find the coefficient of static friction between the ladder and ground.
li
239
p
'"
LLN Ld50N
Strategy Figure 8.J3b is the free-bod)' diagram for O r O d'1 the ladder. The f,rst condition or equilibrium, LF, = O. gives two equations lor three unknowns: (he magnitudes la) Ib) Ie) of' the static friction force Jand the normal force 'Il, both FIGURE 8.13 (illtcractin;: Exarnple 8.8) (a) A ladder leaning acting on the base of the ladder, and the magnilUde of againsl a friction k:ss wall. (b) A free-body Q.iagrarn of the ladder. the force of the wall, P. act.ing on the top of the ladder. (c) Len:1 :Inns fur the force of gravity and P. The second condition of" equilibrium, ~T, = 0, gives a third equation (for P), so all three quantities can be found. The definition of static friction then allows cornputation of the coefficient of static friction. . . ~
,..
~
Solution Apply the first condition of cquilihriu m to the ladder:
(I)
2,F, =
I-
(2)
2,F, =
11 -
P= 0
->
50.0:\ = 0
I=
P
->
11
= 50.0 N
Apply the second condition of equilibrium, computing torques around the base of the ladder, with Tgr'l\' standing for the torque due to the ladder's 50.0-0.' \-veight: The torques due to friction and the normal force are zero about 0 because their moment arms are zero. (tv!oment arms can be found from Fig. 8']3c.)
0+ 0 -(50.0 N)(5.00 m) sin 40.0 0
From Equation (l), we now have I= P= 21.0 N. The ladder is on the verge of slipping, so write an cxpressian for the maximum force of static friction and solve for fL,:
21.0 N =
P~
21.0
+ P(IO.O m)
sin 50.0 0 = 0
~
I= I..",,, =
1'-,11
= 1'-,(50.0 N)
21.0N 1'-, = -.- - = 0.420 ~O.O N
~
~
_
P
Remarks Note that torques \-"ere computcd around an axis through the bottom of the ladder so that only and the force of gravity contributcd nonzero torques. This choice of axis reduces the complexity of the torque equation, often resulting in an equation with only onc unknown. QUESTION 8.8 If a 50.0 N monkey hangs from the middle rung, \'.,'Quld the coefficient of static friction be (a) doubled, (b) halved, or (c) ullchangcd? EXERCISE 8.8 If thc coefficient of static friction is 0.360, and the same ladder makes a 60.0° angle \vith respect to the horizontal, how far along the length of the ladder can a 70.0-kg painter climb before the ladder begins to slip?
Answer
8.5
6.33 m
RELATIONSHIP BETWEEN TORQUE AND ANGULAR ACCELERATION
\,Vhen a rigid object is subject to a net torque, it undergoes an angular acceleration that is directly proportional to the net torque. This result, \..: hich is analogous to NeWlon's sccond law, is derived as follows.
240
Chapter 8
Rotational Equilibrium and Rotational Dynamics
F,
"
,
,... ... ----
--"
The system shown in Figure 8.14 consists of an object of mass 111 connected to a very light rod of length r. The rod is pivoled at the point 0, and its movement is confined to rotation 011 a frictionless horizort{rd table. Assume that a force J~ acts perpendicular La the rod and hence is tangent La the circular path of the object. Because there is 110 force tu up pose this tangential force, the ubject ulHlergocs a tangential acceleration {l, in accordance with Nevy·ton's second law:
m.
'\ O~r,
,".. ......... __ .- ... ..."
"
FI = rna l
Multiply both sides or th is equation by r: FIGURE 8.14 An object of Illass 11/ atlached 10 a 11ghl rod ofkngth r moves ill a circular path 011 a frictionless horizontal surface while "langential force F, acts 011 it.
Fir =
"UTl t
Substituting the equation a, = -ra relating tangential and angular acceleration into the above expression gives [8.4]
The lert side of Equation 8.4 is the torque acting on the object about its axis of rotation, so we can rewrite it as T=
" mr-a
[8.5]
Equation 8.5 shows that the torque on the object is proportional to the angular acceleration of the object, \vhere the constant of proportionality mr 2 is called the moment of inertia of the object of mass m. (Because the rod is very light, its moment of inertia can be neglected.)
QUICK QUIZ 8.1
Using a screwdriver, you try to remove a sere'.. .· from a piece of furniture, but can't get it to turn. To increase the chances of success, you should use a screwdriver that (a) is longer, (b) is shorter, (c) has a narrower handle, or (d) has a wider handle.
Torque on a Rotating Object Consider a solid disk rotating about its axis as in Figure 8.15a. The disk consists of many particles at various distances from the axis of rotation. (See Fig. 8.15b.) The torque on each onc of these panicles is given by Equation 8.5. The net torque on the disk is given by the sum of the individual torques on all the particles:
2;, = (2;",)"2)",
[8.6)
Because the disk is rigid, all of its panicles have the same angular acceleration, so a is not involved in [he SUIll. If the masses and distances of the panicles are labeled with subscripts as in Figure 8.15b, then 2:1nr 2
= In(r]2
+
1n 2T2'2
+
1I/.:3r3 2
+ ..
This quantity is the moment of inertia, I, of lhe whole body: Moment of inertia
[8.7]
~
FIGURE 8.15 (n) A solid disk rotating abolll its axis. (b) The disk consists of man~' panicles. all with the same angular acceleration,
(b)
8.5
Relationship between Torque ond Angular Acceleration
241
The moment of inertia has the 51 units kg· m 1 . Using th is result in Equation 8.6, we see that the net torque on a rigid body rotating about a fixed axis is given by
2> '0 =
[8.8]
Equation 8.8 says that the angular acceleration of an extended rigid object is proportional to the net torque acting on it. This equation is the rotational analog of Newton's second Ia,..· of motion, with torque replacing force, moment of inertia replacing mass, and angular accelerat.ion replacing linear acceleration. Although the moment of inertia of an object is related to its mass, there is an important dif~ ference bet,veen thern. The mass m depends only 011 the quantity of matter in an object, ,vhercas the moment of inertia, I. depends on both the quantity of rnatter and its distribulion (through the /1. term in 1= 17111';1) in the rigid object.
QUICK QUIZ 8.2
A COIlS[antnet torque is applied to an object. vVhich one of the following will not be constant? (a) angular acceleration, (b) angular velocity, (c) moment of inertia, or (d) center of gravity. QUICK QUIZ 8.3 The two rigid objects shown in Figure 8.16 have the same mass, radius, and angular speed. If the same braking torque is applied to each, lvhich takes longer 1.0 stop? (a) A (b) B (c) more information is needed The gear system on a bicycle provides an easily visible example of I.he relalionship between torque and angular acceleration. Consider first a five-speed gear system in which the drive chain can be adjusted 1O wrap around any of five gears attached lo the back wheel (Fig. 8.17). The gears, vvilh different radii, arc concentric with the wheel hub. ''''Then the cyclist begins pedaling from rest, the chain is auachedto the largest gear. Because il has the largest radius, lhis gear provides lhe largestlorquc 1.0 the drive ~vheel. A large torque is required initially, because the bicycle starts from rest. As the bicycle rolls faster, the tangential speed of the chain increases, eventually becolTling too fast for the cyclist to maintain by pushing the pedals. The chain is then moved 1O a gear with a smaller radius, so the chain has a smaller tan~ genl.ial speed lhat the cyclist can more easily maintain. This gear doesn't provide as much torque as the first, but the cyclist needs to accelerate only [Q a somewhal higher speed. This process continues as the bicycle moves faster and faster and the cyciistshiflS through all five gears. The fifth gear supplies the Im...est torque, bUlllm.. · the main function of that torque is to counter the frictional torque from the rolling tires, v.'hich tends to reduce lhe speed or the bicycle. The small radius of the fifth gear allows dle cyciistLO keep lip with the chain's movement by pushing the pedals. A 15~speed bicycle has the same gear structure on the drive ""heel, but has three gears on the sprocket. connected to the pedals. By combining differenl positions of the chain 011 the rear gears and the sprockct gears, 15 different lorques are available.
~ Rotational analog af Newton's
sp.cono low
A
B
FIGURE 8.16
(Quick Quiz 8.3)
APPLICATION Bicycle Gears
FIGURE 8.17 The drive wheel and gears of a bicyc:1e.
More on the Moment of Inertia As wc havc seen. a small object (or a particle) has a moment of inenia equal to mr 2 about some axis. The moment of inenia of a com/Josile ol~ject about some axis is just the sum or the moments of inenia of thc object·s components. For example, suppose a Ill<'!iorcue twirls a baton as in Figure 8.18. Assume Lhat the baton can be modeled as a very light rod of length 2f with a heavy object at each end. (The rod of a real baton has a significanllll
m
FIGURE 8.18 t\ batol! oflenglll 2[ alld m;ISs 2111. (The mass Oflhc conneClillK rod is neglcClcd.) The Illoment ofillenia abOUllhe axis through the balon's cenler and perpendicular to its lenglh is 2mt~.
242
Chapter 8
Rotational Equilibrium and Rotational Dynamics
Because this system consists or two objects with equal masses equidistant from the axis or rotaLion, 1" = for each object, and the sum is
e
1=
Lmr2 = me 2 + mC 2 =
I f the mass of the rod "vcre not negleCled,
\NC
2mt 2
would have to include its moment of
inertia to find the total moment ofinenia of the baLOn. vVe pointed out earlier th
EXAMPLE 8.9 The Baton Twirler Goal
Calculate a moment of inertia.
0.20 kg
0.30 kg
0.30 kg
0.20 kg
Problem In an effort LO be the star of the helir-time 5hO\.. ., a majorette twirls an unusual baton made lip of four spheres fastened to the ends of' very light rods (Fig. 8.19). Each rod is 1.0 In long. (a) Find the moment of inertia of the baLOn about an axis perpendicular to the page and passing through the point where the rods cross. (b) The majoreue tries spinning her su"ange bal.on about the axis 00', as shown in Figure 8.20. Calculate the moment of inertia of the baton abolillhis axis.
Strategy In Figure 8.19, all four balls contribute to the moment ofinenia, ""hereas in Figure 8.20, with the new axis, only the two balls on the left and right contribute. Technically. the balls on the lOp and baLlam still make a small contribution because they're not really point particles. However, their moment of inertia can be neglected because the radius of the sphere is much smaller than the radius formed by the rods. ~
-
,
FIGURE 8.19 C)1~jccts
(ExampleS.gal Four
COllnected to
li~ht rod~ rot'lt-
ing ill the plalle of the page.
..,
.
Solution (a) Calculate the 1110mel1l of inertia of the baton when oriented as in Figure 8.19.
Apply Equation 8.7, neglecting the mass of the connecting rods:
Relationship between Torque and Angular Acceleration
a
Remarks The moment of inertia is smaller in pan (b) because in this configuration the O.20-kg spheres are esscntially located on the axis of rotation.
0.20 kg-
QUESTION 8.9 Tf one of the rods is Icngthened, which onc would cause the larger change in the moment of incrtia, the rod connecting bars one and three or thc rod connecting balls two and four? EXERCISE 8.9 Yel. another bizarre baton is created by taking (our identical balls, each with mass 0.300 kg, and fixing t.hem as before, except that one of the rods has a length of 1.00 m and the other has a length of] .flO m. Ca Icu~ latc the moment of inertia of this baLOn (a) when oriented as in Figure 8.19; (b) ,.. . hen oriented as in Figure 8.20, wit h the shoneI' rod vertical; :and (c) whcn oricnted as in Figure 8.20, but with longer rod vertical.
Calculation of Moments of Inertia for Extended Objects The method used for calculating moments of inertia in Example 8.9 is sirnple when only a few small objects rotate about an axis. "Vhen t.he ol~ject is an extended one, such as a sphere, a cylinder, or a conc, techniques of calculus are often required, unless some simplifying symmetry is presenL. One such extended object amenable to a simple solution is a hoop rOlating about an axis perpendicular to its planc and passing through its center, as shown in Figure 8.21. (A bicycle tire, for example, would approximately fit. into this category.) To cvaluate the moment of inertia of the hoop, we can still use the equation J = ~1nT2 and imagine that the mass of the hoop 1\1 is divided into n small seg~ mcnts having masses '111'1' "ln 2 , }I/.;l' • • , "Ill", as in Figure 8.21, with M = m l + 11/2 + m,3 +. . + mil" This approach isjust an extension of the baton problem described in the preceding cXClmples, exccpl that now '",,'C have a large number of small masses in rotation instead of only four. We can express the sum for I as 1=
2:11/(2
= mI'}
+
1!l'!. r2 'l
+
11/:'''3'1
+ ... +
FIGURE 8.21
A uniform hoop can be divided inLO a large Humber or small scgmems I hal are cquidiswnt from lhe ccmcr of lhe hoop.
m'lr,/l,
All of the segments around the hoop are £II the saml' rlistaure R from the axis of rotation, so we can drop the subscripts on the distances and factor out R'l. to obtain J=
(tfl. l
+
1n2
+
m:1 +
... +
17l,,) R2
= AtTR'.!
[8.9]
This expression can be used for the moment of inertia of any ring-shaped object rotating abollt an axis through its center and perpendicular to its plane. Note that the rcsult is strictly valid only if the thickness of the ring is slTlall relative 1.0 its inner radius. The hoop we selected as an example is unique in that wc were able to find an expression for its moment of inertia by using on Iy simple algebra. Unfonunately, for most extended objects the calculation is much marc difllcult because the mass elements are not all located at the same distance from the axis, so the methods of integral calculus are required. The moments or inertia lor some other comlllon shapes are given without proor in Table 8.1 (page 244). You can usc this table as needed to determine the moment of inertia of a body having anyone of the listed shapes. [I' mass clements in an object are redistributed parallel 1.0 the axis or rotation, the moment of inertia of the object doesn't change. Consequently, the expression f = lV1R2 can be uscd equally well to find the axial moment of inertia of' an embroidery hoop or of a long sewer pipe. Likewise, a door turning on iLs hinges is described by the same moment-oF-inertia expression as that tabulated for a long thin rod rotating about an axis through its end.
Tip 8.3 No Single Moment of Inertia :\1omCnL of inertia is analogous lo mass, bUl there are major dirrerences. Mass is an inherent propcrt)' Or~l1l object. The momenL of inertia of all object depends on lhe shape of the ol::!iecl, its mass, and the choice ofrOlation axis.
244
Chapter 8
Rotational Equilibrium and Rotational Dynamics
TABLE B.1 Moments of Inertia for Various Rigid Objects of Uniform Composition I loop or thin cylil\(t.-ical shell I"" .\In'!
Solid sphc.-c
1= ~ 5 .,\lR' II
Solid cylinder ordi ..k 1=
Thin spherical
shell
~ lvlR 2
1=
2. 3
MN'!.
R "
Long thin rod
I
Long thin rod with rotation axis through end
with rOlatioll ax~.s ~
through center 1= ~ML~
,~
1=
EXAMPLE 8.10 Goal
Warming UI'
/
t
MI.':!
"
Find a moment of inertia and apply the rotational analog of Newton's
second law.
Problem A baseball phlycr loosening lip his ann before a game tosses a 0.150kg baseball, using onl~' the rotation of his forearn'"! to accelerate the ball (Fig. 8.22). The ("rearm has a mass of 1.50 kg and a length of 0.350 m. The ball starts at rest and is released '..·ilh a speed of 30.0 m/s in 0.300 s. (a) find the constant angular acceleration of the arm and ball. (b) Calculate the momcnt of inenia of the system consisting of the foreanll and ball. (c) find the torque excned on the system that results in the angular acceleration round in pan (a). Strategy The angular acceleration can be found with rotational kinematic equations, while the moment of inenia of the system can be obtained by summing the separate moments ofincrtia of the ball and forearm. Multiplying these [\...· 0 results together gives the torque.
FIGURE 8.22
(Example 8.10) :\ ball
being tossed by a pilcher. The roreann is lIsed to accelerate lhc ball.
,.
• • • • • • • • • • • • • • • • • • • • •"'"'IlI
Solution (a) Find the angular acceleration of the ball.
+ crt
The angular acceleration is constam, so use the angular velocity kinematic equation with Wi = 0:
w =
The ball accelerates along a circular arc with radius given by the length of the forearlll. Solve "(J = rw for w and sub~ stitutc:
0=-=-=
W,
w
v
f
rf
~
w
0=(
30.0 mls (0.350 m)(O.300 s)
286 rad/s'
8.5
Relationship between Torque and Angular Acceleration
245
(b) Find the moment of inertia of the system (forearm plus ball). Find the moment of inertia of the ball about an axis thaI passps tlll'Ollgh rhe f'lhow, pf'l-pf'ndiclilar 10 the arm:
11>,,11 = "Ill"? =
Obtain the moment of inertia of" the [orcaI'm, modeled as a rod, by consulting Table 8.1:
1'0,,""'" ~
(0.150 kg)(0.350 m)2 = 1.84 X 10-2 kg' m?
*ML' = H1.50 kg)(0.350 m)'
=6.13X IQ-'kg'm'
Sum the individual moments of inertia 10 obtain the mornent ofillertia ofthc syst.em (ball plus forearm): (c) Find the torque exerted on the system.
Apply Equation 8.8, using tbe results ofparlS (a) and (b):
T
= ["",mcx = (7.97 X 10-2 kg'm')(286 rad/s') 22.8N·m ....................
lio.. • • • • • • •
Remarks Not.ice that having a long forearm C;-1l1 greatly increase the t.orque and hence the acceleration of the ball. This is one reason it's advantageous for a pitcher 1.0 be tall-the pitching arm is proportionately longer. A similar advalltage holds in tennis, where taller players can usually deliver raster serves. QUESTION 8.10 Why do pitchers step forward when delivering the pitch? \Vhy is the timing important? EXERCISE 8.10 A cat
Answers
(aJ 40.3 rad/,'
EXAMPLE 8.11
(b) 453 kg' m'
(c) 1.83 X 10 'N· m
The Fallin Bucket
Goal Combine Ne,vton's second 1frw with its rotational analog. Problem A solid, frictionless cylindrical reel of mass 1\1 = kg and radius R = 00400 m is used to dnnv \vater from a well (Fig. 8.23(1). A bucket of mass 1/1 = 2.00 kg is attached 10 a cord that is '''Tapped around the cylinder. (a) Find the tension Tin the cord and acceleration aofthc buckct. (b) If the bucket stans from rest at the top of the well and falls for 3.00 s before hitting the vlo'ater, ho\,\1 far does it fall? ~.OO
Strategy This problem involves three equations and three unknowns. The three equations are :\e't\/Ion's second law applied to the bucket, NUl = "i.j'~; the rotational version of the second law applied to Ihe cylinder, ler = ~7'I; and the relationship bctwecn linear and angular acceleration, (l = 'tn, \Vh ich connects the dynamics of the bucket and cylinder. The three unknowns are the acceleration a of the bucket, the angular acceleration a of the cylinder, and the tension T in the rope. Assemble the terms of the three equations and solvc for t.he three unknm.. . lIs by substitution. Pan (b) is a review of kinematics.
t
T
("j
gq Ii
T
Mg (h)
(c)
!
mg (d)
FIGURE 8.23 (Example R.ll) (a) A ,,-aler buckclallachcd LO
246 ~
Chapter 8
Rotational Equilibrium and Rotational Dynamics
...
..........................................
Solution
(a) Find the tension in the corel and the acceleration of the bucket.
-mg+ T
Apply Newton's second law to the bucket in Figure 8.23b. There are two forces: the tension T acting upwards and gravity acting downwards.
(1)
Apply T
£JT = la = ,1'vIl,-a
>no. =
mg
=
~
fa to the cylinder in Figure 8.23c:
Notice the angular acceleration is c1ocbvise, so the torque is negative. The normal and gravity forces have zero moment arm and don't contribute any torque. Solve for T and substitute a = a/ R (notice that both and a are negati\"e):
(2)
I,"
-TR = ~!VIR'a
Ct'
Substitute the expression for Tin Equmion (3) inw Equation (1), and solve for the acceleration:
rna = - mg - ~NTa
Substitute the values for m, l\JI, and g, gelling substitute a inLO EquaLion (3) to get T:
a = -5.60 mis'
(l,
(solid cylinder)
then
a=
T=
mg
m + ttv!
8.40 N
(b) Find the distance the bucket falls in 3.00 s. - H5.60 m/s')(3.00 sF
Apply the disphlcemellL kinematic equation for constant acceleration, \-"ith 1= 3.00 sand V o = 0: trrr... • • • • • • • •
Remarks Proper handling of signs is very important in these problems. All such signs should be chosen ini~ tially and checked mathemaLically and physically. In this problem, for example, both the angular acceleration (l' and the acceleration a are negative, so a = tt/I?. applies. If the rope had been wound the other wayan the cylinder, causing coul1lerclock\'.:ise rotation, the wrque woule! have been positive, and the relationship would have been a = -a/ R, with the double negative making the righthand side positive, just like the len-hand side.
EXERCISE 8.11 A hollow cylinder of Illass 0.100 kg and radius 4.00 cm has a string wrapped several times around it, as in Figure 8.23d. J f the string is attached to a rigid support and the cylinder allowed La drop from rest, find (a) the acceleration of the cylinder and (b) the speed of the cylinder ,"vhen a meter of string has ullwound off of it. Answers
(a) -4.90 mis'
(b) 3.13 mls
QUESTION 8.11 How would the acceleration and tension change if most of the reel's mass were at its rim?
8.6
ROTATIONAL KINETIC ENERGY
In Chapter 5 we defined the kinetic energy of a particle moving through space with a speed vas the quantity !mv 2. Analogollsly, an o~iect rotating about some axis with an angular speed w has rotational kinetic energy given bytlw2 . To prove this, consider an object in the shape of a thin rigid plate rotating around some axis perpendicular to its plane, as in Figure 8.24. The plate consists ormany small particles, each of Illass m. All these particles rotate in circular paths around the
8.6
Rotational Kinetic Energy
axis. If r is the distance of one of the particles from Ihe axis of rotation, the speed of that particle is v = rw. Because the total kinetic energy of the plate's rotation i., the slim of all the kinetic energies Clssociated with it'i particles, we have £..J ra . "'(' ~,-=
') = "'('
~mw-
~
'1 ") ~ml·w-
247
wt
t-
'" ") " =:!' ( ~II/Iw-
In the last step, the w'!. term is faClored out because it's the same for every particle. the quantity in parentheses on the right is the moment of inenia of the plate in the limit as the particles become \·anishingly ...mall, so ~ow,
I
,
'"
o [8.10]
where I = ~m,.~ is the moment of inertia of the plate. A system such as a bowling ball rolling down a ramp is described by three types or energy: gravitationaJ potential energy PEg' translational kinetic energy KEI , and rotational kinetic energy KE,.. All these forms of cnerg), plus the potential energies of an)" other conservative forces, must be included in our equation for the conservation of mechanical energy of an isolated sy...tcm: (KE,
-i-
KE,.
+
PEl, = (KE,
+
Klc',
+
PElf
[8.11]
FIGURE 8.24 A rigid plate rotating abollt the :.-.Ixis with angular speed t'. fhe kinelic energy of a panicle ofma..s In is I,InT'"!. The toTal kinetic (,,,erg)' ofth;' plat~ is 1/W'!.
f- Conservotion of mechanical energy
where i and frefer to initial and final values, respectively, and PE includes the potelllial energies of all conservative forcc~ ill a giycn problem. This relation is true onl)' if we ignore dissipative forces !-luch as friction. In thM case, it's necessary to resort to a generalization of the work-energy (heorem: [8.12]
f- Work-energy theorem including rototionol energy
PROBLEM-SOLVING STRATEGY ENERGY METHODS AND ROTATION
I. Choose two points of interest, one where all necessary information is kllO\\ n, and the other where information is desired. 2. Identify the conservative and nonconscn'ath'e forces acting on the system being analyzed. 3. Write the general work-energy theorem, Equation 8.12, or Equation 8.11 if all forces are conseryati\·e.
4. Substitute general expressions for the terms in lhe equation. fl. Use v == rw to eliminate either w or v from the equation.
6. Solve for the unknown.
EXAMPLE 8.12 Goal
A Boll Rolling Down on Incline
Combine gnl\'itational, translational. and rotational energy.
Problem r\ ball of mass J\l and radius N slarts from rest al a height of 2.00 m and rolls down a 30.0° slope. as in Figure 8.25. What is the linear speed of the ball when illeavcs thc incline? Assumc lhat the ball rolls withoul slipping. Strategy The two points of in I.erest are the LOp and hOltom of the incline, ,,,,jth lhe bottom acting as the lero point of gravitational pOl('l1tial energy. As the ball rolls down the ramp, gravitational potential energy is convcned into both translational and rotational kinctic energy without dissipation, so cOllservation of mechanic£11 cnergy can be applied with the lise of Equation 8.11.
r
2.00111
L
3Q.O''(
FIGURE 8.25 (E.x:lInplc8.12);\ billl sian" frum rc~t at the top oran iJl(]ill(' (llId rnlh to Ihe bOllom WilhOul ~Iippillg.
248
Chapter 8
Rotational Equilibrium and Rotational Dynamics
....
.........................................
Solution Appl)' conservation of energy with PE = P/~" the potential enerRY associated with gravity: Substitute the
o+ 0 +
Mgh ~ ~Mv'
+ H~MR2)W' + 0
The ball rolls without slipping, so Rw = v, the "no-slip condition," can be applied:
_t
Ogh _ v- \ -7--
Solve for v, noting thal Mcancels.
....
10(9.80 m/s')(2.00 m) 7
5.29 m/s
......................
~
QUESTION 8.12 Rank from faslcslLO slowest: (a) a solid ball rolling down a ramp without slipping, (b) a cylinder rolling down the same ramp without slipping, (c) a block sliding down a frictionless ramp with the same height and slope. EXERCISE 8.12 Repeat this example for a solid cylinder of the same mass and radius as the ball and released from the same height. In a race ben.. .een the t\Vo ol~jeC[s on the incline, which one would win? Answer
v = Y4gh/3 = 5.11 m/s; the ball would win.
QUICK QUIZ 8.4 Two spheres, one hollow and one solid, are rotating with the same angular speed around an axis through their centers. Both spheres have the same mass and radius. vVhich sphere, if either. has the higher rotational kinetie energy? (a) The hollow sphere. (b) The solid sphere. (e) They have the same kinetic energy.
EXAMPLE 8,13
Blocks and Pulle
Goal Sake a system requiring rotation concepts and the workenergy theorem. Problem Two blocks with masses }}/.I = 5.00 kg and 1112 = 7.00 kg are attached by a string as in Figure 8.26£1, over a pulley \'lith mass M = 2.00 kg. The pUlley, which turns on a frictionless axle, is a hollow cylinder with radius 0.050 0 m over which the string moves without slipping. The horizontal surface has coerFicient of kinetic friction 0.350. Find the speed of the system when the block of mass 171'1 has dropped 2.00 m.
m,
Strategy This problem can be solved with the extension of the work-energy theorem, Equation 8.12. [f the block of mass 11[2 falls from height hto 0, then the block or mass 111 1 moves the same distance, 6.x = h. Apply the work-energy theorem, solve for v, and substitute. Kinetic friction is the sole nonconservativc lorce.
(al
(b)
FIGURE 8.~~ (a) jExamplc 8.13) (h) (Exercise 8.13) In both cases. T] and T~ exert torques on the pullC}".
8.7
,..
249
Angular Momentum
. ....
.
~
Solution Apply the work-energy theorem, with PH = PEg, the potential energy associated \,,"ith gravity: Substitute the frictional work for H~lr' kinetic energy changes for the two blocks, the rotational kinetic energy change for t.he pulley, and the potential energy change for the second block:
-iL,n!1>·
=
-iL,,(m,g)ilx
+
(~Iw' - 0)
= (~IIIIV' -
0) + (tn,"v' - 0)
+ (0 - lII,gh)
For a hoop, 1= A1r 'l. so (//102) = M. Substitute this quantity and v = rw:
Solve for v:
lIt'lgh ~
Substitute III, ~ 5.00 kg, "', ~ 7.00 kg, M = 2.00 kg, g ~ 9.80 l11/s2 , II ~ 2.00111, and ~ 0.350:
m,
lrrrr.. •
v=
J1-k (
_1 2 2U1 JV
m Jg) h -
I 2 I 2 + "21n2V + '2A1v
3.83 m/s
.
••
Remarks In the expression for the speed v, the mass In l of the firsL block and Lhe l11ass M of the pulley all appear in the denominator, reducing the speed, as they should. In the numerator, 111'2 is positive while the friction tcrm is ncgative. Both assertions are reasonable because the force of gravity on 1112 increases the speed of' the system while the force of friction on "fn J slows it dmvn. This problem can also be solved ,vith Newton's second law together with T = la, a good exercise. QUESTION 8.13 How would increasing the radius of the pulley affect the final answer? Assume the angles of the cables arc unchanged and the mass is the same as before.
~
EXERCISE 8.13
Two blocks with masses "Ill, = 2.00 kg and "'" = 9.00 kg are attached over a pulley with Illass 1Vl = 3.00 kg, hanging straight down as in Atwood's machine (fig. 8.26b). The pulley is a solid cylinder with radius 0.050 0 Ill, and there is some friction in the axle. The system is released from rest, and the string moves without slipping over the pulley. If the larger mass is traveling at a speed of 2.50 m/s \vhcn it has dropped] .00 Ill, how much mechanical energy was lost due to friction in the pulley's axle? Answer
29.5.1
8.7 ANGULAR MOMENTUM In Figure 8.5.7, an object of mass 1n rotmes in a circular path of radius ,; acted on by a net force, F net . The resulting nct torque on the object increases its angular speed from the value W o to the value w in a time interval 6./. Therefore, we can wTilc
2> ~ 10:
~w = 1 -= 1 ill
(w
-illwo) = Iw - Iw o ill
,.
as the angular momentum of the
2:
T
= change
ol~jecl.
[8.13]
then we can write
angular momentum time interval
ill
'"
/
If we define the product L= Iw
~/"
ilL ill
[8.14]
FIGURE 8.27 An object of mass 111 rotating in a circular path under the action or a COnSUHlt lorque.
250
Chapter 8
Rotational Equilibrium and Rotational Dynamics
Equation 8.14 is the rotational analog of Nc\v'[on's second law, which can be \\iTilten in the form F= 8Jl/!:1t and states that the net torque acting on an object is equal to the time rate of cbange of the object's angular momentum. Recall that this equation also parallels the impulse-momentum theorem. \Vhcn the net extcnwl torque (~'T) aCling
011 3
system is zero, Equation 8.l4
gives that 6.L/6.t = 0, which says that the lime raLe of change o[ the system's angular momentum is zero. ",,,",c then havc the following important result: Conservation of angular momentum ~
Lct Lj and L{be the angular momenta ofa system at two differcllllimes, and suppose there is no net external torque, so ~T = O. Then [8.15]
L;=0 and angular rnomentllm is said to be L'onserved.
Equation 8.l5 givcs liS a third conservation law to acid to our list.: conservation of angular momentum. \Ve can now st.ate lhat the mechanical energy, linear momentum, and angular momentum of an isolated system all remain constant. If the moment of inertia or an isolated rotating system changes, the system's angular speed will change. Conservation ofangulal' momentum then requires that. if
N APPLlC:...:AT:...:I.=c 0 :...:
_
Figure Skating
APPLICATION Aerial Somersaults
~>=o
[8.16]
Not.e that conservation of angular momentum applies to macroscopic objects such as planets and people, as '.. .ell as to atoms and molecules. Thcre arc many examples or conservation of angular momentum; one of the most dramatic is that of a flgure skater spinning in the finale of her act. In Figure 8.28a, the skater has pulled her arms and legs close to her body, reducing their distance from her axis of rotation and hence also reducing her moment of inertia. By conservation of angular momentum, a reduction in her moment of inerlia must increase her angular velocity. Coming out of the spin in Figure 8.28b, she needs to reduce her ang·ular velocity, so she extends her arms and legs again, increasing her moment of inertia and t.hereby slowi ng her rotation. Similarly, \-"hen a diver or an acrobat 'wishes to make several somersaults, she pulls her hands and feet close to the trunk of her body in order LO rotate at a greater angular speed. 1n this case, the cxternal force clue to gravit)' acts through her center of gravity and hence exerts no torque about her axis or rotation, so the angular momentum about her center of gravity is conserved. For example, when a diver wishes to double her angular speed, she musl reduce her mOfnellt of inertia to half its initial value. An interesting astrophysical example of conservation of angular momentum occurs when a massive star, at the end of its lifetime, uses up all its fuel and collapses under the influence of gravitational forces, causing a gigantic outburst of energy called a supernova. The best-studied example of a remnant. of a supernova explosion is the Crab Nebula, a chaotic, expanding mass of gas (Fig. 8.29). In a
FIGURE 8.28 Michelle Kwau controls her momcnt orincnia. (a) By pulling in her arms anrl legs, she reduces her momellt ofillcrtia and increases her angular velocity (rate of spi 11). (b) Upon lallding, eXlending her arms and legs increases hel 1ll0rnenl orillerlia and helps sial\' her spill
TighTly curling her body, a diver decreast:s her moment of inertia, increasing her <Jllgular vc1ocilY.
(bl
8.7
"
~
] .::
£ ~ ~
(a)
(b)
Angular Momentum
251
FIGURE 8.29 (a) The Crab Nebula in lhe constellation Tam·us. This neblda is llw remnant ora supernova secn on Earlh in A.D. 1054. Ir is 10cHed some 6 300 liKht-years away and is approximately 6lighl-ycars in diameter, still expanding outward. A pubarclccp inside lhe nebula (]asllf>s 30 times every second. (b) Pulsar orr (c) Pulsar on.
(c)
supernova, part of the star's mass is ejected into space, where il even wally condenses into new stars and planets. Most of what is left behind typiGdly collapses into a neutron star-an extremely dense sphere ofmal.ler with a diameter of about 10 k m, greatly reduced from the 10"-km diameter of the origi nal slar and containing a large fraction of the star's original mass. [n a neutron star, pressures become so great that atomic electrons combine with proLOns, becoming neutrons. As the moment of inertia of the system decreases during the collapse, the star's rotational speed increases. More than 700 rapidly rotating neutron stars have been identified sillce their first discovery in 1967, with periods of rotation ranging from a millisecond to several seconds. The neutron star is an amazing- system-an o~ieCl with a mass greater than the Sun, fitting com forI ably ,vithin the space of a small county and rotating so fast that the tangential speed of the surface approaches a sizable fraction of the speed of light!
APPLICATION Rotating Neutron Stars
QUICK QUIZ 8.5 A horizontal disk with moment of" inertia I[ rotates with angular speed w[ about a venical frictionless axle. A second horizontal disk having moment of inertia 12 drops onl0 the first, initially not rotating but sharing the same axis as the first disk. Because their surfaces are rough, the two disks evenLUall>' reach the same:.: angular speed UJ. The ratio w/w l is eqllal to (a) '/1, (b) J,lI, (c) 1,/(/, + '2) (d) 12/(/, + '2) QUICK QUIZ 8.6 [[global \-I.'arming continues, it's likely that some ice from the polar ice caps of the Earth will melt. and the water will be distributed closer to the equator. If th is occurs, would the length of I he day (one revolution) (a) increase, (b) decrease, or (c) remain the same?
EXAMPLE 8.14 The S innin Stool Goal
Apply conservation oranguhtl-momentlll11 to a simple system.
Problem A student sits on a pivoted stool while holding a pair of weights. (See Fig. 8.30.) The sLOol is free to rOlate about. a vertical axis with negligible friction. The moment of inertia of sludent, weights, and stool is 2.25 kg ·m 2 . The student is set in rotation with arms out.stretched, making one completc turn every 1.26 s, arms outstretchcd. (a) \Vhat is the initial angular speed of the system? (b) As he rotates, he pulls the weights inward so that the new moment of inertia of the system (sLudent, objects, and stool) becomes 1.80 kg' m 4 . What is the new angular speed of Ihe system? (c) Find the work done by the swclent on the system while pulling in the \veights. (Ignore energy lost through dissipation in his muscles.)
Wr
I~ I
(a)
(b)
FIGURE 8.30 (Example H.14) (a) The sllldenl is g-ivcn all inilial angular spced while holding lWO
Strategy (a) The angular speed can be obtained frolll the frequency, weights OUl. (b) The ;lIlgU!;lf specrl increases as which is the inverse of the period. (b) There are no external lorques act- the student draws the weights inwards. ing on the system, so the new angular speed can be found with the principle of conservation of angular mOlllenlUm. (c) The work done on the system during this process is the same as the systern's change in rotational kinetic energy.
252
Chapter 8
Rototional Equilibrium and Rotational Dynamics
,.. ..
..................................
~
Solution (a) Find the initial angular speed of the system. Invert the period to gCllhe frequency, and multiply by 27T:
W, ~
2"'f= 2.,,/1'= 4.99 rad/s
(2)
(2.25 kg' m")(4.99 rad/s) = (1.80 kg' m')wj
(b) After he pulls the weights in, what's the system's new angular speed? Equate the initial and final angular momenta of the system:
Substitute and solve {or the Filla I angular speed
w/
w! =
6.24 rad/>
(c) Find the work the student docs on the system.
I'VSllIf!t:1H -"K-llw'-!lw" U r - 2 f j 2' I
Apply the ''lork-energy theorem:
= H1.80 kgm 2)(6.24 rad/s)2
- H2.25 kg· m 2 )( 4.99 rad/5)'
.....
.....
. . .. .. . . . . . . . . . . . . . . . . . .
Remarks Although the angular momentum the student does work on the system.
,
.
.....................
-.
- .......
or the syst.em is conserved, mechanical energy is not conserved because
QUESTION 8.14 If the student suddenly releases the weights, ,...·ill his angular speed increase, decrease, or remain the same? EXERCISE 8.14 A SLar "vith an initial radius of l.0 X l08 m ancl period of 30.0 days collapses suddenly to a radius or l.0 x 10 4 m. (a) Find the period ofrmation after collapse. (b) Find the work done by gravity during the collapse if the mass of the SLar is 2.0 X 10:{O kg. (c) "Vhat is the speed oran indestructible person standing on the equator of the collapsed star? (Neglect any rclativistic or thermal effects, and assume the star is spherical before and after it collapses.) Answers
(a) 2.6 X 10- 2 S
EXAMPLE 8.15
(b) 2.3 X 10.12 J
(c) 2.4 X 10' m/s
The Merry-Go-Round
Goal Apply conscl'vmian of angular momentum ,.. . hilc combining two moments ofinenia. Problem
A mcrry-go-round modeled as a disk of mass
J\t1 = 1.00 X 10 2 kg and radius R = 2.00 m is rotating in a
horizontal plane about a frictionless vertical axle (Fig. 8.31). (a) Arter a student with mass m = 60.0 kg jumps on the rim of the merry-go-round, the system's angular speed decreases La 2.00 rad/s. If Lhe studenL 'walks slowly from the edge toward the center, find the angultJr speed of the systern when she reaches a poim 0.500 m from the center. (b) Find the change in the system's rotational kinetic energy caused by her movement to the center. (c) find the work done on the slUdent as she walks to ,. = 0.500 111.
FIGURE 8.31 (Example 8.15) As the student walks tuward the cen1cr of the rotating platform, the 1fI01llCllLOf inertia of the ~~'stern (stlldcl'll plus plalform) decrcases. Because angular momentum is consen'cd, 1hc 'lng-IlIaI' speed of 111(; system must increase.
Strategy This problem can be solved with conservation of angular momentum by equating the system's initial angular momentum ,,,,,hen the student stands aLthe rim to the angular momentum when (he student has reached r = 0.500 m. The key is to fi nd the different moments of inertia.
8.7
,..-'
.
Angular Momentum
253
..........................................
~
Solution (a) Find the angular speed when the student reaches a point 0.500 111 from the centcr.
Iv = ~MR' = ~(1.00 X 10' kg)(2.00 m)'
Calculate the moment of inertia of the disk, Tf):
= 2.00 X 10' kg· m'
Calculate the initial momcnt of inertia of the s1.lldenl. This is the same as the moment ofincnia ora mass a distance R from the axis:
Is = mR' = (60.0 kg)(2.00 m)' = 2.40 X 10' kg· m'
Surn the t\VO moments of inertia and multiply by the initial angular speed to nnd L i , the initial angular momenturn of' the system:
Li = (I"
+
{,)w i
= (2.00 X 10' kg· m' = 8.80 X 10' kg·
mr/ =
Calculate the student's final moment of inertia, f v ' when she is 0.500 m from the center:
1.\/=
Thc mornent of inertia of the platform is unchanged. Add it to the student's Hnal moment of inertia, and multiply by the unknown final angular spced [Q nnd Li
/1= (If)
Equate the initial and final angular momenta and solve for the final angular speed of the system:
+
+ 2.40 X 10' kg· m')(2.00 rad/s)
m'js
(60.0 kg)(0.50 m)2 = J5.0 kg· m 2
ISJ}w/= (2.00 X 10 2 kg·m'!
+
]5.0 kg·m 2)wJ
L, = Lf
(8.80 X 10' kg· m'/s) ~ (2.15 X 10' kg· m')wf WI =
4.09 rad/s
(b) Find the change in the rotational kinetic energy of the system.
Calculate the initial kinetic energy of thc system:
KE,
=
~/,w,'
= H 4.40 X 10' kg
111
2
)(2.00 rad/s)'
= 8.80 X lO' J Calculate the final kinetic energy of the system:
KI~= ~/fw/ ~ H215 kg·m')(4.09 rad/s)' ~ 1.80 X 10'J
Calculate the change in kinetic energy of the system:
faj -
KE, =
920)
(c) Find the work done on the student. The sLUdentunclergoes a change in kinetic energy thal equals the work done on hcr. Apply thc work-l'nergy theorem:
HI - A K" -.u. L:.,tu
-- 1f, 2 2 SfWf
_
1{
'!
SWi
2
" = 2'( I~.O kg·m-")( 4.09 rad/s )"-
- ~(2.40 X lO' kg·m')(2.00 rad/sf IV= -355.1 ~
.
.
~
The angular momentum is unchanged by imernal forces; hmvc"er, the kinctic energy increases because the studcnt must per[orrn positivc \\lork in ordcr to walk to\vard thc ccnlcr of the platform.
Remarks
QUESTION 8.15 Is energy conservation violatcd in this exalnple? Explain why there is a positive net change in mechanical energy. \'VhClt is the origin of" this cnergy?
254
Chapter 8
Rotational Equilibrium and Rotational Dynamics
EXERCISE 8.15 (a) Find the angular speed of the merry-go-round before the stlldcntjumped on, assuming the SLudcl1l didn't trans-
fer any momentum or energy as she jumped on the merry-go-round. (b) By how much did the kinetic energy of the system change when the sludentjulllpcd on? Notice that energy is lost in this process, as should be expected, since it is essentially a perfectly inelastic collision.
Answers
(a) 4.4 rad/s
---~-
,
(b) KEf - KE, = -1.06 X 10 3J.
--
SUMMARY ~l~~~~-
8.1
•
Torque
F
Let be a force acting on an object, and ICl.r be a position veclor from a chosen poinl 0 to the point of application of thc--->forcc. Then the magnitude of the torque -:; of the fm"ce F is given by T = rFsin 0 [8.2] where r is the length or the position veclor, F till: magnilude orlhe force, and uthe angle bcn...ccn F and The quantity d = r sin 0 is called the lever arm of the
r.
force. 8.2
Torque ond the Two Conditions for Equilibrium
An object in mechanical equilibrium lllust satisfy the following two conditions:
I. The net external force must be zero: tF = O. 2. The net eXlcrnallOrquc must be zero: ~T = O. These two conditions, Llsed in solving problems involving rotation in a plane-result in three equations and three unknowns-n..·o from the first condition (corresponding to the x- anel y-componenls of the force) and olle from the second condition, on torques. These equations must be solved simultaneously.
8.6 Rotational Kinetic Energy If a rigid ol~ject rotates about a fixed axis with angular speed w, its rotational kinetic energy is KE, = !Iw'
[8.10]
where f is the moment of" inertia of the object around the axis of rotation. A s)'stem involving rotation is described by three types or energy: potential energy PE, translational kinetic energy Klc,'" and rotational kinetic energy KE,. All these forms of energy must be included in the equation for conservation or mechanical energy for an isolated system: (KE,
+
KF,
+
PEl, ~ (KE,
+
KE,
+ PElf
[8.11]
,...here i and I refer to initial and final v
8.7
Angulor Momentum
TIle angular momentum of a rotating object is gi,'en by [8.13]
L= Iw
Angular momentum is related to torque in the following equation:
8.5 Relationship between Torque ond Angular Acceleration
"7 __ change in angular momentum __ !1L
The moment of inertia of a group of particles is
.::. [8.7]
If a rigid object free w rotate about a fixed axis has a net extcrn
~T=~ ~.~ This equation is the rotational eqlli\'alcnr of the second law ormotion. Problems are solved by lIsing Equation 8.8 together with I'\ewwn's second law ,lnd solving lhe resulting equations simulLaneollsl)'. The relation a = 1ll' is often key in relating the translational equations to the rotational equations.
time interval
D.l
[8.14]
If the nel externai torque acting on a sySlem is zero, lhe total angu Jar momentum of the system i3 consrant,
L;
~
Lr
[8.15]
and is said to be conserved. Sol.... ing problems usually involves substituting into the expression I,w,
= Irwr
[8.16]
and solving for the unknown.
FOR ADDITIONAL STUDENT RESOURCES. GO TO WWW.SERWAYPHYSICS.COM
1. A wrench 0.500 III long is applied to ,'ard at an angle of 60.0° with respect 10
a line from the boll. through the end of the wrench. How much torque is applied to the nut? (a) 34.6 N· m (b) 4.56 1\:. m (c) JI.8l'· m (el) 14.2 N· m (0) 20.0 N· m
Problems
2. A horizontal plank 11.00 111 long and having 111<15S 20.0 kg rests on two pivots, one at the left end and a second 1.00 III from the right end. Find the magniUlde of the force exertcd on the plank by the second pivot. (a) 32.0 N (b) 45.2 N (c) 112 N (d) 131 N (e) 98.2 N
3. What is tile magnitude of the angular acceleration of a 25.0-kg disk of radius 0.800 m when a torque of magnitude 40.0 N· m is applied to it? (a) 2.50 rad/s 2 (b) 5.00 rad/s' (e) 12.5 rad/s'
(c) 7.50
rad/s'
(d) lO.O
rad/s'
4. Estimate tile rotational kinetic energy of Earth by treating it as a solid sphcre with uniform dcnsity. (a) 3 X 1019 kg' m 2 /s 2 (b) 5 X ]0 2i kg·m 2 /s 2 (c) 7 X 10 3o kg' m 2 /s 2 (d) 4 X lO" kg 'm'/,' (e) 2 X 10'·; kg 'm'/s'
5. Two forces arc acting on an ol~ject. Which of the I"ollowing st.atements is correct? (a) The object is in equilibrium irthe forces arc cqual in magnilude and opposite in direction. (b) The object is in equilibrium if the net torquc on the o~icct is zero. (c) The ol~ject is in equilibrium if" the forces act at the same point on the object. (d) The ol~ject is in equilibrium if"t.hc net force and the net torque on rhe objcct arc both zero. (e) The o~jCCt cannot be in equilibrium because more than one force acts on it. 6. A disk rOGues about a fixed axis that is perpendicular 10 the disk and passes Ihrough its center. At any instant, does every point on the disk have the same (a) cenu-ipetal acceleration, (b) angular velocity, (c) tangential acceleration, (d) linear velocity, or (e) LOtal accclerat.ion? 7. A constant nel nonzero torque is exerted on an objecl. Which of the following quantities (0111101 be constant for this ol~jeCl? fI.·lore than one answer may be correct. (a) angular acceleration (b) angular vdocity (c) moment of inertia (d) center of mass (c)
8. A block slides down a frictionless ramp, while a hol1m,,' sphere and a solid ball roll 'withotlt slipping down it second ramp with the same height and slope. Rank the ,-\rrivaltimes at the bottom from shortcstLO longest. (a) sphel~e, ball, block (b) ball, block, sphere (c) ball, sphere, block (d) block, sphere, ball (e) block, ball, sphere
255
9. A solid disk and a hoop arc silllulraneously released from rest at the LOp of an incline and roll dO'wn withoLlt slipping. 'Which o~iect reaches the bottolll first? (a) The one that has the largest mass arrives first. (b) The one that has the largest radius arrives first. (c) The hoop arrives first. (d) The disk arrives first. (e) The hoop and the disk arrive at the same time.
10. A solid cylinder of mass lVJ and radius R rolls down an incl inc wilhout sl ipping. I t.s moment of inertia about an axis I.hl"OlIgh its center of lTIass is MR'!.j2. At any instant whilc in mOlion, its rotational kinetic energy about its centlT of mass is whal fraction of its total kinetic energy? (a) : (b) t (c) ~ (d) ; (e) None of these 11. The cars in a soapbox derby have no engines; they sim~
ply coasl downhill. \.\rhich of the following design criteria is best from a competitive point of view? The car's wheels should (a) have large moments of inertia, (b) be massive, (c) be hoop-like wheels rather than solid disks, (d) be large wheels rather than small wheels, or (e) have small moments of inenia.
12. Consider two uniform, solid sphercs, a large, massive sphere and a smaller, lighter sphere. They are released frolll res I simultaneously from the top of a hill and roll down withal It slipping. \lVhich one reaches the bottom of thc hill fi rst? (a) The Ia rge sphere reaches the bottom first.. (b) The small sphere reaches the boltom ftrst. (c) The sphere with the greatest density reachcs the bottom first. (d) The spheres reach the bottom at the same time. (e) The answer depends on the values of the spheres' masses and radii. 13. A mouse is initially at rest
011 a horizontal llIrntable mounled on a frictionless, vertical axle. As the mouse begins to walk clockwise around the perimeter, which of the following statements must be true of the turntable? (a) It also turns clockwise. (b) It turns counterclockwise wilh the same angular velocity as lhe mouse. (c) It remains stationary. (d) It t urns counterclockwise bccallse angular momentum is conserved. (e) It turns cOlllllerclockwise because mechanical energy is consen'-eel.
CO"lCEPTUAL QUESTIO"lS 1. Why can't you put your heels firmly against a wall and then bend over withol\t falling?
2. 'Why does a tall athlete have an advantage over a smallclone when the two are competing in the high jump? 3. Both torque and work are products of force alld distance. How are they different? Du they have the sallie units? 4. Is il possible to calculate the LOrqllc acting- on a rigid objcct without specifying an origin? Is Ihe torque independenl of the location of the origin?
5. Can
256
Chapter 8
Rotational Equilibrium and Rotational Dynamics
hold the door closed. If the paleontologist is pressing at the center of the door, and the botanist is pressing at the edge about 8 em from the hinge, est.imate how far the paleontologist would have to relocate in order [0 have a greater effect on keeping the door closed than bOlh of them pushing together have in their original positions. (Question 6 is courtesy of Edward r. Redish. For more questions of this type, sec physics.umd
w,,,,.. .
.edujpcrg/.l 7. In some motorcycle races, lhe riders drive over small hills and the motorcycle becomes airborne ror a short lime. If the motorcycle racer keeps the throttle open while leaving the hill and going into the air, the mOlorC)'cle's nose tends to rise upwards. 'Why does this happen?
n.
In a tapc recorder, the tape is pulled past the read-write heads at a constant speed by the drive mechanism. Consider the reel from which the tape is pulled: As the tape is pulled off, the radius of the roll of remaining tape decreases. How does the torque on the reel change with time? How does the angular speed of the reel change ..vith time? If the tape mechanism is suddenly turned on so that the tape is quickly pulled with a large force, is the tape more likely to break when pulled from a nearly full reel or from a nearly empty reel?
]2. (a) Give an example in which the net force acting on an object is zero, yet the nellorque is nonzero. (b) Give an example in which the net lorque acting on an object is zel'O, yet the net force is nonzero.
8.Uyou toss a textbook into the air, rotating it each time
13. A ladder rests inclined against a wall. Would you feel
about one of the three axes perpendicular to it, you will find that it will not rotate smoothly about one of those axes. (Tq' placing a strong rubber band around the book before the toss so that it will stay closed.) The book's rotation is stable about those axes having the largest and smallest moments of incl-tia, but unstable about the axis of intermediate moment. Try this on your own to find the axis that has this intermediate momelll of inertia.
safer climbing up the ladder if you were told that the 0001' was frictionless, but the wall ,...-as rough, or that the wall was frictionless, bUl the floor was rough? Justify your answer.
9. Stars originale as large bodies of slowly rotating gas. Because of gravity, lhese clumps of gas slowly decrease in size. Whell happens to the angular speed ora Slar as it shrin ks? Explain.
14. A cat usually lands on its feet regardless of the position frolll which it is dropped. A slow-motion film of a cal falling shows that the upper half of its body twists in one direction while the lower half twists in the opposite direction. (See Fig. CQ8.14.) Why does this type of rOlatjon occur?
10. Tf a high jumper positions his body correctly when
going over the bar, the center of gravity of the athlete may actually pass under the bar. (See Fig. CQ8.l0.) Explain ho",.. this is possible.
FIGURE CQ8.14 A falling, u\·jsting cat.
FIGURE CQ8.10
~
PROBLEMS ~
•
WebAssIQl'l
I,:!,
m
~~
=
rEI =
II =
The Problems for this chapter may be assigned online
ell
straightforward, intc:rmediale, challet1~it1g denotes guided problem denotes enhanced content problem
=
biomedical application
0= denotes full solution a\'ailablc in Student Solillio/l.,\ Monual/ Sindy GuidI'
SECTION 8.1 TORQUE
WcbAssign.
1. A grinding wheel of radius 0.350 III rotating on a frictionless axle is broughl to rest by applying a constant friction
force tangential to its rim. The conslant torque produced by this force is 76.0 N· Ill. rind the magnitude of the friclion force.
Problems 2. According to the manual of a certain Cllr, a lIlaxill1l!m torque of magnirude fi5.0 N . m should be applied ",hen tiglllening the lug nuts on the vehicle. 11')'011 lise a wrench or length 0.330 III and you apply the force.:' at the end of the wTellch at an angle 01'75.0° with respect 10 a linc going from the lug nUl through the end of the handle, whal is the magnitude of [he maximum force you can exen on the handle withou! exceeding the recommend<1lion?
SECTION 8.2 TORQUE AND THE TWO CONDITIONS FOR EQUILIBRIUM SECTION 8.3 THE CENTER OF GRAVITY SECTION 8.4 EXAMPLES OF OBJECTS IN EQUILIBRIUM
[]Jill
Tile ann in Figure P8.7 weighs 41.5 N. The force of gravity acting on the ann acts through ..goiJlI. A. Determine the magnitudes o.!lhe tension force F, in the deltoid muscle and the force F, exerted by the shoulder on lhe 11lImerus (lipper-arm hone) to hold the arm in the position shown.
3. Calculate the net torque (magnitude and direction) on the beam in Figure PS.3 abolll (a) an axis through 0 perpendicular to the page and (b) an axis through Cperpclldicular to the p:lge.
I I I
25 N
130
~~5°\~
l20
o
0
:~ I
m \'
200
JK~ 10
A .,.--------
o
;..J
F
-4.0rn
1<-----O.290m
30 N
n A steel band exerts a horizolltal force of 80.0 N all a tooth at point B ill Figure P8A. ,,,,'hat is the torque on the root of the tOoth abOtll point A?
_I
Ii
FIGURE P8.7
FIGURE P8.3
4.
257
8.
E
A nniform beam of length 7.60 m and weight 4.!10 X 1O~ N is carried by twO workers, Sam and Joe, as shown in Figure P8.8. (a) Determine the forces thaI each person exerts on lhe beam. eb) Qualitativel)" ho\·... would the answers change ifS<11l1 moved doser LO the midpoint? (c) What would h
FIGURE P8.4
5. _ A simple pelldulunl consists of a small object 01" mass 3.0 kg hanging at the end ofa 2.0~m-long light string th<11 is connected 10 a pivot point. (,I) Calculate the magnitude orthe LOrque (due to the force of gravity) abolll this pivot point when the string makes a 5.0 0 angle with the \'erlieil!. (b) Does lhe torque increase or decrease as the angle increases? Expl
1-1.00 III
L-2.00 m-
1<------- 7.60 m --------+J FIGURE P8.8
9.111
A cook holds a 2.00-kg canon of milk at mm's length (Fig. PS.g). What force FlJ must be exerted by the biceps muscle? (tgnore the weight of the forearm.)
Wrile lhe necessary equations of equilibrium of 6. _ Ihe object shown in Figure P8.5. Take the origin of the torque equation aboll[ an axis perpendicular lo [he page through the point O.
75.0°
Fg
-:.!5.0 CIll_!
I+--
8.00 em FIGURE P8.9
FIGURE P8.6
10. A meterslick is found [0 balance at the 49.7-crn mark when placed on a fulcrum. ,'\'hen a 50.0-gnIlTI mass is attached
Chapter 8
258
Rotational Equilibrium and Rotational Dynamics
at the 1O.0-em mark, the fulcrum mll~l be moved 10 the 39.2-cm mark for balance. Whal is the mass of the meterslick? [I[]Find the x- and y-coordinates of the center of gnwil.), of a 4.00-1', by 8.00-1', uniform sheet of plywood with t he upper right quadranl removed as shm.. . n in Figure P8.11.
vc'-gc of lipping, find symbolic expr"essiollS for (
e.
15. "
e
Many of the clcmeIHs in horiwilial-bar exercises call
be modeled by represellting the gymnast by fOil I' segy(rL)
4.00
ments consisting of arms, tOrso (including the he'lel), thighs, and lower legs, as shown in Figure P8.l5a.lnenial parameters for a particular gymnast arc as follows:
2.00
Segment
o
x(~)
o
2.00
4.00
6.00
H.OO
Length (m)
reg (m)
[(kg'm')
6.87
0.548
0.239
0.205
Torso
33.57
0.601
0.337
1.610
Thighs
14.07
0.374
7.54
Legs
FIGURE P8.11
12.
Mass (kg)
Arms
Ell A beam resting on IWO pivols has a length of L = 6.00 III and mass !\tJ == 90.0 kg. The pivol. under lhe left end exerts a normal force ilIon the beam, and the second pivot placed a distance = 4.00 In from the left end exerts a normal force 1/,;1' A woman of mass m = 55.0 kg steps onto the left end of the beam .wel begins walking LO the right as in Figure P8.12. The goal i:-; to find the woman's position when the beam begins to tip. (a) Sketch a free-body diagr;tnl, labeling the gravilational and normal forces acting on the beam and placing the woman x meters to the right of the first pivOl, which is the origin. (b) Where is the woman when the normal force 'n l is the g-realesl? (c) What is III when the beam is about. to tip? (d) Use the force eqllation of equilibrium to find the value of 1/,2 when the beam is abollt to tip. (e) Using the result of pan (c) and the torque eClllilibriunl equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. (0 Check the answer w pan (c) by computillg lorques around the first pivot point. Except for possible slight elif:' fercnces due 10 rounding, is the answer the same?
0.151
0.173
0.227
0.16'1
that in Figure P8. ISa Tell" is the d istanct:: to the cenler of gravity measured [rolll the joint dosest to the bar and the masses for the arms, thil;hs, and legs include bmh appendages. I is thc momcnt of inertia of each segmclll abollt its cenler of gravity. Dctermine the distance from the bar to the center of gravity of the gymnast for the t\VO positions shO\<,:n in Figures P8.15b and r8. I5c. .'\'Ote
e
(b)
(e)
FIGURE P8.15
16.
,-----~L-----__>I
II
Csing the data given in Problem 15 and the coordinate s}'stem shown in Figure P8.16b, ca1cuhnc the position or the ccnter of gravity of (hc gym nasi shown in Figurc P8.16a. Pay close attcmion t.o the definition of r".1'; in the table.
M
lI.
leg
.1
f
x
FIGURE P8.12
(a)
13. Consider the following mass distribution, where x- and y-coordinates are given in mcters: 5.0 kg at (0.0, OJ)) Ill, 3.0 kg at (0.0, 4.0) In, and '1.0 kg at (3.0,0.0) In. Where should a founh objccr 0[8.0 k~' be placed so that the center" of gravity of the four-objcct arrangement will be at (0.0,0.0) Ill? 14.
m
r\ bea m of length L :mc1 mass M rests 011 two pivots. The first pivot is at the left end, taken as the origin, and the second pival is at a distance from the left end. A woman or mass m starts at the left end and walks toward the riglH end as in Figure P8.12. \rVhcn the heam is on the
e
(b)
FIGURE P8.16
17.
II
A person bending lorward [Q liha load "with his back" (Fig. P8.l7a) rathcr than "\"Iith his knees" can be iruured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the ercc[Qr spinalis muscle in the back. To sec the magnitude of the forces involved, and to understand why l);Ick problems are COIllman among humans, consider the model shown in Figure P8.l7b ora person bending forward to lifta 200-~ object.
Problems The spine ,md upper body arc represented as a llnif(}rm horizonwl roel of weight 350 N, pivoted at the base of the spine. The erector spinalis muscle, attached at a point lwo·thirds of the wa)' up the spine. maintains the posilion 01" the back. The angle between the spine and this muscle is 12.0°. Find the tension in the back muscle and the COlllpressional force in the spine.
Figure P8.21. Find the tcnsion ill each rope when a person is 0..1)00 m from the left end.
259 700-~
!lack tllll"c1c
\ 1',
R,
l'
Pi\OI
12.0°
1+------- 2.00 In ------~ (b)
(al
FIGURE P8.21
FIGURE P8.17
18. • When a person stands on tiptoe (a strenllou~ po~iti()ll'. the position of the fOOl is as shown in Figure P8.18:\. The LOlal gravittllional force on the body. is supponed b)' t)-It: force exened by the noor 011 the toes of one foo!. A mechanical model of the situation is shown in Figure P8.18h. where T is the force exerted by the Achilles tell· dOli Oil the foot and Ii is the force exerted by the libia on the foot. Find the values of I?, and 0 whell I~ = iOO r\.
Fr.
n
7:
Achilles tendon
O~l' ,
,: J!'J.Oo 1-
f
iB 0
Ii :\"
go.{)O
c~ ::::::.J.
25 n em
D ' (a)
FIGURE P8.18
22. t-\ hllngl-y 700-N hear walks am on ;1 beam in an atlemplto retl-ie\'e some "goodies" hanging at Ihe end (Fig. P8.22). The beam is uniform. weighs 200 N, and is 6.00 m long; lhe goodies weigh 80.0 !\!. (a) Draw a free-body diagram of the Goodies beam. (b) 'When thl' bear is at x = 1.00 m, lind the tension ill FIGURE P8.22 Ihe wire and tile components of the reaction force mlhe hinge. (c) Iflhe wire can withstand a maximum tellsion 01"900 N, what is the maximum distance lhe bear call walk before the wire breaks?
(h)
-
l::!!JA 500·l'\ ulliform reclangular sign 4_00 III wide and 3.00 m high i.. suspended from a .---7" hori.lOntal. 6.00-m-long, uni:W.O' form, 100-N rod as indicated in figure P8.19. The left end of the rod is supported by a Ia.uu:.,,-" $1I0r hinge, and the right end is supported bv tI thin Glble making FIGURE P8.19 a 30.0° angle with the \'ertical. (a) Find the tension T in the cable. (h) Find the horil.omal and \'~nical components of force exerted on the left end of the rod by the hinW" 20. A window washer is standing on a scaffold supported h\ a vertical rope at cach cnd. The scallclld weighs 200 Nand is 3.00 m long. What is the tension in each mpt: whcn till' 700·N worker sl
23. An B.OO-m, 200·1\' uniform ladder rests againsl a smooth wall. The coefficient of static friction between the lcldder ,Illel the ground is 0.600, and the lad del- makes a 50.0° angle with the ground. I-low blr up the ladder can an 800-:"J person climb before the lad del' begins to slip? 24.
A stnlt oflc:ngth L = 3.00 III and ImlSS III = 16.0 kg is held br a cable at an angle of 0 = 30.0° with respeclto the hori/ontal as shown in Figure P8.24. (a) Sketch a free-bod)' diagram, indicating all the forces and thci,· placemelll on the strut. (b) \\'h> is the hinge a good place to use for calculating tor(lllcs? (c) Write (he condition for rotational equilibrium srmbolically, calculating the torques around Ihc hingc_ (d) Lse the LOrque equation to calculate the tension in the cable. (e) \"'rile the .'(- and )-compOnenLS of NeWlon's second law for equilibrium. (f) ese the force equation to find the x- and )'-componems of the force on Ihe hinge. (g) A .. suming the stnn position is to remain the same, would it be advantageolls to auach the cable higher llP on (he W
EI
e
FIGURE P8.24
260
Chapter 8
Rotational Equilibrium and Rotational Dynamics
25. A student gels his car stuck in a snowdrift.. r\oL at a loss, having studied physics, he attaches one end ora SlOlil rope to the car and the olher end to the trunk ora ncarb}' UTe. allowing for a small amount of slack. The student then exerts a force F on the ccntn or the rope in the direction perpendicular LO the car-tree linc
r'---- 12.0", ----+1'1
Tree
28, A I 200-N uniform boom is supported by a cable pnpcndicular to the boom as in Figure P8.28. The booll1 is hinged at the bottom, and a 2000-N weight hangs from iu; top. Find the tension in the supponing cable and the components of Ihe reaction force exerted on the boom by the hinge.
FIGURE P8.28
129.111
The large quadriceps muscle in the upper leg termi~ nates at its lower end in a tendon attached 10 the upper end of the tibia (Fil{. P8.29a). The forces on the lower leg when t~e leg is extended are modeled as in Figure PS.29b, where T is the force of tension in the tendon, is the force of gravity acting on the lower leg. and F is the force of gravity acting on the ]'001.. Find T when the tendon is at an angle 01"25.0° with the tibia, assuming that TV = 30.0 N, F= l2.5l\1, and the leg is extended at an angle () 01"40.0° with the venical. Assume that the cenler or gnl\"ity of the lowel' leg is at its center and that the tendon attaches to the lower leg at a point one-fifth of the way down the leg. ~
FIGURE P8.25
26.
m::J
A uniform beam oflcngth L ;ll1d mass 111 shown in Figure P8.2fi is inclined at an angle 8 to the horizontal. Its llpper end is connecled to a wall by a rope, and its lower end rests on a rough horizontal slll'facc. The coefficient of static fricFIGURE P8.26 tion between the beam and surface is 11-" ASSllJl1c the angle is such thal the static friction force is at its maximum value. (a) Draw ,I free-body diagram for the beam. (b) Using lht: condition of rotational equilibrium, find an expression for lhe tension T in the rope in terms of m, g, and 8. (c) Using Newton's second law for equilibrium, rind a second expression for Tin terms of J.l-" tn, (tnd g. (d) Using the foregoing results, obtain a relationship inmlving only IJ- and the angle e. (e) What happells il" the angle gets smalkr? Is this equation v
w
_C:::a.. Quadriceps
""l':icildon
'1. _-_--
Tibia
./ (a)
25.0°
T
\40.0
j
27.
I'JI
The chewing muscle, the masseter, is one of the strongest in the human bod)'. It is auached to the mandible (lower jawbone) as shown in Figure PS.27a. Thejawbone is pivoted abom a socketjusl in front of the auditory canal. The forces acting 011 the jawbone arc equiv~erH to those acting on the curved bar in Figure PS.27b. Fr is the force exerted by the food being chewed againsltl".!£jm",'bone, T is the force of tension in the masseter, and R is the force exerted by the socket on the mandible. Find T an(! R for a person who bites down on a piece of steak with a force of 50.0 N. ~
3.50 cm
~
(b) FIGURE P8.29
30. One end of a uniform '1.0-111long rod ol'weight TV is supported by a cable. The ot her end rests against a wall, where it is held by friction. (Sec Fig. PS.30.) A. IJ The coefficient of static friction between I he wall (tnd the rod is IJ-, = 0.50. Determine the miniFIGURE P8.30 mum distance x rrom point A at which an fldditionfll weight 7V (the same as the \\'eight of the rod) can be hung wit haUl causing the rod to slip at point A.
SECTION 8.5 RELATIONSHIP BETWEEN TORQUE AND ANGULAR ACCELERATION
7.50 em
31. Four objcC(s are held in position at the corners of a Manrlibli..: (h)
FIGURE P8.27
rCCl(lngle by light rods as shown in Figure P8.31. Find the moment. of incrti'l of the system about (a) the x-axis, (b) the )'-axis, and (c) ,1Il axis through Gand perpendicular to the page.
Problems
llulllcric value for the system's accelerationr (i) What is (he tension in the string? U) How long docs it take the system to drop 1.00 III from rest?
)'
~.OO
kg
2.00 kg-
Il.OO m
V-
x
0
1.0Om.
2.00 kg
FIGURE P8.31
4.00 kg
(Problenls:ll and
~~2)
32. If the system sho\·.. n in Figure PH.3! is sCt in ralation aboUl each of the xes mClllioncd in Prohlem 30, find the lorque thal will produce an angular acceleration of 1.50 rad/5 2 in each casco
,I
33. A large grinding wheel in the shape of a solid cylinder of radius 0.330 III is free to rotale on a frictionless, venical axle. A constant tangential force of 250 ~1 applied lO ils edge causes the wheel La ha,"e an angular accelenllion of 0.940 rad/s:!. (a) What is the moment of inertia of the wheel? (b) Whal is the mass of the wheel? (e) I f the \,·heel starts from rest, whal is its angular H'locit) afler 5.00 shave elapsed, assuming the force is acting during that time? 3-1. _ An O\'crsil.cd yo-yo is made from two identical ,>olic! disks each of mass M == 2.00 kg and radius R == 10.0 Clll. The two disks arc joined by a solid C)linder of raclim r== 4.00 cm and mass In. == l.On kg as in Figure PR.3 i l. Take the CClllcr of the cylindcr as 1 he axis of the sys(eTll. with positive..: torqucs dircctcd to the left along this axis. All torques and angular variables arc lo be calculated around this axis. Light stl-ing is wrapped around the cylinder, <1l1fl the s}'stem is then allowed 10 drop from re.. l. (a) \\lhat is the momcnt of inertia of the s)'stcm? Give it symbolic "'nswcr. (b) What wrqlle docs gravit}' exert 011 the system with respect La the givcn ",xis? (c) Take down· ward as the negativc coordinatc direction. As depicted in Figure P8.34, is the torque exerted by the tension po<;itive or negativc? Is the angular accclcf
7:
FIGURE .8,34
261
35. _ A rope of neg-ligihlc mass is wrapped around a 225-kg solid C)'lindcr of radius 0.400 m. The qlinder is su,>pcllded scveral meters off the ground with ilS ~Lxis or·i· cntecl horizontally, and turns on thm axis without friction. (a) If a i5.0-kg man takes hold of the free end of the rope and falls under the force of gravit)" what is his acceleration? (h) \Vhat is the angular accelerat.ion of the cylinder? (c) If the mass of the ropc wcre not neglected, what would happen to the angular acceleration of the cylinder as the man falls? 36. A pOller\ wheel having a radius of 0.50 111 and a momelll of inenia or 12 kg·m 2 is rotating [reely at 50 rev/min. The potter can st.Op the wheel in 6.0 s by pressing a wet rag against the rim and exerting a radial1\ inward fOI'ce or 70 N. Find the effecti\'c coefficient of kinetic fricuon hetween the wheel and the wet rag. 37. A model airplane with mass 0.750 kg is tethered by a wire so lhat it flies in a circle 30.0 III in radius. The airplane (:ngine providcs a net thrust of 0.800 N pnpendicular to thc tethering wire. (a) Find the torque the net thrust produces abom the center of thc circle. (b) Find the anguhlr acceleration of the airplane whcn it is in level night. (c) Find the linear acccleration or the airplane tangem to ils f1ighL path. 38. A bicycle wllcel has a diamcter of 64.0 (Ill and a mass or 1.80 kg. ASSlllnC thaI (hc wheel is a hoop with all the mass COnCClllnHed on the olltside radius. The bicycle b placed on a ::>tal ionary stand, and a resistive force of 120 N is applie(1 tangent to the rim of the tire. (a) What force Illust be appliefl by a chain passing O\'er a 9.00-cl11diameter sprocket in order to give the wheel an acccleration of4.50 rad/s'2? (b) \",'hm force is rcquired if you shift to a 5.60-cm-diamctcr sprocket? 139,1..-\ 150-kg mcrq·g-o-round in the shape ofa uniform, solid. horizontal disk of radius 1.50 III is set in motion by wl'apping a rope abom the rim of the disk and pulling on the rope. \\'hat constant force must be exerted 011 the rope to bring the mcrq-go·round from resL to an angular speed of 0.500 rcv/s ill 2.00 s?
-to.
mil
An Atwood's machine consists of blocks of masses == 10.0 kg and t11 2 == 20.0 kg llttached b)' a cord run· ning- over a pulley as in Figure P8.40. The pulley is a solid cylinder wiLh mass M == 8.00 kg and radius r::: 0.200 m. The block of Illass 1/1../, is allowed 10 drop, and the cOHI 111 1
FIGURE .8,40
262
Chapter 8
Rotational Equilibrium and Rotational Dynamics
{lIrns the plllley Wilholll slipping. (a) Why must the len· sion T';! be greater than the tension T 1? (b) What is the
147.IA solid, uniform
disk of radius 0.250 m and mass 55.0 kg rolls down a ramp of lenglh 4.50 m that makes an angle of 15.0" with the horizontal. The disk stans from rest from the top of the ramp. Find (a) the speed of the disk's center of lIlass when it reaches the hOllom of [he I"amp and (b) the angular speed of the disk at the bOllom of the ramp.
acceleration of Ihe system, assuming the pulle)' axis is frictionless? (c) Find the tensions T 1 and 7"'1' 41. An airiinCl'lands \\,jth a speed of50.0 Ill/S. Each wheel of
the plane has a radius of 1.25 III and a moment of inertia of 110 kg' Ill:!, At LOllchdown, the wheels begin to spin under the action offriclion. Each wheel supports a weigh I of 1.10 X 10-1 N, and the wheels alLain their angulal' speed in 0.480 s while rolling without slipping. What is the coefficielll of kinetic friction between the wheels and the nlll\\'a~'? Assume thaI the speed orthe plane is constant.
SECTION 8.6 ROTATIONAL KINETIC ENERGY 42. A GIl" is designed LO get its energy from a rotating flywheel with a radius of 2.00 III and a mass of 500 kg. Before a trip. the flywheel is allached to an eleClrie mOLor, which brings the nrwhcers rotational speed lip 10 5 000 rc\/min. (a) Fine! U1C kinelic energy stored in the flywheel. (b) I[the fI)'wheel is to supply energy to the car ;:IS a 1O.0-l1p mowr would, find the lenglh of time the car could rUIl before the flywheel would h,wc to be broughr back up to speed. 43. A hori/.onLal 800-N merry-go-round of radius 1.30 m is starLcc! from rest b)' a constant horizontal force of
50.0 N applied tangcl1lially
(0
the merry-go-round. Find
the kinetic energy of the merrY'-go-roulld after (Assllllle il is a solid cylinder.)
44.
~.OO
s.
Four o~jcCl<;-a hoop, a solid cylinder, a solid sphere, and a Ihill, spherical shell-each has a mass of' 4.80 kg ~lIld a radius of 0.230 m. (a) Find lhe momenl of inertia for each objecl as it rOlales aboulthe axes shown in Table 8.1. (b) Suppose e
E
45. A light rod 1.00 III in length rotates about an axis perpendicular to ils lenglh and passing through its ccntCI" as in Figure PR.45. T"ro panicles of masses 4.00 kg and 3.00 kg are connected to Ihe ends of the rod. (a) ~eglccting the mass of the rod, what is the system's kinetic enel"g} when its angular speed is 2.50 rad/s? (b) Repeal the problem, assuming the mass of the rod is taken to be 2.00 kg.
48.
mil
A solid uniform sphere of mass 111 and radiu<; R rolls withoul slipping down an incline of height Ii. (a) What forms of mechanical energy arc associated with the sphere at any point along the incline when ilS angular speed is w? Answer in words and srmbolically in terms of thc quantities m. g, )'. I. w. and v. (b) What force acting on the sphere causes it 1.0 roll ralher than slip down the inclinc? (c) Determinc I he ralio of lhe sphere's rolational kinctic energy lO its total kinetic energy a[ any instant.
49. The top in Figure P8.49 ha'§ a moment of inertia of 4.00 X IO-~ kg- m 2 and is iniliall}' F at rest. It is free to rotatc .';"'~--7d!!", ... abom a stalionary axis IL-\'. A sl.l"ing wrappcd around a peg along the axis of the tOP is pulled in such a manner as 10 maintain a conslant lension of 5.57 N in the string. If [he string docs not FIGURE P8.49 slip while wound around the peg. what is [he angular speed of thc top aftcr 80.0 em of string 11<1s becn pulled off the peg? Hint: Consider [he work that is done. 50. A conSlant torque of 25.0 N· III is applied 10 a grindstone whose moment of inertia is O.I~O kg·m 2• Using encrgy principles and neglecling friction, fi nd the angular speed aftcr the grindstone has made 15.0 rcvolutions. f-1i1l1:The angular equivalent of \~lel = FD..\' = ~'/1lVl- 4mv/' is HI;,CI = Tan = ~/wi You should convince yourself that this relationship is correct.
- ilwr
51. A IO.O-kg cylinder rolls wilhout slipping on a rough surface. At an instant when its center" of gra\'it}' has a speed of 10.0 mis, determine (a) the translational kinetic energy of its center of gnwity. (h) the rotational kinctic energr aboul its celUer of gravity. and (c) ilS total k.inelic energy.
52. L"sc conser"vation of encrgy 1.0 detcrmine the angular speed of the spool shown in Figure PS.52 after rhe
J 3.00 kg
------;<.t--r- x ;.(10 kg
FIGURE P8.45
Problems 45 and 57
46. A 240-N sphere 0.20 In in radius rolls wilhollt slipping 6.0 m down a ramp th'lt is inclined at 3io w'ilh lhe horizont:ll. What is lhe angular speed of the sphere al the bottOlll of the slope ifilstarts from rest?
FIGURE P8.52
Problems 3.00-kg bllcket has fallen 4.00 m, starting from rest. 'I he light string
153.lm
A giant swing at an ;lmusement park consists of a 365-kg uniform arm 10.0 1ll long, with two seats of negligible mass conncCLed at the lower end of the ann (Fig. P8.53). (a) How far from l.he Lipper end is lhe cemer of mass of the arm? (b) The gravit,ltional potenl ial encl"g"y of the arm is the same as if
10.0
I'll
J
263
57. A light rigid rod 1.00 III in length rotates abotll an axis pcrpendicular 10 its length and through its ccnter, as shown in Figure P8A5. 1\\"0 particles of masses 4.00 kg and 3.00 kg are connected to the ends of the rod. What is Llle angular momentum of the system if the speed o[ each p,lnicle is 5.00 lll/s? (Neglect the rod's mass.) 58. 1-l
I m
FIGURE P8.53
-t I.n m m
SECTION 8.7 ANGULAR MOMENTUM 54. Each of the following ol~jeCls has ;1 radius or 0.180 m and a mass 01'2.40 kg, and each rotates about an axi~ through its center (as in Table 8.1) with all angular speed of 35.0 f<\d/s. find the magnilude or the angular momentum of each obje.:cl. (a) a hoop (b) a solid cylimkr (c) a solid sphere (d) a hollow spherical shell 55. (a) Calculate the ang-ular momenium of Eanh that arises f"rom its spinning motion on its axis, treating- Eanh as a lmirol'm solid sphere. (b) Cakulalc the ;mglllar munwntum of Eanh that arises from ilS orbilalmotion aholltlhe Sun, treating Earth as a poim parlicle. 56.
mil
A O.005-kg bulkttravcl[tinge ing horizontally with a speed of 1.00 X 10:\ m/s enters an IS.O-kg dool', imhedding itsdf 10.0 cm [rom thc side f--1H.O kg opposile the hinges as in Figure P8.56. The I.OO-m-widc 0.0011 kg door is free to swing on its ~ hinges. (a) BeCore it hits the door, docs tile bullet have FIGURE P8.56 anglilar momentum relativc lhe door's axis of rotalion? Explain. (b) Is mechanical encl'gy conscrvcd in this collision? Answer wilhom doing a calculation. (c) AL what angulal' speed docs the door s\.. . ing open immediately after the collision? (The ciaOI' has tbe same moment of incrtia as a r'od with
I I
=
•
-x m
-1.Om-
m
FIGURE P8.59
60. A playground merry-go-round of radius 2.00 III has a momcnt of inertia 1= 275 kg' m 2 and is rotating about a frictionless vertical axle. As a child of" mass 25.0 kg stands at a distance of 1.00 m from thc axle, the system (merrygo-round and child) rotates at the rate of 14.0 rcv/min. The child then proceeds to walk LOward the edge of lhe mCITy-go-round. \Nhat is the angular speed of the system when the child reaches the edge? ffiIJA 'io1id, horil.omal cylinder of mass 10.0 kg and radius 1.00 m rotatcs with an angular speed of 7.00 rad/s about a fixed vertical axis through its center. A 0.250·kg piece of pUll}" is dropped verticall)' onto the cylinder at a point 0.900 III from the center of rotation and sticks to the cylimler. DeLcrmine the fillal angular speed of the system. 62. A studellt sils on a rotating srool holding two 3.0-kg ohject'i. When his al-ms arc e.:xtcnded horizontally, the objccts are.: 1.0 m from the axis of rotation and he rotates \\-·il.h an angular speed 01'0.75 rad/s. The moment of inertia of the sludem plus stool is 3.0 kg· m~ and is assumed to hc constanl. The student thcn pulls in thc objects horizontally to 0.30 III from the rotation axis. (a) Find the new angular specd of the student. (b) Find the kinetic energy of the student before and after the objects al'e pulled in.
Chapter 8
264
Rotational Equilibrium and Rotational Dynamics
63. The puck in Figure P8.63 has a mass of 0.120 kg. Its original distance from the cenler ofrolalion is 40.0 em, and it Illoves with a speed of 80.0 em/so The string is pulled dOWllw;lrd 15.0 CIll through the hole in the frictionless table. Determine the work done on the puck. lIinl: Consider the change in ki netic energy of the puck.
/.,,~~
1...-_ _ :'
(b) Calculate the change in kinctic cnergy due 10 the man's movement. How do YOll account for this change in kinetic eneqjY? --'
FIGURE P8.63
64. A space station shaped like a giant wheel has a radius of 100 III and a moment of inertia of 5.00 X lOR kg' Ill:!, A crew of 150 li\'cson the rim, and the station is rot::uing so thal the crew experiences an apparent accclenilion of 19 (Fig. P8.64). When 100 people move lO the cenlt~r of the station for a union meeting, the angular speed changes. \'\'hal apparelll acceleratIon is experienced by the m3nag· ers remaining at the rim? Assume the
167.IA 60.0-kg woman stands at the rim of a horizontal lUrnlable having a moment of inertia of 500 kg- m 2 and a radius of 2.00 m. The tunuablc is initially at rest and is fl-ee to rotate ",boUL a fl-icLionless, \'ertical axle through its center. The woman then start~ walking around the rim clockwise (as \'ic\',:ed from above the ~ystem) at a constant speed of L50 m/s relati\'c to Earth. (a) In what direclion and with what angular speed does the turntable rotate? (b) Ilow much work does t he woman do to <;et herself and the turntable into motion?
ADDITIONAL PROBLEMS 68. Figure P8.68 shows a c1awhammer as it is being used to pull a nail out of a horizontal board. If a force of magnitude 150 N is exencd horilOntalh as shown, find (a) the force excrted by the hammer claws on the nail and (h) the force exerted by the surface at the point of confact with the hammel- head. Assume that the force the hammer excrts on the nail is pan-dlc1 to the nail and perpendicular to the position vcctor from the poinl ofcolltacL
:lO.O ern
j
FIGURE P8.64 I
65.
ail
A cylinder with moment or inertia II rotates with angular \'elocity Wo abolll a friction less \'ertical axle. A second cylinder. with moment ofincnia I"!. initially not rotal.· iog, drops onlo the first cylinder (Fig. PS.65). Because the surfaces arc rough. the rwo cylinders eventually reach the same angular speed w. (a) Calculate w. (h) Show that kinetic encrg" is lost in this situation. and calculate the ratio of the final to the initial kinetic energy.
Before
:~O.Ool I
: Singk poillt I or confaCI
~ 5.00clll
--:
FIGURE P8.68
69.
E
A 40.0-kg child slands at one end of a IO.O-kg boat IhaL is ·1.00 In long (Fig. P8.69). The boat is inili:llly 3.00 m from the pier. The child notices a tunIc on a rock beyond lhe fal- end of the boal and procecds to walk to that end 10 carch the tunic. (a) Neg-leering friction betwcen the boal and water. describe the motion of the system (child
After FIGURE P8.65
66. A mCIT}'-go-rolind rOlalcs at the rate or 0.20 rcv/s with an 80·kg man standing ala point 2.0 m from the axis of rola· tioll. (a) What is the new angular speed when [he man walks 10 a point 1.0 III from the center? Assume Iha[ thc mL'rf)-go-rollnd is a solid 25-kg cylinder of radius 2.0111.
FIGURE P8.69
Problems plus boat). (b) \'\'herc will the child be relative to the pier when he reaches the rar end of the boat? (c) Will he catch the llIrtle? (Assume that he can reach out 1.00 m from the end of the boat.) 70. A 12.0-kR ol~eCl is attached waCOI'd that is wrapped around a wheel or radius T= 10.0 cm (Fig. P8.70). Tile acceleration or the object down the rriction1c,~s inclille is measured to be 2.00 I11/s 2. Assllming lhe axle or the wheel to be frictionless, determine (a) the tension in the rope. (b) lhe moment of inertia of the wheel, and (c) the angular speed of the wheel 2.00 s aftcr it begins rotating. starting from rest.
~ooz /p' 12.()k~ ~
265
pulling on the rope, the astronaUls shorten the distance between them to d/2. (c) What is the new angular momentum or the system? (d) Whm are t.heir new speeds? (c) \Vhat is the new rotational energy of the system?
345-N scarJold supported by two cables attached to its ends. Bob weighs 750 N and stands 1.00 III from the left end, as shown in Figurc P8.74. Two melers from the len. end is the !iOO-N washing equipmcnl.Joe is 0.500 In from the righl end and ""cighs I 000 f\. Given that the scaffold is in rotaLional and translat..ional ef]uilibriuITl, whal arc the forces on each cable?
37.0 0
FIGURE P8.70
[lIJA
uniform ladder of length Land ...v eight 1V is leaning against a veniC<11 wall. The codTicienl. of statk friction between ,he ladder and the noor is the same as that between the ladder and the wall. If this coefficienl of static friction is I-L., = 0.500, determine the smallest
72. Two astronauts (Fig. P8.72) , each h,wing a mass of75.0 kg, arc connected by a IO.O-m rope of negligible mass. They are isolated in space, movin~ in circles around the point halfway bet.ween them at a speed of 5.00 m/s. Tre
FIGURE P8.72
I
1.00 m _ _l
I~
1<:===~9:_ . <"OO~m'::3:oo~~.:,__()5(10 m r3.00 III FIGURE P8.74
75. A star with mass 3.00 X 10:\0 kg and radius 1.50 X 10!) m rotates on its axis at a rate of 0.010 0 rev/d. If the star suddenly collapses to a neutron star of radius 15.0 km, find (a) lhe angular speed of the star and (b) the tangential speed or all indestructible astronaut standing on the equator.
76. _ A light md 01" length 2L is free 1.0 rotate in a verli((it plane about a frictionless pivot through its center. A particle of mass 111 1 is attachnl at one end of the rod. and a mass 1112 is at the opposite end, where 11/ 1 > "'2. The system is released from rest in the vertical position shown in Fig-uiT P8.76a, and al some later time the system is rOl.ating in lhe position shown in Figure P8.76b. Take the reference point of the gravitational potcntial energy to be at the pivot. (a) Find an expression [or the system's LOLal mechanical energy in [he vertical position. (b) Find an
(Problems 72 and 7:3) I.
73. _ Two astronaUls (Fig. P8.72), each having a mass M. arc connected by a rope of length d having neg-ligible mass. Thcy are isolatcd in spacc. moving in circles around the point halfway bet.ween them at a specd fl. (a) Calculate the magnitudc of the angular 1ll0menl.U1ll of the SystCIH by trC
I,
(b)
(a) FIGURE P8.76
266
Chapter 8
Rotational Equilibrium and Rotational Dynamics
expression for the lotal mechaniOl1 energy in the rOI,lled position shown in Figure P8.76b. (c) Using I he faeL L1ull the mechanical energy of the system is conserved, how would you determine the
77. In Figure Il S.i7, the sliding block has a mass of 0.850 kg, the cOlll1lcrwcight has a mass of 0.420 kg, and the pulley is a uniform solid cylinder with a mass of 0.350 kg and an outer ",dius of 0.030 0 m. The coefficielll of ki netic friction between the block and the hnri/ontal surface is 0.250. The pulley turns witham friction on its axle. The light cord c10es nOt stretch and docs not slip all the pulle}'. The block has
FIGURE P8.77
178.1 (a)
Without the wheels, a bicycle f"ame ha.s a mass of 8.44 kg. Each of the wheels can be roughly modelecl as a uniform solid disk with a mass of 0.820 kg and a radius of 0.343 ITI. Find lhe kinetic energy of the whole bicycle when it is moving forward at 3.35 m/s. (b) Befo,-c the invention ofa wheell.urningon an axle, ancient people moved heavy loads by placing rollers under them. (Modern people me rollers, too: Any hilrdwarc store will sell you <.1 roller bearing for a lazy Susan.) A stone block of mass 844 kg moves forward at 0.335 mis, supported by twO uniform q-lill(lr-i· cal tree trunks, each of Illass 82.0 kg and radius 0.343 m. There is 110 slipping between thc block and the rollers or betwcen the rollers and the ground. Find the totall..inetic energy of I he moving objects.
i9.
~---
2.00
III
---->/
FIGURE P8.79
80. In a circus performance, a la'-ge f).O-kg hoop of radius 3_0 m rolls withollt slipping. If the hoop is given an angular speed of 3.0 rad/s while rolling on the horilOl1lal ground and is then aIlO\...·ed to roll lip a ramp inclined al 20° with the hOI-izomal, how far along the incline does the hoop roll? 81. m:J A unifoml solid cylinder of mass M and radius R rotates on a frictionless horizontal axle (Fig. PS.81). Two objects with equal masses In hang from light corrls wrapped around the cylinder. If the system is released from rc~t. lind (a) the tension in each cord and (b) the accclerauon of each ohject "fter the objects have descended a distance h_
FIGURE P8.81
82. m:J A paimer climbs a ladder lcaning against a smooth wall. At a ccrtain height, the ladder is on the ....erge of slipping. (a) Explain why the force cxerled by the vertical wall on the la(lder is horil.Onlal. (1)) If the ladder of length I. leans "Lt an angle () with the horil.ontal, what is the lever arm for this horizontal force with lhe axis of rotation taken at the base of Ihe ladder? (c) I r the ladder is uniform, what is the lever arm lor the force of gravity acting on the ladder? (d) Let the mass of the painter be 80 kg, L = 4.0 111, the ladder's mass be 30 kg, 0 = 53°. 'lnd the coefficient of friction between ground and ladder be 0.45. Find the maximum dislance the painter can climb up the ladder.
83. A war-wolf, or trelmchet, is a device used during the Middle Ages to throw rocks at castles and 1I0W sometimes used to fling pumpkins and pianos. A simple tfebuchct is shown in Figure P8.83. Model it as a stiff rod of negligible mass 3.00 m long and joining panicles of mass 60.0 kg and 0.120 kg at its ends. It can turn 011 a frictionless hori7ontal axle perpendicular to the rod and 14.0 cm from the panicle of larger mass. The rod is released from rest in a hOI-iLomal or-lem3tion. Find the maximum speed that the object of smaller mass attains.
III
In exercise physiology sllIdics, it is sometimes importallt to determine the location of a persoll's cemcr of gravity. This can be done with the arrang-clIlem shown in Figure P8.79. A light plank rests on [WO scales Ihal rcad Fj.:l = ~80 Nand I'(}. = 320 N. The scales arc separated b}' a distance of~.OO Ill. Ilow far from the woman's fcc I. is her cenlcr of gravily?
no
FIGURE P8.83
267
Problems
184.IA sLJ-ing is wrapped around a ttniforlll cylinder of mass lH and radius R. The cylinder is released fmm rest with the string vertical and its top end tied to a fixed bar AI (Fig. P8.84). Show that (a) the tension in the string is one-third the weight of the cylinder, (b) the magFIGURE PB.B4 lIilllde of the acceleralion of I.he center of gravity is 2g/3, and (c) the speed of the center of gravity is (4gh/3)1/2 after the cylinder has descended Ihrollgh distance h. Verify your answer to P
86. " Swinging on a high bar The gymnast shown in Figure P8.86 is performing a bach..·arcls gianl swing on the high bar. Slarling from reSl in a ncar-\'ertical oriel1l.alion, he rotates around the bar in a counterclockwise direction, keeping his body and anns straight. Friction between the bar and the gymnast's hands cxerts
II
The Iron Cross \Vhen a gymnast weighing 750 N executcs the iron cross as in Figure P8.85a, lhe primary muscles involved in supporting this position arc the latissimus dorsi ("'lat5") ami the pectoralis n1-U0r ("pees"). The rings exert an upward force on the arms and suppOrt the weight of thc gymnast. The fot-:s.e exerted by the shoulder joint on the art~ is labeled F, while the two Illuscles exert a £oI.al forc~ F", on the ann. Estimate the magnitude of the force F m . Note thaI one ring supports half the weight of the gymnast, which is 375 jJ as indicated in Figure P8.85b. Assumc that the force F~J acts at an angle of 45° below the horizontal at a distance of 4.0 cm l'mm the shoulderjoinl. In your eSlim;-lle, take the dislance from lhe shoulderjoinl to the hand [Q be 70 em and ignore the weight of the arm.
® FIGURE PB.B6 87. A 'I.OO-kg tmlSS is connccted by a
light cord [Q a 3.00-kg mass on a smooth sllrf~lCe (Fig. P8.87). The pulley rotates about a frictionless axle and has a momenl.ofinertia of 0.500 kg· m 2 and a radius of' 0.300 Ill. Assuming that the cord docs not slip on the pullcy, find (a) the acceleration of the (wo masses and (h) the tensions 7; and 7;.
3.00 kg '1
2
•
4.00 kg
FIGURE PB.B7
A 10.O-kg Illonkcyclimbs a uniform ladder with weighl w=:I .20
88. _
(a)
37!"i !'\
(b)
FIGURE PB.BS
X 10 2 N and length L =: 3.00 III as shown in Figure P8.8S. The ladder rests (lgainst. the wall at an angle of 0 =: 60.0°. The upper and lower ends of I he ladder rest. on frictionless surfaces, ,·... ith the lower end fastened 10 the wall by a horizoJltal FIGURE PB.BB rope that is frayed and that can suppOri. a m
In the Dead Sea, a lake between Jordan and Israel, the high percentage of salt dissolved in the water raises the fluid's density, dramatically increasing the buoyant force. Bathers can kick back and enjoy a good read, dispensing with the floating lounge chairs.
9.1
States of Matter
9.2 9.3 9.4
The Deformation of Solids
9.5 9.6 9.7 9.8
9.9
9.10
Density and Pressure Variation of Pressure with Depth Pressure Measurements Buoyant Forces and Archimedes' Principle Fluids in Motion Other Applications of Fluid Dynamics Surface Tension, Capillary Action, and Viscous Fluid Flow Transport Phenamena
SOLIDS AND FLUIDS There are four known states of matter: solids, liquids, gases, and plasmas. In the Universe at large, plasmas-systems of charged particles interacting electromagnetically-are the most common. In our environment on Earth, solids, liquids, and gases predominate. An understanding of the fundamental properties of these different states of matter is important in all the sciences, in engineering, and in medicine. Forces put stresses on solids, and stresses con strain, deform, and break those solids, whether they are steel beams or bones. Fluids under pressure can perform work, or they can carry nutrients and essential solutes, like the blood flowing through our arteries and veins. Flowing gases cause pressure differences that can lift a massive cargo plane or the roof off a house in a hurricane. High-temperature plasmas created in fusion reactors may someday allow humankind to harness the energy source of the sun. The study of anyone of these states of matter is itself a vast discipline. Here, we'll introduce basic properties of solids and liquids, the latter including some properties of gases. In addition, we'll take a brief look at surface tension, viscosity, osmosis, and diffusion.
9.1
STATES OF MATTER
I'vlatter is normally classified as being in one of three states: solid, liquid, or gas. Often this classification system is extended to include a fourth state of malter, called a plasma. Everyday experience tells us that a solid has a definite volume and shape. A brick, for example, maintains its familiar shape and size day in and day out. A liquid has a definite volume but no definite shape. When you fill the tank on a la\vn mC)l;ver, the gasoline changes its shape from thal of the original container to that of the tan k on the mower, but the original volume is unchanged. A gas differs from solids and liquids in thc.t. it. has neither definite volume nor definite shape. Because gas cc.n flow, however, it shares many properties with liquids. All matler consists of some distribut.ion of atoms or molecules. The atoms in a solid, held together by forces that are mainly electrical, are located at specific positions with respect to one another and vibrate about those positions. At low
268
9.1
States of Motter
269
temperatures, the vibrating motion is slight and the atoms can be considered essentially fixed. As energy is added to the material. the amplitude of the vibrations increases. A vibrating atom can be viewed as being bound in its equilibrium " .; position by springs attached to neighboring atoms. A collection of such atoms and ill'laginaf)' springs is shown in Figure 9.1. \Ve can piclllre applied external forces c as compressing these tiny internal springs. When the external forces are removed, ~ the solid tends to rellirn to its original shape and size. Consequently, a solid is said ~ 9 .. ~ _....... to have elasticily. Solids can be classified as either crystalline or amorphous. In a crystalline solid Cr)'slals of nalUral quartz (SiD), the atoms have an ordered structure. For example. in the sodium chloride crysone of the most common minerals on E.arth. QuanL crystals are used to tal (common table salt), sodium and chlorine atoms occupy alternate corners of make special lenses and prisms and a cube, as in Figure 9.2£1. In an amorphous solid, such as glass, the atoms arc arc cmplon"d in cerlain electronic arranged almost randomly, as in Figure 9.2b. applications. For any given substance, the liquid state exists at a highel- temperature than the solid state. The intermolecular forces in a liquid aren't strong enough to keep the molecules in fixed positions. and they wander through the liquid in random fashion (Fig. 9.2c). Solids and liquids both have the pl"Operty ,hat when an attempt is made to compress them, strong repulsive atomic forces act internally to resist the compression. (0 the gaseous state. molecules are in consl.ant random motion and exert only weak forces on each other. The average distance bet\.,,·cen the molecules ofa gas is quite large comparcd with the size of the molecules. Occasionally the molecules collide with each other, but most of the time they move as nearly free, noninteracting particles. As a result, unlike solids and liquids. gases can be easily compressed. We'll say more about gases in subsequent chapters. When a gas is heated to high temperature, many of the electrons surrounding each atom are freed from the nucleus. The resulting system is a collection of free, electrically charged particles-negatively charged electrons and positively charged ions. Such a highly ionized stale of matter containing equal amounts of positive and negative charges is called a plasma. Unlike a neutral gas, the long-range electric and magnetic forces allm.... the constituents of a plasma to interact with each other. Plasmas are found inside stars and in accretion disks arOlllld black holes, for example. and are far more common than the solid, liquid, and gaseous Su"ltes FIGURE 9.1 A model ofa portion of :1 solid. The aLOms (spheres) are imagbecause there are far more stars around than any other form of celestial matter. ined as being aLlached to each other Normal matter, however, may constitute only aboul 5% of all maller in the by springs. which represent the elastic Universe. Observations of the last several years point to the existence of an invisnalllrc of the inleratomic forces. A solid consists of trillions of segments ible dark matter, which affecL" the motion of stars orbiting the centers of galaxies. like this, with springs connecting all Dark matter may comprise as much as 25% of the matter in the Universe, several oflhem. times larger than the amount of nonnal matter. Finally, the rapid acceleration of the expansion of the Universe rna)' be driven by an even more m)'sterious form of matter. called dark energy. which rna)' account for 70% of all maller in the Universe.
i ~
(al
(b)
FIGURE 9.2 (a) The l'\aCI struclLlre, with the Na I (gray) and (;1- (~TC(·tl) IIIllS at alternate curners ofa cuhe. (h) In
(e)
270
Chopter 9
Solids and Fluids
9.2
THE DEFORMATION OF SOLIDS
Although a solid may be thought of as having a definite shape and volume, it's possible to change its shape and volume by applying external forces. A sufficiently large force ,...,ill permanently deform or break an object, but othenvise, when the external forces are removed, the o~ject tends to return to its orig"inai shape and size. This is called elastic behavior. The elastic properties of solids are discussed in lenns of stress and strain. Stress is the force per unit area causing a deformation; strain is a measure of the amount of the deformation. For sufficiently small stresses, stress is proportional to strain, with the constant of proportionality depending on the material being deformed and on the nature of the deformation. \'\Te call this proportionality constant the elastic modulus: stress = elastic modulus X strain
[9.1]
The elastic modulus is analogous to a spring constant. It can be taken as the stiffness of a material: A material having a large elastic modulus is vcry stiff and difficult to deform. There are three relationships having the Corm of Equation 9.1, corresponding to tensile, shear, and bulk deformation, and all of them satisfy an equation similar to Hooke's law for springs:
F = -kt..< where F is the applied force, k is the spring constant, and amount by \vhich the spring is stretched or compressed.
[9.2]
ax is essentially
the
Young's Modulus: Elasticity in Length Consider a long bar of cross-sectional area A and length Lo, clamped at one end (Active Fig. 9.3). vVhen an external force F is applied along the bar, perpendicular to the cross section, internal forces in the bar resist the distortion ("stretching") that tends to produce. Nevertheless, the bar attains an equilibrium in wruch (.I) its length is greater than Lo and (2) the external force is balanced by internal forces. Under these circumstances, the bar is said to be stressed. \JVe define the tensile stress as the ratio of the magnitude of the external force F to the crosssectional area A. The 'word "tensile" has the same root as the word "tension" and is used because the bar is under tension. The 51 unit of stress is the newton per square meter (N/m 2 ), called the pascal (Pa):
F
ACTIVE FIGURE 9.3 A long bar clamped at one end is slretched bv the allloun\ i!:.L under action ofa force F.
The pascal
[jl('
-+
1 Pa;::::::; 1 N/m 2 The tensile strain in this cai>e is defined as the ratio of the change in length IlL to the original length La and is therefore a dimensionless quantity. Using Equation 9.1, we can ,",,'Tite an equation relating tensile stress to tensile strain:
F -
A
~
t.L y-
Lo
[9.31
In this equation, Y is the constant of proportionality, called Young's modulus_ Notice that Equation 9.3 could be solved for Fanel put in the form.F = k LiL, where k = YA/ L n, making it look just like Hooke's law, Equation 9.2. A material having a large Young's modulus is difficult to stretch or compress. This quantity is typically used LO characterize a rod or wire stressed under either tension or compression. Because strain is a dimensionless quantity, Yis in pascals. Typical values are given in Table 9.1. Experiments show that (1) the change in length for a fixed external force is proportional 1.0 the original length and (2) the force necessary to produce a given st.rain is proportional to the cross-sectional area. The value of Young's modulus [or a given material depends on whether the material is stretched or compressed. A human femur, for example, is stronger under tension than compression.
9.2
It's possible LO exceed the elastic limit of a substance by applying a sufficiently great stress (Fig. 9.4). At the elastic limit, the stress-strain curve departs from a straight line. A material subjected to a stress beyond this limit ordinarily doesn't return to its original length ,-"hen the external force is removed. A... the sU'ess is increased further, it surpasses the uJtimate strength: the greatest stress the substance can withstand without breaking. The breaking point for brittle matcrials is just beyond the ultimate strength. For ductile metals like copper and gold, after passing the point of ultimate strength, the metal thins and stretches at a lower stress level before breaking.
The Deformation of Solids
271
Stress (MP.)
'100 300 200
o
I!-':"-'>-,-,--,-!,....-:":c-:--:L,,--- Su-ai n O,{X)2 0.004 OJXl6 O.COO 0.01
FIGURE 9.4 Siress-versus-slrain curve for all clastic solid.
Shear Modulus: Elasticity of Shape
-.
Another typc of deforrnation OCCUI'S when an object is subjected to a force F pamltpl to one of its faces while the opposite face is held fixed by a second force (Active Fig. 9.5a). 1f the object is originally a rectangular block, such a parallel force results in a shape with the cross section of a parallelogram. This kind of stress is called a shear stress. A book pushed sideways, as in Active Figure 9.5b, is being subjected to a shear stress. There is no change in volume with this kind of deformation. It's important to remember that in shear stress, the applied force is pamllel to the cross-sectional area, whereas in tensile stress the force is !Jer/JentLiculfl;T to the crosssectional area. 'Ne define the shear stress as F/A, the ratio of the magnitude of the parallel force to the area A of the face being sheared. The shear strain is the ratio tJ..x/h, where ~xjs the horizontal distance the sheared face moves and h is the height of the object. The shear stress is related to the shear strain according to F tJ.x -=sA h
Fixed face (a)
[9.4]
where S is the shear modulus of the material, with units or pascals ([orce per unit area). Once again, notice the similarity to Hooke's law. A material having a large shear modulus is difficult to bend. Shear moduli for some representative materials are listed in Table 9.1. (b)
Bulk Modulus: Volume Elasticity characterizes the response of a substance to uniform squeezing. Suppose the external forces acting on an object are all perpendicular to the surface on which the force acts and are distributed uniformly over the surface of
Th'"
h.d'" ...... ")dulus
TABLE 9.1
Typical Values for the Elastic Modulus Substance Aluminum
Young's Modulus (Pa) 7.0 X 10")
Shear Modulus (Pa) 2.5 X IOlU
Bulk Modulus (Pa) 7.0 X 1010
Bone
1.8 X JOIO
8.0 X IOlU
Brass
9.1 X Will
3.5 X IOIll
Copper
11 X lOW
4.2 X lOw
14 X 10'0
Steel Tungsten Glass
20 X 10 10
16 X 10'0
35 X lO 11l
8.4 X 10 10 14 X 1010
20 X 10 10
6.5-7_8 X 10 10
2.6-3.2 X 10 10
5.0-5.5 X 10 10
5.6 X 10 111 1.2 X 10i
2.6 X 10 10
2.7 X 10 10
Quartz Rib Cartila~e Rubber
Tendon W
6.1 X 1010
0.1 X 10i 2 X IOi O.2J X 10'" 2.8 X 1011)
ACTIVE FIGURE 9.5 (a) A shear deformation in which a rectangular block is disLOrled b}' forces applied LangClH to two of jls faces. (b) A book under shear Slress.
272
Chapter 9
Solids and Fluids
ACTIVE FIGURE 9.6 \A,ihcn a solid is under uniform prcssun.:, it undergoes a change in \'01lime, but 110 change in shape. This cube is compressed on all sides by forces normal 10 its six races.
the object (Active fig. 9.6). This occurs when an object is immersed in a fluid. An object subject to this type of deformation undergoes a change in volume but no change in shape. The volume stress .1.Pis defined as the ratio of the magnitude of the change in the applied force I1F to the surface area A. (In dealing with fluids, we'll refer to the quantity J~/A as the pressure, to be defined and discussed more formally in the next section.) The volume strain is equal to the change in volume 6. V divided by the original volume V. Again using Equation 9.1, we can relate a volume stress to a volume strain by the fannula
Bulk modulus
"'v
",p= - 8 -
~
If
[9.5]
A material having a large bulk modulus doesn't compress easily. Note that a negative sign i.s included in this defining equation so that B is always positive. An increase in pressure (positive (j"P) causes a decrease in volume (negative (j" V) and vice versa. Table 9.1 lists bulk modulus values for some materials. If you look up such values in a different source, you may find that the reciprocal of the bulk modulus, called the compressibility of the material, is listed. Notc from the table that both solids and liquids have bulk moduli. There is neither a Young's modulus nor shear modulus for liquids, hmvever, because liquids simply flow \-vhen subjcucd to a tensile or shearing stress.
EXAMPLE 9.1 Goal
Built to Last
Calculate a compression due to tensile stress, and maximum load.
Problem A vertical steel beam in a building supports a load of 6.0 X 10.-1 N. (a) If the length of the beam is 4.0 III and its cross-sectional area is 8.0 X 10-3 m 2 , find the distance the beam is compressed along its length. (b) "Vhat maximum load in newtons could the steel beam support before failing? Strategy Equation 9.3 pertains to compressive stress and strain and can be solved for ~L, rollowed by substimtion of kI10Wll values_ For part (b), set the compressive stress equal to the ultimate strength of steel from Table 9.2. Solve for the magnitude of the force, ,vhich is the total ,..'eight the structure can support. ~
Solution (a) Find the amount of compression in the beam. Solve Equation 9.3 for Li L and substitute, using the value of Young's modulus from Table 9.1:
F
A
"'L
y-
L" FLo
YA
(6.0 X 104N)(4.0 m) (2.0 X 10" Pa)(S.O X 1O-:{ m')
1.5 X 10- 4 m
9.2
The Deformation of Solids
273
(b) Find the maximum load that the beam can support. Set the compressive stress equal to the ultirn<-lte compressive strength from Table 9.2, and solve [or F:
F A
F ----'-~~~ = 5.0 X 10' Pa
KO X 10-·\
F~ ~..........................................
.
m'
4.0 X 106N ~
Remarks In designing load-bearing structures of any kind, it's always necessary to build in a safety factor. No one would drive a car over a bridge that had been designed to supply the minimum necessary strength LO keep it from collapsing. QUESTION 9.1 Rank by the amount of fractional increase in length under increasing tensile stress, from smallest to largest: rubber, tungsten, steel, aluminum. EXERCISE 9.1 A cable used 1.0 lift heavy materials like sted I-beams must be strong enough to resist breaking even under a load of ].0 X 10 6 N. For safety, the cable must support twice that load. (a) \Vhat cross-sectional area should the cable have if it's to be made of steel? (b) By how much will an 8.0-m length of this cable stretch when subject to the 1.0 X l06-N load?
Problem Adef-ensive lineman of mass fl.:l = 125 kg makes a Oying tackle at 1); = 4.00 m/s on a stationary quarterback of mass m = 85.0 kg, and the lineman's helmet makes solid contact with the quarterback's femur. (a) 'What is the speed VI of t.he two athletes immediately after contact? Assume a linear inelastic collision. (b) If the collision lasts lor 0.100 s, estimate the average force exened 011 the quanerback's femur. (c) If the cross-sectional area of the quarterback's femur is 5.00 X 10-4 m 2 , calculate the shear stress exerted on the bone in the collision. ~
.
Solution (a) What is the speed of the system immediately after contact?
Apply momentum conservation to the system:
.
Strategy The solution proceeds in three well~defined steps. In part (a), use conservation of linear momentum to calculate the final speed of the system consisting of the quarterback and the lineman. Second, the speed found in part (a) can be used in the impulse-momentum theorem to obtain an estimate of the average force exerted on the femur. Third, dividing the average force by the cross-sectional area of the femur gives the desired esti~ mate of the shear stress. ."
274
Chapter 9
Solids and Fluids
Substitute expressions for the initial and final momenta:
Solve for lhe final speed
JWV i
=
(1\1
+ 1ll) V.r (125 kg)( 4.00 m/s)
v/
125 kg
2.38 m/s
+ 85.0 kg
(b) Obtain an estimate for the average force delivered to the quarterback's femur.
Apply the impulse-momenLUOl theorem:
Solve for the average force exerted on the quarterback's femur:
M(v/-
vJ
F = ---'---'-.IV
at
(125 kg)(4.00 m/s - 2.38 m/') 0.100 s
2.03 X 10 3 N
(c) Obtain lhe average shear stress exerted on the quarterback's femur.
Divide the average force found in part (b) by lhe cross-sectional area of the femur: ...................
The ultimate shear strength ofa femur is approximately 7 X break the quarterback's leg.
Remarks LO
=
10 7
~
Pa, so this collision would not be expected
QUESTION 9.2 \tVhat kind of stress would be sustained by the lineman? ''''hat parts of his body "','Quld be affected? EXERCISE 9.2 Calculate the diameter of a horizontal steel bolt if it is expected to support a maximum load having a mass of 2.00 X 10 3 kg but for safety reasons must be designed to support three times thalloacl. (The ultimate shear strength of slee! is aboul 2.5 X lOs Pa.)
Answer
1.73 crn
EXAMPLE 9.3 Goal
Stressing a Lead Boll
Apply the concepts of bulk stress and strain.
A solid lead sphere of volume 0.50 m 3 , dropped in the ocean, sinks to a depth of 2.0 X 10 3 m (about I mile), where the pressure increases by 2.0 X 10 7 Pa. Lead has a bulk modulus of 4.2 X 1010 Pa. What is the change in volume of the sphere?
Problem
Strategy
Solve Equation 9.3 [or .6.\1 and substitute the given quantities.
Solution Start with the definition of bulk modulus:
B=
Solve for
t!.V=
Substitute the known values:
t!.V=
L_ • • • • • • _ _ • . • . . • . . . . . . ,
t!.P t!.VIV V
=
-2.4 X 10- 4 m' ....
9.2
The Deformation of Solids
275
Remarks The negative sign indicates a deaease in volume. The following exercise shows that even V'later can be compressed, although not by much, despite the depth.
QUESTION 9.3 Rank the following substances in order of the fractional change in volume in response to increasing pressure. from smallest to largest: copper, steel, water, mercury. EXERCISE 9.3 (a) By what percentage does a similar globe of water shrink at that same depth? (b) What is the ratio of the new radius to the initial radius? Answer
(a) 0.95%
(b) 0.997
Arches and the Ultimate Strength of Materials As we have seen, the ultimate strength of a material is the maximum force per unit area the material can withstand before it breaks or fractures. Such values are of great importance, particularly in the construction of buildings, bridges, and roads. Table 9.2 gives the ultimate strength of a variety of materials under both tension and compression. Note that bone and a variety of building materials (concrete, brick, and marble) are stronger under compression than under tension. The greater ability of brick and stone to resist compression is the basis of the semicircuLar arch, developed and used extensively by the Romans in everything from memorial arches to expansive temples and aqueduct supports. Bel"ore the development of the arch, the principal method of spanning a space was the simple post-and-beam construction (Fig. 9.7a), in which a horizontal beam is supported by two columns. This type or construction \vas used to build the great Greek temples. The columns of these temples were closely spaced because of the limited length of available stones and the low ulLimale tensile strength of a sagging stone beam. The semicircular arch (Fig. 9.7b) developed by the Romans \.. .' as a great technological achievement in architectural design. It effectively allowed the heavy load of a wide roof span La be channeled into horizontal and vertical forces on narrow supporting columns. The stability of this arch depends on the~compression between its wedge-shaped stones. The stones are forced to squeeze against each other by the uniform loading, as shown in the figure. This compression results in horizontal outward forces at the base or the arch where it starts curving away from the vertical. These forces must then be balanced by the stone walls shmvll on the sides of the arch. It's common to use very heavy walls (buttresses) on either side of the arch to provide horizontal stability. If the foundation of the arch should move, the compressive forces between the wedge-shaped stones may decrease to the extent that the arch collapses. The stone surfaces llsed in the arches constructed by the Romans were cut to make very tightjoints; mortar \...as usually not used. The Gothic arch Flying-
buttress
I
r'ost and beam (a)
FIGURE 9.7
Semicircular arch (ROlnall) (b)
Pointed arch (Cothic) (e)
(a) A simple post-anti-beam structure. (b) The semicircular arch developed by the
Romans. (c) Cothic arch with flying bUll resses 10 provide laleral SUpp0rl.
APPLICATION Arch Structures in Buildings
276
Chapter 9
Solids and Fluids
resistance to slipping between stones was provided by the compression force and the friction between the slone faces. Another important architectural innovation was the pointed Gothic arch, shown in Figure 9.7c. This type of structure \vas first used in Europe beginning in the ]2th century, follmved by the construction of several magnihcent Gothic cathedrals in France in the 13th century. One of the most striking features of these cathedrals is their extreme height. For example, the cathedral at Chartres rises to 118 ft, and the one at Reims has a height of 137 ft. Such magnificent buildings evolved over a very shon l.ime, \.vithout the benefit of any mathematical theory of structures. However, Gothic arches required flying buttresses Lo prevent the spreading of the arch supported by the tall, narrow columns.
9.3
DENSITY AND PRESSURE
Equal masses of aluminum and gold have an important physical difference: The aluminum takes up over seven times as much space as the gold. Although the reasons tor the difference lie at the atomic and nuclear levels, a simple measure of this difference is the concept of density. Density
~
The density p of an object having uniform composition is its mass M divided by its volume V: p=
M
[9.6]
V
SI unit: kilogram per meter cubed (kg/m')
The most common unit.s Llsed for density are kilograms per cubic meter in the Sf system and grams per cubic centimeter in the cgs system. Table 9.3 lists the densities of some substances. The densjtics of most liquids and solids vary slightly with changes in temperature and pressure; the densities of gases vary greatly with such changes. Under normal conditions, the densities of solids and liquids are aboutl 000 times greater than the densities of gases. This difference implies that the average spacing bct\vcen molecules in a gas under such conditions is abom tcn times greater than in a solid or liquid. The specific gravity of a substance is the ratio of its density to the density of water at 4"C, which is ].0 X 10' kg/m 3 . (The size of the kilogram was originally defined to make the density of water 1.0 X )0' kg/m' at 4"C.) By definition, spe-
T BLE 9.3 Densities of Some Common Substances Substance
I' (kg!m')"
Substance
p (kg!m')'
tee
0.917 X 10'
\Vater
1.00 X 10' 1.26 X 10' 0.806 X 10'
Aluminum
2.70 X LOB
Glycerin
Iron
7.86 X 10"
Ethyl alcohol
Copper
8.92 X 10'
Benzene
Silver
10.5 X 10'
Mercury
Lead
11.3 19.3 21.4 ]8.7
Ai,
Gold Platinum Uranium
X X X X
10' 10' 10' 10'
Oxygen Hydrogen Helium
0.879 X 10' 13.6 X 10' 1.29 1.43 8.99 X 10-' 1.79 X 10-]
"All values are al standard atlllospheric Icrnpcratufc and pressure (STP), defined as DOC (273 K) and 1 alill (1.013 X 105 1'<1). To convert LO graHlS per cubic ccntimctl:r, multiply by 1O~~
9.3
..,
~
f---
~f
~
277
FIGURE 9.8 (a) Thl; force cxcr1cd h)' a fluid on a submerged objecl at all)' point is pcrp<:ll(licular 10 the surface of tIll.': ohjecl. Tlte force exerted by the fluid 011 lIn.: walls or the container is perpendicular 10 lhe I,'alls at all poinl~ and inCl'e:l~cs with depth. (b) :\ simple device for measuring: pressure in a fluid.
---
-~
Density and Pressure
-----.h
(a)
(b)
cific gravity IS a dimensionless quantity_ For example, if the specific grc.lVity substance is 3.0, its density is 3.0( 1.0 X J03 kg/Ill') ~ 3.0 X 10' kg/Ill'.
or a
QUICK QUIZ 9.1 Suppose you have one cubic meter of gold, two cubic meters of silver, and six cubic meters oralulllinum. Rank them by mass, from smallest to largest. (a) gold, aluminum, silver (b) gold, silver, aluminum (c) aluminum, gold, silver (d) silver, aluminum, gold
Fluids don't sustain shearing stresses, so the only stress that a fluid can exert all a submerged object is one thaltcnc!s to compress it, which is bulk su'ess, The force exerted by the fluid on the object is always perpendicular to the surfaces of the objen, as shown in Figure 9,8a, The pressure at a specific point in a fluid can be measured \.. . ith the device pictured in Figure 9,Sb: an evacuated cylinder enclosing a light pisLOn connected to a spring that has been previously calibrated \vith known weights, As the device is submerged in a fluid, the fluid presses down on the top of the pistoll and compresses the spring until the inward force exerted by the fluid is balanced by the outward force exerted by the spring. Let F be the magnitude of the force on the piston and A the area of the top surf~lce or the piston. Notice that t.he ltH'cc that compresses the spring is spread 0111 over the entire area, motivating our formal defi n ition of pressu re: If F is the magnitude of a force exerted perpendicular to a given surface of area A, then the pressure P is the (orce divided by lhe area: p=
F
A
TIP 9.1 Force ond Pressure
=----
Equal ion 9.7 makes a deal' distinction bel.\,'een forcc and pressure. A nother important distinction is that/ora is {J ver~ lurand /JrFUU7"(! is (l scalar. The.'c is no direction associ,lled with pressure, but the direction of the force associated with the pressure is perpendicular to the surface of
interest.
~ Pressure
[9.7)
SI unit: pascal (Pa)
Because pressure is defined as force per unit area, it has units of pascals (newtons per square meter). The English customary unil for pressure is the pound per inch squared. Atmospheric pressure at sea level is 14,7Ib/in. 2 , which in 51 units is J.01 X 10' Pa. As we see from Equation 9.7, the effect ofa given force depends criLicalJy on the Clrea to which it's applied. A 700-N man can stand on a vinyl-covered floaT' in regular street shoes without dan1aging the surface, but if he wears golf shoes, the metal cleats protruding from Lhe soles can do considerable damage lo the floor. "\lit.h the cleats, the same force is concentrated into a srllaller area, greatly elevating the pressure in those areas, resulting in . 1 greater likelihood of exceeding the ultimate strength of the floor material. Snowshoes use the same principle (Fig. 9.9). The snow exerts an upward normal force on the shoes to support the person's weight.. According to NeWlon's third law, this upward force is accompanied by a downward force exerted by the shoes on the snm·\!. If the person is wearing snowshoes, that force is distributed over the very large area of each snowshoe, so that the pressure at any given point is relatively low and the person doesn't penetrate very deeply into the snow.
FIGURE 9.9 Snowshoes prC\'Cnt the perSall from sinking into the soft snow because the force on Ihe snow is spread over a larger arC'I, reducing the pressure on the snow's surface.
278
Chapter 9
Solids and Fluids
APPLYING PHYSICS 9,1
BED OF NAILS TRICK
After all exciting but exhausting lecture, a physics professor stretches out for a nap on a hed of nai Is, as in Figure 9.10, suffering no il~iury and only moderate discomfort. How is this possible? Explanation If yOlI try to support your entire weight. on a single nail, the pressure on your body is your weight divided by the very small area of the end of the nail. The resulting pressure is large enough to penetrate the skin. [fyou distribute your weight over several hundred nails, however, as demonstrated by the professor, the pressure is considerably reduced because the area that supports your weight is the total area afall nails in contact with your body. (VVhy is lying on a bed of nails more comfonable than sitting on the same bed? Extend the logic to show that it
would be more uncomfortable yet
LO
stand on a bed
of nails wil"hOlJl shoes.)
FIGURE 9.10
(Applying Physics 9.1) Does anyone have a pillow?
EXAMPLE 9.4 The Water Bed Goal
Calculate a density and a pressure from a weight.
Problem A water bed is 2.00 m on a side and 30.0 cm deep. (a) Find its weight. (b) Find the pressure that the water bed exerts on the floor. Assume the entire lower surface of the bed makes contact with the floor.
Strategy Density is mass per unit volume: first, find the volume of the bed and multiply it by the density of water to get the bed's mass. Multiplying by the acceleration of gravity then gives the weight of the bed. The weight divided by the area of floor the bed rests upon gives t.he pressure exerted on the floor. fr' . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • • . . .....,
Solution (a) Find the weight of the water bed, First, find the volume of the bed: Solve the density equation for the mass and substitute, then multiply the result by g to get the weight:
v = {wli. =
(2.00 111)(2.00 111)(0.300 Ill) ~ 1.20 Ill'
M p~
V
M ~ p\l~ (1.00 X 10 3 kg/1ll 3 ) (1.20 111 3 ) w
= Mg =
= 1.20 X 10 3 kg
(l,20 X 10 3 kg) (9.80 Ill/s') ~
U8 X 10" N
(b) Find the pressure that the bed exerts on the floor. Use the cross-sectional area A = 4.00 m 2 and the value of w from part (a) to get the pressure:
Remarks Notice that the answer to part (b) is far less than atmospheric pressure. ""Vater is heavier than air for a given volume, but the air is stacked up considerably higher (100 km!). The total pressure exerted on the floor would include the pressure of the atmosphere. QUESTION 9.4 Is the pressure inside a rubber balloon greater than, less than, or equal to the ambient atmospheric pressure? EXERCISE 9.4 Calculate the pressure exerted by the water bed on the floor if the bed rests on its side.
Answer
1.97 X 10' Pa
9.4
9.4
Variation of Pressure with Depth
279
VARIATION OF PRESSURE WITH DEPTH
When a fluid is at rest in a container, all portions of the fluid must be in static equilibrium-at rest with respect to t.he observer. Furthermorc, all points at the same depth must be at the same pressure. If this were not the case, fluid \,,'ould now from the higher pressure region 1.0 the lower pressure region. For cxamplc, consider the small block oflluid shown in Figure 9.lla. lftJ,c pressure were greater on the left side of the block than on the right, F I would be greater than F::!, and the block would accelerate to the right and thus ,vould not be in equilibrium. Next, let's examine the fluid contained wilhin the volume indicatcd by the da,rker region in Figure 9.llb. This region has cross-sectional area A alld cXlcnds from position)', [0 position y'.!. bclow the surface of the liquid. Three external forces act on this volume ofrluid: the force ofgravit)', Nlg; the upward force P'l.A cxened by the liquid belmv it; and <'1 dowmvard force PIA exerted by the fluid above it. Becausc the given volume or rJuid is in cquilibrium, these forces must add to zero, so we get ~
1J'l.A - PIA - A1g= 0
~
J (a)
y
o
r
[!l.8]
From the definition of density, ,..'e have M = pV = pA(YI - y,)
[!l.!l]
Substituting Equation 9.9 into Equation 9.8, canceling the area A, and rearranging terms, we get P, = P,
+
pg(YI - .h)
I~l
+
pgh
J) (b)
[!l.10]
Notice that (YI - Y2) is positive, because)'2 < )1' The force P'.!A is greater than the force PIA by exactly the weight of watcr ben\leen t.he two points. This is the same principle experienced by the pcrson at the bottom ora pileup in football or rugby. Atmospheric pressure is also caused by a piling up of fluid-in this case, the fluid is the gas of the atmosphere. The weight of all the air from sea level to the edge of space results in all atmospheric pressure of Po = 1.013 X 10;) Pa (cquiv
)1
[!l.I1]
According to Equation 9.11, tbe pressure Pat a depth h below tbe surface of a liq· uid open to the atmosphere is greater than atmospheric pressure by the amount pgh. Moreover, the pressure isn'l. affected by the shape of the vessel, as shown in Figure 9.12.
FIGURE 9.11 (a) If the block ofOuid is to be in cquilibrium,iJle lorce must balance the force F2 • (b) The net force on the volume ofliquid within the darker region must be zero.
F\
~
c
ij ~,
~
QUICK QUIZ 9.2 The pressure at the bottom of a glass filled with water (p = I 000 kg/m 3 ) is P. The water is poured out and the glass is fi lied with ethyl alcohol (p = 806 kg/m:\). The pressure at the bottom of the glass is now (a) smaller than P (b) equ"l! (0 P (c) larger than P (d) indeterminate.
EXAMPLE 9.S Goal
'"
FIGURE 9.12 This photograph illustrates the fan that the pressure in a liquid is the same at all points lying <11 the same elevation. ~ole thaI the shape of the vessel does nUL aIle, 1 the prcl>surc.
="---------
Oil and Water
Calculate pressurcs created by layers or different nuids.
Air
Po
Problem Tn a huge oil t.anker, salt water has flooded an oil tank to a depth of 5.00 m. On top of the water is a layer oroil8.00 III deep, as in the cross-scnional view of" the tank in Figure 9.13. The oil has a density 01"0.700 g/cm 3 . Find the prcssure at the hOllom of the tank. (Take 1 02f> kg/m 3 as the density of salt water.) Strategy Equation 9.11 must be used twice. First, use it to calculate the pressure PI at the bottom of the oil layer. Then usc [his pressure in place of Pu in Equation 9.11 and calculate the pressure 1-\0<'1 at the bouom of the water layer.
hi I
t
I~
Oil P, Water
P"", FIGURE
9.13
(Example 9.5)
280
Chapter 9
Solids and Fluids
,..-'
..
"
Solution Use EquaLion 9.11 to calculate the pressure at the botlorn of the oil layer:
(1)
P, ~ Po
+ P, Now adapt Equation 9.11 to the new starting pressure, and use it to calculate the pressure at the bottom of the v."ater layer:
+ pgh,
= 1.01 X 10' Pa
(2)
~
(7.00 X 10' kg/m') (9.80 m/s')(8.00 Ill)
1.56 X 10" Pa
P bm = P,
+ pgh,
= 1.56 X 10' Pa
+ (1.025 P hm
~
X
103 kg/Ill') (9.80111/5') (5.00 m)
2.06 X 10' Pa
~
~
Remark The weight of the atmosphere results in Po at the surface of the oil layer. Then the weight of the oil and the weight of the water combine Lo create the pressure at the bOltom. QUESTION 9.5 Why does air pressure decrease ·with increasing altitude? EXERCISE 9.5 Calculate the pressure on the top lid of a chest buried under 4.00 meters of mud with density 1.75 X 103 kg/m 3 at the bottom of a lO.O-m-deep lake.
Answer
2.68
X
10" Pa
EXAMPLE 9.6 Goal
A Pain in the Ear
Calculate a pressure difference at a given depth and estimate a force.
Problem Estimate the net force exerted on your eardrum due to the waLeI' above when you are swimming at the bottom of a pool that is 5.0 III deep. Strategy Use EquaLion 9.11 La find the pressure difference across the eardrum at the given depth. The air inside the ear is generally at atmospheric pressure. Estimate the eardrum's surface area, then use the definition of pressure to get the net force exerted on the eardrum. ,.,
~
,
~
Solution Use Equation 9.11 to calculate the difference between the water pressure at the depth hand the pressure inside the ear:
tJ.P= P- P" = pgh
Mutliply by area A to get the net force on the eardrum associated with this pressure difference, estimating the area of the eardrum as 1 CIn 2.
Filet
~ (1.00 X 10 3 kg/m 3 ) (9.80
m/s') (5.0111)
= 4.9 X 10" Pa = An.P= (1 X 10- 1 m 2 ) (4.9 X ]04 Pal =
Remarks Because a force on the eardrum of this magnitude is uncomfortable, swimmers often "pop their ears" by swallowing or expanding theirjavlo's 'while underwater, an action that pushes air from the lungs into the middle ear. Using this technique equalizes the pressure on lhe two sides of the eardrum and relieves the discomfort.
QUESTION 9.6 Why do water containers and gas cans oflen have a second, sma]]er cap opposite the spout through ·which fluid is poured? EXERCISE 9.6 An airplane takes off at sea level and climbs to a height of 425 m. Estimate the nct outward force on a passenger's eardrum assuming the densily of air is approximately constant at 1.3 kg!m~ and that the inner ear pressure hasn't been equalized.
Answer
0.54 N
9.4
Variation af Pressure with Depth
281
I
£ .....__..... (b)
(a)
FIGURE 9.14 (a) Dial~T,HIl ofa h~'dl-;!lIlic prt:ss (Exampl{' 9.7). Because (he pres~lIre is the ,,:\lIlt':U Ih" len and righ, <;ides. a small foret: F 1 at the kf( produces :llIIuch larger fon"e F:! at (he riKhl. (Il) A "ehide tinder rcp:lir is suppoJ"l(-'d Il\ a h~c1r.lUlic Iill ill a ganlgc.
Because the pressure in a fluid depends on deplh and on lhe value of Po, any increase in pressure at the surface Illust bc transmiued to every point in the fluid. This was first recognized by the French scicmist Blaise Pascal (1623-1662) and is called Pascal's principle: A change in pressure applied to an enclosed fluid is transmitted ulldiminished to evcry point or the flu id and 10 thc walls of the container. An important applica~on of Pascal's principle is the hydraulic press (Fig. 9.14a). A downward force F I is applied to a slllall piston ofar-ea AI' The pressure is transmitted through a fluid to a larger piston of area A 2• As the pistons move and the fluids in the left and right cylinders change their relative heights, there arc slight differences in the pressures at the input and output pistons. Neglecting these small differenccs, the fluid pressure on each of the pistons may be taken to be the same; PI = P'2' Frorn the definition of p~essure, il then follows that Fl / AI-:=' F'2/ A2 · Therefore, the magnitude of the force F~ is larger than the magnitude ofF I by the factor A:/I"\,. That's why a largc load, such as a car, can be moved on the large piston by a much smaller force on the smaller piston. Hydraulic brakes, car lifb, hydraulic jacks. forklifts, and other machines make use of this principle.
EXAMPLE 9.7 Goal
APPLICATION Hydraulic Lifts
The Car Lift
Apply Pascal's principle
LO
a car lift, and show that the input work is the same as the outpUl work.
Problem In a car lift lIsed in a service station, compressed air exerts a fOl"ce on a small piston of circular cross section having a radius of ", := 5.00 ern. This pressure is transmitted by an incomprcssible liquid to a second piston of radius r't = 15.0 cm. (a) ""hat force must the compresscd air exert on the small piston in order lO lift a car \\;eighing 13300 N? Neglecl the weighLl\ of the pistons. (b) \\That air pressure will produce a force of that magnitude? (c) Show that the work done by the input and output pistons is the samc.
Strategy Substillltc into Pascal's principle in part (a), while recognizing thaL the magnitude ofthe output force, F'!., must be equal to the car's weight in order to suppOrt it. Use the definition of pressure in part (b). In part (c). use \,\1= F6.x to find the ratio lV,/\·V2 , showing that it must equal I. This requires combining Pascal's plincipic \vilh the fact that the input. and output pistons move lhrough the same volume.
282
Chapter 9
,..-'
Solids and Fluids
.
,
.
..,
,
Solution (a) Find the necessary force on the small piston. Substitute known values into Pascal's principle, using A = 7T r2 for the area of each piston: 2 _ "-(5.00 X 10- m}' ( '1 J) - 7T (1-.j. 0 X 1 0-'m) '- 1.33 X ION
1.48 X 10'\ N
=
(b) Find the air pressure producing Fl' Substitute into the definition of pressure:
1.48X10'N
F
P = ---..!.- = --;cc-:-c----;,,-"" AI "-(5.00 X 10 2 m}'
1.88 X 10' Pa
(c) Show thallhe work done by the input and output pistons is the same. First equate the volumes, and solve [or the ratio of 1\2
V 1 =V2
to AI:
Now use Pascal's principle
to
get a relationship for 1"/1'2:
--7
A r Llx l =A 2 Llx2
;1~
ax]
Al
LlX2
FI
F';!
FI
A,
A,
F,
11-\ FILl.XI (FI)(Ll.XI) (A )(A~2) = ] F Llx2 = A J
Evaluate the '.Nork ratio, substituting the preceding two results:
W2 = F'l. Llx,:! =
2
2
WI = W2 ~
,
,
~
Remark In this problem, we didn't address the effect of possible differences in the heights of the pistons. If the column of fluid is higher in the small piston, the fluid weight assists in supporting the car, reducing the necessary applied force. If the column offluid is higher in the large piston, both the car and the extra fluid must be supported, so additlonal applied force is required.
QUESTION 9.7 True or False: If the radius of t.he output piston is doubled, the output force increases by a factor of 4. EXERCISE 9.7 A hydraulic lift has pistons with diameters 8.00 COl and 36.0 cm, respectively. If a force of 825 N is exerted at the input piston, what maximum mass can be lifted by the output piston? Answer
1.70 X 103 kg
APPLYING PHYSICS 9.2
BUILDING THE PYRAMIDS
A corollary to the statement that pressure in a Huid increases with depth is that ,vater always seeks its own level. This means that if a vessel is filled with water, then regardless of the vessel's shape the surface of the water is perfectly flat and at the same height at all points. The ancient Egyptians used this fact La make
the pyramids level. Devise a scheme showing how this could be done. Explanation There are many '·.. ays it could be done, but Figure 9.15 shows the scheme used by rhe Egyptians. The builders cul grooves in the base of the
9.5
Pressure Measurements
283
pyramid as in (a) and partially filled the grooves with water. The height or the "vater was marked as in (b), and the rock was chiseled down to the rnark, as in (c). Finally, the groove was filled with crushed rock and gravel, as in (d). (0)
FIGURE 9.15
9.5
(b)
(d)
(c)
(Applying Physics 9.'1)
PRESSURE MEASUREMENTS
A simple device for measuring pressure is the open-lube manometer (Fig. 9.16a). One end of a U-shaped tube containing a liquid is open to the atmosphere, and the other end is connected to a system of unknown pressure P The pressure at point B equ<1ls Po + pgh, where p is the density or the fluid. The pressure at B, however, equals the pressure at A, which is also the unknown pressure P. \Ne conclude thal P = Po + pgh. The pressure P is called the absolute pressure, and P - Po is called the gauge pressure. I r Pin the system is greater than atmospheric pressure, II. is positive. If Pis less than atmospheric pressure (3 partial vacuum), h is negative, meaning that the right-hand column in Figure 9. L6a is lower than the lefL-hand column. Another instrumcnt used LO measure pressure is the barometer (Fig. 9.16b), invented by Evangelista Torricelli (1608-1647). A long tube closed at one end is rilled ,...~th rnercury and then inverted into a dish of mercLII)'- The closed end of the LUbe is nearly a vacuum, so its pressure can be taken to be zero. It follows that Po = pgh, where p is the density of the mercury and II is the height of the mercury colUtllll. Note that the barometer measures the pressure of the atmosphere, whereas the manometer measures pressure in an enclosed fluid. One atmosphere of pressure is defined to be the pressure equivalent of a column of merclll)' that is exactly 0.76 III in height at DoC ,\lith g = 9.806 65 m/s 2 . At this temperature, mercul)' has a densil)' of 13.595 X 103 kg/m:\ therefore, P n = pgh
=
p u
T " 1
p
• IJ
(a)
(13.595 X 10 3 kg/m 3 )(9.806 65 m/s") (0.7600 m) (b)
= .1.013 X 105 Pa = 1 atm
It is interesting to note that the force of the atmosphere on our bodies (assuming a body area of 2 000 in. 2 ) is extrcmely large, on the order of 30 000 lb! If it were not for thc rJuids permeating our tissues and body cavities, Ollr bodies would collapse. The fluids provide equal and opposite forces. In the upper atlTIosphere or in space, sudden decompression can lead to serious injury and death. Air retained in the lungs can damage the liny alveolar sacs, (lnd intestinal gas can even rupture internal organs.
FIGURE 9.16 Two devices for measuring pressure: (a) all 0pelllube nJ31lQnlClcr and (b) a mercUl")' baromclcr.
QUICK QUIZ 9.3 Several cOlllmon barometers are built using a variety of fluids. For which fluid will the column 01" fluid in the barometer be the highest? (Refer to Table 9.3.) (a) mercury (b) water (c) ethyl alcohol (d) benzene
Blood Pressure Measurements
APPLICATION
A specialized manometer (called a sphygmomanometer) is often used to rneasure blood pressure. In this application, a rubber bulb forces air into a cuff wrapped tightly around the upper arm and simultaneollsly into it manometer, as in Figlire 9.17 (page 284). The pressure in the cuff is increased until the now of blood
II Measuring Blood Pressure
284
Chapter 9
Solids and Fluids
through the brachial ,ulery in the arm opened, and lhe measurer listens with a below the cuff. V\'hen the pressure in the maximum value produced by the heart
Manometer
~ Rubber hlllh
is slopped. A valve on the bulb is then stethoscope La the artery at a point just cuff and brachial artery isjusL below the (the systolic pressure), the anery opens
momentarily on each beat of the heart. At this point, the velocity of the blood is high and turbulent, and the f1mv is noisy and can be heard Wilh the stethoscope. The manometer is calibrated to read the pressure in millimeters of mercury, and the value obtained is about J20 mm for a normal heart. Values of 130 l1un or above arc considered high, and medication La Imver the blood pressure is often prescribed for such patients. As the pressure in the cuff is lowered further, intermittent SOUJlds are still beard UJltilthe pressure falls just below the minimum heart pressure (the diastolic pressure). At this point, continuous sounds are heard. Tn the normal heart, this transition occurs at about 80 mm of mercury, and values above 90 require medical intcrvention. Blood pressure readings are usually expressed as the I-alia or the sysLOlic pressure to the diastolic pressure, '\ihich is 120/80 {or a heallhy heart.
QUICK QUIZ 9.4 Blood pressure is normally measured 'with the cuff ohhe sphygmomanometer around the arm. Suppose the blood pressure is measured with the cuff around the calf of the leg of a standing person. ''''ould the reading of the blood pressure be (a) the same here as it is for the arm, (b) greater than it is lor the arm, or (c) less than it is for the arm?
FIGURE 9.17 Asph}'gIllOltlanOmetcr can he lIsed LO measure blood pressun:.
APPLYING PHYSICS 9.3
BALLPOINT PENS
In a ballpoint pen, ink moves clown a tube to the tip, ",here it is spread on a sheet of paper by a rolling stainless steel ball. Near the top of the ink canridge, there is a small hole opcn to the
decrease as the ink was used. Consequclllly, atmospheric pressure exerted against the ink at the botlorn of the cartridge \-"ould prevent: some of' the ink from flowing oul. The hole allows the pressure above the ink LO remain at atmospheric pressure. Why does a ballpoint pen seem LO run out of ink when you ,..Tite on a vertical surface?
Explanation [f lhe hole 'were sealed, or if it were not presenl, the pressure of the air above the ink would
9.6
BUOYANT FORCES AND ARCHIMEDES' PRINCIPLE
A fundamental principle affecting o~jects submerged in nuicls was discovered by Greck madlematician £Ind natural philosopher Archimedes. Archimedes' principle C£l n be stated as follmvs: Archimedes' principle
-+
Any object completely or partially submerged in
9.6
Buoyant Forces and Archimedes's Principle
support to any object placed ill it. 'v\le often say that an object placed in a fluid is buoyed up by the fluid, so 'love call this up,-vard force the buoyant force. The buoyant force is nol a mysterious new force that arises in lluicls. In fact, the physical cause of the buoyant force is the pressure difference between the upper and lower sides of the object, which can be shov.'n to be equal to the weight of the displaced fluid. In Figure 9.18a, the fiuid inside the indicated sphere, colored darker blue, is pressed on all sides by the surrounding fluid. Arrows indicate the forces arising from the pressure. Because pressure increases with depth, t.he arrows on the underside are larger than those on top. Adding them all up, the horizontal components cancel, but there is a net force l.lp\varcls. This force, rille to differences in pressure, is the buoyant force B. The sphere of water neither rises nor falls, so the vector sum of the buoyant force and the force of" gravity on the sphere of" fluid must be zero, and it follows that B = Mg, where 1\1 is the mass of the fluid. Replacing the shaded fluid with a bowling ball of the same volume, as in Figure 9.]8b, changes only the mass on which the pressure acts, so the buoyant force is the same: B = Mg, where .fvf is the mass of the displaced fluid, nol the mass of' the bowling ball. The force ofg1'3vity on the heavier ball is greater than it \,vas all the fluid, so the bowling ball sinks. Archimedes' principle can also be obtained frolll Equation 9.8, relming pressure and depth, using Figure 9,l1b. J--:lorizontal forces frolll the pressure cancel, but in the vertical direction P2A acts upwards on the bOllom of the block or fluid and PIA and the gravity force on the fluid, l\1g, act downward, giving
285
~
[9.12.]
where the buoyancy f-orce has been identiried as a difference in pressure equal in magniLUde (Q the "..'eight of the displaced fluid, This buoyancy force remains the samc regardless of the material occupying the volume in question because it's due to the surrounding fluid. Using the dct-inition ofdcnsit)/, Equation 9.l2a bccomcs [9.12b]
where PlIlIid is the density of the fluid and l~lllid is the volume or the displaced fluid. This result. applies equally to all shapes because any irregular shape can be approximated by a large number of infinitesimal cubes. It's instructive to compare the forces on a totally submerged object with those on a floating objecl.
Case I: A Totally Submerged Object. \Vhcn an object is totally submcrged in a fluid of density Plluid' the upward buoyant force ,Kling on the object has a magniulde of B = Pnuid Vot~jg, where ll"bj is the volume of the objecl. If the o~jcct has density Pohj' the downward gravitational force acting on the object has a magnitude equal to w = mg = Pobj Vobjg, and the net force on it. is B - 1IJ = (Pftllid - Pobj) Vphjg. Therefore, if' thc denSity of the object is less than the density of the fluid. the net force exerted on
ARCHIMEDES Greek mathematician, physicist, and engineer (287-212 B.C.) Archimedes was perhaps the greatest scientist af antiquity. He is well known for discovering the nature of the buoyant force and was a gifted inventor. According to legend, Archimedes was asked by King Hieron to determine whether the king's crown was mode of pure gold or merely a gold alloy. The task was to be performed without damaging the crown. Archimedes allegedly arrived at a solution while taking a bath, noting a partial loss of weight after submerging his arms and legs in the water. As the story goes, he was so excited about his great discovery that he ron naked through the streets of Syracuse, shouting, "Eureka!" which is Greek for "I have found it."
Tip 9.2 Buoyont Force Is Exerted by the Fluid The huoyant force on an object iscxened by lhe nuid and is the saine, regardless of the densil\' of the objecl. Ol~jecls marc dense lhan the fluid sink; ol~jecLs le~s dense risco
FIGURE 9.18 (a) The arrows illdicate fClrCt;S on lilt; sphen: of fluid due to pressure, largcrollllw Undl.:ISide bccallsc preS~lllT increases wit h depth. 'I he net upward force i~ the buoyant forct'. (b) The hUOpl.lll!')]"ce, which is call~(;d by thc511JTlJUndillg fluid. is t be sallIe on am objecl of lht: same \·(J!llme. illcilldinR" lhis ho..... ling ball. The mag"llilude 01 the buoY
displaced fluid.
(a)
(h)
286
Chapter 9
Solids and Fluids
ACTIVE FIGURE 9.19 (a) A lOI,dly submerged o~ieclilwi is It.:s:,> dense than the nuid in which it is subllll:rged is acted upon by a llclllpward f\'Il'Ce. (b) :\ l'llally sllbll\erged ol~jeCl that is c1ell~cr 111:ln lhe nuid sitlks.
less dense th;)n cold air. there is a upward forCl' on rhe balloons.
c Ii
a• •
\C--
J ---------" " 1-IOl-air balloons. Because hot air is
~
'--
(a)
.J
\C--
/Jig
.J
(b)
lIel
the object is !JoSilive (up\vard) and the oqject accelerates 1.l!J1vard, as in AClive Figure 9.19a. I r the densit)' of the object is greall'r than the density of thc nuid, as in AClivc Figure 9.19b, the net force is JI(!galive and the object accelerates downwards.
ACTIVE FIGURE 9.20 All objeci f!u
and the buo)'anl
force B. These tWO forces are equal ill magnitude and opposite in direction.
APPLICATC'-IO;::.:..:N
II
_
Cerebrospinal Fluid
Tubing La draw muifreelc (rOIll tile radiator
lblls of difle-rem densities
FIGURE 9.21 The number of balls thaI floal in this device is a measure of I he densil) of tile antifreeze solutiOll in a vehicle's radiaLOr :Iud. consequently. a measure orthe temperature at which freezing Ili1I OCClll".
Case D: A Floating Object. Nmv consider a partially submerged object in sUltic equilibrium floating in (\ nuid, as in Active Figure 9.20. Tn this case, the upward buoyant force is balanced by the downward force of gravity acting on the objecL If V;luid is the volume of the fluid displaced by the object (which corresponds to the volume of the pan of the object beneath the fluid level), then the magnitude of the buoyanl force is given by B = Pflllill Vnuidg. Because the weight of the oqject is iV = mg = Pobj VI)I~ig~ and becausc w = 13, it follows that Pl1l1irl Vflllidg = P1J1~i V~'I~jg, or Pollj
\~luid
Pnuid
Vol~i
[9.13]
Equation 9.13 neglects the buoyant force of the air, \\ihich is slight because the density of air is only 1.29 kg/m 3 at sea level. Under normal circulllstances, the average density of a fish is slightly greater than the density of water, so a fish would sink if it didn't have a mechanism for adjusting its density. By changing the si7.e of an internal s\\"im bladder, fish maintain neutral buoyancy as they swim to various depths. The human brain is immersed in a nuid (the cerebrospinal fluid) of density I 007 kg/m 3, '¥"hich is slightly less than the average densit)' ofthe brain, I 040 kg/m:\. Consequently, most of the weight of the brain is supported by the buoyant force of the surrounding fluid. J n some clinical procedures, a portion or this fluid must be removed for diagnostic purposes. During such procedures, the nen'es and blood vessels in the brain are placed under great strain, which in turn can calise extreme discomfort and pain. Great care must be exercised with such ptltients until the initial volume of brain fluid has been restOl"Cd by the body. \Vhen service station attendants check the antifreeze in your caT or the condition of your bauery, they often use devices that apply Archimedes' principle. Figure 9.21 shO\,,'s a coml11on device that is used to check the antifreezc in a car radiator. The small balls in the enclosed tube vary in density so that all of them float when the tube is fitled with pure \vater, nOIlC float in pure antifreeze, one floats in a 5% mixture, nvo in a 10% mixture, and so lOrth. The number of balls that float is a measure of the percentage of' antifreeze in the mL,\ture, which in turn is llsed to determine the 100\'est temperature the mixture can withsL.:1.nd wit.hout freezing. Similarly, the degree of charge in some car batteries can be determined with a so-called magic-dot process that is built into the battery (Fig. 9.22). Inside a viewing port in thc top of the battel)', the appearallce of an orange dot indicates that the battery is sufficiently charged; a black dot indicates that the battet)' has lost its charge. If the batteql has sulTicient charge, the density of the battery fluid is high enough to cause the orange ball to float. As the battel)' loses its ch"u"ge, the densit}! of the bauery fluid decreases and the ball sinks bencath the surface 01" the fluid, lea\~ng the dot to appe,lr black.
9.6
Buoyant Forces and Archimedes's Principle
FIGURE 9.22 The orange.: hall in the pla~tic tuhe ill~ide rhe hatH.T)" serve~ a~ all indicator ofwh('lher the harr('!")" is (a) charged or (1)) discharged. As the haltl:I)·los"s ib lharge.:. the dellsit)" 01 tilt" ball"rv fluid decn',lses, arH.! til(' hall
287
APPLICATION Checking the Battery Charge
sinks Ollt ufsi!{hr.
Baller)'
lluid
-
Charged battery
QUICK QUIZ 9.S Atlllospheric pressure varies [rom day to day. The level of a Ooating ship on a high~pressure day is (a) higher (b) lower, or (c) no diffcr~ entthan on a Imv-pressure day.
QUICK QUIZ 9.6 The densil)' or lead is grealer lhan iron, and both melals are denser than water. Is the buoyant force on a solid lead object (a) great.er than, (b) equal to, or (c) less [han the buoyant force acting on a solid iron object of the salllt: dimensions?
EXAMPLE 9.8 Goal
A Red-Ta
Most of lhe vulume of III is iceberg is beneath the \\'ater. Can you determille \\'hat fraction of the total volume is under water?
Sp.ecial an Crowns
Apply Archimedes' prinCiple to a submerged object.
Problem A bargain hunter purchases a "gold" crm.. . n at a rlea market. Af'ter she gelS home, she hangs it from a scale and fmels its weight 1.0 be 7.8 l l l\ (Fig. 9.23a). She then weighs the crown while it. is immersed in wat.er, as in Figure 9.2~b, and now the scale reads 6.86 N. Is the crown made of pure gold? Strategy The goal is to find the density of the crown and compare it to the density of gold. 'A'l' already have the weight of' the crown in air, so we can gel [he mass by dividing by the acceleration of gravity. If we can find t.he volume of the crown, we can obtain the desired density by dividing the mass by this volume_ When the crown is fully immersed, the displaced water is equal [Q the volume of the crm,,"n. This same volume is used in calculating the buoyant fi:Jrce. So our strategy is as follows: (I) Apply Newton's second law to the crown, both in the water and in the air to find [he buoyant force. (2) Usc [he buoyant force to find the crm"'11'5 volume. (3) Divide the crovm's scale vvcight in air by the acceleration of gravit.), to gel t.he mass, then by the volume to gctthe density of the crown. ,... .....
r
r ,,,'====!-) (b)
(a)
(Example 9.8) (a) \Vhell the CroWll is sll.~pclldcd irl air. the scale reads T.,;, = mg, the crown's true \\'l:ighl~ (b) VI/lren the crown is immersed in water, rhe buoyanr forcc B reduccs the scale reading by lhe magnitude of rhe buoyani force, "I'...."n = mg- B. FIGURE 9.23
.
Solution mg = 0
Apply Newton's second la,... to the UO\'1!1l when it's weighed in air. Thel"c arc twO forces on the crown-gravity rind T:1ir> the rorcl; cxerted by the scale on the crown, with rnagnitude equal to the reading on [he scale.
(1)
7~lil -
~hcn
(2)
7;",1<"1" - mg
~
m.g
the crown is immerst:d in wat.er, the scale rorcc is with magnitude equ,lI to theJicale reading, and there is an upward buoyant rorce B and the rorce or gravity. TW;[I"P
+ B=
0
...,
288
Chapter 9
Solids and Fluids
Solve Equation (1) 1'01' mg, subSlillllC inw Equation (2), and solve for the buoyant force, which equals the difference in scale readings: Finn tlw VOlllllW nf,he dispbred
w~ltcr,
llsing the fact
that the magnilllde of the buoyant force equals the weight of the displaced water:
13 = T,;, - 7;""".
~
7.84 N - 6.86 N = 0.980 N
R = PwalcrgVW
(
gp"""
9.80
mid' ( 1.00
X
,
10' kg/m')
1.00 X 10-·' m" 7~lir
The crown is totally submerged, so VCTOKI1 = V"~l(er' From Equation (I), the mass is the crown's weight in air, T air, divided by g:
III ~ -
Find the density of tile crown:
Pcrnwil
g
7.84- N 9.80 mis'
= - - - - - " = 0 800 kg
III
~n'Wll
0.800 kIT c:-::-::-----:-:-'~;__:;_. = 1.00 X 10 ·1 ny'
Remarks Because the density of gold is 19.3 X 1O~{ kg/m 3 • the crown is either hollow, made of an alloy, or both. Despite the mathematical complexity, it is certainly conceivable that this was the method (hal occurred to Archimedes. Concepwally, it's a matter of realizing (or guessing) that equal weights of gold and a silver-gold alloy would have difTerent scale readings when immersed in \vater because their densities and hence their volumes arc different, leading to differing buoyant forces.
...
QUESTION 9.8 True or False: The magnitude of the buoy:ant force on a completely submerged object depends on the object's density. EXERCISE 9.8 The weight of a metal bracelet is measured to be 0.100 N in air and 0.092 N \vhen immersed in water. Find its density.
Answer
Goal
1.25 X 10·l
kg/lll~
Floating Down,-t""h;.;;e""R""i""v,;;;,e,-r
_
Apply Archimedes' principle to a partially submerged object.
Problem A raft is constructed of 'wood having a density of 6.00 X 102 kg/m 3 . Its surface area is 5.70 Ill:!, and its volume is 0.60 m~. "Vhen the raft is placed in fresh water as in Figure 9.24, to what depth II. is the bottom of the raft submerged? Strategy There are two forces acting on the raft: the buoyant force of magnitude E, acting up"vards, and the force of gravity, acting downwards. Because the raft is in equilibrium, the SlIlll of these forces is zero. The buoyant force depends on the submerged volume V\'~lt"r = All. Set up Newton's second Ia\V and solve for fl, the depth reached by the bOLLom of the raft.
"t FIGURE 9.24 (Exalliple 9.9) A raft partially submerged in waLcr.
.
. ••••••••••• '""11
Solution Apply Newton's second law to the raft, which is in equilibrium: The volume of the raft submerged in water is given by \!w:ltcr = Ali. The magnilllde of the buoyant force is equal to the weight of this displaced volume ofw
B=
Now relvrite the gravity Loree on the raft using the raft's density and volume:
m.1~lrl g = (p raI'l \l1~lrl) g
IIIwalcrg
=
(Pwrncl,Vwal"r)g
= (PII':l,crilh)g
9.6
Substitute t.hese two expressions into Newton's second law, B = ntr.lrrg, and solve for h (nole Ihat gcancels):
Buoyant Forces and Archimedes's Principle
(PwaterAh)g
=
II =
289
(Pmft VraJi)g Pr;dt
\~-aft
PW;l1<.~IA
(6.00 X 10" kg/m:l)eO.600 m') (1.00 X 10 3 kg/m 3 )(5.70 m 2 ) 0.063 2 m
.
~
.........
~
Remarks I-Im-" low the raft. rides in the \vaLer depends on the density 01" the rafL. The same is true of the human body: Fat is less dense than muscle and bone, so those with a higher percentage of body fat fioat better. QUESTION 9.9 If the rafL is placed in saiL water, which has a density gre;Her than fresh water, \-vould the value of h (a) decrease, (b) increase, or (c) not change? EXERCISE 9.9 Calculate how much of an iceberg is beneath the surface of the ocean, given that the density afice is 9]7 kg/m 3 and sail water has density 1 025 kg/m:l .
Answer
89.5%
EXAMPLE 9.10
Floating in Two Fluids
Goal Apply Archimedes' principle La an object floating in a fluid having two layers with different densities. Problem A 1.00 X 10 3 _kg cube of aluminum is placed in a tank. "Vater is then added La the tank until half the cube is immersed. (a) \"'hat is the normal force on the cube? (See Fig. 9.2.5a.) (b) Mercury is now slowly poured into the tank until the normal force on the cube goes to zero. (See Fig. 9.25b.) 1-:10\\1 deep is the layer of mercury?
!
IM:'\I
g
(a)
Strategy Both pans of this problem involve applications of NewlOn's second law for a body in eqllilibrillrn, together with the concept or a buoyant force. In part (a) the normal, gravitalional, and buoyant force of waler act on the cube. In part (b) there is an additional buoyant force of mercury, while the normal f-orce goes to zero. Using VII~ = All, solve for the height of mercury, II. ~
FIGURE 9.25
(Example9.10)
.....
,
Solution (a) Find the normal force on the cube when half-immersed in v.'ater. 1.00 X 10 3 kg
Mil
Calculate lhe volume vor the cube and the length d of one side, for future reference (both quantities will be needed for \vhat follows):
II"
\"'rite Ne\vton's second law for the cube, and solve for the normal force. The buoyant force is equal to the weighl of the displaced water (halfthc volumc of the cube).
11 - M'\lg
= -
-
PAl
d = v:~{3
11,=
=
+
'j
3
2.70 X 10' kg/Ill'
= 0.370 m'
0.718 m IJ.val
= 0
k!t\lg- B.. .~ll = l\1Alg- Pw;" (V/2)g
= (1.00 X J03 kg) (9.80 mjs2)
- (1.00 X 10 3 kg/m') (0.370 m 3 /2.00) (9.80 m/,2) IJ
= 9.80 X 10' N - 1.81 X JO' N =
7.99 X 10 3 N
290
Chapter 9
Solids and Fluids
(b) Calculate the level It of added rnercury.
Apply NewLOn's second law to the cube:
n - Ai.",]g + Bwat +
/JIll; =
0
Set
11 = 0 and solve for the buoyant force of mercli ry:
Solve for
11"
h=
noLing that A = d'!.;
h=
........
M"g - /3".,,,
7.99 X 10'?,<
PII,Ag
( 13.6 X 10" kg/m')(0.718 m)'(9.80 m/s')
0.116 m ............................~
,
Remarks Notice that the buoyant. force of mercury cal~ culalcd in part (b) is the same as the normal force in part (a). This is nawrally the case, because enough mercury was added La exactly cancel oul the normal force. We could have used this facl to Lake a shorLCIl1.. simply writing Bllg = 7.99 X lO~ N irnmedialely, solving for 11., and avoiding a second lise of Newton's law. Most of the time, however, we won't be so lucky! Try calcuhiting the normal force when the level of mercury is 4.00 CIll.
9.7
QUESTION 9.10 \Nhat v.,'Quld happen to the aluminum cube ifmore mercury were poured into the tank? EXERCISE 9.10 A cube of alurnilllllll 1.00 111 on a side is immersed onethjrd in ",vater and two-thirds in glycerin. What is the normal force on the cube? Answer
1.50 X 10"' N
FLUIDS IN MOTION
..s
\'\Then a fluid is in mOLion, its Om·" call be charaClcrized in one of two ways. The now is said Lo be streamline, or laminar, i r every panicle lhal passes a particular ~ point. moves along exactly the same smooth path followed by previous panicles ~ passillg that point. This path is called a sl-reamline (Fig. 9.26). Diffcrcmstrcarnlines tl Clln't cross each other under this steady-now condition, and the streamline at any i point coincides with the direction of the velocity of the Ouid at that point. _ _ _' " <: FIGURE 9.26 An illuslf in velocity. Irregubile in;\ lest wim!lullJlel. The Slreallllar Illotions of the fluid, called f'dd)' currents, are characteristic in turbulcnt flow, as lines in Ihe airflow are made \,jsihh' hy slrioke pani<-ks. shown in Figure 9.27. 1n discussions of fluid flow, the term viscosity is L1sed for the degree of internal friction in the fluid. This internal friction is associated with the resistance between tvm a(~jacent la}/ers of the fluid moving relative to each other. A fluid such as kerosene has a lower viscosity than docs crudc oil or molasses. Many feattlres of fluid motion can be understood by considering the behavior ~ g oran ideal fluid, which satisfies the following conditions: CJ
i j I. The fluid is nonviscous, ,,,,'hich means
I here is no internal friction (-orce between adjacent layers. ~ ~ 2. The fluid i~ incompressible, which means its density is constant. ~ 3. The fluid motion is steady, meaning that the velocity, density, and pressure at ~g each point in the fluid don't change with lime. ~ ~ 4. The fluid moves withoUl turbulence. This implies that each element of the fluid ~~ has zero angular velocity about its center, so there can't bc any eddy currents ~~ present in the moving fluid. A small \-"heel placcd in the nuid would translate but not rotatc. ~E
~;f
i
!i
,.;;:33
FIGURE 9.27 TlIrblilellL flow: The Lip ofa rowling blade (llie dark region'll (lie lOp) rorms a vortex ill
Equation of Continuity
air thai is being heall.:d b~' an alcohol lamp. (The \dck is at the hOLlom.) "ott: the air l.nrbult:nce (Ill bOlh sides of the blade.
Figure 9.28a represents a fluid flowing through a pipe of nonuniform size. The particles in the fluid move along the strcamlines in steady-slale 00\·... In a small lime interval D.t, the fluid entering the bottom end of the pipe moves a distance
9.7
Fluids in Motion
291
FIGURE 9.28 (a) A fluid moving '\'ith streamline (Jo\\·throllgh a pipe of\·arrillg cro5s·seniullill area. The volume offlllid flowing through AI in a time inlo.;rval!:J.t Illust equal the vulume flo\\'ill~ through At ill the same time illler....aL Therefore, Arv J = AtV~. (b) Water f1o\\ing slowly OUl or a fallceL The width of the strt:am narrows as lht:" water falls and spl::cds up in
@
I
accord with tht: COntinllilY cqmltioll.
(al
(h)
.6.x[ = VI ~L, where v J is the speed of the fluid al that local ion. If A J is the cross~ sectional area in this region, then the mass contained in the bottom blue region is AM] = PlA I 6.x[ = PIAIVj nt, where PI is the density of the fluid at AI' Similarly, the fluid lhalll10ves out of the upper end of the pipe in the same time interval !1t has a mass of tlA1::! = P~A2v2 D.l. Hov·:cvcr, because mass is conserved and because the flow is steady, the lllass that flows into the bouom of the pipe Lhl'ough A I in the till1C tl/lllllsl equal the mass that flows oUllhrough 1\2 in the same illlcrvai. Thererore, .6..M 1 = 6.M 2 , or
[9.14] For the case of an incompressible fluid, PI = p:.! and Equation 9.14 reduces to [9.15] This expression is called the equation of continuity. From this result, we see that the product of the cross-sectional area of the pipe and the fluid speed at that cross section is a constant. Therefore, the speed is high ",,'here the tube is constricted and low where the tube has a larger diameter. Thc product Av, which has dimensions of volumc per unit time, is called the flow rate. The condition Av = constant is equivalent to the fact that the volume of fluid that enters one end of the tube in a given time interval equals the volume of fluid leaving the tube in the same interval, assuming that the fluid is incompressible and there are no leaks. Figure 9.28b is an example of an application or the equation of continuity: As the stream of ""ate I' flows cOlll.inuously rrom a faucet, the width or the stream narrows as it falls and speeds lip. There arc many instances in everyday experience t.hat involve the equation or continuity. Reducing the cross-sectional area of a garden hose by pULling a thumb over the open end makes the water spray oul with greater speed; hence the stream goes farther. Similar reasoning explains why smoke from a smoldering piece or wood first rises in a streamline pallern, gelling thinner with height, eventually breaking up into a swirling, turbulent paLtern. The smoke rises because it's less dense than air and the blloyant force of the air accelerates it upward. As the speed or the smoke stream increases, the cross-sectional area of the stream decreases, in accordance with the equation or continuity. The stream soon I-caches a speed so great that slreamline flow is nOl possible. ''''e will study the relationship between speed of fluid flmv and lLirbulence in a lat.er discussion on the Reynolds number.
EXAMPLE 9.11 Goal
f- Equation of continuity
Tip 9.3 Continuity Equations The I'ate of flo\\' of fluid inl,o a system cC'jllals the nUt: of flO\\' OUI of the S)'Slcm. The incoming fluid occupies a cenain ,"olumc llnd can enter the system only if the fluid already inside goes alit, thereby making room.
Nio oro Falls
Apply the equation of continuity.
Problem Each second, 5 525 111 3 or water flows over the 670-m-wide c1ifr of the Horseshoe F~llls portion or Niagara Falls. The water is approximately 2 III deep as it reaches the cliff. Estimate its speed at that instant.
292
Chapter 9
Solids and Fluids
Strategy This is an cSlimalc, so only olle significant figure will be ret.ained in the answer. The volullle flow rate is given, and, according to the equation of" continuity, is a constant equallo Av. Find the cross-sectional area, substitute, and solve for the speed.
r'········ . ,
~
Solution Calculmc the cross-sectional area of t.he waleI' as it reaches the edge of the cliff:
A
Multiply this result by t.he speed and set it. equal to the now rate. Then sake for v:
=
(fi70 m)(2 m)
Av
=
I 340 m'
= volume now rate
(1340 m')v = 5 525 m"/s
.....................
v=
--7
4 m/s
.
.J
QUESTION 9.11 "Vhat happens to the speed of blood in an artery when plaque starts to build up on the artery's sides? EXERCISE 9.11 The Garfield Thomas water tunnel at Pcnnsylvania St.atc University has a circular cross scction that consu-icts from a diameter of 3.6 m to the test section, wh ich is 1.2 In in diamcter. 1ft he speed of flow is 3.0 m/s in the larger-diameter pipe, detcrmine the speed of rIm\-' in thc tesl section.
Answer 27 m/s
EXAMPLE 9.12 Goal
Watering a
arden
Combine the equation of continuity with concepts of rIOh' rate and kinematics.
Problem A water hose 2.50 em in diameter is used by it gardener to fill it 30.0-liter bucket. (One liter = L000 cm 3 .) The gardener notices that it takes 1.00 min to fill the bucket. A nozzle with an opening of cross-sect.ional area 0.500 cm 2 is then attached to the hose. The nozzle is held so that waleI' is projected horizont
,...
.........................
.
~
Solution
volume flow rate =
Calculate thc volume flow rate into the buckel, and convert to m 3/s:
= .30.0 L
1.00
= 5.00
III in
(1.00X 10" ern")( 1.00 m )"(1.00 min) 1.00 L 100.0 em 60.0,
X 10-" m'/s
Solve the equation of continuity for l)u.'I:' the x-componen 1 of the initial velocity of the strcam cxiLing the hose:
5.00 X 10 '4 m'l' 0.500 X 10 "m'
10.0 Ill/'
Calculate the time for the stream to fall 1.00 m, using kinematics. Initially, the stream is horizontal, so v o)' is zero: Sct v Oy == 0 in the preceding equation and solve for noting that 6.) = - 1.00 Ill:
It's interesting that the motion of fluids can be treated with the same kinem
QUESTION 9.12 By what. facwr would the range by changed if the flow rate were doubled? EXERCISE 9.12 The nozzle is replaced with a V-shaped IItting that splits the flow in hair. Garden hoses are connected 10 each end of the Y, with each hose having a 0.400 cm~ nozzle. (a) How fast does the water come out of one of the nozzles? (b) How far would one of the nozzles squirt \vater if both ",..'ere operated simultaneously and held horizontally 1.00 m oIT the ground? Hilll: Find the volume flmv rate through each 0.400-CI1l~ nozzle, then follow the same steps as before.
Answer
(a) 6.25 m/s
(b)
2.~2
m
Bernoulli's Equation As a fluid moves through a pipe of varying cross section and elevation, the pressure changes along the pipe. In 1738 the Swiss physicist Daniel Bernoulli (1700-L782) derived an expression that relates the pressure ofa fluid to its speed and elevation. Bernoulli's equation is nOt a freestanding law of physics; rather, it's a consequence of energy conservation as applied to an ideal fluid. In deriving Bernoulli's equation, we again assume the rJuid is incompressible, nonviscous, and flows in a Ilonwrbulent, steady-state manner. Consider the flow through a nonuniform pipe in the time Ilt, as in Figure 9.29. The force on the lower end of the fluid is PjAj, '''''here PI is the pressure at the lower end. The work done on the Imver end 01" the rJuid by the fluid behind it is
where Vis the volume 01" the lower blue region in the figure. In a similar manner, the work done on the fluid on the upper portion in the time!J.1 is
CD
lux1
I~I~
)'I!
~
y':!.
L
FIGURE 9.29 A fluid 110Willl{ lhrough a cOlIslriCled pipe willi sln:amlinc !lo\\'. The fluid ill Ilw scnioll \\'ith
The volume is the same because, by the equation of continuity, the \'olullle of fluid lhat passes through Al in the time III equals the volume that passes through A~ in the same inten..al. The work W2 is negative because the force on the fluid at the top is opposite its displacement. The net work done by these forces in the tirne !J.l is
Pan of this w·ork goes into changing the fluid's kinetic energy, and part goes into changing the gravitational potential energy of the fluid-Earth system. If 1ll is the mass of the (luid passing through the pipe in the time interval Ill, then the change in kinetic enerb'Y of lhe volume of fluid is IlKE = ~mv'!.t
-
~m1JI'!.
The change in the gravitational potential energy is tJ.PH = mgy:i. - mg)'l
Because the net ,,,,'ork done by the rJuid on tIle segment. of r1uid shown in Figure 9.29 changes t.he kinetic energy and the pOlential energy of the nonjsolated sysLCIll, ,ve have
DANIEL BERNOULLI Swiss physicist and mathematician
(1700-1782) In his most famous work, Hydrodynamica, Bernoulli showed that, as the velocity of fluid flow increases, its pressure decreoses. In this some publication, Bernoulli olso attempted the first explanation of the behavior of gases with changing pressure and temperature; this was the beginning of the kinetic theory of gases.
294
Chapter 9
Solids and Fluids
(a) The pressure PI is than the pressure P".!. becausc v J < 11'1' This dcvi<:e GIn be used to measure the speed of lIuid flow. (b) A Venruri lube. located at the top of the photograph. TIle higher h::vcl of nuid in the middle column shows thaL lhe pressure al the LOp of the column, which is in the constriCted regiull of the Venturi Lube. is lower thall/he pressure elsewhere in tilt" column.
FIGURE 9.30
greal~r
lIT
" E " ....._---" (b)
(:t)
The three terms in this equation are those we have just evaluated. Substituting expressions for each of the terms gives
Tip 9.4 Bernoulli'. Principle for Gases Equation 9.16 isn't strictly trlle for
gases because they aren't incom-
PI
V - p"!. V = ~mV2'2
-
~ntVl"!.
pressible. The qualitative behavior
Ifwe divide each term by Vand recall that p =
is the sainI.:, hm,'cvcr: As the speed orthe gas increases, its pres~llrc decreases.
PI -
P'1 = ~ pv/ - ~ Plll'2
+
rr1g)'2 -
JlI/\~
m.g;YI
this expression becomes
+ Pg)':!. - pgYI
Rearrange the terms as 1"01l0\vs:
[9.16] This is Bernoulli's equation, often expressed as Bernoulli's equation '7
P
I
,
+ 'iPlr + pgy
= conSWllt
[9.17]
Bernoulli's equation states that the sum of the pressure P, the kinetic energy per unit volume, !pv'2, and the potential energy per unit volume, pgy, has the same V"dllle at all points along a streamline. An imporLant consequence of Bernoulli's equation can be demonstrated br consid~ ering Figure 9.30, which shows ",:ater flowing through a horizontal constricted pipe from a region of large cross-sectional area into a region of smaller cross-sectional area. This device, called a Venturi tube, can be Llsed to measure the speed offluid 110\\1. Because the pipe is horizontal.)l1 = )'2' and Equation 9.16 applied to points 1 and 2 gives
1,I+ 1'iPVI '2_/:) "!.
I 2 + 'iPv':!.
[9.18]
Because the water is not backing up in the pipe, its speed V2 III tile constricted region must be greater than i\.s speed VI in the region of greater diameter. From Equation 9.18, we see that P2 must be less than PI because V2> VI. This result is often expressed by the statement that swiftly moving fluids exert less pressure than do slowly moving fluids. This irnportant i~lCt enables us to understand a wide range of everyday phenomena. QUICK QUIZ 9.7 You observe \.wo helium balloons floating next to each other aL Lhe ends of stri ngs secured to a t.able. The facing surfaces of the balloons are separated by 1-2 em. You blow Lhrough the opening between the balloons. '''That happens to the balloons? (a) They move toward each otiler. (b) They move a,\'ay from each other. (c) They are unallected.
EXAMPLE 9.13 Goal
Shoot-Out at the Old Water Tank
Apply Bernoulli's equation to fllld the speed ora fluid.
Problem A nearsighted sheriff fires at a caule rustler with his trusty six-shooter. Fortunately for the rustler, the bullet misses him and peneLrates the town water tank. causing- a ]c,lk (Fig. 9.3]). (a) lfthe top of the t<'lnk is open to Lhe
9.7
aUl1osphere, determine the speed at . .v hich the water lca\·cs lhe hole when the watcr le\·cl is 0.500 m abO\"c the holc. (b) ''''here docs the stream hit the ground ifthc hole is 3.00 In aboyc [he ground?
Fluids in Motion
295
I.cro
len:! for
Strategy (a) Assume the tank's cn>ss-scctional "rea is large compared lo the hole's (A:.! » AI), so the water level drops vcry slowl>' and V 1 = 0. Apply Bernoulli's cquation to points Q) and @ in Figurc 9.:~ I, notin~ lhat PI equals atmospheric pressure PI' at the hole and is approximately the samc al thc top of the waler tank. Part (h) can be soh·cd with kinematics,just as if the waleI' wcre a ball thrown horil.ontally.
gr;", il
potential (.' Ilcr ro
,
•
FIGURE 9.31 (Example 9.1:{) The w,lIer speed tIl from the hole in the side of the container is gin:-n b) VI V2glt.
.
~
....
Solution (a) Find the speed of the walcr le..w ing the hole. Substitutc P, = P'l = Po and v'l = 0 into HtTnoulli's equalion, and solve for VI:
= V2ih
VI
= V2gCr2 -
VI
~ V2(9.80 m/5')(0.500 111)
)'1)
=
3.13 m/5
(b) Find where the stream hits the ground. Use the displaccmcllt equation to lind the timc orthe fall, noting that the stream is initially horizontal, so vu> = o.
-3.00
III
= -(4.90 m/5')I'
1= 0.782 s Compute the horizontal distancc thc slrearn travels in this tin1e:
....
,
x = vo~/ = (?.13 l11/s)(0.782
s)
2.45 m
.
... ~
Remarks As the analysis or pan (a) shows, the speed of the water emerging from the hole is equal to thc speed acquired by an object falling freely through the vertical distance It. This is known as Torricelli's law. QUESTION 9.,13
As time passes, what happens to the speed of the water lca\·ing the hole? EXERCISE 9.13
Suppose, in a similar situation, the watcr hit~ the ground 4.20 m from the hole in the tank. If the hole is 2.00 m above the ground, how far above the hole is the watcr le\·el? Answer
2.21 m above the hole
EXAMPLE 9.14
Fluid Flow in a Pip.e
Gool Solve a problem combining Bernoulli's equation and the equation of continuity. Problem A large pipe with a cross-sectional area of 1.00 m'l descends 5.00 m and narrows to 0.500 m~, where it terminatcs in a valve at point Q) (Fig. 9.32). If the pressure at point ® is atmospheric pressure. and the \·alve is opened \vide and water allowed to flow freely, [md thc speed of the water leaving the pipe. Strategy The equation of conlinuit), together with Bernoulli's equation, constilUte twO equations in two unknowns: the speeds v and v 2 . Eliminate v'2. from I.kr· noulli's equation with the equation ofconlinllily, and solve for VI. J
® "11 vt
"
,
",---0)
-----------~J~l
VI FIGURE
9.32
(F.xample9.14)
296 ~
Chapter 9
...
Solids and Fluids
.
~
Solution ',Vrite Bernoulli's equation: Solve the equalion of continuity for v2: (2)
In Equation (1), sel PI = P2 = Po, and substitute the expression for v 2' Then solve for v j ,
(3)
SubstlLUtc the given values:
V, =
11.4 m/s
.
.
~
,
...
Remarks This speed is slightly higher than the speed predicted by Torricelli's law because the narrowing pipe squeezes the fluid.
QUESTION 9.14 Why does setting AI = A 2 give an undefined answer for the speed VI? Hinl: Substitute A 1 = verifywhethcr or not a contradiction is obtained.
1\2
into Equation (3) and
EXERCISE 9.14 Water flowing in a horizontal pipe is at a pressure of 1.40 X 105 Pa at a point where its cross-sectional area is 1.00 m 2 . "\Then the pipe narrows to OAOO m 2 , the pressure drops [Q 1.16 X 105 Pa. Find the water's speed (a) in the wider pipe and (b) in the narrmvcr pipe. Answer
(a) 3.02 m/s
(b) 7.56 m/s
9.8
OTHER APPLICATIONS OF FLUID DYNAMICS
In this section we describe some common phenomena that can be explained, at least in pan, by Bernoulli's equation. In general, an object moving through a fluid is acted upon by a net upward force as the result of any effeu that causes the fluid to change its direction as it flows past the object. For example, a golf ball struck with a club is given a rapid backspin, as shown in Figure 9.33. The dimples on the ball help entrain the air along the curve of the ball's sufi-ace. The figure shows a thin layer of air wrapping parnvay around the ball and being deflected downward as a result. Because the ball pushes the air down, by Newton's third 1£1\\1 the air must push up on the ball and cause it to rise. \Vithout the dimples, the air isn't as well entrained, so the golf ball doesn't travel as far. A tennis ball's fuzz performs a similar funct.ion, though the desired result is ball placement rather than greater distance.
=
FIGURE 9.33 A spinning go[fbal[ is
•
9.8
Other Applications of Fluid Dynamics
FIGURE 9.34 A stream aLlir passill~ over a LUbe dipped ill a liquid GHlSeS the liquid LO ri~e in tIll.: 11Ibl::. This I::ffecl is used ill pcrfllllle baltksand pailll spra)'ers.
297
I'I
FIGURE 9.35 Bloou must travel faster than nonn
Many devices operate in the manner illustrated in Figure 9.34. A stream of air passing over an open tube reduces the pressure above the tube, causing the liquid to rise into the airstream. The liquid is then dispersed into a fine spray of droplets. You might recognize that this so-called atomizer is used in perfume bottles and paint sprayers. The same principle is used in the carburetor ora gasoline engine. Tn that case, the IO\\,'-pressufc region in the carburetor is produced by air drawn in by the piston through the air mter. The gasoline vaporizes, mixes with the air, and enters the cylinder of the engine for combustion. In a person with advanced arteriosclerosis, the Bernoulli effect produces a symptom called vascular flutter. Tn this condition, the anery is constricted as a result of accumulated plaque on its inner walls, as shown in Figure 9.35. To maintain a constant f!m,,' rate, the blood must travel f~lster than normal through the constriction. If the speed of the blood is sufficiently high in the constricted region, the blood pressure is low, and t.he artery may collapse under external pressure, causing a momentary interruption in blood now. During the collapse there is no Bernoulli effect, so the vessel reopens under arterial pressure. As the blood rushes through the constricted artery, t.he internal pressure drops and the artery closes again. Such variations in blood flow can be heard with a stethoscope. If the plaque becomes dislodged and ends up in a smaller vessel thai delivers blood to the heart, it can cause a heart attack. An aneurysm is a \·...,eakened spot on an artery where the artery walls have ballooned outward. Blood nows more slowly though this region, as can be seen from the equation of continuity, resulting in an increase in pressure in the vicinity of the aneurysm relative to the pressure in other parts of the artery. This condition is dangerous because the excess pressure can cause the artery [0 rupture. The lift on an aircraft wing can also be explained in part by the Bernoulli effect. Airplane wings are designed so that the air speed above the wing is greater than the speed below. As a result, the air pressure above the wing is less than the pressure below, and there is a net upward force on the wing, called the l{ft. (There is also a horizontal component called the drag.) Another factor inlluencing the lift on a wing, shown in Figure 9.36, is the slight upward lilt oflhe ,.. . ing. This causes air molecules striking the bottom to be deflected downward, producing a reaction force upward by Newton's third law. Finally, turbulence also has an effect. If the wing is tilted too much, the flow of air across the upper surface becomes turbulent, and the pressure difference across the wing is not as great as that predicted by the Bernoulli effecl. In an extreme case, this turbulence Illay cause the aircraft lo stall.
EXAMPLE 9.15 Goal
APPLICATION "Atomizers" in Perfume Bottles and Paint Sprayers
APPLICATIONC-..__
III
Vascular Flutter and Aneurysms
APPLICATION Lift on Aircraft Wings Drag I
• • • 4
• •
-
--•
~
~
~
1
FIGURE 9.36 Streamline flow around an airplanc wing. The pres+ sure ab(J\'c is less than lllc pressure below, and there is a dynamic lIpward lift forLc.
lift on on Airfoil
Usc.:: Bernoulli's equation to calculate the lift 011 an airplane wing.
Problem An airplane has wings, each with area 4.00 m 2 , designed so that air flows over me lOp of the wing at 245 m/s and underneath I.he wing at 222 m/s. Find I.he Illass 01" the airplane such that the lift on the plane will suppan its weight, assuming the force [rom the pressure dilTerence across the wings is directed straight upwards. Strategy This problem can be solved by substituting values into Bernoulli's equation [Q find the pressure difference between the air under the wing and the air over (he "...ring, followed by applying Nev'llon's second law to find the mass the airplane can lift.
298
Chopter 9
Solids and Fluids
.
r ..
-
..,
-
Solution Apply Bernoulli's equation to the air flowing under the wing (point I) and O\'er the wing (point 2). Gravitational potential energy terms are small compared with the other terms, and can be neglected. Solve this equation for the pressure difference:
H1.29 kg/m')(245' m'/s' -
SubstittllC the given speeds and p = 1.29 kg/m\ the density of air:
tJ.p =
Apply ewton's second law. To Supporllhe plane's weight, the sum of the lift and gravity fOl'ces mllst equal zero. Solve (or the mass l1l of the plane.
2,1 tJ.p - mg
.....
222' m'/s')
tJ.p = 6.93 X 10' Pa
=0
-->
m = 5.66 X 10' kg
. .....
..
Remarks Note the factor of two in the last equation, needed because the airplane has two wings. The density of the <:ttmospherc drops steadily \....ith increasing height, reducillg the lift. As a result, all aircraft have a maximum operating altitude. QUESTION 9.15 Why is the maximum lift affected by increasing allillldc? EXERCISE 9.15 Approxirnately what size wings would an aircraft need all Mars if its engine generates the same differences in speed liS in the example £lnd the toto:ll mass or the crart is 400 kg? The density orail" on the surface or Mars is approximately one percent Earth's densil)' at sea level, and the acceleration of gravity on the surface of Mars is abou13.8 m/s2. Answer Rounding to olle significant digit, each wing would have to have an area or about 10 m 2 . There hm'c been proposals for solar-pO\'\iered robotic \.1ars aircraft, 'which \\'ould have to be gossamer-light with large \'Y·ings.
APPLYING PHYSICS 9.4
SAILING UPWIND
How can a sailboat accomplish the seemingly impossible task ofsailing inLo the \"'ind? Explanation As shown in Figure 9.37, the wind blowing jn the direction of the afrow causes the sail to billow out and take on a shape similar to thal oran airplane wing. By Bernoulli's equation,jusl as for an airplane wing, there is a force on the sail in Ihe direction shown. The component of for"ce perpendicular [Q the boat tends to make the boat 1TI00"C sideways in the watcr, bUllhe keel prevents this sideways mOlion. The component of the force in the forward direction drives the boat almost against the wind. The word almo\'t is used because a sailboat can move ff)rwarcl only when the wind direction is about 10 to 15° with respect to the forward direction. This means Ihatto sail directly against the wind, a boat must rollow a zigzag path, a procedure called ladling, so that the wind
is always at some angle with respect to the direction of tra,·el.
Wind
i- Keel axi...
FIGURE 9.37
(Arpl~ing
Physics 9.4)
9.9
APPLYING PHYSICS 9.5
Velll-
Si~
II \ (API'I}"jll~ Physics
.-
Sewer plpL'
9_5)
in the sevver pipe. "Vhat is the purpose of the vent, which is open to I"1w air ahove the roof of the house? In which dircction is air moving at the opening of the vent, upwards or downwards? Explanation Imagine that thc vent isn't present so that the drainpipe for the sink is simply connected through the trap to the se\Ver pipe. As watel- from the dishwasher moves LO the left in the se\'Ier pipe, the pressure in the sewer pipe is reduced below atmospheric pressure, in accordance with BenlouJJi's principle. The pressure at the drain in the sink is sUll at atmospheric pressure. This pressure difference can push the plug of\vater in the water trap of the sink down the drainpipe and into the sewer pipe, removing it as a barrier to sewer gas. With the addition of the vent to the roof, the reduced pressure of the dishwasher water will result in air entering the vent pipe at the roof. This inflow of air will keep the pressure in the vent pipe and the right-hand side of the sink drainpipe close to atmospheric pressure so thal the plug of water in the ,",vater trap will remain in place.
The exhaust speed of a rocket engine can also be understood qualitatively with Bernoulli's equation, although, in aetual practice, a large number of additional variables need to be taken into account. Rockets actually work better in vacuum than in the atmosphere, contrary to an c~Hly Nf1u YOTIl Times article criticizing rocket pioneer Robert Goddard, which held that they wouldn't work at all, having no air to push against. The pressure inside the combustion chamber is P, and the pressure just outside the nozzle is the ambient atmospheric pressure, Pmll1' Dillcrences in height betwecn the combustion chamber and the end of the nozzlc result in negligible contributions of gravitational potenlial energy. In addition, the gases inside the chamber now at negligible speed compared to gases going through the nozzle. The exhaust speed can be found from Bernoulli's equation, v•. = >(
This equation shows that the exhaust speed is reduced in the atmosphere, so rockets arc actually more dTective in the vacuum of spacc. Also of interest is the appearance of the density p in the denominator. A lower density working nuid or gas will give a higher exhaust speed, which partly explains why liquid hydrogen, which has a very low density, is a fuel of choice.
9.9
299
HOME PLUMBING
Consider the portion ora home plumbing syslem ShO\'.'I1 in Figure 9.38. The water trap in the pipf' below the sink captures a plug of water that prevents sewcr gas from finding its way from the sewer pipe, lip the sink drain, and inlO the home. Suppose the dishwasher is draining and the \V
FIGURE 9.38
Surface Tension, Capillary Action, and Viscous Fluid Flow
SURFACE TENSION, CAPILLARY ACTION, AND VISCOUS FLUID FLOW
If yOLi look closely at a dewdrop sparkling in the morning sunJig-ht, you will rind that the drop is spherical. The drop takes this shape because or ,I property of liquid surfaces called surface tension. In order to understand the origin or surface tcnsion, consider a molecule at point A in a container of water, as in Figure 9.39 (page 300). Although nearby molecules exert forces on this molecule, the net. force on it is zero because it's completely surrounded by other molecules and hence is
APPLICATI0.o...N'---_ Rocket Engines
300
Chapter 9
Solids and Fluids
f1
FIGURE 9.39 The /let force on a molecule: at A is zero uccausc SlIcli a molccu Ie is cOInpletely slirrolllluni b)' other mulecules. The tlCI force on a surf:ace molecule ;u B i~ downward because it isn't completely surrounded hy other moleclllc.~.
attracted equally in all directions. The molecule at B. however, is nOl attracted equally in all directions. Because there arc no molecules above il La exert upward forces, tlle molecule at B is pulled toward the interior of the liquid. The contraction at the surface of the liquid ceases when the inward pull exerted on the surface molecules is balanced by the outward repulsive forces that arise from collisions with molecules in the interior of the liquid. The net efrect of this pull on aLI the surface molecules is to make the surface of the liquid contract and, consequently, to make the surface area of the liquid as small as possible. Drops of water take on a spherical shape because a sphere has the smallest surface area for a given volume. If you place a sewing needle very carefully on the surface ofa bowl of water, you will find that the needle noars evcn though the density of steel is about eight times that of waleI'. This phenomenon can also be explained by surface tension. A close examination of the needle shmvs thal it aCLUally rests in a depression in the liquid surf~lce as shown in Figure 9.40. The WaLeI' surface acts like an elastic membrane under tension. The weight or the necdle produces a depression, increasing the surface alTa of the film. Molecular forces now act at all points along the depression, tending to restore the surface [Q its original horizontal position. Thc venical componcnts of thesc forces act to balance the force of gravity on the needle. The floating net:dle can bc sunk by adding a liltle detergenlto the water, which reduces the surface tcnsion. The surface tension)' in a film of liquid is defined as the magniwcle of the surrace tension force I" divided by the length L along \vhich the force acts: F
[9.19]
')'=-
L
The SJ unit of surface tension is the newton per mcter, and values for a few represcntative materials are given in Table 9.4. Surface tension can be thought of as the energy content of the fluid at its surface per unit surfacc area. To see that this is reasonable, we can manipulate the units of surface tension)' as follows: N FIGURE 9.40
End vic\\' oLi l1(;edle
resting 011 the surface of \~.tlc;r. The components of surface len~itln halance the force ofgra"il)'.
J11
=
J
N'm
In gcneral, in any equilibrium configuration of an object, the energy is a minimum. Consequcntly, a fluid will lakc 011 a shape such that its surface area is as small as possible. For a given volume, a spherical shape has thc smallest surface area; therefore, a drop of water takes on a spherical shape. An apparatlls used to measure the surface tension of liquids is shown in Figure 9.41. A circular wire with a circumference L is lifted from a body ofliquid. The surface film clings to the inside and outside edges of the wire, holding back the w·ire and causing the spring 1.0 stretch. If the spring is calibrated, the force required to overcome the surrace tcnsion of the liquid can be measured. In this case the surface tension is given by F ')' =
TABLE
2L
__
9.~4
Surface Tensions for Various Liquids Surface Tension
(N/m)
Li
FIGURE 9.41 All apparatus for measuring the surface tension of
liquids. The force on the ....·ire ring is mcasun:d jllSt before the ring breaks free orlhe liquid.
Ethyl alcohol Mercury Soapy water \t\'ater Water
20 20
20 20 100
0.022 0.465 0.025 0.073 0.0.09
9.9
Surface Tension, Capillary Action, and Viscous Fluid Flow
301
\Ne use 2L for the length because the surface film exerts forces on buth the inside
and outside of the ring. The surf~lCe tcnsion of liquids decreases with increasing temperature becausc t.he faster moving molecules of a hot liquid aren't bound togcr.her as strongly as are those in a cooler liquid. In addition, certain ingredients called surfactants decrease surface tension when added r.o liquids. For example, soap or detergent decreases the surface t.ension of water, making it easier [or soapy \vater to penetrate the cracks and crevices OfyOUT clothes to clean thcm beller than plain water does. A similar effect occurs in the lungs. The surface tissue of the air sacs in r.he lungs contains a fluid that has a surface t.ension ofaboul 0.050 N/m. A liquid with a surface tension this high would make it very difficult [or the lungs r.o expand during inhalation. However, as the area of' the lungs increases with in halation, the body secretes into the tissue a substance 1.hat gradually reduces the surface tension of" the liquid. At full expansion, the surface tension ot't.he lung fluid can drop to as
APPLICATION
II
Air Sac Surface Tension
low as 0.005 N/m.
EXAMPLE 9.16 Goal
Walking on Water
Apply the surface tension equation.
Problem Many insects can Iilerally walk on water, using surface tension for their support. To show this is feasible, assume the insect's "foot" is spherical. \'Vhen the insect steps onto the water I;vith all six legs, a depression is formed in the wat.er around each fool., as shown in Figure 9.42a. The surface tension of t.he water produces upward forces on the water that tend 1.0 restore the water surface to its normally Oat shape. If the insect's mass is 2.0 X 10-5 kg and t.he radius of each foot is 1.5 X 10- 4 m, find the angle O.
/
' 0 1 1
1 1
e
:
Ii
Strategy Find an expression [or the magnitude of the net (a) (b) force F directed tangentially to the depressed part of the water surface, and obtain the part that is acting vertically, FIGURE 9.42 (Example Y.16) (a) One.:: 1001 uran inseCl restin~ on in opposition to the downward force of gravity. Assume the Ihl: surface ufwaler. (b) This .....ater strider resting: on the ~Ilrfacc ora lake rClnainsoli the sllrl"ace, ralher than sinking, because an Ilpwan! radius of depression is the same as lhe radius of the insecr.'s surface tension force aets on ..:
,.
..,
Solution Start wirh the surface tension equation: Focus on one circular foot, substituting L = by cos to get the vertical component /'·v:
e
F= yL 21TT.
Multiply
F" = y(27TT) cos 0
\-"rite Newton's second law for the insect's one foot., which supports one-sixth of the insect's weight:
L F- 1':1 -
Solve Cor cos 0 and substitut.e:
(I)
F!{rav - 0
y(27Tr) cos II - ;mg = 0
mg
cosO = - - 127TTY
(2.0 X 10- 0 kg)(9.80 m/s') 127T(1.5 X 10 4 m)(0.073 N/m)
_ ' - c -_ _-.,-"-'-,'-c-_'-c---'---c
=
0 47 .
302
Chapter 9
Solids and Fluids
Take the inverse cosine of both sides to find the angle (}:
e=
CO,-I
(0.47) =
tion for
f)
0
...................
Itr... • • • • • • • • • •
Remarks
62
~
If the weight of the insect were great enough to make the riglu side of Equation (1) greater than 1, a soluwould be impossible because the cosine of an angle can never be g_-calCI" than j. In this circumstance the
insect would sink. QUESTION 9.16 True or False: ""'arm water gives more SUppOfl to walking insects than cold waleI'. EXERCISE 9.16
A typical sewing needle floats on \vater when its long dimension is parallel to the water's surf'ace. Estimate the needle's maximum possible mass, assuming the needle is two inches long. Hint: The cosine of an angle is never larger than 1. Answer
0.8 g
The Surface af Liquid [f you have ever closely examined the surface of water in a glass container, you may have noticed that the surface of the liquid near the walls of the glass curves upwards as you move from the center to the edge, as shown in Figure 9.43a. However, if mercury is placed in a glass container, the mercury surface curves downwards, as in Figure 9.43b. These surface effects can be explained by considering the forces between molecules. [n particular, we must consider the forces that the molecules of the liquid exert on one another and lhe forces that the molecules of the glass surface exert all those of the liquid. ]n general terms, forces between like molecules, such as the forces between waler molecules, are called cohesive forces, and forces between unlike molecules, such as those exerted b}' glass on water, are called adhesive forces. \-Vater tends to cling to the walls of the glass because the adhesive forces between the molecules ofwatcr and the glass molecules are gntaterthan the cohesive forces between the waler molecules. In eflect, the water molecules cling (Q the surface of the glass rather than fall back into the bulk of the liquid, When this condition prevails, the liquid is said to "wet" the glass surface. The surface of the mercury curves downward near the walls of the container because the cohesive forces between the mercury atoms are gremer than lhe adhesive forces between mercury and glass. A rnercury atom near the surface is pulled morc strongly toward other mercury atoms than toward the glass surface, so mercury doesn't wet the glass surface. The angle 4> betv./een the solld surface and a line drawn tangent to the liquid at the surf"ce is called the contact angle (Fig. 9.44). The angle is less than 90 0 for
¢, "
,,
'\
\,Vater
,, Glas...; ,
w
~
W
FIGURE 9.43 A liquid ill C0l11aCl wilh
9.9
WaLer drop
, "'----.. \ 4'
~
\,Valt·" drop
\
Surface Tension, Capillary Action, and Viscous Fluid Flow
any substance in which adhesive forces are stronger than cohesive forces and greater than 90° if cohesive forces predominate. For example, ira drop of\.. . atcr is placed on paraffin, the contact angle is approximately 107° (Fig. 9.44a). II" certain chemicals, caJled wetting agems or dctergents, are addcd 1.0 the \\later, the contact angle becomes less than 90°, as shown in Figure 9A4b. The addition of such substances 1.0 water ensures that the WaleI' makes intimate contact with a surface and penctrates it. For this reason, detergents are added to water to wash clothes or dishes. On the other hand, it is sometimes necessary to keep water from making intimate contact with a surface, as in \-val.erproof clothing, \·"here a situation somev"'lat the reverse of that sho\\ln in Figllre 9.44 is called for. T'he clothing is sprayed with a watcrproofing agcnt, which changes cf) from less than 90° to greater than 90°. The waleI' beads up on the surface and doesn't easily penetrate the clothing.
APPLICATION Detergents and Waterproofing Agents
Capillary Action In capillary tubes t.he diameter of the opening is very small, on the order ofa hundredth of a centirneter. In fact, the \-vord mjJiUm) means "hairlike." If such a tuhe is inserted into a fluid for which adhesive forces dominale over cohesive forces, the liquid rises into the tube, as shown in Figure 9.45. The rising of the liquid in the tube can be explained in terms or the shape of the liquid's surl"ace and surl"ace tension effects. At the point of" contact. between liquid and solid, the upward force 01" Sll rface tension is eli rCCled as shown in the Figure. From Equat ion 9.19, lhe ITlagnitudc of this force is
F = yL
~
T
y(27TT)
('r\le use L = 217r here because the liquid is in contact with the surface of the t.ube at all points around its circumference.) The vertical component of this force due 1.0 surface tension is F" ~ y(27Tr) (cas cp)
[9.20]
FIGURE 9.45 /\ liquid rises in a Ilarra\\' lube because orc.tpillar), anion, a result or surface tension and adhesive forces.
For the liquid in the capillary tube to be in equilibrium, this upwLlrd force must be equal to the weight of the cylinder 01" water of height II. inside the capillary tube. The weight of this water is tv = l\!Tg = pVg = PffTTT'1./t
Equating J'~, in Equat.ion 9.20 to lor equilibrium), we have
11)
[9.21 ]
in Equation 9.2J (applying Newton's second law
y(27T1-)(cas ¢) = pg7T1"h
Solving for hgives the height to which water is drawn into the tube: 2y It = -cas¢
pgr
[9.22]
If a capillary tubc is inserted into a liquid in which cohesive forces dominate over adhesive forces, the level of the liquid in the capillar), tube \vill be below the surface of the surrounding fluid, as shown in Figure 9.46. An analysis similar to the above would show that the distance It to the depressed surface is given by Equation 9.22.
FIGURE 9.46 \\'hel1 cohesive Jorces betwccn molecules ora liquid exceed adhesive..: forces, lhe level orlhe liquid inLhe capillary lube is below tJ1C surface of lhe surrouJlding fluid.
Chapter 9
304
Solids and Fluids
Capillary tubes arc often llsed to draw small samples of blood from a needle prick in the skin. Capillary aCLion must also be considered in the construction
or concrete-block buildings because waleI' seepage through capillary pores in the blocks or the mortar Illay cause damage to the inside of the bujlding-. To prevent u.(lI11ag~, tile blucks ~In:: usually cuated with a lvatcrproofillg agent either outside or inside the building. \VaLer seepage through a wall is an undesirable effect of capillary action, but there are many useful effects. Plants depend on capillary action La lranspon waleI' and nutrients, and sponges and paper LOwels use capillary action to absorb spilled nuids.
such
EXAMPLE 9.17 Goal
Risin Water
Apply surface tension to capillary action.
Problem Find the height to which water would rise in a capillary lUbe with a radius equal to 5.0 X 10-[' m. Assume the contact angle betwecn the water and the material of the tube is small enough t.o be considered zero. Strategy
This problem requires substituting values into Equation 9.22. .
~
Solution Substitute the known values into Equation 9.22:
2(0.073 N/m) (1.00 X 10" kg/m')(9.80 m/,2)(5.0 X 10 ,. m)
...............................................
0.30 m
..........................
~
QUESTION 9.17 Based on the result of this cCllculation, is capillary action likely to be the sole mechanism of"wat.er and nutrient transpon in plants? Explain. EXERCISE 9.17 Supposc ethyl alcohol ri!oics 0.250 m jn a thin lUbe. Estimate the radius of the tube, assuming the contact anglc is approximately zero. Answer
2.23 X 10-; m
Viscous Fluid Flow
(al
(h)
FIGURE 9.47
(a) The
partide~ ill
all ideal (nomiscolIs) fluid allmovc
through the pipe willl lhe same
\'docilY. (b) In a viscous fluid. lhe \'e1ocilY orlhe fluid particles is lera at 1he surface orthe pipe and increases to a maximum value althe cenler of lhe pipe.
IL is considerably easier Lo pour water Ollt of a container than to pour honey. This is because honey has <1 hig'her viscosity than water. In a general sense, viscosity refers to the internal friction of a fluid. It's very difficult [or layers or a viscous Auid to slide past onc another. Likewise, it's difficult for one solid surface to slide past another if" there is a highly viscous rJuid, such as soft tar, between them. ',Vhen an ideal (nonviscous) fluid flows through a pipe, the fluid layers slide past one another with no resistance. If the pipe has a uniform cross section, each layer has the same velocity, as shown in Figure 9.47<1. rn contrast, the layers of a viscous fluid have different velocities, as Figure 9.47b indicates. The fluid has the greatest velocity at the center of the pipe, 'whercas the layer next to the wall doesn't move becausc of adhesive forces betwcen molecules (lnd the wall surface. To better understand the concept of' viscosity, consider a layer of liquid between t\Vo solid surfaces, as in Figure 9.48. The lower surface is fixed in position, and the top surface moves to the right with a velocity under the action of an external force F. Because of this motion, a portion of the liquid is distorted from its original shape, ABeD, at one instant to the shape AEFD a mon1cnt later. The force requircd Lo move the upper plate and diSLOfll.he liquid is proportional to both the area A in contact \'\1th the Allid and the specd v of the Auid. Further, the force is
v
9.9
Surface Tension, Capillary Action, and Viscous Fluid Flow
TABLE 9.5 Viscosities of Various Fluids 'f ("e)
Water \-V
20
1.0
X
100
O.~
X 10 :<
B
2.7 X 10 :<
Glycerin
20
I 500 X 10
:<
10-wl. mowr oil
~o
250 X 10
:<
d I
/
.4.
inversely proponional to the distance d between the two plaLes. We can express these proportion ali ties as Frx. !Iv/d. The force required Lo move the upper plate at a fixed speed v is therefore Av d
c
,i 1 I 1
;
1/ I;
;
;
--.... ~
f)
FIGURE 9.48 A la}'ernfliquid betwcen twO solid surfaces in which tl\l' lower surfacc is fixL'd and Ihe LIpper surface moves LO Ihe right with a v(:locily v.
[9.23]
1'="1-
/
E
-t! / J f/
10 :\
~7
Whole blood
H
Viscosity 11 (N ·s/m 2 )
Fluid
305
where TJ (the lowercase Greek letter ftO.) is the coefficient of viscosity of the fluid. The 51 units of viscosity are N· 51m 2 . The units of viscosity in many reference sources are often expressed in dyne· s/C1n 2 , called 1 poise, in honor of the French scientistJ. L. Poiseuille (1799-1869). The relationship bet\vcen the Sl unit of viscosity and the poise is I poise = 10- 1 N . 5/01 2
~
Coefficient of viscosity
~
Poiseuille's law
[9.24]
= 10 2 poise. The coefficients of viscosit)' for some common substances are listed in Table 9.5.
Small viscosities are often expressed in centipoise (cp), where I cp
Poiseuille's Law Figure 9.49 shows a section of a tube of length J, and radius H cOlllaining a fluid under a pressure P, at the left end and a pressure P2 at the right. Because of this pressure difference, the fluid nmvs through the tube. The.: rate of flow (volume pCI' unit time) depends on the pressure difference (PI - P'1), the dimcnsions of thc lube, and the viscosity of thc fluid. The result, knm·vn as Poiseuille's law, is I\. V
Rate of Omo\' = = 1\./
7TR"(P, - R,) 8T/L
•
[9.25]
\vhere TJ is the coefficient of viscosity of (he fluid. \·Ve won't attempt to derive this cquation here because the methods of integral calculus are rcqllired. Howcver, it is reasonable that the rate of flow should increase if the pressure difference across the tube or the tube radius increases. Likewise, the flow rate should decrease il"the viscosity of the fluid or the length of the LUbe increases. So the presence of Rand the pressure difference in the numerator of Equation 9.25 and of I~ and TJ in the denominator make sense. From Poiseuille's law, we see that. in order to maint.ain a constant flow rate, the pressure difference across the tube has to increase if the viscosity of the fluid incre~lses. This fact is imponant in understanding the f1m.. , of blood throllgh the circulatory system. The viscosity of blood incre
R)
I
, ,
I·
P,
-.-v_----I-
P2
·1
FIGURE 9.49 Velocity profile ofa fluid llowing Ihrouj.{h a uniform pipe of circular cross seclion. The rille or flow is given by Poiseuille's I,m'. NOll' thallhe rJuid veJocil}' is greatcsl at the middle of I he pipe.
Chapter 9
306
EXAMPLE 9.18 Gool
Solids and Fluids
A Blood Transfusion
Apply Poiseuille's law.
Problem A patient receives a blood transFusion through a needle of radius 0.20 mm and IcnRlh 2.0 em. The density of blood is I 050 kg/m:l . The bottle supplying the blood is 0.50 III above the patient's arm. What. is the rate of flow through the needle? Strategy Find the pressure difference between the level of the blood and the patient's arm. Substitute into Poiseuille's law, lIsing the value for the viscosil:)' of whole blood in Table 9.5. r" , , '""'lI
Solution 1', - P2 = pgh = (I 050 kg/m') (9.80 m/s 2 ) (0.50 m)
Calculate the pressure difference:
~
Substitute into Poiscuille's law:
5. 15 X 10' Pa
tl V
71' K ' ( 1', - 1',)
tll
8TJL 71'(2.0 X 10- 4 m)·'(5.15 X 10 3 Pal 8(2.7 X 10 'N·s/m')(2.0 X 10 'm) =
~
6.0 X lO-R m'/s ~
_
Remarks Compare this LO the volume flow rate in the absence of any viscosil}'- Using Bernoulli's equation, the calculated volume flow rate is approximately rive Limes as great. As expected, viscosity greatl}' reduces Dow rate. QUESTION 9.18 If the radius ora lube is doubled, by \vhat hlcLor will the flow raLe change for a viscolls fluid? EXERCISE 9.18 A pipe carrying water from a tank 20.0 m L
Answer
0.118
ITl
Reynolds Number At sufficiently high velocities, fluid now changes [rom simple streamline flow to wrbulent flow, characterized by a highly irregular motion or the nuid. Experimentally, the onset o[ lUrbulcnce in a tube is determined by a dimensionless factor called the Reynolds number. RN, given by Reynolds
p1Jd IIN=-
number ~
TJ
[9.26]
where p is the density of the nuid, v is the average speed of the fluid along the direction of now, d is the diameter o[ the lUbe, and TJ is the viscosity of the Ouie!. If RN is below about 2 000, U1C now of fluid through a tube is streamline; turbulence occurs if RN is above 3000. In the region betwecn 2 000 and 3 000, the flow is unstable, meaning thaL the fluid can move in sU'c
EXAMPLE 9.19 Goal
Turbulent Flow of Blood
i'l
Use thc Reynolds number to dctcrminc a speed associatcd with the onset of turbulence.
Problem
Determine the speed at which blood flowing through an artery of diameter 0.20 em wiJl become lurbulent.
Strategy The solution requires only the substitution ofvalucs into Equation 9.26 giving the Reynolds number and then salvi ng it for the speed TJ.
9.10 Transport Phenomena
r-
...............................,
.
Solution Solve EquaLion 9.26 for v, and substitute the viscosity and density of blood from Example 9.18, the diameter d of the artery, and a Reynolds number of" ~.OO X 10~: 110.. • •
(2.7 X 1O-:l N·s/m')(3.00 X 1O:l) T/(t~) v=---= (1.05 X 1O:l kg/m")(0.20 X 10 'm) lui 11
=
3.9 m/s ............
__ • • • • • • • •
Remark
307
The exercise shows I hal rapid ingestion of soda through a straw may create a turbulent state.
QUESTION 9.19 True or False: If the viscosit.y of a fluid flowing through a lUbe is increased, the speed associated with the onset of turbulence decreases. EXERCISE 9.19 Determine the speed vat which water at 20°C sucked up a straw would become turbulent. The straw has a diameter
of 0.006 0 m. Answer
9.10
/
v = 0.50 m/s
TRANSPORT PHENOMENA
',"hen a fluid flows through a tube, the basic mechanism that produces the flow is a difference in pressure across the ends of the tube. This pressure difference is responsible [or the transport of a mass of fluid from one location to another. The fluid may also move from place to place because of a second mechanism-onc that depends on a diffcrence in rona"ntmtion between [\\10 points in the fluid, as opposed to a pressure difference. When the concentration (the number of molecules per unit volume) is hig·hcr aL one location than at anothcr, molecules 'will flow from the point where the concentration is high to the point where it is lowcr. The two fundamental processes involved in nllid transport resulting from concent1-ation differences arc Gilled diffusion and osmosis.
Diffusion In a diffusion process, molecules move from a region where their concentration is high to a region where their concentration is lower. To understand why difTusian occurs, consider Figure 9.::'0, \vhich depicts a container in which a high concentration of molecules has been illtroduced inLO the left side. The dashed line in the figure represcllts an imaginary barrier separating the two regions. Because the molecules arc moving with high speeds in random directions, many or lhcm will cross the imaginary barrier moving from left to right. Very few molecules will pass through moving frolll right to left, simply because there are very few of thcm 011 the right side of the comainer at any instant. As a result, there 'will ah\lays be a net movement [rom the region with many molecules to the region with fewer molecules. For this reason, lhe concentration on the left side of the container will decrease, and that on the right side will increase with time. Once a concentration equilibrium has been reached, there will be no net movement across the crosssectional area: The rate of movement of molecules from left to right will equal the rate [rom right to len. The basic equation lor diffusion is Fick's law, mass aM Diffusion rate = - - = - - = DA time
at
(C' -
L
19.27J
where D is a constant of proportionality. The left side of this equation is called the diffusion roll! and is a measure of the mass being transported per unit timc. The equation says that the rale of dillus ion is proportional to the cross-sectional area A and to the change in concentration per unit distance, (C2 - C 1) / L, which is called
FIGURE 9.50 When lhe conCClllralion of gas molecules 011 lhe lefl side of the conlainer exceeds the COllcCIl~ lratioll on the right side, lhere will be "nel motion (diffusion) of molecules from IeI'I 1.0 rig:ht.
~
Fick's low
308
Chopter 9
Solids and Fluids
TABLE 9.6 Diffusion Coefficients of Various Substances at 20"C Substance
D (m'!s)
6.4 X 10-'
Oxygen through
10- 11
Oxygen through tissue
1X
Oxygen til rough water
1 X 10- 9
Sucrose through water
:1 X 10- 10
Hemoglobin through water
76 X 1O- 11
APPLlCA"'T.:..:IO""N-'--
1'1
Effect of Osmosis on Living Cells
_
the conant-ration gradient. The concentrations C J and C':! are measured in kilograms per cubic meter. The proportionalitY' constant D is called the diffusion coefficient and has units of square meters per second. Table 9.6 list') diffusion coefficients for a few subslances.
The Size of Cells and Osmosis Diffusion through cell membranes is vital in carrying oxygen to the cells of the body and in removing carbon dioxide and other \vaste products from them. Cells require oxygen for those metabolic processes in which substances are either synthesized or broken down. In such processes, the cell uses up oxygen and produces carbon dioxide as a by-product. A fresh supply of oxygen diffuses from the blood, where its concentration is high, into the cell, where its concentration is low. Likewise, carbon dioxide diffuses from the cell into the blood, where it is in lower concentration. Water, ions, and other nutrients also pass into and out of cells by diffusion. A cell can function properly ani)' if it can transport nutrients and waste products rapidly across the cell membrane. The surface area of the cell should be large enough so that the exposed membrane area can exchange materials effectively while tbe volume should be small enough so thm materials can reach or leave particular locations rapidly. This requires a large surface-area-to-volume ratio. Model a cell as a cube, each side with length L. The total surface area is 6L 2 and the volume is L3. The surface area to volume is then surface area
6L2
6
volume
C
L
Because L is in the denominator, a smaller L means a larger ratio. This shows that the smaller the size of a body, the more efficiently it can transport nutrients and v,:aste products across the cell membrane. Cells range in size from a millionth of a meter to several millionths, so a good estimate of a typical cell's surface-to-volume ratio is 10 6 . The diffusion of material through a membrane is partially determined by the size of the pores (holes) in the membrane wall. Small molecules, such as water, may pass through the pores easily, ,,,,hile larger 1110lecules, such as sugar, may pass through only with difficulty or not at all. A membrane that allows passage ofsomc molecules but not others is called a selectively permeable membrane. Osmosis is the diffusion of water across a selectively permeable membrane from a high water concentration to a low water concentration. As in the case of diffusion, osmosis continues until the concentrations on the two sides of the membrane are equal. To understand the effect of osmosis on living cells, consider a particular cell in the body that contains a sugar concentration of 1%. (A 1% solution is 1. g of sugar dissolved in enough water to make 100 ml of solution; "ml" is the abbreviation for milliliters, so 10-3 L = ] cm 3 .) Assume this cell is immersed in a.l5% sugar solution (.15 g of sugar dissolved in enough water to make 100 ml). Compared to the 1.% solution, there are five times as many sugar molecules per unit volume in the 5% sugar solution, so there must be fewer water molecules. Accordingly, '.. . ater will diffuse from inside the cell, where its concentration is higher, across the cell membrane to the outside solution, where the concentration of water is lower. This loss of water from the cell ""ould cause it to shrink and perhaps become damaged through dehydration. If the concentrations were reversed, watcr would diffuse into the cell, causing it to swell and perhaps burst. If solutions are introduced into the body intravenously, care must be taken to ensure that they don't disturb the osmotic balance of the body, else cell damage can occur. For example, if a 9% saline solution surrounds a red blood cell, the cell will shrink. By contrast, if the solution is about 1%, the cell will eventually burst. In the body, blood js cleansed of impurities by osmosis as it -flows through the kidneys. (See Fig. 9.51a.) Arterial blood first passes through a bundle of capillaries
9.10
Transport Phenomena
309
FIGURE 9.51 (a) Diagram ora sing-h' nephroll in the hlllllOill cxcrelory system. (b) All artificial kidllCY. (:lomcru!lIs
-
)1
Collecting duct
To
COlllpressed Fn.::-.h Con.~wnL Used COt ,md :Iir dial)'Sale temper:llure di:dysalc h
rcnal pelvis
(a)
known as a glomerulus, where most of the waste products and some essential salts and minerals arc removed. From the glomerulus, a Ilarrow lUbe emerges thal is in intimate contact with other capillaries throughollt its length. As blood passes tIl rough tIle t llbules, most of thc essential elemcnts are relUrned 10 it; waste prodUCL~ are not allowed LO reenter and are eventually removed in urine. Ir the kidneys fail, an artificial kidney or a dialysis machine can filler the blood. Figure 951 b shows how this is done. Blood frolll an artery in the arm is mixed with heparin, a blood thinner, and allowed to pass through a tube covered \\Tith a semipermeable membrane. The tubing is immersed in a bath or a dialysate fluid with the same chcmical composition as purified blood. Wastc products from thc blood cnter the dialysate by diffusion lhrough the mernbranc. The filtercd blood is thcn returned LO a vein.
APPLICATION "
Kidney Function and Dialysis
Motion through a Viscous Medium ''''hen an ol~iect ralls through air, its motion is impeded by the force of air rcsistancc. Tn general, this force is dependelll on thc shape or the Calling object and un its velocity. The lorce of air resistance acts 011 all falling ol:~jects, but the exact dctails of the motion can be calculated only for a feh' cases in which the object h"1S a simple sh~lpe, such as a sphere. In this scction we will examine the motion of a tiny spherical objcct f~_dling slowly through a viscous medium. In 184.5 a scientist namcd George Stokes found that the magnitude of the resistive force on a very slllall spherical object of radius r falling slmvly through a fluid of viscosity 1] with spced v is given by
[9.28] This equation, called Stokes's law, has many important applications. For example, it describes the sedimentation of particulate matter in blood samples. It was used by Robert Millikan (IR86-19!)~) Lo calculal.e the r..ldills of charged oil dropleL'i failing through air. From this, Millikan was ultimatcly able 1.0 del.crmine the chargc on the eleclron, and was awarded the Nobel prize in 1923 for his pioneering work on elemental charges. As a sphere falls through a viscous medium, three forces act on il, as shown in Figure 9.52: p the force of friction; ii, thc buoyant force of the nuid; and the lorcc of gravity acting on the sphcre. The magnitude ofw is given by
w,
F
lI!
= pgV = pg (
34 7fT"01)
FIGURE 9.52 A sphere falling lllrOllgh a \·l:-.ClllIS medilltll. The forcc1'o
B.
310
Chapter 9
Solids and Fluids
where p is the dcnsil)' of the sphere and 17TT:1 is its volume. According to Archimedes's principle, the magniLUde or the buoyant force is equallo the weight 01" the fluid displaced by the sphere,
where PJ is the densit)' of the fluid. Al the instant the sphere begins to Fall, the force of friction is zero because the speed of the sphere is zero. As the sphere accelerates, its speed increases and so does Fr. Finally, at a speed called the terminal speed v" the net force goes to zero. This occurs 'when the net upward force balances the downward force of gravity. ThclTfore, the sphere reaches lcrrninal speed ,,,",hen ~
1~+B=w
or
When this equaLion is solved for Terminal speed
VI'
we get 2r:!g
-+
lI,
= - - (p - p)
97)
[9.29]
Sedimentation and Centrifugation If an object isn't spherical, we can Slill use the basic approach just described LO dClcrmine its terminal speed. The only difference is that we can't usc Stokes's law for the resistive forcc. Instead, "...e assume that the resistive force has a magnitude given by F r = kv, where II. is a coelTicient that must be determined experimentally. As discussed previollsly, the object reaches its terminal speed when the downward force of gravity is balanced by the nClupward force, or
w=B+Fr
[9.30]
where B = PJgV is the buoyant force. The volullle Vol" the displaced r1uid is related La the density p of the falling o~jccl by V= 'In/p. Hence, we can express the buoyanI. force as
PI B= -mg P We substitute this expression for 13 and F,. condition):
PI
ml{ = -
P
=
mg
kv/ into Equation 9.30 (terminal speed
+
IW I
or v = II!g(l _ PI) I k P
[9.31]
The terminal speed for panicles in biological samples is usually quite small. For example. the terminal speed for blood cells falling through plasma is about 5 cm/h in the gravitational field of the Earth. The terminal speeds [or the molecules that make up a cell arc many orders of magnitude smaller than this because of their much smaller mass. The speed at which materials fall through a fluid is called the sedimentation rate and is important in clinical analysis. The sedimentation rate in a fluid can be increased by increasing the effective acceleration g th"t appears in Equation 9.31. A fluid containing various biologi-
Summary
cal molecules is placed in a centrifuge and whirled at very high angular speeds (Fig. 9.53). Under these conditions, the panicles gain a large radial acceleration a r = '(I2/ r = w2.,. that is much greater than the free-fall acceleration, so we can replace gin Equmioll 9.31 by w~.,.and obtain
311
w
[9.32) This equation indicates thatlhe sedimentmion rate is enormously speeded up in a centrifuge (w 2r» K) and that lhose panicles with the greatest mass will have the largest tcrminal specd. Consequently the most massive particles will settle out on the buttolll ofa testlllbe first. FIGURE 9.53 a ccmrifllg:c
Silllplifkd diaJ..j:rarll of
(top view).
SUMMARY 9.1
States of Matter
9.4
Malter is normally classi fled as bei ng in one of t.h ree states: solid, liquid, or gaseolls. The fourth state oflnatter is called (l plasma, which consists ofa neulral syst.em of charged particles interacting electromagnetically.
9.2
The Deformation of Solids
The clastic properties ofa solid can be described using the concepts of" stress and strain. Stress is related to lhe force per unit area producing a deformation; strain is a measure of the amount of deformation. Stress is proportional lO strain, and the constant of proponionality is the elastic modulus: Stress
= clastic modulus
X strain
[9.1]
Three COlllmon types or deformation are (I) lhe ITS1Slance ()f a solid to elongation or compression, characterized by Young's modulus Y; (2) the l'csistance to displacement or rhe h\Ces of a solid sliding past each oUler, characterized by the shear mocltilus S; and (3) lhe resistance of a solid OJ" liquid to a change in volume, characterized by the bulk modulus B. All three types or deformatioll ohey laws similar to Ilookc's law for springs. Solving problems is usually a matter of identif)~ng the given physical variables and solving- for the ullknown variable.
9.3
Variation of Pressure with Depth
The pressure in an incompressible fluid val-ies with depth It according to the expression p
=
Po + pgh
[9.11]
where PI) is atmospheric jJl"essllre (1.013 X 10-') Pa) and pis the density of the fluicl. Pascal's principle states that when pressure is <:Ipplied to an enclosecl fluid, the pressure is transmitted undiminished to every p()int of the fluid and to the walls of 1he containing vessel.
9.6 Buoyant Farces and Archimedes' Principle 'Alhen an ol~jecl is partially or fully submerged in a fluid, the fluid exerts an upward force, called the buoyant force, on the objecl. This force is, in fact, just the net difference in pressure between the top and boltom of the objecl. It can be shown thaI I he magnitude of lhe blloyant force B is equal to the weight or the nuid displaced by the object. 01" [9.12b] Equation 9.12b is known as Archimedes' principle, Solving a buoyancy problem usually involves putting the buoyant force into Newton's second law and then proceeding- as in ChaplCl" 4.
Density and Pressure
The density p of a substance of uniform composition is its mass pCI' un it vol ume - kiloglams per cubic meter (kg/ m:·\) in the SI sysl.em: M V
19.6]
The pressure P in a fluid, measured in pascab (Pa), is the fOl-ce per unit area that the fluid exerts on an object immersed in it:
p-
F
[9.7]
9.7
Fluids in Motion
Certain aspects of a fluid in motion can be understood by assuming t.he Ottid is nonviscolls and incompressible and that ilS motion is in a steady statc with 110 turbulence: 1. The rlO\.... rate through the pipe is a constanl., which is equivalenl to stating that the product of the cross· sectional area A and the speed 11 at any point is COIlstant. At any two points, therefore, we have
[9.151 This relation IS referred to as the equation of continuity.
312
Chapter 9
Solids and Fluids
2. The slim of the pressure, the kinetic energy per unit volume, and the potential energy per unit volurnc is the same at any Iwo points along- a streamline:
problems with the work-enerb1J' theorem, whereby
[,'i0
poinL<; arc chosen, one point where a quantit), is unknown
and anouler where all quantities are known. Equation 9.16 is then solved for the unknown quantit)'_
[9.161 Equation 9.16 is known as Bernoulli's equation. Solving problems with Bernoulli's equation is similar to solving
FOR ADDITIONAL STUDENT RESOURCES. GO TO WWW.SERWAYPHYSICS.COM
. -"._._. MULTIPLE·CHOICE QUESTIONS I. ,.vhat is the mass or a sol id gold recta ngula r bar that has dimensions of 4.50 CIll X ] 1.0 em X 26.0 em? (a) 24.8 kg (b) 40.6 kg (c) 1l.4 kg (d) 33.2 kg (e) 19.5 kg
La the weight of the boat, (c) less than the weight of the boat, (d) equal to the weight of the displaced waLeI', or (e) equal to the buoyant force 011 the boat?
2. A 66.0-kg man lies on his back on a bed of' nails, with I 208 of the nails in contact with his body. The end of each nail has area 1.00 X 10- 6 m 2 . "Vhat average pressure is exerted by one nail on the man's body? (a) 2.21 X 10' Pa (b) 3.09 X 10' Pa (c) 1.65 X 10 6 Pa (d) 5.30 X 10' Pa (e) 4.11 X 10' Pa
10. Three vessels of different shapes arc ftlled 1.0 the same level with water as in Figure MCQ9.IO. The area orthe base is the same for all three vessels. \""hich of the following statements is valid? (a) The pressure at the top surface of vessel A is greatest because it has the largest surface area. (b) The pressure at. the bottom of vessel A is greatcst because it contains the most water. (c) The pressure at the bOltom of each vessel is the same. (d) The Loree on the bottom of each vesscl is not the same. (e) At a given depdl below the surface of each vessel, the pressure on the side of vessel A is greatest because of its slope.
3. A hydraulic jack has an inpLH piston of area O.O~O m 2 and an omput piston of area 0.70 m 2 . How much force on the input piston is required to lift a car weighing 1.2 X lO' N? (a) 42:-; (b) 68 N (c) 86 N (d) 110 N (e) J30 N 4. A lead bullet is placed in a pool of mercury. What fractional part of the volume of the bullet is submerged? (a) IlA55 (b) 0.622 (c) 0.71'1 (d) 0.831 (e) 0.930
\\
5. Whal is the pressure at the botwm of Loch Ness, ,.. . hich is as much as 754ft deep? (The surface of the lake is only 15.8 m above sea level; hence, the pressure there can be taken to he 1.013 X 10' Pa.) (a) 1.52 X 10' Pa (b) 2.74 X 10' Pa (e) 2.35 X 10 6 Pa (d) 7.01 X 10' Pa (e) 3.15 X \0' Pa 6. Hurricane winds or 95 mi/h are blm....ing over the flat mol' ofa well-sealed house. \O\'hat is the difference in air pressure between the insidc and outside of the house? (a) 1.2 X 10" Pa (b) 2.4 X 10" Pa (c) 3.4 X 10 3 Pa (d) 4.0 X 10" Pa (e) 5.3 X 10" Pa 7. A horizontal pipe narrows from a radius of 0.250 III to 0.100 m. If the speed of the water in the pipe is 1.00 m/s in the larger-radius pipe, what is the speed in the smaller pipe? (a) 4.50 m/s (b) 2.50 m/s (c) 3.75 m/s (d) 6.25 m/s (e) 5.13 m/s 8. Bernoulli's equation can be Llsed to explain, in part, which of the following phenomena? (a) the lift on an airplane wing in flight (b) lhe curve ofa spinning baseball (c) vascular nuuer (d) reduction in pressure of moving fluids (e) all these answers 9. A boaL develops a kak and, after its pas~engers (Ire rescued, eventually sinks to the bottom of a lake. \O\'hen the boat is at the bottom, is the normal force on the boat (a) great.er than the weight. or the boat, (b) equal
(a)
(b)
Ie)
FIGURE MCQ9.10
11. An ideal fluid flows through a horizontal pipe having a diameter that varies along its length. Does the sum of the pressure and kineric energy per unit volume at different sections of the pipe (a) decrease as the pipe diameter increases, (b) increase as the pipe diameter increases, (c) increase as the pipe diameler decreases, (d) decrease as the pipe diameter decreases, or (e) remain the same as the pipe diameter changes. 12. A hose is pointed strLlight up, w'ilh '.. . ater flowing from it at a steady volume flow rate and reaching a maximum height of II. Neglecting air resistance, which of t.he following acljusllllents 1.0 the nozzle will result in the water re~lchillg a height of 4h? (a) Decrease the area of the opening by a factor of 16. (b) Decrease the area by a factor of 8. (c) Decrease the area by a factor of 4. (d) Decrease the area by a factor of 2. (e) Give up because the waler cannot reach a heighl of4h.
Conceptual Questions
313
CONCEPTUAL QUESTIONS I. Why do baseball home run hiuers like to play in Denver, but curvcball pitchers clo nOl?
2. The density or air is 1.3 kg/m:~ at sea level. From your knowledge of air pressure at ground level, estimate the height of the atmosphere. As a simplifying assumplioll. lake lhe allnospherc to be of" lin i for III dcnsil}' up La some hcigllL, after which the density rapidly falls to zero. (In rcality, the density of the atillosphere decreases as we go up.) (This question is courtesy of Edward F. Redish. For more questions of this type, St:C hu p://W\.... w.physics. II mel .edu/perg/.) 3. A woman wearing high-heeled shoes is invited ililO a home in which the kitchen has vinyl floor cO\'cring. Why should the homeowner be concerned? 4. Figure CQ9.4 shows aerial views from directly above' Iwo dams. BOlh dams are equally long (the vertical dimcnsion in the diagram) and equally deep (into the page in the diagram). The dam on thc left holds back a ,"cry large lakt:.', while the dam un the right holds back a narrow river. "'fhich dam has to be built more strongly?
~] Dam
controlling exhalalion are relaxed. Under water, the body equalizes inlernal and external P,"essures. Discuss the cOlldition or tile muscles ir a person under water is breathing til rough a snorkeL Would a snorkel work in deep water? 7. Suppose a damaged ship jllst barely floats in the ocean after a hole in its hull has been scaled. h is pulled by ~l tugboat toward shore and into a rive,", heading toward a dry dock for repair. As the boat is pulled up lhe river, it sinks. Why?
8. \L.111Y people believe that a vacuum created inside a vaCUUlll cleaner callses particles of dirt to be drawn in. Actually, the dirt is pushed in. Explain. 9. A pound of Styrofoam and a pound of lead have the ~ame \\leighl. I [they are placed on a sensitive equal-arm balance, will the scales balance? 10. An ice cube is placed in a glass of\\later. ""hal happens to the level of the water as the ice melts? 11. Place two cans of soft drinks, one regular and one diet, in a container ofw
/
12. Will all ice cube noat higher in ,,'ater or in an alcoholic beverage?
FIGURE CQ9.4
13. Tornadoes and hurricanes often lift the roofs of houses. Use the Bernoulli effect to explain \-vhy. "Vh)' should yOIl keep your \vindows open under lhese conditjons?
Dalll
5. A typical silo on (1 rarlll has many bands wrapped around its perimeter, as shown in Figure CQ9.S. \'Vhy is I he spacing between successive bands smaller at I he lower portions of the silo?
14. P]-,lirie dogs live in underground burrows with al. least two entrances. The)' ventilate their burrows by building a mound over one entrance, as shown in Figure CQ9.14. This entrance is open to a stream of air when a breeze blows from any direction. A second enUance at ground level is open to almost stagnant air. How does this conslruction crecHe an airflow through the burrow?
FIGURE CQ9.5
6.11
During inhalation, the pressure in the lungs is slightly less than external pressllre and the Illuscles
FIGURE CQ9.14
Chapter 9
314
Solids ond Fluids
---------~
PROBLEMS _ _ _ a"
~_'"'"-
~
_
The Problems ror this chapler mar he assil;IH:.:d online al WchAssigll. 3 = stl-aiglnro!"ward, illlernll'di'lll.:, lhallengillg
bone) is 0.50 m long, and calculate the amount of compression this bone can withstand bclore breaking.
WebAssign
I,~,
m
=
m:J =
denoles guided problem dellotes cnh;l1lccd content problem
II = biOllll::dical applicatioll 0= denoles full solution available in
Slud,'11! Sollllimn iVlallll{/lj
S11/d)1 {;lIide
SECTION 9.1 STATES OF MATTER SECTION 9.2 THE DEFORMATION OF SOLIDS 1. If the elastic limit of steel is 5.0 X lOl' Pa, determine the minimulll diameter <J steel wire can have ifil is to SlIpport a 70-kg circlls ped(wlllcr without irs clastic limit being exceeded. 2. _ Comic-book superherues arc:.: sometimes able to punch holes through steel walls. (a) If the ultimate shear strength or steel is taken La be 2.50 X 10M Pa, what forcc is required to punch lhrough a sted plalc 2.00 cm lhick? Assumc the superhcro's fist has cross-sectional area of 1.00 X 10 2 Clll~ and is approximately circular. (b) Qllalitatively, what \\'ould happen to the superhero on delivery of the punch? \\'hat physical law applies?
8. A high-speed lifting mechanism suppOrts an 800-kg oqjecl with a steel cable thai is 25.0 III long ami 4.00 cm 2 in cross-sectional area. (a) Dctenninc the elongation of the cable. (b) By whal additional amount docs the cable increa!':>e in length if [he o~jecl is accelerated upwards at a rate of 3.0 m/s 2 ? (c) Wh
9. A child slides across a floor in a p"ir or rubber-soled shoes. The friction force acting on each foot is 20 1'\, lhe 1'001 print <.\I·ea ofeaeh foot is 14 em'l, and the thickness of the sales is 5.0 mm. Find the horiwlI1.al distance traveled by the sheared face of the sole, The shear modulus of the rubber is :to X 10 6 Pa.
10. The distortion of Earlh's crustal plates is an example of shearon a large scale. A particular crustal rock has a shear lTlOdlllus of I.!i X lOw Pa. What shear stress is involved when a IO-km layer of this rock is sheared through a distance or 5.0 m? [I[]Determine the elongation of" the rod in Figure P9, II if it is under a tension of 5.8 X 10:1 N.
3. A plank 2.00 (Ill thick and 15.0 em wiele is firmly attached La the railing of a ship by clamps so that the rest of the board extends 2.00 III horizontally over the sea below. :\ man of mass 80.0 kg is forced to stand on the vcry end. If the end of the board drops by 5.00 cm hecause of the man's weight, find the shear modulus of the wood.
0.20 em
I.-- 1.3 12.
5. For safety in climbing, a mountaineer uses a nylon rope that is 50 m long and 1.0 cm in diameter. When support~ ing a 90-kg c1imbcr, the rope elongates 1.6 m. Find its Young's modulus. 6.
i'J!
A stainlcss~steel orthodontic wire is applied 10 a lOoth, as in Figu,'c P9.6. The wire has an unslretched length 01'3.1 em and a diameter of 0.22 mm. -'I' lhe wire is stretched 0.10 llllll, find the magnitude and direction of the force on the LOath. Disregard lhe width of the tooth and
FIGURE P9.6
7.
i1
Bone has a Young's modulus of about 18 X lOY p~l. Under compression, i[ can withstand a stress of about 160 X lQi; Pa before breaking. Assume Ihat a fernill' (lhigh-
III
Copper
~.-/.I~,--- ~J) 111 --->Ill
FIGURE P9.11
[]J\r\7hen water freezes, it expands abollt 9.00%. 'What would
be the pressure increase inside your automobile engine block if the water in it frozc? The bulk modulus of ice is 2.00 X lO~t :-.11m 2 .
Aluminum
fIJI
The lOlal cross-sectional area of the load-hearing calcified portion of the two forearm bones (radius and ulna) is approximately 2.4 cm~, During (l car crash, lhe forearm is slammed ag-ainst the dashboard. The arm comes to rest from an initial speed of 80 km/h in !i.0 ms, '[the arm has an effective milSS 01"3.0 kg and bone material can withstand a maximum compressional Slress of 16 X L0 7 Pa, is lhe arm likely to wilhs[and lhe crash?
SECTION 9.3 DENSITY AND PRESSURE 13.
mJ
Suppose two worlds, each having mass M and radius N, coalesce into a single world. Due to gravitational contraction, the combined world has a radius of only ~R. What is the
14. The British gold sovereign coin is an alloy of gold and copper having a lotal mass 01'7.988 g, and is 22-karat gale!. (a) Find the mass or gold in the so\'ereign in kilograms using t.he fact [hat lhe number or k"rats = 24 X (mass of gold)/(total mass). (b) CalcuhHe the volumes of gold and copper, respectively, used to manufacture the coin, (c) C'llculate the density of the Brilisll sovereign coin.
315
Problems
15.
um
Four acrobats ofm
I.:l~ X 1O~ Pa, what IIlllst be the millimum heiglll h of the baA' in order to infH!'>c glucose into the "ein? Assume the 'ipecific gra"itv of Ill(' solution is 1.02.
Hi. A large man sits on a four-legged chair with his ((.oN off the 0001". The combined mass of the mall and chair is ~)5.0 kg. If the chail" legs arc cil"Cular and h;I\'(' a nldius of 0500 cm at the bottom, \\'hat pressure docs each kg- exert on the floor? 17.
The nllclclI~ of an awm can he modded a!'> "C'vera1 protons ;llld neutrons closely packed togel her. E;llh particle ha'i a mass of 1.67 X 10- 27 kg and radius on lhe order of 10 I,", 1lI. (a) L'se this model .md infonnalion LO estimate the densit)' of the nucleus 01" an atom. (b) Compare your re"lIlt with the density of a material such a, iron. What do )Ollr result and comparison sllg-g:est about the sl rllClli rc of malter?
alii
18. The [our lires of an automobile arc inflated to a g-auge pressure 01'2.0 X 10" Pa. Each tire hCls an arca ofO.O~ 11l1~ ill COlllact with the ground, Determine the ",eight of the Clillomohile.
[J]J I f 1.0 m'l of' concrete weighs 5.0
X lOiN, wh,tt is I h(' h('igln
of'the lallesl q-lindrieal concrete pillar Illal will 110t collapse uncleI' its OWI1 weight? The compression strengt h of concrele (the maximum pressllre th'll Gln he exerted on Ihe base of'the SII"IIClllre) is 1.7 X lOi Pa,
SECTION 9.4 VARIATION OF PRESSURE WITH DEPTH SECTION 9.5 PRESSURE MEASUREMENTS 20. The spring of the pressure gauge shown ill Figure 9.Rb ha'i a force constant of I 250 1'\/111, and Ihe pislon has a radius of 1.20 em. As the gauge is lowered inlO water, "'hat change ill depth causes the piston 10 move in b' O.iSO em?
21. Calculate the ahsolule pressure
the boltom of a fn.·shwater lal..c at a depth of27.5 m. ASMlmc tlw density of the water is 1.00 X IO~ kg/Ill:! and the air abcwe is ill a pres'illre of 101.3 kPa. (b) \""hat force is exened hy th .., water on the window of an underwater "ehicle al Ihis depth if I he window is circulal' and has a diameler 01" :l~,O elll:
22.
al
FIGURE P9.23
24, The deepe!'>t point in the ocean is in the Mariana Trench, aholll J I km deep. The pressure at the ocean noor is hug-e, ahollt 1.13 X IO~ :\'/m 2 . (a) Calculate the change in volume of 1.00 Ill' of water carried from the surface to the bottom of the Pacil-ic. (h) The densitr of water at the surfacc is 1.03 X 10 3 kg/Ill:I , Find its densit~ at the bot· tom. (c) 1<; il a good approximation 10 think or water as incompressible?
25 . ..-\ container is filled
to a depth of 20.0 em with water. On lOp of the waler nOfits a ~O,O·cm-thick layer of oil with specific gravitv 0.700. \AThal is the ahsolmc pressure at the hOI tom oflhe comainer?
26. Blaise PasGll duplicated Torricclli's harometer lIsing- a red Bordemlx wine of density 984 kg/m:'! as Ihe work· illg liquid (Fig. P9.26). What was the height 11 of the winc column for normal atmospheric pressure: Would "ou expect the vacuum abo\'e the column to he as good as for mcrcun'?
FIGURE P9.26
!27.lFigure P9.27 shows the essential parts of a hydraulic hrake system. The area of the piston in the master o')inder is 1.8 cm:'! and that ofthc piston in the brake cylinder is 6.4 cm~. The coefficient of friction between shoe and
Whct'!
drum
III
Wh..'ll you sllddenh' stand lip "ftcr l,inA" down for ,1 while. \,0111' bod" may nOI compellsate CJllickh ('l1ough for the P"cssure changes and yOH might red dio)' fn .. a momenl. If the gauge pressure of the blood at rour hean if' 1:\.:\ kPa ;'Ind your bod)" doesn't compensate.'. «I) what would Ihe pressure be ,II your head, 50.0 em ahtlVt.' rour hean? (Il) What would it he at your feel. I.:~O X IO:! nn below \'OUr' Ileart? llilll:The density or blood is I 060 kg/m~.
A collapsible plastic bag- (Figure P9.2~) COlll.lins;1 glllcosc solution.lflhe average gaUR!' pressllre in the vein i<;
h
PCflOlI
M
cylinder
123.1"
FIGURE P9.27
316
Chapter 9
Solids and Fluids
wheel drum is 050. If the wheel has <1 radius or 34 CIll, determine the frictional torque about the axle when a force of 44 N i<; exened on the brake pedal.
librium and the top of the wooden block is at the level of the water, (;-1) What is the densilY or the wood? (b) \Vhat happens to the block when the steel object is replaced by a second steel object with a mass less than 0.310 kg? \<\'hat h,tppens t.o the block when the steel ol~ect is replaced by yet a nother steel ol~icct with a mass greater I han 0.310 kg?
CD in Figure P9.28 has a diameter of O.2.tl in.; pis® has a diameter of 1.5 in. In the absence of' fr"ictioll,
28. Piston ton
determine the force weight.
F necessary
to
SIIPP0r!.
the 500lb :{4.
&I
35,
E
36.
E
FIGURE P9.28
SECTION 9.6 BUOYANT FORCES AND ARCHIMEDES' PRINCIPLE 29. A rubber ball filled with air hasa diameter 01'25.0 em and a mass 01'0.540 kg. vVhat force is required to holrlthe ball
am The average human has a density of945 kg/m 3 after inhaling and I 020 kg/Ill: l after exhaling. (a) Without making any swimming movcmelllS, what percentage of
the human body would be abO\'e (he surface in the Dead Sea (,I lake with a water densit)' of abolll 1 230 kg/rn 3) in each of these cases? (b) Gil'en that bone and Hlllscle arc denser than fat, what physical characteristics difrercnuate "sinkers" (those who tend \.0 sink in water) frorn "noaters" (those who readily float)? 31. A small ferryboat is 4.00 m wiele and 6.00 m long. When
a loaded truck pulls ontO it, the boal sinks an additional 4.00 Clll imo the river. \'Vhat is the weight of the truck? 32.
EI
A 62.0-kg s1lrvimr oCa cruise line disaster rests al.Op a block ofSt}Tofoam insulation, using it as a raft. The Styrofoam has dimensions 2.00 m X 2.00 m X 0.090 0 111. The bottom 0.024 III of the raft is submerged. (
33. _ A woodell block of\'olurne .1).24 X 10-.1 IH:\ floats in watcr. and a small steel object of mass III is pl~\Ccd on lOp o[ till: block. When '1/1 = 0.310 kg, tile system is in equi-
A spherical weal her balloon is filled with hydrogen until its radius is 3.00 m. Its total mass including the instruments it carries is 15.0 kg. (a) Find the buoyant force acting on the balloon, assuming the densit), of air is 1.29 kg/m~, (b) 'What is the net force acting on the balloon and its instruments after the balloon is released from the ground? (c) Why docs the radius of lhe balloon tend to increase as it rises to higher altitude?
A man of mass IJI := 70.0 kg and having a density of 1050 kg/mel (while holding his breath) is completely submerged in water. (a) Write Newton's second law for this situation in terms o[ the man's mass Ill, the density of water p"" Ilis \'olume \~ and g. Neglect any viscous drag of the water. (b) Substitute 'IJ/:= pliinto Newlon's second law alld solve for the acceleralion 0, canceling common factOrs. (c) Calculate the numeric value of the man's acceleration. (d) How long does it take the man to sink 8.00 III to the bouom of the lake?
p
in eqllilibrium immediately belOV"lhc surface ol"watcr ill a swimming pool?
30.
A large balloon of mass 226 kg is filled with heliulll gas until it:-. volume is 325 1ll 3 , Assume the density of air is 1.29 kg/rn:~ and the density of helium is 0.li9 kg/m:\. (a) Draw a rree-body diagram for the balloon. (b) C,lIculare the buoyant force acting on the balloon. (c) Find the net force on the balloon and determine whether the b;-llioon \vill rise or fall after it is released. (d) What maximum additional mass can the balloon suppOrt in equilibrium? (e) What happens to the balloon if the mass of the load is les.\·than the value calculated in part (d)? (f) \Vhat limits the height to which the balloon call rise?
37.
:=
mil On
October 21. 2001, Ian Ashpolc of the ['nitcd Kingdom achieved a record altitude of3.35 km (II 000 ft) powered by 600 wy balloons filled with helium. Each filled balloon had a radius of aboul 0.50 m and an estimated mass of 0.30 kg. (a) Estimate the total buoyant force on the 600 balloons. (b) Estimate the net upward force on all 600 balloons. (c) Ashpole parachuted to Earth after the balloons began to burst at the high altilllde and the system lost huoyanc}'. Why did Ihe balloons burst?
38. m::I A 10.O-kg block of metal is sllspended from a scale and irnmel'sed in water, as in Figure P9.38. The dimensions of the block are 12.0 cm X 10.0 cm X 10.0 em. The 12.0~cm dimension is vert ical, and the top of the block is 5.00 cm below the surface of' the water. (a) What are the forces exerted b}' the FIGURE P9.38 water on the top and bottom of the block? (Take l.Jo := 1.013 0 X 10" :\'/m 2.) (b) What is the reading orthe spring scale? (c) Show that the buoy,tnt force equals the diffcn:nce between the forces at the top alld bottom of the block.
Problems
]39.IA bathysphere used (or deep sea explonllion has a radius of 1.50 ll'l
'"
317
the w,lIer leaving the end of the hose? (b) A nozzle is now
II
(a) Calculate the mass rio\\" rate (in gram~ per second) of blood (p = 1.0 g/cm 3 ) in an aorta with a cross-sectional area of 2.0 em:! if Ihe flow speed is 40 cm/s. (b) AssulTle that the aorta branches to form a brge I1lllnher of capillal-it.:s with a combined cross-sectional ,Irea of 3.0 X 10 3 cm:!. What is the flow speed in the capillaries?
46. A liquid (p = 1.65 g/cm:l ) flows through twO horizontal sections or luhingjoined end La end. In [he first section, the cross-sectional area is 10.0 em:!, the flow speed is ~75 cm/s, and the pressure is 1.20 X to:, Pa. In the second sect;oll, tile cross-sectional arc<-I is 2.50 cm:!. Calcul<:ltc the smaller section's (a) flO\...· speed and (b) pressure.
147.10
A hypodermic syringe contains a medicine with the density ofwmer (Fig. P9.47). The harrel of the syringe has a cro.'is-sectional ,lrea of 2.50 X 10- 5 m 2 . In the absence of a !(lITe on the plun~cr, the pressure everywhere is of magnitude 2.00 N is exerted on 1.00 atm. A force the plunger, making medicine squirt from the needle. Determine the medicinc·s flow speed through the needle. Assullle the prt:ssure in the needle remains equal to 1.00 atm and that lhe syringe is horizomal.
F
k
(aj
(hi
FIGURE P9.40
41. A sample of an unknown material appc(lrs 1.0 \\lcigh 300 N in air and 200 N when imlnersed in alcollOl ofspecifte gravity 0.700. What arc (a) the volume and (b) the density of the material? 42. An object weighing 300 N in air is immersed in water after being tied to a string connected La
-F
FIGURE P9.43
48.
II
49.
mil
SECTION 9.8 OTHER APPLICATIONS OF FLUID DYNAMICS 44. W;ller rIowing through a garden hose ofdiametcr 2.74 cm fills a 25.0-L bucket in LilO min. (
~-
\,Vhen a person inhales, air moves down the bronchus (windpipe) aILS cm/s. The a\'erage flow speed of the air doubles through a constriction in the bronchus. Assuming incompressible now, determine the pressure drop in the constrict ion.
A jet airplane in level night has a mass of 8.66 X 10 4 kg, and thc two wings have <-In eSlimated tOlal area of ~)O.O m:!. (a) \""hat is t.he pressure difference between the lower and upper surfaces of the wings? (b) If the speed of air Iinder the wings is 225 mis, what is the speed of the air over t.he wings? Assume air has a density of 1.2Q kg/lll~. (c) Explain why all aircraft have a "ceiling," a maximum operational altitude.
50. _ An airplane has a mass M, and t.he two wings have a tala I area A. During level flight, tht.: pressure on the lower wing surface is Pl' Determine the pressure P'!. on the upper wing surface. 51.
SECTION 9.7 FLUIDS IN MOTION
p"
v2 : ~- ~\UL':"""",,,:,,,~ ""I'Ib.."",-
FIGURE P9.47
143.IA
1.00-kg beaker containing 2.00 kg or oil (density = 916 kg/m ll ) rests on
ii:::--.
r
EII
In a w,lter pistol, a piston drives watcr through a larger tuhe of radius 1.00 em imo a smaller tube ofr~ldius 1.00 mm as in Figure P9.51. (a) If the pistol is fired hori+ l.omaH), (11,1 height of l.!lO Ill, usc ballistics 1.0 determine the time it takes water 1.0 travel from the noa.le to the g-round. (I'\cglect air resistance and assume
318
Chapter 9
Solids and Fluids
nozzle? (c) Given the areas of the nozzle and cylinder, lise Ihe equation ofcominuity LO calculate the speed at which the plunger must be moved. (e1) 'What is the pressure at the nozzle? (e) U,')c Bernoulli's equation La find the pressure needed in the larger cylinder. Can gravity tcrms be neglected? (0 Ca1culmc the force that must be exened on the trigger to achieve the desired r'mge. (The force that must be exencd is dlle 1O pressure over .mel above atmospheric prcssllIT.)
v,
I?=====±====~~-
FIGURE P9.S1
52. Water moves lhrou~h a constricted pipe in steady. ideal flow. At the lower point shown in Figure 9.29, the pressure is 1.75 X 10" Pa and the pipe radius is ~.OO CIlI. At another painl ~.50 III higher, the pressure is 1.20 X 10:' Pa and the pipe radius is 1.50 em. Find the speed of flow (a) in the lower section
_I IQt..1 lJ
'00
~m
FIGURE P9.SS
56. \Vater is pumpcd through a pipe 01" diameter }fl.O em from the Colorado River lip to Grand Canyon Village, on the rim of the canyon. Tilc river is at 564- III devation and the village is at 2 096 m. (a) At what minimum pressure lTlust t.he wmer be pumped to arrive at the village? (b) If 4 500 m:~ arc pumped per day, what is the speed of the wat.er in tht:: pipe? (c) "Vhat additional pressure is necessaJ)' to deliver this 1I0w? Note: YOli mar assume the frce-fall acceleration and the density of air are constant over the given range or elevations. (57.lold F
58. The \felHuri tube shown in Figure 9.30 may be uscd as a nuid flowmeter. Suppose thc device is used at a service
h
-1
I.OOm
_ _ _ _I,
FIGURE P9.S3
staljon to measure the flow nHe of gasoline (p = 7.00 X 102 kg/m 3 ) through a hose having an outlet radius of 1.20 Cln. Il' the difference in pressure is measured to be PI - P'!. = 1.20 kPa and thc radius of thc inlet. tube to the meter is 2.40 em, find (
SECTION 9.9 SURFACE TENSION, CAPILLARY ACTION, AND VISCOUS FLUID FLOW 59. A square metal sheet 3.0 em on a side and of negligible t.hickness is attached to a balance and inserted into a COIl-
54. A large storage tank, open to the
(a)
(h)
FIGURE P9.S9
Problems LaineI' of fluid. The contact angle is found to he zero, as shown in Figure P9.~9a, and lhe balance 10 which Ihe l1\elal shect is attached reads 0.'10 N. A thin veneer of oil is then spread over the sheel. and the CUlltact angle becomes 180°, as shown in Figure P9.59b. The balance nm.. .· r"-',lds O.~9 N. \",'hat is the sur"face tension or the fluid?
60.11 To lift a wire ring of" radius 1.75 CtTl from the surf~lCe of"a container or blood plasmil, a vertical force of 1.01 X 10 2 N g-reater than the weight of the ring is required. Calculate the surface lension of blood P];.ISI11<1 from this information. 61. A certain fluid has a density of I 080 kg/m 3 and is observed to rise to a heig-Ill of 2.1 un in a I.O-Inm-di"lmeter lube. The contact angle bet\...· een the wall and the f1l1id is /cro. Calculate the sllrface lcnsioll oft.he fillid.
62.
needle so that the flow rate or water through it will be I g/s? (Use 1.0 X 10- 3 Pa' s as the \~scosity ofwmer.) 69.
III
What diameter needle should be used to inject a volume of 500 CI11:\ of a solutloll into a patient in 30 min? A'i:-iUlllC tit ..... kngllJ of the Ileedle is 2.5 em and thc solutjon is e!c\'ated 1.0 m above the point of injcction. Furt.her, assume the viscosily and density of 1J1e solution are those of pure \.. .;Iter, and that t.he pressure inside the vein is atmospheric.
70. \Vater is forced out of a Hre eXlinguisher by air pressllre, as shown in Figure P9.70. \-"hat gauge air pressure in the lank (above atmospheric pressure) is required for the water to have ajet speed of 30.0 m/s when the water level in the lank is 0.500 m below the nozzk'i-
II
) ?rt==it "'" 0500 m
--..t
FIGURE P9.70
F
71.
/_1.20 Ill--¥ /~
0.800 III ~
( FIGURE P9.63
66.
II The pulmonary ane'1', which connects the hean to the lungs, has an inner radius 0[2.6 mm and is 8,4 crn long. If the pressure drop between the hean and lungs is 400 Pa, what is the avcrage spced ot blood in the pldrnOllal)' ancry?
167.ISphel-ical panicles of a protein of dcnsity J.8 g/crn'l are shaken up in a solution of 2ffC water. The 'iOllllioll is allowed to stand for 1.0 h. If the deplh of water in the tube is 5.0 cm, rind the radius of t.he largest panicles lhat remain in solution at lile end orthe hour. 68.
IIJI
A hypodermic needle is 3.0 cm in length and 0.30 111m in diameter. \-\'hal excess pressure is l'cquired along the
II The anrta
in humans has a diameter of about 2.0 em, and at certain times the blood speed through it is about 5~ cm/s. Is the blood now turbulent? The density of whole blood is I 050 kg/m\ and its coefficient of \'iscosit)' is 2.7 X 10 't 1'" s/m~.
72. A pipe carrying 20°C water has a diameter of 2.5 em. Estimatc the maximum flow speed if the flow must be slreamline.
SECTION 9.10 TRANSPORT PHENOMENA 73.
III Sucrose
74.
III
64. A thin 1.5-mm coat.ing of glycerinc has Ill'cn placed bctwecil two microscope slides of"width 1.0 em and length 4.0 cm. Find the force requin.::rl to 131111 one 01" the microscope slides at a constanl speed 01"0.30 m/s relative to the other slide. 65. A straight horizontal pipe with a diameter of 1.0 cm ami ;1 lengt.h of 50 m carries oil with a coefflcient of viscosity 0["0.12 N· s/m 2 . AI. the output of the pipe, tlle flow rate is 8.6 X 10 :; m'\/s and the prcssure is 1.0 atm. Find lhe g"
v
Whole blood has a surface tension of O.05R X/m and a density of I 050 kg"/m 3 . To \... h;\1 height can whole blood ri'ic in a capillary blood vessel that has a radius of 2.0 X 10 b m if the contact angle is zero?
63. Tile block of icc (lel1lpCraturc DOC) shown in Fig:urc P9.63 is drawn (Aer a level surf(lce lubricated by a layer (If" water 0.10 mm thick. Determine the mag-niludc of 1he [
319
is allowed to diffuse along a IO-cm length of tubing filled with water. The tube is 6.0 cm 2 in crossscction<J\ ;'1I"ea. The diffusion coefficient is equal to 5.0 X H)-jl) m~/s, and B.O X 10- 14 kg is lransponed along the t.ube in 15 s. \\'hat is the difference in the concentration levels of sucrose at. the t.wo ends of the tube? Glycerin in water diffuses along a horizontal column thaI. has ;1 cross-sectional area of 2.0 cm 2 . The concentration gradient is ~.O X 1O-:.! kg/m 4 , and the diffusion rate is found 10 be 5.7 X 10- 1:) kg/s. Determine the diffusion (( )effi Cieill.
75. The viscolls forcc on all oil drop is mea'iured to he cCJlIal to ~.O X JO-l:l N when the drop is falling through air with 1
76. Small spheres of diameter 1.00 mm f,lil through 20 C water with a terminal speed of 1.10 em/so Calculate the density 01" the spheres. ADDITIONAL PROBLEMS
177.IAn iron hlock of volumc 0.20 m)l is slispended from a spring scale and immersed in a flask or water. Then lhe iroll block is removed, and an aluminum block of the
320
Chapter 9
Solids and Fluids
same volume replaces il. (a) (11 which case is the buoyant force the gTC,ltest, for the iron block or the ~11Llmin1l1l1 block? (b) In which case does the spring scale read the largest value? (c) Csc the known densities of these materials 10 calculale the quantities requested in pans (a) and (b). Arc your calcu1
lap. A hollow tube is inserted into the spinal column, and
the height to which the fluid rises is observed, as shown in Figure P9.83. If the fluid rises to a height of 160 mm, \\'e wrile its gauge pressure as 160 mm H:p. (a) Express this pressure in pascals. in atmospheres, and in millimeters of mercury. (b) Sometimes it is necessary to dclcrmine whether an accident viclim has suUt.:red a crushed vertebra that is blocking the flo\\' of cerebrospinal fluid ill the spinal column. In other cases a physician may suspect Ih
78. A steel ball is lossed into the ocean and cOllles to rest at a depth 0[2.40 kill. Find its fractional change in \'olulnc, assuming the density of seawater is 1.025 X 10:1 kg/m:l ,
79. As a first approximation. Earth's contincl'lls Illay be thought of as granite blocks noating in a denser rock (c<-liled peridolite) in the same way Ihal ice no,Us in waler. (a) Show thata formula describing Ihis phenomenon is p/!l= p"d where Pg is IJ1C densit), of granite (2.8 X lOi! kg/mS), P" is thc density ofperidotitc (3.3 X 103 kg/m 3 ), I is the thickness of a continent, and d is the depth to which ,1 continent nO<.lLsin the peridotite. (b) Ira continent sinks 5.0 kill into the peridotite layer (this surface may be thought of a... lhe ocean 0001'), what is the thickness oCthe cominent? 80.
II
Take the demit)' of blood La be p and the dist...'1ncc belween the feet and the hean to he hf-{' IgnOl'e lhe flow of blood. (a) Show that the difference in blood pressure between the feet and the heart is given by PI-" - P,-I = pgh". (b) Take the density of blood to be 1.05 X I O:~ kg/m 3 and the dislance bctween thc heart and the feel LO be 1.20 m. Find the difference in blood pressure bet,vecli these two points. This problem indicates that pumping blood from thc cxu'emitics is very ditlicult for the heart.. The veins in the legs have valves in them th~lt open when blood is pumped toward the hear' and close when blood Oows away from the heart. Abo, pumping action produced by ph}!sicaJ activities such as walking and breathing assists the heart.
Is1.ll1
The approx.imate inside diameter of the aorta is 0.50 cm; that of a capillary is JO p,m. The ,lpproxim
82. Superman atlcmpts to drink water through a vcry long vertical straw. VV'ith his greal strengul. he achieves maximum possible suction. The walls of the straw don't collapse. (a) Find tJle ma..x.imum height through which he can lift the water. (h) Still thirsty, the Man of Steel repeats his attempt on the Moon, which has no atmosphere. Find the difference between the water levels inside and outside lhc Slldw. 83.
FIGURE P9.83
84.
II
Determining the densit), of a Ouid has many important applicaLions. A car baltery comains sulfuric acid, and the bauery will not function properly if the acid density is too low. Similarly, lhe effectiveness ofanlifreeze in your car's engine coolant depends on the density of the mix~ lure (usually ethylene glycol and water). \·Vhell you donate blood to a blood bank, its screening includes a determination of the density of the blood bec
T
II
The human brain Clnd spinal cOl-d are immersed in the cerebrospinal fluid. The fluid is normally continuous between the cranial and spinal olVities and exerL" a pressure of 100 to 200 n1l11 of H 20 above the prevailing atmospheric pressure. In mediGII work. pressures arc oftt:n measured in units of mm of' 1-1 2 because body fluids, including the cerebrospinal fluid, typically hm/e nearly the same densit), as waler. The pressure of the cerebrospinal lIuid can be measunx! by means of a sjJinnl
°
L
1 •
FIGURE P9.84
321
Problems
immersed in the liquid of' density p. A lcnl-\"th "of the rod protrucks above the ~urraee of lhc liquid. Show Ihal the densilY of the liquid is gi\'en by
PilL P~f--"
from any air motion while air is blown across the top of Ihe left arm umil the surfaces of the two liquids are at t.he same height (Fig. 9.80c). Dercrm;/lc {he speed of the air being blmvll across the leI"! arm. Assume the density of air is 1.29 kg/m:''.
-,.
85. Figure P9.85 shows a w:llcr tank with a v:l\ve. H the valve is opened, what is the maximum height
r
ShiL:ld
A
\\',lIer
"~-l
(al
(h)
(c)
FIGURE P9.88
h Val"e r-l_ _
,..
/)
l'-~E=~=' fin
89. A I.O-kg hollow ball with a radius of 0.10 III and filled with
'<;;.,/
flil" is released paoloI' water. shom upward? changes in the merged.
FIGURE P9.85
186.IA hdillm-fil1cd balloon is tied to a 2.0-m-long, O.050-kg string. He The balloon is spherical wilh a radius or 0.40 m. When released, il lirts a length It o! the sIring and then remains ill equilihrium, as in Figure P9.86. Dctermine the value of It. 'When FIGURE P9.86 deflated, the balloon has a mas.~ of 0.25 kg-. Ihnl: Only Ih:11 part of the string abo\'e the floor contributes to the IU:ld being held up by the balloon.
11n
87. A 600-kg weather balloon is designed to lift a -lOOO-kg package. What VOllllTIe should the balloon h:l\'e alter being inflated with helium
from rest at the hOllom of:-l 2.0-m-dcep How high above Ihe water docs the ball Neglect :-111 rl'ietional effecls, and neglect ball's motion when il is only partially sllb~
90. Oil having a density of ~130 kg/m~ r10ats on water. A rectangular block of" wood 4.00 (m high and with a cll.:nsit)' of 960 kg/m 3 floats part I}' in the oil and partly in the W;Her. The oil completely covers thl: block. Ilow far below the interrace between the two liquids is lhe bottom of the block?
[ill"
water lank open to the
92. A walkway suspended across a hOlCllobby is supported at Ilumerous points 'lIang its cdges by a vertical c
High temperatures inside a volcano turn water into a high pressure steam. Unless the steam and other gases vent into the atmosphere, pressure can build until a catastrophic explosion results.
10.1
Temperature and the Zeroth Law of Thermodynamics
10.2
Thermometers and Temperoture Scoles
10.3
Thermal Expansion of Solids and Liquids
10.4
Macroscopic Description of an Ideal Gas
10.5
The Kinetic Theory of Gases
THERMAL PHYSICS How can trapped water blow off the top of a volcano in a giant explosion? What causes a sidewalk or road to fracture and buckle spontaneously when the temperature changes? How can thermal energy be harnessed to do work, running the engines that make everything in modern living possible? Answering these and related questions is the domain of thermal physics, the study of temperature, heat, and how they affect matter. Quantitative descriptions of thermal phenomena require careful definitions of the concepts of temperature, heat, and internal energy. Heat leads to changes in internal energy and thus to changes in temperature, which cause the expansion or contraction of matter. Such changes can damage roadways and buildings, create stress fractures in metal, and render flexible materials stiff and brittle, the latter resulting in compromised O-rings and the Challenger disaster. Changes in internal energy can also be harnessed for transportation, construction, and food preservation. Gases are critical in the harnessing of thermal energy to do work. Within normal temperature ranges, a gas acts like a large collection of non-interacting point particles, called an ideal gas. Such gases can be studied on either a macroscopic or microscopic scale. On the macroscopic scale, the pressure, volume, temperature, and number of particles associated with a gas can be related in a single equation known as the ideal gas low. On the microscopic scale, a model called the kinetic theory of goses pictures the components of a gas as small particles. This model will enable us 10 understand how processes on the atomic scale affect macroscopic properties like pressure, temperature, and internal energy.
10.1
TEMPERATURE AND THE ZEROTH LAW OF THERMODYNAMICS
Temperature is commonly associated with how hot or cold an object feels when \\'e touch it. \lVhilc our senses provide us with qualitative indications 01" temperature, Lhey are unreliable and often misleading. A metal icc tray feels colder La the hand, (or example, than a p;'lCk<1ge of ['rozen vegetables at the same Lempcrature,
322
10.1
Temperature and the Zeroth Low of Thermodynamics
323
because metals conduct therm(ll energy more rapidly th"ln a cardboard package. \"/hat \ve need is a reliable and reproducible method of making quantitative measun.:rnents th
~ Zeroth law of thermodynamics
This statcment is importall1. because it makes it possible to define temperature. VVe can t.hink of ten1perature as the property that determines whether or not an ol~ject is in thermal cquilibl"iurn with other objects. Two objects in tbermal equilibrium with each other are at the same temperature.
QUICK QUIZ 10.1
T"vo objcC1.s with diflt.'rcnt sizes, masses, and temperatures arc placed in thermal contact. Choose the best ans\ver: Energy travels (a) from the larger object to the smaller object (b) frum the object with more mass to the one WirJl less Inass (c) from the object at higher temperature to the ol~iect atIO\'\'cr lemperal.Ure.
FIGURE 10.1
The Il'roth 1;1\\' or
1\
A (a)
(b)
(e)
324
Chapter 10
Thermal Physics
10.2 THERMOMETERS AND TEMPERATURE SCALES
FIGURE 10.2
Schematic di'lg-ram
ora 11lL:ITlJt"y IllcrrnOlllcter. Bl'callsc
of thermal expansion, the Il'w'l of 'he mercury rise~ ;l~ the [cmperallln' 01 the Illercury challge~ from 0"<: (LIlt: ice point) to
wrre
(ill(' steam point).
T h
o
1
I'vkrr I II)'
reservoir
\.. envinlOmenl to be measl I red
Flexible hose
FIGURE 10.3 :\ (onstalll-voillme gas thermomcter measures the pre~ sure of the ga.~ contained in dll.: fl
ofg", illlhc flask
j,
!-CptCOIISI,lTIl
b\' r
keep the
llH'flury
!t'H:1 constant.
Thermometers arc devices used LO measure the Wlllperature of an object or a system. \·Vhen a thermometer is in thermal contact with a system, energy is exchanged unt.il the thermometer and t.he system arc in thermal equilibrium with each other. For accurate readings, the thermometer must be much smaller than the system, so that. the energy the thermometer gains or loses doesn't signific3mly alter the energy contenl of (he S}lstelll. All thermometers make use of some physical property that changes '''lith temperature and can be calibrated to rnake the tempera lU re Ineasu rable. Some of t he physical properties llsed arc (1) the volll me of a liquid, (2) the length of a solid, (3) the pressure of a gas held at conSlant volume, (4) the volume of a gas held at constant pressure, (5) the electric resistance ora conductor, and (6) the color ora very hot object. One com mon thermometer in everyday lise consists of a mass orliquicl-usuall)' mercury or alcohol-that expands into a glass capillary tube \'·:hen its temperature rises (Fig. JO.2). In this case the physical property that changes is the volume of a liquid. To serve as an effective thermometer, the change in volume of the liquid \\lith change in temperature must be vcry nearly constant over the temperature ranges of inl.cresl. \l\·11en the cross-sectional area of the capillary LUbe is constant as well, the change in volume of the liquid \'arie~ linearly with its length along the tube. \Vc can then define a temperature in tcrms of the length of the liquid column. The thermometer can be calibrated by placing it in thermal cont<'lct with environments that remain at constant temperature. One such environment is a mixLUre of water ;-lnd ice in thermal equilibrium at atmospheric pressure. Another cOllllllonly lIsed system is a miXl.Ure o[watcr and steam in thermal equilibriulll at atlllospheric presstlre. Once we h;-we marked the ends of the liquid colurnn for our chosen environment on our thermometer, '''o'C need to define (\ scale of numbers associated \\~th various temperatures. An example of such a scale is the Celsius temperature scaJe, formerly called the centigrade scale. On the Celsius scale, the temperaLUre of the icc-,v;;ll.er mixtLlrc is defined to be zero degrees Celsius, ,vritten O°C and called the ice point or freezing point of water. The temperature or the water-Steam mixture is defined as 100°C, called the steam point or boiling point of water. Once lhe ends of the liquid column in the thermometer have been marked at these two points, the distance between rnarks is divided into 100 equal segments, each corresponding to a change in temperature of one degree Celsius. Thermometers calibrated in this way present problems when extremely accurate readings are needed. FOI- example, an alcohol thermometer calibrated at the ice
The Constant-Volume Gas Thermometer and the Kelvin Scale \Ve Gl n construct pract ical t hc:rmol11cters such as the mercury thermometer, but these types of t.hermometers don't define temperature in a fundamental way. One thermometer, however, i!l' more fundamental, and offers a way to define temperature and relate it directly to internal energy: (he gas thermometer. In a gas thermometer, the temperature readings are nearly independent of the substance used in the thermometer. One type o[ gas thcrmometel' is the constanH'olurne unit 5ho""n in Figure JO.3. The behavior observed in this de\'ice is the variation of pressure with temperatlllT ofa nxed volume orgas. vVhen the constant-volume gas thermomeLeI- was developed, it was calibrated Llsing the icc and steam poims of water as follows (a different calibration procedure, 10 be discussed shortly, is now Llsed): The gas nask is insert.ed into an iCC-\\'aler bath, and mercury reservoir B is
10.2
Thermometers and Temperature Scoles
raised or lowered until the volume of the confined gas is at some value, indicated by the zero point on the scale. The height 11, the dilTerence between the levels in the reservoir and column A, indicates the pressure in the flask at DOC. The flask is inserted into water at the steam point, and reservoir B is readjust.ed until the height in column A is again brought to zero on the scale. ensuring that the gas volume is the same as it had been in the icc baLh (hence the designation "constant-volume·'). A measure of the new val LIe for II, gives a value [or the pressure at I Oa°c. These pressure and temperature values arc then plottcd on a graph, as in Figure 10.4. The line connecting the two points serves as a calibration CltrYC for mcasuring unknown temperatures. If we ,·vant to measure the temperature of a substance, vvc place the gas flask in thermal contact with the substance and adjust the column of mercury until the level in column A returns LO zero. The heighl of the mercury column tells LIS the pressure or the gas, cUld we could then flnd the lemperature of the substance f-i'om the calibration ClHye. Now suppose that temperatures are measured \vilh various gas thermometers containing different gases. Experiments show that the thermometer readings are nearly independent of the type of gas used, as long as the gas pressure is 1m\' and the temperature is well above the point at which the gas liquefies. We can also perform the temperature measurements ,.."ith the gas in the Oask at different starting pressures at aoc. As long as the pressure is low, we will generate straight-line calibration curves for each starting pressure, as shown for three experimental trials (solid lines) in Figure 10.5. If the curves in Figure 10.5 are extended back toward negative temperatures, \ve find a startling result: rn every case, regardless of the type of gas or the value o[ the low starting pressure, the pressure extrapolates to zero when the temperature is -273.15°C. This fact suggests lhat this particular temperature is universal in its importance, because i1.doesn't depend on the substance used in the thermometer. In addition, because the lowest possible pressure is P= 0, a perfect vacuum, the temperature -273.l5°C must represent a lower bound for physical processes. We define this temperature as absolute zero. Absolute zcro is used as the basis for the Kelvin temperature scale, ",.. hich sets -273.15°C as its zero point (0 K). The size o[ a "degree" on the Kelvin scale is chosen to be identical to the size of a degree on the Celsius scale. The relationship bctween these two temperature scales is
e
7 = T- 273.15
325
p
.L-
o
-,-'.,----
T lOG)
[00
FIGURE 10.4 A l)'pical graph of pressure \'er~us ternpcrallire taken \\·ith
[10.1]
where Tr; is the Celsius temperature and T is the Kelvin temperature (sometimcs called the absolut.e temperature). Technically, Equalion 10.1 should have units on t.he righI-hand side so that it reads = TOC/K - 273.15°C. The lInils are rather cumbersome in this context, so wc \·vill usually suppress thcm in such calculalions except in the final answer. (This will also be the case when discussing the Celsius and Fahren heit scales.) Early gas thermometers made usc of ice and steam points according to the procedure juS! described. These points arc cxpcrimemally difficult 1.0 duplicate,
'Ie
I'
Tri,tl2 Trial 3
-273.1:') -200
-100
o
IOU
200 T (0C)
FIGURE 10.5 Pn'ssllfc \"(;1"\15 lernpCra\llfl' rorcxperimeillaltrials ill which g.lses h;\\'(; dil"ft:n':llt pres~lIres in a cOlIstanl+\,olullle g;t" thermullletCI'. l'ote lhat for aliliut:e trial .. the prt:<;Sllrc exnapolales LO 7ero allhe temperalure -273.15°C.
Chapter 10
326
Thermal Physics
The kelvin -7
TempcranliT (K)
10" 10"
Ilydrng"cll homb
IOi
-llllcrioroflhc S\l1l
lOt;
Solar corona
however, because they arc pressllrc~sensitive.Consequently, a procedure based on two new points was adopted in 1954 by the International Committee on Wciglus and ]\·lcflsurcs. The first point is absolute zero. The second poinl is the triple point of water, which is the single temperature and pressure at which water, water vapor, and ice can coexist in equilibrium. This poinl is ;.1 conVf>nip.nl and rc~rro ducible reference temperature for the Kelvin scale; it occurs at a temperature of O.OloC and a pressure of 4.58 mm of mercury. The temperCllure althe triple point of waleI' on the Kelvin scale occurs at 273.16 K. Therefore, the 51 unit of tempera~ ture, the kelvin, is defined as 1/273.16 of the temperature of the triple point of water. Figure 10.6 shows the Kelvin temperatures for various physical processes and structures. Absolute zero has been closely approached but never achieved. \'Vhal would happen to a substance ifits temperature could reach 0 K? As Figure 10.5 indicates, the substance \·,:ould exert zero pressure on the ' . . alls of its container (assuming the gas doesn't liquefy or solidify on the ,...'ay to absolute zero). rn Section 10.5 ..." e show thal the pressure of a gas is proportional to Lhe kinetic energy of the molecules of that gas. According La classiG11 physics, therefore, the kinetic energy of the gas would go to zero and there ,...,ould be no Illation al all of the individual cOlllponents of" the gas. According to quantulll theory, however (to be discussed in Chapter 27), the gas would £llways retain some residual energy, called the zero-jJoinl pnergy, at [hat low temperallire.
J05
The Celsius, Kelvin, and Fahrenheit Temperature Scales 10 1
J O~
J>-~t~
Surface of rhe Sun Copper melts \,Va1(,1" frccles Liquid nitrogen
achic,'cd -10-; K FIGURE JO.6 Ah~ulllle lemp('f;lwhich varioll<; selected phr<;ical processes LIke place. Note Ihal lilt" scale is logarithmic.
Equation JO.l shows that the Celsius t.emperature 7(; is shifted from the absolute (Kelvin) temperawre Tby 273.15. Because the size ofa Celsius degree is the same as a kelvin. a temperature difference of 5°C is equal to a temperature difference of 5 K. The twO scales differ onl)' in the choice of zero point. The ice point (273.15 K) corresponds LO O.OO°C, and the steam point (373.15 K) is equivalent to 100.OO°C. The most common temperawre scale in use in the United States is the Fahrenheit scale. I.L sets the lC11lperat.ure of the ice point at 32"F and dle temperalllre of the steam point at 212°.F. The relationship between the Celsius and Fahrenheit temperature scales is [IO.2a]
HilTS al
For example, a temperature of 50.0°F corresponds to a Celsius temperature of lO.O°C <md an absolute temperature 0[283 K. Equation 10.2 can be inverted to give Celsius temperatures in terms of Fahren· heit tern pera tu res:
'Ie = #(7;. A c(1Il1p:lrison urlhe Celsius. Faht"(~llhejl. and Keh"ill lell1pt:"ratUl"l'scales.
32)
[IO.2b]
FIGURE 10.7
$leall1 poinl
100°~
212°=-
373.15 -
-
~
~ ~
0 0 ~
-
~
0 0 ~
lc~
poilll
0°
32°
Cehius
F
27~.15-
Kch-in
10.2
Thermometers and Temperature Scales
327
Equation 10.2 can also be used to rind a relationship between changes in ternperature on the Celsius and Fahrenheit scales. In a problem at the end of the chapter you will be asked to show that ifthc Celsius temperature changes by !J.Tc• the Fahrenheit tcmperature changes by the amollnt !J.7F=~!J.7(.
[10.3]
Figure ]0.7 compares the three lempcraLUre scales we have discllssed.
EXAMPLE 10.1 Goal
Skin TemJ:>erature
Apply the tempcrature conversion formulas.
Problem The tcmperature gradient between the skin and thc air is regulated by cutaneous (skin) blood now. If the cutaneous blood vessels are constricted, the skin temperature and the temperature of the environrnent will be about the same. "Vhen the vessels are dilated. IT10re blood is brought to the surface. Suppose during dilation the skin warms from 72.0°F to 84.0°F. (a) Convert these temperatures to Celsius and find the difference. (b) Convert the ternperatures to Kelvin, again finding the difference. Strategy This is a matter of applying the conversion formulas, Equations 10.1 and LO.2. For pan (b) it's easiest to usc t.he answers for Celsius rather than develop another seL of conversion equations.
.
'-' . . . . . . .. . .....
~
Solution (a) Convert the temperatures from Fahrenheit to Celsius and find the difference. Convert the lower tempcrature, using Equation 10.2b:
7; = BO;. - 32) = B(72 - 32) =
Convcrt the upper temperature:
T" =
Find the difference of the two temperatures:
f17(; = 29"e - 22"C =
(b) Convert the temperatures from Fahrenheit Kelvin and fllld their difference.
:;0;, -
32) = B(84 - 32) =
29"e
7"e
La
Converlthe lower tempenlture. using the answers for Celsius found in pan (a):
T = 22
Convert the upper tcmperamre:
7
Find the difference of the two temperatures:
t;T = 292 K - 285 K = 7 K
7,;=T-273.15 -> T=7';+273.15 + 273.1 5 = 295 K
= 29 + 273.15 =
302 K
.............. Remark
22"e
• • • • • • • • • • .-<11
The change in temperature in Kelvin and Celsius is the same, as it should be.
QUESTION 10.1 Which represents a larger temperature change, a Celsius degree or a Fahrenheit degree? EXERCISE 10.1 Core body tcmperature can rise from 98.6°F to 107°F during extrcme exercise, such as a marathon run. Such elevated temperaturcs can also be caused by viral or bacterial infections or LUmors and arc dangerous if sustained. (a) Conven the given temperatures to Celsius and fincllhe difference. (b) Convert the temperatures to Kelvin, again finding the difference.
Answer
(al 37.0"C, 'IL7"C, 4.7'C
(b) 310.2 K, 3149 K, 4.7 K
328
Chapter 10
EXAMPLE 10.2 Goal
Thermal Physics
Extraterrestrial Temeerature Scale
Understand how to relate different. temperature scales.
Problem An extraterrestrial scicmist invents a I.CmperaLilre scale such lhal v.'ater freezes at - 75 c E and boils at 325°E, where E stands for an cxtrcU<.:rrcsLrial scale. Find an equatioll t!Jdll'e1alcs temperature in "E [Q tcmperalUre in nCo Strategy Using the given data, rind lhe r,llio or the llumber 0[°£ bct\·\,cen the !VI'D temperatures to Lhe number of "C. This ratio will be the same as a similar raLio for any other such process-say, frol11 the freezing point La an unkno"..' !1 temperature-corresponding' to TI~' and 7~;. Sell iIlg the 1.\\'0 ratios equal and solving for ·l~.; in terms or 7 yields the desired relationship.
c
r··
.
• • • • • • • • • • • • • • • • • • • ""111
Solution Find the change in tcmperalure in "E bCl,,,'cen t.he freezing and boiling poims o[water:
/1 Tt. = ~25°E - (-75°E) ~ 400 E 0
Find the ch;lIlgc in temperature in °C bet\veen the freezing and boiling points of,.. . ater:
/17(. = 100°C - O°C = lOO°C
Form the ratio of Illest: two quaJllities.
-=--=4-
This ratio is tht: same between any other 1.\...' 0 temperatures-say, [roln the f"reeL.ing point LO an unknown final tcmperalUre. Set the twO ralios equal to each other:
400 E
°E
/17(.
lOWC
°c
0
7,- (-75°E)
'Ie 'Ii, - (-75°E)
Soke [or 7,.:'
.... Remark
/1 Tf,
OOC
~
4(OEI'C)(T(' - DOC)
..................... -
..
~
The relationship between any other two temperalllres scales can be derived in the same way.
QUESTION 10.2 True or False: Finding the relationship between two temperature scales using knowledge of the freezing and boiling point of water in each system is equivalent to finding the equation ora slraightline. EXERCISE 10.2 Find the equation convening "F to "E. Answer
7~.
= W7j: -
146
10.3
THERMAL EXPANSION OF SOLIDS AND LIQUIDS
Our discussion of" the liquid thermometer made use of one of the best-known changes that occur in most substances: As temperature orthe substance increases, its volume increases. Thi;;; phenomenon, knmvn as thermal expansion, plays an important role in l"Iurnerous applications. Thermal expansionjoims, for example, must be included in buildings, concrete highways, and bridges to compensate for changes in dimensions with variations in temperature (Fig. 10.8). The overall thermal expansion of an object is a consequence of t.he change in the average separation between its constituent atoms or molecules. To understand this idea, consider how the atoms in a solid substance behave. These a(QJ1lS are
10.3
Thermal Expansion of Solids and Liquids
329
located at fixed equilibrium positions; if an atom is pulled Cl\\'ay from its position, a restoring force pulls it back. \·Ve can imagine that the atoms arc particles connected by springs LO their neighboring atoms. (See Fig. 9.1 in the previous chapter.) Han atom is pulled away from its equilibrium position, the distortion of the springs pr()vides a restoring force.
At ordinary temperatures, the atoms vibnue around their equilibrium positions with an amplitude (maximum distance from the centerof"ibration) ofabo lit to-II m, with an average spacing bct\\-'cen thc atoms of about] 0- 10 111. As the temperature of the solid incre~lses, the atoms vibrate with greater amplitudes anclthe average separation between them increases. Consequently, the solid as a whole expands. If the thermal expansion of an object is sufficiently small compared with the ol~ject's initial dimensions, then the change in any dimension is, to a good approximation, proportional to the first power of the temperature change. Suppose an object has an initial length L o along some direction at some temperaUlre 10- Then the leng·th increases by Li L for a change in temperature d T So for small changes in temperaLUrc,
a~ ~
~
(a)
[lOA]
or
where L is the object's final length, Tis its final temperature, and the proportionality constant a is called the coefficient of linear expansion for a given material and has units of (OC) -I. Table 10.1 lists the coefficients of linear expansion for various materials. Note th~lt for these materials a is positive, indicating an increase in length 'with increasing lemperaLUre. Thermal expansion affects the choice of glassware used in kitchens and laboratories. If hot liquid is poured into a cold cOluainer made of ordinary glass, the container may well break due to thermal stress. The insicle surface of the glass becomes hot and expands, while the outside surface is at room temperature, and ordinary glass may not withstancl the difference in expansion without breaking. Pyrex® glass has a coefficient of linear expansion of about one-third that of ordinary glass, so the thermal stresses are smaller. Kitchen measuring cups and laboratory beakers are often made of" Pyrex so they can be used with hot liquids.
TABLE lQ.1 _ _
Average Coefficients of Expansion for Some Materials Near Room Temperature
Material
Average Coefficienl of Linear Expansion [('C)-I]
E.thyl alcohol
1.12 X 10- 4
19 X 10- 6
Benzene
1.24 X 10-.1
17 X 10- 6
Acetone
1.5 X 10-4
X 10- 6
Glycerin
4.85 X 10- 4
3.2 X 10- 6
Mercury
1.82 X 10-.1
Aluminum
24 X 10
Brass and bronze
Copper Glass (ordinary) Class (P>'rex$)
"
9
6
Lead
2Y X 10
Steel
11 X 10- 6
Inv3r (Ni-fe alloy) Concrete
Material
Average Coefficienl of Volume Expansion [("C)-I]
0.9X
10- 6
12 X W- 6
Turpentine
9.0 X 10-"
Gasoline
9.6 X 10- 1
Air IIelium
3.67 X 10-" 3.665 X 10-3
(h)
FIGURE 10,8 (a) Thermal expansion joints are used to separate sections of rO
Tip 10.1 Coefficients of Expansion Are Not Constonts cocfficicnt~ of expansion can vary sOlllcwhat with temperature.
The
so the given coefficient.-'\ arc aeLu,
allyaverage'l.
Chapter 10
330
EXAMPLE 10.3 Goal
Thermol Physics
Ex ansion of a Railroad Track
Apply the concept of linear expansion and relate it to stress.
Problem (a) A steel railroad track has a length or 30.000 III when the tcmper.atul'c is oDe. ''''hat is its length on a hot day ,.. · hen the temperature is 4G.D C C? (b) Suppose the track is nailed down so that it can't expand. What. stress results in the track clue to the temperature change? Strategy (a) Apply the linear expansion equaLion, llsing Table 10.1 and Equation ]0.4. (b) A track thal cannot expand br 6,L due to external constraints is equivalent to compressing the track by ~L, creaLing a stress in the track. Using the cquaLion relating tensile stress to tensile sU'ain together vV1Lh the linear expansion equation, the amount of (compressional) stress can be calculated using Equation 9.3.
(Exalilple 10.3) Thcrmal expat1~ion: The t:Xlrerne Ilcat of aJul}' day in Asl..HIr}' Park, NewJersl:}'. caused lht:se railroad tracks lO huckle.
,... ...
......,
Solution (a) Find Lhe length of the track
al
...,
40.0'C.
Substiwte given quantities into Equation 10.4, finding the change in length:
!1L = aLc,!1T= [11 X 1O- r>('C)-IJ(30000 m)(40.0'C) = 0.013 m
Add the change to the original length to find the final length:
L = L" + !1L =
30.013 m
(b) Find the stress if the track cannot expand. Substitute into Equation 9.3 LO find the stress:
!1L
A
) ' - = (2.00 X 10 1\ Pa )(0.013 m) 30.0 m I"
8.7 X lO' Pa
... Remarks Repealed heating and cooling is an important pan of the weathering process that gradually \vears things out, \,..-eakening structures over time.
QUESTION 10.3 v\'hm happens to the tension ofv.'ires in a piano when the temperawre decreases? EXERCISE 10.3 ',\'hat. is the length or the same Answer
r~lilruad
track on a cold \Villtcr day \V'hen the telllperat.ure is Dot?
29.994 m
APPLYING PHYSICS 10.1
BIMETALLIC STRIPS AND THERMOSTATS
How Gill different coefficicnts of expansion for metals be used as a temperature gauge and control electronic devices such as air conditioners? Explanation When the ternperawres OLI brass rod and a sleel rod or equal length are r<1iscd by the same amount from sOl11e COl11lnon initial value, the brass rod expands Illorc than I he steel rod bccallsc brass has a larger coeJTlcienr or cxpansion than steel. A
simple device thaL lIses this principle is a bimelallic strip. Such strips can be found in the thermostats of certain home heating systems. The st.rip is made by securely bonding t\VO differcnt metals mgclher. As the temperature of the strip increases, the twO mct~ als expand by different amounts and the strip bends, as in Figurc 10.9. The change in shape can make or break an electrical connection.
10.3
Thermal Expansion of Solids and Liquids
(Appl}'inJ,:" I'hy~ics 10.1) (a):\ bimetallic ~trip bClldsas the tcmperature changes because till' two metals ha\'t~ differenl coerflcicllt<; ofcxpan~ion. (b) A bimetallic strip used in a thermostat to brcak or make electrical contact. (c) The illledor ofa thermostat. showing [he coiled bimewllic su-ip. do yOll .<;llppOSC the sll'ip is coiled?
FIGURE 10.9
Steel
--- -~Bra~s -/
331
I ligher temperaturc
Room temperature
(a)
""hy
(h)
(c)
lt may be helpful to piclUre a t.hermal expansion as a magnification or a phot.ographic enlargemcnt. For example, as the tempcrature ora metal washer increases (Active Fig. 10.10), all dimensions, including the nldius of thc hole, increase according to Equation 10.4. One practical application of thermal expansion is the comlllon technique of using hot water to loosen a metal lid stuck on a glass jar. This works because the circumference of the lid expands more than the rim ofthcjar. Because the linear dimcnsions of an objcct change due t.o variations in temper~ ature, it follows that surface arca and volume of the object also change. Consider a square of material having an initial length L o on a side and therefore an initial area A o = L 02 . As the temperature is increased, the length of each side ill creases to
The new area A is ACTIVE FIGURE 10.10
The last term in this expression contains the quantity au T raised to the second power. Because au T is much less th<:ln one, squaring it makes it even smaller. Consequently, we can neglect this term LO get a simpler expression: 1\ =
Thermal expansion ofa homogcnt:ous metal ,... a~hcr.:-\s the washer is healed. all dimensions increase. (;-.laIc lhal the expansion is exagger,ned in this figure.)
L n2 + 2aL(l2 n,T
A = Au
+ 2aA ll tiT
so thal [10.5]
where 'Y = 20'. The quantity 'Y (Greek letter gamma) is called the coefficient of area expansion.
EXAMPLE 10.4 Goal
Rin sand Rads
Apply the equaLion of area expansion.
Problem (a) A circular copper ring at 20.0°C has a hole with an area of9.98 cm~. \.vhat minimum temperature must it have.so that it can be slipped onto a stcel metal rod having a cross-sectional area of 10.0 cm 2 ? (b) Suppose the ring and the rod are heated simultaneously. \i\Thatminimum change in temperalllre ofbolh v.,rill allow the ring to be slipped onto the end of the rod? (Assume no significant. change in t.he coefficients of linear cxpansion over this temperature range.) Strategy In part (a), finding the necessary temperature change iSjllst a matter ofsubslill.lting given values into Equation 10.5, the equation of area expansion. Remember lhat'Y = 2a. Part (b) is a lil1.le harder because now the rod is also expanding. Uthe I;ng is to slip onto the rod, however, the final cross-sectional areas ofbol.h ring and rod must be equal. vVrite this condition in mathematical terms, using Equation 10.5 on both sides of the equation. and solve for IJ.T.
332
Chapter 10
Thermal Physics
............
" ...
,
..,
_
Solution (a) Find the temperature of the ring that will allow it to slip onlO the rod. Write Equat.ion 10.5 and substitute known valucs, leaving n.Tas the sole unkllown:
Solve for tJ.7; then add this change to the initial tcmperalUre to gel the final tcmperaLUre:
~A = yAo~T
0.02 em' = [34 X 10-6 ('C)-'](9.98 ~T =
em')(~T)
58-9°C
T = 7;'
+
~T =
20.0'C + 58.9°C =
78.9'C
(b) IfbOlh ring and rod arc heated, find the minimum change in temperature that \vill allow the ring to be slipped QIlLO the rod.
Set the final areas of the copper ring and s[Cel rod equal to each other: Substitute for each change in area, L\A: Rearrange terms to get n.J" 011 one side facwr it out, and solve:
all Iy,
"YcAc ~T - "y\A,~T = A, - A e ("Y,Ar; - ysA s ) ~T ~T
A. ,
-
= AI -
fl c
A'
I c = --''-''------'-'''--,--
y(;A(. -
".'lAs 10.0 em:'! - 9.98
cm~
(34 X 10 6,C ')(9.98 em') - (22 X 10 r,°C ')(lO.Oem') ~T=
168'C
........................
""Remark
_
,
...
Warming and cooling strategies arc sometimes useful for separating glass parts in a chemistr)' lab, such as rhe glass sLOpper in a bottle of reagent.
QUESTION 10.4 If instead of healing rhe copper ring in part (a) the steel rod is cooled, would the magnitude of the required temperature change be larger, smaller, or the same? "Vhy? (Don't calculate it!) EXERCISE 10.4 A steel ring ,vith a hole having area of 3.99 cm~ is LO be placed 011 all aluminum rod with cross-sectional area of 4.00 cm 2, 130th rod and ring are initially at a tcmperaLUre of35.0°C. AI. what COllllTlon temperature can the steel ring be slipped onto one end of the aluminum rod? Answer
- 61 'c
vVe can also show that the ;ncrpwH'in vol.'ll1ne of an object accompanying a change in temperature is [10.6]
\"here {3, the coefficient of volume expansion, is equal to 30'. (Note that 1" = 20:' and f3 = 30' only if the coefficient of linear expansion of the object is the same in all dircclions.) The proof of Equation 10.6 is similar to the proof of Equation 10.5. As Table 10.1 indicates, each substance has its own characteristic coefficients of expansion.
10.3
Thermal Expansion of Solids and liquids
The thermal expansion of water has a profound influence 011 rising ocean levels. Al current rates of global warming, scientists predict that about one-hal f of the expected rise in sea level will be caused by thermal expansion; the remainder wi II be due to the melting of polar ice.
333
APPLICATION Rising Sea Levels
QUICK QUIZ 10.2 If you quickly plunge a room-temperaLUre mercury thermometer into vcry hot water, the mercury level ,.. . ill (a) go up briefly before reaching a final reading, (b) go dov'ill briefly before reaching a final reading, or (c) not change. QUICK QUIZ 10.3 LfYOll are asked to make a very sensitive glass thermometer, which of the following working fluids would you choose? (a) mercur}' (b) alcohol (c) gasoline (d) gLycerin
QUICK QUIZ 10.4 Tv·..'O spheres are made of the same metal and have the same radius, but one is hollow and the other is solid. The spheres are taken through the same temperature increase. \Vhich sphere expands more? (a) solid sphere, (b) hollow sphere, (c) they expand by the same amount, or (d) not enough information to say.
EXAMPLE 10.5 Goal
Global Warmin and Coastal Floodin
Apply the volume expansion equation together with linear expansion.
Problem (a) Estimate the fractional change in the volume of Earth's oceans due to an average temperature change of 1°C. (b) Use the fact that the average depth of the ocean is 4.00 X JO~ 111 1.0 estimate the change in depth. Note that {3"", = 2.07 X 10-·,(oq-l Strategy In pan (a) solve the volume expansion expression, Equation 10.6, for 6. V/V. For part (b) lise linear expansion La estimate the increase in depth. Neglect the expansion of landmasses, which would reduce the rise in sea level only slightly. " . . ... -
.......
~
Solution (a) Find the fractional change in volume. Divide the volume expansion equation by \1;1 and substitute:
il\l= {3\f"ilT
(b) Find the approximate increase in depth. Use the linear expansion equation. Divide the volume expansion coefficient oh-\'ater by 3 to get the equivalent linear expansion coefficient: ~
.
ilL = (6.90 X 1O- 5 (oC)-')(4000 m)(loC) =
0.3
III
.........................................~
Remarks Three-tenths of a meter may not seem significant, but combined with increased rnelting of the polar ice caps, some coastal areas could experience flooding. An increase of several degrees increases the value of .1Lseveral times and could Significantly reduce the value of\'\>'atel'front property. QUESTLON 10.5 Assuming all have the same initial volume, rank the following substances by the amount of volume expansion due to an increase in temperalllre, ["rom least to most: glass, merCUr}', aluminum, ethyl alcohol. EXERCISE 10.5 A I.OO-liter aluminum cylinder at :J.oooe is filled to the brim with gasoline at rhe same temperature. If the aluminum and gasoline are warmed to 55.aoC, hO\.v much of the gasoline spills out? !-lint: Be sure to account for the expansion
334
Chapter 10
Thermal Physics
of the cOlllainer. Arso, ignore the possibility of evaporation, aile! assume the \'olume coefficients are good LO three digils. Answer The volume spilled is 53.3 cm:\. Forgcuing to take into account the expansion of the cylinder results in a (wrong) answer of 57.6 cm 3 .
The Unusual Behavior of Water
APPLICATION Bursting Water Pipes in Winter
Liquids generally increase in volurne with increasing temperature and have volume expansion coefAcicl1ls aboUl len Limes greater than those of solids. Over a slllall temperature range, \vater is an exception to this rule, as we can see from its c1ensity-verSlls-tcmperamrc curve in Figure 10.] I. As the temperature increases ("rom D"C to 4"C, water COnlr<1C1S, so its density increases. Above 4°C, water exhibits the expected expansion \-v[th increasing temperature, The density ofwarer reaches its maximum value of 1000 kglm 3 at ~lcC. \Ve can use this unusual thermal expansion behavior of water [Q explain \vhy a pond freezes slowly from the top dm\'I1. \'\Then the atmospheric temperature drops from 7°C (0 6°e, say, the water at the su rface of the pond also cools and consequently decreases in volume. This means the surface "rater is more dense than the watcr belm.. .' il, ,,,'hich has not yet cooled nor decreased in volume. As a result, the surface waleI' sinks and warmer water frorn below is forced to lhe surface to be cooled, a process called u/Jwelling. \Nhel1 the allnospheric temperature is between 4°C and ooe. how'ever, the surface ' . .· ater expands as it cools, becoming less dense than the water below it. The sinking process stops, and eventually the surface water freezes. As the water freezes, the icc remains on the surface because ice is less dense than water. The ice continues (0 build up on the surface, and water ncar the bot(Om of the pool remains at ·loC. Further, lhe ice forms an insulating layer that slows heat loss from the underlying water, offering thermal protenion for ma ri ne life. \'Vithout buoy'ancy and the expansion of water upon freezing, life on Earth mar not have been possible. I r ice had been more dense than waler, it would have sunk [Q the bottom of the ocean and built lip over timc. This could have led 10 a freezing of lhe oceans, turning Earth into an icebound 'world similar 10 I-loth in (he Star 'Vars epic The Empire Slrilws l3(lck. The same peculiar thermal expansion properties of water sometimes cause pipes (Q burst in 'winter. As energy leaves the \vater through lhe pipe by heal and is transferred to the olltside cold air, the outer layers of water in lhe pipe freeze nrs1. The continuing energy transfer causes ice to form ever closer to the center of the pipe, As long as there is still an opening through the ice, the \.,ratcr can expand as its lcmperawre approaches or as it freezes into marc ice, pushing itself into another part of the pipe. Evenlllally, however, the ice will freeze to the center
aoe
FIGURE 10.11 Thcdell~ityorll'ater as a function of tellJperalure. TIle inset al (he rig-Ill s!lm,'s dlat the maximum dellsit\' ufll'
l.onoo 0.9999 0.9998 0.9997
0.9996 0.95
0.0995
o
20
40
60
Tcmperat\lre (OC)
80
100
o
2
.,
6
8
10 12
10.4
Macroscopic Description of an Ideal Gas
somewhere along the pipe's length, forming a plug of ice at [hat point. If there is still liquid watcr bet",,'cen this plug and some other obstruction, such as another ice plug or a spigot, then no additional volume is available [or further expansion and freezing. The pressurc in the pipe builds and can ruplllre the pipe.
10.4
MACROSCOPIC DESCRIPTION OF AN IDEAL GAS
The propcnies of gases arc imponanL in a nurnber of therrnodynamic processes. Our wemher is a good example of the types of processes that depend on t.he bchavior of gases. I f,,,'e introduce a gas into a container, it expands to fill the container uniformly, with its pressure depcnding on the size of the comainer, the temperature, and the amount of gas. A largcr container results in a lower pressure, whereas higher temperatures or larger amOUllts orgas result in a higher pressure. The pt"cssure I~, volume \~ temperature 7~ and alllount 11 of gas in a container are relatcd to each other by an (''I lUi/ion oj slale. The equation of state can be very complicated, but is round experimentally to be relatively simple if the gas is maintained at a low pressure (or a low density). Such a low-density gas approximates what is called an ideal gas. Most gases at room temperature and atmospheric pressure behave approximately as ideal gases. An ideal gas is a collection of atoms or molecules that move randomly and exert no long-range forces on each other. Each particle of the ideal gas is individually pointlike, occupying a negligible volume. A gas usually consists of a vcr)' large number of panicles, so it's convenient to express the amount of gas in a given volurne in terms of the number of moles, 11. A mole is a number. The same number of particles is found in a mole of helium as in a mole of iron or aluminum. This number is knO\\;n as Avogadro's nWllherand is given by
+-
Nt = 6.02 X 10 23 particles/mole Avogadro's llumber and the definition or a mole are rundamental 10 chemistry and related branches of physics. The number of moles of a substance is related to its mass iii bv the expression lit
n = --:-"''-----mobr mass
[10.7]
where the molar mass of the substance is defined as the mass of one mole or that substance. usually expressed in grarns per molc. There are 100S of <-l!OI11S in the world, so iI's natural and convellient to choose a ,-ery largc number like Avogadro's number v.then describing collections or atoms. At the same time, Avogadro's number must be special in somc way because olher,"vise why not just count things in terms of some large pmver ort.en, like 10~N? II turns OUl that Avogadro's number was chosen so that the mass in grams of one Avogadro's number of an element is numerically thc same as the mass 01" one iltom of the clement, expressed in atomic mass units (u). This relationship is ,·ery convenient. Looking at the periodic table of the clemenls in the back of the book, we nnd that carbon has an atomic mass of" 12 u, so 12 g of carbon consisls ofexaclly 6.02 X 102:~ atoms of carbon. The aLOmic mass of." oxygen is 16 u, so in 16 g of oxygen there are (Igain 6.02 X 10 2:1 moms 01" oxygen. The same holds true for molccules: The molccular mass of molecular hydrogen, H:!, is 2 u, and there is an Avogadro's number of molecules in 2 g of molecular hydrogen. The technical definition of a mole is as follO\vs: One mole (mol) of any substance is that amount of the substance that contains as many particles (atoms, molecules, or other particles) as there are atoms in 12 g of the isotope carbon-12.
Avogadro's number
335
336
Chapter 10
Thermal Physics
Taking carbon~]2 as a l.estcase, let's rind the mass oran Avogadro"s number of carbon-12 moms. A carbon-12 atom has an atomic Jllass of l2u, or]2 atomic Illass units. One atomic mass ullit is equal to 1.66 X lO-~·N g, abollt the same as the mass or a neutron or proton -panicles lhat make lip alomic nuclei. The mass III of an Avogadl'o's number or GU'bon-12 atoms is then given by
, (1.66 X 10-'-\ 11/= N.,(12u) =6.02 X IO"(12u) LI
ACTIVE FIGURE 10.12 A gas confined 10 a cylinder whose \'olume can be \aried with a mov,lble pblOll.
g)
= 12.0g
So we sce thm Avogadro's number is deliberately chosen LO be the ilwerse of the number of grams in an atomic Illass uniL In this wa,' the atomic mass of an atom expressed in atomic mass units is numerically the same as the mass of an Avogadro's number of that kind of atom expressed in grams. Because there are 6.02 X 10 23 panicles in one mole or any element, the mass per atom for a given elcment is molar mass '\':.\
For example, the mass ora helium atom is mil.., =
Tip 10.2 Only Kelvin Works! Tcmpenllurcs used in the ideal gas law must always be ill kelvin:,.
Equation of state for an ideal gas ~
4.00 g'/mol "3
6.02 X ]0-' atollls/mol
= 6.64 X 10- 24 g/atom
Now suppose an ideal gas is confined [0 a cylindrical conLainer ,.. . ith a volume that can be changed by moving a piston, as in Active Figure 10.12. Assume that the cylinder doesn'l leak, so the number of moles remains constant. Experimcms yield the following observations: First, when the gas is kept at a constant temperature, its pressure is inversely proportional to its volume (Boyle's law). Second, \',..hen the pressure of the gas is kept constant. the volume of the gas is directly proportional to the temperature (Charles's law). Third, \-\lhen the volume of the gas is held COI1S[aIU, the pressure is directly proportional to the lemperawre (Ga)'-Lussac's law). These observations can be summarized by the following equation of Slate, knm,,'n as the ideal gas law: PI! = l1RT
[10.8]
Tn this equation R is a constant for a specific gas that must be determined from l:xperirnents, ,·.. hereas T is the temperature in kelvins. Each point on a !.J~'er~us V diagram would represent. a different state of the system. Experimcnts on several gases shov.. - that., as the pressure approaches zero, t.he quantity PV/nTapproaches the samc value of R for all gases. For this reason, R is callcd thc universal gas constant. In 51 llllits, ,,,,'here pressure is expressed in pascals and volume in cubic meters, The universal gas constant
Tip 10.3 Standard Temperature and Pr~ssure
EXAMPLE 10.6 Goal
[10.9]
If the pressure is expressed in atlllospheres and the volume is given in liters (recall that 1 L = 10''3 cm 3 = ]O-~ m 3 ), then
r
Chemi~ts oflen define standard tcmperatun: and pressure (STP) to be 20"C and].O
R= 8.3IJ/mol·J(
~
R~
0.0821 L·alJl1/lllo1·J(
Using this value of Rand Equation lO.8, the volume occupied by 1 mol orany ideal gas at atmospheric pressure and at O°C (273 K) is 22.4 L J
An Ex anding Gas
Use the ideal gas law to analyze a system of gas.
Problem An ideal gas at20.0°C and a pressure 01" 1.50 X 10:; Pa is in a container having a volume ofl.00 L. (a) Determine the number ofllloles of gas in the container. (b) The gas pushes against a pist.on, expanding to twice its original volume, while the pressurc falls to atmospheric pressure. Find the final temperature.
10.4
Macroscopic Description of an Ideal Gas
337
Strategy In part (a) solve the ideal gas equation of state [or the number of moles, n, and substitute the known quantities. Be sure to convert. the temperature horn Celsius to Kelvin! \"'hen comparing (wo states of a gas as in pan (b) it's often most convenient to divide the ideal gas equation of the final state by the equation of the initial state. Then quantit.ies that don't change can im mediately be cancelled, simplifying the algebra.
. .. ..,
.
~
Solution (a) Find the number of moles of gas.
+ 273
Convert the temperature to kelvins:
T = Tr:
Solve the ideal gas 1m.. . for n and substitute:
PV= nRT
+
273 = 293 K
PV
(1.50 XIO'Pa)(l.OO X 1O- 3 m")
HT
(8.31.J/moIK)(293 K)
11.=-= ~
= 20.0
6.16 X 10-2 mol
(b) Find the I.cmperatun: after the gas expands to 2.00 L. Divide the ideal gas 1m\' for the final state by the ideal gas 13\,,' for the initial state:
Cancel the number of moles and solve for '/j:
'/1
and the gas constant R,
Pr\~
IIK1 f
PI V;
nR7;
ljVr
·Ii
J:
PI\~
(1.0J X 10" Pa)(2.00 L) ( ,) _ 293 K (1.50 X 10' Pall 1.00 L) =
395 K .-"
Ilo... • • • •
Remark Remember the trick used in P~lrt (b); it's often useful in ideal gas problems. Notice that it wasn't necessary to convert units from liters to cubic meters because the units were going to cancel anyway. QUESTION 10.6 Assul11ing constant temperature, does a helium balloon expand, contract, or remain at constant volume as it rises through the air? EXERCISE 10.6 Suppose the tempcrature of 4.50 L of ideal gas drops from 375 K to 275 K. (a) If the volume remains constant and the initial pressure is atmospheric pressure, find the final pressure. (b) Find Ihe number of moles of gas. Answer
(a) 7.41 X 10·' Pa
EXAMPLE 10.7 Goal
(b) 0.146 mol
Message in a Bottle
Apply the ideal gas law in tandem with 1\ewton's second I;n.. .'.
Problem A beachcomber finds a corked bottle containing a message. The air in the boule is m atmospheric pressure and a temperature of 30.0 C. The cork has a cross-sectional area or 2.30 cm~. The beachcomber places the bottle over a fire, figuring the increased pressure v.till push out the cork. At a tcmpera\.llre of 99°C the cork is ejccted [rom the bottle. (a) '"\That was the pressure in lhe bottle just before the cork left it? (b) \-\That Corce of friction held the cork in place? Neglect any change in volume of the bottle. D
Strategy In part. (a) the number of moles of air in the bottle remains the same as it warms over the fire. Take the ideal gas equation for the final state and divide by the ideal gas equation for the initial state. Solve for the final pressure. In part (b) there are three forces aCling on the cork: a friction force, the exterior force of the atmosphere pushing in, and the force of lhe air inside U1C bottle pushing out. Apply Newton's second law. Just before the cork begins to move, the three forces are in equilibrium and Lhe static friction force has iLs maximum value.
338
Chapter 10
Thermal Physics
.~
Solution (a) Find the final pressure. Divide the ideal
ideal gas
];.lW
g~IS la\'>! ;\(
the final poim by lhe
(I)
at the initial point:
Cancel 71, R, and \.; which don't changc, and solve for Pi
SubstilUte known \'ailles, obtaining the final pressure:
I'f
1',
"IOi II In;
I>I\~
PI \/:
'Ii 7;
7"
P =]' ~
-->
J
I
'I;
" 372 K Pr = ( 1.0 I X 1O"Pa)--=
303 K
1.24 X W', Pa
(b) Find the magnitude or the friction force
acting all the cork. Apply Newton's second law to the cork just before it leaves the boule. Pin is the pn:ssure inside the bOILle, and J~,Ul is the pressure Olllside.
2: F= 0
PillA -
~IUIA
1'~'lilti()1I = PillA - ~)lIIA =
-
(Pill -
l~riLlion
=
0
j~HlI)A
(1.24 X 10' I'a - 1.01 X 10' l'a)(2.30 X 10- 1 m') j'lliai"ll
.
---?
=
5.29 N .~
Remark Notice the use, once again, of the ideal gas la\\' in Equation (I). \,yhencver compi1ring the state ofa gas at different points, this is the best ,vay to do the mat h. One other poi Ill: Heating the gas blasted [he cork out of the boule, ,,'hich meant the gas did ,..:ork Oil the cork. The work done by an expanding gas-dri\'ing pistons and generators-is one of the foundations of modern technology and ,.. . ill be studied extensively in Chapter 12.
(\"'0
QUESTION 10.7
As the cork begins to move, what happells to the pressure inside tile bottle? EXERCISE 10.7
A tire contains air at a gauge pressure of 5.00 X 10 1 Pa at ,l temperature of 30.0°C. After nightfall, the temperalllre drops to -I a,Q°c. Find the new gauge pressure in the lire. (Recall lhat gauge pressure is absolut.e pressure minus aunospheric pressure. Assume constant volume.) Answer
3.01 X 10.1 Pa
EXAMPLE 10.8 Submer ing a Balloon Goal
Combine the ideal gas law with the equation of hydrostatic equilibrium and buoyancy.
Problem A sturdy balloon with volume 0.500 m:~ is attached to a 2.50 X 102_kg iron weight and tossed overboard into a freshwater lake. The balloon is made of a light material of negligible mass and elasticity (although j[ can be compressed). The air in the balloon is initially at atmospheric pressure. The system fails to sink and there aiT no more '.. .·eights, so a skin diver decides to drag it deep enough so that the balloon will rcmain submerged. (a) Find the volumc oflhe balloon at the point where the system will remain submerged, in equilibrium. (b) vVhat's U1C balloon's pressure at that point? (c) Assuming constant tempcralUre, 1.0 ,·",hat minimum dept.h must the balloon be dragged?
Strategy As the balloon and weight are dragged deeper into t.he lake, the air in the balloon is compressed and the volume is reduced along with the buoyancy. At SOllle depth 11 lhe total buoyant force acLing on the balloon and weight, 8 ba1 + 13 Fe , will equal the total weight, W haJ + w F,., and the balloon will remain at that depth. SubstiLLHe these forces into Newton's second law and solve for the unknowtl volume of the balloon, answering pan (a). Then use the ideal gas law to find the pressme, and the equation of hydrostatic equilibrium to nnd the dept.h.
10.4
Macroscopic Description of on Ideal Gas
339
..
,.. ... Solution (a) Find the ,'olume of the balloon at. dlC equilibrium poinL. mho
Find the \"olume of the iron, VFc : Find the maijS of the balloon, \vhich is equal to the mass of the air if\\'e neglect thc mass of the balloon's material:
\
~< = -
Pr,·
/II'M'
-
....
2.50 X I O~ kg = -::-,--,----cc;-:--"---;;- = 0.031 8 m'
Apply ~ewLOn's second law to the systcm when it"s in equilibrium: Substitute the appropriate expression for cach term: Cancel the g's and solve lor the volume of the balloon, Vb;! I: 0.6-15 kg
+ 2.50
X
10' kg - (1.00 X 10" kg/m')(0.031 8111') 1.00 X 10' kg/ Ill'
(b) \o\'hat's the balloon's pressure at the equilibrium point?
Now use the ideal gas 1m\-' LO lind the pressurc, assuming constanttemper(lture, so that 71 = 7 . J
/',,"J II/Oj --=--= J->/\'; nR1~, V,
p =-1',=
Ij
I
=
0500 m 3 ( , 1.01 0.219111'
2.31 X 10' Pa
(c) To what minimum depth musl the balloon be dragged'
Use the equation of hydrostatic equilibrium to find the depth:
P, = I~ilm
+
pgh
2.31 X 10" Pa - 1.01 X 10" Pa pg
=
....
(1.00
X
lO'kgfl11')(9.80 m/5')
13.3111
Remark Once again, the ideal gas law was used to good effect. This problem shows how e\'en answering a fairly simple question can require Ihe application of ~e\'cral different physical concepts: density. buoyancy. the ideal gas law, and hydrostatic equilibrium.
QUESTION 10.8 Ira glass i'i turned upside down and then submerged in water. what happens to the \'olume of the trapped air as the glass is pushed deeper under \\Iatcr? EXERCISE 10.8 A boy takes a 30.0-cl11:\ balloon holding air at 1.00
£lUll al the surface or a freshwater lake down to a depth of 4.00 Ill. Find the volume orthe balloon atthi'i depth. Assume the balloon is made orJight material of little elasticity (although it can be compressed) and Ule temperature of the tr~lpp('d air remains constant.
Answer
21.6 cm:1
340
Chapter 10
Thermal Physics
As previously slated, tile flU !TIber of molecules contained in one mole of any gas is Avogadro's number, tV..I = 6.02 X 10 23 particles/mol, so N
n=-
[10.10]
l\/A
,",,"here n is the number of moles and N is t.he number of molecules in the gas. With Equation ]0.10, we can rewrite the ideal gas law in terms of the total number of molecules as
PV = nRT =
li RT N...
or Ideal gas law
-+
[10.11]
PV= NIi"T where
Boltzmann's constant
'J" Jill = - R = 138 X 10-·" j/K
~
N"
.
(10.12]
is Boltzmann's constant. This reformulation of the ideal gas law will be used in the next seClion to relate the temperature of a gas to the average kinetic cllcrg}' of panicles in the gas.
10.5
THE KINETIC THEORY OF GASES
In Section 10.4 we discussed the macroscopic properties of an ideal gas, including pressure, volume, number of moles, and temperature. In this section we consider the ideal gas model from the microscopic point o[view. \
-+
1. The number of molecules in the gas is large, and the average separation between them is large compared with their dimensions. Because the number or molecules is large, we can analyze their behavior statistically. The large separation between molecules means that the molecules occupy a negligible vol~ ume in the container. This assumption is consistent with the ideal gas model, in which we imagine the molecules to be pointlike. 2. The molecules obey Newton's laws of motion, but as a whole they move randomly. By "random I}'" we mean that any molecule can move in any direction w·ith equal probability, with a wide distribution of speeds. 3. The molecules interact only through short-range forces during elastic collisions. This assumption is consistent with the ideal gas model, in which the molecules exert no long-range forces on each other. 4. The molecules make elastic collisions with the walls. 5. All molecules in the gas are identical. Although we often piclLlre an ideal gas as consisting of single atoms, molecular gases exhibit ideal behavior at low pressures. On average, effects associated with molecular structure have no effect on the motions considered, so we can apply the results of the following development. to molecular gases as well as (Q monatomic gases.
Molecular Model for the Pressure of on Ideal Gas As a first application of kinetic theory, ''ie derive an expression for the pressure of an ideal gas in a container in terms 01" microscopic quantities. The pressure of the
10.5
gas is the result of collisions between the gas molecules and the ,\'alls of the container. During these collisions, the gas molecules undergo a change oFmolllentum as a result of the force exerted on them by the walls. Vile now derive an expression for the pressure of an ideal gas consisting of N molecules in a comaincr of volume V. In this section h'e lise m LO represent the mass of one molecule. The container is a cube with edges of" length d (Fig. 10.13). Consider the collision of one molecule moving with a velocity -v.>: to\'V'ard the Icfthand face of" the box (Fig. 10.14). After colliding elastically with the wall, the moleculc moves in the positivc x-dirccljon wilh a velocity +vx ' Because the momentum of the molecule is - m.vx before the collision and + llIVx afterward, the change in its momentum is
.a.fJx
= mv x
(-mv.J = 2mv x
-
IfF I is the magnitude of the average force exerted by a molecule on the wallin the time 6.t, then applying NeWLon's second law to the ".. 'all gives
F = .a.jJ.\ I
The Kinetic Theory of Gases
341
d
d
" A cubical box with sides or length rl containing an ideal gas. The molecule shown 1110\'es \,'ith Vt.."locit}' V.
FIGURE 10.13
2mv x D./
D./
For the molecule to make two collisions \'lith the same wall, it must travel a distance 2d along the x-direction in a time af. Therefore, the time interval between I.\·\,o collisions with the same wall is!1t = 2d/vx ' and the force imparted to the wall by a single molecule is , mv x 2mv~. 2mv x F --I D./ d 2d/v ~
Before collision
x
The LOtal force F exerted by all the molecules on the wall is found by adding the forces exerted by the individual molecules: ,
l' =
' d'Ill ( Vl x ~
+
, v:!x -
+
'l
'1.
=
+V,
Aftcr collision
In this equation v lx is the x-component of velocity of molecule 1, v2~ is the .,,component of velocity or molecule 2, and so on. The summation terminates when we reach Nmolecules because there arc Nmolecules in the container. Note that the average value orlhe square oflhe velocity in the x-direction for N molecules is
Vx 2
.-
FIGURE 10.14 A molccult: moving along the x-axis Illake~ elastic collisiulls I\'ilh the Ivalls orlhe cOllLailll::r. 111 colliding \\ ilh a wall, the molecule's 1ll1)11lClltum is reversed, and the molecule exerts a force on lhe \'1<\11.
')
_V-,I",._+_ _v,,~.~,-.--,-+:-__+ _ _v".'::...'x~ N
where
vx :! is the average value of v/. The total force on the wall can then be \vrincn Nm-.,
F =d- v,>:
Now we focus on one molecule in the container traveling in some arbitrary direction with velocity ~ and having components V x ' v~" and v:. Tn this case we must express the total force on the wall in terms of the speed of lhe molecules rather than just a single component. The Pythagorean theorem relates the square of the speed to the square of these components according to the expression v'!. = v/ + v/ + v:'l. Hence, the average value of v 2 for all lhe molecules in the container is related to the average values V x 2, Vy~, and v: '/. according to the expression v'J. = v} + v/ + v,,2. Because the Illotion is completely random, the average values 2 V x 2, VI 2, and v: are equal 10 each other. Using this fact and the earlier equation for V x 2, we find thal -') 1----;; v~,3" vThe total force on the wall, then, is F=
1~
(m;')
The glass vessel <:olllains dry icc (solid carbon dioxide). Carbon dioxide gas is denser thall air, hence falls when poured from the cylinder. The gas;s colorless, but is made visible hy Ihe formation of tiny ict.: C1"ystab from waler vapor.
342
Chapter 10
Thermal Physics
This expression allows us 1.0 lind the Lotal pressure exerted on the wall by dividing the force by the area:
P=
F= A
F= '3 (N""""",,) d,1 mv- =:i (N) V " I
d2
Pressure of an ideal gas ~
p:::::
I
1llV-
.'3(VN)(I " " " ) "2 mv -
[10.13]
Equation 10.13 says that the pressure is proportional to the number of molecules per unit volume and to the average translational kinetic energy of a molecule, !mJ. \'\lith this simplified model of an ideal gas, '"e have arrh-ed at an imponant result thal relates the large-scale quantity of pressure to an ilLOmic qualuity: the aycrage value of the square of the molecular speed. This relationship prO\1.des a key link be [ween the aLomic world and the large-scale world. Equalion 10.13 captures some familiar fealUres of pressure. One war LO increase [he pressure inside a comaincr is LO increase the number of molecules per unil \'olume in [he cOl1laincr. YOLI do lhis when you add air lO a lire. The pressure in the lire can also be increased by increasing the average lranslalional kinetic energy of the molecules in the tire. As we will see shorllr, this can be accomplished by increasing the lemperalure of the gas inside the tirc. Thm's why the pressure inside a tire increases as the lire warms up during long lrips. The cominuous Oexing of the lires as they mo\'e along the road lransfers energy lO the air inside lhem, incrcasing the air's lempcrat.ure, which in t.urn raises the pressure.
Molecular Interpretation of Temperature Having relalccl [he pressure of a gas to the average kinetic encrgy of the gas molecules, we now relate temperature LO a microscopic descriplioll of thc gas. \'\'e can obtain some insighl into the meaning of I.Clllpel'Cllllre by mulliplying Equation 10.13 by the \olumc:
.. .(' ") P\'=JN'lmvComparing lhis equalion with the equation of state for an ideal gas in the form of Equation 10.11, P\'= NJ:/11~ we note that the Jerl-hand sides of the twO equations arc identical. Equating Ihe right-hand sidcs, \,·e obtain Temperature is proportional to average kinetic energy ~
T= - 2
3/; IJ
I ~ (ijlllV-)
[10.14]
-
This mcans thal the temperature of a gas is a direct measure of the average molecular kinetic energy of tbe gas. As the temperaLHre of a gas increases, the molecules move with higher a\'erage kinetic energy. Rearranging Equalion 10.14. we can relate the translational molecular kinetic energy lO thc lemperalllre: Average kinetic energy per molecule ~
[10.15]
So the a\'erage translational kinelic cnergy per moleculc is ~1:8T. The LOtal translational kinclic energy of N molecules of gas is simply .'\' times the average energ} per molecule, Totol kinetic energy of N molecules ,)
KFW!;,I
=
lv'(~mv:!)
=
~:VhI1T= ~nRT
[10.16]
where we havc used /(/1 = HIIV, for BolLzmann's conSlant and 11 = Nlt\~" for the nurnbcr of moles of gas. From lhis result, we see that the total tr,U1slational kinetic energy of a system of molecules is proportional to the absolute temperature of the system.
10.5
The Kinetic Theory of Gases
343
For a monatomic gas, translational kinetic energy is the only type of energy the molecules 0111 ha\'e. so Equation 10.16 g-ives the internal energy UfoI' a monatomic gas:
u=
~nRJ' (monatomic gas)
[10.17]
For diaLOmi<.' and polyawmic molecules, additional possibililies for energy storage arc available in the vibration and rotation of the molecule. The square roOl of if! is called the root·mean-square (nl1s) speed of the molecules. From Equation 10.15, we get, for I he nils speed, [10.18] where j\;[ is the molar mass in !(iLogmms j)('" 1lI0h', if R is given in 51 llllits. Equation 10.18 sh0'''·s that, a1. a given temperature, lighter molecules tend to move faster than heavier molecules. For example. if gas in a \'essel cOllsisLS of a mixture 01" hydrogen and oxygen, the hydrogen (I-1;!) lIlolecules, with a molar mass of 2.0 X 10-:1 kg/mol, move fOllr times faster than lhe oxygen (O:!) molecules, with Illolar mass 32 X 10-3 kg/mol. If \..'e calculate the rms speed for hydrogen at room temperature (-300 K), we rl11d 3(8.3] 1/1110\' K)(300 K ) , . , =1.9XlO·m/s 2.0 X \0 . kg/mol
+-
Root-mean-square speed
Tip 10.4 Kilogroms Per Mole, Not Grams Per Mole 111 t..he eqll of molar m,,~s 1\1 11lUSt be consistent wllh the unit:>
orlhe gasco11S1allt N. In particu· lar, it His in 51 unils . .\1 must be expressed in kilograms per mole. 1101 gl"amo; per mole.
This speed is abuut 17% of the escape speed for Eanh. as calcult:lIcd in Chapter 7. 13ecilllsc it is an average speed, a large number of molecules have llluch higher speeds and can therefore escape frulll Earth's atmosphere. This is \"'hy Earth's atmosphere doesn't currently contain hydrogen: it has all bled ofT into spa<.:e. Table 10.2 (page 344) lists the rms speeds for variolls molecules at 20°C. A system of gas at a given temperature will exhibit a variety or speeds. This distribution of speeds is known as the lHfl.\"WfU Vl'loril)' dislriulIfion. An example of such a distribution {or nitrogen gas at two different temperawres is given in AClive Figure 10.15. The horizontal axis is speed, and the \"CTtical axis is the nllmber of molecules per unit speed. Notice lhal. three speeds arc of special interest: tlte IllOSI probable speed, corresponding to the peak in the graph; the average speed, ,.. . hich is found b}1 averaging over all the possible speeds; and the rillS speed. For every gas, note lhal. VillI' < Vav < vrll1~' As the temperatllre rises, these three speeds shift to the right.
QUICK QUIZ 10.5 One conwiner is filled \,,..ith argon gas and another with hclillm gas. BOlh containers arc at the same temperature. \Vhich atoms have the higher rms speed? (<1) argon, (b) helium. (c) they have the same speed, or (d) not enough information to say.
200 Curves
160
x=
Gllctllalt'd
for
IO'>11it..roy;cll IllOleCllle~
120
80
40
200
'100
liOO
!:IOO
v (lll/S)
I 000
I 20U
! 400
I GOO
ACTIVE FIGURE 10.15 The Maxwell speed distribulloll fIX JO" nilrogell molecules at 300 K and yon K. The IOlal :H"e~l under either CllIYC l..'quals the totaillumbe." 01 11101el"ldc.~. The most prc,babh.: speed urnI" th(: l\\'('r:age spel'd 1',1\' alld the rool11le;ll\-~qLlare speed 11,,,,. .In:- illdi("ilted for I he ~IO(J-" Cllr\'c.
344
Chapter 10
Thermol Physics
TABLE 10.2 Some rms Speeds Gas
Molar Mass (kg/mol)
1-1,
2.02 X 10-:1
He
4.0 X 10-'
1-1,0
18 X 10-:1 20.2 X 10-"
Ne
APPLYING PHYSICS 10.2
1902 I 352 637 602 51l
N:! and CO
28.0 X 10-5\
NO
30.0 X 10-'
0,
32.0 X 10-'
CO,
44.0 X 10-:1
SO,
64.1 X 10-"
494 478 408 338
EXPANSION AND TEMPERATURE
Imagine a gas in an insulated cylinder with a movable piSLOIl. The pistoll has been pushed imvard, compressing the gas, and is now reJcased. As the molecules of the gas strike the piston, they move it outward. E.xplain, from the point of vic\\' of the kinetic theory, how the expansion of this gas causes its lernperature Lo drop.
piston to move \vilh some velocity. According to the conservation of momentulll, the molecule must rebound with less speed than it had before the collision. As these collisions occur, the average speed of the collection of molecules is therefore reduced. Because temperature is related to the averagc speed of the molecules, the tcmperaturc of the gas drops.
Explanation From the point of view of kinetic theory, a molecule colliding with the piston causes the
EXAMPLE 10.9 Goal
A C linder of Helium
Calculate the illlernal energy ora system and the average kinetic energy per molecule.
Problem A cylinder contains 2.00 mol o[ helium gas at 20.0°C. Assume the helium behaves like an ideal gas. (a) Find the to[
... .
.
...
Solution (a) Find the lotal imernal cnergy of the system. SubSliwte vallles into Equation ]0.17 with n = 2.00 and '1'= 293 K:
u~ ~(2.00l11ol)(8.3IJ/l11ol·K)(293 K)
= 7.30 X 1O'J
(b) ''''hat is the avcrag'c kinetic energy per molecule?
SubSl;wtc gi\len valucs illlo Equation 10.15:
~Jnlf' = ~IiIJT = ~(1.38 X 1O-'''j/K)(293 K) =
6.07 X 10-21 J
Summary
345
(c) Ilow much energy must be added to double the nns speed? From Eqmllion 10.18. doubling the rms speed n:qllin:s quadrupling T C;,]ru];,IP Ihe '''PfJlIireri change OrilHCrnal energy, which is the energy that IIIUSI be put illlo the ')ysLCm:
Remark
Computing changes in
QUESTION 10.9 True or False: At the same
intern~ll
temper~lLllrt'. J
'['. -- ['')
J.
-
L'!, -_:l21/ /'T "\ f -
/Y/', _:l - ~ 11 /'('/' \ I
:l ~ 11"\
-
'I') •
.'> [' = ;(:LUU mol)(H.3J .IImol· ~)[(~,OO X 293 K) - 293 K] 2.19 X 10'J
energ\' will be important in understanding engine cycles in Chapter 12. mole or heliulll gas hali the same inlernal energy as I mole of argon gas.
EXERCISE 10.9 The tcmperalure of5.00 moles of argon gas is !cl\\crcd I"rom 3.00 X IO~ K to 2.-1-0 X 1O~ K. (a) Find the change in the inlernal energy, nU, of the gas. (b) Find thc challgl' in the averag-e kinetic energy per atom. Answer
(a) tJ.U= - 3,74 X lO:lJ
(b) - 1.2~ X 10
"J
SUMMARY 10.1 Temperature and the Zeroth Law of Thermodynamics Two systcms arc in thermal contact if CIlc:q..D' Gill be l'xc!J
The C'xchange of energy betw4.:en [wo OI~ic('I' becau,e of difTercllct's in their tempcl"atures is called heaL. The zeroth Jaw of thermodynamics staiC's that if l\.,."O objecls A and Bare separal.e1y in lhennal e<"]uilibriul11 wilh a third object, then .\ and B arc in t11t·nll'al equilihrium with each mher. EquivalcnLlr, if the third ol~ieci is a thermometer. then the temperature it measures for A and B will be the same. Two objects in thermal equilibriul11 are al the same temperature.
10.2
Thermometers and Temperature Scales
Thermometers measure tcmpcramre and ~lrc hased on physical properties. such as the temperatllre·dependcllt expansion or conu"action of a solid. liquid, or g-ao;. Thcsl' chan~es in volume arc related (0 a line.lr scale, the most cOlllmon being the Fahrenheit, Celsius. and Kelvin scales. The Kehin temperature scale takes its 1£.'1"0 poinl as absolute zero (0 K = -273.15 C), the point at which, b) extrapolation, the pressure orall ga",es f~llls to lero. The relationship bctween the Celsiu" temper;lture '/~ and the Kch'ill (absoltlle) temperature 1'is C
T(:= T- 273.15
110,11
ThL' relationship between the Fahrenheit and Cdsiu'i IcmpcraLllrcs is
[10,201
10.3 Thermal Expansion of Solids and Liquids Ordinarily a substance cxpands when hcated. If an o~iect has an inil iallenglh L o al some tCmperalUre and undergoes ;) ch.mge in tcmpcrature ~ its linear dimension changes by tile amount ~/., which is proponional to the object's initial length and the LCmperaturc change:
7:
[10.11 Th(' parameter a is called Ihe coefficient of linear expan· sian. The change in area or a sublitance with change in temperaLllrc is given by
[10,5] where y = 2a i'i Ihe coefficienl of area expansion. Similad). the change in volume with temperature of most substances is proportional to the initial \"olume '~l and the temperallirc chang-e ~T:
.'>\!= {3\",tJ.T
[10,6]
f3 = 3a is the coefficient of \'oJume expansion. The expansion and comraClion of material due to
where
changes in tcmperature creates stresses and strains, ...onll'timcs surtlciclll to cause fracturing.
10.4 Macroscopic Description of an Ideal Gas A\'ogadro's number i~ 1\~\ = 6.02 X 10~:l panicles/mol. A mole of anything. by definition, consist'\ of an A\'ogadro's lIumber of panicles. The number is derined so Ihat one mole of carbon-12 atoms has a mass of exactly 12 g. The rnas~ of one mole 01" a pure substance in grams is the same, numerically, as that <;ubst.ance's atomic (or molecIllar) mass.
346
Chapter 10
Thermol Physics
An ideal gas obeys the cqumion P\I
=
[10.8J
nRT
where P is the pressure of the gas, V is its VOlli llle, n is Ihe number of moles of gas. R is the universal gas constant (8.31 J/mo!- K), and T is the absolute temperature in kelvins. A real gas at ver), 1m..· pressures behaves approxi-
mately as an ideal gas. Solving problems usually entails comparing twO different states of" the same system of gas, dividing the ideal gas equation for the final slate by the ideal gas equation for the initial SLate, canceling facLOrs thal don't change, and SOlving for the unknown quantity. 10.5 The Kinetic Theory of Gases The pressure of N molecules or an ideal gas contained in a volu me V is g"iven by P =
.'(N)(I V T .I
-
<)
~
~)
tJl,V 2
Thl: average kinetic energy of the molecules ofa gas is directly proportional to the absolute temperature of the
gas: [10.15]
"ll
The quantity is Boltzmann's constant (1.38 X IO-23 J/K). The internal energy of 12 moles of a mOllawlllic ideal
gas is
u=
~'lIRT
[10.17J
The root-mean-square (rms) speed of the molecules of a gas is
l),-",~
_)3"I
-
-
111
J
31iT 111
[10.18]
[10.13]
where &mv'2 is the average kinetic energy per molecule.
FOR ADDITIONAL STUDENT RESOURCES. GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE·CHOICE QUESTIONS 1. On a vcry cold dar ill upslate :\lew York, [he temperatllre is -25°C, which is equivalent to what temperawre in Fahrellheil? (al -46°F (b) -77"F (c) 18°, (d) 298 K (e) -13°,
2. Convert 162"F to the equivalent temperature III Kelvin. (a) 373 K (b) 288 K (c) 345 K (d) 201 K (e) 308 K 3. The Statue of" Liberty is 93 In tall 011 a summer morning when the temperalUre is 20 c C. I r the temperature of the st;uue rises from 20"C to 300C, what is the order of magniwde of the statue's increase in height? Choose lhe best estimate, treating the statue as though it were solid copper. (a) 0.1 I11Il1 (b) I 111111 (c) 1 C111 (d) 10 CI11 (e) I 111
4. A hole is drilled in a metal plat.e. \oVhcll the metal is heated, ,,,,,hat happens to the dial1lct~r of the holc? (a) it decreases. (b) It increases. (c) It remains the same. (d) The answer depends on the initial temperature of the metal. (e) None of these 5. A cOlltai ncr holds 0.50 nl~ of oxygen at all absolute pressure.: of 4.0 atm. A valve is opened, allowing the gas to drive a piston, increasing the volume of" the gas until the pressure drops to 1.0 atm. I f the tempera~ ture remains constant, what ne'.. . volume does the gas occupy? (a) 1.0 mll (b) l.5 ln~ (c) 2.0 rn 3 (d) 0.12 m 3 (e) 2.5 111 3
6. II' the volullle of an ideal gas is doublc(! while its tCIl1pcrawre is quadrupled, does [he pressure (a) remain the same, (b) decrease by a factor of2, (c) decrease by a factor of 4, (d) increase by a factor of 2, or (e) increase by a factor of I.l? 7. One way to cool a gas is to let it expand. \o\'hen a certain gas under a pressure of 5.00 X \06 Pa at 25.0°C is allO\ved to expand to 3.00 times its original volume, its final pressure is 1.07 X lOr, Pa. What is its final temperalure? (
10. Which of the assu mptions below is "/lol made in the kinetic theory of gases? (a) The number of" molecules is very small. (b) The molecules obey NeWLOn's laws of motion. (c) The collisions between molecules are clasI ic. (d) The gas is a pure substance. (e) The average separation between molecules is large compared with their dimensions. 11. Suppo:-.;e for a brier lllomclltthe gas molecules hiu,ing a wall stuck to the \""all instead or bouncing off the wall. How would the pressure 011 [he wall be affected during th,tl brier lime~ (a) The pressure would be I.ero.
Problems (b) The pressure would be halved. (c) The pressure would remain unchanged. (d) The pressure ''lould double. (e) The answer would depend on t.he area of the wall.
347
molecules? (a) Il increases by a faclurof3, (b) It remains the same. (c) It is one third of the original speed. (d) It is v3 limcs the original speed. (e) Il increases by a rae· tor of 6.
12. If the temperature of an ideal gas is illueased frolll 200 K LO 600 K, what happens [Q the rms speed of the
CONCEPTUAL QUESTIONS 1. Wh>, does an ordinary glass dish llsually break whcn placed on a hoI. stuve? Dishes made of Pyrex glass don't break as easily. What characteristic of Pyrex pre\'enLs breakage? 2. A rubber balloon is blown up ,md Ihe end tied. Is the pressure inside the balloon greater than, less thall, ur equal to the ambient atmospheric pressure? Explain. 3. Cornman thermometers are made of a mercury column in a glass tube. Based on t.he operation of t.hese common thermometcrs, which h
7. Some picnickers stop al a cOI1\·cnience store to buy food, including bags of potato chips. They then drive up into the mountains to their picnic site. \-\then they unload the l"ood, they not.ice l.har the bags or chips arc puffed up like balloons. vVhy did this happen? 8. t\:farkings to indicate lenglh are placed on a steel tape in a roOIll that is at a temperature of 22"C. .vleasure· ITlenL<; are then made with tbe same tape on a day \V'hen the temperature is 27 c C. Are the me<-lsuremcnlS too long, roo short, or accurate?
9. Why do vapor buhbles in a pot of boiling water get larger as they approach the surface? ph1l1eL~
4. \-V'hy is a pm...er line more likely LO break in winter than in summer, e\'en if it is loaded with the same ''leight?
10. Why do slmtll sphere?
5. ObjecL<; deep beneal.h the surface of the ocean are subjected [Q extremely high pressures, as we sa\\' in Chapter 9. Some bacteria in these em'ironmcnts ha"e adapted to pressures as milch as a lhousand limes atmospheric pressure, How might such bacteria be alIened if they were rapidly moved [() the surface of Ihe ocean?
Ll . .'vie wi lids on glass jars can often be loosened by running hOl waler over them. \-\'hy does that work?
tend to have little or no aUno--
12. Suppose lhe volume oran ideal gas is doubled wllile the pressure is reduced by hair Does the internal enert,')' of the gas increase, decrease, or remain the same? Explain.
6. After food is cooked in a pressure cooker, why is it "ery importalllto cooll.he conlainer with cold ' . . ater before attempting to remove the lid?
" PROBLEMS ~
The Problems fOl" thi~ chaptl:r mav be assignl:d ouline at WehAssign. 1,~,:.I, = stl'aigillforward, illl(."l"lllcdble,1 h
WebAssign
II
Sindy Guide
SECTION 10.1 TEMPERATURE AND THE ZEROTH LAW OF THERMODYNAMICS SECTION 10.2 THERMOMETERS AND TEMPERATURE SCALES DJFor each of the foHowing temperatures. find the equivalent temperaturc on the indicated scale: (a) -273.15°C
tcmperatun.' whcn the pressure is 0.0400 aun? (b) ,,,that is the pressure at 4500C? 3. COllver-t the following temperatures to their values Oil lhe Fahrenheit. and Kelvin :.cales: (a) thc boiling point of
liquid nilrogcn, -196"C (b) human body temperature, 37.WC 4. Death Vallc)' holds thc record for the highest recorded lCmperalurc in the United States. On Jtlly lO. 1913, at a place called Furnace Creek Ranch, the temperature rose to 13 le F. The lowest U,S. temperature ever recorded occtilTed at Prospecl Creek Camp in Alaska on January 2~, 1971, whcn the tcmperalUrc plumrneted to -79.8" F. (a) COrl\ien these temperatures to the Celsius scale. (h) Con\,cr·t lhe Celsius temperatures to Kelvin. L
on Ole fahTen))e'n scale, (b) 9S.6°F' on tile Celsius scale, and (c) 100 K on the Fahrenheit scale.
QJShow that the tcmperature ~40" is unique in lhal it has the same ntJlllcrical value on the Celsius and Fahrenheit scales.
2. The pressure in a consl:.,nt-volullIe gas thermometer is 0.700 aLlll at lOO°C and 0.512 atm al O"C. (a) ''''hat is the
6. A constanl-volume gas thermometer is calibrated in d11' icc (-SO.O°C) and \11 boiling cth) I alcohol (7S.D"C).
Thermal Physics
Chapter 10
348
COI1-
Lead has a density of 11.3 X I03 kg/lll'~ at We. (a) \'\'hal is the density of' lead at 90°C? (b) Based on your answer to pan (a), now consider a situation in which you plan to invest in a gold bar. \'\Iould you be bettcr otf buying it on a warm da};? E..xphtin.
7. Shmv l1l<11 if the tenlpcrallirc 011 the Celsius scale chang-cs by ~ 'l~ .. the Fahrenheit lClnpcralure dlanges by ~ T,.. = *67'(;.
18. A copper wire WiUl length 10.0 III and cross-sectional arca 2. ' 10 X 10-" nl 2 is strelched tallt between two poles undcr a tension of 75.0 N. What is the tension in the \\~re when the Lcmperallll'C falls by IO.O°C?
The rcspcClivc pressures arc O.~IOO aUll and 1.635 all11. (a) Wll;tl vallie of absolute zero docs the caJibration
[ill-
)~c1d?
(b) \'Vhat pressures would be fOLilld at the rrecL~ ing and boiling points of waLe.:r? (Note that we have the
linear relationship P= A
+ B'I:
where A and IJ are
slanl~.)
8. The temperature dilTcrencc between the inside and the OllL'iidc of it home on a cold winter day is 57.0 F. Express this difference on (a) the Ct'lsius scale and (b) the Kelvin scale. OO
9. _
19. All underground gasoline tank can hold 1.00 X 10'" gal-
lons of gasoline at 52.0°F. If the h'lnk is being filled on a day when lhe outdoor temperawre (and the gasoline in a tanker tnlck) is 95.UoF, how many gallons from ule truck can be poured into the lank? Assume thc tcmperature of the g-
A nlJl"se measures the lClllpcralurc ora P:ll:iCllllo be
43°C. '-\-'hal is this temperature on lhl; Fahrenheit scale? Do yOLl think the p
10. _
Tempel,nure differences on the Rankine scale are identical LO dia'crenccs on lhe Fahrenheit scale, bUl absolU1.e zcro is givcn as OOR. (a) Filld a relalionship convertinglhe lemper;;uures 'Ii, of the FahrenheiL scale La the corresponding tempcraturcs '/~{ of the Rankine scale. (b) find a second relationship convening wmperarures T u of Lhe Rankine scale to the temperaLUrc~ T" or the Kelvin scale.
SECTION 10.3 THERMAL EXPANSION OF SOLIDS AND LIQUIDS 11. The New River Gorge bridge in \Vest Virgini;-l is a 518-111-
long sLcel arch. How milch will iLs length c11
12. _
by T = 27TV LI g, where L is its length. does the ch'lnge in lengul of the rod cau:>c the clock to run J~lst ur sIO\...-? pair of eyeglass frames are made or epoxy plastic (cocf.l"iciCIll or linear expansion = 1.~O X 10-<1 °c- \). AI room temperature (~O.O°C). Lht' frames have circu!;lr lens holes 2.20 cm in radius. To what lempCr(lwre must the rrarlH.':s be hcmed if lenses 2.2\ CIll in rac1iu~ arc [0 be in.o,ened into them?
[:}1JA
20. _ Show that the coetTlciell1 or volullle expansion, {3, is related to the coerticient of linear expansion, (l', through the ('xprei:iSiOll {3 = 3a. 21. A gold ring has an inner diamcter of 2.168 cm at a temperature of 15.0°C, Dctermine il.~ inner diameter at IODoe 50 (O'g"l([ = 1.42 X 1O- C I).
22. A construction worker uses a steel tape 1.0 measure the lengul of an aluminum support column. If the lllca~ured length is 18.700 Il1 \.... hen the temperature is 21.2°C, what is []lC measured length when the temperature rises to 29.4°C? (Nol.f': Don't neglect 1he expansion oflhc tape.) 23. The band ill Figllre P10.23 is stainless steel (cocITicicn t of linear expansion = 17.3 X 10- 6 °e- l; Young's 1ll0dUlrlS = 18 X lOw N/m:!). It is essentially circular with an illiti,d lucan radius of 5.0 III 111 , a height of 4.0 mill, and a Thickness of (UiO mill. If the band JUSt fits snugly over the FIGURE Pl0.23 tooth when he'ltcd to a Lelllperature of 80°C, what is the tension ill the band whcn it cools to
24.
14. A spherical steel ball bearing has a diameter of 2.540 cm at 25°C. (a) What is iL~ diameter ",hell its temperature is raised to loaoe? (Il) \'Vhat temperature change is required LO increase its volume by 1%? 15. A brass ring or di
thill its densit)' at thc
hj~he,.
lemptTil!ure
Po
P~J+f3tJ.T
i~
given by
mJ
Thc Trans-Alaskan pipeline is I 300 kill long. reaching from Prudhoe Bay to the port of Valdez. and is subject to temperat.ures ranging from -73°C to +35°C. How IllllCh docs the steel pipelinc cxpand due to [he di1ference in l.emperalUre? How (an this expansion be compensated for?
125.IThc (Iveragc coefficient of volume expansion for carbon tcu'acll loride is 5.81 X \0-4 (0C) -I. If a 50.0-gal steel container is filled complelely wit.h carbon letrachloride when t.he tcmperatlll"C is 10.0°C, how much will spill over wl](.:n thc temperature rises to 30.0°C? 26.
&I
The density of gasoline is 7.30 X lOt kg/ms at O°c. Its aver
Problems would you gel if )'011 bouglll 10.0 gal of gasolinc al O°C rather than at ~O.O°C from a pump thai is IlOI lempel-ature compt.:nsated? 27. Figure P10.27 ~hows a circular steel casting with a gap. If lhe casting is heated, (a) does the width of the gap incrca.se or decrease? (b) The gap width is 1.600 em when the lcmperallirc is 30.0°C. Determine the gap widtJl when the temperalure is I ~O°c.
FIGURE Pl 0.27
SECTION 10.4 MACROSCOPIC DESCRIPTION OF AN IDEAL GAS 129.IOne mole of oxygell gas is at a pre~sure of 6.00 aIm and a tt:mperatlll"C of27.0°C. (
&I
31. (a) An ideal gas occupies a \'oluTlle of 1.0 cm:1 at 200C and aUl10spheric pressUl·c. Determinc the !lumber of lllulecules of gas in the container. (b) If the pressure of lhc I.O-cm:\ volume is reduced to 1.0 X 10- 11 Pa (an extremely good vacuum) while Ihe temperature remains cunSlant.. how mallY moles of gas rcmain in the container:-
3;~.
10
atlll.
weather balloon is designed to expand to a maximum radius of ~O rn at iL'i working altitude. where tJle air presSllre is O.O~O atm and tJle tcnqJerawre is 200 K. If the halloon is filled al auno~pheric pressure and 300 K, what is its radius atlil"tofl?
~O.O°C.
:~2.
lower the pressure of the g
Ihe tank
J.OO
I:{5.IA
28. On a day whell thl' temperature is
30.
349
A umk having a volume orO.IOO m J contains helium gas al 150 allll. How lll
34. A rigid lank conlains 150 molc~ orall ideal gas. Delermine the number of moles of gas lhal IllllS1 be:.: withdrawn fnun
.%. The cknsily of helium gas at T = we is PI) = 0.179 kg/m:~. The temperatun:, is Ihen raised to '1'= 100~C. but the pressure is kept comlalH. l\ssllmillg- tJ1C helium is an ideal gas. calculale tile new density Pj of the gas.
:H. An air bubble has a \olumc of 1.50 cm 3 when il is rele,lsed b)' a snbmarine 100 In bdow the surface of a lake. What i~ the volulnc of' the bubble when it reaches the .surf~lce? Assullle the tempel'
38. _ Tile ideal g-as law can be recast. in term~ of the densilY or a gas. (a) LSC dimcnsional <:lnaly:-.is to find an expression I"or the density p of;l gas in lerms or the number or moles /I, the volume \I, and the molc(ular \....eighl M in kilograms per mole. (11) \Vith the expression found in pan (a), show Ih"t
p
= .!!- RT M
I'lr an ideal gas. (c) Find the densil}'ol"the carbon dioxidc atmosphere at the sllrface of Venus, where the pressure is 90.0 atm and the temperature is 7.00 X LO:! K. (d) Would all evacuated steel shell of radius 1.00 III and mass 2.00 X 1O~ kg rise or fall in such an allllosphere? \·Vhy?
SECTION 10.5 THE KINETIC THEORY OF GASES 39. \·Vhal i:-, t.he average kinetic energy of a molecule of oxygen at a lem perature of 300 K? 40. A seal<:d cubical cOlHainer 20.0 cm on a side contains three Li Illes A\'ogadro's lllllllber of molecules at a tCl1lpera~ lure of 20.0°e. Find the force exerted by the gas on one of the \valls or the container. [TIJUse Avogadro's numher 10 find the lllass of a helium alUm. 42. Tile rlllS speed 01" an oxygen molecule (O~) in a container of oxygen gas is 625 II1/S. \o\-1lat is thc lcmperature 01" the gas? 43. All ideal gas in ;1 container is at a tcmpcrature of 77.We. \·Vllat is the average translational kinetic energy of a gas molecule in Ihe container? 144.1_ ..-\ 7.00-L vessel contains 350 moles of ideal gas at a prTSSUlT of 1.60 X -I ali Pa. Find (a) the temperature of Ihe gas and (b) the ;)vcl-"ge kinetic energy of a *IS molecule in I he vessel. (c) \'\.'hal additional illlormaLion would you need if you were asked 10 lind the average speed of a gas molecule? kap~ in /i·ont of Lois Lane to Sil\'e her from a volley of bullets. In a I-minut.e illlerval, an automatic weapon fires 150 bullets, each of mass 8.0 g_ at 400 m/s.
45. Superlllan
Chapter 10
350
Thermal Physics
The bullcLs strike his mighty chest, which has an area of O. 75 lll~. Find the average torce exerted on Superman '5 chest if the bullets hOllllCC back after an elastic, head-on collision. 46. In a period or 1.0 s, ,S.O x ](f~ nilmgcn molecules st.'ike a wall of area 8.0 cm 2 , If tJ1C molecules move at 300 m/s and strike the wall head on in a perreCLI)' elastic collision, find the pressure exened on the wall. (The lllass of one N 2 molecule is '1.68 X 10-:;6 kg.) ADDITIONAL PROBLEMS
47. Inside the wall of a hOllse, an Lshaped section of hot-waler pipe consists of
54. _ A venical cylinder of crosssectional area A is fitted with a tigh l-fi Lli Ilg, fdctionless piston of mass Jll (Fig. PI 0.54-). (a) I r 11 moles of all ide.d gas at"e in tJ1C cylinder at a temperature of T. use .\l ewton's second law for equilibriull1 to show that the height h.al which the piston is in equilibrium under its own weight is gi\'en by 11
FIGURE Pl0.47
48. The
150-L Lank at -45.0"C. what is Lhe final pressure in the t
miiJ
On the scale invelHed br French scienlisl R. A. F. de Reaumer (1683-1757), the freezing point of water is 00 butthc boiling poim i.-. 80°. (a) Find a formula converting temperaLUres "j~.- ill Fahrenheit to the tcmpcramrcs Till, of this scale. (b) What is 98.6°F on de Reaumer's scale?
51. A bicycle tire is inflated to a gauge pressure or 2.5 attn when the temperature is l.5"C. While a man is riding the bicycle, the temperluure of the Lire incrcases La 4!1"C. Assuming the volume of the tire docs not change. what is the gauge pressure in the tire at the higher tempcraturc? 52. A l.5~ln-long glass tube that is closed at one end is weighted and lowered 1.0 the bottom of a freshwater lake. Vv'hell thc tubc is recovered, an indicator mark show., thai water rose ro within OAO III of" U1C c1o'ied cnd. Determine the depth of Lhe lake. Assume conSlant temperature. 53. Long-term :.pace missions require reclamat.ion of" the ()xy~ gen in the cmbon dioxide cxh'llcd by the crew. fn olle method of reclamation, 1.00 mol of carbon dioxide produces 1.0011101 of oxygen, with 1.00 mol of methane as a b}'~prodliCl. The methane is stored ill a tank under pressure and is available 1.0 control the '. If" the methane generaLed in the recycling or three astronauts' rcspiralion dllrin~ one week or flighL is ."tared in om originally empty
m
1 11
1 FIGURE PIO.54
nR'J'
= ---'=IJIK +
J~IA
where Po i.<; atmospheric pressure. (I» Is Lhe presslwc inside the cylinder less than, cqll
55. A flask lll~lde of Pyrex is calibrated at 20.0"C. It is filled to the IOO~mL mark on Lhe flask with 3.5.0"C acetone. (a) \·\,11at is the volume of1.he acetone when bOlh it and the flask cool to 2U.DOC? (b) Would the temporary increase in the Pyrex f-lask's volume make an elppreciablc difTercnce in the answer? Why or why not?
56. _ The pressure gauge on a cylinder of gas registers the gauge pressure. which is the differcllce between dle interior (llId exterior pressure, P'1I111' '·\'hen the cylinder is full, the mass of the gas in it is 'm; at a gauge pressure of Pi' Assuming the temperature of Lhe cylinder remains const.ant. lise the ideal gas law and el relationship between moles and mass to show lhatthe J1l
( P,11++ I~"m) I~n",
57. m:J A liquid with a coefficicnt of volume expansion of f3 just fills a spherical lIask of \'olume \";, at temperature T (Fig. PID.57). The flask is made ora rnatcriallhat has a coefficient of linear expansion of Q. The liquid is free 1.0 expand into a capillary of cross-sectional area A at the top. (a) Show that if the lemperature increases by d7: the liquid rises in the capillary by the a11l0UIH dli = (\~/A)(f3 - 30') dT (b) For a typical system. such as a mercury thermometer, why is it a good approxirnalion 1O neglectl.he expansion of the flask?
1
b.h
T,+ b.T FIGURE P10.57
Problems 158.IBefore beginning" long trip on a llot day, a driver inflates an autoillobile tire to a gauge pressure of 1.80 atm at 300 K. At the end of the lIip, the gauge presSUI"e has increased to 2.20 atTn. (a) Assuming the \'OIUlllC has remained constant, whal is lhe l.emperaltlrc of the air inside the tire? (b) What percenwge of the orig·inal mass of air in the tire should be released so thc pressure retllrns to its original valuc? Assume the temperature remains at the value found in part (a) and the volume of the tire remains conswnt as ail' is released. 5~'.
351
(Fig. PJO.61). (a) Derive an expression for lhe angle of bcnding, fJ, as a function of the initial length of the su-ips, their average coefficients oflincar expansion, the change in temperature, and the separation of the centers of the strips (li.r= T2 - 1'1)' (b) ShO\.,.- that the angle of bending goes [0 zero when li. Tgoes 10 Lero or when the two coefficients of expansion become equal. (c) "'hat happens if the bar is cooled?
Two concrete spans of a 250-rn-Iong bridge ,Ire placed end to end so thm no room is allowed for expansion (Fig. PI 0.59a). [f the temperalllre increases by 20.0°C, what is the height.'r to which the spans rise when they buckle (Fig. rIO.59b)' FIGURE P10.61
(a)
(h)
FIGURE P10.59
tio.
All cxpandable cylinder has its top connected to a spring with force COllStant ~.OO X 10 3 N/m (Fig. P10.60). The cylill(ler is tilled v·:ith 5.00 L of gas with the spring relaxed at a pressure of 1.00 aUn and ;1 temperature of 20.00C. (a) If the lid has a cross-sectional are" oro.o I 0 0 Ill:! and negligible mass, how high \\'ill the lid lise when the tcmp<:'!"(Itul·e is raised to 250"'C~ (b) \'\'11<11 is the pressw·c of the gas at 2500C?
"
FIGURE P10.60
6). i\ bimcLallic bar is made of two thin strips of dissimilar met-
als bonded togetller. As lhey are heated, the one with dIe larger average coefficient of expansion expands more than the other. forcing the bar iuto an arc, with lhe ollter strip having both a larger radill~ and a larger circumference
62. A 250-m-long bridge is improperly designed so thaI it can~ nOt expand with temperature. II. is made of concrete with a = 12 X 10 I> "'C- l . (a) Assuming the maximum change in temperature at the site is expecled to be 20°C, lInd the change in length the span would undergo if it were free to expand. (b) Show thallhe stress on an object wilh Young's modulus Y\\'hen raised by llTwith its ends fim11y fixed i~ given by aYllT. (c) If the ma.ximum stress the bridge can wilhstancl withom crumbling is 2.0 X 10 7 Pa, will it cnlmble because of this tcmperature increa~e? Young's modulus ror concrete is abollt 2.0 X 10 11I Pa. 63. When lhe hOl waler in a cenain upstairs bathroom is turned 011, (I series of 18 '"ticks" is heard as the copper hot-water pipe slm\!l)' heats up and increases in length. The pipe runs vertically from lhe hot-water heater in the basemem, through a hole in the floor 5.0 m above lhe waler hemer. Tile '"ticks" are caused by the pipe sticking in the hole ill the floor until the tension in the expanding pipe is great enough to unslick lhe pipe, enabling it 10 jump a shon distancc through the hole. Jr the hot· watcr temperature is 46~C and the room temperature is 20"'c:, determine (a) the dislance the pipe moves with each '"tick" alld (h) the force required to UllSlick the pipe if the cross-sectional area of the copper in the pipe is 355 X 10 :. m"'. 164.IT,.. .o small containers, each with a volume of 100 cm~" contain helium gas al D"'C and 1.00
Glacier fragments fall into the sea. Global worming could melt enough ice to swell the oceans and threaten coastal cities around the world.
11.1
Heat and Internal Energy
11.2
Specific Heat
11.3
Colorimetry
11.4
Latent Heat and Phase Change
11.5
Energy Transfer
11.6
Global Warming and Greenhouse Gases
ENERGY IN THERMAL PROCESSES When two objects with different temperatures are placed in thermal contoct, the temperature of the warmer object decreases while the temperature of the cooler object increases. With time they reach a common equilibrium temperature somewhere in between their initial temperatures. During this process, we soy that energy is transferred from the warmer object to the cooler one. UntiJ about 1850 the subjects of thermodynamics and mechanics were considered two distinct branches of science, and the principle of conservation of energy seemed to describe only certain kinds of mechanical systems. Experiments performed by English physicist James Joule (1818-1889) and others showed that the decrease in mechanical energy (kinetic plus potential) of on isolated system was equal to the increase in internol energy of the system. Today, internal energy is treated
as a form of energy that can be transformed into mechanical
energy and vice versa. Once the concept of energy was broadened to include internal energy, the Jaw of conservation of energy emerged as a universal law of nature. This chapter focuses on some of the processes of energy transfer between a system and its surroundings.
11.1
HEAT AND INTERNAL ENERGY
A Ill<~jor distinClion must be made between heat and internal energy. These terms are not interchangeable: Heal involves a Ira1lJjer or internal energy from one locatioll to anothCI". The follm... ing rorrnal dcrinitions will make the distinction precise.
352
11.1
Heat and Internol Energy
+-
Internal energy U is t.he energy associated with the atoms and molecules of the systcm. The internal energy includes kinet.ic and potential energy associated ,,·..ith the random u'anslational, rotational, and vibrational motion of the particles that make up the system. and any potential energy bonding the particles together.
353
Internal energy
In Chapter 10 '\Ie showed that the internal energy of a monatomic ideal gas is associated with the translational motion of its atoms. In this special case, the internal cner~y is the Lotal translaLional kinetic energy of lhc atoms; the higher the tempcralUre or the gas, the greatcr the kinctic energy of the atoms and the greater the internal energy of' t.he gas. For more complicated dialomic and ~ polyal.Omic gases, internal energy includes other forms of molecular energy, ~ such as rotational kinetic energy and the kinetic and potential encrb'J' asso- ~ ~ ciated ,..· ith molecular \'ibrations. Internal energy is also associated with the ~ ~ intcrmolccular potcnl.ial energy ("bond energy") hct\veen molecules in a liquid ~ jor solid. ~~ I-Ieat was introduced in Chapter 5 as one possible method of transferri ng energy j between ,1 system and its environment, and we provide a formal definition here: J; 3
i
JAMES PRESCOTT JOULE British physicist (1818-1889)
Heat is the transfer of" energy bel\,,'een a system and its environment due to a ternperaturc difference between them.
The symbol Q is used to represent the amount of energy transferred by heat bctween a system and its elwirollmenl. For brevity, we \\'ill often use the phrase "the energy Q I ransferrcd to a systcrn . . ." rather than "the energy Q transferred by hca t to a system. I r a pan of water is heated on the burner of" a stove, il's incorrect to sa)' more heat is in the water. Heal. is the t?an~Ieror thermal energy,jusl as work is the transfer of mechanical energy. '''''hen an object is pushed. it doesn't have more work; rather, it has more mechanical energy tr<:Jnsfcl"fcd by work. Similarly, the pan of water has more tJlermal energy u'ansferred by heat
Joule received some formal education in mathematics, philosophy, and chemistry from John Dalton, but was in large port self-educated. Joule's most active research period, from 1837 through 1847, led to the establishment of the principle of conservation of energy and the relationship between heat and other forms of energy transfer. His study of the quantitative relationship among electrical, mechanical, and chemical effects of heat culminated in his announcement in 1843 of the amount of work required to produce a unit of internal energy.
Units of Heat Early in the dcvelopmenl of thermodynamics, before scientists realized the connection between thermodynamics and mechanics, heat. was defined in t.erms of the lemperaLilre changes it produced in an ol~jeCl, and a separate unit of energy, the calorie, was uscd for heat. The calorie (cal) is defined as the energy necessary to raise the temperature of 1 g of water from 14.5°C to 1.~.5°C. (The "Calorie," with a capital "C," used ill describing the energy content of foods, is actually a kjlocalorie.) Likewise, the unit of heat in t.he U.S. customary system, the British thermal unit (Btu), was defined as the energy required to raise the temperature of 1 Ib of waler from 63'F to 64'F. III 1948 scientists agrccrllhal. because heat (like work) is a measure orlhe t.ransfer of energy, its 51 unit. should be the .joule. The calorie is nm\' defined to be cxaclly4.186J:
1 cal
~
4.186.1
[ ll.l]
+-
Definition of the calorie
+- The mechanical equivalent of heat
This definition makes no re(erence to raising the temperature of water. The calorie is a general energy unit, introduced here for historical reasons, although ,'\'e will make little use of it. The definition in Equation 11.1 is known, from t.he historical background ,ve have discussed, as the mechanical equivalent of heat.
354
Chopter 11
EXAMPLE 11.1 Goal
Energy in Thermal Processes
Workin Off Breakfast
Relate caloric energy to mechanical energy_
Problem A sludclll eals a brcakl~1st consisting of twO bmvls of" cereal and milk, containing a lOlal of 3.20 X 10 2 C~l!01-ics of eller"gy. He wishes 1.0 do an equivalent. amount of work in the gymnasium by performing clIrls with a 25.0-kg barbell (Fig". 11.]). How many times must he raise the ,..'eight (0 expend that IllllCh energy? Assume he raises it through a venical displacernenl of 0.400 III each lime, the distance from his lap (Q his upper chest. Strategy Convert the energy in Calories to joules, then equate thal energy La the work necessary to do 'lI repet.iLions of the barbell exercise. The \vark he docs lifting the barbell can be found from the work-energy theorem and the change in potential energy of the barbell. He does negative ' . .·ork on the barbell going down, to keep it from speeding up. The net work on the barbell during one repetilion is zero. but his muscles expend the same energy both in raising and lowering.
In
r-,'
CL3~ FIGURE 11.1
(Example 11.1)
r································
Solution
1
E= (3.20 X 10'.Ga1) (1.00 X 10 £<11)(4.186 J) 1.00 Qrj ",11
Convert his breakfast Calories, E, to joulcs:
= 1.34 X
J06.1
Use the work-energy theorelll to find the ""ork 1Iecessary Lo Liftlhe barbell up to its maximum height:
W= a KE + aPE = (0 - 0) + ("'gh - 0) = 1/lgh
The slUdclll must expcnd the same amounL of energy lowering the barbell. making 2mgh per repetition. Multiply this amount by n repetitions and SCi it equal LO I he food energ~' E:
n(21/lgh)
Sake for n, substit.uting the [c)od energy lor 1:.":
11
I:
_.
E
= -- =
21/lgh
= lrrrr... • • • • • •
~
1.34 X JO".1
---c---~------cc-c----c
2(25.0 kg)(9.80 m/s')(0400 Ill)
6.84 X 10."1 times
.
.
• • ..o1li
Remarks II" the studcnt does one repetition every 5 seconds. it will t.ake him 9.5 hours LO work off his breakfast! In exercising, a large fraction or energy is lost through heat, however, due to the inefficiency of the boely in doing work. This u'ansfer of" cnergy elram~l.ically reduces the exercise requirement by at least ulree-quarters. a little over two hours. All the same, it. might be best. to forgo thill. second bowl or cereal!
QUESTION 11.1 From the poinl of"vie,v of physics, does the answer depend repetitions affect human met.abolism?
011
how faslthe repetitions are performed? How do [aster
EXERCISE 11.1 How mall)' sprints from rest to a speed of 5.0 m/s would a 65-kg woman have LO complete to burn off 5.0 J"ics? (Assume 100% elTiciency in convening food energy to mechanical energy.) Answer
X
102 Calo~
2.6 X 10 3 sprints
APPLICATION Physiology of Exercise
Gelling proper exercise is an import.ant part of staying healthy and keeping weightllnder control. As seen in the preceding example, the body cxpends energy when doing mechanical '·.. . ork. and these losses are augmcmeel by the inefficiency of converting lhe body's internal stores of encrgy int.o useful work, with threequarters or more leaving the body t.hrough heaL In addition, exercise lcnds LO e1evat.c the body's general met.abolic rat.e, which persists even after the exercise
11.2
is over. The increase in metabolic rate due to exercise, more so than the exercise itself, is helpful in weight reduction.
11.2
SPECIFIC HEAT
The historical definition of the caloric is the amOUJ1[ of energy nccessary to raise the temperature of one gram of
tcmperature by tJ.T =
7j -
'J~,
the specific heat rof the substance is defined by r
Q ==-mllT
TABLE 11.1
Specific Heats of Some Materials at Atmospheric Pressure Substance
SI unit: Joule per kilogram-degree Celsius (J/kg' ·C)
Silver
Table 11.1 lists spccific heats for several substanccs. From the definition of the caloric, the specific heat of water is 4 JH6J/kg' °C. The values quoted are typical, but vary depcnding on the tcmperature and v..'hether t.he matter is in a solid, liquid, or gaseous Slate. From the definition of spccific heat, we can express the energy Q needed r.o raise the temperature ora system of mass 1!l by ~Tas [11.3]
The energy required lO raise the temperaturc of 0.500 kg of \vater by 3.00 c C, lor example, is Q ~ (0.500 kg)(4 186J/kg' ·C) (3.000C) = 6.28 X 10'.1. Note that when tJle tcmperaulre increases, !1T and Q are jJOsitive, corresponding to energy flowing into the system. W'hen (he temperature decreases, Ll T and Q are l1ep:o{ive, and energy nows out of thc systcm. Table 11.l shows that ",·ater has the highest specific heat relative to most other common subslClnces. This high specific heal is responsible lor the moderate tem~ peratures found in regions near large bodies of water. As lhe temperature of a body of water decreases during winter, the ",ater transfers energy to the air, which carries the energy landward when prevailing winds arc to\vard the land. orr the western coast of the United States, the energy liberated by the Pacific Ocean is carried to the east, keeping coastal areas much warmer than they would otherwise be. Winters arc gencrally colder in the castern coastal states, because the pre\'ailing winds tend to carry the energy away from land. The fact that the specific heat of water is higher than I he spcci fic heat of" sand is n..:sponsible for the pattern of airnow at (l beach. During thc day, the Sun adds roughly equal amounts of energy to the beach and the waler, but the Im"icr specific heat of sand causes the beach lO reach a higher tempcrature than the water. As a result, the air above the land reaches a higher lcmperature (han the air above the water. The denser cold air pushes the Icss dense hot air upward (due to Archimedes's principle), result.ing in a brecze from ocean to land during the day. Because lhe hOl air gradually cools as it rises, it subsequently sinks, setting up the circulation pattern sh0'''''n in Figure 1 1.2. A similar e1Tect produces rising layers of (lir called tJU"nnals lho-lt can help eagles soar higher and hang gliders stay in flight longer. A thermal is created when a portion of the Eartll reaches a higher temperature than neighboring regions. Thermals often OCClll' in plowed fields, \,,'hich arc warmed by the Sun to higher temperatures
129
Gold Iron
[11.2]
355
Specific Heat
Steam Tin "Valer
2090 448 128 138 703 23·1 2010 227 4 186
Tip 11.1 Finding ilT In I~Cluation 11.3, beSlll-e LO remember lhal J.Tis always the final LemperaLUre minus lhc inilialLcmperal11l"e: ilT= 7~-- '{~.
Waler
FIGURE 11.2
Circulation of air at Lhe beach. 011 a hOl day. the air "bO"(' lhe salld wanllS faster lhan lhl: air above Lhe cook .. h·ater. The wanner air f1uats up\\'ard due lo Archillledcs's principle. resldting in lhe movement of cook:r air Loward thl' beach.
APPLICATION Sea Breezes and Thermals
356
Chapter 11
Energy in Thermol Processes
Lhan nearby fields shaded by vegetation. The cooler, denser air over the vegetaliollcovered fields pushes the expanding air over the plowed field upwards, and a thermal is fonned. QUICK QUIZ 11.1 SUIJpu.sc yULI have 1 kg each uf iron, glass, and water, and all three samples arc at lOoC. (a) Rank the samples frolll lowest 10 highest temperature after J 00 ] or energy is added La each by heal. (b) Rank them [rom least to greatest aillount of energy transferred by heat if enough energy is transferred so thaL each increases in lcmperawre by 20°C.
EXAMPLE 11.2 Goal
Stressing a Strut
LSC the energy transfer equation in the COlllext or linear expansion and compressional stress.
Problem A steel strut neal" a ship's furnace is 2.00 In long, with a mass of 1.57 kg and cross-sectional area of l.00 X ]0- 4 m 2. During' operation of the furnace, the strut absorbs a net thennal energy of' 2.50 X ]0".1(a) Find the change in tempcraturc of the strut. (b) Find the increase in length or lhe strul. (c) If the strut is not allowed to expand because it's boIled at each end, find the compressional stress developed in the strul.
Strategy This problem can be solved by substituting given quantities into three differcnt equations. In pan
(a), the change in tcmperature can be computed by substituting into Equation 1] .3, which relates temperat.ure change to the energy transferred by heat. In part (b), substituting the result or part (a) into the linear expansion equation yields thc change in length. If that change of length is th,,\'artcc! by poor design, as in pan (c), the result is compressional stress, found with the compressional stress-strain equation. Note: The specific heat or steel may be taken to be the same as that of iron.
...............................,
Solution (a) Find the change in temperawrc.
Solve Equation] 1.3 for the change in temperatllre and subSlirutc:
(250 X lO"J) flY = .,--'--c-c---"-'--7
(1.57 kg)(448J1kg·oC)
(b) Find the change ill length of the strut if it's allm.. .ed to expand.
Substitute into the linear expansion equation:
!H = aLo!!.T= (11 X 10- 6 °C- I )(2.00 m)(355°C) =
7.8 X 10-3 m
(e) Find the compressional stress in the strut if it is not allm'\'ec! to expand. Substitute imo the compressional stress-strain equation:
.
F
A
!!.L II )7.8 X 10-3 m y - = (2.00 X 10 Pa L 2.01 m 7.8 X 10" Pa
.
Remarks Notice thc lise of 2.01 111 in the denorninaLOr of the last calculation. rather than 2.00 m. This is because, in effect, the strut was compressed back to the original ICllgth from the length to which it \vould have expanded. (The difference is negligible, however.) The answer exceeds the ultimate compressive strength of' steel and underscores the importance of allmving 101' thermal expansion. Of course, iI's likely the strut \\'ould bend, relieving some or the stress (creating some shear stress in the process). Finally, if 1 he strut is attachcd al both ends b)' bolts, thermal expansion and contraCl.ion
.........................
~
would cxert sheer stresses on thc bolts, possibly weakening or loosening them over time. QUESTION 11.2 Which of the following combinations of properties will result in the smallest expansion ofa substance due to the absorption of thermal energy? (a) sma\! specific heat, large coefficient of' expansion (b) small specific he.H. small coefficientofcxpansion (c) large spccific heat, small coefficient of' expansion (d) large specific heat, large coefficient of l:xpallsion
11.3
EXERCISE 11.2 Suppose a sted strut having a cross·sectional area of 5.00 X 10-" m Z and length 2.50 m is bolt.ed between two rigid bulkheads in the engine room of a submarine. Assume the density of the steel is ,he same as that of iron.
11.3
Calorimetry
357
(a) Calculate the change in temperaturc of the Strut if it absorbs 3.00 X IO!lJ of thermal energy. (b) Calculate thc compressional stress in the strut.
Answers
(a) 68.2'C
(h) 1.50 X 10' Pa
CALORIMETRY
technique for measuring t.he specific heat of a solid or liquid is to raise the temperature of the substance to some value, place it into a ,"esseJ containing cold waLer of known Ill
QuAd
= -
QhOi
[11.4]
QLuid is positive because energy is flowing into cooler ol~jcct.s. and QhOI is negative because energy is leaving the hot object. The negative sign on the right-hand side of Equation 11.4 ensures that the right-hand side is a positive number, consistent \'vith the left-hand side. The equation is valid only when the system it describes is isolated. Calorimetry problems involvc solving Equation 11.4 lor an unknown quantity, usually either a specific heal or a temperature.
EXAMPLE 11.3 Goal
Findin a Specific Heat
Sake a calorimetry problem involving only two substances.
Problem A 125-g' block of <:1Il un known substance with a tempcrature or gO.ooe is placed in a St)'rofoam cup containing 0.326 kg of,.,..ater at 20.0°C. The system reaches an equilibrium tcmpcrarure of22.4°C. ''''hat is the specific heat, c;.., or ule unknown substance if the heat capacity of the Clip is neglected? Strategy The water gains thermal energy Q,olrl while the block loses thermal energy stilute expressions into Equation 11.4 and solve for the unknown specific heat, (,.. ~..
.
QhOI'
Using Eql'<:llion 11.3, sub-
.
Solution Lei Tbe the nnaltcmperature, and let 7~jJ and T.-: be the initial temperatures of the water and block. respectively. Apply Equations 11.3 and J 1.4:
111,,,1",,.('1'-
Solve for ("x and subst.itute nllmerical values:
-"',c..(T- '1',)
T -1~,.) ,",(7: - '1')
1}/
c;.. =
7~J =
".rlll (
(0.326 kg)(4 190J/kg·oC)(22.4'C - 20.0'C) (0.125 kg)(90.0'C - 22.4°C) 3881/kg· o C 110.- • • • • •
358
Chapter 11
Remarks
Energy in Thermol Processes
Comparing our results to values given in Table J 1.1, the unknown substance is probably copper.
QUESTION 11.3 Objects A, B, and Care at different temperatures, It lowest and C highest. The three objects are put in thermal conlaCl with each olher. '''''i'holll mal energy?
doin~
a calculation, is it possible LO deterllline \-..: hether object Ihvill gain or lose ther-
EXERCISE 11.3 A 255-g block of gold at 85.0°C is immersed in 155 g o[walcr at 25.0<>C. Find the equilibrium temperature, assuming the system is isolated and the heat capacity of the cup can be neglected. Answer
27. goC
Tip 11.2 Celsius Versus Kelvin -,---,---,-.,.,-;::--
In equations in which Tappcar~. .'>uch as the ideal gas law, the Kel-
vin tempt:raLUre mUSl be used. In equations involving !iT, such as calorimetry equal ions, it's pos~i blc to use either edsins or f(ch'in temperatllrcs because a change in tcmperalUrc is the same on both scales. ''''hen in doubt, USc
Kchin. ~--------_/
As long as [here are no more than two substances involved, Equation 11.4 can be used to solve elementary Gllorimelry problems. Sometimes, however, there may be three (or more) substances exchanging thermal energy, each at a different temperature. I f the problem requires finding the final temperature, it may not be dear whether the substance ,"vith the middle temperature gains or loses thermal energy. In such cases, Equation 1-1.4 call't be used reliably. For example, suppose "ie ,"vant to calculate the final temperature of a system consisting initially of a glass beaker at 25°C, hot water at 40"C, and a block of aluminum at 37°C. \Ve know that after the three are combined, the glass beaker warms up and the hot water cools, but we don't knm\, for sure whether the aluminum block gains or loses energ}' because the final temperature is unknown. Fortunately, we can still solve such a problem as long as it's set up correnl}'. With an unknmvn rinal temperature T1, the expression Q = Jllc(7j - 1:) will be positive if 7j > 7~ and negative if 7f < 7:. Equation 11.4 call be written as [11.5]
where QA- is the energy change in the kth object. Equation 11.5 says [hal the sum of all the gains and losses of thermal energy must add up to zero, as required by the consen'ation o[ energy for an isolated system. Each term in Equation 11.5 will have the con:ect sign automatically. Applying Equation] 1.5 to the water, aluminum, and glass problem, we get
Q",+ Q,,,
+
Qg~O
There's no need to decide in advance whether a substance in the system is gaining or losi ng energy. This equation is sirnilar in style to [he conservation of mechanical energy equation, ,,,'here the gains and losses of' kinetic and potent.ial energies sum to zero for an isolaled system: uK + dPI..; = O. As will be seen, changes in thermal energy can be included on the left-hand side of this equation. \-Vhen more than two substances exchange thermal energy, it's cas)' to make errors substituting numbers, so it's a good idea to cOllstructa table to organize and assemble all the data. This strategy is illustrated in the next example.
EXAMPLE 11.4 Goal
Calculate on Equilibrium TemJ!eroture
Solve a calorimetry problem involving three substances at three different temperatures.
Problem Suppose 0.400 kg or water initially at 40.0°C is poured into a O.300-kg glass beaker having a temperawre of 25.0°C. A O.500-kg block of aluminum at 37.0°C is placed in the water and the system insulated. Calculat.e the final equilibrium temperamre of the system. Strategy The energ)' transfer for the water, aluminum, and glass will be designated Q"" Qal' and QgJ respectively. The sum of these transfers lllust equal zero, by conservation of energy. Construct a table, assemble the three terms frolll the given data, and solve for the rinal equilibrium temperature, 'I:
(1.68 X 1O'.J/"C + 4.50 X 10' J/"C + 2,51 X 10' jl'C) T T=
= 9.01 X 10 4
.1
,
...
37,9"C
Irr.. • • • • • • • • • •
Remarks The answer turned out tu be vcry close to the aluminum's initial temperature, so it \vould have been impossible LO guess in advance whether the aluminulll would lose or gain energy. Notice the way the table was organized, mirroring the order of [actors in the diFferent terms. This kind of organization helps prevel1l substitution errors, which are common in these problems. QUESTION 11,4 Suppose thermal energy Q leaked from the syslem. How should the right side of Equation (l) be adjusted? (a) :'olo change is needed. (b) +Q (c) -Q, EXERCISE 11.4 A 20,0-kg gold bar al35,0'C is placed in a large, insulaled 0,800-kg gl
Answer
26,6"C
11.4
LATENT HEAT AND PHASE CHANGE
A substance usually' undergoes a change in t.emperature when energy is transferred between the substance and its environment. In some cases, however, the transfer of" energy doesn't result in a change in tcmpt;ralUre, This can occur when the physical characteristics of the substance change from one form to another, com+ manly referred lO as a phase change. Some common phase chc1nges (Ire solid to liquid (melting), liquid to gas (boiling), and a change in the cl")lstalline structure of a solid. Any such phase change involves a change in lhe internal energy, but no dWlI,!!,Y' in Ihi' tem/JPm.luTP, The energy Qneeded to change the phase ofa given pure substance is Q~-+:'mL
f- Latent heat [11.6]
where L, called the latent heat of the substance, depends on the nature of the phase change as \..·ell as on the substance. The unit of latent heat is the joule per kilogram U/kg). The word lalent means "lying hidden \\,ithin a person or thing," The positive sign in Equation 11.6 is chosen when energy is absorbed by a substance, as when icc is melting. The negative
Tip 11.3 Signs Are Critical Fot phase changes, u~c the cor-rect c:-:plicit ~ign in Equation /1.6, positive ifyoll are adding energy to the slibslanct:. neg.1I he if you're laking it away,
j
Chapter 11
360
Energy in Thermal Processes
TABLE 11.2
Latent Heats of Fusion and Vaporization
Substance
Latent Heat of Fusion (J/kg) cal/g
Melting Po;n! Cc)
1.25
-268.93
2.09 X
6.0~1
-195.81
3. ~O
-182.97
2.01 X 10' 2.13X 10'
-269.65
5.23 X
~ilrogen
-209.97
2.55 X 10 4
Oxygen
-218.79 -1]4
X 1.04 X 3.~~ X 3.81 X 1.~8
0.00
\Vater
Sulfur
119
Gold
660 960.80 I 063.00
Copper
1083
Silver
10' 10 5
24.9
78
10'
79.7
100.00 444.60 I 750
9.10
10'
2.45 X 10' 3.97 X 10)
327.3
Lead Aluminum
Vapol'"izaLion
(J/kg)
103
Helium
El hyl alcohol
Latent Heal of
Boiling Point re)
5.85
8.5'1
X
10 4
I1J-'
2.26 X 10" 3.26 X 10" 8.70 X 10.5 1.14 X 107 2.33 X loti
cal/g
4.99
48.0 50.9 204 340 77.9 208 2720 538
8.82 X 10'
9+.8 21.1
2193
6A4 X 10'
15.4
2 ()6{)
1.58 X 1ali
377
1.34 X 105
32.0
I 187
5.01iX 10"
1210
2450
sign is chosen when energy is rClllo\Tcd [rolll a substance, as when steam condenses to water.
The latent heat of fusion L/ is lIsed when a phase change occurs during melting
or freezing, whereas lhe latent heat of vaporization Lv is used when a phase change occurs during boiling or condensing. J For example, at atlllospheric pressure the latent heat of" fusion for \vater is ~.33 X 10!i .I/kg and the latent heat or vaporization for \.. . ater is 2.26 X IOI;.I/kg. The latent heats of different SUbSU'lIlCeS vary considerably, as can be seen in Table 11.2. Another process, sublimation, is the passage from the solid to the gaseous phase without going through a li(IUic1 phase. The ruming of dry ice (frozen carbon dioxide) illusu-ates this process, which has its own latent heat associated wilh it, the heat of sublimation.
EXAMPLE 11.5 Goal
Boilin
Li'luid Helium
Apply the concept orlatent heal ofvaporiztltion to liquid helium.
Problem Liquid helium has a very low boiling point, <1.2 K, as \vell as a low latent heat of vaporization, equal to 2.09 X 10 lJ/kg. If energy is Lransferred to a container of liquid helium at the boiling point rrom an immersed electric heater al a rale of 10.0 'N, hO\\I long docs it take t.o boil away 2.00 kg of the liquid? Strategy Because L;, = 2.09 X lO,I.I/kg, boiling a\vay each kjlogram of liquid helium requires 2.09 X lO'IJ of energy. Joules divided by \.,'atts is Limc, so find the LOtLl! energy nceded and divide by the power Lo find the time.
,... .....
.......................................
~
Solution
Q = mL" =
Find the energy needed to vaporize 2.00 kg or liquid helium at its boiling point: Divide this resulL by the power
............................................. 1V.'hl'n a gas cool" it
I'n~nluall}
returns Iu Ihl' liquid phasl', or
((}lI
energ:)
~
\){,I" unit 1Il.L~\!;:"in~1l up during-
the pt<>ccss is t:alkd the hrtll ojmlli!I'l!st/!iol/. and it cqll,t1s Ihe he,ll of\ap"riLation. \\' len a 11fplil1 ("OllIs, it e\Clll11' all)' s"lidilics, ;11111 Ihe lum! IJ{.lolidijil.ll!iol/ eqllal" the he;lt ofhl."i"ll.
11.4
Latent Heat and Phose Change
361
Remark i\'otice thal no change of temperature was involved. During such processes, the transferred energy goes into changing the state of the substance involved. QUESTION 11.5 A puddle ofwalcr evaporates after a few hours. \\'here does the energy causing the evaporation come from? EXERCISE 11.5 If 10.0 \·V of power is supplied to 2.00 kg of water at L.OO X lO:!°C, how long will it take the water away?
Answer
La
cornpletely boil
126 h
To better understand the physics 01" pha~c changes, consider the addition of energy to a 1.OO-g cube of ice at - 30.0°C in a container held at constant pressure. Suppose this input of energy turns the ice to ~tcam (water vapor) at 120.0"C. Figure 11.3 is a plOt of the experimental measurement of temperature as energy is added to I he s)'stem. \Ve examine each portion of the curvc separatelr. Part A During this portion of the cun·e. the temperature of the ice changes from -30.0·C to 0.0·c. Because the specific healofiee is 2 090J/kg· 'C, we can calculate the amount of energy added from Equalion 11.3:
Q=
"'fir,
tl T = (1.00
X
10·' kg) (2 090 l/kg· 'C) (30.0·C) ~ 62.71
Part B \,Vhen the ice reaches DOC, the ice-waLeI' mixLure remains at LhaL tcmperaturc-even though energy is being added -until all the ice melts [Q become water at O°C. According to Equation 11.6, the energy required Lo mclL I.OU g or icc at DOC is Q= 11l1'f~ (1.00 X \0' kg)(3.33 X 105.1/k~) = 333J
Part C Bctwcen DoC and 100°C, no phase change OCClirs. The energy added to the water ;s used to increase its ternperature, as in part A. The amount oj' energy necessary to increase the temperalUre from O°C to 100°C is
Q = lIle"..." tlT = (1.00 X 10-' kg) (4.\9 X \ O".J!kg· 'C)( 1.00 X 10 2 'C) Q = 4.19 X 10 2 .1 Part D :\t 100°C, another phase change occurs a ... the water changes to steam at IODoC. As in Part B, the water-steam mixture rcmains at constant temperature, this time at 100°C-even though energy is being added-ulltil all the liquid has been converted to steam. The energy required to convcn 1.00 g of water at 100°C La steam at 100°C is Q= IIIL"= (1.00 X 10 "kg)(2.26 X 10"J/kg) = 2.26 X 10:11 FIGURE 11.3 A plm orlcmpcraLUrc ,·crsu-" CIIl:rg}· added" hen 1.00 g or icc, iniliall}' at -30.0 C, i'l cOllw:rted
J)
,, ,"", : : Steam , 'I
!)It
\\,lIcr + -"tcam
" """ """ I WaleI' " - 30>P.,---'--=L.,--.L-:-:",----.,-:,O-::----::-:':-:::----:-:"::---c:-:-:'::-'+_ (j I
Icc
I ~OO
1000
I
I
1
62.7
396
8J5
}!)(IO
21JOO
25lJU
3000,:
I (
Ellcrg" added (j)
30703110
10
-"tcam at 120 C.
362
Chapter 11
Energy in Thermal Processes
Part E During this portion of the Clirve, as in parts A and C. no phase change OCClIrs, so all the added energy goes into increasing the temperature of the steam. The energy that mllst be added lO raise the ICmperalure of the steam to 120.0°C is
Q ~ II/("",,,n tlT = (1.00 X 10-' kg) (2.0 I X 10" .J/kg· "C) (20.00C)
= 40.2J
The LOtal amount or energy lhal must be added to change 1.00 g of ice at - 30.00e to sleam at 120.0°C is the sum of the results from all five parts of the cur\'c, 3.11 X ] O~I.J. Conversely, lO cool 1.00 g of stearn at 120.0°C to Ihe pain I at !''''Ilich it becomes icc al -30.DoC, 3.1 I X lO:~ J of energy must be removed. Phase changes can be described in terms of rearrangements or molecules when energy is added to or removed from a subSl.ance. Consider first the liquid-w-gas phase change. The molecules in a liquid are close LOguher, and the forces between them are stronger than the forces between the more widely separated molecules of a gas. \'\lark must therefore be clone on the liquid against these attractive molecu~ lar forces so as to separate the molecules. The latent heat of vaporizmion is the amount of energy lhat must be added to the one kilogram o[ liquid to accomplish this separation. Similarly, at the melting point o[ a solid, the amplitUde of vibration of the aWlns aboUl their equilibrium positions becomes great enough to allow the aLOms to pass the barriers of acljacent alOms and move to their nc\,; positions. On average, these new positiuns arc less symmetrical Ulan the old ones and u1ercfore have higher energy. The Iatenl heal of fusion is equal to the work reqllired at the molecular level to transform the mass from the ordered solid phase to the disordered liquid phase. The average diswnce between atoms is much greater in the gas phase u1an in either the liquid or the solid phase. Each aLQm or molecule is removed [rern its neighbors, overcoming the attractive forces of nearby neighbors. Therefore, more vmrk is required at. the molecular level to vaporize a given mass ofa substance than to melt it, so in general the latent heat of vaporization is much greater U1al1 the latent heat affusion (see Table] 1.2). QUICK QUIZ 11.2 CalculaLe the slopes ror the A, C, and E portions of Figurc ]1.~. Rank the slopes [rom least to gl'catesl and explain whC:ltyour ranking means. (a) A, C, E (b) C, A, E (c) E, A, C (d) E, C, A
Make a table for all data. Include separate rows Lor different phases and [or any transition bct\'\'een phases. Include columns for each quantity used and a final column for the combination of the quantities. Transfers of thermal energy in th is last. column are given by Q = mr 6. T, whereas phase changes are given by Q = .±: mLf I'DI' changes bet\veen liquid and solid and by Q = ± mL l for changes bet-ween liquid and gas. !
2. Apply conservation of energy. Ir Lhe sySlem is isolaLed, Lise 2,Q" = 0 (Eg. 11.5). For a non isolated system, the nel energy change should replace the zero on the right-hand side of that equation. Here, l(,!;, iSjllst t.he sum orall the terms in the lasl colu m n of the table. 3. Solve for the unknown quantit}'.
EXAMPLE 11.6 Goal
Ice Water
Solve a problem involving heattransf"er and a phase change from solid to liquid.
Problem At a pany, 6.00 kg or ice at - 5.00°C is added l.emperawre of lhe water when it comes to equilibrium?
10
a cooler holding 30 liLers of water at 20.0=C. \\-1ut is the
11.4
Latent Heat and Phose Change
363
Strategy In this problem, it's best to make a table. vVith the addition of thermal enerb'J: Qice u1e ice ,.,'ill \varm to O°C, thcn melt at O°C with the addition or energy Q,"dl' Next, the melted ice will warm to some final temperaturc T by absorbing energy Qiee-water. obtained from the energy change of the original liquid water, Qwalci' By conservation of energy, these quantities must sum to zero. .
~
.
~
Solution Calculate the mass of liquid watcr: = ( 1.00 X 103
kg/m'')( 30.0 L)
LOOm'3 1.00 X 10' L
kg
= 30.0
Write the equation of thermal equilibrium: Construct a comprehensive tablc:
Q
(kg)
III
600 6.00 6.00 30.0
Qino Ql1leh
Qin'-"~ucr QW'lter
c (J/kg' "C)
L (J/kg)
2090 3.33
X
10'
Tf ("C)
T, ("C)
0 0
-5.00 0 0 20.0
4 190
T
4190
T
Substitut.e all quantities in the second through sixth columns into the last column and sum. which is the evaluation of'Equalion (l), and sake for T:
6.27 X 10".1
T=
Expression In;c('Cice (7j
}//iccLj llI;n J
"';uClj -
'/~)
m"~HcW;l1 (7( - 7~)
+ 200 X 10"j + (2.51 X 10"jrC)('f- O°C) + (1.26 X 10'jrC)('f- 20.0°C)
= 0
3.03°C
....
.
~
- 7~.)
Remarks Making a table is optional. Howe\'cr, simple substiLUtion errors arc extren1ely common, and the table make:s such errors less likely. QUESTION 11.6
Can a closed system containing differcm substances at different initial temperaLUres reach an equilibrium temperature t.hat is lower than all the inilialtcmperalures? EXERCISE 11.6
What mass of icc at -.I0.0°C is needcd to cool a \·vlnlc's water tank. holding 1.20 X 10,1 more comfortable Ia.o°C?
Answer
of water, from 20.0°C to a
1.27 X lO' kg
EXAMPLE 11.7 Goal
m~
Partial Melting
Undersl:<:Ind how to handlc an incomplete phase change.
Problem A S.OO-kg block of ice at DoC is added to an insulatcd container partially f-illed with 10.0 kg of 'vater at 15.0°C. (a) Find lhe final temperal.llre, neglecting thc heat capacity of" t.he container. (b) Find the mass of the icc that was meltcd. Strategy Part (a) is tricky beGlllse the ice docs not entirely melt in this example. "-Vhen there is any doubt concerning whether there will be a complete phase
change, some preliminary calculations are necessary. First, find the total energy ]"cquirccllO melt the ice, Qmdl' and then find Q""w:r' the maximum energy that can be deli\'cred by the water above DoC. If the energy delivered by the water is high enough, allthc ice melts. Ifnot, lhere will usually be a final mixlllre of ice and water at DOC. unless the icc starts at a temperat.ure far below oDe, in which case all the liquid water freez.cs.
364
Chapter 11
Energy in Thermal Processes ..
"
Solution (a) Find the equilibrium temperature, First, compute the arnount of energy necess;1ry to COlll-
Q",,,,, = m".,.LJ = (5.00 kg) (3.33 X 10'.J/kg)
pletely melt I he ice:
= 1.67 X 1OIi .I
:'\exl. calculate the maximum energy that can be lost by the initialm
(10.0 kg)(4 190J/kg"C)WC - 15.0°C) = -6.29 X 10'.1
This result is less than half the energy necessary to melt all the ice, so the final state of the system is a rnixLUI-e of" water and iet:: at the freezing point:
T=
O'C
(b) Compute the Illass of icc melted. SCllhe tolal available energy equal to the heat of fusion of III grams of icc, lIlL!, and solve for m:
....
6.29 X lOr, J = mL! ~ 11/.(3.33 X 10'.I/kg) m =
1.89 kg
.
~
Remarks If this problcITl is solved assuming (wrongly) [hat all the ice melts, a finaltcmperalUre of T= -16.5°C is obLained. The only way that could happcn is if the system \Vcre not isolated, contrary [Q the statement of the problem. In Exercise 11.7, you must. also compute t.hc therm;-ll energy needed La warm the icc La its melting point QUESTION 11.7 What effecl would doubling the initial amount. of liquid '...· mer have on the amount of ice melted? EXERCISE 11.7 If 8.00 kg of ice at -5.00°C is added to 12.0 kg of water at 20.0"C, compute the final t.emperature. How much ice remains, if any?
Answer
7' ~ O°C. 5.23 kg
Sometimes problems involve changes in mechanical energy. During a collision, for example, some kinet.ic energy can be transformed to the illlernal energy of lhe colliding objects. This kind of transformation is illustrated in Example II.S, which involves a possible irnpact of a comet on E;-lrth. In this example, a Ilum· bel' of liberties will be t;-lken in order LO estimate the magniLUde or the destructive power of such a catastrophic event The specific healS depend on temperature and pressure, for example, but that will be ignored. Also, the ideal gas law doesn't appl)1 at the temperatures and pressures attained, and the result of the coil is ion wouldn't be superheated steam, but a plasma of charged panicles. Despite allthese simplifications, the example yields good order-of-magnitude results.
EXAMPLE 11.8 Goal
Armo eddon!
Link mechanical energy
La
thermal energy, phase changes, and the ideal gas law to creme an estimate.
Problem A comet half a kilometer in radius consisting of ice at 273 K hits Earth at a speed or 4.00 X 10'1 m/s. For simplicity, assume all the kinetic cllerb,y converts to thermal energy on impact and that al1the thermal energy goes into warming the comel. (a) Calculate the volume and mass of the ice. (b) C"se consen1alion of energy to find
the I1nal tempcralUre of the comet material. Assume, contrary to fact, that the result is superheated steam and that the usual specific healS are valid, although in facl they depend on both temperalUre and pressure. (c) Assuming the steam retains a spherical shape and has the same initial volume as the comet, calculme the pressure of the
11.4
steam using the ideal gas law. This la.. v actually doesn't apply to a system at such high pressure and ternpcralLire, but can be L1sed [() get all estimate. Strategy Pan (a) requires the volume formula for a sphere and the dennition or density. In pan (b) conservation of energy can be applied. There are four processes involved: (l) melting the ice, (2) warming the ice ,,,,,ater to the boiling point, (3) convening the boiling water to
Latent Heat and Phase Change
365
stearn, and (4) warming the steam. The energy needed for t.hese processes \.. ·ill be designated Qlllt'll' QW
r····························
Solution (a) Find the volume and mass of the ice. Apply the volume formula for a sphere:
4
=
Apply the density formula to fi nel the mass of the ice:
4
\1= -.".r' = -(3.1'1)(5.00 X 10'1ll)' 3 3
fit
5.23
]08
X
111 3
~ p\l~ (917 kg/m')(5.23 X
=
10 8 m')
4.80 X ]0" kg
(b) Find the flllal temperature of the cometary material.
Use conservation of energy:
+
Qmdt
111.1.[ + m.rl.. ..l. ICr D.1~~dl~r
+ The first three terms are negligible compared to the kinetic energy. The steam term involves the llnknO\vn final temperature, so retain only it and the kinetic energy. canceling the mass and solving for T:
+ Qvapof + Q~I~alll + dK= 0 + mL v + 'lIICst~am a ~~l~alll
(1)
(2)
mCSle.l1n
Qwall'(
(0 - ~mv~) = 0
(T- 373 K)
'VI
-1/1lv 2 =
.,
0
, 4.00 X 10 I(
]' ~ - - + 373 K =
'1
m/s )"-
20l0J/kg·K
<;"."",
+ 373 K
T= 3.98 x 10" K (c) Estimate the pressure of the gas, using the ideal gas law.
First, compute the number ofrnolcs of steam:
n = (4.80 X 10"
Solve lor the pressure. using PV = llRT:
l1RT p=-\I
(2.67
j'= ~
I
= 2.67 X 10l'lllOi
x 101'mol)(8.31//11101·K)(3.98 5.23 x iO'm'
1.69 X
1011
X
10"K)
Pa
. ..
.
The estimated pressure is several hundred times greater than the ultimate shear stress ofsteel1 This high-pressure region would expand rapidly, destroying everything \-vithin a very large radius. Fires would ignite across a continent-sized region, and tidal \\laves would wrap around the world, wiping Ollt coastal regions evcrywhere. The Sun would bc obscured for at least a decade, and numerous species, possibly including }-{011lU sapiens, would become extinc\.. Such extinction events arc rare, Remarks
kg)( 0.Ql8llloikg )
but in thc long rUll represent a signincant threat on Earth.
10
~
liFe
QUESTION 11.8 \Nhy would a nickel-iron asteroid be more dangerous than an asteroid of" the same size made mainly of ice? EXERCISE 11.8 Suppose a lead bullet \,.,.ith mass 5.00 g and an initial temperature of 65.0°C hits a wall and completely liqucfies.
366
Chapter 11
Energy in Thermol Processes
''''hat minimulll speed did it have before impact? (J-lint: The minimum speed corresponds LO the case where all the kinetic energy becomes internal energy of the lead
11.5
and the final temperature or the lead is at its melting poilll. Don't neglect. any terms here!) Answer
341 m/s
ENERGY TRANSFER
For some applications iL's necessary to know the rate at which energy is transferred benveen a system and its surroundings and t.he mechanisms responsible for the transfer. This information is particularly important \vhen weatherproofing buildings or in medical applications. such as human survival Limc when exposed to lhe elements. Earlier in this chapter we defined heat as a transfer of energy benveen a system and its surroundings due to a temperat.ure difference between them. In this sect.ion \..'e take a closer look al heat as a means of energy transfer and consider the processes or thermal conduction, convection, and radiation. 2
! f " ~
~
J B
FIGURE 11.4 COllc!uuioll 11Iakl:~ Ihl: metal handle ora cooking pan hal.
A
Encrb')' transfer for Th>T~
FIGURE 11.5 Energy transfer through a conduCling slab of crosssectional area A alld thickness I.. The opposite racc.~ arc at diffen:lll ltlllperatures '1;:l11d T/;.
Tip 11.4 Blankets and Coats in Cold Weather VdH;n }O\J sleep under a blanket in the winter or wear a warm coal outside, the blanket or coat serves as a layer of material with 1(11\' thcrm:1J condUCli\,jt)' lhal reduces tht.; transfer of energy
Thermal Conduction The energy transfer process most closely associated with a temperalllre di ffercnce is called thel-mal conduction or simply conduction. In this process the transreI' can be viewed on an alomic scale as an exchange of kinetic energy between microscopic particles-molecules, atoms, and electrons-with less energetic particlcs gaining cnergy as they collide with more energetic part.icles. An inexpensive pot, as in Figure 11.4-, may have a metal handle with no surrounding insulation. A. 7~ (Fig. 11.5). The slab allows energy to transfer rrom Lhe region of higher temperature to t.he region of lower temperanlre by thermal conduction. The rate of energy transfer, qp = Q/tJ.I, is prop0rlional to the cross-sectional area of the slab and the temperaLUre differ· CJ1ce and is inversely proportional to the thickness of the slab:
a\vot)' from )our bod)' br heaL The
pdmary insulaling mediulll is the air trapped in small pod.els within the material.
).late that qp has units of wallS when Qis injoules and D.l is in seconds.
11.5
Suppose (\ substance is in the shape oCa long, uniform rod orlcngth I~. as in Figure 11.6. We assume the rod is insul 7~. \,\'hen a steady state is reached, the tt:mpel3ulre at each point along the rod is constant in time. In this case !J..]' = Til - 'J: and ii.x = L, so
DoT
T,,-1~
,1x
I.
II
Thermal Conductivity U/s· m .0C)
Substance Metals ('it 25"C) Aluminulll
238 397 314 79.r,
Copper Gold Iron
( T - T)
Lead
[II. 7.1
(
L
3'1. 7 '127
Sil\'er
where Il, a proponionality constanl that depends on the material, is called the thermal conductivity. Substances that arc good conductors have large thermal conductivities, whereas good insulators have low thermal conductivities. Table 11.3 lisL'i the thermal conduClivities ror various substances.
Gases (at 20°C)
0.0234 0.138 0.172 0.0234 0.0238
Air
Helium Hydrogen
r\ilrogen
QUICK QUIZ 11.3 '''lill an ice cube \vrappcd in a wool blanket remain 1'1'0zen lor (a) less time, (b) the same length ortirne. or (c) a longer time than an identical ice cube exposed to air at room lemperaturc?
Oxygen Nonmetals AsbesLO~
QUICK QUIZ 11.4 Two rods of the same length and diameter arc made from diffcrent matcrials. The rods arc to connect two rcgions of different temperature so lhat energ~/ will transfer through the rods by hcat. They can be conneCLed in series, as in FigTlrc 11.7<-1, or in parallel, as in figure 11.7b. In which cast;' is the rate of energy transfer by heat larger? (3) When the rods are III series (b) "Vhen the rods are in parallel (c) The nne is the same in both cases.
Ice
Rubber Watt:r \Vood
"" > '(.
Rod 1
(a)
FIGURE 11.6 COlldllUiol1 01 clll.:rgr through a Illlif'll"lll. rod ofkngth L. The: oppo~ile t:l1Cls arc ill therl1lal <.:Olllacl wilh energy n.:ser\'oir~ ,n different lelllJx;ralLlres.
Goal
Rod 2
Ii.
Energy Transfer
: .(
Rod 2
InSlIlalicHi
ill~IlI,lled
EXAMPLE 11.9
7;
'Ii,
'/~
0.25 1.3 0.84 1.6 0.2 0.60 0.10
Concrele Glass
R'le[ I
Ene!"!.,')" tran:-.fcr
367
TABLE 11.3 Thermal Conductivities
The rate of energy transfer by conduction through the rod is given by
CZJ>=kA
Energy Transfer
(h)
FIGURE 11,7 (Qllick Quiz 11.·1) In which case is lhe raIl' ot l:IH::rgy lran-;fcr larg\T?
Problem Find the energy transferred in 1.00 h by conduClion through a concrete wall 2.0 m high, 3.65 m long, and 0.20111 thick if one side or rhc \,,"'all is held at 20°C and the other side is at 5°C. Strategy Equation 11.7 gives the rate of encrgy transfer by cOllduction ill joules per second. Multiply by the time and subSliwtc given values to find the total thermal energy transferred.
"
. . . . . . .....
Solution Mull iply Equation] 1.7 by !J.t 10 fi nd all expression the total energy Qtransferred through the wall:
fOl-
SllbslilltLe the numerical values 1.0 obtain Q, consul1ing Table 11.3 for h:
Q ~ qp "'/
Q= =
... ..
,
(1.3
.
= hA
J/s·
T) "L ' "'/ (T-
III . °G)
l-oG ) (7.3 Ill') ( - " - (3600 s) 0.20 III
2.6X 106.! ,
,
,
.
,
.
,
..
368
Chapter 11
Energy in Thermol Processes
Remarks Early houses \'iere insulated with thick masonry walls, which restriCl energy loss by conduClion because k is relatively low. The large thickness L also decreases energy loss by conduction, as shown by EquaLiull 11.7. ThcJC arc much belter insulating materials, howcYer, and layering is also helpful. Despite the low thermal conductivity of masonry, the amount of energy lost is still rather large, enough to raise the temperature of 600 kg orwatcr by more than] Cle.
EXERCISE 11.9 A wooden sheller has walls constructed of wooden planks 1.00 em thick. If the exterior temperature is -20.0°C and the interior is 3.0aoe, find the rate of energy loss through a \,,:all that has dimensions 2.00 111 by 2.00 1l1. 1.00 X 10 3 W
Answer
QUESTION 11.9 True or False: Materials having high thcrmal conductivities provide bener insulation than matcrials having 10\\1 thermal conductivities.
Home Insulation To determine whether to add insulation to a ceiling or some other p,lrl ora building. the preceding discussion of conduction must be extended for twO reasons: 1. The insulating properties of materials used in buildings are usually expressed in engineering (l.J.S. customary) rather than Sf units. Measurements stamped on a package of fiberglass insulating board will be in units such as British therlTIal units, feet, and degrees Fahrenheit. 2. In dealing ,vith the insulation ofa building, conduction through a compound slab must be considered, \\lith each portion of the slab having a certain thickness and a specific thermal condu([ivity. A typical \,:all in a house consists of an array of materials, such as vmod paneling, drywall, insulation, sheathing, and wood sid i ng.
The rate or energy transfer by conduction through a compound slab is
Q /1/
11(1/, - T,)
[11.81
LL/Ii;
where 7~1 and "'r;. are the ternper<1tures of the olller extremities of the slab and the summation is over all portions of the slab. This formula can be deri\'ed algebraically, using the facts that the temperature at the interface between two insulating matcrials must be the same and that the rate of energy transfer through one insulator must be thc same as through all the other insulators. If the slab COI1sists of three different materials, the denominator is the sum of three terms. In engineering practice, the 1.enTI LIII for a particular substance is referred to as the R-valuc of the material, so Eqllation U.S reduces to
Q
Ae "Ii, - 7~)
/1/
LN,
[11.91
The R~vailies for a few common building materials are listed in Table 11.4. Note thc unit of Ii and the [aet I hal the R-values are defined for specific thickness.es. ~lCXl to any vertical out.side surface is a very thin, stagnant layer orair that must be considered when the tOlal R-value for a wall is computed. The thickness of this stagnant layer depends on the speed of the ,.. . incl. As a result, energy loss by conduction from a house on a day w'hen the ,-vinci is blowing is greater than energy loss on a day when the wind speed is zero. A represent.Jl.ive R-valuc for a stagnant air layer is given in Table 11.4.
11.5
Energy Transfer
369
TABLE 11.4 R-VaJues for Some Common Building Materials Rvalue (fl" of· b/B,u)
Material
Hardwood siding (1.0 in. thick)
0.91
\Vooc! shingles (lapped)
0.87
Brick (4.0 in. thick)
4.00
Concrete block (tilled cores)
1.93
5.0
St}'rofo
fiberglass balling (3.5 in. thick)
10.90
Fiberglass bal.ling (6.0 in. thick)
18.80 4.35
Fiberglass board (1.0 in. thick) Cclilliose libel' (J.O in. thick)
3.70
Flat glass (0.125 in. thick)
0.89 space)
1.5-1
VertiG1[ air space (3.5 in. thick)
1.01
Stagnant layer orair
0.17
Drywall (0.50 in. thick)
0.'15
Sheathing (0.50 in. thick)
1.32
Insulaling glass
(O.25~in.
EXAMPLE 11.10
.-\ worker inslalling f1b~rglas" insulation ill a home. The ma"k prOleCl.~ Ihe worker agaill.Sl the illllalati(Hl of microscopic fihers, wh ich could be haz;lrdull~ to his healLiI.
The R-Value of a Tical Wall
Goal Calculate the R-value of a \'·.'al1 consisting of several layers of insulating mat.erial.
On'wall
/'
Air
Insulalion
Problem Calculate the to[(l\ R-value for a wall construcled as shown in Figure 11.8£1. Starting oUlside the house (to the left in the figure) and moving imvard, the wall consists of4.0 in. of brick, 0.50 in. of sheathing, an air space 3.5 in. thick. and 0.50 in. of drywall. Sheathing
Strategy Add all the R-values together, remembering lhe.: stagnant air layers inside and olltside the house.
,.
(a)
(bl
FIGURE 11.8 (Example 11.10) A cro:-sscnional view oran eXlerior wall containing (;1) all air space and (b) insu1
.
.
Solution Refer to Table 1].4, and Slim. All quantities arc in units offt 2 • "F' h/Ew.
Convection, presented in the nex( section, can reduce the effectiveness or the outside air layer.
QUESTION 11.10 Does creating insulation with a /-i'-value require (a) a smaller thermal conductivity and larger thickness, (b) a larger thermal conductivity and large thickness, or (c) a smaller thermal conductivity and smaller thickness_ EXERCISE 11.1 0 U a \a)'e.r of ll.berglass insulation 3.5 in. thick is placed inside the wall to replace the air space, as in Figure I L8b, what is the new LOlal R-value? By whal factor is the rate of energy loss reduced? Answer
R~
17 ft,·oF·h/HLU; 2.4
Chapter 11
370
EXAMPLE 11.11 Goal
Energy in Thermal Processes
Sta in Warm in the Arctic
Combine two layers of insulation.
Problem An arctic explorer builds a wooden shelter out of \\'ooden plank~ thal are 1.0 Clll t!lick. Tu illlJJI-UVl: Llle
insulation, he covers the shelter with a layer of ice 3.2 em thick. (a) Compute the R-\'alues for the wooden planks and the ice. (b) JJ the temperature outside the shelter is -20.0°C and the temperature inside is 5.00°C, find the rate of energy loss through one or the walls, if the wall has dimensions 2.00111 by 2.00 m. (c) Find the temperature
at the interface bctw"cen the wood and the ice. Disregard Sl(l~lI~lIll ~ljl-
lay<:rs.
Strategy After finding the R-\ralucs, subsutute il1lo Equation 11.9 lO get the nHe of energy transfer. To answer part (c), lise EquaLion 11.7 for one orthe layers, setling il equal to the rate found in part (b) and SOh~llg for the tcmpcralure.
..... Solution (a) Compute the R-valucs using the data in Table U.3. Find the
(c) Find the temperature in bet\'¥"ccn the ice and \vood.
Apply the equation of heal conduction to the "iood:
k",,,udA( J~I ~ T,.) = rzp
L
(0.10 JIs·Ill· 'C)(4.00 m')(5.00'C O.OIOm
-'Ii
-'---~"----':-':-:-:---"------'~
830 \\'
Solve for the unknO\\rll temperature: ~
....
Remarks The Diller side of the wooden wall and the inner surface or the ice must have the same temperature. and the rale of energy transfer through lhe ice must be the S;'lrne as through the \...· ooden wall. Using Equation 11.7 for icc instead of wood gives the same anS\\ler. This rate of energy transfer is on ly a modest improvement over the thousandwalt rate in Exercise 11.9. The choice or insulating material is imponant! QUESTION 11.11 'Wolllen hayc an extra Ia>'er of subcutalleous fat than men. ',Vhat t\VO survival advantages does this additional layer confer? EXERCISE 11.11 Rather Ihan use ice to cover the \vooden shelter, the explorer glues pressed cork with thickness 0.500 em to the outside of his woodell shelter. Find the new rate of energy loss through the same wall. (NOle l.11ac kcork =
O.0~6./1s· Ill'
Answer
0C.)
480 II'
11.5
Energy Transfer
371
Convection \,Vhen yOll warm your hands over an open flame, as illustrated in Figure 1 J .9, the air directly above the f1arne, being warmed, expands. As ~l result, the density of this air decreases and the air rises, \\icHllling your hands as it flows by. The transfer of energy by the movement of a substance is called convection. ""hell lhe movement results from differences in density, as \.. . ith air around a fire, it's referred to as natural convection. Airflm-v at a beach is an example of natural convection, as is the mixing that occurs as surface water in a lake cools and sinks. \,Vhen the substance is forced to move by a fan or pump, as in some hot air and hot \vater heating systems, the process is called Jorred convation. FIGURE 11.9 \-Varming a hand by con\cction. Convection currents assist in the boiling of v·,Iater. In a teakettle on a hot stove· lOp, lhe lower layers of water are warmed first. The warmed \.. . aler has a lower density and rises to the top, while the denser, cool waler allhe surface sinks to the bOllom of the kctlie and is warmed. The same process occurs when a radiator raises the temperature of a room.l The hot radiator warms the air in the lower regions of the room. The warm air ~ ~ expands and, because of its lower density, rises to the ceiling. The denser, cooler ~ ~ air from above sinks, selling lip the continuous air current pattern shown in Figure ~ a ILiO. ~~ An aUlomobile engine is maintained at a safe operating tcmperalUre by a j:f combination of conduction and (orced cOlweclion. "Vater (actually, a mixture PhOlo~r;lph ofa tcakculc. showing steam <\lId turbulent convectiOIl air Or \'Vater and antifreeze) circulates ill the interior of the engine. As the metal or Cllrn.:nl~. the engine block increases in temperature, energy passes from the hot metal to the cooler water by thennal conduction. The water pump forces water out of the APPLICATION engine and into the radiator. GlITying energy along with it (by forced convection). Cooling Automobile Engines In the I'adiator the hot water passes through metal pipes that are in contacl ..vith the cooler outside air, and energy passes inl.O the air b}1 conduction. The cooled \V.Her is then returned to the engine by lhe water pump to absorb more energy. The process orair being pulled past the radiator by the ran is also forced convection. The algal blooms often seen in temperate lakes and ponds during the spring APPLICATION or fall are caused by cOlweclion currents in the water. To understand this process, I1JI Algal Blooms in Ponds consider Figure 11.11. During lhe summer, bodies of water develop temperature and Lakes gradients, with a warm Lipper layer or \"".Her separated from a cold lower layer by a burfer zone called a t.hermocline. In the spring and rail temperature changes in the \·vater break dm.. : n this thermocline, setling up convection currents that mix the water. The mixing process transports nutrients [rom the botlom lo lhe
FIGURE 11.10 COllveclioll currcms are set lip in a room warmed by a ra(]i'llor.
FIGURE 11.11 (a) During llw ~UHImer. a warm upper layer of waleI' is separated from a cooler lower layer hy a thermocline. (b) COll\ecuon CllrrenlS dnring the spring and fall mil'
(h) Fall and
~prin~
llpwelling
the ""Her ,I lid can cause algal blooms.
372
Chapter 11
Energy in Thermal Processes
surface. The nutrient-rich waleI' [orming at the surface can cause a rapid, temporary increase in the algae population.
APPLYING PHYSICS 11.1
BODY TEMPERATURE
The body temperature of mammals ranges from about 35°C 1.0 38°C, whereas that of bi rcls ranges [rom aboul40°C to <13"C. How can these narrmv ranges of body tcrnpcrature be maintained in cold wcather? A nawral method of maintaining bod}' temperaLUre is via layers ofhlt beneath the skin. Fat protects against bOlh conduClion and convcct.ion because of its 1m\,' thermal condunivity and because there are few blood vessels in fal to carry blood to the
Explanation
II surface, ,,,here energy losses by convection can occur. Birds ruffle their feathers in colel weather LO trap a layer of air with a low thermal conductivity bCl''o'cen the feathers and the skin. Bristling the fur produces the same effecl in Cur-bearing animals. Humans keep warm wiLh ""00] sweaters and down jackets thal [rap the ' . .· armer air in regions close to their bodies, reducing enerST)' loss by convection and cand union.
Radiation
FIGURE 11.12
Another process or transferring energy is through radiation. Figure ]1.12 shows how your hands can be warmed aL an open Oamt: through radiation. Because your hands aren't in physical contact with the flame and the conductivity of air is very low, conduction can't account for thc energy transfer. Nor can convect.ion be responsible for any transfer of energy because your hands aren't above the name in the path of convection currents. The \,varml.h felt in your hands must therefore comc from the transfer of energy by radiation. All objects radiate energy continuously in the form of electrom~lgnetic waves due 10 thennal vibrations of their molecules. These vibrations create the orange glm'l! of an electric stove burner, an electric space heater, and the coils of a toaster. The rate at which an object radiates energy is proportional to the founh power of its absolute temperature. This is known as Stefan's law, expressed in equation form as
Warming"h
radiation.
Stefan's Jaw
~
'!P
=
uAir-'
[11.10]
where '!P is the power in watts (or joules per second) radiated by the object, (J is the Stefan-Boltzmann constant, equal to 5.669 6 X 10- 8 W/m 2 • K4, A. is the surface area of the oqjen in square meters, P is a constant called the emissivity of the ol~jcct, and T is the ol~ject's Kelvin temperanlre. The value of P can vary between zero and one, depending on the properties of the ol:~ject's surface. Approximately l340J of electromagnetic radiation from the Sun passes through each square mcter at the LOp of the Earth's atmosphere every second. This radiation is primarily visible light, accompanicd by Significant amounts of infrared and ultraviolet light. \OVe will study these types of radiation in detail in Chapter 21. Some of this energy is reflected back into space. and some is absorbed by the atmosphere. but enough alTives at the surface of the Earth each day to supply all our energy needs hundreds of times over, if it could be captured and used efficiently. The grmvth in the number of solar houses in the United States is one example oran attempt to make use of this abundant energy. Radiant energy from the Sun affects our day-to-day existence in a number of ways, influencing Earth's ;werage temperatlIre, ocean currents, agriculture. and rain patterns. ft can also affect beha\~or. As another example of the effects of energy transfer by radiation, consider what happens 10 the atmospheric temperature at night. If there is a cloud cover above the Earth, the water vapor in the clouds absorbs part of'the infrared radiation emitted by the.; Earth and re-cmits it back to the surl"ace. Consequently. the temperature at the surface remains at moderate levels. Tn the absence of cloud cover, there
11.5
Energy Transfer
373
FIGURE 11.13 This thermogr'1Il1 of a IlOllse. made during cold wt.:ather, show~ colors ranging from while aud yellow (areas of greatest (:lIcrgy loss) to hltll:: and purple (areal> ork
is nothing to prevent the radiation from escaping into space, so the temperature drops more on a clear night than on a cloudy night. As an object radiates energ}' at a rate given by Equation 11.10, it also absorbs radiation. If it didn't, the object \voulcl evcnlUally radiate all its cllcrg}' and its temperature \"'ould reach absolutc zero. Thc energy an object absorbs comes from its environment, which consists or other bodies that radiate cnergy_ If an ol~jcct is at a temperature T and ito;; surrollndings are at a temperature To, the net energy gained or lost each second by the object as a rcsuh of radiation is ~IWI = UAf(T" - 7()')
[11.11]
When an oqject is in equilibrium with its surroundings, it radiates and absorbs energy atlhe same rate, so its temperamre remains constant. \,Vhen an object is hotter than its surroundings, it radiates more energy than it absorbs and therefore cools. An ideal absorber is an object that absorbs all the light radiation incident on it, including invisible infrared and ultraviolct light. Such an object is called a black body because a room-temperaturc black body would look bbck. Because a black body doesn't reflect radiation at
Thenllogralll ora woman's bre;ISts. ller len breast is diseased (red and orange) and her ri~ht brca.'>l (blue) is hcahhy.
APPLICATION Light-Colored Summer Clothing
APPLICATION Thermography
APPLICATION
II Radiation Thermometers for Measuring Body Temperature
374
Chapter 11
Energy in Thermal Processes
with a rectal thermometer, risks such as bO\vel perforation or bacterial contamination. The instrument measures the intensity of the infrared radiation leaving the eardrum and surrounding tissues and cooverts this information to a standard numerical reading. The eardrum is a particularly good location to measure .body temperature because it's near the hypothalamus, the body's temperature control center.
FIGURE 11.14 A radiatiolllherlTlOlneter Illeasures a paliclll\ !cm~
peralllre Iw monitoring the iJHen~il" of infrared r,Hlialiul) lC:\\'jng the car.
APPLYING PHYSICS 11.2
QUICK QUIZ 11.5 Stars A and B have the same temperature, but star A has twice the radius of star B. (a) \,Vhat is the ratio of star A's power output to star B's output clue to electromagnetic radiation? The emissivity or both stars can be assumed to be 1. (b) Repctlt the question if the st.ars have the same radius, but star A has I,.... ice the absolute temperature of sLu B. (c) 'What's the ratio if star A has bmh twice the radius and twice the absolute temperature of star B?
THERMAL RADIATION AND NIGHT VISION
How can thermal r(1diation be used to see objects in near total darkness? Explanation There are two methods of night vision, one enhancing a combination of\'cry faint visible light and infrarcd light, and another using infrared light only. The latter is valuable for creating images in absolute darkness. Because all ol:~jects above absolute zero emit thermal radiation due to the vibrations of their atoms. the infrared (invisible) light can be
EXAMPLE 11.12 Goal
focused by a speciallcns and scanned by an array or infrared detector clements. These elements create a thermogram. The inforrnation from thousands of separate points in the field ofvic'\' is convened 1O electrical impulses and translated by a microchip into a form suitable for display. Di[ferelll temperature areas are assigned different colors, which can then be easily discerned on the display.
Polor Bear Club
Apply Sldan's law.
Problem A member of the Polar Bear Club, dressed only in bathing trunks of negligible size, prepares lO plunge into the Baltic Sea from the beach in 51.. Petersburg, Russia. The air is calm, \vith a temperature of 5°C. Tfthe s\ovimmer's surface body temperature is 25 C, compute the net rate of energy loss from his skin due to radiation. How much energy is lost in \0.0 min? Assume his emissivity is 0.900 and his surface area is 1.50 m 2 . Q
Strategy Use Equation 11.U, the thermal radiation equation, substituting the given infonllatiol1. Remember to convert temperatures 10 Kelvin by adding 273 to each value in degrees Celsius! r
.
- ..
~
Solution
Convert tempcralHres from Celsius to Kelvin:
l~.oc
=
T". c = Compute t.he net rate of energy loss, using Equation ll.ll:
'!PIll:1
+ 273 = 5 + 273 = 278 K Tr; + 273 ~ 25 + 273 ~ 298 K 1~.
= r.rAf'(T" -
To I)
(5.67 X lO-8 IN/m 2 . K') (1.50 m') X (0.90) [(298 K)" - (278 K)']
Multiply the preceding result by the time. ]0 minutes, to get the energy lost in that time due to radiation:
Remarks Energy is also lost from the body through conveclion and conduction. Clothing- traps layers of air next to the skin, which are warmed by radiation and conduction. In still air these warIll layers are morc readily retained. Even a Polar Bear Club member enjoys sorne benefit from the still air, better retaining a stagnant air layer next to the surface of his skin. QUESTION 11.12 Suppose that at a given temperature the rate 01" all ol~jeCl's energy loss clue to radiation is equal to its loss by conduction. When the object's temperature is raised, is the energy loss due to radiation (a) greater than, (b) equal to, or (c) less than the rate of energy loss due to conduction? (Assume the temperature of the environment is constant.) EXERCISE 1112 Repeat the calculation when the man is standing in his bedroom, \\'ith an ambient temperature of 20.0 c C. Assume his body surClCc temperature is 27.0°C, w·ith emissi\'it) of 0.900. Answer
55.9 IN, 3.35 X 10'J
The Dewar Flask The Thermos boule, also called a Dewar Oask (after its invenlor), is designed to min imize energ}/ transfer by conduction, convection, and radiation. The thermos can store either cold or hotliguids for long periods. The standard vessel (Fig. ] 1.15) is a double-walled Pyrex glass with silvered walls. The space between the walls is evacuated to minimize energy transfer by conduction and convection. The silvered surface minimizes energy transfer by radiation because silver is a very good renector and has very 1m..· emissivity. A further reduction in cnergy loss is achieved by reducing the size of the neck. Dewar flasks are commonly used to store liquid nitrogen (boiling point 77 K) and liquid oxygen (boiling point 90 K). To confine liquid helium (boiling point 4.2 K), which has a very 1m,,· heat of vaporization, it's often necessary to usc a double Dewar system in \vhich the DC\'v'ar flask containing the liquid is surrounded by a second Dewar flask. The space between the two flasks is filled with liquid nitrogen. Some of the principles of the Thermos bOl.tle arc used in the protcct.ion of sensitive electronic instruments in orbiting- space satellitcs. In half of its orbit around the Earth a satellite is exposed to intense radiation frolll the Sun, and in the other hair it lies in the Earth's cold shadov,:. \,Vithout protcction, its interior \vol-dd be subjected to tremendous extremes of temperature. The interior of the satellite is wrapped with blankets of highly reflective aluminum foil. The foil's shiny surface reflects away much of the Sun's radiation while the satellite is in the unshaded part of the orbit and helps retain interior energy while the satellite is in the Earth's shadow.
11.6 GLOBAL WARMING AND GREENHOUSE GASES Many of the principles of energy transfer, and opposition to iI, can be understood by slUdying the operation of a glass greenhouse. During the day, sunlight passes into the greenhouse and is absorbed by the walls, soil, plants, and so on. This absorbed visible light is subsequently reradiated as infrared radiation, causing the temperature of I he interior to risco In addition, convection currents arc inhibited in a greenhouse. As a result, warrncd air can't rapidly pass over the surf,Kes of the green house that are exposed to the outside air and thereby cause all energy loss by conduction through those surl'aces. Most experts now consider this restriction to be a more important warming effect than the trapping of in rrared radiation. Ttl Fact, experiments have shmvn that \vhen the glass over a greenhouse is replaced by a special glass known to transmit infrared light, the temperature inside is lowered onl~· slightly. On the basis of this evidence, the primary mechanism that raises the temperature ora greenhouse
Silvered surfaces Hot or cold liquid
FIGURE 11.15 A cros",-scClional view of
APPLICATION Thermos Bottles
376
Chapter 11
Energy in Thermol Processes FIGURE 11.16 The cOllcelllratioll of alillospheric carboll dioxide in parts per millioll (ppm) of dry air as a funniolJ of lime during UIC laUel' part or the 201h cellLury. These data were recorded
370
:160
alions (red curve:;) coincide with growing
. a
seasons because vegetal ion absorbs c;lrb(Jn dioxide from the air. Tile stead}' illrrea~f.: (black curve) is of concern lu sricnlisLs.
8'
0
~
~0
:'l!'>O
" 0 " "0 ~
.'130
-
320
0
:1-40
IH60
1~lj!)
19,1) 1975 1980 19.'3,j [9!JO 199[, 2000
is nOllhe trapping of infrared radialion, bUl the inhibition of' airnow that occurs under any roof (in an attic, for example). A phenomenon commonly known as the greenhollse effect can also playa major role in determining the Earth's tcmperaLUrc. First, note that the Earth's atmosphere is a good transmitter (and hence a poor absorber) of visible radiation and a good absorber of infrared radiation. The visible light that reaches the Earth's surface is absorbed and reradiated as infrared light, which in turn is absorbed (trapped) by the Earth's atmosphere. An extreme case is the \,;armest planet, Venus, v.·hich has a carbon dioxide (C0 2 ) atmosphere and temperatures approaching 850°F. A!oi fossil fuels (coal, oil, and natural gas) are burned, large amounts of carbon dioxide are released into the atmosphere, causing it to retain more energy. These emissions are or great concern to scientists and governments throughout the world. Many scientists are convinced that the 10% increase in the amount of atmospheric carbon dioxide since 1970 could lead to drastic changes in '''orld climate. The increase in concentration or atlllospheric carbon dioxide in the laner part of the 20th century is shmvn in Figure] l.16. According to one estimate, doubling the carbon dioxide content in the atmosphere will cause temperatures 1.0 increase by 2"C. In temperate regions such as Europe and the United States, a 2°C temperalUre rise would save billions of dollars per year in fuel costs. Unforlunately, it \\-'ould also melt a large amount of land-based ice from Greenland and Antarctica, raising [he level of the oceans and destroying many coastal regions. A 2°C rise would also increase the frequency of drought.s and consequently decrease already low crop yields in tropical and subtropical countries. Even slightly higher average temperatures might make it impossible for certain plants and animals LO survive in their customary ranges. At present, about 3.5 X J0 11 tons of CO 2 are released into the atillosphere each year. Most of this gas results from human activities such as the burning of fossi I fuels, the cutting of forests, and m<1nufacturing processes. Another greenhouse gas is methane (Cl-I,,). which is released in the digestive process of cows and other ruminants. This gas originates from th,u part of the animal's stomach called the rumen, where cellulose is digested. Termites are also major producers of this gas. Finally, greenhouse gases such as nitrous oxide (N 2 0) and sulfur dioxide (502) are increasing due to automobile ~l1ld industrial pollution. \Vhether the increasing greenhouse gases are responsible or not, there is con\·jncing evidence that glob<-ll warming is under way. The evidence comes from the melting of ice in Antarctica and the reu"eat of glaciers at widely scattered sites throughout the world (see Fig. 1] .l7). For exarnple, satellite images of Antarctica show.Jamcs Ross Tsland completcly surrounded by water for the first time since maps were made, about 100 years ago. Previously, t.he island was connected to the mainland by an ice bridg·e. In addition, at various places across the cominent. ice shelves are retreating, some at a rapid rate. Perhaps at no place in the world are glaciers monitored with greater interest than in Switzerland. There, il is found that the Alps have lost aboul 50% of their
Summary
377
FIGURE 11.17 JJrtll!I(Jjtwlas!lr/f. The image ill (a). taken ollJ Ross Island (spider\'-shapcd.ju" off ct:lltcr) befnre the iceberg caln'd. hut afler til(" disilllegr'llion or III(' icc shelf bctwccn.lanu-s Ru.... Island and the Alll;tretic peninsula. In the image in pan (b), t;lkell 011 Febman 12, 1995, die.: icebcrg ha... cahed ,1IIe! b<:gUll mlldng ;l\\'a} from blld. The iccbcrg is aliom 78 kTll hI' 27 kill and 200 III thick. A celllllr}' ;lJ.:"O Janles Ross Islaud ,,'as COlllpkld\' ,urrOlll,ded in ice Ih:njoincd illO AmarCLica. (h)
(a)
glacial ice compared to 130 rears ago. The retreat of glaciers on high-altilude peaks in the tropics is even more severe I han in Switzerland. The Lewis ~Iacier on ~1011nt Kenya and the snows of Kilimanjaro are two examples. I n certain regions of the planet where g-Iacicrs are near large bodies of watcr and arc led by large and frequent snows, however, glaciers continue to advance, so the m'cl,tll piclllre of a catastrophic global-warming sccnal-io rna) be premature. III about,jO rears. though, the amount of carbon dioxide in the atlllosphere is expected to be about l\'lice what it was in the preindustrial era. Because of the possible Gllastrophic consequences, most scienUSL'i \'oice the concern that reductions ill greenhouse gas emissions need to be made now.
SUMMARY 11.1
Heat and Internal Energy
Internal energy is associated \\lith a S}'Slem's microscopic components. Internal energy includes I he kinetic cner~ of translation, rotation. and vibration of molecules, as \\'ell as potential energy. Heat is the transfel- of cnergy acro~s the boundary of a system resulting frolll .1 temperature differcnce between the system and its surroundings. The symbol Q represents I he alllOllnl of t:nerg)' Ira nsfcrred. The calorie is the amount or encrgy neccssaq to raise the lemperaturc of I g of waleI' from 1~1.:)oC [0 1!1..5°C. The mechanical equivalent of heat is 4.IR6 J/cal.
11.2
Specific Heat
11.3
Colorimetry
The energy required to change the tcmperatuft.' of a suhstance of mass m by an amoul1l.lTis
Q
~
/Il,J.T
[11.3J
wh('re r is the specific heat of the "iubst<'lIlcc. In calonm('u-y problems 1he specific heal of a substance c
when.' Q. is Ihe energy change in the kth object in the sysH.oIll. This equatioll G1Jl be soh'ed for the unknown specific heat. or used to determine an equilibrium tcmpcfatUl-e.
11.4
Latent Heat and Phose Change
The encl-gy required stance of mass 111 is
10
change the phase of a pure sllb-
Q ~ =/IlL
[l1.6J
where Lis [he latent heat of the substance. The latent heat of rll~i()n, L r, describes an energy transfer during a ch:1nge f\,()l1l a solid phase to a liquid phasc (or vice vcrsa), while the la[cnt hea[ or vaporil.aion, L", describes an encrbT)' trallsfer dlll-illg a change from a liquid phase to a gaseous phase (or \'ice versa). Calorimeu)' problems ilwoh'ing phase changes are handled wilh Equation 11.5, with latent heal terms added 10 the specific heatterllls.
11.5
Energy Transfer
Energy can be transferred br se\'eral diITerenl processes. including work. discussed in Chapter 5, and by conduction, convection, and radiation_ Conduction can be viewed as an exchange of kinetic energr between colliding molecules or electrons. The I-ate at which energy lran'ifer~ by conduction through a slab of al-ea A ancllhickncss I. is
( T - T) r 'lP= kA;'
r.
[11.7]
Chapter 11
378
Energy in Thermol Processes
where /( is I he thermal conductivity of the malcri ..ll making lip the slab. Energy is tr(l!1sferred by convection as a substance moves fr0111 one place LO another. All objects emil radiation from their surfaces in the form of electromagnetic waves at a nct rale of
W'""
~ (TAI'('I'" -
'I;,')
\,.. here T is the lemperalLlrc or the object and 7~ is the temperature of" the surroundings. An object that is hotter than its surroundings radi.ues more energy than it absorbs, whereas a body that is cooler than its surroundings absorbs more energ}' than it radiates.
[ILl I]
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
- - --,
-,
-
~~-
- ........,-::-
MULTIPLE·CHOICE QUESTIONS 1. COl1vcn 3.fiO X 1O:~ cal LO an equivalent number of
joules, (a) ~.74 X 10' J (b) 1.'17 X (e1) 5.33 X 10'J (e) 7.20 X JO"J 2.
I(J"J
(c) 3.2'l X to' J
COIWCrI 7.80 X 10:' J to the equivalent number or Calories (I Cal = I 000 cal). (a) 186 Cal (b) 238 Cal (c) 325 Cal (eI) 418 Cal (c) 522 Cal
3. How much energy is required to raise the lClnperarure of 5.00 kg of Icad from 20.0"C lO its melting point of 327"C? (a) 4.04 X IO"J (b) 1.07 X 10' J (c) 8.15 X IO'J (d) 2.J3 X 10"J (e) 1.96 X IO'J
4. If 9.30 X [oj J of energy arc transferred to 2.00 kg of ice at we, what is (he final temperature of tile system? (a) 224"C (b) 14.2'C (c) :ll.5"C (d) 18.0°C (e) O°C 5. A wall made of wood 4.nO em thick h(ls area of 48.0111 2 , If the temperature inside is 25""C and the temperature olllside is 14°C, at what ratc is thermal energy transported through the wall by conduClion? (a) 82 \V (b) 210 IV (c) 690 IV (eI) 1.3 X lO' W (e) 2.1 X 10' IV 6. A granite ball of radius 2.00 In and emissivity 0.450 is healed La 135"C, whereas I he ambicllI tcmperalUre is 25.0"C. What is the net power radiated from the ball? (a) 425 IV (b) 3.55 X 10' IN (c) 145 W (d) 2.01 X 10" W (e) 2.04 X 10' IV 7. How long would i, take a 1.00 X 103 vV hcating element to melt 2.00 kg or ice at -20.0"C, assuming all the energy is absorbed by the ice? (a) 4.19 s (b) 419 s (c) 555 min (eI) 12.5 min (e) 2.00 h 8. An unknm.. . n element of mass 0.250 kg, initially at 9S.0°C. is dropped into 0.400 kg of water at 20.0°C contained in an insulated cup of negligible mass .md specific heal. If the equilibrium temperature is 36.0"C,
~~..--
-
-
~
~
~-
.........
which of the following is the most likely identity of the substance? (a) aluminum (b) beryllium (c) cadmium (e1) iron (e) gold 9. An amount of energy is added to ice, raising its temperature from ~10°C to -SoC. A larger amollnt of energy is added to the same mass of water, raising its temperature from lye to 20°C. From these results, what can we conclude? (a) Overcoming the Iatem heat of fusion 01" ice requires an input of energy. (b) The latent heal of fusion of ice delivcrs some energy to the system. (c) The specific heat of ice is less than that of water. (d) The specillc heat of ice is greater than that of water. (e) J\,lore information is needed 1.0 draw any conclusion. 10. Star A has twice the radius and tv,rice the ahsolute tCIllperature of star B. Wh.n is the ratio of the po\,,ieroutput of star A to that of star B? The emissivity of both stars can be assumed to be I. (a) 4 (b) 8 (c) J6 (e1) 32 (e) 64 1l. A person shakes a sealed, insulated bottle containing
coffee for a few minutes. \,Vhat is the change in the temperature of the coffee? (a) a large decrease (b) a slight decrease (c) no change (d) a slight increase (e) a large increase 12. A poke I" is a still, nonflammable rod used to push burning logs around in a fireplace. Suppose it is to be made of a single material. For best functionality and safely, should the poker be made from a material with (a) high specific heat and high thermal conductivity, (b) low specific heat and low thermal conductivity, (c) low specific heat and high thermal conductivity. (d) high specif'ic heat and low thermal conductivilY, or (e) low specific heat and low density?
-
CONCEPTUAL QUESTIONS . . 1. Rub the palm of rour hand on a metal surface for 30 to 4!1 seconds. Place the palm of your other hand on an 11l1rl.lbbed portion of the surblce and I hen on the rubbed portion. The rubbed portion \-vill feel warmer. Now repeat this process on a wooden surface. \'\lh), does the temperature difference bct\.. .ccn the rubbcd and unrubbed portions of the wood surface seem larger than for the metal surface?
2. Pioneers stored fruilS and vegetables in underground cellars. Discuss fully this choice for a storage site. 3. In lIsually warm climates that experience an occasional hard freeze, fruit growers will spray the fruit trees with water, hoping that a layer of icc will form on lhe fruit. Vlhy would such a layer be advantageous?
Problems 4. In winter, why did the pioneers (mentioned in Question 2) store an open barrel of ''later alongside their produce?
enormously larger volullle of air by I.O°C. Estimate that volume of air. The specific heat of air is approximately 1.0 ~J/kg' Take the density of air to be 1.3 kg/m J .
5. Cups of watcr [or co flee or tea can be \Narmecl with a
9. A tile floor Illay feci uncomfortably cold to your bare feel, but a ca'-peted noor in an {lcljoining I"oom :at lhrsame tcmpcrature feels warm. \-Vhy?
coil that is immersed in the water and raised to a high Lernperature by means of electricity. ,·rhy do the instructions warn users not to operate the coils in the absence of ,>vater? Can lhe immersion coil be lIsed to warm lip a Clip of stew?
6. The U.S. penny is now made of copper-coated zinc Can a calorimetric experiment be devised to tesl for the mewl content in a collection of pennies? If so, describe the procedure.
7. On a clear, cold night, why docs frost tend to form on the tops, raliler than ule sides, of mailboxes and cars?
8. The air lemperalure above coastal areas is profoundly inrluenced by the large specific heat of ,'>'arer. One rea-
0c.
10. On a
VlT)' hot day, it's possible to cook an egg on the hood of a car. V/ould you select a black car or a white car on which to cook your egg? \Vhy?
11. Concrete has a higher specific heat than does soil. L'se this faci to explain (panially) why a city has a higher average temperature than thc surroll nding cou ntryside. ""'ould you cxpect evening breezes to blow from cit)' to coulllry or from country to city? Explain.
)2. You need to pick LIp a vcr)' bot cooking pOL in your kitchen. You have a pair of hot pads. Should yOLl soak them in cold waleI' or keep them dry in order to pick up
son is that the energy released when I cubic meter of waleI' cools by l.O°C will raisc Ihc temperature of an
I
379
[he pot llloSt comfortabl)'?
PROBLEMS
~
•
The Problems for this chapter assigned onlin!' al HiebAssign. = straiglnrorwal"cl, internwdiatl', (·h.l\lcll~;l\g
11I<1V
jOllIes of enerb,)' i:lfC the equivalent of onc jelly doughnut? (b) How many stairs must the woman climb to perform an amount 01' mechanical '...·ork equivalent. to the rood energy in one jelly dOtlghJlul? Assume the height ora single stair is It} cm. (c) If the human body is only 25% efTicient in converting chemical eneqw to mechanical energy. how man)' stairs lllust the woman climb to work off her brcakfa'lt?
he
WebAsstgn
I, :!,
;~
lSI =
m
delloles guided probkm
= denotes enhanced rOIHenl probkm
= bioTlledical application 0= c\clloles fnll solution ,wailable in SllIdmt
•
S(Jlllfif!lI~ Mauuall
S/1/(I)' Gl1ide
SECTION 11.1 HEAT AND INTERNAL ENERGY SECTION 11.2 SPECIFIC HEAT 1. The highest recorded waterfall in the world is fOllnd at Angel FaJls in Venezuela. Its longest single waterfall has a hciglll ofRDi In. rfwaler at lhc LOp Or the f~lIls is at15.0"C,
what is the maximum temperature of the water at the bottom of the falls? Assume all the kinetic energy of the water as it reaches the boltom goes into raising the waler's tcmpcrawre.
2. How much energy is requir"Cd to raise the temperaLUre of 1.50 kg of cadmium from 20.00C to 150"C? [[]Lakc Erie conlains roughly 4.00 X 1011 m·~ or waler. (a) !-low much enerb'1' is required LO raise the temperature of that volulllc of water frolll II.DOC LO 12.0"C? (b) Ilow many years would it take 10 supply this amounl of cneq:.,'"Y by using the I OOO-~f\'V cxh(lust energy Or an electric pO\..-cr plant? 4. All ::lluminum rod is 20.0 cm long at 200C and has a mass or 350 g. If 10 OOO.J of enerln' is added LO the rod by heaL. whal is the change in length of the md;: 5. A 75-g sample of silicon is at 25"C. If 750 cal of energy is lransfcrred to the silicon, what is its final tcmperature?
6.
II A
55-kg woman cheals on her diet a.nd eats a 540-
C,\\oric t540 kca\) je\\y donut ror breakfast. (a) How lTlanv
7.
II E
A 75-kg sprinter accelerates from rest to a speed of J 1.0 m/s in 5.0 s. (a) Calculate the mechanical work done by the sprinter during this lime. (b) Calculate the average power lhe sprinter must generate. (c) If the sprinter converts food enerf,'1' to mechanical energy with an efftciency of 25%, at what average rat.e is he burning Calories? (d) What happens to the other 75% of the food encr b'1' being used?
8. " mil A sprimer of mass '/1l accelerates ullilormly from restLO velocity 'V in (seconds. (..I) Write a symbolic cxpres .'lion for the inslantaneous mechanical power ':P required by the sprinter in terms afforce Fand velocity v. (b) Use Newton's second law and a kinematics equation for the velocity at any time to obtain an expression lor 1 he instantaneous power in lerms of 111, n, and I onlv. (c) If a 72">.O-kg sprinter reaches a speed of 11.0 m/s in 5.00 s, calculate the sprinter's acceleration. assuming it to be constant. (d) Calculate the 7S.0-kg sprimer's instantaneous mechanical power as a fUllction of time I and (e) give the maximulll r<:lle at which he burns Calories during the "print. assutlling 25% efficiency of conversion form food cnCl"gy to mechanical energy. 4
9. A !1.00-g lead bullet traveling at 300 m/s is stopped by a large tree. If half the kinetic energy of lhe bullet is transformed into internal energy (lnd remains WiUl ule bullet while the other ha\f is transmitted to the lrce, whal is the increase in lemperature of the bullet?
Chapter 11
380
Energy in Thermal Processes
10. The apparatus shown ill Figure r 11.\ 0 was llsed by Joule to measure UIC mechanical equivalent of heaL. \,\-'ork is done on [he water by a rotcuing paddle wheel, \'ihich is driven b}' t..."o blocks falling at a consh'lnl speed. The temperature of the stirred water increases due 10 the frinio!1 bcn","cen the waler and the paddles. If lhc energy lost in the bc,lI'ings and through the walls is neglected, then the loss in potenrial energy associated with the blocks equals the work done by the paddle wheel on the water. If each block h;l,s a mass of 1.50 kg and the insulated lank is filled with 200 g of water, what is the increase in temperature of Illc water after the blocks fall through a disL
ing a symbolic equation for thermal energy transfer Q, (c) Compute thc mass of the concrele slab g-ivcn {hal it., density is 2.40 X 10 3 kg/m 3 . (d) Conerele has an ultimatc compressive slrenglh of 2.00 X 10 7 Pa, specific heat of 880 J/kg-· "c, and Young's modulus of 2.1 X 10 '0 Pa, How much thermal energy musl be transferred to the slall to reach this compressive stress? (e) What temperature change is refJuired? (0 If the sun delivers 1.00 X 10~ ,·vor power to the LOp surface of the slab and if half the energy, on the avcrage, is absorbed alld retained, how long docs it take the slab to reach the point at which it is in d;wger of cracking due 10 compressive stress?
SECTION 11.3 CALORIMETRY 15. ,,,,'hat mass of wat.er at 25.00C must be allowed to come 1.0 Lhennal equilibriulll with a 1.85-kgcubc ofaluminulll initiallyal 1.50 X I02"C to lower the temperature of the aluminum to 65.00C? Assume
Tllennal il1sulatnr FIGURE Pl1.10 The rallil,g \"cjghl~ rot:lle t hl.: padtlks. GHlsing the lemper
ffiJA 200-g aluminum cup contains 800 g of "',ller in thermal equilibrium WitJl the cup at 80°e. The combination of cup and watcr is cooled uniformly so th
E!I
In the SlllTlmcr or 195H in St. Petersburg, Florida, a new sidewalk was poured Ilear the chi lei Ilood home of aile of fhe
How many pellets must he added to 500 g of water lh(it is initially at 20.0"C to make the equilibrium wmperature 25.0"C? Neglect an)' energy transfer to or from the container.
01].-\ n a lu mi 1ll11l1 cup contai llS 225 g of water and a 40-g copper stirrer, all at 27"C. A 400-g sample of silver at an initial temperature of Sioe is placed in the water. The stirrer is IIsed to stir the mixture llOLil it re
21. A student drops two metallic objcct'i inlo a 120-g steel cont.ainer holding 150 g of water
Problems kincuc energy to thermal energy. If a I 500-kg automobile traveling at 30 IlI/S comes 10 a hail, how much docs Ihe temperature rise in e~lch of the fOLlr 8.0~kg iron brake drums? (The specific heat of iron is 448 .I/kg·" C.)
23.
30.
E
Equal OAOO-kg masses of lead ami tin a\ 60.00e arc placed in 1.00 kg ofwaler at 20.00C. (a) What is Ihe cCJuilibriul1l temperature of the system? (Il) If" an allo), is hedl' Icad and half tin by mass, ".:hat specific heat would )'011 anticipate for the allo),? (c) I low man)' atoms of lin IVs" arc in 0.400 kg of till, and how many atoms of lead N,,!> arc in 0.400 kg or lead? (d) Divide Ihe number 1\1",,, of tin atoms by the number J\ll'h of lead aLOms and cumpare this ratio with the spccific heat of lin divided by the specific heat of lead. Wh,it (;onclusion can be drawn?
124.IAn unknown substance has a mass of 0.12S kg and all initial temperature of 95.0"C. The substance is thell dropped into a calorimeter made of alumillum containing 0.285 kg of \vater initially at 25.OCC. The mass of the aluminum container is 0.150 kg, and lhe temperawre of the calorimeter increases 1O a fimJ1 equilibrium temper,llure or~2.00C. Assuming no thermal energy is trallSkrred to the environment, calculatc the specific hem of the unknown substance.
nee is placed in
32.
c
What is the finaltcmperatllrc of the mixture?
26. A 50-g ice cube Cit O"C is heated until 45 g has become waleI' at 100°C and 5.0 g has become steam al lOWe. Ilow milch energy was added to accomplish the transfol"mat.ion?
28. How much encrgy is required to change a 40-g ice cube from ice al -I D"C (0 steam at. 11 O"C?
29. A 7!1-kg cross-country skicr glides over snow as in Figure PI J .29. The coefficient of friction between skis and snO\...· is 0.20. Assume all the snow beneath his skis i.~ al oce and that all lhe inlernal cllert-f)' generated by f'riction is added to SIlOW. which sticks to his skis until it melts. How far would he have to ski to melt 1.0 kg of snow?
Illto a 0.500-kg aluminum conuiner at 20.00C is placed 6.00 kg of cthyl alcohol al 30.DoC and 1.00 kg ice (It -I O.O"'c. A<;sume the ~ystem is insulated from its environment. (a) Jelent ify all live thermal energy u-ansfers that occur as the system goes to
cal/g'"C
825 g of'water at 25 C.
27. A lOO-g cube of icc at o"e is dropped into \.0 kg of" water that was originally at 80"C. ""hal. is the final temperature of the water arter the ice has mclted?
m
[TIJA 40~g block of ice is cooled to -7S"C and is then a{lded \0 560 g or water in an 80-g copper calorimeter at a lemperature of 25"C. DeLCrmine the final tempenllUl'e of the system consisting of t.he ice, water, and calorimeter. Of nOI allihe ice melt.<;, determine how much ice is left.) Rememher that the ice mllst first warm to O"C, melt, and then continue \varllling as wat.er. (The specillc heal of ice is 0.500
SECTION 11.4 LATENT HEAT AND PHASE CHANGE 25. A 75-g- ice cube at
381
= 2 090 J/kg· 0G.)
II
\""hen you jog, most nf the food energy you burn above your basal metabolic I<\te (BM R) ends up as internal cnerh')' that would raise your body temperature if it wcre not eliminated. The evaporation of perspiration is the primal;" mechanism (-ill' eliminaung this energy. Determille the amount ofwatcr you luse to evaporation when running for 30 minutes at. a rate that uses 400 kcal/h above your BMR. (That amoul1l is often considered to be the "maximum fatburning" energy output.) The metabolism of I gram of' rat generat.es approximately 9.0 kcal of cneq::>,)' and produces approximately] gram or water. (The h)'drogen atoms in the fat molecule are transferred LO oxygen to form water.) \.vhat fraction of your nced for water will be providc{1 by I~lt metabolism? (The lalem heat of vaporization of waleI' at room temperaturc is 2.5 X lOG J/kg.)
33. A high-end gas stove usuetlly has at least one burner rated at 14000 BllI/h. IfYOli place a O.25-kg aluminum pOl containing- 2.0 liters ofw
34. " A 60.0-kg runner expends ~OO W of power while running ;'1 maral.hon. Assuming 10.0% of the energy is delivered to the muscle tissue and that the excess energy is removed from the body primarily by sweating, determine the volume of bodily fluid (assume it is waleI') lost per hour. (AI 37.0°C, the latent heat ol'vaporization ofw-ater is
2.41 X 10fi.J!kg.) 135.1SLCam at IOO"C is added to ice at O"C. (a) Find the
~ '--------~FIGURE P".29 A cross-country skier.
II
The excess internal energy of metabolism is exhausted through a val-icry of channels, such as through radiation and evaporaliol1 of perspiration. Consider another pathway for em:q:...'J' loss: moisUire in exhaled breath. Suppose you breathe flut 22.0 breaths per minute. each with a volume of 0.600 L. Suppose also thai you inhale (II)' air and exhale air at ;"7°C containing water vapor wilh a \-aPlW pressure of 3.20 kPa. The vapor comes from the evaporation of liquid
382
Chapter 11
Energy in Thermol Processes
water in your body. Model the water vapor
37.
mm A ~tOO-g lead
bullet at 30.D"C is fired at a speed of 2.40 X 10 2 m/s into a large, fixed block of ice ,n O"c. in which it becomes embedded. (a) Desuibc the energy transformations that occur as the bullet is cooled. What is the r-inal temperature or the bullet? (11) What qtlantil~' of icc mehE?
SECTION 11.5 ENERGY TRANSFER 38. A glass windowpane in a home is 0.62 CIll thick and has dimensions of 1.0 III X 2.0 Ill. On a certain day, the indoor tCl11per,uurc is 25"C and the olltdoor temperature is DOC. (a) 'What is the rate at which energy is transferred by heal through the Rlass? (b) How much energy is lost through the window in aile day, assuming the temperatures inside and outside remain constant? 39. A concrete slab is 12 Clll thick and has an area of 5.0 m 2 . Electric heclting coils are installed under the slab to melt the ice 011 lhe surface in the winter months. Wh;:lllllinimum po\\'er must be supplied to the coils to nl
40.11 The Ihermal conductivities of human tissues vary greatly. Fat and skin have conductivities or about 0.20 W/m' K and 0.020 "VI'/m' K, respectively, while other tisSHes inside the body have conduclivities of about 0.50 \'\-'/m' K. Assume that bet\\..-cen the core region of lhe bod)' and the skin surE-lee lies a skin layer of 1.0 mill, f~lliayer 01'0.50 cm, and 3.2 Clll of other tissues. (a) Find lhe li'-facLOr for each of lhese layers, and the equivalent R-factor for all layers taken together, retaining two digit.s. (b) Find the rate of" energy loss when the core temperature is 3TC and the exterior temperature is We. Assume that both a protective layel" or clothing and (Ill insulating layer of unmo\'ing air al"e absent, and a body area of 2.0111 2 . ~A steam pipe is covered with 1.50-cm-thick insulating material of lhennal conductivity 0.200 cal/em' "C, s. How milch energy is lost I.:very second when the steam is at 200aC and the surrounding air is at 20.0"C? The pipe has a circumference of 800 em and
42. The average thermal conducti"ity of the walls (including windows) and roof of a house in Figure P-11.42 is 4.8 X
10- 1 kW/m' "e, and their averagc thickncss is 21.0 COl. The house is heated with n,llllral gas, with a heat of combustion (energy released per cubic meter of gas burned) of 9 ~)OO kcal/m 3 . How many cubic meters of gas must be burned each day to maintain an inside temperalure ul" 2:i.O"C if the uUlside telllper'llltn.: i~ O.O"C? Di.':iI-egal-ti radiation and energy loss by heat through the ground. 43. Determine the H-value [or a wall constructed as follows: The outsidc of lhe house consist.s or lapped wood shingles pl O.O"C, determine the rate of energy Iransfer through 1.0 m:! of the window. COIllpare your answer with the rate of enerblJ' transfer through 1.0 lll~ of a single I.O-cm-thick pane of gh-lss. 45. A copper rod and an aluminuIll rod of equal diamcter are joined end 10 end in good thermal contact. The temperature of the free end of the copper rod is held constant al loo"e and lhatofthe br end of the aluminum rod is held at O"c. If the copper rod is 0.15 m 101lg. what must be the length of the aluminum rod so that the temperature at the junction is 50"C? 46. A Styrofoam box has a sluface arca of 0.80 m 2 and a wall thickness of 2.0 cm. The tcmperaltlre of the inner surface is 5.0"C, and U1C outside temperature is 25"C. If it takes 8.0 h for 5.0 kg of ice to melt in the container, determine the thermal conductivity of the Styrofoam. 47. A sphere Ihat is a perfect blackbody radiator has a radius of 0.060 m and is at 20acC in it room where ule temperature is 22"C. Calculate the net I"all.: at which the sphere radiates energy.
148.11\ solar sail
is made or aluminized Mylar having an emissivit}' of 0.03 and renecting 97% of the light that falls on it. Suppose
49. t\'leasuremenls on two st,'ll'S indicate that Star X has a surface temperature of ,I) 727"C and Slar Y has a surface temperature of 11 727 c C. 1r both smrs have the same radius. what is the ratio of the lurninosily (IOtal powel" outPUI) of Star Y to t.he luminosity or Star X? Both stars can be COIlsidered to have an emissivity of 1.0. 50. Calculate the temperature at which a tungsten filament lhal has an emissivity of 0.90 and a surface area of 2.5 X 10-'''> Ill~ will radiate cnergy at tbe rate of 25 in a room ""herc the tcmperaLUre is 22 c C.
"V
ADDITIONAL PR08LEMS
51. The bol.tolll of a copper kettle has a 1O,clll radius and is 2.0 mill thick. The temperature of tile outside surface is
J02'C, and the waler imide
the KrIll" 1'\ I '/' " • )0)
/11K ill
J
Will
Problems
383
of pressure. Find the rale <\1 which energ)' is being lral1SI"erred Lhrollgh the bottom of the kettle.
what rate must the remaining excess cnergy be eliminated through conduction and convectilll1?
52. A l~unil)' comcs homc [rom a long vaGuion WitJl Immdry to do and showers to t;lke. The w,uer heater has bf'cn turned off dllring 111(' ";lr~lIiol1" II" rl\(' IlI''llf'!" h'l" ;, f';lparil), of 50.0 gallons and a 4 800-\V heating clemcm, how much lime is reqllil'cd La raisc Ihe lemperacure of the wmer from 20.0°C to tiD.naC? Assume the heater is w('l1 insulated
I59.I\oVaLcr is being boiled in an open kenlc that has;l 0.500 cmthick circular aluminum bottom with a r;ldius of 12.0 cm. If Ilw \.. ."tel· hoil" 71w"y ;11 ,\ l";It(' of 0500 kg/min, what is the temperature oflhe lower ~ud~lce of the boltom of Lhe kcule? A,,~ume llle lOp surface of the bOllom of the keltle is ;It 100~C:.
53. A '1O-g ice cube lIoaL" in 200 g of water in a 100-g" copper cup; all are at a tempcrat.ure of A piece of lead at 9SOC is dropped into the Clip. and the final eqtlilibrilllll lemperature is t2°e. \'\'hat is the mass of lhe lead?
ace.
54. A WaleI' heatcr is operated by solar power. If the solar collector has an area 01" 6.00 rn~ and the imensity deli\"ered b}' sunlighl is 550 W/m~, how long does it lake 1.0 increase the temper;uurc or 1.00 m~ of \.. ."ter from 20.0~C to 60.ace? 155.IA 200-g block of copper "II a lemperalure or 90 C is dropped inlo 400 g of WaleI' al 27 c C. The water i~ contained in a 300-g g-Iass container. VI/hat is the lintll tt'mpcr;lture or the nJixUire?
60. A <.:lass 0("10 student.s taking an exam has a power output per stlldent of abOllt 200 W. Assume lhe initial temperature of the room is 20°C and that its dimensions are 6.0 m by 15.0 m by 3.0 m. What is the tempcrawre of the room at the end of l.O h if all the enerb,)' remains in the air in the room and nOlle is added by an outside sOlirCl.:? The specif-ic he(Jt of air is 837 J/kg' °e, and iLS densily is about 1.3 X 10 :\g/cm~. 61. A bar of gold (Au) i~ ill thermal contaCt with a bar of sil\'er (Ag) of the same length ,mel area (Fig. PI1.61). One end of' the compound bar is maintained at 80.0 o e, and the opposilc end is al30.0°C. find the temperalllre at the JUHctioll when the energ}' flow reaches a steady state.
56. _ Liquid nit.rag·en has a boiling point of77 K and a latenl heal or vaporil.ation of ~.O I x Hf' J/kg-. A 25-W electric heatin~ elcmem i~ immersed in an inslllau:d vessel cOIlL-,ining 25 l. or liquid nit.rogen al iLS boiling poin L (a) Describe the energy u
57. mil A studenl measures lhe following clata ill a calorim-
All
M;lSS
or WaleI':
1
FIGURE Pll.61
62. An iron plate is held against an iron wheel so that
70.OOC 0.400 kg-
""Iass of calorimeter:
0.040 kg
Specific heat of cllorimcter:
0.6:; kJ/kg . 0c:
Initial temper
27.0°C
t\oJass of aluminum:
0.200 kg
FillallemperaUire of mixlIlre:
6fi.goC
Use lhese data to delcnnine 1h(' speciric heal 01 alll~ minum. Explain wheLher your result is within 15% of' the valuc listed in Table 11.1. 58.
III
O\'erall, 80%01" thc energy used by the body mll~1 be <.:liminated as excess thermal ellerb')' and needs LO be di~ sipaled. The mechanisms or elimination are radialion. evaporation of swe;H (2 4~0 ~J/kg). eV
Ag
IllsulallOll
('try experiment designed lO detennille the specific heat of aluminum: Inilialtemperatllre of water and calorimeter:
I
(l:~.
:'\n automobile has a lllass of I 500 kg, and its aluminum brakes have an overall Illass of 6.0 kg. (a) Assuming all lhe internal energ-y lransfonned by friction when the car SLOpS is deposited in the brakes and neglecting energy transfer. how many Limes could the car be braked 1.0 reSl slaning from 25 m/s (56 mijh) before the brakes \...· ould begin 10 melt? (Assume an inilialtemperawre of 20°C.) (b) Idemil"y some efrects thaI are neglected in pan (a), bUl <1I'e likely 10 be important in a more realistic
16-t.IThree liquids are at temperature:>, or woe, 200e, and ?oooe. respectively. Equal masses of the first tWO liquids are mixed, allCl the equilibrium lempCraLure is 17°C. EfJual masses of the second and third are then mixed, and the equilibrium lemperature i~ 28"C. Find the equilibrium temperaulI-e when equal masses of the flrsl and Ihird arc mixed. 65. _ A 1/01/1 caloril!ll?lt'1" is an apparatus used to measure Ihe specific heal of'a liquid. The technique is to measure
384
Chapter 11
Energy in Thermal Processes
the temperature difference between the input and Ollt· put points of a nowing stream of the liquid while adding cncrg)' at a known rate. (a) Stan with the equations Q = mr(uT) and m = pl~ and ~how thai the rale ,-Il \\>hich energr is added to the liquid is gi\-cn b}' the expre~..ion ~Q/~t
tOI". Then it is plll into the box LO kecp the tesL samples warm as it solidifies. The heat of rusion of tJle phasechange mal.CI-ial is 205 ~J/kg. \Iodel the insulation as a panel with surface area 0.490111 2• Ihickness 9.50 em. and conducti\-ity 0.012 0 W/moC. Assume lhe exterior LemperaLure is 23.00C ror 12.0 hand 16.O"C ror 12.0 h. (a) What mass of the waxy material is rcquil-ed to conduCl Ihe bactcliological lest? (b) Explain \'t'hr youl" calculaLion can be done wi.thout knowing the mas'i of the test &"lmples or of the insulation.
= pr(j,'I)(ilVj.:.\t). (Il) In a panicllla," cxpc';lIlcm.
a liquid of density 0.72 g/cm 3 nows through the GI.1Cll-imetcl" allhe rate of3.5 cm 3/s. At steady :.latc, a ICmperal.tll'C difference of 5.SCC is established between the inpm and QUlplil pointS when energy is supplied aL the rate of 40 Jh. What i:-. the specific hen of the liquid? 66. A wood Slove is llsed to heal a single room. The stove isC\'lindtical in shape. \\ilh a diameter of 40.0 em and a length of 50.0 em. and opernLes aL a temperaLUre of 400°F. (a) If the LemperaLure of the room is 70.0°F. deLcnninc lhc amount of radittlll encrb'Y delivered to the room by the SlOve each second if the emissivity is 0.920. (b) If the room is a ~quarc \\itb walls that are 8.00 ft high and 25.0 ft wide, determine the N-value needed in Lhe walls and ceiling to maintain the inside tempermure at 10.O°F if the oUlside temperlllUre is 32.O"F. NOIf' that we are ignoring ;my heat conveyed by the stO\e via colweclion and an}' ene'"!,"" lostthmugh the walls (and windows!) via convecLion or radiation. 67. A ~')olar cooker" consislS of a curved rerlecting mirror thai foclIsc<; sunlight onto the object to be heaLed (Fig. PI 1.67). The solar power per un iL area re;'lching lile Earth
69. What mass of steam initiallr at 130~C is needed to wann 200 g of water in a IOO-g g:lass conminer from 20.0"C to ,;O.O·C?
70.
The evaporation of perspinltion is the plimal) mechanism for cooling the human body. Estimate tJ1C amount of watC!" you will lose when )'ou bake ill the sun 011 tJ1C beach for an hour. se a value of I 000 W/m:! for the intensil}' of :o,unlight and note Lhat lhe energy rcquired La C\"3poratc a liquid at a partinilar tempel-alure is approximateh' equal 10 the sum of the cncrf,'Y required 10 raise its Icmperalure 10 the boiling point aun the latent heal of \'aporization (determined at the boiling poinl).
,=
[1[]Attime 0, a \'e.s:o.el contains a mixture or lO kg of water and an unknown Illass of ice in equilibrium at Occ. The temperaturc of the mixture is mcasured over a period of hour. wilh the following results: During the firsl.::'O min, the mixl.Ure remains at O°C; from 50 min to 60 min. the temperature inneases steadily from O°C 102°C. Neglecling tJ1C heal. capacity oj" the vessel, deternlinc the mass ar icc that was initially placed in it. Assulllc :1 cansl,ant power input to the container.
,Ill
72.
am An
icc-cube tnty is filled with 73.0 g" of water. After lhe filled tray reaches an equilibrium temperature 20.0°C. it is placed in a freezer SCI al -8_00°C LO make icc cubes. (a) Desnibe the processes that occur as energy is being removed from the waLeI' to make ice. (b) Calculate lhe energ} thaL must be l-cmo\cd from the water to make icc cubes aL -8.00°C.
73. An aluminum rod and an iroll rod arc joined end to end in good thennal conL'lC!. The two rods have equal length... and radii. The free end of the aluminum rod is maintained at a temperature of JOO"'C. and the free end of the iron rod is mainL,lined at O°c. (a) OeLermine the tcmperalure of the intel-facc betl.. .een Lhc twO rods. (b) If cach rod is 15 cm long and each has a cross-sectional area of 5_0 cm~. \dlat quantity of energy is conductcd across the combination in 30 min?
Lance Armstrong is an engine: he requires fuel and oxygen to burn it, and the result is work that drives him up the mountainside as his excess woste energy is expelled in his evaporating sweat.
THE LAWS OF THERMODYNAMICS
12.1
Work in Thermodynamic Processes
12.2
The Fi rst Law of Thermodynamics
12.3
Thermol Processes
12.4
Heat Engines and the Second Law of Thermodynamics
12.5
Entropy
12.6
Human Metabolism
According to the first low of thermodynamics, the internal energy of a system can be increased either by adding energy to the system or by doing work on it. This means the internal energy of a system, which is just the sum of the molecular kinetic and potential energies,
can change as a result of two separate types of energy transfer across the boundary of the system. Although the first law imposes conservation of energy for both energy added by heat and work done on a system, it doesn't predict which of several possible energy-conserving processes actually occur in nature. The second law of thermodynamics constrains the first law by establishing which processes allowed by the first low actually occur. For example, the second law tells us that energy never flows by heat spontaneously from a cold object to a hot object. One important application of this law is in the study of heat engines (such as the internal combustion engine) and the principles that limit their efficiency.
12.1
WORK IN THERMODYNAMIC PROCESSES
Ent:rgy can be transferred to a system by heat and by work done on the system. In musl cast:s of intcresll.reated here, the syslem is a voluJlle of gas, which is importarll in understanding engines. All such systems of gas will be assunH.:d to be in thermodynamic equilibriulll, so that. every pan of the gas is at the same temperature and pressure. If that. were not the case, the ideal gas law wouldn't apply and most of the results prescnted ht:re wouldn't bc valid. Consider a gas contained by a cylinder rltted with a lIlovable piston (Active Fig. l2.la) and in equilibriulll. The gas occupies a volume \land exerts a uniform pressure Pan the cylinder walls and the piston. The gas is compressed slowly enough so the system remains essentially in thermodynamic equ\\ibriulll at all times. As the pistoll is pushed down\'vard by an external force Fthrough a distance Ii.:v, the work done on the gas is
,. (a)
(b)
ACTIVE FIGURE 12.1 (a) A ga~ in a q'linder on upying a volume \!ata pre.~sure I J• (b) PII~hing the pislon down comprt·s:>t.:s the gao,;.
385
Chapter 12
386
The Laws of Thermodynamics
where we have set the magnitude F of the external force equal to /.)1\, possible because the pressure is the same everywhere in the system (by the assumption of equilibrium). Note thal if the piSLOil is pushed downward, 6.y = Yf-)" is negati\'c. so we need an explicit negative sign in the expression for l\1to make the work posi[ive. The change in \'olume of the gas is 6V= A 6.)', which leads to the following definition: The work W done on a gas at constant pressure is given by
w=
-p~v
[12.1]
\\'here Pis the pressure throughout the gas and 6.\'is the change in volume of the gas during the process. If the gas is compressed as in Active Figure 12.lb. aVis negaliYc and the work done on the gas is posiLivc. If the gas expands, LlVis positive and the work done on the gas is negative. The work done by the gas on its environmem, Well'" is simply the negative of lhe work done on lhe gas. In the absence of a change in volume, the work is zero.
EXAMPLE 12.1 Goal
Work Done by' on EXJ1anding Gas
Apply the definition of work
rH.
constant pressure.
Problem In a system similar to that shown in Active Figure 12.1, the gas in the C)'lincier is at a pressure equal to 1.01 X I(f' Pa and the piston has an area ofO.100 111~. As encrgy is slowly added to the gas by heat, the piston is pushed up a disl;lnce of 4.00 em. Calculate Ihe \\lark done by the expanding gas on the surroundings, V\~II\' assuming the pressu rc rcma i I1S constant. Strategy The \\Iork done on the environment is thc negative or the work done on the gas given in Equation 12.1. Compute the chang-c in vohunc and lTluhiply by the pressure. . ..., r'
Solution
= A ~y =
Find the change ill volume of the gas, ~V; which is the cross-sectional area times the displacement:
~v
Multiply this result by the pressure, getting the work the gas docs on the clwironment, \1~0I:
We",
(0.100111')(4.00 X 10-2 Ill)
= 4.00 X 10- 3 Ill'
= =
P ~\I = (1.01 X 10; l'a)(4.00 X 10 -'Ill')
-l04J •••••
..rI
Remark The volume of the gas increases, so the work done on the ellYironment is positi\"e. The work done on the system during this process is lr= -404j. The energ)" required to perform positive work on the em'ironment must come from the energy of the gas. (See the next section for more details.) QUESTION 12.1 If 110 energy were added to the gas during the expansion, could the pressure remain constant? EXERCISE 12.1 Gas in a cylinder similar to Figure 12.1 moves a piston with area 0.200 m:? as energy is slowl) added to the system. If 2.00 X 103 J of work is done on the environment and the pressure of the gas in the cylinder remains constam at 1.01 X 10!> Pa, find the displacement or the piston.
Answer 9.90 X 10-2
In
Equation 12.1 can bc used to calculate the work done on the system only when rhe pressure of the gas remains constant during the expansion or compression. A pl"OCeSS in ,·vhich t.he pressure remains constant is called an isobaric process.
12.1
Work in Thermodynamic Processes
The pressure \IS. volume graph, or PV diagram, of an isobaric process is shmvn in Figure ]2.2. The curve on such <'1 graph is called the !m/.II taken between the initi~t1 and final states. with the arrow indicating the direction the process is going, in this case from smaller to larger vululllc. The area under lhe graph is Area =
pC \} -
387
p
I"
P
_
-_
Vi) = IJ ~ V
The area under the graph in a PV diagram is equal in magnitude to the work done on the gas. This is true in general, whet her or not the process proceeds ~lt constant pressure. JUSt drmv the PF diagram of the process, find the area underneath the graph (and above the horil.ontal axis), and that arca will be the cqual to the magnilllde of the work done on I he gas. I r the arrow on the graph points toward larger volumes, the \vork done on the gas is negative. I f the arrow on the graph point:-:i to\vard smaller volumes, the work done on the gas is positivc. \o\'hencvcr negative work is done on a systcm, positive work is done by t he system on its cllvironment. Thc negative work done on the system represellls a loss of energy from the system-the cost of doing positive work on the environment.
--':-_ V
L-~
\j
\',
FIGURE 12.2 The Pl"diagr;t111 !<1I. a gal> b<:in~ compressed al COllSlalll prc~sllro.:. Tile sh;lckd area n:pre~cllls tllo.: \\ork nOlle on the ga....
QUICK QUIZ 12.1 By visual inspection, order the PV diagrams shown in Figure l2.3 from the most negative work done Oil the system to the lllost positive work done on the systelll. (a) a, b, c, d (b) a, c, b, d (c) d, b, c, a (d) d, a, c, b P(I05 pa)
Notice thatthc graphs in Figure 12.3 all have the same endpoints, but thc areas bencath lhe curves arc different. The work done all (\ s}'stem depends all the path laken in the PVdiagram.
EXAMPLE 12.2 Goal
Work and PV Diagrams
Calculatc work from a PVdiagram.
Problem
Find the numeric value of the work done all the gas in (a) Figure 12.3a and (b) Figure 12.3b.
Strategy The regions in question are composed of reclangles and triangles. Use basic geometric formulas to find the area underneath each Cltrve. Check the direction of the arrow to determine signs. r-"
-
Solution (a) Find the work done on Ihe gas in Figure
12.3~1.
Compute the areas Al and A~ in Figure 12.3a. AI is a rectangle and 1\2 is a triangle.
A, = height X width = (1.00 X lO' Pa)(2.00 m") ~
2.00 X 10 5 .1
A~ = ~ base X height = H2.00 m')(2.00 X 10-' Pal = 2.00 X 10-'.1
Sum l.he areas (the arrows point to increasing volume, so the work done on thc gas is negativc):
Area
~
i'\r~
11,
+
A, = 4.00 X 10'.1
-4.00 X 10 5 .1
388
Chapter 12
The Laws of Thermodynamics
(b) Fi nel the work done on the gas in Figu re 12.3b. Compute the areas or the two rectangular regions:
AJ ~
height X width 1.00 X 105J
(1.00 X 10' Pall 1.00 on')
A 2 ~ height X width ~ (2.00 X 10" Pa)(l.OO on') ~
Sum the areas (the arrows point to decreasing volume, so the work done on the gas is positive):
Area
2.00 X 10'J ~
w~
A,
+ A z ~ 3.00 X lO' J
+3.00 X 10'J
.
~
• • • • • • , • • • • • • • • • • • • • ...oiI
Remarks Notice that in both cases the paths in the PV diagrams start and end at the same points, bUllhe answers are dirferent. QUESTION 12.2 Is work done on a system during a process in 'which its volume remains constant? EXERCISE 12.2 Compute the work done on the system in Figures 12.3c and 12.3d.
Answers
-3.00 X 10 5 J. +4.00 X 10' J
12.2
THE FIRST LAW OF THERMODYNAMICS
The first law of thermodynamics is another energy conservation law that relates changes in internal energy-the energ'y associated with the position and jiggling of all the molecules of a syst.em-to energy transfers due to heat and work. The first law is universally valid, applicable to all kinds or processes, providing a connection bet.ween the microscopic and macroscopic \-vorIds. There are two ways energy can be transferred between a system and its surroundings: by doing work, ,vh ich requires a m;Kroscopic displacement of an ol~ject through the application of a force; and by heat, which occurs through random molecular collisions. Bot h mechanisms result in a r:ltal1gr'in intenwl enrrgy, flU, of the system and therefore in measurable changes in the macroscopic variables of the system, such as the pressure, temperature, and volume. This change in the internal energy can be summarized in the first law of thermodynamics: First Law of Thermodynamics
~
If" syst.em undergoes a change from an initial state to a final st.ate, where Q is the energy transferred to the system by heat and Wis the work done on the system, the change in the internal energy of the system, 6.U, is given by [12.2] The quant.ity Q is positive when energy is transferred into the system by heat and negarive when energy is transferred out or the system by heat. The qualHity 11' is positive when work is done on the system and negarive when the system does work on its environment. All quantities in the first law, Equation 12.2, Illust helVe the same energy units. Any change in the internal energy or a system-the positions and vibrations of the molecules-is due to the transfer of energy by hear or work (or both). From Equation 12.2 we also see that the internal energy or any' isolated system must remain constant, so lhat6.U = O. Even when a system isn't isolated, the change in internal energy \vill be zero if the system goes t.hrough a cyclic process in which all the thermodynamic variables-pressure, volume, temperature, and moles of gas-retlll-n La t.heir original values. It's important to remember that rhe quantities in Equation 12.2 concern a Jyslem, notlhe effect on the system's cilvirolllllenlthrough \\Iork. I[[he system is hot steam
12.2
The First Low of Thermodynamics
expanding against a piston, for example, the system work ltV is negative because the pistol1 can only expand at the expense of the internal energy of the gas. The work \r\lem' done by the hot steam 011 the p'nvimnmenl-in t.his case, moving a pisLOn which moves the train-is positive, but that's not the work Win Equation 12.2. This \\1:(lY of defining \,,'ork in the flrsr hlW Ill::lkf's it ('ollsisl('llI with the concept of work defined in Chapter 5. There, positive 'work done on a system (for example, a block) increased its mechanical energy, whereas negative work decreased its energy. In this chapter, positive work done on a syst.em (typically, a volume of gas) increases its internal energy, and negative work decreases that. intermll energy. In bQ[h the mechanical and thermal cases, the effect on the system is t.he same: positive work increases the system's energy, and negative work decreases t.he system's energy. Some textbooks identify Was the work done by the gas on its environ menlo This is an cquivalent. formulation, but it means that \oV must carry a minus sign in the First law. That convention isn't consistent with previous discussions of the energy of" a system, because 'when \Vis positive t.he system loses energy, whereas in Chapter 5 positive \Vmcans l.hatthe system gains energy. For thal reason, the old convention is not used in this book.
EXAMPLE 12.3 Goal
389
Tip 12.1 Duol Sign Conventions Many physin and cnginee"ing textbooks pl"CSem the firSl law
cxtraCLcd from the enginc.
Heating a Gas
Combine the first law of thermodynamics with work clone during a constant pressure process.
Problem An ideal gas absorbs 5.00 X lO:~J of energy while doing 2.00 X 10:1 J of work on lhe environment during a constant pressure process. (a) Compute the change in the int.ernal energy of the gas. (b) If the internal energy now drops by 4.50 X 10:~ J and 7.50 X 10 3 J is expelled f"rom the system, find the change in volume, assuming a constant pressure process all.OI X 105 Pa. Strategy Part. (a) requires substitution of the given information into the first law, Equalion 12.2. Notice, however, that the given work is done on the environment. The negative of this amount is the work done on the s)'slem, rcpn.:senting a loss of internal energy. Part (b) is a mat.ter of subst.ituting the equation I-or work at constant pressure into the first law and solving for the change in volullle. r····································
• • • • • • • • • • • • """111
Solution (a) Compute the change in internal energy of the gas. Substitute values into the Arstlaw, noting that the work donc on thc gas is ncgativc:
!1U = =
(2 +
W= 5.00 X 10".1 - 2.00 X 10 3 .1
3.00 X 10 3 J
(b) Find the change in volume, noting that aUand Qare bOth ncgalive in this case. Substitute thc equation for work done at constant prcssu re into the fi rst law: Solve ror t.he change in volume,
-4.50 X
]03
=
J=
Q + W = Q - P!1\f
-7.50 X 10' J
-
(1.01 X
]05
Pa)!1\f
av: .
~
!1U
• . .J
Remarks The change in volume is negative, so the system contracts, doing negative \vork on the environment, whereas the work Won the system is positive. QUESTION 12.3 True or False: When an ideal gas expands at constant pressure, the change in the internal energy must be positive. EXERCISE 12.3 Suppose the internal energy of an ideal gas rises by 3.00 X 10.''1 J al a constant pressure of 1.00 X 10:' Pa, whilc the systcm gai ns 4.20 X -10 3 J or energy by heal. Pi nd 1he change in volu me of the system.
Answer
1.20 X
1O-2 111 3
390
Chapter 12
The Laws of Thermodynamics
Recall that an expression lor the internal energy oran ideal gas is U = ~11RT
[l2.3a]
This expression is valid only lor a monatomic ideal gas, which means the particles of the gas consisl of single aloms. The change in the internal encl-gy. !lV, [or such a gas is given by
!::.u= ~nR!::.T
[l2.3b]
The molar specific heat at constant volume of a monatomic ideal gas, ClO ' is defined by [12.4] The change ill internal energy cran ideal gas can thell be written !1U= nC"f1T
[12.5]
For ideal gases, this expression is always valid, c\'cn when the volume isn't constant. The value of the molar specific heat, however, depends on the gas and can vary under dirfercnl conditions of"ternpcraturc and pressure. A gas with a larger mol(l.r specific heat requires more energy to realize a given temperature change. The size of the molar speciflc heat depends on the structure of the gas molecule and ho\\! many different ways it can store energy. A monatomic gas such as helium can store energy as motion in three differelll directions. A gas such as hydrogen, on the other hand, is diatomic in normal temperature ranges, and aside from moving in three directions, itcan also tumble, rotating in two differcntdirections. So hydrogen molectlles can store energy in the form or translational motion and in addition can store energy through tumbling. Further, molecules can also store energy in the vibrations of their constituent atoms. A gas composed of molecules \vith more ways to s[Qre energy ,,,'ill have a larger molar specific heat. Each different way a gas rnolecule can store energy is called a rfegree offreedom. Each degree or freedom contributes ~R to the molar specifk heat. Because an atomic ideal gas can move in three directions, it has a molar specific heat capacity C'O = ~(~R) = ~R. A diatomic gas like molecular oxygen, O~, can also tumble in two different directions. This adds 2 X ~R = It to the molar hcat specific heat, so CT' = ~R fer diatomic gases. The spinning about the long axis connecting the two atoms is generally negligible. Vibration of the atoms in (l moleculc can also contribute to the heat capacity. A full analysis of a giycll system is often complex, so in general, molar specific heats must be determined by experiment. Some representative values of C;.can 1)(.: found in Table 12.1.
12.3
THERMAL PROCESSES
Engine cycles can be complex. Fortunately, they can ohen be broken down into a series of simple processes. In this section the four most common processes will be studied and illuSlrated by their effect on an ideal gas. Each process corresponds to making one of the variables in the ideal gas law a constant or assuming onc of the three qll
12.3 Thermal Processes
TABLE 12,1
Molar Specific Heats of Various Gases Gas
Molar Specific Heat U/mol- K)"
Monatomic G.lses
20,8 20,8 20,8 20,8
He
AI' Nc Kr
12.5
83~
12,5 12,7 12,3
8,~J3
1.67 1.67
8,12
1.6'1 1.69
8A9
Diatomic Gases
1-1,
28,8
N:!
29.1
20.<1 20,8
O~
~9A
21.1
CO
29,3
CI~
~4.7
8.~3
8,33
1.41 1.40
~1.0
833 8,33
lAO lAO
25.7
8.96
1.35
~7,0
28.5
8.!jO
40.4
31.(1
g,oO
11:.:0
~5.4
27.0
8.~7
CI-I"
~5.5
27.1
8A1
1.30 1.29 1.30 1.31
Poly"tomic G.lses
CO 2 S0,
"AII V"IIIl.:S excepl 11131 Ii,]' water Wf:re l,bl;lillerl ;l\ ~HHJ K.
ronment, given by W'll\" = pav. The PV diagram of an isobaric expansion is given in Figure 12.2. As previously discussed, the magnitude of the work done on the gas isjustthe area under the path in its fYVdiagram: height times length, or PD..\!. The negative of this quantity, W= -P6ov, is the energy lost by the gas because the gas does work as it expands. This is the quantity that should be substituted into the (] l'st I:-)\.\,. The work done by the gas on its environment mllst come at the expense o[ the change in its intcrnal energy, tJ.u. Because the change in the int.ernal energy of an ideal gas is given by 60U = 11C,,607; the temperature of an expanding gas must decrease as the internal energy decreases. Expanding volume and decreasing temperature means the pressure must also decrease, in conformity with the ideal gas law, P\!= nR7: Consequcntly, the only way such a process can remain al constant pressure is if thermal energy Q is transferred into the gas by heat. Rearranging the Arst hn-l.', ,\'e obtain Q~
!J.U- W= !J.U+ I'!J.\I
Now we can substitute the expression ideal gas law to substitute Pd.\!= nRAT:
Q=
III
Equation l2.3b for AU and lise the
~l1H!J.7' + IlR!J.T = ;nR!J.J'
Another '·...ay to express this transfer by heat is [12.6]
where ell = ~H. For ideal gases, the molar heat capacity at constant pressure, Cli , is the sum of the molar heat capacity at constant volume, ell' and the gas constant Ii: [12.7]
This can be seen in the fourth column of Table 12.1, where Cp - elJ is calculated for a number 0[" different gases. The difference works oUl to be approximately R in virtually every case.
391
392
Chapter 12
EXAMPLE 12.4 Goal
The Lows of Thermodynamics
Ex anding Gas
Usc molar specific heats and the first law in
constant pressure process.
Problem ~_UO
Suppose a system of monmomic ideal gas at X 1U5 Pa and an initiallcmpcrailire of 293 K slowl)"
expands at constant pressure from a volume of 1.00 L to 2.50 L. (a) Find the work done on the cl1vironmenl. (b) Find the change in internal energy o[ the gas. (c) Usc the first la\...· of thermodynamics 1.0 obtain the thermal energy absorbed by the gas during the process. (d) Use the molar heal capacity at constant pressllre to find the thermal energy absorbed. (c) How would the answers change for a diaLOlllic ideal gas?
equation for work at constant pressure to obtain the answer to part (a). In part (b) lise the ideal gas law twice: to f1ncll.he temperature when \1= 2.00 Land [0 find the Ilumber of moles of t.he gas. These quantities can then be used to obtain the change in iJ1lerna] energy, D.u. Part (c) can then be solved by substituting into the first law, yielding Q, the answer checked in part (d) with Equation 12.6. Repeal. these steps for part (e) artcr increasing lhe ll10lar specilic heats by H because of lhe extra two degrees of freedom associated with a diawlllic gas.
Strategy This problem mainly involves substituting values inLO the appropriate equations. Substitute into the
.......................... Solution (a) Find the work done on the environlllent.
Apply the definition ofv,:ork at constant pressure:
W;", = N),v = (2.00 X ] 0' Pa)(2.50 X 10-' 111·' - 1.00 X 10-' 111')
,1;,,,,.
=
3.00 X III' J
(b) Find the change in the internal energy of the gas. First, obtain the final temperature, using the ideal gas law. noting that P, = Pi
Pi I]
.'j
~;~.
'I:
7j=
--->
V{ (950 X 10- 3 111') T ~ T - ~ (293 K) -;--_.:-:----ccco-~
{
, II,
3
111 ')
733 K (2.00 X 10·' Pa)( 1.00 X 10-3 111 3 )
Again using til<.': ideal gas law, obtain the Ilumber moles gas:
n =
Use these results and given quant.ities to calculme the change in internal cnergy, ilU:
/';U= 11(;,,/';'(= ~I1R/,;T
or
(LOO X 10
(8.31J/K·11101)(293 K)
or
= ~(8.21 X ]0 '11101)(8.3] JlIC 11101)(733 K - 29~ K)
/';U =
4.50 X 10' J
(c) C"se the firSl.law to obtain the energy lransrerred by he~lt. Solve the [-irst. law lor Q, and substitute !1U and W~
-W;",.
~
-3.00 X 10']:
(d) use the 1110lar heat capacity to obtain (1.:
I:;U = Q + W -->
Q
~ /';u -
W
Q= 4.50 X IO'J - (-3.00 X 10'./) al
~
7.50 X 10'./
constant pressure
Substitute values into Equation 12.6:
Q=
nC,,/';T = ~I1R/,;T
= ;(8.21 X W-' 11101)(8.3] JlK· 11101)(733 K - 293 K) =
7.50 X 10']
~
12.3
(e)
1-f0\\'
Thermal Processes
393
would the answers changc [or a diatomic gas?
tJ.u= nC"tJ.T= (~+ l)nRtJ.T
Obtain the new change in illternal energy, DoLl, noting that C" = ~R for a diatomic gas:
= i(8.21 X 10-' rnol)(8.31 j/ K· mol)(733 K - 293 K)
tJ.U = Obtain the new energy transferred by heat. Q:
7.50 X 10'.1
Q = "CI,tJ.T =
(i + l)nRtJ.T
=;(8.21 X 10 'mol)(8.3Ij/K·mol)(733K-293K)
Q. = 1.05 X 10'.1
......
.
~
Remarks Notice that problems involving diatomic gases are no harder than those with monatomic gases. It'sjust a malleI' of" ;')cUu.'iting the molar specific heats. QUESTION 12A True or False: During a constant pressure compressioll, the tcmperallire of an ideal gas must always decrease, and the gas mllst always exhaust thermal energy (Q < 0). EXERCISE 12A Supposc an ideal monatoIT1ic gas at an initial temperature of 475 K is compressed frolll 3.00 L to 2.00 L while its pressure remains constant at 1.00 X JOJ Pa. Find (a) the work done on the gas, (b) the change in internal energy, and (c) the energy transrerred by heat, Q.
Answers
(a) 1.00 X 10'j
(1)) -1.'iOI
(c) -250.1
Adiabatic Processes In an adi;:lbatic process, no energy enters or leavcs the system by heat. Such a system is insulated, thermally isolated from its environment. In general, hm,.. . evcr, the system isn't mechanically isolated, so it can still do work. A sufficiently rapid process m~lY be considered approximatel}' adiabatic because there isn't time [or any sign ificant transfer of energy by hcat. For adiabatic processes Q = 0, so the flrst 1m·\' becomes
tJ.U= HI
(adiabatic processes)
The work done during an adiabalic process Girl be calculated by finding the change in the internal encl-gy. Alternately. the work can bc computed from a P\I diagram. For an ideal gas undergoing an adiabatic pl-ocess, it can be shown that. p\f"t = constant
[I2.8a]
where [12.8b] is called the (uliabfltir i'11df'x of the gas. Values or the adiabatic index for several different gases are given in Table 12.J. After computing the COllstant on the righthand side of ECJll(llion 12.8a and solving for the pressure P, [he area under the curve in thcPFdiagral1l call be found by cOllnting boxes, yielding the work. If a hot gas is allowed to expand so quickly that there is no time for energy to enter or leave the systern by heal, rhe work done on the gas is negative and the il1\ernal energy decreases. This decrease occurs because kinetic energy is transferred from the gas molecules to the moving piston. Such an adiabatic expansion is of praClical importance and is nearly n:alil.cd in all internal combustion engine when a gasoline-air mixture is ignited and expands rapidly against a piston. The following example illust.rates this process.
394
Chapter 12
EXAMPLE 12.5 Goal
The Lows of Thermodynamics
Work and an Engine Cylinder
Use the firsllaw
1.0
find the work done in an adiabatic expansion.
Problem In a car engine operating at a frequency or 1.80 X l03 rev/min, the expansion of hot, high-pressure gas againsla piswn occurs in abollt]O inS. Because energy transfer by heal typically takes a time on the order of minutes or hours, iCs safe La assume little energy leaves the hot gas during the expansion. Find the work done by the gas on the piston during this adiab~llic expansion by assuming the engine cylinder contains 0.100 moles or
an ideal monatomic gas that goes from J.200 X 10 3 K to 4.00 X 102 K, typical engine telllperalllres, during the expansion. Strategy Find the change in internal energy using the given temperatures. For an adiabatic process, this equals the work done on the gas, which is the negative of the work done on the environment-in this case, the piston. .
~
Solution Start with the first law, laking Q = 0: Find 6.Ufrol11 the expression for the internal energy oran ideal monatomic gas:
Remarks The work clone on the piston comes at the expense of the internal energy of the gas. In an ideal adiabatic expansion, the loss of internal energy is completely converted into useful work. In a real engine, there are ahva}!s losses.
QUESTION 12.5 In an adiabat.ic expansion of an ideal gas, why must the change in temperature always be negative? EXERCISE 12.5 A monatomic ideal gas ,,,·it.h volume 0.200 L is rapidly' compressed, so the process can be considered adiabatic. If the gas is in itially at 1.0 I X lO,i Pa and 3.00 X 10 2 K and the final temperature is 477 K, find the work done by the gas on the envi ron ment, H~'m'.
Answer
-17.9 J
EXAMPLE 12.6
An Adiabatic EXJ>ansion
Goal Use the adiabatic pressure vs. volume relation to find a change in pressure and the work done on a gas. Problem A monatomic ideal gas at an iniLial pressure of 1.01 X 10 5 Pa expands adiabatically from an initial volume of 1.50 m3 , doubling its volume. (a) Find the nev.' pressure. (b) Sketch the PV diagram and estimate the work done on the gas. Strategy There isn't enough information to solve this problem with the ideal gas law. Instead, usc Equation 12.8 and t.he given information to find the adiabatic index and the constam C for the process. For part (b), sket.ch the P\! diagram and count boxes to estimate the area under t.he graph, which gives the work.
1.00 0.80
0.60 0.40 0.20
, 1.00
2.00
3.00
If (m )
FIGURE 12.4 (Example 12.6) The jJVdiagram of an adiabatic expansion: the graph of p~ CV.,., where Cis a COllstMlt ;lIld y ~ C,/C".
12.3
Thermal Processes
395
. ..
~.
.,
Solution (a) Find the ncw pressure.
e'i
First, calculate the adiabatic index:
~R
5
y=-=-~-
c=
Use Equation 12.8a to find the constant C:
CIt
ttR
P,II,"
= (loOI
~ lo99
The constant C is fixed for the cntire process and can be used to rind IJ'1:
3 X 10' Pa)(l.50 111')5/"
X 10" Pa·m'
C = 1',11," = P,(3.00
111")5/:;
1.99 X 10"Pa·I11" = P,(6.24 111')
P, =
3.19 X 10' Pa
(b) Estimate the work done un the gas [rom a PV diagram.
Count the boxes betwcen VI = 1.50 Ill:~ and V2 = 3.00 m:l in the graph of P= (1.99 X 10!'> Pa' m!"')V-5/:I in the IJV diagram shmvn in Figure 12.4:
Number or boxes = 17
Each box has an 'area' 0["5.00 X 1O:~.J.
W= -]7'5.00 X
....
IO"J = -8.5 X lO"J .
~
Remarks The exact answer, obtaincd with calculus, is -8.43 X 10'lJ, so our result is a very good estimate. The answer is negative becausc thc gas is expanding, doing positive work on the environment, thereby reducing its own internal energy. QUESTION 12.6 For an adiabatic expansion benveen two given volumes and an initial pressure, which gas does more work, a monalomic gas or a diatomic gas? EXERCISE 12.6 Repeal the preceding calculations for an ideal diatomic gas expanding adiabatically from an initial volume of 0 ..500 m~~ LO a final volume of 1.25 m'\ starting at a pressure of PI = 1.01 X 10 5 Pa. Use the same techniques as in the example.
Answers
P, = 2.80 X 10' Pa, W= -4 X 10 '.I
Isovolumetric Processes An isovolumetric process, sometimes called an isorhoTic process (\vhich is harder to remember), proceeds at conSlant volurne, corrcsponding to vcrtical lines in a PVdiagram. If1.he volume doesn't change, no work is done on or by the systcm, so H'= 0 and the nrstla\\' of thermodynamics reads
t1U= Q
(isovolumclric process)
This rcsult tells LIS that in an isovolumetric process, the change in internal energy of a system equals the energy transferred to the system by heat. From Equation 12.3, the energy I ransfcrred by heal in constant volume processes is given by
Q= "C"f}."!"
EXAMPLE 12.7 Goal
[12.9]
An Isovolumetric Process
Apply thc firstlmv
La
a constant-volulllc proccss.
Problem How much t.hermal encrgy must be added to 5.00 moles ofmonal.Omic ideal gas al3.00 X 10'lKand with a constant. volume of 1.50 L in ordcr 1.0 raise thc tcmperallll"C of the gas (Q ~.80 X 102 K?
396
Chapter 12
The Laws of Thermodynamics
Strategy The energy transferred by heat is equal to the change in the internal energy of the gas, which can be calculated by substitution imo Equation 12.9. ,-
Remark Constant volume processes arc the simplesllo handle and include such processes as heaLing a solid or liquid, in which the work of expansion is negligible. QUESTION 12.7 By what f-actor would the answer change if the gas were dimomic? EXERCISE 12.7 Find the change in temperature 0[22.0 mol of a monatomic idea] gas if it absorbs 9 7ElO.J at constant volume. Answer
35.6 K
Isothermal Processes During an isothermal process, the temperature of a systcm doesn't change. In an ideal gas the internal energy Udepends only on the temperature, so it foIlO\..·s that J1U = 0 because J1T = O. In t his case, the first law of thennodynamics gives Isotherm'll c:xpallsion
Ent:rgy reservoir
OIL
H:lllperalllfe 'Ii,
FIGURE 12.5 The g:IS in lhe cylinder expands isolhermally whik in contacl WiLh a reservoir at lempl:ralUre 7~.
tv = -Q (isolhcrmal process) \'Ve see that i[ the system is an idcal gas undergoing all isothermal process, the work done on the system is cqua I to the negative of the thermal energy transferred to the system. Such a process can be visualized in Figure 12.5. A cylinder filled with gas is in contact ",,lith a large energy reservoir thal can exchange energy with the gas without changing it.s temperall.lre. For a constant temperature ideal gas, l1RT
p=~
If
where the numerator on the right-hand side is constant. The PV diagram of a typical isothermal process is graphed in Figure 12.6, contrasted with an adiabatic pro~ cess. When the process is adiabatic, the pressure falls off more rapidly. Using methods of calculus, it. can be shown that the work done on the environment during an isothermal process is given by
We",. = nRT In
('1)
[12.10]
\~
The symbol "In" in Equation 12.10 is an abbreviation for the natural logarithm, discllssed in AppendiX A. The work Wdone on the gas is just the negative of \~nv· FIGURE 12.6 The J'>Vdiagram oLm i~othermal expansion, graph or P= CV-l, where C i~ a cOllslanl, compan~d 10 an adiabatic" expansion. P= C,\V-1'. r.A is a con..,lanl equal in magnitude to Cin this case blll carrying different units.
P(JOf'Pa) 1.00
0.75 0.50
0.25
1.0
2.0
3.0
4.0
\1(01 3)
12.3 Thermol Processes
EXAMPLE 12.8 Goal
397
An Isothermoll Ex onding Balloon
Find the work done during an isothermal expansion.
Problem A balloon contains 5.00 moles of a monatomic ideal gas. As energy is added to the system by heat (say. by absorption rrom the Sun), the volume increases by 25% at a constant t.emperature of 27.0 uC. Find the work \t\t;~n\l done by the gas in expanding the balloon, the thermal energy Q transferred to the gas, and the work W done on the gas. Strategy Be sure to convert temperatures to kelvins. Use the equation for isothermal \-"ork La find the '\lark done on the balloon, which is the work done on the environment. The JaILer is equal La the thermal energy Q transferred to the gas, and the negative or this quantity is the work clone on the gas. r-. . .....
Solution Substitute into Equation 12.10, finding the work done during the isothermal expansion. Note that T = no°c = 3.00 X 10' K.
nRT In
(Yr) Vi
(5.00 mol)(8.31 jlK' mol)(3.00 X 10' K)
Vo)
1.25Xln ( -
v"
W,"'. =
2.78 X 10' J
Q = \¥,.",. = 2.78 X \03 J
w= -W,,,,. =
The negative of this amount is the work done on the gas:
-2.78 X J03.1
.
~
.• .A
Remarks Notice the relationship between the work done on the gas, the work done on the environment, and the energy transferred. These relationships are true of all isothermal processes. QUESTION 12.8 True or False: In an isothermal process no thermal energy transfer takes place. EXERCISE 12.8 Suppose that. subscquent 1.0 this heating, J .50 X 10"' J of thermal cnergy is removed from the gas isothermally. Find the final volume in terms 01" the initial volume of the example, ~l' (Hint: Follow the same steps as in the example, but in reverse. Also note thaI. the initial volume in I.his exercise is ].25Vo')
Answer 0.375 Vo
General Case When a process follows none of the four given models, it's still possible to lise the first law to get information about. it. The work can be computed from the area under the curve of the PV diagram, and if the tcmperatures al the endpoints can be found, .6.Ufo11ow5 from Equation 12.5, as illustrated in the following example.
EXAMPLE 12.9
A Generol Process
Goal Find thermodynamic quantities for a process lhat doesn't fall into any of the four previously discussed categories. Problem A quantity of 4.00 moles ora monatomic ideal gas expands from an initial volume 01'0.100 m 3 to a nnal volume or 0.300 m 3 and pressure of' 2.5 X 10" Pa (Fig. 12.7a). Compute (a) the work done on the gas, (b) the change in internal energy of the gas, and (c) the thermal energy transferred to the gas.
P(ltPPa)
3.00 ~.OO
1.00
3.00
IJ I
A,
A,
~.oo
1111 I
__ l __ J__
I
1\9
1.00
:119
- V (Ill:s) 0.100 0.200 0300
(a)
FIGURE 12.7
(a) (Example
A
I I I
lA,
~__
B
: A2
'----'---,..-,'----,-'---:- V (m 3 ) 0.100 0.200 0.300 (b)
1~.9)
(b) (Exercise 12.9)
Strategy The work done on the gas is equal to the negative of t.he area under the curve in the PV diagram. Use the ideal gas law to get the temperature change and, subsequently, the change in internal energy. Finally, the firsllaw gives the thermal energy transferred by heaL
398
Chapter 12
The Lows of Thermodynamics .
~
..................................
..,
,
Solution (a) Find the work clone on the gas by computing the area uncleI' the curve in Figure 12.7a. Find £1 1• the area ofLhe triangle:
A, = ~bh, ~ )(0.200 m')( l.50 X 10' Pal = 1.50 X 10'.1
Find A 2 , the area of the rectangle:
A, = bh, = (0.200 m')(1.00 X 10" Pal = 2.00 X 10'.1
Sum the two areas (the gas is expanding. so the work done un the gas is negative and a minlls sign must be supplied):
W= -(A,
+ A,)
~
-3.50 X 10".1
(b) Finclthc.: change in the internal energy during the process. p V (1.00 X 10" Pa)(O.lOO m') l' - ~ - -'------cc----'-'-'------'---'- = 301 K ,,- nR - (4.00 mol)(8.31 .1/K'moJ)
Compute the temperature at points A and B with the ideal gas law:
.. PB \/IJ(,2::c..:::.5.:::.0_X---,-I.:::.0'~P.:::.'.:::.')",(.:::.0.:::3.:::.0.:::.0.:.:m'-..!..') III = - - . = = 2.26 X 10" K
I/N
(4.00 mol)(8.31.J/K· mol)
t1U= ~71Rt1T
Compute the change in il1\.crnal energy:
= ~(4.00 mol)(8.3I.1/K·mol)(2.26
X 10'K - 301 K)
t1U= 9.77 X 10".1 (c) Compute
Qwith
Q = t1U - 1'1'= 9.77 X 10 4 .1 - (-3.50 X lO"'J)
the first law:
= ~
,."
,
1.33X]0'.1
,
,
~
Remarks As long as it's possible to compute the \\lork, cycles involving these more exotic processes can be completely analyzed. Usually, howcver, it's necessary to use calculus.
QUESTION 12.9 For a curve with lowcr pressures but the same endpoints as in Figure 12.7a, would the thermal energy transferred be (a) smaller than, (b) eqllal to, or (c) greater than the thermal energy transfer or the straight-line path? EXERCISE 12.9 Figure 12.7b represents a process involving 3.00 moles or a monatomic ideal gas expanding from 0.100 m:~ to 0.200 111 3 . Find the 'work done on the system, the change in the internal energy of t.he system, and the thermal energy transferred in the process. Answers
W= -2.00 X 10".1, t1U~ -1.50 X 10 4 .1, Q = 5.00 X 10".1
p
Given all the different processes and formulas, it's easy to become confused when approaching one of these ideal gas problems, although most of the time only substitution inl.O the correct formula is required. The essential facts and formulas are compiled in Table] 2.2, both lor easy reference and also to display the similarities and differences between the processes. L FIGURE 12.8 (Quick Quiz [2.2) IdellLify lite nalurcOrpalhsA. B, C, and n.
v QUICK QUIZ 12.2 fdcnLify the paths A, B, C, and D in Figure 12.8 as isobaric, isothermal, isovolumetric, or adiabatic. For path B, (2 = o.
12.4
Heat Engines and the Secand Law of Thermodynamics
TABLE 12.2 The First Law and Thermodynamic Processes (Ideal Gases) Process D-.U Q W Isobaric
Adiabatic Isovolumctric Isothermal General
12.4
nC"AT nC1,tJ.T
I1C/J AT
-P~\f
0
ilU
net/liT
f>U
0
0
-w
ne" t>T
tJ.U- W
- nR'fln
399
11m reser....oir.at Ii,
\t; CJ)
r_~
Engillt:
(PV Area)
HEAT ENGINES AND THE SECOND LAW OF THERMODYNAMICS
A heat engine takes in energy by heal. and partially convcrts it to other forms, such (IS electrical and mechanical energy. In a typical process for producing electricity in a power plant, for instancc, coal or some other fuel is burned, and the resulting internal energy is used to convert V'later to steam. The steam is thcn directed althe blades of a turbine, selling it rOlating. Finally, the mechanical energy associated with Ihis rotation is used to drive an electric generator. In another heat engincthe internal combustion engine in an automobile-encrgy enters the engine as flle) is injected inLO the cylinder and combusted, and a fraction of this energy is converted to rnechanical energy. In general, a heat engine carries somc working substance through a cyclic process! during which (1) energy is transferred by heat [rom a source at a high temperature, (2) work is done by the engine, and (3) energy is expelled hy the engine by heat to a source atlo".·cr temperature. As an example, considcr the operation of a sleam engine in which the working substance is water. The water in the engine is carried through a cycle in which it first evaporatcs into steam in a boiler and then expands against a piston. After the steam is condensed with cooling water, it rt;turns to the boiler, and the process is repe~l[ed. It's useful to draw a heat engine schcmatically, as in Active Figure 12.9. The engine absorbs energy Qh from the hot reservoir, does \,,'ork ltV<.:lIg' then gives up energy (L to the cold reservoir. (Note that negotivp work is done on lhe engine, so L~IUL -,,- - -H(l1g') Because the 'working substance goes through a cycle, always returning to its initial thermodynamic state, its initial and final internal energies are equal, so !::J.U= O. From the first law ofthcrmodynamics, therefore,
Cold reservoir
aL
Tr
ACTIVE FIGURE 12.9 A schematic repn:sentalioll ora heat enginc. The ellginc recei\·cs en erg}' Qh from Lhe hOI resenalr. expels energy Qr to the cold resen'oir, and docs work W
f- Cyclic process
I1U= 0 = Q + W
The last equation shows that the work Wel1g done by a heat engine equals the net energy absorbed by the engine. As wc can sec from Active Figure 12.9, Qnel = 1(1,1 - IQJ Therefore,
[12.11] Ordinarily, a transfer of thermal energy Q can be cithcr positive or negative, so the use of absolute value signs makes the signs of Qh and Qrcxplicil. If the working substance is a gas, then the work done by the engine for a cyclic process is the area enclosed by the curve representing the process on a PV diagram. This area is shown for an al-bitrary cyclic proccss in Figure 12.10. The thermal efficiency e of a heat engine is defined as the work done by the engine, ~ng' divided by the energy absorbed during one cycle:
112,,1
IQ"I - IQ,I IQ"I
I
IQ,I 10.1
[12.12)
ISIr·ictly spc.l1...ing. the imcl"I1al comhuslioll eng:inc i, 110t ,I heal engine aCClIrding to till' dncriptiun 01 lhe qdic process. bccallse lllc air-fuel mixl111"C Illlrkrgra:s onl}' onl· {"~dl· and i, I hell expl'lled IllrOlIg-11 tll(' cxll:luSI sr.'>ltlll.
p
' - - - - - - - - - - If
FIGURE 12.10 The PVdiagram lor an arhitrary cyclic process. The arca enclosed by the curve c.:quals the 11Ct work done.
Chapter 12
400
The Laws of Thermodynamics \~'e can think of thermal efficienc)' as the ralio of the benefit received (work) to the cost incurred (energy lransfer at the higher temperature). Equation 12.12 shows that a heal engine has 100% efficienc), (p = I) onl}' if Q,- = 0, meaning no energy is expelled to the cold reservoir. In other words, a heat engine with perfect cflicicncy would have lU lise all Lile illjJlIl clleq;y rUl duing IllcclJ
EXAMPLE 12.10 Goal
The Efficiency of on Engine
Apply the efficienc)' formula
10
a heaL engine.
Problem During one cycle, all engine extracts 2.00 X 10 3 .1 of energy from a hot reservoir and transfers 1.50 X 10 3 J to a cold reservoir. (a) Find the thermal efficienC}' of the engine. (b) J-1O\v rnuch work does this engine do in one cycle? (c) How much power docs the engine generate ifitgoes through four cycles in 2.30 s? Strategy Apply Equation -12.J21.O obtain the thermal cfficiency, then use the first law, adapted to cngines (Eq. 12.11), LO find thc work done in one cycle. To obtain the powcr generated, di\'idc the \'wrk clone in four cycles by the time it takes to run those cycles.
(b) How milch work docs (his engine do in one cycle? Apply the first law in l.he form of Equation -12.11 to find the work done b}' the engine:
H'~"g
= IQ"I=
5.00
IQ~
= 2.00
X
la' .I -
1.50 X
to 3.1
X 1O~.1
(c) Find the power output of the enginc. IV
Multiply the answer of pan (b) by [our and divide by timc:
.
.
~
'!P=- = !1t
4.00 X (5.00 X I O~J) 2.50 s
'I
= 800X to-w .
~
Remark Problems like this usually rcduce to solving t\\'o equal ions and two unknowns, as here, where tilt two equations are the c1liciellcy equation and the first law and the unknowns arc the efficiency and the work done by the engllle.
QUESTION 12.10 Can the cfhciency or an enginc always be improved by increasing the thermal energy PLit in[Q the system during a cycle? Explain.
EXERCISE 12.10 The energy absorbed by an engine is three times as great as the work it performs. (a) What is its thermal efficiency? (b) 'o\/hat fraction of the energy absorbed is expelled LO the cold reservoir? (c) ,"Vhat is the power output of the engine if the energy input is 1 650 J each cycle and it goes through two cycles every 3 seconds?
Answer
(a) ';"
(b) ';"
EXAMPLE 12.11 Goal
(c) 367 W
Anal zing an Engine Cycle
Combine several concepts to analyze an engine cycle.
Problem A heal engine contains an ideal monatomic gas confined to a cylinder by a movable piston. The gas starts at A, where T= 3.00 X 102 K. (Sec Fig. 12.11a.) The process 8-7 Cis an isothermal expansion. (a) Find the number 11 of moles of gas and the temperature at B. (b) Find !J.U, Q, and W for the isovolulTIctric process A -7 B. (c) Repeal for the isothermal process 13-7 C. (d) RepC
12.4
nal cnergy for the complete cyclc. (f) Find the thermal cncrgy Qh translCrrcd into the systcm, the thermal cnergy rejected, (2(, the thcrmal efficiency, and net work on the environment perfonned by the engine.
Strategy I n pan (a) 11, T, and V can be fOllnd from the ideal gas law, which connects t.he equilibrium values of P, \~ and T Once the tcmpenllure Tis knmvn at the points A, 11, and C, t.he change in internal energy, I:1U, can he cOlllpllted frol11 the formula in Table 12.2 for each process. Qand \\Tcan be similarly computed, or deduced from the first law, using the techniques applied in the single-process examples.
Computel:1UtW ' noting that C" = frH = 12.!)J/mol· K:
ill/"B
~ I1C"ilT~ (0.20~
11101)(12.5.J1mol· K)
X (9.00 X 10' K - 3.00 X 10' K) t!.L~1IJ =
L\.V = 0 for isovolumetl'ic processes, so
110
l. 52 X 10' J
work is done:
\Ve can find QMlfrol11 the t-irst la\-\':
Q'/l = ilUIIJ = 152 X 103 .1
(c) Find I:1Uw., Quo and W/Jc for the isothermal process 13 --> C. This process is isothermal, so the temperaLUrc doesn'l chang'c, and thc change in internal energy is zero:
Compute the work clonc on the system, lIsing the negative ofEquatioll ]2.10:
tJ.U/lc= "lIC,.I:1T=
0
e WI/(; = -nRT In (V ) VIJ = -(0.20~ mol)(8.3l)/mol· K)(9.00 X 10' K)
1.50 X 10-'
X In ( !).OO X 10
W"e =
:I
Ill') m3
-l.67 X 10'.1
Computc (6w [rom the f-irstlaw: (d) Find t1UCI\' Qc.-\" and C --> A.
WCI\
for the isobaric process
Compute the work on [he SyStCI11, with pressure constant:
I'(L)
.
=
(b) Find 1:1u,.\II' cess A ---7 B.
"I~(
1.00
I'(L)
1.-,
10
:")
IJ
2.00
.
Solution (a) Find nand T fj \ ...·ith the ideal gas law:
401
HrCA = -P.6\!= -(1.01 X l05 Pa)(5.00 X ]0-3 m :\
- l.50 X 10-2 m") We" =
1.01 X 10'.1
.~
402
Chapter 12
The Lows of Thermodynamics
Find the change in internal energy,
aUe/!:
!:iU,,, = ~TlR!:iT = ~(0.203 mol)(S.31 ]lK·lI1ol)
X (3.00 X 10' K - 9.00 X 10' K) !:iUcA =
Compute the thermal energy,
QGI'
[rom the Hrsllaw:
QGl
~ !:iU", - WCA = -1.52 X 10 3 .1 -
= (e) Find the net change in internal energy, 6.UI1l "I' for the cycle:
-1.52 X 10'J l.01 X 10'.1
-253 X lO'J
= l.52 X L03j
+0
- 1.52 X 10'.1 = 0
(f) Find the energ)' input, Q,.; the energy rejected, Qr; the thermal efficiency; and the net work performed by the engine: Sum all the positive contributions to find Qh:
Q" = Q,1f< + Que ~ 1.52
X
10'.1
+
1.67 X 10'.1
3.19 X 103 J Sum any negative cOlltributions (in this case, there is only one):
Find the eng"inc efficiency ancllhe net work done by the engine:
Q, =
-2.53 X 10'.1
253 X 10'J
1'=1_IQ"=l IQ"I WCIIg- = - (\t\~1B
+
H/nc
3.19
+
lO'J
0.207
WCA )
= -(0 - 1.67 X 10 3 .1 ~
X
+ 1.01
X
10'J)
6.60 X JO'J
~
~
Remarks Cyclic problems are rather lengthy, but the individual steps are often shon substitutions. Notice that the change in internal energy for the cycle is zero and that the net work done on the environmenL is identical to the net [hennal energy transferred, both as they should be. QUESTION 12.11 If BC\'~'ere <.I slraight~line path, would the work done by the cycle be affected? How? EXERCISE 12.11 4.05 X 10-2 mol of monatomic ideal gas goes through the process shown in Figure 12.l1b. The temperature at point A is 3.00 X 10' K and is 6.00 X 10' K during the isothermal process B-> C. (a) Find Q, !:iU, and Wforlhe constant volume process A ~ B. (b) Do the samc for the isothermal proccss B ---? C. (c) Repeat, for the constanL pressure process C -> A. (d) Find Q", Q" and the efficiency. (e) Find We",. Answers (a) Qcli1 = !:iUAB = 151J, W;lIl = 0 (b) !:iUilC = 0, QBC = -Wile = 1.40 X 10'/ -151.1, WC,\~ 101.1 (d) Q,,~29IJ,Q,= -252.1,e=0.13'1 (e) We", = 39.1
(c) QC,'
= -252 J,
!:iUC..' =
Refrigerators and Heat Pumps Heat engines can operate in reverse. In this case, energy is ir~jecled into the engine, modeled as \",ork IV in Acti\'e Figure 12.12, resulting in energy being extracted from the cold reservoir and transferred to the hot reservoir. The system now operates as a heat pump, a comrnon example being a refrigerator (Fig. ]2.13). Energy Q{ is extracted from the interior of the refrigerator and del ivered as energy Q" to the warmer air in the kitchen. The work is done in the compressor unit of the refrigerator, compressing a refrigerant such as freon, causing its temperature to increase.
12.4
Heat Engines and the Second Low of Thermodynamics
A household air conditioner is another example of a heat pump. Some homes are both heated and cooled by heat pUlllpS. In \'linter, the heat pump extracts energy Qr from the cool outside air and delivcrs energy QI! to the warmer air inside. ln summcr, encrgy Qr is removed from tile cool inside air, while energy Q/i is ejected (Q the \",arm air olltside. For a refrigerator or an air conditioner-a heat pump operaling in cooling rnode-work Wis what you pay for, in terms 01" electrical energy running the compressor, whereas Qr is the desired benefit. The most efficient refrigerator or air conditioner is onc that removes the greatest amount or energy from the cold rescrvoir in exchange for the least amount of work.
403
Ikal pump
The coefficient of performance (COP) for a refrigerator or an air cOllclitiOiler is the magniLUde of thc energy extractcd from thc cold reservoir, !Q,.I, divided by the work Wpcrfonned by the device:
.
COP (cooling mode
) =IQ,I -W
Cold reservoir at T£
[12.13]
51 unit: dimensionless The larger this ratio, the bettcr the performance, because morc energy is being removcd for a given amount of '.. .' ork. A good refrigerator or air conditioner \vill have a COP of 5 or 6. A heat pump operating in healing mode warms the inside ora house in wintcr by extracting energy from the colder outdoor air. This statemelll may seem paradoxical, but. recall that. this process is equivalent to a refrigerator rcmoving energy from its interior and ejecting it int.o the kitchen.
ACTIVE FIGURE 12.12 Sc!lemalic diagram ora heal pump, which takes in energy Q; > () from
The coefficient of performance of a heal pump operating in thc healing Illode is the magnitude of the energy rejected to the hot reservoir, IQ,J, divided by thc work Wdone by the pump: COP(heaLing mode) ~
[12.14]
51 unit: dimensionless In effect, Lhe COP of a heaL PUIllP in the heating mode is the ratio of what you gain (energy delivered to the interior of your hUllle) La what you give (work input). Typical valucs for this COP arc greater than 1, because IQ,J is usually greater than l"~ In a grounchvater heat pump, energy is extracted in the ,vinter from \v.Hcr deep in the ground rather than from the outside air, while energy is delivered to that water in the summer. This strategy incrcases the year·round efliciency of the heating and cooling unit because the groundwatcr is at a higher temperat.ure t.han the air in winter and at a cooler temperalUre than I hc air in slimmer.
FIGURE 12.13
The
COi!H>l1
the hack
ora refrigerator transfer energy h}
heal to the air.
EXAMPLE 12.12 Goal
Cooling the Leftovers
Apply the coefficient of performance or a rcfrigeraLOL
Problem A 2.00-L cont.ainer of leftovcr soup tlt a tcmperature of 323 K is placed in a refrigerator. Assume Lhe spe· cific heal of the soup is the same as that of water and the density is l.25 X 10 3 kg/m 3 . The refrigerator cools the soup to 283 K. (a) If the COP of t.he n..:frigerator is 5.00, find the energy needed, in lhe form of work, to cool the soup. (b) Jf the compressor has a power raling ofO.250 hp, [or what minimum length of time must it. operate to cool the SOllP to 283 K? (The minimum time aSSlIllles the SOllP cools at the same raLe that the heat pump ejects thermal energy from t.he refrigerator.) Strategy
The solution to this problem requires three steps. First, find the total mass
III
of the soup. Second, using
Q = m.c6.7; where Q = Qr' find the energy transfer required to cool the soup. Third, substitute Qrand the COP into Equation 12.13, solving tor \tv. Divide the work
b)1
tlle power to gel an estimate o[ I he time required to cool the soup.
Solution (a) Find the work needed to cool the soup.
= p\1=
(1.25 X 10 3 kg/m 3 )(2.00 X 10-3 m')
Calculate the mass of the soup:
m
Find the energy transfer required to cool1.he soup:
Qc= Q= mctJ.T
= 2.50 kg
= (2.50 kg)(4190.1/kg' K)(283 K - 323 K)
= -4.19 SubSliUlte Q[and the COP into Equation ]2.13:
COP
~
IQcI
X 10 5 .1
--
11'
=
'U9Xl0'.1
11'
= 5.00
11'= 8.38 X 10".1 (b) Find the lime needed to cool the soup. Conven horsepm.. .er to watts:
(ij>
=
Diyide the work by the power to find the elapsed time:
tJ.1
= ""
(0.250 hp)(746 WI] hpj
11'
"
~
8.38 X
10".1
187 W
=
= 187 W
448 s
~
~
Remarks This example illustrates how cooling different substances requires differing amounts ofv·,'ork due to differences in specific hems. The problem doesn't take into account the insulating propenies ohhe soup container and of the soup it.self, which retanl the cooling process.
QUESTION 12.12 lIthe refrigerator door is len open, does the kitchen become cooler? ""'hy or ",tby not? EXERCISE 12.12 (a) How much work must a heat pump \vith a COP of 2.50 do to extract J .00 MJ o[ thermal energy [rom the outdoors (the cold reservoir)? (b) If the unit operates at 0.500 hp, how long will the process take? (Be sure lO use the correct COP!)
Answers
(a) 6.67 X 10 5 .1
(b) 1.79 X 10' s
The Second Law of Thermodynamics There are limits to the efficiency of heat engines. The ideal engine would convert all input energy into useful work, but it turns out that such an engine is impossible to construct. The Kelvin~Planck formulaLion of the second law of thermodynam~ ics can be stated as follows: No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely for the performance of an equal amount of work.
This form of the second lew.' means that the efficiency e = H~llg/IQhl of engines must always be less than 1. Some energy Qr. must always be lost to lhe environment. In other \Vords, il'S theoretically impossible to construct a heat engine with an efficiency of 100%. To summarize, the first law says we can't get a greater amount of energy out of a cyclic process than we put in, and the second law says we can't break even. No matter \,,'hat engine is used, SOllle energy must be transferred by heat. to the cold reservoir. [n Equation 12.11, the second law simply means IQ,J is al'ways greater than zero.
12.4
Heat Engines and the Second Law of Thermodynamics
405
There is another equivalent statement of the second law: If two systems are in thermal contact, net thermal energy transfers ously by heat from the hotter system LO the colder system.
spontane~
Here, spontaneous means the energy transfer occurs naturally, with no work being done. Thermal energy naturally transfers from hotter systems to colder systems. ''''ark must be done to transfer thermal energy from a colder system to a hotter system, hmyever. An example is the refrigerator, which transfers thermal energy from inside the refrigerator to the warmer kitchen.
Reversible and Irreversible Processes No engine can operate with 100% efficiency, but different designs yield different efficiencies, and it turns out that one design in particular delivers the maximum possible efficiency. This design is the Carnot cycle, discussed in the next subsection. Understanding it requires the concepts of reversible and irreversible processes. In a reversible process, every state along the path is an equilibrium state, so the system can relUrn to its initial conditions by going along the same path in the reverse direction. A process that doesn't satisfy this requirement is irreversible. Most natural processes are known to be irreversible; the reversible process is an idealization. Although real processes are always irreversible, some are aluwsl reversible. If a real process occurs so slowly that the system is virtually always in equilibrium, the process can be considered reversible. Imagine compressing a gas very slowly by dropping grains of sand onto a frictionless piston, as in Figure 12.14. The temperature can be kept constant by plaCing the gas in thermal contact with an energy reservoir. The pressure, volume, and temperature of the gas are well defined during this isothermal compression. Each added grain of sand represents a change to a new equilibrium state. The process can be reversed by slowly removing grains of sand from the piston.
LORD KELVIN British Physicist and Mathematician (1824-1907)
Born Wi/fiam Thomson in Belfast, Kelvin was the first to propose the use of on absolute scale of temperature. His study of Carnot's theory led to the idea that energy cannot pass spontaneously from a colder object fa a hot ter object; this principle is known as the second law of thermodynamics. w
The Carnot Engine In 1824-, in an effort to understand the eHiciency of real engines, a French engineer named Sadi Carnot (1796-1832) described a theoretical engine now called a Carnol. engine that is or great importance from both a practical and a theoretical viewpoint. He showed that a heat engine operating in an ideal, reversible cyclenm\' called a Carnat cycle-between nvo energy reservoirs is the most efficiem engine possible. Such an engine establishes an upper limit on the efficiencies of all real engines. Carnat's theorem can be stated as follows: No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs. [n a Carnot cycle, an ideal gas is contained in a cylinder with a movable piston at one end. The temperaLUre of the gas varies between ~ and '/~" The cylinder walls and the piston are thermally nonconducting. Active Figure 12.J5 (page 406) shows the four stages of the Carnot cycle, and Active Figure 12.16 (page 407) is the PV diagram for the cycle. The cycle consists of two adiabatic and two isolhermal processes, all reversible: 1. The process A ~ B is an isothermal expansion at temperature 7}. in which the gas is placed in thermal contact with a hot reservoir (a large oven, for example) at temperat.ure 7~1 (Active Fig. 12.15a). During the process, the gas absorbs energy Q" from t.he reservoir and does work 1~1ll in raising the piston. 2. In the process B ~ C, the base of the cylinder is replaced by a thermally nonconduct.ing \.. . all and the gas expands adiabatically, so no energy ent.ers or leaves the system by heat (Active Fig. 12.15b). During the process, the temperature falls rrom 7~1 to 7~ and the gas does work \.vnf: in raising the piston.
Encq,')'
re~er\'oil'
FIGURE 12.14 i\ gas in thermal COlllact wilh an cncrg>' resel"voI" is compressed slow'>' by grains of sand dropped OntO a piston. The COIllprcs!'>ioll is isolhermal and reversible.
406
Chapter 12
The Laws of Thermodynamics
3. In the process C----t D, the gas is placed in thermal contact \",'ith a cold reservoir attemperat.ure 7; (Active Fig. 12.15c) and is compressed isothermally at temperaLUrc 'I;. During this time, the gas expels energy Qr to the reservoir and the work done on the gas is Weu. 4. In the final p,'ocess, D ----t A, the base of the cylinder is again replaccd by a thermally nonconducting wall (Active Fig. 12.15d) and the gas is compressed adiabatically. The temperature of the gas increases to 7~" and the work done on the gas is l¥o:t.
For a Car not engine, the following relationship bet,,·:een the thermal energy transl-ers and the absolute tcmperatures can be derived:
IQ,I IQ"I
7~.
[12.15]
Til
Substituting this expression inLO Equation 12.12, we find that the thermal efllciency of a Carnot engine is
(arnot is considered to be the founder of the science of thermodynamics. Some of his notes found after his death indicote that he was the first to recognize the relationship between work and heat.
[12.16]
er; = .1
SAD I CARNOT French Engineer (1796-1832)
where Tmust be in kelvins. From this result, we see that all Carnot engines operating reversibly between the same two temperatures have the same efficiency. Equation 12.16 can be applied to any working substance operating in a Carnot cycle bctween two encrgy reservoirs. According to that equation, the eIliciency is zero if T, = 7~1· The efflciency increases as 7; is lowered and as 7~1 is increased. The efficiency can be one (100%), however, only if 7; = 0 K. According LO the third law of thermodynamics, it's impossible to lower the temperature of a system to absolute zero in a finite number of stcps, so such reservoirs are not available and the
Isothermal expansion
J)-)A
B-+C
Adiab:uic compression
.-\di,\balic expansion Encrgy rescryoir at T" (,)
Cycle
Q=O ACTIVE FIGURE 12.15 The Carnot q·ck. In process A ~ B, the gas expal1ds iSOl herlllall~' Idlile
Q=O (b)
(d)
wilh a rescrvoiral 7~.111 proct:ss n----'; C, the gas expands ad i-
in
cont~H.:l
abaticalJ) (Q = 0), In process C ----'; D, the gas is compressed isothermally whilc in contact wil h a resen'oir at T < '/h' In proccssJ)----';/I.ll1ega... is cornpre.'~cd
sions, and thl: downward arrows inrlicate till; adclilioll ofs
C-)f) Isothermal c(llnprcssion
Energy reservoir at (e)
1~
J2.4
ACTIVE FIGURE 12.16 The I~\I diagTam for lhe CarnOl cycle. The llel work dOlle, equals Lhe Ilel energy received by heat ill olle eyc1c.IQhl -
I'
A
L-
Heat Engines and the Second Law of Thermodynamics
407
~\~nl('
IQJ
'I; V
__
~SL
maximum eHkiency is alv,,'ays less than J. In most practical cases, the cold reservoir is near room temperalUre, about 300 K. 50 increasing the efficiency requires raising the temperature of the hot reservoir. All real engines operate irreversibly, due to friction and the brevity of their cycles, and are therefore less efficient than the Carnot engine.
Tip 12.2 Don't Shop for Cornot Engine
The Carnal engine is ollly an idealization. If a Carnol engine were de"e1oped in an effon to maximize erfickncv. il would h,,\'c zero power OU'I.PUL because for all of the processes LO be reversible. lhe engine would have 10 rUIl
QUICK QUIZ 12.3 Three engines operate benveen reservoirs separated in tcmperature by 300 K. The reservoir temperatures are as follows: Engine A: T h Engine E:
'/~
0
infinilely slol\'ly.
= I 000 K, 7; = 700 K = 800 K, T, = 500 K
Engine Co T h = 600 K, 7; = 300 K Rank the engines in order of their theoretically possible efficiency, from highest to 10wesL (a) A, B, C (b) E. C, A (e) C, B, A (d) C, A, B
EXAMPLE 12.13 The Steam Engine Goal
Apply the equations of an ideal (Carnot) engine.
Problem A steam engine has a boiler that operates at 5.00 X 10 2 K. The energy from the boiler changes water to steam, which drives the piston. The temperature of the exhaust is that of the outside air, 3.00 X 10 2 K. (a) What is the engi nets efficiency i r it's an idea I engi ne? (b) I f the 3.50 X 1O:~ J of energy is supplied from the boi leI', fi nd the energy transferred to the cold reservoir and the work done by the engine on its environment. Strategy This problem requires substitution into Equations ]2.13 and 12.16, both applicable to a Carnol. engine. The first equation relates the ratio QrlQ" to I he ratio 7~/Th' and I he second gives the Carnot engine efficiency. .................... , _ -., -
~
Solution (a) Find the engine's efficiency, assuming it's ideal. Substitute into Equation 12.16, the equation for the efficicncy of a Carnot engine: (b) Find the energy transferred to the cold reservoir and the work done on the environment ir3.50 X 103 is delivered to the engine during one cycle.
Pc;
3.00 X 10' K 5.00 X J02 II.
0.400, or 40%
J
The nuio of energies equals the rat.io of temperal1lres:
Substitute, finding the energy transferred to the colel reservoir:
1-
= I
IQ,I/~ 10,1 'Ii, IQ,I
(3.50
1Q,j X
=
IQ;,I
.r'T
, )(3.00 X 10' LO·J
I,
K)
5.00 X 10' K
=
2.10
X
10'J
408
The Lows of Thermodynamics
Chapter 12
Use Equation 12.IJ
10
find the work done by the engine:
IV","
= IQ"I - IQ,I =
3.50 X 10"./ - 210 X 10'./
1.40 X 103J
... .
....
Remarks This problem differs from the earlier examples on work and efficiency because we lIsed the special Carnot relationships, Equations ]2.15 and 12.16. Remember thaL these equations can only be used when the cycle is identified as ideal or a Carnal. QUESTION 12.13 True or False: A non ideal engine operating between the same temperature extremes as a Carnal the same input thermal energy will perform the same amount of\vork as the Carnol enginc.
~ngine and
having
EXERCISE 1213 The highest theoretical eHiciency of a gasoline engine based on the Carnal cycle is 0.300, or 30.0%. (a) If this engine expels its gases into the atlllosphere, which has a temperature or 3.00 x IO~ K, what is the temperature in the C)'lindel' immediately after combustion? (b) If the heat engine absorbs 837.J of energy [rom the hOl reservoir during each cycle, how much work can it perform in each cycle? Answers
(a) 429 K
(b) 251.1
12.5
ENTROPY
Temperatll rc and intern~ll energy, associated with the zeroth and first laws of thermodynamics, respectively, are both state variablcs, mcaning they can bc llsed to describe the thermodynamic state oCa system. A statc variable called the entropy S is related [Q the second law of thermodynamics. vVc define entropy on a macroscopic scale as German physicist Rudolf Clausius (1822-1888) first expressed it III 1865: Let Q,. be the energy absorbed 01' expelled during a rcversible, constant temperature process between two equilibriurn States. Then the change in cnt.ropy during any constant temperature process connccting the (1."0 equilibrium statcs is defined as ~s
-
Q, T
[12.17]
SI unit: jOllies/kelvin (J/K)
"
8
" >.
~
.=;
~ z
RUDOLF CLAUSIUS German Physicist (J822-18BB)
Born with the name Rudolf Gottlieb, he adopted the classical name of Clausius, which was a popular thing to do in his time. "I propose. . ta call 5 the entropy of a body, after the Greek word 'transformation.' I hove designedly coined the word 'entropy' to be similar to energy, for these two quantities are so analogous in their physical significance, that an analogy of denominations seems to be helpful."
"
A similar formula holds ",.'hen the tcmperature isn't constant, but its derivation entails calculus and won't be considered hcre. Calculating the change in entropy, oilS, during a transition bdween two equilibrium states requircs finding a reversible path that. connects the states. The entropy change calculated on t.hat revcrsible path is taken La be!1S for the ac/ual path. This approach is nccessary because quantities such as the temperature of a system can be defincd only for S)lstems in cquilibrium, and a reversible path consists of a sequence of equilibrium srates. The subscript ron the term Qremphasizes thatthc path chosen [or the calculation must be reversible. The change in entropy 6.S, like changes in internal encrg)' ~U and changcs in potcnti;:l1 energy, depends only on the cndpoims, and not on the path connecting them. The concept of entropy gaincd wide acceptancc in pan because it. provided another variable [() describe the stat.e ofa system, along with pressure, volumc, and tcmperature. Its signiflcancc \vas enhanced ,vhen it was {ound that tbe entropy of the Universe increases in all natural processes. This is yet another way of stating the second law of thermodynamics. Although the entropy of the Universe increases in all nalliral proccsses, the entropy o[ a s)'stf'>1n can decreasc. For cxample, if system A trans[Crs encrgy Q to sysl.em B by hCaL, the entropy of system A decreases. This transfer, howcver, can
12.5 only occur if the temperature of system B is less than the temperature of system A. Because temperature appears in the denominator in the definition of entropy, system B's increase in entropy will be greater than syslcm A's decrease, so taken togcl her, the entropy of the Un iverse increases. For cell wries, individuals have attempted to build perpelll::!l mOlion machines 1hal operate continuously wil.hOllt any input of energy or i nCl"case in entropy. The laws of thermodynamics preclude the i nVClllioll of any such macllincs. The concept of entropy is saLisfying because it enables liS 10 present the second
Ent,opy
TiP 12.3 Entropy
409
'* Energy
r
Don't confuse CIl(~l'g'}' a lid cntrop}. Allhough the 11,1Ine~ 'iolllH.1 similar the conccpt'i are different.
law of lhennodynamics in the [onn of a mal hematica} stat.cmcn1. In the next section we find that entropy call also be interpreted in terms of probabilities, <.l relationship that has profound implications. QUICK QUIZ 12.4 Which of the followin!; is trlle [or the entropy change a syst.em that undergoes a reversible, adiahatic process? (a) L\.S < 0
o[
(b) !is = 0 (c) !is> 0
EXAMPLE 12.14 Goal
Mellin a Piece of Le:::;od
_
Calculate I he change in entropy due to a phase change.
Problem (a) Find the change in entropy of" 3.00 X IO~ g of lead when it melts at 327°C. Lead has a latent heat of fusion of 2.4.5 X 104.1/kg. (b) Supposc the same amount of" enerp;y is used to melt part of a piece of silver, which is alrcady at its melting point of"961°C. Find the change in the entropy orthe silver. Strategy sca\e.
This problem can be solved by substitution into E.quation 12.17. Be sure
10
lise the Kelvin tempenllure
.
,.. .....
~
Solution (a) Find the entropy change of the lead. Find the energy necessary to melt the lead:
Q=
Convcrt the tcmperature in degrees Celsius to kelvins:
T= 7;: + 273
Substitute [he quantities found into the entropy c()lI
!is =
IIILr = (0.300 kg)(2.'15 X W.J/kg) = 7.35 X 1O'j
Q
T=
= 327 + 273 = 6.00
7.35 X ]0") (;.00 X 10tK =
X Jot I<
]2.3j/1<
(b) Find the entropy change or the silver. The added energy is the same as in pan (a), by supposition. Substitute into the entropy equation, after first converting the melting point or silver to kelvins:
T
~
7;: + 273
!is = Q = T
..........
~
961 + 273 = 1.234 X JO" K
7.35 X 1O'j 1.23'1 X 10'K =
5.96.J!K
. ..... ~
Remarks This example sllm..'s that adding a given amount of energy to a system increases its disorder, but. adding lhe same amount of energy to another system at higher temperature results in a smaller increase in disorder. This is because the change in entropy is inversely proportional 10 the tcmpcrawre. QUESTION 12.14 Jr the same amount of cnergy were used to melt ice at O°C to water at O°C, rank the entropy changes for ice, silver, and lead, from smallest to largest. EXERCISE 12.14 Find thc change in entropy ofa
Answer
96.4 J/K
2.00~kg block
or-gold at
j
063°C when it melts to become liquid gold at] 063°C.
410
Chapter 12
EXAMPLE 12.15 Goal
The Lows of Thermodynamics
Ice, Steam and the Entro
of the Universe
Calculate the change in entropy for a system and its environmenL.
Problem A block of ice al 273 K is put in thermal contact. with a container of steam at 373 K, converting 25.0 g of ice LO waLer at 273 K \.. . hile condensing some of the steam to water at 373 K. (a) Find the change in entropy of the ice. (b) Find the change in entropy of the Sleam. (c) Find the change in entropy of the Universe. Strategy First, calculate the energy t.ransfer necessary 1.0 melt the ice. The amount of energy gained by the ice is lost by the Sleam. Compute the entropy change for each process and slim to gel the entropy change or the Universe. ".
,
..
Solution (a) Find the change in entropy orthe ice. Use the latent heal of fusion, Lp to compute the thermal energy needed to melt 25.0 g of ice:
Q,,,.
= 1111"/ = (0.025 kg)(3.33 X 10'.P = 8.33 X 10 3 .1
8.33
Calculate the change in entropy of the ice:
X
]O'J
273 K
30.5.1 IK
(b) Find the change in entropy of the steam.
By supposition, the thermal energy lost by the steam
LlS
=
Slcal1,
is equal to the thermal energy gained by the ice:
Q~lcalll
= -8.33 X 10'J
7~te'lll1
373 K
-22.3.1/K
(c) Find the change in entropy of the Universe.
Sum the two changes in entropy: ~ ~
+ 8.2.1/K ~
,
Remark
Notice that. the entropy of the Universe increases, as it must in all nawral processes.
QU ESTION 12.15 True or False: For a given magnitude of thermal energy transfer, the change in entropy is smaller for processes that proceed at lower Icmperature.
EXERCISE 12.15 A 4.00-kg block of ice at 273 K encased in a thin plastic shell of negligible mass mells in a large lake al293 K. At the instant the ice has complctely melted in the shell and is still at 273 K, calculate the change in cntropy of (a) the ice, (b) the lake (which essentially remains at 293 K), and (c) the Universe.
Answers
(aj 4.88
X
EXAMPLE 12.16 Goal
lO'J/K
(b) -4.55
X
1O']lK
(cj +3.3
X
[02.1/K
A Falling Boulder
Combine mcchanical energy and entropy.
Problem A c1ll1nk of rock of mass 1.00 X 10.1 kg at 293 K falls rrolll a eli fr of height 125 In into a large lake, also at 293 K. Find the change in entropy or the lake, assuming all the rock's kinetic energy upon elllering the lake converts to thermal encrgy absorbed by the lake. Strategy Gravitational potential energy ""hell the rock is at. the top of the cliff converts to kinetic energy of the rock before it enters the lake, then is transferred to the lake as thermal energy. The change in the lake's temperawre is negligible (due to its mass). Divide the mechanical energy of the ,'ock by the temperature of the lake to estimate the lake's change in entropy.
12.5
,..
Entropy
411
.
Solution Calculat.e the gravitational potential energy associated with the rock at the LOp of the cliff:
Th is energy is transferred 1.0 1 he lake as therma I energy, resulting in an entropy increase of the lake:
PE= 1IIglz = (1.00 X 10' kg)(9.80
m/s~)(]25 rn)
1.23 X 10'; J
t. s
~
-Q = T
1.23X10"J
---::-:c:-c:--"-
293 K
4.20 X
1O'J!K .
l...
Remarks entropy.
....
This exam pic shm\!s how e\'cn simplc mechanic-d processes can bring about increases in the Universe's
QUESTION 12.16 If you carefully remove your (very heavy) physics book [rom a shelf and place it on the ground, what happens to the entropy of the Cniverse? Does it increase, decrease, or remain the same? Explain. EXERCISE 12.16 Estimate the change in entropy of a tree trunk at 15.0°C \Nhcn a bullet uf mass 5.00 g traveling at 1.00 X 103 m/s embeds itself in it. (Assume the kinetic energy of the bullet transforms to thermal energy, all of wh ich is absorbed by the tree.)
Answer
8.68J/K
Entropy and Disorder A large element of chance is inherent in nal.Ural processes. The spacing bet\.. . een trees in a natural forest, for example, is random; if you discovered a forest 'where
all the trees \vere equally spaced, you would conclude that it had been planted. Likewise, leaves fall to the ground with random arrangements. rt'would be highly unlikely to find thc leaves laid out in perfectly straight rows. Vve call express the results of such observations by saying that a disorderly arrangement is much more probable than an orderly one if the laws of nature are allowed to act without interference. Entropy originally lound its place in thermodynamics, but its importance grew trcmendously as the field of statistical mechanics developed. This analytical approach cmploys an alternate interprctation of entropy. [n statistical mechanics, the behavior of a substance is described by the statistical behavior of the atoms and molccules contained in it. One of" thc main conclusions of the statistical mechanical approach is tint isolated systems tend toward greater disorder, and entropy is a measure of that disorder. In light of this new vic\-\' of entropy, Boltzmann found another method [or calculating entropy through use of the relation
s=
k"ln W
[12.18]
where lin = 1.38 X 1O-:!3 J/K is Boltzmann's constam and W' is a number proporlional to the probability thal the system has a particular configuration. The symbol "In" again stands for naLUrallogarithm, discussed in Appendix A. Equation 12.18 could be applied to a bag of marbles. Imag-ine that you have 100 marblcs-50 red and 50 green-stored in a bag. You are allowed to draw four n1arbles from the bag according to the following rules: Draw one marble, record its color, return it 1.0 the bag, and draw again. Continuc this process until four marbles have been drawn. Notc that because each marble is returned LO the bag before the next one is drawn, lhe probability of drawing a red marble is always the same as the probability of drav.ling a green aile. The results of all possible drm.. . ing sequences are shown in Table 12.3. For example, t.he result RRGR means that a red marble was drawn rirst, a red one second, a
Tip 12.4 Don', Confuse the W's The f;}'mbol IVlIscd here is~ IJro/Jabilir)", nOllO he confused wilh lhe same symhol used for work.
412
Chapter 12
The Laws of Thermodynamics
TABLE 12.3 Possible Results of Drawing Four Marbles from a Bag
APPLICATION The Direction of Time
Total Number of Same Results
End Result
Possible Draws
All R
RRRR
I
IG,3R
RRRG, RRGR, RGRR. GRRR
2G,2R
RRGG, RGRG, GRRG, RGGR. GRGR, GGRR
4 6
3G,IR
GGGR. GGRG, GRGG, RGGG
4
All G
GGGG
I
green one third, and a red one fOlLrth. The table indicmcs thalthere is only one possible ,val' to draw four red marbles. There are four possible sequences that produce one green and three red marbles, six sequences that. produce two green and t\'.'o red, four sequences that produce three green and one red, and one sequence thaI produces all green. From EquaLion 12.18, \!·,'C see thal the state with the greatest disorder (two red and n\"o green marbles) has the highest entropy because it is most probable. In contrast, the most ordered states (all red marbles and all green marbles) are least likely Lo occur and are states of lmvest entropy. The outcome of the draw can range between these highly ordered (Iowestentropy) and highly disordered (highest-entropy) states. Entropy can be regarded as an index or hm\,' rar a system has progressed from an ordered lO a disordered statc. The second law of thermodynamics is really a statement of what is most probable rather than of what must be. Imagine placing an ice cube in contact with a hot piece of pizza. There is nothing in nature that absolutely forbids the transfer of energy by heat from the ice to the much warmer pizza. Sti.ltistically, it's possible for a slo\'·,'-moving molecule in the icc to collide with a faster-mo\'ing molecule in the pizza so that the slmv one transfers some of its energy to t.he Elsler one. ',Vhen the great number of molecules prcsent in the icc and pizza are considered, however, the odds are ovenvhelmingly in favor of the transfer of energy from the fastermoving molecules (0 the slm,,'er-moving molecules. Furthermore, this example demonstrates that a system naturally tends to move from a state of order to a state of disorder. The initial state. in which all the pizza molecules have high kinetic energy and £111 the ice molecules have lmver kinetic energy. is much more ordered than the final state after energy transfer has taken place and the ice has melted. Even more generally, the second Im·v of thermodynamics defines the direction of time for all events as the direction in which the entropy orthe universe increases. Although conservation of energy isn't. violated ir energy flows spontaneoLlsly from a cold object (the ice cube) to a hot object (the pizza slice), that event violates the second law because it represents a spontaneous increase in order. Of course, such an event also viohltes cveryday experience. If the melting ice cube is fihTled and the film speeded up, the difference between running the film in forward and reverse directions ,vould be obvious to an audience. The same \'Iould be true of filming any event involving a large number of particles, such as a dish dropping to the 0001' and shattering. As another example, suppose yOll were able to measure the velocities of all the air molecules in a room at some instant. It's very unlikel}' that you would find alt molecules moving in the same direction with the same speed; that would be a highly ordered stale, indeed. The Illost probable situation is a s)'stem of molecules moving haphazardly in all directions with a wide distribution of speeds, a highly disordered state. This physical situation can be compared to the drawing of mar· bles from a bag: lfa container held ]0 23 molecules ofa gas, the probability of finding all the molecules moving in the same direction with the same speed at some instant "vould be similar to that of drawing a marble from the bag 10 23 times and get ting a red marble on every draw, clearly an unli kely set of events.
12.6
The tendency of nature to move toward a state of' disorder arfecls the ability or a system to do work. Consider a ball thrown toward a wall. The ball has kinetic energy, and its state is an ordered one, which means that all the atoms and molecules of the ball move in unison at the same speed and in the same direction (apan from their random internal mOlions). When the ball hits the wall, however, part of the ball's kinetic energy is transformed into the random, disordered, internal motion of the rnoleculcs in the ball and the "vall, and the temperaLUrcs of the ball and the wall both increase sliglllly, Before the collision, the ball was capable of n doing work. It could drive a nail into the \V'all, (or example. \.\lith the trans[orma- ~ lion of part of the ordered energy into disordered internal energy, this capability of doing work is reduced. The ball rebounds with less kineLic energy than it originally had, because the collision is inelastic. c: Various forms of energy can be cOlwert.ed to internal energy, as in the collision ~ between the ball and the wall, but the reverse t.ransformation is never complete. Tn ~ ~ general, gi\"en twO kinds of energy, A. and B, if 1\ can be completely converted to B 9 and vice versa, we say that A and B are of the same grade. However, if A can be completely converted to B and the reverse is never complete, then A is of a higher grad(J of energy than B. In the case of a ball hitting a wall, the kinetic energy of the ball is of a higher grade than the internal energy contained in the ball and the wall arter the collision. ',Vhen high-grade energy is convened to int.ernal energy, it can never be fully recovered as high-grade energy. This conversion of high·grade energy to internal energy is referred to as the degradation of energy. The energy is said to be degraded because it takes on a form thal is less useful for doing \vork. In ot.her words, in aU real processes, the energy available for doing work decreases. Finally, note once again that the statement that entropy must increase in all nal.Ural processes is true only for isolated systems. There 3rc instances in which the entropy of some system decreases, but with a corresponding net increase in entropy for some other system, "\Then ~dl systems are taken together to form the Un ivcrse, the entropy of the Universe always increases. Ultimately, the entropy of the Universe should reach a maximum. \'Vhen it does, the Universe will be in a state of uniform temperature and density. All physical, chemical, and biological processes will cease, because a state of perfect disorder implies no available energy for doing work. This gloomy state of affairs is somelimes referred to as the ultimate '"'heat death" of the Universe.
Human Metabolism
413
i
QUICK QUIZ 12.5 Suppose you arc throwing two dice in a friendly game of craps. For any given throw, the two numbers that are face up can have a sum of 2, 3.4. 5, 6, 7, 8, 9, 10, II, or 12. Which outcome is most probable? vVhich is least probable?
12.6
HUMAN METABOLISM I'IJI
Animals do work and give off energy by heat, and this lead us to believe the first law of thermodynamics can be applied to living organisms to describe them in a general way. The internal energy stored in humans goes into other forms needed for maintaining and repairing the major body organs and is transferred out of the body by work as a person walks or lifts a heavy object, and by heat ,.. . hen the body is warlTIt:r than its surroundings. Because the rates of change of internal energy, energy loss by heat, and energy loss by work vary widely with the intensity and duration or human aCtivity, it.'s best to measure the time r<1tes of change of !!J.U, Q, and \Iv. Rewriting the first la",:, these time rates or change are related by
I1U 11/
Q I1t
HI
-=-+11/
[12.19]
(a)
(b)
(a) :\ royal flush is a highly ordered poker hand with ;llow probability or occurrence. (b) A disordered and worthless poker hane!' The probabil· ity or Ihis partirularhand occurring is lh~ same as thatoflhc roval nu~h. There arc so many wonbless hands, however, that the probability 01" being dealt a \,·onhless hand is much higlH:r than that of being dealt a royal flush, Can }'t)1l calculate the probability of being dealt a full Iiouse (a pail" and three ofa kind) from a slalldard deck 01"52 canis?
414
Chapter 12
The Lows of Thermodynamics
On average, energy Q flows aul of the body, and work is done b)' the body 011 its surroundings, so both QI~l and 1-1'l'6..t are negative. This means that D-.U/l:1t would be negative and the internal energy and body temperature would decrease \Vilh Lime if a human were a closed system \·...ith no way of ingcsling malleI' or replenishing inlcrnal energy stores. Because all animals are aCLUally open systems, the>' acquire
internal cncrg-y (chemical potential energy) by eating and breathing, so their interllal energy and temperature arc kept constanl. Overall, the energy from the oxidation a[fooel ultimately supplies the \'Iork done by the body and energy lost from the body by heat, and this is the interpretation we give Equation 12.19. That is, tlU/ilt is lhe rate at which internal energy is added to our bodies by food, and this term just balances the ratc of energy loss by heat, Q!d/, and by work, W/b.f. Finally_ if we ha\'e a ,'\'ay of measuring ilU/D.f and \i\tjut for a human, we can calculale Q/!J..t from Equalion 12.19 and gain useful information on the efficiency of" the body as a machine.
Measuring the Metabolic Rate AU/At The value of W/Lil, the work done by a person per unit lime, can easily bc dctermincd by mcasuring the power output supplied by the person (in pedaling a bike, for example). Tbe metabolic rate I1U/Ii.J is the rate at which chemical potential energy in food and oxygen are transformed into internal energy to just balance the body losses of internal energy by work and heat. Although the mechanisms of rood oxidation and energy release in the body are complicated, involving many intermediate reactions and enzymcs (organic compounds that speed lip thc chemical reactions taking place at. "low" body temperatures), an amazingly simple rule summarizes these processes: The metabolic rate is directly proportional to the rate of oxygen consumption by volume. It is [ound that for an r
1
~~
6. U
-
~ ~ Oc~
~~ 9 is: where FIGURE 12.17
Thishike rider is
being: lTIolliLQred fOI oxygclI consumption.
6./
6. 110 .,
= 4.8--'
6./
[12,20]
the mctabolic rate !1[J/D.1 is measured in kcal/s and LiVo/ Lit, the volume rale of oxygen consumption, is in Lis. !\1eclsuring the rate of oxygen consulllption during various activities ranging from sleep to intense bicyclc racing errectivcly measures the variation or metabolic rate or the variation in the total power the body generates. A simultaneous Illcasu rClllct1l of the work pCI' llnitli me done br a person along \vith the metabolic rate allows thc efficiency of t.hc body as a machine to be determined. Figure 12.-17 shows a person monitored for oxygen consulllption while riding a bike atLached to a dynamometer. a device [or measuring pm.. .er output.
Metabolic Rate, Activity, and Weight Gain Table l2.4 shows the measured ratc of oxygen consumption in milliliters per minute per kilogram of' body mass and the calculated met.abolic rale lor a 65-kg fnalc engaged in various activities, A sleeping person uses abollt 80 \V of powcr, (he basal metabolic rate, just to maintain and run different body organs such as the heart, lungs, liver, kidnc>,s, brain, ~nd skeletal mllscles.!\.'lore intense activit), increases the metabolic nlte to a maximum of about 1 600 W for a superb racing cyclist, although such a high raLe can only be maintained for periods ofa few seconds. \'Vhcn we sit watching a riveting film, \ve give off about as much energy by heal as a briglH (250-W) lightbllib. Regardless of level of activity, the daily food intake shouJdjust balance the loss in internal energy if" a person is not 1.0 gain weight. Further, exercise is a poor substitute [01' dieting as a method of losing weight. For example, the Joss of I pound of body f~u requires the muscles La expend 4 100 kC<-l1 of cnerg)'. If the goal is to lose: 1 pound' offat in 35 days, ajogger could run an cxtra rnile a day, because a 65kg jogger lIses abollt 120 kcallojog I mile (35 days X 120 kcal/day = 4200 kcal),
12.6
Human Metabolism
415
TABLE 12.4 Oxygcn Consumption and Metabolic Ratcs for Various Activities for a 65-kg Male' Activity
Ilea,,)' activit), (basketball, swimming a t~lst breaststroke)
30
600
700
Extreme anivily (bicycle racing)
70
I lIDO
L 600
'.')UI4«<::
A (;mllJlflllion
10
Metabolic Rate (W)
80
Medical S/udhs. 'lIe. R. l'assllHU",·. Philadl·lphi.l, F. \. Ib.\ i~. 19(;8.
An easier way LO lose the pOimd of fat would be to diet and eat two fc\...· cr slices or bread per day for 35 days, because bread has a caloric contenl of 60 kcal/slice (35 clays X 2 slices/clay X 60 kcal/slice = 4 200 kcal).
EXAMPLE 12.17 Goal
Fi hting!.-'F""a"'t
_
Estimate human energy usage during a typical day.
Problem Tn the course of24 hours, a 65-kg person spends 8 h ata desk, 2 h puttering around the house, 1 hjoggillg 5 miles, 5 h in moderate activity, and 8 h sleeping. \,Vhal is the change in her internal energy during this period? Strategy The time rate of energy usage-or power-multiplied by time gives rhe amount of energy used during a given activity. Use Table 12.4 to find the power CZP j needed for each activity, multiply each by the lime, and slim them all up. ,. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
Remarks If this is a typical day in the ,\'oman's life, she will have 1.0 consume less than 5 000 kilocalories on a daily basis in order to lose weight. A compliGHion lies in the factthal human metabolism tends LO drop when food intake is reduced. QUESTION 12.17 How could completely skipping meals lead lO weight gain? EXERCISE 12.17 If a fiO.O-kg woman ingests 3 000 kcal a day and spends 6 h sleeping, 4 h walking briskly, 8 h sitting at a desk job, ] h swimming a fast breaststroke, and 5 h watching action movies on TV, about how much 'weight 'NiJ1 the woman gain or lose every day? (Nolf': Recallthatllsing abollt4 100 kcal of energy will burn orfa pound of fat.)
Answer
She'll lose a little more than one-haifa pound offat a day.
Physical Fitness and Efficiency of the Human Body as a Machine One measure of a person's physical fltness is his or her maximum capacity to lise or consume oxygen. This "aerobic" fitness can be increased and maintained with regular exercise, but falls ,,,,,hen training SlOpS. Typical maximum rates of oxygen consumpt1on and corresponding fitness levels are shown in Table 12.5 (page 416);
The Lows of Thermodynamics
Chapter 12
416
we see that the maximum oxygen consumption rate varies from 28 mL/min'kg of buely 11l,-lSS for poorly conditioned subjects to 70 rnL/min-kg [or superb mhJctcs. \Ve have already pointed out that. the first law of thermodynamics call be rewritten \.0 relate the metabolic rate ~U/D.I to the rale at 'rvhich energy leaves the body
TABLE 12.5 Physical Fitness and Maximum Oxygen Consumption Rate Maximum
by work and by heat:
Oxygen Fitness
Level Very poor POOl"
Consumption Rate (mL/min' kg)
28 34
Fair
42
Good
52
Excellent
70
t1V t1/
W
t1/
Now consider the body as (j machine capable of supplying mechanical power LO the oLltside world and ask [or its efficiency. The body's efficiency eis defined as the ratio or the mechanical power supplied by a human to the metabolic nHe or the total pO\\>·er input to the body:
WI t1/ I (' = body's erriciency = It1 VI
~."i(>IIrr(': Ann/lin,
Huol..~.
Q
t1/
-=-+-
K. II. Cuop"r. l\alll;ll11 :-.!("I' Yt,rk. 1~1l;8.
[12.21]
t1/ In t.his definition, absoltlle value signs arc used to show that e is a positive number and to avoid explicitly using minus signs required by our definitions of Wand Q in the first law. Table 12.6 shows the efflcienc}' of" workers engaged in different activities for several hours. These values were obtained by rneasuring the power output and simultaneolls oxygen consurnption of mine workers and calculating the metabolic rale from their oxygen consumption. The table shows that a person can steadily supply mechanical power for several hours at about 100 \V with an efficiency of about 17%. It also shows the dependence of efficiency on activity, and that (' can drop to values as low as 3% for hig'hly inefficient activities like shoveling, which involves lrIallY slans and stops. Finally, it is inlercsting in comparison 10 the average results of Table 12.6 that: a superbly conditioned athlete, efficiently coupled to a mechanical device for extracting power (a bike!), can supply a pO\'..-er of around 300 \'V for about 30 minutes at a peak efficiency of22%.
TABLE 12.6 Melabolic Rale, Power OutpUl, and Efficiency for Differenl AClivilies a Metabolic Rate
Power Output
I1U 11/
TV 11/
(watts)
(waUs)
Efficiency e
Cycling
50S
96
0.19
Pushing loaded coal GIl'S in a mine
525
90
0.17
Shoveling
570
17.5
0.(13
Activity
"Sol/rrl': "Inter- anrl Intra-lndh-idll;11 niITt:n:nCl'~ In Energ"~ Expcndiwn' and Mechanic-al Efficil'l1C\"~ C. H. Wy lid h,lIIl cl ~d .. 1:'r~(IIlIJllIifS !). 17 (1 ~)(ili).
~~MARY 12.2
12.1 Work in Thermodynamic Processes The work clone on a gas at a constant pressure is w~
-Pt1\1
[12.1]
The work done on the gas is positive if the gas is compressed (J.Vis negative) and negative if the gas expands (liVis positive). In general. the work done on a gas that lakes it from some initial state to some final state is the negative of the area under the curve all a PVdiagram.
The First Law of Thermodynamics
According to Ihe first law of thermodynamics. when a system undergoes a chllnge from one state (0 anal her. the change in ilS intel-nal energy liUis
[12.2] where Q is the energy transferred into lhe system by heat and W is the work done on the system. Q is positive when energy enters the system by heat and negative when the sys·
Multiple-Choice Questions lelll loses energy. W is positivt> when work is done on thc system (for example, by compression) and ncgati\'c when the system does positive work on its environment. The change of the intcrnal energy, 6.U, of an ideal gas is givcn by
D.U=
11
[12.51
G',,6.T
where Ct , is the molar specific hem at constant volume.
12.3
IQ,I IQ"I
W;·lIg I!==--~
IQ"I
[12.12]
Ileat pumps are heat engines in reverse. In a refrigerator the heat pump removes t.hermal eller~n' [rom inside the refrigerator. I-leal pumps operaling in cooling mode have coefficient of performance given by COP (cooling mode)
Thermal Processes
An isobaric process is one thai occurs at constant pressure. Tile work done 011 the system in such a process is -p 6.\~ whereas the thermal energy transferred by heat is given by
417
=
IQ,I
W
[12.13]
A heat pump in heating mode has coefficient of perfor-
mance
[12.6]
COP(hcating mode)
=
IQ"I W
[12.14]
with the molar heat capacity at constanl pressure given by C,,~C"+n.
In an adiabatic process no energy is transferred by heat bet.vl.'een the systcm and its surroundings (Q = 0). in this case the f-lrst law gives !i.U = \'11; which means the internal energy changes solely as a consequence of work being done on t.he systelll. Thc pressure and volume in adiabatic processes are related by
pvr =
constant
112.801
where'Y = epiC" is the adiabatic index. In an isovolumetric process the volume doesn'l change and no work is done. For such processt:s, the first law gives /1U= Q All isothermal process occurs:ill constant lempcraLurc. The work done by an ideal gas on the environment is \;11<:11\ =
nRT In
(II), \:
112.IOJ
12.4 Heat Engines and the Second Law of Thermodynamics r11 a cyclic process (in wh ich the system retu rns to its ill iLial stale), D.U = 0 and therefore Q = H('ng' Illeaning the energy tr t'>, transferred into the engine per cycle:
Real processes proceed in an order governed by the second law of thermodynamics, ""hich can be slated in [\'10 ways:
1. Energy will not rIow spontaneously by heat frorn a cold o~icct 1.0 a hot object. 2. No heLlt engine operating in a cycle can absorb energy from a reservoir and perform an equal amount of work. No real heat eng-ine operaling between the Kelvi n temperatures 7~, and ~r;. can exceed the efficiency of an engine operating between the same two t.emperatures in a Carnot cycle, given hy ('I.'
=1
112.16]
Perfc.::cl eff-icicncy or a CarnO!. engine requires a colel reservoir of 0 K. absolute zero. According to U1C third law of thermodynamics, ho,,·:evcr, it is impossible to lower thc telnperalurc ofa system to absolute zero in a finite number of steps.
12.5
Entropy
The second law can also be stated in terms of" a quamit)' called entrop,., (S). The change in entropy of a system is equal to the energy Q~ flowing by heat into (or out of") the system as t.he system changes from one state to another by a reversible process, divided by the absolulc temperature: /18
~ Q,
[12.17]
T
One of the primary findings of stat.istical mechanics is thaI systems tend to\vard disorder, and entropy is a meaSllre of that disorder. An alternate statement of the second law is thal t.he entl"opyofthe Universe increases in all natural processes.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
,~MULTIPLE.CHOICE QUESTIONS 1. An ideal gas is maintained at a constant. pressure of 70.0 kPa during an isobaric process while its volume decreases by 0.20 Ill:!. What is the work done by the sys-
tem on its environment? (a) 14 kJ (b) 35 kJ (c) -14 kJ (d) -35 l;j (e) -72 kJ
418
Chapter 12
The Laws of Thermodynamics
2. A 2.0~lllole ideal gas system is maintained at a constant volume of 4.0 liters. If 100 J of thermal energy is transferred LO the system, \.. . hat is the change in the internal energy or the system' (a) a (b) 400.1 (c) 70.1 (eI) 100.1 (e) -0.100.1
to the gas by heat. (b) No work is done on the gas. (c) The temperature of the gas increases. (d) The internal energy 01" the g,lS remains constant. (e) The pressurc rcmains constant.
3. A monatomic ideal gas expands from l.OO m 3 to 2.50 m:\ at a constant pressure of2.00 X 10" Pa. Find (he change in the internal energy of the gas. (a) 7.50 X 10 5 .1 (b) 1.05 X JO".1 (c) 4.50 X 10'.1 (eI) 3.00 X 10'.1 (e) -4.50 X 10".1
12. \f\.'hen an ideal gas undergoes an adiabalic expansion, which of the following statements is true? ((I) The temperature of the gas doesn't change. (b) No work is done by the gas. (c) No energy is delivered [0 lhe gas by heaL (d) The int.ernal energy of the gas doesn't change. (c) The pressure increases.
4. An ideal gas drives a piSLOI"l as it expands frolll l.00 m 3 to 2.00 111 3 at a constant temperature of 850 K. I r there arc 3~O moles orgas in the pistoJl, how much work docs the gas do in displacing the piston? (a) 1.9 X lOt' J (b) 2.5 X 10';.1 (c) 4.7 X 10 6 .1 (eI) 2.1 X 10'.1 (c) 3.5 X la' .I
13. J r an ideal gas undergoes an isobaric process, which of the following statements is true? (a) The temperature of Lhe gas doesn't change. (b) Work is done on or by the gas. (c) No energy is transferred by heallO or from the gas. (d) The volume of the gas remains the same. (e) The pressure of the gas decreases uniformly.
5. A diaLOmic ideal gas expands adiabatically /"rOlll a volume of 1.00 m 3 to a final volume of 3.50 m 3 . If the initial pressure is 1.00 X 10 5 Pa, what is the final pressure? (a) 6.62 X 10" Pa (b) 1.24· X 10" P" (c) 3.54 X 10"' Pa (eI) 2.33 X 10' Pa (e) 1.73 X 10" Pa
14. Of the follmving, wh icll is rlol a statement of the second law of thermodynamics? (a) No hcat engine operating in a cycle can absorb energy ["rain a reservoir and lise it entirely to do work. (b) :'\Jo real enginc operating between twO energy resen'oirs can be more efficienl than a Carnot engine operating between the same two reservoirs. (c) \Vhen a system undergoes a change in Slale. the change in the interna I energy of the system is the sum of t.he energy transferred to the system by heat and the work done on the system. (d) The entropy of the Universe increases in all natural pro(e~ses. (e) In all real processes, the resulting energy available ror doillg work decreases.
6. How much net work is done by the gas undergoing the cyclic process illustrated in Figure MCQ12.6? Choose the bcsI estimate. (a) 1 X 10'.1 (b) 2 X 10'.1 (c) 3 X 10'.1 (eI) 4 X 10'.1 (e) 5 X W Pa P(JU" Pal -tOO 3.00 2.00 1.00
'-----'------'.--.'-- V (1ll 3) 1.00 2.00 3.00 FIGURE MCQ12.6
7. An engine does 15 kJ of work while rcjectinR 37 kJ to the cold reservoir. \'\-'hat is the efficiency or the engine? (a) 0.15 (b) 0.29 (c) 0.33 (eI) 0.45 (e) 1.2 8. A rcfrigeraLOr does 18 q of work wlJile moving 115 ~J of thermal energy from inside the rcfrigeraLOr. \Vhat is its coefficient of performance? (a) 3.4 (b) 2.8 (c) 8.9 (eI) 6.4 (e) 5.2 9. A steam turbine operates at a boiler temperature of" 450 K and an exhaust temperature of 3.0 X IO~ K. Vii/hat is the maximum theoretical efficicncy of this system? (a) 0.24 (b) 0.50 (c) 033 (eI) 0.57 (e) 0.15 10. A I.OO-kg block of ice at DoC and 1.0 atm melts COI11pler.cly to wafer at We. Calculate the change of the emrop)' of the icc during the melting process. (For ice, LJ ~ 3.33 X 10' .I/kg.) (a) 3 340.J/K (b) 2 170jlK (c) -3340.J/K (d) 1 220.J/K (e) -1 220.J/K 11. If an ideal gas is compressed isothermally, which of the follQ',ving statements is tnle? (a) Energy is transferred
15. If an ideal gas is compressed to half its initial \'olume, which of the following- sr.atcmcnts is true regarding the work done on the gas? (a) The isothermal process involves the most work. (b) The adiabatic process involves the lIlost ' . . ork. (c) The isobaric process involves the most wurk. (d) The isovolumC1.ric process involves the most work. (e) The work done is independent of the process. 16. A window air conditioner is placed on a table inside a well-insulated apartment, plugged in and turned 011. ""'hat happens to the average lemperamre of the apanment? (a) It increases. (b) It decreases. (c) It remains constant. (d) It increases until the unil warms lip and then decreases. (e) The ans'.. .er depends on the initial rcmperatlirc of the apartmenr. 17. The second hnv of thermod}'lllunics implies that the coefficient of performance of a refrigerator must be ",.·hat? (a) less than I (b) less than or equal to I (c) g-reater than or equal to I (d) finite (e) great.cr than 0 18. A thermodynamic process occurs in ,..·hich the enlropy of a system changes by -6 J/K. According LO the second law of thennodynamics, what can you conclude about the entropy change ol'tlle environment? (a) [t must be +6.J/K or less. (b) It IllUst be equal to 5.J/K. (c) It must be between 6J/K and O. (d) It lllust be O. (e) Itmllst be +6 .11K or more.
Problems
CONCEPTUAL
419
Q~E_STIONS
1. Whal arc some factors that. afrecl the efficiency OfatlLOmobile engines?
2. If yOll shake a jar full of jelly beans of different. sizes, the larger beans tend to appear near the lOp and Ihe smaller ones lend to fall to 1.he bOLtom. Why docs that occur? Does this process violate the second law of thermodynamics?
3. For
7. When a sealed Thermos bottle full of hot coffee is shaken, what changes, if' any, take place in (a) the temperature orthe coffee and (b) its internal energy?
8. In solal" ponds constructed in Israel, the Sun's enerb'"Y is concentrated near the bottolll of" a sally panel. With the proper layering of salt in the water. convection is prevented and tcmperawres of 100°C Illay be reached. Can you guess the maximum efficiency with which usefulmechanical ",lUrk can be extracted from the pond?
9. Is i1. possible to construct a heat engine that creates no thermal pollution?
4. Clearly distinguish alllong temperatllre, heal, and internal energy.
10.11' a supersaturated sugar solution is allowed to evapo-
5. Consider the human body performing a strenuous exercise, such as Ii f"Ling weights or riding a bicycle. Work is being done by the body. and energy is leaving by conduction from the skin into the surrounding air. Accarding to the n rst 1a\,,1 of thcrlllodynaill ics, the temperature of the body should be steadily decreasing during the exercise. That isn't \"..hat happens, however. Is the first law invalid for [his siluation? E.xplain.
11. The nrst law of thermodynamics says we can't getlllore out of a process than we put in, but the second law says that ,...'e can't. break even. Explain this statement.
6. A steam-driven turbine is one major component of an deClric power plant. Why is it advantageous to increase the temperature or the steam as much as possible?
rate slowly, sligar crystals form in the container. Hence, sligar molecules go from a disordered form (in solution) to a highly ordered, crystalline fOl"m. Does this process violate the second law of thermodynamics? Explai n.
12. Give some examples of irreversible processes that occur in nattlre. Give an example of a process in nature that is nearly reversible.
PROBLEMS ~
The Problems for this chapter Illay bc assigned online at We·bAssig-n. ~~ = straightf'o,"ward, illlermediatl', t"halll'llging = denotes gu ided problem = denotes enhanced content problclll
WebAssign
l,~,
m _
1'1 =
biomedical applicatioll 0= dellotcs full solution ,wailablc in Student Sol"li(JlH Alal/ual/ 5:turl)' (;1lirh·
SECTION 12.1 WORK IN THERMODYNAMIC PROCESSES l.
mil A gas changes in volume from 0.7['lO
m~ to 0.250 m~ at a constant pressure of 1.50 X 10'-' Pa. (a) How much work is done 011 the gas? (b) I-low much work is done by the gas on its environment? (c)\'Vhich of Newton's \<""s hest explains why the work done on the gas is the [leg-alive of the wOI-k done on the environment?
[}Jeas in a container is at a pressure of 1.5 atm and a vollime of' 4.0 m:~. What is the work done on the gas (a) if it expands at COn!:ilanl pressure to twice its initial volume, and (b) if it is compressed a\. constalll pressure to onequarter its initial volume? 4. A 40.0-g projectile is launched by the expansion of hot gas in all arrangement shown in Figure PI2.4a. The cross-sectional area ofl.hc launch lube is 1.0 em 2 , and the length thal.t.he projectile travels down the tube after starting from rest is 32 cm. As the gas expands, the pressure varies as shown in Figure P12.4b. The values for the initial pressure and volume are P, = 11 X .\05 Pa and V. = 8.0 cm:1 while the final values are Pf = 1.0 X 10 5 Pa and Vf = 40.0 cml'l. Friction between the projectile and the launch tube is negligible. (a) [f the projectile is launched into a vacuum, what is the speed of the projectile as p
2. SkclCh a PV diagt"arn and lind the wOI-k done by the gas during the following stages. (a) A gas is expanded from a volume of 1.0 L La ~.O L al a constant pressure of :1.0 atm. (b) The gas is then cooled at constalll voillme until the pressure falls to 2.0 all11. (c) The gas is then compressed at. a constant pressure of 2.0 alm from a volume of ~.O L to 1.0 L. (No/": Be careful of sig-ns.) (d) The gas is heated until il$ pressure increases from 2.0 atm to 3.0 Cltm at a c.onstant volume. (e) Find the net '.. .ark done during the complete cycle.
P, ~11----;l2 cm---~'I
--I
Hem
LV
L--L-
10--
(h)
(a) FIGURE P12.4
420
The Laws of Thermodynamics
Choptcr 12
it leaves the launch lube? (b) If instead the prc~jeclile is launched into air at a pressure of 1.0 X 10 5 P
5. A gas expands from lro Falong the three paths indicated in Figure P12.5. Calculate the work done on the gas along paths (a) L4/'; (b) II': anti (e) IBF.
,
P(aun)
SECTION 12,3 THERMAL PROCESSES 11. _ The only ronn orenergy possessed by molecules ora monatomic ideal gas is translational kinetic energy. Using the rcsults from the discussion of kinetic theory in Section 10.5, show t.hatthe intcrnal energy of a monatomic ideal gas at pressure P and occupying volume V may be writtcn as U = ~l">V. 12.
A
2
+--;F V (I i t (; rs) o'------!---:---!c---c2 3 ,1 FIGURE P12.S
SECTION 12,2 THE FIRST LAW OF THERMODYNAMICS
(Problems:l and 13)
m
A cylinder of volume 0.300 m:1 contains 10.0 mol of neon gas at 20.0° C. Assume neon behaves as an ideal gas. (a) What is the pressure of the gas? (b) Find the il1lernal energy of the gas. (c) Suppose the gas expands at constant pressu,'c to a volume of 1.000 m 3. f-low much work is done on the gas? (d) What is the tcmperature of the gas at the new volume? (e) Find the internal energy of the gas when its volume is 1.000 m 3 . (f) Compute the change in the internal energy during the expansion. (g) CompUle AU - \IE (h) Must thermal energy be transferred to the gas during the constant pressure expansion or be taken away? (i) Compute Q, [J1C thermal energy transfer. (j) 'What symbolic relmionship between Q. I1U, and Wis suggested by the values obtained?
6. Sketch a PV diagram of the following processes: (a) A gas expands at constant pressure PI from volume VI to volume \12- It is then kepI al constant volume while the pressure is reduced La P,;!, (Il) A gas is reduced in pressure [rom PI to P2 while its volume is held const
[illA gas expands from J to F in Figure PL2.5. The energy added to lhe gas by heat is 418 J when the gas goes from Ito Falong the diagonal path. (a) What is the change in internal energy of the gas? (b) How much energy must be added to the gas by heat for the indirect paLh {!\Fto give the same change in internal energy?
[]JA sample or helium behaves as an ideal gas as it is heated atconstanl pressure from 273 K.w 373 K. jf20.0J of work is done by the gas during this process, what is the mass of helium presel1l?
14. A thermod}'namic system undergoes a process in which its intemal energy decreases by 500 J. If at the same time 220J of work is done on the system, find the energy transferred to or from it by heat.
8. One mole of an ideal gas initially at a temperature of 1.50 X 10 2 °C is compressed at a constant pressure of 2.00 ;lUll to lwo-thirds its initial volumc. (
15. A gas is compressed at a conslalll pressure of 0.800 atm from 9.00 L to 2.00 L. In the process, 400J of energy leaves (hc gas by heal. (a) 'What is the work done on the gas? (b) What is the change in its internal energy?
9. One mole of an ideal gas initially at a temperature of T, = DOC undergoes an expansion at a constant. pressure of 1.00 aLI1l to four times its original volume. (a) Calculate the new temperature 7[01' the gas. (b) Calculate the work done on the gas during the exp.msion. 10. (a) Detcnnine Ihe work done un a fluid that cxpands from ito Jas indicated in Figure PI2.1O. (b) How much work is done on the fluid if it is compressed from JLO i along t.he same path?
16. A qualllity of a monatomic ideal gas undergoes a process in which both its pressure and volume are doubled as shown in Figure PI2.16. \Vhat is the energy absorbed by heat into the gas dluing th is process? (Hint: See Problem 11.)
------71 Po
----(
;
I I
I I
I
:
l-_...l,--~~-v V{J ~Vo
'''>(Pa) FIGURE P12,16
6xlOfi
---
1 I
4 X 106
: I
I
__ ~---IL-_
2x lOG
I
,f I
I
~--c---:':---:----'-- V (m")
o
2
~
FIGURE P12,lO
4
IJ1JA gas is enclosed in a container fitted with a piston of cross-sectional area 0.150 m'.!. The pressure of the gas is maintained at. 6000 Pa a~ the pistoll moves inward 20.0 em. (a) Calculatc the work clone by the gas. (b) 11" the internal energy of Lhe gas decreases by 8.00 J, fmd the amount of heal. removed from the system by heal. during the compression.
Problems 18. A monaLOmic ideal gas undergoes the thermodynamic process shown in the PVdiagram of Figure 1'12.18. Determine whether each 01" the values ~U, Q. and \VIol' the gas is posit.ive, negative, or zero. (I-lint: See Problem 11.) p
2~: ~~~~t\ I I
I I
I
I
~A 5.0-kg block of aluminum is heated from 20°C 1.0 90°C a\ atmospheric pressure. Find (a) the work done by the aluminum, (b) the amount of energy transferred to it by heat, ;)nd (c) the increase in its internal energy.
26. One mule uf gas initii'lly at a pn;ssulc uf 2.00 al.m and a volume of 0.300 L has an internal energy equal to 91.0J. ]n its rinal state, the gas is at a pressure of 1.50 at.m and a ,·ohllne orO.800 L, and it.s internal cnerg) equals 180J. For the paths 111/;; fBI,; and IFin Figure P12.26, Gllculate (a) the work done on the gas and (0) the net energy transferred 10 the gas by heat in t he process.
FIGURE PI2.TS
19. An ideal gas in a cylinder is compressed very slowly to one-third it!'. original volume while its temperature is held constant. The work required to accomplish this task is 75J. (a) \'Vhat is the change in the internal energy of the gas? (b) How much energy is transferred to the gas by heat in this process? 20. An ideal gas in a cylinder is compressed :-ldiah:Hically to one-half its original \'olume. The work required to compress the gas is 125.). «I) How much energy is transferred into or Oul of the gas by heat in this process? (b) What is the change in the imernal energy of the gas? 21. An ideal monatomic gas exp:-lOds isothermally from 0.500 m 3 to 1.25 m:\ at a constant temperature of 675 K. If the initial pressure is 1.00 X 10':' Pa. And (a) the wor-k done on the gas, (b) the thermal energy 1ransrer Q, and (c) the ch:mge in the internal energy. 22. _ An ideal gas expands ,ll constant pressure. (a) Show t.hat P .6.\1 = III?!:J.T (b) Ir the gas is monatomic. start from the definition of internal energy and show that il U = ~ Wen", ",,'here H(nv is the work clone b)' the gas on its environment. (c) For t.he same monatomic ideal gas. show with the first law' that. Q = S\'\~.I'" (d) Is it possible for an ideal gas to expand al. constant pressure while exhausting t.hermal energy? E.xplain.
23. One gram of "",cr change, to ice at a constant pressure of 1.00 atm and a constant. temperature of O°C. In the process, the volume changes from 1.00 cm 3 to 1.09 cm:~. (a) Find the work done on the water and (b) the change in the internal energy of the water. 24. Consider the cyclic process described b}' Figure P12.24. II" Qis negative for the process BC<1nd !1Uis negative for the process CA, det.ermine the signs of Q, W; and MJ associ,Hed with celch process.
P (aun)
: -2]-----
2.00
I -~---,B
150
_A,+--l.....->! ,F I
'-c,;:-,."30"'(;:-'--;;-0.""8""00 II (liters) FIGURE P12.26
27. miJ Consider the Universe to be an adiabatic expansion of" atomic hydrogen gas. (a) Use the ideal gas law and Equation 12.8a to show that TV)'-l = C, where C is a const
SECTION 12.4 HEAT ENGINES AND THE SECOND LAW OF THERMODYNAMICS 29. A gas increases in pressure from 2.00 aun to 6.00 aUll at a constant. vulume of 1.00 m:J
liOO B
initial slate as shown in Figure Pl2.29. How much \\lork is done in one cycle?
30. An ideal gas expands at a constant pressure of' 6.00 X 10" Pa from a mlume of 1.00 m:l [0 a volume of 4.00 m 3 and then is compressed lO one-third llut pressure and a volume of 2.50111 3 as shown in Figure P12.30 before rCllIrning to its initial Slate. How much work is clone in laking a gas through one cycle of the process shm',;n in the figure? P(IO"Pa) 6.00 4.00
2.00 '____'___~'____'____.L_
V (m 3 )
l.Oa 2.00 3.00 4.00 FIGURE P12.30
31. A heal engine operates hetween a reservai," at 25°C and one at 375°C. '",,'hal is the maximulll efficiency possible for this engine? 32.
am A
heat engine is being designed to have a Carnal efficiency of 65% when operating between two heat reservoirs. (a) 1f the tcmpcratUl"c of the colel reservoir is 200C, what must be the tcmperature of the hot reservoir? (b) Can the actual efficiency of the engine bc equal to 65%? Explain.
33. The work done by an engine equals one-fourth lhe energy it absorbs from a reservoir. (a) What is its thermal efficiency? (b) vVhat fraction of' the energy absorbed is expelled to the cold reservoir? 34. A particular engine has a power output of 5.00 kW and an efficiency of25.0%. Iflhe engine expels 8 000 J of energy in each cycle, fmd (a) the energy absorbed in each cycle and (b) the time required to complete each cycle. 135.IOne 01" the mOst eflicient engines ever built is a coal-fi red steam turbine engine in the Ohio River valley, driving an electric generator as it operates between 1 870°C and 430°C. (a) 'Vhal is its maximum theoretical efficiency? (b) Its actual efficiency is 42.0%. How much mechanical power does the engine deliver if it absorbs lAO X 10!i J of energy each second from the hOl rescrvoi r? 36. A gun is a heat engine. In particular, it is an imernal combustion piston engine that does not operate in a cycle, but comes apan during its adiabatic expansion process. A cel"tain gUll consists of 1.80 kg of iron. It fires onc 2.'W-g bullet at 320 m/s with an energy cf"licienc}' of 1.10%. Assume the body of' the gUll absorbs all the energy exhaust and increases uniformly in temperaltlre for a short time before it loses any cnerg'y by heat int.o the environment. Find its temperature increase. 37. An engine absorbs I 700J from a hot reservoir and expels I 200.1 to a cold reservoir in each cycle. (a) ',Vhat is lhe engine's efndency? (b) How much work is done in each cycle? (c) What is the power OLltplll of the engine if each cycle I... sts 0.300 s?
38. A heat pump has a coefficient 01" performance of 3.80
8.00 h' 39. A freezer has a coefticiel1l of performance or 6.30. The freezer is advertised as using 457 HV-h/y. (a) On avcrage, how much cnergy docs the freezer use in a single day? (b) On average, how much thermal energy is removed from the freezer each da),? (c) ,,,Ihat maximum amount of water at 20.0°C cou lei the freezer freeze ina singlc day? (One kilowatt-hour is an amoum of energy equal to running a l-kW appliance for one hour.) 40. _ Suppose an ideal (Carnot) heat pump could be constructed. (a) Using Equation 12.15, oblain an expression lor the coefficient of performanct: for such a heat pump in terms or 1~. and ~. (b) "Vould such a heat pump work belter if the difference in the operating temperatures were great<.:r or were smaller? (c) Com pille the coefficient of performance for such a heal pump if the cold reservoir is 50.0°C and indoor temperature is 70.0°C.
[±[Jln one cycle a heat engine absorbs 500.1 from a hightemperaturt: reservoir and expels 300 J to a low~ lemperature reservoir. Lf the efficiency of this engine is 60% of tht: efficicnc)' or a Carnol engine. what is the ratio of the low temperature to the high temperature in the Camot engine? 42.
mil
A pm'ocr plant has been proposed that would make use of the temperature gradient in the ocean. The system is to operate between 20.0°C (surface water temperawre) and 5.00°C (water temperature al a depLh of about 1 km). (a) What is the maximum efficiency of such a system? (b) If Lhe useful power output of the plant is 75.0 MW, how much energy is absorbed per hour? (c) In vicw of your answer to pan (a), do you think such a system is wonlw,,·hile (considering that there is no charge for fuel)?
43. A nuclear pO\,,,'cr plant has an electrical power output of 1 000 MW and operates with an efficiency of 33%. If excess energy is carried away from the plant by a river with a flow rate of 1.0 X lOfi kg/s, whal is lhe rise in temperature or 1Jle nowi ng water? 44. A heat engine operates in a Carnat cycle between SO. DoC and 350°C. It absorbs 21 000 J of energy per cycle [rom the hot reservoir. The duration of each cycle is 1.00 s. (a) '-\'hal is the mechanical power output of this engine? (b) How much energy does il expel in each cycle by heat?
SECTION 12.5 ENTROPY 45. :\ Styrofoam cup holding 120 g of hOI water at 1.00 X 1.0 2 °C cools to room lemperawre, 20.0°C. ''''hat is Lhe change in entropy of the room? (Neglect the specific heal of the Ctlp and any change in temperature of the room.) 46. Two 2 OOO-kg cars, both traveling at 20 mis, undergo a head-on collision and Slick together. Find the change in entropy of"the Universe resulting from Lile collision if the temperatll re is 2~oC.
Problems
147.1/\ frecLer
is used to freeze 1.0 L of water completely iuto ice. The water and the freezer remain at a constanttelllperature of T= DoC. Determine (a) the change in the CIllWpy of the water and (b) the change in the entrap)' of the freezer.
48. What is the change in entropy of 1.00 kg of liquid water at IOOGC as it changes to steam at 100 0 e? 49. A ?D-kg log falls from a height of25 m inLO a lake. If the log, the lake, and the air are a1l at 300 K, find the change in entropy of the Universe during this process. 50. IrYOli roll a pair of dice, what is the total number of wa~!s in which you can obtain (a) a 12? (b) a 7? []IJThe surface of the Sun is approximately at 5 700 K, and the temperature of the Earth's surface is approximately 290 K. Whatenl.ropy change occurs \.. . hen I 000 J of"encrgy is transferred by heat from lhe Sun to the Earth? 52.
m
\""hen an aluminulll b,.\I- is temporarily connected between a hot reservoir at 725 K and a cold reservoir at 310 K, 2.50 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir. In this irreversible process, calculate the change ill emropy of (a) the hot reservoir, (b) the cold reservoir, and (c) the Universe, neglecting an)' change in entropy of the aluminum rod. (e1) Mathematically, why did the result for the Universe in p:Ht (c) have to be positive?
53. Prepare a table like Table 12.3 for the following occurrence: You toss four coills into the air sinmltaneol,sly and record all the possible resull,s of the toss in terms of the numbers of heads and tails t hat can result. (For example. HIITH and HTI-Il-1 are two possible ,\'ays in which three heads and one tail can be achieved.) (a) On the basis of your table, what is the most probable result of a LOSS? In terms of entropy, (b) what is the most ordered state. and (c) what is the most disordered? 54.
miJ When a metal bar is temporarily connected between a hot reservoir at 'J~, and a cold reservoir at 7;. the energy transferred by heal from the hot reservoir to the cold resc)'voir is Qh' In this irreversible process, find expressions lor the change in entropy of (a) the hot reservoir, (b) the cold reservoir, and (c) the Universe.
SECTION 12.6 HUMAN METABOLISM 55.
II
56.
am II A
Energetically, 1 Ib or fat is equivalent to 1.7 X lO'.J. How much extra weight would you lose each year if you suhstituted one hour of physics study a day (considered desk work) for onc hour of sleep?
weightlifter has a basal metabolic rate of 80.0 Vll. As he is working out, his metabolic rate increases by about 650 ,.\,. (a) How many hours does it take him to work off a 450-Calorie bagel if he stays in bed all day? (b) How long does it take him if he's working om? (c) Calculate the amount of mechanical work necessary to lif[ a 120-kg barbell 2.00 m. (d) L-Ie drops the barbell to the floor and lifts it repeated 1),. How many times per minute must he repeat this process to do an amount of mechanical work equivalent [Q his metabolic rate increase of 650 W during exercise? (e) Could he actually do repeti-
423
tions at the rate found in part (d) at the given metabolic kvcl? Explain. 57.
II
Sweating is one of the main mechanisms with which the bod}' dissipates heat. Sweat evaporates with a latent heat of 2 430 k:J/kg at body tcmpcmtul'c, and t.he bod)' can produce as much (IS 1.5 kg of sweat per hour. If sweating were the only heat dissipation mechanism, what would be the maxilllulll sustainable metaholic rate, in watts, if 80% of the energy used by the body goes into waste heat?
ADDITIONAL PROBLEMS 58. A CarnOl engine operates bet:\veen the temperatures 7"h = IOD GC and c~ = 20°C. By what factor docs the theoretical efficiency increase if the temperature of the hot reservoir is increased to 550°C?
159.IA 1 500-kW heal. engine operales at 25% erficienc)'. The heat energy expelled at the low temperature is absorbed b)' a stream ofwatcr that enters the cooling coils at 20°C. If 60 L flows across the coils per second, determine lhe increase in temperaw rc of the water. liO. A Carnol engine operates bc[\vcen 100°C and 20°C. How much ice can the engine melt from its exhaust after it has done 5.0 X 10'1 J of work? 61. A substance undergoes [he cyclic process shown in FigIlre P12.61. Work ourpmoccurs along path AB\\'hile work input is required along path Be, and no work is involved ill Lhe constant volume process CA. Energy transfers by heat occur during each proccss involved in the cycle. (a) What is Lhe work output during process AB? (b) How Illuch work input is required during process Be? (c) What is the net energy input Qduring this cycle?
P(aun) 5.00
1.00
A
~B
"'-L..c----,,-L'--- V (lilers) \0.0
50.0
FIGURE P12.61
62. ""hen a gas follows path 123 on the PVdiagram in Figure PI2.62, 418J or energy nows into the system by heat and -167.1 of" work is done 011 til(' gas. (a) What is the change in the internal energy of the system? (b) How much energy Q nows into lhe system if the gas follows path 143? The work done on the g-as along this path is -63.0 J. What net work would be done Oll or by the system if the system folI'
2r----,3 ~----l.4
'----------v FIGURE P12.62
424
Chapter 12
The Laws of Thermodynamics
and the lempcrature at painl @. (b) How much work is done by the gas during this expansion? (c) "Vha{ is the change in internal energy of tile gas? Cd) Find the energ}' transferred to the gas by heat in lhis process.
lowed (c) path 12341 and (d) path 14321? (c) What is Ihe change in internal energy of the system in the processes described in parL~ (c) and (d)? 63 . .-\ lOO-kg steel support rod in a building has a length of 2.0 m ala temperature of 20°C. The rod supportl:l a hanging load of 6000 kg. Find (a) the work done on the rod as the temperature increases to 4(Y'C, (b) the energy Q added La the rod (assume the specific heat of steel is the same as thm for iron), and (e) the change in internal energy of the rod.
64.
68.
Two moles of" 1l10lcculal' hydrogen (l-I~) react with ] mole of molecular ox)'gen (02) to produce 2 molcs of "..· ater (T-1:!O) lOgcther with an energy release or 241.8 kJ/mole of waleI'. Suppose a spherical \'essc1 of radius 0500 III cOlilains 14A moles of 1-1 2 and 7.2 moles of 02 at 20.0°C. (a) \'Vhat is the initial pressure in thc vessel? (b) Wh,lt is lhe inilial internal energy oftllc gas? (c) Suppose a spark ignites the mixture and the gases burn completely into water vapor. How much cnerg} is produced? (d) Find thc temperature and pressure of the slearll, assuming it's an ideal gas. (e) Find lhe mass of stcam
69,
II
jO.
III
mil
An iclc;d gas initially at pressure Pn, volume Vo, and temperature To is taken through the cycle described in Figure P12.64. (a) find the net work done by the gas per qcle ill LenllS of Po and \t~). (b) VVhal is the net energy Qad(lt:d to the system per cycle? (c) Obtain a numerical value ror the net work done pCI' cycle for 1.00 mol of Ras iniLiallr al DOC. (flint: Recall that thc work done by the systcm equals the ;Ire
c
FIGURE P12.64
65. One mole of neon gas is heated from 300 K to 420 K at constant pressure. CaJclllaLC (a) I he energy Q tra nsfelTed to the gas, (b) the Ck\llge in the internal energy of the gas, and (c) t he work done 011 the gas. Nore that neon has a molar specific heal of c:::: 20.79 J/mol' K far a conSlantPI"CSSll re process. 66. Ever)' second at \liagar
167.IA
cylinder containing 10.0 moles of (I monatomic ideal gas expands from @ 10@along the path shown in Figure PI2.67. (a) Find the temperature of the gas at poilll @
Suppose you spend ?JO.O minutes on a stair-climbing machine, climbing at a rate 01'90.0 steps per minute, with each SICP 8.00 inches high. If YOll weigh ]50 Ib and the machine repons thai 600 kca! have been burned at the end of the workout, \Vh
Hydrothermal vcnts deep on the ocean floor SPOLlt ,,,';:lter at temperalllres as high as 570°C. This lemperaI.llfe is belm.. . the boiling point or water because of the immense pressure at thai depth. Because the surrounding ocean temperature is at 11.0°C, an organism could lise thc temperatul-e gradient as a source of energy. (a) Assuming the specific heal. of wmcr under these conditions is 1.0 cal/g' °C. how much energy is rclea"ed when 1.0 liter of WaleI' is cooled rrom 570°C to 4.0°C? (b) What is the maximum usable energy an organism can extract rrom th is energy source? (Assilme the organism has some i\ltcrnal t}'pe or heat engine ,\cling bct\\'een the two lemperature extremes.) (c) Water from these vents contains hydrogen sulfide (H:!S) at a concentration orO.90 mmolc/liter. Oxidal.ioll of 1.0 mole or H:!S produces 310 kJ or energy. lIow much energy is available through H 2S oxid,lI ion of 1.0 L of water?
[1[]An electrical power plant has an overall efficiency of 15%. The pia III is to del iver 150 M \V of electrical power to a city, and its t.urbines usc coal as ruel. The burning coal produces steam <11 190°C. which dri\'es lhe turbines. The steam is condensed into water at 25°C by passing through coils tlMt. arc in contact with river water. (a) How man} metric tons or coal docs the plant consume each da)' (1 metric ton = 1 X 10:1 kg)? (b) \'Vllm is the total cost of the fuel per year if' the de!ivc::r}' price is $8 per metric lOn? (c) I[ the river water is delivered al 20°C, at what minimum rate must it rlow over the cooling coils so that its lemperaltlre docsn \ exceed 25°C? (lv'o[f': The heat of com· bllstion of coal is 7.8 X 103 cal/g.)
Ocean waves combine properties of both transverse and longitudinal waves. With proper balance and timing,
Q
surfer con capture some
of the wave's energy and take it for a ride.
". . . . . _---=;;;.............
_-=---------~-----'"
VIBRATIONS AND WAVES
13.1
Hooke's Law
13.2
Elastic Potential Energy
13.3
Comparing Simple Harmonic Motion with Uniform Circular Motion
13.4
Position, Velocity, and Acceleration as a Function of Time
13.5
Motion of a Pendulum
13.6
Damped Oscillations
13.7
Waves
13.8
Frequency, Amplitude, and Wavelength
13.9
The Speed of Waves on Strings
Periodic motion, from mosses on springs to vibrations of atoms, is one of the most important kinds of physical behavior. In this chapter we take a more detailed look at Hooke's law, where the force is proportional to the displacement, tending to restore objects to some equilibrium position. A large number of physical systems can be successfully modeled with this simple idea, including the vibrations of strings, the swinging of a pendulum, and the propagation of
13.10 Interference of Waves
waves of all kinds. All these physical phenomena involve periodic motion.
13.11
Reflection of Waves
Periodic vibrations con cause disturbances that move through a medium in the form of waves. Many kinds of waves occur in nature, such as sound waves, water waves, seismic waves, and electromagnetic waves. These very different physical phenomena are described by common terms and concepts introduced here.
13.1
HOOKE'S LAW
One of the simplest types of vibrational motion is that of an o~ject attached to a spring, previously discussed in the context of energy in Chapter 5. vVe assume the o~ject moves on a frictionless horizontal surface. If the spring is stretched or compn.:ssed a small distance ;r from its unstretched or equilibriulll position and then released, it exerts a force all t.he object as shmvn in Active foigllre 13.1 (page lI26). prom experirnent the spring force is found to obey the equation [ 13.1]
+-
Hooke's law
where xis the displacement ortlle object from its equilibriulll position (x = 0) and h is a posit ive const.anl called the spring constant. This force law for springs was discovered by Roben Hooke in 1678 and is known as Hooke's law. The value of II is a rneasurc orlhe stiffness oft.he spring, Stiff springs have large hvailles, and soft spri ngs ha\'c small Ii values. The negative sign in Equation 13.1 means thalthe force exerted by the spring is always direClcd olJI/mile the displacement of the object. "Vhen the object is to the right of the equilibrium position, as in Active Figure J3.la, x is positive and F is j
425
426
(a)
Chapter 13
Vibrations and Waves
~f!!:!tLtLx ~x-~
X=
0
F,=O
I
(hi
~tE;I':'--.X
(e)
..J-......_ _
-~x
x~~ X=
0
ACTIVE FIGURE 13.1 The force exerted by a spring 011 all objl:cl varies with the disp!a«:mcm of the object from the C'fjllilibrium position, x = O. (a) \Vhen xis positive (Lhe spring' is ~lrelched). the spring' force is 10 the left. (b) Whell x is zero (the spring is unstretdwc!). die ,<,prillg forct:: is 7el"O. (c) \Vhen x is
I1c~ali\'e
(Lhe spring is compn::sscd), the spring force is 10 the right.
negative. This means that the force is in the negative direction, to the left. When the o~ject is to the left of the equilibrium position, as in Active Figure 13.1c, x is negative and J~ is positive, indicating that the direction of the force is to the right. Of course, when x = 0, as in Active Figure 13.1b, the spring is unstrelched and PI = O. Bel:awse the :sprillg furce always <:IUS £award Lhe equilibrium position, it is sometimes called a restoring force. A restoring force always pushes or puUs the object toward the equiJibrium position. Suppose the object is initially pulled a distance A to the right and released from rest. The force exerted by the spring on the o~ject pulls it back toward the equilibriulll position. As the object moves toward x = 0, the magniLUde of the force decreases (because x decreases) and reaches zero at x = O. The object gains speed as it moves toward the equilibriulll position, however, reaching its maximum speed when x = O. The momentum gained by the object. causes It to overshoot the equilibrium position and compress the spring. As the object moves to the left of [\le equilibrium position (negative x-values), the spring force acLS on it to the right, steadily increasing in strength, and the speed of the object decreases. The object finally comes brieOy to rest at x = -A before accelerating back tm\'ards x = 0 and ultimately returning to the original position at x = A. The process is then repeated, and the object continues to oscillate back and forth ovcr the samc path. This type of motion is called simple harmonic motion. Simple hannonic motion occurs when the net force along the direction of motion obeys Hooke's law-when the nel force is proportional to the displacement from the equilibrium point and is always directed toward the equilibrium point. Not all periodic Illotions over the same path can be classified as simple ha.rmonic motion. A ball being tossed back and forth between a parent. and a child moves repetitively, but the motion isn 'l simple harmonic motion because the force acting on the ball doesn't ulke the form of Hooke's law, Equation 13.1. The motion of an object suspended from a vertical spring is also simple harmonic. In t.his case the force or gravit.y acting on the attached object. stretches the spring until equilibrium is reached and the o~jecl is suspended at rest. By defmition, the equilibrium position of the objeCl is x = O. When the object is moved away from equilibrium by a distmce x and released, a net force aClS lOward the equilibriulll position. Because the net force is proportional lo x, the motion is simple hannonic. The follO\\iing three concepts are important. in discussing any kind of periodic mot.ion: • The amplitude A is the maximum distance of the object frolll its equilibrium position. In t.he absence of friction, an object in simple harmonic motion oscillates bet\veen the positions x = - A and x = + A. • The period Tis the t.ime it takes the object to move through one complete cycle of motion, from x = A to x = ~ A and back t.o x = A. • The frequency f is the number of complete cycles or vibrations per unit of time, and is the reciprocal oflhe period (J~ liT).
EXAMPLE 13.1
Measuring the Spring Constant
Goal Use Newton's second law lOgether with Hooke's law to calculate a spring constant. Problem A common technique used to measure a spring constant is illustrated in Figure 13.2. A spring is hung vertically (Fig. 1~.2a), and an ol~ject of mass 171 is attached to the lower end of the spring and slowl), lowered a distance rllO the equilibrium point (Fig. 13.2b). Find thc value of the spring constant if the spring is displaced by 2.00 em and t.he mass is 0.550 kg.
(, (a)
(hi
(e)
FIGURE 13,2 (Example l3.1) Detcrmining I hc ~pril1g <:onstanl. Thc elongation d of the spring is due to lhe suspended wcight mg. Became I he upward sprillg" force balances the weighl when the sy~lell1 is in equilibrium, it follow.~ Ib.11 11:: mg/d.
13.1
Hooke's Law
427
Strategy This example is an application of Newlon's second law. The spring is stretched by a distance d from its initial position uncleI' the action of" the load mg. The spring force is upvvard, balancing the dmvm-vard force ofgravit}' mgwhen the system is in equilibrium. (Sec Fig. 13.2c.) The suspended mass is in equilibrium, so set the sum of the forces equal to zero. . ..
...
Solution Apply the second law (with constant h:
II
= 0) and solve for the spring
2:,1"= Fg + F' =
k
~
mg
-
d
~
-/fW+ b
kd= 0
(0.550 kg)(9.80 m/s') 2.00 X 10 2 m
=
., 2.70 X 10- N/m
..........
. .....
~
Remarks In this case the spring force is positive, because it's directed upward. Once the mass is pulled down from the equilibrium position and released, it oscillates around the equilibriuln posit.ion,just like the horizontal spring. QUESTION 13.1 If this spring is mounted horizontally, what magnitude force does it exert when stretched from equilibrium by 2.0 cm? EXERCISE 13.1
A spring with constant k = 475 N/m stretches 4.50 cm when an spring. Find the acceleration of gravity in this location. Answer
ol~ject
of mass 25.0 kg is auached to the end of the
0.855111/52 (The location is evidently an asteroid or smallll:1oon.)
The acceleriltion of an object moving \vith simple harmonic motion can be found by using Hooke's law in the equation for Newton's second 1<:1\\1, F= ma. This gives ma=F=-lrx
'i
a = --x 1ll
[13.21
Equation 13.2, iln example of a ha:nnonir osrillalorN/llGfiol1, gives the acceleration as ~ runetion of position. Because the maximum n11uc or x is de filled to be the amplitude A, the acceleral ion ranges over (he values - hA/ III to +kA/11/,. Tn the next section we will find equations for velocity as a function or position and For position as a function of time. QUICK QUIZ 13.1 A block on the end ofa horizontal spring is pulled from equilibriulll at x = 0 to x = A ancl released. Through what total distance does it travel in one full cycle of its mOLion? (il) A/2 (b) A (c) 2A (d) 41\
~ Acceleration in simple
harmonic motion
Tip 13.1 ConstantAcceleration Equations Don't Apply The aCcdcr:llioll (lofa partid~ in simple harmonic motion is 110/ con<;lant: itchang-es, varying Wilh x, so
\\'C
can'l
apply the consLant
acceleration kinemalic equ
QUICK QUIZ 13.2 For a simple harmonic oscillator, which of the following p~lirs 01" vector quantities can't both point in the same direction? (The posit.ion vectOr is the displacement from equilibrium.) (a) position and velocity (b) velocity and acceleration (c) position and acceleration
Calculate forces and accelerations for a horizontal spring system.
Problem A O.350-kg object attached to a spring of force constant 1.30 X 10 2 N/m is free 1.0 move on a frictionless horizontal surface, as in Active Figure 13.1. If the object is released frolll rest at x = 0.100 m. find the force on it and its acceleration at x = 0.100 111, x = 0.0500111, .x: = 0 Ill, X = -0.0500111, and x = -0.100 m. Strategy Substitute given quanlilies into Hooke's law to find the forces, then calculale the accelerations with Newton '5 second law. The alllplitllde A is I he same as the point or release from rest, x = 0.100 111.
428 r"
Chapter 13
Vibrations and Waves
..-
.
"
Solution WI"ite Hooke's force law:
I~ =
Substitute the value for It and lake x = A = 0.1 00
-I<x
I~"", ~
111,
-iIA = - (1.30 X 10' N/m) (0.100 m)
finding the force althat point.: Solve Newton's second law ror the acceleration aL x = A:
-13.0 N (l
and substitute to find
'rna =
F'IlIOlX
I~llax
-13.0 N
?It
0.350 kg
-37.1
a=--~
Repeat the same process for the other (our points, assembling a table:
mis'
Position (m)
Force (N)
Acceleration (m/s 2 )
0.100 0.050
-13.0 -650
-37.1 -18.6
o -0.050 -0.100
o
-.
~
o
+6.50 +13.0
+18.6 +37.1
.
~
Remarks The table above shows LhaL when the initial position is halved, the force and acceleration are also halved. Further, positive values of x give negative values of the force and acceleration, whereas negative values or x give positive values of the rorce and acceleration. As the objectllloves to the left and passes the equilibrium point, the spring force becomes positive (for negative values of x), slmving the object down. QUESTION 13.2 \Alill doubling a given displacement always result in doubling the magninlde or the spring force? Explain. EXERCISE 13.2 For the same spring and mass system, rind the force exerted by thc spring and the position x when the object's aecel· eration is +9.00 Ill/5 2.
Answers
3.15 N, -2.42 em
13.2
ELASTIC POTENTIAL ENERGY
[n this section we review the material covered in Section 4 of Chapter 5. A system of interaCling objects has potential energy associated with the configuration of the system. A compressed spring has potential energy that, when allowed to expand, can do work on an object, transforming spring potential energy into the object's kinetic energy. As an example, Figure 13.3 shows a ball being projectcd [rom a spring-loaded toy gun, where the spring is compressed a distance x. As the gun is Fired, the compressed spring does work on the ball and imparts kinetic energy to it. FIGURE 13.3 A b;111 projcctcd I'rom a spring-loaded gUll. Tbc elastic potemial energy slOred in the spring: is transforllled into thc kinetic cllcrgy of the ball.
Ener b')' = c1asLic
Pt.~ _ _....~:::::;::::;l
-
13.2
Elastic Potential Energy
429
FIGURE 13.4 A block sliding 011 a friclionless hori.r.Olltal ~llrraCl; collidt:.~ with a light:.pring. (a) Initially, ,he lnechaniclI energy is <:Iltin.:ly l,he kinetic energy or the block. (],) Tht.' mechanical energy at ~ome arbitr;lI·y position is lhe . . lllll of lhe kinl:tic eJlerg-y of the block alld the dastic potclltia! energy :stored in the spdng. (c) When lhe block comes to rest, the mechani:::;11 energy is clllirely e1
Recall thm the energy stored in a stretched or compressed spring or some othcr elastic material is callcd clastic potcntial encrgy, PEs. given by [13.3]
of- Elastic potential energy
Recall also that the law of conservat.ion of energy. including buth gravitational and spring potential energy, is givcn by [13.4]
If nonconservative forccs such as friction are present. then the change in mechanical energy must equal the \,Iark done by the nonconservative forces: [13.5] Rotational kinetic energy must be included in buth Equation 13.4 and Equation 13.5 for systems involving torques. As an example of the energy conversions that take place when a spring IS included in a system. consider Figure 13.4. A block of mass In slides on a frictionless horizontal surface with constant velocity V, and collides with a coiled spring. The description that I-ollows is greatly simplified by assuming the spring is very light and therefore has negligible kindic energy. As the spring is compressed, it exerts a force La the left on the block. At maximum compression, the block,comes to rest for just an instant (Fig. 13.4c). The initial tot"l energy in the system (block plus spring) before the collision is the kinetic cnergy of the block. After the block collides \vith the spring and t.he spring is partially compressed, as in Figure 13.4b, the block has kinetic energy ~nll/1 (where v < v.) and the spring has potential energy ~/IX~. When the block stops for an instant at the point of" rnaximuITI compression, the kinetic energy is zero. Because the spring force is conservative and because thcre are no external forces that can do 'Yvork on the s)'stem, the total mechanical energy of the system consisting of the block and spring remai.ns constant. Energy is transformed from the kinetic energy or lhe block to the potential energy stored in the spring. As thc spring expands, the block moves in the opposite direction and regains all it.s initial kinetic energy, as in Figure 13Ad. When an archer pulls back on a bowstring, elastic potential energy is stol"ed in both the bent bow and stretched bo\...· slring (Fig. 13.5). When the arrmv is released, the potential energy stored in the system is transformed into the kinetic energy of the arrow. Devices such as crossbO\'V's and slingshots work the same way,
FIGURE 13.5 Elastic potl:lI1ial energy is storl:rl in this drawll bo\\'.
APPLICATION Archery
430
Chapter 13
Vibrations and Waves
QUICK QUIZ 13.3 When an object moving in sirnple harmonic Illotion is at its maximum displacement from equil ibriulll, which of the following is at a maximum? (a) velocity, (b) acceleration, or (c) kinetic energy.
EXAMPLE 13.3
Sto That Car!
Goal Applyconservatioll of energy and the work-energy theorem with spring and gravitational potential energy. Problem A 13 aOO-;\! car starts at rest and rolls dO\\i"n a hill [rom a height of lO.O 111 (Fig. 13.6). It then moves across a level sud~lCe and collides with a light spring-loaded guardraiL (a) Neglecting any losses due La friClion, and ignoring the roLational kinetic energy of the wheels, find the maximum distance the spring is compressed. Assume a spring constant of 1.0 X IO Ii N/m. (b) Calculate the maxjmulll acceleration of the car after contact ,,\~th the spring, assUIlling no frictional losses. (c) If the spring is compressed by only 0.30 Ill, find lhe change in the mechanical energy due to friction.
Strategy Because friction losses are negleClcd, lise conservation of energy in the form of Eqwllion 13.4 to sake f'or lhe spring displacement in part (a). The initial and final values of the car's kinetic energy are zero, so the initial potential energy of the car-spring-Earth syslem is completely converted to elastic potential energy ........... . , - .. . .
FIGURE 13.6 (EXililiple 13.3) A car st
collides \\'ith a spring-loaded guardrail.
in the spring at the end of the ride. In 1',\1'\ (b) apply Newton's second law, substituting the answer to part (a) for x because the maximum compression will give the maximum acceleration. In part (c) friction is no longer neglected, so use the ,.. 'ork-ent:rgy theorem, Equation 1.3.5. Tht: chang'e in mechanical energy must equal the mechanical energy lost due to friClion. ,
~.
,
'~
Solution (a) Find the maximum spring compression, assuming no energy losses due LO friction.
Apply conservation of mechanical encrg}'. Initially, there is only gravitat.ional potential energy, and at maximum compression of the guardrail, Lhere is only spring potential energy.
(KJi
Solve (or x:
x=
+
PI'"
+
pE,), = (KE
o + mgil + 0 = 2mgh
+ P/~~ +
,
pt;JJ
"
0 + 0 + ,'''-"
2(13000 N)(lO.O m) 1.0XlO"N/m
h
0.51 m
(b) Calculate lhe maximum acceleration of lhe car by the spring, neglecting Friction.
Apply Newton's second law:
'IfI,(J.
Substiwt(; values:
a=
= -/(,x
(LO
~
X
hx m
a =
!<xl'
/(,xg
'/Ilg
w
lOr, N/m)(0.51 m)(9.8m/s') 13 000 N
-380 mis' (c) If the compression of the guardrail is only 0.30 m, find the change in the mechanical energy due to friction. Use the work-eneq:D' theorem:
.........
W = (Kt'+ PEg 1I(
PF) PEg ....\ J - (KF+ ...
a + ~kx')
=
(0 +
~
H1.0 X
W", =
+
- (0
+
mgil
+
pF) ...s I
+ 0)
10" N/m)(0.30f - (13000 N)(lOO Ill)
-8.5 X 10" J
....... , ... ~
13.2
Elastic Potential Energy
431
Remarks The answer to part (b) is about 40 times greaLer than the acceleration of gravity, so \\'e'd hener be wearing our scat belts. Note that the solution didn't requirc calculation of" the velocity of the car. QUESTION 13.3 Truc or False.: In th(' ahsence of f'lwrgy 10ssC's chIC' In friclion, dOllbling the height of the hill doubles thc maximum acceleration delivered by the spring. EXERCISE 13.3 A spring-loaded gun IIres a O.lOO-kg puck along a tabletop. The puck slides up a curved ramp and flies straight up illlo the air. If the spring is displaced 12.0 cm from equilibrium and the spring conStant is 875 N/m, how high does the puck rise, neglecting friction? (b) If instead it only rises to a hcight of 5.00 m because of friction, what is the change in mechanical energy?
Answers
(a) 6.43
III
(b) -1.40.r
In addition to studying the preceding example, it's givcn in Section 5.4.
~l
good idea to review those
Velocity as a Function of Position Conservation of energy provides a simple mcthod of deriving an expression for the velocity of an object undergoing periodic Illotion as a function of position. The ol~ject in question is initially at its l1l
2 _
I
2"ItA - '!mv
t
+
I
,;!/(x
:-- ,\---1
~III (a)
: ,
E=--.L/u\2 ~
'!
I I I
I I
I
~x--1
~;-m-,--_=v_-'
2
,
Solving for v. we get
(h)
v= ::t
-"(" A- - x'")
[13.61
I/!
This expression shows that the o~jeCl's speed is a maximum at x = 0 and is zero at the cxtreme positions x = ::tA. The right side of Equation 13.6 is preceded by the ::t sign bccause the square root of a number can be either positive or negative. If the object in Figure 13.7 is moving to the right, v is positive; if the object. is I1"lOvillg to the left, v is negative.
FIGURE 13.7 (a) An objcu:Ju:lched to:1 sprill~ on a frictionlc:>s surface is 1'c1c:lscd from rest with the spring extended a distance A.Just befol'c the object is released, the total energ}' is the elastic potclItial cncr~y luflJ2. (b) Whell the objccl reaches posilion x. it has kinctic ("nergy mv"/2 and lhe c1a..tic pOlcmial energy has deCI'cased lO
kx'l/2.
_
Apply the time-indepcndent velocity expression, Equation 13.6, to an
Problem A O.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates all a frictionless horiLolltal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 COl. (b) What is the velocity of the object when the displacement is 2.00 em? (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.
E= tkx2+tmif
o~ject-springsystem.
Strategy The tOlal energy of the system call be found most casily at the amplitude x = A, where the kinetic energy is zero. There, the potential energy alone is equal to the total energy. Conservation of energy then yields the speed at x = O. For part (b), obtain the velocity by sub· stituting the given value of x into the lime-independent velocity equation. Using this resuh, the kinetic energy asked for in part (c) can be found by substitution, and the potential energy can be found by substitution into Equation 13.3.
Solution (a) Calculate the lata I encrg'y and maximum speed i r the amplitude is 3.00 em. Substitute x = A = 3.00 em and h = 20.0 N/m into the equation [or the toLal mechanical energy E:
Use conservation of energy with Xi = A and XI = 0 to compute the speed oCthe object
)' X 10 -"-m-
(KE+ PEl! + I'li\·= (KE+ PI'- '+g pe) sli 'f
o+ 0
+ !
~/u\2 = ~mV~lax :.!
-
2J11Vllu"" -
+0+0
9.00 X J 0 -;I J
(
J8.0 X 10-3J 0.190 m/s
0.500 kg (b) Compute the velocity of the displacement is 2.00 em.
ol~iect when
the
Substitute known values directly into Equation 13.6:
= 0:
20.0 _ N/m [( 0.0300 m )'- - ( 0.0200 m )'] 0..000 kg
0:0.141 m/s (c) Compute the kinetic and pOlential energies when the displacemclll is 2.00 em. I " = 'iI ( -) = ,1/IVO.~OO kg( 0.141 m/s )"-
4.9/- X 10- '.j J
SubstitUlc into the equat.ion for kinetic energy:
KE,
Substitute int.o the equation for spring potential energy:
PE, = ~h.\'2 = ~(20.0 N/m)(2.00 X 1O- 2 m)' =
~
4.00 X 1O- 3 .J
~
~
Remark With the given information, it. is impossible to choose between the positive and negative solutions in pan (b). Notice thalthe sum Kl!,' + P/~'f in part (c) equals the total energy E found in part (a), as it should (except lor a small discrepancy due LO rounding). QUESTION 13.4
True or False: Doubling the initial displacement doubles the speed of the object allhe equilibrium point. EXERCISE 13.4
For whal values of' x is the speed of the ol=!.jecl 0.10 m/s? Answer
:!: 2.5!) cm
13.3
COMPARING SIMPLE HARMONIC MOTION WITH UNIFORM CIRCULAR MOTION
'We can better understand and visualize many aspects of simple harmonic motion along a straight line by looking at its relationship to uniform circular motion. Active Figure 13.8 is a top vic\v o[an experimental arrangement that is lIseful for this purpose. A ball is attached to the rim of a turntable of radius A, illuminated [rom the side by a lamp_ \Ve II nd that as the turntable rotates with constant angular speed, the shadow of the ball moves back and forth with simple harmonic motion.
13.3
an
433
Comparing Simple Harmonic Motion with Uniform Circular Motion
This fact can be understood from Equation ]3.6, ,·vhich says that the velocity of" ol~jcctmovingv,"ith simple harmonic Illotion is related to the displacement by
where C is a constant. To see that the shadow also obeys this relation, consider Figure 13.9, ,.. ., hich shQ\.\'s the ball moving with a constant speed V/I in a direction tangent to the circular path. Atl.his instant, the velocity of the ball in the xdirection is given by v = 110 sin 0, or sin
(J
=
Q
•
Ball
~.-/ A
P
TUfIl\.Llblt.:
"
/ ·--73\:,ccn
~
"0
Shadow of ball
From the larger triangle in the figure we can obtain a second expression for sin 0:
Equating the right-hand sides of the tv..·o expressions for sin O. we find the folJowing relationship between the velocity v and the displacement x: v
VA' - x'
ACTIVE FIGURE 13.8 All exp<.'rilllcllLal SCLUp lor demOllstrating the connection between simpk harmonic motion
II or V
110 "- /
= -
A
" 'J v A-'J - x-'J = C ...v /A- x-
;
The velocity of the ball in the x-direction is related to the displacement x in exactly the same way as the velocity of an object undergoing simple harmonic motion. The shadow therefore moves wilh simple harmonic motion. A valuable example of the relationship between simple harmonic motiun and circular motion can be seen in vehicles and machines lhal use the back-and-Iorrh motion ora piston 1.0 create rotational motion in a wheel. Consider the drive wheel of" a locomotive. ]n Figure 13.10, the curved housing at the left contains a piston th(lt moves back and forlh in simple harmonic motion. The piston is connected to an arrangement of' rods that transforms its back-and-fonh motion into rotational motion of the wheels. A similar mechanism in an automobile engine transfurms the back-anel-forth mOlion of the pistons to rotational Illotion of the crankshaft.
,
I I
,, ,
,,
'.
FIGURE 13.9 The ball rot,ul'S \\'ith constant speed Un' The x-component oflhe balrs \'elocity equals lhe pro.jec· lion of tI on the x-a.xi.~.
v
APPLICATION Pistons and Drive Wheels
Period ond Frequency The period T of the shadmv in Active Figure 13.8, which represents lhe time required for one complete trip back and forth, is also the time it takes the ball to make one complete circular trip on the lllrntable. Because the ball moves through the distance 27TA (the circumference of the circle) in the time 7~ the speed V o of the ball around the circular path is ~1TA
"0 = - -
T
and the period is
27TII
T=-~
v"
[13.7]
l11"'lagine that the ball moves from P to Q, a quarter of a revolution, in AClive Figure 13.8. The motion of the shadow is equivalent to the horizontal motjon of an ol~je<.:t on Ihe end ora spring. For this reason, the radius A of the circular motion is the same as the amplinlde A of the simple harmonic motion of the shadow. During the quarter of a cycle shovm, ule shadow moves from a point where the energy of
FIGURE 13.10 The drive ,\'heel mechanism oran otd locomotive.
434
Chapter 13
Vibrations and Waves
the system (ball and spring) is solely elastic potential energy to a point where the energy is solely kinetic energy. By conservat.ion or energy. we have
which can be solved lor A/vo:
Substituting this expression for A/vo in Equation 13.7, we flnclthallhe period is The period of on object-spring system moving with simple harmonic motion -?
[13.8]
Equation 13.8 represents the time required for an object of mass m attached to a spring with spring constant k to complete one cycle of" its motion. The square roOl or the mass is in the numerator, so a large mass v·,rill mean a large period, in agreement with intuition. The square root of the spring constant k is in the denominator, so a large spring constant will yield a small period, again agreeing with intuition. It's also interesting that the period doesn't depend on the amplitude A. The inverse of the period is the frequency of" the lIlotion: I
[13.9]
j~ T
Therefore, the frequency of the periodic motion of a mass on a spring is I Frequency of on object-spring
system -?
j= 27T
{k \j;;
[13.10]
The units of frequency are cycles per second (5- 1), or hertz (Hz). The angular frequencyw IS Angular frequency of on object-
spring system -?
Tip 13.2 Twin Frequencies The JrcqllF.1ICJ givt:s the numhcr of cycles pcr second, whereas the angttlarJrequencygives the number of radians per second. T1lcse L\"O physical cotlcepts are nearly identical and arc linked by the conversion factor 2", fad/cycle.
W = 27Tj=
~ -
i
[l3.11]
1Il
The frequency and angular frequency are actually closely related concepts. The unit of frequency is cycles per second, 'where a cycle may be thought of as a unit of angular measure corresponding to 27T radians, or 360°. Viewed in this way, angular frequency is just a unit conversion or frequency. Radian measure is used for angles mainly because it provides a convenient and natural link between linear and angular quantities. Although an ideal mass-spring system has a period proportional to the square root of t.he object's mass ?n, experiments show that a graph of T2 versus Tn doesn't pass through the origin. This is because the spring itself has a mass. The coils of the spring oscillate just like the object, except the amplitudes are smaller for all coils but the last. For a cylindrical spring, energy arguments can be llsed to show thal the t:Uecliveadditional mass ora light spring is one-third the mass of" the spring. The square of the period is proportional to the total oscillating mass, so a graph of T2 versus total mass (the mass hung on the spring plus the dlective oscillating mass of the spl-ing) would pass through the o,·igin. QUICK QUIZ 13.4 An object of mass m. is attached to a horizontal spring, stretched to a displacement A from equilibrium and released, undergoing harmonic oscillations on a frictionless surface with period 10- The experi~ ment is then repeated with a mass of 4m. What's the new period or oscillation? (a) 2To (b) 70 (c) To/2 (d) 7;'/4
13.3
Comparing Simple Harmonic Motion with Uniform Circular Motion
435
QUICK QUIZ 13.5 Consider the situation in Quick Quiz 13.4. [s the subsequent total Illechan ica] energy of the object with mass 4m (a) greater than, (b) less than, or (c) equal to thc originall.OtaJ mechanical energy?
APPLYING PHYSICS 13.1
BUNGEE JUMPING
A bungee cord can be modeled as a spring. Hyou go bungcejulllping, you will bounce up and down at the end of the elastic cord after your dive off a bridge (Fig. 1:\.11). Suppose you perform a dive and measure the frequency of}'our bouncing. You then move to another bridge, but flnd that the bungee corel is too long {or dives off this bridge. \Nhat possible solutions might be applied? In terms of the original frequency, what is the frequency of vibration associated with the solution? Explanation There are two possible solutions: Make the bungee cord smaller or fold it in half. The laltcr would be the safer of t.he two choices, as we'll see. The force exerted by the bungee cord, modeled as a spring, is proponional to the separation of t.he coils as the spring is extended. First, '·...e extcnd the spring by a given distance and measure t.he distance between coils. 'vVe then cllt1.he spring in half. Trone of the half-springs is now extended by the same distance, t.he coils will be twice as far apart as they were for the complete spring. Therefore, it takes twice as much force to stretch the half-spring through the same displacement, so the half-spring has a spring constant twice t.hat of the complete spri ng. The folded bungee cord can then be modeled as two half-springs in parallel. Each half has a spring constant that. is twice the original spring- constant of' the bungee cord. In addition, an ol~ieet hanging on the folded bungee cord ,,,,ill experience two forces, one fTom each half-spring. As a result., t.he required forcc f'or a given extension
FIGURE 13.11 (Applying Physics 13.1) Bungcejulllpillg from a bridgr-.
will be four times as much as for the original bungee corel. The efTcctive spring constant of the folded bllngee cord is therefore [our times as large as the original spring constant. Because the frequency of oscillation is proportional to t.he square root. of the spring constant, your bouncing frequency all the folded cord will be l1vice what it was on the original cord. This discussion llegleCL'i the fact lhatlhe coils of a spring have an initial separation. It's also imporLantto remember that a shorter coil may lose elasticity more readily, possibly even going beyond the elastic limit. for the material, \vith disastTous reSUll'i. Bungeejumping is dangerous; discreuon is advised!
EXAMPLE 13.5 That Car Needs Shock Absorbers! Goal
Understand the relationships bet\veen period, frequency, and angul(;lr frequency.
Problem A 1.30 X I03-kg car is constructed 011 a frame supported by four springs. Each spring has a spring constant of2.00 X 10 1 N/m. If l.\vo people riding in the car have a combined n1ass of 1.60 X 102 kg, find the frequency of'vibration of the car when it is driven over a pothole in the road. Find also the period and the angular frequency. Assullle the wcight is cvenly distributcd. Strategy Because the weight is evenly distributed, each spring supports one-fourth of the mass. Substitute this value and the spring constant into Equation 13.10 to get the frequency. The reciprocal is the period, and multiplying the frequency by 27T gives the angular frequency.
,..
.
Solution Compute one-quarter of the total mass:
. '" ~ ,;(m,;oc + mp;,J = 365 kg
= HI.30 X 103kg
'-"
+ 1.60 X 10'kg)
436
Vibrations and Waves
Chapter 13
2.00 X 10" N/m
Substitute into Equation 13.lO La lind the frequency:
Remark Solving this problem didn'L require any knowledge of lhe size of the pothole because the frequency doesn't depend on Lhe amplitude of the motion.
QUESTION 13.5 True or False: The frequency of vibration of a heavy vehicle is greater than that of a lighter vehicle, assuming the two vehicles arc supponed by the same set of springs. EXERCISE 13.5 A 45.0-kg boy jumps on a 5.00-kg pogo SLick with spring consLanL3 650 N/m. Find (a) Lhe angular frequency, (b) Lhe rrequency, "nel (c) the period of the boy's motion. Answers
(a) 8.54 rad/s
y
(h) 1.361-1z
w p
, ,,
"
1\ ,/ : \
8 \ \
---4...-L-J.-;'x
,,
I
Xx
()
x
Q ,;
,
'
ACTIVE FIGURE 13.12 A rcfc]'('nce circle. A~ the hall at f> rotale~ in a circle \\,;th uniform angular spcecl, its projection Qalong the x~axis moves with simple harmullic moti()I1.
13.4
(c) 0.735 s
POSITION, VELOCITY, AND ACCELERATION AS A FUNCTION OF TIME
\·Ve can obtain an expression for the position of an ol~ject moving with simple harmonic Illotion as a fUI1Clion of time by returning to the relationship between simple harmonic motion and uniform circular Illotion. Again, consider a ball on the rim of a rotating turntable of radius A, as in Active Figure 13.12. 'We refer {Q the circle made by the ball as the referena! cirde for the Illotion. \Ve assume the lUrntablc revolves at a (;O'nsl.al1t angular speed w. As the ball rotates on the reference circle, the angle 0 made by the line OP with the x-a.xis changes with time. Meam·vhile, the prqjection of P on the x-~lXis, labeled point Q, moves back and forth along the axis with simple harmonic motion. From the right triangle OPQ, we see that cos 8 = xl A. Therefore, the x-coordinate ofthe ball is
x= AcosO Because the ball rOlateS with constant angular speed, it follows that 8 = wt (see Chapler 7), so we have
x
~
A cos (wi)
[13.12]
In one complete revolution, the ball rotates through an angle of21T fad in a time equal to the period T In other '.. .ords, the motion repeats itself every T seconds. Therefore, [13.13]
INhere fis the frequency of the motion. The angular speed of the ball as it moves ,-lround the reference circle is the same as the angular frequency of the projected simple harmonic mOlion. Consequently, Equation ]3.12 can be written x = A cos (2",{/)
[13.]4a]
13.4
Position, Velocity, and Acceleration as a Function of Time
This cosine function represents the position of an object moving with simple harmonic mOL ion as a (uncl.ion of time, and is graphed in Active Figure 13.13a. Because the cosine function varies between 1 and -I, x varies between A and - A. The shape of the graph is called sinusoidal. Active Figul'cs I3.13h and 1~{.13c represent curves for velocity and acceleration as a function of time. To find the equation for the velocity, usc Equations ]3.6 and 13.I4a l.Ogether with the identity cos2 e + sin 2 e = 1, obtaining
v= -Awsin(21Tjl)
[13.14b]
where we have used the fact that w = V!(/ nt. The ::t sign is no longer needed, because sine can Lake both positive and negative values. Deriving an expression for the acceleration involves substituting Equation 13.1421 into Equation 13.2, Newton's second law for springs:
,,= - Aw' cos (21T{I)
x=
x
437
II cosw/
(a)
v
(h)
n
«)
[13.14c] (/=-w'lAcoswt
The detailed steps of these derivations are left as an exercise [or the studelll. Notice that ,\'hen the displaccment x is at a maximum, at x = A 01" X = -A, the velocity is zero, and when x is zero, the magnitude of the velocity is a maximum. Further, when x = + A, its most positive value, the acceleration is a maximum but in the negative x-direction, and when x is at its most negative position, ". = -A, the acceleration has its maximum value in the positive x·dircction. These facts are consistent with our earlier discussion of the points at which TJ and a reach their m::Lximum, minimum, and zero values. The maximum values of the position, velocity, and acceleration are always equal to the magnitude of the expression in front of the trigonometric function in each equation because the largest value or either cosine or sine is 1. Figure 13.14 illustrates one experimental arrang-emellt that demonstrates the si nusoidal nature of simple harmonic motion. An object. connected to a spring has a marking pen attached to it. While the object vibrates venically, a sheet of paper is moved horizontally with constant speed. The pen traces out a sinusoidal pattern.
ACTIVE FIGURE 13.13 (a) Displacement, (b) \'elocil)'. and (c) acceleration \"ersus time far an object 1ll0\
MOl-ion ~
QUICK QUIZ 13.6 If the amplitude ora system moving in simple harmonic motion is doubled, which of the following quantitics r!o(',sn't change? (a) total energy (b) maximum speed (c) maximum acceleration (d) period
EXAMPLE 13.6 Goal
orpaper
~
FIGURE 13.14 An eXI>crimClll
The Vibrating Object-S
Identify' the physical parameters ofa harmonic oscillator from its rnathematical description.
Problem (a) Find the amplitude, frequency, and period of motion ror an object vibrating at the end of a horizontal spring if the equation for its position as a runetion of time is
(b) Find the maximum magnitude of the velocity and acceleration. (c) What are the position, velocity, and acceleration of the o~jeCl aft.er 1.00 s has elapsed?
Strategy In pan (a) the amplitude and frequency can be found by compa,-ing the given equation with the standard form in Equation 13.14a, matching up the numerical values with the corresponding terms in the standard form. In pan (b) the maximum speed will occur when the sine function in Equation 13.14b equals 1 or -1, the extreme values of the sine function (and similarly for the acceleration and the cosine function). In each case, find the magnitude of the expression in front of the trigonometric function. Pan (c) is just a mattel' of substituting values into Equations 13.14a, b, and c.
Solution (a) Find the amplitude, Frequency, and period. ""rile the standard [arm g-ivcn by Equation 13.14a and underneath it ,vrite the given equation:
Equate the factors in frolll of [\lC cosine find the amplitude:
rllnctjon~
La
(I)
x = A cos (21T(1)
(2)
x = (0.250 m) cos
A =
0.2.00 m 1T
(-..!!..t) 8.00
The angular frequency w is the factor ill frollt of I in Equations (1) and (2). Equate these factors:
W= 91TJ= - rad/s = 8.00
Divide w by 2rr to get the frequency J:
J= -
The period Tis the reciprocal of the frequency:
.,,-~ ,- 16.0 s J
w
27T
=
0. 39~. rad/s ,
0.062.0 Hz
(b) Find the maximum magnitudes of the velocity and the acceleration. Calculate the maximum speed from the of the sine function in Equation] 3.14b:
f~1ctor
in front
Calculate the maximum acceleration from the factor in front of the cosine function in Equation ]3.14c:
v""" = Aw = (0.2.00 m) (0.393 rad/s) = 0.0983 Ill/s
(/.m",
0.0386 Illis'
= Aw' = (0.250 m) (0.393 rad/s)' =
(c) Find the position, velocity, and acceleration of the object aflcr 1.00 s.
Substitute I. = LOO s in the given equation:
x = (0.2.00111) cos (0.393 rad) =
Substitute values into the velocity equation:
v
=
0.231
III
-Aw sin (wI.)
-(0.250111)(0.393 rael/s) sin (0.393 rad/s·I.OO s) v=
Substitute values into the acceleration eqllation:
-0.0376 Ill/s
a = -Aw2 cos (wi) ~
a=
............... -
- (0.250 m) (0.39~ rad/s')' cos (0.393 rad/s ·1.00 5) -0.0357 m/s:'! -
.
.......................~
Remarks Tn evaluating the sine or cosine Function, the angle i~ in radians, so you should either set your calculator to evaluate trigonometric functions based on radian measure or convert from radians to degrees. QUESTION 13.6 If the rnass is doubled, is the or (c) ullchanged?
m~lgnitude of
the acceleration 01" the system at any position (a) doubled, (b) halved.
EXERCISE 13.6 If the object-spring system is described by x = (0.330111) cos (1.50/), find (a) the amplitude, the angular frequency, the frequency, and the period, (b) the rnaxjmurn magnitudes or the velocity and acceleration, and (c) the position, veloci[)', ,mc! acceleration \.\'I1en 0.250 s.
,=
Answers (a) II = 0.330 m, w = 1.50 rad/s, J= 0.239I-1z, (c) X = 0.307 m, V = - 0.181 mis, a = - 0.691 mis'
T=
4.19 s
(b)
v"""
= 0.4%
mis, am."
= 0.743
mis'
13.5
439
Motion of a Pendulum
13.5 MOTION OF A PENDULUM A simple pendulum is another mechanical system that exhibits periodic motion. II consists of a small bob of' mass fit slispended by a light string of length L Axed at its upper end, as in Active Figure 13.15. (By a light st.ring, we mean that the string's Illass is assumed to be vcry small cornparcd with the mass of the bob and hence can be ignored.) "Vhen released, the bob swings to and fro over the same path, but is its motion simple harmonic? Answering this question requires examining the resLOrlllg force-the force or gravity-that acts on the pendulum. The pendulum bob moves along a circular arc, rather than back and forth in a straight line. When the oscillations arc small, however, the mOlion of the bob is nearly straight, so Hooke's law may apply approximately, In Active Figure 13.15, s is the displacement of the I)ob from equilibrium along the arc, Hooke's law is F= -Irx, so we arc looking for a similar expression involving ~, l~ = -hs, where l'~ is the force acting in a direclion tangent to the circular arc. From the figure, the restoring force is l'~
Since s = UJ, the equation for
I~
= - mgsin 0
can be written as I~ =
-mgsin
(-,)
,e
: \:T
L:,
I I --~---\
mgsll1 B \
r m
'\ e
--
_IIIKcosB
mg
ACTIVE FIGURE 13.15 A simple pendulum consists ora bob orma-;s 111 suspcll(kd b' a light sLring of IengLh 1.. (I, is the di:.Lalicc from tIll.: pivot to tht, n'nler of ma~s or Lhe bub.) The I"eswring rorce Lhal causcs the pendulum ro lJl1dcrgo simple harmonic Illotion is the componenl orgravilational force tangelllio the path of mati on. mgsin O.
~
This expression isn'tofthe form FI = -ks, so in general. the motion ofa pendultllll is Twl simple harmonic. For small angles less than abotll 15 degrees, however, the <.Ingle 0 measured in radians and the sine of the angle are approximately equal. For example, 0 = 10,0° = 0.175 rad, and sin (10.0°) = 0,174, Therefore, if we restrict the motion to small angles, the approximation sin 0 = 0 is valid, and the reslOring force can be wri tten FI = -mgsin 0 = -lIlg0
Substituting 0 = jjJ., we obtain F, =
-('~)\
This equation follml,1s the general form of Hooke's force 1<1\1,1 j~ = -hs, with h = ·mg! L. vVe are justified in sa)~ng that a pcndulum undergoes simple harmonic Illotion only when it swings back and forth at small amplitudes (or, in this case, small values of 0, so that sin 0 ~ 0), Recallthal for the object-spring system, the angular frequency is given by Equa.ion 13.11: w =
27T/=
~- '
Tip 13.3 Pendulum Motion Is Not Harmonic Remember that rhe pendululll dOPs I/O/exhibit true -;imple harmonic mOlion for an)' angle, If Ihe allgle is less than about 15", the mOLion can be modeWllas approximately simple harmonic.
111
SubSlituting the expression of Il for a pendulum, 'we obtain
w=
mg! L m
=
fi.
"Vi.
This angular frequency can be subslituted into Equation 13.12, \vhich then malh~ crnalically describes the motion ofa pcndulum, The freqltency is just the angular frequency divided by 27T, while the period is the reciprocal of the frequency, or
[13.15]
f- The period of a simple pendulum depends only on L and 9
This equation re"eals the somewhat surprising result thtlt the period of a simple pendulum doesn't depend on the mass, but only on the pendulum's length and on
440
Chapter 13
Vibrations and Waves
ACTIVE FIGURE 13.16 Simple harmollic Illotion for;ll1 ()1~jeCl-spring~}'SlCI1l. and its analogy, the motiun oLl simple pendulum.
•
•
alll;tX
I
~WKKNJ-• •
-...;..-...... \';"ax
----t-------'-----+-I- - x -A
APPLICATION Pendulum Clocks
APPLICATION Use of Pendulum in Prospecting
o
"
the free-fall acceleration. FUrLhermore, the amplitude of the motion isn'l a faclor as long as it's relatively small. The analogy between the motion ora simple pendulum and the object-spring system is illustrated in Active Figure ]3.16. Galileo nrst noted that the period of a pendulum was indcpendcm of its amplitude. He supposedl}' observed this while aucnding chufch services at the cathedral in Pisa. The pendulum he studied was a swinging chandelier that was set in motion when someone bumped it while ligilling candles. Galileo was able LO measure its period by timing the swings \,lith his pulse. The dependence of the period of a pendulum on its length and on the free-fall acceleration allows us to use a pendulum as a timekeeper for a clock. A number of clock designs employ a pendulum, with the length
QUICK QUIZ 13.7 A simple pendulum is suspended from the ceiling 01'£1 stationary elevator, and the period is measured. If the elevator moves with constant velocity, does the period (a) increase, (b) decrease. or (c) remain
13.5
Motion of
0
Pendulum
441
the same? If the elevator accelerat.es upward, docs t.hc pcriod (a) increase, (b) decrease, or (c) rcmain the same?
QUICK QUIZ 13.8 A pendulum clock depends on the period or a pendulum to keep correct time. Suppose a pendulum clock is keeping con-eeL tirnc and then Dennis the tvJenace slides the bob of the pendulum downward 011 the oscillat.ing roc!. Does the clock run (a) slo\\', (b) fasl, or (c) correcLly? QUICK QUIZ 13.9 The period of a simple pendulum is measured to be T on the E.arth. If t.he same pendululll were sct ill motion on the Moon, would its period be (a) less than 1; (b) greater t.han T, or (c) equal to T?
EXAMPLE 13.7 Goal
Measuring the Value of
Detcrmine g [rom pendulum motion.
Using a small pendulum of" length 0.1.71 \"'hat is the value orgin this location?
Problem
lll,
a geoph),!,iicist. counLs 72.0 complet.e swings in a time of 60.0 s.
Strategy First. calculate the period of the pendulum by dividing the lotal time by the number of complete swings. Solve Equation 13.J5 for gand substitute vailles. ..., ..... ~-
Solution
Calculate the period by dividing the total elapsed time by I he number of complete oscillations:
r
2.0
# of" oscillations
T~
Solve Equation 13.15 for gand substilllte values:
60.0 s = 0.833 s - = -7-
time
~
f;,.
27T\O'
->
~
.'1
,)
L g
1~=4iT--
(39.5)(0.171 Ill) (0.833
sf
\1.73
mis'
.....
lrrrr... • • • • • •
Remark
.\1easuring- such a vibration is a good way of determ ining the local value of' the accelcration of gravity.
QUESTION 13.7 True or False: A simplc pcndulum of length 0.50 length 1.0 m.
III
has a larger frequency of vibration than a simple pendulum of
EXERCISE 13.7 \"'hat would be t.he period of the 0.17t-m pendulum on the .\100n where the acceleration of gra\~t)' is 1.62 m/s~? l
Answer
2.04 s
The Physical Pendulum The simple pendulum discussed thus far consists of" a mass attached to a string. A pendulum, hm...ever, can be made from an ol~ject or an)' shape. The general case is called the physilXll jJfmduJum. In Figure 13.17 a rigid object is pivoted at point 0, which is a distance L from the oqjeCl's center of" mass. The center of Illass oscillates along a circular are, just like the simple pendulum. The period ofa physical pendululll is given by T= 27T
~ mgl. /
l'ivol
[13.16]
where I is the objecl's moment of" inertia and m is the object's mass. As a check, notice thal in the special case of a simple pendulum with an arm of" length J. and
mg FIGURE 13.17 pivoted at 0.
A physical pendulum
442
Chapter 13
Vibrations and Waves
negligible mass, the moment ofincnia is 1= 1llL2. Substituting into Equation 13.l6 results in
7' = 27T
ml} mgL
= 27T ~ g
'which is the correct period for a simple pendulum.
13.6
APPLICATION Shock Absorbers
Oil or other viscous fluid
I
Piston with holes
(a)
. . Shock absorbe r Coil spring ~_
DAMPED OSCILLATIONS
The vibrating motions we have disclIssed so far have taken place in ideal systems that oscillale indefinitely' uncleI' the action of a linear restoring force. In all real mechanical systems. (orces of friCLion retard the motion. so the systems don't oscillate indefin itely. The friction reduces the mechanical energy of the system as time passes, and the motion is said to be damped. Shock absorbers in automobiles (Fig. 13.18) are one practical application of damped motion. A shock absorber consists of a piston moving through a liquid such as oil. The upper part of the shock absorber is firmly attached to the body or the car. vVhen the car travels over a bump in the road, holes in the piston allow it to move lip and down in the fluid in a damped fashion. Damped mOlion varies with the fluid used. For example. if the fluid has a relatively low viscosity. the vibrating motion is preserved but the amplitude of vibration decreases in time and the motion ultimately ceases. This process is knO\..·n as underdamped oscillation. The position vs. time curve for an object undergoing such oscillation appears in Active Figure 13.19. Figure 13.20 compares three types of damped motion, with curve (a) representing underdamped oscillation. ffthe fluid viscosity is increased, the object returns rapidly to equilibrium after it's released and doesn't oscillate. Tn this case the s>'stem is said to be critir:ally dmnjJed. and is shown as curve (b) in Figure ]3.20. The piston returns to the equilibrium position in the shortest time possible \vithout once overshooting the equilibrium position. [f the viscosit}, is made greater still, the system is said to be overdamped. In this case the piston relUrns to equilibrium without ever passing through the equilibrium point, but the time required to reach equilibrium is great.er than in critical damping, as illustrated by curve (c) in Figure 13.20. To make automobiles more comfortable to ride in, shock absorbers are designcdLO be slightly underdamped. This can be demonstrated by a sharp downward push on the hood of a car. After lhe applied force is removed, the body of the car oscillates a few times about the equilibrium position before returning to its fixed posir.ion .
x
(b)
FIGURE 13.18 (a) A shock absorher consists ora piston oscillating in a chamher filled with oil. As the piston oscillates, the oil is sqllccl.ed through holes between the pistoll ::tnd the chamber, Glllsing a damping of rhe piston's oscillations. (bl One lype of aulomotilT: suspcnsioll s}'stcm. in which a shock ahsorber is placed inside a coil spring at each IdH:eL
ACTIVE FIGURE 13.19 A graph ordispbccll1elll verSllS lime.; I{ll" all llnderdarnpcd oscillator. Note Ihe.; decrease in illllplitude with limt'.
FIGURE 13.20 l'lmsofdisplace· IlIcnt vcrsus lime for (a) an llnderdamped oscillator. (b) a critically damped oscillator, and (c) an overdampcd oscillalor.
13.7
Waves
13.7 WAVES The world is full of'tvavcs: sound waves, waves 011 a sl1'ing, scism ic waves, and electromagnetic waves, such as visible light, radio waves, television signals, and x-rays. All these waves have "lS their source a vibrating object, so we can apply the concepts or-simple harmonic motion in describing them. In the case of sound \\:aves, the vibrations that produce \vaves arise frol11 sources such as a person's vocal chords or a plucked guitar string. The vibrations of" electrons in an antenna produce radio or television waves, and the simple upand-down motion of a hand can produce a wave on a string. Certain concepts are common to all \\;ayes, regardless of their naLUrc. In the remainder orthis chapter, we focus our attention on the general propenies of waves. In later chapters we will sLUdy specific types of" waves, such as sound waves and electromagnetic \,·:aves.
What Is a Wave? \'''hcn you drop a pebble inl.O a pool of water, t.he disturbance produces \vater waves, which move away from the point where the pebble entered the water. A leaf noming near the disturb(lJ1ce moves up and down and back and forth abollt its origi nal position, but doesn't undergo any net displacement attrihutable 1.0 the disturbance. This means that the water wave (or disturbance) moves from one place to another, butlh(1 waler isn't mrri(1d with if. When we observe a \vater wave, we see a rearrangement of the water's surface. Without the water, there wouldn't be a wave. Similarly, a wave traveling on a string wouldn't exist without the string. Sound \v;-wes travel through air as a result of" pressure variations from point to point. Therefore, we can consider a \"lave to be lhe motion ofa dis/lIrbanCf. In Chapter 21 we discuss electromagnetic waves, which don't require a medium. The mechanical \'I.'aves discllssed in this chapt.er require (J) some source of disturbance, (2) a medium that can be diswrbed, and (3) some physical connection or mechanism through which adjacent portions of the medium can innuence each other. All waves carry energy and momentum. The amount of energy transmitted through a medium and the mechanism responsible for the transport of energy differ from case to case, The energy carried by ocean waves during a storm, [or example, is much greater than the energy carried by a sound wave generated by a single human voice.
APPLYING PHYSICS 13.2
BURYING BOND
At one point in On !-Ier /\!lajesty\ Secret Sp-rvirt', aJames Bond film ["rom the 1960s, Bond was escaping on skis. Hc had a good lead and was a hard-to-hit moving target. There was no point in \..· asting bullets shooting at him, so why' did the bad guys open fire? Explanation These misguided gentlemen had a good understanding of the physics or waves. An impulsi\'e sound, like a gunshot, can calise an
acoustical disturbance that propagates through the air. Hit impacts a ledge of snow that is ready to break free, an avalanche can result. Such a disaster occurred in 1916 during World \Var I when Austrian soldiers in the Alps were smothered by an avalanche caused by cannon fire. So the bad guys, who have never been able to hit Bond with a bullet, decided to use the sound ofgunrire to start an avalanche.
Types of Waves One or the simplest ways to demonstrate \'V'ave Illolion is to flip one end ofa long string that is under tension and has its opposite end fixed. as in figure 13,21 (page 444). The bump (called a pulse) travels to the right Wilh a definite speed. A dis~ lu rbance of th is type is called a traveling wave. The figure shows the shape of the string at three closely spaced times.
443
444
Chapter 13
Vibrations and Waves
FIGURE 13.21 A W;l\'C pulse traveling along a sm':lched ~Iring. The shape or LltC pulse is approximatch ullchallgcd as it tra\'d~.
As such a ,',lave pulse travels along lhc string, each segment of the string that is disturbed moves in a direction perpendicular to the wave motion. Figure 13.22 illustrates this poilu for a particular tin)' segment P. The string never moves in the direction of the 'wave. A traveling wave in which the panicles of the dis· turbed rnedillm movf' in ;:J dirr.c:rion perpf'ndic.ular 1.0 the "\lave velocity is called a transverse wave. Figure 13.23a illustrates the formation of transverse waves on a long spri ng. I n another class of waves, called longitudinal waves, the elements of the medium undergo displacements parallel to the direction of wave motion. Sound waves in air are longitudinal. Their disturbance corresponds to a series of high- and 1m,,·· pressure regions that may travc1through air or through any material medium with a certain speed. A longitudinal pulse can easily be produced in a stretched spring, as in Figure "13.23b. The free end is pumped back and forth along the length of the spring. This action produces compressed and stretched regions of the coil that travel along the spring, parallel to the \.. . ave motion. ""aves need not be purely transverse or purely longitudinal: ocean waves exhibit a superposition of both types. 'Vhen an ocean wave encounters a cork, the cork executes a circular motion, going up and dmvn ,,,,,hile going fanvard and back. Another type of wave, called a soliton, consists of a solitary wave front that propagates in isolation. Ordinary water 'Naves generally spread Ollt and dissipate, but solitons tend to maintain their form. The study of solitons began in ]849, when Scottish cngineerJohn Scott Russell noticed a solitary wave leaving the turbulence in front of a barge and propagating forward allan its own. The wave maintained its shape and traveled down a canal at about 10 rni/h. Russell chased the wave t\,..o miles on horseback before losing it. Only in the ]9605 did scientists take solitons seriously; they are now \videly used to model physical phenomena, from elcmen tar)' panicles to the Giant Red Spot ofJupiter. 4
FIGURE 13.22 A pulse IraH:,ling on a SlI'cLChcd sirilll; is a tr;\1lsvcrsc waH:". :\ 11)' element I~ on the rope mon~s
(brlle arrows) in a direction perp(:lJ(licnl
Picture of a Wave Active Figure 13.24 shows the curved shape ofa vibrating string. This pauern is a sinusoidal curve, the same as in simple harmonic mOlion. The brown curve can be thought of" as a snapshot of a traveling wave taken at some inSL
J'
-Tf---A:--\+--f'\,x Compressed
COlllpl'esscd
1=0
ACTIVE FIGURE 13.24 .r\ one·dimcllsiOlI;IJ
.~i!lllsojdal 1I'
traveling [0 lhe right with a speed I'. The brown curve is a snap~llOl of Ihe \,'a\"t~ at 1= O. and the bllH' cune is anOlhcr snapshot:1t some later [ilm.: I.
• •
Strelched (b) Longitudinal wavc
FIGURE 13.23 (a) A transvcrse wa\'e is St:1 tip ill a spring br mm·ing one cnd of the spring perpendicular to ils length. (b) :\ longitudinal pulse along a streIch cd spring. The displacement ofLhe coils is in lhe direction ol"lhe wave motiun. FIll" theslartinl-{ Illotion describerl ill the lext, Ihe cOlllprc~s(;d region is followed by a stretched region.
13.8
Frequency, Amplitude, and Wavelength
The same v.laverorm can be used La describe a longitudinal wave, even though no up-and~down motion is taking place. Consider a longitudinal wave traveling on a spring. Figure 13.25£1 is a snapshot of this wave at some instant, and Figure 13.25b shows the sinusoidal curve that represents the wave. Points where the coils of the spring ;'IIY" cnrnprpsscrl cOITPspnnd tn the crests of Ihe waveform, and stn~lThf"d I'egions correspond La troughs. The type of wave represented by the curve ill Figure 13.25b is often called a density w(wcor !Jressure wave, because the crests, where the spring coils arc compressed, are regions of high density, and the troughs, where the coils are stretched, are regions of low density. Sound 'waves are longitudinal waves, propagating as a series of high- and low-density regions.
13.8
~(a)
11
(b)
FIGURE 13.25 (a) A longitudinal W;l\'e on a spring. (b) The cresL~ of the waveform correspond to compressed regions of the spring. and the troughs correspond to strelchcd regions of the spring.
A
FREQUENCY, AMPLITUDE, AND WAVELENGTH
Active Figure 13.26 illustrates a method or producing a continuous wave or a steady stream of pulses on a very long string. One end of the string is connected to ,,1 blade that is set vibrating, As the bbde oscillates vertically with simple harmonic motion, a traveling wave moving to the right is SCI. up in the string. Active Figure 13.26 consists of views of the wave at intervals of one-quarter of a period. Note that each small segment of the string, such as P, oscillates vertically in the y.-direction with simple harmonic motion. That must be the case because each segment follows the simple harmonic motion of the blade. Evel)' segment of the string can therefore be treated as a simple harmonic oscillator vihrating with the same frequency as t.he blade that drives the string. The frequencies of the waves studied in this course will range from rather low values for waves on strings and waves on water, to values for sound waves between 20 Hz and 20000 Hz (recall thaL 1 Hz = ] s-'), to much higher frequencies for electromagnetic waves. These waves have different physical sources, but can be described with the same cOnCepl"i. The horizontal dashed line in Active Figure 13.26 represents the position of the string when no wave is present. The maximum distance the string moves above or below this equilibrium value is called the amplitude A of the wave. For the waves we work with, the ampliwcles at the crest and the trough will be identical. AClive Figure 13.26b illusu-ates another characteristic of a wavc. The horizontal arrows show the distance between t\lI'D successive points that behave identically. This distance is called the wavelength A (the Greek letter lambda). \¥e can lise these definitions to derive an expression fur the speed of a wave. We st.art with the defining equation for the wave speed v: ~x
v=~l
Vibrming blade
p (d)
ACTIVE FIGURE 13.26 One medwr! for producing traveling waves Oil a continuous string. The left end of the strillg is connected to a blade that is set vihrating. E\!ery part oflhe strillg, such as point P, oscillates venically with simple harmonic motion.
The wave speed is the speed at which a particular part of the wave-say, a creslmoves through the medium. A wave advances a distance of one wavelength in a time interval equal 1.0 one period of the vibralion. Taking.6.x = i\ and at = 7~ we see that A
v= -
T
Because the frequency is the reciprocal of the period, we have v = fA
[13.]7]
This important general equation applies to many different types of waves, such as sound waves and electromagnetic waves.
445
f- Wave speed
446
Chapter 13
EXAMPLE 13.8 Goal
Vibrations and Waves
A Traveling Wave
Obtain information about a wave direcLly from its graph.
Problem A w~l\'e traveling in the positive x-direction is pictured in pigurc 13.27a. Find the ampliwdc, wavelength, speed, and period of the ",,'ave if it has a frequency of 8.00 Hz. In Fig-
ure ]3.27a, D.X
=
40.0 em and 6.)'
=
15.0
y
CIll.
)'
++--"'------,I-- x
Strategy The amplitude and wavelength can be read directly from the figure: The maximum vertical displacement is the amplitude, and the distance from one crest Lo Lhe next is the \vaVelenglh. Multiplying lhe wavelength by the frequency gives the speed, whereas the period is the reciprocal of the frequency.
----I:--¥----l--x
(h)
(aj
FIGURE 13.27
(a) (Example 13.8) (b) (Exercise ]3.8)
.
,.-' ...
~
Solution
= b.y ~ 15.0 cm = 0.150 m
The maximum wave di.splacement is the arnplitude A:
A
The distance from crest to crest is the wavelength:
A = b.x = 40.0 cm = 0.400 m
Muiliply the wavelength by the frequency to get the speed:
v
= JA = (8.00 Hz) (0.400 m)
Take the reciprocal of the frequency to get the period:
T
=J
.....
I
~
3.20 m/s
1
- - s = 0.125 s 8.00
.
~
Remark It's important not to confuse the wave \-\lith the medium it travels in. A wave is energy transmitted through a medium; some \,,raves, such as light waves, don't require a mediulll.
QUESTION 13.8 Is the frequency of a wavc affected by the wave's amplitude? EXERCISE 13.8 A wave traveling in the positive x-direction is pictured in Figure 13.27b. Find the (lmplitude, wavelength, speed, and period of the wave if it has a frequency of 15.0 Hz. In the figure, D.x = 72.0 cm and 6.)' = 25.0 em.
Answers
A = 0.25
EXAMPLE 13.9 Goal
In,
A = 0.720
In,
v = 10.8 mis, T= 0.0667 s
Sound and Li ht
Perform elememary calculations using speed, wavelength, and frequency.
Problem A wave has a wavelength of 3.00 m. Calculate the frequency of the wave if it is (a) a sound wave and (b) a light ·wave. Take the specd of" sound as 343 Ill/s and thc speed of light as 3.00 X lOR m/s.
,..-' Solution (a) Find the frequency ora sound wave with A = 3.00
.. .. . .....,
Ill.
Soh'e Equation 3.17 for the r'requency and substitute:
(1 )
v A
I~-~
343 m/s 3.00 m
Il4 Hz
(b) Find the frequency ora light ,"vave with A = 3.00 Ill. Substitute into Equation (1), using the speed of light lor c: Iil...• • • • • •
J= -c
A
=
3.00
X
to" m/s
.'-'-'_~~
3.00 m
1.00 X 108 Hz
............................~
Remark The same equation can be uSl'd to nnd the frequency in each case, despite the great difference between the physical phenomena. Notice hmv nlUch hll'gcr freC(uencies of light waves are than frequencies of sound waves.
13.9 The Speed of Waves on Strings
447
QUESTION 13.9 A wave in one medium encounters a ne\v mediulll and enters it. Which of the following wave properties will be affected in this process? (a) wavelength (b) frequency (c) speed EXERCISE 13.9 (a) Find the wavelength of an electromagnetic wave with frequency 9.00 GHz = 9.00 X 10~ Hz (C = giga = 10!.!), which is in the microwave range. (b) Find the speed ora sound wave in an unknown fluid medium if a frequency of 567 Hz has a \'\-'avclength 01"2.50 Ill.
Answers
13.9
(a) 0.033 3 III
(b) 1.42 X 10" llI!s
THE SPEED OF WAVES ON STRINGS
In this section we focus our aLlention on the speed of a transverse wave on a stretched string. For a vibrating string, there are two speeds to consider. One is the speed of the physical string that vibrates up and down, transverse to the string, in lhe y-direction. The other is the wave sjJeed, which is the rate at which the disturbance propag-ates along the length of the string in the x-clircClion. \tVe wish to find an expression for the wave speed. If a horizontal string under tension is pulled vertically and released, it starL<; a1 its maximum displacement, )1 = A, and takes a certain amount of time to go to )1 = - A and back to A again. This amount of time is the peliod of the waye, and is the same as the Lirnc needed for the \-"ave to advance h01izonlally by one wayciength. Dividing the wavelength by the period of one transverse oscillation gives the wave speed. For a fixed wavelength, a su'ing under greater tension F has a greater wave speed because the period of vibration is shoner, and the wave advances one wavelength during one period. It also makes sense th~ll. a string \\~th greater mass per unit length, fL, vibrates more slm\'ly, leading LO a longer period and a slowel" wave speed. The wave speed is given by [13.18] where Fis the tension in the string and 1.L is the mass of the string per unit length, called the bum!" density. From Equation ]3.18, it's clear that a lal"ger tension F results in a larger \\lave speed, whereas a larger linear density J.l.- gives a slower wave speed, as expected. According to Equation 13.l8, the propagation speed ofa mechanical wave, such as a \..'ave on a string, depends only on the properties of the string through which 1he disturbance travels. 1L doesn't depend on the amplitude of the vibration. This turns out to be generally true ofwavcs in various media, A proof of Equation 13.18 requires calcullis. but dimensional analysis can easily verify that the expression is dimensionally correcL Note that the dimensions of F are ML/T:!, and the dimensions of J.l are MIL The dimensions of FIJ.l arc therefore L:!/T:!, so those of \IFIlL arc LIT, giving the dimensions of speed. No other combination of Pand f.L is dimensionally correct, so in the case in \,,..hich the tension and mass density are the only relevant physical faclOrs, we have verified Equation \3.18 up to an overall constant. According to Equation I3.l8, we can increase the speed of a wave on a stretched string by inCl"casing the tension in the string. Increasing the lIlass per unitlenglh, on the other hand, decreases the wave speed. These physical facts lie behind the metallic windings on the bass strings of pianos and guitars. The windings increase the !nass per unit length, fL, decreasing the waye speed and hence the frequency, resulting in a lower tone. Tuning a suing to a desired frequency is a simple matter of changing the tension in the string'.
APPLICATION Bass Guitar Strings
448
Chapter 13
Vibrations and Waves
EXAMPLE 13.10 A Pulse Traveling on a String Goal
Calcu!;:nc the speed of a wave on a string.
Problem A uniform sIring has a mass M of' 0.030 a kg and a length L of 6.00 m. Tension is maintained in the string by sllspending a block of mass JII = 2.00 kg from one end (Fig. 13.28). (a) Find the speed of ~l transverse wave pulse on this string. (b) Find the time it takes the pulse to travel [rom the wall LO lhe pulley. Neglect the mass of the hanging part of t.he string. Strategy The (ension Fcan be obtained from Ncv.'ton's second Imv for equilibrium applied to the block, and the mass per Ilnit length of the string is J.L = I\;l/ L. '·Vith these quantities, the speed at" the transverse pulse can be found by substitution into Equation 13.18. Pan (b) requires the formula d = vi. ~
.
,
lti=
5.00 '"
--------i r\.00 m
L
2.00 kg FIGURE 13.28 (Exal1lpk 13.10) 1'h<: tellsion Fin the Hrillg is mainlail1<:d hy lite ~lIspelided block. The W,I\'t: spccd is l{ivell b}' the cxprcssioll v =
Solution (a) Find the speed of the wave pulse. Apply the second law to the block: the tension F is equal and opposite to the force of gravity.
L,F= F- mg= 0
---7
F= mg
Substitute expressions for Fand f-t into Equation .13.18:
!
(2.00 kg)(9.80 m/s') J96 N (0.0300 kg)/(6.00 m) = \ 0.00500 kglm 62.6 m/s (b) Find the time it takes the pulse to travel from the wall LO the pulley. Solve the distance formula for time:
I =
....
d 5.00 In = v 62.6 m/s
~
0.0799 s
.
...
Remark Don't confuse the speed of the wave on the string with the speed oFLhe sound wave produced by the vibrating string. (See Chapter 14.) QUESTION 13.10
If the mass of the block is qlladruplcd, what happens to the speed of the \<,Iave? EXERCISE 13.10
To \,,:hat tension must a string with mass 0.010 0 kg and length 2.50 speed of 125 m/s? Answer
III
be tightened so t.hat \""aves will u"avcl on it at. a
62.5 N
13.10
INTERFERENCE OF WAVES
)'. 1 any interesting wave phenomena in nalLlre require n\'o or more waves passing through the same region or space at the same time. Two traveling waves can meet and pass through each ot.her without being destroyed or even altered. For inst.ance, when two pebbles are thrmvn into a pond, the exp~lJ1ding circular \·...aves don't destroy each other. In faet, the ripples pass through each other. Likewise, ,..,hen sound waves [rom two sources move through air, they pass through each other. In the region of" overlap, the resultant. wave is found by adding the displacements of the individual waves. For stich analyses, the superposition principle applies:
13.11
V\'hen two or more traveling waves encounter each other while moving through a medium, the resultant V'lave is found by adding together the displacements of the individual V'laves point by point. Experiments show that the superposition principle is valid only ,·vhen the individual waves have small amplitudes of displacelll en I, which is an assumption we make in all our examples. Figures 13.29£1 and 13.2gb shm'\-' two waves of the same amplitude and rrequency. IT at some instant of time these two waves were traveling through the same regiun of space, the resultant wave at that instant ,.. .ould have a shape like that shown in Figure 13.29c. For example, suppose the 'waves are water waves ofamplilude 1 m. At the instant they overlap so that crest meets crest and trough meets trough, the resultant 'wave has an amplitude of2 m. \"'aves coming together like that are said to be in jJ/lase and to exhibit constructive interference. Figures 13.30a and 13.30b show two similar 'Naves. In this case, however, the crest of one coincides with the trough of the other, so one wave is inverfl'{! relative to the other. The resultant wave, shown in Figure 13.30c, is seen La be a state of complete cancellat.ion. If these were ,valeI' ,,'aves coming together, one of the waves would exert an upward force all an individual drop of water at the same instant the other wave was exerting a downward force. The resuIL would be no motion of the water at all. In such a situation the two waves are said to be l80° out of phase and to exhibit destructive interference. Figure ]3.31 illustrates the interference of V·later waves produced by drops of"watcr falling into a pOllel. Active Figure 13.32 shows constructive interference in two pulses moving toward each other along a stretched string; Active Figure 13.33 (page 450) shows destructive interference in two pulses. Notice in both figures that when the two pulses separate, their shapes are unchanged, as if they had never met!
13.11
449
Reflection of Waves
~ Superposition principle
[rv\f\!
(e)
FIGURE 13.29 Constructive imerference. [[two waves having lhe same frequency aud amplitude are in phase. as in (a) and (b), the resultanI wave when the)' combine (c) has the same frequency as the individual waves, but twice their amplitude.
t/\JV"I (a)
REFLECTION OF WAVES
In our discussion so I~u, we have assumed waves could travel indefinitely withollt striking anything. Oft.en, such conditions arc not realized in practice. \Vhenever ,,1 traveli ng wave reaches a boundary, part or all of the wave is reflected. For example, consider "I pulse traveling on a string thm is fixed at one end (Active Fig. 13.34, page 450). \o\'hen the pulse reaches the wall, it is reOected. Note that the reOecred pulse is inverted. This can be explained as follows: \"'hen the pulse meets the wall, the string exerts an upward force on the wall. According to Newton's third law, the wall must exert an equal and opposite (downward) reaction force on the string. This downward force causes the pulse 1.0 invert all reflection. Now suppose the pulse arrives at the string's end, and the end is attached to a ring or negligible mass that is free to slide along the post without friction (Active
1-
(a)a=?~ (b)~
I
I
(e)
FIGURE 13.30 encl'. \Vhcn
twO
Destructive imerferwaves with the same
frequency and ampliLUde are 180" out of phase, as in (a) and (b), the result when the)' combine (c) is complete canccllal ion.
(d)~ !~
-
(l')~~~
(C)~-jJ/f\~,~ ACTIVE FIGURE 13.32 Two wave pulses traveling on a sln:lchcd string ill opposito.: dirccl ions pass through each olher. When the PLtI.~cs ovcrlap. as ill (b). (c), and (d). the nel.displaccmcnt of the strillg equ:lls the SLl11l of the displacclllellt~ produced by each pulse.
FIGURE 13.31 Interference pal/crus produced by olllward-spreading waves from marlY drops of liquid falling into a body of water.
Chapter 13
450
Vibrations and Waves
---(b)J~ --
ACTIVE FIGURE 13.33 Two wave pulses Lr;wc1ing in opposite directions wil h displacemellts lhal are im'erted relative to each other. ""hell the twO overlap, as ill (c), their displacements SubtraCl from c:1Ch allier.
(a)~
--
(d)~
-- --
(e)~
(e)~
Fig. 13.35). Again the pulse is reflected, but this lime it is not inverted. On reaching the post, the pulse exerts a force on the ring, causing it to accelerate upward. The ring is then returned to its original position by the downward component of
Incident
pulse
--..
the tension in the slI"ing. An alternate method of showing that a pulse is reflected without inversion when it strikes a free end of a string is to send the pulse down a string hanging vertically. \,Vhen the pulse hits the free end, it's reflected without inversion, just as is the pulse in Active Figure 13.35. Finally, ,""hen a pulse reaches a boundary, it's partly reflected and partly transmitted past the boundary into the new medium. This effect is easy to observe in the case of L\VO ropes of different density joined at some boundary.
~I
':"(b7')===~
====#1 (e)
~=d=)===~ ACTIVE FIGURE 13.35 Thc rdlcetioll of a traveling wa\'e at the free end of a stretched string. in this case the reflected pulse is not invenccl.
~Clc~;~1 ......-
pulse
ACTIVE FIGURE 13.34 The rcncctioll of a Ira\"clilll{ wave at the fixed end of
=7tf
= = (b)
Reflected
==~~Ulse,
==~o:!:J
(d)
SUMMARY 13.1
Hooke's Low
Simple harmonic motion occurs when the net fOI"Ce on an object along the direction of motion is proportional to the object's displacement and in the opposite direction: F,
~
-h"
[13.1]
This equation is called Hooke's law. The time required for one complete vibration is called the period of the motion. The reciprocal of the period is the frequency of the motion, which is the number of oscillations per second. When an object moves with sil'nple harmonic motion, its acceleration as a function of position is (J
13.2
=
k
--x m
[13.2]
Elastic Potential Energy
The energy stored in a stretched or compressed spring or in some other elastic millerial is called elastic potential energy: [13.3]
The velocity of an object as a function of position, when [he object is moving with simple harmonic motion, is TJ
= ±
'V/1:-(11' m -
.<')
[13.6]
13.4 Position, Velocity, and Acceleration as a Function of Time The period of' an objen of mass 111 moving with simple harmonic Illation while attached to a spring of spring constant k is
T= 27T
r;;; YT
[13.8]
",'here Tis independent of the amplitude A. The frequency of an object-spring system is J= l/'T The angular frequency w of the system in rad/s is w ~ 2",/=
,f-f;'
-. '/fI
[13.111
Multiple-Choice Questions
When an object is moving with simple harmonic mOlion, the position, velocity, and acceleration of the object as a function of time are given by
.v = t\ cos (2r.P)
[13.14a]
v = -"-\w sin (2-rrft)
[13.14b]
a = -Aw:? cos (2r.Jl)
[13.14e]
451
This relationship holds for a wide variety of different waves. 13.9
The Speed of Waves on Strings
TIlf' SPf'f'rl of;'! wavf" tnl\'f'ling on ,I sIrf'lchf'rl ,'I'-jng of mass per unit length J1. and under tension Fis
[13.18] 13.5 Motion of 0 Pendulum A simple pendulum of length L moves v",ith simple harmonic motion for small angular displacements from the vertical, with a period of T=
2n~~
[13.15]
13.7 Waves In a transverse wave the elements of the medium move in a direction perpendicular to the direction of the wave. An example is a wave on a stretched string. In a longitudinal wave the elements of the medium move parallel to the direction of the wave velocity. An example is a sound ' . . ave.
13.10 Interference of Waves The superposition principle states that if two or more traveling ,..:aves are moving through a medium, the resultant wave is found by adding the individual waves together point by point. \,Vhen waves meet crest to crest and I rough to trough, they undergo constructive interference. \,Vhen crest meets trough, the waves undergo destructive interference. 13.11 Reflection of Waves \"lhen a wave pulse reflects from a rigid boundary, the pulse is inverted. \'''hen the boundary is free, the reflected pulse is Ilot inverted.
13.8 Frequency, Amplitude, and Wavelength The relationship of the speed, wavelength, and frequency of a wave is v~p.
[13.17]
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE·CHOICE QUESTIONS 1. The distance between the crest of a water wave and the next trough is 2 m. If the frequency of a particular wave is 2 Hz, what is the speed of the wave? (a) 4 m/s (b) I mls (e) 8 m/s (d) 2 ml' (e) impossible to delermine from the information given
5. If an object of mass '171. attached to a light spring is replaced by one of mass 9m, the frequency of the vibrating system changes by what multiplicative factor? (a) ~ (b) ~ (c) 3.0 (d) 9.0 (e) 6.0 .
2. The position of an object moving with simple harmon ic motion is given by.\: = 4 cos (6rr/), where x is in meters and I is in seconds. What is the period of the oscillating system? (a) 4 s (b) :; s (c) }\ s (d) 6rr s (e) impossible to determine from the information given
6. An ol~ject of mass 0.10 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic motion. "'!hat is the magnitude of the acceleration of the object ,.,,·hen it is at its maxi IT'll1m displacement 01" 0.10 m? (a) 0 (b) 0.45 mis' (e) 1.0 mis' (d) 2.0 rnls' (e) 2.40 mis'
3. A block.spring system vibrating on a frictionless, horizonlal surface with an amplitude of 6.0 cm has a total energy of 12 J. If the block is replaCl:d by one having twice the mass of the original block and the amplitude of the motion is again 6.0 Clll, what is the energy of the more massive syslem' (a) 12 J (b) 24 J (e) 6 J (d) 48 J (e) 36 J 4. A mass of 0.40 kg, hanging from a spl"ing with a spring constant of 80.0 NITTI, is set int.o an up-allCl-down simple harmonic motion. If the mass is displaced frolll equilibrium by 0.10 m and released from rest, what is its speed when rnoving through the equilibrium point? (a) 0 (b) 1.4 mls (c) 2.0 mls (d) 3.4 m/' (e) 4.2 m/'
7. A runaway railroad car with mass 3.0 X 10 5 kg coasts across a level track at 2.0 mls when it collides elastically with a spring-loaded bumper at the end of the track. If lhe spring constant of the bumper is 2.0 X 10 6 N/m. what is the maxi mUlTI compression of the spring during rhe collision? (a) 0.77 m (b) 0.58 m (c) 0.34 m (d) 1.07 m (e) 1.24 m 8.1fa simple pendulum oscill;Hes with a small amplitude and its length is doubled, what happens to the frequenc)' of its mOlion? (a) [tdoubles. (b) It becomes \12 limes as large. (c) I, hal\'es. (d) 1L becomes J/V2 as large. (e) 1L rema ins lhe sa me.
Chapter 13
452
Vibrations and Waves
9. A simple pendulum has a period or 2.5 s. Whal is its period if its length is made four times as large? (a) 0.625 s (b) 1.25 s (e) 2.5 s (d) 3.54 s (e) 5.0 s 10. A particle on a spring moves in simple harmonic motion ,dong the x-:n:is helwcen IlIrning points at Xl = 100 em and x 2 = 140 eln. At which of the following positions does the panicle have its maximum kinetic energy? (a) 100 elll (b) 110 elll (e) 120 em (d) 130 em (e) 140 em
11. 'Which of the following statements is '1101. true regarding a mass-spring system that moves with simple harmonic motion in the absence of friction? (a) The total energy of the system remains constanl. (b) The energy of the system is continually transformed between kinetic and
potential energy. (c) The tolal energy of the system is proportional to the square or the amplitude. (d) The potential energy stored in the system is greatest when the mass passes through the equilibrium position. (e) The velocity of the oscillating mass has its maximum value 'when the mass pClsses through the equilibrium position. 12. A block is attached to a spring hanging vertically. After being 5Im.. . l)' lowered, it hangs at rest with the spring stretched by 15.0 em. If the block is raised back up and released from rest. wilh the spring unstretched, what maximum distance does it fall? (a) 7.5 cm (b) 15.0 em (c) 30.0 (m (d) 60.0 em (e) impossible to determine without knm.. . ing the mass and spring constant
CONCEPTUAL QUESTIONS I. An object-spring system undergoes simple harmonic moLion with an amplitude A. Does the total energy change if the mass is doubled but the amplitude isn't changed? Are the kinetic and potential energies at a given point in its motion affected by the change in mass? Explain.
8. If a grandfather clock were running slow, how could we
2. If a spring is Slant?
9. A grandfather clock depends on t.he period of a
CUl
in half, what happens to its spring con-
7, A pendulum bob is made from a sphere filled with water. \""hat would happen to ule frequency of vibra~ tion of this pendulum if the sphere had a hole in it that allov.:ed the water to leak out ~10\\'1)'? a(~jllst
the lengtJl or ule pendulum to correct the time? pen~
dululTI to keep correct time. Suppose such a clock is calibrated correctly and then the temperature of the room in ,...hich it resides increases. Does Ule clock nlll slow, fast, or correctly? Hint: A material expands \'i·hen its temperature increases.
3. An object is hung on a spring, and the frequency of oscillation of the system, f, is measured. The object, a second identical object, and the spring are carried to space in the space shuttle. The [WO objects are attached to the ends of the spring, and the system is taken out into space on a space walk. The spring is extended, and the system is released to oscillate ,.. . hile floating in space. The coils of the spring don't bump into one another. v\11at is the frequency of oscillation for this system in terms off?
10. If you st.retch a rubber hose and pluck it, you can obsel\lc a pulse traveling up and down the hose. What happens to the speed of the pulse if YOli stretch the hose more tightly? What happens to the speed if you fill the hose ,,,ith water?
4. If an object-spring system is hung venicaJly and set into oscillation, why docs the motion eventually stop?
11. Explain v,.·hy the kinetic and potential energies of an object-spring system can never be negative.
5. Is a bouncing ball an example of' simple hClrmonic motion? Is the daily movement of a student from home to school and back simple harmonic motion?
12. '''That happens to the speed ofa wave on a suing when the frequency is doubled? ..\ ssume the tcnsion in the string remains the samc.
6. If a pendulum clock keeps perfect time at the base of a mountain, will it also keep perfect time when it is moved to the top of the mountain? Explain.
PROBLEMS ~
WebAssign
The Problems for this chapter Illay be assigned online ill WebAssign.
I, 2. J = slraightfor\l"ard, illlcrmcdialc, challeng:illg
m
mil (I
= denotes guided problem = denotes enhanced content problem
= biomedical application
0= denoll.:s. full solution available in SllIdeJIl Solutions Manual/ StudJ Guide
SECTION 13.1 HOOKE'S LAW
rnA O.60-kg block attached
to a spring with force constalll 130 N/m is free to move on a frictionless, horizontal sur face as in Figure 13.7. The block is released from rest aftcr the spring is stretched 0.13 ll'l. At that installl, find (a) the force on the block and (b) ilS acceleration. 4
2. When a 4.25~kg object is placed on top ofa venical spring. thc spring compresses a distance of2.62 cm. "Vhat is the force constanl of the spring?
453
Problems
3. A ball dropped from a height of ·LOO III makes a pcrfc("L1~ elastic collision with the ground. Assuming no mechanical energy is lost due LO air resistance, (a) show that the motion is periodic <md (b) deLermine thc period of the motion. (c) Is the motion simple hannonic? Expl'lin. 4. A load of 50 :'IJ attached to a spring hanging vt"rtically stretches the spring 5.0 cm. The spring is nO\, placed horil.omall}' on a table and stretched 11 CI11. (a) What force is required 10 stretch the spring by that amount? (b) Plot a graph of force (on the J-axis) \'ersus spring displacement from the equilibrium position along the x-axis. 5. A spring is hung from a ceiling, and an object attached to its lower end stretches the spring by a distance of 5.00 cm from its unstretched position when the system is in equilibrium. If the spring constant is 47.5 t\ 'm, determine the mass of the objecl. 6. An archer must exert a force of 375 N on the bowstring shown in Figure PI3.6a such that the Siring makes an angle of 8 = 35.0" with the vertical. (a) Determine the tCllsion in the bowstring. (b) If the applied force is replaced by a stretched spring as in Figure Pl3.fib and the spring IS stretched 30.0 cm from its unstretched length, what is the spring constant?
k
III
(al
of these bands 1.0 em. (a) What is the potential energy stored in the two bands together when a 50-g SLOne is placed in the cup and pulled back 0.20 m from the equilibrium position? (b) With what :..peed does the stone lea\'e the slingshol? 10. An ing (a) (b)
11. A child's to\' consists of a piece of plastic attached to a spring (Fig. P13.11). The spring is compl'essed against the floor a distance of 2.00 cm, and the to)' is released. If u1e toy has a mass of 100 g and rises 1.0 a maximum height of 60.0 cm, estimate the force conSl.:'l1H of the spl·ing.
FIGURE P13.11
ill]A 1O.Q.-g bullet is fired into, and embeds itself in, a 2.00-kg block attached to a spring with it force constant of 19.6 ~/m
am An
15.
EI
16.
am A O.25Q-kg block
FIGURE P13.6
SECTION 13.2 ELASTIC POTENTIAL ENERGY 8. A spring-loaded pellet gun is designed to fire 3.00·g projectiles horil.onl.ally at a speed of 45.0 m/s. (a) If the spring is compressed to its maximum design difference of 8.00 cm, what spring constant is required? (b) What maximum force is required to load the gun? rnA slingshot consists of a light leather cup conl
t
12, An automobile having a mass of I 000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring \\~th constant 5.00 X 101; r\'/m and is compressed 3.16 cm as the car is brought 10 rest. \\'hat wa£ the speed of the car before impact., assuming no energy is lost in the collision \\~th the wall?
(b)
7. _ A spring 1.50 m long with force conslant 47.~ l\'/m is hung from the ceiling of an ele\'alor, and a block of mass 10.0 kg is attached to the bottom of the spring. (a) By how much is the spring stretched when the block is slowly lowered to its equilibrium poim? (b) Jr the cleva· tor sub:..equentlr accelerates upward at 2.00 m's~, what is the position of the block, taking the equilibrium position found in part (a) as)' = 0 and upwards as the positive y-direClion. (e) ff the elevator cable :..naps during the accelermion, describe the subsequent motion of the block relative to the freelr falling elevator. What is tht amplitude of its motion?
archei' pulls her bowsu·ing back 0.400 m by exerta force then increases uniformly (rom I.ero to 230 N. What is the equi\'alem spring constant of the bow? How much work is done in pulling the bow?
o~ject-spring system moving with simple harmonic motion has an amplilude A. (a) \'\That is the total energy of the syslCm in terms of k and A onl)'? (b) Suppose at. a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this situation, using only me \'al;ables for ule mass Ill, velocity v, spring constant /(, and position -". (c) Using the results of pan.s (a) and (b) and the consen':ttion of energy equation, find the positions x of the object \,'hen its kinetic energy equals tWlce the potential energy stored in ule spring. (The answer should in terms of A only.) A hOI;zomal block-spring system with the block on a frictionless surface has total mechanical energy E = ·li.O J and a ma.ximulTI displacement from equilibrium of 0.240 m. (a) \'\Thal is the spling constant? (b) What. is the kinetic energy of the system at u1e equilibrium point? (c) If the maximum speed of u1e block is 3.45 mis, what i:.. irs mass? (d) What. is the speed of the block when ill) di.splacement is 0.160 m? (c) Find the kinetic energ}' of the block at x = 0.160 m. (f) find the potcntial energy stored in U1C spring when ."( = 0.160 m. (g) Suppose the same system is released from rest at x = 0.240 III on a rough surface so that it loses 14.0J by the time it reaches its first turning point (after passing equilibrium at x = 0). \Vhat. is ib position at thai. insl,ant?
resting on a frictionless, hOI'izontal slll-face is attached to a spring having force constant 83.8 N/m as in FiljUl"e PI~.16. A hOI;l.ol11al force causes
F
454
Chapter 13
Vibrations and Waves
the spring to sU'erch a distance of 5.46 em from it'i equilibrium position. (a) Find [he value of F. (b) \,Vhat FIGURE P13.16 is the tOlal energy stored in the s)'stem when the spring is stretched? (c) Find the magnillJde or the
23. Consider the simplified single~piston engine in figure Pl 3.23. Tf the wheel rotates at a conS(;Hll angular speed w, explain v..'hy the piston rod oscillates in simple harmonic molion.
w Piston
I
I
x=-A
x(t)
other information would you need to know to answer?
SECTION 13.3 COMPARING SIMPLE HARMONIC MOTION WITH UNIFORM CIRCULAR MOTION SECTION 13.4 POSITION, VELOCITY, AND ACCELERATION AS A FUNCTION OF TIME 17. A OAD-kg object connected LO a light spring with a force COnStllJll of 19.6 N/m oscillates on a friClionless horizOlllal surface. If the spring is compressed 4.0 cm and released from rest, determine (a) the !TI(lximum speed of the o~ject, (b) the speed of the object when the spring is compressed 1.5 COl, and (c) the speed of the object when the spring is stretched 1.5 cm. (d) For what value of xdocs the speed equal onc~half the maximum speed? o~ject-spring system oscill,nes with an amplitude of 3.5 em. If the spring consh"1nt is 250 N/m and the object has a mass orO.50 kg, determine (a) the mechanical energy of the s)'stem, (b) lhe maximum speed of the object, and (c) the maximum acceleration of the o~jCCI.
18. All
c::![]At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 20.0 cm on a spring with a force constant of 16,0 N/m. It is observed tJlatthe maximum speed of' the bunch of bananas is 40.0 cm/s. What is the weight of the bananas in newtons? 20. A 50.O-g object is attached to a horizontal spring \...·ith a force constant of 10.0 N/m and released from rest with an ampliwde of 25.0 cm. What is the velocity of Lhe object when it is halfway 1O the equilibrium position if the surface is frictionless? 21. While riding behind a car traveling at 3.00 mis, you notice that one of the car's tires has a smaJl hemispherical bump on its rim, as in Figure P13.21. (a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion. (b) If the radius of the car's tires is 0.30 m. ",hal is tJle bump's period of oscillation?
FIGURE P13.23
24. The period or motion of an o~ject-springsystem is 0.223 s when a 35.4-g o~ject is auached to the spring. \Vhat is the force constant of the spring? 25. A spring stretches 3.9 cm when a lO-g o~ject is hung from it. The object is replaced with a block of mass 25 g thal oscillates in simple harmonic motioll. Calculate the period of mOlion. 26. When four people with a combined mass of 320 kg sit down in a car, they find thallhe car drops 0.80 em lower on its sprillgs. Then lhey get Out of the car and bounce it up and down. What is the frequency of {he car's vibration if its mass (when it is emptr) is 2.0 X 10 3 kg?
127. fA can of Illass 250 g is placed on a frictionless horizontal air track. A spring having a spring constant or9.5 N/m is auached between the cart and the left end of the track. \""hen in equilibrium, the cart is located 12 cm from the left end of the track. If the c;1rt is displaced 4.5 em from ils equilibrium position, find (a) Lhe period at which it oscillates, (b) its maximum speed, and (c) its speed when it is ]4 cm from the left end of the track. 28. The position of an o~jec{ connected to a spring varies with time according to the expression x :::: (5.2 em) sin (8.0 'Tit). Find (a) the period of this motion, (b) the frequency of the motion, (c) the amplitude of lhe motion. and (d) the lirsttime after l :::: 0 Lhatthc object reaches the position x = 2.6 em. 29. A 326~g object is (luached to a spring and execures simple harmonic motion with a period of 0.250 s. If the tOtal energy of the system is 5.83 j, fmd (a) the maximum speed of the object, (b) the force constant of the spring, and (c) the amplitude of the moL ion. 30.
FIGURE P13.21
22. An ol~ject moves uniform I)' around a circular palh of radius 20.0 cm, making one complete revolution e....ery 2.00 s. \""hat are (a) the lransL:nional speed of the ol~jeCl, (b) the frequency of motion in henz, and (c) the angular speed of the object?
aD
An object executes simple harmonic motion with an amplitude A. (a) At what values of its position does its speed equal hall' its maximum speed? (b) At whal values of ils position does its potential energy equal half the total energy?
31. A 2.00-kg object on a frictionless horizontal track is ;'lltached to lhe end of ,1 horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.00 In {Q the right from its equilibrium position and then released. initiating- simple harmonic motion. (a) \l\lhat is the force
Problems
(magnitude and direction) acting on the object 3.fiO s after it is released? (b) How many times docs Ihe ol~jcct oscillate in 3.50 s? 32.
(b) What is its period if the elcvator is accelerating downward at 5.00 lll/s2? (c) \,Vhat is the period of simple harmonic motion for the pendulum if it is placed in a u'lIck thal is accelerating horizontally at 5.00 Ill/S2?
EJI
A spring of negligible Illass stretches 3.00 cm from its t'elaxed length when a fOl"ce of 7.50 N is applied. A O.SOO-kg panicle rests on a friclionless horizontal surface and is attached to the free end of lhe spring. The panicle is displaced from the origin LO x = 5.00 cm and released from rest at t = O. (a) \"!hat is dle force constant of the spring? (b) ""'hat arc the angular frequency w, the frequency, and the period of the motion? (c) \OVhat. is the tota] energy of the system? (d) \Vhat is the amplitude of the motion? (e) Whal. arc the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from t.he equilibrium position at t = 0.500 s. (g) Determine the velocity and acceleration of the panicle when 1= 0.500 s.
455
SECTION 13.6 DAMPED OSCILLATIONS SECTION 13.7 WAVES SECTION 13.8 FREQUENCY, AMPLITUDE, AND WAVELENGTH 41. The sinusoidal wave shown in Figure P13.41 is traveling in I.he positive x-direction and has a frequency of 18.0 Hz. Find the (a) amplitude, (b) wavelength, (c) period, and (d) speed of the wave.
133.ICiven thaI. x = J1 cos (wl) is a sinusoidal function of time, show that v (velocity) and a (acceleration) arc also sinusoidal functions of time. Hillt: Use Equations 13.6 and 13.2.
-+-\---+---\-8.26 em
~y_ _---,l 5.20 em
SECTION 13.5 MOTION OF A PENDULUM 34. A man enters a tall tower, needing to know its height. He notes that a long pendulum eXlends from the ceiling almostLO the floor and that its period is 15.5 s. (a) How tall is the to\\>'er? (b) If this pendulum is taken LO lhe !\lloon, where the free-fall acceleration is ] .67 lll/s2, what is the period there?
FIGURE P13.41
-12. An o~iect attached to a spring vibrates with simple harmonic motion as described by Figure P13.42. For this motion, find (il) the amplitude, (b) the period, (c) the angular frequency, (d) the maximum speed, (e) the maximum acceleration, and (f) an equation for its position x in terms ofa sine function.
35. A simple pendulum makes 120 complete oscillations in 3.00 min at a location where g = 9.80 Ill/52 . Find (a) the period oftlle pendulum and (b) its length.
.>:(em)
36. A "seconds" pendulum is one that moves through its equilibriulll position once each second. (The period of the pendulum is 2.000 s.) The length of a seconds pendululll is 0.9927 m at Tokyo and 0.994 2 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations? 137.IA pendululll clock that works perfenly on the Eanh is taken to the ~'Ioon. (a) Does it run f~lst or slow there? (b) If the clock is started at 12:00 midnight, wh,ell will it read after one Eanh day (24.0 h)? Assume the free-fall acceleration on lhe 1'\'100n is 1.63 Ill/S2. :lR. An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0"C. (a) When placcd in a room at a temperature of -S.U"C, will it gain lime or lose time? (b) HO\.. . milch lime will it gain or lose eve I)' hour? Hint: See Chaptcr 10. 39. The free-fall acceleration on Mars is 3.7111/52. (a) \·Vhat length of pendulum has a period of I s on Earth? \'Vhal length of pendulum would ha"e a 1-s period all Mars? (b) An object is suspended from a spring with force constant 10 N/m. Find the mass suspended from this spring that would rcsult in a period of I s on Earth and on Mars. 40. A simple pendululll is S.OO m long. (a) What is thc period of simple harmonic motion for this pendulum if il is located in an elcvator accelcrating upward at 5.00 mjs2?
f----'--+--!;--f-----:--+- I
(,)
FIGURE P13.42
43. A certain I'M radio station broadcasts jazz music at a frequcncy of 101.9 MHz. Find (a) the wave's period and (b) its wavelength. (Radio waves are electromagnetic Wilves lhilt travel at the speed of light, 3.00 x 10 8 m/s.) 44. The dist.ance between twO successive minima of a transverse wave is 2.76 m. Five crests of the wave pass a given point. along- the direction of travel e\'ery 14.0 s. Find (a) the frequency of the \'i·ave and (b) the wm'e speed. 45. A harmonic wave is travcling along a rope. It is observed that the oscillalor Ihal gencrates the wave complctes 40.0 vibrations in 30.0 s. Also, a given maximum travels 425 cm along thc ropc ill 10.0 s. What. is the wavelength? 46.
II
A bat can detect small objects, such as an insect, whose siJ;c is approximately equal to one wavelength of the sound the bat makes. Ifbats emita chirp ata frequcnc)' of 60.0 kHz and the speed of sound in air is 340 mis, what is the smallest insect a bat Gill detect?
4S6
Chapter 13
Vibrations and Waves
47. A cork on t.he surface of a pond bobs up and down two times per second on ripples having a wavelength of 8.50 em. lfthe cork is 10.0 III from shore, how long does it take a ripple passing the cork to reach the shore? 11.8.IOcean waves an: traveling to the east at ~I.O m/s with a dis-
tance 0[20 III between crests. With whal frequency do the waves hillhe from of a boat (a) when the boat is at anchor and (b) when the boat is moving \.. .csnvarcl at 1.0 m/s? ~.O
kg
FIGURE P13.57
SECTION 13.9 THE SPEED OF WAVES ON STRINGS 49. A phone cord is 4.00 In long and has a mass of 0.200 kg. A transverse wave pulse is produced by plucking one end of the laut corel. The pulse makes [our trips down and back along lhe cord in 0.800 s. What is the tension in the cord? 50. A circLis performer ~tretchcs a tightrope bem'cell tWO towers. He strikes one end of Lhe rope and sends a wave along it toward the other tower. He notes thaI it takes the wave 0.800 s to reach the opposite [Ower, 20.0 Tn away. If a I-m length of the rope has a massorO.350 kg, find the tension in the tightrope.
5!. A transverse pulse moves along a stretched cord of length 6.30 II'l having a mass of 0.150 kg. If Lhe tension in the cord is 12.0 N, find (a) the wave speed and (b) the time it takes the pulse to trave! the length of the core!. 52. A taUl clothesline is 12.0 III long and has a mass of 0.375 kg. A transverse pulse is produced by plucking one end of the clothesline. If the pulse takes 2.96 s to make six round trips along the clothesline, find (a) the speed of the pulse and (b) thc tension in the clothesline. 153.ITransversc w;wcs with a speed of 50.0 mls arc to be produced on a sll'elched string. A 5.00-m length orstring ""'ith a tOI.<"llmass of 0.060 0 kg is Llsed. (a) 'What is the required tension in the string? (b) Cctlcl.llate the \.,,·ave speed in the stdng ifthe tension is 8.00 :-.;, 54, An aslronaul on lhe :'vloon wishes to measure the local value of g by timing pulses traveling down a wire thm has a large oqject suspended from it. Assume a wire of lllass 4,00 g is J ,60 m long and has a g,OO-kg object suspended from it. A pulse requires 36.1 ms to traverse the length of the wire. Calculate gMoon from these dala. (You ma)' neglect the mass of the wire when calculming the tension in it.) 55. A simple pendulum consislS of a ball of lllass S.OO kg hanging from a uniform string of mass 0.060 a kg and length L If the period or oscilhlliOll or the pendulum is 2.00 s, determine the speed of a u'ansverse wave in tJ1e string when the pendulum hangs vertically, 56. A string is 50.0 cm long and has a mass of 3.00 g. A wave u"avels at 5.00 mls along this string, A second string has the same length, but half the mass of the first. If the two stJings are under the same tension, what is the speed of a wave along the second string? 57. Tension is maintained in a string as in Figure 1'13.57, The observed wave speed is 24 mls when Lhe suspended mass is 3.0 kg. (a) V\,llat is the mass per unit length of U1C string? (b) \,Vhat is the wave speed when lhe suspended mass is 2.0 kg?
IS8.IThe clastic limit of a piece of steel wire is 2.70 X 1O!1 Pa. \Vhat is the maximum speed at which transverse wave pulses can propagalc along the wire withoLiI exceeding its clastic lim it? (The density of steel is 7.86 X 10 3 kg/lll~.) 59.
mil
60.
mil
A 2.65-kg power line running between two lOwers has a length of 38.0 m and is under a tension of 12.5 N. (a) What is the speed of a transverse pulse set up on the line? (b) If the tension in the line was unknown, describe a procedure a worker on the ground might use to estimate the tension,
A taut clothesline has length L and a mass M. A transverse pulse is produced by plucking one end of the clothesline. If the pulse makes n round trips along the clothesline in f seconds, find expressions [or (a) the speed of the pulse in terms of '11, L, and I and (b) lhe tension Fin lhe clothesline in terms of the same variables and mass 1\'1,
SECTION 13.10 INTERFERENCE OF WAVES SECTION 13.11 REFLECTION OF WAVES 61. A wave of amplitude 0.30 III interferes with a second wave or amplitude 0.20 m traveling in the same direclion. What are (a) the largest and (b) the smallest resultant amplitudes that can occur, and under what conditions willlhese maxima and minima arise? 62. The position of a 0.30-kg object attached LO a spring is described by x =
(0.25 m) cos (0.47T/)
Find (a) the amplitude of the motion, (b) the spring constant, (c) the position of lhe object at f = 0.30 S, and (d) the object's speed at l = 0.30 s,
ADDITIONAL PROBLEMS 63. An object of mass 2.00 kg is oscillating freely on a vertical spring with a period of 0.600 s. Another object of unknown mass on the same spring oscillates Wilh a period o[] .05 s. Find (a) the spring constant Ii and (b) the unknown mass, 64. A certain lUning fork vibrates at a frequency of 196 Hz while each tip of ils two prongs has an amplitude of 0.850 mm. (a) "Vhat is the period of this motion? (b) Find the wavelength of the sound produced by the vibrating fork, taking the speed of sound in air to be 343 m/s. 65. A simple pendulum has mass 1.20 kg and length 0,700 m. (a) What is the period of the pendulum near the surface
Problems of Eanh? (b) If the same mass is attached to a spring, what spring constant would result in the period of motion found in parT (a)? 66. A 500-g block is released from rest and slides down a frictionless track thaL begins 2.00 m above the hOl'izontal, :as shown in Figure PI3.66. At the hallam orthe tr(lck, where the surface is horizontal, the block strikes and sticks 1.0 a light spring with a spring constant of20.0 N/m. Find the maximum distance the spring is compressed.
457
tem is as shown in the figure, and the incline is smooth. The 25-kg block is pulled 20 em down the incline (so that the 3D-kg block is 40 cm above the floor) and is released from resl. Find the speed of each block when the 3D-kg block is 20 cm abo\'e the floor (th'lt is, when the spring is unstretched).
25k g
sao g
.1 30 kg
2.00 m
j
FIGURE P13.69 FIGURE P13.66
67. A 3.00-kg object is fastened to a light spring, \.. . ith the intervening cord passing over a pulley (Fig. PI3.67). The pulley is friclionless, and its inertia may be neglected. The ol~ject is released from rest when the spring is unstretehed. I f the object drops 10.0 em before stopping. find (a) the spring constant of the spring and (b) the speed of the object when it is 5.00 CIll below its starting poinl.
70. A spring in a toy gun has a spring constant of 9.80 N/m and can be compressed 20.0 cm beyond the equilibrium position. A 1.00-g pellet resting againstlhe spring is propelled fon.. . arcl when the spring is released. (a) Find the muzzle speed of the pellel. (b) If the pellet is fired horizontall}' from a height of 1.00 m above the floor, what is its ranger 3.00 kg
Im-
A light balloon filled ,.. . ith
helium
of density
0.180 kg/m 3 is tied to a light string of length L = 3.00 m.
FIGURE P13.67
68. A 5.00-g bullet moving with an initial speed of 400 m/s is fired into and passes through a I.OO-kg block, as in Figure PI3.68. The block, initially at reS! on a frictionless horizontal surface, is connected to a spring with a spring constant of900 N/m. Jfthe block mm'cs 5.00 cm to the right after impact, find (a) the speed at which the bullet emerges from the block and (b) the mechanical energy lost in the collision.
The string is tied to the ground, fanning an "inverted" simple pendulum (Fig. P13.71a). If the balloon is displaced slightly from equilibrium. as in Figure PI3.71b, show that the Illation is simple harmonic and delcrmine the period of the motion. Take the density of air to be 1.29 kg/m~. Hhl/: Use an analogy with the simple pendu. IUln discussed in the text, and see Chapter 9. He
Air
Air
I I I
He
I
g.
L
t
I
it
:0
L
400 m/s -~
•
(a)
(b) FIGURE P13.71
5.00ClTl-:
r
,.
"""'+-
=-
72. _ An object of mass m is connected to two rubber bands of length L, each under tension F. as in Figure P13.72. The object is displaced vertically b)' a small dislance y. Assuming the tension docs not change, show that (a) the restoring force is -(2F/L)y and (b) the system exhibits simple harmonic motion with an angular frequency w = V2F/ mL.
FIGURE P13.68
69. A 25.kg block is connected (Q a 30-kg block by a light Siring that passes over a frictionless pulley. The 30-kg block is conneClcd to a light spring of force constant 200 N/m, as in Figure Pl~.69. The spring is unstretchcd \.. .·hcl1 the sys-
FIGURE P13.72
458
Chapter 13
Vibrations and Walles
73. Assume i1 hole is drilled through the center of the Eanh. h can be sI1O\..·n thal an object of mass 111 aL a distance r from the centcr of the Earth is pulled (Oward the cenler only by the material in the shaded portion of Figure P13.73. Assume Eilnh h;u a uniform density p. Write down ;\"cwlon's law of gravilatioll for FIGURE P13.73 an object ilt a distance l' from the ccnlCf of the Eanh and show that the force on it is of the [ofm of Hooke's law, F= -h, \\;I.h an effective force constant of k = C07TpGm, where G is the graviu'uional conStant. 74.
76,
mil
A s}'stem consisls of a venical spri ng with force constant k = I 250 11m, length L = 1.50 m, .and object of mass 111 = 5.00 kg attachcd to the end (Fig. PI3.i6). The object is placed at the level of the point of auachment with the spring ul1Stretchcd, at position )'1 = '-, and then It is released so that it swings like a pendulum. (a) \\Trite Ncwton's second law symbolically for the system as the object passes through its lowest point. (~Ole Ihat at the lowest point, r = L - )).) (b) Write the conservation of enel-gy equation symbolically, equaling the tOlal mechanical energies at the initial point and lowesl poinl. (c) Find the coordinate position of the lowes1 poinl. (d) Will [his pendulum's period be greater 01" less than the period ora simple pendulum with the same mass 111 and length mass L? Explain.
Figure P13.74 shows a crucle model aran insect wing. The mass m represcllLs the entire mass of the wing, which pivots aboUl the fulcnam /~ The spring represents the surrounding COIHlcctive tissue. MOlion of thc wing con'csponds to vibration of the spring. Suppose the mass of the wing is 0.30 g and the effecti,'c spring constant of the tissue is '-1.7 x IO~4 ~/m. If the mass m moves up and down a distancc 01"2.0 mm from i,s position of equilibrium, what iii the maximum speed of tile outer-tip orthe ,.. . ing?
y
)'=
-~)
3.00 mill
I. .
J.50
Cill
1
0
)j-
_v FIGURE P13.76
FIGURE P13.74
177.IA large block P executes horizontal simple harmonic motion as it slides across a frietionless surface ",ith a frequency J= 1.501-fz. Block B rests on it, as shown in Figure PI3.77, and the coefficient of static friction between the two is IL, = 0.600. '''!hat maxilllulll amplitude of oscillation can the system have if block JJ is notlQ slip?
i5. A 2.00·kg block hangs without vibrating at the end of a spring (k = 500 N/m) that is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of gl3 when the acceleration suddenly CC
p,
'"
IJ p
I--_~-------l.._...
FIGURE P13.n
Sound can be used to creal-e strik~ lng images of the human body's interior, as in this three-dimensional ultrasound of twins in utero.
14.1 14.2
Producing a Sound Wove Characteristics of Sound
Waves
SOUND
14.3
The Speed of Sound
14.4
Energy and Intensity of Sound Waves
14.5
Spherical and Plane Waves
14.6
The Doppler Effect
14.7
Interference of Sound Waves
14.8
Standing Waves
14.9
Forced Vibrations and Resonance
Sound waves are the most important example of longitudinal waves, In this chapter we dis-
cuss the characteristics of sound waves: how they are produced, what they ore, and how they
14.10 Standing Waves in Air Columns
travel through motter. We then investigate what happens when sound waves interfere with
each other. The insights gained in this chapter will help you understand how we hear.
14.1
PRODUCING A SOUND WAVE
Whether it conveys the shrill whi ne of ajct engine or the soft melodies of a crooner, any sound \V
14.11
Beats
14.12
Quality of Sound
14.13 The Ear
(a)
Low-dell~it"
region
(b) FIGURE 14.1 A vibraling tuning fork. (a) As the right tinc: of Ihe fork mo\'t~s 10 the right, a high-dcllsil), regiol1 (compression) of air is formed in frOIll of it:. rnovemclII. (b) As the right tine moves to the left. a loh'densit), region (rarefaction) of air is formed behind it.
459
Chapter 14
460
Sound
FIGURE 14.2 (a) As the luning fork \'ibratcs, a series of compressions and rarefacLions moves outward, away from the furk. (b) The crests of the wave correspond to compressions, the troughs 10 rarefacliolls.
(b)
to represent a sound wave. as in Figure 14.2b. Notice that there are crests in the sinusoidal wave at the points where the sound wave has compressions and troughs where the sound wave has rarefactions. The compressions and rarefactions of the sound ",,'aves are superposed on the random thermal motion of the atoms and molecules of the ail- (discllssed in Chapter 10), so sOllnd wa\'es in gases travel at about the molecular rms speed.
14.2
CHARACTERISTICS OF SOUND WAVES
As already noted, the general motion of elements of air near a vibrating object is back and forth between regions of compression and rarefaction. This back-andforth motion of elements of the medium in the direction of the disturbance is characteristic of a longitudinal wave. The motion of the elements of the medium in a longitudinal sound wave is back and forth along the direction in which the wave travels. By contrast, in a transverse wave, the vibrations of the elements of the medium are at right angles to the direction of travel of the wave.
Categories af Sound Waves Sound ,,"'aves F(lll into three categories covering different ranges of frequencies. Audible waves are longitudinal waves that lie within the range of sensitivity of the human ear, approximately 20 to 20 000 Hz. Infrasonic waves are longitudinal waves v",ith frequencies below the audible range. Earthquake waves are an example. Ultrasonic waves are longitudinal waves with frequencies above the audible range for humans and are produced by certain types of whistles. Animals such as dogs can hear the waves emitted by these whistles.
Applications of Ultrasound APPLICATION Medical Uses of Ultrasound
Direclion of \;bnttion
!
Elecu'ical
'-c7""::::;2:,--co_ni.~,-e_cI ions
Crystal
FIGURE 14.3 All alternating vollage applied to Ihe faces ora pie7.0cleClric crystal callses the cl"}'stal 10 \'ibraTe.
Ultrasonic waves are sound waves with frequencies greater than 20 kHz.. Because of their high frequency and corresponding short wavelengths, ultrasonic waves can be used to produce images of small objects and are currentl}' in wide use in medical applications, both as a diagnostic tool and in certain treatments. Internal organs can be examined via the images produced by the reflection and absorption of ultrasonic waves. Although ultrasonic waves are far safer than x-rays, their images don't always have as much detail. Certain organs, howe\'er, such as the liver and the spleen, are invisible to x-rays but can be imaged with ultrasonic waves. ~'ledical workers can measure the speed of the blood now in the body with a de"ice called an ultrasonic flow meter, which makes use of the Doppler effect (discussed in Section 14.6). The rIow speed is found by comparing the frequency of the '\'aves scattered by the nowing blood with the incident frequency. Figure 14.3 illustrates the technique that produces ultrasonic waves for clinical use. Electrical contacts are made to the opposite faces of a crystal, such as quartz or strontium titanate. fF an alternating voltage of high frequency is applied to these contacts, the crystal vibrates at the same frequency as the applied voltage, emining a beam of ultrasonic waves. At one time, a technique like this was used to produce sound in nearly all headphones. This method of transforming electrical energy into mechanical energy, calleclthe piezoelectric effect, is reversible: lfsome exter* nal source causes the cryst.al to vibrate, an alternat:ing voltage is produced across
14.3
The Speed of Sound
461
it. A single crystal can therefore be used to both generate and receive ultrasonic waves. The primary physical principle that makes ultrasound imaging possible is the fact that a sound wave is partially reflected whenever it is incident on a boundary bCI\Veen t"-!O matcri<.lls having different densities. If a sound wave is u'a,'eling in a material of density Pi and strikes a material of density P" the percentage of the incident sound wave intensity refleCted, Pl? is given by
PH =
(p, - P')' X p,
100
+ PI
This equation assumes that the direction of the incident sound wave is perpendicilIaI' to the boundary and that the speed of sound is approximately (he same in the two materials. The latter assumption holds very well for the human body because the speed of sound doesn't vary much in the organs of the body. Physicians commonly use ultrasonic waves to observe fetuses. This technique presents far less risk than do x-rays, which deposit more energy in cells and can produce birth defects. First the abdomen of the mot.her is coated ,.. . ith a liquid, such as mineral oil. If that wcre not done, most of the incident ultrasonic waves from the piezoelectric source would bc reflected at the boundary between the air and the mother's skin. Mineral oil has a densil)l similar to that of skin, and a very small fraction of the incident ultrasonic wave is reflected when Pi = P" The ultrasound energy is emitted in pulses rather than as a continuous wave, so the same cl)fsl.al can be used as a detector as well as a transmitter. AJ1 image 0[-" the fetus is obt.ained by using an array of' transducers placed on the abdomen. The reflccted sound waves picked up by the transducers are converted to an electric signal, which is used to form an image on a fluorescent. screen. Difficulties such as the likelihood or spontaneous abortion or of' breech birth are easily detected with this technique. Fetal abnormalities such as spina bifida and water on the brain are also readily observed. A relativel),> ne,v medical application of ultrasonics is the lXlVitmn ultrasonic 511rgiwi aspirator (GCSA). This dcvice has made it possible to surgically remove brain tumors that were previously inoperable. The probe of the CCSA emits ultrasonic waves (at about 23 kHz) at its tip. '·\·11en the tip touches a tumor, the part of the tumor near the probe is shattered and the residue can be sucked up (aspirated) through the hollow probe. Lsing this technique, neurosurgcons are able to remove brain tumors ,.. ~thout causing serious damage to healthy surrounding tissue. Ultrasound is also used to break up kidney stones that are otherv"ise roo large to pass. Previously, invasive surgcry ''ias more often required. Another interesting application of ultrasound is the ulLrasonic ranging unit used in some cameras to prO\~de all
14.3
AllultrasoLlnrl image ol";t human fc1tls in the \\·Ilmb.
Thc speed of a sound wave in a fluid depends on the f1uid's compressibility and inertia. If the fluid has a bulk modulus B and an equilibrium density p, the speed or sound in it is
v=~
[14.1]
f- Speed of sound in a fluid
462
Chapter 14
Sound
Equation 1.4.1 also holds true for a gas. Recall from Chapter 9 that the bulk modulus is defined as the ratio of the change in pressure, D.P, LO the resulting fractional change in volume, 6V/V: !lP ---!lV/V
[14.2]
B is ah'iays positIve because an lIlcrease III pressure (positive UP) results in a decrease in volume. Hence, the ratio D,Pj,1,Vis ahvays negative. It's interesling to compare Equation 14./ with Equation 13.18 for the speed or transverse waves on a su"ing, v = VFj j..L, discussed in Chapter 13. In both cases the wave speed depends on an elastic property of the medium (B or F) and on an inertial property of the medium (p or J..L). Tn fact, the speed orall mechanical waves follows an expression of the general form
,---,---...,,-----elastic property
11=
\ inertial property
Another example or this general form is the speed of a longitudinal wave in a solid rod, \.. . hich is
v=jf Speeds of Sound in Various Media :Medium
Gases Air (O°C) Air (JOO°C) Hydrogen (O°C) Oxrgen (DoC) Helium (DoC)
v (m/s)
33] 386 1290 317 972
where V is the Young's modulus of the solid (see Eg. 9.3) and p is its density. This expression is valid only for a thin, solid rod. Table 14.1 lists the speeds of sound in various media. The speed of sound is much higher in solids than in gases because the molecules in a solid interact more strongly with each other than do molecules in a gas. Striking a long steel rail \.. . ith a hammer, tor example, produces two sound waves, one moving through the rail and a slower \,,'ave moving through the air. A student "... ith an ear pressed against the rail r-irst hears the faster sound moving through the rail, then the sound moving through air. In general, sound travels faster through solids than liquids and faster through liquids than gases, although there arc exceptions. The speed of sound also depends on the temperature of the medium. For sound traveling through air, the relationship between the speed of sound and temperature is
Liquids at 25°C \o\'ater
Melhyl alcohol
Sea water
1490 1140 1530
Solids Aluminulll
5100
Copper Iron Lead Vulcanized rubber
3560 5130 1320 54
APPLYING PHYSICS 14.1
[l4.3J
v = (331 m!s) )
27~ K
[14.4]
~vhere 331 mls is the speed of sound in air at O°C and T is the absolute (Kelvin) temperature. Using this equation, the speed of sound in air at 293 K (a typical room temperature) is approximately 343 m/s.
QUICK QUIZ 14.1 Which of the following actions will increase the speed of sound in air? (a) decreasing the air temperature (b) increasing the frequency of the sound (c) increasing the air temperature (d) increasing the amplillide of the sound wave (e) reducing the pressure of the air
THE SOUNDS HEARD DURING A STORM
How does lightning produce thunder, and causes the extended rumble?
~vhal
Explanation Assume you're at ground level, and neglect ground reflections. \'\!hen lightning strikes, a channel ofionizecl air carries a large electric current from a cloud to the ground. This results in a rapid
temperature increase of the air in the channel as the current moves through it, causing a similarly rapid expansion of the air. The expansion is so sudden and so intense that a tremendous dislllrbance-thunderis produced in the air. The entire length of' the channel produces the sound at cssentially the samc instant
14.4
of time. Sound produced at the bottom of the chanlIel reaches YOll first because that's the point closest to you. Sounds from progressi,"ely highel" portions of the channel reach yOlI at later times, resulting in an extended ,"oar. If the lightning channel were:a per·
EXAMPLE 14.1 Goal
Energy and Intensity of Sound Woves
463
feetly straight line, the roar might be steady. but the zigzag shape of the path results in the rumbling variation in loudness. with different quantities of sound energy from different segments arriving at any given instant.
Ex losion over an Ice Sheet
Calculate time of travel for sound through various media.
Problem An explosion occurs 275 III abO\"c an 867-m-thick ice sheet that lies over ocean \..'ater. I f the air temperature is -7.00°C, how long does it take the sound to reach a research "essel 1 250 m below the ice? Neglect any changes in the bulk modulus and density with temperature and depth. ( se Bkc = 9.2 X 10 9 Pa.) Strategy Calculate the speed of sound in air with Equation 14.4 and use d = vIto find the time needed for the sound to reach the surface of the ice. Use Eqmllion 14.1 to compute the speed of sound in icc, again finding the time with the distance equation. Finally, use lhe speed of sound in salt watel" to find the time needed to traverse the water and then sum the three times. .
otice that the speed of solind is highest in solid icc, second highest in liquid water, and slowest in air.
QUESTION 14.1 Is the speed of solind in rubber higher or lower than the speed of sound in aluminum? \Vh)'? EXERCISE 14.1 Complllc the spced orsound in thc rollowing substances at 273 K: (a) lead (Y = 1.6 X 10'0 Pa). (b) mercury (E X 10'" Pa), and (c) air at -15.0°C.
Answers
14.4
(a) 1.2 X 10' mls
(b) 1.4 X 10' mls
(c) 322 m/s
ENERGY AND INTENSITY OF SOUND WAVES
As the tines of a tuning fork mm'e back and forth through the air, they exert (\ force on
= 2.8
464
Chapter 14
Sound
TllC average intensity I of a wave 011 a given surface is defined as the ratei::U which energy nmvs through t.he surface, 6.£.;11/, divided by the surface area A: [~
1 tlE
-A ill
[14.5]
where lhe direction of energy flow is perpendicular to the surface at every poilll. S1 unit: walt per meter squared (W/m')
A rate of energy transfer is power, so Equation 14.5 can be written in the alternate form Intensity of a wave
-+
power
(i/'
area
J1
[14.6]
[~--=-
where gp is the sound pow'er passing through the surface, measured in wans, and the intensity again has units of'\laLts per square meter. The faintest sounds the human ear can detect at a frequency of 1 000 Hz have an intensity of about I X lO~I~ W/m 2. This intensitY' is called the threshold of hearing. The loudest sounds the ear can tolerate have an intensity or about I W/m 2 (the threshoJd of pain). At the threshold of hearing, the increase in pressure in the ear is approxirnately 3 X 10-5 Pa over normal atmospheric pressure. Because aUTIOspheric pressure is aboutl X ]0 5 Pa, this means the ear can detect pressure fluctuations as small as about 3 parts in 101Uj The maximum displacement of an air molecule at the threshold of hearing is about I X 10-11 tn, which is a remarkably small number! If we compare this displacement with the diameter of a molecule (about 10~1O m), ,·..e see that the ear is an extremely sensitive detector of sound waves. The loudest sounds the human ear can tolerate at I kHz correspond to a pressure variation of about 29 Pa away from normal atmospheric pressure, with a maximum displacement of air molecules of 1 X J 0- 5 m.
Intensity Level in Decibels The loudest tolerable sounds have intensities about 1.0 X 10 12 times greater than the faintest detectable sounds. The most intense sound, however, isn't perceived as being LO X 10 12 times louder than the faintest sound because the sensation of loudness is approximately logarithmic in the human ear. (For a review of logarithms, see Section A.3, heading C, in Appendix A.) The relative intensity of a sound is called the intensity level or decibeJ level, defined by Intensity level
-+
Tip 14.1 Intensity Versus Intensity Level Don'L confuse intcnsitywith intcnsil)' level. Intensity is a physical quantity with units o[watlS per meter squared; intensity lc\'cl, or decibel level, is a convenient mathem:l.licaltransformalion of inH:nsily to a logarithmic
scale,
(3
~
10 log
(-f)
[14.7]
The constant If) = J.O X J o~ 12 W/m 2 is the reference intensity, the sound intensity at ule threshold of hearing, lis the intensity, and f3 is the con-esponding intensity level measured in decibels (dB). (The ' . . ord decibel, which is one-tenth ofa be4 comes from lhe name of the invenlor of the lelephone, Alexander Craham Bell (1847 -1922)). To get a feel [or various decibel levels, we can substitute a few representative numbers into Equation 14.7, starting with 1 = 1.0 X 10- 12 W/m 2 :
l.0 X 1O- w/m') 10Iog ( I" ., ~ 10 log (1) 1.0 X 10 - W/m12
(3
=
= OdB
From this result, \ve see that the lower threshold of human hearing has been chosen to be zero on the decibel scale. Progressing upward by powers often yields 11
W/m') =
1.0 X 10(3 = 10 log ( I"" 10 X 10 -W/m10
W/m')
1.0 X 10(3 = 10 log ( I'" 1.0 X 10 - W/m-
10 log (10)
= ]0 dB
= 10 log (l00) = 20 dB
14.4
Energy and Intensity of Sound Walles
Notice the pattern: MU/lif)/),ing a given intensity by ten addj 10 db to the intensity level. This pallern holds lhroughout the decibel scale. For example, a 50-dB sound is ]0 times as intense as (\ 40*dB sound, \vhereas a 60*dB sound is 100 times as intense as a 40*dB sound. On this scale, the threshold of pain (I = 1.0 VV/Ill::!) corresponds to an intensir-y Icvelof{3= 1010g(l/1 X 10-") =IOlog(IO") = 120 clB. Nearby jet airplanes can creaLe intensity levels of 150 dB, and subways and riveting machines have levels of 90 LO 100 dB. The electronically amplil1ed sound heard al rock coneens can attain levels of up to 120 dB, the threshold of pain. Exposure to such high intensity levels can seriously damage the ear. Earplugs are recommended whenever prolonged intensity levels exceed 90 dB. Recent evidence suggesls that noise pollution, which is common in most large cities and in some industrial environments, may be a eon* tribuling factor to high blood pressure, anxiety, and nervousness. Table L4.2 gives the approximate intensity levels of various sounds.
TABLE 14.2 Intensity Levels in Decibels for Different Sources Source of Sound
i\'earby jCt airplanc Jackhammer, machine
Goal
(3 (dB)
150 130
gun Siren, rock cOllcert
120
Subway, power mower Busy tra rfic Vacuum cleaner Normal cOlwcrsalion
100
MosquiLO buzzing \fhisper Ruslling leaves Threshold of hearing
EXAMPLE 14.2
465
80 70 50 40 30 10
o
Machine
',Vorking with watts and decibels.
Problem A noisy grinding machine in a factory produces a sound intensity of 1.00 X 10-5 \'V/m 2 . Calculate (a) the decibel level or this machine and (b) the new intensit)'lcyel when a second, identical machine is added to the factory. (c) A cenain number ofaddit.ional such machines arc put inlo operation alongside these two machines. \-Vhen all the rnachines are running at the same time the decibel level is 77.0 dB. Find the sound intensity. Strategy Pans (a) and (b) require substituting into the decibel formula, Equation 14.7, with the intensity in part (b) lwice the intensity in part. (a). In pan (c), the intensity level in decibels is given, and it's necessary to work backwards, using the inverse of the logarithm runetion, to get the intensity in wallS per meter squared.
r···
,
,_
~
Solution (a) Calculate the intensity level of the single grinder. Substitute the intensity into the decibel rormula:
1.00 X 10-5 W1m') {3= 1010 ( g 1.00 X 10 "W/m'
10 log (10 7 )
70.0 dB (b) Calculate t.he ne\,; intensity level when an additional
machine is added.
2.00 X 10-5 W 1m' ) =1010 ( . {3 g 1.00 X 10 12 W/m'
Substitute l\viee the intensity of part (a) into the decibel formllla:
73.0 dB
(c) Find the intensity corresponding to an intensity level of 77.0 clB
Substitute 77.0 dB into the decibel formula and divide both sides by 10:
{3
=
77.0 clB
770 = log ( Make each side the exponent of ]0. On the righl*hand side, 1010gll = /1., bydeRnition or base 10 logarithms.
107.,<) = 5.01
= I"
10 log I
,,)
10 -W/mX
]0 7
I
~ -----;;,---."
1.00 X 10 1= 5.01 X 10- 5 W
~
,
_.-
"._........
(~)
.
,'-,
I' W/m'
1m' ~
Remarks The answer is nve times the intensity of the single grinder, so in pan (c) there are five such machines operal iog simultaneously. Because of the logarithmic definition of intensity level, large changes 1n intensity correspond to small changes in i nrensity level.
466
Sound
Chapter 14
QUESTION 14.2 By how many decibels is the sound intensity level raised '.vhen the sound intensity is doubled? EXERCISE 14.2 Suppose a mal111[aclUring plant has an average sound intensity level of 97.0 dB created b~' 25 identical machines. (a) Find the LOtal intensity created by a[1 the machines. (b) Find rhe sound intensity created by UII~ ~uLh IlJachine. (c) \\'hat's the sound intensity le\'el irtl\'c such machines are running?
Answers
(a) 5.01 X 10-' W/m'
APPLICATION
I/JI
OSHA Noise-Level Regulations
Spherical wave fronL
• \ t
I~
I '2
FIGURE 14.4 A spherical wave propagating radially oUlward from an oscillating sphere. The inlensilY of the W;1\'1;: varies as IIr~.
(1)) 2.00 X 10-" W/m'
(c) 90.0 dB
Federal OSHA regul€ltions now demand that no office or factory worker be exposed to noise levels that average more than 85 dB over an 8-h day. From a management point of view, here's the good news: one machine in the factory may produce a noise level of 70 dB, but a second machine, though doubling the total intensity, increases the noise level by only 3 dB. Because of the logarithmic nature of intensity levels, doubling the intensity doesn't double the intensity level; in fact, it alters it by a surprisingly sm(lll amount. This means that equipment can be added to the factory without appreciably altering the intensity level of' the environlllcnt. ~ow here's the bad news: as you remove noisy machinery, the intensit}, level isn't lowered appreciably. In Exercise 14.2, reducing the intensity! level by 7 dB would require the removal of 20 of the 25 machines! To lower the level another 7 dB would require removing 80% of the remaining machines. in 'which case only one machine ,',,'ould remain .
14.5
SPHERICAL AND PLANE WAVES
If a small spherical object oscillates so thal its radius changes pcriodically ,vith time, a spherical sound \-vave is produced (Fig. 14.4). The \-vave moves outward from the source at a constant speed. Because all points all the vibrating sphere behave ill the same way, we conclude that the energy in a spherical ".'ave propagates equally in all directions. This means that no one direction is preferred over any other. IfCZPav is the average pm...·er emitted by the source, then at any distance r from the source, this power must be distributed over a spherical sllrface of area 4-rrr 2 , assuming no absorption in the medium. (Recall that 4-rrr 2 is the surface area of a sphere.) Hence, the intensity of the sound at a distance r from the source is I~
"
Wave front ......
average power area
rJPa ,.
czpa \
A
4-rr1"~
~-=--
[14.8]
This equation shows that the intensity ofa wave decreases with increasing distance from its source, as you might. expect. The fact that Ivaries as l/r 2 is a result of the assumption that the small source (sometimes called a point source) emits a spherical wave. (In f~lct, light W;:lves also obey this so-called inverse-square relationship.) Because the average power is the same through any spherical surface centered at the source, we see that the intensities at distances '1'\ and 1"2 (Fig. 14.4) from the center of the source are
The ratio of the il1lensities at these two spherical surfaces is [14.9] FIGURE 14.5 Spherical w
It's llseful to represent spherical waves graphically with a series of circular arcs (lines of maximum intensity) concentric with the source representing part ora spherical surface, as in Figure 14.5. We call such an arc a wave front. The distance between adjacent wave fronts equals the wavelength A. The radial lines pointing ourward from the source and perpendicular to the arcs are called rays.
14.5
PI.lIle
467
)'
wave Iron 1 " / 1 \
FIGURE 14.6 Far.twm from a poilll sourcc. tht.· h
Spherical and Plane Waves
I
FIGURE 14.7 \ representation 01 a plant' wan: moving: in thl' po!>itiw x-direction \\·jth a speed 1'. The wa\l· frolllS are planes parallel La Ihe J-:-plane.
No\\'" consider a small portion of a ""we from t hat is at a great distance (relati\"c to A) from the source, as in Figure 14.6. fn this case the rays are nearly parallel to
each other and the wave fronts are very close to being planes. At distances from the source that are great relative lO the wan~length, lherefore, we can approximate the wave front with parallel planes, called plane waves. Any small portion ofa spherical wave that is far from the source can be considered a plane w~we. Figure 14.7 illustrates a plane wave propagating along the x-axis. If the positive x-direction is taken 1.0 be the direction of the 'Nave Illation (or ray) in this figure, then the wave fronts are parallcl to the plane containing the y- and z-axes.
EXAMPLE 14.3 Goal
Intensity Variations of a Point Source
Relatc sound imensities and lheir di.lilances frolll a point source.
Problem A small source emits sound waves with a power outpul of 80.0 \,\t. (a) Find the intensity 3.00 In from the source. (b) At whaL distance would the intcnsity be one-fourth as much as it is at r = 3.00 m? (c) Find the distance at which the sound level is 40.0 dB. Strategy The source is small, so the emitted wa\'es are spherical and the intensity in pan (a) can be found by substituting values into Equation 14.8. Pan (b) invol\'es solving for rin Equalion 14.8 followed by substitution (although Eq. ILl.9 can be used instead). In part (c), con\"crt from the sound intensity level 1.0 the intensity in W/m'!., using Equation 14.7. Then subslitute into Equation 14.9 (although Eq. 14.8 could be used instead) and soh'e for r2 . ~
. . . . ..
.
. ....
Solution (a) Find the inlensil)' 3.00 m from the source. Substitute ?P'l\ = 80.0 \,,- and r = 3.00 minto Equalion 14.8:
1= '!P.,
+rr,.'
= _8::.:0:.:.'0=--.:.:\'\_',-, = 41T(3.00 m)'
0.707W/m'
(b) At what distance would the intensity be one-fourth as much as it is at r::= 3.00 Ill? Take 1= {O.707 \'\'/m'!.)/4, and sol\'e for rin Equation 14.8:
,.=
(::'J'
80.0W ]" [ 4'11 ( 0.707 W/m')/ ~.O
(c) Find the distance at. which the sound level is 40.0 dB. Convert the intensity level of 40.0 dB to an intc:nsit>1 in W/m' by solving Equation 14.7 fa,. I:
40.0 = 10 log
(.!.-) II)
-->
4.00
= log (.!.-) II)
6.00 m
Chapter 14
468
Sound
Solve Equation 14.9 [or J}, substitute the intensity and the result of part (a), and take the square rooL:
"
I,
r~-
I,
I"r
2
--. "2-"
(3.00m)'(
1'2
......
=
=
, I,
1'1- -
1'2,
0.707W/m' ) 1.00 X 10- 8 W/m'
2.52 X 10"
Jl)
..........................~
Remarks Once the intensity is known at one position a certain distance away from the source, it's easier to use Equation 14.9 rather than Equation 14.8 to find the intensity at any other location. This is particul~1rly true for pan (b), where, llsing Equation 14.9. we can see right away that doubling the distance reduces the imcnsiLy to oncfourth its previous value.
QUESTION 14.3 The power aUlpLlL of a sound system is increased by a [aeLOr of 25. By what factor should from the speakers so lhe sound inten~ity is the same?
yOll
adjust your distance
EXERCISE 14.3 Suppose a cenainjet plane creates an intensity level of 125 dB ala distance 0[5.00 m. "\That intensity level does it creale on the gTound directly underneath it when Oying at. an altitude of 2.00 km? Answer
73.0 dB
14.6
THE DOPPLER EFFECT
I f a car or truck is moving while its horn is blmving, the frequency of the sound yOll hear is higher as rhe vehicle approaches you and lower as it moves away from you. This phenomenon is one example of the DopJJler effect, named for Austrian physicist Christian Doppler (1803-1853), who discovered iL. The same effect is heard if you're on a motorcycle and the horn is stationary: the frequency is higher as you approach the source and Im,,'cr as yOll move 3\vay. Although the Doppler efTect is most often associated "vith sound, it's common to all waves, including light. In deriving the Doppler effect, we assume the air is stationary and that all speed measurements are made relative to this stationary medium. The speed Vo is the speed of the observer, V s is the speed of' the source. and v is rhe speed of sound.
Case 1; The Observer Is Moving Relative to a Stationary Source tn Active Figure 14.8 an observer is moving with a speed of IJO w"iard the source (considered a point source), which is at rest. (vs = 0). '/lie take the frequency of the source to be Is, the ,,,'avelength of the source to be As, and the speed of sound in air to be v. If both obsen'er and source are stationary, the obsen·cr detcclS Is wave fronlS per second. (That is, ,,,hen V o = 0 and V s = 0, the obsen'cd frequency fo equ<-1ls the source frequency J~.) \\'11en 11l00~ng tm\~lrd the source, the observer moves a disrance of Vol in l seconds. During this interval, the observer det.ects an additional number of wave fronts. The number of extra \vave fronts is equal to the diSlance o"aveled, Vol, divided by the wavelength As:
-'-U(J
Source
.........L---"'''''-If----+-+--o S
o
iI.~ = 0
ACTIVE FIGURE 14.8 ,\n observer l1\ovillg \,'ith a speed Va to'L'l/rda SIrth<Jll the ~OLlrcc frcf]t1ellcyJI"
Additional ,"'/ave fronts detected = vot As Divide this equation by the time t to gel the number of additional W"we fronts detected /Jer second, volAs. I-Icnce, the frequenc)' heard by the observer is increased to
14.6
The Doppler Effect
SubstiLUling A,\ = v/fsinto this expression forjowe obtain V
+
469
----""i'O
va)
[14.10]
10= fs ( - v -
Source
vVhen the observer is 77lovingawtl)'from a stational'y source (rig. ]4.9), the observed
rrequency decre;'Jses. A derivation yields the same result as Equation 14.10, but with v - V o in the numerator. Thererore, ,"vhen the observer is moving away from the source, substitute -vo for V o in Equation ]4.10.
Case 2: The Source Is Moving Relative to a Stationary Observer Now consider a source moving toward an observer at rest, as in Active Figure 14.10. Here, the '.. . ave fronts passing observer A are closer together because the source is moving in the direction of the outgoing ,,,,'ave. As a result, the wavelength "-0 measured by observer A is shorter than the wavelength A.\. of the source at rest. During each vibration, which lasts for an interval T (the period). the source moves a distance vsT = v.,/f~· and the wavelength is shortened by that amount. The observed wavelength is therefore given by vs Ao = AI - ).,
Because As =
v/J.~,
.
Ao
A. _ .1
v Vs } \
Tip 14.2 Doppler Effect Doesn't Depend on Distance
v
v
FIGURE 14.9 An obsener moving wilh a speed or DO awa) from a st,llionarr source hears a frequencY./(J lhal is I01lln-lhan the: source freq\lencY'/I'
The sou nd from a sou .-cc approaching at constant speed will increase in intensity, bIll the observed (elevated) frequency will remain unchanged. The Doppler effect doesn't depend on distance.
the frequency observed by A ;s
}o=-=
S~---l---+_--+"""-----"-L Vs'" 0
v
vs
1s
h
or
}n =
I\(v ': vJ
[14.11]
As expected, the observed frequency increases when the source is moving toward the observer. \,Vhen the source is moving awa) from an observer at rest, the minus sign in the denominator must be replaced 'with a plus sign, so the factor becomes (v + v s ).
General Case \,Vhcn both the source and the observer are in motion relative to Earth, Equations Jl l.l0 and 14.11 can be combined to give V
+
va)
!n=fs ( - .v-
[14.12]
v.~
of- Doppler shift equotionobserver and source in motion
In this expression, the signs for the values substituted for va and V s depend on the direction of the velocity. When the observer moves toward the source, a !Jositive ACTIVE FIGURE 14.10 (a) A SOllrce Smo\"ing IVilh speed TiS 100\'ard Sl;\lionary observer A and awav (rom stationary observer B. ObS(:r\'cr A hears an incrl'{/scd frequency, and observer /3 hears a detTer/sed frequency. (b) The Doppler effect in water, observed in a ripple tank. The source producing the water wa\(~s is moving to lhe right.
(al
(bl
470
Chapter 14
Sound ~peed
is substituted for Yo; when the observer moves away from the source, a nega.tive speed is subsLituted for vo. Similarly, a jJositive speed is slibsliUllCd for Vs ,vhen
the source moves t.oward the observer, a l1rgativespeed when the source moves away from the observer. Choosing incon-ecL signs is the most common mistake made in \.. .orking a Dop-
pler effect problem. The following rules may be helpflll: The word lowanl is associated with an il1creaseill the observed frequency; the words away/rom are associated \'\"ith a decrease in the observed frequency. These t,vo rules derive from the physical insight that when the observer is moving [O\varcl the source (or the source LOward the observer), there is a smaller observed period between wave ClTsts, hence a larger frequency, ",,"ith the reverse holding-a smaller observed frequency-when the observer is moving away from the source (or the source away from the observer). Keep the physical insight in mind whenever you're in doubt about the signs in Equation 14.12: Acljust them as necessary to get the correct physical result. The second most common mistake made in applying Equation l4.12 is to accidentally reverse numerator and denominator. Some find it helpful to remember the equation in the follO\ving form:
Io 11
+
V()
v -
Vs
The advantage of this form is its symmetry: both sides are very nearly the same, with D's on the left and S's on the righl. Forgetting which side has the plus sign and ,,,'hich has the minus sign is not a serious problem as long as physical insight is used to check the ans\-"er and make adjustments as necessary.
QUICK QUIZ 14.2
Suppose you're on a
hOl
air balloon ride, carrying a
buzzer that emits a sound of frequency f If you accidentally drop the buzzer over the side while the balloon is rising at constant speed, what can you conclude about the sound you hear as the buzzer falls l.m\"ard the ground? (a) The frequency and imensity increase. (b) The frequency decreases and the intensity increases. (c) The frequency decreases and the intensity decreases. (d) The frequency remains the same, but the intensity decreases.
APPLYING PHYSICS 14.2
OUT-OF-TUNE SPEAKERS
Suppose you place your stereo speakers far apart and run past them from riglH to left or left to right. If you run rapidly enough and have excellent pitch discrimination, you may norice that the music playing seems 10 be ow of tune ,.."hen you're bel\.. . een the speakers. ',\Thy? Explanation When you are between the speakers, you are running away frolll one of them and LOward
EXAMPLE 14.4 Goal
the other, so there is a Doppler shift downward for the sound from the speaker behind you and a Doppler shift upv.mrd For t.he sound from the speaker ahead oryou. As a result, the sound from the two speakers ''''ill not be in tune. A calculation shows that a \vorld-c1ass sprinter could run fast enough to generate about a semitone difference in the sound from the two speakers.
Listen, but Don't Stond on the Trock
Solve a Doppler shift problem when only the source is [noving.
Problem A train moving at a speed of 40.0 mls sounds its whistle, which has a frequency of !i.OO X 102 Hz. Determine the frequency heard by a stationary observer as the train a!tproac!les the observer. The ambient temperature is 24.0'C.
14.6
The Doppler Effect
471
Strategy Usc Equation 14.4w gel the speed of sou lid at the ambient temperature, then subSlitlHC values into Equa· tion 14.12 for the Doppler shift. Because the train approaches the observer, the observed frequency will be larger. Choose the sign of V s LO rencet this fact ,... .
'~
Solution use Equation 1.... .4 to calculate the speed of sound in air at T= 24.0°C:
v
The obscn'cr is slmionary. so va = O. The Irain is moving toward the observer, so V s = + 40.0 m/s (Po.\iliT!e). Substitute these values and the speed of sound into the Doppler shift equation:
= (331
m/s)
= (331
m/s)
Jo
=
J" (
V
+
11 -
r-T \/2731( (273
.,
(
3<15 mls ) 345 mls - 40.0 mls
566 Hz
......................................... Remark
If the u-ain were going away frol11 the obscncr.
345 mls
va) v.,
= (5.00 X 10- Hz) ~
+ 2'kO) K 273 K
. Vii
= - 40.0
~
m/s would ha,te been chosen instead.
QUESTION 14.4 Docs thc Doppler shift change due to temperature variations? If so, wh),? For lypical daily \'ariations in temperature in a moderate climat<..:, ,.. .ollid any change in thc Doppler shift be best characteriLed as (a) nonexistent, (b) small, or (c) large? EXERCISE 14.4 Dctermine the frequency heard by thc stationary observcr as the train r('cedes from the observcr.
Answer
448 Hz
EXAMPLE 14.5 The Nois Siren Goal
Solve a Doppler shift problem when both the source and observer are moving.
Problem An amhulance travels down a highway at a speed of 75.0 mi/h, its siren emitting sound at a frequency of ll.OO X IO~ lIz. \\'11m frequency is heard b}' a passenger in a car traveling at 55.0 mi/h in the opposite direction as the car and ambulance (a) aJ)pmarh each other and (b) pass and move aWllY from each other? Take the speed of sound in air to be 11 = 345 m/s. Strategy Aside from convening mi/h to miSt Ihis problem only requires substitution into the Doppler formula. but two signs must be chosen correctly in each part. In pan (a) the obsen'er moves tm..· ard the source and the source moves toward the obseryer, so both Vo and iI.\ should be chosen to be positive. Switch signs after they pass each other.
,..
.
Solution Conven the speeds from mi/h to m/s:
. .... ..., 0.447 m/s)
Vs
= (75.0 milh) ( . 1.00 1111/h
Vo
=
Jo
=
J,
=
HI (345 mls + 24.6 ( 4.00 X LO-<') 345 mls - 33.5 m/s
0.447 m/s) (55.0 milh) ( . I.00I111/h
= 33.5
mls
= 24.6 mls
(a) Compute the observed frequency as the ambulance and car approach each other.
Each vehicle goes toward the olher, so substilutc V o = +24.6 mls and Vs = +33.5 mls into the Doppler shirt formula:
+
V ( 1/ -
1/ 0 )
Vs
Ill/s)
~
475 Hz
472
Chapter 14
Sound
(b) Compute the observed frequency as the ambulance and car recede from each other. Each vehicle goes away from the other, so substitute va = -24.6 m/s and V s = -33.5 m/s ;1ll0 the Doppler shift formula:
Remarks Notice hm.. · the sig'ns were handled. In part (b) the negative signs were required on the speeds because both observer and source were moving £1\\1'£1)' from each other. Sometimes, of course, one of the speeds is negative and the other is positive.
QUESTION 14.5 Is the Doppler shift affected by sound intensity level? EXERCISE 14.5 Repeal this problem, but assume the ambulance and car are going the same direction, with the ambulance initially behind the car. The speeds and the frequency of the siren are the same as in the example. Find lhe frequency heard by the observer in the car (a) before and (b) after the ambulance passes the car. Note: The highway paLrol subsequently gives the driver of the car a ticket for not pulling over for an emergency vehicle!
Answers
(a) 411 Hz
(b) 391 Hz
Shock Waves ''''hal happens when the source speed V s exceeds the wave velocity v? Figure l4.l1a describes this situation graphically. The circles represem spherical wave fronts emitted by the source at various times during its motion. At l = 0, the source is at point So, and at some later lime " the source is at point Sil' Tn the intervalt, the wave front. centered at SO reaches a radius of vi. In this same inlelval, the source tra\'els to Sill a distance of vit. At the instant the source is at SII' the waves just beginning to be generated at this point. have wave fi'onts of zero radius. The line drawn [rom 5" to the wave front centered on So is tangent to all other v·,'ave fronts generated at intermediate Limes. All such tangent Jines lie on the surface ofa cone. The angle 8 between one of these tangent lines and the direction of u'avel is given by v
sin () = -
VJ
The ratio v./v is called the Mach number. The conical ,...ave front produced \".hen vs > 1.J (supersonic speeds) is knov·,'I) as a shock wave. Figure 14.11 b is a photograph of a bullet traveling at supersonic speed t.hrough the hot air rising above a candle.
FIGURE 14.11 (a) A represent'll ion of a shock wave, produced when a source moves from Si} to \ with a .~peed lJ. I hal is grro(f'r lhan I he wave speed v in that medium. The envelope of the wave fronts forms a cone with half~angle of sin 0 = vlv•. (b) A stroboscopic pholograph of a bllllcL moving aL sllpersonic speed throllgh the hot air above a candle.
Conical wave fronl
o
I
(a)
•
(bl
14.7
"Pressure
Interference of Sound Waves
473
•
~
Atmospheric pressure (bl
(a)
FIGURE 14.12 The V-shaped how wave ofa boat is fanned because the bmu travels at a speed greater than I he speed of the water waves. A bow w;·lve is
FIGURE 14.13 (a) The two shock wavcs produced by the nose and tail ofajet airplane traveling at supersonic speed. (b) :\ shock wave due La ajet tnl\'eling at thc speed ofsollnd is made visihlt: as a fog ofwatcr vapor. The largt: pressure \'ariation in the shock wave causes Lht: warer in the air to condense into water droplets.
Notice the shock waves in the \~cinity or the bullel. Another interesting example or a shock ' . . ave is the V-shaped \..'ave front produced by a boac (the bow wave) ,,\·hen the boat's speed exceeds t.he speed of the \vater waves (Fig. 14.12). Jet aircraft and space shut.tles traveling at supersonic speeds produce shock waves that are responsible for the loud explosion, or sonic boom, heard on the ground. A shock wave carries a great deal of energy concentrated on the surface of" the cone, with correspondingly great. pressure variations. Shock waves are unpleasant to hear and can damage buildings when aircrafL fly supersonically at low altitudes. In fact, an airplane flying at supersonic speeds produces a double boom because two shock waves are formed: one from the nose of the plane and one from lhe lail (Fig. 14.13).
'J nlL
QUICK QUIZ 14.3 As an airplane flying with constant velocity moves from a cold air Illass into a warm air mass, docs the .\1ach number (a) increase, (b) decrease, or (c) remain the same?
14.7
INTERFERENCE OF SOUND WAVES
1
Sound wa\,es can be made to interrere with each other, a phenomenon that can be demonstrated with the device sho\'vn in Figure ]4.14. Sound from a loudspeaker at S is sent into a tube at P, ,.. · here there is a T-shaped junction. The sound splits and follows t\VO separate pat.hways, indicated by the red arrows. Half of the sound travels upward, half downward. Finally, the two sounds merge at an opening where a listener places her ear. 11" the two paths r\ and 1"2 have the same length, \','aves Lh..l l enter the junction \,,'ill separate into [wo halves, travel the t"\\'o paths, and then combine again at the ear. This reuniting of the two waves produces f01Htructive inler!prnnp, and the listener hears a loud sound. If t.he upper path is adjusted to be one rull wavelength longer than the lower pat.h, const.ructive int.erference of the two waves occurs again, and a loud sound is detected at the receiver. lyVe have the following result: If the path difference T2 - '-1 is zero or some integer multiple of wavelengths, then constructive interference occurs and (n= 0, 1,2, ...)
[14.13]
Suppose, however, LhaL one or the path lengths, "(2' is adjusted so Lhat the upper path is half a wavelength longer than the Imver pat.h fl' In this case an entering sound \·\lave splits and travels the two paths as before, bUI now t.he wave along the upper path must travel a clisLo1.nCe equivalent to half a wa\'elength farther than Lhe ",'ave traveling along the lower path. As a result, the crest of one wave meets the trough or the othe,- when they merge at the receiver, causing the two \"laVeS 10 cancel each other. This phenomenon is called lotally rleslnu"tivf inlmference, and
1l L) jrC~ivec
Speaker
FIGURE 14.14 An acoustical system ror dernonstratillg inl.errert:ncc of sound waves. Souno hom the speaker enters rhe tube ,md SpJiL" into two parts at P. The two waves combil1e al Ihe opposite side and are detected at N. The upper path length is v,wicd by the sliding section.
+-
Condition for constructive interference
Chapter 14
474
Sound
Condition for destructive interference ~
APPLICATION Connecting Your Stereo Speakers
Tip 14.3 Do Waves Really
Interfere? ]n popular usage, [0 iTltl!lf(!fe means "to come illlo conOiCl ....'ilh" or "to intervene [0 affect all otUcomc.~ Thi~ differs from its lise in physics, where \,'aves pass through each other and interfere, but don't affect each oLher in any way.
no sound is detected at the receiver. in general. if the path difference r 2 wavelengths, destructive interference occurs and
~, l~,
r, - " = (11 + ~)A
(11
= 0,
1,2,.
r, is
-
24 ...
[14.14]
.)
Nature provides many other examples of interference phenomena, most notably in connection ,..· ith light "vaves, described in Chapter 24. In conncning the wires bcr"vcen your stereo system and loudspeakers, you may noLice that the ,,,ires are usually color coded and that the speakers have positive and negative signs on the connections. The reason for this is that the speakers need to be connencd with the same "polarity." H they arcn'l. then the same electrical signal l'cd to both speakers will result in one speaker cone moving outward at the same time that the other speaker cone is moving inward. In this case, the sound leaving the two speakers will be 1.80 0 out of phase with each other. If you are sitting michvay bel\.. .een the speakers, the sounds from both speakers travel the same distance and preserve the phase difference they had ,,,,'hen they left. In an ideal situation, for a ]800 phase difference, you would get completc destructive interference and no sound! In reality, the cancellation is not complete and is much more signi fkant [or bass notes (which have a long wavelength) than for the shaner 'wavelength treble notes. Nevertheless, to avoid a significant redl.lcrion in the illlensit)' of bass notes, the color-coded wires and the signs on the speaker connections should be carefully noted.
EXAMPLE 14.6 Two Speakers Driven by the Same Source Goal
Use the concept of interference to cOl11pute a frequency'.
Problem Two speakers pll'lced 3.00 111 apan are driven b}' the same oscillalOf (Fig. )4.15). A listener is originally at point 0, which is located 8.00 m from the center of the line connecting the £\...·0 speakers. The listener then walks to point P, which is a perpendicular distance 0.350 In from O. before reaching the first minimum in sound intensity. "'That is the frequency of the oscillator? Take the speed of sound in air to be V s = 343 m/s.
~~ ----~-__ I
~
a.350m
----~if. - - .J
r
FIGURE 14.15
(Example 14.6) Two loudspeakers dril'en by the same
Strategy The position of the fIrst minimurn in sound source can produce interference. intensity is given, which is a point of destructive interference. ''''e can find the path lengths r\ and 1'2 with the Pythagorean theorem and then use Equation tive interference to nnd the wavelength A. Using v = fA then yields the frequency. ~
1~1.]4 [or
destruc-
,
~
Solution Use the Pythagorean theorem to find the path lengths 1'\ and 1'2:
Substitute these values and solving for the wavelength:
(J.l5 m)'
= 8.08 In
r, = V(8.00 m)' + (1.85 m)' = 8.21
= 0 into Equation ]4.14,
11
'; -
1", = (n
.
I=
11
-=
A
34·3 m/s 0.26 m
In
+ VA
8.2] m - 8.08 m = 0.13 m = A/2
Solve v = Ai for the frequency f and substitute the speed of sound and the wavelength: ~
" = V(8.00 m)' +
-->
A = 0.26
In
1.3 kHz
............
,
Remark For problems involving constructive interference, the only difference is that Equalion H.l3, r2 would be used instead of Equation 14.14.
~
-
1'1
= nA,
QUESTION 14.6
True or False: In the same conlext, smaller ,,,,'avelengrhs of sound would create more interference maxima and minima than longer wavelengths.
14.8
Standing Waves
475
EXERCISE 14.6 If lhe oscillmor frequency is acljusled so that the location of the first minimum is at a distance of 0.750 m from 0, what is the new frequency?
Answer
0.624 kHz
14.8
STANDING WAVES
St£lIlding ,,,"ayes can be set up in a stretched string by connecting one end of the string to a stationary clamp and connecting the other end to a vibrating object. such as the end ofa tuning fork, or by shaking the hand holding the string up and down at a stcady rate (Fig. 14.16)_ Traveling wavcs then reOect from thc ends and move in both directions on the string. The incidcllL and reflccted waves combine accord ing to the superposition principle. (Sec Section] 3.] 0.) I f the stri ng vibrates at cxactly thc right frcquency. the "'a,·e appears to stand still, hence its name, standjng wave. A node occurs wherc thc twO travcling wa"es always ha\'e the same magnitudc of displaccment. but the oppositc sign, so the nCt displaccmcnt is zero at th'l1 point. Therc is no motion in thc string atlhc nodcs, b1lt midway between nvo acljacent nodes, at an antinode, the so;ng vibrates with tJ,c largest ampliLUdc. Figure 1.<1.17 shows snapshots of the oscillalion of a standing wm-e during half of a cyclc. The pink arrows indicate the direction of motion of different pans of Ihe string. Notice thal all points on the string oscillate together vertically with the same frequency, but different points have different amplitudes of motion, The points ofal.lachment to the wall and all olhcr slalional'y points on the string are called nodes, labeled N in Figure 14.l7fl. From the ngure, obsen-e that the disLance betwcen a(ljacelll nodes is one-half the wa\'elen~lh of the wave:
Vibrating blade FIGURE 14.16 Standing wa\'es can be SCt up in a stretched string by connecting ouc end or the :>lring to a \'ibrating blade_ When the blade vibrates at Olll" of Ihe nallll""] frcqucncies or the string, largc·ampliUlde ~landing \\'a\'c:> an- Cl-caled.
d~~ = ~A Consider a string of length L thai is fixed al both ends, as in AClive Fig·ul'e 14.18, For a string, we can set up standing-,vave p,Htcrns at many frequencies-the morc
1+---- L ---+1
(a}
,\
FIGURE 14.17 .\ st:lI\ding·wa\e p:ltIITn itl;l Slrclched Sli ing, shoWl1 or ~llap~hOl~ ollhe 'ilring dill ing olle-hit/fora ncJe. III pan (a) N dl'll('ICS
(h)
ACTIVE FIGURE 14.18 (a) Standing \\ a\"n in ,1.\1 retched :>1 ring or length L lixetl at both C'l1d~. TIl(' ch~lrach;rislic freqllt:ncies clr\,illl'al ion llll"m ;L harlllonic series: (b) the flindalllenta] fl"equcllCr. or first harmonic; (c} the second h:ll"l'n,)nic; and (d) the thirrl harmonic. I\'ole lh:1l N
476
Chapter 14
Sound
loops, the higher the frequency. Three such patterns arc showl1 in Active Figures 14.18b, 14, 18c, and 14.18d. Each has a ch;uacterisLic frequency, w"hich we will now calculate. First, tbe ends of the string must be nodes, because these points are fixed. If the string is displaced at its midpoint and released, the vihr;:ltioll shown in Artivp Figure 14.18b can be produced, in which case the center of the string" is an anlinode, labeled A. Note that I'rolll end to end, the pattern is )J-A-N. The distance [rom a node to its acUacent Clntinode, N-A, is ahvays equal co a quarter v"'-Clvelenglh, ',/4. There are two such seglllems. N-A and A-i':, so L = 2(A/4) = ',/2, and AI = 2L. The frequency of this vibration is therefore
(a)
v
Ji
A;" =
=
v 2L
[14.15]
Recall that the speed ofa wave on a string is v = \/F/J.L, where Fis the tension in the string and J..L is its mass per unit length (Chapter 13). Substituting imo Equation 14.15, we obtain [14.16] (c)
Multillash phntographs of standillgwa\e pancrns in a cord driven by a \"ibralOr ai, the left end, The singleloop pattern in (a) represcliis the
fundamemal frequency (n
=
I). the
two-loop paHern in (b) the second
harmonic (n:= 2), and Ihe Ihree-loop paHern in (e) the lhird h,lnTlonic (II = 3).
This lowest frequency or vibration is called the fundamental frequency of the vibrating string, or the first harmonic. The first harmonic has nodes only at the ends: the points of attachment, with node·anlinode paLLerll ofN-A-N. The next harmonic, called the second harmonic (also called the first overtone), can be constructed by inserting an additional nodeanlinode segment between the endpoints. This makes the pattern N-A-N-A-f\T, as in Active Figure 14.18c. We count the node-antinode pairs: N-A, A-N, N-A, and A-N, four segments in all, each representing a quarter ' I. . avclength. \'Ve then have L = 4(A:/4) = A2 , and the second harmonic (first overtone) is
h
= -
v Lv 2 (9_Lv)
A2
= -
=
=
2Ji
This frequency is equal to lwice the fundamental frequency. The third harmonic (second overtone) is constructed similarly. Inserting one Illore J\'-A segment, "ie obtain Active Figure 14.18d, the patt.ern of' nodes reading K-A-N-A-N-A-N. There are six node-anti node segments, so L = 6(A 3 /4) = 3(A:/2), which means that A~ = 2L/3, giving
fl
v A,
3v 2L
=-=-=
3J;
All the higher harmonics, it turns out, are positive imeg·er multiples of the fundamental:
1" The frequencies!l>
2/1,
=
nJi
n = 2L
(F-
V;
11.
= 1,2,3, ...
[14.17]
3.il, and so on form a harmoll.icseries.
QUICK QUIZ 14.4 Which of the following frequencies are higher harmonics of a slring wilh fundamental frequency of 150 Hz? (a) 200 Hz (b) 300 Hz (c) 400 Hz (d) 500 I-Iz (e) 600 Hz \Alhen a stretched string is distorted to a shape that corresponds to anyone orits harmonics, after being released it vibrates only at the frequency of that harmonic. If the string is struck or bm.. .ed, however, the resulting vibration includes different amounts of' various hannonics, including the fundamental frequency. Waves not in the harmonic series are quickly damped out on a string fixed at both ends. In effect, when disturbed, the string "selects" the standing-wave rrequencies. As "ie'll
14.8
see later, the presence of several harmonics on a string gives stringed instruments thcir characteristic sound. which enables LIS to distinguish one from anothcr cven when they are producing identical fundamental frequencies. The frequcncy of a string on a musical instrument can be changed b>1 varying either the Icn>iion or the Ipllgth. TIl(' tf'nsion in glJit:.:l:r ::Inc! violin slrings i,1; V::lJ"ied by turning pegs on the neck of the instrumclll. As the tension is increased, the frequency of the harmonic series increases according to Equation 14.17. Once the instrument is tuned, the musician varies the rrequency by pressing the strings againslthe neck at a variely of positions, thereby changing the effective lengths of the vibrating portions of the strings. As the length is reduced, the frequency again increases, as [ollo\'.'s from Equation 14.17. Finally, Equation 14.17 shows that a string of fixed lcngth can be made to vibrate at a lower fundamental frequency by increasing its mass per unit length. This increase is achieved in the bass strings of guitars and pianos by wrapping the strings with metal windings.
EXAMPLE 14.7 Goal
Standing Waves
477
APPLICATION Tuning a Musical Instrument
Guitar Fundamentals
Apply standing-wave concepts to a stringed instrumcnt.
Problem The high E string on a certain guitar measures 64.0 em In length and has a fundamental frequcncy of 329lh. \Vhen a gllitar~ ist presses down so that the sU-ing is in contact with the first fret (Fig. 14.19£1), the string is shortencd so that it plays an F note that has a frequency 0[349 Hz. (a) How far is the fret [rom the nut? (b) Overtones can be produced on a guitar string by gently placing the index finger in the location of a node of a higher harmonic. The string should be touched, but. not depressed against. a fret. (Given the width of a finger, pressing l.00 hard v,,Iill damp out higher harmonics as well.) The fundamental Frequency is thercby suppressed, making it possible to hear overtones. Where on the guitar string relative LO the nut shoulclthe finger be lightly placed so as to hear the second harmonic? The fourth harmonic? (This is equivalent to finding the location of the nodes in each case.)
(a)
bridge
~nd
frel
Strategy For pan (a) lise Equation ]4.15, corresponding LO the funda(b) mcntClJ frequency, to find the speed of waves on the string. Shortening lhe string by playing a higher note doesn't affect the wave speed, ,...· hich FIGURE 14.19 (Example 1'1.7) (a) Plaring an F note all a guitar. (b) Some pans ora guilar. depends only on the tension and linear density of the su-ing (vl.'hich are unchanged). Solve Equat.ion 14.15 for the ncw length L, using the new Fundamental frequency, and subtract this length [1'0111 the original length to find the distance frol11 the nut to the first frer. In part (b) remember that. the distance from node to node is haifa v,:avelength. Calculate the wavelength, divide it in two, andlocC\te the nodes, "vhich are integral numbers of half-,vavelengths [rom the nuL ""oip: The nut is a small piece of wood or ebony at the LOp of the fret hoard. The distance from the nut 10 the bridge (belm\' the sound hole) is the length of the string. (See Fig. 14.19b.)
Solution (a) Find the distance from the nut to the first frct. Substiwtc L(( = 0.640 m and}; = 329 Hz into Equalion 14.15, lInding the wave speed on the string:
It ;
v 9L -
0
v; 2 Lof, ; 2(0.640 m)(329 Hz) ; 421 m/s 421 m/s 2 ( 349 Hz )
Solve Equation 14.-15 for the length L, and substitute the wave speed and the frequency of an F note. Sllbtractthis length from t.he originallengt.h L o to find lhe distance from the nul to the firSl fret.:
D.X;
Ln - L
~
64.0
~
. 0.603
elll -
III ;
60.3
60.3
CIll;
CIll
3.7
CIll
478
Chapter 14
Sound
(b) Find the locations of nodes for the second and fourth harmonics.
The second harmonic has a \'\'avelength A:? = 1.... 0 = 64.0 em. The distance from nullo node corresponds to haIfa wavelength. The fourth harmonic, of wavelength ;\4 = ~Lo = 32.0 CIll, has three nodes bet\vecn the endpoints:
Remarks Placing a finger at the position Ax = 32.0 em damps out the fundamental and odd harmonics, but not all the higher even harmonics. The second harmonic dominates, however, because the rest of the string is free to vibrate. Placing the finger at .1.x = 16.0 (Ill or 48.0 em damps Ollt the first through third harmonics, allowing the founh harmonic 1O be heard.
QUESTION 14.7 True or False: If a guitar string has length L, gently placing a thin object at the position (~)II twill always result in the sounding a higher harmonic, where 11 is a positive intcger. EXERCISE 14.7 Pressing the E string dm.. .' n on the fret board just above the second fret pinches the string firmly against the fret, gi\'· ing an F-sharp, which has frequency 3.70 X 10' Hz. (a) Where should the second frel be localed' (b) Find lwo locations \,'here you could tol/ch thc open E string and hear the third harmonic. Answers (a) 7.l cm from the nut and 3,4 cm from the first fret. )Jote that the distance from the first to the second fret isn't the same as from the nut to the first fTeL (b) 2].3 Clll and 42.7 cm from the nut
EXAMPLE 14.8 Goal
Harmonics of a Stretched Wire
Calculate string harmonics, relate them to sound, and combine them ' . .· ith tensile stress.
Problem (a) Find the frequencies of the fundamental, second, and third harmonics of a steel wire 1.00 III long ,,,,,ith a mass per unit length of 2.00 X 10-3 kg/Ill and under a tension ofSO.O N. (b) Find the wavcJengths of the sound 'Naves created by the vibrating wire for all three modes. Assume the speed of sound in air is 34-5 m/s. (c) Suppose the wire is carbon steel with a density of 7.80 X 10 3 kg/nl"'I, a cross-sectional area A = 2.56 X 10-7 m 2 , and an elastic limit or2.S0 X 108 Pa. Find the fundamental frequency if the wire is tightened to the elastic limit. Neglect an)' stretching or the wire (which would slightly reduce the mass per unit length).
Strategy (a) It's easiest to find the speed of waves on the wire then substitute into Equation H.15 to find the first harmonic. The next tVI'O are Illultiples of the first, given by Equation 14.J7. (b) The frequencies of the sound waves are the same as the frequencies of the vibrating wire, but t.he wavelengths al-e different. Use 11£ = fA, ,.,.'here u£ is the speed of sound in air, LO find the wavelengths in air. (c) Find the force corresponding to the elastic limit and substitute it into Equation 14.16.
Solution (a) Find the first three harmonics at the given tension. Use Equation l3.ISLO calculate the speed orthe \\lave on the wire:
v=
!?:' I -~
/.L
v 2L
Find the \-"ire's fundamental frequency from Equation 14.15: Find the next two harmonics by multiplication:
\
I,
=
2I,
\
80.0 N , =2.00XIO-m!s 2.00 X 10 3 kg! m
2.00 X 10' mls 2(1.00 m)
= 2.00 X 10' Hz ,j; =
1.00 X 10' Hz
3I,
= 3.00 X 10' Hz
14.9
Forced Vibrations and Resonance
479
(b) Find the wavelength of the sound waves produced. Solve v" = JA for the wavelength and substitute 1he frequencies:
~
vif,
= (345 m/s)/(1.00 X 10' Hz) =
3.45 m
A, =
v/};
= (345 m/sl/(2.00 X 10' Hz)
1.73 m
A" =
vi;;
~ (345 m/sl/(3.00 X 10' Hz)
l.l5 m
A,
(c) Find the fundamental frequency corresponding LO the elastic limit. Calculate the tension in the wire from the elastic limit:
Suhstitute the values of F, f.l, and L into Equation 14.16:
F - = clastic limit
A F
=
Ii
= 2L \ / ;
Ii
= 2(1.00 m) \
---+
F = (elastic limit)"'\'
(2.80 X 108 Pa)(2.56 X 10-7 m')
= 71.7 N
IfF I
!
71.71\ 200 X 10 'kg/m
94.7 Hz • _ •• _.A
.
~
Remarks From the ans\ver to pan (c), it appears we need to choose a thicker wire or use a better grade of steel with a higher elastic limit. The frequency corresponding to the elast.ic limit is smaller than the fundamental! QUESTION 14.8 A string on a guitar is replaced with one of Im·ver linear densil)'. To obtain the samc frequency sound as previously, must the tension of the ne\v string be (a) grcater than, (b) less than, or (c) equalLO the tcnsion in the old string? EXERCISE 14.8 (a) Find the fundarnental frequcncy and second harmonic if the tension in dle wire is increased to 115 N. (Assume the wire doesn't stretch or break.) (b) Using a sound speed 0[345 mis, find the \v;welengl.hs of the sound waves produced.
Answers
14.9
(a) 1.20 X 10' Hz, 2.40 X 10' Hz
(bl 2.88
Ill,
1.44 m
FORCED VIBRATIONS AND RESONANCE
In Chapter 13 we learned that the energy of a damped oscillator decreases over time because of friction. It's possible to compensate for this energ'y loss by apply~ ing an external force that. does positive work on the system. For cxample, suppose an objccr-spring systcm having some naUiral frequency of vibration.fo is pushed back and fonh by a periodic force \\~th frequency! The system vibrates althe frequency/of the driving force. This type or motion is referrcd to as a forced vibration. ll.s amplitude reaches a maximum when the frequency of the driving force eqllals the natural frequcnC)' of tJ1e system ic"l' called the resonant frequency of the system. Under this condition, t.he systcm is said to be in resonance. In Section 14.8 we learned that a stretched string can vibrate in one or more of its natural modes. Here again, if a periodic force is applied to t.he string, the amplitude or vibration increases as the frequency of the applied force approaches one of the su-ing's nault'al frequencies of vi braLion. Resonance vibrations occur in a wide variety of circumstances. Figure 14-.20 illustrates one experiment tJlat demonsu·ates a resonance condition. Se\'eral pendulums of different lengths are suspended from a flexible beam. If one of them, such as A, is set in mOlion, the others begin to oscillate because ofvibrmions in the flexible beam. Pendulum C, the same length as A, oscillates with the greatest amplitude because its natural frequency matches that of pendulum A (the driving force). Another simple example of resonance is a child being pushed on a s\ving, which is essentially a pendulum with a natural frequency that depends on its length. The swing is kept in motion by a series of appropriately limed pushes. For its amplitude 10 increase, the swing 111ust be pushed each time it returns to the person's hands.
AV B FIGURE 14.20 I{csOIl,l11ct-·.lfpendulum 1\ is SCI. ill oscillation, ani)' pcndullim C, wilh a lenglll matching lhal of 1\, will e,cmualh' oscillate with ,I large ;ImpliLud..:. or resonale. The :!lTOh'S indicl\c motion perpcnrliClllar 10
the page.
480
Chapter 14
Sound
APPLICATION Shattering Goblets with the Voice
FIGURE 14.21 Standing-wave pattern in a vibrating wineglass. The glass will shatter if the :l.Inplilude of vibration becomes too large.
This corresponds to a frequency equal to the natural frequency of the swing. If the energy put into the system per cycle of mOlion equals the energy lost due LO friction, the amplitude remains constanl. Opera singers have been known to set crystal goblets in audible \'ibration with lhr.ir pnwe.rflll voices, as shown in Figure 14.21. This is yet another example of resonance: The sOllnd waves emill.ed by the singer can set up large-amplitude vibrations in the glass. If a highly amplified sound wave has the right frequency, the amplilUde of forced vibrations in the glass increases to the point where the glass becomes heavily su"ained and shaLters. The classic example of structural resonance occurred in 1940, \.. . hcn the Tacoma Narrm.. . s bridge in the slate of \VashingLOn \.. .as set in oscillation by the wind (Fig. 14.22). The ampliulde of the oscillations increased rapidly and reached a high value until [he bridge ultimately collapsed (probably beG1USe of met<:ll faligue). In recent years, hm-veve1', a number of researchers have called this explanation into question. Gusts of\vind, in general, don't provide the periodic force necessary for a sustained resonance condition, anclthe bridge exhibited large twisting oscillations, rather than the simple up-and-dov.:n oscilhnions expected of resonance. A more recent example of" destruction by su·ucLUral resonance occurred during the Lama Prieta earthquake near Oakland, California, in 1989. In a mile-long section of the double~deckerNimitz Freeway, the upper deck collapsed onto the Im...er deck, killing several people. The collapse occurred because that particular section was built on mud fill, whereas other parts were built on bedrock ..-'\.s seismic waves pass through mud fill or other loose soil, their speed decreases and their amplitude increases. The section of the free\vay that collapsed oscillated at the same frequency as other sections, but at a much larger amplimde.
14.10
STANDING WAVES IN AIR COLUMNS
Standing longitudinal waves can be set up in a tube of air, such as an organ pipe, as the result of interference bet\'\'een sound waves traveling in opposite directions. The relat.ionship bet\"o'een the incident wave and the renected wave depends on ""'hether the renecting end of the tube is open or closed. A portion of the sound wave is rerlected back into the lUbe even at an open end. If one end is closed, a node must exist at that end because the movement of air is restricted. If the end is open, the elements of air have complete freedom of motion, and an antinode exists. Figure ]4.23a shows the first three modes of vibration of a pipe open at both ends. ''''hen air is directed against an edge at the left, longitudinal standing waves are formed and the pipe vibrates at its nalUral frequencies. Note that, from end to end, the pattern is A-~-A, the same pattern as in the vibrating sU"ing, excepr node and antinode have exchanged positions. As before, an anLinode and its adjacent node, A-N, represent a quarter-v,:avelengt.h, and there are 1.'\'0, A-N and N-A, so L = 2(A,/4) = A,/2 and A, = 2L The fundamental frequency of the pipe open at both ends is then.lt = v/ Al = v/2L The next harmonic has an additional node and antinode betv.'een the ends, creating the pattern A-N-A-N-A.
FIGURE 14.22 The collapse ortlle Tacoma Narrows suspcllsiolJ bridge in 19'10 has been cited as a demollstration ofrnecbanical resonance.lligh winds sel up standing waves in the bridge, causing it to oscillate alone of its nalural frequencies. Once cstahlished, Ihe rCSOllance lIlay have led to the bridge's collapse. (Mathcmalicians and physicists arc curn:nlly challeng-ing this inlcrpret'ilion.)
14.10
:\
r L-1 ~ ~XX ~
A
\l
A
N
T'
Ii = ~:=~, ,\~ = /.
fn.'f!lll·l\cics arc {,.
:~I,.
,·111' and .\01111.
f.! =i='2fl
Tip 14.4 Sound Waves Are Not Transverse
~ ==?1 ~XJ ~XXJ
A
FIGURE 14.23 {;II SI:lnding longiwdin
Al = '2/ T'
481
Standing Walles in Air Columns
AI
Ii
I
41
",2!..=2!.. .:II
The standing Jongiuldinal
First harmonic
wa\'c~
Fig-Lin' 14,2~ are dr~l\\'n ;tS tranS\Cl"SC wa\'e5 on Iy bec~llse il\ difficliltio draw longitudinal di ..plaCCIlW1l1s: l1wr're in the sallle dllCCllOll as the \\,\\e propagallOtl III till: rigul c, 1111' \ClllcaJ axl" I cpIcsenLS Cllhel pt CSSIII C or hOIIZ()Ilml (!l:,>placclllclll oflhe e1t'lllCnt\ !..hL' llwdlUIll ill
-II.
[ Third h,lIllWllic
-= I
or l----------_/
(h) Clo:-cc!.1t one clld, upcn ;11 lhc OLlJ(.T
,,,le COlinL the pairs: A-N, N -A, A-N, and N -A, making lour segments, each WiLh Icngth A:/4. \!\7 C have I, = 4(A 2 /4) = A:!. and the second harmollic (first overtone) = v/A:.! = v/L = 2(v/2L) = 2/i. All higher harmollics. it turns OUl, are positi\"c integer lllulLip]e~ of the fundamenl.al: is./~
J,Ii
"
~ 1 1 - = /1(,
2L
11
.II
=
I, 2. 3, .
rI4.18]
+-
Pipe open ot both ends; all harmonics are present
w/1ere 11 b Lhe speed of sound in air, Notice the similarity 10 Equation 14-.17, which also involves Illulliples of l.he fundamental. Ira pipe is open al 011(' end and closed allhe other, lhe Opt'll cnd is .1n antinode ;llld the closed end is (l node (Fig'. 14.23b). In such a pipe. the fundamental I"rectllellc)' consists of;1 single aminodc-lIode pair, A-~, so 1. = A[/~l and Al = ~1L. The fundamental harmonic for a pipe closed at one end is then.ft = TI/ AI = v/4L. The first m"ertone has another node and anti node bet\\'eell the open end and closed end. making the pattern A-N-A-N. There arc three alllinode-node segIllCllts in this paLlerll (A-N, ~-r\., ~lJ1d ;;'\-N) , so L = ~(.A:J-I) and A:l = 41./3, The first overtone therefore has li'equelley j~ = v/ A:\ = :iv/4L = 3/i. Silllilarly. I, = 5]... III ("otltl"asL to the pipe open
J,,,QUICK QUIZ 14.5
TJ
II - .
41-
= n r,
A pipe open
"'
ill
n = I, :1,5..
[14.19]
both ends resonales at a fundamental
fn.::qllencyJ;"lpl'n' \'\"hen olle end is covered
describes how these 1\\'0 reson;lnt frequcncies COlll)Xlre? ((I) j~l'''''d = J..,pen (h) .J:1o'~l'rl = 1)."'1)(·" (c) .r.I",~.d = 2j;'1l("1 (d) /'1<,,,·,1 = ~.f,'P"" (t·) 110n(' ort.hesc
+-
Pipe closed at one end; only odd harmonics ore present
482
Chapter 14
Sound
QUICK QUIZ 14.6 Balboa Park in San Diego has an outdoor organ. When tilL air t.emperature increases, the fundamental frequency of one of the organ pipes (a) increases, (b) decreases, (c) stays the same, or (d) is impossible to determine. (The thermal expansion of the pipe is negligible.)
APPLYING PHYSICS 14.3
OSCILLATIONS IN A HARBOR
\Vh)' do passing ocean waves sometimes cause the ,,'mer in a harbor to undergo ycry large oscillations, called a seir/u' (pronounced sflysh)? Explanation Water in a harbor is enclosed and possesses a natural frequency based on the size of the harbor. This is similar to the natural frequency of the enclosed air in a boule, which Gill be excited by blowing across the edge of the opening. Ocean waves pass
APPLYING PHYSICS 14.4
WHY ARE INSTRUMENTS WARMED UP?
VI-hy do the sU'ings go flal and dle wind instrumCIHS go sharp during a performance iran orchestra doesn'l \\"arm up beforehand? Explanation Without warming up, all the instruments \,·ill be at room temperature (It the beginning of the concert. As the wind instrulllents arc played, they fill with warm air [rolll the player's exhalation. The increase in wlllpcralure of the air in the instruments causes an increase in the speed of sound,
APPLYING PHYSICS 14.5
Explanation Songs lor [he bugle are limited to harmonics o[ the fundamental frequency because there is no control over f"rcqucncics withoLit valves, keys, slides, or finger hules. The player obtains different notes by changing the tension in the lips as the bugle
Goal
which raises Lhe resonance frcql.lcnciesofllJeair colullllls. As a result. the instruments go sharp. The strings all the stringed instrlllllcms also increase ill telllperatll re clue to the friction of rubbing with the bow. This results in thermal exp~lIlsion, which causes a decrease in tension in the strings. With the decrease in tension, the wave speed on the strings drops and the fundamental frequencies decrease, so the stringed instruments go nat.
HOW DO BUGLES WORK?
.-\ bugle has no ,·al\'es. keys. slides, or finger holes. I-low can il be llsed to playa song?
EXAMPLE 14.9
by the opcni ng of tIle harbor at a certain frequency. [fthis frequency matches [Ilal of the enclosed harbor, then a large standing wan' Gin be set up in the water by resonancc. This situation can be simulatcd by carrying (\ fish t~lnk fillcd with water. Ify"Oul"\\I'alking frequency matches the natural frequency of the ,,·ater as it sloshes back and lorth, a large standing \nwe deyelops in the fish tank.
is played, exciting different harmonics. The normal playing range ofa bugle is alllong the third, fourth, firth. and sixth harmonics of the fundamental. '"Rc\,eille," for example, is played with just the three nOles G, C, and F, and "Taps" is played WiLh these three notes and the G one octave above the lower G.
Harmonics of a Pi e
Find frequencies of open and closed pipes.
Problem A pipe is 2.46 11"1 long. (a) Dctermi ne the frequencies 01" the first I hree harmonics i r the pipe is open
14.l0
Standing Waves in Air Columns
483
. .. ...,
,..-' ..... Solution (a) find the frequencies iflhe pipe is open at both ends. Substitute imo Equalion 14.18, with n = 1:
.It
.\hlltiply La find the second
f,
343111/s 2(2.46 Ill)
"
2/. = 21, =
139 H,.
69.7 lIz
I, = 3Ji
209 J-ll.
(b) 1-I0\\' many harmonics lie between 20 liz and 20 ODD Hz ror I his pi pel V 1 / - = 1/
SeLl.he frequency in Equation 14.18 equal to 2.00 X 10 I Hz and solve for II:
2L
343 Ill/s 2·2.46
= 2.00 X 10' Hz
III
n = 286
This works out La 11 = 286.88. which lllllst be trullcated down (11 = 287 gi\'es a frequency o\'cr 2.00 X 10 I Hz): (c) Finc1the frequencies for lhe pipe closed al aile end.
Apply Equation 14.19 with
'!7
J,
= I:
The next twO harmonics
I, =
11
343 m/s
4L
'1(2.46 m)
3f,
= [05
Hz
34.9 Hz
f,
~
5f, =
175 H,
.... . .
.........
QUESTION 14.9 True or false: The fundamental \\"
Answer
(a) OA64 111,7.40 X 10' H,
EXAMPLE 14,10
(b) IS!') Hz, 5.'i5 Hz
(c) 54
Resonance in a Tube of Variable Length
Goal Understand resonance ill llIbes and perform elcmentary calculal ions. Problem Figure l-J.2--1a shows a sil11ple apparatus for demonstrating resonance in a lube. A long tube open at both ends is plll·tially submerged in a beaker of W3[er, and a \'ibratillg lUning fork or llnkno\...·n frequency is placed ncar the lOp of the Lube. The length of [he air column, I"~ is acljusted by rnm'ing Lhe tube \'crticall). The sound ,,'aves generated by the fork are n:inforred when the length of the air colul11n corresponds [() one or the resonant. rrefJuencies of the tube. Suppose the sl11<1l1est \'alue of L for which a peak occllrs in the sound inlensity is 9.00 em. (a) \'Vith this l11ea~t1rel11ent, determine the frefJuellc), of the tuning [ark. (b) Find the wa\elenglh and the next 1.\\'0 air-column lengths giving resonance. Take the speed 01" sound 10 be 345 mh.
Fir'it
reSOllanct.' Sl'COlld n::s0t1iJI1Ce
\\'aler
(third
harmonic) (a)
Third n':S(lllann: ( fif1l1 h;u'lllllnic)
(h)
(FX
FIGURE 14.24
Strategy Once the Lube is in Ihe ,\'ater. Ihe setup is the same as a pipe closed at one end. For pan (a). subs[ itllle \'alues for v
484 ro
(a) Find the Crequcncy of the Illlling fork. SubstillllC n = J, 11 = 345 mis, and into Equation 14.19:
1...,1
= 9.00 X IO-:?
345 III
4(9.00
X
Ill/ S
10 'm)
958 Hz
(b) Find the wavelength and the next two ' . . . ater levels gh'ing resonance.
0.360 m
Calculate the wavelength, using the faCl the-H, for a tube open at one encl, ,\ = 4L for the fundamental. Add a half-wavelength or distance resonance position:
101- 1
to gel the next
Add another half-H'
L, = L,
+ A;2
=
0.0900 m
L, = L,
+ A/2
=
0.270 m
+ 0.180 m = 0.270111
+ 0.l80 m = 0.450 m
resonance position:
....
• • • • • • • ...011
Remark This experimclllal arrangement is often lIsed to Illca~lIre lhe speed of sound, in which case the Frequency of tile tuning fork must be knO\nl in advance, QUESTION 14.10 True or False: The rcsonanl Crequcncy of' an ~\ir column depends on the length of the colulllll and the speed of sound. EXERCISE 14.10 An unknown gas is introduced inlo the aforementioned apparatus using L1le same tuning lurk, and the first resonance OCCllrs v,,'hell the air column is 5.R4 (m long, Find thc speed of sOllnd ill the gas.
Answer
224 Illis
14.11
BEATS
The interfercncc phcnomena \\ie have been disclIssing so fal' have invoked the superposition of t\\'o or morc ",a\'cs with the sallle frequency. tfcwcJing in opposite directions. Another type of interference effect results from the superposition of two \-vavcs \,.. ith slighlly differcnt frequencies. III such a situation, the \\im'es at some fixed point are periodically in and out of phase, corresponding to an alternation in time between cOllstnlClive and destructive interference, To understand this phenomenon, consider Active Figure 14.25. The twO waves shown in Acri\'e Figure 14.2.~a were cmitted by two tuning forks having slightly different frequencies: Active Figure 14,25b shO\·\ls thc sllperposition of these waves. At sOllie time the wan:s arc in phase and cOllst.ructi\"e ill(Crfcrence occurs, as demonstrated by (he resultant curve in Active Fig'ure 14.25b. AI. some lalcr lime, however. Lhe vibrations of the t\
'II
ACTIVE FIGURE 14.25 BealS afc fOfllll:d lw the Clllnhill;lliun of 1\\'0 \\;II'L'S of ~lightl\' dillerclIl ffcquencie:. lr;l\'elillg" in Ihe sam(' dilCClIIJ!l, (;I) The imli\idllal \\,l\es lH::nd b) ;]11 "bscrlt·r al a fixcd poinl ill 'pan·. (bl The rOlllbilll'd 1I".lH' !la, all ,lrnplillllk (dashed lilll') tll"l osril1;]It" in lime.
mon' out or phase, then illlO phase again. and so on. As a conscClucncc, a listener al some fixed point hears an aliernation in loudness. knowil as beats, The IlllmbCT or beats per second, or the bmt jiHjlil'Jlf)', equells the difference in rrequency between the two sources: };, =
II, - I,l
~
[14.20]
where J" is the beal frequency and I\ and j~ are the t,,'o frequencies. The absolute ,'alue is used because the beat rref]uenc), is a pusiti,'c quantity and will occur regardless of the order or subt raction. A stringed instrument stich as a piano can be tUlled by beating a note on the instrument against a note or knm"'11 frequency. The string can then be wned to the desired frequenc), by a(~jll.stinp; llle tension LllllilllO healS
Beat frequency
APPLICATION Using Beats ta Tune a Musical Instrument
QUICK QUIZ 14.7
YOLI are luning a guitar!w cumparing the sO\l1lcl of the string ,,'ith that or a standard wning fork. You notice a heat frequency of :=> H'l ,,'hen both souncls are present. As you tighten Ihe guitar string. the heat rrequency rises steadil~·to 8 F-Iz, To tune Ihe string exactly to the wning lark, yOll should (a) fominlle to lighten the string, (b) loosen the string, or (e) impossible to determine from the gi'T'n infurmation,
EXAMPLE 14.11 Goal
Sour Notes
Apply Ihe beat frequency concept.
Problem A certain piano string is supposed to "ibrate at a frequellcy uf 4.40 X IO:! Hz. To check its frequency. a tuning fork known to vibrCllc
ning, "'hat beat frequency is obsenc:d by the
assista11l~
Noh': Assume: all llumbers are accurate to t,,;() decimal
places. llL:CeSsary J'or this last cakulal iUIl. Strategy (a) The beat frequency is equal to the
.
.
Solution (a) Filld the two possible frequcncies.
Case I: J:! - ./; is alreacly positin·. ahsoillte-,'alue signs:
Sf)
just drop the
=
j, =
Case 2: j~ - /1 is negativc, so drop the (lb"olute-",dlle signs. but apply an m'erall negalive sig'n: (b) Fi lid the speed of the ob~erver ir rllnll i ng the piano results ill a beal frequency of~ H,..
ft,
j, -
I,
~
4 H,.
~
}, - 4.40 X 10' Hz
444 Hz
/; ~ -(j,
-I;)
~
4 H,
= -(j; -
-lAO X 10' HI)
.J; = -136 Hz
toW~1 I'd
Apply the Doppler shif't lO the case ,,'here frequency of t ht.' pi,lllO string heard by lht: rll1llling ob . . en-er is /0 = 4~8 Hz:
I() = -~38
/t ~ ~::)
,_ .) (:~43Illh + HI = ( 4.\b liz
V())
343 Ill/s
"" ~ (
'138 I-Iz - 43.(; HZ) 436 Hz
(3.J:~ Ill/s)
J .57 m/s
..
486
Chapter 14
Sound
(c) \\-hat beal frequency does the assistant observe?
. f~ (vv-+ VO)
Appl~
Equation l4.12. Now the source is the LUlling fork, so JI ~ 4AO X lO' Hz,
.to =
=
(4.40 X 10-" I-Iz) (
ii, = J, - i;
Compute the beat frequency:
343 343
Ill/S -
= 442 Hz - 436 Hz =
Ill/S
)
1.57 m/s
~ 442 Hz
6 Hz
.
.
~
11,\
~
Ilnd the tuning fork arc struck at the same time. (a) Find the I.\VO possible frequencies of the string. (b) Assume the aCLllal string frequency is the higher frequency. If the piano llIncr runs m.. . ay from the piano at 4.00 m/s while holding the \'ibraling luning fork. \vhal beat frequency does he hear? (c) \,Vhat beal frequency does lhe assistant on (he bench hel r? Use 34:\ rn/5 for the speed of sound.
Remarks The assistant on the piano bench and the tuner running with the fork observe different beat frequencies. tVlany physical obserVrllions depend on the statc of Illot ion of Ihe observer, a subject discllssed morc fully in Chapter 26, on relativity. QUESTION 14,11 \I\'hy aren't bears heard when two different notes are played 011 the pi(lno~
Answers
(al 438 Hz, 442 I-Iz
(b) 3 I-Iz
(c) 7 Hz
EXERCISE 14,11 The assistal1lacljusls the tension in the same piano string, and a beat r'requency of 2.00 Hz is heard when the note
14.12
!\f\!\
IOIVVVV Tuning: lurk
Ibl
-r-D-+-t-D--+--+-O-t-V V V V FluLl;
1<1VAAClarinet
FIGURE 14.26
prodll("l'd
W;\H.'lorllh
b\ (a) .1 LUlling" fork. (b);\ Ihue. ,lilt! te) a clarillc1. all :ll.lpprnxilll:llt'!\, rll ..
:<.une freCjucnc\', Prl'ssure \erti(,:lllv.
LillII'
i~
QUALITY OF SOUND
The sOllild-wa,'c patterns produced by lllOSll1lusical instruments are complex. Figure 14.26 shows characteri.'itic wavcforms (pressure is ploued on the \'enical axis. time all the horizontal axis) produced by a lUning fork, a flute, and a clarinet. each playing the same steady note. Although each instrument has its own characteristic pattern, the figure reveals lhat each of the \'iavefol'ms is periodic. .\lore that the tuning fork produces on 1),' one harmonic (the fundamental frequency). but the two instrulllenlS emit mixtures of harmonics. Figure 14.27 graphs the harmonics 01" the waveforms of Figure 14.26. \Vhen the nOle is phlyccl on rhe flute (Fig. 1L1.26b), pa rt of' the sou nd consists or a vi bralioll ;u the fundamental frequency, an C\'en higher intensity is contributed by I he second harmonic, the fOllrth harmonic produces about the same intensity as the fundamental. and so all. These sounds add together according tu the principle ofsuperpositioll to give the complex waveform showll. The clarinet emits a ("ert,lin intensit.y at
plotled
hurifOJl[;llll'.
ClarilH.'1
Tuning: lork
~
FIGURE 14.27
Flutl'
II
1!arnHlllil'~
:\ 'I :1 h Ilarlll()11 ics
(a)
(hi
~\
4
:1
~
(i
1brmonin oflhe
\\,n-d\,rlll~ ill I~igllrt:
2
i
1-1.26. !'\Oll' thl'ir\':lrialion ill
~
4 n-
11
-
HanW)llics re) illlCll.~il~.
I
II
8 9
j
<
14.13
so ["orlh. The resultant superposition of Ihese rrequcncies produces the pauern shO\vn in Figure l4,26c. The wning fork (Figs. H.26a and 14.27a) emits sound only at ()1e frefluenq: 01" lile fi rst harmonic. In Illusic, lhe characteristic sound of any instrument is n:ferred LO as the qua/i/), or lilllbre. of the sound. The quality depends on the mixture of harmonics in tile sOLlnd. \·\'e say IhaL the nOle C on a lIute differs in quality from the same C on il clarinel. InstrlllllCIl1.S such as the bugle. trulllpet, ,"iolin. and tuba are rich in harmonics. A mu..,ici;-ln playing a wind inslrument can emphasi/e one or anothl'r of tllese harmonics by changing the configuration of the lips. thereby playing difTererll musical notes ,,>,itll the same valve openings.
APPLYING PHYSICS 14.6
The Ear
487
Tip 14.5 Pitch Is Not the Same as Frequency Ahhough pilch is related ll1ostll' (but 1l0lcompletdy) to l1'eqllcncy, the two terms arlO nOI the S,lmt:, A phra<,c such as Hth~ pilch of the sound" is incorrect bcclllse pildl is not;, physical properly (lfLll(' ~O\lnd. Fn:qucncy b til(' phr<,ical lneasurcmClll nflhe number of oscillation.,. pl'l" second of lilc sound. PilCh i~ a psychological rcaclioll LO sOllnrlLh
WHY DOES THE PROFESSOR SOUND LIKE DONALD DUCK?
A professor performs a demonslralion in \,'hich he brellhes helium and then speaks with a comical voice. Onl.' st udem ex pia ins, ''The vl.'loci 1)' of sou nd in helium is higher than in air, so the fundamental frequency of the standing wavl.'s in 1he mouth is increased." Anothcr ';tudenl says, "1\'0, the fundamellLal frequency is dctermined by \ he \'ocal folds and cannot be changed. Only the qualily of the "oice has challged.'· vVhich studellt is correct? Explanation The sccond student is correct. The fundamcIltal frequency orthe cumplex lOlle from the
voice is determined by the \'ibration of the ,"ocal folds and is not changed by substituting a different gas ill the lllouth. The illiroduclioll of" the heliulll into the mouth restllt.s in harmonics ofhighcr frequencies bei ng excited more I.h'-l n in the normal "Dice, but the rundalllent;-t1 frequency or the \'oice is the same. only Ihe qllalit)' has chang-ed. The unusual inclusion of Ihe higher frequency hanT\onics results in a COllllllon description of I.hi~ efTen as a "high-pitched" "oice, but thai description is incorrect. (It is really a "quack)''' ti III brc.)
14.13 THE EAR 111 The human ear is di\'ided into three regions: the outer ear, the middle ear, and the inner ear (Fig. 14.2H), The (Juln ('(fJ" consists of the ear canal (which is open 10 the
Seillicircular l<\llals
FIGURE 14.28 The ~tlllClllrl' uf tlw hllmall C;lr. The Ihn'e lilly hOlll;S (ossicle,,) l\1a1 COllllcCllhe t"ardrUlll to Ill": WilldO\\ orllle cochl..:" 'H·t
488
Chapter 14
Sound
down the ear canal to the eardrum, which vibrates in and out in phase with the pushes and pulls caused by the alternating high and 10\\' pressures of the waves. Behind lhe eardrum arc three small bones of the middlr- em; called lhe hamIller, the anvil, and lhe stirrup because of their shapes. These bones transmit the vibration to lhe inner em; which cOl1lains the cochlea, a snail-shaped tube aboul 2 ( I l l long. The cochlea makes contact with the stirrup at the o\'al window and is divided along its length by the basilar membrane, which consists of small hairs (cilia) and nerve fibers. This membrane varies in mass per unit length and in tension ;llong its Icngth, and different portions of it resonate at difTerclll frequencies. (Recall that the nawral frequency of a sU-ing depends on its mass per unit lcngth and on the tension in it.) Along the basilar membrane are numerous IH::r\'e endings, which sense the vibration of the membrane and in lurn tnlllsmil impulses La the brain. The brain ilHerprels the impulses as sounds of"arying frequcncy, depending on the locations along the basilar membrane of the impulsetransmiuing nerves and on the rates at \\:hich the impulses are transmittcd. Figure 14.29 shows thc frequency response CUITes of an average human car for sounds of equal loudness, ranging 1'1'0111 0 La 120 dB. To interpret this series of graphs, take [he bOllom curvc as the t11l"esholcl of hearing. Compare the intcnsity level on thc vertical axis for the t\\·o frequencies 100 Hz and I 000 I-h.. The \'crtical a..xis shows that the 100-Hz sound mllst be about 38 dB greater than the lOOO+Lo: sound to be at the threshold of hearing, which means that the threshold of hear· ing i~ vcry strongly dependenl on frequency. The easicst frequencies to hear arc around 3 300 Hz; those above 12000 Hz or belm,,: about 50 Hz must be relati\el~ intense to be heard. Nm..· consider the curve labeled 80. This curve uses a 1 aOO-Hz LOne at an inlensity level of HO dB as its reference. The curye shows that a tone of frequency 100 f-lz would have to be about!.j dB louder than the SO-dB, 1 OOO-I-h LOne in order wsound as loud. Notice that the curves nallen out as the il1lellsilies levels or tile sounds increase. so when sounds arc loud, all frequencies Gill be heard equally well. The slll~,ll bones in the llliddle ear represent an intricate lever system Ihal increases the force on the oval \\'indmv. The pressure is greatly magnified because [he surface arca of the cardrum is aboul 20 times [hat of the O\·al window (in anal· ogy with a hydraulic press). The middle ear, together \','ith the eardruill and oval window. in eHeet acts as a lIlalching Iletwork between the air in the OIHer car and the liquid in the inner car. The o\"enlll energy transfer between the outer car and [Ile inner ear is highly cillciem, \vith pressure amplificalion bcrors or se\'eral thousand. In other words, pressure variations in lhe inner ear arc much greater than those in the ollter ear. The car has its own built-in protection againsl loud sounds. The: muscles COI1nect.ing the three middle-car bones to the walls conu'ol the volume of the sound
FIGURE 14.29 Clllles"rililensilr le\d U'lSllS frerlllellCl for ,ulIlle!:. lhal perccin.:d III be uf equal )oudlles;;;. Note lhat till" ear is lllO,1 ~ellsi\in' ,H a Ireqll<~llry "f ,tbolH:l :~OO Hz. rlll'lowl',l (unc cc'ITt'sponds to the threshold ufh\'
-'-12~O~_-2T2h~,·e<eslioldot paill
----
120
;trt.~
100 100
lilt' pUptllalion.
.~
60
;,
..:; --to 2fJ
20
Threshold ot
hearing
o L~o
u .:::",_~-:::--_,,?__.
Inn
:l(ll)
:10(1() 10 000
1000
rrl'qLH'11C\
(I
[I.)
Summary
by changing the tension 011 the bones as sound builds up. thtls hindering their ability to translllit vibrations. In addition, the eardrulll becomes stiffer as the sound inlellsil)' increases. These two events make the ear less senSili\'c to loud incoming SOllnds. There is a time dda>' between the onset ora loud ~ollnd and the ear's protecti\'(.' reaction, howevcr, so a vcry sudden IOILd sOllnd can still damage the car. The complex stnlclure or the human cal" is believed to be related to the Ltct then mammals evol\"ed from seagoing creatures. In comparison, insect cal"s are considerably simpler in design bccause insects hm'(' always heell bnd residents, A typical insect ear consists of an C
orr
The speed or sound in a medillll1 of Im\k modltlus Band density p is
14.5
Sphericol ond Plone Waves
The intensity of a s/J!leriml wm", produced by a point source is propOri iOllal to the average pO\\'er emined and inversel\' pl'oponiollal LO the square 01" tile distance ['rom the source:
The Doppler Effect
The change in freqllency heard by an obsen'er ,,\'hellt.'ver there is relari\'c mOl iOIl between a ~OllITl' or sound and the ObSCl \'er is called the Doppler effect. if the ob:-'l.'rver i~ 1ll0\·illg with ~pet.'d lIO and the source is ll1(wing with speed l'~, thl' ubscrn-d frequcncy is
Jo ~ F (vv + "0)
whel"e Ti.., the absolute (Kelvin) [emperalure and :)31 mls is I he speed or sound ina ir at O°C,
Energy and Intensity of Sound Waves
The average intensity or sound incidenl ddined b) power
(-j'
a rea
1\
011
a surface is
[14.6]
where Ihe power .f i., the energ) per ullit time flowing lhrollgh the :-.url'acc, which has area .1. The intensity level or
=
III lug
Cochlear Implonts
The cUllsLlnl 10 = 1.0 X 10 l~ W/Ill~ is a reference illtenSil}'. lIS1l
14.6
[14.4]
f3
rnI
114.1)
The speed uf sound also dcpt:ndo;, on the temperature or tIle tlll'dium. The relationship beTwecn tempcrature and the 'ilx:ct! ufsollnd in air is
1= - - =
APPLICATION
[14.8]
,,~~~
14.4
489
(~) I"
[14.7]
~
114.12]
liS
wilen.'. v is L111' speed or sound. A positi\'e speed i!'> ~lIbsti luled {or I/O when the ob~el"ver mO\'cs toward the ..,ource, a neg
14.7
Interference of Sound Waves
When wave" interfere, the resultalll wavc is found by adding lhl' individual W,l\'CS together point by point. When cre:-.I 11H'l'ls cresl alld IrOllgl1 llleelS trough, tile waves undergo constructive interference. with pat It length difference r, -
1'1
= Jli\
/I
= 0,1. 2"
114.13J
490
Chapter 14
Sound
\\Thell crest meets trough. destructive interference occurs. with p
14.8
= O. 1,2,.
[14.14]
14.10
Standing Waves
j"
Standing 'waves
~ 2L
Standing Waves in Air Columns
St
IIF
\1;
= 1,
'2, 3..
I' JI -
2L
=
n=
njj
I,~,
3.
114.IHJ
If" t he lUbe is rlosn! at the reflecti ng end, only the odd harmonics are present ellld the natural frequencies of yibl'ation are
.f" 11
=
[14.17]
I'
= II-=}~{; 41.
1/
= 1,3,5,.
[14.19)
where F is the tension in the string and f.L is its mass per
14.11
unit !eng-til.
The phenomenon of beats is
14.9
Forced Vibrations and Resonance
A S\"Slelll capable of oscillating is said to be in resonance with SOllle driving- force ,dlcnevcr Ihe frequency of Ihe ciriying force matches olle of the natural frequencies or the sysrem. \,,",'hen the system is resonating, it o . . ci!latcs with maximum amplitude.
Beats
j" = where J'1. and
II, - jil
[14,20]
j; arc the two source frequcncies.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS 1. A sound W,I\"e traveling in air has a ["l'cquency J and wavclength A. A second sOllnd wa\'e tra\'cling in air ha~ \\'(-l\'elenglh AI?. What is the frequency or the second sOllnel wa"c' (a) 4/(b) 2/(e) I(d) ;/(c)
:1
2. "'hat is rhe speed of a longillidinal \,'(-l\"C in a bar of aluminum? (a) :140 m/s (b) 570 Ill/s (c) J 400 lll/S (cl) ~ 200 m/s (e) 5 100 m/s 3. Compule the speed of sound in ethyl alcohol. (The bulk modulus or ethyl alcohol = 1.0 X 1O~' Pa.) (<1) I JOO l11/s (b) :>40 mls (e) 820 Ill/s (d) '\50 m/s (e) 1 :;00 m/s 4. The temperalllrc al Furnace Creek in Dealh Valley reached 114°F on.July 10. 1913. \,Vhat is the speed of sound in air al this temperature? (a) 321 m/s (b) g43 Ill/S (c) 3114 Illis (eI) 37!i m/s (e) 41l!i m/s 5. A poinr source broadcasts sound into it Ilniform mediulll. If lhe distance from Ihe source is tripled, how does tht: intensity change? (a) il becomes oneninth as large. (b) It becomes one-third as large. (c) It is unchanged. (d) It becomes lhree times larger. (c) It becomes nine limes larg-er.
6. The sound intensity level ofajet plane going down lhe runway as obser\"ed from a certain location is 105 dB. What is the intensin: of lhe sound at lllis location? (a) ~.45 X 10-2 W/m; (b) 3..'>'1 X 10- 3 \VIm' (e) 8.25 X 10-3 W/m' (eI) 3.16 X 10-' W/Ill' (e) 1.05 X 10-' \VIm' 7. 1f a I.OO-k I-IL SOli nd source moves at a ,"'pet:d of 50.0 m/s loward
frequency heard by lhe listener? (The velocity 01" sounel is 340 m/s.) (a) 937 Hz (h) 947 Hz (e) I 060 Hz (d) 1 070 liz (e) I ~~O I-Iz 8. A nute has a length or 5H.O em. rr the speed of sound in air is 3-l~ mis, what is the fundamental frequency of the nute. assuming it is a wbe closed alone end and open at the mhcr? (a) 1~18 Hz (b) 29f1 1-17 (c) ·144 HI. (e1) 591 Hz (c) ~40 I-Iz 9. When t",,·o tllning forks are sounded at lht: same time. a beat. rrefJul:l1cy of" 5 Hz OCelli"'>. If one of the lUning forks has a fn::quency of 2-15 Hz. whal is lhe frequenc), of the other llllling fork? (a) 240 Hz (b) 2-12.:'l HI. (e) 2·17.5 HI (d) ~50 Hz (c) \'Iore than one answer could be correct.
10. \Vhat happens Lo a sound W3\'C lrare! when it tr
Problems
491
sounding its siren at a frequency of !)OO I-h. \\"hich sLilemCnt i~ correct? (a) VOLI hear a frccl'.Iency less 1han 500 }-I/, (b) You hear a frequency equal to ~100 Hz, (c) You hear a frequency greater [han .~OO liz. (d) Vou hear a frequenc}" greater than 500 Hz, whereas the ambulance drivt::r hear.s"l frequency I{l\\·er than 300 lIz. (e) You hear a frequcncy less [han ,":100 Hz, \dlCl"eas the arnbul:mce driver he;trs a frequency of 500 Hz.
14. ;\ hollo\\· pipe (such as an organ pipe open at bOl h ends) is made to go into resonance at frequency Iopel) , One end of" the pipe is 1I0W cO\·ered and the pipe is
13. Two sirens tI and B are sounding so that the frequency from .\ is twice the frequency from B. Compared Wilh
15. Doubling the po\\'er Output from;) sound source emir-
'i!.'l'l." ting a single frequency will result in what increase in ckcibel bel? (a) 0.50 dB (b) 2.0 dB (c) 3.0 dB (d) "dB (e) above 20 dB
the speed of sound from A, is lhe speed of sound from 13 (a) twice as fast, (h) half as fasl.. (c) four times as f
CONCEPTUAL QUESTIONS I. (a) YOll arc driving dmvn the higlHvay in YOllr GIl' when a police car sounding- its siren overtakes YOtl and pa~scs you. Hits f"requency at re.st is j;), is the frequency )'Oll hear while [he Car is GHching tip to you higher or lower than };l? (b) \Vilat :elbollt the frequency )'011 hear al"ler the car has passed YOII?
7. YOII
2. A nude Illodel of the human throat is rhat or a pipe
8. The radar systems lIsed by police
open at both ellds with a \'ibrating source [0 in[rodl1cf'" lhl' sound into the pipe at one end. A'isunling lhe vibrating source producc!o> a range of fl·eqllenc;es. dis("[ISS the eDen of" changing the pipc·s Icnglh.
1.0 detect speeders are sensitive to the Doppler 5hift of a pulse of radio w;wes. Discuss hm\" this sensitivilY can be used to measure the speed of a car.
9. All archer shoots all arrow from a bo\\". Docs the string
3. All aUlOfocllS cClmera sends out ~\ p111:-;c 01" wund and measures lhe time taken for the pulse LO reach an ol?jccl. reflect ofT of iI, and return to be detected, Can the temper:ltUH: affect the camera's focus?
of lhc bow exhibit standing wavcs after the alTOW Ic;l\·e5~ If so, and if the bow is perfectly sylllllletric 50 that the arrm.".1cavcs from the center of the string, what harmonics are excited?
4, Explain ho\\, the distance to a lig-htlling bolt can be determined by counting Ihe scconds between the rbsh and the sOllnd or thunder.
10 . ..-\ son drink boltlc resonates as air is blown across il1i top, \,Vhat happens [0 the resonant frequcncy as the 1('\"(,1 orrluicl in the boltlc decreases?
5. Senel agenL<; in the mo\·ie3
11. An airplane mechanic notices that the sound I"rom a twill-eng-inc ain:rafL varies rapidly in loudness when both engincs arc running. V\'ha[ cOllld be causing this yariation fi"om loud to soft?
6, Of lht' folhm'ing SOllllds, stalt: which is mo..,r likel~ to have an intensity level or 60 dB: a rock conccrt, the turning of a page in this texl, a normal conw'l"s,-lIifJll, a cheering crowd at a football g-Clllle. or background Iloise at a chllrch?
PROBLEMS ~
WebAssign
l. 1,
m
m:J
~~
12. Why dncs :l \·ibrating guitar string sound louder when placed on IIle instrulllent than it would ir allowed to vibrat(' in the air while on'the instrument?
. The I'roblem:> for Ihi~ ch;qJlcr Ina\ bl' assignl'd oltlinl: at \\'l:bA~"igll.
SECTION 14.2 CHARACTERISTICS OF SOUND WAVES
= :.tJ";liglllforward, inlnnH'di
= d('lllHt"~
guided prohlem
=
ellh
ell'nDll':'
II = bi\llllcdical applicaLi()Il o dl...·lllltl:~ lull S,)IULion aV;libble in Stlldl'lIl SolulilJ/!I '\!(Il/I/a{j SII/d) (;lIidp
SECTION 14.3 THE SPEED OF SOUND UIIII'H oflle1"1I'i.51' I·fafnl, lI~e )-15 m!s
1.
fIJ
fllf' s/Jl'l'd of.mund in ail:
mil Suppose yOlI hear a clap of thunder 16.2 s after seeing tlw associated lighllling stroke. The speed of sound waH;.s ill air is ~H:~ mh.
492
Chapter 14
Sound
qllcncy of 2!)0 HI. at ,lll intensity o[ 3.0 X 10- 11 W/m':!. What is the intensity delivered 10 the eardrum?
Do YOH need to know the value of the speed of light to
answer? Explain.
2. Earthquakes at fault lines in Eanh's crust create seismic waves, which arc longitudinal (P-W3\'CS) or transverse (5\\",\\"e5). The P-W;l\"CS have a speed of about 7 klll/s. Estimate the a\,(;rage bulk modulus of Eal'lh"s crllst given that the density of rock is aboul 2 500 kg/m: l .
14,
to produce \'ery loud sOLinds. The lLIbe iii its swim bladder. llsed as an alllplirlcr. The sound 1c\'el oflhis crealllre ha'i been mC
4. "
5. A group of hikers hears an echo 3.00:-. after shaming. [f
lhe lClllperillllre is 22.0°C, how far away is the t11011lHain Ihat n:fleclcd lhc sound \\'~\\'c?
6.
n The range of human hearing extends from
approximately 20 lIz to 20 000 Hz. Find the \vavelengths of Ihese ('xtrt::lnes al a temperature of 27°C.
[}]You are watching
8. A stolle is dropped ti'om rest into a well. Tile sOlll1d of the splash is heard exactly 2.00 s later. Find thl' depth of the well if the air temperature is I a.D°c.
9..-\ sound wan: tr;\\'t::ling in air at 65°C has ;1 frequency of R45 Hz. Find (a) the wave speed ,Iud (b) the wavelength.
SECTION 14.4 ENERGY AND INTENSITY OF SOUND WAVES SECTION 14.5 SPHERICAL AND PLANE WAVES 10.
am The
illtcnsily level produced by a jet airplane at a certain location is I!lO dR. (a) Cdcuble the intensity of the sound 1I"
11. One of the louelL''it SOllncls in recent h iswry was that made by Ihe explosioll of Krakatoa on August 26-27, 1883. According to harolllctric lllcaSliremt.'IllS. the sound had ;} dcrihel k\'c\ of 180 dB at a distance of 161 km. Assuming the intensit} fall,; off as the inverse of the disl:i\llce :-.quarcd, whal \\'as the dccibel le\'cl 011 Rodriguez lsland, -I ROO km :-l\\'a\'? 12. A sound \\';n'e from a 'iil'en has an intensity of 100.0 W/m':! ;It a certaill point. and
01]111 :\ person wears a hearing aid
Ihal uniformly increases Ihe illlensily 1('\'cl of all alldiblc rrequencies of sound by :10.0 dB. The hearing aid picks up sOlilld ha\'ing a fre-
The area ofa typical eardrum is abotll5.0 X 10-[' m:? Calculate the sound po\'·...er (the energy per second) incidellt on an eardrum at (a) the threshold of hearing and (b) the threshold of'pain.
15. • The toadrlsh makes lise of resonance in a closed wbl.'
3. The coldest recorded temperature of air on Earlh, -128.6c F, occurred on July 21. 1983, at Vosl.Ok,;1 Rus~ian 3latioll in Amarclica. What is the _"peed of sound in air;oH this lCmper:1111re? A dolphin located in seawater al a IClllpCral.llrc of 25°C emits a sound directed toward the bollorn of the ocean 150 m below. Ho\\' much time passes before it hC:Jrs an echo?
III
16.
m
A trumpet creates a sound ilHensilr lnel of 1.15 X JO':! dB at a distance 01" 1.00 m. (a) \,\;hat is the sound imensil}' of a trumpet at this distance? (b) What is the sound intens.ity or fj\·c trumpets al Lhis distance? (c) Find the sound intensity of live trumpets at the location of the firsl row or an audience. 8.00 III away. assLlming. for simplicity, the sOHnd ('nergy propagates lllliforll1l~' in all direclions. (d) Calculate the dt'cibcl len:l of the five trumpels in the t-irst row. (e) If lhe trumpels arc being pla}cd ill an outdoor auditorium, ho\\' far away. in thcory, call lheir combined soulld be heard? (0 In pr,iCtice such a :-.ound could not be heard once the li~tener was 2-3 kill away. \Vhy can'l the sound be heard al the dislance found in pan (c)? J-lin!: In a \'cry qlliet room the ambienL sound intensity level is about 30 dB.
17. There is cvidence that clephity Icvel of these sounds can reach 103 dB. measured a distance of 5.0 III from the source. Determine the intensity le\-e! of the infrasound 10 km from the source, assuming the souncl cllerg)' rildi;:lles ulliformly ill illl directiolls. lB. A family ice show is held at an enclosed arena. The skaters perform lo music playing at :llevcl of80.0 dB. This intcnsit)' level is too loud for your baby, who yells at 75.0 dB. (a) What tOlal sound il1lell'iil~' eng-ulfs you? (b) Whal is the combined sOllnd lc\"CI? 19. A Irain sounds its horn as it approachc" an imer:-.ection. The h01"11 Cill jusl be heard al ,t le\'cl of 50 dB by ,Ill ohscl"\'er 10 km 'Iway. (a) \\'hat is lhe
20. An outside loudspeaker (considered a small source) emits SOlllld waves with a power output of 100 W. (a) Find the intensity lO.n III I'rom the source. (b) Find the ilHensit~ levd in decibels al thilt di~lancc. (c) At what distance would yOll experience lhe sound
!mE
Show lhat lhe dilTel-ence in decihel len.:ls (31 and f3'! of a sOlll1d source is related 1.0 the ratio of ils distances 'I and I~ from lhe I"ecei\'cr~ by the [ormula
{3, - {3,
~
21llog
('l) J,!
Problems l'xplocle~ 100 m abme the groulld (Fig, Three obscn'crs arc spaced 100 III apart. wilh Ihe nrH (A) direnly under the explosioll. (a) \Vltat is the ralia of Ihl' sound il1lcnsit)' heard by ()bst:rn'r A LO that heard by obst:rn:T B? (b) What is thc ratio of the inll:nsily he
the tuning fork when Wil\'eS of fn:q\lency 48.~ HI reach the release point? Take the speed of sound in air to be
22. A "kyrockl't PI4,~2),
100
A I_ ".~."'---]00
f
J
I B C m--r--100 ----I III
FIGURE P14.22
SECTION 14.6 THE DOPPLER EFFECT 23 .. \ COlllllluter train passes a passenger plailorm al a COII'itant speed of -10.0 mh. The u-ain horn is sounded
24. An airplane Ira\'c1ing
ilt half the speed (If sound (v = 1721ll/s) emits a sound of frequency 5,00 kll/. Al whal frequency does a stationary listener hear the sOllnd (a) as Ihe plane ilpproaches? (b) .\fttT il passes?
125, IT\\'o trains on separate tracks mO\'e tQ\\'anl t'ach other. Train I has a speed of IgO km/h, train 2 (l speed of ~lO.O kill/h. Train 2 blows its horn, emilling
3'10 mis, 30.
tl1
1*
1'1
A bal fh ing- al 5,00 Ill/s is chasing a in"cct tl~ ing ill Lht: same direCliol1. Iflhe bat emits a -lO.O-kHz chirp and rccei\'e!') h;ICk all ccho at -IDA kllz. whitt is Ihe speed olthe imeet:: (Take t he speed of sound ill ;1 ir l(l be ;)<] /) m/s,)
129.1:-\ III n ing fork vibr;lt.ing al 312 J-1"1 fa lIs from resl and a(celer
493
n
Expenallt parcms arc thrilled LO heal" their unborn baby\ heartbeat, revealed by all ultrasonic Illotion detecLOr. SllPP0,sC the fetus's \cntricular Willi moves in simple harmonit: motion witll amplitude 1,80111111 and frequency 11.'1 per minute. (a) Find the Illaxinlltm lincar speed of" the heart wall. Suppose the Illotion detector in contact with the mother's abdomen produces ~ollnd ,ll prccisely 2 i\[I-I/, which travels thl"Ough tissue ,ll l,,~O klll/s. (b) Find the maximu1ll fl'equency at which sound arri\cs
31. Thc now-discominllcd Conrorde new al !\fach 1.5, which meanl that the "peed of tile plane was 1.5 timcs the speed OfSOllild in air. What wa~ the angle bet\\'ecn lhe direction of" propagal ion of the ~hock wa\'e and the direction of Ihe pLtne\ wlocity?
32.
EI.\
yellow submarine tr,\\'eling hori/ontally at 1,0 mls uses sonar with a frequency of 5.2i X 10'1 I Lo:. A red subnl
SECTION 14.7 INTERFERENCE OF SOUND WAVES 33. A pair of spe;1kers connected to the S,lllle sound system face each 01 her, one
by a speaker emining sound of frequency i56 HI. (a) If cunslructive interference ()ccur~ ilt a paniculal- ilbtanl. by whal minimum amouill should the path length ill the Ilppt:r U-shapcd lUbe be increased so tllat destrllCli\'c interference occurs inslead? (b) \\"h
494
Chapter 14
Sound
mKJThc ship in Figure P14.35 travels along a straight line parallel to the shore and 600 m from il. The ship's radio receives Sil1lllILancoLl.~ signals of the same frequency from anlennas A and B. The signals interfere consLl'tlclively at point C, which is cquidist
,r---- 800 m
,,
-------I
I
I
600
:
:
I
L
~
~
C
.-------------
III
J)
FIGURE P14.35
36.
1'\\'0
loudspeakers are placed above and belo\\' each
other. as in Figure 14.1,1), and dri\'cn by the same source al a frequency of 4 ..~O X \()2 HI',. An observer is in fronl of the speakers (La the right) al paim 0, at the same distance from c<\Ch speaker. If [he speed of sOlilld is 345 m/s. what minimum venic-lI distance up\vard should lh~ top speaker be rnov<.:d to create destl"uClive intc:rft:rence at point Q? 37. A pair ofspe,lkers separated by 0.700 1ll aloe driven by the sallle oscillawr at a frequency of 690 liz. An obsclTcr originall) positioned at Olle of t he speakers begins to walk along a line perpendicular to the line joining the speakers. (a) How far must the observer . .v alk before reaching a relmive maximum in intcnsity~ (b) Tlow far will the observer he from the speaker when the nr."t relative minimUIll is detected in the intcnsity?
SECTION 14.8 STANDING WAVES 38. A sted wire in a piano has a length of 0.700 U m and a mass of '1.300 X 10 {kg. To \vhal tension mUSl this wire be stretched so th;tt tile fundamenlal vibration corrcspomls to middle C Ck = 261.6 Hz 011 the chromatic musical sealck
ima of t.he amplitude or the prcssure would be expected. (Usc v
= 343
lI1/s.)
42. Two pieces 01" sted wire wilh identical cross sections have lengths of f. and 2 L. The wires are each fixed at both ends .0 g and lenglh 1.35 m is strung on a ba.. . s so that the distance from the nullO the bridge is 1.10 Ill. (a) Complnc the linear density of U1C string. (b) \\11at \"elocit) wave on the string will produce the desired fundamental frequency of the E t sU'ing, 41.2 Hz? (c) Calclllrlte the tension required to obtain the proper frequenc)'. (d) Cdculatc the \\'avelenglh of the string's vibration. (e) \Vhat is lhe wavelength of the sound produced in air? (Assume lhe speed of sound in air is ~43 m/s.) 44. _ :\ standing wave is set up in a string ofvariablc length
,
,
1-11---·,
,
39. A stretched ~lrjng fixed at each elld has a mass of -10.0 g and a length of H.OO m. The tension in the string is -I~J.O N. (a) Determinc the posiljons or the nodes and anti nodes fOl" thL' third harmonic. (b) What is the \'ibration fl"equellcy ror this harmonic? 40. Resonance or sound Wilves can bc produced within an alumi11um rod by holding the rod at ilS midpoint alld stroking il with all alcohol-satllrmed paper I.owel. In this resonance Illode, the middle of tile rod is a node ".,-hile the ends arc anti nodes; no other nodes or antinodcs ~lre present.. What is lhL' frequency or the rcsonance ir lhe rod is 1.00 III long? ffiJTwo spctlker~ an.; driven by a common oscillator at 800 }-I/ <mel face each other at a distance of" 1.25111. Locate the points along- a linejoining the speakers where relative I11ln-
,
,, - - r ! - - I
,Jr:>Yj -I g'"t / , / m (b) FIGURE P14.45
46. In the arrangement shown in Figure PH.46. an objeCl of n13S~ III = 5.0 kg- hangs from a cord around a light pullc\-. The length or Ihe cord between poillt Pand {he pulley i~
495
Problems
= ~.O m, (a) When the \'ibrcuor is set to a fleqllCllC~ 01 LjO 117. a ~tanding wa\'e with six luops i" formed. What mllSl be the lillear mass demity of the (01'& (b) 110\\ m;\l1\ loops (if all}) will result if III i~ chanj.\'ed 10 \.') kg? (c) I It)\\ mall\' loops (if all}) \\ill result if 11/ i~ (hanged LO L
10
5.... Two ;tdjacclll natlll'
k~'
SECTION 14.11 BEATS 55. In certain ranges of a piano kl'\boarcl. morc than aile \ ibr ,1Ior
I
J.
------->\
slrillg i:) tUlled to the ~amc nOte to prO\ide extra loudness. For example. the note at 1.10 X ID! HI has twO 'ltrings at this frequenq, If one ,>tring slip~ from it-. normal tension or 6,00 x 101 1': to 5..10 x 10 2 ?,\, "hal heat f1-cf]ucno' is heard when the hammer 'itri .... e~ the t\\'O ~tring .. simulta-
~>-
nl.'ol1sh~
'" FIGURE P14.46
[If]
56. The G strin~ nn a \ iolin has a fundamental frequcnC\' of 196 I-b. It is 30.0 cm long and ha" a Illa ..s of 0.500 g. While lhi" '>trin~ i-; ..ounding, a ncar/)\' \io/inis, effcctiyely :)hortcns Ihe G stl'ing: 011 her idC'llticll \iolin (b\ 'iliding her finger dO\\'1l lhe ~trillg) until a beat freqlleu(\" of 2.00 Hz b heard bCl\\ccn the two ::.trin~.,. \\'hen lhat occurs. \\'hat is the dlectiw length of her strillg~
r\ 60.00-(m guitar sIring Hilder a tt'nsioll 01' 50.000:\1 ha" a mass pn IInit length of 0,100 00 ~/(,1l1. What i.. the highe...t rC<jOll
SECTION 14.9 FORCED VIBRATIONS AND RESONANCE "'S, SI;lIHling-w<\\{' \'ibrmions are set lip in a crystal goblel wilh IClIlr nodes and four ,wtinodes (qllally 'ipaced around lhe 20.0-cm circumferencc or its rim, Iltr:lll,>\('rsc \\'1l?
!'ii, Two [rain \\'hi"t1es ha\'e idt'nticli lrequcncie~ of 1.80 X IO~ liz. When one train is
SECTION 14.10 STANDING WAVES IN AIR COLUMNS 49. '!'ht' \\ inclpipe ora Ivpical whooping crane is "b
50, The o\'erall length of a piccolo is 32.0 cm. Th(' re,sOIl;Ilin~ air column \'ibrates as in . (h) Opening- hole:-. in the side eneni\l~h !)hOlten~ the.' !eng-Ih of ,hc re'lOIl,tlll colulllll, If Ihe hi,!{h<"SI nOl(' il pit,colo call "'ound is oj- 000 HI. find Iht· dist;\I1te bt't\\l'ell adia<':clll ;HHinodes for this marie of \ihratioll,
51.
II Tht'
human e.'ar canal i", abolll 2$ elll IOIll4'. If il i.., a tube thai i.. open at one cnd ,HId closed at Ihe eardnnll, whal i'l Ihc fUlldamclltal fl'ccjUt'!1n ,llOlllld which we would expect hea"ing to be mo~t 'll'll'lili\e? 1 ;t.... c the .. ,)('t'rI of .,ound to he ~HO m/s. rq.~arckrl ,,b
52.
mJ
A t\lnnd under a l"in'T is 2.00 kill long:. (<1) Al wh,tl fn'f]lIellCies{'anlhe air in th(' wllllel re..onate.'? (bl Explain whclhel il would be good to make a rille a~aillst blowingvour (ill' horn \\hell \'ou are in til(" tunnel.
53. A pipe open at bOlh ends ho°C?
Two pipes of equal length are carh open at one end, Each has :1 fundamental rrcqlll'nC~ of ·ISO III. at 300 K. In one pipe the air tcmpcrature is increa~ed lO 305 K, If the IwO pipes are soullded together. what heat frequency resllIL'i?
15!1.].\ "WdCll! holds a lunillg fork o\cillatillg at 2561-11.:. lie \\ alb wward a wall at a constanl speed of 1.33 mis, (a) What beal rrequeuc\' doC's ht' observe bc[ween Ihe tuning fork and its echo? (b) 110\\ fast must he walk away from Ihe wall to ob'iel'\'t' a heat frcqut'nn of :l.00 liz?
SECTION 14.13 THE EAR 60,
If a hum,lll ('ar canal
Gin
bt'
tholl~ht
ofa:-'I'csembling
all organ pipe. do 'led at one cnd, that resonates
a fUll-
dallll'lllallrequelKy of:l 000 I'll. whal is the length of the canal: L'st' a normal bod~ temperature of 3i"C for raul' delenninalioll of the "peed or sound in the canal. 61,
Some 'iLudies suggest lhat the uppn frequency limit of hearing is determined bv the diaml'ter of the eat"dnlln. The \,an:'lcng'th or tht, sound \\~,l\e and the diameter of the {'ardrum arf' apprOXilllal('" eqlluJllc a bod\ tl'mperallln: of37~C.)
ADDITIONAL PROBLEMS 62. Thc intensil\ lewl (Iran orchc.',>Lra is W> dB. A single \ iolin reaches a !t'\-e! of 7.0 X In ' dB. \'\'hal is the ratio of the "/lund ililellSil) of tht' filII orchestra to the illlC'llsit)' 01 a sing-Ie \'iulill:
496
Chapter 14
Sound
63. ASSlIllle a IOlldspeaker broadcast> soulid equ"ll)' ill "II directions and produces ~ollnd with a level of 103 dB at a distance of 1.60 III from its ccmer. (
;e
i
: 4.00
III
I I
~
(~L -----.-" 3.00 ill
:
~
'~~i)
-:--
2.00
!'lk::.-
FIGURE P14.64
65. An intcrslalC highway has heen huilt though a poor neighborhood in a cit),. In lhe aflernooll, the ~ound level in a rented room is 80.0 em as 100 GlrS pa . . s outside the Willdol';,evc.:ry milllilc. LaiC at tli~'ht, the traffic flow is only five cars per minute. Wh,tt is L1le average law-night s(nlnd level?
oscillation. so lhe qUClrtz is described as pi{'!of!('((ric. An dcctric circuit feeds in energy to maintain the oscillal ion and also COUI1IS the voltage pulses to keep time. 70. A IlowerpOl. is knocked off a balcony 20.0 III abO\e the sickw,dk and f,lIb tUWill d illl 1l1l~1I.... pet..li[Jg I./J-nJ-lilJI lIIall who is st.anding below'. How close to the sidewalk call the flowerpot fall before it b too late for a warning shouted 1'1'0111 t he balcony 10 reach the man in time? Assume the man below requires O.~O(J s 10 respond to the warning.
[1[]On a workday, the a\'er3ge decibcllcvel 01" a busy street is 70 dB. with 100 cars passing a given point c\'err minule. If the number or cars is reduced to 25 evcry minule 011 a weekend, what i~ the decibel kvel or the 51 re(.'t~ i2. A Ilule is designed so that il plays a frequenc)'of261.6 I-I/., middle C:, v,'hen all the holes are covered and thc tcmperature is 20.()OC. (a) Consider the nUll' to be a pipe open at both ends ;md find its length, asstllllin~ the middle-C frequency is the rUlldalllcnt;-t1 frequency. (b) A second player. nearby in a colder room, also alLClllpts w pia} middle C on an identical flute. A beat frequency of 3.00 beal.'>/s is heard. \Vh,lt is the temperawre of the room? 73. A block with
166.IA stm!cnt uses an ;ludio oscillator or acUustablc frequency to measure the depth ora water well. He reports hearing two successive reSOIUllces at :")2.0 Hz and 60.0 HI:. If the: speed of sound is ~'15 mis, how deep is the \vell? 6i. \·"'hen;\t rest, t\\'o trains ha\l~ :-irens that emit a frequency of :WO H,. The Irains travel 100~'arri each olller and to\\'ard an ohserver stationed bCI\H.'en them. Olle of the trains move:-. at 30.0 mls. and the observer hears :l beat freqtlellcy 01 ~.O bealS per second. \Vhat i:- tile speed of' the "ccond lrain, which travels faster than 30.0 mist 68. A commuter train blows its horn as it passes tI p'lssen· gel' platform al (\ COnsl;lI1t speed or ·10.0 m/s. The horn soumis at a trequency or 3~0 1-1/ when L1le train is ttt rest. "\'hal is the freqUl..:ncr obscn'cd by a person on t he plat. form (a) as the t.rain approilches and (b) :IS the train recedes from him? (c) \-"hat wavelength docs the obsen'el' Jind in each case? 69. A qual'\Z walch contains a crystal oscilbt.or ill the form or a block or qtlartz Ihal vihrates by contracl ing' and expanding. Two opposite races or the block, 7.05 mill apart, arc antinodes, movillg alternately toward and awa} from each orlH'\'. The plane halfw:l)' betwecn these 1\1"0 EKl"S is a node or thc \:ibratioll. Tile :-peed or sound in qU<Jrlz is ~.70 km/s. Find lIlt' frequency of the vibration. All oscillating electric \'oltagc accompanies tile mechanical
FIGURE P14.73
\74.IA studenl stands several meters in front 01" a slllomh rcOl'cting wall, holdinl-?: a board all which a wire is fixed at each t'nd. The wirc, vibrating in its third harmonic. is 75.0 em long-, h,ts it lIlass of 2.2!l g. and is under a lension of 400?\. A sl.:colld siudent. mm'ing towards tht: wall, hears H.30 heats per second. Whal is lhe speed o["the "ltldClll approaching the wall? Use 340 m/s as the speed 01 sound in ail". 75. B)' proper excital1011, it is possible to produce both longitudinal and transvcrse waves in a long metal rod. In a particll[;1r case, the rod j<; 150 cm long and 0.200 cm in r:l.dius and ha" a mass of 50.9~. Young's modulus for the material is 6.~O X lOl(l Pa. Delermine the required Icn:-iOll in rhe rod so that lhe ratio orthe speed of long it lid iIlal \..·a\'e~ lO the :-peed of t fanS\ erse waves is 8.
This nighttime view of multiple bolts of lightning was photographed in Tucson, Arizona. During a thunderstorm, a high concentration of electrical charge in a thundercloud creates a
higher-than~normal
elec·
tric field behveen the thundercloud and the negatively charged Earth's surloce. This strong electric field creates on eledric discharge- -on enormous spark-between the charged cloud and the ground Other discharges observed in the sky include c1oud-to-cloud discharges and the more frequent introcloud discharges.
ELECTRIC FORCES AND ELECTRIC FIELDS
15.1
Properties of Electric Charges
15.2
Insulators and Conductors
15.3
Coulomb's Law
Electricity is the lifeblood of technological civilization and modern society. Without it, we revert to the mid-nineteenth century: no telephones, no television, none of the household
15.4 The Electric Field
appliances that we toke for granted. Modern medicine would be a fantasy, and due to the
15.5
Electric Field Lines
lack of sophisticated experimental equipment and fast computers-and especially the slow
15.6
Conductors in Electrostatic Equilibrium
15.7
The Millikan Oil-Drop Experiment
15.8
The Von de Groaff Generator
15.9
Electric Flux and Gauss's Law
dissemination of information-science and technology would grow at a glacial pace. Instead, with the discovery and harnessing of electric forces and fields, we can view
arrangements of atoms, probe the inner workings of the cell, and send spacecraft beyond the limits of the solar system. All this has become possible in just the last few generations
of human life, a blink of the eye compared to the million years our kind spent foraging the savannahs of Africa. Around 700 s.c. the ancient Greeks conducted the earliest known study of electricity. It all began when someone noticed that a fossil material coiled amber would attract small objects after being rubbed with wool. Since then we have learned that this phenomenon is not restricted to amber and wool, but occurs (to some degree) when almost any two nonconducting substances are rubbed together. In this chapter we use the effect of charging by friction to begin an investigation of electric forces. We then discuss Coulomb's law, which is the fundamental law of force between any two stationary charged particles. The concept of on electric field associated with charges is introduced and its effects on other charged particles described. We end with discussions of the Von de Graaff generator and Gauss's law.
15.1
PROPERTIES OF ELECTRIC CHARGES
An cr ru n n ing a plastic comb l.h rough yOLi r ha i r, yOll wi 11 fi net that [he comb attracts bilS or paper. The attractive lorce is often strong enough to sllspend lhe paper
497
498
Chapter 15
Electric Forces and Electric Fields
FIGURE 15.1 (.1) A negatively charged rubber rod, suspellded by a thread, is attracted 10 a pOtlilivc1y charged glass rod. (b) A negatively charged rllbber rod is repelled b)' another IIcg-atively charged rubber rod.
(a)
8ENJAMIN FRANKLIN (1706~1790)
Franklin was a printer, author, physical scientist, inventor, diplomat, and a founding father of the United Stoles. His work on electricity in the late 17405 changed a jumbled, unrelated set of observations into 0 coherent
science.
Like charges repel;
unlike charges attract.
~
Charge is conserved ~
rrom the comb, defying the gravitational pull of the entire Earth. The same effect occurs with other rubbed materials, such as glass and hard rubber. Another si mple experiment is to rub an innatcd balloon against wool (or across your hair). On a dry day, the rubbed balloon \vill then stick to the wall ofa room, often lor hours. These materials have become electrically charged. Vou can give your body an electric charge by vigorously rubbing your shoes on a wool rug or by sliding across a car scat. YOll can then surprise and anlloy a f"riend or coworker with a light touch on the arm, deliveri ng a slight shock to both yourself and your victim. (If the coworker is your boss, don't expect a promotion!) These experiments work best on a dry clay because excessive moisture can bcilit3te a leaking away of the charge. Experiments also demonstrate that there arc t,.. . o kinds of electric charge, which Her~jamin Franklin (1706-1790) named positive and negative. Figure 15.1 illustrates the interaction of the two charges. A hard rubber (or plastic) rod that has been rubbed \·I!ith fur is suspended by a piece of string. ,",Vhen ;L glass rod that has been rubbed with silk is brought near the rubber rod, the rubber rod is attracted lm..: ardthe glass rod (Fig. 13.]a). If two charged rubber rods (or two charged glass rods) ~lre brought near each other, as in Figure 15.lb, the force between them is repulsive. These observations may be explained by assllming the rubber and glass rods have acquired different kinds of excess charge. VVe use the convention suggested by F1-anklin, \vhere the excess electric charge on the glass rod is called posiLive and that on the rubber rod is called negmive. On the basis of such observations, we conclude that like charges repel one another and unlike charges attract one another. Objects usually contain cqual amounts of positive and negatlve charge; electrical forces betwecn objects arise when those objects have net negat.ive or positive charges. Nallire's basic carriers of positive chargc are protOns, which, along with ncutrans, are locatcd in the nuclei of atoms. The nucleus, about 10- 15 In in radius, is surrounded by a cloud of negat.ively charged electrons aboUl ten thousand times larger in extent. An electron has the same magnitude charge as a prolan, butlhe opposite sign. In a gram of matter there are approximately IO~:1 positively charged protons and.just as many negatively charged electrons, so Ihc net charge is zcro. Because the nucleus of an atom is held nrmly in place inside a solid, protons never move from one material to another. Electrons are far lighter than prolons and hence more easily accelerated by forces. Further, they occupy the outer regions of the atom. Consequently, ol~jects become charged by gaining or losing electrons. Charge transfers readily from one type of m
15.2
a glass rod is rubbed ''''ith silk, as in Figure J:1.2, electrons are transferred from the rod LO the silk. As a result, the glass rod carries a net positive charge, the silk a net l1cg~llive charge. Likewise. whl:11 rubber is rubbed \vith fur, eleClrons are transferred from the fur r.o the rubber. In 1909 Ruben tVlillikan (l886-195~) disconTed that if an object is charged, ilS charge is ahvays a multiple of a fundamental unit of charge, clt:signated by the symbol e. In modern tcrms, the charge is said In be quantized, meaning that charge occurs in discrete chunks that can't be further subdivided, An object may h~1ve (l charge of ±f', ±2e, ±3e, (lnd so on, but neveI'I a fractional charge of ::to.Sf' or ±O.22/'. Other experiments in \1illikan's time showed thm the electron h;lS a charge or ~() (lnd the proton has all equal <:md opposite charge or +('. Some particles, such as a neutron, have no nel charge. A neutral atom (an atom \vith no net charge) contains as many protons as electrons. The \'alue of f' is no\\' known to be 1,602 19 X 10 19 C. (The SI unit of electric charge is lhe coulomb, or C.)
15.2
INSULATORS AND CONDUCTORS
Insulators and Conductors
FIGURE 15.2
499
When a glass roel
is ruhhL'd Wilh sill.... eknrolls art' 1l";\Ilsh-rrcd from lite gbss ttl lilt, silk. lkcau'l: 01 ronscn';uiml 01 chargl'. L'
Substances can be classified in terms of their ability to conduct electric charge.
In conductors, electric charges move freely in response to an c1eClric lorce. All other materials are called insulators, Glass and rubber are insulators. \Vhen stich mmerials are charged by rubbing, only 1he rubbed area becomes charged, and there is no tendency lor the charge to move into other regions of the material, Tn contrast, materials such as copper, aluminum, and silver are R"0od conductors. ''''hen such materials are charged in some small reg·ion, the charge readily distributes itself over the entire surface of the material. If yo II hold a copper rod in your hand and rub the rod with wool or fur, it will not auract a piece of paper. This might suggest thai ;1 meta] can't be charged. Hcw,icver, if you hold the copper rod with an insulator and thc:n rub it with wool or fur, the rod remains charged and aUraClS the paper. In the first casc, the electric charges produced br rubbing readily move from the copper through your body and finally LO ground. In the second case, the insulating handle prevents the flow of" charge to ground, S'em;ro1lduclOfS are a lhird class of materials, and their cleClrical properties are somewhere bet\vecn those of insulators and [hose or conduClors. Silicon and germanium are well-known semiconductors that are widely tlsed ill the fabrication of a variety of electronic devices.
(
Charging by Conduction Consider a negatively charged rubber rod brought into COlllaCl \',Iith an insui
(ll) COlllaCt
FIGURE 15.3 (J1~eClI)\'
Charging by Induction An
ol~ieCl connected to;-I
conducting wire or copper pipe buried in t.he Earth is said
to be grounded. The Earth call be considered ,In inlinitc reservoir for electrons; 'Thvi (" i~ -trulIg" ("\'id~ll(e I'll IhI' ~XiS("11Le 01 hll1dall1~1l1.11 pal"l id.., {,l1k{l {juarks lhal han:. d,arg~s of ::!:d3 01 ::!:'.!.,./ :~. ThL' ch;lrg:e IS ~tlll qllanli/ell, hIli in ((nib "I =e/3 ralht'r liiall -P. _\ IlIUIT ("!llplt'lt· dl"lls,itlll or qll"rb "I,d
11[('lr pr"pclli", is pr"Sl:lll,'d ill C:llapl'T
:~O,
Ch:lrgiuj.{ a Illt.:lalli(
conduction. (a) .Imt hdore CO!ll3ll, the l1egatil'l' rod repl'b thl' sphL'n:·~ t.:lcetrolls, illchlling a lo,ali/cd po~iti\'c charg"(.'. (b) After COllliHl, c1t'ClrOlis froll) the rod flow OIlIO t he sp"l'n~, Ill'IIi raliLing I Ill.:' local pt,sitil'(' chalg-es, (I") VI-h.. nll11' rod is fL'IIW\cd, lhe :.phere i' left with:l nl'g-at ilL' chal ge,
500
Chapter 15
Electric Forces and Electric Fields
in erfect, it can accept or supply an unlimited number of electrons. With this idea in mind, we can understand I.he charging or a conductor by a process known as induction. Consider a ncgatively charged rubber rod brought near a nCUlral (uncharged) conducting sphere that is insulated, so there is no conducting path to ground (Fig. 15.4). Initially the sphere is elcctrically neutral (Fig. 15.4a). ''''hen the negatively charged rod is brought close to the sphere, the repulsive force bet\veen the electrons in the rod and those in the sphere causes some electrons r.o move to the side of the sphere farthest ~nvay from t.he rod (Fig. 15.4b). The region of the sphere ncarest the negatively charged rod has an excess of positive charge because of the migration of electrons away from that location. If a grounded conducting \...'ire is then connect.ed to the sphere, as in Fig'ure 15.4c, some of the electrons leave the sphere and travel to ground. If the wire to ground is then removed (Fig. 15.4d), the conduCling sphere is left with an cxcess of induced positive charge. Finally, when the rubbcr rod is rcmovcd [rom the vicinity of the sphere (Fig. 15.4-e), the induced positive charge remains on the ungrounded sphere. Even though the positively charged atomic nuclei rCITI
la)
(h)
(e)
(d)
Ie) FIGURE 15,4 Charging a metallie object hy hulur(ilili. (a) A neutral metallic spht:re, with equalllllllllll'l"s of positive alld !leg-arive ch;rrgcs. (b) The elect rOllS
UII
a llclltral sphere
art: rt::distriblltcd whcll a charged I-ubbcr rod is placed Ilear till.; sphere. (c) When lht; sphere isgrollnded. some of its c1ectrolls leave it through the gmund wire. (d) When Lhe ground connection is rellloved. Ill(' sphere has excess positil'e charge Ihal
QUICK QUIZ 15.1 A suspended object i\ is attracted to a neutral wall. It's also attracted to a positively charged o~jcct B. \Vhich of the following is true abollt object A? (a) It is uncharged. (b) II. has a negative charge. (c) It has a positive charge. (d) It may be either chargcd or uncharged.
is nUlluniforml)' disrril.JllI('d. (e) ,",Vhen tht: rod is rClllll\"C(1. the rt.:lllainill~
15.3
eleclJolls redistribute uniformly ,uld there is a net unifi,rm distribution positive charA"c UII Ihe sphere.
In L785 Charles Coulomb (1736-1806) experimentally established the fundamen· lal law of' elcClric force between t\,,.o stationary charged panicles.
or
COULOMB'S LAW
An electric force has the following properties: 1. It is directed along a line joining the two particlcs and is inversely proportional to the squarc of the separation distance t; between them. 2. It is proportional to the product of the magn iludes of the charges.lqll and Iq21, of the two particles. 3. It is attractive if the chargcs are of opposite sign and repulsive if the charges have the same sign.
From these observations, Coulomb proposed the following mathematical form [or [he electric force bet\'\-'cen t\..IO charges:
15.3
Coulomb's Law
501
FIGURE 15.5 (a) TIlt' char~cd objecl on the len induce'S charge.; Oil the surface of an in';lIlatnr. (h) A dlarg<-'d comb atlracts hiLS of paper because charg-cs art.: di ...placcd ill lhe p:lper.
Wall
\
Charg;ed balloon (h)
(a)
The magnilUdc of the electric force F bctween charges 1\ and q'1. scparated by a distance,. is given by
,
l=/lr
1'1,11'1,1
,.-
<)
f- Coulomb's low
[15.1]
where hI' is a constant called the Coulomb ronslrl11t. Equation 15.1, known as Coulomb's law, applies exacLly only to point charges and to spherical distributions of charges, in which case r is lhe distance between the
t,\'o ccnters or charge. Electric forces bet'·.. ecll unmoving charges are called elel'lros/fllicforces. Moving chargcs, in addition, create magnetic forces, studied in Chap-
tcr 1Y. The value oCthe Coulomb constant in Equation lEi.l depends 011 the choice of units. The 51 unit of charge is the coulomb (C). From experiment, wC' knO\v that t.he Coulomb constant in 51 units has the valuc Ii,
= 8.9875
X 10" N· m'/C'
[15.2]
This number can be rounded, depending on the accuracy of other quantities in a given problem. \,Vc'lIuse either two or three significant digits, as usual. The chargc on thc proton has a magnitude or e = J.6 X IO- l!) C. Therefore, it would takc 1/1' = 6.3 X 101.'1 prolOns to create a lotal charge or +1.0 C. Likewise, 6.~ X 101.'l eleclrons would have a total charge of -1.0 C. Compare t.his charge widl the number or free elcc£rons in J cm:-l of copper, which is on t.he order of 10 23 . Even so, 1.0 C is a very large amount of charge. In typical electrostatic experiments in which a rubber or glass rod is charged by friction. [here is a net charge on the order of10- 6 C ( = I ILC). Only a very small rraClion orthe total aV
CHARLES COULOMB (1736-1806)
TABLE 15.1 Charge and Mass of the Electron, Proton, and Neutron Particle
Charge (e)
Mass (kg)
Electron Proton Neutron
-1.60 X LO '" +1.60 X IO-EI 0
9.1] X 10- 31 1.67 X JO- 27 1.67 X J 0-27
Coulomb's major contribution to science was in the field of electrostatics and magnetism. During his lifetime, he also investigated the strengths of materials and identified the forces that affect objects on beams, thereby contributing to the field of structural mechanics.
502
Electric Forces and Electric Fields
Chapter 15
force of repulsion between two positively charged particles. Like other forces, c1eclric forces obey Newton's thircllaw; hence, the forscs F 1:! and F2! arc equal in mag· nittlcle but opposite in direction. jThe notation Fl~ denotes the force exencd by particle I on particle 2; likewise, F~l is the force exerted by panicle 2 011 panicle 1.) FrullI NewwlI's thinJ hlW, F I2 alld F:!l ale s of whether 1ft and q'!. have the same magnitude. ~
~
QUICK QUIZ 15.2 Object A has a charge or +2 p.C, and objeci B has a charge of +6 j..Le. Which statement is true? (a)
(h)
sign. the 1"01'('(' is rcplllsiH>. (11) \Vht:ll
Ihe charges arc ofopposill:
the
altf;H"liVl:.
EXAMPLE 15.1 Goal
~ign.
Forces in
-FR.'
=
(c)
3FAB = - FnA
The Coulomb force is sirnilar to the gravitational force. Bmh act at a distance without direct. contact. Both (Ire inversely proportional to the distance squared, "vith the force directed along a line connecting the two bodies. The mathematical fonn is the same, with the masses !lit and 1n'1. in Newton's law replaced by fli and (1'2 in Cou 10mb's law and ,vith ~c',\·ton's constant (; replaced by Coulomb's constant Itr· There are t\vo il11portant differences: (I) eleClric forces can be either attractive or repulsive, but gravitational forces are ahvays attractive, and (2) the c1eclric force between charged elementary panicles is far stronger than the gravitational force bct\veen the same panicles, as the next example shows.
ACTIVE FIGURE 15.6 Two poill! .harges ~t:paralcd h)" a distance "ext"rt a rurteol1 t;;lch oIlier given b\' Coulomb's law. The lorte un f}1 is egu:.ll ill magniwdl' and oppo:-ile ill din'clinn to the forcl' 011 (h (a) \'"hen lhe chargl'~ arc .)[Ihe saille
fore<.· i1>
FAI \ = -3F\H (b) FAil
0
Hydro en Atom
Contrast the magnitudes oran electric force and a gravil<1lionaJ force.
Problem The electron and proton of {\ hydrogen atom are separated (on the average) by a distance of about 5.3 X 10 II Ill. (a) Find [he magnitudes of the electric force and the gravitational force that each particle exens on the other, and the ratio of the c1ectl·ic force }~. Lo the gravitational force (b) Compute the accelcration caused by the electric force of the proton on the electron. Repeat for the gravitational acceleration.
I'>
Strategy Solving {his problem is just a matter of substituting known quantities into the t\\'o force laws anclthen llnding the r~ltio.
....
.
~
Solution (a) Compute the magnitudes of the electric and gravitational forces, and llnd thc ration 1~!Fg. Substitute If/II = Iq21 = r and the distance into Coulomb's law to llnd the electric force:
,
", =
I,d"
k,--., ,-
=
(
8.99 X
10 .jN.m") (1.6 X10- C)' 10' --.,- (_ _")' C-
~.3
X 10m -
8.2 X 10-8 N
Substitutc the masses and distance into Newton's law of gravitr 1.0 find the gravitational force:
(b) COl11pllll' lhe accelermion or the electron caused by the electric force. Repeal I()I' tIle graviwlional acccleratiOlI. U... e "ewton'" ,econd law and the electric l(nee lOll nd in pa rt (a):
III fl
U~e
mrfl~
NC'\'lon\ second law and the gravitational force found in pan (a):
r
r
= }.:
= f~
->
F,
8.2 x 10 Il:\
1It r
9.11 X 10 " kg
a, = - =
,
F
->
(ll!=-=
m,
3.6 X 10
9.0 X IO~2
Ill/S 2
17~
9.11 X 10 ," kg
4.0 X 10
17
mis'
.......... Remarks The graVilaLional force between the charged constituents of the atom is negligible compared with the dcclric force between them. The electric force is so SHang, however, that any net charge on an object quickl)' atlraCls nearhy opposite charges, neutral il.ing the ol~jecl. As a rc\uh. gravity plays a greater role in the mechanics of 1l'1Oving ol~ecls ill cvcryday life. QUESTION 15.1 I I" the distance between two charges is doubled, by Wllilt facLor is the magnitude of the clectric force changcd? EXERCISE 15.1
Find I hc magnitude of the electric force bet\',ieen t\\10 protons separated by I fcmLol11eter (10- 1:' 01), approximately the di.."tance hetween two protons in the nucleus of a helium atom. The answer may not appear large, bUl if not for the 51 rang nuclear force, the two prolOns \vottld accelerate in opposite directions at over I X 10~9 Ill/S2 ! Answer
2 X 10'2 ~
The Superposition Principle When a number of" separate ch:.uges act on the charge 01" interest, each exerts an electric force. These electric forces call all be computed separately, one
EXAMPLE 15.2 Goal
Findin Electrostatic Equilibrium
Apply Coulomb's law in one dimensioll. (----2.0111---
Problem Three charges lie along the x-axis as in Figure 15.7. The positi\'c charge fJJ = !;') p.C is at x = 2.0 111, and the positi\-e charge fJ'.! = 6.0 p.C is at the origin. \\'herc must a negative charge (/:I be placed on the x-axis so that the rc.:SIl II a Ilt ciCCI ric force on it is zero? Strategy If '1:1 is to the right or left of the other two charges, the net force on (h call't be zero because then FI :1 and F'.!:~ act ill the same direction. ConsequeJ1lI)', 'I';'. rnust lie between the twO other charges. \\'ritc F1:1 and F'2:l in terms of the unknown coordinate position x, then ~um (hem and set them equal to zero, "ioh-ing for the unknown. The solution call be obtained with the quad raLic formu la,
-x
FIGURE 15.7
2.11 I l l - X
(Example E",.2) Tim.'..'
poilll ChMgC~ are plaet-°d along lhe x-'lxis. The charge,,\ is ncgali\t.', whereas 11 and il":! arc P()~ili\"\.'o Ifl~le rc..ullanl force on q] i~ Il'I'Oo lhe fOl'cc fl'clo,toned by q, on 1l mUSI bt' eqllal,illlllagnitudc and opposilt, lilt" I(,,"("e
F'~(.'lI.crtt.'d
b' 11011/h,.
...
~
Solution WI'ite the x-component of F\:l=
\,Vrite J he x-component of F'.!",,:
SCi lhl'
slim equal to zero:
+ k,
(1.0 X 10-· C) 'III ( 2,0 III - X )"-
F'.!.:h =
- hr
(6.0 X 10 "c)I'I11 _, = 0 x-
~
504
Chapter 15
Electric Forces and Electric Fields
Cancel hr , 10- 6 , and (13 [rom the equation and rearrange terms (explicit significant figures ~lI1d units arc temporarily suspended lor clarity):
(I)
6(2 - x)' = l5x'
Put this equation into sLand;Hd quadratic fOI-m,
6(4
4x I ).2) = J5~
mil
+ UX + f = 0:
3,-2
+ 8x -
x=
On I)' the positive root makes sense:
x= 0.77 m
4x I .\2)
2(4
=
5.\.2
8 = 0
-8:': v'64
Apply the quadratic fornlllh,:
-).
(4)(3)( 8) 23
.....
. • , • • • • • • • • • .A
Remarks Notice thal physical reasoning \"'35 required to choose between the t\\lO possible ansv,:ers for x. which is nearly always the case when quadratic equaLiolls are involved. Use of the quadratic formula could have been avoided by laking the square root or both sides ofEqua1.ion (1), hmv'cvcr this short cut is OrlCn unavailable. QUESTION 15.2 If ql has (hc same magnitude as before but is negative, in what region along the x-axis would it be possible for the net electric force on q3 LO be zero? (a) x < 0 (b) 0 < x < 2 III (c) 2 III < X EXERCISE 15.2 Three charges lie along the x-axis. A positive charge (11 = 10.0 J.Le is at.\" = 1.00 m, and a negative charge tj'!. = -2.00 p,C is at the origin. "Vhere must a positive charge fl.", be placed un the x-axis so that the resultant force on it is zero?
Answer
x
= -0.809 m
EXAMPLE 15.3 A Charge Triangle Goal
Appl)1 Coulomb's law in two dirncnsions.
J
Problem Consider three point charges at the corners of a triangle, as shown in Figure 15.8, \\ihere 'II = 6.00 X 1O-9 C, If" = -2.00 X 10-" C, and If, = 5.00 X 10- 9 C. (a) Find th~ componcllls of the lorce F~.", e~erted by (/2 on q:~. (b) Find the components of the force FI:~ exerted by 'II on q3- (c) Find the resultant force on q"" in telTns of' components and also in terms of magnitude and direction. Strategy Coulomb's law gives the magnitude of each force, which em be split with right-triangle trigonometry into x- and J-components. Sum the vectors componenl\visc and lhcn l-ind the magnitude and direction orl.he resultant vec(Qr.
'I'!. -
:I.OO1Jl
'II
+J----------------.,
FIGURE 15.8 (Example 15.3) TIl(: force exerted hv 'II Oil 13 IS Thl.: force ('xencd by q.. . Oil q~l i~ F.. ~. The rr..',IIf1mltjofo,'F 3 exerted on 1:1 is Ihe VI'florSlIlll FI:I + F~_i'
.
FI~'
~
Solution (a) Find the compollcl1ls orthe (orce exerted by q2 on q:\. Find the magniLUde or F:!:I \vith Coulornb's hl\.. . : _ (
-
1'2:\
"2
8.99 X 10 N'm
= 5.62 x J O-!) N
IC
')
(2.00 X
[0-"
C)(500 x 10-' C)
( )" 4.00 III -
15.4
505
The Electric Field
Bl.:calisc F~l:\ is horizontal and points in the ncgmive x-direction, lh<.: negative of lhe magnitudc gives the x~componcn1.. and the y-componcnt is zero: (b) Find the components of the forcc exerted by (11 on fJ:\·
1'1,11,'1,,1
Ii,
~
Find the magnitude ofF,.\\:
r" l ( 8..19 X
,
10 N'1ll
,2)(6.00 X 1O-"C)(5.00 X 10-"C) IC ( )' 5.00111 "
F", = 1.08 X 10-' N
Fu ., = I~" cos
Use the giyen triangle to find the components of F1:l :
e=
(1.08 X 10-':\) cos (37')
= 8.63 X 10-" N i 1 ", = F u sin
e=
(1.08 X 10-' N)sin (37')
(c) Find the componems of the resliltani veClOL I'~
Sum the x-components 1.0 find the resultant Fx :
= -5.62 X 10-" N - 8.63 X 10-" N 3.0l X 10-":\
I'; = 0 +
Sum the y-components to find Ihe resllltam ~\:
6.50 X 10 "N = 6.50 X 10-" N
, , IFI=VF'+F'
Find the magnilUde of' the resultant Corce on the charge Q3' using the Pythagorean 1 heorcm:
= V(3.01 X 10 "N)' + (6.50 X 10 "N)' =
7.16 X 1O-"N
(F) =
() = tall-I ~ I'.,
Find the angle the resultant force makes \ViLh respect 1.0 lhe positive x-axis:
lan-I
(63.01'0 X X .:J
.
~.
Remark
..
~
The methods llsed here (Ire just like Ihose llsed with NeWlon's law of gTavity in t\VO dimensions.
QUESTION 15.3 Without actually calculating lhe electric force
011 f]'!.,
determine the quadrant into which the electric rorce veClor poinLS.
EXERCISE 153 Using the same triangle, find the vector compon<':llIs of' the e1eClric force on qJ and the vector's magnitude and c1in.:ction.
Answers
15.4
F, = -8.63 X 10-<' N, I'; = 5.50 X 10 "N.F
~
1.02 X 10-':\.
e=
141'
THE ELECTRIC FIELD
The gravitational force and the electrostatic !orce arc both capable of acting through space, producing an erfcCl even when there isn't any physical conLKt between the ol:~jeCls involved. Field forces can be discussed in a variety or \~aysJ but. an approach developed by Michael Faraday (J791-1867) is the most practical. In this approach an electric field is said to exist in the region of space aroltnd a charged ubject. The electric field exerts an electric force un any other charged object within the field. This differs from the Coulomb's 1m.. . concept or a {-orce
506
Electric Forces and Electric Fields
Chapter 15
'In
J
E
FIGURE 15.9 A small objtCl with charge 'I" placed Ileal' all objeCl wilh itin~
exerted at a distance in that the force is no\\' exerted by something-the nelclthat is in the sarnc locaLiol1 as the charged object. Figure 15.9 shows 3n objcCl with a small positive charge flo placed near a second o~jcC[ with a lllllch larger positive charge Q.
E
The electric field produced by a charge Q at the location of a sillall "tcst" charg'c flo is defined a~ the electric force exened by Qon flo divided by the lest charge 10:
F
-)E""-F
[15.3]
flo
51 unit: newton per coulomb (N/C)
Concepwally and experimentally, the test charge 10 is required to be very small (arbitrarily small, in fact), so it do~n't cause any significant rearrangement of the charge creating the electric rield E. lvlathematically, however, the size of the test charge makes no diff<:-rence: the calculation comes Ou[ the same, regardless. In vie\'", o('this, using qn = I C in Equation ]5.3 can be convenient ifnOL rigoroLis. \Vhcn a positive tcst charge is used, the electric field ahvays has the same direction as the electric force on the test charge, which follows from Equation 15.3. Hence. in Figure 15.9, the direction 01" the electric fIeld is horizontal and to the right. The electric field at point A ill Figure 15.10£1 is venical and downward because at that point a positive tesl charge would be aLLractcd to\.. · ard the negatively charged sphere. Once the c1eclI'ic ncld due to a given arrangement of charges is known al some point, the force on an)' panicle with charge fJ placed at that point can be calculatcd [-"rom a rearrangemelll. of Equation ]5.3: [15.4] Here '10 has been replaced by q, which need nOt be a mere: test charge. As 5hO\'\'n in Active Figtlre 15.11, thc direction ofE is the direction of the force that acts on a positive test charge q() placed in the field. We say th,u an electric field exists at a point if a test charge at that point is subject to an electric force there. Consider a point charge q located a dislance T from a test charge quo According to Coulornb's law, the magnitude of tht; eketl'ic force of the charge q on the test charge is [15.5] Because the rnagnitude of the electric field ill the position of the [cst charge is deri ned as E = F/qo, we sec that the magnillideor the electric field due to the charge q at [he position of ql) is [15.6] Equation 15.6 points out an important properl)' of electric fields that makes them useful quantities for describing electrical phcnornena. As the equation indicates, FIGURE 15.10
(a) The l:leClric fidd ,11
A due lO the: nel.{all\L'l} chargc:d splic:n: is
downward. tuward llll' nel.{:llivl: charge:. (b) Thl: electric fleld al Pduc (n Lhe po.'>il i\·c1)' ch:lrged conduuing sphere i:. upward. away frum the positi\c c!J;lrgc. (c) A Lest cll:lrge 1/1, placed 81 JJ\"ill (
P I'
(h)
10
+
Ie)
p
15.4
an electric field at a given point depends only on the charge q on the o4jcct setting up the field and the distance r from that object to a specific point in space. As a result, we can say that an electric field exists at point Pin Active Figure ]!'i.ll whether or not there is a test charge at P. The principle of supcrposition holds when the elcctric field due to a group of point charges is calcu1
The Electric Field
+
507
(aJ
(b)
QUICK QUIZ 15.3 A test charge or +~!-'C is at a point Pwhere the electric field due to other charges is directed LO the right and has a magnitude of 4 X lOti N/C. If the \.est. charge is replaced "Iilh a charge of -~ p,C, the electric field at P (a) has t.he same m<'lgnitude as before, hut changcs direct.ion, (b) increases ill magniutde and ch~nges direct.ion, (c) rcmains the same, or (d) decreases in magnitude and changes direct.ion.
ACTIVE FIGURE 15.11 A test charge fjlj at I'is a di!>lallce ,-frum radially ml11l/onlJ rrom
q.
(h) 11' q is
negalive, lhe electric fidd at !"
J~poillb
QUICK QUIZ 15.4 A cireular ring of charge or radius b has" tOlal charge q uniformly distributed around it. Find the magnitude to the electric field in the cenler of" the ring. (a) 0 (b) 1:,.'1/IJ' (c) h,'1'/IJ' (e1) k,.'1'/1J (e) None of lhese answers is correct.
QUICK QUIZ 15.5 A "free" electron and a "free" prolOn arc placed in an identical electric field. v\Jhich of the following st.atcmelllS are true? (a) Each particle is acted upon by the same electric force and has the samc (\ccelcra~ tion. (h) The electric force on the proton is greater in magnilUde t.han the clectric force on the electron, bllt in the opposite direction. (c) The electric force on the proton is equal in lnagnitude LO the electric force on the electron, but in the opposite direction. (d) The magnitude of t.he acceleration or the electron is greater than that of the proton. (c) Both particles have the samc acceleration.
EXAMPLE 15.4 Goal
Electrified Oil
Use electric forccs and fields toget.her with NeWlOn's second 1;:\\\1 in a one-dimensional problem.
Problem Tiny droplets of oil acquire a small negative charge \\lhile dropping through a vacuum (pressure = 0) in an cxperiment.. An electric field of ITlagnitude 5.92 X 10" N/C points straight dm.. . n. (a) One particular droplet is observed to rcmain stlspendcd againstgravit.y. Ift.he mass oft.hc droplet is 2.93 X ]0- 1:, kg, find the charge carried by the droplet. (b) Another droplet of the same mass falls 10.3 Clll from rest in 0.250 s, again moving through a vacuum. Find the charge carried by t.he droplet. ~
Strategy ''''e use Newton's second law with bOI h gr;witCltional and electric forces. In hot.h pans the electric field E is puintingciown, t.aken as t.he negative direction, as llsual.ln part (a) the acceleration is equal to zero. In pan (b) the acceleration is uniform, so the kinemat ic equations yield, he acceler"ll.ioll. Newton's law can thell be solved for q.
"
.,
..
.
Solution (a) Find t.he charge on the suspended droplet.
Apply Newton's second law to t.he droplet in the vertical direction:
(I)
ma=2:F=-mg+l'.'q
508
Chapter 15
Electric Forces and Electric Fields
(2.93 X 10- 15 kg)(9.80 m/,')
E points downward. hence is negative. Sct a = 0 in Eqmnioll (I) and solve for 'I:
-592 X 10" N/C =
-4.85 X lO-"'C
(b) Find the charge all the raIling droplet. Use the kinematic displacclllcl1l equation La And the acceleration:
Substiwte lJ.y = -0.103 m, I = 0.250 s, and V u = 0:
-0.103
Solve Equation (1) for q and substitute:
'1=
III
= ,I
fI (
0.250 s)"-
->
0.=
-3.30m/s'
1Il( a + Ii) /i
(2.93 X 10-" kg)( -3.30 mis'
+ 9.80 m/s')
-5.92 X 10' N/C -3.22 X 10- 19 C .
Irrr... • • • • • • • • • • • • • • • • •
...
Remarks This example exhibits features similar to the Millikan Oil-Drop experiment discllssed in SCClioll ]5.7, which determined the value of the fundamental electric charge t. Notice thai in both paris of the example, the charge is ver), nearly a lllultiple of' P. QUESTION 15.4
\Vhat would be the accelenllion ufthe oil droplet in part (a) if the electric field suddenly re\ierscd direction without changing in magnitude? EXERCISE 15.4 Suppose a droplet of unknown mass remains suspended against gravity when E minimum mass of tlie droplet?
Answer
= -2.70
X lOr; N/C. ''''hat is the
4.41 X 10-1'> kg
T
Y
CALCULATING ELECTRIC FORCES AND FIELDS
The following procedure is used to calculate electric forces. The same procedure can be used to calculate an electric rield, a simple matter of replacing the charge of interest, q, \.. · ith a convenient test chargc and dividing by the tcst charge at the end: 1. Draw a diagram of the charges ill the problem. 2. Identify the charge of interest, q, and circle it. 3. Convert all units to 51, with ch;ugcs in coulombs and distances in meters, so as to be consistent with the Sf value of the Coulomb constam k,. 4. Apply Coulomb's law. For each charge Q, find the electric force on the charge of interest, q. The magnitude of the force can be found using Coulomb's 1£1\.. '. The vector direction 01" the electric force is along the line of the t\•..-o charges, directed away frolll Q if the charges have the S;.llOe sign, toward Q if the charges have lhc opposite sign. Find the angle (J this vector makes with the positive x-axis. The x-component of the electric force excITed by Q on q will be Feos 0, and the J-cornponem will be Fsin O. 5. Sum all the x-components, getting the x-component of [he resulwlll electric force.
15.4
n.
The Electric Field
509
Sum all the y-components, gelling the y-componcnt oCthe resultant c1ectri<.: force.
7. Use the Pythagorean theorem and trigonometry to find the magnitude and direction of the resultant force if desired.
EXAMPLE 15.5
Electric Field Due to Two P"'o"'in"'t""C:.;.h"'o"'rg"e"'s"-
"
.'
Goal Use the superposition principle LO calculate the electric field due 1.0 t\'-.'O poin1 charges. Problem Charge fJJ = 7.00 fLC is at the onglll, and charge (/2 = -5.00 fLC is all the x-axis, 0.300 ITI fmlll Lhc origin (fig. 15.12). (a) Find the magnitude and direction of thc electric field at point P, which has coordinates (0,0.400) m. (b) Find the ((:Jrce on a charge of2.00 X 10-.'1 C placed aL P. Strategy Follow the problem-solvillg strategy, finding the electric field at point P due to each individual charge in terms of x- and )'-COlllponcnts, then adding the components of each type to get the x- and y-componellLs of the resultant electric field at P. The magnitude of Lhc /l)fce in parr (b) can be found by simply multiplying the magniLudc of the electric field by the charge.
,... ..
"
"
0.·100
m
FIGURE 15.12
(Example I!;.!'i)
The rcsultalll c1(:cli'ic field E al P e.qllals the Vl'etor sum E1 + E~. wilen.: E 1 is the field 4llc to the p<Jsitiv(" ch.lrgc ql ,l1ul E~ is Ihl' lield dill' In
x
lhl' m:galive charge 'h.
....
"..,
.
Solution (a) Calculate the electric field
ZIt
P. " 10 N
Find the magnitude ofE l '.",jth Equation 15Ji:
0
" " ) (7.00 X 10- C) /C 'm (0.400 m)"
= 3.93 X 10" N/C
The veClor El is vertical, making Zln angle 0['90 with respect to Lhe positivc x-axis. Use this fact to find ilS components:
E"
f\lext, find the magnilUde ofE2, again with Equation I Z>.6:
£';1 =
0
= }o', cos (90') = 0 I,;" = Ie', si II (900) = 3.93
",.1'1',1
= (8.99 X 10" N'm'/C') (5.00 X 10 ':,C)
ri 1.80
Obtain Lhe x-component of E:!, Ilsing the triangle in Figllre 15.12 to find cos 8:
X 10" N/C
(0500 m)-
w" N/C
X
acU 0.300 cos (} = = - - = 0.600 hyp 0500 Ii" = E, cos 0 = (1.80 X ~
Obtain I he y-componcnt in the same way, hUl a miJius sign has to be provided for sin 8 because this component is directed downwards:
1.08
X
10" l'\/C) (0.600)
lO" I\;/C
. opp 0.400 SII'O = - - = - - = 0.800 h)'p 0.500
E" = E, sill 0
~
(1.80 X 10" N/C)(-O.800)
= -1.44 X JO" N/C
Sum t.he x-components the resultant vector:
1.0
get the x-component of"
lo~ =
E"
+
Ii," ~ 0
+
1.08 X 10" N/C ~ 1.08 X 10" N/C
510
Electric Forces and Electric Fields
Chapter 15
SUIll the )' -components to get the y-component or the result.ant vector: Use the Pythagorean theorem to rll1d the magnitude of the resuhalll vector:
EJ
= E,y + E" = 0 + 3.93
1':, = 2.49 X 10' N/C
E=VE}+Jo.~'= 2.71 X10'N/C
The inverse tangent function yields the direction of the resultant vector:
(b) Find the force placed at P.
all
(all
-I
(2.49 X 10'• N/C) = 66.6 108 X 10" N/C
0
a charge of2.00 X 1O-R C
Calculate the magniLUde of the force (the direclion is the same as that ofE because the charge is POSiliyc): ~
F= Eq = (2.71 X 10' N/C)(2.00 X 10-' C) 5.42 X 10-3 N
.
.
~.-.-
X 10" N/C - 1.44 X 10' N/C
......
Remarks There were numerous steps to this problem, but each \·...as vcry short. ''''hen attacking such problems, it's important La focus on one small step at a time. The solution comes not frolll a leap of genius, bm [rom the assembly of a number of rclativdy easy parts. QUESTION 15.5 Suppose q'!. were moved slowly to the right. vVhat \·.:auld happen to the angle c/J? EXERCISE 15.5 (a) Place a charge of -7.00 j.LC al point Panel find the magnilude and direction of the electric field at the location o[ q'1 due to 'II and the charge at P. (b) Find the ITIZlgnitude and direction or I he force 011 (h.
Answer
(a) 5.84 X
JO' N/C. = 20.2° (b) F ~ 2.921'1, = 200.°
15.5
ELECTRIC FIELD LINES
A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the electric field vector at any point. These lines, introduced by Michael Faraday and called electric field lines, are related 1.0 the electric field in any region of'space in the following way: ~
1. The electric field vector E is tangent to Ihe electric field lines at each point. 2. The number of lines per unit area through a sllrface perpendicular to the lines is proportional Lo the strength of the electric field in a given region. Note that E is large \".. hen the field lines are close together and small when the lines arc far apart. Figure 15.13a shows some reprcsentative electric field lines for a single positive poinl charge. This t\\'o~dimensionaldrawing contains only the field lines that lie in the plane containing the point charge. The lines are actually directed radially outward [rom the charge in all directions, some"...hatlikc the quills of an angry porcupine. Because a positive test charge placed in this field would be repelled by the charge fj~ the lines are directed radially away from the positive charge. The electric fidd lines for a single negative point charge are directed [Oward the charge (Fig. 15.13b) because a positive lest charge is attracted by a negauve charge. fn either case the lines are radial and extend all the \-va)' to infinity. Note that the lines are closer together as they get near the charge, indicating that t.he strength of the flcld is increasing. Equation 15.6 verifies that this is indeed the case. The rules for drawing electric field lines ror any charge distribution loHmv directly from the relationship between electric field lines and electric field vectors:
15.5
Electric field Lines
Sl1
.~
(a)
Ie)
(b)
FIGURE 15.13 The den!"i, lidd lil1('~ rnr a poitllcliargl". (,1) For.l po~ilin:' pOilllch;lrgt..:, lh~ linc~ radiale Oll1\\'ard. (b) For a Ilt'g"
1. The lines for a group of point charges must begin 011 posit.ive charges and end 011 negative charges. In the case of an excess of charge, some lines will begin or end infinitely far away. 2. The !lumber of lines dnn,vll leaving a positive charge or ending on a negative charge is proponional LO the magnitude 01" the charge. 3. No t\\'o fidd lines can cross each other.
Figure 15,14 shows the bealltifully s}'lllmetric electric field lines for lW'O point charges of equal magnitude but opposite sign. This charge configuration is called an electric dipole. Note that the number of lines that begin at the positive charge mllsl equal the number that terminate at the negative charge_ At points very near eilher charge, the lines arc nearly radial. The high densily of lines bel\·...een the charges indicates a strong cleclI'ic J'ield in this region. Figllre 15.15 (page 512) shows the electric field lines in the vicinil)" oft\Vo equal posit ive point charges. Again, close to either charge the lines arc nearly' radial. The same number of lines emerges from Ci.lch charge because the charges
TIP 15.1 Electric Field Lines Aren't Paths af 1 Particles ElcClric fidd lines are 1101 material Ol~jccl~. The) art-' used only as a piclorial rcprc,,>cmat ion of I he electric ri.ckt at variolls locations. Excepl in special cases. they do lIol represent 1he pal h of
a charged parlic1e released in
FIGURE 15.14 (a) Thc ckctric field lilll.:~ ror (\\,0 l'qu;ll alld OPPOSill' poilll
"kelt k dipole), ;..lUll:: lhe Ilumher orlil1l'~ leal'inA" the POSili\"e charge equah (he 1HlmhL'r It'Tll1illalillg
chal ~es (;Lll
lhal
(a)
(hi
512
Chapter 15
Electric Forces and Electric Fields
~-q
(b)
ACTIVE FIGURE 15.16 The cleClric field lines for a point charge of +:!qand a second poill! charge of -g. :\lotI..' that tll'O lines !caw the charge +2'1 for cvery lille that lcrmillales un -If.
FIGURE 15.15 (a) The c!eclril· field lilll'~ fOI Lll'O pusitive point cllargc~. The paim... A, B. and Care di.'Cllsscd in Quick Qui; 15.11. (b) TIlt: dark linl'1'> are .. mall picles nflhrc'ad su~pended ill oil, I\hich aligllll'illl the dCClric fidel produced by 11110 rharged Ctlliductors.
QUICK QUIZ 15.6
Rank the magnilUdes of the electric field at poims A., B, and Cin figure 15.15, with the largest magnitude lIrsl. (aJ A, /3, c: (b) A, C, IJ (c) C, A, IJ (el) The answer can't be determined by visual inspection.
APPLYING PHYSICS 15.1
MEASURING ATMOSPHERIC ELECTRIC FIELDS
The electric fielel near the surface of' the Earth in fair weather is about lOO N/C downward. Under a thundercloud, the electric field can be very large, on the order 0[20 000 I\jC. How are these electric fields measuITd?
Explanation A device for measuring these flelds is called the Jield mil/.. figure 15.17 sho\vs the fundamental componclHs ofa field mill: two metal plates parallel to the ground. Each pi
is suddenly moved over the lower plate, as in Figure 15.-17b. This plate is also connected to ground and is also negatively charged, so the rield lines now end on the upper platc. The negative chargcs in the lower plate are repelled by those on Lhe upper plate and must pass through the ammeter, registering a now of charge. Thc amount of charge that was on the lower plate is related to the strength or the electric field. In this way, the no\\' of charge through lhc ammeter can be calibratcd 1.0 measure I he electric field. The plates are normally designed like the blades ora fan, v,iLh the upper plme rotating so that the lower plate is alternately covered and uncovered. As a result, charges flow back and forth continually through the ammeter, and the reading can be relatcd to thc electric Aeld strength.
II I III III d I
~~ ~ ~
(a)
(b)
FIGURE 15.17 (Applying Phy.. ics lEU) III (a) e1eclI·ic field lille~ elld on neg;ui\·c charges all the lower plale. In (b) the secund plaIt.: i~ l1\o\'cd ah(l\e the.; lower plate. I::lectric field lilIes HOI\' end all the l1pper plate. and the l1l:gative charges in tho:: lower plate ;\re I"t'pclkd Lhrolq;h lhl.: ammeLer.
15.6
15.6
Conductors in Electrostatic Equilibrium
S13
CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM
A good electric conductor like copper, ~ll.hollgh electrically ncutral, contains charges (electrons) lha\. aren't bound 10 any a\.om ,1Ild are free to rnnvf' ahOlll within the malerial. \-Vhen no net motion ofcharg-e occllrs within a conductor, the conductor is said to be in electrostatic equilibrium. An isolated conducl.Or (one Ihal is insulated from ground) has the following properties: 1. The eleClric field is zero everywhere inside the conducting material.
2. Any excess charge on an isolated conducl.Or resides entirely on its surface. 3. The electric field just oUlside a charged conductor is perpendicular LO the
f- Properties of an isolated conductor
conductor's surface. 4. On an irregularly shaped conductor, the charge accumulates at sharp points, where the radius of curvature of the Sllr1~ICC is smallest. The first property can be underslood by examining what would happen if it \vcre not true. If there were an electric field inside a conductor. the free charge I here vl'Ould rna\"(:: and a flow of charge, or current, \vould be created. 1f there were a net movement of charge, however, the conductor would no longer be in electrostatic equilibrium. Property 2 is a direct result of the ]/r 2 repulsion between like charges described by Coulomb's law. If by some means an excess 01" charge is placed inside a COI1ducwr, the repulsive forces between the like charges push them as far apart as possible, causing them \.0 quickly migTatc to the surface. (\Ve wan', prove it here, but the excess charge resides on the surface because Coulomb's la\v is an invcrsesCJuarc law. \Vith any other power law. an excess of charge would exist on the surface, bUlthere would be a distribution of' charge, of either the same or opposite sign, inside the conducl.Or.) Property 3 can be understood by again considering what would happen if it were not true. If the electric field in Figure 15.l8a were not perpendicular to the surface, it. would have a componell1. along the surface, \.. . hich \vould cause the free charges of the conductor to move (to the lel't in the figure). J f the charges mo\"ed, however, a current \voldd be created and the conductor would no longer be in electrostatic equilibrium. Therefore, E must be perpendicular to the surface. To see why propeny 4 must be true, consider Figure 15.19£1 (page 514), \dlich shmvs a conductor that is fairly flat at one end and relatively pointed
I'
f
•
~
~ ~
~
0 5
I
0
f~ ~§
u
(a)
>
~
] -~ ~
(h)
• I Electric fidel paucrn ofa ch;lrgL:d coneiliCLing plalL: llcar all oppo~ild~· charged p,)inteel condliclOL Small pieces nflhread sllspL:ndcd in oil align with the dcnric field lillL:S. 1\olice th;lllhe electric field is IIlost illlense near the poililed pan of llie condllclor. \I'llere the radius of CIII ....alnre is the smallest. ..\lso.lhe lincsar(' perpendicular \0 the C011dllcton,.
FIGURE 15.18 (a) !\t:galivt:cllargc"al lhesurfa,cofacQnduClor.lfllll' t:leclric f-ield \l'l:r(' al ,Ill allgle lO the ~1l1 face, as ~hOWll. an deetde force wOlild ilL: eXCl"l<-'d 011 lile dlilrg('S ;tlUllg Ill(' sllrfac(' and 1I1C\ would moHo' In ll1e Iell: BCClILl:.C Lhe conduClor is assumed to be in ekctro~latic c:quilihriulll. E ClilllOl h,II'<-" a COmpOlll'lIl along lhe ,urlace ;lIld hCllce Illu:-.L he perpcndicular 10 it. (h) Tile eke\! ic field pallcrll 01',1 charged condll(l,i Ilg pI ,Ill' near :Ill oppo.sildy charged UlilduCI ill,!.!; cvlilld('1". Small pil·CC' or thread sllspt:ndcd ill oil ;llig-I) witll L!1l' t:kclric field lines. :--,:roli(t: thai (I) rht: el"clrir hl·lrllillcs arc P(·qwlldkillar LO rhe condllcrur~alld (2) then' are nil line:. in:.idc rhe cylinder (E = 0).
514
Chapter 15
Electric Forces and Electric Fields
FIGURE 15.19 (a) A COIlc/ueLm with a f!aller ('llll:1 :Illd a rd:llivcly -;harp end n. Excess charge pl:H'ecl on thi .. conduclor resides emirely at liS .. urface al\d is disH ibnled <;() lhal (bl t1lr..Tf'" is lc% charge pcr unil an:a on Ill..: flatter ('lid ;1 lid (c) there is ,. largt· cliarg'p per IInit area on the ... bou-per cud.
t ~
fj
IJ
(a)
U
_I
(a)
- +
+ -
+
+
-+
+-
-++
++ +-
W'
-+ + 1 -~++~-~-(b)
(e)
l-JP W
(d)
I
FIGURE 15.20 An experimenl showing Ihal ;Hl> charge transferred 10 a condlillor r<:~idcs on i\s surface ill c!ectro'lalic cqllilibrillrn. The Ilol\m\conducLor is insulated fwm grolilld, and the sillall mel'll ball is <;lIpponeri bV'lll inslILliing: thread.
APPLICATION Lightning Rods
(b)
(e)
per unit area is greater than at the flat end. The cumulative effect of many such outward forces from nearby charges at the sharp end produccs"1 large resultant force directed m...ay from tile surface that can be great enough to cause charges to leap from the surface into the surrounding air. Many experiments have shown that the net charge on a conductor resides on its surf-ace. One such experiment was first performed by tI,·1ichaci Faraday and is referred to as Farr{(laY:~ ice-Imil exl)('rinwnl. Faraday 100\'cred a metal ball having a negative charge at the end of a silk thread (an insulator) into an uncharged hollow conductor insulated from ground, a metal ice-p;]il as in Figure 1.5.20a. As the ball entered the pail, the needle on an electrometer attached to the outer surface of the pail was observed to deflect. (An elcctrometer is a dcvice used to measure charge.) The needle deflected because the charged ball induced a positive charge on the inner "'all of the pail, which left an equal negative charge on the outer wall (Fig. 15.2Gb).
Faraday next touched the inner surface o[ the pail with thc hall and noted that the deflection of the ncedlc did not change, either when the ball touched the inner surface of the pail (Fig. 15.20c) or when it was removed (Fig. 15.20d). Further, he found that the ball was now uncharged because when it touched the inside of the pail, the excess negative charg·e on the ball had been drawn oJ{ neutralizing the induced positive charge on the inner surface of the pail. 1n this way Faraday discovered the useful result that (til the excess charge all an object can be transferred to an already charged metal shell if the object is t.ouched to the insielf' of the shell. As we will see, this result is the principle of operation of the Van de Graafr generator. Faraday concluded thaI because the denection of the needle in the electromClCr didn't. change "lhen the charged ball \.Quched the inside of thc pail, the positivc charge induced on the inside surface of the pail was just enough to neutralize the negative chargc on tht: ball. As a result of his investigations, he concluded that a charged object suspended inside a rnetal container rearranged the charge on the container so thal the sign of the charge on its inside surface "'liS ojJIJOsite the sign of t.he charge on the suspended ol~ject. This produced a charge on the outside surface of the cOlltainer of the same sign as that on the sllspended ol~jecL Farada~· also found thal if the electrometcr was connected to the inside surface o[ the pail after lhe experiment had been rlln, the needle showed 110 dencClion. Thus, the pxress charge acquired by the pail when contact was made bel\\'cen ball and pail appeared on the aliter surl~lCe of the pail. If a metal rod having sharp points is attached to a hOllse. most of any charge on the house passes through these points, eliminating lhe induced charge on the house produced by storm clouds. In addition, a lig·htning discharge striking the house passes through the metal rod and is safely carried to thc ground lhrough ",,·ires leading I"rom t.he rod to the Earth. Lightning rods llsing Ihis principle were first dcveloped by Benjamin Franklin. Some European countries couldn't accept the fact that such a v,,'orthwhile idea could have originated in the New 'World, so they "impro'"cd" the design by eliminating the sharp points~
15.7
APPLYING PHYSICS 15.2
Explanation \,Vhen the spherical shell is placed around the charge, the charges in the shell rcarrange 1.0 satisfy the rules for a conductor in equilibrium. A net charge of -Q 1n00'es to the interior surface or the
conducLOr, so t.he electric flcld inside lhe conductor becomes zero. This mcans the field lines originating on the +Qcharge now terminate on the negative charges. The movement of the negative charges LO the inner surface of the sphere lea\"es a net charge or +Qon the ouler surface orthe sphere. Then the field lines olltside the sphere look just as before: the only change, overall, is the absence offield lines within the conductor.
DRIVER SAFETY DURING ELECTRICAL STORMS
Why is it safe to stay inside an automobile during a lightning storm? Explanation Many people believe that staying- inside lhe car is safe because of the insul'lting characteristics of the rubber tires, but in fact that isn't true. Lighl11ing Clln travel through several kilomcters of ai r, so it ca n cerl ai n Iy penetrate a cenl imeter of rub-
15.7
515
CONDUCTORS AND FIELD LINES
Suppose a poilH charge +Qis in ernpt)' space. \-Veal.. . ing rubber gloves, you proceed to surround the charge with a concentric spherical conducting shell. "'ThaL effect docs that have on the field lines from the charge?
APPLYING PHYSICS 15.3
The Millikan Oil-Drop Experiment
ber. The safety orrcrnaining in [)le car is due to the racl. that charges on the metal shell of the car will reside on the outer surface of the car, as noted in property 2 discussed earlier. As a result. an occupant in the automobile touching the inner surfaccs is not in danger.
THE MILLIKAN OIL-DROP EXPERIMENT
From ]909 to ]913, Robert Andrews \1; lIikan (1868-1953) performed a brilliant set of" experiments at the University of Chicago in which he measured the elementary charge po[ the electron and demonstrateclthe qllantized nature of the electronic charge. The apparatus he used, diagrammed in Active Figure 15.2], contains 1\'1"0 parallcl metal plates. Oil droplets lhat have been charged by friction in an atomizer arc allowed to pass through a small hole in the upper plate. A horizolltal lig-Ill beam is used to illuminate the droplets, which are \"iewed by a telescope with axis at right angles to the beam. The droplets then appeal' as shining stars against a clark background, and the rate or fall of individual drops can be determined. We assume a single drop having a mass or 111 and carrying a charge of q is being viewed and its charge is negative. Irno electric field is present bel\.. . een the plates, the two forces acting on the charge arc the force of gravity, mg, acting downward, and an upward viscolls drag force D (Fig. 15.22<1. pagc 516). The drag force is proportional to the speed of the drop. When the drop reaches its terminal speed, v, the tv,,'O forces bCllance each other (JI/g = D). ~
Oil droplets
!
Tckscnpe wilh scale ill eyepiece
Pinhole
ACTIVE FIGURE 15.21 A schl·tlIalic \'icw orMillik::lll'S oildrop ;lpP;lrallls.
516
Chapter lS
Electric Forces and Electric Fields
FIGURE 15.22
The forces un;\
nCI-;":1.-
lively charged oil droplcl in r.,·lillibll's cxperirnellt.
(3) Field off
(b) Ficld
011
)Jaw suppose an electric field is setup between the plates by a baltery t.9nncclcd so that the upper plate is positively charged. In this case a third force, qE, acts on t.he charged drop. Because q is negative and E is downward, the electric Coree is upwmrJ as in Figure 15.22b. rr this force is great enough, the drop moves upward and the drag force D' aels downward. \,Vhen the upward electric force, qE, balances the sum of the force of gravity and the drag force, both acting downward, the drop rcaches a new tcrmincl! speed v'. VVith the llelc1 tllrned on, a drop moves slowly upward, typically at a rate of Itundredr.hs of a centi meter per second. The rate of [all in the absence of a field is comparable. I-rence, a single droplet with constant mass and radius can be followed lor hours as it alternately rises and falls, simply by turning the electric field on and off. /\. rter maki ng measurcments on thousands of droplets, Millikan and his coworkers Found that, LO w'ithin about 1% precision, e\'cr}' drop had a charge equal to some positivc or negativc integer multiple of the elementary charge e, ~
~
~
q = ne
Metal dOllle
FIGURE 15.23
A diagram ofa V~11
rotating bell. TIl(' c1largl' is depllsiled \l111h<: bell al poilll A and tr;ulskrn,d [0 the dOlm: at pnilll B.
= 0,
::tl, ::::!:2, ::::!:3, .
[15.7]
where I' = 1.60 X 10- 19 C. It was later established that positive integer multiples of (' would arise when an oil droplet had lost one or more electrons. Likewise, negative integer multiples of e \','Quld arise ".fhen a drop had gained one or more eleC[rons. Gains or losses in integral numbers provide conclusive evidence thaI charge is Cluantized. In -1923 Millikan was awarded the Nobel Prize in Physics for this work.
15.8
de CraaffgeneraLOf. Ch;lrgf' is rran~ ll:rred to lhe dome by me,iIlS 01" a
1/.
THE VAN DE GRAAFF GENERATOR
In 1929 RobenJ. Van de Graafr (1901-1967) designed and builL an electrostatic generator t.hat has been used extensively in nuclear physics research. The principles of its operat ion can be understood with knmvledge of the propenies of electric fields and charges already presented in this chapter. Figure 15.23 shows the basic const.ruction of' this device. A motor-driven pulley P moves a belt past positively charged comb-like metallic needles positioned at A. Negative charges are allraCl.cd to these needles from lhe belt, leaving the lefr. side of the belt with a net positive charge. The positive charges attracr. electrons onto the belt as it moves past a second comb of needles at. 13, increasing the excess positive charge on the dome. Because the ciCCI ric neld inside the metal dome is negligible, the positive charge on it can easily be increased regardless of how much charge is alread)' presenl. The result is that t he dome is left with a large amount of' positive charge. This acclllllulation of charge on the dome can't continue indefinitely. As more and more charge appears on the surface of the dOllle, the magnitude of the electric Ficld at thatsllrface is also increasing. Finally, t.he sr.rength of the field becomes great enough to partially ionize the air near lhe surface, increasing lhe conductivit.y of the air. Charges on the dome nov.' have a pathway to leak oiTinto 1he air, producing some spectacular "lightning bolts" as the discharge occurs. As noted earlier, charges find it easier to le£lp orra surL\Ce at points where the C\lrvaturc is great. As a result, one way 10 inhibit the electric discharge, and to increase the amount of charge thal can be stored on the dome, is to increase its radius. Anolher method
15.9
Electric Flux and Gauss's Law
517
for inhibiting discharge is to place the entire systcm in (l cOlltainer r-illcd \vi1.h a high-pressure gas, ,,~'hich is signillcantly more difficult to ionil.c than air at almospheric pressure. Ir prawns (or other charged particles) are introduced into a tube al.lachcd to lilt:: dOllie, tile lalge elect! ic field of [he donlc exell:j a Iepllbive rUl ce 011 t1le protons, causing them 1.0 accelerate to encrgies high enough [0 initiate nuclear reactions between the protons and various target nuclei.
15.9
ELECTRIC FLUX AND GAUSS'S LAW
Gauss's law is cssel1lially a technique for calculating tile iwcrage electric lield on a closed surface, developed by Karl Friedrich Gauss (1777-1855). vVhen the elecu-ic field, because of its symmetry, is constant everywhere on that surface and perpendicular to it, the exactdectric field GlIl be found.ln such special cases, Gauss's la\,\,' is far c(lsier 1.0 apply than Coulomb's law. Causs's law relates the clec\.ric nux through a closed surf-ace and the tOl;·11 charge inside that surface. A r1os('(l .w11fatt has an inside and all outside: all example is a sphere. 1~·lecJ.ri(.lI/l.x;s a measure of how mllch the electric AcId vectors penetrate through a given surface. 11" the electric field vectors arc tangent to the surface at all points, for example, they don't penetrate the surface and the electric flux [hrollgh [he surface is zero. These concepts \.. . ill be discllssed more flllly in the next t\V() subseci ions. As \,,'c'1I sec, Gauss's law states that 1he elect ric flux lh rough a closed sllrfacc is proportional to the charge contained i.nsidethc surface.
FIGURE 15.24 Ficld Iinc~ ot a 11Iliform dcnri(- field pCIll:lraling a plane of area A PCI pelldicula'·lo lht· field. The dcetric t1\lX iIl/., 1hroug-h 111is area is cquai to Ell.
Electric Flux Consider an electric lleld that is uniform in both magnitude and direction, as in Figure 15.24. The electric Field lines penetrate a surr~lcc of area A, which is perpendicular 10 the field. The lechnique lIsed ror clra\'lIing- a figure SHcil as Figure 1:").24 is that the number of lines per unit area, :V/A, i.., proportional 1.0 the magnilllde of the electric Field, or Er:x !'vIA. VIle can rewrilc this proportion as /\' 0'.: HA, which means that the number offield lines is proportional to lhc !Jroductof Eancl A, called the electric flux and represented by I.he symbol q\:
[15.8] Note (hat (1)/, has Sf units of N· m~/C and is proportional lo the number of field lilies that pass through some area A oriented perpendicular to the field. (It's called nllX by analogy with the term .llux in fluid flow, which is the volume ofliqllid flowing through a perpendicular area per second.) If the surface under consideration is not perpendicular to the llcld, as in Figure 15.25, the exprcssion for the electric flux is
<1>1! =
','A cos e
[15.9]
where (j vcc1.or perpendicular to the area A is at all angle 0 \\lith respect to the field. This vector is often said 10 be JW'1J/.rilto thc surface, and we will refer to it as "the nonnal vector 1.0 tile surface." The number of lines that. cross 1 his arca is equal to lhe llulllbcr lhat cross the projected area A', which is pcrpcndiclllar to the field. \rVc sec that the 1.\..'0 areas arc relatcd by AI = /\ cos 8. Prom Equation ]5.9, we see that t he flux through a surface of fixed area has the maxi mum value EI\ when the surface is perpendicular to the field (when 0 = 0°) and thai the flux is t,cro when the surf~\Ce is parallel \.0 the field (when () = 90°). By convention, for a closed surface, the flux lines passing into the interior of the volume are negative and those passing out of the interior of the volume are positive. This convenlion is equivalent [0 requiring the normal vector of the surface 1.0 point. outward when computing the flux through a closed surface.
Nt~rlll;tl
A
,,
~~
~~~ /v ~
A'=AcosO
FIGURE 15.25 Fidd lines for a uniform electric lidd through all arc.. i\ that is al an
of (90 C
-
0)
10
IhL:
ficld. Because the Ilumbcr of lilIes lhal go lhrough the ~had(·d area A' is the same as the /lllmher Iha1 go through .4., \lif"conclude Iha1llw flux thr<Jlll-{h .4' i:. equal la lht.: flux lhrough A alld is givl:lI by
~
Electric flux
518
Chapter 15
Electric Forces and Electric Fields
QUICK QUIZ 15.7 Calculate the magnitude of the flux of a constant electric field of" 5.00 N/C in the z-dircClion throug"h a rectangle with area '1.00 111 2 in the xy-plane. (a) 0 (b) 10.0 N 'm'/C (c) 20.0 N· m'/C (d) More information is needed QUICK QUIZ 15.8 Suppose the electric fIeld of Quick Quiz 15.7 is lilted 60 m.. . a)' from d1C positive z~direction. Calculate the magnitude of the flux through rhe same area. (a) 0 (b) 10.0 N ·m 2 /C (c) 20.0 N 'm'/C (d) More information is needed 0
EXAMPLE 15.6 Goal
Flux Through
0
Cube
Calculate the electric nux through a closed surface.
Problem Consider a uniform electric field oriented in the :I:-direclion. Find the eleClrie nux through each surface ora cube \\lith edges L oriented as shown in Figure 15.26, and the net flux. Strategy This problem involves substituting inLO the ddinition of electric flux given by Equation IS.9. In each Celse E and A = L2 are the same; the only difference is the angle f) that the electric field makes with respect [Q a vector perpendicular to a given surface and pOinting olll\vard (the normal vector to the Slld~lCC). The angles can be deLermined by inspection. The flux through a surface parallel to the A")'-plane \-vill be labeled cV,; and fUrl her designated by position (front, back); others will be labeled similar]y: ): left or righL. ~
FIGURE 15.26 (bampk 15.6) A hypolhcliGl1 slirEll:e ill dw shapf' ora cuhe ill <\ uniform electric field paral]el to tile x-axis. The Ilet flux through I lit: Sllrfacc is I.ero whell the llt.:t charge insirlt.: I hc cube is lero.
. . . . . . . . . . . . . . . . . ....
-
, , ,
...
~
Solution The normal veClOr to the xy-plane points in the neg;lth'e z-direction. This, in turn, is perpendicular to E, so () = 90°. (The opposite side works similarly.)
(V.n = 6i\ cos (90°)
0 (back and from)
The normal vector to the xz-plane points in the nega.... tive y-direetion. This, in turn, is perpendicular to E, so again fI = 90°. (The opposite side works similarly.)
<1>" = Ell cos (90°)
0 (top and boltom)
The normal "ector LO surface CD (the y.z-plane) points in the negative "\-~direction. This is anl.iparallel to E, so 8 = 1S00.
<1.\ =
Surf~l.Ce
® has normal vector pointing in the positive x-direction, so B = 0°.
el'" = Efl cos (00)
\,Ve calculate the net flux by summing:
...
Efl cos (ISOO)
-EL' (surfaceQ))
EL' (su rface
0 + 0 + 0 + 0 - t'L':l +
CV)
1:.''-.,2 =
0
.............
-~
Remarks In doing this calculation, it is necessary to remember [hal the angk in the definition of nux is measured from Lhe normal \'ector to the surface and thal this vector !TIust point ollt\-vards for <1 closed surface. As a result, lhe normal vector for the )'z-planc on the left points in the negative x-direction, anclthe normal "eemf to the plane parallellO the )'z-planc on the right points in the positive x-direction. Notice that there aren't allY charges in (he box. The net electric flux is always zero for closed Slit· races that don't contain nct charge. QUESTION 15.6 If the surface in Figure ]:").26 were spherical. would the answer be (a) greater than, (b) less than, or (c) the same as the net electric flux found for the cubical surface?
15.9
Electric Flux and Gauss's Law
519
EXERCISE 15.6 Suppose the constant electric fIeld in Example 15.6 points in the positive y-dircction instcad. Calculate the flux through the xz-planc and the surface parallel to it. \Vhat's the net electric nux through the surface of the cube? Answers
(flx~ =
-El} (bollom), ep~~ = +EJ} (lOp). The net flux is still zero.
Gaussian surface
Gauss's Law
/
Consider a point charge fJ surrounded by a spherical surface of radius,. centered on the charge, as in Figure 15.27a. The magnitude of the electric field evcrywhere on the surface or the sphere is
q
(a)
Note Lhat the clectric field is perpendicular LO the spherical surface at all points on the surl"tlce. The elcctric flux through the surface is therefore l~A, where A = 47i""~ is the surface area or the sphere:
It's sumetimes convenient to exprcss he in terms of" another constant, Eo, as he = 1/(47TE u). The constant En is called thc permittivity of free space and has the value E
o
I 41Th"
1')
'/
'/
= - - = 8.85 X 10- - C-/N'I11-
[15.10]
"r
The use of or Ell is strictly a matter of tasle. The electric nux through the closed spherical surface thal surrounds the charge q can now be expresscd as
q
(b) FIGURE 15.27 (a) The flux through a ~pherical SUI face ofradills rsurrounding a poi III charge 'I is 11\ = q/€n' (b) The lIux through allY arbitrar~" surface <;urrounding the charge is ,1!sO eqllalLO q/€o'
This result says that the electric flux through a sphere that surrounds a charge q is equal to the charge divided by the constant Eo. Using calculus, this result can be proven for any closed surface that surrounds the charge fJ. For example, if the surface surrounding q is irregular, as in Figure 15.27b, the flux lhrough that surface i~ also qlEiJ. This leads to the following general result, known as Gauss's law: The electric nux q)E through any closed slIrface is equal to the net charge inside the surf~1Ce, Qillsidl" divided by E(I: QlllSide
I = ---
()E
~ Gauss's Low
[15.11]
EO
Although it's not obvious. Gauss's law describes how charges create electric fields. In principle it c(tn always be used to calculate the electric field of a systcm of charges or a continuous distribution of charge. In practice, the technique is useful only in a limited number of cases in which there is
. '1 Q~ -
-
En
[15.12]
TIP 15.2 Gaussian Surfaces Aren't Real Camsian SIll'f';\Ct' is an imaginary sllrf
A
facilitate a lllatl1emalical c
520
Chapter lS
Electric Forces and Electric Fields
as \..· ill be seen in the examples. Although Gauss's la\.. . in this form can be used LO obtain the electric field only for problems with a 10l of symmetry, it Gln n/wa)1S be used to obtain the averagfclcclric field on any surface.
-'2C
+
.,..2C
QUICK QUIZ 15.9 Find the electric flux through the surface in Active Figure 1528. (a) -(3 C)/E" (b) (3 C)/E" (c) 0 (d) -(6 C)/Eo -oC
+
J
C
QUICK QUIZ 15.10 Fo!" a closed surface through 'which the net flux is zero, each of the [ollm.. . ing four statements muld be true. \iVhich of the statements must be true? (There may be more lila 11 olle.) (a) There are no charges inside the surface. (b) The net charge inside the surface is zero. (c) The electric field is zero c\'cr)'where on the surface. (cI) The number of electric f1elcllincs entering the surface equals the Ilumber leaving the surface.
ACTIVE FIGURE 15.28 (Quick QuL~ 1:),9)
EXAMPLE 15,7
The Electric Field of
0
Chorged S hericol Shell
Goal Use Gauss's ]a\\i' 10 determine electric Relds v.:hen t he symmetry is spherical. Problem A spherical conducting shell of inner radius a and outer radius b carries a total charge +Q distributed on the sllrf.~lCe of a conducting shell (Fig. 15.29a). The quantit} Q is taken to be positive. (a) Find the elcClric field in the interior of the conduCling shell, for r< a, and (b) the electric field outside the shell, for r> b. (c) Ifan additional charge of -2Qis placed at the center, find the electric field for r> b. (d) ,~rhat is the distribution of charge on t.he sphere in part (c)?
Strategy For each pan, dra\\' a spherical Gaussian surface in the region of illlerest. Add up the charge inside the Gaussian surface, substitute it and the area illLO Gauss's la\'/, and solve for the electric field. To find the distribution 01" charge in pan (c), LIse Gauss's law in reverse: the charge distribution must be such that the electrostatic field is zero inside a condUClOr.
,
Caussian
t I
e",·,·r,,,e
If
,f/
/
b
--'Z:.-:::'V --....
I
GalJssian surface
" '
+
+
+
\ (,)
(hi
(el
FIGURE 15.29 (Exaillplc 15.7) (a) Tht.· electric field in"ide a ulli~ forml)' charg·ed spherical shell i~ zero. It is also zero for tht: conducting material in the regioll a < r < b. Tlie fidd olll"ide is tht: same as LhaL of a POillt charge Ilaving a total charge Q located :ll the center of the shell. (b) '["he cOllstruclioll of a GauI;si:'l11 surface for calculating the electric lield insirlea spheriral shell. (c) The cOlbtruction ora G;mssi;lll surfan: for e
1'"' •
.~
Solution (a) Find the electric field for 1·<
{f.
/01
£(417r') =
Apply Gauss's law, Equation 15.12, to the Gaussian surface illustrated in Figure 15.29c:
/~.\ =
£(4"r')
Divide by the area:
£=-41TE or:?
:\pply GclUSS'S law, Equ:nion 15.-12, LO the Gaussian surface illustrated in Figure 15.2gb (note that there isn't any charge inside this surface): (b) Find the electric held for
Qill,;idc
= 0
En
r> b.
Q
Q
-->
1:'=0
15.9
Electric Flux and Gauss's Low
521
(c) Now an additional charge of -2Qis placed at the center of the sphere. Compute the new electric field olltside the sphere, for,.> b, Appl}' Gauss's law as in part (b), including' he ne,,' charge ill Qin.;;id,.:
+Q-2Q
/';;\
F=
Q
(d) Find the charge dislribution on the sphere for part (c). \~rritc
Gauss's Jaw for the intcrior of the shell:
Fincllhe charge on the inner surface of the shell, noting that the electric field in the conductor is zcro:
Find the charge on the ollter surface, noting that the inner and OLlter surface charges must Slim to +Q: ~
Q(}lIl<'rSlIrfilU'
+
Qillllcrslirfarc
QOIIT"r
=
-Q"'lIcr'llrLlct"
= Q
+ Q = -Q
.....
..
Remarks The important thing to notice is that in each case, the charge is spread out m"cr a region with spherical symmctry or is located atlhc exact center. That is what <111O\\ls the computation ora value [or the electric field.
QUESTION 15.7 If the charge at the center of the sphere is lllade positive, how is the charge on the inner affected?
SUrl~lCC
of the sphere
EXERCISE 15.7 Suppose I he charge at the center is now increased to +~Q, while I he surface ofLhe conductor still retainsa charge or +Q. (a) Find the eleci ric l1eld exterior to the sphere, lor r > b. (b) \Vhat's the electric field inside theconduc[Qr, for a < r < b? (c) Find the charge distriblllion on the conductor. Answers (a) E = ~Q/47TEur2 (b) t; = 0, which is always the case when charges are nol moving in a conductor. (c) Inner surface: -2Q; outer surface: +3Q
In Example] 5.7, not much was said about the distribut ion of charge on the conductor. Whenever there is a net nonzero charge, the individual charges will try 1.0 gel as far away from each 01 her as possible. fIence, charge will reside eithcr on the inside surrace or on t.he olltside surface. Because thc electric field in the conductor is zero, there will always be enough charge on the inll(:r surface to cancel \"'haleVtT charge is at the center. In pan (b) there is no charge on the inner surface and a charg<: or +Q on th<: outer surl"ace. In pan (c), with a -Q charg<: at the center, +Q is on the inner surface and 0 C is on the ollter surf~lCe. Finally, in the exercise, with -2Q in the center, there musl be +2Q on the inner sllrl~lce and -Q on the outer surface. In each case the t.Otal charge on the conductor remains t.he same, +(2; ii's just arranged differently. Problems like Example 15.7 arc oftcn said to have "thin, nonconducting shells" GlIT)'ing a uniiorlllly distributed chargc. In these cases no distinction need be made between the outer surface ;-lnd inner surblce of 1 he shell. The next example makes that implicit assumption.
522
Chapter 15
Electric Forces and Electric Fields
EXAMPLE 15.8 Goal
Appl), Gauss's law to
Problem
~l
problem with plane symmetry.
Find the electric field above and below a nonconducting infinite plane sheet of charge with uniform posi-
tive charge per unit area
(T
(Fig. 15.30a).
Strategy By symmeli')', the electric field must be perpendicular to the plane and directed away from it on eit.her side, as shown in Figure 15.30b. For the GaussiClIl surface, choose a sIllall cylinder with axis perpendicular to the plane, each end having area 1\0> No electric Aeld lines pass through the curved surface of the cylinder, only through the t,..,o ends, which have LOtal area 2A Il . Apply Gauss's Imv, llsing Figure 15.30b.
Solution (a) Find the electric fidd above and below a plane of uniform charge. Apply Gauss's law, Equation 15.12:
The total charge inside the Gaussian cylinder is the charge densily times the cross~scctional area: The electric flux comcs entirely from the two ends, each having area 1\0' Substitutc A = 2A(I and Qill,idl' a nd solve lor 1:.'. This is the magnillldf' of the cleclric field. Find t.hc z.-compotlcnt or the field above and below the plane. Thc elcctric ficld points a\vay from thc plane, so it's positivc above the plane and negative below t.he plane. ~
E, =
•< 0 .
.
~
Remarks NOlice here that the plate '.vas taken 1.0 be a thin, nonconducling shell. Ifit's made ol'melal, of course, the eleclric field inside il is zero, wilh half" lhc charge on the upper surface <'lnd half on the lowcr surface. QUESTION 15.8 In reality, the shect carrying charge woulcllikely be metallic ilnd have a small but nonzcro thickncss. If"it carries the sarne charge per unit area, what. is the c1ectJ'ic field inside the sheet betwecn the two surfaces?
Summary
523
EXERCISE 15.8 Suppose an infinite nonconducting plane of charge as in Example 15.8 has a Find the electric field above and below the plate. Sketch the field.
Answers
-IT
},' = - ~~
2€l)'
z'>
o· Ez =
-,
IT
- , z <. 2€n
o. Set.:
Ull
iform negative charge density of"
~cr.
Figul'c 15.30c for the sketch.
An imponant circuit element that will be studied extensively in the next chapter is the parallel-plate capacitor. The devicc consists of a plate 01" positive charge, as in Example 15.8, with the negative plate of Exercise 15.H placed above it. The sum of these two rields is illustrated in Figure 15.31. The result is an electric field with double the magnitude in bel\.. .een the twO plates: [15.13] Outside the plates, the electric fields cancel.
t t t t t t~ +
FIGURE 15.31 CJ"OS~ ~cclioll ol'
SUMMARY 15.1
Properties of Electric Chorges
The magnitude of' lhe electric field due to a point (/wrge
q £II a distance .,-frolll the poin1 charge is
Electric charges have the lollowing pmpt.'rlies: l. Unlike charges attract one another and like charges repel one another. 2. Electric charge is ahva):s conserved. 3. Charge cOllles in discrete packets Ihal are i1ll.egrallllultiples or the basic electric charge e = J.6 X 1O-1~1 c. 4. The force between two charged particles is proportional to the inverse square or the distance between them.
15.2
Insulators and Conductors
Conductors are materials in which charges move freely in response to an electric f.e1d. All other materials are called insulators.
15.3
Coulomb's Low
Coulomb's law states that the electric force between two slationary charged particles separated by a dislancc r hilS tile magnilude
F~
1",11",1
Ii "
r-
.)
[15.1]
where If/II and khl an: the magnillldcs of the charges on the parlicks in coulombs and h,= 8.99 is the Coulomb constant.
15.4
X
10 9 N'm'jC'
[15.2]
[15.61
15.5
15.6
The Electric Field
-.E =F-
"9
Conductors in Electrostotic Equilibrium
.-\ conductor in electrostatic equilibrium has the following propcrties: I. The electric field is zero everywhere insicle the conducting mat.erial. 2. Any excess charge on an isolated conductor must reside cntirely 011 its surrace. 3. The eleci ric fIeld jusl outside a charged conductor is perpendicular to the conductor's surface. 4. On an irregularly shaped conductor, charge accumulates ""here the radius of curvature of the sllrf~lCe is smallest, al sharp points.
15.9
An c1eclric field E exists at some point ill space if it .'!Imall t('st charg.s qu placed at that point is acted upon by an electric lorce F. The electric field is defincd as
Electric Field Lines
Electric field lines are lIseful lor visuali7.ing the e1ec:!.ric field in any region of space. The electric field vector E is tang-ent to the electric field lines at every point. Further, the number or elcnric [-Ield lines per unit area through a surface perpendicular to the lines is proportional to lhe strengl h ( ) f the clectric field althal surface.
Electric Flux and Gauss's Low
Gauss's law states that []lC:: electric nux through any c10scd surface is eCJllal 1.0 the lIet charge Q inside the surElce divided by the pcrmiuivily ofrree space, EO: EA =
[ 15.3]
The dir·ection of" thc electric field ;u a point in sp"ice is defined to be the direction of the elcctric force thaI would be excrlcd on a small positive charge placed al.thal poin!.
rl
'-j)/~
Qillsirk = ---
[15.12J
'0
For highly symmetric distributions of' charge, Gauss's Illw can be used La calculate electric fields.
524
Chapter lS
Electric Forces and Electric Fields
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS 1. The magnitude of the electric force between two protons is 2.3 X "10-26 N. How far ,lpart arc Lhey? (a) 0.10 III (b) 0.022 m (e) 3.1 m (d) 0.005 7 m (e) 0.48 III 2. Estimate the magnitude of the electric held strength due to the proton in a hydrogen atom at a distance of 5.29 X /0-11 Ill, the Bohr radius. (a) [0 II N/C (b) 10' 'J/C (e) 1O"'1'1C (d) 10" 'J/C (e) 10" N/C 3. A \'cry small ball has a mass of 5.0 X 10-'\ kg and a charge of 4.0 }Le. \Vllat magnitude electric field directed upward will balance the weight of I he ball? (a) 8.2 X Ill' N/C (b) 1.2 X ]0' N/C (e) 2.0 X 10-' !'I/C (d) ".I X 10" NIC (e) 3.7 X 10" N/C
4. An electron \.,;ilh a speed of 3.00 X lOr, m/s moves into a lin dorm electric field of magn illlclc 1.00 X lO:~ N/C. The Held is parallel 10 the e1curoll's motioll. Hmll far does the electron travel before it is brought to rest? (a) 2.56 cm (b) !).l2 em (e) 11.2 em (d) 3.34 m (e) 4.24 III 5. Charges of 3.0 nC, -2.0 IIC. -7.0 nC, and 1.0 ne are contained inside
Ill'
6. A uniform electric field of 1.0 N/C is sct up by a uniform distribution of charge ill the .\)'-plane. What is the deClric field inside a metal ball placed 0.50 m abO\'c the Xl-plane? (a) 1.0 N/C (b) -1.0 'J/C (e) 0 (eI) 0.25 :-.l/C (e) It v
7. A charge of -4.00 nC is localed at (0, 1.00) 111. \Vhat is the x-component or the electric field at UI.OO, -2.00) Ill? (a) I.IS N/C (b) -2.2-1 NIC (e) 3.91 !'I/C (eI) -115 N/C (e) 0.863 N/C
8. Two point charges auraet each other with all electric force or magnilUde F II" one charge is reduced to Ol1Cthird its original value and the dist.ance between the charges is doubled, whaL is the rt..'suILing magnilUde of the eleClric force between them? (a) /'/]2 (b) F/3 (e) FI(; (d) 31'/4 (e) 3/'/2
9. ""hat happens when a charged insulator is placed near an uncharged metallic object? (a) They repel each other. (b) The}' auract each other. (c) They may attract or repel each other, depending on whether the charge on the insulator is positive or negat.ive. (d) They exert no electrostatic force on each mIK!'. (e) The charged insulator always spontaneously discharges.
10. 1n which of the following contexts Gill Gauss's law not be readily applied to I-inc! [he electric field? (a) ncar a long, uniformly charged wire (b) above a large uniformly charged plane (c) inside 1I uniformly charged ball (e1) outside a uniformly charged sphere (e) Gauss's law Gin be readily applied 1.0 fi nd t.he electric field in all these contexts. II. What prevents gravity from pulling yOll through the ground 1.0 the center of Earth? Choose the best answer. (a) The density or malleI' is too great. (b) The positive nuclei 01" your body's atoms repel the positive nuclei of the aloms of" the ground. (c) The densiry of the ground is greater than the density of your body. (d) Awms are bound together by chemical bonds. (e) Electrons on the ground's surface and lhe surface or your {-eeL repel one another. 12. A merallic coin is given a positive electric charge. Does its mass (a) increase measurably, (b) increase by an alllount too small to measure directly, (c) stay unchanged, (d) decrease by an ,1lTIount too slllall to measure directly, or (e) decrease measurably?
]3. Three charged particles are arranged on corners of a square as shown in Figure MCQI5.13, with charge ~Q on both the particle at the upper left corner and the particle at the lower right corner, and charge +2Q on the particle at. I he lower left corner. What is the direc· tion of the electric field at lhe Lipper right corner, which is a point in empty space? (a) upward and to the right (b) LO the right (c) dmvmvard (d) downward and to the lefr (e) The field is exactly zero at Lhat. point.
~
al
-Q
(bl
(d)
+2Q
(c)
-Q
FIGURE MCQ15.13
14. Suppose [he +2(2 charge at the lower left corner of" Figure MCQ15.13 is removed. Which statemenl is trUt:· about the rnagn it ude of the electric field al rhe upper right corner? (a) It becomes !;.ll·ger. (b) It becomes srnaller. (c) It stays the same. (d) It changes unprediclably. (e) It is zero.
Problems
525
CONCEPTUAL QUESTIONS 1. A glass object is charged to +3 nC b>' rubbing it with
UIJjt;Cl 01
have elcci rOllS been removed from it?
2. \'Vhy must hospital personnel wear spl'cial conducting
shoes \'ihite working around oxygen in an operating rt)om? What might happen if the personnel \\'llrt: shoes wit h rubber soles? 3. A person is placed in a large, hollow metallic sphere (hal is insulated from ground. Ira large charge is placed 011 the sphere, will the person be harmed upon LOllChilig the inside of the sphere? 4. E.xplain from an atomic viewpoirll why c11
5. Explain how a positively charged ol~jecl. call be used 10 leave another metallic ()~ject with a net neg-ative char~('. Discuss the motion or charp;es during the proces~. 6. I r a sLlspendcd o~iect A is attractcd to a charged 13, com W(; cOllchldc thaLl\ is charged? Explain.
ol~jeC'l
7. If a nlt'tal objcCl receives
8. Consider poinl A in Figure CQ15.H. Does charge exist al Ihis point? Does a force exist at this poin1? Does flcld exist <Jtthis point? Explain.
9. In fair weather there is an electric field at the surface of the Earth. pointing down into the ground. vVhat is the electric ch
It. There are great similarities hetween electric and gra\'italional fields. A room can be electrically shielded so that there are no electric fields ill the room by surrounding it \\lith a conductor. Can a room be gravitationally shielded? Explain. 12. Wily should a ground \\lire be connected to the metal support. rod for;'l television antenna?
13. A charged comb often attracts small bils of dry paper that t1wn ny away when they touch the comb. Explain. 14. A spllerical surface surrounds <-l point charge q. Describe what happens to the total Flux through rhe surface if (a) tile charge is tripled, (b) the volume oft.he sphere is doubled, (c) tlte surface is changed to a cube, (el) the charge is Illoved to another location inside Ihe slirJ~lce, and (c) the charge is llloved outside I he surface.
it
'I,
15. Iflllorc electric field lines leave a Gaussian surface thaI! enter it, what CClll you conclude about the net charge enclosed by t.hat surface?
FIGURE CQ15.8
PROBLEMS ~
..
WebAsslgn
I.
~. :~
The
Probkms lill' I hi1'> chaptl:r Ill
= 1'>lraighlforw
Illll'lll]('di~'I{', lhall(,lI~ill.~
m
= d(,llOlC~
_
= (\(,llott:~ cnh;tnced C(lnlCnl prohlem
II = 0=
guided problem
hlOlllCdicll applicallon
dellr'ICS full SOIUCIOll availallle in Slndl'III.'lolutiolls !'I(wlIlIII
SInd" Guidi'
SECTION 15.3
3. _ Two metal halls A and fj of negligible radius arc float~ ing
COULOMB'S LAW
I. A 7.!J-nC charg'c i'i loeaLed 1.8 III Crom
2. A c1wrgcd panicle /\ excrts a force 01'2.62,uN to the riRht 011 charged particle IJ when the panicles are 13.7 lllm
am
(a) Find the e1eclrostatic force bCt'.. .eell a Na-i ion and i-\ Cl- ion scpar;\led by 0.50 nm. (b) Would the answer change if the sodium ion were replaced by Lj+ and the Chloride ion by Br-? E.xplain.
[ilrhe ltUdellS orHBe. which consisls of4 protons and 4 neuIrons. is very unslable and spolltanl'ollsly break~ inti) 1\\'0 alpha particles (helium nuclei, each consisling of 2 protons and ;Z ncutrons). (a) \·Vhm is the force belwecll the t\,·o alpha particles when the)' arc 5.00 X 10 Jf'm apart. and (h) what is the iniliallllagnitude of the acceleration ()f1he alpha particles due 1.0 this force? Note that the mass nfall alph;;\ pa'·licle is 4.0026 u.
526 6.
Chapter 15
II A
Electric Forces and Electric Fields
Illolccllle of DNA
(deoxyribollucleic
acid)
3.00
is
r
2.17 P.1ll long. The ends of the molecule become singlr ioni71'd: negative on one end, positi'"c on the other. The helical molecule aelS like a spring and compresses 1.00% upon becoming charged. Determine the effective spring conSlanl of the molecule. 7. Suppost.' 1.00 g of Il)drogcn is separated into electrons and pruLOIlS. Suppose also the pmwils al"e pla<:cd .u the EarLh's North Pole and the electrons al"C placed at the South Pole. What is the resulting compressional force on the Eanh? 8.
m:iI FOllr point ch
as shown in Figure P15.S. Determine the magnire~uhanL electric force on q. with k~. q, and a left in ~ymbolic form. II
LUcie and direction of the
lUlOO
III
0.500
III
lie 0.500
'!!..j
6.00nC
1 2.00
lie
FIGURE P15.12
13. Three poin[ charges are localed al the comers of an equilalel-al triangle as in Figure P15.13. Find the magnitude and direction of Ihe Ilt't electric force on the 2.00 JLC charge. )"
7.nOpe
"
'1q.-----~'1
n
n
@-'-----Q.... x ~.oo
,Lie
FIGURE P15.13 FIGURE P15.8
[]JT\\'o small identical conducting spheres arc placed with their eeillen, 0.30 III apart. One is given ;'1 charge of 12 X 10-\1 C. lhe olher a charge of -18 X 10-~J C. (
-4.DOpe
(Prohlems
J:~alld
24)
!illA charge of -3.00 nC and a charge of -5.80 nC are 'ieparated by a distance of 50.0 CIlI. Find the position at which a third charg-e of +7.50 nC can be placed so that [Jle net e1cctrostal ic force on it is zero. 15. Two small melallic spheres. each or milSS 0.20 g, arc suspended as penduhlllls b)' light strings from a common puint as shown in Figure PI!'"I.15. The spheres arc givcn the samc c1cnric charge. and it is found lhal thcy come to equilibrium when each string is alan angle 01'5.0° with the vcnical. If each string is :\0.0 Clll long. what is lhe magnitude of the charge all each sphere?
PI5.IO.
6.0U pC
1.50 pC
---------
~;\.oocm FIGURE P1S.l0
-2.001IC -----~
~2.00 em
(Problems 10 and 18) 0.20 ~
11. Three charge;; are arranged as shown in Figure PI!'">. I I.
Find the magnitude and direction of the elcnrostatit· force on the charg-c .at the origin.
FIGURE P15.1S
16. )
s.oo n(.
(->'00 nC
0.300 III -0--"==-'::'1--.'
O.HJO III -3.0u nC
FIGURE P15.11
12. Three charges are arranged a~ shown in Fi~llre P15.12. Filld the magni1.11de and direction of the eleclrostalic force on I he 6.00-nC charge.
0.20 g
EiI
Particle A of charge 3.00 X 10 ~ C is at lhe origin. panicle B of charge -6.00 X 10 I C is ;u (4.00 m, 0) and particle C of charge 1.00 X 10 I C is at (0. 3.00 m). (a) What is the x-component 01 the electric force exerted by A on C? (b) What is the y·compollent of the force exerted b) A. on C? (c) Find the nWb'l1itude of the force exerted by Bon C. (d) Calculate the x-component of the force exerted by Bon C. (e) Calculate the )L.component of the force exerted by Bon C. (f) SUlllthe two x-components to obtain d1e resultant x-component of the electric force acting all C. (g) Repeat pan (0 for lhe y-componcnL (h) Find [J1C magnitude and direclioll of the resultant electric force acting all C.
Problems
SECTION 15.4 THE ELECTRIC FIELD 17. A small ol~jeci of mass :1.80 g and cha"gc -lR p'e "floats" in a uniform c1eclric field. \Vhat is the magnilude :lnd direction orthe c1enric field?
527
28. Three charges arc al the corner~ of' an equilater;:ll triangle, as shown in Figure PIEi.28. Calculate the electric field at CI point midway hetv,,'ecn the two charges all lhe x-axis. J
3.00 nC
18. «(\) Delermine the clcCLric field strcll14th al a point l.00 ( I l l to the left of the middle charge shown in Figure PIS.IO. (b) If a charge of-2.00 J-LC is pl
19. All airplane is flying through a thllilclercloud at
Iit-L----~x
8.00 nC
-S.OO nC
FIGURE P1S.28
29. Three idcmical charges (q = -5.0 J-LC) lie alung a circle of radius 2.0 m at angles of' 3OC, 150", and 270", as shown in Figure Pl!1.29. Whm is the rcsullant electric fielcl at the center of the ci rcll'? J
21. A chargc 01" -!).O nC is at Ihe origin and a second c1nrge of 7.0 nC is at x = 4.00 Ill. Find the magnit.udc and direction of the electric field halfway in bctween Ihe 1\\'0 ch'lrges. 22. Loch 01 the prolans in " parlicle bcam has a kinetic energy of 3.2!) X 10-1:'.1. '·\fhat arc the magnit.ude and direction of thc cleClric field th;ll will SlOp t.hes(; protons in a distance 01" 1.25 m?
123.1/\ prown
-~'I FIGURE P1S.29
SECTION 15.5 ELECTRIC FIELD LINES SECTION 15.6 CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM 30. Figure P15.30 shows t.he clecu"ic field lines for tWO point charges separated by" small distance. (a) Determine the riltio '1';q~. (b) What are the signs of q\ and (h.?
25. An
.
r-I.o,n
I ~ 6.0 f../C
FIGURE P1S.27
FIGURE P1S.30
31. (a) Sketch lhe electric field lines around ,Ill isolated point charg-e q> O. (b) Sketch the c1eclric fidel pattern around all isolaled nl'gati\'c point charge oflllagnitude -2q. 32. (a) Skelch the eleclric fielel pallcrn al"Ound LWO posiliw: point charge.'i of magnitude I f..LC placed close togeLher. (h) Skel.ch the electric fidel pattern arollnd two negative poinl charges of -2 ,uC, placed close together. (c) Sketch the pal1ern around two point charges of +1 p,C and -2 p,e, placed close lOKether.
528
Chapter 15
Electric Forces and Electric Fields
133.IT\'.:o poi III charges are ,I small distance apart. (a) Sketch the elcci ric field lines for the two if Ollt.: has a charge four times that of the other and both charges arc positive. (b) Repeal for the case ill which both charges arc negative. 34. (a) Skelf"h the f'lertrir fif'lrl
p,lltPnl
sel
lip
hy;:) ]lO<;ili"f'lr
charged hollow sphere. Include regions inside and regions oUlsidc the sphere. (b) A conduCling cube is given a positive charge. Skelch the ckClric field pallcrn bmh inside and olltside 'he cube.
35. Refer to figure 15.20. The charge lowered inLO the cenleI' of the hollow conductor has a magnitude of 5 Me. Find the magnilUdc and sign of the charge on the inside and olltside of 1he hollow conductor when the charge is as shown in (a) Figure 15.20<1, (b) Figure J5.20b, (c) Figure 15.20c.
ImAlI electric j'ield of intensity 3.50 kN/C is applied <dong the .\"-axis. Calculate the electric flux through a rectangular plane 0.350 m \\,'ide and 0.700 III long if (a) the plane is parallel 10 the y:::-pi
42.
mJ
The electric (ield everywhere on the surface of a charged sphere of' radius 0.230 m has a magnitude of 575 t\/C and points radi(llly out.ward £i'om the center or the sphere. (a) \'Vhat is the net charge on lhe sphere? (b) What Gill you conclude about the n<1wre and distribution of charge inside the sphere?
43. FOll1- dosed surfaces, 51 throug'h Sl' together Wilh the charg-cs -2Q, Q. and -Q, are sketched in Figure PI5.43. (The colored Ii nes are lhe inrersections of the ~urfaces with lhe page.) Find the electric flux through each surface.
SECTION 15.8 THE VAN DE GRAAFF GENERATOR 36. The dome ofa Van de Gr
FIGURE P15.43
44. ;\ \'crtical electric field of magnitude 1.80 X 10 1 N/C exist.s above Earth's surbce on a slormy eLly. A car with a rcctangubr size of 5.50 m by 2.00 m is tra\'cling along
,I horizontal roadway. find the magnilude of the electric flllX
through the bmtolll of the car.
145.IA
point charge q is 10catc(1 at the center of a spherical shell of radius a thai has a charge -'1 uniformly clistribmed on its surface. Find lhe electric field (a) for all points outside the spherical shell and (b) for a point inside the shell a diswl1ce rf"rom the center.
46.
m:J
A ch
or edge
47. Suppose the conducting spherical shell of Figure 1!1.29 carries a charge or 3.00 ne alld that a charge of -2.00 nC is
SECTION 15.9 ELECTRIC FLUX AND GAUSS'S LAW 40. A fbt surface having an area of ?J.2 111 2 is rotated in i-I uniform electric field or magnitude Ii. = 6.2 X 10 5 N/C. Determine the electric flux through this area (a) when the electric Aeld is pel-pendiclliar to the surrace and (b) when Liw electric field is parallello the surface.
148.IE A
very large nonconducting' plate lying in lht.: x)"plane GIlTies a charge per unit area or cr. A second such plale localed at z = 2.00 cm and oriented parallel to the .\y-plalle carries a charge per unit area of -2cr. Find the elecl,-ic field (a) for z. < O. (b) 0 < z. < 2.00 CIn, and (c) ::. > 2.00 em.
Problems
lowered to rc,lc!l its equilibrium position? (b) The ball is given a charg"c of 0.050 0 C. II" an eleClric field directed upward is applied. increasing slowly to a maximum value of 355.0 N/C. how far below the unstrctched position is the new equilibrium position or the ball?
ADDITIONAL PROBLEMS 49.
[n deep space two sphcrc~ each of radius :).OU III are conneclcd by a 3.00 X IO~ III nonconduCling core!. If a uniformly di'ilribulcd charge of 35.0 me resides on the surface of each sphere, calculate the lension ill the corel.
.~O.
mil
A tlonconducting. thin plane sheet of ch
56.
.I)J. Three poinl charges are aligned along the x-axis as shown
in Figure PI5.51. Find the electric field at the pl)sition x +2.0m.)'=O.
=
O.ROllli
529
ED
A 2.00-mC charged 1.00-g cork ball is suspended vertically on a O.500-m-long light string in the prcscnce of a uniform downward-directed electric field or magnilUde E = 1.00 X IO£' N/C. If the ball is displaced slightly from the vertical. it osci Ilates Ii ke a si mpJe pendu lum. (a) Determine the period of the ball's oscillation. (b) Should gra\'it)' be included in the calculalion for pan (a)? Explain .
57. Two 2.0-g spllt.~res are suspended by lO.O-cm-long light strings (Fig. PI5.57). A uniform electric tield is applied in the x-direction. If the spheres have charges or -5.0 x IU-R C and +5.0 X 10-1> C, dctermine the electric field intensity that enables the spheres to be in equilibrium at 0 = 100.
-""""'.---~;)--------O---x
-
5.0 nC
3.0 nC
FIGURE.15.51
52. A small, 2.00-g plastic b,dl is sLlspended b}' ,\ ~O.O~cm~ long string in a uniform electric field, as showl1 in Figure P15.52.1l"lhe ball is in equilibrium when the string makes a 15.0° angle wilh the vertical as indicated. what is the net charge 011 the ball?
• E FIGURE.15.57
58. I', '" 1.UU X 103 :J/C
I!I'"
2.00 g
FIGURE.15.52
153.I(a) Two identical point dlar~es +q arc loclted on the y-axis al )' = +a and y = -a. What is the electric field along the x-:lxis ar x = b? (b) A circular ring of charge of radius a has a LOtal positive charge Q diqribllted uniformly around it. The ring is in the x = 0 plane with its cemer at the origin. \'\.'l1a1 is the electric field along the x-axis at x = b due 10 the rin~ 01" charge? (J lilll: Consider the charge Q to consist of" m;my pairs of idelllical point charges positioned at the cnds of diameters or Ihe ring.)
&
A POilll charge of magnitude 5.00 j1.C is at thc origin of a coordinate system, and a charge of -4.00 MC is at the point x = 1.00 m. There is a point on the x-axis. at x less than infinity. where the elcctric fidd goes to zero. (a) Show by conceptwll arguments that this point cannol be located between the charges. (b) Show b} concepLUal arg-uments tha\. the point cannot be <11. any location between .\- = 0 and negative illlinity. (c) Show by cOllceptual arg-uments that (he point must be between x = 1.00 Jl1 and x = positi\'e infinity. (d) Use the values given 1O find (he point and show that it is consistent with }'our conceptual argumenl.
159.IT\\'0 hard rubber spheres or Illass 15 g are rubbcd vigorously with fur un a dl'y day and arc then suspended from a rod with t\VO insulating strings of length 5.0 em. They are observed to hang at equilibrium as sho,.. . n in Figure P1S.59_ each at an angle of lac with the vertical. Estimme the amoulll ol"cbarge that is found on each spherc. (Problem 59 is coul'lesy ofE. F. Redish. For more problems of this type, visit www.physics.ulllcl.edu/perg/.)
54. _ The cketrons in a particle beam each havc a kinetic energy 1\.. Find the magnitude of the electric field Ihat ",ill SlOp these eleclrons in a distance d, expres~ing the answer symbolicall}' in terms of K. e, and d. ShOllld the electric field point in the direclion of the motion or the electron. or should it point in the opposite direction? 55. A venical spring with constant 845 N/m h:l:-' a ball of mass 4.00 kg attached to tile boltom of it. which i~ held with the spring unstrel.ched. (a) How far mu~t the ball be
FIGURE.15.59
530
Chapter 15
Electric Forces and Electric Fields
60. Two small siker spheres, each with a mass of 100 g,
ffi[]Each of the eleclrolls in a panicle beam has a kinetic cncrg\ of 1.60 X 10 17 J. (a) What is the magnitude the uniform electric field (pointing in the direction the electron')' movement) lhal will SLOp these c1eClrons in a di<;t:lllcc of 10.0 em? (h) I low long will it take to stop the ele('l rOllS? (e) After Ihe electrons stop. what will they dor Explain.
or or
64. Protons arc projected with an initial speed l\1 = 9550 ntis imo a region where a uniform electric fidd E = i20 N/C is pn:M'1l1 (Fig. PI5.6..J). The protons ;m: to hit a target that lies a hori7ontal distance of 1.27 mill from the point where the prowns arc launched. Find (a) the two pro· jectioll angles 8 lhal ".'ill resull in a hil and (b) the lotal duration or f1iglll 1'01' each of the two lr
62. Three identical poim charges, each or Illa~~ 111 = 0.100 kg', hang rrom Ihree strillg~. as shown in Figllre PI5.62. Irlhe leng-th.'i of the lert and righl strings ;m.: t::lCh L = 30.0 ClI! and if the angle 0 is lI5.00, dClcrminc Liu.:: valllc of q.
,
8 8
/ / Proton
L
"
beam FIGURE P15.64
+q
''I no
no FIGURE P15.62
The world's largest air-insulated Von de Grooff generator produces bolts of lightning indoors at the Museum of Science in Boston. The discharges occur when the voltage difference gets large enough to ionize the air, on instance of dielectric breakdown.
16.1
Potential Difference and Electric Potential
16.2
Electric Potential and Potential Energy Due to Point Charges
16.3
Potentials and Charged Conductors
16.4
Equipotential Surfaces
16.5
Applications
servative forces of gravity and springs. By using the principle of conservation of energy, we
16.6
Capacitance
were often able to ovoid working directly with forces when solving problems. Here we learn
16.7
The Parallel-Plate
ELECTRICAL ENERGY AND CAPACITANCE The concept of potential energy was first introduced in Chapter 5 in connection with the con-
Capacitor
that the potential energy concept is o[so useful in the study of electricity. Because the Coulomb force is conservative, we con define an electric potential energy corresponding to that
16.8
Combinations of Capacitors
16.9
Energy Stored in a Charged Capacitor
16.10
Capacitors with Dielectrics
force. In addition, we define an electric potential-the potential energy per unit chorge-corresponding to the electric field. With the concept of electric potential in hond, we con begin to understand electric circuits, starting with an investigation of common circuit elements called capacitors. These simple devices store electrical energy and have found uses virtually everywhere, from etched circuits on a microchip to the creation of enormous bursts of power in fusion experiments,
16.1
POTENTIAL DIFFERENCE AND ELECTRIC POTENTIAL
Electric potcntial cnergy and electric potclllial are closely related concepts, The electric potential turns out 1.0 be just the electric potential energy per unit charge. This relationship is similar [0 that between clcClric force and the e1ecu-ic field, which is the electric force per unit charge.
Work ond Electric Potential Energy Recall from Chapter 5 that the work done by a consen'ative force F on an object depends only on the initial and nnal positions of the object and not on lhe path t.aken between Ihose two points. This, in turn, means that a potential energy function (-Jf exists. As we have seen, potential cnergy is a scalar quantity with the change in potcl1lial energy equal by deflllition to the negative of the work done by the conservative force: 6.P/:.' = P£I- Pt.:; = -'VI;'
531
Chapter 16
532
High PE
Electrical Energy and Capacitance
Low PE
E
+ +
-+- + o
A
~x
•
IJ
• -x
+
+ FIGURE 16.1 When a charge q moves in a uniform electric nekl E from point A to poim B. the work done on the charge br lite eleclric force is qE~ fix.
Both the Coulomb force Jaw and the universal law of gravity are proportional [Q 1/r 2 , Because they have the same mathematical form and because the gravity force is conservative, it follows that the Coulomb force is also conservative. As with gravity, an electrical potential energy function can be associated with this force. To make these ideas morc quantitaEve, imagine a small positive charge placed
at point A in a uniform electric fIeld E, as in Figure 16.1. For simplicity, we first consider only constant electric fields and charges that move parallel to that field in one dimension (taken to be the x-axis). The electric field between equally and oppositely charged parallel plates is an example of a field that is approximately constant. (See Chapter 15.) A~the charge moves from point A to point B under the influence of the electric field E, the work done on the charg·e bv, the electric field is equal to the part of the electric force qE acting parallel to the displacement times the displacement 6.x = X.r - Xi: ~
~1~B
= Fx 6.x = qE... (x/ -
Xi) ~
In this expression r; is the chargs and E.>: is the vector comp~lenl of E in the x-direction (noL the magnitude of E). Unlike the magnitude ofE, the component b~ can be positive or negative, depending on the direction ofE, although in Figure 16.1 Ex is positive. Finally, note that the displacement, like q and Ex, can also be either positive or negative, depending on the direction of the displacement. The preceding expression for the work done by an electric field on a charge moving in one dimension is valid for both positive and negative charges and for constant electric fields pointing in (m)' direction. "Vhen numbers are substituted with correct signs, the overall correCl sign automatically results. In some books the expression Ml = q/~'r1 is used, instead, \vhere E is the magnitude or the electric field and rl is the distance the particle travels. The \\o'eakness of this formulation is that it. doesn't allow, mathematically, for negat.ive electric work on positive charges, nor for positive electric 'work on neg;:nive charges! Nonetheless, the expression is easy to remember and useful for finding magn itudes: the magnitude of the work done by a constant e1eClric field on a charge moving parallel to the field is always given by 11-1'1 ~ IqIKI. We can substitute our dellnition of electric \\lark into the work-energy theorem (assume other forces are absent): ~
The electric force is conservative, so the electric work depcnds only on the end~ points of the path, A and B, not all the path taken. Therefore, as the charge acceleratcs to the right in Figure 16. L, it gains kinetic energy and loses an equal amount of potential energy. Recall from Chapter 5 that the work done by a conservative force can be reinterpreted as the negative of the change in a potential energy associated with that force. This interpretation motivates the dcfinition of the change in electric potential energy: Change in electric potential energy -+
The change in the electric potential energy, 6.PE, of a system consisting of an object of charge q moving through a displacement 6.x in a COl1stam electric field If is given by ;"'PE = - ~IIJ ~ -qE,;"'x
where Ex is the x-component of the electric field and 8x = xf placement of the charge along the x-axis.
[16.1] Xl
is the dis-
51 unit: joule (J)
Although potential energy can be defined for any electric field Equation 16.1 is valid only for the case of a uniform (Le., constant) electric field, for a particle that undergoes a displacement along a given axis (here called the x-axis). Because the electric field is conservative, the change in potential energy doesn't depend 011
16.1
the path. Consequcntly, it's unimponant whether 01" not thc charge remains on the axis at all times during dle displacemem; the change in potential energy \vill be the same. In subsequent. sections \\'e \'vill examine siwations in which t.he electric f-Ield is not lIniform. E.lectric and gravitatioll
!:.KE + !:.PEd = !:.KE + (0 - 1'I[I:"d) = 0
->
6.KE = l'IIEd
The absolute-valuc signs on qare there only to make explicit that the charge is positive in this case. Similarly, the ol~jeCl in Figure 16.2b gaills kinetic energy equal in magnitude to the loss of gravitational potential energy:
I1KI," + !:.PI", ~ I1KI," + (0 - mgd) = 0
~
533
Potential Difference and Electric Potential
!:.KE ~ mgd
A'
'1
",g~8'" J (a)
(b)
FIGURE 16.2 (a) WhclI the e1euric lield it i~ din:c.:Il:d downward. PUilll 8 i~ at a lower dt::ctric pOlcluiallhan poillt A. As a posit in' test charge 1ll0VCS from A tu B, the t:kctric pulential energy decrt:ases. (h) As an object or mass m 11100e~ in the dircclion or Lhe gra\"ilal ional field g, Ille gra\,it:'l' t iOllal pOlent i:ll cnergy d('creases.
So for positive charges, elcctric potcntial energy works very much like gravitational potential energy. In both cases moving an o~ject oppositc the direction of the ficld results in a gain of potential energy, and upon release, the potential energy is converted to the object's kinetic energy. Elcctric potential energ·y differs significantly from gravitational potential energy. however, in that there are I wo kinds of eknrical charge-positive and negalive-whereas gravity has only positive "gravitationaJ charge" (i.e. mass). A negalively charged panicle at rest at point A in Fig-Lire 16.2a would have to be IJ'llShPr{ clown to point B. To see ""hy, apply the work-energy theorem to a negative charge at rest at point A and assullled to have some speed von arriving at point 13:
W = !:.KE + !:.PF,", ~ (!mv' - 0) + [0 - (-Iq Ed)] W = !mv' +
I'll Ed
Notice that the negative charge, -[ ql. unlike the positive charge, had a posit.ive change in electric potential energy in moving from point A to point B. If the negative charge has any speed at point /3, the kinetic energy corresponding to that spccd is also positive. Because both terms on the right-hand side of the work-energy equation are positive, there is no way of gelling the negative charge from point A to point Hwithout doing positive work Hfon it. In fan, if the negalive charge is sirn· ply released at point A, it will "fall" upwards againsl t.he direction of the fidel!
QUICK QUIZ 16.1 If an electron is released from rest in a uniforrn elect ric field, docs the electric potential encrgr of the charge-field system (a) increrlse, (b) dccrea::ie, or (c) remain the same?
EXAMPLE 16.1 Goal
Potential Ener y Differences in an Electric Field
Ilhlstrate the concept of elcClric potential energy.
Problem
prolon is released fl-om rest at x = in a constant electric field with magnilude 1.50 X 10:~ N/C, pointing in the positive x-direction. (a) Calculate the change in the electric potentia] energy associated \vith the proton when it reaches x = 5.00 Clll. (b) An electron is now fired in the same dircction from the same position. \,Vhar is its change ill electric potential energy associated with the electron if it reaches x = 12.0 cm? (c) If the direction of -~.OO Clll
A
the electric held is reversed and an electron is released from rest at x = 3.00 cm, by how lllllch has lhe electric pOlential encrgy changed when the electron reaches .r: = 7.00 cm? Strategy This problem requires a straightforward substitution of given values into the definit.ion of elect.ric potential energy, Equation 16.1.
534
Chopt.. 16
Electrical Energy and Capacitance
.....
.
,.
.
~
Solution (a) Calculate the change in the electric potential energy associated with the proton.
Appl) Equiltion 16.1:
-qJ-.:,('i - x,J
l1PIi = -'lEx l1x =
= -(1.60 X
10
X
19
C)(1.50 X 10' iVC)
[0.0500 rn - (-0.0200 rn)]
-1.68 X 10- 17 J (b) Find the change in electric potential energy associated "ith an electron fired frorn x = -0.0200 In and reaching x = 0.120 Ill. Apply Equalion 16.l, but in this case notc 1hal the eleClric charge q is negative:
;:"PE = -qEx;:"x
=
= -(-1.60 X X
-qEx(''j -
x,l
10- 19 C)(1.50
X
10' 'Ie)
[(0.120 rn - (-0.0200 rn»)
+3.36 X 10-".1 (c) Find the change in potential energy associated ,,,,ith an electron traveling from x = 3.00 em to x = 7.00 em if the direction of the electric field is reversed. SubsliulIe, but now the electric flcld points in the negathe x-dircClioll, hence carries a minus sign:
l1P/-: = -'IE, l1x = -'lEx (') - x,)
= -(-1.60 X X (0.070
1O- 19 C)(-I ..:;O X lO'N/C)
III -
0.030 rn)
-9.60 X 10- 18 .1
.
trrr... • • •
~
Remarks Notice thallhe proton (actuall)' the prown-field system) lost potential energy \il.'hen it moved in the posiLi\e .\-din:clion, whereas the clecu"on gained potential energy when it moved in the same direction. Finding changes in potential energy with the field rc\"crsed was only a maLlc.- of supplying a minus sign, bringing the total number in this case 10 three! It's imporwlll not to drop an)' tile signs.
or
QUESTION 16.1 True or False: \\'hen an electron is released from rest in a constant electric field, the change in the electric pOlemial energy associated with the electron becomes mOfC negmh"c with lime. EXERCISE 16.1 Find the change in eleClric pOlenlial energy associated with the eleClron in pan (b) as it goes on from x = 0.120 III to .\' = -0.180 ITI. (Note that the electron must turn around and go back at some poinl. The location of the turning pOilH is unimporlalll because changes in potential energy depend only on the endpoints of the path.) Answer
-7.20 X 10
17
J
EXAMPLE 16.2 D namics of Char ed Particles Goal
Use cleClric pOlcmial energy in conservation of encrgy problems.
Problem (ill Find the speed of the proton at x = 0.0500 rn in pan (a) of Example 16.1. (b) Find the iniliilj speed of the electron (at x = -2.00 em) in pan (b) of Example 16.1 given that its speed has fallen by half when it reaches x= 0.120 n1.
16.1
Potential Difference and Electric Potential
535
Strategy Apply conservation of energy, solving [or the unknown speeds. Part (b) involves two equations: the conservation of energy equation and the condition v[ = ~v, for the unknown initial and final speeds. The ch,-lI1ges in electric potential energy have already been calculat.ed in Example 16.1. .... _ r' . . . . . ~
Solution (a) Calculate the proton's speed at x = 0.050
Ill.
+ 1';1'1( = 0
Usc conservation of energy, Wilh an initial speed of zero:
1';1(£
Solvc for v and substitute the change in potcntial energy [ound in Example l6.la:
v- = --!lPlc
(1~mfiv , - 0) +
-->
UA
1'1" - 0 J-
~
.>
'm ,
l
2 (l .67 X lO
T
'kg-
)(-1.68XIO
17
J)
1.42 X 10' Ill/s
(b) rind lhe electron's inilial speed given lhal. its spccd has fallcn by halfal x = 0.120 Ill. Apply conservation or energy once again, substituting expressions for the initial and final kinetic energies:
1';KH + 1';1'£ = 0
(~m.~v/ - ~m ..v,2)
+
tiPE = 0
Substitutc the condition vI = ~Vi and subtract the change in potential energy from both sicles: -1';1'1£
Combine terms and solve for v" the initial speed, and subsl illite the change in potential energy found in Examplc 16.11,:
81';1'£ Vi=
3m"
8(3.36 X 10
".Il
3(9.11 X lO " kg)
9.92 X 10(; m/s ~
-
_-
.
_
..........
Remarks Although the changes in potential energy £lssociated with the protoll and electron were similar in magnitude, the effect on their speeds differed dramatically. The change in potential energy had a proponionalely much greater effect on the much lighter electron than 011 the proton. QUESTION 16.2 True or False: lfa proton and electron both Illove t.hrollgh the same displacement in an electric field, the change in potential energy associated with the proton must be equal in magnitude and opposite in sign to the change in pOlential energy associated with the electron. EXERCISE 16.2 Refer to Exercise 16.!. Find the electron's speed at. x
Answer
1.35 X 10
7
lll/S
= -0.180111.
The ans\\'cr is 4.5% of" the speed of light.
Electric Potential In Chapter ]5 iL was convenienl to define an electric field E relat.ed to the electric force = (iF.... In this 'tva)' the propert.ies 01" nxed collections or charges could be easily studied, and the force on any particle in the electric field could be obtained simpl}' by multiplying by the panicle's charge 'I. For the same reasons, it's
F
Chapter 16
536
Electrical Energy cnd Capacitance
useful to deAne an eleclrir IJo/en/iat dijfPTf'ure !:iP/': = q!:iV: Potential difference between
two points
-+
~lf reLued
to the potential energy by
The elcClric pOlential difference 8\f between points i\ and 13 is the change in electric potential cnel-gy as a charge 1 moves from A to B divided by the
charge 'I: !:iPE !:iV=V,,-VA = - -
q
[16.2]
51 unit: joule per coulomb, or volt O/C, or V) This definition is completely general, although in Illany cases calculus ,\-'ould be required to compute the change in potential energy of the system. Because electric potential energy is a scalar quantity, electric potential is also a scalar quantity. From Equation 16.2, we see that electric potential difference is a measure of the change in electric potential energy per unit charge. Ahernatcly, the electric potential diff-erence is the work per unit charge that would have [Q be done by sornc force to move a charge from point A to point B in the electric held. The SI unit of electric potential is the joule pCI' coulomb, called the volt (V). From the definition ofthat. unit, 1 J of work must. be done to move a J-C charge bctwecl1two points that are at a potential di f"lerence of 1 V. In the process o[ moving through a potential difference of I V, the l-C charge gains] J of energy. For the special case of a uniform electric Held such as that. between charged pantllel plates, dividing Equation 16.l hy qgives !:iP/,'
- - = ~E.:D.x
q
TIP 16.1 Potential and Potential Energy Electric potential is characteristic
of the fi~kl only. independent or a Lest charge thaI mar be placed in that field. On lhe other hand, potential energy is a characteristic of the charge-field system due to an interaction between the field and a charge placed in the field.
,.
FIGURE 16.3 & ll'iA)
Comparing this equation \·...ith Equation 16.2, we find that !:i\!= -E,tix
[16.3]
Equation 16.3 shows that potential difference also has units of electric lIeld times distance. It then l"ol1o\-v& that the 51 unit of t.he electric field, the newton per coulomb, can also be expressed as VOllS per meter:
J N/C = I Vim Because Equation 16.3 is directly related to Equation 16.J, remember that it's valid only [or the systcm consisting of a unil-orm electric field and a charge moving in one dimension. Released from rest, positive charges accelerate sponL,lIleous!}' from regions of high potential to low potential. If a positive charge is given some initial velocity in the direction of high potential, it can move in that direclion, but will slow and finall>' turn around, just like a ball tossed upwards in a gravity field. ~egative charges clo exactly the opposite: released from rest, they accelerate from regions of [ow potential l.oward regions of high potential. 'Nork must be done on negative charges to makc them go in thc direction of lower electric potential.
QUICK QUIZ 16.2 If a negatively charged panicle is placed at rest in an electric potential field thaI. increases in the positive x-direction, will the panicle (a) accelerate in the positive ;\"-c1irection, (b) accelerate in the negative x-direction, or (c) remain al rest?
(Quick Qui7.:f.es 16..1
QUICK QUIZ 16.3 Figure 16.3 is a graph of an electric potential as a function of" position. If a positively charged particle is placed at point 1\, what will its subsequent motion be? \'Vill it (a) go to the right, (b) go to lhe left, (c) remain at point A, or (d) oscillale around poim B?
16.1
Potential Difference ond Electric Potential
537
QUICK QUIZ 16.4 If a negatively charged particle is placed al poilH B in Figure 16.3 and given a very slllall kick to the right, ,·\'hat will its subsequent Illotion be? \Vill it (a) go to the riglll and nOI rewrn, (b) go to the left, (c) remain at point 13. or (d) oscillate around point B? A n application of potential di ffercnce is the 12-V battery 10Ulld in an automobile. Such a battery maintains a potential difference across its terminals, with the positive terminal 12 V higher in potential than the negative terminal. Tn practice the negative terminal is usually connected to the metal body of the car, ,vhich can be considered to be at a potcntial of zero volts. The battery provides the electrical current necessary to operatc headlights, a radio, pO\\'>er \.. . indows, motors, and so forth. Now consider a charge of +1 C, to be moved around a circuit that contains the battery connectcd to some of these external devices. As the charge is moved inside the bauer)' frolll the negative terminal (at 0 V) to the positive terminal (at 12 V), the work done on the charge by the battery is 12 J. Every coulomb of positive charge that leaves the positive terminal of" the battery carries an energy of 12 J. As the charge moves through the external circuit I.Oward the negative terminal, it gives tip its 12 J of electrical energy to the external devices. \'Vhen the charge reaches the negative terminal, its electrical energy is zero again. At this point, the battery takes over and restores 12.J of energy to the charge as it is moved from the Ilegativc 1.0 the positive terminal, enabling it to makc another transit orthe circuit. The actual amount of charge that lea\'es the bauery each second and trayerses the circuit depends on the properties of the external devices, as seen in the next chapter.
EXAMPLE 16.3 Goal
APPLICATION Automobile Batteries
TV Tubes and Atom Smashers
Relate electric potential to an electric field and conservatiun of energy.
Problem In atom smashers (also known as cyclotrons and linear accelerators) charged particles are accelerated in much the same way they are accelerated in TV tubes: t.hrough potential differences. Suppose a proton is injected at a speed of 1.00 X IOfi m/s between two plates 5.00 cm apart, as shown in Figure 16.4. The proton subsequently accelerates across the gap and exit.s through the opening. (a) What must the electric potential difference be if the exitspeed is LO be 3.00 X 10 6 m/s? (b) \·Vhat is the magnitude of the electric field between the plates, assun1ing it's constant? Strategy Csc conservation of energy, writing the change in potential energy in terms of the change in electric potential, Ln~ and solve for ~E For part (b), solve Equation 16.3 [or the electric fielcl.
fi:)v.+
+
+
,
+ + LOI\' pote1ltial
5.00 cm FIGURE 16.4 (Example 16.3):\ proton enters a ca\'iry and accclenllcs from one charged plate 1,£\I'ard the other in ;HI e1eclric field E .
. . . . . . ...,
" .. Solution (a) Find the electric potential yielding the desired exit speed of the proton. Apply conservation of energy, writing the potential energy in terms of the electric potential:
t!.KE+ t!.PE= t!.K1:'+ qt!.v=
Solve I he energy equation lor the ch<-mge in potential:
t!.\!=
Substilllte the given values. obtaining lhe necessary potential difference:
i:11' =
a
i:1KE
q
111 11
q
.)
- - (vJ- - Vi') 2'1
(J.67 X 10-" kg) 2(160 X 10 "C) [(3.00 X 10"m/s)' - (1.00 X 10" m/s)'J
t!. \!~ -4.18 X 10' V
538
Chapter 16
Electrical Energy and Capacitance
(b) ",'hat electric field must exist between the plates? Solve Equation 10.3 [or the electric field and substitute:
E=
~V
4.18 X 10'V
~x
0.050 0 m
8.36 X 10'> N/C
...
• • • • • • • • • • • • • • • • ...011
Remarks Systems of ~uch cavities, consisting or allernaLing positive and negaLivc plates, are used LO accelerate charged particles to high speed before smashing them into targets. To prevent a sIO\,,-ing of, say. a positively charged particle after it passes through the negative plate of one cavity and clllers t lie next, the charges on the plates are reversed. Otherwise, the particle "'.'Quid be traveling from the negative plate to a positive plate in the second caviLy. and the kinetic energy gained in the previolls cavity would be lost in the second.
EXERCISE 16.3 Suppose electrons in a TV tube are acceleraLCd through a potential differencc of 2.00 X 10" V from the heated cmhode (neg€ll.ivc electrode), where they are produced, roward the screen. which also seryes as the anode (positive elcctrode), 25.0 em av.. . ay. (a) At what speed would the eleClrons impact the phosphors on the screen? Assume they accelerate from rest and ignore relmivistic effects (Chapter 26). (b) \'Vhat's the magnitude of rhe electric field, ifit is assumed constant?
QUESTION 16.3 True or False: A more massive panicle gains less energy in Iraversing a gi\-en potential difference than does a lighter particle.
Answers
1.00
- P; ill vollsjm -Vinvolts
0.800 0.600 0.400
~;;--L..!~""':L::;::s;='7;;', r (Ill) 0.0 4.00 li.OO FIGURE 16.5 E.leclric fleld and electric putentialversus dislanc(; from ,1 poim charge of l.11 X 10- 10 C. Nolc lhal l'is propunionallo l.1I; where" . . E is proportional to l/r~.
Electric potential created by a point charge ~
Superposition principle ~
16.2
(a) 8.38 X 10' m/s
(b) 8.00 X 10' Vim
ELECTRIC POTENTIAL AND POTENTIAL ENERGY DUE TO POINT CHARGES
In electric circuits a point or zero electric potential is often defined by grounding (connecting to the Earth) some point in the circuit. For example, if the negative terminal of a 12-V bauery were conneCled to ground, it would be considered to have a potemial of zero, \'v'hereas the positive termin<'ll would have a potcntial o[ +12 V. The potential difference created by the battery, however, is only locally defined. In this section we describe the elcctric potential ora point charge, \',;hich is defined throughout space. The e1cClric field of a point charge extends throughout space, so its elcctric pOlential docs, also. The zero point o[ electric potential could be taken anywhere, but is usually taken to be an infinitc distance from the charge, far from its influence and the innuence o[ any other charges. V,,'ith this choice, the methods of calculus can be llsed to shm... that the electric potential created by a point chargc qat any distance rfrom the chargc is given by
q
V= h., r
[16.4]
EqU<'llion 16.4 shows that the electric potential, or work per unit charge, required to move a test charge in [rom infinity to a dist.. \I1ce r [rom a positive point charge q increases as the positive test charge moves closer to q. A plot of Equation 16.4 in Figure 16.5 shows that the potential associated with a point charge decreases as 1/r with increasing ,; in COlltrasttO the magnitude of the charge's electric field, ",;hich decreases as 1./l. 2 . The e1eClric potential of 1.\",'0 or more charges is obtained by applying the super· position principle: the total electric potential at some point P due to several point charges is the algebraic swn of the electric potentials due to the individual charges. This method is similar to the one used in Chapter 15 to find t.he resultant electric field at a point in space. Unlike elcctric field superposition, \vhich involves a sum of \'ectors, the superposition of electric potentials requ ires cvaluati ng a sum of scalars. As a result, it's much easier to evaluate the electric potential at some point due 1.0 several charges than to evaluate the electric field, which is a yector quantity
16,2
Electric Potential and Potential Energy Due to Point Charges
FIGURE 16.6
539
rhe eleClri(' potclltial
(ill ;lrbilran'llI1Jts) JUllie p[;lllC nlll-
taining all electric dipole. Potential is plottt'd in the n.'nical dinwtlsiOll.
~.O
=
1.0
"
~
.="
0
~
W -J.O
-~,U
Figure 16,6 is a compmer-generated plot of the electric potential associalCd with electric dipole, which consists of two charges of equallllagniwdc but opposite in sign, The charges lie in a horizontal plane al the center of the potential spikcs. The value of the potential is plolled in the "ertical dimension. The computer program has added the potential of each charge to arrive at total ,'allies of the potential. Just as in the case of constant eleClric fields, there is a relationship bc(\.veen electric potemial and electric potential energy. 1f FI is the electric pOLential due 1.0 charge fll at. a point P (Active Figure l6.7a) the work required to bring charge fI':! from infinity to Pwithout acceleration is fj'!.VI • By definition, this \,mrk equals the potential energy j-Jf of the two-panicle system when the particles are separated by a distance,. (Active Fig. 16.7b). v..'e can therefore express the electrical potential energy of the poiror charges as ,Ill
PJ~'
=
f.1"- VI
=
fJlfJ?,
,.
l:f--
[16.5]
f- Potential energy of a pair of charges
I r the charges are of the salflP sign, PI; is posit.ive, Because likt: charges repel, positive \-vork must be done on the system by an external agent to force the t.\\IO chargt:s near each other. Conversely, if the charges (lrc 01" o/J/JOSilesign, the lorce is attractive and PE is negative. This means that negative work must be done to prevent unlike charges from accelerating toward each other as the)' are brought. close together.
QUICK QUIZ 16.5 Consider a collection of' charges in a givcn rcgion and suppose all othcr charges are distant and have a negligible effect. Further. the electric pot.ential is taken to be zero at infinit),. II" the electric potential at a gi\'en point in the region is zero, which of the following statements must be true? (a) The electric field is zero at that point. (b) The electric potential energy is a minimum at that point. (c) There is no nct charge in thc region, (d) Sornc charges in the region are positive. and some are negali\·e. (e) The charges have the same sign and are symmetrically arranged around the given poi nt. QUICK QUIZ 16.6 A spherical balloon comains a pusiti\'cJy charged particle at its centcr. As the ba Iloon is in fla ted to a 1([ rger volume ",'h iIe the eha rgcd panicle rernains at the center, which of the following are true? (a) The electric potential at the surface of the balloon increases. (b) The magnitude of the electric field at the surface of the balloon increases. (c) The electric flux through the balloon remains the same. (d) none of these.
'I, (a)
",.",.",.",. I'E= k~qJ{l:!
,.
'I, (b)
ACTIVE FIGURE 16.7 (a) The electric pOlcmial \'t at Pduc tht· point ch::lr~e 11 is \'1 = 'I/It II', (h) JLI second charge. (b i" brouglll from illfinil\ to P. the pUlemial cllcrg'Y orlhe pair is P/~ = Ji/!l{h/I: EO
540
Chapter 16
Electrical Energy and Capacitance
PROBLEM-SOLVING STRATEGY ELECTRIC POTENTIAL
1. Drawa diagram orall charges and circle the poilll of interest. 2. Cdculate the distance from each charge to the point ofinLerCSl, labeling it 011 the diagram.
",'I
3. For each charge q, calculate the scalar quantity V = - . Till' sign of Mr"h rhargf'
lfI-U,\t
r
be 7l/rluded in 'Your mlculationsf
4. Sum all the numbers fOllnd in the previous slep, obtaining the c1ecu"ic pOlcnlial al the point of interest.
EXAMPLE 16.4 Finding the Electric Potenti.:;:o;.1 Goal Calculate the electric potential due to a collection of point charges.
."
~ )' (Ill)
(fl.4.flO) P
Problem A 5.00-JLC point charge is at the origin, and a poim charge f('!. = -2.00 p,C is on the x-axis at (3.00, 0) 111, as in Figure 16.8. (a) If the electric potential is taken LO be: zero at infinity', find the lotal electric potcllli,1! duc LO these charges at point P with coordinates (0,4.00) m. (b) How much \\lOrk is required to bring a third point charge or 4.00 J1.C [1'0111 infinity Lo P?
,,
,,
, ,, ,
rl
' \ r':!
,,
fJj
FIGURE 16.8 (b,:'UTlple 16.'1) The t'kClrk pOlel1lial al poim Pdlle 10 lhe poinl charges ql and (h is lhe algebraic SUlll Oflhc POll'[llial~ due LO the illdividllal ch;uges.
1~
\
--ul-----GI--x (Ill) o (3.00.0)
Strategy For part (a), the electric potential at Pdue 1.0 each charge can be calculated from V = "///1: The LOtal cJec~ lric potential at P is the sum or these two numbers. For pan (b), usc the work-energy theorem, together \"lith Equalion 16.5, recalling that the pOlenti~1l at infinity is taken to be zero. .
~
.
~
Solution (a) Find the electric potemial
Calculate the electric potential at Pclue to the 5.00-ILC charge:
X 10 6 C ) \,; -_ Ii, "-'- -_ ( 8.99 X 10" N .Ill')(5.00 " 1", C4.00 III
1.12 X 10'Y Find the electric pOlcntial at Pduc: to the -2.00-p.C charge:
_
V, - Ii, ~
'I, _ ( - 8.99
X
T~
10
,,~.Ill')(-2.00 X 1O-6C) " C~ 5.00 m
~ -0.360 X 10" Y
Sum the two numbers to flnd the total electric potential at P:
(b) Find the work needed from infinily to P.
La
bring the
v"
+ V,
=
V,
=
7.6 X IO"Y
= L.l2 X 10' Y
+
(-0.360 X 10" V)
~1.OO-,uC charge
Apply the work-energy theorem, with Equation 16.5:
W= I1PE = '1,111'= '1,(1'" - V.) (4.00
X
10-6 C)(76
X
10' V - 0)
W= 3.0 X 10- 2 J ~
......•........ -
..
,
.
.
~
16.2
Electric Potential and Potential Energy Due to Point Charges
increase. (b) It would decrease. (c) Tt would remain the same.
Remarks Unlike the electric field, where vector aclrtilion is required, the electric potential due to more than one charge can be found with ordinary addition ofscalars. Further, notice that the \.. .ork required to move the charge is equal to the change in electric potential energy. The slim of the work done moving the paniciL plus the work done by the electric field is I.ero (H~'lht"f + Wch-nric = 0) because the particle starts and ends at resl. Therefore, \r\'utlll'r = -H~lcl.:lri( = .6.lJ.'leddc = q .6.\~
EXERCISE 16.4 Suppose a charge of -2.00 IJ..C is at the origin and a charge of 3.00 Me is aLLhe poim (0, 3.00) ITI. (3) Find the electric potential at (4.00, 0) 111, assuming the electric potential is zero at infinity, and (b) find the work necessary to bring a 4.00 p.C charge from infinity to the point (4.00,0) ITI.
QUESTION 16.4 If (1'2 were moved to the right, \.. . hat would hClppen to the electric potential ~, at point IJ? (a) It wOlild
EXAMPLE 16.5
541
Answers
(3) 8.99 X 10' V
Electric Potential Energy an:::d:..::;DLn:.:;a:::om=ic;;;:s
(b) 3.60 X 10-3J
_
Goal Apply conservation of" energy and electrical potential energy to a configuration of charges.
Problem Suppose three protons lie on the x-axis, at rest relative to one another at a g'iven instanl of time, as in Figure 16.9. If proton (/3 on the right is released while the others are held l-ixed in place, find a symbolic expression for the prolan's speed at inl-inity and evaluate this speed when to = 2.00 f"m. (Note: "I frn = 10- 15 m.)
FIGURE 16.9
(Example 16.!i)
Strategy First calculate the in itial electric potential energy associated with the systern of three particles. There will be three terms, one for each interacting pair. Then calculate lhe final electric potential energy associated with the system when the proton on the right is arbitrarily far away. Because the electric potential energy falls ofras 1/r,. two of" the terms will v;;'Inish. Using conservation of energy then yields the speed of the panicle in question. "...
.
.
...
Solution Calcu late the electric potential energy associated with the initial configuration of charges: Calculate the clcnric potential energy associated with the final configuration of" charges: \-Vrite the conservation of energy equation: Substitute appropriate terms:
1 (, 'i11l lJ - -
0
- ~:\ I
.,
:!.m:~v3~ ~
hl'e 2
+ -'0- -
(h,e' + -.h,e') -.2ft)
Solve for v:~ after combining the twO rcmaining potential energy terms:
3(8.99 X 10" N'm'/C')( 1.60 X 10- 1" C)' (1.67 X 10 27kg)(2.00 X]O 15 m )
.
1.44 X 10' m/s ~
542
Chapter 16
Electrical Energy and Capacitance
Remarks The difference in the initial and fi nal kinetic energies yields the energy available for motion. This calculation is some,.. . hat contrived because jt would be difficult, alt.hough not impossible, to arrange such a configuration of prowns; it could conceivably occur by chance inside a star. QUESTION 16.5 If a fourth proton were placed to the right of fJ",. how many additional potential energy terms would have lated in the initial configuration?
[Q
be calcu-
EXERCISE 16.5 Starting from the initial configuration of three proLOns, suppose the end ['<.:0 particles are released simultaneously and the middle panicle is fixed. Obtain a numerical ans\.. .er for the speed of the twO particles at infinity. (Note that their speeds, by symmeli")', must be the same.)
Answer
1.31 X 10' m/s
16.3
POTENTIALS AND CHARGED CONDUCTORS
The electric pot.ential at all points on a charged conductor can be determined by combining Equations 16.1 and 16.2. From Equation 16.1, v..e see that the work done on a charge by electric forces is related to the change in elcctrical potclllial energy of the charge by
W= -!!.PE From Equation 16.2, we see that the change in electric potential energy between two points A and B is related to the potential difference between those points by
!!.PE = q(V/i - V,) Combining these two equations, we find that [16.6] Using this equation, we obtai n the following general result: No net work is required to move a charge between two points that are at the same electric potential. Tn mathematical terms this result says that \'\!= 0 whenever VII = \~\' In Chapter 15 we found that when a conductor is in electrostatic cquilibriurn, a net charge placed on il. resides entirely on its surface. Further, \,"'e showed that the electric field just outside the Stll-f~lCe of a charged conductor in electrostatic equilibrium is perpendicular to the sur1~lce and thaI. thc field inside the conductor is zero. We now show that all points on the surface of a charged conductor in electrostatic equilibrium are at the same potential. Consider a surface path connecting any points A and Bon a charged conductor, as in Figure 16.10. The chargcs on the conductor are assumed to be ~ equilibrium with each other, so none are moving. In this case the electric field E is ahvays perpendicular to the displacement along this path. This must. be so, for otherwise the pan or the electric fielcl tangent to the surface \',;oldd move the charges. Because E is perpendicular to the path, no work is done by t.he electric field if a charge is moved between the given two points. FI·om Equation 16.6 we see thaI. if the work done is zero, the difference in electric potential, lIJI - VA' is also zero. It fol1mvs t.hat the electric potential is a constant everywbere on the surface of a charged conductor in equilibrium. Furlher, because the electric field inside a conductor is zero, no work is required 1.0 move a charge bet.ween two poi illS inside the conductor. Again, Equation ]6.6 shows that if the volOrk done is zero, the difference in clectric potential between any two points inside a conductor must also be zero. 'Ne conclude that the electric potential is constant everywhere inside a conductor. Finally, because one of the points inside the conductor could be arbitrarily close to the surbce of' the conductor, we conclude that the electric potential is constant ~
FIGURE 16.10 All ;lrbitrarily~haped conductor \\'hh an exn:ss positi\"(: charge. \Vhl:ll the conductor i~ ill e1cctrostatic equilibriurn, alilh e charge resicle.:s aL til\' surface, E :::: O. in~id<;, lht:.' t:onductur, ;md liH.: decLric field JUSt outside lhe t:unduCI,nr is perpendicular LO lht: surface.:. The POlclltial is \:II11SL
~urfJ.ce.
16.4
Equipotential Surfaces
543
everywhere inside a conductor and equal to that same value at the surface. :-\s a consequence, no work is required to move a charge from the interior of a charged conductor to its surface. (II's important La realize that the potenlial inside a conductor is not necessarily zero, evell though the interior eleci ric field is zero.)
The Electron Volt An appropri;uely sized unit ofcnergy comrnonly llsed in atomic and nuclear physics is the electron VOIL (eV). For example, electrons in norm"l aLOms typically have energies of tens of eV's, excited electrons in aLOrns emitting x-rays ha"e energies of thousands of eV's, and high-energy gamma rays (electromagnetic waves) emilled by Lhe nucleus have energies of millions of eV's. The electron volt is defllled as the kinetic energy that an electron gains when accelerated through a potelllial difference of 1 V. Because I V = l.1/C and because the magnitude of the charge on the electron is 1.60 X 10- 19 C, we see that the electron volt is related to the joule br 1 eV = 1.60 X 1O~19 C' \' = 1.60 X JO~I".J
[16.7]
QUICK QUIZ 16.7 An electron initially at rest accelerates th rough a potential difference of I V, gaining kinetic energy K..L~. whereas KE" (d) The answer can't be dClermined rrom the given information.
16.4
EQUIPOTENTIAL SURFACES
A surl~lCe on \-"hich all poil1l.s are at the saine pOlellLial is called an equipotential surface. The potential difference benvccn any two points on ,1l1 equipotential surface is zero. l-Ience, no work is required to move a charge at constant speed on an equipotential surface. Equipotential surfaces have a simple relationship to the electric field: The electric field at every point of an equipotential surface is perpendicular to the surface. If Lhe electric field E had a component parallel to the surface, that component would produce an electric force 011 a charge placed on the surface. Thi.s force would do work on the charge as it moved from onc pain! to another, in cont.radiction to t.he defll1ition of an equipotential surface. Equipotential surfaces can be represented on a diagram by drawing equipotential contours, which are two-dimensional views of the imersections of the equipotent.ial surfaces with the plane of the drawing. These equipotential contours are generally referred to simply as equipotentials. Figure 16.11a (page 544) shows I he equipotentials (in blue) associatcd \\lith a positi\'c point charge. Note that the equipotentials are perpendicular La the e1cctric Field lines (in red) at all points. Recall that the electric potential created by a point. charge q is given by V = k"q/I: This relation shows that, for a single point charge, the potential is constant on any surface 011 which 'ris constant. It follow's that. the equipotcntials ofa point charge arc a r~11l1i1)' or spheres centered on the point charge. rigure 16.11b shmvs the equipotcntials associated with two charges of equal magn ilude but opposite sign.
~
Definition of the electron volt
544
Chapter 16
FIGURE 16.11
Electrical Energy and Capacitance
Equipolcllti:ll<;
(dashed blue lines) and ,,'leetTi, field lines (red lines) for (a) iI posi· th-c point charge and (b) twO poilll charga::s of equ.d magnilUde alld opposiu: sign. In all ca~s the cquipotcmials are pnpnulirlliarlo lhe clec· I ric" field lines;:11 c\cl) point.
EIl'etric liclcllinc
,,
' \ EquipolcllliaJ
(0)
Ib)
16.5 APPLICATIONS The Electrostatic Precipitator One imponant application or electric discharge in gases is a de\'ice called an eleelrostalic jnl'ci/JilaI01: This dc\"ice removes particulate maller from combustion gases, thereby reducing air pollUlion. Il'S especiall)" lIseful in coal-burning power plants and in industrial operations lhal. gcner<:lle large quantities of smoke. Systems cur· rently in usc can climinate approximmely 90% by mass of thc ash and dust from the smoke. Unfortunatel)", a very high percentage of thc lighter panicles still escape, and they contribute significantly to smog and haze. Figure 16.12 illustrates the basic idea of the electrostatic precipitator. A high Yoll.age (lypically 40 kV to 100 k\') is maintained between a wire running down the cenLCr ofa duct and the ouler wall, which is grounded. The wire is mail1lained at a negative electric potcntial with respcClto the wall, ::;0 thc elcClric field is directed toward the wire. The electric field near the wire reaches a high enough value to calise a discharge around the wire and the [ormation of positive ions, eleCLrons, and negative ions, such as O 2 . As the electrons and negative ions are acceler-
the outer wall by the nonuniform electric l-ield, the dirt particles in the streaming gas become charged by collisions and ion caplure. Because most of t he charged dirt particlcs arc negative, they are also dra\'~"n to the outer wall by the electric field. When the duct is shaken, the particles Elll loose and are collected at the bOlLOIll. In addition to reducing the amounts orharmful gases and paniculmc matter in the atmosphere. the electrost.atic precipitator recovers valuable metal oxides frorll t.he SLack. A similar dcvice called an elerlrosfalic air rleallf'r is used in homes to relieve the discomfort of" allergy sul-Terers. Air laden v~'ilh dust and pollen is drawn illw t.he device across a positively charged rnesh screen. The airborne particles becomc positively charged when they make contact with the screen, and then lhey pass through a second, negati\"c1y charged mesh screen. The electrostatic force of attraction between the positively charged particles in the air and the negatively charged screen causes the panicles to precipitate om 011 the surface o[ t.he screen, removing a very high percenlage of contaminants from the air stream.
545
Applications
aLeel LOwarel
APPLICATION The Electrostatic Air Cleaner
Xerogrophy ond Loser Printers
APPLICATION
Xerography is widely used to make phol.Ocopies of printed materials. The basic idea behind the process \.. . as developed by Chester Carlson, \vho was granted a pal.entror his invention in 1940. In 1947 the Xerox Corporation launched a full-scale program LO develop automated duplicating machines using Carlson's process. The huge success of that development is eviden t: today, practically all offices a nd libra ries have one or more duplicating machines, and the capabilities o[these machines continue to evolve. Some features of the xerographic process invoke simple concepts from electrostatics and optics. The one idea t.hat makes the process unique, however, is the use ofphoLOconductive materialLO form an image. A phoLOconductor is a material that is a poor conductor of electricity in the dark, but. a reasonabl)' good conductor when exposed to I ighl. Figure 16.l3 illustrates the steps in the xerographic process. First, the surface ora plate or drum is coated \~iith a thin film of the photoconductive material (usually selenium or some compound ofseleniurn), and the photoconductive surface is given a positive electrostatic charge in the dark (Fig. 16.13a). Thc page LO be copied is then pr~jected onto t.he charged surface (Fig. 16.13b). The phoLOconducting
Xerographic Copiers
....- Lens
;; ....'.:;'.
iJ~"l ~
6Ir
Selcnium-coated druin
(a) Ch:lrging the drulll
I.aser beam
Neg-ativdy charged toner
(b) Imaging the document
(c) Appl)'illg the tonel·
(d) Transrerring-thc
Loner to the paper rhe xerographic process. (a) The pholocondllnin: ~urfact.: i~ positively ("har~ed. (b) Through the lIseofa lighl source and a len,. a hidden image is formed 011 the charg:l.:d surface ill llu: form of posilive charges. (c) The slirface COli win ing the image is cOI'lTcd with a neg.nin·]}' charged pOl"der, \\'llich adheres only to the image area. (d) A piece of paper i, placed mer the surface and gi\'cn a charge. This lr;lllsfers Lhe image to llie paper, which is then healed to "lix-lhe powder to Lhe paper. (f:) The image olllhe drum ofa laser print...:r is produced b}' LUrning a lasl'r beam on and offas it sweep:o, ;)cr(lSS the selelliulll-coated drnlll.
FIGURE 16.13
(e) Laser printer dntm
546
Chapter 16
Electrical Energy and Capacitance
surface becomes conduCling only in areas where light slrikes: there the lighr produces charge calTiers in I he pholoconductor thalncutralize the positivel)' charged surface. The charges remain on those areas of the phowconducwr not exposed to light, however, leaving a hidden image of the object. in the Form ora positive distri· bution of surface charge. Next. a negatively charged powder called a loner is dusted onto the photocondllcting surface (Fig. 16.J3c). The charged powder adheres only to the areas that. contain the positively charged image. At this point, the image becomes visible. It is then lransierred to thc sLlrJ~lce of a sheet of' positively charged paper. Finally, the toner is "fixed" to thc surface of the paper by heal (Fig. 16.13d), resulting in a permanent copy of the original. The steps for producing' a documcnt on a laser primer are similar to thosc used in a photOcopy machine in that parts (a), (c), and (d) of Figure 16.13 remain essentially the same. The difference bct ween thc two tech niques lies in the way the image is formed on the selenium-coated drum. In a laser primer the command to print. the lell.er O. (or instance, is sent to a laser {'rom the memory or a computcr. A rotating mirror inside the printer causes the beam of the laser LO sweep across the selen ium-coatcd drum in an interlaced pattern (Fig. ·16.13e). Electrical signals generated by the printer turn the laser beam on and offin a pattern th<1ttracesout the letter 0 in the form of positive charges on the selenium. Toner is then applied to the drum, and the transfer to paper is accomplished (IS in (l photocopy machine.
APPLICATION Loser Printers
16.6 -Q
+
FIGURE 16.14 A parallel-plaIt' capacitor consists of two par-alIt:! plates. each of area .1. sep:lrated by a distance d. The pl;ncs carry equal and
OPPOSill: charges.
Capacitance of
0
CAPACITANCE
A capacitor is a device used in a variety o[ electric circuits, such as to tlIne the frequency of radio receivers, eliminate sparking in automobile ignition systems, or store short-term energy [or rapid release in elect.ronic flash units. Figure ]6.14 shows a typical design for a capacitor. [t consists of t\-vo parallel metal plates separated by a distance rI. Csed in an electric circuit, the plates are connected to the positive and negative t.erminals of a battc,"y or some other voltage source. 'When this connection is made, electrons are pulled ofT one of the plates, leaving it with a charge of +Q, and are transfcrred through the batt.ery to the other plate, leaving it with
c= Q
pair
of conductors
~V
~
51 unit: farad (F)
TIP 16.2 Potentiol Difference Is Jl.y, Not V Use the symbol ill'for the poten~ tial difference across:J circuit elemCllt or a de\'ice (many other books use :-illlply Vror potent ial difference). Till; dual use of I'LO repn:scnl pOlcnlial in one place and a potential difference ill another can lelf! to llllllcccssary confusion.
[16.8]
= coulomb per volt (C/V)
The quantities Q and .l\i are always laken to be positive when used in Equation 16.8. For example, ira 3.0-,uF capacitor is connected to a 12-V battery, the magnitude of the charge on each plate of the capacitor is
Q=
C ~\I = (3.0 X 10-'; F) (12 V) = 361'C
From Equation 16.8, we sec thal a large capacitance is needed to store a large amount or charge ror a given applied yoltage. The farad is a \'ery large unit of capacitance. In pn1ctice, most typiC£l1 capacitors have capacitann:s ranging from microfarads (I 1'1' = 1 X 10 ,; F) [0 picofarads (1 pI' = J X 10-1' F).
16.7
The Parallel-Plate Capacitor
547
FIGURE 16.15 (a) The electric field hetween the plalC!i ofa panl1Ici-phllC capacitor is uniform near thc ce1lter, but nOlllUlrorm nea.· the edges. (b) Electric fidd pattern of LWo nppo~irel)' charg.'d
cnndllCl illg p:lnll_
lei plaLes. Small pieces of thread on an oil surface aligll with Iht' declric Held.
(h)
(0)
16.7
THE PARALLEL-PLATE CAPACITOR
The capacitance of a device depends on the geometric arrangernelll of the conductors. The capacitance of a parallel-plate capacitor with plates separated by air (see Fig. l6.14) can be easily calculated frolll three facts. First, recall from Chapter 15 that the magnitude of the electric field bet,,,,'cen two plates is given by h: = (T/~I' where if is the magnitude of the charge per unit area on each plate. Second, \\lC round earlier in this chaptcr that the potential difference between two plates is av = Ed, where d is the distance bcn...· een the platcs. Third. the charge on one plate is given by if = erA, where A is the area of the plate. Substituting these three facts into the definition of capacitance gives the desired result:
If i\.V
ITA
Ed
«T/Eo)d
c=-=-= Canceling the charge per unit area,
fT,
Ke)'----_
FIGURE 16.16 When the kev of one type of keyboard is pressed, the capacitance ofa parallel-plate capacilor increases as the plale spacing decreases. The Sllbst;lncc l.. bcled ~dielcctric" i~ an insulating matnial. :lsnescribcd in Section 16.10.
yields
c=
;\ En-
d
[16.9]
wherc A is the area of one of the plates, d is I he cI istance between llJe plates, and EO is the permittivity of free space. From Equation l6.9, we see that plates with larger area can store more charge. The same is true for:a small plate separation d because then the positive charges on one plate exert a stronger force on the negative charges on the other plat.e, allowing more charge to be held on the plates. Figure 16.15 shmvs the electric field lines ofa more realistic parallel-plate capacitor. The electric field is very ncarly constant in the center between the plates, but becomes less so \vhen approaching the edges. For Illost purposes. however, the field Inay be taken as constant throughout the region between the plates. One practical device thal uses a capacitor is the nash attachment on a camera. A battery is used 1.0 charge the capacitor, and the stored charge is then released when the shutter-release bunon is pressed 10 take a picture. The stored'charge is delivered to a nash tube very quickly, illuminating the subject at the instant more light is needed. Computers make use of capacitors in many ways. For example, one type of C0111puter keyboard has capacitors at the bases orits keys, as in Figure 16.16. Each key is connected l.0 a movable plate. which represents one side of the capacitor; the fixed plate on the bottom of the keyboard represel1ns the othel" side of the cap
~
Capacitance of a poralielplate capacitor
APPLICATION Camero Flash Attachments
APPLICATION Computer Keyboards
APPLICATION Electrostatic Confinement
548
Chapter 16
EXAMPLE 16.6 Goal
Electrical Energy and Capacitance
A Parallel-Plate Cap.acitar
Calculate fundamental physical properties ofa parallel-plate capaciLOr.
Problem A parallel-plate capacitor has an area A = 2.00 X 10- 4 m 2 and a picHe separation d = 1.00 X 1O-:~ m. (a) Find its capacitance. (b) How much charge is on the positive plate irthe capacitor is connected to a 3.00-V bauery? Calculate (c) the charge density on the positive plate, assuming the density is uniforrn, and (d) the magnillide 01" the electric field between the plates. Strategy Parts (a) and (b) can be solved by substituting into the basic equations lor capacit£lllce. [n part (c) usc the definition of charge density, and in pan (d) use the faCLLhat the voltage difference equals the electric field times the distance.
..
,... Solution (a) Find the capacitance.
c=
Substitute into Equation ]6.9:
c
=
A = rl
€lJ~
I
( 8.85
1.77 X J 0
12
II1')
X 10- I'"·C-/N·m ,)(2.00 X jQ- 3 1.00 X 10 . III F
=
1.77 pF
(b) Find the charge on the positive plate after the capacitor is connected to a 3.00-V battery.
c=R t.1f
SubstiLUte into Equation 16.8:
Q~ Ct.If~ (1.77 X 10-" F)(3.00 V)
= 5.3 J X HI
12
C
(c) Calculate the charge density on the positive plate. Charge density is charge divided by area:
10- 12 C
Q
5.31
A
2.00 X 10
X
a-=~=
4
m'
2.66 X 10-' C/m'
(d) Calculate the magnitude of the eleClric field between the plates. t. If L,·= =
Apply t.V = Ed:
rl
~
3.00 V = 1.00X10'm
,
1
3.00 X 10: Vim ,
~
Remarks The ans'..·cr to pan (cl) could also have becn obtained from the electric field derived for a parallel plate capacitor, Equation 15.13, E = a-/€r)'
QUESTION 16.6 How do the answers change if' the distance between the plates is doubled? EXERCISE 16.6 Two plates, each o[ area 3.00 X ]0- 4 m~, are lIsed to construct a parallel-plate capacitor with capacitance l.00 pF. (a) Find the necessary separation distance. (b) II' the positive plate is to hold a charge of' 5.00 X 10- 12 C, find the charge density. (c) Find the electric field bet,..'een the plates. (d) \Vhat voltage batlery should be attached to the plate to obtain the preceding results?
Answers
(a) 2.66 X 10-:1 m
(b) 1.67 X 10-8 elm·
(c) 1.89 X 10:1 N/C
(e1) 5.00 V
Symbols for Circuit Elements and Circuits
r-----
The symbol that is commonly used to represent a cap,leito!" in a circuit is -----1 or sometimes ~r-----. Don't confuse eilher of these symbols with the circuit symbol,
~C-- which
is used to designate a battery (or any other source of
direct current). The positive (crill inal of the battery is at the higher potential and is represented by the longer vertical line in t.he battery symbol. ]n the next chapter
16.8
we discuss anothcr circuit clemen!., called a resisl.Or, rcpresentcd bY' the symbol \Vhen wires in a circuit c1on't havc appreciable resistance compared \,vith the resistance of" othcr c1cmellts in the circui!., the wires arc represcllted by straight lines. It\ important L0 realize lhat a circui!. is a collection or real objects, usually contain ing a source of electrical energy (such as a battery) connected to clements lhat convert electrical energy to other forms (light, heat. sound) or st.ore the energy in electric or !nagnctic fields [or later retrieval. A real circuit and its schematic dia~ gram are sketched sidc by side in Figure 16,17. The circuit symbol for a lightbulb
Combinations of Capacitors
549
I:!V
~.
shown in Figllre -16.17b is
1"\vo or more capacitors can be combined in circuiLs in several ways, but most rcduce LO two simple configuraLions, called parol/f)/ and sPI'it's. The idea, then. is 10 And the single cquivalem capacitance due to a combination of" sevt:ral different capacitors that arc in parallel or in series with each other. Capacitors are m
Capacitors in Parallel Two capacitors connected as shown in Active Figure 16.18£1 are said to be in f)(fral~ If'!. The lefL pl,lte of each capaciLor is connected to the positive terminal of the battery by a conducting wire, so the left plaLes are at the same potential. In the same way, t he right. plates, boLh connected 1.0 the negative terminal oj' the battery, are also m the same potenLial. This mcalls that capacitors in parallel both have the same potential difference &.Vacl'oss them, Capacitors in parallel are illustrated in Active Figure 16.18b. When the capacitors arc firsL cOllnecLed in the circuit, ciCCI rOilS arc transferred from the left plates throug'h the batLery to the right plates, leaving the left pl.Hes positively charged and the right plaLes negatively' charged. The energy source for !.his transfcr of charge is the inLcrnal chemical energy stored in the baLtery, which is cOllvened to electrical energy. The flow of charge stops when the voltage across lhe Glpacitors equals the voltage of the baw::ry, at which timc [he capaciLOrs have their maximum charges. If the maximulll charges on the two capacitors are Ql and Q~, respectively, the lolrti rharg(', Q, stored b)' the two capacitors is
Q=Q,+Q,
0.1/1 =~V:!=.1.JI
+
1
:IlTn."" I hi" CotP;ll'ilnrs ;11 (' I he S
L'';!.
o.:quhal<:'IH capacil;lllcc" is
(-
+
[16.10]
ACTIVE FIGURE 16.18 (a) A par. (b) Tltl' CilTlliLdi:tgrallllol the p'lrallc.:1 combillaliun. (c) rll{' potenLial diffen:llcts
~-
1
+
(a)
(b)
(e)
(a)
12V +
r
COMBINATIONS OF CAPACITORS
(;1
~
®
If you are not familiar \,;ith circuit diagrams, trace [he p
16.8
ResisLor
c.:~.
=
(."1
and llie C2 .
+
(b)
FIGURE 16.17 (a) A real circuit and (h) it~ equivalent circllil diag-ralll,
550
Electrical Energy and Capacitance
Chapter 16
\r\'e can replace these tWO capacitors ,-\lith one equivalent capacitor having a capacitance of C(:(r This equivalent capaciLOr must have exactly the same external effcC( on the circuit as the original two, so it must slore Q units of charge and have the same potential difference across it. The respective charges on each capacitor are
Q, = C1
TIP 16.3 Voltoge Is the Some os Potentiol Difference
The charge
011
Q2 = C,
the equivalcm capacitor is Q= C",
A voltage (Jcmss a dc\"icc. stich as a capacitor, has the same meaning as the pOlcl1lial difference acros~ the c1c\"icc. For example, if\,'e sa)' lhal the vollage across a C
Substituting these relationships into Equation 16.10 gives
Ceq
[16.11]
I f we extend this tremment to t.hree or more capacitors connected in parallel, the equivalent capacitance is found to be
parallel ) ( combination
[16.12]
\Ve see that the equivalent capacitance of a parallel combination of capacitors is larger than any of the individual capacitances.
EXAMPLE 16.7 Goal
Four CORacitors Connected in Parallel
Analyze a circuit ,.. . ith several C<:lpacilOrs in parallel.
Problem (a) Determine the capacitance of the single capacitor that is equivalent to the parallel combination of capacitors shown in Figure 16.19. Find (b) the charge all the 12.0-,uF capacitor and (c) the total charge contained in the configuration. (d) Derivc a symbolic expression for the fraction of the total charge contained on one of the capacitors. Strategy For part (a), add the individual capacilances. For part (b), apply the formula C = QI
,... .....
,
300 jlF
6.00 jlF
II II II
II
r-----
,12.0 jlF
II
II
24.01'1'
II
-
+
L, J
FIGURE 16.19 (Example 16.7) FoUl" capacitors connected in parallel.
18.0V
~
....,
.
Solution (a) Find the equivalclIL capacitance. Apply Equation 16.12:
(b) Find the charge on thcl2-fLF capacitor (designated C,). Solyc the capacitance equation for Qand substitute:
Q ~ C3
X
10-6 F) (18.0 V) = 216
X
10- 6 C
16.8
Combinations of Capacitors
551
(c) Find the total charge contained in Lhe conngurmion. Q~ C,qLl.V = (45.0/LF)(18.0V)
L'se the equivalent capacitance:
8.IOXIO"/LC
(d) Derive a symbolic expression [or the f"raction or the total chargc comainccl in one of the capacitors. \,Vrite a symbolic expression for the fractional charge in the il.h capacitor and use Lhe capacitor dennition: • • • • • • • • ...OiI
Remarks The charge on anyone oCthe parallel capacilOrs can be fOllnd as in part. (b) because the potential difference is the same. Notice that finding the LOtal charge does not require finding the charge on each individual capacitor and adding. lL's easier to use the equivalent capacitance in the capacitance definition. QUESTION 16.7 If all four capacitors had the same capacitance, what fraction of the total charge would be held by each? EXERCISE 16.7 rind the charge on thc 24.0-p.F capacitor.
Answer
432 /LC
Capacitors in Series Now consider two capacitors conJl(.:cted in serif'S, as illustratcd in Activc Figure 16.20a. For a series combination of capacitors, the magnitude of the charge must be the same on all the plates. To understand this principle, consider the charge transfer process in some detail. \Vhcn a battery is connected to the circuit, electrons with total charge ~Q are translerrecl from thc left plate or C1 to the right plate of C~ through thc battery, leaving the lcft plate of C1 wil h a charge or +Q. As a consequence, the magniludes of the charges Oil the left plate of C\ and the right plate of C~ must be the same. :'\0\\' consider the right plate or Ct and the lerl. plate of' C~, in the middle. These platcs are not connected to the b£1uery (because of the gap across the platcs) and, taken together,
c!
(.'1
~\'I
.1.\':!
r
J
+Q -Q
+(2
+
-Q
lI +
(b)
~
Q is the same for all capacitors connected in series
ACTIVE FIGURE 16.20 :\ seri('~ LombinatiOIl of u,'o capaci. LOrs. The charges 011 lhe cap"dtol's arc the ...ame, and the equh'alcnl capacilance C
G,I.
552
Chapter 16
Electrical Energy and Capacitance
After an equivalent capacitor for a series of capacitors is fully charged, the equivalent capacitor must end up with a charge of -Q on its right plate and a charge of +Q on its left plate. Applying the definition or capacitance to the circuit in Active Figure 16.20b, we have
"v=JL C e
,·..here ~V is the potential difference bCl\\i'cen the terminals of the bartery and Ceq is the equivalent capacitance. Because Q = C ilV can be applied to each capacitor, the porential differences across them arc given by
"V, =C,QFrom Active Figure ]6.20a,
\,,'C
see lhaL il\'=
dl~
+ LiV~
[16.13)
where LillI and LlV2 are the potential differences across capacitors C1 and C2 (a consequence of the conservation of energy). The potential difference across any number of capacitors (or other circuit ele~ ments) in series equals the sum of the potential differences across the individual capacitors. SubstiLUting these expressions into Equation 16.13 and noting that Li.V:;:: Q/Ceq , \VC have
JL= Q+R Ct'q
Canceling Q, we arrive
aL
C1
C2
the following relationship:
I
1
1
Ceq
C\
C2
-=-+-
series ) ( combination
[16.14]
If this analysis is applied to three or more capacitors connected in series, the equivalent capacitance is found to be
l
1
J
1
CC(l
C1
C2
C:\
-=-+-+-+
series ) ( combination
[16.15)
As \.. 'e \-vill shO\v in Example 16.7, Equation ]6.15 implies that the equivalent capacitance of a series combination is always smaller than any individual capacitance in the combination. QUICK QUIZ 16.8 A capacitor is designed so lhal one plale is large and the other is small. If the plates are connected to a battery, (a) the large plate has a greater charge than the small plate, (b) the large plate has less charge than the small plate, or (c) the plates have equal, but opposite, charge.
EXAMPLE 16.8
Four Capacitors Connected in Series
Goal Find an equivalent capacitance of capacitors in series, and the charge and \'ollage on each capacitor. Problem Four capacilOrs are connected in series ,..· ith a battery, as in Figure J6.2l. (a) Calculate the capacitance or- the equivalent capacitor. (b) Compute t.he charge on lhe 12~,uF capacitor. (c) Find the voltage drop across the 12~,uF capacitor. Strategy Combine all [Jle capacitors into a single, equivalent capacitor using Equa~ Lion J6.15. Find the charge on this equivalent capacitor lIsing C:;:: Q/~V This charge is the same as on the individual capacitors. Usc this same equation again to find the voltage drop across the 12-fLF capacitor.
+ 18 V
FIGURE 16.21 (Example 16.8) Four capacitor~ connccled ill !'cries.
16.8
.
,...
Combinations of Capacitors ..................... -
553
...
~
Solution (a) Calculate the eqllivalent capacitancc of thc series. 1
1
1
1
I
Coq
~.o fLF
6.0 fLF
12 fLF
24 fLF
-=---+--+--+--
Apply EquaLion 16.15:
C~q =
1.6 fLF
(b) Compute thc charge on the 12-ILF capacitor. The desired charge equals the charge on the equivalent capaciwr:
Q= (;,." "'V = (1.6X 10-hF)(18V)
=
29JLC
(c) Find the voltage drop across the 12-p.F capacitor.
Notice that the equivalent capacitance is less than that of any of the individual capacitors. The relationbe used to find the "oltage drops on the other cap~1CiLOrs,jLlstas in pan (c).
QI~V can
QUESTION 16.8 Over which capacitor is the voltage drop thc smallest? The largest? EXERCISE 16.8 The 24-,uF capacitor is removed from the circuit, leaving only three capacitors in series. Find (a) the equivalent capacitance, (b) the charge on the 6-p.F capaciLOr. and (c) t.he voltage drop across the 6-JLF capacitor.
I. Combine capacitors that are in series or in parallel. following' the derived formulas. 2. Redraw the circuit aftcr every' combination. 3. Repeat the first. t.wo steps until there is only a single equivalent capacitor. 4. Find the charge on the single equivalent. capacitor, using C =
(V~V.
5. Work backwards through the diagrams to the original one, finding the charge and voltage drop across each capacitor along the \\fay. To do this, use the follm\'ing collection of facts: Q/~V
A. The capacitor equation: C =
B. Capacitors in parallel: Ce'l = C1
+
C'-!
C. Capacitors in parallel all have the same voltage difference, DoV, as does their equivalent. capacitor.
.
D. CapacItors
.
.
111 SCt"lCS: -
1
= -
1
+
1
~
C"q Cl C'1 E. CapaciLors in series all have Lhe same capacitor.
char~e,
Q, as docs Lheir equivalent
554
Electrical Energy and Capacitance
Chapter 16
EXAMPLE 16.9 Goal
Equivalent Cap.acitance
Solve a complex combination ofscries and parallel capacitors.
Problem (a) Calculate the equivalent capacilance between (f and b lor the combination of capacitors shown in Fig"lIre 16.22a. All capacilances arc in rnicror'.lrads. (b) Ira 12-V battery is conncClcd across the system between points a and b, find the charge on the 4.0-jLf capacitor in the first diagram and the voltage drop across it.
:!.o
{}
---!f---
a 6.0 b
4.0
Strotegy For part (aJ. usc Equations 16.12 Ie) (al Idl Ihl and 16.1.'> to reduce the combination step by FIGURE 16.22 (Exalllple 16.9) To flild the cquivalclIl capacitance orthc circuit ill step, as indicated in the figure. For part (b), (a). usc the series and paralld rult;s described ill the lcx110 sLlccessively rc(lucc the to find the charge on the 4·.0-p,F capacitor, cirCllit as indicated ill (Il), (c), i.lIld (
.
Solution (a) Calculate the equivalelll capacitance. Find the equivalelll capacitance of the parallel l.O-p,F and 3.0-JLF capacitors in Figure 16.22a:
Ceq
=
Find the equivalent capacitance of the parallel 2.0-IlF and 6.0-p,F capacitors in Figure 16.22a:
C,q
= C, + C2 = 2.0 iLl' + 6.0 iLl' = 8.0 iLl'
Combine the t\.. .o series 4.0-,uF capacitors in Figure 16.22b:
-=-+-=---+--
C,
+ C, =
1.0 iLl'
+ 3.0 iLl' = 4.0 iLl'
I
I
I
j
1
Ceq
C,
C,
4.0 iLl'
4.0 iLl'
I
2.0 ILl'
2.0 iLl' 1
I
1
C
C,
C,
1
1
-~-+-=---+-8.0 8.0 e"
Combine the two series 8.0-p.F capacitors in Figure 16.22b:
ILl'
ILl'
1 '1.0 iLl' Finally, cOtnbilie the two parallel capacitors in Figure 16.22c to find the equivalent capacitance between a and b:
Ce"
=
C,
+
C2
= 2.0 iLl' + 4.0 ILl' =
6.0 ILl'
(b) Find the charge on the 4.0-.uF capacitor and the ,"oltage drop across it. Compute the charge on the 2.0-,uF capacitor in Figure L6.22c, which is the same as the charge on the 4.0-pJ' capacitor in Figure 16.22a: Use the basic capacitance equation to f-i nd the voltage drop across the 4.0-,uF capacitor in Figure 16.22£1: ~
-,
C= Q £>11
-->
Q = C£> 11 = (2.0 iLl') (I 2 V)
c= -Q
-->
£>11
£>11
.
6.0V • • . • • . . • • . . . • • • • . . • . •.A
Remarks To rind the rest. of the charges and voltage drops, it's just a matter of using C = Q/8V repeatedly, together \"·:ith ('acts 5C and 5E in the Problem-Solving Strategy. The voltage drop across the 4.0-,uF capacitor could also have been round by noticing, in Figllre 16.22b, that both Glpacitors had the sarne value and so by symmetry would split the total drop of 12 volts between t.hem.
16.9
QUESTION 16.9 Which capacitor holds more charge, the
I.O~,uF capacitor
Energy Stored in a Charged Capacitor
555
or the 3.0-J.LF capacitor?
EXERCISE 16.9 (a) Tn Example 16.9 find the charge on the R.O-,uF capacitor in Figure 16.22a and the voltaRc drop across it. (b) Do t.he same for the 6.0-j.LF capacitor in Figure ]6.22a.
Answers
16.9
(a) 48 ,.,.C. 6.0 V
(b) % ,.,.C. 6.0 V
ENERGY STORED IN A CHARGED CAPACITOR
Almost everyone who works with electronic equipment has at some time verified that a capacitor can store energy. I f the plates of a charged capacitor are connected by a conductor such as a wire, charge transfers rrorn one plate La the other until the lwO are uncharged. The discharge can oftcn be observed as a visible spark. If you accidentally touched the opposite plates of a charged capacitor, your fingers would act as a patlw..:ay by \..r hich the capaciLOr could discharge, inflicting an electric shock. The degrec of shock would depend on the capacitance and voltagc applied to the capacitor. When' high voltages a.nr/.lruge qua II lilies of dwrge are pr('sen.I as J
in thl' !Jown' SIlIJI}I)' of{[ it'levis-ion .'lei, such a shoch can he falal.
Capacitors store electrical energy. and that energy is the same as the work required to movc charge onto the plates. I r a capacitor is initially uncharged (both plates are neutral) so that the plates are at the same potential, very little work is required to transfer a small amount of charge LlQ from one plate to the other. Once this charge has been transferred, however. a small potential difference ilV = 6..QlC appears between the plates. so work must he done to transfer additional charge against this potel1l.ial difference. From Equation 16.6, if the potential difference at any instant during the chargin~ process is ill~ the work LHVrequircd to move marc charge LlQ through this potential difference is givcn by
ilW=ilVilQ \Ve know thal .6.V= Q/Cfor a capacitor that has a LOtal chargc of Q. Therefore, a plot of voltage versus total charge gives a straight line with a slope of I/e, as shown in Figure 16.23. The work Lll-V, for a particular ilV is the area of the blue rectangle. Adding up all the rectangles gives an approximation of the total \\lork needed to fill the capacitor. In the limit as LlQ is taken to be inllnilcsimally small, the total work needed to charge the capacitor to a final charge Q and voltage LlV is the area under Lhe linc. This isjllst Lhe area ora triangk. one-half the base times the heighl, so it follm\ls that
al'
[16.16] As prcviollsly slaled, Wis also the energy stored in the CilpaciLor. From the definition of capacilance, '.. . e have Q = C ~.v; hence, we can express the energy stored three different ways: Energy stored = tQ il V = tC (i'l I')' = Q' . . 2C
Q
[16.17]
For example, the amount of energy slored in a 5.0-j.LF capacitor ,.. . hen it is con-
nected across a 120-V battery is Energy stored = :C(i'lV)' =
aQ FIGURE 16.23
Hs.o
X 10 6r)(120V)' = ~.(; X 10- 2
J
fn practice, there is a limit to the maximulll encrg-y (or charge) thal can be stored ;n a capacitor. At some point, the Coulomb forces between the char~es on t.he plates become so strong thaL electronsjulllp across the gap, discharging thc capacitor. For this reason, capacitors are llsually labeled with a maximum operating voltage. (This physical fact can actually be exploited to yield a circuit wilh a regularly blinking light).
t\ plOl of\Ohork n.:quircd lo charge 1he capacitor to a fin"ll ch;:lJ"ge I I I Q is the area IInder tht· slr;ligh1 tine, whidl cfjltals Q A~/2.
Chapter 16
556
Electrical Energy and Capacitance
APPLICATION
II
Defibcillatars
Large capacitors can slore enough electrical energy to cause severe burns or even death iF they are discharged so that the flow o[ charge can pass through the hearL. Under the proper conditions, hmvevcr, they can be used to susLain life by slopping cardiac fibrillation in heart attack victims. \'\Then fibrillation occurs, the heart produces a rapid, ilTcgular pauenl of beats. A fast discharge of eleClrical
energy through the heart can relurn the organ to its normal betH pattern. Emergency medical teams usc port.able ddibriliaLOfs that contain bauerics capable or charging a capacitor to a high voltage. (The circuitry aCluall)' permits the capacitor to be charged to a much higher voltage than the baner)'.) fn this case and others (camera nash units and lasers used lor fusion experiments), capacitors serve as energy reservoirs that Gill be slowly charged and then quickly discharged to provide large amounts of energy in ,l short pulse. The stored electrical energy is released through the hean by conducting electrodes, called paddles, placed on both sides of the victim's chest. The paramedics must wait between applications of electrical energy because of the time it takes the capaciLors to become fully charged. The high voltage on the capacitor can be obtained from a low-voltage battery in a portable machine through lhe phenomenon of electromagnetic induction, l:O be studied in Chapter 20.
EXAMPLE 16.10 Goal
Apply energy and power concepts to a capacitor.
Problem A fully charged defibrillator contains 1.20 kJ o[ energy stored in a 1.10 X ]0- 4 F capacitor. In a discharge through a patient, 6.00 X -10~.1 of electrical energy are delivered in 2.50 ms. (a) Find the voltage needed to store 1.20 ~r in the unit. (b) ',Vhat average power is delivered to the patient? Strategy Because we know the energy stored and the capacitance, we can use Equation 16.17 to I1nd the required voltage in pan (a). For part (b), dividing the energy delivered by the time gives the average pO\..·er.
,..
.
• • • • • • ••• , • • • • • • • • • • • • •""'1
Solution (a) Find the voltage needed to store L.20 kJ in the unit. Solye Equation 16.17 for L\V:
Energy stored = ~C Li. V 2 I'>v=
2 X (energy stored) C
2(1.20 X lO"J) 1.l0X10"F
4.67 X 10' V (b) What average power is delivered to the patient? Divide the energy delivered by the time:
2P av
energy delivered
= --=----
1'>1
6.00X10'J 2.50 X ] 0 3,
2.40 X 10"W
.....
.............. , .. the study of Re circuits in Chap-
~
Remarks The power delivered by a draining capacitor isn't COl1slanL as we'lll1nd in ter 18. For thal reason, we were able to nnd only an average power. Capacitors are necessary in defibrillators because they can deliver energy far more quickly than batteries. Batteries provide current through relatively slow chemical reactions, whereas capacitors can quickly release charge that has already been produced and stored. QUESTION 16.10 If the voltage across the capacitor ,,,'ere doubled, would the energy stored be (a) halved, (b) doubled, or (c) quadrupled?
16.10
Capacitors with Dielectrics
557
EXERCISE 16.10 (a) Find the encrg·ycOIuained in a 2.50 X ]0-:' F parallel-plate capacitor ifit holds 1.75 X 10 3 Cofcharge. (b) \r\'hat's the.: voltage between Ihc plates? (c) \Vhat new voltage will result in a doubling of the sLOred energy? Answers
(a) 6.13 X 10-2 J
(b) 70.0 V
APPLYING PHYSICS 16.1
(cJ 99.0 \'
MAXIMUM ENERGY DESIGN
Hmo\! should three capacitors and two batteries be connected so that the capaciLOrs will sLOre the maxiIllUIll possible energy?
potential difference, so we would like to maximize each of these fJuantities. I f the three capacitors are connected in parallel, their capacitances add, and if the batteries are in series, their potential differences, similarly, also add together.
Explanation The eneq{Y stored in the capacitor is proportiollal to the capacitance and the square uf the
QUICK QUIZ 16.9 A parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance farther apart. Do the foIIO\-\'ing quantities increase, decrease, or stay the same? (a) C (b) Q (c) I.ibetween the plates (d) ,,11 (e) energy stored in the capacitor
16.10
CAPACITORS WITH DIELECTRICS
A dielectric is an insulating material, such as rubber, plastic, or \.. . axed paper. \'\Ihcn a dielectric is inserted between the plates of a capaciLOr, the capacitance increases. fr the dielectric completel), fills the space between the plates, the capacitance is multiplied by the factor K, called the dielectric constant. The following experiment illustrat.es 1 he eHcct 01" a dielectric in a capacitor. Consider a parallel-plate capacitor uf charge Qo and capacitance Co in the absence or a dielectric. The potential diHerence across the capacitor plates can be measured, and is given by a\~) = Qo/Cil (Fig. 16.24a). Because the capacitor is not COI1necLed La an external circuit, thcre is no pathway for charge to leave or be added lO the plates. If a dielectric is nuw inserted between the plales as in Figure 16.24b, the voltage across the plates is Teduc('(/' by the !"acLOr K to the value
"\\.
"11= -
K
Dit:l~etri(
/
+
+
6\~)
~
(a)
(J \
,It'
(h)
FIGURE 16.24 (3) With air between the plates, the vollage across the capacitor is .1.1'0' the capacil;lncc is e,), and the charge is Q(I. (b) \Vith a dielectric hetween the plates. lhe {·harl4c n':lllains at '20. bllltlH: vo!laj.\·c ,llld capacit,lI)cc change.
558
Chapter 16
Electrical Energy and Capacitance
Because K > 1, .1Vis less than 6..\10 _ Because the charge Qo on the capacilOrdoesll't change, we conclude that. the capacitance in the presence of the dielectric must change to the value
or [16.18]
According lO this result, the capacitance is mullijJlied by the [aclor K when the dielectric fills the region bcnveen the plates. For a parallel-plate capacilOr, where the capacitance in the absence of a dielectric is Co = EnA/d, we can express (he capacitance in the presence ofa dielectric as C=
/~. I
"" I
··,.,··, , .. W .'
I
T··
...
........
FIGURE 16.25 Dielectric breakdO\\ II in air. Spark!> an: produn:d when a large "It('mating voltagr: is applied ,tCI"U;"S tIll.: \\'irc~ by a hig-h\"ohage induclioll coil power sllpp1r.
t\
[16.19]
KEO-
rl
From this result, it appears that the capacitance could be made very large by decreasing d, the separaLioll bCl\veen the plates. In practice the lowest value of d is limited by the electric discharge that can occur through the dielectric material separating the plates. For any given plate separation, there is a maximum electric field that can be produced in the dielectric before it breaks down and begins La conduct. This maxilllulll electric field is called lhe dielectric strength, and for air it.s value is about 3 X lOll Vjm. Most insulating mat.erials have dielectric strengths greater than that of air, as indicated by the yalues listed in Table 16.1. Figure 16.25 sho\Vs an instance of dielectric breakdown in air. Commercial capacimrs are ortcn n1ade by using mctal foil intcrlaced with thin sheets of paraffin-impregnated paper or Mylar®, which serves as the dielectric material. These alternat.e layers of metal foil and dielectric arc rolled into a small cylinder (Fig_ 16.26a). One type of a high-voltage capacitor consists of a number of interwoven metal plates immersed in silicone oil (fig. 16.26b). Small capacitors are often constructed [rom ceramic materials_ Variable capacitors (typically 10 pF t.o 500 pF) usually consist or two inten\loven sets of metal plates, one fixed and the other movable, with air as the dielectric. An electrolytic capacitor (Fig. 16.26c) is often used to store large amounts of charge at relatively low voltages. 1L consists of a metal foil in contact with an e1ec·
TABLE 16.1 Dielectric Constants and Dielectric Strengths of Various Materials at Room Temperature Dielect.-ic Dielectric Constant K Strength (Vim) Material Air
1.00059
Bakclilc®
24 X 106
Neoprcne rubber
4.9 3.78 6.7
Nylon
3.4
14 X 101;
Paper
3.7 2.;)6
J() X lOr,
24 X 10'
5.6 2.;)
15 X
Fused quartz
Polystyrene Pyrex® glass Silicone oil
Strontium l.itanate Teflon® Vacuum \-V;llcr
233 2.1 1.00000
80
3 X 10'
8 X 10' 12 X 10'
1'1 X lO" IOtl
8X lOt, 60 X IO€>
16,10
""0
•
I
559
FIGURE 16.26 I hret· colllllleiTial tapadtHI fiesig'n~. (a) A tubular c;IP~1( itor whCJ\c plate..
( :;lse
Me.:tal foil
Capacitors with Dielectrics
COlllaCl."
\It,t.tlli(' foil + uxide Ia\ef Papcr (b)
(aj
(t")
trolyte-a solution that conducts charge by virtue.: of the motion of the ions contained in it. \\'hen a \'oltage is applied beL\\'Cen the foil and [he electrolyte, a thin la)'er of metal oxide (an insulaLOr) is formed on the foil, and this layer scn·cs as 1he didcctric. Enormous capacitances can be attained because the dielectric layer is very thin, Figure 16_27 shows a variet)' of commercially <'l\'ailablc capacitors, Variable capaciLOrs are used in radios to adjust the frequency. ""hen electrolytic capacilOrs are used in circuits, the polarity (thc plus and minlls signs on the dcvice) musl be observed, If the polarity of the applied voltage is opposite that intended, the oxide layer will he rCIllO\'cd and the capacitor will conduct rather than store charge, Further, reversing the polarity can result in such a large current thalthe capacitor lIlay cithl'r burn or produce steam and explode.
•
I
(a) A wllcnioo of in.1 variety oj applicatiom_ (b) :\ \'ariabk capacitor. \Vhcll OIiC "l't 01 metal plalcs i.. rOlaled so a:-. to ]iI' bell"e.:cll ,I fixed st't ofplall-:-., the t-aparit:\Ilcc or the dl'\ irc chan~l.:s, u\pa("itor,~ Ilsed
!
"~ ]
,.t
~
~
~
" ~
~
j
'",
'"
FIGURE 16.27
.
ii.
(al
APPLYING PHYSICS 16.2
"
(h)
STUD FINDERS
Iryoll ha\'e ever tried to hang a picture on a wall securely, you knO\\- that it can be difficult to locate ~I \\louden stud in which LO anchor )'our nail or screw. The principles discussed in this section ('an be lI')cd to detect a stud electronically. The primary clement or an electronic stud finder is a capacitor wit h it!\ plates arranged side by side instead of racinJt on(" another, as in Figure 16_28. How does this de\'icc work?
Capacitnl Illau.::-.
0;;1 lid
Explanation As the detector is moved alollg' a \\all, its capacitance changes when it passes across a swd because the dielectric constant of lhe material "bel ween" the plates changes. The change in capacilance can be used to cause a lightLO cOllle un, signaling I he presence of the stud,
' " fimkr \\;.lIb(l,lnl (al
Stud
(b)
FIGURE 16.28 (Apph il\~ I'ln,,,ics lti_2J .\ stud finder, (a) The.: materiab helwl"en lhe pl.lll'\ of lhe capacilOr :11'(' lhe drywall and lht, air behind it. (I» Thl" 1Il,\lt'l"iOIls hecome fir}'''';111 :\I1d \~(Jnd when lhe detcctor II1m·e~ .lCross a ~llld in 1Il<' \\,;,11, T!Il': c1ldngc in thedit'lf"tlric COII"t<1Il1 C<1U"es iI "ignallig"lllIO ilhllllin;tlt'.
560
Chopter 16
Electrical Energy and Capacitance
QUICK QUIZ 16.10 A fully charged parallel-plate capacitor remains connCCled to a bauery while a dielectric is slid between the plates. Do the following quantities increase, decrease, or st.ay the same? (a) G' (b) Q (e) E bcnveen the plates (d) .1V (e) energy stored in the capacitor
EXAMPLE 16.11 Goal
A PaRer. Filled Capacitor
Ca1culme fundamental physical properties of a parallel-plate capacitor with a dielectric.
Problem A parallel-plate capacitor has plates 2.0 em by 3.0 em. The plates are separated by a LO-l11m thickness of paper. Find (a) the capacicallce of this device and (b) the maximum charge that can be placed 011 the capacitor. (e) After the fully charged capaciLOr is disconnected from tIle baltery, the dielectric is subsequently removed. Find the ne\v electric field across the capacitor. Does the capacitor discharge? Strategy For pan (a), obtain the dielectric conSlant for p(lper [rom Table ]6.1 and substiwle, with other given quantities, into Equation 16.19. For part (b), not.e that Table 16.1 also gives the dieleClric strength of paper, which is the maximum electric field that can be applied before electrical breakdown occurs. Use Equation 16.3, 1.\1= Ed, to obtain the maximum voltage and substitute into the basic capacitance equation. For part (c), remember that disconnect.ing the bat.tery traps the extr~l charge on the plmes, '.vhich must remain even after the die]ecn·jc is removed. Find the charge density on the plates and use Gauss's law to find the new' electric field bet\veen the plates.
,..
"
.
.
,
,
~
Solution (a) Find the capacitance of this device. SubstilLlte into Equation .I6.J9:
C=
A KEn-
r1
4Ill')
)(6.0 X 10- , = 3.7 ( 8.85 X 10- 12 -C' -0. N 'm1.0 X 10 . = 2.0 X 10
II
111
F
(b) Find the maximum charge that call be placed on the capacitor. Calculate the maximum applied voltage, using the dielectric strength or paper, EIl\
Solve the basic capacitance equmion for Qillax and substitute d~'I
Q",,,,
= C "'I~"" = (2.0
X 10- 11 F)(1.6 X 10' V)
0.32/-,C
(c) Suppose the rully charged capacitor is disconnected from the battery and the dielectric is subsequently removed. Find the new electric fIeld between the plates of the capacitor. Does the capacitor discharge? 3.2 X 10- 7 C
Compute the charge density 011 the plates:
Calculate the electric field from the charge density:
6.0 X 10 "' m~
E
= 5.3
= '!.. = __5_.3_X_l_0-c-;-4_Ci l_ITI-;;'_ EO
8.85
X
10
12
C2 /m 2 N
X
10-" C/ITI'
6.0
X ]07
N/C
Because the electric field withoul the dielectric exceeds the value of the dielectric strength of air, the capacitor discharges across the gap. ~
Remarks Dielectrics allow K times as much charge to be swred on a capacitor for a given ,·ollage. They also allow an increase in the applied voltage by increasing the threshold or electrical breakdown. QUESTION 16.11
Subsequent La pan (c), the capacil.Or is reconnected (b) smaller than, or (c) the same as found in pan (b)?
1.0
the baLtery. Is the charge on the plates (a) larger than,
EXERCISE 16.11
A parallel~plate capacitor has plate area of' 2.50 X 10-:1 m:! and distance between the plates of 2.00 111m. (a) Find the maximum charge that can be placed on the capacitor ifair is between the plates. (b) Find the maximum charge if the ai r is replaced by polystyrene. (a) 7 X 10-8 C
Answers
(b)].4 X 10-" C
o
An Atomic Description of Dielectrics The explanation of why a dielectric increases the cap(lcitance of a capacitor is based on an awmic description of the material, which in turn involves a property of some molecules called polarization. A molecule is said 1.0 be polarized when there is a separation between the average positions of its negative charge and its positive charge. In some molecules, such as water, this condition is ahvays present. To see why, consider the geometry of a water molecule (Fig. 16.29). The molecule is arranged so that the negative oxygen atom is bonded to the positively charged hydrogen atoms with a 105 angle bet\veen the two bonds. The center of negative charge is al. the oxygcn atom, and the center of positive charge lies at a point midway along the line joining the hydrogen atoms (point x in the diagram). Materials composed of molecules that are permanently polarized in this '....ay have large dielectric constants. and indeed, Table 16.1 shows that the dielectric constant of water is large (K = 80) compared with other common sllb~ stances. A symmetric molecule (Fig. 16.30£1) can ha\'e no permanent polarization, but a polarization can be induced in it by an ext.ernal electric field. A field directed to the len, as in Figure 16.30b, v..'Quld cause the center of positive charge to shin to the left from its initial position and the center or negative charge to shift to the right. This il1duad jJolarization is the effect that predominates in most materials used as dielectrics in capacitors. To understand why the polarization of a dicleclric can affect capacitance, con~ sider the slab of dielectric shown in Figure 16.31. Before placing the slab between the plates of the capacitor, the polar molecules are randomly oriented (Fig. 16.318). The polar molecules are dipoles, and each creates a dipole electric Held, but because or their random orientation, this AcId avcrages to zero. After insertion of the dielectric slab into the electric field Eo between the plates (Fig. 16.31b), the positive plate attracts the negative ends of the dipoles and the negative plate attracts the positive ends of the dipoles. These forces exert a torque on the molecules making up the dielectric, reorienting them so that on average the negative pole is more inclined to\vard the positi"e plate and the positive pole is more aligned toward the negative plate. The positive and negative charges in the 0
" x
,
-• -
+
+ -
., +
I
-
"
x
EI) (a)
+
(b)
+
Ell
:I
Eind (cj
-
H
H
+
+
FIGURE 16.29 The water molecule, H:!O. has a permanent polarization result ing from ilS bem gCOlllclIy The !,oim labeled xis lhe cemer of pusili,t' cllarg(;.
-
+
--
--
+-
(a)
• -
+ -
FIGURE 16.30 (a) A s\'llIlllctric molecule has no perlllnnent polarization. (b) All external electric field induces a polarization in the molecule.
FIGURE 16.31 (a) In dIe absence Ufalll:xternal declric field. polar molecules are randomly orir.:l1led. (h) \Vhen an external elcctric field is applied, tlie molecules partially align ,dth the field. (c) The charged edges of the dielectric call be modeled as an additional pair of parallel plates cSlablishing an electricjield E'''d in the direction opposite Eo.
562
Chapter 16
Electrical Energy and Capacitance
middle still cancel each ol.her, but there is a net accumulation of negative charge in the dielectric next [Q the positive plate and a net accumulation of positive charge next to the negative plate. This configuration can be modeled as an additional pair of charged plates, as in Figure 16.31c, creating an induced electric field E ilUl that pal"tly cancels the o,-iginal electric fleld Eo- If the battery is not connected ",:hen the dielectric is insened, the potential difference .:lVo across the plates is reduced LO Li1~/K. If the capacitor is still connected to the bauery, however, the negative poles push more electrons off the positive plate, making illllore positive. Meanwhile, tile positive poles £lllract more electrons onto the negative plate. This situation continues until lhe potential difference across the battery reaches its originalmagnitude, equal to the potential gain across the battery. The nct efree[ is an increase in the amount of charge stored on the capacitor. Because the plates can store more charge for a given volLage, it follows from C = QLl\lthat the capacitance must increase. ~
~
QUICK QUIZ 16.11 Consider a parallel-phue capacitor with a dieleClric material bet""een the plates. If the temperature of the dielectric increases, does the capacitance (a) decrease, (b) increase, or (c) remain thc same?
SUMMARY 16.1 Potential Difference and Electric Potential The change in the electric pOlerHial enel-gy of a system consisting of an o~ject or charge q mov0g through a displacement Dox in a constant eleC(ric field E is given by
These equations can be used in the solution or consen·alion of energy problems and in the work-energy theorem.
[16.1]
16.4 Equipotential Surfaces Every point on the surface or a charged conductor in electros[~lIic equilibrium is at the same potential. Funhcr, the potential is constant everywhere inside the conductor and equals its v£llue on the surf~lce. The electron volt is defined as the energy that an electron (or proton) gains when ,1cceleralcd through a potential difference of I V. The conversion between electron volts and joules is
where E:r. is the componcnt of the electric field in the xdirection and UX = xf - X,. The difference in electric potential between two poi illS A and B is
.6. F = FB
-
b.PE
.-' = -q-
\/
[16.2]
where fiPE is the chauw'in electrical potential energy as q moves bel ween A and B. The units of potential difference arejoulcs per coulomb, or volts; lJ/C = I. V. The electric potential differ~nce between two points A and 13 in a uniform electric field E is
16.3
a charge
Potentials and Charged Conductors
1 eV
~
1.60 X
10~"
C- V
~
1.60 X
10~'''J
[16.7]
An)' surface on which the potcmial is the same at every point is called an equipotential surface. The electric field is always oriented perpendicular to an equipotemial surface.
[16.3]
where ax = x - x, is lhe displacement between A and f Band E:r. is the x-component of the electric field in that region.
16.2 Electric Potential and Potential Energy Due to Point Charges The electric potential due to a poi nl charge q at. distance ·r from the poi1ll. charge is V= Ii , 'i
,.
[16.4]
The electric potential energy of a pair of point. charges separated by distance ris [16.5]
16.6
Capacitance
A cap~lcitor consists of two meral plate:; with charges that
are equal in magniwde but opposite in sign. The capacitance C or any capacitor is the ratio of the magnitude of the charge Qon either plate to the magnitude o[potentia! difference a\lbetween them:
c=
[l6.8J
Capacitance has the units coulombs per volt, or farads: 1 C/V = 1 F.
16.7 The Parallel-Plate Capacitor The capacitance of l''''O parallel metal plates oLuea A separaled by distance d is C=
A EO-
,{
[16.9]
Multiple-Choice Questions
where EO = 8.85 X ]0-['2 C~/N' m 2 is a constant called the permittivity of free space.
16.8
same voltage drop, and that series capacitors h ..w e the same charge.
16.9
Combinations of Capacitors
The equivalent capacitance of a parallel combination of
capacitors is [16.12] If two or more capacitors are cOllneCl.cd in series, the equivalent capacitance of the series combination is
Energy Stored in a Charged Capacitor
Three equivalem expr~ssions f'or stored in a charged capacitor are Energy stored .
16.10 [l6.15J Problems involving a combination of capacitors can be solved by applying Equations 16.12 and 16.13 repeatedly to a circuit diagram, simplifying it as much as possible. ThIS step is followed by working backwards to the original diag-ram, Clpplying C = Q/Li\~ that parallel capacilors ha"e the
563
r~klll(liing
1 1 ( = ;;Q .::i \1 = ;;C .6. V)"-
= -Q'
2C
the energy
[16.17]
Capacitors with Dielectrics
\Vhen a nonconducting material, called a dielectric, is placed between the plates of a capacitor, the capacitance is multiplied by the factor K, which is called the dielectric constant, a property of the dielectric material. The capacitance ora parallel-plate capacitor filled with a dielectric is [16.19]
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS 1. A proton is released at. the origin in a constant electric field of 850 N/C acting in the positive x-direction. Find the change in the elecrric potential energy associated ,,,'ith the proton after it travels to x = 2.5 Ill. (a) 3.4 X 10- 16 J (b) -3.4 X 10- 16 J (e) 2.5 X 10-16 J (d) -2.5 X 10- 16 J (e) -1.6 X 10- 1<1 J
2. An electron in a TV picLUre tube is accelerated through a potential difference of 1.0 X 10-1 V before it hits the screen. What is t.he kinetic energy of lhe electron in electron volts? (a) l.0 X to" eV (b) 1.6 X 10-15 cV (e) 1.6 X 10-" eV (el) 6.25 X 10" eV (c) 1.6 X 10- 1<1 cV 3. A helium nucleus (charge = 2e, mass = 6.63 X LO-~7 kg) I.raveling at a speed of' 6.20 X 10:; m/s enters an electric fleld, traveling from point ®, at. a potcmial of 1.50 X 10 3 V, to point ®, at 4.00 X 10 3 V. '·Vhat is its speed at point ®? (a) 7.91 X 10' Ill!' (b) 3.78 X 10·' m!, (e) 2.13 X 10' m!s (el) 2.52 X 10" Ill!' (c) 3.01 X 10' Ill!s 4. The electric potential at x = 3.0111 is 120 V, and the electric potential at x = 5.0 III is 190 V. \'\That is the electric Aeld in this region, assuming it's constant? (a) 140 N/C (b) -140 N!e (e) 35 N!e (d) -35 I'!C (e) 75 WC 5. An electronics technician wishes to construct a parallel-plate capacitor llsing rutile (K = l.00 X 10 2 ) as lhe dielectric. If the cross-sectional area of the plates is 1.0 cm 2 , ,vhat is the capacitance if the rutile t.hickness is 1.0 111m? (
7, If three unequal capacitors, initially uncharged, are connected in series across a batter)', which of the following statements is true? (a) The equivalent capacitance is greater than an)' of the individual capacitances. (b) The largest voltage appears across the capacito!· ''''ith the smallest. capacitance. (c) The largest voltage appears across the capacitor with the largest capacitance. (d) The capacitor with the largest capacitance has t he greatest charge. (e) The capacitor with the smallesl capacitance has the smallest charge. 8. A parallel-plate capacitor is connected to a bauery. \,Vhat happens if the plate separation is doubled while the capacitor remains connected to the battery? (a) The stored energy remains the same. (b) The stored energy is doubled. (c) The stored energy decreases by a factor of 2. (d) The stored energy decreases by a factor of 4. (e) The stored energy increases by a factor of 4. 9. A parallel-plate capacitor filled with air carries a charge Q. The battery is disconnected, and a slab of material with dielectric constant K = 2 is inserted between the plates. Which of the following statements is correct? (a) The voltage across the capacitor decreases by a factor 0("2. (b) The voltage across the capacitor is doubled. (c) The charge on the plates is doubled. (d) The charge on the plates decreases by a factor of 2. (e) The electric field is doubled. 10. After a parallel-phue capacitor is charged by a battery, it is disconnected from the battery and its plate separation is increased. Which of the following st.at.ements is correct? (a) The energy stored in the capacitor decreases. (b) The energy stored in the capacitor increases. (c) The electric field between the plates decreases. (d) The potential difference between the plates decreases. (e) The charge on the plates decreases.
564
Chapter 16
Electrical Energy and Capacitance
11. A battery is attached to several different capacitors connected in parallel. 'Which of the following statements is true? (a) All the capacitors have (he same charge, and the equivalent capacitance is greater than the capacj~ Lance of any of the capacitors in the g'roup. (b) The capacitor with the largest capacitance carries [he smallest charge. (c) The potential difference across each capacitor is the same, and the equivalent capacitance is greater than any of the capacitors in the group. (d) The capaciwf with the smallest capacitance carries the largest charge. (e) The potential differences across the capacitors
12. A battery is auached across several different capacitors connected in series. Which orthe following statements are true? (a) All the capacitors have the same charge, and the equivalent capacitance is less than the capacitance of any of the individual capacitors in the group. (b) All thc capacitors have the same charge. and the equivalelll capacitance is greater Lhan any of the indi~ vidual capacitors in the group. (c) The capacitor with the largest capaciLallce carries the largest charge. (c) The pQ[emial difference across each capacitor must be the samc. (d) The largest pmcntial difference appears across the capacitor having the largesL capacitance. (e) The largest potemial difference appears across the capaciLOr with the smallest capaciLance.
CONCEPTUAL QUESTIONS 1. (a) Describe the motion of a proton after it is released from rest in a uniform electric field. (b) Describe the changes (if' any) in its kinetic cncrg'y and the electric potential energy associated with the proton.
2. Describe how yOll can increase the maximum operating voltage ofa parallel-plate capacitor for a flxed plate separation. 3. A parallel~plale capacitor is charged by a bauer)', and the bauery is then disconnected frolll the capacitor. Because the charges on the capacitor plates are opposite in sign, the)' attract each other. I-!ence, it takes positive work La increase the plate separation. Show that the external work done \·,;hel1 the plate separation is increased leads to an increase in the energy slored in rhe capacitor. 4. Distinguish between electric potential and electrical potential energy. 5. Suppose yOLl are sitting in a car and a 20-kV pO'wcr line drops across the car. Should you stay in the car or get out? The power line potemial is 20 kV cornparcd LO the pOlel1l ial or the ground. 6. ,.vh)' is it important to avoid sharp edges or points on conductors used in high-voltage equiplllcm? 7. Explain why, under static conditions, all points in a conductor must be at the samc electric potential.
~
The Problems for this chapLer may be 011 line at WebAssign. 1,2,3 = slraigblforwarcl. interlllediate, rhalknging
m
m!I
WebAssign
assign eo
= denotes guided
problem
= denotcs enhanced conlcnt problem
II = hiomedical application o = denotes fuJI solUlitlll available in Student Solulion\ Ma/wall Sind) G'uidr
8. If yOH are gi\'en three differem capacilOrs CI , C2 , and Cj , how many different. combinations of capacitance call yOll produce, llsing all capacitors in your circuits? 9. "Vhy is it dangerous to louch the terminals of a highvoltage capacitor evcn after the voltage source that charged the battery is disconnected from Lhe capacitor? What can be done to make the capaciwr safe to handle after the voltage source has been removed? 10. The plates or a capacitor are connected to a battery. "Vhal happens to the charge on the plates if the connecting wires are removed from the bauery? What hap· pens to the charge if the wires are removed from the bauery and connected LO each other? 11. Can e1ect.ric field lines ever cross? Why or why not? Can equipotentials ever cross? Why or why not? 12. Is it always possible to reduce a combination of capacitors to one equivalent capacitor wiLh lhe rules developed in t.his chapter? Explain. 13. I r you were asked to design a capaciLOr ror which a small size and a large capacitance v,,'ere required, what factors would be important in your design? 14. Explain ,"\ihy a dielectric increases the maximum opel'·
SECTION 16.1 POTENTIAL DIFFERENCE AND ELECTRIC POTENTIAL 1. A uniform electric field of magniLude 375 N/C poillling
in the positive x-direction aets on an eleCtron, which is ini· Lially al rest.. rU1.er lhe electron has moved 3.20 cm, what is (a) the work done by the field on Lhe electron, (b) the change in potential energy associated wilh the electron, and (c) the velocity of tile clccu'on?
Problems 2. A uniform electric field of magnitude 32i N/C is directed along the +)'-axis. A 5AO-p,C charge 11lOvCS from the origin to the point (x. y) = (-15.0 cm, -32.0 cm). (a) \Vhat is the changc in the potential cnergy associmed with tllis charge? (b) Through what potcntial difference did the charge move?
mil A
potential difference of 90 mV exists between the inner and outer surfaces of a eell membrane. The inner surface is negative relat ive to t he outer surf
4. An ion accelerated through a potential difference of60.0 V has its potential energy decreascd by 1.92 X 10-17 .J. Calcul(lte the charge on the ion. 5. The potential difference bCl\\'cen the accelcrating pI ales of a TV set is
7. Oppositely charged parallel plates are separated by 5.:\3 mm. A potcnlial difference of 600 V exists between thc plates. (a) What is the magnitude of the electric ficld between the plates? (b) What is the magnitude of Ihe force on an electron betwecn the plates? (c) How milch work must bc done 011 thc clenron La move it LO the negative plate if it is initially positioned 2.90 mm from lhe positive plate?
8.
am
(a) Find lhe potential difference d\l~ required to
SLOp
an electron (callcd a "sLOppingpotcntial") moving with an initial speed of2.85 X t0 7 m/s. (b) \·Vould a proton lraveling at the same specd rcquire a grcater or lesser magniwc1e pOlcntial difference? Explain. (c) Find a symbolic expression for the ratio of thc proton stopping potential and thc elcctron stopping pOlcntial, d\~/dv,.. The answer should be in terms of the proton mass J1l!J and electron mass m~. 9.
~r,.0 /LC is connected to a spring for which k = 78.0 N/m. The block lies 011 a frictionless, horizontal surface and is immersed ill a uniform electric field or magnitude £ = 4.86 X IOl N/C directed as shown in Figure PI6.9. Tfthe block is released from rest when the spring is unstrctched (x = 0), (a) by what maximum distance does the block move from its initial position? (b) Find the subsequent equilibrium position of the block and the amplitude of its motion. (c) Using conservation of energ}'. find a symbolic rela-
tionship giving the potential diffcrence bctween its initial position and the point of maximum cxtension in terms of" the spring constant Jr. the amplitude A. and the charge Q.
[IQJOn planct Tchar. the fret-fall accc1eralivn is the Silllle as that on the Earth, but there is also a strong downward eleClric field Lhat is uniform close LO the planct's surface. A 2.00-kg ball having a charge of5.00 /LC is thrown up""'ard at a speed of 20.1 m/s. It hits the gTolind after all interval of 4.10 s. What is the potential difference between the starting poilll ;-md the top point orlhe trajecLOry?
SECTION 16.2 ELECTRIC POTENTIAL AND POTENTIAL ENERGY DUE TO POINT CHARGES SECTION 16.3 POTENTIALS AND CHARGED CONDUCTORS SECTION 16.4 EQUIPOTENTIAL SURFACES 11.
am An electron is 'H the origin. (a)
Calculate tile electric potential \fA at point A, x = 0.250 em. (b) Calculale thc electric potcntial \lJia!. point B, x = 0.750 cm. Whal is the potential difference \'8 - lfA? (e)'.·Would a negatively charged particle placed at point A necessarily go through this same potential difference upon reaching point B? Explain.
12. Two poinl ch;1rges arc on the )'~axis. A 4.50-p,C chargc is locatcd at)' = L.25 em. and a -2.24-f.LC charge is locatcd at y = -1.80 cm. Find the Lotal electric potential at (a) the origin and (b) the point having coordinates (1.50 cm, 0). 113.I(a) Find the electric potential, taking zero at infinity, at the upper right corner (the corner without a ch;-lrge) of the rectangle in Figure P16.13. (b) Repeat iflhe 2.00-.uC charge is repl;lCed with a charge of -2.00 /Le. B.OO pC
~.oo
mJ A 74.0-g block carrying a charge Q =
Ill,
Q
-
..J-_.....
FIGURE P16.9
600 em
em
..
\. 2.00 pC FIGURE P16.13
·1.00.uC
(Problems 13 and 14)
14. Three charges are siLuatcd at corners ora rectangle as in Figure PI6.13. How much energy would be expended in moving the 8.00-ILC charge to infinity? 15.
am Two
point charges QI = +5.00 ne and Q2 = -3.00 nC are separatcd by 35.0 cm. (a) Whal is the electric potential at a POilll mirlwa}' between the chat'ges? (b) \"'hal is thc potential energy of the pair of charges? What is the sign irIca nce of the 1gehraic sign of rour answer?
,I
Ii
L
565
~x
16. A point charge of 9.00 X IO-!} C is loc
Chapter 16
566
Electrical Energy and Capacitance
P16.23. How close does thc alpha panicle get to lhe gold nucleus before turning around? Assume the gold nucleus rcmains stationary. 24.
mil
Four point charges each having charge Qare located Ihe ('orner<; of
SECTION 16.6 CAPACITANCE FIGURE P16.17
SECTION 16.7 THE PARALLEL-PLATE CAPACITOR
t8. An electron stans from rest 3.00 em [rom the center of a uniformly charged sphere of radius 2.00 em. [[the sphere carries a total charge of 1.00 X 10-9 C, hO\... fast will the electron be moving v,,'hen it reaches the surface of the sphere? 19.
DI
A prown is located at the origin, and a second prolon is located on (he x-axis at x = 6.00 fm (1 fm = 10-1:; m). (a) Calculate the electric potential cn<.:rgy aSSOCi
~mJ A proLOn and an alpha particle (charge = 2e mass = 6.64- X 1O<>~ kg) arc initially at rest, separated by 4.00 X lO-lI; m. (a) If they are bmh released simultaneously, explain why you can'l find their velocities at infinity using only cOllScn'ation of cnergy. (b) What other conservation law call be applied in this case? (c) Find the speeds o[the proton and alpha particle, respectively, at infinity. J
21. A small spherical o~jcct carries a charge o[ 8.00 nCo At what distance from thc center of the object is the potential equal to 100 V? 50.0 V? 25.0 V? 1s the spacing of the equipotentials proportional to the change in voltage? 22.
mil
Starting with the definition of work, provc Lhatthe local electric field must be everywhere perpendicular to a surface having the same potemial at every pain!.
23. in Rlllherlord's famous scattering experimellls that led to the planctary model of the atom, alpha panicles (having charges of +2e and masses of 6.64 X 1.0-27 kg) ",,·ere fired toward a gold nuclcus with charge +7ge. An alpha particle, initially very far from the gold nucleus, is fired al 2.00 X 10 7 mls direClly toward the nucleus, as in Figure
v=o +
~d-I FIGURE P16.23
25. Considcr the Earth and a cloud layer 800 m above the planct to be the plates ora parallcl~platc capacitor. (a) r[ the cloud layer has an area of 1.0 km 2 = 1.0 X 10 6 m 2 , what is the capacit.<'l1lcc? (b) If an electric field strength greater than 3.0 X 10 6 N/C causes the air to break down and conduct charge (Iig'htning), what is the maximum charge the cloud can hold? 26. (a) When a 9.00-V bauery is connected to the plates of a capacit.or, it stores a charge of 27.0 /-Le. What is the value of the capacitance? (b) 1f the same capacitor is can nected to a l2.0-V bmtcry, what charge is slored? 27. An air-filled parallel-plate capacitor has plates of arca 2.30 Cl11 2 separated by 1.50 mm. The capacitor is connected to a 12.0-V baucry. (a) Find the value of its capaci~ tance. (b) What is the chargc on the capacitor? (c) '",,'hat is the magnitude of the uniform electric ficld between the plates? 28. (a) How much charge is on each plate ofa 4.00-JLF capacitor when It is connected to a 12.0-V baner}'? (b) If this samc capacitor is connectcd to a l.50-V battery, what charge is stored? 129.1A n air-fi lIed capacitor consists of two parallel plates, each ,,,,ith an area of 7.60 cm 2 and separated by a distance of l.80 mill. If a 20.0-V potential difference is applied to these plates, calculate (a) the electric Aeld betwcen the plates, (b) the capacitance, and (c) the charge on each platc. 30. A J-megabit computer mcmory chip contains many 60.0 X 1Q-1!"-F capacitors. Each capacitor has a plate area of 21.0 X 10- 12 m 2 • Determine the plate separation of such a capacitor. (Assume a parallel-plate configuration.) The diameter of an :ltorn is on the order of lO-IO m = 1 A. Express the plate separation in angstroms. 31.
mil
A parallel-plate Glpacitor with area 0.200 m 2 and plate separalion of 3.00 mill is connected LO a 6.00-V battery. (a) ,'\'hal is the capacitance? (b) Ho\'I.' much chargc is stored on the plates? (c) \Vhat is the eleCl.ric field betwcen the platcs? (d) Find thc magnitude of the charge density on cach platc. (e) \Vithout disconnecting the baHcry, the plates are moved farthcr apart. Qualilatively. what happells to each of the prcvious answcrs?
32. A small object with a mass of 350 mg carries a charge of 30.0 nC and is sllspcndcd by a thrc.:ad between the vertical plates of a parallel-plate C
Problems
SECTION 16.8 COMBINATIONS OF CAPACITORS 33. Given a 2.50-,uF capacitor, :l 6.25-,uF capacitor, and a 6.00-V battery. tind the charge 011 each capacitor if you connect them (a) in series across the bauery and (b) in
39. Filld Lhe charge
011
567
each of the capacitors ill Figure
PI6.~Cj.
n
.u. 8.011)/.0 4.00 ~F 1.00
IJarillld iicross Lht: balLet-yo
34. Find the equivalent capacitance of a 4.20-p,F capacitor and an 8.50-p.F C
series and (b) in p;lrallel.
)/F
5.00
FIGURE P16.39
135.IFind (a) the equivalent capacitance of the capaciwrs in
Figure PI6.35, (b) the charge on each capacitor. <mel (c) the ]JOlemial difference across each capaciLOr. 6.00 III"
""""f~:'~
f"OO""
9.00 V
FIGURE P16.35
36. Two capaciLOrs give an equivalent capacitance 01'9.00 I)F when connected in parallcl and an equivalent capacitance or 2.00 pF when connected in series. What is the capacilance of cach capacitor? 37. For thc system of capacitors shown in Figure Plfi.~H, find (a) t.he cquivcl1ent capacitance or the syst.em, (h) the charge 011 each capacitor, and (c) the potential dirferencc across each capacitor.
40. A 1O.0-/lF capaciwr is fully charged across a 12.0-V battcry. The capacitor is then disconncCled [rom thc ballery and connecterl across an initially uncharged capacilOr with capacitance C. Thc resulting voltage across each capacitor is 3.00 V. What is the valuc of Q 41. A 25.0-/J..r capacitor and a 40.0-,:.LF capacitor are charged by being connected across sepClnile 50.0-V batteries. (a) Determinc the resulting charge on each capacitor. (b) The capacitors are then disconnected from their batterics and connected 10 each other, with each negativc plate connected to the other positive plate. What is the final charge of' each capacitor, and what is the final potential difference across the 40.0-,:.LF capacitor? 42. (a) rind the equivalelll capacitance between points a and b for the group of capacitors connected as shown in Figure P16.42 if C! =: 5.00 J.LF, Co? =: 10.00 j.tr, and C3 = 2.00 .uF. (b) I r the potcntial between points a and b is 60.0 V, what charge is stored on C:,-?
90.U V FIGURE P16.37
38. _ Consider the combination of capacitors ill Figure PL6.38. (a) Find the cquivalcnL single GlpacitcHlcC of the two capacitors in series and redraw Lhe diagram (called diagram J) with this equivalcnL capacitance. (b) In diagram I fI nd thc eqllivalclll Glpacitance of the th ree cap
1 I I 1-
36.0 V
~.oo ~F ~.oo~.
-.l
J..:
FIGURE P16.38
I··LL 1.\ I.OO-,uF capacitor is charged by being connected auoss a 10.0-V bauer),. It is then disconnected from the bauer}' and connected across an uncharged 2.00-,uF capacitor. Determine the resulting charge on each capacitor. 44. Find the equivalent capacitance between points a and bin the combination ofcapacilOrs shown in Figure PI6.44.
4.0 pF
T
·o .uF
"
b
5.0 ,uF
6.0 JIF
FIGURE P16.44
r'11I11F T-.JH.OO .'IF
SECTION 16.9 ENERGY STORED IN A CHARGED CAPACITOR 45. A 12.0-V battery is connected to a .... 50-j.tF capacitor. How much cnergy is stored ill the capacitor?
568
Chapter 16
Electrical Energy and Capacitance
46. _ Two capacitors. C1 = 18.0 ,uF and C2 = 36.0 ,uF, arc connected in series. and a 12.0-V balter}' is connected across them. (a) Find the equivalent capacitance, and the energy contained in this eqllivalclll capacitor. (b) Find the energy stored in each individual capacitor. Show that the sum of these two energies is the same as the energy found in part (a). Willlhis equality ahvays be true, or docs it depend on the number of Cap(lcllOrs and their capaci£ances? (c) If the same capacitors were conncCled in parallel, whaL potential difference would be required across them so that. the combination Slores the same energy as in part (a)? Which capacitor slores more energy in this situation, C1 or C/ 147.1.-\ parallel-plate capacitor has capacitance 3.00 J.LF. (a) How much energy is stored in the capacitor if it is conllecled to a 6.00-V banery? (b) If the bau.ery is disconnened and the distance between the charged plates doubled, ,.. . hat is the energy stored? (c) The battery is subsequently reattached to the capacitor, but the plate separation relTtains as in pan (b). How much energy is slOred? (Answer each part in microjoules.) 48. A certain storm cloud has a potential difference of 1.00 X lOll V relative to a tree.lr. during a lightning storm. 50.0 C of charge is transferred through this potcntial difference and 1.00% of the energy is absorbed by thc tree. how much watcr (sap in the lrce) initially at 30.acC can be boiled away? Water has a speciric heal. of 4 186J/kg C. a boiling point of 100°C, and a heat of vaporization of 2.26 X lOti J/kg.
before the capacitOr is rolled up? (Use the parallel-platc formula. Adding a second strip of paper and rolling up the capaciror doubles its capacitance by allowing both sllrf'aces of each st rip of foil 10 store charge.) A model 01" a I-cd blood cell pOI-tra)'s. the cell as a shpcrical capacitor, a positively charged liquid sphere of surface arca A separated from the surrounding negatively charged rluid by a membrane of thickness I. Tiny electrodes illlroduced into lhe interior of the ccll show a potential difference of 100 mV across the membrane. The membrane's thickness is estimated to be 100 nm and has a dielectric con Slant of 5.00. (a) [f an average red blood cell has a mass 01" 1.00 X 10- 12 kg, estimate the volume of the cell and thus fmd its surface area. The density of blood is I 100 kg/Ill:". (b) Estimatc the capacitancc of the cell by assuming the membrane surfaces act as parallel plates. (c) Ca1culme the charge on the surface of thc membrane. How many electronic charges docs t.he surface charge reprcsent?
53. "
ADDITIONAL PROBLEMS
54. _ Three parallel-plate capacitors arc constructed, each having the same plate spacing d and with C1 having platc area AI' C 2 having area .4 2, and C3 having area A 3 • Show that the LOtal capacitance C of the three capacitors connected in parallel is the same as that ofa capacitor ha"ing plate spacing dand plate area A = AI + 1\2 + As·
Cl
SECTION 16.10 CAPACITORS WITH DIELECTRICS 49. _ The voltage across an air-filled parallel-plate capac ilOr is measured to be 85.0 V. When a dielecLl-ic is inserted and completely rills the space between the plates as in Figure 16.24, the \'oltage drops to 25.0 V. (a) \·Vhat is the dielectric constant of the inserted material? Can you identify the dielectric? (b) If the dielectric doesn't completely fill the space betwecn the plates, what could rOll concludc about the \'oltagc across the platcs? 50. A parallel-plate capacitor in air has a plate separation of 1.30 cm and a plate arca of 25.0 cm 2 • The plates are charged to a pOlential difference of 2.50 X 102 V and disconnected from the source. The capacilOr is then immersed in distilled wat.er. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference aftn immersion, and (c) the change in energy stored in the capacitor due LO immersion. Assume the distilled water is an insulator.
55. _
Three parallel-plate capacitors are constructed, each having the same plate area A and with Cl having plate spacing d l , C2 having pla(e spacing~, and C3 having plate spacing ((", Show that the LOlal capacitance Cofthe three capacitors connccted in series is the same as a capacitor of plate area A and wit h plate spacing d = d l + d'l + d?o.
56, _ For the system of four capacitors shown in Figure PI6.37, find (a) the total energy stored in the system and (b) the energy stored by each capacitor. (c) Compare the slim of the answers in part. (b) with your result to part (a) and explain your observation. ~ _ A parallel-plate capaciLOr with a plale separation d
has a capacitance Co in the absence of a dielectric. A slab or dielectric male rial of dielectric constant K and thick· ness d/3 is then inserted between the plates as in Figure PI6.57. Show that the capacitance of this partialJ)' filled capacitor is given by
C (2}: l)C =
(Hint: Treat the system as t\\'o capacitors connected in series, one with dielectric in it and the othcr one emplY.)
[]I]Oelcrmine (a) the capacitance and (b) the maximum voltage that can be applied LO a Tenon-filled parallel-phllc capacitor haying a plate area of 175 cm 2 and all insulation thickness of 0.040 0 mm. 52. A com mercial capacitor is constructcd as in Figure 16.26a. This particular capacitor is made from a strip of aluminum foil separated by two strips of paraffi n-coated paper. Each st.rip of foil and paper is 7.00 em wide. The foil is 0.00400 mm t.hick, and the paper is 0.0250 mIn thick and has a dielectric const.ant of 3.70. \Nhatlength ShOldd the strips be if a capacitance of 9.50 X 1O-l-; F is desired
O
til.Ll===;ii--=-...il C,
A
..*dl !
(0)
(b)
-
FIGURE P16.57
~
Problems 58.
E
Two capacitors givc an equivalent capacitance of C" when connected in parallel and an cqui\'alclllcapacit3nce of Cs when connected in series. What is the capacitance of each capaciwr?
59. An isolated capacito," of unknown capacitance has been charged to a potential difference of 100 V. \''-'hen the charged capacitor is disconnected from the battery and then connected in parallel (Q an ullch
60. Two charges of 1.0 J.LC and -2.0 J.LC are 0.50 m apart at twO vertices of an equilateral trianglc as in Figure 1'16.60. (a) What is the electric potential due to the 1.0-J-LC charge at lhe third vertex, point P? (b) What is the electric potential due to the -2.0-J.LC charge at P? (c) Find the total elcctric potential at P. Cd) \-Vhat is the work required to move a 3.0-J-LC charge from inftn ity to P.
0.50
0.50
III
III
FIGURE P16.60
61. Find the equivalent capaciulnce of the group of tors shown in Figure PI6.6t.
capaci~
5.00 JJF
~:300IlF
4.00;'F-t"~OOUF 1 3.00 r---l
III'
II
7.00 JIF II
L -_ _- - - \
tric shock to the chest over a time of a few milliseconds. The de\'ice contains a capacitor of a few microfarads, charged to several thousand vallS. Electrodes called paddIes, abolll8 em across and coated with conducting paste, are held against lhe chest on both sides of the hean. Their handles are insulated to preycnt injury to the operator, who calls "Clear!'· and pushes a button on one paddle to discharge the capacitor through the patiem·s chest. Assume an energy of 300 W' s i:-. to be deli\'ered from a 30.0-,u.F capacitor. To whtlt potential difference muSt it be charged?
64. \-\Then a certain
air~filled parallel-plate capacitor is conneCled across a battery, it acquires a charge of 150 J1..C on each plate. While the battery connection is maintained, a dielectric slab is inserted into, and I1l1s, the region between the platcs. This results in the accumulation of an additional charge 0[200 p,C on each plate. 'What is the dielectric constant of the slab?
165.IC
1.0 pC
6.00uF .
569
'I--:c-,--,-,----' ·18.nV
FIGURE P16.61
162.IA spherical capaciwr consists of a spherical conducting shell of radius band charge -Qconccntric with a slnaller conducting sphere of radius (I. and chaqse Q. (a) Find the capacitance of this devicc. (b) 5hO\...· that as thc radius b of the outer sphere approaches infinity, the capacitance "pproaches the value a/h.. :::: ~11i"EI)(/"
63.11 The immediate c
66. The energy stored in a 52.0-,uF capacitor is used to melt a 6.00-mg sample of lead. To what "ohage Illustthe capacitor be initially chargcd. assuming the initial temperature of the lead is 20.0°C? Lead has a specific hcal of 128 J/kg'''C, a melting point of 327.3"C, and a latcnt heat of fusion of 24.5 kjlkg. 67. Metal sphere A of radius 12.0 cm carries 6.00 J.kC of ch
)'
-I
, I'atho[ the electrOll
~
j__II,~=ell=_-,,=F =="J~I' FIGURE P16.68
These power lines transfer energy from the power company to homes and businesses. The energy is transferred at a very high voltage, possibly hundreds of thousands of volts in some cases. The high voltage results in less loss of power due to resistance in the wires, so it is used even though it makes power lines very dangerous.
17.1
Electric Current
17.2 A Microscopic View: Current and Drift Speed 17.3
Current and Voltage Measurements in Circuits
17.4
Resistance, Resistivity, and Ohm's Law
17.5
Temperature Variation of Resistance
17.6
Electrical Energy and Power
17.7
Superconductors
17.8
Electrical Activity in the Heart
CURRENT AND RESISTANCE Many practical applications and devices are based on the principles of static electricity, but electricity was destined to become an inseparable part of our doily lives when scientists learned how to produce a continuous flow of charge for relatively long periods of time using batteries. The battery or voltaic cell was invented in 1800 by Italian physicist Alessandro Volta. Batteries supplied a continuous flow of charge at low potential, in contrast to earlier electrostatic devices that produced a tiny flow of charge at high potential for brief periods. This steady source of electric current allowed scientists to perform experiments to learn how to control the flow of electric charges in circuits. Today, electric currents power our lights, radios, television sets, air conditioners, computers, and refrigerators. They ignite the gasoline in automobile engines, troveI thraugh miniature components making up the chips of microcomputers, and provide the power for countless other invaluable tasks. In this chapter we define current and discuss some of the factors that contribute to the resistance to the flow of charge in conductors. We olso discuss energy transformations in electric circuits. These topics will be the foundation for additional work with circuits in later chapters.
17.1
ELECTRIC CURRENT
In Figure 17..1 charges move in a dircClion perpendicular LO a surface of area A. (The area could be the cross-sectional area of
Suppose 6.Q is the amount of charge th . l .l flows t.hrough an area A in a time interval 6.t and that the direction of now is perpendicular to the area. Then the average current Ia\o is equal to the amount of charge divided by the time inlerval: J a\
Ll.Q
~-
6. f
Slunit: coulomb/second (C/s), or the ampel'e (A)
570
[17.1a]
17.1
Current is composed of individual moving charges, so for an extremely low current, it is conceivable that a single charge could pass through area A in one instant and no charge in the next instant. All currents, then, are essentially averages over time, Given the very large number of charges usually im·olved. hm,:ever, it makes sense to define an instantaneous current. The instantaneous current I is the limit of the average current as the time interval goes to zero: , I' t.Q I = Ilin II\" = JlllJ./----j.O·
J./----j.o
Electric Current
571
TIP 17.1 Current Flow Is
Redundant The phrases flow of(Urrelll and ,urrl'11/ }low are common ly used, but hCl'C the t\'ord flow is l'edundallt bcc
[l7.lb]
n.t
51 unit: coulomb/second (C/s), or the ampere (A) When the current is steady, the average and instantaneous currents are the same. Note that one ampere of current is equivalent LO one coulomb of charge passing through the cross-sectional arC~l in a time interval of 1 s. \Vhen charges flow through a surface as in Figure 17.1, they can be positive, negative, or both. The direction of conventional current used in this book is the direction positive charges flow. (This historical com'ention originated about 200 years ago, \v'hen the ideas of positive and negative charges were introduced.) In a common conductor such as copper, the current is due to the motion of negatively charged electrons, so the direction of the current is opposite the direct.ion of" motion of the electrons. On the other hallet. for a beam of positively charged protons in an accelerator, the current is in the same direction as the motion of the protons. ]11 some cases-gases and electrolytes. for example-the current is the result of the flows of both positive and negative charges. f\:loving charges, whether positive or negatjve, are referred to as rhmge mrrinJ. In a metal, for example, the charge carriers arc electrons. In electrostatics, \-vhere charges are stationary, the electric potential is the same everywhere in a conductor. That is 110 longer true for conductors carrying current: as charges move along a wire, the electric potential is continually decreasing (except in the special case of superconducLOrs). EXAMPLE 17.1 Goal
+-
Direction of current
FIGURE 17.1 Charges in Illotion an area A. The lime rale of flow of charge through the area is defined as the current I. TIl(' rlirection of the current is the direction of flow of positive charges. I hrollgh
Turn On the Light
Apply the concept of current.
Problem The amount of charge that passes through the filament of a certain lightbulb in 2.00 sis 1.67 C. Find (a) the average current in the lightbulb and (b) the number of electrons that pass through the filament in 5.00 s. Strategy Substitute into Equation 17.1<:1 Cor part (a), then multiply the answer by the time given in part (b) to get the LOt'll charge that passes in that time. The total charge cquals the number ;Vof electrons going through the circuit times the charge per electroll. ,.. . . • . • . , •• , . , . , • , • • . • . , . , . , . • • . . . . . - . . . , • • • . . . • . . , . , , ••• , , • , - . • • • . • • . • . • • • • . . . , •. '''ll
Solution (a) Compute the average current. in the lightbulb. 1.67 C 2.00 s
Substitute the charge and time into Equation 17.la:
0.835 A
(b) Find the number of electrons passing through the filamcnt in 5.00 s.
Z"
The total number . V of' electrons times the charge per electron equals the total charge, In.\ n./:
(1)
Substitute and solve forN:
N(1.60 X 10- 19 C/electron)
N ~
_
,
,.,
,
,.
=
IVq =
t./ (0.835/\.)(5.00 s)
2.61 X 10 19 electrons
.
."'
572
Chapter 17
Current and Resistance
Remarks In developing the solution, it was important La use units LO ensure the correctness or equations such as Equation (1). Notice the enormous number ofe]eclrons passing through a given point in a lypical circuil. QUESTION 17.1 Is it possible La have an instantaneous current of ('/2 per second? Explain. Can the average current take this "aluer EXERCISE 17.1 Suppose 6.40 X 10 21 electrons pass through a wire in 2.00 min. Find the average current. Answer
8.53 A
QUICK QUIZ 17.1
Consider positive and negalivc charges moving horizontally through the rour regions in Figure 17.2. Rank the magnitudes of the currents in these four regions [rolll lowest to highesl. (f(l is the current in Figure 17.2a, I b the current in Figure l7.2b, etc.) (a) ~i~ 1(1, (, 10 (b) la' (J IbJ lrl (c) I" In' 1.1, I, (eI) 1.1' I b, I" In (c) In. I" I" J" (f) None of these
FIGURE 17.2 (QllickQlliz17.I)
(h)
(a)
17.2
FIGURE 17.3 A section ofallniform conduCHlf of cross-sectional ;11'C
by 1IA.1I,!~I. where 11 is lhe lllllllhcr of' mobile carricrs per lInil \lJIIIIIlC.
(e)
(d)
A MICROSCOPIC VIEW: CURRENT AND DRIFT SPEED
Macroscopic currents can be related to the Illotion of the microscopic charge carriers making up the current. It turns out that current depends on the average speed or the charge carriers in the direction of the current, the number of charge carriers per unit volume, and the size of the charge carried by each. Consider identically charged panicles moving in a conductor of cross-sectional area A (Fig. 17.3). The volume of an element of length of the conductor is it Ir n represents the number of mobile charge carriers per unit volulllc, the number of carriers in the volume elemem is 1/1l The mobile charge LlQ in this element is therefore
ax
ax.
ax.
aQ = number of carricrs X charge per carrier
= (nil D.x)q
where q is the charge on each carrier. If the carriers move with a constant average speed called the drift speed vr/. the distance they move in thc tirnc interval ill is ax = vr/ D.l. ·'We can therefore write
oQ =
(nAvd 01)"
11' we divide both sides of this equal ion by ill, and take the limil as ill goes to zero, we see that the Cllrrenl. in the conductor is
11--ACTIVE FIGURE 17.4 A
~chcrll:Hic
repreSenl
zigzag mol ion 01';] charge c:lrricr ill
I =
oQ
lim , i l - ()
D.t
= ?If/VilA
[17.2]
To understand the meaning of drift speed, consider a conductor in which the charge carriers are free electrons. If the conductor is isolated, these electrons undergo random Illotion similar to the motion of the molecules of a gas. The drift speed is normally llluch smaller than the free electrons' average speed between collisions with the fixed atoms of the conductor. \'Vhen a potemial difference is
17.2
A Microscopic View: Current and Drift Speed
appl ied bctv,reen the ends of the conductor (say, with a bauery), an electric field is set up in the conductor, creflling an electric force on the elect.rons ..1 nd hence a current. III reality, the electrons don't simply move in straight lines along the conductor. Instcad, they undergo repeated collisions with the atoms of the metal, and the result is a complicated zigzag motion with on1r a small average drift speed along the wire (Active Fig. 17.4). The energy transferred from the eleClrons to the metal atoms during a collision increases the vibrational cnergy of the atoms and causes a corresponding increase in the temperature of the conductor. Despite the collisions, howevcr, the electrons move slowly. along the conducLOr in a direction opposite E with the drift velocity Vii'
/~-----
~
EXAMPLE 17.2 Goal
573
TIP 17.2 Electrons Are Everywhere in the Circuit Electrons don'l han> 10 Iravel from the ligh1 switch to the lig-hlbulb lor the liglllbulb lU op..:r:llc. Electrons alreadv in the filament of the lightbulb move in response to the c1ec1rk Held set up by 1he bauer}'. Also, the baltcl)' dClcs not provide electrons to the circllit; it provides e//('rf!J' to the CXiS1 illg electrons.
Drift SReed of Electrons
Calculate a drift speed and compare it with lhe rms speed of an electron gas.
Problem A copper wire of cross-sectional arca 3,00 X 1O-i ; m~ carries a current of 10.0 A. (a) Assuming each copper atom contributes one free electron to the metal, find the drift speed of the electrons in this wire. (b) Use the ideal gas model to compare the drif't speed with the randall) rms speed an electron would have at 20.0°C. The density of copper is 8.92 g/cm 3 . and its atomic mass is 63.5 ll. All the variables in Equation 17.2 are known exccpt [or 17, the number of' frec charge carriers per unit volume. 11 by recalling that one mole of copper contains an Avogadro's number (6.02 X 1023 ) of atoms and each atom contribut.es one charge carrier to the mctal. The volume of one mole can be found from copper's known density and atomic mass. The atomic mass is the same, numerically. as the number of grams in a mole of the substance. . Strategy
\·Vc can find
~........
~
Solution (a) Find t.he drift speed of the elcctrons. Calculate the volume of one mole of copper [rom its density and its atomic
In
63.5 g
p
8.92 gl Cln'
\1= - =
= 7.12 cm~
mass:
Convert the volume from cm 3 to m 3 :
7.12CI11~(~)3 10-
7.12 XlO-Gm:1
(Ill
Divide Avogadro's number (the number of electrons in onc mole) by the volume per mole to obtain the number density: Solve Equation 17.2 for the drift speed and substitute:
11
=
6.02 X lO~n electrons/mole 7.12 X 10 flm~/molc
= 8.46 X 102R eleClrons/m,1
I v" = - IIqll
10.0 Cis (8.46 X 10" eleclrons/m')(1.60 X 10 19 C)(3.00 X 10
v" =
2.46 X 10
~·I
fi
111 2)
mls
(b) rind the rms speed of a gas of electrons at 20.0°C. Apply Equalion 10.18:
Convert the tcmperature to the Keh'in scale and substitut.e valucs:
v
IS( 1.38
= noh
\
X 10 '''J/K)(293 K) 9.11 X 10 " kg
~ 1.15 X 10'
................................ - .....
mls
.
...
574 Remark
Chapter 17
Current and Resistance
The drift speed of an electron in a wire is very small, only about one-billionth of its random thermal
speed.
QUESTION 17.2 True or False: The drift velocity in a \\"ire of a given composition is inversely proponional to the number density of charge carriers. EXERCISE 17.2
,,\That current in a copper wire \vith a cross-sectional area of 7.50 X 10-7 m::! would result in a drift speed equal to 5.00 X 10- 4 m / 5?
Answer
5.08 A
Example 17.2 shows that drift speeds are typically "cry small. In fact, the drift speed is much smaller than the average speed between collisions. Electrons traveling at 2.46 X 10-'1 mis, as in the example, ,,,'ould take about 68 min to travel J Ill! 1n view of this 10''1' speed, ".. hy does a ligillbulh LUrn on almost inst3maneously ,,:hen a switch is thrown? Think of the flow of water through a pipe. If a drop of water is forced into one end of a pipe that. is already filled ""ith ,",,'atcr, a drop must be pushed ouL the other end of the pipe. Although it may take an individual drop a long time to make it through the pipe, a flm'\' initiated at one end produces a similar flmv at the other end very quickly. Another familiar analogy is the motion of a bicycle chain. \'Vhen the sprocket moves one link, the other links all move rnore or less immediately, even though it takes a given link some time [0 make a complete rotation. Tn a conductor, the electric field driving the free electrons travels at a speed close to that of light, so ,vhcn yOll flip a light s,,,·itch, the message for the electrons to start moving through the wire (the electric field) reaches them at a speed on the order of 101'1 m/s!
QUICK QUIZ 17.2 Suppose a current-carrying wire has a cross·sectiollal area that gradually becomes smaller along the wire so that the wire has the shape ora very long, truncated cone. I-Im,,' does the drift speed vary along the wire? (a) It slows down as the cross section becomes smaller. (b) It speeds up as the cross section becomes smaller. (c) It doesn't change. (d) More information is needed.
17.3 CURRENT AND VOLTAGE MEASUREMENTS IN CIRCUITS To study electric current in circuits, we need to understand how to measure currents and volt.ages. The circuit shown in Figure 17.5£1 is a drawing of the actual circuit necessary for measuring the current in Example 17.1. Figure 17.5b shov..s a stylized figure called a circuit diagram that represents the actual circuit of Figure Ii.5a. This circuit consists of only a bauery and a lightbulb. The word circuil.means "a closed loop of some sort around which current circulates." The battery pumps charge through the bulb and around the loop. i\o charge would flow ,..ithout a complete conducting path from the positive terminal of the battery into one side ofthe bulb, out the other side, Clnd through the copper conducting wires back to the negative terminal of the battery. The most important quantities that characterize how the bulb works in different situations are the current / in the bulb and the potential difference IlVacross the bulb. To measure the current in the bulb, we place an ammeter, the device for measuring current, in line '"\lith the bulb so there is no path for the current La bypass the meter; all the charge passing through the bulb lllust also pass through the ammeter. The voltmeter measures the potential difference, or volt-
17.4
Resistance, Resistivity, and Ohm's Law
Halu.:ry
+
~+ Bulb
+
~.. OmR
+
Ammeter
~
-
-'--. + V01LIlH.:ler (c)
(11)
FIGURE 17.5 (a) A sketch ofal!
age. between the two ends of the bulb's fllament. If \'ie use I.V.'O meters simultaneously as in Figure 17.5a, 'we can remove the voltmeter and see if its prescnce arfects the currel1l reading. Figure 17.5c shows a digital llluilimeter, a convenicnt device, with a digital readout, that can be used to measure voltage, current, or resistance. An advanlage of using a digiL
QUICK QUIZ 17.3 Look at the foul' "circuits" shown in Figure 17.6 and select those that wi\llight the bulb. +
+
. D 'PS
• -
(0)
FIGURE 17.6
17.4
(b)
(e)
(Quick Quit. li.g)
RESISTANCE, RESISTIVITY, AND OHM'S LAW
Resistance and Ohm's Law \Vhen a voltage (potential dilTerence) ~V is applied across the ends of a metallic conductor as in Figure 17.7 (pagc 576), the current in the conductor is found to be
(d)
575
576
Chapter 17
Current and Resistance
proponional to the applied voltage; [rx.1v. If the proportionality holds, ,";e can write L\V = IH, ,.,.'here the proponionality constant R is called the Tes£slauce of the conductor. ln fact, we define the resistance as the ratio of the voltage across the conductor to the current it carries: Resistance
...
l
r
1\,'
,
[17.3] Resistance has 51 units of volts per ampere, called ohms (0). If a potential difference of 1 V across a conduclOr produces a current of 1 A, the resistance of the con~ ductar is 1 n. For example, iran electrical appliance connected ta a 120-V source carries a current of 6 A, its resistance is 20 n. The concepts of electric current, volt.age, and resistance can be compared to the flo",.. of,..'ater in a river. As \.. .'<:lter no\',,'s dmvnhill in a river of constant width and depth, the flow rate (water cun-em) depends on [)le steepness of descent of the river and the effects of rocks, the riverbank, anel other obstructions. The voltage difference is analogolls to the steepness, and the resistance to the obstructions. Based on this analogy, it seems reasonable that increasing the voltage applied to a circuit should increase the current in the circuit,just as increasing the steepness of descent increases the watcr current. Also, increasing the obstructions in the river's path ,"vill reduce the water current,just as increasing the resistance in a circuit will lower the electric current. Resistance in a circuit arises due to collisions between the electrons carrying the current \",'ith fixed atoms inside the conductor. These collisions inhibit the movcment of charges in much the same ,-vay as would a force of friction. For many materials, including most metals, experimellls show that the resistance remains constant over a wide range of applied voltages or currents. This statement is known as Ohm's law, after Georg Simon Ohm (1789-1834), who \.,ras the first (0 conduct a systematic slUdy of electrical resistance. Ohm's law is given by
t~
'\ IA I
r-~
Ifb .._ _
, \
-+
~=:.:=~_"'"' II" Ii
FIGURE 17.7 A uniform COIlc!uctor oflenglh land cross-sectioJlal area A. The currellt Jill the conductor is proponiollallo the applied ~ILage 1l.1' = It - \~. The electric field E set up ill the cOilduClor is also proporTional 10 the current.
L'. \f= !R
[17.4]
where Ris understood to be independent of
GEORG SIMON OHM
(1787-1854) A high school teacher in Cologne and later a professor at Munich, Ohm
formulated the concept of resistance and discovered the proportionalities expressed in Equation 17.5.
§
! ~
c:
~~
QUICK QUIZ 17.4 In Figure 17.8b does Lhe resisLance or the diode (a) increase or (b) decrease as the positive voltage ~Vincreases?
An assortmc.:lll of n:sistors used for a ,·ariet\' of applicatiolls in e1eClronic circuits.
0
QUICK QUIZ 17.5 All clenric devices are required to have identirying plates that specify their electrical characteristics. The plate on a certain steam iron states that the iron carries a current of 6.00 A when COll-
17.4
Resistance, Resistivity, and Ohm's Low
necred to a source of 1.20 X 10 2 V. ''''That is the resistance of the stearn iron? (a) 0.0500 fl (b) 20.0
n (c)
36.0
n
e
e
R=p-
[17.5]
A
where the constant of proportionality, p, is called the resistivity or the material. Every material has a characteristic resistivity that depends on irs electronic structure and on temperature. Good electric conductors have very low resistiviries, and good insulators have very high resistivities. Table 17.1 lists the resis[ivities of various materials at 20°C. Because resistance values are in ohms, resistivity values must be in ohm-meters (.0.' m).
TABLE 17.1 Resistivities and Temperature Coefficients of Resistivity for Various Materials (at 20°C) Resistivity
Copper Gold AlutTlinunl Tungslen Iron Platinum Lead Nichrome Carbon Germanium Silicon Glass I-lard rubber
Sulfllf Quartz (fused) .,:\ nickcl-chromitllll
~"e=~
T-~v
Electrons don't move in straigill-line paths through a conductor. [nstead, they undergo repeated collisions with the metal at.oms. Consider a conductor with a voltage applied across its ends. An electron gains speed as the electric force associ· atecl with the internal electric AcId accelerates it, giving it a velocity in the direction opposite that. of the clcClric field. A collision with an (ltom randomizes t.he electron's velocity, reducing it in the direction opposite the field. The process then repeats it.self. Together, these collisions affect the electron somewhat as a force of internal friction ,.. .ould. This step is the origin of a material's resistance. The resist
Silver
I
I
Resistivity
Material
577
(n'm)
1.59 X 10 8 1.7 X 10- 8 2.44 X 10-8 2.82 X 10-8 11.6 X IQ-!l
10.0 X 10-8 II X 10- 8 22 X 10-:'1
150X 10-8 3.5 X 10.:0
Temperalure Coefficient of Resistivitv [('C)-I] .
3.8 X 10" 3.9 X 10-" 3,4 X 10-" 3.9 X 10-" 4.5 X 10-:1 5.0
X
10'1
3.92 X 10 " 3.9 X 10 :~ 0.4 X 10 "
-0.5
X 1O-:i
0.46
-48 X 10-'"
640
-75X 10<\
10lU _IO H
=]0 1:\
10 1;; 75 X lOw allo~' cummonly u~ed
in
he;l(in~ eIeIllCtlL~.
(a)
(b) FIGURE 17.8 (a) The currCllLvoltage curve for an ohmic material. The cun-c: is linear. and lhe slope gives the resistance of 1 he conductor. (b) A nonlinear currelll-\,oltllge cun'e for a semiconducting diode. This device doesn't obey Ohm's law.
578
Chapter 17
Current and Resistance
APPLYING PHYSICS 17.1
DIMMING OF AGING LIGHTBULBS
As a lightbulb ages, ,
Illt::ans that the filament will carry less current for the
when new?
sallle applied voltage. \'Vith less currelll in the r-J1a-
Explanation There are two reasons for the lightbulb's behavior, one clectrical and one optical, but both are related to the same phenomenon occurring within the bulb. The filamclll oran old lightbulb is made of a tung-sten wire that has bcen kept at a high temperature for many hours. High temperatures evaporate tungsten from the filament, decreasing its radius. From R = pel!\, \YC see that a decreased crosssectional area leads LO an increase in the resistance of the filamenl. This increasing rcsistance ,,,,'ith age
melll, there is less light output, and the fllamcntglows morc dimly. At the high operating temperature of the 1I1amcnt, tungsten atoms leave its surface, much as waleI' molecules evaporate from a puddle or ,,·atcr. The atoms arc carried a.. v<.1Y by convection currents in the gas in lhe bulb and arc deposited on the inner surface of the g·lass. 111 lime, the glass becomes less transparent because of the tungsten coating. which decreases the amollnt of light that passes through the glass.
EXAMPLE 17.3 The Resistance af Nichrome Wire Goal
Combine the concept of resistivity with Ohm's law.
Problem (a) Calculate the resistance per unit length or a 22-gauge Nichrome wire of radius 0.321 III Ill. (b) If a potential difference of 10.0 V is maintained across a l.OO-m length of the Nichrome wire, what is the current in the wire? (c) The wire is melted down and recast with twice it.s original length. Find thc new resistance R,vas a multiple of the old resistance Ro . Strategy Pan (a) requires substitution into Equation 17.5, after calculating the cross-sectional area, whereas pan (b) is a maLterofsubstitlllion into Ohm's la\,,'. Part (c) rcquires somc algcbra. The idea is to take the exprcssion for the new resistance and substitute expressions for ("and A,\r, the new lengt.h and cross-sectional area, in terms o[the old length and cross-section. For the area substit.ution, remember that the volumcs ol"thc old and new wires are the same. ,...-'
Remarks From Table 17.1, the resistivity of Nichrome is about 100 times that or copper, a typical good conductor. Thererore, a copper wire of the same radius would have a resistancc per unit length of onl)' 0.052 Dim, and a 1.00-m length of copper \"ire ofthc S(1me radius \,'ould carry the same current (2.2 A) with an applied voltage of only O.115Y.
Becausc of its rcsistance heatcrs.
La
oxidation, Nichrome is often used for heating elemcnts in toasters. irons. and elcctric
QUESTION 17.3
Would replacing the Nichrome with copper result in a higher current or 100,,'er current? EXERCISE 17.3
',Vhat is the resistance of a 6.0-m length or Nichrome \\:ire that has a radius 0.321 mm? calTY \vhen connected to a 120-V source? Answers
I-IO\\'
much current does it
28 fl; 4.3 A
QUICK QUIZ 17.6 Suppose an electrical wire is replaced with one h;'H'ing every linear dimension doubled (i.e., the length and radius have twice their original values). Does the wire now have (a) more resistance than berore, (b) less resistance, or (c) the same resiswncer
17.5
TEMPERATURE VARIATION OF RESISTANCE
The resistivity p, and hence the resistance, of a conductor depends on a number of [actors. One 01" the most impon£lnt is the temperature o[ the metaL For most metals, reSistivity increases \"ith increasing temperature. This correlation can be understood as rollows: as the temperature of the material increases, its constituent atoms vibrate \"\'ith greater amplitudes. As a result, the electrons find it more difficult to gel by those atoms,just as it is more difficult to weave through (I crowded room when the people are in motion th
+ a(T -
7;,)]
[17.6]
where p is the resistivity at some temperature T (in Celsius degTees), Po is the resistivity at some rererence temperaLUre 70 (usually taken to be 20°C), and a is a parameter called the temperature coefficient of resistivity. TemperalUrc coefficients for various materials are provided in Table ]7.1. The interesting negative values of a ror semiconductors arise because these materials possess weakly bound charge carriers that become free to move and contribute to the current as the temperature rises. Because the resistance of a conductor with a unirorm cross section is proportional to the resistivity according to Equat.ion 17.5 (R = pI/A), the temperature variat.ion of resistance can be written R = R.,[l
+ a(T-
7;,)]
[17.7]
Precise temperature measurements are ol"t.en made Llsing this property, as shown by the following example.
In an old-fashiOllcd carbon fibmeJlf illcllldcscent billp. the electrical resl<;tancc is l)'picallv 10 n, hilt changes with temperaturc.
580
Chapter 17
Current and Resistance
EXAMPLE 17.4 A Platinum Resistance Thermometer Goal
Apply the temperature dependence of resistance.
Problem A resistance thermometer, 'which measures temperature by measuring the change in resistance of a conductor, is !lude of platinum and has a resistance of 50.0 n at 20.0°C. (a) When the device is immersed in a vessel comaining melting indium, its resistance increases to 76.8 n. From this information, find the melting point of indium. (b) The indiulll is heated further Ilntil it reaches a temperature of' 235°C. Vlhar
is the ratio of the new current in the platinum to the current imp allhe melting point? Strategy For parl (a), solve Equation 17,7 lor 1'- 70 and get a for platinum from Table 17,], substituting known quantities. For part (b), use Ohm's law in Equation 17.7.
...,
~-
Solution (a) Find the !TIehing point ofindiurn. 76,8!1 - 50,0 fl [3.92 X 10-' (oCt'][50.0 il]
1'- 70 =
Solve Equation 17,7 lor T- To:
137°C Substitute 7~) = 20.0°C and obtain the Illelting point of indium:
r
=
157°C
(b) Find the ratio of the ne\\! currenlt.o the old v,:hen the tempcrature rises from 157°C La 235°C.
Write Equation 17.7, with No and 7~) replaced by R11Ip and Tmp ' l he resistance ;mel tempcraLUre at the melting point.
If = R mp [ I -\- a(T - T,np)]
According to Ohm's la",,,, R = DV/land R llIp = ~V/llIlr' SubstilUle these expressions into Equation 17.7: 1
Cancel the voltage differences, invert the lWO expressions, and then divide bOlh sides by Imp: Substitutc T= 235°C, '(np = 137°C, and the value for 0., obtaining the desired ralio:
As t.he lcmperawre rises, both the rills speed of the electrons in the metal and the resistance increase.
QUESTION 17.4
,.vhat happens to the drift speed of the electrons as the temperalUre rises? (a) It becomes larger. (b) It becomes smaller. (c) It remains unchanged. EXERCISE 17,4
Suppose a wire made of an unknmvn allo)' and having a temperature o[20.0°C carries a current of 0.450 A. At 52.0°C the current is 0.370 A for (he same potential di [ference. find the temperature coefficient of resistivity of the alloy.
Answer 6,76 X
lQ-3
eC)-'
17.6
ELECTRICAL ENERGY AND POWER
I r a bauery is lIsed to establish an electric current in a conductor, chemical energy stored in the baltery is continuously transformed into kinetic energy of the charge carriers. This kinetic energy is quickly lost as a result of collisions between the charge carriers and fixed atoms in the conductor, causing an increase in the tem-
17.6
perature of the conduct.or. In this \,"ay the chemical energy storcd in the bauery is continuously transformed into thermal energy. To understand the process or energy transfer in a simple circuit, consider a balLery WiLh terminals connecteel La a resistor (Active Fig. 17.9; remcmber that the positive lerminal of the bauery is always at the higher potential). No\\-' imagine following a quantity of positive charge ~Q around the circuit from point fl, through the bauery and rcsistor, and back to ...\. Point A is a reference point that is grounded (the ground symbol is ~), and its pOlential is taken to be zero. As the charge LiQ moves [rom A to B thro7l g h l.he bau.er)", the electrical potential energy of the system increases by the amount LiQ L\Vancl the chemical potcntial energy in the bauery decreases by the same amount. (Recall from Chapter 16 thal. LiPE = fJ 6,\() As the charge movcs from C to D through the resistor, hm.. . ever, it loses this electrical potential energy dllring collisions with atoms in the resistor. In the process the energy is transformed to internal energy corresponding to increased vibrational motion of those atoms. Because we can ignore the very small resistance of the interconnecting \vires, no energy transformation occurs for paths Be and DA. \rVhen the charge returns to point A, the net result is th
.it
581
Electrical Energy and Power
1/ D
ACTIVE FIGURE 17.9 A cirCllit consist illg of a ballery ~nd a resistance N.. Positive charge flow:-. clockwise frOIll the positive l{) the:Ileg:ati\re terminal ofthc bane!")'. Poilll 1\
is grounded.
~\'
where I is the current in the resistor and .!iV is the potential difference across it. Of course, the charge regains this energy ,vhen it passes t.hrough the battery, at the expense of chemical energy in the banery. The rate at ,,,'hich lhe system loscs pot.ential energy as the charge passes through the resistor is equal to the rate at which the system gains internal energy in the resistor. Therefore, the power CZP, representing the rate at \vhich energ}' is delivered to the resistor, is [17.8]
'8'=fli\l
~
Power
~
Power delivered to a resistor
Although this result ,"\,as developed by considering a battery delivering energy to a resistor, Equalion 17.8 can be used to determine the power transferred from a voltage source to any device carrying a current land having a potential difference LiVbetween its terminals. Using Equation 17.8 and the [act that.iV = IR f<)]" a resistor, we can express the power delivered to the resistor in the alternate (orms li \I' '8'= f'R=--
R
[17.9]
\,Vhen I is in amperes, /1Vill volts, and H in ohrns, the Slunit of power is the waU (introduced in Chapter 5). Thc po\',;er delivered to a conductor of resistance R is oftcn referred to as an I'!.R loss. Note that Equation 17.9 applies only to resistors and nolto nonohmic dcvices such as lightbulbs and diodes. Regardless of the ,...ays in which you LIse electrical energy in your home, yOll IlltimCltely IllUSt pay for il or risk having your power turned oIT. The unit of cncrgy Llsed by elcctric companies to calculate consumption, the kilowalt-hour, is defined in terms of the unit of power and the amount of time it's supplied. Onc kilowallhour (kvVh) is the cnergy convened or consumed in 1 h at the constant ratc of 1 kW. lL has the numerical value 1 kWh = (lO' W)(3 600 s) = 3.60 X 10 6 j
[17.10]
On an electric bill, the amount of electricity L1sed in a given period is usually stated in mulliples ofkilowau-lJours.
TIP 17.3 Misconception About Current Current is 1/ul "used up- in a
n:sis~
LOr. Rather, some of the cllerg)' the charges have rccei\cd frOI11 I he voltage source is ddi\'ered 10 llie rl:sisLor. making it hoI and causing it to radiate. Also, the current doesn't slow down when going through lht' I"e~i~tor: il\ the S,llllC throughout the circuit.
582
Chapter 17
Current and Resistance
APPLYING PHYSICS 17.2
L1GHTBULB FAILURES
\Vhy do liglubutbs fail so often immediately af'tcr they're turned on? Explanation Once the s\,vilch is closed, the line voltage is applied across the bulb. As the voltage is applied across the cold filament when tbe bulb is first turned on, the resistance of the filament is 1m..·• the currcm is high, and a relatively large amount of
QUICK QUIZ 17.7 A voltage t.V is applied across the ends of a Nichrome hearer \\lire having a cross-sectional area A and length L. The same \'oltage is applied across the ends of a second Nichrome heater ,-"ire having a crosssectional area A and length 2L. ',\'hich wire geLS hOller? (a) The shorter ,,,:ire does. (b) The longer \"ire does. (c) More in formation is needed.
30W
-v A
J
QUICK QUIZ 17.8 For the two resistors shown in Figure 17.10, rank the currents at points a through 1frol11 largest to smallest.
{~(~ ~v B
d
(a) I" = I" (c) I, =
FIGURE 17.10
lr
> I, = IJ > I, = I" (b) J"
'r> I, = 1" > I" ~
I
"
power is delivered to the bulb. This current spike at the bcg'illning of operation is the reason lightbulbs ofLen fail immediately after they are lllrned on. As the filament 'varms, its resistance rises and the current decreases. As a result, the power delivered La the bulb decreases and the bulb is less likely to burn Ollt.
~ J"
> I, = J" > I, = If
I,
b
~I'
(Quick Quiz
17.~)
QUICK QUIZ 17.9 T\vo resistors, A and B, arc connected in a series circuit with a battery. The resistance of A is twice thal ofB. \Vhich resistor dissipates more pm,vcr? (a) Resistor A does. (b) Resistor B does. (c) More informat.ion is needed. QUICK QUIZ 17.10 The diameter of wire A is greater than the diameter of wire B, bUllhcir lengths and resistivities are identical. For a given voltage difrerence across the ends, what is the rel£lLionship between fPA and 2Jl B , the dissipated power ["or wires A and B, respectively? (a) 'J'., ~ (YlB (b) (Yl,\ < 01'B (c) 01'A > 01'"
EXAMPLE 17.5 The Cost of Lighting Uj) Your Life Goal
Apply the electric power concept and calculate the cost. of power usage lIsing kilowatt-hours.
Problem A circuit provides a maximum current of 20.0 A at an operating voltage of 1.20 X lO~ V. (a) How many 75 "V bulbs can operate with this voltage source? (b) At $0.120 per kilov'iau-hour, how much does it cost to operate these bulbs for 8.00 h'
r
Strategy Find the necessary power with QP = .6Vthcn divide by 75.0 '''' per blllb to gCllhe total number of bulbs. To find the cost, converL pm.. . er to kilowatts and multiply by the number of hours, then multiply by the cost per kilowatt-hour. ,...-
~
Solution (a) Finclthe number of bulbs that can be lighled. Substiwte into Equat.ion l7.8 to get the Lota] power:
(Yl"" .., ~ I L'.V = (20.0 A)(L20 X 10' V)
Di\-ide the total power by the pm-vel' per bulb Lhe number of bulbs:
Number of bulbs
1.0
geL
([PIOlal
- --
QPbulb -
= 2.40 X
2.40 X 10,1 \V
75.0 W
=
10' W
32.0
17.6
Electrical Energy and Power
583
(b) Calculatc the cost of this electricity for an 8.00-h day. Enero"\' = '!PI = (2.'10 X 10' W)(
Finclthe energy in kilowatt-hours:
~.
1.00 kW )(8.00 h) 1.00 X 10'W
= 19.2 kWh
Cost = (19.2 kWh)($0.12/kWh)
Multiply the energy by the cost per kilmvatt-hour:
$2.30 • ' , •••• , ' , •...A
Iro.. • • • , • • • • •
Remarks This amount of cnergy might correspond to what a small orflcc uscs in a working day, t<'lking into account all pm'\'er requiremcnts (notjustlighling). 1n general. resistive devices can have variable power output, depending on how the circuit is wired. Here, power outputs wcre specified, so such considerations \...· ere unnecessary,
QUESTION 17.5 Considering how hot the parts of an incandesccl1llight bulb get during operation, guess what fraClioll of the encrgy Clllitlcc! by an incanc1escentlightbulb is in the form ofvisiblc light. (a) 10% (b) 50% (c) 80% EXERCISE 17.5 (a) How many Christmas tree lights dr<'l\"'ing 5,00 W of power each could be run on a circuit operating at 1,20 X IO~ V and providing 15.0 A of current? (b) Find the cost to operate one such string 24,0 h per day for the Ch ristmas season (two weeks), lIsing the rate $0. 12/kvVh.
Answers
(a) 3.60 X 10 2 bulbs
EXAMPLE 17.6
(b) $72.60
The Power Converted by on Electric Heoter
Goal Calculate an electrical power Olilput and link dynarnics.
1.0
its effect 011 the environment through the first law of thermo-
Problem An electric hemer is operated by applying a potcntial difference of 50,0 V to a Nichrome wire of total resistance 8.00 n. (a) Find the current carried by the wire and the pmver rating or the healer. (b) Using this heater, how long would iltakc to heat 2.50 X!O:l moles of diatomic gas (e.g,. a mixture of oxygen and nitrogen, or air) from a chil1y IO,O°C to 25,0"C? Tlike the molar specific heat at constant volume crail' to be ~R. For part (a), find the CUlTClll with Ohm's law and substitute illlO the expression for power. Part (b) is an isovolllmCl.ric process, so the thermal energy provided by the heater all goes into the change in internal energy, au. Calculate this quantity llsing the first law or thcrrnodynamics and divide by the power to get the lime. Strategy
~
,
"-,
..
,
,."
,
,.,.,
..
,.,
..
,
,
.. .. ,
"
,
~
Solution (a) Compute the currem and power output, ,; \I 50.0 V / = = -- =
Apply Ohm's law to get the current:
R
8.00 fl
6.25 A
QJ> ~ /'R = (6.2.0 A)'(8.00 fl)
Substitute into Equation 17,9 to find the powcr:
313 W
(b) How long does iLLake La heaLLhc gas?
Calculate the thermal energy transfer [rom the n.rstla\.,.', ~ote that W= 0 because the volume doesn't change,
Q=
,; U = lie,,'; T
= (2.50 = 7.79
Divide the thel"lTlal energy by the power gel. the time: ~,
'
..
'-' '.,'
La
, ,
I
..
Q
x I 0' 1ll01)(~·8.31.J/1ll01·K)(298 K x
283 K)
10' J
7.79
10'J -"13 W X
= g; = -3-c---:-
2.49 X 10' s • •••• ...A
584
Chapter 17
Current and Resistance
Remarks The number of moles orgas given here is approximaLelywhat would be found in a bedroom. "Varming the air with this space heater requires only about 40 minutes. The caiculCllion, however, doesn't take into account conduction losses. Recall that a 20-cm-thick concrete \\1<,111, as calculated in Chapter 11, permiued the loss of more than 2 rnegajoulcs an hour by conduction! QUESTION 17.6
If the heater wire is replaced by a wire with JO\-I!cr resistance, is the time required to heat the gas (a) unchanged, (b) increased, or (c) decreased? EXERCISE 17.6
A hOHvatcr heater is rated at 4.50 X 10:1 \·V and operates at 2.40 X 10 2 V. (a) Find the resistance in the heating element and the current. (b) Hal..' long does it take to heat 125 L of water from 20.0°C to 50.0°C, neglecting conduction and other losses? (a) 12.8 fl, 18.8 A
Answers
(b) 3.49 X 10 3 S
17.7
R(D)
0.15
I I I I
0.10
,, ,
0.05
I
7~
I I
0.00
4.2
4.0
4.4
T(K)
FIGURE 17.11 Resistance n:rslls temper.llurc for a sample of mercllry (Hg). The g-r;tph follOl'·s thatof:lllormal m<.'lal above the critical temperature 7~. The resistallce drops to zero al the criticallelllpcrature, which is 4.2 K for mercury. and rClllain~ ill zero for lower LempCraltlre~.
TABLE 17.2
Critical Temperatures for Various Superconductors Material
Zn AI Sn
Hg Pb :\b
Nb 3 Sn ::\bJGe
YBa ZCu g 07 Bi-$r-Ca-Cu-O Tl-Ba-Ca-Cu-O HgBa 2 Ca::CI1 3O S
Tc(K)
0.88 1.19 3.72 4.15
7.18 9.46 18.05 23.2 90 105 125 134
SUPERCONDUCTORS
There is a class of metals and compounds ""ith resistances thm fall vinuall}' to zero below a certain temperalUre 7~ called the critical tempera/lI.Te. These materials are knm.. . n as superconductors. The resistance vs. temperature graph for a superconductor follm'Y's that of a normal mClal aL temperalUres above ~ (Fig. 17.11). \Vhen the tempera lUre is at or below 7~, however, the resistance suddenly drops LO zero. This phcnorncnon was discovered in ]911 by Dutch physicist H. Kamerlingh Onnes as he and a graduate student worked with mercury, which is a superconductor below 4.1 I.e Recent measurements have shmvn that the resistivities of superconductors below 7~ are less than 4 X .10- 25 n· m, around 10 1; times smaller than the resistivity of copper and in practice considered to be zero. Today thousands of sllperconductors are known, including such common metals as aluminum, tin, lead, zinc, and indium. Table 17.2 lists the critical temperatures of several superconductors. The value of ]~ is sensitive to chemical composition, pressure, and crystalline structure. Interestingly, copper, silver, and gold, which are excellent conductors, don't ex hibit superconductivity. A truly remarkable feature of superconductors is that once a current is set up in them, it persists without any n/lplied voltage (because R = 0). In fact, steady currents in superconducting loops have been observed to persist for years with no apparent decay! An imporLant development in physics that created much excitement in the scientific cOllllllunity \vas the discovery of high-temperature copper-oxide-based superconductors. The cxcitement began with a 1986 publication by J. Georg Bednorz and K. Alex M(dler, scientists at the IB.vI Zurich Research Laboratory in Switzerland, in which they reponed evidence [or superconductivity at a temperature near 30 K in an oxide of barium, lanthanum, and copper. Bednorz and M(Hler were awarded the Nobel Prize in Physics in ]987 for their important discovery. The discovery \vas remarkable because the criticaltemperatllre was significantly higher than that orany previously known superconductor. Shortly thereafter a new family of compounds was investigated, and research activity in the field of superconductiyity proceeded vigorously. In early 1987 groups at the Universit)' o[ Alabama at Huntsville and the University of Houston announcecl the discovery of superconductivity at about 92 K in an oxide of yttrium, barium, and copper (YBa2C1I307)' shown as the gray disk in Figure 1.7.12. Late in 1987, teams of scientists from Japan and the United Slates reported superconductivity at 105 K in an oxide of bismuth, strontium, calcium, and copper. More recently, scientists have reponed superconductivityat lcmperawres as high as ]50 K in an oxide conlaining mercury. The search for novel superconducling materials continues, with the hope o[ someday obtaining a room-temperature superconducting material. This research is impOl'· tallt bOlh (or scientific reasons and [or practical applications.
17.8
An important and useful application is the construction of sliperconducting magnets in v·..hich the magnetic field intensities are abom ten times greater than those of thc best nonnal electromagncts. Such magnets are being considered as a means of storing energy. The idctl of using superconducting power lines to trans-
'•·..... •. r ..'
mit power efficiently is also receiving seriolls consideration. Model-n sUpct-conducting eleClronlc devices consisting of two thin-film superconducLOrs separaLed by a
ELECTRICAL ACTIVITY IN THE HEART
e
Electrocardiograms Evcry action invol\"ing the body's muscles is initiated by electrical activity. The voltages produced by muscular action in the heart are particularly important to physicians. Volt.age pulses cause the heart to beat, and the waves of electrical excitation that sweep across the heart associatcd with the heartbeat are conducted through the body via the body fluids. These voltage pulses are large enough to be detccted by suitable monitoring equipment. attached to the skin. A sensitive voltmcter making good electrical contact ,.. . ith [he skin by means of contaCLS attached with conduct.ing past.e can be used to measure hean pulses, which are typically of the order of 1 mV at the surface of the body. The voltage pulses can be recorded on an instrument called an electrocardiograph, and the pattern recorded by this instrument is called an electrocardiogram (EKG). To understand the information contained in an EKG pallern, it is necessary first to describe the underlying principles concerning electrical activit)' in the heart. The right atriulll of the heart contains a speci.lIizcd set of llluscle fibers called the SA (sinoatrial) node that. initiates the heartbeat (Fig. 17.13). Electric impulses that originate in these fibers gradually spread from cell to cell throughout the right and left atrial muscles, causing them to contracL. The pulse that passes through the muscle cells is often called a depolarization wave because or its effecL on individual cells. Ifan individual muscle cell were examined in its resting state, a double-layer elcctric charge distribution would be found on its surface, as shown in Figure 17.14a. The impulse generated by the SA node momentarily and locally allows positive charge on the outside of the cell to now in and neutralize the negative charge on the inside layer. This effect changes the cell's charge distribution to that shown in Figure 17.14b. Once the depolarization ,.,.ave has passed through an individual heart Illuscle cell, the cell recovers the resting-state charge distribution (positive out, negative in) shown in Figurel7.14a in about 250 illS. \Vhen the impulse reaches the atrioventricular (AV) node (Fig. 17.13), the muscles of the atria begin to relax, and t.he pulse is dirccted to the ventricular muscles by the AV node. The muscles of the ventricles contract as the depolarization wave spreads through the ventricles along a group of fibers called the PUTkinjt} fibers. The vent.ricles then relax after the pulse has passed through. At this poinl, the S/.\ node is again triggered and the cycle is repeated.
+
+
+
+
+
~
+ + '-
+
+ +
+
~
+
+
~
Dcpolaril
(al
(bl
~ ".~
.........
"'k,,
t.hin insubtor have been constructed. Among these devices are magnctometers (mtlgnetic-neld measuring devices) and various microwa,"c devices.
17.8
585
Electrical Activity in the Heart
"
--------'
FIGURE 17.12 A small permanent magnet floal.s freely abo,e a ceramic disk made of the superconduclor YBa 2Cu:i 0 7• cooled bv liquid nilro~(:n at 77 K. The superconductor has zero electric resisfallce at temperatures bela....· 92 K :lnd expels any applied magnctic field.
APPLICATION "
Electrocardiograms
Purkinjc fibers
Atrioventricular (AV) node FIGURE 17.13 The electrical COtlduction system of the human heart. (RA: right atrium: LA: left atrium: RV: right vcntricle; LV: Icft ventricle.)
FIGURE 17.14 (a) Charge di:>tribution ofa musck cell in lht:: ;Hriul1l beforc a depolarization wave has passed through the cell. (h) Charge di~tributioll a~ Ihe W:il\-e passes.
586
Chapter 17
1.0
R
1°·5 " ~
~
Current and Resistance
T
I' 0
Q
- 0.5
S 0
0.2
004
0.6
Time (s)
FIGURE 17.15 An EKG response for a normal heart.
APPLICATION
II Cardiac Pacemakers
A sketch of the electrical activity registered on an EKG for one be(ll of a normal hean is shown in Figure 17.15. The pulse indicated by Poccursjusl before the atria begin to contract. The Q!~SpLllse occurs in the ventricles just before theycomracl, and the T pulse occurs \-"hen the cells in the ventricles begin to recover. EKGs for an abnormal hean are shown in Figur"c 17.16. The QRS portion of the patlern shQ\vll in Figure 17.16a is H'ider than normal, indicating thal the patient may have an enlarged hearl. (\Vhy?) Figure 17.l6b indicates that there is no constclm relationship between the P pulse and the QI?S pulse. This suggests ~l blockage in the electrical conduction path between the SA and AV nodes which results in the atria and ventricles beating independently and inefficiem heart pUlllping. Finall)', Figure 17.16c shows a situation in which there is no P pulse and an irregular spacing bet'\'een the QRS pulses. This is symptOmatic of irregular atrial contraction, which is called fibril/ation. In this condition, the atrial and ventricular contractions arc irregular. As noted previously, the sinoatrial node directs the heart to beatalthe appropriate rate, usually about 72 bems per minute. Disease or the aging process, however, can dam
An Emergency Room in Your Chest APPLICATION
II
Implanted Cordioverter Defibrillators
InJune 2001 an operation on Vice President Dick Cheney focused all.cntion on thc progress in treating heart problems with tin)' implanted electrical devices. Aptly called "an emergency room in your chest" by Cheney's attending physician. devices called Implanted C
FIGURE 17.16
Abnormal EKGs.
I?
R
P
J;: * ~ P
I'
•
s
R
(Z
Q
(b)
(a)
R
R
II
(e)
I?
R
17,8
Electrical Activity in the Heart
FIGURE 17.17 (a) A dual-chamher ICD wilh leads in the hean. One lead monitors and stimulales the right atrium, and the olher monitors alld slimulales lh(' right vel1ll"icle. (b) Medlronic Dual Chamber leo.
Dual-chamber leO Blowup or deflbrillalor! llIonitor lead
(a)
(b)
2. storing about a half hour of heart signals that can easil~' be read out by a physician 3. being easily reprogrammed with an external magnetic wand 4. performing complicated signal analysis and comparison 5. supplying either 0.25- to 10-V repetitive pacing signals to speed up or slow clown a malfunctioning heart, or a high-voltage pulse of about 800 V to halt the potentially fatal condition ofventricubH'llbrillation, in which the heart quivers rapidly rather than beats (people who have experienced such a high-voltage jolt say that it fcels like a kick or a bomb going off in the chest) 6. automatically acljusting the number of pacing pulses per minute to match the patient's activit}' ICDs are powered by lithium batteries and have implantcd lifetimes of 4 to 6 years. Some basic properties of these adjustable leDs are given in Table 17.3. In the table tachycardia means "rapid heartbcat" and bradycardia means "sIO\\" heanbeaL" A key [actor in developing tiny electrical implants that serve as defibrillators is the development ofcapaciLOrs with relatively large capacitance (125 p.f) and small physical size.
TABLE 17.3
Properties of Implanted Cardioverter Defibrillators Physical Specifications YI ass (g) Size (em)
85
Antitachycardia Pacing
leo del j"ers a burst of criticall}' timed
Number of bursts
I-If;'
Burst cycle length (ms)
Pulse width (ms)
200-552 2-20 7.5 or 10 1.001' 1.9
High-Voltage Defibrillation Pulse energy (j) Pulse ampliUlde (V)
801
7.3 X 6.2 X 1.3 (abom fi"e stacked silver dollars)
A dual-chamber leD can steadily dclivcr repetitive pulses to bOlh the atrium and the vcntricle 40-100
0.25-7.5 0.05,0.1-1.5.1.9
lO wW"I.... phoionicd,com/spct:s,html.
587
588
Current and Resistance
Chapter 17
SUMMARY 17.1
Electric Current
The average electric current 1 in a conductor is ddincd as
i1Q
If a conductor has length resistance is
e and cross-sectional area It, its
[17.10]
[17.5]
,..· here uQ is the charge that passes through a cross section of the conductor in time .:it. The SI unit of current is the ampere (A); 1.:-\ = I Cis. By con"cllliol1, the direction of current is 'he direction of flow of positive charge. The instantaneous current I is the limit of the average current as the time interval goes (Q zero:
where p is an intrinsic property or the conductor called the electrical resistivity. The 81 unit: of resistivity is the ohm· meter (n· m).
1
/ =
.w
lim u/---!'(l
17.2
~-
ill
la~' =
lim tlQ .l1-.4 (I
D,t
[17.1b]
A Microscopic View: Current and Drift Speed
The current in a conductor is related to the motion of the charge carriers by [17.2] where n is the number of mobile charge carriers per unit volul11c, q is the charge 011 each carrier, V,I is the dri ft speed of the charges, and A is the cross-sectional area of the conductor.
17.4
17.5
Resistance, Resistivity, and Ohm's Law
The resistance R of a conduclor is defined as the ratio of the potential difference across the conducLOr La the currentin it: [17.3]
Temperature Variation of Resistance
Over a limited temperature range, the resistivity of a conductor varies with temperature according to the expression p
= Po[l + a.(1'-
[,\1=
m
[17.4]
[17.6]
where a is the temperature coefficient of resistivity and Po is the resistivity at some reference temperature 70 (usually taken to be 20°C). The resistance of a conductor varies with temperature according to the cxpre:-;sion R = Ro[l
17.6
+ a.(T-
7,)]
[17.7]
Electrical Energy and Power
Ira potential difference .6.\1 is maintained across an elecu'ical device, the power, or rate at which energy is supplied to the device, is
(ip=ji1V
[17.8]
Because the potential difference across a resistor is dV= JR, the power delivered to a resistor can be expressed as L\ V:'> '!J'= j-R=-,.,.
The 51 units or resistance are volt.s per ampere, or ohms (fl); I n = I VIA. Ohm's law describes many conductors for which the applied voltage is directly proportional to the current it causes. The proportionality constanl is the resistance:
7,)]
'J
R
[17.9J
A kilowatt-hour is the amount of energy convened or consumed in one hour by a device supplied with power at the rate of I k\'V. II. is equivalent to 1 kWh ~ 3.60 X 106J
[17.10]
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE·CHOICE QUESTIONS 1. A wire carries a current or 1.6 A. How many elect.rons per second pass a given point in the ,..· ire? Choose the besl estimate. (a) 10" (b) JOIS (c) JO'" (d) 10'" (0) 10"
R, \.. . hat is the resistance of" wire B? (a) 4R (b) 2R (c) R (d) R/2 (0) 11/4
2. \Iv'ire A has lhe same length and l.wice the radius of wire B. Both wires are made of the same material and carry the same current.. Which or the following equations is t.rue concerning the drift velocities VA and VII of electrons in the wires? (a) V..1 = 2vB (b) v.4 = v/I (c) V,I = v,j2 (d) V tl = 4v li (e) V._\ = v,J4
4. Three wires are made of copper having circular cross sections. Wire 1 has a length L and radius 1: Wire 2 has a length L and radius 2'1". ·Wire 3 has a length 2L and radius 3,: \'Vhich wire has the smallest resistance? (a) \\'ire 1 (b) wire 2 (c) wire 3 (d) All three wires have the same resistance. (e) Not enough information is given to answer t.he question.
3. Wire B has the same resistivity, twice the length, and twice lhe radius of wire A. If wire A has a resistance
5, \Vhich of the following combinations of units has unit.s of energy? (0) Cis (b) kWh (c) kW (d) A· h (e) W/s
Conceptual Questions
6. The current versus \'oltage behavior of a certain elec· tl'ical device is showll in Figure rVICQJ7.6. When the potential difference across the de\'ice is 2 V, what is the resistance? (a) I n (b) ~ n (c) n (d) undefi ned (c) none of these
i
2
"
.~
234 \'oltage (V)
FIGURE MCQ17.6
9. A color television set draws about 2.5 A when connected to a 120 V source. What is the cost (with electrical energy at 8 cents/kWh) of running the set for 8.0 hours? (a) 2.0 cents (b) 4.0 cents (c) ]9 cellls (d) 40 cents (e) 62 cents 10. A potential difference or l.0 V is maintained across a 10.0-11. resistor for a period of 20 s. "Vbat LOtal charge passes through the ,.. . ire in this time inren1al? (a) 200 C (b) 20 C (c) 2 C (d) 0.005 C (e) 0.05 C
~
$ 2
589
] I. Three resistors, A, B, and C, are connected in parallel and
Multiple-Chuice Questions Gand 7
7. The current vs. vollage behavior of a certain eleClrical device is shO\"n in Figure MCQI7.6. \,Vhell the potentia] difference across the device is 3 V, what is the resistance? (a) 08~ n (b) 1.2 n (c) 3.0 n (d) 1.33 n (e) 2.3 n 8. A metal wire has a resistance or 10.00 {l at a temperature of 20.0°C. Ir the same wire has a resistancc of 10.55.n at 90.0°C, what is the resistance of this same wire when ilS temperature is -20°C? (a) 0.700 11. (h) 9.69 n (c) 10.3 n (d) 13.8 n (e) 6.;;9 n
12. Three resistors, A, B, and C, are connected in series in a closed loop wilh a battery, with the resistance of A being the smallest and the resistance of C the greatest. Across which resistor is the voltage drop the greatest? (a) A (b) B (c) C (d) The \'oltage drops arc the same for each. (e) [V[ore information is needed to answer the question. 13. The power delivered to a resistor is 4.0 \,V when a certain voltage is applied across it. How much power is delivered to t.he resistor when the \'oltage is doubled? (a) 2.0 VI' (b) 4.0 W (c) 8.0 W (d) 16 VI' (e) 32 W
CONCEPTUAL QUESTIONS 1. Car baueries are often rated in ampere-hours. Does this unit designatc the amount of current, pO\'I"er, energy, or charge that can be drawn from the battery?
7. When I he voltage across a certain conductor is doubled, the current. is observed to triple, ,",Vhat can you conclude about the conductor?
2. We have seen that <1n electric Ileld lllUst exist inside a conductor that carries a curren\. How is that possible in view of the fact that in electrostatics we concluded that the electric field lllllSl bc tero inside a conductor?
8. There is an old admonition given to experimenlcrs to "keep one hand in the pocket.·' when working around high voltages. Why is this warning a good idea?
3. Why don't the free electrons in a metal fall 1.0 the bot10m of' the metal due to gravity? And charges in a COI1dllClOr are supposed to reside on the surface; why don'l the free electrons £111 go to the surface? 4. III an analogy between lrarfic flow and electrical current, what would correspond to the charge Q? What would correspond 10 t.he current. n 5. (\lc'.... spaper articles often have st.atements such as "10000 volts of electricity surged through the yictim's boely." What is wrong wilh this statcment? 6. Two lightbulbs are each connected to a voltage of 120 V. One has a pOWCI' 01'25 \'\', the other 100 W. Which lighlbulb has the higher resistance? \""hich Iightbulb carries more current?
9. When is more power delivered to a lightbulb, immedi· ately after il is turned on and the glow oflhe filament is increasing or after ir has been on for a few seconds and t.he glow is steady? 10. Some homes have light dimmers that. are operated by rotating a knob. What is being changed in the electric circuit ,vhen the knob is rotated? 11. ,",Vhat could happen to I he drift velocity of the electrons in a ""'ire and to Lhe current in lhe wire if the electrons could move through it freely without resistance?
12. Use the atomic theory of maner to explain why the resistance of' a material should increase as its tempera1U re increases.
590
Chapter 17
..
Current and Resistance
_~---.
~
PROBLEMS' •
~
WebAssign
I, :!. :3 -
m
mJ
•
~.v;:~'
(b) Starling with the density of iron and lhe resull of part (a), compute the molal' density of iron (the nUlllbel' of lllOIeS of iron per cubic meter). (c) Calculate the number density or iron atoms using Avogadro's number. (d) Obti"lin lhe number density of conduction electrons given that there are two conduction electrons per iron ,110m. (e) I f" the wi re carries a current of 30.0 A, calculi"ltc the drif"t speed of conduction elect.rons.
The Problems for this chaplcr Illay he assigned online OIL WebAssign.
;,ll itigillful \~,tld. lll{l"lllll"l!ialt:, l h.lllen~in~
= denotes guided problem =
denotes enhanced conlellt problem
II = biomedical application o = denolt.:s full solUlion available in Student Soluliolls Mmlllol/ Study Guide
SECTION 17.1 ELECTRIC CURRENT SECTION 17.2 A MICROSCOPIC VIEW: CURRENT AND DRIFT SPEED
[I]lf a
current of 80.0 mA exists in a metal wire, how many eleClrons flo\\' past a given cross section of the wire in 10.0 min? In what direction do the electrons travel with respecl to the current?
2.
mil A copper wire has a circular cross section with a
SECTION 17.4 RESISTANCE, RESISTIVITY, AND OHM'S LAW 10. An eleCTric heater G1ITies a current of 13.5 A when operating at a voltage of 1.20 X 102V. Whal is the resistance of the hearer? 11.
II A person
notices a mild shock if" the current along a path through the thumb and index nnger exceeds 80 J1.A. Compare the maximum possible voltage without shock across the thumb and index finger with a dry-skin resistance of -4.0 X 10" n and a wet·skin resistance of2 000 n.
radius of 1.25 !TIm. (a) If the wire carries a current of 3.70 A, find the drifL speed of the electrons in the wire. (See Example 17.2 for relevant data on copper.) (b) All other things being equal, what happens to the drift speed in wires made of metal h;wing a larger number of condm;tion electrons per atom than copper? Explain.
J 2. Suppose you wish to f"abricate a uniforlll wire out of 1.00 g of copper. If the wire is to have a resistance a = 0.500 n, and if all the copper is to be used, what will be (a) the length and (b) the diameter of the wil'e?
3. In the Bohr model of the hydrogen atom, an electron in the lowest energy state moves at a speed of 2.19 X lOt> mls in a circular path having a radius of 5.29 X 10- 11 Ill. What is the effenin,: currenL associated with this orbiting electron?
13. Nichrome wire of" cross-sectional radius 0.791 mm is 1O be L1sed in winding a heating coil. If the coil must carry a current of9.25 A when a voltage of 1.20 X 10 zV is applied across it.s ends, find (a) the required resistance ohhe coil i"lnd (b) the length of wire YOli musl use to lI'ind the coil.
4. A prOlan beam in an accelerator carries a current of 125 J1.A. If the beam is incidel1l on a I.ilrget, how man}' protons slrike the target in a period of 23.0 s?
14. Eighteen-gauge wire has a diameter of 1.024 mm. Calculate the resistance of]5 III of IS-gauge copper wire at
5. If the curreJ1L carried by a conductor is doubled, what happens to (a) the charge carrier density? (b) The elec· tron drift velocit}'?
[I[]A potential difference of 12 V is found to produce a current of 0040 A in a 3.2~m length of wire with a uniform radius of 0.40 em. What is (a) the resistance of the wire? (b) The rcsistivity ohhc wire?
6. If3.25 X 10-;\ kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.78 h. what is
the currem in the cell during that period? Assllllle the gold ions «-lIT)' one elementary unit of positi ....e charge.
[1JA
20'C.
16. ami Aluminum and copper wires of equal length are found 1.0 have the same resisti"lnce. What is the ratio of their radii?
200-km-long high-voltage transmission line 2.0 cm in diameter carries a steady current of" I 000 A. If the conducLOr is copper with a free charge density of" 8.5 X 10~8 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable?
17. A wire 50.0 m long and 2.00 mm in diameter is connected to a source widl a potemial difference of9.1 I V. and the current is found Lo be 36.0 A. Assume a temperalure of" 20°C and, using Table 17.l, identify the metal au( of which the \vire is made.
8. An aluminulll wire carrying a current of 5.0 A has a cross· sectional area of 4.0 X 10- 6 m 2 . find the drift speed of the electrons in the wire. The density of aluminum is 2.7 g/cm 3• (Assume three electrons are supplied by each atom.)
18. A rectangular block of copper has sides of length ]0 em. 20 cm, and 40 cm. If the block is conneCLcd to a G.O-V source across two of its opposite faces, what are (a) lhe maximum current and (b) the minimum currenl the block can carry?
9.
Ell
An iron wire has a cross-sectional area of" 5.00 X 10-[,111 2 . Carry oul steps (a) through (e) to cOlllpUle the drift speed of the conduction e1eClrol1s in the ",,·ire. (a) How mi"lny kilograms arc there in I mole of iron?
[I[]A wire of" initial length LlJ and radius 1"0 has a measured resistance of 1.0 n. The wire is drawn under tensile stress to a new uniform radius of r = 0.251'0' ',Vhat is the new resistance of the wire?
Problems
1'1
20. _ The human body can exhibit a wide range of resistances LO currellt depending on the path of the (lll-· rent, contact area, and sweat iness of I he ski n. Suppose I he resisl
m
Starting from Ohm's law. show that E = Jr. where E is the magnilllde 01' the electric field (assullIed constant) and) = f/A is called the current density. The resuh is in E:lct trlle in gcneral.
SECTION 17.5 TEMPERATURE VARIATION OF RESISTANCE 22. I I' a certain silvcr wire has"1 resistance of6.0U fl at 20.0"C. what resistance will it have at 34.0°C? 123.IWhile taking photographs in Death Valle}' on a day when the temperature is 58.W' C, Bill Hikcr finds tlut a certain voltage applied lO a copper wire produces a current of 1.000 A. Bill then travels 10 Antarctica and applies the same voltage lO the same wire. \Vhat current does he registcr therc if the tcmperature is -88.DOC? Assumc no change occurs in the wire's shape and size. 24. _ A length of aluminum wire has a resistance of 30.0.n ;:H 20.00C. V\:l1en the wire is warmed in an o\'en and reaches Ihermal equilibrium, the resio;;tancc or the wire increases to 46.2 .n (a) Neglecting thcrmal expansion, find the temperature of the oven. (b) Qualitatively. how would thermal expansion be expected lO arfect the answer? 25. A certain lightbulb has a tungsten f-ibmelll h"l\'ing a resistance of 15 n at 20°C and 160 fl when hoI. Assume E.quation li.7 can be used over the large temperature range here. Find the temperature of the filament when it is hoI. 26. At what tempcrature will aluminum have a rcsistivity lhat is threc times t.he resistivity or copper at room temper<\ture? 27. At 20°C. the carbon resistor in an electric circuit connected to a !i.O-V baucr}' has a resisLlI1ce or 2.0 X 10 2 n. \'Vhal is the current in the circuit \.. . hen the temperature of the ca rbon rises lO 80°C? 28. A wire 3.00 III long and 0.'150 mm:! in cross-sectional area has a resistance or 41.0 f1 al 20°C. If its resistance increases l.0 41.4 n at 29.0 c C, \\'11,11 is t.he tCllll)erawre coefficient of resistivity? 29. (a) A 34.5-m length or copper wire at 20.O"C has a radius of 0.25 mm. I I' a potcntial differencc of 9.0 V is applied across the length of the wire. determine the current ill Ihe wirc. (b) If the wire is heated 10 30.0°C while the 9.0-V potential difference is maintaincd. whal is the resulting current in the wire? 30. A toaster rated at 1 050 'N operates on a 120-V household circuit and a ll.OO-m length of Nichrome wire as its heat-
591
ing element. The operaling tcmpenuure of this elcmel1l is 320°C, What is the cross-sectional area of the wire?
31.11 Tn one form of plethysmograph (a device for measuring- volume), a rubber capillary tube with an inside c1i"IlWlt'I' of 1.00 111m is fillt'rl wiLh Hlcrcur}' at 20°C. The resistance of the mercury is measured with the aid of electrodes sealed into the ends of thc wbe. If 100.00 cm or the lube is wound in a spiral around a patient's uppcr arm, the blood flow during a heartbcat causes the arm 10 expand, stretching the tube to a length of 100.04 cm. From this observation, and assuming cylindrical symmetry, you can find the change in \'ollime of the arm, which g-i\'es an indicat.ion of blood flow. (a) Calculate the resistance of the mercury. (b) Calculate thc fractional change ill resistance during the heartbeat. (/-lint: The fraction by which the cross-scctional area of" the mercury t.hread decreases is the fraction by which lhe length increases, since the volume of mercury is cOnSl.anL) Take Pllg = 9.4 X 10-7 fl·m. 132.IA platinum resistance thermometer has resistances of 200.0 fl when placed in a DoC icc bath and 253.8 n when immersed in a crucible containing melting potassium. \'\'hal i~ the melting point of pOlassiu1ll? (Hill I: First dctermine the resistance or the platinum resistance thcrmomdel' at room temperature, 20°C.)
SECTION 17.6 ELECTRICAL ENERGY AND POWER 33. Suppose your waffle iron is rated at 1.00 k\V \.. . hcn connected 1.0 a 1.20 X 102 V source. (a) \-Vhat currenl does the waffle iron carry? (b) What is its resistance? 34. If' electrical energy costs 12 cents. or 0.12. per kilowatl!lour. how much does il cost to (a) burn a 100-W light.bulb for 2--1- h! (b) Operate an eleclric oven for 5.0 h ifit carries ;1 current of 20.0 A at 220 V?
35. A certain compact disc player draws a current of 350 mA at 6.0 V. I-:Iow much power is requircd player?
10
operate the
36. A high-voltage t.ransmission line with a resistance of 0.31 fi/km carries a currenl of I 000 A. The linc is al a potcntial or iOO kV af the power station and G\lTies the current to a cit}' located 160 km from the station. (a) \-Vhat is the power loss due (0 resistance in the line? (b) 'Vh'lt fraClion of the transmitted power does this loss represellt? [lliThe heating clement of a colTeemakcr operates at 120 V and carries
592
Chapter 17
Current and Resistance
39. ,,'hal is the required resistance of an immersion heater thai will increase the temperawre of 1.50 kg of water from IO.acc to 50.aoC in lO.O min while operating at 120 V' 40. A certain toaster has a healing e1emenllllade 01" Nicluomc
resistallce wire. When the toaster is I-irsl connected to a 120-V source ofpolcmial difTerence (and the wire is at a temperature of 20.0"C), the initial current is 1.80 A but rhe current begins to decrease as the resistive clemen! warms up. 'When the toaster reaches its final operating temperature, the current has dropped to 1.53 A. (a) Find the power the toaster com'ens when it is at its operming temperature. (b) What is Ihe final temperature of the heating element?
GOA copper cable is designed to carry a current or300 A with a power loss of 2.00 W/m. "Vhat is the rcqllil-ed radius of this Cable? 42. Baucrics arc rated in terms of ampere-hours (1\' h). For example, a ballery that can deliver a currem of 3.0 A [or 5.0 h is rated aL 15 A·h. (a) What is Ihe total ellergy, in kilowau-hollrs, stored in a 12 V bauery rated al 55 A· h? (b) AI $0.12 per kilowau-hour, what is the \'alue of the electricity that can be produced by this bauery?
43.11 The poremial difference across
44.
The cost of e1eclricity varies widely throughoUI! he United Slates; $0.120/"-Wh is a typical valuc. At this unit price, calculate the cost of (a) leaving a 40.0-\'\" porch liglu on for 2 weeks while you are on vacation, (b) making a picce of dark toast in 3.00 min with a 970-W waster, and (c) drying a load of clothes in 40.0 min in a 5 200-\V dryer.
4,5. An ll-W energy-efficient rluorescelll I
146.1_
An office worker uses an immersion heater to warm 250 g of 'vace I' in a liglu, covered, insulated cup from 20~C to lOWe in 4.00 minutes. The heater is a Kichrome resistance wire connected to a 120-V pO\'v'er supply. Assume the wire is at \Ooce throughom the 4.00-min time inter\-al. Specify a diameter and a length the wire can have. Can it be made from less lh
47. The heating coil of a hot-water heater has a rcsistance of 20 n and operates at 210 V. If electrical energy costs SO.OSO/kWh, wh
linearly over this temperature range. (d) \Vhat . . o ltage drop is required across the wire? (e) Why are LUngsten lightbulbs energetically inefficient as light sources? ADDITIONAL PROBLEMS
49, If a battery is rated at 60.0 A· h, how much total charge GlI1 it deliver before il goes "dead"? 50. A car owner forgcts to turn off the headlights of his car \',:hilc it is parked in his garage. If the 12-V battery in hili car is ratcd at 90 A· h and each headlight requires 36 \,V of power, how long will it take the bauer}' 1.0 completcly discharge? 51. An electronic device requires a power of 15 W when con· nccted to a 9.0-V baltery. How much power is delivered to the device if it is connccted to a 6.0-V battery? (Neglect the resistances o[ the batteries and assume the resistance of the de\"ice does not change.) 52, Detcrmine the temperature at which the resistance of an alulllillllll1 wire will be twice its value at20 c C. (Assume its coefficient of resistivit)' remains constanl.)
153.IA particular wire has a resistivity or3.0 X 10-t1f!'m and cross-sectional are", of 4.0 X JO-ti m 2 . A length of this wire is to be used as a resistor that will develop 48 \·v of power when connected across a 20-V batter}'. What length of wire is required?
54. Birds resting on high-voltage power lines are a common sight. The copper 'wire on which a bird stands is 2.2 cm in diameter and carries a current of 50 A. If Ihe bird's feet arc 4.0 cm apart, calculate the potential difference across its body. 55. An experiment is conducted to measure the electrical resistivity of Nichrome in the form of wires with differem lengths and cross-sectional areas. For one set of measuremellIs, a student uses 30-gauge wire, which has a crosssectional area of 7.30 X IO- B m 2 . The student measures the potential difference across the wire and the currelll in the wire with a voltmeter and an ammeter, respecrively. For each of' the measurements given in lhe following table takcn on wires or three different lengths, calculate the resistance of the wires and the corresponding value of Ihe resistivity. L (ml
"'v (V)
I (Al
0.5'W 1.(128
5.22
0.500
5.82
0.276
1.543
5.94
0.187
R(fl)
P (fhn)
'What is the average value of the resistivity, and how does this \'alue compare with the value given in Table I7.I? 56. A c
a I 200-\\' waffle iron and a 500-\"1' coffeepot. Usually, you
Problems operate these appliances from a 110-V outlet for 0.500 h each day. (a) At 12 cems per HVh, how much do you spend LO cook breakfast during a 30.0-day period? (b) You find yourseifaddicle(lto warnes and would like LO upgrade to a 2 400-W waffle iron that '..·ill enable you to cook twice as many warnes during a h
58. The current in a conductor varies in lime as shown in Figure P17.58. (a) How many coulombs of charge pass through a cross section of the conductor in the imerval from I. = 0 to 1= 5.0 s? (b)\o\'l1al constant current would transport the same LOtal charge during the 5.0-s interval as does the actual current?
593
current of 4.00 A. Is this system adequatcl}' protected against overload? 63. A resistor is constructed by forming a material of resistivity 3.5 X 10'" n· m illl.o the shape of a hollow cylindel" of length 4.0 cm and inner and outer I'adii 0.50 cm and 1.2 em, respectively. In use. a potential difference is applied between the ends of the cylinder. producing a current parallel to the length of the cylinder. Find the resistance of the cylinder. 64. A 50.0-g sample of a conducting material is all that is 'lvailable. The resistivity of [he material is measured to be II X \0-8 n· 111, and the density is 7.86 g/cm 3 . The material is to be shaped into a solid cylindrical wire that has a total resistance of 1.5 n. (a) \\1nat lenglll of wire is required? (b) What must be the diameter orthe wire?
165.111
An x-ray tube used for cancer therapy operates at 4.0 /\:rv, with a beam current of 25 mA striking the metal t
5 ~ 91-__-1 UooL---"---'c-..Jc--'----!-,2 3 4 Time (s) FIGURE P17.58
59. An electric car is designed to run off a bank of 12.0-V baLleries with
When a straight wire is heated. its resistance changes 66. _ according to the equation R~
Roll +,,('1'-70)]
(Eq. 17.7), where a is the temperalllre coefficient of resistivity. (a) Show that a more precise result, which includes the length and area ora wire change when it is heated, is
R [l + ,,(T- 7~)][1 + ,,'('1'- 70)] R= o [I + 2,,'('1'- 70)] where a' is the coefficient of linear expansion. (See Chapter 10.) (b) Compare the two results for a 2.00-m-long copper v,.·ire of radius 0.100 mm, starting at 20.0°C and heated to 100.OOC. 67. :\ man wishes to vacuum his car with a canister vacuum cleaner marked 535 \·V at 120 V. The car is parked far from the building, so he uses an extension cord 15.0 mlong to plug the cleaner inw a 120-V source_ Assume the cleaner has COmitant resistance. (a) If the resistance of each or the two conductors of the extension cord is 0.900 n, what is the actual power delivered to the cleaner? (b) If. instead, the power is to be at least 525 W. what Illust be the diameter of each of two identical copper conductors in the cord the young man buys? (c) Repeat part (b) if the power is to be at least 532 \\1. (Suggfslion: A symbolic solution can simplify the calculations.)
This vision-imporied patient is wearing on artificial vision device. A comero on the lens sends video to
0
portable computer. The com-
puter then converts the video to signals that stimulate implants in the patient's visual cortex, giving him enough sight to move around on his own
18.1
Sources of emf
18.2
Resistors in Series
18.3
Resistors in Parallel
18.4
Kirchhoff's Rules and Complex DC Circuits
18.5
RC Circuits
18.6
Household Circuits
18.7
Electrical Safety
18.8
Conduction of Electrical Signals by Neurons
o
"
z &
DIRECT-CURRENT CIRCUITS Batteries, resistors, and capacitors can be used in various combinations to construct electric circuits, which direct and control the flow of electricity and the energy it conveys. Such circuits moke possible 011 the modern conveniences in a home: electric lights, electric stove tops and ovens, washing machines, and a host of other appliances and tools. Electric circuits are also found in our cars, in tractors that increase forming productivity, and in all types of medical equipment that saves so many lives every day. In this chapter we study and analyze a number of simple direct-current circuits. The onaly+ sis is simplified by the use of two rules known as Kirchhoff's rules, which follow from the principle of conservation of energy and the low of conservation of charge. Most of the circuits are assumed to be in steady state, which means that the currents ore constant in magnitude and direction. We close the chapter with a discussion of circuits containing resistors and capacitors, in which current varies with time.
18.1
SOURCES OF EMF
A ell rrenl. is m£1 inl.a ined ina closed ci rClI it by a source of cm f.l Among such sources are any devices (for example, baueries and generators) that increase the potel1lial cnergy of the circulating charges, A source of emf can be thought of as a "charge pump" thal forces electrons to move in a direction opposite the electrostatic field inside the source. The emf B of a source is the work done per unit charge; hence Ihe Slunit o[cmfis the volt. Consider the circuit in Active Figure 18.1a consisting of a bauery connected to a resisLOr. \,Ve assumc the connecting ",..ires have no resistance, If we neglecl Ihe internal resistance of the battery, t.he potential drop across the battery (the tcrmin<11 voltage) equals Ihe emf of the bauery. Because a real battery ah•..-ays has some inlernal resis£ance ,~ however, the terrninal \'ollagc is nol equal LO the emr. The circuit of Active Figure 18.la can bt: described schcmatiGllly by the diagram ITlie terlll ..... a~ origitully an abb. (" i,ltiOI1 t"l ("olIJrag:cll.
594
~{l'(lrollloli,'"{i't"re.
bill <-'mf is nOI rt:ally a filHT, s" Ille IUllg" form is db-
18.2
Resistors in Series
in Active Figure 18.1b. The battery, represented by the dashed rectangle, consists o[ a source o[ emf in series with an internal resistance I: ~ow imagine a positive charge movi ng through the battery [rom {f La h in the figu reo As the charge passes from the negative to the positive terminal or the bauery, the pot.ential of the charge increases by 8. As the charge moves through the resist.ance }~ hm"'evel', its potent.ial decreases by t.he amount lr, where I is t.he current in the circuit.. The terminal voltage of the bauery, 6,\1= \lb ~ \~. is therefore given by
Baltery +
e
D.\I~e-Jr
[18.1]
From this expression, '...-e see that e is equal to the terminal voltage when the cur.'ent is zero, called t.he open-circuit voltage. By inspect.ing Figure 18.lb, we find that. the terminal Yoltage 1\1 must also equal the potential difference across the ext.ernal resist.ance N, often called the load resistance; that is, 6.V = IR. Combining this relationship with Equation 18.\, we arrive at
(a) iE;-----~-' (l
e~IR+h
595
[18.2]
1-
;
,I + '/
1
",
---~
I
Solving for t.he current gives
b
II 13 I~- R+r
d
The preceding equation shows that the current in this simple circuit depends on both the resistance external to the bauery and the ill1.ernal resistance or the batlery. Tf R is much greater lhan r, 've can neglecl r ill our analysis (an option ,ve usually select). If'we multiply Equation 18.2 by the current I, we get
[13
~
['R+
(h)
ACTIVE FIGURE 18.1 (a) A drcuil consisling of a resislor conneCied LO the lenninals ofa bal· lei;". (b) A cireuil diagram of a source of em r £ h'lVing illlernal resistance r cOllnected to an external resislor R.
['1'
This equation tells us that. t.he total pm..· er outpUl IS of the source of emf is con~ "erted at the rate J'lR al which energy is delivered to t.he 10:c:ld resistance, plu.s the rate 12r at which energy is delivered to the internal resistance. Again, if r« R, most. of the power delivered by t.he battery is t.ransferred to the load resist.ance. Unless otherwise stated, in our examples and end-Dr-chapter problems we will assume the internal resistance o[ a battery in a circuit is negligible.
QUICK QUIZ 18.1 True or False: \Vhile discharging. the terminal voltage of a battery can never be greater t.han the emf of the baltery. QUICK QUIZ 18.2
18.2
Why docs a baller)' get warm while in lise'
An assortment of baitcries.
RESISTORS IN SERIES
\,Vhen two or more resistors are connected end to end as in ACliye Figure 18.2, they are said to be in s(!ries. The resistors could be simple devices, such as lightbulbs or heating elements. \~7hcn 1\\"0 resistors R] and R'2 are connected to a battery as in Active Figure 18.2 (page 596), the current is the same in the two resistors because any charge that flows through R1 must also flow through R2 • Th is is analogoLis to water flowing through a pipe '\'ith t.wo constrictions, corresponding to R[ and R 2 . Whatever volume of water flows in one end in a gi,·en time imerval must exit the opposite end. Because the potential difference between a and bin Acti"e Figure IS.2b equals IN] and t.he potential difference between band c equals JR 2 , the potential difference between (J and cis D.V~
IR, + IR,
~
I(R, + R,)
Regardless of hm.,,· many resistors we ha"e in series, the sum of the potential differences across the resistors is equal to the total potential difference across the combination. As we will show lat.er, this result is a consequence of the conservation
TIP 18.1 Whot's Constont in a Battery? Equalion 18.21>how.. thaI lhe current in a circuit depends on lhe resistance of the bauery, so a ballel·y can 'I be considered a source of constant current. Even the terminal "oltage of a bauery givt;n by Equalion 18.1 can't bl.' considered conSlant because I he internal resistance can ch;lIlge (due LO "'anning, for example, during the operation ofthc baIter)'). A banery is, however, a source of constant emr.
596
Chapter 18
Direct-Current Circuits
'" ~I
(h)
(a)
(01
ACTIVE FIGURE 18.2 t\ series COllllcction of 111'0 resi.~lors, HI ,[Ild U:!.. The <:lllTCllls ill the resistors are the same, ,lllrllhe cCjuivI?~q
a!ellt n:sisl.IIlCC" oflbe cOlllbination isgi,"en by
= HI + l?:!..
of energy. Active Figure IR.2c sho\vs an equivalent resisLOr Ref] that can replace the two resistors of the original circuit. The equivalent resistor has the same effect on the circuit because it results in the same current in the circuit as the two resistors. Applying Ohm's la\'l to this equivalcnt resistor, \ve have L'>V= lReq
Equaling thc preceding two expressions, \ve have
or Rcq
Equivalent resistance of a series combination of resistors ~
= R 1 + R'J,
(series combination)
[18.3]
An extcnsion of the preceding analysis shows that the equivalent resistance of three or more resistOrs connected in series is [18.4]
Therefore, the equivalent resistance of a series combination of resistors is the algebraic sum of the individual resistances and is always greater than any individual resistance. Note that if the filament. of one lightbulb in Active Figure 18.2 \vere t.o fail, the circuit \voldd no longer be complete (an open-circuit condition would exist) and the second bulb would also go out.
APPLYING PHYSICS 18.1
CHRISTMAS LIGHTS IN SERIES
A new design for Christrllas lights aIlO\,,'s them to be connected in series. A failed bulb in such a string- would result in an open circuit, and all the bulbs would go out. I-1m-\>' can the bulbs be rcdesig"ned to prevent that
from h<'lppening?
d II
Explanation Jfthe string oflighls contained the usual kind of bulbs, a failed bulb would be hard to locate. Each bulb \vallle! have to be replaced with a good bulb, one by one, umil the failed bulb was found. Tf there happened LO be t\\'o or more failed bulbs in
FIGURE 18.3 (Applying Physics 18.1) (a) Schclllmicdiagram of ;llllodern "llliniallire" holiday liglllhulb, wilh ajnlllperconnection LO pro\'ide a current palh if the fihtlllenl breaks. \Vhcn the fdanll.:llt is inlacl, charges flow in the filanlcll1. (b) A holida~ lighlbulb with a hrob.:ll tilamellt. In this case, charge:. flow in the
Filallwnl
Jump!.:!" CI;l:>S
(a)
insulator
(hi
jltlllpCr COllll('Cli'JII.
18,2
the string of" lights, finding them would be a lengthy and annoying task. Christmas lights use special bulbs that have an insuJ:ued loop of ,·.. ire (ajumper) across the conduc[~ lllg supports to the bulb filaments (Fig. 18.3). If the f1Iamcnl breaks and the bulb fails, the bulb's resis~
Resistors in Series
597
tance increases dramatically. As a result, mOst of the applied voltage appears across the loop of wire_ This voltage causes the insulation around the loop of 'v ire to burn, causing the metal wire ro make electrical COil tact with the supports. This produces a conducting path through the bulb, so the other bulbs remain lit.
QUICK QUIZ 18.3 In Figure 18.4 the current is measured with the ammeleI' at the bottom 01" the circuit. ","'hen the switch is opencd, docs the reading on the ammeter (a) increase, (b) decrease, or (c) llot change? QUICK QUIZ 18.4 The circuit in Figure ]8.4 consists of £\'\'0 resistors, a s\vil.ch, an ammeter, and a bauery. \Vhen the switch is closed, power ~<.:. is delivered to resistor R\. \-"hen the s\vitch is opened, which of the following statements is true about the power q]>" deli\-ered to RJ ? (a) lP" < qpc (b) f!Pu = 7P(
FIGURE 18.4 and 18.4)
(c)'iJ',,>'iJ',
EXAMPLE 18.1 Goal
(Quick QuiLZ<:S lH.3
Four Resistors in Series
Analyze several resistOrs connected in series.
~.O
n
4.0
n
:>.0 fl
i.a
n
18.0n
Problem Four resistOrs are arranged as shown in Figure ]8.5£1. find (a) the equivalent resistance of the circuit and (b) the current in the circuit if the emf' of the battery is 6.0 V. Strategy Because the resistors are connected in series, summing their resistances gives the equivalent resistance. Ohm's law can then be used to find the current.
6.0 V
(a)
(b)
6.0 V
FIGURE 18.5 (Ex,llllple 18.1) (a) Four resiSlors connected in (h) The equivalcnl resislancc oflhe circuit ill (a) .
Solution (a) Find the equivalent resisLance of the circuit. Apply Equation 18.4, summing the resistances:
R,,,
=
II, + R, + R, + H., = 2.0
~
18.0
n + 4.0 n + 5.0 fl
T
7.0 fl
n
(b) Find the current in the circuit. Apply Ohm's law to lhe equivalent resistor in Figure 18.5b, solving for the current:
...... -
6> II
6.0 V
R,."
18.0 fl
I~-=---= ~A
.
_
.
_
~
Remarks A common misconception is that the current is "used up" and steadily declines as it progresses through a series of resistors. T'hal would be a violation of conservation of charge. QUESTION 18.1 The illternal resistance of lhc batLery is negleCled in this example. How would il affect the final current, if taken into account? EXERCISE 18.1 Because the current in the equivalent n:sistor is A. find t.he \Tollage drop across each resistor.
1
Answers
Ll Vln =
*V;
*A, the currenl in each resisLOr of the original circuit must also be
~ V"n = ~ V; ~ ViH = ~ V; ~ Vi!! = ~ V
598
Chapter 18
Direct-Current Circuits
RESISTORS IN PARALLEL
18.3
Nov{ consider two resistors cOllnected in parallel, as in Active Figure 18.6. In this case the potential differences across the resistors are the same because each is connected directly across the battery terminals. The currents are generally not the same. vVhen charges reach point a (called ajunclion) in Active Figure 18.6b, the current splits into two pans: f l , flowing through Rj ; and [2' Om...ing through Rc.!.. If R 1 is greater than R'2' then 'I is less than 1'1. In general, more charge travels through the path with less resistance. Because charge is conserved, the current I that enters point a must equal the tot.al current II + 12 leaving that point. ~'1ath ematically, this is written
The potential drop must be the same for the t\-\'o resistors and must also equal the potential drop across the battery. Ohm·s 1m.,: applied to each resistor yields
Ohm's law applied to the equivalent resisLOr in Active Figure 18.6c gives ~v
{=Re(l
"Vhen these expressions for the currents are substituted into the equation' = II 12 and the .:lV's are cancelled, we obtain
+
[18.5]
(parallel combination)
An extension of this analysis to three or more resistors in parallel produces the following general expression for the equ iva lent resistance: Equivalent resistance of a parallel combination of resistors ~
1
1
1
1
R,q
R1
H,
R,
--=-+-+-:-+
[18.6]
From this expression, \...·e see thal the inverse of the equivalent resistance of two or more resistors connected in parallel is the sum of the inverses of the individw ual resistances and is always less than the smallest resistance in the group.
ACTIVE FIGURE 18.6 (a) A parallel connection of (Wo lightbulbs with resistances RI and R.2" (b) CircuiL di;Jgram for the twOresistor cirellil. The pOlclIlial differ-
RI
ences across HI and R2 arc the same. (c) The equivalent resistance uf the combinatioll i~ gi\"cn by the reciprocal fdaliamhip l/R~'l = IIR I + IIR.!"
ilV J =.6.lf2 =.6.V
1
R,
Re ,!
..
d a
-
'
(a)
..
Balt!.:l)"
" "F
T
I' (b)
I
R;- R'2
R';!
I,
It
'.
I
'"
b
It I."F
'-----j+I' (c)
18.3
Resistors in Parallel
599
QUICK QUIZ 18.5 In Figure 18.7 the current is measured v,lith the ammeter on the right side of the circuit diagram. \,Vhen the switch is closed, does the reading on the ammeter (a) increase, (b) decrease, or (c) remain the sallle? QUICK QUIZ 18.6 \'Vhen the switch is open in Figure 18.7, power 'd'" is delivered to the resistor Rl. When the s\..·itch is closed, which of" the following is tfue about the power ~<: delivered to R]? ("Neglect the internal resistance of lhe banery.) (a) '!Pc < '!Po (b) '!P,. = '!Po (e) '!Pc > '&"
FIGURE 18.7
(Quid. Quizzes 18.5
and 18.6)
EXAMPLE 18.2 Three Resistors in Parallel Goal
-
Analyze a circuit having resistors connected in parallel.
Problem Three resistors are connected in parallel as in Figure ]8.8. A potential difference of 18 V is maintained between points (l alld b. (a) Find the current in each resistor. (b) Calculate the power delivered to each resistor and the total power. (c) Find the equivalent resistance ol"the circuit. (d) Find the total power delivered to the equivalent resistance.
a
I, j
-=-
18 V
Strategy To get thc currcnt in each resistor we can use Ohm's la\\' and the fact1.hat the voltage drops across parallel resistors arc all the same. The rest of the problem jusl requires substitution into the equation for power delivered to a resistor, ~ = /'2R, and the reciprocal-sum law for parallel resistors.
,..
I
1,
,
•
> 3.0 fl
j I,!
6.0 fl
b
.
I,
• • '"""11
~v
ISV
R,
3.0 D
~V
ISV
~\!
IS V
6.0A
10 = = - - = 3.0 A H, 6.0 D I, = HI = 9.0 D =
2.0 A
(b) Calculate the power delivered to each resistor and lhe l.01.al power. Apply qp = ['2R to each resistor, substitutinp; the results Crom part (a):
Sum to get the lotal power:
3 D:
'&,
=
I,'R,
= (6.0 A)'(3.0 fl) ~ 11 0 W
6 fl:
'&,
=
I,' H,
= (3.0 A)'(6.0 fl) =
9 fl:
'&3
=
l,'H, = (2.0 A)'(9.0 fl) = 36 W
'!l'eo'
= 110 W
54 vI'
+ 54 W + 36 W ~ 2.0 X
]Q' W
(c) Find the equivalent resistance of the circuit. Apply the reciprocal-sulll rule, Equalion 18.6:
9.0,0
FIGURE 18.8 (Example 18.2) Thrct-· resislOrs connected in parallel. The ,"oil age across each re<;islOr is 18 V.
Solution (a) find the current in each resistor. Apply Ohm's law, solved for the Cllrrenl [delivered by the battery to find the current in each resistor:
'>
>
1
I
I
I
Ref]
R]
R'l.
R?,
-=-+-+I
I
]
I
II
He',
3.0 fl
6.0 fl
9.0 fl
IS fl
He, =
11
-=--+--+--~--
IS
D
=
1.6 fl
600
Chopter 18
Direct-Current Circuits
(d) Compmc the power dissipated by the equivalent resistance. (if
Remarks There's something imponanl to notice in pan (a): the smallest 3.0 [l resistor carries the largest current, whereas the other, larger resistors of' 6.0!1 and 9.0 [l carry smaller currents. The largest current is ah'\'ays found in the path orIeast resistance. In pan (b) the power could also be [ouncll,.vith r;p = (.~V)2/R. Note that @\ = 108 VIi', but is rounded to 110 W because there are only two significant figures. Finall)', notice that the total power dissipated in the equivalent resistor is the same as the sum of the power dissipated in the individual resistors, as it should be.
TI P 18.2 Don't Forget to Flip It! The mOSI COlllmon llIistake in calculating the equivalent resislance for resistors in parallel is to rorgel to inverl the answer afrer summing the reciprocals. Don't rorgetlo nip it:
QUESTION 18.2 If a founh resiswr were: added in parallel to the other three, haw would the equivalelH resistance change? (a) It would be larger. (b) II would be smaller. (c) More information is required to determine the effect. EXERCISE 18.2 Suppose the resistances in the example are 1.0 fl., 2.0 0, and 3.0.n, respectively, and a new voltage source is provided. If the current measured in the 3.0-0 resistOr is 2.0 A, find (a) the potential difference provided by the ne\v battery and the cllrrems in each of the remaining resistors, (b) the pO\.. .' er delivered to each resistor and the LOlal power, (c) the equivalem resistance, and (d) the total current and the power dissipated by the equivalent resistor. (a) £ = 6.0 V, I, = 6.0 A, J, = 3.0 A (b) '!J'"q = 66 W
Answers
(if,
= 36 W, (if, = 18 W, '!J',
~ 12 W, (ifw,
= 66 W (c) ff f1
(d) 1=11 A,
QUICK QUIZ 18.7 Suppose )'ou have three identical lightbulbs. some wire, and a bauery. YOll connect one Iightbulb to the batter)' and take nOte of its brightness. You add a second lightbulb, connecting it in parallel with the previous lightbulbs, and again take note of the brightness. Repeat the process ,·... ith the thi rei lightbulb, connecting it in parallel ,.. .Iith the other two. As the lightbulbs are added, what happens 1.0 (a) the brightness of the lightbulbs? (b) The individual currents in the lightbulbs? (c) The power delivered b), the bauery' (d) The lifetime of the balter)'? (l\eglect the battery's internal resistance.) QUICK QUIZ 18.8 Trthe lightbulbs in Quick Quiz 18.7 are connected one by one in series instead of in parallel, what happens to (a) the brightness of the lightbulbs? (b) The individual currents in the lightbulbs? (c) The pm...· er delivered by the battery? (d) The lifetime of the battery? (Again, neglect the battery's internal resistance.)
APPLICATION Circuit Breakers
Household circuits are always wired so that the electrical devices are connected in parallel, as in Active Figure 18.6a. In this way each device operates independently of the others so that if one is switched off, the others remain on. For example, if one of the lightbulbs in Active Figure 18.6 '.. ·ere removed from its socket, the other would continue to operate. Equally important is that each device operates at the sallle voltage. If t.he devices 'were connected in series, the voltage across any one device would depend on hm..· lllany devices were in the combination and on their individual resistances. In many household circuits, circuit breakers are used in series with other circuit elements for safety purposes. A circuit breaker is designed to switch off and open the circuit at some maximUll1 value of the current (typically 15 A or 20 A) that depends on the nature of the circuit. IT a circuit breaker were not used, excessive currents caused by operating several devices simultaneously could result in exces· sive ,,,ire temperatures, perhaps causing a fire. In older home construction, fuses
18.3
601
Resistors in Parallel
\vere llsed in place of circuit breakers. "\Then the current in a circuit exceeded some value. the conductor in a fuse melted and opened the circuit. The disadvantage of fuses is that they are destroyed in the process of opening the circuit, whereas circuit breakers can be reset.
APPLYING PHYSICS 18.2
L1GHTBULB COMBINATIONS
Compare the brightness of the four identical lightbulbs shown in Figure 18.9. \·Vhat happens if bulb A fails and so cannot conduct current? \'Vhat irC fails? \'\'hat ifD fails? Explanation Bulbs A and B are connected in series across the emf of t.he bauery, whereas bulb C is cOllneClcd by itself across the battery. This means the voltage drop across C has the same magnitude as the bat.tery emf, whereas this same emf is split between bulbs A and B. As a result, bulb C will glow more brightly than either of bulbs A and B, which will glow equally brightly. Bulb 0 has a wire connected across it-a shon circuit-so the potcntial difference across bulb D is zero and it doesn't glow. )fbulb A fails, B goes out, but C stays lit. lrC fails, there is no effect on the other bulbs. If D fails, the e\"ent is undetectClble because 0 was not gIO\...· ing initially.
APPLYING PHYSICS 18.3
A
~
B
~
~c
::
-f FIGURE 18.9
D
(Applying Physics
18.2)
THREE-WAY LIGHTBULBS
Figure 18.10 illustrates how a three-way lightbulb is constructed to provide three levels of light intensity. The socket or the lamp is equipped with a three~way switch for selecting diffcrentlight intensities. The bulb contains twO filaments. \Nhy arc the f1laments connected in parallel? Explain hm.. . the two filaments are used to provide the three dirferentlight intensities. Explanation If the fllaments were connected in series and one of them were to fail, there would be no current in the bulb anclthe bulb would not glow, regardless of the position orthe switch. \'\Then the fllaments are connected in parallel and one of them (say, the 75-W filament) fails, howc\Ter, the bulb will still operate in one of the s\"itch positions because there is current. in the other (100·\'\') filament. The three light. intensities arc made possible by selecting one of three values offilament resistance, lIsing a single valuc of 120 V for the applied voltage. The 75-\V filament offers Olle value of resistance, the 100-\0\' filament offers a second value, and the third resistance is obtained by combining the two filaments in parallel. \'\'hen switch 51 is closed and s\vitch 52 is opened, only the 75-'W filament carries current. \,Vhen s\.. . itch 5 J is open cd and swit.ch 5:! is closed, only the 100-\'\' filament carries current. \,Vhen both S\'I,jlches are closed, both filaments carry current and a total illumination corresponding to 175 \0\' is obtained.
PROBLEM-SOLVING STRATEGY
SIMPLIFYING CIRCUITS WITH RESISTORS l. Combine all resistors in series by summing the individual resistances and
draw the new, simplified circuit diagram. Usefu I facts:
= R1 + R'J, + R:~ + The current in each resistor is the same.
Rt,Xl
lOO-W filament
FIGURE 18.10 18.3)
(Applying Physics
602
Chapter 18
Direct-Current Circuits
2. Combine all resistors in parallel by summing the reciprocals or the resistances and then laking the reciprocal of the resull. Dnlw t he new, simplified circuit diagram. 1 Useful facts: - - =
Ro,
I
I
I
R,
H,
+- +- +
~
R,
The potential difference across each resisLOr is the same.
3. Repeallhc first two steps as necessary, until no further combinations can be made. If there is only a single battery in the circuit, the result ""ill LlsuaUy be a single equivalent resistor in series 'with the bancry. 4. Usc Ohm's law, 8.\1= JR, to determine the current in rhe cqlliv~dent resistor. Then work back\varcls through the diagrams, applying the lIseful facts listed in step 1 or step 2 to find the currents in the other resisLOrs. (In morc complex circuits, Kirchhoff's rules \",'ill be needed, as described in the next section).
EXAMPLE 18.3
Equivalent Resistance
Goal Solve a problem involving both series and parallel resistors. Problem Four resistors are connected as shown in Figure 18.lla. (a) Find the equivalent resistance between points a and c. (b) \tVhat is the current in each resisLOr if a 42-V battery is connected between a and c? Strategy Reduce the circuit in steps, as shm'\'n in Figures 18Jlb and l8.llc, lIsing the sum rule for resistors in series and the reciprocal-sum rule for resistors in parallel. Finding the currents is a malleI' of applying Ohm's law \"hile ' . .· orking backwards through the di'1grallls.
G.on
-I
8.0 II (a)
"
4.0 II
I
I,t ,
b
1,1 ~.O
II
!
12.0 II
2.0 II
(b)(~
FIGURE 18.11 (Example 18.3) The rOllr resisLOrs shown ill (a) call be reduced in sLeps LO all c(!llivalcllt I'l-fl resisLOI".
14.0 II (0)
~
" ~
.
• • • • • • • • • • • ' ' '111
Solution (a) Find the equivalent resistance of the circuil. The 8.0-0 and 4.0-D resistors are in series, so use the sum rule to Ilnd the equivalent resistance between (f. ancl b: I
1
I
1
1
I
1<,,,
H,
1<2
6.011
3.0 n
2.011
The 6.0-0 and 3.0-f! resistors arc in parallel, so use the reciprocal-sulll rule to find the equivalent resistance between band r (don't forget to invert!):
-=-+-=--+--=--
Ln the new diagram, 18.llb, there arc no\v twO resistors in series. Combine lhem with [he sum rule to find [he equivalent rcsislance of the circuit:
R o " = H, + 1<, ~ 12.011 + 2.0 11 = 14.011
1(" ~ 2.011
18.4
Kirchhoff's Rules and Complex DC Circuits
603
(b) Find thc current in each resistor if a 42·V batter)' is connected between points a and c.
a \~I"-
Find the ClllTcnt in the equivalent rcsisLOr in figure 18.llc, which is the LOtal current. Resistors in series all carry the same current, so the \"alue is the current in the 12-0. resistor in Figure 18.11b and also in the 8.0-0 and 4.0-0 resistors in Figure 18.1Ia.
1= -
Appl)' thejllnction rule to point b:
(1)
R~I
~
I = I,
42 V - - = 3.0 A 140
+
I~
The 6.0-0 and 3.0-fl resistors are in parallel. so the \'oltage drops across them are the same: 3.0A = I, + 21, = 31,
Substilllte this result into Equation (1), with 1= 3.0 A:
-->
I,
1.0 A
I, = 2.0 A
....
• • . • • . . • .A
Remarks As a final check, note that .'>1'" = (6.00)1, = (3.00)1, = 6.0 V and .'>V" = (120)1, = 36 V; therefore, a"f1f = aVab + .lV/x = 42 V, as expected. QUESTION 18.3 'Which of the original resistors dissipates energy at the grealesl rate? EXERCISE 18.3 Suppose the series resistors in Example 18.~ are now 6.00 fl and 3.00 n \vhile the parallel resistors are 8.00 fl (top) and 4,00 n (bottorn), and the battery provides an emf of 27.0 V. Find (a) the equivalent resistance and (b) the currents /, 11, and '2'
Answers
18.4
(a) 11.70 (b) 1= 2.31 A, I, = 0.770 A,
I~
= 1.54 A
KIRCHHOFF'S RULES AND COMPLEX DC CIRCUITS
As demonstrated in the preceding section, we call analyze simple circuits using Ohm's law and (he rules for series and parallel combinations of resistors. There arc, howc\'cr, man)' ways in \"'hich resistors can be connected so that the circuits formed can't be reduced to a single equivalent resistor. The procedure for analyzing mOl-c complex circuits can be facilitated by the use of t\.. .o simple rules called Kirchhoff's rules: (a)
I. The sum of the currents entering an)' junction mllSt equal the Slim of the
currellls leaving that junction. (This rule is often referred to as the junction rule.) 2. The slim of the potential differences across all the elements around an)' closed circuit loop must be zero. (This rule is usually called the loop rule.) Thejulletion rule is a statement of (o1l.'wnJation ofclwrgi'. \\'hate\"cr current enters a gi\'en point in a circuit must leave that poilll because charge can'l build up or disappear at a point. If we apply this rule to the junction in Figure 18.12a, we gel I, =
I~
+
I,
Figure 18.12b represents a mechanical analog or the circuit shown in Figure 18.12£1. In this analogwmer nows through a hranched pipe with no leaks. The now rate inlo the pipe equals the total now rate out of the two branches. The loop rule is equivalent to the principle of C01JSI'f1.H1lion oj nJergy. Any charge thaI. rnoves around any closed loop in a circuit (st:lrting and ending at the same
-
FJowin
(b) FIGURE 18.12 (a) A schematic diagram illuslraling Kirchhofrsjullclion rule, COllserva, ion of charge requires I hal \\ hatc\er current elller'i ajllllc,ioll musllt'a\"c ,haljunctioll. 111 this case. therefore. II = I'!. + 13, (b) A mechanical analog of the jllllcLion rule: lhe 11l;L flow in mUSl equallhc nt:L f1owOlll.
604
Chapter 18
Direct-Current Circuits
-
I
(a)
•
fI
•
'N'I
.:1\'= \'I,-I~= -1/1
-
b
I
(b) fI
~\'= \'.- \'<:= .LIU
b
8
(c)
-, +
•
a
\~=
..11'= \'1.-
+8
,-
b
8 (d)
+ a
"\1'= li.- l~'" -8
•
b
FIGURE 18.13 Rllk~ for dClcrmining the pOlCllti.d diffcn'lIces :Kro~, a resistor ami a hanel \', asstllllillg the batl<:r) has 110 illlertlal n:sistiltlCC.
point) must gain as much energy as illoses. Ilg£lins energy as it is pumped through a source of emf. ILs enc.::rg}/ may decrease in the form ora potential drop -IR across a rcsistor or as a rC811lt of nowi ng backward th rough a sou rce of em Cfrom the positive to the negative terminal inside the bau.ery. In the lauer case, eleClrical energy is convened to chemical energy as the battery is charged. \Vhen applying Kirchhoff's rules, you must make two decisions at the beginning of the problem: 1. Assign symbols and direclions to the currents in all branches of the circuit. Don't wOITY about gucssing the direction of a current incorrectly; the resulting answer will be negati\'c, but its magnitude will be rmTPct. (Because the equations are Iinearin the currents, all currents are to the first power.) 2. \Vhen applying the loop rule. you must choose a direction for traversing the loop and be consistent in going eithel" clockwise or cOllnterclock,v,.·ise. As rou tra\crse the loop, record \'ollage drops and rises according to the following rules (summarized in Fig. 18.l3, where it is assUll"lcd that mO\'emcnt is from point a toward point b):
(a) If a resistor is traverscd in the direction o[ thc currcnt, the change in electric potential across the resistor is -IR (Fig. 18.13a). (b) If a resistor is traversed in the direction opposite the current, the change in electric potential across the resistor is +IR (Fig. 18.13b). (c) If a source of emf is traversed in the direction o[ the emf (from - to + on the tcrrnimlls), the changc in electric potential is +E: (Fig. IS.ISc). (d) Ir a source of emf is traversed in the direction opposite the emf (from + to - on the tcrminals), the change in electric potential is -E: (Fig. J8.13d). There arc lirnits lU the number of times thejullction rule and the loop rule can be llsed. You can lise l.hejunclion rule as often as needed as long as, ~ach time you write an equal ion, you include in it a cUlTenlthat has not been lIsed in a previous junction-I'll Ie equ<Jtion. (I r l.h is procedu re isn't followed, the new equation will just be a combinalion of twO other equations that yOll already have.) In general, the number of times the junction rule can be used is one fewer than the number of junction points in the circuil. The loop rule can also be used as often as needed, so long as a new circuit element (resistor or battery) or a net..· current. appears in each new equation. To solve a particular circuit problem, you need as many independent equations as you have unknowns.
PROBLEM-SOLVING
SiJTR:MA;rT~EG~YW55555==55555555§t
APPLYING KIRCHHOFF'S RULES TO A CIRCUIT 1. Assign labels and symbols to all the known and unknown quantities. 2. Assign directiolls to the currents in each part of the circuit. Although the assignmem of current directions is arbitrary. you must stick with your original choices throughollt the problem as you apply Kirchhoffs rules. 3. Apply the junction rule to any junction in the circuit. The rule may be applied as many times ~lS a new current (one not used in a previousl}' found equation) appears in the resulting equation. GUSTAV KIRCHHOFF
German Physicist (1824-1887) Together with German chemist Robert Bunsen, Kirchhoff, 0 professor at
Heidelberg, invented the spectroscopy that we study in Chapter 28. He also formulated another rule that
states, "A cool substance will absorb light of the some wavelengths that it emits when hot."
4. Apply Kirchhoff's loop rule to as many loops in the circuit as are needed to solve for the unknowns. To apply this rule, you must correctly identify the change in electric potential as rou cross each elcmcnt in traversing the c10scd loop. ''''atch out for signs~ 5. Solve the equations simultaneollsly [or the unknown quantities, using substiwtion or any other method familiar to the student. 6. Check your answers by substituting them into thc original equations.
18.4
Kirchhoff's Rules and Complex DC Circuits
605
EXAMPLE 18.4 Ap'p'lying Kirchhoff's Rules Goal Use Kirchhofrs rules to find currents in a circuit wiLh three currents and one baLlery.
5.0n
Problem Find the currents in the circuit shO\...n ill Figure 18.14 by using KirchhoWs rules. There are three unknown currents in this circuit. so we must obtain three indcpcndelll equations. which then can be solved b) substitlHion. \\'e can find the equations with one application orthejullClion rule and two applications orthe loop rule. \,Ve choose junction r. UUIlction d gives the same equation.) For the loops, we choose Ihe hallam loop and the top loop, both shown by blue arrows, which indicate the direction ,,'e are going to traverse the circuit mathematically (not necessarily the FIGURE direction of the current). The third loop gi,'es all equation thaI can be obtained by a linear combination of the other [",,'0, so it prmides no additional information and isn't lIsed.
c
Strategy
,..
_ n
b
ti.O \' (Example 1R.4)
18.14
.
. ..... -....
Solution Apply thejullction rule to point c. /1 is directed into the junclion, /'1 and /3 are directed out of the junction. Select the bOI tom loop and tnl\'erse it clockwise starting at point a., generating an equation with lhe loop rule: Sclec[ t hc top loop and traverse it clockwise from point c. Notice the gain across the 9.0-0 resistur because it is traversed ag(finslthe direction or the current!
L L\. V =
Li. \lllal
+ n. \I~.()n + 6. \I~.un
+ (9.00) /, = 0
- (5.0 0)1,
/,=/,+/"
(2)
4.0/,
(3)
-5.0/,
Solvc Equation 3 for I:!. and substilute imo Equa· lion (I):
/,
1.8/3
Sub3tilllte the lauer expression into Equation (2) and solve for 13 :
4.0(2.81,) + 9.0/, = 6.0
Substitute
-5.0/,
/3
back into Equation (3)
SubstitUlC /'j inlO Equation (2)
LO
LO
gel I'!.:
gCl/l:
0
6.0 V - (4.00)/, - (9.00)/, = 0
(I)
Rewritc the three equations, rearranging terms and dropping units for the moment, lor convenience:
=
~
/, = /, +
4.0/,
~
+ 9.0/,
I,
= 6.0
+ 9.01, =
0
+
l,
=
1.8/"
-->
/" = 0.30 A
=0
-
/2
= 6.0
-
/, = 0.831\
9.0(0.30 A)
+ 9.0(0.30 A)
= 2.8/,
=
0.54 A
ll.. • • • • • • •
Remarks Substituting these values back into the original cquations verifies that they are ('orreCl, with any small discrepancies due LO rounding. The problcm can also be soh-ed by firsL combining resistors. QUESTION 18.4 How would the answers change if the indicated directions of the currents in Figure IRI4 werc allrcverscd? EXERCISE 18.4 Sllppose the 6.0-V bauery is replaced by a ballery of unkl10wn t'lllfand an ammeter measures /t = 1.5 A. Find the other two currents a nd tIle emf of the battery.
Answers
/2::;
U.96 A, 1:\ = 0.54 A,
e ::; ]1 V
TIP 18.3 More Current Goes in the Poth of less Resistance You ma> have heard lhe ..t..'ttCmt,:1ll -Current lakc~ Ihe palh of kast rcsi ..tancc.- For it parallel comhinalion ofrc'\i~lors, lhi .. slatement i~ inllccurate hcc
...
606
Chapter 18
EXAMPLE 18.5
Direct-Current Circuits
Another ApP'licotion of Kirchhoff's Rules
Goal Find the currents in a circuit with three currents and two batter-
14 V
I
e
+I'
ies when sorne current directions
a re chosen wrongly. Problem Find fl. 101 , and 13 in Figure 18.15a. Strotegy Use K..irchhoffs two rules, lhejullclion rule once and the loop rule l",rice. to develop three equations for the three unknown currents. Solve the equations simultaneousl)'.
Solution Apply Kirchhoff'sjullcLion rule Lojunction c. Because of the chosen current directions, II and 12 are directed into the junction and 13 is directed out of thejunClion. Apply Kirchhoff's loop rule to the loops abcda and bifcb. (Loop aejdngives no new information.) Tn loop befch, a positive sign is obtained when the 6.0-0 resistor is traversed because the direction of the path is opposite the direction of the current II' Using Equation (1), eliminate 13 from Equation (2) (ignore units for the moment):
+
(I)
I" = I,
(2)
Loop abed{/.:
(3)
Loop beJcb:
I,
10 V - (6.0 f1)I, - (2.0 fl)I, = 0
-]
V
10 - 6.01, - 2.0(1,
+ I,)
= 0
10 = 8.01,
Divide each term in Equation (3) by 2 and rearrange the equation so that the currents are on the right side:
(5)
-12 = -3.0£,
Subtracting Equation (3) from Equation (4) eliminates [2 and gives 11:
22
Substituting this value of [I into Equation (5) gives 12 :
2.01,
= Ill, ~
I, = ['2
into
I,
~
I,
-->
- IOV = 0
+ 2.01,
(4)
Finally, substitute the values found for /1 and Equation (l) to obtain 13 :
Remarks The fact that /2 and 13 are both negative indicates that the \\>Tong directions were chosen for these currents. Nonetheless, the magnitudes are correct. Choosing the right directions or the currents
Is it possible for the current. in a battery to be directed from the positive terminal to\'\'ard the ncgative tcrminal? EXERCISE 18.5
Find the three currents in Figure IS.15b. (Note that the direction of one current \,,'as deliberately chosen wrongly!) Answers
I, = -1.0 A, 12 = l.0 A, I, = 2.0 A
18.5
607
ACTIVE FIGURE 18.16 (ill :\ capacitor in M'rit·s wit II a n'~is lor, ,I ballery, and iI swilch. (b);\ plm of Iht' charg-e on Ihe capilciwr versus lim/.' after lhe switch on lhe circuil i.;
cE: O.(;3~CS
~fI
RC Circuits
clo.;,.d. AfT"1" OIH' lilllP ,01'';';1111 T, tl1l' i~ 6:{% of tht: m
(·harge
CEo Till' charge approacht.·s ils Illaxiilium value as I approacht.'s illiinity.
S (b)
(a)
18.5
RC CIRCUITS
So Car, we have been concerned ,'vith circuits ,.. . i'h constant currents. \Ve now consider direct-current circuits cOlltaining capacitors, in which the (urrellls vary wil.h lime. Consider the series circuit in Active Figure IS.If). \·Ve assume the capaciLOr is initially uncharged with the switch opened. After lhc switch is closed, the battery begins to charge the plates or the capacitor anclthe charge passes through the resistor. As the capacitor is being charged. the circuit carries a changing currenL. The charging process continues until lhe capacitor is charged to its maximulll equilibriulll value, Q = ee, where 8 is the maximum vohage across the capacitor. Once the capacitor is fully charged, the current in the circllit is zero. If\-ve assume the capacitor is ullcharged before the switch is closed, and if the switch is closed at L = 0, \ve fmc! that the charge on the capacitor varies with time according to the equation
q = Q(J - ,-'/f"')
=
He
(a)
[18.7]
where t> = 2.718. . is Euler's constant, the base of the natural logarithms. Active Figure L8.16b is agraph of this equation. The charge is zero at /. = 0 and approaches its maximum value, Q, as (approaches infinity. The voltage 6.Vacross the capacitor at. any time is obtained by dividing the charge by the capacitance: 6V = qlC. As you can see from Equation 18.7, it would lilke all infinite amount of timc, in this model, for the capacitor to become fully charged. The reason is mathematical: in obtaining that equation, charges are assumed to be infinitely small, whereas ill reality the smallest charge is that of an electron, with a magnitude or ] .60 x ]0- 1\) C. For all practical purposcs, thc capacitor is fully charged after a finite amount of' time. The term RCthat appears in Eql13tionl8.7 is called the time constant T (Creek letter tau), so T
[/
[18.8)
The time constant n.:presents the time required for the charge to increase frolll zero to 6~.2% of its maxilllulll equilibrium valuc. This means that in a period of time equal to one t.ime constant, the charge on the capacitor increases from zero to 0.632Q. This can bc sct.:n by subslituting t = T = RC into Equation 18.7 and solving for q. (Note that I - ,,-I = 0.632.) It's imponanl La note lhal a capacitor charges very slO\vl)' in a circuit with a long time constant., whereas it charges vcry rapidly in a circuit with a shon time constant. After a time equal to t.en time const.anIS, the capacitor is more than 99.99% charged. Now consider the circuit in Active Figure 18.17a, consisting of a capacitor with all initial charge Q, ;] resistor, and a s\vitch. Before the switch is closed, the potential differcnce across the charged capclcitor is Q/C. Once the switch is c1oscd, the charge begins to flow through the resistor [rolll one capacitor platc to the other until the capacitor is fully discharged. 1r-the switch is closed at t = 0, it can be shmvn thallhe charge q on the capacitor varies with time according to the equation [18.9] The charge clecreases exponentially with time, as shown in Active Figure 18.17b. In the intervalt = 'T = HC, the charg-e decreases from its initial value Q to O.368Q.
(b) ACTIVE FIGURE 18.17 (US lime ;tlkr the swilch is closed.
E- Time constant of on RC circuit
608
Chapter 18
Direct-Current Circuits
In other word:), in a lime equallo one time constant, the capacitor loses 63.2% of its initial charge. Because av = q/C. the voltage across the capaciLOr also decreases exponentially with time according to the equation .6. V = Be-I.'m:, where e (,..·hich equals Q/C) is the initial voltage across the fully charged capaciLOr.
APPLYING PHYSICS 18.4
TIMED WINDSHIELD WIPERS
~Ian}' automobiles are equipped with windshield wipers thal can be used intermittently during a liglu rainfall. How does the operation of this feature depend on the charging and discharging of a capacilOr?
ferent values of R through a multi position switch. The brief time thal the wipers remain on and the lime they are off are determined by the value of the time constant of the circuit.
Explanation The wipers are part of an RC circuit ",·ith lime constant 'hal can be varied by selcCling dif-
APPLYING PHYSICS 18.5
BACTERIAL GROWTH
In biological applications concerned with popUlation growth, an equation is used that is simil(1r to the exponential equations encountered in the analysis of RCcircuits. Applied to a number of bacteria, this equation is !\~-=
N1 2"
where Nris the number ofbactcria present aft.er n doubling times, iVj is the number present initial1y, and n is the number of gro\,.lth C}'cles or doubli ng times, Doubling times vary according to the organism. The doubling time for the bacteria responsible for leprosy is abo1ll30 days. and that for the salmonella bacteria
APPLYING PHYSICS 18.6
III responsible for food poisoning is about. 20 minutes. Suppose only 10 salmonella baCleria find their wa}' onto a turkey leg after your Than ksgiving meal. Four hours later you come back for a midnight. snack. How l11<=lny baeteri<=l are present now? Explanation The number or doubling limes is 240 min/20 min = 12. Thus,
1"J =
iV,2" = (10 bacteria) (2") = 40960 bacteria
So your system will ha\'e to deal with an invading host of about 41 000 bacteria, which are going to cominue lO double in a ver), promising environment.
ROADWAY FLASHERS
Many roadway construction sites have flashing >'ellow lights to warn motorists of possible dangers. What causes the lights to flash' Explanation A lypical circuit for such a flasher is shown in Figure 18.18. The lamp L is a gas-filled lamp that acts as an open circuit until a large potemial difference causes a discharge, which gives offa brightlighl. During this discharge, charge flows through t.he gas between the electrodes of the lamp. When the switch is closed, the battery charges the capacitor. At the beginning, the current is high and the charge on the capacitor is 10\\', so most of the potelllial difference appears across the resistance R. As the capacitor charges, more potential difference appears across it, reflecting the lower currelll and lower potencial dirference across the resistor. E\'entually, lhe potelHial difference across the capacitor reaches a \'~llue at which the lamp will conduct, causing a flash. This flash discharges the capacitor through the lamp, and the process of charging begins again. The period between flashes can be adjusted by changing the time const<=lnl of the RC circuit.
FIGURE 18.18 18.6)
(Appl~illg
Phtsics
18.5
QUICK QUIZ 18.9 The switch is closed in Figure 18.19. After a long time compared with the time constant of the circuit, \..· hat will the current be in the 2-fl resistor? (a) 4 A (b) 3 A (c) 2 A (d) I A (e) More information is needed.
RC Circuits
L
609
IF
2fl Ifl
Ifl FIGURE 18.19
EXAMPLE 18.6 Goal
Char in a Capacitor in a
Calcublc elementary propenies ora simple RCcircuit.
Problem An uncharged capacitor and a resistor are con~ neeted in series to a battery, as in Active Figure 18.16a. If 8 = 12.0 V, C = 5.00 f.LF, and R = 8.00 X 10" fl, Gnd (a) the time constant of the circuit, (b) the maximum charge on the capacitor, (c) the charge on the capaciLOr after 6.00 s, (d) the potential difference across the resislOr after 6.00 S, and (e) t.he current in the resistor allhat 1i me. Strategy Finding the lime constant in part (a) requires substitution into Equation ]8.8. For part (b), the maxi~
(Quick Quiz 18.9)
mum charge occurs after a long time, when the current has droppeclto zero. By Ohm's law, LI\!= JR, the potential difference across the resisLOr is also zero at that time, and Kirchhoff's loop rule then gives the maximum charge. Finding the charge at some particular time, as in part (c), is a matter of substituting into Equation 18.7. Kirchhoff's loop rule and the capacitance equation can be used to indirectly find the potential drop across the resistor in pan (d), and then Ohm's law yields the current.
Solution (a) Find the time constant of the circuit. Use the definition of the time constant, Equation 18.8:
T =
RC = (8.00
X
10' fl)(5.00
X 1O~"
F)
(b) Calculate the maximum charge on the capacitor.
Apply Kirchhoff's loop rule to the RC circuit, going clockwise, \...· hich means that the voltage dirt"erence across the battery is positive and the differences across the capacitor and resistor are negative:
( 1)
From the definition of capacitance (Eq. 16.8) and Ohm's law, we have 8Vc = -q/Cand nVN = -JR.. These are voltage drops, so they're negative. Also,
(2)
.a.V,);l1
=
If 8---JR=0 c
+8.
\,Vhen the maximum charge q = Qis reached, 1= O. Solve Equation (2) (or the maximum charge: Substitllte to find the maximum charge:
Q=
ce
Q = (5.00 X 10- 6 F)(12.0 V)
(c) Fi nd the charge on the capacitor after 6.00 s. Substitute into Equation 18.7: = 46.6 f.LC
~
60.0 f.LC
4.00 s
~
610
Chapter 18
Direct-Current Circuits
(d) Compute the potential difference across the resistor after 6.00 s.
Compute the ,·uILage drop thal time:
~VC
q G
46.6 JLC
-- =
across the capacitor at
-9.32 II
5.00 JLl'
Ll\l" = -1'.11,,,,, - I'.\le = -12.0 - (-9.32 V)
Solve EqU
-2.68 V
=
(e) Find the current in lhe resistor after 6.00 s.
-I'. \III /=--=
Apply Ohm's law, lIsing the resulls or pan (d) (remember that ~\fR = -fR here):
II
-( -2.68 V)
(ROO X 10' 0)
3.35 X 10- 6 A ~..
.
-
Remark In solving this problem, we paid scrupulous attention (Q signs. These signs must always be chosen when applying Kirchhoff's loop rule and must rernain consistent lhroughoUl the problem. Alternately, magnitudes can be used and the signs chosen by physical intuition. For example, the magnitude or the potential difference across the resistor lllust equal the magnitude of the potential difference across the battery minus the magnitude ofthe pOlential difference across the capacitor. QUESTION 18.6 In an RC circuit as depicted in I'igurc 18.16£1, what happens to the time required for the capacitor to be charged
EXAMPLE 18.7 Goal
....................~
.
to half its maximum value if either the resistance or capacitance is increascd? (a) It increases. (b) It decreases. (c) TI remains the samc. EXERCISE 18.6 Find (a) the charge on the capacitor after 2.00 shave elapscd, (b) the rnagn itude of the potential difference across the capacitor arter 2.00 S, and (c) the magnitude of' the potential difference across the resistor at that same time.
Answers
(a) 23.6 JLC
(h) 4.72 V
(e) 7.28 V
Discharging a Cap.acitar in an RC Circuit
Calculate some elementary properties of a discharging capacitor in an HC circuit.
Problem Consider a capacitor C being discharged through
,..
Strategy This problem requires substituting given values into variolls equations, as well as a few algebraic manipulations involving t.he natural logarithm. In part (a) scI. q = tQ in Equation 18.9 {or a discharging capacitor, where Q is the initial charge, and solve for time t. In part (b) substitute into Equations 16.8 and ]8.8 to find (he initial capaciLOr charge and time constant, respectively. In part (c) substitute the results of pan (b) and q = 1.60 XIO- 1H C into the discharging-capacitor equation, again solving for time.
...............................
.
~
Solution (a) How long does it take the charge on the capacitor LO reduce to one-fourth its initial value? Apply Equation 18.9:
Substitute q(t) = Q/4 into the preceding equation and cancel Q: T<1ke natl.lrallogarithms of both sides and solve for the ti me I:
In
W=
-liRe
I = -IIG In
W=
1.39RG = 1.397
18.6
Household Circuits
611
(b) Compute the initial charge and time constant from the given data. Use the capacitance equation charge:
La
find the initial
C=
i\.~
---7
Q = C i\. \i = (3.50
X
10-6 F)(l2.0 V)
Q= 4.20 X 10-5C Now calculate the time constant:
T
=
RC = (2.00 11) (3.50
X
10-6 F)
7.00 X 10- 6 S
(c) Hm\.· long does it take to drain all but the last quanLUI11 of charge? Apply Equation 18.9, divide by
Q, and take natural
q( I)
e
logarithms of both sides:
-tiT
=-q Q
Take the natural logs of both sides:
Substitute q = 1.60 X 10- 19 C and the values for Qand T found in part (b):
1.60 X \0-19 C) 1 = -(7.00 X 10-6 s)ln ( _ 4.20 X 10 ' C =
Remarks Part (a) shows how useful information can often be obtained even when no details concerning capacitances, resistances, or voltages are known. Part (c) demonstrates that capacitors can be rapidly discharged (or conversely, charged), despite the mathematical form of Equations 18.7 and 18.9, which indicate an infinite time would be required. QUESTION 18.7
Suppose the initial \'oltage used to charge the capacitor were doubled. \'Vould the time required for discharging all butLhe last quantum of' charge (a) increase, (b) decrease, (c) remain the same? EXERCISE 18.7
Suppose the same type of series circuit has R = 8.00 X 10'1 n, C = 5.00 j,LF, and an initial voltage across the capacitor of 6.00 V. (a) How long does it take the capacitor to lose half its initial charge? (b) How long does it take to lose all but the last 10 electrons on the negative plate? Answers
18.6
(a) 0.277 s
(b) 12.2 s
HOUSEHOLD CIRCUITS
Household circuits are a practical application of SOllle of the ideas presented in this chapter. In a typical installation the utility company distributes electric pm\'er to individual houses \\lith a pair of wires, or power lines. Electrical devices in a house are then connected in parallel La these lines. as shown in Figure 18.20. The potential difference between the 1.\'\10 wires is about 120 V. (These currents and voltages are actually alt.ernating currents and voltages, but for the present discLlssion \\le will assurne they are direct currents and voltages.) One of the wires is connected to ground, and the other \\lire, sometimes called the "hot" wire, is at a potential of 120 V. A meter and a circuit breaker (or a fuse) are connected in series with the wire entering the house, as indicated in the figure. In modern homes, circuit breakers are used in place of fuses. ',Vhen the current in a circuit exceeds some value (lypically 15 A or 20 A), Ihe circuit breaker acts as a switch and opens the circuit. Figure 18.21 shows one design for a circuit breaker. Current passes through a bimetallic strip, the top ofv.!hich bends to the left \vhen excessive current heats it. If the strip bends far enough La the left, iL settles into a
~.- - - - , - - - - 1 2 0 V Live .k.., ~-y---< l'.'lcter Neutral
Il, '--..L-,-J-...L- 0 V FIGURE 18.20 A wiring diagram for a household circuiL. The resistances RI , R.2 • and Rj represelll appliances or other electrical dl:\'ices lhal operate at an appl iec! \'oltage of 120 V.
612
Chapter 18
Direct-Current Circuits
groove in the spring-loaded metal bar. \Vhen this scLLling occurs, the bar drops enough to open the circuit ~lL the cont.act point. The bar also nips a switch that indicates that the circuit breaker is not opermional. (After the overload is removed, t.he S\\liLch can be flipped back 011.) Circuit breakers based on this design have the
ContacL
palms /MCI:l[ h(lf
/l\imClatlic strip
disadv.-ltl1age that some lime is .'equirecl for the heating of the strip, so the circuit Illay not be opened rapidly enough when it is overloaded. Therefore, lllany circuit
hreakers are nov,' designed to use electromagnets (discussed in Chapter 19). The ,,,,,'ire and circuit breaker are carefully selected to meet the current demands ora circuit. lrthe circuit is to carry currents as large as 30 A, a heavy-duty wire and an appropriate circuit breaker must be used. Household circuits that are normally used I.e) power lamps and small appliances often reCJuire on I)' 20 A. Each circuit has its own circuit breaker to accomlTIodme its maximum safe load. As an example, consider a circuit that powers a toaster, a microwave oven, and a heater (represented b), RI , R"L' and R3 in Fig. 18.20). Using the equation W> = J ~f~ we can calculate the CUITCIll carried b}' each appliance. The toaster, rated at] 000 \"1, dra'\IS a CUlTent of I 000/120 = 8.33 A. The micrO\-\,ave oven, rated at800 \·V, dra\...,s a current of 6.67 A, and the heater, ratcd at 1 300 \·V, draws a currCll[ of 10.8 A. If thc three appliances are operated simultaneously, they drm.. · a LOtal current of 25.8 A. Therefore, the breakcr should be able to handle at least this much current, or elsc it will be tripped. As an alternative, the toaster and microwave oven could operale on one 20-A circuit and the heater on a separate 20-A circuit. Many heavy-dUly appliances, such as electric ranges and clothes dryers. require 240 V 1.0 operate. The power company supplies this voltage by providing. in addition to a live wire that is 1.20 V above ground potential, another wire, also considered live. that is 120 V below ground potential (fig. 18.22). Therefore, the potential drop across the n.l,'o live wires is 240 V. An appliance operating from a 240-\1 line requires hal r the current of one operating from a 120-V line; conscquently, smaller ,.. ·ires can be used in the higher-voltage circuit withollt becoming overheated.
FIGURE 18.21 A circuit breaker lhal uses a binLt:lallic sLrip for il~ operation.
APPLICATION Fuses and Circuit Breakers
+120 V
-120 V
I
(a) ~
J
~
" FIGURE 18.22 I'O\"CI" conncctiOllS for a 240-V appliallce.
APPLICATION Third Wire on Consumer Appliances
ELECTRICAL SAFETY
A person can be electrocuted by tOuching a live wire ,,,'hile in contact '\J"ith ground. Such a hazard is often due to frayed inslllation that exposes the conducting ,·...ire. ~ The ground contact might be made by touching a water pipe (which is normally aL ground potential) or by standing on the ground with wet fect because impure water is a good conductor. Obviously, such situations should be avoided at all costs. Electric shock can resLdt in fatal burns, or it can C<-luse the muscles of vital organs, such as the heart, to malfunction. The degree of damage to the body depends on [he magnitude of the current, the length of time it acts, and the part of the body through which it passes. Currents of 5 mA or less can cause a sensation of shock, but ordinarily do little or no damage. If the current is larger than about 10 mA, the hand muscles contract. and the perSall may be unable to let go of the li\'e wire. If a current of about 100 mA passes through the body ('or just a few scc~ onds, it can be [atal. Such large currents paralyzc the respiratory muscles. In some cases, currents or about] A through the body produce serious (and sometimes fatal) burns. As an additional sa('cty feature for consumers, electrical equipment manufacturers now use electrical cords that have a third \-vire, called a case ground. To undersland how this works, consider the drill being used in Figure 18.23. A n'lO-wire device that has one wire, called the "hot" wire, is connected to the high-potelllial (l20-V) side of' the input power linc, and the second wire is connected to ground (0 V). If the high-voltage wire comes in conlact with the case of the drill (Fig. 18.23a), a short circuit occurs. In this undesirable circumstance, the path'\J"ay for the currcnt is from the high-voltage 'wire through the person holding the drill and to Earth. a pathway that can be ratal. ProteClion is provided by a third wire, connCClcd to [he case of the drill (Fig. 18.23b). In this case, if a shon occurs. the path of" least resistance for the current is from the high-voltage wire through the case
E
S
(b)
18.7
18.8
Conduction of Electrical Signals by Neurons
"Ouch!"
613
FIGURE 18.23 The: -hot" (or -live") wire. ;It 120 V, alwa}'s includl-'s a circuit br..:aker for safel}'. (a) \-Vhen the drill is operated with t\\"o win's, the normal current path it' from the "hot" wire. through lhe moto!' conneclions. ilnd backlo groulld lhrough tlw -nentral" wire. Herc. ho\\·c"er. the high-\'ohagc side has come ill contact with the drill case, sn, the person holding t he drill I'f'{"civcs an elcclrical shock. {h} Shock call bl: prc\'cilled b)" a third wire rUllnillgfronl Ih..: drill case to lh..: grollnd.
"Ground'- ~
I
]~ L--... 120 \' CirCllit breaker Ground "':"
(b)
~l11d back to ground through the third wire. The resulting high currelll produced will blow a fuse or trip a circuit breaker before the consumer is injured. Specicd power owlets called ground-fault interruptcrs (GFTs) are nm.. . being used in kitchens, balhrooms, bascments, and other hazardous areas of ne\v homcs. Thcy ,He designed to protect people from electrical shock by sensing small currents-approximately 5 mA and grcater-Ieaking to ground. \·Vhen current above this level is detected, the device shuts off' (intcrrupts) the current in less than a millisecond. (Ground-fault interrllpters will be discussed in Chapter 20.)
18.8 CONDUCTION OF ELECTRICAL SIGNALS BY NEURONS2 JIll The most remarkable use of electrical phenomcna in living organisms is ['ound in the nervous system of animals. Specialized cells in the body called neurons form a complex nctwork that receives, proccsscs, and translTlits inf'orrnation from one part 0[' the body to another. The center of this network is located in the brain, which has the ability 1.0 store and analyze information. On the basis of this inforrnaLion, thc nervous systcm controls parts of the body. The ncrvous systen1 is highly complex and consists of about 1010 interconnected ncurons. SOll1C aspccts of the nervous system are well known. Over the past several decades the method of signal propagation through the nervous system has been establishcd. The messages transmittcd by neurons are voltage pulses called artion /JO(fl1tials. \'\lhen a neuron receives a strong enough stimulus, it produces identical \'oltage pulses that ~lre actively propagated along its structurc, The strength of the stimulus is conveyed by the number of" pulses produced. vVhcn the pulscs reach the end of the neuron, they activate eithcr musclc cells or other neurons. Thcre is a "flring threshold" for neurons: action potentials propagat.c along a nellron only i r the stimulus is Sll fficiently strong. Neurons can be divided into three classes: scnsory neurons, motor neurons, and interneurons. The sensory neurons receive stimuli ['rom sensory organs thal monitor the external and internal environment of thc body. Depending on t.heir specialized functions, the sensory neurons convey messages about factors such as
614
Chapter 18
Direct-Current Circuits
light, temperawre, pressure, muscle tension, and odor to higher cemers in the nervous system. The mowr neurons carry messages that cOl1u'ol the muscle cells, The Smapsc Dendrites messages are based on the information provided by the sensory neurons and by the brain. The interneurons transmit information from one neuron to another. ~uclellS Each nCU1"On consists of a cell body to ,,\,hich al"C attached input ends called Cell body dendrites and a long tail called the axon, ,vhich transmits the signal a\'..·a)" from the cell (Fig. 18.24). The far end of the axon branches into nerve endings that neurOll Axon---~t transmit the signal across small gaps to other neurons or to muscle cells. A simple sensorimotor neuron circuit is shO\,;,n in Figure 18.25. A stimulus from a muscle produces nerve impulses that travel to the spine. Here the signal is transmitted to / Nen.'c motor neurOll, 'which in turn sends impulses to control the muscle. Figure 18.26 a - .........._U endings 5hO\'I'5 an electron microscope image of neurons in the brain. FIGURE 18.24 Diagram ora neuron. The axon, which is an extension of the neuron cell, conducts electric impulscs away from the cell body. Some axons are extremely long. In humans, for example, the axons connecting the spine \-\'ith the fingers and toes are more than 1 m long. The neuron can transmit messages because of the special active electrical characSensory neuron teristics of the axon. (The axon acts as all active source or energy like a battery, rather than like a passive stretch of resistive wirc.) Much or the informacion about Interneuron J the electrical and chemical propenies of thc axon is obtained by inserting small needlelike probes into it. Figure 18.27 shows an experimental setup. Spine Note tbat the outside of the axon is grounded, so all measured volLages are with respect to a zero potential on the outside. "Vith these probes it is possible to inject current into the axon, measure the resulting action potcntial as a function of time i\fOlor neuron at a flxed point, and sample the cell's chemical composition. Such experiments Muscle are usually difficult to run because the diameter of most axons is very small. Even FIGURE 18.25 A simple IH.:ur;11 the largest axons in the human nervous system have a diameter of only abom 20 X circuit. 10--1 cm. The giant squid, hov,:ever, has an axon with a diameter of about 0.5 mm, which is large enough for the convenient insertion of probes. :Much of the information about signal trallsmission in the nervous system has come rrom experiments ,\'ith the squid axon. In the aqueous environment of the body, salts and other molecules dissociate into positive and negative ions. As a result, body fluids are relatively good conSi ductol"s of electricity. The inside of the axon is filled with an ionic fluid that is ~ .§. separated from the surrounding body fluid by a thin membrane that is only about ~ ~ 5 nm to 10 nm thick. The resistivities of the internal and external fluids are about j ~ the same, but thcir chemical compositions are substantbllv different. The external fluid is similar to se;:l\val.cr: its ionic solutes are mostly p~sitive sodium ions and g, ~negative ch loride ions. Inside the axon, the positive ions are mostly potassium ions ~ and thc negative ions arc mostly large organic ions. ~~ Ordinarily. thc concentrations of sodium and potassium ions inside and outsidc the axon ,.. 'ould be equalized by diffusion. The axon, however, is a living cell with FIGURE 18.26 Stellate neuron [rom an energy supply and can change the permeability of its membranes on a time human cortex. scalc of milliseconds.
-
•
}1 i ..., <:...
Oscilloscope
~;;;;;;;;;;;;r""""Q"01301Ol
Current pulse g-cncrawr
FIGURE 18.27 An axon :>timulalt..:d e1eclricallr. The lert probe injects a
short pulse o[ current. and the right probe measures rhe resulLing
Axon
CurrenLinjcning
probe
VoltageIllcasudng Pl'Obe
Summary FIGURE 18.28
Actiun POICllli;\! (mV)
(a)
Trpical
615
pUle11lial as a function 01 time.
+20
(b) Cllfa'll[ in the axon /IleJllhr-anc
of----'--+-!:--\----!:---'------::Time (111~) :')
wall as a function of lime.
-20 -'10
-60
-8u (a)
CU1Tellt
in
lllcll1hr~\lle wall
in arbitrary
o I-~~--~--;;---S",.~- Tinlt' ~')
(illS)
K~Ollt
(h)
When the axon is not conduCling an electric pulse, 1 he axon membrane is highly permeable to potassium ions, slightl), perrneable to sodiulll ions, and impermeable to large organic ions. Consequently, although sodium ions cannot easily elller the axon, potassium ions can leave it. As the potassium ions lcavt: the axon, however, they lemle behind large negative organic ions, ,·\'hieh Gll1not lollow thcm through the membrane. As a result, a negative potential builds up inside the axon with respect to the outside. The final negative potential reached, which has been measured at abollt -70 mY, holds back the outflow ofp0l.assium ions so that at equilibrium, the concentration of ions is as we have statcd. The mechanism for the production oran electric signal by the neuron is concepwally simple, but was experimellwlly diffi.cult to son oul.. ""'hcn a neuron changes its resting potcntial bccause oran appropriate stimulus, the propenies of its mcmbrane change locally. As a result, there is a sudden flow of sodium iOlls inlo the cell that lasts for about 2 illS. This flow produces the +~O mV peak in the action potential shown in Figure 18.28a. lmmcdialely arter, there is an increase in potas siu11l iOIl rim\' out of the cell that restorcs the resting aClion pot.ential or -70 mV in ~ln additional 3 ms. Both the Na+ and K+ ion flows have been measured by using radioactive Na and K tracers. The nerve signal has been measured to tnl\'cl along the axon at speeds of 50 m/s to about 150 m/s. This flow of charged panicles (or signal transmission) in a nene axon is ul/li/u'signal transmission in a rnelal wire. In an axon charges move in a direction perpendicular to the direction of t.ravel or the nen'e sign,ll, and the l1er\'e signal moves much more slowly than a voltage pulse traveling along a metallic wire. Although the axon is a highly complex strucl.Ure and much of how Na~ and It' ion channels open and close is not understood. standard electric circuit concepts of Cllrrent and capacitance can be used to analyze axons. It is left as a problem (Problem 41) LO show that the axon, having equal and Opposile charges separated by a thin dielectric membrane, acts like a capacitot-. w
SUMMARY 18.1 Sources of emf Any device, such as a battery or generator, that increases til(; electric potential energy 01' charges in an electric circuit is called a source of emf. Batteries convert chemical
energy into electrical potelltial energy, and generators con\'crt mechanical energy inlo electrical potential encl-g)'. The lerminal vollage LlVofa baltery is gi\"cn by t>V=8-1r
[18.1]
Chapter 18
616
Direct-Current Circuits
e
where is the emf of the baltery, T is the current, and r is the imernal resistance of the ba(tery. Generally, the internal resistance is slllall enough to be neglected.
18.2
Resistors in Series
The equivalent resistance of a sel of resiswrs con nee red ill
The first rule, called the junction rule, is a statement of conservation of charge. The second rule, called the loop rule, is a statement of conservation of energy. Solving problems involves using these fules to generate as lllany equations as there are unknown currents. The equations can then be solved simultaneously.
series is [18.4] The current remains at a constant value as it passes through a series of resistors. The potential di rFerence across an)' (',":0 resistors in series is different, unless the resistors have rhe same resistance. 18.3 Resistors in Parallel The equivalent resistance of a set of resistors parallel is ill R]
R2
[18.7] The term RC in Equation 18.7 is called the time constant T (Greek letter tau), so
COIl
]
-~-+-+-+
R cq
18.5 RC Circuits In a simple RCcircuit with a battery, a resistor, and a capacitor in series, the charge on the capacitor increases according to the equation
R3
nee ted in
[18.6]
The potential difference across any two parallel resistors is the same; the current in each resistor, hmvever, will be different unless the two resistances are equal. 18.4 Kirchhoff's Rules and Complex DC Circuits Complex circuits can be analyzed using Kirchhoff's rules:
1. The sum of the currems entering an)' junction must equal the sum of the currents leaving that junction. 2. The sum of the potential differences across all the elements around any closed circuit loop must be zero.
T
= RC
[18.8]
The time constant represents the time required for the charge to increase from zero to 63.2% of its maximum equilibrium value. A simple RC circuit consisting of a charged capacitor in series \-\,ith a resistor discharges according to the expression [18.9] Problems can be solved by substituting into these equations. The volt.age 6.Vacross the capacitor at any time is obtained by dividing the charge by the capacitance: D.\!= q/C. Using Kirchhoff's loop rule yields the potential difference across the resistor. Ohm's la\\I' applied to the resistor then gives the current.
FOR ADDITIONAL STUDENT RESOURCES. GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE·CHOICE QUESTIONS 1. The terminals ofa banery are connected across two different resistors in series. \'Vhich of the following statements are correct? (There may be more than one correct statement.) (a) The smaller resistOr carries more current. (b) The larger resistor carries less current. (e) The current in each resistor is the same. (d) The voltage difference across each resistOr is the same. (e) The voltage difference is greatest across the resistor closest to the positive terminal.
2. The terminals of a battery are connected across t\\'o different resistors in parallel. \,\'hich of tbe following statements are correct? (There may be more than one correct statement.) (a) The larger resistor carries more current than the smaller resistor. (b) The larger resistor carries less current than the smaller resistor. (c) The voltage difference across each resistor is the same. (d) The voltage difference across the larger resistor is greater than the voltage difrerence across the smaller resismr. (e) The voltage difference is gre4:Her across the resistor closer to the battery. 3. A 1.00-0 and a 2.00-0 resistor are in parallel. \Vhat is the equivalent single resistance? (a) 1.50 n (b) 3.00 (c) 0.667 n (d) 0.333 n (e) ] .33 n
n
4. Two lightbulbs are in series, one operating at 120 \'\r and the othel" operating at 60.0 W. If the voltage drop across the series combination is ] 20 V, \vhar is the current in the circuit? (a) ].0 A (b) 1.5 A (c) 2.0 A (d) 2.5 A (e) 3.0 A 5. When connected to a 120 V source, an electric heater is rated at 1 200 \·v and a toaster at 6.0 X 102 W. If both appliances are operating in parallel on a 120-V circuit, what total current is delivered by an external source? (a)i 2 A (b) 24 A (c) 32 A (d) 8.0 A (e)i 5 A
6. '·Vhat is t.he current in the 1.O.O-f! resistance in the circuit shown in Figure fo..'lCQI8.6? (a) 0.59 A (b) 1.0 A (e) 11.2 A (d) 16.0 A (e) 5.3 A 2.00
4.0 il
6.0 il
100
FIGURE MCQ18.6
7. Which of the following equations is not a consequence of Kirchhoff's laws, when applied to figure ;"'ICQ18.7?
Conceptual Questions
(a) 9 - " - 21, = 0 (b) I, + I, - f, = 0 (c) 4 - 2', 0 (d) 5 - II - 3[:\ = 0 (e) All 1hesc equations are correct.
c
3(~ =
-
5V
- ;]
I '1 V
Bauery FIGURE MCQ18.1 0
I,
I
-
] I. Several resistors arc connected in parallel. Which of
3n
I,
FIGURE MCQ18.7
8. If the terminals of a battery are connected across lIW identical resistors in series, the total pm-vcr delivered by the baltery is 8.0 \1\,'. I r the same banery is cOllnected across the resistors in parallel, what is the total pmver delivered by the bauery? (a) 16 W (b) 32 W (c) 2.0 II' (d) 4.0 W (e) 8.0 W 9. \Vhal is the time constant of the circuit shown in Figure MCQI8.9? Each of the five resistors has resistance R, anel each of rhe five capacitors has capacitance C. The internal resistance of the battery is negligible. (a) Re (b) 5RC (c) IORC (d) 25RC (e) None of these answet's is con-eeL.
R
\5
R
R
R
R
CT 1
~I
.:~
C2>----,
1n
I,
617
the following statements are true of the correspond~ ing equivalent resistance? (There may be more than one con-en statement.) (a) It is greater than the resistance or any of the individual resistors. (b) It is less than the resistance of any of the individual resistors. (c) It is dependent on the volt<'lge applied across the series. (d) It is equal to the sum of the resistances of all the resistors in the series. (e) II. is equal to the reciprocal of the sum of the inverses of the resistances of al I the resistors, 12. Several resistors are connected in series. \·Vhich of the following statements are true of the cOlTesponding equi\·alent resistance? (There may be more than one correct sl atement.) (a) It is greate,- than the resistance of any of the indi,,'idual resistors, (b) It is less than the resistance or em}' of the individual resistors. (c) It is dependent on the voltage applied across the group. (d) It is eql.lalto the sum of the resistances of all the individual resistors. (e) It is equal to the reciprocal of the sum of the inverses of the resistances oEall the resistors. 13. The capacitor in Figure MCQ18.l3 is initially uncharged. \Vhen the switch is closed at I = 0 S, what is the voltage drop across the capacitor after one time constan! has elapsed) (a) 2.03 V (b) 3.25 V (c) 6.63 V (eI) 3.79 V (e) 4.43 V
J(;
FIGURE MCQ18.9
10. The capacitor in Figure MCQIS.10 has a large capacitance and is initially uncharged. After t.he switch is closed, what happens to the lighlbulb? (a) It never glows because the capacitor represents an open circuit. (b) It glows only after the capacitor is fully charged. (c) ltglows for a very shan time as the capacitor is being charged. (d) It glows cominuously. (e) II g!m,,·s intermittently.
2.00 F
5.00 n
FIGURE MCQ18.13
CONCEPTUAL QUESTIONS I. Is the direction of current in a bauery always from the negative terminal to the positive one? Explain.
5. if you ha\'e your headlights all while you start your car, why do Lhey dim while Lhe car is starting?
2. Given three lightbulbs and a battery, sketch different circuits as you can.
6. If electrical power is transmitted over long distances, the resistance of the wires becomes significant. Why? \,Vh ich mode of tra nsm ission wou Jet result i n less energy loss, high current and low voltage or low current and high voltage? Discuss.
(IS
many
3. Suppose the energy transferred to a dead bauer)' during charging is W The recharged bauery is then llsed ulltil [ully discharged again, Is the total energy trans~ fen-ed out of the battery during use a Iso \V? 4. A shan circuit is a circuit comaining a p
7. Connecting baueries in series increases the emfapplied to a circuit. \-\-'hat advantage might there be to connecting them in paral1c1? 8. Two sets of Christmas lights are available. For set A, when one bulb is removed, the remaining bulbs remain illuminated. For sel B, when one bulb is removed, the
Chapter 18
618
Direct-Current Circuits
remaining bulbs do not operate. Explain the difference in wiring for lhe (wo sets. 9. Suppose a parachutist lands on a high-voilltgc wire and grabs lhe ",,'ire as she prepares to be rescued. \-Vill she be electrocuted? If the " . ire t.hen breaks, should she continue to holel onto the wire as she falls 10 the ground? 10. (a) Two resisLOrs arc cOllllcncd in series ZlUOSS a bat· Lcry. Is the po\,.:er delivered to each resistor (i) the same or (ii) not necessarily the same? (b) T,.. .o resistors are connected in parallel across a battery. Is the power deliyercd to each resisLOf (i) the same or (ii) nor necessarily the same? 11. \Vhy is it possible for a bi rei to Sil on a h igh-vollage \vi re
12. A ski resort consists ora fc\..' chairliCrs and several interconnected dmnlhil1 flillS on the side of a mountain, with a lodge at the bOllOlTI. The lifts arc analogous to batteries, and the rUIlS are analogous to resisLOrs. Describe how two runs can be in series. Describe how three nms can be in parallel. Sketch ajunction of one lin and two runs. One of the skiers is carrying an altimeter. Slate Kirchhoff's junction rule and Kirchhoff's loop rule for ski resorts.
13. Embodied in Kirchhoff's rules are two conservation laws. What are they?
14. \O\'hy is it dangerous to turn on a light when you arc in a bathtub?
without being electrocuted? (Sec Fig. CQ18.11.)
FIGURE CQ18.11 volt,lge wire.
~
Birds 011 a high-
The Problems for lhis chapler Ilia)' be: assigned oilline at WebAssign. Sl ra ig-lllJorward, i 111(") lllcdia(e, l·h;1llell~i11g-
WebAssign t, :2. J
B
m:1 = "
o
=
= den01es guided problem
del\otl;s enhanced conte III problem
= biomedical applicalion = denot(~s ful1 solution available in Sflldell( SO/If/ions Malltlal/
Stud)' Guide
4. (a) Find the current in ,1 8.00-fl resistor connecrcd to a balleI')' that has an internal resistance of 0.15 n if the voltage across the b
SECTION 18.1 SOURCES OF EMF
9.00 fl
4.00n
SECTION 18.2 RESISTORS IN SERIES 10.0
SECTION 18.3 RESISTORS IN PARALLEL
n
a
1. A batlery having an emf 01" 9.00 V delivers 117 rnA when connected to a 72.0-fl load. Determine the internal resislance of thc bauer),. 2. Three 9.0-0 resistors are conneClcd in series with a 12-V bancr)'. Find (a) the equivalem resistance of the circuil and (b) the currenl ill each resisLOr. (c) Repeal for lhe case ill which all lhree resisLOrs are connected in parallel across the bauer)'.
FIGURE P1S.S
6. Consider the combination of resistors shown in Figul-c P1S.6. (a) Find the equivalent resistancc between point a and b. (b) If a \,olt;'-Ige of35 V is applicd between points a and b. rind the current in each resistor. 4.0 fl
12J1
[]JA
liglHbulb markcd "75 W [at] 120 \!" is screwed illlO a socket alone end of a long extension cord in wh ich each of the two conducLOrs has a resistance of 0.800 n. The other end of the extension cord is plugged illlO a 120-V oullet. Draw a cirCllil diagram, and And the actual pm.. .er of tile blilb ill the circuit described.
b
"
5.0n 6.0 J1
8.0 FIGURE P1S.6
n
Problems
619
7. What is the equivalent resistance of the combination of identical resistors between points {[ and bin Figure PIH.i? R
~ R
R
R
FIGURE P1S.10 FIGURE P1S.7
8.
m
Consider the circuit shown in figure PI8.8. (a) Calculate the equi\alent resistance of the 10.0-0 and 3.00-0 "esistors connected in parallel. (b) L~ing the result of pan (a). calculate the combined resistance of the 10.0-0. 5.00-0, and 4.00-0 resistors. (c) Calculate the equivalent I-esistance of the combined resistance found in part (b) and the parallel 3.00·[! resislOr. (eI) Combine the equivalent resistance found in pan (c) with the 2.00·n resi~tor. (e) Calculate the total current in the circuit. (0 What is the voltage drop across the 2.00-,0 resistor~ (g) Subtract· iug the result 01" pan (f) from the batter)' '·ollage. find the \oltage across the 3.00-0 resistor. (h) Calculate the current in the 3.00-0 resistor.
..
10.0 11
•
2,O.0.11
..
\00 11
,
5,~0 11 vVV
"v
[I[]The resistance between terminals (l and bin Figure PIS.II is 750. If the resistors labeled R ha'-e the same value. determine R.
R
1200~
R
b <>-_ _-'------'_ _- - '
FIGURE P1S.11
12. mil Three identical resistors are connected as shown in Figure P18.12. The maximum power thal can safel}' be delivered by a battery connected between a and b is 24 W. (a) What is the equivalent resistance between points a and b? (b) Find an expression for the maximum \'oltage that can be applied between a and b. (c) \""hat is the power dcli"cred to each resistOr when the maximum voltage is applied between a and b?
..
3,0,Op
1/
a
I,
S'fO I'
yy
R
I'
FIGURE P1S.12
FIGURE PIS.S
[]JConsider the circuit shown in Figure PIB.9. Find (a) the current in the 20.0-n resistor and (I» the pOlCl1tial differ· ence between points (l and b.
13. Find the current in the 12·0 resistor in Figure PI8.I3. 6.0 11 3.00
10.0
a
a
~
5.000
~.011
25.0 \' b
I
I~
60a
~ZP
:
~yyy--
5.000
20.0n
~t--I
-
FIGURE P1S.9
10. mJ Consider the two circuits shown in Figure P18.1O in which the lightbulbs and batteries are identical. The re:-,istance of each Jiglubulb is R. (a) Find the currents in each lightbllib. (b) How does the brightllcss or B compare with that of C? E.xplain. (c) How does the brightllcss of A COlllp
18 V
FIGURE P1S.13
14. mil (a) Is it possible to reduce the circuit shown in Figure P18.14 (page 620) to a single equivalelll resistor connected across the battery? Explain. (b) Find the currel1l in the 2.00-.0 resistor. (c) Calculate the pov,,'er delivered by the bauer)' to the circuit.
620
Direct-Current Circuits
Chapter 18
19, Figure P18.19 shows a circuit diagram. Determine (a) the current. (b) the potential of wire A relative to ground, and (c) the voltage drop across the I 500-0: resistor.
:won
6.00
3.00 II
n
~I
-1
4.00n FIGURE P18.14
15. (a) YOll need a 45-0 resistor, but the sLOck,-ool1l has on I)' 20-11 and 50-f! resistors. How can the desired resistance be achieved under these circumstances? (b) What can VOll do ifYOll need a 35-0 resistor?
1 -1
15000
0000 311.0\'
1
2Uoon
+
25.0\'
+
jOQn
20.0 V
SECTION 18.4 KIRCHHOFF'S RULES AND COMPLEX DC CIRCUITS
FIGURE P18.19
For some circuits, the currellls are nO( necessarily in the dircClion shown.
20. In the circuil of Figure P18.20, the current '1 is 3.0A and the \"alues of E and R a,'e unknown. \\'hal arc the currellls I';! and ,3 ?
SOil':
16.
mil
(tI) Find the (urrem in each rc:-isLOr of Figure P 18.16 b, using the rules for resislOfS in series and parallel. (b) 'Write three indepcndcm equations for [he Ihree currents using Kirchorrs laws: one wilh the node rule; a second using the loop nde
through
Ihe
b,ltlcry.
-11" 12!l -
e R
,1I11I11CI('1"
b
42 V
FIGURE P18.20
the
FIGURE P18.16
GJJ-n resistor, and the 24.0-0 resistor; and the third lIsing the loop rule throngh the J2.0-0 and 24.0-11 rcsisLOrs. Soh.c to check the answcr~ found in pan (a).
[]I]The
21V
21. (a) In Figure PJ8.21 find the current in each resistor and (b) the pmver delivercd LO each resistor. 24.0 V
2B.on
It.OV
12.0n
shawl! in Figure P18.17 reads 2.00 A. Find
'I' '2' and E.
-I,
~~1_5_.0_\_'_~ I,
II
16.0ll
500 II A
I"
11 .. "" ~
~F---FIGURE P18.17
18. Dctclllline the potcntial difference j.\~~ for the circllit in Figure PIS. IS.
FIGURE P18.21
22. m:J FOll!' resistors arc connected to a hattery with a terminal \'oltage of 12 V, as shown in Figure PIS.22. (a) How would }'Oll reduce lhe circuit to an eClui,-alem single resi:'Jotor connected to the batter",? Use this procedure [0 find the cqui\'alclll resiSl
,..
30.0
Ion
12V
a
s.on
t-1
..
3.00
+
n
50.0.0
6.0 Q
1
(Y'
·1.0 V
+
S.O\!
FIGURE P18.18
80.0
2.00
"
n
-'1+
L-AJv,,\,-----1,i20.0.0
12V
FIGURE P18.22
Problems
123.IUsing Kirchhoff's niles. (a) find the current in each resistor shown in Figure 1'18.23 and (b) find the poten[ial difference between points rand!
~o.o
621
n
5.00 fl Ii
v:: ~3
( 4 . 0 0 kfl
---'k,
'-8-,
d
60.0 V
iO.O V
20.0n
HO.O V
FIGURE P18.27
28. R~
3.flO k!!
2.00 kf!
I FIGURE P18.23
24. Two 1.50-V batteries-with their positive tcrminals in the same direction-arc inscrLed in series into the barrel of a flashlight. One hauery has ,111 internal resistance of 0.255 fl, Ihe other an imernal resistance or O.15~ n. When the switch is closed, a current of 0.600 A passes through the lamp. (a) What is the lamp's resistance? (b) \\'haL fl"aCLion of Ihe power dissipated is dissipated in I he batteries?
25.
am
E!I
For the circuit shown in Figure PI8.28, use Kirchhoffs rules to obtain equations for (a) the upper loop. (b) the lower loop, and (c) the node on the left side. In each case suppress units for clarity and simplify, com· bining like terms. (d) Solve the node equation for 136 (e) Using the equation found in (d), eliminate 1.% from the equation found in pan (b). (0 Solve the equations found in part (a) and pan (e) simultaneousl}' for the t\·,:o unknowns [or 11R and /12' respectivcl)'. (g) Substitute [he answers found in pan (f) into the node equation found in part (d), sulving for 13(" (h) What is the signifiGll1ce of the negative answer lor 112 ? 18 V
-
8.00 n
(a) Can the circuit shown in Figure P18.25 be reduced to a si ngle resistor cOllnected to the batteries? Explain. (b) Calculate each of the ullknown currents iI' I'J.' and r." for the circuit.
- /12
I 1.0 n
4.0 n
-
jI,
I;
3.0n 1.0 n
n
j
12
26. A dead bauery is charged by connecting it [0 the liye billtery or another car wit h jumper cables (Fig. 1'18.26). Determine the current in the starler and in the dead battery.
+
'%
36 V
12.0\,1
-.Eoov IlS.OV 4.00 il
2.00 n
3.on il FIGURE P18.29
SECTION 18.5 RC CIRCUITS
1.00 n 0.06 n
+
-
[1[]Find the potential difference across each resistor in Figure P18.29.
.5.000
FIGURE P18.25
0.01 n
iOOn .
FIGURE P18.28
2.0 n
5.0
I~V+
+ 5.00 fl
~4V
5.00 n
'IR
+
Slarter
30. Show I hat T = RC has units of time.
FIGURE P18.26
31. Consider the series RC-circuit shown in Active Figlire 18.16 for which R = 75.0 kfl, C = 25.0 .uF, and e = 12.0 V. Find (a) the time constant or the circuit and (b) the charge on the capacitor one time constant after the swilch is closed.
27. _ (a) Can the circuit shown in figure P18.27 be reduced to a single resistor connected La the ball.eries? Explain. (b) Find the magnitude of the current and iu.. direction in each resi~tor.
32. An uncharged capacitor and a resistor arc can nected in series to a source of emf. If e = 9.00 V. C= 20.0 ,uF, and R = 1000., flnd (a) the time constam of' the circuit, (b) 1he maxi mum charge on the capacitor, and (c) the charge 011 t.he capacitor after one lime conSlanl.
Live
bauery
]0 V
Dead bauery
622
Chapter 18
Direct-Current Circuits
33. Consider a series RC circuit for which R = 1.0 Mfl, C = 5.0 j..LF, and e = 30 V. Find the charge on the capacitor 10 s after the switch is closed.
a D.IO-m piece of axon when it is not conducting an electric pulse. How Illany K+ ions are on the outside of the axon assuming an initial potential difference of 7.0 X 10-2 V? Is this a large charge per unit area? [Hint: Calculate the charge per unit area in terllls of the llulllbcrofangstroms (..\2) per electronic charge. An atom has a cross section of about 1}...'2 (l A = 10-10 111).] (b) How much positive charge must flow through the cell membrane to reach the excited state of +3.0 X 10-2 V from the resting state o[ -7.0 X LQ-2 V? How many sodium ions (:--J"a+) is this? (c) If it takes 2.0 illS for the Na+ ions [Q enter the axon, what is the average current in the axon wall in this process? (d) How much energy does it take [Q raise the potential of the inner axon wall to +3.0 X 10-2 V. starting from the resting potential of -7.0 X 10-2 V?
134.1:\ series combination ora 12-kf! resistor and an unknown capacirof is connected La a 12-V bauer)'. One secOlld after the circuit is completed. the voltage across dlC capacitOr is IO V. Determine the capacitance of the capacitor. 35. A charged capaciwf is connected to a resistor and a slvilCh as in Active Figure 18.li. The circuit has a tirnc constant of 1.5 s. After the s\\.. itch is closed, the charge 011 the capacitor is 75% orits initial charge. (a) Find the lime it takes to reach this charge. (b) I [R = 250 kO, what is the value of C? 36. A series He circuit has a lime conslam of 0.960 s. The bauery has an emf of 48.0 V, and the maximum current in the circuit is 0.500 mAo What are (a) the value of the capacitance and (b) the charge stored in the capacilOr 1.92 s arter the s\\'itch is closed?
External fluid
+ Positive charge { byer
SECTION 18.6 HOUSEHOLD CIRCUITS 37. \01/hat minimum number of 75-VI/ lightbulbs must be connected in parallel to a single 120-V household circuit LO trip a 30.0-A circuit breaker?
x \()-4lm
Negative { charge
layer
38. A lamp (R = 150 fl), an electric heater (R = 25 D), and a fan (R = 50 fl) are connected in paraJlel across a 120-V line. (a) What total current is supplied to the circuit? (b) What is the voltage across the fan? (c) \Vhat is the current in the lamp? (d) What pmvcr is expended in the heater?
(39.1/\ heating clement in a slove is designed 1.0 dissipate 3000 ,"V when connected to 240 V. (a) Assuming the resistance is cOnStanl, calculate the current in the heating element ifit is connected to J20V. (b) Calculate the power it dissipates at that volmge. 40.
mil
A coffee maker is rated at I 200 W, a toaster at ] 100 \N, and a waffle maker at I 400 W. The three appliances are connected in parallel to a common 120-V household circuit. (a) \\'hal is the Cllrrelll. in each appliance when operating independent.ly? (b) "Vhal total current is delivered to the appliances when all are operating simultaneously: (c) Is a 15-A circuit. breaker sufficient in this situation? Explain.
SECTION 18.8 CONDUCTION OF ELECTRICAL SIGNALS BY NEURONS 41.
II Assume
a length of axon membrane of about 0.10 m is excited by an action potential (length excited = nerve speed X pulse duration = 50.0 mls X 2.0 X 10- 3 S = 0.10 m). In the resting stat.e, the outer surface of the axon wall is charged positively with K+ ions and the inner wall has an equal and oppositc charge of negative organic ions, as shown in Figure 1'18.41. I'vlodel the axon as a parallelplate capacitor and take C = KEoA/d and Q = C n,VIO investigate the charge as follows. Use typical values [or a cylindrical axon of cell wall thickness d. = 1.0 X 10-8 Ill, axon radius r= 1.0 X 101 tLlll, and cell-wall dielectric constant K = 3.0. (a) Calculatc the positive charge on the oLllside of
Consider the model of the axon as a capacil.Or from Problem 41 and Figure P18.41. (a) How much energy does it take to restore the inner wall of the axon to -7.0 X 10-2 V, starting from +3.0 X 10-2 V? (b) Find the averagc current in the axon wall during this process.
43. " L"sing Figure 18.28b and the results of Problems 18Ald and 18.42a, find the po\',:er supplied by the axon per action potential.
ADDITIONAL PROBLEMS 44. How man)' different. resistance values can be conslructed from a 2.0-fl, a 'l.0-n, and a 6.0-£! resistor? Show how yOll would get each resistance value either individually or by combining them. 45. Find the equivalent resistance between points Figure PI8.45. 2A.{l
5.1 II
ao-----"Nv----,-~WV_-7
1.811
3.5 !l
3.611 FIGURE P1B.45
(l
and bin
Problems
S
146.\_ For thc circuit shown in Fig-lire PI8.46, the voltmeter reads 6.0 V and thl.: ammcter rcads 3.0 mAo Find (a) the value or R, (b) the emf of the battery, and (c) the \'oltage across the 3.0-kV resistor. (d) What assumptions did you have to make to solve this problem?
g.OO \
~2
623
120kll
10.0 IfF =IS 0 kn 3.00 kn
FIGURE P18.50
51. Ii
3.0 kD:
FIGURE P18.46
47. Find (a) the equivalent resistance of the circuit in Figure
"
-
,
X
)"
J:t
(i.On
A
c
B
2.4D:
")
15V--
G.on
"
A circuit consists of three identical lalTlps, each of reSiSI
PISA?, (b) each current in the circuit. (c) the potential rlifference across each resistor, and (d) the power dissipated by each resislOr.
-"
mil
G.O n
d
(j.on
g.on
:r
FIGURE P18.47
FIGURE P18.51 {l and b in Figure PIS,52 drops to onc-halr its original value when SWiLCh S is closed. Determine the value of R.
52. The resistance bctwcen points
4S. Three 60.0-\'\/, 120-V lightbulbs ,He connected across a 120-\' power source, as shown in Figure P18A8. Find (a) the LOtal power delivered 10 the three bulbs and (b) the potcntial difference across each. Assume the resistance of elCh bulb is const,lnt (even though, ill reality, the resistance increases markedly with current).
/I
90
Ion
{1
a
s
b
Ion
gon
120 V
FIGURE P18.52
FIGURE P18.48
49. An automobile bauery has an emf of 12.6 V and an internal resi~tance ofO.OSO n. The hcadlig'lllS ha\'e a total resistance 01" 5.00 n (assumed COl1st(1111). vVhat is the potential difference
50. III Figure P18.50 suppose the switch has been c1o~ed for a length of time ~llffici(;ntly long lor the capacitor La become fully charged. Find (a) the steady-sLate current in each resistor and (h) the charge on the capacitor.
s:t
A generator has a terminal voltage of 110 V when it delivers 10.0 A and 106 V when it delivers 30.0 A. Calculate the emrancl the imernal resislance ofLhc generator.
54. An emf of 10 V is connected to a ..e .. ic~ RCcircuit consist·
ing of" ,1 resistor of ::!.O X lOll fl and a capaciLOI' of ::S.n,uF. Find thc time required for the charge on Ihe capacitor to reach 90% or its fin"l vaillc. 155.IThe student cngillt;cr or a campus radio station wishes to veri f"} I he clTectivcllcss of the Iightn ing- rod on the a mcnna lllast (Fig. PI8.55, page 624). The unknown resistance RiC is betwecn points Cand J.:. Point Eis a "true gl"Ound," but is inaccessible for direct measurcmenl because the Stn1tum ill which it is located is sevl.:ral meters below Earth's
624
Chapter 18
Direct-Current Circuits
surface. Two identical rods are driven illlo the ground at A and B, introducing an unknown resistance Ry The procedure for finding the unknown resistance R.~ is as follo\\'s. i\'Ieasurc resistance R l between points A and B. Then (OnneCl A and B with a heavy conducting wire and measure resistance R'l. between points A and C. (a) Derive a formula for R" in terms of the observable resistances R 1 and R2 - (b) A smisfaclOf)' ground resistance would be R x < 2.0 n. Is the grounding of the station adequate if measurcmems give R j =: 13 nand R 2 = 6.0 D?
~,
r',
,r/J
Rx
R\
-
Ii
•
ill a simple circuit diagram as a resiswr in series with the battery.) (a) Graph the power dissipated in I he load resistOr as a function of the load resistance. (b) With your graph, demonstrate the following important theorem: Tltp power delivered to a load is a. maximum if the load resistance eql/als the i1'l1('(l/a[ -resistanfe of the SOltrte. 160.IThe circuit in Figure PIS.GO contains two resislors, R] = 2.0 kfl and R';!. = 3.0 kn, and two capacitors, C] = 2.0,u.F and C.! = 3.0 ,uF, connected to a bauer)' with emf e = 120 V. If there arc no charges on the capacilOrs before S\',iitch S is closed, determine the charges fJl and (12 on capacitors C 1 and C2 ' respectively, as functions of time, after the S\.... iICh is closed. (Hi-nt: First reconstruct the circllit so that it becomes a simple RC circuit colHaining a single resislor and single capacitor in series, connected to the battery, and then determine the total ch
!?J
-
-
FIGURE P18.SS
56. The resistor R in Figure PIS.56 dissipates 20 \V of power. Determine the value of R.
5.un
FIGURE P18.60
40
75 V
~o
a
a
61. Consider the circuit shown in Figure Pl8.61. Find (a) lhe
potential difference bet\.. .cen points a and b and (b) the current in Ihe 20.0-D resistOr.
II
FIGURE P18.56
58. For the network in Figure PlS.5S, show that the resistance bet\.. .een points a. and bis RUb = -H fl. (Hint.: Connect a battery with emf e across points a and b and determine Sf I, where lis the currelll in the bauery.) 1.0
a
1.0
1.0
a
n
3.0n
"V, "
lOon
57. A voltage !1Vig applied to a series can IIguration of n resistors. each of resistance H. The circuit componems arc reconnected in a parallel configuration, and voltage L\.V is again applied. Show that the power consumed by the series configuration is 1/!I~ times the power consumed by the parallel configuration.
"
e
s
~ 1---------!1
a
5.0
b
10.00.
a
5.000.
9 0 \-'
1~5
b
"
.
.
soon
20.0
n
FIGURE P18.61
62. In Figure PI8.62, N[ = 0.100 fl, I{l = 1.00 n. and Rj = 10.0 fl. Find the equivalent resistance of the circuit and the current in each resistor when a 5.00-V power supply is connected bct\.. .een (a) points A and B, (b) points A and C. and (c) points A and D.
lOon
n R,
II,
II,
FIGURE P18.58
59. A battery wilh an internal resistance of 10.0 n produces an open-circuit voltage of l 2.0 V. A variable load resis~ lance with a range from 0 to 30.0 n is connected across the baltery. (Note: A battery has a resistance that depends on the condition orits chemicals and that increases as the bauer)' ages. This internal resistance can be represented
A
13
c
D
FIGURE P18.62
163.1What arc the expcCled readings of the ammeter and \'oltmeIer for the circuit in Figure PI8.53?
Problems 6.0011
625
remains on (a) the 3.00-f.LF capacitor? (b) The 2.00-f.LF capacitor? (c) What is the CUITCIll in the resil)tor?
Aj----Vl/lr-----, 10.011
::~
5.0011
6.0011
~
FIGURE P18.63
5.00
64. Consider [he two arrangcmcms of bauci-ies and bulbs shown in Figure PI8.64. The two bulbs are identical and have resistance R, and the two batteries arc idclllical with output voltage 1V (a) [n case I, with the two bulbs in series, compare the brightness of each bulb, the currCnt in each bulb, and the power delivered to each bulb. (b) In case 2, with the twO bulbs in parallel, compare the brightness of each bulb, the current in each bulb, and the power supplied to each bulb. (el Which bulbs arc brighter, those in case 1 or those in case 2? (d) In each case, ifone bulb fails, will the other go om as well? If the other bulb doesn't fail, will it gel brighter or Slay the same? (Problem 64 is counes)' of E. F. Redish. For other problems of lhi~ Iype, visit htlp://www.physics.umd.cdu/perg/.)
LJ
al' -
AI
-f-al'r 1
Il.~
C..
CaM' 2
FIGURE P18.64
65. The gi\cn pair of capacitors in Figure 1'18.65 arc fully charged by a 12.0-V bauery. The bauer)' is di~conncCled and thc circuit dOl)cd. After 1.00 ms, how much charge
IO~
n
FIGURE P18.65
66.
E
What is the equivalent resistance of the collection of resistors shown in Figure PI8.66~
R
1/
1/
FIGURE P18.66
67.
A'l-
x
II
An electric eel genenltcs elccu'ic currents through it~ highly specialized Hunter's organ, in which thousands of disk-shaped cells called electroc}'tcs are lined up in series, very much in the same way baucries flrc lined up inside a f1ashliglu. When activat.ed. each eleclroc}'tc can maintain a potcnt.ial difference ofabou!. 150 mVat.a currentofl A for abom 2.0 ms. Suppose a grown electric eel has 4.0 X 10 3 eleClrocytcs and can delh-er up to .!300 shocks in rapid series over about I s. (a) Whall1laximum electrical power can an e1eClric eel generate? (b) Approximately how llluch energy docs it rcleasc in one shock? (e) How high would a mass of I kg havc to be lifted so that its gravitational potcntial encrg}' cquals lhc energy released in 300 such shocks?
Aurora borealis, the northern lights. Displays such os this one are caused by cosmic ra.... portides trapped in the magnetic field of Earth_ When the particles collide with atoms in the atml iphere, they
cause the atoms to emit visible
I;ght
19.1
Mognets
19.2
Earth's Mognetic Field
19.3
Magnetic Fields
19.4
Magnetic Force on a Current-Carrying Conductor
19.5
Torque on a Current Loop and Electric Motors
19.6
Motion of 0 Charged Particle in a Magnetic
F;eld
19.7
19.8 19.9
19.10
Magnetic Field of a Long, Straight Wire and Ampere's Law Magnetic Force Between Two Parallel Conductors Magnetic Fields of Current Loops and Solenoids Magnetic Domains
MAGNETISM In terms of applications, magnetism is one of the most important fields in physics. Large electromagnets are used to pick up heavy loads. Magnets are used in such de.... ices as meters, motors, and loudspeakers. Magnetic tapes and disks are used rou1inely in sound- and videorecording equipment ond to store computer data. Intense magnetic fields are used in magnetic resonance imaging (MRI) devices to explore the human body with better resolution and greater safety than x-rays can provide. Giant superconducting magnets are used in the cyclotrons that guide particles into targets at nearly the speed of light. Roil guns (Fig. 19.1) use magnetic forces to fire high-speed projectiles, and magnetic bottles hold antimatter, a possible key to future space propulsion systems. Magnetism is closely linked with electricity. Magnetic fields affect moving charges, and moving charges produce magnetic fields. Changing magnetic fields can even create electric fields. These phenomena signify on underlying unity of electricity and magnetism, which James Clerk Maxwell first described in the 19th century. The ultimate source of any magnetic field is electric current.
19.1
MAGNETS
MOSl people have had experience wilh some form of magnet. YOLI arc most likely familiiuwilh the cOllllllon iron horseshoe magncllhat can pick up iron-containing objccls such as paper clips and nails. 1n the discussion thal follo\\'s, we assume the magnet has the shape of a bar. fron objects arc most strongly altracted to either end of such a bar magnel, called its poles, One end is called the north pole and lhc olher the south pole. The names come from the bcha,'ior of a magnet 111 the presence or Earth's magnetic field, If a bar magnet is suspcnded from its midpoint by a piece of su+ing so that il can swing freely in a horiZOnLal plane, il will roratc lIJ1lil its north pole points to the north and ilS south pole points to the sOlllh, The 'iamc idea is used to construct a simple compass, \1agnctic poles also exert attractive or repulsive forces 011 each olher similar to the electrical forces bel",:een
626
19.1
Magnets
627
FIGURE 19.1 Rail guns launch projectiles at high speed using magnetic force. Larger versions could scnd pa>'lo"ds into space or be llsed as thrusters to move metal-rich aSlcroids frum deep space to Eilrth orbil. In lids photo a rail ~1I1l at Sandia l'\ational LaboraTOries in Albuquerque, ;\lew Mexico fires a pr~iectile at over three kilometers per second. (For morc information 011 this, sec Appl),ing Ph~'sics 20.2 011 p;lgC 671.)
charged obje([s. in fact, simple experiments with two bar magnets show that like poles repel each other and unlike poles attract each other. Although the force bcnveen opposite magnetic poles is similar to the force between positive and negative electric charges, there is an importa11l difference: positive and negative electric charges can exist in isolation of each other; north and south poles don't. No maHer how rnany times a permanent magnet is cut, each piece always has a north pole and a south pole, There is some theorctical basis, however, for the speculation that magnetic monopoles (isolated north or south poles) exist in nature. and the attempt to detect them is currently an active experimental field of investigat.ion. An un magnetized piece of iron can be magnet.ized by stroking it ,... it.h a magnet. M~lgnetisl1l can also be induced in iron (and other materials) by other means. For example, ira piece of unmagnetized iron is placed near a strong permanent magnet., the piece of iron eventually becomes magnetized. The process can be acceler· ated by heat.ing and then cooling the iron. Naturally occurring magnetic materials such as magnetite are magnetized in th is way because they have been subjected to Earth's magnetic field for long periods of time. The extcnt to which a piece of material retains its magnetism depends on whether it is classiRecl as magnetically hard or soft. Soft rnagnetic materials, such as iron, are easily magnetized, but also t.end to lose their magnetism easily. In contrast, hard magnetic materials, such as cobalt and nickel, are difficult 10 magnetize, buttcnd to retain their magnetism. In earlier chapters we described the interaClioll between charged objects in terms of electric fields. Recall that an electric field surrounds any stationary electric charge. The region of space surrounding a moving charge includes a magnetic Reid as well. A magnetic Aeld also surrounds a properly magnetized magnetic material. To describe any type or vector field, we must deAI2,.e its magnitude, or strength, and it.s direction. The direction or a magnetic field B at any location is the direction in which the north pole of a compass needle poillls ell that location. Active Figure 19.2a shows how the magnetic field ora bar magnet can be traced ,..· ith the aid of a compass, defining a magnetic field line. Several magnetic fleld lines of a
ACTIVE FIGURE 19.2 (a) Tracing the magnelic field ofa bar magnet with compasses. (b) Scveral magnetic field lines ora bar magnct.
(aj
(b)
628
Chapter 19
Magnetism
FIGURE 19.3 (a) The magnetic field pattern of a bar magnet, as displayed with iron filings 011 a sheet of paper. (b) The magnetic field paltern between 1I11like poles of 11m bar magnets. as displayed ....,jlh iron filings. (c) The magnetic field pauern between l\,'O like poles.
(,)
APPLICATION Dusting for Fingerprints
bar magnet traced out in this way appear in the lwo-c1imcnsioJ1ClI represemation in Active Figure 19.2b. Magnetic field pallerns can be displayed by placing small iron filings in the vicinity ora magnet, as in Figure )9.3. Forensic scientists use a technique similar to that shmvn in Figure 19.3 to find fingerprints at a crime scene. One way to fincllatcnt, or invisible, prints is by sprinkling a powder of iron dust on a surface. The iron adheres to any perspiration or body oils that are present and can be spread around on the surface "lith a magnetic brush that never comes iIHo contaCl with the powder or the surface.
19.2 TIP 19.1 The Geogrophic North Pole Is the Mognetic South Pole The nonh pole ora magnet in (l compass poims norlh because it's attracted to Eanh's magntlie sOlllh pole, located ncar Earth's
l7ogmphirnonh pole.
FIGURE 19.4 Earth's magnetic field lines. Note that magnetic south is at the north geographic pole ami magnetic north is at the south geograpli ic polc.
Icl
(h)
EARTH'S MAGNETIC FIELD
A small bar magnet is said to have north and south poles, but it's more accurate to say it has a "north-seeking" pole and a "south-seeking" pole. By these expressions, we mean that if such a magnet is used as a compass, one end will "seek," or point 1.0, the geogrclphic North Pole of Eanh and the other end will "seck," or point lO, the geographic South Pole of Earth. We conclude lhal the geographic North Pole of Earth corresponds to a magnetic south pole, and the geographic South Pole of Earth corresponds to a magnetic north pole. In fact, the configuration of Earth's magnetic field, pictured in Figure 19.4, very much resembles whal would be observed ira huge bar magnet were buried deep in the Eanh's interior. II' a compass needle is suspended in bearings that allow it to rotate in the verti~ cal plane as 'well as in the horizontal plane, the needle is horizontal 'with respect to Earth's surface only near the equator. As the device is moved northward, the
Magnetic axis
~ ~\f2~:' ,\x;, of
South magnctic
rotation North
\
South gcogr
magnetic pole
19.2
needle rotates so that it points more and more toward the surface of Earth. The angle between the direction of the magnetic field and the horizontal is called the dip angle. Finally, at a point just north of Hudson Bay in Canada, the north pole of the needle points directly downward, with a dip angle of90 c . That site, rirst fOlilld in 1832, is considered to be the location of the south magnetic pole of Earth. It is approKimately 'I 300 miles from Eanh's geographic North Pole and varies \.. . ith time. Similarly, Earth's magnetic north pole is about 1200 miles [rom its geographic Sout.h Pole. This means that compass needles poilll only approximately north. The difference betv·:een true north, den ned as the geographic North Pole, and nonh indicated by a compass varies from point [Q point on Earth, a difference referred to as magnetic rife/ina/ion. For example, (lIang a line through South Carolina and the Great Lakes a compass indicates true nonh, w'hereas in \·Vashington slale it aligns 25" east 01" lrue north Wig. 19.5). Although t.he magnetic field pattern of Earth is similar to the pauern that would be set tIp by a bar magnet placed at its center, the source of Earth's field can't consist of large masses of permanently magnetized material. Earth does have large deposits of iron ore deep beneath its surface, but the high temperatures in the core prcvent the iron from retaining any permanent magnetization. It's considered more likely lhat the true source ofEanh's magnetic field is electric current in the liquid part of its core. This current, which is not well understood, may be driven by an intcraction between the planet's rotation and convection in the hot liquid core. There is some evidence t.hat the strengt.h of a plane['s magnetic field is related to the planet's rate of rotat.ion. For example, Jupiter rotates faster than Earth, and recent space probes indicate that Jupiter's magnetic field is stronger than Earth's, even though Jupiter lacks an iron core. Venus, on the other hand, rotates more slmvly than Earth, and it.s magnetic field is weaker. Investigation into the cause of Earth's magnetism continues. An interesting fact concerning Earth's magnetic field is that its direction reverses every few million years. Evidence for this phenomenon is provided by basalt (an iron-containing rock) that is sometimes spewcd forth by volcanic activity on the ocean floor. As the lava cools, it solidifies and retains a picture of the direction of Earth's magnetic field. \·Vhen the basalt deposits are dated, they provide evidence for periodic reversals of the magnetic field. The cause of these neld reversals is sti II nOt underst.ood. ft has long been speculated that some animals, such as birds, use the magnetic ncld of Earth to guide their migrations. Studies have shown that a type of anaerobic bacterium that lives in swamps h€ls a magnetized chain of magnetite as part of its internal structure. (The term anaerobic means that these bacteria live and grow without oxygen; in fact, oxygen is toxic to them.) The magnetized chain acts as a compass needle that enables the bacteria to align \\lith Earth's magnetic field. When t hey find themselves out of the mud on the bottom of the swamp, thc)' return to their oxygen-free environment by following the magnetic field lines of Earth. Further evidence r-or their magnetic sensing abilit), is that bacteria found in the northern hemisphere have internal magnetite chains that are opposite in polarity to those of similar bacteria in the southern hemispherc. Similarly, in the nonhern hemisphere, Earth's field has a downward component, whereas in the southern hemisphere it has an up\\lard component. Recently, a mcteorite originating on Mars has been found to contain a chain of magnetite. NASA scientists believe it Illay be a fossil of ancient Martian bacterial life. The magnetic field of Earth is used to label runways at airports according to their direction. A large number is painted on the end of the runway so that it can be read by the pilotoran incoming airplane. This number describes the direction in which the airplane is traveling, expressed as the magnetic heading, in degrees measured c10cbvise from magnetic nonh divided by 10. A runway marked 9 would be directed toward the east (90 divided by lO), whereas a runway marked 18 would be directed toward magnetic south. 0
Earth's Magnetic Field
629
f:J 15"E • \',
j";
FIGURE 19.5 A map or the con~ tinciltal L:lliH~d States showing the
declination ora compas<; rrom true north.
APPLICATION Magnetic Bacteria
APPLICATION Labeling Airport Runways
o
630
Chapter 19
Magnetism
APPLYING PHYSICS 19.1
COMPASSES DOWN UNDER
On a business trip to Australia, >'QU take along your that you may have used on a camping trip. Does this compass work correctly in AustraEa?
American~madecompass
Explanation There's no problem with using the compass in Australia. The north pole of the magnet in the compass will be attracted to the south magnetic pole near the g'eographic North Pole,just as it
was in the United States. The only difference in the mag-netic field lines is that they have an upward component in Australia, whereas they have a dQ',vl1ward component in the United States. Held in a horizontal plane, your compass can't detect this difference, however; it only displays the direction of the horizontal component of the magnetic field.
19.3 MAGNETIC FIELDS Experiments show that a stationary charged panicle doesn't interact with a static magnetic field. When a charged particle is moving through a magnetic field l however, a magnetic force acts on it. This force has its maximum value when the charge moves in a direction perpendicular to the magnetic field lines, decreases in value at other angles, and becomes zero when the panicle moves along the field lines. This is quite different from the electric force, '\vhich exerts a force on a charged particle whether it's moving or at rest. Further, the electric force is directed parallel to the electric field 'whereas the magnetic force on a moving charge is directed perpendicular to the magnetic field. In ollr discussion of electricity, the electric field at some point in space was defined as the electric force per unit charge acting on some test charge placed at that point. Tn a similar 'way, \V'e can describe the properties of the magnetic field B at some point in terms of the magnetic force exerted on a test charge at that point. Our test o~ject is a charge q moving with velocity It is found experimentally that the strength of the magnetic force on the particle is proportional to the magnitude of the charg~/b the magnitude of the velocity V, the strength of the external magnetic fleld B, and the sine of the angle () bet\\'een the direction of and the direction of B. These observations can be summarized by 'writing the magnitude of the magnetic force as ~
v.
v
~
F~
qvBsin
e
[19.1]
This expression is lIsed to define the magnitude of the magnetic field as
B=
F
qv sin
[19.2]
e
If F is in newtons, q in coulombs, and v in meters per second, the 51 unit of magnetic field is the tesla (T), also called the weber (Wb) per square meter (1 T = I \·Vb/m 2 ). If a l-C charge moves in a direction perpendicular to a magnetic field of magnitude] T with a speed of 1~n/s, the magnetic force exerted on the charge is 1 N. \·Ve can express the units ofB as
[B]
=
T
=
'~~ m'
=
_N_ ~ ~ C'm/s
[19.3]
A'ill
Tn practice, the cgs unit for magnetic field, the gauss (G), is often used. The gauss is related to the tesla through the conversion
IT = la' G Conventional laboratory magnets can produce magnetic fields as large as about 25000 G, or 2.5 T. Supercondllcting magnets that can generate magnetic fields as great as 3 X 10 5 G, or 30 T, have been constructed. These values can be compared with the value of Earth's magnetic field near its surface, lvhich is about 0.5 G, or 0.5 X 10-' T.
19.3
Magnetic Fields
631
From Equation 19.1 we see that the force on a charged particle moving in a magnetic field has its maximum value when the particle's motion is !Jf!"11Jendicu[arLo the magnetic field, corresponding to () = 90°, so that sin () = I. The magnitude of this maximum force has the value l~lI,t:< = gvB
[19.4]
v
Also from Equation 19.1, F is zero when is parallel to Ii (corresponding to () = 0° or 180°), so no magnctic force is exerted on a charged panicle "",hen it moves in the direction of the magnetic field or opposite the field. Experimcnts sho~ that the direction of the magnetic force is ahvays perpendiciliaI' to both and B, as shown in Figure 19.6 {or a positively charged panicle. To determine the direction of the force, we employ right-hand rule number 1:
v
1. Point the flllgers orY'our right hand in the direction of the velocity 2. Curl the fingers in the direction of the magnetic field
v.
B, moving through
FIGURE 19,6 The direclioll of the magnetic force on a posilivdy charged particle moving with a "docily ~ in lIlt.: prt:scnce of it tn;tgtlClic field. When ./ is at an ang-l<: (J wilh respeclto Ii, lhe magnelic fOI'.se is perpendicular LO hOlh v and B.
the smallest angle (as in Fig. 19.7). ~
3. Your thumb is now pointing in the direction of the magnetic force F exerted on a positive charge.
F
If the charge is negative rather than positive, the force is directed o/J/JOsile that shown in Figures 19.6 and 19.7_ So if' q is negative, simply lise the right-hand rule to ftnd the direction for positive q and then reverse that direction for the negative charge.
QUICK QUIZ 19.1
A charged particle moves in a straight line I hrough a region of'space. vVhich of the following answers mu.st be true? (Assume any other fields are negligible.) The magnetic field (a) has a magnitude of zero (b) has a zero component perpendicular 1.0 the particle's velocity (c) has a zero component parallel to the particle's velocity in that region. QUICK QUIZ 19.2 The north-pole end of a bar magnet is held near a stationary positively charged piece of plastic. Is the plastic (a) attracted, (b) repelled, or (c) unaffected by the magnet?
EXAMPLE 19.1 Goal
FIGURE 19.7 RighI-hand mJe number I for delermining the direc· tion oflhc magnetic force 011 a positive charge tlIoving~'ith a velocity V in a magnctic field B. Poi1l1 the fingers ofyout rig'" hand ill lhe directioll orv 'lIlr!.thcn curllhcm ill lhc--. direnioll ofB. The mag-uNic force F points ill the direction 01 y01l1" riglll thumb.
A Proton Travelin in Earth's Ma netic Field
Calculate the magnitude and direction or a magnetic force.
Problem A proton moves with a speed of 1.00 X 10:' m/s through Earth's magnetic field, which has CI value o[ 55.0 1.1..T at a particular location. When the proton moves eastward, the magnetic force acting on it is directed straight upward, and when it moves northward, no magnetic force acts on it. (a) What is the direction of the magnetic field, and (b) what is the strength of the magnetic force when the proton moves eastward? (c) Calculate the gravitational force on the proton and compare it with the magnetic force. Compare it also with the electric force if ~
-v·
there 'were an electric field w'ith a magnitude equal to I:; = 1.50 X 10 2 N/C at that location, a cOlllmon value at Earth's surface. Note that the mass of the proton is 1.67 X 10" kg. Strategy The direction of the magnetic field can be found from an application of the right-hand rule, l.ogether with the fact that no force is exerted on the proLan when it's traveling north. Substilllting into Equation 19.J yields the magnitude of the magnetic field.
....
• • '""IIl
Solution (a) Find the direction of the magnetic field. No magnetic force acts on the proton when it's going north, so the angle such a proton makes with the magnetic field direction must be either 0° or 180°. Therefore, the ma~netic field 8 ITlUst point either north or south. ~ow apply the right-hand rule. vVhen the panicle ~
travels east, the magnetic force is directed upward. Point thu III bin the d ireCl.ion of the force and you r fi ngel's in the direction of" the velocity eastward. ,"Vhen you curl your fingers, they point north, which mllstlherefore be the direction of the magnetic ficld. yOll r
632
Chapter 19
Magnetism
(b) Find the magniwde of the magnetic force. Substitute the given values and the charge of a prolon into Equation 19.1. From pan (a), the angle between the \'clocilY of the proton and the rnagnclic field Ii is 90.00.
F ~ quB sin
(c) Calculate the gravitational force 011 the prolon and compare it with the magnetic force
Fgm =
Remarks The information regarding a proton moving north was necessary to fiX the direction of the magnetic field. Otherwise, an upward magnetic force on an eastward-moving proton could be caused by a magnetic field pointing anywhere northeast or northwest. I\'otice in pan (c) the relative strengths of the forces, with the electric force larger than the magnetic force and both much larger than the gra\'itational force, all for typical field values found in nature.
QUESTION 19.1 An electron and proton moving with the same \'elocit)' enter a uniform magnetic field. In what two ways does the magnetic field affect the electron differel1lly from the proton? EXERCISE 19.1 Suppose an e1eClron is moving due wcst in the same magnetic field as in Example 19.1 at a speed of 2.50 X 10 5 m/s. Find the magnitude and direction of the magnetic force 011 the electron. Answer
2.20
X
10- 18 N, straight up. (Don't forget, the electron is negatively charged!)
EXAMPLE 19.2 A Proton
M:::o:.:.vi:.:.;n::c:..:....:=~=~~~~
,
Goal
Calculate the magnetic force and acceleration when a panicle moves at an angle other than 90" to the field.
I
Problem A proton moves at 8.00 X 106 mls along the x-axis. It enters a region in which there is a magnetic field of magnitude 2.50 T, directed at an angle of 60.0" with the x-axis and lying in the X)I-plane (fig. 19.8). (a) Find the initial magnitude and direction of the magnetic force on the proton. (b) Calculate the pralOn's initial acceleration. Strategy finding the magnilUde and direction of the magnetic force requires substituting values into the equation for magnetic force, Equation 19.1, and llsing the righl-hand rule. Applying Newton's second law solves pan (b).
.J FIGURE 19.8 (Exa.!.~lple 19.2) The magnetic force F on a proton is in tl~ positive z-direnion when "nd B lie in the x)'·pl"ne.
v
Ii
k-----y
x
r'
'"'II!
Solution (a) Find the magnitude and direction of the rnagnetic force 011 the proton. Substitute 1.J = 8.00 X 10 6 mis, the magnetic field strength B = 2.50 T, the angle, and the charge o[ a prown into Equation 19.1:
F= qvBsin (]
(1.60 X 10- 19 C)(8.00 X 1O"m/s)(2.50 T)(sin 60") F= 2.77 X 10- 1' N
19.4
Apply rigllt~hand rule number 1 La find the initial direction of the magnetic force:
Magnetic Force on a Current-Carrying Conductor
633
Point the fingers of' the right hand in the x-direction (the direction orV') and then curl them toward B. The thumb points upward, in the positive z-direClion.
(b) Calculate the proton's initial acceleration.
ma = F
Substitute the force and the mass of a proton into Newton's second law:
-->
(1.67 X 10- 27 kg)" = 2.77 X 10- 12 N
...............
lrr.. • • • . . • •
Remarks The initial acceleration is also in the positive z-direction. Because the direction ofV' changes, however, the subsequent direction orthe magnetic force also changes. In applying right-h£lnd rule number 1 to find the direction, it was imponantLO take into consideration the charge. A negatively charged panicle accelerates in the opposite direction. QUESTION 19.2 Can a constant magnetic Aeld change t.he speed ofa charged particle? Explain. EXERCISE 19.2 Calculate the acceleration oran electron that moves through the same magnetic field as in Example 19.2, at the same velocity as the proton. The mass of an electron is 9.11 X 10- 31 kg.
Answer
3.04 X 10 1M m/s~ in the negalive z-direcrion
19.4
MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR
Ifa magnetic field exerts a (orce on a single charged particle when it moves through a magnetic field, it should be no surprise that magnetic forces are exerted on a current-carrying \\lire as well (see Fig. 19.9). Because the current is a collection of many charged particles in motion, the resultant force on the \'lire is due to the SUIll of the individual forces on the charged particles. The force on the panicles is transmitted to the "bulk" of the 'wire through collisions with the atoms making up the wire. Some explanation is in order concerning notation in manv of the figures. To indicate the direction orB, we lise the following conventions: ~
If B is directed into the page, as in FigurS 19.10, we use a series of green crosses, representing the t.ails of arrows. If B is directed out of the page, we use a series of green dots, representing the tips of arrows. If B lies in the plane of the page, we usc a series of green field lines ·with arrowheads. ~
The force on a current~Glrrying conductor can be demonstrated by hanging a wire between the poles ora magnet, as in Figure ]9.10. In this figure, the magnetic
;;
]
..._-
FIGURE 19.9 Thisapparaws demOnStrales the force on a currentcarrying conductor in an external magnelic field. Why does the bar swing {Jwa_~ from the magnet afler lhe SI\"ltch is closed?
FIGURE 19.10 A segmell1 ofa flexible \'ertical wire partially stretched beth'cen thc poles ofa magnet, with LiH; field (green crosses) direcled imo the page. (a) When there is no current in the wire, it remains vertical. (b) When [he current is upward. the wire deflects to the left. (cl ·When the current is downward, the wire deflects to the right.
(b)
(el
634
Chapter 19
Magnetism
A
field is direcled into lhe page and covers the region within the shaded area. The v.'ire denects to the right or left when it carries a current. \Ve can quantify this discussion by considering a straight segment of wire of _ . 0 Icng,th and cross-sectional area A carrying current I in a uniforJTI external maunetic field B, as in Figure ]9.11. \!Ve assume that the magnetic field is perpendicular (Q the wire and is directed into the page. A force of magnitude 1\11.1), = qvdB is exerted on each charge carrier in the wire, \,,!here v" is the drift velocity of the charge. To find the total force on the wire, \ve Illultiply the force on one charge carrier by the number of carriers in the segment. Because the volume of the segIllent is A[, the number of carriers is nAt', where 11. is the number of carriers per unit volume. Hence, the magnitude of the total magnetic force on the wire of length .f is as follows: Tota I force = force all each charge carrier X total number of carriers ~
FIGURE 19.11 Aseetiollora wire conwining moving ch~Hres in an eXlernalmaglictic field B-
TIP 19.2 The Origin of the Magnetic Force on a Wire When a magnetic field is applied at some angle to a wire carrying a current,
e
l~"", ~ ('1v"B)(nAC)
From Chapter J7, however, we knO\v that the current in the wire is given by the expression 1= nqvdl1, so [19.5] This equation can be used only when the current and the magnetic field are at right angles to each other. If the wire is not perpendicular to the field but is at some arbitrary angle, as in Figure 19.12, the magnitude of the magnetic force on the wire is F~
B
FIGURE 19.12 A wire carrying a current I in the prssence of an external magnetic field B that makes an angle owith the wire.
APPLICATION Loudspeaker Operation
cone
B
N
FIGURE 19.13 <;peaker.
:\ diagram of
a
[19.6]
where (J is the angle bel\veen and the direction of the current. The direction of this force can be obtained by the use of right-hand rule number 1. In this case, however, you must place your fingers in the direction of the positive current ,. rather than in the direction of~. The current, naturally, is made up of charges moving at some velocity, so this reall}' isn't a separate rule. In Figure 19.12 the direction of the magnetic force on the wire is out of the page. Finally, \,,'hen the current is either in the direClion of the field or opposite the direction of the field, the magnetic force on the 'wire is zero. A magnetic force acting on a current-carrying \\lire in a magnetic field is the operating principle of rnost speakers in sound systems. One speaker design, shO\\ln in Figure 19.13, consists 0[£1 coil of wire called the voice coil, a flexible paper cone that acts as the speaker, and a permanent magnet. The coil of wire surrounding the north pole of the magnet is shaped so that the magnetic field lines are directed radially out\"ard from the coil's axis. \Vllen an electriG1J signal is sent to the coil. producing a current in the coil as in Figure 19.13, a magnetic force to the left acts on the coil. (This can be seen by applying right-hand rule number 1 to each turn of wire.) "Vhen the current reverses direction, as it would [or a current that varied simlsoidall}', the magnetic force on tbe coil also reverses direction, and the cone, \·,rhich is attached to the coil, accelerates to the right. An alternating current through the coil causes an alternating force on the coil, which results in vibrations of the cone.
,====="Current Paper
Blt sin
FIGURE 19.14 A silliple eleClromagnetic pump has 110 Illoving pans to damage a condllcting fluid, such as blood. passillg Lhrough. i\pplication of right-hand rule Ilumber 1 (right fingers in the direci iOIl of the c.!:inent I. curl them in the direction of B. thumb points ill the direction of the force) shows that the fmcc all the current-carrying segment of the fluid is in the direction of the velocilY.
19.4
Magnetic Force on a Current-Carrying Conductor
The vibrating cone creates sound waves as it pushes and pulls on the air in front or it. In this way, a I-kHz electrical signal is converted LO a I-kHz sound wave. An application of the rorce on a current-carrying conductor is illustrated by the electromagnetic pump shown in Figure 19.14 (page 634). Artificial hearts require a pump to keep the blood Ao\\,'ing, alld kidllt::)' dialysis machines also requi,"e a pUlllp to assist the heart in pumping blood that is to be cleansed. Ordinary mechanical pumps create problems because they damage the blood cells as they move through the pump. The mechanism shmvn in the figure has demonstrated some promise in such applications. A magnetic field is established across a segment of the tube contain ina the blood, nowing in the direction of the velocitY, V. An electric current passing through lhe ntlid in the direction sho\\'n has a magnetic force acting on it in the direction of V, as applying the right-hand rule shows. This force helps to keep the blood in Illotion.
635
APPLICATION
rtI
Electromagnetic Pumps for Artificial Hearts and Kidneys
~
APPLYING PHYSICS 19,2
LIGHTNING STRIKES
In a lightning strike there is a rapid movement of negative charge from a cloud 1.0 the ground. In what direction is a lightning strike deOeCled by Eanh's magnetic field?
EXAMPLE 19. Goal
A Current-Carr in
Explanation The dowm...· ard flow of negalive charge in a lightning stroke is equivalent to a current rnoving upward. Consequent.ly, we have an upward-mm'ing current in a nonhward-directed magnetic field. According to right-hand rule number 1, the lightning strike \"iotild be deflected toward the west.
Wire in Earth's Magnetic Field
Compare t.he magnetic force on a current-carrying wire with the gravitational force exerted on the wire.
Problem A wire carries a current of" 22.0 A from west to east. Assume the magnetic field or Earth at this location is horizontal and directed from south 1.0 north and it has a magnilude 01'0.500 X 10- 4 T. (a) Find the magnitude and direction of the magnetic force on a 36.0-m length of'wire. (b) Calculate the gravitational force on the same length of wire if it's made of copper and has a cross-sectional area 0[2.50 X 10-(; m 2 . r································
Solution (a) Calculate the magnetic force on the wire.
e=
Substitute into Equation 19.6, using the fact that the magnetic field and the current are at right angles to each other:
F = Blf sin
Apply right-hand rule number I 1.0 fmd the direction of the magnetic force:
\Vith the fingers oryour right hand pointing west to east in the direction of the current, curl them north in the direction of the magnetic field. Your thumb points upward.
=
(0500 X 10-' T)(22,0 A)(36.0 111) sin 90.0'
3.96 X 10-' N
(b) Calculate the gravitational force on the wire seg·ment.
First, obtain the mass of the wire from the density of copper, the length, and cross-sectional area of the wire:
III
= pll = p(Af) = (8-92 X 10' kg/m 3)(2,50
X 10-6 111'· 36.0111)
= 0,803 kg
To gel the gravitational force, multiply the mass by the acceleration of gravity: ..........................................
Remarks This calculation demonstrates that under normal circumstances, the gravitational force on a currentcarrying conductor is much greater than the magnetic force due La Earth's magnetic field.
QUESTION 19.3 ''''hat magnetic force is exerted on a wire carrying current parallel La the direction of the magnetic field? EXERCISE 19.3 ''''hat current would make the magnetic force in the example equal in magniwde to the gravitational force? Answer
4.37 X 10 3 A, a large current that would very rapidly heat and melt the ,.. . ire.
19.5
TORQUE ON A CURRENT LOOP AND ELECTRIC MOTORS
Tn the preceding section we showed how a magnetic force is exerted on a currentcarrying conductor when the conductor is placed in an external magnetic field. With this starting point, \'Ve nm.. . ' show that a torque is exerted on a current loop placed in a magnetic Held. The results of this analysis will be of great practical value when we discuss generators and motors in Chapter 20. Consider a rectangular loop carrying current J in the presence of an external uniform magnetic field in the plane of the loop, as shown in Figure 19.15a. The forces on the sides of length a are zero because these wires are parallel to the field. The magn itudes of the magnetic forces on the sides of length b, however, are
F, = F, = BIb ~
The direction of F I , the force on the left side of the loop, is out of the page and that of F 2 , the force all the right side of the loop, is into the page. Lf we view the loop from the side, as in Figure 19.15b, the forces are directed as shown. Ifwe assume the loop is pivoted so that it can rotate about point 0, we see that these two forces produce a torque about 0 that rotates the loop clockwise. The magnitude of this torque, T max ' is ~
a
T",,,
a
= F, 2 + 1<:, 2 ~ (BIb)
it
a
2 + (BIb) 2 = Blab
where the moment arm about 0 is a/2 lor both forces. Because the area of the loop is A = ab, the torque can be expressed as To""
= BIA
[19.7]
This result is valid only when the magnetic fleld is jJarallel to the plane of the loop, as in Figure 19.15b. If the fleld makes an angle 0 with a line perpendicular to the
---
-
I
Ii
(al
1 1
F]
I~l
b
I·
a
.\
-
F,
T (b)
F2
a/2
, ,,
; 8
Ii
;
-
flsin(}/
2'
(e)
/
,
F2
-
FIGURE 19.15 (a) Top viewofa lcct,tngular loop in a lmirarm magnetic field B. ~o magnetic forces act on the sides of leng! h (). parallel to ii, bl!} rorce~do act 011 the sides ofJenglh b. (b) /\ side vicw of the rectangular loop 5110\\';; tlWL l!le forces F] alld F~ on the sirles of ll.:ngdl bcreale a torque that tends to twist Iht: loup clockwise. (c) IFB is alan angle 0 with a line perpendicular to Ihe plane oflhe loop, the torque is A"ivcn hy BfA f',ill tJ.
19.5
Torque on a Current Loop and Electric Motors
637
plane of the loop, as in Figure 19.15c, the moment arm for each force is given by (0/2) sin 8. An analysis similar to the previous one gives. for the magnitude of the torque, T = BfA sin e [19.8] This result shows that the torque has the maximum value BfA when the field is parallel to the plane of the loop (8 = 90°) and is zero when the field is perpendicular [Q the plane of the loop (8 = 0). As seen in Figure 19.]5c, the loop tends to rotale to smaller vailles of 8 (so that the normal 10 the plane of the loop rotates toward the direction of the magnetic fleld). Although the foregoing analysis \.. 'as for a rectangular loop, a more general derivation shows that Equation 19.8 applies regardless of the shape of' the loop. Further, the torque all a coil with N turns is T
= BIAN sin
e
[19.9a]
The quantity p.. = JANis defined as the magnitude ofa vector i1 called the magnefic moment of the coil. The vector i1 always points perpendicular to the plane of the loop(s) and is such that if the thllmb of the right hand points in the direction of 11. the fingers of the right hand point in the direction of the current. The angle () in Equations 19.8 and 19.9 lies between the directions of the magnetic moment Jl and the magnetic field B. The magnetic torque can then be written T
p,B sin 8
=
[19.9b]
QUICK QUIZ 19.3 A square and a circular loop with 0.1e same area lie in the X)l-plane, where there is a uniform magnetic field B pointing at some angle 0 with respect to the positive z-direction. Each loop carries the sarne current, in the same dircction. 'Which magnetic torque is larger? (a) the torque on the square loop (b) the torque on the circular loop (c) the torqucs are the same (d) more information is needed
EXAMPLE 19.4 The Torgue on Goal
0
Circulor Loop in
a~M=a~g!.:.n:::e:.:.t:::ic;.;F~i~e.:.:ld:....
_
Calculate a magnetic torque on a loop of" current.
Problem A circular wire loop of radius 1.00 III is placed in a magnetic field of magnitude 0.500 T. The normal to the plane of the loop makes an angle of 30.0° with the magnetic field (Fig. 19.16a). The curren' in the loop is 2.00 A in the direction shown. (a) Find the magnetic moment of the loop and the magnitude of the lorCJue at this instant. (b) The same current is carried by the rectangular 2.00-m by 3.00-m coil with lhree loops shown in figure 19.16b. Find the magnetic 1ll0mentofthe coil and the magnitude orthe torque acting on the coil at that instant.
.' (h)
Strategy For each part, WCjllst have to calculate the area, use it in the calculation urlhe magnetic moment. and multiply the result by B sin O. Altoget.her, this process amounts to substituting values into Equation 19.9b.
I
(e)
FIGURE 19.16 (Ex;unplc 1.9.'1) (a) A circular curn:nt loop in an eXlernal magnetic licld B. (h) A rectallgular current loop ill tlte ~ame lIeld. (c) (Excrci~e IY.'I)
638 ~
Chapter 19
Magnetism
...
..
.
Solution (a) Find the magnetic moment of the circular loop and the magnetic lOrque exerted on it.
= ",r' = ",(1.00 m)' = 3.14 m'
First, calculate the enclosed area of the circular loop:
A
Calculate the magnetic moment of the loop:
JL= fA,\'= (2.00A)(3.J4m')(I) = 6.28A·m'
1'0\,- substitute values for the magnetic moment, magnetic field, and (J into Equation 19.9b:
T
= JLBsin =
e ~ (6.28A·m')(0.500T)(sin 30.0°)
1.57 N'm
(b) Find the magnetic moment of the reCLangular coil and the magnetic lOrque exened on it.
= LX
=
= 6.00 m'
Calculate the area of the coil:
A
Calculate the magnelic moment of the coil:
JL = JAN ~ (2.00 A)(6.00 m')(3) = 36.0 A·m'
SuhstitU[c values into Equation 19.9b:
T
= JLB sin =
.
~
H
(2.00 m)(3.00 m)
e~
(0.500 T) (36.0 A· m')(sin 30.0°)
9.00N·m
.
...
Remarks [n calculating a magnetic torque. it's not strictly necessary to calculate the magnetic moment. 1nstead, Equation "19.9a can be used directly.
QUESTION 19.4 What happens to the magnitude of the torque if the angle increases to\\.'ard gOO? Goes beyond 90 0 ? EXERCISE 19.4 Suppose a right triangular coil with base of2.00 rn and height 3.00 111 having twO loops carries a current 0[2.00 A as shown in Figure 19.16c. Find the magnetic moment and the torque on the coil. The magnetic field is again 0.500 T and makes an angle of30.0° with respect to the normal direction. Answers
JL = 12.0 A· m',
T
= 3.00 N· m
Electric Motors APPLICATION Electric Motors
DC Power source ACTIVE FIGURE 19.17 Simplified skelch of:l DC eleclric motor.
It's hard to imagine life in the 21st century without electric motors. Some appliances that contain motors include computer disk drives, CD players, VCR and DVD players, food processors and blenders, car starters, furnaces, and air conditioners. The motors convert electrical energy to kinetic energy of rotation and consist of a rigid cUlTent-carrying loop that rotates when placed in the field ofa magnet. As we have just seen (Fig. 19.15), the torque on such a loop rotates the loop to smaller values of 0 until the torque becomes zero, when the magnetic field is perpendicular to the plane of [he loop and 8 = O. If the loop turns past this angle and the current remains in the direction shown in rhe figure, the torque reverses direction and LUrns the loop in the opposite direction, that is, counterclockwise. To overcome this difficulty and provide continuous rotation in one direction. the current in the loop must periodically reverse direction. 1n alternating current (AC) motors, sllch a reversal occurs naturally 120 times each second. 1n direct current (DC) motOrs, the reversal is accomplished mechanically with split-ring contacts (commutators) and brushes, as shown in Active Figure 19.17. Although actual mOtors contain many current loops and commutators. for simplicity Active Figure 19.17 shows only a single loop and a single set of split-ring contacts rigidly attached to and rotating with the loop. Electrical stationary contacts
19.6
Motion of a Charged Particle in a Magnetic Field
639
FIGURE 19.18 Till: clll-{inl-" COlllj>ilrIJ11l:llt of thl: Tovola I'riu~, a h)'hl id \'chicle,
ca lied bms!lt!s are mai nl a ined in c1cctrica I contact with the rotali ng spl i t ri llg. These brushes are usually made or graphite because it is a good electrical conduct.or as well as a good lubricant. Just as the loop becomes perpendicular to the magnetic field and the torque becomes 7.ero, inertia carries thc loop forward in the c1ock't\'isc direction and the hrushes cross lhe gaps in the ring, causing the loop current to reverse its dircClioll. This rcversal provides another pulse of torque in the clockwise direction for another 180°, the current reverses, and the process repcats itself. Figure 19.18 shows a modern motor used to power a hybrid gas-electric GU,
19.6
Bill
v·
MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD
Consider the case of' a positively charged panicle moving in a uniform magnctic ('ield so that the direction of' II~ panicle's velocity is perpendintla!; 10 flip Jidd, as in Active Figure 19.19. The label Bill and the crosses indicatc that B is dirccted inl0 the LJd~e. ADplication of the right-hand rule at point Pshows lhat the direClioll of the magnetic force F at lhat location is upward. The force causes thc particle 10 alter its direction of travel and 1.0 follow a curved path. Application of' the righthand rule at any poill( shows that the magnetic force is always directed toward the center of the circular path; therefore, the magnelic force causes a ccmripetal acceleration, which changes only the direclion of and not its magnilllde. Because F produces the centripetal a~celeration, we can equate its magnitude. qvB in this case, to the mass of Ihe panicle multiplied by the centripetal accclcrcnion v'l./1: from Newton's second law, we find that
v
q ACTIVE FIGURE 19.19 I,.VIII.:11 rhe \'dod!, ofa charged parLiclc is perpcndicular to a uniform magllctic field, the panicle lTlo\'e's ill :I cird~. whose plane is perpendicular to B. \,'hich is directed into the page. (The crosses represent rile lails Oflhl: magnelicj1e1d \'ecLOrs,) The magnclic force F on lhe ch;\rge is alwiI\'s directed toward the ccnler of lhe cil'ck.
fIIV-"
F= '1v8=-T
wh ich gives 1'=
mv '1 B
+q
Helical paLh
[19.10]
This equ
ACTIVE FIGURE 19.20 A (hargt.:e! panicle having a vdocit)' dircctl:d al an angle wilh a tlniform lTIagrlt'lic field 1t\00'cs in a helical path.
640
Chapter 19
Magnetism
APPLYING PHYSICS 19.3
TRAPPING CHARGES
StOring charged particles is important for a variet), of applications. Suppose a uniform magnetic field exists in a finite region of space. Can a charged particle be injected into this region from the olltside and remain trapped in the region by magnetic force alone?
Explanation It's beslLO consider separately the components of the particle "elacity parallel and perpendicular to the field lines in the region. There is no magnetic force on the particle associated with the velocity component parallel to the field lines, so thal ,"clacity component remains unchanged. Now consider dle component of velocity perpendicular to the field lines_ This component will result in a magnetic force that is perpendicular to both the field lines and the velocity component itself. The path of a panicle for ,,,'hich the force is always perpendicular to the velocity is a circle. The particle therefore follows a circular arc and exits the field on the other side of the circle, as shown in Figure 19.21 for a particle with constant kinetic energy. On the other hand, a particle can become trapped if it loses some kinetic energy in a collision after entering the field, as in Active Figure 19.20.
Particle motion
FIGURE 19.21
(App1ring
rh~:.ic~
19.3)
Particles can be injected and contained if, in addition to the magnetic field, electrostatic fields are involved. These fields are lIsed in the Penning trap. \t\'ith these devices, it's possible to Store charged particles for extended periods. Such traps are useful, for example, in the storage of antimatter, which disintegrates completely on contact l'l'ith ordinary matter.
QUICK QUIZ 19.4 As a charged particle moves freely in a circular path in the presence of a constam magnetic field applied perpendicular La the particle's velocity. the panicle's kinetic energy (a) remains constant, (b) increases,
or (c) decreases.
EXAMPLE 19.5 Goal
The Mass S ectrameter: Identif ing Particles
Use the cyclotron equation to identify a particle.
Problem A charged particle enters the magnetic field of a mass spenromeler at a speed of 1.79 X 10 6 m/s. It subsequentl}' 1TI00'es in a circular orbit with a radius of 16.0 (111 in a uniform magnetic field of magnitude 0.350 T having a direction perpendicular to the particle's velocity. Find the parlicle's mass-to-charge ratio and identify it based on the
table on page 641. Strategy
After finding the mass-to-charge ratio with Equation ]9.10, compare it with the \'allles in the table, identi-
fying the particle. ~
.
.
Solution \"'rite the cyclotron equation:
Soh'e this equation for the mass divided by the charge. m/q, and subst.itute values:
~
1llV '}'=-
1B
11/.
IB
(0.160 m)(0.350 T)
1
v
J.79 X lOOmis
kg
8313 X 10. . C
19.6
fdentify the particle from the rable. All panicles are completely ionized.
Remarks The mass spectrometer is an important tool in both chemistry and physics. Unknown chemicals can be heated and ionized, and the resulting particles passed through the mass spectrometer and subsequently identified. QUESTION 19.5 ',Vhat happens to the momentum of a charged particle in a uniform magnetic field? EXERCISE 19.5 Suppose a second charged particle enters the mass spectrometer at the same speed as the panicle in Example 19.5. If it travels in a circle with radius 10.7 em, find d1e mass·to-charge ratio and identify the panicle from the table above.
Answers 2.09
X
10-8 kg/C; deuterium
EXAMPLE 19.6 Goal
The Moss Spectrometer: Seporot"'in"'9...I"'s""o.:.:t0::.tp."'eg;s
_ APPLICATION
Apply the cyclotron equaLioll to the process of separating isotopes.
Moss Spectrometers
Problem Two singly ionized atoms move out of a slit at point S in Figure 19.22 and into a magnetic field of magnitude 0.100 T pointing into the page. Each has a speed of1.00 X 10 6 m/s. The nucleus of the first atom contains one proton and has a mass of 1.67 X 10- 27 kg, whereas the nucleus of the second atom contains a proton and a neutron and has a mass of 3.34 X 10-27 kg. Atoms \\lith the same number of protons in the nucleus but different masses are called isotopes. The two isotopes here are hydrogen and deuterium. Find their distance of separation \vhen they strike a photographic plate at P. Strategy Apply the cyclotron equation to each atom, finding the radius of the path of each. Double the radii to find the path diameters and then find their difference.
xxxxxxx
FIGURE 19.22 (Example 19.6) TI\'o isotopes leave the slit at point Sand travel indifferent circular paths before striking a photographic plate at P.
. .. ..,
r·········· .
Solution
10- 27 kg)(1.00 X 106 m/s) (l.60 X 10 to C)(O.lOO T)
(1.67
Use Equation 19.10 to find the radius of the circular path followed by the lighter isotope, hydrogen:
X
= 0.104 m
Use the same equation to calculate the radius of the path of deuterium, the heavier isotope:
"11l<jV
19
-
= --- =
(3.34 X 10- 27 kg)(1.00 X 10 6
'IE
m/s) (1.60 X 10 19 C)(O.lOO T)
= 0.20901
Multiply the radii by 2 to find the diameters and take the difference, getting the separation x between the isotopes: 1Irr.- • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • . . • • •
.
~
Remarks During \Vodd \Nar II, mass spectrometers were used LO separate the radioactive uranium isotope U-235 from its far more common isotope, U-238.
642
Chapter 19
Magnetism
QUESTION 19.6 Estimate the radius of the circle traced
OLlt
by a singly ionized lead atom moving at the same speed.
EXERCISE 19.6 Use the same mass spectrometer as in EXJmpJeI9.6 to find the separation between two iSQ(opes of helium: normal helium-4, which has a nucleus consisting of two protons and two neutrons. and helium-3, which has two protons and a single neUlron. Assume both nuclei, doubly ionized (having a charge of 2e = 3.20 X 10-1\) C), enter the field at 1.00 X lOti m/s. The helium-4 nucleus has a mass of 6.64 X 10- 2i kg, and the helium-3 nucleus has a Illass of 5.01 X 10-27 kg.
Answer
0.102 m
ACTIVE FIGURE 19.23 (a) \\'hen there is no CUlTcnl ill the venical wire, all compass ncecl1L:s point in the same dirCClioll. (b) \'\'hcn the I,'ire carries ~I strong cunent, the comp
I'
II>I
"
z
HANS CHRISTIAN OERSTED Danish Physicist and Chemist {l777-18511
Oersted is best known for observing fhat a compass needle deflects when
placed near a wire carrying a current. This important discovery was the first evidence of the connection between electric ond mognetic phenomena. Oersted was also the first to prepare pure aluminum.
"TIP 19.3 Raise Your Right Hand! We have introduced two righthand rules in this chapter. He sure to use only your right hand when applying t.hest'" rules.
19.7
MAGNETIC FIELD OF A LONG, STRAIGHT WIRE AND AMPERE'S LAW
During a lecwre demonstration in 1819, Danish scientist Hans Oersted (1777]851) found that an electric current in a wire deflected a nearby compass needle. Th is momentous discovery, lin king a magnetic field with an electric current for the first Lime, \·\,a5 the beginning of our understanding of the origin of magnetism. A simple experiment first carried out by Oersted in 1820 clearly demonstrates that a current-carrying conductor produces a magnetic field. In this experiment, several compass needJes are placed in a horizontal plane near a long vertical wire, as in Active Figure 19.23a. \\'hen there is no current in the wire, all needles point in the same direction (that of Earth's field), as one v\'Ould expect. \A/hen the wire carries a strong, steady current, however, the needles all deflect in directions tangent to the circle, as in Active Figure 19.23b. These observations shml;' that the direction orB is consistent with the following convenient rule, right-hand rule number 2: Point the thumb of your right hand along a wire in the direction of positive current, as in Figure 19.24a. Your fingers then naturally curl in the direction of the magnetic field B. ""hen the current is reversed, the filings in Figure 19.24b also reverse.
FIGURE 19.24 (a) Right-hand rule Ilumbcr 2 lor detcITllillillg the dircc1ion of the magllc1ic Held dllc to a long. straiglH wire cHf)'ing a currellt. Kote that the magnelir field lines form circles arouud the wire. (b) r:ircular magnetic field lines:;urrouncling a curren I-carrying wire. displayed by iron filings.
IOJ
Ib)
19.7
Magnetic Field of
0
Long, Straight Wire and Ampere's Law
~
643
~
Because the filings point in t.he direclioll orB, we conclude that the lines orB form circles about the ,..'ire. By symmetry, the magnitude of is the same every''''here on a circular path centered on the ,\lire and lying in a plane perpendicular to the wire. By varying the~current and distance from the wire, it can be experimentally determined that B is proponional La the current and inversely proportional to the dist<\I1ce from the wire. These observations lead to a mathematical expression for the strengt.h of the magnetic field due to the current J in a long, straight wire:
B
,"of
B=-
[19.11]
21Tr
~ Magnetic
field due to a long,
straight wire
The proportionality constant JJ.nJ called the permeability of free space, has the value [19.12]
tI
T
Ampere's Law and a Long, Straight Wire
..\ .rbiuaf}' closed path
I
Equation 19..11 enables us to calculate the magnetic field due to a long, straight wire carrying a current. A general procedure for deriving such equations was proposed by French scientist Andre-Ma"ie Ampere (177S-]836); it provides a relation between the current in an arbitrarily shaped wire and the magnetic field produced by the wire. Consider an arbitrary closed path surrounding a currelll as in Figure 19.25. The path consists of many shon segments, each of length .6.e. Multiply one of these lengths by the component of the magnetic field parallel to that segment, where the product is labeled BII.d.t. According to Ampere, the sum of all such products over the closed path is equal to f..Lo times the nel current rthat passes through the sur{-~1Ce bouncled by the closed path. This statcrnem, known as Ampere's circuital law, can be \Hillen
FIGURE 19.25 An arbilrar)' closed path aroulid a Cllrrelll is used 10 caludale lhc magnetic field of the current b~' the usc of Ampere's rille.
[19.13]
Ile
where 8 1 is the component of B parallel to the scgmel1l of length and 2.BII n.t' means that \\'C take the sum over all the products Ell ;ie around the closed path. Ampere's Jaw is the fundamental law describing how electric currents create rnagnetic fields in the surrounding empty space. ''\'e can use Ampere's circuital law to derive the magnetic field due to a long, straight wire carrying a current I. As discussed earlier, each of the magnetic field lines of this configuration forms a circle with the wire at its centers. The magnetic field is tangent to this circle at every point, and its magnitudc has the same valuc B over the entire circumference ofa circle of radius r, so B = B, as shmvil in Figure 19.26. In calculating the sum Bu tie over the circular path, notice that. BII can be removed from the slim (because it has the same value H for each clement on the circle). Equation 19.13 then gives
L
FIGURE 19.26 A closed. circular p,lIh of radius raround
ANDRE.MARIE AMPERE (1775-1836) Ampere, a Frenchman, is credited with the discovery of electromagnetism, the relationship between electric currents and magnetic fields.
644
Chapter 19
Magnetism
Dividing both sides by the circumference 27TT, we obtain
B = !-Lo' 21Tr This result is identical to Equation 10.11, which is the rnagnctic lleld due to the
current f in a long, straight wire. Ampere's circuital law provides an elegant and simple method for calculating the rnagnetic Fields of highly symmetric currelll configurations, but it can't easily be used to calculate magnetic fields for complex current configurations that lack symmetry. In addition, Ampere's circuital lavI' in this form is valid only when the currents and nelds don't cbange with time.
EXAMPLE 19.7 The Magnetic Field of a Long Wire Goal Calculate the magnetic flcld ofa long, straight wire and the force that the field exerts on a particle. I
Problem A long, straight wire carries a current of 5.00 A. At one instant, a proton, 4.00 mm from the wire, travels at a speed of 1.50 X 10 3 m/s parallel to the wire and in the same direction as the current (Fig. ]9.27). (a) Find the magnitude and direction of the magnetic ficld created by the wire. (b) Find the magnitude and direnion of the magnctic force the wire's magnetic field exerts on the prown.
•
r
FIGURE 19.27
(Example 19.7) The
magnetic field due to the current is illlu the page
1
Strategy First use Equatioll 19.1 J to find the magnitude of the magnetic field at the given point. Use right-hand rule number 2 to find the direction of the field. Finally, substitute into Equation 19.1, computing the magnctic forcc on the proton. , .. -
~
.
.
'-'
Solution (a) Find the magnitude and direction of' the wire's magnetic field.
Usc Equation 19.1l LO calcubte the magnitude of the magnetic field 4.00 mll1 from the Voiire:
B
!-Lol
=-
27T)"
=
Apply right-hand rule number 2 to find the direction --> of the magnetic field B:
(47T X 1O-'T·m/A)(5.00A) 27T(4.00 X 10 'm)
= -'------'-------'----','---,----"-
2.50 X 10- 4 T
With the right thumb pointing in the direction of the current in Figure 19.27, the fingers curl into the page at ths.location of the proton. The angle 0 between and B is therefore 90°.
v
(b) Compute the magnetic force exerted by the wire on the proton. Substitute into Equation 19.1, which gives the magnilUde of the magnetic force on a charged particle:
F= qvBsin 0 = (1.60 X 10-" C)(1.50 X J03 m/ s) X (2.50 X 1O-'T)(sin900) =
Find the direction of the magnetic force with righthand rule number I: trr... . . • • • • • • .
6.00 X 1O- 211 N
Point your right fingers in the direction orv, curling them into the page toward B. Your thumb points to the left, which is thc direction of the magnetic force .
. . . . . . . . . . . . . . . ..
... . .. . . . . ......
Remarks The location of the proton is important. On the left-hand side, the wire's magnet.ic field points outward and the magnetic force on the proton is 1.0 the right.
19.8
Magnetic Force Between Two Parallel Conductors
645
QUESTION 19.7 How ,.. .' ould the magnetic force differ if it acted on an electron moving with the same velocity? EXERCISE 19.7 Find (a) the magnetic field created by the wire and (b) the magnetic force on a helium~3 nucleus located 7.50 mill to the left of the wire in Figure 19.27, traveling 2.50 X 103 m/s opposite the direction of the current. (See the data table presented in Example 19.5 on page 641). Answers
(a) 1.33 X 10-.1 T
(b) 1.07 X 10 '" N, direcled
lO
the ler, in Figure 19.27
19.8 MAGNETIC FORCE BETWEEN TWO PARALLEL CONDUCTORS As we have seen, a magnetic force acts on a current-carrying conductor when the conductor is placed in an external magnetic field. Because a conductor carrying a current creates a magnetic field around itself, it is easy to understand lhal two current-carrying wires placed close together exert magnetic forces on each other. Consider two long, straight, parallel wires separated by the distance d and carry· ing currents 11 and I'.! in the same direction, as shown in Active Figure 19.28. \\fire I is directly above wire 2. \Vhat's the magnetic force 011 one '...· ire due to a magnetic field set up by the other wire~ In this calculation we are finding the force on wire 1 due to the magnetic field of wire 2. The current''!. sets up magnetic field B'!, at wire 1. The direction ofB:! is perpendicular to the wire, as shown in the figure. Using Equation 19.11, we nnd that the magnitude of this magnetic field is J-Lo
'2
IJ, = 27Td According to Equ~ion 19.5, the magnitude of the magnelic force on wire I in the presence of field U:! due to ' 2. is
F = 13,' I
.! I
e=
(1'-"")' e 27Td
I
=
2
--"'!I'-i-";;:~~ 1 \
ACTIVE FIGURE 19.28 Two parallel wires, oriented \'enically. carry .'>1 cady currents and exerl forces 011 each other. The field B;- al wire I duc to wire 2 produces a force all win: I givcn by 1"1 :: B.,. i1f:. The force i~ .Ill racli,c if the current~ ha\'e lhe !;amc direction, as !;hO\\,ll, alld rt.:plllsivc irthe t\\'o currents have 0PP0l>ilC directions.
l'-oJ,',e 21Td
\Ale can rewrite this relationship in terms of the force per unit length:
FI
e=
}Loll I'},
27Td
[19.14]
The direction of Fl is downward, toward wire 2, as indicated by right-hand rule nurnber 1. This calculation is completely symmetric, which means thal the force on wire 2 is equal to and opposite F\, as expected from Newton's third law of action-reaction. We have shown thal parallel conductors carrying currents in lhe same direction attract each olher. You should use the approach indicated by Figure 19.28 and lhe steps leading to Equation 19.14 to show that parallel conductors carrying currents in Opposile directions 1-e/JP[ each other. The force bel ween two parallel wires carrying a current is used to define the SI unit of currem, lhe ampere (A), as follows:
F\
If two long, parallel wires I m apart carry lhe same current and the magnetic force per unit length on each wire is 2 X 10 7 N/m, the current is defined to be I A.
~
Definition of the ampere
The Siunito( charge, the coulomb (C), can now be defined in lerms of the ampere as follows: If a conductor carries a steady current of 1 A, the quantity of charge that flows through any cross section in I sis 1 C.
f'- Definition of the coulomb
646
Chapter 19
Magnetism
QUICK QUIZ 19.5 Which of the following actions would double the magnitude of the magnetic force per unit length between [".;0 parallel currentcarrying wires? Choose all correct answers. (a) Double one of the currents. (b) Double the distance behveen them. (c) Reduce the distance between them by half. (d) Double both currents. QUICK QUIZ 19.6 [f, in Figure 19.28, f l ~ 2 A and f, = 6 A, which of the following is true? (a) F I = 31'2 (b) F I ~ f'2 (c) F I ~ F2 /3
EXAMPLE 19.8 Goal
Levitating a Wire
Calculate the magnetic force of one current-carrying 'v'ire on a parallel currenl-calTying wire.
Problem Two wires, each having a weight per unit length of 1.00 X 10- 4 N/m, are parallel with one directly above the other. Assume the ,.. .,ires carry currents that are equal in magnitude and opposite in direction. The wires are 0.10 m apart, and the sum of the magnetic force and gravitational force on the upper \.. . ire is zero. Find the current in the wires. (Neglect Earth's magnetic field.) Strategy The upper wire must be in equl1ibrium under the forces of magnetic repulsion and gravity. Set the sum of the forces equal to zero and solve for the unknovvn current, J. , _ ...
~
Solution
~
F gra...
Set the sum of the forces equal to zero and substitute the appropriate expressions. Notice thaL the magnetic force between the wires is repulsive.
The currents are equal, so II = /2 substitutions and solve for 12 :
= I.
~
+ F mag
0
1 2 = -'-(2_,,_d-,-)(,--"-,,,gl,--C.:...)
Make these
/La
,
f- =
Substitute g'iven values, finding 1 2 , then take the square root. Notice that the weight per unit length, mg/ is given.
(2,,·0.100 m)(l.OO X 10- 4 N 1m) _
~
.,
50.0 A-
(4" X 10 'T·m)
e,
~.-
=
1= 7.07 A ,
-_.~
Remark Exercise 19.3 showed that using Earth's magnetic field to levitate a wire required extremely large currents. Currents in wires can create much stronger magnetic fields than Earth's magnetic field in regions near the wire. QUESTION 19.8 Why can't cars be constructed that can magnetically levitate in Earth's magnetic field? EXERCISE 19.8 If the current in each \vire is doubled, hmv far apart should the wires be placed if the magnitudes of the gravitational and magnetic forces on the upper wire are to be equal? Answer
0.400 m
19.9
MAGNETIC FIELDS OF CURRENT LOOPS AND SOLENOIDS
The strength of the magnetic field set up by a piece of wire carrying a current can be en hanced at a specific location if the wire is formed into a loop. You can understand this by considering the eHen of several small segments of the current loop, as in Figure 19.29. The small segment at r.he top of the loop, labeled LlX I , produces a magner.ic field of magnitude Bj at the loop's center, directed out of the page. The
19.9
direction ofB can be \"erified using right~hand rule number 2 for a long, straight wire. Imagine holding- the wire with your right hand, \vith rour thumb pointingi!) the direClion of' the cllrrent. Your tingers then curl around in the direClion ofB. /\. segment 01" lenglll 6.x 2 at the bottom orth<..: loop also contributes to the field at the center. incl"casing its slreng·th. The field produced al the center by lhe segment ilx1 has the same magnimde as 13 1 and is also directed oul of the pagt.~. Similarly. all other such segments of the current loop contribute to the field. The net effect is a magnetic field for Ihe current loop
APPLYING PHYSICS 19.4
FIGURE 19.29 :\11 sep;111eflls of" Iht" ("[liTem loop produce a magnet ic fidd Oil Ihe n'llu.:r ofllll.' loop, directed 011/ of t!lp /HlKI'·
TWISTED WIRES
In electrical circuits it is often the case that insulated wires carrying' currents in opposite directions arc Iwisted together. """hat is the advantage of doing this? Explanation If the h'ires are not twisted together, the combination of the two wires forms a currenl
+
647
Magnetic Fields of Current Loops and Solenoids
loop. which produces a relatively strong magnetic field. This magnetic Iielel generated by 1he loop could be strong enough to affect acljacent circuits or com· pOllents. \'Vhen the wires are twisted together, t.heir magnetic fields tend to cancel.
FIGURE 19.30 (a) ~I;lgn("ljl- lidd HUl'S for a rUITell1 loop. Note thaL til(' lines resemble Ihuse of a bar mag-llt.:l. (b) The magnelic f-ield of a bar magnt.:l is ~illlilar lO thaI ofa current loop. (c) Field lilll'~ ofa CIlITCII( loop. displayt.:d bv iron nlings.
I
(a)
(h)
(e)
The magniwde of the magnetic field at the center of a circular loop carrying current! as in Figure 19.30a is gi\'cn by flo! B=-
2R
I /
This equation mllst be deri\·ed with calculus. It can be shown, however, to be rca· son(lhle by calculaling the field at1.he Ct:nter of four long wires, each carrying currelll / and forming a square, as in Figure 19.31. with a circle of radius R inscribed \v"ithin it. Intuitively, this arrangt:ment should give a magnetic field at the cenler thai is similar ill magnitude to the field produced by the circular loop. The currenl in the circular wire is closer 1.0 the center. so that wire would have a magnetic field somewhat stronger than just the four legs of the rectangle, but the lengths or the siraight wires beyond the reclangle compensate for it. Each \·vire contributes Ihe same magnetic field at tile exaCl center, so the total field is given by
This result is njJjJl"oximalply the same as the fidel produced by the circular loop 01" current.
I~
t '\
-
._E __
/
I
FIGURE 19.31 The field ora (irloop carr}'illj.{ Cllnl'lll 1 Coil) he ;lpproxilll,ued by Ihe fidel due 10 fUlir slraiglll wln.'s, padl earn lng C"llrrenL I. CiliaI'
648
Chapter 19
Magnetism
\"'hen the coil has N loops, each carrying current 1, the magnetic field at the cenler is given by 1
fJ.o B= N -
[19.15]
2R
Magnetic Field of a Solenoid Ha long, straight ,,,,,ire is bent into a coil of several closely spaced loops, the resLlll~ ing device is a solenoid, often called an electromagnet. This device is important in many applications because it acts as a magnet only when it carries a current. The magnetic field inside a solenoid increases with the current and is proportional to the number of coils per unit length. Figure 19.32 shows the magnetic field lines of a loosely wound solenoid of length and total number or lUrns N. ~otice that the field lines inside the solenoid are nearly parallel. uniformly spaced, and close together. As a result, the field inside the solenoid is strong and approximately uniform. The exterior field at the sides of the solenoid is nonuniforrn, much weaker than the interior field, and opposite in diredion to the field inside the solenoid. If the turns are closely spaced, the field lines are as shown in Figure ]9.333, entering at one end of the solenoid and emerging at the other. One end of the solenoid acts as a north pole and the other end acts as a south pole. If the length of the solenoid is much greater than its radius, t.he lines that leave the north end of the solenoid spread our over a wide region before returning to enter the 50Ulh end. The more \..·iclel)' separated t.he field lines are, the weaker rhe Field. This is in contrast to a much stronger field inside the solenoid, where the lines are close together. Also. the Field inside the solenoid has 3 constant magnitude at all points far from its cnds. As will be sho\,.'n subsequently, these considerations allo\'.' the application of Ampere's law to the solenoid, giving a result of
e
FIGURE 19.32 The magnetic field lines for a loosely woulld solenoid.
The magnetic field inside
a solenoid
~
[19.16]
B = fJ.onI
for the field inside rhe solenoid, ",·here 17 = Nle is the number of turns per unit length of the solenoid. So-callcd steering magnets placed along the neck of the picture tube in a television set, as in Figure 19.34, are used 1.0 make the electron beam move to the desired locations on the screen, tracing out the images. The rate at \vhich the electron beam sweeps over the screen is so f~\st that to the eye it looks like a picture rather than a sequence of dots.
Electromagnets
TV
Elecu.".o;":.c::};il:;(,1 gun ::
wbe
~_~--+_Path of
beam
<--'-Screen
(bl FIGURE 19.33 (3) ~Iagnctic ficlcllines far a tightlywaulld solenoid oflinitc length carrying a steady currellt. The fi.eld inside lhe solenoid is Ilearl)" uniform and stl"Ollg. Note that the field lines resembk lhose ofa bar magnet. so the solenoid efrectively has north and south poles. (h) The magnerir field pallcrn of" bar mag-net, displ;l~·ed by small iron filings Oil a sheet of paper.
FIGURE 19.34 Electl"Olllagnets arc used to deflect elect rons to desired positions on lhe screen ora te!c\·isioll tube.
19.9
EXAMPLE 19.9 Gool
The Ma netic Field Insid
Magnetic Fields of Current Loops and Solenoids
649
Solenoid
Calculate the magnetic field of a solenoid from given data and the momentum of a charged panicle in this
field. Problem A certain solenoid consists of 100 llIrns of wire and has a length of 10.0 cm. (a) Find the magnitude of the magnetic field inside the solenoid when it carries a current of 0.500 A. (b) \Vhat is the 1l"1Omentum ofa proton orbiting inside the solenoid in a circle with a radius of 0.020 Ill? The axis of the solenoid is perpendicular to the plane of the orbit. (c) Approximately how much wire would be needed to build this solenoid? Assume the solenoid's radius is 5.00 em. Strategy In part (a) calculate the number of LUrns per meter and substitute that and given information into Equation 19.16, getting the magnitude of the magnetic field. Pan (b) is an application ofXewton's second law. . . . . . . . . ...... -
~
Solution (a) Find the magnitude of the magnetic field inside the solenoid when it carries a current of 0.500 A. 100
N
tlInlS
Calculate the number of turns per unit length:
n
= - = -,----,-=-- = t 0.100 III
Substitute 11. and Jinto Equation 19.16 to find the rnagnitude of the magnetic field:
B
~
l.00 X 10' turns/m
MOll1
= (4". X 10-7 T· m/A)(l.OO X 10' lurns/m)(0.500 A) = 6.28 X 10- 4 T
(b) Find the momentum ofa proton orbiting in a circle of radius 0.020 m near the center of the solenoid. \,\Trite 1\'ewton's second law for the prOton:
mel = F= quE
Substitute the centripetal acceleration a = v'l/'r:
v-" m,. = ovB ,
Cancel olle [acwr of v on both sides and multiply by J~ getting the momenmm mv:
mv ~ ,.qB = (0.020 m)(1.60 X 10- 19 C)(6.28 X 10-" T) /' =
/ltV
=
2.01 X 10- 24 kg'm/s
(c) Approximately hm,,- much ''''ire would be needed to build this solenoid? Multiply the number of turns by the circumference
DrOne loop:
Length ofw·ire = (number ofturns)(27ir)
= (1.00
X 10 2 turns)(2".· 0.0500 m) = 31.4 m
.
Irrrr... • • • •
~
Remarks An electron in part (b) would have the same momentum as the proton. but a much higher speed. It would also orbit in the opposite direction. The length of wire in part (c) is only an estimate because the wire has a certain thickness, slightly increasing the size of each loop. In addition. the wire loops aren't perfect circles because they wind slowly lip along the solenoid. QUESTION 19.9 \,Vhat would happen to the orbiting proton if the solenoid were oriented vertical Iv? EXERCISE 19.9 Suppose yOll have a 32.0-m length of copper wire. 1r the wire is wrapped into a solenoid 0.240 III long and having a radius of 0.04-0 0 m, how strong is the resulting magnetic field in its center when the current is 12.0 A?
Answer
8.00 X 10-' T
650
Chapter 19
Magnetism
Ampere's Law Applied to a Solenoid
• • • • • • • • • • •
""Ve can use Ampere's 1m\' to obtain the expression for the magnetic field inside a solenoid carryi ng a current f. A cros2. section taken along the length of part of our solenoid is show~ in Figure 19.35. B inside the solenoid is uniform and parallel Lo the axis, and B outside is approximately zero. Consider a rectangular path of length Land 'l,.vidth W, as shown in the figure. \Ve can apply Ampere's law to this path by evaluating the sum of Ell Ll-t over sach side of the rectangle. The comribution along side 3 is clearly zero because B = 0 in this region. The contributions . from Sides 2 and 4 are both zero because B is perpendicular to 6.t along these paths. Side 1 of length L gives a contribution BL to the sum because Ii is uniform along this path and parallel to Lle. Therefore, the sum over the closed rectangular path has the value ~
x x
FIGURE 19.35 A cross-sectional "iell' ofa tightlrwollnd solenoid. If the solenoid is long rebli\"c to ils .-adius. we call aSSUllle the lTI
The right side of Ampere's ]a,.... involves the total current that passes through the area bounded by the path chosen. In this case, the total current through the rectangular path equals the current through each turn of the solenoid, multiplied by the number of turns. If lv' is the number of turns in the length L, then the toml current through the rectangular path equals Nf. Ampere's law applied to this path therefore gives
2:
BII
tie
~ BL = /huNT
or N B = 1"0 L f ~ /hun!
where n
19.10
FIGURE 19.36 Classical model ofa spinning denron.
=
Ivl L is the number of turns per unit length.
MAGNETIC DOMAINS
The rnagnetic field produced by a current in a coil of wire gives us a hint as to what might cause certain materials to exhibit strong magnetic properties. A single coil like that in Figure 19.30a has a north pole and a south pole, but if that is true for a coil of wire, it should also be true for any current confined to a circular path. In particu lar, an individu.al atom should (lct as a magnet because of the motion of the eleclmns abou.t lhe nucleus. Each electron, with its charge of 1.6 X 10- 19 C, circles the atom once in about "10- 16 s. fr we divide the electric charge by this time interval, we see that the orbiting electron is equivalent to a current of1.6 X 10-3 A. Such a current produces a magnetic field on the order of 20 T at the center of the circular path. From this we see that a very strong magnetic field would be produced if several of these atomic magnets could be aligned inside a material. This doesn't occur, hO\,,·ever, because the simple model we have described is not the complete story. A thorough analysis of atomic structure shows that the magnetic field produced by one electron in an atOI11 is often canceled by an oppositely revolving electron in the same atom. The net result is thal the magnetic effect produced by the electrons orbiting the nucleus is either zero or very small for most materials. The magnetic properties of many materials can be explained by the fact that an electron not only circles in an orbit, but. also spins on its axis like a top, ,\lith spin magnetic moment as shown (Fig. ]9.36). (This classical description should not be taken too literally. The property of electron sjlin can be understood only in the context of quantum mechanics, which '\Ie will not discuss here.) The spinning electron represents a charge in motion that produces a magnetic field. The fleld due to the spinning is generally stronger than the rield due to the orbital motion. In atoms containing many electrons, the electrons usually pair up with their spins oppo-
19.10
I
'-
--
\-
--
(a)
(b)
-- '" , ,,\
"
/
/
"-
..-
\
/
/
.....
I
_8 (c)
651
FIGURE 19.37 (a) Random orientation of domains in an unmagnetilcd sub:.tance. (b) When ~Il external magnetic field Ii is ~pplied. the domains lend to align with the magnetic field. (c) As the field is made e"en stronger. the domains not aligned whh the external field become "er) small.
I
/
/
Magnetic Domains
-
•8
site each other so that their fields cancel each other. That is why most substances are not magnets. In certain strongly magnetic materials, such as iron, cobalt, and nickel, however, the magnetic fields produced by the electron spins don't cancel completely. Such materials are said to be ferromagnetic. Tn ferromagnetic materials strong coupling occurs between neighboring atoms. forming large groups of atoms with spins thal are aligned. Called domains, the sizes of these groups typically range from about 10- 4 (Ill to 0.1 em. In an unmagnetized substance the " domains are randomly oriented, as shown in Figure 19.37a. ',Vhen an external field i is applied, as in Figure }9.37b, the magnetic field of each domain tends to come ~ nearer to alignment with the external field, resulting in magnetization. ~ In what are called hard magnetic materials, domains remain aligned e\'en after the external field is removed; the result is a permanent magnet. In soft magnetic materials, such as iron, once the external field is removed, thermal agitation proFIGURE 19.38 The pennanent duces motion of the domains and the material quickly returns to an unmagneIll:lgnet (redl tempOl'ari1r maguetizes some paperclips. which then cling to tized state. each 01 her through magnetic forces. The aligl1lTlent of domains explains why the strength of an electromagnet is increased dramatically by the insertion of an iron core into the magnet's center. The magnetic field produced by the currelll in the loops causes the domains to align, thus producing a large net external field, The use of iron as a core is also advantageous because it is a soft magnetic material that loses its magnetism almost /"TIP 19.4 The Electron '\ instantaneously after the current in the coils is turned off. Spins, but Doesn't! The formation of domains in ferromagnetic substances also explains \\Thy such E.ven though we use the word substances are attracted to permanent magnets. The magnetic field of a permaSPill, the electron, unlike a child's lOp, isn't physicall}' spinning in nent magnet realigns domains in a ferromagnetic object so that the object becomes this sense. The electron has an t.emporaril)' magnetized. The object's poles are then attracted to the correspondintdnsic angular momentum ing opposite poles of the permanent. magnet. The object can similarly attract other that causes it to act as iJil were spinning, bUI the concept of spin ferrornagnetic objects, as illustrated in Figure 19.38.
J
,
j
"
angular momentum is actually a rc1alh-islic quanwm effecl.
~---------_/
Types of Magnetic Materials
Magnetic rnaterials can be classified according to how they react to the application of a magnetic field. In ferromagnetic materials the atoms hayc permanent magnetic moments that align readily with an externally applied magnetic field. Examples of ferromagnetic mat.erials are iron, cobalt, and nickel. Such substances c; can retain some of their magnetization even after the applied magnetic field is ~ i remm·ed §"j Para~agnetic materials also ha,·e magnetic moments that tend to align with an ~ externally applied magnetic field, but the response is extremely weak compared ~ with that of ferromagnetic materials. Examples of paramagnetic substances are ~ ,~ aluminurn, calcium, and platinum. A ferromagnetic material can become para- ~ ~ magnetic when warmed to a certain critical temperature, the Curie tenlperature, ~ ~ that depends on the material. FIGURE 19.39 Diamagnetism. A · . . I ) frog is le\ itated in a 16-T magnetic rn d lamagnetlc matena s, an externa ) I)· a))) ied magnetic field induces a \'en.· field at the NUmcgen High Field Magweak magnetization that is opposite the applied field. Ordinarily diamagnetism lIet Laboratoll' in the Netherlands. isn't observed because paramagnetic and ferromagnetic effects are far stronger. The le\'itatioll force is excned on the _· 19 ,39 ,10We\'er, l I : diamagnetic waLer molecules in the I n 1'lgure aver)' e ligh magnetic neld exerts a levitating force on the rrog·s body. The frog suffered JlO ill diamagner.ic warer molecules in a rrog. effects frOIll Ihe levitation experience,
j
i
652
Chapter 19
19.3
Magnetic Fields
Magnetism
UIV r~-
The magnetic force that aet~on a charge q moving with velocity in a magnetic f-icld B has magnitude
v
F= qvRsine
[19.tJ
v
where e is the angle between and B. To find the direction of this force, use right-hand rule number 1: point the fingers of your open right hand in t'1e direction of and then curl them in the direction of B. Your thumb then points in the direction of the magnetic force F. If the charge is negative rather than positive, the force is directed opposite the force given by the right-hand rule. The 51 unitofthe magnetic field is the tcsla (T), or weber per square meter (VlTb/m 2 ). An additional commonly used unit [or the magnetic field is the gauss (G); 1 T = 10'] G.
v
19.4 Magnetic Farce an a Current-Carrying Conductor
e
If a straight conductor of length carries current I, the magnetic force on that conduct,2r when it is placed in a uniform external magnetic field B is F~
BIe sin ()
[19.6]
where fJ is the angle between the direction of the current and the direction of the magnetic field. Right-hand rule number 1 also gives the direction of the magnetic force on dle conductor. In this case, however, you must poim your fingers in the direction of the current rather than in the direction ofv. 19.5 Torque on a Current Loop and Electric Motors The torqu~T on a current-carrying loop of wire in a magnetic field B has magnitude T
= BfA sin ()
[19.8]
where lis the current in the loop and J1 is its cross-sectional area. The magnimde of the magnetic moment of a currentcarrying coil is defined by}J- = JAN, where N is the number of loops. The magnetic moment is considered a vector, fl, that is perpendicular to the plane of the loop. The angle ~ ~ between Band J.t is e.
19.6
qB
where m. is the mass of the particle and q is its charge.
19.7
Magnetic Field of a Lang, Straight Wire
and Ampere's Law The magnetic field at distance -r from a long, straight wire carrying current [has the magnitude
[19.11] where J.Lo = 4rr X lO-7T'mjA is the permeability of free space. The magnetic field lines around a long, straight wire are circles concentric with the wire. Ampere's law can be used to find the magnetic field around certain simple current-carrying conductors. It can be written [19.13] ~
where 8 n is the component of B tangent to a smalt current element of length f1e t.hat is pan of" a closed path and! is the total current. that penetrates the closed path.
19.8 Magnetic Farce Between Twa Parallel Conductors The force per unit length on each oflwo parallel wires sep,Hated by the distance d and carrying currents 1] and 1'1. has the magnitude [19.14] The forces are attractive if the currents are in the same direction and repulsive if they are in opposj(C directions.
19.9
Magnetic Field of Current Loops and Solenoids
The magnetic field at the cenLer of a coi I of N circular loops o/"radius R, each carrying current I, is given by fL0 1 N-
Motion of a Charged Particle in a Magnetic Field
If a charged particle moves in a uniform magnetic field so that its initial velocity is perpendicular to the field, it will move in a circular path in a plane perpendicular to the magnetic field. The radius 'I" of the circular path can be found from Newton's second law and centripetal acceleration, and is given by
[19.10]
2R
[19.15]
The magnetic field inside a solenoid has the magnitude [19.16] where n length.
=
Nje is the number of turns of wire per unit
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
Multiple-Choice Questions
653
MULTIPLE-CHOICE QUESTIONS I. An electron moves across Earth's equator at a speed of
2.5 X 10(; m/s and in a direction 35 0 N of E. At this pOilll, Eanll's magnetic held has a din.:uiuIJ tlue nOI til, is parallel to the surface, and has a magnitude of 0.\0 X 10--1 T. What is the force acting on the e1ec~ tron due to its interaction with Earth's magnetic field? (a) 5.\ X IO-IH N due west (b) 4.0 X 10- 18 f\' toward Earth's surface (c) 2.4 X 10- 18 N away from Eanh's surface (eI) 3.:\ X to- 18 N [O\\lard Eanh's surface (e) 4.0 X 10- 18 N away from Earth's surface 2. A wire of length 0.50 m carries a current of 0.10 A in the positive x~direClion, parallel to the ground. 11' the wire has a weight of 1.0 X 10-2 N, what is the minimum magnilUclc magnetic field that exerl.s a magnetic force on the wire equal LO the wire's ' . . eight? (a) 0.20 T (b) 0.30 T (e) 0.40 T (d) 0.50 T (e) 0.60 T 3. A rectangular coil ofwirc consisting or ten loops, each with length 0.20 m and width 0.30 Ill, lies in the xyplane. If the coil carries a current of 2.0 A, what is the LOrque exerted by a magnctic field of magnitude 0.010 T directed at an angle of 30.0'" with respect to the pOsili,-e z-axis? (a) 1.2 X 10 2 N'1ll (b) 2.4 X 10-2 N· III (e) 6.0 X 10-' N'1ll (d) 4.0 X 10-' N'1ll (e) 3.0 X 10-' N'1ll 4. A proton enters a constant magnetic held of" magnitude 0.050 T and traverses a semicircle of radius 1.0 IlltTI before leaving the field. What is the proton's speed? (a) 1.6 X 10' Illis (b) 2.3 X 10' rnls (e) 2.8 X 10' rn/s (d) 1.8 X 10 2 mls (e) 4.8 X 10' m/s
a force that is perpendicular to the direction of motion. (e) It doesn't change the mag-niwde of the momentum 01 the panicle. 10. A proton moving horizontally enters a region where a un iform magnetic field is eli n:cteel perpendicular to the proton's velocity, as shown in Figure MCQI9.1O. After the proton enters the field, c10es it (a) deflect down' . . ard, with its speed remaining const.ant; (b) conI inue to move in the horizontal direction with constant velocity; (c) move in a circular orbit and become trapped by the ficld; (d) deriect upward, moving in a semicircular path with constant speed, and exit the field moving LO the left; or (e) deflect out of the plane of the paper~
,
v
7. What is the magnitude of the magner.ic field at the core of a 120-turn solenoid of length 0.50 m carrying a eurrem or 2.0 A? (a) 2.4 X 10-4 T (b) 4.8 X 10" T (e) 1.2 X ]()-. T (d) 3.6 X 10-" T (e) 6.0 X 10-" T 8. A uniform magnetic held cannot exert a magnetic force on a particle in \\'hich of the following circumstances? (There may be more than one correct statcmenl.) (a) The particle is charged. (b) The panicle moves perpendicular to lhe field. (e) The panicle moves parallel to the held. (d) The magnitude of the magnetic ficld change'~ with time. (e) The panicle is at resl.
x
,
FIGURE MCQ19.10
I I. Two long, parallel wires each carry the same current as in Figure MCQ19.11. Is the total magnetic held at lhe point Pmidway between the wires (a) zero, (b) directed into the page, (c) directed OUI of the page, (d) directed to the left, or (e) directed to the right?
5. A long wire carries a current of I A. Find the magnit.ude or the magnetic field 2 m away from tlte wire. (a) 1 X 10' T (b) I X 10-' T (e) 1 X 10-· T (d) 1 X 10- 7 T (e) I X ]()-IO T 6. Estimate the magnilUde of the magnetic force per unit length between a pair of parallel wires separated by 2 rn if they each carry a current of 3 A. (a) 1 X 10- 1 N/m (b) 1 X ]()4 N/m (e) I X IIJ ., N/m (d) 1 X 10- 0 N/m (e) 1 X 10-' 1\/111
x
-/
.p -/
FIGURE MCQ19.11
12. Two long, straight wires cross each other at right angles, and each carries the same current as in Figure .'vICQI9.12. vVhich of the following statements are true regarding the total magnetic field at. the various points due to the two wires? (There lllay be more than one correct statemenL) (a) The lleld is strongest at points Band D. (b) The field is st.rongest at points A and C. (c) The field is Ollt of the page at point Band inLO the page at point D. (d) The field is out of the page c:lt point C and nUl of the page at point D. (e) The held has the same magnitude at all four points. B
•
A
•
-/
9. A charged panicle is t.raveling through a uniform mag-netic ficld. Which or the following statements are true of the magnetic fielclr (There may be more than nne correct statement.) (a) It exerts a force on the panicle that is parallel 10 the held. (b) It exel"lS a force on the panicle along the direction of its motion. (c) It increases the kinet.ic energy 01" the panicle. (d) It exerts
c
•
D
•
FIGURE MCQ19.12
13. Two long, parallel wi res carry currents of 20 A and 10 A in opposite directions, as in Figure MCQI9.13. Which
654
Chapter 19
Magnetism
of the following statements must be true? (There may be morc than one correct statement.) (el) In region I the magnetic field is into the page and is never zero. (b) rn region 11 the field is into the page and can be zero. (c) In region III it is possible for the field to be zero. (d) In region I the magnetic field is Ollt of the page and is never zero. (e) There arc no points where the field is zero. I
20A
n -lOA
III
FIGURE MCQ19.13
magnitude ClllTent, and the uniform magnetic field points in the positive x-direction. Rank the coils by the magnitude of the torque exerted on them by the field, from largest 10 smallest. (0) A, B, C (b) A, C, B (e) B, A, C (eI) B, C, A (0) C, A, B
15. A long, straight wire carries a current J as in Figure MCQ19.15. \o\'hich of the following statements are true regarding the magnetic field due to the wire? (There may be more I han one correct st
14. A magnetic field exens a torque on each of the current· carrying single loops of wire 5110""11 in Figure M.CQ19.14. The loops lie in the .\)'-plane, each carrying the same -{
)' (m)
'I
3
2
B
FIGURE MCQ19.15
•
DCD ~ 2
3
4
.;
:o:(m)
6
FIGURE MCQ19.14
16. Solenoid A has length L and I'" turns, solenoid B has length'lL and lv' turns,
CONCEPTUAL QUESTIONS 1. In your home television set, a beam or electrons moves from the back of the picture tube to the screen, where it strikes a nuorescent dO[ that glm.,,·s with a particular color when hit. Earth's magnetic field at the location of the television set is horizontal and toward the north. in which direetion(s) should the set be OI-iented so t.hat the beam undergoes the largest deflection? 2. Can a constant magnetic AcId set a proton at rest into motion? Explain your answer. 3. How can the motion of a charged pacticle be used to distin~uish between a magnetic field and an electric field in a certain region?
FIGURE CQ19.5
4. 'Which way \\'ould a compass point if you were at Earth's nort h magnetic pole?
6. A magnet attracts a piece of iron. Thc iron can then attract another picce of iron. On the basis of domain alignment, explain whal happens in each piece of iron.
5. 'Why does the picture on a television screen become distOrted when a magnel is brought ne..l r the screen as in Figure CQI9.5? Cautio//.: You should not do this experiment at home on a color television set because it Illay permanemly affect the picture qualit.y.
7. A Hindu ruler once suggestcd that he be entombed in a magnetic coffin with the polarity arranged so that he could be forever suspended between heaven and Earth. Is such magnClic levitation possible? Discuss.
Problems
nets from this observation? (d) If the upper magnet were inverted, what do you suppose would happen?
9. Suppose you Illove along a wire at the same speed as the drift speed of the electrolls in the wire. Do you now measure a magnetic field of zero?
*
~ c
10. Is the magnetic field created by a current loop uniform? Explain.
t
11. Can you usc a compass to detect the currents in wires in the walls near light switches in your home?
~
!9 "'-
12. Why do chargcd particles from outer space, calleel cosmic rays, strike Earth more frequently at the poles than at the equator? 13. Parallel wires exert magnetic forces on each other. "\'hat about perpendicular wires? Imaginc t,,,,,o wires oriented perpendicular to each other and almost touching. Each wire carries a current. Is there a force between lhe wires? 14. Ilow can a currenlloop be used to determine the presence of a magnetic field ina given region of space?
~
8. Will a nail be attracted to eithcl- pole of a magnet? Explain what is happening inside the nail when it. is placed ncar the magnet.
FIGURE CQ19.16
17. The electron beam at the top of the photograph in Figure CQ19.l7 is projected to the right. The beam deflects downward in the presence of a magnetic field produced by a pair or current-carrying coils. (a) ',Vhat is the direction ofthe magnetic field? (b) What would happen to the beam if the magnetic field \..·ere reversed in direction?
15. A hanging Slinky@; toy is aU3ched to a pmverful battery and a s,..· itch. When the s\vitch is closed so that the toy now carrics current, does the Slinky compress or expand?
16. Figure CQ19.16 shows four permancnt magnets, each having a hole through its center. :'Jotice that the blue and yellow magnets arc levitated above the red ones. (a) How does this levitation occur? (b) ""Vhat purpose do the rods serve? (c) "Vhat can you S3Y about the poles of the mag-
FIGURE CQ19.17
'. PROBLEMS
v
The Problems for this ch,lptcr may be assigned online al 'VebAssign. I, '2,:1 = straightforward, inll'lmediall", challenging, I!II = denOI(:s guided problem
WebAssign
m;J
denotes en hanced content prohlem biomedical application D = denotes full solution available in Stut/nlt SolntioliS Manllfl!!
B,"
1
=
Stud)' Guide
SECTION 19.3 MAGNETIC FIELDS I. A prown is projected into a magnetic field that is di rCCl.ed along the positive x-axis. find the direction of the magnetic force excrtcd on the proton for cach of the following directions of the pl"Oton's velocity: (a) the positivc y-dil-ection; (b) the negative y-dircclion; (c) the positive x-direction. 2, (a) Find the direction orthe force on a proton (a positively charged particle) moving through the magnetic fields in Figu"e P19.2, as shown. (b) Repeat pan (a), assuming the moving panicle is an electron.
(a)
J II
!!! B
(e)
Ie)
-1-' ------v
~
II =
655
IbJ
B
oUl
• • • • • • • • vII • • • • • • • • (d)
(t)
FIGURE P19.2 (Problems 2 and 12) For Problem velocit)' n.:ctor wil h a current i II that direCl ion.
12, replace Ihe
656
Chapter 19
Magnetism
[]JFind the direction of the magnetic field acting on the positively charged particle moving in the various siruaLions shown in Figure P19.3 if the direction of the magnetic force acting on it is as indicated.
lr
F ....f----@
9. An electron moving in a direct.ion perpendicular to a uniform magnetic field at a speed ofl.5 X 10 7 mls undergoes an acceleration of 4.0 X 10 16 m/s 2 to the riglu (the positive x-direction) when its velocit), i~ upward (the positive J-direction). Determine the magnitude and direction of the field. 10.
v(in)
v
®'V(out)
(a)
(b)
(e)
FIGURE P19.3 (Problems 3 and 15) For Problem 15, replace lhe velocity \"eetor wilh a currenl in thal dirCCl ion.
4. Determine the initial direction of the deflection of charged particles as they enter the magnetic fields, as shown ill Figure PI9A. (a)
Bill
1-------
+
)(
X
,
(c)
B"p
(b) 1
x x I I xxx >:: I x x x x : : x x x X I I
Tnl~
,1
_
,
Bright
1------- 1 I ,
I
I 1
1 1--: I
r
1 I
.-!
+
FIGURE P19.4
5.
E
A laboratory electromagnet produces a magnetic field of magnitude 1.50 T. A prawn moves through this field willl a speed of 6.00 X 10 6 rnls. (a) Find the magnitude of the maximum magnetic force th;.u could be exerted on the proton. (b) "Vhat is the magnitude of the maximum acceleration ohhe proton? (c) \,Vould the field exert the same magnetic force on an electron moving through the field with the same speed? ',",ould the electron undergo the same acceleration? Explain.
6. The magnet.ic field of Earth at a certain location is directed vertically downward and has a magnitude of 50.0 ,u.T. A prot.on is moving horizontally wward the west in this field with a speed of 6.20 X 10 6 m/s. "\/hat arc the direction ancl magnitude of the magnetic focce the field exens on the prown? L1J'\11at velocity would a proton need to circle Earth I 000 km above the magnetic cquawr, where Earth's magnetic field is direCled horizolllally nonh and has a luagnitmle of 4.00 X 10-8 T? 8. An electron is accelerated through 2400 V from rest and [hen enters a region where there is a uniform 1.70-T magnetic field. \Vhat are (a) the maximum and (b) the minimum magnitudes of the l1l
II Sodium
ions (Na+) move at 0.851 mls through a bloodstream in the arm of a person standing near a large magnet. The magnetic field has a strengt.h of 0.254 T and makes an angle of 51.0 0 , .... ith the motion of the sodium ions. The arm contains 100 emil of blood with a concentration of 3.00 X 10 20 Na + ions per cubic centimeter. If no other ions were presem in the arm, what would be the magnetic focce 011 the arm?
11. A.t the equator, near the surface of Earth, the magnetic field is approximately 50.0 p.,T northward and the electric field is about 100 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron with an inst
SECTION 19.4 MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR 12. Tn Figure P19.2 assume in each case the velocity vector shown is replaced with a wire carrying a current in the direction of t.he velocity vector. For each case, find the direction of the magnetic force acting on the wire. 13. A current T= 1.5 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length of 0.12 :\lIm acts on the conductor in the negative )'~direc[ion. Calculate the magnitude and direction of the magnetic field in the region through which the current passes. 14.
mil
A straight wire carrying a 3.0-A current is placed in a uniform magnetic field of magnitude 0.28 T directed perpendicular [Q the ,,,,,·ire. (a) Find the magnitude of the magnetic force on a section of the wire having a length of 14 cm. (b) Explain why you can't determine the direction of the magnetic force from the information given in the problem.
15. In Figure Pl9.3 assume in each case the velocicy vector shown is replaced with a wire carrying a current in the direct.ion of the velocity vector. For each case, find the direction of the magnetic field that will produce the magnetic force shown. u::h]A wire having a mass per unit length of 0.500 glcm carries a 2.00-A current horizontally to the somh. \'\fhat are the direction and magnitude of lhe minimum magnetic field needed to lift this wire venical1y upward? 17. A wire carries a cunent of JO.O A in a direction that makes an angle of 30.0 with the direction of a magnetic field of strength 0.300 T. Find the magnetic force on a 5.00-111 length of the wire. 0
18. At a certain location, Earth has a magnelic field of 0.60 X 10- 4 T, point.ing 75 0 below the horizontal in a north-south plane. A 10.O-m-long straight wire carries a J f)-A current. (a) If the currem is directed horizomall)'
Problems
657
toward the east, what are the magnitude and direction of the magnetic force on the wire? (b) '''''hat are the magnitude and direction of the force if the current is directed vertically upward? !iliA wil'e with a mass of 1.00 g/cm is placed on a horizontal surface with a coefficient of friction of 0.200. The wit'e carries a current of 1.50 A eastward and moves horizontally to the north. 'What are the magnitude and the direction of the smflllest venicalmagnetic field that enables the wire La mm'e in this fashion? 20. A conductor suspended by t,vo flexible wires as shown in Figure 1'19.20 has a mass per unit length of 0.040 0 kg/tn. 'What current must exist in the conducLar for the tension in the supporting wires to be zero when the magnetic field is 3.60 T into the page? What is the required direction ror the current? 21.
----1' x
x
x
x
x
x
X
XBi"
X
X
X
X
x
x
x
x FIGURE P19.23
24. A horizontal power line of length 58 m calTies a current of 2.2 kA as shown in Figure P19.24. Eanh's magnetic field at this location has a magnitude of 5.0 X 10-5 T and makes an angle of 65" wilh the power line. Find the magnitude and direction of the magnetic force on the power line.
FIGURE P19.20
am
Consider the system pictured in Figure P19.21. A l5-cm length of conductor of mass 15 g, free La move vertically, is placed between two thin, vertical conductors, and a uniform magnetic field acts perpendicular (0 [he page. When a 5.0-A current is directed as shown in the figure, the horizontal wire moves upward at constant velocit), in the presence of gravity. (a) \Vhat forces act on the horizontal \.. . ire, and under what condition is the wire able to move upward at consta11l velocity? (b) Find the magnitude and direction ofthe minimum magnetic field required to move the wire at constant speed. (c) 'What haplJens if the magnetic field exceeds this minimum value? (The wire slides with om friction on the two venical conductors.)
FIGURE P19.24
SECTION 19.5 TORQUE ON A CURRENT LOOP AND ELECTRIC MOTORS 25. A 50-turn coil of radius 5.0 cm rotates in a uniform magnetic field having a magnitude of 0.50 T. lrthe coil carries a current of25 rnA, find the magnitude oflhe maximum torq lie exerted 011 the coil. 26. A current of l7.0 mA is maintained in a single circular loop with a circumference of2.00 m. A magnetic field of 0.800 T is directed parallel to the plane of the loop. \"'hal is the magnitude of the l.Orque exerted by l.he magnetic field 011 the loop?
-15clll-
22. A wire 2.80 III in length carries a current of 5.00 A in a region where a uniform magnetic field has a magnitude of 0.390 T. Calculate t.he magnitude of the magnetic force on the wire, assuming the angle bet.ween the magnetic field and the current is (a) 60.00, (b) 90.0°, and (c) 12OC.
127.IAn eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor 40.0 em axis of 30.0 em (Fig. P19.27). ~ The coil lies in the plane of the page and has a 6.00-A r-30.0 current flow·ing clockwise around it. If the coil is in a FIGURE P19.27 uniform magnetic field of 2,00 X 10- 4 T directed toward the left of the page, what is the magnitude of the torque on the coil? (Hint: The area of an eHipse is A = 7Tav, where a and bare, respecti\'ely, the sel11im~or and semiminor axes of the ellipse.)
23. Tn Figure P19.23 the cube is 40.0 cm on each edge. Four straight segments of wire-ab, lx, cd, and da-form a closed loop that carries a current r = 5.00 A in the direction shown. A uniform magnel.ic fielcl of magnitude B = 0.0200 T is in the positive y-dil·ection. Determine [)le magnitude and direction of the magnetic force on each segment.
28. A rectangular loop consists of 100 closely wrapped turns and has dimensions 0.40 m by 0.30 m. The loop is hinged along the y-axis, and the plane of the coil makes an angle of 30.00 with the x-axis (Fig. P19.28, page 658). What is the magnit.ude of the torque exened 011 the loop by a uni+ f"(mll magnetic field of 0.80 T directed along the x-axis \..,,·hen the current in the windings has a value 01" 1.2 A in
5.0A
f
~
5.0 A
FIGURE P19.21
T
""=cm"1
1
658
Chapter 19
Magnetism
the direction shown? What is the expected dir"cction of rotation of the loop? y
T~2(JA o.~oo.J1 1 ., ,
SECTION 19.6 MOTION OF A CHARGED PARTICLE IN A MAGNETIC FJELD
x
~
0.300 m
::>0.0
""I FIGURE P19.28
29. A 20D-turn rectangular coil having dimensions of 3.0 em by 5.0 em is ph-lCed in a unitorm magnetic field of magnitude 0.90 T. (a) find the currem in the coil if the maximum torque exened on it by Ihe magnetic field is 0.15 N· m. (b) Find the magnitude of the torque on the coil when the magnetic field makes an angle of 25° with the normalLo the plane of the coil.
30. A copper wire is 8.00 In long and has a cross-sectional area of l.OO X 10--1 m~. The wire forms a one-turn loop in the shape of square and is then connected to a battery that applies a potential difference of 0.1 00 V. If the loop is placed in a uniform magnetic field of magniwde DADO T, what is the maximum torque that can act on it? The resistivity of copper is 1.70 x 10- 8 {1. m. 31. A long piece of wire with a mass of 0.100 kg and a total length of 4.00 m is used to make a square coil with a side of 0.100 m. The coil is hinged along a horizontal side, carries a 3.40-A current, and is placed in a venical magnetic field with a magnitude of 0.010 0 T. (a) Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium. (b) Find the torque acting on the coil due to the magnetic force at equilibrium. 32.
EI
x-axis, will the loop rotate clockwise or counterclockwise? (f) Compute the magnitudc of the magnetic moment of the loop. (g) V\That is the angle between the magnetic moment vector and the magnetic field? (h) Compute the LOI'que on the loop using the values found for the magnetic moment and magnetic field.
,
A rectangular loop has dimensions 0.500 m bv 0.300 m. The loop is hinged along the x-axis and lies in the x)'-plane (Fig. Pl9.32). A uniform magnetic field of y 1.50 T is directed at an angle x of 40.0 0 with respect to the FIGURE P19.32 positive J-axis and lies parallel everywhere to the yz-plane. The loop carries a currelll of 0.900 A ill the direction shown. (Ignore gravitation.) (a) In what direction is magnetic force exerted on wire segmem ab? \lv'hat is the direnion of thc magnctic torquc associattd with this force, as compliled with respect to the x-axis? (b) What is the direction ofthc magnetic force exerted on segment cd? 'What is thc direction of the magnetic torque associated with this lorce, again computed with respect to the x-axis? (c) Can the forces examined in par"ts (a) and (b) combine to calise the loop to rotate around the x-axis? Call thc)' affect the motion of the loop in any way? Explain. (
33. An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of2.0 mT. If the speed of the electron is 1.5 X 10 7 m.!s, delennine (a) the radius of the circular path and (b) the time it takes to complete one revolution. 34. A proton travels \'liith a speed of 5.02 X Jon mls at an angle of 600 with the direction ofa magnetic field of magnitude 0.180 T in the positive x-direction. What are (a) the magnitude of the magnetic force on the prolan and (b) the prown's acceleration? 35. Figure P19.35a is a diagram of a device called a velocit)' selector, in which panicles of a specific velocity pass through undeflected while those with greater or lesser' velocitics are deflected either upwards or downwards. An electric field is directed perpendiculal' to a magnetic field. producing an electric force and (l magnetic force on dle charged particle that can be equal in magnitude and oppositc in direction (Fig. P19.35b) and hence cancel. Show that particles ,.. . ith a speed or v = fiB will pass through the velocity selector undeflected. Bill X
X
+
SOlllce
X
±
•
X
X
SIll
X
X
x
x x
x x
x x
X
X
X
x
x x
V
x
X
X
.Hl~ l x x
X
•x
+
.. x x
-
+
qvB
------
+q
X X
'IE
(a)
(bl
FIGURE P19.35
36. Consider thc lllass spectrometer sho\'m schematically in Figure 1'19.36. The electric field benveen the plates or Photographic plate
-x -
x
x
x
x
"
x
x
x
X
x\x
x
X
X
X
,x
X
x
x
x
x
:~"5<-- x
x
X
x
p
x
,. ,
Bill X
X
X
E-:rR
X
X
x
x
x
x
x1 x
x
x
x
x
v
x x
Vt:locit>· selector
X
X
f :- . x
x
x
f
I
,
x
X
x
X
x
x x
x
x
x
x
x
x
x
x
x
FIGURE P19.36 (Problcm~ 36 and 41) :\ mass spectrometer. Charged particles are l1rst ~etll through.:.' velocity selector. They then cllter a regioll where a magnclic l1eld BI> (directed inward) callse~ positive ions to Illo\:e in a semicircular path and strike a photographic film at P.
659
Problems
the velocity selector is 950 Vim, and the magnetic fields in both the velocity selector and the denection chamber have magnillldes of 0.930 T. Calculate the radius of the path in the system for a singl), charged ion with mass /IJ = 2.18 X 10-26 kg. (Hinf: Sce Problem 35.)
x
x
x
x
x
X
x
•
137.IA singl), charged positive ion has a maSS of 2.50 X 10-26 kg. After being acceler
•
• •
•
difference of 250 V, the ion emers a magnetic field of 0.500 T, in a direction perpendicular to the field. Calcu· late the radius of the path of the ion in the field.
40.
•
44. 111 each of pans (a), (b), and (c) of Figure P19.44, find the direction of the current in t.he wire that would produce a magnctic field directed a!:i shown.
• BOlli • • •
•
Bin
•
T
+
T
•
...
+
+
+
(e)
FIGURE P19.44
45.
II
Neurons in Ollr bodies carry weak cUlTems that produce detcctable magnetic fields. A technique called magnetoencephalogmpll)'J or MEG, is used to study e1ecu·ical activit}' in the brain llsing lhis concep£. This technique is capable of detecting magncLic fields as \\-'eak as 1.0 X 10- 1'-; T. ~fodel the neuron as a long wire carrying a current and find Lhe currcnt itl11USL carry to produce a field of this magnitude at a distance of 4.0 cm from the neuron.
146.lln 1962 measuremcnts of the magnetic field of a large tornado were made iltthe Geophysic.al Observatory in Tulsa. Oklahoma. If t.he magnilude of the tornado's field was B = 1.50 X 10-8 T pointing north when the tornado was 9.00 km east of the observatOl-}', what current was carried up or down the funnel of the tOI-nado? ~lodel the vortex as a long, straight wire carrying a ClllTcnt.
47.
II A cardiac
pacemaker can be affected b)' a static magnetic field as small as 1. j mT. How close can a pacemaker ,,'earer come to a long. str"ight wire carrying 20 A?
48. The [.\'''-0 wires shown in Figure P19.48 calTY currents of 5.00 A in opposite dil·ections and are separated by 10.0 cm. Find the dircction and magnitude of the net magnetic field (a) at a point midway between tlle wires; (b) at point PI' JO.O cm to the right of the wire on the right; and (c) at poilll P2' 20.0 cm to the left of the wire on the left.
SECTION 19.7 MAGNETIC FIELD OF A LONG, STRAIGHT WIRE AND AMPERE'S LAW 43. A lightning bolt may carry a currem of 1.00 X 10"" A for a short time. Whm is the resulting magnetic field 100 m from the bolt? Suppose the bolt ext.ends far above and below the point of observation.
•
•
J
42. A cosmic-I-a}' proton in interstellar space has an energy of 10.0 MeV and executes a circular orbil having a radius equal to that of Mercury's orbit around the Sun, which is 5.80 X 1010 m. What is the magnetic field in that region of space?
(b)
•
J
41. _ A particle passes through a mass spectrometer as illustrated in Figure P19.36. The electric field between the plates of the ,·elocity selector ha!:i a magnitude of 8250 Vim, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.0931 T. In the deficClion chamber the panicle strikes a photographic plate 39.6 crn I·ernm'ed from its exit point after traveling in a semicircle. (a) ,,'hat is the mass-tocharge ratio of the particle? (b) What is the mass of the panicle if it is doubly ionized? (c) 'Vhat is its identity. assuming it's an element? .
~/
BOllI
•
EiJ
A panicle widl a charge q and kinetic energy K Iravels in a Ilniforl11 magnetic field o[magnituc!e B. lfthe particle llloves in a circular path of radius R find an expre!:ision for its speed in terms of its kinetic energy K charge q, the magniwde of the magnetic field B, and the radius R.
• •
~~
(a)
38. A mass spectrometer is used to examine the isotopes of l1I·anium. Ions in the beam emerge from the velocity ~elector at a speed of 3.00 X 105 mls and entcr a uniform magnetic field of 0.600 T directed perpendicularly to the velocity ofthe ions. \Vhat is U1C distance between the impact poims formed on the photographic plate b)' singly charged ions of 235C and 23SU? 39. A proton is at rest at the plane vertical boundary of a region containing a uniform ,·ertical magnctic field B. An alpha panicle mO\'ing horizontally makes a head·on elastic collision \,..· ith the proton. Immediately after the collision, both particles enter the magnetic field, moving perpendicular to the direction of the ficld. Thc radius of the proton's trajectory is R. Find the radius of the alpha panicle's tr~ectory. The mass of the alpha particle is lour timcs that ofthc proton, and its charge is twice Ihal or the proton.
~B
Bm
i
5.00 A
r-20.0cm
I -10.0
5.00 A
~I
eml-IO.O em-j
FIGURE P19.48
660
Chapter 19
Magnetism
A0-----------0 , ,c
49. Four long, parallel can· ductors carr}" equal currems of J "'" 5.00 A. figure P19.49 is an end vicw of the conducLOrs. The dircClion of the current is into the page al points A and B (indicated h)' the crosses) and Olit of the page at Ganci D (indicmed
br
,,
tance is it 0.100 J.LT? (b) At one instant, the two conduclors in a long household extension cord carr)' equal 2.00-A currents in opposite direClions. The twO wire'i are 3.00 mm apart. Find the magnetic field 40.0 cm awa), f!"Om the middle of the su'aight cord, in the plane of the two wires. (c) At what distance is it one-tel1lh as large? (
,, , ,,:0.200 III ,,
p•
ncb------------0D 0.200
III
FIGURE P19.49
the dot.s). Cdcul
the magnetic field at point P, lOC
50. The two wires in figure P19.50 carry currClib of 3.00 A and 5.00 A in the direClion indicated. (a) Find the direction and magnitude of the magnetic field al a point midway between the wires. (b) Find the mag-nilude and dircClion orthe magnetic field at point P, loe-lled 20.0 em abO\c the wire carrying the 5.00-A currelll. I'
54.
mil
Two long, parallel wires .\ separated by a dislance 2dcarry eqllal currents in the sarnt; 1" t((J,y) clircction. An end view of the two wires is shown in Figure d -';d+--~~- x P19.54, where the currents arc x P (x, 0) alit of the page. (a) What is the • direClion of the magnetic field FIGURE P19.S4 at P on the .~-a.xis 'let up by thc two wires? (b) Find an expression for the lTlagnitude of the field at P. (c) From your result 10 part (b), determine the field at a point midway bctween the two wires. Doco; your remit meet with your expcctation? Explain.
•
20.0 em
I
3.00 r\ ,.}- - - - - - - \.~ 5.00 A
1.--20.0 cm-./ FIGURE P19.S0
ISI.IA wire carries .1 7.00-;\ currclll along the x-axis, and another wire carries a 6.00-A curTelll along the y-axis, as shO\...' n in Figure 1)19.51. Whal is the magnetic field at point P, located at x = 4.00 Ill, J = 3.00 m? .>
•
SECTION 19.8 MAGNETIC FORCE BETWEEN TWO PARALLEL CONDUCTORS 55. Two parallel \vires arc separatcd by 6.00 cm, each carrying 3.0 A of current in same direction. What is the rnagnillIdc of the force per un it length between the wires? Is the force attractive or repulsive? 56. _ Two parallel wires separated by 4.0 cm repel each other with a force per unitlcnglh of 2.0 X 10' Nm. The current in one wire is 5.0 A. (3) Find the current in the other wire. (b) Arc the currents in the ~allle di,-ection or in opposite directions? (c) \-Vhat would happen if the direction of one ClllTcnl were reversed and doubled?
157. IA wire 6.00 . \
f
I
~--T
(4.00,3.00) on
1',
.~x
7.00 A fiGURE P19.S1
52.
iii
A long, straight wire lies Oil a horizolltaltablc in the xy.. planc and carric'i a current of 1.20 J.1A in the positive x-direction along- the x-axis. A proLan is traveling ill the negative ;\"-dircction at speed 2.30 X 10 4 m/'i a di:-.tance d abo\'e Ihe wire (i.e. 4 = d). (a) \0\1 hat is the direction ofthc magnetic field of the wire aL the position or Ihe proLOn? (b) What is the direclion or the magnetic force acting on the prolOn? (c) Explain why the direClion of the proton's motion doesn't chan~c. (d) Using Newton\ :-.econd law, find .1 symbolic expression for d in terms of the acceleration of gravity g, the prolOn mass 111, its speed v, charge q, and the current I. (c) Find the numeric an,,;wer for the di'itance d using the results of part (d).
53. The magnetic fidel .110.0 cm away from a lon~, straight "'ire carrying current 2.00 A is 1.00 JLT. (a) At what dis-
\"'ith a weight per unit length of 0.080 N/m is suspended direclly above a second wire. The lOp wire carries a current of 30.0 A, and the bOllum wire carries a current of 60.0 A. Find the distance of separation between the wires 'iO that the tOP wire will be held in place by magnelic repulsion.
58. In Figure P19.58 the current in lhe long, straight wire is 11 = 5.00 A, and lhe wire lies in the plane or the rectangular loop, which carries 10.0 A. The dimcn'lions shown arc r = 0.100 m, a = 0.150 m, and t = 0.450 rn. Find the magnilUde and direction of the net force exerted by the mag-netic field due 10 the straight wire on the loop.
I,l
ll,
•
I
t
r-+-u-J FIGURE P19.S8
1
Problems
SECTION 19.9 MAGNETIC FIELDS OF CURRENT LOOPS AND SOLENOIDS 59. A 3D-turn circulal' coil of length 6.0 em produces a magnetic field of magnitude 2.0 mT at its cenler. Find the CUfrem in the loop. 60. A certain superconducling magnet in the form of a solenoid of length 0.50 111 can generate a magnetic field of 9.0 T in its core when its coils carry a current of 75 A. The windings, made of a niobium-titanium allo)', must be cooled La 4.2 K. Find the numbn of LUnlS in the solenoid. 0ITJ1l is desired to construct a solenoid thaL will h~l\'e a resislance of 5.00.0 (at 20 C) and produce a magnetic field of 4.00 X 10 2 T at its (Clller when it carries a current of 4.00 A. The solenoid is to be constructed (rol11 coppel" v\o'ire having a diameter of 0.500 mm. If the radius of the solenoid is 1O be 1.00 em, determine (a) the number of turns of wire needed and (b) the length the solenoid should have. C
H2. A single-turn square loop of wire 2.00 em on a side carries a counterclockwise current of 0.200 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30 turns per centimeter and carries a coulllcrclockwise current of 15.0 A. Find the force on each side of' the loop and the torque acting on the loop. ()~.
carries a current of 0.30 A, and a mass of 20.0 g mu~t be added to the right side to balance the system. \-Vhat r~ the magnitude of the magnetic field? 65. Two coplanar circular loops 01 wil'c calTY CUITcnts of /1 - 5.0 A and I,! = 3.0 A in opposite direc· tions as in Figure PI9.65. (a) If r = 9.0 cm, what is the magnitude and direction of the net magnetic field at the cemer of the two loops? (b) Determine the value of l' such that the net field at the cel1lcr is zero.
magnitude of the magnetic field al point P, \.... hich i~ In the same plane as the two wires. (b) Find the magnetl( field at a point 30.0 ern above the point of intersectIon (30.0 em out of the page, toward you). y
,
t I
~.oo.-\
II~fP 40.0 em
~
__
=~~~!..-_. -x 5.00 A
FIGURE P19.67
68. A O.200-kg metal rod carrying a current of lO.n A glides on two horizontal rails 0.500 III apart. \\'hat \'ertical mag· netic field is required to keep the rod mO\'ing at a constanl speed if the coefficient of kinctic friction bcu\cell the rod and rails is 0.1 DO?
169.1
Csing an eleClromagnetic nO\\,'meter (Fig. P19.69), a heart surgeon monitors the flow rate of blood through all artery. Eleclrodes A and B make COlllaCl with (he outel Artery
s
R FIGURE P19.64
FIGURE P19.65
6i. Two long, straight wires cross each other at right angles as shown in Figure PI9.6i. (a) Find the direction and
ADDITIONAL PROBLEMS
coil
I,
66. An electron enters a region of magnetic field of magnitude O.OlD 0 T, traveling pcrpendicular to the linear boundary of the region. The direction of the field is per pendicular to the \'e1ocity of the electron. (a) lJetermlllc the time it takes for the electron to lea\e the "ficld-filled" region, noting that its path is a semicircle. (b) Filld the kinetic energy of the electron if the radill~ of its sCllIicir cular path is 2.00 cm.
An electron is moving at a speed of 1.0 X 10" rn/s in a ci r· clIlar path of radius 2.0 em inside a solenoid. The magnetic rteld of the solenoid is perpendicular to the plane of the electron's path. Find (a) the strength of the magnetic field inside the solenoid and (b) the current in the wlenoid if it has 25 turns per centimeter.
64. _ Figure PJ9.64 is a setup that can be used to measure magnctic fields. A rectangular coil of wire contains N turns has a width w. The coil is attached to one arlll of a balance and is suspended between the poles or a mag· net. The field is uniform and perpendicular to the plane or thc coiL The system is first balanced . . . hen the currcl1l in the coil is zero. When the switch is closed and thc coil carries a currem I, a mass m must be added to the right side to balance the system. (a) Find an expression for the magnitude of the magnetic field and determine its direction. (b) Wh}' is the result independent of the vertical dimension of the coil? (c) Suppose the coil has 50 turns and width of 5.0 em. \Vhen the switch is closed, the coil
661
FIGURE P19.69
662
Chapter 19
Magnetism
surface of the blood vessel. which has illlerior diamcLer 3.00 1111ll. (a) For a magnetic field magnitude of 0.040 0 T, a potential difference of 160 J.LV appears between the electrodes. Calculate the speed of the blood. (b) Verify that electrode A is posit ive, as shown. Does the sign of the emf depend on whether the mobile ions in the blood are predominantly positively or negatively charged? Explain.
70. A uniform horizomal wire with a linear Illass density of 0.50 g/m carries a 2.0-A currelll. II is placed in a constant magnetic fle1d with a strength of 4.0 X 10- 3 T. The field is horizontal and perpendicular LO the wire. As the wire moves upward starting from rest, (a) whal is its acceleraLion and (b) how long does illtlkc LO !"ise 50 em? Neglect the magnetic field of Earth. 71. Three long. parallel conductors carry currents of [= 2.0 A. Figure P19.71 is an end view of the conductors, \.... ith each currem coming out of the page. Given that a = 1.0 em, determine the magnitude and direction of the magnetic field at points il, 13. and C.
(.)/
,
, /Y\ , , // I,,', , , ,
,
,.\..: - _2_ "
, ....
,
'
-n+---,,--~~c ,I
(II
,/
'
--(1--
B!
173.IProtons having a kinetic energy of 5.00 .\-leV are moving in the positive x-direction and enter a magnetic field of 0.050 a T in the z-direction, out of the plane of the page, and extending from x = a ta x:::: 1.00111 as in Figure PI9.73. (a) Calculate the y-componelll of' the protons' momentum as Lhe)' leave the magnetic field. (b) Find the angle a between the initial vclocity vector of the proton beam and the velocity yectar aftcr the beam emerges from the field. (Hin/.: Neglect relativistic effects and note lhall eV = 1.60 X 10-19 J)
··· ·· ·· ··· ··• ·· ·· ·· ···· ··
·· ·· ··· · •
FIGURE P19.73
74. A straight wire of mass 10.0 g and length 5.0 cm is suspended from two identical springs that, in t.urn. form a closed circuit (Fig. PI9.i4). Thc springs stretch a distance of 0.50 em under the weight of the wire. The circuit has a total resistance of [2 fl. \'\Then a magnetic field direCled out of the page (indicated by the dots in the figure) is turned on, the springs are observed to streich an additional 0.30 em. What is the strength of the magnetic field? (The upper portion of the circuil is fixed.)
/
2'1 V
\l/ ~/
FIGURE P19.71
72. TV.lo long, parallel wires, each with a mass per unit length of -to g/m, are supported in a horizontal plane by 6.0-cmlong strings, as shown in Figure PI9.;2. Each wire carries the same currenl I, causing the wires to repel each other so that the angle 0 between the supporting strings is 16°. (a) Are the currcllls in tJle same or opposite directions? (b) Determine the magnitude of each current. y
FIGURE P19.72
•
•
1•
•
•
•
.1
~S.Ocm~
•
FIGURE P19.74
75. A I.DO-kg ball having net charge Q = 5.00 I-/-C is thrown out of a window horizontally at a speed v = 20.0 m/s. The \·... indow is at a height Ii = 20.0 m above the ground. A uniform horizontal magnetic field of magnitude B = 0.010 a T is perpendicular to the plane of the ball's trajecLOry. Find the magniwde of the magnetic force acting on the ball just before it hits the ground. (f-lint: Ignore magnetic forces in finding the ball's fiml1 yelocit},.) 76. Two long, parallel conducLOrs separated by 10.0 cm carry currents in the same direction. The firsl wire carries a current I] = 5.00 A, and the second carries 1'2 :::: 8.00 A. (a) \'Vhal is the magnitude of the magnctic field created by [\ at the location of I,) (b) '-Vhat is the force per unitlengLh exerted by 1\ 011 1ii (c) 'What is thc magnitude of the magnetic field created by 1'2 at the location of II? (d) What is the force per length exerted by I::. on [I?
A forest of fluorescent lights, not wired to any power source, ore lighted by electromagnetic induction Changing currents in the power wires overhead create limedependent magnetic flux in the vicinity of the tubes, indLlcing a voltage across them.
INDUCED VOLTAGES AND INDUCTANCE In 1819 Hans Christian Oersted discovered that an electric current exerted a force on a magnetic compass. Although there had long been speculation that such Q relationship existed,
20.1
Induced emf and Magnetic Flux
20.2
Faraday's Low of Induction
20.3
Motional emf
2004
Lenz's Law Revisited (The Minus Sign in Faraday's Law)
20.5
Generators
20.6
Self-Inductance
20.7
RL Circuits
20.8
Energy Stored in a Magnetic Field
Oersted's finding was the first evidence of a link between electricity and magnetism. Because nature is often symmetric, the discovery that electric currents produce magnetic fields led sci~ entists to suspect that magnetic fields could produce electric currents. Indeed, experiments conducted by Michael Faraday in England and independently by Joseph Henry in the United States in 1831 showed that a changing magnetic field could induce an electric current in a circuit. The results of these experiments led to a basic and important law known as Faraday's law. In this chapter we discuss Faraday's law and several practical applications, one of which is the production of electrical energy in power generation plants throughout the world.
20.1
INDUCED EMF AND MAGNETIC FLUX
An experiment first conducted by F"H3day demonstrated thaL a current can be produced by a changing magnetic field. The apparatus shown in Figure 20.1 (page 664) consists of a coil connected to a switch and a battery. We call this coil the jJri'/!uuy mil and the corresponding- circlJit the primary circuit. The coil is wrapped around an iron ring to intensify the magnetic field produced by t.he current in the coil. A second coil, called the semndW)l roil, at. the right, is ""rapped around the iron ring and is connect.ed to an an1meter. The corresponding circuit is called the secondary circuit. It's important 10 notice that there is no battery in the sec~ ondary circuit. At first glance, you might guess that no currenl ""auld ever be detected ill the secondary circuit. "'''hen the s,,,'iLch in t.he prima!")' circuit in Figure 20.1 is suddenly closed, however, something amazing happens: the ammeter measures a current. in the secondary circuit and then returns to zero! ""hen the s\,vitch is opened
663
664
Chapter 20
Induced Voltages and Inductance
FIGURE 20.1 Farada~"s expCrimellL. \Vhcn the switch in the primar}' circuit
A.. lllmeler
r; aaa'R
.,d
.0'
PI'jma]"y
Baltery
coil
l
MICHAEL FARADAY
again, the arnmeter reads a current in the opposite direC[ion and again returns to zero. Finally, whenever there is a steady current in the primary circuit, the amme~ ~ tel' reads zero. ~ From such observations, Faraday concluded that an electric current could be ~ _ produced by a changing magnetic ReId. (A steady magnetic ReId doesn't produce ~ ~ a current unless the coil is moving, as explained belmv.) The current produced in ~ ~ the secondary circuit occurs for only an instant \",'hile the magnetic field through !i the secondary coil is changing. In effect, the secondary circuit behaves as though ~ ~ a source or emf \·"ere connected to it for a short tirne. It's customary to say that an c:: " induced emf is produced in the secondary circuit by the changing ~agD~lic field.
British physicist ond chemist
(l791-1867) Faraday is often regarded as the greatest experimental scientist of the 1800s. His many contributions to the study of electricity include the inventions of the electric motor, electric
generator, and transformer, as well as the discovery of electromagnetic induction and the laws of electrolysis. Greatly influenced by religion, he refused to work on military poison gas for the British government.
Magnetic flux ~
Magnetic Flux To evaluate induced emfs quantitatively, we needlo understand what factors affect the phenomenon. Although changing magnetic fields ah..'ays induce electric fields, in other situations the magnetic field remains constant, yet an induced electric field is still produced. The best example of this is an electric generator: A loop of conductor rotating in a constant magnetic field creates an elecrric current. The physical quantity associated with magnetism that creates an electric field is a changing magnetic flux ..Magnetic nux is defined in the same \.. .ay as electric nux (Section ]5.9) and is proporlionallo both the strength of the magnetic field passing through the plane ora loop of wire and the area of the loop. The magnetic flux «)nthrough a loop of wire with area A is defined by
"
=
I3 j A = I3A cos
e
[20.1]
where 81. is the component of Ii perpendisular to the plane of the loop, as in Figure 20.2a, and 8 is the angle between B and the normal (perpendicular) to the plane of the loop. SI unit: weber (Wb)
FIGURE 20.2 (a) A uniform magnetic field Ii making an angle 0 I,·jth a dire-Clion flonllal (0 the plal1\' of a wire loop of area A. (b) An edge view of the loop.
Loop or area :l
~~ '/ (b)
20.1
Induced emf and Magnetic Flux
From Equation 20.1, it follows tha...!.. 13.1. = B cos 8. The magnetic nux, in other words, is t.he rnagnitude of the pan of B t.hat is perpendicular to the plane of the loop times the area of the loop. Figure 20.2b is an edge ,·iew of the loop and the penetrating magnetic field lines. \Vhen the field is perpendicular to the plane of the loop as in Figure 20.3a, e = 0 and ~)IJ has a maximum "alue, (1)ll.max = BA. \Vhen the plane of the loop is parallel to B as in Figure 20.3b, 8 = 90° and <~H = o. The nllx can also be negative. For example. when () = 180:>. the flux is equal to -BA. Because the 51 unit of B is the tesla, or weber per square meter, the unit of nux is T· m', or weber (lI'b). \!Ve can emphasize the qualitative meaning of Equation 20.1 by first drawing magnetic field lines, as in Figure 20.3. The number of lines per unit area increases as the field strength increases. The value of the magnetic flux is proportional to the total number of lines passing through the loop. \Ve see that the most lines pass through the loop when its plane is perpendicular to the field. as in Figure 20.3a, so the flux has its maximum value at that tirne. As Figure 20.3b shows, no lines pass through the loop when its plane is parallel to the field. so in that case
APPLYING PHYSICS 20.1
Explanation Greenland (hltitudc 60:> north to 80:> north) is closer to a magnetic pole than Argentina (IaLillide 20° south to 50° south), so the magnetic
Gool
(1)/1.
"'.l"
=
8= 90'
Btl
(aJ
(bJ
FIGURE 20.3 An edge ,ie\,· ofa loop ill a uniform magnelic field. (a) When the field lines are perpcndiClllar lO the plane of the loop, lhe magnelic flux through Ihe loop is a maximum and equal 10 <J>B = Rt. (b) When Ihe held lines are parallel 10 lht.· plane of the 100p,Ihe magnetic flux Ihrough I he Joop is zero.
FLUX COMPARED
Argentina has more land area (2.8 X 10l> km 2 ) than Greenland (2.2 X JO° km 2). Wh)' is the magneLic nux of Earth's magnetic field larger through Greenland than th rough Argentina?
EXAMPLE 20.1
8= O·
665
field is stronger there. That in itself isn't sufficient to conclude that the magnetic flux is greater, but Greenland's proximity to a pole also means that the anglc magnetic field lines make ",,"ith the vertical is srnaller than in Argent.ina. As a result, more field lines pCIletl'ate the surface in Greenland, despite Argentina's sligllll), larger area.
Mo netic Flux
Calculate magnetic flux and a change in nux.
Problem A conducting circular loop of radius 0.250 1ll is placed in the .'9'-plane in a uniform magnetic field of 0.360 T that points in the positive .:-direction, the same direction as the normal to the plane. (a) alculate the magnetic nux through the loop. (b) Suppose the loop is rotated clockwise around the x-axis. so the normal direc-
tion now points at a 45.0:> angle with respect to the z-axis. Recalculate the magnetic flux through lhe loop. (c) \Vhat is the change in flux due to the rotation of the loop? Strategy After finding the area, substitute "alues into the equation for magnetic flux for each pan.
,.. .....
.
..... ...,
Solution (a) Calculme the initial magnetic flux through the loop. 7Or 2
First. calculale the area of the loop:
A =
Substitute A, B, and 8 = 0 into Equation 20.1 to find the initial rnagnetic flux:
q'/J =
0
= 7T(0.250 mf = 0.196 m'
ABcos 8
=
(0.196 m')(0.360 T) cos (0') 0.070 6 Wb
= 0.0706 T· m 2 =
(b) Calculate the magnetic flux through the loop afler it has rotated 45.0 around the x-axis. 0
Make the same substitutions as in part (a), except the angle between Ii ~ll1d the normal is 110\'18 = 45.0 0
(c) Find the change in the magnetic Allx due to the rOlalion of the loop. SubtraCllhe result of pan (a) from the result of
i'I (I'" = 0.0499 Wb - 0.0706 Wb = -0.020 7 Wb
pan (b): ..........
lrrr... • • • • • •
,
~
Remarks 0Jotice that the rmation of the loop, nOI any change in the magnetic field, is responsible for the change in nux. This changing magnetic nux is essential in the [uncI ioning or electric motors and gcncraLOrs.
QUESTION 20.1 True or False: Tf the loop is rotated in the opposite direction by the same amount, the change in magnetic flux has the same magn illlcle but opposite sign.
EXERCISE 20.1
The loop, having rotated by 45°, rOtates clockwise another 30°, so the normal LO t.he plane poims at an angle or 75 with respect to the direction of the magnetic ficld. Find (a) the magnetic flux through the loop when 0 = 75 0 and (b) the change in magnetic nux during the rotation from ~15° to 7~0. 0
Answers
(a) 0.0183 Wb
(b) -0.031 6 Wb
20.2
Tip 20.1 Induced Current Requires a Change in
Magnetic Flux The existellce of magnetic flux through all area is not sufficient 10 create all indllccd elllf. A dumgein the magnetic flux o\'cr some time interval al must occur for an cmho be induced.
FARADAY'S LAW OF INDUCTION
The usefulness of the concept or magnetic flux can be made obviolls by another simple experiment that demonstrates the basic idea of electromagnetic induction. Consider a wire loop connected to an anl111eter as in Active Figure 20.4. Ifa magnet is moved toward the loop, t.he ammeter reads a currenl in one direction, as in Active Figure 20.4£1. ""hen the magnet is held stationary, as in Active Figure 20Ab, the ammeter reads zero current. If the magnet. is moved away from the loop, the ammet.er reads a current in the opposite dircction, as in Active Figure 20.4(. If the magnet is held stationary and the loop is moved either toward or away [rom the magnet, the alTImeter also reads a current. From these observations, it can be concluded that a current is set up in the circuit as long as there is relative motion between the magnet and the loop. The same experimental results arc found whethcr the loop moves or the magnet moves. \Nc call such a currenl an induced current because it is produced by all induced emf. This experiment is sim itar to the Faraday experiment discussed in Section 20.l. In each case, an emf is induced in a circuit when the magnetic nux through the circuit changes ".. . ith time. It turns out that the instantaneous emf induced in a circuit equals the negative of the rate of change of' magnetic flux with respect to time through the ci rcuit. This is Faraday's law of magnetic induction. ACTIVE FIGURE 20.4 (a) '·\'hCII a magnet is moved toward a wire loup ("011llc("ICd LO <111 a III Illt.:t er. lhe alllmcter reach a curn:lIt as
shown, indicaLing lhat a currenl lis illduced in till.: loop. (1)) ''''hen Ihe mag-lid is held stationary. IlO CLlITClll is ill<Juced ill lhe loop. ('ven Wh<:ll the Jllagnct is inside the 'uop. (c) When the magnet is llluved :lway from th(' loop. Ihl' lImm('ll'r reads.1 CUlTell' ill llle opposi1e dirt.:t'lillll. illdicuillg an indllt'{'d CllrreliL goillg oppusite tIll' direction orthc Cllrrem ill pan (a).
20.2
If a circuit contains lv' tightly wound loops and the magnetic flux through each loop changes by the amount llC]>/J during the interval 6./, the average ernfinduced in the circuit during time llt is 6 cl)/J 8= - N -
Faraday's Law of Induction
~
Faraday's law
~
Lenz's law
667
[20.2]
6.1
Because cDn = BA cos 0, a change of any of the factors B, A, or f) with time produces an emf. \Ve explore the effect of a change in each of these facIal's in the following seClions. The minus sign in Equation 20.2 is included to indicate the polarity of the induced emf. This polarity determines \vhich of two directions current will flow in a loop, a direction given by Lenz's law: The current caused by the induced emf travels in the direction that creates a magnetic field with flux opposing the change in lhe original flux through the circuit. Lenz's law says that if the magnetic flux through a loop is becoming more positive, say, then lhe induced emf creates a current and associated magnetic field that produces negative magnetic nux. Some misl]kenly think this "counler magnetic field" creat.ed by the induced current, called Bind ('-ind'-jor induced), will always point in a direction opposite the applied magnetic field B, but thaI is only Irue half the time! Figure 20.5a shows a field penetrating a loop. The graph in Figure 20.5b shows thalthe n~.~lgniLUde of the magnetic field B shrinks withJime, which me<1ns that the nux orB is shrinking \vith time, so the induced field Bind will actually be in the same direction as B. In effect, Bind "shores up" the field B, slowing the loss of flux through the loop. The direction of the current in Figure 20.5a can be determined by right-hand rule number 2: Point your right thumb in the direction that will.c:.ause the fingers on your right hand [Q curl in the direction of the induced field Bind' In this case, that direction is counterclockwise: with the right thumb pointed in the direction of the current.. your fingers curl dmvn outside the loop and around and up through the inside of the loop. Remember, inside lhe loop is \vhere it's important for the induced magnetic field to be pointing up. ~
~
QUICK QUIZ 20.1 Figure 20.6 is a graph of the magnitude B versus time for a magnetic
B
field that passes through a Fixed loop and is oriented perpendicular t.o the plane of the loop. Rank the magnitudes of the ernf generated in the loop at the three instants indicated from largest to smallest.
a
FIGURE 20.6
EXAMPLE 20.2
(al
~
IJ
~, (b)
EIGURE 20.5 (a) The magnetic lleld B becomes smaller with lime, reducing the flux, so current is induced in a direction that creates all induced magnetic lleld Bmd opposing the change in magnetic flux. (1)) Graph or tile magnitude of the magnetic field as a functiun of time.
b (Quick Quiz 20.1)
Faraday and Lenz to the Rescue
Goal Calculale an induced emf and currenl with Faraday's 1m,,' and apply Lenz's Im'i when the magnetic field changes with lime. Problem A coil with 25 turns of"wire is wrapped on a frame with a square cross section 1.80 cm on a side. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 0.350 n. An applied uniform magnetic field is perpendicular to the plane of the coil, as in Figure 20.7. (a) If the field changes uniformly from 0.00 T to 0.500 Tin 0.800 s, what is the induced emf in the coil while t.he field is changing? Find (b) the magnitude and (c) the direction of" the induced current in the coil \.. . hile the field is ch<mging.
x )'
FIGURE 20.7
(Example 20.2)
668
Chapter 20
Induced Voltages and Inductance
Strategy Part (3) requi res substitut.ing into Faraday's la..\', Equation 20.2. The necessary information is given, except. for 6,/:l' the change in the magnetic flux during the elapsed time. Compute the initial and final magnetic fluxes with Equation 20.1, find the difference, and assemble all terms in Faraday's law. The current can then be found with Ohm's law, <'"Ind its direction with Lcnz's law.
.
~
Solution (a) Find the induced emf in the coil. To compule the flux, the area of the coil is needed:
JI = L' = (0.0180 m)' = 3.24 X 10-" m'
The magnetic flux ([)B.llhrough the coil all = 0 is zero because B = O. Calculate the flux at 1= 0.800 s:
<[>,,;= BA cos 0 = (0.500 T)(3.24 X 10-' Ill') cos (0') = 1.62 X 10-" Wb
Compute the change in the magnetic flux through the cross section of the coil over the 0.800-s interval: SubstilUte into Faraday's law ofindl.lction to find the induced emf in the coil:
e= =
- NtiQ:>ll - - = - ( 25 lUrns )(1.62 X IO-'Wb) tit 0.800 s -5.06 X W- 3 y
(b) Find the magnitude of the induced current. in the coil.
Substitute the voltage difference and the resistance into Ohm's law:
tiV R
/=-=
5.06 X 1O- 3 y 0.350 fl
1.45 X 10- 2 A
(c) Find the direction of'the induced current in the coil.
The magnetic field is increasing up through the loop, in the same direction as the normal to the plane; hence, the flux is positive and is also increasing. A do..vnward-poiming induced magnetic field will create negative nux, opposing the change. IfYOll point
.....
Remark
your right thumb in the clockwise direction along the loop, your fingers curl down through the loop, Wh1ch is the correct direction for the counter magnetic field. Hellce the current must proceed in a clockwise direction. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • .<11
Lcnz's law can best he handled by first sketching a diagram.
QUESTION 20.2 What average emf is induced in the loop if, instead, the magnetic field changes uniformly from 0.500 T to O? How would that affect the induced current? EXERCISE 20.2 Suppose the magnetic field changes unirormly rrom 0.500 T to 0.200 T in the next 0.600 s. Compute (a) the induced emf in the coil and (b) the magnitude and direction of the induced current.
Answers
(a) 4.05 X 10-3 Y
FIGURE 20.8 Essential components of a ground faull interrupter (conterns of the g-ray hox in Fig. 20.9a). III Ilewcr humcs such devices ar(" built directly into wall ollliets. The purpose of the ~ensing coil and circllil. breaker is to cut off till' CUITCIIL belore damage is
APPLICATION Ground fault interrupters
(b) 1.16 X 10-2 A, counterclockwise
The ground fault interrupter (GFI) is an illlercSling safety device that proteCls people against electric shock when they touch appliances and power tools. Its operation makes use or Faraday's law. Figure 20.8 shows the essential parts of a ground fault interrupter. Wire "I leads from the I rOil wall outlet to the appliance to be protected, and ring wire 2 leads from the appliance back to the wall outlet. An iron ring surrounds the twO wires to confine the magnetic field set up by each v,"ire. A sensing coil, which can activate a circuit breaker ,·."hen changes in magnetic flux occur, is wrapped around part of the iron ring. Because the currents in the ,.. ., ires are in opposite directions, the net
20,2
Faraday's Law of Induction
669
I I
•
Wall fIllllel
Cround l~lUll intetTuplcr
120 V
r
~
1
(h)
(aj FIGURE 20.9 (a) This hair dryer has been plugged into a ground rault inlerrupter lhal is in turn phl/.mc·d illto all unprotected wall outkt, (bl YOII likdyhave scen this kind or ground fault inlerrupter ill a 1lOId balhroolll, where hair dryers alld electric shavers are orten usc(l by people jU:o>1 alii of I he shower or Whll might tollch a water pipe, providing
magnetic field through the sensing coil due to the currents is zero. If a shan circuit occurs in the appliance so lhatthere is no returning current, however, the net magnetic field through the sensing coil is no longer zero. A short circuit call happen if, for example, one of the wires loses its insulation, providing a path through you LO ground if you happen to be touching the appliance and are grounded as in figure 18.23a. Because the current is alternating, the magnetic flux through the sensing coil changes with time, producing an induced voltage in the coil. This induced volLage is used to trigger a circuit breaker, stopping the current quickly (in about 1 ms) before it reaches a level that might be harmful to the person lIsing the appliance. A ground fault interrupter provides raster and rnore complete protcction than even the case-ground-and-circuit-breaker combination shown in Figure 18.23b. For t.his reason, ground fault interrupters are commonly found in bathrooms, where electricity poses a hazard to people. (See rig. 20.9.) Another interesting application of Faraday's law is the production of sound in an electric guitar. A vibrating string induces an emf in a coil (Fig. 20.]0). The pickup coil is placed ncar the vibrating guitar string, which is made ofa metal that can be magnetized. The permanent magnet inside the coil magnetizes the portion of' the stri ng nearest the coil. When the guitar string vibrates at some frequency, its magnetized segment produces a changing magnetic nux through the pickup coil. The changing nux induces a voltage in the coil; the voltage is fed to all amplifier. The OUtput of the amplifier is sent to the loudspeakers, produci ng the sound waves that we hear. Sudden infant. death syndrome, or SIDS, is a devastating affliction in which a baby sucldenl)' SLOpS breathing during sleep \vithoUl an apparent causc. One type of' monitoring device, called an apnea monitor, is sometimes used to alert caregiv~ ers of the cessation of breathing. The device uses induced currents, as shown in
Pickup coil
Magnetil.ed porlion or string
Electric guitar pickups
APPLICATION
II!
Apnea monitors
FIGURE 20.10 (,,) Ill::tn dt:Ut ie- gUI tar;\ vihratin!-{ strillg illdll«'~ a voltage in the pickup coil. (b) Stwral pj(kllp~ allow llw \ ibratioll lO Ik ddectNI frOUI dirferellt portion, of lhe .~llillg
Magnd
To amplifier Cuitar string (aj
APPLICATION
(h)
Induced Voltages and Inductance
Chapter 20
670
FIGURE 20.11 This infanl Is\\,(;;\ring a monitor designed to alert caregivers ifbrcJ.lhing SlOpS. ~olicc the twO wires attached to opposite sides of I he chest.
FIGURE 20.12 r\ straight conductor of length (' lllo\,jng with velocity ~ Ihrough a uniform magnetic field B directed perpendicular to The venor F", is til(' magnetic lorce all an electron in the conductor. :\11 Clnfof BCv is indllced between the ellcls (If
v.
tIll: bar.
-1 F",
R
-'---===tJ::-1.-x~1
-
MOTIONAL EMF
In Section 20.2 ,.,'e considered em[s induced in a circuit when the magnetic field changes with time. In this seclion we describe a particular application of Faraday's law in which a so-called motional emf is produced. It is the emf induced in a conductor moving through a magnetic field. First consider a straight conductor of length .e moving with constam velocity through a uniform magnetic field directed into the paper, as in Figure 20.12. For sirnplicity, we assume the conductor moves in a direction perpendicular to the field. A magnetic force of magnil.Ude Fill = flvB, directed downward, acts on the electrons in the conductor. Because of this magnetic force, the free electrons move to the lower end of the conductor and accumulate there, leaving a net positive charge at the upper end. As a result of this charge separation, an electric field is produced in the conductor. The charge at the ends builds up until the do,,·,'ll,..'ard magnetic force qvB is balanced by the upward electric force qE. At this point, charge stops flowing and the condition for equilibrium requires that
or
E= vB
Because the electric field is uniform, the field produced in the conductor is related to the potential difference across the ends by 6.V= E£, giving
---
IEI= Bf"
(b)
ACTIVE FIGURE 20.13 (a) A cOlldllcling bilr sliding wilh velocity v along 111'0 conducting rail~ ~nder the.; ael iOll of :Ill apelied force F"l'p' Tht: magnetic force F .. opposes the motion. and a cOUlHerclock-
\\ ise currell1 is induced in the loop. (b) The equi\';denl (:irctlit of thaI in (a).
20.3
qE= qvB
J
-
Figure 20.11. A coil of wire attached to one side of the chest carries an allernating current. The varying rnagnetic flux produced by this current passes through a pickup coil attached to the opposite side of the chest. Expansion and contraction of the chest caused by breathing or movement change the strength of the voltage induced in the pickup coil. ]f breathing stops, however, the pattern of the induced voltage stabilizes, and external circuits monitoring the voltage sound an alarm to the caregivers after a momentary pause to ensure that a problem actually does exist.
Ll.V= E£ = Bev
[20.3]
Because there is an excess of positive charge at the upper end of the conductor and an excess of negative charge at the lower end, the upper end is at a higher potential than the lower end. There is a potential difference across a conductor as long as it moves through a field. ]f the motion is reversed, the polarity of the potential difference is also reversed. A more interesting siwation occurs if the moving conductor is part of a closed conducting path. This situation is particularly useful for illustrating how a changing loop area induces a current in a closed circuit described by Faraday's law. Consider a circuit consisting of a conducting bar of length sliding along l'\"O fixed, parallel conducting rails, as in Active Figure 20.13a. For simplicity, assume the moving bar has zero resistance and the slationary pa-rt of the circuit has constant resistance R. A uniform and constant magnetic field B is applied perpendicular to
e,
20.3
v
the planc or the circuit. As the bar is pulled to the right wilh velocity under the influence of an applied force ap !', a magnetic force along the length of the bar <Jets on the free charges in [he bar. This force ill turn sets up an induced current because the charges are free to move in a closed conducting path. In this case, tlie changing magnctic flux t.hrough the loop and the corresponding induced emf across the moving bar arise from the chang£' in an){l of till-' lOOI) as the bar moves throllgh the magnetic field. Assume the bar moves a dist.ance 6.)." in time t1t, as shown in Figure 20.14. The increase in flux 6.
F
e
671
Motional emf
,----11, Ii
'-------I
IJ
FIGURE 20.14 :\~ lhe bilr move... La dle rig-hl. lhe area orlbe loop ill("reasc~ by Ihe ;Hnounl e.o.x
Using Faraday's law and not.ing that there is one loop (N = 1), we find t.hat the magnitude of the induccd emf is
tl c]>" [81 = - - = tll
tlx l3e= Bev
[20.4]
tlt
This induced emf is oftcn called a motional emf because it arises from the motion or a conductor through a magnet.ic field. f'urlher, if" the resistance of the circuit is R, thc magnit.ude of the induced current in thccircuitis
181
uev
R
R
1=-=-
[20.5]
Activc Figurc 20. L3h shows the cquivalent circuit diagram fat" th is example.
APPLYING PHYSICS 20.2
SPACE CATAPULT
Applying a force all the bar \\"ill result in an induced ern!" in the circuit shown in Active Figure 20.13. Sup' pose we remove the external magnetic field in t.he diagram and rcplace the resistor 'with a high-voltage source and a ~witch, as in Figure 20.15. ,",Vhat will happell when t.he switch is closed? \,Vill the bar move, and do(:s it malleI' which \\lay we connect the high-voltage source? Explanation Suppose t.he source is capable of cstablishing high curren!. Then the t.wo horizontal conducting rods 'will create a strong magnetic field in t he area between then'l, directed into the page. (The movable bar also creates (l magnetic field, but this Ill"
C()I)dllcling -
harfn:e \.0 slid!"
1==
C~ldUC(illg rails
FIGURE 20.15 20.2)
(Applying I'hysin
field cCln't exert. force on the bar itself.) Bccause the moving bar carries a downward current., a magnetic force is exerted on the bar, directcd t.o the right. !-Ience, the bar accelerates along the rails away from the power supply. If the polarity of the power were reversed, the magnetic field would be out of the page, the current in the bar \'\fould be upward, and the force on the bar would still be directed to the right. The Bie force exerted by a magnetic field according t.o Equation 19.6 causes the bar to accelerate away from the voltagc sourcc. Studies have shown that it's possible to launch payloads into space with this technology. (This is the working principlc ofa rail gun.) Very large accelerations can be obtained with currently available technology, with payloads being accelerated to a speed of several kilolll<.:ters per second in a franion ora second. This acceleration is larger than humans can tolerate. Rail gUlls have been proposed as propulsion systems [or moving asteroids int.o more useful orbits. The matcrial of the asteroid could be mined and launched ofT the surface by a rail gun, \vhich would act. like a rocket engine, modifying the velocity and hence the orbit of the asteroid. Some asteroids contain trillions of dollars worlh of valuable 111etals.
Chapter 20
672
Induced Voltages and Inductance
QUICK QUIZ 20.2 A horizontal metal bar oriented east-west drops straight down in a location \.. . here Earth's magnetic field is due north. As a result, an emf develops bct\-veen the ends. ',Vhich end is positivel}' charged? (a) the east end (b) the \.. .est end (c) neither end carries a charge
QUICK QUIZ 20.3 YOli intend to move a rectangular loop of wire into a region of uniform magnetic field at a given speed so as to induce an emf in the loop. The plane of the loop fnUS[ rernain perpendicular to the magnetic field lines. In which orient.ation should you hold the loop \vhile you move it into the region with the magnetic field to generate the largest eme? (a) wiLh the long dimension of the loop parallel to the velocity vector (b) with the short dimension of the loop parallel to the velocity veCLOr (c) either way because the emf is the same regardless of orientation
EXAMPLE 20.3 Goal
A Potentiol Difference Induced Across Air lone Win s
Find the emf induced by motion through a magnetic field.
Problem An airplane with a wingspan of 30.0 III flies due north at a location where the dOl'·;nward component of Earth's magnetic field is 0.600 X 10- 4 T. There is also a component pointing due north that has a magnitude of 0.470 X 10- 4 T. (a) Find the difference in potential beLween the wingtips when the speed of the plane is 2.50 X 10' m/s. (b) Which wingtip is positive?
Strategy Because the plane is flying north, the northern component of magnetic field W'on'l have an)' effect on the induced emf. The induced emf across the \',..ing is caused solely by the downward component of the Earth's magnetic field. Substitute the given quantities into Equation 20.4. Use right-hand rule number] to find the direction positive charges would be propelled by the magnetic force. JI"""' • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • .,
Solution (a) Calculate the difference in potential across the wingtips. ''''rite the motional emf equation and SubStllUte the given quantities:
e= =
BCv = (0.600 X 10-' T)(30.0 m)(2.50 X 10' m/s) 0.450 V
(b) \Vhich wingtip is positive? Apply right-hand rule number 1:
Point the fingers oryollI' right hand north, in the direction of the velocity, and curl them dO\\!J1, in the direction of the magnetic field. Your thumb points west.
An induced emf such as this one can cause problems on an aircraft.
QUESTION 203 In what directions are magnetic forces exerted on electrons in the metal aircraft. if it is flying due west? (a) north (b) south (c) east (d) west (e) up (D down EXERCISE 20.3 Suppose the magnetic field in a given region of space is paI'alleilO Earth's surface, points north, and has magnitude 1.80 X lO~4 T. A metal cable attached Lo a space stalioll stretches radially outwards 2.50 kill. (a) Estimate the potential difference that develops bet\..· een the ends of the cable if it's traveling eastward around Earth at 7.70 X 10 3 m/s. (b) 'Vhich end of the cable is positive, the 100A,'cr end or the upper end?
Answers
(a) 3.47
X
to'V
(b) The upper end is positive.
20.3
EXAMPLE 20.4 Goal
Motional emf
673
Where Is the Ener y Source?
Use motional emf to find an induced emf and a cun-enl. Strategy For pan (a), substiwte imo Equation 20.4 for the motional cmf. Once the emf is found, substitution illlO Ohm's 1m" gives the curreill. In pan (c), Lise Equation 19.6 for the magnetic force on a current-carrying conductor. In part (d), usc the fact that the power dissi~ pated by the resistor multiplied by the elapsed time Illust equal the work done by the applied force.
Problem (a) The sliding bar in Figure 20J 3a has a length of 0.500 m and moves at 2.00 mls in a magnetic field of magnitude 0.250 T. Using the concept of motional emf, find the induced voltage in the moving rod. (b) If the resistance in the circuit is 0.500 fl, find the currel1l in the circuit and the power delivered to the resistor. (Noll'; The current in this case goes counterclockwise around the loop.) (c) Calculate the magnetic force on the bar. (d) Use the concepts of work and pm·vcr to calculate the applied force.
..,
,.. Solution (a) Find the induced emfwith the concept of motional emf. Substitute into Equation 20.4 to find the induced emf:
t: = BCv = (0.250 T)(0.500 m)(2.00 m/s) = 0.250 V
(b) Find the induced currellt in the circuit and the power dissipated by the resistor.
t:
0.250 V
R
= -'-O-.5-'--oo=-n::-
0.500 A
Substitute the emf and the resistance into Ohm's la,,,' to find the induced current:
I
SubstiLuLe f = 0.500 A and t: = 0.250 V into Equation 17.8 to find the power dissipated by the 0.500-n resistor:
'J' = f LlV= (0.500 A)(0.250 V) = 0.125 W
=
(c) Calculate the magnitude and direction of the magnetic force on the bar.
e
!Be =
Substitute values for J, B, and into Equation 19.6, wiLh sin 8 = sin (900) = I, to find the magnitude of the force:
I~" =
Apply right-hand rule number 2 to find lhe direction of the force:
Point the fingers of your right hand in the direction of the positive current, then curl them in the direction of the magnetic field. Your thumb points in the negative X~(lireClion.
(0.500 A)(0.250T)(0.500 m) = 6.25 x lO-'N
(el) Find the value of f',ppo the applied force.
Set the work done by the applied force equallO the dissipated power times the elapsed time: Solve lor
I~pp
and substitute d = v At:
F
app
'!f'tJ, t
'!f'tJ,t
'!f'
d
vAt
v
~-=-
0.125 W = 2.00 m/s
-:'-c_ _
6.25 X 10'N
......................................................................................~ Remarks Pan (d) could be solved by using Neh'LOn's second law for an object in equilibrium: T,vo forces act. horizontally on the bar and the acceleration of the bar is zero, so the forces must be equal in magnitude and opposite in direction. Notice the agreement between the answers for Fm and I\\PP' despite the very different concepts used. QUESTION 20.4 Suppose the applied force and magnetic field in Figure 20. L3a are removed, but a battery creates a current in the same direction as indicated. \""hat happens LO the bar?
Chapter 20
674
Induced Voltages and Inductance
EXERCISE 20.4 Suppose the current suddenly increases to 1.25 A in the sallle direction as before due La an increase in speed of the bar. Find (a) the emf induced in the rod and (b) the new speed orthe rod.
Answers
(a) 0.625 V
(b) 5.00 m/s
20.4 I
I
ii> ,.
.
__ Fm
•
(al
Ibl
FIGURE 20.16
(a) As the COlldUeling bar slides Ull 1IH: 1\\'0 II xed conduclillg" rails, the magnetic nux
through the luop il1crea,es ,,'jth timl'. By Lent's law, the im!lKed Cllr~ rent must be ('(lImlf'l't{ockJU;w so as tIl produce ,t COlllllccH:Ling nux UI/I oj 011' palm: (b) When the bar moves to the left, tilt::: illduced Cllnent Illust bt rlocRwiJI'. \Vh)'?
LENZ'S LAW REVISITED (THE MINUS SIGN IN FARADAY'S LAW)
To reach a bellcr understanding of Lenz's law, consider the example ora bar moving to the righl on l\VO pandlel rails in the presence of a uniform magnetic field directed into the paper (I'ig. 20.16a). As the bar moves to the right, the magnetic flux through the circuit increases \,"ith time because the area of the loop increases. Lenz's law says that the induced currcntmust be in a direction such that the flux it produces opposes the change in the external magnctic flux. Because the flux due to the external rield is increasing into the paper, the induced current, to oppose the change, must produce a flux ouf or the paper. Hence, the induced current must be counterclock\"'ise when the bar moves to the right. (Use rig-ht-hand rule number 2 from Chapter 19 to verify this direction.) On the other hand, if the bar is moving to the left, as in Figure 20.16b. the magnetic flux through the loop decreases with time. l3eGlUSe the flux is into the paper, the induced current has to be clockwise to produce its 0"'1'11 flux into the paper (which opposes the decrease in the external flux). Tn either casc, the induced current tends to maimain the original flux through the circuit. Novv we exami ne this situation from the viewpoint of energy conservation. Suppose the bar is given a slight push to the right. In the preceding analysis, we found that this motion led to a counterclockwise current in the loop. Let's see what would happen if we assume the current is clockwise, opposite the dircction required by Lenz's law. For a clockwise current /, the direction of the magnetic force Blf on the sliding bar is to the right.. This force accelerates the rod and increases its velocity. III turn, the arca of' the loop increases more rapidly, thereby increasing the induced current, ,,,,hich increases t.he force, which increases the current. and so forth. rn effect, the system acquires energy with zero input energy. This is inconsistent with all experience and ,.. . ith the law of conserv;Hion of energy, so we're forced to conclucle that the current must be counterclockwise. Consider another situation. A bar magnet is moved LO the right toward a stationary loop of wire, as in Figure 20.17a. As the magnet moves, the magnetic flux through the loop increases with time. To counteract this rise in flux, the incluced current produces a flux to the left, as in Figure 20.17b; hence, the induced current is in the direction shown. Note that the magnetic field lines associcltcd with the induced current oppose the motion of the magnet. The left face of the current loop is therefore a north pole, and the right face is a south pole.
lal
(bl
Ie)
(d)
Whell thl:; mag-I let is lTIo\'I:;d toward tile ~liltiollary condliCling loop. a CUlTClll is illCluccd in tlw dircl.:lioll :-hO\\'Il. (h) This induced ClilTClll produccs i~ own tllIX to till; left 10 COllllleraCI Ihe iIH:l"ea~ing t;xtern;ll flllx lO the rig-In. (c) \'\'IJcn tile magllctis moved
(a)
20.4
Electromagnet
Lenz's Low Revisited (The Minus Sign in Faraday's Low)
FIGURE 20.18 Lenl's law.
Coil
\
la)
(b)
675
An example or
(e)
On the other hand, if the magnet were moving to the left, as in Figure 20.17c, its nux through the loop, which is tov,,'ard the right, 'would decrease \.. . ith time. Under these circumstances, the induced current in the loop would be in a direction to set up a field directed fr0111 left to right through the loop, in an effon to maintain a constam number of flux lines. Hence, the induced current in the loop would be as shown in Figure 20.17d. ] n this case, the left face of the loop would be a south pole and the right face would be a nonh pole. As another example, consider a coil of wire placed near an electromagnet, as in Figure 20.18a, "Ve wish to find the direction of the induced current in the coil at various times: at the instant the switch is closed, after the switch has been closed for several seconds, and when the switch is opened, "Vhen the s\vitch is closed, the situation changes from a condition in which no lines of nux pass through the coil to one in v,,'hich lines of flux pass through in the direction shown in Figure 20.l8b. To counteract this change in the number of lines, the coil must set up a field frolll left to right in the figure, which requires a currcnt dire((ed as shown in Figure 20,18b. After the switch has been closed for several seconds, there is no change in the number of lines through the loop; hence, the induced current is zero. Opening the switch causes the magnetic field to change from a condition in which flux lines thread through the coil from right to left to a condition of zero !lux. The induced current must then be as shown in Figure 20.18c so as to setup its m... n field from right to left,
Tip 20.2 There Are Two Mognetic Fields to Consider '''''hen <\pplying Lcnz's la\\', there are two mecond is the magnetic field produced by the induced currem in the loop.
'----------
QUICK QUIZ 20.4 A bar magnet is falling toward the center of a loop of wi re, with the north pole oriented downwards. Vie\ved from the same side of the loop as the magnet, as the north pole approaches the loop, \·vhat is the direction of the induced current? (a) clockwise (b) zero (c) counterclockwise (ell along the length of the magnet
QUICK QUIZ 20.5
Two circular loops are side by side and lie in the '<Jplane. A s......itch is closed, staning a counterclockwise current in the left-hand loop. as viewed fro111 a point on the positive z-axis passing through the center of the loop. "Vhich of the following statements is true of the right-hand loop? (a) The current remains zero. (b) An induced current moves counterclockwise. (c) An induced current moves clockwise.
Tope Recorders One common practical use of induced currents and emfs is associated \'lith the tape recorder. 1v1any different types of tape recorders are made, bur the basic principles are the same for all. A magnetic tape moves past a recording and playback head, as in Figure 20.19a (page 676). The tape is a plastic ribbon coated with iron oxide or chromium oxide. The recording process makes use of a current in an electromagnet producing a magnetic field. Figure 20.l9b illustrates the steps in the process. A sound wavc sent into a microphone is transformed into an electric current, amplified, and allO\\"ed to pass dwough a wire coiled around a doughnut-shaped piece afiron. which functions
APPLICATION Magnetic tape recorders
676
Chapter 20
Induced Voltages and Inductance
FIGURE 20.19 (a) i\oL~ior pans of a rlIa~JI~l ic tar}/' recorder. If
From
Rccnrd/pla}'back head
Bulk era::w head
amplifier
(a)
(b)
as the recording head. The iron ring and the wire constitute an electromagnet, in v,:hich the lines o[the magnclic rield are contained completely inside the iron except at the point where a slot is cut in the ring. I-[ere the magnetic field fringes out of the iron and magnetizes the small pieces of iron oxide embedded in the tape. As the tape moves past the slot, it becomes magnetizcd in a pattcrn that reproduces both the frequency and the intensity orthc sound signal entering the microphone. To reconstruct the sound signal, the previously magnetized tape is allowed to pass through a recorder head operating in the playback mode. A second wirewound doughnut-shaped piece of iron with a slot in it passes close to the tape so that the varying magnetic fields on the tape produce changing field lines lhrough the wire coil. The changing f1llx induces a current in the coil that corresponds to the current in the recording head that originally produced the tape. This changing electric current can be amplified and used to drive a speaker. Playback is thus an example of induction ofa current by a moving magnet. Hard drives in computers also use induction to write data, though for faster access the data is written to a coated, spinning disk rather than tape. The read heads, however, now use the giam magnct.oresistance effect, or GM R, rather than induction. The greater sensitivity of the new read heads led to larger storage capacity.
20.5 APPLICATION Alternating-current generators
GENERATORS
Generators and motors are imponant practical devices that operate on the principle of electromagnetic induction. First, consider the alternating-current (AC) generator, a device that converts mechanical energy to electrical energy. In its simplest form, the AC generator consists of a wire loop rotatcd in a magnetic field by somc external means (Active Fig. 20.20a). In commercial power plants, the energy required to rotate the loop can be derived from a variety of sources. In a hydroelectric plant, for example, falling water directed against the blades of a turbine produces the rotary motion; in a coal-fired plant, heat produced by burning coal
ACTIVE FIGURE 20.20 (a) A Sf III::m,Hir diagram ufan AC gl·lwrator. All t'l11fis imluct'd ill a roil, whit h I nl<J,tcs Ily SOlllC cx(erllalmc:ans 111 a mag-netic field. (II) A plul orthe "ltt'l nar1l1g emf induced in I he loop Vf'rsu-. lime.
(a)
(h)
20.5
vsin 0 H
MagllClic field lint:s
Generators
677
FIGURE 20.21 (a) A loop rotating al a constallt allgular velocity in an extcnlal magnetic fkld. The clnl induced in lhe loop varies sinu~oi dall>' with tinlt'. (bl An o.:dge vic...., of lht: rotating: loop.
(h)
(a)
is llsed lO convert water to steam, and this steam is direcled against the turbine blades. As the loop rotates, the magnetic nux through it changes with time, inducing an emf and a currelll in an external circuit. The ends of the loop arc connected to slip rings that rotate with the loop. Connections to the external circuit arc made by stationary brushes in contact with the slip rings. We can derive an expression for the emf generated in the rotating loop by making use of the equation for motional emf, 8 = Bev. Figure 20.2Ja shows a loop of wire rotating clockwise in a uniform magnetic Held direcled to the right. The magnetic force (q11B) on the charges in \I,,'ires A B and CD is not along lhe lenglhs of the \-vires. (The force on thc electrons in these wires is perpcndicular to the wires.) Hence, an emf is generated only in wires BC and AD. At any instant, ,·vire Be has velocity vat an angle v\"ith the magnetic field, as shown in Figure 20.21b. (Note that the component of velocity parallel to the rield has no effect on the charges in the wire, whereas the component or-velocity perpendicular to the field produces a magnetic force on the charges that moves electrons from C to 8.) The emf generated in wire BCequals Bevl.' where is the length of the wire and 11.1. is the componenl ofvclocily perpcndicular to the field. An emfof 13£11_ is also generated in wire DA, and thc sense of this emf is the same as that in wire BC Because vl. = v sin e, the total induced emf' is
e
e
t: = 2BfvL = 2 Btu sin
e
[20.6]
If lhe loop rotates with a constant angular speed w, we can use the relation () = wI. in Equation 20.6. Furthermore, because every poilll on the wires BC and
OA rotates in a circle about the axis of rotation ,·vith the same angular speed w, we have 11 = 'fW = (a/2)w, where a is the length of sides AB and CD. Equation 20.6 therefore reduces to
t:
=
2Bf
(~)w sin wi =
Turhint;S LUrn electric geller<.llor~ at a hydroelectric power pl,lnt.
Bfrtw sin wi
If a coil has N turns, the emf is IV times as large because each loop has the same c..:rnf induced in it. Further, because the area of the loop is A = ta, the total emf is [20.7]
S = NI3Aw sin wI.
This result shows lhat the emf varies sinusoidally with time, as planed in Active Figtlre 20.2Gb. NOle thaI the maximum emf has t.he valuc [20.8]
e
which occurs when wI = 90° or 270°. Tn other words, = 8 11lilx when the plane of the loop is parallel to the magnetic rield. Further, the emf is zero when wI. = 0 or ISO°, which happens whenever the magnetic field is perpendicular to the plane or 1he loop. rn the Uniled SLa1.es and Canada the frequency uf rotal ion for commercial generators is 60 Hz, whereas in some EUr()pe~1n countries 50 Hz is used. (Recall tllat w = 2,"./; where Jis the frequency in hertz.) The direct-current (DC) generator is illustrated in Active Figure 20.223 (page 678). The componems are essentially the same as those of the AC generator except
APPLICATION Direct-current generators
678
Chapter 20
Induced Voltages and Inductance
ACTIVE FIGURE 20.22 (a) A schemafic diagram of a DC generaLOr. (b) The ernfflucHlales in magnitude, btU always has the same polarilv.
e
_kvwv (b)
that the contacts to the rotating loop are made by a split ring, or commutawr. rn this design the output voltage a]v.:ays has rhe same polarity and the current i!\ a pulsating direct current, as in Active Figure 20.22b. Note that the contacts LO the split ring reverse their roles every half cycle. Allhe same time, the polarity of the induced em f reverses. Hence, the polarity of the split ring remains the same. A pulsming DC current is not suitable for most applications. To produce a steady DC current, commercial DC generators use many loops and commutators distributed around the axis of rotation so that the sinusoidal pulses from the loops overlap in phase. When these pulses are superimposed. the DC output. is almost free of fluctuations.
EXAMPLE 20.5 Goal
emf Induced in
n AC Generator
Understand physical aspects of an AC generator.
Problem An AC generator consists of eight turns of wire, each having area A = 0.090 0 m 2 , , ... ith a tOlal resistance of 12.0 n. The coil rotates in a magnetic fleJd of 0.500 T at a constant frequency of 60.0 Hz, with axis of rotation perpendicular to the direction of the magnetic ficld. (a) Find the maximum induced emf. (b) \,Vhat is rhe maximum induced currem? (c) Dcterminc the induced emf and current as functions of time. (d) 'Nhat maximum torque must be applied to keep thc coil turning? Strategy From the given frequency, calculate the angular frequency wand substitute it. LOgether ,·.. ith given quantities, into Equation 20.8. As functions of time, the emf and current have the form A sin wt, where A.. is the maximum eInf or current, respectively. For part (d), calculate the magnetic torque on the coil when the current is at a maximum, (See Chapter 19.) The applied torque must do work against this magnetic torque to keep the coil turning. ~
,~
Solution (a) Find the maximum induced emf'.
First, calculate the angular frequency of the rotational Illotion:
w = 2",/= 2",(60.0 I-Iz) = 377 radls
Substitute the values for N, A, n, and w into Equation 20.8, obtaining rhe maximulll induced emf:
(b) ''''hat is the maximum induced current? Substitute the maximum induced emf B max and the resistance Rinto Ohm's law La And the maxirnum induced currenr:
f
. = Bmax
max
R
136 V 12.0 D
11.3A
(c) Determine the induced emf and the current as functions of time. Substitute B m •Lx and w il1lo Equation 20.7 to obtain the variation of B with time t in seconds:
8 = 8 m . , sin
wi
= (136 V) sin 3771
20.5
The time \'ariation orthe currcllllooksjllst like this expression, except \'Vith the maximum CUITenl Ollt in rront: (d) Calculate the maximulll applied to keep the coillUrning.
1.00'qllC
l~
Generators
(1l.3A) sin 377/
necessary = p.Bsin fJ
Write the equation for magnetic torque:
T
Calculate the maximum magnetic moment of the coil. p.:
JL = I"".JIN = (11.3 A)(0.090 m')(8) = 8.14 A· m'
SubstilUte into the magnetic torque equation, with 8 = 90° to lind the maximul1l applied torque: Remark
679
l,lldX
= (8.14-A·m 2 )(O.500T) sin 90° = 4.07 N'rn
The nUlllber of loops, N, can't be arbitrary because there must be a force strong enough
QUESTION 20.5 What effect docs doubling the frequency have
Oil
10
turn lhe coil.
lhe maximum induced emf?
EXERCISE 20.5 An AC generator is to have a maximum outpllt of 301 V. Each circular turn of \vire has an area of 0.100 m 2 and a resistance of O.BO n. The coil rotates ill a magnetic field of 0.600 T with a frequency oC 40.0 Hz, with the axis of rot<,u ion perpendicul'lr to the direction of the magnetic field. (a) How many turns of wire should the coil have to produce the desired emf? (b) Find the maximum currelll induced in the coil. (c) Determine the induced emf' as a function of lillle. Answers
(a) 20 turns
(b) 18.8 A
(c) 8 = (301 V) sin 2511
Motors and Back emf Motors are devices that COllvert electrical energy to mechanical energy. E.ssentially, a motor is a generator run in reverse: instead 0[' a current being generated by a rotating loop, a current is supplied to the loop by a source or eml~ and the magnetic torque on the cUlTenl-carrying loop causes it to rO(;·lte. A rnolor can perform usel'ul mechanical work when a shaft connected to its rotating coil is attached to some external device. As the coil in the 1110101' rotates. however, I he changing magnetic f1tlX t.h rOllgh it induces an elll r that acts to reduce the current in the coil. Ifit illo"(](fsed the current, Lenz's law would be violated. The phrase back emf is used for an emf that. tends to reduce the applied current. The back emf" increases in magnitude as the rot.uional speed of the coil increases. VIle can picture this state of affairs as lhe equivalent circuit in Figure 20.23. For illustrative purposes, assulTle the external pO\\'er source supplying current in the coil of the illataI' has a voltage of 120 V, the coil has a resistance 01']0 fl, and the back emf induced in the coil '1\.lhis instant is 70\1. The voltage available to supply current equals the difference between the applied voltage and llle back emf, or 50 V ill this case. The current. is always reduced by the back emr. \"'hen a motor is turned on, there is no back emf initially and the current is very laq..;'e because it's limited only by (he resistance of the coil. As the coil begins to rotate, the induced back emf opposes the applied voltage and the C\llTem in the coil is reduced. II' the mechanical load inCl'eases, the motor slows down, which decreases the back cm[ This reduction in the back emf increases the current in the coil and lherefore also increases the power needed from the external volt.age source. As a result, the power requirements lor starting a motor and for running il under heavy loads are greater t.han lhose for running the motor under ;l\'crage loads. If the motor is allowed to run under no mechanical load, the back emf reduccs the current. to a value just large enough 1.0 balance energy losses by heat and friction.
APPLICATION Motors
lO !leoil n:siSlance
70V back emf
'--------I:J 120 V cxtl..:rnal sOllrce
FIGURE 20.23 rt'~ellLcd
emr.
A mOlor com he f('p' as a ]"esiS"lIlCl' plu~ a back
Chapter 20
Induced Voltages and Inductance
EXAMPLE 20.6
Induced Current in a Motor
680
Goal
Apply the concept ora back emf in calculating the induced current in a motor.
Problem A motor has coils with a resistance of 10.0 fl and is supplied by a voltage of 6.11= 1.20 X 10 2 V. "Vhen the mOLOr is running at its maximum speed, the back emf is 70.0 V. Find the current in the coils (a) when the motor is firsllurned on and (b) when the motor has reached its maximum rotation rate. Strategy For each part, find the net voltage, which is the applied voltage minus the induced emf. Divide the net voltage by the resistance to gel the current. ,.. •• - • • • • • • • • . • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •"'111
Solution (a) Find the initial current, when the motor is first turned on. if the coil isn't rotating, the back emf is zero and the
I = _£_-_£---"fJ,,,,e",'
R
current has its maximum value. Calculate the difference between the emfancl the initial back emf and divide by lhe resistance R, obtaining the initial current:
1.20 X 10'\1 - 0 10.0 [l
12.0 A
(b) Find the current ,.. . hen the motor is rotating at its maximum rate.
Repeat the calculation, using the maximum value of the back emf:
The phenomenon of back emfis one ,,,"ray in ,.. . hich the rotation rale of electric mOLOrs is limited.
QUESTION 20.6 As a mawr speeds up, what happens to the magnitude of the magnetic torque? (a) lL increases. (b) it decreases. (c) It remains constant. EXERCISE 20.6 If the current in the motor is 8.00 A at some instant, what is the back em f at that time?
Answer 40.0 V
20.6
FIGURE 20.24 Afrcr tlte switch ill the circuit is closed, thc current produces its 01'111 lllagn<:tic flux through the loop. As the current incre;\ses toward il~ eqllilibriulll valuc, th<: flux changes in time and induces all emr ill the loop. The balter~' drawn wirh dashed lilies is a s~ mbol ror lhe sell: induced <.:Illf".
SELF·INDUCTANCE
Consider a circuit consisting of a switch, a resistor, and a source of emf, as in Figure 20.24. When the switch is closed, the current doesn't immediately change from zero to its maximum value, ciR. The law of electromagnetic induction, Faraday's Imv, prevents this change. "Vhat happens instead is the following: as the currem increases with time, the magnetic flux through the loop due to this current also increases. The increasing flux induces an emfin the circuit that opposes the change in n1agnelic flux. By Lenz's law, the induced emf is in the direction indicated by the dashed battery in the figure. The net potential difference across the resistor is the emf of the battery minus. the opposing induced emf. As the magnitude of the current increases, the mte of increase lessens and hence the induced emf decreases. This opposing emf results in a gradual increase in the current. For the same reason, when the switch is opened, the current doesn't immediately fall to zero. This effect is called self-induction because the changing flux through the circuit arises from the circuit itself. The emf that is set up in the circuit is caIled a self-induced emf. As a second example of self-inductance, consider Figure 20.25, which shows a coil wound on a cylindrical iron core. (A practical device would have several hundred turns.) Assume the current changes with time. ''''hen the current is in the direction shown, a magnetic field is set up inside the coil, directed from right to left. As a result, some lines of magnetic flux pass through the cross-secLional area
20.6 LClIz'" 1,,\\, emr
'.
-
Lcnz .... lawemr
Self-Inductance
681
FIGURE 20.25 (a).\ current in (he coil produces a m;lgnl·tic field
-!J t-
dirc:cu'r\ to dlL' lefl. (b) Irlhe currclll inCl"ea~e~. ,he coil an.'> as a source of cmfdilccled a.. shO\,·nll\ the dashed haller\,. (c) The induced emfin the coil change" il.'t polaril\ ifiluc: currelll
dccrea"l':t. I increasing
(b)
('I
I decreasing (e)
of the coil. As the current changes with time, the nux through the coil changes and induces an emf in the coil. Lenz's law shows thalthis induced emf has a direction that opposes the change in the current. If the current is increasing, the induced emf is as pictured in Figure 20.25b, and if the current is decreasing, the induced emfis as shown in Figure 20.25c. To e\'aluatc self-inductance quantitatively, first note that. according to Faraday's law, the induced emf is given by Equation 20.2:
8=
a'D II
-;\I--
at
The magnetic nux is proportional to the magnetic field, which is proponional to the current in the coil. Therefore, the self-induced emf must be proportional to the rate of change of the current with time, or
M 8= - L -
at
JOSEPH HENRY
[20.9]
where L is a proponionalit)' constclnL called the inductance of the de\'ice. The negative sign indicates that a changing current induces an emf in opposition to the change. I n other \'lords, if the current is increasing (8/ positive), the induced emf is negative, indicating opposition to the increase in current. Likewise, if the current is decreasing (81 negative), the sign of the induced emf is positive, indicating that the emfis acting to oppose the decrease. The inductance of a coil depends on the cross-sectional area of the coil and other quantities, ,vhich can all be grouped under the general heading of geometric factors. The Slunit of inductance is the henry (H), which, [rom Equation 20.9, is equal to 1 volt-second per ampere: I 1-1 = I\,· s/.-\
American physicist (1797-1878) Henry became the first director of the Smithsonian Institution and first president of the Academy of Natural Science. He was the first to produce an electric current with a magnetic field, but he failed to publish his results as early as Faraday because of his heavy teaching duties at the Albany Academy in New York State. He improved the design of the electromagnet and constructed one of the first motors. He olso discovered the phenomenon of self-induction. The unit of inductance, the henry, is named in his honor.
III the process of calculating self-inductance, it is often convenient to equate Equations 20.2 and 20.9 to find an expression for L:
.l"'/I .It
;\1--=
:ll L.ll
L=
APPLYING PHYSICS 20.3
[20.10]
f- Inductance
MAKING SPARKS FLY
In some circuits a spark occurs between the poles of a switch when the switch is opened. ""hy isn"t there a spark when the switch for this circuit is closed? Explanation According to Lenz's la\v, the clirec· tion of induced emfs is such thm the induced lllagnClic field opposes change in the original magnetic nux. When the switch is opened, the sudden drop
in the magnelic field in the circuit induces an emf in a direction that opposes change in the original current. This induced en1f can cause a spark as lhe current bridges the air gap belween Ihe poles of the s,\·itch. The spark doesn't occur when the switch is closed. because the original current is zero and the induced emf opposes any change in that current.
Chapter 20
682
Induced Voltages and Inductance
fn general, determining the inductance of a given current element can be challenging. Finding an expression for the inductance ora common solenoid. however, is straightforward. Let the solenoid have N llIrns and length Assume is large compared with the radius and the core DC the solenoid is air. We take the illlcrior magnetic flcld La be uniform and given by Equation 19,16,
e.
N
= !-,,,nl = !-,,, f
B
e
J
v.:here n = Nle is the number of turns per unit length. The magnetic nux through each turn is therefore
where A is the cross-sectiollal area of the solenoid. From th is expression and Equation 20.10, "ve find that [20.11a]
e
This equation shows thal L depends on the geometric faclors and A and on /-Lo, and is proportional LO the square of the number of turns. Because N = nf, we can also express the result in the form
("f)'
e
1=" • ,-0 - - J I =" r-""'O ,,'Jlt ~",,'\I ,-0
[20.11b]
where \1= Ai is the volume of the solenoid.
EXAMPLE 20.7 Goal
Inductance, Self-Induced emf, and Solenoids
Calcubte the inducrancc and self-induced emfofa solenoid.
Problem (a) CalcuLue the inductance of a solenoid containing 300 turns if the length or the solenoid is 25.0 cm and its cross-sectional area is 4.00 X ]O~4 m 2 . (b) Calculatc the self-induced elllf in the solenoid described in part (a) if the curreIll in the solenoid decreases at the rate of 50.0 A/s. Strategy Substiwting given quantities into Equation 20.11a gives the inductance L. For part (b), substitute the result of part (a) and M!tJ./, = -30.0 A!s into Equation 20.9 to gctthc sclf-induccd cmf. .. - .
.,
,...
Solution (a) Calculate the inductance of the solenoid. Substitute the number 1\/ of turns, the area A, and the length into Equation 20-l1a to find the inductance:
L =
N'J.A _r_"__ II.
f
e
= ( 477
X 10
_;
T'm/A
) (300)'(4.00 X 10- 4 m') <J 2S.0X IO-m
= 1.81 X 10-" T·m'! A =
(b) Calculate the
selJ~induccd emf
0.18Iml-1
in the solenoid.
Substitute Land 6.1/6,t = -50.0 A/s inw Equation 20.9, finding the self~indllced emf:
B
=
M -Lt!d
=
-(1.81 X 10-·'H)(-SO.OA!s)
= 9.0S mY
...................... ~
10..
Remark
Notice thaL 1:11/6,1 is negative because the currenL is decreasing with time.
QUESTION 20.7 If the solenoid were wrapped into (I circle so as to become a toroidal solenoid, ,... hat ,vould be truc of its self-inductance? (a) It would be the same. (b) h would be larger. (c) It ,...· ould be smaller.
20.7
RL Circuits
683
EXERCISE 20.7 A solenoid is to have an inductance of 0.285 mH, a cross-sectional area of 6.00 X 10-'1 m 2 , and a length of 36.0 cm. (a) How many turns per unit length should it have? (b) If the self-induced emf is -12.5 mV at a gi\'en time, at what rate is the current changing at that instant? Answers
20.7
(a)
l.O~
X 10' turns/ill
(b) 43.9 A/s
RL CIRCUITS
A circuit element that has a large inductance, such as a closely wrapped coil of many
turns, is called an inductor. The circuit symbol for an inductor is ~. \'Ve will always assume the self-inductance of the remainder of Lhe circuit is negligible compared with thar of the inducLor in the circuit. To gain some insight into the effecL oran inductor in a circuit, consider the t,.. . o circuits in Figure 20.26. Figure 20.26a shows a resistor connected to the terminals ora battery. For this circuit, Kirchhoff's loop rule is S ~ IR. = O. The voltage drop across the resistor is
1 (a)
[20.12]
L uuu
[n this case, we interpret resistance as a measure of opposition to the current. Now consider the circuit in Figure 20.26b, consisting of an inductor connected Lo the terminals ofa battery. At Lhe instant the swiLch in this circuit is closed, because fR = 0, the emf of the battery equals the back emf generaLed in the coil. I-fence, we have
6.f E:L=-L ilt
L
=-
R
E: (1 - ,-'I') R
L---II-----l e
(b)
[20.14]
Using methods of calculus it can be shO\,,'n that the current in such a circuit is given by l= -
1
[20.13]
From this expression, we can interpret L as a measure of opposition to the rate of change of current. Active Figure 20.27 shows a circuit consisting of a resistor, an inductor, and a battery. Suppose the s\",itch is closed at t = O. The current begins to increase, but the inductor produces an emf that opposes the increasing current. As a result, the current can't change from zero to its maximum \"alne of SIR instantaneously. Equation 20.13 shows that the induced emf is a maximum when the current is changing mOst rapidly, which occurs when the switch is first closed. As the current approaches its steady-state value, the back emf of the coil falls off because the current is changing more slowly. Finally, when the current reaches its steady-state value, the rate of change is zero and the back emf is also zero. Active Figure 20.28 plots current in the circuit as a function of time. This plot is similar to that of the charge on a capacitor as a function of time, discussed in Chapter 18. In that case, we found it convenient to introduce a quantity called the time constant of the tifT/dl, vvhich told us something about the time required for the capacitor to approach its steady-state charge. In the same ,vay, time constants are defined for circuits containing resistors ~lnd inductors. The time constant T for an RL circuit is the time required for the current in the circuit to reach 63.2% of its final "alue SIR; the time constant of an RL circuit is given by T
s
[20.15]
This equation is consistent with our intuition: when the switch is closed at t = 0, the current is iniliallyzero, risingwirh time to some maximum value. :"Jotice the mathematical similarity between Equation 20.13 and Equation 18.7, which features a
FIGURE 20.26 A cOlllparisun of the effect ofa resistor with that of an inductor in a simple circuit.
t
n
R
+
-~,
+ L
S
Ii
ACTIVE FIGURE 20.27 A series IlLcircuil. As the current illcrease~ toward its maximum value, the inductor produces an emf that opposes the increasing current.
+- Time constant for an RL cir-
684
Chapter 20
Induced Voltages and Inductance
capacitor instead of an inductor. As in the case ofa capacitor, the equation's form suggests an infinite amount of time is required for the currenl in the inductor to reach its maximum value. That is an artifact of assuming current is composed of moving charges thaL arc in frnitcsimal, as will be demonstrated in Example 20.9. QUICK QUIZ 20.6 The switch ill the circuit shown in figure 20.29 is closed, and the liglllbulb glows steadily. The inductor is a simple air-core solenoid. An iron roel is inserted into the interior of the solenoid, increasing the m(lgnitudc of the magnetic field in the solenoid. As the rod is inserted, the brightness of the lightbulb (a) increases, (b) decreases, or (c) remains the same.
ACTIVE FIGURE 20.28 A plm of ClJlTf'lll versus lime for Ihe RI. (;rcllit ~ho\\'n in Figure 20.27. The switrh is clused all = 0, and the curITnt Illcreases LOward its maximum valu(' e/R. Thl:; lime comtallt (is Tilf time illakcs the current to reach 632'1 01 its maximu1ll value
EXAMPLE 20.8 Goal
An
/I
FIGURE 20.29
(Quick Quiz 20.6)
RL Circuit
Calculatc a timc constant and relate it to current in an RL circuit.
Problem A 12.6-V battery is in a circuit with a 30.0-mH inductor and a 0.]50-0 resistor, as in Activc Figure 20.27. The switch is closed at 1= O. (a) Find the time constant of the circuit. (b) Find the current after onc time constant ha~ elapsed. (c) Find t.he voltagc drops across the resistor when I. = 0 and t = one time constant. (d) What.'s the rate of f
hange of the current after one time constant?
Strategy Pan (a) requires only substitution into the e1efi nition of time constant. With this value and Ohm's law, the (urrent afr.er one time constant can he found, and multiplying this current by the resistance yields the voltage drop across the resistor after one time constant. "Vith the voltage drop and Kirchhoff's loop law, the voltage across the Indllctor can be found. This value can be substituted into Equation 20.13 \.0 obtain the rate of change of the current. ............................................................
Solution (a) \Vhat's the time conSlant of the circuit? ';llhstitUlc I he inductance L and resistance R into E'luation 20.14, finding the time constant:
T
L
30.0 X 10-' H
R
0.1500
0.200 s
(b) Find the current after one time constant has f"iap,,·d. ]2.6 V = 84.0 A 01500
First, use Ohm's 1m-\' to compute the final value of the (I1rH'nt after many time conStants have elapsed: /\flf"r one time constant, the current rises ItS final value:
[Q
63.2% of'
J,.
= (0.632)J",,, =
(0.632)(84.0 A)
=
53.1 A
(c) Find the \'oltage drops across the resistance when 0 and I = one time constant..
I
Initially, the current in the circuit is zero, so, from Ohm',,; I
Nf·XI. using Ohm's law, find the magnitude of the volt"-Re drop across the resistor after one time constant:
(d) \,Vhat's the rate of change of the current after one lime constant? IJsing Kirchhoff's voltage rule. calculate the voltage drop across the inducl.Or
DoVe = fR DoVR (t = 0 s) = (0 A)(0.150 OJ = 0
D.V" (l = 0.200 s) = (53.1 A)(0.150 D) = 7.97V
~
20.7
= -8 -
I'.VI. Now solve Equation 20.13 for
a'/~l
and substitute:
,
I'.VII = -12.6V - (-797V)
-4.6V
M 1'.1
-
-4.6V 30.0 X 10 'H
I'. VI.
L
150 A/s
..
,
Remarks The values used in this problem were taken from actual components salvaged from the starter system of a car. Because the current in such an RL circuit is initially zero, inductors are sometimes referred to as "chokes" because they temporarily choke ofT the current. In solving pan (d), \-ve traversed the circuit in the direction of positive current, so the voltage difference across the battery was positive and the differences across the resistor and inductor were negative.
685
I'. V, = - L -
1'.1 I'.t ~
RL Circuits
....
EXERCISE 20.8 A 12.6-V battery is in series with a resistance of 0.350 n and an inductor. (a) After a long time, what is the current in the circuit? (b) \Vhat is the current after one tirne constant? (c) \'\Ihat's the voltage drop across the inductor at this lime? (d) Find thc inductance if the time constant is 0.130 s.
Answers
(a) 360 A
(b) 22.8 A
(e) 462 V
(d) 4.55 X lo-' H
QUESTION 20.8 Find the current in the circuit after two time constants.
EXAMPLE 20.9 Goal
Formotion of a Magnetic Field
Understand the role or time in seui ng up an inductor's magnctic ficld.
Problem Given the RL circuit of Example 20.8, And the time required for thc currentLO reach 99.9% of its maxivalue after the s\vitch is closed.
mum
Strategy The solution requires solving Equation 20.-15 for time followed by substitution. Notice that the maximum current is 'IlHl)', = 8/ R. ." , . , .. _ ,._ .
~
~.-
Solution ""rite Equation 20.15, with 'isubstilUted for the currenL: Divide bot.h sides by
8(1_-'h)=1 (l--
'ma)',:
I
Subtract 1 from both sides and then multiply both sidesby-I:
- - f- = 111\
e-lIT
Take the natural log of both sides:
Solve for' and substiLUlC the expression ror 7" from Equation 20.14:
1=
Substitute values, obtaining the desired time:
1=
~
-_.
,-"
,,--
_!C
In (1 _
R
-.!L) 'max
30.0 X 10-:1 H ---=-=-:,.,-:---In(] - 0.999) = 0.150 "
1.38 s
.........
Remarks From this calculation, it's found that forming the magnetic field in an inductor and approaching the maximum current occurs relatively rapidly. Contrary to what might be expected from the mathematical form of Equation 20.15, an infinite amount of time is not actually required.
686
Chapter 20
Induced Voltages and Inductance
QUESTION 20.9 If the inducwnce were doubled, by what factor would the length of time found be changed? (a) 1 (i.e., no change) (b) 2 (c) ~ EXERCISE 20.9 Suppose a series RL circuit is composed of a 2.00-.0 resistor, a 15.0 H inductor, and a 6.00 V banery. (a) \-Vhal is the time constant for this circuit? (b) Once the switch is closed, how long does it take the current to reach half its maximum ,'allle?
Answers
(a) 7.505 (b) 5.20 s
20.8
ENERGY STORED IN A MAGNETIC FIELD
The emf induced by an inductor prevents a battery from establishing an instantaneous current in a circuit. The balter)' has to do \-"ark Lo produce a current. \Ve can think of this needed work as energy stored in the inducLor in its magnetic field. In a manner similar to that used in Section 16.9 to find the energy stored in a capacitor, we find that the energy stored by an inductor is Energy stored in an inductor
~
PEr.
Notice that the result is similar in form charged capacitor (Eg. 16.18):
Goal
the expression for the energy sLOred ina
Ma netic Ener
Relate the storage of magnetic energy to currents in an RL circuit.
Problem A 12.0-V bmtery is connected in series to a 25.0-0, resistor and a 5.00-H inductor. (a) Find the maxi~ mum current in the circuit. (b) Find the energy stored in the inclucwr at this time. (c) Hmv much energy is stored in the inductor when the current is changing at a rate of 1.50 A/s? Strategy In part (a) Ohm's law and Kirchhoff's voltage rule yield the maximum current because the voltage
,..
LO
[20.16]
PEr = ~C(1l V)'
Energy stored in a capacitor ~
EXAMPLE 20.10
= ~U'l.
across the inductor is zero when the current is maximal. Substituting the current into Equation 20.l6 gives the energy stored in the inductor. In part (c) the given rate of change of the current can be used Lo calculate the voltage drop across the inductor at the specified time. Kirchhoff's voltage rule and Ohm's law then give the current I at that time, which can be used [Q find the energy stored in the inductor.
......... ,
.
~
Solution (a) Find the maximum current in the circuit. Apply Kirchhoff's voltage rule to the circuit:
.6.~)att
+ .6.Vn + u\ft
= 0
M £-[R-L-=O
III
When the maximum current is reached, 6.1/6.1 is zero, so the voltage drop across the inductor is zero. Solve for the maximum current Jllla~:
£
12.0 V
J
= -R = -= 25.0 n
PEL
= ~Ll;,~ = &<5.00 H)(0.480 A)' =
max
0.480 A
(b) Find the energy stored in the inducwl- at this time.
Substitute known values into Equation 20.16:
0.576J
687
Summary
(c) Find the energy in the inductor \vhen the current changes at
Apply Kirchhoff's volt
1=
~ (EO R
L Cl/) Cli
I
= - - n [12.0 V - (5.00 H)(L50 A/s) ] = O.IRO A
25.0
Finally, substitute the value for the current into Equation 20.15, finding the energy stored in dle inductor:
PI', =
H
bU' =
b(5.00 1l)(O.180A)' = 0.081 OJ
,.-
...
Remarks Notice 11m" important it is to combine concepts from pre\'ious chapters. Here, Ohm's bw and Kirchhoff's loop rule were essential to the solution of the problem.
QUESTION 20,10 True or False: The larger the value oflhe inductance in such an RLcircuit, the larger the maximum currenl. EXERCISE 20,10 For the same circuit, Rnclthe energy stored in the inductor \...· hen t he rate of change of the current is 1.00 A/s. Answer
0.196J
SUMMARY 20,1
20,5
Induced emf and Magnetic Flux
The magnetic rlliX <1)/31hrough a closed loop is defined as (1)11
=
BA cos ()
[20,11
where B is (he strength of the uniform magnetic field, A is the cn~s-seclional area of lhe loop, and e is the an~1c between B and a direction perpendicular 10 the plane 01" the loop,
20.2
Faraday's Law of Induction
[20,2] where N is the number or loops in the circuit. The mag~ nc.:lic fltlX (!)/Jcan change with time whenever the magnetic ficld fi, the area 11, or the angle () changes \'lith time. Lenz's law o;tates that the current from lhe induced emf" creates a magnetic field with flux opposing the rhf/l/!{e in magnctic rlliX thrOllgh
20,3
Motional emf
If (l conducting bar of Icngt~ (i 1l100'es through a magnetic held with a "peed 11 so that B is perpendicular to the bar, the emf" induced in the bar, often called a motional emf, is
lei
~
Be"
e
(20.4]
= NUAw
sin wI
[20,7]
Such generators naturally produce alternating current (AC), which changes direction with frequency W/27T, The AC current can be transformed to direct current.
20.7
Faraday's law of induction slatcs that the instantaneous elllf induced in a circuit equals the negative 01" the rate of change ofmagnelic flux through [he circllit,
Generators
When ;-l coil of wire with N tUnlS, each of area A. rotates with cunstant angular speed w in a uniform magnetic field B.the elllfinduced in the coil is
RL Circuits
,"Vhen the current in a coil changes with time, an emf is induced in the coil according to Far
e
= -" -ClI Clt
[20,9]
where I. is the inductance of the coil. The 51 unit for inductance is (he henry (1-1); I H = I V· s/A, The inductance of a coil can be found from lhe expression
120,10] when; N is the number oftlll"lls on Lhe coil. f is the currellt in the coil, and
[20,l1aJ
688
Chapter 20
Induced Voltages and Inductance
If a resistor and induclOr arc connected in series to a baltery and a S,,,... ilCh is closed at I. = 0, the current in rhe circuit doesn"t rise inslantly to its maximum value. After one time constant T = LIR, the current in the circuit is 63.2% of its final value ej R. As the CUlTcnl approaches its final. maximum value, the voltage drop across the inductor approaches zero. The currenl lin such a circuit any time I after the eircl/it is completed is
e
1 = - (J
R
20.8
Energy Stored in
0
Magnetic Field
The energy stored in the magnetic field of an inductor carrying current I is [20.16] As the current in an RL circuit approaches ils maximum v,duc, the s[Qred energy also approaches a maximum value.
[20.15]
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.CDM
MULTIPLE-CHOICE QUESTIONS 1. \Vhal is the magnetic flux through a rectangle lying in the .\)-plane with lengrh 2.0 m and width 1.0 m if a uniform magnetic field has magnilUcie 3.0 T and makes an angle of 30.00 with respect to the -"y-plane? (a) 1.4 \'\-'b (b) 2.2 Wb (c) 3.0 "lib (d) 4.1 "lib (e) 5.2 Wb 2. A lO-lUrn square coil 0.5 In on a side lies in the xy-plane. A uniform magnetic field with magniLUde 1.0 T in the negative z-direction changes steadily to 3.0 T in the posili"e z-direction, a process that takes 2.0 s. \t\'hal is the magnitude of the induced emf in the coil? (a) 2.0 V (b) 5.0 V (c) 3.0 V (d) 4.0 V (e) 6.0 V
respect to the field. (d) Rotating the coil. (e) Removing the coil from {he field. 9. A rectangular loop is placed near a long wire carrying a current /J as shown in Figure MCQ20.9. If I decreases in time, what can be said of the current. induced in t.he loop? (a) The current. has a direclion that depends on the size of the loop. (b) The current is clockwise. (c) The current. is coullterclockwise. (d) The current is zero. (e) The magnitude of the current depends on the size of the loop. /I
3. A small airplane with a wingspan of]2 In flies horizontally and due north at a speed of 60.0 mls in a region where the magnetic fIeld of Eanh is 60.0 Jl,T directed 60.0° below the horizontal. What is the magnitude of the induced emf between the tips of the airplane's wings? (a) 51 mV (b) 31 mY (c) 37 IllV (d) 44 mV (e) 22 IllV 4. A generalOr cOlllains a 100-turn coil that rotates 10.0 times per second. If each turn has an area of 0.100 m 2 and the magnetic field through the coils is 0.050 0 T, \\'hatis the maximum eml'induced in the coil? (a) 31.4 V (b) 62.8 V (cj 0.314 V (d) 6.28 V (e) 3.14 V 5. The current in a 5.00-H inductor decreases uniformly at a rate of2.00 A/s. \Nhat is the voltage drop across the inductor? (aj 2.50 Ii (b) 5.00 V (cj OAOO V (eI) 10.0 Ii (e) 7.50 V 6. A 5.0~H inductor is connected in series with a IO.O~V battery, a switch, and a 2.5~n resistor. After closing the switch and completing the circuit, what is t.he estimated length ofrimc required for the current to reach half its maximum value? (Select the closest answer.) (a) 200 s (b) 50 s (c) 20 s (d) 2 s (e) 0.2 s 7. The current is doubled in an illductor. By what factor does the stored energy change? (a) ~ (b) 1 (c) 2 (d) 4 (e)
*
8. ·When a coil is placed in a conS[(lnt. external magnetic field directed perpendicular to the plane of the coil, which of the rollO\'v'ing actions does nol induce an emf in the coil? (a) Collapsing r.he coil. (b) Changing the strength of the magnetic field. (c) ;vloving the coil through {he field without changing its orienr.alion ,... itb
LJ -I
FIGURE MCQ20.9
10. Two rectangular loops of" "... ire lie in the same plane, as shown in Figure MCQ20.l0. If the current [in the ollter loop is counterclockwise and increases with time, what is t.rue of the current induced in the inner loop? (a) Ir is zero. (b) It is clockwise. (c) It is counterclockwise. (d) Its magnitude depends on the dimensions of the loops. (e) Its direction depends on the dimensions of the loops.
D FIGURE MCQ20.10
1 J. A bar magnet is held above the center of a wire loop lying in the horizontal plane, as shown in Figure MCQ20.11. The south end of" the magnet is tm...ard the loop. i\fter the magnet. is dropped, what is true of the current. in the resistor as viewed from above? (a) It is clockwise as the
db /I
FIGURE MCQ20.11
ConceptLJol Questions
magnet falls LOward the loop. (b) It is cOllmerclockwise as the magnet falls toward the loop. (c) It is c10cbvise arIel' the magnet has moved through the loop and moves away from it. (d) It is always clockwise. (e) It is first counterclock,,'ise as the magnet approaches the loop and then clockwise after it has passed through Ihe loop. 12. A circuit consists of a lightbulb conncctcd to two conducting rails and a conducting movable bar, as shown in Figure MCQ20.12. An external mag-netic field is dirccted perpendicular to the planc of the circuit. During which of the follmving actions will the bulb lighl up? (More than one statemem may apply.) (a) The bar is moved to the left. (b) The bar is moved to the righL
1. A circular loop is located in a uniform and constant magnetic fIeld. Describe how an emf Gill be induced in the loop in this situation.
2. Docs dropping a magnet down a copper tube produce a current in the tube? Explain.
3. A spacecraft orbiting Eanh has a coil of wire in iL An astronaut measurc.;s a sm"ll current in the coil, although there is no battery connected to il and there are no magnets in the spacecraft. "Vhat is causing the current? 4. A loop or wire is placed in a unifonll magnetic field. For what orientation of the loop is the magnetic nux a maximum? For what orientation is the flux zero?
5. As the conducting bar in figure CQ20.5 moves to the right, an electric field directed downward is set up. If the bar were moving to the left, explain why the electric field would be upward.
X
B;ll +
+
x
X
X
X
x
FIGURE CQ20.5
-
',l
FIGURE MCQ20.12
9. Eddy currents are induced currents set up in a piece of metal when it moves through a nonuniform magnetic tield. For example, consider the nat metal plate swinging at the end of a bar as a pendulum, as shown in Figure CQ20.9, At position 1, the pendulum is 1110\'ing from a region where !.!lere is no magnetic field into a region where the field Bin is directed into the paper. Show that at position I the direction of the eddy currenl is counterclockwise. Also, at position 2 the pendulum is movi ng out of the field into a region of zero field. Show that the direction of the eddy current is clockwise in this case. Use right-hand rule number 2 LO show that these eddr currents lead to a magnetic force on the plate directed as shown in the figure. Because the induced eddy current always produces a retarding force when the plate enters or leaves the field, the swinging plate quickly comes to rest.
X
X
X
X
X
X
"
" "" "" "" "" \\
v
E x
(c) The magniwde of the magnetic f-ield increases wilh time, (d) The magnitude of the magnetic field decreases w'ith lime. (e) The magnetic field remains constant in time.
\ PiVOl X
+ X
689
x
x
x
x
<)
\~
(Collceptual Questions'::' and 6)
6. As the bar in Figure CQ20.5 moves perpendicular to the ficld, is an external force required 1.0 keep it moving with constant speed? 7. \IVearing a metal bracelet in a region ofstrong magnetic field could be hazardous. Discuss this statement.
8. How is electrical energy produced in dams? (That is, hmv is the energy of motion of the water convened to AC electricity?)
FIGURE CQ20.9
10. A bar magnet is dropped toward a conducting ring lying on the noor. As thc magnet falls toward the ring, does it move as a freely falling o~ject? 11. A piece of aluminum is dropped vertically downward between the poles of an electromagnet. Does the mag-
nctic field affect the vclocil:)' of the aluminum? Hinl: Sce Conceptual Question 9.
690
Chapter 20
Induced Voltages and Inductance
12. Is it possible ro indllce a constant emf for an infinite amount of time: ]3. "Vhy is the induced emf that appears in an indllcwr called a back (collmer) emf?
14. ;\ magneto is used to cause the spark in a spark plug in many lawn mowers today, A magneto consists ofa permanent magnet mounted on a f1yv,"heel so that it spins past a fixed coil. Explain how t.his arrangemcm genera(es a large enough potential diffen~n('e to cause the spark.
-
PROBLEMS ~
Th~ Problcnl~ for this chapter mar be assigned online at \I\'ebAssigll. = straightforward, intcrmediate, ch~Jlkn~in!.\
y
WebAssign I.
2.:~
m
=
_
= (\enoll.;s enhanced (;on\ent problem
II = 0=
denotes guided problem
,
biomedical application denotes full solution available in Slur/rILl SOllllioll.1 ,'H(II/11f11/
,
S/lIdy GuidI'
e
SECTION 20.1 INDUCED emf AND MAGNETIC FLUX 1. A uniform magnetic field of l11agnillide 0.50 T is directed perpendicular LO the plane or a rectangular loop having dimensions 8.0 em by 12 em. Find the magnetic flux through the loop.
2. Find the flux of Earth's magnetic field of magnitude 5.00 X 10~:; T through a square loop of area 20.0 CIl1~ (a) whell the field is perpendicular to Ihe plane of the loop, (b) when the field makes a 30.0 0 angle with the normalLO the plane of the loop, and (c) whcn lhe f-ield makes a 90.00 angle with the normallO the plane. 3. A circular loop of radius 12.0 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane of the loop and the magnetic flux through the loop is 8.00 X 10-:1 T· m:!, what. is thc strength of the magnetic field? (b) If the magnetic field is directed parallel to the plane of the loop, what is Ihe magnetic flux th rough the loop?
4. A long. >;traight wire carrying a current of2.00 A is placed along the axis or a cylinder of r:ld illS 0.500 m ,I nd a length of3.00 m. Determine the total magnetic flux through the cylindcr.
rnA long, straight wire lies in the plane ofa circular coil \...·ith a radius of 0.010 m. The wire carries a currCIll of 2.0 A and is placed along a diameter of the coil. (a) What is the net flux through the coil? (b) If the wire passes through the center of the coil and is perpendicular to Ihe plane of' the coil, what is the net flux through the coil?
6. A ~IOO-turn solenoid oflenglh 36.0 em and I'adills 3.00 cm carries a current of 5,00 A. Find (a) the magnelic field strength inside thc coil at its midpoint and (b) the magnetic flux through a circular cross~sectiollal area of Ihe solenoid at its midpoint. 7. A cube of edge length f = 2.5 ( I l l is positioned as shown in F'iRure P20.7. There is a ullifonn Illagnctic fielel throughout the region with components B", = +5.0 T, R1 = +4.0 T, and B: = +3.0 T. (a) Calculate (he flux throuRh the shaded face of the cube, (b) ''\-'hat is (he total nux emerging frolllthe volume enclosed by the cuhc (i.e" the total nux through all six faces)?
I ,,
FIGURE P20.7
SECTION 20.2 FARADAY'S LAW OF INDUCTION 8. " Transcranial magnetic stimulalion (TMS) is a noninvasive t.echnique llsed LO stimulate regions of lhe human hrain. A small coil is placed on the scalp, and a brief burst of current in the coil pl"oduces a rapidly changing magnetic field inside the brain. The induced emf call he sufficielllLO slimulate neuronal aClivily. Olle sHch device generates a magnetic field within lhe brain lhat rises from Lcro to 1.5 T in 120 illS. Determinc the induced emf within a circlc of tissue of I-ad iliS 1.6 mm and that is perpendicular LO the direction or tile field.
9.
A rectangular coil is loc
10. The flexible loop in Figure P20.JO has a radius of 12 cm and is in a magnetic field 01" strength 0.15 T. The loop is grasped at poillls A and Band slretched tllltil its area is nearly zero. If iL takes 0.20 s to close the loop, what is the magllilHde of the average induct'd emf in it during this time?
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
FIGURE P20,lO
x
x
x
x
Problems 10. 12. alld 18
ImA wire loop or radius 0.30 III lics so thal an external magnetic field of magnitude 0.30 T is perpendicular to the loop. The field reven;es its direction, and its magnitude changes to 0.20 T in 1.5 s. Find the magnitude of lhe average induced emfin the loop durin~ this lime.
Problems
691
12. A circular loop of ,·.. ire of radius 12.0 em is placed ill a magnetic field directed perpendicular to the plane of the loop. as shown in Figure P20.1O. If the field decreases at the rate of 0.0500 Tis in some time imerval, what is the magnitude of the emf induced in the loop during this inten'al? 13. The plane of a rectangular coil, 5.0 Clll by 8.0 Cl~ is perpendicular to the direction of a magnetic field B. If the coil has 75 turns and a total resistance of 8.0 il, at what nIle must the magniLUde ofB change to induce a current of 0.10 A in the windings of the coil? 14. A square, single-turn wire loop 1.00 cm on a side is placed inside a solenoid that has a circular cross section of' radius 1.00em 3.00 cm, as shown in Figure P20.I4. The solenoid is 20.0 cm 1.00 ( I l l long and wound with 100 turns ol'wire. (a) If the current in the solenoid is 3.00 A, what is the FIGURE P20.14 flux through the loop? (b) [f the current in the solenoid is reduced to zero in 3.00 s, what is Lhe magnitude of the average induced emf in the loop?
~oo1
[7
15.
18.
A 300~tllrn solenoid ,.. . ith a length of 20.0 em and a radius of 1.50 em carries a currentof2.00 A. A second coil or four turns is ''''Tapped tightly around this solenoid, so it can be considered to have the same radius as the solenoid. The current in the 300-turn solenoid increases steildily to 5.00 r\ in 0.900 s. (a) Use Ampere's law to calcubte the initial magnetic field in the middle of the 300-wrn solenoid. (b) Calculatc thc magnetic field of the 300¥wrIl solenoid after 0.900 s. (c) Calculate the arC71 of the 4-turn coil. (d) Calculate the change in the magnetic flux through the 4-turtl coil during the same period. (e) Calculate the average induced emf in the 4-turn coil. Is it equal to the instantaneous induced emf? Explain. (f) \Vhy could contributions to the magnetic field by the current in the 4-turn coil bc neglected in this calculation?
II
To monitor the breathing or a hospital patient. a thin belt is girded around the patient's chest as in Figure P20.17. The belt is a 200-turn coil. \Nhen the patient inhales, the area encircled by the coil increases by 39.0 cm 2 . The magnitude or Earth's magnetic field is 50.0 pT and makes an anglc of 28.00 with the planc of thc coil. Assuming a patient takes 1.80 s to inhale, find lhe magnitude of the average induced emf' in the coil during that time.
mil An N¥LUl'l1 circular wire coil of radius r lies in the .\)'-plane. as shown in figure P20.lO. A uniform magnetic field is turned on, increasing steadily from 0 LO Bo in the positive z-direction in I seconds. (a) Find a symbolic expression ror the emf, e, induced in the coil in terms of the variables given. (b) Looking down on at the x)'-ph1l1e from the positive z-axis, is the direction of the induced CUITCI1l clockwise or counterclockwise? (c) If each loop has resistance R, Fllld an expression for the magnitude of the induced current, f.)
m
CI.[]A circular coil enclosing an area of 100 em:? is made of 200 turns of copper wire. Tile wire making up the coil has resistance of 5.0 fl, and the ends of the wire arc connected to form a closed circuit. Initially. a 1.1-T uniforrn magnetic field points perpendicularly upward through the plane of the coil. The direction of the field thcn reverses so that the final magnetic field has a magnitude of l.l T and points downward through the coil. If the lime required for the field to rncrsc directions is 0.10 s, what average current flows through the coil during that time? 17.
FIGURE P20.17
SECTION 20.3 MOTIONAL emf 19. A pickup truck has a width of 79.8 in. If it is traveling north at. 37 m/s through a magnetic field with vertical cornponem of 35 ,uT, what magnitude emf is induccd bet'..· een the driver and passenger sides of the truck? 20. A 2.00-111 length of wire is held in an east-west direction and moves horil.ontally to the north with a speed of 15.0 m/s. The vertical componelll of Earth's magnetic field ill this region is 40.0 ,uT directed downward. Calculatc the induced emf between the ends of the wire and determine which end is positive. [1IJAn automobile has a venical radio antenna 1.20 m long. The aUlomobile travels at 65.0 km/h on a horizontal road where Earth's magnetic field is 50.0 ,uT, directed toward the north and downward at an angle of 65.0° below the horizontal. (a) Specify the direction the automobile ~hould move so as to generate the maximum motional em r in the 7lntenna, with the LOp of the antenna positive relati,'e \.0 the boltom. (b) Calculate the magnitude of this incluced emf. 22.
mil An astronaut is connected to her spacecraft
by a 25-m-long tether cord as she and the spacecraft orbit Earth in a circular path al a speed of 3.0 X 10;1 m/s. At one instant, the vollage me..lsurcd betwcen the ends of a wire embedded in the cord is measured to be 0.45 V. Assume the long dimension of the cord is perpendicular La the vertical component of Eanh's magnetic field at that instant. (a) \·Vhat is the magnitude of the VertiGll component of Earth's field al this location? (b) Does the mea· sured voltage change as the system moves from one localion to another? Explain.
692
Chapter 20
Induced Voltages and Inductance
23. A Boeing-7l li jet with a wingspan of60.0 m i,." flying horifontally
29. Find the direclion of tile currenl in the resistor Rshown in Figure P20.29 tlfter each of the following sleps (takcn in (he order given): (a) The switch is closed. (b) The \'ariable resistance in sel'ies with the batten is decreased. (c) The cil'cuit containing resistor I? is moved to the left. (el) The switch is opened.
e
H
FIGURE P20.29
FIGURE P20.24
30.
Prohlems 24,32. ami 59
26.
v·
5
A conducting rectangular loop of mass M, resistance dimensions IV b~ l falb from rest ilHo a magnetic field B. as shown in Figure P20.30. During Ihe time interval before the lOp edge of th(: loop reaches the field. the loop approaches a terminal speed v.,. (a) Sho\\' thai
R.
SECTION 20.4 LENZ'S LAW REVISITED (THE MINUS SIGN IN FARADAY'S LAW) 125.IA hal' ma~llel is positioned ncar a coil or wire, as shown in Figure P20.25. What is the direction of the currenl in the resistor when the m
E
ar~
11.,. =
N
11
MglI B"!. 7II 'l
(b) \.vhy is 1), proportional proportional to If.??
FIGURE P20.25
(0
m (c) Wh}' is it inversely
-If/--
(11 Fignre P20.26 what is the direnion or the current induced in the resistor a( the instant the switch is closed?
1
L-,..... v· FIGURE P20.30
FIGURE P20.26
27. A copper bar is mo\ed lO the right while its axis is maimained in a direction perpendicular to a magnetic ficld, as ~hown in Figure:: P20.27. If the top of the bar becollles po~iti\'e rehltive to the boltom. what is the direction of the magnct ic field?
+
(1[]A I'eclangular coil ""ilh rcsislancc J? ha~ N lurn~. each of length and widlh w, as shown in Figure P20.31. The coil moves into a uniform magnetic field BWilh constam \'e1ocit)' v. What are the magnitude and direction of the tOlal magnelic force on lhe coil (a) a~ il enter:, the magnetic field, (b) as it moves wilhin the field, and (c) as it leaves lhe field?
e
+
- .,. FIGURE P20.27
B,n
28. Find the
{CG~C[
s~
e
FIGURE P20.31
e
11 FIGURE P20.28
•
32. m:J A conducting bar of length 1ll00'cs LO the right on t\\'o friClionlcss rails, as shown ill FiJ.{lll'C P20.2-1. A uniform magnelic field directed into the page has a magnitude of 0.30 T. Assume l = :\5 cm and N = 9.0 n. (a) At
Problems what const
693
voltage induced in the coil if the axis of rotation of the coil is along its m<~jor axis and is aligned (a) perpendicular to Earth's ma~netic field and (b) parallel LO Earth's magnetic field? (Nole: The area of an ellipse is g-iven by A = 7illb, where II is the length flf the JP1l1im'Uor axi~ and b is the length of the snmminor axi..;.)
SECTION 20.5 GENERATORS 33. A generator produce", 24.0 V when turning at 900 rc\·imin. What potemial difference does it produce \\hen turning at 500 rev/min?
SECTION 20.6 SELF-INDUCTANCE 40. A technician wraps wire aroulld a tube of length 36 cm h'H"in~ a diameter of R.O C111. When the windings are evenly spread over the full It'ngth of the LUbe. the result i'i a solenoid containing- 580 turns of wire. (a) Find the self.. inductance of thi~ solenoid. (b) If the current in this solenoid increases al the rate of 1.0 A r",. \\ hat is the selfinduced emf in the solenoid?
34. A 100-tlirn square wire coil of area 0.040 m~ rotate'i about a venical axis at I !l00 rev/min. as indicated in Fig-ure P20.34. The hori.lontal component of Earth's magnetic field at the location of the loop i'i 2.0 X 10 -!) T. Calculate the maximum emf induced in the coil by Earth\ fide\.
41. TIlt:' Cllnent in a coil change'i from 3.5 A
10 2.0 A in 0.50 .... If the average emf induced in the coil is )2 m\'. whal is the sdf-inductance of t he coil?
42,
E
Show that the twO expressions for inductance given
b) A/(Il 11 L=-1
and
L=
-I';
:'1/:'1
have the same units. FIGURE P20.34
35.
lit!
COllsickrahle scientific work is currend)' under way to determine whether weak oscillating magnetic fields
sitch
I :17. lin
a model AC gcneraLOr. a 500-turn r('ctangular coil 8.0 em h, 20 cm rotates at 120 re\/min ill a uniform magnetic field of 0.60 T. (a) What i') the maximum emf induced in the coil? (b) What is the instantaneous value of the emf in the coil at t = (1T/32) s? AS'iume the ('mf i'i .lero at t = O. (c) What is the smallest value of I for which I Ill' emf will ha\'e its maximum value:-
38. A mOlor has coils with a resistance of 30 nand opnate'i from a voltag-c of 240 V. \'\'hen the motor is oper;iling al it~ maximum speed. the back emf is 145 \'. Find the: cur· rent in the coils (a) when the motor is first wrned on and (b) when the motor has reached maximum ...peed. (c) If lht' current in the motor is 6.0;\ at some inslant. what i... the back emfal thai time: 39, A coil of 10.0 turns is in the shape of all ellipse haying a Im~jor axi'i of 10.0 cm and a minor axis 01" 4.00 cm. The coil l"OLal.esatlOO rpm in a region in which th(" magnitude of Earth's maglletic field is 55 J.LT. What is the maxi ilium
43, A solenoid of radius 2.5 CIll has 400 turns and a length of 20 elll. Find (a) its inductance :lnd (b) the rate at which Clirrent must change through it to produce an emf of 75 mV. QTIAn emf of 24.0 mV i'i induced in a 500-turn coil when the current is changing at a ratt: of 10.0 Ah. What is the magnetic nux through each turn orthe coil at an instal1l whell the current is 4.00 A?
SECTION 20.7 RL CIRCUITS
45. A battery is connected ill series with a .:to-O resistor and a 12-mH inductor. The maximum cuncnt in the circuit is 150 mAo Find (a) the time constant of this circuit and (b) the emf of the battery.
46. An RL circuit wiLli L = 3.00 J-I and an RC circuit with C = 3.00,u.F ha\'e tht: same time comtant. If the two circuits ha\'e the same resistance (a) what i\ the value of Rand (b) what is this common time consl ..U1t?
n,
QI]A batten' is connected in series with a 0..30-0 resistor and an inductor, as shown in Figure 20.27. The 'iwitch is closed at I = O. The time constant of the circuit is 0 . 25 l), and the maximum current in the circuit is R.O A. Find (a) the emf of the batten', (b) lhe inductance of the circuit. (c) the current in the circuit after one time comtal1l hal) elapsed, and (d) rhe voltag-e across the resi'iLOr and the vollage aeros .. the indunor arlt'r one lime con ..tant has elapsed. 48. A 25-mll inductor. all R.o·n resiHor, ,md a f').O-\' batlery arc connected in series. The switch is closed at t = O. Find, he mILage drop acros\ the re..islo," (a) at t = 0 and (b) after aile time constalll. has pil'ised. Also, find the voltage drop across the inductor (c) at 1= 0 ;lnd (d) after one time constant has elapsed. 49. Calculate the resistano..' in an RL circuit in which L = 2.50 H and the current increases to 90.0% of it.. final value in 3.00 s.
694
Induced Voltages and Inductance
Chapter 20
50. Consider the cirellil shown in Figure P20.50. Take e = 6.00 V. L ~ 8.00 m H, and R ~ 4.00 n. (a) What is the inductive lime conSLalll of the circuit? (b) Calculate the current in the circuit 250 J.LS after the switch is closed. (e) What is the value of the final steady-slate currclll? (d) How long does it take (he current lO reach 80.0% of its maximum value? 5
L
R FIGURE P20.S0
SECTION 20.8 ENERGY STORED IN A MAGNETIC FIELD
56. A 500-turn circular-loop coil [5.0 cm in diameter is ini~ tially aligned so that its axis is parallcllo Earth's magnetic field. In 2.77 inS the coil is Ripped so that its axis is perpendicular LO Earth's magnetic field. Ifan average voltage 0[0.166 V is thereby induced in the coil, \\lh
51. (a) If an inductor carrying a 1.70-A current SI01"eS an energy of 0.300 IlD. what is its inductance? (b) How much energy does the same inductor store if it carries a 3.0-A elilTent?
24.0 !1 820 turns I
IM!1
152.IA 300-ulrll solenoid has a radius of 5.00 em and a length of 20.0 em. Find (a) the inductance of the solenoid and (b) the energy stored in the solenoid when the Cl.lrrelll in its windings is 0.500 A. 53. A 24-V battery is conncacd in series with " resistor and an inductor, with R = 8.0 nand L = 4.0 H, respectivcly. Find the energy slored in the inductor (a) when the current reaches irs maximulll value and (b) one tin)c constant af"ter the switch is closed. 54.
60.0 V
58.
mill
59.
mil
to
ADDITIONAL PROBLEMS
55. In Figure P20.55 the bar m
'lOii::
S
r-.lotion tow,lrd the loop
11 b
FIGURE P20.SS
[JG' "on"
12500
FIGURE P20.57
Ell
A 60.0-m length of insulated copper wire is wound form a solenoid of radius 2.0 cm. The copper wire has a radius of 0.50 111m. (a) Whal is the resistance of the wire? (b) Treating each turn of the solenoid as a circle, how many turns can be made with the wire? (c) How long is the resulting solenoid? Cd) What is [he self-inductance of the solenoid? (c) If' the solenoid is attached to a batter)' with an emf of 6.0 V and internal resistance of 350 mfl, compute the time constant of the circuiL. (f) What is the maximum current attained? (g) How long would it take to reach 99.9% or its maximum current? (h) 'What maximum energy is stored in [he inductor?
•
A spacecraft is in a circular orbit of radius 3.0 X ]0-1 kill around a 2,0 X 10:\0 kg pulsar. The magnetic field or the pulsar at that radial distance is 1.0 X 10 2 T directed perpendicular to the velocity of the spacecraft. The spacecraft is 0.20 km long with a radius of 0.040 km and moves counterclockwise in the X)-planc around the pulsar. (a) 'What is the speed of the spacecraft? (b) If the m'lgnetic field points in the positive ~-direClion, is the emf induced from the back w the front of the space~ craft or from side LO side? (c) Compute the induced emf. (d) Describe the hazards for astronauts inside any spacecraft moving in the vicinityofa pulsar.
e
A conducting rod of length moyes on two horizontal frictionless rails, as in Figure P20.24. A cons!
60. \Vhen the coil of a mOLOr is rotating at maximum speed, the current in the windings is 4.0 A. \-"'hen the motor is first turned on, the current in the windings is 11 A. Tf"the motor is operated at 120 V, find (a) the resistance of the windings and (b) the back emf in the coil at maximum speed, lIDThe bolt of lightning depicted in Figure P20.61 passes 200 m from
Problems in 10.5 /.lS, what is the average voltag-e induced in the coil? Assume the distance 10 the center of the coil determines the <wcntgc magnetic field at the coil"<; position. Trealtht.' lightning boll as a long, \'cnical wire.
FIGURE P20.61
62. The square loop in Figure
.1
P20.62 is made of wires with
" -----"7""":------ IJ 2.00 em FIGURE P20.63
64. An aluminum ring of radius 5.00 em and rcc;ist
/
0;.00 un
--
J
1 •
FIGURE P20.64
1 ---.1 3.00 em
-I
695
noid is increasing at
FIGURE P20.65
66. GEl An iV-turn square coil with side [and resistance R is pulled lU the rig"l11 at constant speed v in the posilive x-direction in the presence ora uniform magnetic field B acting perpendicular to the coil. a~ shown ill Figtll'e P20.66. At 1= 0, the right side of the coil is althe edge of the ficld. After a time I ha" elapsed, the entire coil is in the region where B = O. In lerms of the (Iuantitics lV, 8, [, ll, and R. find ~\ mbolic exp"cssions for (a) the magnitude ofthe induced emfin the loop dul'ing the time intcr\"<\1 t, (b) the magnitude of the induced currcnt in the coil. (c) the power delivered to the coil. and (d) the force required LO remO\'e the coil fmm the field. (e) What i.. the direction of the induced currem in the loop? What is the direction of the magnetic force on the loop \\hile it i" beinf.4 pulled out of the ficld~
8".
I
x
x
x
x
x
X
x
X
X
X
X
X
X
X
X
x
x
x
x
x
x
X
8,0
x
FIGURE P20.66
F.• I'!'
Arecibo, a large radio telescope in Puerto Rico, gathers electromagnetic radiation in the form of radio waves. These long wavelengths pass through obscuring dust douds, allowing astronomers to create images of the core region of the Milky Way galaxy, which can't be observed in the visible spectrum.
21.1
Resistors in on AC Circuit
21.2
Capacitors in an AC Circuit
21.3
Inductors in on AC Circuit
21.4
The
21.5
Power in an AC Circuit
21.6
RLC Series Circuit
Resonance in a Series
RLC Circuit
21.7
The Transformer
21.8
Maxwell's Predictions
21.9
Hertz's Confirmation of Maxwell's Predictions
21.10
Production of Electromagnetic Waves by on Antenna
ALTERNATINGCURRENT CIRCUITS AND ELECTROMAGNETIC WAVES Every time we turn on
0
television set, a stereo system, or any other electric appliances, we
call on alternating currents (AC) to provide the power to operate them. We begin our study of AC circuits by examining the characteristics of a circuit containing a source of emf and one
22.11
Properties of Electromagnetic Waves
other circuit element: a resistor, a capacitor, or an inductor. Then we examine what happens
21.12
The Spectrum of Electromagnetic Waves
to simple series configurations of the three kinds of elements.
21.13
The Doppler Effect for Electromagnetic Waves
when these elements are connected in combination with each other. Our discussion is limited We conclude this chapter with a discussion of electromagnetic waves, which are composed of fluctuating electric and magnetic fields. Electromagnetic waves in the form of visible light enable us to view the world around us; infrared waves warm our environment; radiofrequency waves carry our television and radio programs, as well as information about processes in the core of our galaxy; and X-rays allow us to perceive structures hidden inside our bodies and study properties of distant, collapsed stars. Light is key ta our understanding of the universe.
21.1
RESISTORS IN AN AC CIRCUIT
An AC circuit consisls of combinations of circuit elemems and an AC generator or an AC source, which provides the alternating current. \Ve have seen that the out~ put oran AC gcneraLor is sinusoidal and varies with time according to
llV = dV;nax sin 27TJt
[21.1]
where Llv is the instantaneous voltage, dl111lax is the maximum voltage or the AC generator, andJis the frequency at which the voltage changes, measured in hertz (Hz). (Compare Equations 20.7 and 20.8 with Equation 21.1.) VI/e first consider a
696
21.1
-e--),
simple circuit consisting of a resistor and an AC source (designated by the symbol
Resistors in an AC Circuit
697
1"'----1<."1<-----I
as in Active Figure 21.1. The current and the voltage versus time
across the resistor are shown in Active Figure 2] .2. To explain the concept of alternating cun-ent, we begin by discussing the current versus time curve in Active Figure 21.2. At poi III (J. on the curvc, the currcnt has a maximum valuc in one direction, arbitrarily called the positive direction. Between points a and b, the Cllrrent is decreasing in magnitude but is still in the positive direction. At point b, the current is momentarily zero; it then begins to increase in the opposite (negative) direction between points band c At point c, the current has reached its maximum value in the negative direction. The current and voltage are in step with each other because they vary identically with time. Because the current and the voltage reach their maximum values at the same time, they are said to be in phase. Notice that the average value of the current over one cycle is zero because the current. is maintaincd in one direction (the positive direction) for the same amount of time and at the same magnitude as it is in the opposite direction (the negative dircction). The dircClion of the current, howevcr, has no effect on the behavior of the resistor in the circuit: the collisions between e1eclrons and the fixed atoms of the resistor result in an increase in the resistor's temperature regardless of the direction of the current.. We can quantify this discussion by recalling that the rate at which electrical energy is dissipated in a resistor, the power CZP, is
L -_ _--{~)_--~
Av =n V,lla-'t sin 2Trfl ACTIVE FIGURE 21.1 A series circuit consisting of a resislor Nconncclcd La an AC generator, designated by the symbol
--e---
'!P = i'R where i is the instantaneous current in the resistor. Because the heating effect of a current is proportional to the squrt11!ofthe current, it. makes no difference whether the sign associated with the current is posil.i,·e or negative. Thc hearing effect produced by an alternating currcnt with a maximum value of Iula >: is no/. thesameas that produced by a direct current of the same value, however. Thc rcason is that the alternating current has this maximum value for only an instant of time during a cycle. The important quantity in an AC circuit is a special kind of average value of current, called the rms current: the direct current that dissipates the same alllount of energy in a resistor that is dissipated by the acwal ailernating current. To find the rms current, we first square the current, then find its average value, and finally take the square root of this average value. Hence, the rms current is the square root of the average (mean) of the square of the current.. Because i 2 varies as sin 2 27rft, the avcrage value of i 2 is 4J~1il){ (Fig. 21.3b, page 698),l Thercfore, the rms current 'rills is related to the maximum value of the alternating current JlllilX by [21.2]
This cquation says that all alternating current with a maximulll value of 3 A produces the same heating effect in a resistor as a direct current of (3/V2) A. We can therefore say that the average power dissipatcd in ,,1 resistor that carries alternating current fis
IWe call show lhal (i~)", "" J';",./'t. ;IS follows: Till: currenl in Ihe circuil \ with lime accordillg In the c:xpressil,n i I",., sin 2rrJI, ~o j'! I;:',,~ sin 2 2rrJt. Then:fore, \,·c can lind Ihe average value of t~ bl' calculating lhe a\'eragt: value of sin 2 211Jt. :--Iale thai a graph 01" cos 2 2..JI versus time is identical to a g:raph of ,in 2 2rrjlvcrsu" limc, cxcepl Ihal Ihc poi III s arc sh iftcd "n lhe lim(; axis. Th liS, Ihe Ii mc ;;Vl:ragt: or .~i n 2 '!.'rrfl is equal tu I he f i lIle avcrag-e of cu.,,~ 2-;;JI, wken 0\·(:1" aile or mort: cycles. Th:u is,
=
=
(Sill~ 27TJt)~"
Wil h Lhis fact and lhe I ri~onome, ric illeillill'
Sill~
(sill'! ?:"jlJ,.-
+
= (cos'! 2TrJI)',
0 I- cos 2 0 = I. we gel (cos'! '2.Trft)., = 2(.~iIl2 ':!.7Tfl)." = I
(siTl~ 271ft)., = ~ W\1('11 tlli, re.'>1111 i~"llhsli1\llcd illl" llw expression ,~= I~,., sin~ 2rrfl, we get (1"2) .., =
1"",,1 v'Q, where 1,,,,, is ,llc rmSClirrenl.
I;,,,,
=1~,,,,1'!..
or 1m,. =
ACTIVE FIGURE 21.2 A pIal of current and voltage across a resistor versus time.
698
Chopter 21
Alternating-Current Circuits and Electromagnetic Waves
FIGURE 21.3 (a) 1"10lofthe currell! in it resistor as ;t funnioll uf lime. (b) Plot of the sfjuarc or the nlrn:::lll in a resistor as a fU1lction of time. Notice thaI lhe grar shaded rcgilills UlIdP.Tlhe Cllrve and nbrltll!th(: dashed Jille for {~,~./2 have the same area as the gray sh,Hlt;d regions above Ihe curve and below the dashed line Illr 1~,..:J'l..
ThIIS, the average value of /2
,-.<,
is [~,.J2.
(h)
Alternating voltages arc also best discussed in terms of rills voltages. with a relationship identical La the preceding one, rms voltage
-+
T~!.li..
Notation Used in This
Chapter Voltage Current
Instantaneous value
dV
Ma.ximum
dV,ll""
J",,",,"
value rms value
~\'rn"
l'IIls
[21.3]
where av,-m.~ is the rm.'i volLage and 6.lt;lIa~ is the maximum value of the alternating voltage. When we speak of mcasuring an AC voltage of 120 V from an electric outlet, we aelUall}' mean an rms voltage of L20 V. A quick calculation using Equation 21.3 shows that such an AC voltage actually has a peak value of about 170 V. 1n this chapter we usc rms values when discussing alternating currents and voltages. One reason is that AC ammet.ers and voltmeters are designed to read rms values. Further, if we use !"ms values, many of the equations for alternating currem will have the same form as those used in the study 01" direct-current (DC) circuits. Table 21.1 summarizes the notations Iisecl throughout this chapter. Consider the series circuit in Figure 21.1, consisting of a resistor connected to an AC gt;nerator. A resistor impt;des the current in an AC circuit,just as it does in a DC circuit. Ohm's law is therefore valid [or an AC cil-cuit, and ,ve have [21.4a] The rms voltage across a resistor is equal to the rms current in the circuit times the resistance. This equation is also true i[ maximllm values of current and voltage are used:
[21.4b]
QUICK QUIZ 21.1 W·hich of the following statements can be true for a resistor connected in a simple series circuit to an operating AC generatOl-? (a) {jP'J\. = 0 and iav = 0 (b) fJP.". = 0 and i av > 0 (c) CZPa ,. > 0 and (\I" = 0 (d) CJPav > 0 and lav> 0
EXAMPLE 21.1 Goal
What Is the rms Current?
Perform basic AC circuit. calculations lor a purely resistive circuit..
Problem An AC voltage source has an output of llv = (2.00 X 102 V) sin 27TJI. This source is connected 1.00 X 10 2 n resistor as ill Figure 21.1. Find the nns voltage and nns current in the resistor.
LO
a
Strategy Compare the expression for the voltage output just given with the general form, llv = 6.V:l1
21.2 Capacitors in on AC Circuit
.
-
~
Solution Obtain the maximum \'oltage by compadson of the gi\'cn expression for the output with the general expression:
............
.lv = (2.00 X 102 V) sin 27iJI
2.00
Next, substitute into Equation 21.3 to find thc rms voltage of the source: Substitute this result into Ohm's law current:
10
hndthe nns
699
X
dv = dt: nax sin 27fjt
]0' \'
v'2 141 V 1.00 X 10 2 [)
141 V
1.41 A
............ _ ...
Il.. • • • • • •
~
~
Remark ),Jotice how the conccpt of rms values allows the halldling of an AC circuit quantitatively in much the same way as a DC circuit. QUESTION 21.1 True or False: The rms current in an AC circuit oscillates sinusoidally \·. . ith time. EXERCISE 21.1 Find the maximum current in the circuit and the a\'erage power delivered to the circuit.
Answer
2.00 A; 2.00
X ]0'
W
APPLYING PHYSICS 21.1
ELECTRIC FIELDS AND CANCER TREATMENT
Cancer cells rnul1iply far more frequently than most normal cells, spreading throllghoutthe body, using its resources and interfering wilh normal functioning. Most therapies damage both cancerous and healthy cells, so finding methods that target cancer cells is important in dc\'eloping bcttcr treatments for the disease. Because cancer cells multiply so rapidly. it's natural to consider treatments that prevcnt or disrupt cell division. Treatments such as chemotherapy interfere with the cell division cycle, but can also damage healthy cells. It has rccently been found that alternating electric j"ields produced by AC currents in the range of 100 kHz can disrupt the cell division cycle, either by slOWing the division or by causing a dividing cell to disintegrate. Healthy cells that divide at only a very slow rme arc less vulnerable than the rapidlydividing cancer cells, so such therapy holds out promisc for certain types of cancer.
21.2
III
The alternating electric fields are thoughtlO affect the process of mitosis, which is the dividing of the cell nuclells into two sets of identical chromosomes. ear the end of the first phase of mitosis. called the prophase, the mitotic spindle forms. a struClllre of fine filaments that guides the two replicated sets of chromosomes into separate daughter cells. The mitotic spindle is made lip of a polymerization of dimers of LUbulin, a protcin with a large electric dipole moment. The alternating electric field exerts forces on these dipoles, disrupting their proper functioning. Elcctric field therapy is especially promising for the treatment of brain tumors because healthy brain cells don't divide and lherefore would be unharmed by thc alternaling electric fields. Research on such therapies is ongoing.
CAPACITORS IN AN AC CIRCUIT
To understand the cffect of a capacitor on the behavior of a circuit containing an AC \"oltagc source, we first revicw what happens ,.. . hen a capacitor is placed in a circuit containing a DC source, such as a battery. ""hen the s,.".itch is closed in a series circuit containing a batl.ery, a resisl.or. and a capachor, the initial charge
700
Chapter 21
Alternating-Current Circuits and Electromagnetic Waves
1-1'- - - ~"('--~'I ,-------j"f----, r;
'---~~}---~
FIGURE 21.4 ,\ series circlIit consisting of a capacitor (; cOlllleClcd to an AC generator.
The voltage across a capacitor logs the current by 90 0 ~
Capacitive reactance
on the plates of the capacitor is zero. The Illotion of charge through the circuit is therefore relatively free, and there is a large current in the circuit. As more charge accumulates on the capacitor, the voltage across it increases, opposing the current. After some time interval, which depends on lhc time constant Re, the current approaches zero. Consequently, a capacitor in a DC circuit. limits or impedes the current so that it apprw.lches zero arter a brief time. NO\v consider the simple series circuit in Figure 21.4, consisting ofa capaciwf connected 1.0 an AC generalOr. Vie sketch curves of CUITent versus time and voltage versus time, and then attempt to make the graphs seem reasonable. The curves arc shown in Figure 21.5. First, notice that the segrnent of the current curve from a to b indicates that the Cllrrent starts out at a rather large value. This large value can be understood by recognizing that t.here is no charge on the capaciwr at I = 0; as a consequence, there is nothing in the circuit except the resistance of the wires to hinder the flow of charge at this instant. The current decreases, however, as the voltage across the capacitor increases from eta d on the voltage curve. When the voltage is at point d, the current reverses and begins to increase in the opposite direcrion (from b to I' on the current curve). During this time, the voltage across the capacitor decreases t"rom d to fbecause the plates are now losing the charge they accumulated earlier. The remainder of the cycle for both voltage and current is a repeat of what happened during the first half of the cycle. The current reaches a maximum value in the opposite direction at point e on the current curve and then decreases as the voltage across the capacitor builds up. In a purely resistive circuit, the currcnt and voltage are always in slep with each other. That isn't [he case when a capacitor is in the circuit. In Figure 21.5, when an alternating voltage is appllcd across a capacitor, the voltage reaches its maximum value one-quarter of a cycle after the current reaches its maximulll value. We say that the voltage across a capacitor always lags the current by 90°. The impeding effect ofa capacitor on the current. in an AC circuit is expressed in terms ora factor called the capacitive reactance Xc> dcfined as
~
Xc -
1 27rJC
[21.5)
\,\'hen Cis in farads and lis in hertz, the unit of" Xc is the ohm. Notice that 27iJ= w, the angular rrequency. From Equation 21.5, as the frequency of the voltage source increases, the capacitivc reactance Xc (the impeding effect of the capacitor) decreases, so the current increases. At high frequency, there is less time available to charge the capacitor, so less charge and voltage accumulate on the capacitor, which uanslates into less opposition to the flow or charge and, consequently, a higher current. The analogy between capacitive reactance and resistance means that we can write an equation of the same form as Ohm's law to describe AC circuits containing capacitors. This equation relates the nns voltage and rills current in the circuit to the capacitive reactance:
.r
FIGURE 21.5
Pluts of curn:nt and
voltage ,lcross a rapaCilOl" \'I'rsus lilTW in an .. \C circuit. The milage lags I he currem by 90~.
EXAMPLE 21.2 Goal
[21.6)
A Purel Co ocitive AC Circuit
Perform basic AC circuit calculations for a capacitive circuit.
Problem An 8.00-JLF capacitor is COlll1eCleclto the terminals of an AC generator with an rills voltage of 150 X 10' V and a frequency of60.0 I-Iz. Find the capacitive reaClance and the rms current in the circuit. Strategy Substitute values into Equations 21.5 and 21.6.
21.3
Inductors in an AC Circuit
.
r"' . • . . • . .
Solution
Substitute the values ofJand C into Equation 2 I.5:
Solve Equation 2l.6 for the current and substiwte the values ror X(:and the nns volLage to find the rms current:
I Xc = - . 21'JC
l
= -::--;-::-::-:--cc-c-:-::-:-,--------,o--,21'(60.0 l-lz)(8.00 X 10 "F) 1.50 X 10' V 332 fl
,
332
..
~
n
0.452 A
.
11.. • • • • • • • •
Remark
701
~
Again, notice how similar the technique is to that oranalyzing a DC circuit with a resistor.
QUESTION 21.2 True or False: The larger the capacitance ora capacilOr, the larger the capacitive reactance. EXERCISE 21.2 If the Crequency is doubled, what happens
LO
the capacitive reactance and the rms current?
Answer
Xc is halved, and
21.3
INDUCTORS IN AN AC CIRCUIT
'nils
is doubled.
No'lo\' consider an AC circuit consisting only of an inductor conneCted to the terminals oran AC source, as in Active Figure 21.6. (In £Illy real circuit there is some resistance in the ,,,,'ire Conning the inductive coil, but 'lo\ie ignore this consideration for now.) The changing current output of the generator produces
.
=
f!,.[
1.f!,.l
[21.7]
The effective resistance of the coil in an AC circuit is measured by a quantity called the inductive reactance, XI,: [21.8]
1+'
llv,.
·>l1
I.
'----{~}-----'
dv =6. ~llax sin 271ft ACTIVE FIGURE 21.6 A series circuit consisting or an inducLOr Lconllcctcd to an ACgcllL:rator.
When lis in hertz and L is in henries, the unit of XI. is the ohm. The inductive reactance increases with increasing frequency and increasing inductance. Contrast these facts v.'ith capacitors, where increasing frequency or capacitance decreases the capacitive reactance. To understand the meaning of inductive reactance. compare Equation 21.8 with Equation 21.7. First, llote [rOll] Equation 21.8 that the inductive reactance depends 011 the inductance L, which is reasonable because the back emf (Eq. 21.7) is large for large values of /.,. Second, note that the inductive reactance depends on the frequency f This dependence, too, is reasonahle because the back emf depends 011 1:11/l:1l, a quantity that is large when the current changes rapidly, as it \\'0111d for high frequencies. vVith inductive reactance defined in this \Vay, we can ,,,'rite an equation of the same form as Ohm's law for the voltage across the coil or inductor: [21.9]
where I:1VLrrns is the rms voltage across the coil and Inns is the rills current in the coil. Active Figure 21.7 shows the instantaneous voltage and instantaneous current across t.he coil as functions of time. vVhen a sinusoidal voltage is applied across all inductor, the voltage reaches its maximum value one-quarter of an oscillation period before the current rea.ches its maximum value. In this situalion '.. .e say that the voltage across an inductor always leads the current by 90°.
ACTIVE FIGURE 21.7 I'lols of currellt alld voltage across illl inductor versus time in an AC cirnlit. The voltagc leads' Iw currem by 90.
Chapter 21
702
Alternating-Current Circuits and Electromagnetic Waves
To sec ·.vhy there is a phase relationship between voltage and current, we examine a few points on the curves of Active Figure 21.7. At poim a. on the current curve, the curre11l is beginning to increase in the positive direction. At this instant the rate of change 01" current. !J.l/8.1 (the slope of the current curve), is at a maximum, and \\Ie see from Equation 2].7 thaL the voltage across the induclor is consequently also al a mllximulll. As the current rises between poims a and b on the curve, I1J/l1t gradually decreases until it reaches zero at point b. As a result,
the voltage across the inductor is decreasing during this same time interval, as the segment bCl\veen c and d on the voltage curve indicates. Immediately after point b, the current begins to decrease, although it still has the same direction it had during the previous quarter cycle. As the current decreases to zero (from b to eon the curve), a voltage is agai 11 induced in the coil (frol11 d to J), bm the polarity of this voltage is opposite the polarity of the voltage induced bctween ("and d. This occurs because back cmfs always oppose the change in the current. "Ve could continue to examine other segments of the curves, but no new information would be gained because the current and voltage variations arc repetitive.
EXAMPLE 21.3 Gool
A Purely Inductive AC Circuit
Perform basic AC circuit calculations for an inductive circuit.
Problem In a purely inductive AC circuit (sec Active Fig. 21.6), L = 25.0 mH and the rillS voltage is .1.50 X 102V. Find the inductive reactance and rms current ill the circuit if the frequency is 60.0 Hz.
,..
.
• • • • • • • • • • • • • • ' '111
Solution Substitute L and limo Equation 21.8 to get the inducti\'c reactance:
Solve Equation 21.9 for the rms current and substitute:
X" = 2rrJL = 2rr(60,0 5- 1)(25.0 X 1O-3 H) = 9.42 n Li VLnns 1.50 X ]0' V I m,s = -X-- = 9.42 n I.
Remark The analogy with DC circuits is even closer than in the capacitive case because in the inductive equivalent of Ohm's law, the voltage across an inductor is /JrojJortionalto the inductance L.just as the voltage across a resistor is proportional to R in Ohm's law.
QUESTION 21.3 True or False: A larger inductance or frequenc), results in a larger inductive reactance. EXERCISE 21.3 Calculate the inductive reactance and rms current in a similar circuit if the frequency is again 60.0 Hz, but the rms voltage is 85.0 V and the inductance is 47.0 m H. Answer
XL = 17.7 n, 1= 4.80 A
21.4 /1
L
c
ACTIVE FIGURE 21.8 A series circuit consisting ora resistor, an inductor, ;mcl a capacitor connected to an AC generator.
THE RLC SERIES CIRCUIT
In the foregoing sections we exarnined the effects oran inductor, a capacitor, and a resistor \vhen they are connected separately across an AC voltage source. \OVe no\\! consider v•.'hat happens ,-.. . hen these elements are combined. Active Figure 21..8 shows a circuit containing a resistor, an inductor, and a capacitor connected in series lIcross an AC source that supplies a total voltage ~v at some instant. The current in the circuit is the same at all points in the circuit at any instant and varies sinusoidally with time, as indicated in Active Figure 21.9a. This fact can be expressed mathematically as
i = J lllax sin 27Tfl
21.4
Earlier, we learned that the voltage across each clement mayor may not be in phase with the current. The instantaneous voltages across the three elements, shown in Active Figure 21.9, have the following phase relmions to the instantaneous current:
The RLC Series Circuit
la)
;M", V, A ,
703
/'
'CT'
(hi
1. The instantaneous voltage !1VJI across the resistor is in PhflSIJ with the instantaneous currell!.. (Sec Active Fig, 21.9b.) 2. The instantaneous voltage dvi. across the inductor Leads the current by 90°. (See Active Fig. 21.90.) 3. The inslantancous voltage !1v(;across the capacitor lflgs the current by 90°. (See Active Fig. 21.9d.)
The net instantaneous voltage 111J supplied by the AC source equals the sum of the instantaneous voltages across the separate elements: ~v = dvu + ~1JC + drJ1,' This doesn't mean, however, that Lhe voltages rneasured with an AC voltmeter across R, C, and L Slim to the measured source voltage! In faet, the measured voltages don"t sum 10 the measured source voltage because the voltages across R, C, and L all have different phases. To account for the different phases or the voltage drops, \\le use a t.echniquc involving vectors. We represent the voltage across each clement with a rotating vector, as in Figure 21.]0. The rotating veClOrs are referred to as phasors, and the diagram is called a phasor diagram, This particular diagram represents the circuit voltage given bY' the expression ~v = ~V:II'IX sin (27T'ft + eM, where ~V:llax is the maximum voltage (the magnitude or length of the rotating vector or phasor) and 4> is the angle between the phasor and the positive x-axis when 1= O. The phasal' can be vie\'...ed as a vector of magnitude 6.\J;rm rOlating at a constant frequency [so that its projection along the y-axis is the instantaneous voltage in the circuit. Because 4> is the phase angle between thc voltage and current in the circuit, the phasor lor the currcn! (not shown in Fig. 21.10) lies along the POSilivc x-axis when l = 0 and is exprcssed by the relation i = 'm;\l< sin (27rJl). The phasor diagrams in Figure 21.] I are Llseful for analyzing the .w~rir~s RLC circuit. Voltages in phase \·...ith t he current are represented by vectors along the positive x-axis, and voltages out of phase with the current lie along other directions. ~VR is horizontal and 1.0 t.he right because it's in phase v.lith the current. Likewise, ~VL is represented by a phasor along the positive y-axis because it leads the current by 90°, Fin:: ":. AI~.:- along the negative )'-axis because it lags the ClilTent~ by 90°. If the pha501-5 are added as vector quantities so as to account for rhe different. phases of the voltages across R, L, and C, Figure 21.11a shows that the only x-component lor the voltages is d,Vu and the net y-component is dVI . - AVe. \,Ve now add the phasors vectorially 1.0 find the phasor AV:nax (Fig. 21.1]b), which represents the maximum voltage.
(e)
Id)
(e)
ACTIVE FIGURE 21.9 I'hase relations in dw series ILI,CcirCUil shown in Anivc Figure 21.8.
I(Hz)
----1iL---l..-----x
FIGURE 21.10 A phasordiagr:tlll ror the \'olla~l" in all ACcirclIit, \\'here
!ill
is the instan·
voltage.
FIGURE 21.11 (a) A phasor diagram for lhe ULCcircllil. (b) Additi01l of the ph~lsors as nXlOrs giw·s .i \~".~ = Vol V,,? + (.i \I, -
.I' ~v,
I------<---x (b)
Ie)
la)
~t\ 11ll1Cmnl1;C 1O help you n:ll1cllIbcr Ih~ ph,l~l" rdaliullships in lirc circuits is "HI./lhc fCE rnarl.~ I~ rcpresl'lllS the "'011:\1-\(' 8, 1111e CIl'TCIIL. I.tll\' illllul-lallu', .\11(1 (; Ihe r;ap:,cilanCt', TIlliS. Ihe 11:111\(' J:/.lll1,'alls that ill :In i'ulllCli\(' ,ircuil, Ihe v\,ltagl· 8 lC;\fh the Cl11'r..111 In ,I capacitive circuit ICE (llcal'S that Ihe 111rn·111 I,'ads lh,' vnlragl',
r,
704
Chapter 21
Alternating-Current Circuits and Electromagnetic Waves
The right triangle in Figure 21.11b gives the following equations for the maximum voltage and the phase angle qJ between the maximum voltage and the current: [21.10] [21.11] In these equations, all voltages are maximum values. Although we choose to use maximum voltages in our analysis, the preceding equations apply equally well to rms voltages because the two quantities are related to each other by the same fae· tor for all cirellil elements. The result for the maximum voltage .IlV:llax given by Equation 21.10 reinforces the fact that the voltages across the resistor, capacitor, and inductor are not in phase, so one cannot simply add them to get the voltage across the combination of element or to get the source voltage.
QUICK QUIZ 21.2
For Lhe circuiL of Figure 21.8, is Lhe inSLa11laneOus voltage of the source equal to (a) the sum of the maximum voltages across the elements, (b) the sum of the instantaneous voltages across the elements, or (c) the sum of the rms voltages across the elements?
We can write Equation 21.10 in the form of Ohm's law, using the relations ~VR = fmaxR, .1VL = fm ,·... here ~nax is the maximum current in the
circuit: [21.12] ]t's convenient LO denne a parameter called the impedance Z of the circuit as Impedance
-+
[2I.l3]
so that Equation 2] .12 becomes [2I.l4]
Equation 2] .J4 is in the form of Ohm's law, 6.V = JR., with R replaced by the impedance in ohms. Indeed, Equation 2"1.]4 can be regarded as a generalized form of Ohm's law applied to a series AC circuit. Both the impedance and therefore the. current in an AC circuit depend on the resistance, the inductance, the capacitance, and the frequency (because the reactances are frequency dependent). It's useful to represent the impedance Z v.'ith a vector diagram such as the one depicted in Figure 21.]1c. A right triangle is construned with right side XL - Xc, base R, and hypotenuse Z. Applying the Pythagorean theorem to this triangle, we sec that
which is Equation 21.13. furthermore, we see from the yeetor diagram in Figure 21.1lc that the phase angle
X"-L_-_X-,,C
tan
~
R
[21.15]
The physical significance of the phase angle ,viII become apparent in SeClion 21.5. Table 21.2 provides impedance values and phase angles for some series circuits containing different combinations of circuit elements. Parallel alternating current circuits al-e also useful in everyday applications. \Ve won't discuss them here, however, because their analysis is beyond the scope of this
book.
21.4
The RLC Series Circuit
70S
TABLE 21.2
Impedance Values and Phase Angles for Various Combinations of Circuit Elements Circuit Elements
Impedance Z
Phase Angle
tP
I? o----J..M~
L ~
II
e
R
L
-wv----j~ ~ R
L
e
~I--
II
0°
Xc
_90 0
Xl.
+900
..JR'l. +Xl:
NChrau\c. between _90 0 and 0°
..J R 2 + X 2L
Positive, hetv..ecn 0° and 900
~R'
+ (X, -Xc)'
Negative if Xc > XL Positive if X c < Xl.
Not,.; III each cOlse ,," AC \"Ollagl: (1101 sI1O\\,.) is applied acruss the: combi'Mlion of dClllcnt~ (IIml is, ..crOss the dots).
QUICK QUIZ 21.3 If switch A is closed in Figure 21.12, what happens to the impedance of the circuit? (a) It increases. (b) It decreases. (e) It doesn't change. QUICK QUIZ 21.4 Suppose X,. > Xc. If switch A is closed in Figure 21.12, what happens to the phase angle? (a) It increases. (b) 1L decreases. (e) It doesn'L change. QUICK QUIZ 21.5 Suppose XL> Xc. If switch A is left open and switch B is closed in Figure 21.12, what happens to the phase angle? (a) It increases. (b) It decreases. (c) It doesn't change.
FIGURE 21.12 21.3-21.6)
(Quick Quiucs
QUICK QUIZ 21.6 Suppose XL> Xcin Figure 21.12 and, with both switches opell, a piece of iron is slipped into the inductor. During this process, whal happens to the brighuless of the bulb' (a) It increases. (b) It decreases. (c) It doesn't change.
PROBLEM-SOLVING STRATEGY
RLC CIRCUITS The following procedure is recommended for solving series RLC circuit problems: 1. Calculate the inductive and capacitive reactances, XI. and >4.."
2. Use Xl and Xc together with the resistance I? to calculate the impedance Z of the circuit.
3. Find the maximum current or maximum voltage drop with the equivalent orOhm's law, .11~11aJ( = Ima"Z. 4. Calculate the voltage drops across the individual elements with the appropriate variations of Ohm's law: ~"/l.max = Jm;l>.!?-' .6.Vf .,max = 'nHlxX", and AVe,In,IX = 'l1l:lxXe· 5. Obtain the phase angle using tan q> = (X,. - Xd/R.
NIKOLA TESLA (1856-1943) Teslo was born in Croatia, but spent most of his professional life as on inventor in the United States. He was a key figure in the development of alternating-current electricity, high-voltage transformers, and the transport of electrical power via AC transmission lines. Tesla's viewpoint was at odds with the ideas of Edison, who commiHed himself to the use of direct current in power transmission. Teslo's AC approach won out.
706
Chapter 21
EXAMPLE 21.4 Goal
Alternoting~Current Circuits and Electromagnetic Waves
An RLC Circuit
Analyze a series NLC AC circuit and find the phase angle.
Problem A series RLC AC circuit has resistance R = 2.50 X ]02 n, inductance L = 0.600 H, capacitance C = 3.50 j.LF, frequency I= oU.O Hz, and maximum voltage liV;IMS = 1.50 X IO~ V. Find (a) the impedance of the circuit, (b) the maximum current in the circuit, (c) t.he phase angle, and (d) the maximum voltages across the clements. Strategy Calculate the inducLivc and capacitive reactances, which can be llsed with the resistance impedance and phase angle. The impedance and Ohm's lalA,' yield the maximum current.
.
-
~
...........
La
calculate the
_
~
Solution (a) Find the impedance of the circuit. First, calcuJ;;ue the inductive and capacitive reactances:
X,. = 27TJL = 226 n
SubstilUte these results and the resistance R into Equation 21.13 to obtain the impedance of the Circtlit:
z = VII' + (X,. -
Xc = 1/27TJC = 758 Il
Xc)'
= V(2.50 X 10' 0)'
+ (2260 - 7580)' = 588 Il
(b) Find the maximum current in the circuit.
Usc Equation 21.12, the cqulvalentofOhm's law, to find the maximum current:
r.
,=
rna);
1.50 X 10 2 V 5880
1l'v:11.1:" Z
0.255 A
(c) Find the phase angle. X,. - Xc
Calculate the phase angle between the current and the voltage 'with Equation 21.15:
~ tan-I (2260 - 758 Il) = -64.8" 2.50 X 10'Il
R.
(d) Find the maximum voltages across the elements.
Substitule into the ';Ohm's law" expressions fol' each individual type of' current element: ~VJ."",,, ~ {",,,,X,. = (0.255 A)(2.26 X 10 2 Il) ~ 57.6 V ~Vc.",,, lrrrr... • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
Remarks Because the circuit is morc capacitive than inductive (Xc> X,), ep is negative. A negative phase angle means that the current leads the applied voltage. Notice also that the sum of the maximum voltages across the elements is 6.VR + .6.V,. + 6.vc = 314 V, which is much greater than the maximum voltage 01' the generator, 150 V. As we saw in Quick Quiz 2].2. the sum or the Ill:aximum voltages is a meaningless quantity because when alternating voltages are added, both (Iuti:,. mnjJl'iludes and theirtJhases must be taken into account. We know that the maximum voltages across the various c1ernents occur at different limes, so it docsn'l make sense to add all the maximum values. The correct way to "add" the voltages is through Equation 21.10.
=
!",,,Xc
=
(0.255 A)(7.58
X ]02
Il) = 193 V
.
~
QUESTION 21.4 True or False: In an RLC circuit, the impedance must always be greater than or equal to the resistance. EXERCISE 21.4 Analyze a series RLC AC circuit for which R = 175 n, L = 0.500 H, C = 22.5t.tF, J= 60.0 Hz, and ~V""" = 325 V. Find (a) the impedance. (b) the maximum current, (c) the phase angle, and (d) the maximum voltages across I he elements.
Answers (a) 1890 (b) 1,72 A (c) 22.0" (d) 301 V, .6. \11,.max = 324 V, .6.'vC.max = 203 V
~VJ<."'", =
21.5
21.5
Power in on AC Circuit
POWER IN AN AC CIRCUIT
No power losses arc associated with pure capacitors and pure inductors in an AC circuit. A pure capacitor, by definition, has no resistance or inductance, \",hercas a pure inductor has no resistance or capacitance. (These definitions are idealizations: in a real capacitor, for example, inductive effects could become important at high frequencies.) vVe begin by analyzing the power dissipated in an AC circuit thal contains only a generator and a capacitor. When the current increases in one direction in an AC circuit. charge accumulates on the capacitor and a voltage drop appears across it. When the voltage reaches its maximum value, the energy stored in the capacitor is
This energy storage is only momentary, however: ""hen the current reverses direction, the charge leaves the capacitor plates and returns to the voltage source. During one-half of each cycle the capacitor is being charged, and during the other half the charge is being reLUrned to the voltag-e source. Therefore, the average power supplied by the source is zero. In other \.\Iords, no power losses occur in a capacitor in an AC circuit. Similarly, the source must do work against the back emf of an inductor that is carrying a current. Vlhen the current reaches its maximulll value, the energy stored in the inductor is a maximum and is given by
\!Vhen the current begins to decrease in the circuit, this stored energy is returned 1.0 the source as the inductor 3uempts to maintain the current in the circuit. The average power delivered to a resistor in an RLC circuit is [21.16]
The average power delivered by the generator is converted to internal energy in the resistor. No power loss occurs in an ideal capacitor or inductor. An alternate equation for the average power loss in an AC circuit can be found by substiLUting (from Ohm's law) R = 6.\lR,rm/frillS into Equation 2] .16:
lI.'s convenient to refer to a voltage triangle that shows the relationship among aVU,fms, and 6.VL ,nns 6.VC,rms' such as Figure 21.1Ib. (Remember t.hat Fig. 21."11 applies to both maximum and rIllS voltages.) From t.his figure, we see that the voltage drop across a resistor can be \vritten in t.erms of the voltage of the source, 6.\I;'I11S'
6.l~'Ill~:
Hence, the average pO\'ier delivered by a generator in an AC circuit is [21.17]
where the quantity cos cP is called the power facto... Equation 21.17 shows that the power delivered by an AC source to any circuit depends on the phase difference between the source voltage and the resulting current. This [act has many interesting applications. For example, factories often lise devices such as large motors in machines, generators, and transformers that have a large inductive load due to all the windings. To deliver greater power to such devices without using excessively high voltages, factory technicians introduce capacitance in the circuits to shift the phase.
~
Average power
APPLICATION Shifting phase to deliver more power
707
708
Chapter 21
EXAMPLE 21.5 Goal
Alternating-Current Circuits and Electromagnetic Waves
Average Power in an RLC Series Circuit
Understand power in RLCseries circuits.
Problem
C
Strategy After fil1Cling the nns current and rills voltage with Equations 21.2 and 21.3, substiLulC into Equation 21.17, using the phase angle found in Example 21.4. .....................
r·································· Solution First, use Equations 2].2 and 2L3 to calculate the
~
0.255 A V2 = 0.180 A
rillS
current and rms voltage:
1.50 X ]02 V
C="--V2~2'-=-'- = J06 V
Substitute these results and the phase angle (~ = -64-.8° imo Equ;uion 21.17Lo find the average power:
qp,.. = =
... Remark
l,m,~~'"" cos c/.> = (0.180 A)(106 V) cos (-64.8°)
8.12 W
.
~
The same result can be obtained from Equation 21.16, 9f;". = 1;lllsR.
QUESTION 21.S Under what cilTumstance can the average pmvcr of an RLG' circuit be zero? EXERCISE 21.5 Repeal this problem, lIsing the system described in Exercise 21.4.
Answer
259 W
1~t1IS
21.6
RESONANCE IN A SERIES RLC CIRCUIT
In general, the rillS currenl in a series RLC circuiL can be written I
1<,
I
FIGURE 21.13 A plO! Ofeufr(:1ll amplillld(; in a series HI.C(:ircllit versus frequency orlhe goener-alar voltage. !'\olice that Ihe current reaches its maximum value allhc resunallce frequencY.lII·
Resonance frequency
APPLICATION Tuning Your Radio
-+
nns
n l'rms a v,'ms = -Z- = -:r=~=o====Cii VR 2 + (XL - Xr:J2
[21.18]
From this equation, we sec that if the frequency is varied, the current has its maximum value when the impedance has its minimu'm, value, which occurs when XL = Xc' 1n such a circumstance, the impedance of the circuit reduces to l = R. The frequency 10 at which this happens is called the resonance frequency of the circuil. To find/o, we set XL = Xc> which gives, from Equations 2l.5 and 21.8,
27Tfr,L
fu
=
1
=--
27TfoC
1 27TVLC
[21.19]
Figure 2-1.13 is a plot of current as a fUIlCllon of frequency for a circuit containing a fixed value f'or both the capacitance and the inductance. From Equation 2-1.18, it must be concluded that the current would become infinitc at resonancc when N = O. Although Equation 21.18 predicts this result, real circuits always have some resistance, which limits the value of the current. The tuning circuit of a radio is an in"lportant application of a series resonance circuit. The radio is tuned to a particular station (.."hich transmits a specific radiofrequency signal) by varying a capacitor, \I.lhich changes the resonance frequcnc)' of the lUning circuit. 'A'hen this resonance frequency matches that of the incoming radio wave, the current in the tuning circuit increases.
21.6
APPLYING PHYSICS 21.2
Resonance in a Series RLC Circuit
709
METAL DETECTORS AT THE COURTHOUSE
When you walk through the doorway ora courthouse metal detector, as the person in Figure 21.14 is doing, you are really walking through a coil of many turns. How mighllhe metal det.ector \",'ork? Explanation The mctal detect.ur is essentially a resonant circuit. The portal you step through is an inductor (a large loop of conducting wire) that is pan of the circuit. The frequency of the circuit. is lUned to the resonant frequency of the circuit when there is no rnetal in the inductor. \-Vhen you walk through with metal in your POCkCL, you change the effective inductance of the resonance circuit, resulting in a change in the currcnt in the circuit. This change in cllrrenl is detected, and an elect.ronic circuit causcs a sOllnd to bc emitted as an alarm.
EXAMPLE 21.6 Goal
~
FIGURE 21.14
(Apply-
ing Ph)'Sics 21.2) A
COUl"t-
hOlISt:: lTIt::1al dClcclor.
~
"• ...._---~
A Circuit in Resonance
Understand resonance frequcncy and its relat.ion to inductancc, capacitance, and the rms current.
Problem Consider a series RLC circuit for which R = 1.50 X 102 .0, j, = 20.0 mH, 6.Y;'I1lS = 20.0 Y, and J= 796 s I (a) Determine the value of t.he capacitance for which the rms current is a maximum. (b) Find t.he maximum nns cur.. rent in the circuit. Strategy The current is a maximum at the resonance frequency Io, which should be set equal to the driving frequency, 796 S-I. The resulLing equation can be solved for C. For part (b), substitute into Equation 21.18 to get the maximum rms current. . • • • • • • • • • - - • ""'II!
~
Solution (a) Find t.he capacitance giving the maximum current. in the circuit (the resonance condition). Solve the resonance frequency for the capacitance:
I
fc'=9-rpC _1T v LL1 C = -47T-;;-'Ji,:7.,cL
Insertlhe given values, substiwting the source frequency for the resonance frequenc)\L:
I
C = '17T'(796 Hz)'(20.0 X 10
3
H)
2.00 X 10- 6 F
(b) Find the maximum rms current in the circuit. The capacitive and inductive reactances are equal, so Z = Ii = 1.50 X 10' fl. Substitute into Equation 21.18 to find the rms current: lrr.. • • • • • •
20.0 V 1.50 X 10' n
0.133 A
.
~
Remark Because the impedance Z is in the denominator of Equatiun 21 J8, the maximum current will always occur when XI. = Xc because that yields the minimum value of Z.
QUESTION 21.6 True or False: The magnitude of thc current in an RLCcircuit is never larger than the rms cunent.
710
Alternating-Current Circuits and Electromagnetic Waves
Chapter 21
EXERCISE 21.6 Consider a series RLC circuit l'or which R = 1.20 X 10' fl, C = 3,10 X 10-5 F, t;v.,,,,, = 35,0 Y, and I= 60,0,-1, (a) Determine the value of the inductance for which the rms current is a maximum. (b) Find the maximum rInS current in the circuit. Answers
(a) 0,227 H
(b) 0,292 A
s
z\
Primary (inplll)
l=:==:'~
Secondary (ompul)
FIGURE 21.15 An ideal transformer consists oft\l'o coils wound on the same soft iron core. An AC \'ollagc d( 'r is applied 10 the primary coil, and the output voltage ~\l~ is observed across lhc load resistance Rafter Ihe switch is dosed.
21.7
THE TRANSFORMER
IL's often necessary to change a small AC voltage Lo a larger one or vice versa. Such changes are effected with a device called a transformer. In its simplest form the AC transformer consists of two coils of wire wound around a core of soft iron, as shown in Figure 2] .15. The coil on the left, \\'·hich is connected to the input AC voltage source and has N] turns, is called the primary winding, or the pri·"U.L't)'. The coil on the right, which is connected to a resistor R and consists of N 2 turns, is the secoru.ia-ry. The purposes of the common iron core arc to increase the magnetic flux and to provide a medium in \",-hich nearly all the flux through one coil passes through the other. V\'hen an input AC voltage L\VI is applied to the primary, the induced voltage across it is given by t;1l
-N-I
t;t
[21.20]
where (1)8 is the magnetic flux through each lllrll. If we assume that no nux leaks from the iron core, then the flux through each turn of the primary equals the nux through each turn of the secondary_ Hence, the voltage across the secondary coil is [21.21] The term 6.(I>Il/6.t is common ro Equations 21.20 and 21.2.1 and can be algebraically eliminated, g"iving N,
t; If., = - - t; V, N,
[21.22]
vVhcn N'2 is greater than N I • 6.\12 exceeds L~.vl and the transformer is referred to as a sle/J-up transfonna ''''hen N2 is less than N I , making aV2 less than dVI , \'ie have a ste/J-down transfonne):
In on ideol transformer, the input power equals the output power ~
By Faraday's law, a voltage is generated across the secondary only when there is a change in the number of flux lines passing through the secondary. The input current in the primary must therefore change with time, which is what happens when an alternating current is used. "Vhen the input at the primary is a direct current, however, a voltage output occurs at the secondary only at the instant a switch in the primary circuit is opened or closed. Once the current in the primary reaches a steady value, the output voltage at the secondary is zero. rt may seem that a transformer is a device in which it is possible (Q get something [or nothing. For example, a step-up transformer can change an input voltage from, say, 10 V to 100 V. This means that each coulomb of charge leaving the secondary has 100 J of energy, whereas each coulomb of charge entering the primary has only 10 J of energy. That is not the case, hovvever, because the power input to the primary equals the power output at tbe secondary:
[, t;v, = [, t;v,
[21.23]
Although the voltage at the secondary may be, say, ten times greater than the vol[age at the primary, the (,'urre7ll in the secondary will be smaller than the primary's
21.7
current by a f'actor of' ten. Equation 21.23 assumes an ideal transformer in which there are no power losses between the primary and the secondary. Real transformers typically have power efficiencies ranging from 90% to 99%. Power losses occur because of such facLOrs as eddy currents induced in the iron core of the trallSfonner, which dissipate energy in the fon11 of j2R losses. When electric power is transmitted over large distances, it's economical to use a high voltage and a low current. because t.he pm-vcr lost via resistive heating in the transmission lines varies as /2R. ff a utility company can reduce the current b)1 a factor of ten, [or cxamplc, thc power loss is reduced by a factor of one hundred. In practice, the voltage is st.epped up to around 230 000 Vat the generating station, thcn stepped down to around 20000 Vat a distribution station, and finally stepped down to 120 V at the customer's utility pole.
EXAMPLE 21.7 Gool
Distributin Power to
0
The Transformer
711
APPLICATION Long-Distance Electric Power Transmission
Cit
Understand transformers and their role in reducing power loss.
Problem A generator at a utility company produces 1.00 X \0' A of current al 4.00 X 10 3 V. The vollagc is stepped up to 2.40 X 10'"' V by a transformer bcfolT being sent 011 a high-voltage trl\llsmission line across a rural area to a city. Assume the efTective resistance of the power line is 30.0 n and that the transformers are ideal. (a) Determine the percentage 0[' power lost in t.he transmission line. (b) What percentage of" the original power would be lost in the transmission lille if the voltage were llot stepped up? ".
.
Strategy Solving this problem is just a matler of substitution into the equation [or transformers and t he equation for power loss. To obtain the fraction of power losl., it's also necessary to compute the power output of the generator: the current times the potential difference creat.ed by the generator.
""
.
Solution (a) Determine the percentage of power lost in the line. (1.00 X 10' A)(4.00 X LO'V)
Substitutc into Equation 21.23 to find the current in the transmission line:
1.67 A
2.40 X 105 V
\'1',,,,,
= I,'R = (1.67 A)'(30.0 D) = 83.7 W
Now lise Equation 21.]6 to find the power lost in the transmission line:
(1)
Calculate the power output of the generator:
\'I' = I, Jl.V, = (1.00 X 10' A)(4.00 X 10' V) ~ 4.00 X 10" W
Finally. divide Q"Plo.'il by the power outpHt and multiply by 1001.0 rind the percentage or power lost;
% power 10Sl
=
83.7 W ) X 100 ( 4.00 X \0',W
=
0.0209%
(b) ''''hat percentage of the original power would be lost in the transmission line if t.he vohage were not stepped up? = I'R = (1.00 X 10' A)'(30.0 D) = 3.00 X 105 W
Replace the stepped-up current in Equation (1) by the original currentofl.OO X 10 2 A:
\'1',,,,,
Calculate the percentage loss, as before:
% power
\OSl =
X W)
3.00 10" ( 4.00 X 10-',Wx 100 = 75%
712
Chapter 21
Alternating-Current Circuits and Electromagnetic Waves
This cylindrical step-down transformer drops the voltage from 4000 V LO 220 V lor dclh'ery La a group of residences.
Remarks This example illustrates the advantage of high-voltage transmission lines. At the city, a transformer at a substation steps the voltage back down La about 4 000 V, and this voltage is maintained across utility lines throughout the city. When the power is to be used at a home or business, a transformer on a utility pole near the establishment reduces the volt-
age
LO
240Vor 120V.
QUESTION 21.7 If the voltage is stepped lip to double the amount in this problem, by what [actor is the power loss
changed? (a) 2 (b) no change (c) ~ (d) ~
-1l
~
i?'....._ ~
EXERCISE 21.7 Suppose the sallle generator has the voltage stepped up to only 7.50 X 10 4 V and the resistance of the Iinc is 85.00. Find the percentage of power lost in this case.
Answer
~c ~
!
Q
0.604%
21.8
MAXWELL'S PREDICTIONS
During the early stages of their study and development, electric and magnetic phenomena were thought to be unrelated. In -1865, however, James Clerk Maxwell (1831-1879) provided a mathematical theory that showed a close relationship between all electric and magnetic phenomena. In addition to unifying the formerly separate fields of electricity and magnetism, his brilliant theory predicted that electric and magnetic fields can move through space as waves. The theory he developed is based on the following [our pieces ofinformation:
1. Electric field lines originate on positive charges and terlllinate on negative charges. 2. Magnetic field lines always form closed loops; they don't begin or end anywhere. 3. A varying magnetic field induces an emf and hence an electric field. This fact is a statement or Faraday's law (Chapter 20). 4. Magnetic fields are generated by moving charges (or currems), as sUlllmarized in Ampere's law (Chapter 19).
JAMES CLERK MAXWELL Scottish Theoretical Physicist (1831-1879)
Maxwell developed the electromagnetic theory of light ond the kinetic theory of gases, ond he explained the nature of Saturn's rings and color vision. MaxwelJ's successful interpretation af the electromagnetic field resulted in the equations that bear his nome. Formidable mathematical ability combined with great insight enabled him to lead the way in the study of electromagnetism and kinetic theory.
The first statement is a consequence of the nature of the electrostatic force bcn.. .een charged particles, given by Coulomb's law. It embodies the fact that free charges (electric monopoles) exist in nat.ure. The second statement-that magnetic fields form continuous loops-is exemplified by the magnetic field lines around a long, straight wire, which are closed circles, and the magnetic field lines or a bar magnet, which form closed loops. It says, in contrast to the first statement, that free magnetic charges (magnetic monopoles) don't exist in nature. The third statement is equivalent to Faraday's law of induClion, and the fourth is equivalent to Ampere's law. In one of the greatest theoretical developments of the 19th century, Maxwell used these four statements within a corresponding mathematical framework to prove that electric and magnetic fields play symmetric roles in nature. It was already known [rom experimenLS that a changing magnetic field produced an electric field according LO Faraday's law. Maxwell believed thalnature was symmetric, and he therefore hypothesized that a changing electric field should produce
21.9
Hertz's Confirmation of Maxwell's Predictions
713
a magnetic field. This hrpmhesis could not be proven experimentally at the time it \vas developed because the magnetic fields generated by changing electric fields are generally very weak and therer-ore difficult 1.0 detect. Tojustify his hypothesis, :\1axwcll searched for other phenomena [hat might be explained by it. He turned his attention to the Illotioll of rapidly oscillating (accel~ erating) charges, such as those in a conducting rod connected to (Ill alternating voltage. Such charges are accelerated and, according to Maxwell's predictions, g"enerate changing electric and magnetic fields. The changing fields cause electromagnetic disturbances that travel through space as waves, similar to the spreading water waves created by a pebble thrown into a pool. The waves sent out by the oscillating charges are Auctuating electric and magnetic fields, so they are called electromagnetic waves. From Faraday's law and from Max\,,·e1l's 0\\'11 generalization of Ampere's law, MaK\,,·e11 calculated the specd of the waves to be equal to the spced of light, c = 3 X ]0 8 m/s. He concluded that visible light and other electromagnetic waves consist of fluctuating electric and magnetic fields traveling through empty space, with each varying field inducing the other~ His was truly one of the greatest discoveries of science, on a par with NeWlon's discovery of the laws of motion. Like Newton's laws, it had a profound innuence on later scientific developments.
21.9
HERTZ'S CONFIRMATION OF MAXWELL'S PREDICTIONS
In 1887, after Maxwell's death, Heinrich Hertz (1857-1894) was the first to generate and detect electromagnctic waves in a laboratory sctting, using I.e circuits. Tn such a circuit a charged capacilor is connected to an inductor, as in Figurc 21.16. 'VVhen the switch is closed, oscillations occur in the current in the circuit and in the charge on the capacilOr. If the resistance of the circuit is neglected, no energy is dissipated and the oscillations continue. In the following analysis, we neglect the resistance in the circuit. Vve assume the capacitor has an initial charge of Qmax and the switch is closed at t = O. W'hen thc capacitor is fully charged, the total energy in the circuit is stOl'cd in the electric field of the capacitor and is equal to Q711:1)/2C. At this time, the current is zero, so no energy is stored in the inductor. As the capacitor begins 10 discharge, the energy stored in its electric Aeld decreases. At the same time, the current increases and energy equal to L1'2/2 is now stored in the magnetic field of the inductor. Thus, energy is transferred from the electric field of the capacitor to the magnetic field of the inductor. \Vhen the capacitor is fully discharged, it stores no energy. Al this time, the current reaches its maximum valuc and all the energy is stored in the inductor. The process then repeats in the reverse direct.ion. The cnergy continues to transfer between the inductor and thc capacitor, corresponding to oscillations in the current and charge. As we saw in Section 21.6, the frequency of oscillation of an LC circuit is called the Tesonancejrequf'1/c)' of the circuit and is given by
j"
=
+ C
FIGURE 21.16 A simpk' LCcircuit. The capacitor has an inilial charge of QrnH' and the switch is closed ;ll I = O.
TnduClion coil
1 27TVLC
The circuit Hertz used in his in\"estigations of electromagnetic waves is similar to that just discussed and is shown schemat ically in Figure 21.17. An induction coil
(a large coil of wire) is connected 1.0 two meta) spheres with a narrow gap between them to form a capacitor. Oscillations are initiated in the circuit by shon \"oltage pulses sent via the coil to the spheres, charging one positive, the other negative. Because L and Care quite small in t.his circuit, the frequency of oscillation is quite high./= 100 MHz. This circuit is called a transmitter because it produces electromagnetic waves. Several meters from the transmiuer circuit, Hertz placed a second circuit, the receiver, which consist.ed ofa single loop of wire connected to two spheres. It. had its own effective inductance, capacitance, and natural frequency 01" oscillation.
FIGURE 21.17 A schemalic diagram of Hertz's app3ratus for gcnenHing :l11d detecti]lg e1cnromagnl:tic wan:s. The transmitter C011sist~ of l\~'O spherical den rodes connected to an induction coil. which prO' ides short \'olla~e sl1rgc.~ to the ~phert's. sc.:ujug up o~cil1adol1s i1l1hc discharg(:. The rtcei,"er is a !lead)\" single loop of wire ('ol1l:linillg a second spark gap.
714
Chapter 21
Alternating-Current Circuits and Electromagnetic Waves
HEINRICH RUDOLF HERTZ German Physicist (1857-!894) Hertz made his most important discovery of rodio waves in 1887. After
finding that the speed of a radio wave was the same as thot of light, Hertz showed that rodio WQves, like light
waves, could be reflected, refracted, and diffracted. Hertz died of blood poisoning at the age of 36. During his short life, he mode many contributions to science. The hertz, equal
Hertz found that energy \vas being sent from the transmit.ter LO the receiver when the resonance frequency of the receiver was adjusted to match that. of the transmitter. The energy translcr was detected when the voltage across t.he spheres in the receiver circuit became high enough to produce ionization in the air, which c
to one complete vibration or cycle per second, is named after him.
21.10
APPLICATION Radio-Wave Transmission
PRODUCTION OF ELECTROMAGNETIC WAVES BY AN ANTENNA
In the previoLls section we found that the energy stored in an LC circuit is continually transferred bcnveen the electric field of the capacitor and the magnetic Field of" the inductor. This energy transfer, hm\lcver, continues for prolonged periods of time only ,..,hen the changes occur slowly. If the current alternates rapidly, the circuit loses some of its energy in the form of electromagnetic '\laves. In faet., electromagneLic waves are radiated by (In)' circuit carrying an alternating current. The fundamental mechanism responsible for this radiation is the acceleration ofa charged panicle. Whenever a charged particle accelerates, it radiates energy. An alternating vall age 'lppljed to the wires of an antenna forces electric charges in the antenna to oscillatc. This common technique for accelerating charged par· ticIes is the source of the radio waves emitted by the broadcast antenna of a radio station. Figure 21.18 illustrates the production of an elcctromagnetic wave by oscillating electric charges in an antenna. Two metal rods afe connected 10 an AC source, which causes charges to oscillate between the rods. The output voltage of the generator is sinusoidal. At l = 0, the upper rocl is given a maximum positi\'e charge and the bottom rod an equal negative charge, as in Figure 21.18a. The electric Held Ileal' t.he antenna at this instant is also shown in the figure. As the charges oscil·
FIGURE 21.18 All clecu"ic field set up by uscillating charges ill an antenna. The field moves aw,ly from the antenna ,tl the speed oflig-ht.
(a) 1=0
(bl
(c) l=
~
(d) (==T
21.11
Properties of Electromagnetic Waves
late, the rods become less charged, the field near the rods decreases in strength, and the dm"-'Il\..· ard-directcdlll
21.11
TIP 21.1 Accelerated Charges Produce Electromagnetic Waves Stalionary charges produce only electric fields. whe.-eas charges in uniform motion (Le., constam \-elocity) produce electric and magnetic llelcb, but no cleCtro~ magnetic Il'avcs. III contrast, accelerated charges produce eleclromagnetic \\"a\'cs as well as electric and llwgnctic fields. An accclerati ng charge also radiales energy.
FIGURE 21.19 1\1agnclic field lines around an a/Henna carrying a c1mng· ing currenl.
PROPERTIES OF ELECTROMAGNETIC WAVES
"vVe have seen that Nlaxwcll's detailed anal}/sis predicted the existcnce and propert.ies of clectromagnetic waves. In this scction we summarize what we know about clectromagnetic wm'es thus far and consider some additional properties. In our discussion here and in future sections, '\ie will oftcn make reference to a typc ofw'ave called a plane wave. A plane electromagnetic wm'e is a wave travcling from a very d ista nt sourcc. Active Figu re 21.20 pictu res such a wave at a given iIlslant of time. In this case the oscillations of the electric and magnetic fields take place in planes perpendicular to the x-axis and are therefore perpendicular to the direction of travel of the wave. Because of the latter pl2perty, electromagnetic waves are transverse ~avcs. In the figure the electric field E is in thc y-direction and the magnetic field B is in the z-direction. Light propagates in a direction perpendicular to these two fields. That direction is determined by yet a~other right-hand rule: (I) point tl~e r-ingers oryour right hand in the direclion orE, (2) curl them in the direction ofB, and (3) lhe right thumb then points in the direclion of propagation of the wave. Elcctromagnetic waves travel with the speed of light. In fact, it can be shown that the speed of an electromagnetic wave is related to lhe permeability and perlllittivity of the mediulll through which it tr.wels. Maxwell found this relationship for free space to be 1 c~---
V fLo'o
[21.24]
715
~
Speed of light
716
Chapter 21
Alternoting~Current Circuits and Electromagnetic Waves
ACTIVE FIGURE 21.20 An elccLromagnL:tic \\'ave scm (Jut by oscillating charges in all alllenna, represented at elIte instant of time and far from the anlt:lllla, mewing ill the positive x-direction with speed r. Notice that the electric field is perpendicular tu the magnclit" field,
y
and both arc perpendicular LU the dil"eetion OfW
x
where c is the speed of light, JJ-o = 4iT X ]0- 7 N's2/C 2 is the permeability constam of vacuum, and €() = 8.854 19 X 10- 12 C2 /t\:'m:! is the perllliuivit}: of free space. Substiltlling these values into Equation 2] .24, \VC find that t = 2.99792 X 10"
mls
[21.25]
Because electromagnetic waves travel at the same speed as light in vacuum, scienconcluded (correctly) that light is an electromagnetic wave. Maxwell also proved the [allowing relationship for electromagnetic waves:
TIP 21.2 E Stronger thon B?
t ists
The relationship E = Be makes the dectric th.:lds assochned with light are nlllch larger than the magnetic fields. That is nOllhe case: The \llliL~ arc different, so the quantities can', be din:ctly compared. The tWO fields colllribmc equally to lhe energy of a ligh! wave.
[ it appear lh:'11
E
B
[21.26]
c
which states that the ratio of the magnitudc of the electric field to the magnitude of the magnctic ficld equals the specd of light. Electromagnetic \vavcs carry energy as they travel through space, and this cnergy can be transferred to objects placed in their paths. The average rate at which energy passes through an area perpendicular 1.0 the direction or travel of a \vave, or the average power per unit area, is called the intensity I of the wave and is givcn by 1=
Eilia:>, Bm,IX
[21.27]
2JLo
where h:llla " and Billa" an: the 'JI/.axirn'll.fn values or E and B. The quamity I is analogous to the intensity of sound '...·avcs introduced in Chapter 14. From Equation 21.26, \-vc see that Ema " = (Bmax = BIIl
Light is an electromagnetic wave and transports energy and momentum
-+
I!.:~,a" C =- = -2JLoc
2JLo
')
B~nax
[21.28]
Note that in these expressions we use the avemgf power per unit area. A detailed analysis would show that the energy carried by an electromagnetic \',.-avc is shared equally by the electric ancl magnetic fields. ElcClromagnetic v.'avc::; have an average intensity given by Equalion 21.28. When the waves st.rike an area A 01" an object's surface [or a given time IJ.t, energy U= 1.'\ 6.1 is transferred to the surface. Momentulll is transferred, as well. Hence, pressure is exerted on a surf~lce when an electromagnetic wavr.:: impinges on it. In \-vhat follows, we assume the clcClromagnelic wave transports a total energy U lO a 5ur-
21.11
Properties of Electromagnetic Waves
717
face in a timc ,11. I f the surface absorbs all the incident energy U in this time, 1\·laxwell sho\l,Icd that the total momcntum p deli\'ered to (hi!'. surface has a magnilllde U jJ = (
(complete
~Ibsorption)
[21.29]
If tht.: surface is a perfect reflector. then the momel1lum transferred in a time 8( for normal incidence is twice that gi\'cn b) Equation 21.29. This is analoguus to a molcCllle of gas bouncing off the wall of a container in a perfectly elastic collision. If the molecule is initially traveling in the positi\'c x·direction at \'elocity vand after Ihe ('ollision is lI"a\'eling in the negative x-direction at \'e1ocity -v. its change in momentum is given by n/J = 11lV - (-11lv) = 2ml1. Light bouncing off a perfect reflector is a similar process, so for complete reflection.
2U
p=r
(completc reflenion)
[21.30]
Lighl
~lirror
FIGURE 21.21 .\n apparatus for lIIl',hlll ing lh{· racliatiun pn:ssurc: uf light. In pranin·. lhe 'i\Slt:m is COil, 1.lined in a high \dCUIIIll.
Although radiation pressures are very sinall (abuut 5 X 10-1; :'\/m~ fOI" direct sunlight), they ha\'e been measured with a device such as the one shown in Figure 21.2J. Light is allowed to strike a mirror and a black disk that are connected to each olher by a horizontal bar suspended from a fine.:' fiber. Lig-ht striking the black disk is completely absorbed. so all the momentum of 1he light is transferred 1.0 the disk. Light striking the mirror head-on is totally reflected; hence, the momentum transfer to thc mirror is Iwice that transn'1iucd 10 tile disk. As a result. the !Jori1.0nl<11 bar supporting the disks twists counterclockwise as st.:en from above. The bar comes to equilibrium at some angle under the aCtion of the torques caused by radiation pressure and the twisting 01" the libel'. TIlL' radiation pressure can be determined by me<\suring the angle at which equilibrium occurs. The apparatus must be placed in a high vacuulll to e1imillate the effect:-. of air currents. It's inlcrcsti ng that ::;imilar experiments demunsl ral c thaI electromagnetic waves ((Irry angular momentum, as well. III summary, electromagnetic \.. . avcs Iraveling through free space havc the following prapenie::;: 1. Electromagnetic ,..·a\·e5 travel at the speed of light. 2. Electromagnetic waves are transverse wC\\es because Ihe elcctric and magnctic fields are perpendicular to the direction ofpropag-ation afthe wave and to each olher. 3. The nllio of the electric field to the magnetic field in an electromagnetic wa\'e eqtlal~ the speed of light. 4. Electromagnetic wa\'es carr) b01h energ) and momentum, \\·hich can be delivered to a surface.
APPLYING PHYSICS 21.3
~
Some properties of electromagnetic waves
SOLAR SYSTEM DUST
In the il1lcrplanetaq space in the Solar S)stcm. there is a large amount of dust. Although inLerplanelan dust can in theory ha\'e a ,'aricty of si/es-from molecular size upward-why are there vel") few dU:'lL panicles smaller than about 0.2 J.un in the solar sn· tem? Hint: The solar system originally contained dusl panicle~ of all sizes. Explanation Dust particles in the solar sy~tem are sul~jecl 10 two furces: the gravitational force toward the Sun and the force from radial ion prcs..,lIre, which is dircclcd away from the Sun. The gravitational force
is proportional to the cube of the radius of a spherical dU'it particle because it is proportional to the mass (pV) of the particle. The nldiation pressure is proportional to the square of the radius because it depends on the cross-sectional area of the particle, For large particles, the gra\'ilational force is larger than the furce of radiation pressure, and the weak attraction to thc Sun causes such particles to move slowly toward it. For small panicles, less (han about 0.2 ,u.m, the larger force from radiation pressure sweeps Ihem out afthe solar system.
Chapter 21
718
Alternating-Current Circuits and Electromagnetic Walles
QUICK QUIZ 21.7 In an apparatus such as that in Figure 21.21, suppose the black disk is replaced by one with half the radius.\'Vhich of the following are different after the disk is replaced? (a) radiation pressure on the disk (b) radiation force on the disk (c) radiation momenlUm delivered to the disk in a given time interval
;:====::::::::::;:;;::::=::::=====:::
EXAMPLE 21.8 A Hot Tin Roof (5010r- Powered ",H~o~m~e~s!-l Goal Calculat.e some basic properties of light. and relate t.hem to thermal radiation. Problem Assume the Sun delivers an average power per unit area ofahout1.00 X 10' W/m' to Earth's surface. (a) Calculate the towl power incidelll 011 a flat tin roof 8.00 m by 20.0 Ill. Assume the radiation is incident normal (perpendicular) to the roof (b) Calculate the peak e1eetl'ic field of the light. (c) Compute the peak magnetic field of the light. (d) The tin roof reneClS some light, and convection, conduction, and radiation transport the rest of the thermal energY' away until some equilibrium temperature is established. If the roof is a perfect blackbody and rids itself of one-half of the incident radiation through thermal radiation, what's its equilibriulll temperawre? Assume the 3mbiclll temperature is 298 K. ,
Solution (a) Calculate the power delivered to the roof. = 1.4 = (1.00 X 10 3 W/m')(8.00 m X 20.0 m)
'!P
\1ultip1r the intensity by the area to get the power:
=
1.60 X JO'W
(b) Calculate Lhe peak cJeetl'ic field 01' the light. Solve Equation 21.28 for Ema ,,:
E",,~
= V2(4.". X 10 ':>:·s'/C')(3.00 X 10'm/5)(1.00 X ]O'W/m') =
868 V /m
(c) Compute the peak magnetic field of the light.
B
Obtain Bl1lax lIsing Equation 21.26:
Emax
=
ma."
=
C
868 V /m 3.00 X 108 m/s
(d) Find the equilibrium temperature of the roof.
Substitute into Stefan's law. Dnly one-hal f the incident power should be substituted, and twice the area of the roof (both the LOp and the underside of the roof cOllnt).
""-.:.r -
'\(T'-'I"') ()
(TeJ
T'=T"+~ o uf'A 4
(2981Z)
+
(0.500)(1.60 X 10'W/m') (5.67 XIO 'W/m'·K4 )(1)(3.20 X 102 m')
T= 333 K = 6.0 X 10'oC Irrr.... • • •
.
-
-
~
21.11
Remarks If the incident power could aLL be converted to electric power. it would be more than enough for (he average home. Unfortunately. solar energy isn'l easily harnessed, and the prospects for large-scale conversion are not as bright as they may appear from this simple calculation. For example. the conversion efficiency from solar to electrical energy is far less than 1000/<:: 10% is typical for phoLOvoltaic cells. Roofsystems for lIsing solar energy (a raise the temperature of ,vater with dTiciencies of around 50% ha\·e been builL. Other pranical problems must be considered, however, such as overcast days. geographic location, and energy storage.
Properties of Electromagnetic Waves
719
EXERCISE 21.8 A spherical satellite orbiting Earth is lighted on one side
by the Sun. with intensit) 1 340 \V/m 2 . (a) If the radius of the satellite is 1.00 m, whal power is incident upon it? Noll': The satellite effectively intercepts radimion only m·er a cross section, an area equal to that of a disk, 7rr 2 . (b) Calculate the peak electric field. (c) Calculate the peak magnetic field. Answers (a) 4.21 X 10' \\' (b) 1.01 X 10' \'1m (c) 3.35 X 10-6 T
QUESTION 21.8 Does the angle the roof makes with respcCl to the horizontal affect the amount of power absorbed by the roof? Explain.
EXAMPLE 21.9 Gool
eli Poer Ships of Space
Relate the intensity of light
10
its mechanical
~rfcct on
Problem Aluminized Mylar film is a highly reflective, lightweight material that could be llsed to make sails for spacecrafl driven by t.he light. of t.he SUIl. Suppose a sail with area 1.00 kl11~ is orbiting the Sun at a dislancc of 1.50 X lOll 111. The sail has a mass of 5.00 X 10!~ kg and is telhered to a payload 01" mass 2.00 X 10 1 kg. (a) I I" the intensity of sunlight is 1.34 X 103 Vvr alld the s~lil is orienlcc! perpendicular to the incident light, \'Ih",! radiltl force is exerted on the sail? (b) Abollt how long would it
matter.
take to change the radial speed of the sail by 1.00 krn/s? Assume the sail is perfectly reflecting. Strategy Equation 2] .30 gives the momentum imparted ,,,,,hen light strikes an object and is totally reOectcd. The change in this momentulll with tirne is a force. For part (b), lise Newton's second law to obtain the acceleration. The "elocity kinematics equal ion then yields the necessal") time to achieve the desired change in speed.
.,
".. Solution (a) Find the force exerted on the sail. Write Equation 21.30 and substiLUte U for the energy delivered to the sail:
=
Divide both sides by ill. obtaining the force il/J/ill exerted by the Jighl on the sail:
2Mill :1p= -2U =2'!J>ill --=-c
,
ill>
/- = -
~
2M
-
ill =
c
c
2(1 340 W/m')(1.00 X 106 m 2 )
= -'-----'----'-'---;:-------"-
3.00X10"m/s
c
8.93:\
(b) Find the time it takes to change the radial speed b), 1.00
km/s. F
=- =
SubSliwtc the force into NeWlOn's second law and soh'e for the acceleration of the sail:
II
Apply lhe kinematics velocit)' equation:
v = al
Solve for t:
...
111
+
8.93:\ 2.50 X 10' kg
X Hr I m/s 2
DO
1= _V_-_V_" =
a
= 3.57
.,-:.1,,0,,0,-,-X-=..1::.0·.,-;"","1/"s,,, 3.57 X 10 1 mis'
2.80 X 106 s
.
..
720
Chopter 21
Alternating-Current Circuits and Electromagnetic Waves
Remarks The answer LO parL (b) is a little over a 111onth. vVhile the acceleration is very 10\...·• there are no fuel costs, and within a few momhs the velocity can change sufficiently [Q allow the spacecraft to reach any planet in the solar system. Such spacecraft may be useful for certain purposes and are highly economical, but require a considerable amount of patience.
pose you were flaming in space and pointed the laser beam away from you. What v·muld your acceleration be? Assume your tolal mass, including equipment, is 72.0 kg and the force is direCled through your center of mass. !-lin/.: The change in momentum is the same as in the nonrefieClive case. (c) Compare the acceleration found ill pan (b) \\Iith the acceleration of gravity ofa space station of mass 1.00 X lOG kg, if the station's center of mass is 100.0 III away.
QUESTION 21.9 By what facLor will the force exerted by the Sun's light be changed when the spacecraft is twice as far from the Sun' (a) no change (b) ~ (c) t (d) k EXERCISE 21.9 A laser has a power of 22.0 \V and a beam J'adius of 0.500 mm. (a) Find the imensity of the laser. (b) Sup-
21.12
...
Answers (a) 2.80 X 10' Wlm' (b) 1.02 X 10-9 mis' (c) 6.67 X 10-9 m/s2 . I r you were planning 1.0 use your laser welding torch as a thruster [Q get you back La the station, don't bother because the force of gravity is stronger. Better yet, get somebody to toss you a line.
THE SPECTRUM OF ELECTROMAGNETIC WAVES
All electromagnetic waves travel in a vacuum with the speed o[ light, t. These waves transport energy and momenl.Urn [rol11 some source La a receiver. In 1887 Hertz successfully generated and detected the radio-frequency electromagnetic \'laves predicted by Maxwell. Maxwell himself had recognized as electromagnetic g \-,'aves both visible light and the infrared radiation discovered in 1800 by William E ~ Herschel. ft is now known that other forms of electromagnetic waves exist that are ~ disl.ing·uished by their frequencies and wavelengths. Because all electromagnetic waves travel through free space with a speed c, their ~III:"d <= frequency Janel wavelength A arc related by the important expression
\Io/earing sunglasses lacking tthraviolet (UV) proleetioll is worse for yOllr eyes than wearing llO sllnglasses at all. Sunglasses \\'ithow proteClinn absorh some \'isibk light, causing [he pupils to dilate. This allows more UV tiglll to enler the ere, increasing [he damage lO the lens of lhe c)e over time. \\'ithOlll the sllnglasses. thl' pupils constrict, redllcing l:wlll visible aod dangerous CV radimion. Ik cool: wear sunglasses with U\' pruteclion.
J
r;=
fA
[21.31]
The various types of electromagnetic waves are presented in Figure 21.22. Notice the \,\-'ide and overlapping range of frequencies and wavelengths. For instance, an AM radio wave with a frequency of 1.50 M Hz (a typical value) has a wavelength of _ c _ 3.00 X 108 m/s _ 6 1 - 2.00 1.50 X 10 s-
A- - -
.r
X
, 10
111
The following abbreviations are often used to designate shon wavelengths and distances: I micrometer (,u..rn) ::;:: ]0- 6
In
I nanometer (nm) = 10-9 m 1 angstrom (A) = 10- 10 m
The wavelengths of visible light, for example, range from O.4,urn to 0.7,um, or 400 nm LO 700 nm, or 4 000 A La 7 000 A.
QUICK QUIZ 21.8 Which of the following statements are true about light waves? (a) The higher the rrequency, the longer the wavelength. (b) The lower the frequency, the longer the wavelength. (c) Higher frequency light travels raster than lower frequency light. (d) The shorter the wavelength, the higher the frequency. (e) The lower the frequency, the shorter the wavelength. Brief descriptions of the \.. . ave lypes follow, in order of decreasing wavelength. There is no sharp division beh.. . een one kind of' electromagnetic w~we and the next. All forms of electromagnetic radiation are produced by
21.12
The Spectrum of Electromagnetic Waves
Frequcncy, Hz 10'.!2 10~1
10 20 101'J 10 18 10 17
10 16 10Ei 10 14
FIGURE 21.22
lOll
lOW
The clcctromag-
llelic spCClrum. NOlice the ovcrlap
I
,,
=1------
hClwcen acljaCCnllYpCS ofw~ves. The expanded \'icw 10 lhe right shO\\'s delails of the \'isiblc spectrum.
Ipm
X-r.ays
t]~l~~~~v!'i~~;I!>I~ctli!g~h!' ~
Inm
1,1m
Violel Blue Green ycllO\v
10 1:l 10 12
721
Il-----
Imm t em
Microwaves
10!1
I
ID"
TV,FM
10 7
1m
Radio waves
AM
lkm
Longwavc
I Radio waves, which were discussed in Section 2IJO, are the result of charges acceleraLing through conducting ,vires. They are, of course, used in radio and television communication systellls. Microwaves (short-wavelength radio waves) have wavelengths ranging between about 1 mill and 30 CIll and are generated by electronic devices. Their short wavelengths make them well suited for the radar systems used in aircraft navigation and for the slUdy or atomic and molecular properties of matter. Microwave ovens are an interesting domestic application of these waves. It has been suggested that solar energy might be harnessed by beaming microwaves to Earth from a solar collector in space. Infrared waves (sometimes incorrectly called "heat waves"), produced by hot objects and molecules, have wavelengths ranging frolll about 1 mill to the longest wavelength of visible light, 7 X 10-7 Ill. They are readily absorbed by most materials. The infrared energy absorbed by a substance causes it to get warmer because the energy agitates the "Homs of the object, increasing their vibrational or translational motion. The result is a rise in temperature. Infrared radiation has many practical and scientific applications, including physical therapy, infrared photography, and the study of the vibrations of atoms. Visible light, the most familiar form of electromagnetic waves, Illay be defmed as the part of the speclrum that is detected by the human e)'e. Light is produced by the rearrangement of electrons in atoms and moleCldes. The wavelcngths of visible light are classified as colors ranging from violet (A = 4 X 10-7 m) to red (A = 7 X ]0-7 m). The cye's sensitivity is a function of 'v ave length and is greatest at a wavelength ofabout5.6 X -1O-7 III (yellow green). Ultraviolet (UV) light covers \vavelengths ranging from about 4 X -W-7 m (400 nm) down to 6 X 10- 10 m (0.6 nm). The Sun is an important source of ultravioletlig'ht (which is the main cause orsuntans) ..1\10st of the ultraviolet light from the Sun is absorbed by atoms in the upper atlllosphere, or stratosphere, which is fortunate, because UV light in large quantities has harmful effects on humans. One important constituent of the stratosphere is ozone (03)' produced from reactions
722
Chapter 21
Alternating-Current Circuits and Electromagnetic Waves
i
Ii
~ ii
0
<
~
"~
jj 0
...i
z
(b)
(al
(c)
0
g
;i
""
<
<0
(d)
FIGURE 21.23 ObS(.'n.llion'io ill diffcf{'1lI pans of the c1eclfomagnclic Spt'etrum "hO\\" different realUre, o[ L1w Crab ~ehilla. (;1) X-I'ill image. (b) Optical image. (e) Polaril.cd \'i\iblc light. (d) Radio image.
of oxygen with ultraviolet radiation. The resulLing Olone shield causes lethal highenergy ulLraviolcl r~tdiaLjon La warm the slraLOspherc. X-rays are clcClromagnclic W3\"CS with wavelengths from aboUl 10 -R III (10 nm) down to 10- 1:\ m (10 I nm). The mOSt common source of x-rays is the acceleration of high-energy electrons bombarding a metal target. X-rays are used as a diagnostic [001 in medicine and as a treatment for certain forms of cancer. Because x-rays easily penetrate and damage or destroy living tissues and organisms, care must be taken to avoid unnccessary exposure and overexposure. Gamma rays-electromagnetic waves emilled b) radioactive nuclei-have wavclengths ranging from abollt 10-11) m to less than 10- 1•1 m. ThcYtHC high I}' pcnetrating and cause serious damage when absorbed by living tissues. Accordingly, those working ncar such radiatiun must be protected by ganncms contllining heavily absorbing materials, such as layers or lead. \OVhen astronomers observe the saine celestial object lIsing detecLOrs sensitive to different regions of the electromagnetic spectrulll, striking variations in the object'S femures can be seen. Figure 21.23 shows images of the Crab Nebula made in three different wa\!elcngth ranges. The Crab Nebula is the remnant of a supernova explosion that was seen on Earth in 1054 A.D. (Compare with Fig. 8.28.)
APPLYING PHYSICS 21.4
THE SUN AND THE EVOLUTION OF THE EYE
The «('Iller 01 sensiti\'it)' of our eyes coincides \,..-ith the «"mel of the wm'clcngth distribution of the Sun. Is this an amazing coincidence? Explanation This fact is nQL a coincidence; rather, i('(j the result of biological evolution. Humans have ('volved with vision must sensitivc to wavelengths that
21.13
are strongest from the Sun. Ifaliens from another planet evcr an-i\'ed at Eanh, their eyes would have the center of sensitivity at wavelengths differcm from ours. lftheir sun \.. .ere a red dwarf, for example. the alien's eyes would be most sensitivc to red lighl.
THE DOPPLER EFFECT FOR ELECTROMAGNETIC WAVES
As we saw in Section 14.6, sound wavcs exhibit the Doppler effect when the observer, the source, or both are moving relative to the medium of propagation. Recall that in the Doppler effect, the observed frequency of the wave is larger or smaller than the frequency emitted by the source of the wave. A Doppler effect also occurs for electromagnetic waves, blll it differs from the Doppler effect for sound \'1'aves in two ways. First, in the Doppler cffecl for sound waves. motion relative to the medium is mOSl important because sound waves
"
Summary
723
require a medium in which to propagate. In contrast, the medium of propagation plays no role in the Doppler effect for electromagnetic waves because the waves require no medium in which to propagate. Second, the speed of sound that appears in the equation for the Doppler effect for sound depends on the reference frame in which it is measured. In contrast, as we see in Chapter 26. the speed of eleClromagnetic waves has the same value in all coordinate systems that. are either at rest or moving at constant velocity with respect to one another. The single equation that describes the Doppler effect for electromagnetic waves is given by the approximate expression if u«
c
[21.32J
where fo is the observed frequency, j~ is the frequency emitted br the source, u is the relative speed of the observer and source. and {' is the speed of light in a vacuum. Note that Equation 2] .32 is valid only if u is much smaller than r. Further, it can also be used for sound as long as the relative velocity or the source and observer is much less than the velocity of sound. The positive sign in the equation must be used \.. . hen the source and observer are moving toward each other, whereas the negative sign Tllust be used \vhen they are Tlloving away [rom each other. Thus, we anticipate an increase in the observed frequency if the source and observer are approaching each Other and a decrease if the source and observer recede from each other. Astronomers have made important disco\'eries using Doppler obser\'ations on light. reaching Earth from distant stars and galaxies. Such measurements have shm'\!11 that most distant galaxies are moving away [rom Earth. Thus. the Universe is expanding. This Doppler shift is called a tn[ shift because t.he observed wavelengths are shifted to\\'ard the red portion (longer wavelengths) of the visible spectrum. Further, measurements show that the speed of a galax)' increases \\'jth increasing distance from Earth. l'vlore recent Doppler effect measurements made with the Hubble Space Telescope have shmvn that a galax), labeled M87 is rotating, with one edge moving toward us and the other moving away. Its measured speed of rotation was used to identify a supermassive black hole located at its center.
SUMMARY 21.1 Resistors in on AC Circuit I fan AC circuit consists of a generator and a resislOr, the current in the circuit is in phase with the voltage, which means that the current and voltage reach their maximulll values at the same time. In discussions of volt.ages and currents in AC circuits, rms values of voltages are usually used. One reason is that AC ammeters and voltmeters are designed to read rillS values. The rills values of currcnts and voltage (Inm and Ll\~.Ill.I) are related to the maximum \·alues of these quantities (fill:", and Ll~ll
=
1m •lx
v2
and
Xc
= -1-
[21.5] 2"'jC \vhere Jis the frequency of the AC voltage source. The rIllS voltage across and the rills current in a capacitor arc related by [21.6]
[21.2,21.31
The rms voltage across a resistor is related to the rms current in the resistor by Ohm's law: 6VR . rrm
voltage reaches its maximum v defined as
= {rm,R
[21.4aJ
21.2 Capacitors in an AC Circuit )f(lll AC circuit consists ofa generator and a capacito'·, the volt.age lags bchind the current by 90"". This means thallhe
21.3 Inductors in an AC Circuit If an AC circuit consists of a generator and an inductor, the vollage leads the current by 90°. This means the voltage reaches its maximum value onc·quaner of a period before the currelll reaches its maximulll value. The effective impedance of a coil in an AC circuit is measured by a quantity called the inductive reactance X" defi ned as XI
= 2",jL
121.8]
724
Alternating-Current Circuits and Electromagnetic Waves
Chapter 21
The rills voltage across a coil is rebtcd to the rills current in the coil by
21.4
The RLC Series Circuit
111 an RLCserics AC circuit, the maximulTl applied voltage av is related to the maximum vo1Lages across the resistor (8.VR ) , capacitor (LlVd, and inductor (6VJ by l21.1O] If an AC circuit. contains a resistor, an indllcwr, and a capacitor connected in series, the limit they place on the current is given by the impedance Z or the circuit,
defined as
21.8-21.13
Electromagnetic Waves and Their
Properties Electromagnetic waves were predicted by .lames Clerk ~1axwell and experimentally confirmed by Heinrich Hertz. These waves are created by accelerating electric charges and have the lallowing properties:
1. Electromagnetic waves are transverse waves because the electric and magnelic fields arc perpendicular to lhe direction of propagation of the waves.
2. Electromagnetic waves travel at the speed of light. 3. The ratio of the electric field to the magnetic field at a given point in an electromagnetic ",.·ave equals the speed of light: 10"
121.13] The rdationship between the maximum voltage supplied ro an NrC series AC circuit and the maximum currem in the circuit, which is the same in evt:ry clement, is 121.14] Tn an R/Lscries AC circuit, the applied rills voltage and current dl-e out of phase. The phase angle tP between the current and volta
¢ =
x -
Xc
-,,'-~_'o /{
21.5
[21.15]
Power in an AC Circuit
The average power delivered by the voltage source in an RLCseries AC circuit is @a\ = lrnl~llVnu~ cos
Il \~ lib
[21.181
The current Il
10 = - - - ; = 2r. \f[C
N; "'V,
U
(complete absorption)
[21.29]
If the surface is a perrect reflecLOJ", the momentum delivered in time 6.f at normal incidence is twice thal given by Equation 21.29:
l'
2U c
=-
(complete reflection)
[21.30J
6. The speed c, rrequcncy.f, <:lnd wavelength Aoran electromagnelic wave are related by [21.31]
the secondary winding consists of IV'!. turns and then an input AC voltage 1\/1 is applied to the primary, the induced voltage in tile second(lq' winding is given by 1\':,
where Em,o< and B rnax arc the maxi lTIum values of the electric and magnctic fields. 5. Electromagnetic ","'aves l.ranspon linear and angular momentum as v·,'ell as energy. The momentum IJdelivered in time dt at normal incidence to an object that completely absorbs light energy U is given by
[21.19J
21.7 The Transformer If the primary winding of a transformer has /'1/1 llIrns and
"'1', =
[21.27,21.28]
[
Resonance in Q Series RLC Circuit In general, the rms currel1l. in a series RLC circuit can be written
.0. Vrlll~
_ 1 J= 10""''''0<1",_13-"."",1:1",
/) = -
21.6
--Z = ~Vr.J""I'=c+=r(X#'L~=CX;=;,.)""
[21.26]
(
4. Electromagnctic ,vaves carry energy as they travel through space. The average power per unit area is the intensily I, given by
l21.17]
\v!lel"c rhe (onSlalll cos ¢ is called the power factor.
I'm' =
B
[21.22]
\"'hen N'2 is greater than N i , LiV2 exceeds LiV1 and the Il"ansformer is relerred to as a slr.jJ-up fmm!ormn: \'\'hen N 2 is le:::.~ than N" making ~\f2 less than .0.\11 , we have a stell-down lransformn: In an idealtransfonner, the power output e(lllals the power input.
The electromagnetic spectrum includes waves covering a bro£ld r£lnge of frequencies and \'\'(Ivelengths. These waves have a variel)' of" applications and characteristics, dcpendin~ on theil· f'l'equencie."i or wavelengths. The l'requency of" a given wave can be shi fled by the relative velocit}" of"observer and source, with the observed l"rcqucncyj()given by if It
«
c
[21.32]
\·vhcre f\ is the frequency of the source. "II is the relative speed of the observer and source, and (is the speed of light in a V£lClltllll. The positive sign is IIsed when the source and observer approach each other, the negative sign when they recede frolll each mher.
Multiple-Choice Questions
725
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE.CHOICE QUESTIONS l. A sinusoidally varying voltage has a maximum valuc of 170 V. What is its rills value? (a) 24-0 V (Il) 170 V
(e) 120V(d)0(e) -120V
2. ,",Vhen a particular inductor is connected to a source of sinusoidally varying voltage with constant amplitude and a frequency of 60.0 Hz, the nns current is 3.0 A. What is the rms current if the source frequency is doubleel' (a) 12 A (b) 6.0 A (e) 4.2 A (d) 3.0 A (e) 1.5 A
3. What is the rms voltage across a capacitor wilh capacitance (1.00/27T) .uF carrying 1.OO-mA rms AC current al60.0 Hz? (a) 16.7 V (b) 13.2 V (e) 32.2 V (eI) 0.560 V (e) 4.43 V 4. Find the voltage across a (1.0/21T)-]-J inductor when it carries 2.0 A of rl1lS current at 60.0 liz. (a) 160 V (b) 140 V (e) 120 V (eI) 95 V (e) 85 V 5. A series RLCcircuit contains a resistor of20 fl, a capacitor 01"0.75 ,uF, and an inductor of 120 1111-:1. Ira sinusoidally varying nns voltage or] 20 V is applied across this combination of' elements, what is the nns current in the circuit when operating at its resonance frequency? (a) 2.4 A (b) 6.0 A (e) 10 A (eI) 17 A (e) 8.2 A 6. A 6.0-V battery is connected across the primary coil ora transformer having 50 turns. If the secondary coil ofthc transformcr has 100 turns, what voltage appears across the secondary? (a) 24 V (b) 12 V (e) 0.0 V (d) 3.0 V (e) none of these
7. An electromagnetic \NaVe with a peak magnetic field component of magnitude 1.5 X lO- i T has an associated peak electric field cornponcnt or what value? (a) 0.50 X 10-" N/C (b) 2.0 X 10-' N/C (e) 2.~ X 10 I N/C (eI) 45 N/C (e) 22 N/C 8. An inductor and a resistor are connected ill series across all AC generator, as shm.. . n in Figure MCQ21.8. lmmediatcly after the switch is closed, which of the following statements is t.rue? (a) The current is AF/H. (b) The volt(.Ige across the inductor is zcro. (c) The currcnt in the circuit is zero. (d) The voltage across the resistor is 6.v. (e) The voltage across the inductor is half its maximum valuc.
R
L
s
the switch is closed, which of the following statements is true? (a) The voltage across the capacitor lags I he current by 90°. (b) The voltage across the resistor is out of phase . .vith the current. (c) The voltage across the capacitor leads the current by 90°. (d) The current decreases as the frequency of the generator is increased, bm its peak voltage remains the same. (e) none of these
R
c
s
'-------{~ }----~
FIGURE MCQ21.9
10. A resistor. capacitor, and inductor are connected in series across an AC generator. Wh ich of the following statements is f'alse? (a) The instantancous voltage across the capacitor lags the current by 90°. (b) The instanwncous voltage across the inductor leads the current by 90°. (c) The instantaneous voltage aCI'Oss the rcsistor is in phase wit.h the current. (d) The voltages across the resistor, capacitor, and inductor are not in phase. (e) The rms \'olL:"1ge across the combination of' the three elements equals the algebraic sum of the rms voltages across each element separately. 11. A resistor, capacitor, and inductor are connected in
series across an AC generator. \'Vhich 01" the following statements is true? (a) All the power is lost in the inducLOr. (b) All the power is lost in the capacitor. (c) All the power is lost in the resistor. (d) Power is lost in all th"ee clements. (e) The power delivered hy the generator dues not depend on the phase difrerence between the generator voltage and current. 12. I f the voltage across a circuit element has its maximum
value when the current in the circuit is zero, does it follow that (a) the circuit elemtlll is a resiswr, (b) the circuit element is a capacitor, (c) the circuit element is an inductor, (c1) the current and voltage are 90° OUI of phase, or (e) the Cllrrent and voltage are 180° OUI or phase? 13. \"'hal is the phase angle in a series RLCcircuitat resonance? (a) 180" (11) 90' (e) 0 (eI) -90' (e) :t45'
'--------1 ~ }----~ FIGURE MCQ21.8
9. A capacitor and a resisLOr are connected in series across an AC generator, as shown in Figure MCQ21.9. Aftcr
14. A series RLC circuit contains a 20.0-fl resistor, a O.75-f.LF capacitor, and a 120-mll inductor. If a sinusoidally varying rms voltage or 120 Vatf= 5.0 X 10 2 1-11. is applied across this combination of clements, what is the rillS CllJTcnt in {he circuit? (a) 2.3 A (b) 6.0 A (c) 10 A (eI) 17 A (e) 4.8 A
Alternating-Current Circuits and Electromagnetic Waves
Chapter 21
726
CONCEPTUAL QUESTIONS 1. Befor'e the advcm of cable television and satellite dishes, homeowners either mounted a television antenna on the roof or used "rabbit ears" alop their selS. (See Fig. CQ21.1.) Cenain oriel1lations of the receiving antenna on a television set gave better reception than Olhers. FUTlhermore, the best orienr
this process:\ f or F'~I? \\'hat is the carrier frequency? \-\'hat is the signal frequenC}'? \Vhat is the broadcasting amenna? \Vhal is the receiving amennar
6. \\'hen light (or other electromagnetic radiation) travels across a given region, ",hal is it ,hat oscillates? What is it that is transported? 7. Tn space sailing, which is a proposed alternative for transport [Q the planets, a spacecraft carries a \'ery large sail. Sunlight striking the sail exerts a force, accelerating the spacecrllfl. Should the sail be absorptive or reflective to be mos! effective?
8. How can the average value of an alternating current be zero, ret the square root of the a\'crage squared \'alue nOt be zero? 9. Suppose a creature from another planet had eyes that were sensitive to infrared radiation. Describe what it would see if it looked around the room that rou are now in. That is, ",..hat would be bright and what would be dim? FIGURE CQ21.1
10. \Yhy should an infrared photograph of a person look different from a photograph taken using visible light?
2. "That is the impedance of an RLC circuit at the ,"eso!lallce frequenc)/?
3. Receiving radio ,llllcnnas can be in the form of conducting lines or loops. 'What should the orientation of each of these antennas be relative to a broadcasting anlenna lhal is venical? 4. If the fundamental source of a sound wave is a \'ibrating object. what is the rllndamental source or an electromagnetic wa\'e? 5. In radio transmission a radio wave serves as a carrier wa\'c, and the sound signal is superimposed on the carrier ,..-a\'e. In amplitude modulation (A:\'I) radio, the amplitude of the carrier ,..rave varies according to the sound wave, The C,S. t\avy sometilnes uses flashing lights to send t-.lorse code between neighboring ships. a process that has similarities to I-adio broadcasting. Is
-
-.--
-. -
", PROBLEMS· ~
,-
'T"'
'--r:r-~
..
11. Does a wire connected to a battery emit an electromagnetic wave? 12. J fa high-frequency current is passed through a solenoid containing a metallic core. the core becomes warm due (0 induction. Explain why the temperature of the maLerial rises in this situation. 13. If the I-esistance in an RLC circuit remains the same, but the capacitance and inductance are each doubled. how will the resonance frequenc), change? 14. Why is the sum of the maximum voltages across each of the elements in a series RLC circuit llsually greater than the maximum applied voltage? Doesn't this \·iolate Kirchhoff's loop rule? 15. \·"'hat is the advanlage of lransmitting power at high voltages?
-
The Problems for this chapter may be :1ssigned online at WebAssigl1. l. 2. :\ = st..,\ightforward, ;lI\('rllu:di;tl.t:. (h;tlkll~,\In~
generator, (c) the maximum current in the resistor, and (d) the avcrage powel' delivered to the resistor.
•
WebAssl9n
m B
= denotes guided
problem
= denotes enhanced content problem
= biomedical applic,llion 0= denotes full solutioll av,lilable in Silldi'lll 5011lli01l5 Mallual/ Slud)' GuidI'
SECTION 21.1 RESISTORS IN AN AC CIRCUIT 1. When an AC generator is connected across a t2.0-fi resis~
LOr. the rms CUlTClll in the resistor is 8.00 A. Find (a) the nils voltage across the resistor, (b) t he peak voltage or the
2.
mil A certain lighlbulb is raled at 60.0 ,0\, when
opcrat~
ing at an nns \'ohage or 1.20 X 1O:! V. (a) What is the peak \'oltage applied auoss the bulb? (b) "'hal is the resistance of t he bulb? (c) Does a 1.00 X lOt \f bulb ha\'c grealer or less resistance than a 60.0-''\: bulb? Explain. 3. An AC power supply that produces a maximum voltage of ill';".,,, = 100 V is connected to a 24.0-!l resistor. The current and the resistor voltage arc respective!)' measured with an ideal AC ammeter and an ideal AC '·oltmcler, as shown in Figure P21.3. Whal does each meier read~ Note
Problems lhat an ideal alllmeter has zero rcsistanu: VolUllCI.Cr has infinite resistance. ~\'",.lX =
Cllld
all ideal
100 V
[]JWhen a 4.0~,uF capacitor is connected to a gcnerator whose nns output. is 30 V, the current in thc circuit is obsCI"ved 1O be 0.30 A. \Vhat is the frequencr of the source? 10.
R= 24.0n
L----{\,}--' FIGURE P21.3
4.
Fi~lIre P21A shows three lamps cOlllleCted 10 a 120-V AC (nns) household supply ,'oltage. Lamp~ I and 2 have 150-W bulbs; lamp 3 has a 100-W bulb. Find the nns c'urrent and the resistance of each bulb.
727
mil
An Af: gt'nl T
11. What muSl be the capacit311ce ofa capacitor in ...ened in it 60-H7 circuit in snies with a generator of liO-V maximum output. voltage to produce an rms current outptll of 0.75 A? 12, A generator dcli"crs an AC voltage of the fonn .lv = (98.0 V) sin (807i"l) to a capacitor. The maximum ClllTent in the circuit is 0.500 A. Find the (a) nns voltage of the g-eneralOr, (h) f"equenC) oj the generator, (c) rills cur-
rem. (d) reactance, and (e) \ "Iut:.' of the capacitance. 120 \'
SECTION 21.3 INDUCTORS IN AN AC CIRCUIT 13. Show that the inductive reactancc X, has 51 units of ohms. FIGURE P21.4
IIIf\1l
FIGURE P21.S
6.
The output voltage of an AC ~eller3tor is gi,en b~ (170 \') sin (60../). The generator is conneCted across a 20.0-n '·csistor. By inspectioll, what an: the (a) ma,illllllll \'OIt
Ell tiv
=
SECTION 21.2 CAPACITORS IN AN AC CIRCUIT 7. Show lhat the 51 unit of capacitivc reactance ,'( i... thc ohm,
8. Whal is the maximum currenl deli\erecl
to a Cil"Cllit COil· laining a 2.20-,uF capacitor when it is conlleclt:d across (a) a i'\onh American oUllct having .:i\:'u" = 120 V and J= GO.n 1'-1/ and (b) a European outlet having £i\~n" = ~40 V ,u,dJ= 50.0 I-b'?
14. An AC ~cnerator has an OUlPUt rillS vollage of78.0 V at a rrequenc}" of 80.0 1-1"1. I r I he gcncrawr is can nectcd across
ing a maximlllll output voltage of 4.00 V al a fre(lucncy of 300.0 Hz. What inductance is needed to keep the rillS cllrrcntles~ than 2.00 mA? 16,
&I
The Olllplii. vollage of an AC generator is given by .lv = (1.20 X IO:! V) sin (301Tt). The generator is connected acro~s a 0.500-1-1 inductOr. Find the (a) frequency of the generator, (b) rms voltage acro~s the inductor. (c) incillcti"e re.lctance. (d) ,·m .. currcnt in the inductor. (c) maximulll (urrem in the inductol', and (0 average power c1di"cred to the inductor. (g) Find an cxpl'essioll for the instantaneous (un-ent. (h) At what time after 1 = o docs the instantancous current first reach 1.00 A? (Use the im'erse sine function.)
[J1JDetcrmillc the min.il11111ll magnetic flux through an inductor connecled to a standard OtILlet (.l\~m" = 120 \', f~ 60.0 liz).
SECTION 21.4 THE RLC SERIES CIRCUIT ~v = (80.0 V) sin (1501) is applied to a series RLCci"cuit with I. = 80.0 mil. C = 125,0.u.F. and If = 40.0 n, (a) What is the impedance of thc circuit? (h) ';\'hat is the maximulll ClIlTelll in the circuit?
18. A 'iinusoirlal ,"oltage
19. A -I-O.O-p.F capacitor is connected to a 50.0-fl. resistor and a generahK who)'>e rills output is 30.0 V at 60.0 liz. Find (a) the nns current in the circuit. (b) the rms ,'oltage drop anuss the resistor, (c) the rillS voltage drop across the capacitor, and (d) lht' phase angle for Ihe ci,·cuil. 20. An inductor (I" = 400 nil I), a capacitOi (C= 4.431J.f), and a resistor (U = !iOO fl) ;lrc connected in series. A
728
Chapter 21
Alternating-Current Circuits and Electromagnetic Waves
50.0-Hz AC generator connected in series to these elements produces a maximum cuncnL of250 rnA in the circuit. (a) Calculate the required maximum voltage u\~nax' (b) Determine the phase angle by which the current leaels
or Jags the applied voltage.
(d) Does thc voltage lead or lag the current? How will plltling an inductor in series with the existing capacitor and resistOr affect the current? Explain. 27.
I1DA resistor (Ii = 9.00 X 10' fl), a capacitor (C =
0.250 I'F), and an inductor (L = 2.50 H) are connected in series across a 2.40 X lO2~HI. AC source for which ~\~llax = lAO X [0 2 V. Calculate (a) the impedance of the circuit, (b) the maximum current delivered by the source, and (c) the phase angle belween the current and voltage. (d) Is the current leading or lagging the voltage?
22. A 50.0--0 resistor, a 0.100-1-1 inductor, and a IO.O-j.LF capacilOr are conneueel in series to a 60.0-l-Iz source. The rms current in the circuit is 2.75 A. Find the rms voltages across (
24. An AC source operating at 60 Hz with ;1 maximum voltage of 170 V is connected in series wit.h a resistor (R = 1.2 ko.) ltnd an inductor (L = 2.8 H). (a) What is the maximum value of the currellL in the circuit? (b) What are the maximum values of the potential difference across the resistor and the inductor? (c) \
28.
ED
A series RLC circuit comains the fo[IO\\'ing components: R = 1.50 X 102 n, L == 2.50 X ]02 ml-I, C = 2.00 ,uF, and a generator with 6\~n:ox = 2.10 X 102 V operating at 50.0 HI:. Calculate the (a) inductive reactance, (b) capacitive reactance, (c) impedance, (e1) maximum currem, and (e) phase angle between the current and generator voltage. (I) Calculate the individual maximum ....oltages across the resistor, inductor, and capacitor.
mm Consider the RLC circuit in
Problem 21.27. When the voltage across the resistor is a maximum, what are the individual voltages across the capacitor and induclor? Explain.
29. An AC source with a maxinlum voltage of 150 V and
I== 50.0 Hz
is connected between points II and d in Figure P21.29. Calculate the nns voltages berween points (
bed
~f---. 40.0 n 185 mH 65.01lF FIGURE P21.29
SECTION 21.5 POWER IN AN AC CIRCUIT
Iso. IAn
AC source opel"
31. A multi meter in an HL circuit records an rms current of 0.500 A and a GO.D-Hz rms generator voltage of 104 V. A
wattmeter shows that the average power delivered to the resistor is 10.0 \1\", Determine (a) the impedance in the circuit, (b) the resistance R, and (c) the inductance L Transformer
5000 V
I
32. An AC voltage of the form {),.V = (90.0 V) sin (350l) is applied to a series RLe circuit. If R = 50.00., C = 25.0 i-tF, and L = 0.200 1-1, find the (a) impedance of the circuit. (b) rms current in the circuit, and (c) average power delivered to the circuit. 33. Calculate the average power delivered desnibed in Problem 21.
to
the circuil
FIGURE P21.2S
26.
r:mI
A 60.0-!1 resistor is connccled in series with a 30.0-p,F capacitor and a generator" h~wing a maximum voltage of 1.20 X 10 2 V and operating al 60.0 Hz. Find the (a) capacitive rcaClance of the circuit, (b) impcdance of the circuil, and (c) Ill
34. A series RLC circuit. has a resistance of 22.0 11 and an impedance of 80.0 n. If the rms voltage applied to the circuit is [60 V, what average power is delivered to the circuit? 135.IAn inductor and a resistor are connected in series. When connected to a fiO-Hz, gO-V (rms) source, [he voltage
Problems
drop across the resisLOr is (-ound to be 50 V (rms) and the power dclivcred to the circuit is 14- W. Find (a) the value of the resistance and (b) the value of the induct
36. _ Consider a series RLC circuit with R = 25 n, L = 6.0 mH, and C= 25,u.-F. The circuit is connected to a lO-V (rms), 600-Hz AC source. (iI) Is the sum of the volt
SECTION 21.6 RESONANCE IN A SERIES RLC CIRCUIT 37. _ A resonant circuit in a radio receiver is tuned to a certain station when the inducLOr has a value of 3.00 J-LH and the capacitor has a value of 2.50 pF. The resistance of the circuit is 12 fl. (a) Find the frequency of the radio station. (b) Is there any information given in the problem thal. is not needed to solve it? Explain.
38. The LC circuit or a radar transmiller oscillates at 9.00 CHz. (a) What inductance will resonate with a 2.00~pF capacitor at this frequency? (b) 'What is the inductive reactance of the circuit at this frequency? 139.IThe AM band extends frolll approximately 500 kHz to l 600 kHz. Ira 2.0-,uH inductor is used in a luning circuit for a radio, what are the extremes that a capacitor must reach to cover the complete band of frequencies?
40. Consider a series RI.Ccircuit with R = 15 fl, L = 200 ml-l, C = 751LF, and
42. A series circuit contains a 3.00-H inductor, a 3.00-J,LF capacitor, and a 30.0-fl resistor connected to a 120-V (r111s) sourcc of variable frequency. Find lhe power deliv· ered to the circuit when the frequency of" the source is (<1) the resonance frequency, (b) one-half the resonance frequency, (c) one-[ounh the resonance frequency, (d) two times the resonance frequency, and (e) four times the resonance rrequenc),- From your calculations, can you draw a conclusion about the frequency at which the maximum power is delivered to the circuit?
SECTION 21.7 THE TRANSFORMER 43. The primary coil of a transformer has N I = 250 turns, and its secondary coil has N2 = I 500 turns. If the input voltage across the primary coil is.au = (170 V) sin wl, \\'hat rms voltage is developed across the secondary coil?
729
44. A slep-down transformer is used [or recharging the bat· teries of ponable devices. The (Urns ratio N 21NI for a particular transformer used in a CD player is 1:]3. When L1sed with 120-V (rms) household service, lhe transformer draws an rms current of250 rnA. Find the (a) rms output voltage of the transformer and (b) power delivered to the CD player. 145.IAn AC power generator produces 50 A (rrns) at ~ 600 V. The voltage is stepped up to 100000 V by an ideal transformer, and the energy is transmilled through a long. distance power line that has a resistance of 100 D. What percentage of the power delivered by the generaLOr is dis~ sipated as heal. in the power line? 46. A transformer is to be used to provide power for a computer disk drive that needs 6.0 V (rms) instead of the 120 V (rms) from the wall outlet. The number of turns in the primary is 400, and it delivers 500 mA (the secondary current) at an output voltage of 6.0 V (rms). (a) Should lhe transformer have more turns in the secondary compared with the primary. or fewer turns? (b) Find the current in the primal")'. (c) Find the number of turns in the secondary. 47. A transformer on a pole ncar a factory steps the vollage down from 3600 V (rms) to 120 V (rms). The transformer is to deliver 1 000 kW to the factory at 90% efficiency. Find (a) the power delivered to the primary, (b) the current in the primary, and (c) lhe current in the secondary. 48. A transmission line lhal has a resistance per unitlenglh of 4.50 X 10-'1 film is to be used to transmit 5.00 M'\oV over 400 miles (6.44 X 10" m). The outpUl voltage of the generator is 4.50 kV (rms). (a) What is the line loss if a transformer is used to step up the vollage to 500 kV (rms)? (b) \Vhat fraction of the input pown is lost to the line under these circumstances? (c) What difficulties would be encountered on altempting to tranSmitlhe 5.00 MW at the generator voltage of 4.50 kV (rms)?
SECTION 21.10 PRODUCTION OF ELECTROMAGNETIC WAVES BY AN ANTENNA SECTION 21.11 PROPERTIES OF ELECTROMAGNETIC WAVES 49. The U.S. Navy has long proposed the construction of extremely low frequency (ELF waves) communications systems; such waves could penetrate the oceans to reach distant submarines. Calculate the length of a quartcrwavelength antenna for a transmiuer generating ELF waves of frequency 75 Hz. How praCtical is this antenna? 50. (a) The distance to Polaris, the North Slar, is approximately 6.44 X 10 18 m. If Polaris were to burn out today, how mallY years would it take to see it disappear? (b) How long docs it take sunlight LO reach Earth? (c) How long does it take a microwave signal to travel from Earth l.0 lhe Moon and back? (The distance from Earth to the Moon is 3.84 X 10' km.) 51. An electromagnetic wave in frce space has an electric field of amplitude 330 Vim. Find the amplitude of the corresponding magnetic field.
730
Chapter 21
Alternating-Current Circuits and Electromagnetic Waves
52. ExpcrimenLcrs at the National Institute or Standards and Technology have made precise measurements or the speed of Jig-hl using the fact that, in vacuum. the speed of e1ectrom
frequency 2.45 GHz. The micmwaves enter thc oven and are rencC(cd by the walls. The standing-wavc p'll.lern produced in lhe oven can cook food unevenly, with hot spots in the food at anti nodes and cool spots al nodes, so a turntable is oflen uscd to rotate the food and distribute the energy. If a microwavc oven is used with a cooking dish in a fixed position, the anti nodes can appear as burn m,H-ks on fi)ods such as carrot strips or cheese. The separation distance between the bums is measured to be 6.00 cm. Calculatc the speed or the microwaves from these data.
Uf3JII
Oxygemltcd hemoglobin absorbs weakly in the red (hence its I"cd color) and sirongly in the ncar infrared, whel'ca~ dcoxygenau:d hemoglobin has the opposite absorption. This hlct is llsed in a "pulse oximeter'" to measure oxygel1 saturatio1l in ancrial blood. The device clips onto thc' end ofa per~on's fillger and ha~ two light-em iui ng diodes-a red (660 nm) and an infrared (940 nm)-and a photocell that detects the amOlllll of liglu transmiued thmugh the fingel al each wavelength. (a) Determine the freqllen<:y of each of thc~e light sources. (b) I r 67% of the e!lelgy of the rcd source is absorbed in the blood, by what lilClor doe.~ thc amplitude of the c1cClromagneti<: wave change? I-lint: The intensity of the wave is equal to the averagt' power per unit area as given by Equation 21.28.
54.
II
Operation of the pulse oximeter (see previous pToblem).l·he Lransmission of light energy as it passes thmllgh a solution of1ight-absorbing molecules is described by the Beel-I.ambert law
,= '0
10 cu.
or
log", ( ') I"
~
-.C/.
which gives Liw dccn:asc in intensity I in tenns of the distancc 1. LtH..: light ha~ tr
iii
A la~er beam is used to levitate a metal disk against LIlt' force of Earth's gravit.y. (a) Derive an equation giving thl' l-equircd im<.:nsity of light, I, in terms of the mass tTl 01 the disk, the gra"itat ional acceleration g, the speed of lig-ht" and the cross+set:tional area of the disk A. Assume the disk is perfcctly rellcLling and the beam is directed perpendicular LO th<.: disk. (b) If the disk has mass 5.00 g and radius 4.00 cm, find the necessary light intensity. (c) C;iw two reasons why using light PI'cSSlIrc as propulsion near Earth's SUrratT i~ impnlctit:al.
57. A microwave oven is powered by an electron tube called a ma~·Ill·lron lhal generales e1eetromagnetic wavcs of
58. Assume Lhe solar radiation incident on Earth IS I 340 Wim 2 (at the top of Earth's atlllosphere). Calculate the total power radiated by the Sun, taking the average separation between Earth and the Sun to be 1.49 X lOll m.
SECTION 21.12 THE SPECTRUM OF ELECTROMAGNETIC WAVES 59.
II
The eye is most sensitive to light of wavelength 5.50 X 10 7 m, which is in the green-ycllow region of the visible c1c<.:tromagnctic spectrum. \'\Ihal is Ihe frequency of this light?
60.11 A diathermy machine, used in physiotherapy, generates c1eClromagnetic ['adiatinn th,u gives the efTect of "deep heat" when absorbed in tisslle. One assigned frequency for diathermy is 27.33 MHz. What is [he wavelength of t his radiation? 61. \-Vhat are the wavelength ranges in (a) lhe AM radio band (540-J 600 kHz) and (b) the FM radio band (88108 MHz)?
62. An important news announcement is transmiued by radio waves 1.0 people who are JOO kill awa>', sitting neXI to their radios, and by sound waves to people sitting across the newsroom, 3.0 III from the newscaster. Who receives the news first? Explain. Take the speed of sound in air La be ~43 m/s. J63.llnfrared spectra al-e used by chemists to help idenliry an unk.nown substance. Atoms in a molecule that arc bound LOgctJler by a particular bond vibrate at a predictable frequency, and light at that. frequency is absorbed strongly by the atom. In the case of the C=O double bond, for example, the oxygen atom is bound to the carbon by a bond thal has an effective spring constant of 2800 N/m. If we assume the carbon atom remains stational-y (it is attached to other atoms in the molecule), determine the resonant rrequency or this bond and the wavelength of light that matches that frequency. Verif-)' that this wave~ length lies in the infrared region of the spectrum. (The mass of an oxygen atom is 2.66 X 10-21 ) kg.)
21.13 THE DOPPLER EFFECT FOR ELECTROMAGNETIC WAVES 64. A spaceship is approaching a space station at a speed of 1.8 X 10'-' tn/so The space station has a beacon that emits green light with a frequency of 6.0 X 10 14 Hz. What is the frequency of the beacon observed on the spaceship? What is the change ill frequency? (Carr)' five digits in lhese calculations.)
Problems
65. While driving at a constant speed of 80 km/h, yOll are passed b)' a car traveling at 120 km/h. If the freqncnC)' of light emittcd b)' lhe taillights of the car that passes you is <1.3 X lOH Hz. what frequency will yOll obscrve? What is the change in frequency?
731
as shown in Figure P21.73. The radio signal is a continuous sinusoidal wave with amplitude Em;... = 0.20 J.LV/m. Assume the antenna absorbs all the radiation that falls on the dish. (a) \·Vhat is the amplitudc of the magnetic field in this wavc? (b) 'What is the intensity of the radiation rcceivcd by the antenna? (c) What is the power recei,-ed b)' the antenna?
66. A spceder tries to explain to [he police that the yellow warning lights on the side of the mad looked green to her because of the Doppler shift. How fast would she have been traveling if rellow light of wavelength 580 nm had been shifted to green with a wavelength of 560 nm? Note: For speeds less than 0.03c, Equation 21.32 will lead to a \'alue for the change of frequenC)' accurate to approximately two significant digits. ADDITIONAL PROBLEMS
67. A 50.0-fi resistor is connected in series with a 15.0·p.F capacitor and a 60.0-Hz, 1.20 X 102·V (rms) source. Find the (a) impedance of the circuit and (b) rms current in thc circuit. (c) What is the value of the inducLOI-that must be insened in the circuit to reduce the current to one· half thai found in pan (b)? 68. The intensil)' of solar radiation at the top of Earth's atmosphel-e is 1340 W/m 3 . Assuming 60% of the incoming solar energy reaches Earth's surface and assuming yOIl absorb 50% of the incident energy, make an order-of:' magnitude estimate of the amount of solar encrgy yon absorb in a 60-minmc sunbath. 69. A 200-!1 resistor is connccted in series with a 5.0-.u.1i' capacilor and a GO-Hz, 120-V rms linc. lf eleclrical energy costs 0.080/kVlh. how much does it cost to leave this circuit connected for 24 h? 70. A series RLC circuit has a resonance frequcllcr of 2000/17" HI.. When it is operating at a frequenc), of W > w o, XL = 12!1 and Xc = 8.0!1. Calculate the values of Land Cfor the circuit. [TIJAs a way of determining tJle inductance of a coil used in a research project, a sllldem first connects the coil to a 12.0-V battcq' and measures a CUlTcnt of 0.630 A. The studell1. then connects the coil to a 24.0-V (rms). 60.0-llz gcnerator and measures an rms current of 0.570 A. What is the inductance? 72. (a) What capacitance will resonate with a one-turn loop of inductance 400 pH to gi\'e a radar wave of wavelength 3.0 cm? (b) If the capacitor has square parallel plates separated by 1.0 mm of air, what should the edge length of the plates be? (c) What is the common reactance of thc loop and capacitor at resonance? 73. A dish antenna with a diameter of 20.0 m recehes (at normal incidence) a radio signal fron"! a distant soul-ce.
FIGURE P21.73
74. A particular inductor has appreciable resistance. When the inductor is connected \.0 a 12-V bauer)', the current in the inductor is 3.0 A. When it is connected 1.0 an AC source with an rms output of 12 V and a frequency of 60 Hz, the cUITentdrops to 2.0 A. \Vhalare (a) the impedance at 60 Hz and (b) the inductance of the inductor? 175.IOne possible means of achieving space flight is to place a perfectly renecting aluminized sheet into Earth's orbit and to lise the light from the Sun to push this solar sail. Suppose such a sail, of area 6.00 X 10-1 m 2 and mass 6000 kg, is placed in orbit facing the Sun. (a) What force is excrted on the sail? (b) What is the sail's accelcration? (c) I-low long docs it take this sail to reach the Moon, 3.84 X 108 m away? Ignore all gravitational effects and assume a solar intensity of 1 340 W/m 2 • Hint: The radiation pressure by a reflected wave is given b)' 2 (average powel" per unit areal/Co
76.
II am
The U.S. Food and Drug Administration limits the radiation leakage of microwa\'e ovens to no more than 5.0 mW/cm 2 at a distance of 2.0 in. A trpical cell phone, which also transmits microwm'cs, has a peak Output power of about 2.0 W_ (a) Approximating the cell phone as a point source, calculate the radiation intensil)' of a cell phone at a distance of 2.0 in. How does the answer compare with the maximum allowable microwave m'en leakage? (b) The distance from your ear to your brain is about 2 in. What would the radiation imensity in rour brain be if you used a Bluetooth headset, keeping the phone in your pocket, 1.0 m away from your brain? Most headsets are so-called Class 2 devices with a maximum Output power of 2.5 mW.
Light is bent (refracted) as it passes through woter, with different wavelengths bending by different amounts, a phenomenon called dispersion. Together with reflection, these physical phenomena lead to the creation of a rainbow when light passes through small, suspended droplets of water,
22.1
The Nature of Light
22.2
Reflection and Refraction
22.3 The Law of Refraction 22.4
Dispersion and Prisms
22.5 The Rainbow 22.6
Huygens' Principle
22.7 Total Internal Reflection
REFLECTION AND REFRACTION OF LIGHT Light has a dual nature. In some experiments it acts like a particle, while in others it acts like a wove. In this and the next two chapters, we concentrate on the aspects of light that are best understood through the wove model. First we discuss the reflection of light at the boundary between two media and the refraction (bending) of light as it travels from one medium into another. We use these ideas to study the refraction of light as it posses through lenses and the reflection of light from mirrored surfaces. Finally, we describe how lenses and mirrors can be used to view objects with telescopes and microscopes and how lenses ore used in photography. The ability to manipulate light hos greatly enhanced our capacity to investigate and understand the nature of the Universe.
22.1
THE NATURE OF LIGHT
Until the beginning of the 19th century, light was modeled as a stream of particles emitlccl by a source that stimulated the sense of sight on entering the eye. The chief architect of the particle theory of light was Newton. \·Vith this theory, he provided simple explanations of" some known experimental racts concerning the nature of light, namely, the laws or reflection and rcfractioll. Most scicntists acccpted :"Jc\.. · ton's particlc theory of light. During NeWlon's lifetime, however, another theory was proposed. In ]678 Dutch physicist .1I1d astronomer Christian Hu)'gens (1629-](j95) showed that a wave theory oflighl could also explain the 1,1\'\'5 of reflection and refraction. The \\'ave theory didn't receive immediate acceptance, for several reasons. first, all the waves known at the time (sound, water, and so on) traveled through sOlne sort ofmcdium, but light from the Sun could travellO Earth through empty space. Funher, it was argued that if light were some form or wave, it would bend around obstacles; hence, we should be able to see around corners. II. is now known that
732
22.2
Reflection and Refraction
light does indeed bend around the edges of o~jccts. This phenomenon, known as diffrartion, is dirficult to observe because light waves have such short wavelengths. Even though experimental evidence [or the diffraction of tight was discovered by Francesco Grimaldi (1618-1663) around 1660, for more than a century most scientists rejected the wave theory and adhered to NC\..,LOn's particle theory, pl'obabl)' due 10 Newton's great repmalion as a scientist. The first clear demonstration of the wave mlturc of light was provided in 1801 by Thomas Young (1773-1829), who showed that under appropriate condit.ions, light exhibits imerf"erence behavior. Light waves emitted by a single source and traveling along two different paths can arrive at some point and combine and cancel each other by destructive interference. Such behavior couldn't be explained at that time by a particle model because scientists couldn't imagine how twO or more particles could comc together and cancel one another. The most important development in the theory o[ light was the work of "tvlax"...ell, who predicted in ]865 that light was a form of high-frequency electromagnetic wave (Chapter 21). His theory also predicted that these \.. . aves should have a speed of 3 X 108 mis, in agreement \vilh the measured value. A Ithough the classical theory of electricity and magnetism explained most known properties of light, some subsequent experiments couldn't be explaincd by the assumption that light was a wave. The most striking experiment \\'as the jJ/wloelectric effect (which we examine more closely in Chapter 27), discovered by Hertz. Hertz found that clean metal surfaces emit charges v.ihen exposed to ultraviolet light. In 1905, Einstein published a paper that formulated the theory of light quanta ("particles") and explained the photoelectric effect.. He reached the conclusion that light was composed o[ corpuscles, or discontinuous quanta or energy. These corpuscles or quanta are now called !J/totOTIS to cmphasize their particle-like Il
Astronomer
[22.1]
~ Energy
E= hJ
where h = 6.63 X 10-:H .J' sis /JlanDk:\· rons/anl. This theory retains some features of both the \\'ave and panicle theories of light. As we discuss lat.er, the photoelectric effect is t.he result or energy transfer from a single pholon to an electron in the metal. This means the electron interacts with one photon of light as if the electron had been struck by a particle. Yet the photon has wave-like characteristics, as implied by the fact that a frequency is llsed in its definition. In view of these developments, light must be regarded as having a dual na.ture: In some experiments light acts as a wave and in others it acts as a particle. Classical electromagnetic wave theory provides adequate cxplanations of light propagation and of the effects of interference, whereas the photoelectric effect and other experiments involving the interaction of light with matter are best explained by assuming light is a panicle. So in the final analysis, is light a wave or a particle? The answer is neither and both: light has a number of physical properties, some associated "'lith waves and others with particles.
22.2
REFLECTION AND REFRACTION
\"'hen light traveling in one medium encounters a boundary leading into a second medium, the processes of reflection and refraction can OCCUl". In reflection part of the light encountering the second medium bounces off that medium. In refraction the light passing into the second medium bends through an angle with respect to the normal to the boundary. Often, both processes occur at the same time, with pan of the light being renected and pan refracted. To study reflection and refraction we need a way of thin king about beams of light, and this is given by the ray approximation.
733
CHRISTIAN HUYGENS (1629-1695), Dutch Physicist and Huygens is best known for his contributions to the fields of optics and dynamics. To Huygens, light was a vibratory motion in the ether, spreading out and producing the sensotion of light when impinging on the eye. On the basis of this theory, he deduced the lows of reflection and refraction and explained the phenomenon of double refraction.
of
Q
photon
734
Chapter 22
Reflection and Refraction of Light
---
Rays
Wave fronls FIGURE 22.1 A plane wave tr;wclingLa the righl. Note thallhe ra),s, corresponding to the direction of\\',wc mQLioll, are 51 ..aight lines perpclldiclllar to the \,'<\vc fronts.
The Ray Approximation in Geometric Optics An important property of light thal can be understood based on common experience is the following: light travels in a straight-line path in a homoge· DeOllS medium, until it encounters a boundary bem'een two different materials. \t\'hen light strikes a boundary, it is reflected from that boundary, passes into the material on the other side of the boundary, or partially does both. The preceding observation leads us to use what is called the ray approximation to represent beams of light. As shO'..·n in Figure 22.1, a ray oflight is an imaginary line drawn along the direction of travel of the light beam. For example, a beam of sunlight passing through a darkened room traces out the path of a light ray. \i\,'e also make use of the concept of wave fronts of light. A wave front is a surface passing through the points of a wave that have the same phase and amplitude. For instance, the \.. . ave fronts in Figure 22.1 could be surfaces passing through the crests of waves. The rays, corresponding to the direction of wave Illotion, are straight lines perpendicular to the wave fronts. When light rays travel in parallel paths, the wave fronts are planes perpendicular to the rays.
Reflection of Light
APPLICATION Seeing the Road on a Rainy Night
When a light ra)' traveling in a transparent medium encounters a boundary leading into a second medium, part of the incident ray is reflected back into the first medium. Figure 22.2a shows several rays of a beam of light incidem on a smooth, mirror-like refleCling surface. The reflected rays are parallel to one another, as indicated in the figure. The reflection of light from such a smooth surface is called specular reflection. On the other hand, if the reflecting surface is rough, as in Figure 22.2b, the surface reflects the rays in a variety of directions. Reflection from any rough surface is known as diffuse reflection. A surface behaves as a smooth surface as long as its variations are small compared with the wavelength of the incident light. Figures 22.2c and 22.2d are photographs of specular and diffuse reflection of laser light, respectively. As an example, consider the two types of reflection from a road surface that a driver might observe while driving at night. vVhen the road is dry, light [rom oncoming vehicles is scattered off the road in different directions (diffuse reflection) and the road is clearly visible. On a rainy night ",,then the road is wet, the road's irregularities are filled \vith \vater. Because the wet surface is smooth, the light undergoes specular reflection. This means that the light is reflected straight ahead, and the driver ofa car sees only what is directly in front of him. Light from the side never reaches the driver's eye. Tn this book we concern ourselves only with specular reflcC1.ion, and \ve use the term -reflection to mean specular reflection.
FIGURE 22.2 A schcmaLic representation of (<1) specular rdlection, where the reflcCled rays are all paralleila Olle another. and (b) diffuse reflection. where the reneclef! rays travel in random directions. (c, e1) PhOlographsofspecubrimd dirru"e reflection, Illadc wil h laser light. (n)
(e)
(b)
(d)
22.2
QUICK QUIZ 22.1
Reflection and Refraction
735
Which part of Figure 22.3, (il) or (b), beller shows spec-
ular reneclion of light from the roadway?
(b)
(a)
FIGURE 22.3
(Quick Qui7 22.1)
Normal
Consider a light ray traveling in air and incidCnL at some angle on a nat,
unit enters the eye and is reflcClcd back along ils original path from the retina. This type of rcflection back along the original direction is called rf'l'rorifln;lion. If the flash unit and lens are close together, retrordlected light can elllcr the lens. Most or thc light rcflectcd from the retina is red due to the blood vessels at the b;(lck of the eye, giving the reel-eye effect in the photograph.
APPLYING PHYSICS 22.1
\
.
81
: OJ ",---" I
I
ACTIVE FIGURE 22.4 According LO lhe law or rdkclioll, 0 1 = 0;.
APPLICATION
lIlI
Red Eyes in Flash Photographs
THE COLORS OF WATER RIPPLES AT SUNSET
An observer on the west-facing beach ora large lake is 'watching the beginning ofa sunset. The water is \'ery smooth except for some areas with small ripples. The observer notices that some areas of the water are blue and some are pink. ''''hy does the ,vater appear to be different colors in different areas?
APPLYING PHYSICS 22.2
Rcncclcd ray
I I I I I I
m)'
[22.2] You may have noticed a COllllllon occurrence in photographs of individuals: lhcir eyes appear to be glowing red. "Red-eye" occurs when a photographic flash device is llsed and the flash unit is close to the camera lens. Lig'ht from the flash
,
Incidenl
smooth surface. as in Active Figure 22.4. The incident and rcneCled rays make angles 01 and 0;, respectively, with a line perpendicular to the surface at the point where the incident ray strikes the sllrl~lce. \o\le call this line the normal to the surface. Experiments show lhal the angle of reflection equals the angle of incidence:
Explanation The different colors arise from specular and diffuse reflection. The smooth areas of the water will specularly reflect the light from the west, which is the pink light from the SLinset. The areas with small ripples will rdlectthe light diffusely, so light from all parts orlhe sky will be reflected into the observer's eyes. Because most of the sky is still blue at the beginning of the sunset, thcse areas will appear to be blue.
DOUBLE IMAGES
Whcn standing outside in the sun close to a single pant: window looking to the darker interior ora building, why can you often see two images of yourself, one superposed on thc otht:r? Explanation Reflection occurs whenever there is an interface between two different media. For the
glass in the window, there are two such surfaces, the window surface facing outdoors and the window surface racing indoors. Each of these interfaces results in an image. You will notice that one image is slightly smaller than the other, because the reflecting surface is farther away.
736
Chapter 22
EXAMPLE 22.1 Goal
Reflection and Refraction of Light
The Double-Reflecting Light Ray
Calculate a resultant angle from two reflections.
Problem Two mirrors make an angle of 120 0 with each other, as in Figure ~~.5. A ray is incident on mirror M 1 at an angle o[ 65° to the normal. Find the angle the ray makes with the normal LO M 2 after it is reflected from both mirrors. FIGURE 22.5 (E.xample 22.1) Mirrors fI'l] and M? make all angle or 120 0 with each other.
Strategy Apply the law of reflection twice. Given the incident r<'ly at angle fJ illC ' find the final resultant angle, I3r~f'
.
.
".
Solution Apply the law of reflection reflection, Brei:
Lo
~
M I to find the angle of
Find the angle cP that is the complement of the angle 0rcf:
(~ =
Find the unknown angle a in the triangle of MI' .M 2• and the ray traveling from M1 to M 2 • lIsing the fact that the tJ,ree angles sum to 180":
180"
90" - 0", = 90" - 65" = 25" ~
25"
+ 120" +
fX
-->
fX ~
35"
The angle a is complemenwfy to the angle of incidence, f3il1C' for M 2: Apply the law ofrefleClion a second time, obtaining {3rcf: lrrrr... • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
Remark
f3ref = f3inc = 55°
.
~
NOlice the heavy reliance on elementary geometry and trigonometry in these renection problems.
QUESTION 22.1 Tn general, what is the relationship between the incident angle (Jille and the final reflected angle f3refwhen the angle benveen the mirrors is 90.OO? (a) 8 in e: + f3ref = 90.0° (b) 8 ine - f3 ref = 90.0" (c) 8ine + {3rcf = 1800 EXERCISE 22.1 Repeat the problem if the angle of incidence is 550 and the second mirror makes an angle of 100 0 with the first mirror.
Answer
45"
Refraction of Light '''lhen a ray of light traveling through a transparent medium encoumers a boundary leading into another transparent medium, as in Active Figure 22.6a, part of the ray is reflected and part enters the second medium. The ray that cillers the second medium is bent at the boundary and is said to be reJ'ncted. The incident ray. the reflected ray. the refracted ray, and the normal at the point of incidence all lie in the same plane. The angle of refraction, 82 , in Active Figure 22.6a depends on the properties of the 1:\\'0 media and on the angle of incidence, through the relationship sin 8,)
~~-
sin 8 J
V9
=
-=-
= constant
[22.3]
VI
where VI is the speed of light in medium I and v'!. is the speed of light in medium 2. Note that the angle of refraction is also measured v,lith respect to the normal. In Section 22.7 we derive the laws of reflection and refraction llsing Huygens' principle.
22.3
Incident rd.Y
l\'ormal
.\
V,
t\ir Glass
737
ACTIVE FIGURE 22.6 (a) A. rayobliqllel) incidcnI on an air-glass illlcrfacc. The refracted r"y is bClllloward lhe normal because 1.':/< lit, (b) Light incidcnl 011 lhe Lucite block bend.. bOlh \,'hen il enler.. the bloc-I.. and \\hen it leaves the block.
Rdlecled ray
I
The Law of Refraction
I
~. :821
, I
B Rcfr.lCled
ray (a)
(hI
Experiment shows that the path of a light ray through a refracting surface is reversible. For example, the ray in Acti\'c Figure 22.6a travels from point A to point B. If the ray originated at B, it would follow the same path to reach point A, but the reflected ray would be in the glass.
QUICK QUIZ 22.2 rfbearn 1 is the incoming beam in Active Figure 22.6b, \Vh ich of the other four beams are due to refleClion? \Vhich arc due to refraction? When light moves [rom a material in which its speed is high to a material in which its speed is 100'l"er, the angle of refraction 9'1 is less than the angle of incidence. The refractcd ray thercfore bcnds tov./arc! t.he normal, as shown in Acti\'c Figure 22,7£1. If the ray moves from a material in which it travels slov./Iy to a material in which it travels more rapidly, 8 2 is greater t.han 9 1, so the ray bends away from I.he normal, as shown in Active Figure 22.7b.
Normal "I
I I I
(JI >(J'1.
01 :
Air
Glass
'<"1, :.0
~2,
I'\!
< til
I I
(a)
22.3
THE LAW OF REFRACTION
Normal
When light passes [rom one transparel1tmedium to another, it's refracted because the speed of light is different in the two media. 1 The index of refraction, n, of a medium is deAned as the ratio clv; speed of light in 7/=
r
\,3CllUITI
speed of light in a medium
v
I I
\ Class Air
[22.4]
From this definition, we see that the index of refraction is a dimensionless number that is greater than or equal to 1 because v is always less than c. Further, n is equal to one for vacuum. Table 22.1 (page 738) lists the indices of refraction for various substances. As light travels from one medium to another, its frequency doesn't change. To see wh)" consider Figure 22.8. \Vave froills pass an obscn·er at point A in medium
8 J < 82
~I: r , I
I.......
1'2> VI
: O2
"
(h)
ACTIVE FIGURE 22.7 (a) When the light beam moves from air into glass, ilS path is bent toward lhe normal. (b) When the beam moves from glass inlO air, its path is hem away from the normal.
r
1/1=-
"I
A
'"
1
2 /J II.! '"
FIGURE 22.8 As the wa\e mo\'('~ rrom /1u'diull1 I to mediulU 2, its w:l\'dCUJ.'1I1 challges, but ilS frequency remains constant.
r_ 1~
I-I he 'pced orlighl "aries hCl\\(:Cll medii' bCGlliSC the lin1l' ];18' UllL!tcd h}' Ihe absOl pliorl II 11';1\'1,1, frellll alnm 10 alom (l(;p~l1d on I he p:lrLiClllar c!cclrOllic
SlrllC(lln'
Chapter 22
738
Retlection and Refraction of Light
TIP 22.1 An Inverse Relationship The index or refraction i~ iJlvrrsely proponional to [he .....ave speed. Therefore, HS I he: wan: speed v
Substance
decreascs.lhe index ofrdraclion. 11,
TIP 22.2 The Frequency Remains the Some doc~ not passes from
Thejrl'qul'1/r)'ofa wave \\';I\'C
Index of Refraction
Substance
Solids at 20°C
iJiUNlSes.
change as the
TABLE 22.1 I ndices of Refraction for Various Substances, Measured with Light of Vacuum Wavelength '\0 = 589 mn
one medium to anOlher. Both th<.:
wave speed and the wa\'elength do change, bUllhe freqllency remains the same.
Index of Refraction
Liquids at 20°C
Diamond (C)
2.419
Benzene
1.511 1
Fluorit.e (CaF 2)
1.434
Carbon disulfide
1.628
Fused quartz (SiO:!)
1.458
C-I rbon tctrach Imide
1.461
Class, crown
1.52
Ethyl alcohol
1.361
Glass, flilll
1.66
Glycerine
1.473
Ice (H,o) (a'O°C) Polystyrene
1.309
Water
1.333
Sodium chloride (Nael)
1.544
Gases at O°C, 1 atm
Zircon
1.923
Air
1.000293
Clrhon dioxide
1.00045
1.49
"I with a certain frequency and are incident on the boundary between medium I and medium 2. The frequency at which the 'Nave fronts pass an observer at point B in medium 2 must equal the frequency aL which they arrive at point A. If not, the wave fronts would eit.her pile up at the boundary or be destroyed or created at the boundary. Because neither of these events occurs, the frequency must remain the same as a light ray passes from one medium into another. Therefore, because the relation v = fA must be \'alid in both media and because !, = !, = I we see ,ha, (lnd = fA,
v,
Because VI ¥- v2 • it follows that AI c.F A~. 1\ relationship between the index of refrac~ tion and the wavelength can bc obtained by dividing these two equations and mak· ing lIse of the deHnition of the index of refraction given by Equation 22.4:
"I
112
A,
V,
c/ nl
1'12
'\2
V,
C/112
1/. 1
[22.5]
which gives Aln l =
A211.2
[22.6]
Let medium I be the vacullm so thaI. 'Ill = 1. It follows frolll Equation 22.6 that the index of refraction orany mediulll can be expressed as the ratio n = FIGURE 22.9
low inckx 01" refraction to UIl(' will! a higller index of refranioll.
Snell's low of refraction
[22.7]
A"
A schematic diagral1l
orlhc mlu,lioll ill wa\"t:lell~lh \,'hell light travels from a mediulll with;\
Au
where Ao is the wavelength of light in vacuulll and All is the wavelength in the mediulll havi ng index of refraction 17. Figure 22.9 is a schematic representation or this reduction in \'I'avelength \I/hen light passes from a vacuum into a transparent medium. We arc now in a position to express Equation 22.3 in an alternate form. If \I/C substitute Equation 22.5 into Equation 22.3, we get
[22.8]
~
The expcrimenlal discovery of this relationship is usually credited to \Villcbr0rcl Snell (1591-1626) alld is t.herefore known as Snell's Jaw of refraction. A material has an index or refraction that increases con~ tinuollsly frolll LOp 1.0 bottolTl. Of the three paths shown in Figure 22,10 (page 739), which path will a light ray follow as it passes through the Jll
QUICK QUIZ 22.3
22.3
" ",,----
FIGURE 22.10
(Quick Quiz
The Low of Refraction
739
22.~)
"
\ (b)
(aJ
Ie)
QUICK QUIZ 22.4 As light (ravels [rom a vacuum (n = I) 10 a medium such as glass (n > ]), which of the following propenies remains the same, the (a) \'lavelength, (b) wave speed, or (c) frequency?
EXAMPLE 22.2 Gool
Angle of Refraction for Glass
Apply Snell"s law
(0
a slab of glass.
Incidenl
I\'ormal
, ,
T
FIGURE 22.11 (Example ~~.~) Refraction of light by glass.
,
Problem A light ray o[wavelength 589 nm (produced by a sodium lamp) traveling through air is incident on a smooth, flat slab of crown glass at an angle of 30.0° to t.he normal, as sketched in Figure 22.11. Find the angle of refraction, 82 ,
" :\00°:
\;--:, ,,
Air
t
~Class
o
Strategy Substitute quantities inLO Snell's law and solve for the unknown angle of refraction, f)2'
From Table 22.1, find ", ~ 1.00 [or air and "2 = 1.52 for crown glass. Substitute these values into Equation (1) and take the inverse sine of both sides:
sin 8 2 =
(l.~O)(Sin 30.0°) 1.~2
= 0.329
8 2 = sin-I (0.329) = 19.2°
~
~
Remarks Notice that the light ray bends LOward the normal when it enters a material of a higher index of refraclion. If the ray left the material following the same path in reverse, it would bend away from the normal. QUESTION 22.2 If t.he glass is replaced by a transparent material with smaller index of refraction, "",ill the refract.ion angle (a) smaller, (b) larger, or (c) unchanged?
{)2
be
EXERCISE 22.2 J r the light. ray moves from inside the glass toward the glass-air interface at an angle of 30.0° to the normal, determine the angle of refraction.
Answer
The ray bends 49.5° away from the normal, as expected.
EXAMPLE 22.3 Gool
Li ht in Fused
uartz
Use the index of refraction to determine the effect of a medium on light'S speed and wavelength.
Problem Light of wavelellgth 589 nm in vacuum passes through a piece of fused quartz of index of refraction n = 1..158. (a) Find t.he speed of light in fused quartz. (b) "VhaL is the \vavelength of this light in fused quarLL? (c) ,,"Vhal is the frequency of the light. in Fused quartz?
Solution (a) Find the speed of liglH in fused quartz. c
3.00 X 108 m/s
11
] .458
v= - =
Obtain the speed from E.quation 22.4:
2.06 X 108 mls
(b) ''''hat is the wa\"clength of this light in fused quartz?
.1 0
A, = 11-
Use Eq. 22.7 to calculate the wavelength: (c) ,,\That is the frequency of' the light in [used quanz?
c
j=-=
The frequency in quarLz is the same as in vacuum. Solve r = fA [or the frequency:
A
.189 nm
404 nm
1.458
3.00 X 108 m/s 589 X 10
5.09 X 10'" Hz
9 111
..............................
a...••..•.••.•. _ .....•••••.
QUESTION 22.3
Remarks It's interesting to notc that the speed of light in vacuum, 3.00 X 1011 mis, is an upper limit for the speed of material o~jccts. In our treatment of relativity in
True or False: If light with wavelength A in glass passes into waleI' \vith index 11 w ' the new wavelength of the light is A/nw '
Chapter 26, we will find that this upper limit is consistent ,..;ith experimental observations. However, it's possible for a particle moving in a medium to have a speed that exceeds the speed of light in that medium. for example, it's theoret.ically possible for a particle to travel through fused quartz at a speed greater than 2.06 X lOR mis, but ill11ust slill have a speed less than 3.00 X 108 rn/s.
EXERCISE 22.3 Liglll with wavelength 589 nm passes through crystalline sodium chloride. In this medium, find (a) the speed of light, (b) the wavelength, and (c) the frequency of the light. Answer
(a) 1.94 X 108 mls (b) 381 nm (c) 5.09 X 1O'4I-lz
EXAMPLE 22.4 light Passing Thraugh a Slab Goal
Apply Snell's law when a ray passes i [lto and oul of another medium.
Problem A light beam traveling through a transparclltllledi\lm of index of refraction 11. 1 passes through a thick transp;uelll slab with parallel faces and index of refraction 1l~1 (Fig. 22.12). Show that the emerging beam is parallel to the incident beam.
FIGURE 22.12 (l::xarnple2~.4) \Vhen lig-ht passes through a naT slah of ll1aterial, rIle emergillg" be'11ll is parallel 10 the illcidcllt beam; therefore, OJ = 8s·
'"
~ :8 ------1-', _-!',
I
2\ :
Strategy Apply Snell's Jaw twice, once al the upper -' surface and once at the lower surface. The two equa1/1 j ~ (ions will be related because lhe angle of refraclion ,r 83 ." al the upper surface equals the angle of incidence at the lower surface. The ray passing through the slab makes equal angles with lhe normal at. the entry and exit points. This procedure will enable us to compare angles fJ 1 and 8". ~
.
'""
Solution Apply Snell's law to the upper surface:
(I)
Apply Snell's law to the lower surface:
(2)
'fl.]
sin O'!. = -
n':! 1·/.<)
Substitute Equation (I) into Equation (2):
sin 0 3 =
sin 8 1
-=sin 8 2 11.1
22.3
The Low of Refraction
741
Take the inverse sine ofhoth sides, noting that the angles are positive and less than 90 0
;
...
trr... • • • • • •
~
Remarks The pl'eccding result proves that the slab doesn'l alter the direction of the beam. It docs, however, produce a lateral displaccment of the beam, as shown in Figurc 22.12.
QUESTION 22.4 Suppose an additional slab with indcx 11:\ ' ....ere placed below the slab of glass. \'\'ould the exit angle at the botlOm surrace of this second slab still equal the incident angle at the upper surface of the first slab? (a) yes (b) no (c) depends on the mcdia EXERCISE 22.4 Suppose the ra)'. in air with n = 1.00, enters a slab with n = 2.50 at a 45.0° angle with respect to the normal, then exits the bOllom of the slab into water, \'lith n = 1.33. At what angle to the normal does the ray Icave the slab? Answer
32.Jo
EXAMPLE 22.5 Goal
Refraction of Laser Li ht in a Digital Videodisc (DVD)
Apply Snell's law together \vith geometric constraints. 1.20111111 and <:111 index of refraction 11 = 1.55. Note that this syslern is relatively immune lO small dust particles degrading the \'ideo quality because panicles would have 1.0 be as large as 0.700 mm 1.0 obscure the beam at lhe poil1l \..· here it Cl1lcrs the plastic.
Problem A DVD is a video recording consisting of a spiral lrack about 1.0 J.Lm wide with digital information. (See Fig. 22.13£1.) The digital information consists 01" a ."cries of pits that are "read" by a lascr beam sharpl)' focused on a track in the information layer. If the width (l of I he beam al. the in formation layer must cqual 1.0 J.lm 10 distinguish individual tracks and the width w of the beam as it enters the plaslic is 0.700 a mm, find the angle 0, at which the conical beam should ent.er t.he plastic. (See Fig. 22.13b.) Assume the plastic has a thickness
--- -
("l
A
A
("1
(i'l
~
~
(.)
r-,
,......
fil ~
Pl r"'
-
('"\
I""
~
ft)
11""
,...... m
(a)
,.
L=
Strategy Use right-triangle trigonometry to dcwnn inc the angle 02 and then apply Snell's law 1.0 obtain the angle 0,.
-Infonmninn la)'cr
,, , ,,, ,, , ,,
Plaslic 1/:1.35
~2.
0,
i' ,
w
FIGURE 22.13
'il., .. ~\rr
(Exalllpk 22.5):\
111 icrogTaph of a DV 0 "urfacc showillg: Lracb and piLS along each Iral·k. (Il) Cros~ seclion of a Calle-shaped la~cr 1lt..:
,0,
,
(h)
.
Solution From the top and bOllom of Figure 22.13b, obtain an equation relating 70, b, and a;
. IV
= 2b
Solve Ihis equal ion for band substitute given values:
b=
Now lise the langem function to fi nd e'!.:
tan
+
a
w- a
-~-
2
e.} -
....
b
=
700.0 X 10
6
m -
2 349.5!Lm
= - = ----'-00-I 1.20 X 10'!Lnl
1.0 X IW 6 m
= 349.5 !Lm
742
Reflection and Refraction of Light
Chapter 22
Finally. use Snell's law to find 6 1:
n'2sin 8'2
sin 0 1 = -n',,-":::si:::I1-=8-=.,
..................................... ~
.................... Remark
1 , 5 ~- Sin ' 16 ,-90 = 0.433 1.00
Despite its apparent complexity, the problem isn'L that. different from Example 22.2.
QUESTION 22,S Suppose the plastic were replaced by a material with a higher index ofrefracLion. How vl'Ould the width of the beam allhe information layer be affected? (a) rt would remain the same. (b) It would decrease. (c) It would increase. EXERCISE 22,5 Suppose you wish LO redesign the system to decrease the initial width of the beam from 0.700 0 mm LO 0.600 0 mill but leave the incident angle 8] and all other p:aramcters the same as before, except the jndex of refraction for the plastic matcrial (n2) and the angle 02' "'Vhat index of refraction should the plastic have? Answer
1.79
22.4 "
l.54
Crown glass
1.52 1.50
1.'18
1.46
400
500
600
700
A.,nm FIGURE 22.14 \'ariationsofindcx of refraction ill rhe visible spl.:ctrum with respeCllO \-aCIlUlll wave1crlgLh for lhree rnaLerials.
DISPERSION AND PRISMS
In Table 22.:1 we presentcd values for the index of refraction of various materials. If \,ve make careful rneasurements, however. we find that the index of refraction in anything but vacuum depends on the ".,'avelcngth of light. The dependence of the index of refraction on wavelength is called dispersion. Figure 22.14 is a graphical representation of this variation in the index of refraction with wavelength. Becallse n is a function of wavelength , Snell's 1m·\' indicates that the angle of refraction made when light enters a material depends on the wavelength of the light. As seen in the figure, the index of refraction for a material usually decreases with increasing wavelength. This means that violet light (A == 400 nm) refracts more than red light (A =0 650 nm) when passing from air into a material. To understand the effects of dispersion on light, consider what happens when light strikes a prism. as in Figure 22.15£1. A ray of light ofa single wavelength that is incident on the prism from the left emerges bent away from its original direction of travel by an angle called the angle of deviatioD. Now suppose a beam of ",hile light. (a combinat.ion of all visible wavelengths) is incident on a prism. Because of dispersion. the different colors refract through different angles of deviation and
a,
\Vhite li~lll
"""
Red
(a)
(e)
FIGURE 22.15 (a) A prism rerracl.~ a light ray and deviates the light through the
22.4
Dispersion and Prisms
743
\"'hile lighl
(a)
(b)
FIGURE 22.16
(a) Dispersion of whiLe light by a prism. Because n varies wilh wavelength. the prism r1isperst:s the white lighl 11110 its various spectral components. (h) DilTercnt nllon. of" light Lhat pass I hrollgh a prism arc refracted al diffc'·ClIl all~1t.:s bet·Ollise the index of rt'fme lion of I he ghss depends nil wavelength. Violet light bends the mosl. red lighl the least.
the rays lhat emerge from the second {';:'lce of the prism spread out in a series of colors known as a visible spectrum, as shown in Figure 22 ..16. These colors, in order of decreasing wavelength, are red, orange, yellow, green, blue, and violet. Violet light deviates the most, rcd light the least, and the rcmaining colors in the visible spectrum fall between thcse extremes. Prisms are often lIsed in an instrument known as a prism spectrometer, the essential clements of which are shown in Figure 22.17a. This instrument is COlllmonly used to study the wavelcnglhs emittcd by a light source, such as a sodium vapor lamp. Light from the sourcc is sent through a narrow, adjustable slit and lens to produce a parallel, or collimated, beam. The light then passes through t.he prism and is dispersed inLO a speClrum. The refracted light is observed through a telescope. The cxperimenter secs different colored images of the slit through the eyepiece of the telcscope. The telescope can be moved or the prism can be rotated 10 view the various wavelengths, which havc differcm angles of deviation. Figure 22.17b shows one type of prism spectrometer uscd in undergraduate laboratories. All hot, low-pressure gases emil1.heir own characteristic spcctra, so one use ora prism specLromCler is La identify gases. For example, sodiulll emits only tVI.'O \\'avclengths in the visible spectrum: two closely spaced yellow lincs. (The bright line~ I ike images of the slit secn in a spectroscope are called .'jJaimllines.) A gas em itting Ihese, and only thesc, colors call be identified as sodium. Likewise, mercury vapor has its own characteristic spectrum, consisting or four prominent wavelengthsorange, grecn, blue, alld violctlines-along with some wavelengths of lower intensity. The particular wavelengths emitted by a gas serve as "FlIlgerprints" of that gas. Spectral analysis, which is the measurement of Ihe wavelengths emitted or
APPLICATION Identifying Gases with a Spectrometer
FIGURE 22.17
(-.) A diagr'lIll t i L l prislll spectl"ome'ter. The' col(lrs III 1he "penrulll arc vl<'we~d through a 1l:lc ..,e:ope. (h) A pri!'1ll "'peCIl"lIIllI·ter ,,'illl illtc>n:h'lngeablc cUlllpnrlent ....
Slil
Tekscope (a)
(h)
744
Chapter 22
Reflection and Refraction of Light
absorbed by a substance, is a powerful general lOol in many scientific areas. As examples, chemists and biologists use infrared spectroscopy to identify molecules, astronomers lise visible-light spectroscopy to identify elements on distant stars, and geologists lise spectral analysis to identify minerals.
APPLYING PHYSICS 22.3
DISPERSION
When a beam of light enters a glass prism, which has nonparallel sides, the rainbow of color exiting the prism is a testimonial to the dispersion occurring in the glass. Suppose a beam of light enters a slab of material with parallel sides. "Vhen the beam exits the other side, traveling in the same direction as the original beam, is there any evidence of dispersion?
Explanation Due to dispersion, light at the violet end of the spectrum exhibits a larger angle ofrefraction on entering the glass than light at the red end. All colors oflight return to their original direction of propagation as they refract back out into the air. As a result, the outgoing beam is white. The net shift in the position of the violet light along the edge of the slab is larger than the shift of the red light. howevel-, so one edge of the outgoing beam has a bluish tinge to it (it appears blue rather than violet because the eye is not very sensitive to violet light), whereas the other edge has a reddish tinge. This effect is indicated in Figure 22,]8. The colored edges of the outgoing beam of white light are evidence of dispersion.
EXAMPLE 22.6 Goal
TraJ1SparCIll $lab
FIGURE 22.18
(Applying Physics
2V~)
Light Throu h a Prism
Calculate the consequences of dispersion
Problem A beam of light is incident on a prism of a cenain glass at an angle of OJ = 30.0", as shown in Figure 22.19. H the index of refraction of the glass for violet light is 1.80, And (a) 0" the angle of refraction at the air-glass interface, (b) <1>" the angle of incidence at the glass-air interface. and (c) 4)1' the angle or refraction when the violet light exits the prism. (d) What is the value of t1)l, the amount by which the violet light is displaced vertically?
Strategy This problem requires Snell's law to find the refraetjon angles and some elementary geometry and trigonometry based on Figlll"c 22.19.
60· FIGURE 22.19
,..
.
........... -
Solution (a) Find O'!., the angle of refraction at the air-glass interface.
Use Snell's law to find the first angle of refraction:
n J sin OJ = sin 0, --> 0,)
-
(1.00) sin 30.0 = (1.80) sin 0,
0.500) = sin- l ( -= 1.80
16.1°
~
22.5
The Rainbow
745
(b) Find tP2' the angle ofincidcllce at the glass-air interface. Compute the angle {3:
f3 = 30.0' - 02 = 30.0' - 16.1' = 13.9'
Compute the angle Q' using the factlhatlhe sum or the interior angles ofa triangle equals l800:
180'
The incident angle tP2 at the glass-air interface is complementary to a:
=
13.9' + 90' + a --> a
= 76.1' 13.9'
(c) Find tPl' the angle of refraction when the violet light exits the prism. Apply Snell's law:
(~)Sin-I
(",sin
c.~o) sin-I [(1.80) sin
25.6'
(d) What is the value of 6.y, the amount by which the violctlight is displaced vertically? Use the tangent function to finclthe vertical displacement:
lI.y tanf3=II. x
-->
lI.y=II.."tanf3
lI.y = (6.00 em) tan (13.9') =
IA8 em
....
.. ~
Remarks The same calculation for red Iighl is left as an exercisl:. The violet lighl is bent more and displaced farther dm'\'n the face of the prism. Notice that a theorem in geomctry about parallel lines and the angles created b)1 a transverse line give cP 2 = f3 immediately, ,,:hich v.,'Quld have saved some calculation. In general, however, this tactic miglu not be available.
QUESTION 22.6 On passing lhrough the prism, will yellow light bend through a larger angle or smaller angle than the violet light? (a) Yellow light bends lhrough a larger angle. (b) Yellow Jighl bends lhrough a smaller angle. (e) The angles are the same. EXERCISE 22.6 Repeat pans (a) through (d) of the example for red light passing lhrough the prism, given that the index of refraction for red light is 1.72.
Answers
22.5
(a) 16.9' (b) 13.1° (e) 22.9' (d) 1.40 em
THE RAINBOW
The dispersion of light into a spectrum is demonstrated most vividly in nature through the formation of'a rainbow, often seen by an observer positioned between the Sun and a rain shower. To understand how a rainbO\.. . is formed, consider Active Figure 22.20. A ray of light passing overhead strikes a drop of water in the atlllosphere and is refracted and rcnected as follO\\"s: Lt is ftrst refracted at the front surface or the drop, with the violet light deviating the most and the red light the least. At. the back surface of the drop, the light is reOected and returns to the frollt surface, where it again undergoes refraction as il moves from water into air. The rays leave the drop so that the angle between the incident white light and the reUlrning violet ray is 40° and the angle between the while light and the returning red ray is 42°. This small angular difference between the returning rays causes us to see the bow as explained in the next paragraph. Now consider an observer viewing a rainbow, as in Figure 22.21a (page 746). If a raindrop high in the sky is being observed, the red light returning from the drop can reach the observer because it is deviated the most, but. the violet light passes
~ghl
~ V II
ACTIVE FIGURE 22.20 Refractioll of sunliglll by a sphericli railldrop.
746
Chapter 22
Reflection and Refraction of light
FIGURE 22.21 (a) The formatioll ofa rainbow. (b) Thi~ phowgraph of a rainbow ShOWS;1 distinct sl.:Cfllldary rainbow with 11w colors re\-l;rst:;d.
(a)
(h)
over the observer because it. is deviated the least. Hence, the observer sees this drop as being reel. Similarly, a drop lower in the sky would direct violcllighl tOward the observer and appear 1.0 be violet. (The red light from this drop ,..·ould strike the ground and not be seen.) The remaining colors of the spectrulll would reach the observer from raindrops lying between these two extreme positions. Figure 22.2] b shows a beautiful rainbow and a secondary rainbow with its colors reversed.
22.6
Huygens' principle
~
HUYGENS' PRINCIPLE
The laws or reflcClion and refraction can be deduced using a geometric method proposed by Huygens in 1678. l-Iuygens assufned light is a form of wave motion raj her than a stream of particles. He had no knO\vledgc of the IlaW re of light or of its electromagnetic character. Nevertheless, his simpliAcd wave model is adequate for understanding many practical aspects of the propagation of light.. Huygens' principle is a geometric construction for determining at some lnstant the position of a ne\.. . \vave front from knowledge of the wave front that preceded it.. (A wave front is a surface passing through those points ora wave which have the same phase and amplitude. For instance, a wave fronL could be a surface passing through the crests of waves.) [n Huygens' construction, all points on a given wave front are taken as point sources for the production of spherical secondary waves, called wavelets, tbat propagate in the forward direction with speeds cbaracteris~ tic of waves in that medium. After some time has elapsed, the new position of the wave front is the surface tangent to the wavelets. Figure 22.22 illustrates t\·..-o simple examples of Hu}/gens' construction. First, consider a plane lNave moving through free space, as in Figure 22.22a. At t = 0, the wavc front is indicatcd by lhe planc labcled AA'. 1n l-Iuygcl1s' construclion, each point on this wave from is considered a point source. For darity, only a few points on AN arc shown. "Vith these points as sources [or the wavelets, we draw circles of radius c !J.t, \vhcre cis [he speed oflight in vacuum and ~f.is the period ofpropagation from one wave from to the next. The surface drawn tangent to these wavelets is the pl
FIGURE 22.22 J-1u)'gens' construeliom for (~t) ,1 plane wave propag-alill~ to the right ,llld (h) a spherical wave.
B
A
Old
Oldwnl
W;1Ve
froll!
A' (a)
(h)
Nr:w \'{aVl: fronL
22.6
2 I
3 ~ ~
A'D
A'
f)
!J
). I' ,I JJ ,I
\..-\
6,,
,,f\
,
8'
8,
C
(hi
(a)
Huygens' Principle Applied to Reflection and Refraction The laws of reflection and refraclion were stated earlier in the chapter \v'ithol.lt proof. vVe nm,,· derive these laws using Hllygens' principle. Figure 22.23£1 illustrates the law of reflection. The line AA' represents a wave front of the incident light. As ray 3 travels from A' to C, ray 1 reflects from A and produces a spherical wavelet of radius AD_ (Recall that the radius of a Huygens W3\-elCt is v til.) Because the twO wavelets having radii A'Cand AD are in the same medium, they have the same speed v, so AD = A'e. \'feamvhilc. the spherical \·vavelct centered at B has spread only halfas far as the one centered at A because ray 2 strikes the surface later than ray 1. From J-luygens' principle, we find that the reflected \va\-e [rant is CD, a line tangent \.0 all the outgoing spherical \vavelets. The remainder or our analysis depends on geometry, as summarized in Figure 22.23b. Note that the righl triangles ADC and AA'Care congruent because they have the same hypotenuse, !le, and because AD = A'e. From the figure, we have
A'C sin 8 1 = AC
and
,
AD AC
smO' = I
The rigl1l-hand sides are equal, so sin 8 = sin 0;, and it follows that 0 1 = 8;, which is the law of reflection. Huygens· principle and Figure 22.24£1 can be llsed to deri\'e Snell's law ofrefraction_ 1n the time interval til, ray 1 moves from A. to B and ray 2 moves from A' to C. The radius of the outgoing spherical wavelet centered at A is equal to v2 dr. The distance A'e is equal to VI 6./. Geon1etric considerations 5hm.. . that angle A~4C equals 8] and angle ACE equals 82 . From triangles AA'Cand ACB, we find that and
2
i\kdiuml, speed of liglH
v,,6.l sin O'J = - - -
-
AC
VI
A'
--r-,Icdiulll 2. spc.;ed or light t'2
\
2
(a)
FIGURE 22.24
747
FIGURE 22.23 (a) J-111ygCI1S· COllstruel ion for prm·ing (IIC law of reflecI ion. (b) Triangle _\DC is congr (ICIlI to triangle AA'C.
~
....""-
Huygens' Principle
(b)
(a) I-Illygens' conslruction for proving lhe la\\' ofrefraclion. (II) Overhead view ora band rolling from concrete onto grass.
Chapter 22
748
Reflection and Refraction of Light
Hwe divide the first equation by the second,
\\-'C
sin 8 1
VI
sin 0':1.
v2
From Equation 22.4, though, we know that
VI
gel.
= r/tl. and v 2 = c/n 2 . Therefore,
sin 8)
clnj
112
sin 8'1
cll1'!.
11)
and it foIl0\"·5 that
which is the law of refraction. A mechanical analog of refraction is shown in Figure 22.24b. When the left end of thc rolling barrel reaches the grass, it slows down, while the right end remains on the concrete and moves at its original speed. This difference in speeds causes the barrel to pivot. changing the direCLion of its motion.
22.7 TOTAL INTERNAL REFLECTION
,
.~
/I' . This pll(,tograpll ,hows Ilonparallellighl rav, ('IHering a glass prism. TIl(' bOl\um
IWO I';\\'$
undcq..(o 10t,11
iJllt'rnal,'efknloll alThe IOllgcsl side of the prism. Thl' top three ra~'S an: refracted al the longest side as tlwy le;l\e the prism.
An interesting effect called lolal internal reflection can occur when light encounters the boundary bClween a medium with <:t higher index of refraction and one with a lower index of refraction. Consider a light beam traveling in medium 1 and meeting the boundary between medium L and medium 2, where 11 J is greater than 112 (Active Fig. 22.25). Possible directions or the bealTI are indicated by rays I through .'1. Note that t.he refracted rays are bent away from the normal because 11 1 is greater than 12 2. At some particular angle of incidence 8" called the critical angle, the refracted light ray moves parallel La the boundary so that (J':!. = 90° (Active Fig. 22.25b) . .For (I1/,glps oJ inridence greater Own 8" the beam is enLirely reflected at the boundary, as is ray 5 in Active Figure 22.25a. This ray is reflected as though it had struck a perfectly rdlecling surface. It and all nlys like it obey the law ofreflcction: the angle of incidence equals the angle of reflection. ",'e can use Snell's law 1.0 find the critical angle. \Vhen 8. = O( and 8'1 = 90°, Snell's law (Eq. 22.8) gives
[22.9] Equation 22.9 can be used only when 11 1 is greater than II;! because total internal reflection occurs only when Light attempts to move from a medium of higher index of refraction to a medium of lower index of refraction. If '11 were less than 7'-2, Equation 22.9 would give sin 8( >], which is an absurd result because the sine oran angle can never begrealer than J. ACTIVE FIGURE 22.25 (... , Ra\-s from a medium with index of rdractioll I'I tnwd 10 a medium with indc:'l. ofr<:fl,telioll 1I!. \\hen: 11, > 112" lh th(' allgk of incidcllce illcre"..cs, the all~lc nf rerranion (11 increases until O~ is ~O (1';1\' I). For ('\'('11 larger angles "f illddcllrc, total internal reflection nccur~ (ra\ :». (b) The
;'IIormal
Nor·mal
n.>ll'! 2
.11
3
en/lml (nljfll'lJ r
(aJ
(b)
22.7
Total Internal Reflection
FIGURE 22.26
749
Intern;tl reflection
ill a prism. (il) The ra} j" de\ i:llCd by 90°. (b) The r1ircClion oflhe fa)' is revcrsed. (c) Two prism:> ll."e
perisctlpc.
(h)
(a)
When mediulll 2 is air, the critical angle is small for substances with large indices of refraction, such as diamond, where n = 2.42 and Or = 24.0", By comparison, for crown glass, n = 1.52 and Or = 41.0°. This propeny, combined with proper faceting, causes a diamond to sparkle brilliantly. A prism and the phenomenon of LOLal internal relleClion can alter the direction oftra\'cl ora light beam. Figure 22.26 illustrates two such possibililies. In one case the light beam is deflected by gO.a" (Fig. 22.26a), and in the second case the path of the beam is reversed (Fig. 22.26b). A common application ofLOtal internal reOeclioll is a submarine periscope. In this device two prisms arc arranged as in Figure 22.2fic so that an incidelll beam of light follows the path shm.. . n and the user call "see around corners."
APPLYING PHYSICS 22.4
f
.
e
(i\ppl}'il1g I'h}'sin. 22.4)
EXAMPLE 22.7
Submarine Periscopes
TOTAL INTERNAL REFLECTION AND DISPERSION
A beam ofv,!hite light is incident on the curved edgc ora semicircular piece of glass, as shown in Figure 22.27. The light enters the curved surface along the normal, so it shows no refraction. It encounters the straight side of" the glass
FIGURE 22.27
APPLICATION
normal to the curved side and encounters the straight. side at. the center of curvature of the curved side. ,",Vhy does the refraC1.ed beam become redder as it approaches a direction parallel to the straight side? Explanation \""hen the outgoing beam approaches the direction parallel to the straight side, the incident angle is approaching the critical angle for total internal reflection. Dispersion occurs as the light passes out or the glass. The index of refraction for light at the violet end of thc visible spectrum is larger than
A View from the Fish's E e
Goal Apply the concept of total imernal reflcction. (a) Find the critical angle lor a water-air boundary. (b) Use I.he result or pari (a) to predict what a fish will see (Fig. 22.28) if it looks up LOw'ard the water surface al angles of 40.0°. 48.6°, and 60.0°.
FIGURE 22.28 (E.x~lInplc 22.7) A fi!>h Inok~ Ilpward loward lhe \~'att:r's surface.
Chapter 22
750
Reflection and Refraction of Light
Strategy After finding the critical angle by substitution, usc that the path of a light ray is reversible: at a given angle, where,"er a light beam can go is also where a lighl beam can come from, along the same path.
Solution (a) Find the critical angle for a \"'
SubstilUlC into Equation 22.9 to flfld the critical angle:
sin
11, 1.00 e, = ---=. = - - = 0.750 n\ 1.333
e, =
5In-\ (0.750) = 48.6'
(b) Predict whata fish will see ifit looks up lO\\"arci the water surface at angles of 40.0°, 48.6°, and 60.0°, At an angle of 40.0°, a beam of light from underwat.er will be refracted at the surface and enter the air above. Because the path of a light ray is reversible (Snell's law works bOlh going and coming), light from above can follow the same path and be perceived by the nsh. At an angle of 48.6°, the critical angle for water, light frolll underwater is bem so that it travels along the surface. So
light following the same path in reverse can reach the fish only by skimming along the water surface before being refracted to\vard the fish's eye. At angles greater than the critical angle of 48.6°, a beam of light shot toward the surface "viII be completely reflected down toward the bOltom of the pool. Reversing the path, the fish sees a reflection of some object on the bottom .
...... •
•
•
•
•
• • ••• • ••••• • • • • • •• • •
• •
•
•
•
•
•
•
•
• •
•
• ••
• •
• •
• •
•
•
• •
•
•
•
•
•
•
•
•
•
• •
•
• •
• •• •
• • • • • • • • • • • • • • ",j
QUESTION 22.7 If the water is replaced by a transparent nuid \'lith a higher index of refraction, is the critical angle of the fluid-air boundary (a) larger, (b) smaller, or (c) the same as for water? EXERCISE 22.7 Suppose a layer of oil with n = 1.50 coals the surface of the water. \,\That is the critical angle for total internal refleclion for light traveling in the oil layer and encountering the oil-\\'ater boundary? Answer
62.7°
Fiber Optics
FIGURE 22.29 Light travels in a cun'ed transparent rod br multiple internal reflections.
APPLICATION
(I
Fiber Optics in Medical Diagnosis and Surgery
APPLICATION Fiber Optics in
Telecommunications
Another interesting application of total internal reflection is the use of solid glass or transparent plastic rods to "pipe" light from one place to another. As indicated in Figure 22.29, light is confined (Q traveling within the rods, even around gemle curves, as a result of successive internal reOeclions. Such a light pipe can be quite flexible if thin fibers are Llsed rather than thick rods. If a bundle of parallel fibers is used to construct an optical transmission line, images can be transferred from one point t.o another. Very lillIe light intensity is 10s1. in these fibers as a result of reflections on the sides. Any loss of intensity is due essentially to reflections from the twO ends and absorption by the fiber material. Fiber-optic devices are particularly useful for viewing images produced at inaccessible locations. Physicians often use fiber-optic cables to aid in the diagnosis and correction of cenain medical problems without the intnlsion of major surgery. For example, a fiber-optic cable can be threaded through the esophagus and into the stOmach to look for ulcers. In this application the cable consists of two fiber-optic lines: one to transrnit a beam of light into the stomach for illumination and the other to al10w the light to be transmitted out of the stomach. The resulting image can, in some cases, be viewed directly by the physician, but rnore often is displayed on a television Illoniwr or saved in digital form. In a similar way, flber-optic cables can be used to examine the colon or to help physicians perform surgery \'lithout the need for large incisions. The field of fiber optics has revolutionized the entire communications industry. Billions of kilometers or optical fiber have been installed in the United States to carry high-speed Internet traffic, radio and television signals, and telephone calls. The fibers can carry mLlch higher volumes of telephone calls and other forms of communication than electrical wires because of the higher frequcllcyofthe infra-
red light used LO carry the information on optical fibers. Optical fibers are also preferable to copper ",Ii res because Ihey arc insulaLOrs ,lOci don't pick up stray cIccI ric and magnetic fields or electronic "noise."
APPLYING PHYSICS 22.5
DESIGN OF AN OPTICAL FIBER
An optical f'iber consists ofa transparellt core surrounded by cladding, \·,,rhich is a l1Iau:rial with a Im\:er index ofrcfraction than the core (Fig. ~2.;)O). A cone ofang-Ies, called the acceptance cone, is at the el1lrance 1.0 the llber. Incoming light at angles within this cone will be transmittcd through the fiber,
1II········~Jach;' I -
Cladding
\vhcreas light entcrinl{ the corc from anglt:s oUbide the cone will not be transmittcd. Thl' figure 'sh()w~ a light ray entering the fiber-just within the aneptance cone and undergoing total internal reneClioll al the interface betwecn the core and the cladding. If it. is technologic£llly difficult to produce light so that it enters the fiber [rom a small range of angles. how could you a(~jusl. the indices or refranion of t.he cnrt: and cladding to increase the size of the acceptance cone? \Vould you design the indices to be farlhel apart or closer together?
Corl'
FIGURE 22.30
Explanation The acceptance cone would become larger if the Cl-ilical angle (0,. in the figure) could be made smaller. This acHustmenl can be donc by making the index ofrefraclinn of the cladding material smaller so lhatl.lle indices of refraction of the u,re and cladding material would be fanher apan.
(Applying l'hysics 22.5)
SUMMARY 22.1
The Nature of Light
122.2]
Lighl has a dual nature. In some experiments it acts like (\ W;IVt::. in others like a p;trticle, called a photon by Einstein. The energy ora phOlon is proportional to its frequency, I,·
where 11
22.2
=
6.63 X 10
~
,11.1' sis
/1
[22.11
Plallck\- col1Jlrml.
Reflection and Refraction
In the reflection of light otT a flat. smooth surface, the angle of incidence, 0 1, with respect to a line perpendicular LO the surface is cqualLO rhe angle ofrefleetion, 0;:
,t
I.ight that passes inlo lranspa"enll1lediulll i . . b('lll al tht: boundary and is said 10 be I"Ij·'nded. The angle of refractioll is the angle the ray makes with respect to a lint' pt:rpcndiLular to the surface afler it has entered till' IH'W mediulll.
22.3
The Law of Refraction
The index of refraction of" a material. n=
r
v
II,
is defined a.'s
122.41
Chapter 22
752
Reflection and Refraction of Light
where cis the speed ofliglH in a vacuum and lJ is the speed of light in the mmcrial. The index of refraction of a material is also
[22.7]
"0
where is the waye!cngth of the light in vacuum and A" is its wavelength in the material. The law of refraction, or SocII's law, stales thal
[22.8] where n j and 11'2 are the indices of refraction in the two media. The incident ray. the reflected ray. the refracted ray. and the normal to the surface all lie in the same plane.
22.4
Dispersion ond Prisms
22.5
The Roinbow
The index of refraction ora material depends on the wavelength of the incidenlliglu, an effect called dispersion. Light at the violct end of the spectrum exhibilS a larger angle of refraction on elltering glass than light at the red end. Rainbows are a consequence of dispersion.
22.6
Huygens' Principle
Huygens' principle states that all POilll5 on a w3\·e frolll are pointsourc~sfor the production of spherical secondal'y waves called wavelets. These wavelets propagate forward ata speed characteristic of waves in a particular medium. After some time has elapsed, the new position of the wave front is the surface tangent to the \,,'avelets. This principle can be used to deduce the laws of reflection and refraction.
22.7
Totol Internol Reflection
Total imernal reflection can occur when light, traveling in a rnedium with higher index of refraction, is incidclll on the bounda/-y of a material with a lower index of refraction. The maximum angl.e oJ incidence 0l for \vhich light can move from a medium wilh index 11\ into a medium with index 112 , where 11] is greater than 1l2 , is called the critical angle and is given by 11.,
sin Or =--=.
[22.9]
'" Total internal reflection is used in the optical fibers that carr)' data at high speed around the ,.. .orld.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE·CHOICE QUESTIONS 1. How many SOO-nm photons does it take to have the same Lotal energy as four 200-nm photons? (a) -I (b) 2 (c) 4 (d) 8 (e) 16 2. Carbon disulfide (11 = 1.63) is poured into a container made of crown glass (11 = 1.52). \oVhal is the critical angle for imernal reflection of a ray in the liquid when it is incident on the liquid-to-glass surface? (a) 89° (bl 69' (c) 2]· (d) 4.00 (c) 33' 3. A monochromatic light source emits a wavelength of 495 nm in air. When passing through a liquid, the wavelength reduces to 434 nm. \,,"'hat is the liquid's index of refracLion' (a) 1.26 (bl 1.49 (e) 1.14 (d) 1.33 (e) 2.03 4. What happens to a liglll wave when it travels from air imo glass? (a) Its spced remains the same. (b) lts speed increases. (c) Its wavelength increases. (d) lts wavelength remains the same. (e) Its frequency remains the same. 5. The index of refraction lor water is about4/~·t What happens to light when it lnwels [rom air into water? (3) Its speed increases to 4c/3, and its frequency decreases. (b) Its speed decreases to 3c/4, and its wavelength decreases by a facLOr of 3/4. (c) lls speed decreases to 3c/4, and its wavelength increases by a [acLOr of 4/3. (d) Its speed and frequency remain the same. (e) Its speed decreases to 3(/4, and its frequency increases. 6. Light can travel from air into water or from water into air, with some possible paths as shown in Figure MCQ22.6. Which path will the light most likely follow? (a) II (b) B (e) C (d) D (el E
Air
Waler A
8
c
D
E
FIGURE MCQ22.6
n,
7. Light traveling in a medium of index of refraction is incident on another medium having an index of refraction 112 , Under which of the following conditions can LOtaI reAcction occur at the interface of the t""·o media? (a) 11.2 > 11 1 (b) 'ILl> rl 2 (c) Light travels slm.. ·cr in the second medium than in the first. (el) The angle of incidence is less than the critical angle. (c) The angle of incidence lTILlSl equall.he refraction angle. 8. A light ray containing both blue and red wavelengths is incident at an angle on a slab of glass. Which of the sketches in Figure MCQ22.8 represents the most likely oULcome? (al A (b) B (c) C (d) 0 (e) none ofLhese
B
C FIGURE MCQ22.8
Problems 9. 'Which color light is bent the most \",hen entering crown glass frol11 air at some positive angle 0 with respect to Lhe normal?
subsequcntly passes into a second slab of m'llerial with
753
index of refraction n 1 before passing back into vacuum again. The surfaces of the different materials are all parallel to one another. As the light exits the second slab, whal can be said of the final angle 4J that the liglll makes with the normal? (a) 8 (b)
CONCEPTUAL QUESTIONS 1. A ra~ of light is mm-ing from a material having a high index of refraction inw a material with a lowcr index of refraction. (a) Is the ray bentLOward the normal or away from it? (b) If the wm'elenglh is 600 nm in the material with the high index of refraction, is it g-reater, smaller. or the same in the material with the lower index of rerraction? (c) How does the frequenc), change as the light moves between the two materials? Do('s il increase. decrcase. or remain the same?
2. Why docs the al-c ofa rainbow appear "lith red on LOp and violet on the bOllom?
a body of water warmer than the air. "-hat is seen is the hill and an ilwerted image directly below it. Explain how these mirages are fonned.
7. Explain why a diamond loses most of its sparkle when submerged in carbon disulfide. 8. Suppose YOll are told that onl) two colors of light (X and Y) are sent through a glass prism and that X is benl more than Y Which color tra\els more slowly in the prism?
9. The level of water in a clear, colorless glass can easily be
3. A scientific supply catalog advenises a material having an index of refraction of 0.85. Is that agooe! prOdUClI.O buy? Why or \\'h}' not?
obseryed with the naked cye. The level of liquid helium in a clear glass vessel is extremely difficult to see with the naked eye_ Explain. Hint: The index of refraction of liquid helium is close to that of air.
4. Under what conditions is a mirage formed? On a hot day, what arc we seeing when we observe a mirage ora w,lIer puddle on the road?
10. Is it possible to have t.otal internal reflection for light
5. I n dispersive male rials, the angle of refraction for a light ray depends on the wavelength of the light. Does thc angle of reflection from the surrace orlhe material depend on 1he wavelength? Why or why lIot?
6. A type of mirage called a pillgo is oftcn observed in Alaska. Pingos occur when the light from a small hill passes to an observer by a path lhat takes the light ovcr
incident from air on waler? Explain. 11. ',Vhy does a diamond show flashes of color when
observed under white light? 12. Explain why an oar partially submerged in water appears to be bent.
13. Why do astronomers looking al distam galaxies talk abotll looking backward in lim6
PROBLEMS ~
The )}mblems for this chapter m;l\ bl· assigned online al WchAssigH_ J. 2.:1 = sllaiglufofW;lfd. inllfllll·diall·, rhallt·llglllJ.\ = dcnole!oo gnided problem _ = denotes enhanced content problem
WebAssign
m
- bimncdical applicalion 0= denotes full solution available in Slut/rol Solutro"s .\lmillai/ Sillily
(;1J.1t/~
SECTION 22.1 THE NATURE OF LIGHT I. During lhe Apollo XI Moon landing, a rClrordlC:Cling
pal1el "'a3 erected on the Moon's surrace. The spc:ed of lighl c
2. miJ (a) \'\'hat is the encrg)' in joules of an x-ray phoLOn with wa,·e1ength 1.00 X 10-10 m? (b) Convert the energy LO electron volts. (c) If more penetrating x-rays are desired, should the wa,·c1ength be increased or decreased? (d) Should the frequency be increased or decreased? [IJrind the energy of (a) a photon having a frequenn of 5.00 X 10li 1-17 and (b) a photon having a ,,'a,'clength of 3.00 X 102 nm. Express }:our answers in units of eleclron '-Oils, nOling that I eV = 1.60 X 10-19 J. 4. _ (a) Calculalc the wa"c1ength or liglll in vacuum that has a frequency of 5.45 X 101-1 Hz. (b) \\Ihm is its wave1englh in benzene? (c) Calculate the energy of one phoLOn of such light in vacuulll. Express the answer in electron volts. (d) Does the energy of lhe photon change when it enters the benzene? Explain. 5. Find the speed of light in (a) water, (b) CrQ,,·n glass, (lnd (c) diamond.
Chapter 22
754 6.
Reflection and Refraction of Light
aa
(a) Find a symbolic expression for the wavelength A ofa photon in terms of its energy E. Planck's COnSl
7. A ray of light travels from air into another medium, making an angle of 450 with the normal as in Figure P22.7. Find the angle of rerraClion 8',! if the second medium is (a) quanz, (b) carbon disuHide, and (c) water.
Air
"
pendicular to the plane in which the light moves. Find the (a) speed, (b) frequency, and (c) ,,,·a,·elength of the light in the Lucile. Suggestion: Usc a protractor. 15. The light emincd by a helium-neon laser l1<1s a wavelength of 632.8 nm in air. As lhe light travels from air into zircon, find its (a) speed, (b) wavelength, and (c) frequency, all in the zircon. 16. A flashlight on the bottom of a 4.00-m-deep swimming pool sends a ray up\\'ard and at an angle so that lhe ray st.rikes the surface of the water 2.00 m from the point directly above the flashligill. What angle (in air) docs the emerging ray make with the water's surface? []L]How many times will the incident beam shown in Figure P22.l7 he reflecled by each of the parallel mirrors?
P
SECTION 22.2 REFLECTION AND REFRACTION SECTION 22.3 THE LAW OF REFRACTION 8. The 1\\'0 mirrors in Figure P22.8 meet at a right angle. The beam of liglH in the vcrueal plane P strikes mirror 1 as shown. (a) Determine the distance Ihe rdlecled light beam travels before striking mirror :l. (b) In what direction does the light beam Lnwel after being reflected from mirror 2?
~
Mirror
MirTOI'
FIGURE P22.7
1.00
l
~----, Light
I
Mirror: <) ~
I --X, 9
I
jFbC;lm (
1 -IO/JO 1I <' 0-
-
._:I~
1 1 I ,
1',... -_
MilTor
FIGURE P22.8
9. An underwater scuba cliver sees the Sun (It an apparclll angle of 4-5.0° from the vertical. What is the actual direction of the Sun? 10. A beam oflightlra\'cling in air strikes the surface of minera] oil at an angle 01'23.1 ° with the nonnall.O the surface. If the light tra\'cls at 2.17 X 10 8 m/s in the oil, what are the ("j index ofrcfraction o[miner",l oil and (b) angle of refraction?
[TIJA laser be;llll is incident at an angle of 30.0° l.o the venical onLo a solution of com syrup in water. If the beam is rdracted to 19.24° Lo the venical, (a) whm is the index of refraction of the syrup solution? Suppose the light is red. with wavelength 632.811111 in a vacuum. Find its (b) wavelength, (c) frequency, and (d) speed in the solution.
12. Ughl containing wavelengths of 400 nm, 500 nm, and 650 nm is incident ('rom air on a block of crown glass at an angle 01" 25.0°. (a) Are all colors l"efracted alike, or is one color bent morc than the others? (b) Calculate [he angle of rcfraction ill each case to ,·crify yow· answcr. 13. A ray or-light is incident on the surface ofa block of clear ice fit an angle of <10.0° with the normal. Pan of the light is reflected, and pan is refracted. Find the angle between the reflected ;Ind refracted light.. 14. A laser beam with vacuum wavcleng·th 632.8 nm is incident [rom air OlltO a block of" Lucil.e, as shown in Active Figure 22.Gb. The line of sight of the phoLOg-raph is per-
ill
sooo ___
~
~n~J(~er~1 ~l:~n~
L- 1.00 m-----J FIGURE P22.17
IS.
Bail A ray of light strikes a fLu. 2.00-cm-thick block of glass (11 =- 1.50) atan angle of30.0" with respect to the nor· mal (Fig. P22.18). (a) Find the angle of refraction at lhe top surface. (b) find the
2.00 em I
FIGURE P22.18
[illThe light beam shown ill Figure 1)22.19 makes an angle of ~O.W with the normal line NN' in the linseed oil.
~ \
Ait
N
Linseed
: fOOO ~
'0:
I
N' FIGURE P22.19
oil
Water
755
Problems
Determine the angles linseed oil is ].48.)
and 0'. (The refractivc index for
(J
20. A laser beam is incident on a 45°_<15°_90° prism perpendicular to one o[its faces. as shown in Figure P22.20. The transmitted beam that exits the h}'putcllu~C uf the pI iS1l1 makes an angle of 15° with the direction of the incident beam. Find the index of refraction of the prism.
FIGURE P22.20
21. Two light pulses are emiucd simultaneousl}' from a source. The pulses take parallel paths to a deteclOr 6.20 m away. but one moves through air and the other through a block of icc. Determine the difference in the pulses' times of arrival at the detector. 22.
III
A narrow beam of ultrasonic waves reflects ofT lhe liver tumor in Figure P22.22. 1r the speed of the wave is 10.0% less in the liver than in the surrounding medium, determine the depth of the tumor. I 1............. 12.0
em
I
:~
,
I
l
"'\
25. A beam of" light both rencers and refracts at the surf~lce between ail" and glass, as shown in Figure P22.2S. If the index of refraction ohhe glass is l1g , find the angle of incidcnce, 0t, in the air that would result in the reflected ray and the refracted ray beillg perpendicular to each olher. 111:111: Remember lhe identit}' Sill (90° - e) = cos e.
I
I I
~
I
:>
FIGURE P22.25
26. A cylindrical cistern, conslruCled below ground level, is 3.0 m in diameter and 2.0 III deep and is filled to the brim with a liquid whose index of refraction is 1.5. A small ol~iect rests on the bottom of the cistern at its center. I [ow far from the cdge of the cistern can a girl whose eycs are 1.2 m from the ground stand and still see the object? 27. An opaque cylindrical tank with an open top has a diametef of 3.00 III and is completely filled with W;Hef. Whcn the afternoon Sun reaches an angle of 28.0° above the horizon, sunlight ceases to illuminate the bonom of rhe tank. How deep is the rank?
I I
\
beam is sent from the submarine so that the beam strikes the surface 01" the watcr 2.10 X 10 2 m from the shorc. A building stands on the shore, and the lascr beam hits a target ar the top of the building. The goal is to r-ind the height of the target above sea level. (a) Draw a diagram of the silLlation, identifying the two triangles that are imponant LO finding lhe solution. (b) Find the angle of incidence of the beam srriking the water-air interface. (c) Find the angle of refraction. (e1) \o\That angle does the refracted beam make v"'ith respect to the horizontal? (e) Find the height of the target above sea level.
Tumor
FIGURE P22.22
123.1_ A person looking inlO an empty container is able to sec the far edge of the container's botLOm, as shown in Figure P22.23a. The height of the container is h, and its width is d. When the container is completely filled with a nuid of index of refraction II, the person call see a coin at the middle of the container's bonom, as shown in Figure P2~.23b. (a) Show that rhe ratio hid is given by
~= d
' JB ~ 4 - 11'2
(b) Assuming the container has" w'idth of 8.0 Clll and is filled with W
I
h
~~l
f--d--l (a)
28. A certain kind of glass has all index of refracrion of 1.6.50 [or blue lightofwavclength 430 nm and an indcx of 1.6]5 for red light of wavelength 680 11m. If a beam containing t.hese two colors is incident at an angle of 30.00° on a piece of this glass, what is the angle between Ihe two beams inside the glass? 129.1The index of refraction for red light in water is 1.331 and t.hat for blue light is 1.340. If a ray of white light enters the water at an angle of incidence of 83.00°, whal arc the underwater angles of refraction for the blue and I-cd components of the light? 30. The index of refraction lor crown glass is 1.512 at a wavelength of 660 nm (red), whereas its index of refraclion is 1.530 at a wavelength of 410 nm (violet). If both wavelengths are incident on a slab of crown glass at the same angle of incidence, 60.0°, what is the angle of refraction for each wavelengt h?
1
\
SECTION 22.4 DISPERSION AND PRISMS
(b)
FIGURE P22.23
24. _ A submarine is 3.00 X 10 2 m horizontally from shore (\ nd 1.00 X 102 m beneath the surface of the water. A laser
31. A light beam containing red and violet wavelengths is incident on
756
Chapter 22
Reflection and Refraction of Light
32. The index of refraction for violcllight in silica fillll glass is 1.66 and that for red light is 1.62. What is the angular dispersion of visible light passing through an eqllilat~ eral prism of apex angle 60.0" if the angle of incidence is 50. 0°? (See Fig. P22.32.)
result in total internal reflection at the water-air interface? (c) Does the thickness of the water layer or glass aJfeCl the result? (cJ) Does the index of refraction of the intervening layer affect the result? 41. A room contains air in which the speed of sound is
343 m/s. The walls of the room are made of concrCte, in which the speed of sound is I 850 m/s. (a) Find the critical angle for total intermd reflection of sound at the concreteair boundary. (b) In which medium must the sound be traveling in order to undergo LOtal internal reflection? (c) ",", bare concrete wall is a highly efficient mirror for sound." Give evidence far or against this statement.
Devialion of _ yellow light
42. Three adjacent faces (tJlat all share a corner) of a plastic cube of index or rerraction n are painted black. A clear spO[ althe painted corner sen'es as a source of diverging rays when light comes through it. Show lhal a ray from this corner to the center of a clear face is totally reflected if n ;;:: "\.If.
FIGURE P22.32
lli]A ra}' of light strikes the midpoinl of one face of an equiangular (60°_60"_60°) glass prism (11 :::: 1.5) at an angle of incidence of 30", (a) Trace the path of the light ray through the glass and find the angles of incidence and refraction at each Surf:KC. (b) If a small fracLioll of liglll is also reneclccl at each surface, whal are the angles of reflection at the surfaces?
143.IThe light beam in Figure 1'22.43 strikes surface 2 at the crilical angle. Determine the
Surface I
SECTION 22.7 TOTAL INTERNAL REFLECTION 34. For liglll of wavelength 589 nm, calculate the critical angles 1'01' the following substances when ~L1rmuncled by air: (a) fused quartz, (b) palyslyrene, and (c) sodium chloridc. 35. Repeat Problem 34, bm this time assume the quarlZ, polystyrene, and sodium chloride are surrounded by water.
36. A beam 01" light is illcident fram air on the surface af a liquid. If the ::lllglc of incidence is 30.0° and the angle of refraction is 22.00, find the critical angle for the liquid when surrounded by air.
FIGURE P22.43
44. Ajewel thief hides::l diamond by placing it on the bOllom of a public swimming pool. He places a circular raft on the surface af the water directly above and centered over the diamond, as shown in Figure P22.44. If the surface of the water is calm and thc pool is 2.00 m deep, find the minimum diameter oflhe rafl that would prevent the diamond from being seen.
37. A plastic light pipe has an index of refraction of 1.53. For LOtal internal reflection, what is the minimum angle of incidence if the pipe is in (a) air and (b) watcr?
38. Dctermine the m'-lximull1 angle
e for
which the light rays inciclem on the end of the light pipe -0\ in Figure P22.38 ;Ire su~ject LO !"~~.--~ total internal reflection along the walls of the pipe. Assume the FIGURE P22.38 light pipe has an index of refraction of 1.36 and the outside mcdium isair.
39. A light ray is incident normally LO the long face (the hypotenuse) of a 45°-45°-900 prism sunou nded by air, as shown in Figure 22.26b. Calculate the minimum index ofrcfrac· tion of the prism for which the ray will LOtally imernally reflect at each of the two sides making the right angle. l40.1_ A beam of laser light with wavelcngth 612 nm is directed through a slab of glass having index of n.:frac· tion 1.78. (a) For what minimum incidelll angle would a ray of light undergo total internal rerleetion? (b) If a layer of wat.er is placed over the glass, what is the minimum angle of incidence on the glass-water jntert';-lce thm will
2.00
III
Diamond FIGURE P22.44
ADDITIONAL PROBLEMS
45. A layer of ice having p
am A ray of light is incident at an angle 30.0° on a plane slab of flint glass s\lrrounded by water. (a) Find thc refrac-
757
Problems
lion angle. (b) Suppose the index of refracti 011 orlhe surrounding medium can be 3cljusted, but thc incidclH anglc of the light remains the same. As Ihe index of refraClion of thc medium approaches thal of the glass. what happens LO Ihe rcrraction angle? (c) What happens LO lhe refnlction angle when the medium's index of refraction excceds that of the glass?
47. A ray of light is incidcnt on a flat surface 01 a block of crown glass that is sUlTounded by waler. The angle of rcfraction is 19.6". Find the angle of reflection. 4g. A liglll ray of wa\·e1englh 5S9 nm is incidem at an angle 0 on the LOp ,0 surface of a block of polystY'·ene surrounded by air, as shown in Figurc P22.48. (a) Find the maximum valuc of 0 for which Lhc rcfracted ray will undergo total illlcrnal reflcctjon at the left vcnicill race FIGURE P22.48 of the block. (b) Repeat Ihe calcllialion for Ihe case in which the pol)sr)rclH.· block is immcrsed in water. (c) What happens if the block i't immersed in C
o
149.11\,<; shown in Figure P22.49, a lig-ht ray is incidcllt normal to one face of a 30 0 -60"-90" block of dcnsc flint glass (a prism) that is immcrsed in waler. (a) Dctermine the exit angle 04 of the ray. (b) A substance is dissoh·cd in the watcr to increase the index of refraction. Al what value 01" 11~ docs total internal reOection cease at point/ J ?
-
52. An optical fiber with index of refracT d tion 11 and diameter d is surrounded 1by air. Light is sent into the fiber along its axis, as shown ill Figure P22.52. (a) Find the sm.lllest outside radius R pcrmined for a bend in the fiber I I I • if no light is to escape. (b) Does thc result for part (a) pl·edict reasonable FIGURE P22.S2 behavior as d approaches 7.cro~ As Ii increases? As n approaches unity? (c) Evaluate R. assuming the diameter of the fibcr is 100,um and its index of refraction is lAO.
153.1.-\ piece of wire is bent through an angle 0. The bent wire is partially suhmerged in benzene (index of refraction = 1.50) so that, to a person looking along the dry part, the wire appears to be straight and makes an angle of 30.0" with the hori7ontal. Determine the value of8. 54. A light ray tra\'eling in air is incidem on one face ofa rightangle prism with index of refraction 11 = 1.50, as shown in Figure P22.5-4, and the ray follows the path shown in the figure. Assuming 0 = 60.0" and the base of the prism is mirrored, determine the angle rJ> made by the outgoing ray ' . ·. ith thc normal to the right face of the prism.
,
60.0 0
pI
00", '"
\ O;J., .' 30.0
FIGURE P22.49
50. A narrow beam of light is incidclll from air OIlLO a glass surface with index of refraction 1.56. Find the angle of incidence for which the corrcspollding
FIGURE P22.S4
55. A transparent cylinder 01" radius R = 2.00 m has a mirrored surface on its righl half, as shown in Figure P22.55. :\ light ra}' traveling in air is incident on the left side of the cylinder. The incident light I'ay and the exiting light ra)' are parallel, and (1 = 2.00 m. Dctcrmine the index of refraction of the material.
,
Incident r
,---,'>S-. ,
, ,'C
/..;. E..xiLi ng ray --,_~..'"
,,
~tirrorcd
surface
"
FIGURE P22.SS
56. A I;lser beam strikes one end of ;1 slab of material, as in Figure P22.56. The index of refraction of the slab is 1.48. I,
I
12.0 em
·_~
- - - - -L".-__
.;0.0'>
FIGURE P22.S1
,,
11/
by 11= 2A.
,
,,
'_'_=_I_.'_18
FIGURE P22.S6
....-J }
3.] 0
1111ll
758
Reflection and Refraction of Light
Chapter 22
Determine the number ofinlcrnal reflections of the beam before it emerges from the opposite end of the slab. 57. A light ray enters a rectangular block of plastic at an anglt; 0 1 = 45.0 0 and emerges ,u an angle B,! = 76.00, as shown in Figure P22.57. (a) Determine [he index of refraction of the plastic. (b) If the light ray enters [he plastic at a palm L = 50.0 COl from the botlom edge, how long does illake the light ray to travel through the plastic?
\I
159.IFigure P22.59 shows the p,nh of a beam of light through sevnal laycrs with different indices of'refraction. (a) If 8 1 = 30.0 what is the angle 02 of the emerging beam? (b) 'What must the incident angle 8 1 be to have tOlal imer· !1fll rertection at the surLlce between the mediulll with n = 1.20 and the medium with n = LOO? 0
,
,8,
"
.~,
I.
I
Use the data to verify Snell's law of refraction by plolling thc sine of the angle of incidence versus t.he sine of the angle of refraction. From the resulting plot, deduce the index of refraction of water.
,
11 = 1.60
,
,, e,.
11=1.40
'It,,. ,
FIGURE P22.57
,
58. StuciclllS allo\\' a narrow beam of laser light to strike a waleI' surfilce. They arrange to measure the angle of refraction for selected angles of incidence and record the data shown in the following table: Angle of Incidence (degrees)
Angle of Refraction (degrees)
10.0
7.5
20.0 30.0 40.0 50.0 60.0 70.0 80.0
15.1
'"
,
tl = 1.20 ~
1/=
FIGURE P22.59
60. Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on LOp of shcet 2, and a laser beam is directed onlo the sheets from abovc so that it strikes the interface at an angle of 26.5 with the normal. The refracted beam ill sheet 2 makes an angle of 3l.r with the normal. The experiment is repe
22.3
28.7 35.2 40.3 45.3 47.7
1.00
This beautiful photograph shows a raindrop suspended from a leaf. The raindrop acts as a lens. It refracts light twice to produce a real, inverted image of the foliage beyond
23.1 23.2
Flat Mirrors Images Formed by
Concave Mirrors
MIRRORS AND LENSES
23.3
Convex Mirrors and Sign Conventions
23.4
Images Formed by Refraction
23.5
Atmospheric Refraction
23.6
Thin Lenses
23.7
Lens and Mirror Aberrations
The development of the technology of mirrors and lenses led to a revolution in the progress of science. These devices, relatively simple to construct from cheap materials, led to micro-
scopes and telescopes, extending human sight and opening up new pathways to knowledge, from microbes to distant plonets. This chapter covers the formation of images when plane and spherical light waves foil on
plane and spherical surfaces. Images can be formed by reflection from mirrors or by refraction through lenses. In our study of mirrors and lenses, we continue to assume light travels in straight lines (the ray approximation), ignoring diffraction.
23.1
FLAT MIRRORS
\tVe begin by examining the flat mirror. Consider a point source of light placed
at 0 in Figure 23.1, a distance p in front of a flat mirror. The distance p is called the object distance. Light rays leave the source alld are reflected from the mirror. After renection, the rays diverge (spread apart), but the)' appear to t.he viewcr t.o come from a point I behind the mirror. Point J is called the image of the ol~jeCl at 0. Regardless of t.he system undcr study, images are formed at the point where rays of light actually intersect or where they appear to originate. Because the rays in the figure appear to originate at f, which is a distance q behind the mirror, t.hat is t.he location ofl-he image. The distance q is called the image distance. Images are classified as real 01" virtual. In the format.ion ora real i"1l/ag~), light actu ally passes through the image point. For a vl1-fu(f.{ image. light. doesn't pass through the image point, but appears to come (diverge) from there. The image formed by the flat In irror in F'igu re 23. L is a vi rtu"ll image. ] n fact, the images seen in flat mi rl'ors are always virLual (for real objects). Real images can be displayed on a screen (as at a movie), but virtual images cannoL. "vVe examine some or the properties of the images formed by nat mirrors by using the simple geometric techniques. To find Ollt where an image is formed, it's
1--/' -4-1---'1-1
,;/1
,;:// ~,o...;-./
4
FIGURE 23.1 All image fnrmo.:d hy I"enec[ion frolll a fhu mirror. The im;\gc poim I is b(>hind tilt' mil for ilt dislance £I. which is equal ill m'lgniluck to [he ol~jecl distanco.: I).
759
760
Chapter 23
Mirrors and Lenses
ACTIVE FIGURE 23.2 A geometric cOllHrucrion 10 locale tbe image of all ol~j<:n placed in fronl ofa flat mirror. Because the triangles
PQRand P'QR'lre idl:lllical, p = and h
necessary to follO\\' at least twO rays of light as they reflect from the mirror as in Active Figure 23.2. One of those rays starts at P, follows the horizontal path PQ (Q the mirror, and reflects back on itself. The second ray fo110\-\'5 the oblique path PR and reflects as shown. An observer to the left of the mirror \...· ould trace the two reflected rays back (Q the point from '.vhich they appear to have originated: point P'. A continuation of this process for points other than Pon the objecL would result in a virtual image (drawn as a yellow arrmv) La the right of the mirror. Because triangles PQR and P'QR are identical, PQ = P'Q. I-renee, we conclude that the image formed by an object placed in front of a flat mirror is as far behind the mirror as the object is in front of the mirror. Geometry also shows that the object height h equals the image height h'. The lateral magnification !vI is defined as
If/I
= II'.
TIP 23.1 Magnification Enlargement
*
Note lhal the \\'ord IlIflFJlIijiClltioll, as used in optics, doesn't always mean enlargell/e1lt hecause the image coilld be sm,lller than the
objecl.
image height
h'
object heighL
h
,11=
[23.1]
Equation 23. L is a general definiLion of the laLeral magnification of' any type of mirror. For a flat mirror, lH = 1 because It' = h. In surnmary, the image formed by a flat mirror has the following properties: 1. The image is as far behind the rninor as the object is in front. 2. The irnage is unmagnified, virtual, and upright. (By uj)right, we mean thal if the objecL alTow points upward, as in Active Figure 23.2, so does the image arrow. The opposite oran upright image is an inverted image.)
Finally, nOLe Lhat a flat mirror produces an image having an a.pparent left-right reversal. You can see this reversal standing in front of a mirror and raising your right hand. Your image in the mirror raises the left hane!. Likewise, your hair appears 10 be parted on the opposite side, and a mole on your right cheek appears to be on your image's left cheek. QUICK QUIZ 23.1 J n Lhe overhead view of Figure 23.3, Lhe image of rhe stone seen by observer 1 is at C. "Vhere does observer 2 see the image: aL A, at B, at C, at D, at E, or nol at all? A
IJ
c
f)
Ii
2 FIGURE 23.3
EXAMPLE 23,1 Goal
(Quick Quiz 23.1)
"Mirror, Mirror, on the Wall"
Apply the propenies ora flat rninor.
Problem A man 1.80 m tall stands in front ora mirror and sees his full height, no morc anelno less. Ifhis eyes are 0.14 III from the LOp of his head, \,,,hat is the m inimulll height or the mirror?
23.1
Strategy Figure 23.4 shm'l's two rays of light, Olle from the man's feet and the other from the top of his head. renecting off the mirror and entering his eye. The rar from his feet just st.rikes the bOLlom of the mirror, so if the mirror were longer, it would be too long, and if shorter, the ray would not be reflected. The angle of incidence and the angle of renection are equal, labeled e. This means the two triangles, £lBD and DBC, are identical because they are right triangles with a common side (DB) and twO identical angles e. Use this key fact and the small isosceles triangle FEG to solve the problem. ~
..
,
,.,
"
"
-=
E
T e( -----------eS---- 1
761
FIGURE 23.4 (Example 23.1)
d
B
.. , . ,
Solution vVe need to find BE, ,...·hich equals d. Relate this length to lengths on the man's body:
Remarks The minor must be exactly equal to half the height of the man for him to see only his full height and nothing more 01' less. Notice that the ans\ver doesn't depend on his distance from the mirror. QUESTION 23.1 \"'ould a taller man be able to see his full height in the same mirror? EXERCISE 23.1 How large should the mirror be if he wants
Answer
0.30
1.0
see only the upper third of his bod)'?
III
Most rearview mirrors in cars have a clay setting and a night setl.ing. The night setting greatly diminishes the intensity of the image so that lights from trailing cars will not blind the driver. To understand how such a mirror ''I'orks, consider Figure 23.5. The mirror is a wedge of glass with a reflecting metallic coaling on the back side. \'''hen the mirror is in the day setting, as in Figure 23.5a, Iighl frol11 an ol~jeCl behind the car strikes the mirror at point l. l'vlost of the light enters the
~tting
(a)
Day and Night Settings for Rearview Mirrors
FIGURE 23.5 Cl'Oss-\ectional views ora rearview mirror. (a) With the da)' selling, t he silvered back surface of the mirror refleclS a briglll ray 11 into the drivcr's eves. (b) ,,,'ith tht' nighl setting. the glass of the un<;ilvercd frollt surface oCthe mirror reflects a dim ra)' J) into the driver's cyes.
Reflccling side of mirrol"
Daytime
APPLICATION
l\"ighltime (b)
~ettillg
762
Chapter 23
Mirrors and Lenses
wedge, is refracted, and reflects from the back of the mirror Lo return La the front surface, \",here it is re[raCled again as it reenters the air as ray B (for bright). In addilion. a small ponioll of rhe light is renected at the front surface, as indicated by ray D (lor dim). This dim reflected light is responsible for the image observed when the min-or is in the night setting, as in Figure 23.5b. Now the wedge is rotated so
[hal Lhe path lollowed by Lhe brightlighL (ray B) doesn'Llead to the eye. InSlead, the dirn light reflected [rom the front surl"ace travels La the eye, and the brightness oflrailing headlights doesn't become a hazard.
APPLYING PHYSICS 23.1
ILLUSIONIST'S TRICK
The professor in the box shown in Figure 23.6 appears LO be balancing himself on a few Hngers v... ith both of his fect elevated from the noor. He c
23.2
FIGURE 23.6
(Appl)'ing Php.ics 23,1)
IMAGES FORMED BY CONCAVE MIRRORS
A spherical mirror, as its name implies, has the shape ofa segmentofa sphere. Figure 23.7 shows a spherical mirror with a silYcred inner, conGl"e surface; this type of mirror is called a concave mirror. The mirror has radius or curvature R, and llS center of curvature is ~lt point C. Poinl Vis the center of the spherical segment, and (t line drawn from eta Vis called the principal axis ohhe mirror. Now consider a point source of light placed at point 0 in Figure 23.7b, on the principal axis and outside point C. Several diverging rays originating at 0 are shown. After reflecting from the mirror, these rays converge to meet at I, called the image point. The rays then conlinue and diverge from 1 as if there were an objec! there. As a result, a real image is formed. \"'henever reflected light actually passes through a point, the image formed there is real. \Nc often assume all rays that diverge from the object make small angles with the principal axis. All such rays renect through the image point, as in Figure 23.7b. FIGURE 23.7 (it) A concave mirror of radius R. The Cl;lller ofcurv;\lLlrl', C. i.. locatcd 01\ the principal axis. (b) :\ point ol~jecl placed at 0 in frOlll of a COrlG]\"{· spherical milTor of radius U, where 0 is ;lllY POillt Oll tho.:: principal ;]xis fanhl:'- lhall R frolll the surface orlhe mirror. forms a rca} ill1;lgc al I. Hlhe rays di\'(~rg,; frolll Oal small angles. they all reflect through lbe S;IIlH' imagl: poin!.
Mirror
Ct.:lller or
(a)
:Odinor
(bl
23.2
Images Formed by Concave Mirrors
Rays that make a large angle with the principal axis, as in Figure 23.8, converge to other points on the principal axis, producing a blurred image. This ellen, called spherical aberration, is present to some extent with any spherical mirror and \.. ,ill be discussed in Section 23.7. ''''e Gill liSt: lhe geumelry ~ilo\'''''11 ill FiglilC 23.9 Lv c£tlt.:1I1me tlie image distance q from the object distance /) and radius of curvature R. By convention, these distances are measured from point \~ The figure shmvs two rays of light leaving the tip of" the oqjeet. One ray passes through the center of curvature, C, of" the mirror, hitting the mirror head-on (perpendicular to the mirror surface) and reflecting back on itself. The second ray strikes the mirror at point Vand reflects as shown, obeying the la\\! of ref-icClion. The image of the tip of the arrow is at the point where the two rays intersect. From the largest triangle in Figure 23.9, we see that tan (J = hlp; the light-blue triangle gives Lan (J = -h'/q. The negative sign has been introduced to satisfy our convention that h' is negative when the image is inverted with respect to the object.. as it is here. From Equation 23.1 and these resulLs, we find that the magnification of the mirror is 1Vf=
11.'
~
q
=
., FIGURE 23.8 Ra}'s at huge ;lIIgles from the horizOnlal axis rcflecl from a spherical, concave mirror 10 inter· sect the principal axis at diflcrCIlI points. resulting in a blurred ilnagc. This phenomenon is called splm'1C{/1 aberration.
[23.2]
I'
11.
763
From two othcr triangles in the figure, we get t<:111
11. n = ---
and
/' - R
tan
Q'
=
11.'
R-q
from \\.'hich we find that
R-q
Iz'
----
[23.3]
/J - R
Iz
If we compare Equation 23.2 ' . . irh Equation 23.3, \\'e see thal
R-q /J-R
q /'
Simple algebra reduces this equation to 1
1
2
/'
q
R
-+-=-
[23.4]
~
Mirror equation
This expression is called the mirror equation. Iflhe object is very far [rom the mirror-if the object. distance p is great enough compared with R that !Jcan be said to approach infinity~thcn 1//) = 0, and we see from Equation 23.4 that q = R/2. In other words, when the objcct is very far from thc mirror, the image point is halfway between the center of curvature and the center of the mirror, as in Figure 23.10a (page 764). The incoming rays are essentially parallel in lhat figure because the source is assumed to be very far from the
FIGURE 23.9 The image formed by a splll~ric:l1 conCide Ibe center of curv:llure, C.
o
e
f
c
a
h'
9
Principal axis
1-----/1---. I---------p------~
764
Chapter 23
Mirrors and Lenses
FIGURE 23.10 (a) Light .-ays from a distant object (jJ = (0) rdlect frol11 a conca\'e mirror thruugh the focal point F In this case the image distance q = f?/2 = J where [is the local length oCthe mirror. (b) A phoLOgraph ortlle n:l\eClion of parallel r
c
from a concave mirror.
(b)
(a)
mirror. In this special case we call the image poinllhe focal pointFand the image distance the focal length f where H
f=-2
Focal length -)
The
In iITOI'
[23.5]
equation can therefore be expressed in terms of the focal length: I
l
I
-+-=jJ q f
[23.6]
Not.e that rays [rom objects;H inf·inity are always focused
TIP 23.2 Focol Point '" Focus Point The focal point is
flot
at which light rays
[OCIlS
the point to form
an image. The focal point ora mirror is determined solely by its ClIl-vature; ildocSIl'1 depend on the location ofan)'objecl.
23.3
~lt.
the focal point.
CONVEX MIRRORS AND SIGN CONVENTIONS
Figure 23.11 shm\'s the formation of an image by a convex mirror, which is silvered so that light is reflected from the out.er, convex surface. It is sometimes called a diverging mirror because the rays from any point on the object diverge after reflection, as though they were coming from some point behind the mirror. The image in Figure 23.11 is virtual rather than real bccause it lies behind the mirror at the poim the renecl.ed rays appear to origin'lte. In general, the image formed by a convex mirror is upright, virtual, and smaller than the object. \,Ve won't derive any equations for convex spherical mirrors. If we did, we ""ould lind that the equations dcveloped for concave mirrors can be llsed with convex mirrors if particular sign conventions are used. V~'c call the region in which light rays move the front side of the mirror, and the other side, where virtual images are formed, the back sidf'. For example, in Figures 23.9 and 23.11, the side to the lefL of the mirror is the front side and the side to the right is the back side. Figurc 23.1.2 is helpful for understanding the rules for o~jeet and image distances, and Table 23.1 slImmarizes the sign conventions for all the necessary quantities. Notice that when the quantities !l, '7, and f(and R) are located where the light is-in front of the mirror-t.hey are positive, whereas when they are locaLed behind the mirror (where the light isn't), they are negative.
Ray Diagrams for Mirrors \,Ve can conveniently determine the positions and sizes of images [armed by mirrors by constructing ra.y diagrams similar to I he ones we have been using. This kind of graphical construction tells us the overall naLUre of the image and can be used to check parameters calculated from the mirror and magnification equaFIGURE 23.11
Formation of all
Front
Back
image by a spherical, cUI1\,ex mirror. )Jole thill Lht' image is 1';rllJa/ and IIp."ight.
,..
C
23.3
Convex Mirrors and Sign Conventions
TABLE 23.1 Sign Conventions for Mirrors Quantity
Symbol
Object local ion
FronL, or real, sirle
In Front
q
Focal length
I
Image height
h'
Magn i Ilcation
M
Upright Image
Inverted Image
+ + +
"
Image location
In Back
+ +
palld qpositi\'e
[ncidenL light• • Reflected light
765
Bilck,or \rirlual, ~jdL'
I
!Jand '1" c g-.u i \'e
No light
Convex or concave mirror
FIGURE 23.12 A diagram desCI'ibing I he signs ofP and q for COllV!.:X and concave mirrOl·s.
tions. Making a ray diagram requires knowing the position of" the ol~ject and the location of the center of curvature. To locate the image, three rays arc constructed (rat.her than only the I.vw \,,·c have been constructing so far), as shown b)' the examples in Active Figure 23.13. All three rays stan from the same object point; for these examples. the tip of the arrow \vas chosen. For the concave mirrors in Active Figures 23.13a and b, the rays are drawn as f"ollows: J. Ray] is drawn parallel to the principal axis and is renected back through the focal point F. 2. Ray 2 is drawn through the focal point and is reflected parallclLO the principal axis. 3. Ray 3 is drawn through the center of curvat ure, C, and is reflected back on ilself.
3
! ....----=---~:-'------=:'_f
0 Principal axis
Back
(a)
2
~,---~---),
1
3 Bad;
(b)
o
c
(c)
ACTIVE FIGURE 23.13 Ray diagrams for spherical mirror~ and corresponding" photographs of lhe images of ('andles. (a) \Vhen an object is outside the cellter of CUI"\'3tun~ ofa concave mirror, the image is real, inverted, alld reduced ill size. (h) \o\'hcn an ol~iecl is betwt:en :1 concave mirror and the focal point. the image is virtual. upright, and magnilied. (c) When an objccL i" in frOllt of a convex mirror, the image is virtual. lIpriglll, and reduced in si/.e.
766
Chapter 23
Mirrors and Lenses
Note that rays actually go in al1 directions from the object; we choose to follmv those moving in a direction that simplifies our drawing. The intersection of any two of these rays at a point locates the image. The third ray serves as a check of our construction. The image point obtained in this fashion must a]\,"ays agl-ee with the \'alue of q calculated from the mirror formula.
FIGURE 23,14 A (omcx side-view mirror on a vehicle produces (111 upriglH image Ihal is smaller than the objecl. The smaller image means that lhe object is doser tban its apparent distance as ObSL:fVL:d in the mirror.
APPLYING PHYSICS 23.2
In the case ora concave mirror, note what happens as the objectis moved closer to the mirror. The real, inverted image in ACli\'e Figure 23.13a moves lO the left as the object approaches the focal point. ''''hen the object is at the focal point, the image is infinitely far to the left. ,,\Then the object lies betwecn the focal point and the mirror surface, as in Active Figure 23.13b, hm.. ·cvcr, the image isvirtual and upright. "Vith rhe convex mirror shown in Active Figure 23.13c, the image of a real object is always \'inual and upright. As the object distance increases, the virtual image shrinks and approaches the focal point as 1) approaches infinity. You should construct a ray diagram to verify these statements. The image-forming characteristics of curved mirrors obviously determine their lIses. For example, suppose you want to design a mirror that will help people shave or apply cosmetics. For this, you need a concavc mirror that puts the user inside the focal point, such as the mirror in Active Figure 23.13b. With that mirror, the image is upright and greatly enlarged. In contrast, suppose [he primary purpose ora mirror is to observe a large field of view. In that case you need a convex mirror such as the one in Active Figurc 23.13c. The diminished size of the image means that a fairly large field of view is seen in the mirror. Mirrors like this one are often placcd in stores to help emplo)'ees watch for shoplifters. A second use of such a mirror is as a side-vie\v mirror on a car (Fig. 23.14). This kind of mirror is usually placed on the passenger side of the car and carries the \varning "Objccts arc closcr than thcy appeal-." \'Vithout such warning, a driver might think she is looking into a flat mirror, which doesn't alter the size of the image. She could be fooled into belicving that a truck is far away because it looks small, when it's actually a large semi very close behind her, but diminished in size because of the image formation characteristics of the com'ex mirror. CONCAVE VERSUS CONVEX
A virtual image can be anywhere behind a concave
mirror. \'Vhy is there a maximum distance at which the image call exist behind a convex mirror? Explanation Consider the concave mirror first and imagine two different liglH rays leaving a tiny object and striking the mirror. If the o~ject is at the focal point, the light rays reflecting from the mirror will be parallel to the mirror axis. They can be interpreted as forming a rinual image infinitely br av.'ay behind the mirror. As the object is brought closer to the mirror, the reflected rays will diverge through larger and larger angles, resulting in their extcnsions converging closer and closer to the back of thc mirror. When the
APPLYING PHYSICS 23.3
o~jeCl.
is brought right up to the mirror, the image is right behind the mirror. ""Then the object is much doser to the mirror than the focal length, the mirrors acts like a flat mirror and the image isjust as far behind the mirror as the object is in front of it. The image can therefore be anywhere from infinitel)' far away to right at the surface of the mirror. For the convex mirror, an objcct at infinity produces a virtu ..d image at the focal point. As the object is brought closer, the reflected rays diverge more sharply and the image moves closer to the mirror. As a result, the virtual image is restricted to the region betwecn the mirror and the focal point.
REVERSIBLE WAVES
Large trucks often have a sign on the back saying, "lfyou can't see my mirror. 1 can't see you." Explain this sign. Explanation The trucking companies are making use of the principle of the reversibility of light rays.
For an image of you to be form cd in the driver's mirror, there must be a pathway for rays oflight to reach the mirror, allO\ving the driver to see your image. If you can"t see the mirror, this patll\vay doesn't exist.
23.3
EXAMPLE 23.2 Goal
Convex Mirrors and Sign Conventions
767
Images Formed by a Concave Mirror
Calculate properties 01" a concave milTor.
Problem Assume a cenain concave. spherical mirror has a focal length of 10.0 cm. (a) Locate the image and find the rnagnification for an object distance of 25.0 cm. Determinc whcther the image is real or \'inual, inverted or upright. and larger 01" smaller. Do the samc for objcct distances of (I» 10.0 cm and (c) 3.00 cm.
For each part, substiwte into the mirror and magnification equations. Part (b) in\'olvcs a limiting process because the answers are infinite.
Strategy
• • • • • • • • . • • • '''ll
r···························
Solution (a) Find the image position for an object distance of
25.0 em. Calculate the magnification and describe the image. I
I
I
Use the mirror equation to find the image distance:
-+-=P 'I J
Substitute and solve lor q. According to Table 23.1. jJ and fare positive.
~:---
I
1
+-
= ~'-
'!
25.0 em
10.0 em
'1= 16.7em BeC
M=
16.7 cm 25.0 em
'I
/'
-0.668
The imClge is smaller than the oqjeCl because IA'11 < 1, and it is inverted because 1\11 is negal.i\'e. (See Fig. 23.13a.) (b) Locate the image distance when the object dis-
tance is 10.0 cm. Calclliate the magnifletltion and describe the image. The
is at the focal point. Substiwl.e I} = ~n"".!= 10.0 cm into the mirror eqtlCltion:
ol:~ject
10.0 ....~
--c-:---
10.0 em
I
+ - = --c-:--q
10.0 em
I -=0
q
-->
q=
00
Because 1\1 = -qll}, the magnificalion is also infinite. (c) Locate the image distance when the o~ject dis-
tance is 5.00 em. CCllculate the magniflcation and describe the illlClgC. Once again, substitute into the mirror equation:
I
+-= q 10.0 em 5.00 em I
'!
10.0 em
5.00 em
10.0 elll
q ~ -JO.O em The image is virtual (behind the mirror) because q is negative. Usc Equation 23.2 La calculClle the magnification:
.111=
q jJ
_(-lo.oem) = 2.00 5.00 em
The image is larger (magnified by a faclor or 2) because 1"11 > I, and upright because M is posit ive. (Sec Fig. 23.13b.) lrrr... • • • • • • • • • • •
Note the characteristics oran image formed by a concave, spherical mirror. \tVhen the ObjCCl is olltside the focal point, the image is inverted and real; al the focal point, the image is formed at infinity; inside the focal point, the image is upright and virtual.
Remarks
QUESTION 23.2 ''''hat location does the image approach as the object gels arbitraril}' far a\vay from the mirror? (a) infinity (b) the focal point (c) the radius ofcurvature of the mirror (d) the mirror itself EXERCISE 23.2 Tr the object distance is 20.0
Answer
nnd the image distance and the magnification of the mirror.
q = 20.0 em, M = -1.00
EXAMPLE 23,3 Goal
CIll,
Images Formed by a Convex Mirror
Calculate propenies ofa convex mirror.
Problem An objecl 3.00 (Ill high is placed 20.0 em from a convex mirror with a focal length of magnilUde 8.00 em. Find (a) the position oft.he image, (b) the magnification of the mirror, and (c) the height of the image.
Strategy This problem again requires only substitution into the mirror ;lI1d magnification equations. rv{ultiplying the object height by the magnification gives the image heighl. ,
~
~
Solution (a) Find the position of the image. Because the mirror is convex, its focal length is negative. Substitute into the mirror equation:
I
1
1
P
'I
I
-+-~-
--'----- + -I 20.0 em
'I
= ---'---
-8.00 em
q = -5.71 em
Solve [or q: (b) Find the magnification of the mirror.
1\1=
Substitute into Equation 23.2:
q
/)
_(-5.71 em) = 0.286 20.0 em
(c) Find the height o[the image.
Multiply the
ol~ject
height by t.he magnification:
h' ~ hM
= (3.00 el11)(0.286) = 0.858 em
~
~
Remarks The negative value of q indicates the image is virtual, or behind the mirror, as in Figure 23.13c. The image is upright because A1 is positive. QUESTION 23.3 True or False: A convex mirror can produce only virtual images. EXERCISE 23.3 Suppose t.he o~ject is moved so that it is 4.00 em from the same mirror. Repeat parts (a) through (c).
Answers
(a) -2.67 em
EXAMPLE 23.4 Goal
(b) 0.668
(c) 2.00 em; the image is upright and vinual.
The Face in the Mirror
Find a focal length from a magnification and an object distance.
Problem When a \voman stands with her face 40.0 em from her face. \,Vhat is the focal length of the mirror?
Col
cosmetic mirror, the upright image is twice as tall as
Strategy To nndfin this example, we must first find q, the image distance. Because the problem stat.es that the image is upright, the magnificat.ion must be positive (in this case, 1\1 = +2), and because .iVl = -ql!), we can delcr!TIme q.
Images Formed by Refraction
23.4
,..
.
.
Solution
-
769 _
~
'I --=2
Obtain q from the magnification equation:
'I Because q is negative, the image is on the opposite side of the mirror and hence is virtual. Substitute q and jJ into the mirror equation and solve rorJ:
~
P -21' ~ -2(40.0
UII) - -80.0 elll
1
40.0 em
80.0 em
f
f= 80.0 em
..........
.
..
Remarks The positive sign for the focal length tells us that the mirror is concave, a fact \..'e already knew because the mirror magnified the object. (A convex mirror would have produced a smaller image.)
QUESTION 23.4 If she moves the mirror closer to her face, what happens to the image? (a) It becomes inverted and smaller. (b) It
ren1ains upright and becomes smaller. (c) It becomes inverted and larger. (d) It remains upright and becomes larger.
EXERCISE 23.4 Suppose a fun~house spherical mirror makes you appear to be one-third your normal height. lfyou are 1.20 m away from the mirror, find its £-ocal length. Is the mirror concave or convex? Answers
23.4
-0.600 m, convex
IMAGES FORMED BY REFRACTION
In this section we describe hm"-' images are formed by refraction at a spherical surface. Consider t\.. . o transparent media with indices of refraction t1 1 and 11 2 , \vhere the boundary between the two media is a spherical surface of radius R (Fig. 23.l5). \,Ve assumc the medium 1.0 the right has a higher index of refraction than the one to the left: 112 > n l • That would be the case for light entering a curved piece of glass from air or for light entering thc water in a fishbowl from air. The rays originating at the objeCt location Dare refracted at the spherical surface and then converge to the image point I. vVe can begin with Snell's law of refraction and use simple geometric techniques to show that the object distance, image distance, and radius of curvalllre are related by the equation
[23.7] Further, the magnification ofa refracting surface is 11.'
nllj
II
n'l/J
M=-= - -
[23.8]
As with mirrors, certain sign conventions hold, depending on circumstances. First note that real images are [armed by refraction on the side of the surface 0/)/)0sil" the side from which the light comes, in contrast to mirrors, where real images arc formed on the S(J.JJ11' side of lhe renecting surface. This makes sense because light reflects off mirrors, so any real images must form on the same side the light comes from. \'\7ith a transparent medium, the rays pass through and naturally form real images on the opposite side. \Ve define the side of the surface \\,'here light rays originate as the front side. The other side is called the back side. Because of the difference in location of real images, the refraction sign conventions for q and R are the opposite of those for reflection. For example, p, q, and R are all positive in Figure 23.15. The sign convcntions for spherical refracting surfaces are summarized in Table 23.2 (page 770).
FIGURE 23.15 An image formed by refraction at a spherical surface. Ra)'s makillg small angles with the principal axis dh'erge from a pOilU objecl at o and pa~~ lhrough the image point!.
770
Chapter 23
Mirrors and Lenses
TABLE 23.2 Sign Conventions for Refl'acting Surfaces Symbol
In Front
ObjeCllocalioll
/J
+
[mag"c location
q
Radius
R
Image height
I,'
Quantity
APPLYING PHYSICS 23.4
Upright
Inverted
Image
Image
+ + +
UNDERWATER VISION
,I
''''hy does a person with normal vision see blurry image if the eyes are opened underwater h'ith no goggles or diving Illask in use?
Explanation The eye presents a spherical refraction surface. The eye normally functions so that light entering from [Jle air is refractcclto form an image in the retina located at the back of the eyeball. The difference in rhe index of refraction bctween ""ater and the eye is smaller than the difference in the index of refraction ben·...een air and rhe eye. Conse-
quently. light entering the eye from the water doesn't undergo as much refraction as does lighlentcring from the air, and the image is formed behind the retina. A diving mask or swimming goggles hayc no optical action of their own; they eIre simply flat pieces or glass or plastic in a rubber lTlounL. Theydo, however, provide a region or air ;,}cUacclllLO the eyes so that the correel refraction relationship is established and images will be in focus.
Flat Refracting Surfaces If the refracting surf'ace is rIat, then R approaches inAnity and Equation 23.7 reduces to /
o~ ,,
1/,
17'1
/J
q 1/2
'1= --I) II,
!----!'ACTIVE FIGURE 23.16 Thc image form cd bra nat refraCting surface is virtual and 011 the S;Jmc side of the surface as the object. \'ote that iflhe Iighl rays are re\"cr~ed ill direction, we h:we the situation (k~crihcd ill Example 22.7.
[23.9]
Frolll Equation 23.9, \ve see that the sign of q is opposite that of /J. Consequently. the image formed by a flat refracting surface is on the same side of the surface as the object. This statemel1L is iliustr<'lI.CcI in Active Figure 23.16 for the situatioll in which 'Ill is greater than 11'2' where a virtual image is formed between the object and the surface. ~otc that the refracted ray bends away from the normal in this case because 11 1 > 'l.:!..
QUICK QUIZ 23.2 A person spearfishing from a boat sees a fish localcd ~ m from the boat at an apparent depth of 1 m. To spear the fish, should the person aim (a) at, (b) above, or (c) below [he image or the fish? QUICK QUIZ 23.3 True or False: (a) The image of an object placed III front ora concave mirror is ahvays upright. (b) The height of the image oran object placed in front of a concave mirror must be smaller than or equal to the height of I he object. (c) The image of an object pl
EXAMPLE 23.5 Goal
Goze Into the Cr stol Boll
Calculate the properties of an image created by a spherical lens.
23.4
Images Formed by Refraction
Problem A coin 2.00 cm in diameter is embedded in a solid glass ball of radius 30.0 em (Fig. 23.17). The index of refraction or the ball is 1.50, and the coin is 20.0 cm from the surface. Find the position of the image of the coin and the height or the coin's image.
771
FIGURE 23.17 (Example 2::1.5) A coin embedded ill a glass ball forms a vinllal image bcrween lhe c.;Oill and tht: surface of the glass.
Strategy Because the rays are mm'ing from a medium of high index of refraction (the glass ball) to a medium of lower index of refraction (air), 20.0 cm the rays originating at the coin are refracted away from the normal at the surface and diverge OUtward. The image is formed in the glass and is virtual. Substitute into Equations 23.7 and 23.8 for the image position and magnification, respectively.
,..
.
Solution Apply Equation 23.7 ane!take ", jJ = 20.0 em, and R = -30.0 em:
• • • • • • • • • • • • • • • • ' ' '11
nI
/t2
-+-= jJ q
.50, II, = 1.00,
_1.,,-5-,-0_ 20.0 em
+ _l._0_0
= 1.00 - 1.50
q
Solve for q:
q=-17.1cl11
To find the image height, use Equation 23.8 for the magn ification:
111=
R -30.0 em
1.50(-17.1 em) 1.00(20.0 em)
h'
h
ii' = 1.28h = (1.28)(2.00 em) = 2.56 em
.
.
~
..
~
Remarks The negative sign on q indicates that the image is in the same medium as the object (the side of incident light), in agreement with our ray diagram, and therefore mll.':it be virtual. The positive value for .M means that the image is upright. QUESTION 23.5 How would the final answer be affected if the ball and observer ,,,ere immersed in water? (a) It ",ould be smaller. (b) It would be larger. (c) There woule! be no change. EXERCISE 23.5 A coin is embedded 20.0 cm from the surface of a similar ball of transparent substance having radius 30.0 cm and
unknown composition. If the coin's image is virtual and located 15.0 em from the surface, find the (a) index of rcfraClion of the substance and (b) magnifIcation. Answers
(a) 2.00
XAMPLE 2,6
(b) 1.50
The One That Got Awa FIGURE 23.18 (Example 23.6) The ,tpplll'ent depth q or t he fish is less than the true depth d.
Goal Calculate the properties of an image created by a nat refractive surface. Problem A small fish is swimming at a depth rJ below the surface of a pone! (Fig. 23.18). (a) What is the ap!Jorent depl.h of the fish as viewed frolll directly overhead? (b) Tf the fish is 12 cm long, how long is its image? Strategy In this example the refracting surface is flat, so R is infinite. Hence, "ie can use Equation 23.9 to determine the location of the image, which is the apparentloc3tion of the fish.
Solution (a) Find Lhe apparent depth of Lhe ush. Substitute 23.9:
11 1 =
1.33 for water and 1) = d
inlO
Equation
-0.752<1
'1=
(b) ''''hat is the size oflhe fish's image?
'/0
Use Equation 23.9 [Q eliminate q from the Equation 23.8, the magnification equation:
.
~
n2l1
h
h' = h =
)
nj ( -~jJ
h' M=-=
12 em
.
~
Remarks Again, because 'I is negative, the imag'c is virtual, as indicated in Figure 2:1.18. The apparent depth is three-fourths the aClUal depth. For instance, if d = 4.0 lll, then q = -3.0 m. QUESTION 23.6 Suppose a similar experiment is carried Olll. \\lith an objcCl immersed in oil (n = 1.5) the same distance belo".. the surface. How c10es the apparent depth of the object compare with its apparent depth when immersed in water? (a) The apparent depLh is unchanged. (b) The apparent depth is larger. (c) The apparem depth is smaller. EXERCISE 23.6 A spear fisherman estimates that a trout is 1.3
Answer
III
below the water's surface. \tVllat is the actual depth of the fish?
2.0 m
23.5
FIGURE 23.19 Because light is refracted by Earth's atmosphere. an observer ,ll ()sccs the Sun c\'en though il has fallell below the horizon.
ATMOSPHERIC REFRACTION
images formed by refraction in our atlTIosphere lead to some interesting phenom~ ena. One such phenomenon that occurs daily is the visibility of the Sun at dusk even though it has passed belm-\' the horizon. Figure 23.]9 shows why it occurs. Rays of light from the Sun strike Earth's atlllosphere (represented by the shaded area around the planet) and arc bent as they pass into a medium that. has an index of refraction different. from that of the almost empty space in which they have been traveling. The bending in this situation differs somewhat from the bending we have considered previously in that it is gradual and continuous as the light moves through lhe almosphere toward an observer at point 0. This is because the light
(al
(b)
FIGURE 23.20 (a) A mirage is prudnced by lile bending oflighl rays in the almospherc when lhere arc large tcmpCr
23.6
Thin Lenses
773
moves through layers or air that have a continuously changing index of refraction. \,Vhen the rays reach the obscrvcr, the e}:e follows them back along the direction from which thcy appear to have corne (indicated by the dashed path in the figure). The end result is that the Sun appears to be above the horizon even after it has fallen below it. The mirage is another phenomenon of nature produced by refraClion in the atlllosphere. A mirage can be observed \.. . hen the ground is so hot that the air directly above it is warmer than thc air at higher elevations. The desert is a region in which such circumstances prevail, but mirages arc also seen on heatcd road\vays during the summer. The layers of air at diJTerent heights above Earth have different densities and different refractive indices. The effect these differences can have is pictured in Figure 23.20a. The observer sees the sky and a tree in two different ways. One group of light rays reaches the observer by the straight-line path A, and the eye traces these rays back to see the tree in the normal fashion. In addition, a second group of rays travels along the curved path 13. These rays are directcd to\vard the ground and are then bent as a result of refraction. As a conscquence, thc observer also sees an inverted image of the tree and the background of the sky as he traces the rays back to the point at which they appear to have originated. Because an upright image and an inverted image are seen when the image ofa tree is observed in a reflccting pool of water, the observer unconsciollsl}' calls 011 this past experience and concludes that the sky is reflected by a pool of water in front of the tree.
23.6
THIN LENSES
A typical thin lens consists or a piece of glass or plastiC, ground so that each of its t",.'O refracting surfaces is a segment of either a sphere or a plane. Lenses are commonly used to form images by refraction in optical instruments, such as cameras, telescopes, and microscopes. The equation that relates object and image distances for a lens is virtually identical to the mirror equation derived earlier, and the method lIscd to derive it is also similar. Figure 23.21 shows some representative shapes of lenses. Notice that we have placed these lenses in two groups. Those ill Figure 23.2la are thicker at the center than at the rim, and those in Figure 23.21b are thinner at the center than at the rim. The lenses in the first group are examples of converging lenses, and those in the second group are diverging lenses. The reason for these nallles will become apparent shortly. As we did for mirrors, it is convenient to dellne a point called the focal point ror a lens. For example, in Figure 23.22£1 (page 774), a group of rays parallel to the axis passes through the focal point Faftcr being cOIwerged by the lens. The distance from the focal point to the lens is called the Cocallength! The focal length is the image distance that corresponds to an infinite object distance. Recall that we arc considering the lens to be very thin. As a result, it makes no difference whcther we take the focal length to be the distance from the focal point to the surface of the lens or t.he distance [rom t.he focal point to the center of the lens because the difference bCl\v'een these two lengths is negligible. A t.hin lens has fwo focal points, as illustrated in Figure 23.22, one on each side oft.lle lens. One focal pointcol'rcsponds to parallel rays traveling from the left and the other corresponds to parallel rays traveling from the right. Rays parallel to the axis di\'erge after passing through a lens ofbiconc<:1ve shape, shown in Figure 23.22b. In this case the focal point is deAned to be the point \vhere the diverged rays appear to originate, labeled Fin the figure. Figures 23.22a (page 774) and 23.22b indicate why thc names convf'1gillgand divergillgare applied to these lenses. Now consider a ray of light passing through the center of a lens. Such a ray is labeled ray 1 in Figure 23.23 (page 774). For a thin lens, a ray passing through the
Bicollvex
Convcxconcave
Planoconvex
(a)
Biconcave
Convcx-
Plano-
concave
concave
(b) FIGURE 23.21 Various lens shapes. (a) COll\erging lenses have posith'c focallellgth .. and arc thickest at the middle. (h) Divcrgillg lenses have llegative focal lengths and are thickl:SI allhe edges.
774
Chapter 23
Mirrors and Lenses
F
tal §
5 E
-0
•
~-:::..
g-
...
-
1-;<:::;::::---
-~
•
_-:,-:-tJe F
F
F
I
S
~
~-J
J-~
"
~
(b) FIGURE 23.22 (!-pjt) PhOlographs of t he effects of cOlln~rging and diverging lenses 011 parallel rays. (Righ!) The focal poims of the (a) bicolH'cX lens and (b) biconcave lens.
center is llildenected. Ray 2 in the same fIgure is parallel to the principal axis of the lens (the horizontal axis passing through 0), and as a result it passes through the focal point Farter refracLion. Rays I and 2 intersect at the point that is the tip of the image arrow. \'\'e first note that the tangent of the angle Q' can be found by llsing the blue and gold shaded triangles in Figure 23.23: Lan
II.
a = -
p
or
Lan 0' =
h'
'I
From this result, "ie find that 11.'
q
h
P
M~-=
[23.10]
The equation for magniFIcation by a lens is the same as the equation for magnification by a mirror. 'Iti'e also note from Figure 23.23 thal
PQ
tan fJ ::: -
FIGURE 23.23 A geomelric COllstruCtion for developing rhe rhill·lens equation.
I
or
tan
" 0
L,
h'
---
'I-I Back
Front
2
e :::
p Principal axis
fa
-~-
~!--j
h'
23.6
The height PQused in the first of these equations. l1o\\'e\'cr. is the same as h, the height of the object. Therefore, II
r
II' ---
r
q-
Back
I' po"iti\'c q ncgative
p nc~alive
1 'I -
FIGURE 23.24 \ di,lgnull ror ohtai"inK llll' siKIlS or panel q ror a thin lellS or.1 rcrracling "urr"ce.
1
1
/J 1
I
1
/'
q
1
-+-=-
[23.11]
This equation. called the thin·lens equation, can be used with both converging and divcrging lenses if we adhere to a sct of sign cOlwcntions. Figure 23.24 is useful for obtaining the signs of p and q, and Table 23.3 gives the complete sign cOI1\·en· tions for len~es. Note that a converging lens has a positive focal length under this convention and a diverging lens has a negative focal length. Hence, the names /JO.fiilivI>and npKati1Jeare offen given to these lenses. The focallengl.h for a lens in air is relalcd to the curvatures of its front and back surfaces and to the index of refraction 11 of the lens material by I (I -I ) -=(,,-1) I H, H,
[23.12]
where HI is the radius of CUl'vallire of the front surface or the lens and /1...2 is the radius of curvature of the back surfacc. (As \,,'ith mirrors, we arbitrarily call the side from which the light approaches the fnml of the lens.) Table 23.3 gives the sign cOlwcmions for R 1 and R 2 . Equation 23.12, called the lens-maker's equation, enables us to calculate the focal length from the known propenies of the lens.
Ray Diagrams for Thin Lenses Ra)' diagrams are essential for understanding the overall image formation by a thin lens or a system of lenses. They should also help clarify the sign conventions already discllssed. Active Figure 23.25 (page 776) illustrates this method for three sinRle-lens silll
2. The second ray is drawn through the center of the lens. This ray continues in a straight line. 3. The third ray is drawn through the other focal poim and emerges from the lens parallel to the principal axis.
Objl.·ct loctllion Image location I.ens radii
Focal lengl h
p q HI' H'2
J
•
Refractcd light
which reduces to
TABLE 23.3 Sign Conventions for Thin Lenses Qu:.lIltity Symbol In Front
775
q positive
Incidenlli~hl
Using the latter equation in combination with Equation 23,10 gi\'cs
q
Front
•
'1-1
II' It
Thin Lenses
In Back
Convergent
+ + + +
Divergent
~
Thin-lens equation
776
Chapter 23
Mirrors and Lenses
Back
-F]
Front
o
0 Back
Front
(bl
(a)
F]
[
3
Fz
2 Back
From (e)
ACTIVE FIGURE 23.25 R"y diagrams for locating the image of an ohject. (<1) The object is outside the focal poi III ora cOlwerg' illg lens. (b) The object is inside rhe focal point ofa converging lens. (c) The object is oLllside the focal point ora diverging lellS.
TIP 23.3 We Choose Only a Few Rays AltJ10Ugh om ra)' diagrams in Figure 23.25 only show three rays Icaving all object. an infinite number of rays can bc drawn belwcen t.he ol~ject and its image.
A similar conslruClion is lIsed LO locate the image formed by a diverging lens, as 5hm-\'0 in AClive Figure 23.25c. The point of intersection of any two or the rays in these diagrams can be used lO locate the image. The third ray serves as a check on construction. For the converging lens in AClive Figure 23.25a, where the object is outside the front focal point (P > J), the ray diagram shows that the image is real and inverted. \Vhen the real object is inside the front focal point (/) < j), as in Active Figure 23.25b, the image is virtual and upright. For the cliverging lens of Active Figure 23.25c, the image is virtual and upright.
QUICK QUIZ 23.4 A clear plastic sandwich bag filled with water can act as a crude converging lens in air. If the bag is filled ,vith air and placed under ',,';'Her, is the effective lens (a) converging or (b) diverging?
QUICK QUIZ 23.5
In Aetive Figure 23.25a the blue object arrow is replaced by one that is much taller than the lens. How many rays from the object will strike the lens?
QUICK QUIZ 23.6 An object is plaeed to the left of a converging lens. ",Vhich of the follm.. . ing statements are true, and which are false? (a) The image is alvvays to the right of the lens. (b) The image can be upright or inverted. (c) The image is always smaller or the same size as the object. Your success in vo.'Orking lens or mirror problems will be determined largely by whether you make sign errors ,.. . hen substituting into the lens or mirror equations. The only way to ensure yOll don't make sign errors is to become adept at using the sign conventions. The best ways to do so are to work a multitucle of problems on your m,m and to construct confirming ray diagrams. ,,,,Tatching an instructor or reading the example problems is no substitute for practice.
APPLYING PHYSICS 23.5
VISION AND DIVING MASKS
Diving masks often have a lens built inLO the glass faceplate for divers who don't have perfect vision. This lens allows the individual to dive without the necessity or glasses because the faceplate perfonns the necessary refraction to produce clear vision. Normal glasses have lenses that are curved on both the front and rear surfaces. The lenses in a diving-mask faceplate often have curved surfaces only on the inside oUhe glass. \rVhy is this design desirable? The main reason for curving only the inner surface of the lens in Lhe diving-mask face-
Solution
plate is to enable the diver to see clearly while underwater and in the air. Il"there were curved surFaces on both the front and the back of the diving lens, there would be two refractions. The lens could be designed so that these twO refractions would give clear vision while the diver is in air. \'\then the diver went underwaLer, however, the refraction between the water and the glass at the first interface would differ because the index of refraction oh..· ater is different from that of air. Consequently, the diver's vision wouldn't be clear underwater.
23.6
EXAMPLE 23.7 Goal
Thin Lenses
Ima es Farmed b a Converging Le"'ngs:...
777
_/
~
Calculate geometric quantities associated with a converging lens.
A converging Jens or local length 10.0 cm forms images of an object siLUated at various distances. (a) If the object is placed 30.0 em from the lens, locCite the image, state whether it's real or virtual, and I-ind its magnitica(jon. (b) Repeat the problem when the object is at 10.0 cm and (c) again when the object is 5.00 Clll from the lens.
Problem
All three problems require only substillltion into the thin-lens equat.ion and the associated magnification equation, Equations 23.10 and 23.11, respectively. The conventions of Table 23.3 must be followed.
Strategy
~
. ... '..,.
.
Solution (a) Find the image distance and describe the image when the ol~ject is placed at 30.0 em. I
l
1
-+-=/) q J
The ray diagram is shown in Figure 23.26a. SubstilUte values into the thin-lens equation to locate the image:
J I +30.0 em q
I
= ---,--,-----
~,-----
10.0 em
Solve for q, the image distance. It's positive, so the image is real ancl on the far side of the lens:
q = +15.0 em
The magnification of the lens is obtained fro111 Equation 23.JO. M is negative and less than 1 in absolute valuc, so the image is inverted and smaller than the object:
M=
q
15.0 cm 30.0 em
/)
-0.500
(b) Repeallhe problem, when lhe objeel is placed al 10.0 em. 1
]
10.0 em
q
Locate the image by substituting into the thin-lens equation:
---,--,----- + -
This equation is satisfied only in the limit as q becomes infinite.
q -->
~
1
---,--,----10.0 em
1 -=0 q
(c) Repeal lhe problem when lhe object is placed 5.00 em from the lens.
1
]
5.00 em
q
See the ray diagram shown in Figure 23.26b. SubstiLUte into the thin~lens equation to locate the image:
----,-,----- + -
Solve for 'I, which is negllt.ivc, meaning the image is on the same side as the object. and is virwal:
q = -10.0 em
1
= ---,--,-----
10.0 em
FIGURE 23.26
a 1--+---->1 10.0 em
15.0
{'Ill
J
10.Oem
:).00 em
~30.0cm
10.0 em (al
(b)
(Example 23.7)
Chapter 23
778
Mirrors and Lenses
q
M=
SubstilUte the values of j) and q into the magnification equation. iV1 is positive and larger than I, so the image is upright and double the object size:
l'
em) em
_(-10.0 5.00
+2.00
~
~
Remark scopes.
The ability of a lens to magnify objects led to the inventions of reading glasses, microscopes, and tele-
QUESTION 23.7 If the lens is lIsed
LO
form an image of the sun on (\ screen, hm...., far [rom the lens should the screen be located?
EXERCISE 23.7 Suppose the image o[an object is upright (lnd magnified 1.75 times when t.he object is placed 15.0 cm from a lens. Find the location of [J1C image and the focal length of the lens.
Answers
o~jeet)
(a) -2fi.3 em (virtual, on the samc side as thc
(b) 34.9 em
EXAMPLE 23.8 The Case of a Diver ing Lens Goal
Calculate geometric quantities associated \',1ith a diverging lens.
Problem
Repeal the problem of Example 23.7 ror a divergi-nglens of focal length 10.0 cm.
Strategy Once again, substitlllion into the thin-lens equation and the associated magnification equation, together with the conventions in Tablc 23.3, solve the various parts. The only difference is the negative rocallength.
.
"-'
.
~
Solution (a) Locate t.he image and ils magnification if the object is at 30.0 cm. I
I
1
-+-=!J (/ I
The ray diagTam is given in Figure 23.27a. Apply the thin-lcns equation with j) = 30.0 cm to locate the image:
1
1
---+-= ~o.o
Solve for 'I, which is ncgative and hence virtual:
q=
Substitute into Equation 23.10 to get the magnification. Because J11 is positive and has absolute value Icss than I, lhe image is upright and smaller than the objeCl:
A1 =
elll
~7.50
'I
10.0 em
cm
'I
!J
em)
-7.50 ( 30.0 elll
+0.250
(b) Locate the image and flnd its magnification irthc o~icet is
Remarks Notice that in every case the image is virtual, hence on the same side of the lens as the object. Further, the image is smaller than the objecl. For a diverging lens and a real object. this is always the case, as can be provcn mathcmatically. QUESTION 23.8 Can a divcrging lens be used as a magnifying glass? Explain. EXERCISE 23.8 Repeat the calculation, finding the position of the image and the magnification if the object is 20.0 cm from the lens.
Answers
q = -6.67 em, M = 0.334
Combinations of Thin Lenses Many useful optical devices require two lenses. Handling problems involving two lenses is not much different from dealing with a single-lens problem n...,ice. First, the image produced by the first lens is calculated as though the second lens were not present. The light then approaches the second lens a~ iJit had come from the image formed by the first lens. Hence, the image formed by the first lens is treated as the object for the second lens. The image formed by the second lens is the final image of ule system. 11' the imagc formed by the f,rSl.lens lies on the back side of the second lens, the image is treat.ed as a virtual object for the second lens, so /) is negative. The same procedure can be extended to a system of three or more lenses. The overall magnification ofa system of thin lenses is the product of the magnincations of t.he separate lenses.
EXAMPLE Goal
23.9 Two Lenses in a Row
Calculate geometric quantities for a sequential pair of lenses.
Problem Two converging lenses are placed 20.0 cm apart, as shown in Figure 23.28a (page 780). with an object 30.0 em in front of lens 1 on the lefl. (a) If lens 1 has a focal length of 10.0 em, locate the image formed by this lens and det.ermine its magnification. (b) If lens 2 on the right has a focal length of20.0 crn, locate the final imagc formed and find the LOtal magnification of the system. Strategy ',Ve apply the I hin-Iens equation to each lens. The image formed by lens I is treated as the object for lens 2. Also, v.re use the fact thalthe total magnification of the system is the product of the magnifications produced by the separate lenses.
780
Chapter 23
FIGURE 23.28
Mirrors and Lenses
(Example 23.!1)
Objea
11 '"
I 0.0 em 12 = 20.0 em
(a)
I.l;lls 2
Lens 1 (0)
...... , .....
r·························
~
Solution
(a) Locate the image and determine the magnifIcation of lens 1. I
I
30.0 em
q
I
See the ray diagram, Figure 23.28b. Apply the thinlens equation to lens l:
.,..,--,------ + -
Solve for q, wh ich is positive and hence to the riglll of the first lens:
q = +15.0 em
Compute the magnification or lens 1:
M,
= -,--,------
10.0 em
q
15.0 ( I l l 30.0 em
/]
-0.500
(b) Locate the final image and find the lOlal magnification.
The image fOfmed by lens 1 becomes the ol~jecl for lens 2. Compute the object distance for lens 2: Once again apply the thin-lens equation to lens 2 La locate the flnal image:
/] = 20.0 em - 15.0 em = 5.00 em
I c:-::-:-+- = 5.00 em q
-,--,------
20.0 em
q = -6.67 em Calculate lhe magnification of lens 2:
Multiply the two magnillcations to get the overall magnification of the system: ~
M,= M
q
/J
= M,NI,
~
(-6.67 ern) 5.00 em
+ 1.33
(-0.500)(1.33)
= -0.665 ~
Remarks The negative sign for !vi indicates that the final image is inverted and smaller than the object because the absolute value of lVl is less than 1. Because fJ is negative, the final image is virtual.
23.7
Lens and Mirror Aberrations
781
QUESTION 23.9 If lens 2 is moved so it is 40 cm away from lens 1, ,.. ' ould the final image be upright or inverted? EXERCISE 23.9 If the two lenses in Figure 2:\.28 arc separated by 10.0 em, locate the final image and find the magnification of the system. Hint-· The object lor the second lens is virtual!
Answers
4.00 cm behind the second lens, 1\1 = -0.400
(a)
(b)
(e)
FIGURE 23.29 Lenses call produce variolls furms of aherratiolls. as shown by these blurred phOIOgraphic images of ,I point sourcc. (a) Spherical aberration occurs whell light passing thrOuKh the lens a1 difrcrclll distances frolll t hl: principal axis is focused at different poi illS. (b) Astigmatism is an aber~ l";lli(ln th:11 occurs when Ihe object is not on the prillcipal axis of the It'll.... (c) Coma. This aberration ncclIrs wilen lighl passing through lhc Jens far frolll the principal axis focuses at a different part ortllc fllcal plane than light passing IIcar [he cellter of the lens.
23.7
LENS AND MIRROR ABERRATIONS
One of the basic problems of systems containing mirrors and lenses is the imperfect quality of the images, ,...hich is largely the result of defects in shape and form. The simple theory of mirrors and lenses assumes rays make small angles with lhe principal axis and all rays reaching lhe Jens or mirror from a point source are focused at a single point, producing a sharp image. This is nol. always true in the rcal world. Where lhe apprOXimations used in this theory do not hold, imperfcct images arc formed. 1r onc wishes to analyze image formation precisely, it is necessary to trace each ray, lIsing Snell's law, at each refracting surface. This procedure shows that there is no single point image; instead, the image is blurred. The departures orreal (imperfect) images from the ideal predicted by Lhe simple theory arc called aberrations. Two common types or aberrations are spherical aberration and chromatic aberration. Photographs of three forms of lens aberrations are shown in Figure 23.29.
~
--
~-
Spherical Aberration Spherical aberration results from the fact thal the {-ocal points of light rays passing far ['·Olll the principal axis of a spherical Icns (or mirror) are different from the focal points of rays with the same wavelength passing neal- the axis. Figure 23.~O illustrates spherical aberration for parallel rays passing through a converging lens. Rays ncar the middle of the lens arc imaged farther from the lens than rays al the edges. Hence, there is no single focal length for a spherical lens. Most C<1meras are equipped with an acUustable aperture to control the light intensity and, when possible, reduce spherical aberration. (A 11 aperture is an open-
~
•
"-
"
.....~
"
-",
....
'"
FIGURE 23.30 Spherical aberralion produced by a converging lens. Does a diverging lens pruduce spherical aberration? (A ngles are greatly cxaggerated for clarity.)
782
Chapter 23
Mirrors and Lenses
ing thal cOlllrols the amount of lighL transmitted through the Jens.) As the aperture size is reduced, sharper images arc produced because only the central porlion of the lens is exposed to the incident light when the aperture is very small. At the same time, however, progressively less light is imaged. To compensate for this loss, a longf'r exposlIrF limp is 1ISf'd. An f>xrlmplp (}flhr. n>sltlrs oh'::lined with small
apertures is the sharp image produced by a pinhole camera, with an aperture size of approx imaLcly 0.1 1111ll. In the case of mirrors used for very distant objects, one can eliminate, or at least minimize, spherical aberrat.ion by employing a parabolic rather than spherical surface. Parabolic surfaces are not used in many applications, however, because they are very expensive to make with high-quality optics. Parallel light rays incident on such a surface focus at a common point. Parabolic reflecting surfaces are used in many astronomical telescopes to enhance the image quality. They are also used in flashlights, in which a nearly parallel light beam is produced from a small lamp placed at the [ocus of the refleCLing surface.
Chromatic Aberration Different wavelengths of light refracted by a lens focus at different poims, which gives rise to chromatic aberration. In Chapter 22 we described how the index of refraction ofa material varies Wi111 ""avclength. vVhen "vhite light passes through a lens, for example, violet light rays are refraCLcd more than red light rays (see Fig. 23.31), so the focal leng·th for red light is greater thall for violet light. Other wavelengths (not shown in the figure) would have intermediate focal poims. Chromatic aberration for a diverging lens is opposite that for a converging lens. Chromatic aberration can be greatly reduced by a combination of converging and diverging lenses.
FIGURE 23.31 ChrolllaLie aberration produced by a comcrging lens. R
_
~
__ -
'~T'
SUMMARY 23.1
Flat Mirrors Images are formed where rays of light intersect or where they appear to originate. A real image is formed when light intersects, or passes through, an image point. In a virtual image the light doesn't pass through the image point, but appears to diverge from it. The image formed by a rIat. mirror has the following properties:
1. The image is as hir behind the mirror as the object is in front.ofil. 2. The image is unmagnified, virtual, and upright.
23.2
Images Farmed by Concave Mirrors
23.3
Convex Mirrors and Sign Conventions
1
1
l'
q
J
Images Formed by Refraction
1/.j
-
/)
n2
+-
iH
h' q [23.2] 111=-= Ii I' The object distance and image distance for a spherical mirror of radius Rare related by lhe mirror equation:
1
23.4
An image can be formed by refraction al a spherical surface of radius R. The object and image distances for refraction ['rom such a surface are related by ~
q
n2-ul
--"---'/I
[23.7]
The magnification of a refracting surface is
The magnification M ofa spherical mirror is defined as the ratio of the image height 11,' to the object height h, which is the negative of the ratio of the image distance q to the object dislance /1:
-+-~-
where J= 1?/2 is the focal length of the mirror. Equations 23.2 and 23.6 hold for bmh concave and convex mirrors, suqject to the sign conventions given in Table 23.1.
[23.6J
11.'
11 I
q
= -= ---
II.
[23.8]
1/.211
where the object is located in lhe medium with index of refraction n l and t.he image is formed in the medium with index of refraction 'H2 . Equations 23.7 and 23.8 are subjecl to the sign conventions of Table 23.2.
23.6
Thin Lenses
The magnification of a tbin lens is h'
'I
II
I'
,\;1=-=
[23.101
Conceptual Questions
The object and image distances of a thin lens are rdaled b\ the thin-lens equation: I
1
I
-+-=/' q J
[23.11]
783
from the focal points of lighl ravs far from Ihe principal axis ofa spherical lens or mirror being differelll fl'om those of nl}S passing through the center. Chromatic aberration arises becausc light rays of di ffcrclll wavelengths focus at different poinL'i when refracted by a lens.
Eqwu:ions 23.10 and 2:\.11 al'e subjectLO the sign conventions of Table 23..~.
23.7 Lens and Mirror Aberrations Aberrations are respom,ible for the formation of imperfect images by lenses £lnd mirrors. Spherical aberration results
FOR ADDITIONAL STUDENT RESOURCES. GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS J. When an image ora real objecl is formed br;' flat mirror, which of Ihe following statcments are always true? (a) The ima):{e is larger Ihan the ohject. (h) The image is the same size as the object. (c) The image is virtual. (d) The image is smaller than the object. (e) The image is upright.
2. An object is placed 16.0 Clll away from a conCan~ mirror with focal length 6.00 em. Find the location of the image. (a) 9.60 em in frol1\. of the mirror (b) ~1.:~6 cm in fl'ont of the mirror (c) 9.60 em behind the mirror (d) 4.36 em behind the mirror (c) 10.0 em in front of lhc mirror
7. \'\Then (he image of a real object is formed by a conca\·e mirror, which of the following statements are true? (a) The image: is always ,'cal. (b) The image is always virtual. (c) If lhe object dist~lJ)ce is less than the focal length. the image is real. (d) If the object distance is greater than tbe local length, the image is real. (e) If q < 0, the image is virtual.
8.:\ convex mirror forms an image of a real object. Which
3. All object is phlced 16.0 em 3\\'a)' from a convex mirror wilh 3 focal length of magniwde 6.00 cm. \"'hm is the location of the image? (3) 9.60 ell1 in frontoflhc mirror (b) 4.36 em in from of the mirror (c) 9_60 cm behind fhe mirror (d) 4.36 em behind the mirror (e) 10.0 Clll in fronlofthe mirror 4. A cOlwel-ging lens has a Yirwal object located :l fillite distance ;:m:ay. 'Which of t.he following must be Irue? (a) The image is virtual. (b) The: image is real. (c) /) < 0 (e1) /) > 0 (e) J> 'I
5. r\ I hin, convergcnl lens has a focal length of H,OO em. Ir~l real, im·erted image is located 12.0 cm to Ihe right of Ihe lens, where is the object located? (a) 12.0 em to the left of the lens (b) 24.0 CIl1 10 the righl of the lens (c) 24.0 cm to the left of the lens (d) 18.0 Clll lo 1he right of the lens (e) 18.0 em 10 the left of the lens 6. A real object is 10.0 cm to the left of a thin, lells hewing a focal length of magnitude 16.0
is the localion of the image? (
di\'eq~ing Clll.
of the following 'itatemelllS is always true? (a) The image is rc~d and upright. (b) The image is vinual and larger than the ol~iect. (c) The image is virtual and upright. (d) The image is larger than the object. (e) The image is virtual and inverted.
9. An ol:~jccl is located 50 cm from a converging lens h3\·ing a focal length of 15 Cln. \Vhich of the foJlO\,;ing is true regarding the image formed by the lens? (a) [t is virtual, upright, and larger than the object. (b) It is virtual, inverted, and smaller than the object. (c) Il is real, inverted, and smaller than the objecl. (d) It is real, ilwcned, and larger than the objeCi. (e) h is real. upright. and larger than the object. 10. For a real ol~j('ct. is the image formed by a diverging lens always (a) dnual and inverted, (b) virtual and upright. (c) real and upright, (d) real and ilwertcd, or (e) real and magnified:
\,Vhal
CONCEPTUAL QUESTIONS 1. Tape a picture of yourself on a bathroom mirror. Siand s<:vcral centimeters away from the min'or. Can YOll fOClIS your e)'es on both the picture taped La the mirror ami your image in lhe mirror at thl' same tit1U'? So where is the image of )'OUrsclI?
2. Explain wh) a mirror cannol give rise to chromatic ~lb('rralion.
3. A flat mirror creates a virtual image of your face. Suppose the flat mirror is combined with another optical
Chapter 23
784
Mirrors and Lenses
element. Can the mirror form a real image in such a combination? 4. \-Vhy does a clear stream always appear to be shallm\'cr than it aClllally is?
to. H a cylinder of solid glass or clear plastic is placed above the words LEAD OX fDE and viewed from the side, as shown in Figure CQ23.l0, the word LEAD appears inverted, but the word OXIDE does not. Explain.
5. A \,jrtual image is often described as an image through which light nl)'s don't actually travel. as they do for a I-cal image. Can a virtual image be photographed? 6. A cOlllmon mirage is formed when the air gelS gradually cooler as the height abo\'e the ground increases. ''''hat might happen if the air grm\iS gradually warmer as the height increases? This often happens over bodies ofwalcr or snow-covered ground; the effecL is called looming. 7. Suppose you wanl to lise a converging lens to project the image of 1\\'0 trees onto a screen. One tree is
""~
,o,.~
I
,
Screen
\1
I
(b)
FIGURE CQ23.10
11. In a Jules Verne novel, a piece of ice is shaped into a magnifying lens to focus sunliglll to s[art a fire. ls that possible?
12. Light from an o~jcct passes through a lens and forms a visible image on a screcn. If the screen is removed, would you be able to see the image (a) if you remained in your present position (lnd (b) if)'oll cotlld look at lhe Icns along the its axis, beyond the original position of the screen?
Far trce
I
+------- x ,I Image of J ncar u·ce ... , - - - - - - - 2x -------~.
~
rEVD OXIDE
LEAD OXIDE
FIGURE CQ23.7
8. Lenses used in eyeglasses, ,.. . hether converging or diverging, are fllways designed such that the middle of the lens curves away from lhe eye. Why? 9. Can a converging lens be made La diverge light if placcd in a liquid? How aboll! a converging min-ol·?
13. \-Vhy does t.he local length of a mirror nOl depend on
the mirror material when [he focal length of a lens does depend on Lhe lens matcrial?
14. An invcrted image of an objecl is view·ed on a screen. rrom lhe side facing a converging lens. An opaque card is Lhen illlroduced covcring only the upper half of the lens. "Vhal happens to the image on the screen? (a) Half the image would disappear. (b) The entire image would appear and remain unchanged. (c) Half the image would disappear and be dimmer. (d) The entirc image would appear, bUL would be dimmer.
-~
PROBLEMS ~
The Problem~ for {his chapter m:i1v be assigned online at \'Veb.--\ssign. ' L, ~,j = straightforward, illICllllCdi,llt',
WebAssign
m
II =
biomedical application D = dcnotes full solution available in Stllde1ll SO/Illiuta Mall1/(/1!
Sindy Gnide
yOIl older or younger than your actual age? Compllle an order-or.magnitude eSlimate for [he age difference, based on da[a you specify.
3. A person walks into a room lhat h
2.lm Two plane mirrors sland facing each other. 3.0 In apan, and a woman slands between lhem. The woman faces one of the mirrors [rom a dislance of 1.0 111, with the
4. In a church choir lofl.. lWO parallel walls are 5.30 m apan. The singers Hand against the north waiL The organist faces the south wall, silting 0.800 m awa~' (rom it.. So
SECTION 23.1 FLAT MIRRORS l. Docs your bathroom mirror show
I
palm of her left hand facing lhe closer mirror. (a) ,,\-'hat is lhe apparem position of the c10sesl image of her lefl hand, measured from the surface oflhe mirror in front of her? Does i[ show thc palm of her hand or the back of her hand? (b) ,,"Vhal is lhe posi[ion of the neXl image? Does il show the palm of her hand or the back of her hand? (c) Repeat for the lhird image. (d) Which of the images are real and which arc vinual?
Problems that she can sec the choir, a flat mirror 0.600 m wide is lllounted on the south wall, straight in front of the organist. \Vhm width of the north wall can she see? Hint: Draw a lop-view diagram to juslify your answer. 5.
m
Use Active Figure 23.2 to givf' £1 gronwtri, proof that the virtual image formed by a plane mirror is the same distance behind the mirror as the object is in frolll of it.
of the hubcap. (a) How br is his face from the hubcapr (b) What is the radius of curvature of the hubcap? 13. A concave makeup mirror is designed so that a person 25 cm in [rollL or' it sees an upright image magnified by a factor of two. What is the radius of curvature of the mirror? 14.
SECTION 23.2 IMAGES FORMED BY SPHERICAL MIRRORS SECTION 23.3 CONVEX MIRRORS AND SIGN CONVENTIONS !l1t/l('following problems, olgebmic signJ arp nol.givl'll. We team' it to yolt to determine fhe con1'cl sign to use with each quantity, basfd on a11 a1/al:)'si.~ of Ihe pmblt'ln and the sigu emwenliolls in Table 23./.
6. A delllist uses a mirror to examine a tooth that is l.00 em in front of the mirror. The image of the tooth is formed 10.0 cm behind the mirror. Determine (a) the mirror's radius of curvature and (b) the magnification of the image. 7. _ An object is placed 40.0 cm from a concave mirror of radius 20.0 cm. (a) Find the location of the image. (b) What is the magnification of the mirror? Is the image real or virtual? Is the image upright or inverted? 8.
II To fit a contact lens to a patient's eye, a !<eratlJllIcferCan be used to measure the cun'ature of the cornea-the [ront surface of the eye. This instru ment places
9.
mm
10.
mm
An object of height 2.0 cm is placed 30.0 cm from a convex mirror of focal1cngth having magnitude 10.0 cm. (a) Find the location of the image. (b) Describe the properties of the image.
While looking at her image in a cosllletic mirror, Dina notes that her bee is highly magnified vy'hen she is close to the milTor, buL as she backs away from the mirror, her image first becomes blurry, then disappears when she is about 30 cm [rom the mirror, and then inverts ' . . hen she is beyond 30 cm. Based on these observations, what can she conclude about the properties of the mirror?
OI]A 2.00-cm-high ol~ject is placed 3.00 Clll in front ora concave mirror. I[the image is 5.00 cm high and virtual, what is the focal length of the mirror? 12. A dcdicated spans car enthusiast polishes the inside and omside surfaces of a hubcap that is a seClion of a sphere. 'Nhen he looks into one side of the hubcap, he sees an illlage of his face 30.0 (In in back of it. I-Ie then turns the hubcap over, keeping it the same distance from his face. He now sees an image of his face 10.0 em in back
785
ol:~ject is placed JO.O em from a convex mirror having a focal length of magnit.ude 15.0 Clll. (a) Use graph paper to construct an accurate ray diagram for this siwation. From your ray diagram, determine the location of" the image. its magnification, and whether it is upright or inverted. (b) Check your ansv,:ers to part (a) using the mirror equation.
mil An
!KIA Illan standing 1.52 m in front of a shaving mirror produces an inverted image 18.0 Clll in from ofil. How close to the mirror should hc stand jf he wants 10 form an upright image of his chin that is twice the chin's actual size? 16. A convex mirror has a focal length of magnitude 8.0 cm. (a) If the image is virtual, what is the object location for which the magnilUde of the image distance is one third the magnitude of the object distance? (b) Find the magnification of the image and state whether it is upright or inverted. Ii. A woman holds a tllbe of lipstick 10.0 cm from a spherical mirror and notices that the image of the tube is upright and halfits normal size. (a) Determine the position of the image. Is it in front of or behind the mirror? (b) Calculate the focal length of the mirror. 18.
mil A concave mirror has a radius of curvature of24 cm. (a) Determine t.he object position for which the resulting image is upright and three times the size of the object. (b) Draw a ray diagram to determine the position of the image. Is the image real or vinual?
II9.IE A spherical mirror is to be used to form an image, fivc timcs as t
SECTION 23.4 IMAGES FORMED BY REFRACTION 21. A cubical block of ice 50.0 cm on an edge is placed on a le\'cl floor over a speck of dust. Locate the image of the speck, when viewed from directly above, if the index of refraction of ice is 1.309. 22. One end ofa long glass rod (1'1 = 1.50) has the shape ofa convex surC'lcc or radius 8.00 em. An object is positioned in air along the axis of the rod in front of the convex surface. Find the image position that corresponds 1.0 each of the follovY'ing object positions: (a) 20.0 cm, (b) 8.00 cm, (c) 4.00 cm, (d) 2.00 cm. 23. A paperweight is made of a solid glass hemisphere with index of refraction 1.50. The radius of the circular cross
786
Chapter 23
Mirrors and Lenses
section is 4.0 em. The hemisphere is placed on its flat surface, with the center directly over a 2.5-1ll1ll-long line drawn on a sheet of pareL \'\-'hat length of line is seen by someone looking vertically down on the hemisphere?
(lnd (c) 10.0 Clll. For each case, state whether the image is real 0" virtual and upright or invened, and flnc! the magnification. 34.
24. The top ora swimming pool is atgrOllnd level. l(the pool is 2 m deep, ho\\' far below ground level does the bottom of [}lC pool appear to be located when (a) the pool is completely filled with water and (b) the pool is filled halfway WitJl water? 125.IA transparent sphere ofullknown composition is observed to form an image of rhe Sun on its surface opposite the Sun. What is the refractive index 01' the sphcl"c m;:tlcrial? 26. A nim glass plate (n = 1.66) rests on the bonam of ,m aquarium tank. The plate is 8.00 cm thick (vcnical dimension) and covered with water (n = 1.33) to a depth of 12.0 cm. CaJcuhlle the apparent thickness of the plate as viewed from above the water. (Assume nearly normal incidence of light rays.)
am ;\ diverging lens has a focal
length of 20.0 CIll. Use graph paper to construct accur
35. A transparellt photographic slide is placed in from of a converging lens \"... ith a focal length of 2.44 cm. The lens lonns an im
36. The nickel's image in Figure P23.36 has twice the diameter of the nickel when the lens is 2.84 cm fl"Omthe nickel. Determine the focal length of the lens.
27. _ A jellyfish is Ooating in a wmer-filled aquarium 1.00 In behind a Oat pane of glass 6.00 cm thick and having an index ofrcfraetioll of 1.50. (a) \,,"'here is the image of the jellyfish located? (b) Repeal. the problem when the glass is so thin that its thickness can be neglected. (c) How does the thickness of the glass affect the answer to pan (a)?
28. A goldllsh is swimming at 2.00 cm/s toward the front wall of a rectangular aquarium. What is the appan..:nl speed 01" the fish as measured hy an observer looking ill from outside the front \,'all of the tank? The index of rcf,'acLion of water is 1.333.
SECTION 23.6 THIN LENSES /29.111 A contact lens is made of plastic with an index of refraction of 1.50. The lens has an outer radius of cur~ vature of +2.00 cm and an inner radius of curvature of +2.50 Clll. \",,'hat is the focallenglh or the lens?
30. The left 1~lcc ofa hiconVt.:x lens has a radius of curvature of 12.0 cm, and the right face has a radius of curvature of 18.0 Clll. The index of refraction of the glass is 1.44. (a) Calculate the {(Kal length of' the lens. (b) Calculate lhe focal length if lhe radii 01" curvature of lhe (\',.-0 faces are interchanged. 31. A converging lens has
(l focal length o[ 10.0 cm. Locate the images for object distances of (a) 20.0 CIll, (b) JO.O cm, and (c) 5.00 Cill, if they exist. For each case, st.ate whether t.he image is real or virtual, upright or inverted, and find the magnificalion.
32.
m
A converging lens has a focal !eng·th of W.O cm. Usc graph paper LO construct accurate ray diagrams for object distances of (a) 20.0 cm and (b) fl.O crn. From your ray diagrams, detcrminc the location of the image and its properties and compare your results Wil h the values found algebraically in Problem 31. (c) Name at leastth"ee errors in construcling [he graph that could lead to differences in the final answer.
133.IA diverging lei IS has a focallcngth 0[20.0 Clll. Locate the images lor o~jeet dist.ances of (a) 40.0 em, (b) 20.0 cm,
FIGURE P23.36
37. A certain LCD projector contains a single thin lens. An object 24.0 mm high is 10 be pr~jected so thal ils image fills a screen 1.80 m high. The objecl-to-screen distance is 3.00 lll. (a) Determine the local length of the pr~jcC(ion lens. (b) How f
139.IA converging Jens is placed 30.0 cm to the right of a diverging lens of focal length 10.0 CIll. A beam of paralIcl light enlers the diverging lens from lhe left, and the beam is again parallel when it emerges from the converging lens. C
40.
am
(
41. Two converging lenses, each of focal length 15.0 cm, are placed 40.0 em apart, and an ot~ject is placed 30.0 cm in from of t.he firsl lens. 'Where is the final image formed, and what is the magnification of the system?
Problems
42.
Ol~ject 0 1 is
15.0 em LO Ihe left of a converging lens with a lO.O-cm focal length. A second lens is positioned 10.0 cm LO the right of the first lens and is observed to form a final image at the position of the original objecl 01' (a) What is the focal length of the second lens? (b) What is the overall magnification or this system? (c) What is Ihe nature (i.e.. real or virtual. upright or inverted) of the final image?
43. A I.OO-em-high o~jecl is placed 4.00 em LO the left of a converging lens of focal length 8.00 elll. A diverging lcns of focal length -16.00 em is 6.00 em 10 the right of the converging lens. Find the position and hcighl of the final image. Is the image invcrted or upright? Real or virtual?
44. Two converging lenses having focal lengths of 10.0 em and 20.0 em are placed 50.0 em apart, as shown in Figure P23.44. The final image is LO be located between the lenses, at the position indicalcd. (a) How far LO the left or the first lens should the O~jCCl be positioned? (b) \Nha1. is the overall magnification of the system? (c) Is lhe final image upright or inverted? Objecl
it I.
I,
(IACnl)
V
h
"m'm~
(2
cm)
fj A
-+-i:-,,"~J-J
50.0 em - - - - - ,
FIGURE P23.44
145.1 Lens L1 in Figure P23.45 has a foca 1length of J5.0 em and is located
Film
787
and lens. (c) Calculate the magnification of the firsl. lens. (f) Calculate the magnification of the second lells. (g) What is the total magnification for the system? (h) Is Ihe final image real 01' .... irtual? Is il upright or inverted?
ADDITIONAL PROBLEMS 47. An o~ject placed 10.0 em from a concave spherical mirror produces a real image 8.00 em from the mirror. If the o~iecl is moved to a new position 20.0 em from the mirror, what is the position of the image? Is the final image real or vinllal?
A real o~ject's distance from a eOlwerging lens is five times the foeallenglh. (a) Determine the location of the image qin t.ermsofthe l-ocallenglhf (b) Find the magnification of the image. (e) Is the image real or virlual? Is il upright or inverted? Is the image on the same side of the lens as the object or on the opposite side?
48. _
49. A convergent lens ·with a 50.0-mm focal length is llsed LO focus an image of a very distant scene onto a nat sneen 35.0 mm wide. What is the angular width a of the scene included in the image on the screen? 50. A diverging lens (n = 1.50) is shaped like that in Active Figure 23.25c. The radius of the first surface is 15.0 em, and that of the second surface is 10.0 em. (a) Find the food length of the lens. Determine the positions of the images for object distances of" (b) infinity, (c) 3111, (d) III, and (e) 1/112 51. The lens and the mirror in Figure P23.51 are separated by 1.00 m and have focal lcngths of +80.0 em and -50.0 cm, rcspcClively. If an object is placed 1.00 III to Ihe left. of the lens, where will the final image be 10caLCd? State whether the image is upright or inverted. and determine the overall magnification. Object
MilTOI'
Lens
1--1.00 Ill-~·I-,-FIGURE P23.51
152.IThe object in Figure P23.52 is midway between the lens and the mirror. The mirror's radius of curvature is FIGURE P23.45
46.
Ell
I\n object is placed l5.0 em from a first converging lells of focal length 10.0 em. ;\ second converging lens with focallcng1.h 5.00 em is placed 10.0 ( I l l to the rig-Ill or the f-i.rsl converging lens. (a) Find the position tIl of the image formed by the first converging lens. (b) How far from the second lens is the image of the first. lens? (e) What is the valuc of 1)'1' the object position [or the second lens? (d) Find the position f{'.!. of the image fanned by the sec-
Lens
Mirror
Objcel
_~_R_ ~25.0cm FIGURE P23.52
788
Mirrors and Lenses
Chapter 23
20.0 em, and the: lens 11;\5 a focal Icng\h of -16.7 em. Considering onl}' the light that leaves the object and travels firslLOward the mirror, !ocmc the final image formed by this system. Is the image real or virtual? Is jt upright or im"cned? 'Nhal is the overall magnification o[the image? 53. A parallel beam of light el1lers a glass hemisphere perpendicular 10 the Hat [ace, as shown in Figure P23.53. The radius of the hcmi.sphcre is R= 6.00 em, and the index of refraction is n = 1.560. Determine the point at which the beam is foclIsed. (Assume paraxial rays; i.e., assume all rays arc located close LO the principal axis.) Air
157.IAn object 2.00 cm high is placed 40.0 cm to the left of a converging lens having a focal length of 30.0 cm. A di"erging lens having a focallenglh of -20.0 em is placed 110 em to the right of the converging lens. (
FIGURE P23.53
54. A converging lens of focal1ength 20.0 em is separated hy 50.0 em from a cOllverging lens of local length 5.00 em. (a) Find the position of rhe final image of an ol~ject placed 40.0 em in front of the first lens. (b) If the height of the object is 2.00 em, what is the height of' the final image? Is lhe image real or \'irLual? (c) [f the two lenses arc now placed in contact with each other and the object is 5.00 em in fronl. of this combination, where will the image be located? (See Problem 56.)
r'5.
To work this problem, use the fact that the image U 2.00 cm 4.00 em/\. formed by the first surface -S.OOcm-/ becomes the object for the second surface. Fig1.00 ure P23.55 shoh's a piece FIGURE P23.55 of glass with index or refraction 1.50. The ends are hemispheres with radii 2.00 em and 4.00 em, and the celllcrs of the hemispherical ends are separated by a distance of8.00 em. A point object is in air, 1.00 em from the left end of the glass. Locate the image or the object due to refraction at the two spherical surfaces.
j
56.
SU
I
~-------J
mJ
Consider two thin lenses, one of focal lengt.h j; and the other of local length i'2' placed in contact with each other, as shown in Figure P2~.56. Apply the thin-lens equation to each of these lenses and combine the results to show tl1at this combination of lenses behaves like a thin lens havinga focallcngthigiven by 1//= Vii + I/j~. Assume the lhicknesses of the lenses can be ignored in comparison to the Dlher distances involved.
FIGURE P23.58
59. Figure P23.59 shows a converging lens Witll radii III = 9.00 em and R'l = - 11.00 Clll, in front ofa concave spherical mirror of radius R = 8.00 cm. The focal points (F l and F'.!.) for the thin lens and the center of curvature (C) of the mirror are also shown. (a) If the focal points F l and /0"1, are 5.00 cm from the vertex of"t.hc tJlinlcns, what is the index of refraction of the lens? (b) If the lens and mirror are 20.0 em apart and an object is plac.:ed 8.00 cm to the left of the lens, \\lhat is the position of the fInal image and ils magnification as seen by the eye in the figure? (c) Is the final image inverted or llpright? Explain.
c F
Ji
,
h
FIGURE P23.56
FIGURE P23.59
Problems 60.
m
Find the object dislances (in terms of j) for a thin converging lens of focal length fif (a) the image is real and the image distance is four times the focal length and (b) the image is virtual and the image distance is three times the focal length. (c) Calculate the magnification of the lens for cases (a) and (b).
l!IJThe lens-maker's equ£nion for a lens with index in a medium wilh index n.1 takes the form
III
789
times larger than the object. The lens has a focal length of 10.0 em. The lens and minor are separated by 40.0 cm. Determine the focal length of the mirror. (Don't assume the figure is drawn to scale.) Minor
Lcn~
immersed
1 ("',,,-1 )( R:-R, J I) 7= A thin diverging glass (index = 1.50) lens with R] = -3.00 m and H2 = -6.00 m is surrounded by air. An arrow is pl
FIGURE P23.62
63. The lens-maker's equation applies 1.0 a lens immel'sed in a liquid if n in the equation is replaced by n l /1I2 • Here 111 refers to the refracti\'e index of the lens material and n,.! is that of the medium surrounding the lens. (a) A certain lens has focal length of 79.0 em in air and a refractivc index of 1.55. Find its local length in watcr. (b) A certain mirror has focal lengLh of 79.0 em in air. Find its focal length in water.
The colors in many of a
humming~
bird's feathers are not due to pigment. The iridescence that makes the brilliant colors that often appear on the bird's throat and belly is due to on interference effect caused by structures in the feathers. The colors vary with the viewing angle.
24.1
Conditions for Interference
24.2
Young's Double-Slit Experiment
24.3
Change of Phose Due to Reflection
24.4
Interference in Thin Films
24.5
-g
" :-_---------------~ ~
Using Interference to
Read CDs and DVDs
24.6
Diffraction
24.7
Single-Slit Diffraction
24.8
The Diffraction Grating
24.9
Polarization of Light Waves
WAVE OPTICS Colors swirl on a soap bubble as it drifts through the air on a summer day, and vivid rainbows reflect from the filth of oil films in the puddles of a dirty city street. Beachgoers, covered with thin layers of oil, wear their coated sunglasses that absorb half the incoming light. In laboratories scientists determine the precise composition of materials by analyzing the light they give off when hot, and in observatories around the world, telescopes gather light from distant galaxies, filtering out individual wavelengths in bonds and thereby determining the speed of expansion of the Universe. Understanding how these rainbows are made and how certain scientific instruments can determine wavelengths is the domain of wave optics. Light can be viewed as either a particle or a wove. Geometric optics, the subject of the previous chapter, depends on the particle nature of light. Wave optics depends on the wave nature of light. The three primary topics we examine in this chapter are interference, diffraction, and polarization. These phenomena can't be adequately explained with ray optics, but can be understood if light is viewed as a wave.
24.1
CONDITIONS FOR INTERFERENCE
In our discussion of interference of mechanical waves in Chapter 13, we found that two waves could add together either constructively or destrllctively. Tn constructive interference the amplitude of the resultant wave is greater than that of either of the individual 'Naves, whereas in destructive interference, the resultant amplitude is less than that of either individual \vave. Light waves also interfere with one another. Fundamentally, all interference associated with light waves arises \vhen the electromagnetic fields that constitute the individual waves combine. Interference effects in light waves aren't easy to observe because ofthe short ,'V'avelengths involved (about 4 X ]0-7 m to about 7 X 10-7 m). The follm,,'ing two conditions, however, facilitate the observation of interference beGveen two sources oflight: Conditions for interference ~
790
1. The sources are coherent, which means that the waves they emit must maintain a constant phase with respect to one another. 2. The waves have identical wavelengths.
24.2
Young's Double-Slit Experiment
791
Two sources (producing two traveling wavcs) arc needed to create interference. To produce a stable interference pattern, the individual waves must maintain a constant phase with one another. ''''hen this siwalion prevails, the sources are said to be coherent. The sound waves cmitled by twO side~by-sidc loudspeakers driven by a single amplifier can produce interfer"ence because the t,..,o speakers respond to the amplifier in the same way at the same time: they are in phase. If two light sources are placed side by side, however, no interference effecls are observed because the light waves from one source are emitted independenLly of the waves from the other source; hence, the emissions from the two sources don't maintain a constant phase relationship with each other during the time of observation. An ordinary light source undergoes random changes about once every 10-8 s. Therefore, the conditions for construClive interference, desrructi\'e interference, and intermediate states have durations on the order of 10-8 s. The result is that no interference effects are observed because the ere can't follow such short-term changes. Ordinary light sources are said to be incoherent. An older method for producing two coherent light sources is to pass light [rom a single \\lavelengl.h (monochromatic) source through a narrow slit and then allow the light to fall on a screen containing two other narrow slits, The first slit is needed LO create a single wave front that illuminates both slits coherently. The light emerging from the two slits is coherent because a single source produces the original light beam and the slits serve only LO separate the original beam into two parts. Any random change in the light emitted by the source will occur in the two sep<:lrate beams at the same time, and interference effects can be observed. Currently it's much more common to use a laser as a coherent source to demon· strate interference. A laser produces an intcnse, coherenl, monochromatic beam over a width of several millimeters. The laser may therefore be used to illuminate multiple slits directly, and interference effects can be easily observed in a fully lighted room. The principles of operation or a laser are explained in Chapter 28.
24.2 YOUNG'S DOUBLE·SLlT EXPERIMENT Thomas Young first demonstrated interference in light waves from two sources in 1801. Active Figure 24.1a is a schematic diagram of the apparatus used in this expcriment. (Young used pinholes rather than slits in his original experiments.) Light is incident on a screen containing a narrow slit So' The light waves emerging from this slit arrive at a second screen that contains two narrow, parallel slits SI and 52' These slits serve as a pair of coherent light sources because \..'aves emerging ACTIVE FIGURE 24.1 (a) A diagram of Young's doubleslit experiment. The narrow slits aCl as sources of \\·
max
min
~'T
max
5,
min
max
•
fie
min
u
max
Firsl screen
SC('ond screen
g
j
min
"~
""
~ Screen (a)
(b)
792
Chapter 24
Wove Optics
FIGURE 24.2 An interference patlern involving water waves is produced by 1\\,0 vibr
Path difference
from them originate from the same \',lave front and therefore are always in phase. The light from the two slits produces a visible pattern on screen C consisting of a series of bright and dark parallel bands called fringes (Active Fig. 24.lb). When the light from slits SI and S2 arrives at a point on the screen so that constructive interference occurs at that location, a bright fringe appears. V·/hen the light [rom the two slits combines destructively at any location on the screen, a dark fringe results. Figure 24.2 is a photograph of an interference pattern produced by two coherent vibrating sources in a water tank. Figure 24.3 is a schematic diagram of some of the ways in which the two waves can combine at screen C of Figure 24.1. Tn Figure 24.3a two waves, which leave the t\vo slits in phase, strike the screen at the central point P. Because these waves travel equal distances, they arrive in phase at P, and as a result, constructive interference occurs there and a bright fringe is observed. In Figure 24.3b the t\'I"O light waves again start in phase, but the upper wave has to travel one ,"..avelength farther to reach point Q on the screen. Because the upper 'wave falls behind the lower one by exactly one ".:avelength, the t\vo waves still arrive in phase at Q, so a second bright fringe appears at that location. Now consider point R, midway bet\veen P and Q, in Figure 24.3c. At R, the upper wave has fallen half a 'wavelength behind the lower ,..'ave. This means that the trough of the bottom 'wave overlaps the crest of the upper wave, giving rise to destnlCtive interference. As a result, a dark fringe can be observed at R. \Ve can describe Young's experiment quantitatively with the help of Figure 24.4. Consider point P on the viewing screen; the screen is positioned a perpendicu~ lar distance L from the screen containing slits SI and S2' which are separated by distance d, and T J and T2 are the distances the secondary waves travel from slit to screen. vVe assume the waves emerging [rom SI and S2 have the same constant frequency, have the same amplitude, and start out in phase. The light intensity on the screen at P is the result of lig'hl from both slits. A wave from the lower slit, however, travels farther than a wave from the upper slit by the amount d sin 8. This distance is called the path difference 0 (lowercase Greek delta), where
o=
~
T2 -
TI
= d sin fJ
[24.1]
Equation 24.1 assumes the h\iO waves travel in parallel lines, 'which is approximately true because L is much greater than d. As noted earlier, the value of this path difference determines whether the two waves are in phase when they arrive at P. If the path difference is either zero or some integral multiple of the wavelength, the two ,-vaves are in phase at P and constructive interference results. Therefore, the condjtion for bright fringes, or constructive interference, at Pis Condition for constructive interference (two slits) ~
o=
d sin 8 LJri f,(hl
=
mA
[24.2]
m = 0, :t:l, ::t:2,.
s, p
---~--
Slits
p
R Dark area
Q Bright Q
area
Screen (h)
(a)
FIGURE 24.3
(0)
(a) Constructive imerference occurs at Pwhen the wavcs combine. (b) CO!lstJ'ucth'e illterfcrence also occurs at Q (c) Destrllctive interfercnce occurs al R. when the wave from the upper slit falls half a wavelength behind the wave from the lower slit. (These figures are not drawn to scale.)
24.2
Young's Double-Slit Experiment
FIGURE 24.4 A geometric construction that describes Young's double-slit experiment. The path difference between the two rays is 0 = r2 - r l = d sin B. (This figure is not drawn to scale.)
p
1 o
793
j
(b)
(a)
The number m is called the order number. The central bright fringe at 8brig:ht = o (m = 0) is called the zeroth-order maxim.um. The first maximum on either side, \·vhere 'In = .::t::], is called the first-order maximwn, and so forth. "Vhen 8 is an odd multiple of A/2, the two waves arriving at Pare 180 out of phase and give rise to destructive interference. Therefore, the condition for dark fringes, or destructive interference, at Pis 0
tn= 0, ±1, ±2,.
[24.3]
~ Condition for destructive
interference (two slits)
[f m = 0 in this equation, the path difference is 8 = A/2, which is the condition
for the location of the first dark fringe on either side of the central (bright) maximum. Likewise, if In = 1, the path difference is 8 = 3A/2, which is the condition for the second dark fringe on each side, and so forth. It's useful to obtain expressions for the positions of the bright and dark fringes measured venically from 0 to P. In addition to our assumption that L » d, \.\'e assume d» A. These assumptions can be valid because, in practice, L is often on the order of] m, d is a fraction of a millimeter, and A is a fraction of a micrometer for visible light. Under these conditions () is small, so V·le can use the approximation sin 8 ~ tan 8. Then, from triangle OPQ in Figure 24.4, 'we see that )' = Ltan 0 = Lsin 8
TIP 24.1 Small-Angle Approximation: Size Matters! The small-angle approximation r;jn 0 == tan 0 is tl"lle 10 three-digit precision only fOI' angles less than abolll 4".
[24.4]
Solving Equation 24.2 for sin 8 and substituting the result into Equation 24.4, we fi nel that the positions of the brighl fringes, measured from 0, are m = 0, :::!::l, ::!:2, .
[24.5]
Using Equations 24.3 and 24.4, we find that the dark fringes are located at
Yclark
=
~L(m. +~)
111
= 0,
± 1, ::!: 2, ...
[24.6]
As ,ve will show in Example 24.1, Young's double-slit experiment provides a method for measuring the wavelength of light. In fact, Young used this technique to do just that. In addition, his experiment gave the wave model of light a great deal of credibility. It \"as inconceivable that particles of light coming through the slits could cancel each other in a "Tay that would explain the dark fringes.
Reflection, illierference, and diffraction can be seen in this aerial photograph of waves in the sea.
794
Chapter 24
Wave Optics
APPLYING PHYSICS 24.1
A SMOKY YOUNG'S EXPERIMENT
Consider a double-slit experiment in which a laser beam is passed through a pair ofyer)' closely spaced slits and a clear interference pattern is displayed on a distant screen. 'ow suppose you place smoke panicles between the double slit and the screen. \Vith the presence oC the smoke particles, wi II YOll see the effeclS of interference in the space between the slits and the screen, or will you see only the effects 011 the screen?
APPLYING PHYSICS 24.2
Explanation You will see the interference pattern bOlh on the screen and in the area filled with smoke between the slits and the screen. There will be bright lines directed LOward the bright areas on the screen and clark lines directed LOward the dark areas on the screen. This is because Equations 24.5 and 24.6 depend on the distance to the screen, L, ,...-hich can take any value.
TELEVISION SIGNAL INTERFERENCE
Suppose you are watching television by means of an antenna rat.her than a cable system. If an airplane flies near your location, you may notice wavering ghost images in the tele,·ision picture. \r\'hat might cause this phenomenon? Explanation Your television antenna receives twO signals: the direct signal from the transmil.ling
antenna and a signal reOected frolll the surface of the airplane. As the airplane changes pOSition, there are some times v.{hen these t\\'o signals are in phase and other times when they are out of phase. As a result, the intensity of the combined signal received at your antenna ""ill vary. The wavering of the ghost images of the picture is evidence of t.his variation.
QUICK QUIZ 24.1 In a two-slit interference pattern projected on a screen, are the fringes equally spaced on the screen (a) everywhere, (b) only for large angles, or (c) only lor small angles? QUICK QUIZ 24.2 H· the distance between the slits is doubled in Young's experiment, '.vhat happens to the width of the central maximum? (a) The width is doubled. (b) The width is unchanged. (c) The width is halved. QUICK QUIZ 24.3 A Young's double-slit experiment is performed with three different colors of light: red, green, and blue. Rank the colors b)' the distance between acljacent bright fringes, from smallest to largest. (a) red, green, blue (b) green, blue, red (c) blue, green, red
EXAMPLE 24.1 Goal
Measuring the Wavelen th af a Li ht Saur.:.c;;,e
_
Show how Young's experiment can be lIsed to measure the wavelength of coherent light.
Problem A screen is separated from a double-slit source by 1.20 m. The distance between the two slits is 0.030 0 mm. The second-order bright fringe (m = 2) is measured to be 4.50 cm [rom the centerline. Determine (a) thc wa\-clength of the light and (b) the distance between adjacent bright fdnges. Strategy Equation 24.5 reJates the positions of the bright fringes to the other variables, including the wa,·elength of the Light. Substitute into this equation and solve for A. Taking the difference between )',"_1 and )'", results in a general expression for the distance between bright fringes. r-. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .... . ..........
Solution (a) Determine the wavelength of the light. Solve Equation 24.5 for the wavelength and substitute the values ?n = 2')'2 = 4.50 X 10-2 m, L = ] .20 m, and d = 3.00 X 10-" 111:
d = -'-(4.::.=-50::....c..X.::l.:..:O.::-,..,'.::m.::)-':'(3:c..O"'O'--X--=I--=O_-'--=m=) A = _fl_ /ilL 2(1.20 m) = 5.63 X 10- 7 m =
563 nm
24.3
(h) Determine the distance between rringes.
Change of Phase Due to Reflection
795
a(~jacenl bright
Use Equation 24.5 to find the distance between any acUacent bright fringes (here, those characterized by 'til. and m. + I):
AL !'>y=y m- \-)' III =-(1Il+l) d
AL
AL d
--1n=-
d
(5.63 X 10- 7 Ill)( 1.20 Ill) = 2.25 elll 3.00 X 10 r, III
...
..............
~
Remarks This calculation depends on the angle 8 being small because the small-angle approximation ,vas implicitly used. The measurement of the position of the bright fringes yields the wavelength of light, which in turn is a sigIlature of atomic processes, as is discussed in the chapters on modern physics. This kind of measurement therefore helped open the world of the a 1.0 Ill. QUESTION 24.1 True or False: A larger slit creates a larger separation between interference fringes. EXERCISE 24.1 Suppose the same experiment is run with a different light source. If the first-order maximum is found at 1.85 from the centerline, what is the wavelength of the light?
Answer
463 n m
24.3
CHANGE OF PHASE DUE TO REFLECTION
Young's method of producing two coherent light sources involves illuminating a pair of slits with a single source. Another simple, yet ingenious, arrangement for producing an interference pattern with a single light source is known as Lloyd's mirrm: A point source of light ;s placed at. point S, close to a mirror, as illustrated in Figure 24.5. Light waves can reach the viewing point Peither by the direct path SP or by the path irl\'olving reflection from the mirror. The reflected ray can be treated as a ray originating at the source S' behind the mirror. Source 5', which is the image of S, can be considered a virtual source. At points far from the source, an interf-erence pattern due to 'waves from Sand 5;' is observed, just as for tWO real coherent sources. The positions of the dark and bright fringes. however, are revl'J"sed relative to the pattern obtained from tWO real coherent sources (Young's experiment). This is because the coherent sources Sand S' differ in phase b)' 180°, a phase change produced by reflection. To illustrate the point further, consider pi, the point where the mirror intersects the screen. This point is equidistant from Sand S'. If path difference alone were responsible for the phase difference, a bright fringe would be observed at P' (because the path dirf"erence is zero for this point), corresponding to the central fringe of the t\vo-slit interference pall.ern. Instead, we observe a dark fringe at P, from "ihich we conclude that a 180° phase change must be produced by reflection from the mirror. In general, an electromagnetic wave undergoes a phase change of 180° upon reflection from a medium that has an index of refraction higher than the one in which the wave was traveling. An analogy can be drawn between renected light waves and the reflections of a transverse wave on a stretched string when the wave meets a boundary. as in Figtire 24.6 (page 796). The reflected pulse on a string undergoes a phase change of 180° when it is reflectcd from the boundary of a denser string or from a rigid barrier and undergoes no phase change when it is rerJected frolll the boundary of a less dense string. Similarly, an electromagnetic wavc undergoes a 180° phase change \,,'hen reflected frolll the boundary of a mediulll with index of refraction higher than [he one in \vhich it has been traveling. There is no phase change "..· hen the wave is renected from a boundary leading to a medium of lower index of refraction. The transmitted 'wave that crosses the boundary also undergoes no phase change.
CIll
Viewing screen
Real source
s P' / /
Virtualc,"'''' source S'
MiJTor
FIGURE 24.5 L1oyd's mirror. All imerference pattern is produced Oll a screen at Pas a result ofcbe combinalion of the direcl ray (billel and the reflected ray (brown). The reflected ray undergoes a phase change of 180".
796
Chapter 24
Wave Optics
1800
pha~c
change
No phase change
~ Incident wave
Incident wave
-~
III
III
-
?l.2
<
Rdlcclcd wave
liZ
Rigid supporl
11 I
>
Free suppon
1/2
(h) FIGURE 24.6 (a) A ray rellct:lillg from a mediulll of higher refractive index undergoes a 180' phase change. The "ighl side shows the analogy with a renecled pulse on a slring. (b) A ray reflecting from a mediulll of lower rcf.-active index undergoes no phasc change.
180 0 phase
Ch"ng~
f
.Air
No phase
c"""Re
Film with index /I
/
Air
24.4
INTERFERENCE IN THIN FILMS
I.nterferencc effects arc commonly observed in thin Alms, such as the thin surf~lce of a soap bubble or thin layers of oil on water. The varied colors observed \vhen incoherent white light is incident on such films result from the interfercnce of""aves rerJected from the two surfaces of the film. Consider a film of uniform thickness t and index of refraction n, as in Figure 24.7. Assume the light rays traveling in air are nearly normalLO the (\'\'0 surfaces of the film. To determine ,.. . hcther the rcrJcned rays interfere constructively or destructively, we fi rst note the follm\'ing facts: 1. An elcClromagnctic ,..'ave traveling from a medium ofindex of refraction Jl 1 toward a medium of index or refraction n2 undergoes a 180 phase change on reflecLion ,<\'hen 112 > n l . Therc is no phase change in the reflectcd '',lave if n"'l < n l . 2. The wavelcngth of light ,1.1/ in a medium ,\!ith index of refraction n 15 0
FIGURE 24.7 Interferel1<:e ohservcd in light reflected from a thin film is due to a combination arrays reflected from the llppcr and 100\<:r surfaces.
A"
A
[24.7J
='II
,,:here A is the 'wavelength of light in vacuum. We apply these rules to the film of Figure 24.7. According to the first rule, ray 1, which is reflected from the upper surface A, undergoes a phase change of (80 0 with respect to the incident wave. Ray 2, which is reflectcd from the lower surface B, undergoes no phase change with respect to the incident ,,·ave. Therefore, ray l
~.-.~
b
.
'-.
"\'f4IIJ" ,.. - ,"" "
~ ~
"~ 'J l -..-,,'T. ~~-
;.'
f"
__
-
Soap bubbles Ull w,ucr. The colors are due to itltcrfcrence bClWCCIl light nl)'S rdleClcd from the from .111(1 back of the thin fdm of soap making lip the
bubble. Till: color depclld.~ Oil the thickness orlhc film, ranging from black when: L1le film is at its thinnest to mag<:llla where it is thickest.
is 180 0 out or phase with respect to ray 2, ,\'hich is equivalent to a path difference of Aj2. We must also consider, though, that ray 2 travels an extra distance of 2l before the ,,:aves recombine in the air above the surface. For example, if21 = A7/2, rays 1 and 2 recombine in phase and const.ructive interference results. In general, the condition for ronslruclive inlclference in thin lllms is 2l = (m
+ ~)AII
m = 0, 1, 2,.
[24.8]
This condition takes into account two factors: (1) the diffcrence in path length for the two rays (the term mA) and (2) the 180 phase change upon reflection (the term An /2). Because All = A/'11, we can ,vrite Equation 24.8 ;n the [orm 0
2nt = ('/II + ~)A
111
= 0, 1,2, ' , '
[24.9]
If the extra distance 2ltraveled by ray 2 is a multiple of An. the twO waves combine out of phase and the result is destructive interference. The general equation for dcslrucliv(] inle1/ermu;e in th i n fi 1illS is 2nt = mA
111
= 0,1,2,_
[24.10]
24.4
Equations 24.9 and 24.10 for constructive and destructive interference are valid when there is onJy one phase reversal. This will occur when the media above and below the thin film both have indices of refraction greater than the film or "'when both have indices of refraction less than the film. Figure 24.7 is a case in ~ point: the air (n = 1) Lhat is uoLli ,1I.JOve and below the film has an index of refraction less than thaL of the film. As a result, there is a phase reversal on reflection off the top layer of the film but not the bottom, and Equations 24.9 and 24.10 apply. If the film is placed between two different media, one of lower refractive index than the film and one of higher refractive index, Equations 24.9 and 24.10 are reversed: Equation 24.9 is used for destructive interference and Equation 24.10 for constructive interference. In this case either there is a phase change of 1800 for both ray I reflecting [rom surface A and ray 2 reflecting [rom surface B, as in Figure 24.9 of Example 24.3, or there is no phase change for either ray, v,lhich ""ould be the case if the incident ray came from underneath the film. Hence, the net change in relative phase due to the reflections is zero.
QUICK QUIZ 24.4 Suppose Young's experiment is carried oul in air, and then, in a second experiment, the apparatus is immersed in water. In \...-hat way does the distance between bright fringes change? (a) They move farther apart. (b) They move closer together. (c) There is no change.
Interference in Thin Films
797
~\
thin film of oil on I,'ater displays interference, evidenced by the patlern of colors when white light is incident on the film. Variations in the film's thickness produce the intersecting color pattern. The razol- blade gives you an idea oflhe size of the colored hands.
Newton's Rings Another method for observing interference in light waves is to place a plano-convex lens on top ora flat glass surface, as in Figure 24.8£1. ''\lith this arrangement, the air fi 1m between the glass surfaces varies in thickness from zero at the point of contact to some value l at P If the radius of curvature R of the lens is much greater than the distance l'and the system is viewed from above light ofwavelengrh A, a pattern o[ light and dark rings is observed (Fig. 24.8b). These circular fringes, discovered by Newton, are called Newton's rings. The interference is due Lo the combination of ray], reflected from the plate, lNith ray 2, reflected from the lower surface of the lens. Ray 1 undergoes a phase change of 180" on reflection because it is reflected from a boundary leading into a medium of higher refractive index, v\:hereas ray 2 undergoes no phase change because it is reflected from a medium of Im'ller refractive index. Hence, the conditions for constructive and destructive interference are given by Equations 24.9 and 24.]0, respeclively, with n = 1 because the "film" is air. The contacL point at a is dark, as seen in Figure 24.8b, because there is no path difference and the total phase change is due only to the 180 phase change upon reflection. Using the geometry shown in Figure 24.8a, \\'e can obtain expressions for the radii of the bright and dark bands in terms of the radius of curvature Rand vacuum wavelength A. For example, the dark rings have radii ofr= VmAR/n. 0
(a)
(bl
FIGURE 24.8 (a) The combination of rays reflected from the glass plate ,\lid the curved surf~lce of the Icnsgivcs rise to an interference pattern known as \lewton's rings. (b) A phOlograph of Newton's rings. (c) This asymmetric interference paT.1ern indicates imperfections in the lens.
(e)
TIP 24.2 The Two Tricks of Thin Films Be sure
to include both errenspath length and phase changewhen }'ou analyze an imerfen:nce pattern from a thin film.
798
Chapter 24
Wove Optics
APPLICATION Checking for Imperfections in Optical Lenses
One imponanlllse of Newlon's rings is in the testing of optical lenses. A circular pauern like that in Figure 24.8b is achieved on I}' when the lens is ground to a perfectly spherical curvature. Variations from such symmetry might produce a pattern like that in Figure 2~l.8c. These variations give an indication of how the lens must be reground and ,"epolished to remuve imperfeCLions.
OBLEM-SOLVING STRATEGY THIN-FILM INTERFERENCE The following steps are recommended in addressing thin-film interference problems: 1. Identify the thin film causing the interference, and the indices of refraction in the film and in the media on either side ofiL
2. Determine the number of phase reversals: zero, one, or t\\'o. 3. Consult the following table, which contains Equations 24.9 and 24.10, and select the con'eet column for the problem in question: Equation (111 - 0, 1, ...)
1 Phase Reversal
0 or 2 Phase Reversals
= (m + ~)A [24.9] 2nt = m). [24.10]
Constructive
DestruClive
Destructive
Conslnlclive
2nl,
4. Substitute values in the appropriate equations, as selected in the previous step.
EXAMPLE 24.2 Goal
Interference in a SoaR Film
Study constructive interference effects in a thin film.
Problem (a) Calculare the minimum thickness ora soap-bubble film (n = 1.33) that will result in constructive interference in the renected light if the film is illuminated by tight .."ith \"avelength 602 nm in free space. (b) Recalculate the minimum thickness for constructive interference when the soap-bubble film is on top of a glass slide with n = 1.50. Strategy [n pan (a) there is only one inversion, so the condition for constructive interference is 2nt = (111 + ~)A. The minimum film thickness for constructive interference corresponds to m = 0 in this equation. Pan (b) involves two im'ersions, so 21lt = mA is required. r' . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . • . . . . • . , , . . . . . . . . . . . . . • . . . . . . . . . , ,
..,
Solution (a) Calculate the minimum thickness of the soap-bubble film that will result in constructive interference. Soh·e 2nl = A/2 for the thickness I and substitute:
A
(= -
4"
602 11m
= .::..:;.:'-':':'~ 4(1.33)
113 11m
(b) Find the minimum soap-film thickness "..· hen the fi IIn is on top of a glass slide with n = 1.33. Write the condition for constructive interference, when two inversions take place:
Goal Study destruct.ive inlerference effects in a t.hin film when there are t.wo inversions. Problem Semiconductors such as silicon are used to fahricate solar cells, devices that generale electric energy when exposed to sunlight. Solar cells (lrc orten coated \vilh a transparent thin film, such as silicon monoxide (SiD; n = 1.45), to minimize reflective losses (Fig. 24.9). A silicon solar cell (11 = 3.50) is cOaled with a thin l-ilrn of silicon monoxide for this purpose. Assllming normal incidence, determine the minirnllm thickness of the film that will produce the least reflection at a wavelength of 552 nm.
FIGURE 24.9 (Exampk 24,3) Reflectivc lo.~ses fl·om a <;ilinlll ~olar cell arc millimin>d by CO,lIillg il \,'ith a thin film nf<;ilicon monoxide (SiD).
If}O° phase ch;11l~e
1\ " 4 Air 11=
5iQ
Si
11=
11
1.45
1RO° ph;!<;t;
i~changc
/
I
_
t
\~
I
j
1
= 3.50
Strategy Reflection is least: when rays 1 and 2 in Figure 24.9 meet the condition [or destructive interference. Note that uolh rays undergo 180 0 phase changes on reflect.ion. The condit.ion for a reflection minimum is therefore 21'11= V2.
,..
.
.
Solution
A 4'11
552 nm
I~-=:::":;-::"":':::;-
Solve 2nl = A/2 for I, the required t.hickness:
.
~."
4( 1.45)
~
95.2 nm ~
"
Remarks Typically, such coatings reduce the rdlective loss from 30% (\.. . ith no cOelting) to 10% (with a coming), thereby increasing the cell's efficiency because more ligl1l is available to create charge carriers in the cell. In reality the coating is never perfectly nOllreflcct.ing because the required thickness is wavelengt.h dependent and t.he incident 1ig·ht covers a wide range of wavelcllgt hs. QUESTION 24.3 To rninimize reflection of a smallcr wavelength, should the thickness of" t.he coat.ing be thicker or thinner? EXERCISE 24.3 Class lenses lIsed in cameras and other opt.ical instrument.s are usually coated with one or more t.ransparent thin films, such as magnesium fluoride (MgF 2 ), to reduce or eliminate ul1\\'anted reflection. Carl Zeiss developed this method; his first coat.ing was 1.00 X 10 2 nm thick, on glass. Using '11 = 1.~8 for Mgf'2' \.. . har visible waYelength would be eliminated by destructive interference in the reflected light?
Answer
552 nm
EXAMPLE 24.4
Interference in a Wedge-Shaped Film FIGURE 24.10 (Example ~/l.4) Inrerference bands ill n"flecred light call be observed by illulUillalillg a h'edge-shaped film with monochromatic li,g-hl. Thc· dark area.~ illlhe inrerferencc pallern cOITespnlld (0 positions or elcsr rucriH" inrerfen::nce.
Goal Calculat.e interference effects when the fllm has variable thickness. Problem A pair of glass slides 10.0 Clll long and with n = 1.52 are separated on one end by a hair, fortning a triangular wedge of air, as illustrat.ed in Figure 24.10. \,Vhen coherent. light from a helium-neon laser with
x
L
800
Chapter 24
Wove Optics
wavelength 63311111 is incident on the film froll,,) above, 15.0 dark fringes per centimeter are observed. How thick is the hair?
Strategy The imcrfcrcnce pattern is created by the thin film orair having variable thickness. The p3ltern is a sCI-ies ofaltcrnating bright and dark parallel bands. A
dark band corresponds to dcstl'uCli\'c illlcrference, and there is one phase reversal, so 211( = 111). should be used. We can also use the similar triangles in Figure 24.10 to obtain the relation I/x = DIL. \,Ve can Ilnd the thickIIC:S~ [UI ,"I)' m, and if the position x can also be found, this last equation gives the diameter of the hair, D.
,. ..
....
.
Solution Sake the deslruClire·imcrfercnce equation for the thickness of the film, t, with n = 1 for air: If d is the distance from one dark band LO the next. then the x-coordinate of the 1fllh band is a multiple of d:
By dimensional
anal~'sis, d
mA 2
1=-
x= md
( 15.0
bands)-I -c;;;-
mA/2 md
A 2d
isjusl the inverse of the number of bands per ccmimcter.
d
Now lise similar triangles and substilUtc all the information:
-=--=-=-
Solve for D and substitute gi\'cn values:
D= - = 2d
x
=
AI-
=
6.67 X 10
_~~ band
D
L
(633 X 1O- 9 m)(0.100 rn) 2(6.67 X 10 • rn)
= 4.75 X 10- 5 m
~
~
Remarks Some may he concerned about interference caused by light bouncing off {he top and ballom of, say. the upper glass slide. It's unlikely, however, that the thickness of the slide will be half an illleger multiple o[ the wavelength of the helium-neon laser (for some very large value of 111). In addition, in contrast to the air wedge, the thickness of the glass doesn't vary. QUESTION 24.4 If the air \\-'edge is filled Wilh water, how is the distance between dark bands affected? Explain. EXERCISE 24.4 The air wedge is replaced with water. with laser light hits the glass slides.
Answer
5.02 X 10- 4
11
= 1.33.
Find the distance between dark bands when the helium-neon
III
24.5 APPLICATION The PhY5ics of CDs ond DVDs
USING INTERFERENCE TO READ CDs AND DVDs
Compact discs (COs) and digital \'ideodiscs (DVDs) have revolutionized the computer and entcnainment industries by providing fast access; high·densit}1 storage of text. graphics, and movies; and high-quality sound recordings. The data on these videodiscs arc slored digitally as a series of zeros and ones. and these zeros and ones arc read by laser light reflected from the disc. SU'ong reflections (constructive interference) from the disc are chosen to represent zeros, and weak reflections (destructi\'e interference) represent ones. To see in more detail how thin-film intci-ference plays a crucial role in reading CDs, consider Figure 24.1 I. This figure shows a photomicrograph or several CD tracks which consist or a sequence of pits (when viewed from the LOp or label side ohhe disc) of varying leng·th formed in a reflcCling-mctal information layer. A cross-sectional view of a CD as shown in Figure 24.12 reveals that the pits appear as bUlllps LO the laser beam, which shines on the metallic layer through a clear plastic coaling ["rom below.
24.5
Using Interference to Read CDs and DVDs FIGURE 24.12
801
Cross section ora CD
showing meLallic pits of depth (and a laser beam detecdng the edge ora pit.
lranspan'nl plastic, 1/ '" 1.6
(2) (I) ~ !-/ reflected beam incident laser beam
FIGURE 24.11
A phololllicrograph uf ;l(!j,lCent 1racks 011 a cOlllpacl disc (CD). The infOrll1alioll ellcuded in these piLs and Sillool 11 area~ is read hr a laser beam.
As the disk rotates, the laser beam reflects off the sequence of bumps and lower areas into a phol.Odetector, which converts the fluctuating reflected light intensity into an electrical string of zeros and ones. To make the light fluctuations more pronounced and easier to detect, the pit depth l is made equal to one-quarter of a wavelength of the laser light in the plastic. \'Vhen the beam hits a rising or falling bUlllp edge, part of the beam reflects from the top of the bump and pan from the lower adjaccnt area, ensuring destructive interference and very low intcnsit)' \-"hen the reflected beams cornbine at the detector. Bump edges are read as ones, and flat bump tops and intervening flat plains are read as zeros. In Example 24.5 the pit depth for a standard CD, using an infrared laser of wavelength 780 nm, is calculated. DVDs lise shaner \\"avelength lasers of 635 nm, and the track separation, pit depth, a,ndminimum pit length are all srnaller. These differences allow a DVD to store about 30 times more information than a CD.
EXAMPLE 24.5 Goal
Pit Der.th in a CD
Apply interference principles to a CD.
Problem Find the pit depth in a CD that has a plastic transparent layer with index of refraction of 1.60 and is designed for use in a CD player llsing a laser with a wavelength 0[7.80 X -10 2 nm in air. Strategy (See Fig. 24.12.) Rays 1 and 2 both reneet from the metal layer. \·...hich acts like a mirror, so there is no phase difference due to reflection between those rays. There is, hmvevcr, the usual phase difference caused by the extra distance 2t traveled by ray 2. The wavelength is A/n. where n is the index ofrc[raction in the substance. . . . . .. . - .. .. .... . -
~.
~
Solution
A
Cse the appropriate condition for deslructive interference in a thin film:
2/= 2n
Solve for the thickness I and substitute:
1=-=
~
,........
A
411
.
7.80Xl0~nm,
()( 4 1.60 )
=
<}
1.22 X ]0- 11m
-............
.
~
Remarks Different CD systems have diflerenttoleranees for scratches. Anything that changes the renective properties of the disk can affect the readability of the disk. QUESTION 24.5 True or False: Given two plastics "lith different indices of refraction, the material with the larger index of refraction will have a larger pit depth. EXERCISE 24.5 Repeat the example for a laser \\lith wavclength 635 nm.
Answer
99.2 nm
802
Chapter 24
Wove Optics
FIGURE 24.13 (a) Ifligilldid nol spread out arle.. pil.~sing through lhe 1'1il~. no illtcl'ferencc would OCClir. (h) Tilt: light from lhl: IwO slits overlaps :I.' il spr<:ads out, filling the \;XpeClL'd sltadowcd regions with light alul producillg illlcrfcrcnc(~ fringes.
+rt~
Ii - +0--
-
I+:
~.
~
1+0--1
I
(a)
24.6
FIGURE 24.14 The rliffraclioll paltern that appears on a scrc(~n whcn light passes through a lIarrow H~rtical ~liL The paltern consists of a bro
balld~.
FIGURE 24.15 The diffraction pattern of a penny placerl midw
}
(b)
DIFFRACTION
Suppose a light beam is incident on t\\lO slits, as in Young's double-slit experimenl. Ir the light truly traveled in straight-line paths after passing through the slits, as in Figure 24. [3a, the '...aves wouldn't overlap and no interf"erence pattern would be seen. fllstcad, Huygens' principle requires Ihal I he ,..'aves spread out [rom the slits, as shown in Figure 24. L3b. In other words, t.he light bends from a straight-line path and enters the region that would otherwise be shadowed. This spreading out of light frolll its initial line of travel is called diffraction. In general, diffraClion occurs when waves pass through small openings, around obstacles, or by sharp edges. For example, when a single narrow slit is placed between a distant light source (or a laser beam) and a screen, the light produces a dilTracl.ion pattern like that in Figure 24.14. The pattern consists of a broad, intense central band flanked by a series of narrO'wer, less illlense secondary bands (called secondary maxima) and a series of dark bands, or minima. This phenomenon can't be explained within the frame'Yvork ofgeomctric optics, ·which says that light 1"
ACTIVE FIGURE 24.16 (a) The Fr,Lllllhnfer diffraction pallerll or;, !'ii IIg"1c !'ilit. Tlw parallel raY's 'l1"l." broug-hl illin [(-Jeus un the screen \,'ilh a convergillg lens. The pauern l"OIISisIS ofa <:('1111'<11 bright region tbrlkl·d by IllllCh weaker maxima. (Thisdr;'~l'ill~is notlo scale.) (b) A pholOgraph Or,l sjll~le-sli1 Frallnhofer difTracliol1 pallerll.
Slil
.: ~ ~g~-
Lcm
:-~--
d
J
(
Incomingwa\"l,;
Sen.-en (a)
(1))
24.7
24.7
803
Single. Slit Diffraction
SINGLE-SLIT DIFFRACTION
Until now \ve have assumed slits have negligible \vidth, acting as line sources oflighl. I n this section we determine how their nonzero widths are the basis for undersLanding the nature of the Fraunhofcr diffraction pal.l.crn produced by a single slil. We can deduce some imporlant features or this problem by examining waves corning from various portions of the slit, as shown in Figure 24.J7. According Lo Huygens' pri nciple, each portion of the slit acts as a source of waves. Hence, light from one portion of the slit can interfere with light from another portion, and the resultant intensity on the screen depends on the direction O. To analyze the diffraction pattern, it's cOIwcnienl to divide the slit into halves, as in Figure 24.]7. All the waves that originate at the slit are in phase. Consider waves I _md 3, which originate at the bottom and cel1Ler of the slit, respectively. 'Nave 1 travels farther than wave 3 by an amount equal [Q the path difference (a/2) sin 8, where a. is the width of the slit. Similarly, the path difference between waves 3 and 5 is (a/2) sin e. If this path difference is exactly halfofa wavelength (corresponding to a phase difference of .180°), the two wavcs cancel each other and destructive interference results. This is true, in fact, [or any two \vaves that. originate at points separated by half the slit width because the phase difference ben....een two such points is 180°. Therefore, ,.. 'aves from the upper half of the slit intcrfere destntctivd)' \vith ",'aves from I he lower half of" the slit when
!!2
:.ill
8
FIGURE 24.17 Diffraction of light b}' a narrow slit orwidth fl. Each portioll ofthc slit acts as a poiltl source of waves. The palh difference bel ween waves I and 3 or between wave:. 2 and 4 is equal to (a/2) sin O. (This drawing is not LO scale, and LI\(; waves are assumed to converge al :I distant poii'll.)
A
a
2"
2 sin 0 = or when
A
sin
(J = -
n
Ifwe divide the slit into four parts rather than two and use similar reasoning, we find that the screen is also dark when
2A
sin 0
a
Continuing in this way, we can divide the slit inlo six pans and show that darkness occurs on the screen when sin
3A
(J = fl
TIP 24.3 The Same, but Different Although ECJuaLions 24.2
patlern, whereas Equation 24.11 describes lhe dark rcgion:-. in a single·slil interference palLern.
Therefore, the general condit.ion for destructive interference [or a single slit of width ais
sin
A n
0dark = 11/-
11/=
±1,±2,::!:::3 •...
[24.11]
Equation 24-.11 gives the values oro for which the diffraction pattern has zero il1Lensity, where a dark fringe forms. The equation tel1s us nothing about the variat.ion in intensity along the screen, however. The general features of the intensity distribution along the screen are shown in Figure 24.18. A broad cemral bright
sin 0riark
=
2,11 a
sin OLlar\;' = A/a
-)'1
sin Odark
==
-AI n
--~.J ~)':1
sin Oclark
==
-2,l1 II
Viewing scn.:cn
FIGURE 24.18 Posilionsofthc minima for the FI';1unhofer diffraction pallcrn ora single slit of width (I. (Thi:. d."awing is not to scale.)
f- Condition for destructive interference (single slit)
804
Chapter 24
Wave Optics
fringe is flanked by much ,,,,'cakeI' bright fringes allCrllaLing with dark fringes. The various dark fringes (points of zero intensity) occur at the values of that satisfy Equation 24.1l. The points of constructive interference lie approximately halfway between the dark fringes. Note that the central bright fringe is twice as wide as the
e
weaker maxima having m> 1.
QUICK QUIZ 24.5
In a single-slit diffraction experiment, as the width of the slit is made smaller, does the width of the central maximum of the diffract.ion pattern (a) becomes smaller, (b) become larger, or (c) remain the same?
APPLYING PHYSICS 24.3
DIFFRACTION OF SOUND WAVES -
of the slit, so sound is effectively diffracted by the opening and the central maximum spreads throughout the roolll. Light \v3vclcnglhs are much smaller than the slit width, so there is virtually no diffraction for the light. You must have a direct line of sight to detect the light waves.
Ifa classroom door is open even a small amount, you can hear sounds coming from the hal1way, yet you can't see ,vhal is going on in the halhvay. f-Iow can this difference be explained? Explanation The space between the slightly open door and the wall is acting as a single slit for waves. Sound waves have wavelengths larger than the width
EXAMPLE 24.6 Gool
A Single-Slit Experiment
Find the positions ofl.he dark fringes in single-slit diffraction.
Problem Light of wavelength 5.80 X IO~ 11m is incident on a slit of width 0.300 mm. The observing screen is placed 2.00 m from the slit. find the positions of the first dark fringes and the width or t.he central bright fringe. Strategy This problem requires substitution into Equation 24.11 to find the sines or the angles of the first dark fringes. The positions can then be founc! with the tangent function because for small angles sin (J = tan fJ. The extent of the central maximum is defined by these two dark fringes.
.
.
~
Solution
5.80 X 10- 7 m
A
The first clark fringes that flank the central bright fringe correspond to III = ::::!: I in Equation 24.]]:
sin
e
tan
e~
)',
Use the triangle in Figure 24.18 to relate the position of the fringe to the t.angent [unction: Because (J is very small, \·...e can lise the approximation sin () = tan (J and then solve for )'1:
.Sill
.
e=
tan () = -L
Compute the distance between the positive and negative first-order maxima, which is the width waf the central maximum:
-=.:.=..:--'-'--'-'c--,::.:.. 0.300 X 10- 3 m
::::!:-= +
a
~
:t 1.93 X 10- 3
L
y,
), = L sin e =
(2.00 m)(:t 1.93 X 10-') = :t3.86 X lQ-3 m
w= +3.86 X lQ-:lm- (-3.86 X 1O-3 m) = 7.72 X 10-3 m
lrrrr... • • • • •
Remarks Note that this value of w is much greater than the width of the slit. As the width of the slit is increased, however, the diffraction pattern narrows, corresponding to smaller values of e. In facl., for large values of a, the maxima and minima are so closely spaced that the
only observable pattern is a large central bright area resembling the geometric image of the slit. Because the width of the geometric image increases as the slit width increases, the narrmvcst image occurs when the geometrjc and diffraction widths are equal.
24.8
The DiHroction Grating
805
QUESTION 24.6 Suppose the entire apparatus is immersed in \vater. If the same \\"a\'clength of light (in air) is incident on the slit in1mersed in water, is the resulting central maximum larger or smaller? Explain. EXERCISE 24.6 Dctermine the width of the fina-order bright fringe in the example, when the apparallls is in air.
Answer
3.86 mm
24.8
THE DIFFRACTION GRATING
The diffraClion grating. a useful device for analyL..ing light sources, consists of a large number of equally spaced parallel slits. A grating can be made by scratching pandlel lines on a glass platc with a precision machining technique. The clear panes between scnll.ches act like slits. A lypical grating contains se\'eral thousand lines pCI' cenlifneter. For example. a grating ruled with 5 000 lines/em has a slit spacing d equal to the reciprocal of that number; hence, d = (1/5000) cm = 2 XIQ-·I cm. Figure 24.19 is a schematic diagram ofa section ofa plane diffraction grating. A plane wa\·e is incident from the left, normal to the plane of the grating. The intensity of the paltern on the screen is the result of the combined effects of interference and diffraction. Each slit causes diffraction, and the diffracted beams in llIrn interfere with one anal her 10 producc the pauern. Morem'er, each slit acts as a sourcc of waves, and all \,,';wes stan in phase at the slits. For some arbitrary dircction (J measured from the horizontal, howe\·cr, the waves must nayel dijJl'r('nt path lengths before reaching a particular point Pon the screen. In Figurc 24.19, not.e that. t.he path differencc belw~en waves frorn any t\\lO adjaccnt slits is d sin O. [rlhis path difference equals one \,...avclcngth or some integral multiple ora ,"'ave· length, waves from all slits will be in phase al. P and a bright linc \...·ill be observcd at that point. Therefore, the condition lor maxima in the interference pal.tern at t he angle (} is lit
= 0, ±l, ±2,.
[24.121
+-
Condition for maximo in the interference pattern of a diffraction grating
Incomin~
plane wavc of light
p
Fir~t-ordcr
maximum
~
....
•
......
9
p-
~
~
(m= I)
C'.enl.rd.1 (or 7croth-orrkT) ma."imum
Ie
~
....
......... ~,
(m:: 0)
~
First-order maximum
p
(m::-I)
DiITrauion gr.lting
.,-
....
... \
FIGURE 24.19 A ~ide \icll' ora djfIT,lctiol1 gr:uing. The slit scparation is fl. and the path difference between adj,ICt'lll slil~;,!, dsin 8.
806
Chapter 24
-~
'"
-1
Wove Optics
0
2 CollimalOr
o sin
,\
2'\
d
d
e -----..
ACTIVE FIGURE 24.20 Imcnsil)' vcrsus sill e for the dilTrac~ lion gratillg. The zerOlh-, firsl-, amI second-order princip
(}
"1 ~~gt=~
""'·1 /
I'
Grating
ACTIVE FIGURE 24.21 A diagram of
shown.
Light emerging from a slit at an angle other than thal for a maximum interferes nearly completely destructively with light [1'0111 some other slit on the grating. All such pairs will result in little or no transmission in that direction, as illustrated in AClive Figure 24.20. Equation 24.12 can be used to calculate the wavelength from the grating spacing and the angle of deviation, O. The integer 111 is the order number of the diffraction paLLern. If the incident radiation contains several wavelengths, each wavelength deviat.es through a specific angle, which can be found from Equation 24.12. All \vavelengths are {ocused at 0 = 0, corresponding to m = 0. This point is called the u-roJ.h-urder '/flClxi'llt'll:m. The first-urrln maxim.um, corresponding (Q tn = I, is observed at an angle Lhat satisfies t.he relationship sin (j = Aid; the second-order maximum, corresponding to 1Jl = 2, is observed at a larger angle e, and so all. Active Figure 24.20 is a sketch of the intensity distribution for some of the orders produced by a diffraction grating. Note the sharpness of the principal maxima and the broad range of the dark areas, a pattern in direct cont.rast to the broad bright fringes charaetcrisLic of Lhe tW'o-slil interference patLern. A simple arrangement that can be used to measure the angles in a diffraction panern is shown ill Active Figure 24.21. This setup is a form of a c1iffractiollgrating spectrometer. The light to be analyzed passes through a slit and is formed into a parallel beam by a lens. The light then strikes the grating at a 90° angle. The diffracted light leaves the grating at angles that satisfy Equation 24.12. A telescope is lIsed to view the image of the slit. The wavelength can be determined by measuring the angles at which the images of the slit appear for the various orders_
QUICK QUIZ 24.6 If laser I19h1 is reflec[ed rrom a phonograph record or a compact disc, a diffraction pattern appears. The pattern arises because both devices contai n parallel tracks of in formation that act as a reflection diffraction grating. "Vhich device. record or compact disc, results in diffraction maxima Lhat are farther apart?
APPLYING PHYSICS 24.4
PRISM VS. GRATING
'·\'hen white light enters through an opening in an opaque box and exits throug'h all opening on the other side of the box, a spectrum of colors appears on the \"'£111. From t.his observation, how \,,'ould yOll be able to determine whether the box contains a prism or a diffraction grating? Explanation The determination could be made by nOlicing the order of the colors in the spectrum rela-
rive to the direction oft.he original beam orwhite light. For a prism, in which the separation of light is a result or dispersion, the violellight \,... ill be refracted more t.han the red light. I-Ience, the order of t.he spect.rum from a prism will be from red, closest to the original direction, to violet. For a diffraction grating, the angle of diffraction increases with wavelength, so the spectrum from the diffraction grating will have
24.8
colors in the order from violet, closest to the original direction, to reel. Furthermore. the difTraction grating \vill produce two first-order spectra Oil either side
APPLYING PHYSICS 24.5
The Diffraction Grating
of the grating, whereas the prism \,,:ill produce only a single spectrum.
RAINBOWS FROM A COMPACT DISC
\Vhite liglll n.:f1eCLcd from the surface ora CD has a multicolored appearance, as shown in Figure 24.22. The observation depends on the orientation of the
FIGURE 24.22 (.·\pp1ring Physi<-~ 2,1.5) ComraCl disc:,> act a .. din raction gratiu).;"<; \,·hell ob:.-cn·ed Ulld(·,·whilt.: liglll.
disc relative to the eye and the position of thc light sourcc_ Explain how all this \\lorks. Explanation The surface or a CD has a spiralshaped track (with a spacing of approximately 1 JLIl1) that acts as a refleClion grating. The light scattered by these closely spaced parallel tracks interferes con~ structivcly in cenain directions that depend on both the wavelength and the direction of the incident light. Anyone section of the disc sen'es as a diffraction grating for \\lhite light, scnding beams of constructive interference for differcnt colors in different dircctions. The different colors yOll see when viewingone scction of the disc change as the light source, the disc. or you move to changc the angles of incidence or diffraction.
Use of a Diffraction Grating in CD Tracking
APPLICATION
If a CD player is 1.0 reprodllce sound faithfully, the laser beam must follmv the spired track of"infnl"mation perfectly. Sometimcs the laser beam call drift orftrack, however, and without a feedback procedure tu let the player know that is happening, the fidelity of the music can be greatly reduced. Figure 2-1.23 shows hm..· a diffraction gr
Tracking Information on a CD
Compact disc:
"'"
-
FIGURE 24.23 Tllt· !
.- Diffranioll
graling
I.aser
807
808
Chapter 24
EXAMPLE 24.7 Goal
Wove Optics
A Diffraction Grating
Calculate different-order principal maxima for a diffraction grating.
Problem :YIonochromatic light from a helium-neon laser (A = 632.8 nm) is incident normallv on a diffraction grating containing 6.00 X 103 lines/clll. Find the angles at which one would observe the first-order maximum, the second-oreler maximum, and so forth. Strategy Find the slit separation by inverting the number of lines per centimeter, then subsLitulc values imo Equation 24.12.
,..
.
..,
.
Solution Invcn the number of lines per centimeter to obtain the slit separation:
1.67 X In-"em = 1.67 X 10'nm
A d
Substitmc m = I inlO Equation 24.12 to find the sine of the angle corresponding to the first-order
632.8 nm
~==~i'-= 0.379 3
.67 X 10 11rn
maXIlllum:
Take the inverse sine lind 0,:
or the preceding result to
()I
= sin- 1 0.379 =
22.3 0
2A 2(632.8 nm) sin 0, = - = -"-----.,---'= 0.758 d 1.67X103 nlll
Repeat the calculation for m = 2:
O2 ~ 49.3°
3A d
si nO, = -
Repeat the calculation for m = 3:
.
3(632.8 nm) 1.67 X IO'nm
= --=-'-'-'"---"---='-
~
1.14
Because sin 8 can't exceed 1, there is no solution for 83 , ,
~
.
.............................
,
~
Remarks The foregoing calculation shmvs that t.here can only be a finite number of principal maxima. In this case only zeroth-, nrsl-, and second-order maxima \voulcl be observed. QUESTION 24.7 Does a diffraction graling wilh more lines have a smaller or larger separation ben-veen adjacenl principal maxima? EXERCISE 24.7 Suppose light wit.h wavelength 7.80 X 102 nm is used instead and the diffraction grating has 3.30 X 103 lines per centimeter, find the angles orall the principal maxima.
Answers
0', J4.9', 31.0', 50.6'
24.9
POLARIZATION OF LIGHT WAVES
I n Chapter 21 we described the transverse nalUl'e of eleClromagnetic waves. Figure
....c x
FIGU RE 24.24 A schemal ic diagram of a polarilcd elec! romagnclic wal'e propagating in the ~dircC1ioll, The eleClric field \'eClOr E vibr,ltl'S in the X)'-p1al'!St ,,'hereas Ihe magnellc fidd vector B vibrates ill the x:-phllle.
24.24 shmvs thm the elect.ric and magnetic field vectors associated with an electromagnetic wave are at right angles to each other and also to the direction of wave propagation. The phenomenon of polarization, described in this section, is firm evidence of the transverse nature of electromagnetic waves. An ordinary bearn of light. consists of a large number of electromagnelic \vaves emitted by the atoms or molecules of the light source. The "ibr~uing charges associated with the moms act as tin)' antennas. Each atom produces a wave \vith its own orientation of E, as in Figure 24,24, corresponding to the direction of atomic vibration. Because all directions of vibration are possible, however, the resultant electromagnet.ic wave is a superposition of "laves produced by the individual
24.9
atomic sources. The result is an unpolarized light wave, represenled schematically in Figure 24.25£1. The direction of wave propagation shown in the figure is perpendicular to the page. Note that aU directions of the electric field vector are equally probable and lie in a plane (such as the plane of this page) perpendicular to the direction of propagation. --7 A wave is said to be linearly polarized if the resultant electric field E vibrates in the same direction at aU timesata particular point, as in Figure 24.25b. (Somcti mes such a wave is descri bcd as plane polarized or si mpl), IJolruized.) The wave in Figure 24.24 is an example of a \\lave thatls linearly polarized in the y-direcl.ion. As the wavt::..Yropagates in the x-direClion, E is always in the y-direction. The plane formed by E and the direction of propagation is called the IJlane of jJola.rization of the wave. In Figure 24.24 the plane of polarization is the xy-plane. It's possible to obtain a linearly polarized beam from an unpolarized beam by removing all waves from the beam except those \\Iilh electric field vectors that oscillate in a single plane. \Ve now discuss three processes for doing this: (1) seleClive absorption, (2) reflection, and (3) scattering.
Polarization of Light Waves
809
, (a)
(b)
FIGURE 24.25 (a) An lInpo!ariLcd lighl bealli viewcd along the direction oFpropagatioll (perpendicular tu the page). The transverse electric ridd vector Gin vibrate in any dirccl ion with equal prubability. (b) :-\ lincarl>' polarized light beam with Ihe eleclric field vcclor \'ibraling in the vertical directioll.
Polarization by Selective Absorption The most COlllmon technique for polarizing light is w use a material that transmits waves having electric field vectors that vibrate in a plane parallel to a certain direction and absorbs those waves v.'ith electric ricld vectors vibrating in directions perpendicular to that direction. In 1932 E. H. Land discovered a material, \vhich he called Polaroid, that polarizes light through selective absorption by oriented molecules. This material is fabricated in thin sheets of long-chain hydrocarbons, 'which are stretched during manufacture so that the molecules align. After a sheet is dipped into a solution containing iodine, the molecules become good electrical conductors. Conduction takes place primarily along the hydrocarbon chains, however, because the valence elecll'ons of the molecules can move easily only along those chains. (Recall that valence electrons are "free" electrons that can move easily through the conductor.) As a result, the molecules readily absorb light having an electric field vector parallel to their lengths and transmit light v·:ith an electric field vector perpendicular to their lengths. It's common to refer to the direction perpendicular to the molecuR lar chains as the transmission axis. [n an ideal polarizer.1ll light with E parallel 1.0 the transmission axis is transmitted and all light with E perpendicular to the transmission axis is absorbed. Polarizing material reduccs the intensity of light passing through it. In Active Figure 24.26 an unpolarized light beam is incident on the first polarizing sheet, called the polarizer; the transmission axis is as indicated. The light that passes th~pugh this sheet is polarized vertically, and the transmitted electric field vector is Eo. A second polarizing sheet, called the analyzer, intercepts this beam \'\'ith it.s .!:l'ansmission axis at an angle of 0 to the axis of the polarizer. The component of Eo that is perP.sndicular to the axis of the analyzer is completely absorbed. The component 01" En that is parallel to t.he analyzer axis, Eo cos 0, is allowed to pass through the analyzer. Because t.he intensity of the transmitted beam varies as the ~
ACTIVE FIGURE 24.26 Two polari/.ing sheets whose lrallsmis-
sion ax(:s make an angle 8 with each other. Oldy a Fraction of lhe pulaI" ized lighl illCidcllt 011 the analyzer is transmiu.cd.
810
Chapter 24
Wove Optics
FIGURE 24.27 The llllensity ()f light transmitted through two polari"crs depends
011
the relative orientations
oflhcir Irall~1Tlissioll axes. (a) The transmitted light has ma.till/llm intensity when the transmission axes arc aligned with each olher. (b) The Iransmilled light intensity diminishes when the transmission axes ,ire al ,1Il allgle or45" with each other. (c) Thl: lransmilted light intensit)' is a milli-
(h)
(a)
I')
mum when the transmission axes an:
at ri1{hI angles to each other.
square of it.s amplitude E, we conclude that the intensity of the (polarized) beam transmitted through the analyzer varies (IS Malus's law
-+
[24.13] where 10 is the intensity of the polarized wave incident on the analyzer. This expression, known as Malus's Jaw, applies LO any two polarizing ITImcrials having transmission axes at an angle of 8 LO each other. Note [rom Equation 24.13 that the transmitted intensity is a maximum when the transmission axes are parallel (0 = 0 or 180°) and is a minimum (cornplete absorption by the analyzer) when the transmission axes are perpendicular to each other. This variation in transmitted intensity through a pair of polarizing sheets is illustrated in Figure 24.27. "\Then unpolarized lighl of intensity 10 is sent through a single ideal polarizer, the transmitted linearly polarized light has intensity 10 /2. This fact follows from Malus's law because the average value of cos 2 0 is one-half.
APPLYING PHYSICS 24.6
POLARIZING MICROWAVES
A polarizer for microwaves call be made as a grid of parallel metal wires about 1 cm apan. Is the electric field vector for microwaves transmitted through this polarizer parallel or perpendicular to the mete'll wires?
Explanation Electric field veCLOrs parallel to the metal wires cause electrons in the metal to oscillate parallel to the wires. Thus, the energy [rom the waves
EXAMPLE 24.8 Goal
with these electric field vectors is transferred to the metal by accelerat.ing the elect.rons and is eventually tre'll1sformed t.o internal energy through the resistance o[ the metal. \'''aves with electric field vectors perpendicular 1.0 the metal wires arc not able to accelerate eleCl.rons and pass through the wires. Consequently, the electric field polarization is perpendicular to the mctal wires.
Polorizer
Understand how polarizing materials affect light intensity.
Problem Unpolarizecl light is incident upon three polarizers. The first polarizer has a verticallransmission axis, the second has a transmission axis rotated 30.0° with respect to the first, and the third has a transmission axis rotated 75.0° relative to the f-irs1.. IT the initial light intensity of the beam is ' M calculate the light intensity after the beam passes through (a) the second polarizer and (b) the third polarizer. Strategy After the beam passes through the first polarizer, it is polarized and its intensity is cut in hall'. Malus's law can then be applied to the second and third polarizers. The angle llsed in Malus's law must be relaLive to the imme~ diatel)' preceding transmission axis. ~
.
.
..
Solution (a) Calculate the intensity of the beam after it passes through the second polarizer. The incident intensity is Ib /2. Apply Malus's lavv to the second polarizer:
(v'3)' 222
h = -I, cos' (30.0°) = -I -
3
-[
8 '
24.9
Polarization of Light Waves
(b) Calculate the intensity of the beam after it passes through the third polarizer.
, I, = I,cos-" e = 3 ,c05-(45.00) = 3 1b
8/
The incident intensity is now 3tb /S. Apply Malus's law 10 the third polarizer.
...............................
,
,
.,
,.......
.
..
(v'2)"
8 2
811
3 16 '
-I
~
,
Remarks Notice that the angle used in part (b) was not 75.0°, but 75.0° - 30.0° = 45.0°. The angle is always with respect LO the previous polarizer's transmission axis because the polarizing material physically determines what direction the transmitted electric fields can have. QUESTION 24.8 At what angle relative to the previous polarizer must an additional polarizer be placed so as light?
LO
completely block the
EXERCISE 24.8 The polarizers are rotated so that t.he second polarizer has a transmission axis of 40.0° with respect to the first polarizer and the third polarizer has an angle of90.0° with respect to the first. If lb is the intensity of the original un polarized light, what is the intensity of the beam after it passes through (a) the second polarizer and (b) the third polarizer? (c) What is the (-inal transmitted intensity if"the second polarizer is removed?
Answers
(a) 0.293/b
(b) 0.J21/,
(c) 0
Polarization by Reflection \'Vhen an unpolarized light beam is reflected from a surface, the reflected light is completely polarized, partiallr polarized, or unpolarized, depending on the angle of incidence. If the angle or incidence is either 0 0 or 90 0 (a normal or grazing angle), the reflected beam is unpolarized. For angles of incidence between 00 and 90°, hmvever, the reflected light is polarized 1.0 some extent. For one particular angle of incidence the reflected beam is completely polarized. Suppose an un polarized light beam is incident on a surface, as in Figure 24.28a. The beam can be described by t\.. .'o electric field components, one parallel to the surrace (represented by dots) and the other perpendicular to the first component and to the direction of propagation (represented by brown arrows). It is found that the parallel component reflects more strongly than the other components, and the result is a pa,rtially polarized beam. In addition, the refracted beam is also partially polarized. Nm.. .' suppose the angle of incidence, 01' is varied until t he angle between the reflected and refracted beams is 90° (Fig. 24.2Sb). At. this particular angle of incidence, called the polarizing angle Op. the rerIected beam is completely polarized, with its electric field vector parallel to the surface, while the refracted beam is partially polarized. An expression relating the polarizing angle LO the index of refraction of" the reflecting surface can be obtained by the use of" Figure 24.2Sb. From this figure.
Incidellt beam
Rerlccted beam
,,
;
•
OJ
----
OJ
,, • • nJ
Incident btam
"'-...' ,
RerJeCled be
Of'
'.
'" Refracted htam (aj
Rcfractl:d beam (b)
FIGURE 24.28 (a) When tlllpol'ing the equation 11 '=' tan Of"
812
Chapter 24
Wave Optics
we sec that. al the polarizing angle, Op + 90 0 + 0'2 = 180 0 , so 02 = 90 Snell's law and laking 11, = nair = 1.00 and 112 = 11 yields sin OJ
sin
sin 8 2
sin (Jz
0 -
0p. Using
(Jp
1/=--=--
Because sin Brewster's low
(J'2 =
sin (90
Polaroid Sunglasses
-
0) = cos 0P' the expression ror sin (J p n = - - - = tan 0 cos Op P
-7
APPLICATION
0
11
can be wriuen [24.14]
Equation 24.14 is called Brewster's law, and the polarizing angle Op is sometimes called Brewster's angle ancr its discoverer, Sir David Brewster (178]-1868). For example, Brewster's angle for crown glass (where 11. = 1.52) has the value 81> = tan- I (1.52) = 56.7°. Because nvaries \-vith wavelength for a given substance, Brewster's angle is also a function ofwavclengl.h. Polarization by refleClion is a common phenomenon. Sunlight reflected from watcr, glass, or sIlO''''' is partially polarized. If the surface is horizol1lal, the electric field vector of the reflected light has a strong horizol1Lal component. Sunglasses made of polarizing material reduce the glare, which is the renected light. The transmission axes of the lenses are oriented vertically to ahsorb the strong horizontal component of the reflected light. Because the reflected light is mostly polarized, lllost of the glare can be eliminated ,...irhout removing most of the normal light.
Polarization by Scattering
Unpolarizcd
light
,
-
•,
~
Air molecule
FIGURE 24.29 The scaucring or unpolarilcd sunlight b~ air ll1ol~ eculcs. The 1i~hl ob~ef\'cd ill right angles is linear!} polarilcd because the vibt
When light. is incident on a system of panicles, such as a gas, the electrons in the !l1t.:dium can absorb and reradiate part of the light. The absorption and reradiation of light by the medium, called scattering, is what causes sunlight reaching an observer on Eart.h from straight overhead to be polarized. You can observe this effect by looking directly up through a pair of' sunglasses made of polarizing glass. Less light passes through at certain orientations of the lenses than at others. Figure 24.29 illustrates how the sunlight becomes polarized. The left side of the figure shows an incident unpolarized beam of sunlight on the verge of striking an air molecule. 'When the beam strikes the air molecule, it sets the elcctrons of the molccule into vibration. These \"ibrating charges act like those in an anLenna except that they vibrate in a complicated pattern. The horizontal part of the elecu'ic field ,·ector in the incident wave causes the charges to vibrate horizontally, and the vertical part of the \"ector simultaneously causes them to vibrate vertically. A horizontall}' polarized wave is emitted b)' the electrons as a result of their horizontal motion. and a \"crtically polarized ''''avc is emitted parallel to Earth as a result of their ,·cnical motion. Scientists have found that bees and homing pigeons use the polarization of sunlight as a navigational aid .
Optical Activity Many imponant practical applications of polarized light involve the lise of certain matcrials that display the property of optical activity. A substance is said to be optically active if it rotales the plane of polarization of transmilled light. Suppose unpolarized light is incident on a polarizer from the left, as in figure 24.30a. The transmittcd light is polarized vertically, as shown. If this light is then incident on an analyzer with its axis perpendicular to that of the polarizer, no light emerges from it. If an optically active material is placed between the polarizer and analyzer, as in Figure 24.30b, the material causes the direction of the polarized beam to rotate through the angle 8. As a result, some light is able to pass through the analyzer. The angle th rough which the light is rotated by the material can be found by rotating the polarizer until the light is again extinguished. IL is found that the angle of rotation depends on the lengLh of the sample and, if the substance is in solution,
24.9
.
Unpolal-izl:d li~hl
...
~
~
(a)
813
FIGURE 24.30 (a) When crossed polarizers arc used, none or Ihe polarind light can pass through the analyzer. (b) All optically acti\'c material rotatcs Ihe direction ofpolariz
A.nalFer Polarizer
Polarization of Light Woves
Polan/l:
i ~ Optical I>, activc material
... ... (b)
...
, , ..
on the concentration. One optically active material is a solution of common sugar, dextrose. A standard method for determining the concelllration of a sugar solution is to measure t.he rotation produced by a fixed lengt.h of the solution. Optical activity occurs in a material because of an asymmetry in the shape of its constituent molecules. [tor examplc, some proteins arc opt.ically active because or their spiral shapes. Other materials, such as glass and plastic, become optically act.ive when placed under stress. Jf polarized light is passed t.hrough an unstressed piece of plastic and then through an analyzer with an axis perpendicular to that of the polarizer, none of the polarizcd light is transmitted. If the plastic is placed under stress, hmvever, the regions of greatest stress produce the largest angles of rotation or polarized light, and a series of light and dark bands are observed in the transmiucd light. Engineers oftcn use this property in the design of structures ranging from bridges to small tools. A plastic model is built and analyzcd under dilTerent load conditions to determine positions ofpotel1tial \,,·eakness and failure under stress. If the design is poor, p(llterns of light and dark bands will indicate the points of' greatest weakness, and I he design can be corrccted at an early stage. Figure 24.31 shO\vs examples of stress patterns in plastic.
APPLICATION Finding the Concentrations of Solutions by Means of Their Opticol Activity
Liquid Crystals
APPLICATION
An crfectsimilar LO rotation of the plane of polarization is used t.o create the familiar displays on pocket calculators, wristwatches, notebook computers, and so forth. The properties of a uniquc substance called a liquid crystal make these displays (called LeOs, for liquid C'r'ys(f.ll displays) possible. As its namc irnplies, a liquid crystal is a substance \.\'ith properties intcrrnediate between those ofa crystalline solid and those ora liquid; lhat is, the molecules of the substance are more orderly than lhose in a liquid, but less orderly th'.11l those in a pure crystalline solid. The forces
Liquid Crystal Displays (LCDs)
(b)
FIGURE 24.31 (a) Straill dislribution in a pbstic model of a replacement hip lIscd in a medical resl:arch laboralor)·. Tilc pattern is produced when the Illude I is placed hClwl:cll a polarizer and Hll analyzl:r oriclll.ed perpendicular 10 each other. (I» A plastic model oran arch structure under load condiliolls obscncd belwcell pcq)cltdicular polari~.ers. Such pallcrlls are lIseful ill lhe optimum design of archi ICC lura I componclllS.
814
Chapter 24
Wave Optics
that hold the molecules together in such a state are jusl barely strong enough to enable the substance to maintain a definite shape, so it is reasonable to call it a solid. Small jnputs of mechanical or electrical energy, hmvcvcr, can disrupt these weak bonds and make the substance flow, rotate, or lwisl. To see how liquid crystals can be used to create a display, consider Figure
24.32a. The liquid crystal is placed between two glass plates in the pattern shown, and electrical contacts, indicated by the thin lines, are made. "Vhen a voltage is applied across any segment in the display, that segment turns dark. In this fashion any number between 0 and 9 can be formed by the pal.Lern, depending on the voltages applied to the seven segments. To see why a segment can be changed from dark to light by the application of a voltage, consider Figure 24.32b, which shows the basic construction of a portion of the display. The liquid crystal is placed between h\lo glass substrates thal are packaged between two pieces of Polaroid material with their transmission axes perpendicular. A reflecting surface is placed behind one of the pieces of Polaroid. First consider what happens when light falls on this package and no voltages are applied LO the liquid crystal, as shown in Figure 24.32b. Incoming light is polarized by the polarizer on the left and then falls on the liquid crystal. As the light passes through the crystal, its plane of polarization is rotated by 90°, allowing it to pass through the polarizer on the right. It reflects from the reflecting surface and retraces its path through the crystal. Thus, an observer to the left of the crystal sees the segment as being bright. When a voltage is applied as in Figure 24.32c, the molecules of the liquid crystal don't rotaLe the plane of polarization of the light. In this case the light is absorbed by the polarizer on the right and none is reflected back to the observer to the left of the crystal. As a result, the observer sees this segment as black. Changing the applied voltage to the crystal in a precise pattern at precise times can make the pattern tick off the seconds on a watch, display a letter on a computer display, and so forth.
Jbdt
FIGURE 24.32 (a) The lighlsegment pattern of a liquid crystal display. (b) Rotation ofa polarized light beam by a liquid Cll'stal when the applied voltage is zero. (c) .\101eculcs of the liquid crystal align II'ith the electric field when a milage is applied.
P
(
*
Liquid crystal
Incoming liglH
Pol,u"izer
(b)
!lV= 0 \'Vithout \'oltage
t ! "'=:r.==?Y Wilh voltage (e)
Polarizer
Reflecting surface
106 !
Light absorbed here
815
Summary
SUMMARY 24.1
Conditions for Interference
Interference occurs when two or more 1ig-ln \'laves overlap at a ~i\'ell point. A sustained interference paucrn is observed if (1) the sources (lTC coherent (that is, Ihey maintain a canst
24.2
Young's Double-Slit Experiment
In Young's double-slit experiment two slits separated by dislance dare illllll1inatcd by a single-wavelength light source. An interference pallCrn consisting of bright and dark fringes is observed on a screen a diswnce L from I he slits. The condit ion for bright fringes (constructive interference) is d sin
0hrigln
= '1II.A
11/,=
0, ±l, ±2,.
+ ~)A
111
= 0, ± I, ::::!:::2,.
[24.3]
The position J", of the bright fringes on Lhe screen can be determined by lIsing the relation sin 6 = tan 0 = )'"/1,, which is lrue for small angles. This rel
Diffraction
24.7
Single-Slit Diffroction
Diffraction occurs when \'iaves pass through small openings, around obswc1es, or by sharp edges. The diffraction pattern produced by a single slit on a distant screen consists of a central bright maximum flanked by less bright fringes alternating "'llh dark regions. The angles (J at which the diffraction pattern has tero intensity (regions of destruclive interference) arc described by sin Delar \.. = •
24.4
Interference in Thin Films
An dectrolllagncric wave undcrgoes
A" = -
[24.7]
"
where A is the wavelength of Lhe light in frce space. Light cncountering a thin film of thickness l wil1 rdleCl off the top and bOllom of the rilm, each ray undergoing a possible phasc change as described above. The two rays recombine, and bright and dark fringes will be observed, with the COllditions ofintcrfercnce givcn b}' the following table: Equation (m;;;;; 0, 1, ...) '211/ = (m '2u/ = mA
+
~),\
[24.9] [24.IOJ
= ± I, ±2, ::::!:::3, ..
[24.11]
24.8
The Diffraction Groting
1\ diffraction grating consist.s of many equally spaced, idcntical slits. The condition [or maximum intensity in the interferelH:C pauern ora diffraction grating is d sin
0bright = 11/A
m= 0, .:!:I, ±2,
[24.12]
wherc d is the spacing between adjacent slits and m is the order number of the diffraction patt.ern. A diffraction grating can be made by putLing a large llumber of cvenly spaced scratches on a glass slide. The number of such lines per centimeter is the inverse of the spacing d.
[24.5]
A similar expression can be derived for the dark fringes. This equation can be used either to localc the maxima or 10 determine the \vavelength oflighL by measuring y",.
Change of Phose Due to Reflection
111
where A is the wavelength of the light incident on the slit find a is the width of the slit.
24.9
24.3
A '1!1.fl
[24.2]
The number III is called the order number oUhe fringe. The condition for dark fringes (deslrucLivc interferencc) is dsin O,I.,rl.: = (fit
24.6
I Phase Reversal
o or 2 Phase Reversals
Constructive Dest'·llctivc
Desll"llclivc Consl fllctive
Polorization of Light Woves
Unpolarized lig-Ill can be polarized by selecLive absorption. reflection, or scauering. A material can polarize light if it transmits waves having electric field vectors that vibrate in a plane pllrallel1.o a certain direction and absorbs waves with electric field vectors vibrating in directions pel"pcndicular to that direction. 'When unpolarized light passes through a polarizing sheet, its intensity is reduced by half and Lhe light becomes polarized. When this light passes through a second polarizing sheet with transmission axis at ('tn angle of' 0 with respect to the transmission axis of the first sheet, the transmitted intensity is given by 1 = 10 cos 2 8
[24.13]
\",,'herc 10 is tile illl.ensity or the light after passing- through the first polarizing sheet. In general, light reflected from an amorphous material, such as glass, is partially polarized. Reflected light is completely polarized, with its electric field parallel 10 the sllrface, v.'hen the angle of incidence produces a 90 angle between lhe reflected and refracted beams. This angle of incidence, called the polarizing angle 6", satisfies Brewster's law, given by 0
n=wn~
~4.~
\-"here n is the indcx of refraction of the reflecting medium.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
816
Chapter 24
Wave Optics
~MULTIPLE-CHOICE QUESTIONS ~.=.;j!~'---_T ~ __ ~
1. A monochromatic light beam having a wavelength of 5.0 X 10 2 11m illuminates a double slit having a slit ;:)cpal aliull uf 2.0 X 10-:' til. \Vltal i~ the angle uf tlie second-order bright fringe? (a) 0.050 rad (b) 0.025 rad (e) 0.10 rad (d) 0.25 rad (e) 0.010 rad
2. A thin layer of oil (11 = 1.25) is floating on water (n = 1.33). Whm is the minimum nonzero thickness of the oil in the region that strongly reflects green light (A = 530 11m)? (a) 500 11 III (b) 313 11m (e) 404 11m (d) 212 11m (e) 285 11m 3. A Fraunhofer diffraction pattern is produced on a screen located 1.0 m from a single slit.. If a light source o[ \vavelength 5.0 X 10- 7 m is used and the distance from the center of the cent.ral bright fringe to the first dark fringe is 5.0 X 10-3 Ill, what is the slit ,.. · idth? (a) 0.010 mm (b) 0.10 mm (e) 0.200 mm (d) 1.0 mm (e) 0.005 Il1Il1 4. If plane-polarized light is scm through two polarizers, the first polarizer at 45 0 to the original plane of polarization and the second polarizer at 90° to the original plane of polarization, what fraClion of the original polarized imensitygets through the last polarizer? (a) a (b) 025 (e) 0.50 (d) 0125 (e) 0.10 5. A plane monochromatic light wave is incident on a double-slit as illustrated in Figure 24.4. As the slit separmioll decreases, what happens to the separation between the interference fringes on the screen? (a) It decreases. (b) It increases. (c) It remains the same. (d) It may increase or decrease, depending on the wavelength of the light. (e) More information is required. 6. A plane monochromat.ic light wave is incident on a double-slit as illustrated in Figure 24.4. If the viewing screen is moved away from the double slit, what hap-
pens to the separation between the interference fringes on dle screen? (a) It increases. (b) It decreases. (c) It remains the same. (d) It may increase or decrease, depending on the wavelength of the light. (e) -'1ore information is required. 7. In one experimelll light of wavelength AJ passes through a single slit and forms an interference pattern on a screen. In a second experiment light ofwavelength '\2 passes through the same slit and forms another interference pat.tern on the screen. If '\2 > AI' what is the separation between bright lines in the two experiments? (a) It's the same for both experiments. (b) It's greater in the first experimenl. (c) It's greater in the second experimcnt. (d) It's dependent on the intensity of the light. (e) ~tore informat.ion is required. 8. W'hat causes t.he brightly colored patterns sometimes seen on wet streets covered with a la)'er of oil? Choose the best answer. (a) diffraction and polarization (b) interference and deflection (c) polarization and reflection (d) refraction and diffraction (e) reflcnion and interference 9. 'What happens to light when it is reflected ofT a shiny surface on a sunny day? (a) h undergoes refraction. (b) It is wtally polarized. (c) It is unpolarized. (d) It tends to be partially polarized. (e) More information is required. 10. Consider a wave passing through a single slit. What happens 1.0 t.he width of the ccntralm3ximuIll of its diffraction pat.tern as the slit is made half as wide? (a) It becomes one-fourth as wide. (b) It becomes one-half as ",..· ide. (c) Its width does not change. (d) It becomes twice as wide. (c) It becomes four times wider.
il).CONCEPTUAL QUESTIONS ~
1. Your automobile has two headlights. What sort of interference pattern do rOll expectLO see from them? 'Nhy? 2. Holding your hand at arm's length, you can readily block direct sunlight from your eyes. Why can you not block sound from your ears this \~'ay?
3. Consider a dark fringe in an interference pattern at which almost no light energ)' is arriving. Light from both slits is arriving at this point, but the wavcs cancel. '",'here docs the energy go? 4. If Young's doublc~s]it experiment were performed under water, how would the observed interference pattern be a frected? 5.ln a laboratory accident, you spill two liquids onto \,·ater, neither of \vhich mixes with the water. They both form t.hin films on the water surface. As the films spread and become very thin, yOll not.ice that one film becomes bright and the other black in reflected light. \Vhy might. that be?
6. If white light is used in Young's double-slit experiment rather than monochromatic light, how docs t.he interference pattern change? 7. A lens with ourer radius of curvature R and index of refraClion 11 I-ests on a flat gla~s plate, and the combination is illuminat.ed from white light [rom above. Is there a dark spot or a light spot at the center of the lens? ""Vhat does it mean if the observed rings are noncircular? 8. Often, fingcrprints left on a piece of gla~s such as a windowpane show colored spectra like that from a diffraction grating. ""hy? 9. In everyday experiencc, why are radio waves polarized, whereas light is not? 10. Suppose reflected \vhite light is used [Q obsen;e a thin, t.ransparent coating on glass as the coating material is gradually deposited by evaporation in a vacuum. Describe some color changes that might occur during the process of bu ilei ing up the th ickness of the coating.
Problems
817
11. Would it be possible t.o place a nonreflecth'e coating 011 an airplane to cancel radar waves orvo.'avelengt.h 3 cm?
12. Cert.ain sunglclsses use a polarizing material to reduce the intensity orlight reflected from shiny surfaces, such as \-vat.er or the hood of a car. \'\That orientat.ion or the transmission axis should thc material ha\'c [Q be most effectivc?
13, \,yhy is it so much easier to perform int.erference experiments with a laser than with an ordinary light source? 14. A soap film is held vertically in air and is viewed in reflected light as in Figure CQ2L1.14. Explain why lhe film appears to be dark at.thc top.
FIGURE CQ24.14
PROBLEMS ~
WebAssign
I.~.:l
m
mil
II = 0=
The Problems for this chapter may be assigned online at \A/ebAssign.
= straightfol"ward,
ilHermccli,ltc.
dl
= denoles guided problem denotes enhanced content pl'oblelll biomedical ilpplic3lioll denotes ruB solution available in Stllt/I'lll So!lll;olls Maul/al!
=
Stud) Guide
SECTION 24.2 YOUNG'S DOUBLE-SLIT EXPERIMENT I. A laser beam is incident on tWO slitS with a separation
of 0.200 mm, and a screcn is placed 5.00 m from lhe slits. If the bright interference fringes on thc screen are separated by 1.58 em. \.. . hat is the w,n'clength of the laser light? 2. In a Young's double-slit experiment, a set of parallel slits with a separation of 0.100 111m is illuminated by light having a wavclenglh of 589 nm, and the interferencc pauern is observed 011 a scrcen 4.00 m from the slits. (a) What is the difference in path lengths from each of the slits to the location ora third-order bright rring'e on the screen? (b) What is the differcnce in path lengths from the (WO slitS to the location of the third dark fringe on thc screen, away from the cemcr of the patteI'll?
rnA pair of narrow, parallel slits separated by 0.250 mm is illuminated by the green component from a mercury vapor lamp (i\. = 546.1 11m). The illlederence pallern is observed on a screen 1.20 m from the plane of the panl1lei slits. Calculate the distance (a) rrom the ccntralmaxi~ Tllum to the firS\. bright rcgion on either side of the central maximuTll and (b) bet.ween the first lind sccond dark bands in the interference pattern. 4. Lighl of wavelength 5.30 X [0 2 nm illuminates a pair of slits separat.ed b)' 0.300 111m. If a screen is placed 2.00 III from the slits. detcrmine the dist(lnce between the firSt and second clark Fringes. 5. In a location where the speed of sound is 354 m/s. a 2 OOO-Hz sound wavc impinges on (\''1'0 slitS 30.0 em apan. (a) At what angle is the first maximum located? (b) If the sound wave is replaced by 3.00-em microwaves, what slit scparation gives tht same angle ()t. the /"irst maximum?
(c) If' lhe slit separation is 1.00 jlm, what frequcnc)' of light gives the same Ilrst maximum angle? 6. Two slits separated by 0.050 0 mm and located 1.50 m from a viewing screen arc illuminated with monochromatic light. The third-order bright fringe is 5.30 cm from the zeroth-order bright fringe. Find the (a) wave::length or the light and (b) separation between 3cUacent bright fringes.
7. Two radio antennas separated by 300 m. as shown in Figure P24.7, simultaneously transmit. identical signals of the same wavelength. A radio in a car tra\'e1ing due north recei\'es the signals. (a) lf the car is at the position of the second maximum, what is the wavelength of the signals? (b) How much farther must the car tra\'el to encounter the neXt minimum in reception? /-linl: Determine the path difference between the lWO signals at the two locations of the car.
r-®
:-IOom
30~ 1l~--~IOOO'~---~-!
.®
FIGURE P24.7
8. Light of v,'a\'clength 6.0 X 10 2 nm f
Chapter 24
818
Wave Optics
the centra/maximum of this pattern. (a) What is the path difference 8 for the two sliLS a1 the location .'\'? (b) Express this path difference in terms of the wavelength. (c) Will the i11lcr[erence correspond (Q a maximulll, a minimum, or an illlcnnediatc condition? 11. A riverside warehouse has two open doors, as in Fif{lIrc P2<1.I1. Its interior is lined with a sound-absOl"bing l'nalcrial. A boat on tJ1C river sounds its horn. To person A, the sound is loud and clear. To person Il, the sound is barely
audible. The principal wavelength
or
the sound
WatTS
is
3.00111. Assuming person B is at the position of the first minimum, determine the distance between the doors, cenler LO center.
~
--U).,\ r-::~~---
III
A
U
150mFIGURE P24.11
12.
mil
A student sets up a double-slit experiment using monochromatic light of wavelength A. The distance between the slits is equal to 25A. (a) Find the angles ,u which the III = 1,2. :lnd 3 maxima occur on the viewing' screen. (b) At what angles do the first three dark fringes occur? (c) Why arc the answers so nell ly spaced? Is the spacing even for all orders? E.xplain.
U1JRadio waves from (l star, of wavelength 250 m, reach a radio telescope by two separate paths, as shown in Fig-lire P24.13. One is a direct path to the receiver, which is situated on the edge of a dill by the Ocean. The second is bv reflection off the water. The first millimum of destructi,'c interference occurs when thc star is 25.0" above the horizon. Find the hci~'ht of the clifT. (Assume no phase chang-e on reflection.)
path Reflcned
\
FIGURE P24.13
14.
EI
SECTION 24.4 INTERFERENCE IN THIN FILMS 16. _ A soap bubble (11 = 1.33) ha,'ing a wall thickness or 120 nm is noating ill air. (a) What is the wavelength of the visible light that is most strongly reflected? (b) Explain how a bubble of differcm thickness could also strongly reneel light 01" this same wavclcngt.h. (c) Find the two smallest fi 1m thicknesses larg·er than the aile given lhal call produce strong-I), reflected light ot' th is same wavelength.
= 1.756) is sandwiched between two nat, parallel plates ofgla:is (n = 1.50). '",,'hat is the minimum t.hickness of the liquid laycr if normally incidellL light with A = 6.00 X IO'! Ill'll ill air is {Q be strongly reflected?
17. A thin layer of liquid methylene iodide (II
18.
A t.hin film of oil (11 = 1.25) is loealcd on smooth, wei pavement. \Vhen viewed from a direction perpcndicul:-lr to the p;WCIllCllL, the film rdlects most strongly red light at 640 nm and reflects no green light at 512 Ill'll. (a) What is the minimum lhickness of the oil film? (b) Let tnl correspond to the order of lite constructive interference and /11'2 to the order of the destructive interference. Obtain a relationship between /III and m't thai is consistent with lhe given data.
mil
19. A coming- is applied t.o a lens to minimii'c rdleClions. The index of refraction or the coating is 1.55 and that of the lens is 1.48. if the coating is 177.4 nm thick, what \\'avc+ length is minimally reflectcd [or normal incidence ill the lowest order?
Di~
~"fj , / -/
15. \Vavcs from a radio station have a wavelength of 300 m. The)' travel by two paths to a home receiver 20.0 km from the transmitter. One path is a direct p
SECTION 24.3 CHANGE OF PHASE DUE TO REFLECTION
B
20.0
first-order bright fringe :Hld comptlLe lhe sine of that angle. Are the sine and tangent of the anglc comparable in value? Doe." your answer always hold true? (eI) Calculate the wavelength ofthc light. (e) Complltc the angle of the firlh-order bright fringe. ([) Find its position on the screen.
Monochromatic light ofwavclcllglh A is incident on a pair of slits separated by 2.-1-0 X 10- 4 J'Il. and forms an imerference pauern all a screen is pbced l.SO m away from the slits. The first-orelcr bright fringe is 4.52 mm from the cenLer of the cenu'a] maximulll. (a) DnH\' a pic· ture, labeling the angle 0 and the legs ol'thc right tri,lIlgle associated with the first-order bright fringe. (b) Compute the tangcm of the angle 0 associated wit.h the fil'st-order briglll fringe. (e) Find the angle corresponding to the
20. A transparent oil with index of refraction 1.29 spills on the surr(lce o[wmer (index ofrcfranioll I.R3), producing a maximum of reflection with normally incidelH orange light (w;n:elengdl 600 nm in air). Assuming the maximum occurs in thc first oreler, determine the Ihickness of the oil slick. []IJA possible means for making an airplane invisible to radar is to coat the plane with an anti reflective polymer. If radar waves have a wavelength of 3.00 cm and the index orrerraction of the polymer is 11 = 1.50, how thick would yOll make the coating? 22.
An oil film (n = 1.45) floating on waleI' is illuminated by white light al nonnal incidellce. The film is
mil
Problems
2.80 X IO:.! nm thick. Find (a) the wavelength and color of the light in the visible spectrum most strongly reflected and (b) the w
mil
Astronomers obscn'c Ihe chromosphere of the Sun with a filter that p;1sses the rcd hydrogcn spectrallinc of wC\,"elength 656.3 nm, called the H a line. The filter con~ sists of a transparellt dielectric of Ihick ness d held bel ween two partially aluminized glass plates. The filter is kept at a constant temperaturc. (a) Find the minimum value of d that will produce maximum trtll1smission ofperpcndicular H(l' light if thc dielectric has an index of refraction of 1,378. (b) If lhe temperature of the filter increases above the normal value increasing its thickness, \.. . har happens to the lnll1smitted wavelcngth? (c) The dielectric will also pass what neal--visible wavelength? One or the glass plates is colored red to absorb this light.
24. Two rectangular optically flat plates (11, = 1..:')2) are in cOlltacl
819
Icngths in the visible spectrum intensified in the renected light? 30. A beam of light ofwavc1cngth 580 11m passes through twO closely spaced glass plates, as shown in Figure P24.30. For \vhal minimum nonzero value of" the plate separation d will the transmitted light be brighl? This arrangement is orten used 1.0 measure the wavelength of light and is called a Fabry-Pcrot interferomctcr.
FIGURE P24.30
SECTION 24.7 SINGLE-SLIT DIFFRACTION 31. Light of wavelength 5.40 X 102 nm passes through a slit of width 0.200 Illm. (a) Find the width of the central maxi· mum on a screen located 1.50 m from the slit. (b) Determine the \ ic!lh of the first-order bright fringe. 32. Light of \ avc1ength 600 nm falls on a 0.40·1l1111-widc slit and forms a diffraction p,Htern on a screen 1.5 m a'·iily. (a) Find the po."ition of thc first dark band 011 each side 01" the cemral maximum. (b) Find the width of Ihe central maximum.
FIGURE P24.24
(Problems 2'1 ,Hid 25)
25. An air wedge is formcd between two gla:-;s plates separated
133,ILight of ,vavclengLh 587.5 11m illuminates a slil of width 0.75 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is 10 be 0.85 mm from Lhe central maximum? (b) Calculate the width of the cenlral maximum. 34. :Microwavcs of wavelength 5,00 cm enter a long, narrow window in a building that is otherwise essentially opaquc to thc incorning '.. . <.1\'es. Il"the ''''indo\\' is 36.0 em wide, whal is Ihe distance from the central maximum to Ihe firslorder minimum along a wall 6.50 111 from the windm...·? 35. A beam of monochromatic light is diffracted by a slit of width 0.600 llUll. The diffraction pattern forms on a wall 1.30111 beyond the slit. The width of the central maximum is 2,00 111m. Calculate the wavelength of the light. 36. A screen is placed 50.0 cm from a single slit thal is illuminated with light ofw<\"c1ength 680 nm. If the distance between the first and third minima in [he difTraction pattern is 3.00 min, what is the width of the slit? 37. A slit of width 0.50 mm is il\umin
SECTION 24.8 THE DIFFRACTION GRATING 38. A helium-neon laser (A = 632.8 nm) is used to calibrate a diffraction graling. Jr the first-order maximum occurs at 20.5 c , what is the spacing belween acljacentgrooves in the grating?
820
Chapter 24
Wove Optics
139.IThree discrete spectral lines occur at angles of 10.1°, (3.T. and 14.8 c , respecl.i\'c1y, in the first.-order spectrum of a diffraction-grating spectrometer. (a) I r the grating has 3 660 slits/em, whal arc the wavelength!) of the light? (b) At what angles are these lines found in the second~ order spectra?
40. Intense while light is incident on a diffraction grating that has 600 lines/mm. (a) \-\'hat is the highest order ill which the complete visible ~pcclrum can be seen with this grating? (b) Whm is the angular separation between the \'iolet edge (400 11m) and the rcd edge (700 nm) of the first-order spectrum produced by llle gt""lling? 41. The hydrogen spcnfum has a red line at 656 11m and a violet line at 434 nm. What angular separation between these two spectral lines is obtained with a diffraction grating thai has 4500 lines/cm?
length or sodium. (c) Find the position of the secondary maximum on the sneCI1. (d) Repeat pans (b) and (c) for the second-order maximum of tile 589.6-nm wavelength of sodium, finding the angle and position on the screen. (e) What is the distance between the two secondary maxima on the screen? (f) Now find the distance between the two secondary maxima without rounding, carrying all digits in calculator memory. How many significant digits are in agreement? What can you conclude about rounding and lhe use ofirltcrmediate answers in this case? 49. A beam of !,)'ll-nm light is incident on a diffraction grati ng that has 4·00 Ii l1(,:s/111m. (a) Determine tile angle of the second-order ray. (b) If the entire aplJaratus is immersed in water, determine the nc\,;, second-order angle of dif.. fr
50. Light containing two 42. A grating v,'ith 1500 slit.s per centimeter is illuminated ,,·ith light of wavelength 500 nnl. (a) What is the highest-order number that can be observed with this grating? (b) Repeat for a grating or 15 000 slits per cClllimeler. 43. A light source emits two tm~jor speclral lines: an orange line of wavelength 610 nm and a blue-green line ofwa,-elength 480 nm. If the speClrulll is resolved b}' a diffraction grating having 5000 lines/cm and viewed un a screen 2.00 rn from the grating, whm is the distance (in ceillimetel's) between the two spectral lines ill the second-oreler spectrum?
44. White light is spread out imo its spectral components by a diffract.ion grat.ing. If the grating has 2000 lines per centimet.er, at what angle does red ligln of wavelength 640 nm appear in the f-irst-order spectrum?
45. Light from an argon laser strikes a diffract.ion grating that has 5 310 grooves pCI' cent.imeter. The central and first-order principal maxima arc separated by 0.488 m on a wall 1.72 m from the grat.ing. Determine t.he wavelength of the laser lighL.
46.
am Take red light at 700 nm and \"iolet at400 nm as the ends of the visible speClrulll and consider I he continuous spect.rum of white light formed b}' a difTractioll grating "..· ith a spacing of d meters between ,,(~j(\cent lines. Show that the intervale,,! ~ 0 ~ O,',! of t.he continuous spectrum in second order must overlap the interval O,,:,! S 0 S Or'!> of the third-order spectrum. Note: 0,1-1. is I he angle of the \'iolet light in second order, and 0r2 is the angle made b)' red light in second ordcr.
!47.)Sunlight is incident on a r1iffractioll grating that has 2750 lines/cm. The seconrl~order spectrum over the visible range (400-700 nm) is to be limited to 1.75 cm along a screen that is a dist.ance L from the gratillg. \·Vhat is Ihe required value of L? 48.
m
A diffraClion grating has 4.200 X I(p rulings per centimeter. The screen is 2.000 m from the grat.ing. In parts (a) through (e), round each result to {our digils, using the rounded "alues for subsequent calculations. (a) Compute the value of d, the distance bct\'..· ccn (lCUacellt rulings. Express thc answer in meters. (b) Calculate the anglc of the second-order maximum made by the 589.0-nm wave-
different wavelengt.hs passes through a diffraction grating with f 200 slits/cm. On a screen 15.0 Clll from lhe grating, the third-order maximum of Ihe shorter wavelength falls midway between lhe central maximum and the first side maximum for the longer wavelength. I f' the neighboring maxima of the longer wa\'elength arc 8.44 IllITl apan on the screen, what are the wavelengths ill the light? !lin/: Csc the small-angle approximation.
SECTION 24.9 POLARIZATION OF LIGHT WAVES [K[]Thc angle of incidence of' a light beam in air OIllO a reflecting surface is continuously variable. The reflected ray is found to be compktcly pohll"ized when the angle of incidence is 48.0°. (a) What is the index of l'efraction of the reflect.ing material? (b) If some of" the incident fight (at an angle of 48.00) passes int.o the material below the surface, what is Ihc angk of refraction?
52. L-npolarized light passes through t.wo Polaroid sheets. The t.ransmission axis of the analyzer makes an angle of 35.0° \...· ith thc axis of" the polarizer. (a) What fraClion of the original unpolarized light is transmitted through the analyzer? (h) What fraction of the original light is absorbed by the analyzcr? 53. The index of refraction of a glass plate is 1.52. What is the Brewster's angle when the plate is (a) in air and (b) in watu? (Sec Problem 57.) 54. At. what angle abovc the horizon is (he Sun if light from it. is completely polari:ted upon rellcct.ion from water? 55. A lighl bearn is incident on a piccl:: 01" fused quartz (n = 1.458) at. the Brewster's angle. Find the (a) value of Brewster's ang·le and (b) t he angle or refraction for the transmilled ray.
56. The crilical angle for tot'11 internal reflection for sapphire surrounded by air is 34.4°_ Calculate the Brewster's angle for sapphire if the light i'i incident from the air. 57. Equation 24.14 aSSUlllCS the incident light is in air. 1f the light is incidelll from a medium of index 1/1 onto a medium of index ll t , follow the procedure used to derive Equation 24.14 to sho\\' that Ian 0:, == n~';IlI'
Problems 58. Plane-polarized light is incident 011 a single polarizing disk, with the dircction of J~'o parallel to the direction of the transmission axis. 'l'hrough whal angle should the disk be rotated so thelt the inlcnsity in the transmitted beam is reduced hya factor of (a) 2.00, (b) 4.00, and (c) 6.00? 159.IThree polarizing platcs whose planes arc parallel are centered on a common axis. The directions of the transmission axes relative 10 the common vertical directioll arc shown in Figure P24.59. A linearly polarized beam of lig-ht with plane or pol:, ril.:ltion parallel to the vcrtical refercncc dircction is incident. from the left onto thc first disk with intensily I, = 10.0 unit.s (arbitrary). Ca1cul,-1te the transmitted intensity II when 1 = :!O.OC. 2 = 40.00, and (}.~ = 60.0°. /-lint: Make repeated lISC or Malus's law.
°
°
821
screen, wilh a cent.imeler I"oler below it. Did the light pass through one slit or twO slits? Explain how you call tell. II' the an:-.wcr is Olle slit, find it.s width. If the answer is two slits, find the distance between their centers. !64.1_ III a YOllllg's illterftTCnce experimcllI, the IWO slits arc separated by 0.150 mm and the incidcm light includes two wavelength~: Al = 5-10 nm (green) and ,\~ = 450 11m (blllc). The overlapping intel-ference patterns are observed 01l ,\ screen 1.40 m from the slits. (a) Find a relatioJ1ship belween the orders IlI l and 111'2 that determines where a brig-Ill rringe of the green light coincides with a bright fringe ()rtl1c blue light. (The order In l is associaled with AI' ;llId 'I!!'!. is associated \vilh ,\~.) (b) Find lhc rniniIlltltll vallles or IfI I and mA2 such that the overlapping of the bright fringcs will occur and find the position of the over· lap on the screen.
65. Light ofwavcJcngth 546 nm (the intense green line [rom a mercury source) produces a Young's interference pattern in which the second minimum from the central maximum is along a direction that makes an angle of 18.0 min of arc with Ihe axis through the central maximum. Whal is the dislance between the parallel slits?
FIGURE P24.S9
(['rohll"1Jl~ 59
alld 70)
liO. L.ight of intensit)' II) and polaril.C:d parallel to the transmission axis ofa polarizer is incidclltoll an analyzer, (a) lfthc transmission axis of the analyzer makes an angle of 45° with the ,lxis of the polarizer, wh,tl is Ihe intensity or the transmitted light? (b) WhatshOldd Ihe
II
Liglll with a wavelength in vaclium or 546.1 nm perpendicularly on :I hiolog-ical specimen that is l.OOn ,ulTl thick. The light splits illto two beams polarized at right angles, for which the indices orrefl'action are 1.320 and 1.333, respecl i\·c1y. (a) Calcillale the wa\Tlength of each componenl. of the light while il is traversing the specimen. (b) Calculate Ihe phase difference between the two beams when t.hey emer~e from the specimen. 1~\lIs
66. The two speakers arc placed 35.0 cm apart. A single oscilbtor makes the speakers vibrate in phase at. a frequency of 2.00 kl-Il. AI. what angles, measured from the perpcndicllhlr bisector or the line joining the speakers, would a disl
P---. Source
ADDITIONAL PROBLEMS
, :"
•
62. Light from a helium-neon laser (,\ = 632.H nm) is incidem on a single slit. What. is the maxinlllm width of the slit for which no diiTrauion minim~1 arc observcd?
63. _ Laser lighl with a wavcJellgth 0["632.8 nm is directed through one slilol- two slits ..Ind allO\ved to rail 011 a screen 2.60 1TI beyond. Figure P2'1.63 shows Ihe llall.erll on Ihe
5
6
7
8
9
10
FIGURE P24.63
11
12
13
I
J
o
IvlilTOI" FIGURE P24.67
68.
II
]\,1.111Y cel Is arc Iransparent and co101"lcss. Structures of" great interest. in biology and medicine can be practicall)' invisihle t.o ordinary rnicroscop}'. An intel/crena mirrovope reveals a differencc in rel'l'active index as a shift in interference fringes to indicate the size and shape of cell struc· ulr('.r". The idea is excmplified in the following problem: An air wedge is formed bctween two glass plmes in contact along one edge and slightly separated at the opposite edge. \,\'hell t.hc plalcs arc illuminated wilh monochro~ matic lighl from above, lhe reflected light has 85 dark rringes. Catcillate t.hc number of dark fringes thai appear ifwatcr (11 = 1.33) replaces the air between the plates.
822
Chapter 24
Wove Optics
169.IFigure P24.69 shows
71.
mil The transmiuing antenna on a submarine is 5.00 m above the water when the ship surfaces. The captain wishes to transmit a message to a receiver 011 a 90.0-m-tall cliff at the ocean shore. If the signal is to be completely polarized by reflection off the ocean surface, how far must the ship be [rom the shore?
72. _ A plano-convex lens (Oat on onc side, convex on the other) with index of refraction 11 rests with its curved side (radius o[ curvature R) on a flal glass surface of the same index ofrefraclion with a film of index nfi1m between them. The lens is illuminated from above by ligluofw3velength A. Show that the dark Newton rings that appear have radii of
where m is an integer.
173.IA diffraction pattern is produced on a screen 140 em Receiver
Transmitter FIGURE P24.69
70. Three polarizers, centered on a common axis and ' . . ith their planes parallel to one another, have transmission polariLCd with its plane or polarization oricmed vertically, is incident from the left OlllO the first polarizer. ·What is the ratio 'ill;ofthe final transmitted intensity to the incident intensity if (a) 01 = 45°, 2 = 90", and ()g = OO? (b) OJ = 0", 82 = 45°, and OJ = gO"?
°
[rom a single slit, using monochromatic light of wavelength 500 nm. The distance [rom the center of the centr
The twin Keck telescopes otop the summit of Hawaii's Mauna Keo volcano ore the world's largest optical and infrared telescopes. Each telescope has a primary mirror
lam in diameter comprised of 36 hexagonal segments that function together as a single piece of reflective gloss. For from city lights, high on a dormont volcano in dry, unpolluted air, the telescopes are ideally placed to probe the mysteries of the Universe.
OPTICAL INSTRUMENTS 25.1 We use devices made from lenses, mirrors, and other optical components every time we
put on a pair of eyeglasses or contact lenses, toke a photograph, look at the sky through a
The Camera
25.2 The Eye
telescope, and so on. In this chapter we examine how optical instruments work. For the most
25.3
The Simple Magnifier
part, our analyses involve the laws of reflection and refraction and the procedures of geomet~
25.4
The Compound Microscope
25.5
The Telescope
25.6
Resolution of Single-Slit and Circular Apertures
25.7
The Michelson Interferometer
ric optics. To explain certain phenomena, however, we must use the wave nature of light.
25.1
THE CAMERA
The single-lens photographic camera is a simple optical instrument having the features shown in figure 25.l. It consisls of an opaque box, a converging lens that produces a real image, and a photographic fdlll behind the lens to receive the image. Digital can1eras differ in that the image is formed on a charge-coupled device (CCD) or a complementary melal·oxidc semiconductor (CI\:JOS) sensor instead of on film. Both the ceo and the C~·10S image sensors convert the image into digital form, which can then be stored in the camera's memory. rocusing ofa camera is accomplished by varying the distance bet\ovcen the lens and film, with all <.)cljustable bellows in antique cameras and other mechanisms in contemporary models. For proper focusing. which leads to sharp images, the lells-to-filn1 distance depends on the object dist.ance as well as on the focal length of the lens. The shutter, located behind the lens, is a mechanical de\'ice that is opened for selected time intervals. \'\lith this arr<1ngement, moving objecls can be photographed by using shon exposure times and dark scenes (with low light levels) by using long exposure limes. If I.his aqjuslInent were not available, it would be impossible to lake s1.op-ac1.ion photographs. A rapidly moving ych iclc, for example, could move far enough ",,,hile the shutter was open La produce a blurred image. Another m ..~or cause of blurred images is movement of the rauwm while t.he shutter is open. To prevent such movemenl, y01l should mount the camera on a I.ripod or lise short exposure times. Typical shutler speeds (lhat is, exposure times) are 1/30 s, 1/60 s, 1/125 s, and 1/250 s. St.alioJ1<1ry objects are often shot with a shutter speed of 1/60 s.
Shutter
{
,,'
eel)
~~, L"."~~ Y~;~'OS ....L
,.~<::
I
0
~
sensor
Image
Apenun'
~ f>-"f.-- (/
FIGURE 25.1 Cross-sectional \"ic\\ ora ~illlple digital camera. The CCD or C~IOS illlag-c sen sur i" IIll': lig-hlSCllSil ivc compollenl uf lile callH.·ra. 111;1 llondigital camera the JighllrolTl the Il.:ll~ fall~ 011[0 pholographic lilln. III n:alil)' /»> If.
823
824
Chapter 25
Optical Instruments
Most cameras also have an aperture of acljustable diameter [0 further control the intensity of the light reaching the Him. ',Vhen all aperture of small diameter is used, only light from the central portion of the lens reaches the film, so spherical abernllion is reduced. The iIlLen:siL)' J uf the light reaching the film is proportional to the area of the lens. Because this area in turn is proportional LO the square of the lens diameter D, the intensiL)' is also proponional to D?. Light intensity is a measure of the rate at v.. hich energy is received by the film per unit area of the image. Because the area of the image is proporLional La rl in Figure 25.1 and q = ! (when fJ » f, so that p can be approximated as infinite), we conclude that the illlensity is also proportional to 1//'2. Therefore, ]-:x D2/f2. The brightness of the image formed on the film depends on the light intensity, so 'we see that it ultimately depends on both the focallengthfand diameter D of the lens. The ratio flD is called theI-number (or focal ratio) of a lens:
f
f-number == D
[25.1]
The I-number is oftcn given as a dcscription of the lens "speed." A lens \..·ith a low.f-number is a "fa~H" lens. Extremely fast lenses, v\'hich have an I-number as low as approximately 1.2, are expensive because of the difficulty of keeping aberrations acceptably small v.lith light rays passing through a large area of the lens. Camera lenses are often marked with a range of j-numbers, such as 1.4,2,2.8,4, 5.6,8, and 11. Anyone of these scltings can be selected by acljusting the aperture, which changes the value of D. Increasing the sening from one I-number to the next-higher value (for example, from 2.8 to 4) decreases the Clrea of the aperture by a [actor of 2. The 1m-vest I-nu mber scui ng all a camera corresponds to a \vide open aperture and the use of the maximum possible lens area. Simple cameras lIsually have a fixed focal length and fixed aperture size, with an f-number of abollt 11. This high value for the f-number allmvs for a large depth of field and means that objects at a wide range of distances from the lens form reasonably sharp images on the film. In other words, the camera doesn't have to be focused. i\tlost cameras \\lith variable I-numbers acUust them automatically.
25.2 THE EYE HI Like a camera, a normal eye focuses light. and produces a sharp image. The mechanisms by which the eye controls the amount of light admitted and adjusts to produce correctly focused images, hm,\'cver, are far more complex, intricate, and effective than those in even the most sophisticated camera. In all respects the eye is a physiological wonder. Figure 23.2a shows the essential parts of the eye. Light entering the eye passes through a transparent structure called the cornea, behind which are a clear liquid (the aqueous humor), a variable aperture (the jJupil, which is an opening in the in's), and the C1)'stalline lens. Most of the refraction occurs at the outer surface of the eye, where the cornea is covered with a film or tears. Relatively little refraction occurs in the crystalline lens because the aqueous humor in contact with the lens has an average index of refraction close to that of the lens. The iris, which is the colored portion of the eye, is a muscular diaphragm that controls pupil size. The iris regulates the amount of light entering the eye by dilating the pupil in low-light conditions and contracting the pupil under conditions of bright light. The I-number range of the eye is from about 2.8 to 16. The cornea-lens system focuses light onto the back surface of the e>'e-the ret~ i17a-'..vhich consists of millions of sensitive receptors called mds ancl (;Ones. ',Vhen stimulated by light, these structures send impulses to the brain via lhe optic nerve, converting them into our conscious view or the world. The process by which the brain performs this conversion is not \\"ell understood and is the su~ject of much
FIGURE 25.2 (a) F.ssenlial p
speculation and research. Unlike filrn in a camera, the rods and cones chemically ac!just their sensitivity according to the prevailing light conditions. This aqjustment, \'vhich takes about 15 minutes, is responsible for the experience of ;'geLling used to the dark" in such places as movie theaters. Iris aperture control, which lakes less than 1 second, helps protect the retina from overload in t.he acljustment process. The eye focuses on an o~ject by varying the shape of the pliable crystalline lens through an amazing process called accommodation. An important. componelH in accommodation is the ciliary muscle, which is silllated in a circle around the rim of the lens. Thin filaments, called zonllles, fun from this muscle to the edge of the lens. \"'hen the eye is focused on a distant. object, the ciliar}' muscle is relaxed, tightening the zonules that auach the ciliary muscle to the edge of the lens. The force of the zonules causes the lens to flatten, increasing its focal length. For an ol~ject distance of infinity, the focal length of the eye is equal to the fixed distance between lens and retina, about 1.7 cm. The eye focuses on nearby objects by lensing the ciliary muscle, \vhich relaxes the zonules. This action a11mvs the lens to bulge a bit and its focal length decreases, resulting in the image being focused on the retina. All these lens acUustmcnts take place so swiftly that we are not even aware of the change. In this respect even t.he finest electronic camera is a toy comparcd with the eye. There is a limit to accommodation because objects that are vcry close to the eye produce blurred images. The near point is the closest distance for which the lens call accomrnodate to focus light on the retina. This distance usually increases \'·,ith age and has an average value of 23 cm. Typically, at age 10 the Ileal' point of the eye is about 18 em. This increases to about 25 cm at age 20, 50 cm at age 40, and 500 cm or greater at. age 60. The far point of the eye represents the farthest distance for which the lens of the relaxed eye can focus light on the retina. A person wilh normal vision is able to sec ve!")' distant objects, such as the Moon, and so has a rar point at infinity.
Conditions of the Eye \t\'hcn the eye suffers a mismatch belween the focusing power of the lens-cornea system and the length of the eye so that light rays reach the retina before they converge to form an image, as in Figure 25.33 (page 826), the condition is knmvn as farsightedness (or h)1}(>wjJia). A farsighted person can usually see faraway ol~jects clearly but not nearby objects. Althollgh the near point or a normal eye is approximale]y 25 Clll, the nea r poi Ilt or a farsighted person is much fal'l her tha n that. The
The Eye
825
826
Chapter 25
Optical Instruments
COlllraCled
1ll11scle
COllvcrgil1g
•
Ncar puillt
I\'car
~~===E~~I',\ Object
point
[mage
Ie",
-
Relaxed lem c'_
\
Image at retina
behind retina (a)
(~
FIGURE 25.3 (a) A farsighted eye is slig-lltly shorl(~r 1hall normal; hence, the image or d Wilh a ("(JIl\'crging lens. (1'11(' \l~ieCl is assullIed to he vcr)' ~Jllall in Lllest.: figures.)
APPLICATION "
Using Optical Lenses to Correct for Defects
eye ora farsigl1lcd person tries to focus by accommodation, by shortening its fOC"11 length. Accommodation \vorks for distant objects, but because the focallengrh of []lC r
25.2 The Eye
827
Relaxed [enl'
01~it.:CI
.,.~.~:~~"C=--~
;~
Farpoilll
Di\"erg:illg lens \
Ohjl:Cl
/'-'"
Far point
Image in frOJ'lIOfrCl.ina
(al
(h)
(a) A lIearsighlcd eye is ~lightl)' longer Ih;lll normal: henn', the ill1;\ge ofa distanl object !llCIISt.:S il/FolI1 of till: retina. (b) The conditioll ,an be (IllTeeted \,·jlb a diverging' lens. (The object is assumed LO Ill: very :.mall in these figures.) FIGURE 25.4
EXAMPLE 25.1 Goal
Prescribing a Corrective Lens for a Forsig"h.;.:t"'e"'d..:P..:o"'t"'ie"'n"'t_B....,.
Apply geometric optics
La
"
correct farsightedness.
Problem The near point of a patient's eye is 50.0 em. (a) \Vhat focal length must a corrective lens have to enable the eye to see clearly an ol~jec[ 25.0 Cill zl\.\,ay? Neglect the eye-lens distance. (b) "V hal is the power of this lens? (c) Repeat the problem, taking into account that., for typical eyeglasses, the corrective lens is 2.00 cm in front of the eye. Strategy This problem requires substitution into the thin-lens equation (Eq. 23.11) and then using the dell nition of lens pmver in terms or diopters. The object is at 25.0 cm, but the lens must form an image at the patient's near point, 50.0 cm, thc closest point at which the patient's eye can see clearly. In pan (c) 2.00 cm must be subtractcd from both the object distance and the image distance to account [or the position of the Icns. ".'
.....
. . . • . • • • • • . • . , ''''Il
Solution (a) Find thc focal length of the corrcClive lens, neglccting its distance from the eye. I
I
I
Apply the thin-lens equation:
-+-=!J q f
Substitute /) = 25.0 em and q = -50.0 em (the latter is negative because the image must be virtual) on the same side of the lens as the oqject:
--'-- + --,-:-':--
Solve for f The focal length is positive, corresponding to a converging lens.
f= 50.0 em
25.0 em
-50.0 em
f
(b) 'What is the power of this lens?
The power is the reciprocal or the focal length in meters:
'!P
1
1
= - = -,---,-,----
f
0.500
+2.00 diopter,
III
(c) Repeat the problem, noting that the corrective lens is aet ually 2.00 em in front of the eye.
Substitute the corrected values of p and th in-lens equation:
Remark Notice that the calculation in pan (c), which doesn't neglect thc eyc-lens distance, results in a difference of 0.26 diopter.
828
Chapter 25
Optical Instruments
QUESTION 25.1 True or False: The larger the distance
lO
a ncar point, the larger lhe power of the required corrective lens.
EXERCISE 25.1 Suppose a lens is placed in a device lhat determines its power as ~.75 c1ioptcrs. Find (a) the focal leng-I.h of the lens and (b) the minimum distance at ,...· hich a patient ,dll be able to [OCLIS 011 an ol~jccl if the paticlll's Ilcar point is 60.0 em. Neglect the eye-lens distance.
Answers
Goal
(
(b) 22.7 em
Apply geometric optics
10 CDlTcet
nearsightedness.
Problem A particular nearsighted patient can't see objects clearly when Lhey are beyond 25 em (tile far poil1l or the eye). (a) What focallenglh should the prescribed COlllact lens have to correct this problem? (b) Find lhe power of the lens, in diopters. Neglecl the distance between the eye anclthe corrective lens. Strategy The purpose of the lens in this instance is to take ol:~jects at infinity and create an image of t.hem at. the patient's ral' point.. Apply the thin-lens equation.
,.. ..
...
.
Solution (a) Find the focal length of the corrective lens. 1
I
- +- = /J q
Apply the thin-lens equation for an object at infinity and image at 25.0 em:
I= 1
(b) Find the power ofthL: lens in dioplcrs.
(fj> = -
I
=
-
I
oc
1
+ --=----(-25.0 em)
I
-25.0 em
-0.250 m
= -4.00 diopters
.
t.. • . . . .
~
Remarks The focallenglh is negativc, consistent \vith a diverging lens. Noticc that the power is also ncgati\'c alld has the same numeric v'llue as llle sum on tllC lert side of the thin-lens equation. QUESTION 25.2 True or False: The shorter the distance to a patient's far point, the morc negative the power of the required corrective lens. EXERCISE 25.2 (a) \Nhat pmver lens ,vould you prescribe for a patient with a rar point of 3.1.0 cm? Neglect the eye-lens distance. (b) Repeat, assuming an eye-correcLivc lens distance of2.00 ern.
Answers
(a) -2.86 diopters
APPLYING PHYSICS 25.1
(b) -3.03 diopters
VISION OF THE INVISIBLE MAN
A classic science fiction story, The Invisiblf' Man by H. G. \'Vells, tells ora man wllo becomes invisible by changing the index of refraction of his body to that of air. SLUdellls who know how the eye works have criticii'.cd this story; tile}' claim thallhe invisible man ,,,,'ould be url
Explanation He vo/Ouldn't be able to sec. For the eye sec an ol~jecl, incoming light must be refracted at the cornea and lens to form an image on the retina. If the cornea and lens have the same index of refraction as air, refraction can't occur <'ll1el all image wouldn't be Jormcd. LO
25.3
The Simple Magnifier
829
QUICK QUIZ 25.1 Two campers wish to starl a fire dllring the day. One camper is nearsighted and one is farsighted. vVllose glasses should be lIsed La focus the Sun's rays onlo some paper to starllhe fire? (a) either camper's
(b) the ncarsigillcd camper's (c) the farsighted camper's
25.3
THE SIMPLE MAGNIFIER
The simple magnifier is one of lhc most basic of all optical instruments because it consists only uf a single converging lens. As the name implies, this device is lIsed 1.0 increase the apparent sil'.c or an object. Suppose an object is vie\vcd aL some distance /) from the eye, as in Figure 25.5. Clearly, the size orlhe image formed at the retina depends on the angle 0 sub1.cncled by lhe object at the eye. As the object moves closer La the eye, 8 increases and a larger image is observed. A normal eye, hm",ever, can't [oells on an o~jeCl closer than about 25 cm, the near point (Fig. 25.6a). (Try ill) Therefore, 0 is a maximum at the ncar poim. To further increase the apparent angular size of an object, a converging lens can be placed in from or the eye with the object positioned at point a,just inside the roeal point of t.he lens, as in Figllre 2!).6b. At Ihis lucation, the It::ns forms a virtual, upright, and enlarged image, as shown. The lens allows the object to bc viewed closer 1.0 the eye than is possiblc without the lens. \'Vc define the angular magnification m as the ratio of the angle sublended by a small object when the lens is in use (angle in Fig, 25,6b) to the angle sublended by the object placed ;:u the near point ,..· ith no lens in usc (angle 00 in Fig. 25.6a):
FIGURE 25,S Thl' si/.\' of lhc image fonlH.:d 011 the n:lina depends 011 lhc ,mg-k 0 subtellckd at Ill<:: eyc,
e
'/11
= -88
[25.2]
f- Angular magnification with the object at the near point
0
For Ihe case in which the lens is held close to thc eyc, the angular magnification is a maximum ,vhen the image formed by the lens is ,Il the ncar point of the eyc, which corresponds to q = ~2!'i cm (see Fig. 25.6b). The ol~iect dist
1 _-,--_ -+-
P
~25
cm
[25.3]
I
2"I /' = 2"
+I
Here,Jis the focal length of the magnifier in centimeters. From Figures 25.6a and 25.6b, the small-anglc approximation givcs h
tan 0 0
= O{J = 25
Ii
tan 8 = I) = -
and
l'
[25.4]
Equation 25.2 lhercfore becomes
e
11./ I)
25
0"
h/2.'>
/'
=-=--=-=
111
m."
25 25j/(25
+ I)
so thal
25 cm
+---
125.5]
I
r ~"":...__ ~.I)cllil '~n,-",,-:,-::o.-~k. ~-----
an 'lI1g-lc
2:') cm ---'-----I;lJ' (al
FIGURE 25.6 (a) An object placed ,lllhe near point (/J = 2:) cm) slIbtcmb ,Ill ;lIlgle ul"0o = 11/25 "llhe l:)"l:. (b) An ol~j('CI placed Ill-'ar lilt.: focal poinl ora t'oll\'{'q~ing 1\'lIs prot!lln:s a m'lgnified image. \,'hich su!ltcilds
oro = 11'/25
lhal in Ihis ~illlalion q = -2!"", CIll,
(bl
Chapter 25
830
Optical Instruments
The maximum angular magnification given by Equation 25.3 is the ratio of the angular size seen with the lens to the angular size seen without the lens, with the object at the near point of the eye. Although the normal eye can focus on an image formed anywhere between the near point and infinity, it's Illost relaxed when the image is at infinity (Sec. 25.2). for the image Cormed by the magnifying lens to
appear at infinity, the object must be placed allhe [-ocal point of the lens so that jJ = f In this case Equation 25.4 becomes
oo =
-
II.
II.
and 0=-
25
!
and the angular magnification is
8
23 em
00
!
[25.6]
1n=-=---
\Vith a single lens, it's possible w achieve angular magnifications up to about 4 ".. ithout serious aberrations. Magnifications up to about 20 can be achieved by using one or two additional lenses to correct for aberrations.
EXAMPLE 25.3 Goal
Magnification of a Lens
Compute magnifications ofa lens when the image is at the near point and when it's at infinity.
Problem (a) What is the maximum angular rnagnificatioll ora lens with a focal length of ]0.0 cm? (b) What is the angular magniflcation o[this lens when the eye is relaxed? Assume an eye-lens distance of zero. Strategy The maximum angular magnification occurs \vhen the imCige formed by the lens is at the near point of the eye. Under these circumstances, Equation 25.5 gives us the maximum angular magnification. In part (b) the eye is relaxed only if the image is at inrinity, so Equation 25.6 applies. ,...-' , , ....,
Solution (a) Find the maximum angular magnification of the lens. 25 em
9" _:) Clll
I
10.0 em
1+--=1+
Subslitute into Equation 25.5:
3.5
(b) Find the magnification or the lens when the eye is relaxed.
''''hen the eye is relaxed, the image is at infinity, so substitute into Equation 25.6:
11/
25 ( I l l
25 em
I
10.0 em
= --- =
.........................
110..-• • • • • • • • • • • • • • • • • • • •
QUESTION 25.3 For greater magnification, should a lens \vith a larger or smaller focal length be selected? EXERCISE 25.3 'o\'hal focal length would be necessary if the lens ,,;ere to have a maximum angular magnification of 4-.00? Answer
8.3
Clll
25.4
THE COMPOUND MICROSCOPE
A simple magnifier provides only limited assistance with inspection of the minute details of' an object. Greater magnification can be achieved by combining two lenses in a de\'ice called a compound microscope, a schematic diagram ofwhich is shmvn in Active Figure 25.7£1. The instrument consists of two lenses: an objective with a very shon focal length Irl (where I" < 1 em) and an ocular lens, or eyepiece,
25.4
The Compound Microscope
831
OI~jeCli\'e
(b)
(a)
ACTIVE FIGURE 25.7 (a) A diagram of a compound rniClTlM:opt·. whirh cOll~i.'>b 01 all ol~jeCliH' ,md all eyepiece. or ondar Il"ns. (b) A CtlillpOlllld llIiuo.'>cope. The lhree-o]~it'cti\l' turret allow., lht' l11>er 10 switch 10 ,c\"(,;ral difrert..'lIt pnw('rs of m of ('\'epicl't's Will) diffl'lcnt local kllglh, and difl"t.:rt:lI1 objecli\'es CIIl produce;\ wid(' rangt.: ofmagnific
\\I'ith a focallengthJ;ofa few centimeters. The two lenses are separated by distance L that is much greater thClll eithcr 1, ur j". The basic approach used LO analyze the image formation properties of a microscope is lhat of two lenses ill a row: the image formed by thc first becomcs the object for the second. The object 0 placed just olltside the focal length of the objective forms a real, invertcd image at I] that is at or just inside the focal point oCthe eyepiece. This image is much cnlarged. (For clarity, the enlargement of Jj is not shmvn in Active Fig. 25.7£1.) The eyepiece, \.. . hich serves <15 a simple magnifier, IIses the image at II as its object. and produces an image at I'J.' The image seen b): Ihe eye at 1'1. is virtual, inverted, and vcry much enlarged. The latcral magnification !\1[ urthe first image is -qll/),. Note that fli is approximalely equal LO L because the object is placed close to the focal point of the objective lens, which ensures that the image ('"armed \.. . ilJ bc I"eJ.r from the objective lens. Further, because the ol~jecl is very close to the focal point or the objective lens, PI = );,. Therefore, the lateral magnification of the objective is
'I,
Nil = - -
/J,
L
=
J,
From Equation 25.6, the angular magnification of the eyepiece for an object. (corresponding to the image at II) placed at the focal point is found to he
2!) cm
=---
111
"
.f.,
The O\'erall magnification of the compound microscope is defined as the lIet of' the lateral and angular magnifications: III
=
lH I 1fl r
(25
= - -L - -CIT' -) j;, I..
prod~
[25.7]
Thc negati\'e sign indicates that the image is inverted \,'ilh respecl 10 the o4ject. The microscope has extended our visiun into the prc\'iollsly Ilnknown realm of incredibly small objects, and the capabilities of this instrument. have increased steadily with improved techniques in precision grinding of lenses. A natural question is whether there is (lny limit to how powerful a microscope could be. For example, could a microscopc be made powerful enough to allow us LO see an atom? The answer to this question is no, as long as visible light is used to illumimuc the object.. To bc seen, the object uncleI' a microscope must be at least as large as a wavclengl h or lig·hl. An atom is many times smaller than thc wavelength of visible light, so its mysteries must be probed via other techniques.
~ Magnification of a microscope
832
Chopter 25
Optical Instruments
The "'<1VClenglh dependence of the "seeing" ability of a ",:ave can be illustrated by \vaLer waves set lip in a balhLUb in the rollO\"'ing way. Imagine that you vibrate your hand in the watcr until waves with a wavelength of aboUl 6 in. arc moving along the surface. If you fix a small object, such as a roolhpick, in the path of the waves, yOlI will find thal the waves arc not appreciably disturbed by the toothpick, but continue along their p£lth. Now suppose you Ax a larger ol~jeet, such as a toy sailboat, in the path of the waves. In this case the waves arc considerably disturbed by the object. The toothpick '·...as much smaller than the \vavelength of the waves, and as a result the \V'aves didn't "see" it. The toy sailboat, however, is about the same size as the wavelength of the waves and hence creates a disturbance. Light waves behave in this same general way. The ability of an optical microscope to view an oqject depends on the size of thc o~jccl rehllivc to the wavelength of the light lIsed La observe it. Hence, it will never bc possible Lo observe atoms or molecules \.. . ith Stich a microscope because their dimensions are so small ("""0,] nm) relative to the wavelength of the light (=500 11m).
EXAMPLE 25.4 Microsco e Magnifications Goal
Understand the critical factors involvcd in determining the magnifying powcr ora microscope.
Problem A certain microscope has t\\lO interchangeable o~jcctives. One has a foeallength of2.0 cm, and the other has a focal length of 0.20 cm. Also available are two eyepieces of focal lengths 2.5 CI11 and 5.0 cm. If the length of the microscope is 18 cm, compute the magnifications for the [oHO\'ling combinations: the 2.0-cm objective and 5.0-cl11 eyepiece, the 2.0-cm objective and 2.5-cm eyepiece. and the O.20-cm Objective and 5.0-cm eyepiece.
Strategy The solution consists ofsubstiLuting into Equation 25.7 for three different combinations of lenses.
...
• • • • • • • • . . • • • ••• •• •• "''''111
Solution Apply Equation 25.i and combine the 2.0-cm objective \vith the 5.0~cm eyepiece: Combine the 2.0-cm objective \\"ith the 2.5-cm eyepiece: Combine the O.20-cm eyepIece:
.........
ol~jective
with the 5.0-cm
m. =
__ 18cm I. (25 em) __ - - - (25cm) --- 2.0 eln 5.0 cln I" I,
Remark t\·luch higher magnifications can be
Answer
-9.0 X 10'
25.5
THE TELESCOPE
There arc two fundamentally different types of telescope, both designed to help us vie\.\' distant objects such as the planets in our solar system: (1) the refracting telescope, which uses a combination of lenses to form an image, and (2) the reflecting telescope, \vhich uses a curved mirror and a lens to form an image. Once again, we can analyze (he telescope by considering it to be a system of tWO optical e1e-
25.5
833
ACTIVE FIGURE 25.8 (a):\ diagram ofa rdraCling telescope, with lhe objecr at infinity. (b) A refracting telescope.
E)'cpit:ce lens Objective lens
1';1.
The Telescope
(a)
(b)
ments in a row. As before, the basic technique followed is that the image formed by the first element becomes the object for the second. 1n the refracting telescope 11\'0 lenses are arranged so that the objective forms a real, inverted image of the distant ol~ject very near the focal point of the eyepiece (Active Fig. 25.8a). Further, the image at [I is formed at the focal point of the objective because the object is essentially at infinity. Hence, the two lenses are separated by the distance 1, + fe, which corresponds to the length of the telescope's tube. Finally, at /2' the eyepiece forms an enlarged image of the image at [I' The angular magnification of the telescope is given by fJ/8 u> where 80 is the angle subtended by t.he object at t.he objective and 8 is the angle subtended by the final image. From t.he triangles in Active Figure 25.8a, and for small angles, we hm'c h' 8=-
and
f"
Ii
The Hubble Space Telescope enables liS 10 see both funhcr inlO space and further back in lime than ever before.
8=-
"
fo
Therefore, the angular magnification of the telescope can be expressed as
8
h'lf,.
;;,
8,
10f"
f"
11/=-=--=-
[25.8]
This equation says that the angular magnification of a t.elescope equals the ratio of the objective focal length to the eyepiece focal length. Here again, the angular magnification is the ratio of the angular size seen wit.h t.he telescope to the angular size seen with the unaided eye. In some applicat.ions-for instance, the observation of relatively nearby objects such as the SUIl, the Moon, or planets-angular magnification is imponant. Stars, hm·\.'cver, ;ue so far away that they always appear as slllall points of light regardless of how much angular magnificat.ion is used. The large research telescopes used to swell' very distant objects must have great diameters to gather as much light as possible. It's difficult and expensive to manufacture such large lenses for refract· ing telescopes. In addition, the heaviness of large lenses leads to sagging, which is another source of aberration. These problems can be partially overcome by replacing the objective lens with a reflecting, concave mirror, usually havjng a parabolic shape so as to avoid spherical aberration. Figure 25.9 shows the design of a typical reflecting telescope. Incom· ing light rays pass down the barrel of the telescope and (Ire reflected by a parabolic mirror at the base. These rays converge toward point A in the figure, where an image would be formed on a photographic plate or another detector. Before this image is formed, however, a small, flat mirror at 1"1 reflects the lig·ht toward an opening in the side of the tube that passes into an eyepiece. This design is said t.o have a Newtonian focus, after its developer. Note that in the reflecting telescope the
~
Angular magnification
of a telescope
.\
1
.
\
En:pie<.:c
Parab(llic " - - - - - " - - mirrur FIGURE 25.9 t\ refiel'ling tel<.:scope with a NeWlonian rOCtls.
834
Chapter 25
Opticollnstrurnents FIGURE 25.10
The I-laic
lek.~(C)pc
f
at Moult\. Palomar Observatory. .!IISL
~
bt:lure laking lhe de\"atur up 10 the prime-rocllS cage, a firsl.-limc ()h~(~rvt:r j~ alw,lys laid, ~Cood ricwiug! And, ifroLi should fall. try to miss the
~ o
"~ ~
fllirror.~
light never passes through glass (except for the small eyepiece). As a result, problems associalcd wilh chromatic aberration are virtually eliminated. The largest optical telescopes in the world arc the two lO-Ill-diamclcr Keck reflectors 011 Malina Kea in Hawaii. The largest single-mirrored renccting telescope in the Cnitecl Staws is the 5-m-diamCl.cr instrument on Mount Palomar in California. (See Fig. 25.10.) In comrast, the largest refracting telescope in the world, atlhe Yerkes Observatory in 'Villiams Bay, vVisconsin, has a diameter of only 1m.
EXAMPLE 25.5 Goal
Hubble Power
Understand magnification in telescopes.
Problem The Hubble Space Telescope is 13.2 m long, but has a secondary mirror thaL increases its effective focal length to .:17.8 m. (See Fig. 25.11.) The telescope doesn't have an eyepiece because various instruments, not a human ere, record lhe colleeled light. It can, however, produce images several lhollsand times larger than they would appear \..· jth the unaided human eye. 'What focal-Iengt.h eyepiece used wit.h the Hubble mirror system would produce a magnil'iccuion of8.00 X IQ:\?
Strategy Equation 25.S for telescope magnification can be solved [or I he eyepiece focal length. The equation lor finding the angular magnification ora reflector is the same as that for a refractor.
FIGURE 25.11 A schclllatic of the Hubhle Sp'lCt· Telescope.
-
Aperilire doo[,
ldary __ :0-=, mirro
SClI)!
,.
Secondary mirror ba.llk
,
f\:laill banlc
Cenll"d bamc
~I
Prima lllirro
,""
Focal plane ~
.
,... ..... Solution Solve forIrin Equation 25.8 and substitute values:
.....
.
j,; 1/1.=-
j
j
I = ;; =
'"
....
57.8 rn 8.00 X 10'
................
Remarks The light-gathering power of a telescope and the Icngth of the baseline over which liglll is gathered are in fact more important than a telescope's magnification, because these two factors contribute to the resolmion of the image. A high resolution in1agc can always be rnagnilJed so its dctails can be examined. A low resolULion image, however, is oflen fuzzy ,.. . hen magnified. (See Seerion 25.6.)
QUESTION 25.5 Can greater magnification of a telescope be achieved by increasing the focal length of t he mirror? What effect will increasing the focallengrh of the eyepiece have on the magnification?
25.6
Resolution of Single-Slit and Circular Apertures
835
EXERCISE 25.5
The Hale telescope on Maull[ Palomar has a focal length of 16.8 m. Find the magnificaLion or the telescope in conjunction with an eyepiece having a focal length of 5.00 111m. Answer
3.36 X 10'
25.6
RESOLUTION OF SINGLE-SLIT AND CIRCULAR APERTURES
The ability of an optical system such as the eye. a microscope. or a telescope to distinguish bet\lleen closely spaced objects is limited because of the wave nature orJight. To understand this difficulty. consider Figure 25.12, which shows two light sources far from a naITOw slit of width a. The sources can be taken as two point sources 51 and 52 that are not coherent. For example. they could be two distant stars. If no diFfraction occurred, two distinct bright spots (or images) '\lould be observed on the screen at t.he right in the figure. Because of dim-action, however, each source is imaged as a bright central region flanked by ,'.:eaker bright and dark rings. \"'hat is obsen'ed on the screen is the sum oft\".'O diffraction patterns, one from 51 and the other from 52' If the two sources are separated so thaI their central maxima don't overlap, as in Figure 25.l2a, their images can be distinguished and are said to be -resolved. If the sources are close together. however, as in Figure 25.12b, t.he two central maxima may overlap and the images are not resolved. To decide whether two images are resolved, the following condition is often applied to their diffraction patterns:
When the central maximum of one image falls on the first minimum of another image. the images arc said to be just resolved. This limiting condition of resolution is known as Rayleigh's criterion.
+-
Rayleigh's criterion
Figure 25.J3 (page 836) sho\\ls diffraction patterns in three situations. The images are just resolved when their angular separalion satisfies Rayleigh's criterion (Fig. 25.13a). As the objects are brought closer togcther, their images are barely resolved (Fig. 25.13b). Finally. when the sources are very c10sc to each olher, their images are not resolved (Fig. 25.13c). From Rayleigh's crit.erion, we can determine the minimum angular separation ()ll1in subtended by the source at the sljt so that the images will bejust resolved.]n Chapter 24 we found that the first minimum in a single-slit diffraction pattern occurs at the angle t.hat satisfies the relationship A
sinO = a
FIGURE2S,l2 Twopointsources far rrom a narrow slit each produce a diffraction panern. (a) The angle subtended by the sources at the aperture is large enough SQ thai the diffraction paltCI'1IS arc distinguishable. (b) The angle subtcnded by the sources is so small that Ihe diffraction patterns are not distinguishable. (Nole thal the anglc~ are p;rcatly exaggerated. The drawing is IIOl to scale.) Slit (aj
Viewing screen (b)
836
Chapter 25
Opticollnstrllments
FIGURE 25.13 Thediffrat:lioll patlerns of two POilil sources (solid CLlf\'CS) and till' n . :sullam pallcrn (dash~d cun'~) for lhree angular separal ions of the sntlrces. (a) The Vlurcc" are <;ep arc so rlose togelher lhat their pattcrlls ;lrc nul rcsolved.
I ,
"
,
\ I
,
I \
r
\
J!L (a)
(b)
(c)
\\;here II is the ""idth of the slit. According to Raylcigh's critcrion, this expression gives the smallest angular separation for which the two images can be resolved. Because A « a in Illost situations, sin () is small and we can use the approximation sin 0 = O. Therefore, the limiting angle 01" resolution for a slit of'\i'idth fI is Limiting angle for a slit
-+
[25.9] ,,,,'here 0lllin is in radians. Hence, the angle subtended by the t\'/o sources at the slit be grealf;"rthan A/a if" the images arc to be resolved . .,.§ rvlany optical sYstems use circular apertures rather than slits. The diffraction ~ ~pattern of a circ~t1ar aperture (Fig. 25.14) consists of a central circular briglH ~ g,region surrounded by progressively fainter rings. Analysis shows that the limiting :, ~ angle of" resoilltion of the circular aperture is ~ Illust
.
if ~
A
1.22D
~.
[25.10]
g~where
D is the diameter of the aperture. Note that Equation 25.10 is similar to 25.9 except for the factor 1.22, \-vhich arises from a complex mathematical analysis of diffractioll frolll a circular aperture.
~ ~EquaLion
FIGURE 25.14 Th(' dilTr
APPLYING PHYSICS 25.2
QUICK QUIZ 25.2 Suppose you <:Irc observing a binary star with a telescope and are having difficulty resolving the two stars. \Vhich color ftlter will better help resolve the stars? (a) blue (b) red (c) neither because colored fi Iters have no effect on resolution
CAT'S EYES
Cats' eyes have vertical pupils in dim light. \Vhich would cats he 1110st successful at resolving at night, headlights on a distant car or vertically separated running lights on a distant boat's masl having the same separation as the car's headlights?
Explanation The dTectivc slit width in the vertical direction of the cat's eye is larger than that in the horizontal direction. Thus, it has more resolving power for lights separated in the vertical direction and would be more effective at resolving the l11asl.lighl.s on the boaL
25.6
EXAMPLE 25.6 Goal
Resolution of a Microscope
837
-'
SLUdy limitations on the resolution of a microscope.
Problem Sodium light of wavelength 589 11m is used [Q view ~lll object under a microscope. The aperliire of the objccti\'e has a diameter of 0.90 CI11. (a) Find the limiting angle of resolutioll for this microscope. (b) Using visible light of any wavelength you desire, find the maxilllulll limit of resolUlion for this microscope. (c) \'Vater of index or refraction 1.33 now rills the space between lhe ol~ject and the objecti\·e. \\'hal. erfect would this water have on the resolving power of the microscope, using 58~)-nm light? JI""
Resolution of Single-Slit and Circular Apertures
Strategy Parts (a) and (b) require substitution into Equation 25.10. BeG-lUse t.he wavelength appears in the numerator, violet light, with t.he shortest \'isible wavelength, gives the maximum resolution. In part (c) the only difference is that the wavelength changes 10 A/n, where 11 is lhe index ofref"raction of water.
.........
••••
Solution (a) Find the limiting angle of' resolution for this mICroscope.
589
'J'J ~ 0 ._D = 1.22 ( 0.90 X 10 - III
Substitute into Equation 25.10 to obtain the limiting angle of resolmion:
8.0
X jQ-9 III )
10-' rad
X
(b) Calculate the lnicroscopc's maximum limit of resolution.
To obtain the rnaximull1 resolution, sllbsti1.llte the shortest visible wavelength available, which is violet light, of wavelength 4.0 X lQ
A
() mill
99._J)
=
7
99 ( 4.0 X 10- III ) 1.__ <> 0.90 X 10 - III
5.4 X 10- 5 rad
"",,'hat effect docs \",',Her bel.wec.;n the object and the have on the resolution, with 589-11 m
ol~jecl.ive lens
lif\hl'
Calculate the wavelength oft.he sodiulIllighl in the w~lter:
Subsl il.UIC this wavelength into Equal ion 25.10 to get the resolut.ion:
.................
589 nm
443
:--::-:- =
1.33
1.22 (443 0.90
X X
I1Ill
1O-~, III ) ~ 10 - III
6.0 X 10-5 rad
................................... ~
Remarks In each case any t\VO point.s on t.he object. subt.ending an angle of' less than the limiting angle Hillin at the objeci ive cannOl be dislinguished in the irnage. Consequently, it may be possible to see a cell bUlt.hen be unable to clearly see smaller Sl ructures within the cell. Obtaining an increase in resolmion is the motivation behind placing a drop of oil on the slide for certain ol~jcctive lenses.
QUESTION 25.6 Does having t\.. .o eyes instead of one improve the hllman ability to resolve distant widely spaced eyes increase visual resolving power? Explain.
ol.:~jeet.s?
In general, would more
EXERCISE 25.6 Suppose oil \..· ith n = 1.50 fills the space bet.ween t.he object and the objective for this 111icroscape. Calculate the lin1iting angle ()mill rar sodium light. or wavclenglh 589 nm in air. Answer
5.3 XW-·-, rad
838
Chapter 25
Opticollnstruments
EXAMPLE 25.7 Goal
Resolvin Craters on the Moon
Calculate the resolution of a telescope.
Problem The Hubble Space Telescope has an apenure of diameter 2.40 Ill. (a) \·Vhat is its limiting angle of resolution at a wavelength of 6.00 X 102 11m? (b) ',\]hat's the smallest crater it could resolve on the Moon? (The 1\1.o011'S dislance from Earth is 3.84 X .10 8 m.) Strategy After substituting into Equation 25.10 to find the limiting angle, lise s = TO to compute the minimum size of craLcr that call be resolved. ,....
,
...,
"
Solution (a) \Vhat is the limiting angle of resolution at a wavelength of6.00 X 10'nm? Substitute D = 2.40 m and A = 6.00 Equation 25.10:
x 10-; III into
()min
= 1.22
~ -_ 1.22 (6.00 X 10D
7Ill)
2.40 m
3.05 X 10- 7 rad (b) \"'hat's the srnallesl lunar crater the Hubble Space Telescope can resolve? The twO opposite sides of the crater must subtend the minimum angle. Use the arc leng[h formula: ~
,
.
s = .. 8 = (3.84 X 10" 1ll)(3.05 X 10-; rad)
.
117m ~
Remarks The distance is so great and the angle so small that using the arc length of a circle isjustified because the circular arc is very nearly a straight line. The Hubble Space Telescope has produced several gigabytes of data e\'ery day since it first began operation. QUESTION 25.7 Is the resolution of a telescope betLcr at the red end of the visible spectrum or the violet end? EXERCISE 25.7 The Hale telescope on Mount Palomar has a diameter of 5.08 In (200 in.). (a) Find the limiting angle of resolution for a wavelength of6.00 X 10 2 nrn. (b) Calculate the smallest crllter diameter the telescope can resolve on the Moon. (c) The answers appear better than what the Hubble can achieve. \Vhy are the answers misleading?
Answers (a) 1.44 X 1O~7 rad (b) 55.3 m (c) Allhough the numbers are beu:er than Hubble's, the Hale telescope must contend with the effects of mmospheric turbulence, so the smaller space-based telescope actually obtains far bctter results.
IL's intercsting to compare the resolution of thc Hale telescope with that of a large radio telescope, such as the system at Arecibo, Puerto Rico, which has a diameter of 1 000 ft (305 m). This telescope detects radio waves at a wavelength of 0.7.:S m. The corresponding minimum angle of resolution can be calculated as 3.0 X 10-3 racl (10 min 19 s of arc), w'hich is more than 1O 000 times larger than the calculated minirllum angle [or the Hale telescope. W'ith such relatively poor resolution, \vhy is Arecibo considered a valuable astronomical inst.rument? Unlike its optical counterparts, Arecibo can see through clouds of dust. The center of our Milky "Vay galaxy is obscured by such dust clouds, which absorb and scatter visible light. Radio waves easily penetrate the clouds, so radio telescopes allow direct observations of the galactic core.
Resolving Power of the Diffraction Grating The diffraction grating studied in Chapter 24 is mOSt useful for making accurate wavelength measurements. Like the prism, it can be used to disperse a spectrum into its components. or the two dC"ices, the grating is better suited to distinguish-
25.6
Resolution of Single-Slii and Circular Apertures
839
ing between twO closely spaced wavelengths. We say thalthe grating spectrometer has a higher H'.wlution than the prism spectrometer. If AI and A2
R= Nm
[25.12]
~
Resolving power of a grating
So, the resolving power R increases with the ordtT number", and is large for a grating with a great numbt:l· of illuminated slits. Note that [or III = 0, R = 0, which signifies that all wavflengths are indistinguisho.ble for the zeroth-order maximum. (All wavelengths fall at the same point on the screen.) Consider, hmvcver, the second-order diffraction pl\ttern of a grating that has 5000 rulings illuminated by the light source. The resolving power of such a gratillg in second order is H = 5 000 X 2 = 10000. Therefore, the minimum wavelength separation between two spectral lines that can bcjust resolved. assuming a mean \'vavelength or 600 nm, is calculated from Equation 25.12 LO be l.iA = AIR= 6 X lO-~ nm. For the thirdorder principal maximum, R = 15000 and 6..\ = II X 10 ~ nm, and so on.
EXAMPLE 25.8 Goal
light from Sodium Atoms
Find the necessary resolving power to distil\guish spectral lines.
Problem Two bright lines in the spectrum of sodium have \\:avelengths of 589.00 11m and 589.59 nm, respectively. (a) \,Vhat Illust the resolving power ofa grating be so as LO distinguish these wavelengths? (b) To resolve these lines in the second-order spectrum, how man)' lines ol"l.he grating must be il1l1minated? Strategy This problem requires little more lhl\n substilliting ilHO Equations 25.n and 25.12 . ..........
• • • • • • • • • • • • • • '""'11I
Solution (a) What must the resolving power ora grating be in order to distinguish the given wavelengths?
(b) To resolve these tines in the second-order spectrum, how many lines of the grating must be illuminated? Solve Equation 25.12 for N,,111d substitute:
Ii N = m
~
1.0 X 10" 2
.:.::'--'-'--'-:...
5.0 X j 0' lines
......
Iro.. • • • • • •
Remark
~
The ability to resolve spectral lines is particularly important in experimental atomic physics.
QUESTION 25.8 True or False: II' two diffraction gratings are identical except for the number of" lines, the grating with the larger Ilumber of lines has the greater resolving power. EXERCISE 25.8 Due to a phenornellon called electron spin, when the lines OL1 spectrurn
840
Chapter 25
Optical Instruments
having wavelengths of 656.272 11m and 656.285 nm. (a) \Vhallllusl be the resolving pmver of a grating so as to distinguish these wave1cng[}ls? (b) How many lines of the grating must be illuminat.ed to resolve these lines in the thirdorder spectrum? Answers
(a) 5.0 X 10'
(b) 1.7 X 10·' lines
25.7
THE MICHELSON INTERFEROMETER
The Michelson interferometer is an optical instrument having great scientific importance. Invented by American physicist A. A. Michelson (1852-1931), it is an ingenious device that splits a light beam into two parts and then recombines them to form an interference pattern. The interferometer is used to make accurate length measurements. Active Figure 25.15 is a schematic diagram of an interferometer. A beam of light proyided by a monochromatic source is split into two rays by a partially silvered mirror IVI inclined at an angle of 45 0 rebuive LO the incident. light beam. One ray is reflected vertically upward to mirror M J , and the other ray is transmitted horizontally through mirror M t.o mirror M:!. I-Ience, the two rays travel separat.e paths, L J and '--'2' After reflcCling from mirrors M I and M2 , the two rays eventually recombine to produce an interference pattern, which can be viewed through a telescope. The glass plate P, equal in thickness to mirror M, is placed in the path of the horizontal ray to ensure that the two rays travel the same distance through glass. The interference paltern for the two rays is determined by the difference in their path lengths.VVhen the two rays are vie'.. . ed as shown, the image ofM z is at 1\>1;, parallel to MI' Hence, the space between i\tl; and M 1 forms the equivalent ofa parallel air film. The effective thickness of the air film is varied by using a finely threaded screw to move mirror M J in the direction indicaled b)' the arrows in Active Figure 25.J5.lfone of the mirrors is tipped slightly with respect to the other, the thin film between the two is wedge shaped and an interference pattern consisting of parallel fringes is set up, as described in Example 24.4. NO'.. . SllPpose we [OCLIS on one of the dark lines with the crosshairs ofa telescope. As mirror 1·1 J is moved LO lengthen the path L JJ the thickness of the wedge increases. ',Vhen the thickness increases by A/4, the destruClive interference that initiall)' produced the dark fringe has changed to constructive interference, and we now observe a bright fringe at the location of the crosshairs. The term fringe shift is used to describe the change in a fringe from dark Lo light or from light to dark. Successive light and dark fringes are formcd each time tvl J is moved a distance of A/4. The \\ravelcngth of light can be measured by counting the number of fringe shifts for a measured displacement of M J' Conversely, if the wavelength is accurately known (as ,.. . ith a laser beam), the mirror displacement can be cleter-
~II ~t Adjustable mirror
ACTIVE FIGURE 25.15 A diagram of I he Midwlson ililerfcrOlllcler. A single beam is split i11lo two rays by the half·~il\'ered mirror ~1. The p,lrll difference between I he two rays i~ varied with lhe adjuslable mirror .\1 1,
~·12 i=...1
t
1. 1
Light
source~
__
Image oD1 2
L M
;;
.ll''-.....;......?/n-L _ .....-
h
Beam spliuer
Telescope
Summary
841
mined to \,,'ithin a fraction of the \vavclength, Because the intcrferometer can measure displacements precisely, it is oftcn used to make highly accurate measurements of the dimensions of mecha n ical components. If the mirrors are perfectly aligned rather than tipped with respect to each other, the path difTercnce differs slightly (01- different angles of vicw. This <:1lTangement results in an interference pauern that resembles Newton's rings, The pattern can be used in a fashion similar to that for tipped mirrors. An observer pays attention to the center spot in the interference pattern. For example, suppose the spot is initially dark, indicating that destructive interference is occurring. If M] is no\\' moved a distance of A/4, this central spot changes to a light region, corresponding to a fringe shift.
SUMMARY 25.1
The Camera
The light-concentrating power of a lens 0[" local length 1" and diameter D is determined by theJ-number, defincd as . Iler ~ j-nuill
/
0
III
[25.1]
The smaller the fnul11ber of a lens, the brighter the imagc formed.
25.2
25.5
=-
1, C
5 m ) /
The Telescope
.r;
m=j;
8
111
=80
25.6
Resolution of Single-Slit and Circular Apertures
Two images arc said to be just resolved ,""hen the central maximum of the diffraclion panern for one image falls on the fjr~l minimum or the other image, This limiting condition of resolution is kno\'v"n as Rayleigh's criterion. The limiting angle of resolution for a slit of width a is A
fl min
[25.2J
where () is the angle subtended by an object at the eye with a lens in use and ()o is the angle subtended by the object when it is placed at the near point of the eye and no lens is used. The maximum angular magnification of a lens is IJI rnax
=
1
25 cm
+ -/--
[25.5J
== -;;
m=---
/ 25.4
[25.6]
The Compound Microscope
The overall magnification of a compound microscope of length L is the product of the magnification produced by
[25.9]
The limiting angle of resolution ora circular aperture is Omin
=
A J.22 D
[25.10]
where D is the diameter of the aperture. If AI and '\2 are two nearly equal wavelengths between which a grating spectrometer can jusL barely distinguish, the resolving power R of the grating is defined as
\-\-'hen the eye is relaxed, the angular magnification is
25 em
[25.8]
where.£ is the f'ocallength of the objective and It is the focal length of the eyepiece.
The Simple Magnifier
The angular magnification of a lens is defined as
[25.7J
The angular magnification of a telescope is
The Eye
Hyperopia (farsightedness) is a defect. of the e}'e that occurs either when the eyeball is too shon or when the ciliary muscle cannOt change the shape of" the lens enough to form a properly focLlsed image. Myopia (nearsightedness) occurs either when the eye is longer than norm<\\ or when the maximum focal length of the lens is insufficient to produce a clearly ["ocused image on the retina. The power of a lens in diopters is the invcrse of the focal length in meters.
25.3
the objective, orfocallengthfo, and the magnification produced by the eyepiece, of focal length j,.:
[25.IIJ where A == A1 == A2 and u,\ = A2 - AI' The resolving power ofa difTraction grating in the Ullh order is
R= N'm
[25.12]
wherc N is the number of illuminated rulings on the grating.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
842
Chapter 25
Optical Instruments
MULTIPLE·CHOICE QUESTIONS l. ''\Thy is it advantageoLls 10 lise a large-diameter objective lens in a telescope? (a) It diffracts the light more
effectively than smaller-dialllcccr objective lenses. (b) It increases its magnification. (c) It increases its resolution. (d) It enables yOll to see morc objects in the field of view. (c) It rcflcns unwanted wavelengths. 2. A Jens has a focal length or 25 CIll. "Vhat is the power of the lens? (a) 2.0 '!iopters (b) 4.0 diopter, (e) 6.0 dioptel'S Cd) 8.0 dioplcrs (e) nUllc of these 3.lf you increase the aperture diameter of a camera by a factor of 3, hm.. . is the intensity of the lighl striking the 111m affected? (a) It increases by a faclor or 3. (b) It decreases by a factor of 3. (c) It increases by a factor of 9. (d) It decreases by a faclor of 9. (c) Incrcasing the apenure doesn't affeCI the ilHensily.
4. If a mall has eyes that arc shorter than normaL how is his vision affected and how can it be correet:ed? (a) The man is Llrsightcd (hyperopia), and his vision can be conccted with a diverging lens. (b) The man is nearsighrcd (myopi'l), and his vision can be corrccted with a diverging lens. (c) The man is farsighted, and his vision can be corrected with a converging lens. (d) The man is ncarsig-htcd, and his vision can be conceted \.. . ith a converging lens. (e) The man's vision is nor corrcctible.
5.U a woman's eyes are longer than normal, how is her vision affected and how can her vision be corrected? (a) The \\lOlllan is farsighted (hypcropia), and her vision can be corrected with a diverging lens. (b) The wOlllan is nearsighted (myopia), and her vision can be cOITeCl.cd with a diverging lens. (c) The wOlllan is far-
sighted, and her vision can be corrected with a cOlwerginf.{ lens. (d) The wOlllan is nearsighted, and her vision can be cOITcetcc! with a converging lells. (c) The woman's vision is not corrcctible. 6. A telescope can be used as a camera if the 111m is placed al the image position of the ol~jcctive lens (or mirror); so the object ivc lens becomes 1he lens of the camera. If a 2.00-m f()Cal length objecti\·e lens 01" a telescope is llsed to photograph the image of two stars that arc 20.0 mill apart on the film, what is the angular separalioll or the two sLars' (a) 0.010' (b) 0.10' (e) 0.286' (d) 0.573' (e) 0.353' 7. The near poilll of a given individual's cye is 55 cm. What focal length corrective lens should be prescribed so dun an O~jCCl can be clcarly seen when placed 25 cm in from of" the eye? (a) -25 Clll (b) 17 cm (c) 46 cm (d) ~2 cm (e) 54 Clll
8. A compound microscope has objective and eyepiece lenses offocallcngths 0.80 CIll and4.0 cm, respectively. II" the microscope length is 15 cm, what is the approximate magnification of' the microscope? (a) 3.2 (b) 6.3 (e) 48 (d) 120 (e) 160 9. Estimate the size aperture required of (I lelescope to resolve nvo sIal'S [hat arc 200 Iy from Earth and orbiting within 10-:' Iy of each other. (a) 10 cm (b) I III (c) 10 In (d) 50 In (e) 100 In 10.;\ diffraction grating Illust distinguish wavelengths that are 0.1 nm apan in the visible bam!' \Vhat minimum resolving- power must the grating have? (a) 3000 (b) 4 000 (e) 5 000 (d) 6 000 (e) 7 000
CONCEPTUAL QUESTIONS ~
.~
-
I. A lens is used 1.0 examine an objecl across a room. Is
the lens probably being used as a simple magnifier? 2. V\Thy is it difficllit or impossible to on an ol~jeet across a room?
fOCllS
a lnicroscopc
3.11 The
opric nerve and the brain invert the image formed on the retina. "Vhy don't we see cverything upside down?
4.lfyou wantLO examine the fine detail of
7. Large telescopes are usually reflecting rather than refracting. List some reasons for this choice. 8. If"you',Nantto use a converging lens to set f'ire LO a piece of paper, why should the light SOlUTe be farther from the lens than its focal point? 9. Explain why it is theoretically impossible [0 see all ol~ject as small as an atom regardless of" the quality of' t.he light microscope being used. 10. Which is most irnpon;ll"It in Ihe use ofa ClIllera photoflash unit, the intcnsity of'the light (the energy per unit area per unit time) or the product of the intensity and the time ol"the flash, assuming the time is less than the shutler speed? JI.ll A patient has a ncar point of" 1.25 m. Is she ncar-
sigllled or farsighted? Should the corrective lens be converging or diverging? 12. A lens with a certain power is used as a simple magnifier. If the pm\ier of the lens is doubled. does the angular magnification increase or decrease?
Problems 13. During LA.SI K eye surgery (laser-assisted in situ keratomileusis), the shape of the cornea is modified by vaporizing some of its material. II' the surgery is per-
843
formed to correct for nearsightedness, how does the cornea need 10 be reshaped?
PROBLEMS ~
The Problcms for lhis chaplcr may be assigllc.:d onlinc al \'VebAssign. straightforward, inlermt:diaL~,(h~dlt:lIgill).:
WebAssign
1,~,:1 =
III = denoles gui(lL:d problem m:I = denOles enhanced conlenl
problem
II = biomedical applic,llion
0= denolcs full "ohllion available in Student SolutioTis ,Haul/fI!!
SECTION 25.2 THE EYE 9. " Thl: near poil1l of a person's cye is 60.0 cm. To see oqjects clearly at a distance of 25.0 cm, what should be lhe focal length and power or the appropriate corrcctive lens? (l"<eglecl the distance from the lens to t he eye.) 10.
Slud)' Guide
SECTION 25.1 THE CAMERA I. A lens has a focal Icngth of" 28 em and a diameter of" 4.0 em. ,,,That is thef-numbcr of the lens?
2. A certain camera Ius f-numbcrs thm range from 1.2 to 22. If the focal length of the lens is 55 mm, what is the range of apenure diameters for tbe camera? [[JA photographic image of a building is 0.092 0 m high. The image was made ""'ith a lens with a focal length of !'>2.0 mlTI. If the lens was 100 m from the building when the photograph was made, determine [hc heiglu of the building. 4. The image area of a typical 35 mm slide is 23.,~ mm by 35.0 mm. If a camera's lens has a focal length of 55.0 mm and forms an image of the constellation Orion, which is 20 c across, will the full image fit on a 35-mm slide? 5. A camera is being lIsed with the correct exposure at an fnumbcr of '1.0 and a shutler speed of 1/32 s. Tu "SlOp" a fast-moving sul:~ject, the shutter speed is changed to 1/2,~6 s. Find the new f-number that should be used to maintain satisfactory exposure, assuming no change in lighting conditions. 6. (a) esc conceplllal arguments to show that the intensit), of light (energy per unit area per unit time) reaching the film in a camera is propor[ional to the square of lhe reciprocal of the f-number as I
Ix - - -
(pO)'
(b) Thc correct exposure time for a camera set to f/1.8 is (1/500) s. Calculate the correct exposure time if the f-number is changed to f/4 under the same lighting conditions. Noll': "f/4," on a camera, means "an f-numbcr of 4:'
rnA certain type offilm requires an exposure t.ime of 0.010 s with an j/ll lens sctting. Another type of film "equires twice the light energy to produce the same level of exposure. What fnumber does the second type of film need with the 0.010-s exposure timc? 8. A certain camera lens has a focal length of 175 mill. Its posit.ion can be adjusted to produce images \"'hen the lens is between 180 mm and 210 mm from the plane of the film. Over what range of object. distances is the lens lIsef"ul?
III
Ell A paticlll can't see objects closer than 40.0 cm and wishes to clearly sce objects that are 20.0 cm rrom his ere. (a) Is the patient nearsighted or farsightcd? (b) If the eye-lens distance is 2.00 em, what is the minimulll object distance f) f"rom thc lens? (c) ""hat image position with respect to the lens will allow the patientlo see the object? (d) Is lhe imClgc real or virtual? Is the image distance q positive or negati\'c? (e) Calculate the required focal length. (f) Find the power of the lens in diopters. (g) If a cOlltClCllens is to bc prescribed instead, find /1, 'I, andl, and the power of dIe lens.
o:::IJllJI
The accollllllodation limits for Nearsighted Nick's eyes arc 18.0 Clll and 80.0 cm. When he wears his glasses, hc is able 1.0 see faraway objects clearly. At what minimum distance is he nble to see objects clearly?
12.
II
A certain child's near point is 10.0 cm; her far point (with eyes relaxed) is 125 Clll. Each eye lens is 2.00 em from the retina. (a) Between v,,'hat limits, measured in diopters, does the power of this lens-cornea combination var),? (b) Calculat.e the power or the eyeglass lens the child should lise for relaxed distance \·ision. Is t.he lens converging or diverging?
13.11I An individual is nearsighted; his ncar point is 13.0 cm and his far point is 50.0 COl. (a) What lens powel' is needed to correct his nearsightedness? (b) \Vhen the lenses are in use, what is this person's near point?
aD. DiiJ A patient has
a near point of 45.0 cm and far poim of85.0 em. (a) Can a single lens correct the patient's vision? Explain the patient's options. (b) Calculate the power lens needed 1.0 correct the near point so that the patient can see objects 25.0 cm away. Neglect the eye-lens distance. (c) Calculate the power lens needed to correct the patient's rar point, again neglecting the eyc-Iens distance.
15.1lJI An artificial lens is implanted in a person's eyc to replace a diseased lens. The distance betv,,'een the artificial lens and the retin(\ is 2.80 cm. In the absence of the lens, an image of a distant ol~jeet (formed by refraction at the cornea) falls 2.53 em behind the retina. The lens is designed 1.0 put the image of the distant object on the rctina. What is [hc power of the implanted lens? Hint: Consider (he image formed by the cornea to be a virtual object. 16.11 A person is to be fi([ed with bifocals. She can see clearly when the object is between 30 em and 1.5 III from
844
Chapter 25
Optical Instruments
the eye. (a) The upper portions of the hifoGlls (Fig. P25.16) should be designed to enable her to sec distam o~jecls clearly. What power should they have? (h) The lower portions of the hilueals should enable her \0 sec ObjcCls comfonably at 25 em. What power should ther hayc?
Leaf
t-------H--------I--Cr::-· I~.----r·I·~-~ 71.0em
1.26m
FIGURE P25.23
24. A lens having a rocal length of 2.» cm is used as a simpic m
SECTION 25.4 THE COMPOUND MICROSCOPE 17. " A nearsighted woman can't seC' objects clearly beyond 40.0 em (her far poi Ill). II" she has no astig'matism and CQ11lacl lenses ;Irc prescribed, what power and type of lens (Ire required to correct her vision?
am
18. " A person sees clearly wearing eyeglasses that have a power of ~4.00 diopLcrs when the lenses arc 2.00 em in front of the eyes. (a) What is the focal length or the lens? (b) Is the person nearsighted or Ltrsightcd? (e) If [he pnson w;lllls to switch to contact lenses placed dirccdy on the eyes, wh,ulcns power should be prescribed?
SECTION 25.3 THE SIMPLE MAGNIFIER 19...-\. stamp colleclor uses a lens with 7.5-cm focal length as a simple magnifier. The virtual illlage is produced allhe normal near point (21') em). (a) 1-10\\1- ron rrom the lens should the stamp be placed? (b) \"fhm is the expected angular magnification? 20. A magnifier ha.'> a maximum an~ular magnification or +6.0. Find the (a) focal length or the len~ and (b) angular m,q:~nification when lhe eye is relaxed.
21. A biology stude lit uses a simple lI1agnifier [0 examine the sU'uctural features or an insect's wing. The wing is held 3.50 em in front of the lens, and the image is rormed 25.0 Clll [rom the eye. (a) \''''hat is the roeallength or the lens? (b) \'Vhat angular magniflcat ion is achieved? 22. Ajewclcr's lells offocallength 5.0 cm is used as a magnifier. \"Iith the lens held ncar the eye, determine (a) the angular magnification when the object is at the rocal point or the bls and (b) the anglilar magnificat ion when the image formed by the lens is at the ncar point or the eye (25 cm). (c) \\ThaL is the ol~ject distance giving the m
SECTION 25.5 THE TELESCOPE 25. The desired overall magnification ora compound microscope is 140X. The objective alone produces a lateral magnification or 12X. Determine the rcquired focallcngLh of [he eyepiece. 26. The distance bct\.. .c(;n the eyepiece and the objc([ive lens in a certain compound rnil:roscope is 20.0 cm. The focal length or the objcctive is 0.500 CJll, and that of the eyepiece is 1.70 Clll. Find the overall magnific'llion or the III icroscope. 27. Your instructor asks rOll to dcsign and constnleta refracting telescope having a magllif-icatioll or 45 using an available cyepiel:c ofrocallength 4.0 cm. How long should }'Oll make the telescope lube? A microscope has an ol~jectivc lens with a focal or 16.22 mill and an eyepiece with a local length or 9.50 nllll. \Vith the length of the barrel sct at 29.0 em, the diamet(;r ofa rcd blood cell's image subtends an angle of 1.43 mrad \\'i1.h the eye. If the linal image distance is 29.0 em from the eyepiece, whar is the actual diameter of the red blood cell?
128.111
len~th
29. The length of a microscope tube is 15.0 Clll. The focal length or the ol:~jecti"e is 1.00 cm,
wilen ac!illstcd ror viewing a dismnl. ol~ject \vith minimum eyestrain. The angular magnification produced b)' lIlt: telescope is 45. COll1pUle the rocallength or each lens.
Problems
32.}\ certain tele'icope ha:-. an ol~jeclive 01" fOGll length I 500 cm. If the t\toon is used as an ohject. it I.O-cm-lollf.{ ill1;1f4(' fOI med by the objective corre:.pom!:-. to what di:--tallce. in miles. on the l\.foon? Assullw ~.8 X 10~ III for the Eanh-t\lllOIl distance.
33.
~4.
E
,htronolllt.:rs often take phoLO~raphs\... ith the ohjective lens 01' mirror of a telescope alone. withollt an evepiece, (a) Show that the image size It' for a lelesnlpe used in thi" manner is gi\'en by II' = III/(J - Pl. where" j-; the object si/e. li-:-. the objective fOGd length, and!J i:. the object distanet:, (b) Simplify the cxpr('!:lsion in part (a) if the object distance is much gre'ltCl" than lhe objectivc focal length. (c) The -wingspan~ olthl' Internatiollal Spac(' Station is lO8.6 m. the o\'erall width of ilS ..olar panel configuration. When it is orhiting at all altillide 01 107 "'m. lind the width of the image formed by a teleSUlpe ohjective of fm'allength '1.00 m. An elderly sailor is shipwrecked on a (lcsl,rt i"iland, but manages to S
135.IA
person dccide~ 1O lise an old pair of ('yep;la~ses to Illake sorne optical instruments. He knows lhat the near point in hi\ left eyc is 50.0 ern and the near poilll in his rig-Ill C)e is 100 Clll. (a) ,""'hm is the maximum angular magnification 1l(' Gin prodllCl' in ;-1 telescope? (b) [f'Ill: p1
:~6.
EiJ
Galileo devised a simple tcrrestrial telescope that produces an upright image. It consists of a cOll\erging objective lens and a diverging eyepiece at oppo:--itt' cnds of the IClcscope LUbe. For distant ol~iects, I,he lUbe lell!"ph is the objective focal length Ic~s the ab"olllie vallie of the e\epi('ce focal length. (a) Docs the uscI' of the telest.-ope st'e a real or virtual image? (b) \-"here i.., tht, final illlage~ (c) If a telescope is to be constnKtcd \... ith a tube of length 10.0 Clll and a magnification of 3.00, what arc the 100:al kngths of the objective and C}epicce?
SECTION 25.6 RESOLUTION OF SINGLE-SLIT AND CIRCULAR APERTURES 37. A cOllverging lens \\'ith ~ diameter of ~\O.O cm forlll\ an image ofa satellile passing m'crhead. The satellite has two green light ... (wavelength 500 nm) ~paccd 1.00 III apart. If lhe li~hts canjllsl be re...olved according 10 thc Ra\leigh criterion. ",hat is the aitilllde of the satdl ite? 38. The pupil of a cat's ere narrow... to a vntical slil of widlh 0.:,00 mm in daylight. What i.., the angula,' rl.'~ollltion for a pair ofhorizomall)' separated mice? (USl' !lOO-nm lighl in rOlll calculation.) 39. To increase the resolving power of a microscope, the ol~eCl and the objcctive arc imn"!ersed ill oil (11= 1.5). I" I hc limil ing angle of resolution \\ ithout lilt..' oil i" 0.60 fLrad. whal is the limiting ,lIlg1e of resolution with t he oil? Hint: The oil changes the wavekngl hoI' lhe light.
40.
845
n (a)
Calculate the limiting angle of resolution for tile eye. a'isllming a pupil diameter 01'2.00 mm, a wavclength or 500 llm in air, and an index of refraction for the cye of 1.33. (b) \'Vhat is the maximum distance from the eye at which two poi Ill... separated by 1.00 em could be resolvedr
Iill.\ \ehicle with headlights ~eparated b} 2.00 m approaches an ohserver holding an infrared detector sensitive to radiation ofw
42. :\ helium-neon laser emits liglH thai has a wa\'c1ength of 632.8 nm. The circular apertlll'e through \\hich the beam emerges ha~ a diameter of 0.200 cm. ESLimate the diameter of thc beam 3.00 km frolll the laser. --43, Sllppo~e a 5.00-m-diamcter telescope were constructed on the 1\.'loon, where the ab...cnce of atmospheric dis-
lortion "'ould permit excellcnt ,·icwing. If obscrvation'i were madt' using 500-nm light, what minimum separation hel\""een two object'i cOllldjust be J'esolved on Mars al closest approach (whell 1\.lars is 8.0 X 107 km from lhe f\toon)? 44. A spy satcllite circles Earlh at an
SECTION 25.7 THE MICHELSON INTERFEROMETER 47. Light o[wa\'c1ellglh 550 11m is used to calibrate a rvlichcl· son interferometer. With the use of a micrometer suew, the platform on which olle mirrOt' is mounted is mo\'cd 0.180 mm. I low many fringe shifts are counted? 48.
~Ionochromatic light is beamed into a Michelson interferomeLer. The movable mirror is displaced 0.382 mm. causing lhe C('l1tral spot in the imerferometer' pattern Lo change from brighl 10 dark and hack to bright .v = I 700 times. (
mJ
49. " An interferometer is used to measure the length of a bacterium. The wav('leng'Lh or the light u....;'d is 650 nm, As one
Chapter 25
B46
Optical Instruments
powers because they are at. different distances from the cye.
of the cellw the other, 310 fringe shifts arc coullt.ed. How long is the bllCleriulll? 50. Mirror M] in Active Figure 25.15 is displaced a distance fiL. During this displacemellt, 250 fringe shifts are counted. The light being used has a wavelength of
632.8
11m.
157.111
The neal' point of an eye is 75.0 ern. (a) \Nhat should be the pmvcr of a corrective lens prescribcd 1.0 enable thc erc to see an objcCl deadr at 25.0 em? (b) If, using dte correcti\'c lens, the person can sec an oqject clearly at 26.0 em but not at 25.0 em, by how many diopters did the lens grinder miss the prescription?
Calculate the displaccmcm AI...
51. A thin sheet of transparent material has an index of refraction of 1.40 and is 15.0 J.L1TI thick. When it is inserlcd in the light path along one
58.11 It" a typical eyeball is 2.00 em long and has a pupil opening that can range from about 2.00 mm to 6.00 mm, what are (a) the focal length of the eye when it is focused on objects 1.00 m mvay, (b) the smallest f-number or the cye when il is focused on objects 1.00 m away, and (c) the largest f-number of t.he cyc when it is focused on objects 1.00 m away?
152.IThe Michelson imerfcromclcr can be llsed to measure the index of refraction of a gas by placing an evacuated transparent tube in d1C light path along one arm of the dc,·ice. Fringe shifts occur as the gas is slowly added to the tube. Assume 600-nm light is uscd, the tube is 5.00 em long, and 160 fringe shifts occur as the pressure of the gas in the tube increases to armospheric pressure. \""hat is the index of refraction of the gas? Hin{: The fringe shifts occur because the \.. .;lvclength of the light changes inside the gas-filled tube.
59.
II
60.
II
ADDITIONAL PROBLEMS
53. The Yerkes refracting telescope has a I.OO-m-diametCl" objective lens of focal length 20.0 m. Assume it is L1sed with an eyepiece of focal length 2.50 cm. (a) Determine the magnification oflhe planet rvlars as seen through the telescope. (b) Are the ohserved Martian polar caps right side up or upside down? 54. Estimalc the minimum angle subtended at the eye of a hawk flying at an altillldc of 50 m necessary to recognize a mouse on the ground.
55.
The wavelenglhs of the sodium spectrum are A1 = 589.00 nm and A2 = 589.59 nm. Determine the minimum !lumber of lines in a grating thai will a11m.. . resolution of the sodium spectrum ill (a) tl1<: first order and (b) the third order.
56.
III
A person with a nearsighted eye has near and f~lr points of 16 em and 25 em, respectively. (a) Assuming a Icns is placed 2.0 cm from the eye, \vhat power must the lens have to correct this condition? (b) Suppose contact lenses placed directly on the cornea are lIsed to COITect the pel·son's eyesight. \\'hat is the power of the lens required in this case. and what is the new· near point? Hint: The contact lens anc! lhe eyeglass lens require slightly different
A cataract-impaired lens in an eye may be surgically removed (lnd replaced by (l manufactured lens. The focal length required for t.he ncw lens is determined by the lens-to-retina distance, which is mcasured by a sonarlike device, and by the requirement thai the implant provide for correct distance vision. (a) If the distance from lens to retina is 22.4 mm, calculalx the power of the implanted lens in diopters. (b) Since there is no accommodation and the implant allows for correct distance "ision, a corrective lens for close work or reading must be used. Assumc a reading distance of 33.0 CIll, and calculate the power of the lens in the reading glasses. rf the aqueous humor of the cye has an index of refraction of 1.34 and the distance from the \"ertex of the cornea to the retina is 2.00 cm, what is the radius of curvature of the cornea for which distant objects will be focused on the retina? (For simplicity, assume all refraction occurs in thc aqucous humor.)
ffiLJA Boy Scout starts a fire by using a lens from his eyeglasses to focus sunlighton kindling 5.0 cm from the lens. The Bo~' Scout has a near point of 15 em. Vv'hen the lens is used as a simple magnifier, (a) what is the maximum magnification dlat can be achieved and (b) what is the magnification \,,'hen the eye is relaxed? Cautioll: The equations derived in the text for a simple magnifier assume a Hnormal" eye. 62. A laboratory (astronomical) telescope is used to vic\\' a scale Ihat is 300 em from the objective, which has a [ocal length of' 20.0 em; the eyepiece has a focal length o[ 2.00 cm. Calculate the angular magnification when the telescope is
RELATIVITY Most of our everyday experiences and observations hove to do with objects that move at speeds much less than the speed of light. Newtonian mechanics was formulated to describe
Albert Einstein revolutionized modern physics. He explained the random motion of pollen grains, which proved the existence of atoms, and the photoelectric effect, which showed that light was a particle as well as a wove. With Satyendra Nath Bose he predicted 0 new form of matter, the Bose-Einstein condensate, which was only recently discovered in the laboratory. His theory of special relativity made dear the foundations of space and time, and established a key relationship between mass and energy. His theory of gravitation, general relativity, led to 0 deeper understanding of planetary motion, the structure and evalution of stars, and the expanding universe. The equation in this photo, loosely translated, says that the average curvature of spacetime is zero in empty space.
the motion of such objects, and its formalism is quite successful in describing a wide range of phenomena that occur at low speeds. It foils, however, when applied to particles having speeds approaching that of light. Experimentally, for example, it's possible to accelerate an electron to a speed of O.99c (where c is the speed of light) by using a potential difference of several million volts. According to Newtonian mechanics, if the potential difference is increased by a factor of 4, the electron's kinetic energy is four times greater and its speed should double to 1.98c. Experiments,
however, show that the speed of the electron-as well as the speed of any other particle that has mass-always remains less than the speed of light, regardless of the size of the acceler· ating voltage. The existence of a universal speed limit has far-reaching consequences. It means that the usual concepts of force, momentum, and energy no longer apply for rapidly moving objects. A less obvious consequence is that observers moving at different speeds will measure different time intervals and displacements between the same two events. Relating the measurements made by different observers is the subject of relativity.
26.1
26.1
Galilean Relativity
26.2
The Speed of Light
26.3
Einstein's Principle of Relativity
26.4
Consequences of Special Relativity
26.5
Relativistic Momentum
26.6
Relativistic Energy and the Equivalence of Mass and Energy
26.7
General Relativity
GALILEAN RELATIVITY
To describe a physical event, it's necessary to choose afm11l(> of refere-no'. For example, when you perfor111 an experimcm in a laboratory, you select a coordinate s)'stem, or frame of reference, that is at rest with respect to 1he Iabo.-
847
848
Chapter 26
Relativity
FIGURE 26.1 (a) The observer on lhe airplane sees Lhe ball lnm"c in a yenical palh when lhrO\\'n upward. (b) The observer on Earth vie\l's lilt: palh oflhe ball to be a parabola.
(0)
(b)
those reference frames in which Ne\.. . ton's laws arc valid. In these frames, objcns move in straight lines at constant speed unless acted on by a nOll-zero net force, thus the name "inertial frame" because objects observed from these frames obey Ne\,tton's first law, the la" . of inenia. For the situation described in the previous paragraph, the laboratory coordinate system and the coordinate system of the moving car arc both inertial frames of reference. Consequently, if the laws of mechanics arc found to be t.rue in the laboratory, then the person in the car must also observe the same laws, Consider an airplane in flight, moving with a constant velocity, as in Figure 26.1a. If a passenger in the airplane throws a ball straight up in the air, the passenger observcs that the ball moves in a vertical path. The mOlion of the ball is precisely the same as it 'would be if the ball \vere thrO\vn \'\'hile at rest on Earth. The law of gravity and the equations of motion under constant acceleration are obeyed whethcr the airplane is at rest or in uniform rnotion. Now consider the same experiment when vie\ved by another observer at rest on Eart.h. This st.ationary observer views the path of the ball in the plane to be a parabola, as in Figure 26.1b. Further, according to this observer, the ball has a velocity to the right equal to the velocity of the plane. Although the two observers disagree on the shape of the ball's path, both agree that the motion of the ball obeys the law of gravity and ~ewton's laws of motion, and they even agree on hm...· long the ball is in the air. \Ve draw the following important conclusion: There isno preferred frame of reference for describing the laws of mechanics.
26.2
THE SPEED OF LIGHT
It's natural to ask whether the concept of Galilean relativity in mechanics also applies 1.0 experiments in electricity, magnetism, optics, and other areas. Experiments indicate that the answer is no. Further, ir we assume the la\,·/s of electricity and magnetism are the same in all inertial frames, a paradox concerning the speed of light immediately arises. According to electromagnetic theory, the speed or light always has the fixed value of 2.99792458 X 10H m/s in free space. According to Galilean relativity, however, the speed of the pulse relative to the stationary observer S outside the boxcar in Figure 26.2 should be c + v. Hence, Galilean relativil)' is inconsistent with .Maxwell's well-tested theory of electromagnetism. Elect.romagnetic theory predicts that light \-,'aves must propagate through free space with a speed equal to the speed oflighL. The theory doesn't require the presence ora medium for wave propagation, hO\"¥'ever. This is in contrastlO other types of waves, such as water and sound waves, that do require a medium to support the dislUrbances. I.n the 19th century physicists thought that electromagnetic waves also required a medium to propagate. They proposed that such a medium existed
26.2 The Speed of Light
5'
5
•
•
v
and gave it the name luminiferous ether. The ether was assumed to be presclll everywhere, even in empty space, and light wa,"es were viewed as ether oscillations. Further, the ether would hm"c to be a massless but rigid medium with no effect on the morion of plancLs or other objects. These concepts are indeed strange. In addition, it was found thal the troublesome laws of elcClricity and magnetism would take on their simplest forms in a special frame of reference at rest with respect La the ether. This frame was called the absolute frame. The laws of electricity and magnetism would be ,"aliel in this absolute frame. but lhe)' \,"ould have to be modified in any reference frame 1ll00-ing with respect to the absolute frame. As a result of the importance attached to the ether and the absolute frame, it becarne of considerable interest in physics to prm'e by experiment that they existed. Because it was considered likely that Earth was in motion through the ether, from the view of an experimenter on Earth, there was an "ether wind·' blm...·ing through the laboratory. A direct method lor deteCting the ether wind would use an apparatus fixed to Earth 10 measure the wind's innuence on the speed of light. If v is the speed of the ether relative to Earth, then the speed of light should have its maximum value, (+ V, when propagating dm'iI1wind, as shown in Figure 26.3a. Like""ise, the speed of light should have its minimum value, ,. - v, when propagating upwind, as in Figure 26.3b. and an int.ermediate value, «('2 - y2)1/2. in the direction perpendicular to the ether wind, as in Figure 26.3c. If the Sun were assumed to be at rest in the ether, then the velocity of the ether wind would be equal to the orbital velocity of Earth around the Sun. which has a magnilUde of approximately 3 X 10·' m/s. Because ( = 3 X 108 mis, it should be possible lO detect a change in speed of about I part in 10 1 for measurements in the upwind or downwind direction.
FIGURE 26.2 .-\ pulse 01" light is sent by a person in a mo,·ing boxcar. According to Galilean relati,-ity, U1C ~peed of the pulse should be + \' rclati\'c to a stationary observcr. Ollt
c
•
v
•
c+v (a) Downwind
-
v
c-v
The Michelson-Morley Experiment The most famous experiment designed to deteCt these small changes in the speed of lighl was first perfOl-med in 1881 by Alben A, Michelson (1852-1931) and laler repealed under ,·arious conditions b) Michelson and Edward W. Morley (1838-1923)_ The experiment was designed to determine the velocity of Earth relatiye to the hypothetical ether. The experimental tool used was the Michelson interferometer, shown in Acti,'e Figure 26.4 (page 850). Arm 2 is aligned along the direction of Earth's motion through space. Earth's moving through the ether at speed y is equivalent 1.0 the ether nowing past Earth in the opposite direction with speed v. This ether wind blowing in the direction opposite the direction of Eanh's motion should cause the speed of light measured in the Earth frame to be c - vas the light approaches mirror M2 and (. + v after rerJecrion, where c is the speed of light in the ether frame. The two beams renected from .\1 1 and M2 recombine. and an interference pattern consisting of alternating dark and bright fringes is formed. The int.erference pall.ern was observed while the interferometer ,...as rotated through an angle or 90°. This rotation supposedly would change the speed of the ether wind along the direction of arm 1. The effect of such rotation should have been [Q cause the fri nge pattern to shift slightly but. measurably, but measurements failed to show any change in the intelference pattern! E\-en t.hough Michelson-Morley experiment was repeat.ed
849
(b) Upwind
(c) Across hind FIGURE 26.3 Trthe speed ohhe ether wind relative to Earth is v dnd (is the speed of light relative to lhe elhc:r, the speed oflighl relatil-e 10 Earth is (a) (+ v in the downwind direction, (b) (- v in the IIp\\'ind direction, and (e) V c2 - vI! ill the direction perpendicular to the \'Iilld_
850
Chapter 26
Relativity
ACTIVE FIGURE 26.4 Accurding to till: ether wind lhc()rr, the speed of Ii gill slloukl be c- vas tile bealll
ALBERT EINSTEIN German-American Physicist
Telescope
(1879-19551 One of the greatest physicists of all time, Einstein was born in Ulm, Ger-
many. In 1905 at the age of 26 he published four scientific papers thot revolutionized physics. Two of these
popers introduced the special theory of relativity, considered by many to be
his most important work. In 1916, in on exciting race with mathematician
David Hilbert, Einstein published his theory of gravity, coiled the general theory of relativity. The most dramatic prediction of this theory is the degree to which light is deflected by a gravitational field. Measurements
made by astronomers on bright stors in the vicinity of the eclipsed Sun in
1919 confirmed Einstein's prediction, and as a result Einstein became a warld celebrity_ Einstein was deeply disturbed by the development of quantum mechanics in the 1920s despite his own role as a scientific revolutionary. In particular, he could never accept the probabilistic view of events in nature that is a central feature of quantum theory. The lost few decades of his life were devoted to on unsuccessful search for a unified theory that would combine gravitation and electromagnetism.
Postulates of relativity
~
at different times of the year when the ether wind ,vas expectcd to change direction, the rcsults ",ere ahvays the same: no fringe shift of the magnitude required was ever observed. The negative results of the Michelson-Morley experiment not only cOI1lradicted the ether hypothesis, but also shmved that it was impossible to measure the absolute velocity of Earth with respect to the ether frame. As v·:e will see in the next. section, hmvever, Einstein suggested a postulate in the special theory of relativity that places quite a different interpretation on these negative results. In later years, when more "vas known about the naLUre of light, the idea of an ether that permeates all space \-vas discarded. Light is now understood to be an electromagnetic wave that requires no medium for its propagation.
26.3
EINSTEIN'S PRINCIPLE OF RELATIVITY
In L905 Albert Einstein proposed a theory that explained the result of the Michelson-1\:lorley experiment and completely altered our notions of space and time. He based his special theory of relativity on two postulates: 1. The principle of relativity: All the laws of physics arc the same in all inertial frames. 2. The constancy of the speed of light: The speed of liglu in a vacuum has the same value, ( = 2.99792458 X 108 mis, in all inertial reference frames, regardless of the velocity of the observer or the velocity of the source emitting the light. The first postulate asserTs that all the lav.ls of physics are the same in all reference frames moving with constant velocit.y relative to each other. This postulate is a s\.. .eeping generalization of the principle of Galilean relativity, which refers only to t.he laws of mechanics. From an experimental point of view, Einstein's principle of relativity means that an)' kind of experimenL-mechanical, thermal, optical, or electrical-performed in a laboratory at rest must give the same result when performed in a laboratory moving at. a constant speed past the first one. Hence, no preferred inertial reference frame exists, and it is impossible to detect absolute motion. Although postulate 2 was a brilliant theoretical insight on Einstein's part in 1905, it has since been confirmed experimentally in many ways. Perhaps the most. direct demonst.ration involves mcasuring the speed of photons emitted by particles traveling at 99.99% of the speed oflighL The measured photon speed in (his case agrees to nve significant. figures wilh the speed of light in empty space.
26.4
Consequences of Special Relativity
851
The null result of the Michelson-Morley experiment can be readily understood within the framework of Einstein's theory. According to his principle of relativity, Ihe premises of the Ylichelson-Morley experiment '.vere incorrect. Tn the process of trying to explain the expected results, 'ove stated that when light traveled against the ether wind its speed \.,:;)s c - v. If', hmvevcr, the state of motion of the observer or of the source has no influence on the ,"alue found for the speed of light, the rneasured value must ah,,·ays be c. Like,vise, the light makes the return trip aftcr rcflection from the mirror at a spced of c, not at a speed of (" + v. Thus, the motion of Earth does not influence the fringe pattern observed in the \1ichelson-\1.orley experiment, and a null resulr should be expected. I f we accept Einstein's theory of relativity, we must conclude that uniform relative motion is unimportant when measuring the speed of light. Atthe same time, we have to adjust our commonsense notions of space and time and be prepared for some rather bizarre consequences.
QUICK QUIZ 26.1
True or Falsc: If you wcre traveling in a spaceship at a speed of c/2 relalive to Earth and yOll fired a laser beam in the direction of the spaceship's motion, the light from the laser 'ovould lravel at a speed of 3c/2 relative to Earth.
26.4
CONSEQUENCES OF SPECIAL RELATIVITY
Alrnost everyone who has dabbled even superficially in science is aware of some of" the startling predictions that arise because of Einstein's approach La relaLive
motion. As \'Ie examine some of the consequences of relativity in this seclion, we'll find lhat they conflict with some of our basic notions of space and time. \'\'e restrict our discussion to the concepls of length, time, and simultaneity, which are quite different in relativistic mechanics from what they arc in Newtonian mechanics. For example, in relativislic mechanics the distance belween two points and the time intelTal bel'.. .een t\\'o events depend on the frame of reference in which they are measured. [n relativistic mechanics there is no such thing as absolute length or absolute time. Further, events at different locations that are observed to occur simultaneously in one frame are not observed to be simultaneous in another frame moving uniformly past the first.
f- Length and time measurements depend on the frame of reference
Simultaneity and the Relativity af Time A basic premise of Newtonian mechanics is thal a universal time scale exists that
is the same for all observers. NeWlOn and his fo1lowers simply LOok simultaneity for gTanted. In h is special theory of relativity, Einstein abandoned lhat assltmption. Einstein devised the follmving lhought experiment to illustrate this point. A boxcar mm"es wilh uniform velocity, and t,,,"o lightning bolts strike its ends, as in Figure 26.5a (page 852), leaving marks on the boxcar and the ground. The marks on the boxcar are labeled A' and /J', and those on the ground are labeled A and 15. An observer at 0' moving with the boxcar is midway between A' and 8 ' , and an observer on the ground at 0 is midway between A and B. The events recorded by the observers are the striking of the boxcar by the t\vo lightning bolts. The light signals recording the instant when the two boILS struck reach observer O(ltthe same time, as indicated in Figure 26.5b. This observer realizes that the signals have traveled at the sallle speed over equal distances and so rightly concludes that the events at.4 and B occurred simultaneously. Nm'\' consider the sarne events as vie",·ed by observer 0'. By the time the signals have reached observer O. observer 0 ' has moved as indicated in Figure 26.5b. Thus, the signal from 8 ' has already s,"vept past 0', but the signal from A' has not yet reached 0'. Tn Other words, 0' sees the signal from 8' before seeing the signal from A'. According to Einstein, Ihe li.oo
r-------~\
TIP 26.1 Who's Right? Which persoll is correct concerning the simultaneity of the tWO e\"ents? Both aloe correct hecau<;e the principle of relativit)' stales lhal no inCrlial frame orl'cfer~ ence is prcfcn-cd. Although the two observers rna" reach diffcrcnl conclusions, both' arc COITCCf in their own reference frame. An)' uniformly moving frame of referencc can be used to CkS(Tibe c"cnts and do physics_
852
Chapter 26
Relativity
-
v
v
.•\ ()
rv+-° ..... (Il)
(a)
FIGURE 26.5 'Iwo lij.{hUling: holtssldke the ends ofa IIH,\·jng- hose:l1. (:I) I he l"\Clll~ appear to be simullaI1C()II~ wille sl;lltUna ....' (lb~cl'vcr at O. who is miclwa}' between A .1Ilri It (b) 111l' C;\clltsdon"l appear 10 hI.' :.illltIII,UlI.-·Ull~ to til(' Olht'l"w:r at 0', who cl;lim~ thai till: from of thL" ILliu is strllck lJ,Jorr I he rear.
o!Jlierun:'i 1/ltl.\1 fi lld thaI lighllravels at the sam" spl'i'fl. Therefore, observer 0' concludes that the lightning struck the front of the boxcar before it struck the back.
This thought cxperin1cnt clearl)' demonstrates that the 1,,,'0 CycntS thal appear LO be simultaneous to obsenter Odo not appear 10 be simuILaneolis to absencr 0'. In Olhcr words, T'wo events Ihm are simultaneous in olle refcrence frame arc in general not sirnultancolls in a second frame moving relative to the firsL. Simultaneity depends on the state of motion of the observer and is therefore not an absolutc concept. At this point, rOll mighl \fonder which observer is right concerning the two events. The al1s\-"er is thal/XJil! are correct because the principle orrebtivity slates thaI there is no pref'elTed inertial frame of reference. Although Ihe twO observers reaeh different conclusions, both are con-eet in I heir O\\ln reference frames because the concept of Sil11111lancity is nol absolute. I n fact., this is the central point o[reJativity Any incrtial frame 01" reference can be used 10 describe c,·cnu, and do physics.
Time Dilation
'-"c can illuo;;t.rat.e that observers in different incnial frames may measure differem Iime intervals bet weell a pair of cyents by consideri ng a 'eh ide moving to the right with a speed II as in Active Figure 26_6a. A mirror is fixed to the ceiling of the vehi-
-
v
v
:\Iirror
-'
O'
~~~x'
(a)
o
(h)
(c)
ACTIVE FIGURE 26.6 (a) :\ Illirrvr i~ fixerilO a moving vehicle. il11d a light pulse leaves 0' at rnl ill the \ehicle. (b) ReI,lli,e lO a ~lalinn,tl)- ob~er\(:l 011 Earth, lhe mirror ,md 0' move Wilh a ~pl'(:d II. Note that Lhe dislallce lhe pulse (rawb i.~ gn:;tter 1hall '1ft .l~ measured by the stal innary obscl"\cr. (l) The I ighl 1Liang-Ie ror calculaling lhl' n:lalion:.llip 1>clW('('11 :'\lalld Ill!"
26.4
Consequences of Special Relativity
cle, and an observer 0' at rcst in this system holds a laser a distance d below [he mirror. At sorne instant, the laser emits a pulse of light directed I.Oward the rninor (event 1), and at some later time after reflecting from the mirror, the pulse arrives back at the laser (event 2). Observer 0 1 Glrrics (t clock and uses it 1.0 measure the lillie interval 0.1 , berwcen these two events, \vhich she views as OCCUlTing at the 1 safne place. (The subscript I) stands for jJro/lN, as \\.c'll sec in a moment.) Because the light pulse has a speed t, the Lin1e it takes it LO travel from point A 1.0 the mirror and back to poim /1 is distance
2d
Lr~1\'ded
speed
[26.1]
(
T'he time interval 11/.1! measured by 0' requires only a single clock located ai. tlte same place as the laser in this frame. Now consider the same scI. events as viewed by Oin a second frame, as shown in Active Figure 26.6b. According to this observer, the mirror and laser are mm'ing 1.0 the right with a speed v, and, as a result, I he sequence of events appears differcm. By the timc the light from lhe laser reaches the mirror, the mirror has moved l.0 the right a distance v 111/2, \.. . here 6.1 is the t.illle it takes the light pulse to travel from point A to the mirror and back to poim A as measured b)' 0. (n other words, o concludes that because of' the motion of the vehicle. if the light is to hit the mirror, it must leavc the laser at an ('Ingle with respect to the vertical direction. Comparing Active Figures 26.6a and 26.6b, we see that the light ITllisttravei ranher in (b) than ill (a). (Note that neither observer ;'knows" that he or she is moving. Each is at rest in his ur her O\.... n inenial frame.) According to the second poswlatc of the special theory of' relativity, both observers mllst measure c lor the speed or light. Because the light travels fanher in the frame of 0, it follows that. the time interval (J,,/ measurcd by 0 is longer than lhe time intcrval ill" rneasured by 0'. To obtain a relationship bct\veen these two l.ime intervals, it is convenient 1.0 examine the right triangle shO\vn in Active Figure 26.6c. The Pythagorean theorem gives
or
C:J v:J = (
+
d'
Solving for!:ll yields
2d Ilccause I1tp = 2d/c, we can express this result as
[26.2] where y =
VI -
v'/ "
[26.3]
Because y is ahv'£lys greater t.han ], Equation 26.2 says thaL the time interval at between two events measured by an observe.· moving with respect to a dock I is longer than the time interval alp between the same two events measured by an observer at rest with respect to the clock. Con~equently, (J,,/ > illp' and the proper time interval is expanded or dilaled by the faclor y. Hence, Ihis effect is known as time dilation. For example, suppose the uhserver at rest with respect. to I he clock measures the time required lor the light nash to leave the laser and return. We assume the measured time interval in this frame of reference, !YJ.I1" is I s. (Th is would require a \'ery IAn 1l'llIy. Figun: 26.6 ~IHI\\'~ tIlt" clock mndlll{ ;llld nOlllu: oh_l"nl"l. bUllhat is {'CJlli\;delll tu Ob'lT\'cr 0 modul-; 111l: lefl with velocity v with respect Lo lit,· clock.
to
f- Time dilation
853
854
Chapter 26
Relativity
A clock in motion runs more slowly than an identical
stationary dock
~
;--
TIP 26.2 Proper Time
"'
Jntervol ,"on must be able to corrcn).,.. identify rhe observer who . measurc,~ the proper lime inter\'al. The proper time interval between twO events is the time imerval measured by an observer for whom the two e\'ent~ take place at Ihe same position.
----_/
l'vluon is created
I" 111 6.6 x 10I Muon decays
(al
Muon is created
-l.8 X 103 m
~1UOll
decays (b)
FIGURE26.7 (a) IfnOlfortimedilalion, a milan moving at 0.99, relative 1,0 an observer on Earth would lra\'e! ani) 6.6 x IO~ m. on average. before decaying. (b) Because of time dilalion, the observed lifetime is longer. <;0 such a muon lnl\'cls a distance of aboul4.80 x 10 3 Ill, on average, a~ measured hy an ollscl"\'cr on Earth.
tall vehicle.) Now we find the time interval as measured by observer 0 moving with respect to the same clock, If observer 0 is traveling at half the speed of light (v = 0.500c), then l' = J.15 and, according to Equation 26.2, 1'.1 = l' 1'.lp = 1.15(1.00 s) 1 = L15 s. Therefore, when observer 0 claims that 1.00 s has passed. observer 0 claims thal 1.15 s has passed. Obsel"ver 0 considers the clock of 0' to be reading too Iowa value for the elapsed urne between the two e\'ents and says that the clock of 0 1 is "running slow." From this phenomenon, we may conclude the following: A clock moving past an observer at speed v runs more slowly than an identical clock at rest with respect to the observer by a factor of y-l. The time interval D.lp in Equations 26.1 and 26.2 is called the proper time. In general, proper time is the time interval benveen two events as measured by an observer who sees the events occur at the same position. Although you may have realized it by now, it's important to spell ou[ that relativity is a scientific democracy: the view of 0' that 0 is actually the one moving with speed v to the left and that the clock of 0 is funning more slowly isJUSt as yalid as the view of O. The principle of relativity requires that lhe views of tWO observers in uniform relative motion be equally valid and capable of being checked experimentally. \Ve have seen that moving clocks run slow by a factor of y-I. This is true for ordinary mechanical clocks as well as for the light clock just described. In fact we can generalize these results by stating that all physical processes. including chemical and biological ones, slow down relative to a clock when those processes occur in a frame moving with respect to the clock. For example, the heartbeat of an astronaut n10ving through space \",'ould keep time with a clock inside the spaceship. Both the astronaut's clock and heartbeat vvould be slowed down relative to a clock back on Earth (although the astronaut would have no sensation of life slowing down in the spaceship). Time dilation isa vcry real phenomenon that has been verified byvariousexperiments involving the ticking of natufal clocks. An interesting example of time dilation involves the observation of muons, unstable elementary particles that are very similar to electrons, having the same charge, but 207 times the mass. Muons can be produced by the collision of cosmic radiation with atoms high in the atmosphere. These particles have a lifetime of 2.2 /-LS when measured in a reference fI-ame at rest with respect to them. If we take 2.2 /-LS as the average lifetime of a muon and assume that their speed is close to the speed of light, we find that these particles can travel only about 600 m before they decay (Fig. 26.7a). Hence, they could never reach Earth from the upper atmosphere "..· here they are produced. Experiments, however. show that a large number of muons do reach Earth, and the phenomenon of time dilation explains how. Relative to an observer on Earth, the muons have a lifetime equal to 'YTp' where Tp = 2.2 J1..S is the lifetime in a frame of reference traveling with the muons. For example, for v = 0.99c, y:::::::: 7.1 and yTp = 16/1-s. Hence, the average distance muons travel as measured by an observer on Earth is YUlp = 480001, as indicated in Figure 26.7b. Consequently, muons can reach Earth's surface. In J976 expel-iments with muons were conducted at the laborarory of the European Council for )\uclear Research (CERN) in Geneva. Muons were injected into a large storage ring, reaching speeds of about 0.999 4c. EleCLrons produced by the decaying muons were detected by counters around the ring, enabling scientists to measure the decay rate and hence the lifetime of the muons. The lifetime of the moving muons '\.'as measured to be about 30 times as long as that ofstationary muons to within (\\'0 parts in a thousand, in agreement with the prediction of relativity. QUICK QUIZ 26.2 Suppose you're an astronaut being paid according to the time you spend traveling in space. You take a long voyage traveling at a speed near that of light. Upon your return to Earth, you're asked how you'd like to be paid: according to the time elapsed on a clock on Earth or
26.4
Consequences of Special Relativity
855
according to your ship's clock. To maximize your paycheck, which should you choose? (a) the Earth clock (b) the ship's clock (c) either clock because it doesn't make a difference
EXAMPLE 26.1 Goal
Pendulum Periods
Apply the concept of time dilation.
Problem The period ofa pendululll is measurecllo be :1.00 s in the inertial frame of the pendulum at Earth's surface. \,Vhat is the period as measured by an obsen'cr mO\'ing at a spced ofO.950rwith respect to the pendulum? Strategy Here, we're given the period of the clock as measured by an observer in the rest fre-lmc of the clock, so Ihat's a proper time interval Atr Vile want to know hm'" much I ime passes as measured by an observer in a frame mO\'~ ing relative to the clock, which is D.t. Substiltllion imo Equation 26.2 then solves the problem. ~
. ....
.....
Solution Substitute the proper time and relative speed into Equation 26.2:
3.00 s
~
9.6] s
(O.950r)'
'.
r"
....... Remarks The moving observer considers the J}f'1ldlllnm to be moving, and moving clocks are observed to run more slowly: while the pendulum oscillates once in 3 s for an observer in the rest frallle of the clock, it takes nearly 10 s to uscillate once according the moving observer. QUESTION 26.1 Suppose a mass-spring system ,.. . ith the sallle period as the pendululll is placed in the observer's spaceship. When the spaceship is traveling at a speed ofO.95r relative to an observer on Earth, what is the period of the pendulum as measured by t.he Earth observer? EXERCISE 26.1 What is the period of the pendululll as measured by a third observer moving at 0.900,.?
Answer 6.88 s
Confusion arises in problems like Example 26.1 because movement is relative: 1'1'0111 the point of vic,... of someOlle in the pendulum's rest frame, the pendulum is standing still (except, of course, for lhe swinging Illotion), "... hercas 10 someOIlC in a framc thaI is moving with respect to the pendulum, it's the pendulum that's doing t.he moving. To keep it straight, ahva)'s focus on the observer making the measurement and ask yourself whether t.he clock being measured is 11l0\"ing with respect to that observer. I I' the answer is no, then the observer is in the rest. frame of the clock and measures the clock's proper tillie. If the answer is yes, then the time Hleasllred by the observer will be dilated, or larger than the clock's proper ti me. This con fusion or perspectives led to the famous "t,·... in paradox."
The Twin Paradox An intriguing consequence ort ime dilation is the so~callcd twin paradox (Fig. 26.8). Consider an experiment involving a scI or t\vins named Specdo and CosIo. \rVhen they are 20 years old, Speedo, the more advellluresome of the two, sets Ollt. on an epic journey to Planet X, located 20 light-yeCll's from Eanh. Further, his spaceship is capable of reaching a speed of' 0.95<: relative LO the inertial frame of his twin brother back home. After reaching Planet X, Speedo becomes homesick and immediately reLUrns to Earth at the same speed of 0.95(, l!POI1 his return, Spcedo
856
Chapter 26
Relativity
FIGURE 26.8 (a) 1\s the twins depan, they're the sallie age. (b) When Spcedn returns from hi.~jollrncr to Planet X. he's ~ounger than his lldn Goslo, who remained on Earth.
(a)
(b)
is shocked LO discover Lhat Cosio has aged 2D/v ~ 2(20 ly)/(O,95 Iy/yr) = 42 years and is now 62 years old. Speedo, on the other hand, has aged only 13 years. Sorne wrongly consider this the paradox; that t\vins could age:H different rates and end up ~Irter a period of time having very different ages. AJthough contrary [Q our common sense, that isn'l the paradox at all. The paradox is that, from SpeedD's point of view, he was at rest \·vhile CosIo (on Earth) sped a\-\'ay [rom him at O.95cand returned later. So Cosio's clock ,vas moving reb.nive to Speedo and hence running slow compared with Speedo's clock. The conclusion: Speedo, not Cosio, should be the older of the twins! To resolve this apparent paradox, consider a third observer moving ata constant speed of O.,l')c relative to Goslo. To the t.hird observer. Goslo never changes inertial frames: his speed relative to the third observer is always the same. The third observer notes, however, that Speeda accelerates during his journey, changing refermct frames in Ihe /Jrocess. 1'1'0111 the third observer's perspective, it's clear that there is something very different about the motion of Goslo when compared with that of Speedo. The roles played by Goslo and Speedo are not symmetric, so it isn't surprising that time flows differently for each. Further, because Speedo accelerates, he is in a noninenial frame of reference and is technically outside the bounds of special relativity (although there are methods for dealing wiLh accelerated motion in relativity). Only Goslo, who is in a single inertial frame, can apply the simple Lime-dilation formula to Speedo's Lrip. CosIo finds Lhat instead of aging 42 years, Speedo ages only (l - V'/C 2)1/2(42 years) = 13 years, or Lhese 13 years, Specdo spends 6.5 years traveling to Planet X and 6.3 years returning, for a LOtal traye] time of 13 years, in agreement with our earlier statement. QUICK QUIZ 26.3 True or False: People t.raveling near the speed of light relative to Earth ,,,'ould measure their lifespans and find them, all the averag'e, longer than the average human lifespan as measured on Earth.
Length Contraction
TIP 26.3 The Proper Length You must be able to corrcclly
identify the obscn'crwho measures the proper length. The proper length bCl\,'een two poi illS in space is l.he length measured by all obser\'er ,H rest with respect to the kngth. Very often. the proper time interval and the proper length are not measured by the sallle observer.
The measured distance between two points depends on the frame of reference of the observer. The proper length L p of an ol~ject is the length of the object as measured by an observer at rest relative to the object. The length of an object measured in a reference frame that is moving with respectLO the object is always less than the proper length. This effect is known as length contraction. To understand length contraction quantitatively, consider a spaceship traveling with a speed v from onc star to al10Lher as seen by two observers, one on Earth and the other in the spaceship. The observer at rest on Earth (and also assumed to be at rest with respect to the two stars) measures the distance between the stars to be L p. According Lo this observer, the time it t.akes the spaceship to complete the voyage is III = Lp/v. Because of time dilation, the space traveler, using his spaceship clock, measures a smaller time of travel: 6.11! = 6.1/y. The space traveler claims to be at rest and sees the destination star moving toward the spaceship with speed v. Because the
26.4
Consequences of Special Relativity
857
space traveler reaches the star in time atp' he concludes that the distance L between the stars is shorter than Lv The distance measured by the space traveler is t;{
L=vt;{!'~v-
y
Hecause Lli =
11
dl, it follO\vs that ~ Length contraction
[26.4]
According to this result, illustrat.ed in Active figure 26.9, ifan observer at rest. with respect to an objectmcasures its length to be L/I , an observcr moving at a spccd 11 relative to the o~ject "vill find it to be short.er than it.s proper length by the factor V [ ~ v 2/ r:!. Note lhallength contraction takes place only along the direction of motion. Time-dilation and length contraction effects have interesting applications for future space travel to distant Slars. For the star to be reached in a fraction of a human lifetime, the trip must be taken at very high speeds. According to an Earthbound observer, the time for a spacecraft to reach t.he destination star will be dilat.ed compared with t.he time int.erval measured by travelers. As discussed in the treatmcnt. of thc twin paradox, the travelers will be younger than their twins when they ret.urn to Eanh. Therefore, by the time the travelers reach the star, they will have aged by some number of years, while their partners back on Earth will have aged a larger number or years, the exact ratio depending on the speed of the spacecraft. At a spacecraft speed ofO.94r, t.his ratio is about 3:1.
y'
QUICK QUIZ 26.5
Lp
---->II
I I
I I
x'
0'
QUICK QUIZ 26.4 You are packing lor a trip La another star, and on your journey yOll will be traveling at a speed of 0.99(. Can you sleep in a smaller cabin than usual, because you will be shoneI' when you lie down? Explain your answer.
i
I' I I
(a)
o
(b)
.,
ACTIVE FIGURE 26.9 A meterstick moves to the righl with il speed u. (a) The meterstick as viewed by an obsen'er ,H rest wil h respect 10 the meterstick. (b) The meterstick as seen by an observer moving Idlh a speed 11 with respect to it. The moving melerstick is always measured to be short!'r than in its own rest frame by a faclorOrVl - v 2/c 2•
observe a rocket moving away from you. Compared with its length when it was at rest on the ground, ""ill yOll measure its length to be (a) shaner, (b) longer, or (c) the same? Compared to the passage of time measured by the watch on your wrist, is the passage of time on the rocket's clock (d) fast.er, (e) slower, or (f) the same? Answer the same quesLions if (he rocket turns around and cOllleS to\vard >'ou. YOll
EXAMPLE 26.2 Goal
Apply the concept of length contraction to a distance.
Problem (a) An observer on Earth sees a spaceship at an altitude of 4350 km moving dO\\'n""ard toward Earth with a speed ofO.970c. ''''hat is the distance from the spaceship to Earth as rneasured by the spaccship's captain? (b) After firing his engines, the captain Illeasures her ship's ahitude as 267 km, ""hcreas the observer on Earth measures it to be 625 km. What is the speed of the spaceship althis instant? Strategy To the captain, Earth is rushing toward her ship at O.970c; hence, the distance between her ship and Earth is contracted. Substitution into Equation 26.9 yields the answer. In part (b) use the same equation, subsLilllting the distances and solving for the speed. "-'
.
....
Solution (a) Find the distance from the ship to Earth as measured by t.he capt.ain. Substitute into Equation 26.4, getting the altitude as measured by lhe captain in the ship:
L
=
Lp'/1
=
1.06
X
v'/c' 10' km
=
(4350 km)VI ~ (0.970c)'/c'
"'"
858
Chapter 26
Relativity
(b) Whal is the subsequent speed or tile spaceship if the Earth observer Illeasures the distance from the ship to Eanh as 625 kill and the captain measures it as 2fi7 km?
Apply the length-contraction equation: Sq\larC both sides of Illis equation and solve for v:
, "(
")
L = L,,- 1 - v-Ie II
~
(v I -
(LI Ly
->
1 - v'l (2
=
(0L)'
= ("Ill - (267 km/625 km)'
v = 0,904c
.
~.
Remark
~
The proper length is always the length measured by an observer at rest with respect to that length.
QUESTION 26,2 As a spaceship approaches an observer at nearly the speed oflighl, the captain directs a beam of yellow light at [he observer. \-Vhat would tlie observer report upon seeing the light? (a) Its wavelength \vollid be shifted toward the red end of the spectrulll. (b) Its \\'<1velength v','Quld correspund to yellow light. (c) Irs wavelength would be shifted toward the blue end of the specrrulTI. EXERCISE 26.2 Suppose the observer on the ship mcasures the distltllCe [rolll Earth as 50.0 km, \·\:hereas the observer on Earth measures the distance as 125 km. At. what. speed is t.he spacecraft approach ing Eanh?
Answer
0,917"
Length contraction occurs only in the direclion of the observer's motion. No cOlltl"action occurs perpendicular 1.0 t hal. direction. For example, a spaceship at rest relative to an observer may have t.he shape of an equilatcraltrianglc, but if it passes the observer at relativistic speed in a direction parallc1 to its base, the base will shorten while the height remains the same. Hence the observer will report that the craft has the form of an isosceles triangle. An observer traveling with the ship will still observe itlo be cquilateral.
26.5
RELATIVISTIC MOMENTUM
Properly describing the Illation of particlcs within the framework of special relativit.y requires generalizing Newton's laws or motion and the definitions of momentum and energy. These generalized definitions reduce 10 the classical (nonrelativisLic) definitions vl'llcn v is rnuch less thall c. First, recall t.hat conservation or mOlllenlUm states that when two ol~jects collide, the total momentum of the systcm remains constant, assuming thc objects are isolated, reaning only v,"ith each other. V\'hen analyzing such collisions from rapidly movi ng inertial frames, howcver, it is found thaI momentum is not consened if the classical definition or momentum, p = 'IIIV, is uscd. To have momentum conserva1ion ina 11 iner1.ia I f"ra mes-even I hose movi ng at a n appreciable fraction of c-thc definition of momentum must be modified to read Relativistic momentum '7
/1::::::;:
mv
' '111 -
='Y IIIV
[26.51
v'I'"
whcre v is the speed of the panicle and "III is its Ill
Relativistic Energy and the Equivalence of Moss and Energy
26.6
EXAMPLE 26.3 Goal
859
The Relativistic Momentum of on Electron
Contrast the classical and relativistic definitions ormomentulTI.
Problem An electron, which has a mass or 9.11 x 10 :~l kg, moves with a speed or 0.750c. Find the classical (nonrelativistic) momentum and compare it with its relati\'istic counterpart !J rd . Strategy Substitute into the classical definition to get the classical momentum, then multiply by the gamma {'actor to obtain the relativistic version.
.
".
.
Solution First, compute the classical (non relativistic) momentum with v = 0.750r:
-_
I' = =
mv
= (9.11 X 1O~" kg)(0.750 X 3.00 X 10' m/s)
2.05 X 1O~" kg·m/s 2.05
Multiply this result by y to obtain the relativistic momentum:
~
.. -
~
X 1O~"
kg·m/s
yI] - (0.750clc)' ...........
.
~
_
Remarks The (corrcct) relativistic result is 50% greater than the classical result. In subsequent calculations no notational distinction will be made betwcen classical and relativistic momentum. For problems involving relative speeds ofO.2c, the ans\.,..er using the classical expression is about 2% below the correct answer. QUESTION 26.3 A particle with initial momentum /Ji doubles its speed. 1-10\.,.. does its final momentum momentum? (a) /Jr> 2p, (b) /Jr = 2p, (c) PI < 21',
PI compare
EXERCISE 26.3 Repeat the calculation for a proton traveling at 0.600r.
Answers
26.6
p = 3.01
X 1O~19
kg· mis, /',." = 3.76 X
10~19
kg·m/s
RELATIVISTIC ENERGY AND THE EQUIVALENCE OF MASS AND ENERGY
\Ve have seen that the definition of momentum required generalization to make it compatible with the principle of relativity. Likcwise. the definition of kinctic energy requires modification in relativistic mechanics. Einstein fOllnd that the cOlTecl expression for the kinetic energy of an object is
KE =
ymc'!. -
me'l
[26.6]
~
Kinetic energy
~
Rest energy
The constant t.erm Hll.· 2 in Equal ion 26.6, which is independent or thc speed of I he object, is called the rest energy or the object, EN: Ell. = me'2
[26.7]
The term ymr'l in Equation 26.6 depends on the object's speed and is the sum of the kinetic and rest energies. \·Ve definc yn/(;'2 to be the total energy I:,;, so Total energy = kinetic energy + rest energy or, I1sing Equation 26.6, E:::: KE+ mc 2 = ytllc'!
[26.8]
with its initial
860
Chapter 26
Relativity
Because'Y :::::: (l - v'll c2 )-Ji'!., we can also express the Total energy
-+
lOlal
energy Eas [26.9]
This is Einstein's famous mass-energy equivalence equation.'!. The relal ion E = 'Y17u:'2 = KF + nil''.!. shmvs the amazing result that a stationary particle with zero kinetic energy has an energy proportional to its mass. Further, a small mass corresponds to an enormous amount or energy because the proportionality constant between mass and energy is large: ['2 = 9 X 10 16 m'!./s'!.. The equation L'u = '111(2, as Einstein first suggested, indireclly implies that the mass ofa panicle may be completely convcniblc to energy and that pure energy-ror example, electromagnetic energy-may be convened to particles having mass. That is indeed the case, as has been shown in the laboratory mallY times in interactions involving matter and anlimaller. On a larger scale, nuclear power plants produce energy by the fission of ura11 iu m, ,,,'h ich involves the conversion of a smal J alllQU nt of the mass of the uran ium into encrgy. Thc SUll, too, convens mass into energy and continually loses mass in pOll ri ng out a tremendous a IllOll nt of" elecl romagnetic energy in all d ireclions. I t's extremely intcresti ng 1hat cllthollgh \ve have been tal king' about the imerconversion of mass and encrgy for panicles, the expression E = rtI.r 2 is universal and applics to all objects, processes, and systCJlls: a hot object has slightly more mass and is slightly more difficult to accelerate than an identical cold ol~jecI because it has more thermal energy, ancl a stret.ched spring has more clastic potential energy and more Jllass than an identical ul1streLched spring. A key point, however, is that these changes in mass arc orten rar too small to meaSllre, Our best bet for measur~ ing mass changes is in nuclear transformations, where a measurable fraction of the Illass is converteel into energy,
QUICK QUIZ 26.6 True or F"lse: BeC"LlSe Lhe speed or a panicle cannot exceed the speed of light. there is an upper limit to its momentum and kinetic energy.
Energy and Relativistic Momentum OfLen the momentum or energy of a panicle rather than its speed is measured. so it's useful to find an expression relating the total energy E to the relativistic momellLUIll fl. \Ve can do so by llsing the expressions E = 'YIJlC2 and /) = 'Yrrtv. By squaring these equations and subtracting, we can eliminate v. The result, after some algebra, is [26.10]
""'hen the particle is at rest, /) = 0, so E = Eu = 1IIr 2 . In this special case the [Otal energy equals the rest energy, For the case of panicles that have zero mass, such as photons (massless, chargcless particles of light), we set Tn = 0 in Equation 26,10 and lind that
E=
JJC
[26.11]
This equation is an exact expression relating energy and momentulll for photons, wh ich always travel at the speed of light. tn dealing 'with subatomic panicles, it's convenient to express their energy in electron volts (cV) because the particles arc given energy when accelerated through <;lll electrostatic potential difference. The conversion factor is ~r\11 hallAh thi, l:){p]'c.~sion Iloesll'l lou\, t:xacll}" like I be f:1111011S (''"Illatiull E = 111('2, it us("c1to bl' nllnlllOll tn write //I = "1m" (Einstdn himsc1fwr"u' if that wa~'), when' 11/ is the cflcetiw: Illa" "Llt1 object \I1o\'ing at speed vand m" i, tlll' mass of Ih:tl0b.ll'ct as 111l'a.~tlrl'tlll}":I1l ollM·rvcr al n's! with respect LO Ill!.' ol~il'CI, Tllt'n 0111" E= "'f/1lr hecomes tin: I"amiliar F. = Il is nllT1'1I11~ 1Il1f:l.~hillnable \() US(' III = ""(111 0,
mr.
26.6
Relativistic Energy and the Equivalence of Moss and Energy
861
I cV = 1.60 X 10-1".1 For example, the mass orCIn electron is 9.1 I X 10 -31 kg. Ilence, the rest energy of the electron is (9.11 X 1O-:~1 kg)(3.00 X 101'\ m/s)~ - H.20 X IO- H J
11I/ 2 -
Converting
lO
eV, we have
III,e'
= (8.20 X 1O-H/J(l
eV/1.60 X 1O-1u/J
= 0.5] I MeV
li
where I MeV = -IO eY. llecallse we frequenlly use the expression J':= )'11/(·2 In nuclear physics and because "1/1 is usually in alOmic mass units, u. it is useful tu have the conversion factor I u = 931.494 MeV/r:'. Using this factor makes it cas}', for example, to find the rest. energy ill \1eV of the nucleus ora uranium alom with a mass of235.043 924 11:
ER =
lIIe' =
(235.043924u)(931.494 MeV/II'''')('') = 2.18942 X 10"MeV
QUICK QUIZ 26.7 A photon is reflecled from a mirror. Trllc or Falsc: (a) Because a photon has zero mass, it does not exert a force on lhe mirror. (b) Allhough the photon has energy, it can't transfer any energy to the surface because it has zero mass. (c) The photon carries momentum, and when it reflects off the mirror, it Ilndergoes a change in momentum and excns a force on the mirror. (d) Although the photon carries Illomentum, its chang-e in momentulll is zero when it reflects from the mirror, so it can't exert a force un the mirror.
EXAMPLE 26.4 Goal
AS eedx Electro:.:.n
~
~
_J
Compute a total energy and a relativistic kinetic energy.
Problem An electron moves "vith a speed v = O.850r. Find its total energy and kinetic energy in mega electron volts (\1cV) and compare the lauer to the classical kinetic energy (10 1' eV = l.rvleV).
Strategy Substit.utc into Equation 26.9
to gCl.the LOtal ellergy and subtract the rest mass energy to obtain the kinetic
energy.
,..-'
.
''1
Solution Substitute vallles into EqU
(9.] I X 1O-:1I kg)(3.00 X 10' m/s)'
VI
(_ -13)( 1.601.0oeV) 1.06 X 10'.1 X 10 '''.I =
The kinetic energy is obtained by subtracting the rest energy from the total energy:
- (0.850cl C)'
0.975 MeV
Klo = E -
11I/ =
0.975 MeV - 0.511 MeV = 0.464 MeV
Calculate the classical kinetic energy: ~ ~ (9.11 X 10-" kg)(0.850 X 3.00 X 10' m/s)' ~ ~
Remark
2.96 X 10-H] = 0.185 :VleV
.
.
...
Notice the large discrepancy bel ween the relativistic kinetic energy and the classical kinetic energy.
QUESTION 26.4 According to an observer, the speed v of a p;-lrticle with kinetic energy KJ".'i increases to 2v. How docs the final kinetic energy KEjcompare with the initial kinetic energy? (a) KJl'l> 4Kf.. .j (b) 4KEj (c) Kl.j< 4KEj
KJ'1=
Chapter 26
862
Relativity
EXERCISE 26.4 Calcu]alc the lotal energy and the kinetic energy in MeV ora proLOIl traveling at 0.600e. (The rest. energy ora proton is approximalely 938 MeV.)
Answers
E = 1.17 X 10' MeV, KE = 2.3 X 102 MeV
EXAMPLE 26.5 Goal
The Conversion of Mass to Kinetic Energ in Uronium Fission
Understand the production of energy from nuclear sources.
Problem The fission, or splitting, of uranium was discovered in 1938 by Lise Mcitncr, who slIccessfully interpreted some curious experimental results found by Ouo Hahn as due to fission. (Hahn recei,,·ccl the Nobel Prize.) The fission of 2§~C begins with the absorpLion of a slowITIm,jng neutron that produces an unstable nucleus of 236U. The 2~6U nucleus then quickly decays into two heavy fragments moving at high speed, as well as several neutrons. Most of the kinetic energy released in such a fission is carried off by the two large fragments. (a) For the typical fission process,
,..
I~n
+ 2~~U
-7
l~~Ba
+ ~~Kr + 36n ....
calculate the kinetic energy in MeV carried off by the fission fragments, neglecting the kinetic energy of the reactants. (b) '>\That percentage of the initial energy is converted inw kinetic energy? The atomic masses involved, given in atomic mass units, arc
bn
'§~U
= 1.008 665 u
'~bBa = J40.903 496 u
= 235.043 923 u
i~Kr = 91.907936 u
Strategy This problem is an application of the conservation of relativistic energy. ''''rite the conservation law as a sum of kinetic energy and rest energy, and solve for the final kinetic energy.
... ..-
..,
Solution (a) Calculate the final kinetic energy for the given process. Apply the conservation of relativistic energy equation, assuming KEillitial = 0: Solve for KEfill :tI and subsrillltc, converting to ~'leV in the last step:
KE o",,'
= (1.008665 u + -
235.043923 L1)c2
[140.903496 u
+ 91.907936 u + 3(1.008665 u)]c'
= (0.215161 u)(931.494 MeV/u'c')(c')
=
200.421 MeV
(b) ''''hat percentage of the initial energy is converted into kinetic energy? Compute the LOtal energy, which is the initial energy:
~ (1.008665
1I
+ 235.043923 u)c'
= (236.05259 u)(931.494 MeV/lI'c')(c') = 2.19882 X 10' ~1eV
Diyiclc the kinetic energy by the total energy and Illulliply by 100%:
200.421 MeV -====.o---x 2.19882 X JO"MeV
~
100% = 9.1J5 X 10-'% ,
,
.. ,
~
Remarks This calculation shows that nuclear reacrions liberate only about one-tenth of 1% of the rest energy of the constituent particles. Some fusion reactions result in a percent yield several times as large. QUESTION 26.5
",Thy is so little of the mass converted 1.0 other forms of energy?
26.7
General Relativity
863
EXERCISE 26.5 In a fusion rcaction light clemcllts combine to form
+ ¥O
-7
~T
+ IH
(a) Calculate the energy releascd in the form of kinetic encrgy, assuming for simplicity the initial kinetic energy is zero_ (b) \lVhat percentagc of the rest mass is convened to energy? The atomic masses involved arc ;D = 2.014 102 Answers
(a) 4.033 37 MeV
\I
:;T = 3.016049 u
:H = 1.007825 u
(b) 0.1075%
26.7 GENERAL RELATIVITY Special relativity relates observations of inertial observers. Einstein sought a more general theory that would address accelerating systcms. His search was motivated in part by the following curious fact: mass determines the incrtia of an ol~jeet and also the strength or the gravitational neld. The mass involved in inertia is called incrl ial mass, 'J1I" whereas the mass responsible for the gra\"itational field is called I he gravitational mass, "IlI/{ These masses appear in Newton's law of gravitation and in the second law of motion: Gravitational property
Fg =
Inertial property
F= ,
The value for the gravitational constant G was chosen to make thc magnitudes of '111~ and Ill, numerically equal. Regardless of how G is chosen, howe\"er, the strict proponionality of JI1}rand U1.i has been established experimentally to an extremely high degree: a few parts in 10 12 . It appears that gravitat.ional rnass and inertial mass may indeed be exact.ly equal: 1/1, = mit In Einstein's view the remarkable coincidence that lllg and Ul i were exactly equal was evidence for an intimate connect.ion between the two concepts. He pointed out that no mechanical experiment (slich as releasing a mass) could distinguish between the t.wo silllations illust.rated in Figures 26. lOa and 26.1Gb. fn each casc a FIGURE 26.10 (a) The obscn'er ill Lhe Lubicle is at n:SL in a unilill·m g-ra\ iralional ficld He expericllccs a normal force n. (b) NO\\' llle obscn'cr is ill a region whl'l"c gravit)' ~ IIcglig-iblc. but all eXICl"nal force F aels on ,he frame of refcrcllce. producing all accderatioll WiLh magnitude l!- Again. rhe l1lan experiences a norm,t! force 'Ii lhat acceh:ratcs him along with lhe cubicle. According ro Einstein, ,he frsed hall would. (d) Ci,'cn the equivalencl: or the frames, the same phenomenon should be ousenn! in rhe presellce of a gravity field.
g.
(a)
(bl
(c)
(d)
864
Chapter 26
Relativity
mass released by the observer undergoes a downward acceleration of g relative to the floor. Einstein carried this idea further and proposed thaL noexperimcnt, mechanical or otherwise, could distinguish betweell the two cases. This extension to include all phenornena (noljust mechanical Diles) has interesting consequences. For exam-
ple, suppose a light pulse is sem horizontally across the box, as in Figure 26.10c. The trajectory of the light pulse bends dmvnwarcl as the box accelerates upward to meet il. Einstein proposed that a beam of light should also be bent downward by a gravitational field (Fig. 26.lOd). The two posLUlalCs or Einstein's general relativity are as follows: 1. All the laws or nature have the same form [or observers in any frame of reference, accelerated or not. 2. In the vicinity of any given point, a gravitational field is equivalent to an accelerated frame of reference without a grayitational field. (This is the ItriTiciple oj eqll.ivalence.)
The second postulate implies that gravitational mass and inertial mass are completely equivalent, not just proportional. \'\That were thought to be two different types of mass are actually identical. One interesting effect predicted by gencral relativity is that time scales are altered by gravity. A clock in the prescnce of gravity runs more slowly than one in \.. . hich gravity is negligible. As a consequence, light emitted from atoms in a strong gravit.y field, such as the Sun's, is observed LO have a lower frequcncy than the same light emitted by atoms in the laboratory. This gravitational shift has been detected in spectral lines emitted by atoms in massive stars. It has also been \'erified on Earth by comparing the frequencies or gamma rays emitted from nuclei separated vertically by :abol.lt20 ffi. The second postulate suggests a gravitational field may be "transformed away" at any point if we choose an appropriate accelerated ffame of reference: a freely falling one. Einstein developed an ingenious method of describing the acceleration necessary to make the gravitational field "disappear." He specified a certain quantity, the CU1'7..HllUTff ofsjJ{/.cetime, that describes the gravitational effect at every point. In fact, the curvature of spacetime completely replaces Ne\..· ton's gravitational theory. According to Einstein, there is no such thing as a gravitational force. Rathel', the presence of a rnass causes a curvalure of spacetime in the vicinity of the mass. Planets going around the Sun follow the natural contours of the spacetime, much as marbles roll around inside a bowl. Tn 19i9 John 'Vheeler summarized Einstein's general theory of relativity in a single sentence: "~'1ass one tells spacetime how to curve; curved spacetime tells mass n'lO hm.. . to move:' The fundamental equation of general relativity can be roughly stated as a proportion as follows: Average curvature of spacetime
ex:
energy density
Einstein pursued a new theory of gravity in large part because of a discrepancy in the orbit of Mercury as calculated from Newton's second law. The closest approach of ~lercl1ry to the Sun, called the perihelion, changes position slO\l/ly over time. Newton's theory accounted for all but 43 arc seconds per century; Einstein's general relativity explained the discrepancy. The most dramatic test of general relativity came shonly after the end of \,vorld \'Var I. Einstein's theory predicts that a star would bend a light ray by a certain precise amount. Sir Arthur Eddington mounted an expedition to Africa and, during a solar eclipse, confirmed that starlight bent on passing the Sun in an amount matching the prediction of general relati\1ty (Fig. 26.11). "\'hen this discovery was announced, Einstein became an international celebrity. General relativily also predicts that a large star can exhaust its nuclear fuel and collapse to a vcry small volume, turning into a black hole. Here the curvature of spacetime is so extreme that all matter and light \vithin a certain radius becomes trapped. This radius, called the Sl"hwflrzsdzild radius or pvent horizon, is about 3 km
Summary
..
FIGURE 26.11 Dcnection of starlight passing near the Sun. Because of lhis effect. the SlIll and othel' remOle objects call ael aS;I gmviral;ollallrns. In his gencrallheor) ofreJativit)'. Einstein calculated that starlightjuSl gr:lling Ihe Sun's surface should be deflected by all angle of 1.75".
Apparent direClioll LO Slal DL:neeted path of light [romslar
---
=-
__
~
-
1
1.7:>' To star
SUII
865
(actual direction)
Earth
for a black hole with the mass of our Sun. At the black hole's center may lurk a sil1guIm'ity, a point of infinite density and curvalLlre where spacetime comes to an end. There is strong evidence for the existence of a black hole havi ng a mass of millions of Suns at the center of our galaxy.
APPLYING PHYSICS 26.1
FASTER CLOCKS IN A "MILE-HIGH CITY"
At.omic clocks are extremely accurate; in fact, an error of 1 sin 3 million years is typical. This error can be described as aboUl one pan in 10 1.'. On t.he other hand, the awmic clock in Boulder, Colorado, is often 15 ns faster than the atomic clock in 'Vashington, D.C., after only one day. This error is about one pan in 6 X ]0 12 , which is about]7 t.imes larger than the lypical error. Ifatomic clocks are so Clccurate, why docs a clock in Boulder not remain synchronous \\o'ilh one in Washington, D.C.?
Explanation According to the general theory of relativity, the passage of time depends on gravity: clocks rUIl more slowly in strong gravitational fields. "Vashington, D.C., is at an elevation very close to sea level, whereas Boulder is about a mile higher in altitude, so the gravitational field at Boulder is weaker than at vVashington, D.C. As a result, an atomic clock runs more rapidly in Boulder than in vVashinglOn, D.C. (This effect has been verified by experiment.)
26.3 Einstein's Principle of Relativity The two basic postulates of the special theory of relativity a re as follows:
2. The length of an object in motion is con/raeted in the direction of mation. The equation for length contraction is
1. The laws of physics are the same in all inertial framcs of reference. 2. The speed of light is the same for all inertial observers, independently of their motion or of the motion of the source of light.
[26,4]
26.4 Consequences of Special Relativity Some of' the consequences of the special theory of relativity are as follows: 1. Clocks in motion relative 10 an observer slow down, a phenomenon known as time dilation. The relationship between time intervals in the moving and at-rest systems is [26.2] where 6.1 is the time interval measured in the system in relative motion with respect to the clock, 1
y
= -:r=~=;;
VI - v'I.'
[26.3]
and u.t p is the proper time interval measured in the system moving with the clock.
\vhere L is the length measured by an observer in motion relative to t.he object and Lp is the proper length measured by an observer for \\'h0111 the object is at rest. 3. Events that are simultaneous for one observer are not simultaneous for another observer in motion relative to the first.
26.5 Reiativistic Momentum The relativistic expression for the momentum of a panicle moving \.. . ith velocity v is
/J
mv ::::=:
v'1 -
v'J.; ('1
= ymv
[26.51
26.6 Relativistic Energy and the Equivalence of Mass and Energy The relativistic expression (or the kinetic energy of an o~jeCl is KE = ymr'2 - mr'l [26.6) where mc'1 is tbe rest energy of the object, ER
<
866
Chapter 26
Relativity
The total energy of a panicle is 'lflC~
E=--r=~=
VI -
The relativist ic momentum is related to the LOlal energy through the equation
[26.9]
[26.10]
11'2/ (;'2
This is Einstein's famous mass-energy equivalence equation results when v = O. FOR ADDITIONAL STUDENT RESOURCES. GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS 1. \,"hich of the following stalcll1cl1ls are fundamental poslulates of the special Iheory of relativity? (a) Light moves through a subsLan<.:e called lhe ('ther. (b) The speed of light depend:-. all the inertial reference frame in which it is measured. (c) The laws of physics depend on the inenial reference frame in which they are lIsed. (d) The laws of physics arc the sallle in all incnial reference frames. (c) The speed of light is independcm of the incnial reference framc in which it is measured_ 2. A spaCt:ship moves at one-quancr the speed or light relative to Earth in a dircClion perpendicular to the tine of sight of an observer at rest with respectLO Earth. If the spacesh ip has a length L when at rest, wh ich stau::ment is true concerning the moving ship's length C as measured by the observer? (a) L' > L (b) L' < L (c) r: = L (eI) r: « L (e) C » I. 3. A car's headlights come all, and an observer on the ground measures the speed of the light beam to be c = 3.00 X 10 8 m/s. As the driver heads dowll the highway traveling al a speed 1) away from the ground observer, which 01" the following statements are true ahoUl. the measured speed of the light beam? (a) The ground observer measures the light speed to be c + v. (b) The driver rneasures (Ile lighr speed LO be c. (c) The ground obser\'(;r measures the light speed to be c. (d) The driver measures the Iight's speed to be slightly less than c. (e) The ground observer measures the light'S speed to be slightly less than c. 4. An astronaut is tnl\'c1illp'" in
5. A pendululll in a spaceship traveling through space ,;ll a speed of O.ge reicH ive to Eanh has a period P as measured by an aslronaUl in the spaceship. \Vhat is true aboul thc period T as measured by an observer on Earth? (a) P> T(b) p~ T(c) P< T(eI) P< Tor p>.r; depending on the direction or motion of the spacecraft (e) None of these 6. A cube measured at rest has ,'olume E An observer then passes the cube pal-allel to one of its sides at 0.981', so)' = 5. What is the volume of the cube as measured by the moving observer' (a) 11/125 (b) W25 (c) W5 (eI) V(e).oV 7. Suppose a photon, proton, and electron all have the sallle lotal energy t.:. Rank their momenta from smalleSlto greatest. (a) photon, elcnron, proton (b) proton. photon, electron (c) cleClroll, photon, proton (d) electron, proton, photon (c) proton, eleClron, photoll 8. \Vhal is the y factor of an electron having a kinetic energy or2mc'? (a) I (b) 2 (c) 3 (eI) 4 (e) 5 9. A mass-spring system moving with simple harmonic mOlion has a period T when measured by a ground observer. rr lhe same systcm is then placed in an iner~ tial frame of reference that moves past the ground observer al a speed of 0.50c, b)' what faclor should 7' be multiplied to give the system's period as measured by the gronnd observer' (a) 0..00 (b) 0.87 (c) 1.00 (eI) 1.15 (e) 1.35
10. The pm',:er Olllput of the Sun is 3.7 X lO~li \V. What is rhe rale at which the Sun's mass is converted ilHo energy' (a) 4.1 X 10' kg/s (b) 6.3 X 10" kg/s (c) 7.4 X 10" kg/s (d) 3.7 X 10' kg/s (e) 2.5 X 10" kg/s
CONCEPTUAL QUESTIONS 1. A spacecraft with the shape of a sphere or diameter D moves past an observcr on LHth with a speed of 0.5c What shape does t.he observer measure for the spacecraft as itlllO\'CS past?
2. What twO speed measurements will two observers in relative motion always agree upon? 3. The speed of light in water is 2.30 X 10 8 m/s. Suppose an elcctron is moving through water at 2.50 X 10/1 rn/s. Does this panicle speed violate the principle of relat ivitv?
4. "Vith regard to reference frames, how docs general,-e1ali"ity differ from special relativity? 5. Some distantslar-like objects. called quasars, are receding from us at half the speed oflight (or greater). What is the speed or the light we receive from these quasars? 6. It is said that Einslein, ill his teenage years, asked the question, "What would T see in a mirror if I carried il illllly IWllds and rail at a speed near that of light?" How would you answer I his question?
Problems
867
7. List some ways our day-la-day lives would change if the speed of light were only 50 m/s.
9. Photons of lighl have zero mass. How is it possible thai
8. Two identically constructed clocks are synchronized. One is put into orbit around Earth, and the other remains on Earth. \rVhich clock runs morc slowl}'? When the moving clock returns to Earth, will the two clocks still be synchronized? Discuss from the standpoints of both special and general relativity.
10. Imagine an astronaut on a trip to Sirius, which lies 8 light-years from Earth. Upon arrival at Sirius, t.he astronaut finds that the lrip lasted 6 years. If the u'ip was made at a constant speed of 0.8r, how can Ihe 8-1ight-year dist.ance be reconciled ",,'ith the 6-year duration?
the)' have moment urn?
PROBLEMS ~
The Problems for Ihis chapter may be WebAssignassigncd online al \'VebAssign. I, :!, :\ = straiglHforward, il\lerlnt'di~lte, cil,llienging
m
= denotes guided problem
II =
denotes ellhanced content problem biomedical application denotes filII solution available in Student Solulions Almwal/
mil = 0=
Stud)' Guide
SECTION 26.4 CONSEQUENCES OF SPECIAL RELATIVITY OJlfastronams could travel at tJ = 0.950c, we on Earth would say it takcs (4.20/0.950) = 4.42 ycars to reach Alph
relative to Earth respcct to Earth. according to the the answer change Earth in the oppo-
3. Al what speed docs a clock move if it is measured to run at a rale one~halrthe rate of an identical clock at rest with respect to an observer?
4.11 An
astronaut at rest on Earth has a hean rate of 70 beats/min. 'When the astronaut is traveling in a spaceship at 0.90c, what will this rate be as measured by (a) an observer also in the ship and (b) an observer at rest on Earth?
WThe average lifetime of a pi meson in ils own frame of reference (i.e., the proper lifetime) is 2.6 X IO~8s. If the meson moves with a speed of 0.98c, what is (a) its mean lifetime as measured by an observer on Earth. and (b) the average distance it travels before decaying, as measured by an observer on Earth? (c) What. distance would it travel iflimc dilation did not occur? 6.
II An
astronaut is traveling in a space vehicle t.hat has a speed ofO.500crelmive to Earth. The astronaut measures his pulse rate 75.0 per minute. Signals generated by the astronaut's pulse are radioed to Earth when the vehicle is moving perpendicularly to
,u
Earth ohserver mcasure? ·What would be the pulse rate if the speed of the space vehicle were increased to O.990(~
7.
m
A muon formed high in Earth's atmosphere 1J"
8. A deep-space probe moves away from Earth with a speed ofO.80c. An antenna on the probe requires 3.0 s, in probe time, 10 rolate through 1.0 rev. How much time is required for 1.0 rev according to an observer on Eanh? [[]The proper length of one spaceship is threc times that of another. The twO spaceships are traveling in the same direction and, while both are passing overhead, an Earth observer measures the two spaceships to ha\e the same length. If the slower spaceship has a speed of 0.350r with respect to Earth, determine the speed of the faster spaceship. 10. An atomic clock moves at 1 000 km/h for 1.00 h as measured by ;:t,n observer llsing an identical clock on Earth. How many nanoseconds slow will the moving clock bc at the end of" the I.OO-h interval? Hinl: The following approximation is helpful:~' = (I ~ x/2) for x« 1. 11. A supcrtrain of proper length 100 m tra\'eJs at a speed of 0.95r as it passes through a tunnel having proper length 50 m. As seen by a trackside observer, is the u'ain ever completely within the tunnel? If so, by how much? 12. A box is cubical with sides of proper lengths L I = L 2 = L:1' as shown in Fig~ ure P26.12, when viewed in its own L"l. rest fnune. If lhis block moves parallel to one of' its edges with a speed of L, 0.80r past an observer, (a) what shape FIGURE P26.J2 does it appear to have to this observer? (b) \"'hat is the length of each side as measured by the observer?
868
Chapter 26
Relativity
SECTION 26.5 RELATIVISTIC MOMENTUM 13. (3) Find the change in classical momentum oran electron thal accelcrate~ from O.900e LO 0.940c, ncg-Iccling relati,'istic CffCClS. (b) Repeat the calcul;uion, lI"iing the expression for relati\'jo;tic 1ll01llf"1l11l1ll. 14.
E
Cllcul
115.IAIl unstable particle ,n rest brc
SECTION 26.6 RELATIVISTIC ENERGY AND THE EQUIVALENCE OF MASS AND ENERGY 16. A pmton mo\'cs with a speed of 0.950r. Calculate (a) ils rest en erg}'. (b) ils tOlal energ)', and (c) it.. kinctic energy. thro\\'~ a O.15-kg baseball al a speed of 0.9001'. Find the haseball\ (a) rcsl energy, (h) kinelic energy" and (c) lotall.ncrgy.
17. Superman
~leV of work LO accelerme speed of 11, c1clennine lJ.
18. If it lakes 3 750 rest to
(l
~l
pm IOn [rom
[I[]\Vhal speed must a panicle atwin before it!'; kinetic energy is douhle tht: value predicted hy the llonrelmivist ic expre,~sion 1\../.: = ~mll?
20. Detcrmine Ihl' cllcrl:D' rCCjuircd to accelerate an e1eclron from (a) 0.5001' to 0.9001' and (h) 0.900e to O.!l90r. 21. A cube of ~led has a volume of 1.00 cm:~ and
123.1_ Starting with the deJinilions of relali\'istic encl"gy and mOl1lellllllTI, show I.hat/~'1 = fP,'! + m2r l (Eq. 26.1O). 24.
EIJ ~lBr
Consider the rC
equation symbolically in terms of the rest energy and the kinelic energv. seuing the iniliallOtal energy to (he final lOtal energy. (b) L:sing values r!"Om Appendix B. find the lOlal mass of the inilial panicles. (c) Using the valucsgivell below, find the IOlal mass of the panicles afler the reaction takes place. (d) Sublract the filial particle mass from the initial parlicle mass. (c) COllvcn lhe an~wcr to part (d) to Me\'. obtaining thc kinetic ellcrg\ of the daughtcr particles. ne~leCling Ihe kinetic energy of the reactanls. Note: Lanlhanum-148 has alomic maSs 147.932236 u; bromine-87 has alomic lllass 86.920711 19 1I. ADDITIONAL PROBLEMS
25. Consider electrons acceknue(1 10 a tolal t.:nergy of 20.0 GeV ill the 3.00-km-long Stanford Linear Accclera-
tor. (a) \Vhal is the l' faclOr" for ule electrons? (b) How long does (he accelerator' appear to the eleclrons? Electron mass energy: 0.511 MeV.
26. An c1ecu"on has a specd of O. 750c. (a) Find lhe speed of a P"OIOIl lhat has the samc kinctic energy as the electron. (b) Find lbe speed of a prOlon thai has the same momentum as the electron.
t27.IThe reSl energy of an electron is 0.511 Me\'. The rest energy ora protoll i~ 938 fvleV. Assume bOlh panicles have kinetic energies of 2.nO Me::V. Find the speed of (a) lhe c1eclron and (b) the prOlon. (c) By how much docs lhe speed of the electron exceed that of the prolon? Note: Per~ form the calculalions in MeV; don·t. convert the energies tojoulcs. The ansv.lcr is sensitive to rounding-. 28.
am A spring of for"ce conSlalH k is compressed b}' a distance x from ils equilibrium lenglh. (a) Docs lhe mass of the spring change when the spring is compressed? Explain. (h) Find an expl·cssion for the change in mas'i of the ~pring in terms of k, x, Clnd c. (b) \,'hat is the change in mass if the force constant is 2.0 X 10 2 N/m and x = 15 cm?
29. Whal is Lhe speed of a prolon thal has becn accelcratc(1 from rest through a difference of potel1li,l1 of (a) 500 V and (b) 5.00 X lOR \'?
30. An elcctron has a lOtal energy equal lO five limes ils rcst energy. (a) IvVllal is iUi momentum? (b) Repeat for a proton. !mAn astmnaut wishes lo visil lhe Andromeda galaxy, making a one-way trip thaL will take 30.0 years in the spaccship's frame of refcrence. Assume lhe galax}' is 2.00 million light-years away and his speed i~ conslant. (a) How fasl must hc lravel rehui\'e to Earth? (b) \Vhat will be the kinetic energy orhis sp:lcecraft, which has lllass of 1.00 X 10" kg? (c) \fhal is the COSl of this energy if it is purchased al a Lvpical COnSHITler price for eleclric energ\, 13.0 cenls per kWh? The following approximation will prove lIseful: _~~I=
,~
., =)-2
forx«
1
32. An alarm clock is sel to sound in 10 h. Al 1= 0, the clock is placed in a spaceship moving with a speed ofO.i5r (relative lo E.anh). \Vhm di~lancc, as delermined bV;:1I1 Eanh observer. docs tile spaceship lravel before the alarm clock sOllnds? 33. A sp;lCcship of proper 1cng"th 300 In lakes 0.75 /.LS to pass an CHlll observer. Determine the speed of this ~paceship as measured by 1he Ea rt h ob'iervcr.
34, The cosmic rays of highest e!lerg)' are prolons lhat have kinetic energy on the order of 1013 MeV. (it) I low long would il take a pl"OLOn of this energy lo lravel across the Milk}" "Vay galaxy, having a diameter - lO'i light-years, as measured in the proton's frame? (b) From the paim oh,.;ew of the pmton. how man} kilometer~ across is lhe gala..x}? 135.IThe 1I0nr'e1alivistic expression [or the lllomelHum of a panicle, p = lJI'V. em he lIsed if v« c. For whal speed docs the usc of lhis formula gi\'e an error in the mOl11e11.LUm of
36. (a) Show Ihat a potcntial difference of 1.02 X 106 V would be sufficient to give an electron a speed equal 10 Lwice the
Problems
speed of liglll if Newtonian mechanics remained valid at high speeds. (b) What speed would an electron actually acquire in falling thwllgh a pOlential dilTercllce equal to
1.02 ~7.
X
10" V'
The llluon is ~m llnstable panicle that spontaneously decays into an e1eclron and two neutrinos. In a reference frame in which [he Illuons arc stationary, if the number of' muons at t = 0 is lV'n, the number at time I is given by N = NI)['/T, where T is the mean lifetime. equal to 2.2 J1.s. Suppose lhe muons move at a speed of 0.95r and thcl·e (Ire 5.0 X 104 muons at I = O. (a) \o\,'hat is the observed lifetime of the muons? (b) How lllallY muons remain after travel ing a distance of ~.O k m?
38. Imagine that the entire Sun collapses to a sphere of radius HI: such lhat the work required to remove a small mass
111
from the surface would be equal to its rest energy JI/('.!. This radius is called the gmvilolionnl mdius for the Sun. rind Ug • (It is believed (hatlhe ultimate l~t[e arvel"}' massive stars is to collapse beyond their gravitational radii into black holes.)
velocity, 70.0% or the speed of light. Jts nucJear~powered battcries supply the cner~n' LO keep its data transmittcr active continuously. The batteries have a lifetime of" 15.0 years as me::l.slIl'ed in iI rest fr'lmc. (a) How long do the battcries on the space probe la'il as measured by mission control all Earth? (b) How far is the probe from Earth when its batteries fail a:s measured by mission conLrol? (c) 1-10\,,,· far is the probe frolll Earth as mcasured b}' its built-in trip odometer when its baw:rie:s fail? (d) For what lotal time after launch (Ire data received from the probe by mission control? Note that radio wavcs travel at the speed of light and fill the :space between the probe and Earth at the time the baltery fails. 41. An observer 1ll00·ing at a speed of 0.99ft( relative to a rod (Fig. P26.41) mea3ures its length to be 2.00 III and sees its length 1.0 be oricntcd
39. The identical lwin:s Speedo and Cosio join a migration from Earth to Planet X, which is 20.0 light-years away in 01'. Go:slo·s at 0.750r. Calculate the age difference between tile l",,'ills ,Ifter Cosio's spacecraft lands on Planet X. Which twi 11 is the older? [iQJAn intcr:stellar space pmbe is launched from Eanh. After a brief period 01" accdcratioll, it moves with a constant
869
J\'lotion of ohsen-et·
FIGURE P26.41 rig-lit.
Vit:;w or rod as SeCI) br an
oh~cn'('r moving \0 lilt::
Color-enhanced scanning electron micrograph of the storage mite Lepidoglyphus destructor. These common mites grow to 0.75 mm and feed on molds, flour, and rice. They thrive at
2YC and high
humidity and can trigger allergies.
27.1
Blackbody Radiation and Planck's Hypothesis
27.2
The Photoelectric Effect and the Particle Theory
of Light 27.3
X-Rays
27.4
Diffraction of X-Rays by Crystals
27.5
The Compton Effect
27.6
The Dual Natu,e of Light and Motter
27.7
The Wave Function
27.8
The Uncertainty Principle
QUANTUM PHYSICS Although many problems were resolved by the theory of relativity in the early port of the 20th century, many others remained unsolved. Attempts to explain the behavior of matter on the atomic level with the laws of classical physics were consistently unsuccessful. Various phenomena, such as the electromagnetic radiation emitted by a heated object (blackbody radiation), the emission of electrons by illuminated metals (the photoelectric effect), and the emission of sharp spectral lines by gas atoms in an electric discharge tube, couldn't be understood within the framework of classical physics. Between 1900 and 1930, however, a modern version of mechanics called quantum mechanics or wove mechanics was highly successful in explaining the behavior of atoms, molecules, and nuclei. The earliest ideas of quantum theory were introduced by Planck, and most of the subsequent mathematical developments, interpretations, and improvements were mode by a number of distinguished physicists, including Einstein, Bohr, Schr6dinger, de Broglie, Heisenberg, Born, and Dirac. In this chapter we introduce the underlying ideas of quantum theory and the wave-particle nature of motter, and discuss some simple applications of quantum theory, including the photoelectric effect, the Compton effect, and x-rays.
27.1 TIP 27.1 Expect to Be Confused Your life experiences lake place in the macroscopic \\'orld, where quantum effects arc not evident. Quantum dTects can be even morc bizarre t.han relaLivistic effects, but. don·t.despair: confusion is normal "ilnd expecled. As l\'obel Prize-winning physicisl Richard Feynman once said, "Nohod)' understands quantulll mechanics."
870
BLACKBODY RADIATION AND PLANCK'S HYPOTHESIS
An object at an)' temperature emits electromagnetic radiation, called thermal radiation. Stefan's hw..·, discussed in Section 11.5, describes the total pm\;er radi,Heel. The spectrulll of the radiation depends on the temperature and properties of the object. At low temperatures, the \.. . avelengths of the thermal radiation are mainly in the infrared region and hence not observable by the eye. As the telll~ perature of an object increases, the object eventually begins to glow red. At suffIciently high temperatures, it appears to be white, as in the glmv ofl.he hOL tungsten filament. of a lightbulb. A careful study of thermal radiat.ion shows that it consists of a continuous distribution ofwavelenglhs from the infrared, visible, and ultraviolet portions of the spectrum.
27.1
Blackbody Radiation and Planck's Hypothesis
From a classical vic\vpoint, thermal radiation originates from acceleratcd ch
871
FIGURE 27.1 An opr.:nill~ in lhe Gl\·ilY Or;l lwei" i.. a good appll)ximaI ion or a hlackbod),. As li~hl ('lllers lh(' caviL}' lhrollgh Ihl' small 0pl'llill~, pan is reneCled ;111(\ pan is ahsorlwd on each reflf'Clioll from lhe illl('1 ior walls. Afler Olan\· I('f1pnions. CS.Sl·lllilllly all the incidt"lIt l'1It'l"g-\ is ab.,orlwd.
where Amax is the wavelength at which the curve peaks and T is the absolute temperature of the o~ject cmiuing the radiation.
APPLYING PHYSICS 27.1
STAR COLORS
If you look carerull)' at stars in the night sky, yOll can distinguish three main colors: red, white, and blue. \Vhat caUSeS these particular colors? Explanation These colors result ['rom the different surface telnperaturcs of stars. A relatively cool star, with a surface LCmperalure or~ 0001\., has a radiation curve similar to the middle curve ill Active Fig-lire
27.2. The peak in this cur\'c is above the visiblc wavelengths, 0.4 ,u,n1 100.7 ,um, bcyond thc wavelength of red light, so significantly morc radiation is emillcd \vithin the visible range '-\tthe red end than the blue end of the spectrum. Consequently. the star appcars rcddish in color, similar to the red glow from the bll rner of an electric rangc.
A hOller star has a radiat.ion curve more like lhe upper curve in Active Figlire 27.2. In t.his case tile star emits significant radiation throughollt the visible range, and the combination of all colors causes 1 he star to look white. Such is the case with our own Sun, with a surface temperature of 5 800 K. For much hotter stars, the pe
In
K
T = ==-=--=--,----"'------'-"--"'Allldx
0.2898 X 10 ~ m·K 2.30 X 10
7
m
t
1.26 X 10' K
Attempts to use classical ideas to explain the shapes of" 1he curves shown in Active Figure ~7.2 failed. Active Figure 27.3 (page 872) shows an expcrimental plot of the blackbody radimion spectrum (red curve), together with the theoretical picture of\-\'hatlhis curvc should look like based on classical theories (blue curve). At long wavelengt.hs, cl<'lssical theor)' is in good a~reement \vith the experimental clata. At short \'I-'avelcngths, however, major disagreement exists bel.ween classical theory and experiment. As A approaches zero, classical theory erroneously
4 \Vavd('ng:th (pill)
ACTIVE FIGURE 27.2 tlllcnsily ot hlackhod\ radial i(Jl1 \·C'rSllS \"a\c!l'Ilg"lll :Illhrcc dif"lt.:n.:nl (l'llIpe..:ralllrl·~. NoLC lhat lIlt· lnl:d radiaLiOll
cmilll·d (lhe..: an-a ulldcr
curve) innTa~,-~ with inCll';lsillg lcmperatlln'_
:1
872
Chapter 27
Quantum Physics
ClassicI} ulcor}'
Experimental daLa
predicts that the intensity should go to infinity, when the experimental data show it should approach zero. In 1900 Planck developed a formula for blackbody radiation that was in complete agreement with experiments at all wavelengths, leading to a curve shown by the red line in AClive Figure 27.3. Planck hypothesized that blackbody radiation was produced by submicroscopic charged oscillators, which he called resonators. He assumed the \valls of a glowing cavity were composed of billions of these resonators, although their exact nature was unknm-vn. The resonators were allowed to have only certain discrete energies E,l' given by [27.2]
\Vavelcnglh ACTIVE FIGURE 27.3 Comparison ofexpcrimclllal dolla with the classical theory of blackbod~' radiation. Planck's thCOl'~ matches the experimental data perfectl)'.
where 11 is a positive integer called a quantum number,fis the frequency of vi bration of the resonator, and It is a constant known as Planck's constant, which has the value [27.3J
Because the energy or each resonator can have only discrete values given by Equalion 27.2, v..'e say the energy is qlumtizuL. Each discrete energy value represents a dirferent quantum stale, with each value of n representing a specific quantum slate. (\·"'hen the resonator is in the n = 1 quantum Slale, its energy is 1If, when it is in the n = 2 quantulTI state, its energy is 2 Iif; and so on.) The key point in Planck's theory is the assulTIption of quantized energy states. This radical departure from classical physics is the ';quantum leap" that led to a totally new understanding of nature. It's shocking: it's like saying a pitched baseball can have only a fixed number of different speeds, and no speeds in between those nxecl values_ The fact that energy can assume only certain, discrete values instead of any one ofa continuum of values is the single most imponantdifference beth'cen quantum theory and the classical theories of NC\''t''ton ancl.\tlaxweIL
MAX PLANCK
German Physicist (1858-1947) Planck introduced the concept of a "quantum of action" (Planck's constant h) in an attempt to explain the spectral distribution of blackbody
radiation, which formed the foundations for quantum theory. In 1918 he was awarded the Nobel Prize in Phys-
ics for this discovery of the quantized nature of energy.
Photoeleclrons
~c~
E
V
II Variable power stlppl}'
ACTIVE FIGURE 27.4 A circuit diagram for smdying the phoLOelcctric effect. '<\'hen light strikes plate £ (the emiuer), pholOelectrons arc ejected from the plate. Electrons moving from platc E to plate C (the collector) crcatc a current in the cirelli .. registered allhe ammeter. A.
27.2
THE PHOTOELECTRIC EFFECT AND THE PARTICLE THEORY OF LIGHT
In the latter part of the 19th century, experirncnls showed that light incident on certain metallic surfaces caused the emission of electrons from the surfaces. This phenomenon is knm,,'11 as the photoelectric effect, and the erniued electrons arc called photoelectrons. The first discovery of this phenomenon was made by Hertz, \'.'ho was also the first to produce the electromagnetic waves predicted by Max\ovell. Active Figure 27.4 is a schematic diagram of" a photoelectric effect apparatus. An evacuated glass tube known as a photocell contains a metal plate E (the emitter) connected to the negative terminal of a variable power supply. Another metal plate, C (the collector), is maintained at a positive potential by the power supply. \r\'hen the tube is kept in the dark, the ammeter reads zero, indicating that there is no current in the circuit. 'When plate E is illuminated by light having a wavelength shorter than some particular wavelength that depends on the material used to make plate E, however, a current is detected by the ammeter, indicating a f!m.. . of charges across the gap between E and C. This current arises from photoelectrons emitled from the negative plate E and collected at the positive plate C. Figure 27.5 is a plot o[ the photoelectric current versus the pOlenti;JI difference.nV between E and C for two light intensities. At large values of .n1~ the current reaches a maximum value. In addition, the current increases as the incident light intensity increases, as you might expect. Finally, when .n.Vis negative-that is, when the power supply in the circuit is reversed to make E positive and C negative-the ClllTcnt drops to a low value because most o[ the emitted photoelectrons are repelled by the now negative plate C. In this situation only those electrons having a kinetic energy greater than the magnitude of e aVreach C, where e is the charge on the electron.
27.2
The Photoelectric Effect and the Particle Theory of Light
'-\-'hen t.\V is equal to or more negative than -t.\\~, the stopping potential, no electrons reach C and the current is zero. The stopping potential is indejJenrlent or the radiation intensity. The maximum kinetic energy of the photoelectrons is related to the stopping potential through the relationship
873
Current High intensity
Low intensity
[27.4] Several features of the photoelectric effect can't be explained \vith classical physics or with the \vave theory of light: • No electrons are emitted if the incident light frequency falls below some cutoff frequency 1, also called the threshold frequency, which is characteristic of the material being illurninated. This fact is inconsistent with the \,,·ave theory, 'tvhich predicts that the photoelectric efTect should occur at. nny frequency, provided the light intensity is surtlciently high. • The maximum kinetic energy of the photoelectrons is independent of light int.ensity. According to wave t.heory, light of higher int.ensity should carry more energy into the metal per unit time and therefore eject photoelectrons having higher kinetic energies. • The maximum kinetic energy of Ihe photoelectrons increases with increasing light frequency. The wave theory predicts no relationship between photoelectron energy and incident light frequency. • Electrons are emitted from the surface almost instantaneously (less than 10- 9 s after the surface is illuminated), even at low liglu intensities. Classically, we expect the photoelectrons to require some time to absorb the incident radiation before they acquire enough kinetic energy to escape from the metal.
Applied voltage FIGURE 27.5
PhoLOdectric currerll
verSllS applied potelllial difren'nce fOJ"
two light intensities. The current
increases Idlh intensit)'. but reaches a salurationlevel for large valucs of ~F At voltages equal to or less I harl -.:l\~,
where .:lV, is the SLOpping potential, the currcm is zero.
A successful explanation of the photoelectric effect. was gi\'en by Einstein in 1905, the same year he published his special t.heory of relativity. As pan of a gen-
eral paper on electromagnetic radiation, for which he received the Nobel Prize in Physics in 1921, Einstein extended Planck's concept of quant.ization to electromagnetic waves. He suggested that a tiny packet of light energy or photon would be emitted when a quantized oscillator made ajump from an energy state En = nil! to the next lower state E n- l = (/1 - l)lif Conservation of energy would require t.he decrease in oscillator energy, /?f, to be equ
E = lif
[27.5]
~
Energy of a photon
~
Photoelectric effect equation
where h is Planck's constant and fis the frequency of the light, which is equal to the frequency of Planck's oscillator. The key point here is that the light energy lost. b}' the emiuel', 11j; stays sharply localized in a tiny packet or panicle called a photon. In Einstein's model a photon is so localized that it can give all its energy It/to a single electron in the metal. According to Einstein, the maximum kinetic energy for these liberated photoelectrons is KEIl1ilX = hf - ¢
[27.6]
where 4> is called the work function of the metal. The work funct.ion, \,,rhich represents the minimum energy with which an electron is bound in the metal, is on the order ora few electron volts. Table 27.1 lists work functions [or various mctals. ""Vilh the photon t.heory of light, we can explain the previously mentioned feal.ures of the photoelectric effect. that cannot be understood using concepts of classical physics: • Photoelectrons are created by absorption of' a single phot.on, so the energy of that photon must be greater t.han or equal 1.0 the work function, else no photoelectrons will be produced. This explains the cUIOf'f' frequency. • From Equation 27.6, KElIlax. depends only on the frequency of the light anclthe value of the work function. Light int.ensit.y is immaterial because absorption of a single photon is responsible for the clectron's change in kinetic energy • Equation 27.6 is linear in the frequency, so KEIll ;O( increases with increasing frequency.
TABLE 27.1 '''lark Functions of Selected Metals Metal
'" (eV)
Ag
4.73
AI
4.08
ell Fe
4.70 4.50
:\la
2.46
Pb P,
4.14 6.35
Zn
4.31
874
Chapter 27
Quantum Physics
• Electrons arc cmiuecl almost instantaneously. regardless of intensity. because the light energy is concentrated ill packets rather than spread Oul in waves. [[ the frequency is high enough, no time is needed for the electron 10 gradually acquire sufficient energy LO escape the metaL
FIGURE 27.6 A skClcb of K/·;", .• ~ lhe rrequency of incich.:nl light for pholoekclrons ill a typical photoelectric effccl experiment. I'hOlOllS with frequency less I him h d01l'1 have surficient cncrp;y 10 eject an clcct ron from the IlIl.::taL
Experimentally, a lincar relationship is observed betwecnJand KEmax • as sketched in Figure 27.6. The intercept on the horizontal axis, corresponding to KEllla ..... = 0, gives the cutoff frequency belmv ,vhich no photoelecll'ons are em ilted, regardless of light intensity. The cutoff wave/('ngth A, can be derived from Equation 27.6:
n:~I'SUS
ftc
A=,
[27.7]
,.. ., here c is the speed of light. \,Vavelengths greateT than A( incident on a material with \vork function cP don't result in the emission of photoelectrons.
EXAMPLE 27.1 Goal
Photoelectrons from Sodium
Understand the quantization of light. and its role in the photoelectric effect.
Problem A sodiulll surface is illuminated ",.'ith liglu ofwavelcngth 0.300 /L1ll. The work fuoClion for sodium is 2.46 cv. Calculate (a) the energy of each photon in electron volts, (b) lhe maximum kinetic energy of the ejected photoelectrons, and (c) the cULOfTwavelength for sodium. Strategy
Parts (a), (b), and (c) require substitution of va lues into Equations 27.5, 27.6, and 27.7, respectively.
. ..... Solution (a) Calculate the energy of each photon. Obtain the frequency from the wavelength:
Use Equation 27.5 to calculate the photon's energy:
JA
J=
1.00 X lO'" Hz
E= hJ= = 6.63
-4
J = A = 0.300
(6.63 X 1O-'·'j·s)(1.00 X 10"[-[z)
X 10- 19 j
_(
-
3.00 X 10' Ill/' X 1O-'; III
c
c=
6.63 X JO
-'9)( .J
1.00 eV )_ 1.60 X 10 '
(b) Find the maximum kinetic energy of the photoelectrons.
SubstilUte into Equation 27.6:
KE",,, = hl-
(c) Compute the cutoff wavelength.
Convert qJ from electron volts to joules:
4)
= 2.46 eV = (2.46 eV)(1.60 X 10- 1" jleV) = 3.94 X lO- I ".J
Find lhe clILOfTwave!cngth using Equation 27.7.
he
(6.63 X 1O-"j·,)(3.00 X lO'm/s)
3.94 X 10 "'j
A=-= ,
= 5.05 X JO~7
In
=
........... ~
........ Remark
505 nm
The cutoffwaYelength is in the yellow-green region of the visible spectrum.
27.3
X-Roys
875
QUESTION 27.1 True or False: Suppose in a given phoLOclecLric experiment the f requenc), of light. is larger than the cutoff frequency. The magnitude of the stopping potential, then, is larger than the energy of the incident photons. EXERCISE 27.1 (a) What minimum-frequency light will ejeCt photoelectrons from a copper surface? (b) If this frequency is tripled, nnd the maximum kinetic energy (in eV) of the resulting photoelectrons. Answers
(a) 1.13 X 10" Hz
(b) 9.40 eV
Photocells The photoelectric effect has many interesting applications lIsing a device called the jJhotocell. The photocell 5hO\'I.'11 in Active Figure 27.4 produces a current in the circuit when light of sufficiently high frequency falls on the cell, but it doesn't allow a current in the dark. This device is used in streetlights: a photoeleClric control unit in the base of the light activates a switch that turns off the streetlight when ambient light strikes it. ~v1any garage-door systems and elevators use a light beam and a photocell as a safety feature in their design. ''''hen the light beam strikes the photocell, the electric current generated is sufRciemly large to maintain a closed circuit. Vlhen an object or a person blocks the light beam, the current is interrupted, \.. . hich signals the door to open.
27.3
X-RAYS
X-rays were discovered in 1895 by \Vilhelm Rontgen and much later identified as electromagnetic waves, rollowing a suggestion by l\'Iax yon Laue in 1912. X-rays have higher frequencies than ultraviolet radiation and can penetrate most materials with relative ease. Typical x-ray wavelengths are about 0.1 nm, which is on the order of the atomic spacing in a solid. As a result, they can be diffracted by the regular atomic spacings in a crystal lattice, which act as a diffraction grating. The x-ray diffraction pattern of NaCI is shown in Figure 27.7. X-rays are produced when high-speed electrons are suddenly slowed down, such as when a meta] target is struck by electrons that have been accelerated through a potential difference of several thousand volts. Figure 27.8a shows a schematic diagram of an x-ray tube. A current in t.he filament causes electrons to be emitted, and these freed electrons are accelerated toward a dense metal target, stich as tungsten, which is held at a higher potential than the filament. Figure 27_9 (page 876) represents a plot of x-ray intensity versus wavelength for the spectrum of radiation emitted by an x-ray lUbe. NOle that the spectrum has are two distinct components. One component is a continuous broad spectrum that depends on the \'oitage applied to the tube. Superimposed on this componenl. is a series of sharp, intense lines that depend on the nature of the target material. To
\'oltage
-.
------
Tungslen targel
Filament
~
'*
:; I,••II:=:::!, ~
" ~
_
ekclrons
~..rv-..
x-ri'lVS
(a)
•
•
•
•
•
• FIGURE 27.7
•
•
•
• X-ra~'diITraclioll pat-
tern of NaCI.
x-ra)' tube.
V
Filamenl
•
•
(a) Diagram of an x-ray tube. (b) Photograph of an
.-------IIII-+--~ ~::E==,. --
•
•
FIGURE 27.8
lligh voltage
~
APPLICATION Photocells
o
~
• ~
(b)
Chapter 27
876
Quantum Physics
observe these sharp lines, v\'hich rcprcselll radial ion emitted by the target atoms as their electrons undergo rearrangements, lhe acceleraling voltage must excecd a ccrt.Jin value, called the threshold voltage. \Ve discuss lhreshold voltage funher in Chaplc,- 28. The continuous radiation is sometimes called bremsstrahlung,
K
"
•
I
'=
a Gennan \'lord meaning "braking radimion," because elcClrons emit radiation
"<;=
1(/.1
¥
v""---
/ 30
40
50
60
70
RO
90
A.pmFIGURE 27.9 Till': x-nw ~rCetnllTl of a llletalwfgct nmsisrs 01';1 broad continuous \peClrlllTI pillS a l1\lHlber
of sharp lines, which
when they undergo an acceleration inside the t.arget. Figure 27.10 illustrates how x-rays are produccd when an electron passes near a charged target nucleus. As the electron passes close to a positi,'ely charged nucleus contained in the target material, il is deflected from its path because of its e1eClrical attraction to the nucleus; hence, it undergoes an acceleration. An analysis from classical physics 5hO\,,·s that any charged particle ,.. . ill emit electromagnetic radiation when it is accelerated. (An example of this phcnomenon is the production of electrom;'lgnetic '\laves by accelerated charges in a radio antenna, as described in Chapter 2-1.) According to quantum theory, this radiation must appear in the form of photons. Because the radiated photon shown in Figure 27.10 carries encrgy. the electron musl lose kinetic energy because of its encounter ,,,,·ith [he target nucleus. An extreme example consists of the electron losing all of its energy in a single collision. In t.his case the initial energy orthe electron (fLiV) is transformed completely into the energy of the photon (I~rll1;IX). In equation [arm, [27.8]
Nucleus of lMgCt atom
IOdlCCh:r1 1"lo1WCI" energy ./1
electron
..-.--- ~
where e LlV is rhe energ}' of the elcctron after it has been accelerated through a potential difference of aVvolts and e is the charge on the electron. This erruation says that the shortest wavelength radiation that can be produced is
Incoming ekclron
lu; Amill
Emilled ' " phOlon
FIGURE 27.10 All electron passingne.tI a charl{l'd IClrg-et atom e:qwrian acceleration. and a photon is emilled in the process.
ClltCS
APPLICATION Using X~Rays to Study the Work of Moster Painters
tube Collimitlor
Photographic film
FIGURE 27.11 Schematic diagram ollhe ICClllliquc Llsnl to ubserw: lhe dilTracr ion ofx-rars h~ (\ single crystal. 1'110.: ;lrT:\~' of ~POl!' formed 011 Ihe film by Ihe diffr,lcted he,lm" is called a I.aue pattern. (See Fig. ~i.7.)
('6. V
[27.9]
:\'ot all the radiation produced has this part.icular w;,welength because many of the electrons aren't. SLOpped in a single collision. The result is the production of the con ti n UOllS speclru III of wavelcngths. Interesting insights into the process of painting and revising a masterpiece arc being revealed by x-rays. Long-wavelengt.h x-rays are absorbed in varying degrees by some paints, such as those having lead, cad mium, chromium, or cobalt as a base. Thc x-ray interactions v"ith the paints give contrast because the differelll elements in the paints have different electron densities. Also, thicker layers will absorb more than thin layers. To examine a painting by ;,In old master, a film is placed bchind il while it is x-rayed fro1l1thc front. Ghost outlines of earlier paintings and earlier forms of the final mastcrpiece are sometimes revcaled when the film is de,-eloped.
27.4
X-rd)'
=
DIFFRACTION OF X·RAYS BY CRYSTALS
]n Chapter 24 we described how a diffraction grating could be used to measure the wavelength of light. In principle the wavelength of any electromagnetic "'ave can be measured if a grating having a suitablc tine spacing can be f()lllld. The spacing bct\·\,ecn lines must be approximately equal to the wavelength of the radiation to hc mcasured. X-rays are electromagnetic waves ""ith wavelengths on the order of 0.1 nm. It would be impossible LO construct a grating with such a small spacing. As noted in the previous section, hO\vever, the regular array of atOllls in a crystal could act as a three-dimensional grating for observing the diffracrion or x-rays. One experimental arrangemcnt for observing x-ray diffraction is shown in Figure 27.1]. A na1"1"O\\" beam of x-rays ,-vith a continuous wavelength range is incident on a crystal such as sodiurll chloride. The diffracted radiation is . . ery intense in certain directions, corresponding 10 constructive interference from wayes ref-leCled from laycrs of atoms in the crystal. The diffracted radiation is detected by a photogTaphic nJrn and forms an array of spots known as a Lrwf !Jallern. The
27.4
t.r .? ). ,I,(
....,
/'
./
I/"
-
,.( I:
IncidenL beam
,I
-
ReflecLed hcam
.)'"
7'11,1
,/
I/'"'
Diffraction of X~Rays by Crystols
.~
\.> J ,/
~
Lowe.- plane
dsin
FIGURE 27.12 A Illudel ufLhe cubic cr}stal1ine SlrllClUt"r.: or sudium chloride. The blue ~phen's repres~'llt Ihe Cl iUlIS, and the red spheres represent the N;ll iOlls. 'I"ltc lellgtll ofthc cube edge i~ (/ = O.5t13 nllL
e
FIGURE 27.13 :\ two-
crystal structure is determined by analyzing the positions
e.
2rlsin 0 = mA
111
= 1,2,3,.
[27.10]
This condition is known as Bragg's law, after 'N. L. Bragg (1890-1971), who first derived the relationship. 1ft he wavelength and diffraction angle are measured, Equation 27.10 can be used to calculatc the spacing bctween aLomic planes. The techniquc of x-ray diffraction has been lIsed 1.0 determine the atomic arrangement of complex organic molecules such as proteins. Proteins arc large molecules containing thousands of al.Oms t.hat. help regultue and speed up chcmi· cal ld"c processes in cells. Some proteins are amazing catalysts, speeding up the slow room-temperature react.ions in cells by 17 orders of magnitude. To underst.and t.his incredible biochemical reactivity, it is import.ant to deterrnine t.he strllc~ lUre ot" these intricate molecules. The main technique used to determine the molecular st.ructure of prot.eins, DNA, and RNA is x-ray diffraction using x-rays of wavelength of about I A (1 A = OJ nm = I X 10- 10 m). This technique allows t.he experimenter to "sec" individual atoms that are separated by about t.his distance in molecules. Because the biochcrnical x-ray diffraction sample is prepared in crystal form. the gPOllIetlJ (position of the bright Spots jn spacc) of the diffraction pattern is det.ermined by the regular three-dimensional crystallall.ice arrangemenl of molecules in thc sample. The inLensities of the bright diffraction spots are determined by the atoms and their elcctronic distributions in t.he fundamcntal building block or the crystal: the unit cell. Using complicated computational techniqucs, investigators can essentially
f- Bragg's law
877
Chapter 27
878
Quantum Physics
-= deduce the molecular structure by matching the observed intensities of diffracted g beams with a series of assumed aLOmic positions that determine the acomic struc~ j lUre and electron density of the molecule. _ Crystallizing a large molecule such as D~A and obtaining its x-ray diffraction
i
g .~
bS
FIGURE 27.14 An x-raydifTraclion phOlOgr.lph of DNA taken by Rosalind Franklin. TIle cross pattern OfSPOIS \\':1S ,} clue that DNA h;lS a helical structure.
FIGURE 27.15 The double-helix Structure of DNA.
EXAMPLE 27.2 Goal
pattern is highly challenging. Ln the case of DNA, obtaining sufficiently pure crystals is especially dirficull because there exist two crystalline forms, A and E, which
arise in mixed form during preparation. These two forms result in diffraction patterns that can't be easil,' deciphered. In 1951 Rosalind Franklin, a researcher at King's College in London, developed an ingenious method of separating the two forms and managed to obtain excellent x-ray diffraction images of pure crystalline DNA in B-form. 'Vith these images she determined t.hat the helical shape of DNA consisted of two inten.. ·oven strands, \...· ith the sugar-phosphate backbone on the olltside of the molecule, refuting prior models t.hat had the backbone on the inside. One oEher images is shown in Figure 27.14. James v\Tatson and Francis Crick used Franklin's work to uncover further details of the moleCll1e and its function in heredity, particularly in regards to the internal struCLUre. Attached 1.0 each sugar-phosphate unit. of each strand is one of four base molecules: adenine, cytosine, guanine, or thymine. The bases are arranged sequentially along the strand, with patterns in the sequence of bases acting as codes for proteins that carry out various functions for a given organism. The bases in one strand bind to those in the other strand, forming a double helix. A model of the double helix is shm"n in Figure 27.l5. In 1962 \'\latson, Crick, and a colleague oFFranklin's, Maurice vVilkins, received lhe Nobel Prize for Physiology and Medicine for their work on understanding the st.ructure and function of D1\'A. Franklin would have shared the prize, but died in 1958 of cancer at the age of thirty-eight. (The prize is not awarded post-humously.)
X-Ra Diffraction from Calcite
Understand Bragg's law and apply it to a crystal.
Problem IJthe spacing between certClin planes in a crystal of calcite (CaCO g) is 0.314 nm, fll1d the grazing angles at which first- and third-order interference will occur for x-rays of\vavelength 0.070 0 11m. Strategy
Solve Bragg's la,,,,, for sin
e and substitute, lIsing the inverse-sine function .
~.......
Solution Find the grazing angle corresponding to m = 1, [or first-order interference:
to obtain the angle.
1/1). (0.0700 nm) 5;n8=-= =0.111 2d 2(0.314 nm)
8 = sin- I (O.I11) = 6.37 Repeat the calculalion for third-order interference (1/1
= 3):
~
0
===--=-="-'-
. iliA 3(0.0700 nm) sm 0 = = 2d 2(0.314 nm)
8 = 5in- I (0.334) = 19.5 ~
Remark
0.334
0
~
,
Notice there is little eli fference benveen this kind of problem and a Young's slit experiment.
QUESTION 27.2
True or False: A smaller grazing angle implies a smaller distance bel\veen planes in the crystal lattice, EXERCISE 27.2
X-rays of wavelength 0.060 0 nm are scattered from a cryst.alwith a grazing angle of 1J .7°. Assume cess. Calculate the spacing between the cl"}'stal planes.
Answer 0.148 nm
111
= 1 for this pro-
27.5
27.5 THE COMPTON EFFECT
Recoiling
Furthcrjuslific;uioll for the photoll nature of light carne from an experiment conducted by Arthur H. Compton in 1923. In his experiment Compton directed;-tll xray beam of wavelength Ail to\varcl a block of graphite. He found that the scaLlcred x-rays had a slightly longer wavclcnglh A th(lll Ihe incident x-rays and hence the energies or the scattered rays were 100,,·cr. The amount of energy reduction depended on the angle at which the x-rays were scallereel. The change in wavelength 6./\ between ~l scattered x-ray and an incident x-ray is called the Compton sbift. To explain this dfecl, Compton assumed if a photon behaves like a panicle, its collision with other panicles is similar to a collision bcl'.. .ecn t\vo billiard balls. Hence, the x-ray photon carries both measurable fling)! and momentum, and thesc t\-\lO quantities must be conserved in a collision. If Ihe incident photon collides with an electron initially at rest, as in Figure 27.16, the photon transfers some of its energy and momcntum to the c1eClron. As a consequcnce. the energy and frc· qucncy of t he scattered photon are lowered and its wavelength incn.:ases. Applying relativistic energy and momentum conservation to the collision described in Figure 27.16, the shifl in \vavclength of the scallered photon is given by ~A = A - An = -
Ii
In,,r
The Compton Effect
(l - cos 0)
[27.11]
879
CI~Clrnrl~ /
/<1> ~~ U-------
FJ \
FIGURE 27.16 OiagnHll repres('flling CornplOll sClIl{.'ring ora pholon br all dl:Clroli. The scaltl'n.:d photoll has le~s energy (01 a longer \~'a\'{' kllgLh) Lhan rhe it!tidelll phlll(ln.
~
The Compton shift formula
where Ill, is the mass of the electron and 0 is the angle bet\vcen the directions of the scaucred and incident phot.ons. The quantity II/m."c is called the Compton wavelength and has value or 0.00243 nlll. The Compton wavelength is vcry small relativc to the wavelcnglhs of visible light, so the shift in \.. .avelcngth would be difficult to detect if visible light were used. Further, note t.hat. the Compton shift depends on the scaucring angle f) and not on the wavelength. Experirncntal result.s for x-rays scattered [rom various targets obey Eqllation 27.11 and strongly support. the photon concept.
,l
QUICK QUIZ 27.1
True or False: "Vhen a photon scatters off;.lll electron, the photon loses energy.
QUICK QUIZ 27.2 All x-ray photoll is scattered by an electron. Docs the rrequency of the scallcred phot.on relative to that of the incident phot.on (a) increase, (b) decrease, or (c) remain lhe same:-
QUICK QUIZ 27.3 A pholon of energy Eu slrikes a free electron, with lhe scattered pholon of energy E moving in the direction opposite lhat of the incident pholan. In this Complon effecl interaction, what is the resulting kinetic energy of the electron? (a) E" (b) /; (c) En - I' (d) Eo + E
EXAMPLE 27.3 Goal
Scattering X""-R"'a"'y:.;;s
ARTHUR HOLLY COMPTON American Physicist (1892-1962) Compton attended Wooster College and Princeton University. He become director of the laboratory at the University of Chicago, where experimental work concerned with sustained chain reactions was conducted. This work was of central importance to the construction of the first atomic bomb. Because of his discovery of fhe CompIon effect and his work with cosmic rays, he shored the 1927 Nobel Prize in Physics with Charles Wilson. ~
___'
Understand Compton scattering and its effect on the photon's encrgy.
Solution (a) Calculate the wavelength of the x-rays.
h !'.A=-(I-cos6)
Substitute into Equation 27.11 to obtain the wavelength shift:
//1 eC
6.63 X 10- 34 J S (1 - cos 45.0·) (9.11 X 10 ,\ kg)(3.00 X 10' m/s)
-;---~~--.::..,~-"--~-;-----;-
= 7.11 X 10-\' m = 0.000 711 nm
Add this shift to the original \.. . avelength to obtain the \va\'clcngth of the scattered photon:
A/=!'.A
+ A, = 0.20071] nm
(b) Find the fraction of energy lost by the photon
in the collision. Rewrite the enerb'l' E in terms of v.ravelcngth. using e= fA: Compute
~/E using
t
E= hf= h-
A
I:.ithis expression:
E !'.E E
Cancel he and rearrange terms:
E,
Ei
hel AI - hel A, hel Ai
1 I_A-c' = Ai _ 1 = _A,_"- _AI = _!'.A _1_1A-'.I_-_ _ 1/A i AI AI AI
0.000711 nm -3.54 X 10- 3 0.200711 nm ~ -.~ Remark It is also possible to find this answer by substiwting inLO the energy expression at an earlier stage, but the algebraic derivation is more elegant and instructive because it shows hO\<\! changes in energy are relatedlo changes in wavelength. !'.E E
Substitute values frolll part (a):
QUESTION 27.3 The incident photon loses energy. \,Vhere does it go? EXERCISE 27.3 Repeal the example [or a photon with \vavelength 3.00 X 10~2 nm thaL scaUers at an angle of60.0°.
Answers
(a)
3.12 X 10-2 nm
(b) !'.EIE = -3.88 X 10-'
27.6
De Broglie's hypothesis
~
THE DUAL NATURE OF LIGHT AND MATTER
Phenomena such as the photoelectric effect and the Compton effect offer evidence that when light (or other [arms of electromagnetic radiation) and maHer interact, the light behaves as if it. were composed of particles having energy hJand momentum hiA. In other contexts, however, light acts like a \..·ave, exhibiting interference and diffraction effects. This apparent duality can be panly explained by considering the energies of photons in different contexts. for example, photons with frequencies in the radio wavelengths carry very lilLIe energy, and it may take 10 10 such photons to create a signal in an antenna. These photons therefore act LOgether like a wave to create the effect. Gamma rays, on the other hand, are so energetic that a single gamma ray photon can be detected. Tn his doctoral dissertation in ]924, Louis de Broglie postulated Lhat because photons bave wave and particle characteristics, perhaps an forms of matter have both properties. This highly revolutionary idea had no experimcmal confirmation at that time. According 1.0 de Broglie, electrons, just like light, have a dual p"l rticle-\vave nature.
27.6
The Dual Nature of Light ond Matter
881
In Chapter 26 we found that the relationship between energy and momentum for a phoLOn, which has a rest energy of zero, is /) = f/c vVe also know [rom Equation 27.5 that the energy or a photon is he E = IzJ=A
[27.12J
Consequently, the momentum ofa photon can be expressed as E
he
c
cA
h A
/)=-=-=-
[27.13]
~ Momentum of 0 photon
From this equation, 'we see that the photon wavelength can be spcciAed by its momentum, or A = h/f}. De Broglie suggested that all material particles with momentum /J should have a characleristic wavelength A = h/t). Because the Illomentum of a panicle of rllass IJI and speed v is mv = I), the de Broglie wave length of a particle is M
h mv
[27.14]
fo- De Broglie wavelength
Further, de Broglie postulated that the frequencies of matter waves (waves associated with panicles having nonzero rest energy) obey the Einstein relationship for phoLOns, E = hf, so that [27.15]
~ Frequency of matter waves
The dual nature of maner is quil.e apparent in Equations 27.14 and 27.15 because each contains both panicle concepts (mv and I.:.') and wave concept.s (A and J). The fact thal these relationships had been established experimentally [or photons made the de Broglie hypothesis t.hatmuch easier to accept. The Davisson-Germer experiment in 1927 connnned de Broglie's hypothesis by showing that electrons scattering orr crystals fOrlTI a diffraction pattern. The regularly spaced planes of atoms in crystalline regions of a nickel target act as a diffraction grating for electron matter waves.
QUICK QUIZ 27.4 True or False: As the momentum ofa panicle o[mass m increases, its wavelength increases. QUICK QUIZ 27.5 A nonrclativistic electron and a non relativistic proton are moving and have the same de Broglie wavelength. vVhich of the following are also the same for the I.wo panicles? (a) speed (b) kinetic energy (c) momenLum (d) frequency
lOUIS DE BROGLIE French Physicist (1892-1987) De Broglie attended the Sorbo nne in Paris, where he changed his major from history to theoretical physics. He wos awarded the Nobel Prize in Physics in 1929 for his discovery of the wove nature of electrons.
EXAMPLE 27.4 The Electron Versus the Baseball Goal
Apply the de Broglie hypothesis to a quanlulll and a classical object.
Problem (a) Compare the de Broglie v\'avelength for an electron (IlI r = 9.lJ X 10-31 kg) moving at a speed equal to 1.00 X 10 7 m/s with that. of a baseball of mass 0.l45 kg pitched at 45.0 m/s. (b) Compare these wavelengths with that oran electron trClveling at 0.999(;. Strategy This problem is a matter ofsubstit.ution into Equation 27.14 for the de Broglie wavelength. In pan (b) the relativistic momentulll must be used.
882 "..
Chapter 27
Quantum Physics
..
...
,
,
..,
Solution (a) Compare the de Broglie wavelengths of the electron and the baseball. Substitute data for the electron into Equation 27.14:
A,.
It
(9.] 1 X 10 '" kg)(1.00 X ]0 7 m/s)
m,.v
7.28 X ]0-\\ m
Repeal the calculation \.vith the baseball data:
A - b -
h
-
1II 0l! -
6.63 X 1O-3.\ J's
-;---cc------''------c
(0.145 kg)(45.0 m/s)
(b) Find the \\iavelength for an electron traveling at 0.999[. Replace the momentum in Equation 27.14 'Vilh t.he relativistic lllOlllcmum:
m,.v
(6.63
X
10-'\'\ J s)V I - (0.999r)'/c'
A=-;-'------~...o!-'---------'--'----~
Substilllte:
,
(9.]] X lO-31 kg)(0.999·3.00 X lO'm/s) 1.09 X lO-13 m
. ...............•..
lrrrrr.- • • • • • • • • • • •
Remarks The electron \vavelength corresponds LO that of x-rays in the electromagnetic spectrum. The baseball, by contrast, has a wavelength much smaller than any apcrLure through which the baseball cOlild possibly pass, so we couldn't observe any of its diffraction effects. It is generally true that the wave properties or large-scale objects can't be observed. Notice thaL even at extreme relativistic speeds, the eleC[ron wavelength is still far larger than the baseball's.
~
QUESTION 27.4 How does doubling the speed of a panicle
3.97 X ]O-H m
Application: The Electron Microscope APPLICATION Electron Microscopes
A practical device that relics on the wave characteristics of electrons is the electron IniCl"oscope. A lnm.smission electron microscope, uscd for viewing flat, thin samples, is shown in Figure 27.17. In Illany respects it is similar to an optical microscope, but the electron rnicroscope has a Illuch greater resolving power because it can accelerate electrons to very high kinetic energies, giving them very shon wavelengths. No microscope can resolve details that are signiricantly smaller than the wavelength of the radi<,uion Ilsedlo illuminat.e the objecL. Typically. the ,.\'avelengths of electrons in an electron microscope are smaller than the visible wave~ lenglhs b}' a Factor of about JO-5 . The electron beam in an electron microscope is controlled by electrostatic or mag-neLic deflection, ,.. . hich acts on the electrons to focus the beam LO an image. Due 1.0 limitations in the e1eclromagnclic lenses used, however, the improvement in resolution over light microscopes is only about a factor of I. 000, two orders of magniwde smaller than that implied by the eleClron '.. . avelength. Rather than examining the image through an eyepiece as in an optical microscope. the viewer looks at an image {ormed 011 a r1uorescent screen. (The viewing screen must be fluorescent because ol.hcn..!ise lhe image produced wouldn"!. be visible.)
27.7
The Wave Function
883
Electron gun \~lcuum
AnOdC-:::-==U~~~~~
Eleclromagnelic lem
Core Coil
Eleclrornagnelic condenser lens
Electron beam Specimen goes here
Scrccn-------..
Specimcn chamber door
E
Projeclor lens
j
Visual transmission
~
•"
Photo cham bel· (b)
(0)
FIGURE 27.17 (a) Diagram of a transmission electron microscope fOI" viewing a Ihin. seclionerl ~'lmple. The "Ienscs~ that comrol the elecll'oll beam are magnetic deneClion coil". (b) An electron microscope.
APPLYING PHYSICS 27.2
X-RAY MICROSCOPES?
Electron microscopes (Fig. 27.17) take advantage of the \\lave naLUre of particles. Electrons are accelerated to high speeds, giving them a shon de Broglie '\'avelength. lmagine an electron microscope using electrons with a de Broglie ,vavelength of 0.2 nm. "Vhy don't we design a microscope using 0.2-nm photons to do the same thing?
27.7
Explanation Because electrons are charged particles, the)' interact electrically with the sample in the microscope and scaUcr according to the shape and density of various portions of the sample, providing a means of viewing the sample. Photons ofwaveleng1.h 0.2 nm are uncharged and in d1e x-ray region of the spectrum. The)' tend La simply pass through the thin sample without interacling.
THE WAVE FUNCTION
De Broglie's re"olutionary idea that panicles should hm·e a wave nature soon moved out of the realm of skepticism to the point where it was viewed as a necessary concept in understanding the subatomic world. fn 1926 Austrian-German physicist Erwin Schr6dinger proposed a ,",,'ave equation that described how matter \""aves change in space and time. The Schr6dinger wave equation represents a keyelement in the theory of quantum mechanics. It's as imponant in quantum mechanics as Ne\l,Iton's la,vs in classical mechanics. Schr6dinger's equation has been successfully applied to the hydrogen atom and to many other microscopic systems. Solving Schrodinger's equation (beyond the level of Ihis course) determines a quantity \(, called the wave function. Each panicle is represented b)' a wave function "I' that depends both on position and on time. Once \)1 is found, '\1/2 gives LIS information on the probability (per unit volume) of finding the particle in
884
Chapter 27
Quantum Physics
ERWIN SCHRODINGER Austrian Theoretical Physicist (1887-1961) Schrodinger is best known as the creator of wave mechanics, a less cumbersome theory than the equivalent
matrix mechanics developed by Werner Heisenberg. In 1933 Schr6dinger
left Germany and eventually settled at the Dublin Institute of Advanced Study, where he spent 17 happy, creative years working on problems
in general relativity, cosmology, and the application of quantum physics
any given region. To understand this, we return to Young's experiment involving coherent light passing through a double slit. First, recall from Chapter 2J that the intensity ora light beam is proportional to the square or the electric field strength E associated ,,,'ith the beam: I 0:: 1~·2. According to the wave model of light, there are certain points on the vicwing screen where the net electric field is zero as a result of destruct.ive interference of waves from the two slits. Because E is zero at these points, the intensity is also zero and the screen is dark there. Likewise, at points on the screen at which constructive interference occurs, Eis large, as is the intensity; hence, these locations are bright. NO'v consider the same experiment when light is vie\-ved as having a panicle nature. The number of photons reaching a point on the screen per second increases as the intensity (brightness) increases. Consequently, tile number or photons that strike a unit area on the screen each second is proportional to the square of the electric field, or A'a. IJY. From a probabilistic point of view, a photon has a high probability of striking the screcn at a point where the intensity (and /:.''l.) is high and a low probability ofs1.riking the screen where the intensity is low. '·Vhen describing particles rather than photons, qr rather than 1~: plays the role of the amplitude. Using an analogy with the description of light, we make the folIm,,'ing interpretation 0['(1 for particles: If \1' is a wave function used to describe a single particle, the value of \1,2 at some location at a given time is proportional to the probabi lily per un it VOltlllle of' finding the particle at that locat.ion 31 that time. Adding all the values of '1,2 in a given region gives the probability of finding the particle in that region.
to biology. In 1956 he returned home
to Austria and his beloved Tyrolean mountains, where he died in 1961.
27.8 THE UNCERTAINTY PRINCIPLE I f you \vere to measure the position and speed of a particle at any inst<1lll, you ".;Quid always be f-~lced ''lith expcrimemallJncertainties in your measuremenl.s. According to classical mechanics, no fundamental barrier to an ultimate refinement of the apparat1-IS or experimental procedures exists. In other words, it's possible, in principle, to make such measuremcnts 'Ni1.h arbitrarily small uncenainly. Quantulll theory predicts, howcver, thal such a barrier does exisL.ln 1927 "Verner Heisenberg (1901-1976) introduced this notion, which is now knQ\vn as the uncertainty principle: If a measurement of the position of a panicle is made with precision Il.\: and a simultancous measurement of linear momentum is made with precision IllJx' the product of the two uncertainties can never be smaller than h/47r: WERNER HEISENBERG
[27.16]
German Theoretical Physicist
(1901-1976) Heisenberg obtained his Ph.D. in 1923 at the University of Munich,
where he studied under Arnold Sommerfeld. While physicists such as de Broglie and Schr6dinger tried to develop physical models of the atom, Heisenberg developed an abstract mathematical model called matrix mechanics to explain the wavelengths of spectral lines. Heisenberg made many other significant contributions to physics, including his famous uncertainty principle, for which he received the Nobel Prize in Physics in 1932; the prediction of two forms of molecular hydrogen; and theoretical models of the nucleus of an atom.
1n other words, it is physically impossible to measure simultaneously the exact position and exact linear momentum of a particle. If D.X is very small, then 6./)x is large, and vice versa. To understand the physical origin of the uncertainty principle, consider the following thought experiment. introduced by Heisenberg. Suppose YOll \-..,ish to measure the position and linear momentum of an electron as accurately as possible. You might be able to do so by viewing the electron with a powerfullightll1icroscope. For you 1.0 sce the electron and determine its location, at least one photon of light musl bounce off the electron, as shown in Figure 27.18a, and pass through the microscope into your eye, as shown in Figure 27.18b. ',Vhen it strikes the electron, however, the photon transfers some unknown amollntofits momentum to the electron. Thus, in the process of locating the electron very accurately (that is, by making very small), the light that enables you to succeed in YOllr measurement changes Ihe electron's lllomenLUIll to some undeterminable extent (making 6.px . .'cry large).
ax
27,8
colli'iion
Incidcnl photon ScalltTCcl
"".--_.
885
FIGURE 27.18 A ulOught experimem for viewing an electron ''lith a p()\\'crful lllicroS<'Opt" (
..-\rIel
Before collision
The Uncertainty Principle
photon ..j'"
sJ
Ekclron
Recoiling electroll
(b)
The incoming photon has momentum Ii/A. As a result of the collision, the photon transfers pan or all of its momentum along the x-axis to the electron. Therelore, the lt1U:ertainl)' in the electron's lllomentulll after thc collision is as great as the momcn1.Um of the incoming photon: 6./).;; = Ii/A. Further, because the photon also has \\:ave properties, we expect to be able to determine the electron's position = A. Multiplying to w"ithin one wavelength of the light being used to view it, so these l\oVO uncertainties gives
ax
The value Ii represents the minimum in the product of the uncertainties. Because the uncertainty can always be greater than this minimuJll, we have D.x I:1jJ,? Ii
Apart [rom the numerical factor 1/4-7T introduced by Heisenberg's more precise anl1lysis, this inequality agrees with Equation 27,]6. Another form of'the uncertainty relationship sets a limit on the accuracy v.'ith which the energy E of a system can be measured in a finitC' time interval ~t: Ii D.I' D.l ? 4rr
[27.17]
It can be inferred from this relationship that the encrgy of a particle cannot be measured with complete precision in a very short interval of lime. Thus, \vhen an electron is viewed as a particle, the uncertaiIlly principle tells liS that (a) its position and veiocilY canllOt both be knowil precisely at the same time and (b) its energy can be ullcertain for a period given by 6.[ = 11/(41' DE),
Problem The spl:cd of' an electron is measured to be 5.00 X 10:1 m/s to an accuracy of 0.00300%. Find the minimum unccnaim}' in determining the position of this eleClron. Strategy After computing the momentum and its uncertainty, substillitc into Heisenberg's uncertainty principle, Equation 27.16. ~
.
• •••••••••••••••
Solution Calculate the momentum of the electron:
jJ, = "',v ~ (9.11 X 10
:>1
kg)(5.00 X 10 3 111 /s)
= 4.5fi X 10 "kg· m/s
The uncertainty in jJ.;; is 0.003 00%
01" 1 his
value:
D./J,
= 0.000030 OjJ, = = 1.37 X 10
31
(0.0000300)(4.56 X 10- 27 kg· m/s)
kg· m/s
'"""'111
886
Quantum Physics
Chapter 27
Now calculate the uncertainty in position lIsing this
c,x 1:>.p<
2: -
value of !1px and Equation 27.17:
1:>.x
h
471
--->
1:>.x
Ii
2: - - -
47T !1jJx
6.626 X 10-" 2:
47>-(1.37 X 10
J. s .
31
kg·m/s)
~ 0.384 X
10- 3 m
0.384 mm
...........
.
,..-'
~
Remark Notice that this isn't an exact calculation: the uncertainty in position call take any value as long as it's greater than or equal to the value given by the ullcenainl)' principle. QUESTION 27.5 True or False: The uncertainty in the position ofa proton in the helium nucleus is, on average, less than the uncertainly ora prown in a uraniulll aLom. EXERCISE 27.5 Suppose an electron is found somewhere in an atom of diameter 1.25 X 10- 10 m. Estimate the 1I11certailllY in the electron's mOmellllll1l (in one dimension). Answer
!1p ~ 4.22
_'_T.~--
X
10- 25 kg' m/s
-~---
SUMMARY 27.1 Blackbody Radiation and Planck's Hypothesis The characteristics of blackbody radiation can't be explained with classical concepts. The peak of" a blackbody radiation curve is given by "rien's displacemeotlaw, Ama),T = 0.2898 X 10-2 m· K
[27.1]
where t\nax is the wavelength at which the curve peaks and T is the absolmc temperature of" the o~ject cmitting the radiation. Planck first introduced the quantum conccpt when he assumed the subatomic oscillators responsible for blackbody radiation could have only discrete amounts of energy given by [27.2]
27.3
X-Rays
27.4 Diffroction of X-Rays by Crystals X-rays are produced when high-speed electrons are suddenly decelerated. W'hen electrons have been accelerated through a voltage \~ the shonesHvavelength radiation that can be produced is he Amin = {' ~ \f
[27.9]
The regular array of atoms in a crystal can act as a diffraction grating for x-rays and [or electrons. The condition for construct.ive int.erference of the diffracted rays is given by Bragg's law: 2d sin fj
= 11/11.
m = 1,2,3, .
[27.10]
where 11 is a positive integer called a quantum number and fis the frequency of vibration of the resonator.
Bragg's law bears a similaritr to the equation for the dif:' fraction pattern of' a double slit.
27.2 The Photoelectric Effect and the Particle Theory of Light The photoelectric effect is a process whereby c!cnrons are ejected from a metal surface when light is incident on t.hat surface. Einstein provided a successful explanat.ion of this effect by extending Planck's quantum hypol hesis to electromagnetic waves. In this model, light is viewed as a stream of panicles called pholOns, each \vith energy E = hf, where Jis the light frequenc)' and h is Planck's constant. The maxi~ mum kinetic energy of the ejected photoelectrons is
27.5 The Compton Effect X-ravs from an incident beam are scauered at ,'arious angl~s by electrons in a target such as carboll. In such a scauering event, a shift in 'wavelength is observed for the scattered x-rays. This phenomenon is knO\""n as the Compton shift. Conservation of momentum and energy applied to a pholon-eleclron collision yields the following expression for the shift in \-vavelength of the scaucred x-rays:
KEma ), =
where
h!- (I)
q) is the work function of' the metal.
[27.6]
Ll.A
=
Ii A - Ao ~ (I - cos 0) 11/,C
Here, 11/1' is the mass of the electron, and fj is the scattering angle.
C
[27.11]
is the speed of light,
Multiple-Choice Questions
27.6
The Dual Nature of Light and Matter
27.8
Light exhibits both a particle and a wave.:' nature. De Brog-lic proposed that al/malter has both a panicle and a wave nature. The de Broglie wavelength of any particle of mass JlI :-lIlel Sr(~('rl 71 is
It II. A = -=/)
The Uncertainty Principle
According to Hcisenberg's uncerlainly principle, it is impossible to measure simultaneously the exact position and cxact rnornenlllm of a panicle. If.:lx is the uncertainty in the measured position and ~/J" the uncerlainty in the momC'ntum, the product ~x ~Px is given by
127.14J
!If V
De Broglie also proposed thaI the frequencies of the w.wcs associated with particles abc)' the Einstein relationship 1,'= hf
27.7
887
The Wave Function
I n the theory of quantum mechanics, each panicle is described by ~l quantity 'V called the wave function. The probability per unil volume offinc1ing the panicle ala particular point at sOllle instant is proportional to 'It:!. Qllanlum mechanics has been highly successful in describing the behavior of atomic and molecular systems.
[27.16] Also, [27.17] where ill:.' is the uncertainty in the energy or the panicle and ~I is the u ncenainly in (he time it ta kcs to meaSLl re the cnerg}'.
FOR ADDITIONAL STUDENT RESOURCES. GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS I. What is the surface temperature ora distant star ha\'ing a peak wavelength of 475 nm? (a) 6100 I< (b) G 6~O K (c) 5510 K (e1) 6350 K (e) 6560 K
2. In a photoelectric experimcnt, a metal is irradiated with light of cnergy 3.56 eY. 1f a SlOpping porcnt.ial or 1.10 V is required, what is the work function of the metal? (a) 1.83 eV (b) 2.46 eV (c) 3.20 eV (e1) 4.66 cV (e) 0.644 eV
3. An electron is acceleraled through a potemial diflcrcnce of :1.00 V before colliding ,.. . ith a metal target. What minimum-wavelength light can such an dcctron emit? (a) 204 Hm (b) 352 nm (c) 4H nm (d) 536 nm (e) 612 11m 4. What is the de Broglie wavelength of an electron ,Kccleratcd from rcstthrou~h a potential difference ofSO.O V? (a) 0.10011111 (b) 0.17411111 (c) 0.139 11111 (
6. Which of the following statements are t rue according to the uncenaint)' principle? (a) It is impossible 10 simultaneously determine both the position and the momentum ofa particlc with arbitrary accuracy. (b) 11. is impossible to simultaneously dctermine both the energy and the mOmcJll.lll11 of a particle with arbitrary accuracy. (c) It is impossible to determine a particle's energy with
arbitrary accuracy in a finite amount of time. (e1) II. is impossible to measure the position of a panicle with arbitrary accuracy in a Ilnitc amount of time. (e) It is impossible to simultaneously measure both the energy and position ora panicle with arbitrary accuracy. 7. The first order x-ray scattering angle of crystal A is greater than the corresponding scallering angle of crystal B for x-rays of a given energy. What can be said of the separation distance dtl of crystalline planes in crystal A as compared with the separation distance dJj in B? (a) dA = d/l (b) tiLl> dB (c) dt! < d n (d) The answer depends on the energy or the x-rays. (e) The answer depends on the intensity of the x-rays. 8. A proton, electron, and a helium nucleus all move at speed v. Rank their de Broglie wavelengths from Ion· gesl to shonesl. (a) prolan, helium nucleus, electron (b) helium nucleus, proton, electron (c) proton, electron, helium nucleus (d) heliulTI nucleus, electron, proton (e) eleclron, proton, heliulll nucleus
9. \-\'hich one of the following phenomena most clearly dcrnonsLral.cs the particle nature of light? (a) diffraction (b) the photocleClric etTcCl (c) polarizal ion (d) intcrfercnce (e) refraction 10. \·"'Ilich one of the loIIO\...· ing phcnomena most clearly demonstrates the wave nature of electrons? (a) the photoclectric effect (b) Wien's law (c) blackbody radiation
(d) the CompLOn effect (e) diffraction of eleclrons by crystals
888
Chapter 27
Quantum Physics
-~-
~
-
,
CONqPTUAL QUESTlqNS
_
1. Tryon observe objects inside a vcry hot kiln, why is it difficulL to discern the shapes of the objects? 2. V{hy is an electron microscope more suitable t.han an optical microscope for "seeing" objects of atomic size?
3. Are blackbodies black? 4. ''''h)' is it impossible to simultaneously measure the position and velocity ora particle \.\'ilh infinite accuracy? 5. All ol~jecls radiate energy. ""hy, then, arc we sec all the objects in a dark room?
110t
able
LO
6. Is light a W
8. '''Thich has more energy,
9. 'Why does the existence of a clIwff freq uency in the photoelectric effect favor a particle theory or light rather than a wave theory? ~._-
,
~.
10. \·\lhat effect, if any, would }'Otl expect the l.emperal.Ure of" a material to have on the ease with which electrons can be ejected from it via the photoelectric effect? 11. The cutorf" rrequency of a material is 10' Are eleCLrOl1S emilled from the material when (a) ligln of frequency greater than .fo is incident on the material? (b) Less than for
12. The brightest star in the constellation Lyra is the bluish star Vega, whereas the brightest star in Bootes is the reddish SUII' ArclUrus. How do you account for the difference in color of the two stars? 13. If the photoelectric effect is observed in one metal, can you conclude that. the effect. will also be observed in another metal under the same conditions? Explain. 14. The aLQllls in a crystal lie in planes separated by a few lcnths of a nanometer. Can a crystal be used LO produce a diffraction pattern with visiblc light as it does for x-rays? Explain your answer with reference to Bragg's law.
.
. ~RQ_~~EMS ~
WebAssign
J. 2, 3 =
m
E
The Problems for this chapter ma)' be
assigned online at WebAssign. straightforward, inlcrlllcdialc, lh;lllelli-{illg"
= denOtes guided problem = denOles enhanced content problem
=
biomedical applicalion
0= dcnoles l"ull solution available in
Sllull'nt SolutioJls !\:Imll/alj
Stud)" Guide
SECTION 27.1 BLACKBODY RADIATION AND PLANCK'S HYPOTHESIS
[JJ (a)
What is the surface tcmper:lt.ure of Betelgeuse, a red gianl star in the constellation of Orion, which radiates with a peak wavelength of about 970 nm? (b) Rigel, a bluish-while Slar in Orion, rad iat.es widl a peak wavelength of 145 nm. Find the temperature of Rigel's surface.
2. (a) Lighlning produces a maximum air temperature on the order of 10 1 K, whereas (b) a nuclear explosion produces a temperature on the order of 10 7 K. Use "Vien's displacement law to find the order of magnitude of the wavelength of the thermally produced photons radimed with greatest intensity b}' each of these sou rces. :\lame the part or the electromagnetic spectrum where you would expect each to radiate most strongly. 3.
III
4.
E
The human eye is most sensitive to 560-nm (green) light. \rVhal is the temperature of a blackbodr that would radj~dle most intensely at this wavelength?
A tungsten filament of a glowing lightbulb has a temperature or approximately 2 000 K. (a) Assuming the operating filament is a blackbody, det.ermine the peak wa"c1ength or its emitted radiation at this temperature.
(b) \-Vhy does your answer to pan (a) suggest that more energy from a lightbulb goes illlO heat than imo light? 5. Calculate the energr in electron vaILS ofa photon having a wavelength (a) in the microw(lve range, 5.00 em, (b) in the \'isible light range, 500 11In, and (c) in the x-ray range, 5.00 nm. WSupposc a star with radius 8.50 X lOll m has a peak wtlyelength of 685 nm in the spectrum of its emitted radiation. (a) Find lhe energy of a photon with this wavelength. (b) What is surface tempera1.Ure or the slar? (c) Al whm rate is energy emitted from the star in lhe form of radiation? Assume the star is a blackbody (e = I). (d) Using the answer to pan (a), estimate the rate al which photons leave the surface of the star. 7. An F\1 radio transmiuer has a power output of 150 kW and operates at a frequency 01'99.7 MHz. I-Iow Illany photons pCI' second does the transmitter emit? 8. II The threshold of dark-adapted (scotopic) vision is 4.0 X 10- 11 W/I1l~ at a central wavelenglh of SOD nm. If liglll with lhis intensity and wa\'elenglil enters the eye when the pupil is open to its maximum diameter of 8.5 mm, how man)' photons per second enter the cye?
SECTION 27.2 THE PHOTOELECTRIC EFFECT AND THE PARTICLE THEORY OF LIGHT []JWhen light. of \vavelengl.h 350 nm falls on a potassium surface. electrons having a maximum kinetic energy of 1.31 eV are emillcd. Find (a) the work function of potassium, (b) the cUloff wavelength, and (c) the frequcncy correspolldi llg 1.0 the cutoff wa\·elcngth. 10. Electrons are ejected from a certain metallic surface with speeds ranging lip to 4.6 X 10 5 mls when light with a
Problems wavelength of A = 625 nm is used. (a) What is the work function of the metal? (b) What is the ClllOff frequellc), for this metal?
II.
12.
D
The work fUllction for platinulll is 6.35 eV. (a) Conven the value of the work function [rom eleClron volts to joules. (b) Find the cutoff frequency for platinulll. (c) What maximum wavelength of light incident all platinum releases photoelectrons from the platillum's surrace? (d) If light of energy 8.50 eV is incident 011 zinc, what is the maximum kinetic encrgy of the ejected photoelectrons? Give the answer in electron volts. (e) For photons of energy 8.50 cV, what stopping potential would be required to arrest thc current ofphotoelcclrons?
mil
Lithium, beryllium, and mercury have work functions of 2.30 eV, 3.90 eV, and 4.50 e\'. respeeti\'cly. 1.ight with CI waveleng111 of 4.00 X 10 2 nm is incident on cach of these metals. (a) \-\'hich of these metals emit photoelectrons in response to the light? Why? (b) Find the maximum kinetic cnergy for the phUlOelcctrolls in each case.
~When light of wavelength 254 nm
falls all cesium. the required stopping potential is 3.00 V. Iflight ofw:welength 436 nm is used, the stopping potemial is 0.900 V. Use this information to plot a graph like thaI shown in Figure 27.6, and rrom the graph determine the cutoff frcquency for cesium and its work function.
14. Ulu'aviolct lighl is incident normally 011 the surface of a certain substance. The binding energy of the electrons in this substance is 3.44 eV. The incident light has an intensity of" O.055\,\I/m 2 • The c1ectrons arc phOlocleClricall}' emittcd with a maximum speed of 4.2 X 10 5 m/s. How many electrons are emitted from a square centimeter of the surface each second? Assume the absorption or every photOn ejects an electron.
SECTION 27.3 X-RAYS 15. The extremes of the x-ray portion of the electromagnetic :-;pectrum range from approximalc1y 1.0 X 10- 8 m to \.0 X 10- 1:'1 m. Find the minimum accelerating voltages required to produce wavelengths at these two extremes. 16.
mil
Calculate the minimum-w
17. What minimull1 accelerating voltage is required to produce an x-ray with a wavelength of70.0 pm?
SECTION 27.4 DIFFRACTION OF X-RAYS BY CRYSTALS 18. Whcn x-rays of wavelength of 0.129 nm arc incident Oil the surface of (1 crystal having a struCLUre similar to thaI of NaCI. a first-ordel" maximum is observed al 8.15°. Calculate the interplanar spacing of I. he crystal based on this information. [][]Potassium iodide has an illlerplanar spacing of d = 0.296 nm. t\ monochromatic x-ray beam shows a first-
889
order diffraction maximum when the grazing anglc is 7.6°. Calculate the x-ray wavelength. 20.
mil
The first-order diffraction maximum is observed at 12.6° [or a crystal having an illlcrplanar spacing of 0.240 nm. How many other orders can be obscl"\'cd in the diffraction pattcrn, and al what angles do they appear? 'Why is there an upper limit to the number of observed oroers?
21. X-rays of wavelength 0.140 nm arc reflected from a certain crystal, and the first-order maximum OCClirs at an anglc of 14...1°. What value does this gi\'e for the illlerplanflr spacing of the cryst.al?
SECTION 27.5 THE COMPTON EFFECT 22. X-rays are scattered from a target at an angle of 55.0° with the direction of the incident beam. Find the wavelength sh ift or the sC
123.IA 0.001 6-nm phOlOll scauers from a free electron. For whaL (photon) sC
SECTION 27.6 THE DUAL NATURE OF LIGHT AND MATTER 27. (a) [f the wavelength or an electron is 5.00 X 10- 7 m, how Fast is it moving? (b) If the electron has a speed equal to 1.00 X 10 7 m/s, what is its wa\'elength? 28. Calculate the de Broglie wavelength of a proton moving al (a) 2.00 X 10' m/s and (b) 2.00 X 10' m/s. [illDc Broglie postulated that the relationship A = h/ll is valid for relativistic particles. What is the dc Broglie wavelength for a (rc1aLivistic) electron having a kinetic energy
01" 3.00 MeV' 30. (a) Calculatc the momentum or a photon having a wavelength of 4.00 X 10~ nm. (b) Find the speed of an electron having the same momentum as the photon in part (a). 31. The resoh"ing powerofa microscope is proportional to the wavelength lIsed. A resolution of 1.0 X 10- 11 III (0.010 11m) would be required in order to "see" an atom. (a) If electrons were used (electron microscope), what minimum kinetic energy would be required of the electrons? (b) If photons were used, what minimum photOn energy would be needed I.oohtain 1.0 X 10- 11 m resolmion?
890 32.
Chapter 27
Quantum Physics
am
A panicle of mass If/. and charge q is accelerated from rest through a potential difrcrcnce ~v. (a) Use conservation of cncrg}' (0 find a symbolic expression for the momentum orllle particle in lcrmsof Ill, q, lind aV Assume the panicle's speed isn't relali"istic. (b) Write a symbolic expression for the de Broglie wavclenglil \Ising the result of part (a). (c) If an electron and proton gn throllgh the same pOlcmial difference bm in opposite directions, which particle will have the shorter wavelength?
grazing angle is 2.5°. What is lhe spacing bctween ing planes in the crysral?
reflcct~
Figure P27.4~ shows the spectrum of light emittcd by a firerly. Determine the tcmperature of a blackbody that would emit radiation peaked at the same frequency. Based on your result, would you say that fircfly radiation is blackbody radiation?
43. _
1.0
."C' u.8
SECTION 27.7 THE WAVE FUNCTION
0
.~ ."" <:"
SECTION 27.8 THE UNCERTAINTY PRINCIPLE 133.IIn the ground state of hydrogen, the ullcenainty in the position of the elcclron is roughly 0.10 nm. If the speed of the electron is approximately the same as the uncertainty in its speed, about how fast is it moving?
34. A O.50-kg block rests on the icy surface of" a frozen pond, which )'OLl can assume to be frictionless. If the location of the block is measured to a precision of 0.50 cm, what is the minimum uncertaimy in the block's speed, assuming I he mass is known cxactly?
35. Suppose optical radiaLioll (,\
= 5.00 X 10-' Ill) is \lsed 1O determine the position of an electron to within the wavelength of the lighl. \-Vhat will be the resulling uncenaimy ill the e1ectron's velocity?
36. Suppose Fuzzy, a quanwm mechanical duck, lives in a world in which II = 21TJ '5. Fuzzy has a mass of" 2.00 kg and is initially known to be within a pond 1.00 III , ...· ide. (a) Wha( is Ihc minimum uncertainty in tile duck's speed? (b) Assuming this uncertainty in speed to prevail for 5.00 s, determine the uncertainty in Fuzzy's position after this timc.
I 37.IThe
average lifctime of a muon is aboUl 2 J.ls. Estimate the minimum 11l1certailllY in the cnergy ofa muon.
38. mil (;-1) Show that the kinctic encrgy of a nonrc1ativistic particle can be written in tcrllls 01" its momcntum as KE = p'l./2m. (b) Usc the results of part (a) to find the minimum kinclic energy ora proton confincd within a nucleus having a diameter of 1.0 X 10 15 Ill.
0.6
-
0.4
'"
0.2 400
500
fiOO
Wavelength (11m) FIGURE P27.43
44 .. Iohnny .Julllper's favorite trick is to step out of his 16tl1stor}' "lindow and r~l1150.0 Jll into a pool. A news reporter takes a piclllre of 75.0-kg Johnny just before he makes a splash, llsing an exposure time of!i.OO ms. Find (a) Johnn)"s de Broglie wavclength at. this moment, (b) the uncertainty of his kinetic energy measurcment during such a pcriod or time, and (c) the percent error caused by such an uncenaint}'. 145.IPhotons of wa.. . elength 450 nm are incident on a metal. The most cnergetic electrons ejected from the met.al are bent into a circular arc of radius 20.0 cm by a magnetic field with ;l magnil ude of" 2.00 X 10 -f> T. \Vhal is the work function of' the meLal? 46. A 200-.\1e\l photon is scattered <1t40.0° by a free proton that is initially al. rest. Find the energy (in MeV) of the
scattered phOto11.
47. A light source of wavelength ,.\ illuminatcs a metal and cjects photoelectrons with a maximum kinetic energy or 1.00 e\'. A secolld light sourcc or wavelength ,.\/2 ~jects photoelectrons with a maximum kinelic energy of 4.00 eY. What is I.he work ['unction of the metal?
ADDITIONAL PR08LEMS
39. A microw<we photon in the x-band region has a wavelength of 3.00 CIll. Find (a) the momentum, (b) the frequency, and (c) thc energy or the phoLan in electron \"Olts. 140.IFind the speed of an electron having a de Broglie wavelength equal to its COlllplOll wavelenglll. Hinl: This electron is relativistic. 41. A 2.0-kg object falls from a height of 5.0 m to the gTound.
IIallthc gravitational potential energy of lhis mass could be cOllvcned to visiblc light of wavelength 5.0 X 10-7 nl, how man)' photons wOll1d be produced?
42. An x-ray tube is operated at 50 000 V. (a) Find the
mini~
mum wavelength of the radiation emiued by this tube. (b) Il'the radiation isdirectcd ;It a crystal, the first-order maximum in the reflccted radiation occurs when dle
48. Red light ofwavelcngth 670 nm produces photoelectrons from a certain photoemissive matcrial. Green light or waveleng-lIl 520 nm produces pholoe1ectrons from the same Jllalerial wilh 1.50 times the maximum kinetic energy. What is the material's work function?
149.11 low
fast Illust an electron be moving if ;lll il.s kinetic energy is lost to
50. Show that if an elt:ctron wcre conrined inside an atomic nucleus of diamcter 2.0 X 10-15 nl. it would have to bc moving relativistically, whcreas a proton confined to lhe same nucleus can be moving at less than olle-temh the speed of light.
Burning gunpowder transfers energy to the atoms of colorproducing chemicals, exciting their electrons to higher energy states. In returning to the ground state, the electrons emit light of specific colors, resulting in spectocular fjre works displays. Strontium produces red, and sodium produces yeilowl orange. 4
ATOMIC PHYSICS A hot gas emits light of certain characteristic wavelengths that can be used to identify it, much as a fingerprint con identify a person. For a given atom, these characteristic emit-
ted wavelengths can be understood using physical quantities coiled quantum numbers. The simplest atom is hydrogen, and understanding it can lead to understanding the structure of other atoms and their combinations. The fact that no two electrons in an atom can have
28.1
Early Models of the Atom
28.2
Atomic Spectra
28.3
The Bohr Model
28.4
Quantum Mechanics and the Hydrogen Atom
28.5
The Exclusion Principle and the Periodic Table
28.6
Characteristic X-Rays
28.7
Atomic Transitions and Lasers
the same set of quantum numbers-the Pauli exclusion principle-is extremely important in
understanding the properties of complex atoms and the arrangement of elements in the periodic table. Knowledge of atomic structure can be used to describe the mechanisms involved
in the production of x-rays and the operation of a loser, among many other applications.
28.1
EARLY MODELS OF THE ATOM
The model of the atom in the days of Newton was a tiny, hard, indestructible sphere..r\lthough this model was a good basis [or the kjnetic theory of" gases, new models had to be devised when later experiments revealed the electronic nature of" atoms. J. J. Thomson (1856-1940) suggested a model of the atom as a volume of positiye charge with electrons embedded throughout the \'olume, much like the seeds in a watermelon (Fig. 28.1). 1n 191J Ernest Rutherford (187]-1937) and his students Hans Geiger and Ernest ~llarsden pcrfonlled a critical experiment shm.. . ing that Thomson's model couldn't be correct. In this experiment a beam of positively charged alpha particles was projected against a thin metal foil, as in Figure 28.2£1 (page 892). Most of the alpha panicles passed through the foil as if it were empty space, bllt a few panicles ,.. 'ere scaLtered through large angles, some even traveling backwards. Such large deflections ,veren'[ expected. In Thomson's model a positively charged alpha panicle would never come close enough 1O a large positive charge to cause any large~angle deflections. Rutherford explained these results by assuming the positive charge in an atOm was concentrated in a region called the nucleus that was small relative to the size of the atom. Any electrons belonging to the atom
EleClron
• • •
• • • • •
FIGURE 28.1 Thomson's mudd or tIll; alom. with the electrons embed, <1<-'d inside Ih...: positiw~ charge like ~('...:ds in a \,';llel"meloll.
891
892
Chapter 28
Atomic Physics
FIGURE 28.2 (a) Geiger and Marsden"- H.:<:hniqut: for obst:l"ving lht:" scallcring of alpha particles from a
lhill
foillargt:l. The source is a
Viewing screen
natll-
I"all\' occurrillg radioactive SUbsl;lllce, such as radilllll. (h) Rllrllf'r1ilrCl\ ]11;mCl
~~
SOllrce.~
1Ji~
SIR JOSEPH JOHN THOMSON English Physicist (1856-1940)
Thomson, usuolfy considered the discoverer of the electron, opened up
the field of subatomic particle physics with his extensive work on the deflection of cathode rays (electrons) in an electric field. He received the 1906 Nobel Prize in Physics for his discovery of the electron.
, ,, I I I I
-----
~
~
,
,,
~
~
•
\
,
,,
,
•,
SCI"Cell
(a)
, ,,, ,
,,
~
,
~
~
I I
r
,,
(b)
were visualized as orbiting the nucleus, much as planets orbit the SUll, as shm'ill in Figure 28.2b. The alpha panicles used in Rutherford's expcrimcnLs were bter identified as the nuclei o[ helium atoms. There \.. . ere two basic difficulties with Rutherford's planetary model. First, an atom emits certain discrete characteristic frequencies of eleclromagnetic radiation and no others; the Rutherford model was unable to explain this phenomenon. Second, the electrons in Rutherford's model undergo a celllripctal acceleration. According to !'vlaxwell's theory of electromagnetism, centripetally accelerated chargcs rcvolving with frequency lshuuld radiate clectromagnetic waves of the same f'requency. As the electron radiates energy, the radius of its orbit steadily decreases and its frequency of rcyolution increascs. This process leads to an everincreasing frequency of emitted radiation and a rapid collapse of the atom as the electron spinds into the nucleus. Rutherford's model or the atorn gave \vay to that of' Niels Bohr, which explained the characteristic radiation emill.ed I'rom atoms. Bohr's theory, in turn, \.. . a5 supplanted by quanwlll mechanics. Both the lattcr theories arc based on studies of atomic spectra: the special pattern in the \.. .avclengths of emittcd light that is un iq lie for every d incren t clement.
28.2 ATOMIC SPECTRA Suppose an evacuated glass tube is fliled with hydrogen (or some other gas) at 1m.... pressure. If a voltage applied between metal electrodes in the tllbe is grcat enough to produce an electric current in the gas, the tube cmits light having a color that depends on the gas inside. (That's how a neon sign works.) When the emitted light is analyzed \-vith a spectrometer, discrete bright lines are observed, each having a different wavelength, or color. Such a series of spectral lines is called an emission spectrum. The wavelengths comained in such a spcctrum are characteristic orthe element emitting the light (Fig. 28.3). Because no two elements emitthc same line spcctrum, this phcnomcnon represents a reliable technique for identifying elements in a gaseous substance. The emission spectrum of hydrogen shown in Figure 28.4 includes four prominent lines that occur at wavelengths 01" 656.3 nm, 486.1 n111, 434.1 11m, and 4]0.2 11m. In 1885 Johann Balmer (1825-1898) found that the wavelengths of these and less prominent lines can be described by the simplc crnpirical equatioll Balmer series ~
[28.11 where n may have integra] values of 3, 4, 5,. ., and RH is a constant, called the Rydberg constant. If the wavelength is in meters, then RH has the value
Rydberg constont
-+
[28.2]
28.2
~oo
A (nm) 400
FIGURE 28,3 \'isible "Pt'("[I':I. (a) Line ~p(·t.-Lra produc-.:d h)' emission in tht: vi.~ihle raJlgc for the elements hydrogen, mercury. :\I1d IIcon. (Il) The ab~orption spect rlllll li)r
70U
600
893
Atomic Spectra
I
hydrogf'll. ThL' dark absorpti('ll
line~
occur at lhe same wa\"e!clIgths as til{: emission lillt·" for hydro~l'll shown in (a).
(a)
.
.
.
,
I A(nlll) 400
Gon
.:;00
700
(hi A.(llln )
The first line in the Balmer series, at 656.3 11m, corresponds LO 11 = ~ in Equation 28.1, thc line at 486,1 lllll corresponds Lo n = 4, and so on. In addition LO the Balmer series of speclral lines, tile Lyman series ,vas subsequently discovered in the Far ultraviolet. with the radiated wavelengths described by a silllilar equalion, with to! in Equation 28.1 replaced by}2 and the integer 11 greater than I. The Paschen series corresponded to longer ,..'avelengths than 1he Balmer series, with the 2~ in Equation 28.1 replaced by 32 and 11 > 3. These models, together with man)' other observations, can be combined 10 yield the Rydberg equation,
+= (J" - -!,,) II"
11
ur
[28.3J
Ir
""here m and "II are positive integers and n> m. In addition to emitting light at specific wavelengths, an clement can absorb light at specific wavelengths. The spenral lines corresponding to this process Corm what is known as all absorption spectrum. An absorption spectrum can be obtained by passing a continuous radiation spectrum (one containing all h'avelengths) through a vapor orthe element being 'lllalyLed. Thc absorption spectrum consists of a series of dark lines superimposed on the olhen"ise bright, continuous spectrum. Each line i tl the: a bsorptioll spectrulTI of a givell clement coi ncides wi Lil aline in the emission spectrulll orlhe element. Irhydrogcn is thc absorbing vapor, for example, dark lines will appear at the visible wavelengths 656.3 11m, ll86.l nm, 4:14.1 nm, anc1410.2 nm, as shown in Figures 28.3b and 28.4. The absorption spectl'urn of all c1cment has many practical applications. For example, thc continuous spectrum of radiation emitted by the Sun lllust pass through the cooler gases 01" the solar atmosphere before reaching Earth, The vari~ ous absorption lines observed in the solar spectrum have been used to identify ele· IIlcnts in the solar atlllosphere, including heliulll, which was previously unknm.. . n.
APPLYING PHYSICS 28.1
~(H.6
I 410.2
\
4R6.1
fi~6.;{
434.1
FIGURE 28.4 The Balmcr ~I:ries Ofspcxlrallilles for ,\tomi(' hydrogen, with sC\"t:rallines lll
+-
Rydberg equation
APPLICATION Discovery of Helium
THERMAL OR SPECTRAL
On observing a yellow candle flame, your laboratory partner claims that the light from the flame originates from excited sodium atoms in the flame, You disagree, stating that bCG.llIse the' ctlndle flame is hOl, the radiation 1Tlllst be thermal in origin. Before 1he
disagreement becomes more intense, how could you dctermine who is correct? Explanation A simple determination could be made by observing the light from the candle flame throllgh
894
Chapter 28
Atomic Physics
a spectrometer, which is a slit and diffraction grating combination discussed in Chapter 25. I r lhe spectrum of the light is continuous, it's probably thermal in origin. Uthe spectrum shows discrete lines, it's atomic
APPLYING PHYSICS 28.2
AURORAS
At extreme northern IatiLUdes, the aurora borealis provides a beautiful and colorful display in the night sky. A similar display, called the aurora australis, occurs near the southern polar region. What is the orlgin of the various colors seen in the auroras? Explanation The aurora results from high-speed panicles interacting with Eanh's magnetic neld and entering the atmosphere. \·Vhen these particles col-
-,
, ,, I I
,
,I
in origin. The results or the experiment show that the light is indeed thermal in origin and originates from random molecular motion in the candle name.
28.3
lide \vith molecules in the atmosphere, they excite the molecules just as does the voltage in the spectrum tubes discussed earlier in this section, Tn response the molecules emit colors of light according to the characteristic spectra of their atomic constituents. For our atmosphere, the primary constituents are nit.rogen and oxygen, which provide the red, blue, and green colors of the aurora.
THE BOHR MODEL
At the beginning of the 20th cenlUry, it wasn't understood why atoms of a given element emitted and absorbed only certain wavelengths. Tn 1913 Bohr prO\,jded an explanation of the spectra of hydrogen that includes some features of the cur· remll' accepted theory. His model of the hydrogen atom included the follmving basic assumptions:
I
FIGURE 28.5 Diagram representing: Bohr's model or the h~'drogen atom. The orbiting electron is allowed only in specific orbits of discrete radii.
1. The electroll Jlloves in circular orbits about the proton under the influence of the Coulomb force of attraction, as in Figure 28.5. The Coulomb force produces the electron's centripetal acceleration. 2. Only certain electron orbits are stable and allO\ved. In these orbits no energy in the form of electromagnetic radiation is emitted, so the LOtal energy of the atom remains constant. 3. Radiation is emitted by the hydrogen atom when the electron '~UI1lPS" from a more energetic initial state to a less energetic state. The '~ump" can'l be visualized or treated classically. The frequency fof the radiation emitted in the jump is related to the change in the atom's energy, gi"en by [28.4]
1':, - I'J= hi
where E; is the energy of the initial st.ate, t./is the energy of the final state, It is Planck's constant, and Ei > Et The frequency of the radiation is indejJendenl of Ihe.!1"equenc)' o.!the electron:~ oTbital molion. 4. The circumference of an elecn'on's orbit lllust contain an integral number of de Broglie \vavelengths, 1'1 = 1,2,3,. 27TT = nA (See Fig. 28.6.) Because the de Broglie wavelength or an electron is A = II/meV, we can wTite the preceding equation as tn"vr = nfl.
FIGURE 28.6 Standing-wave pattern for an electron wave in a stahle orbit of h\·drogen. There arc three full wavelengths ill this orbit.
[28.5]
n=J,2,3,.
where 17 = h/27T. vVith these four assumptions, we can calculate the allowed energies and emission wavelengths of the hydrogen atom using the model pictured in Figure 28.5, in \""hich the electron travels in a circular orbit of radius r\'\:ith an orbital speed v. The electrical potential energy of the atom is Q,Q2
PE=k-=k r r r
(-e)(e) r
e-"
,.
-k , -
28.3 The Bohr Model
895
where kf is the Coulomb constant. Assliming Ihe nucleus is at rest, the lOtal energy I'~ or the aLOm is the sum or tile ki netic and pOlcl1lial energy: _
E=
_
+
Kl~
I ') <j11l,.v- -
PE =
"
.,
e-
,.
ki'-
[28.6]
By Newton's second lav,', the electric force of" tlllraction on the electron, h,.f''J./r'J., equal w"a,. 'where (If = v~/ris the centripelal acceleration or the electron, so
ll1usl
'I
,)
v-
p-
m,,~=
r-
Multiply both sides of this eqwllion by 42 energy: J
1.0
get an exprc.'Jsion for the kinetic
128.8]
= -
Combining Illis result ,,:ith Eqll
E=
[28.9]
2,-
where the negative value of the energy indicates that the electron is bound to the proton. An expression for T can be obtained by solving Equations 28.5 and 28.7 for v 2 and equating the resuhs: ')
1r
1/2 fI
= --.,-') mf-r-
NIELS BOHR Danish Physicist (1885-1962)
k,P'J. 21'
"
""',V" -
[28.7]
1".---;;
r
Bohr was on active participant in the early development of Quantum mechanics and provided much of its philosophical framework. During the 1920s and 1930s he headed the Institute for Advanced Studies in Copenhagen, where many of the world's best physicists come to exchange ideas. Bohr was awarded the 1922 Nobel Prize in Physics for his investigation of the structure of atoms and of the radiation emanating from them.
hr(''J.
=-111,1"
'I1'!.fI 2 I"'j
= ---., JIl
k
r r
1/=
('-
1,2,3,.
[28.10]
This equation is based on t.he assurnptioll that the electron can exist only in certain allowed orbits determined by the integer II. The orbit with the smallesL radius, called the Bohr radius, flO' corresponds to n::.... : ~... ..:. :'
,," "
ftn =
~--'J
= 0.052 9 nm
[28.11]
111" "eA general expression [or the radills of any orbil in the hydrogen atom is obt<:lined by substituting Equation 28.[ 1 into Equation 28.10: ACTIVE FIGURE 2B.7
[28.12] The first three Bohr orbiLs [or hydrogen art" shm.. . n in AClive Figure 28.7. Equation 28.10 can then be subslituLed into Equation 28.9 to gi\'e the following expression for the energies orlhe quantum statcs: n
=
1,2,3,.
[28.13]
Il"we substitutc numerical values into Equation 28.13, we obtain
13.6
l!;11 = - -.,- cV n"
[28.14]
The Im.. . est-energy state, or ground state, corresponds to 11 = I and has an energy f] = -ml'k/,e l /2 'J. = -13.6 eV. The nexL state, corresponding to 1/ = 2, has an energy E~ = 1'~]/4 = -3.40 eV, and so on. An energy Ic\'cl diagram showing thc
n
The firM lhl'ee circular orbits prt.:dieted hy lhc Bohr model of till,: hydrogen atom.
t- The energy levels of hydrogen
896
Chapter 28
Atomic Physics
ACTIVE FIGURE 28.8 An energy level diagram lor hydrogen. Quantum lIulllbcrs are given on the lcfl. and energies (in electron \'ohs) arc gi\cll on the riglu. Venical arrows n:pl-CSCtll the four lowe~l-enel"g}' I nll1<;ition"
10(,1')
"
~~~~~i;~ ]+I-'-U1. -
(1.(10
for eaell of tile spec-
3 ~C+j-I.-l
lral series shown. Till: colored arrows for the 13almer series indicate lilal this series results illl'isible liglu.
-0.5442 I ! 11! -O.~504 Paschen 1.::11 2 series
2 ~1·+I-'-'C!.1'-_ -3.401 Balmer series
Lyman series
ll.ll.~
-13.606
energies of these stat.ionary states and the conesponding qua mum numbers is given in Active Figure 28.8. The uppermost level shown, corresponding to E = 0 and 11 ---+ x, represents the state for which the electron is completely removed from the aLOm. In this state the electron's KE and PE are both zero, which means that the electron is at rest infinitely far m.. . ay from the proton. The minimum energy required to ionize the atom-that. is, t.o completely remove the electron-is called the ionization energy. The ionization energy [or h)'drogen is 13.6 eV. Equations 28.4 and 28.13 and the third Bohr postulate shm.. ' that if the electron jumps from one orbit with quantum number 11; to a second orbit with quantum number 11[, it emits a photon o[ frequency fgiven by
I=
[28.15]
where 11[ < nj • To converlthis equation into one analogous to the Rydberg equation, substitute f= (ilt and divide both sides by c, obtaining -J =
A
R1j ( -1-
n}
-
I )
-.-,
n,-
[28.16]
where [28.17]
...TIP 28.1 Energy Depends on n Only for Hydrogen Because all other quantities in EqlL
moms, the energy 1e\'cls depend on other well.
qUClnllllll
numbers as
'I
Substituting the known values of Ill" If,. P, C, and fi verifies that this theoretical value for the Rydberg constant is in excellent agreement with the experimelllally derived value in Equations -12.1 through 12.3. \,Vhen Bohr demonstrated this agreement, it was n.:cognized as a major accomplishment of h is theory. We can use Equation 28.16 to evaluate the wavelengths for the various series in the hydrogen spectrum. For example, in the Balmer series, 1'1 = 2 and l1 i = 3, 4, 5, . (Eq. 28.1). The energy level diagram for hydrogen shown in Active Figure 28.8 indicates (he origin of the spectral lines. The transitions between levels are represented by vertical arrows. NOle that whenever a transition occurs between a state designated by '11, to one designated by 11/ (where n i > n/), a photon with a frequency (Ej - E/)/h is emiued. This process can be interpreted as foJlows: the lines in the visible part of the hydrogen spectrulll arise when the electron jumps from [he third, fourth, or even higher orbit to the second orbit. The Bohr theory SlICcessfully predicts the wavelengths or all the observed spectral lines of hydrogen.
28.3
EXAMPLE 28.1
The Balmer series for the hydrogen
E (cV)
II
OO~~i~o.oo G -0.38 5 -0.54
corresponds to electronic transitions that terminate in lhe slale with qU
" ~
-0.85
-,.--t--t-++- - 1.s I
2 _.L-.L--'----'----'---_ Balmc.·
~1.-1 0
series
Strategy This problem is a maller of substillning values intO Equation 28.16. The frequency can then be obt;'linecl from c =.fA and the energy from E = Iif. The longest-wavelength photon corresponds La the one that is cmiLLed \-"hen the electron jumps from the n i = 3 state to the III = 2 state. The shortesH...· avelcngth photon corresponds 10 the one that is emitted when the eleclronjumps from 'n/ = ::0 to the 111 = 2 state.
,.
897
The Balmer Series far H)ldra en
Goal Calculate the wavelength. frequency, and energy ofa photon emitted during an electron transition in an ;HOIll. Problem
The Bohr Model
FIGURE 28.9
(Example 28.1) T"allSiLiolls rel>pollsible for the B;dlller series for the hydrogen atom. Al1lransitiollS terminate at the /I = 21c\'cl.
...,
.
Solution (a) Find the longesHvavelength photon emitted in
the Balmer series and determine its frequency and energy. Substitute into Equation 28.16, \-"ith n j = 3 and 'IIf = 2:
I
A
36 ') = 6.563 5(1.097 X 10 7 m
Take the reciprocal and substitute, finding thc wavelength:
X
656.3 nm
=
c
2.998 X 10' m/5 6.3fi~ X 10- 7 m
i':O\v use c = fA to obt.ain the frequency:
J= -
Calculate the pholon's energy by substituting into Equation 27.5:
E = hf= (6.626 X 10-3< J. 5)(4.368 X 10" Hz)
(b) Find the shortest-wavelength photon emitted in the Balmer series. Substitute into Equat.ion 28.16, with] /n; ---10 0 as "lI,~ccand 11/ =2:
Take the reciprocal and substilUte, finding the wa\'elength:
A
1\
R"
X 10-'"J
=
(_I" "/1/ - -.!,)
4
I1 j -
4.368 X 10'" Hz
1.892 eV
=
R"
(21"- 0) -
R" 4
4
A= = -,----------;c--"o-) = 3.646 X 10- 7 m R" (1.097 X 10'm =
ltr.... • • • _ • • • • • • • • •
=
= 3.027
+~
10- 7 111
364.6 nm
.
~
Remarks The firsl \\la\·elength is in the red region of the visible spectrum. \Ne could also obtain lhe energy of the photon by usi ng Equation 28.'1 in the form 11/= F.] - £2' \.. . here £2 and £3 are the energy levels of the hydrogen al.Ol11, calculated rrom Equation 28.14. Note t.hat this phot.on is the lowest-energy phol.Ol1 in t.he Balmer series because it involves the smallest energy change. The second photon, the 1110st energetic, is in the ultraviolet region. QUESTION 28.1 What is the lipper-limit energy ora photon thaI can be emill.cd rrolll hydrogen due to the transition of an electron between energy levels? Explain.
898
Chapter 28
Atomic Physics
EXERCISE 28.1 (a) Calculate the energy of the shoncsl-w3\'clcngLh photon emirted in the Balmer series for hydrogen. (b) Calculate the wavelength of the photon emitlcd when an eleclI-on transits from 11, = 4 to n = 2. Answers
(a) 3.40 eV (b) 486 nm
Bohr's Correspondence Principle In QLlr study of relativity in Chapter 26, we [ound that Newtonian mechanics can't be used to describe phenomena that occur at speeds approaching the speed of lighl. Newtonian mechanics is a special case of relativistic mechanics and applies only when 11 is much smaller than c. Similarly, quantum mechanics is in agreement with classical physics when the energy differences between quantized levels are very small. This principle, Ilrst set fonh by Bohr, is called the correspondence principle.
Hydrogen-like Atoms The analysis used in the Bohr theory is also successful when applied to h)'drogen/i/(p atollls. An aLOin is said to be hydrogen·likc when it contains only one e1ectrOll. Examples are singly ionized helium, doubly ionized lithium, and triply ionized beryllium. The results of the Bohr theory ror hydrogen c
E
"
=
11
= .I, 2, 3, ...
[28.18]
and ~ = ,
1\
1lt r
h}Z2 e'I(_I -1".'-\ ., 7iCl( 1//
4
1_) 'J
11
t
[28.19]
Although ..nany attempts were made LO extend the Bohr theory to more complex, multi-electron atoms, the results were unsllccessful. Even today, only approximate methods are available for treating multi-elect.ron atollls.
QUICK QUIZ 28.1 Consider a hydrogen a!.Olll and a singly ionized helium atom. ''''hich atom has the lower ground state energ)'? (a) Hydrogen (b) Helium (c) The ground state energy is the same for both.
EXAMPLE 28.2 Goal
Singly Ionized Helium.:...
~
_
Apply the modified Bohr theory to a hydrogen-like atom.
Problem Singly ionized helium, I-Ie+, a hydl'ogen-like system, has one electron in the Is orbit when the atom is in its ground state. Find (a) t.he energy of the system in the ground state in electron volts and (b) the radius of the groundstate orbit. Strategy Part (a) requires substitution into the modified Bohr model, Equation 28.18. In pan (b) modify Equation 28.10 [or the radius of the Bohr orbits by replacing (''2 by Ze 2 , where Z is the number of protons in the nucleus. ~
Solution (a) Find the energy of the system in the ground state. ""rite Equation 28.18 lor the energies ora hydrogen1i ke s)'stem:
~
28.4
Substitute the constants and convert
lO
E
electron volls:
"
Quantum Mechanics and the Hydrogen Atom
~
Substitute Z = 2 (the atomic number or helium) and n = 1 to obtain the ground state energy:
Z'( 13.6 eV)
,
n-
For our case,
11
=
10
a
hydrogen~like
-54.4 eV
-4( l3.6 eV)
(b) Find the radius of the ground state. Generalize Equation 28.10 substituting Ze'1 for e'1:
1lli?-
alom by
1 and Z = 2:
m,.Ii"Ze'l 1"1
=
899
n'L =
n'1
Z (a,,) = Z (0.0529 11m)
0.0265 nm
........
ltrr... • • • • • • • • • • • • •
Remarks Notice that for higher Z, the energy of a hydrogen-like atom is lower, which means that the electron more tightly bound than in hydrogen. The result is a smaller aLOm, as seen in part (b).
~
IS
QUESTION 28.2 \Nhen an eleClron undergoes a transition from a higher to lower state in singly ionized heliurn, how will the energy of the emined photon compare ,,,,ith the analogous transition in hydrogen? Explain. EXERCISE 28.2 Repeat the problem for the first excited state of doubl)1 ionized lithium (Z = 3, n = 2).
Answers
(3)
E,
= -30.6 eV (b) r, = 0.0705 11m
Bohr's theory was cxtcnded in an ad hoc manner so as to include further details of atomic spectra. All these modifications were replaced with the theory of quantum mechanics, developed independently by "Verner Heisenberg and Erwin Sch rod inger.
28.4
QUANTUM MECHANICS AND THE HYDROGEN ATOM
One of the first great achievements or quantum mechanics ,vas the solution of the wave equation for the hydrogen atom. Although the deLaiis of the solution are beyond the level of this book, the solution and ilS implications for atomic structure can be described. According to quantum mechanics, the energies of the allowed states are in exact agreement with the values obtained by the Bohr theory (Eq. 28.13) when the allowed energies depend only on the principal quamum Ilumber n. In addition to the principal quantum number, two other quantulll numbers emerged frorn the solution of the Schrodinger ,,,,,ave equation: the orbital quantum number, and the orbital magnetic quantum number, JII i . The effect of the magnetic quantum number 11I( can be obserYed in spectra when magnetic fields are present, wh ich results ina splitting of individual spectral lines into several lines. This splitting is called the Zeema.u f1fifl'l. Figure 28.10 shO\'\lS a single spectral line being split into three closely spaced lines. This indicates that the energy oran electron is slightly modified when the atom is immersed in a magnet ic field. The allmvcd ranges orthe values or these quanwm numbers are as follows:
e,
• The value of n can range from I to 00 in integer steps. • The value of can range from 0 (Q n - 1 in integer steps. • The value of mf can range frolll -f (Q [ in intcger steps.
e
From these rules, it can be seen that. for a given value of n, there arc 11 possible + I possible val lies of 111(. val lies of f, whereas for a given value of thcre are
e,
2e
.\
IJ
FIGURE 28.10 A single line (A) call split into lhrec separate lilll:S (13) in a magnet ic field.
900
Chapter 28
Atomic Physics
TABLE 28,1 Three Quantum Numbers for the Hydrogen Atom Quantum Number I
Allowed Values
Number of Allowed Stales
Principal quamum number
1,2,3, "
.!\..ny number
Orbital quantum number
0, 1,2, ... , /1- I -t,-I:+I,_ 0, ... , I.C
2e + I
Name
Orbital magnetic quantum number
e, e
TABLE 28,2 SheU and Subshell Notatiou 11
Shell Symbol
2
L
° I
P
3
Iv!
2
d
4
N
3
f
5 6
0 P
4
g h
K
e-
"
(
Subshell Symbol
s
5
rnr example, if n = 1, t.here is only I value of = O. Because 2C +1 = 2·0 + 1 = I, t.here is only olle value of me, which is m'e = O. I r n = 2, the value of f may be or I; ii"C = 0, then 11/.(. = 0; but ire = I, then 1I1e ma)' be 1,0, or -1. Table 28.1 slllllmarizes the rules for determining the allm\ied vailles of and me [or a given valuc of n. For historical reasons, all states with the same principal quantum number fl are said to form a shell, Shells are identified by the letters K, L, M, ... ,which designate the states for \.,.,hich n = 1,2,3, and so fonh. The states with given values of 11 and are said to form a subshell. The letters s, j), el, J, g, ... are lIsed to designate the states for I;vhich t = 0, 1, 2, 3, 4, . . . These notations are sUlllmarized in Table 28.2. States that violate the rules given in Table 28.1 can't exist. One state that cannot exist, (or example, is the 2d state, which \'.'Quld have n = 2 and = 2. This state is nOt allowed because the highest allowed value of" tis 11 - I, or 1 in this case. So for n = 2, 2s and 2j) are allmved states, but 2d, 2f, ... are not. for n = 3, the allowed states are ~s, 3/), and 3d. In general, ror a given value of 11, t.here are 11.'2 Slales with distinct pairs ofvalucs of' e ~lnd 'lilt"
°
e
e
e
QUICK QUIZ 28.2 \'Vhen the principal quantum number is n = 5, how many different values of (a) and (b) mf arc possible? (c) How many states have distinct. pairs of values of eand m,t?
e
Spin Spin lip
/
/'
,
/--------' , t :\fucleus
"
I ---+--(a)
-------Nucleus
---(b)
Spin down
FIGURE 28.11 Asan electron rno\'l:s in its orbit aboul the Iluclells. its spin can be either (a) up or (b) down.
In high-resolution spectrometers, close examinat.ion of one of the prominent lines of sodium vapor shm.. .s lhal. it is, in [act, t,.. .o very closely spaced lines. The wavelengths of these lines occur in the yellow region of the spectrum, at 589.0 nm and .189.6 nm. This kind of splitting is referred to as fine structure. rn 1925, ,.. . hen this doublet. ,,,'"as ill'S!. noticed, atomic theory couldn't explain it, so Samuel Goudsmit and Georgc Uhlenbeck, following a suggestion by Austrian physicist '~Tolfgang Pauli, proposed the introduction ofa rourth quantum number to describe atomic energy lcvels, '1lI" called the spin magnetic quantum number. Spin isn't found in the solutions of" Schrodingcr's equations; rather, it naturally arises in the Dirac equation, derived in 1927 by Paul Dirac. Thi~ equation is important in relativistic quantum theory. In describing the spin quantum number, it's convenient (but technically incorrect) to think of the electron as spinning on its axis as it orbit.s the nucleus, just as Earth spins on its axis as it orbits the Sun. Unlike the spin of a ,.. .orld, however, therc arc only twO ways in which the electron can spin as il orbits the nucleus, as shown in Figure 28.11. If the direction of spin is as shown in Figure 28.11£1, the electron is said to have "spin up." If the direction of spin is reversed as in Figurc 28.llb, the electron is said to have "spin down." The energy of the electron is slightly diff"erent (or thc two spin dircctions, and this energy difference accounts for the sodiulll doublet. The quantum numbers associated with electron spin are
28.4
Quantum Mechanics and the Hydrogen Atom r~-----
~ for the spin-up state and 111, = - ~ for the spin-down state. As v,.'e see in Example 28.3, this new quantulll number dOllbles the number of allowed Slates specified by the quantulll numbers 11, C, and JII(. For each electron, there ,He two spin states. A subshell corresponding to a given Llctor of e can contain no more than 't(2e + I) electrons. This number is used because electrons in a sllbsheli Illust have uniquc pairs of the quantum \lumbers (m.t. mo). There are 2C + I differentmaglletic qll<:lnLUIll numbers /II f and two diffcrent spin quantum numbers 171." making 2(2C + 1) unique pairs (m,e. '11/.,). For cxampic, the /) subshell (C = I) is filled when il contains 2(2' I + I) = 6 clcclrons. This fact can bc extended 10 include all four quantulll numbers, as will be important to us later when ,..Ie discuss the Pauli ('xdusiun /Jrillri/)/e. III,
=
EXAMPLE 28.3 Goal
= 2 Level of H dro
TIP 28.2 The Electron Isn't Actually Spinning The dcClron is nol ph)':->icall)'
spinning. Elcclron spin is
spinning.
en
Coullt and tabulate distinct quantum st,ltes and detcrmine their energy based
Problem 1/
The n
901
011
e, '111(, and
(a) Determine thc number of states with a uniquc set of valucs for
atomic energy level. m$ in the hydrogen atom for
= 2. (b) Tabulate the distinct possible quantum states, including spin. (c) Calculate the energics of these Slates in
lhe absence ora magnetic fteld, disregarding small differences causcd by spin. Strategy This problcm is a matter of counting. following thc quant.um rules for n, e. me, and 'tn,. "Uniquc" means that no other quantum state has the same sel of numbers. The energies-disregarding spin or Zeeman splitting in rnagnetic ftelds-are all t.he same because all states have the samc principal quantum number, 11 = 2. ..........
• • • • • • '""III
Solution (a) Dctermine the number of states \-vith a unique set of values {or and 111(' in the hydrogen atom ["or n = 2.
e
Detcrmine the differcnt. possible \'allics of II = 2:
e for
Find the diffcrent possiblc values of" 1Jl t for
e=
List the distinct pairs of (e. nit) for Find the differcnt possible values List the distinct pairs of" (e,
"171 1)
Sum the results for C = 0 and
e=
o('"I7lt
lor
e=
0::::;;
0:
0: for
e::::;;
-e::::;;
Jn t
= 2,0::::;; e : : ;
n - I, so for n
::::;;
e, so
-0 ::::;;
1111' $:
There is only one: (C, m,)
e=
I:
1:
e
= I and mult.iply by 2 LO account for the two possible spins of each slate: (b) Tabulatc the different possible sC1.s of quallnumbers.
Ulm
Usc the results oepart (a) and recall lhat thc spin qllantum number is always +~ or ~.
-f::::;;
me::::;;
e, so -I::::;;
1 and
e=
0 or I
0 implies that me = 0
= (0,0).
Ul('::::;;
1 implics that}llt = -1,0, or
There are lhree: (C, m,) = (I, -1), (t, 0), anel (I, I). Number 01" slales
=
2(1
e
11/(
nl,
2 2 2
1
-1
-2:
1
-1
1
-\f
2
1
0 0
2 2
1
1
-'2"
I
1
2
0
0
-"2
0
0
/I
2 2
I
I I
"I I I
+ 3) = 8
902
Chapter 28
Atomic Physics
(e) Calculate the energies of these
sl;llCS.
The common energy of all the slates, disregarding Zeeman splitting and spin, can be fonnd with Equation 28.l4:
E
"
=
13.6 eV
13.6 eV
n"
22
.,
.
.
-3AGeV
~
Remarks Although these states normally hm"c the same energy. the application of a magnetic field causes them to take slightly different energies celltered around the energy corresponding La 11 = 2. In addition, the slight difference in energy due lO spin state was neglected. QUESTION 28.3 Which of the [OUf quantum numbers arc nc\'er negmi\'c? EXERCISE 28.3
(a) Determine the number of stales with a unique pair of values for t, mt, and 11l f in the 11 (b) Determine the energies of those SLates, disregarding any splitting effects. Answers
(a) 18
(b)
I::, =
= 3 Ic\"e! of hydrogen.
-1.51 eV
Electron Clouds The solution of the wave equation. as discussed in Section 27.7, yields a wa\'e function \1' that depends on the quanLUm numbers 11, and mi' Recall thal if p is a poi nt and Vp a very small volume containing that point, then \J12~1 is approximately the probability of flllding the electron inside the volume ~J' Figure 28.12 gives the probability per unit length of finding the electron at various distances from the nucleus in Lhe Is state of hydrogen (n = 1, = 0, and me = 0). Note that the curve peaks at a value of,. = 0.0529 nm, the Bohr radius for the first (11 = 1) electron orbit in hydrogen. This peak means that there is a maximum probability of finding the electron in a small interval o[ a given, fixed length cel1lered at that distance from the nucleus. As the curve indicates, however, there is also a probability of finding the electron in such a small interval centered at any other distance from the nucleus. In quantum rnechanics the electron is not confined to a particular orbital distance from the nucleus. as assumed in the Bohr model. The electron may be found at various distances from the nucleus, but finding it in a small interval centered on the Bohr radius has the greatest probability. Quantulll mechanics also predicts thatlhe \..·a\·e function for the hydrogen atol11 in the ground state is spherically symmetric; hence, the electron can be found in a spherical region surrounding the nucleus. This is in conu'ast to the Bohr theory, which confines the position of the electron to points in a plane. The quantul11 mechanical result is often illlerpl'eted by \·ie\•..-ing the electron as a cloud surrounding the nucleus_ An attempt at picturing this cloud-like behavior is shown in Figure 28.13. The densest regions of the cloud represent those locations where the electron is most likely [Q be found. If a similar analysis is calTied out for the n = 2, C = 0 Slate of hydrogen, a peak of the probability curve is found at 4ao' whereas for the n = 3, = 0 state, the curve peaks algar,. In general, quantum mechanics predicts a most probable electron distance to the nucleus that is in agreement with the location predicted by the Boh r theory.
e,
e
(10
=
0.0529
r
11m
FIGURE 28.12 The probability per unit length of finding the electron ,ersus dislance from the 1l1lc!CllS for the hvdl'Ogcn atom in the h (ground) Slate. :\'ote lhat the graph has its maximum value when requals the fir'lol Bohr radius, ti~.
x
FIGURE 28.13 The spherical electron cloud for the hydrogell atom ill iL~ 15 Slate.
e
28.5
THE EXCLUSION PRINCIPLE AND THE PERIODIC TABLE
The state or an electron in a hydrogen atom is specified by [our quantum numbers: n, t, 11I l , and ms. As illUrns Ollt, the state of any electron in any other atom can also be specified by this same set of quantulll numbers.
28.5
The Exclusion Principle and the Periodic Table
903
I-1m·... many electrons in an atom can have a particular set of quantum numbers? This important question was answered by Pauli in 1925 in a powerful statement known as the Pauli exclusion principle: No two electrons in an atom can ever have the same set of values for the set o[ quantulll numbers'll, 111. 1> and nt,.
e,
The Pauli exclusion principle explai ns the electronic structure of complex atoms as a succession of filled levels with different quantulll numbers increasing in energy. where the outermost electrons are primarily responsible for the chemical properties of the clement. If this principle weren't valid, every electron ,.. .' ould end up in the lowest energy slate o[ the atom and the chemical behavior of the elements would be grossly different. Nature as ' . . e know it would not exist, and wewould not exist to wonder about it! As a general rule, the order that electrons /ill an atom's subshell is as follows. Once one subshell is filled, the next electron goes into the vacant subshell that is lowest in energy. If the atolll were not in the lowest energy state available to it, it "lIould radiale energy umil it reached that state. A subshell is filled when il contains 2(2£ + 1) electrons. This rule is based on the analysis of quantum numbers to be described later. Following the rule, shells and subshells can contain numbers of electrons according to the pattern given in Table 28.3. The exclusion principle can be illustrat.ed by examining the electronic arrangement in a fev.,' of the lighter atoms. HydroW'1I has only one electron, which, in its grollnd state, call be described by either of two sets 01" qualllllm numbers: 1.0,0,! 01" 1,0,0, -~. The electronic configuration of" this atom is orten designated as 151. The notation Is refers to a state for which 11 = I and = 0, and the superscript indicates that one electron is present in this level. Neutral helium has two electrons. fn the ground state. the quantum Ilumbers for these two electrons are 1, 0, 0, ~ and 1, 0, 0, -~. No other possible combinations of' quantum numbers exist for this level, and we say that tile K shell is filled. The helium eleclronic configuration is designated as ls~. Neutral lithium has three electrons. In the ground state, two of them arc in Ihe ls subshell and the third is in the 2s subshcll because the Itttter subshell is lower in energy than the 2p subshell. Hence, the electronic configuration for lithium is
~ The Pauli exclusion principle
TIP 28.3 The Exclusion Principle Is More General Tht.: t.:xclusion principle :;lated hCI'c is a limited rOm1 of Ihe more gCI\('I"al exclusion principle, which slales that no lwofi'rm;om. (panicles wilh spin ~. ~ ...) can be in the same quantum Sl:lle.
e
I s22.\·I.
A list. of electronic ground-state configurations for a number of atoms is provided in Table 28.4 (page 904). In 18il Dmilry Mendeleyev (1834-190i), a Russian chemist, arranged the clements known at that time into a table according to their atomic masses and chemical similarities. The first table Mendeleyev proposed
TABLE 28.3 Number of Elecll'Ons iu Filled Subshells and Shells
Shell
SubshelJ
Number of Electrons in Filled Subs hell
K (II
~
I)
.'iC
~
0)
2
L (n
= 2)
.,(C
~
0)
2
W'~I)
6
M (n
~
3)
N (11. = 4)
,(f
= 0)
2
I'll
~
I)
6
dlC
~
2)
10
,(I: ~ 0)
2
/'(I:
= I)
6
d(C
~
}(C
2)
= 3)
10 14
Number of Electrons in
Filled Shell 2
I }
)
8
18
32
WOLFGANG PAULI Austrian Theoretical Physicist
(1900-1958) The extremely talented Pauli first gained public recognition at Ihe age af 21 with 0 masterful review article on relativity. In 1945 he received the Nobel Prize in Physics for his discovery of the exclusion principle. Among his other major contributions were the explanation at the connection betvveen particle spin and statistics, the theory of relativistic quantum electrodynamics, the neutrino hypothesis, and the hypothesis of nuclear spin.
904
Chapter 28
Atomic Physics
TABLE 28.4 Electronic Configurations of Some Elements Z
,Voir; Tilt:' bracket notation is llsed as a ~h()rl h:\lld lIlo.:lhod to ,!\'e ,if I n; pCI it ion in i nd icat ing inn<;r-~hell eleci rOils. TillIS, f IleJ reprcSCllb I J 2, [l'\e) represents I.f~2s22f1', fArllTpresems Is~2J!2p63s~3/"',and '0 (111.
contained many blank spaces, and he boldly stated that the gaps were there only because those clements had not yet been discovered, By noting the column in \'vhich these missing elements should be located, he was able to make rough predictions about their chemical properties, ',\Iithin 20 years of this announcement, those elements were indeed discovered. The elements in our current version of the periodic table are still arranged so thaL all those in a vertical column have similar chemical properties. For example, consider the elements in the last column: He (helium), Ne (neon), AI' (argon), Kr (krypton), Xe (xenon), and Rn (radon). The outstanding characteristic of these elements is that they don't normally take part in chemical reactions,joining with other a!.Oms to form molecules, and are therefore classified as inert. They are called the noble gases. ''''e can partially understand their behavior by looking at the electronic configurations shown in Table 28.4, The element helium has the electronic configllration 152 , Tn other words, one shell is filled. The electrons in this lliled shell are considerably separated in energy from the next available level, the 2s level, The clectronic configuration for neon is Ls'.!.2SL2JF Again, the outer shell is filled lind there is a large difference in energy bCl\Veen the 2jJ level and the 35 level. Argon has the configuration ls'l2s'!.2f/'3s 2 3fP. Here, the 3f) subshell is filled and there is a wide gap in energy bet\",'een the 3f} subshell and the 3d subshell. Through all the noble gases, the pattern rernains the same; a noble gas is formed when either a shell or a subshell is filled, and there is a large gap in energy before thc next possible level is encountered. The clements in the first column of the periodic table are called the alkali metals and are highly active chemically. Referring to Table 28.4, we can understand wh)' these elements interact so strongly with other elements. These alkali memls all have a single outer electron in an ,', subshelL This electron is shielded from t.he nucleus by all the electrons in the inner shells. Consequently, it's only loosely
28.6
Characteristic X-Rays
905
bound to the atom and can readily be accepted by other atoms that bind it more tightly to form molecules. The elements in the seyenth column of the periodic table are called the halogens and are also highly active chemically. All these elements are lacking one electron in a subshell, so they readily accept electl'ons f1-om othel- atoms to form molecules.
QUICK QUIZ 28.3 Krypton (atomic number 36) has how many electrolls in its next-to-outer shell (n = 3)? (<1) 2 (b) 4 (c) 8 (el) 18
28.6
CHARACTERISTIC X-RAYS
X-rays are emitted when a metal target is bombarded with high~cnergy electrons. The x-ray spectrum typically consists of a broad continuous band and a series of intense sharp lines that arc dcpendenl on the type of metal used ror the target, as shown in Figure 28.14. These discrete lines. called characteristic x-rays, werc discovered in 1908, but their origin remained unexplained until the details of atomic structure were developed. The First step in the production of characteristic x-rays occurs when a bombarding electron collides with an electron in an inner shell of a target atom ,.. . ith suI":' ficient energy to remove the electron [rom the atom. The vacancy created in the shell is R1Ied when an eleclron in a higher le\'el drops down into the lower-energy level containing the vacancy. The time it takes for that to happen is very short, less than 10-9 s. The transition is accompanied by the emission of a photon \/o.'ith energy equaling the difference in energy between the two levels. Typically, the energy of such transitions is grealer {han I 000 eV, and rhe emiLLed x-ray phot.ons have \vavelcngths in the range 01'0.01 nm l.0 1 nm. vVe aSSUIllC the incoming electron has dislodged an atomic electron ["1'0111 the innermost shell, the I( shell. If the vacancy is filled by an elcctron dropping from the next-higher shell, thc L shell, the photon emitted in the process is refcrreclto as the K£r linc on the curve of Figure 28.1~1. If the vacancy is filled by an eleClron dropping from the M shell, the line produced is callcd the Kf3line. Other characteristic x-ray lines are rormed when electrons drop [rom upper levels to vacancies other than those in t.he K shell. For example, L lines are produced when vacancies in the L shell arc filled by electrons dropping [roIll higher shells. An La line is procluced as an elcctron drops frorn the ~\:1 shell to the L shell, and an L,B line is produced by a transition from the i\ shell to the L shell. vVe can estimate thc cnergy of the emil.t.ed x-rays as [ollm.. . s. Consider twO electrons in rhe K shell of an atom whose atomic number is Z. Each electron partially shields the other [rom the charge of the nucleus. Ze, so each is subject to an effective nuclear charge ldf = (2 - l)e. vVe can now use a modified form of Equation 28.18 to estimate the energy of either electron in the K shell (with 11 = 1). vVe have ,
I~K
=-
r"
k,":!.e..j
JII.,.lz,n-.-"
=-
211\'r'here £1) is the ground-state energy. Substituting
i
30
4{J
50
60 70 80 A,pm-
90
FIGURE 28.14 The x~l"ay spectrum ofa metal targd consists of ,1broad comilllloHs spectrum (brcmsslrahlLing) plus a number ofshal-p lines Ihat are due to characlaislir x-rays. The data shown were obtained whell ~5-keV electrons bombarded a mol) bdenulll target. Note that I pm :: IO-J~ III = 0.001 nm.
r" Z;/TEO It.ff
Ii" = -(Z-1)'(13.6eV)
= Z - ] gives
[28.20]
As Example 28.4 shows, \ve can estimate t he energy of an electron in an L or an tvl shell in a similar fashion. Taking- the energy difference between these lvm levels, we can then calculate the energy and wavelength of the emitted photon. In 19]4 Henry G. J. l\:loselcy plotted the Z values for a number of clements against Vf7I, where i\ is thc \...avclcngt.h oft.he K£r line for each element. He found that such a plot produced a straight linc, as in Figure 28.15, which is consistent. with our rough calculations of"the energy levels based on Equation 28.20. From his plot., Moseley was able to determine the Z values o[ other elements, providing a periodic chart in excellent agreement. with the known chemical properties of the clcmcl1l.s.
z FIGURE 28.15 A J\'foselcr plotor ~ \·ersus Z. where A is the W
Chapter 28
906
Atomic Physics
EXAMPLE 28.4 Goal
Characteristic X-Ra s
Calculate the energy and wavelength of characteristic x-rays.
Problem Estimate the energy and wavelength of the characteristic x-ray emitted from a tungsten target \vhen an electron drops from an M shell (11, = 3 state) to a vacancy in the K shell (n = 1 state).
tion 28.20. For the M-shell estimate, we need a new equation. There is 1 electron in the K shell (because one is missing) and there are 8 in the L shell, making 9 electrons shielding the nuclear charge. Therelore Zdf = 74 - 9 and EM = -Z~n1~73' ,·",here £.'" is the energy of the n = 3 level in hydrogen. The difference EM - EK is the
Strategy Develop two estimates, one for the electron in the K shell (n = ]) and one for the eleClrol1 in the M shell (11. = 3). For the K-shell estimate, we can use Equa".. , " .. , .
energy ofthe pholon.
Solution Use Equation 28.20 to estimate the energy of an electron in the K shell of tungsten, atomic number
E K = -(74 - ])'(13.6 eV) = -72500 eV
• • • • • • • • • • • • • • • • •"""11
Z= 74:
-Z;"E, =
Estimate tbe energy of an electron in the M shell in the same way:
-(Z - 9)'
~g
=
-(74 _ 9)'
(13~ eV)
-6380 eV Calculate the difference in energy between the M and K shells:
EM - E K = -6380 eV - (-72500 eV) = 66100 eV
Find the wavelength of the emitted x-ray:
M= h[= h-
c
A
-->
he
A~
t;E
8 (6.63 X 1O-''']-s)(3.00 X 10 m/5) --"---'---C.-_ _~-'---'c (66] X 10" eV)(1.60 X 10-" ]leV)
These estimates depend on the amount of sh ielding of the nuclear charge, which can be difficult to dcter-
1l11ne.
QUESTION 28.4 Could a transition from the L shel ( to the K shell ever result in a more energetic photon than a transition from the M to the K shell? Discuss. EXERCISE 28.4 Repeat the problem [or a 2/) electron transiting [rom the L shell to the K shell. (fol- technical reasons, the L shell electron must have = 1, so a single Is electron and two 2s electrons shield the nucleus.)
e
Answers
(a) 5.54 X 10" eV
1',=======
£,------£,-------
1',------FIGURE 28.16 Energy level dia· gram of an alOin wilh various allowed states. The lowesL-cnergystalc, £1' is the ground stall:. All olhers ;"Ire excited stales.
(b) 0.0224 nm
28.7
ATOMIC TRANSITIONS AND LASERS
'v\'e have seen that an atom will emit radiation only at certain frequencies that correspond to the energy separation between the various allm...·ed states. Consider an atom with many allmvcd energy states, labeled E I , E2 , E"" .. , ,as in Figure 28.16. \'\-'hen light is incident on the atom, only those photons with energy lif matching the energy separation f1l!, between two levels can be absorbed by the atom. A schematic diagram representing this stimulated absorption process is shown in Active Figure 28.17. At ordinary temperatures, most of lhe atoms in a sample are in the ground state. If a vessel containing Illany atoms of a gas is illuminated with a light beam containing all possible photon frequencies (that is, a continuous spec-
28.7
Atomic Transitions and Lasers
ACTIVE FIGURE 28.17 Diagram rt;'presenting the process of stilllil/alnl (".lso/l;tion of a phmon bv all ,Horn. The blut;' dot represellt~ all electron. The electron is tran~ferred from lhl' ground Slalc to the excited stale \\'hen the alom absorbs a pholon of energy I!f= I:'~ - £1'
Atom ill ground Slale
I': I
Before
Before
LrUIl1), only those photons 0(" energies E'2 - 1:"1' I~':\ - El , £,1 - E I , (Ind so on can be absorbed. As a result of this absorption, some aLoms are raised to various al1m,,'cd higher-energy levels, called excited states. Once an aLom is in an excited state, there is a constant probability thaL it will jump back to a Im,,'er level by emitting a photon, as shown in Figure 28.18. This process is known as spontaneolls emission. Typically, an atom will remain in an excited state for only about 10-8 s. A third process that is important in lasers, stimulated emission. was predicted b)1 Einstein in 1917. Suppose an atom is in the excited state E2 , as in Figure 28.19, and a photon with energy hI = E2 - £1 is incident on it. The incoming photOn increases the probability that the excited atom will return to the ground state and thereby emit a second photon having the same energy h! Note that two identical photons result from stimulated emission: the incident photOn and the emitted photOn. The emilled photon is exat:tlJ in phase wilh (he incident j)hofon. These photons can stimulate other atoms to emit photons in a chain of similar processes. The intense. coherent (in-phase) light in a laser (light mnpliflcation by stimu* lated emission of radiation) is a result of stimulated emission. [n a laser, voltages can be used to put more electrons in excited sL;Hes than in the ground statc. This process is called population inversion. The excited state of the system must be a mf'laslable slalf, which means that its lifetime must be relatively long. \Vhen that is the case, stimulated emission will occur before spontaneous emission. Finally, the photons produced must be retained in the system for a while so that they can sLimulate the production of still more photons. This step can be done \\lith mirrors, one of which is partly transparent. Figure 28.20 is an energy Icvel diagram for the neon at.om in a helium-neon gas laser. The mixture of helium and neon is confined to a glass tube sealed at
Scientist checking the performance of an experlmcmallascr-clluillg dC\'ice mounted on a robol arm. The laser is being USL:d 10 Cut lhrnugh a melal plate.
E,* -.-L,-----
~
,
A =632.8 nm
hJ ~
Olllino energy
~
•
lif
InpUI enerb'") E I ...L._ _- ' -_ _
After
FIGURE 28.19 Diagram n.:presenling- the process of stimulated I'mission ofa photon br an incoming: photon of energy Iif. Initially, the atorn il> in the excited Slale. The incoming" pholon ~l.imillates the awm [Q emit a second photon of cllergy '1= I'.~ - £1'
AIleI'
FIGURE 28.18 Diagram reprcsl'IHing the process Offl)(mtallN.IIH emiJsi(j// ofa photon br an alom thaI is iniLially in tht;' excited Sl<11l' E",. \,'hen thl' L:!t-clron falls to thc grOlllld Slate, the al0m erniL~ a photon Df energy 11~ L", - 1':1'
\klastable Slate
Before
907
FIGURE 28.20 ElIcrg)" level diagram for lhe neon atom in a Ilclilll11-11COll laser. The atom emils 63~.8-llm photons through slimulated cmission in I he lransillon £3* - I'.'t. This trall~itiOI\ is lhe l>ource ofcohen:n1ligIH in the laser.
908
Chapter 28
Atomic Physics
the ends by mirrors. A high voltage applied LO the lube causes electrons to sweep through it, colliding with the mOllls of the gas and raising them imo excited states. Keon atoms arc excited to slate E~r through this process and also as a result of collisions with excited helium atoms. "Vllcn a neon atom makes a transition to state
II'J.' it stimulates emission by neighboring excited atoms. The result is the production of coherent light at a wavelength of 632.8 nm. Figure 28.21 summarizes the steps in the producLion ora laser beam. Lasers that coyer wavelengths in the infrared, visible, and ultraviolet regions of the spectrum arc 110'\! available. Applications include the surgical "welding" of detached rctinas, "lasik" surgery, precision survcying and lcngth measurement, a potcntial sourcc for inducing nuclear fusion rcaClions, prccision cmling of metals and other matcrials, and telephone communication along oplical flbers.
APPLICATIO""N"'--_ _ Loser Technology
SpOlllanc.:OlIS el1li~sion
;3
ralldorn din:etioll~
n::= °0°°0068'6=:0::: ==:n UT ° ° -U 0 0 00-
MlllOl
one\
~ ~
L"e, OUlPUl
MIlIOI t1\o
~ .~
o
~"
~ ....II&'~ ~
Q
Stimulating h
•
~
Encq...')' illplll
(b)
(a)
FIGURE 28.21 (a) SICPS in the production ofa laser bC:lln, TIle 111be contains,ll,oms, which represenl Ilic activt: IlH.:dillln. A n external source of energy (optical. electrical, t:tc.) is needed to "pump" the allJll1S U) eXCiled ener~r stares. The parallel end mirrors provirle llie feedback of the stimulating wave. (h) Ph'Hograpll of the firSI I'llhy laser, showing the flash lamp .~llITOllllding Ihe ruby rod.
SUM¥ARY 28.3 The Bohr Model The Bohr model of the atom is successful in describing the spectra of atomic hydrogen and hydrogen-like ions. One basic assumption of the model is thai the electron can exist only in certain orbits such that its ang·ular momentum 11mr is an integral multiple of Ii, whcre Ii is Planck's constant divided by 21i. Assuming circular orbits and a Coulomb force of attraction between e1cctron and proton, the energies of lhe quantum states for hydrogen arc , __ m,h/p·t (~)
E" -
'2fi'!
n'!
n
=
I,
2,~,.
E, f~
I~
[28.15]
It
Bohr's correspondence principle stares thal quantum t'nechanics is in agl-eement with classical physics \vhen the quanlum numbers for (l system are very large. The Bohr rheol'y can be gcneralized to hy(lrogen-like atoms, such as singly ionized helium or doubly ionized lithium. This rnoclirk
[28.13]
where Ire is the Coulomb const.ant, I' is the charge on the electron, and n is an integer called it quantum number. If the electron in the hydrogen al.Om jumps from an orbit having quantum number 11" to an ot'bit having quantum number nf , it emils a phol.On of frequcncy J. given by
28.4 Quantum Mechanics and the Hydrogen Atom One of the many sliccesses of quantum mechanics is that t.he quantllm numbers 11, and 71Ir associated with atomic structure arise directly from the malhematics of the theory. The quantum Ilumber 'It is called the principal quan· tum number, is the orbital quantum number, and me is
e,
e
Multiple~Choice Questions
the orbital magnetic quantum number. These quantum numbel-s can take only certain valucs: I :s 11 < 00 in integer stcps, a :s e:s 11 - 1, and -C s '/fl'c:S C. In addition, a fourLh quanLum number, called the spin magnetic quantum number 'In" is needed to explain a fine doubling of lines in atomic spectra, with 'ln = j
:±:1.
28.5 The Exclusion Principle and the Periodic Table An understanding of thc periodic table of the elements became possible when Pauli formulated the exclusion principle, which states that no two electrons in the same atom can have the same values for the set of quantum numbers n, -e, m." and m. A particular set of these quantum numbers is called a qua11lum state. The exclusion principle explains how different energy levels in atoms are populatcd. Once one subshell is filled, the next electron goes into the vacant subshell that is lowest in encrgy. Atol11s with similar configurations in their outermost: shell have similar chcmical properties and are fOLlnd in the same column of the pcri~ odic table. j
•
28.6 Characteristic X-Rays Characteristic x-rays are produced when a bombarding elcctron collides \vith an electron in an inner shell of an atom \vilh suHicient energy to remove the electron from
909
the aWIll. The vacancy is filled when an electron from a higher level drops down into the level cOlllajning the vacancy, emitting a photon in the x-ray pan of (he spectrum in the process.
28.7
Atomic Transitions and Lasers
\t\-'hen an atom is irradiated by light of all different wavelengths, it will only absorb only wavelengths equal to the difference in energy of two of its energy levels. This phenomenon, called stimulated absorption, places an atom's electrons into excited states. Atoms in an excited stat.e have a probability of returning to a lower level of excitation by spontaneous emissjon. The wavelengths that can be elllit~ ted are the same as the wavelengths that can be absorbed. Iran "'Will is in an excited state and a photon with energy JIj = E'2 - £1 is incident on it, the probability of emission of a second photon of this energy is greatly enhanced. The emitted photon is exactly in phase with the incident phoLOn. This process is called stimulated emission. The emit~ ted and original photon can then stimulate more emission, creating an amplifying effect. Lasers are monochromatic, coherent light sources that work on the principle of stimulated emission of radiation from a system or atoms.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS 1. An electron in the n = .5 energy level of hydrogen undergoes a transition to the n = 3 energy level. \'\That wavelength photon does the atom emit in this process? (a) 1.28 X 10- 6 III (b) 2.37 X 10- 6 Jll (c) 4.22 X 10-' III (d) 3.04 X 10-'; 01 (e) 5.92 X 10-' m
2. A beryllium atom is stripped of all but one or its electrons. Vv'hat is the energy of the ground state? (a) -]3,(, eV (b) -218 eV (c) -122 eV (d) -40.8 eV (e) 0 3. An elcctron in a hrdrogen atom is in the 17 = 3 energy level. lim'll many different quamurn states are available to it? (a) 1 (b) 2 (c) 8 (d) 9 (e) 18
7. \t\-fhich of the following electronic configurations arc not allowed' (a) 25'21''' (b) 35'3/)' (c) 3d'45' (d) 3d,04s'41''; (e) 1.~~2s22d1
8. What can be concluded abollt a hydrogen atom ,vith an clectron in the d state? (a) The atom is ionized. (b) The orbital angular momeJ1ltllll of the atom is zero. (c) The orbital angular momentum of the atom is not zero. (d) The atom is in its ground state. (e) The principal quantum number is n = 2. 9.lf an electron had a spin of~, its spin quantum number could have the following four values: 111, = +~, +~, I . . -2", and -"2. If that \\7ere true, whIch one of the following elements 'oNith a filled shell would become the first of the noble gases? (a) He with 2 clectrons (b) Li with 3 elecrrons (c) Be with 4 eJecrrons (el) B with 5 electrons (e) C with 6 electrons ~
4. Consider an atom having four disti net energy levels. If an electron is able to make transitions bctween any two levels, how many different wavelengths of radiation could the atom emit? (a) 2 (b) 3 (c) 4 (d) .') (e) 6 5. The periodic table is based mainly 011 which of the follo\ving principles? (a) It is baseci on the uncertaillty principle. (b) All electrons in all atolll must have the same set of quantum numbers. (c) No two clectrons in an atom can have the same set of quantum nUIllbel'S. Cd) All electrons in an atom are in orbitals having the same energy. (e) Energy is conserved ill all interactions. 6. Tf an elect.ron in an atom has the quantum numbers 17 = 3, = 2, "IIle = I, and IU, = &, what state is it in? (a) 35 (b) 3/) (c) 3d (d) 4d (e) 3J
e
~
10. In relating Bohr's theory to the de Broglie wavelength of electrons, why does the circumference of an electron's orbit become nine times greater when the electron moves from the ·n = 1 level to the n = 3 level? (a) There arc nine times as many wavelengths in the new orbit. (b) The \\7avelength of the electron becomes nine times as long. (c) There are three tillles as many wavelengths, and each wavelength is thl-ee times as long. (c1) The electron is moving nine times faster. (e) Thc atom is partly ionized.
910
Atomic Physics
Chap'" 28
CONCEPTUAL QUESTIONS I. In the hydrogen mom, the quantum number'll. can increase withom limit. Because of this fact, does the frequency of possible spectral tines from hydrogen also increase without limit?
8. If matter has a wave nature, \-vhy is that not observable in our dady experience?
2. Does the light emluccl by a neon sign constitute a continuous spectrum or only a few colors? Defend your answer.
10. Can the electron in the ground state of hydrogen absorb a photon of energy less than 13.6 eV? Can it absorb a photon of energy greater than 13.6 eV? Explain.
3. In an x-ray tube, if the energy with which the electrons strike the meta] target is increased, the '.. . . avelengths of the characteristic x-rays cia not change. 'Nh)' nor?
11. \·Vhy do lithium, potassium, and sodium exhibit similar chemical properties?
4.
Ylu~t an atom first be ionized before it can emit light? Discuss.
5. Is it possible for a spectrum from an x-ray tube to show the continuous spectrum of x-rays without the presence of the characteristic x-rays? 6. Suppose the electron in the hydrogen atom obeyed classical mechanics rather than quantum mechanics. '"\Thy should such a hypothetical atom emit a continuous spectrum rather than the observed tine spectrum? 7. \,Vhen a hologram is produced, the system (including light source, objcn, beam splitter, and so on) must be held mmionless within a quarter of the light's wavelength. 'Why?
~
•
WebAsslgn
9. Discuss some consequences of the exclusion principle.
12. List some ways in which quantum mechanics altered our view of the atom pictured by the Bohr theory. 13. It is easy to underst.and how two electrons (one with spin up, one with spin clown) can fill the Js shell for a helium atom. How is it possible that eight more electrons can fit into the 2s, 2p level to complete the Is2.\2.2pti shell for a neon atom? 14. The ionization energies for Li, :'\la, K, Rb, and Cs arc 5.390,5.138,4.339,4·.176, and 3.893 eV, respectively. Explain why these values are to be expected in terms of the atomic structures. 15. W-hy is stimulated emission so important in the operation of a laser?
force between an electron and a proton separated by this distance. (b) Detcrmine (in eV) the electrostatic potential energy of the atom.
The Problems for this chapter may be assigned online at WcbAssign.
denotes guided problem denotes enhanced content problem
biomedical application
= denotes full solution availahle in Studenl Solulions Manual! Slud)1 Guide
SECTION 28.1 EARLY MODELS OF THE ATOM SECTION 28.2 ATOMIC SPECTRA 1. _ The wavelengths of the Lyman series for hydrogen are given by
+= RL~
(1 -
1~.2)
n
= 2, 3, 4" .
(a) Calculate the wavelengths of the first three lines in this series. (b) ldentify the region of the electromagnet ic spectrum in which these lines appear. 2. The wavelengths of the Paschen series (-or hydrogen are given by n
~
4,5.6,.
(a) Calculate the wavelengths of the first three lines in this series. (b) Identity the region of the eleClromagnetic spectrum in which these lines appear. [TIThe "size" of the alom in Rutherford's model is about 1.0 X lO- LO m. (a) Determine the attractive electrostatic
4. The "size" of the nucleus in Rutherford's model of the atom is about 1.0 fm = 1.0 X 10-\;' m. (a) Determine the repulsive electrostatic force bet",.-een two protons separated by this distance. (b) Determine (in MeV) the electrostatic potential energy of thc pair of prolons. 5.
mJ
The "size" of the atom in Rutherford's model is about 1.0 X 10-10 m. (a) Deter'mine (he speed of an electron moving about the proton using the altractive electrostatic force betwccn an electron and a proton sepanued by this diswnce. (b) Does this speed suggest that Einsteinian relativity must be considered in studying the atom? (c) Compute the de Broglie wavelength of the electron as it moves about the proton. (d) Does this wavelength suggest that ",.·ave effects, such as diffraction and interference, must be considered in studying the atom?
6. In a Rutherford scattering expel"iment, an a-particle (charge:::: +2e) heads directly toward a gold nucleus (charge = +7ge). The a-panicle had a kinetic energy of 5.0 MeV when "cry far (-I' --.,. (0) [rom the nucleus. Assuming the gold nucleus to be fixed in space, determine the distance of closest approach. I-lint: Use conservation of energy with PE = k r qjq2/?:
SECTION 28.3 THE BOHR MODEL
WA hydrogen atom is in its first excited state (n::::
2). Using the Bohr theory or the mom, calculate (a) the radius
Problems of the orbit. (b) the linear momentum of the electron, (c) the angular momentum of the electron, (d) the kinetic energy. (e) the potential encl-gy, and (0 the total encrgy. 8. For a hydrogen atom in its gl-ound slate, lise lht:: Bohl model to compute (a) the orbital speed of thc electron, (b) the kinetic energy of the c1ectmn, and (c) the eleClrical potential energy of the atom.
9. _
Show that the speed of the electron in the mh Bohr orbit in hydrogen is given by
911
18. Consider a large number of hydrogen atoms, with clec· trons all initially in the n = 4 state. (a) How many different wavelengths would be obsen'ed in lhe emission spectrum of these atoms? (b) What is the longest wavelength that could be observed? To which series does it belong? 19. Two hydrogen atoms, both initially in the ground state. undergo a head-on collision. If both atoms are to be excited to the 11 = 21cvel in this collision, what is the minimum speed each atom can haw befOl·c thc collision?
10. A photon is emitted when a hydrogen atom undergoes a tnlnsition fmm the 11 = 5 state to the 11. = 3 state. Calculate (a) I he wavelength, (b) the frcclucncy, and (c) the energy (in electron volts) of the emitted photOlI.
20. (a) Calculate the angular momentum of the l\:loon due to its orbital motion aboUl Earth. In your calculation use 3.84 X 108 III as the average Earth-~toon distance and 2.36 X lOti s as the period of the Moon in its orbit. (b) If the angular momentum of the Moon obeys Bohr's quanti. zation rule (I... = nil), determine the value ofthe quantum number fl. (c) By whal fraction would the Earth-Moon radius haye to be increased to increase the quantum number by I?
OLJA hydrogen atom emits a photon ofwavelenglh 656 nm. From what energy orbit to whatlower-energ)' orbit did the clcClron jump?
12LIAn electron is in the second excited orbit of hydrogen. corresponding to It = 3. Find (a) the radius of the orbit and (b) the \\>·avelength of the electron in this orbit.
12. Following are four possible transitions for a hydrogen atom
22. _ (a) Write an expression relaLing the kinetic energy KE of the electron and the potential energy PE in the Bohr model of the hydrogen atom. (b) Suppose a hydrogen atom absorbs a photon of energy E, resulting in the transfer of the electron tu a higher-energy level. Express the resulting change in lhe potential energy of' the system in terms of E. (c) Wh
= 2; IIf = 5 III. 11., = 7; 11[= 4 I.
11,
H.
11.,
= 5; 11, = 3
IV.
"It;
= 4; 'I1 = f
7
(a) Which transition will emit the shonest-w;lvelength photon? (b) For which transition \·vill the atOlll gain the most energy? (c) For which transition(s) docs the atom lose energ)'?
23. The orbital radii of a hydrogen-like atom is given by the equation
13. ,..vhat is the energy of a photon that, wlien absorbed by a hydrogen atolll, could cause an electronic transition from (a) the n = 2 state to the n = 5 state and (b) the 11 = 4 state to the n = 6 state? [KIA hrdrogen atom initially in its ground state (u = t) absorbs a photon and ends up in the state for which It = 3. (a) What is the energy of the absorbed photon? (b) If the atol11 eventually returns to the ground state, what photon energies could the atom emit? 15. Ilow much energy is ,·equi,·ed to ionize a hydrogen atom when it is in (a) the ground sLate and (b) the n = 3 sLate:16.
E
A panicle of charge q and mass m. moving with a consumt speed iI, perpendicular to a constant magnetic field follows a circular path. If in this case the angular momentum about the center of this circle is qualllized so that mVT = 211", show that the allowed radii for the partide are
n,
where
'11
= 1,2,3,.
01J((l) Ifan electron makes a transition from the 11 = -I Bohr orhit to thc n = 2 orbit, detel"l11ine the wavclength of the photOn created in the process. (b) Assuming thaI the ,1l0m was initially at rest, determine the recoil speed of the hydrogen ,Hom when this photon is emitted.
r=
Zm~h~e'2
What is the radius of the first Bohr orbit in (a) He+, (b) Li'2+. and (c) BeH ? 24,
EI
Consider a Bohr model of doubly ionized lithium. (a) \hite an expression similar to Equation 28.14 for the energy levels of the sole remaining e1ecuon_ (b) Find the energy corresponding to n = 4. (c) Find (he energy corresponding to 11 = 2. (d) Calculate the energyo[the photon emitted when thc electron transits from the fourth enel-gy Inel to the second energy le\'el. Express the answer both in electron volts and in joules. (e) Find the frequenc)' and wavelength of the emittcd photon. (f) In what part of the spectrum is the emitted light?
25. Detennine the wavclength of an elecu-on in the third excited orbit of the hycll"Ogen atom, with 11 = 4. 126.1_ sing the concept of standing wayes, de Broglie was able to derive Bohr's stationary orbit postulate. He assumed a confined electron could exist only in states where its de Broglie waves form st
912
Chapler 28
Atomic Physics
SECTION 28.4 QUANTUM MECHANICS AND THE HYDROGEN ATOM 27. List the possible SeLS of quantum numbers for electrons in the 3d subshell.
28. When the principal quantum number is n = 4, how many different values of (a) and (b) Inf are possible?
e
29. The p~ll1cson has a charge of -e, a spin quantum number of 1, and a mass 1 507 times that of the electron. If the electrons in atoms were replaced by p-mesons, lisllhe possible sets of quantum numbers for p-mesons in the 3d subshell.
SECTION 28.5 THE EXCLUSION PRINCIPLE AND THE PERIODIC TABLE 30. (a) Write out the electronic conAguration of the ground state [or nitrogen (Z = 7). (b) Write out the values for lhe possible seL of quantum numbers n, "m!.', and 1n, for the electrons in nitrogen.
e,
e
[]IJTwo electrons in the same atom have n = 3 and = l. (a) List the guanwlll numbers for the possible states of the atom. (b) How many states would be possible if the exclusion principle did not apply to the atom? 32. How many different sets ofguantum numbers are possible [or an elecn'on for which (a) 1/ = I, (b) n = 2, (c) n = 3, (cl) n = 4, and (e) n = 5? Check your resuils to show that they agree with the general rule that the number of different sets of quantum numbers is equal to 2n2 . 33. Zirconium (Z = 40) has two electrons in an incomplete €I subshell. (a) ,-Vhar are the values of nand for each electron? (b) Vv"hat are all possible values of -m,c and m} (c) 'Vhat is the electron configuration in the ground state of zirconium?
e
SECTION 28.6 CHARACTERISTIC X-RAYS 34. The K-shel1 ionization energy of copper is 8 979 cv. The L-shell ionization energy is 95\ eV. Determine the wavelength of the KG. emission line of copper. \Vhat must the minimum voltage be on an x-ray rube with a copper target to see the KG. line? 35. The K" x-ray is emitted when an electron undergoes a transition frolll the L shell (n = 2) to the K shell (n = I). Cse the method illustrated in Example 28.4 to calculate the wavelength of the Ko x-ray from a nickel target (Z ~ 28). 36. ''''hen an elect.ron drops from the M shell (n = 3) to a vacancy in the K shell (11 = 1), the measured wavelength of the emitted x-ray is found to be 0.\01 nm. Identify the element. []1JThe K series of the discrete speClrulll of tungsten contains wavelengths of 0.0185 nm, 0.020 9 nm, and 0.021 5 nm. The K-sheU ionization energy is 69.5 keV. Determine the ionization energies of the L, rVl, and N shells.
ADDITIONAL PROBLEMS 38. In a hydrogen atom, what is the principal quanwm number of the electron orbit with a I'adius closest to ].0 }Lm? 39. (a) How much energy is required to calise an electron in hydrogen to move from the n = ] smte to tile n = 1 state? (b) If the electrons gain this energy by collision bel:ween hydrogen atoms in a high-temperature gas, find the minimum temperature of the heated hydrogen gas. The thermal energy of the heated atoms is given by 3klJT/2, where II l1 is the Boltzmann constant. 40. A pulsed ruby laser emits light at 694.3 nm. For a l4.0-ps pulse containing 3.00 J of energy, find (a) the physical length of the pulse as it travels throug'b space and (b) the number of photons in it. (c) If the beam has a circular cross section 0.600 em in diameter. what is the number of photons per cubic millimeter? [![JAn electron in chromium moves from the '/J. = 2 state to the 1/ = 1 state without emitting a photon. Instead, the excess energy is transferred to an outer electron (one in the 11 = 4 state), which is then ejected by the atom. In this Auger (pronounced "ohjay") process, the ejected electron is referred to as an Auger electron, (a) Find the change in energy associated with the transition from n = 2 into the vacant 11 = 1 state using Bohr theory. Assume only one electron in the K shell is shielding part of the nuclear charge. (b) Find the energy needed to ionize an 11- = 4 electron, assuming 22 electrons shield the nucleus. (c) Find the kinetic energy of the ejected (Auger) electron. (All ans,>vers should be in electron volts.) 42. (a) Construct an energy level diagram for the He+ ion, for which Z = 2. (b) \Vhat is the ionization energy for He+? 43. A laser used in eye surgery emits a 3.00-mJ pulse in 1.00 ns, focused to a spot 30.0,u..m in diameter on the ret~ ina. (a) Find (in Sl units) the power per unit area at the retina. (This quantity is called the il'radiance.) (b) What energy is delivered per pulse to an area of molecular size (say, a circular area 0.600 nm in diameter)? 44. An electron has a de Broglie wavelength equal to the diameter of a hydrogen atom in its ground state. (a) 'Vhat is the kinetic energy of the electron? (b) How does this energy compare ,vith the ground-state energy of the hydrogen atom? 145. IE Use Bohr's model of the hydrogen atom to show that w'hen the atom makes a transition from the state n to the state 11 - 1, the frequency of [he emiued light is given b}'
46. Suppose the ionization energy of an atom is 4.100 eV In this same atom, we observe emission lines that have wavelengths of 310.0 nm, 400.0 nm, and] 378 nm. Use this information 1.0 construct the en erg}' level diagram with the least Humber of levels. Assume that the higher energy levels are closer together.
Aerial view of a nuclear power plant that generates electrical power. Energy is generated in such plants from the process of nuclear fission, in which a heavy nucleus such as 235U splits into smaller particles having a large amount of kinetic energy. This surplus energy con be used to heat water into highpressure steam and drive a turbine.
29.1
Some Properties of Nuclei
29.2
Binding Energy
29.3
Radioactivity
29.4 The Decay Processes
NUCLEAR PHYSICS
29.5
Natural Radioactivity
29.6
Nuclear Reactions
In this chapter we discuss the properties and structure of the atomic nucleus. We start by
29.7
Medical Applications af Radiation
describing the basic properties of nuclei and follow with a discussion of the phenomenon of radioactivity. Finally, we explore nuclear reactions and the various processes by which nuclei decoy.
29.1
SOME PROPERTIES OF NUCLEI
All nuclei arc composed of two lypes of panicles: protons and neutrons. The only exception is the ordinary hydrogen nucleus, which is a single prolon. Tn describing some of the properties of nuclei, such as their charge, mass, and radius, we make usc of lhe follm.. . ing quantities: • • •
the atomic number Z, ,,,,hieh equals the number of prolOns in the nucleus the neutron number N", which equals the number of neutrons in the nucleus the mass number A, ,·,'hich equals the number of nucleons in the nucleus (n1.lcleon is a generic term used to refer to either a proton or a neutron)
iX,
The symbol ,..Ie use to represent nuclei is where X represents the chemical symbol for the element. For exarnple, HAl has the mass number 27 and the atomic number 13; therefore, it contains 13 protons and 14 neutrons. \Vhen no confusion is likely to arise, we often omit the subscript Z because the chemical symbol can always be used to determine 1. The nuclei of all atoms of a particular element must contain the same number of proLOns. but they may contain different numbers of neutrons. :\uclei thal are related in t.his way are called isotopes. The isotopes of an element have the same Z value, but different N and A values. The natural abundances of isotopes can differ substantially. for example. IbC, !~C, I~C, and liC are four isot.opes of carbon. The natural abundance of the I~C isotope is about 98.9%, \'I.'hereas that of the InC isotope is only about L 1%. Some isot.opes don't occur naturally. but can be produced in t.he IrtboraLOry through nuclear reactions. Even the simplest element. hydrogen, has isotopes; \B, hydrogen; deuterium; ancllH, tritium.
TH,
ERNEST RUTHERFORD New Zealand Physicist (l871-1937)
Rutherford was awarded the 1908 Nobel Prize in Chemistry for studying radioactivity and for discovering that atoms can be broken apart by alpha rays. "On consideration, I realized that this scattering backward mllst be the result of a single collision, and when I made calculations 1saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater port of the mass of the atom was concen~ trated in a minute nucleus. It was then that 1hod the ideo of on atom with a minute massive center carrying a charge."
913
914
Chapter 29
Nuclear Physics
TABLE 29.1 Masses of the Proton, Neutron, and Electron in Various Units Mass
kg
u
MeV/c 2
Neutron
1.672 6 X LO-27 1.5750 X 10-27
Electron
9.109 X 10-31
1.007 276 1.008665 5.486 X 10-"
938.28 939.57 0.511
Particle Proton
Charge and Mass
Definition of the unified
mass unit,
U ~
TIP 29.1 Mass Number Is Not the Atomic Mass Don't confuse the lIlilS~ Ilumber A with the atomic mass. .\fass number is an imegcr that specifies
age of the masses of the isotopes of a given ekmem and has unils
ofu.
The proton carries a single positive charge +e = 1.602 17733 X lO-lfJ C, the electron carries a single negative charge -e, and the neutron is electrically nClilral. Because the neutron has no charge, it's difficult to detect. The prown is abollt I 836 times as massive as the electron, and the masses of the proton and the neutron are almost equal (Table 29.1). For i:lLOmic masses, it is convenient to define the unified mass unit, 11, in such a way thm the mass of one atom of the iSOLOPC J2C is exactly "12 Lt, where I 1I = 1.660559 X ]0- 27 kg. The proton and nelltron each have a mass oraboml u, and the electron has a mass that is only a small fraction of an alOlllic mass unit. Because the rest energy of a particle is given by EN = mc 2 , it is often convelliel1l to express the particle's Illass in terms of its energy equivalent. for one atomic Illass unit, we h::lVC an energy equivalent of Ell
= me' = (1.660559 X 10-" kg)(2.997 92 = 1.492431 X 10- 10 .1 = 931.494 MeV
X 10' m!s)'
In calculations nuclear physicists often express mass in terms of the unit MeV/t:;!, '""here I LI ~ 931.494 McV/r'
The Size of Nuclei The size and strUClUre of nuclei were rirst investigated in the scattering experiments of Rutherford, discllssed in Section 28.1. Using the principle of conservation of energy, Rutherford fOllnd an expression for how close an alpha panicle moving directly toward the nucleus call come to the nucleus before being LUrned around b}' Coulomb repulsion. rn stich a head-Oil collision. the kindic energy of the incoming alpha panicle must be convened completely to electrical potential energy when the particle SLOpS at the point of closest approach and 1\lrnS around (Active Fig. 29.1). [f we equate the initial kinetic energy of the alpha particle to the maximum e1ectric,,1 potential energy of the system (alpha panicle plus target nucleus), we have I"
'!I'!'
'i m:1r = It"-r- = It~
(2e)(Ze) d
where rl is the distance or closest approach. Solving for rI, we get 4h"Z e'l. d = - -2-
mv
ACTIVE FIGURE 29.1 All alpha panicle UJI a head-oll colli3i'ln course wil h a lludcllS of charge Xf'. Because oflhe r.olJlom\) n:plllsiun between lhe like charges. the :llpha particle will Slup in~I"Ullalleousl>'al a di:.l'lIln- d from the lIucleus. called tilt' rlist;ln("(' of closest appruach.
From this expression, Rutherford [ound that alpha particles approached to ",Iithin ~.2 X 10- 14 III ora Ilucleus when the foil was made of gold, implying that the radius of" the gold nucleus must be less than this value. For silver atOllls, the distance or closest approach was 2 X 10- 11 111. From these results, Rutherlord concluded that the positive charge in an atom is concentrated in a small sphere, which he called t.he nucleus, wilh radius no greater than about 10- 11 Ill. Because such small lengths arc common in nuclear physics, a convcnientlillit of" length is the ]em/omeler (fm), sometimes called the fermi and defined as
29.1
1 fm
='
Some Properties of Nuclei
915
10- 15
Since the time of' Rutherford's scanering experiments, a mu\tilude of other experiments have sho\\'11 that most nuclei arc approximately spherical and have an average radius given by T = foA 1':1 l29.IJ where I() is a constant equal to 1.2 X 10- 15 III and A is the total number of nucleons. Because the volume of a sphere is proponional to the cube of its radius, it follO\vs from Equation 29,1 that the volume of a llucleus (assumed to be spherical) is directly proportional to A, the lotalnumbcr of Ilucleons. This relationship then suggests all nuclei have nearly the same density. Nucleons combine to [arm a nucleus (IS though they \',"ere tighlly packed spheres (Fig. 29.2).
FIGURE 29.2 A nuclells Catl be visualized a~ a duster of tightly p,lcked ~pheres. each of which is a llucleon.
Nuclear Stability Given that the nucleus consists of a closely packed collection of proLOns and neutTons, you might be surprised thm it can e\'en exist. The very large repulsive electrostatic forces between proLOns should caLise the Ilucleus to fly apart. Nuclei. however, arc stable because of the presence of another, short-range (about 2-fm) force: the nuclear force, an attractive force that acts bct\,·;een £111 nuclear particles. The proLOns attract each other via the nuclear force, and at the same time they repel each other through the Coulomb force. The attractive nuclear force also aCts between pairs of neutrons and between neutrons and protons. The nuclear attractive force is stronger than the Coulomb repulsive Iorce \vithin the nucleus (at short ranges). If it were not, stable nuclei would not exist. Moreover, the strong nuclear force is nearly independent of charge. In other words, the nuclear forces associated with proton-proton. proton-neutron, and neutronneut.ron interactions are approximately the same. apan from t.he additional repulsivc Coulomb force [or the prown-proton interactioll. There are about 260 stable nuclci; hundreds of others have been observed, but are unstable. A plot of N versus Z for a number of stable Iluclei is giycn in Figure 29.3. l~O
120
MARIA GOEPPERT.MAYER German Physicist (1906-1972) Goeppert-Mayer is best known for her development of the shell model of the nucleus, published in 1950. A similar model was simultaneously developed by Hans Jensen, a German scientist. Mario Goeppert-Mayer and Hans Jensen were awarded the Nobel Prize in Physics in 1963 for their extraordinary work in understanding the structure of the nucleus.
110 ]00 ~lO
'"
80
~
~= 70 9 60
"
Z
50 40 30 FIGURE 29.3 A pial nfllIe 1Il.;utrOIl Humber Nver.'lls the proton lillmbel
20
10
20
30
40
~O
6U
Proton number Z
70
SO
90
Zl"or the siable nuclei (solid points). The dashed straight line cOITespond~ to the conditiotl lv' = Z. The lluclei ,He centered on Ihe so-called lint: ofstabilit)'- The ~hadcd area shows f"dioaet i\(~ (Lln~tablc) lIuclei.
916
Chapter 29
Nuclear Physics
NOlL: that lighl. nuclei arc most stable if they contain equal numbers of protons and neulrons so that N = Z, but heavy nuclei are more stable if N > Z. This difference can be panially understood by recognizing thalas the number ofprototls increases, the strength of the Coulomb force increases, which lends to break the nucleus apart. As a result, more ncutl"OIlS arc needed to keep the nucleus stable because neutrons are afl-ected only by the attractive nuclear forces. fn effect, the additional neutrons "diILllC" the nuclear charge density. Eventually, when Z = 83, the repulsive forces between prolOns cannot be cornpens3lcd for by the addition of Ilclltrons. Elements that contain more than 83 protons don't hayc st<:tble nuclei, but, rather, decay' or disinl.egr<-lte into other paniclcs in various amounts of time. The masses and some other propcnies of selected isotopes arc provided in Appendix B.
29.2
BINDING ENERGY
The lOtal l1l<Jss of a nucleus is always less than the sum of the masses or its nucleons. Also, because mass is anot.her man ifcstation of energy. the total energy of the bound system (the nucleus) is less than the combined energy of the separated nucleons. This difference in energy is called the binding energy of the nucleus and can be thought of as the encrgy that must be added to a llucleus to break it apan into its separated ncutrons and protons.
EXAMPLE 29.1 Goal
The Bindin
of the Deuteron
Calculate the binding energy ora nucleus.
Problem The nucleus of the deuteriulll atom, cal1ed the deuteron, consists of a proton and a neutron. Calculate thc deuteron's binding energy in '-leV, givcn that its atomic m<-lSS, thp mass of fl deufpriurn nudeus !Jlus an "lec/ron, is 2.014 102
L1.
Strategy Calculate the sum of the masses of the individual particles and subtract the mass of the combined particle. The masses of the neutral alOJ11S can be used instead of the nuclei because the electron masst:S cancel. Usc the values from Appendix B. The mass of an atom given in Appendix B includes the Illass of Z electrons, where Z is the atom's atomic number.
,.. ...
.
Solution To find the binding energy, first sum the masses of the hydrogen al.Om and ncutron and subtract the mass of the deuteron:
6.m =
(nip
+ m,) -
(1.007825 u
~
1n d
+
1.008665 u) - 2.014 102 u
= 0.002388 u
Convert this mass diffcrence
1O
its equivalent in IvleV:
Ef• = (0.002388 u)
... Remarks This result tells us Ihat to separate a deuteron into a proton and a neutron, it's neccssary to add 2.224 MeV of energy 1.0 t.he deut.eron to overcome the attractive nuclear force between the proIon and the neutron. One way to supply the deuteron ,·.. ith this energy is bombard it ,.. . ith energctic panicles. If the binding energy of a nucleus were zero, the Ilucleus ,.. .ould separate into its constituent prawns and nelllrons withoUI the addition of any ellerg>'; that is, it would spontaneollsly break aparl.
931.5 MeV Iu
=
2.224 MeV . • . . . . . • • . . • . .J.
QUESTION 29.1
Tritium and helium-3 have the same number of nucleons, but. tritium has one proton and two neutrons whercas helium-3 has two protons and one neutron. \'Vithout doing a calculation, which nucleus has a greater binding energy? Explain. EXERCISE 29.1
Calculate the binding energy ofiHe. Answer
7.718
~tcV
29.2
Binding Energy
917
It's interesting LO examine a plOt orbilldiIlg energy per nucleon, EbIA, as a function of mass number lor various stable Iluclei (Fig. 29.4). Ex<..:ept for the lighter Iluclei. the average binding energy per nucleon is abo1ll8 \Ie\'. NOLl' that the cune peaks in the dcinil)' of ...l = 60, which mC~IIlS that nuclei \\ ilh mass numbers greawr 0'· less than 60 are nOl as strongly bound as lho~<-' ncar the middle of the periodic table. As we'l\ sec latcr, this fan allows cner!:o to be released ill fission and fusion reaClion~. The curve is slowly varying for A > 40. which sugge~ts the nuclear force ~aturates. In other ,,,,'ords, a particular nucleon can illlcraet with only a limited number of other nucleons. which can be viewed as Ihe "nearest neighbors" in the close-packed struClure illustrated in Figure 29,2.
APPLYING PHYSICS 29.1
BINDING NUCLEONS AND ELECTRONS
Figure 29.4 shO\\'s a graph of the amount of energ) required 1.0 remove a nucleon from the nuclells. The figure indicmes that an approximate!) COllstaJ1l amount of energy is necessary to remmc a nucleon abO\·c A = 40. whereas we sc=w; in Chapter 28 lhat widel)' \'arying- <1 1110U lllS of energy are required lO rCl11Q\'c an electron from the aLom. ,,vllill aCCOllnts for this difference? Explanation In the case of Figure 29.4, the approxill1ately constant \'alue of the nuclear binding cnerj.{y is a result 01" the short-nlnge nat ure of t.he nuclear force. A given nucleon interact.s only "",ith its Ii-'\\' Ileal'· est neighbors rather lhan with alilhe Ilucleons in the nucleus. Consequelllly, no matter how man)' nucleon":i arc prescJl1 in the nucleus, removing all)1 nucleon
I~C
I
~
~,"F
6
:l.;...
.. , ~.=
..-:.
"= 5
b
~'~"
I7~Ge
\
'l,~
~l()
6~"i
2iU.
I
l:r.
j
V,'I
Th
pb
'!It>Ra
PI;
Au
W;Ag
l:loIl
U
''''Fe
B
,Be
U
~:,;
~g
iii
C1
~\~"
7
c
c 'S5
9
"0
Region (If g-rcat(':-l Slahilit\ ~'O
"lie
in\'oh'es separating it only from its nearest neighbors. The energy to do so is therefore approximately independent of how man~ nucleons arc present. For the clearest comparison with the eleCtron, think of averaging the energies required to rel1lO\'e all the e1ect.rons from an atOm, from the outermost valence electron to the innermost K-shell elenron. This avcr~lge increases with increasing atomic number. The electrical force binding the electrons 1.0 the nucleus in an atom is a long-range force, An electron in an .Hom interacts with all the proLOns in the nuclells. \<\-'hen Ihe nuclear charge increases, there is a SlrOIlgel' attraction bel.,..een the nucleus and the cleClrons. l'herclore, as the nuclear charge increases, more energy is necessary to remove an average eleCtron.
3
2 ""II
0 0
~o
,0
60
~U
100
120
'-HI
Ma..., l1umher A
160
I~O
~oo
220
2'10
FIGURE 29.4 Billrling em.'I·g\ per lludt:ol1 H:Pilh Ihl' ma" numbt.·!" A for ll11cki thilt :In' alollj:;" the line 01 ~tahilit" ,110\\'11 ill Figure 29.3. Some repre~elll;uiH~ nuclei ,lpp";\1 as blue dOl, with taheb. (~uc1('i to the righl of 2UII Ph are Ullslahle. The ('u,",'(' repre,('IllS lht.' hillding cm:rg\ for tit" lllu,t st,thk iSOIOpt.·.... )
918
Chapter 29
Nuclear Physics
29.3
RADIOACTIVITY
tn 1896 Becquerel accidentally discovered that uraniulll salt crystals emit an invis-
MARIE CURIE Polish Scientist (1867-1934) In 1903 Marie Curie shared the Nobel Prize in Physics with her husband, Pierre, and with Antoine Henri Becquerel for their studies of radio-
active substances. In 1911 she was awarded a second Nobel Prize, this time in chemistry, for the discovery of radium and polonium. Marie Curie died of leukemia caused by years of
exposure to radioactive substances. "I persist in believing that the ideas
that then guided us are the only ones which con lead to the true social progress. We cannot hope to build a better world without improving the
individual. Toward this end, each of us must work toward his own highest
development, accepting at the some time his share of responsibility in the generollife of humanity."
Detectur alTa)'
ible radiation thal can darken a photographic plate even if the plate is covered to exclude lighl. After several such observations under controlled conditions, he concluded that the radiation emitted by I he crystals was or a new type, one requiring no external stimulation. This spontaneous emission or radiation was soon called radioactivity. Subsequent experirnents by other scientists showed that other substances were also radioactive. The most significant investigations of" this type ".'ere conducted by .Marie and Pierre Curie. After several years of careful and laborious chemical separa· Lion processes on tons 01" pitchblende, a radioactive are, the Curies reported the discovery of two previollsly unknown elements, both of ,.. . hich ,vere radioactive. These elements were named polonium and radium. Subsequent experiments, including Rutherford's famous \\lark on alpha-particle scattering, suggested that radioactivity ,...as the result or the decay, or disintegration, of unstable nuclei. Three types of radiation can be emitted by a radioactive substance: alpha (a:) particles, in ,vhich the emilted particles are p-Ie nuclei; beta ({3) particles, in ".'hich the emitted particles are either electrons or positrons; and gamma (y) ra}'s, in '\'hich the emitted "rays" are high-energy photons. A positron is a particle similar to the electron in all respects except that it has a charge of +e. (The positron is said to be the antiparticle of the electron.) The symbol e- is used to designate an electron, and e+ designates a posilron. It's possible to distinguish these three forms of radiation by using the scheme described in Figure 29.5. The radiation from a radioactive sample is directed into a region \,,'ith a magnetic field, and the beam splits into three components, two bending in opposite directions and the lhird not changing direction. From this simple observation, il can be concluded that the radialion of the undeOected beam (the gamma ray) carries no charge, the component. denected upward cOlllains positive!y charged particles (alpha panicles), and the component deflected dov..' nward contains negatively charged panicles (e-). If the beam includes a positron (e+), it is deflected upward. The three types of radiation have quite different penetrating powers. Alpha particles barely penetrate a sheet of paper, bela particles can penetrate a few millimeters of aluminum, and gamma rays can penetrate several centimeters of lead.
The Decoy Constant and Half-Life Observation has shm"'n that if a radioactive sample contains N radioactive nuclei at some instan!., the number of nuclei, D.N, that decay in a small time interval n.t is proportional to 1\/; mathelllat.ically,
l..ead
I:1N 1:1/
Radioactil'c source
or FIGURE 29.5 The radi,lIion from
Decoy rote
-+
I:1N = -AN 11/
[29.2]
where A is a constant called the decay constant. The negative sign signifies that N decreases with lime; that is, D.N is negative. The value of A for any isotope determines the rate at which that isotope will decay. The decay rate, or activity R, of a sample is defined as the number of decays per second. From Equation 29.2, we see that the decay rate is
I1N
R=
I 1:1/ I =
AN
Isotopes wilh a large A value decay rapidly; those with small A decay slowly.
[29.3]
29.3
A general deca)' curve for a radioactive sample is shown in Active Figure 29.6_ It be shown [rom EquaLion 29.2 (using calculus) that the number of nuclei present varies with time according to the equation
Radioactivity
919
NU)
Gill
[29.4a] where N is Ihe number of radioactive nuclei present al time L, NIJ is the number prcsent at timc 1= 0, and (, = 2.718 . . is Euler's constant. Processes lhal. obey EfJH
(I)"
I I I I
{No
---i-I I I
[29.4b]
where 11 is the nurnberofhalf:-lives. The number nGlll lake allY nonnegative value and need not be an integer. From the definition, it follows that 11 is related to time I and the half-life 7;/'1 by {
n=--
[29.4e]
7'1 / 2
I =
.J
I
N=l\~'2
Setting IV = No/2 and
1\1=}V,ll,-ltI
ACTIVE FIGURE 29,6 1'101 of the exponcl1lial (\(ocay 1,IW 101' l'
'/1/2 in Equation 29Aa gives -Nl = i\/ (' _,\."I·,
2
(J
\·Vriting this expression In the forlll /,Tu = 2 and taking the natural logarithm of both sides, we get
In 2
,,
0.693
129,5]
1, / 2 = -A- = -A-
Equation 29_5 is a convcnicnt expression relating the half:-life to the decay constant.. Note that afler an elapsed time of" OIlC haIr-life, i'vl(1/2 radioactivc nuclei remai n (by definition); after two half-lives, half of those will have decayed and .No/4 radioact.ive nuclei will be left; after three half-lives, iVI /8 will be left; and so on_ The unit of activity H is the curie (Ci), defined as
== 3,7 X 10 10 decays/s
TIP 29.2 Two Half-Lives Don't Make a Whale Life
[29.6]
A h;llf~lifc is the lime it lakes for half of a gi\-cn number of nuclei
This IInit \',las selected as the original activity unit because it is the approximate activity of I g ofradiulll_ The SI IInil of activity is the becquerel (Bq):
to decay. Dlll-ing a second hall~ life. half the remaining !luclei decay, so in IWO ha[I~lhes, thrcequarters of the original material h-a" decayed, not all of il.
1 Ci
I Bq
~
[29.7]
1 decay/s
Therefore, 1 Ci = 3.7 X 10 10 fiq. The most commonly used units millicurie (10- 3 Ci) and the microcurie (10- 6 Ci).
0["
activit)' are the
QUICK QUIZ 29.1 True or False: A radioactive atol11 ahv£\)'s decays after two half·lives have elapsed.
QUICK QUIZ 29.2
\,Vhat fraction ora radioactive sample has decayed after lhree half-lives have elapsed? (a) 1/8 (b) 3/4 (e) 7/8 (d) none oflhese
QUICK QUIZ 29.3
Suppose the decay constant of radioacti\'c substance A is twice the decay conSLant of radioactive substance B. If substance B has a halfIife of 4 h, whal's 1he hair-life of subslance A? (a) R h (b) 4 h (c) ~ h
Calculate the activity ora radioactive substance at different limes.
Problem The half-life of the radioaClivc nucleus 2§~Ra is 1.6 X lO:~ yr. lr a sample initially contains 3.00 X 1016 such nuclei, determine (a) the initial aClivit)' in curies, (b) the number of radiulll nuclei remaining after 4.8 x 10 3 yr, and (e) the activity at this later time.
Strategy For pans (a) and (c), find the decay constant and multiply it b}' lhe number of nuclei. Pan (b) requires multiplying the iniLiaJ number of nuclei b}: one-half for every elapsed half-life. (Essentially, this is an application of Eq.29.4b.)
.
.
~
,.,
~
Solution (a) Determine the initial activity in curies. Convert the half-life to seconds:
1~/' = (lo6 X 10 3 yr)(3.l56 X 10 7 s/yr) = 5.0 X 10 ' " s
Substitme this value into Equation 29.5 to get the decay constant:
0.693 0.693 A= -- = 10 = 1.4 X 10- 11 Til' 5.0 X to s
Calculate the activit}" of the sample all = 0, using R.o = ANa, \"'here Ro is the decay rate at t = 0 and No is the llumber of radioactive Iluclei present at l = 0: Convert
clIries to obtain the activit)' at l = 0, llsing J Ci = 3.7 X 10 10 decays/s: to
R o = ANn
=
(1.4 X
[0-11
,-I
,-1)(3.0 X 1Olfinliclei)
= 4.2 X 10 5 decays/s
Ci - ' - - -) Ro = (4.2 X 10'c\ccays/s) ( - - - - -1' 7 3.7 X 10 '0 decays/'
(b) How maTI}' radium nuclei remain after 4.8 X 10' yr? ~
4.8 X 10' yr -::-::-:-=--,-,-'-:-::-c= 3.0 half-lives 1.6 X 10 3 yr/haUClife
Calcuhue the number ofha\{-:'lives, n:
"II.
MulLiply the initial number of nuclei by the number of factors of one-half:
(1)
Substit ute No = 3.0 X 10 16 and
N = {3.0 X 10 '6
11
= 3.0:
"2
N= N" ( 1)"
nllclei)(~)'"
3.8
X
10 15 nuclei
(c) Calculate the activity after 4.8 X 103 yr. ]\·tulliply the number of remaining Iluclei by the decay constant to find the activity R:
R= AN= (1.4 X 10- 11 ,-1)(3.8 X lO"nliclei) = 5.3 X 10' decays/s
= 1.4 !LCi ~
,
~
The activity is reduced by half every half:'lifc, \,...hich is naturally the case because activity is proportional to the number of remain ing nuclei. The precise number or nuclei at any tin1e is never truly exact because particles decay according to a probability. The larger the sample, hmvever, Ihe more accurate the predictions [rom Equation 29.4. Remarks
QUESTION 29.2 Hm,,' would doubling the initial mass of radioactive material aff'ect the initial activity? How would it affect the half-life? EXERCISE 29.2 Find (a) the number of remaining radium nuclei after 3.2 X 10 3 yr ancl (b) the activity at lhis time.
Answers
(a) 7.3 X 10" nuclei
(b) 2.8 !LCi
29.4
29.4
The Decay Processes
921
THE DECAY PROCESSES
As stated in the previous section, radioactive nuclei decay spontaneollsly via alpha, beta, and gamlT'la decay. As we'll see in this section, these processes are very different from one another.
Alpha Decay If a nucleus ern its an alpha panicle (~l-Ie)J it loses two protons and ('wo neutrons. Therefore, the neutron number N of a single nucleus decreases by 2, Z decreases by 2, and A decreases by 4. The decay can be writlcn symbolically as [29.8] where X is called the parent nucleus and Y is known as the daughter nucleus. As examples, 2~RU and 22GRa are both alpha emitters and decay according to the schemes [29.9]
and [29.10] The half-life for 238 U decay is 4.47 X 10 9 years, and the half-life for 226Ra decay is 1.60 X 103 years. In both cases, note thaL the mass number A of the daughter nucleus is four less than that of the parent nucleus, and the atomic number Z is reduced by t\vo. The differences are accounted for in the emitted alpha panicle (the 41-le nucleus). The decay of 226Ra is shown in Active Figure 29.7. "Vhen one element changes into another, as happens in alpha decay. the process is called spontaneous decay or transmutation. As a general rule, (1) thc sum of the mass numbers A must be the same on both sides of the equation, and (2) the sum of the atomic numbers Z must be the same on both sides of thc equation. rar alpha emission to occur, the mass or the parent must be greater than the combined mass of the daughter and the alpha particle. In the decay process, this excess mass is converted into energy of other fOfms and appears in the form of kinetic energy in the daughter nucleus and the alpha particle. Most of the kinetic energy is carried away by the alph
Ii 2m
KE=-
Because momentum is conserved, the twO particles emitted in the decay of a nucleus at rest must h£we equal, but oppositely directed, momenta. As a result, the lighter particle, with the smaller mass in the denominator, has more kinetic energy than the more massive particle.
APPLYING PHYSICS 29.2
KE~=O
PRa:: 0 Before deca)'
KERn
, K! 'a a"'--~
After deca}' ACTIVE FIGURE 29.7 The alpha deca)' of radium-226. The radium nucleus is initially at resi. After the decay. ,he radoll nucleus hal> kinetic energy KERn and mornelltUJIl PR,,, and tile alpha panicle has kinetic energy KEn and momelllum p".
ENERGY AND HALF-LIFE
In comparing alpha decay energies from a number of radioactive nuclides, why is it round that the halI-life of the decay goes down as the energy of the decay goes up? Explanation It should seem reasonable that the higher the energy of the alpha particle, the more
likely it is to escape the confines of the nucleus. The higher probability of escape translates La a faster rate of decay, \vhich appears as a shorrer half-life.
922
EXAMPLE 29.3 Goal
Nuclear Physics
Chapter 29
Decaying Radium
Calculate the energy released during an alpha decay.
Problem "Ve shm.. .ed thal the 2~~Ra nucleus undergoes alpha decay to ~~~Rn (Eq. 29,]0). Calculate the amoutH of energy liberated in this decay. 'J'ake I he Illass of 2~~Ra to be ~26.025 40't tI, lhal of 2~~Rn Lo be ~~~.O 17 571 tI, and
that ofjHe Lo be 4.002 602
Lt,
as found ill Appendix B.
Strategy The solution is a matter or subtracting the ncutralmasses of the daughter particles from the original mass of the radon atom. ,
~
~
Solution
+
+ 4.002602 u = 226.020 L73 u
Compute the sum of the mass or the daughter particle. mil' and the mass orthe alpha particle, 1/1.n :
111"
Compute the loss ofmtlss, tim, during the decay by subtracting the previous result [rom M/I' the mass of the original panicle:
ti'll1 = 111" - ('111" + "'a) = 226.025 402 u - 226.020 173 u
COIwcn the loss of mass D:m to its equivalent energy in MeV:
Ie·
~
.
..
-
-
m. = 222.017571 u
= 0.005 229 u
=
(0.005 229 u) (931.494 MeV/u)
=
4.871 MeV
-
~
Remark The potential barricr is Lypically higher than th is vC=lluc of the energy, but qllantum tunneling permits the event to occur anyway.
QUESTION 29.3 COI1\'en the Rnal answer LOjoulcs and estimmc the energy produced by an Avogadro's number ofslIch decays. EXERCISE 29.3 Calculate the cnerg>' released when ~Be splits into two alpha particles. Beryllium-8 has an atomic mass 0[8.005 305
1I.
Answer 0.094 I MeV
Beta Decay \·Vhen a radioactive nucleus undergoes beta decay, the daughter nucleus has the same number of nucleons as the parent nucleus, but the aLOmic number is changed by]: [29.11] [29.12] Again, note that the nucleon number and total charge are both conserved in these decays. As we will see shortly, however, these processes are not described completely by these expressions. A t}Vical beta decay event is
[29.13] The cmission of electrons [rom a nucleus is surprising because, in all our previous discussions, we staled thm the nucleus is composcd or protons and neutrons only. This apparent discrepancy can be explained by noting that the emitted electron is created in the nucleus by a process in \'ihich a neutron is transformed into a proton. This process can be represented by [29.14] Consider the energy of the system of Equation 29..1 3 before and after decay. As with alpha decay, energy must be conserved in beta decay. The next example illustrates how to calculate the amount of energy released in the beta decay of 1;\C.
29.4
EXAMPLE 29.4 Goal
923
The Decoy Processes
The Beta Decay' of Carbon-14
Calculate the energy released in a beta decay.
Problem Find the energy liberated in the beta decay of 1;~C to IiI\'. as represented by Equation 29.13. That equation refers to nuclei, whereas Appendix 11 gives the masses of neutral atoms. Adding six electrons to both sides of Equalion 29.13 yields
Strategy As in preceding problems, finding the released energy involves computing the difference in mass between the resultant panicle(s) and the initial particle(s) and converting to .\1eY. _._
~
~
Solution Obt.ain the masses of It,C and IjN from Appendix B and compute the difference between them:
tJ.:m = me - mN = 14.00324-2 u - 14.003074 u = 0.000 168 1I
Convert the mass difference to MeV:
E ~ (0.000 168 u)(931.494 MeV/u)
~
0.156 MeV
........ Remarks The calculated energy is generall)' more than the energy observed in this process. The discrepancy led to a crisis in physics because it appeared that encrgy wasn't conserved. As discussed below, this crisis was resolved by the discovery that another particle \'..-as also produced in the reaction. .
~
QUESTION 29.4 Is the binding energy per nucleon of the Il itrogen-14 nucleus greater or less than the binding energy per Ilucleon Jar the carbon-14 nucleus? Justify your ans\-\'cr. EXERCISE 29.4 Calculate the maximum energy liberated in the beta decay of radioactive potassium to calcium:
Answer
i8K ~ ~8Ca
+ e-.
1.31 MeV
From Example 29.4, \I,'C see that the energy released in the beta decay of 14C is approximately 0.16 rVleY. As with alpha decay, we expect the electron to carry away virtually all this energy as kinetic energy because, apparently, it is the lightest panicle produced in the decay. As Figure 29.8 sho\\;5, however, only a small number of electrons have this maximum kinetic energy, represented as KEmax on the graph; most of the electrons emiued have kinetic energies lower than that predicted value. If the daughter Ilucleus and the electron aren't carrying away this liberated energy, where has the energy gone? As all additional complication, further analysis of beta decay shows that the principles of conservation of both angular momentum ancllincar momentum appcar to ha\'c been violated! In 1930 Pauli proposed that a third particle must be present to carry away the "missing" energy and to conserve momentum, Later, Enrico Fermi developed a complete theory of beta decay and named this particle the neull'ino ("little neutral one") because it had to be electrically neutral and have little or no mass. Although it eluded detection for many years, the neutrino (I)) \vas finally detected experimcntally in ]956. The nClltrino has the following properties: • Zero electric charge • A mass much smaller than that of the electron, but probably not zero. (Recent experiments suggest that the neuu-jno definit.ely has mass, but the \'alue is uncertain, perhaps less than 1 eV/c~.) • A spin of ~ • Vcry weak interaction with mauer, making it difficult to detect \,Vith the int.roduction of the neut.rino, we can now represent the beta decay process of Equation 29.13 in its correct form: [29.15]
Kinetic energy (a)
Kinctic energy (bl FIGURE 29.8 (a) DistributioJl of bela p.tnicle energies in a typical bela decay. All cllergh:s are observed up to a maximum \·alne. (Il) In COllll"(lst.lhe energies of alpha particles from ,1.11 alpha dCc
924
Chapter 29
Nuclear Physics
TIP 29.3 Moss Number of the Electron Another nOl,lLioli that is ~ome limes u::-cd for an electron i:. _fe. This notation docs not imply that the electron has I'ero rest energy_ The masl'o of the t:!ectron is lllLlch smaller than that orthe ligllleSI nucleon. so we e
I
---
The bar in the symbol v indicates an antineutrino. To explain what an anlineutrino is, we first consider the (ollowing decay: [29.16] we sec that when l:.'N c1ecCl)'s into I~C, a particle is produced that is idcmical to the electron except that it has a positive charge of +e. This particle is called a positron. Because it is like the electron in all respects except charge, the positron is said La be the antiparticle or the electron. \·Ve discuss
Gamma Decay Vcry often a nucleus that undergoes radioactive decay is left in an excited energy SlOHe. The nucleus can then undergo a second decay to a lower energy state-perhaps even to the ground state-by emitting one or more high-energy photons. The process is similar to the emission of light by an atom. An atom emits radiation La release some extra energy \...· hen an electron "jumps" from a state of" high energy to a state of IO\'..·er energy. Like\''''ise, the nucleus uses cssentially the same medlOd to release any extra energy it may have I-allowing a decay or some other nuclear event. In nuclear dc-excitation, the "Jumps" Lhat release energy are made by pro· tons or neutrons in the nucleus as they move from a higher energy level to a lower level. The photons emitted in the process are called gamma rays, which have very high energy relative to the energy ofvisiblc lighl. A nucleus may reach an excitcd stille as the result of a violent collision with another particle. IL's more C01111110n, however, lar a nucleus to be in an excited state as a result of alpha or beta decay. The following sequence of events t}'pines the gamma decay processes: I~B
If,C*
ENRICO FERMI Italian Physicist (1901-1954)
l~C~:
---7
---7
+ e- + 17
[29.17]
+ 'Y
[29.181
IlC
Equation 29. L7 represents a beta decay in which 12B decays to 12C·, where the asterisk indicates that the carbon nucleus is left in an excited state follmving the decay. The excited carbon nucleus then decays to the ground sLate by emiuing a gamma ray, as indicated by Equation 29.18. Note that gamma emission docsn't result in any change in either Z or 11.
Fermi was awarded the Nobel Prize in Physics in 1938 for producing the
Practical Uses of Radioactivity
transuronic elements by neutron irradiation and for his discovery of
CARBON DATING The beta decay of 14C given by Equalion 29.15 is common I)' used
nuclear reactions bought about by slow neutrons. He mode mony other outstanding contributions to physics, including his theory of beta decoy, the free~electron theory of metals, and the development of the world's fjrst fission reactor in 1942. Fermi was truly a gifted theoretical and experimental physicist. He was also well known for his ability to present physics in a cleor and exciting manner. "Whoteyer Nature has in store for mankind, unpleasant as it may be, men must accept, for ignorance is neyer better than knowledge."
to dale org,:l1lic samples. Cosmic rays (high-energy particles from outer space) in the upper atmosphere cause nuclear reactions that create He [1'0111 14N. In fact, the ratio of I
29.4
SMOKE DETECTORS Smoke detecwrs are rn:quenlly L1sed ill homes and industT}' ror fire prQ(eetion, Most or the common ones an..: the ionization-type that usc radioactive materials. (See Fig. 29.9.) A smoke detector cunsists of" an ionization chamber, a sensitive current detector, and all alarm. A weak radioactive source ionizes the
925
RADON DETECTION Radioactivity can also affect our daily lives in harmful ways. Soon after the discovery of radium by the Curies, it was found that the air in contact with radium compounds becomcs radioacti"e. [t was then shO\\:n that this raclioactivit), came from the radium itself, and the product was therefore called "radium emanation." Rutherford and Frederick Soddy succeeded in condensing this "emanation," confirming that it. was a real substance: the inert, gaseous element. now called radon (Rn), Later, it was discovered that the air in uranium mines is radioactive because of" the presence of" radon gas. The mines must therefore be well vcntilatcd to help prOlcctthe miners. Finally, the fear of radon pollution has moved from uranium mines into our own homes. Because certain types of rock, soil, brick, and concrete contain small quantities or radium, some of the resulting r"ldon gas finds its way into our homes and olher buildings. The most serious problems arise from leakage of radon from the ground into the struCllIrc. One practical remedy is to exhaust the air through a pipe just above the underlying soil or gravel directly to the outdoors by means ora small fan or blower. To usc radioactive dating techniques, \..'c need to recast some of the equations already introduced. \'\le start by multiplying both sides or Equation 29.4 by A:
AN
FIGURE 29.9 All ionizalion-I)pl:' sll1okt: delt:ctor, Smoh' enteritlg the chamber redures lhe detened curl"t'n!. causing the alarm to sound.
APPLICATION Radon Pollution
= ANne-AI
From Equation 29_3, we have AN = R and ANn = HI). Substitute these expressions into the above equation and divide through by Rq : -H= ('
AI
lIu
where R is the present activity and Ro \.. ' as the activity when the ol~ject in quest.ion was P<:lrt of a living organism. \,Ve call solve for time by taking the natural logarithm of both sides oft.he foregoing cquation: In
(~) =
In
(e''')
-AI
II"
In 1=
EXAMPLE 29.5 Goal
(~) II"
A
[29.19]
Should We ReRort This Skeleton to Homicide?
Apply the technique of" carbon-I.4 dating.
Problem A 50.0-g sample of carbon taken from the pelvic bone or a skeletoll is found to have a Glrbon-14 dccay rale of 200.0 decays/min. It is known that carbon from a living organism has a decay rme of 15.0 decays/min' g and that I'IC has a ha\l~life or5 730 yr = 3.01 X 10\) min. Find the age orthe skeleton. Strategy Calculatc the original activity and lhe decay const.ant and then substitute those numbers and t.he currenl activit), into Equation 29.19.
Chapter 29
926
Nuclear Physics
.....
.........................................
Solution Calculate the original activit}' Ro [rom the decay rate and the mass of the sample:
RI)
~
In
R is given, so now we substiwte all values into Equa-
1=
, decavs 7.50 X 10'--'-
deCayS) .-'- (500 g) ( 15.0 ITIm'g
min
__==C-_
0.693
Find the decay constant from Equation 29.5:
~
0.693 = 2.30 X 1O- lO mu,-1 3.01 X 10' min
(li) No
lion 29.19 to find the age ofLhe skeleton:
200.0 decays/ min ) In ( " 7.50 X 10- decays/min 2.30 X 10 10 min '
l.32
= 5.74 X 10" min =
.....
1.09
X
10" yr
••••••••••••••••••••••••••••••••••••••••••
Remark mates.
...ס1lII
For much longer periods, other radioactive substances with longer halr·lives must be llsed to develop esti-
QUESTION 29.5 Do the results of carbon dating depend on the Illass of the original sample? EXERCISE 29.5 A sample of carbon of mass 7.60 g taken [rom an animal jawbone has an activity of 4.00 decays/min. How old is the ja\\<-bone? 2.77 X 10 1 yr
Answer
29.5 .\' 228Ra
140
c-
e
135
228Ac
a
Z2~h
a
220Rn a
130 20~Tl
125
c208Pb
80
85
90
z
FIGURE 29.10 Dccaperics beginning Wilh ~32Th.
NATURAL RADIOACTIVITY
Radioactive nuclei are generally c1assined into two groups: (J) unstable nuclei found in nature, which give rise to what is called natural radioactivity, and (2) nuclei produced in the laboraLOry through nuclear reactions, which exhibit artificial radioactivity. Three series of naturally occurring radioactive nuclei exist (Table 29.2). Each starts with a specific long-lived radioactive isotope with half·life exceeding that of any of its descendants. The fourth series in Table 29.2 begins with 23iNp, a transuranic element (an element having an aLOmic number greater than that of uranium) not found in nature. This element has a half-life of "only" 2.14· X 10 6 yr. The t\vo uranium series are some\',,'hat more complex than the 232Th series (Fig. 29.10). Also, there are several other naturally occurring radioactive isotopes, such as He and 4oK, that are not part of" either decay series. NalUral radioactivity constantly supplies our environment with radioactive elements that would otherwise have disappeared long ago. tor example, because the solar system is about 5 X lOY years old, the supply 0[226Ra (with a half-life of only 1600 yr) would have been depleted by radioactive decay long ago were it not for the decay series that starts with 238 U, with a half-life of 4.47 X 101 yr. TABLE 29.2 The Four Radioactive Series Series
Starting Isotope
L'ranium
2~~L~
AClinium
23~U
Thorium
~~~Th
:"Jeptuniull1
~n~Np
Half~life (years)
Stable End Product
4.47 X 10" 7.04 X 10"
2g~Pb
1/11 X 10 10 2.14X 10 6
2g~Pb
2g~Pb 2g~Pb
29.6
29.6
Nuclear Reactions
927
NUCLEAR REACTIONS
It is possible to change the strtlClurc ofnllclei by' bombarding them \vith energetic particles. Such changes are called nuclear reactions. Rutherford was the first to observe nuclear reactions, using naturally occurring radioactive sources for the bombarding particles. He found thai protons were released \vhen alpha particles were allowed to collide with nitrogen atoms. The process can be represented symbolicallyas [29.20] This equation says that an alpha particle (~l-Ie)strikes a nitrogen nucleus and produces an unknovm product nucleus (X) and a prOlan (:I-i). Balancing atomic numbers and mass numbers, as we did lor radioactive decay, enables us to conclude that the unknQ\.\'n is characterized as lAX. Because the element with atomic number 8 is oxygen, we sec that the reaction is [29.21] This nuclear re
QUICK QUIZ 29.4 (a) 6n + :?;~~U --7 (b)
i)n + ~ij~U
(c) An
+
~~1Pll
--7
IjSXe + ~~Sr + 20)n) 1~~Sn + '~t4Mo + 3(6n)
---7
1~~1
EXAMPLE 29.6 Goal
vVhich of the following are possible reactions?
+ !lfNb + 3(bn)
The Discovery of the Neutron
Balance a nuclear reaction to determine an unknown decay product.
Problem A nuclear reaction of significant note occurred in 1932 when Robert Chadwick, in England, bombarded a berylli1lm target with alpha panicles. Analysis of the cxperiment indicated that the following reaction occurred: JHe
+
~Bc
---..,
I~C
+ )x
\J\'hal is iX in this reaction?
Strategy
,," ...
"
Balancing mass Ilumbers and atomic numbers )'ields the answer.
.
.
Solution ',Vrite an equation relating the aLOmic masses on either side:
4
Write an equation relating the atomic numbers:
~+4=6+Z
Ielen ti fy rhe panicle:
~x
~
Remarks
"
- ..
+ 9 = 12 + A -; A = ]
=
-;
Z=o
bn (a ncutron)
. This was the fi rsl. experiment
. 1.0
.....
"
provide positive proof of the existence of" neutrons.
QUESTION 29.6 \'\Ihere in nature is the reaction bct\veen helium and beryllium comrnonly found?
-
....
928
Chapter 29
Nuclear Physics
EXERCISE 29.6 Identify the unknown particle in the reaction ~He
Answer
+
IjN
~
IJO
+
ix
2X = \H (a neutral hydrogen aLOm) Q Values ~re have just examined some nuclear reactions for which mass numbers and atomic numbers must be balanced in the equations. "Ve will now consider the energy involved in these reactions because energy is another importam quantity that must be conserved. \·Ve illustrate this procedure by analyzing the nuclear reaction
;H
+ IjN
-->
I~C
+ iHe
[29.22]
The LOtal mass on the left side of the equation is the sum ofl.he lllassofTH (2.014 102 u) and Lhe mass of IjN (14.003074 u), which equals 16.017 176 u. Similarly, the mass on the right side of tile equation is the sum of the mass of l~C (12.000000 u) plus Lhe mass ofp-Ie (4.002602 u), for a Lotal of 16.002 602 u. Thus, the LOLal mass before the reaction is greater than the total mass after the reaction. The mass difference in Lhe reacLion is equal to l6.017 176 u - 16.002 602 u = 0.014574 u. This "lost" mass is converted to the kinetic energy oCtile nuclei prcselll after t..he reaction. 1n energy units, 0.014574 u is equivalent to 13.576j\·leV of kinetic energy carried away by the carbon and helium nuclei. The energy required to balance the equation is called the Qvalue of the reaction. rn Equation 29.22, the Q value is 13.576 NieV. Nuclear reactions in which there is a release of energy-that is, positive Qvalues-are said to be exothermic reactions. The energy balance sheet isn't complete, however, because we must also consider the kinetic energy of the incident particle before the collision. As an example, assume the demeron in Equation 29.22 has a kinetic energy of5 MeV. Adding this value La our Q value, v·:e find thaL the carbon and helium nuclei have a total kinetic energy of ]8.576 J'vleV [ol}m..· ing the reaction. Now consider the reaction
lAO + IH
[29.23]
Before the re~lCtion, the total Illass is the SUIll of the masses of the alpha particle and the niLrogen nucleus: 4.002 602 u + 14.003074 u = 18.005676 u. After the reaction, the total mass is the sum of the masses of the oxygen nucleus and the proton: 16.999133 u + l.007 825 u = 18.006 %8 u.ln this case the LOral mass afLer Lhe reaction is greater than the total Illass before the reaction. The mass deficit is 0.001 282 ll, equivalent to an energy deficit of L194 MeV. This deficit is expressed b}' the negative Q value of the reaction, -l.J94 ~1eV. Reactions with negative Qvalues are called endothermic reactions. Such reactions won't take place unless the incoming panicle has at least enough kinetic energy to overcome the energy deficit. At first it might appear that the reaction in Equation 29.23 can take place if the incoming alpha particle has a kinelic energy of 1.194 MeV. In practice, however, the alpha panicle must have more energy than that. H it has an energy of only 1.194 MeV, energy is conserved; careful analysis, though, shows that momentum isn't, \',1hich can be understood by recognizing that the incoming alpha particle has some momenUl1ll before the reaction. If its kinetic energy is only L.194 MeV, however, lhe products (oxygen and a proton) \volild be created \'lith zero kinetic energy and thus zero momentum. [t can be shown that to conserve both energy and momentulll, lhe incoming particle must have a min imum kinetic energy given by
KE",;" =
(1+ Mm) IQI
[29.24]
29.7
Medical Applications of Radiation
where 111 is the mass of the incident particle, /'vI is the mass of the target, and the absolute value of the Q value is used. For the reaction gi\·en by Equation 29.23, we find that
KE m ;" ~ ( 1
4.002602 )
+ 14.003074 1- J.I94 Me"l - 1.535 MeV
This minimum value of the kinetic energy 01" the incoming particle is called the threshold energy. The lluclear reaction shown in Equation 29.23 won't occur if the incoming alpha panicle has a kinetic energy of less than 1.535 MeV, but. can occur if its kinetic energy is equallO or greater 1hall 1.535 MeV. QUICK QUIZ 29.5 Ifthe Qvalue of an endothermic reaction is -2.17 "leV, the minimum kinetic energy needed ill the reactantllllclei for the reaction to occur must be (a) equal to 2.17 MeV, (bl greater than 2.17 MeV, (c) less than 2.17 MeV, or (d) cxactly halfof2.17 McV.
29.7
MEDICAL APPLICATIONS OF RADIATION III
Radiation Damage in Matter Radiation absorbed by matter can cause severe damage. The degree and kind of damage depend on several factors, including the type and energy of the radiation and the properties of the absorbing material. Radiation damage in biological organisms is due primarily to ionization effects in cells. The normal function ofa cell may be disrupted when highly reactive ions or radicals are formed as the result of' ionizing radiation. For example, hydrogen and hydroxyl radicals produced from water molecules can induce chemical reactions that may break bonds in proteins and other vital molecules. Large acute doses of radiation are especially dangerous because damage to a great number of molecules in a cell may cause the cellLO die. Also, cells thaL do survive the radiation may become defectivc, which can leadlO cancer. In biological systems it is common 1.0 separate radiation damage into two categorics: somatic damage and genetic damagc. Somatic damage is radiation damage to any cells except the reproductive cells. Such damage can lead to cancer at high radiation levels or seriously alter the characteristics of specific organisms. Genetic damage affects only reproductive cells. Damage 10 the genes in reproductive cells can lead to defective offspri ng. Clearly. we must be concerned abollt the effect of diagnostic treatments, such as x-rays and other forms of exposurc to radiation. Several units are used to quantify radiation exposure and dose. The roentgen (R) is defined as the amount of ionizing radiation that will produce 2.08 X 10 9 ion pairs in 1 cm 3 of air under standard conditions. Equivalently, the roentgen is the amount of radiation that deposits 8.76 x 10-3 J of energy into] kg of air. For most applications, the roentgen has been replaced by the rad (an acronym for 1ildial.ion absorbed dose), defined as follows: One rad is the amount of radiation that deposits 10-2 J of energy into 1 kg of absorbing material. Although the rad is a pcrfectly good physicalullit, it's not the beSlunit for measuring the degree of biological damage produced by radiatioll because the degree of darnagc depends nol only all thc dose, but also on the fJJle of radiation. For example, a given dose of alpha panicles causes about ten times more biological d8111age than an equal dose of x-rays. The RBE (1-e1ative biological effectivcness) facLOt" is defined as the number of rads of x-radiation or gamma radiation that produces the same biological damage as 1 rad of the radiation being used. The RBE factors for different types of radiation are given in Table 29.3 (page 930). NOle that the values are only approximatc because they vary with panicle energy and t.he form of damage.
929
930
Chapter 29
Nuclear Physics
TABLE 29.3 RBE Factors for Several Types of Radiation Radiation
RBE Factor
1.0 1.0-1.7 10-20 4-5 LO 20
X-rays :Oll1d gam ma 1""yS
Beta particles
Alph
Finally. the rem (roentgen '>quivalenl in man) is den ned as the product of the dose in rads and the RBE factor: Dose in rem
APPLICATION
II
Occupational Radiation Exposure Limits
APPLICATION Irradiation of Food and Medical Equipment
'=
dose in rads X RBE
According to this definition, 1 rem of any two kinds of radiation will produce the same amount of biological damage. fo'rom Table 29.3, we see lhat a dose of 1 racl of fast neutrons represents an effective dose of 10 rem and lhal. 1 r(ld of x-radialion is equivalent to a dose of 1 rem. Lmv-levcl radiation 1'1'0111 natural sources, such as cosmic rays and radioactive rocks and soil, delivers a dose of abollt 0.13 rem/year per person. The upper limit or radiation dose recolllmended by the U.S. government (apart from background radiation and exposure related to medical procedures) is 0.5 rem/year. Many occupations involve higher levels of radiation exposure, (tnd for individuals in these occupations, an upper limit of 5 rem/rear has been set for whole-body exposure. Higher upper limits are permissible for certain parts of the body, such as the hands and forearms. An acute whole-body dose of 400 to 500 rem results in a mortality rate of about 50%. The most dangerous form of exposure is ingestion or inhalation of radioactive isotopes, especially those elemellls the body retains and concentrates. such
Tracing APPLICATION
II
Radioactive Tracers in Medicine
Radioactive particles can be used 1.0 trace chemicals participating in various reactions. One of the most valuable uses of radioactive tracers is in medicine. For example, l:tl [ is an artif"icially produced isotope of iodine. (The natural. nonradioactive isotope is 1271.) Iodine. a necessary nutrient for our bodies, is obtained largely through the intake of seafood and iodized sail. Thc thyroid gland plays a m~ior role in the distribution of iodine throughout the body. To cvaluatc the performance of the thyroid, the patient drinks a small amount or radioactive sodium iodidc. Two hours latcr, thc amount o[ iodine in (hc thyroid gland is dctermined by measuring the radiation intensity in the neck area.
Summary
A medica I application of the use of radioactive tracers occurring in emergency situations is that of locating a hemorrhage inside the body. Often the location of the site cannot. easily be determined, but radioactive chromium can identify the location with a high degree of precision. Chromium is taken up by red blood cells and carried uniformly throughout the body. The blood, hmvever, will be dumped at a hemorrhage site, and the radioactivity of that region will increase markedly. Tracing techniques are as wide ranging as human ingenuity can devise. Current applications range from checking the absorption of nuorine by teeth to checking contamination of" food-processing equipment by cleansers 1.0 monitoring deterioration inside an automobile engine. In the last case a radioactive material is used in the manufacture orthe pistons, and the oil is checked for radioactivity t.o determine the amount of,l,Iear on the pistons.
931
(a)
Magnetic Resonance Imaging (MRI) The heart of magnet.ic resonance imaging (MRI) is that ,,:hen a nucleus having a magnetic momeni is placed in an ext.ernal magnetic field, its moment. precesses about the magnetic field with a frequency that is proportional to the field. For example, a proton, ,"',lith a spin of &, can occupy one of two energy states when placed in an external magnetic field. The lower-energy state corresponds to the case in which the spin is aligned with the field, whereas the higher-energy state corresponds to the case in which the spin is opposite the field. Transitions between these t,,,",o states can be observed v.'ith a technique known as nuclear magnetic resonance. A DC magnetic field is applied to align the magnetic moments, and a second, weak oscillating magnetic field is applied perpendicular to the DC field. vVhen the frequency of the oscillating field is adjusted 1.0 match the precessional frequency of the magnetic moments, the nuclei will ;'flip" between t.he l'\I'o spin states. These transitions result in a net absorption of energy by the spin system, which can be detected electronically. In MRf, image reconstruction is obtained lIsing spatially varying magnetic fields and a procedure for encoding each point in the sample being imaged. Two MRf images taken on a human head are shown in Figure 29.1]. In practice, a computer-controlled pulse-sequencing technique is used LO produce signals that are captured by a suitable processing device. The signals are then subject.ed to appropriate mathematical manipulations to provide data for Lhe final image. The main advantage of MRI over other imaging techniques in medical diagnostics is that it causes minimal damage to cellular structures. Photons associated with the radio rrequency signals used in MRI have energies of only about 10-7 eV Because molecular bond strengths arc much larger (on the order of" I eV), the rf photons cause little cellular damage. fn comparison, x-rays or ')'-l'ays have energies ranging [rom 10'1 to 10" eV and can cause considerable cellular damage.
(b) FIGURE 29.11 Complller·enhanct:d t..., RI images of (a) a normal hllmall bl'
APPLICATION "
Magnetic Resonance Imaging (MRI)
SUMMARY 29.1
Some Properties of Nuclei
29.2 Binding Energy Nuclei are represented symbolically as QX, where X represents the chemical s}'lllbol for t.he clemem. The quantity A is the mass number, ,·,;hich equals the total number of nucleons (neutrons plus protons) in the nucleus. The qU<1l1lity Z is the atomic number, which equals the number of protons in the nucleus. l\:uclei that contain the sallle number of" protons bm different numbers of neutrons arc called isot.opes. In other words, isotopes ha,'e the same Zvalues but dirferem A values.
Most nuclei are approximately spherical, with an average radius given by
r=r"A'·'
[29.1]
where '1"0 is a constant equal to 1.2 X to-I.'> m and A is the mass number. The total mass of a nucleus is always less than the sum of Ihe masses of its individual nucleons. This mass difference ISm, multiplied by c 2, gives the binding energy of the nucleus.
932
Chapter 29
29.3
Nuclear Physics
Radioactivity
The spomaneolls emission of radiation by certain nuclei is called radioactivity. There are three proces~es by which a radioactive substance can decay: alpha (a) decay, in which the emillNI particlps art' ~Hf' 111ldf'i; hel,l (~) den,y. in which the emiucd particles are c1ecu'ons 01" po~ilrons; and gamma (y) dCGlY. in ,..·hich the emiucd panicles are hiHhenergy photons. The decay rate, or activity, R, of a sample is ~iven b)' R
:;,-"1 = AN = 11
W'hcn a nucleus undergocs beta decay,
[29.3]
1
where Si~ the llumberofradioaCli'"c nuclei aLsomc in';;l
where thc asterisk indicates that Lhe carbon nucleus was in an excited state before gamma emission.
29.6
Nuclear Reactions
Nuclear reactions can occur when a bombarding panicle strikes another nucleus. A typical nuclear reaction is [29.21]
[29.4aJ
\"'here lV is the number of radioacti\'e nuclei present at time f, '\(1 is the Ilumber altime ( = O. and (' = 2.718 ... is the base of the lIaLurallogarithms. The half-life '/~:,! ofa radioaClive substance is the time required for half ora given number of radioactive nuclei to dccay. The half-life is related to the deca)' const.ant b)' ,. IIi2 =
0.693 -A-
29.4 The Decay Processes If a nucleus decays by alpha emission. it loses and
lWO
neutrons. A typical
~lIpha
129.5J
t\.... o
prowns
In thi!'> reaction, an alpha particle stri kcs ~l nitrogen nucleus, producing an oxygen nuc1eu:-. and a proton. As in radioactive decay, atomic numbers and mass nlllnbers balance on lhe two sides of the arrow. ;\judear reactions in which en erg)' is released are said to be exothermic reactions and :II-e characterized by posiLive Q \'alues. Reactions with negative Q valucs, called endothermic reactions, canll0t OCCIII' unles~ the incoming panicle h",s 31. least enough kinelic energy 1.0 overcome lht' energy deficil. To conserve both cnerg-y and momentum, t.he incoming particle must have a minimum kinetic energy. called the threshold energy, given by
decay is
KE min = 129.9J
~ote
that in this decay, as in all radioactive decay processe!'l, the sum of the Z\'alues on the left equals lhe sum of the Z values on the right the same is true for the A va lues. A typical beta decay i~
( I + -AI'Ill) IQ!
[29.24J
where m is the mass of the incidel1l panicle and t\I is the mass of the tal-get atom.
[29.151
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS 1. The atolllic mass of oxygen is 1:>.999 ll. Com en th is maSli
to kilogr.lInli. (a) 1.28 X 1O-2,i kg (b) ~.37 X 10- 21; kg (c) 1.67 X 10-27 kg (d) 4.72 X lO-'?fo kg (e) 2.66 X 10 -". kg
2. COfn'erl the atomic mass of an oxygen atom to units of ~leV/c'.
(a) 1.88 X 104 ~leV/e2 (b) 2.47 X 10' ~le\'/c2 (d) 9.21 X 10" ~leV/e'
(c) 7.37 X 10' ~le\'1e2 (e) 1.49 X 10' _lleV/e'
3. EqimalC the ratio of the volume of a neon nucleus 10 lhat of a helium nucleus. (a) I (b) 2 (c) 4 (d) ;) (c) 7 4. Rank chlorine. nickel, and gold by binding- energy per nuckon, from smallest to largest. (a) gold, chlorine,
rine (d) gold, nickel. chlorine (e) chlorine. nickel, gold 5. Substance A has a deGI)' constant A that is three times that of B. Find the ratio of the half-life of substance A to the half-life or substance B. (a) j (b) I (c) 3 (d) 2 (e) ~ 6. The half-life of radium-224 is about 3.6 d. \\'halapproximaLe fraCl 011 of a :-.ample rern
(b): (c)
*(d) t'r. (e) h
Multiple-Choice Questions
7. Docs a neutral atom designated as I~X comain (a) 20 nClllrons find 20 protollS. (b) 22 proton:" and 18 nl..'l1Irons. (c) 18 protons ~l1ld 22 neutrons, (el) IH proton~ and 40 neutrons. or (e) 40 protons and 18 neutrons? 8. \'Yhen the ~iKr nuc!<.:us undergoe:-. beta dCGI}. docs the daughtcr nucleus (Rb) comain (a) 58 neutrOIlS and 37 protons, (b) 58 protons and 37 neutrons. (c) 54 ncu· lions and 41 protons. (d) 59 neutrons and ~16 protons, or (c) 55 neutrons and 40 protons? 9. Which of the following statements is truc regarding a radioactive isotope ~X that deca}s by t"ll1iuing a g-amllla ra)'? (a) The resulting isotope has a difrerc11I Z value. (h) The resulting isotope has the same A \'£tluc and the ~all1C 7. ,allle. (c) The resulting isotope has a differcllt
933
A value, (d) Both r\ and Z are different. (e) The atom jumps lO a state or higher encrg}'.
10. III the decay :!i~6Th ~ 1Ra + Jlle. idcmif\ the mass number and lhc alomic number of the Ra nucleus. (a) A = 230. Z = 92 (b) " = 238, Z = 88 (e) II = 230, Z ~ 88 (d).1 = 234, l= 88 (el A = 2:18. Z = 86 II. When ::~p deGIY~ to 7~S. which of the follO\\ ing panicles are emiued? (a) a protoll (b) an alpha panicle (c) an electron (d) a gamma ray (e) an antineutrino 12. What is the - n' Ca) 8.4 (e) 4.2 ~leV
for the reaClion liRe + a ~ I:!C Cb) 7.3 ~leV (e) 6.2 ~Ie\' (d) 5.7 ~Ie\'
Q value ~leV
CONCEPTUAL QUESTIONS positron. A neUlron, though. has a larger rest energy lhan a proton. How is thal possible?
L. \Vhy do i..otopes ofa gi,"cn element han: difTercnl ph)s·
ical properties, such as mass, but the same c1H.'mical properties?
8. An alpha particle h;-IS lwice the charge ora bela particle. Why does the former deflect less than the latter when passing between electrically charged plates, assuming they both have the salllC specd?
2. If a heavy nllclell!'! that is initially at rC'lt lI11dcr!-{ot's
alpha decay, which has more kinetic energy after lht decay, Ihe alpha panicle or the daughter nucleus?
9. Can carbon-L4 dating be used to meaSure the age of a Slone?
3. A Sl.lldCIll claims thaI a heavy larm of hydl"o~('n decay:-by (11pl1a emission. Huw do you respond?
4. Explain the 1l1ain differences bel ween alpha, bela, lind g-limma rays. 5. In beta decay, the energy of lhl' electron or positron (.'milted from the nucleus lies somc\\'llere in a relatively larp;e range of possibilities. In alpha decay ho\\'('ver, the alpha particle energy can only have dilicn:te values. \\'liy is l here I h is d i fTerencc? 6. I f Ii IIII is kept in a box, alpha particles from a radio.u··
the SOlll"CC outside the box cannot bt.'ltt particles can, Explain,
expo~e
the film, blll
10. Pick an}' bet<-l-deGlY process and sliow Ihat the neutrino must have zero charge. 1J, \,\'hy do heavier e1emcms require more neult-ons to Inaintain stability? 12. Suppose it could be shown that the inlensil.r of cosmic rays was much grealer 10000 years ago. How would that affect the ages we as~ign to ancicm samples of onceliving mailer?
13. Compare and contrast a photon and a neutrino.
7. In posilron deca)' a prolon in the nucleus he(·oll1t·s a nClIlron. and the posith'e charge is carried awa) b, lht.·
PROBLEMS ~
The Prohlems for Ihi~ chapler assig-ned onlinc:u \\'d>.\!'sij.{ll. I. 2, \ = slr'aigllLfon\<\nl, il1ll·IIllUIi.lt'. Ch'llh.:'llglllj:, •
(a) DClcrmine the mass number ofa nucleus having a radius approximalcl}' equal lO two·thirds the radius of :!~~Ra. (b) Identifr lhc:: elemclll. Are any other answer~ possible? Explain.
3. L'sing 2.3 X 10'7 kg/m\ a'l the dell:-,ily of nuclear nUtuer, lind the radius ofa sphc:n..' ofslIch mailer lhat would have a mass equal to that 01 E«rlh. Eanh ha'i a mass equ;ll lO 5.98 X 10~N kg
'i/ml.\' (;/lul,. A IH(
BiiJ
Af)pendix Il
4. Consider the ~r:cll nudeus. Find approximale value,> Ii-.n-
ils (a) radius. (b) volume. add (c) dcnsit}.
SECTION 29.1 SOME PROPERTIES OF NUCLEI 1. Compare Ihe nllclear radii of the following Iluclides: YII. ~~c(), 1~~Au, 1~1Pll.
I
alpha panicle (1. = 2, rna".'; = 6.64 X 10-27 kg) approaches to wilhin 1.00 X 10 11 m ofa Glrboll nucleus (Z = 6). Whal ~He (
S.IAn
934
Chapter 29
Nuclear Physics
the alpha particle, (b) the acceleration of the alpha particle at this time, and (c) the potclllial energy ofdlC
16.
a A drug tagged with
J7.
m
18.
mm Tritium has a half-life of 12.33 years. (a) \~/hal frac-
am
Singly ionized carbon aLoms are accelerated through 1 000 V ,llld jMs:::.td intu a mass speetl-omClcr to determine
the isotopes present. (See Chapter 19.) The magnetic field strength in U1C spectrometer is 0.200 T. (a) Determine the orbital radii for the 12e and the DC isotopes as they p,BS through the field. (b) Show that the ratio of the radii may be wriuen in the form 1"1
~= \
~ 1111
and verify Ihat)'our radii in part (a) satisfy this formula.
~~Tc (half~lile = 6.05 h) is preparcel for a patient. If the original activity of the sample was 1.1 X 10 4 Bq, what is its activit)' after it has been on the shelf for 2.0 h?
The ha1f~lilc of 1:111 is 8.04· days. (a) Convert the halflife to seconds. (b) Calculate the decay constalH for this isotope. (c) ConvenO.500 J.LCi to the Slunitthe becquerel. (d) Find the number of 131( nuclei necessary 10 produce a sample '\iith an
7. (a) Find the speed an alpha panicle requires to come within 3.2 X 10-1-1 In of a gold nucleus. (b) Find the energy of the alpha panicle in \-1eV.
t.ion of the nuclei in a tritium sample will remain after 5.00 )...? (b) 10.0 yr' (c) 123.3 yr' (d) According to Equation 29Aa, an infinite amount. of time is required for the entire sample to decay. Is that realist.ic? Discuss.
8. At the end of its life, tI star with a mass of twO times the Sun's mass is expected to collapse, combining its prowns
~rvlany smoke detectors Lise small quantities of the iso-
and electrons to form a neutron star. Such a star could be thought of as a gigantic atomic nucleus. If a star of mass 2 X 1.99 X 10 30 kg collapsed into neutrons (m" = 1.67 X 10-:n kg), what would its radius be? Assume r = rnA 1'3.
SECTION 29.2 BINDING ENERGY 9, Calculate the average binding energy pcr nucleon of (a) WNa (sodium), (b) ~~Co (cobalt), and (c) 1~4Ag (silver). 10. Calculate the binding energy per nucleon [or (a) (b) 'IHe, (c) "I'Fe, and (d) 23RU.
2(-1,
QIJA p
tope 141Am in their operation. The ha1i~lire of 'lIIAm is 432 yr. How long will it take the activity of this material to decrease to 1.00 X 1O-?1 of the original activit)'? 20.
II
After:l plam 01" animal dies, its 1.le content. decreases with a half-life of 5 730 yr. If an archaeologist finds an ancient firepil conu1ining partially consumed fIrewood and the HC conte III of the wood is only 12.5% that of an equal carbon sample from a present-day tree, what is the age of the ancient site?
[lliA
freshly prepared samplc of a certain radioacti\'c isoLOpe has an activjty of 10.0 mei. After 11.00 11, t.he activity is 8.00 mCi. (a) Find the decay constant and half-life of [he isol.ope. (b) How many at.olTls of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity 30 h after it is prepared?
22. A building has become accidentally contamin"ued wit.h radioact.ivity. The longest-lived material in the building is strontium-90. (The atomic mass oQUSr is 89.9077 tI,) If the building initially contained 5.0 kg of this substance and the safe level is less than 10.0 counls/min, how long willl.he building be unsafe:
SECTION 29.4 THE DECAY PROCESSES
13. "Eva nuclei having the same mass number are knO\\in as iso/mrs. Calculate the difference in binding energy per nucleon for the isobars TtNa and gl\·lg. How do you accoulll for this difference? (The mass or i~Mg = 22.994127 u.)
23. Identify the missing nuclides (X) in the following decays: (a) X ~ ~~Ni + 'Y (b) ~~:;Po -7 X + a (c) X ---7 ~~Fe + e+ + v
14. CalCllla1.c the binding energy of the last neutron in the iilCa nucleus. Hint: You should compare the mass of ~gCa with the mass of ~3Ca plus the mass of a neutron. The mass oq~Ca = 41.958 622 u, whereas the mass of ~6Ca = 42.95877011.
25. The mass of ·%Fe is 55.9349 u, and the mass of 5GCO is 55.9399 ll. Which isotope decays into the other and by what process?
SECTION 29.3 RADIOACTIVITY [[[JThe half~life of an isotope of phosphorus is 14 days. If a sample conlains 3.0 X 10lG such lluclei, determine its aClivit)', Express your ans",,"er in curies.
24. ldenti fy the missing panicles (X) in the following reactions: (a) I~Cd + X ~ 1~~Ag + v (b) I~~' + ~J-1e --. X + I~O
26. Find t.he energy released in the alpha decay of ~~~U. The following mass value ","ill be useful: 2~JTh has a mass of 23·\-.043 583 tl. 127.10ctermine which of the follov,,'ing suggested decays can occur spontaneously: (a) ~gCa ~ e"'"
+
13K
(b) 1I,;\INd -+ ~He
+
I~Ce
Problems 28. ~~Ni (mass = 65.929 1 u) undergoes bt.'la dt'Gl\ to ~~Cu (mass = 65.9289 u). (a) Writc the completc decar formula for this process. (b) Find the maximuill kinctic energ-y of the emerging electrons.
29. An 3H nucleus beta decays illlo :'I-Ie b, cre~Hillg- .111 ek·t-tron and an antineulrillo according 10 the r('action :1'11
-)0
~Hc
35. (
mil
+ e- + "
Consider twO reactioll'i:
iH ~ 111 PI + il--l -7 ~lIe
(l) n
Use Appendix B 10 determine the tOlal enng} released in t his reaction. 30. A piece of charcoal used for cooking is found at the remains of an ancient campsite. A I.OO-kg sample of carhall from the wood has an activit}' of 2.00 X IQ:{ decays pcr minute. Find the age of the charcoal. Hint: Li"in~ material has all ani\ it}' of 15.0 decaYS/ill inute pt:r gram of carbon prcst:nL.
l1IJA wooden artifact is found in an ancient tomb. lis carhonII (lite) actidt) is mea'iurcd to be 60Jl% of that in a fre'ih sample of wood from the same region. Assuming the same amount of lie was initialJ}' prcse11l in the wood fronl which lhe artifact was made, delerminc ,he a~e of th(' artifact.
(2)
+
(a) Computc the Q val lies for thesc reactions. Identify wherher each reaction is eXOlhenHic or endolhermic. (b) Which rC;lction rc~ults ill more released energy? Wh),? (c) Assuming Ihe difference is primarily dul.' 10 Ihe work clone by tbe electrit.· force. calculate the distance bel ween the ["'0 prolons in helillm-3.
137.INawral gold ha.. on I} One isotope. 1~~AlI. If gold is bomharded with .. Io\\, ncutrons, e~ panicles are emilled. (a) \,'rite the appropriatc reaction eqlJalion. (b) Calcuhue the maximum energy of the emilled bela panicles. The mas'i of I~~I-IA" i.. 197.96675 u. 38. The following reactions are obselTcd:
Q = 6.R12 Me'·
SECTION 29.6 NUCLEAR REACTIONS
Q= -1.665
32. A beam of 6.61-MeV protons is incidellt all a target of nAI. Tho'ic protons that collide with the t
p + ~~AI
--)0
nSi +
11
(i1Si has a mas~ of 26.986 721 u.) Negleeti lIg any rccoi I of the produClllUdclIS, determine the kinetic cnergy orthe clnerging nCUlrons. 133.Udenlif" the unknown panicles X and X' in the following: nuclear reactions:
One method of producing nClllfOns for expcrlmental use is to bombard ~Li with prolons. The neutrons ;UI.' emilled according to the reaction
\11 -I- ~Li
-)0
;Be
of 8 Be
Calculate the nl
~IeV
and ,nBe in atomic mass units
39. When 18 0 is slruck b}' a proton, IHF and anolher particle are produced. (a) What is thc other pari icle? (b) The re
SECTION 29.7 MEDICAL APPLICATIONS OF RADIATION 40.
III
In tcrms of biological
41. (I A person whose mass is 75.0 kg is expQ~ed to a wholebod\' dose of 25.0 rad. How Illany joules of energy are deposited in the person's body? 42.
+ ,I,n
(a) Calcul,lte the mass in atomic mass llnits of the particles on the left side of Ihe eqll reaction in tcrlll~ of the masse~ 111". 11111", lit" • • mel the ,·c1ocitics. (e) Writc all expression relating: the kinetic energies of panicles beforc and after togelhcr with Q. ('1 Whal minimum kinctic cnergy must thc incidellt proton have if I h is reaci ion is 1.0 nccu r?
935
II
A 200-rad clost, of radiation is adm inistered to a pat ient in an effort to combat. a cancerous growth. Asslllning all the eneq.,'")' deposited is absorhed by the ~rowt.h, (a) Gdclilatc Ihe ;unOIlIlI. of energy delivered per unit mass. (b) Assuming Ihe growth has a mass of' 0.25 kg 311d a ~pccillc lie~ll cqualw lhat o("water, calculate its temperature rise.
143.IA "c1e"cr" technician decides to heal. some water for his cofTee with an x-ray machine. If the machine produces 10 rad/s, how long will it lake to raise Ihe temperature ofa cup ofwatcr by 50"'C? Ignore heat losses during this time. 44.
II
45.
II
An x-ray technician v,'-orks 5 dars per we('k. 50 week!> pCI' "ear. Assullle Ill(' technician take!'. an ayel
Chapter 29
936
Nuclear Physics
of I4.3 days. The average kinetic energy of the emiucd electrons is 700 keV. If the initial activity of the sample is 1.31 MBq, determine (a) the number of electrons emitted in a lO-day period, (b) the lotal energy deposited in the body during the 10 days. and (e) the absorbed dose if I he electrons arc completely absorbed in 100 g of tissue. 46.
III
A particular radioactive source produces 100 mrad of 2-MeV gamma rays per hour at a distance of 1.0 m. (a) How long could a person sland at this diswncc before accumulating an intolerable dose of I rem? (b) Assuming the gamma radiation is emitted uniforrnly in all direc~ lions, at what dislance would a person receive a dose of 10 mrad/h from this source?
ADDITIONAL PROBLEMS
47. A radioactive sample contains 3.50 J.Lg of pure IIe, which has a half-life of20.4 min. (a) How many moles of IIC are present initially? (b) Determine the numbel- of nuclei prcsent initially. What is the activit)' of the sample (c) initiallyand (el) after 8.00 h? 48. How much time elapses before 90.0% of the radioactivity ora sample oq§As disappears, as measured by its (\nivity? The half-life of ~~As is 26 h. 149.IA 200.0-mCi sample ofa radioactive isotope is purchased by a medical supply house. If the sample has a half-life of 14.0 days, how long will it keep before its activity is reduced to 20.0 mCi? 50.
mJ
(a) Why is the beta decay p ----7 11 + e+ + II forbidden for a free proton? (b) Why is the same reaction possible if the proton is bound in a nucleus, as in the reaClion I~N ----7 InC + e+ + v? (c) How much energy is released in thc reaction given in part (b)? Suggestion: Add seven electrons 10 both sides of the reaction and write il for ncutral atoms.
51. _ In a piece of rock from the :vfoon, the 8iRb contenl is assayed to be 1.82 X 1010 atoms per gram of material and the SiS r content is found t.o be 1.07 X 'IO!I atoms pcr gram. (The relevant decay is s7Rb ----7 87S r + e-. The half-
life of the decay is 4.8 X 1010 yr.) (a) Determine the age of the rock. (b) Could the material in U1C rock actually be much older? What assumption is implicit in using the radioactive-dating method? 52.
Em (a) Can 5iCO decay by e+ emission? Explain. (b) Can IIe decay by e- emission? Explain, (c) If either answer is yes, what is the range of kinetic energics available [or the beta particle? Atomic masses: 5iCO: 56.936294 u; 5iFe: 56.935396 u
153.11'1 A medicallaboratolJ' stock solution is prepared with an initial activity due to 2·tNa of 2.5 mCi/ml, and 10.0 ml of the st.ock sollllion is diluted at to = 0 to a working solution whose t.otal volume is 250 ml. After 48 h, a 5.0-ml sample of the working solution is monitored wilh a counter. What is the measured activity? Notf': 1 ml = I milliliter. 54. " After the sudden release of radioactivity from the Chernobyl nuclear reactor accident in 1986, the radioactivity of milk in Poland rose to 2000 Bq/L due to iodine· 131, with a half-lifcof8.04-days. Radioactivciodine is particularly hazardous becausc the thyroid gland concentrat.es iodine. The Chernobyl accident caused a measurable increase in thyroid cancers among children in Belarus. (a) For comparison, find the aClivityofmilk due lO pOlassium. Assume I liter of milk cOlltains 2.00 g of potassium, of v..' hich 0.011 7% is the isotope 'loK, which has a hall:'life of 1.28 X 10') yr. (b) Afler what lengLil of time would the activity due to iodinc fall below t.hal due to pOtassium? 55. During t.he manufacture of a steel engine component., radioactivc iron (5~Fe) is included in the total mass of 0.20 kg. The component is placed in a test engine when the activity due to the isotope is 20.0 p..Ci. After a I OOO~h test period, oil is removed from the engine and is found to contain enough 59Fe to produce 800 disinte~ grations/min per liler of oil. The tolal \'olull1c of oil in t.he engine is 6.S L. Calculate lhe LOla I mass worn from t.he enginc component per hour of operalion. (The halflife of "'9Fe is 45.1 da)'s.)
This photo shows scientist Melissa Douglas and part of the Z machine, an inertial-electrostatic confinement fusion apparatus at Sandia National Laboratories. In this device giant capacitors discharge through a grid of tungsten wires finer than human hairs, vaporizing them. The tungsten ions implode inward at a million miles an hour. Braking strongly in the grip of
0
"Z.pinch," the ions emit
powerful x-rays thot compress a deuterium pellet, causing collisions between the deuterium atoms that lead to fusion events.
NUCLEAR ENERGY AND ELEMENTARY PARTICLES
30.1
Nuclear Fissian
30.2
Nuclear Fusion
30.3
Elementary Particles and the Fundamental Forces
30.4
Positrons and Other Antiparticles
30.5
Classification of Particles
30.6
Conservation Laws
30.7
The Eightfold Way
all matter in nature is constructed from only two families of particles: quarks and leptons.
30.8
Quarks and Color
Finally, we describe how such models help us understand the evolution of the Universe.
30.9
In this conduding chapter we discuss the two means by which energy can be derived from nuclear reactions: fission, in which a nucleus of large moss number splits into two smaller nuclei, and fusion, in which two light nuclei fuse to form a heavier nucleus. In either case,
there is a release of large amounts of energy that con be used destructively through bombs or constructively through the production of electric power. We end our study of physics by examining the known subatomic particles and the fundamental interactions that govern
their behavior. We olso discuss the current theory of elementary particles, which states that
Electroweok Theory and
the Standard Model
30.1
30.10 The Cosmic Connection
NUCLEAR FISSION
30.11
Nuclear fission occurs \-\'hen a heavy nucleus, such as 23:>U, splits, or fissions, into IWO smaller nuclei. In such a reaction the total mass of the products is less than the original mass of the heavy nucleus. The fission of 235C by slow (low·energy) neutrons can be represented by the sequence of events -----?
X
+ Y + neutrons
Problems ond Perspectives
[30.1]
where 236U' is an intermediate state that lasts only for about 10- 12 s before splitting into nuclei X and Y, called fission fragments, Man)' combinations of X and Y satisfy the requirements or conservation of energy and charge. In the fission of uraIlium, about 90 different daughter nuclei can be formed. The process also results in the production of se\'eral (typicall)' two or th ree) neutrons per fission event. On the average, 2,47 neutrons are released per event. A lypical reaction of th is type is [30.2]
937
938
Chapter 30
FIGURE 30.1
A !ludear
Nuclear Energy and Elementary Particles
r.5~ioll
y
l..'\crll:l<;; d(:scribcd by the liqllid·drop
lIludd ofillc nucleus. (
approaches a ~"'U lIude.. ".
o
(h) Tlw lIeutrOll is absorbed b)' lhe :.o"-'L Illl(lcll~. changillg iliO !\l;U·. whidl is.1 :!~hC lI11c1clIS ill.1Il cxciled <"':1((:. lcl The llucleus deform" :Ind o ..cillale'ilil..c a liquid drop. (ell The llllc!ell
(aJ
Sequence of events in a
nuclear fission process
~
Ie)
(d)
The fission fragmcms. bariulll and krypton, and the released neutrons have a great deal of kinetic energy following the fission event. )Jolicc thaL the sum of the mass numbers, or number of Ilucleons, on the left (l + 235 = 236) is the same as the total numberofnucleonson the right (l41 + 92 -;- 3 = 236). The LOtal number of protons (92) is also the same on both sides. The energ), Qreleased through [he disintegration in Equation 30.2 can be easily calculated using the dara in Appendix B. The details of this calcuhuion can be found in Chapter 26 (Example 26.5), with an answcr of Q = 200.422 MeV. The breakup of the uranium nucleus can be compared to what happens to a drop of water whell excess ellergy is added to il. All the atoms in the drop ha\'e energy. but no!. enough to break up the drop. If enough energy is added LO set the drop vibrating, however, it. "'ill undergo elongation and compression ulllil the amplilllde of vibration becomes large enough to cause the drop LO break aparl. In the uranium nucleus a similar process occurs (Fig. 30.1). The sequence of C\'CIllS is as follows:
1. The 2:l5U nuclells captures a thermal (slow-moving) neutron. 2. The capt1lre results in the formation of 230U*, and the excess energy of this nuclcus GUises itLO undergo violclll oscillations. 3. The 2:lliU· IIllcleus becomes highly elongated, and the force or repllision bet\vecll protons in the 1\\'0 halves of the dumbbell-shaped nucleus tends 10 increase the distortion. 4. The nucleus splits into twO fragments, cmilling se\'eral ncutrons in the process. T),pically. the a mou nt of energy released by the fission of a si nglc he
APPLYING PHYSICS 30.1
UNSTABLE PRODUCTS
If a hC3\'Y nucleus \,,·ere LO fission into only two product nuclei. they \"ould be \'ery unstable. \"Thy? Explanation According LO Figure 29.3. the ratio of the number of neutrons LO the number of prowns increascs wilh Z. As a result. when a hea\'}' nuclells
EXAMPLE 30.1 Goal
splits in 3 fission reaction LO two lighter nuclei. the lighter nuclei tend to have LOO man) neutrons. The resull is instability because the nuclei return to the cune in Figure 29.3 by decay processes that reduce the numbcr of ncutrons.
A Fission-Powered World
Relate raw l1lalcrial to energy output.
Problem (a) Calculate the lOlal energy released if 1.00 kg of 2:l."iU undergocs fission, taking the disilllcgralion energy per e,"eIllLO be Q= 208 t\-fcV. (b) 1-10\.. . many kilograms of 2:~~JU would be needed to satisfy the ''''orld's annual energy consumption (abolll -I X J0 20 .P? Strategy In pan (a), lise the concept of a mole and Avog
30.1
Nuclear Fission
.
,....
939
• • • • • • • • • """II
Solution (a) Calculate the total energy released frol11 1.00 kg of'2:15U. Find the LOla I nurnber ofnuc1ei in 1.00 kg of uranium:
N = (
6.02 X 10" n uclei/ 11101) ( 1.00 X 10' g) 235 g/l11ol
= 2.56 X 10" nuclei
Multiply Nby the energy yield per nucleus, obtaining the total disintegration energy:
E = NQ = (2.56 X 10" nUclei)(208 ---".'vl:..:e\-,-I_) nucleus =
5.32 X ] 0'" MeV
(b) l-low many kilograms "..'Quld provide for the world's annual energy needs?
Set the energy per kilogram, Ekg , times the number of kilograms,lVkg , equal to the total annual energy consumption. Solve for Nkv,:
EkgNkg = E lOI 4 X 10 20 J
E
'Vk =~
··S
Ekl!-
(5.32 X 10" eV Ikg)( 1.60 X 10 "J/eV)
5 X 10"kg ~
..
"
,
.
• • • . ..i
Remarks The calculation implicitly assumes perfect conversion to usable power, which is never the case in real systems. Enough uranium deposits are knm.,.. n so as to provide the \.. . orld's current energy requirements [or a few hundred years. Breeder reactor technology can greatly extend those reserves.
QUESTION 30.1 Estimate the average mass of'B5U needed to provide pO'ver for one family for one year. EXERCISE 30.1 How long can 1 kg ofuranium-235 keep a 100-wattlightbulb burning ifall its released energy is convened to electrica I energy?
Answer
- 30 000 yr
Nuclear Reactors The neutrons emiued when 2:\5U undergoes fission can in turn trigger other nuclei to undergo fission, with the possibility ofa chain reaction (Actiyc Fig. 30.2). Calcu· huions shO'\l that ir the chain reaction isn't controlled, it will proceed LOO rapidly and possibly result in the sudden release of an enormous amount of energy (an explosion), even from only 1 g 0[25'15U. If the energ}' in 1 kg of 235L'" were released, it would equal that released by the detonation of about 20 000 tons of TNT! An uncontrolled fission reaction, of course, is the principle behind the first nuclear bomb. A nuclear reactor is a system designed to maintain ""hal is called a self·sustained chain reaction, first achieved in 1942 by Enrico Fenni. !'vlost reactors in operation today also lise uranium as fuel. Natural uranium contains only about 0.7% of the 215U isotope, with the remaining 99.3% being the 238U isotope. This fact is important to the operation of a reactor because 238U almost never undergoes fission. Instead, it tends to absorb neutrons, producing neptunium and plutonium. For this reason, reactor fuels must be
940
Chapter 30
Nuclear Energy and Elementary Particles
ACTIVE FIGURE 30.2 A nuclear chain reaction initiated by the CilpllJre ora lIClJlron.
as the average number of neutrons from each fission event that will cause another event. A sell:'sustained chain reacLion is achieved \".'hen K;:;:::: I. Under this condition, the reaClor is said to be critical. ""hen K is less than 1, the reacmr is subcritical and the reaction dies oul. ,,\,'hen Kis gre~uer than 1, the reactor is said to be supercritical and a runa,vay reaction occurs. In a nuclear reactor used to furnish power 1.0 a utiliL)] company, it is necessary to maintain a Kvalue close to l. The basic design ofa nuclear reactor is shmvn in Figure 30.3. Thc ruel elemcl1ls consist of ell riched uranium. The size of the reactor is important in redtlcing neutron leakage: a large reactor has a smaller surface-lo-volume ratio and smaller leakage than a smaller reactor. It's also imponal1tto regulate the neutron energies because 510\'..' neutrons are far more likely to cause fissions than fast neutrons in ~:l5U. Further, 23 RU doesn't absorb slow neutrons. For the chai n reaction to continue, the neutrons must, thercfore, be slov..· ed claVI-'ll. This slowing is accomplished by surrounding the fuel with a substance called a moderator, such as graphit.e (carbon) or heavy water (D 20). ]\'1ost modern reactors use heavy water. Collisions in the moderator slow the ncutrans and enhance the fissioning or2:~:)u. The power output ofa fission reactor is cOl1lrolled by the controll'Ods depicted in Figure 30.3. These rods are made of materials like cadmium that readily absorb neutrons. Fissions in a nuclear reactor heat molten sodium (or '.. . ater, depending on the system), which is pumped through a hear exchanger. There, the thermal energy is transferred to water in a secondary system. The water is converted to steam, which drives a turbine-generator to create electric power. Fission reactors are extremely safe. According to the Oa./t Ridgf' Na.tional Labomfory Revil'lv, "The heal1h risk of living within 8 km (5 miles) of a nuclear reactor for 50 years is no greater than the risk of smoki ng 1.4 cigarettes, drin king 0.5 L of wine, traveling 240 km by car, flying' 9600 km byjer, or having one chest x-ray in a hospital. Each activil y, by itself, is cst.i mated 1.0 increase a person's chance o[ dying in any given year by one in a million." The safet.), issues associated w'ith nuclear power reactors are complex and often emotional. All sources or energy have associated risks. Coal, [or example, exposes "vorkel's to health hazards (including radioactive radon) and produces atmospheric pollution (including greenhollse gases). In each case the risks must be weighed against the benefits and the availabi lity of the energy source.
30.2
30.2
Nuclear Fusion
NUCLEAR FUSION
When two light nuclei combine to form a heavier nucleus, the process is called nuclear fusion. Because the mass of the final nucleus is less than the slim of the masses of the original nuclei, there is a loss of mass, accompanied b}' a release energy. Although fusion power plants have nOl ret been developed,
or
Fusion in the Sun All stars generate their energy through fusion processes. About ~)O% of stars, including the Sun, fuse hydrogen, whereas some older stars fuse helil.llll or other heavier elcrnents. The energy produced by fusion increases the pressure inside (he star and prcvcllls its collapse due to gravity. Two conditions Illllst be met before fusion reactions in the star can sustain its energy needs. First, the temperature must be high enough (about 107 K for hydrogen) to allow the kinetic energy or the positively charged hydrogcn nuclei to ovcrcome their mutual Coulomb repulsion as the)' collide. Second, the density of nuclei must be high enough 10 ensure a high nHe of collision. It's interesting to nute thata quantulll effect is key in making sunshine. Temperatures inside stars like the Sun are not high enough (0 al1m" colliding protons to overcome the Coulomb repulsion. In a certain percentage of collisions, however, the nuclei pass through the balTier anyway an example of qllrlJltwn fill/neling. The proton-proton cycle is a series ofthlTe llilclear reactions that are believed to be the stages in the liberation of energy in the Sun and other stars rich in hydrogen. An overall vie"..' of the proton-proton cycle is t.hat four protons combine to form an alpha panicle and two posit.rons, with the release of 25 rvleV of energy in the process. The specific steps in the proton-prOLOn cycle are
:T-I +
IH
~
~D
+eI +
V
and :1-1
+
~D
~
~He
+ 'Y
[30.3]
where D st.ands for deuterium, the isotope of hydrogen having- one proton and one neutron in the nucleus. (IL can also be written as ~H.) The second re3Clion is followed by eilher hydrogen-helium fusion or helium-helium fusion: :1-1
+ ;I-Ie
-->
iHe
+ c+ + "
or ~He
+
~He
~
~He
+ 2CH)
Thc energy liberated is carried primarily by gamma rays, positrons, and neutrinos, as can bc seen from the reactions. The gamma rays are soon absorbed by the dense gas, raising its terr"lperature. The positrons combine \vith electrons to prodllce gamma rays, which in turn are also absorbed by the gas wit.hin a fe\\' cenI imeters. The neutrinos, howevcr, a1mos1 neyer int.eracl with mat.tcr; hence, they escape from the star, carrying about 2% of the encrgy generated \\lith thclll. These energy-liberating fusion reactions are callcd thermonuclear fusion reactions. The hydrogen (fusion) bomb, first. exploded in 1952, is an example of an uncontrolled thermonuclear fusion reactioll.
Fusion Reactors A great deal of effort is under way 1.0 develop a sustained and controllable fusion power reactor. Cont.rolled fusion is often called the ultimate energy source becallse of the availability of water, its fuel source. For example, if" deuteriulll, the isotope of hydrogen consisting ofa proton and a neutron, \,'ere used as the fuel, 0.06 g of it could be extraClcd rrom I gal or\,v·ater at a cost. OrabOUl four cents. Hence, the fuel
APPLICATION Fusion Reactors
941
942
Chopl.,JO
Nuclear Energy and Elementary Particles
costs of even an inefflcient reaCLQr would be almost illsignificam. An additional advant.age of Fusion reactors is Lhal comparatively few radioactive by-products are formed. As noted in Equation 30.3, the end product of the fusion of hydrogen nuclei is safe, nonradioactive heliulll. Unfortunately, a thermonuclear reactor that can delivc,' a net power output over a reasonable time interval is not yet a ,'cality, and rnany problems fnust be solved before a successful device is constructed. The fusion reactions that appear most promising in the construction of a fusion power reactor involve deuterium (0) and tritium (T), which areisoLOpes of hydrogen. These reactions are
TO + ~D ;0 + TD
--> -->
~He
+ 611
?T +
11-1
Q ~ 3.27 MeV Q ~ 4.03 MeV
[30.41
and
lD
Lawson's criterion ~
Vacuum
/
Plasma FIGURE 30.4
Diagram ora tokamak
used in the magnetic confinement scheme. Tile plasma is trapped within
the spiralillg magnetic fidd lines as shown.
+
iT
-->
lHe + &11
Q = 17.59 MeV
".'here the Qvalues refer to the amount of energy released per reaction. As noted earlier, deuterium is available in almost unlimited quantities from Ollr lakes and oceans and is very inexpensive to extract. Tritiulll, however, is radioaClh'e (Tv:.! = l2.3 yr) and undergoes beta decay [Q 3He. For this reason, tritium doesn't occur naturally to any great extent and must be artificially produced. The fundamental challenge of Iluclear fusion power is to give the nuclei enough kinetic energy to overcome the repulsive Coulomb force benveen them at close proximity. This step can be accomplished by heating the fuel to extremely high temperawres (about .10:) K, far greater than the interior temperawre of the Sun). Such high temperatures are not easy to obtain in a laboratory or a power plant. At these high temperatures, the atoms are ionized and the system then consists of a collection of electrons and nuclei, common ly reI-erred to as a !Jlasma. In addition to the high temperature requirements, two other critical factors determine whether or not a thermonuclear reactor will function: the plasma ion density 11 and the plasma confinement time T, the time the interacting ions are maintained at a temperature equal to or greater than that required for the reaction to proceed. The density and confinerncnl time must both be large enough to ensure that more fusion energy will be released than is required to heat the plasma. Lawson's criterion states that a net power output in a fusion reactor is possible under the follmving conditions: 1"17
2:::
101'1 s/cm 3
Deuteri u m-tritiu min Leraction
117
2:::
IO IG s/cm:\
Deuteriu m-deuteriulll interaction
[30.5]
The problem of plasma confinement time has yet to be solved. How can a plasma be confined at a temperature of 10 8 K for tirnes on the order of 1 s? Most fusion experiments use magnetic field confinement Lo contain a plasma. One device, called a tokamak, has a doughnut-shaped geometry (a toroid), as shmvn in Figure 30.4. This device uses a combination of t\·,;o magnetic fields to confine the plasma inside the doughnut. A strong magnetic field is produced by the current in the windings, and a weaker magnetic field is produced by the current in the toroid. The resulting magnetic field lines are helical, as shown in the figure. In this configuration the field lines spiral around the plasma and pre\'ent it from touching the walls of the vaCUUlll chamber. There are a number of other methods of creating fusion events. In inertial laser confinement fusion, the fuel is put into the form of a small pellct and then collapsed by liltrahigh-pm'ler lasers. Fusion can also take place in a device the size of a TV set and in fact was invented by Philo Farnsworth, one of the fathers of' electronic television. In this method, called inertial electrostatic confinemcnt, positively charged particles are rapidly attracted to\vard a negativcly charged grid. Some of the positivc particles then collide and fuse.
30.3
EXAMPLE 30.2 Goal
Elementary Particles and the Fundamental Forces
Astrofuel on the Moon
Calculate the energy released in a fusion reaction.
Problem
Find lhe energy released in the reaClion of" helium-3 with deuterium: ~I-Ie
Strategy
+
~D
~
JHe
+ :II
The energy released is the difference between the mass energy of the reactants and the products. . ....
,.. ... - ....
Solution Add the masses on the left-hand side and subtract the masses on the right. obtaining llln in atomic mass units:
Convert the mass difference
~
3.016029 u
"
+ 2.014 102 u - 4.002603 u - 1.007825 u
= 0.019703 u 10
an equivalent amount
I~=
(0.019703u) (
of energy in MeV:
931.5 MeV) = 18.35 MeV 1 II
........ - ......
lrro... • • • • • •
Remarks The result is a large amount of energy per reaction. Helium-3 is rarc 011 Earl h but. plentiful on the Moon, where it has become trapped in the nile dust of the lunar soil. Helium<1 has the advantage of producing morc protons than neutrons (some neutrons are still produced by side reactions, such as D-D), but has the disadvantage ora higher ignition tempcralUre. Lrfusion power plants using helium-3 became a reality, slUdies indicate that it would be economically advantageous to minc hclium-3 robotically and return it to Earth. The energy reLUrn per dollar would be far greater than for mining coal or drilling for oil~
30.3
943
QUESTION 30.2 f-]ow much energy, in joules, could be obtained from an Avogadro's number of helium-3-deuterium fusion reactions? EXERCISE 30.2 rind the energy yield in the fusion of t\O,IQ helium-3 nuclei: ~He
Answer
+
~j-Jc
12.9 MeV
ELEMENTARY PARTICLES AND THE FUNDAMENTAL FORCES
Besides the constituent.s of atoms-proLOns, electrons, and neutrons-numerous other particles can be found in high-energy experiments or observed in nature. subsequent to collisions involving cosmic rays. Unlike the highly stable protons and elcctrons, these particles decay rapidly, wit h halr·lives ranging from 10 13 S to 10 Ii s. There is very strong indirect cvidence that most of these particles, jnclud~ ing neutrons ancl proLOns, arc combinalions of more elemental'}' particles called quarks. Quarks, leptons (the electron is an example), and the particles that convey !'orces (the photon is an example) are nnw thoughl Lo be the truly fundamental particles. The key to understanding the propenies of elemenlary particles is the description of the forces of nature in which they pari icipate. All particles in nature are subject to four fundamental forces: the strong, elect rOlllagnetic, weak, and gra\'it<'\tion(ll forces. The strong force is responsible for the tight binding of quarks to form nculrons and protons and for the nuclear force, a son of residual strong force, binding neutrons and proLOlls into nuclei. This force represents the "glue" that holds the nucleons togcther and is thc strongesl of all the fundamental forces. It is a very short-range force and is negligible for separalions greater than about 10- 1:, m (the approximate size of the nucleus). Thc electromagnetic force, \.. . hich is about LQ-2 times the strength of the strong force, is responsible for the binding of ato1"n5 and molecules. It's a long-range force that decreases in strength as the inverse square of the separation between interacting
-->
iHe
+ 2(1 H)
944
Nuclear Energy and Elementary Porticles
Chapter 30
~
0;
§
l
~
~
6(
"RICHARD FEYNMAN American Physicist (1918-1988)
Feynmon, together with Julian S. Schwinger and Shinichiro Tomonogo, won the 1965 Nobel Prize in Physics for fundamental work in the prin-
ciples of quantum electrodynamics. His many important contributions to physics include work on the first atomic bomb in the Monhattan project, the invention of simple diagrams to represent particle interactions graphically, the theory of the weak
interaction of subatomic particles, a reformulation of quantum mechanics,
ond the theory of superfluid helium. Later he served on the commission investigating the Challenger tragedy, demonstrating the problem with the spoce shuttle's O-rings by dipping a scale-model O-ring in his gloss of ice water and then shattering it with a hammer. He also contributed to physics education through the magnificent three-volume text The Feynmon Lectures on Physics.
particles. The weak force is a short-range nuclear force thal is exhibited in the instability of certain nuclei. h's involved in the mechanism of beta decay, and its strength is only about 10-6 times that or the strong force. Finally, the gravitational force is a long-range force vvith a strength on Iy about] 0- 43 times that of the strong force. Although this familiar interaction is the force that holds the planets, sw,·s, and galaxies together, its effect on elementary particles is negligible. The gravita+ tional force is by far the ,,,"'eakest ofalllhe fundamental forces. Modern physics often describes the forces bet·ween particles in terms of the actions of field particles or quanta. In the case of the familiar electromagnetic interaction, lhc field particles are photons. In the language of modern physics, the electromagnetic force is mediated (carried) by photons, which are the quanta of the electromagnetic field. The strong force is mediated by field particles called glu.ons, the '',leak force is mediated by particles called the \V and Z bosom, and the gravitational force is thought to be mediated by quanta of the gravitational field called gravitons. torces between two particles are conveyed by an exchange offield quanta. This process is analogous to the covalent bond between two atoms created by an exchange or sharing of electrons. The electromagnetic interaction, for example, involves an exchange of photons. The force between two particles can be undersLOod in general with a simple illustration called a FeJl1ma.n diagram, developed by Richard P. Feynman (19181988). Figure 30.5 is a Feynman diagram for the electromagnetic interaction between t\'.'O electrons. In this simple case, a photon is the field particle that medi+ ates the electromagnetic force between the electrons. The photon tra nsfers energy and momentum hom one electron to the other in the interaction. Such a photon, called a virtualtJll.Oton, can never be detected directly because it is absorbed by the second electron very shortly after being emitted by the first electron. The existence of a virtual photon might be expected to violate the law of conservation of energy, but doesn't because of the time-energy uncenaimy principle. Recall that the uncertainty principle says that the energy is uncertain or not conserved by an a mall n t dE for a ti me I:1l such that 1:1£ L\ l = Ii.. I f the exchange of the vi rtual photon happens quickly enough, the brief discrepanc)' in energy conservation is less than the minimum uncertainty in energy and the exchange is physically an acceptable process. All the field quanta have been detected except for the graviton, which may never be found directly because of the '.. .' eakness of the gravitational field. These interactions, their ranges, and their relative strengths are summarized in Table 30.1.
30.4
POSITRONS AND OTHER ANTIPARTICLES
In the 19205 theoretical physicist Paul Dirac (j902-19S4) developed a version of quantum mechanics that incorporated special relativity. Dirac's theory accounted for the electron's spin and its magnetic moment, but had an apparent flaw in that it predicted negative energy states. The theor}' was rescued b}' positing the existence of an anti-electron having the same mass as an electron but the opposite charge,
Time e
" e-
e-
Virtual photon
•
' ... ",I
TA BLE..,3~O~.I
_
Particle Interactions
",
..... 1
Interaction e
(Force)
Range of Vorce
Mediating Field Particle
1 10- 2
Short (= 1 fll1)
Eleetroll1agncl ic
Long (ex: l/r 2)
Cluoll PhoLon
\'\'eak
10-li
Short (= 10-3 fll1)
\o\':tand Z
Gravilational
10-43
Long (" 1/,.')
Craviton
Strong
FIGURE 30.5 Feynman diagram represcnl ing 11 pholOn mediating Lhe electrOmagneLic rorcc between lWO electrons.
Relative Strength:!
aFor
(1'·0
quarks separated by ~ X 10- 1,
Ill.
b050n5
30.5
called a jJOSilm11. The general and profound implication of Dirac's theory is that for every particle, there is an antiparticle with the same mass as the particle, but the opposite charge. An antiparticle is usually designated by a bar over the symbol [or the particle. For example, 15 denotes the antiproton and v the anti neutrino. Tn this book t.he notation c+ is pl-eferrcd for the positron. Practically every known elementary panicle has a distincl antiparticle. Among the exceptions are thc phoLOn and the neutral pion (71'), which are their own (Intipanicles. In 1932 the positron was discovered by Carl Anderson in a cloud chamber experimen1. To discriminate between positi,·c and negati\"c charges, he placed the c1011d ch.unber in a magnetic lleld, causing moving charges to follow curved paths. T-Ie noticed that some of the electron-like tracks deneclcd in a direction corresponding 1.0 a positively charged panicle: positrons. When a particle meets its antiparticle, both panicles arc an nih dated, resulti ng in high-energy photons. The process of electron-positron a.nnihilation is used in the medical diagnostic technique ufpositron-elnission tomography (PET). The patient is injected with a glucose solution containing a radioactive substance that decays by positron cmission. Examples or such substances are oxygcn-15, nitrogen-13, carbon11, and fluorine-IS. The radioactive material is carried to the brain. \,Vhen a decay occurs, thc emitted posilron annihilates with an electron in the brain tissue, resulting in two gamma ray photons. \\lith the assistance ora computer, an image can be cre~lted of" the sitcs in thc brain where the glucose accumulates. The images from a PET scan can point 10 a wide variety 01" disorders in the brain, including Alzheimer's disease. Tn addition, because glucose metabolizes more rapidly in active areas or the brain than in other parts or the body, a PET scan can indicatc which areas or the brain are involved in various processes such as language, music, and vision.
30.5
Classification of Particles
945
TIP 30.1 Antiparticles An Hlltiparticle is nm idemificd soleI)' on 1he basis of (lpposilc charge. Evcn ncmral parlic1c~ ha\'(: antipaniclel:>.
APPLICATION
III
Positron-Emission Tomography (PET) Sconning
CLASSIFICATION OF PARTICLES
All particles othcr than t.hose thatlransmit forces can be classified into two broad catcgories. hadrons and leptons, according to their intcractiuns. The hadrons are composites of quarks, whereas the leptons are thought Lo be truly elememary, although there hm'e been suggestions that they might also have internal structure.
Hodrons Particles that interact through the strong force are called hadmlls. The two classes of hadrons, known as IJ/f'sons and baryons, are distinguished by their masses and spins. All mesons are knm... n to decay finally illlo electrons, pOSiU'OIlS, neutrinos, and photons. A good examplc of a meson is the pion (rr), the lightest of the knO\.. . n mesons, with a mass of." about 140 MeV/c 2 and a spin of O. As secn in Table 30.2 (page 946), the pion comes in three varieties, corresponding to three charge states: 'Tr~, rr~, ancl rro. Pions are highly unstable panicles. rorexample, the ']T-, which has a lifetime of about 2.6 X lO-H s, decays into a Illuon and an antincutrino. The J.Lmuon, essentially a heavy electron with a lifetime of 2.2 fJ..S, then decays into an electron, a neutrino, ancl an a11lineutrino. The sequence or decays is [30.6] fJ..
--+
e-
+
v
+ II
Baryons have masses equal to or greater than the proton mass (the name baryon means "heavy" in Greek), and their spin is always a nonintcger ntlue (~ or ~). Protons and neutrons are baryons. as are many other particles. vVil.h the exception of the proton, all baryons decay in such a \\lay thal thc end products includc a proton. For example, the baryon called the:: hyperon first decays to a N' in about 10- 10 s. The AO then decays to a proton and a 7T- in about. 3 X lO- IU s.
PAUL ADRIEN MAURICE DIRAC British physicist (1902-1984) Dirac was instrumental in the understanding of antimatter Gnd in the unification of quantum mechanics and relativity. He made numerous contributions to the development of quon~ tum physics and cosmology. In \933 he won the Nobel Prize in Physics.
946
Chapter 30
Nuclear Energy and Elementary Particles
TABLE 30.2 Some Particles and Their Properties
Category
Leptons
Particle Name
Symbol
Antiparticle
Mass (MeV/e')
Electron
c
(: +
Electron-ncutri no
",
V,.
f....l uon
I"
1'"
J05.7
fIlL
,,"
<
Muon-ncu(rino
Tau Tau-neutrino
0.511
< 7eV/r'.!. o.~
T+
I 784
",
",
< 30
,,'
7T
7
B
L,
L"
L,
0 0 0 0 0 0
+1 +1 0 0 0 0
0 0 +1 +1 0 0
0 0 0 0 +1 +1
0 0 0 0 0 0
0 0
0 0
0
0 0
0
S
Principal Decay Modes a
Lifetime(s)
Stahle
Stable 2.20 X IO-fi Stable < 4 X 10- 13
e-
IJcVj.l
e- V,l/7
J.L- VjtlJ p
Stahle
Hadrons Mesons
Self
139.6 135.0
K+
K-
493.7
0
0
0 0
KU,
KOS
K"I.
KOL
497.7 497.7
0 0
0 0
Eta
7)
548.8
Prawn
TJ p
Self Self
p
Neutron
11
n
0 0 +1 +1
Lambda
:\1)
1\0
Sigma
~+
Xi
L il LEll
0 0 0 0 0 0 0 0 0 0
Pion
1T
Kaon
Baryons
Omega
1l
"f+
958 938.3 939.6 1 115.6 I 189.4 I 192.5 I 1973
2"0
I 315
:::.
:=+
1321
n-
fl+
IfU
1672
+1 +1 +1 +1 +1 +1 +1
0
2.60 X 10-' 0.83 X ]()-Ifi
j1. ... J'J.l
1.24 X JO-'
11- vji'"
7T:!:c:::v ,37To
0
0 +]
0 0
0 0
+1 +1
0.89 X 10- 10 5.2 X "10-8
0 0 0 0
0 0 0 0
0 0 0
< 10- 18 2.2 X 1O-~]
0 0 0 0 0 0
0 0 0 0 0 0
0
0
0 -1 -]
+
-
(I
'7T
7T+1i-,27TO
c 7T:!:J.L+Vp.
2y,37T 7Ff
, r.
-
S(able
920 2.6 X 10 0.80 X 10
6
'"
10
1O-2lJ
-1 -1 -2 -2
2.9 X 10- 10 1.64 X 10- 10
-3
10- 10
"Nolalions ill Ihis column. ~lIch as p1T . n7fi, 1l11';ln (1,'0 pllssible dC('ay Illodes. In I hi.~ case 1)1l: two po~~ible decays lire 11.° ~ P
2y
X
1.5 X 10-10
0.82 X
pc-v" pro - • Ilr.
"
+ p7T , 1l7T 'J
:\Oy
"" I\
0 O 1T
1\D7T -
Su1To, AOK-
+ rr- and /\Il --+ 11 - 1TO.
Today it is believed that hadrons arc composed of' quarks. Some tant properties of hadrons arc listed in Table 30.2.
or the impor-
Leptons Leptons (from the Greek {eiJlOS, meaning "small" or "light") are a group of particles that participate in the weak interaction. All leptons have a spin of~. Included in this group are electrons, muons, and neutrinos, which are all less massive than the lightest hadron. A muon is identical to an electron except that its mass is 207 times the electron mass. Although hadrons have size and strucLUre, leptons appear to be truly elementary, with no structure down to the limit or resolution of experiment. (about 10- 19 m). Unlike hadrons, the number of knmvn leptons is small. Currently, scientists believe that there are only six leptons (each h~lVing an antiparticle): the electron, the muon, the tau, and a neutrino associated with each:
The taLI lepton, discovered in 1975, has a mass about twice that of the proton. Allhough neutrinos have masses of ~lbout zero, there is strong indirect evidence thal the electron neutrino has a nonzero mass of aboul3 eV/c 2 , or 1/180 000 oflhe
30,6
Conservation Laws
947
electron mass. A firm knowledge of the netltrino's mass could have great significance in cosmological models and in our understanding of the fuwre of the Universe.
30.6
CONSERVATION LAWS
A number of conserv
Baryon Number The law of conservation of baryon number means that whenever a baryon is created in a reaction or decay, an antibaryon is also created. This information can be qU<1ntined by assigning a baryon number: 13 = 1 for all baryons, B = -1 for all antibaryons, and 13 = 0 for all other panicles. Thus, the law of conscl'vation of baryon Dumber st.ates that whenever a nuclear reaction or decay occurs, the sum or the baryon numbers before the process equals the sum of" the baryon numbers afler the process. Note that if the baryon number is absolutely conserved, the proton must be absolutely stable: if it ,,,'ere not for the law of conservation or baryon number, the proton could decay into a positron and a neutral pion. Such a decay, however, has never been observed. At present, we can only sa)' that the proton has a half-life of at least lO·'H years. (The estimated age of the Universe is about 10 lD yeal-s.) ]n one version ora so-called grand unified theory, GUT, physicists predined that the proton is actually unstable. According to this theory, the baryon number (sometirnes called the ba'I)lol1ir: charge) is not absolutely conserved, whereas electric charge is always conserved.
EXAMPLE 30.3 Goal
~ Conservation
of baryon number
Checking Baryon Numbers
Lse conservation of' baryoll number to determine whet her a given reaction can occur.
Problem
Determine whether the following reaction can occur based on the law of conser\'ation of bal-yon number: p+n
~
p+p+n+p
Strategy COLtIll t.he baryons on both sides of lhe reaction, recalling- thal B = 1 for baryons and 13 = -] for al1tibaryons. .
~
.
Solution Coul1tthe baryons on the left:
~
The neutron and prolan are both baryons; hence, 1 + 1 = 2.
Count the baryons on the right:
There are three baryons and one antibaryon, so j
+ I + 1+(-1)=2. .
lrr... • • • • • •
...
Remark Baryon number is conserved in this reaction, so it can occur provided that the incoming proton has sufflcient energy. QUESTION 30.3 True or False: A proton can't decay into a positron plus a neutrino. EXERCISE 30.3 Can the following reaction occur. based on the law of conservation or baryon number?
p+n Answer
-->
p+p+p
No, (Compute the baryon numbers on both sides and show thal they're not equal.)
Chapter 30
948
Nuclear Energy and Elementary Particles
Lepton Number Conservation of lepton number ~
There arc three conservation laws involving lepton numbers, one for each variety of lepton. The law of conservation of electron-lepton number states that the sum of the elcClron-leplon numbers before a reaction or deca)' must equal the sum of the electron-lepton numbers after the reaction or decay. The electron and the electron neutrino arc assigned a positive elcnron-Iepton number Lc = 1, the anlileptons C' and 17c are assigned the electron-lepton number Le = -1, and all other particles have L c = O. For example, consider neutron decay: n
Neutron decoy -io
p+ + e- +ve
---*
°
Before the decay, the electron-lepton IlU mber is L e = 0; after the decay. it is + ] + (-1) = 0, so the electron-lepLOn number is conserved. It's important to recognize that baryon number lUust also be conserved. This can easily be seen by noting that before the decay B = 1, v.:hereas aftel- the decay B = 1 + + 0 = 1. Similarly, ,..,'hen a decay involves muons, the muon·lepron number LIJ. is conserved. The p.- and the l)jL are assigned LI-t = +1, the antimuons p,- and vlJ. are assigned LIJ. = -1, and all other particles have LJL = O. Finally, the tau-lepton number LT is conserved, and similar assignments can be made for the 'T lepLOn and its neutrino.
°
EXAMPLE 30.4 Checking Lep.ton Numbers Goal
Usc conservation of lepLOIl number to determine ,·.. hether a given process is possible.
Problem
Determine which of the [ollowing decay schemes can occur on the basis of conservation of lepton number: (1)
(2) Strategy
P.
"
+
~
e-
+ 17(: +
~
f.l+
+
J)J.t
vlJ.
+
Vc
COllnt the lepwns on each side and see if the numbers are equal.
Solution Because decay I im,olves both a muon and an electron, Lp. and L e must both be conserved. Before the decay, LIJ. = + 1 and Lc = O. After the decay, LJ.L = 0 + 0 + ] = +1 and 1...(: = - 1 - 1 + 0 = O. Both lepton numbers are conserved, and on this basis, the decay mode is possible. Before decay 2 occurs, Lp. = 0 and Lc = O. A fter the decay, LJl. = -] + ] + 0 = 0, but L(. = +1. This decay isn't possible because the eleclron-lepLOIl number is not conserved. 1Ir... • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • , •• , • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • ..OiI
QUESTION 30.4 Can a neutron decay into a positron and an electron? Explain. EXERCISE 30.4 Determine whether the decay"'Answer
----t J.L-
+ vp. can occur.
No. (Com pUle lepton numbers on both sides and show that they're not equal in this case.)
QUICK QUIZ 30.1 (a) p + p --> p (c) p.+ --> e +
QUICK QUIZ 30.2 (a) p (c)
W·hich of the follmving reactions cannot occur?
+P+P ", + v"
n --> K+
+
(eI) "- --> p.-
e-
+ V,
+ v"
\'\Thich of the following reactions cannot occur?
+ p --> 2y
rr'l +
(b) n --> p
(b) y
+ :£-
+ P --> n + "0 + P --> K+ + ~+
(eI) "+
30.7
The Eightfold Woy
949
Conservotion of Strangeness The K, 1\, and 2: panicles exhibilullusual properties in their production and decay and hence are called stmnge particles. One unusual property of strange panicles is that they are alwrl)'s produced in pairs. For example. when a pion collides v'lith a proton, two neutral strange particles are produced \\dth high probability (Fig. 30.6) following the reaction:
e-\
~ ~
On the other hand, the reaction 7r- + p+ ~ KU + n has never occurred, even though it violates no known conservation laws and the energy of the pion is sllf~ ricicllllO initiate the reaction. The second peculiar feature o[ strange panicles is that although the}' are proe1uced by the strong interaction at a high rme, 1 hey don', decay into particles that interact. via the strong force at a very high rate. Instead, they decay very slowly, which is characteristic of the weak interaction. Their half-lives are in the range of 10- 10 S [010-8 s; most other panicles that interact via the strong force have much shorter lifetimes on the order of 10-2:i s. To explain these unusual properties of strange particles, a law called conseroali.on oj strangeness \vas imroduced, lOgcther with a new quantulll number S called strangeness. The strangeness numbers for some particles are given in Table 30.2. The production of strange panicles in pairs is explained by assigning S = +1 to one of the particles and S = -1 to the other. All nonstrange panicles are assigned strangeness S = O. The law of conservation of strangeness states thai whenever a nuclear reaction or decay occurs, the sum of the strangeness numbers before the process must equal the sum of the strangeness Ilumbers after the process. The slow decay of strange panicles can be explained by assuming the strong and eleetromagnetic interactions obey the law of conservation of strangeness, 'whereas the weak interaction does no\.. Because the decay reaction involves the loss of one strange panicle, it violates strangeness conservation and hence proceeds slowly via the "veak interaction. 1n checking reactions for proper strangeness conservation, the same procedure as with baryon number conservation and lepton number conservation is followed. Using Table 30.2, counL the strangeness all each side. rf the two results are equal, the reaction conserves strangeness.
APPLYING PHYSICS 30.2
~
~
I ,;.
p\
I a:;
~ ~
~
FIGURE 30.6 This drawing rcpre· SCIII.S tracks of many events obtained
by analyzing a bubble-chamber photograph. The strange particles AO and KO are formed (at the bottom) as the ...- imcracls with a prawn according to the interaction 7T + P -+ Nl + KO. (7'\ole that the lIeutral particle,s leave no Iracks, as indicated by the dashed lines.) The AO and K(I then decay according to lhc interactions All -+ 7r +- P and fell -+ 7r + JL + /)'1'
BREAKING CONSERVATION LAWS
A student claims Lo have observed a decay o[an electron into two neutrinos traveling in opposite directions. \,Vhat conservation h1\\"s would be violatcd by lhis decay? Explanation Several conservation laws would be violated. Conservmion of electric charge would be violated because the negative charge of the eleclroll has disappeared. Conservation of electron lepton number \.. · ould also be violated because there is one lepton before the decay and two afterward. If bOlh neutrinos were electron neutrinos, electron lepton number conservation v.:culd be violated in the final
30.7
"0
state. If one or the product neutrinos were other than an electron neutrino, however, another lepton conservation law would be violated because there were no other leptons in the initial state. Other conservation la\\ls would be obeyed by this decay. Energy can be conscrved; the rest energy of the electron appears as thc kinetic energy (and possibly some small rest energy) of lhe neutrinos. The opposite directions of the two neutrinos' ,"clodties allow for the conservation of momentum. Conservation of baryon number and conservation of other lepton numbers v,,'ould also be upheld in this decay.
THE EIGHTFOLD WAY
Quantities such a5 spin, baryon number, lepton number, and strangeness are labels we associate with particles. Many classification schemes that group panicles into
950
Chapter 30
Nuclear Energy and Elementary Particles
FIGURE 30.7 (a) The hexagonal eightfold.way paltcrll for the eight spin-~ baryolls. This strangeness \er"US charge plm uses a horizolllal axi~ for the str
MURRAY GELL·MANN American Physicist {b. 1929} GelJ-Mann was awarded the 1969 Nobel Prize in Physics for his fheoretical studies dealing with subatomic particles.
families based on such labels have been proposed. First, consider the first eight baryons listed in Table 30.2, all having a spin of~. The family consists of the proton, the neutron, and six other panicles. If we plot their strangeness \,crsus their charge using a sloping coordinate system, as in Figure 30.7a, a fascinating pattern emerges: six of the baq:ol1s form a hexagon, and the remaining two are at the hexagon's center. (Particles with spin quanLUm number! or ~ are called Jermions.) Nm\' consider the family of mesons listed in Table 30.2 with spins of zero. (Panicles with spin quanllllll number 0 or "I are called bosons.) If we count both particles and (lmiparticles, there are nine such mesons. Figure 30.7b is a plot of strangeness versus charge for this Family. Again, a fascinating hexagonal pattern emerges. In this case the particles on the perimeter of the hexagon lie opposite their anti~ panicles, and the remaining three (which form their own antiparticles) are at its ~ center. These and related symmetric patterns, called the eightfold way, were pro.~ posed independently in 1961 by Murray Cell-Mann and Yu"all'.'e·eman. ~<:: The groups or baryons and mesons can be displayed in many other symmetric ~ patterns ''''ithin the [ramevl'Ork of the eightfold way. For example, the family of spin-~ baryons contains ten particles arranged in a pattern like the tenpins in a bowling alley. After the pattern was proposed, one of the panicles was missing; it had yet to be discovered. Gell-.\'lann predicted that the missing panicle, ""hich he called the omega minus (n-), should have a spin of~, a charge of -I, a strangeness of -3, and a mass of about 1 680 1\1cV/c 2 . Shortly thereafter, in ]964, scientists at the Brookhaven National Laboratory found the missing particle through careful analyses of bubble-chamber photographs and confirmed all its predicted properties. The patterns of lhe eightfOld way in the AcId of panicle physics have much in common '',lith the periodic table. \Vhcncver a vacancy (a missing particle or element) occurs in the organized patterns, experimentalists have a guide for their investigations.
30.8
QUARKS AND COLOR
Although leptons appear to be truly elementary particles without measurable size or structure, hadrons are more complex. There is strong evidence, including the scattering of electrons off nuclei, that hadrons are composed of more elementary particles called quarks.
The Quark Model According to the quark model, all haclrons are composite 5)'stems of tWO or three of six fundamental constituents called quarks, \vhich rhymes with "sharks" (although some rhyme it with "forks"). These six quarks are given the arbitrary names ut), down, strange, charmed, bollom, and lOp, designated by the letters ll, d, s, c. b, and t.
30.8
TABLE 30.3
Quark Composition of Several Hadrons
Quarks Bar)'on Symbol Spin Charge Number Strangeness Charm Bottomness Topness 9
Up
+if'
II
Down
I
d
"1
Strange
s
I
Charllled
c
Bottom
b
'i I
"1
-~1'
I
j
, I
0
0
0
0
0
0
0
0
-~t'
-I
0
0
0
+i..
0
+1
0
0
-*",
0
0
+1
0
()
0
0
+1
+~p
Top
Antiquarks
Name
I
II
:!
Anti-down
d
:!
I
Anti-botLOm
-*e
+~e
+;..
Anti-strange
f\llli-l.Op
Particle
r
b
I
"2
, ., I
"2
-~('
+~l'
.,
-~('
, I
Quark Composition
Mesons
,,'
dll
1T
lid Sll
us sci
Baryons
p
Baryon Symbol Spin Charge Number Strangeness Charm Bottomness Topness
Anti-up
Anli~chanllec1
951
TABLE 30.4
Properties of Quarks and Antiquarks
Name
Quarks and Color
uud mid lids
()
0
0
0
I
0
0
0
0
uds
I
+1
0
0
0
dds
0
-I
0
0
llSS
dss
, , ,, .,, .,,
0
0
-I
0
0
0
0
-]
UliS
n
sss
Quarks have fractional electric charges, along with other properties, as shmvn in Tahle 30.3. Associated with each quark is <'In antiquark of opposite charge, bar)lon number, and strangeness. Mesons consist of a quark and an amiquark, whereas baryons consist of three quarks. Table :lOA lists the quark compositions of several mesons and bar)lons. )Jote that jusltwO of the quarks, u and cI, are contained in ~dl hadrons encountered in ordinary matter (protons and neutrons). The third quark, s, is needed only to constnlct stnlnge particles with a strangeness of either + I or -I. Active Figurc 30.8 is a pictorial represcntation of the quark compositions of several particles. The charmed, bottom, and top quarks are more massive than the other quarks and OCClIr in higher-energy interactions. Each has its O\\ln quantum number, called charm, boltomness. and topness, respectivel)'. An examplc of a hadron fOfmcd from lhesc quarks is theJ/'J! panicle, also called charmoniulll, which is composed Or~l charmed quark and an anticharmcd quark, ce.
APPLYING PHYSICS 30.3
CONSERVATION OF MESON NUMBER
We havc seen a law of consen'ation of lepton numher and a law of conservation of baryon number. Why isn't there a law of consen'ation of meson number? Explanation ',Ve can answer this question from the poin! of"icw of creating panicle-antiparticle pairs from available energy. If energy is cOlwerted to the rest energy ora lepton-antilepLOn pair, there is no net change in lepton number because the lepton has a lepton number of +1 and the antilepton -I. Energy could also be transformed into the rest energy ora
baryon-anti baryon pair. The baryon has baryon number + l. the antibaryon -1, and there is no net change in baryon number. Now suppose energy is transformed into the rest energy of a quark-antiquark pair. By definition in quark theory, a quark-ar'lliquark pail" is a meson. 1n this reaction. therefore, the number of mesons increases from zero to one, so meson number is nOL conserved.
952
Nucleor Energy and Elementary Particles
Chapter 30
Color TIP 30.2 Color Is Not Reolly Color lVhcn we usc the word colono describe a quark, it has BUllting (0 do with visual sensation from light. It is simply n (:Olwcnicnl name for a property analogolls LO electric charge.
Baryons
~[esons
n+
p
- - -, ,
, I
\
,,
d
U
, , I
-, -"- -, , --
, I
-"- \
I
, ,
,
-- -
I
-- -
ACTIVE FIGURE 30.8 Quark compositiOlIS of [wo ml:son~ and t\\"O baryons. \'ole that L1IC mesons on the left cOlllaill two quarks and thallhc baryons on the righl <:011lain three quarks.
,
,,
....
-- -
I I
I
Quarks have another property called color or color charge. This propen)1 isn't color in the visual sense; rather, it's just a label for sornething analogous to elecrric charge. Quarks are said to come in three colors: red, green, and blue, Amiquarks have the properties antired, anrigrecn, and antibluc. Color was defined because some quark combinations appeared to violate rhe Pauli exclusion principle. An example is thc omega~l11inus particle (n-), which consists of three strange quarks, sss, that ,He all spin up, giving a spin o[~. Each strange quark is assumed to have a different. color and hence is in a distinct quantum stat.e, satisfying the exclusion principle. Tn general. quark combinations must be "colorless." A meson consists ora quark of one color and an amiquark of the corresponding anticolor. Baryons must consist of' one recl, one green, and one blue quark, or their anticolors. The theory of how quarks intcract with one another by means or color charge is called quantum chromodynamics, or QCD, to parallel quanLUlll clectrodynamics (t.he theory of intcractions among elecrric charges). The strong force between quarks is often called the color force. The force is carried by massless particles called gluons (\"vhich are ana1ogolls to photons for the electromagnetic force). According to QCD, there arc eight glLLons, all \vith color charge, and their allligluons. \rVhcll a quark emirs or absorbs a gluon, its color changes. For example, a blue quark that emits a g[lIon may become a red quark. and a red quark that absorbs this gluon becomes a bluc quark. The color force between quarks is analogous to the electric force betwccn charges: like colors repel and opposite colors attract. Therefore, two red quarks repel each other, but a red quark \\o'ill be attracted to an <'llllired quark. The attraction between quarks of opposite color to form a meson (qq) is indicated in Figul·e 30.9". Different-colored quarks also attract one another, but with less intensity than opposite colors of quark and antiquark. For example, a cluster of recl, blue, and green quarks all attract. one another to form baryons, as indicated in Figure 30.9b. Every baryon contains three quarks of three different colors. AILhough the color force bet\<\'een two color-neutral hadrons (such as a proton and a neutron) is negligible at large separations, the strong color force between thei r constilUent quarks does not exactly cancel at small separations of about I fll1 . This residual strong force is in fact the nuclear force that binds protons and neutrons to form nuclei. It is similar to the residual electromagnetic force that binds neutral atoms into molecules.
~'Icson
\
,,
\
30.9 ... - -
- ....
(01
... - -
-
I
, I
I
Baryon
,
\
\
,, (bl
FIGURE 30.9 (a) A green qllark is atlraclcd to all 311ligrccn qll
(qq). (b) Three dirrercll1-culored
quarks attract one another baryon.
to
farll! a
ELECTROWEAK THEORY AND THE STANDARD MODEL
Recall that the \·...eak int.eraction is an extreme1)' shorr range force having an interaction distance of approximately 10- 11" m. Such a short-range interaction implies that the quantized panicles that carry the \.. . eak field (the spin 1 Wi-, Vti'-, and ZO bosons) are cxtremely massive, as is indeed the case. These amazing bosons can be thought of as struclUrelcss. point-like particles as massive as krypton atoms! The weak interaction is responsible for the decay of the c, s, b, and r quarks into lighter, more stable u and d quarks, as well as the decay of the massive Ji and 7lepro!1s into (lighter) electrons, The weak interaction is very important because it governs the stability of the basic particles of matter. A mysterious fealUre of the weak lnleraction is its lack of symmetry, especially when compared \·/ith rhe high degree o[ symmetry shovm by the strong, electromagnetic, and gravitational interactions. For example, the \.veak interaction, unlike the strong interaction, is not symmetric under rnirror reflection or charge exchange. (/\I[irrol" nfleclion means that all the quantities in a given pC'lrticie reaction are exchanged as in a mirror reflection: left for right, an inward motion toward the mirror for;:'lll out\'':..l rd motion, and so forth. Cluuge exc!wng(' means that all the
30.9
Electroweok Theory and the Standard Model
953
FIGURE 30.10 The Sl
:vIATI'ER AND ENERGY
Chlon Elcclro-
PhOIOIl
I~a~n~t~c
\\'cak
Wand Z bo:-,ons
b
Leptons r
era\ iLOn
v,
eleClric charges in a panicle reaction are converted to their opposites: all positives 1.0 negatives and vice versa.) Not symmetric means t.hat the reaClion wit.h all quantities changed occurs less rrequently than the direct reaction. For example. thc decay of the KfJ , which is govcrncd by the weak imeraction, is not symrnetric under charge exchange bccause thc reaction KO ---) -rr- + e+ + Vtc occurs much morc frequently than t.he rcaction KQ ---t -rr+ + e~ + V,.. The electroweak theory uniAes the electromagnelic and weak interactions. This theory postulates that thc weak and electromagnctic interactions havc t.he same strength at very high panicle energies and are differciltmanifestations of a single unifying e1ectroweak interaction. The phot.on and the three massive bosons (\.\7± and ZO) play kcy roles in the e]ectroweak thcory. The theory makes many concrete predictions, such as t.he prediction or the masses of the \N and Z panicles at about 82 CeV/c 2 and 93 GeV/c1 . respectively. These predictions have been experimentally verified. The combination of" the clectroweak theory (tlld QCD (or the strong interaction rorms what is referred to in high-energy physics as the Standard Model. Although the details or the Standard Model are complex, its essential ingredients can be summarized with the help of Figure 30.tO. Thc strong force, mediated by gluons, holds quarks together to rorm compositc particles stich as protons, neutrons, and mcsons. Leptons participate only in the electromagnetic and weak interactions. The electromagnetic force is mediatcd by phot.ons, and the weak force is mediated by VV and Z bosons. ~ote that all fundamental f"orees are mediated by bosons (panicles with spin I) having properties given, to a large extent, by symmctries involved in the theories. Thc Standard Model, however, doesn't answer all questions. A major question is ,"vhy the photon has no mass although the VV and Z bOSOIlS clo. Because of this mass difference, the electromagnetic and weak forces are ,'ery different at low energies bllt become similar in nature at very high energies, where the rest energies of the Wand Z bosons are insignificant fraClions of their tOlal energies. This behavior during lhe transition from high to low energies, called symmetry breaking. doesn't answer thc question or the origin or panicle masses. To resolvc that problem, a hypothetical particle called the Higgs boson, which provides a mechanism for breaking the elcctroweak symmetry and bestowing different panicle masses on different. particles, has been proposed. The Standard Model, including the Higgs -~u· ll1cchanism, provides a logically consistent explanat.ion of the massive nal.Ure of "0 the 'A' and Z bosons. Unfortunately, the Higgs boson has not. yet been found, hut j physicists predict that its mass should be less than 1 TeV/r 1 (]()12 eV). Following the success of t.he electroweak theory, scientists allempled (a combine it wit.h QCD in a grand unification theory (GUT). In this model the electrowcak
A dew from inside lhe Large rJeclrnll~Po~ilroll (LEI') collideI' 11111IIC!.
\~'hi("h
is 27 kill in cil'Cumrercllce.
954
Chapter 30
Nuclear Energy and Elementary Particles
force was merged with the strong color force to [arm a grand unified force. One version of the theory considers leptons and quarks as members of the same f~\m ily that are able to change into each other by exchanging an appropriate particle. Many GUT theories predict that protons arc unstable and \\Iill decay with a lifetime of about 10 31 years, a period LlI- g,-eater than the age of the Cniverse. As yet, proton decays have not been observed.
30.10 THE COSMIC CONNECTION According to the Big Bang theory, the Universe erupted from an inflnitely dense singularity about 15 billion to 20 billion years ago. The first few minutes after the Big Bang saw such extremes of energy that it is believed that all {-our interactions of physics were unified and all matter was contained in an undifferentiated "quark SOllP
~
]
I ]' ~
~
"t:: t
3 GEORGE GAMOW Russian Physicist (1904-1968) Gomow and two of his students, Ralph Alpher and Robert Herman, were the first to take the first holf hour of the Universe seriously. In a mostly
overlooked paper published in 1948, they made truly remarkable cosmo-
logical predictions. They correctly cal-
culated the abundances of hydrogen and helium ofter the first half hour (75% Hand 25% He) and predicted that radiation from the Big Bang should still be presenl and have an apparent temperature of about 5 K.
FIGURE 30.11
The evolution of the lour fundamental forces from the Big Bang to the present is shown in Figure 30.11. During the first ]0- 43 s (the ultrahot epoch, with T< 10 32 K), the strong, elccu-oweak, and gravitational forces were joined to form a compJetely unified force. In the first 10-;\5 s following the gig Bang (the hot epoch, with T < ] O~9 K), gravity broke free of th is unification and the strong and clectroweak forces remained as one, described by a grand unification theory. During this period panicle cnergies were so great (> lOIt] CeV) that very massive particles as ,,,,'e11 as quarks, leptons, and their antiparticles existed. Then, after 10-35 s, the Universe rapidly expanded and cooled (the warm epoch, with T< 10 2!lto ]0 15 K), the strong and c1eetroweak forces parted company, and the grand unification scheme '"as broken. As the Universe continued to cool, the electroweak force split into the weak force and the electromagnetic force about 10-10 s after the Big Bang. After a few minutes, protons condensed out of the hal. soup. For half an hour, the Universe underwent thermonuclear detonation, exploding like a hydrogen bomb and producing most of the helium nuclei now present. The Universe cominucd to expand, and its temperaulre dropped. Until about 700 000 years after the Big Bang, the Universe was dominated by radiaLion. Energetic radiation prevented matter from forming single hydrogen atoms because collisions would instantly ionize any atoms that might form. Photons underwent continuous Compton scattering from the vast number or free electrons, resulting in a Universe that was opaque to radiat.ion. By the time the Universe was about 700 000 years old, it had expanded and cooled to about 3 000 K. Protons could now bind to e1ect.rons to form neutral hydrogen atoms, and the Universe suddenly became transparent to photons. Radiation no longer dominated the Universe, ,md clumps of neutral matter grew steadily: first atoms, ~oIlO\. . 'ed by molecules, gas clouds, stars, and finally galaxies.
Protons and
:\ briefhislOl"yof Quarks and leplons
t.he Universe from the Big Bang LO Lhe prcselll. The fOllr forces became dislinguishable during the first micrusecond. Then, all the quarks combined to form panicles that inleraCI via the SI1'Ong rorcc. The leplOlls remained scparale, however, and exist as individually observable partide~ 10 this day.
GraviUttiun:11
can form
(
Nuclei can form Atoms can form
I1CllLrons
,/
~
I I
:
Gravilahon
I
,
I
~
I
I
/ Sln)[lg and eleCll"oweak Unified
Two
force
rOl"ces
I I
II
Stroll!' r6rce 0 I
I
"Veak rorce
I
, •
/
ElccU'owcak
I
, I
I
I I
F.lecl romagnetic '1rcc \ Four rorcco, !
Thn:c forces
I
10-20 Big Bang
I
I
I
I
'
,
IOU Age oflhc Univcl"."iC (s)
Present age of the 1.1 ni\'erse
30.11
Problems and Perspectives
955
Observation of Radiotion from the Primordial Fireball ~
In 1965 Arno A. Penzias (b. 1933) and Raben W. Wilson (b. 1936) of Bell Labora- ~ l.orics made an amazing discovery while testing a sensitive microwave receiver. A 'S~ pesky signal producing a faint background hiss was imerfering with their satellite ~ communications experiments. Despite all their cffons, the signal remained. Ulli- ~ ma1.el)', it became clear th.u they were observing microwave backgTound radiation i (at a wavelength of7.35 elll) represellting the leftover "glow" fro111 the Big Bang. The lllicrmvavc horn that served as their receiving antenna is shown in Figure 30.12. The intensity or the detected signal remained ullchanged as I.he (lntenna was pointed in different directions. The radiation had equal strength in all direclions, \\'hich suggested thal the ent.ire Universe \".'as the source of this radiation. Subsequent experiments by other groups added intensity data at different wavelengths, as shown in Figure 30.13. The results confirm that the radiation is that of a blackbody £112.9 K. This figure is perhaps the most clear-clll evidence for the Big Bang theory. The cosmic background radiation \o\'as fOLlnd to be too uniform to have led to the development of galaxies. In 1992, following a study using the Cosmic Background Explorer, or COBE, slight irregularities in the cosmic background were found. These irregulal'ities are t.hought to be the seeds of galaxy formation.
'" FIGURE 30.12
30.11
Robt"rl \V. WilSOl1 (lrft) and Arno A. PCll/ias (rilfht). with
Bell T\'lcphone Laboral0rics' lIofllrcncclor antenna.
PROBLEMS AND PERSPECTIVES
While panicle physicisls havc becn exploring the realm of the vcry small, cosmologists have been exploring cosmic history back to the first microsecond of" the Big Bang. Observation of the events that occur when two panicles collide in an accelerator is essential in reconstructing the early moments in cosmic hisLOry. Perhaps the key LO understanding the early Universe is first to understand t.he worlel of elementary particles. Our understanding of physics at shan and long distances is far frorn complete. Particle physicists are bced with many unanswered questions. \'\'hy is there so little antimatter in the Universe? Do neutrinos have a small mass, and if so, hmv much do they contribute 1.0 the "dark matter" holding the Uniyerse together gravitationally? How can \ve understand the latesl astronomical measurements, which show t.hat Ihe expansion of the Universe is accelerating and that there Illay be a kind of "anrigTavil)' force," or dark energy. acting between \,·idel)' separated galaxies? Is it possible LO unify the strong and electroweak theories in a logical and consistent manner? "Vhy do quarks and leptons forlll three similar but distinct families? Are
FIGURE 30.13 Theorctical bl
--..E
"c
;;-
, ~
10'
;:
~
lOb
" C
"E" I(rl ~
PCll/.iasand Wibon
0.1
I \<\:an~lell~lh
10 (cln)
100
956
Chapter 30
Nuclear Energy and Elementary Particles
muons the same as electrons (apart from their different masses), or do they have subtle differences that have not been detecLed? \0\711y are some panicles charged and others neutral? vVhy do quarks carry a fractional charge? What determines the masses of the fundamental particles? The questions go on and on. Because of the "arid advances and nevI' discoveries in the related fields of particle physics and
cosmology, by the lime you read this book some of these questions may have been resolved and olhers may have emerged. An imponant question that remains is whether leptons and quarks have a substructure. Many physicists bel ieve that the fundamental quantities arc not infinitesimal points, but extremely tiny vibrating strings. Despite more than three decades of string theory research by thousands of physicists, however, a final Theory of Everything hasn't been Found. Whether there is a limit to knowledge is an open question.
--
--_._~
~
SUMMARY 30.1
Nuclear Fission
In nuclear fission the total mass of the products is ah\-'ays less than the original mass of the reaClams. Nuclear fission occurs when a hem'}' nucleus splits, or fissions, into two smaller nuclei. The lost mass is t.ransFormed into energy, electromagnetic radiation. and the kinetic energy of daughter particles. A nuclear reactor is a system designed to maintain a self-sustaining chain reaction. Nuclear reactors using controlled fission events are currentl}' being used to generate electric power. A useful parameter for describing the level of reactor operation is the reproduction constant K, which is the average number of neutrons from each fission evem that will cause another event. A self-sustaining reaction is achieved when K = 1.
30.2
Nucleor Fusion
In nuclear fusion two light nuclei combine to for III a heavier nucleus. This type of nuclear reaction occurs in the Sun, assisted by a quantum tunneling process that helps particles get through the Coulomb barrier. Controlled fusion events oner the hope 01" plentiful supplies of energy in the future. The nuclear fusion reactor is considered by many scienLists to be the ultimate energy source because its fuel is water. Lawson's criterion states that a fusion reactor will provide a net output power if the product of the plasma ion density n and the plasma confinement time T satisfies [he Following relationships:
nT
2:::
10 14 siemS Deuterium-tritium interaction
nT
~
IOIt's/em 3 Deuterium-deuterium interaction
[30.5]
30.3 Elementary Porticles and the Fundamentol Forces There are four fundament.al forces of nature: the strong (hadronic), electromagnetic, weak, and gravitational forces. The strong force is the force between nucleons that keeps the nucleus together. The weak force is responsible for beta decay. The electromagnetic and weak forces arc now considered to be manifestations of a single force called the electrowcak force.
Every fundamental interaClion is said to be mediated by the exchange of field panicles. The electromagnetic interaction is mediated by the photon, the weak interaction by the 'V:t and ZO bosans, the gravitational interaction by gravitons, and the strong interaction by gluons.
30.4
Pasitrons and Other Antiporticles
An antiparticle and a particle have the same mass, but opposite charge, and may also have other propenies with opposite values, such as lepton number and baryon number. It is possible to produce particie-alllipanicle pairs in nuclear reactions if the available energy is greater than 211/(2, where m. is the mass of the particle (or antiparticle).
30.5
Classification of Particles
Panicles other than photons are classified as hadrons or leptons. Hadrons interact primarily through the strong force. They have size and structure and hence are not elementary particles. There are two types of hadrons: ba'}ons and mesons. Mesons have a baryon nlllllberofzeroand have either zero or integer spin. Bar·yons, which generally are the most massive particles, have nonzero baryon numbers and spins of ~ or ~. The neutron and proton are examples of baryons. Leptons have no known structure, down to fhe limits of current resolution (about 10- 19 m). Leptons interact only through the weak and electromagnetic Forces. There are six leptons: the elect.ron, e~; the muon, J.I.-; the tau, T-; and their associated neutrinos, TJ" vI" and v"
30.6
Conservation Laws
In all reactions and decays, quantities such as energy,linear momentum, angular momentum, electric charge, baryon number, and leptOn number are strictly conserved. Certain particles have properties called strangeness and charm. These unusual propert.ies are conserved only in those reactions and decays that occur via the strong force.
30.8
Quarks ond Color
Recent theories postulate that all hadrons are composed of smaller units known as quarks, which have fractional electric charges and baryon numbers of ~ and come in six ·'f1a~ vors": up, down, strange, charmed, t.op, and bottom. Each
Conceptual Questions
baryon contains three quarks, and each meson contains one quark and one amiquark. According to the theory of quantum chromodynamics, quarks have a property called color, and the strong force between quarks is referred to as the color force. The color force increases as the distance between particles increases, so quarks are confined and arc never observed in isolalion. When two bound quarks are widel)' separated, a new quark-amiquark pair forms between them, and the single panicle breaks imo n,'o new panicles, each composed 01" a qllark-antiquark pair.
957
30.10 The Cosmic Connection Observation of background lllicrowavc radiation by Penzias and Wilson st.rongly confirmed that the Universe started with a Big Bang about 15 billion years ago and has been expanding ever since. The background radiation is equivalent 10 that 01" a blackbody at a temperature of about 3 K. The cosmic micrm,'ave background has very small irregillarities, corresponding to temperature variations of 0.0003 K. \·Vithout these irregularities acting as nucleation SiICS, particles would never have clumped together to form galaxics and stars.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
MULTIPLE-CHOICE QUESTIONS 1. Tn the first atomic bomb, ~1 lest carried Ollt in l\'cw rvlexieo called Trinity. the energy released was equivalent to approximately 17 kilotons of TNT. ESl:im
2. When a nClilron decays
1.0 a proton, electron, and ncuu'ino, how much energy is released? (Neglect the energy of the neutrino.) (a) 0.781.1 MeV (b) 0.352 MeV (c) 4.31 MeV (eI) 2.22 y[eV (c) UO MeV
3. \-Vhy can't. a proton decay into a positron and a neutrino? (a) Charge conservation is violated. (b) Baryon number conservation is violated. (c) Conservation or energy is violated. Cd) Lepton number conservation is violated. (c) Conservation of strangeness is violated.
J.
4. The n~ panicle is a baryon with spin Which of the following is truc of Ihis particle? (a) It has three possible spin st.ates in a magnetic ficlcl. (b) tt has four possible spin slates. (c) It has three times the charge or a spin-~ particle. Cd) It has three timcs the mass of a spin-~ particle. (e) It can't decay to a spin-~ particle. 5. When an electron and a positron meet at low speed in empty space, the)' annihilatc each other to produce two O.511-MeV gamma rays. What law would he violatcd if t.hey produced one gamma ray with an energy or 1.02 MeV? (a) conservation of energy (b) conservation of momentum (c) conservat.ion of charge (d) conservalion of baryon number (e) conservation of electronlepton number. 6. I r t.he moderator were suddenly removed from a nuclear reac[Qr in an electric generating station, what
is Lhe most likely consequence? (a) The reactor would go supercrilical, and a runaway reaction would occur. (b) The nuclear reactions would proceed as before, bllt the reactor \\,'ould overheat. (c) The reactor would become subcritical, and the rate at which reactions OCCUlTed would diminish. (d) No change would occur in the reactor's operation. (e) The rate at which reactions occurred would increase, but the reaCI.OI- wouldn't go supercritical. 7. \·Vhich parlicle is most likely to be captured by a 235U nucleus and cause it to undergo fission? (a) an energetic prawn (b) an energetic neutron (c) a slow-moving alpha particle (d) a slow-moving neutron (e) a fastmoving electron 8.ln a fission reaction a 235U nucleus captures a neutron, splitting the nucleus into 13il and 96v. How many free neutrons result. from this process? (a) 1 (b) 2 (c) 3 (d) 4 (c) 5 9. \·Vhich of the following particle reactions can't occur? (a) p + n -> p + P + P (b) n --+ p -I- e - -I- Ii, (c) P, -
--+
(d) rr
--+
(c)
--+
T-
-I- Ii, -I- vJL p,- -I- vJL e -I-I- V,
c
v:
10. \Vhich of the following particle reactions can't occur? (a) p + P -> 2y (b)rrfl-l-n (c) Y -I- P
(eI) fl (e)
7T+
--+
K~~-I-~()
n -I- rro
--+
22- -I- c'" -I- v,
+ P
--+
K+ -I-
~+
CONCEPTUAL QUESTIONS I. If high-energy e1ect.rons with de B,-oglie wavelengt.hs
smaller than the size of the Iwcleus are scattered from nuclei. the behavior of thc electrons is consistclll with scattering from very massive structures much smaller in size than the nucleus, namely, qllarks. Ilow is this behavior similar to a classic experil11elll that detected small structures in an atom?
2. What factors make a fusion reaction difficull to achieve?
3. Doubly charged baryon~ are knm..'n to exist. \'Vhy are there no doubly charged mesons? 4. \Vhy would a fusion reactor produce less radioactive was[C than a fission reacwr?
958
Chapter 30
Nuclear Energy and Elementary Particles
5. 'Vhy didn't at0111s exist until hundreds of thousands of years aflcr the Big Bang?
9. Describe the properties or baryons and mesons and the important differences between them.
6. Panicles known as resonances have ,'en short half-lives, on the order of 1O-2~ s. \Vould rOll gu~ss that they are haurulI'i Ul lepwlIs?
to. Identify the panicle decays in Table 30.2 thal occur by the electromagnetic interaction. Justifr your ans\\·er.
11. Kaons all decay into final slales that contain no protons
7. Describe the quark model of hadrons, including the
propenies of quarks. 8. In the theory of quantum chromodynamics, quarks come in three colors. How would }'Oll justify I he slate~ menL, "All bar}"ons and mesons afC colorless"?
01'
nemrons. \Vhat is lhe bar)'on number ofkaonsr
12. \·rhy is a neutron stable inside the nucleus? (In free space the neutron deca}'s in 900 s.)
PROBLEMS ~w" L..A~irtn The Problems for this chapter t. ...
lll'I\'
ship lI'avel per kilogram of fuel if the fue) consists of enriched uranium containing 1.7% of the fissionable iso· tope 2 35 U and the ship's engine has an efficiency 01'20%;: Assume 2081\:[eV is released per fission en::I1!.
SECTION 30.1 NUCLEAR FISSION 1. Burning I mctric Ion (I 000 kg) of coal call yield an energ}' of 3.30 X 10 10 J. Fission of one nucleus of 'Iran i'lm235 yields an :werage energy of approximately 208 f\"cv. Whal lllass of uraniulli produces the same energy as 1 metric ton of coal?
2. Find the energy released in the fission reaction n + '.!~U
--+
:mZr + I~Tc + 3n
The atomic masse" of the fissioll producLS are 97.912 0 u for 1SZr and 134.908711 for 1~~'Te. 3. Findlhe energy released in the fission reaction
4. The nldioactivc isotope strontium-90 is a panicular1)' dangerous fi~sion product of '.!:I."U because it SlIbstillltes for calcium in bones. What other direct fission prodllCls would accompany it in the neutron-induced fission of '.!_v·U? .vole: This reaction may release two, three. or four free IH:'lltrons.
[]JAssullle ordinar~ soil contains nawl'al uranium in amoullls of I pan per million by mass. (a) How much ura· nium is in the top 1.00 III of soil on a l·acre (43 560-ft'.!) plot 01 grounel, assuming the specific gravit} of soil is
6. A t} pic~11 Iluclear fission power plant produces aboul 1.00 CWoI' electrical power. Assume the planl hali an overall eff-iciency o[ 40.0% and each I1ssion produces 200 .\Ic" of Ihermal energy. Calculate thc mass of '.!1[,U consumed cach day, 7. Suppose the water exerts an average [rictional drag of 1.0 X JO" N on ,I nuclear-powered ship. How far can the
8.
iii According to one estimate, there are 4.4 X 10" metric Ions of world lll'anium reserves extractable at 130/kg or less, About 0.7% of naturally occurring ufClIlium is the fissionable isotope ~3:·U. (a) Calculatc the mass of 2j5 Ll in this reserve in grams. (b) Find the Humber of moles of 23:,U and convert to a number of iltoms. (c) Assuming 208 MeV is obtained from each reaction and all this energy is captured, calcuh\lc the total energy thal C'Ill be extracted from the reserve in joules. (d) Assuming world power consulllption 1.0 be const
9. An all-electric home uses approximately 2000 kWh of electric energy per mOlllh. Ilow much uraniulll-235 woulfl be required to providc this house wilil its energy necds for one year? Assume 100% conversion efllciency and 208 Me\' released per fission.
to. E
Seawater contains 3 mg of uf
SECTION 30.2 NUCLEAR FUSION [IIJ\-\'hen a star has exhausted its hydrogen fuel, it ma}' fuse other nuclear fuels. At temperatures abo\'c 1.0 X 10.'1 K. helium fusion can occur. Write lhe equations for the [ollowing processes. (a) Two alpha particles fuse to produce a nucleus A and a gamma ray. \",'hat is nucleus A? (b) I\UCICllS A absorbs an
m
Problems
12. Find the energy released in the fusion reactioll
IH + jl-l
--7
~He
+ "Y
13. If an all-electric home uses approximately 2 000 k\Vh 01" electric cnergy per month, how many fusion events dcscribed by the reaction ~H + {J-I --7 ~He + bn would be required to keep this home running for one year?
14.
EII
Another series of nuclear reactions that can prodllCC energy in the interior of stars is the cycle described below. This cycle is most efficicm when the central temperalllre in a star is above 1.6 X 107 K. Because the temperature at the center of the Sun is only ].5 X 10 7 K, the following cycle produces less than 10% of the Sun's energy. (a) A high-energy proton is absorbed by 1:!C. Another nucleus, A, is produced in the reaction, along with a gamma ray. Identify nucleus A. (b) Nucleus A decays Ihrough positron emission to form nucleus B. Identify nucleus B. (c) ~UclCIlS 13 absorbs a proton to produce nucleus Cancl a gamma ray. ldcntify nucleus C. (d) ~Ilclells C absorbs a proton to produce nucleus D and a gamma ray. Jdentify nucleus D. (c) Nucleus D decays through positron emission to produce nucleus E. Identify nucleus E. (0 Nucleus E absorbs a proton to produce nucleus F plus an alplla particle. \Vhat is nucleus F? Nole; ffnucleus Pis 1101 l?C_ that is, the nucleus you startcd with-you h:we made an error and should review the sequence of events.
[IiJAssul1'le a deuteron and
iI-I + ~;r-l
~
1He +
(lIn
16. _ A reauion lhat has been considered as a source of energ)' is the absorptioll ofa proton by a boron-II nuclcus to produce til rec alpha panicles:
:1-1 +
'~B
-7
SECTION 30.6 CONSERVATION LAWS 20. For the following two reactions, the first may occur hut the second cannot. Explain. (can occur) (cannot occur) ImEach of the following reactions is forbidden. Determine a conservation law that is violated for each reaction. (a) p + p --> ",- + c(b) ro + p --7 P + ro+ (c)p+p --7 p+ro l (d) p + P --> P + P + n (e) "y + p --> n + n"
22. Determine the type of neutrino or antinelll.rino involved ill each of the following- processes. (a) rot- --7 roO + e+ + ? (b) ? + P --7 J.L- + P + rot(c) IV l --7 P + J.L +? (d) T t --7 11-+ + ? + ? 23. Identify the unknown particle on the left side of the reaction ?+P
3(iJ-le)
This reaction is an attractive possibilit)' because boron is easily obtained from Earth's crust. A disadvantage is th,lt the protons and boron nuclei must have larg"e kinetic energies for the reaction to take placc. This requirement contrasts to the initi'ltion ofuraniulll fission by slow \leuI rollS. (a) How much energy is releascd in each reaction? (b) Why must the reaCl.ant particles have high kinetic cnergies?
SECTION 30.4 POSITRONS AND OTHER ANTIPARTICLES 17. A photon produces a proton-antiproton pair according to the reaction y --7 P + p. What is the minimum possible I"reqllellcy of the photon? What is its wavcleng-Ih?
[ffiJA photon with an energy of 2.09 CeV creates a protonantiproton pair in '.. . hich the proton has a kinetic energy of95.0 MeV. What is the kinetic energy of the antiproton? 19. A neutral pion at. rest deca)'s into two photons according10
Find the energy. momentum, and frequency of each pholon.
--7
n
+ J.L+
24. mil (a) Show that baryon number and charge are conserved ;n the following reactions of a pion with a proton:
+ 17.6 .\tleV
Neglecting relativistic corrections, dctcnnine the kinetic energy acquired by Ihe IlClltrOll.
959
(I)
rr'
+
p
--.
K-
+
(2)
rr" + p
--7
ro+
+ 1+
I+
(b) The first reaction is observed, blll. the second never occurs. Explain these observations. (c) Could the second reaction happen ifit created a third particle? Usa, which panicles in Table ::S0.2 might make it possible? Would the reaction require less energy or morc energy than the reaction of Equatioll (J)? Why?
25. _
Identify the consen'cd quantities in the following pn)cesses. (a) 2- --> (b) 1(0 (c) K
-7
+P
--7
2.0 + n
(d) In --> 1\.0 + "y (e)e-+e- --7 11-+
(f)
p+
n
-->
+ 11--
Afl + 1-
SECTION 30.8 QUARKS AND COLOR 26. The quark composilion of the proton is uud, whereas that of the ncULron is mid. Show that the charge, baryon number, and strangeness of these particles equal the sums of these !lumbers fOI" their quark constituents. 127.lrind the number of dectrons, and of each species of quark, in I L of water. 28. The quark compositions of the KO and AU particles arc (15 and mis, respectively. Show that the charge, baryon number. and strangeness of these panicles equal the sums of these numbers for their fJuark constituems.
29. Identify the particles corresponding (a)
SUll,
(b) tid, (c) sd, and (d) ssd.
to
the quark st.ates
960
Chapter 30
Nuclear Energy and Elementary Particles
30. \"'hat is the elcCLrical charge or the baryons with the quark compositions (,,) ulid and (b) ueld? What are these baryons called?
ADDITIONAL PROBLEMS
[]I):\.
~o panicle tr;weling through malleI' strikes a prown and a k+, and a gamma rny, as \\'ell as a third parlicle, emerges. Use the quark model of each 10 determine the identity of the third panicle.
32. :\ame at least one conservation law that prevents each or the following rC;:lcrions from occurring.
+ P --. s- + "" ~ ",- + v~ p ~ 7[+ + IT! + ",-
(a) ",-
(b) (c)
J.L~
133.Wind the: energy released in the Cusioll reaction I-I
34.
35.
+ ~H
~
~Hc
+ e- + "
Occ.:asjonaJl~',
high-energy muons collide with electrons and produce two neutrinos according to the rcaClion j.L+ + e- ----7 2/1. \'\'"hat kind of neutrinos are they?
am
Each or the following decays is forbidden. For each process, determine a cansen'atian law that is violalCd. (a) JL(b) n
--. .....-7
(c) AO .....-7 (eI) P ~ (c)
Sil
~
+~ p T e- + v.. P + 1Jo e+ + 11(1 n + ...H e-
36. Two protons approach each other with 70.4 MeV of kinetic energy and engage in a reactjon in which a proton and a positive pion cmerge at rest. \lVhat third parlicle, obviously uncharged and therefore difficult to detect, must have been created? 13i.IA 2.0-MeV neUlron is emitted in a fission reactor. If it loses one-half its kinetic energy in each collision with a modeL1LOr atolll. how lll
mm The fusion reaction iD + iD
.....-7 ~I-Ic + /In releases 3.27 MeV of ellcrg}'. If a fusion reaClor operates strictly on the basis or Ihis reaction, (
(b) At eight cents a kilowall-hour, how milch would t.he produced energy be wonh? (c) Heavy water (J)zO) costs about $300 per kilogr
-II. A ,,-meson
at
rest decays accordi ng to 7T
.....-7
/-L-+v,..
\,vhal is the energy carried off by the neutrino? Assume the neutrino has no lllass and moves offwilh the speed of light. Take 1II.71' C2 = 139.6 MeV ,md m!,.r~ = 105.7 MeV. Note: Use relativity; sec Equation 26.10.
42. The reaction ,,- + P --t ....'"0 + AO occurs with high probability, whereas t.he reaction 1J- + P -4 I0 + n never occurs. Analyze these reactions at t.he quark level. Show that the first reaction conserves the total number or each type of quark and the second reaction docs not. 143.IThe Sun radiates energy at t.he rate of 3.85 X IO~6 V",'. Suppose lhe net rcaclion 4p
+ 2e-
--t
(l'
+ 2v" + 6y
accounts for all the energy released. Calculate {he number of protons fused per second. ,"'v'ote: recall that an alpha panicle is a helillm-4 nucleus.
P E D A G O G IC A L C O LO R C H A R T M ec ha ni cs Di sp la ce m cl 1l an d , po sI tIo n I'e et or ' ,
- -...
Li ne ar (v ) an d an gu la r (w ) vc lo cit ) I'c cto rs \'c 1o ci n co m po ne nt Ic ct or s
- -..
,
Fo rc e ve ct or s (F ) Fo rc e co m po ne nt \'l 'el or ,
...
--
Ac ce lc ra tio n \'l 'ct or s (a )
Li ne ar (p ) an d an gu la r (L) m om en tu m l'eClors
..
--
- -.. Li ne ar or ro ta tio n, d m ot io n di rl' ui ol l, ~pring,
I'u lll '\ ,
Ac cc le ra tio n co m po ne nt I'l 'el or s
Electricity an d M ag ne tis m El ec tri c field. ~lagnetic
•
In dl lll l)l S (o tl, )
fic ld s
\ oh nH 'tl 'I'
\
. leg ali ye ch ar gn
\1I 11 11 l'lt T'
,\
Re sis to rs
\(
Po :it i\' t' ch ar ge ,
Ba lle rie s an d ot he r DC po we r up pl ie , wi tc he
- - If + -
"O Ul tl' S
(,r ou nd Sl m ho l (:UITellt
Light an d O pt ic s Li gh t ra y
Ob je ct s
Le n e an d pr ism
Im ag es
r.l ilT or -
•
ApPENDIX A Mathematics Review
A.l
MATHEMATICAL NOTATION
Nially mathematical symbols are lIsed throughout this book. These symbols arc described here, with examples illuSll-ating their use.
Equals Sign: = The symbol = denotes the mathematical equality of two quantities. In physics, it also makes a statement about the relationship of different physical concepts. An example is the equation I:.' = me'.!., This famous equation says that a given mass m, UuxlsllTed in kil.ogmms, is equivalent La a certain amount of energy, E, nUYIS'lITed in joules. The speed of lighl squared, [2, can be considered a constant of proportiol1al~ ity, neccessary because the units chosen for given quantities are rather arbilralJ'. based on hisLOrical circulllstances.
Proportionality:
IX
The symbol rx denotes a proporLionaliL)'. This symbol might be llsed when rocusing on relationships rather than an exact mathematical equality. For example, we could write E 'X. m, ,""hich says in \.. .Grels that "the energy E associated with an object is proportionalLO the mass mol' the objec!." Another example is found in kinetic energy, which is the energy associated with an object's motion, defined by KJ}.' = ~IIlV2, where 111 is again the mass and v is [he speed. Both 1n and v are variables in this expression. Hence, the kinetic energy KE is proportional to Ill, KE IX JII, ([lid at the same time KE is proportional to t.he speed squared, KE rx v 2 . Another term used here is "directly proponional." The density p of an object is related LO ilS mass and volume by p = UI/ V: Consequently, the density is said to be direcLly proportional to mass and inversely proponional to volume.
Inequalities The symbol < means "is less than," and> means "is greater than." For example, PFe > PAl means that the density of iron, PF",' is greater than the density of aluminum, PAl' If there is a line underneath the symbol, there is the possibility of equality: :::; means "less than or equal to," whereas?: means "greater than or equal lO." Any particle's speed v, for example, is less than or equal LO the speed of light, r: v:::; c,
Sometimes the size of a given quantity great.ly differs from the size of another quantity. Simple inequality doesn't convey vast differences. For such cases, the symbol « means "is much less than" and » means "is much greater than." The rnass of the Sun, Msull ' is much greater than the mass of the Earth, lylE: J\rlslIl1 » ME' The mass of an electron, 111 1" is much less than the mass of a proLan, mil: '/!I'I'« mIl'
Approximately Equal: = The symbol = indicates thatlwO quantities are approximately equal LO each other. The mass of a proton, 111/), is approxirnately the same as the mass of" a neutron, lIll/' This relationship can be wrillcn mil = 11/".
Equivalence: == The symbol == means
"is defined as," which is a different statement than a simple
=, It means that the quantity on the left-usually a single quantity-is another way
A.I
A.2
Appendix A
Mathematics Review
of expressing the quantity or quantities on the right. The classical momentum of an object, II, is defined to be the mass of the object m times its velocity v, hence!J == mv. Because this equivalence is by definition, there is no possibility of /) being equal to something else. Contrast this case \\rith that of the expression for the velocity vof an object under constant acceleration, which is v = at 1 vo' This equation ,.. .oldd never be written with an equivalence sign because v in this context is not a defined quantit-y; rather it is an equality that holds true only under the assumption of constant acceleration. The expression for the classical momentum, hmvever, is always true by definition, so it would be appropriate to write p == rnv Lhe first time the concept is introduced. After the inu'oduction of a term, an ordinary equals sign generally suffices.
Differences:
~
The Greek letter.6. (capital delta) is the symbol Llsed to indicate the difference in a measured physical quantity, usually at t\VO different times. The best example is a displacement along the x-axis, indicated by ::lx (read as "delta x"). Note that ax doesn't mean "the product of Li. and x." Suppose a person out lor a morning stroll starts measuring her distance away from home when she is 10 III from her door\vay. She then continues along a straight-line path and stops strolling 50 m from the door. Her change in position during the walk is dx = 50 m - 10 In = 40 ITI. In symbolic form, such displacements can be \vritten 6.x
=
xJ -
Xi
In this equation, "i is the final position and x; is the initial pOSItiOn. There are numerous other examples of di.fferences in physics, such as the difference (or change) in momentum, !J.jJ = Pf - jJi; the change in kinetic energy, !J.K= Kf - ~; and the change in temperawre, !J.T= Tf - T i .
Summation: ~ In physics there are often contexts in which it's necessary to add several quantities. A useful abbreviation for representing such a sum is the Greek letter 1- (capital sigma). Suppose we wish to add a set of five numbers represented by XI' -'=2' x3 ' x4 , and x 5 . III the abbreviated notation, \ve would write the slim as 5
Xl
+ X2 +
X3
+
X4
+
X5
=
L.
Xi
i=l
where the subscript i on x represents anyone of the numbers in the set. For example, if there are five masses in a system, m l , n~, 1n?>, In,,, and 1ns, the total mass of the system 1\1 = 1n 1 + Jn 2 + "TIt'!, + 1n4 + 'ms could be expressed as 5
M=
2: m i i=l
The x-coordi nate of the center of mass of lhe five masses, meanwhile, could be written
"
2:
;=1
XCM
\<,Iith similar expressions for the
Absolute Value:
)1-
= -'---'-:-,---
lvI
and z-coordinares orthe center of mass.
II
The magnitude ofa quanLil-y x, written Ixi. is simply the absolute value ofrhatquanlity. The sign of Ixl is always positive, regardless of the sign of x. For example, if x ~ -5, then Ixl = 5; if x = 8, then Ixl = 8. In physics this sign is useful whenever
A.2
the magnitude of a quantity is more important than any direction that might be implied by a sign.
A.2
SCIENTIFIC NOTATION
\!Ian)' quantities in science have very large or very small yalues. The speed of light is about 300 000 000 m/s, and the ink required to make the dot over an i in this textbook has a mass of about 0.000 000 00 I kg. ] t'S very cumbersome to read. '''Tite, and keep track of such numbers because the decimal places have to be counted and because a number with one significant digit may require a large number of zeros. Scientific notation is a ,...'ay of representing these nurnbers withollt having to ''''rite out so many zeros, which in general are ollly used 1.0 establish the magnitude oCthe number, not its accuracy. The key is to use pm.. . ers of 10. The nonnegative pO'wers of 10 are I0" ~ 1 10' = 10
10'
=
10 X 10
=
100
10" = 10 X 10 X 10 = 1 000 10' = 10 X 10 X 10 X 10 = 10000 10" = 10 X I0 X 10 X lOX 10 = 100 000 and so on. The number of decimal places following the llrst digit in the number and to the left of the decimal point corresponds to the pm.. .er to which 10 is raised, called the exponent of 10. The speed of light, 300 000 000 Ill/s, can then be expressed as 3 X lOR m/s. Notice t.here are eight decimal places La I he right of the leading digit, 3, and to the left of where the decimal point would be placed. In this method, some representative numbers smaller than 1 are 1 10
lO-'~-=O.l
10-' = 10-:<
~
I
10 X 10
~
0.01
I
10
X
10
X 10
~
0.00 I
1 0 000 1 10 X 10 X 10 X 10 = .
10-" = - - - - - ' - - - - -
1 10 X 10 X IOX10XI0
~
0.000 OJ
[n these cases, the number of decimal places to the right of the decimal point LIp to and including only the first nonzero digit. equals the value of the (negative) exponent. Numbers expressed as some power of 10 multipl ied by anOther number between I and 10 arc said to be in scientific notation. For example, Coulomb's constant, which is associated with electric forces, is given by 8 987 551 789 N'm 2 /C 2 and is written in scientific Ilotation as 8.987 551 789 X I O!) N-m 2 /C 2 . Nc,vtoll's cOllstant of gnlvit.atioll is given by 0.000 000 000 066 731 N'rn~/kg2, writtcn in scientific notation as 6.673 I X 10-" N '111 2 /kg'. When nurnbers expressed in scientific notation arc being Illultiplied, ule following general rule is very useful:
1011 X 10 11I = 10 11 + 11I
[A.I]
,v!u":l'e 11 and m can be any numbers (not necessarily integers). For example, 10:! X 10:1 = 10'. The rule also applies if one of the exponents is negative: 10:< X 10-8 = 10-'.
Scientific Notation
A.3
A.4
Appendix A
Mathematics Review
\t\Then dividing Ilumbers expressed in scientific notation, note that lOll
[A.2]
EXERCISES
'With help from the above rules, verify the answers to the following: I. 86 400 ~ 8.64 X [04 2. 9816762.5 = 9.8167625 X 106 3. 0.0000000398 = 3.98 X 10-' 4. (4 X 10') (9 X 10") = 3.6 X 10"
5. (3 X 10 7 )(6 X 10-") = 1.8 X 10-"
6. 7
75 X 10- 11
[) X 10
3
= 1.5 X 10- 7
(3 X 10 6 )(8 X 10- 2 )
. (2 X 10 17 )(6 X 10')
A.3
=9XlO- 18
-
ALGEBRA
A. Some Basic Rules \-Vhen algebraic operations are performed, the laws of arithmetic apply. Symbols such as x, )', and z arc usually used LO represent quantities that arc nOl specified, 'what are called the unknowns. First, consider the equation
8x= 32 If we wish [Q solve [or x, \
32 8
Hx 8
x~4
Next consider t.he cqualion
In lhis type of expression, we can add or subtract the same quantity from each side. lfwe subtract 2 frolll each side. we obtain
x
1n general, if x + II = b, then x = b Now consider the equation
~
6
ll.
x
-=9 5
If we multiply each side by 5, we are left with x on the left by itself and 45 on the right:
G)rSl ~
9 X 5
x = 45
[n all cases, whatever operation is performed on the left side of the equality must also be performed on the right side.
A.3
Algebm
A.5
The following rules for multiplying, dividing, adding, and subtracting fractions should be recalled, where a, b, and rare three numbers: Example
Rule
WW=:
Multiplying
(01 b)
Dividing
mm~
(el d)
ad :t or 0 e -::!:-=--bd b d
Adding
(2)(5) 10 5 (4)(3) 12 6 (2)(5) - (4)(3)
213 4/5
ad be
~
8 15
4 2 - -3 5
(3)(5)
2 15
Very oflen in physics \.. .e are called upon (Q manipulate symbolic expressions algebraically, a process most students find unfamiliar. Tt's vcry important, however, because substituting numbers into an equation too early can often obscure meaning. The following two examples illustrate how these kinds of algebraic manipulations are carried Ollt.
EXAMPLE A ball is dropped from the
Solution
lOp
ofa building 50.0
tall. Hmv long
In
c10es
it take the ball to fall
[0
a height 0[25.0 m?
First, '''Tite the general ballistics equation for this siwation: x = ~a!2
+
vnl
+
Xu
Here, a = -9.80 m/s2 is the acceleration of gravity thaL causes the ball to rail, va = 0 is the initial velocity, and 50.0 m is the initial position. SubstilUte only the initial velocity, va = 0, obL.:-'lining the following equation:
Xu
=
x = 4a!'2 + .Yo This equation must be solved for t. Subtract Xo from both sides:
Multiply both sieles by 21a:
x( ~)( a
xo) =
(~)~a.I' = a -
I'
It's customary to have the desired value on the left, so switch the equation around and take the square root of both sides:
Only the positive root makes sense. Values could now be substilUted to obtain a final answer.
EXAMPLE A block of" mass
111 slides over a frictionless surface in the positive ;\'-dircction. It encounters a patch of roughness having coefficient of kinetic friction /L/(. If the rough patch has length a. x, find the speed of the block after leaving the patch.
Solution
Using the work-energy theorem, we have
Add ~m.vo2 to both sides:
A.6
Appendix A
Mathematics Review
Multiply both sides by 2/m: 11
2
2fJ-J.g nx
2
= v0 -
Finally, take the square root or both sides. Because the block is sliding in the positive x-direction, the positive square root is selected.
v =
Vv02 -
2}J-kg 6.x
EXERCiSES
Tn Exercises 1-4, solve for );' ANSWERS
1- a a
J l.a=-1+ x 2. 3x - 5 ~ i3 3. ax - 5
~
bx
5
:'(=--
.\" = ()
+2 3
4.--~--
2x + 6
4.<
+
8
7 a - !J
;-.;=--
II
x=
7
5. Solve the f'ollmving equation for vI:
Answer:
VI
B. Powers \,Vhen powers ofa given quantity xare multiplied, the following rule applies: [A.3]
For example, x'lx 4 = X 2 +·1 = )..6. "Vhen dividing the powers ofa given quantity. the rule is :.\:/I
-
=
XU-I/I
x"
[AA]
For example, x 8 /x 2 = X:'l-2 = xll . A pO\vcr thal is a fraClion, such as t corresponds to a root as follows: x l/
JI
=
V"X
[A.5]
For example, 4 1/:1 = -V4 = 1.5874. (A scientific calculator is lIseful for such calculations.) Finally, any quantity Xli raised to the Ulth power is [A.6]
TABLE A.I Rules of Exponents
Table A.l summarizes the rules of exponents. EXERCISES
Veri ry the followi llg:
(x")'" = x""
1. 2. 3. 4. 5. 6.
3' X 3 3 ~ 243 xrlx- s = x- 3 xlOj x- 5 = x 15 5 ' / 3 ~ 1.709975 (Use your calculator.) 60 ' /.' ~ 2.783 158 (Use your calculator.) (.\:.')3
= :\,12
A.3
c.
Algebra
A.7
Factoring
The following arc some lIseful formulas for factoring an equation: ax
a'!.
+
+
ay
+
rLZ
= a(x
+ )' +
2ab
+
/)2
= (a
+
b)'1.
perfect square
a' - ,/ = (a
+
b)(a - b)
differences ofsqllares
cOlllmon factor
z)
D. Quadratic Equations The g-eneral form of a quadratic equation is
ax'.?+bx+c=O
[A.7]
when.: x is the unknmvn quantity and a, b, and r: arc numerical factors referred to as coefficients of the equation. This equation has two roots, given by x=
-b:!:Vil-4ac
[A.8]
2a
I r b2 - 4ac> 0, the roots arc real.
EXAMPLE The equation x 2
+
5x
+4
= 0 has the following roots corresponding to the
x=
where sign.
Xl
-5 :!: V5' - (4)(1)(4) 2( I)
V9
-5 :!: 2
(\',10
signs of the square-roollcrm:
-5:!: 3 2
-5 - 3 -5 + 3 -I -4 2 2 refers to the root corresponding to the posiLive sign and x 2 refers to the rool corresponding to the negative
EXAMPLE A ball is projected upwards at 16.0 m/s. Use the quadrat.ic formula hcight or8.00 m above the point of release. Solution
LO
determine the time necessary for it to reach a
From the discussion of ballistics in Chapter 2, we can write (I)
The acceleration is due to gravity, given by a = -9.80 I"n/s2 ; the init.ial velocity is 110 = 16.0 m/s; and thc initial position is the point of release, taken to be Xli = O. Substitute these values into Equation (1) and set x = 8.00 m, arriving at x ~ -4.901'
+ 16.001 = 8.00
where unils have been suppressed for mathematical clarity. Rearrange this expression into the standard form of Equation A.7:
-4.901'
+ ]6.001 - 8.00
= 0
The equation is quadratic in the: time, t. \·Ve have a = -4.9, b = 16, and c = -8.00. Substitutc these values into Equation A.8: -16.0:!: V16' - 4(-490)(-8.00) -16.0:!: v992 1 = ------;---;---'--,--'-'------"2(-4.90) -9.80
v'99.2 1.63:;: - - = 0.614 s, 2.65 s 9.80 Both solutions are valid in Ihis case, one corresponding to reaching the point of' interest all the way up and the other to reaching it on the wa)' back down.
Appendix A
A.S
Mathematics Review
EXERCISES
Solve the rollowing quadratic equations: ANSWERS
1. J.2 + 2x - 3 = 0 2. 2x' - 5x + 2 ~ 0
3. 2x' - 4x - 9 = 0
Xl
= 1
Xl =
~
.<,
I
=
Xt
=
-3
x~ = ~
+ v22/2
x2
=
1 - v'22/2
4. Repeat the ballistics example [or a height of 10.0 rn above the point of release. Answer: I, = 0.S42 s I, = 2.42 s
E. Linear Equations A linear equation has tht: general form
FIGURE A.I
y
0>0 b< 0
where m. and b are constants. This kind of' equation is called linear because the graph of)1 versus ;r is a straight line, as shmvll in Figure A.1. The constam b, called the )'-intercept, represents the value of )' at wh ieh the sLJ~aighl line intersects the )I-axis. The constant m is equal to the slope of the straight line. If any two points on the straight line are specified by the coordinates (x J , )11) and (x 2, Yz), as in Figure A.1, the slope of the straight line can be expressed as Slope =
x
FIGURE A.2
[A.9]
y=m.x+b
x
)'2 -
)' 1
x2
XI
-
il)
[A.10]
Note that m and b can have either positive or negative values. if In > 0, the straight line has a positive slope, as in Figure A.I. If "Ill < 0, the straight line has a negative slope. In Figure A.l, both II/. and b arc positive. Three other possible situations a re shown in Figu re A. 2.
EXAMPLE Suppose the electrical resistance of a metal wire is 5.00.n. at a temperature of 20.0°C and 6.-14 n at 80.0°C. Assuming Lhe resistance changes linearly, what is t.he resistance of the wire aL 60.0°C? Solution Find the equation of the line describing the resistance R and then substitute the new temperature into it. T"..' o points on the graph of resistance versus temperalllre, (20.0°C, 5.00 D) and (So.ooe, 6.14 0), allow computation of the slope: !J.R 6.140 - 5.000 ,> = = 1.90 X 10--fl/"C (1) '/1l = !J. T SO.O°C - 20.0"C Now lise the point-slope formulation ofa line....vith this slope and (20.0°C, .5.00 .0):
R - Ro =m(T -
(2) (3)
R - 5.00 0
~
7~)
(1.90 X 10-2 OJ"C)(T - 20.0°C)
Finally, substitute T = 60.00 into Equation (3) and solve for R, getting R = 5.76 n.
EXERCISES
1. Dra.. " graphs or the follO\ving straight lines: (a)y=5x+3 (b)y=-2x+4 (e)y~-3x-6 2. Find rhe slopes or the straight lines described in Exercise 1. Answers: (a) 5 (b) -2 (e)-3 3. Find the slopes of the straight lines that pass through the following sets of points: (a) (0, -4) and (4.2) (b) (0,0) and (2, -5) (e) (-5, 2) and (4, -2) Answers: (a) 3/2 (b) -5/2 (e) -4/9
A.3
Algebra
4. Suppose an experiment measures the following clisplaccmclllS (in meters) from equilibriulll of a vt:rtical spring due 1.0 allClching weights (in NCWLOIlS): (0.0250 fIl, 22.0 :"J), (0.0750,66.0 N). Find the spring constant, which is the slope oCthe line ill lhe graph ofvY'cight versus displacement Answer: 880 N/m
F. Solving Simultaneous Linear Equations Consider the equation 3x + 5y = 15, \vhich }l;-lS l\"O unknO\,.rns, x and ). Such an equation doesn't have a unique solution. For c:xanlple, note that (x = 0, )' = 3), (x = 5, Y = 0), and (x = 2, y = 9/5) arc all solutions to this equation. rra problem has t\\/O unknowns, a unique SOllllion is possible only ifw'c have L\'/O equations. Tn general, jf a problem has n unknowns, its solution requires 1/ equations. To solve t\vo simultaneous equations involying twO unknowns, x and y. we solve one of the equations for x in terms ofy and SubSI illite this expressiun into the other equation,
EXAMPLE Solve the follmving t\vo simultaneous equations: (1) Solution
!;x
+ y = -8
2x - 2)' = 4
(2)
From Equation (2), we find that x = J + 2. Substitution of this value inl.O Equation (1) giycs
"(y + 2) + Y = -8 6)'=-18 )' =
-3
x=y+2= -I Alternate Solution
Multiply each term in Equation (1) by the [actor 2 and add the result to Equation (2): lOx+2y=~16
2x-2)'=4 J2x=-12 x = -1
Y= x - 2
-3
Two linear equations containing two unknowns can also be solved by a graphical method, If the straight lines corresponding to the t\\'o equations arc plotted in a conventional coordinatc systcm, thc intcrsection of the I.WO lincs represents the solution. For example, consider the two equations
y 5
,,--),=2 x - 2)' = -I
These equations are ploned in Figure A.3. The intcrseCLion of the two lines has the coordinates x = 5, Y = 3, which rcprcsents the sulution to the equations. YOll should check this solution by the analyticaltechniquc discussed abm'c.
FIGURE A.3
A.9
A.10
Appendix A
Mathematics Review
EXAMPLE A block of Illass 1/1 = 2.00 kg travels in the positive x-direction at v, = 5.00 rn/s, while a second block, of Illass iv[ = 4.. 00 kg and leading the first block, travels in the positive x-dircClion at 2.00 m/s. The surface is flicLionless. VVhal. are the blocks' velocities after collision, if that collision is perrenly elastic? Solution As c<'In be seen in Chapter 6, a perfectly elastic collision involves equations for the momentum and energy. With algebra, the cnerg)' equation, \vhich is quadratic in v, can be recast as a linear equation. The two equations are !riven bv,
~
(l)mv; (2)
Substitute the known quantities
vi
= 5.00
+
MV,
~mv/
+ MV/
vi-Vi=-(vJ-lj)
m/s and Vi = 2.00 m/s into Equations (1) and (2): (3) (4)
18 = 2vj
+ 4 \j
3 = -v(+ V(
Multiply Equation (4) by 2 and add to Equation (3): 18 = 21'1+ 41j 6 = -2v(+ 2'1
24 SUbSLinIling the solution for
= 6\j
Ij = 4.00 m/s
-->
' 1 back int.o Equation (4) yields vf =
1.00 Ill/S.
EXERCISES
Solve the following pairs of simultaneous equations involving two unknowns: ANSWERS
J.x+)'=ll x-)'=2 2.98 - T = T - 49 = 3. 6x + 2)' = 8x - 'I" =
x
= 5, J' =
3
lOa 5a
T = 65.3, a = 3.27
6
x=
~,y=-3
28
G. Logarithms and Exponentials Suppose a quantity x is expressed as a power of some quantity a: x= ([.)'
[A.Il]
The number a is called the base number. The logarithm of x with respect to the base a is equal to the exponent to which the base must be raised so as to satisfy the expression x = a J: [A.12] y = loga x Conversely, the antilogarithm of)' is the number x:
x = antilog,,)'
[A.13]
The antilog expression is in fan identical to the exponemial expression in Equation A.11, which is preferable for practical purposes. In pracLice, the two bases most orten used are base 10, called the common logarithm base, and base e = 2.718 . _. , called the nal"UTallogarilhm base. When COlllmon logarithms are lIsed, [A.14]
A.3
Algebm
A.ll
\t\'hcn nalUral logarithms are used, )' ~ In x (or, = e')
[A.I5]
For example. log lO 52 = 1.716, so alltilogllJ 1.716 = IOl.illi = 52. Likewise, 111 .. 52 = 3.951, '\0 antilnl' 3.951 = e~·951 = 52. In general. note lhm yOll can cOllvert bCl\\'cen base 10 and base e with the eqllalil} In x = (2.302 585)log,,,.\
[A.16]
Fin::lll). some useful properties oflogarilhms are log (lib) = log
II
+
log b
In
p
= I
log(alb) = log a - log b log(lI') =
11
I"G) -In
log a
=
a
Logarithms in college physics are used most nOlabl) in the definition of decibel le\'el. Sound imcnsi(y \'aries across several orders of magnitude, making it awkward to compare different intensities. Decibellc\'c1 (onH:'rts these intensities to a morc manageable logarithmic scale.
EXAMPLE (LOGS) Suppose ajcl testing its engines produces a sound intensity of' J = 0.750 'IN at a given location in an airplane hangar, \,Vhat decibel level corresponds 1.0 this sound intensity? Solution
Decibelle\'el {3 is defined by (3 =
where III = I X
10-1~ \'\'/m~
10IOg(.!...) III
is the standard reference il1l.cnsit}', Subsliwte the given information: (3
= 10 log (
0.750 W/m') I" " = 1 19 dB 10 - W/m-
EXAMPLE (ANTILOGS) A collection of four identicalm3chilles create... a decibel lc"c1 of f3 = 87.0 dB in a machine shop, \\'hat sound intensity ,.. .ould be created by only one such machine? Solution \\Te usc the equation of decibel Ic\'cl to find the lOtal sound intensity of the four machines. and then ,,:e divide by 4. From Equation (I):
87.0 dB = 1010g(
,., I ,) 10 - \l'/m-
Di\'idc both sides by 10 and take the antilog of both sides, which means, equivalcl1lly.lO exponentiate: lOX., = IO~f1.:fJ 1 = 10 "·10" = 10
3.:\
iii
'J =
_1_
10- 12 = 5.01 X 10-" W/m'
There are four machines, so this result must be divided b) 4 to get the intensity of" aile machine: 1 = 1.25 X 10
1
W/m'
A.12
Appendix A
Mathematics Review
EXAMPLE EXPONENTIALS) The half-life of tritium is 12.33 rears. (Tritium is the heaviest isoLOpe of hydrogen, with a nucleus consisting of a prown and two neutrons.) If a sample contains 3.0 g of tritium in ilially. how much rell"lains after 20.0 years? Solution
The equaliull giving the 1I11111UCI uf Ilucil:i of a n.tdiuacti,-e substance as a [unction of time is
N=
Ala
(2"I)"
where Nis the number of nuclei remaining. ''\'u is the initial number of nuclei, and 11 is the IlUmbCI- of half-lives. i\'ole lhat this equation is an exponential expression ,.. . ith a base of~. The number of half-lives is gi\'en b}' n = 1/0."2 = 20.0 yr/ 12.33 yr = 1.62. The fractional alTIollnl of tritium that remains after 20.0 )'1" is therefore
(,)'62 = 0.325
-N = No
2
Hence, of the original 3.00 g of tritium, 0.325 X 3.00 g = 0.975 g remains.
A.4
GEOMETRY
Table A.2 gives the areas and volumes [or several geometric shapes used [hrollgh~ oul this texl. These areas and volumes are important in numerous physics applications. A good example is the concept of pressure P, which is the force per unit area. As an equation, it is wriucn P = F/ A. Areas must also be calculated in problems involving the volume rate of fluid flow through a pipe using the equation of continuity, the lensile ~lrcss exerted all a cable by a \vcight, the nne of thermal energy transfer through a barrier, and the densil)' of current through a wire. There arc numerous other applications. Volumes are imponanl in computing the buoyant force excncd by waler on a submerged object, in calculaling densit.ies, and in determining the bulk Slress of" Ouid or gas on an object, which affects its volume. Again, there are numerous other applications.
TABLE A.2 Useful Information for Geometry
Surface area Area =
{w
=41l"T 2
Volume = ~.II"T'
Sphere
Rectangle
Surface area = 2.11"r~ ... 21frt
c
Volume
Are-d = 1ft :1
=.11" rtf
Circumference = 2;rr
Circle
~ b Triangle
./r?
11
i
~
til
l
t Recl:lnglllar box
Surface area = 2(iht (wt /'w) Volume =(wit
A.S
A.5
Trigonometry
TRIGONOMETRY
Some of the most basic facts concerning trigonometry arc presented in Chapter 1, and we encourage yOll to sLUdy me material prcscmed there if you afe having trouble with this branch of mathematics_ The most imponam trigonometric concepts include the P}/thagorean theorem:
[A. I?]
This equation states thal the square distance along the hypotenuse of a right triangle equals the slim of the squares of the legs. It can also be lIsed LO find distances between points in Canesian coordinates and the length of a ,'ceLOr, where .6.x is replaced by the x-compone~ of the veelor and .:lJ is replaced by the )'-compOnenl of the ,'eelor. If the vec[Qr A has components Ax and A~, the magniLUde A of the vector satisfies [A.18]
which has a form completely analogous LO the form of the Pythagorean theorem. Also highly useful are the cosine and sine functions because they relate the length of a veCLOr LO its x- and J-components: [A.19] [A.20j
Ax = A cosO 1\ = AsinO
The direction 8 of a vector in a plane can be dCLcnnined by usc of the Langent fUllction:
e=
tan
A,
~
[A.21]
A.
A relative of the Pythagorean theorem is also frcquclllly uscful:
sin'
e + cos' e =
[A.22]
Details on thc above concepts can be found in the extensive discussions in Chapters 1 and 3. The following are some othcr trigonometric identiLies that can sometimes be useful:
e= cos e =
cos(90· -
sin
sin (90· -
e) e)
sin 28 = 2 sin () cos 8 cos~
cos 20 =
8 -
sill~
()
e cos
sin(O :+:
The following relationships apply to any triangle, as shown in Figure A.4: a
+ {3 +
y = 180·
a':!. = b'l + c'l - 2bccos Q' b'!. = a 2 c'!. = a 2 ft
1". It Coplen. ~Atolllic Wdghts of the Elements I99Y". a tcchttical rcporltu the l1Jlern;ltional Union of Pure and Applied Chemistry, ;111(\ puhlhhcd in PlIrY' all/I Applit'i/ Clu:1Il1JIf)', 73('\), 6fi7-6R3, 2001. Atomic l1l;I,~eS oft he ;50lopes :uc from G. Audi and A, II. \Yapsll-a, "The 1995 Update to lll(~ Atomic r,1ass E,,;II11ali"ll;' Nurll'(ll" "hf.~h"s. A.'l!):>, ....01. '1,409-180, D•.:ccrnln.:1' 25, 1U9:J. Pel'U'nl ablilldan.:e values are frolll K.j. R. Rosmall and I'. D. P. Tad"r, "Isotopic COIupositiollS of I It!' Elements \19\t.~;1 technical repolt to Ill<' Intel"ualional Lllion of Pure aml ..\pplied CIH..:mistr}'. ,lIld puhlished in /'W7' (md AfrPbrd Chemislr")'. 70(1), ~17-236, 1991'1.
C
ApPENDIX
Some Useful Tables
TABLE C.2
TABLE C.l Mathematical Symbols Used in the Text and Their Meaning
Standard Symbols for Units Symbol
Unit
Symbol
Unit
Symbol
..I. ..I.
ampere
leal
kilocalode
ang~lrom
ku
kilog-ram
atm
<-llll1ol'>phcrc
kill
kilomeler
Meaning is equal is
1101
10
equal to
is clcfi ned
> <
= ~
_x,
Bq
hequerel
J...mol
kilomole
is proportiollallO
Bw
B,-ilish lhcrlllalunil
I.
lite"!"
is greater than
C
COU]
Ib
pound
is les.. . lhan
'C
I)
lighl)'car
is milch greater thall
(al
degree Celsius caloriC:
III
meter
is much less than
em
cemimcler
min
minUlc
is approximately eqllal to
Ci
CUriC
mol
mole
is on the orcin of magnilUdc of
d
day
~
newton
chang-c in
deg
degrc(: (angk)
11m
nanometer
pa ...cal
ill'>
\"01'
unccnaint> in x
sum of all quantities x,
eY
dCClrOIl\"oh
~lbsoltl1e
OF
degree Fahrc...·nheit
Pa rad
F
rarael
n'\'
It
loOl
,
second
G
Gauss
T
tcsla
g
g-ram
1I
atomic mass unit
H
henry
\'
\'olt
h
hour
II'
\\'all
hp
hOI"SCpOWcr
lI'b
well<.'I'
I·!;'
henz
\I
\'C'lI
value of x (always a positi\e quantity)
TABLE C.3 The Greek Alphabet Alpha
.1
"
"
radian rCH>lutiOIl
in.
inch
fUll
micrometer"
J
joule
n
ohm
K
kclvill
TABLE C.4
~1I
v
Physical Data Often Used"
Bela
B
i>
Xi
y
Omicron
-0
Average Eanh-\Ioon distance
r
~
Gamma
0
Averag<.: Eartlt-Sun distance
1.496 X
Delta Ep.. . ilun Zeta Eta Tlwla Iota
..I
IS
Pi
II
1T
Equatoria 1 rad ius of Earth
p
Density of;lir
6.3R X 101) m 1.20 kg/m:\
"
Density ofwall'r (20
Kappa
K
L
.1 II
Z
,
II 0
TJ 0
E
E
Rho
P
Sigllla
~
Tall
T
7
(20~C
3.8'-1 X 1Ol'i m
(lnel 1 all11) C (;
and 1 alln)
101 r III
l.00 X 103
k~/m~
9.80 lIl/s~ .;.9R X 10" kg
Up~ilon
Y
u
Free-I'
Phi
(I>
t\lass of ~tOOll
7.36 X IO!:! kg
K
Chi
A
X
1.99 X 10:,n kg
A
P~i
~'fass 01
'I'
Omega
n
'"
Standard atmospheric pre:-,slI re
f'.
J
Sun
1.01:1
X
10" Pa
0> .• Tlles("
,II ,.
r1)('
\';llll("~
nf 1hl'
("OIlSl,llll,
as used i" till'
It·:" I.
A.19
A.20
Appendix C
Some Useful Tables
TABLE C.5 Some Fundamental Constants Quantit),
Symbol
Value"
Atom ic mass unit
1I
1.66053886 (28) X 10 27 kg 931.494043 (80) MeV/e' 6.0221415 (10) X 10" particles/mol
Avogadro's number ef;
Bohr magneton
1J.B = - -
Bohr radius
au
9.27400949 (80) X IO-"j/T
2m, h'l.
= - -..
~
m.,.-k,
5.291 i72 108 (18) X 10- 11 m 1.3806505 (24) X 10-" J/K
Bolll'mann's constant
Compton wavelength
m,c
2.426310238 (16) X 10- 12 m
Coulomb constant
8.987551 788 ... X 10' N· m'/e'
Deuteron mass
3.34358335 (57) X 10-27 kg 2.013 553 212 70 (35) 1I 9.1093826 (16) X 10-31 kg 5.4857990945 (24) X 10-< II
ElcClrOll mass
Electron voll
Elementary charge Gas constant Gravitational ConSlalll
Josephson frequency-voltage ratio
Magnetic nux quantulll
0.510998918 (4'1) MeW" 1.602 17653 (H) X IO-I!I J 1.602 17653 (14) X 10-]9 C
eV e R
8.314472 (15) J/mol· K 6.6742 (10) X 10 II N· m'/kg'
G 2,
4.83597879 (41) X 10" Hz/V
Ii
2e
2.06783372 (18) X 1O- 15 T·m' 1.67492728 (29) X 10- 27 kg 1.00866491560 (55) 1I
Neutron mass
939.565360 (81) :VIcV/c' Nuclear magneton
ell
11 ..
= 2m p
5.05078343 (43) X 10" J/T
PermeabiliL), of free space
4'7T X 10-"; T· Ill/A (exact)
Perm ittivity of free space
8.854 187817.
Planck's constant
6.626069 3 (II) X 10 -" J . S
h ~
h =-
2"
Prown mass
R)'dberg constant Speed oflighl in vacuum
1.054571 68 (18) X 10-" J. S 1.672 621 71 (29) X 10-" kg 1.007 276 466 88 (13) 1I 938.272029 (80) MeV/e' 1.0973731568525 (73) X 107 m -] 2.99792458 X lO~ m/s (exact)
XOIl!:Thes<-' conslams are Ihe \',llues recommended in 2002 b~' CODATA, based Il1l ,1 Icast,sqll"r<-'~ Mljuslmen, of dala frolll diffcrentmeasurcl11o:nts. For a more cOlllplele 1i\t, 'ICe P.]. Mohr and It :-J. Taylor. -CODATA Recommended Vahu" orthe FUlIdamcrllal Physical ConSlalll": 2002.~ Ilro. Mod. 1llt)l. 77:1. 2005. "The 1lI1l1lbcr~ in parentheses lor che \"'alues represenl the ullCcnainlics IIf,l.c I;ISI t\Hl digits.
ApPENDIX D SI Units
TAB LE~D~.l"--_ 51 Base Units Base Quantity
51 Bnse Unit Name Symbol
Lenglh
meter
Mass
kilogram
ka0
Time
second
s
m
Electric currem
ampere
A
TCmper
kelvin
K
AmounL of substance
mole
mol
Luminous imensity
candela
cd
TABLE D.2
Derived Sf Units
Quantity
Expression in Terms of Base Units
Expression in Terms of Other 51 Units
N,lI11C
Symbol
Plane angle
radian
rad
In/Ill
Frequenc),
henz
I-Iz
S-l
Force
newton
N
kg· tHis:!
.11 m
Pressure
pascal
Pa
N/m 2
Energy: work
joule
.J
kg/Ill' s~ kg'1ll 2js:!
Power
waLL
W
kg- mtjs:\
.lis
Electric charge Electric potential (cmf)
coulomb
C
A·s
volt
V
kg- m 2 /A
Capacitance
farad
F
A:!'sl/kR'm 2
C/V
Electric resistance
ohlll
n
kg'm:!/A 2 's 3
VIA
.\ilagnctic flux
weber
Wb
"g-m!/A-s2
V·s,T-rn 2
Mrlgnctic field intensity
lcsla
T
I ncluctance
henr}'
H
2
N·m
' 53
kg/A's kg·m 2 /A2· s2
W/A..J/C
\Vb/m 2 \Vb/A
A.21
ApPENDIX
E
MeAT Skill Builder Study Guide
VECTORS EXAMPLE 1 A hiker walks due Ilonh at 4.00 km/h from his campsite. A grizzly bear, starting 3.00 km due east of the campsite, ,\;alks at 7.00 krn/h in a direction 30° west of nonh. After one hour, hm..- far apart are they? Solution \Vrite the components of the position vcctor of the hiker after one hour, choosing the campsite as the origin:
H ., = 0
H, = 4.00 m
The bear travels at an angle or 30" + 90° = 120 \-vith respect to due cast, where due east corresponds to the positi"c x-direction. 'Vrilc the components of the position vector of the bear after onc hour, again relative the campsite:
B, = 3.00 km
Subtracting the bear's position vector from the hiker's position vector ·will result in a vector Ii pointing I"rol11 the bear to the hiker:
Rx = F-l, - B, = 0 - (-0.500 km) = 0.500 kill
0
Calculate the magnitude ofthe resultant vector, obtaining' the distance between the hiker and the bear:
+ (7.00 km) cos 1200
= -0.500 km
Bo' = (7.00 km) sin 120' = 6.06 km
R, =
I-I, - By = 4.00 km - 6.06 kill = -2.06 kill
R= (R} + R,')Ii2 = [(O.500km)' + (-2.06km)'J'" =
2.12km
EXAMPLE 2 TWO\'eCIOrs Aand Ii are not parallel to each other (nor "'antiparallel"-pointing in opposite directions).lfR = A + \Vh ich of Lhe following mUSL be Lrue of Lhe magn iLl.ldes A, /1, and Hi (a) R> A + B (b) R = A + /1 (e) R < A + B
H,
Conceptuol Solution Because the vectors Aand Ii arc not paraltsl, they can be arranged to form t\VO sides or a triangle. By the geometric la\... ofveclOr addition, the resultant vector R forms the third side of the same triangle. The shortest distance between t\-\'o points is a straight line, so the distance along R is shorter than the sum of the lengths of A and B. Hence, the correct answer is (c).
MULTIPLE·CHOICE PROBLEMS I. A car travels due caSl for a distance of ~ mi and lhcn due nonh for an additional 4 mi bcrore SlOpping. \Vhat is the shortest slraight-Iine dist.ance between U1t starting and ending poinls of this trip? (a) 3 mi (b) 'lll'li (c) 5 mi (d) 7 mi 2. In the previolls probkm, whaL is Ihe angle n of the shortest. paLh relative 1O due north? (a) IX = arc cos 3/5 (b) Q' = arc sin 5/3 (c) Q' = arc sin 4/3 (d) a = arc tan 3/4 3. A vector
A makes
all angle or 60 e with the x-axis of a Canesian coordinale s)'stem.
\-\o'hich of lhe following statements is true of the indicated magnitudes? (a) ,-\... (b) A, > A, (c) A, > A (el) A, > A
> AI'
4. Force is a vector quamit), measured in units of ncwtons, N. What must be the angle bel\.. .een lWO concurrentl;' aClillg forces of 5 :\1 and 3 "\1, respectively, if the resultant veOor has;[ magniwdc of8 N? (a) 0° (b) <15° (c) 90° (d) 180°
A.22
Motion
A.23
5. Two forces £let concurrently on an objcn. Both vectors have the same magnitude of 10 N and act al right. angks to each other. What is the closest estimate of the magnitude of their resultant? (a) 0 ~ (b) 141'\ (c) 20 N (d) 100:\1
Answers 1. (c). The [\\'0 legs of the trip are perpendicular; therefore, the shoncst distance is given b)' the h)'potenuse or the corresponding riglll triangle. The value is given by the Pythagorean theorem: R = (32 + ~J2)1 '2 = .5 mi. 2. (d). The angle given by tan a.
Q'
that gives the direction of" the hypotenuse relative 10 the J-axis is
= opposite side/
~
a
= arc
tan 3/4
3. (b). Choices (c)
4. (a). The only war the vector sum ofa 3-:'\1 and a 5-N force can equalS N is ifboth forces 'let in the sallie direction. The angle bel ween I hem must therefore be zero degrees.
5. (b). The two vectors form the legs ora right triangle. The rcsultalll vector is the hypotenuse. Choices (a) and (c) can be eliminated immediately because they require that the two forces be ailliparallcl or parallel respectively. Choice (d) can also be eliminated because it is too large to be the hypotenuse. The magnitude of the resultant can be calculated with the Pyt hagorcan theorem: /I ~ (10'
+ 10')'12
~
(200)'"
~
10(2)'"
~
10(1,41)
= 14 N
MOTION EXAMPLE 1 A motorist traveling 24.0 m/s slams on his brakes and comcs to rest in 6.00 s. "Vhat is the car's acceleration assuming it's constant, and how far does the Car travel during t.his time? Solution Apply the kinematics equation for velocity: Solvc Equation (1) for the acceleration a and substilute v = 0, l = 6.00 s, ~lnd vI) = 24·.0 m/s:
(1)
v= al+ Vo
v -
0
24,Om/s 0,005
o = --,-----'--a = --V
I
-4,00 mis'
The displacement can then be found with the t.imeindependent equation: Solve Equation (2) for
nx and substitute valucs:
Llx =
o2a
(24,001/5)' = 72.0 m 2( -4,0001/5')
EXAMPLE 2 A student. on top ofa building releases a rock from rest. \Vhcll the rock is halfway to the ground, he releases a golf ball from rest. As both objects fall, does the distance bel\'\'een them increase, decrease, or stay the same? Conceptual Solution At the instant the gall" ball is released the rock is traveling faster, hence will always travel faster before it reaches the ground because the velocitics orboth ol~jects arc changing at the same rate. Because the rock is a/ways falling faster, Lhe distance between it and the golf ball will increase.
A.24
Appendix E
MeAT Skill Builder Study Guide
EXAMPLE 3 A baseball player thrm'V's a ball at an angle 0 above the horizol1lal and at speed Un' Neglecting air drag, whal is the ball's speed when it returns to the same height at which it \-"as thrown? (a) The ball's speed \vill be the same. (b) The ball's speed will be larger. (c) The ball's speed witl be smaller. Conceptuol Solution The ball's speed will be the same, so the carreel answer is (a). Because there is no acceleration in the horizontal direction, the velocity in the x·dircC[ion is conSlanL. and only the y-direclion need be considered. By symmetry. the time taken La reach maximum height is the same time taken to fall back LO Ihe original height. lfthe ball's \"e!aciL)' in the y-direction changes from vo, to 0 on the way to maximum height, then it ,"viII change ["rom 0 to -v(h on returning to its original heigill. Hence the ball's velocity components will be the same as bef()re, except for a signJch~lt1ge in the y~col11poncnL, which doesn't affect the specd.
Quantitative Solution Lct the poim of release correspond to x = 0, ')' = o. Set the equalion of displacement in the y~direction equal to zero:
( 1)
I
It
+
'J
u)' = ~al-
vili
= 0
Sake Eqllation (I), obtaining the I imes when .1y = 0 (i.e., at tile start and end of the trajcctory): Substitute these t\VO Solulions into lhe equation for the velocity in the y~direction:
Case 1: t = 0 Case 2: f
-~l!O") (/. -- +
(
- 2v0 1
= -_. a
fl
The speed at any time is given by the following expression: Hecause lhe J'-cOlTlpOnCIll is squared, the negative sign in Case 2 makes no difference and the speeds at lhe two times in question are the same, again giving alls\'\'cr (a).
EXAMPLE 4 A merry-go-round has a radius of2.00 m and turns once every 5.00 s. \,Vhal magnitude net force is required to hold a 35.0-kg boy in place at the rim?
Solution First. calculate the speed oran ol~iect at the rim, which is the distance around the rim divided by the time for one rotation:
d
27TT
I
I
v=-=--=
~7T(2.00
Ill)
!J.OO s
1)'1.
A body traveling in a circle at uniform speed requires a centripetal acceleration of ac = v'2/r. The force producing this acceleration can be round by substitut.ing this expression into Newton's second law:
F
=
,.
mat = m- = ( 35.0 ku ) o
= 2.51 m/s
(2.5] Ill/S)' 2.00
III
= 1.10 X 10'N
MULTIPLE-CHOICE PROBLEMS 1. A bird rIics ,1.0 rn elite north in 2.0 s and then flies 2.0 III due west in 1.0 s. What is the bird'~ average spet:d? (a) 2.0 m/s (b) 4.0 m/s (c) 8.0 m/s (d) 2v5/3 m/s
Motion
2. Applying the brakes to a car traveling at 45 km/h provides an acceleration of 5.0 m/s~ in the opposite direction. HoI\' long will it take the car to stop? (a) 0.40 s (b) 2.5 s (e) 5.0 s (d) 9.05 3. A ball rolls down a long inclined plane with a uniform acceleration of magnitude 1.0 m/s 2 . If its spced at somc installt of time is 10 mis, what will be its speed 5.0 scconds later? (a) 5 m/s (b) 10 m/s (c) 15 m/s (d) 16 m/s 4. A ball rolls down an inclined plane with a uniform acceleration of 1.00 m/s~. If its initial velocity is 1.00 m/s down the incline, how far will it travel along the incline in 1O.0s? (a) 10.0111 (b) 12.0 In (e) 60.0 111 (eI) 1.00 X 10' m
5. An object with an initial velocity 01" 25.0 m/s accelerates uniformly for W.O s to a final velocity of 75.0 m/s. What is the magnitude of its acceleration? (a) 3.00 m/s2 (b) 5.00 m/52 (c) 25.0 rn/s 2 (d) 50.0 m/s 2 6. A rock is dropped from a height of 19.6 III above the ground. How long does it take the rock to hit the ground? (a) 2.0 s (b) 4.0 s (c) 4.9 s (el) 9.8 s
7. A spacecraft hovering above the surface ofa distant planet releases a probe
to explore the planet's surface. The probe falls freel)' a distance of 40.0 m during the first 4.00 s after its release. What is the magnitude of the acceleration due to gravity on this
8. A ball is dropped [rom the roof of a very tall building. What is its spt:ed aftcr falling for 5.00 s' (a) 1.96 m/s (b) 9.80 m/s (e) 49.0111/5 (el) 98.0 m/s 9. A quarterback lhrows a football wilh a velocit}' of 7.00 m/s at an angle of l.~.O° with the horizontal. HO\\i br m\ia)' should the receiver be to catch the football successfully? (a) 1.25 III (b) 2.50 m (c) 5.00 III (d) 6.25 m 10. A bomber is fl),ing parallel to the ground with a speed of 500 km/h at all altiwde of 1960 m when it drops a bomb. How long does it take the bomb to hit the ground?
(a) 0.5005 (b) 20.05 (e) 50.05 (d) 2.00 X 10' s II. A ball is thrown horizontally with a specd of 6.0 m/s. What is its speed after 3.0 seconds of flight? (a) 30 m/s (b) 1!l.81ll/s (e) 18 m/s (c1) 4.9 III Is 12. A 70~kg woman standing at the equator rotates with the Earth around its axis at a tangeillial speed of abolll 5 X 102 m/s. If the radius of the Earth is approximale1y 6 X lOr. m, which is the best estim,ttc of the centripetal acceleraLion experienced by the woman? (a) 4 X 10~2 m/s2 (b) -1 Ill/S2 (c) 10-4 m/s 2 (d) 24 m/s 2 ]3. A ball rolls with uniform speed around a flal, horizontal, circular track. If the speed of the ball is doubkd, its centripetal acceleration is (a) quadrupled (b) doubled (e) halved (e1) unchanged
Answers I. (a). The average speed is Lhe LOlal distance traveled divided by the elapsed time.
v = tlll1t
= 6.0
m/3.0 s
= 2.0 m/s
(vf - v,)la. The final velocity i~ 0 km/h, and the acceleration is -5.0 m/s 2 . (The negalivc sign appears because the acceleration is antiparallcl to the velocity.) Remember to convert from kilometers to meters and from hours to seconds.
2. (b).
1=
t1v
t = -
3. (c). Ilf =
4. (e). d ~ 5. (b). a =
a
=
() - (45km/h)(10"0l/1 km)(1 h/36005)
, , = 2.5
-5.0 m/s-
= 10 m/s + (1.0 m/s2) (5.0 s) = J5 m/s v;l + ;"1' = (1.00 m/s)(JO.O s) +;( 1.00 m/5')(10.0 sf ~ !1IJldl = (75.0 m/s - 25.0 m/s)/ 10.0 s = 5.00 m/s 2 VI
+
s
fit
60.0 m
6. (a). The rock is in free Fall with zero initial velocity and initial height of )'0 = 19.6 m. The displacement. equation becomes Do)' = )' - )'0 = ~~gI2. Soh'ing for time I with y = 0 gives t
= (2)'u/g)11:! = (2 X 19.6 m/9.8
1lI/s2)112
= (4.0 s:!)l/2 = 2.0 s
7. (b). This question is (\ free-fall problem. The object starts from rest, V o = O. Rearrange l1y = -4gt 2 and solve for the acceleration of gravity, g: g~
-2(y- y,,)/I'
~
-2(-40.0m - 0)/16.05'
~
5.00m/s'
A.25
A.26
Appendix E
MeAT Skill Builder Study Guide
8. (c). The
rrcc~rall
2 Ill/5 )
-(9.8
velocit.y starting with an initial velocity or zero is v = -gl
=
(5.00 s) = -49 m/s
9. (b). The initial velocity in the y-dircclion is TJl)y = V o sin () = (i.OO m/s) sin .15.00 = 1.81 m!s. Maximum height is reached when the velocity in the )'-direction is zero: v = o = -gl + v Uy --+ 0 = -(9.80 lll/s2)t + 1.81 mls - 4 l = 0.185 s. Douhling this J'csult gives the total time or flight. The range is fix = (llo cos O)t = (7.00 m/sHeas 15.00)(0.370 s) =
2,E>(J m. 10. (b). The horizontal and the vertical components of velocity are independent. The ver-
or gravity:
tical componcnt is due only to the acceleration
)' = ~ ~p
---+
1=
(2),/ g)I/2
, ~ [2(1 960 m)((9.80 m/s')]'/2 ~ ('1.00 X 102 s')'/2 = 20.0 s 11. (a). The hariz;ontal companelll of velocity is V x is vJ = gt = (9.80 m/s\!) (3.00 s) = 29.4 mis, so
v
~
(1'/ + 1'/)'/2
~
= 6.00 m/s and the vertical componcllt
[(ii.OO m/s)2 + (29.4 m/s)']'/2 = 3.00 m/s
12. (a). The wornan is moving with constallt speed equal to the rotation of lhe Earth on ils axis in a cil·de equal 10 the radius of the Eanh. The centripctal acceleration is fir
13. (a).
fl,
=
=
BCGlllSe
iJ2/N
=
(500 m/s)2/6 X 1O(i
III
= (25 X
I()'I m\!/s~)/6
X 10 1;
III
= 4 X 10-2 mjs2
1?/R. Thc accelnation and the squarc of the velocil)' arc directly proponional. the velocity is squared, doubling- the velocity quadruples the acceleration.
FORCE EXAMPL:::oE... l
-----------
A net force Faccelerates an object of mass mat 6 m/s2 . The same net force is applied the accelermion of lhe more massive body?
[0
a body "lith mass 3m. \Vhat is
Conceptual Solution According to the l\ewton's second law, 1.he acceleration imparted to an ol~jeet by a force is inversely proportionalLO the object's mass. The acceleration or an object three times as massive is therefore ~(6 m/s 2 ) = 2 m/s2. Quantitative Solution Write Newton's second law for each of the objects:
Divide Equation (2) by Equation (1):
Subslimle M
=
3m into Equation (3) and solve for aM:
(I)
mUm
(2)
i\1a,\J = F
(3)
= F
F F
i\1aM 1JUl III
(3m)a.\1 ma m
EXAMPLE 2 A block of" mass m is attached by a horizont.al string of negligible mass to the left side of a second block of mass 1\.1, with 171 < 1\1. A second such string is au.ached to the right side of mass M and the system is pulled in the positive x-direction by a constant force F. If the surface is frinion less, what can be said about the tension Tin lhe string conncuing lhc two blocks? (a) l' = F (b) T < F (c) T> F J
Conceptual Solution The answer is (b). The acceleration of the block m is the same as lhat of" the system of both, which has IOta} mass m + 111. The tension Tmllst accelerate only the mass m, whereas the force Fmus1. provide the same acceleration to a system ofgrealer mass, 111 + i\1.lt therelore [allows tha1.F> T
Force
Quantitative Solution In this problem the y-component or Newton's second law gives only the normal forces acting 011 the blocks in terms of the gravity force. Because there is no fricLion, these forces don't arfect the acceleration. \,Vrite the x-component. of 'Kewton's second law for the system:
(1 )
{m+M)a=F
Write the X'-component orNcwton's second law for the less massive block:
(2)
ma= T
Divide Equation ("I) by Equation (2), canceling the acceleration, which is the same for both:
EXAMPLE 3 A block \'\'ith mass JfI is started at the top oran incline with coefficient or kinetic friction ILk and allowed to slide to the bott.om. A second block \-"ith mass iVf = 2m is also allowed to slide down the incline. If J.Lk is the same for both blocks, how does the acceleration aM of the second block compare with the acceleration fim of the first block? (a) niH = 2a m (b) fl,\J = am (c) aM = ~am
Conceptual Solution The accelereltions are the same because I he force of grm'ity, the force of friction, and the ma side of:-\ewton's second law when applied in this physical context are all proportional 1.0 the mass. Hence, a different mass should not affect the acceleration of the body, and the answer is (b).
EXAMPLE 4 Planet A has twice the mass and L\vice the radius of Planet B. On \vhich planet would an astronaut have a greater weight? (a) Planet A (b) Planet B (c) The astronaut's weight would be unaffected. Conceptuol Solution ''''eight as measured on a given planet is the magnitude of the gravitational force at the surface of that planet. In 0Jewton's law of gravitation, the gravitational force is directly proportional to the mass of each of two bodies but inversely proportional to the square of the distance between them. l-lence, in considering only the masses, planet A exerts twice the gravitational force of planet B. Planet A also has twice the radius of" planet B, howe\"cr, and the inverse square of 2 is~. Overall, therefore, Planet A has a weaker gravilational acceleration at its surface than Planet B by a facLOr of one-half, and the weight of an astronaut will be greater on planet B [answer (b)]. Quantitative Solution \Vrite Newton's law of" gravitation for the weight of an astronaut of mass m on pl,lnet A: \r\'rite the same law for the weight or the astronaLlt on Planel. 13:
Divide Equalion (2) by Equalion (J):
(I)
(2)
II/,Nl nG
W/J=--,-
'"8
1\1J1TJI:!
lVI..j /",/"
A.28
Appendix E
MeAT Skill Builder Study Guide
MB (2'"n)' (2M,,),} Solve [or wlJ:
W II
=
2
2w, [answer (b)]
MULTIPLE·CHOICE PROBLEMS 1. Which of the following will result from the application of a nonzero net force on an object? (a) The velocity of I he object v·lill remain consl.anL (b) The velocity of the object ' . . . il1 remain conSlam, bUllhe direction in which U1C objecllllo....es will change. (c) The velocity of the object will change. (e1) None of the above.
2. Body A has a mass that is twice as great as that of body B. Ir a force acting on body A is half the value ora force acting on body 13, which statement is true? (a) The acceleralion of A will be twice that of 13. (b) The acceleration of A \.. . ill be half that of B. (c) The acceleration of A will be equal 10 thal of B. (eI) The acceleration of A will be one-fourth that of B.
3. \Nhich of the following is a statement of New·ton's second law of motion? (a) For every action there is an equal and opposite reaction. (b) Force and the acceleration it produces are dircctly proporl ion
may speed up. 5. What is the weight of a 2.00-kg body on or near the surface of the Earth? (a) 4.90 N (b) 15.0 Ib (c) 19.5 N (e1) 54.0 kg' mis' 6. Two objccts of equal mass arc separated by a distance of2 m. If the mass of one object is doubled, the force of grm'ity between the twO o~jects will be (a) halfas great (b) twice
as great (e) one-follrth as gTcat (d) four times as great 7. The distance belween a spaceship and the celller of the Earth increases from one Earth radius to three Earth radii. What happens to the force of gr
9. The measured weight of a person standing on a scale and riding in an elevator will be the greatest when (a) the elcvator rises at a constant velocity. (b) the elevator accelel-ates upward. (e) the elevator falls at a constant velocity. (d) none of the these, because weight is conSlant. 10. A block with mass 25.0 kg rests on a rough level surface. A horizontal force of 1.50 X 10 2 N is applied to it, and an acceleration of 4.00 m/s 2 is subsequently observed. \Vhat is the magnilUde of the force of friction acting on the block? (a) 25.0 N (b) l.00 X 10' N (c) 75.0 N (eI) 50.0 N
Answers 1. (e). Force produces accelna1.ion, which results in a change in velocity.
2. (d). Obtain the rauo of the accelerations from Newton's second law, F = 1TI:1
a"
1St
In,PH
F1J
-->
(2m'/J)a"
al'~J)
In[PII
FII
-->
ll,t
1
lllJ
4
1/lll:
3, (b). Choice (a) is a statement of the third law, and choice (c) is a statement of the first
law. 4. (b). Choices (c) and (d) can be eliminated immediately because changing velocity (either its magnitude or direction) requires the application ofa force. Choice (a) can
Equilibrium be eliminated because the with uniform velocity.
fir~1
law applies to both
5. (e). w= F= IIIg~ (2.00 kR)(9.80 lIl/s')
bodic~
A.29
at rest and those moving
= 19.6 N
6. (b). In the law ofunivcrsal gnwitation, the gravitational force is direcLly proportional to the product of the masses. Doubling one of the masses double~ their product and therefore doubles the force. 7. (a). Choices (b) and (d) can be eliminated because the forcc of gravitational attranion must decrease with increasing distance. Because j'~r;" = G1II]1n<:/r 2 , t.he force decreases by a factor of9 when r is tripled.
8. (c). Sce Example 4. Gravitational force is proportional to mass and inversel}' proporI-ion£1lto the radius squared, so nine times the mass and three times the radius yield.s a factor 01'9/32 = I. There is no discernible difference in weig-ht because the Iwo effects exactly cancel each other. 9. (b). The weight measured on the scale is essentially the magnitude of the normal force. Apply Newton's second la\'..· to the pa~~enger and solve for Ihe normal force: /1/0
= 11 -
mg
-)0
11
= UUl + mg
Inspecting the equation for the normal force. it'~ clear that a positive acceleration results in a larger normal force. hence a larger reading on the scale. 10, (d). Use Newton's second law:
ma = j~rp
(25.0 kg)(4.00
2 Ill/S )
j~ri,
= 1.50
J'~ric
X
102 1'\ -
j'I'ie
= 50.0 N
EQUILIBRIUM EXAMPLE 1 A uniform steel beam w'ith length L and mass ..vI = 60.0 kg rests on two pivots. The first pivot is located ~H the left end of the beam and exerts a normal force 11 1 at that point. A second pivot is two-thirds of the distance from the left end to the right end. "Vhat is the normal force 11~ exerted on the beam by the second pivot set? (a) 441 N (b) 264 N (c) 372 N (d) 188 N Strategy In equilibrium, the SUIll or the torques is zero. Let 'Til, be the torque exerted by the normal force Til associated with lhe lIrst pivot, T'I~ t.he torque exerted by the normal force 11'2 associated with the second pivot, and TAt the torque exerted by t.he force of gravity on the beam. Choose the pivot at the left end as the point around which to calculate torques. Usc 'T = rFsin 8 for each torque, where fJ is the angle between the position vector orr and the force F. The sign ora given torque is positive if the force tends to roLate the beam counterclockwise and negative if the force tends to rotate the beam clockwise. Solution In equilibrium, the slim of the torques is zero: The torque due La the normal force 11 1 is zero because it.s lever arm is zero. The weight of u1e uniform beam may be considered to act at the middle of" the beam, its cent.er of gravity. Equation (1) becomes:
(2)
Notice that for n~ the torque is positive, whereas the gravitational torque is negative (see the St.rategy). Cancel the [actors of L in Equation (2) and solve fo!' "fI~l:
"2
n2(2L/3) sin 90° - Mg(L/2) sin 90°
" = ,Mg = ,'( 60.0 kg )( 9.80 m/s-0) =
441 N [answer (a)]
= 0
= 441 N
A.30
Appendix E
MeAT Skill Builder Study Guide
EXAMPLE 2 Ice skater A has mass M and is at rest on a frozen pond, holding a backpack of mass 1/1. She tosses the backpack to ice skater B. who also has mass Nl and is initially at rest. If the surface of tile pond is considered frictionless, which skater has the larger subsequent speed? (Changes of momenltllll in the venical direction can be ignored, here.) (a) skater A
(b) skater B (c) Both skaters have the same subsequent speed. Conceptual Solution
By conservation of momentum, the final momenwlll of skater A is equal and opposite the momentum of the backpack. Skater B acquires this same momentum when caLching the backpack, but also increases her mass from m LO 1'1 + m. Because momel1lUI1l is proponional to both mass and velocity, skater B, with greater mass, has a smaller subsequent speed. Thus, the answer is (a). Quantitative Solution Apply conservation of momentum to skatcr A. The initial momentum equals the finalmomcntum: The initial total momentum is zero. Skater A's final momentum is M\~\, \',;hcreas the backpack has momentum 'mv: (1 )
II,1 =
Solve for the velocity lin of the backpack-skater B system:
(2)
\Ill =
Divide the magnitude or Equation (I) by Equation (2):
(3)
Soh'e for the \'elocitY V-I of skater A:
mv
M
Apply conscrvation of mumentum to sk<:ller B. The initial momenwm of the backpack, mv, equals the final momenLUIll of the system of skater and backpack:
By Equation (3), I1·:.IiI 11,,1 > 1, so
WII >
mv 'Tn
I 1~,1 I 1111 1
+ lW
1-':1 I171. mv I + i\!1
11/ + 1\1 =--->1
M
[11111 [answer (a)].
MULTIPLE-CHOICE PROBLEMS 1. A nonuniform bar 8.00 III long is placed on a pivot 2.00 In [rom the right end of the bar, which is also the lighter end. The center of gravit), of the bar is located 2.00 m from the heavier left encl. [f a weight of H/ = 5.00 X 10 2 N on the right end balances Ihe bar, what must be the weigh! war the bar? (a) 1.25 X 10:? N (b) 2.50 X 1O:? N (c) 5.00 X 10' N (d) 1.00 X 10";\1
2. A rod ornegligiblc mass is 10 m in length.lfa 3D-kg object is suspended from the left end of the rod and a 20-kg object rrom the right end, where must the pivot poim be placed to ensure equilibrium? (a) 4 m from the 30-kg object (b) ~I III from the 20-kg object (c) 8 m from the 3D-kg object (d) 5 m from the 20-kg object 3. A car with a mass or8.00 X IOC' kg is stalled all a road. A tnlck with a Jll
Equilibrium sion, the two vehicles remain in contact and cominue to move as one unit. \Vhat is the speed O[[!lC coupled vehicles right after impact? (a) 2.00 m/s (b) 4.00 m/s (e) 5.00 m/s (eI) 6.00 m/,
5. A pistol with a mass of 0.20 kg fires a a.50-gram bullet with an initial speed or 1.0 x 10'1 Jll/:-;. \OVhlll is the magnilllCle or the recoil velocity of the pistol? (a) 0.25 m/s (b) 0.50 Ill/' (c) 1.0 mls (eI) 10 Ill/S 6. A O.20-kg ball is bounced against a wall. It hits the wall with a speed of 20.0 m/s and rebounds in an elastic collision. What is the magniLUde of the changc in ll"lomentLIm of the ball? (a) 0 kg' mls (b) 4.0 kg' mls (c) 8.0 kg' m/s (d) 10 kg' m/s 7. A tennis ball is hit with a tennis racket and the change in the momcntulll of the ball is <1.0 kg' m/s. If the collision timc of thc ball and rackct is 0.01 s, what is the magnirude of the force exerted by the ball on the racket? (a) 2.5 X lO-:~ N (b) 4 X 10-2 N (c) 4.0 N (eI) 400 N
8. A 3D-kg can traveling" due north at 5 m/s collides wilh a 50-kg cart that had bcen traveling clue south. Both carts immcdiately corne to rest after the collision. What must have been the speed of thc sOluhbound can? (a) 3 m/s (b) 5 mls (c) 6 m/s (d) 10 m/s
Answers I. (b). At equilihrium,
The torqu<:: "7" ext:l'tcd by the pivot is zero because the torques are computt:d around that point, meaning a zero lever ann for thc force exened by the pivol. The center of gravity is 4.00 III from the pivot pOilU, and the gravitational force exerts Cl counI.crclockwise torquc T/I on the bar as if all the mass were concentrated at (hat poinl. The torque due to the weight 01' the bar is thereforc equal 10 +rF = +(4.00 m)wu' The weight Won the right side givcs a clockwise torque Til" 2(2.00 m) W Hence, from Equation (1),
0+ w,l-I.OO Ill) ~ (5.00 X
]Q:!
N)(2.00 Ill)
:=
0
-----)0
lOU
=
:l.50 X 10:! N
2. (.1). The pivot point must be closer to the heavier weight; mherwise, the torques due [he weights could not be equal in magnilude and oppositc in direction. Givcn this information, choices (b) and (c) can be eliminated immediately. Choice (d) can also be eliminated because it puts the pivot althe center of the rod, which would only give equilibrium if the two \veights were identical. The choice must be (a), which can be confirmed using the second condition of equilibrium. If x is the distance of the pivot from the 30-kg ma~s, then 10 - x is the distance from the pivot to lhe 20-kg mass. Apply the second condition of cquilibrium, kT, = 0: 10
(30 kg)g(x) - (20 kg)g(lO - x)
~
0
~
x
~
4
111
3. (c). Conservation or momentum ensures that the toud momentum afler the collision must equal the 101al momentum before the collision. I I' A is the car of mass 8.00 X 102 kg and B is the I 200-kg truck, then
Because the car is stalled, its initial velocity is zero and Inr\U.\",ili,,1 is zero. After the colli~ sian, the vehicles are joined and v/\ fill,,1 = VB fill,,1 = ur",,,I' The equation bccomes
Solve for vrmal: Vlill.t1
=
nlI\Tln,,,,u,,J/(m/\
=
J2 m/s.
+ Il/.IJ,) ==
(I 200 kg)(20.0 m/s)/(I 200 kg
+ 8.00
X 10:1 kg)
4. (a). Conservation of" momentum gives Pbefurt. = !J"r".!"' The total momentum before the collision is the sum of momenta for the two vehicles. After thc collision, /J is the JIlomenlllm associated with t.heir coupled masscs: (1.00 X 10" kg)(5.00 I11/S)
v,,,,,, =
(5.00 X 10'
+
(3.00 X 10" kg)(1.00 I11/S) ~ (4.00 X lO" kg) v,,,,,,,
+ 3.00 X JO')(kg' m /s)/(4.00 X 10" kg)
~
2.00 Ill/S
5. (a). The momentum is conserved because there are no external forces acting on the pistol-bullet system during the tiring. The initial momentum, however, is zero. The
A.31
A.32
Appendix E
MeAT Skill Builder Study Guide
final momentum is the slim of the momentum of the bullet and the recoil momentum of the pisml: I~erur" = jJa1wr Vpl",,'
0
-
=
IIIpi'lolVpiswl
= -(0.50 g)(I.O X 10' m/s)/0.20 X
+
IIIhl1lktVuullc1
10'g
= -0.25 m/s
6. (c). The change in momentulll is given by D./J = mVr- mv" where viis the initial velocity
and viis the rebound velocity. Because the collision is elastic, the ball rebounds in the opposite direction hut with the same speed with which it hit the waIL Therefore, v = f 2vi , and the change in momcnlUm is 8./) =m(-v,) - mVi
= -2mv,
= -2(0.20 kg)(20.0 m/s)
The magnitude ofa number is its absolUlc value:
= -8.0 kg' m/s
I-S.o kg' Ill/51
= 8.0 kg'l1l/s
7. (d). Force is related to momentum by the impulsc·momclllum equation, Ftlt Therefore,
= t!.jJ/t:.t = (4 kg' m/s)/O.Ol
F
=tl/J.
s ~ 400 kg' mis' = 400 N
8. (a). Momentum is conserved. Because both cans come La rest, the tolal momentum arter the collision lTlust be zero, which means thal the LOtal momentum prior to the collision was also zero. Ir the two cans are labeled A and B, then lnAvr\
=
llI Ut'B
-
VB
= (30 kg)(5 m/s)/50 kg = 3 m/s
WORK EXAMPLE 1 A car traveling;H speed v can stop in distance d due to kinetic [riClion. At twice the speed, what SLOpping distance is required? (a) d (b) 2d (c) 4d (d) 3d Conceptual Solution The answer is (c). Answer (a) can be eliminated immediately because at double the velocity, the car's energy mUSt be greater thllll before and hence the SlOpping distllllce must be greater. Because kinetic energy is proportional to v 2 , doubling the velocity quadruples the energy, "~York done by friction is linear in distance d, so four times the distance is required. This result can more readily be seen through a proportion. Quantitative Solution Apply the work-energy theorem:
(1)
HI = D.KE = KEJ - KE,
(4)
Fhd[ = ~mvJ2
(5)
F"d'.! = ~mv2'2
Assume the car is traveling in the positive x-direction. Substitute expressions for work done by a kinetic friction force with magnitude Fk, performed over a displacement 6.xand for kinetic energy, noting thal KEJ = 0: Multiply botll sides by -) and drop tile zero term: Denote the two possible diA-'erent: displacements by el J and c4, the two velocities by VI and lJ.2 ' Substitute into Equation (3), creating one equation for the first case and a similar equation for the second case:
Now divide Equation (5) by Equation (4), canceling terms Fh, 1ft, and ~: It follows, then, that answer (c) is correct:
J~d'2 l~dJ
1
2
'2'1nV2 1 ,
'2 1nV [-
Work
A.33
EXAMPLE 2 A truck of mass 1\11 and a car of mass 'J1I are traveling at the same speed. They both slam on their brakes, locking them and skidding along the road, which has friClion coefficient J.Lk' The car stops after going a distance d. Assuming J.Lk is the same for both vehicles, which of the following is true of the distance D thallhe [rllck u"avels before slopping? (a) IJ ~ d (b) D> d (c) D < rI
Conceptual Solution Kinetic energy is directly proportional to mass, and because the truck is morc massive than the car, its kinetic energy is greater and hence it might be thought thiu a greater braking distance \vould be required. The friction force, however, is also directly proportional LO mass, so the friction force is proportionately greater lor rhe truck than the car. All other things being equal, the truck and car stop in the same distance, D = d, '...· hich is ans,\'er (a). Quantitative Solution \'\Trite the expression for the force of rriction, where is I he normal force: Now Lise Equation (4) from Example 1:
SubsLitute the expressions for the kinetic friction force from Equation (1):
11
(I)
(2)
F, = iL' n = iL,mg
Fk"rlu:k
D
1~I,r.lrd
1 ., ~A1v~ 1 ? 2mv~
(iL,Mg)J)
~j\1v2
MIJ
111
IJ
(iLl,mg) rI
1 ? 2ml)~
md
m
rI
The answer is (a), as expected:
EXAMPLE 3 A car undergoes uniform acceleration from rest. \·\'hal. can be said about the instantaneous power delivered by the engine? (a) The instantaneous power incn.:ases with increasing speed, (b) The instantancous power remains the samc. (c) The instantaneous power decreases with increasing speed. Conceptual Solution The correct answer is (a). Because the acceleration is uniform, the force is constanl.. The installlaneous power is givcn by qp = Fv, so with increasing velocity the delivered pmver must also increase.
EXAMPLE 4 A circus stuntman is blasted straight up out or a cannon. Irail' resistance is negligible, \.. . hich of the follmving is true abollt his lOta) mechanical energy as he rises? (3) It increases. (b) ft remains the same, (c) It decreases. Conceptual Solution [11 the absence of non conservative forces, mechanical energy is conserved, so it. will remain the same, and ans\",cr (b) is correcL Application Irthe stunl.man has mass 65.0 kg and his initial vclocity is 8.00 m/s, find the maximum height reached. Solution Apply conscrvation of mechanical energy:
flKE+ fll'E= 0
1
.,
The final velocity is zero, and initial height may' be taken as zero:
0-
Solve for II/and substitute values:
hr = - ' =
'I1nVI-
+ mghJ - 0
= 0
v·':!.
(8.00111/s)'
2g
2(9.80111/s')
3.27 111
A.34
Appendix E
MeAT Sk;1I Bu;lder Study Gu;de
MULTIPLE·CHOICE PROBLEMS 1. If the speed at which
\~lhal
is the magnilUde of the force exerted by ai." on a plane if 500 kilowatts of pm....cl·
arc needed to maintain a constant speed of 100 meters per second? (a) 5 t\ (b) 50 N (c) 500 N (d) 5000 N to the speed ofa body ifits kinetic energy is c101lbled? (a) It increases by a ractoror2 1/ 2 , (b) II is doubled. (e) his halved. (d) It increases bya facLorof4.
3. Vlf'hal happens
4. A ball with a mass of 1.0 kg sits at the lOp ora 30° incline plane that is 20.0 meters long. If the potential energy of the ball relative the bottom orthe incline is 98j at the top of the incline, \"'ha1 is its potential energy once it rolls halfway down the incline? (a) O.J (b) 49J (c) 98J (d) 196J 5. Ilo\\' much work is done by a horizomal force of 20.0 N when applied to a mass of 0.500 kg over a distallce or 10.0 on' (a) 5.00 J (h) 1O.0.J (c) 49.0 J (d) 2.00 X 10'.1 6. A body located 10.0 meters above the sUI"E-tce of the Earth has a gravitational potential energy 01' 4.90 X 10::! J relative to the Earth's surface. Whal is the new gravitational potemial cnergy if the body drops to a height of 7.00 m above the Earth? (a) 70.0 J (b) 147J (c) 2811 (d) 343J 7. Cart A has a mass of 1.0 kg ,md a conslant velocity of 3.0 m/s. Can B has a mass of 1.5 kg and a conSlant velocity of 2.0 m/s. Which of the following statements is true? (a) Can A has the greater kinetic energy. (b) Can B has the greater kinetic energy. (c) Cart A has the greater acceleration. (d) Can B has the grc
'!P!v
~
500
X
IO'W!(lOO m!,) = 5 X 10 3 :\
3. (a). Let K'-~i be the initial kinetic energy, and Kl"ilhe Ilnal kinctic energy. ro..lake a proportion: I
KEJ
:?lnv/
KE;
'1 1nV i-
I
Solve for VI' getting VI =
(2KE,)
-->
.,
KE,
I
,
I
"
= 2 1J1Vj
"2-n1 V i-
-->
2
= v/ v,?
Y2Vj.
4. (b). Because the height is decreasing, thc gravitational potential energy is decreasing ..Hid must have a value less than its initial reading of98 J. Choice (d) can be e1iminatcd because its value is higher than the initi;'ll value ol"1.he potcntial energy, as can choicc (c) because its valuc is the salllc as [he initial value of the potential cnergy. Choice (a) can be eliminated because, by the problem statement, the gravilal ianal potelllial energy is zero at the boLLom orthe ramp. That leaves (b) as the only option. Quantital ivcly: PE 5. (d). W= I'D. s
~
= mgh =
(1.0 kg)(9.8 m/s'.!)[(10.0 m) . sin (30°)]
(20.01')(10.0 on)
~
= 49.1
2.00 X 10'.1
6. (d). Thc gravitational potential cnergy and the height of the body above the ground are direclly proportional: PE = 1JIgh. Reducing the heighl by a facLOr of 7/10 must reduce the gravitational potential energy by the samc factor: (4.90 x 102 J)(0.700) = 343.) 7. (a). Because the velocity is constant, the accelcr
KE" = H1.0 kg)(3.() m!s)' ~ 4.5J
KE" ~
H1.5 kg)(4.0 m!s)' ~ 3.01
Motter
A.35
8. (d). By definition, gra\'italional potential energy changes with the height or a bod}' abo\'t~ the ground. 9. (c). The average power is the work done on an ohject divided by the length of' the time intcl-Yal, fJ> = H'l~/. The work done here i" ag-ainst l-{ra\ity. In the period under cOI1:'lider~llion, the woman mllst incrca:'le hel" IlH:ch.ln;cal cncl-g)' hy an amutllll equal to her change in gravitational pOlcntial energy. so.P = IJIKh/1i1 = (50.0 kg)(9.80 m/s:!)
(2.00 on)/1O.0 ,
~
98.0 \\'.
MATTER EXAMPLE 1 Tank A is filled with \\-'ater, and, in £In identical tank B, the boltom half is filled with waler and the LOp half with oil. For which tank is the pressure at the boltom of the tank the g-reatest? (a) Tank A (h) Tank B (c) The pressure in bOlh tanks is the same. Conceptual Solution The oil in lank B floats on LOp of the water because it's less dense lhan water. The lotal weight of the fluid in tank B muSI therefore be less than an equal volume of\\'ater alone. Consequentl)" the fluid weight per unit area is greater at the bot1.Om of" tank A, and the ans\,'er is (a).
EXAMPLE 2 A slee! cable \,,,ith cross-sectional area A and leng-th I,stretches by length 6.L 1 \'\'hcn suspending a given weight. A second steel cable with half the cross-scctional arca and three times the length is used to support the same wcight, and .,'relchc., by an amounl t1L,. Whal is thc ralio t11_1/t1L,? (a) 2 (b) 4 (c) 6 (d) 8 Conceptual Solution Tcnsile Slress, a force per unit area, is proportional to Lensile strain, \vhich is the fractional change in length of an objecl or the change in Icngth diYided by the original length. The amolllltthe cable stretches is thus in\'ersely proportional 1.0 the cross-sectional area and direcll)' proportional to the length. Hence. the second cable, being three times as long, should stretch three times as far IInder a gi\ren load. "\'ith only half the cross-sectional area, it should stretch fanher by a factor of (~t], or 2. Taken logether, the second cable should stretch six times as far as the firSt !'ihoncr and thicker cable, so 1:i.L.1./I:i.L 1 = 6, which is answer (c). Quantitative Solution "\'rite the equation relating tensile stress and strain for the first cable: Rc\vriLC Equalion (1) for the second cable, noling that }', Young's modulus, is cOnStanL for a gi\'cn matcl'ial:
(I)
(2)
F,
A, F,
A,
F';!,
Divide Equation (2) by Equation (I), canceling r:
Cancd I~~ and F] in Equation (3) because they are cqU
(3)
~1.1
rL,
:::'L,
)'--"
L'l
:-'L,
A2
/...'2,
F,
lJ,L,
A,
L,
.....
il/... 2 /...) I~ AI --=--
F1 1\2
A, 3L, ----
~AI L1
dLI L 2
3
,
:!
6
A.36
Appendix E
MeAT Skill Builder Study Guide
EXAMPLE 3 An object floats \yith lhree-quarLers of its volume submerged in water. What is its density? (PII.
~
~
~
2:F = F /j + Fl;'l~l\" = 0
(1)
FB
-
IIIg = 0
By definition, the mass of the object is given by 1Jl = P(,bj V, where Pobj is the ol~jcet's density and V is its ,"olumc. The buoyant force is the weight of the displaced water: F/j = P"~lIcr l\uhg, \...· herc Y~lIlJ is the volume ofwatcr displaced. Substiwle these expressions illLO Equation (I):
Solve Equatiol1 (2) for Pohi' obtaining ansv,rer (c):
PO"I ~ =
EXAMPL.=,E,;:,4
V p,,,,,,., = \(,,"
()(
0.750
1.00 X tOo°1 kg/nr<')
7.50 X 10' kg/Ill"
_
At point I, water Ao","'s smoothly at speed VI through a horizol1lal pipe with radius r l • The pipe then narrows to half that radius at point2. iNhat can be said of the speed VI of the water at point I compared with its speed v2 at point2? (a) VI = v 2 (b) VI > iJ';l (c) VI < v 2 Conceptuol Solution The volume.: flow rate.: is proportional to the velocity and cross-sectional area of the pipe. A larger radius at point 1 means a larger cross-sectional area. Because water is essentially incompressible, the flow rate must be the same in both sections of pipe, so the larger cross-section at point 1 results in a smaller nuid velocity and the ansv.rer is (c). Quantitative Solution Apply the equation of continuity for an incompressible fluid: Substitute an expression for the area on each side of Equalion (1): Solve Equation (2) lor obtaining <:lI1s\,. . er (c):
VI'
and substitute
r2
=
!rl,
(2)
Motter
MULTIPLE-CHOICE PROBLEMS 1. A divcr is swimming 10.0 m helm\! the surface of the water in a reservoir. There is no current, the air 11,)s a pressure of" 1.00 atmosphere, aIle! the density of the water is 1.00 X 10:1 kilograms per cubic meter. \"Ihat is the pressure as measured by the di\'er? (a) 1.10 aim (b) 1.99 X lO' Pa (c) ] 1.0 atm (eI) 1.01 X 10" I'a
2. The aorta of a 70.0-kg man has a cross-scClional area of 3.00 cm 1 and carries blood with a speed of 30.0 cm/s. What is the average volume flow rate? (a) 10.0 CHI/S (b) 33.0 cm'/s (c) lO.O cm'/, (eI) 90.0 cm"/' 3. At 20.0 c C the density of '·...ater is 1.00 g/cm 3 . What is the density of a body that has a weight of 0.980 N in air but registers an apparent weight of only 0.245 N on a spri ng scale \...·l1el1 fully immersed in water? (a) 0.245 g/crn 1 (b) 0.7~5 g/cm 3 (c) 1.33 g/cm 3 (eI) 4.00 gAm' 4. Two insoluble bodies, A and B, appear to lose the same amount of weight v,.·hell submerged in alcohol. V\"hich statement. is mwa applicable? (a) Both bodies have the same mass in air. (b) Both bodies have the same volume. (c) Bo1l1 bodies have the same density. (d) Both bodies have the same weight in air. 5. The boltom or each foOl. or an 80-kg man has an Mea of aboul 400 cm~. What is the crfcCl of his wearing snowshoes with an area of abom DADO m 2 ? (a) The pressure exerted on the snow becomes 10 I imcs as great. (b) The pressure exerted on the snow becomes 1/10 as great. (c) The pressure exerted on the sno\\' remains the same. (d) The force exerted on the snow is IllO as gre;n. 6. In a hydraulic lift, the surface of the input pistoll is 10 cm 2 and that of the output piston is 3 000 cm 2 . "Vhat is the work done if a 100·t\ force applied to the input piston raises the outp'" pistoll by 2.0 m? (a) 20 J;J (b) 30 kJ (c) 40 J;J (eI) 60 kJ 7. Young's modulus for steel is 2.0 X 1011 N/m 1 . What is the stress on a steel rod that is 100 cm long- and 20 mm in diameter when it is stretched by a force of 6.3 X 103 N? (a) 2.01 X 10' N/m' (b) 12.6 X 10" N/m' (c) 3.1!i X 10' "i/m' (eI) 'kO X lO" N/m'
Answers 1. (b). The fluid is at reSl (no cUITems), so this problem is a hydrostatic pressure calculalion. In Sf units I atm = 1.0l X 10:> Pa. Choice (d) can be eliminated bec<\llse the pressure below the water must he greater than the pressure at the surface.
I'd"',,
= "'no« + pgh ~ (1.01 = 1.99 X 10' Pa
X 10' Pa)+ (1.00 X 10' kg/m')(9.80 m/s') (10.0 m)
2. (d). The answer follows from the dellnition: Volume flow rate = vA = (~O c01/s)(3 cm 2 ) == 90 cm:"/s 3. (c). This problem is an application of Archimedes' principle. The apparent loss of weight of a submerged body equals the weight of the fluid displaced. The weight wof the displaced water is w == 0.980 N - 0.245 N = O.i3fl N. Using the definit.ion of dCllsi ty, P".llcrg = w/ \~ so the volume occupied by til is weight or water is V = 0.735 NIPwalc.K The volume of Ihe body Inust equal the volume of the water displaced. The density 01" the body is p == Ul/V= (0.980 N/0.73?J N)PwaLer = 1.33 g!cm:'l. 4. (b). According to Archimedes' principle, the apparent weightlosl is equal to the weight of the displaced nuid: therefore, both mUSl have the same volume because the volume of the fluid equals the volume of the hody. 5. (b). The force exerted by the man is his weight and it is assumed to he constant, which eliminates choice (d). For a constan1 force, pressure and are,l are inversely proportional. The area of the snowshoes is ten times the area of the foot, so the pressure associated with lhe snowshoes is the inverse of 10, or l/lO the pressure exerted by the fool. 6. (d). By Pascal's principle. the pressure a[ the input and output pistons is the same, so /~n = /~Ul ;\0111
;t;n
~
(3 000 c~') lOI) N 10 cor
~
3 X 10" N
Work is a force times a displacement: w~ l'!1s = (3 X 10' N)(2m) ~ 6 X 10"J ~ 60 kJ
A.37
A.38
MeAT Skill Builder Study Guide
Appendix E
7. (a). Stress is force per IlnilJrCa, so neither Young's modulus nor the length of the rod are needed lO solve the pmhlcm. Stress = F/i\ = (6.30 X
I(fi
N)/1T(l X 1O-~ m? = 2.01 X 10i N/m 2
WAVES EXAMPLE 1 A transverse wa\'e travels at speed VI and has twice the frequency and one-quaner the wavelength ora second transverse wave. (They could be waves on two different strings, for example.) How does the speed VI of the first wave compare \\"ith the speed v'!. of the second wave? (a) VI = 1)'1 (b) VI = 2v'!. (c) VI = ~v'!. Solution The speed of a ",rave is proportional ro both the frequenc)' and ' . . avclcllgth:
(1)
v = fA
lIsing Equation (1) and substitute;; ~ 2f, and A, = A,/4:
v, 1)4
JiA, j,A,
Solve for
v,
:! V2
Make a ratio of v'!. to
VI'
lJ]
obtaining ans,,,,,cr (c):
(2j,)(A,/4)
2
;;A,
4
I
-
2
I
EXAMPLE 2 1J1. oscillates at the end of a horizontal spring on
A block of mass
7:
Conceptual Solution An increased mass '.. ·oukl increase the inenia or the system without augmenting the mechanical energy, so a mass of 2m should move more slowly than a mass of m. That in turn "..auld lengthen the period, eliminating choices (b) and (d). The period is proponional to the square rOOt of the mass, so doubling the Illass increases the period by a factor of the square rom of 2, which is answer (a).
Quantitative Solution Apply the equation for the period ora mass-spring system:
(I)
T=
Using Equation (I), make a ratio of the new period 7~e\, to the old period, 7~, canceling common terms and substituting mne(\' = 2m():
(2)
Solving Equation (2) for the nc,v period 7;,0.;\'1 yields ans,,,er (a):
7~,cw =
2"'-,[f
v'2 To
[answer (a)]
EXAMPLE 3 A simple pendulum swings back and forth 36 times in 68 s. \,Vhat is its length if the local acceleration of gravity is 9.80 l11/s" (a) 0.0677111 (b) 0.356111 (c) 1.25 III (d) 0.887 III
Solution First. find the period Tor Lhe Illotion:
T =
68 s 36 = 1.89 s
Waves
T= 27T
Writc thc equation for the period of a pendulum and solve it for the length L, obtaining answer (d):
~ ~
_ gr' _ (9.80 m/s')(1.89 s)'
g
L---,-
47T-
0.887 m [answer (d)]
MULTIPLE·CHOICE PROBLEMS I. A simple pcndululll has a period or 4.6~ s at a place on lhe Earth where the accelera~ Lion ofgnlvily is 9.82 Ill/s~. At ,I diffcrenllocation, the period increascs LO 4.64 s. \-Vhat is the value of gatlhis second point? (a) 9.78 m/s~ (b) 9.82 m/s 2 (c) 9.86 Ill/S2 (d) Canllot be determincd without knowing Ihe length 01" the pendulum. 2. What is the wavelength OLl transverse wave having a speed of 15 or5.0 I-Iz? (a) 3.0111 (b) 10 111 (c) 20 111 (d) 45 m
Ill/S
ami
~l
frequency
3. What is the optimum difference in phase for maximum destructive intcrferencc betwcen LWO waves of the same frequcncy? (a) 3600 (b) 270" (c) 180" (d) ~:IO"
4. SUlnding wavcs can bc formed if coincide/II. w,wes have (a) the same direclion of propagation. (b) the same frequency. (c) different amplitudes. (d) diflerent wavelengths. ~ s. For lhc pendulum to havc a period of 4 s, we must (a) halve the length. (Il) quarter the length. (c) double the length. Cd) quadruple the length.
5. A simple pendulum with a lenglh I. has a period of
6. Ira simple pendululll 12 In long has a frequency of 0.25 1-11., what ,,,'ill be lhe period of a second pendulum
7. A pendulum clock runs too slowly (i.e., is losing lime). Which of tht: following
ments could rectify the problem? (a) The weigh 1 of the bob should be decreased so it can move faster. (b) The Icnglh of" the wire holding tile bob should be shortened. (c) The amplitude of Ihe swing should be reduced so the path covered is shorter. (d) NOlle or the above.
8. A 20.0-kg object placed on a frictionless floor is all
g, ~ pj'g,/T,')
~ (4.63 s)'(9.82 111/>')/('1.64 s)'
= 9.78 I11/S'
2. (a). Wavelength is veloci!y divided by frequency. The formula does nOl depend upon the type 01" wave involved.
,.\ = v/f=
(15 m/s)/5.0 s
1=
3.0 m
3. (e). Two waves are completel)' out of phase when t.heir anti nodes coincide so lhat each crest on one wave coincides with a trough on the other. This sitnation occurs when the W
(d). In a pcndlllulll the period and Ihe square root of the length arc directly proportional:
T= 27T(L/g)l/'t. so that T/O/ 2
= consl.anl
To double lhe period, yOlI must douhle the square root or the length. To double the squarc 1'001 of the lengLh, YOll must qlladruplc thc length: (4L)Ii2
= 4- 1/'.1 1}/'2 = 2/,1/2
A.39
47T
2
A.40
Appendix E
MeAT Skill Builder Study Guide
6. (a). The frequency is the reciprocal of the period, f = 1!7~ so the first pendulul11 h
7. (b). The period of a pendulum is directly rel
8. (c). The force conslanL is k = F/l1x
=
5.0 Njl.O m
5.0
=
T= 2r.(m.jk)'/'!. = 271(20.0 kg/S.Oa N·
[\1.
Ill-I, The period is
m- I )1/2 =
12.6 s
SOUND EXAMPLE 1 When a given sound intensity doubles, by how lllllch does the sound intensity level (or decibel level) increase? Solution
Write the expression for a di fference in decibel level,
!J.[3 = [3, - [3, = 10 log
(J
!J.[3: Factor and app1r the logarithm rule log a - log b = log(a/b) = log(fI.· b-'):
!J.[3 = 10 [lOg = 10 log
SubstilUte 1'2 = 211 and cvaluate the expression:
-10 log (,,) (--=-1,,) I In
!J.[3 = 10 log
G~) ~ log
GJ]
('2 10)
10 log - . In II
C:) 21,) (J:
10 log(2) ~ 3.01 dB
EXAMPLE 2 A sound wave passes from air into water. \,Vhat call be said of its wavelength in water compared '\"ith air? (a) The wavelengths are the same. (b) The wavelength in air is greater than in \,,·ater. (c) The wavelength in air is less than in water. Conceptual Solution
The frequency of the sound doesn't change in going from one type of media to t.he next because it's caused b)' periodic variations of pressure in ti me. The ''\'avelength must change, however, because the speed of sound changes and during a single period thc sound wave will travel a different distance, which corresponds to a single wavelength. The speed of sound in water is greater than in air, so in a single period the sound wave will travel a greater distance. Hence, the wavelength of a given sound wave is greater in '.. .' ater than in rlir, \v'hich is (c). This result call be made quantitative by using v = fA, where v is the wave speed, fthc frequency, and A the wavelength.
EXAMPLE 3 A sound wave emiued from a sonar device reflects offa submarine thal is traveling away [rom the SOllar source. How does the frequency of I he reflected wave, fu, compare with the frequency of the source,!",? «1) In = Is (b) fa < f) (c)
J" > J,
Solution The reflected wave will have a Im·.. . e r frequency [answer (b)]. To see why, consider pressure maximums impinging sequentially on the submarine. The first pressure maximum hits t.he subm(lrine and refleus, blll the second rcaches the submarine when it. has moved farther away. The dislance between consecutive maximums in the reflected waves is thereby increased; hence the wavelength also increases. Because u = fA and the speed isn't affect.ed, tlle rrequenc), must decreasc.
Sound
MULTIPLE·CHOICE PROBLEMS I. The foghorn of a ship echoes off an iceberg in the distance. If the echo is heard
5.00 seconds after the horn is sounded and the air tcmperature is -l5.00C, how far II1/" i,('IJPrg? (;'1) 224 111 (h) RO.') 111 (r) R27 In (d) 930 m
::l\\Jay i"
2. ''''hal is the sound level ora wave with an i1llensily of 10 -:1 W/1ll2 ? (a) 30 dB (b) 60 clB (e) 90 dB (d) 120 dB 3. Al DoC, approximately how 10llg does it take sound La I ravel 5.00 kill through air?
(a) 15 s (b) 3U s (el 4;, s (d) 60
S
4. [fthe speed ora transverse wa\-e Of'l violin string is 12.0 m/s and lhe frequency played is 4.00 Hz, what is the wavelength of" the sound in air? (Use 343 Il1ls for the speed of sound.) (aj 48.0 111 (b) 12.0111 (e) 3.00111 (d) 85.8 111 5. 1[" two ideillical sound waves interact in phase, the resulting wave will have a (a) shorter
6. What is the speed of" a 10ngiUldinai SOlllld \\'ave in a steel roel if Young's modulus for steel is 2.0 X 1011 ;";/m 2 and lhe densily of steel is 8.0 X 10:1 kg/m'l? (a) 4.0 X IO-R m/s (h) 5.0 X J03 m / s (c) 25 X 1O(i m / 5 (d) 2.5 X 10'lm/s 7. If"two frequencies emitted from (\\,0 sources arc 48 Hz and 54 Hz. how many heats per second arc heard? (a) 3 (b) 6 (e) 9 (d) I ~ 8, The frequency registered by a detecLOr is higher thall the frequcnc)' emilled by rhe source. 'rVhich of the statcmc::l1ls below 1!Itu/ be true? (a) The source must be moving ,Hvay from the deleclor. (b) The source must he moving toward the dewcror. (c) The distance bet".;een the source and lhe detector must be decreasing. (d) The delcctor ml/st he moving "way from the source.
Answers
1. (b). The normal speed of sound at OOC in air is various remperaLUres is gi\'en by v =
=
(331
m/s
IT ( ) \/2731( = 331
3~1
m/s. The speed of so I 1I1d in air at
)
IlI/s \
!273K- 15.0K 273 K
322 m/s
Calcu late the disl "nce traveled in hal f of 5 seconds: tI
2. (e). Substitute:
13
=
vi
= 10 log 1110 =
=
(~22
m/s)(2 ..::'0 s) = 80S m
10 log (I X LQ--I W/m:!110
12
= 90 dB.
W/m:!)
3. (a). At Dce, the speed of sound in air i), 331 mis, so / = (5.00 X 10 m)/r~31 m/s) 15 s. 3
4. (d). The frequency is the same for the string as lor the sound wave the srring produces in lhe "uITounding air, blll.thc wavelength differs. The speed ofsoulld is equal to the product ofl.he frequency and wavelength. so A = vl!= (343m/s)/4.00 S-l = 85.8m. 5. (b). Two waves arc in phase if their crests and troughs coincide. The amplitude of lhe resulting wave is the algebraic sum of the amplitudes of lhe two W(lves being superposed at that point, so the amplitude is doubled.
6, (b). The sollllion requires substituling into an expression for the vdodt)' of sound through a rod of solid material having Young's modulus Y:
v
= (Ylp)I/2=
(2.0 X 10'1 kg' II1/s:!)/(8.0 X
= 25 X 106m2/s~)I!~ = 5.0 X 10 3
7. (b) . .ft".." ~
/. - /, = 54
- 48
1O:~
kg/m:l)l ."
m/s
= 6 bealS
8. (c), Movement of a sound source relative to all observer gives rise to the Doppler effect. Because lhe frequency is shirted lo ;-1 higher value. the source and rhe detector must be gelling closer logether, effectively shortening the wavelengLh from the observer's point of view. This observation eliminates choices (a) and (d). Choice (b) can be eliminated because it isn't necessarily true: the same kind of efrect occurs when the source is held steady and the detector moves towards it.
A.41
A.42
Appendix E
MeAT Skill Builder Study Guide
LIGHT EXAMPLE 1 \l\11ich substance will have smaller critical angle for total internal reOeclion, glass with an index of refraction of 1.50 or diamond with an index ofrefraClion of2.42? (a) glass (b) diamond (c) The critical angles are the same for both. Conceptuol Solution A larger index of refraction indicates a slower speed of light inside the material, which in turn means a larger refraction angle, or a larger bending towards the normal on entering the material from air and a larger bending 3W3)' from the normal 011 passing from the material into air. (Recall thal a normal line is perpendicular to the surface of the material.) For LOtal internal reflection to occur, a refraction angle of 90° must be possible, which occurs whenever the refracting medium has a lower index of refraction than lhe incident medium. Diamond bends light more than glass, so the incident ray can be closer to the normal and still be bent enough so the angle of refraction resulLs in LOtal internal reflection. Closer to the normal means a smaller critical angle, so the answer is (b). Quantitative Solution "'Trite Snell's law for the diamond-air interface: Compute the critical angle for diamond, lIsing ll A = ] .00 for air and (JA = 90'\ together with 170 = 2.42:
2.42 sin 0 0 = ] .00
Repeallhc calcuhnion for glass:
1.50 sin 0" = 1.00
The calculation cxplicitly shows that diamond has
_ 01) -
->
Oe
. _I (1.00) _ sIn ?? - 24.4 • _.4-
. -I (1.00) = Sill - - = 41.8
0
1.50
,I smaller critical angle than glass [answer (b)l
EXAMPLE 2 A patient's ncar point is 85.0 cm. ""hat focal length prescription lens will allow the patient to see objects clearly that arc aL a distance of 25.0 em from the eye? Neglect the eye-lens distance. Solution Use the thin lens equation:
An object at distance p = 25.0 Col must form an image at the patient's near point of 85.0 cm. The image must be \"irtual. so q = -85.0 cm:
I
1
]
f
1 =-=-+ ---:---:---:- = 2.82 25.0 em -85.0 cm
-=-+f p q X
/0-2
cm- l
f= 35.4 em
EXAMPLE 3 Two photons tra\"eling in vacuum have different wavelengths. \,Vhich of the following statements is true? (a) The photon with a smaller wavelength has greater energy. (b) The photon with greater wavelength has greater energy. (c) The energy of all photons is the same. (d) The photon wil h the greater w·avelength travels at a lesser speed. Solution Answer (d) can be eliminated immediately because lhe speed of light in vacuulll is the san1e for all wavelengths of light. The energ}1 of a photon, or particle of light, is given by}!.' = lifand consequently is proponionalto the frequency f, which in turn is inversely proportional to the wavelength because f = ciA, where c is the speed of light. A smaller wavelength photon has a greater frequency and therefore a greater energy, so answer (a) is truc.
Light
MULTIPLE·CHOICE PROBLEMS I. Glass has an index of refraction of 1.50. What is the frequency of light lhm ha~ 2 nm in glass? (a) 1.00 liz (b) 2.25 Hz (c) 'l.00 X JOlll1z ;,l wavelength of 5.00 X 10
(d) 9.00 X 10"; Hz 2. Watcr has an index ofrefraClion of 1.33. Ifa plane mirror is submerged in w,ller, what can be said of the angle or reOcclion 0 ifligllt strikes the mirror with an angle of inc idencc of" 30°? (a) fJ < 30 (b) () = 30C (e) SOC < () (eI) No light. is rdlected because 30" is C
Ihe critical angle for water. 3. The index of reFraction for W;;J(cr is 1.33 and 111at for glass is 1.50. A light ray sLri kcs the water-glass boundary with an incident angle of 30.00 on the walcr siele. Which of the following is the refraction angle in the glass? (3) 26.3° (b) 34.7<> (e) 30.0" (el) 60.0°
4. Light is incident on a prism at an angle of gOD rc1mi\'e \0 its surface. The index of refraction of the prism material is 1.50. '",'hich of Ihe following statements is most accurate about the angle of refraction fJ? (a) 00 < 0 < 4S
e
5. White light incident on an air-glass interface is split into
6. A real object is placed 10.0 em from a converg'ing lens lhat has
8. \""hat is the magnification ofa lens that forms an image 20.0 cm 10 its right when a real objecl is placed 10.0 cm to its len? (a) 0.500 (il) 1.00 (c) 1.50 (d) -2.00 9. The human eye can respond to light with a tolal encrgyofas lillie as]O II'} I I" red light has
Answers 1. (c). The vclocil)' of light in glass can be found fwm the definition o[ refractive index. 11 = c!lJ: wavelength and frequency are related (0 velocity by the general wa\'c relation, v = fA. Thercrore f~
"/A
~
(InA
~
(3.00 X 10'm/s)/(l.50)(S.00 X 10- 7 111)
~
cl.OO X 10"Hz
2. (b). The law of reflection is independem of the medillm im·olved. The angle of reflection is ah....ays equal to the angle of incidence. 3. (a). Snell's law is given by /1 1 sin 01 = '/I'lsin 02; hence, 1.33 sin 30.0° = 1.50 sin O':!.. From this expression, we see that sin 02 < sin 30.0", so 0'2 must be less than 30.0°. Therefore. (a) is the on I)' reasonable choice. 4. (c). Snell"s law is "Ill sin 01= Jl 2 sin 02' Because Ihe incident rays arc normal 1.0 the prism surface. 01 = 00 and sin 0" = O. Consequently, 0 := I/';! sin 8',! -l> 8 2 = 0". 5. (b). The greater the frequency of light, lhe greater its energy <md the fastCl" its speed Ihwug-h any material mcdillm. From Snell's law, the velocity and sin 0 with respect to the normal arc i,wersely proporlional. Of the choiccs. violellight has the highest frequency and therefore the highest velocilY and the greatest angle of refraction. 6. (b). By the thin lens equation, Jlf= 1//) + l/q; hence. with /) = 10.n em and/= 6.00 em, substitution results in q = 15.0 cm. The image distance is positi\'e; hcnce, the image is real. The rnagnifical.ion isg'i\'cn by M = -q//) = -15.0 cm/IO.O = -1.50. Because M is negative, the image is im·erted, whereas IMI > I means thaI. the image is enlarged.
A.43
A.44
Appendix E
MeAT Skirt Builder Study Guide
7. (c). The focal length is given by the thin lens equ,uion, 1/1 = lip + 1/1}. Because the image formed is viruwl, the sign of if is negative. Therefore, I
1
J
1
I
/)
q
15.0 em
-~-+-~
+~~~-
-30.0 em
8. (d). Magnification is given by M
= -q/P =
-~u.u
f = 30.0 em
-->
30.0 em
em/W.O em
=
-'LOO. The sign of
q
is positive because the image is real. The negative value of M means that the image is invened.
9. (c). E = hC/A = (6.63 X to-:".1 5)(3.00 X 10' m/s)/(6.00 X 10- 7 m) ~ 3.31 X 1O-'''J This result is the energy of each red phOLOIl. The number of such phOlOl1S needed LO produce a total of 10 ll'lJ of energy is
(10-1'.1)/(3.31 X 1O-".J/phoLOn) - 3 photons 10. (d). Polarization can only occur with transverse waves because the motion must be perpendicular to the direction of propagation.
ELECTROSTATICS EXAMPLE 1
'P_/Jon
Two protons, each of charge 'Ip = e, exert an electric force of magnitude each other when they are a distance 'if apart. A pair of alpha particles, each of charge qa = 2e, exert an electric torce Fl)-a = Fil-/I on each other. 'Vhat is the distance between the alpha particles, 1~. in terms of the diswllce bet\veen the proLOns, ' / (a) rfl' = 4rp (b) I~ = 21 p (e) 1~ = rp (eI) More information is needed. Solution
keftpf/p
Use Coulomb's la\\lLO find an expression for the force between the proLOns:
(1)
Fp _ p = --.-,-
Use Coulomb's law to find an expression for the force between the alpha panicles:
(2)
, _ hrqo,qa Fo - o ra -
,.I'-
t
h"e
2
,.I'0
",(2,)'
<)
r
"
2
4lt ee':!.
,. ,a
,( fe'.!.
Divide Equation (2) by Equation (1) mon terms:
~llld
cancel com-
(3)
F';,-p Fa-a
.,
r I' --"
4",r T cr
N"ow substitUle F{(-II =!4 Ffr-f! and solve for r(r ~.. Take square roOLS, obtaining
I~
in tenns of
2
r. 41'
2
I'
2
r 2 =16?·2 "
p
'i,:
The distance between the alpha panicles is four tlrnes the distance benveen Lhe proLOns, which is answer (a).
EXAMPLE 2 Sphere A has twice the radius ora second, very distant sphere B. LeL the electric pOlCntial at infinity be taken as zero. If Lhe electric potelllial at the surface o[ sphere A is the same as at the surface of sphere 13, what can be said of the charge Q,\ on sphere 11 compared with eha'-gc Q/i on B? (a) Q... = 2Q. (b) Q... = QR (c) Q.., = Qw'2 Conceptual Solution
By Gauss's lav~', a spherical distribution of charge creates all electric field outside the sphere as ifall the charge ,..-cre concentraled as a point charge at the center of the sphere. The elect.ric potential due to a point charge is proportional to the charge Q and inversely proportional to the distance frol11 that charge. Tv,"ice the radius reduces the electric potelllial 3lthe surface of A by a [actor of one-hair. The charge of sphere A. I11Ust be twice t.hat of B so thal the electric potentials will be the same for both spheres. Hence, the answer is (a).
Electrostatics
Quantitative Solution Write the equation for the electric potential of a point charge q:
(I)
Make a ratio ofEqu3tion (I) for charge A and charge 13, respectively:
Substitut.e ~l = VII and rA = 2rfJ into Equation (2) and solve for ~, again obtaining [answer (a»):
(2)
A.45
k,Q
v=,.
k"Q.I' Vt=~ \I'j
!leQ'j
Q.,T/I Q,/2r/l)
--;
Q" =
2Q" [answer (a)]
EXAMPLE 3 How much work is required to bring a proton with charge 1.6 X 10- 19 C and an alpha particle with charge 3.2 X 10- 19 C from rest at a great distance (effectively infinity) to rest positions a distance of] .00 X 10- 15 m away from each other? Solution Use the work-energy t.heorem:
W= t!.KE + t!.PE = KEj- KE; + PJ'r - PEl I:,q/J%
The velocities are zero both init.ially and finally, so the kinetic energies are zero. Substitute values into the potential energy and find the necessary work to assemble the configuration:
W= 0 - 0 + - - - 0 r
(9.00 X 109 kg·m'/C'·s')(l.60 X 1O-'9C)(3.20 X 10- 19 C) LOO X 10- 15 m
EXAMPLE 4 A fixed, conStant electric field E accelerates a proton from rest through a displacement 60s. A fully ionized lithium atom with three times the charge of the proton accelerates through the same constant electric field and displacement. Which of the following is true of the kinetic energies of the particles? (a) The kinetic energies of the two particles are the same. (b) The kinetic energy of the proton is larger. (c) The kinetic energy of the lithium ion is larger. Solution
The work done by the electric field on a particle of charge q is given by \tV = F 60s = qE .6.s. The electric field E and displacement 60s are the same for both particles, so the field does three times as much work on the lithium ion. The work-energy theorem for this physical cont.ext is W = .6.KE, so the lithium ion's kinetic energy is three times that of the proton and the answer is (c).
MULTIPLE-CHOICE PROBLEMS l. What is the potential difference between poilll A and point B if LO.O J of work is required to move a charge of 4.00 C from one POilllLO the other? (a) 0.400 Y (b) 2.50 V (c) 14.0 V (d) 40.0 V
2. How much work would have to be done by a nonconservative force in moving an electron throug"h a positive potential difference of2.0 X I06y? Assume the electron is at rest both initially
A.46
Appendix E
MeAT Skill Builder Study Guide
5. Two poinl charges, A and B, with charges of2.00 X 10- 1 C and -4.00 X 10
'e, respec-
tively, are separated by a distance of6.00 m. Wh,ll is the magnilLlde oftJ1e electrostatic force exntcd on charge A? (a) 2.20 X 1O-!l:\ (b) L.30 N (e) 20.0;'" (d) 36.0;\ 6. Two poin! charges. A ,mel 13, are separated by 10.0 m. IF the distance between them is ITChJrFC! 10 .~.OO Ill, 111(' fOlTC' f'xC'l"Ipri on {,;lrh (a) decre,lses to one-half its original vallie. (b) incrc
7. Sphere A with a lIet c1nrge of" +3.0 X 1O-~ Cis tollched LO a second sphere 13, which has a net charge of -9.0 X LO-:1 C. The two spheres, which arc exactly the same size and composition, are then separated. The net charge on sphere A is now (
8. If the charge on a particle in an electric field is reduced
{O half its original value, the force exerted on the p;lrLicle by the field is (a) doubled, (b) halved. (c) quadrupled. (d) unchanged.
9. In the flgllre below, poims A, B, and C are at various distances from a given poim charge.
'Wh ieh statement is most. accllrate? The c1ecu'jc field strength is (a) greatest at point A. (b) greatest at point B. (c) great.est at point C. (d) the same al all three poims. 10. The electros£atic force henveen two point charges is F I rthe charge of one point charge is doubled and I hat of the ot.her charge is qlladrllplcd, the [orcc becomes which of the following? (aj /-/2 (b) 2F (e) 4F (el) SF
Answers 1. (b). Potential difference bct\\;een two points in an electric field is the work per unit charge required to move a charge bctween the tvm points:
aV= Wlq = 10.0.114.00 C = 2.50V 2. (d). If t.he on I)' effect on the p
aKE+
= -3.2
q~l'=
0
+
(-1.6 X 1O-"C)(2.0 X 10 6 1')
X 10-".1
3. (b). PrOlOIlS are fixed in the nucleus and cannOl be transferred by friction. Electrons can be t.r
F ~ (9.00 X 10"" lll'/C')(2.00 X 10-" C)(-'I.OO X 10-" c)/(6.00 111)' ~ -20.0 N so the magnitude of the force is 20.0 N. 6. (d). From Coulomb's law, F = k,(qAqn)lt·~ and force is inversely proportional to the squal'c or the distance separating the points. Decreasing the distance to half its original value means that the force quadruples. 7. (b). This problem is all application o[ dlC law of conservation of charge. The initial net charge is (+3.0 X 10 -:\ C) + (-9.0 X IO-~\ C) = -6.0 X 10-:'1 C. The samc net charge mllst exist after contact. The -6.0 X 1O-~ emUS! hc cvenly distributed between the two spheres because physically the)' are identical.
Circuits
A.47
8. (b). The electric field strength is the ratio of the «")fCC exerted on a unit charge in the field: J:: = F'/q. Therefore, Fand q (Ire directly proportional and linearly related. 9. (c). From Coulomb's law, (orce varies inversely wilh tlie square of the distance from
the charge. The strength of the electric field ,u a point is the ratio of Ihis force to the eh3.l·ge: 1:,' = Nr'll"'!., Therefore, E and 1"2 al'e inversely pl·opal·tiona!. The smaller the
value of r, lhe smaller the value of r:!and the gTeatcr the valut' of I~·. 10. (d). From Coulomb's law, F = k,(fhfh)/r2, force is directly proportional to the product of the charges. If ql is doubled and q'J.. is quadrupled, the product of the charges increases eighLfold and so docs the force.
CIRCUITS EXAMPLE 1 Three resistors are connected together. How should t.hey be combined so as to minimize the resistance of the combination? (a) They should be connected in series. (b) The two larger resistors should be put in parallel, and the remaining resistor put in series with the first two. (c) All three resistors should be placed in parallel. Conceptuol Solution Resistance is proportional to length and inversely proportional to the cross-sectional area or a resistor. Putting all three resistors in parallel effectively minimizes the overall length 01" the combined resistor and maximizes the effective cross-sectional area, so (c) is the correct answer. Quantitative Solution Let R 1 and R';? be the two larger resistors. Calculatc the resistance R s of the three resistors in series: Calculate the resist.ance !{mmbo series with a third resistor:
0[£1
parallel pair in
1
Rwillbo = ( Ii; +
1
)-1
f{2
Divide the numerator and denominmor of the parallel resistor term by R 1:
Calculate the resistance Rp of" three resistors in parallel:
Divide ule numerator and denominator by R I !?.'!.:
Notice in Equation (2) that Rmmbu is less than R 2 + 1<3 because lhe denominator of" the fil-St term is greater Ulan 1, which in turn means that Rcorllbu is less than R s = H[ + R2 + R?o. Finally, notice in Equation (3) that Rt,is less than R 3 , again because of a denominator larger than I, and so H. p is less than R comho ' The purely parallel combination therefore yields the least resistance, and the ans\ver is (c).
EXAMPLE 2 Two resistOrs are to dissipatc as much energy as possiblc when a fixed voltage difference is placed across them. Should they be installed in parallel or in series? Conceptual Solution The power dissipated by a resistor is proportional Lo thc voltage dirference squared and inversely proportional to the resistance. The smallest possible resistance will therefore result in the largest pmvcr out.put. Consequently, the two resistors should be placed in parallel, which results in the lowest combined resistance.
A.48
Appendix E
MeAT Skill Builder Study Guide
EXAMPL:::,E.,::3
_
A ccna i n off-I he~shelf resistor of resistivity p has resistance R. Suppose it is com mercially desi rable to design a nc\\." resisLOr thal has one-third the length and one-fourth the cross-sectional area of the existing resistor, bllllhe same Q\'crall resistance. What should the resistivity p" of the new resistor be, in terms of the resistivity p of the original device? Solution \\'rite an expression for the resistance R II of the new resisror in terms OrilS resistivity, cross-sectional area A, and its length L:
( I)
R = PilL" Il
All
p"L,. Divide Equation (I) b)' the same expression for the original resisLOr:
(2)
R" R
SubSlillllC An = A/4, Rn = R, and L'1 = L/3 into Equalion (2):
(3)
1
Sake EquaLioll (3) for the new resistivity. POI:
p,/ = ~p
~ pL A
p,),,,A pLA"
p,,(L/3)A 4p" =-pL(A/4) 3p
MULTIPLE-CHOICE PROBLEMS 1. Three rcsi~LOrs of resistance 1.0,0, 2.0 n, and 3.0 n, respectivel)" arc in series. If a potent ial differcnce of' 12 V is applied across the combination, what is the resulting current in the circuit? (a) 0.50 A (b) 2.0 A (c) 6.0 A (d) l2 A 2. If the Icngl h of a conducting wire with resistance R is doubled, what will the resistance 01" Lhe longer wire be? (a) /1/'1 (b) "/2 (c) 2/1 (d) 4 n 3. [fa 2.0-0 rcsistor and a 6.0 fl resistor are connc([ed in parallel, what is their combined resistance? (a) 1.5 n (b) '1.0 f1 (c) 8.0 f1 (d) 12 n 4. rf all the components of an electric circuit are connected in series, which of the following physical quantil ics must bc the same at all poinls in the circllit? (a) voltage (b) current (c) resistance (d) power 5. The current in a conductor is 3.0 A when it is connected acro~s a 6.0N bauer}. How much power is delivered to the conductor? (a) 0.50 W (b) 2.0 W (c) 9.0 W (d) 18 W 6. A 12-f! resistor is connccted across a 6.0-V dc source. I-low much energy is delivered to fhe resistor in halfan hOllr? (a) 1.5 X 1O- 3 kWh (b) 2.0 X 1O-:ikWh (c) 3.0 X ro-3 kWh (d) 12 X 10-' kWh 7, A bauery with an emf of 6.20 \' ca'Ties a cun'enl of 20.0 A. If the internal resistance of the batter'Y is 0.01 n. whal is the lerminal ....oltage? (a) 1.24 V (b) 6.00 \' (c) 6.40 \' (d) 31.0 V
8. Devices A and B are connected in parallel to it battery. [f fhe resistance R I of device A is four times as gl'car as the resistance R B of de\"ice B, what is true of II and /8' lhe currellls in de..ices A and B? (a) I, = 218 (b) I I ~ 1i/2 (c) I, = 4/" (d) I, = 1i/4 9. A wire has re<;istallce R. What is the resistance of a wire of Ihc same substance Ihat has lhe same length but twice the cross-sectional area? (a) 2R (b) R/2 (c) II R (d) R/4 10. Whal must be the reading in ,he ammeter A 1'01" lhe circuit section shown? ·LOA
2.0:\
---:;>::--,----+---GA~--
6.0A
(a) 0 A (b) 6.0 A (c) 8.0 A (d) 12 A
:2
Atoms II. What i~ the curn.'llt in a wire if 5.00 c: or charf.{c (a) 1.00 A (b) 250 A (c) 5.00 A lei) 1o.\) A
pas~cs
A.49
through the wire in O.!lOO s?
Answers I. (b). The three rc~islOrs arc ('oll1lccted in ~erit's, ..,0 HI", ~''/'''I<>I= [101= 12V/6.0!1 = 2.0A.
= :£ n = fl.a n.
From Ohm's law.
2. (c). Resislallcc is dire<.:lIy proportional LO length and in\'ersely proportional to the cross-sectional ~ITa of (he conductor. If the
3. (a). The resistor.., arc in parallel: 1/",
= 11,11,/(1/, +
II,) = 120'/8.0 fl ~ 1.5 fl
In Cael, Ihe equivalent resi.~L<-lnce or p
4. (b). A series circllil has only one path fot· ctIlTenl."'O it 1ll1lSt be the saml' at all the circuit.
5. (d). 71'
= /..1 V~
P()illl~
in
IR W.
6. (a). The energy uscd by a load is the proc!IICl or 1 he power It tl:-.es per unit time alHI the length or I irnl' il is opermcd:
7. (b). The product Ir is the potential drop ol..:cllrring within the batrer~': Ir = (20.0 :-\)(0.01 .0) = 0.20 V. Ikcall~e the bauny is producing current and not being recharg-cd, the Icnnin
L~I",=O
Ilence. one-fourth
1.\
U II
If. II
I
1'1
HI
..fUll
·1
mucl! current goc,; lhrough resistor A.
9. (b). Re~i~tallcc is Inversely proportional rcsistallCl' bv half'.
LO
lhe are,,: doublillg til(' arca reduces Ihe
10. (d). Kirchhorr's ClllTent. rule S
ATOMS EXAMPLE 1 Trilillm is
Solution Calculatc lhe number /I or half.. li\·cs rC'luired. The equation can be soh'cd by inspection (logari[hms are ordinarily required):
= ( ~)" ~
j
20.0 g .60XIO'g
1
800
n=3
A.50
Appendix E
MeAT Skill Builder Study Guide
~hJ!lipl)' the Ilumber of hall:'livcs by the length or the half-life to find the necessary lime interval in question. \'crifying thal-the ans\ver is (c):
lJ.1 = III, , = 3 (12.33 yr)
37.0 yr
EXAMPLE 2 'Vhat is the mass number ofa carbon atom having 6 protons and 8 neutrons? (a) 6 (b) 8 (c) 14 Solution
Don't confuse Illass I1mnber \virh tht: alOmic llurnber, 1:, \.. . hich is the number of prowns in the nucleus. The mass number A is the number ornudeons ill the nucleus. To calculate the mass Ilumber,just add the Illlmhcr of protons and neutrons LOgcLhcr: 6 + 8 = 14, \'·;hich is answer (c).
EXAMPLE 3 vVhich or the following' particles is given by ~.xin the f(Jllowing reaction? (a) ;Li (b) ~Bc (c) liN (d) I\;C
iH = ~X
-7
~He
Solution Equate the sum of the mass numbers on both sides of the reaction:
By charge conservation, the number of protons must also be the same on b(l{ h sides: Based on these two results, the correCt answer is (a),
+ JHe
1+11=4+4=8
--7
11=7
I+Z=2+2~4
--7
2=3
~Li,
MULTIPLE·CHOICE PROBLEMS L III the nuclear reaction below, what particle does X represent?
TT:"\a
---7
Y6>Je +
X+
flo.:
(a) all a-particle (b) a ,t3-panicle (c) a positron (d) a y-pholOll
2. \Vhat is the atomic number oCthe daughter nuclide in the following reaction? ---7
(a) 14 (b) 16 (c)
~O
,'Si Z
+
c+
+
JI,
(el) 31
3. If I~J\ has a half-life of about 10.0 min, how long will it take for 20 decay to 2.5 g-? (a) [) min (b) 10 min (c) 20 min (d) 30 min
g or the isotope to
4. A certain radinnuclidc decays by emitting an a-panicle. What is the difTerence between the atomic numbers of the parent and the daugh1er IIllclidesr (a) I (b) 2 (c) 4 (el) 6
5. \-Vhat is the difference in Illass number between tIle parent
(el) 13
7. \'\-'hich one of the following is an i~olope of 1~5X? (a) I~~X (b) I~~X (c) 1~~\X (dl I~~X
8. III the following lwclear equation, '....h"l is X? I~N
(a)
+
Jlle
---7
I~O
+
X
proton (b) a positron (c) a ,t3-pclrlicle (d) an a-particle
9, \Vhat is the !lumber ol'ncmrons in 1~>tIXe? (al 5"1 (b) 86 (c) 140 (d) 19·1
Atoms 10. Radon-2~2 has a halr·life of (lboUl 4 days. If a 'iample 01" :!~:!Rn gas ill a cOlllaincr i..; inilially doubled, the halr-lile will be (a) h;l1vcd. (b) doubled. (e) quarteret!. (d) unchanged.
11. r\ radiolluclide decays completel}. In the reaction flask. the only gas that is IOIJlld is Iwlil11l1, which was 1101 prcse-l1t whell the flask was ,'iealt.:d. The decay process was prohahly (a) ,B-dcea>'. (h) a-decay. (c) y-ckeav. (d) posilron em ission. 12. Wh;H is the hall:lire of a radiOllllclide if 1/1601" its initial mas.., is present ;\fter (a) Jr) min (b) ~O lllin (c) 4r:. mill (d) 60 min
~
h?
13. The hall~lifc oQiNa is '2.6 y. If Xgrams of this sodilllll isolOpe are iniLially presellt. how ",uch;, len afler I~ yr? (a) X/32 (b) X/13 (e) X/8 (d) X/f, Answers I. (c). ~Olice that the mass llumber is unchanged. whel'L'as Llw atOmic Humber has heen reduced by I, \.. . hich implies that a proton has changed into a nClllron. The emiLling panicle must ha"e;l charge eC(ualto a proton hut ;111 atomic mass number of lero. The positron is lhe only choice that has both these attributes. 2. (a). The ;l1omic number fore, Z = 14.
i~
Z. Conservat.ion of charge llleans lhal I.)
=Z+
3. (d). First. tinct the lHllllbcr or hall~li\'es and Ihell multiply by t.he "duc or the g-eL t.he elapsed time:
~~:=i=G)' U/
1/
=
I, There-
half~lire to
3
= JIll·:! = 3(lO min) = 30 min
4. (b). An a-panicle is a ~I-Ie helium nuclem, III a-decay, two protons arc efTcni\'ely rCIllO\'ed. 5, (b). J3-dcc3Y emit.s a high-energy electron. jle. In dlt' p1"Oce:-.~. a neulron decays into a proton plus Ihe emiucd electron and all alllint"lIl.rino, The number of nuclcons rcmains Ilnchallged, with the proton replacing the neutron ill the sum ornUCleOlIS. 6. (d). The llla..;s number is the
SlIlll
OfllclltrollS and protons: 6 + 7
= 13 lIucleon..;.
7. (c). !,,;oLOpes of all clement. have the same atomic number but dilTcl-ent number:-. of" neutrons. so their mass numbers are different. The atomic number or X is 63. 8, (a). The Illas..; numbers must be the same on hOlh ..,ides 01' the reCloioll. II" A is the Illass number of X, then 14 + 4 = 17 + .-I., so il = L. As for alomic number, i + 2 = 8 + Z. Therefore, Z = I, which describes a proton, 1H. 9. (b). The number of nelll.rons is the mass number minlls llle atomic number: i\ - Z 141l - 54
~
=
86.
10. (d). The hall:lifc is a const.alll that depends on the identity of the nuclide, not. on the . Each a-particle lhen acquires I\m electrons to form a neutral helium alom. 12. (b). In every haIr-life, the mass decreases to halfils previousyaluc: 1/16:::: 1/2-t, It takco; fOllr hall~livcs to decay down to 1/16 the original mass. Each must be 30 min long
hecause the entire process takes two Itouro;. 13. (a). In 13 yr, there ,.,..ill be 5 hall:livcsoI"2.(i yr cacll (5 X 2.6 = 13). The isotope dccreascs
to 1/2:'
= 1/32 orits original amount.
A.S1
ANSWERS TO QUICK QUIZZES, EXAMPLE QUESTIONS, ODD-NUMBERED MULTIPLE CHOICE QUESTIONS, CONCEPTUAL QUESTIONS, AND PROBLEMS CHAPTER 1 EXAMPLE QUESTIONS 1. F'1Isc.." 2. True 3. U.9Sm:!
4. 2X.O rn/ .. =
(2S.0J~)(2.21I1li
.h) = 62.7 lIIi/h
s 1.00 In S The amwer i~ slighlh diffen:nl because l!l(' dilfert'nl ('"ol1\(.'r..ion faClor~ WCIT rounded. leading 10 !>l1lall, tlllp' ('dic. l;lbk c1iffcren{cs inlhe finill.It1.'i\\(:-r~.
5. (60.0 "';")' 1.00 h 6. An ;In""{"f of 101:! Ct'll" is witllin an order 01 mOign illldc: of Ill(' gin'>n answer, COl re ..ponrtillJ.:" 10 sli~htlr dilTcrc!H choicc~ in Ihe HlllllllC ('stimaliol1s. COllsequcntl). !OI:! cells is al~o" rCil<;ol1;1blc estim'lIe.:\1l estimate of ID" Ct.'lIs would be smpcci hCGwse (workiJl~ backward ..) it would imply Ct.'lls 1hal could be !it'CIl wil h lIw naked l'n~! 7. - -I X 1011 .
8. 9.
W\:!
\~'oJ'kin~ backwards,
,. = 4.9!18. which fllrthn rounds LO ,"l.tlO,
= :n.():\c. \\·hich fllrther rounds 10 37.0°. The sliRhl diff('ren("('~ are caused Iw rounding. 10. Yes. The ~ccall{ Il1llLlion.wgelbcr with the distann~ to dw building", gi\'e:. lIlt:.' 1cllglh of' Ihe brpotcnllSl' or the 1riangle. I I. !l·IO kill \\hCI(:8:.
f)
MULTlPLE·CHOICE QUESTIONS I. (a)
7. (e1)
9. (a) II. (e1)
CONCEPTUAL QUESTIONS I. (al -0.1 m (b) -1m (c) Iktween 10 III and 100 III (d) - 10 m (e) - 100 III 3. - 10" s 5. (a) -1010 beats (b) -10'1 b(.'a(:o,
7. TIlt' lellA:lh Of8 hand v'aries from person 10 per...oll. so it i..n', it llseful "andard of length. 9. A dimemionalh cOlTeet equation isn't necess"rih true. For ('xample, Ihe equal ion 2 dogs = j do~~ is dilll{'nsionalh eOl'rect. bill i~n'l trlle. I-Io\\·en:r. ifan equation i~ not dilllell~ioll alh corrCCl, it C;tnnOl be corren. It. An 1'<:lilllah:, en,'n if imprecise b\' an order of magnililde, ~r('atlr Ieduces (he J
,I
pnl\"isioIIS.
PROBLEMS 3. (b) 11,,10\<1," = 7TR~, A ....,!-U,)-.'1,t." rl~ ... = 1c-1Ig:th X width 5. Ill:l / (kg' s'2)
A.52
35. (2.0m. I. 1m)
37, r=2.2m.O=27 39, 5.69 III 41. (a) 6.71 III (h) 0.891 (c) 0.;·16 43.3..11
3. (e1) 5. (e)
7. 22R.H
9. (a) three ~ignilicant riglll"CS (b) fOUl signifiGHlt figlll'es (c) three significant figures (d) (wo ,ignificmt figures lJ. (a) 35.7cm~ (Il) 99.2cm\ (c) 17.7cm:?,99.:lcm' (d) In tht: "otmding proccs..... slllall ;nnOlml~ are either' added 10 or suhtracted fmm an answer to satisfy the rull" of:.ignificanl rigllt"e~. For a ghen rounding. diJTcrelH 'lIlilll
('Ill
III
45. (a) 3.00 (b) :,.00 (e) OBOO (e1) (J.ROO (e) 1.33 47.5.00.17.00; ,-he angle itsdfb 35.:)0 49. 70.0 III 51. (.:1) \JI.T :! (b) kg' m's:? 53. (..) 1.0 X 10'\ kg (b) "'cdl = :l.2 X 10 I.. kg. /film",,, = 0.27 kg. IIlII> = I.~ X 1O-~ "-g 55. The \'alue of k. a dimemionle£:o> constant. cannUl be found b,' dimensional anal\·sis. 57. - 10 '1 m 59. -10' tuners (assnmes one Iliner pel 10 OO(J residents and a pnplIlillion of /.5 million) 61. (a) 3.16 x 10~ s (b) Between lOlU \'1 and lOll VI'
3, The \"e1ocit\ ,'s. timc gnlph (a) ha~ a cnn . . t;\Ilt ~Iop(', indicating a cOIlSUIll ;:Iccelel-ation, which is rl·prc· . . cnted b\ the accd~ l'ralion \'s. lime graph (e). Graph (b) r('preselll~ an ohject with inue.. :.ing speed, but a:. Lime pl·ogl"es~es. Iht'lilll'~ (11;1\\11 langcm to the cun'e ha"e illCl"e:Jsing ~Iope:-,. Sinn' thl' ;Kcell'lation i, cqllallo Ille -",lope 01 Lhe lallgellllillc. ,he i!e
ANSWERS TO QUICK QUIZZES, EXAMPLE QUEST/ONS, ODD-NUMBERED MULTIPLE CHOICE QUESTIONS, CONCEPTUAL QUESTIONS, AND PROBLEMS CHAPTER 1 EXAMPLE QUESTIONS 1. False 2. True 3. -13.98 m t
m)(2.24 mVh) = 62.7 mi/h 4. 28.0 m/s = ( 28.0s 1.00 m/s
The answer is slightly diffcrclll because the different cOllversion faclors wen~ roullrled. leading 1O small, unpl"edic13ble differences in the final ans\\'er~.
5. (60.0
min)'
1.00 h
6. An ,mS\\'N of 101:? cells iswilhin an order ormagnilude ortbe gi\'cn answer, corresponding to slightly different choices in lhe \'OILllllC estimations. (;mlscfJ"ently, ]01'..' cells is also a reasonable estimate. An eSlimate:: of 10~· cells would be sllspeCl because (working backwards) it would i rnply cells I hat could he seen wit.h 1he naked eye! 7.-4XIO" . 8. - 10 11 9. Working backwards. r = 4.99R, which funhcr rounds to 5.00, whereas (} = ~7.03', which further rounds to 37.W. Th~ slight dilrercnc:e~ are causer! by rollnrling. 10. Yes. The seell1! function, together with the distance to the builcling, gives tile leng-ul of the hypotclllisc of the triangle. 11. 540 kill
CONCEPTUAL QUESTIONS I. (a) -O.lm (b) -I III (c) Between lOmalld LOOm (d) -10 m (e) -100 III 3. - lOfl S 5. (a) - 1()6 ocals (/) - 10" bea!.'; 7. The length ofa hand \~.lries from person to person. so it isn'l a useful standard of lenglh. 9. .-\ dimemionalh correct equatioll isn'l necessadly Irue. FOI" example, the equation 2 dogs = 5 dogs is dimensionally cor· reel. bUl isn'llrue. Howe\·er. ifan equlllion is nOl dimensionall" correct. it cannot be correct. 11. An estimate, C\'cn if imprecise b}' an Ol"der of magnittlde, greatly reduces the range of plausible answers 1.0 a gin'n queslion. The eSlimate gi\'es guidance as to \\ hat correCLi"c measures mighl be feasible. FOI" example, if)'oll eSlimate that ·10 000 people in a countr}' will die unless they ha\e food assistance and iflhis number is reliable \\'ilhin an order of magnillide. rOll know Ihal iH lHO~l 400 000 people will need prO\;siollS.
PROBLEMS 3. (b)
A'lli"'h'"
5. 111:1/ (kg' s~) 7. 228.8 e1l1
A.52
=
2 1TR .
Aft'''l:'"~'''I." I'LIII" = length X widlh
9. (a) three significant figlll"es (b) four signHicant figures (e) three significalll figun's (e1) Iv..·O significanl figures II. (a) 35.7 em:! (b) 99.2cm" (c) 17.7cm:!,99.5cm' (d) In the rounding p"ocess, small alllounts arc eilher added 10 or subtracted from an answ'er to satis!)' the Iltle~ ofsignificalll ligures. For a given munding. different small adjustmentS are made. introducing a certain amount of randomness in the Ia.-;l significant digit of Ihe final answer. 13. (a) 797 (b) 1.1 (e) 17.66 15. 2 X 10k fadlOms 17. 0.20-1 m~ 19.1.39 X IOlm:! 21. I X 10 17 ft 23. 6.71 X 10::;' mi/h 25. 3.08 X 104 m:l 27. 9.82 cm 29. - I Of, balls 31. (a) - 10 14 . Olhel" answers are possible. This answt'r assumes an intcstinal1cngth of about 7 III and (tn;ragc diameter of abolll 4 em. There al"e probably more bacteria in )'0111' illlestinalut\Cl than Lhel-e are cells in your body! (b) The large number meallS that they arc probably nOI dangerous. Illlcstinal b<-lclcria help digest food and prO\'idc importanl nmrient<;. Humans and bacteria el~jor <-l mllluall}' Ix:neHcial symbiotic relationship. 33. -lOire\' 35, (2.0 m, 1.4 Ill) 37, r= 2.2 Ill, 0 = 27" 39. !"i.59 III 41. (a) 6.71 m (b) 0.89'1 (elO.746 43. 3.4 1 III 45. (a) 3.00 (b) 3.00 (e) 0.800 (d) 0.800 (c) 1.33 47. !"i.()0/7.00: the angle iL.,e!f is 35.5"' 49. 70.0 m 51. (fl) \ILT-~ (b) kg'm/s:! 53. (a) 1.0 X 10:1 kg (b) m cdl = 5.2 X IO- Iti kg. III kxh " .., = 0.27 kg. m fh = 1.3 X IO-;kg 55. The value of k, a dimensionless constant, cannol be found b\' dimensional analysis. 57. - lfJ-'1 III 59. -103luners (assumes one l!lller per 10000 residents and a population 01'7.5 million) 61. (3) 3.16 X 10' s (b) Bel\\'een 1Oro),r and 1011 )1'
3. The velocily \"S. time graph (n) has a COIl~lalll slope, inrliC
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
Graph (c) depicts an object which lir~t has tl velocilY lhat increases at a conStant nUl:. which mcans lhat the objeCl's accelen:tlioJl is cOllstant. The vdocity Ihen ~IOPS changing. which means that Lhe acceleratioll orthl.' object is Lero. This behavior is best maLched by gnlph (I"). 4. (b)
5. (a) blue graph (b) red l-;"raph (c) grcc.:.'ll graph
6. (e) 7. (e) 8. (a) and (f)
LIle displacement i~ :.'ero n:ganlk~s ortlle total distance eled duri ng tIle giv(-'Jl t iIll('.
[1';.1\'-
2. No. A venicalline in a posi[iull v:>.. [ime graph would mean t haL an object had sOlla'how I.ravCI·~ed all points along lhe gi\'en path instantalleously, which is physically impossible.
3. 1'\0. A \'enical LangCll[ line would corrc~pond to an infinite acceleratioll, which is phr~icall)' illlP{I~sihle. 4. 35.0 m/s i~
Ihe illtt:rseClion ora straight line
6, The coasting displacl:l1lt:'llI would double to [olal displac~menlof 715 Ill.
14~~
m, \,'ith a
7. The acceleration is Icro \\·lten:'vcr the tangent to the \clocity \crsus time graph is horia)]]ta!. Visually, Lhat OCCllPi from - 50 ~ to 0 sand [hf'1l a~,litl at approximatel>' 180 s, 320 S, and 330 s.. 8. The upwalTljump would :>.Iightl), increase the ball's initial \'clocit>', slighLly increasing [lie maximulll height. 9. I) 10. The engine should be Ilrt:-([ agi-lill;1I
CONCEPTUAL QUESTIONS 1. Yes. If the velocil)' oflhe panicle is 1I0n/era, tile panicle is in
5.
7. 9.
3. (a) 52.9 klll/h (h) 90.0 km 5. (a) Boal A Will~ by liO km (b) U 7. (a) 180 knl (b) 6::H klll/h 9. (a) 4.0 m/s (h) -4.0 Il1ls (c; 0 (d) 2.0 m/s II. I .:~2 h 13. 2.80 11,218 kill 17. (a) 5.00 Ill;:" (b) -2..')(1 m/s (c) 0 (d)
I. :'\lo. The objecllllay not he lra\'dill~ in (l sLraighlline. IfLhe initial and fillal positions an; in [he same place. for example.
3,
PROBLEMS I. = 0.02 "i
15. 2i4 kl11/h
EXAMPLE QUESTIONS
5. The graphical ~Ollitioll and a parabola.
A.S3
mOlion. If the acceleration is leI"C), [he velocil> of the panicle is unchanging or is COllst'llll. Yes. Iflllisoc("lIr~, Ille IO\\illg down. (c) The GIl' is mo\"ing to [lit: east at constant speed. (0) The car is 1l10\'illg to Lhe IH'sl btll slowing down. (e) The car is moving to the \"l'.~t and speeding up. (f) The cal" is lllo\"ing to the lVeM:ll COll~[an[ ~pCC:.·d. (g) The car ~Larls frOIll l't',,1 :lnd bcgin~ to speed up LQh'ard lhe ea~t. (h) The C:1r Sl,lrt~ rrOIll r('~t and !wgil1s to ~peed lip toward [he wt'sI.
s.on m/s
19. 0.1811li west or[h(.' flagpole
21, (a) !l2.4 ft/s..l[l.n fils, !):'i.!J IVs, 56.9 fI/s. 57.-1IVs (bl 0<181 lIN 23. ~.7 s 25. (.nllle lime. Wl-iting expressions for the position versus Lilla: for each vchicle and equaling the 1.'\\0 gives a qua(h-at ic equat i(lIl in I whose solution is eiLhel' 1 I A s or 13.6 s. The 'irst solution. IIA s. is the time of the collision. The collision occurs wIlen the van is 212 m from the ol'iginal po~ition of"Suc·:>. car. 37. 200 III 39. (
In
41. 958 III 43, (,1) H.2 S (h) 1.3 X IO:.! III 45. (a) 31.91ll (h) 2.:):' s (c) 2.55 s (r1) -25.0 m/s 47. 38.2 m 49. I [;ll"(lwood Iloor: fl = 2.0 X IO~ ll1/s:.!, 111 floor: ({ = 3.9 X 10:.! m/s:.!, tit = 7.1 illS
=
1.4 ms; carpeted
51. (a) It is ,t I"n'l'ly l:lllin!{ object, so il.~ acceleration -9.80 Ill/~:.! (dowl1ward) while in night. (b) 0 (c) 9.80 lllh (d) '1.90 m
i~
53. (a) It cOlltinlll':>; upward. slO\~ing under the influence of gr.lVit) until it reachf.'s a maximum heighl, and thcn it falls to [anlt. (11) ~OH 1lI (c) 8.51 s (d) 16.4 s 55. 13.0 ~ 57. (a) -~50 X 10" Illh:! (h) 2.86 X 10 4 s 59. (a) 10.0 1ll/.'> upward (b) 4.68 m/s downward 61. (a) 4.0 Ill!.'>, 1.0 1m (11) 0.82 m 63. 0.60 s 65. (:1) 3.00 S (b) -2'1.1) mis, -24.5 m/s (c) 23.5 01 67. (a) 1.21 s anc!' [he lil".~t ball is dropped (b) 7.2 m below the wilHlo\\' 69. No. III [h(' [ime imnval equallo David's reaction time, the $1 bill (a freely railing ol~ject) falls a distance or I{t'!./2 == 20 em, which i~ about twice the distance benvcen the top of llit: hill :Inri its (·enter. 71. (a) i l = !'i.O S, (:.! = 8;') s (b) 200 ft/s (c) 18500 ft from starting point (d) lOs arier slal"ling to slow down (Iolalu'ip time = 100 s) 73. (a) 5.5 X J(J'I fL (Il) :\.7 X IO'!. I"t/s (c) The plane would lJ-a\'el onl)' 0.002 n in tllc time it takes tilt" lighl from the bolt to !"t_'ach the cyt'. 75. (a) 7.82 III (h) 0.782 <;
A.54
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choke Questions, Conceptual Questions, and Problems
CHAPTER 3 QUICK QUIZZES I. (e)
x-component
)L.componcnt
A
+ 3. (bl 4. (al 5. (e)
is alll';l)'o; pnpcndicular to tht: initial ve[ocil~', resulting in a parabolic p:lth. As regard .. Ih(' spacecraft. I he inilial \'clociry plfl),<; the !"Ole of the horil.Otllal velocity ofthl,; projectile, and Ihe It'a ki rig gas plays I he role Ihar gravity pia} ... in rhe case of Ihe pn!jeClile. If t Ill;" orientation of thc spacecraft were to change in response to lhe gas leak (,,'hieh is br far lhe more likel)' rcsull), then the acCclCI":llion would change direction and Ihl' lllotioll could bt:collle vcry complicaled. 9. For :Illgles (J < 45~, the pr(~jeClile thrown
6. (b)
EXAMPLE QUESTIONS
PROBLEMS
1. I f the \'cctors poi11l ill 1he same direct ion, I he Slim of the magn ilucles of the t\\'o wetol's equals 1he Ill
2. Because 11,., B,,, and Jlat"e all known, all)' orlhe trigonomelric flillCI ions Gill be used to lind the angle. 3. The hikers' displacemellt n:ctors arc lhe s'lIne. 4. The initial and final \docit}'veClors arc eqllal in tllflgniltidc because the X-colllponcm doesn't change and the .,'-componc11l changes only by a sign. 5. To Ihe pilot, the package appears to drop straight cloWIl because lhe x-compolICnls of velocity for I.he plane and p;lck~ age arc 1he same. 6. Fal .. €" 7. -t5~ 8. False 9. Faist' 10. To an observer 011 tIll.; g"round, the ball drops straight dowll. II. The angle decreases willI increasing SI)('I,;(I. 12. The 'Illgle is differellt because relative vdud!y depends on bOlh 1he magniLllclt.: ;lnd the direction of Ihe veiociLY "ecton. In Example 3.12. the boat's velocit)" veCLor forms the h) pOle· nllse ora right ll-ian~l(', whereas in Example 3.11. that vcctor form" a leg ofa right triangle.
MULTIPLE CHOICE QUESTIONS
0
0
(b) (v n-). (c) (d)
= 86.6
mi/h, (V' II)y
= 50.0 mi/h
v. II = \:.11' - V \E v jr" = ~.90 X IO~ rni/It, 7.3fC!\ orE,
39. 249 ft upstream 41. (a) 2.02 X J(fl s (b) 1.67 X 10:1 s (c) The lillll' sa\ings going downstream with the ClllTCHI is always less tlml1the extra time n:'qllired to go the same distance against The current. 43. (a) 6.80 kill (b) 3.00 kill \cnicallyabo\'e the impact point (c)
CONCEPTUAL QUESTIONS J. Th(' lll:lgnitildes add \\'11en
1. 43 tlnit~ in Ihe negative y-direnion 3. (a) Appmx;ma,ely 5.0 ,n,;" at -53 (bl Approx;lIIately 5.0 units;u +5~" 5. Approximately 42.1 ft : = 3.00 X 10'!·mi/h. (v.f.J) = 0
A
B
and arc in Ihe same dircclion. The reSUltalll will he 1.('1"0 when the two vectors are equal in magnitude lind opposite in direction. 3. (a) Atlhe top orthe pn~ieClilc's nighl. ils vclocil}' is horizolltal alld its acceleration is downward. This is Ihe only poi lit :H which Lhe velocit)' and ;lcce\eration vectors are perpcndicubu. (b) If the projcClilc is thrown straiglll IIp 01' down, thell the velocity and acceleration will be paralld thmughout the motion. For any Olher kind of projectile Illotion. the vclocit} and acceleration vectors arc never parallel. 5. (a) Thc acceleration is zero. since bOlh the tn::lgniwde and direct ion of the velocily n;ma in consl l'cgards the pr(~jcClilc. gravilY pm\'ides an acceleration thai
v
G6.2~
45, 18.0 s 4,7. (a) ·1O.!l m/s (b) Rathl,;r l.han falling like;1 rock, the skier glides through the air like a bird. proionging-Ihejlllllp. 49. 68.6 km/h 51. (a) 1.52 X 10:1 111 (b) 36.1 s (c) 4.05 X 10: 1 m 53. H"H"'" = 18 Ill. H\l,'" = 7.9 III 55. (a) '12 ntis (b) 3.8 s (c) r.,. = :l4 m/s. Vv = -13 IH/s;
lJ=:17 ll1/s 57. 61. 63. 65.
7.5 mls in the dil'ecuoll the ball was thrown 11/2 7.5 mill 10.8 III g . 1 . 67. (b))' = Ax'! with ;\ ~ - - , where lJ· IS the Illlll.l.lc \'e OCltv 2v,~
1
(c) 1'1.5 mls 69. (a) 3!l.1" 01" 54.9<> (b) 42.2 III 01' 85.4 m. respecti\·t:!y 71. (a) 2U.0 above lhe hOI"ilOlHaf (b) 3.05 s 73. (a) 2.1 nt/~ (b) 360 m horiLOlltall~' from the base of cliff
•
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
7. (b) 8. (h) By exerting an up\\'ard Force conlponent
reduce tlte nOI"mal force ()ITe of k.i net ic friction.
011
Oil the sled. you the gmund and so reduce the
7.
EXAMPLE QUESTIONS 1. Othel' than the forces rnentioned in (he pmblem, the force or gr{l\,ity pulls dowll\\';Hds all the boal. Because the boat doesn't sink. a force exened by the waler all the boat must oppose the gl';wity rorce. (Tn Chapter 9this force will be identified as the buoyancy force.) 2. False. The angle at which lhe forces are applied is also imporI.Jlll in detennining the direclioll of the acceleration n:Cl()]". 3.0.2 N
4.2g 5. The tensions would double. 6. The magnitude of the tension force would be greater. and the magniLllde 01" the normal force would be smaller. 7. Doubling the weight doubles the mass. which halves both the acceleration and displacement. 8. A gentler slope ll1e(lns a smaller angle and hence a smaller acceleration dowll the slope. Consequently, the car would Lake longer to reach the bottom of tile hill. 9. The scale reading is grCatel" than the \...· eight of the I'ish dm· ing the first accdel·,ltioll phase. \'\!hen the velocity becomes constant, the scale reading is equal 10 the weight. When tbe cle,·ator slows dowll, the scale reading is less than the weight. 10. A1tach one end oCthe cable to the object to be lil"ted and the other end Lo a platform. Place liglnel' weights on t.he platform until the total mass of the weights and platform exceeds the mass of the heavy object. II. A larger stalic friction coeHicient would increase lilt:" maximum angle. 12, The coefficiellt ofkinctic friction ,\'ould be larger than in the example. 13, Both the acceleration and the tension increase when In'.! is incI·eased. 14. The top block would slide ofT the back end of the lower block.
MULTIPLE-CHOICE QUESTIONS I. (a)
3. (eI) 5. 7. 9. II. 13. 15.
(b) (b) (a)
(d) (e)
(a)
CONCEPTUAL QUESTIONS I. (a) Two externJI fOI"Ces aCI on the ball. (i) One is a downward gravitational force exerted by Earth. (ii) The second force on the ball is an upward Ilormallorce exerted by the hand. The reactions to these forces are (i) all upward gr'lvitational force exerted by the ball on Earth and (ii) a downward force exerted by the ball on the ham!. (b) Afler the ball leaves the ham!. the only external force act'ing on the ball is the gravi-
9.
11.
13.
A.S5
tational force cxertcd by Earth. The reflction is all upward grm'itational force exerted by the ball on Eanh. The coefficient of static friction is larger than that of k.inctic friction. To start the box moving. you must cOUlllerb,·Jlance the maximum static friction force. This fOITe exceeds the kinetic frict ion force that you must counlerbabnce 10 maint.ain the constant velocity 01" the bo:\ once it start!; mO\·ing. The inertia 01" the su itcase IvOtJid keep it moving forward as the bus stops. There would be no tendellcy fOI' the suitcase to be thrown backward 10ward the passenger. Tile case should be dismissed. The force causing an automobile to move is the friction between the tires and the roadway as the automobile attempts to push the roadway backward. The fOITe driving a propeller ail"j)lanc forwal-d is the reaction force exened by the air on the propeller as the rotating pmpeller pLlslles the air backward (the action). In a rowboat, the rower pushes the water backward with the oars (the action). The water pushes fOr\\-arc! on the oars and hence the boat (the reaction). \Vhcn rhe bus starts moving. Claudette's mass is accelerated by the force exerted by the back of the seat on her body. Clark is standing. however, and the only force acting 011 him is the friction between his shoes and the noor of the bus. Thus, when thc bus starts moying, his !"eet accelel'ate fOI'w'lrd, but the l'est of his body experiences almost no acceleral.ing force (only t.hat due to his being fll.tached to his accelerating feCI!). As a consequence, his body tends 10 stay almost at rest, according to Newton's first law, I'elalive to the ground. Relalive 10 Cbudeue, howe\'er, he is moving towal"d her and falls into her lap. Both performers won Academy Awards. The tension in the rope is the maximum fOITe that OCCllrs in fmlhdircnions. In this case, then. since both are pulling \\'ith a force of magnitildc 200 N, the tcnsion is 200 N. If the rope does nor move, then the force on each athlete mus1 equal zero. Tberefore, each athlete exerts 200 J\ against the grolllld. (a) As the lllan takes lhe step, the anion is the force his fOOl exerts on Earth; the reacrion is tile force exerted by Earth on his foot. (b) I-Jere, the anion is the fOITe exerted by lhe snowball on the girl's back; the reaction is the lorce exel"(ed br the girl's back on the snowball. (c) This action is the fOI'ce exerted by the glove nn the ball; the reanion is the force exerted by the ball on the glove. (d) This action is the force exerted by the air molecules on the window: the reaction is rhe [olTe exerted by the window all the air molecules. In each case, we could equall), well inter'change the terms Maction" and "reaction."
PROBLEMS 1.2XI0 4 r-.; 3. (a) 121\" (b) 3,0 m/s 2 5. 3.71 ~,58.7 N, 2.27 kg 7.9.6N 9. lAX 10J:'\I
II. (a) 0.200 m/s 2 (b) 10.0 m (c) 2.00 m/s 13. (a) 0.2.1] kg/m (b) 7.35 m/s 2 15. l.J X 10-1 N
17. (a) 600 N in \'cnical cable, 997 N in inclined cable. 796 N in hol"izolllal cable (b) Irthe poilll.ofau<1chmelll were lllOH'd higher up on the walL the left cable would have a y-colllponelll that would help support the G1I burglar, thus reducing the tension in the cable on the dghL. 19. 150:\ in venic'll cable, 75 N in l-ighr-side cable. 130 N in leftside cable 21. (a) ~ = 79.8 N, .,~ = ~9.9 t\ (h) 2.3<11ll/S~ 23. 613 N
A.S6
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
25. 641\ 27. (a) i.50 X JO:\ N backward (b) 50.0 m 29. (a) 7.0 m/s:! horizontal (Inc! to (he right (b) 21 N (c) 14 N horiwntal allel LO the ]-iglll 31. (a) TRA N (b) 1051\ 3!l. 2:1 m/s 35. (a) T> 111 (b) T= w (c) T< IV (d) 1.85 X 10"1 N: \'es . (e) 1.47 X 10,1 N; yes (f) 1.25 X 10\ N; yes 37. (a) 2,]5 X IU~ i'\ forward (b) 6401\ forwanl (c) 64.7 N rear-
ward (d) 1.02 X 10 4 N at 74.1° below lloriLOnL
39. 41. 43. 45. 47. 49.
JL, ~ 0.38. 1', ~ U.31 (a) 0.256 (h) 0.509 mjs:!
(a) 14.7 J-Lk
III
(b) Neither mass is necessary
= 0.287
32.1 N (a) 33 m/s (b) :--,10. The object will speed llP to 33 m/s from any lower speed and will 510\\' down to 33 m/s from any higher speed. 51. (al 1.78 m/s:! (b) 0.368 (c) 9.:n N (d) 2.m m/s 53. 3.30 m/s:.' 55. (a) 0.404 (101 45.8110 57. (aj
4 kg
(4 kg)g
(b) 2.31 m/s~, down for Ihe 4.0U-kg object, left for the
59.
61. 63. 65. 67. 69.
71. 73.
1.00-kg ol~jeCl, up for the 2.00-kgobjeCl (c) 30.0 N in the lefl cord, 24.2 N in the riglll cord (d) Without friCI ion. Ihe +-kg block falls more rreely, so the tension '1; in the string anached 10 it is reduced. The 2-kg block iii accelerated upwards at a higher rate, hence Ihe tension force '/~ acting on it must be gTeater. (a) 84.9 N upward (b) 8'1.9 N downward 50 m (a) friction between box and truck (b) 2.94 rnls:! (a) 2.22 m (b) 8.74 m/s do\\'n lhe incline (a) 0.232 m/s 2 (b) 9.68 N (a) 1.7m/s2 , 17 N (b) O.6911l/s:!, 17 N (a) 3.43 kN (b) 0.967 m/s horizontally forward (c) ROI h would increase. (a) 3U.7' (b) 0.8431\
75. 5.51 X IO~ N
77. 72.0;oJ 79. (a) 7.] X lO'l\ (hi 8.] X IO' N (c) 7.1 X J02;oJ (d) 6.5 X 10:.' N 81. (a) 0.408 lll/s2 upward (bl Fl3.3:\J
CHAPTER 5 QUICK QUIZZES 1.(C) 2. (d)
3.
(c)
4. (c)
EXAMPLE QUESTIONS I. As long as the same displacement is produced by the sallle force. doubling the load will not challge lhe amount or work done by the applied force.
2. Doubhng the displacement doubles the a",oum of",mk done in each case. 3. The wet mad would reduce the coefficient orkinClic rriction. so the final velocity would be greater. 4. (el 5. In f'arh <;-lSC lhc n'-lnril>' wOllld h,IVP::ill addition'll horizontal componenr, meaning that the overall speed would be greater. 6. A smaller angle means thai a larger initial speed would be l"Cquired wallow the grasshopper to reach the indicated heigh\.
7. 1n the presence orfriclion a different sh
MULTIPLE-CHOICE QUESTIONS 1. (e)
3. (el)
5. (c) 7. (b) 9. (b) 11. (d) 13. (hi
CONCEPTUAL QUESTIONS 1. Because no motion is taking place, the rope undergoes no displacelllenl and no wOI-k is done on it. For the same reaSOIl, no work is being done on the pullers or the ground. ',Vork is being done only wilhin the bodies orlhe pullers. For example. Lhe heart of each puller is applying forces on tIle blood to move blood through the body. 3. \-\lhen the slide is rriclionless, changing the lenglh orshape of the slide will nOl make any difference in the final speed of the child, as long as the difrerence in the heights of the upper and lower ends of the slide is kept constanLlfrriction must be considered, the path length along which the friction rOl"Ce does negative work will be greater when the slide is made long(~r or given bumps. Thus, Ihe child will arrive at lhe 10\\'el- end with less kinetic energy (and hence less speed). 5. If we ignore an)' effects due to rolling friction all the tires of· the cal·, we find that the same amount of work would be done in driving up tlte switchback alld in driving straight 11]1 the mOllntfl in hecause the weight or the car is mo\·ed upw<11Tls against gravity by the same \'crtical distance in e<Jch case. If we inclur1e friction, there is more work done in driving the switchback because the distance over which the friclion [Ol'ce acts is Illuch lange ... So \\Jhy do we use s\\'itch-
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
hack:-.:- The answer lies in the lorcc rC
\\'here
is Ihe speed of the ball as illca\'es the lOe. Aftci the with the toe, only the gravitational force and the relarding fOITe due to ail" l"csiSlance conlinue 10 (10 work on the ballthrougholll iLS flighl. J I. Yes. the total mechanical cncrg}' of the s}'slcm i~ cOll'ien'ed because the only lorces aning al"e conservali\e: the force of g-ravil)' :ll1d tile spring force. The,"e al'e twO forms ofpotelltifll eIH.:rg} in Ihis cast": grflvitalional potential energy and c::Iaslic potential energy stored in the spring. 13. Lel's a:-'~lIrne )'oulih the book slo"'I>'. III lhis case, there are two forces all the book thal are almos1 eqll:ll in m:lgnitlltlc: the lifting force and the force ofgravil)'. Thus. 1Ill' po~irive work dOIle.: by YOIl and the negative work done hy gr'l\'ity can· eel. Then.~ is no ncr \\"ol'k performed and no nel change in rIll.: kinetic energy, so lhe work-energy lheorem is sal isfied. 15. A~ lhe s..llel1ite moves in a cin:lIlar orbit abolll the Earth, liS displacement during any small lime illlerval i:-. perpendicular 10 the gr
balllose~ conlaCl
Cllllcotanl.
PROBLEMS I. 7001 3.2XIO'J 5. (a) 61.3 J (b) --16.3.1 (c) 0 (d) The work done by gra\ ity would 1101 change, the work dOlle b, the friCTion force would d('crca~e, and the work done by Ihe normal force would nOI change. 7. (a) 79.1 ~ (Il) 1.49 kJ (e) -1.·19 kl 9. (a) 2.00 lllh (b) 200:\i II. (a) 115.1 (Il) 165.1 13. (a) -5.6 X IO:!.! (b) 1.2 m 15. (a) ~.34 X 10' I\' (b) l.91 X 10- 1 s
17. 1.0 m/s J 9. '1.1 11I 21. :).11 I\'/m 23. ~Ui2 X 10" N llll\\ard 25. II = 6.94 III
A.57
Whitt:p, = I~OJ, H~hill"'1' = 290,j. addilional muscles must be invoked 29. (a) 4.30 X 10'.1 (Il) -3.97 X 10'.1 (e) 115m/s 31. (a) The mas'!'. spring, alld E.'Hth (including the \\'all) constilute Ule system. The mass and Earth interact through the :->pring fOiTC. gra\ ily, and Ihe llUlllI'i.l fUILt:. (L) tile [.Juiul or maximum eXlension, x = 0.060 0 m. and the equilibrium point,x=O (c) 1.33Jmx =6.00cm;OJatx=0 (d) ~"LVI'! + ~kx/ = it1ftv/ + ~kx,!'! --+
27.
U'!
= I UI~ + .! (x,:!
V
m
-
xl),
1.75
m/s (e)
J.51
m/s. This
anS'...er is not one-half Ule first answer because the equation is nOI linear. 33. 0.459 m 35. (a) 0.350 III (b) The result would be less than 0.350 III because some of the mechanical encr!,"}' is losl as a result of the force of fliction between the block and track. 37. (a) 10.9 m/s (b) 11.6 m/s 39. (a) Initially, all the energy is stol'ed in ule compressed spring. After the gUll is fired and the projeClile lea\'es lhe gUll, ule energy is transfelTed to the kinelic energy of lhe projeclile, resulting in a small increase in gravit.ational potemial energy. Once Ihe projectile reaches its ma.ximum height, the energy is all associated \'lith its gr'ork done hy the force of friction. (e) 15.3 m/s. 43.2.1 kN 45. 3.8 m/s 47. 289 III 49. (a) 24.5 mls (b) Yes (c) 206 m (d) Unrealistic; the actu X IO~ N 71. (a) 3.1~ m/s (Il) 4.43 m/s (e) 1.00 m 73. (") 0.588.1 (Il) 0.588.1 (e) 2.-12 m/s (d) PE, ~ 0.392.1 (e) KE,.~ 0.196.1 75. (") 0.0204 m (Il) 7.20 X 10' Kim 77. (a) 423 mi/gal (b) 7i6 mi/gal 79. (a) 28.0 m/s (h) 30.0 In (c) 89.0 m be)'"ond Ihe end of tile u'ack 81. (a) 101.1 (b) 0.410 III (c) 2.84 m/s (d) - 9.80 mm (t') 2.85 m/s 83. 9H :\/m 85. \1~1l'1 = 0, Wg ,.., = -2.0 X lOl.J. H~,,,""~l = 0, IVr,;";",, = 2.0 X 10 1 .1 87. (a) 10.2 kW (Il) 10.6 kW (e) 5.82 X 10".1 89. 4.~ m/s 91. hCI\It.:t.:n 25.2 km/h ;"Inti 27.0 klli/h
A.58
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
!l. It will be easiest to catch the medicine ball when its speed (and kinetic energy) is lowest. 'I'he firstoption-ulrowing the medicine ball at. the sallle \'clocity-will be the most dilTicull.. beC
(1/1,,,'II,)
v",= v, -
EXAMPLE QUESTIONS l. 44 m/s 2. When one cal" is ovenaking another, the rei'll ive velociT)' is small, so on impacL the change in momentum is also small. In a head-on collision, hO\l'e"er, the .dative velocity is large because the cars arc traveling ill oppositt'" directions. Consequently, lhe change in momentum ora passenger in a headon collision is greater than when hit from behind, which implies a larger ltl'cragc force. 3. The bow does Ihe same WOI"k on each arrow, so they have the same kinetic energy. By Quick Quiz G. L however. the heavier arrow will then have the greater momentum, so tIle recoil it causes will be greater, as well. 4. The final vclocily would be unaffected, but Ihe change in kinetic energy would be doubled. 5. Energy can be losl due to friClion in the mechanisms and associated thermal energy loss due to air drag and through emission of SOllnd waves. 6. No. lfthat were the case, energy could not be conserved, 7. The blocks cannol both come 10 rest at the same time because Ihen by Equation (l) momentum Ivould llot be conserved. 8. 45° 9. 1//«(/ + g)
MULTIPLE-CHOICE QUESTIONS I. (b)
3. (c) 5. (c) 7. (d)
9. (.) 11. (d)
CONCEPTUAL QUESTIONS 1. (<1) 1\'0. Ii cannot carry more kinetic energy Ihan it possesses. That would violate the law of" energy conservation. (b) Yes. By bouncing from the object it strikes, it can deliver mOI"e momenlum in a collision lhan il possesses in its flight. 3. Ifall1he kinetic energy disappears, there must be no motion of either of the objects arter the collision. If neither is moving. the final momentum of the system is zero, and the initial momentum of the system must also have been zero. A siulation in which this could be true would be the head-on collision 01" two objects ha\'ing momenta of equalmagn itude but opposite direction. 5. Initially, lhe clay has mOmenlLUll directcd tow;u-cl the wall. \\lH.:n it collidcs and st.icks to the wall, neithcl-the clay nor the wall appears to have any momentum. Thus, it is tempting 10 (wrongfully) conclude that momentum is not conse.-ved. However, the "lost" momentum is aclually imparted to the wall and Earth, callsing both to move, Because of EartJI's enormous mass, its recoil speed is too small 10 delect. i. As the water is forced Out of the holes in the <\rIn, the ann imparts a hOlizontal impulse to the \V
The third option. throwing the medicine ball with the kinetic cncrgy, will also n~duce the velocily, btu only by the square root of tile r
S
l 2
..
_./11
V~
I
=
I
it folloll"s that vm
=
v,
~'"
-. lI1'n
The slowest and easiest throw will be made when the momentum is held constant. [[you wish to check this answer, try substituting ill values of VI = I mis, 111/ = 1 kg, and 111,. = 100 kg, Then the SalllC-1ll01llClllUm throw will be c
PROBLEMS 8.35 X 1O-~1 kg' m/s (b) '1.30 kg' m/s 750kg'm/s (d) 1.78 X 10:"'9kg'm/s 31.0 m/s (b) the bullet, 3.38 x I O~J versus 69.7 J 563 kg . m/s (b) 751 N 22.0 m/s: 1.14 kg (a) 10.4 kg' m/s in the direction of the final \'clocit}' of the h.,1l (b) 173 "J J.39N·sup ;:\64 kg 'm/s fon.. . ard, 438 i\' forwaH[ (a) 8.0)1· s (b) 5.3 m/s (c) ~.3 Ill/S (a) 12N's (b) S.ON·s (I..:) 8.0 m/s,S.3 m!s (a) 9.60 X 10-' s (b) 3.65 X 10' N (c) 26.6g 65 m!s (a) 1.15m/s (b) 0.346 m!s directed opposite 10 girl's mOlion (a) J 5<1 111 (b) By Nev,'ton 's Ihird law. when the aSlronaut exerts a force on Ihe tank, the tank exerts a force back on the astron'UIL. This reaction force accelerates the astronaltL towards the spacecraft.
1. (
7. 9. 11. 13.
15. 17. 19. 21. 2:{. 25.
27.
V,hm"<"
29.
(a)
= 2.48 mis, Vi mt
VG'ld,~~ = 2.25 X JO-~ m!s
1----.-
V:, "'2 ........
(b) The collision is best. described as perfenly inelastic because the sk
Answers to Quick Quizzes, Exomple Questions, Odd+Numbered Multiple-Choice Questions, Conceptual Questions, and Problems (c) IJItV I + II'';!.,',;! = (11I 1 T 11I';!.)ll, (d) "J = (11/ 11'1 + IJI';!.H2 )/(1II 1 T 11I'2) (c) 6.3:1m/s 31. 1!l.6 m/~ 33. (a) 1.80 m/~ (h) 'l.ln X 10lJ 35. (a) I 37.1.32m 39. 57 m 41. ::!73 m/s 4:i. (a) - 6.67 cm/s, 133 cm/s (h) O.H:':9 45. 17,1 CIl1h (25.0-g o~ject). 22.1 C1H/S (1 O.O-g ol~jcn) 47. (a) O\cr a \(:ry shon timt: intcn·al. external forces ha\'e no tillle lo impart significant impul:-.e lO the plar('r~ during tht;> collision. The t\\'O players llHl\"l..' tog-ether after tht' tackle. so Ihe collision is completely incla:-.tic. (b) 2.HR 11l/1> at ~23° (c) 7BEl .J. The lost kinetic cncl'gy is tran.'ifornwd into mher form.'i of eller6')' .'iuch as lhennal cnerh'J' and sound. 49. 5.59 m/s llonh 5\. (a) 2.50 m/s ill -60" (b) elaslic collision 53. 1.78 X \0:\ N on truck driver. 8.89 X 1O~;\ on car dri\'er 55. (a) 8/3 Ill/s (incident panicle), 32/3 m/s (target p,Hlicle) (Il) -16/::1 m/s (incidem p:lrlicle). 8/31ll/s (targ-t'l pilnicle) (c) 7.1 X 10 '2.1 in case (a). :tnd 2.8 X 1O-~.1 ill C:I~e (b). The incident panicle loses mOl'C kim·tic energy in ca~(' (:1). in which the larget lll:l~S is 1.0 g. 57. 1.\ X \0:1 N (upward) 59. 71.9 N 61. (a) 3 (b) ~ 63. (a) -2.33 tn/s. '1.67 m/s (b) 0.277 m (c) 2.98111 (e1) IA91ll 65. (<1) -0.667 11l/s (b) 0.952111 (i7. (a) 3.5'-1 Ill/:-. (b) 1.77 III (c) ;:)5·1 X 10-1 :\ (d) No, lilt' normal f(HTe exerted by the rail conlrihllteS upward IllOIllt'lltlltllLO the s\-stem. li9. (a) 0.28 or 214% (b) 1.1 X IU u.1 for lhe neutron. 4.5 X 10 II .1 lor carboll 71. (a) No. Arter colliding, Lhe CMS, 11lm·ing as a unit. would lravel llonhca:-l, . . 0 Ihe")' couldn'l d<1mage properL\' 011 Ihe sOlltheast COnlL'r. (b) .\"-componetlL: 16.3 km/h . .v-component; ~1.17 klll/h, angle: the final velocity of the car i~ IK7 km/h at 29.r nonl1 of caSl, consistent with pan (a). 73. (a) '-I.8ft 11l/s (b) RAI III
75.
71 11
77. (:1)
(~)V2P,{!,d '" 1,1 llI/s at 30~ below lhe pn:-.itivc x-axis
=
(b) O.:t1 or 32%
CHAPTER 7 QUICK QUIZZES I. (c) 2. (b) 3. (b) 4. (b) 5. (a) Ii. I. (e) 2. (a) :1. (h) 7. (c)
8. (h). (c) 9. (e) 10. (eI)
EXAMPLE QUESTIONS I. Yes. The conversion factor is (1800/7r rad). 2. All gi\'en (pl,llliities and :1I1~WerS ,Ire in angul:lr IIllilS. so allel-ing the radius of the \\'IH:d Ita... no effeci 011 IhL' answers. 3. The anSWl:r \\'oHld ha\e been redllced by a factor of one-half.
A.59
4. III Ihis case, dOllhling the angular acceleration doubles lhe :Ingular displacemellt. Thill is true here because the initial angular speed i~ Icro. 5. The angular acn:lcration ora record player during pIa} is LITO. A CD player must have non-zero <1cceleration because lhe angllla.- spcl..·d llluSt change. 6. (b) 7. 1l would be illcl'l..':lsed. 8. The angle ol"lhe hallk :lnd tIle radius of the circle (!<.:Ict"minc lhe minimum safe spe~(l. 9. The nornwl force i~ ~ti1l7ero. 10. Yes. The force of g-r:lVily acting on eilch billiard ball holds lhe balls agaiml lhe table and assisb in creaLing f"ictioll forces thal aIIO\\' Ihe balls to roll. The gravity forces between tIle halls arc illsi~niricalll, however. II. Fil'st, lllostastcroids arc ilTegul:lr in . . h"pe, so Equation (1) will not apply because the acceleration may not be uniform. Second. the asteroid Illay be so small that lhere will be lIO significant or useful region wllcre the acceleration is uniform. In Ihal case. Newlon's more general law of gravitat ion would Iw required. 12. (b) 13. Mechanical energy i.< conserccd in lhis syslcm. Becan.'c lhe potelllial en{''l"gy al perigee is lower. the kinetic energy musl be higher. 14. 3 days
MULTIPLE-CHOICE QUESTIONS J. (e)
:J.
(c)
5. (c) 7. (c) 9. (d) II. (b) 13. (e)
CONCEPTUAL QUESTIONS 1. (a) The head will tend to lean lO\\'ard the righl . . houldcr (lha{ i~, toward the outside ofllle curve). (b) \",'hen there is no .~lrap_lension in Ihe neck muscles must pmduce lhe ccnlripelal acce1enllioll. (c) Wilh a stntp, the lension in Ihe ~\rap performs lhis fllllClion . allowing the neck muscles 10 n:lnain relaxerl. 3. ThL' speedOlllt'tCrwill be inaccurillc. The speedomeler mea~lIres the numln'l" of tire revolutions per .'iecond. so its readings will be LOO low. 5. TIlt" car cannol round a turn al con1>wnt velocity, becclllse "conStanl \'clocit)''' means tht" direction of tile \'c1ociIY is not ch:mging. The slOltement is correCI i["the word veiOcilY" is replaced by lhe worrl "speed." 7. Consider an individu<11 slanding against lhe inside wall of tIle cylinder \\'illl hC'r head pointcd LOward the axis orlhe C\linder. As the evlinder rotates, the person tends to move in ,1 straight-line palh tangelllto IhL' circular p(llh followed by the t:ylinder wall. As a result. lhe person presses againsl the wall, and the normal force exerted on her provirles the radinl force required 10 keep her moving in a circular path. 1ft he rotational speed is acljusted such that this normal force i.. equal ill magnitllde to her weighl on Earth. she will not be ahle to distinglli1>h between lhe artificial gravit\, ollhe colony and ordinar)' gT:I\·it)'. 9. l'he tendcllcy orlhe \\'aler is to [nove in :l straight-lint: path tangclll to the {"il'udar path followed by Ihe containcr. As a result, atlhe top orthe circulal- path, lhe \\';lter pressc'i against the haltom of the pail. anrllhe normal force cxcned k
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
A.60
by the pilil on the water provides the radial force required
LO
keep the water moving in its circu13r path. 11. Any object thaI 1l10\'es such that the diree/ion arits velocity changes has an acceleration. A cal" moving in a circular path will always have a centripetal acceleration. 1. (a) 7.27 X 10-"> rad/s (b) Because ofiL5 rotation ahoUl its axis, Eanh bulges at the equator. 3. (a) 3.2 X lOll rad (b) 5.0 X 1()7 rev 5. (a) 821 rad/5 2 (b) 4.21 x 10:1 rad 7. (a) 3.:) rad (b) The angulal-disp!acemenl increases by a facror of 4 because Equation 7.9 is quadra1 ic in the angular velocities. 9. ~lain rotor: 179 m/... = 0.522TJ"'''''d Tail rotor: 221 m/s = 0.6441i",,,nrl 11. (a) 116 rev (b) 62.1 md/s 13. 13.7 ra<.l/s~ 15. (a) 3.37 X 10- 2 m/s (h) 0 (c) The grm'ilalional fon:e and the normal force 17. (a) 0.35 m/!:;::! (h) 1.0 mls (c) 0.35 m/s::!, 0.94 mfs::!, 1.0 Ill/5 2 at 20° rorward with respect to lhe direction of U r 19. 12.0 III 21. (a) 1.10 kN (b) 2.04 times her lVeight 23. 22.6 m/s 25. (a) 18.0 Ill/5 2 (b) 900 N (c) ] .8<1; this large coefficiem is unrealistic, and she will not be able to stay on the merry-go-round. 27. (a) 9.8 "l (b) 9.8!\ (e) 6.3 m/s 29. (a) The force of static friClion (lCling lOw,u"(1 the road's ccntcr of curvature causes the briefcase's centripetal acceleration. \o\11en I.he necessarl.' cel1lripet
Earth. (a) 9.58 X 10'; m (b) 5..117 h 1.90 X 10"' kg
(a) 1.48 h (b) 7.79 X JO~ m/s (c) 6.,13 X 109 .1 1.50 h or 90.0 min (a) 9.40 rev/s (b) H.I rev/s 2 ; {I, = 2 590 m/s:!; (II = 206111/s:! (c)"-, = 514 \l; r~ = 40.7 \l 51. (a) 2.51 m/s (b) 7.90 m/s:! (c) 4.00 m/s 53. (a) 7.76 X ]0 3 111/5 (b) 89.3 min
55. (a)
II
=
m{ g - i.~~)
57. (a) FK.,m<. =
59. 61. 65. 67.
(b) 17.1 mh;
FK.apPMt'1ll
+
mRfw:!
(b) 732 N (equator), 735:'J (either pole) 11.8 km/s (a) 15.3 km (h) 1.66 X lOw kg (c) 1.13 X 10.1 s (a) 10.6 kN (b) 14.1 m/s (a) 0.71 yr (b) The departure must be timed so that the spacecraft arrives at aphelion \lillen the largcl planet is there.
69. (a) I
~
28w
g
(b) w
~
71. (a) 109 N (b) 56.4 1\' 73. (a) 106 N (h) 0.396
75. 0.131
QUICK QUIZZES I. (d) 2. (h) 3. (h)
PROBLEMS
41. 43. 45. 47. 49.
CHAPTER 8
Vii
Ir.g
4. (a) 5. (e) 6. (a)
EXAMPLE QUESTIONS 1. The revolving dool· begins
2. 3. 4.
5. 6.
1.0 move clockwise instead of cou nterclockwise. Placing the wedge closer to the doorknob increases its eITt"cti"eness. I r the woman leans backwards, the torque she exerts 011 the seesaw increases and she begins tn descend. (h) The system would remain balanced in equilibl·iuJl1. The angle tll(lde by the biceps force would still not vary much from 90° bUL Lhe length of the moment ann would be doubled, so lhe required biceps force would be reduced by nearly half. The Lension in lhe Glble would increase.
7. 8. (e)
9. Lengthening-lhe rod between balls 2 and 411'ould create Lhe lal-ger change in the moment of inertia. 10. SLepping forward Ira nsfcrs the momenLUm 01" Lhe pitcher's body to the hall. Without proper liming, the transfer will not take place or will have less efrect. 11. The magnil.Ude of the acceleration would decrease; that of Lhe Lension wOllld increase. 12. Block, ball, cylinder 13. The final anSlver wouldn't change. 14. Ilis angular speed remains Lhe same. 15. Energy conservaLion is nOL violated. The positive !le( change occurs becausc the sLudent is doing work on the system.
MULTIPLE-CHOICE QUESTIONS J. :1. 5. 7. 9. 1 J. 13.
(a)
(b) (d) (b), (e) (d) (e)
(el)
CONCEPTUAL QUESTIONS l. In order lor you LO r-emain in equilibrium. rour center of
gra\'itr must always be O\'er your point of Slippori, dle feet. If yOLlt· heels are ag'ai nst a wall, your cemcI- ofgraviLy cannOl remain above your feel when you bend forward, su you lose yOlll" balance. 3. There ar-e l.WO rl1<~or differences between I.Orque and work. The primary difference is LhaL the displacement in Lhe expression for \\'ork is direct.ed along lilc rorce, while the important distance in [he torqut: expression is perpendicular to the force. The second difference involves whelher there is motion. In the case of work, work is done only ifl.he force Sllcceeds in causing a displacement of the point of applicaLion of" I he force. By contrast, a force applied at a perpendicular dislance ("I-om a l'Olation axis results in a torque regardless ofwhethet- Lhere is mOl ion. As far as unils arc concel·ned, the maLhemalical expres~ sions lor bmh work and torque are in units tlrat are lhe prodUCI of newtons and meters, btll this product i~called ajoule in the case of" work and remains a neWl.On-metcr in lhe case or torque.
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
5. t\'o. For an object to be in c.:quilibrilltn, the nel extJ:'rn,d lorce 'KLing on it must bc zero. This is no{ possible when on Iv ol1e lorce aCls on (Ill: ol~iecl, unless Lh
As the I-adius of the S()Ul'ce reel decrease:'>. ,l\i\'en a decreasing lension. lhe torque in the source reel \\'ill d("creas(' t:n::n faster. ;IS lhe following eqllation show~:
Thi~ torque will be panly absorhed hy frirtion in tlte feed heads (which we assume to be slTlall): sOllle will be ahsorhed by friction in the source reel. Another sm:lll amount of the torqlle will be absorbed by thc increasing alll;ul
PROBLEMS I. 217 .\J 3. (:I) 30 :\. III (COlllllf'l'clm:kwisc) (b) 36 N· III (COlIIHCI"clockwise) 5. (a) :1.1 N· III (b) The torque inCITa~es. bec;lll~e the torquc is proponional to the nHllllcllt arm. I.."in U. alld litis lanor increases as (j increases. 7. I'~ = 724 :-.J. !'~ = 716 ~
9. ~H 2 N II ..\:~=~3:~rl.-,,,~= l.fi7ft
13. (-1.5
11l.
-1.5
Ill)
A.61
15. l.OI III in Figure I'~U5b; 0.015 m towards the head in Fij.{llre P8.1."Jc' 17. T=2.71 kN.R,,=2.6?)kN 19. (a) +13 N. (b) 222 N (to the right). 21(; N (upwards) 21. '1'1 = 501 N, .J~ = 672 N, T\ = :1R4 N 23. 6.1:1 III 25. 2.861.N 27. H= 107N. T
157N 29. 209 N 31. (a) 99.0 kg' m t (b) 41.0 kg' lilt (c) 14~ kg'm~ 33. (;1) 87.8 kg·1Jl2 (b) I.GI X lO~ kg (e) ·1.70 rad!:; 35. (a) 3.9~ m/s~ (11) 9.RO rad/s~ (c) As llie rope IC:l\es the cylimler. the mass of the cylinder decreases, also decreasing the q'lindel"s momellt of inertia. AI the same lime. the Illass of rope leaving the cylinder wOllld increase the torque exerted on the cylinder. Hence, Ihe accderation or the system would increase in magnitude with time. (The increase would be slil4'ht ill this case. given the large mass 01 the c)'linder.) 37. (a) ~".O N· 111 (b) O.O~.l,5 6 rad/s~ (e) 1.07 m/s~ 39. 41. 43. 45.
47. 49. 51. 53. 55.
57. 59. 61. 63. 65.
177:',J O..')~4 276.J (a) SA7 J (b) :).99 J (,I) 3.90 m/s (b) 1!l.6 rad!s I 'l~) rad!s (a) 500 J (b) 250) (e) 750 J (a) !l.OO III (b) 2.31 X IU1 J (c) 1.79 X 101J (d) 9.28 rn/s (a) 7.08 X IO:\:~.I' s b) 2.67 X IO HI .I· s 175.J·s COllllterclockwise 8.0 rev/s fi.n rad!~ :).~Iq x 10 ~.1 I, ') KI~1 (a) W = ( --'- WI! (b) -.- ~ < IIT/~ KJ~, 1 +1 1
2
67. (;1) 0.160 rad!s COlilltcrclockwi'ic (b) 99.9.1 69. (a) .\s the child walks to the right end of Lhe boat, tile boat Illo\t·s left (Iowards the pier). (b) The boat moves 1.4~1 m closer to tlw pier. so the child will be !'>.5S m fmm the pil·r. (c) Nn. I-k will he short of reaching the lunJe b\· 0.4~) m. 71. 16.Y 73. (a) Mud (h) Mu~ (c) .'h,d (d) 2v (e) 4MlI~ (f) ~~.\111'! 75. (a) 7.27 x 10:1 rad/s (h) 1.09 X lOs IlI/S. abollt ollc-third the speed of liglll 77. (a) 1.1l?! m/s (b) 51.2 rad/s 79. l.m) III 4mg MIliK ~
-,-;--,--'''- (I, I ({I = c-:--,---':-M+ -1m ,\1 + 4/1/ 83. 21}) m/s 85. 1)(/1 = -1-.5 kN 87. (a) ~.12 m!~~ (b) '1'1 = 26.7 N. '/~ = 9.37:\
7. (,) EXAMPLE QUESTIONS 1. TllllKslcn. Sled, aiuminulll, rubber 2. The lineman's skull and neck \"OLlld llll(krgo compressional Slre:-.s.
A.62
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
3. Steel. copper, mercury, \\',\l(:r 4. TIlt' prC'~sure in:,>idc a rubher balloon is greater th,ul ;UlllOspheric pressure became it lllust also excr! a fOITC ,lgainsl the elastic force of the rubber. 5. Al higher aitillldt:. the column orair above a gi\clI al"ea is progrl'ssively shoneI' anrllcss dense, so Ille weighl or the air column is reduced. Prcssurt..: is caused by 1 he weight of the air column, so the presslll-e is also reduced. 6. As fluid pours Oilithroligh a single opening. the air" inside the can above Lilt, fluid t'xpallcls imo a largcr \"ollllnC', reducing the pressure to below ;ltillospheric pressure. Air must then ('mer the same opening going the opposite dircction, resulting in disrupted fluid flow. A separatc openin~ for ait" intake Ill;tint,aills air pressure inside the call without disnlpling tlte flow of lhe fluid. 7. True 8. False:' 9. (a) 10. Tlw aluminum cuhe would float free orthe hOUlllll. 11. The speed Ofl)ll' hlond in tile.: nalTo\\'ed region illcre;lses. 12. A factor of2 13. The speed slo\\l~\\"ilh time. 14. Substitllting AI = rI~ into Eqll
9. The balance will nOl be ill equilibrium: The side with lIlC le;ld will be lower. Despite.: the factI hat the weight." un both sides of the bafance are the same, tIll' Styrofoam. due lO its larg-el- volume. will expericllce a lS lhan an equal \'Ullin Ie of the regular 1I1iX(lJre. 13. Opening the windows reslLlts in a smaller pre~slIre differellce betwecn the exterior and interior of the hOllse and, therefore, less tendl'llcy for se"ere damage..' to lhe structure dut: to the Bernoulli effect.
PROBLEMS I. I.:{
3. 5. 7. 9. Il. 13. 15.
17.
MULTIPLE-CHOICE QUESTIONS I. (a) (e) (c)
3. 5. 7. 9.
(d) (e)
II. (e)
CONCEPTUAL QUESTIONS 1. The density of air is IOWel" ill the mile-Iligh city of Denver thall il is 3tlowe.:r altitudt:::1>. so the effcel orair drag is less in Dt.:nvcr than it would be ill a cilY such as l\"ew York, The reduced air draR" means a well-hit ball will go farther. benefit· ing home-run hillcrs. On the.: other hand, curve ball pitchers prefer to throll' at lower al1ittldes where tiLe highn dcnsity ait" produces greater defleCling forces on a spinning ball. 3. She eXel"lS enough pressure IlI1 the flOOl' to denl or plll1C1I1fC the (Joor covcring. Tlu.: large pressllre is CiIIIS('d by the fact lhat her II'cight is distribuled O\'f'r Ihe "ery sillall crossseClional area of her high heels. Ifyoll are the homeowner, you Illighl want to suggest that she remove her hig-h Ireels and pill all some slippers. 5. If you think orllle grain stored in the silo as a fluid, the pressure the grain exerts 011 the walls of the si 10 incn:ase1> with incrca'iing depth,just as waler pressure ill ,1 lake illcreases with illcreasing depth, Thus, the spacing uet\\'een h
19. 21. 23.
25. 27. 29. 31. 33.
35.
37.
39.
41. 43. 45.
4i.
IIUIl
1.05 X 10 7 Pa R.,) X 10~ Pa 4.<1 111111 0.02"1 mm 1.9cm 4.74Pn (a) '2550 N (b) :-l.00 X 10 1 Pa (c) No. The pressllrl; will be the sallll' on Iv ir l he acr-obats all wear lhc S;:lme-siLe.: ~hoe. (a) - '"' x' I0 17 kg/m~l (b) The demit} of an aLOI11 is ahoUT IOL 1 Iinu.:s greater than th~ dcnsity of iron and othcr common solids and liqll ids. which sugt!;CSIS that an atom is mostly empl r ::;pacc. Liquids and solids a'i well as ga~es are mostly empt}' space. 3.'1 x 10:! m (a) 3.71 X 10'· p" (b) 3.57 X HI'1'i O.IR3 m LOS X 10' I")a 27 N· III 7'I.~J N 9.41 k:\ (a) 40S kg/Ill l (b) If t he qed objecl'1> Illass isjll~t slightly gTealcr lhan O.~IO kg, Ihell the block i1> submerged bUl the steel object and the block do not sink to the holtom, the rcaSOIl being that the st~el objeCl 110\,' starts LO displace water and 1he buoyant force is now illCrcasing. If the steel ol~jL'Cl is solid. il will displace a lillie more than 0.039 kg at" waler when ["ully submerged. The stcel objecl (Ind thc block will sink to llll: bOtLOlll when the stct:1 object's 1llaSS exceeds about 0.350 kg, and 1'01' ntlues less dwn this, bill greater than O.RIO kg, the block will he sllbmerged and the steel object pa 1't ial Iy SUbtlll'l"ged. (a) I A3 kN upward (b) l.iH kf\' upward (c) The balloon c",-pands bCGllISC the extcmal pl'e&1>ure decline1> with increasing altiLude. (a) 4.0 kN (h) i.2 ki'\ (c) The air pressure at this high altitude is much lower than aunospheJic pressure at the surfuce of Earth, so the halloons expanded and e"cl1mall)' burst. 2.67 X I O~ kg (a) 1.46 X 10·:1 Ill' (b) 2.10 X 10:\ kg/1I1~ 173 N (upper scale). 31. i K (Iom.:r scale) (a) 80 g/s (b) 0.27 millis 12.6 m/s
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems 49. (a) 9.-13 X 10: 1 Pa (b) 235 m/s (cl The d('nsit\ 01 air decrease" \\'ith increasing hcight. resulting in a smaller pressure dillerence. Beyond Ilw maxilllllrn operational altitude, the pressure dilTel-ence can 110 longer support lhe aircrafL 51. (a) 0.55~.l, s (b) 14.5 Ill/S (e) O.I·1511l/s (el) I.OI~ X 10" I'a (e) 2.00 X 10;' 1'<1; gr,lvily lenm C:1Il be lIegICClt·d. (I) 3:-\.0 t\ 53. 9.00 elll 55. 1.47 elll 57. (a) 28.0 lllh (b) 28.0 tll/s (c) 2.11 MPa
3.
59. 8.3 X 10 ~ N/m 61. 3.6 X Io-C! N/lll 63. H.G ~
65. 2.1 i\IPa 67. 2.8/oLlIl
69. 0.'11 1ll1l1 71. UN = 4.3 X JO'I; luruu!l:nl flow 73. 1.8 X 10- 3 kg/Ill:1 75.1.4 X IO-"N·,,/ lIl C! 77. (a) The buoyanl fOITes arc th(' sallie hecause the two hlocks displace equal amounts of water. (b) The sprill~ scale rCOlds largest value for the iron block. (c) B = 2.0 X 10 1 N for hOlh blocks. '1';,,,,, = 13X 10~~. T.'I"""I\\I'" = :t3 X 10:1 N.
5.
7.
79. (b) 5.9 kill 81. 2.5 X 10 7 capillal'ies 83. (a) 1.57 kPa. I ..:',!l X JO C! aun, 11.8 mm of Hg (b) The /lilid level in the lap slJould risc. (c) Blocbge or (low of the eercbrospill:lliluid
85. 2.25 In aho\'e lhe le\'el of poinl n I(f 1
87. 4.J.! X m'l 89. 6.4 m 91. 17.0 cm above Lhe floor
2. True 3. 'A'hen Ihe tempt::rature c1<.'crea'ie.s. the ICllsion ill the wire iIlCI"ea<;es. 4. The mag-nitucil- ortl1<:: n.:quircd H.:mperallllT cllHllge would Ix: larger because 1 he I incar expansion coelTiciem of steel is less than Ihal of cop peL 5. Class. aluminulIl, eLhyl alcohol, Illercury 6. The balloon expands. 7. The pre<;~ure is slightl~ reduct..:d. 8. The \'olume of air decreases. 9. True
MULTIPLE-CHOICE QUESTIONS 1. Ie)
3. 5. 7. 9.
(c) (e)
(d) (b) t 1. Ib)
CONCEPTUAL QUESTIONS I. An ordinary gla~s dish will usually bn:ak becall"C of Sl n.·sses Ill
9. 11.
A.63
less lhan the dish of ordinal-y glass and does not normally dc"elop :-.uflicit'm slress 10 cause breakaKl::. l\:lercur}' 1l1iiSt Ila\,(' the larger coefficielll of expansioll. As Ille tenlpcratun.' ora IllerlllOllwlCI' rises, bOlh lile merelll")' a Ildllw Klass expa nd. I r they bot h had! he same cocffkiem of'linCdl t:.x]J
PROBLEMS 1. la) -~60°C (b) 37.OOC (c) -280·C 3. la) -32IT,i71\ (b) 98.lioF,3.10 X IO"I\ 9. 109°F: yes. Thc Ilormal human body tempel"'lLUre i:-.98.6"F, so the pa!it'llt has a high ICver that needs immediate anenlion. II. J I em 13..~:J.OQC
23. '1..7 X IO~ N 25. 0548 gal 27. (a) incn.:(l!-,es (b) l,fi03cIn 29. (:I) 627"(: (Ill 927°C 31. (a) 2.5 X 10 1') molecules (b) 4.1 X 10-;11 mol 33. 1.28
~7kg
47. (J.G63
al 78.2 0 below Ihe horilonml
mt1l
:11
J
49. 3.5!:", L 51. 2.86'1l1ll
53. 6.;wl 7 tvl Pa 55. (;I) 99.8 !HI.. (11) The challf{t.: in volume of the nask is far sl1l:1ller hecause I')'rex has a much smalkr coefficient of
A.64
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
expansion l..hilll acetollC. Hence. the change in volurne of tile flask ha~ a negligible clTecl on the all."wer. 57. The expansion of the mercury is almost 20 lime~ that of the flask (assuming Pyrex glass). 59. 2.74 III (u" - u,)I,,(AT) 61. (a) 0 ~ -'--"--"--"----'-
5,
A,· (c) The bar bends in the opposite direction. 63. (a) 0.12 nllll (b) 96 N
CHAPTER 11
7.
QUICK QUIZZES I. (a) \"r;\lcr, gla~s. iron. (b) Iron, glass, watel'. 2. (b) The slopes arc proportional to I he reciprocal of" the specific heal, so a larger <;pecific heal n.:sults in a smaller slopt-:, meaning more energy is required to achieve a givC'll lempcratlJre change. 3. (c) 4. (bl 5. (a) 4 (bl 16 (e) 64 9.
EXAMPLE QUESTIONS 1. From t he pOint of view of physics, faster repetitions don't affect Ihe fin'll answer; phy~iologic
5. The cIH.:rgy for evaporalion comes from Ihe internal entrgy of the fluid. HowevcL this inlcrnal energy is affected by thermal energy transferred to anc! from its ellvir'onmclll, which in turn "ffeels the ralt: of ('vapormioll.
6. No 7. The Illass of icc melted would double. 8. Nickel-iron :lsteroids have a Ii igher density and 1 hercforc a greater mass, which llleans Lhey can deliver lllon' energy on impart lor a giren speed. 9. F
10. (a) 1 J. The extra layer offal is aclvaillageolls for survivill!J," cold weather and food depri\'ation. 12. (a)
MULTIPLE-CHOICE QUESTIONS I. (b)
11.
PROBLEMS 1. 1(i,9"e
3. (a) 1.67 X 10 1".1 (b) :)~.9yr 5. HfloC 7. (a) 4.5 X 10: 1.1 (b) ~)IO W (c) 0.87 Calls (d) The excess lhermal energy is I r
9. 176T II. 88 W 13. <1.2 X lOt'.!
15. 17. 19. 21. 23.
3. (e) 5. (d)
7. (d) 9. Ie) II. (d)
CONCEPTUAL QUESTIONS 1. \Vhcn you rub the surface, yOll increase tlte temperature of the rubbed region. With the metal surface. some of this energy is transferred away frum the rubbed site hy conduction. Consequelltly.lhe temperature ill the rubbed area is not as high for the metal as it is [01" the wnod, and il feels l'e1alively cooler Ihan thl' wood. 3. The fruil loses enel"gr into the ail" b} radial ion and cOlweclion [rolll ilsSlirfacf". Before icc crystals can form inside the fmil 10 I'uplllrl' cell Willis, atl of the liquid water Oil the skin will ha\'e to freeze. The resulting time delay may pre\'eTll damage withillthc fruit throughoul a frosl)' night. FunheL
a sllrf
25.
27. 29. 31. 33.
35. 37.
39.
0.845 kg HO g 1.8 X IO'J/k~·"e 0.26 kg (' = 18 min lal all ;ce mel", Tr~ 40'e (b) 8.0 g mel IS. 'I;~ () C (a) Till' bullel loses ;111 its kinetic cllergy;ls il is Slopped by rill' icc. Also. lhermal energy must be l'emOl'ed frolllllle bullet 10 cool it from 30.0°C 10 O~c. The SHm ofrhese 1\\'0 el1t'r~ gies eqllals tlie energy it takes to melt part of the icc. The fin;ll tempel'
41. 4021\1\"1' 43. 1<1 fi.'!·oF· h/BIIl
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, ond Problems
45. 9.0 em 47.0.11 kW 49. If): I 51. 12kW 53. 2.3 kg 55. 2~:rc 57. 8.00 X 1O:! J/kg' °C. This value differs from the tabulated valueb~' I-l%,so they agree within 15%. 5!). 109°C 61. 51.2°C 63. (a) 7 ~tops (b) Assumes that no energy is 10s1 to rhe sUITollnclillgs and that all internal cller!:.')· generated Stays with the brakes. 65. (b) 2.7 X LO·'.1/kg·' C 67. 12 h 69. 10.9 g 71. LA kg 73. (a) 75.0'C (b) 3(i k.1
CHAPTER 12 QUICK QUIZZES I. (b) 2. A is iso\'olumetric, B is adiahatic, C is isothermal, 0 is isobaric. 3. (e) 4. (b) 5. The number 7 is t he most probable oulcome. The numbers 2 and 12 arc the least probable olltcomes.
EXAMPLE QUESTIONS 1. No 2. No 3. True 4. True 5. The change in temperature must ,tiways be negative because lhe system docs work on the environmt:l1t at the expense of its internal energy and no thermal energy can be supplied Lo Lhe system to compensale for the loss. 6. A diatomic gas does more \\lork ,mder t hese as~tlmptions. 7. ~ 8. False 9. (a) 10. No. The efficienc)' imprm'es onl}" if Lhe ratio IQ, IQhl becomes smaller. Further, too large an increase in Qh will damage the engine, so there is a limil evcn if QI" remains fixed. 11. If the path from B to Cwere a slraight Ii ne, more work WQllld be done per cycle. 12. No. The compressor does work and warms the kitchen. \~'ith the refrigerator door open, the compl'cssor wuuld run continllotlsly. 13. False 14. Silver, lead, icc 15. F,llse 16. The thermal ener b'")' created by JOur body during the exertion would be dissipated into the environment, increasing Lllf-' enlropy of the Universe. 17. Skipping meals ciH110wcr the basal metabolism, reducing the rate aL which enf.:rgy is llsed. v\'hen a large meal is eaten laler, Ihe lower mctabolism me
MULTIPLE-CHOICE QUESTIONS I. (e) 3. (e)
5. 7. 9. II. 13. 15. L7.
A.65
(e)
(b) (e)
(d) (b) (h) (d)
CONCEPTUAL QUESTIONS 1. First, the elliciency of the automobile engine cannot exceed the Clrnotefficicncy: it is limitf.:d by the temperature of the burni ng fllel and the tempcrature of the cnvironment into which the exhaust is dumped. Second, Ihe engine block c
PROBLEMS l. (a) 7.30 X LO'.1(b) -7.50 X 10'.1 3. 5. 7. 9. 13. 15. 17. 19. 21. 23. 25.
(c) NewtOn's third law of motion (a) -6.1 X 10".1 (b) 4.6 X 10".1 (a) -8L0.1 (h) -"07.1 (e) -203.1 96.~ mg (a) 1.09 X 10" K (b) -6.RI kJ (a) -88.5.1 (h) 722.1 (a) 567) (h) 167/ (a) -180j (b) +188) (a) Zem (b) -75.1 (a) -'1.58 X 10'.1 (h) 4.58 X 10'.1 (e) 0 (a) -9.12 X 10-".1 (h) -333.1 (a) 0.95.1 (b) 3.2 X LO'.1 (e) 3.2 X LcP.1
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple.Choice Questions, Conceptual Questions, and Problems
A.66
27. (b) 1.3 X IO'~ K 29. 405 kJ 31. 0.540 (or 54.0%) 33. (a) 0.25 (or 25%) (b) 3/4 35. (a) 0.672 (or 67.2%) (b) 58.8 kW 37. (a) 0.29-1 (0)' 29.490 (b) 500J (e) 1.67 kW 39. (a) 4.51 X 10'J (b) 2.84 X 10'J (e) 68. I kg 4 I. I '3 43. 0.49'C 45. 140j/K 47. (a) -1.Hj/K (b) I.2kJ/K 49.57j K 51. 3.27 j/K 53. (a) End Resuh All II
EXAMPLE QUESTIONS 1. 5.40 N 2. 'a. If a spring issuclched 100 far, il no longer satisl-ies Hooke's law and can become permanently deformed. 3. False 4. Tnu.>
5. False 6. 7. 8. 9. 10.
(b) True No (a), (e) The speed is doubled.
MULTIPLE-CHOICE QUESTIONS I. (e) 3. (al
(bl (a) (e)
Id)
CONCEPTUAL QUESTIONS 1. No. Because the lOtal energy is £ = ~kA.2, changing the mass of the object while keeping A constalll has no effect 011 the total energ}. \'\"hen the objecl is al a disFlacemcllt x from equilibrium, the potential ellerg} is ~h:C, independent of the mass, and the kinetic energy is KE = f - ~kx:!, also independent of the mass. 3. \rhcn the spring with tWO objecb on oppositc ends is SCt imo oscillation in space, the coil althe exact cemcr of the spring does not move. Thus. we can imagine clamping the cemer coil in place without affeCling thc motion. Ifh'c do this. h'e have [\,'0 separate oscillming systems. one on e4lch side of the clamp. The half-spring on each side of the clamp has lwice the spring constant of the full spring, as shown by the following argument: The force exerted bva spring i.. proportional to the separation of tile coils as the spring is extended. Imagine that we extend a spring by a gh'en distance and measure the distance between coils. \-"e thell cut the spring in half. If onc of the half-springs is now exu~nded by thc same distance, the coils will be twicc as far aran as Ihe}' wcre in the complcle spring. Thus, it lakes twice as much force to slretch the half-spring, from which we conclude Ihat the half-spring has a spring constant which is twice that of the complete spring. Hence, our clamped system of ol~jects all twO half-springs will vibrate with a frequency that is higher than Ib)' a factor of the square root of two. 5. The bouncing ball is not an example ofsimplc hallllonic mOlion. The ball does not [allow a sinusoidal flillction for its position as a ["unction oflimc. Tht: daily movement ofa student is also not simple harmonic Illotion, because the student stays;u a fixed location. school. for a long time. If this motion were sinusoidal, the sludenl would move morc and more slowly as she approached hel' desk. and as soon as she sal down m the desk, she would start to movc back toward home ag-tlin. 7. We assume that the bUO\
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems 19. :19.21'1 21. (a) You obscn'c uniform circular motion projected on a plane perpendicular to lhe mOlion. (1)) 0.628 s 23. The horizomal dil>placemclll is de~cribed by x(1) = A cos wi. where A is lhe distance from the cenler of Ihe wheel (0 the cn"mkpin. 25. 0.63 s 27. (a) 1.0 s (hj 0.28m/' (cj 0.25 m/s 29. (a) 5.98 m/s (b) 206 N/rn (c) O.2~H m 31. (a) 11.0 N toward the left (b) 0.881 oscillaliolls 33. 1/ = ± wA sin Wi. II = - (tPA cos wi 35. (a) 1.:)05 (1)) 0.5.59 III 37. (a) slow (b) 9:47 39. (a) LLmh = 25 cm, L M... , = 9.4 em. (b) "'1..U"lh = 111\\;".. = 0.25 kg 41. {a} 4.13 cm (b) lOA cm (c) 5.56 X IQ-:! s (d) 187 cm '5 43. (
CHAPTER 14 QUICK QUIZZES I. (c) 2, (l') 3. 4. 5. 6.
(h) (h). (e) (
7. (h)
EXAMPLE QUESTIONS I. The speed of ~oulld i.') much less in rubber because the bllik modlilus of nlbbcr b much le~s than th'll of 'llullli11llm.
2. 3.0 dB 3. YOli should illcrea"t-' roul" dislancc from Lhe sound S(Jllrce hy a factor of 5. 4. Ye". It changes he('
9. True 10. True t I. The nOll'S
A.67
MULTIPLE-CHOICE QUESTIONS I. (h)
3. (a) 5. (a) 7. (
9. (e) II. (h)
13. (e) 15. (c)
CONCEPTUAL QUESTIONS 1. (a) higher (b) lower 3. The camera i" designed to operate at an assumc:d speed of sound of 345 mis, Ihe speed of ..ound aL a room LemperalUre of 23°C. If the lempcraLUre should decrease to ....a\". O°C. the speed of sound \\ill aho decrease, and the camera will respond (I) the fact that it tah's longel' for the sound 10 make its round trip. Thlls. it will operate as if the object is fanher away thMI iL really is. h is or imerest to note that bats use echo sounding like this LO locate insects or to avoid obstacles in fnlllt ofthcm, "omething thaI thc\ must do because of poor eyesight and the high ~pecds at which the} fh-. Likewise. blue \"hales use this lechnique to help them a\-oid objecls in thei. path. The need hel'c is ohl iom. because a t} pical \\,hale has a mass of 10-" kg and travels at a relathd\ fast speed of 20 mi h. so il take~ a long time [or iL to stop ilS motion or to change dircnioll. 5. Sophisticated e1eclronic de\ices break the freqllcnC\' range ofabo1ll60 to 1000 Hz used in telephone cOJl\'crsations imo sen:ral frequency bands and lhen mix them in a pl-edetermined pattern so lhallhey become unintelligible. The dcscrambler. of COil I'se. moves Lhe bands back into their proper order. 7. The echo i'i Doppler shift(·d. and Lhe shifl i"like both a 1n0\ing source and a moving obsener. The sound that leaves \our horn in the forward direclion is Doppler shifted to a higher fl"C(luency. because it is coming from a moving source. As the sound rcnect~ back and comes toward you. you are a moving obsen-er. so there is a second Doppler 'ihift to an even higher fn·quency. If the sound reOects from the spacecrall coming toward you_ there i:o. a different moving-source "hift 10 an C\'cn higher frequency. The reOecting "urface of Lhe 'ipacecrafl
PROBLEMS 1. 3.50 k11l. As long as Ihe speed of lighL is much greater than the speed ofsolllld. iLs actll
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
A.68 27. 29. 31. 33. 35. 37. 39.
0.391 m/s 19.3 m ~8°
x = 0.500111, 1.00111,1.50 m. 2.00 111, 2.50 m, 3.00 m. 3.50 III SOD III (a) 0.2-10 III (b) 0.855 III (a) Nodes at O. 2.67 m, 5.33 Ill, and 8.00 m; anti nodes at 1.33 111,4.00 m. and 6.67 m (b) 18.6 Hz 41. At 0.0891 111,0.303111,0.518 m. 0.732 111, 0.947 m. and 1.16 m from one speaker. 43. (a) 1.85 X 10- 2 kg/Ill (b) 90.6 m/5 (c) 152 N (d) 2.20 m (e) 8.33 1ll 45. la) 79 N Ihl 2.1 X 10' Hz 47. 19.976 kHz 49. 58 Hz 51. 3.0 kHz 53. (a) 0.552 111 (b) 317 Hz 55. 5.64 bcats/s 57. 3.88 m/s aW
CHAPTER 15 QUICK QUIZZES I. Ib) 2. Ib) 3. (e) 4. la)
5. (e) and (d) 6. lal
7. (el 8. (bl 9. (d) 10. Ib) and (dl
EXAMPLE QUESTIONS I.
t
2. la) 3. FOllrth quadrant 4. The sllSpendcd droplet would accelerate dowllw
MULTIPLE-CHOICE QUESTIONS I.
lal
3. (b) 5.
(h)
7. Id) 9. (b) I I.
lei
13. (dl
CONCEPTUAL QUESTIONS I. Electrons have been removed fmm (he ol~iecl. 3. No. The charge on the metallic sphere I'csides on irs outer surface, so rhe person is able to louch the surface withoul causing any charge transfer.
5. \'[o\'e an objecl A wilh a net positive charge so It is near, but nOllollching, a neutral metallic object B that is insulated from the ground. The presence of A will polarize 8. causing an excess negative charge to exislon the side nearest A and an excess positive charge of equal magnitude La exist all the side farthest from A. While.A is still nenr B, touch Bwith your hane!. Additional electrons will then flow from ground, lhrough raUl' body and onto B. With A. continuing to be near but not in contact with B, remove your hand from B, thus trapping the excess electrons on B. \I,.'hcl1 A.. is now removed, B is left with excess electrons, or a net negative charge. By means of O1ulllal repulsion, t.his negative charge will now spread uniformly over the emire surface of B. 7, An objecl's mass decreases very slightly (illlllleasmably) when it is given a positive charge, because it loses electrons. \\ihen the object is given a negative charge, its lllass increases slightly because it gains electrons. 9. Electric field lines stan on positive charges and end on negative charges. Thus, if the fair·wealher field is directed i!l1o the ground. the ground must han: a negative charge. 11. The e1eclric shielding effeCl of conductors depends on the fact thatthe.·e al'e two kinds of charge: positive and negativc. As a result, charges can move within the conductor so that the combinat ion of positive and negarive charges establishes an electric field that exactly cancels the external field within the condllctQl' and any c:wil.ies inside the conducLOr. There is onl)' one type of gravitat ion charge. howeYer, because there is no negative mass. As a result. gravitational shielding is not possible. A room cannot be gravitationally shielded because mass is always POSil ive or zero. never negative. 13. \Vhen the comb is nearh}'. charges separate on the paper, and the paper is al.lracted to the comb. After conlact, charges from the comb arc transferred to the paper, so that it has the same type of charge as the comb. The paper is thus repelled. 15. YOII can only conclude that the net charge inside the Gaussian surface is positive.
PROBLEMS I. 8.7 X 10-s N; thc force is repllisive. 3. (a) 2.36 X 10-[' C (b) The chal-ges induce opposite charges in the bulkhc5 J 25. 0.849 m 27. ].8 III to the left of I he -2.S·.uC charge 29. zero 35. (a) 0 (b) [I.ue inside, -5 fLC outside (c) a inside, ~5 fLC outside (d) 0 inside, -5 J1.C outside 37. 1.3 X 10-3 C 39_ (a) 2.54 X lO 15 N (b) 1.59 X 1O~ N/C radiall)' outward 41. lal 858 N· m'/C IblO lei 657 N· m'/C 43. -QI EU for 51; 0 for 52; -2Q1 EO for 53; 0 for 8'1 45. (a) 0 (b) lV/I,,?' outward
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple·Choice Questions, Conceptual Questions, and Problems 47. (a) -7.99 N/C (b) 0 (e) 1.44 N/C (d) ~.OO nC on "'e inner surface; 1.00 ne on the outer surface 49. 115 N 51. 24 N/C ill the positive x-di."ection 53. (a) E:::: 211,qb ({/~ + b~)-3n in the positive x-direClion (b) E= h,Qb«(I~ + b2)-~/2 inthc positive x-direction 55. (a) 4_64 X \0-2 III (b) 2.54 X 10- 2 m 57. 4.4 X 10' N/C 59. - 10-' C 61. (a) 0 (b) 7.99 X 10' N/C (ou,,,,ard) (c) 0 (d) 7.34 X 101; r\/C (outward) 63. (a) l.00 X 103 N/C (b) 3.37 X 10-11 Ii (c) accelerate at 1.76 X 101.' m/s:! in the direction opposite thai of the eleCtric field
7.
9.
CHAPTER 16 QUICK QUIZZES I. (b)
II.
2. (a) 3. (b) 4. (d) 5. (d) 6. (e) 7. (a) 8. (e) 9. (a) Cdecrcascs. (b) Qstays [he same. (c) Esta)'s the same. (d) ,iVincreases. (e) The energy stored increases. 10. (a) Cino-eases. (b) Qincreascs. (c) ESTays the same. (d) ~\ 'remains the same. (e) The energy slOl"ed increases. II. (a)
EXAMPLE QUESTIOHS 1. True 2. Tnle 3. False 4. la) 5. Three 6. Each answer would be reduced by a factor of one-half. 7. One-quarter 8. The voltage drop is (he smallest across the 24-ILF capacitor and largest across the 3.0-j1.F capaciwr. 9. Tlie 3.0-,uF capacitor 10. Ie) II. (b)
MULTIPLE·CHOICE QUESTIOHS I. 3. 5. 7. 9. II.
(b)
(b) la) (b) la) (e)
COHCEPTUAL QUESTIONS I. (a) The protOn moves in a straight line with constalll acceleration in the direction of the electric field. (b) As its vclocil}' increases. its kineLIc energy inCt-eascs and Ihe c1eclI'ic potcnti,t} energy associated \\'ith the prol.on decreases_ 3. The work done in pulling the capacitor plates farther apan is transferred into additional electric energy stored in the capacitol-. The charge is constant and the capacitance decreases, bm the potential difference between the plates increases, which results in an incl-ease in the stored electric energy. 5. If the power linc makes eleCtrical contact with the metal of the car. i[ wi[1 raise the potemial of the cal' to 20 kV. It will <\[so raise the potential OfrOlll- body (Q 20 kV, because YOll
13.
A.69
are in contact with t.he car. In itself, [his is not iI problem_ If YOll step out of the car, however, yOUl" body a[ 20 kV will make contact lI'ith the grollnrl, which is at zero volts. As a result, a current will pass Through yOUl- body and you will likely be irUured. Thus. it is best to stay in the car until help arrives. [[tWO points on a conducting objcct WCI'C at differ-cnt potentials. then free charges in the object would move and we \\'ould not have static conditions, in contradiction to the initial assumption. (Free posiLive charges would migrale from locations of higher to locations of lower potential. Free electrons wOllld ]-apidly mO\'e from locations of lower to locations of higher parential.) All of the charges would continue 10 move 1I1ltilthe pOTe1l1ial became equal e\'cry",hc\-e in [he conductor. The capacitor often l-emains charged long after the voltage source is disconnected. This residual charge can be lethal. The capacitor can be safely handled afTer discharging the plates by short-circuiting the device with a conductor. sl1ch as a screwdriver with '-In insulating handle. Field lines represent the direction of the electric force on a posit.ivc test charge. I f electric field lines were to cross, then. a( the point ofcrossing. There would be an ambiguity regarding the direction of the force on the lest charge, because there would be two possible forces there. Thus, eleCtric field lines cannot cross. [\ is possible for equipotcntial surfaces LO cross. (1Iowe,-er. equipotelllial surfaces at different potentials cannot inTersect.) For example. suppose t\\'o identical positive charges are at diagonally opposite corners of a square and twO negative charges of equal magnitude are aT the other two corners_ Then the planes perpendicular to the sides of I he square at their midpoinTS are equipO[elllial su.-faces" These (WO planes cross each other 3tthe line per penrlicular 10 the square at its celller. VOIl should lise a dieleCtric-filled capacitor whose die!e,tl'ic constant is \'cl") l.uge. Further. you should lll
PROBLEMS l. (a) 1.92 X 1O- 1.'1J (b) -1.92 X 1O- 18 J (c) 2.03 X 106 m/s in
lhe negati\'c x-direClion 3. lAX 1O- 20 J
5. 1.7 X 10" N/C 7. (a) 1.13 X 10'N/C (b) 1.80 X 10-'" N
Ie) 4.38 X 10-".1
9. (a) The spring stretches by 4.36 cm. (b) Equilibrium: x= 2.18cm;A
=
2.IBcm (c) 1t'=
2/u1 2
Q
II. (a) -5.irl X [0-7 V (b) -1.92 X 10-7V, ~F= 3.84 X 10- 7 V (c) :"Jo. L'nless fixed in place, (he electron would move in the opposite direction, increasing its distance fl"om points A and JJ and lowering The potential difference between them. 13. (a) 2.67 X 10 6 V (b) 2.13 X 10 6 V 15. (a) 103 \' (b) -3.85 X 10-7 J; positive \\-ork must be done 10 separaTe The charges_ 17. -11.0 kV 19. (a) 3.84 X 10-".1 Ib) 2.55 X 10-".1 (e) -2.17 X 10-".1 (d) 8.08 X 10 6 mls (e) 1.24 X 10 7 mls 21. 0.719111, 1.4<1 m. 2_88 m. No. The equipotentials are not uniformly spaced. Ins(ead, the radius of an equipolcnial is inversely proportional to the potential. 23. 2.7-1 X 10- 14 III 25. (a)l.1 X IO-'F (b)27C 27. (a) 1.36 pF (b) 16.3 pC Ie) 8.00 x 10" V/m 29. (a) 11.1 kV/m lOwardthc negative plate (b) 3.74 pF (c) 7-1.7 pC and -74_7 pC 31. (a) 5.90 X IO- ro Fib) 3.54 X 10-; C (e) 2.00 X 10' N/C (d) 1.77 X 1O-11 elm:! (e) All the answers are reduced.
A.70
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
011 each capacimr (b) 15.0 p.,C 011 the 2.50-tLF capacitor and 37.5 p..C on thc 6.25-p.,F capacitor (a) 2.67 p,F (b) 24.0,u.C on each 8.00-p.,F capacitor, 18.0 tLC on the 6.00-,u.F capacitor, 6.00 p.,C on the 2.00-J.lF capacitor (c) 3.00 V across each capacitor (
33. (a) LO.7 p..C
35.
37.
39. 41.
43. 45. 47. 49.
5.
7.
9.
85.0 V.
51. (a) 8.13 nF Ib) 2.40 kV 53. (a) volllme 9.09 X 10- 16 m3, ,1I"ea ·1·.54 X W-w m~ (b) 2.01 X 10-" F (c) 2.01 X 10- 14 C. 1.26 X 10" electronic charges 59. 4.29 p.F
61. 6.25!,F 63. 4.47 kV
65. 0.75 me on C1, 0.2., mC on C''.!. 67. Sphere A: 0.800 J-I-C; sphere B: 1.20 p,C
CHAPTER 17 QUICK QUIZZES I. lei)
2. 3. 4. 5.
(b) Ie), (eI) Ib)
(1)) 6. (b) 7. (a) 8. (b) 9. (a) 10. (e)
EXAMPLE QU ESTIONS l. 1\'0. Such a current corresponds to the passage of one elec-
tron c\'ery 2 seconds. The cl\'cragc currenl, however, can have ilnyvalilc. 2. Tl·ue
3. Higher 4. (b)
5. (a) 6. (e)
MULTIPLE-CHOICE QUESTIONS I. (e)
3. (eI) 5. (b)
7. Ib) 9. (e) II. (a) 13. (eI)
11.
\.lIed lhroughollllhe metal. The concept of charges residing on the surface ofa metal is true for a metal with ,Ill cxcess charge. The !lumber of free elect rons in an electrically ncutral piece of melal is the same as the number of positive ions-lhe metal has zero net charge. A voltage is not something that "surges through" a completed cit"cuit. A volulge is a potential difference that is applied across a device or a circuit. It would be more correct to say "1 ampere of electricity surged through the victim's body." Ahhough this amount of current would have disastrous results on the human body, a value of I (ampere) doesn't sound as exciting for a newspaper article as 10 000 ('lolls). A nOI hel- possibility is to ""rile "} 0 000 volts of elect ricit)" \verc applied across the victi Ill'S body," which still doesn't sound quiLe as excit i ng. \Ve would conclude th~H the conductor is nonohmic. Once the switch is closed. the line volLage is <tpplied across the bulb. As the \'oltage is applied across the cold filament when it is first turntcd on, the resistance of the f\1
PROBLEMS I. 3.00 X 1020 electrons move past in the direction opposite to the currcnt. 3. 1.05 mA 5. (a) n is unaffected (b) Vii is doubled 7. 27 yr 9. (a) 55.85 X lO-j kg/mol (b) 1.41 X 10" mol/m'l (c) 8.49 X 10 2/\ iron atoms/m 3 (d) 1.70 X 102!1 conduction elcctl'OllS/m J (e) 2.21 X 10- 1 mls 11. 32 V is 200 times large!' than 0.1 f) V 13. (a) 13.0 fl (b) 17.0 m 15. (a) 30 fl (b) ."1.7 x 10-' fl· m 17. silvcl- (p = 1.!l9 X 1O- Il U'm) 19. 256 23. 1.98 A 25. 2200 'c 27. 26 rnA 29. (a) 3.0 A 11» 2.9 A 31. (a) 1.2 n (b) 8.0 X 10- 1 (a 0.080% increase) 33. (a) 8.33 Alb) 14.'1 fl 35.2.1 W 37. 11.2 min 39. 3'1.'1 fl 41. 1.6cm
n
43. 15 p,\,V
45. 23 cents 47. $1.2 49.2.16 X 105 C
CONCEPTUAL QUESTIONS J. Char,g-c. Bccal/sean ampere isa lIniLofcurrenl (1 A = I Cis) and an hour is a unit of time (1 h = 3600 s), then 1 A· h = 3600 C. 3. The gravitational force plilling t he electron to I he bottom of a piece of metal i~ much smaller Ihan the t:1ectdcal repulsion pushing the electrons "pan. Thus, free electrons Slar dislrib-
51.6.7W 53. 1.1 km ti5. 1.'17 X 10- 6 n . m: differs by 2.0% from value in Table 17.1 57. (a) $3.06 (b) No. The circuit must be able LO handle at least 261\. 59. (a) 667 A (b) 50.0 km 61. 3.77 X I02f''/m'l
Answers to Quick
Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
63. 3i .\lfl 65. 0.48 kg/' 67. (a) 470 \V (b) 1.60 mill or marc (c) 2.93 mill or more
EXAMPLE QUESTIONS 1. It would reduce the final current. 2. (il) 3. The 8.00-n resistor 4. The answers would be negative, but the)" wonld have the same magnitude as before. 5. Yes 6. (a) i. (a)
MULTIPLE-CHOICE QUESTIONS I. (e) (e) (e) (a)
3, 5. i. 9. 11. 13.
(d) (il). (e) (d)
CONCEPTUAL QUESTIONS 1. No. When a ballery sen'es as a SOlOTe and supplies current ..:t. the conventional current flows through the batlery from the negative terminal 10 the positive one. However, when a source hadng a larger emf than the battery is lIsed to charge the bauery. the conventional current is forced to flow through the banery from the posil ive terminal [Q the negative one. 3. The LOtal amount of energ)' delivered by the ballery will be less than l·V. Recall that a baltery can be considered an ideal. resislf\l1celcss batterr in series "'ith the imemal resistance. When the batteql is being charged, the energy deli\'ered to il includes the energy necessary to charge the ideal ballcry. pillS the energy that goes into raising the temperallll'e of the b
A.71
9. She will not be electrocuted if she holds OntO only one highvoltage wirc, because she is not completing a circuil. There is no pOle11lial difference across he,- body as long as she clings to only one wire. However, she should release the wire illlmt:diatel)" ollce il breaks. because she will become part of a closed cil Luil when ~he ,-e,a:hes the ground or comes into contact with another object. 11. The bird is resling on a wire of fixed potential. In order to be elcctmcllted, a large potential diflerence is re
PROBLEMS I. 4.92 fl 3. 73.8 'N. Your circuit diagram will consist of IWO O.SOO-!) resis-
tors in series with the 192-fl resistance of the bulb. 5. (a) Ii.1 fl (il) 1.99 A 10" 4.00 fl and 9.00 fl, 1.17 A lor i.OO fl, 0.818 A lor 10.0 n
7. iR/3 9. (a) 0.227 A (b) 5.68 \' 11.55fl 13. 0.43 A 15. (a) ConneCltWO 50-f} resistors in parallel, and then connect this combination in series h'ith a 20-n resistor. (b) Connect t\\·o 50~n resistors in parallel, connect tWO 20-fi resistors in parallel, and then conneCI these two combina· tions ill series with cach other. Ii. 0.iI4A, 1.29 A, 12.6\' 19. (a) 3.00 mA (il) -19.0V (e) 4.50V 21. (a) I, ~ 0.492 A, I, ~ 0.148 A, I, ~ 0.639 A (il) 6.i8 W to the 28.0-fl resislOr. 1.78 W to the 12.0-fl resistol-, 6.53 W 10 the 16.0-f! \'esistor 23. (a) 0.385 mA, 3.08 111A, 2.69 mA (b) 69.2 \', with {"at the higher potential 25. (a) No. The only simplification is to note that the 2.0-!1 and 4.0-.fl resistors are in series and add to a resistance 01"6.0 I1. Likewise. the 5.0-!1 and 1.Q-.fl resistors are in sel"ies and add [0 a resistance 01'6.0 n. The circuit cannot be simplified any further. Kirchhoffs rules must be used to analyze tJ1e circuit. (b) 11 = 3.5 A. 1:1 = 2.5 A. l"j = 1.0 A 27. (a) No. The 1l11l1tiloop circuit cannot be simplified any furlher. KirchlloWs rules mUSI be used to analy1.e the circuit. (b) 130 = 0.353 A directed to the right, f~ = 0.118 A directed to tlte right, 1'211 = 0.471 A directed to the left 29. J. \''2 = 3.05 V, fi \1,\ = 4.57 \1, fiF., = 7.38 V. n \'5 = 1.62 V 31. (a) 1.88s(il) 1.90X 10-'C
33. 1.3 X 10-" C 35. 37. 39. 41.
(a) 0.43 s (il) 6.0 "I' 481ightbl.llbs (al 6.25 A (il) i50 IV (a) 1.2 X 10- 9 C. 7.3 X 10!1 K+ ions. Not large, only 1 e/290 A~ (b) 1.7 X 1Q-!1 C, 1.0 x 10 10 Na+ ions (c) U.83 /LA (d) i.5 X 10-".1
43. 11 n\'V
45. 7.5 n 4i. (a) 15 fl (b) 1I = 1.0 A, I~ = /: 1 = 0.50 A, r.~ = 0.30 A. and ' 5 = 0.20 A (e) (t> I')", ~ 6.0 V, ('" V)" ~ 1.2 V, (t> I')", ~ (t> I')!' ~ 1.8 V. (~I')" ~ 3.0 V, (t> I'),,, ~ 6.0 V (d) ?f',,, ~ 6.0 w. '!Po ~ 0.60 IV, '!Pm ~ 0.54 W. 71'/, ~ 0.36 W, 'lP"t = 1.5 \0\', CJ1'dh = 6.0 VI 49. (a) 12.4 V (b) 9.65 V
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
A.72
51. (a) R~rn = 3R,R~~''''d
=
e'
e'
2R(b)1J"lloCn = 31?'~d""'d = 21?
(c) Lamps A and B increase in brigillness.lamp C goes oui.
53. 112 V. 0.200 n 55. (a) I? t = R~ - tR[ (b) R" = 2.8 n (inadcqU
59. '3'=-,-
(1/
+ 10.00)'
61. (a) 5.G~ V (b) 0.227 II 63. 0.395 II, 1.50 V 65. (a) 2.41 X 10 ·'C (b) 1.61 X 1O-'C (e) 1.61 X 10-'A 67. (a) 6.0 X lO' \I' (b) 1.2./ (e) 37 m
CHAPTER 19 QUICK QUIZZES I. (h)
2. (e) 3. (e) 4. (a)
5. (a), (e) 6. (b)
EXAMPLE QUESTIONS 1. The force on the electron isopposilC the dir"cetion of the force on the prowl1. and the acceleration of the electron i!l far greatcr than the acceleration on the prown due to the electron's lower mass. 2. A magnetic field always eXCrls a force perpendicular to the \'clocit}' ora charged panicle. so it call change the panicle's direcrion butllot its ~peed. 3. Zcro 4. A.. the angle approaches 90°, the magnilllcJe or the force increases. After going beyond 90°, it decre<\scs. 5. The magnilllde of the momentum remai ns constant. The direction ofmomelllum changes. unless the panicle's velocit)" is parallel or antiparallcl to thc magnel ic ficld.
6. 20m 7. It would have the same magnilude and point to the right. 8. Earth\ magnetic field is extremel)" .... eak. so the ClllTCnt C.\I·· ried by the car \,'ould have to be correspondingly \'cry large. Such a current would he imp'-:Jclical to genel',ue, and if gencratcd. it would heal and mell the Wil'CS carrying iL 9. The proton would accelcratt:; dowllward dnc to gravilY while circling.
MULTlPLE·CHOICE QUESTIONS I. (d) 3. (e) 5. (d) i. (e)
9. (d), (el
II. (a)
13. (e), (d) 15. (a)
CONCEPTUAL QUESTIONS I. Thc St't ~hOllld be oriemed such that tht:; beam i::. mo,'ing
eithl'r toward the cast or toward the \\eSL 3. Tht:; magnetic force on a moving charged panicle is always perpendicular to the panicle's di reel ion of mOl ion, There is no magnetic force on the charge when it moves parallclLo the direction of the magnetic field. Ilowe\"cr. the force on a charged panicle mO\-ing in an eleclric field is ne\'er zero ilnd is always parallel to the direction of the fidd. Therefore. b, projecting the chftrged particle in different directions. it is po'i.!>ible to determine the nalllre oftbe field. 5. The magnetic field produces a magnetic force on the elec· troll.!> moving toward Ihe screen that produce the image. This magnetic force deflects the electrons to regions on the scr(;~n other than the ones 10 which they arc supposed to go, The result is ft dislOrtcd imag"t:. 7. Sitch !c\'itaLion cOllld 1Ic"er occur, Atlhe I\'onh Pole. where Earlh's magnetic field is direcled downward, toward the eClllivalcm ofa buried south pole, a coffin would be repelled ifit'i sottLh magnetic pole were dil'eeted downward. HoweH:-r. eCluilibrium would be only tl-ansit.ol"y, as any slight disturballce would upset the balance belween the magnetic force and the gravilmional force. 9, If VOLI were mo\ing along with the electrons. you would mC;lSlIre a zero current for the electrons, so lherwould not proch Ice a magnct ic field according lO your observations. Ilowever, the fixed positi ....e cbarges in the metal would now be ll1o\'ing backwards relative La rou, crealing a currem cquivalenLLO the forward motion ofille electrons when }'OU \\'cre stationary. Thus, you \\'ould measure the samc magnetic field as when you \\{"I'e stationary, but it would be due LO the po"iLi\"e charges presumed to be moving from your poim of \'icw. n, A compass does nOi deteCt currents in wires ncar light switche~. for twO reasons. The fir~t is that, because the cable to the light switch contains two wircs, one carrying current to the switch and the other carrying it. awar from the switch, the netlllagnClic field would be vcry small and wculd fal! off rapidl} with increasing disLance. The second reason is that the current is alternating
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
It Fl
53. 55. 57. 59. 61. 63. 65. 67.
.)
".
":'-1
Ii • -..!.-;
Ii Z
! F
69. 71.
15. Each coil oflhc Slink)!» will become a magnet., becHI~e a coil acts a.s a ClilTenl loop. The $ense of rot.ation of t.he Cllrrenl is the same in all coils, so each coil becomes a magnet ,,'ith the same orientation of poles, Thus,
73. 75,
(a) 3.0 4.5 3.2
A.73
4.00 III (b) 7.50 nT (c) J.26 m (d) zem X 10 ~ N/m; altraclivc mill
A (a) 920 tllrns (b) 12 em (a) 2.8 JLT (b) 0.89 mA (a) 5.3 JLT into the page (b) 7.2 cm (a) 0.500 p,T OUI of the page (b) 3.89p.T parallel 10 .\)'-plane and <1t59.0" clockwise from + x-direction (a) 1.3:\ m/s (b) the sign of the emf is independclll of lhe charge 53 /LT IowaI'd the bOllom of the page, 20,u.T towanl the hOllom of the page, and 0 (a) - 8.00 X 10-" kg· m/s (h) 8.90° 1.41 X 10- 1; N
CHAPTER 20 QUICK QUIZZES
PROBLEMS I. (a) The negative z~direction (b) The positive z-direction (c) The magnetic force is zero in this case.
3. (,,) inLO the page (b) toward the right (c) t(Mard the bOllom of the page 5. (<1) 1.44 X 1O-1':! \! (b) 8.62 X 10 14 m/s':! (c) A force wOlild be exerled on the electron that had Ihe same magniLUdc as the force on a proton. but in the opposite dircuion because of its negative ch:lrgc. TIle acceleration of the elect ron would be much greater than Ihat of the prolon because the mass of Ihe electron is much smaller. 7. 2.83 X 10' m/s west 9. 1.5 X 10-':! T into the page (the negative z-direclion) II. I'~ = 8.93 X 10 ~o N (downward), J'~ = 1.60 X 10- 17 r\ (upward), J'~n = 4.80 X 10- 17 N (downward) 13. 8.0 X 10-.'1 T in the +z-direction 15. (a) into the page (b) loward the ri~ht (c) toward the bOllom of the page 17. 7.50 N 19. 0.131 T (dO\vfl\\'ard) 21. (a) The magnetic force and the force of gnwil)' bOlh act on the wire. \·Vhen the magnetic force is IIp\',,'ard and bal,mces the dowm\'ard forcc of gradty, Ihe net force on the wire is zero, ;:mel the wire can move llpward
1. 2. 3. 4. 5. 6.
h, C, (al (hi
tI
(e)
(h) (h)
EXAMPLE QUESTIONS I. False
2. 5.06 X 10-3 V: thl.: current would be in the opposite direclion. 3. (a). (e) 4. :\ magnl.:tic force directed to the right will be exerted on the
bar. 5. Doubling the freqllcncydollbles the maximum induced emf. 6. (h) 7. (a)
8. 72.6 A 9. (h)
10. False
MULTIPLE-CHOICE QUESTIONS I. (e) 3. (e) 5. (
9. (h), (e) II. (al
CONCEPTUAL QUESTIONS I. Acco1"din~ to Faraday\ law. an emf is induced in a wire loop
if the magnetic flux tht'ough the loop changes wilh lime. In this situalion. an emf can be induced either by l"Otaling the loop around:-ln ad)itr
A.74
Answers to Quick Quizzes, Example Questions,
Odd~Numbered Multiple-Choice
were nOI a continuous baml. this cmfwould cause high. yoltage
$B,llel
=0
11.9o.JmV 13. 2.7 Tis 15. (a) 3.77 X JO-~T (b) 9.'12 X 1O-3T (e) 7.07 X 10- 1 111 2 (d) 3.99 X IO-oWb (e) 1.77 X 10-5 V; the average induced elllf is equal to the instantaneous in (his case because the ClllTcnl increases steadil}'. (n The induced emf is small, so the curn;lll in tILe 4-turn coil and its magnetic nclcl will also be small. 17. 1O.2p.V 19. 2.6 mV 21. (a) toward lhe east (b) 4.58 X 10- 1 V 23. 0.763 V 25. (a) from lefl to right (b) from righl to lefl 27. into the p~lge 29. (a) from righl 10 lefl (b) from "ight to lcft (cl from left (0 right (d) from left 10 riglH 31. (a) F= ,v'!B'2 w'2 v /Rto the left (b) 0 (e) F= ;V'!/3'lll,'2 vIRto the left 33. 13.3 \' 35. 1.9 x 10 37. (a) 60 V (b) 57 \' (e) 0.13 s 39. (a) IB.Ip.V (b) 0 41. ·l.0 mH 43. (a) 2.0 mH (b) 3BA/s
II,·
45. (a) 4.0 illS (b) 0.-15 V 47. (a) 2.4 V (b) 75 ntH (cl 5.1 A ~I"R ~ -1.5 \'; ~\·I. ~ -0.9 \. 1.92 n (a) 0.20Bmll (b) 0.936 mj (a) 18j (b) 7.2j negali\'e (\~ < \'.) (a) 20.0 ms (b) 37.9 V (el 1.52 IllV (d) 5l.S mA (a) 0.500 A (b) 2.00 \\. (e) 2.00 W
(d)
61. 115 kV 63. (a) 0.157 IllV (end Jj is posili\'e) (b) 5.89 IllV (end A is positivc) 65. (al 9.00 A (b) 10.8 N (c) b is ;ltthe higher pOlenl inl (d) Ko
CHAPTER 21 QUICK QUIZZES I. (e) 2. (b)
7. (b), (c) 8. (b). (d)
EXAMPLE QUESTIONS 1. False
2. 3. 4. 5.
False True True The a,erage pm.,.cr will be zero if the phase angle is 90' or -90", i.e. when R::o O. 6. False 7. (d) 8. Yes. When the roof is perpendicular to the sun's rap the amount of energv intel-cepted by the roof is a maximum. AI OIher angles the amount is smaller.
MULTIPLE-CHOICE QUESTIONS
9. 1.5 IllV
49. 51. 53. 55. 57. 59.
3. (b) 4. (a) 5. (a) 6. (b)
9. (e)
PROBLEMS J. 4.B X JO-'T'm' 3. (a) O. J77 T (b) 0 5. (a) q'B.,..., = 0 (b) 0 7. (a) 3.1 X 10 :iT-ill:! (b)
Questions, Conceptual Questions, ond Problems
I. 3. 5. 7. 9. 11. 13.
(e) (a)
(b) (d) (a)
(e) (e)
CONCEPTUAL QUESTIONS 1. For best reception, the length or tile antenna should be parallel to the orientation of the oscillating electric field. Because of
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems ~hines, as does human skin, the same color for all races, Clothing is dark as a rule, but your seat and the chair seal glow alike after you stand up. YOllr face appears lit from within, like ajack-o'-Ianlern; your nostrils and the opening"s ofyollr eon canals an: briglll; brighler still al-e the pupils of yOllr C)'C:S. 11. No. The wire will emit electromagnetic waves only if the curn;:nt varies in time. The I'adiation is the resultof
PROBLEMS I. 3. 5. 9. II.
(a) 96.0 V (b) 136 V (e) 11.3 A (d) 768 IV 70.7 V. 2.95 A 6.76 IV 4.0 X 10' Hz 17/LF
15, 0.7£)0 II
17. OA50T'1I1~ (a) 0.361 A (b) 18.1 V (e) 23.9 V (d) -53.0' (a) 1.43 kG (b) 0.0979 A (c) 51.l e (d) voltage lead~ Ulrn~nt (a) 89.6 V (b) 108 V
19. 21. 23. 25. 27.
1.88 V (a) 78.5 f! (b) 1.59 kf! (e) 1.52 kfl (d) 138 ",A (e) -8·1.3' ,6.\111 = 20.7 \', ~\fL:= 10.8 V, ~\c = 2.20 X 10:! V (a) 103 V (b) 150 V (e) J27 II (d) 23.6 II
(r)
29. 31. 33. 35. 37.
(a) 208 fl (b) 40.0 fl (e) 0.541 H 4.30 W (a) 1.8 X lO' fl (b) 0.71 H (a) 5.81 X 10 7 Hz (b) Yes; the resistance of the circuit is not needed. The circuit's resonance frequency is found b) equaling the inductive reactance to the capacitive reactance, which leads 10 Equation 21.19.
39,
emi " = 4.9 llF, ema >: = 51nF
41. 43. 45. 47. 49. 51.
0.242J 721 II
53,
55, 57. 59. 61. 63.
65,
0.18% is lost (a) 1.1 X 10' kIV (b) 3.1 X 10' A (e) 8.3 X 10' A I 000 kill; there will al\\'a)'s be better usc for tax money,
1.10x 1O-,iT X J0 14 I Iz,./iR := 3.19 X lOH Hz, E"'d>:/Emu.i:= 0.57 l~',,,,,~ = 1.01 X 10~ Vim, BIIl •• ~ := 3,35 X 10-6 T 1.Yl l X IOllm/s !lAS X 10 11 Hz (a) 188111 to 556 III (b) 2.78 m to 3.'l m 5.2 X 10 1:1 Hz, 5.8 J.L1ll 4.29999984 X W l 'l Hl: -1.6 X 10 7 liz (the frequency decl'cases) (a) 184 fl (b) 1,53 mA (e) 1.44 H l.7cellls
L.,t := 4.55
67. 69. 71. 99.6m1I 73. (a) 6,7 X 10- 11) T (b) rl.3 X 10 17 W/I1l~ (e) 1.7 X 10-1-1 W 75. (a) 0.536 N (b) 8.93 X 10-5 m/s~ (c) 33.9 days
CHAPTER 22 QUICK QUIZZES I. (a)
2. 13ealllS 2 and 4 are reflected; beams 3 and 5 are rcfl'acted.
CONCEPTUAL QUESTIONS I. (a) Away from the normal (b) increases (c) remains the same 3. No, the inrormation in the cat.alog is incorrect. The index of refraction isgivcll by 11:= c/v, where cis the speed of light in a vacuum and vis the speed of light in the material. Because light travels [aste]' in a vacuum than in any Olher material, it is impossible for the index ofrefnletion of any matcri;l! to ha\'c a value less than I. 5. There is no dependence of the angle of reflection on wave· length, because: the light does not emer deeply into the malerial during reflection-itl'eflects rrom the sud'ace. 7. On the;: one hand, a ball covered with mirrors sparkles by reflecting light from ilS surface. On llle other haneL a faceted diamond lets in Jig-Ill at the LOp, reflects it by lotal inLernal reflection in the botTom half, and sends the liglll OUI throllgh the top again. Because or its high index of refraclion, the critical angle for diamond in air for total internal ,-eflcClion, Ilntllely 0r:= sill-I(n"ir!ndi.''''''''d)' is small. Thus. lighl rays cnter thl"Ough a lal"ge area and exitlhrough a ,'er)' slllall area with a much higher intensity. '-Vhen a diamond i~ immersed in carbon disulfide, the critical angle is increased to Or = sin-l(lI'Mboll
PROBLEMS L :\.00 X 1011 m/s 3, (a) 2.07 X 10~ eV (b) 4 14 cV
A.76
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
5. (a) 2.25 X 1U~ m/s (b) J .97 X 108 m/s (c) 1.24 X 1O~ m/s 7. (a) 29' (b) 2li' (e) 32' 9. 19.5° above the horizontal II. (a) 1.52 (b) 417 nm (c) 4.74 X 10 1•1 Hz (d) 1.98 X IO" m/s 13. 111' 15. (a) 1.559 X 108 m/s (b) 329.J nm (c) 4.738 X IO).! Hz:
3.
17. five times from the right-hand mirror and six times from lhe left 19. 8 ~ 30.4', 0' 21. 6.39 IlS 23. (b) 4.7 cm 25. 0 = tan- l (lI..) 27. 3.39 m
(a) FCllse (b) False (c) True (b) An infinite number (a) False (b) True (c) False
EXAMPLE QUESTIONS 1. )Jo 2. (b) 3. True 4. (b)
5. (a) 6. (e) 7. The screen should be placed one focal length away from {he lens. 8. No. The largesl IllClgnificarion would be 1.80 a diverging lens would not make a good magnifying glass. 9. Upriglll
CONCEPTUAL QUESTIONS I. You will not be able to focus your eyes on both the piclllre and your image at the same time. To foclls on the piclUre, you mllSt adjust yOlll- eres so thaI an object several centimeters away (the picture) is in focus. Thus, rOll are focHsing on
13.
the min'or surface_ BUI yOlll' image in the mirror is as far behind lhe mirror as you are in front ofil. ThIIS, you must focus YOllr eyes beyond the mirror, twice as far away as the picture to bring lhe image illlO focllS. A single Oat min'or forms a virtual image of an object due lu twu faLlul~. Filst, Lilt; lighlla)"s flOlnthe object arc necessarily diverging from the object, and second, the lack of curvalUre of the flat mirror canno{ convert diverging rays to converging I-ars. I r another optical element is first used 1O calise light rays LO converge, then the natlllirror can be placed in the region in which lIlC converging rays are pres· ent. and it will change the direction of tile rays so lhat the renl image is formed at n different location. For example, if a real image is formed by a convex lens. and the flat mirror is placed between the lens and the image position. the image formed by the mirmr will be real. Light rays diverge from the position ofa dnual imagcjust as they do from an actual objecl. Thus, a virtual image can be ClS easily photographed as any object. can. Of course, the camera would have [0 be placed near the axis of the lens or mirror in order to intercept the liglH rays. We consider the {wo frees [Q be th'O separate objects. The far tree is an object that is fat·ther from the lens than the near [I·ce. Thus. lhe image of the far tree will be closer LO the lens than the image of the near ll·ee. The screen must be moved closer 10 the lens (Q pUI the far tree in focus. If a cOllverging lens is placed in a liquid having an index of ref"action larger than thaI of the lens material, the direction of refractions at the lens surfaces will be reversed, and the lens will diverge liglH. A min-or depends onl)' on reflection which is independent of the surrounding material, so a COllverging minor will be col1\'el'ging ill any liquid. This is a possible scenario. \\'hen light crosses a boundary between air and ice, it will refn'tct in the same manner as it does when crossing a boundary of lhe same shape between ail' and glass. Thus, a cOlwcrging lens may be made from ice as well ;15 glass. Howevel', ice is such a strong absorber of infrared radiation lhat it is unlikely you will be able to start a fire with a small ice lens. The focal length for a mirror is determined by the law of refleClion fmm the mirror SUI-face. The law ofrefleClion is indcpendclIl of the material of which the mirror is made and of the surrounding medium. Thus. the focal length depends oilly 011 lhe radius of curvature and not on the material. The focal length ofa lens depends on the indices of refractiall of the lens material and surrounding medium. Thus. the focal length of a lens depends on the lens maleri
PROBLEMS 1. on the order of 10-9 s younger 3. 10.0 ft. 30.0 ft, 40.0 ft 7. (a) 13.3 crn in fmllL of the mirror (b) -0.333; the image is real and inverted as in Figure 23.13a. 9. (a) q = -7.50 cm, so the image is behind the mirror. (b) The magnification is 0.25, so the image has a height 01'0.50 cm and is upl"ighl. 11.5.00 cm 13. 1.0 m 15. 8.05 cm 17. (a) 5.0 em behind the milTor (b) -10.0 cm 19. (a) concave with focallengthJ= 0.83 m (b) Object must be 1.0 m in front of the mirl"Or. 21. 38.2 cm below the upper surface of the ice 23.3.8 mm 25. 11 = 2.00 27. (a) The image is 0.790 m fmm the OUler surface oflhe glass pane, inside the water tank. (b) \,,"'ith a glass pane of negligible thickness, lile image is 0.i50 In inside the tank. (c) The
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple+Choice Questions, Conceptual Questions, and Problems
29. 31.
33.
thicker {he gl'lss, the greater [he distance between the final image and the outer surface of the glass. 20.0 em (a) 20.0 em beyond the lens; real, ilwenecl,:VI = -1.00 (b) :\0 image is formed. Parallel rays leave the lens. (c.;) 10.0 un in flOllt oftht:: lens; virtll'll, lIlJ,-ight. M = +2.00 (a) 13.3 cm in front of the lens, virtual, upright. M = +I/:~ (b) 10.0 Clll in front of the lens. viruml. upl"ight. ,\/ = + 1/2 (c) 6.67 elll in front of tile lens, Vil"tllal, upriglu. IH = +2/3 (a) either 9.63 cm 01" 3.27 cm (b) 2.10 em (a) 39.0 111m (b) 39.5 mm 40.0 cm 30.0 em to tile left of the second lens, M = -3.00 7.47 cm in fronl of the second lens; 1.07 cm; virllla!, upright from 0.22l l m to 18.2 III real image, 5.71 em in from of the mirror
35. 37. 39. 41. 43. 45. 47. 49. 38.6° 51. 160 Clll to the left of the lens. inverted. AI = -0.800 53. q = 10.7 cm 55. 32.0 cm to the right of the second surface (real image) 57. (a) 20.0 cm to tlte right of the second lens: I'd = -6.00 (b) inverted (c) 6.67 cm to tile right of rhe second lens: AI = -2.00: inverted 59. (a) J .99 (b) 10.0 cm 10 the left of the lens (c) inverted 61. (a) 5.45 m to the left of the lens (b) 8.24 III 1O the left of dIe lens (c) 17.1 III [Q the left of the lens (d) by slln"ollnding the Icnswith a medium having a refracti"e index greater than that of tlte lens material. 63. (a) 263 cm (b) 79.0 cm
CHAPTER 24 QUICK QUIZZES I. (c) (c) (c)
2. 3. 4. 5. 6.
(bi (b) The compaCl disc
EXAMPLE QUESTIONS I. False 2. The soap film is thicker in the region II1;1t reflects red light. 3. The coating should be thinner. 4. The wavelength is smaller in water than in air. so the dis1;"WCe between cbrk bands is also slllal1e ... 5. False 6. Because the ,,·avclength becomes smaller in woter, the angles 10 the first maxima become smaller. resulting in a smaller central maximum. 7. The separmioll between principal maxima will be larger. 8. The additional polarizer must make rln angle of 90° with respect to the previous polarizer.
MULTIPLE·CHOICE QUESTIONS I. (a)
3. (b) 5. (b) 7. (e)
9. (e1)
CONCEPTUAL QUESTIONS I. YOll will nor see an interference paltern from the automobile headlights, for two reasons. The first is that the headlights
A.77
are not coherent sources and are therefore incapable of producing sustained interference. Also, the headlights 'He so far apan in comparison \0 the wavelengths emitted that. even if I hey were made into coherent SOUI"CeS, the intel"fC,·ence maxima and minima would be too closely spaced to be obsen'abh:::. 3. The result of the double slil is to redistribute the energy aniving 'lt the screen. Although there is no energy at the location ofa (\nrk fringe, there is f'our times as much energ), at the location ofa bright fringe as there would be with ani)' a single narrow slit. The total amount of" energy arriving at the screen is twice as llluch as with a single slit. as it mllst be according 10 the law of conserv,nion of energy. 5. One of the materials has a higher index of refraction than wate!". and the other has a lower index. The material with the higher index will appear black as it approaches zero 1hickness. There will be a 180" phase change for the light reflected from the upper surface, but 110 such phase change for the light reflected from1he lower surface. because the index of refraction for water on the other side is lower than that of the film. Thus, the two reflections will be out of phase and will illlerfere des1ructi\'el)'. The material with index of refraction lower than waleI' will have a phase change for the light reflected from both the upper and lhe lower surfHce. so that the reflections from the zero-thick.ness film will be back in phase and the lilm will appear bright. 7. Becallse the light reflecting at the lower surface of the film undergoes a 180° phase change, while light reflecting from the upper surface of the film does not undergo sHch a change. the central Spot (where (he film has neal" zero thickness) h'ill be dark. lfthe observed rings are not circular, the cllrved surface of tile lens does not have a true spherical shape. 9. For regional communication ell the Earth's surface, radio wa,'cs are typically broadcast from currents oscillating in tall vertical towers. These waves have vertical planes of polarization. Lighl originates from the vibrations ofalOllls or electronic transitions within alOlllS. which represent oscillations in all possible directions. Thus. light generally is not pol;1rized. 11. Yes. Jn order to do this. first measure the radar reflectivit)' of the metal of your airplane. Then choose a light. durable materialthal has approximately half the radar renectivit)' of the metal in your plane. J\·teasurc its index of refraction. and place onto the metal a coating equal in thickness to onequarter of 3 cm, divided by that index. Sell the plane quickly. and then )'0\1 can sell the supposed enemy new radars operating al 1.5 cm. which the coated metal will reOeCl with extrahigh efficiency. 13. Jfrol! wish to perform an interference experiment. YOll need monochromatic coherent light. To obtain it, YOll must first pass light from an ordinary source through a prism or diffraClion grating to disperse differem colors into different directions. Using a single narrow slit, select a single color and make 1hat lighl diffract to cover both slits for a Young's experiment. The pl"OccdHl"c is much simpler with a laser because its output is already monochromatic and coherent.
7. ('l) 55.7m (b) ]24 m 9. 5IS nm 11. 11.3m 13. 148 III 15. 75.0 III 17. 85.4nm
A.78
Answers to Quick Quizzes, Example Questions, Odd~NumberedMultiple-Choice Questions, Conceptuol Questions, and Problems
19. 21. 23. 25. 27.
550 nm 0.500 Clll (a) 238 nm (b) A will increase (c) 328 nm 4.35 ?till 4.75 /-lom 29. No, the wavelengths intensified al'c 276 nm, 138 nm, 92.0 nm, .. 31. (a) 8.10 111m (b) 4.05 mm 33. (a) 1.1 m (h) 1.7 mill 35. 462 nm 37. 1.20 mill. 1.20 mill 39. (a) 479 nm, 647 nm. 698n111 (b) 20.5 ,28.3°,30.7° 41. 5.91° in first order; 13.2° in second ordcl'; and 26.5° in third order 43. 44.5 cm 45. 514 nm 47. 9.13 Clll 49. (aj 25.6' (bj 19.U" 51. (aj 1.11 (b) 42.0" 53. (a) 56.7~ (b) 4H.8~ 55. (aj ';5.6 (bI34.4" 59. 6.89 units 61. (a) 413.7 nm. 409.7 nm (b) 8.6° 63. One slil 0.12 rnm wide. The central maximum is lwice as wide as the other maxirna. 65.0.J56mm 67. 2.50 mm 69. (a) 16.6 III (b) 8.28 m 71. 127m 73. 0.350 mm
CHAPTER 25 QUICK QUIZZES I. (eJ 2. (aj
3.
5.
7.
9.
II. 13.
example is p
PROBLEMS I. 7.0 3. 177 m 5. 1.4 7. 8.U
2. True 3. A smaller focal length gives a greater Ill,lg-nificalion and should be selected. 4. True 5. Ye.~.ltlcreasing the focal lcngth of the mirror increases lhe magnilication.lncreasing the focallengl.h of the eyepiece decreases the magnification. 6. More widel}'-spaced eyes increase visual resolving power by effectively inneasing the aperture siz.~, J). in Eqllalion 25.10. The limiting angle of resolulion is thereby decreased, meaning liner details of dislant objeCfs can be resolved. 7. Resolution is beller a1 the violet end ofille visible spectrum. 8. True
MULTIPLE-CHOICE QUESTIONS
CONCEPTUAL QUESTIONS 1. The obser\'er is nut using the lens as a simple l1l
9. II. 13. 15.
'10 (a) (a) (a) (a) (a)
j.Llll
8.00 X 102 (b) The im(lge is inverted. 1.0 X lO: l lines (b) 3.3 X 102 lines +2.67 dioplcrs (b) 0."16 diopter too low +44.6 diopters (b) 3.03 dioplers m = 4-.0 (b) m = 3.0
CHAPTER 26 QUICK QUIZZES 1. False: lhe speed orJight is r for all obselTers. 2. (a)
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
3. False 4. No. From
perspcClive you're at re,')t wifh respect 10 the cabin, so you will measure yourself as having YOUI' normal length, and will require ,I normal-sited cabin. 5. (a), (e); (a), (e) 6. False 7. (a) False (b) False (c) Truc (d) False )'0111'
EXAMPLE QUESTIONS I. 9.61 s
2. (e) 3. (a) 4. (al 5, Verv lillIe of the mass is convened to other forms of' energy in tl1ese reactions becallse the total III I 111ber ofnClitfons and praLOns doesn't change. The energy liberated is only the energy associated with their imeractions.
CONCEPTUAL QUESTIONS I, An ellipsoid. The dimcnsion in the direction or motion would be measured to be less th
5, The light from the quasar moves at :,.00 X lOil m/s. The speed of light is indepcndent of the 1II00ion of the source or the observcr. 7. For a wonderful fictional exploration of this question, get a "f\lI1'. Tompkins" book by George Gamow. All of the rc1ativit), effects would be obvious in our lives. Time dilation and length contraction \\'ould both occur. Driving home in a hurry, you would push on the gas pedal nO[ 10 increasc your specd "cI'y much, but to make thc blocks shorter, Big Doppler shirts in wave rrequencies would make red lights look green
PROBLEMS L (a) 1.38)T (b) 1.31 lightyears
3
.v'oi ,'
5. (a) 1.3 X 10- 7 s (b) 381l\ (e) 7.6 m 7, (a) 1.55 X 10- 5 s (b) 7.09 (c) 2.19 X 10- 6 S (d) 649 III (e) F"OIll the third obscrvel''s pOint of\'ie\\', the muon is tra\'cling fasler, so according to the third observer, the muon's H(etimc is longer than that measured by the obs<.;rver at rest with respect (Q Earth.
9. 0.950r II. Yes, with 19 m 10 spare 13. (a) 1.09 X 10-23 kg' "lIs (b) 1.89 X 10-" kg' Ollis 15, O.285r
17. (a) 1.35 X IOIfiJ (bl 1.75 X 10"'.1 (el 3.10 X 10'" J
A.79
19. 0.786r 21. ISA g/cm 3 25. (
CHAPTER 27 QUICK QUIZZES I. 2. 3. 4, 5.
True (b) (e) False (e)
EXAMPLE QUESTIONS I, False 2, False 3. Some of the photon's energy is transft:rred to the electron. 4. Doubling the speed ofa panicle doubles its momentum, reducing the particle's wavelength by a factor of one*half. This ans\>Jcr is no longer true when the doubled speed is rel
MULTIPLE-CHOICE QUESTIONS I. (a)
3. (e) 5. (d)
i.
(e)
9. (b)
CONCEPTUAL QUESTIONS I. The shape o[an objeci is normall} determined hy observing the light ITflening from its surface. In a kiln. the o~ject will be very hot and will be glowing reel. The emitted radiarion is far stronger than the reOected radiation, and the lhermal radiation emiued is on Iy slighlly dependent on lhe material fram which the o~iect is made, Unlike reflected light, the emitted light comes from all surfaces with equal intenSity, so contrast is lost and the shape of the object is harder 10 discern. 3. The "blackness" ofa bl
A.BO
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
electron is measured by the sLopping potential. The reverse voltage (sLOpping vollage) reqllired to stop (he Cllrrent is proportional to the frequency of the incoming liglrL More intense light consists of more phOLOIlS 51 ri king a unit area each second, bUI atoms al"e so small thaL one emitted decu"on never gets
"kick" from more than one photoll. Inc."cas-
ing the ilnensi!} of the light will generally illcrease the size of the current, but will nOI change the energy of the individual electrons that
PROBLEMS 1. (a) =3 000
I,
hydmgcn-Iikc ,HOlliS are proponional to l~, where Zis the atomic nllmber. 3. Tile quantum IlUmbel's /I and arc never negative. 4. No. The rvl shell is at a higher energy; hellce, transitions from the 1\,1 shcllw the I{ shell will always ."esult in morc energelic photons than all)' transit ion from I he L sl 1I.:1 I LO tilt: I{ shell.
e
MULTIPLE·CHOICE QUESTIONS I. (al
3. (e) 5. (0) 7. (b). (0)
9.
CONCEPTUAL QUESTIONS I. r f the energ")' of tllC hydrog"cn atOll} were pmporl ionall.O /I (or any powel' of 11), {hen the energy would become infinite
3.
(1)) =20 000 K
3. 8.2 X 10:\ K 5, (a) 2.4~J X 10-,1 t'V (b) 2.'19 cV (c) 249 eV
7. 2.27 X 10'111 pholons/s 9. (a) 2.24 eV (b) 555 nm (c) 5.4J X 10H H L 11. (a) 1.02 X 10- 1"'.1 (b) l.~,g X lOl', liz (c) 19G nm (d) 2.15 eV (0) 2.1., Y
13. 4.8 X lOll Hz, 2.0 eY 15. 1.2 X 1O~ V and 1.2 X 10i V, respectivcly ]7.l7,8kV 19. 0.078 nm 21. 0.28111nl
23. 70° 25. 1.18 X
1O-~3
5.
kg· mis, '178 eV
27. (a) lAG km/s (b) 7.28 X 10-11 29. 3.58 X 10- 13 m 31. (a) ]5 keY (1)) 1.2 x W keY 33. - lOti !HIs
111
116 m/s 3 x W-"'} (a) 2.~1 X 1O-:1~ kg' m/s (h) 1.00 X lOU) Hz (c)
35. 37. 39. 41. 43.
EXAMPLE QUESTIONS I. The energy associated with the quantum 1l1lmbcl- 11 increases with increasing qllClntLtlll numh.·1 n, gOillg LO zero in till: limit of arbitrarily large n. A transilion from a vel')' high energy le"ellO the ground stare therefore results in the emis"ion of photons approaching an energy of 13.6 cV, the same as the ionization energy. 2. The energy difference ill the helium alOm \\'ill be four times thar of the same transition in hydrogen. Energy le\'els in
(e)
13.
15.
as /I grew to infinity. But the energy of the atom is ilwersel}' proponional to 1J'.!. Thus, as 11 groh's to illfinity, the energy or the atom approaches a value that is above the ground state bY:l finite a1110\1nl, namely, Ihe ionization energy 13.6eV A~ tile electron falls from one bound state to another, its energy lo,>s is always less than thc ionization energy. The energ} and fn.:qllency orallY cmiued photon are finite. The characleristic x-rays orip;inate from transitions within the atoms or the target, such as an electron from the L shell making a transition to a vacancy in the K shell. The \'acallq' is caused when an accelerated electron in the x-ray lUbe supplies energy to Ihe K shcll e1eClron to eject it from the alom. If the energy of the bombarding elenrolls were 1.0 be increased, the K shell elcClron will be ejected from the alOlll wit h morc remaining kinel ic energy. Bill the energy difference between the K and I. shell has not changed, so the emiLled x-I'ay has exactly the same wavelength. 1\ continuous speclruIll without characteristic x-rays is possible. At a low accelerating potenti,11 difference for the electron. the electron may not have enough energy to eject all eleclron from a tal'gel aLOIll. As a result, there ",ill be no characteristic x-rays. The change in speed of the electron as it Clllel'S lhe cargcl will result in the cominllous spectrum. The hologram is an interference paLtern between light scattcn::d from the object and the reference beam. [fanything moves by a distance comparable lO the wavelength of the light (or more), the pauern will wash out. The eHeCl isjust like m
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems ton pa~sing b)' an excited aram stimulates that atom to emit phOIOII, the emitted photon i~ in pha~e with the ori~inal phOlon and tr,lVeis in the same direction. As Ihis pm cess is repeated m
2. OOllblingthe inilial mass of radioactive llla£el'ial doubles tht: initial ani\·iry. Doubling the Illas.~ has no effect 011 lhc half-
tl
life.
PROBLEMS 1. (a) 121.5 nm, IO~",:') nm, 97.20 nm (b) F
II. 13. 15. 17. 19. 21. 23.
25. 27.
as thc atom. (a) 0.212 nm (b) 9.95 X W-:!!" kg'm/s (c) ~.11 X (d) 3AO eV (e) -6.80 eV (f) -3.40 eV E=-1.5IeV(n=3)wE=-3A-OcV(11=2) (a) 2.86 cV (b) 0.472 cV (a) 13.6 eV (b) 1.51 e\' (a) '188 nm (b) 0.8H m/s 4.42 X ID'1 m/s (a) 0.-176 nlll ( b) 0.997 11m (;1) 0.0265 11m (b) O.Oli 6nm (1.:) 0.01:-{ 2 om 1.33 11m n t wt Ill,
3 3 3 3 3 3 3 3 3 3
2 2 2 2 2 2 2 2 2 2
3.7.79--1 X 10-1:1.1.4.69 X 1011J 4. The binding cllcrg-y of carbon-ItI <;hould be greatel,than nitrogell-14 because nilrogen ha~ more protons in its llucleus. The mUllial repulsion of the protons means thatlhc nil rogen nuclells WOllld rcquire less energy to brc
MULTIPLE-CHOICE QUESTIONS L (e) 3. (d) 5. (a)
lO~IJ'S
7. (e) 9. (h) I J. (e). (e)
CONCEPTUAL QUESTIONS 1. IsotOpe~ of a ~i\'en element correspond to nuclei with dif-
37. L shell: 11.7 ke\'; .\·1 shell: 10.0 ke\'::'\! shell: 2.::1,0 ke\' 39. (71) 10.2 eV (b) 7.88 X 10.1 K 4 L (a) 5.,10 keV (b) 3.40 eV (e) 5.~9 keV 43. (a) --1.2·1 X 10 1", \\"lm:! (b) 1.20 X 1Q-1:! J
9.
11.
CHAPTER 29 QUICK QUIZZES I. False
2. (e) 3, (e)
4. (a) ..w d (b) 5. (b)
EXAMPLE QUESTIONS I. Tritiulll has lhe grealer binding en("l·~Y. Unlike tritiulll, helillll1 has two prolOns thar exert a rcplllsi\'c electroslatic fOITe on e ligllll}' bound as the lritiullliluclell~.
13.
fcrenl Tlul1lben ofnt::utrons. This h'ill'Tsull in a v,u·iety of differenl physical properties for lhe nllclei, including the obdoll.~ olle of mass. The chemical behavior. howc\·el·, is gm'emed b) til(" elemen"s electrons. All isolopes ofa given element 11;1\'c the .~:lmc nu mbcr or electrons and, thcrefi,re. the ~amc chcmical beh not ali\'e to ulke up carbon from the e1l\"irollIl1CnL Only the ages ofobjecls that were ()I1CC ;llive can be CStitlldled br the electrostatic force. r r enough prornns were pl.) Both must Iravel al Ihe speed of light and are capable oflran:-.fclTing both ellerKY alld lIHJlllel1lllll1. They differ in that the phOlO1l has spin (il1lrillsic aflgutar momenrum) Ii and is involved in
A.82
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
electromagnetic interactions, while the neulrino hets spin
11/2. and is c10sdy related LO bela decays.
PR08LEMS I. A :::: 2, ,. =: 1.5 fill; A == 60. r:::: 4.7 fm; A ::::: 197, r = 7.0 fm;
A:::: 239. r= 7.'1 fm 3. l.8 X 10'.! III 5. (a) 27.61\ (b)4.16 X 10" mls' (c) 1.73 MeV 7. (a) 1.9 X 10 7 m/s (b) 7.1 r-,'!eV 9. (a) 8.1 ~lcV (b) 8.7 \'!cV (el 8.61\-1eV II. 3.5·'1 MeV 13. 0.210 \leV/nucleon greater for n:--Ja, allribulable to less p.-oIOn repulsion 15. 0.46 Ci 17. (a) 6.9:) X 10;' S (b) 9.98 X 10-7 S-I (c) 1.9 X ]0 1 decays/s (d) 1.9 X 10 111 nuclei (c) 5.0.200 mCi 19. 4.31 X 10 3 yl 21. (a) 5.5R X "to '.! h- 1• 1~.'1 h (b) 2.39 X 10 1:1 llnclci (c) 1.9 mCi 23. (a) ~~~i* (b) '.!;JPb (c) ~~Co
25. e+ c1ecar. ~~Co --7 ~~Fe + e+ + JI 27. (a) canllot occur spol\laneously (b) call occur spontaneollsly 29. 18.6 keV 31. 4.22 X 10:1 vr 33. (a) nr\e (b) 1~.lXc (c) X = e+. x' = " 35. (a) lie (b) I~B 37. (a) 1~4Al\ + n --+ I~~I-Ig + c + Ji (b) 7.88 MeV 39. (a) II)n (b) Fluoride mass = 18.000953 u 41. 18.8J 43. 24 d 45. (a) 8.97 X 1011 t'lcLirollS (b) O.lOOJ (c) 100 nld 47. (a) 3.18 X JO 7 mol (b) 1.92 X IO li nuclei (c) 1.08 X 10H Sq (d) 8.96 X JU 6 Bq 49. 46.5 d 51. (a) 4.0 X 10 9 yr (b) II could be no older. The I"Ock could be younger if some Si'Sr were initially present. 53. 54,u.Ci 55. --t.":! X .to II kg/h
CONCEPTUAL QUESTIONS 1. The experiment described is a nice analogy to the Rutherford sCaltcring cxperilllcnt. In the RutherfoH! experiment, alpha panicles were scauered fl"Om atoms and the scanering was consistent with a small structure in Ihe atom containing the positi\'c charge. 3. The largest charge quark is 2r/3, so (l combination or only 1\,'0 panicles. a quark and an aIlliquark forming a meson.
could not have an electric charge of +2e. Only panicles containing Ihree quarks, each with a charge 01'21:'/3, can combine to produce a total charge or2e. Until aboul 700 000 years after the Big gang, the temperature or the Universe was high enough 1'01' any atoms that formed to be ionized by ambient radialion. Once the average radiation energy dropped below the hydrogcn ionization energy or 13.6 eV. hydmgcn atoms could form and rcmain as ncmral atoms for n:lativcly long period or time. In the quark model, all hadrons :11"e composed of smaller units called quarks. Quarks hav(; a fractional e1enric charge and a baryon number of k. There are six flavors of quarks: up (II), clown (d), strange (s), charmed (c), top (t), and hOllom (b). All baryons contain three quarks. and all mesons con1;lin onc qU
29. 31. 33. 35.
126 tvJcV (a) 16.2 kg (b) 117 g 2.9 X 10:1 km (=1 800 mile~) 1.01 g (a) ~Be (b) I~C (c) 7.27 l\lcV 3.07 X 10~2 evcllls/yr H.I MeV (a) 4.53 X 10~:\ Hz (b) 0.622 fm 67.5 MeV, 67.5 MeV/c, 1.63 X lO:.!~ I-It: (a) conservation of electron-lepton number and consen'alion of muon-lepton number (b) conservalioll or charge (c) conservation or baryon number (d) conservation or baryon I1\llnber (e) conservation of charge lip. (a) charge, bal)'on number, L,. L T (b) charge, baryon number, Lt' Lp.' L.,. (c) charge. Lt' L,.., L.,., strangeness number (d) charge, baryon number, L" L,.., LT' strangeness number (c) chClrge, baryon number, L,", L,.." L.,., strangeness number (0 charge, baryon number, Lt , L.. , L T , strangeness number 3.34 X 10 20 electrons. 9.36 X lO~i; up quarks, 8.70 X 102li down quarks (a)~:+ (b) ,,- (c) K" (d) 2
37. ~6 39. (b) 12 days 41. 29.8 MeV 43. 3.60 X 10:18 prolons/s
INDEX
Pa~c numhers foIlO\..<.-d b\Jand II-efer ligures and table... respecti\-e1y_
to
A Ab...olute pre'isure. 28;{ Absolute tCmpcl·;lIllre. 325~326, 326J Absolme zero. 325. :-\26 Ab:loorbcrs. ideal, ~n:-\ Absorption ...('leCli\'e. polarization of Ii gIlt wa\'C!i b), 809-811.
S09/. 810J r\b<;orption spectra, 893 AC. Se,Altemating CUITCnt i\cceleration. 31-33, -47. See also Angular acceleralion: CCllu'ipelal accelel-at ion average (a) in one dimension. 31-32. 47 in IWO dimemion:-.. 61. 74 constant definition of. 3.~ in one dimension. 35-41. 35f,-I7 dccelenniotl lIS. negalive accelc.:ralion,32 lree-fall. ·12, ~081 inSt (a) in one dilllCll . . inn. 32-33, 47 on vclocit) us. time grrlph, 32, 3~/ in IWO dimcnsions, 61, 74 lIegati\e. 32, :1'1 ~e\\'ton's second 1;1\\," ancl. 86~88.
9i-IOI, 108, 109 peak. 10 problem-solving strategies. 36-37 in simple harmonic mOlion as funci ion of posit ion, 427-428.4;')0 ;\5 function of time, 437, 437j, 451 LOI
Adhc.·...i\c lo!'ce::>. 302-~'\W\ ,\diahalic index, 393. 395 ,\di"batic proces:loes. 390. 417 1"'diOigrams, 393-395. 39-1/ 'H)rk done Oil gas in. 393-393, -417 Al'fohic fitness. -115-116,1161 Ail. S/'/' (I1~oAlInosphere ('oll\(.'ction C1IITents in. 371 dic.:If.:(tric (onstal1l, 55fil c1i(·leClric strength. 558t indc.·x of refraction. 7381 spt't:d of :)ound in. -162. 162/. -163 th('rm;}1 conducli\'it~_ 3ti7t Ail bags. and momentum reduction. 166 Air col II III lb. standing \,-a\(':lo in, -480--184.
1%,218 rOlalion;\1 motion under, con ..lant ac('clcl·ation. 19-1-195,218 1Or(llI(' and. 239-241,254 LOlal ;l{"(('!t.·Lttlion. 200-201 ,\nglllar di ..placcmem. 191, 191/ An~uLlI frequency (w) in pendulums. 439 ill :loilllpk hannonic Illotion. --13·1. 1:-\:1-·136.450 :\ngulal lIlag-nilication ofsilllpll' mag-nifier. 829-830. 8·11 of"lcll'scoIW. 833, B41
Ll
1.2
Index
Angular momentum (L), 2-19-234, 2501 conservation of, 250-234 definition of, 249 of electromagnetic wave, 717, 724 torque and, 249-254 Angulal- position, 191 Angular quantities l"e1ation to linear quantities, 194, 196-199,218,228-229 as vectors, 202 Angular speed ClTerage (w~J, 191, 218 instantaneous (w), 192 tangential speed and, 196. 218 Anglllar\'c1ocity (~), direction of, 202,202J Antenna, electromagnetic wave production by, 714-715. 714f,741J Antifreeze testing devices. 286, 2861 Antimaner, sTorage of, 6-10 Antinelltrinos. 924. 932 A ntillode, of standing wave, "175--176, 475J, 'ISO, 4S1J Antiparticles. 918, 924, 944-945. 9461,956 AiHiqu
Astronomy (Cull/intwl) comets, impaCl 011 Earth, 142,364-36:') Crab Nebula. 250, 25]J cun-"llure of spacetime, 864 dark energy, 269 dark mattCI-, 269 escape speeds, planetary, 213-215, 214/ history of. 24-25, 207, 215 Kepler's laws, 215-218, 219 neutron stal"S, 251 observations. eleclromagnetic spectrum and, 722, 722] plasmas, 269 puls:lI"s,25IJ solar system dust in, size of, 717 geocentric model 01", 24, 215 hcliocentdc model of, 24-25, 215 stars COIOl" of, 871 formation of, 213 surface temperature of, 871 supernovas, 250, 231f Atmosphere" See also Air carbon dioxide concentrations in. 376-377,376/ escape speed of molecules and, 214 green house effect, 375-377 hcat radiation from, 372-373 I"cl'raction of light by, 7i2-773, 772] At11l0sphere (atm), 283 :\ tlnospheric electric field. meaSUI"ement oC 5J2, 312] Atillospheric pressure, 27i, 279 Al0m(s). ''IN' (llso Nucleus, atomic ch
Audible waves, 460, 489 Aurora
571,
5~8
Average force (F,,,), impulse-momentum theorem and, 163, ]63f Average power (f¥), definition of. 14-3 Average speed, 26-27, 47 A,·erage velocity (li) in one dimension. 27-28, 30. 47 in constant accelera[ion, 35 from position us. time graphs, 28. 29J in two dimensions, 6J, 74 AV (mriovcntriculal") node, 585, 385J Avogadro's llumber, 335, 340 Axial myopia, 826 Axon, 614, 614J
B Back side of convex III inor, 764, 764/ of spherical refracting surface, 769 oflhinlells.773 Bacleria doubling time of, 608 magnetic navigation in, 629 Bakelite dielectric constant, 558t dielectric strengrh. 558/ Ballislic pendulum. 17L l72f Ballpoint pens. pressme within, 284
Gas(cs). Ser also Ideal gas characteristics of. 268-269 density of, 276, 276t inlcrnal energy of, 343. 3:'13, 390 kinel ic theory of, 340-3,'1.5, 346 molecules, degl"ees 01" fn:edolll, 390 real m. ideal. propenies of. 335 rool-mean-square (rills) speed of, 343, 3441,346 spectral analysis of, 743-7-1-1 speed OfSOLllld in, '161-462, 462t lherrnal condUclivit)', 3n71 work done on, at constant pressure, 3S5-388 (:,lS constant (R), 39J Gas thermometers, 324-326, :124)," 325] Gauge pressure, 283 Gallss (C). 630,652 COlwcrsion to/from lesl
Brcwsler's angle for, 812 index of refraction, 738/, 742] as i nsu laLOr, 499 resisl ivit}'. 5771 specific heat, 355f therm
1.9
Goeppcn-~1aver, Maria. 915/ Gold latent heal of fusion, 3601 l;ltcllt heal of vaporization, 360, resistivity. 577/ specilic hL:al, 355t temperatLi re coefficient of rcsislidl~. 577/ thermal conductlvit}. 367t Golfball, aerodynamics of. 296, 296/ Goudsmit, S'lllluel. 900 Grade, or energy. 413 Grand unil1ed theory (GeT). 947, 953-954 Graviwtiol181 force (/~), 89-90. Sf'I'(f[I'O Frce fall acceleration due to. -12 anillcial,203 as conscn'ati\'e force, 127 dctermining, with pendulum, 440. ·141 as field force, B4 as fundamental fOI-ce. 84, 9"14, 9-1'1/, 954,95-1/956 G'lllSS'S Jaw. 208 in general n:,hnivit)'. 863-865 law of' universal gl-avitation (N('wl.on). 90, 108.207-215, 218-21H and light, bending or, 863j, 864-S65,865/ in mOl ion through viscolls medium. 309-310 net. 209-210 as poinl force, 208 in Standard ~'fodel, 933. 953I and time, 864, 865 llniversal gravitation constant ((;), 90, 207, 208-209, 208J, 219 variation with distance, 207-208. 20Rt V,I. electric force, 502-503 work done by. 127-128. 128/ 150. 211-212 Graviwtionallenses.865] Gravitational potclHial energy (Plil;)' 127-135,150 and conservation of mechan leal energy, 130-133 and gravitational work. 127-128. 128/ 150,211-212 for high objects, 211-213 reference levels for, 129-)30. 129/ in rOlationalmolion. 247, 254 Slunits of, 128 V~. electric pOlelllial energ). 33~. 5?3f work-energy theorem and. 127-128, 133-135, 150 Gravil,llional work (\r1~), gl-avitational potential energy and. 127-128. 12Sj: 150,211-212 Graviwm, 944, 94-11 Greenhollse effect. 375-377 Greenhouse gases. 376-377, 376] Grimaldi. F,·ancesco. 733 Ground, electrical dcfi ned. 499-500 symbol ror, 581. 581! Grollnd-faull. inlerruptel's (GFls), 613, 6oS-069, 668J, 669! Ground Slate, 895
1.10
Index
Guanine, 878 GUT. See Grand unified theory
H Hadron.'>, 945-946, 946/, 956 quark model of, 950-952, 9511,
956-957 Hahn, Ouo, 862 Hale telescope, 834J, 838 Half-life (TlI~)' 919, 921, 932 Halogens, 905 I-Jammer (ear bone), 487f, 488 I-lard magnetic materials, 627, 651 Harmonic(s) in air column waves. 480-481, 4S1J, 482-183,490 and quality (timbre) of sound, 486-487,486J in string waves, 475/ 476-479. 476f Harmonic oscillalOr equal ions, 427 Harmonic series, 476-'177 Hearing, human, 464-466, 465/, 487-488. 487f 48S! I-lean artificial,635 electrical activit), of, 585-587, 585f 586J Heat. See also 'Molar specific heal cOlwection, 371-372, 37IJ, 378 definition of, 323, 353, 377 and energ'Y transfer, 141 ideal absorbers, 373 ideal reflectors, 373 latent (L), 359, 377 of fusion, 360, 360; units of, 359 ofvapodzation. 360-361, 360t, 362 mechanical equivalent of, 353. 377 for phase change, 359-362, 377 radiation, 372-375, 372f, 3731, 378 to raise mass In by tJ.T 355-356 specific heal (c), 355-357 definition of, 355, 377 measurement of, 357-359. 377 of selected substances, 3551 thermal conducrjon, 366-368, 36~f, 377-378 conducLivity of selecLed substances, 367t and home insulalion, 368-3iO units of. 353-355 Heat engines, 399-408, 3991, 417 Carnot engine, 405-408, 4061, 417 cyclic pl"Ocess in, 399, 400-402 Pl1diagram [01", 399, 399f refl'igerators and hear pumps as, 402-404, 403f 417 thermal efficiency of, 399-400, 404, 406-407, 417 work done by, 399-400, 417 Heat pumps, 402-404, 4031, 417 Heisenberg, Werner, 884. 884J, 899 Helicopters, J\'ewton's third law and, 93 Heliocentric model. 24-25, 215 Helium (He) ground-state configuration, 903. 904t helium-neon laser. 907, 908! latenl heaLoffusion. 360t
Helium (I-Ie) (Continued) latent heat of"aporizarion, 360t liquid, storage of, 375 molal" specific heat, ~191 t as noble gas. 904 speed of sound in, 462/ lhermal conductivity. 3671 Henry (1-1), 681. 687 Henry,Joseph. 663, 68] I-lerman, Roben, 954f Herschel, \Nilliam, 720 Hertz (Hz), 434 Henz, I-Ieinl"ich Rudolf. 713-714, 7]3J, 720.724.733.872 Higgs boson, 953 Hilbert, David, 850f Hooke, Robert, 135,425 Hooke's la\\'. 135,425-428, 426J, 450 Horsepower (hp), 143 Household cil"cuits, 600-601, 611-612,612J Hubble Space Telescope, 723, 834 l-Iuman body aneriosclerosis, vascul(lr fluner in. ~97. 297J blood clennsing. by kidneys, 308-309,309J in oxygen diffusion, 308 pressure. measuring of. 283-284. 284I speed of, measuring, 460 viscosity of, 305, 305t brain boxing injlUl·lo. J6~ fluid supporting, 163,286 circulaLOry system and Poiseuille's law, 305~306 and Reynolds number, 306-307 DNA. structure of, determining, 877-878, 878} ear anatomy and function. 464-466, 465/, 487-488, '187;; 488J efficienc)' of. 416. 4161 and exercise energy expenditun~ in, 147 physical filness and oX)'gen consumption. 415-4·]6, 416t physiology of, 354-355 and weight conlroL 414-415 eye. 824-829, 825f accommodation. 825-826 ana tom}'. 824-825, 82~f condirions of, 825-829, 8201, 827;; 841 diving masks wilh vision-colTecling lens, 776 nuid pressure in, 169 image fonnalion in, 824~825 I'cd-eye in photographs. 735 and sunglasses UV protection, 720/ underwaLCr vision, 770 vision cOITection, 826-828. 826f 827J wavelength sensitivity of, 722 heart artificial. 635 electrical aUivity of, 585-587. 585f 586J
I-[uman body (Colltinued) inlravenous soilltions, and osmotic balance. 308 kidneys artificial, 309, 309/ blood cleansing by, 30R-.109, ~09f dialysis, 309, 309f 635 lung nuid surface tension, 301 nervous system, electrical signal conduction in, 613-614, 614f neurons eleclrical signal conduction in, 613-614, 614J srructure of, 614, 614f rypes of, 614-615 physical fitness, 415-416, 416/. pOlassiulll ion channels, 614-615 power omputof, 146-147, Hit sodium ion channels, 614-615 thermal erficiency of, 416, 4161 I-Inman metabolism, 413-416 first law of thermodynamics and. 413-414 metabolic rate, 414-415, 415t oxygen consumption and metabolic rate, 414-415, 415t and physical fitness, 415-416, 41Gl and weighrgain, 414-415 Hummingbirds, feather color in, 790f Hu}'gens, Christian, 732, 733f I-Iuygens' principle, 746-748, 746J, 7471, 752 H}'draulic brakes, 281 Hydraulic press, 281, 281/ Hydrogen (H) atom, size of, 3t, 4 Balmer series, 897, 89~r Bohr model of, 894-898, 894J, 908 emission specrrum, 892, 893j, 894, 896 energy level diagram, 895-896, 896f fusion of, in Sun, 941 ground-state configuration, 903, 9041 ionizalion energy, 896 isotopes, 913 quantum mechanical model of,
899-902.908-909 qua mum stale energies, 895, 908 rms speed of, 343, 344t speed of sound in, 462t thermal COndUCli\'ity of. 367t Hydrogen bomb, 941 Hydrogen-like alOtllS, Bohr model and, 898-899.908 Hyperopia (farsightedness), 825-827, 826(,841
leos, See Implanted Cardiovener Defibrillarors Ice densi[}', 2761, 334 index of refraction, 738t specific heat, 355/ (hermal conducth1ily, 3671 Ice~pail experiment (Faraday), 514, 514f Ice (freezing) point, 324, 325-326, 3261 Ideal absorbers, 373 Ideal fluid, characteristics of, 290
Index
Ideal gas definition of, 335 internal enngy change in, :390 pmpenies macroscopic, 335-340, 3·15-346 mol('('lIl:H models fOl-, 3<10-345, 346 vs" real gas. 335 Ideal gas law, 336. 340, 346 applications, 336-339 Ideal reflectors, 373 ideal transformers, 711, 724 Image b:nely resolved, 833, 836/ definition of. 759 just resoh'cd, R~5, 836j, 841 not resolved, 835, 8351, 836/ real, 759, 762, 7651, 782 resolved, 835, 835/ virtual. 759-760, 765l, 782 Image distance, 759,759/763,769,782 11ll;lge point. of spherical mirror, 762-7(;3, 762f, 763J Impedance (2), 703j, 70-1. 705/, 724 Impending moL ion, 102 Implanted Canlioverter Dcf-ibrillatol-S (ICDs), 586-587. S87j; 5871 [mpulse (1\ 162,181 ' [mpulse-momcnlum theol-em, 162-163,181 Index of refraction, 737, 738t, H2, 742): 751 I nclillm. as superconduclor. 584 Induced currclH, 666 Induced emf applications. 675-680 back emf. G79-680, 679/ definition of. 663-664, 664/ direction of. 667-668. 667/. 674-675, 674f, 675J Fanlday's law of magnetic induction, 666-670,687 in g'enerators, 676-680 Lenz's law. 667-668, 674-67!l, 674); 675/ motional, 670-67..J., 67~f, 6711, 687 self-induced, 680-683, 680,[, 6811, 687 Induced polarizalion, of molecules, in dielecl.l"ic. 561, 561/ Induclance (L), 681, 687 of solenoid, 682-683. 687 Induction, 663-664. 664:1 ;-lpplica!ion~. 675-680 back emf. 679-680, 679J charging by, 499-500. 500/ eliminating charge from houses. 51-I dil"eClion ofinduced emf. 667-668, 667f, 674-675, 674f, 675/ Faraday's law of, 666-670. 687 in generalOrs. 676-680 I.enz:s law, 667-668, 67'1-675. 674j. 675/ motional emf. 670-674, 670j. 671j, 687 self-induced emf, 680-6S~. 680j, 681f, 687 Inductive reaCL;lllce (X,), 701-702. 7011, 723-72-1 Inductol"(s) in AC circllits, 701-702, 701,f, 72~-72<1 Rl, circuils, 705, RLCseries circuits" 702-706, 702j, 703/724
Inductor"(s) (Continued) circllil symbol for. 683, (83) in DC HLcircuits. 683-687, 6831, 684f, 688 deflnilion of. 683 energy slOred in magnetic field of', 686-687.688 impedance, 705t phase angle. 7051 Inelaslic collisions, 169. [76,177-178. 181 perfectly ineiaSlic. 169, 170-173. 181 Inertia, R5-86. 108 Inertial electrostatic confinemenl f'lsion, 9':Hj, 942 Inerl.iallaser confmClllcnl fusion, 9'12 Infrared waves, 721. 72[/ Infrasonic waves. -160, 489 In phase. defined. -149, 4-19/ Insects. cars of, 489 lnstant
1.11
Interference (Continued) of sound wa,"cs. 473-..J.75. 473f, 489-490 ofwavcs, genera II),. .<148-449. 449f,451 beats. 484-'186, 484], 490 cons(ructh-e, 449, 4-19j, 451, 473.489 destructi'"e, 4<19. 449j, 451, 473-474. <190 lmcrferometcr. ~\'lichelson, 840-841, 840f, 849. 830J Imernal combustion engine as heat engine, :\99 work done by. ~93-394 Imernal energy (U) change in in adiabatic processes, 393-395, '117 f'irsllaw ofthcrmodynamirs. 388-390, <\ 16-417 in generic processes, 397-398 for ideal gas. 390 in isobaric processes, 390-393, <117 in isothermal processes. 396-397. ~96J in isO\-olumctric processes. 395-396 conscrralioll of energy and, 141 definition of, 353, 377 degradation or, 413 of isolated system. 388 of monatomic gas, 34-3, 390 of polyatomic gas, 353 Imcrnational Commiuee on \Veigills and Measures. 326 Imcrnellrons, 613-614. 614[ 11llr
1.12
Index
J
L
Jensen. Hans. 915/ Joule U), 120. 124. 128 cOlwersion to electron "olt, 860-86\ Juule.James r., 352. 353J Junction rule, 603-60G, 603t GIG Jupiter escape speed. 214t planetary data. 217/ Just resol"ed image, 835. 836); 841
Lakes. convection currents in, 37 l-372, 372f Lambda (tV)). 946/, 951f Lamin'na III ics L.-.w of l"cf'l"aCljon (Snell), 737~7'12,
Wave fronl oflight wave, 734. i34:j Huygcns' princ.:iple, 74fl-i48, i4(~f, 747f, 752 of spherical wave. 466, "1I16J Wave function ('1').883-884.887.902 Wavelength (..\) ofliglH absorption spectra, 893 cutoff, in photOelectric effect, 874 detennining from diffraction grating, 806 emission spec/rums, 892-894, 893J measurement of, 793-7!N. 838-839, B'IO-841 in medium, 7~16, 815 oflinear wa\'e. 445--146. 445J, 451 ofmauer (de Broglie wavelength), 881-882, 887 of!ipherkal wave, 466, 46(~1 of x-rays charac/eri!ilic x-rays. 90!l-906 Compton shirl, B79-8sa, 879j, 88G emiw.:d by x-ray wbe, 875-876. S75); 876j. 887 Wavelets, 746 Wave optic~, 790. Sre (/150 DilTraClion; Intcrfen:ncc; PolariJ:arioll \rave speed, 445-446, 447-448, 451 W bosons. 9/14, 944t. 952-9fI3 \'\Ieak force Big Bang and, 9,:)4, 954J
Chal"actcristics of, 944. 944[, 946, 9f)6 and electrQwcak theory, 952-953 as fundamental force, 84 in Stand,ml Model. 953, 953J Weber (Wb). 664-665 Weber per s<]uare meIer (Whim:'), 6~\O,
6S2
Weight apparClll. acceleration and. 99-100 gravit
Wilson, Robert \,V., ~)[lS, lJ55I Wind instruments, sound production in, 480-483 \"I'ood, thermal conductivity, 367/ WO"k (110, 119-123. 120j. 121); 150 in adiabfltic processes. 393-395 ro charge capacilOl', 555, 55'51 by constanl force. 120 dcfinilion or, 120 and electric potelll ia I energy.
531-535. 5321 and energy transfer, I,ll frictional, 122-123 on gas at conStant pl-essurc, 385-38S in generic processes. 397-~98 gravitational. 127-128, 128J, 15U, 211-212 by heat engine. ~99-'100, '117 in isobal'ic pmccsscs, 385-388, 39039~.417
in isothermal proccssf'S, 396-397. 3961, 417 in iso\'olllllletric processe!i. 395-396, 4 17 mechanical energy change and, HI, lSI and potential eneIX)', 127-128 from PV diagr3m. 387-388 SI unilsof, 120 b}'spl'ing. 136-138, 148-150 in thermodynamic processes. 385-388, '116 by varying force, 1'17-150, 1'18I \'Vork-energy theon.:rn. 12'1-127, gO with gravitational and clastic potential energy componcilts. J36-138.150 wit h gravitational potential enel'gy componcllt, 127-128, 13~~135, lilO for .-otaTional motion with nonconscrvativc rorccs. 247, 248-249,254 Vv'ol'k fUllction, 87~, 873/, 886
x x-component or glancing collision. 176 of vect.or, 57 Xenon (Xc), as noble gas, 90'1
z Z bosollS, 94'1, 944l, 952-95~ Zeeman effect. 899, 899J
Zero-poil1l energy. 326 Zemth law or thermodynamics. 323. ~2V 345 ZCrDlh-ordcr maxillllllll. 79:l. 806 Zinc (Zn)
ground-!itatc configuration. 904/ as sliperconductor, 584, 58'1/ work function, 871t Zircoll. index of refraction, 738/ Z machine. 937J ZOllllles, 82fl
