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Copyright © 2006, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved. No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher. All inquiries should be emailed to
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ISBN (13) : 978-81-224-2709-7
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PREFACE
The present book is based on my experience, extending over a period of about fourteen years of teaching graduate and postgraduate students of the University of Lucknow. Various universities have continuously been updating postgraduate courses in physics. The courses at senior school level have also been fairly modernized. This book has been prepared for students of B.Sc. (General) and B.Sc. (Hons.) keeping these trends in mind. The subject matter has been selected and developed in such a manner so as to provide a bridge between these advanced and introductory level courses. Mechanics being the vital building-block of physics, the treatment of key concepts has to be fairly exhaustive. Only then the students would be able to comprehend advanced level courses in classical mechanics, quantum mechanics and other areas of physics at postgraduate level. Keeping in view the special requirements of undergraduate students in particular a number of worked-out examples are distributed throughout the book to elucidate and, in some cases, to extend the various concepts as well as to give the students a good grasp of the basic concepts and theories of mechanics. This book is designed as a textbook according to the updated syllabus primarily for the undergraduate physics and applied physics students of the various Indian universities and colleges. This book while being equally useful and inevitable for the students preparing for competitive examinations for admission into engineering and medical colleges will continue to be useful also for the first year engineering students. Some advanced topics like “special theory of relativity” have also been treated in detail in this book. Some of the problems of mechanics which are expected to help the students gain more practical knowledge in the subject have been given as follow-up problems at the end of each chapter. In other words, the presentation is oriented towards introducing the fundamental concepts of mechanics leading to, and associated with, practical applications. The most distinguishing feature of the book is that the material has been treated and presented—and the various results derived—systematically and thoroughly, yet to the point as to make it suitable for self-study by the students. Suitable diagrams are given in the book to illustrate the basic principles.
(vi) Short-Answer Type Questions with Answers have been given in each chapter to help students in the comprehension and appreciation of the finer points of physics. The book has been written in a very simple and lucid way. Every effort has been made to make the treatments simple and comprehensive. The difficult topics are explained with the help of neat and clean diagrams. Throughout the book emphasis has been laid on the physical concepts. I am absolutely sure that all readers, particularly students, will be tremendously benefitted by this book. This book will be extremely useful to teachers and students of various courses—involving studies in fundamental sciences—being pursued in various institutions across the globe. AUTHORS
CONTENTS
Preface
(v)
1. MEASUREMENT
1
1.1
What is Physics
1
1.2
Scope and Excitement in Physics
1
1.3
Measurement
2
1.4
Physical Quantities
3
1.5
Fundamental Units of S.I. System
3
1.6
Reference Frames
4
1.7
Inertial and Non-inertial Frames
5
1.8
Scalars and Vectors
9
1.9
Addition of Vectors
11
1.10
Resolution of Vectors
16
1.11
Multiplication of Vectors
17
1.12
Vectors and the Laws of Physics
21
1.13
Speed and Velocity
22
1.14
Acceleration
24
1.15
Rectilinear Motion
24
1.16
Acceleration of Gravity
25
1.17
Accuracy and Errors in Measurement
25
SUMMARY
29
2. FORCE AND MOTION
30
2.1
Mechanics
30
2.2
Cause of Motion: Force
30
2.3
Newton’s Laws of Motion
31
2.4
Freely Falling Bodies
33
2.5
Motion in a Vertical Plane
33
2.6
Projectile Motion
34
2.7
Equilibrium of Forces
37
2.8
Frictional Forces
37
2.9
Static Friction and Coefficient of Static Friction
38
(viii) 2.10
Dynamic Friction and Coefficient of Dynamic Friction
38
PROBLEM
39
3. DYNAMICS OF CIRCULAR MOTION AND THE GRAVITATIONAL FIELD
86
3.1
Uniform Circular Motion
86
3.2
Centrifugal Force
87
3.3
The Centrifuge
88
3.4
Banking of Curved Roads and Railway Tracks (Banked Track)
88
3.5
Bicycle Motion
89
3.6
Conical Pendulum
90
3.7
Motion in a Vertical Circle
91
3.8
Motion of Planet
92
3.9
Kepler’s Laws of Motion
93
3.10
Derivation of Law of Gravitation
93
3.11
Newton’s Conclusions from Kepler’s Laws
94
3.12
Newton’s Universal Law of Gravitations
95
3.13
Gravity and the Earth
96
3.14
Acceleration due to Gravity
96
3.15
Expression of Acceleration due to Gravity g in Terms of Gravitations Constant G
96
3.16
Difference between Mass and Weight
97
3.17
Inertial Mass and Gravitational Mass
97
3.18
Gravitational Field and Potential
97
3.19
Equipotential Surface
98
3.20
Weightlessness in Satellites
99
3.21
Velocity of Escape
100
3.22
Relation between Orbital Velocity and Escape Velocity
101
3.23
Satellites
101
3.24
Orbital Velocity of Satellite
101
3.25
Orbital Speed and Period of Revolution of a Satellite Very Close to Earth
103
3.26
Artificial Satellites
104
3.27
Black Holes
105
4. WORK, ENERGY AND MOMENTUM 4.1
Work
142 142
(ix) 4.2
Power
143
4.3
Work in Stretching a Spring
143
4.4
Energy
144
4.5
Kinetic Energy
144
4.6
Potential Energy
145
4.7
Gravitational Potential Energy
145
4.8
Work-Energy Theorem
146
4.9
Significance of the Work-Energy Theorem
147
4.10
Conservative Force : First Definition
147
4.11
A Central force is Conservative
148
4.12
Non-Conservative Force
149
4.13
Relation between Conservative Force and Potential Energy
151
4.14
The Curl of a Conservative Force is Zero
152
4.15
Linear Restoring Force
154
5. LINEAR AND ANGULAR MOMENTUM
222
5.1
Conservation of Linear Momentum
222
5.2
Centre of Mass
223
5.3
Cartesian Components of the Centre of Mass
225
5.4
Centre of Mass of a Solid Body
225
5.5
Position Vector of the Centre of Mass
226
5.6
Velocity of the Centre of Mass
226
5.7
Center of Mass Frame of Reference
227
5.8
Motion of the Center of Mass of a System of Particles Subject to External Forces
227
5.9
Linear Momentum in Center of Mass Frame of Reference
228
5.10
System of Variable Mass
228
5.11
Motion of a Rocket
230
5.12
Multi Stage Rocket
232
6. COLLISION
241
6.1
Collision
241
6.2
Elastic Collision in One Dimension
241
6.3
The Ballistic Pendulum
245
6.4
Collision in Two Dimension
246
6.5
Value of the Scattering Angle
247
6.6
Scattering Cross-Section
249
(x) 6.7
Differential Scattering Cross-Section
250
6.8
Total Cross-Section
251
6.9
Impact Parameters
251
6.10
Rutherford Scattering
253
7. ROTATIONAL KINEMATICS
311
7.1
Rigid Body
311
7.2
Moment of a Force or Torque
311
7.3
Angular Acceleration
311
7.4
Relation between Angular Acceleration and Linear Acceleration
312
7.5
Kinetic Energy of Rotation
313
7.6
Angular Momentum
314
7.7
Relation between Torque and Angular Momentum
314
7.8
Conservation of Angular Momentum
315
7.9
Torque Acting on a Particle
315
7.10
Angular Momentum of the Center of Mass of a System of Particles
316
7.11
Precession
318
7.12
The Top (Precession of a Top Spinning in Earth’s Gravitational Field)
318
7.13
The Gyrostat
321
7.14
Man with Dumb-bell on a Rotating Table
322
8. MOMENT OF INERTIA
336
8.1
Moment of Inertia
336
8.2
Role of Moment of Inertia in Rotational Motion
337
8.3
Radius of Gyration
337
8.4
Analogous Parameters in Translational and Rotational Motion 338
8.5
General Theorems on Moment of Inertia
339
8.6
Calculation of Moment of Inertia
342
8.7
Moment of Inertia of a Uniform Rod
342
8.8
Moment of Inertia of a Rectangular Lamina (or Bar)
343
8.9
Moment of Inertia of a Thin Circular Ring (or a Hoop)
345
8.10
Moment of Inertia of a Circular Lamina or Disc
345
8.11
Moment of Inertia of an Angular Ring or Disc
346
8.12
Momnt of Inertia of a Solid Cylinder
347
(xi) 8.13
Moment of Inertia of a Solid Cone
348
8.14
Moment of Inertia of a Hollow Cylinder
350
8.15
Moment of Inertia of a Spherical Shell
351
8.16
Moment of Inertia of a Solid Sphere
351
8.17
Moment of Inertial of a Hollow Sphere or a Thick Shell
352
8.18
M.I. of a Uniform Triangular Lamina
353
8.19
Kinetic Energy of Rotation
354
8.20
A Body Rolling Down an Inclined Plane (Its Acceleration Along the Plane)
356
8.21
Identification of Hollow and Solid Sphere
358
8.22
Compound Pendulum
358
8.23
Fly Wheels
360
9. MECHANICAL PROPERTIES OF MATTER
393
9.1
Rigid Body
393
9.2
Elasticity
393
9.3
Perfectly Elastic and Perfectly Plastic
393
9.4
Stress
393
9.5
Strain
394
9.6
Hook’s Law
394
9.7
Elastic Limit
394
9.8
Young’s Modulus
395
9.9
Bulk Modulus of Elasticity
396
9.10
Modulus of Rigidity (Shear Modulus)
396
9.11
Poisson’s Ratio
397
9.12
Potential Energy in Stretched Wire
397
9.13
Equivalence of a Shear to a Tensile and a Compressive Strain at Right Angles to Each Other and Each Equal to Half the Shear
398
Equivalence of Compression and Equal Perpendicular Extension to a Shear
399
9.14 9.15
Equivalence of a Shearing Stress to an Equal Extensional Stress Plus an Equal and Perpendicular Compressional Stress 400
9.16
Relations Connecting the Elastic Constants
401
9.17
Theoretical Limiting Values of Poisson’s Ratio
404
9.18
Poisson’s Ratio for an Incompressible Material
405
9.19
Twisting Couple on a Cylinder
406
(xii) 9.20
Determination of the coefficient of Rigidity (η) for the Material of a Wire
408
9.21
Torsional Oscillations
411
9.22
Determination of Modulus of Rigidity of a Torsional Oscillations
411
9.23
Beam
412
9.24
Bending Moment
413
9.25
Cantilever
416
9.26
Beam Supported at its Ends and Loaded in the Middle
418
9.27
Determination of Young’s Modulus by Bending of a Beam
420
9.28
Determination of Elastic Constants by Searle’s Method
421
10. FLUIDS
463
10.1
Molecular Forces
463
10.2
Definition of Surface Tension
464
10.3
Explanation of Surface Tension
465
10.4
Surface Energy
465
10.5
Relation between Surface Tension and Work Done in Increasing the Surface Area
465
10.6
Shape of Liquid Meniscus in a Glass Tube
466
10.7
Angle of Contact
467
10.8
Capillary Action
468
10.9
Rising of Liquid in a Capillary Tube of Insufficient Length
469
10.10
Effects on Surface Tension
469
10.11
Ideal Liquid
470
10.12
Steady or Stream Line Flow
470
10.13
Equation of Continuity of Flow
471
10.14
Energy of the Fluid
472
10.15
Bernoulli’s Theorem
473
10.16
Velocity of Efflux
475
10.17
Velocity of Efflux of a Gas
476
10.18
Viscosity
477
10.19
Flow of Liquid in a Tube: Critical Velocity
478
10.20
Velocity Gradient and Coefficient of velocity
478
10.21
Poiseuille’s Equation for Liquid-Flow through a Narrow Tube
479
Poiseuille’s Method for Determining Coefficient of Viscosity of a Liquid
481
10.22
(xiii) 10.24
Rate of Liquid-Flow through Capillaries in Parallel
483
10.25
Poiseuille’s Formula Extended to Gases
483
10.26
Determination of Viscosity of a Gas
484
10.27
Stoke’s Law of Viscous Force
486
10.28
Stoke’s Formula for the Terminal Velocity of a Falling Sphere
487
Velocity of Rain Drops
487
10.29
11. HARMONIC MOTION
523
11.1
Periodic Motion
523
11.2
Simple Harmonic Motion (S.H.M.); As a Projection of Uniform Circular Motion
523
11.3
Displacement Equation of S.H.M.
524
11.4
Conditions for Linear S.H.M.
525
11.5
Equation of Motion of a Simple Harmonic Oscillator
526
11.6
Importance of S.H.M.
527
11.7
Energy of Harmonic Oscillator
527
11.8
Average Values of Kinetic and Potential Energies
529
11.9
Position-Average of Kinetic and Potential Energies
530
11.10
Fractions of Kinetic and Potential Energies
530
11.11
Mass Attached to a Horizontal Spring
531
11.12
Motion of a Body Suspended by a Vertical Spring
532
11.13
Mass Suspended by a Heavy Spring
533
11.14
Oscillations of a Floating Cylinder
535
11.15
Oscillations of a Liquid in a U-tube
536
11.16
Helmholtz Resonator
536
11.17
Composition of Two Simple Harmonic Motions of Equal Periods in a Straight Line
537
11.18
Composition of Two Rectangular S.H.M. of Equal Periods
538
11.19
Composition of Two Rectangular S.H.M.’s of Time Periods Nearly Equal
540
Lissajou’s Figures from Two Rectangular S.H.M. in Frequency Ratio 2:1
542
Uses of Lissajous Figures
544
11.20 11.21
12. WAVE MOTION
576
12.1
Wave Motion
576
12.2
Transverse Wave
576
(xiv) 12.3
Longitudinal Wave
576
12.4
General Equation of Wave Motion
577
12.5
Equation of a Plane Progressive Harmonic Wave
578
12.6
Principle of Superposition
580
12.7
Longitudinal Waves in Rods
580
12.8
Longitudinal Waves in Gases (Pressure Variations for Plane Waves)
581
12.9
Waves in a Linear Bounded Medium
583
12.10
Flow of Energy in Stationary Waves
588
12.11
Characteristics of Stationary Waves
590
12.12
Wave Velocity (or Phase Velocity)
590
12.13
Group Velocity
591
12.14
Relation between Group Velocity and Wave Velocity
592
12.15
Normal and Anomalous Dispersion
593
12.16
Non-dispersive Medium
593
13. STATIONARY WAVES (WAVES IN A LINEAR BOUNDED MEDIUM) 14. DAMPED AND FORCED HARMONIC OSCILLATION
597 608
14.1
Damping Force
608
14.2
Damped Harmonic Oscillator
609
14.3
Logarithmic Decrement
612
14.4
Power Dissipation in Damped Harmonic Oscillator
613
14.5
Quality Factor Q
614
14.6
Forced (Driven) Harmonic Oscillator
615
14.7
Amplitude Resonance
617
14.8
Sharpness of Resonance
619
14.9
Velocity Resonance
620
14.10
Power Absorption
621
14.11
Driven LCR Circuit
623
15. RELATIVITY
643
15.1
Back Ground of Michelson-Morley Experiment
643
15.2
The Speed of Light Relative to Earth
645
15.3
Michelson Morley Experiment
646
15.4
Explanation of Negative Results: Principle of Constancy of Speed of Light
648
(xv) 15.5
The Relativity of Simultaneity: The Relativistic Concept of Space and Time
649
15.6
Postulates of Special Theory of Relativity
651
15.7
Lorentz Transformation Equations
651
15.8
Length Contraction
656
15.9
Time Dilation
659
15.10
An Experimental Verification of Time Dilation
659
15.11
Transformation and Addition of Velocities
663
15.12
Relativistic Doppler’s Effect
666
15.13
Confirmation of Doppler’s Effect
670
15.14
Conservation of Momentum: Variation of Mass with Velocity
671
15.15
Mass-Energy Relation
673
15.16
Relation between Momentum and Energy
680
15.17
Transformation of Momentum and Energy
681
Appendices
685
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1 MEASUREMENT 1.1 WHAT IS PHYSICS Physics is branch of science, which deals with the study of the phenomena of nature. The word ‘physics’ is derived from a Greek word meaning nature. The word ‘Science’ comes from the Latin word scientia, which means ‘to know’. Man has been observing various natural phenomena from time immemorial. He has always been curious about nature and the world around him. The motion of the moon and other heavenly bodies in the sky has aroused owe and amazement in him. The regular repetition of sunrise and sunset and the seasons of the year have fascinated him. Man has observed these and other natural phenomena and responded to them in an orderly manner. The experiences gained over a period of time were transmitted from generation to generation and began to be termed as ‘knowledge’. Each generation added new fact to the knowledge obtained from the previous generation. The systematized knowledge thus gained was termed ‘Science’. Every bit of knowledge is not called science. Only such knowledge as is collected by what is called the scientific method. The scientific method is the basis of scientific development. The method involves four steps: 1.
Observation of the relevant facts,
2.
Proposal of a hypothesis or a theory based on these observation,
3.
Testing of the proposed theory to see if its consequences or predictions are actually observed in practice, and
4.
Modification of the theory, if necessary.
1.2 SCOPE AND EXCITEMENT IN PHYSICS The various sciences may be divided into two broad classes, physical and biological. Physical sciences deal with nonliving matter and biological sciences with living matter. About a hundred years ago, it was possible for one man to master the knowledge of both sciences and many outstanding workers in physical sciences were also competent doctors and biologists. There was no clear-cut division between the several branches of physical sciences, as we know them today. In fact, all were included in the term natural philosophy. Aristotle, Archimedes and even Galileo and Newton called themselves natural philosophers. But today the situation 1
2
Mechanics
is different. The tremendous upsurge of scientific activity and the accumulation of knowledge in the last century have forced scientists to narrow down their field of activity. Not even a genius could hope to keep up with the developments in the various branches of physical sciences as we know them today, namely, physics, chemistry, astronomy, biology, medicine, geology, engineering etc. Those concerned mainly with the application of science to the betterment of human life and environment are called engineers. The invention of the steam engine and the electric motor gave rise to engineering. Physicists and chemists are concerned more with the basic aspects of nonliving matter chemistry deals primarily with molecular changes and the rearrangement of the atoms that form molecules. Physics deals with the phenomena of the non-living world such as mechanics (motion), heat, sound, electricity, magnetism and light. The division of the subject of physics into these and other branches is largely a matter of convenience. These branches are inter-related, as you will discover in the course of your study of the subject. Physics may be defined as that branch of knowledge that deals with the phenomena of non-living matter. In physics, we deal with many physical phenomena and experiences. Merely reading about the experiences and observation of others is not enough. If students are to understand and enjoy physics, they must have same of these experiences themselves. These experiences are not only exciting but also very educative. The swinging hanging lamp in a church led Galileo to a method of measuring time. The fall of an apple and the motion of the moon led Newton to his famous law of gravitation. The rattling (or dancing) of the lid of a kettle led to the invention of the steam engine. The flowing of a flute causes vibrations that produce sound. The light from stars tells us something about stars and their evolution. The study of electricity helps us to design motors and dynamos. The study of semiconductors helps us to design radios, televisions, calculators and even computers.
1.3 MEASUREMENT Observation can be subjective or objective. An observation that varies from individual to individual is subjective. For example, different individuals observing the same thing, e.g. a painting or a flower, feel differently. Physics does not deal with such subjective observations. Physics is a science of objective observation, an observation that is the same for all individuals. An individual observer through his sense of touch or sight, but these senses is not always reliable. To illustrate the inaccuracy of our sense of touch, we consider three pans containing cold, warm and hot water. If you put your finger first in cold water and then in warm water, your sense of touch will tell you that it is hot. But if you put your finger first in hot water and then in warm water, your sense tells you it is cold. This clearly suggests the necessity of making a measurement to arrive at the truth. It is necessary to measure the degree of hotness of water in each pan. In other words, it is not enough to describe a phenomenon in a general and qualitative way. A number must be tied to it. Thus, physics is a science of measurement. Lord Kelvin, a leading physicist of the 19th Century, once said: “When you can measure what you are talking about and express it in numbers, you know something about it; but when you cannot, your knowledge is of a meagre and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely in your thoughts advanced to the stage of a science”.
Measurement
3
1.4 PHYSICAL QUANTITIES Physical quantities are often divided into fundamental quantities and derived quantities. Derived quantities are those whose defining operations are based on other physical quantities. Fundamental quantities are not defined in terms of other physical quantities. The number of quantities regarded as fundamental is the minimum number needed to give a consistent and unambiguous description of all the quantities of physics. All Physical quantities occurring in mechanics can be expressed in the units of ‘length’, ‘mass’ and ‘time’. The units of these three quantities are independent of one another and no one can be changed or related to any other unit. These quantities are called ‘fundamental quantities’ and their units are called ‘fundamental units’. In the same way, to fix the units of physical quantities occurring in electromagnetism, thermodynamics and optics, electric current, temperature and luminous intensity are taken as fundamental quantities and their units as ‘fundamental units’. The system of units based on the units of the seven fundamental quantities (length, mass, time, electric current, temperature, luminous intensity and amount of substance) is called International system of units. It is abbreviated as SI from the French name Le Systeme International d’units. It is based on the following seven fundamental (or Basic) and two supplementary Units: Basic physical Quantity 1.
Name of the Unit
Symbol
length
meter
m
2.
mass
kilogram
kg
3.
time
second
s
4.
Electric Current
ampere
A
5.
Temperature
Kelvin
K
6.
Luminous intensity
candela
Cd
7.
Amount of substance
mole
mol
Supplementary physical Quantity
Name of the Unit
Symbol
1.
Plane angle
radian
rad
2.
Solid angle
steradian
Sr
1.5 FUNDAMENTAL UNITS OF S.I. SYSTEM Basic and Supplementary Units of SI System The seven fundamental and two supplementary units of SI system are defined as follows: 1.
Metre: On atomic standard meter is defined as to be equal to 1,650,763.73 wavelengths in vacuum of the radiation emitted due to transition between the levels 2p10 and 5d5 of the isotope of krypton having mass number 86. Krypton86 emits light of several different wavelengths. The light emitted by Krypton-86 due to transition between the levels 2p10 5d5 in orange red in colour and has wavelength 6057.8021 Å or 6.0578021 × 10–7 m. The number of these wavelengths in 1 m can be counted by using an optical interferometer which comes out to be 1,650,763.3.
4
Mechanics
2.
Kilogram: There is no definition of unit kilogram on atomic standards. Therefore in SI system, kilogram is the mass of a platinum-iridium cylinder kept in the International Bureau of weights and measures at Paris. In practice, the mass of 1 litre of water at 4°C is 1 kilogram.
3.
Second: Unit second can again be defined on atomic standards. One second is defined to be equal to the duration of 9,192,631,770 vibrations corresponding to the transition between two hyperfine levels of Caesium –133 atom in the ground state.
4.
Kelvin: It was adopted as the unit of temperature. The fraction 1/273.16 of the thermodyamics temperature of triple point of water is called 1K.
5.
Ampere: It was adopted as the unit of current. It is defined as the current generating a force 2 × 10–7 Newton per metre between two straight parallel conductors of infinite length and negligible circular cross-section, when placed at a distance of one meter in vacuum.
6.
Candela: It was adopted as the unit of luminous intensity. One candela is the luminous intensity in perpendicular direction of a surface of 1/600,000 meter 2 of a black body at a temperature of freezing platinum (2046.64 Kelvin) and under a pressure of 101,325 N/m2. Candela was redefined in 1979 as below: It is the luminous intensity in a given direction due to a source which emits monochromatic radiation of frequency 540 × 1012 Hz and of which the radiant intensity in that direction is 1/683 watt per steradian.
7.
Mole: It was adopted as the unit of amount of substance. The amount of a substance that contains as many elementary entities (molecules or atoms if the substance is monoatomic) as there are number of atoms in 0.012 kg of carbon-12 is called a mole. This number (number of atoms in 0.012 kg of carbon –12) is called Avogadro constant and its best value available is 6.022045 × 1023.
8.
Radian: It was adopted as the unit of plane angle. It is the plane angle between the two radii of a circle, which cut off from the circumference, an arc equal to the length of the radius. Plane angle in Radian = length of arc/radius
9.
Steradian: It was adopted as the unit of solid angle with its apex at the centre of a sphere that cuts out an area on the surface of the sphere equal to the area of the square, whose sides are equal to the radius of the sphere. Solid angle in steradian = area cut out from the surface of sphere/radius2
1.6 REFERENCE FRAMES The same physical quantity may have different values if it is measured by observers who are moving with respect to each other. The velocity of a train has one value if measured by an observer on the ground, a different value if measured from a speeding car, and the value zero if measured by an observer sitting in the train itself. None of these values has any fundamental advantage over any other; each is equally ‘correct’ from the point of view of the observer making the measurement.
Measurement
5
In general, the measured value of any physical quantity depends on the reference frame of the observer who is making the measurement. To specify a physical quantity, each observer may choose a zero of the time scale, an origin in space and an appropriate coordinate system. We shall refer to these collectively as a frame of reference. Since the space of our experience has three dimensions, we must in general specify three coordinates to fix uniquely the position of an object. The Cartesian coordinates x, y, z are commonly used in mechanics. Thus, the position and time of any event may be specified with respect to the frame of reference by three Cartesian coordinates x, y, z and the time t.
1.7 INERTIAL AND NON-INERTIAL FRAMES A system relative to which the motion of any object is described is called a frame of reference. The motion of a body has no meaning unless it is described with respect to some well defined system. There are generally two types of reference systems: 1.
The frames with respect to which unaccelerated body is unaccelerated. This also includes the state of rest.
2.
The frames with respect to which an unaccelerated body is accelerated.
The frames with respect to which an unaccelerated body is unaccelerated, i.e., is at rest or moving with constant linear velocity are called inertial frames i.e., unaccelerated frames are inertial frames. Let us consider any coordinate system relative to which body in motion has coordinates (x, y, z). If the body, is not acted upon by any external force, then m
d2 x = 0, dt2
m
d2 y = 0, dt2
m
d2 z =0 dt2
Hence,
d2 x = 0, dt2
d2 y = 0, dt2
d2 z =0 dt2
Which gives
dx dy dz = ux = constant, = uy = constant, = uz = constant dt dt dt Where ux, uy and uz are the components of velocity in x, y and z directions respectively. From above equations we see that the components of velocity are constant, i.e., we say that without application of an external force, a body in motion continues its motion with uniform velocity in a straight line, which is Newton’s first law. Hence we may say ‘An inertial frame is one in which law of inertia or Newton’s first law invalid’. The frame with respect to which an unaccelerated body is accelerated are called noninertial frames, i.e., accelerated frames are called non-inertial frames. A frame of reference moving with constant velocity relative to an inertial frame is also inertial. Since acceleration of the body in both the frames is zero, the velocity of the body is different but uniform. Experiments suggest that a frame of reference fixed in stars is an inertial frame. A coordinate system fixed in earth is not an inertial frame, since earth rotates about its axis and also about the sun.
6
Mechanics
Galilean Transformation Galilean transformations are used to transform the coordinates of a particle from one inertial frame to another. They relate the observations of position and time made by two of observers, located in two different inertial frames. Let us consider two inertial frames S and S′. S being at rest and S′ moving with a constant velocity v relative to S. The positions of two observers O and O′ observing an event at any point P coincide with the origin of the two frames S and S′. Then the problem is to transform the data of the event recorded in the first frame to those recorded in the second. Case I: When the second frame moves relative to first along positive direction of x-axis. Let the origins O and O′ of two frames S and S′ coincide initially. Let the event happening at P be denoted by (x, y, z, t) in frame S and by (x′, y′, z′ and t′) in frames S′, if we count time from the instant when O and O′ momentarily coincide, then after a time t, the frame S′ is separated from frame S by a distance vt in the direction of x-axis as shown in figure 1. Then the observations of two observers O and O′ of the same event happening at P may be seen to be related by the following equations:
x′ = x − vt y′ = y z′ = z t′ = t
U| || V| || W
Y′
Y S
...(1)
Vt
P
O′
O
Z
S′
X, X′
Z′
Fig. 1 These equations are called Galilean transformation equations and relate to observations of position and time made by two sets of observers, located in two different inertial frames. Case II: When the second frame is moving along a straight line relative to first along any direction as shown in Fig. 2. Let the second frame S′ be moving relative to first frame S with a velocity v such that →
$ $ + $jv + kv v = iv x y z
Where vx, vy, and vz are the components of v along x, y and z axes respectively. Let (x, y, z, t) and (x′, y′, z′, t)′ be the coordinates of an event happening at P at any instant as observed by the two observers situated at O and O′ of frames S and S′ respectively. If the origins of two systems coincide initially, then after a time t the frame S′ is separated from frame S by a distance vxt, vyt and vzt along x, y and z axes respectively. Then referring to figure 2, we have x′ = x – vxt y′ = y – vy t z′ = z – vzt t′ = t
U| || V| || W
...(2)
Measurement
7 Y Y′
x′
P y′
v xt
x′
O′
v yt
O
z
X
v zt z′
Fig. 2
These are the Galilean transformations, relating to the observations of position and time made by two observers in two different inertial frames. Case III: When the second frame has uniform angular velocity relative to first. Let us consider two frames S and S′, the latter moving with uniform angular velocity ω relative to S about z-axis. Let the origins and axes of two frames coincide initially, i.e., at t = t′ = 0. In Fig. 3, z-axis is taken perpendicular to the plane of the paper. After time t, x′ and y′ axes are rotated by an angle ωt relative to x and y-axes respectively as shown in Fig. 3. The observations (x′, y′, z′, t′) taken by observer in S′ are related to those (x, y, z, t) taken by observer in S of the same event of P by the equations. x′ = Component of x along x′ + component of y along x′ + component of z along x′ y′ y
ωt x′
ωt x
z ′, z
Fig. 3
= x cos ωt + y sin ωt + z cos 90° or
x′ = x cos ωt + y sin ωt
also
y′ = Component of x along y′ + component of y along y′ + component of z along y′
...(3)
= x cos (90° + ωt) + y cos ωt + z cos 90° or
y′ = –x sin ωt + y cos ωt
and
z′ = Component of x along z′ + component of y along z′ + component of z along z′
...(4)
= x cos 90° + y cos 90° + z cos 0° or
z′ = z
...(5)
8
Mechanics
and
t′ = t
...(6)
Equations (3), (4), (5) and (6) are called time dependent Galilean transformations since they are time dependent and were obtained by Galileo. Galilean Transformation of the Velocity of a Particle: Let us consider two inertial frames S and S′, the frame S′ moving with velocity v relative to S, which is given by
$ $ + $jv + kv v = iv x y z
→
Let r and r′ be the position vectors of any particle at time t as observed by observer in frame S and S′ respectively. Then from Galilean transformations, we have r′ = r – vt and
t′ = t
...(1)
Differentiating eq. (1), keeping v constant, we get dr′ = dr – vdt and \
dt′ = dt
...(2)
dr′ dr dt dr −v = −v = dt ′ dt ′ dt′ dt
...(3)
dr′ dr = u′ = velocity of a particle relative to frame S and = u = velocity dt′ dt of particle relative to frame S′ because since
Hence, Eq. (3) gives u′ = u – v
...(4)
Which shows that the velocities measured by the observers in the two frames of reference are not the same. Alternatively, we can say that velocity of a body is not invariant under Galilean transformations. The inverse transformation from S′ to S, is obviously given by u = u′ + v
...(5)
Relation (4) or (5) are known as Galilean law of addition of velocities.
Invariance of Newton’s Second Law Newton’s second law of motion, which also includes the first law, is the real law of →
motion. To test it let a force F be acting on a mass ‘m’ in frame S, then since force is the rate of change of momentum, we have →
→ → d dv (m v ) = m = m a′ F = dt dt ′ →
→
Where the mass has been assumed to be independent of velocity and a is the acceleration produced in the mass Similarly, in frame S′, if F′ be the force acting on m, then →
→ → d d v′ = m a′ (mv′ ) = m F′ = dt ′ dt′ →
Measurement
9
→
where a′ , is the acceleration produced in mass m in frame S′. But since in inertial frames, →
→
a = a′
→
We have
→
F = F′
This shows that force and hence Newton’s second law of motion is invariant under Galilean transformation.
1.8 SCALARS AND VECTORS The physical quantities are of two types: Scalars and Vectors. Scalar Quantities: The quantities which have only magnitude and no direction, are called ‘Scalar Quantities’, e.g., mass, distance, time, speed, volume, density, work, charge, electric current, potential, frequency etc. A scalar quantity can be completely defined by a number and a unit. ‘mass’ of the truck is 200 Kg, the distance of my college is 5 km from my these statements we have given complete information about the quantity. subtraction, multiplication and division of scalar quantities can be done by
For example, the residence. In all The summation, ordinary algebra.
Vector Quantities There are certain physical quantities whose complete description not only requires their magnitude (i.e., a numerical value with appropriate unit) but also their direction in space e.g., velocity of a train. The magnitude of velocity is represented by a number such as 100 Km/ hour. This tells how fast the train is moving. But the description of velocity is complete only when we specify the direction of velocity also. We can represent the magnitude of velocity and the tip of the arrow represents its direction. If a particle is subjected to two velocities simultaneously its resultant velocity is different from the two velocities and is obtained by using a special rule. Suppose a particle is moving inside a long tube with a speed of 6 m/sec and the tube itself is moving in the room at a speed of 8 m/sec along a direction perpendicular to its length. Figure 1 represents the position of the tube and the particle at initial instant and after a time interval of 1 sec. Geometrical analysis gives the result that particle has moved a distance of 10 m in a direction θ = 53° from the tube. Hence, the resultant velocity of the particle is 10 m/sec along this direction. C t = 1 sec
R e su lta n t Ve lo city
8m
A
2 nd velo city
B 6m
1 st ve lo city t=0
Fig. 4
In figure 4, line AB represents the first velocity with point B as the head. Then we draw another line BC representing the second velocity with its tail coinciding with the head of the first line. Thus the line AC with C as head and A as the tail represents the resultant velocity. The resultant may also be called as the sum of the two velocities. We have added two velocities AB and BC and have obtained the sum AC. This rule of addition is called as the triangle law of addition.
10
Mechanics
Thus, the physical quantities which have magnitude and direction and which can be added according to the triangle rule, are called Vector quantities.
Different Types of Vectors 1.
Like Vectors: Two vectors are said to be like vectors if they have same direction, →
→
but different magnitude. Fig. 5 show two vectors A and B which have different magnitude but are paralled to each other. → A
→ B
Fig. 5
2.
Equal Vectors: Two vectors are said to be equal, if they have the same magnitude →
→
and direction. Fig. 6 shows two vectors A and B having the same magnitude and →
→
same direction and therefore, A = B
Fig. 6
For two vectors to be equal, it does not matter, whether the two vectors have their tails at the same point or not. If the scales selected for both the vectors is the same, they are represented by two equal and parallel lines. 3.
Unlike Vectors: The vectors having opposite direction and different magnitude, →
→
are called unlike vectors. Fig. 7 shows two such vectors A and B which have different magnitudes and are antiparallel to each other. → A
→ B
Fig. 7
4.
Opposite Vectors: The vectors having same magnitude but opposite direction, are →
→
known as opposite vectors. Fig. 8 shows two such vectors A and B having the same magnitude and opposite direction and therefore, →
→
A = –B
Fig. 8
5.
Unit Vectors: A vector divided by its magnitude is called a unit vector along the direction of the vector. Obviously, the unit vector has unit magnitude and direction is the same as that of the given vector. → $ and is read as ‘A A unit vector in the direction of same vector A is written as A cap’ or ‘A caret’ or ‘A hat’. Therefore, by definition
Measurement
11 →
→ $ = A $ or A A = AA A Thus, any vector can be expressed as magnitude times the unit vector along its own direction, unit vectors along x, y and z axes are represented by i$, $j and k$ respectively.
or
6.
Co-initial Vectors: Vectors are said to be coinitial, if they have a common initial point. In →
→
Fig. 9, A and B starting from the same point O as their origin are called Co-initial vectors. →
7.
Fig. 9
→
B A Co-linear Vectors: Two vectors having equal or unequal magnitude, which either act along the same line. Fig. 10(a) or along the parallel lines in the same direction Fig. 1-(b) or along the parallel lines in opposite direction. Fig. 10(c), are called colinear vectors. Like, unlike, equal and opposite vectors are collinear. → A
→ B
(a)
→ A
→ B
→ B
→ A
(b )
(c )
Fig. 10
8.
Coplanar Vectors: Vectors lying in the same plane are called coplanar vectors. Fig. 11(a) shows three vectors
→
→
→
x , y and z along mutually ⊥ axis x, y and z
respectively. These vectors are non coplanar but the vectors →
→
→
→
→
x – y , y – z and
→
z – x are coplanar Fig. 11(b). → y
→ y
→ → x–y
→ → y–x → x
→ z
→ x → z
→ → z–x
(a)
(b )
Fig. 11
9.
Null Vector: It is defined as a vector having zero magnitude. It has a direction, which is indeterminate as its magnitude is zero.
1.9 ADDITION OF VECTORS Since vectors have both magnitude and direction, they cannot be added by ordinary →
→
algebra. In Fig. 12 is shown the method of addition of two vectors A and B . Fig. 12(a). For
12
Mechanics →
→
this, we first draw vector A . Then starting from the arrow-head of A we draw the vector →
→
→
B Fig. 12(b). Finally, we draw a vector R starting from the initial point of A and ending →
→
→
→
at the arrow-head of B . Vector R would be the sum of A and B . →
→
→
R = A +B →
→
→
The magnitude of A + B can be determined by measuring the length of R and the →
→
→
direction can be expressed by measuring the angle between R and A ( or B ).
→ → → R=A +B
→ B
→ B → A
→ A → → → R=A +B (a )
(b)
Fig. 12
Geometrical Method of Vector Addition There are three laws of Vectors addition, namely triangle, parallelogram and polygon laws of vector addition. These laws can be used to add two or more vectors having an inclination with each other. (i)
Triangle law of Vector addition: According to this law if two vectors are represented both in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented totally (both in magnitude and direction) by the third side of the triangle taken in the opposite order. F
→ → → R=A +B
→ B → A
D
E → B
→ B
→ A
E
→ A
F
→ → → R=B +A
D
Fig. 13 →
→
Figure 13 shows two vectors A and B . In order to find the sum of these vectors by →
→
→
using triangle law of vector addition, draw vector DE = A . Then move vector B parallel to →
→
itself until its tail coincides with the tip of vector A. Show the arrow head of B by point F. →
→
→
Then vector DF = R drawn from the tail of vector A to the tip of vector B is the sum or →
→
resultant of vector A and B . Thus →
→
→
R = A+B
Measurement
13 →
→
→
→
(ii)
Vector addition in Commutative: From Figure 13, it is clear that A + B = B + A . This property of the vector addition according to which the vector addition is independent of the order in which the vectors are added is called cumulative property of vector addition. A physical quantity having both magnitude and direction is not a vector, if it does not obey commutative law.
(iii)
Parallelogram law of Vector addition: According to this law if two vectors acting simultaneously at a point can be represented both in magnitude and direction by two adjacent sides of a parallelogram, the resultant is represented completely (both in magnitude and direction) by the diagonal of the parallelogram passing through that point. →
→
Suppose we have to find the resultant of two vector A and B as shown in Fig. 14. →
→
−→
In order to find the resultant, draw the vector DE = A. Then move vector B →
parallel to itself, till its tail coincides with the tail of vector A . If we represent it →
−→
arrow head by point G, then vector DG represents vector B . Then complete the parallelogram. → −→
−→
−→
−→
A
G
→
Now EF = DG = B
→ B
→
and GF = DE = A
→ B
→ A
F
→ R
→ B
→ A
D
E
Fig. 14 →
It is clear from figure that diagonal DF of the parallelogram represents the resultant →
→
of sum of vectors A and B . (iv)
Polygon Law of Vector Addition: This law helps us to obtain the resultant of more than two vectors and is just the extension of the triangle law of vector addition. R → D
→ C
→ D
S → A
→ B
→ → → → A+B +C +D O
→ C
Q → → → A+B +C → → B B → + A P → A
Fig. 15
According to this law if a number of vectors are represented in magnitude and direction by the sides of a polygon taken in same order, then their resultant is represented in magnitude
14
Mechanics
and direction by the closing side of polygon taken in the opposite order. Figure 15 represents → → →
→
the addition of four vectors A , B, C and D by this method.
Analytical Method of Vector Addition (i)
→
→
Triangle law of vector addition: Suppose two vectors A and B represent both −→
−→
in magnitude and direction the sides PQ and QR of the triangle PQR taken in →
same order. Then according to triangle law of vector addition, the resultants R is represented by the closing side PR taken in the opposite order. →
Magnitude of the resultant R : Draw a perpendicular RS from the point R, on the side PQ which meets the line PQ at point S when produced forward. Then, from the ∆PSR , we get (PR)2 = (PS)2 + (SR)2 = (PQ + QS)2 + (SR)2 = (PQ)2 + (QS)2 + 2 PQ.QS + (SR)2 But ∴
(QS)2 + (SR)2 = (QR)2 (PR)2 = (PQ)2 + (QR)2 + 2PQ.QS
...(i)
R
→ R
→ B
α P
θ → A
Q
S
Fig. 16
From right angled triangle QRS, we get cos θ = QS/QR QS = QR cos/QR cos = θ
or Therefore, eq. (i) becomes
(PR)2 = (PQ)2 + (QR)2 + 2 PQ. QR. cos θ Now,
PR = R, PQ = A, QR = B
\
R2 = A2 + B2 + 2 AB cos θ
or
R =
[A2 + B2 + 2AB cos θ]
→
→
Direction of the resultant R : Suppose the resultant R makes an angle α with the → direction of A then from right angled triangle PRS, we get from Fig. 16 tan α = Now
RS RS = PS PQ + QS
PQ = A, QS = B cos θ, RS = B sin θ
Measurement
15
tan α =
Hence, (ii)
B sin θ A + B cos θ →
→
Parallelogram Law of Vector Addition: Suppose two vectors A and B inclined to each other at an angle θ be represented in magnitude and direction both by the −→
−→
concurrent sides PQ and PT of the parallelogram. PQUT as shown in figure 17. →
→
Then according to parallelogram law, resultant of A and B is represented both in −→
magnitude and direction by the diagonal PU of the parallelogram. → A
T
U
→ R
→ B θ
→ B
α
θ → A
P
Q
S
Fig. 17 →
Magnitude of the resultant R : Drop a perpendicular from the point U on the line PQ which meets the line PQ at some point S. From right handed triangle PSU, we get PU 2 = PS2+ SU2 = (PQ + QS)2 + SU2 = (PQ)2 + (QS)2 + 2 PQ.QS + (SU)2 But ∴
or
(QS)2 + (SU)2 = (QU)2 (PU)2 = (PQ)2 + (QU)2 + 2 PQ.QS
Now
PU = R, PQ = A and QU = B
Further
QS = B cos θ
Hence,
R2 = A2 + B2 + 2 AB cos θ R =
[A2 + B2 + 2 AB cos θ]
→
→
Direction of the resultant R : Suppose the resultant vector R makes an angle α with → vector A . Then from right angled triangle PUS, we get tan α =
US US B sin θ = = PS PQ + QS A + B cos θ
16
Mechanics
1.10 RESOLUTION OF VECTORS The process of splitting up a vector into two or more vectors is known as resolution of a vector. The vectors into which a given vector is split are called component vectors. The resolution of a vector into two mutually perpendicular vectors is called the rectangular resolution of vector in a plane or two dimensions. →
−→
Figure 18 shows a vector OR = r in the X-Y plane drawn from the origin O. Let the vector makes an angle α with the x-axis and β with the y-axis. This vector is to be resolved into two component vectors along two mutually perpendicular unit vectors i$ and j$ respectively, where i$ and $j are the unit vectors along x-axis and y-axis respectively as shown in figure 15. From point R, drop perpendiculars RP and RQ on x and y-axis respectively. The length OP −→
→
is called the projection of OR on x-axis while length OQ is the projection of OR on y-axis. According to parallelogram law of vector addition →
−→
−→
−→
r = OR = OP + OQ →
Thus we have resolved the vector r into two parts, one along OX and the other along OY, the magnitude of the part along OX is OP = rx = r cos α and the magnitude of the part along OY is OQ = r = r cos β i.e., in terms of unit vector i$ and j$ , we can write y
→ OP = →
and
Thus
→
i$ r cos α = i$ rx →
OQ =
$j r sin α = $j r y
=
$j r cos β = $j r y
→
r = =
Y
Q
θ
R
→
→
i$ r cos α + $j r cos β
ry
^j
→r β
$ + $jr ir x y
^i
α O
P
X
rx
Fig. 18
r If the vector r is not in the X-Y plane, it may have non zero projection along x, y and
z-axes and we can resolve it into three components i.e., along the x, y and z-axes. If α, β →
and γ be the angles made by the vector r with respect to x, y and z-axes respectively, then we can write →
r = i$ r cos α + $j r cos β + k$ r cos γ $ r = i$ rx + $j ry + k rz
→
Where i$, $j and k$ are the unit vectors along x, y and z axes respectively, the magnitude →
(r cos α) is called the component of r along x-axis, r cos β is called the component along y-axis and r cos γ is called the component along z-axis.
Measurement
17
Above equation also shows that any vector in three-dimensional can be expressed as a linear combination of the three unit vectors i$, $j and k$ .
1.11
MULTIPLICATION OF VECTORS There are three kinds of multiplication operations for vectors i.
multiplication of a vector by a scalar,
ii.
multiplication of two vectors in such a way as to yield a scalar, and
iii.
multiplication of two vectors in such a way as to yield another vector.
There are still other possibilities, but we shall not consider them here. The multiplication of a vector by a scalar has a simple meaning. The product of a scalar →
→
k and a vector a , written k a , is defined to be a new vector whose magnitude is k times the →
→
magnitude of a . The new vector has the same direction as a if k is positive and the opposite direction if k is negative. To divide a vector by a scalar we simply multiply the vector by the reciprocal of the scalar. Scalar Product (Dot Product): The scalar product of two vectors is defined as a scalar quantity having magnitude equal to the product of the magnitude of two vectors and the cosine of the smaller angle between them. Mathematically, if θ is the angle between vectors →
→
A and B then → →
A . B = AB cosθ Above equation can also be expressed as below → →
A . B = A(B cos θ) = B(A cos θ)
Fig. 19 →
→
where B cos θ is the magnitude of component of B along the direction of vector A and A cos θ →
→
is the magnitude of component of A along the direction of vector B . Therefore, the dot product of two vectors can also be interpreted as the product of the magnitude of one vector and the magnitude of the component of other vector along the direction of first vector. Dot product of two vectors can be positive or zero or negative depending upon θ is less than 90° or equal to 90° or 90° < θ < 180°.
Properties 1.
Dot product of two vectors is always commutative i.e., → →
→ →
A .B = B . A → →
A . B = AB cos θ →
→
where θ is the angle between A and B measured in anti clock wise direction → →
and B. A = BA cos( − θ) = AB cos θ → →
→ →
Thus A . B = B. A
18
Mechanics → B → B 0
A co s θ θ
θ B co s θ
→ A
→ A
Fig. 20
2.
The dot product of a vector with itself gives square of its magnitude i.e., → →
A . A = A . A cos 0° = A 2
3.
The dot product of two mutually perpendicular vectors is zero i.e., if two vectors →
→
A and B are perpendicular then → →
4.
A . B = AB cos 90° = 0 The dot product obeys the distributive law i.e., →
→
→
→ →
→ →
A . (B + C) = A . B + A . C 5.
Two vectors are collinear, if their dot product is numerically equal to product of their magnitudes i.e., When θ = 0° or 180° → →
A . B = AB → → For example i$ . i$ = $j . $j = k . k = 1 (as θ = 0°) →
→
i$ . $j = $j . k = k . i$ = 0 (as θ = 90° ) .
6.
Dot product of two vectors in term of their rectangular components in three dimensions. → →
A . B = (i$ A x + $j A y + k$ A z ) . (i$ B x + j$ B y + k$ Bz ) = AxBx + AyBy + AzBz Examples of some physical quantities which can be expressed as scalar product of two vectors: (a)
→
→
Work (W) is defined as the scalar product of force ( F ) and the displacement ( S ) i.e., → →
W = F.S (b)
→
→
Power (P) is defined as the scalar product of force ( F ) and the velocity ( v ) i.e., → →
P = F. v
Measurement
(c)
19
Magnetic flux (φ) linked with a surface is defined as the dot product of magnetic →
→
induction ( B ) and area vector ( A ) i.e., → →
φ = B. A Vector Product (Cross Product): The vector product of two vectors is defined as a vector having magnitude equal to the product of the magnitudes of two vectors with the sine of angle between them and direction perpendicular to the plane containing the two vectors in accordance with right handed screw rule or right hand thumb rule. →
→
If θ is the angle between vectors A and B , then →
→
A × B = AB sin θ n$ →
→
The direction of vector A × B is the same as that of unit vector n$ . It is decided by any of the following two rules:
Fig. 21 →
→
Right handed screw rule: Rotate a right handed screw from vector A to B through the smaller angle between them, then the direction of motion of screw gives the direction of →
→
vector A × B . Right hand thumb rule: Bend the finger of the right hand in such a way that they point →
→
in the direction of rotation from vector A to B
through the smaller angle between them, →
→
then the thumb points in the direction of vector A × B . The cross product of two vectors in term of their rectangular components is →
→
A × B = (i$ A x + j$ A y + k$ A z ) × (i$ B x + j$ B y + k$ B z ) = (Ay Bz – Az By) i$ + (Az Bx – Ax Bz) $j + (Ax By – Ay Bx) k$
=
i$
$j
k$
Ax
Ay
Az
Bx
By
Bz
20
Mechanics
Properties of Vector Product 1.
The cross product of the two vectors does not obey commutative law. →
→
→
→
→
→
A × B = − (B × A ) i.e., 2.
→
→
A × B ≠ (B × A )
The cross product follows the distribution law i.e. →
→
→
→
→
→
→
A × ( B − C) = A × B − A × C
3.
The cross product of a vector with itself is a NULL vector i.e., →
→
A × A = (A) . (A) sin 0° n$ = 0
4.
The cross product of two vectors represents the area of the parallelogram formed by them. From figure 22, show a parallelogram PQRS whose adjacent sides PQ and PS are →
→
represented by vectors A and B respectively Now, area of parallelogram = QP × SM = AB sin θ S
R
→ B B sin θ θ P
M
→ A
Q
Fig. 22 →
→
Because, the magnitude of vector A × B is AB sin θ, hence cross product of two vectors represents the area of parallelogram formed by it. It is worth noting that →
→
→
→
area vectors A × B acts along the perpendicular to the plane of two vectors A and B . 5.
The Cross product of unit vectors are i$ × i$ = $j × $j = k$ × k$ = (1) (1) sin 0° n$ = 0
i$ × j$ = (1) (1) sin 90° k$ = k$ Where k$ is a unit vector perpendicular to the plane of i$ and $j in a direction in which a right hand screw will advance, when rotated from i$ to $j . Also Similarly, and
$ $ − $j × i$ = (1) (1) sin 90° ( − k) = k $j × k$ = − k$ × $j = i$ k$ × i$ = − i$ × k$ = $j
Measurement
21
Examples of some physical quantities which can be expressed as cross product of two vectors: →
(a)
The instantaneous velocity ( v ) of a particle is equal to the cross product of its →
→
angular velocity (ω ) and the position vector ( r ) i.e., →
→
→
v = ω× r
(b)
→
The tangential acceleration ( at ) of a particle is equal to cross product of its angular →
→
acceleration ( α ) and the position vector ( r ) i.e., →
→
→
at = α × r
(c)
→
The centripetal acceleration ( ac ) of a particle is equal to the cross product of its →
angular velocity and the linear velocity ( v ) i.e., →
→
→
ac = ω × v
(d)
→
The force F on a charge q moving inside magnetic field is equal to charge times →
→
the cross product of its velocity ( v ) and magnetic induction ( B ) i.e., →
→
→
F = q ( v × B) (e)
→
→
→
The torque ( τ ) of a force ( F ) is equal to cross product of the position vector r and →
the force ( F ) applied i.e., →
→
→
τ = r ×F
(f)
→
→
The angular momentum ( L ) is equal to cross product of position vector ( r ) and →
linear momentum ( P ) of the particle i.e., →
→
→
L = r ×P
1.12 VECTORS AND THE LAWS OF PHYSICS Vectors turn out to be very useful in physics. It will be helpful to look a little , more → →
→
deeply into why this is true. Suppose that we have three vectors a , b and r which have components ax, ay, az: bx, by , bz and rx, ry, rz respectively in a particular coordinate system xyz of our reference frame. Let us suppose further that the three vectors are related so that →
→
→
r = a+ b
...(1)
22
Mechanics
By a simple extension of above equation r x = a x + bx r y = a y + by and
...(2)
r z = az + bz Now consider another coordinate system x′ y′ z′ which has these properties 1.
Its origin does not coincide with the origin of the first, or xyz, system and
2.
Its three axes are not parallel to the corresponding axes in the first system.
In other words, the second set of coordinates has been both translated and rotated with respect to the first. The components of the vectors
→ →
→
a , b and r in the new system would all prove, in general, to be different, we may represent them by ax′, ay′, az′, bx′, by′, bz′ and rx′, ry′, rz′ respectively. These new components would be found, however, to be related in that.
rx′ = ax′ + bx′ ry′ = ay′ + by′ and
...(3)
rz′ = az′ + bz ′ That is, in the new system we would find once again that →
→
→
r = a+ b
In more formal language: relations among vectors, of which Eq.(1) is only one example, are invariant (that is, are unchanged) with respect to translation or rotation of the coordinates. Now it is a fact of experience that the experiments on which the laws of physics are based and indeed the laws of physics themselves are similarly unchanged in form when we rotate or translate the reference system. Thus the language of vectors is an ideal one in which to express physical laws. If we can express a law in vector form the invariance of the law for translation and rotation of the coordinate system is assured by this purely geometrical property of vectors.
1.13 SPEED AND VELOCITY The speed of a moving object is the rate at which it covers distance. It is important to distinguish between average speed and instantaneous speed. The average speed v of something that travels the distance S in the time interval t is v =
s t
Average speed = distance traveled/time interval Thus a car that has gone 150 km in 5 h had a average speed of v =
s = (150/5) km/h = 30 km/h t
Measurement
23
The average speed of the car is only part of the story of its journey, however, because knowing v does not tell us whether the car had the same speed for the entire 5 h or sometimes went faster than 30 km/h and sometimes slower. Even though the car’s speed is changing, at every moment it has a certain definite value (which is what is indicated by its speedometer). To find this instantaneous speed v at a particular time t, we draw a straight-line tangent to the distance- time curve at that value of t. The length of the line does not matter. Then we determine v from the tangent line from the formula v = ∆s/∆t Where ∆s is the distance interval between the ends of the tangent and ∆t is the time interval between them. (∆ is the Greek capital letter delta). The instantaneous speed of the car at t = 40 s is from figure Total distance, (m)
0
100
200
300
400
500
Elapsed time, (S)
0
28
40
49
57
63
V = ∆s/∆t = 100 m/10 s = 10 m/s
D ista nce (m )
5 00 4 00 3 00 2 00 1 00 0 10
20
30
40
50
60
70
Tim e (S )
Fig. 23
When the instantaneous speed of an object does not change, it is moving at constant speed. The speed of a moving object is a scalar quantity. Its unit is meter/second and its dimension are (LT–1).
Velocity The object’s velocity, however, includes the direction in which it is moving and is a vector →
quantity. If the object undergoes a displacement s in the time interval, its average velocity →
v during this interval is →
v = s/t
Instantaneous velocity v is the value of ∆s/∆t at a particular moment, and for straightline motion is found by the same procedure as that used for instantaneous speed v but with the direction specified as well.
24
Mechanics
1.14 ACCELERATION If the velocity of a moving object is changing, then its motion is called ‘accelerated motion’. The change in velocity may be in magnitude (speed), or in direction or in both. If the object is moving along a straight line, then only the magnitude of velocity (speed) changes. The time-rate of change of velocity of an object is called the ‘acceleration’ of that object. That is Acceleration = Change in velocity/time interval →
Acceleration is generally represented by ‘ a ’. Since velocity is a vector quantity, the acceleration is also a vector quantity. Suppose, the velocity of a moving object is v1 at time t1 and becomes v2 at time t2. It means that in the time-interval (t2 – t1), the change in the velocity of the object is (v2 – v1). Hence, the average acceleration of the object in time-interval t2 – t1 is →
a =
v2 − v1 ∆v = ∆t t2 − t1
If the time interval ∆t is infinitesimally small (∆t → 0) then the above formula gives acceleration ‘at a particular time’. This is called ‘instantaneous acceleration’ and is given by →
a = Lim
∆t → 0
∆v dv = ∆t dt
The unit of acceleration is meter /second2 and its dimensional formula is (LT–2). If the velocity of an object undergoes equal changes in equal interval of time, then its acceleration is said to be ‘uniform’. If the magnitude of the velocity (speed) of an object is increasing with time, then the acceleration of the object is positive. If the magnitude of the velocity (speed) is decreasing, then the acceleration is negative and it is called ‘retardation’.
1.15 RECTILINEAR MOTION When a particle moves along a straight line, its motion is said to be ‘rectilinear’. In this type of motion the acceleration of the particle is either zero (velocity constant) or arises from a change in the magnitude of the velocity. Let us consider a particle in rectilinear motion, say along the x-axis, under a constant acceleration a. We can derive relations between the kinematical variable x, v, a and t. Suppose the particle has velocity v0 at the origin (x = 0) at time t = 0, and moving with constant acceleration a, acquires a velocity v at time t. From the definition of acceleration, we have a = or
∆v v − v0 = ∆t t−0
v = v0 + at
...(1)
Since the acceleration is constant, the average velocity vav in time-interval t equal to one half the sum of the velocities at the beginning and the end of the interval. Thus
vav =
v0 + v 2
Measurement
25
Hence, the displacement at time t is x = vav × t =
v0 + v t 2
Substituting for v from Eq.(1), we get x =
v0 + (v0 + at) t 2
x = vo t + ½ at2
...(2)
Eliminating t between (1) and (2), we can get v 2 = v20 + 2ax
... (3)
Eqs. (1), (2) and (3) are equations of rectilinear motion under constant acceleration.
1.16 ACCELERATION OF GRAVITY When a body is allowed to fall over earth’s surface from a certain point above it, it is seen that (a)
its velocity increases continuously and the acceleration is found to be uniform and
(b)
that the acceleration is the same for all bodies.
This uniform acceleration of all the bodies when falling down under the earths action of gravity is called acceleration due to gravity. It is represented by ‘g’. If the resistance offered by air to the motion of body is assumed negligible, then all the bodies falling freely under gravity have the same acceleration ‘g’ acting vertically downwards. The value of g varies from a value 9.781 ms–2 at the equator to the value 9.831 ms–2 at the poles. The equation derived for rectilinear motion along the x-axis can be used for freely falling bodies as well. Let us take y-axis as vertical, here the constant acceleration is g directed vertically downwards. Thus after replacing x by y and a by g in the equation of motion, we obtain equations for vertically downward motion: v = vo + gt y = vot + ½ gt2 v 2 = vo2 + 2gy →
In case of bodies moving vertically upward, the acceleration due to gravity g acts in the opposite direction i.e., as retardation. Hence the equations of motion would be v = vo – gt y = vo t – ½ gt2 v 2 = vo2 – 2gy
1.17 ACCURACY AND ERRORS IN MEASUREMENT In many experiments in the physics laboratory, the aim is to determine the value of physical constant. To determine a physical constant, we have to measure the various quantities
26
Mechanics
which are connected with that physical constant by a formula. For example, to determine the density (ρ) of a metal block, we have to measure its mass (m) and its volume (V) which are related to ρ by the formula: ρ = m/V The accuracy in the value of ρ obviously depends upon the accuracy in the measurements of m and V. Measurements of the quantities in the formula involve errors which are of two types: (a) Random errors and (b) Systematic errors
Random Errors Random errors may be due to (i) small changes in the conditions of a measurements and (ii) the incorrect judgment of the observer in making a measurement. For example, suppose you are determining the weight of a body with the help of a spring balance. You will usually make an error in estimating the coincidence of the pointer with the scale reading or in assessing the correct position of the pointer when it lies between two consecutive graduations of the scale. This error, which is due to incorrect judgment of the observer, is also an example of random errors. Random errors cannot be traced to any systematic or constant cause of error. They do not obey any well-defined law of action. Their character can be understood or appreciated from the illustration of firing shots at a target using a rifle. The target is usually a bull’s eye with concentric rings round it. The result of firing a large number of shots at the target is well known. The target will be marked by a well-grouped arrangement of shots. A large number of shots will be nearer a certain point and other shots will be grouped around it on all sides. These shots which are grouped on both sides of the correct point obey the law of probability which means that large random errors are less probable to occur than small ones. A study of the target will show that the random shots lie with as many to one side of the centre as to the other. They will also show that small derivations from the center are more numerous than large deviation and that a large deviation is very rare.
Method of Minimizing Random Errors If we make a large number of measurements of the same quantity then it is very likely that the majority of these measurements will have small errors which might be positive or negative. The error will be positive or negative depending on whether the observed measurement is above or below the correct value. Thus random errors can be minimized by taking the arithmetic mean of a large number of measurements of the same quantity. This arithmetic mean will be very close to the correct result. If one or two measurements differ widely from the rest, they should be rejected while finding the mean.
Systematic Errors During the course of some measurements, certain sources of error operate constantly or systematically making the measurement. Systematically greater or smaller than the correct reading. These errors, whose cause can be traced, are called systematic errors. All instrumental errors belong to this category, such as the zero error in vernier callipers and micrometer screw, the index error in an optical bench, the end error in a meter bridge, faulty graduations of a measuring scale, etc.
Elimination of Systematic Errors To eliminate systematic errors, different methods are used in different cases.
Measurement
27
1.
In some cases, the errors are determined previously and the measurements corrected accordingly. For example, the zero error in an instrument is determined before a measurement is made and each measurement is corrected accordingly.
2.
In some cases the error is allowed to occur and finally eliminated with the data obtained from the measurements. The heat loss due to radiation is taken into account and corrected for from the record of the temperature at different times.
Order of Accuracy Even after random and systematic errors are minimized, the measurement has a certain order of accuracy which is determined by the least count of the measuring instruments used in that measurement. Suppose we determine the value of a physical quantity u by measuring three quantities x, y and z whose true values are related to u by the equation. u = xα yβ z–γ
...(a)
Let the expected small errors in the measurement of quantities x, y and z be respectively ± δx, ± δy and ± δz so that the error in u by using these observed quantities is ± δu. The numerical values of δx, δy and δz are given by the least count of the instruments used to measure them. Taking logarithm of both sides of Eq. (a), we have log u = α log x + β log y – γ log z Partial differentiation of the above equation gives
δu δx δy δz +β −γ = α u x y z
...(b)
δu . The values of δx, δy and δz may be u positive or negative and in some case the terms on the right hand side of Eq. (b) may counteract each other. This effect cannot be relied upon and it is necessary to consider the worst case which is the case when all errors add up giving an error δu given by the equation: The proportional or relative error in u is
FG δu IJ H uK
= α maximum
δx δy δz +β +γ x y z
...(c)
Thus to find the maximum proportional error in u, multiply the proportional errors in each factor (x, y and z) by the numerical value of the power to which each factor is raised and then add all the terms so obtained. The sum thus obtained will give the maximum proportional error in the result of u. When the proportional error of a quantity is multiplied by 100, we get the percentage error of that quantity. It is evident from Eq. (c) that a small error in the measurement of the quantity having the highest power will contribute maximum percentage error in the value of u. Hence, the quantity having the highest power should be measured with a great precision as possible. This is illustrated in the following example.
28
Mechanics
Example: In an experiment for determining the density ( ρ ) of a rectangular block of a metal, the dimension of the block are measured with callipers having a least count of 0.01 cm and its mass is measured with beam balance of least count 0.1 g. The measured values are: mass of the block (m) = 39.3 gm, length of block (x) = 5.12 cm, Breadth of block (y) = 2.56 cm, Thickness of block (z) = 0.37 cm. Error in m = δm = ± 0.1 g Error in x = δx = ± 0.01 cm Error in y = δy = ± 0.01 cm Error in z = δz = ± 0.01 cm Find the maximum proportional error in the determination of ρ. Ans: The density of the block is given by ρ = m/xyz
39. 3 = 8.1037 g cm−3 5.12 × 2. 56 × 0. 37 Proportional error in ρ is given by the equation Calculating ρ omitting errors, ρ =
δρ δm δx δy δz − − − ρ = m x y z
The maximum proportional error in ρ is given by the equation
FG δρ IJ H ρK
=
δm δx δy δz + + + m x y z
=
0.1 0. 01 0. 01 0. 01 + + + 39. 3 5.12 2. 56 0. 37
max
= 0.0025 + 0.0019 + 0.0039 + 0.0270 = 0.0353 ∴ Maximum percentage error = 0.0353 × 100 = 3.53% Hence the error is 3.53% of 8.1037. Clearly it is absurd to give the result for ρ to five significant figures. The error in ρ is given by δρ = 0.0353 × ρ = 0.0353 × 8.1037 = 0. 286 ≅ 0. 3 Hence the value of ρ = 8.1037 is not accurate up to the fourth decimal place. In fact, it is accurate only up to the first decimal place. Hence the value of ρ must be rounded off as 8.1 and the result of measurements is written as ρ = ( 8.1 ± 0. 3) g cm −3 It is clear that such a large error in the measurement of ρ is due to a large error (=0.027) in the measurement of z, the smallest of the quantities measured. Hence the order of accuracy of ρ should be increased by measuring z with an instrument having a least count which is smaller than 0.01 cm. Thus a micrometer screw (least count = 0.001 cm), rather than a vernier caliper should be used to measure z.
Measurement
29
SUMMARY Physics is a branch of science that deals with the study of the phenomena of nature. The scientific method used in the study of science involves observation, proposal of a theory, testing the consequences of the proposed theory and modification or refinement of the theory in the light of new facts. The applications of physics have played a very great role in technology and in our daily lives. Physics is a science of measurement. All quantities which can be measured either indirectly or directly such as length, mass, time, force, temperature, light intensity, electric current etc. are called physical quantities of numerous such quantities, length, mass and time are regarded as fundamental quantities. The measurement of these quantities involves the choice of a unit. The internationally accepted units of length, mass and time respectively are metre, kilogram and second. The units of all other mechanical quantities are derived from these three basic units. These quantities are measured by direct and indirect methods. The measured values of these quantities, show a very wide range variation. Different physical quantities have different dimensions. The dimensions of a physical quantity are the number of times the fundamental units of mass, length and time appear in that quantity.
30
Mechanics
2 FORCE AND MOTION 2.1 MECHANICS Mechanics, the oldest of the physical sciences, is the study of the motion of objects. The calculation of the path of an artillery shell or of a space probe sent from Earth to mars are among its problems. When we describe motion (or trajectories) we are dealing with that part of mechanics called kinematics, ignoring the forces producing the motion. When we relate motion to the forces associated with it and to the properties of the moving objects, we are dealing with dynamics.
2.2 CAUSE OF MOTION: FORCE From daily experience we know that the motion of a body is a direct result of its interactions with the other bodies around it (which form its environment). When a batsman hits a ball, he interacts with the ball and modifies its motion. The motion of a freely falling body or a projectile is the result of its interaction with the earth. The motion of an electron around a nucleus is the result of its interaction with the nucleus. An interaction is quantitatively expressed in terms of a concept called ‘force’. An intuitive motion of force is derived in terms of a push or a pull. When we push or pull on a body, we are said to exert a (muscle) force on it. Earth which pulls all bodies towards its centre is said to exert a (gravitational) force on them. A stretched spring pulling a body attached to its end is said to exert a (elastic) force on the body. A locomotive exerts a force on the train it is pulling or pushing. Thus every force exerted on a body is associated with some other body in the environment. It is not always that an application of force will result in motion or change in motion. For example, we may push a wall, i.e., there is an interaction between us and the wall, and hence there is a force, but the wall may not move at all. Thus, force may be described as a push or pull, resulting from the interaction between bodies, which produces or tends to produce motion or change in motion. The analysis of the relation between force and the motion of a body is based on three laws of motion which were first stated by Sir Issac Newton. 30
Force and Motion
31
2.3 NEWTON’S LAWS OF MOTION Newton’s laws of motion are the basis of mechanics. Galileo’s version of inertia was formalized by Newton in a form that has come to be known as Newton’s first law of motion. Newton’s first law of motion: stated in Newton’s words, the first law of motion is: “Every body continues in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it”. Newton’s first law is also known as law of inertia and the motion of a body not subject to the action of other forces is said to be inertial motion. With the help of this law we can define force as an external cause which changes or tends to change the state of rest or of uniform motion of a body. Newton’s first law is really a statement about reference frames. We know that the motion of a body can be described only relative to some other body. Its motion relative to one body may be very different from that relative to another. A passenger in an aircraft which is on its take-off run is at rest relative to the aircraft but is in accelerated motion relative to the earth. Therefore we always choose a set of coordinate axes attached to a specified body, relative to which the motion of a given body is described. Such a set of coordinate axes is known as a “reference frame”. The first law tells us that we can find a reference frame relative to which a body remains at rest or in uniform motion along a straight line (i.e., it has no acceleration) when no net external force acts upon it. Such a reference frame is called an ‘inertial frame’. Thus an inertial frame is one in which Newton’s first law correctly describes the motion of a body not acted on by a net force. Such frames are either fixed with respect to the distant stars or moving at uniform velocity with respect to them. A reference frame attached to the earth can be considered to be an inertial frame for most practical purposes, although it is not precisely so, due to the axial and orbital motions of the earth. But if a frame of reference is inertial, then every other frame which is in uniform motion relative to it is also inertial. Newton’s first law makes no distinction between a body at rest and one moving with a constant velocity. Both states are “natural” when no net external force or interaction acts on the body. That this is so becomes clear when a body at rest in one inertial frame is observed from a second (inertial) frame moving with constant velocity relative to the first. An observer in the first frame finds the body to be at rest, an observer in the second frame finds the same body to be moving with uniform velocity. Both observers find the body with no acceleration. Thus both observe that Newton’s first law is being obeyed.
Newton’s Second Law of Motion Newton’s second law tells us what happens to the state of rest or of uniform motion of a body when a net external force acts on the body i.e., when the body interacts with other surrounding bodies. This law states: “The change of motion of an object is proportional to the force impressed, and is made in the direction of the straight line in which the force is impressed”. By “change of motion” Newton meant the rate of change of momentum (p) with time. So mathematically we have
32
Mechanics →
F ∝
d → ( p) dt
→
F = k
or
d → ( p) dt
...(1)
→
Where F is the impressed force and k is a constant of proportionality. The differential operator d/dt indicates the rate of change with time. Now, if the mass of the body remains constant (i.e., neither the body is gaining in mass like a conveyer belt nor it is disintegrating like a rocket), then →
→
→ → d p = d (m v ) = dv m = ma dt dt dt
→
Where a =
→
dv = the acceleration of the body. Thus Eq. (1) becomes dt →
→
F = km a F = km a
This law provide a quantitative definition of force. Thus force is equal to mass times acceleration, if the mass is constant. The force has the same direction as the acceleration. This is an alternative statement of the second law. →
We note that the Newton’s first law is contained in the second law as a special case, if →
F = 0, then a = 0. In other words, if the net force on a body is zero, the acceleration of the body is zero. Therefore, in the absence of a net force a body will move with constant velocity or be at rest, which is the first law.
Newton’s Third Law of Motion A force acting on a body arises as a result of its ineraction with another body surrounding it. Thus any single force is only one aspect of a mutual interaction between two bodies. We find that whenever one body exerts a force on a second body, the second body always exerts on the first a force which is equal in magnitude but opposite in direction and has the same line of action. A single isolated force is therefore an impossibility. The two forces involved in every interaction between the bodies are called an ‘action’ and a ‘reaction’. Either force may be considered the ‘action’ and the other the ‘reaction’. This property of forces was stated by Newton in his third law of motion: “To every action there is always an equal and opposite reaction”. If a body A exerts a force FAB on a body B, then the body B in turn exerts a force FBA on A, such that FAB = –FBA So, we have
FAB + FBA = 0
Force and Motion
33
2.4 FREELY FALLING BODIES It is a matter of common observation that bodies fall towards the ground. Leaves of the tree, fruits from the tree, a body just dropped from the top of a tower, a stone thrown upward etc. all reach the ground. According to Aristotle it was believed that the time taken by a heavier body to reach the ground as less as compared to a lighter body dropped from the same height. This belief continued until the sixteenth century and nobody either challenged it or tried to prove its correctness. In 1590, Galileo disapproved this idea. He showed from his famous experiments at the leaning tower of Pisa that all bodies dropped from the same height reach the ground at the same instant irrespective of their masses. Galileo showed that bodies of different masses when dropped from the top of the leaning tower of Pisa, reached the ground at the same time. A large crowd was present to see the correctness of Galileo’s idea. When a stone and a piece of paper were dropped from the tower at the same time, it was found that the stone reached the ground earlier. This difference in time was explained by Galileo to be due to the resistance offered by air on the paper. He showed that if a paper and a stone are dropped from the same height in vacuum, both take the same time to fall through the same height.
Laws of Falling Bodies 1.
All bodies fall with equal rapidity in vacuum irrespective (Equations of Motion in Free Fall) of their masses
2.
The velocity acquired by a body falling freely from rest is directly proportional to the time of its fall or
3.
v ∝t v = gt
The distance moved by a body falling freely from rest is directly proportional to the square of the time of fall. S ∝ t2 S = ½ gt2
2.5 MOTION IN A VERTICAL PLANE Equations of motion in a vertical plane is given by v = u – gt S = ut – (gt2 )/2 v 2 = u2 – 2 gs
Components of Vectors In mechanics and other branches of physics, we often need to find the component of a vector in a certain direction. The component is the ‘effective part’ of the vector in that direction. We can illustrate it by considering a picture held up by two strings OP and OQ each at an angle of 60° to the vertical. Figure 1(a). If the force or tension in OP is 6 N, its vertical component S acting upwards at P helps to support the weight W of the picture, which acts vertically downwards. The upward component T of the 6 N force in OQ acting in the direction QT, also helps to support W. Figure 1(b) shows how the component value of a vector F can
34
Mechanics
be found in a direction OX. If OR represents F to scale, we draw a rectangle OPRQ which has OR as its diagonal. Now F is the sum of the vectors OP and OQ. The vector OQ has no effect in a direction OX at 90° to itself. So the effective part or component of F in the direction OX is the vector OP. If θ is the angle between F and the direction OX, then OP/OR = cos θ or OP = OR cos θ = F cos θ So the component of any vector F in a direction making an angle θ to F is always given by = F cos θ In a direction OY perpendicular to OX, F has a component F cos (90° – θ) which is F sin θ. This component is represented by OQ in the figure (b). S
T
O 6N
6 0°
6 0°
Y
R
Q
P
F
F sin θ
Q
9 0° θ O
F co s θ
P
X
W (a )
(b )
Fig. 1 Components of vectors.
2.6 PROJECTILE MOTION A body projected in the horizontal direction with a uniform velocity is under the action of (i) uniform velocity in the horizontal direction and (ii) uniform acceleration due to gravity in the vertically downward direction. Such a body is known as projectile e.g., a bomb released from an aeroplane moving with a uniform speed and a bullet fired from a rifle. The path traversed by the projectile is called its trajectory.
A
u
h
Consider a tower AB of height h. A stone is thrown in the horizontal direction from the top of tower. As soon as the stone is released, it is under the action of C B (i) uniform velocity ‘u’ in the horizontal direction and Fig. 2 (ii) acceleration due to gravity ‘g’ in the vertically downward direction. Under the combined effect of the two, the stone does not reach the foot of the tower but it falls at some distance away from the foot of the tower. Suppose the stone reaches the point C. The distance BC is the horizontal range. Suppose the time taken by stone to reach the ground = t
Force and Motion
∴
35
Vertical distance h = ½ gt2
...(1)
Since the initial velocity in the vertical direction is zero, the horizontal distance
A
u +
BC = x = u.t ∴
t = x/u
...(2)
+
h
+
From Eqs. (1) and (2) h = ½ g (x/u)2 x2
or
B
2u2 h = g
C
Fig. 3
This is the equation of a parabola. Therefore the path traversed by the stone is a parabola. Take the case of an aeroplane flying at a height h from the ground. It is to drop a bomb at C. If the bomb is released just at the moment when the aeroplane is vertically above c, the bomb will not fall at C but it will fall at a point some distance away from C. From the knowledge of the height h, time t can be calculated from the relation: h = ½ gt2, or t = and
2h g
A
BC = ut
Therefore the bomb should be released at a point A which is vertically above B and t seconds earlier. Then after traversing a parabolic path the bomb will reach the point C. It is to be remembered that the aeroplane will also reach a point vertically above C just at the moment when the bomb reaches the point C. In calculating the distance BC air resistance is neglected.
+
h
+
+
B
C
Fig. 4
Show that the horizontal range is maximum when the angle of projection is 45° with the horizontal. Suppose a body is projected with a velocity u making an angle θ with the horizontal. Resolve this velocity u into two components. (i)
u cos θ in the horizontal direction,
(ii)
u sin θ in the vertically upward direction. B
u sin θ
u θ A u cos θ
C
Fig. 5
36
Mechanics
The horizontal component remains constant throughout the flight of the body in air, whereas the vertical component of the velocity changes continuously. The acceleration due to gravity is negative. Maximum height: Suppose the maximum vertical height reached is h. From the equation,
v2 – u2 = 2gh v = 0, u = u sin θ and g is negative
∴
0–
u2
sin2
θ = –2gh u2 sin2 θ h = 2g
or
...(i)
After time t, the distance travelled in the horizontal direction, x = (u cos θ)t
...(ii)
Distance travelled in the vertical direction y = (u sin θ) t – From equation (ii),
t = x/(u cos θ) y =
1 2 gt 2
FG H
...(iii)
u sin θ x 1 x − g u cos θ u cos θ 2
FG H
g x = x tan θ − 2 u cos θ
IJ K
IJ K
2
2
...(iv)
This is the equation of a parabola Time in air: Suppose the projectile remains for time t in air: It means y = 0 From equation (iii) 0 = (u sin θ) t – or
t =
1 2 gt 2
2u sin θ g
...(v)
Maximum range: The horizontal range x = (u cos θ) t Putting value of t from Eq. (v), we get x = (u cos θ). =
u2 sin 2θ g
2u sin θ g ...(vi)
The range x will be maximum When sin 2θ = 1 or 2θ = 90° i.e., θ = 45° Hence the horizontal range is maximum when the projectile make an angle of 45° with the horizontal.
Force and Motion
37
2.7 EQUILIBRIUM OF FORCES When a particle is in equilibrium, then the resultant of all the forces acting on it is zero. It then follows from Newton’s first law of motion that a particle in equilibrium is either at rest or is moving in a straight line with constant speed. It is found that for a large number of problems, we have to deal with equilibrium of forces lying in a plane. Therefore, we shall restrict our discussion to the case when a particle is in equilibrium under the influence of →
→
→
a number of coplanar forces F1 , F2 , F3 ... . The required condition is given by →
→
→
F1 + F2 + F3… = 0
...(i)
Since the forces are coplanar, we can resolve them along two mutually perpendicular directions of x and y-axes, O being the particle. So above equation (i) can be rewritten as
eF or
j e bF + F g i$ + dF
j
$ + F $j + F i$ + F j$ + ... = 0 1y 2x 2y
1x i
1x
2x
1y
i
+ F2 y $j = 0
Σ(F1 x + F2 x + ...) = 0 and
Σ (F1 y + F2 y + ...) = 0
Y F2
UV W
F1
(ii) O
Equation (ii) can be expressed in a concise form as Σ Fx = 0, and
Σ Fy = 0
Where Σ denotes summation of the x- or y-components of the forces.
X
F4 F3
Fig. 6
2.8 FRICTIONAL FORCES When one body moves in contact with another, its motion is opposed by a force which comes into play at the plane of contact of the two bodies. This resistive force which tends to destroy the relative motion between the two is called the “force of friction”. Frictional forces may occur between the surfaces of contact even when there is no relative motion between the bodies. Work is done in overcoming the friction and consequently friction produces power losses in moving machinery whose efficiency falls below hundred percent. Friction has many useful aspects also. We could not walk without friction and if somehow we start moving we could not stop. It would have been difficult even to fix nail in the wall. Though annoying, friction is a necessity for us as our everyday activities depend upon friction.
Origin of friction The phenomenon of friction occurring between dry surfaces of two sliding bodies is complicated when it is observed at microscopic level. The force laws for friction do not have simplicity and accuracy of very high order. Large number of surface irregularities is the key reason of friction. When a pull is exerted on a body so as to let it slide over the other surface; the relative motion is resisted on account of large number of surface irregularities and an opposing force is developed which is the force of friction. As the pull is increased, the opposing force also increases and at a certain pull, the body begins sliding.
38
Mechanics
2.9 STATIC FRICTION AND COEFFICIENT OF STATIC FRICTION Suppose a block rests over a horizontal table Fig. 7(a). weight of block is balanced by the normal reaction. Next a very small horizontal force Fa is applied to the block Fig. 7(b). The block still remains stationary. One can say that “frictional force Fs” has come into existence (Fs – Fa). The frictional force preventing a body from sliding over the surface of other body is termed static friction. Again if applied force Fa is increased a little, the block still remains stationary which suggests that frictional force Fs has also increased with Fa. Increasing Fa, a stage is reached when the block begins to move. Till the motion does not start, the Fs is equal and opposite to applied force in all stages. The greatest value of Fs at the plane of contact of surfaces of two bodies in the stage when one body is just to slide over the other is termed “limiting friction”, and is denoted by Fs. Which acts between surfaces at rest with respect to each other is called ‘force of static friction’. Again Maximum force of static friction, Fs = Smallest force needed to start motion FL The coefficient of static friction (µs) is defined as µs =
Magnitude of maximum forces of static friction Magnitude of normal force
i.e.,
µs = Fs/N
or
Fs = µsN N
N Fa Fs
mg
mg
(a ) N o fo rce a pp lied
(b ) A p plie d force exists, b u t n o m o tio n
Fig. 7
2.10 DYNAMIC FRICTION AND COEFFICIENT OF DYNAMIC FRICTION In general a smaller force is needed to maintain uniform motion as compared to start the motion of a body over the surface. The force acting between surfaces in relative motion is called forces of sliding friction (or kinetic friction or dynamic friction) and is denoted by Fd. The kinetic friction is measured by the force necessary to keep the two surfaces in uniform relative motion in contact with one another, coefficient of dynamic friction, (µd) is defined as µd = Magnitude of force of dynamic friction/magnitude of normal force i.e.,
µd = Fd/N
or
Fd = µd N
N U n ifo rm m otion Fo Fd = µd N
mg
Fig. 8
Dependence of Coefficient of Friction Coefficient of friction (µ) , depends upon (i) finishing of the surface, (ii) nature of material, (iii) surface films, (iv) contamination, (v) temperature.
Force and Motion
39
The explanation of the two laws of friction lies in surface adhesion theory. In “rolling”’ the frictional force is proportional to the velocity of the body; i.e., F = –kv Where k = constant of proportionality If, motion of the body is so fast that fluid swirls around (aeroplane moving in air) than frictional drag is given by F = –kv2 For still higher velocities above equation does not hold good.
Angle of Friction and Cone of Friction Let us consider a block A of mass m. HL is a horizontal rough surface. A force P is applied parallel to surface on the block. Fs is frictional force, N normal reaction force. If body be at rest, the resultant of Fs and N is a single force R that makes angle θ with normal to surface, we have N = R cos θ Fs = R sin θ tan θ = Fs/N
and so
N P
FS
θ
N R
A P L
H
θ
Fs
Fig. 9
mg
Fig. 10
As P is slowly increased, Fs also increases and thus θ also increases and θ is maximum when P is such that block is just to move. The friction in this case is termed “limiting friction”. The angle of friction (λ) is defined as the angle which the resultant reaction (R) of the surface makes with the normal when the friction is limiting. Thus, (tan θ)limiting = tan λ =
(Fs )limiting N
FL = µs N = (Fs)limiting
tan λ =
or As
FL
Resultant of FL and N lies on the surface of cone having λ as the semi-vertical angle and direction of N as its axis. This cone is termed “cone of friction” as shown in figure 11.
Fig. 11
PROBLEM Q. 1. Two blocks of masses m1 and m2 are connected by a massless spring on a horizontal frictionless table. Find the ratio of their accelerations a1 and a2 after they are pulled apart and then released.
40
Mechanics
Solution. The blocks are pulled apart by equal and opposite forces. On being released they start moving toward each other under equal and opposite (elastic) force exerted by the massless spring. Since force is equals to mass multiplied by acceleration, we have F = m1 a1 = m2a2 ∴
a1 m2 = a2 m1
Q. 2. A car of mass 1200 kg moving at 22 m/s is brought to rest over a distance of 50m. Find the breaking force and the time required to stop. Solution. As soon as the brakes are applied, the car is decelerated. If a be the (negative) acceleration, then from the relation ν2 = ν2o + 2 ax, we have 0 = (22)2 + 2a (50)
22 × 22 = − 4. 84 m/s2 2 × 50 The braking (retarding) force on the car is, therefore,
so that
a = −
F = mass × acceleration = 1200 × (–4.84) = –5808 nt. The negative sign signifies retardation. Let t be the time the car takes to stop. Then, from the relation v = v0 + at, we have 0 = 22 + (–4.84)t ∴
t = 22/4.84 = 4.55 sec.
Q. 3. A car having a mass of 150 kg is moving at 60 km/hour. When the brakes are applied to produce a constant deceleration, the car stops in 1.2 min. Determine the force applied to the car. [Ans. –347 nt.] Q. 4. A gun fires ten 2 gm bullets per second with a speed of 500 m/sec. The bullets are stopped by a rigid wall. (a) What is the momentum of each bullet? (b) What is the kinetic energy of each bullet? (c) What is the average force exerted by the bullets on the wall? Solution: (a) The momentum of each bullet is P = mv = (2 × 10–3 kg ) (500 m/s) = 1 Kg-m/s. (b) The kinetic energy of each bullet is K = ½ mv2 = ½ (2 × 10–3 kg) (500 m/s)2 = 250 joule. (c) The bullets are stopped by a rigid wall. The magnitude of the change of momentum for each bullet is therefore 1 kg-m/s. Since 10 bullets are fired per second, the rate of change of momentum is dP/dt = 10 kg-m/s2 But this must be equal to the average force exerted on the wall. Thus F = dP/dt = 10 kg-m/s2 = 10 nt.
Force and Motion
41
Q. 5. A body of mass 10 gm falls from a height of 3 meter into a pile of sand. The body penetrates the sand a distance of 3 cm before stopping. What force has the sand exerted on the body? Solution. Let v be the velocity of the body at the instant it reaches the pile of sand. Then from the relation v2 = v2o + 2 gy, we have v 2 = 0 + 2 × ( 9.8 meter/sec2) × 3 meter. = 58.8 (meter/sec)2 This velocity is reduced to zero due to the deceleration ‘a’ produced by the sand. Thus, from the relation v2 = v2o + 2ay, we have 0 = 58.8 + 2a (0.03 meter)
58. 8 = − 980 meter/sec2 2 × 0. 03 The mass of the body is 10 gm = 0.01 kg. Hence the (retarding) force exerted by the sand on it is a = −
F = ma = 0.01 kg × (–980 m/s2) = –9.8 nt 10–31
Q.6. An electron (mass 9.0 × kg) is projected horizontally at a speed of 1.2 × 107 m/sec into an electric field which exerts a constant vertical force of 4.5 × 10–15 nt on it. Determine the vertical distance the electron is deflected during the time it has moved forward 3.0 cm horizontally. Solution. The time taken by the electron to move 3.0 cm (= 0.03 meter) horizontally is given by t =
0. 03 meter = 2. 5 × 10−9 sec. 1.2 × 107 meter/sec
The vertical acceleration of the electron is a =
Force 4 . 5 × 10−15 = = 5. 0 × 1015 nt/kg Mass 9. 0 × 10−31
The vertical distance moved under this acceleration in time t is y =
1 2 at 2
1 × (5.0 × 1015) × (2.5 × 10–9)2 2 = 1.56 × 10–2 meter = 1.56 cm. =
Q. 7. A block of mass m = 2.0 kg is pulled along a smooth horizontal surface by a horizontal force F. Find the normal force exerted on the block by the surface. What must be the force F if the block is to gain a horizontal velocity of 4.0 meter/sec in 2.0 sec starting from rest?
42
Mechanics
Solution. Let N be the normal force exerted by the smooth surface on the block. The net vertical force on the block is N – mg. Since there is no vertical acceleration in the block, we have by Newton second law (net force = mass × acceleration) N – mg = 0 or
N = mg = 2.0 × 9.8 = 19.6 nt
The horizontal acceleration ‘a’ may be obtained from the relation v = v0 + at : 4.0 = 0 + a (2.0) ∴
Fig. 12
a = 2.0 m/sec2
The required horizontal force is therefore, by Newton’s Law, given by F = ma = 2.0 × 2.0 = 4.0 nt. Q. 8. A block of mass M is pulled along a smooth horizontal surface by a rope of mass m. A force F is applied to one end of the rope. Find the acceleration of the block and the rope. Also deduce the force exerted by the rope on the block.
Fig. 13
[Ans. F/(M + m), MF/M + m)] Q. 9. Two blocks of masses m1 = 2.0 kg and m2 = 1.0 kg are in contact on a frictionless table. A horizontal force F = 3.0 nt is applied to m1. Find the force of contact between m1 and m2. Again, find the force of contact if F is applied to m 2. Solution. Let F′ be the force of contact between m1 and m2. The block m1 exerts a force F on m2. and by the law of action and reaction, m2 also exerts force F′ on m1, as shown in the figure.
Fig. 14
Fig. 15
The new force on m1 is F – F′, and that on m2 is F′. Let a be the (common) acceleration produced. Then, by Newton’s second law, we have F – F′ = and
m1a
F′ = m2a Solving:
F′ =
m2 F m1 + m2
F′ =
1. 0 kg (3. 0 nt) = 1. 0 nt. 2.0 kg + 1.0 kg
Putting the given values:
If F is applied to m2., then we would have
F ′ = 2.0 nt.
Q. 10. Three blocks of masses m1 = 10 kg, m2 = 20 kg and m3 = 30 kg are connected as shown in the Fig. 16 on a smooth horizontal table and pulled to the right with a force T3 = 60 nt. Find the tension T1 and T2. (Lucknow, 1985)
Force and Motion
43
Solution. The force T3 is exerted on the entire system. The block m3 exerts a forward force T2 on m3; while by Newton’s third law the block m2 exerts a backward force T2 on m3. Similarly, m3 exerts a forward force T1 on m1 and m1 exerts a backward force T1 on m1. This is shown below.
Fig. 16
Now, suppose all the blocks acquire an acceleration a. First consider the system as a whole. Net force on the system = mass × acceleration T3 = (m3 + m2 + m1)a
...(i)
Again, we consider m2 and m1 together. Net force on m2 and m1 = mass × acceleration. T2 = (m2 + m1)a
...(ii)
Finally, we consider m1 only. T1 = m1 a
...(iii)
Here, T3 = 60 nt; m1 = 10 kg, m2 = 20 kg and m3 = 30 kg. Substitution in eq. (i) gives 60 = (30 + 20 + 10 ) a a = 60/60 = 1 m/sec2 Substituting the value of a in (ii) and (iii), we get T2 = 30 nt. T1 = 10 nt. Q. 11. An inextensible string connecting blocks m1 = 10 kg and m2 = 5 kg passes over a light frictionless pulley as shown. m1 moves on a frictionless surface while m2 moves vertically. Calculate the acceleration a of the system and the tension T in the string. Solution: The diagram shows the forces on each block. The hanging block m2 (due to gravity) exerts through the tension force T on m2. Now, the forces on m1 are: its weight mg vertically downward, normal force N exerted by the smooth surface and the tension T. If a be the acceleration of the block in the horizontal direction then by Newton’s law, we have T = m1a And
N – m1g = 0
...(i) ...(ii)
Since the string is inextensible, the acceleration of the block m2 is also a. The new force on m2 is m2 g – T. Hence by Newton’ second law, we have m2 g – T = m2a Adding (i) and (iii), we get m 2 g = (m1 + m2)a
...(iii)
44
Mechanics
m2 g a = m1 + m2
or
N
a T
m1
Substituting the given value of a in (i) we get T
m1m2 g T = m1 + m2
a
m1g
m2
Substituting the given values: m1 = 10 kg and m2 = 5 kg.
m2g
a = (1/3) g = (1/3) × 9.8 = 3.27 nt/kg and
Fig. 17
T = (10/3) g = (10/3) × 9.8 = 32.7 nt.
Q. 12. An inextensible string connecting unequal masses m1 = 1.0 kg and m2 = 2.0 kg passes over a massless and frictionless pulley as shown in figure 18. Find the tension in the string and the acceleration of the masses. Solution. As the string is inextensible, both masses have the same acceleration a. Also, the pulley is massless and frictionless, hence the tension T at both ends of the string is the same. The heavier mass m2 is accelerated downward and the lighter mass m1 is accelerated upward. The new (upward) force on m1 is T – m1g, and the net (downward) force on m2 is m2g – T. Therefore, by Newton’s second law, we have and
T – m1 g = m1a
...(i)
m2 g – T = m2a
...(ii)
Adding (i) and (ii), we get (m2 – m1) g = (m1 + m2)a or
a =
m2 − m1 g m1 + m2
...(iii)
Substituting for a in eq. (i) and solving , we get T =
2m1m2 g m1 + m2
Thus, for m1 = 1.0 kg and m2 = 2.0 kg, we have a = (1/3) × 9.8 = 3.27 m/sec2 and
T
T = (4/3) g = (4/3) × 9.8 = 13.1 nt.
Q. 13. A uniform flexible chain of length l with mass per unit length λ, passes over a small frictionless, massless pulley. It is released from a rest position with a length of chain x hanging from one side and a length l – x from the other side. (a) Under what condition will it accelerate? (b) If this condition is met, find the acceleration a as a function of x. (Lucknow 1997) [Ans. (a) l ≠ 2x , (b) a = g (1 – 2x/l)]
a
T
m 2g
a m1g
Fig. 18
Q. 14. A block of mass m1 = 3.0 kg on a smooth inclined plane of 30° is connected by a cord over a small, frictionless pulley to a second block of mass m2 = 2.0 kg hanging vertically
Force and Motion
45
as shown in figure 19. Calculate the acceleration with which the blocks move, and also tension in the cord. (take g = 10 m/s2). Solution. Let a be the acceleration of the blocks m1 and m2, and T the tension in the cord, which is uniform throughout as the cord is massless and the pulley is frictionless. The force acting on the block m1 are: (i) its weight m1g which is the force exerted by the earth, (ii) the tension T in the string, and (iii) the normal force N exerted by the inclined surface (the force is normal because there is no frictional force between the surfaces). The net force on m1 along the plane is T-m1g sin θ, and that perpendicular to the plane is N – m1g cos θ. The block m1 is accelerated along the incline while the acceleration perpendicular to the plane is zero. Therefore, by Newton’s second law, we have and
T – m1 g sinθ = m1a
...(i)
N – m1g cosθ = 0
...(ii)
Again, the forces on m2 are : (i) its weight m2g and (ii) the tension T in the string. The net vertical (downward) force is m2 g – T. As the block is accelerated downward, Newton’s second law gives m2 g – T = m2a
...(iii)
Solving (i) and (iii), we get
and
a =
m2 − m1 sin θ g m1 + m2
T =
m1m2 (1 + sin θ ) g m1 + m2 a T
N
T
m1
m
gs 1
i
nθ
m2 θ
o gc
sθ
m1
m 2g
m 1g
θ
a
Fig. 19
Here m1 = 3.0 kg, m2 = 2.0 kg, sin θ = sin 30° = ½ and g = 10 m/s2
FG H
2. 0 − 3. 0 × a =
1 2
3. 0 + 2. 0
FG H
IJ K 10 = 1.0 m/s
(3. 0) (2. 0) 1 + and
T =
3. 0 + 2. 0
2
1 2
IJ K 10 = 18 nt.
Q. 15. A car, mass 1000 kg moves uphill along a smooth road inclined 30°. Determine the force which the engine of the car must produce if the car is to move (a) with uniform motion
46
Mechanics
(b) with an acceleration of 0.2 m/s2. (c) Find also in each case the force exerted on the car by the road. [Ans. (a) 4900 nt, (b) 5100 nt, (c) 4900 √3 in each case] Q. 16. A block of mass m = 2.0 kg is kept at rest on a smooth plan, inclined at an angle θ = 30° with the horizontal, by means of a string attached to the vertical wall. Find the tension in the string and the normal force acting on the block. If the string is cut, find the acceleration of the block. Neglect friction [Ans. 9.8 nt, 17 nt, 4.0 m/s2] Q. 17. A space traveller, whose mass is 100 kg leaves the earth. Compute his weight on earth (g = 9.8 m/s2) and on Mars (g = 3.8 m/s2). Solution: Weight = mass × acceleration due to gravity. Therefore, weight on earth WE = 100 kg × 9.8 m/sec2 = 980 nt, and weight on Mars WM = 100 kg × 3.8 m/sec2 = 380 nt. The weight in interplanetary space would be zero. Of course, the mass will remain 100 kg at all locations. Q. 18. A body has a mass of 100 kg on the Earth. What would be it (i) mass and (ii) weight on the Moon where the acceleration due to gravity is 1.6 m/sec2. [Ans. (i) 100 kg, (ii) 160 nt.] Q. 19. A massless string pulls a mass of 50 kg upward against gravity. The string would break if subjected to a tension greater than 600 Newtons. What is the maximum acceleration with which the mass can be moved upward? (Lucknow 1990, 1996) Solution. The force acting upon the mass are (i) its weight mg and (ii) tension T in the string. The net (upward) force, T – mg, is responsible for the upward acceleration a of the mass. By Newton’s second law (net force = mass × acceleration), we have T – mg = ma Here (maximum) tension T = 600 nt and m = 50 kg. Therefore; the (maximum) acceleration a is given by
T − mg m 600 − (50 × 9. 8) = 2. 2 m/sec2 a = Fig. 20 50 Q. 20. A body of mass 50 kg is hanging by a rope attached to the ceiling of an elevator. Calculate the tension in the rope if the elevator is (i) accelerating upward at 4 m/s2,(ii) downward at 4 m/s2, (iii) falling freely, (iv) moving with uniform velocity of 5 m/s upward or downward. a =
Solution. The forces of the body are its weight mg and the tension T in the rope which is the upward force exerted on the body by the rope. (If the body were suspended by a spring balance, then T would be the reading of the balance. T is also known as the ‘apparent weight’ of the body).
Force and Motion
47
(i) The body is at rest relative, to the elevator, and hence has an upward acceleration ‘a’ relative to the earth. Taking the upward direction as positive, the resultant force on the body is T – mg. Hence from Newton’s law (force = mass × acceleration), we have T – mg = ma ∴
T = mg + ma
Thus the apparent weight T is greater than the true weight mg, and the body appears “heavier” Substituting the given values: m = 50 kg, a = 4 m/s2 T = m (g + a) = 50 (9.8 + 4) = 690 nt. (ii) When the elevator is accelerating downward (‘a’ negative), we may write T – mg = – ma ∴
T = mg – ma
In this case the apparent weight T is less than true weight mg, and the body appears “lighter”. Substituting the given values: T = m(g – a) = 50 (9.8 – 4) = 290 nt (iii) If the elevator falls freely , a = g, then we have T = m(g – a) = m (g – g) = 0 In this case the tension (apparent weight) is zero and the body appears “weightless”. If the body were suspended by a spring balance in a freely falling elevator, the balance would read zero. (iv) If the elevator is at rest, or moving vertically (either up or down) with constant velocity, a = 0. In this case T – mg = 0 T = mg The apparent weight equals the true weight. T = mg = 50 × 9.8 = 490 nt. T
T
T a=0
a
a mg (a)
mg (b )
mg (c )
Fig. 21
Q. 21. A body hangs from a spring balance supported from the roof of an elevator. (a) If the elevator has an upward acceleration of 2.45 m/s2 and the balance reads 50 nt, what is the true weight of the body? (c) What will the balance read if the elevator cable breaks? [Ans. (a) 40 nt (b) 2.45 m/s2 downward, (c) zero]
48
Mechanics
Q. 22. An elevator weighing 500 kg is pulled upward by a cable with an acceleration of 2.0 m/s2 (a) What is the tension in the cable? (b) What is the tension when the elevator is accelerating downward at 2.0 m/s2. [Ans. (a) 5900 nt, (b) 3900 nt] Q. 23. An elevator and its load have a total mass of 800 kg. Find the tension T in the supporting cable when the elevator, moving downward at 10 m/s, is brought to rest with constant acceleration in a distance of 25 m. Solution. Let us take the upward direction as positive. The elevator having an initial downward velocity of 10 m/s is brought to rest within a distance of 25 m. Let us use the relation v 2 = v02 + 2ax Here v = 0 , v0 = –10 m/s and x = –25 m. Thus 0 = (–10)2 + 2a (–25)
− ( −10)2 = 2 m/s2 2 ( −25) The acceleration is therefore positive (upward). The resultant upward force on the elevator is T – mg, so that by Newton’s law or
a =
T – mg = ma T = mg + ma = m (g + a) = 800 (9.8 + 2) = 9440 nt. Q. 24. An elevator is moving vertically upward with an acceleration of 2 m/s2. Find the force exerted by the feet of a passenger of mass 80 kg on the floor of the elevator. What would be the force if the elevator were accelerating downward? Solution. The force exerted by the passenger on the floor will always be equal in magnitude but opposite in direction to the force exerted by the floor on the passenger. We can therefore calculate either the force exerted on the floor (action force) or the force exerted on the passenger (reaction force). Let us take the upward direction as positive, and consider the forces acting on the passenger. These are the passenger’s true weight mg acting downward, and the force P exerted on him by the floor which is acting upward. P is the apparent weight of the passenger. The resultant upward force on the passenger is thus P – mg. If a be the upward acceleration of the passenger (and of the elevator), we have by Newton’s second law P – mg = ma or
P = mg + ma Fig. 22
Thus the apparent weight P is greater than the true weight mg. The passenger feels himself pressing down on the floor with greater force (the floor is pressing upward on him with greater force) than when he and the elevator are at rest (or moving with uniform velocity).
Force and Motion
49
Substituting the given values: m = 80 kg and a = 2 m/s2 P = m (g + a) = 80 (9.8 + 2) = 944 nt. If the elevator is acelerating downward (a negative),we may write P = mg – ma. In this case the apparent weight P is less than the true weight mg. The passenger feels himself pressing down on the floor with less force than when he and the elevator are at rest (or moving with uniform velocity). Substituting the given values: P = m (g – a) = 80 (9.8 – 2) = 624 nt If the elevator cable breaks, the elevator falls freely (a = g). In the case P = m(g – a) = m(g – g) = 0 Then the passenger and floor would exert no forces on each other. Then passenger’s apparent weight would be zero. Q. 25. A 90-kg man is in an elevator. Determine the force exerted on him by the floor when: (a) the elevator goes up/down with uniform speed, (b) the elevator accelerates upward at 3 m/s2, (c) the elevator accelerates downward at 3 m/s2, and (d) the elevator falls freely. [Ans. (a) 882 nt, (b) 1152 nt, (c) 612 nt, (d) zero] Q. 26. The total mass of an elevator with a 80-kg man in it is 1000 kg. This elevator moving upward with a speed of 8 m/s, is brought to rest over a distance of 20 m. Calculate (a) the tension T in the cables supporting the elevator, and (b) the force exerted on the man by the elevator floor. Solution. (a) Let us take the upward direction as positive. The elevator having an initial upward speed of 8 m/s is brought to rest within a distance of 20 m. Let use the relation v 2 = v02 + 2ax Here v = 0, v0 = 8 m/s, and x = 20 m. Thus 0 = (8)2 + 2a (20)
( 8)2 = − 1. 6 m/s2 2 × 20 The acceleration is therefore negative (downward). The resultant upward force on the elevator is T – mg, where mg is the total weight of the elevator. By Newton’s law a = −
T – mg = ma or
T = mg + ma = m (g + a) = 1000 (9.8 – 1.6) = 8200 nt.
(b) Let P be the (upward) force exerted on the man by the elevator floor. If m′ be the mass of a man, its weight acting downward is m′g. Thus the net upward force on the man is P – m′g. Since a is the acceleration of the man (and of the elevator), we have by Newton’s law P – m′g = m′a or
P = m′g + m′a = m′ (g + a) = 80 (9.8 – 1.6) = 656 nt.
50
Mechanics
Q. 27. A bob hanging from the ceiling of a car acts as an accelerometer. Derive the general expression relating the horizontal acceleration a of the car to the angle θ made by the bob with the vertical. Solution. When the car has an acceleration ‘a’ toward the right, the rope carrying the ball makes an angle θ with the vertical. a T co s θ θ
T θ T sin θ mg
Fig. 23
Two forces are exerted on the ball: its weight mg and the tension T in the rope (shown separately). The resultant vertical force is T cos θ – mg, and the resultant horizontal force is T sin θ. The vertical acceleration of the system is zero, while the horizontal acceleration is a. Hence by Newton’s second law, we have T cos θ – mg = 0 and T sin θ = ma Eliminating T, we get tan θ = a/g or a = g tan θ This is the required result Q. 28. Consider free fall of an object of mass m from rest through the air which exerts a frictional force on it, proportional to its velocity. If the proportionality constant is k : (a) Write down its equation of motion. (b) Show that it ceases to accelerate when it reaches a terminal velocity v T = mg/k. (c) Verify that the velocity of the object varies with time as v = vT(1 – e–kt/m). Solution. (a) The forces acting on the object are (i) the force of gravity mg in the direction of motion and (ii) the frictional force kv against the motion. The net force in the direction of motion is therefore mg – kv. This, by Newton’s second law, is equal to mass m multiplied by the acceleration a of the body. Thus mg – kv = ma
...(i) 2
If y represents the (vertical) displacement, then v = equation may therefore be written as mg – k
dy dv d y = 2 . The above and a = dt dt dt
dy md2 y = dt dt2
d2 y k dy + g = − m dt dt2 This is the equation of motion of the object
...(ii)
(b) Let vT be the ‘terminal’ velocity attained by the object when the acceleration a becomes zero. Making this substitution in eq. (i), we have
Force and Motion
51
mg – kvT = 0 vT =
mg k
...(iii)
(c) The equation of motion (ii) can be written as
k dv = − v+ g m dt But g = (k/m) vT from equation (iii). Therefore
k k k dv (vT − v) = − v + vT = m m m dt dv k dt vT − v = m Integrating, we get
− log e (vT − v) = When
k t + C (constant) t = 0m, v = 0. Thus – loge vT = C
Therefore, the above equation becomes
− log e (vT − v) = or
or
log e
k t − log e vT m
−k ( vT − v) t = m vT kt − (vT − v) = e m vT kt
or
or
or
1−
− v = e m vT kt v − m e 1 − = vT
v = vT
F1 − e I GH JK −
kt m
This is the required expression. Q. 29. A block weighing 20 nt rests on a rough surface. A horizontal force of 8 nt is required before the block starts to slide, while a force of 4 nt keeps the block moving at constant speed once it has started sliding. Find the coefficients of static and sliding (kinetic) friction.
52
Mechanics
Solution. The smallest (horizontal) force F required to start the motion is equal to the maximum force of static friction fs , i.e., F = fs But we know that (maximum) fs = µs N = µs W, Where µs is the coefficient of static friction and N is the normal reaction which is equal to the weight W of the block. Thus F = µs W, or
µs =
F 8 nt = = 0. 4 W 20 nt
Again; the force F′ required to maintain a uniform motion is equal to the force of kinetic friction, fk . That is F′ = fk = µk N = µkW or
µk =
F′ 4 nt = = 0. 2 W 20 nt
Q. 30. A block weighing 10 nt is rest on a horizontal table. The coefficient of static friction between block and table is 0.50. (a) What is the magnitude of the horizontal force that will just start the block moving? (b) What is the magnitude of a force acting upward 60° from the horizontal that will just start the block moving? (c) If the force acts down at 60° from the horizontal, how large can it be without causing the block to move? Solution. (a) As shown in Fig. 24, the horizontal force F that will just start the block moving is equal to the maximum force of static friction. Thus F = fs = µsN Where µs is the coefficient of static friction and N is the normal reaction. Also N = W, So that
F = µsW = 0.50 × 10 nt = 5.0 nt.
(b) In this case the forces on the block are shown in Fig. The applied force is inclined at φ upward from the horizontal. Its horizontal and vertical components are Fx = F cos φ and Fy = F sin φ. In this case, we have, for equilibrium Fx = fs = µs N and
Fy = W – N Thus
Fx = µs (W – Fy) F cos φ = µs (W – F sin φ)
or or
F = Here
µ sW cos φ + µs sin φ
µs = 0.50, W = 10 nt, cos φ = cos 60° = 0.50 and sin 60° = 0.866 F =
0. 50 × 10 = 5. 36 nt. 0. 50 + (0. 50 × 0. 866)
Fig. 24
Force and Motion
53
Note that in this example the normal reaction N is not equal to the weight of the block, but is less than the weight by the vertical component of the force F. (c) In this case (Fig. 25), the equilibrium equations will be: F x = fs = µ s N and
Fy = N – W From these we can see that F =
µ sW cos φ − µ s sin φ
F =
0. 50 × 10 = 74. 6 nt 0. 50 − (0. 5 × 0. 866)
Fig. 25
In this case the normal reaction is greater than the weight of the block. Q. 31. A block is pulled to the right at constant velocity by a 10-nt force acting 30° above the horizontal. The coefficient of sliding friction between the block and the surface is 0.5. What is the weight of the block? [Ans. 22.3 nt] Q. 32. A block of mass 10 kg rests on a horizontal surface. What constant horizontal force F is needed to give it a velocity of 4 m/s in 2 sec, starting from rest, if the friction force between the block and the surface is constant and is equal to 5 Newton? Solution. Let a be the acceleration in the block when in motion. From the relation v = v0 + at, we have a =
v − v0 t
Here v = 4 m/s, v0 = 0 and t = 2s. Thus a =
4−0 = 2 m/s2 2
If F be the force applied on the block, then the net force producing the above acceleration would be F – fs , where fs is the static force of friction between the block and the surface. Thus, by Newton’s second law, F – fs = ma or
F = ma + fs Here
m = 10 kg, a = 2 m/s2 and fs = 5 nt.
∴
F = (10 × 2) + 5 = 25 nt.
Q. 33. A body of mass 2000 kg is being pulled by a tractor with a constant velocity of 5 m/s on a rough surface. If the coefficient of kinetic friction is 0.8, what are the magnitudes of the frictional force and of the applied force? Solution. The force of kinetic friction is given by fk = µk N, where µk is the coefficient of kinetic friction and N is the normal reaction which is equal to the weight W (= mg) of the body. Then fk = µk mg
54
Mechanics
= 0.8 × 2000 kg × 9.8 nt/kg = 15680 nt, in the direction opposite to motion. In order to pull the body with a constant velocity, the applied force must be just equal to the frictional force, and applied in the direction of motion. Q. 34. The coefficient of friction between a block of wood of mass 20 kg and a table surface is 0.25. (a) What force is needed to give it an acceleration of 0.2 m/s2, (b) What forces are acting when it is moving with a constant velocity of 2 m/s, (c) If no external force is applied, how far will block travel before coming to rest. [Ans. (a) 53 nt, (b) applied and frictional force each 49 nt. (c) 0.816 m] Q. 35. A block sliding initially with a speed of 10 m/s on a rough horizontal surface, comes to rest in a distance of 70 meters. What is the coefficient of kinetic friction between the block and the surface? Solution. Let us assume that the force of kinetic friction fk between the block and the surface is constant. Then we have a uniformly decelerated motion. From the relation 2 v2 = v0 + 2ax with the final velocity v = 0, we obtain
0 = (10 m/s)2 + 2a (70 m) so that
a = −
10 × 10 = − 0. 71 m/s2 2 × 70
The negative sign indicates deceleration. The frictional force on the block is, by the second law of motion, given by fk = –ma where m is the mass of the block. But we know that fk = µkN where µk is the coefficient of kinetic friction and N is the normal reaction which is equal to the weight W (= mg) of the block. Thus fk = µk mg µk = =
fk − ma − a = = mg mg g
[fk = –ma]
0. 71 m/s2 = 0. 072. 9. 8 m/s2
Q. 36. If the coefficient of static friction between the tyres and the road is 0.5, what is the shortest distance in which an automobile can be stopped when traveling at 72 km/hour? Solution. Let fs be the constant force of static friction between the tyres and the road. The automobile is under uniformly decelerated motion. From the relation v 2 = v02 + 2ax with the final velocity v = 0, we obtain x = −
v02 , 2a
...(i)
Force and Motion
55
where the minus sign means that the acceleration a caused by fs is opposite to the direction of motion. Now, by Newton’s second law, we have fs = – ma, Where m is the mass of the automobile. But we know that fs = µsN = µs mg where µs is the coefficient of static friction and N (= mg) is the normal reaction. From the last two expression, we have –ma = µs mg or
a = –µs g. Then from eq. (i) the distance of stopping is x =
v02 2µ s g
Here v0 = 72 km/hour = 20 m/sec and µs = 0.5 (20 m/s)2 = 40. 8 meter. 2 × 0. 5 × ( 9. 8 m/s2 ) Q. 37. Block A weight 100 nt. The coefficient of static friction between the block and table is 0.30. The block B weights 20 nt and the system is in the equilibrium. Find the friction force exerted on block A. Also find the maximum weight of block for which the system will be in equilibrium. (Lucknow 1991, 1997)
∴
x =
Solution. The forces acting on the knot O are the tension T, T′ and W (weight of B) in the cords. (The reactions to T and T′ which act on A and on the wall are also shown). Since the knot is in equilibrium, we have T = T′ cos 45° and
T′ sin 45° = W
Fig. 26
56
Mechanics
These equations give T = W = 20 nt. Now, the forces on the block A are its weight W, normal reaction N, tangential force of static friction fs and the tension T. Again, since the block is in equilibrium, we have N = W and
fs = T = 20 nt.
As W (and hence T) increases, the friction force fs also increases until it acquires its maximum value, at which fs = µs N
(where µs is coeff. of static friction)
= µs W
(∴ N = W)
= 0.30 × 100 = 30 nt. At this limiting state of equilibrium, W, T and fs all are equal and maximum. Thus Wmax = 30 nt. Q. 38. A 4.0 kg block A is put on top of a 5.0 kg block B; which is placed on a smooth table. It is found that a horizontal force of 12 nt is to be applied on A in order to slip it on B. Find the maximum horizontal force F which can be applied to B so that both A and B move together, and the resulting acceleration of the blocks. Solution. Let m1 and m2 be the masses of the blocks A and B. If µk be the coefficient of sliding friction between A and B, then the frictional force on A exerted by B is fk = µk N = µk m1 g. A force of 12 nt maintains a slip of A on B against the force of friction fk. Thus 12 = fk = µk m1 g Here
m 1 = 4.0 kg.
12 12 3 Fig. 27 = = m1 g 4. 0 × g g B is placed on a smooth table. Hence in order to move A and B together we require a force F on B which just overcomes the frictional force on B exerted by A and that on A exerted by B. Thus
µk =
F = µk m1 g + µk m2 g = µk (m1 + m2) g = 3/g (4.0 + 5.0) g = 27 nt This force is causing both A and B to move on the smooth table. Hence the acceleration produced in them is
F 27 = = 3 m/s2 m1 + m2 4 . 0 + 5. 0 Q. 39. Block A weight 4 nt and block B weight 8 nt. The coefficient of sliding friction between all surfaces is 0.25. Find the force F to slide B at a constant speed when (a) A rests on B and moves with it, (b) A is held at rest, (c) A and B are connected by a light cord passing over a smooth pulley. (Lucknow, 1994) a =
[Ans. (a) 3 nt, (b) 4 nt, (c) 5 nt]
Force and Motion
57
A
A
F
B
A
F
B
(a )
F
B
(b )
(c )
Fig. 28
Q. 40. A block of mass 5 kg resting on a horizontal surface is connected by a cord passing over a light frictionless pulley to a hanging block of mass 5 kg. The coefficient of kinetic friction between the block and the surface is 0.5. Find the tension in the cord and the acceleration of each block. Solution. Let m1 and m2 be the masses of the blocks A and B. The diagram shows the forces on each block. The block B exerts through the tension in the cord of force T on block. A, while A exerts an equal reaction force T on B. Now, the forces on A are: its weight W1 (= m1 g) vertically downward, normal reaction N exerted by surface, tangential frictional force fk exerted by the surface, and the tension T. If a be the acceleration of the block A in the horizontal direction, then by second law of motion, we have T – fk = m1a N – W1 = 0
and or
N = W1 = m1g. But ∴
fk = µkN = µkm1g T – µkm1g = m1a
...(i)
Since the cord is inextensible, the acceleration of the block B is also a. The net force on B is W2 – T. Hence by law of motion, we have W2 – T = m2a or m2 g – T = m2a
...(ii)
Substituting the value of T from (i) into (ii), we get m2g – (m1a + µkm1g) = m2a or
m2g – µk m1g = (m1 + m2)a
or
a =
(m2 − µ km1 ) g m1 + m2
...(iii)
Substituting this value of a in Eq. (ii), we get m2 g – T = m2
(m2 − µ km1 ) g m1 + m2
LM N
T = m2 g 1 −
or
T =
m2 − µ k m m1 + m2
m1m2 (1 + µ k ) g m1 + m2
OP Q ...(iv)
58
Mechanics N
A
T
a
fk
T B
a
W 1 (= m 1 g) W 2 (= m 2 g)
Fig. 29
Substituting the given values in (iii) and (iv); m1 = 5 kg, m2 = 5 kg, µk = 0.5
and
a =
(5 − 0. 5 × 5) ( 9. 8) = 2.45 m/s2 5+5
T =
5 × 5 (1 + 0. 5) (9. 8) = 36.75 nt. 5+5
Q. 41. The masses of A and B in Figure 30 are 10 kg and 5 kg. Find the minimum mass of C that will prevent A from sliding, if µs between A and the table is 0.20. Compute the acceleration of the system if C is removed. Take µk between table and A also to be 0.20. [Ans. 15 kg, g/5] Fig. 30
Q. 42. A 10 kg block is sliding down a plane inclined at an angle of 30° with the horizontal. Find the normal reaction and the acceleration of the block. The coefficient of kinetic friction is 0.5. Solution. The forces on the block (Fig. 31) are its weight W and the normal and the (kinetic) frictional components of the force exerted by the plane, namely, N and fk respectively. The weight W may be resolved into two components, W sin θ parallel to the plane and W cos θ perpendicular to the plane. Thus the net force along the plane is W sin θ – fk and that perpendicular to the plane is N – W cos θ. If a be the acceleration down the plane, we have by second law of motion W sinθ – fk = ma
...(i)
Since there is no motion perpendicular to the plane, we have N – W cos θ = 0 Also
fk = µk N
...(ii) ...(iii)
From eq. (ii), we get N = W cos θ Here
W = mg = 10 × 9.8 = 98 nt and cos θ = cos 30° = 0.866
Force and Motion
59
Substituting this value of N and µk = 0.5 in (iii), we get fk = 0.5 × 84.87 = 42.43 nt
Fig. 31
Now, Eq. (i) a =
W sinθ − fk m
= g sin θ – (fk/m)
[∴ W = mg]
= (9.8 × 0.5) – (42.43/10) = 4.9 – 4.243 = 0.657
[sin θ = sin 30° = 0.5]
m/s2
Q. 43. A piece of ice slides down a 45° inclined plane in twice the time it takes to slide down a frictionless 45° inclined plane. What is the coefficient of kinetic friction between ice and inclined plane? (Lucknow, 1983) Solution. Let a be the acceleration of the ice piece down the first (rough) plane. The forces on the piece are shown in Fig. 31. The net force down the plane is W sin θ – fk and that perpendicular to the plane is N – W cos θ. By second law of motion, we have W sinθ – fk = ma
...(i)
N – W cos θ = 0
and
N = W cos θ
so that
fk = µk N = µk W cos θ
Also,
Putting this value of fk in (i), we get W sin θ – µk W cos θ = ma a = g sin θ – µk g cos θ,
or
...(ii)
because W = mg, here θ = 45°, so that sin θ = cos θ = 1/ 2 . Thus a =
g 2
(1 − µ k )
Let t be the time taken by the piece to slide down the plane. Then from x = v0t +
1 2 at , 2
we have (here v0 = 0) x =
1 g (1 − µ k )t2 2 2
...(iii)
60
Mechanics
For the frictionless plane (µk = 0), we have a = g sinθ = g/ 2 The time is now t/2. Thus
1 g x = 2 2
FG t IJ H 2K
2
...(iv)
Comparing (iii) and (iv), we get 1 – µk = ¼ ´µk = 1 – ¼ = 3/4 = 0.75. Q. 44. A block slides down an inclined plane of slope angle θ with constant velocity. It is then projected up the same plane with an initial speed v0. How far up the incline will it move before coming to rest? Will it slide down again? (Lucknow 1986, 1990, 1993) Solution. Refer to Fig. 31. The net force on the block down the plane is W sin θ – fk which is zero because the block slides down with constant velocity (zero acceleration). Thus W sin θ – fk = 0 fk = W sin θ
or
When the block is projected up (now fk will be down the plane), the net force down the plane becomes W sinθ + fk = W sin θ + W sin θ = 2 W sin θ. Hence the acceleration down the plane would be a =
force 2 W sin θ = = 2 g sin θ mass m
Now, using the relation v 2 = v02 + 2ax with the final velocity v = 0, we have 0 = v02 + 2 (2 g sin θ) x x = −
v02 4 g sin θ
Hence the distance covered up the plane –x =
v02 4 g sin θ
As soon as the block stops, the frictional force becomes static, which is larger than the kinetic. Hence the block will not slide down again. Q. 45. A block of mass 0.2 kg starts up a plane inclined 30° with the horizontal with a velocity of 12 m/s. If the coefficient of sliding friction is 0.16, how far up the plane the block travels before stopping? What would be the block’s speed when it is made to return to the bottom of the plane? [Ans. 11.5 m, 9.01 m/s] Q. 46. A body with a mass of 0.80 kg is on a plane inclined at 30° to the horizontal. What force must be applied on the body so that it slides with an acceleration of 0.10 m/s2 (a) uphill, (b) downhill? The coefficient of sliding friction with the plane is 0.30.
Force and Motion
61
Solution. Let us first consider the body moving uphill (Fig. 32a). The forces acting on the body are its weight W (= mg) acting vertically downward, the normal force N and the kinetic frictional force fk exerted by the plane, and the force F applied uphill. The frictional force fk is always against the motion and hence in this case downhill. The weight W may be resolved into two components, W sin θ parallel to the plane and W cos θ perpendicular to the plane. Thus the net force on the body along the plane directed uphill is F – W sin θ –fk, and that perpendicular to the plane is N – W cos θ. If a be the acceleration up the plane, the second law of motion gives F – W sin θ – fk = ma
...(i)
Since there is no motion perpendicular to the plane, we have N – W cos θ = 0 N = W cos θ
or Also, we know that
fk = µkN = µk W cos θ. F
N
N
W sin θ
W sin θ θ
fk
θ
fk θ
F
W co s θ W
θ
(a )
W co s θ W (b )
Fig. 32
get
Where µk is the coefficient of kinetic friction. Substituting this value of fk in eq. (i), we F – W sin θ – µk W cos θ
= ma
Replacing W by mg, we get F – mg (sin θ + µk cos θ) = ma F = m[a + g (sin θ + µk cos θ)]
or Here m = 0.80 kg, a = 0.10
m/s2,
...(ii)
µk = 0.30 and θ = 30°, thus
sin θ = sin 30° = 0.5 and cos θ = cos 30° = 0.866. Whence F = 0.80 [0.10 + 9.8 (0.50 + 0.30 × 0.866)] = 6.04 nt. (b) When the body is moving down hill, the forces are as shown in Fig. (b). We now assume that the applied force F is downhill. fk is always against the motion and hence now it is uphill. If a be the acceleration down the plane, then the equation of motion is
62
Mechanics
fk = µk W cos θ (as before), therefore
But
F + W sin θ – µk W cos θ = ma F = m[a – g (sin θ – µk cos θ)]
or
...(iv)
Putting the values: F = 0.80 [0.10 – 9.8 (0.5 – 0.30 × 0.866)] = –1.80 nt. The negative sign means the force F is uphill instead of downhill as we had assumed. Still the motion is downhill. Q. 47. Forces required to move a body on a rough inclined plane with a uniform velocity in the upward and downward directions are in the ratio 2:1. Find the inclination of the plane if the coefficient of friction is 3. (Lucknow, 1987) Solution: Let F1 and F2 be the required forces. Putting a = 0 (because velocity is uniform) in eqs. (ii) and (iv) in the last problem, we get F 1 = mg (sin θ + µk cos θ) F 2 = –mg (sin θ – µk cos θ)
and But
F 1 = 2 F2. This gives sin θ + µk cos θ = –2 (sin θ – µk cos θ) tan θ =
or Now
µk 3
µk = 3 tan θ = 1
or
θ = 45°
Q. 48. A block A weighing 100 nt is placed on an inclined plane of slope angle 30° and is connected to a second hanging block B by a cord passing over a small smooth pulley. The coefficient of sliding friction is 0.30. Find the weight of the block B for which the block A moves at constant speed (a) up the plane, (b) down the plane. [Ans. (a) 76 nt, (b) 24 nt]
Fig. 33
Q. 49. Two blocks with masses m1 = 1.65 kg and m2 = 3.30 kg are connected by a string and slide down a plane inclined at an angle θ = 30° with the horizontal. The coefficient of sliding friction between m1 and the plane is 0.226 and that between m2 and the plane is 0.113. Calculate the common acceleration of the two blocks and the tension in the string. [Ans. 3.62 meter/sec2, 1.06 nt]
Fig. 34
Q. 50. A block is at rest on an inclined plane making an angle θ with the horizontal. As the angle of incline is increased, slipping just starts at an angle of inclination θs. Find the coefficient of static friction between block and incline in terms of θs. Solution. The forces on the block are shown in Fig. 35. W is the weight of the block, N is the normal reaction and fs the tangential force of static friction exerted on the block by
Force and Motion
63
the inclined surface. The weight W may be resolved into two components, W sin θ parallel to the inclined plane and W cos θ perpendicular to it. Since the block is in equilibrium, the net force along the plane as well as perpendicular to the plane is zero. That is fs – W sin θ = 0 N – W cos θ = 0
and
fs < µsN.
Also
When we increase θ to θs, slipping just begins. Thus for θ = θs, we can use fs = µs N. Substituting this above, we get µsN = W sin θs N = W cos θs
and Dividing, we get
µs = tan θs N
W sin θ a
θ
fs W co s θ
W θ
Fig. 35
Hence measurement of the angle of inclination at which slipping just starts provides us a method for determining the coefficient of static friction between two surfaces. In a similar way, we can show that the angle of inclination θk required to maintain a constant speed of the block sliding down the plane is given by µk = tan θk where θk < θs and µk is the coefficient of sliding (kinetic) friction. Q. 51. A block rests on a plane inclined at θ with the horizontal. The coefficient of sliding friction is 0.50 and that of static friction is 0.75. (a) As the angle θ is increased, find the minimum angle at which the block starts to slip. (b) At this angle find the acceleration once the block has begun to move. (c) How long is required for the block to slip 8 meter along the inclined plane. [Ans. (a) 37°, (b) 1.96 m/s2, (c) 2.86 s] Q. 52. A particle starting from rest moves in a straight line with acceleration a = (0.5 + 0.1 t) m/s2, where t is the time. Calculate its velocity and the distance travelled after 2 seconds. (Lucknow, 1984, 1989) Solution. Suppose that initial position of particle is at origin i.e., at t = 0, x = 0, v = 0 Given or
a = 0.5 + 0.1t m/s2 dv/dt = 0.5 + 0.1t or; dv = (0.5 + 0.1 t) dt
Integrating we get,
64
Mechanics
v = 0. 5 t +
0.1 t2 + k; 2
where k is constant of integration, determined by boundary condition. At start;
v = 0 and t = 0; so above equation yields; k = 0
So,
v = 0. 5t +
At
t = 2 sec; v = 0.5 × 2 + ½ × 0.1 × 4
or
0.1 t2 2
v = 1 + 0.2 = 1.2 m/sec Again rewriting v,
F GH
I JK
2 dx 0.1 t2 ; i.e., dx = 0. 5 t + 0.1 t dt = 0. 5 t + dt 2 2
integrating, one gets x =
0. 5 t2 0.1 t3 + +k 2 2 3
at start, x = 0, t = 0; so above equation gives; k = 0 so
x =
0. 5 t2 0.1 t3 + 2 2 3
x =
0. 5 0.1 8 ×4+ × 2 2 3
and at t = 2;
= 1 + .133 or
x = 1.133 m
Q. 53. Two blocks of masses m1 and m2 are connected by a massless spring on a horizontal frictionless table. Find the ratio of their acceleration a1 and a2 after they are pulled apart and released. (Lucknow, 1988) Solution. Situation is shown in the Fig. 36 below. Here F1 and F2 are forces in opposite directions on m1 and m2. T1 and T2 are resulting tension in the spring due to F1 and F2. T1, F1 are one action-reaction pair T2, F2 are other action-reaction pair As the system is in equilibrium F1 + F2 = 0 and T1 + T2 = 0 i.e., i.e.,
m1a1 + m2a2 = 0 −m2 a1 = m1 a2
Fig. 36
Force and Motion
65
Thus, the ratio of the acceleration is inversely proportional to their masses. The negative sign suggests that acceleration of the block m1 and m2 are oppositely directed. Q. 54. The displacement (x) of a particle moving in one dimension, under the action of a constant force is related to time (t) by the equation t = x + 3, where x is in meter and t in sec. Find displacement of particle when its velocity is zero. Solution. Since
x+3 x = (t – 3)2 = t2 – 6t + 9
t =
Differentiating v = dx/dt = 2t – 6 When v = 0, t = 3 sec. ∴ Displacement at time t = 3 sec. Will be x = (3)2 – 6(3) + 9 = 0 Q. 55. A bullet of mass 20 gm, moving with velocity 16 m/s penetrates a sand bag and comes to rest in 0.05 sec. Find (i) depth of penetration, (ii) average retarding force of sand. (Lucknow, 1989) Solution. Given, u = 16 m/s, v = 0 and t = 0.05 sec. The retardation in sand is given by 0 = u – at i.e.,
a = 16/.05 = 320 m/s2
(i) The depth of penetration is obtained by using equation v 2 = u2 – 2as or
0 = (16)2 – 2(320)s
or
s =
16 × 16 = 0. 4 m 2 × 320
(ii) The average retarding force of sand is F =
mv − mu t
= m
FG v − u IJ = ma H t K
= (20 × 10–3) × 320 = 6.4 Newton Q. 56. A light string which passes over a frictionless pulley, and hangs vertically on each side of it, carries at each end a mass of 240 gm. When a rider of 10 gm is placed over one of the masses, the system moves from rest a distance of 40 cm in 2 seconds. Calculate g. Solution. Suppose P and Q are two masses each, equal to 0.240 kg. They are hanging from the pulley as shown. when rider R of mass 0.010 kg. is placed on mass Q, the system moves through 0.40 m in 2 sec. The final position is shown by dotted lines.
66
Mechanics
Let T be the tension in string and a be the acceleration either P or Q. Since P moves upward we have for its motion; T – 0.240 g = 0.240 a
...(i)
and as Q moves downwards, therefore (0.250 g – T) = 0.250 a
...(ii)
from (i) and (ii); eliminating T; 0.250 g – 0.240 g = 0.240 a + 0.250 a or
0.010 g = 0.490 a g = 49 a
T
a
To find a, we have
T a
P
s = 0.40 m, t = 2, u = 0, a = ? s = ut + ½ at2
Using the formula
0.40 = 0 +
1 a × 4 = 2a 2
a = 0.20 m/sec2
or
Q P Q
mPg
4 0 cm
mQ g
Fig. 36(a)
g = 49 a = 49 × 0.20 m/sec2 = 9.80 m/sec2
So,
Q. 57. A uniform rope of length D resting on a frictionless horizontal surface is pulled at one end by a force P. What is the tension in the rope at a distance ‘d’ from the end where the force is applied? Solution. Suppose the rope to be divided into two segments A and B of length ‘d’ and (D – d). Let T be the tension at the boundary of A and B. If M is total mass of rope and ‘a’, the acceleration produced a = P/M in system, Now the equation of motion of segment B is T =
=
T =
b g FGH MP IJK L D − d OP PM N D Q L dO P M1 − P N DQ M D−d D
;
FG M = mass /length IJ HD K
Fig. 37
Q. 58. A uniform rod of length L and density ρ is a horizontal acceleration a. Find the magnitude of through the mid-point of rod? Solution. If A is cross sectional area of rod, its ∴ Force acting forward on rod, F = A Lρα If p is stress at the mid-cross section of rod, the of rod will be = F – P.A
being pulled along a smooth floor with stress at the transverse cross-section (I.I.T. 1993) mass M = AL.ρ resultant force on right half length ‘b’
Force and Motion
67
∴ From Newton’s second Law, F – PA = mass of length b × accn A L ρα – PA = i.e.,
P =
FG L A ρIJ α H 2 K 1 L ρα 2
Fig. 38
Q. 59. A mass of 500 gm is placed on a smooth table with a string attached to it. The string goes over a frictionless pulley and is connected to another mass of 200 gm. At t = 0, the mass of 500 gm is at a distance of 200 cm from the end and moving with a speed of 50 cm/sec towards the left (see Fig. 39). what will be its position and speed at t = 1 sec. Solution. Suppose ‘a’ represent the acceleration of both the block at t = 0 The equation of motion of 200 gm is 200 g – T = 200 a ...(i) The equation of motion of 500 gm is T = 500 a ...(ii) Adding equation (i) and (ii) we get 200 g – 500 a = 200 a ...(iii) i.e., a = 2/7 g so a = (2/7) × 980 = 280 cm/sec2
Fig. 39
for displacement of block of 500 gm
1 2 at 2 u = –50 cm/sec, a = 280 cm/sec2, t = 1 sec 1 2 ∴ s = –50 × 1 + × 280 × t 2 = –50 + 140 s = +90 cms Thus, the block of 500 gm has moved 90 cms to the right of its initial position. The velocity v of block of 500 gm is then v = u + at; u = –50 cm/sec, a = 280 cm/sec2, t = 1 sec = –50 + 280 × 1 = 230 cm/sec (to the right) Q. 60. A uniform flexible chain of length L and mass per unit length λ passes over a smooth massless small pulley. At an instant a length x of the chain hangs on one side of the pulley and length (L – x) on the other side. Find the instantaneous acceleration. Also find the time interval in which x changes from L/4 to 0, if the chain moves with a negligible velocity at x = L/2. (Lucknow, 1987) Solution. Suppose instantaneous common acceleration is ‘a’. Weight of chain of length x = x λ g. Net upward force on chain of length x = T – xλg = xλa or, T = xλa + xλg = xλ(a + g) ...(i) weight of chain of length (L – x) = (L – x) λg Using,
s = ut +
68
or or or
Mechanics
Net force acting downward on chain of length (L – x) is (L – x) λg – T = (L – x) λa ...(ii) Putting T from Eq. (i) in Eq. (i), we get (L – x) λg – xλ (a + g) = (L – x) λa Lλg – x λg – a xλ – g xλ = (L – x) λa Lλg – 2xλg = (L – x) λ.a + a.xλ λg (L – 2x) = aλL
(L – x) x T
or
λg (L − 2 x) g = (L − 2 x) a = λL L
or
a =
a
g (L − 2 x) L
–
T
x λg
Fig. 40
Q. 61. Two rough planes, inclined at 30° and 60° to horizontal and of same height are placed back to back. Two masses of 5 kg and 10 kg are placed on the faces and connected by a string passing over the top of the planes. If coefficient of friction is 1/ 3 , find the resultant acceleration. Solution. The 10 kg mass moves down and 5 kg mass moves up the plane. Let; a = acceleration of system, T = the tension of the string and N1 and N2 be the normal reaction of the plane. For 5 kg mass, (ΣF = ma) T – (µN1 + 5 g sin 30°) = 5a
...(i)
5 g cos 30° = N 1
...(ii)
and For 10 kg mass
10 g sin 60° – T – µ N2 = 10 a
...(iii)
10 g cos 60° = N 2
...(iv)
T – 5 g (sin 30° + µ cos 30°) = 5 a
...(v)
(ii) and (i) gives and (iv) in (iii) yields 10g (sin 60° – µ cos 60°) – T = 10 a
...(vi)
T
So, (v) and (vi) give
F 10 g G H
3 1 I − J − 5 g FGH 12 + 12 IJK 2 2 3K
µN 2
N1
a N2
= 15 a
2 3−3 a = g 9
or
T
µN 1
3 0°
1 0g
5g
6 0°
Fig. 41
Q. 62. A block of mass 2 kg slides on an inclined plane which makes an angle of 30° with the horizontal. The coefficient of friction between the block and surface is (i)
3/2 ;
What force should be applied to the block so that block moves down without any acceleration?
Force and Motion
69
(ii)
What force should be applied to the block so that block moves up without any acceleration.
(iii)
Calculate the ratio of power, needed in the above two cases: if the block moves with the same speed in both the cases.
Solution. (i) The figure 42 shows the situation, in which various forces are depicted, Resolving the forces along and perpendicular to the plane; we get µN = P1 + mg sin 30°
...(i)
N = mg cos 30°
N
...(ii)
µN
(ii) and (i) yields
P1
µ mg cos 30° = P1 + mg sin 30° or
P 1 = mg (µ cos 30° – sin 30°) = 2 × 9. 8
=
F 3. GH 2
e
19. 6 3 − 2 2 2
3 1 − 2 2
I JK
sin mg
3
0°
30
mg
30°
m g co s 30 °
Fig. 42
j = 19. 6 × 1. 586 ;
P1 = 11 Newton
2 × 1. 414
(ii) The forces acting in this case are shown in figure 43. Resolving along and perpendicular to the rough inclined plane; one gets; P 2 = mg sin 30° + µ N
...(iii)
N = mg cos 30°
...(iv) N
(iv) and (iii) gives
P2
P 2 = mg (sin 30° + µ cos 30°) or
LM1 + 3 . 3 OP MN 2 2 2 PQ L 2 + 3 OP = 9.8 4.414 = 2 × 9. 8 M MN 2 2 PQ 1. 414 = 2 × 9. 8
sin mg 3 0°
30
°
30 °
µN
mg
m g co s 30 °
Fig. 43
P 2 = 30.6 Newtons (iii) As Power = (force × displacement)/time = force × velocity As the velocity is same in both the cases, the ratio of powers shall be equal to the ratio of forces applies i.e., P1 F 11 = 0. 36 = 1 = P2 F2 30. 6
63. A block weighing 400 kg rests on a horizontal surface and supports on top of it, another block of weight 100 kg as shown. The block W2 is attached to a vertical wall by a string 6 m long. Find the magnitude of the horizontal force P applied to the lower block as shown in figure which shall be necessary so that slipping of W2 occurs.
70
Mechanics
(µ for each surfaces in contact is ¼ ) Solution. Forces on block W2 are shown below: It may be seen, that T cos θ = µN2
...(i)
N2 + T sin θ = 100g
...(ii)
From figure 44(a) : sin θ = From (i)
T =
11 / 6 : cos θ =
5 ; tan θ = 11 / 5 6
µN2 ; then (ii) is, cos θ
N2 + µ N2 tan θ = 100 g N2 (1 + µ tan θ) = 100 g
or or
N2 =
100 g 1 + µ tan θ
...(iii)
T sin θ
A
N2
N2
6m
µN 2
T B
5m
θ C
9 0°
T co s θ
w2 w1
P
w2
w1 µN 2
µN 1
P N1
1 00 g
(a )
(b )
(c )
4 00 g
Fig. 44
Forces on block W1 are as shown in Fig. 44(b) As observed for block W1: P = µN2 + µN1 = (as N2 = N1 + 400 g)
LM µ × 100 g + µ F 100 g + 400 gI OP JK PQ MN1 + µ tan θ GH 1 + µ tan θ LM µ × 100 g + F 100 + 400I OP MN1 + µ tan θ GH 1 + µ tan θ JK PQ O L 1000 + 100OP 9. 8 L 200 × 20 + 400 P = 9. 8 M M 4 MN 20 + 11 PQ N 20 + 3.316 Q
P = µg
=
= 142.8 kg wt Q. 64. A block of mass 3 kg rests on a rough horizontal table, the coefficient of friction between the surface is 0.5. A massless string is tied to the block which passes over a smooth light pulley at the edge of the table and supports a block of 5 kg. Find the acceleration of the system, if any.
Force and Motion
71
Solution. For 3 kg mass, eqn. of motion is, T – µN = 3 a ...(i) For 5 kg mass, 5 g – T = 5a ...(ii) and µN = 0.5 × 3 × 9.8 = 14.7 Newton ...(iii) adding (i) and (ii) and using (iii) 5 g – µN = 8 a or 5 × 9.8 – 14.7 = 8 a or
a =
49 − 14. 7 34. 3 = 8 8
a = 4.29 m/s2 Fig. 45
Q. 65. Find the least force required to drag a particle of mass m along a horizontal surface. Solution. Refer Fig. 46 P = force acting at ∠θ, R = resultant of N and Fs, λ = angle of friction Resolving the forces horizontally and vertically P cos θ – R sin λ = 0
...(i)
P sin θ + R cos λ = mg
...(ii)
from (i)
R =
P cos θ sin λ
...(iii) N
then (ii) is;
R
P sin θ +
P cos θ . cos λ = mg sin λ
θ
P sin θ + P cos θ cot λ = mg
or,
Differentiating (iv) w.r.t. to θ sinθ
P λ
FG H
dP dP + P cos θ + cot λ cos θ − P sin θ dθ dθ
...(iv)
IJ K
Fs mg
Fig. 46
= 0
dP (sin θ + cos θ cot λ ) + P( cos θ − cot λ sin θ) = 0 dθ
...(v)
dP P (sin θ − cot λ − cos θ) = dθ sin θ + cos θ cot λ for P to be minimum; dP/dθ = 0 i.e.,
sin θ cot λ − cos θ = 0; or; tan θ = tan λ, i.e.,
θ = λ
...(vi)
72
Mechanics
Putting (vi) and (iv) P sin λ + P cos λ cot λ = mg Multiplying both sides by sin λ, P sin2 λ + P cos2 λ = mg sin λ or P = mg sin λ Q. 66. A block slides down an inclined plane of slope θ, with a constant velocity. It is then projected up the same plane with an initial velocity v0 . How far will it move before coming to rest. Will it slide down again ? (Lucknow, 1982, 1986) Solution. We have, while sliding as shown in Fig. 47(a) mg sin θ = µN, mg cos θ = N i.e., µ = tan θ ...(i) Now, in upward motion on same plane shown in Fig. 47(b) Retarding Force = mg sin θ + µ N; (N = mg cos θ) = mg sin θ + µ mg cos θ = mg (sin θ + µ cos θ) N
h n w it t i o on tarda i t M o re
N
µN
rm if o un y θ it in lo c ve m g s
θ mg
θ
mg θ
m g co s θ
s in
θ
θ
(a )
µN
mg
m g co s θ
(b )
Fig. 47
= g (sin θ + µ cos θ)
Hence, retardation
= g (sin θ + tan θ . cos θ) = 2 g sin θ using with
v2
= u2 + 2as;
u = v0; a = –2g sin θ, v = 0 0 = v02 − 4 g sin θ x s
or
s =
v02 4 g sinθ
The block will not slide down again since the component of force along the plane downwards will be just balanced by the force of the motion. Q. 67. A uniform rod rests in limiting equilibirum in contact with a floor and a vertical wall (rod in a vertical plane). Supposing the wall and floor to be unequally rough, compute the angle between rod and wall. Solution. Situation is depicted in the figure 48. LK is rod; MS is wall; θ is angle between rod and the wall. Let, 2l = length of the rod; µ = coeff. of static friction between the rod and the wall; µ′ = coeff. of static friction between rod and floor.
Force and Motion
73
For equilibrium in horizontal direction, µ′N′ = N
N′ =
or
N µ′
...(i)
For equilibrium in vertical direction M
N′ + µN = W
N + µN = W µ′
or
or
N =
or,
N
FG 1 + µIJ H µ′ K
FG 1 + µµ ′ IJ = W H µ′ K
µN N
= W
θ
K
l G
...(ii)
Now taking moment about point L, we have
N′ l w = mg L µ ′N ′
W.LD = N.KS + µ N.LS
D
S
Fig. 48
but as LD = l sin θ; KS = 2l cos θ; and LS = 2l sin θ above equation becomes W.l sin θ = N.2l cos θ + µN 2l sin θ and putting the value of N from (ii) W.l sin θ = or
W.l sin θ –
FG H
Wl 1 −
µ′ W µ′ W . 2l cos θ + µ. . 2l sin θ 1 + µµ ′ 1 + µµ ′
2µµ ′ W sin θ l 2 µ ′ Wl cos θ = 1 + µµ ′ 1 + µµ ′
IJ K
FG H
2µ ′ Wl cos θ 2µµ ′ 2µ ′ 2µ ′ sin θ = or tan θ = or θ = tan−1 1 + µµ ′ 1 + µµ ′ 1 − µµ ′ 1 − µµ ′
IJ K
Q. 68. Forces required to move a body on a rough inclined plane with a uniform velocity in the upward and downward direction are in the ratio 2:1. Find the inclination of the plane if the coefficient of friction is 0.3. (Lucknow, 1987) Solution. (i) Uphill uniform motion: Net force in uphill direction is zero because body moves with constant velocity; i.e., F1 – (mg sin θ + µ N) = 0 or
F1 = mg sin θ + µ mg cos θ (or N = mg cos θ)
or
F1 = mg (sin θ + µ cos θ)
(ii) Downhill uniform motion: Again, resultant force in downhill direction is zero as the body moves with constant velocity. If F2 is applied downward force, then (F2 + mg sin θ) – µN = 0 or
F2 = µ mg cos θ – mg sin θ = 0
or
F2 = mg (µ cos θ – sin θ)
74
Mechanics N
Un
ifo
rm
ti mo
mg
s in
N
µN
on
F2 θ
θ mg
θ
m if o r Un F 1
sin mg
m g co s θ
Mo
ti o
n
θ
θ µN
θ
mg
(a )
m g co s θ
(b )
Fig. 49
According to question
F1 2 , F2 = 1 i.e., F1 = 2F2 mg (sin θ + µ cos θ) = 2 mg (µ cos θ – sin θ) sin θ + µ cos θ = 2 µ cos θ – 2 sin θ
or
3 sin θ = µ cos θ
or
tan θ = µ /3 = 0.3/3 = 0.1; θ = tan–1 (0.1)
or
Q. 69. A piece of ice slides down a 45° incline in thrice the time it takes to slide down on a frictionless 45° incline. Find the coefficient of kinetic friction between ice and the incline. (Lucknow, 1983, 1988) Solution. Frictionless incline: Let length of incline be s, initial velocity = 0 downward force = mg sin 45° = ma a =
mg sin 45° g = m 2
Now, if t1 is the time taken to slide down the incline, using S = ut +
1 2 at ; with u = 0, t = t1, a = g/ 2 2
We have, s =
1 2
FG g IJ t H 2K
2 1
...(i) N
N
µN mg
sin
° 45 mg
4 5°
m g co s 45 °
(a )
Fig. 50
sin mg 4 5°
4
5° mg (b )
m g co s 45 °
Force and Motion
75
For incline having friction: Downward force = mg sin 45° – µN = mg sin 45° – µ mg cos 45° ma = mg (sin 45° – µ cos 45°) = mg so, downward acceleration;
a =
FG 1 − µ 1 IJ H 2 2K
g (1 − µ) 2
if t2 is the time the body takes to slide down the incline using, s = ut +
1 2 at ; with u = 0, 2
t = t2 and
a =
g (1 − µ) 2
s = 0+
1 g × (1 − µ)t22 2 2
...(ii)
from (i) and (ii) 1 2
or
FG g IJ t H 2K
2 1
=
g (1 − µ )t22 2 2
t 12 = (1 − µ ) t22
...(iii)
But according to question, t2 = 3 t1 then (iii) is t 1 2 = (1 − µ) ( 3t1 )2 or 1 − µ = 1 / 9 or
µ = 8/9 = 0.889
Q. 70. Two blocks are kept on a horizontal smooth table as shown. Find out the minimum force that must be applied on 10 kg block to cause 6 kg block just to slide. Coeff. of static friction between two blocks is 0.3. (Lucknow, 1994) Solution. Let F be the minimum force applied on 10 kg block, the acceleration produced in system is, α = F/(10 + 6) = F/16. This gives a reactionary force 6 α which will tend to slide the 6 kg block where as frictional force will tend to stop it. Hence for 6 kg not to move relative to 10 kg. 6 α = µs × 6 × g (6F/16) = 0.3 × 6 × 9.8 or
Fig. 51
F = 0.3 × 16 × 9.8 = 47.04 Newton Q. 71. Determine the frictional force of air on a body of mass 1 kg falling with an acceleration of 8 m/sec2 (g = 10 m/sec2).
76
Mechanics
W′ = mg′ = mg – Fr so
Fr = mg – mg′ = 10 – 8 = 2 Newton
Q. 72. Block B in the figure 52 has mass 160 kg. The coefficient of static friction between block and table is 0.25, what is the maximum mass of block A for which the system will be in equilibrium ? (Lucknow, 1988) Solution. Let m be the maximum mass of block A for which system is in equilibrium. Various forces are shown in the figure 52. For equilibrium; T cos 45° = µN
...(i)
N = Mg
...(ii)
and
mg = T sin 45°
...(iii)
(i) and (ii) give, T cos 45° = µ M g T/ 2 = 0.25 × 160 × 9.8 or
T = 0.25 × 160 × 9.8 × Given
M = 160 kg; g = 9.8
2
m/s2;
...(iv)
µ = 0.25
N T sin 4 5 ° µN
T 4 5°
B
T co s 45 °
Mg W a ll
A
mg
Fig. 52
Putting (iv) in (iii) mg = 0.25 × 160 × 9.8 × or
2 ×1/ 2
m = 0.25 × 160 kg or m = 40 kg.
Q. 73. A body rests upon inclined plane and will just slide down the plane when the slope of the plane is 30°. Find the acceleration of the body down the plane when the slope is increased to 60°. Solution. When body just slides down the inclined plane, the inclination of the plane is 30°; so
mg sin 30° = µN
Force and Motion
77
or
µ = tan 30° (as N = mg cos 30°)
When slope is 60°; the body moves down the plane. The force acting down is:
N
mg sin 60° – µN = ma s in mg
(with N = mg cos 60°) so,
m(g sin 60° – µ g cos 60°) = ma
where
a = accn
or
a = g (sin 60° − tan 30° cos 60° ) = 9. 8
F GH
6 0°
60
µN
° 60°
mg
m g co s 60 °
Fig. 53
I JK
3 1 1 − × = 5. 66 m/s2 2 2 3
Q. 74. A body with a mass of 0.84 kg is on a plane inclined at 30° to the horizontal. What force must be applied on the body so that it moves with uniform acceleration of 0.12 m/sec2. (a) uphill (b) downhill; (The coefficient of sliding friction with the plane is 0.32). (Lucknow, 1978, 1982) Solution. (a) Uphill motion: Let F be the applied force so that body moves uphill with an acceleration = 0.12 m/sec2. F – (µ N + mg sin 30°) = ma or
F = ma + (µ mg cos 30° + mg sin 30°) = ma + mg (µ cos 30° + sin 30°) = 0.84 × 0.12 + 0.84 × 9.8 [0.32 × ( 3 / 2) + (1/ 2)]
or
F = 0.1008 + 6.397 = 6.498 Newton N
Mo
t io
n Mo
F θ sin mg θ
θ
µN
mg
θ = 3 0° m g co s θ
n
µN
θ F sin g m θ
(a )
t io
θ mg
m g co s θ
(b )
Fig. 54
(b) Downhill motion: Suppose F is the applied force down the hill so that body moves with an acceleration = 0.12 m/s2 then, F + mg sin θ – µ N = ma or,
F = ma – mg sin 30° + µ mg cos 30° (as N = mg cos 30°) = ma – mg (sin 30° – 0.32 cos 30°) = 0.84 × 0.12 – 0.84 × 9.8 [(1/2) – 0.32 ( 3 / 2) ] = 1.7330 Newtons; i.e., in upward direction.
78
Mechanics
Q. 75. A body of 5 kg weight is just prevented from sliding down a rough inclined plane by a force of 2 kg weight acting up the line of greatest slope. When this force is increased to 3 kg weight; the body just begins to slide up the plane. Prove that coefficient of friction between the body and the plane is
3 /15.
Solution. Body is just prevented from sliding down the plane when 2 kg force acts up the inclined plane; it suggests that friction is limiting and acts up the plane. Resolving the forces on the body, along and perpendicular to the plane; N = 5 g cos θ; and µN + 2g = 5 g sin θ
...(i)
µ 5g cos θ + 2g = 5 g sin θ
i.e.,
5 sin θ – 5 µ cos θ = 2
or
...(ii)
When 3 kg weight is applied; the body just slides up the plane. The friction is again limiting but now will act down the inclined plane. Resolving the forces along and perpendicular to the inclined plane; 3g =
5 g sin θ + µ N and N = 5 g cos θ ...(iii)
3 g = 5 g sin θ + µ 5 g cos θ
i.e.,
5 sin θ + µ 5 cos θ = 3
or
...(iv)
adding (ii) and (iv) 10 sin θ = 5 or sinθ =
1 or θ = 30° 2
then from (iv)
5×
1 + 5µ × ( 3 / 2) = 3 2 1
= 0.115 5 3 Q. 76. A block is released from the rest at the top of a frictionless inclined plane 16 m long. It reaches the bottom 4.0 sec later. A second block is projected up the plane from the bottom at the instant the first block is released in such a way that is returns to the bottom simultaneously with the first block. or
µ =
(i)
Find the acceleration of each block on the incline.
(ii)
What is the initial velocity of the second block.
(iii)
How far up the inclined plane does it travel?
Solution. Taking the displacement down the incline to be positive the acceleration of both the blocks when they return simultaneously is, a = g sin θ; from s = ut + = 0, t = 4 sec., s = 16 m. 1 a 42 2 a = (2 × 16)/16 = 2 m/s2
16 = 0 +
1 2 at with u 2
Force and Motion
79
direction of v0 (up the plane) is taken as positive, then acceleration is negative i.e., a = –2 m/s2 using t = 4, u = v0
1 2 at ; with a = –2 m/s2, t = 4, u = v0 2
then
s = ut +
i.e.,
0 = v0 × 4 −
or
1 × ( −2) × 42 2
v 0 = 4 m/s
(iii) Supposing the second block to travel a distance s, up the incline where it comes to rest, gives; v 2 = u2 + 2as with
v = 0, u = v0 = 4 m/s, a = –2 m/s2
then
0 = 42 + 2 × (–2)s
or
s = 4 m
Q. 77. Two blocks of mass 2.9 kg and 1.9 kg are suspended from rigid support by two extensible wires each 1 meter long. The upper wire has negligible mass and lower wire has uniform mass of 0.2 kg/m. The whole system has upward acceleration 0.2 m/s2 as shown. Find (i) Tension at mid point of lower wire (ii) Tension at mid point of upper wire
(I.I.T., 1989)
Solution. Since upper wire has negligible mass, the tension T1 in upper wire will be same all along its length. If T2 is tension at the upper point of lower wire, for equilibrium of m1. T1 – T2 – m1 g = m1a
...(i)
and taking m3 as mass of lower wire, for equilibrium of m2, T2 – m2 g – m3 g = (m2 + m3) a
T1
...(ii)
Adding the two equations
m 1 2.9 kg
T1 – (m1 + m2 + m3) g = (m1 + m2 + m3) a T1 = (m1 + m2 + m3) (g + a)
i.e.,
T2 0.2 kg /m
= (2.9 + 1.9 + 0.2) (9.8 + 0.2) = 5 × 10 = 50 Newton
T2
This will be the tension at mid point of upper wire. Putting value of T1 in equation (i) 50 – T2 = 2.9 (9.8 + 0.2) = 29 i.e.,
T2 = 21 Newton
1.9 kg m2
Fig. 55
At mid point tension will be less than at upper point since only half length of lower wire will contribute to tension due to its own weight. i.e., or i.e.,
FG H
Tm + 0. 2 ×
IJ K
1 × ( 9. 8 + 0. 2) = 21 2
Tm + 0.1 × 10 = 21 Tm = 20 Newton
80
Mechanics
Q. 78. How can a mass of 100 kg be lowered with a massless rope which can support, without breaking, a mass of 80 kg only. (Lucknow, 1994) Solution. Let, the 100 kg mass be lowered with aceleration ‘a’ then taking T < 80 × 9.8 Newton, the equilibrium of motion for the system will be 100g – T = 100 a taking maximum possible of T, we get 100 g – 80 g = 100 a or
a = (1/5) g Hence, the given mass can be lowered with acceleration of g/5 or more than this.
Q. 79. A stone weighing 2 kg and tied at the end of string of length 2.5 meters, revolves with frequency of 10 revolutions/sec. Find out the force on the stone as measured (i) in inertial frame (ii) in frame rotating with the string. Solution. (i) The actual force which makes body to rotate, in inertial frame, is the centripetal force = −
mv2 = − mrω2 r
(–ve sign shows that force is towards center). So, force on stone, in inertial frame, = –m ω2 r = –2 (2π × 10)2 × 2.5 = –19719.2 Newton force being supplied by tension of string (ii) The frame rotating with the string is non-inertial and in this frame the acceleration of the stone is 0. So total force in this frame = 0. The equilibrium of the stone in rotating frame is obtained with the help of centrifugal force (Fictious force). Q. 80. Two blocks A and B are joined to each other by a string and a spring of force constant 1960 N/m which passes over a pulley, attached to another block C as shown. Block B slides over block C and A moves down along C with same speed. Coefficient of friction between blocks and surface of C is 0.2. Taking mass of block A as 2 kg, calculate. (a)
mass of block B
(b)
energy stored in spring.
Solution. When both block B and A at same speed the spring is in condition of maximum extension. Taking T as tension of spring and the string, and ‘m’ as the mass of block B, we have, from Fig. 56 For motion of B,
T – µ mg = 0
For motion of A,
T = 2g
combining (i) and (ii) we get µ mg = 2 g or,
0.2 m = 2 so, m = 10 kg
...(i) ...(ii)
Force and Motion
81 B
B
µm g
µ = 0 .2
M
T T
µ = 0 .2
C
C A
A
(a )
(b )
2g
K = 1 9 60 N /M
Fig. 56
Putting value of ‘m’ in equation (i) or using equation (ii) we get T = 2 × 9.8 = 19.6 N The elongation of a string is therefore x =
Tension 19. 6 = = 0. 01 m Force − constant 1960
Hence, energy stored in spring U =
1 2 1 kx = × 1960 × ( 0. 01)2 2 2
= 98 × 10–3 = 0.098 joule Q. 81. A body falling freely from a given height H hits an inclined plane in its path at a height ‘h’. As a resutl of impact the direction of velocity of body becomes horizontal. For what value of (h/H), the body will take maximum time to reach the ground. Solution. Let t1 be the time of fall through height (H – h) and t2 the time of free fall through height ‘h’ then H– h = and
h =
1 2 gt1 2 1 2 gt2 (since after impact there is no 2 initial velocity along vertical)
t = t1 + t2 = or
t =
2 g
2(H − h) / g + 2h / g
b H − hg
1/2
for time to be maximum the condition is dt/dh = 0 1
− or
1
− 1 1 − (H − h) 2 + h 2 = 0 2 2
h =
which yields, h 1 = 2 H
(H − h)
+ h1 /2
82
Mechanics
Q. 82. A projectile is launched with an initial speed v in direction θ to the horizontal which hits an inclined plane of inclination β with horizontal (β < θ) and passing through the launching point. Derive an expression for the range R on this inclined plane. When will the projectile hit this plane? Solution. Let the projectile hit the inclined plane at M and OM = R (range). At the instant of hitting the plane, the x and y displacements are ON and NM respectively as shown in Fig. 57. ON = R cos β; NM = R sin β
Now
Clearly time taken to cover horizontal distance ON, t =
ON R cos β = v cos θ v cos θ
...(i)
For vertical distance NM; we have s = NM; u = v sin θ, a = –g, t = R cos β/v cos θ s = ut +
Then
1 2 at gives; 2
Fig. 57
R sin β = v sin θ
R cos β 1 ( R cos β)2 − g v cos θ 2 ( v cos θ )2
1 R2 cos2 β g = R (tan θ cos β − sin β) 2 v2 cos2 θ or
2v2 cos2 θ sin θ cos β − cos θ sin β R = cos θ g cos2 β
or
R =
2v2 cos θ sin(θ − β) g cos2 β
Putting above in (i) t =
2v2 cos θ sin(θ − β) cos β 2v sin(θ − β) = v cos θ g cos β g cos2 β
Q. 83. A block Q (refer figure 58) having mass 0.2 kg. is placed on the top of block P of mass 0.8 kg. Coefficient of sliding friction between P and the table is 0.2 and that between P and Q is 0.5. The pulley is light and smooth. What horizontal force A will maintain the motion of P with uniform speed?
84
Mechanics
Q. 85. The water of 0.5 km wide river is flowing with a velocity of 4 km/hr. A boat man standing on one of the bank of the river wishes to take his boat to a point on the opposite bank exactly infront of his present position. He can row his boat with a velocity of 8 km/hr relative to water. In which direction should he row his boat. Obtain the time to cross the river. Solution. Let the boatman start from P and Q be his destination. Then resultant velocity (relative to earth i.e., bank) should be in direction PQ. So boat should be rowed along PR. Let resultant velocity be v. Sin α = RQ/PR = 4/8 i.e., α = 30° and
PQ =
i.e.,
R
4 km /h r
Q
82 − 42 = 48 = 6. 9
v = 6.9 km/hr
V
8 km /h r
So boat man should row in direction 30° with PQ, upstream, Time to cross the river;
α
PQ 0. 5 P = Speed-in-direction PQ 6. 9 Fig. 61 1 60 hr = min = 4.35 min = 13. 8 13. 8 Q. 86. An aeroplane is flying with velocity 70 km/hr in northeast direction. Wind is blowing at 30 km/hr from north to south. What is the resultant displacement of aeroplane in 4 hours? t =
Solution. Let v1 = velocity of aeroplane; v2 = velocity of wind then resultant velocity of aeroplane, v is, v = v1 + v2 Resolving v1 and v2 in x and y direction; v 1 = v0 cos 45° i + v0 sin 45° j
70 $ $ (i + j ) 2 v 2 = –30 j
v1 =
= 35 2 i$ + (35 2 − 30) $j v = (49.5 i + 19.5 j) km/hr. The result and displacement r = (49.5 i + 19.5j) × 4 = (198 i + 78j)
LM r OP N Q →
and
=
1892 + 782 = 204. 46 km.
if φ = angle made by r with x-axis (i.e., east) tan φ = y/x = 19.5/49.5 or φ = 23° nearly. Q. 83 An aeroplane is flying in a horizontal direction with a velocity of 600km per hour and a height of 1960 meters. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at a point B. Calculate the distance AB. (IIT, 1973) Ans. When a body is dropped, it is acted upon by (i)
Uniform velocity = 600 km/hour =
600000 m/s in the horizontal direction. 3600
Force and Motion
(ii)
85
Acceleration due to gravity in the vertical direction.
Here
h = 1960 m
Suppose the time to reach the ground = t second h = ½ gt2; here g = 9.80 m/s2
9. 80 2 t 2 1960 t2 = = 400 second2 4 . 90 t = 20 second
1960 =
Distance
AB = S = ut S =
600000 × 20 = 3333.33 m. 3600
SELECTED PROBLEMS 1.
What do you understand by a frame of reference. Define inertial frame. Comment on the reference frame attached with earth. (Lucknow, 1994, Agra 1996)
2.
What is a coriolis force? Find an expression for it. Exaplain its importance. (Lucknow, 1995)
3.
4.
Explain (a)
Fictitious force,
(Agra)
(b)
Galilean transformations,
(c)
Coriolis force.
(IAS)
A reference frame ‘a’ rotates with respect to another reference frame ‘b’ with uniform angular velocity ω. If the position, velocity and acceleration of a particle in frame ‘a’ are represented by ‘R’, va and fa show that acceleration of that particle in frame ‘b’ is given by where: fB = fa + 2ω × va + ω × (ω × R) Integrate this equation with reference to the motion of bodies on earth surface.
5.
Show that the motion of one projectile as seen from another projectile will always be a straight line motion. (Agra, 1970)
6.
A plane flies across the north pole at 400 km/hr and follow a longitude 450 (rotating with earth) all the times. Compute what angle does a freely suspended plumb line in the plane makes as it passes over the pole, make with a plumb line which is situated on the earth surface at north pole.
7.
A thin uniform rod AB of mass m and length 2λ can rotate in a vertical plane about A as pendulum. A particle of mass 2 m is also fixed on the rod at a diatance x from A. Find x so that the periodic time of small swing may be minimum.
8.
A person is running towards a bus with a velocity of constant magnitude u and always directed towards the bus. The bus is moving along a straight road with a velocity of constant
→
→
→
→
magnitude v . Initially at t = 0. a is perpendicular to v and the distance between the person and the bus l. After what time will the person catch the bus?
LM Ans. ul OP N v −u Q 2
2
86
Mechanics
3 DYNAMICS OF CIRCULAR MOTION AND THE GRAVITATIONAL FIELD 3.1 UNIFORM CIRCULAR MOTION Centripetal Force: suppose a body is moving on a horizontal circle (radius r) with a uniform velocity v. At any point on the circle, the direction of the velocity is directed along the tangent to the circle at that point i.e., at A the direction of the velocity AX, at B along BY, at C along CZ and so on. If the body is free at any point it will move tangential to the circle at that point. But as the body is moving along the circumference of the circle, there must be a force acting towards the center of the circular path and this force is called the centripetal force. Centripetal force is defined as that force which acts towards the center along the radius of a circular path on which the body is moving with a uniform velocity. In figure 1 let A and B be two positions of the body after an interval of time t. Then
x
y
v
B
v
C r
v
A
r r
z O
Fig. 1
AB = velocity. time = vt
Let o′a and o′b be vectors representing the velocities at A and B respectively in figure 2(ii). Then ∠ao′b = θ (the angle between the tangents equal to the angle between the radii). ab is the change in velocity from A to B. Acceleration =
ab Change in Velocity = t Time
Since A and B are very close to each other, therefore arc AB can be taken as a straight line. ∆OAB and ∆o′ab are similar ∴ ∴
ab AB = o′ a OA
vt ab ab v2 or = = r v t r 86
Dynamics of Circular Motion and the Gravitational Field
∴
87
ab v2 = t r
Acceleration =
a
v B v
α
α r
α θ
A v
r
O θ
α b
V
c
(ii)
(i)
Fig. 2
∴ The acceleration on the body acting towards the center of the circular path = ∴
Centripetal force = mass × acceleration = m F =
v2 r
v2 r
mv2 r
Direction of the Centripetal Force When a body moves along a circular path with uniform speed, the magnitude of the velocity remains the same but its direction changes at every point. It means there is change in velocity whenever there is a change in direction, there must be some acceleration. As a body has mass also, a force must act upon the body. This force must act along such a direction that the magnitude of the velocity does not change. As a force has no component at right angles to its direction, the force must act at every point in a direction perpendicular to the direction of the velocity. As the velocity is tangential to the circle at every point, the force must be acting along the radius and towards the centre of the circular path. Due to this reason it is called centripetal force.
3.2 CENTRIFUGAL FORCE When a body is rotating on a circular path, it has a tendency to move along a tangent. If a body A leaves the circular path at any instant, for an observer A who is not sharing the motion along the circular path (i.e., a body B standing outside the reference circle), the body A appears to fly off tangentially at the point of release. For an observer C, who is sharing the same circular motion as that of the body A, the body A appears to be at rest before it is released. According to C, when A is released, it appears to fly off radially away from the center. It appears to the body C as if the body A has been thrown off along the radius away from the center by some force. This inertial force is known as centrifugal force. Its magnitude is mv2/r. It is not a force of reaction. Centrifugal force is a fictitious force and holds good in a rotating frame of reference. When a car is on turning round a corner, the persons sitting inside the car experience an outward force. This is due to the fact that no centripetal force is provided by the passengers. Therefore to avoid this outward force, the passengers are to exert an inward force.
88
Mechanics
3.3 THE CENTRIFUGE It is an appliance to separate heavier particles from the lighter particles in a liquid. The liquid is rotated in a cylindrical at a high speed with the help of an electric motor. The heavier particles move away from the axis of rotation and the lighter particles moves near to the axis of rotation.
Application (a)
Sugar crystals are separated from molasses with the help of a centrifuge.
(b)
In cream separators, when the vessel containing milk is rotated at high speed, the lighter cream particles collect near the axle while the skimmed milk moves away from the axle.
(c)
In drying machines, the wet clothes are rotated at high speed. The water particles fly off tangentially through the holes in the wall of the outer vessel.
(d)
Honey is also separated from bees with the help of a centrifuge.
(e)
Precipitates, sediments, bacteria etc., are also separated in a similar way.
In an ultra-centrifuge, the speed of rotation is very high and is of the order of 30 to 40 thousand rotations per minute.
3.4 BANKING OF CURVED ROADS AND RAILWAY TRACKS (BANKED TRACK) A car moving on road, or a train moving on rails, requires a centripetal force while taking a turn. As these vehicles are heavy, the necessary centripetal force may not be provided by friction, and moreover, the wheels are likely to suffer considerable wear and tear. In this case, the centripetal force is produced by slopping down the road inward at the turns. Similarly, while laying the railways tracks, the inner rail is laid slightly lower than the outer rail at turns. By doing so, the car, or the train leans inward while taking turn and the necessary centripetal force is provided. This force is produced by the normal reaction of the earth, or the rail. In figure 3, is shown a car taking a turn on a road while is given a slope of angle θ, G is the center of gravity of the car at which the weight mg of the car acts. When the car leans the total normal reaction R of the road exerted on the wheels makes an angle θ with the vertical. The vertical component R cos θ balances the weight mg of the car, while the horizontal component R sin θ provides the necessary centripetal force mv2/r. Thus R cos θ = mg and
R sinθ =
mv2 r
...(i) ...(ii)
R co s θ
R
θ G
Dividing Eq. (ii) by (i), we get 2
v rg From this formula the angle θ can be calculated for given values of v and r. Thus the slope θ is proper for a particular speed of the car. Therefore, the driver drives the car with that particular speed at the turn. Since m does not appear in the formula, hence this speed does not depend upon the weight of the car.
tan θ =
R
θ
CG
mv2 r
θ s in
mg
Fig. 3
Dynamics of Circular Motion and the Gravitational Field
89
If a car is moving at a speed higher than the desired speed, it tends to slip outward at the turn, but then the frictional force acts inward and provides the additional centripetal force. Similarly, if the car is moving at a speed lower than the desired speed it tends to slip inward at the turn, but now the frictional force acts outward and reduces the centripetal force. The hilly roads are made slopping inward throughout so that the vehicles moving on them lean automatically towards the center of the turn and are acted upon by the centripetal force. If these road were in level then, at the turns, the outer wheels would be raised up thus overturning the vehicle. The flying aeroplane leans to one side while taking a turn in the horizontal plane. In this situation the vertical component of the force acting on the wings of the aeroplane balances its weight and the horizontal component provides the necessary centripetal force. In the “well of death” the driver while driving the motor-cycle fast on the wall of the well, leans inward. The vertical component of the reaction of the wall balances the weight of the motor-cycle and the driver, while the horizontal component provides the required centripetal force.
3.5 BICYCLE MOTION As shown in figure 4, a rider taking a turn towards his left hand. Let m be the mass of the rider and the bicycle, v the speed of the bicycle and r the radius of the (circular) turn. The centripetal force mv2/r necessary to take a turn is provided by the friction between the tyres and the road. Therefore, when the bicycle is turned, a frictional force F (= mv2/r) towards the center of the turn is exerted at the point A on the F 1 G G F2 F1 F1 road. Let us imagine two equal and opposite forces F1 and F2, each equal and parallel to F ( F1 = F2 = F), acting mg mg at the center of gravity G of the rider and the bicycle. The force F acting at A can now be replaced by the force F1 θ C e ntre o f acting at G, and the anticlockwise Tu rn A F A F couple formed by F1 and F2 (= F). The R R force F1(= F) is the necessary centripetal force. If the rider remains (a ) (b) straight while turning, then his weight Fig. 4 mg acting vertically downward at G and the earth’s normal reaction R (= mg) acting vertically upward at A would cancel each other. In this situation the anticlockwise couple would overturn the bicycle outward. If, however, the rider leans inward towards the center of the turn, then his weight mg and the normal reaction R ( = mg) of the earth from a clockwise couple which balances the anticlockwise couple (Fig. 5). Hence the rider moves on the turn without any risk of overturning. Suppose the rider leans through an angle q from the vertical for balancing the two couples then: couple formed by mg and R (= mg) = couple formed by F1 and F2 (= F)
90
Mechanics
mg × GA sin θ = F × GA cos θ tan θ =
or But
F =
F mg mv2 r
tan θ =
C .G .
R θ F
r
θ
2
v rg From this formula, the angle θ can be calculated.
∴
m v2 r
mg
C
S
F1
F2
d
d θ Since the centripetal force mv2/r is provided by the frictional force, the value of mv2/r should be less than the limiting frictional force µr otherwise the bicycle Fig. 5 would slip. This is why the rider slow down the speed of the bicycle while taking a turn and follows the path of a larger radius. During rainy season the frictional force decrease appreciably and cannot provide the centripetal force. Hence during rains bicycle rider usually slip on the roads while taking turn.
3.6 CONICAL PENDULUM Suppose a small object A of mass m is tied to a string OA of length l and then whirled round in a horizontal circle of radius r, with O fixed directly above the center B of the circle, If the circular speed of A is constant, the string turns at a constant angle θ to the vertical. This is called a conical pendulum. Since A moves with a constant speed v in a circle of radius r, there must be a centripetal mv2 force acting towards the center B. The horizontal component, T sin θ , of the tension r T in the string provides this force along AB. So
mv2 ...(i) r Also, since the mass does not move vertically, its weight mg must be counter balanced by the vertical component T cos θ of the tension. So T sin θ =
T cos θ = mg
...(ii) O
Dividing (i) by (ii), then
v2 tan θ = rg
T l
A similar formula for θ was obtained for the angle of banking of a track, which prevented side-slip. A pendulum suspended from the ceiling of a train does not remain vertical while the train goes round a circular track. Its bob moves outwards away from the center and the string becomes inclined at an angle θ to the vertical, as shown in figure 6. In this case the centripetal force is provided by the horizontal component of the tension in the string.
θ
T co s θ θ A m
r T sin θ
mg
Fig. 6
B
Dynamics of Circular Motion and the Gravitational Field
91
3.7 MOTION IN A VERTICAL CIRCLE When a body tied to the end of a string is rotated in a vertical circle, the speed of the body is different at different points of the circular path. Therefore, the centripetal force on the body and the tension in the string change continuously. Let us consider a body of mass ‘m’ tied to the end of a string of length R and whirled in a vertical circle about a fixed point O to which the other end of the string is attached. The motion is circular but not uniform, since the body speeds up while coming down and slows down while going up. The force acting on the body at any instant are its weight mg directed vertically downward, →
→
and the tension T in the string directed radially inward. The weight m g can be resolved into tangential component mg sin θ and a radical (normal) component mg cos θ. Thus the body has tangential force mg sin θ and a resultant radial force T – mg cos θ acting on it.
O R
T
θ
m g co s θ θ
m g sin θ mg
Fig. 7
The tangential force gives to the body a tangential acceleration, which is responsible for the variation in its speed. The radial force provides the necessary centripetal force. Thus
T – mg cos θ =
mv2 R
The tension in the string is therefore
LM v + g cosθOP NR Q 2
T = m
...(i)
Let us consider two special cases of the equation: (i)
At the lower point A of the circle, θ = 0 so cos θ =1 T = TA and v = vA (say).
O TA
Equation (i) takes the form TA = m
LM v + gOP NR Q 2 A
vA
A mg
Fig. 8
Which means that the tension must be large enough not only to overcome the weight mg but also provide the centripetal force
mv2A . R
92
Mechanics
(ii)
At the highest point B, θ = 180°, so cos θ = –1, T = TB and v = vB (say) Then we have Equation (i) TB = m
LM v − gOP NR Q 2 B
TB is less than TA, because the weight mg provides a part of the centripetal force. If vB decreases, the tension TB would decrease and vanish at a certain critical speed vC (say). To determine this critical speed, we put TB = 0 and vB = vC in the last expression. Then 0 = m vc =
Fig. 9
LM v − gOP NR Q 2 C
Rg
This speed of the body is called the “critical speed”. In this state the centripetal force is provided simply by the weight of the body. If the speed of the body at the highest point B is
R g , then the required centripetal force would be less than the weight of the body which will therefore fall down (the string would slack).
less than the critical speed
If a bucket containing water is rotated fast in a vertical plane, the water does not fall even when the bucket is completely inverted. This can be explained in the following manner. The bucket rotates in a vertical plane under a changing centripetal force. The water contained in the rotating bucket experiences a centrifugal force which is always equal and opposite to the centripetal force. When the bucket is at the highest point B of this circular path, then the centrifugal force
mvB2 on the water is directed upward. If the speed of rotation R of the bucket is quite fast, the centrifugal force is greater than the weight W of the water. The difference
F mv GH R
2 B
the water pressed to the bottom of the bucket.
−W
I JK
keeps
2
m vB R B
. W
Fig. 10
3.8 MOTION OF PLANET Our solar system consists of a sun which is stationary at the center of the Universe and nine planets which revolve around the sun in separate orbits. The name of these planets are: Mercury, Venus, Earth, Mass, Jupiter, Saturn, Uranus, Neptune and Pluto. The planet Mercury is closest to the Sun and Pluto is farthest. These are certain celestial bodies which revolve around the planets. These are called ‘satellites’. For example, moon revolves around the earth, hence moon is a satellite of the earth. Similarly, Mars has two satellites, Jupiter has twelve satellites, Saturn has ten satellites, and so on.
Dynamics of Circular Motion and the Gravitational Field
93
3.9 KEPLER’S LAWS OF MOTION Kepler found important regularities in the motion of the planets. These regularities are known as ‘Kepler’s three laws of planetary motion’. (i)
Shape of the Orbit: All planets move around the sun in elliptical orbits having the sun at one focus of the orbit. This is the law of orbits.
(ii)
Velocity of the orbit: A line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal speed of the planet remains constant. This is the law of areas. When the planet is nearest the sun then its speed is maximum and when it is farthest from the sun then its speed is minimum. In Figure 11, if a planet moves from A to B in a given time-interval, and from C to D in the same timeinterval, then the areas ASB and CSD will be equal. B
A
C S D
Fig. 11
(iii)
Time periods of Planets: The square of the period of revolution of any planet around the sun is directly proportional to the cube of its mean distance from the sun. This is the law of periods. If the period of a planet around the sun is T and the mean radius of its orbit is r, then or
T 2 ∝ r3 T 2 = Kr3
where K is a constant. Thus, larger the distance of a planet from the sun, larger will be its period of revolution around the sun.
3.10 DERIVATION OF LAW OF GRAVITATION Suppose the mass of the planet A is M1, the radius of its orbit is R1 and time period of revolution is T1. It is assumed that the orbit is circular. The force of attraction exerted by the sun on the planet (centripetal force)
LM 2π OP NT Q
2
F1 =
M1R1ω12
= M1R1
...(i)
1
Similarly for a second planet B of mass M2, Radius R2 and period of revolution round the sun T2 F2 =
M2R2ω22
= M2R2
LM 2π OP NT Q 2
2
...(ii)
94
Mechanics
F1 F2 =
FG M IJ FG R IJ FG T IJ HM K HR K HT K 1
1
2
2
2
1
2
...(iii)
But according to Kepler’s third law,
FG T IJ HT K 2 1
2
FR I = G J HR K
3
2
...(iv)
1
Substituting these values in equation (iii)
F1 F2 =
FG M IJ FG R IJ HM K HR K 1
2
2
1
2
F1R12 F2R22 = = constant M1 M2 ∴ or
F ∝ M/R2
(i)
F ∝ M and
(ii)
F ∝ 1/R2
Thus, the force of attraction exerted by the sun on a planet is proportional to its mass and inversely proportional to the square of its distance from the sun.
3.11 NEWTON’S CONCLUSIONS FROM KEPLER’S LAWS Newton found that the orbits of most of the planets (except Mercury and Pluto) are nearly circular. According to Kepler’s second law, the areal speed of a planet remains constant. This means that in a circular orbit the linear speed of the planet will be constant. Since the planet is moving on a circular path; it is being acted upon by a centripetal force directed towards the center (Sun). This force is given by F =
mv2 r
Where m is the mass of the planet, v is its linear speed and r is the radius of its circular orbit. If T is the period of revolution of the planet, then
∴
v =
linear distance travelled in one revolution 2πr = T period of revolution
F =
m 2πr r T
FG IJ H K
2
=
4 π2mr T2
But according to Kepler’s third law, T2 = K r3 ∴ or
F =
4 π2mr 4 π2 = K K r3
F ∝
m r2
FG m IJ Hr K 2
...(i)
Dynamics of Circular Motion and the Gravitational Field
95
Thus, on the basis of Kepler’s laws, Newton drew the following conclusions: (i)
A planet is acted upon by a centripetal force which is directed towards the sun.
(ii)
This force is inversely proportional to the square of the distance between the planet and the sun
(iii)
FG F ∝ 1 IJ . H rK 2
This force is directly proportional to the mass of the planet (F ∝ m). Since the force between the planet and the sun is mutual, the force F is also proportional to the mass M of the sun (F ∝ M). Now, we can replace the constant
4 π2 in Equation K
(i) by GM, where G is another constant. Then, we have F = G
Mm r2
Newton stated that the above formula is not only applied between sun and planets, but also between any two bodies (or particles) of the universe. If m1 and m2 be the masses of two particles, then the force of attraction between them is given by F = G
m1m2 r2
This is Newton’s Law of Gravitation.
3.12 NEWTON’S UNIVERSAL LAW OF GRAVITATIONS In 1686, Newton stated that in the Universe each particle of matter attracts every other particle. This universal attractive-force is called ‘gravitation’: The force of attraction between any two material particles is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. It acts along the line joining the two particles. Suppose two particles of masses m1 and m2 are situated at a distance r apart. If the force of attraction acting between them is F, then according to Newton’s law of gravitation, we have F ∝
m1m2 r2
F = G
or
m1m2 r2
where G is the constant of proportionality which is called ‘Newton’s gravitation constant’. Its value is same for all pairs of particles. So it is a universal constant. If m1 = m2 = 1 and r = 1, then F = G Hence, gravitation constant is equal in magnitude to that force of attraction which acts between two particles each of unit mass separated by a unit distance apart. From the above formula
Fr2 G = m1m2
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Mechanics
Thus, if force (F) be in Newton, distance (r) in meter and mass (m1 & m2) in kg, then G will be in Newton-meter2/kg2. It dimensional formula is [M–1L3T–2]. In rationalized MKS units G = 6.670 × 10–11 newton-m2/kg2.
3.13 GRAVITY AND THE EARTH In Newton’s law of gravitation, the gravitation is the force of attraction acting between any two bodies. If one of the bodies is earth then the gravitation is called ‘gravity’. Hence gravity is the force by which earth attracts a body towards its center. Clearly gravity is a special case of gravitation. It is due to gravity that bodies thrown freely ultimately fall on the surface of the earth.
3.14 ACCELERATION DUE TO GRAVITY When a body is dropped down freely from a height, it begins to fall towards the earth under gravity and its velocity of fall continuously increases. The acceleration developed in its motion is called ‘acceleration due to gravity’. Thus, the acceleration due to gravity is the rate of increase of velocity of a body falling freely towards the earth. It is denoted by ‘g’. It does not depend upon the shape, size, mass, etc. of the body. If m be the mass of a body then force of gravity acting on it is mg (weight of the body). Therefore, the acceleration due to gravity is equal in magnitude to the force exerted by the earth on a body of unit mass. The unit of acceleration due to gravity is meter/second2 or Newton/kg. The universal constant G is different from
→
g . The constant G has the dimension
→
M–1 L3 T –3 and is a scalar, g has the dimension LT–2 and is a vector, and is neither universal nor constant.
3.15 EXPRESSION OF ACCELERATION DUE TO GRAVITY g IN TERMS OF GRAVITATIONS CONSTANT G Suppose that the mass of the earth is Me, its radius is Re and the whole mass Me is concentrated at its center. Let a body of mass m be situated at the surface of the earth or at a small height above the surface, this height being negligible compared to the radius of the earth. Hence the distance of the body from the center of earth may be taken as Re. According to the law of gravitation, the force of attraction acting on the body due to the earth is given by F =
GMem R2e
...(i)
The acceleration due to gravity g in the body arises due to the force F. According to Newton’s second law of motion, we have F = mg From Eqs. (i) and (ii), we have mg = G
Mem R2e
...(ii)
Dynamics of Circular Motion and the Gravitational Field
g =
97
GMe R2e
This expression is free from m. This means that the value of g does not depend upon the mass of the body. Hence if two bodies of different masses be allowed to fall freely (in the absence of air), they will have the same acceleration. If they are allowed to fall from the same height, they will reach the earth simultaneously. In the presence of air, however, the buoyancy effect and the viscous drag will cause different accelerations in the bodies. In this case the heavier body will reach the earth earlier.
3.16 DIFFERENCE BETWEEN MASS AND WEIGHT Mass is the amount of matter contained in a body. This is fixed for a given body. Its value is the same at all places. Weight is the force with which a body is attracted towards the center of the earth. It is different at different places. It depends on the value of g. A body has the same mass at the poles and at the equator whereas its weight will be more at the poles than at the equator. With the help of a beam balance, the mass of the body is determined. With the help of a spring balance the weight of the body is determined. The units of mass are gram or Kg. The units of weight are dyne or Newton.
3.17 INERTIAL MASS AND GRAVITATIONAL MASS Inertial Mass: According to Newton’s second law of motion, when a force is applied on a body, the body moves with an acceleration F = ma
or
m = F/a
This mass is called the inertial mass of the body. If the force is increased, the acceleration also increases and F/a = constant for a given body. If the same force is applied on two different bodies and the acceleration produced are equal, then the inertial masses of the two bodies are equal. If the same force is applied on two different bodies, the inertial mass of that body is more in which the acceleration produced is less and vice versa. Gravitational Mass: According to the law of gravitation, the Gravitational force of attraction of a body towards the center of the earth is equal to the weight of the body. Let the weight of the body be W W = mg m = W/g This mass is called the gravitational mass of the body. This is determined with the help of a beam balance. It will be the same even on the surface of the moon. If the value of g is less at the moon, the weight of the body will also be less but the gravitational mass is the same.
3.18 GRAVITATIONAL FIELD AND POTENTIAL The intensity of gravitational field at a point due to a mass is defined as the force experienced by unit mass, placed at that point. It is, however, supposed that the introduction of the unit mass at the point does not change the configuration of the field. If m be the mass of a gravitating particle, the intensity of gravitational field at a distance r from it is given by
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Mechanics
E = −
Gm $ r r2
→
→ where r$ = r is a unit vector along r . r Here the negative sign indicates that the direction of force is opposite to r$ i.e., towards the mass m. Thus, the force per unit mass is a measure of the field intensity.
The gravitational potential at a point in a gravitational field of a body is the amount of work, required to be done on a unit mass in bringing it from infinity (i.e., the state of zero potential) to that point. In other words, the potential at a point is equal to the potential →
energy per unit mass. Hence, the potential at a distance r from a mass m is given by
z r
V = −
→
→
E .d r =
∞
z r
∞
Gm Gm dr = − 2 r r
Therefore, the potential energy of a mass m′, placed at that point, is given by U = m′V = −
Gmm′ r
Similar to the case of an electric field, the expression for the intensity of the gravitational field is E = −
dV dr
3.19 EQUIPOTENTIAL SURFACE An equipotential surface is the surface, at all the points of which the potential is constant. The potential due to a point mass m at a distance r is V = −
Gm r
and it is constant on a spherical surface of radius r. This is the equipotenial surface. The intensity of the field E at r is given by B dr
dV $ E = −∇ V = − r dr where r$ =
A m o
→
r is along OA r →
Now if a unit mass moves a small distance AB = dr on the surface, then
E
→
→
E.d r = −
Fig. 12
dV $ r . dr = 0 dr
because vector r$ is perpendicular to the vector dr at the surface. This means that for
Dynamics of Circular Motion and the Gravitational Field
99
equipotential surface, the intensity of the field E is perpendicular to the surface at any point A. If a unit mass is moved from one point to the other on an equipotential surface, the
z
B
amount of work done is equal to zero (because
→
FG IJ H K →
E d r = VB − VA = 0)
A
If we have a mass m′ at A, then force on the mass is given by F = −∇U = − ∇(mV) = − m∇V = mE Thus the force is also directed perpendicular to the equipotenial surface at A. Hence, if any mass is moved on an equipotential surface, no work will be done.
3.20
WEIGHTLESSNESS IN SATELLITES
The weight of a body is felt due to a reactionary force applied on the body by some other body (which is in contact with the first body). For example, when we stand on a plane we feel our weight due to the reaction of the plane on our feet. If under some special circumstances the reaction of this plane becomes the zero then we shall feel as our weight has also become zero. This is called “State of weightlessness”. If the ropes of a descending lift are broken, then persons standing in the lift will feel this state. Weightlessness is also felt by a space-man inside an artificial satellite. Suppose an artificial satellite of mass m is revolving around the earth (mass Me) with speed v0 in an orbit of radius r. The necessary centripetal force is provided by the gravitational force.
G
Mem mv02 = r r2 GMe v02 = r2 r
or
...(i)
If there is a space-man of mass m′ inside the satellite, he is acted upon by two forces: (i) gravitational force
GMem′ , (ii) reaction R of the base of the satellite, in the opposite r2
direction. Thus there is a net force
LM GM m′ − ROP N r Q e 2
on the man. It is directed towards the
center of the orbit and is the necessary centripetal force on the man. That is,
GMem′ m′ v02 R − = r r2 GM R v02 − = m′ r r2
or
Substituting the value of v02 / r from equation (i), we get
GMe R GMe − = 2 m′ r r2 ∴
R = 0
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Mechanics
Thus, the reactionary force on the man is zero. Hence he feels his weight zero. If he stands on a spring-balance, the balance will read zero. In fact, every body inside the satellite is in a state of weightlessness. If we suspend a body by a string, no tension will be produced in the string. Space-man cannot take water from a glass because on tilting the glass, water coming out of it will float in the form of drops. Space-man take food by pressing a tube filled with food in the form of paste. Although moon is also a satellite of the earth, but a person on moon does not feel weightlessness. The reason is that the moon has a large mass and exerts a gravitational force on the person (and this is the weight of person on the moon). On the other hand, the artificial satellite having a smaller mass does not exert gravitational force on the space-man.
3.21 VELOCITY OF ESCAPE It is a thing of common experience that when a body is projected in the upward direction, it returns back due to the gravitational pull of the earth on it. Now, if the body is projected upwards with such a velocity which will just take the body beyond the gravitational field of the earth, then it will never come back. This velocity of the body is called the velocity of escape. Let us consider a body of mass m lying at a distance r (>R) from the center of the earth (or planet or sun). The force with which the earth attracts the body is given by F = −
GMm r2
where M is the mass of the earth. If this mass be moved through a distance dr away from the earth, then the small work done on the body is dW =
GMm dr r2
In order that the body may not return on the earth, sufficient work must be done on it to import the body so much kinetic energy that it moves to infinity. Therefore the work done in moving the body from the surface of the earth to infinity is
z
∞
W =
R
LM N
GMm GMm dr = − 2 r r
OP Q
∞
=
R
GMm R
Evidently, a body of K.E. greater than this GMm/R would escape completely from the earth. Hence the minimum velocity v of projection of the body to escape into space is given by the relation
1 GMm mv2 = 2 R But
g = GM/R2
∴
v =
or
v=
2GM R
2 gR
Expressions (i) and (ii) are for the escape velocity of a body from earth.
...(i)
...(ii)
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101
If at the surface of the earth, we take g = 9.8 m/sec2 and its radius R = 6.4 × 106 m, then escape velocity v =
2 × 9. 8 × 6. 4 × 106
v = 11.2 × 103 m/sec v = 11.2 km/sec Hence, if a body is projected in the upward direction with a velocity 11.2 km/sec or more than this, it will never return to the earth. Escape velocity is the same for all bodies of different masses.
3.22 RELATION BETWEEN ORBITAL VELOCITY AND ESCAPE VELOCITY The orbital velocity of a satellite close to the earth is v0 = for a body thrown from the earth’s surface is v0 =
∴
gR e
v0 ve =
2 gR e
ve =
2 v0
=
gRe , and the escape velocity
2 gR e . Thus
1 2
If the orbital velocity of a satellite revolving close to the earth happens to increase to 2 times, the satellite would escape.
3.23
SATELLITES
In the solar system, different planets revolve round the sun. The radii of the orbits and their time periods of revolution are different for different planets. In these cases, the force of gravitation between the sun and the planet provides the necessary centripetal force. Similarly the moon revolves around the earth and the force of gravitation between the earth and the moon provides the necessary centripetal force for the moon to be in its orbit. Here moon is the satellite of the earth. From 1957, many artificial satellites around the earth have been launched. These satellites are put into orbits with the help of multi-stage rockets.
3.24 ORBITAL VELOCITY OF SATELLITE When a satellite (such as moon) revolves in a circular orbit around the earth, a centripetal force acts upon the satellite. This force is the gravitational force exerted by the earth on the satellite. In figure, a satellite of mass m is revolving around the earth with a speed v0 in a circular mv02 . r Let Me be the mass of the earth. The gravitational force exerted by the earth on the GMem , where G is gravitation constant. As the gravitational force provides satellite will be r2 the required centripetal force, we have
orbit of radius r. The centripetal force on the satellites is
Dynamics of Circular Motion and the Gravitational Field
But
103
GMe = g Re2
∴
(R e + h)3 2 gR e
T = 2π
From Eq. (iii) or (iv) the period of revolution of the satellite can be calculated. It is evident from these equations that the period of revolution of a satellite depends only upon its height above the earth’s surface. Greater is the distance of a satellite above the earth’s surface, greater is its period of revolution. This is why the moon, which is at a height of 3,80,000 km above earth, completes one revolution of earth in nearly 27 days, while an artificial satellite revolving near the earth’s surface completes 10 to 20 revolution in a day. If the height of an artificial satellite above earth’s surface be such that its period of revolution is exactly equal to the period of revolution (24 hours) of the axial motion of the earth, then the satellite would appear stationary over a point on earth’s equator. It would be synchronous with earth’s spin. Such a satellite is known as “geostationary satellite”. It is used to reflect T.V. signals and telecast T.V. programs from one part of the world to another. Now, from Eq. (iv), we have (Re + h)3 =
T2 × gR2e 4 π2
LM T gR OP N 4π Q 2
(Re + h) =
2 e
1/ 3
2
...(v)
Substituting T = 24 h = 24 × 60 × 60 = 86400 s, Re = 6.37 × 106 m and g = 9.8 m/s2 in Eq. (v), we get Re + h = 42,200 Km This is the orbital radius of a geostationary satellite. The height of the satellite above the earth’s surface is h = 42,200 – 6370 = 35,830 km. Thus, for a satellite to appear stationary, it must be placed in an orbit around the earth at a height of 35,830 km from the earth’s surface. This is often called ‘parking orbit’ of the satellite. Artificial satellites used for telecasting are placed in parking orbits. The orbiting speed of the geostationary satellite is given by v =
2π (R e + h) T
v =
2 × 3.14 × 42, 200 km = 11042 km/hr 24 h
3.25 ORBITAL SPEED AND PERIOD OF REVOLUTION OF A SATELLITE VERY CLOSE TO EARTH If a satellite is very close to the earth’s surface (h << Re), then h will be negligible compared to Re. In this case putting h = 0 in Eq (ii), the orbital speed of the satellite is given by
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Mechanics
v0 = Re
g = Re
gR e
Putting g = 9.8 meter/sec2 and Re = 6.37 × 106 meter, we have v0 =
9. 8 × (6. 37 × 106 ) = 7. 9 × 103 m/s ≅ 8 km/s
Similarly, putting h = 0 in Eq. (iv), the period of revolution of the satellite is given by T = 2π
Re = 2 × 3.14 × (6. 37 × 106 ) / 9. 8 g
= 5063 seconds ≈ 84 minutes. Thus, the speed of a satellite revolving very close to the earth’s surface is nearly 8 km/sec and its period of revolution is nearly 84 minutes.
3.26 ARTIFICIAL SATELLITES We have seen above that when a satellite revolves around the earth in an orbit near the earth’s surface then its orbital velocity is about 8 km/s. Therefore, if we send a body a few hundred kilometers above the earth’s surface and give it a horizontal velocity of 8 km/s, then the body is placed in an orbit around the earth. Such a body is called an artificial satellite. Like moon, an artificial satellite also revolves, around the earth under the gravitational attraction exerted by the earth which acts as the centripetal force. One may ask, how is it that the satellite continues to revolve in an orbit at a definite height above the earth, instead of falling towards the centre of the orbit under the centripetal force. In fact, the satellite continuously falls towards the centre of the earth, but due to the curvature of the earth it is maintained at the same height. This we can see in Fig. 14. If there were no centripetal force on the satellite then it would have moved along a straight-line path PQR————. But due to the presence of centripetal force it moves along a circular path PST————. Thus it is continuously falling towards the center of the earth through the distances QS, RT——. An artificial satellite is placed in an orbit by means of a multi-stage rocket. The satellite is placed on the rocket. On being fired, the rocket moves vertically upward with an increasing velocity. When the fuel of the first stage of the rocket is exhausted, its casing is detached and the second stage comes in operation. The velocity of the rocket increases further. This process continues. When the rocket, after crossing the dense atmosphere of the earth attains proper height, a special mechanism gives a thrust to the satellite producing a pre-calculated horizontal velocity. A satellite carried to a height (<< earth’s radius) and given a horizontal velocity of 8 km/s is placed, by earth’s gravity, almost in a circular orbit around the earth, It then continues revolving (without using any fuel). Even due to a slight mistake in calculating the velocity, the orbit of the satellite would change considerably. The satellite is always given such a velocity that its orbit remains outside the earth’s atmosphere; otherwise the friction of atmosphere would cause so much heat that it will burn. The satellite revolves around the earth in an orbit with earth as center, or a focus. If a packet is released from the satellite, it will not fall to the earth but will remain revolving in the same orbit with the same speed as the satellite.
Dynamics of Circular Motion and the Gravitational Field
105
Uses of Artificial Satellite (1)
They are used to study the upper regions of the atmosphere.
(2)
Information about the space of the earth can be obtained.
(3)
Weather forecast can be communicated to the earth.
(4)
Radiation from the sun and outer space can be studied.
(5)
Distant telecasting can be achieved.
(6)
Meteorites can be studied.
(7)
Space flights are possible due to artificial satellites.
O E a rth
Fig. 14
3.27 BLACK HOLES Gravitational attraction occurs throughout the universe. It is used in theories of galaxies and stars in addition to explaining the motion of planets and earth satellites as we have done. Stars appear to have formed after a ‘big bang’ or explosion of matter when the universe first began. Initially, the hydrogen gas atoms produced moved towards each other under gravitational attraction, the speed and energy of the atoms increased and their temperature rose. Many atoms then had sufficient energy to fuse together and form helium atoms. The fusion of hydrogen atoms to form helium atoms produces nuclear energy and the high temperature rise produces light. A star reaches a stable size or radius when the outward pressure of the gas is balanced by the inward gravitational attraction between the atoms. When all the hydrogen in the star is used up in the fusion process, which may occur after by millions of years, the unbalanced gravitational attraction between the atoms will result in the star becoming smaller and smaller. Eventually the star will collapses. From the waveparticle duality of light, light can be considered as particles (photons). So as the star collapses in size and becomes denser, the gravitational attraction on light from its surface increases. This make the light bend more and more towards the inside of the star and soon no light escapes from the star. So a collapsed star has such a strong gravitational field that not even light can escape from it, which is why it appears black. The region round the star from which no light can escape is called a black hole. Fortunately, our star, the sun, is estimated to continue producing nuclear energy for about 5000 million years. In 1994 the Hubble Space Telescope discovered
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Mechanics
one of the biggest black holes in the universe, about 500 light years across and in a galaxy 52 million light years away from the earth. (1 light year = 9.46 × 1012 km approx.).
PLANETS AND SATELLITES Q. 1. The maximum and minimum distances of a comet from the sun are 1.6 × 1012 m and 8.0 × 1010 m respectively. If the speed of the comet at the nearest point is 6.0 × 104 m/sec, calculate the speed at the farthest point. Ans. The speed of a satellite round a planet in elliptical orbit varies constantly. In order →
→
→
to conserve angular momentum. L ( = r × m v ) at these points is →
L = mv1r1 = mv2r2 v 1 r 1 = v2 r 2
or
Putting the given values, we have v1 × (1.6 × 1012) = (6.0 × 104) × (8.0 × 1010) ∴
v1 =
(6. 0 × 104 ) × (8. 0 × 1010 ) 1. 6 × 1012
= 3. 0 × 103 meter /sec Ans. Q. 2. Earth’s revolution round the sun has radius 1.5 × 1011 meters and period 3.15 × second (one year). If the gravitational constant is 6.67 × 10 –11 newton-m2/kg2, deduce the mass of the sun. The formula is to be deduced. 107
Ans. The period of revolution T of the earth round the sun (mass Ms) in a circular orbit of radius r is given by
∴
T2 =
4 π2r3 GMs
Ms =
4 π2r3 GT2
Substituting the given values: Ms =
4 × (3.14 )2 × (1. 5 × 1011 m)3 (6. 67 × 10−11 newton-m2 / kg2 ) (3.15 × 107 s)2
= 2.0 × 1030 kg Ans. Q. 3. Determine the mass of the earth from the moon’s revolution around the earth in a circular orbit of radius 3.8 × 105 km with period 27.3 days, G = 6.67 × 10–11 newton-m2/kg2. Ans. The moon revolves round the earth. If r be the radius of its orbit and T the orbiting period, then the mass of the earth is given by M =
4 π2r3 GT2
Dynamics of Circular Motion and the Gravitational Field
107
Here r = 3.8 × 105 km m 3.8 × 108 meter, T = 27.3 days = 27.3 × 24 × 60 × 60 sec. ∴
M =
4 × ( 3.14)2 × ( 3. 8 × 108 )3 ( 6. 67 × 10−11 ) × (27. 3 × 24 × 60 × 60)2
= 5. 8 × 1024 kg. Q. 4. With what horizontal velocity must a satellite be projected at 800 km above the surface of the earth so that it will have a circular orbit about the earth. Assume earth’s radius 6400 km. What will be the period of rotation? g = 9.8 m/sec2. Ans. The orbital velocity of a satellite revolving round the earth at a height h from earth’s surface is given by v = R
R = 6400 km = 6.4 × 106 meter
Putting the given values and
FG g IJ H R + hK
R + h = 6400 + 800 = 7200 km = 7.2 × 106 meter
we get
6 v = ( 6. 4 × 10 ) ×
9. 8 7. 2 × 106
= 7.5 × 103 m/sec = 7.5 km/sec. The satellite must be given a horizontal velocity of 7.5 km/sec after taking it to a stable orbit round the earth. The period of revolution is
T =
2π (R + h) 2 × 3.14 × 7200 km = v 7.5 km/sec
= 6029 sec = 100.5 min Ans. Q. 5. The period of a satellite in circular orbit of radius 12000 km around a planet is 3 hours. Obtain the period of a satellite in circular orbit of radius 48000 km around the same planet.
3 Ans. The period of a satellite is proportional to the th power of its orbital radius. 2 i.e.,
T ∝ ( r )3 / 2 Thus, if T′ be the period in an orbit of radius r′, thus
T = T′
FG r IJ H r′ K
3 /2
Here T = 3 hours, r = 12000 km and r′ = 48000 km. Therefore
3 = T′
FG 12000IJ H 48000K
3 /2
=
FG 1 IJ H 4K
T ′ = 24 hours Ans.
3/ 2
=
1 8
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109
R from earth’s 2 surface, where R is the radius of the earth, calculate its period of revolution (R = 6.38 × 106 meter). Q. 8. A satellite moves in a circular orbit around the earth at a height
Ans. The period of revolution of a satellite at a height h from earth’s surface is given by
For h =
T =
2π (R + h) R
FG R + h IJ H g K
T =
2π × 1. 5 R R
1. 5 R g
R , we have 2
= 2π
FG R IJ (1. 5) H gK
3 /2
6. 38 × 106 meter × (1. 5)3 /2 9.8 meter/sec2
= 2 × 3.14 ×
= 5067 sec × (1.5)3/2 = 9310 sec = 2 hours 35 min Ans. Q. 9. Calculate the limiting velocity required by an artificial satellite for orbiting very closely round the earth (R = 6.4 × 106 meter, g = 9.8 meter/sec2). What would be the period. Ans. The orbital velocity of a satellite revolving round the earth (radius R) at a height h from earth’s surface is given by v = R
g R+h
For a satellite orbiting very closely the earth’s surface, h << R, so that v = R
g = R =
gR 9. 8 × 6. 4 × 106 = 7. 92 × 103 m/sec
= 7.92 km/sec. 2π (R + h) R
The period is
T =
For h << R
T = 2π
R+h g
R = 2 × 3.14 × g
6. 4 × 106 = 5. 075 × 103 sec 9. 8
= 84.6 min = 1.41 hours Ans.
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Mechanics
Q. 10. A satellite is revolving round near the equator of a planet of mean density ρ. Show that the period T of such an orbit depends only on the density of the planet. Ans. Let m be the mass and v the velocity of satellite which orbits round the planet just above its surface. The radius of the orbit is practically equal to the radius R of the planet. Therefore, the centripetal force acting upon the satellite is mv2/R. If M be the mass of planet, the gravitational force between the planet and the satellite is GMm/R2 and this supplies the required centripetal force. Thus
Mm mv2 = G 2 R R or
GM R
v = The period of the satellite is T =
2πR R3 = 2π v GM
Now, the mass M of the planet (mean density suppose ρ) is M =
4 3 πr ρ 3
Making the substitution in the last expression, we get
3π Gρ
T =
Thus the period depends only on the density of the planet. Q. 11. A satellite is launched into a circular orbit 1600 km above the surface of the earth. Find the period of revolution if the radius of the earth is R = 6400 km and the acceleration due to gravity is 9.8 m/sec2. At what height from the ground, should it be launched so that it may appear stationary over a point on the earth’s equator? Ans. The orbiting period of a satellite at a height h from earth’s surface is T =
2π (R + h) R
Here R = 6400 km, h = 1600 km =
R . Then 4
FG H
2π R + T =
= 2π
R+h g
R 4
IJ FG R + R IJ K H 4K
R
FG R IJ FG1 + 1 IJ H gK H 4K
g 3/ 2
Dynamics of Circular Motion and the Gravitational Field
= 2π
111
R (1. 25)3 /2 g
Putting the given values:
6. 4 × 104 m (1. 25)3 / 2 9.8 m/sec2
T = 2 × 3.14 ×
= 7092 sec = 1.97 hours. Now, a satellite will appear stationary in the sky over a point on earth’s equator if its period of revolution round the earth is equal to the period of revolution of the earth round its own axis which is 24 hours. Let us find the height h of such a satellite above earth’s surface in terms of earth’s radius. Let it be nR. Then T =
2π (R + nR) R
= 2π
R + nR g
R (1 + n)3 / 2 g
= 2 × 3.14
6. 4 × 106 meter (1 + n)3 /2 9.8 m / sec2
= (5067 sec) (1 + n)3/2 = (1.41 hours) (1 + n)3/2 For T = 24 hours, we have (24 hours) = (1.41 hours) (1 + n)3/2 or
(1 + n)3/2 =
24 = 17 1. 41
or
1 + n = (17)2/3 = 6.61
or
n = 5.61 The height of the geo-stationary satellite above earth’s surface is nR = 5.61 × 6400 km = 3.59 × 104 km Ans.
Q. 12. Two satellites of same mass are launched in the same orbit round the earth so as to rotate opposite to each other. They collide in elastically and stick together as wreckage. Obtain the total energy of the system before and just after the collision. Describe the subsequent motion of the wreckage. Ans. The potential energy of a satellite in its orbit GMm GMm . The total , and the kinetic energy is is − 2r r energy is E = K+U=
GMm GMm − 2r r
r
→ v m
→ v m
M
Fig. 15
112
Mechanics
GMm 2r where m is the mass of the satellite, M the attracting mass (earth) and r the orbital radius. When there are two satellites, each of mass m, in the same orbit the energy would be = −
−
GMm GMm GMm − = − 2r 2r r
→
Let v be the velocity of the wreckage after the collision. Then, by the law of conservation of momentum, we have →
→
→
m v + m v = (m + m) v′
or
mv – mv = (m + m) v′ ∴
v′ = 0
Therefore, the wreckage of mass 2m has no kinetic energy but only potential energy. Hence the total energy just after the collision would be
GM (2m) r As the velocity of the wreckage is zero, the centripetal force disappears and the wreckage falls down under gravity. −
Q. 13. A small satellite revolves round a planet in an orbit just above planet surface. Taking G = 6.66 × 10 –11 MKS units and mean density of planet as 8.00 × 103 MKS units, calculate the time period of the satellite. Ans.
T =
=
3π Gρ 3 × 3.14 6. 66 × 10−11 × 8. 00 × 103
= 0.420 × 104 sec = 70 min Ans. Q. 14. Two earth satellites; A and B, each of mass m are to be launched into circular orbits about earth’s centre. Satellite A is to orbit at an altitude of 6400 km and B at 19200 km. The radius of earth is 6400 km. (a) What is the ratio of the potential energy of B to that of A, in orbit, (b) Ratio of kinetic energy, (c) Which one has the greater total energy. Ans. (a) The potential energy of an earth’s satellite in a circular orbit of radius r is
GMm r where M is the mass of the earth and m that of the satellite. U(r) = −
For A:
r = 6400 + 6400 = 12800 km
Dynamics of Circular Motion and the Gravitational Field
For B: ∴
113
r = 19200 + 6400 = 25600 km
UB 12800 1 r = . = A = UA rB 25600 2
(b) The kinetic energy of an earth’s satellite moving with velocity v in a circular orbit of radius r is K = ∴
1 GMm mv2 = 2 2r
rA 1 KB = = KA rB 2
(c) The total energy of the satellite is
GMm GMm GMm + =− r 2r 2r Which is negative. Clearly the farther the satellite is from the earth, the greater (that is, less negative) is its total energy E. Hence the satellite B has the greater (less negative) total energy. E = U+K= −
Q. 15. A stream of α-particles is bombarded on mercury nucleus (z = 80) with a velocity 1.0 × 109 cm/sec. If an α-particle is approaching in head-on direction, calculate the distance of closest approach. The mass of α-particle is 6.4 × 10 – 24 gm and electronic charge is 4.8 × 10–10 esu. Ans. The distance of closest approach of an α-particle to a nucleus is given by ro =
2 ze2 cm ki
where ki is the initial kinetic energy. Here ki = =
1 mv2 2 1 × (6. 4 × 10−24 gm) (1.0 × 109 cm/sec)2 2
= 3.2 × 10–6 erg. ∴
ro =
2 × 80 × ( 4. 8 × 10−10 esu)2 3. 2 × 10−6 erg
ro = 1.152 × 10–11 cm
Ans.
Q. 16. In a double star, two stars (one of mass m and the other of 2m) distant d apart rotate about their common centre of mass. Deduce an expression for the period of revolution. Show the ratio of their angular momenta about the centre of mass is the same as the ratio of their kinetic energies. Ans. The centre of mass c will be at distances
d 2d and from the masses. 2m and m 3 3
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Mechanics
respectively. Both the stars rotate round c in their respective orbits with the same angular velocity ω. The gravitational force acting on each star due to the other supplies the necessary
G (2m) m . If we consider the d2
centripetal force. The gravitational force on either star is
rotation of the smaller star, the centripetal force (mrω2) is m
FG 2d IJ ω . H 3K
Fig. 16
∴
FG IJ H K
G (2m) m 2d ω2 = m d2 3 ω =
or
3G m d3
Therefore, the period of revolution is given by T =
2π = 2π ω
d3 3 Gm
The ratio of the angular momenta is
(Iω)big (Iω )small
=
Ibig = Ismall
FG d IJ H 3K F 2d I mG J H 3K
2
(2m)
2
=
1 2
Since ω is same for both. The ratio of their kinetic energies is
FG 1 Iω IJ H2 K FG 1 Iω IJ H2 K 2
big
2
=
Ibig 1 = Ismall 2
small
which is the same as the ratio of their angular momenta.
2
Dynamics of Circular Motion and the Gravitational Field
115
Q. 17. Estimate the mass of the sun assuming the orbit of the earth around the sun to be circular. The distance between sun and earth is 1.49 × 1013 cm, G = 6.66 × 10–8 egs units and earth takes 365 days to make one revolution around the sun. Q. 18. An artificial satellite of the earth moves at an altitude of 640 km along a circular orbit. Find the orbital velocity of the satellite. (Radius of earth = 6400 km). Q. 19. Show that the time of revolution of a satellite just above the earths surface is 84.4 min. Density of earth is 5.51 × 103 kg/meter3 and G = 6.67 × 10–11 nt-m2/kg. Q. 20. What is the smallest radius of a circle at which a bicyclist can travel if his speed is 7 m/sec and the coefficient of static friction between the tyres and the road in 0.25. Under these conditions what is the largest angle of inclination to the vertical at which the bicyclist can without falling? Ans. The (maximum) force of friction is given by fs = µsN = µsmg This must supply the required centripetal force. If R be the smallest radius, the maximum required centripetal force would be mv2/R. Thus µs mg = or
mv2 R v2 µsg
R =
here v = 7 m/sec, µs = 0.25 and g = 9.8 m/sec2
(7 m / s)2 = 200 meter 0. 25 × (9. 8 m/sec2 ) The (largest) angle of inclination θ is given by ∴
R =
tan θ =
v2 (7 m/sec)2 = = 0. 25 Rg 20 m × 9. 8 m/sec2
θ = tan–1 (0.25) = 14° Ans. Q. 21. A cord is tied to a pail of water and the pail is swing in a vertical circle of radius 1 meter. What must be the minimum velocity of the pail at the highest point of the circle if no water is to spill from the pail? (g = 9.8 m/sec2). Ans. At the highest point of the vertical circle, the centripetal force is supplied by the tension in the cord and also by the weight of the pail plus water. The force must be atleast equal to the weight, otherwise water would spill from the pail under gravity. Hence if vc be the minimum velocity required at the highest point, we have mvc2 = mg R
or
vc = Here
Rg
R = 1 meter vc =
1 × 9. 8 = 3.13 m/sec Ans.
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Mechanics
Q. 22. A circular curve of highway is designed for traffic moving at 15 m/sec. (a) If the radius of the curve is 100 m, what is the correct angle of banking of the road. (b) If the curve is not banked, what is the minimum coefficient of friction between tyres and road that would keep traffic from skidding at this speed? Ans. (a) Let θ be the correct angle of banking. Then tan θ =
v2 Rg
Putting the given values, we have (15 m/s)2 = 0. 23 (100 m) (9.8 m/s2 ) θ = tan–1 (0.23) = 13°
tan θ =
(b) If the road is not banked, then the required frictional force f (say) must supply the entire centripetal force,
mv2 . Thus R f =
mv2 R
µmg =
mv2 R
But f = µN = µmg. Thus
or
µ =
v2 = 0. 23 Rg
Ans.
Q. 23. The radius of curvature of a railway line at a place is 800 meter and the distance between the rails is 1.5 meter. What should be the elevation of the outer rail above the inner one for a safe speed of 20 km/hour? Ans. Let θ be the correct angle of banking. Then tan θ =
v2 Rg
Here v = 20 km/hour = 5.56 m/sec and R = 800 m tan θ =
5. 56 × 5. 56 = 0. 004 800 × 9. 8
If x be the elevation of the outer rail and l the distance between the rails, then tan θ = or
x l
x = l tan θ = (1.5 m) (0.004) = 0.006 m = 0.6 cm.
Dynamics of Circular Motion and the Gravitational Field
117
Q. 24. A mass m on a frictionless table is attached to a hanging mass M by a cord through a hole in the table. Find the condition (v and r) with which its must spin for M to stay at rest. Ans. The mass M will stay at rest when its weight Mg is used up in supplying the required centripetal force
mv2 to the mass m, that is when r Mg =
mv2 r
Mg v2 = m r
or
This is the required condition. The equilibrium will, however, be unstable. Fig. 17 Q. 25. A smooth table is placed horizontally and an ideal spring of spring-constant k = 1000 nt/m and unextended length of 0.5 m has one end fixed to its centre the other end is attached to a mass of 5 kg which is moving in a circle with constant speed 10 m/s. Find the tension in the spring and the extension of the spring beyond its normal length.
Ans. The centripetal force required for the mass m to move in a circle is supplied by the tension T produced in the stretched spring. The streched length of the spring is R, equal to the radius of the circle. If lo be the unextended length of the spring, then the elongation is (R – lo) and the tension produced is given by T = k (R – lo),
...(1)
where the spring-constant k is the tension per unit elongation. But this is equal to the centripetal force mv2/R. Therefore, k(R – lo) =
mv2 R
kR (R – lo) = mv 2
or
Here k = 1000 nt/m, lo = 0.5 m, m = 5 kg, v = 10 m/sec. ∴ or or
1000 R(R – 0.5) = 5 × 100 R(R – 0.5) = 0.5 2R2 – R – 1 = 0
Solving this quadratic eq. we get R = 1.0 m The extension of the spring beyond its normal length is R – lo = 1.0 – 0.5 = 0.5 m Substituting this value in (1), the tension is T = 1000 × 0.5 = 500 nt. Ans. Q. 26. An electron is moving in a circular orbit of radius 5.3 × 10 –11 meter around the atomic nucleus at a rate of 6.6 × 1015 rev/sec. Find the acceleration of the electron and centripetal force acting on it. The mass of the electron is 9.1 × 10 –31 kg. Ans. Let R be the radius of the orbit and f the number of revolutions per second. Then the velocity of the electron is given by
118
Mechanics
v = 2πR f Hence its acceleration is a =
v2 = 4 π2R f 2 R
= 4 × (3.14)2 × (5.3 × 10 –11) × (6.6 × 1015)2 = 9.1 × 1022 m/sec2, towards the nucleus. The centripetal force is Fc = ma = (9.1 × 10 –31) × (9.1 × 1022) = 8.3 × 10 –8 nt towards the nucleus. Q. 27. A smooth table is placed horizontal and a spring of unstretched length lo and forceconstant k has one end fixed to its centre to the other end of the spring is attached a mass m which is making f revolutions per second around the centre. Show that the radius R of this uniform circular motion is klo/(k – 4π2mf 2) and the tension T in the spring in 4π2mklof 2/ (k – 4π2mf 2). Ans. The centripetal force required for the mass m to move in a circle is supplied by the tension T produced in the stretched spring. The stretched length of the spring is R, equal to the radius of the circle. Thus the elongation in the spring is (R – lo) and the tension produced is given by T = k (R – lo)
...(1)
as k (force-constant) is the tension per unit elongation. The (linear) velocity of motion is given by v = 2πRf Therefore, the required centripetal force is
mv2 = 4π2Rf2m R Since T =
...(2)
mv2 , we have by (1) and (2) we get R k (R − lo ) = 4π2Rf 2 m
or
R =
klo k − 4 π2 f 2m
Substituting this value of R in Eq. (1) we get T = k
T =
LM kl − l OP N ( k − 4 π f m) Q o 2 2
4 π 2 f 2mlo k k − 4 π 2 f 2m
o
Ans.
Dynamics of Circular Motion and the Gravitational Field
119
Q. 28. An artificial satellite is revolving round the earth at a distance of 620 km. Calculate the minimum velocity and the period of revolution. Radius of earth is 6380 km and acceleration due to earth’s gravity at the surface of the earth is 9.8 metres/sec2. Ans. Radius of earth’s satellite orbit r = Radius of earth + Distance of satellite from earth’s surface = 6380 + 620 = 7000 km = 7 × 106 m/sec. Radius of earth ∴
R = 6380 × 103 m, g = 9.8 m/sec2
Period of revolution T =
and
2πr R
orbital velocity v = R
2π × 7 × 106 r = 6380 × 103 g g = 6380 × 103 r
7 × 106 = 5775 sec. 9. 8
9. 8 7 × 106
= 7. 55 × 103 m/sec. Q. 29. A stone of mass 1 kg is attached to one end of a string 1 m long, of breaking strength 400 nt, and is whirled in a horizontal circle on a frictionless surface the other end of the string is kept fixed. Find the maximum velocity the stone can attain without breaking the string.
mv2 is to be supplied by the tension in the string R which should not exceed 400 nt. Thus, if v be the maximum velocity the stone can attain, we have Ans. The required centripetal force
mv2 = T = 400 nt R Here m = 1 kg and R = 1 m. Then v 2 = 400 ∴
v = 20 m/sec
Ans.
Q. 30. A mass of 1 standard kg is placed at sea level on the earth’s equator and is moving with the earth in a circle of radius 6.40 × 106 meter (earth’s radius) at a constant speed of 465 m/s. Determine the centripetal force needed. Also find the force exerted by the mass on a spring balance from which it is suspended at the equator (its weight). Assume that the mass would weight exactly 9.80 nt the equator of the earth did not rotate about its axis. Ans. The centripetal force is given by
1 × 465 mv2 = = 0. 0338 nt. R 6. 40 × 106 w′ = w – Fc Fc =
The weight
= 9.80 – 0.0338 = 9.77 nt Ans.
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Mechanics
Q. 31. Escape velocity from solar system. Show that the escape velocity of a body from solar system, launched from the earth is
2GMs/Res where Ms is the mass of the sun and Res is the distance of the earth from the sun. Neglect the earth’s rotation and influence of earth’s gravity. What is the escape velocity, if G = 6.67 × 10–11 S.I. unit, Me = 1.33 × 1030 kg and Res = 1.49 × 1011 m. Ans. In the gravitational field of the sun, the potential energy of a body of mass m situated on the earth is GMsm R es
v = −
If the body is to be escaped from the gravitational influence of the sun, the body should be imparted with kinetic energy
FG 1 mv IJ H2 K 2
equal in magnitude to GMsm/Res. So that
1 GMsm mv2 = or v = 2 R es
2GMs R es
From the given data, the escape velocity (launched from the earth) from the solar system is v =
2 × 6. 67 × 10−11 × 1. 33 × 1030 1. 49 × 1011
= 4.2 × 104 m/sec = 42 km/sec. Q. 32. Compare the root mean square velocity of oxygen molecules at 27°C with the escape velocity from earth’s surface (Boltzmann constant k = 1.4 × 10 – 23 joule/k). Ans. If V is the R.M.S. velocity of the oxygen molecular, then the mean kinetic energy per molecule is given by
1 1 mv2 = kT 2 2 where m [ = 32 × (mass of a proton) = 32 × 1.7 × 10 –27 kg] is the mass of an oxygen molecule and T the absolute temperature.
3 kT = m
3 × 1. 4 × 10−23 × 30° 32 × 1. 7 × 10−27
∴
V =
Q
T = 273 + 27 = 300 k
Now, escape velocity from the earth v =
∴
V = v
2 g Re = 2 × 9. 8 × 6. 4 × 106 3 × 1. 4 × 10−23 × 300 32 × 1. 7 × 10−27 × 2 × 9. 8 × 6. 4 × 106
Dynamics of Circular Motion and the Gravitational Field
121
V ≈ 0. 04 Ans. v Q. 33. What is the potential energy of a mass of 1 kg on the surface of the earth, reference to zero potential energy at infinite distance? Calculate also its potential energy at a distance of 10 5 km from the centre of the earth. Ans. In case of earth (or solid sphere) the potential energy is given by
GMm r (i) For the surface of the earth, r = R v(r) = −
GMm − (6. 67 × 10−11 ) × 5. 98 × 1024 × 1 = R 6. 37 × 106 = –6.23 × 107 joules
v(r) = −
∴
(ii) when
r = 105 km = 108 m v(r) = −
(6. 67 × 10−11 ) × 5. 98 × 1024 × 1 108
= –3.98 × 106 joules Q. 34. If a body is to be projected vertically upwards from earth’s surface to reach a height of 10R, how much velocity should be given? Ans. The body should be supplied so much kinetic energy, so as it can reach a height 10R from the surface of the earth or 11R from the centre of earth. If v is the velocity given to the body initially, then its kinetic energy = Increase in potential energy = − = − ∴ or
1 mv2 2
gR2m gR2m + r2 r1 gR2m gR2m 10 + = mgR 11 R 11 R
1 10 mv2 = mgR 2 11 v =
20 gR = 11
20 × 9. 8 × 6. 4 × 106 11
= 10.6 × 103 m/sec. Q. 35. For the earth-sun system, calculate (i) the kinetic energy and (ii) the work which will have to be done in doubling the radius of the orbit of the earth. Ans. The gravitational potential of the earth due to the sun = –GM s/r ∴ Potential energy of the earth = −
GMsMe r
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Mechanics
1 Mev2 2 In case of earth the necessary centripetal force for its rotation is provided by gravitational attraction i.e., K.E. of the earth =
GMsMe Mev2 = r2 r ∴
1 1 GMsM e Mev2 = 2 2 r
GMsMe 1 GMsMe GMsMe + =− 2 2r r r Evidently negative sign indicates that the earth is bound with the earth with this energy ∴
Total energy of the earth = −
i.e., The binding energy =
1 GMsMe 1 = Mev2 = Mer2ω2 2 2r 2
FG H
1 1. 5 × 1011 × 2 × 3.14 × 5. 98 × 1024 × = 2 365 × 24 × 60 × 60
IJ K
2
= 2.7 × 1033 joules
FGQ r = 1.5 × 10 H
11
m and ω =
2π rad /sec 365 × 24 × 60 × 60
IJ K
when the radius of the earth’s orbit is doubled, then the final binding energy will be =
GMsMe 1 GMsM e = 2 (2 r ) 2 2r
2. 7 × 1033 = 1. 35 × 1033 joules. 2 Therefore the amount of work done in doubling the radius of the orbit = 2.7 × 1033 – 1.35 × 1033 = 1.35 × 1033 joules Ans. Q. 36. Two bodies of masses M1 and M2 are placed distant d apart, show that at the position, where the gravitational field due to them is zero, the potential is given by =
G ( M1 + M2 + 2 M1M2 ) d Ans. Let the gravitational field be zero at a point distant r from the mass M1. Then, the distance of the same point from the mass M2 = d – r. At the point, under consideration, the field due to M1 and M2 is zero, i.e.,
V = −
−
or
GM2 GM1 = − 2 ( d − r)2 r d−r = r
M2 M1
Dynamics of Circular Motion and the Gravitational Field
123
Adding 1 on both sides, we have
M2 +
d = r
∴
M1
or r =
M1 d M1
d – r = d−
M1 +
M2
=
d M1 M1 +
M2
d M2 M1 +
M2
Therefore, potential at the point is V = − = − = − V = −
GM1 GM2 − r ( d − r) GM1 d M1
e
( M1 +
G M1 + d
M2 ) −
GM2 d M2
( M1 +
M1M2 + M2 + M1M2
G (M1 + M2 + 2 M1M2 ) d
M2 )
j
Ans.
Q. 37. If a mass 50 kg is raised to a height 2R from the earth’s surface, calculate the change in potential energy (g = 9.8 m/sec2; R = 6.4 × 106 m). Ans. The potential energy of a mass m distance r from a solid sphere of a mass M will be given by U(r) = Potential × m = −
GMm r
QV = −
GM r
(referring zero potential energy at infinite distance) But
g =
GM , where R is the radius of earth, R2
gR2m r Difference of potential energy of mass ‘m’ between the points at r1 and r2 is
∴
U(r) = −
U(r2) – U(r1) = − here ∴
gR2m gR2m + r2 r1
r 1 = R and r2 = 2R + R = 3R gRM 2 + gRM = gmR U(r2) – U(r1) = − 3 3 2 × 9. 8 × 50 × 6. 4 × 106 = 2.1 × 109 joules Ans. = 3
Q. 38. (a) A small block of mass ‘m’ slides along the friction less track. Calculate the height ‘h’ at which mass must be released on the track to be able to go round the track of radius ‘R’.
Dynamics of Circular Motion and the Gravitational Field
so
125
2mg ( h − 2R) − mg R 2mg (5R − 2R) − mg = R = 5 mg Ans.
NC =
NC
Q. 39. A bead can slide without friction on a circular hoop of radius 10 cm in a vertical plane. The hoop rotates at a constant rate of 2 rev/sec, about a vertical diameter. (a) Find angle θ at which the bead will be in vertical equilibrium, (b) Can bead rise up to the height of the centre of hoop, (c) What will happen if hoop rotates at 1 rev/sec. Ans. For equilibrium along the hoop mrω2 cos θ = mg sin θ or or or
(R sinθ) ω 2 cos θ = g sin θ
cos θ =
g 9. 8 = 2 0.1 (2π × 2)2 Rω
θ = cos–1 (0.62)
Fig. 18
(b) The bead can not rise up to the height of centre of hoop because, then, there will be no vertical component of mrω2 to balance the weight mg of bead. (c) If ω = 2π × 1 rev/sec. 9. 8 98 = , which is greater than 1. cos θ = 2 0.1 (2π × 1) 4 π2 This means no value of θ is possible. Hence bead will remain stationary at the bottom of hoop. Q. 40. A hemispherical bowl of radius R = 0.1 m is rotating about its own axis, which is vertical, with angular velocity ω. A particle of mass m = 10 –2 kg, on frictionless inner surface of bowl is also rotating with same angular velocity. The particle is at a height ‘h’ from the bottom of bowl. (a) Obtain a relation between h and ω. What is minimum value of ω needed to have nonzero value of h? (b) It is desired to measure ‘g’ using this set-up by measuring ‘h’ accurately. Assuming that R and ω are known and least count in the measurement of ‘h’ is 10 –4 m, find minimum ω possible error in measurement of g. Ans. (a) The forces acting on rotating particle P are o (1) Weight, mg acting downwards θ N R -h (2) Normal reaction, N θ (3) Centrifugal force, mrω2 P m rω2 For equilibrium of rotating particle C r N cos θ = mg ...(1) and N sinθ – mrω2 = Centripetal force mg = mrω2 i.e.,
N sin θ = 2 mrω2
...(2) Fig. 19
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Mechanics
From (1) and (2);
But
tan θ =
2rω 2 g
tan θ =
PC r = CO R − h
r 2rω 2 = R−h g
so
g 2(R − h) 9. 8 4. 9 = ω2 = 2 (0.1 − h) 0.1 − h ω2 =
or i.e.,
which is desired relation. For non-zero value of h, ω 2 >
...(3)
4.9 . Hence, ω > 7 radian/sec. 0. 1
(b) Since from Eq. (3), we have g = 2ω2 (R – h) Taking log on both side differentiating, taking h → 0,
log g = log 2ω2 + log (R – h)
∆g ∆h = 0− g R−h ( ∆g)min = g.
∆h 10−4 = 9. 8 × = 9. 8 × 10−3 R 0.1
Q 41. A thread is passing through a hole at the center of a frictionless table. At the upper end, a block of 0.5 kg is tied and another mass 8.0 kg is tied to the lower end which is freely hanging. The smaller mass is rotated about an axis passing through hole so as to balance the heavier mass. If hanging mass is changed to 1.0 kg, what is fractional change in radius and the angular velocity of smaller mass to balance the hanging mass again. Ans. Let r = radius of circular path traced by smaller mass and ω = angular velocity of rotation then tension in string T = mrω2 Since, the string balances hanging mass M, Hence
T = Mg
For equilibrium of system, mrω2 = Mg or
0.5 rω2 = 8 g
...(1)
Fig. 20
if r′ is the new radius and ω′ the angular velocity when 8.0 kg mass is changed to 1.0 kg 0.5 r′ω′2 = 1 g
...(2)
Dynamics of Circular Motion and the Gravitational Field
127
Dividing (1) by (2) we get
FG IJ H K
r′ ω ′ r ω
2
= 8
...(3)
Again, since angular momentum of system will be conserved mr2ω = mr′2ω′
ω = ω′
or
FG r′ IJ H rK
2
putting the value in Eq. (3) we get
FG r IJ H r′ K
δr r ′ − r 8 − 1 = = =7 r r 1
= 8 or
and
δω ω′ ~ ω ω′ = =1 = ω ω ω
or
δω = ω
FG r IJ H r′ K
2
−1=
1 63 ~1 = 64 64
Q. 42. A particle describes a horizontal circle on a smooth surface of inverted cone. The height of the plane of circle above the vertex is 9.8 cm. Find speed of the particle. Ans. The particle while describing circular path along smooth surface of inverted cone is acted upon by two forces along the inclined surface of cone. (1) mg cosθ, component of its weight down the inclined surface.
Hence for circular motion of particle to be possible, without sliding along surface, condition is,
9 .8 cm
mv2 sin θ, the component of reaction of r centripetal force up the inclined plane. (2)
r
θ mg m g cos θ
mv2 r
θ
Fig. 21
mv2 sinθ = mg cos θ; where θ is half above the cone angle r v =
rg cot θ
From figure
r = h tan θ = 9.8 × 10–2 tan θ
Hence
v =
9. 8 × 10−2 tan θ × 9. 8 cot θ
= 0. 98 tan θ cot θ v = 0.98 m/sec
Ans.
Dynamics of Circular Motion and the Gravitational Field
i.e.,
vmax =
µgR =
129
0. 4 × 9. 8 × 30 = 10. 84 m/sec Ans.
Q. 46. Calculate the altitude of an artificial satellite if it is always above a certain place on earth’s surface, assuming its orbit to be circular. (Mean radius of earth = 6400 km, g = 9.80 m/s2). Ans. When the period of satellite in its orbit is equal to the period of earth rotation about own axis, then the satellite shall always be at a fixed point in reference to earth. Thus, period of satellite;
T = 24 hrs
Since
T =
2π ω
ω =
2π 2 × 3.142 = rad/s T 24 × 60 × 60
or For satellite in circular orbit,
GMm mv2 = mrω2 = r2 r r3 =
or where
GM ω2
...(1)
r = satellite to earth distance, M = mass of earth, m = mass of satellite, ω = angular velocity of satellite.
Again on earth’s surface,
GM = g or GM = gR2 R2 putting this in eq. (1) we get r3 = or
gR2 9. 8 × (6400000)2 × (24 × 3600)2 = ω2 (2 × 3.142)2
r = 42400 km height of satellite from earth surface h = r – R
or
h = 42400 – 6400 h = 36000 km Ans.
Q. 47. A satellite revolves round a planet in an elliptic orbit. Its maximum and minimum distances from the planet are 1.5 × 107 meters and 0.5 × 107 meters respectively. If the speed of the satellite at the farthest point be 5 × 103 m/sec. Calculate the speed at the nearest point. Ans. Let, F = Farthest position of satellite from planet situated at focus at the elliptical path
130
Mechanics
r1
M
F
N m
r2 P lan et
Fig. 22
N = Nearest position of satellite, m = mass of satellite, M = mass of planet, r 1 = farthest distance, r 2 = nearest distance, v 1 = velocity at farthest position, v 2 = velocity at nearest position. Assuming that path of satellite nearly circular at farthest and nearest position, we have for farthest position,
mv12 GMm = r12 r1 GM = v12 r1
or
...(1)
at nearest position,
mv22 GMm = r22 r2 GM = v22 r2
or
...(2)
Dividing (2) by (1), we get
v22 r1 v12 = r2 v 2 = v1
or
FG r IJ Hr K 1
1 2
...(3)
2
putting values, we get v 2 = 5 × 103 or
(1. 5 × 107 ) / (0. 5 × 107 ) = 5 × 103 3
v 2 = 8.660 × 103 m/s Ans.
Q. 48. Find the magnitude of centripetal acceleration of a particle on the tip of a fan blade of 0.3 meter diameter, rotating at speed of 1000 rev/min.
132
Mechanics
Ans. Let R = actual distance of earth from the sun and R′ =
R = supposed distance 3
between sun and earth. The gravitational force of attraction provides required centripetal force i.e.,
GMm mv2 = 2 R R But where
...(1) Fig. 23
v = Rω = R2πh M = mass of sun, m = mass of earth, R = Sun to earth distance, v = tangential velocity,
T = time taken by earth to complete one revolution. Since
v =
2πR T
Putting the value of v in Eq. (1), we get
FG H
GMm m 2πR = R2 R T T2 =
or
IJ K
2
4 π2R3 GM
...(2)
Now, if T′ is time of one revolution of earth when distance is R ′ = we have,
4 π2 T ′2 =
or
T′ So
2
=
T′ =
3
...(3)
GM
Dividing Eq. (3) by Eq. (2), we get
T ′2 = T2
FG R IJ H 3K
R , then by analogy, 3
4 π2
FG R IJ H 3K
GM
3
×
1 GM = 4 π2 R3 27
T2 but T = 1 year = 365 days 27 T 365 = = 70. 3 days Ans. 3 3 3 × 1. 732
Q. 52. A large mass M and a small mass m hang at the two ends of a string that passes over a smooth tube as shown in the figure 24. The mass m moves around a circular path which
Dynamics of Circular Motion and the Gravitational Field
133
lies in a horizontal plane. The length of the string from the mass m to the top of the tube is l and θ is the angle which the length makes with the vertical, what should be the frequency of rotation of the mass m so that the mass M remains stationary. Ans. Various forces on masses m and M are as shown in figure 24. For equilibrium of mass m, we have T cos θ = mg T sin θ =
and
...(1)
mv2 = mω2r r
...(2)
For equilibrium of mass M, T = Mg From the diagram
r = l sin θ
putting the value of T and r in Eqn. (2) l T co s θ
T
in Ts
θ
mg
m v2 r
M Mg
Fig. 24
Mg sin θ = mω2 l sinθ or and
ω =
Mg ml
frequency: n =
ω 2π
n =
1 2π
Mg Ans. ml
Q. 53. A boy is sitting on the horizontal platform of a joy wheel at a distance of 7m from the centre. Wheel begins to rotate and when the angular speed exceeds 10 rev/min, the boy just slips. What is the coefficient of friction between boy and platform. (g = 9.8 m/s2). Ans. The boy slips when centrifugal force exceeds the limiting force of friction. If ω is the maximum angular velocity of wheel when boy just slips, µ mg = mrω2 i.e.,
µ =
rω2 g
134
Mechanics
2π × 10 π = rad/sec 60 3
given,
ω =
and
r = 7m
we get
µ =
FG π IJ H 3K
2
×
7 = 0. 78 9. 8
µ = 0.78 Ans. Q. 54. A 40 kg. mass hanging at the end of a rope of length l, oscillates in a vertical plane with an angular amplitude θo. What is the tension in the rope when it makes an angle θ with the vertical. If the breaking strength of the rope is 80 kg, what is the maximum amplitude with which the mass can oscillate without breaking the rope? Ans. The figure shows the position of the oscillating mass at angular amplitudes θ0 and θ. The velocity v at θ (after a desent of distance NM from extreme position) is given by v 2 = 2 gNM
S
NM = SM – SN = l cos θ – l cos θ0
But
θ θ0
v 2 = 2gl (cos θ − cos θ0 )
then
l
If T is tension in the rope at position Q. Then T = mg cosθ +
T
mv2 l
Putting the value of v2 T = mg cos θ +
m × 2 gl (cos θ − cos θ0 ) l
Putting
Q
M
θ
P
T = 3mg cos θ − 2mg cos θ0
or
R
N
m = 40 kg
mg
m g co s θ
Fig. 25
T = 3 × 40 × g cos θ – 2 × 40 × g × cos θ0 T = 40 g 3 cos θ − 2 cos θ0
or
The maximum tension in the rope occurs when θ = 0 (mean position), i.e., Tmax = 40 g [3 − 2 cos θ0 ]
i.e., or
But given
Tmax = 80 kg
So,
80 g = 40 g [3 − 2 cos θ0 ] 2 cos θ0 = 1 cos θ0 =
1 i.e., 2
θ0 = 60°
Ans.
Q. 55. A body slides down from the top of a hemisphere of radius r. The surfaces of block and hemisphere are frictionless. Show that the height at which body loses contact with the surface of the sphere is
2 r. 3
Dynamics of Circular Motion and the Gravitational Field
135
Ans. Let height at which body loses contact be ‘h’, then OR = h and ∠QOR = θ If at position Q, velocity is v, then taking limiting case of equilibrium for circular motion mg cos θ =
P
mv2 r
...(1)
OR h = cos θ = ...(2) OQ r
where
Q
R h
r θ
θ
mg m g co s θ
O
But the velocity ‘v’ at Q, in vertical desent PR = (r – h), is given by
Fig. 26
1 mv2 = mg (r – h) 2 v 2 = 2g (r – h)
or
putting the values of cos θ from Eq. (2) and value of contact upto point Q
mg which gives or
...(3) v2
from Eq. (3) into Eq. (1), we get for
m × 2 g ( r − h) h = r r 3h = 2r h =
2 r Ans. 3
Q. 56. The radius of curvature of a railway track at a place is 800 meter and the distance between the rails is 1.5 meters. What should be the elevation of the outer rail above the inner one for a safe speed of 20 km per hour? Ans. Suppose the outer rail is elevated by an angle θ with respect to inner rail i.e., h is height to which outer rail is raised vertically, then tan θ = sin θ =
h h or tan θ = (as θ is small) d 1.5
The banking angle is related to safe speed by tan θ =
100 v2 where v = 20 km/hr = m/sec 18 rg
putting values, we get
100 × 100 h = 18 × 18 × 800 × 9. 8 1. 5 or
Fig. 27
h = 0.59 cm Ans.
Q. 57. A heavy particle at the end of a tight string (length 20 cm), the other end of which is fixed is allowed to fall from a horizontal position of the string. When the string is vertical, it encounters an obstruction at its middle point and the particle continues its motion in a circle of 10 cm radius. Find the height which the particle will attain before the string slackens.
136
Mechanics
Ans. Let OP be the string in horizontal position with end O fixed, and the particle falls from position P. Then the velocity at position Q when string is vertical is given by v 2 = u2 + 2gh where u = 0 v 2 = 2 × 980 × 20 = 39200
or
...(1)
Let K be mid point of OQ, QM–circular path which particle will describe with K as centre, M–the position where string slackens. r = KQ = KM At M, the component of mg along MK provides the centripetal force, i.e., mg cos θ =
mv12 ; v1 = tangential velocity at M r
Fig. 28
v12 i.e., v1 = rg cosθ r KL QL − QK h − r = = cos θ = KM QK r
g cos θ =
or again we get
v1 =
rg
( h − r) r
Considering the journey of particle from Q to M, we have
v12 = v2 – 2gh putting the value of v1 get rg
FG h − r IJ H r K
= v2 – 2gh
g(h – r) = v2 – 2gh
or which gives
3gh = v2 + rg
v2 + rg 3g 39200 + 980 × 10 h = 3 × 980 h = 16.6 cm Ans.
or
h =
or
Q. 58. A sphere of mass 200 gm is attached to an inextensible string of length 130 cm whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 50 cms. (1) Calculate the time period of one revolution. (2) What is the tension in the string. Ans. Given
OP = l = 130 cm, PC = r = 50 cm r = l sin θ
The vertical component of tension balances the
Dynamics of Circular Motion and the Gravitational Field
137
weight and horizontal component provides the centripetal force; Thus mg = T cos θ and
mv2 = T sin θ r
Dividing we get v2 rg
tan θ = or
v =
again
...(1)
rg tanθ
CP = tan θ = CO
r OP2 − CP2
Fig. 29
=
50 (130)2 − (50)2
=
5 12
Putting values of r, g, tan θ in Eq. (1) we get
50 × 980 ×
v =
5 350 = cm/sec. 12 6
The time period of one revolution, T =
2π 2π r 2πr = . = ω ω r v
T =
2π × 50 = 2. 2 sec. Ans. 350 / 6
(2) Using the condition for vertical equilibrium, T = where
then
cos θ =
T =
mg cos θ
12 12 = 25 + 144 13 200 × 980 = 2.12 × 105 dynes Ans. 12 / 13
Q. 59. A nail is locked at a certain distance vertically below the point of suspension of a simple pendulum. The pendulum bob is released from a position where the string makes an angle of 60° with the vertical. Calculate the distance of nail from the point of suspension such that the bob will just perform revolution with the nail as centre. Assume the length of pendulum to be one meter. Ans. Let SP be the initial position with S as point of suspension then,
PS 1 = 2 2 RQ = SQ – SR = SP – SR SR = PS cos 60° =
and
= 1−
1 1 = m 2 2
138
Mechanics
Now, bob comes from P to Q descending vertically by distance RQ.
S
Using v2 = u2 + 2 gd for this journey with u = 0, g = 9.8
6 0°
m/sec2, S = we get
1 m 2
1m
v 2 = 0 + 2 × 9.8 ×
1 = 9. 8 ...(1) 2
We know that velocity of the particle at bottom of a vertical circle, so that the particle may successfully describe vertical circle, is given by v = or
5rg ; r is the radius of circle.
v 2 = 5 × r × 9.8 Equating Eq. (1) and (2), we get 9.8 = 5 × r × 9.8
or
r =
i.e., ∴
RN = = SN = = SN =
R
P r
N Q
Fig. 30
...(2)
1 = 0. 2 meter = NQ 5 RQ – NQ 0.5 – 0.2 = 0.3 m SR + RN 0.5 + 0.3 = 0.8 m 0.8 m Ans.
Q. 60. A particle of mass 0.2 kg is moving inside a smooth vertical circle of radius r = 50 cm. If it is projected horizontally with velocity v = 4 m/sec from its lowest position, find the angle θ at which it will loose contact with the circle. Ans. Situation is shown in figure 31. Here m = 0.2 kg, r = 50 cm = 0.5 m, g = 9.8 m/sec2 Let h = height from lowest position where particle loses contact with vertical circle. θ = angle with vertical and v = velocity at Q then, component of mg along QO = mg cos θ.
mv2 = T + mg cos θ r T = 0 at Q.
centrifugal force = But So, where
mg cosθ = cosθ =
mv2 r
...(1)
RO h − r = OQ r
Again, velocity at Q is, from v2 = u2 – 2gs, with u = 4 m/sec v 2 = 16 – 2 × 9.8 × h putting value of cos θ in (1), we get
...(2)
Dynamics of Circular Motion and the Gravitational Field
mg
FG h − r IJ H r K
=
or
g (h – r) = Substituting the value of v2 from g (h – r) = or 9.8 (h – 0.5) = or 9.8 h – 4.9 = which gives h =
mv2 r v2 Eq. (2) we get 16 – 2 × 9.8 × h 16 – 2 × 9.8 × h 16 – 19.6 h 0.71 m
139
θ O
θ = cos–1 (0.42)
or
T m g co s θ mg
h
r
h − r 0.71 − 0. 50 = = 0. 42 cos θ = r 0. 50
then
Q
R
v = 4m /s
Fig. 31
Ans.
Q. 61. A smooth table is placed horizontally and an ideal spring of constant k = 1000 nt/m and unextended length of 0.5 m has one end fixed to its centre. The other end is attached to a mass of 5 kg which is moving in a circle with constant speed, 10 m/sec. Find (1) The tension in the spring. (2) Extension of spring beyond its normal length. Ans. Let l = unextended length of spring = 0.5 m x = extended length of spring then, extension in the spring = (x – l) At position P of ball, force on the spring towards centre O, is F = k × extension = centripetal force =
But and Spring force
mv2 x
...(1)
mv2 5 × 102 500 = = ...(2) x x x F = k (x – 0.5) = 1000 × (x – 0.5)
then Eq. (2) gives 500 = 1000 x – 1000 × 0.5 x
or
500 = 1000 x – 500 x
or
500 = 1000 x2 – 500 x
or
2x2 – x – 1 = 0 Fig. 32
140
Mechanics
1 x = 1, x = − ; but negative elongation is not possible, 2 x = 1, Hence tension in string
which gives So, So, and
mv2 5 × 10 × 10 = = 500 Newton 1 x Extension (x – l) = (1 – 0.5) = 0.5 meter F = T=
F = 500 Newton Extension = 0.5 meter
Ans.
Q. 62. A string with a ball is held horizontally as shown in the figure 33. A nail located at a distance ‘d’ vertically below the point of suspension. Show that ‘d’ must be at least 0.61 l. (l being the length of the string) if the ball is to swing completely around a circle centered on the nail. Ans. Ball falls under gravity from P reaches M (vertical descent = OM = l) then it has to swing completely around the circle with centre at C (nail) of diameter MN = h. From v2 = u2 + 2gh, velocity of ball at M is given by v 2 = 0 + 2gl ...(1) Let velocity of ball at N be v1. Considering the ascent MN and using v 2 = u2 + 2hg u = v, v = v1, h = MN, g = –g,
where
v 1 2 = v2 – 2gh or v12 = 2gl – 2gh
we have
Fig. 33
...(2)
For successful completion of circle, condition at N is, Centrifugal force = weight
gh mv12 = mg or v12 = 2 h/2 Equating Eqs. (2) and (3) we get gh or h = 4 l − 4 h 2gl – 2gh = 2 4l = 0. 8 l So MN = h = 5 then O C = d = OM – CM = l – 0.4 l i.e.,
∴
d = 0.6 l
...(3)
Ans.
UNSOLVED PROBLEMS 1.
Show that a centripetal acceleration acts on a particle moving on a circular path.
2.
Explain the statement “Centrifugal force is a pseudo force”.
3.
Explain the principle of cream separating machine.
4.
If a car is suddenly turned to left, a passenger of car strikes with right wall, explain.
Dynamics of Circular Motion and the Gravitational Field
141
5. 6.
A passenger in spaceship uses spring clock-give reason. The speed of a body is constant. Can it have a path other than a straight line or circle.
7.
Why are tracks banked at sharp turns. An automobile overturns when going too fast around a curve. Which wheel leaves the ground first. The handle of the carpenter’s screw driver is much thicker then the handle of a watch makers screw-driver. Explain the reason. A simple pendulum with a bob of mass m swings with an angular amplitude of 40°. When its angular displacement is 20°, then the tension in the string is greater than mg cos 20°. Is the statement correct. As astronaut in spaceship feels weightlessness. Explain why ? If a satellite orbits earth at an altitude of 100 km. Find its time period of revolution in circular orbit. A particle of mass m rotates in a circle of radius a, with a uniform angular speed ω. It is observed from a frame rotating about the z-axis with a uniform angular speed ω0. Find the centrifugal force on the particle. A smooth circular tube is held firmly in a vertical plane. A particle which can slide inside the tube is slightly displaced from rest at its highest position in the tube. Find the pressure between the tube and the particle in terms of its mass ‘m’ and the angular displacement θ from its highest position. Obtain expression for the orbiting velocity of artificial satellite. A satellite is moving in a circular orbit round the earth at a distance of 5R from its centre. Calculate its orbital speed and time period. Find an expression for the total energy of a satellite of mass ‘m’ moving round the earth in circular orbit of radius r. Write short note on (a) Satellite launching (b) Escape velocity (c) Geo-stationary satellite. In an orbit of radius 1.5 × 1011 meter, earth completes one revolution in 365 days around the sun. Calculate mass of sun (G = 6.67 × 10–10 nt-m2/kg2). A motor cyclist with the motor cycle has a mass of 250 kg and travels round a curve in a road of 36 metre radius. Snow and ice on the road reduce the coefficient of friction between the tyres and road to 0.2. If the curved road is banked to 15° in motor cyclist’s favour; Find (a) maximum velocity attainable without slipping (b) the angle, the rider must make with the road surface at this velocity assuming that rider and motorcycle remain in one plane.
8. 9.
10. 11.
12.
13. 14. 15. 16. 17. 18.
19.
20.
21.
A train is travelling at 64 km/h and the diameter of one of the wheels of the engine is 1.5 m. Find the velocities of the two points on this wheel. Which are at a height of 1.2 m above the ground. A fighter plane flying in the sky dives with a speed of 360 km/hr in a vertical circle of radius 200 m, weight of the pilot sitting in it is 75 kg. Find the value of force with which pilot presses his seat when aeroplane is (a) at highest position (b) at lowest position. The moon revolves around the earth in a circle of radius 3.8 × 106 m and requires 27.3 days to make a complete revolution. What is the acceleration of the moon towards the earth.
142
Mechanics
4 WORK, ENERGY AND MOMENTUM 4.1 WORK Whenever a force acting on a body produces a change in the position of the body, work is said to be done by the force. If there is no change in the position of the body, work is not done. We may exert a large force on a wall, but if the wall remains intact in its present position then we have not done any work. If a coolie having a heavy box on his head is standing at a fixed place, he is not doing any work. Work is said to be done only when there is a displacement in the direction of the force. Work is measured by the product of the applied force and the displacement of the body in the direction of the force, that is Work = force × displacement in the direction of the force. If a force F acting on a body produces a displacement ∆s in the body in the direction of the force (Fig. 1a), then the work done by the force is given by W = F × ∆s If the force F is making an angle θ with the direction of displacement of the body (Fig. 1b), then the work done is W = F cos θ × ∆s, because F cos θ is the component of F in the direction of displacement. F
θ
F
∆s
F co s θ
∆s
(a )
(b )
Fig. 1
If θ = 90°, then cos θ = 0 and so W = 0. This means that if the displacement is perpendicular to the force, no work is done. When a satellite revolves around the earth, the direction of the force applied by the earth is always perpendicular to the direction of motion of the satellite. Hence no work is done on the satellite by the centripetal force. 142
Work, Energy and Momentum
143
Force and displacement both are vector quantities but work is a scalar quantity. The unit of work is ‘joule’. If a force of 1 newton produces a displacement of 1 meter in the direction of the force, the work done is called 1 joule: 1 joule = 1 newton × 1 meter
4.2 POWER The rate of doing work by an agent or a machine is called power: Power =
Work time
If W is the work done by an agent in t second, then his power P is given by P =
W t
Since the unit of work is joule, the unit of power will be ‘joule/second’. This is called ‘watt’. If 1 joule of work is done in 1 second, the power is 1 watt. 1 watt = 1 joule/second Another unit of power is ‘horse-power’ 1 horse-power = 746 watts The power of a normal person is from 0.05 to 0.1 horse power. If an agent exerts a force F in the direction of motion, then P =
W Fr = = Fv t t
where v = r/t
4.3 WORK IN STRETCHING A SPRING When a spring is stretched slowly, the stretching force increases steadily as the spring elongates, i.e., the force is ‘variable’. Let one end of the spring be attached to a wall its length being along the x-axis. Let the origin x = 0, coincide with the free end of the spring in its normal, unstretched state. Let the spring be stretched through a distance x by applying a force Fapp at the free end. The spring, on account of its elasticity, will exert a restoring force F on the stretching agent given by (within elastic limit) F = –kx;
(Hooke’s law)
Where k is the force-constant or stiffness of the spring. The minus sign indicates that the restoring force is always opposite to the displacement x. Since the restoring force F is equal and opposite to the applied force Fapp; the latter is given by
x=0
x
Fapp = –F = kx The work done by the (varying) applied force Fapp. in the displacement from x = 0 to x = x is
z
x= x
W =
x=0
→
→
F app. d x =
F app
z
x= x
Fapp dx
x=0
Fig. 2
144
Mechanics →
→
∴ Angle between F app and d x is zero.
z x
=
0
Lx O kx dx = k M P N2Q 2
x
= 0
1 2 kx 2
Similarly if the spring is stretched so that its free end moves from x1 to x2, the work done is
z
x = x2
W =
Fapp dx = k
x = x1
=
z
x2
x1
Lx O x dx = k M P N2Q 2
e
1 2 1 2 1 kx2 − kx1 = k x22 − x12 2 2 2
x2
x1
j
4.4 ENERGY The capacity of doing work is called energy. Energy has various forms such as mechanical energy, heat energy, light energy, magnetic energy, sound energy, chemical energy etc. Mechanical energy has two forms: Kinetic energy and potential energy.
4.5 KINETIC ENERGY The kinetic energy of a moving body is measured by the amount of work which has been done in bringing the body from the rest position to its present position, or which the body can do in going from its present position to the rest position. Let a body of mass m be in the rest position. When we apply a constant force F on the body, it starts moving under an acceleration. If a be the acceleration, then by Newton’s second law, we have a =
F m
suppose the body acquires a velocity v in moving a distance S. According to equation v2 = u2 + 2as (u = 0, since the body was initially at rest) we have v 2 = 2as = 2 ×
F ×s m
F × S = ½ mv2 But F × S is the work which the force F has done on the body in moving it a distance ‘S’. It is due to this work that the body has itself acquired the capacity of doing work. This is the measure of the kinetic energy of the body. Hence if we represent kinetic energy of a body by K, then K = F × S = ½ mv2 Thus, the kinetic energy of a moving body is equal to half the product of the mass (m) of the body and the square of its speed (v2). In this formula, v occurs in the second power and so the speed has a larger effect, compared to mass, on the kinetic energy. It is because of this reason that the bullet fired from a gun injures seriously inspite of its very small mass.
Work, Energy and Momentum
145
Now, suppose a body is initially moving with a uniform speed u. When a force is applied on it then its speed increases from u to v in a distance S. Now, we have v 2 = u2 + 2as v2 – u2 = 2 × (F/m) × s F × S = ½ mv2 – ½ mu2 But F × S is the work done W on the body by the force. Hence W =
FG W = 1 mv H 2
1 1 mv2 − mu2 2 2
IJ K
1 mu2 is the increase in the kinetic energy of the 2 body. Thus when a force acts upon a moving body, then the kinetic energy of the body increases, and the increase is equal to the work done. According to definition,
2
−
4.6 POTENTIAL ENERGY Bodies can do work also by virtue of their ‘position’ or ‘state of strain’. The energy in a body due to its position or state of strain is called the ‘potential energy’ of the body. For example, the water at the top of a water-fall can rotate a turbine when falling on it. The water has this capability by virtue of its position (at a height). Similarly, a wound clock-spring keeps the clock running by virtue of its state of strain. Thus water and wound spring both have potential energy, the former has ‘gravitational’ potential energy and the later has ‘elastic’ potential energy. A system of electric charges also has potential energy which is called ‘electrostatic’ potential energy. If two opposite charges are taken further away from each other, the work done against their mutual attraction is stored in the form of potential energy, so that the potential energy of the system increases. In case of similar charges, the potential energy increases when they are brought closer. The potential energy of a body is measured by the amount of work which has been done in bringing the body from its zero-position to the present position or which that body can do in going from its present position to the zero-position. Therefore, if a body is taken, under the action of a force, from one position to the other, then the work done is stored in the body in the form of its potential energy (provided there is no loss of work against friction, etc.).
4.7 GRAVITATIONAL POTENTIAL ENERGY A mass held stationary above the ground has energy, because, when released, it can raise another object attached to it by a rope passing over a pulley, for example. A coiled spring also has energy, which is released gradually as the spring uncoils. The energy of the weight or spring is called potential energy, because it arises from the position or arrangement of the body and not from its motion. In the case of the weight, the energy given to it is equal to the work done by the person or machine which raises it steadily to that position against the force of attraction of the earth. So this is gravitational potential energy. In the case of the spring, the energy is equal to the work done in displacing the molecules from their normal equilibrium positions against the forces of attraction of the surrounding molecules. So this is molecular potential energy.
146
Mechanics
Suppose a body of mass m is raised to a height h from the earth’s surface. In this process, work is done against the force of gravity (mg). This work is stored in the body in the form of gravitational potential energy U. Thus U = work done against force of gravity = weight of the body × height = mg × h = mgh
4.8 WORK-ENERGY THEOREM Let us consider a body of mass m acted upon by a resultant accelerating force F along the x-axis. Suppose as the body moves from a position x1 to a position x2 along the x-axis, its velocity increases from v1 to v2. The work done by the force in the displacement is
z
x = x2
W =
F dx
x = x1
By Newton’s second law, we have F = ma = m 3
v = dx/dt
z
x = x2
∴
dv dv dx dv =m . = mv dt dx dt dx
W = m
x = x1
LM v OP N2Q 2
= m
dv v dx = m dx
v2
= v1
z
v = v2
v dv
v = v1
1 1 mv22 − mv12 2 2
The quantity ½ mv2 is defined as the kinetic energy K of the body. Thus the above equation may be written as W = K2 – K1 Where K2 and K1 are the final and initial kinetic energies of the body. Thus, if ∆K represents the change in kinetic energy, ∆K = K2 – K1, then we have W = ∆K Thus we conclude that “whenever a body is acted upon by a number of forces such that the resultant force is not zero, then the work done by the resultant force (whether constant or variable) is equal to the change in the kinetic energy of the body. This is known as the work-energy theorem. If the kinetic energy of the body decreases, the work done on it by the resultant force is negative, or the work is done by the body against the resultant force. Therefore, a moving body is said to have a store of (kinetic) energy in it, which it loses in doing work. Hence the kinetic energy of a moving body is defined as the work it can do before coming to rest. The units of kinetic energy and of work are the same. Kinetic energy, like-work, is a scalar quantity.
Work, Energy and Momentum
147
When several forces act upon the body, the work done by the resultant force is the algebraic sum of the works done by the individual forces i.e., W = W1 + W2 + ----------------Hence the work-energy theorem may also be written as W1 + W2 + ------------- = ∆K
4.9 SIGNIFICANCE OF THE WORK-ENERGY THEOREM The work-energy theorem is useful for solving problems in which the work done by the resultant force is easily computed and in which we are interested in finding the particle’s speed at certain positions of greater significance, perhaps is the fact that the work-energy theorem is the starting point for a sweeping generalization in physics. It has been emphasized that the work-energy theorem is valid when W is interpreted as the work done by the resultant force acting on the particle. However, it is helpful in any problems to compute separately the work done by certain types of force and give special names to the work done by each type. This leads to the concepts of different types of energy and the principle of the conservation of energy.
4.10 CONSERVATIVE FORCE: FIRST DEFINITION A force acting on a particle is conservative if the particle, after going through a complete round trip, returns to its initial position with the same kinetic energy as it had initially. v
v=0
A (a )
(b )
Fig. 3
When we throw a ball upward against gravity, the ball reaches a certain height coming momentaily to rest so that its kinetic energy becomes zero. Then it returns to our hand under gravity with the same kinetic energy with which it was thrown (provided the airresistance is assumed zero). Thus the force of gravity is conservative. Similarly, when a block is moved from a position A on a horizontal plane with a velocity v so as to compress on a spring (Figure 3a). It is first brought to rest by the elastic (restoring) force of the spring and loses all its kinetic energy (Figure b). Then the compressed spring re-expands and the block moves back under the elastic force gaining kinetic energy. As the block returns to its initial position A, it has the same velocity v, and hence the same F N kinetic energy, as it had before (provided the horizontal plane F θ is frictionless and the spring is ideal). Thus the elastic force A exerted by an ideal spring is conservative. The electrostatic force is also conservative. A force is termed to be conservative if the work done by it on a particle in moving it from one point to another in space, depends only on the location or position of the particle and not on the path followed by the particle.
→ F
C
M
Fig. 4
148
Mechanics →
Imagine a particle moving from point M to N under action of force F . There may be →
three alternative paths, Path MAN, Path MBN, Path MCN. If the force F is conservative, then
z
N
WMN =
→
→
F. d r =
M
z
N
→
z
N
→
F. d r =
M
→
→
F. d r
M
Path MAN Path MBN
Path MCN
Work done by a conservative force around a closed path →
Consider the closed path MANBM. under action of a conservative force F , a particle moves from M to N along path MAN and returns back to M, along the path NBM. →
Now, work done by the force F in moving the particle from M to N along path 1, is
z
N
WMN =
→
→
F. d r = TN − TM
M
Where TN = K.E. of particle at N and TM = K.E. of particle at M. Similarly the work →
done in carrying the particle from N to M, by the force F is given by
z
M
WNM =
→
→
F. d r = TM − TN
N
→
So, total work done by the conservative force F , along closed path MANBM, is
z
→
→
F. d r =
z
→
→
F. d r +
Path MANBM Path 1(MAN)
z
→
→
F. d r
Path 2(MBN)
WMANBM = TN – TM + TM – TN = 0 or
z
→
→
F. d r = 0
Fig. 5
Thus, work done by a conservative force around a closed path is always zero i.e., the change in kinetic energy of the particle in motion due to a conservative force, over a closed path is zero.
Examples of conservative forces Position dependent forces depend on the instantaneous position of the particle or body (not on its velocity). Central forces are position dependent and in addition, are directed towards a fixed center. Position dependent forces are (a) Gravitational force (b) Electrostatic force (c) Elastic force. Most of the position dependent forces are conservative in nature.
4.11 A CENTRAL FORCE IS CONSERVATIVE A force is termed central force if it acts on a particle in such a way that it is directed towards or away from a fixed point and its magnitude depends only on the distance of particle from the fixed point.
Work, Energy and Momentum
149 →
It may be observed, that central forces are conservative. Consider a central force F , by definition: →
F = FsS →
Where Fs is function of position ‘s’ only and S is unit vector along S . The above force is directed away from the fixed point and acts on a particle. Let the particle move from M to N under action of this force. Now, total work done by the central force in this movement is,
z
N
WMN =
→
→
F. d r
M →
Putting the value of force F ,
z z
N
WMN =
→
FsS . d r =
M
z
N
Fs . dr cos θ
M
N
or
WMN =
Fs dS
Fig. 6
M
As Fs is function of s only, so its integral (say α) is also a function of S, hence
z
N
WMN =
FsdS = α
N M
= α N − αM
M
Thus, the work done by the central force depends only on the position of points M and N (not on path followed). This suggests that central force is a conservative force.
Properties of Work done by conservative force (i)
It is independent of path.
(ii)
It is equal to difference between final and initial values of energy function.
(iii)
It is completely recoverable.
4.12 NON-CONSERVATIVE FORCE A force is non-conservative if the work done by it on the particle moving between two points depends on the path followed by the particle. Thus, if a particle moves from M to N →
under influence of a non-conservative force F , then work done along three possible paths by the force is not same (i.e., it is path dependent). Thus
z
N
WMN =
→
→
M
Path 1
z
N
F. d r ≠
→
→
M
Path 2
z
N
F. d r ≠
→
→
F. d r
M
Path 3
150
Mechanics
Thus, work done by a non-conservative force around a closed path is not zero i.e.,
z
N
WMN =
→
→
F .d r ≠ 0
M
Again, if a conservative and a non-conservative both, forces act upon the particle and WC be the work done by conservative force and WM-C be the work done by non-conservative force, then
Fig. 7
WC + WN–C = Change in K.E. of Particle = ∆T Example of non-conservative force: Frictional forces, viscous forces etc are examples of path dependent forces.
Potential Energy Potential energy of a body is defined as the energy stored in body due to its position, configuration or state of strain. Consider a particle lying in a conservative force field due to force FC. The force FC that acts on the particle may be balanced by applications of suitable applied force Fapp. Now the particle may be moved extremely slowly (so that no K.E. develop) by applied force against the conservative force FC. The work done in moving the particle will appear as potential energy of the particle. Thus, potential energy is measured by the amount of work that it can do when it moves from referred position to some standard position. Potential energy is generally denoted by U.
Potential Energy Difference The difference in potential energy between two positions of a particle is defined as the work done by applied force on the particle in moving it from the first position to second position. Thus, if r2 and r1 are the final and initial position, then by definition
z
r2
U(r2) – U(r1) =
→
→
z
r2
→
→
F app. d r = − F c . d r
r1
(as Fapp = –Fc)
r1
z
r2
→
→
U(r2) – U(r1) = − F c . d r
dr
r1
If first position is at infinity (i.e., r1 = ∞) having potential energy zero there (i.e., U(r1) = 0), then potential energy at position r (i.e., r2 = r) is given by
z z
r2
0 – U(r) =
→
→
Fc . d r
∞
r2
or
– U(r) =
∞
Fapp
FC →
→
Fc . d r
Fig. 8
Work, Energy and Momentum
151
z r
or
U(r) =
→
→
F app . d r
∞
Thus potential energy of a particle at a point r may be defined as the work done by the externally applied force in moving the particle from infinity to that point. The position at which the conservative force acting on the particle is zero, is the reference position with respect to potential energy. In case of gravitational and electrostatic forces the reference position is infinity. In case of springs the normal unstretched length is considered as reference position. In case of earth’s gravitational force field, earth’s surface is the reference position.
4.13 RELATION BETWEEN CONSERVATIVE FORCE AND POTENTIAL ENERGY By definition of potential energy
z r
→
→
U(r) = – F . d r ∞
In rectangular Cartesian system one may write F = i Fx + j Fy + k Fz →
and
→ d r = i dx + j dy + k dz Then the potential energy will be given by →
U( r ) = U (x, y, z)
z z z r
i.e.,
) U(r) = − (i Fx + j Fy + k Fz ).(i dx + j dy + kdz ∞
or
→
y
x
z z
U( r ) = − Fx dx − Fy dy − Fz dz ∞
Differentiating the above equation partially
∞
∞
∂U ∂U ∂U = −Fx , = − Fy , = − Fz ∂x ∂y ∂z →
In view of these relations, the force F can be expressed as →
F = − i
LM N
OP Q
∂ + j ∂ + k ∂ U F = − i ∂x ∂y ∂z →
or
∂U ∂U ∂U − j −k ∂x ∂y ∂z
152
Mechanics →
→
F = −∇U
or
→
∇ = +i
where
∂ ∂ ∂ + j + k ∂x ∂y ∂z
→
Vector operator ∇ is called ‘del’ or ‘nebla’. Thus, a conservative force is the negative gradient of potential energy U. Only for one dimension F = −
dU dx
or
z
U = − F dx
4.14 THE CURL OF A CONSERVATIVE FORCE IS ZERO Now let us see what is the value of curl F for a conservative force: A conservative force may be expressed as negative gradient of potential energy i.e., →
→
F = −∇U →
The curl of F is, →
→
→
→
→
curl F = ∇ × F = ∇ × ( − ∇ U)
i ∂ = − ∂x ∂U ∂x
k ∂ ∂z ∂U ∂z
j ∂ ∂y ∂U ∂y
LM F MN GH
I F JK GH
I F JK GH
∂2U ∂2U ∂2U ∂2U ∂2U ∂2U = − i − − j − + k − ∂y ∂z ∂z∂y ∂x∂z ∂z∂x ∂x∂y ∂y∂x Since, U is perfect differential, so ∂2U ∂2U = and so on ∂x ∂y ∂y ∂x
∴ i.e.,
→ Curl F = − i ( 0) − j ( 0) + k ( 0) = 0 →
→
∇×F = 0
Thus, the curl of a conservative force is zero.
I OP JK PQ
Work, Energy and Momentum
153
The Total Mechanical Energy is conserved in a conservative force field →
Consider a particle to be displaced by a conservative force F from a position M to N. By work-energy theorem, the work done by force is equal to the increase in K.E. of particle, i.e.,
z
N
WMN =
→
→
F .d r = TN − TM
M
one can express the same work done by decrease in potential energy of the particle, i.e.,
z
N
→
→
WMN = − F . d r = UM − UN M
Equating above,
TN – TM = UM – UN
or
TN + UN = TM + UM which suggests that the sum of kinetic and potential energy of a particle under action of conservative force remains constant, i.e., T + U = Total Mechanical Energy (Constant in conservative force field) Sometimes it is convenient to call the quantity E = T + U as the energy function. This energy function is invariant with respect to change in time.
Motion of a Body near the Surface of the Earth Let us consider a body of mass m be situated at a height h in the rest position above the earth’s surface. Suppose that its potential energy is zero at the earth’s surface. Let the x direction be normal to the surface of the earth and directed upwards. If the body starts to fall down and at any instant its height above the earth’s surface is x, then work done by the gravitational force – mg on the body is
z x
W =
( − mg) dx = − mg ( x − h) = mg ( h − x)
h
Where the direction of force is opposite to x. This amount of work done on the body will increase its kinetic energy W = ½ mv2 ∴
½
mv2
∴ initial velocity = 0
= mg (h – x)
mgh = ½ mv2 + mgx
or
...(i)
But a body situated at a height x has the capacity to work mg × x, hence the potential energy of the body at this height is mgx. Thus, initially, the body has no kinetic energy, but only potential energy mgh. ∴
Initial total energy = initial K.E. + initial P.E. = 0 + mgh = mgh
At any height x, total energy = ½ mv2 + mgx = mgh (from Eq. (i))
154
Mechanics
At earth surface (x = 0), total energy = mgh + 0 = mgh Hence, the total energy of a freely falling body is the same initially, finally and in any middle position. In other words, the sum of potential and kinetic energies of a freely falling body remains constant throughout the motion.
4.15 LINEAR RESTORING FORCE When a force acting on a particle is directly proportional to the displacement of the particle from a certain fixed point (called mean position) and is directed oppositely to the displacement, it is called linear restoring force. Thus →
→
Fα − r →
F = − kr r
or →
Where r = position vector of particle at any instant = r r →
r = r r = i x + jy + kz
Now or
→
→
F = −k r Linear restoring force is a conservative force: For a conservative force →
→
∇×F = 0 →
→
) + kz F = − k r = − k ( i x + jy
and →
→
→
→
∇ × F = ∇ × (− k r )
so
=
i ∂ ∂x − kx
j ∂ ∂y − ky
k i ∂ ∂ = −k ∂z ∂x − kz x
j ∂ ∂y y
k ∂ ∂z z
= 0
Thus, linear restoring force is a conservative force.
Massless Horizontal Spring – One End Fixed For small displacement, an ideal compressed or stretched spring produces a linear restoring force (due to elastic properties of the spring in accordance to Hook’s law). Let there be a massless spring whose one end is rigidly attached to a wall and other end be tied to a block which could slide on a horizontal table having no friction.
Normal Position No force is produced in the spring. Let us regard the origin at the normal position of block.
Work, Energy and Momentum
155
Stretched Position: If an external force stretches the spring slowly, then work will be done on the system against the restoring force which constitutes the potential energy of the system. The linear force produced in the spring is in the direction opposite to displacement; i.e., →
F = −kxi where k = force constant of spring (or spring factor) and x = displacement. Now, potential energy, by definition, is
z x
→
→
z x
→
z x
U = − F . d r = − F . dx cos θ = − ( − kx ). dx
z
0
0
x
or
U =
kxdx =
0
∴
0
1 2 kx 2
U = ½ kx2
Which gives the potential energy stored in stretched spring suffering extension x. If ‘a’ is maximum stretching (or elongation) of spring Umax =
1 2 ka 2
...(1)
Motion after release: If body be stretched up to maximum stretching and then released it moves under linear restoring force towards origin. Initial total energy just at the instant of release = ½ ka2 Total energy at position x during return =
1 2 kx + ½ mv2 2
(where v = velocity in position x) Using energy conservation principle, ½ ka2 = ½ kx2 + ½ mv2 ½ mv2 = ½ k (a2 – x2)
or
k 2 ( a − x2 ) m the velocity will be maximum in mean position (x = 0), is given by
v0 = ±
or
∴
vmax = ± a
k m
...(2)
Compressed Position: When spring is compressed then the displacement is –x and, thus, the linear restoring force produced is →
F = − k.( − i x) = i kx
156
Mechanics
The spring linear force in compressed spring is in +x direction. The maximum energy (Umax) and maximum velocity (Vmax) gained in mean position will be given by Equation (1) and Equation (2) respectively. Equation of motion: Newton’s law
Restoring force at stretching x of the spring is F = –kx. By
F = m
d2 x dt2
So equation of motion is m
or
d2 x = –kx dt2
d2 x k + x = 0 2 m dt
Fig. 9
which is the differential equation of motion of spring having time period m k Vertical spring in uniform gravitational field with mass attached to spring: Figure 10 (a) shows the natural length of an ideal massless, vertical spring of force constant k. When a weight mg is gently attached to the free end of spring, an elongation x0 is produced. Figure (b) shows the force in the spring (acting upward) balances the weight mg. Thus
T = 2π
mg = kx 0 or
mg – kx0 = 0
If a further, displacement x is given to the mass, figure 10 (c) then net or effective restoring force = [mg – k(x0 + x)] = kx0 – kx0 – kx = –kx
Work, Energy and Momentum
157
From Newton’s law Force = m m
thus
or
d2 x dt2
d2 x = –kx dt2
d2 x k + x = 0 2 m dt
Fig. 10
which is the equation of motion. Note that gravity has no effect in the motion which is simple harmonic of time period given by T = 2π
m k
or frequency (n) =
1 2π
k m
Potential Energy Curve If the potential energy U of a particle is a function of position (i.e., changes from point to point), then a graph may be plotted to show the variation of potential energy with the position of the particle. Such a graph is known as potential energy curve. In case, if the particle is allowed to move in one dimension only, say along x-axis, the potential energy U will be the function of x-coordinate only, and then the force (conservative in nature) on the particle will be given by F = −
dU( x) dx
...(1)
158
Mechanics Y
D G B
C P U
K
E
Q U0 A
U O
X
x x0 x
Fig. 10(a)
Fig. 10(a) shows a possible potential energy curve for one dimensional motion. The slope (dU/dx) of this curve at any point gives the force (F = –dU/dx), acting on the particle placed at that point. The slope (dU/dx) is positive, whenever U increases with the increase in x (e.g., for the part AB and CD of the curve) and negative whenever U decreases with increase in x (e.g., for the part GA and BC). Therefore the force F is negative or directed to the left whenever the potential energy is increasing with x and positive or directed towards right whenever the potential energy is decreasing with x. This means that the force acting on the particle at any point tries to bring the particle to the region of lower potential energy. At points A, B, C and D, the potential energy U is minimum or maximum, the value of dU/dx is zero. Therefore, if a particle is placed at such a point with zero velocity, it will experience no force (F = –dU/dx = 0) and so it will remain at rest. These points of the curve are known as positions of equilibrium. B, and D are the points corresponding to maximum potential energy. A particle at rest at such a point will remain at rest. However, if the particle is displaced even the slighest distance from this point the force F(x) = –dU/dx will tend to push the particle farther away from the equilibrium position. Hence the points B and D are the positions of unstable equilibrium. A and C are the points corresponding to minimum potential energy. A particle at rest at such a point will remain at rest. If the particle is displaced slightly in either direction by giving a little energy the force will tend to take it back towards the equilibrium position. Hence the points A and C are the positions of stable equilibrium. In case, if U is constant in a region, then the slope dU/dx and hence the force acting on a particle placed at any point of that region is zero. In such a region, if a particle is displaced from one point to the other, it will remain there without experiencing any restoring force. So we call the region of constant potential energy as the region of neutral equilibrium. For example, a book placed on a table anywhere remains in equilibrium. Bounded region or potential well. Now let us discuss about the point A (or C) of stable equilibrium and the region near to it, which is of great interest in physics. A particle, placed at the point A at rest, can be displaced from this point by giving some energy so that the total
Work, Energy and Momentum
159
energy E of the particle becomes more than minimum value of the potential energy U0. The total energy E = K + U of the particle has been indicated.
NUMERICALS Q.1. An object of mass 5 kg falls from rest through a vertical distance of 20 m and reaches a velocity of 8 m/s. Calculate the work done by push of air exerted on the object. Solution. The forces acting on the body are: (i) the weight, acting downward, (ii) the air push, acting upward. The body is getting accelerated downward under influence of resultant force (mg – P), P being the push of air. Now work done by resultant force in 20 m fall = (mg – P) × 20 Joules Gain in KE in this fall = = as,
1 1 mv22 − mv12 2 2 1 1 m(8)2 − m × 0 = 32 m Joules 2 2
work done = change in K.E. ∴
or
(mg – P) × 20 = 32 m mg – P =
or
32m 32 × 5 ; or P = 5 × 9.8 – 20 20
P = 41 Newton So, work done by push of air (force) = –41 × 20 = –820 Joule (–ve sign refers to expense of work).
Q.2. A man pulls a block weighing 10 kg upto a distance of 10 meter on a horizontal rough surface (coefficient of kinetic function = 0.20) at a constant speed. Calculate the work done by the man if the pull exerted by him makes an angle of 30° with the horizontal. Solution. In the figure various forces are shown where
P = pull exerted by man, mg = weight of block, µN = frictional force.
Then; work done by the man
Fig. 11 →
→
W = P . r = Pr cos 30°
(r = displacement)
As block moves with uniform speed; net force is zero ∴ and also, which gives,
P cos 30° – µN = 0
...(i)
N + P sin 30° – mg = 0 N = mg – P sin 30°
...(ii)
160
Mechanics
Putting this value in (i) P cos 30° – µ mg + µP sin 30° = 0 or
P (cos 30° + µ sin 30°) = µ mg
or
⇒ So,
P =
0. 2 × 10 × 9. 8 µ mg = 1 cos 30° + µ sin 30° 3 / 2 + 0. 2 × 2
P =
19. 60 = 20. 3 Newton 0. 966
work done = Pr cos 30° = 20.3 × 10
3 /2
= 175.803 Joule Q.3. A block initially sliding on a rough horizontal surface with velocity of 10 m/sec stops after travelling a distance of 70 m. Find the coefficient of kinetic friction between surface and the block.
1 1 mv2 = m × 102 = 50 m Joules 2 2 Frictional force acting on block = µN = µ mg Solution.
Initial K.E. =
Work done against frictional force in moving 70 m = µ mg × 70 Joule This work done must be equal to the initial K.E. possessed by the block i.e.,
µ mg × 70 = 50 m
or
µ =
50 50 = = 0. 073 70 g 70 × 9. 8
Q.4. A body of mass 40 kg climbs a vertical rope 10 meters in length in 15 sec with constant velocity. Calculate: (i)
Work done by the body,
(ii)
Power output.
Solution. Velocity of the body =
10 = 0. 667 m/sec 15
The body has to exert force to support his own weight. So force exerted = mg = 40 × 9.8 = 392 Newton Thus,
Work = force × displacement = 392 × 10 = 3920 J
W 3920 = = 261. 33 Watt t 15 Q.5. If a net force of 10 Newton acts on a body, initially at rest, of mass 30 kg; then what is the
and
Power =
(i)
Work done by the force in the third second.
(ii)
Instantaneous power exerted by force at the end of third second.
Work, Energy and Momentum
161
Solution. (i) The acceleration of the body, a = i.e.,
a =
Force Mass
10 1 = m /sec2 30 3
Displacement in 1 sec. S = ut + = 0+
1 2 at 2
1 1 1 . (1)2 = meter 2 3 6
Work done at the end of 1 sec. W = F × displacement = 10 ×
1 = 1. 667 Joule 6
Displacement in 2 sec.
1 1 2 1 2 at = 0 + × (2)2 = meter 2 3 3 2 Work done at the end of 2 sec = W = F × displacement S = ut +
= 10 ×
2 = 6. 667 Joule 3
Displacement in 3 sec. S = 0+
FG IJ H K
1 1 9 × × 32 = m. 2 3 6
Work done at the end of 3 sec;
W = F × displacement
9 = 15 Joule 6 Work done in the 3rd sec = Work done in 3 sec – Work done in 2 sec. = 10 ×
∴
= (15 – 6.667) = 8.33 Joules (ii) Velocity at the end of 3rd sec. v = u + at v = 0+
or
1 × 3 = 1 m/sec. 3
So, instantaneous power, P = F× v = 10 × 1 = 10 watt. Q.6. When a 500 kg stone is pushed over a level horizontal rough surface, by a horizontal force of 100 Newton, its velocity is found to be uniform. Calculate the work done in following cases, assuming frictional force to be constant:
162
Mechanics
(a)
Velocity remains constant.
(b)
Velocity increases from 0 to 1 m/sec.
(c)
Velocity decreases from 1 to 0 m/sec. in a distance of 5 meter.
Solution. (a) For uniform velocity over rough surface Fapplied = Ffrictional ∴ Work done in 5 meter distance against frictional force; W = 100 × 5 = 500 Joule (b) Here, the applied force increases the velocity from 0 to 1 m/sec. in addition to overcoming the friction. Using
v 2 = u2 + 2aS = 0 + 2a × 5
1 m/sec2 10 Force required for this acceleration, So
a =
1 = 50 Newton 10 Total force = 100 + 50 = 150 Newton F = ma = 500 ×
So, Work done in 5 meter distance = 150 × 5 = 750 Joule (c) In this case velocity decreases from 0 to 1 m/sec. So,
v 2 = u2 + 2aS 0 = 1 + 2a × 5 a = −
So, ∴ So,
1 m/sec2 (retardation) 10
retarding force = ma = – 500 ×
1 = –50 Newton 10
Total force acting = 100 – 50 = 50 Newton Work done = 50 × 5 = 250 Joules
Q.7. A small block of mass m slides along the frictionless track as shown in the figure 12. Calculate the height h at which the mass must be released on the track to be able to go round the rack of radius R. If h = 5R, what is the reaction force exerted by the track on the block when it is at (i) point A (ii) point B (iii) point C. Solution. Motion is under gravitational force, which is conservative, so mechanical energy is conserved i.e., loss in P.E. in coming from P to C = Gain in K.E.
Fig. 12
Work, Energy and Momentum
∴
163
mg(h – 2R) =
1 mv2 2
...(i) [where v = velocity at C]
Now velocity at C should be such that centrifugal force at C = weight of block for successful completion of loop.
mv2 = mg or v2 = Rg R
i.e.,
...(ii)
From (i) and (ii) Rg = 2g (h – 2R) R = 2h – 4R h =
5R 2
if h = 5R, velocity at A is given by Loss in P.E. = gain in K.E. i.e.,
mgh =
1 mv2A 2
v2A = 2gh = 2g × 5R = 10gR
or At pt. A, centrifugal force,
=
mv2A m × 10 g R = = 10 mg R R
Weight of block at A; acting down wards = mg So, total outward force (action) = 10 mg + mg = 11 mg Thus, reaction by track on block at A = 11 mg Similarly, velocity of block at B will be
or
mg(h – R) =
1 mvB2 2
mg(5R – R) =
1 mvB2 2
vB2 = 8 gR
or i.e., Centrifugal force at B;
=
mvB2 m × 8 gR = = 8mg R R
Since mg acts at 90° to centrifugal force at B; so action by block on the track = 8mg Thus, reaction exerted by track on the block at B = 8 mg
164
Mechanics
The velocity of block at C, is given by
or
mg(h – 2R) =
1 mvC2 2
mg(5R – 2R) =
1 mvC2 2
vC2 = 6gR
Then,
centrifugal force at C =
mvC2 m6 gR = = 6mg R R
weight mg of block acts in opposite direction to centrifugal force; So,
Action force of block = 6 mg – mg = 5 mg
Thus, reaction exerted by track on the block at C = 5 mg. Q.8. A particle slides along a track with elevated ends and a flat central part, as shown in the figure. The flat part has a length l = 2.0 meter. The curved parts are frictionless while for the flat part the coefficient of kinetic friction is 0.20. The particle is released at a point A, which is at a height h= 1.0 meter above the flat part of the track. Where does the particle finally come to rest ? Solution. D
A mg
B
l
C
Fig. 12
P.E. of particle at A w.r.t. B = mgh On reaching B, entire energy is kinetic; so K.E. at B =
1 1 mv2 so; mv2 = mgh 2 2
...(i)
Suppose particle travels distance d in flat part before coming to rest. Thus all K.E. is used against friction, so
1 mv2 = µN × distance = µ mgd 2 1 mv2 = µmgd 2
or (i) and (ii) give
mgh = µmgd d =
h 1 = = 5 meters µ 0. 2
(as N = mg) ....(ii)
Work, Energy and Momentum
165
Thus particle travels B to C (2 meter) then climbs towards D (no energy loss, as path is frictionless), then descends to C, then comes from C to B (2 meters); again climbs towards A (no energy loss, as track is frictionless); comes back to B then travels half of BC (1 meter) and stops. Total distance travelled on flat part = 2 + 2 + 1 = 5 meters. Q.9. A constant force of 5 Newton acts for 10 sec. on a body whose mass is 2 kg. The body was initially at rest. Calculate (a)
Work done by the force,
(b)
final kinetic energy,
(c)
average power of the force.
Solution. (a) Acceleration,
Force 5 = = 2. 5 m/sec2 mass 2
a =
Now displacement is given by
1 2 at 2 1 2 = 0 + × 2. 5 × (10) 2 = 125 meter
S = ut +
So work done by force = 5 × 125 = 625 J. (b) The final velocity v is given by v = u + at v = 0 + 2.5 × 10 = 25 m/sec. So, final K.E. (c)
= Pavg =
1 1 mv2 = × 2 × (25)2 = 625 J. 2 2 Work done 625 = = 62. 5 Watt. time 10
Q.10. A rod of length 1.0 m and mass 0.5 kg fixed at one end, is initially hanging vertically. The other end is now raised until it makes an angle of 60° with respect to the vertical. How much work is required ? Solution. Given
OP = 1 m ∠POQ = 60°, m = 0.5 kg
Rod is moved aside against gravity, from position P to Q. This makes C.G. of rod to move from C to D. The vertical displacement of C.G. = CD′ = OC – OD′ = 0.5 – 0.5 cos 60° = 0.25 meter. So, work done in moving the rod against gravity. = mg (CD′) = mg × 0.25 = 0.5 × 9.8 × 0.25 = 1.225 Joule.
Fig. 13
166
Mechanics
Q.11. A body of mass 0.5 kg starts from rest and slides vertically down a curved track which is in the shape of one quadrant of a circle of radius 1 meter. At the bottom of track, speed of the body was 3 m/sec. What was the work done by the frictional force ? Solution. The motion is under gravity. Frictionless force resist the motion and in doing so it does work. Supposing that curved track were frictionless, the velocity of body in vertical descent of 1 m is, v 2 = u2 + 2gS = 0 + 2g × 1
or v =
2 g = 2 × 9. 8
v = 4.427 m/sec.
1 1 mv2 = × 0.5 × 19.6 = 4.9 Joules. 2 2 1 available K.E. at bottom = × 0.5 × (3)2 2 = 2.25 Joules.
and
K.E. would have been =
but
The difference of K.E. is 4.9 – 2.25 = 2.65 J has been spent in overcoming friction. So work done by frictional force = 2.65 J. → →
P. r where P is constant. r3 Solution. As force is negative gradient of potential energy i.e., Q.12. Find the force described by the potential energy U =
F =
−∇U
F P . r I δ LM P . r OP −∇ G GH r JJK = − δr MMN r PPQ δ F P cos θ I G JK δr H r → →
So force,
F =
=
→
3
3
3
→
→ →
1 2 p . r 2 p . r = 2P cos θ × = = 3 r r3 r4 Q.13. The potential energy of a particle is given by
U = 40 + 6x2 – 7xy + 8y2 + 32 z where U is in joule and x, y and z are in metre. Find the force acting on the particle when it is in position (–2, 0, 5). Solution. Using the relation F = −
dU , we can get components of force along x, y and dr
z directions. Thus
Fx = −
dU = − 12 x + 7 y dx
Work, Energy and Momentum
167
Fy = −
dU = + 7 x + 16 y dy
Fz = −
dU = − 32 dz
In position (–2, 0, 5) we have, Fx = 24, Fy = –14 and Fz = –32. Resultant force on the particle in given position
(24)2 + ( −14 )2 + ( −32)2
F =
= 42.38 Newton. Q.14. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by
a b − 6 x12 x where a and b are the +ve constants and x is distance between atoms, determine Ux =
(a) value of x at which Ux is zero. (b) Value of x at which Ux is minimum. (c) Force between atoms. (d) Dissociation energy of molecules. Solution. (a)
a x12
or
F aI x = G J H bK
or (b) For (Ux)min; or
or
a b − 6 =0 12 x x b = 6 x
Ux =
d (Ux ) = 0 dx d dx
FG a Hx
12
−
b x6
IJ K
1/6
= 0 or
2a b = x6 1
or
−12a 6b + 7 =0 x13 x
FG IJ H K
2a 2a , or x = x = b b 6
(c) Force is –ve gradient of potential energy so; F = −
d d (Ux ) = − dx dx
LM −12a + 6b OP Nx xQ LM12a − 6b OP Nx x Q
= −
or
F =
13
13
7
7
FG a Hx
12
−
b x6
IJ K
1/6
168
Mechanics
(d) Dissociation energy (D) is changed in potential energy from its minimum value (which occurs at equilibrium separation) to zero (which occurs at x equal to infinity). So, D = (Ux )at x = ∞ − ( Ux )min
LM a = O−M MM F 2a I MN GH b JK = − Thus, dissociation energy,
D =
2
OP b P − 2a P b PP Q
b2 b2 b2 + = 4 a 2a 4 a
b2 . 4a
Q.15. Show that following forces are conservative; →
(a) F = yz i + zx j + xy k →
(b) F = ( 2 xy + z2 ) i + x 2 j + 2 xz k →
→
Solution. (a) We know that for conservative force, ∇ × F = 0 . →
Substituting the value of F ,
i → → ∂ ∇ × F = ∂x yz = i
j ∂ ∂y zx
k ∂ ∂z xy
LM ∂ ( xy) − ∂ ( zx)OP − j L ∂ ( yz) − ∂ ( xy)O PQ ∂x ∂z N ∂y Q MN ∂z L ∂ ( xz) − ∂ ( yz)OP + k M ∂y N ∂x Q
= i x − x + j y − y + k z − z = 0 So, given force is conservative.
(b)
i ∂ → → ∇×F = ∂x (2 xy + z2 )
j k ∂ ∂ ∂y ∂z x2 2 xz
Work, Energy and Momentum
169
LM ∂ (2 xz) − ∂ ( x )OP + j L ∂ (2 xy + z ) − ∂ (2xz)O PQ ∂z ∂x N ∂y Q MN ∂z L∂ O ∂ (2 xy + z )P + k M ( x ) − x y ∂ ∂ N Q
= i
2
2
2
2
= i 0 − 0 + j 2 z − 2 z + k 2 x − 2 x = 0 So, force is conservative. Q.16. Calculate the work done in following cases: →
(a)
Force F = 2i + xyj + xz 2k acts on a particle and displaces it from position (2, 3, 1) to (2, 3, 4) parallel to z-axis.
(b)
Force F = ( 2xy + z2 )i + x2 j + 2xzk makes a particle to move from position (0, 1, 2) to (5, 2, 7).
(c)
Force F = Ax + Bx2 acts parallel to x-axis on a particle and moves it from x = 1 to x = 2.
→
→
Solution. (a)
W = =
z z
N→
→
F . dr =
M N
M
z
N
M
(2i + xyj + xz2k ) ( dxi + dyj + dzk)
2dx + xydy + xz2dz (since i . j = i . k = 0)
Since motion occurs along z-axis, so no change in x and y occurs i.e., dx = dy = 0 and we have W =
z
N
M
2
xz dz = x
LM z OP N3Q 3
= 2.
(b)
W = = = =
z z z z
4
=
1
N→
M N
M
M
Lz O z dz = x M P N3Q 2
3
N
M
2 3 [4 − 1] = 42 units . 3
z
N
M
(Fxi + Fy j + Fz k ). ( dxi + dy j + dzk )
Fx dx + Fy dy + Fz dz
N
M N
M
→
F . dr =
z
N
(2 xy + z2 ) dx + x2dy + 2 xz dz (2 xy dx + x2dy) + ( z2dx + 2 xz dz)
170
=
z
N
M
d ( x2 y) + d ( z2 x) =
2 2 = x y+ z x
N
z
Mechanics N
M
d ( x2 y + z2 x)
= x2 y + z2 x
M
( 5, 2, 7) ( 0, 1, 2)
= 295 units.
z
x=2
(c)
W =
z
Fdx
x=1 2
= =
1
( Ax + Bx2 ) dx
LM Ax N2
2
+
Bx3 3
OP Q
2
1
A 2 B (2 − 12 ) + (23 − 13 ) 2 3 3 7 = A+ B 2 3 Q.17. An ideal massless spring can be compressed 2 meter by a force of 200 N. The spring is located at the bottom of a frictionless inclined plane which makes angle of 30° with the horizontal. A 20 kg mass is released from rest at the top of inclined plane which comes to rest after compressing the spring by 4 metres. Calculate =
(a) Distance travelled by mass before coming to rest. (b) Speed of mass before it reaches the spring. Solution. (a) The linear restoring force in the spring is given by, F = –kx here, Let,
k =
200 = 100 N/m. 2
l = distance along inclined plane that mass travels before coming to rest. x = compression of spring = 4 meter.
The initial P.E. of mass will be converted into P.E. of spring during compression of 4 meter.
or or
l′
l
or
1 2 kx 2 1 2 mgl sin θ = kx 2 1 × 100 × 42 20 × 9.8 × l × sin 30° = 2 l = 8.16 meter.
mgh =
4
i.e.,
3 0°
Fig. 14
Work, Energy and Momentum
171
(b) Let l' be a distance travelled when mass just touches the spring and v be its velocity then, l = l' + 4 or
l' = l – 4 = 8.16 – 4 = 4.16 meter. As K.E. of mass at the instant of touching the spring = Initial P.E. of mass i.e.,
1 mv2 = mgl' sin θ 2 v 2 = 2gl' sin θ = 2 × 9.8 × 4.16 ×
or
1 2
v = 6.385 m/sec. Q. 18. An ideal spring (Force constant = 8 × 104 dynes/cm) hangs vertically and supports 0.8 kg mass at rest. Calculate the distance by which the mass should be pulled down so that it may pass through, on being released, the equilibrium position with a velocity of 1 m/sec. Solution. The linear restoring force produced in the stretched spring is F = –kx In stretching the spring, work is done which is stored as P.E. If a vertical downward pull of ‘l’ cm is made, the work done against the restoring force = potential energy =
1 2 kl 2
On releasing the mass after stretching, it moves upwards and passes the mean position with velocity v (say). Then K.E. at mean position =
1 mv2 2
since entire P.E. at extreme position totally converts into K.E. at mean position,
1 2 1 kl = mv2 2 2 as m = 0.8 kg, k = 8 × 104 dynes/cm and v = 1 m/sec. l =
mv2 / k =
800 × (100)2 /(8 × 104 ) = 10 cm
Q.19. A block of mass 2.0 kg is dropped from a height of 0.4 m. onto a spring of force constant K = 1960 N/m. Find the maximum distance through which the spring will be compressed [Neglect Friction] Solution. Situation is shown in the diagram Loss in P.E. of mass = gain in pot. energy by the compressed spring i.e.,
mg (d+ y) =
1 2 ky 2
given m = 2 kg, d = 0.4 m; k = 1960 Nt/m, g = 9.8 m/sec2; y = ? So,
2 × 9.8 × (0.4 + y) =
1 × 1960 × y2 2
172
or
Mechanics
y2 – 0.02 y – 0.008 = 0
2 .0 kg
0. 02 ± (0. 02) − 4 × 0. 008 2 2
or
y =
d
0. 0004 − 0. 032 2 = 0.0999 m (taking +ve value). =
0. 02 ±
Fig. 15
Q.20. The scale of spring balance, reading from zero to 100 Newton is 20 cm long. A body suspended from the balance is observed to oscillate vertically at 2 vib/sec. What is weight of the body ? Solution. Force constant of spring =
100 = 500 Nt/m. 0. 2
Since frequency of oscillation of vertical spring with mass is given by, 1 2π with n = 2, K = 500 Nt/m, π = 3.14 kg
n =
m =
K K ; or m = m 4 π2n2
500 500 = = 3.169 kg 4 × ( 3.14 )2 × 4 157. 75
weight mg = 3.169 × 9.8 = 31.06 Newton. Q.21. An object is attached to a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by an amount ‘d’. If the same object is attached to the same spring but permitted to fall instead, through what distance does it stretch the spring ? Solution. Fig. 16(a) shows the normal unextended spring. In Fig. 16(b), by slowly placing the mass, equilibrium state is shown and elongation is d.
(a ) d d′
(b )
(c)
Fig. 16
Work, Energy and Momentum
So,
173
Restoring force = weight
mg ...(i) d In Fig. 16(c); If mass is permitted to fall, the mass will go down to an extreme, will rise up and thus will execute S.H.M. i.e.,
Kd = mg or K =
Suppose in extreme position elongation is d then; Work done by force mg on the spring = mg × d' P.B. corresponding to extreme position =
...(ii)
1 Kd′2 2
...(iii)
Equating eqs. (ii) and (iii)
Putting for K from (i), or
mgd′ =
1 Kd'2 2
mgd′ =
1 mg . d'2 2 d
d' = 2d i.e., stretching is double of the previous case.
Q.22. Two block A and B are connected to each other by a string and a spring, the string passes over a frictionless pulley as shown. The block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of blocks is 0.2. Force constant of spring is 1960 Nt/m. If mass of A = 2 kg, calculate the mass of block B and energy stored in the spring. Solution. Let m = mass of B; T = tension in the spring For motion of B; T = µN = µmg = 0.2 mg B
...(i)
T
T
C A
Fig. 17
for motion of A from (i) and (ii) again;
T = mg = 2g
...(ii)
2g = 0.2 mg or m = 10 kg T = 2 × 9.8 = 19.6 Newton (from (ii))
if x is the elongation of the spring, then magnitude of linear restoring force; F = kx (upwards –ve) or x =
F k
174
Mechanics
The linear restoring force acting in the present case is T itself i.e., F = T So,
x =
Energy stored in the spring =
T 19. 6 = k 1960 1 2 kx 2
...(iii)
FG H
IJ K
2
1 19. 6 × 1960 × = 0. 098 Joule 2 1960 Q.23. A body of mass 2 kg is attached to a horizontal spring of force constant 8 Nt/m. A constant force of 6 Nt. is applied along the length of the spring. Find =
(a) The speed of the body when it is displaced through 0.5 m (b) If force is removed, then; how much farther shall the body move before reaching the state of rest. Solution. (a) Work done by the constant force = 6 × 0.5 = 3 Joule If v is velocity, K.E. of body in displaced position will be given by
1 1 × 2v2 mv2 = 2 2 Again, potential energy stored in the spring at displaced position of 0.5 m
1 2 1 × 8 × (0. 5)2 kx = 2 2 So, Work done in 0.5 m displacement = K.E. + P.E. i.e.,
W =
1 1 × 2 × v2 + × 8 × (0. 5)2 2 2
W = v2 + 1
or Since
W = 3 Joule, 3 = v2 + 1 or v = 1.4 m/sec.
(b) Total energy of system at the time of removal of force = 3 Joule. Suppose body moves distance ‘l’ still onwards before coming to rest; on removal of force P.E. at the extreme position = total energy of system
1 × 8 × (0. 5 + l)2 = 3 2 3 4 l = 0.36 m.
(0.5 + l)2 =
or So,
Q.24. A 10 lb block is thrust up a 30° inclined plane with an initial speed 16 ft./sec. If it is found to travel 5 ft. along the plane; stop and then slide back again to the bottom. Calculate (a) The force of friction (regarded as constant) on block. (b) Speed of block when it reaches bottom of the inclined plane. Solution. (a) Upward motion: At the top where block momentarily stops; total energy,
Work, Energy and Momentum
175
E = K.E. + P.E. = mg × 5 sin 30° = 10 × 5 ×
1 = 25 ft lb 2
At foot of incline, Ei = (K.E.)i + (P.E.)i Ei =
FG IJ H K
1 10 (16)2 + 0 2 32
Ei = 40 ft-lb Energy lost = Work done by frictional force in 5 ft. distance i.e.,
25 – 40 = –F × 5 or F = 3 lb
(b) Downwards motion: Suppose block returns to the foot of incline with speed ‘v’. At bottom where motion ends: E = K.E. + P.E. =
FG IJ H K
1 10 2 5 2 v +0= v ft - lb 2 32 32
At top where motion starts: total energy = 25 ft-lb again energy lost = work done by frictional force
5 2 v − 25 = –15 or 32
v2 =2 32
or v = 8 ft/sec.
Q.25. Figure shows a vertical section of a frictionless track surface. A block of mass 2 kg is released from position A. Compare its K.E. at the positions B, C and D. (g = 9.8 m/sec2). Solution. Motion occurs under gravitational force which is conservative, so energy is conserved. In coming down block looses its P.E. and corresponding gain in K.E. occurs; at point B:
(K.E.)B = (P.E.)A – (P.E.)B = (mgh)A – (mgh)B = mg (14 – 5) = 2 × 9.8 × 9 = 176.4 Joules.
at Point C:
(K.E.)C = (P.E.)A – (P.E.)C = mghA – mghC = mg (14 – 7) = 0.2 × 9.8 × 7 = 137.2 Joules
at Point D:
(K.E.)D = (P.E.)A – (P.E.)D = mghA – mghD = mg (14 – 0) = 2 × 9.8 × 14 = 274.4 Joules.
Q.26. A point mass m starts from rest and slides down the surface of a frictionless sphere of radius r. Measure angles from the vertical and P.E. from the top. Find (a)
Change in P.E. of the mass with angle.
(b)
K.E. as function of angle.
(c)
Radial and tangential acceleration as function of angle.
(d)
Angle at which mass flies off the sphere.
176
Mechanics
(e)
If friction be present between mass and sphere, shall the mass fly off at greater or lesser angle as compared to situation in (d).
Solution. (a) At top position P (reference point) P.E. = 0 at bottom position at depth, y = 2r ; ∴ P.E. = –mgy
P
m
at a general position M;
M
M′
P.E. = –mgPM' = – mg (PO – M'O)
θ
F M′ O . MOIJ = – mg G r − H MO K
r
O
= – mgr (1 – cos θ) So change in P.E.; (P.E.)M – (P.E.)P = – mgr (1 – cos θ) – 0 ∆(P.E.) = – mgr (1 – cos θ)
B
Fig. 18
...(i)
(b) At top total energy is wholly potential = 0. so from energy conservation At general position M; K.E. + P.E. = 0 i.e., K.E. + [– mgr (1 – cos θ)] = 0 K.E. = mgr (1 – cos θ)
or (c) At general position M; ∴
Radial acceleration =
But K.E. at same position is
...(ii) M
v = tangential velocity
m g co s θ θ O mg
v2 r
1 mv2 = mgr (1 – cos θ) 2
m g sin θ
r
v 2 = 2gr (1 – cos θ)
or
N
...(iii)
Fig. 19
2 gr (1 − cos θ) = 2 g (1 − cos θ) Radial acceleration = r
So,
tangential acceleration = ∴ as
v = rw =
rdθ dθ v ∴ = dt dt r
dv v dv = dt r dθ
so, From (iii)
dv dv dθ = . dt dθ dt
∴
dv 1 2 gr (1 − cos θ) = dθ 2
...(iv) −
1 2
. 2 gr sin θ
then tangential acceleration is
dv v gr sin θ . = g sin θ = dt r v (d) At position where mass flies off the sphere-surface the condition for such event is Centrifugal force = Component of weight along the radius
Work, Energy and Momentum
177
mv2 = mg cos θ r
i.e.,
v 2 = gr cos θ
or
...(v)
equating (iii) and (v); 2gr (1 – cos θ) = gr cos θ 3 cos θ = 2 or θ = cos–1
or
FG 2 IJ H 3K
(e) If friction is present, then velocity at any general point M shall be less. v is less means cos θ is less which suggests θ is large. Therefore mass leaves at greater angle as compared to previous (frictionless) case. Q.27. A body of mass 5 kg moves down from state of rest on an inclined plane of length 5 meters having inclination of 60° with the horizontal. If coefficient of friction be 0.2 find the speed of body at the bottom. How far shall it slide on the rough horizontal surface having same coefficient of friction as the inclined plane. P.E. = mgh = mgd sin θ ; and K.E. = O
Solution. At top:
frictional force = µN = µmg cos θ
Now,
Work done by frictional force = –µmg cos θ × d (–ve sign gives, nature or direction) Since,
change in total energy = work done
i.e., change in P.E. + Change in K.E. = work done ∆T + (–mgd sin θ) = – µ mgd cos θ
So;
∆T = mgd (sin θ – µ cos θ) =
or
1 mv2 2
θ
Fig. 20
v 2 = 2gd (sin θ – µ cos θ ) g = 9.8 m/sec2; d = 5 meters, θ = 60°, µ = 0.2 v 2 = 2 × 9.8 × (sin 60° – 0.2 cos 60°) × 5
or Now, which gives,
v = 8.664 m/sec
At the start on horizontal rough surface the K.E. possessed by body =
1 mv2, 2
This becomes zero finally due to expense in doing work against frictional force; i.e.,
O–
1 mv2 = Work done against friction 2
Supposing that l distance is travelled before coming to rest, then frictional work done = – µ mgl So, as calculated earlier,
–
1 mv2 = – µ mgl or v2 = 2 µ gl 2 v 2 = 75.06 ; µ = 0.2, g = 9.8 l =
75. 06 v2 = = 19.15 meter . 2µg 2 × 0. 2 × 9. 8
178
Mechanics
Q.28. A block of mass 1 kg collides with a horizontal weightless spring of force constant 2Nt/meter. The block compresses the spring 4.0 meter from its normal position. Calculate the speed of block at the instant of collision if kinetic friction between block and surface is 0.25. Solution. Suppose v = speed of block at the instant of collision then its kinetic energy
1 mv2 2 This energy is spent in overcoming the friction and compressing the spring, if x is compression of spring; =
1 1 2 mv2 = kx + µmgx 2 2 given
x = 4 meters, k = 2 Nt/meter, m = 1 kg, g = 9.8 m/sec2, µ = 0.25
1 1 × 1 × v2 = × 2 × (4)2 + 0. 25 × 1 × 9. 8 × 4 2 2
∴ or
v =
51. 6 = 7.18 m/sec.
Q.29. A horizontal force pushes a 10 kg mass up an inclined plane ( θ = 30°) from bottom, by a distance of 3 meters on the inclined plane. If the initial and final speeds of the mass are 1 m/sec and 3 m/s, calculate the work done by the force. Solution. Height in the final position, h = AB sin 30° = 3 × ∴ Change in P.E.,
∆U = mgh = 10 × 9.8 × 1.5 = 147 Joule
Change in K.E. = ∆T = = ∴
1 = 1.5 m 2
1 1 1 mv22 − mv12 = m (v22 − v12 ) 2 2 2
1 × 10 × (32 – 12) = 40 J 2
Change in total energy = (147 + 40) = 187 Joule
which gives the work done.
N
Q.30. A chain of length d and mass m lies on a frictionless horizontal table such that one third of its length hangs over the edge. Calculate the work needed to pull the hanging part back onto table. Solution. Mass per unit length of chain = m/d
E
W
S
m Mass of infinitesimally small length dy of chain = . dy work done in lifting this d segment against force of gravity; by a distance y; is dW =
FG m dyIJ .g.y Hd K
So, work done in lifting or raising the
d length 3
Work, Energy and Momentum
179
W =
z
d/3
0
mg = d
=
mg 2d
FG mg IJ . y dy H dK
z
d /3
0
mg ydy = d
F d I = mgd . GH 9 JK 18
Fy I GH 2 JK 2
d /3
0
2
Q.31. A 0.5 kg block slides from point A on a horizontal track with initial speed 3m/s towards a weightless horizontal spring of length 1m and force constant 2 Nt/m. Part AB of track is frictionless while part BC has coefficient of static and kinetic friction 0.22 and 0.2. If AB = 2m and BD = 2.14m, find total distance block moves before coming to rest (g =10m/s2). Solution. The block will reach B with initial K.E. V1 =
1 × 0. 5 × (3)2 = 2. 25 J 2
In going from B to D work will be done against kinetic friction. ∴
W = µN (BD) = 0.2 × 0.5 × 10 × 2.14 = 2.14 J
Therefore, the K.E. with which block will hit the spring at D = V1 – W = 0.11 J If block compresses the spring through x, this energy will be partly used up in doing work against friction during the compression x and rest will be stored in spring as its P.E. i.e.,
1 2 Kx + µNx = 0.11 2 1 × 2 × x2 + 0. 2 × 0. 5 × 10 x = 0.11 2
or
i.e.,
x2 + x – 0.11 = 0 which gives x =
−1 ±
1 + 0. 44 = 0.1 m 2
It means total distance converted by block before it comes to rest is D = AB + BD + x = 2 + 2.14 + 0.1 = 4.24 m. Q.32. A body of mass m is accelerated uniformly from rest to a speed V in time. Show that work done on the body as a function of time is W =
1 V2 2 m .t . 2 T2
180
Mechanics
Solution. Let ‘a’ be the acceleration in body then, since u = 0
V T
V = 0 + a.T or a = The velocity gained by body in time t will be v = at =
V . t T
Hence K.E. gained in time t, which gives work done W = =
FG H
1 1 V mv2 = m .t 2 2 T
IJ K
2
1 V2 m 2 . t2 2 T
Q.33. The scale of a spring balance reads from 0 to 250 kg and is 25 cm long. What is the potential energy of the spring when a 20 kg weight hang from it ? Solution. As given, 25 cm elongation of spring takes place corresponding to 250 kg. So the force constant of spring is given by 250 × 9.8 = K × 0.25 K = 9.8 × 103 N/m If 20 kg weight hangs from balance, elongation will be obtained by 20 × 9.8 = Kx' i.e.,
x' =
20 × 9. 8 meter K
∴ Potential energy stored in spring will be
F GH
1 1 20 × 9. 8 U = K ( x' )2 = × 9. 8 × 103 2 2 9. 8 × 103
=
I JK
2
1 (20 × 9. 8)2 × 2 9. 8 × 103
= 1.96 Q.34. Calculate the work done by a force F = kx2 acting on a particle at an angle of 60° with x-axis to displace it from x1 to x2 along the x-axis. Solution. The work done is given by W = =
z z
→ −→
F . dr x2
x1
1 k 2
Fdx cos 60° =
LM OP N Q
1 x3 = k 2 3
x2
= x1
e
z
x2
x2dx
x1
j
1 k x23 − x13 . 6
Work, Energy and Momentum
181
Q.35. A cord is used to lower vertically a block of mass M a distance d at a constant downward acceleration of g/4. Find the work done by the cord on the block. Solution. Let T be the tension in the cord acting vertically upward. The net force on the block is T-Mg, acting upward, where Mg is the weight of the block. If a is the downward acceleration of the block, then by Newton’s second law, we have T – Mg = –Ma or
T = M(g – a) Here,
a = g/4
∴
T = M g−
FG H
IJ K
3 g = Mg 4 4
Therefore, work done by the tension T is given by W = tension (T) × displacement in the direction of T = T × (–d) [3 d is downward]
3 Mgd . 4 Q.36. A block of mass 10.0 kg is pulled up at constant speed from the bottom to the top of a smooth incline 5.00 meter off the ground at the top. Calculate the work done by the applied force which is parallel to the incline. (g = 9.8 m/s2). = −
Solution.
Fig. 21
The force acting on the block at any instant are its weight mg, the normal reaction N and the applied force F. The weight can be resolved into a component mg sin θ to the inclined plane and a component mg cos θ perpendicular to it. The net force parallel to the plane is F mg sin θ. Since the block moves on the plane at constant speed, the net force is zero. Thus F – mg sin θ = 0 or
F = mg sin θ = 10 × 9.8 ×
3 = 58.8 Nt 5
182
Mechanics
The work done is W = Force × displacement along the force = 58.8 Nt × 5 m = 294 Joule Q.37. A boy pulls a 5 kg block 10 meter along a horizontal surface at a constant speed with a force directed 45° above the horizontal. If coefficient of kinetic friction is 0.20, how much work does the boy do on block ? Solution. The force acting on the block are its weight mg, normal reaction N exerted by the surface, sliding frictional force fK against the motion and the applied force F at 45° above the horizontal. The net horizontal force is F cos 45° – fK, and net vertical force is N + F sin 45° – mg.
Fig. 22
The block is unaccelerated, so that from Newton’s second law, we obtain and
F cos 45° – fK = 0
...(i)
N + F sin 45° – mg = 0
...(ii)
Also, we know
fK = µKN
...(iii)
where µK is the coefficient of kinetic friction. Substituting the value of N from Eq. (ii) in Eq. (iii), we get fK = µK (mg – F sin 45°) Now putting this value of fK in Eq. (i); we get F cos 45° – µK (mg – F sin 45°)= 0 or
...(iv)
F (cos 45° + µK sin 45°) = µK mg
µK mg cos 45° + µK sin 45° with µK = 0.20, mg = 5 × 9.8 = 49 Nt and cos 45° = sin 45° = 0.707,
or
F =
we obtain F = 11.55 Nt. The block is pulled through a horizontal distance r = 10 meter. Then the work done is W = Fr cos 45° W = (11.55) (10) (0.707) = 81.66 Joule. Q.38. A block of mass m = 3.57 kg is pulled at constant speed through a distance r = 4.06 m along a horizontal surface by a rope exerting a constant force F = 7.68 Nt inclined at θ = 15° to the horizontal. Find (i) the total work done on the block (ii) work done on the block by the rope and by friction, (iii) coefficient of kinetic friction between block and surface. Solution. (i) Since the block moves at constant speed, the net force on it is zero. Hence the total work done on it is zero. (ii) The force exerted by the rope is inclined at 15° to the horizontal . Hence the work done on the block by the rope through a distance r = 4.06 m is given by
Work, Energy and Momentum
183
M = Fr cos 15°´ = 7.68 × 4.06 × 0.966 = 30.1 Joule Since the total work done on the block is zero, the work done on it by friction is –30.1 Joule, i.e., work is done by the block against friction. (iii) From eq. (iv) of the last problem, we can write
F cos15° mg − F sin 15° 7. 68 × 0. 966 = 3. 57 × 9. 8 − 7. 68 × 0. 259
µK =
7. 419 = 0.225 33. 00 Q.39. A running man has half the kinetic energy that a boy of half his mass has. The man speeds up by 1.0 m/s and then has the same kinetic energy as the boy has. Find speed of man and boy. =
Solution. Let m be the mass of man, and m/2 that of boy. Let v1 and v2 be their original 1 m 2 1 2 v1 . Since the energy speeds. Then kinetic energy of man is mv1 and that of boy is 2 2 2 of man is half that of boy, we have
FG IJ H K
LM FG IJ OP N H K Q
1 1 m 2 1 v2 mv12 = 2 2 2 2 or
v12 =
1 2 v2 4
...(i)
When the speed of the man becomes (v1 + 1), his kinetic energy equals to that of the boy. That is,
FG IJ H K
1 m 2 1 v2 m ( v1 + 1)2 = 2 2 2 (v1 + 1)2 =
1 2 v2 2
...(ii)
Solving eqs. (i) and (ii) we get v 1 = 2.41 m/s v 2 = 4.82 m/s. Q.40. A 30 gm bullet initially travelling 500 m/s penetrates 12 cm into a wooden block. What average force does it exert ? Ans. The kinetic energy of the bullet is given by 1 mv2 2 1 (30 × 10−3 ) × (500)2 = 3750 Joule = 2
K =
184
Mechanics
It stops after travelling a distance r = 12 cm. = 0.12 m in wooden block. If F Nt be the (retarding) force exerted by the block, then the work done by the bullet against this force is W = Fr = F × 0.12 Joule By work-energy theorem, the kinetic energy equals this work done, that is 3750 = F × 0.12 F =
3750 = 31250 Nt. 0.12
Q.41. Show with the help of work-energy theorem that the minimum stopping distance for a car of mass m moving with speed v along a level road is v2/2µsg, where µs is the coefficient of static friction between tyres and road. Solution. The kinetic energy of the car is K =
1 mv2 2
It is retarded due to the force of static friction fs (rolling friction neglected), whose maximum value is given by fs = µsN = µsmg N being the normal reaction which equals weight mg of the car on a level road. If x be the minimum stopping distance, then the work done against the force fs is given by W = fsx = µs mgx This, by work-energy theorem, equals the initial kinetic energy K of the car. Thus, µsmgx = x =
1 mv2 2 v2 2µ s g
Q.42. A mass of 0.675 kg is being revolved on a smooth table in a horizontal circle by means of a string which passes through a hole in the table at the centre of the circle. If the radius of circle is 0.50 m and the uniform speed is 10.00 m/s, find the tension in string. If the radius of the circle is reduced to 0.30 m by drawing the string down through the hole, the tension is increased by a factor 4.63. Find the work done by the string on the revolving mass during the reduction of the radius. Solution. The tension T of the string supplies the required centripetal force mv2/R i.e.,
mv2 0. 675 × (10. 0)2 = = 135 Nt R (0. 5) On reducing the radius, the new tension is T =
T' = 4.63 T = 4.63 × 135 = 625 Nt If v' be the new velocity, then we have, T' =
mv′2 R′
Work, Energy and Momentum
185
625 =
0. 675 v′2 0. 30
v' 2 =
625 × 0. 3 = 278 ( m/s)2 0. 675
The increase in kinetic energy of the moving mass is therefore ∆K = =
1 m (v′2 − v2 ) 2 1 × (0. 675) × (278 − 100) 2
= 60 Joule By work energy theorem, this is equal, to the work done on the moving mass. Hence,
W = ∆K = 60 Joule
Q.43. A constant force of 5 Nt acts for 10 seconds on a body whose mass is 2 kg. The body was initially at rest. Calculate the work done by the force, the final kinetic energy and the average power of the force. Solution. Let a be the (constant) acceleration in the body. Then a =
F 5 = = 2. 5 m/s2 m 2
The distance x moved in 10 sec is x =
1 2 1 at = × 2. 5 × (10)2 2 2
= 125 m. The work done by the force F during this distance is W = Fx = 5 × 125 = 625 Joule The initial kinetic energy of the body is zero, By work-energy theorem the final kinetic energy is the same as the work done, i.e., 625 Joule. The body covers distance x (= 125 m) in 10 sec. Then its average velocity is x 125 v = t = 10 = 12. 5 m/s
The average power of the force is therefore →
→
P = Fv
= 5 × 12.5 = 62.5 Watt. Q.44. A boy whose mass is 51 kg climbs, with constant speed, a vertical rope 6 m long in 10 second. How much work does the boy perform ? What is his power output during the climb ?
186
Mechanics
Solution. The body does work against his weight in climbing. This is given by W = force × distance = (51 × 9.8) × 6 = 3000 Joule This work is done in 10 sec. Hence the power output is W t 3000 = 10 = 300 Watt.
P =
Y
θ
θ0 T
l
A F dy
θ dx
O
h
X mg
Fig. 23
potential energy when the bob has been raised to a vertical height h in the x-y plane is U(x,
y)
= U(x,
h)
= mgh
...(i)
At the point A, the kinetic energy is zero i.e., the potential energy is equal to the mechanical energy E. Then E = mgh
....(ii)
If the bob be released at the point A, it returns under the gravitational (restoring) force, and the potential energy begins to convert into the kinetic energy. At any point in its path the sum of the two energies remains equal to the mechanical energy. Thus if v be the velocity and y the vertical height of the bob at any point, we have by the law of conservation of mechanical energy.
But
1 mv2 + U( x, y) = E 2 U(x, y) = mgy
1 mv2 + mgy = E 2 Comparing eq. (ii) and (iv) we have ∴
1 mv2 + mgy = mgh 2
....(iii)
...(iv)
Work, Energy and Momentum
187
1 mv2 = mg (h – y) 2
or or
v =
2g ( h − y)
Thus at O (y = 0); v =
2gh (maximum); the energy is all kinetic. At A (y = h); v = 0 (minimum); the energy is all potential. 1 mv2 varies between mgh and O. This means 2gh and O or 2 that from equation (iii), U(x, y) can never be greater than E i.e., U(x, y) >/ mgh. In other words the bob cannot rise higher than h, its release point A. Thus v varies between
(iv) Magnetic Potential Energy. Let τ be the (restoring) couple acting upon a magnet of moment M when its axis makes an angle θ with a magnetic field H. Then we have τ = –MH sinθ Let U(0) be the potential energy in the θ = 0 position. The work in turning the magnet from θ = 0 to θ position is the potential energy further acquired by the magnet. Thus if U(θ) be the potential energy at position θ, then U(θ) – U(0) = −
z
θ
0
τ dθ
= MH
z
θ
0
sin θ dθ
= MH (1 – cos θ) If U(0) be assumed to be zero, then U(θ) = MH (1 – cos θ) This is the required expression. If the magnet is released from the position θ, its potential energy is converted into the kinetic energy of rotation and in position θ = 0 the entire energy becomes kinetic. It I be the moment of inertia of the magnet about the axis of suspension and ω the angular speed attained at θ = 0 position, then the kinetic energy is
1 2 Iω . Therefore, 2
1 2 Iω = MH (1 – cos θ) 2 ω =
2MH (1 − cos θ) I
→
Q.45. If a force F = ( 2 xy + z2 ) i + x 2 j + 2 xzk , then show that it a conservative force. Determine its potential function. Solution. Let us find the curl of given force.
188
Mechanics
i ∂ curl F = ∂x Fx
k i ∂ ∂ = ∂z ∂x Fz 2 xy + z2
j ∂ ∂y Fy
→
j k ∂ ∂ ∂y ∂z x2 2 xz
= i (0 − 0 ) − j (2 z − 2 z) + k (2 x − 2 x) = 0 →
→
As curl F is zero, the force F is conservative. →
The potential energy function of F is given by. U = − = −
z z z z
→ −→
F . dr = −
z
(Fx dx + Fy dy + Fz dz)
(2 xy + z2 ) dx + x2 dy + 2 xz dz
= − (2 xy dx + x2dy) + ( z2dx + 2 xzdz) = −
d ( x2 y) + d ( z2 x) = −
= − ( x2 y + z2 x)
z
d( x2 y + z2 x)
→
Q.46. Show that the force represented by F = yz i + zx j + xy k is conservative. →
Q.47. The position of a moving particle at an instant is given by r = A cosθ i + A sinθ j . Show that the force acting on the particle is conservative. Solution. The position vector of the particle at an instant t is given by →
r = A cos θ i + A sin θ j = A (cos ωt i + sin ωt j )
where ω is the angular velocity of the particle and ωt = θ The linear velocity and acceleration at t are given by →
v =
and
→
dr = Aω ( − sin ωt i + cos ωt j ) dt
→
→
→ dv = − Aω2 (cos ωt i + sin ωt j ) = − ω2 r dt If m be the mass of the particle, then the force acting upon it is given by
a =
→
→
→
F = m a = − mω 2 r →
Since curl r = 0, we have →
curl F = 0 and so the force is conservative.
Work, Energy and Momentum
189
Q.48. What is the potential energy of an 800 kg elevator at the top of a building 380 m above street level ? Assume the potential energy at street level to be zero. What would happen to this energy if the elevator comes down to the street level ? Solution. The (gravitational) potential energy of the elevator at a height h when its value at street level is zero, is given by U(h) = mgh Here, ∴
m = 800 kg and h = 380 m U(h) = 800 × 9.8 × 380 = 2.98 × 106 Joule.
When the elevator comes down to the street level, the potential energy is first converted into kinetic energy and then into heat when the elevator has come to a stop. Q.49. A rod of length 1.0 meter and mass 0.5 kg is fixed at one end and is initially having vertically. The other end is now raised until it makes an angle of 60° with the vertical. How much work is required ?
Fig. 24
Solution. The weight mg of the rod acts at its centre of gravity G. As the lower end of the rod is rotated through 60°, G moves to G', i.e., it is raised through a height h, where, h = CG = OG – OC = OG – OG' cos 60° = OG (1 – cos 60°) because
CG = OG'
Here OG = 0.5 meter and cos 60° = 0.5 ∴
h = 0.5 (1 – 0.5) = 0.25 meter
Thus; the gain in the potential energy of the rod is U = mgh = 0.5 × 9.8 × 0.25 Since the work done W has been stored as potential energy, we have W = U = 1.225 Joule
190
Mechanics
Q.50. A uniform chain is held on a frictionless table with one-fifth of its length hanging over the edge. If the chain has a length l and a mass m, how much work is required to pull the hanging part back on the table ? Solution. l/5
Fig. 25
The mass of the hanging part of the chain is m/5. The weight mg/5 of this part of the chain acts at its centre of gravity which is distance l/10 below the surface of the table. Hence the gain in potential energy in pulling the hanging part on the table is
mg l mgl = 5 10 50 This is also the work done in pulling the chain. Thus U =
W = U = mgl/50 Q.51. The potential energy function for the force between two atoms in a diatomic molecule can be expressed as follows: U(x) =
a b − 6 x12 x
where a and b are positive constants and x is the distance between the two atoms. Derive an expression for the force between the two atoms and show that the two atoms repel each other for x less than x0 and attract each other for x greater than x0. What is the value of x0 ? Solution.
U(x) =
a b − 6 12 x x
We know that the force is the negative gradient of the potential energy. Therefore the force between the two atoms is given by F = − = − =
dU( x) dx
LM N
d a b − dx x12 x6
OP Q
12 a 6b − 7 x13 x
Let the force F be zero, when x = x0, then we have 12 a 6b − 7 = 0 x13 x0 0 2a x06 = b
x0
F 2a I = G J HbK
1/6
Work, Energy and Momentum
191
This is the equilibrium point (a point of stable equilibrium). Now, we may write F =
=
12 a 6b − 7 x13 x
LM N
OP Q
6b 2a / b 6b −1 = 7 7 6 x x x
R|F x I S|GH x JK T 0
6
−1
U| V| W
Clearly, when x < x0 then F is positive and the atoms repel each other. When x > x0 then F is negative and the atoms attract each other. Q.52. The potential energy function for the force between two atoms in a diatomic molecule may be expressed as follows: U(x) =
a b − 5; 10 x x
where a and b are positive constants and x is the distance between the atoms. (a) Calculate the distance x at which P.E. is minimum. (b) Assume that one of the atoms remains at rest and that the other moves along x. Describe the possible motion. (c) The energy needed to break up the molecule into separated atoms (x = ∞ ) is called the dissociation energy. What is the dissociation energy of the molecule ? Solution. (a) The value of x at which U(x) is a minimum is found from
so that
or
d dx
FG a Hx
10
−
−
b x5
IJ K
d U( x) = 0 dx
= 0
10 a 5b + 6 = 0 x11 x
or
x5 =
or
x =
2a b
FG 2a IJ H bK
1/5
(b) The force between the two atom is F = −
The force is zero when x =
FG 2a IJ H bK
d 10 a 5b U( x) = 11 − 6 dx x x
1/ 5
. When x is less than (2a/b)1/5, the force is positive
and the atoms repel each other. When x is greater than (2a/b)1/5, then force is negative and the atoms attract each other. If one of the atoms is fixed, then the other atom would oscillate about the equilibrium separation (2a/b)1/5. (c) The dissociation energy D is equal to the change in potential energy from its minimum value at equilibrium separation x = (2a/b)1/5 to the zero value at x = ∞ .
192
Mechanics
∴
D = U(x = – ∞ ) – U{x = (2a/b)1/5} = D–
RS a T (2a/b)
−
5
b 2a/b
UV W
= b2/4a If the kinetic energy at the equilibrium position is equal to or greater than its value, the molecule will dissociate. Q.53. The potential energy function of the force between two atoms in a diatomic molecule can be expressed approximately as: U(x) =
a b − 6 x12 x
where a and b are positive constants and x is the distance between atoms (i) At what values of x is U(x) equal to zero and U(x) a minimum ? (ii) Determine the force between atoms (iii) Calculate the dissociation energy of the molecule.
LM Ans. (i) F a I GH b JK MN
1/6
or ∞ ,
FG 2a IJ HbK
1/6
; ( ii)
12a 6b − 7 , ( iii) b2 /4 a x3 x
OP PQ
Q.10. The potential energy between the protons and neutrons inside a nucleus is given by U(r) = −
r0 U0 e− r/r0 r
Find the corresponding expression for the force of attraction and compute the ratio of this force at r0 , 2r0 , 4r0 and 10r0 to the force at r = r0. What conclusion you draw from the results? Solution. The force is the negative gradient of potential energy. Thus
RS r U e UV Tr W R|F 1 I + FG 1 IJ e e j F − 1 I U|V r U SG − J e GH r JK |W H rK T|H r K FG 1 + 1 IJ −r U e r Hr r K
F(r) = −
=
=
d d U( x) = dr dt
0
0
− r / r0
0
0
0
− r / r0
2
0
− r / r0
0
− r / r0
0
This is the expression for the force of ‘attraction’. At r = r0 the force is Fr0 = − U0 e−1
FG 2 IJ Hr K
F2r0 = − U0 e−2
3 4 r0
F4r0 = − U0 e−4
5 16 r0
0
Similarly,
Work, Energy and Momentum
193
F10r0 = − U0 e−10
11 100 r0
Therefore we have
F2r0 Fr0 F4 r0 Fr0 F10r0 Fr0
=
3 −1 e ~ 0.14 ~ 1.4 × 10–1 8
=
5 −3 e ~ 0.0078 ~ 7.8 × 10–3 32
[3 e3 = 20. 09 ]
=
11 −9 e ~ 0.0000067 ~ 6.7 × 10–6 200
[3 e9 = 8108 ]
[3 e = 2. 718]
The result shows that the force is short range. Q.54. The potential energy of a body is given by U = 40 + 6x2 – 7xy + 8y2 + 32z where U is the joule and x, y, z in meter. Deduce the x, y, z components of the force on the body when it is in position (–2, 0, +5). Solution.
U = 40 + 6x2 – 7xy + 8y2 + 32z
The negative gradient of the potential energy with respect to the potential variables gives the intrinsic force. Therefore, the (x, y, z) components of the force at the position (–2, 0, +5) are given by Fx =
− ∂U = − 12 x + 7 y = + 24 Nt [putting x = –2, y = 0] ∂x
Fy =
− ∂U = 7 x − 16 y = − 14 Nt [putting x = –2, y = 0] ∂y
Fz =
− ∂U = − 32 Nt ∂z
Q.55. If the potential in a plane is given by U = 4x2 – 10xy + 2y2, deduce the x and y components of field at the point x = 2, y = 4. Solution.
U = 4x2 – 10xy + 2y2
The x and y components of the field are the negative gradients of U with respect to x and y respectively. Thus
and
Fx =
− ∂U = − 8 x + 10 y ∂x
Fy =
− ∂U = 10 x − 4 y ∂y
at x = 2, y = 4 we have Fx = 24 and Fy = 4. Q.56. The electric potential in a region of space is given by V = 5x – 7x2y + 8y2 + 16yz – 4z volt.
194
Mechanics →
Deduce an expression for the electric field E . Calculate the y-component of the field at the point (2, 4, –3). V = 5x – 7x2y + 8y2 + 16yz – 4z volt
Solution.
...(i)
→
The electric field E may be written in terms of its cartesian components as →
E = Ex i + E y j + E z k Since the electric field in a direction is the negative potential gradient in that direction, we may write →
E = −
∂V ∂V ∂V i − j− k ∂x ∂y ∂z
...(ii)
By partial differential of eq. (ii) we get
∂V ∂V ∂V = − 7 x2 + 16 y + 16 z and = 16 y − 4 = 5 – 14 xy, ∂y ∂z ∂x Putting these values in eq. (ii), we obtain →
E = ( −5 x + 14 xy ) i + (7 x2 − 16 y − 16 z) j + ( −16 y + 4) k This is the required expression. The y-component at the point (2, 4, –3) is Ey = 7x2 – 16y – 16z = 12 volt/meter. Q.57. The electric potential in a system is given by V(x, y, z) = 20 + 6x2 – 5xy + 4y2 + 3z2 joule/coulomb. where x, y, z are in meter. Deduce in i, j , k notation the force on a 2 × 10 – 15 coulomb charge placed at position (2, 0, –3) meter. Solution. The electric field is the negative derivative of the potential and is given by →
E =
− ∂V ∂V ∂ V i− j− k ∂x ∂y ∂Z
= − (12 x − 5 y) i − ( −5 x + 8 y) j − 6 zk Nt/Coul. At position (2, 0, –3) meter, the field is therefore →
E = − (24 − 0) i − ( −10 + 0) j − ( − 18 ) k or
→
E = −24 i + 10 j + 18 k Newton/Coulomb The force on a charge q = 2 × 10–15 coulomb at this point is →
→
F = q E = ( −48i + 20 j − 36 k ) × 10−15 Newton This is the required expression.
Work, Energy and Momentum
195
Q.58. A scalar potential field is given by V = 6x + 8y – 12xy2 + 7yz2 – 5y2 joule/coul. where x, y, z are in meters. Calculate (a) the work done on a 5 coulomb charge in moving it from position (2, 0, 0) to (2, 5, 0). (b) the force on a 4-coulomb charge placed at the origin (0, 0, 0). Solution. (a) The potential at positions (2, 0, 0) and (2, 5, 0) are obtained by the given expression as V2, V2,
0, 0
= –12 joule/coulomb and V2,
5, 0
= –697 joule/coulomb
Therefore the work done in carrying a charge q = 5 coulomb from a point V2, 5, 0 is given by (work = charge × potential difference) W = q (V2,
0, 0
~ V2,
0, 0
5, 0)
= 5 × 685 = 3425 Joule (b) The expression for the electric field is →
E = −
∂V ∂V ∂V i− j− k ∂x ∂y ∂z
= − ( 6 − 12 y2 ) i − ( 8 − 24 xy + 7 z2 − 10 y) j − 14 yz k At (0, 0, 0) we have →
E0 = – 6i − 8 j Newton/coulomb.
The force on a charge q = 4 coulomb is therefore →
→
F = q E0 = − 24 i − 32 j Newton. The scalar magnitude of this force is F =
(24)2 + ( −32)2
= 40 Newton. Q.59. Show that for the same initial speed v0 , the speed v of a projectile will be the same at all points at the same elevation regardless of the angle of projection. Solution. In the absence of air resistance, the only force on a projectile is its weight, and the mechanical energy of the projectile remains constant. Fig. 26 shows two trajectories of a projectile with the same initial speed (and hence the same total energy) but with different angles of departure. Now, at all points at the same elevation the potential energy is the same; hence the kinetic energy is the same and so the speed is the same.
X
v0
v0 X
Fig. 26
to
196
Mechanics
Q.60. The bob of a 2m long pendulum has mass m = 0.5 kg. It is pulled to a side to make θ = 30° angle with the vertical. Calculate the change in potential energy, the work done and the speed of the bob when it passes the lowest point after being released. Solution. As the bob of mass m is moved from A to B, it rises through the height h, where h = CA = OA – OC = l – l cos θ = l (1 – cos θ) Thus the gain in potential energy is U = mgh
O θ l
l co s θ l C h
m B
m A
Fig. 27
= mgl (1 – cos θ) Here m = 0.5 kg, l = 2 meter and cos θ = cos 30° = 0.866 ∴
U = 0.5 × 9.8 × 2 (1 – 0.866) = 1.31 Joule
All the work done is stored as potential energy. Therefore Work done = 1.31 Joule On being released at B, when bob passes the lowest point A, it has lost all its potential
1 mv2 , where v is the velocity at A. No work is done 2 by the tension in the string because it is always perpendicular to the displacement. Hence by the conservation of mechanical energy, we have energy which appears as kinetic energy
mgh = or
v =
1 mv2 , 2 2 gh
Here h = l (1 – cos θ) = 2 (1 – 0.866) = 0.268 meter ∴
v =
2 × 9. 8 × 0. 268 = 2. 3 m/s
Q.61. (a) The bob of a simple pendulum of length l is released from a point in the same horizontal line as the point suspension and at a distance l from it. Calculate the velocity of the bob and the tension in the string at the lowest point of its swing. (b) If the string of the pendulum is catched by a nail located vertically below the point of suspension and the bob just swings around a complete circle around the nail, find the distance of the nail from the point of suspension.
Fig. 28
(c) If the string of the pendulum is made of rubber then show that it will stretch by 3 mg/ K (where K is the force constant) on reaching the bob at the lowest point.
Work, Energy and Momentum
197
Solution. (a) Let S be the point of suspension and A the point from which the bob of mass m is released (SA = l). The point A is at a vertical height l above the lowest point B of the swing. Hence is the potential energy mgl. On being released, the bob swings along the dotted arc, reaching the lowest point B where its entire energy is kinetic,
1 mv2 (v being the 2
velocity at B). By the conservation of mechanical energy, we have mgl = ∴
v =
1 mv2 2 2gl
This is the velocity at the lowest point. As the bob moves toward B, the tension in the string increases and becomes maximum at B. If T be this (maximum) tension, the net vertically upward force on the bob is T-mg, and this provides the required centripetal force mv2/l to the swinging bob. Thus T – mg = But at B,
mv2 l
v 2 = 2gl (proved above). T = mg +
mv2 l
T = 3mg (b) Now, suppose a nail N is located at distance d vertically below S. As the bob reaches the point B, the swing of the pendulum is caught by the nail and the bob swings around a complete circle of radius r(say). C is the highest point of this circular swing. Clearly the bob will do so provided its velocity at C, say vc, is such that the required centripetal force
mv2c r
(downward) is provided by its entire weight mg. That is mv2c = mg r
or
v2c = gr
1 m ( v2 − vc2 ). This appears 2 as gravitational potential energy mg(2r) of the bob at the point C. Thus
The decrease in kinetic energy as the bob goes from B to C is
1 m ( v2 − vc2 ) = mg (2r) 2
Substituting the values of v2 and v2c from above, we get 1 m (2 gl − gr) = mg (2r) 2
2gl – gr = 4gr
198
Mechanics
2gl = 5gr r =
2 l 5
∴ The distance of the nail from the point of suspension is d = l – r = l –
2 3 l= l 5 5
(c) If the string is made of rubber, then on reaching the bob at B the string is stretched by an amount ∆l (say). The string would therefore experience an upward elastic force K(∆l). This minus the (downward) weight mg would provide the centripetal force mv2/l at B. Thus K(∆l) – mg =
mv2 l
B; v2 = 2gl
at ∴
m(2 gl) = 2mg l ∆l = 3mg/K.
K(∆l) – mg =
Q.62. A light meter stick, pivoted about a horizontal axis through its centre, has a 2kg body attached to one end and a 1 kg body to the other. The system is released from rest with the stick horizontal. What is the velocity of each body as the stick swings through a vertical position ? Solution.
Fig. 29
Initially the bodies are at A and B, having gravitational potential energies mgh only, where h is the vertical height from the lowest point A'(h = 0.5 meter). The mechanical (potential) energy of the system is = 2 × 9.8 × 0.5 + 1 × 9.8 × 0.5 = 3 × 9.8 × 0.5 Joules On being released, the heavier body comes down in the position A' and the lighter one 1 mv2 goes upto B'. Let v be the velocity of each. Now the body at A' has kinetic energy 2
FG H
IJ K
Work, Energy and Momentum
199
FG 1 mv IJ and the potential energy (mgh); H2 K h being now 1 meter. Thus the mechanical energy of the system 1 F1 I × 2 × v + G × 1 × v + 1 × 9. 8 ×1J = H K 2 2 alone, but the body at B' has both the kinetic energy
2
2
2
3 2 v + 9. 8 2 By the law of conservation of mechanical energy, we have 3 2 3 × 9.8 × 0.5 = v + 9. 8 2 3 2 v = 3 × 9.8 × 0.5 – 9.8 = 0.5 × 9.8 or 2 2 × 0. 5 × 9. 8 = 1. 81 m/s ∴ v = 3 Q.63. A light rod of length l and with a mass m attached to its end is suspended vertically. It is turned through 180° and then released. Calculate the velocity of the mass and the tension in the rod when the mass reaches at lowest point. =
If the system be released with the rod horizontal, at what angle from the vertical the tension in the rod would be equal to the weight of the body ? A Solution. Let S be the point of suspension and A the point from which the mass m is released. The point A is at a vertical height 2l above the lowest point B of the swing. Hence at A the mass has gravitational potential energy mg(2l). On being released, the mass swing along the dotted semicircle, reaching the lowest point B where its energy is
1 entirely kinetic, mv2 (v being the velocity at B). By energy 2 1 conservation, we have mg (2l) = mv2 2 ∴
v = 2 gl
S T
l
B v mg
Fig. 30
As the mass is moving in a circle or radius l, it must be acted upon by a centripetal force. At the point B, the net upward force on the mass is T-mg, where T is the vertical tension in the string. This provides the required centripetal force. Thus mv2 l = 4gl
T – mg = But
v2
4 mgl = 5 mg l Now, suppose the mass is released from the point A'. On reaching the point B', it descends vertically through a distance l cos θ, thus losing potential energy by mgl cos θ. It, 1 mv′2 , where v' is the velocity at B'. however, gains kinetic energy 2 ∴
T = mg +
200
Mechanics
Thus
l
1 mv′2 mgl cos θ = 2 ∴
S
T – mg cosθ =
B′ θ m g co s θ m g
B
Fig. 31
mv′2 l
(2 gl cos θ) = 3 mg cos θ l Let the angle θ be such that T = mg, then we have mg = 3mg cos θ cos θ =
1 3
θ = cos−1
∴
FG 1 IJ = 71° H 3K
v1
Q.64. A ball is tied to a cord and set in rotation in a vertical circle prove that the tension in the cord at the lowest point exceed that at the highest point by six times the weight of the ball. Solution. Let v1 be the velocity of the ball as, it passes the highest point. The force acting on it are its weight mg and the tension T1 in the cord both acting downward. The resultant force is thus T1 + mg, which provides the centripetal force R
l co s θ
T = mg cos θ + m
or
mv12
θ
T
v′ 2 = 2gl cos θ
At the point B', the net radically inward force on the mass is T – mg cosθ, which provides the required centripetal force mv′2/l. Thus
or
A′
T1
mg R O T2
v2 mg
Fig. 32
. Thus
mv12 R Similarly, at the lowest point we shall have T1 + mg =
mv22 R m 2 (v − v12 ) + 2 mg T2 – T1 = ...(1) R 2 At the highest point, the ball has gravitational potential energy mg (2R) and the kinetic 1 1 mv12 . At the lowest point it has entirely kinetic energy mv22 . By the law of energy 2 2 conservation of energy, we have
T2 – mg =
mg (2R) +
1 1 mv12 = mv22 2 2
Work, Energy and Momentum
or
201
m 2 (v − v12 ) = 4 mg R 2 Substituting this value in eq. (1) we have T2 – T1 = 4 mg + 2 mg = 6 mg.
Q.65. A body slides down a curved frictionless track which is one quadrant of a circle of radius R. Find its speed at the bottom of the track. Point out the usefulness of the energy method of solving dynamical problems. Will the conclusion regarding the speed at the bottom hold if friction were present ?
1
m
N
2 mg
R e feren ce
Ans. Motion on a frictionless inclined le vel plane: Let m be the mass of the body which Fig. 33 starting from rest at the point 1, slides down a frictionless curved track to the point 2. As there is no friction the forces acting on the body are (i) its weight mg, and (ii) the normal force N exerted on it by the track. Since the force N is always normal to the direction of motion of the body, it does no work. In other words, only the gravitational force mg does work. Since the force is conservative, the mechanical energy of the body is conserved. Thus we can write. E = K1 + U1 = K2 + U2
1 mv 2 + mgh = 1 mv2 + mgh 2 2 1 1 2 2 Here v1 = 0, h1 = R, h2 = 0, v2 = ? O + mgR = v =
1 mv22 + 0 2 2gR
The speed is therefore the same as if the body had fallen vertically through a height R. This proves the fact that the work done by a conservative force in moving a body through a distance is independent of the path chosen. Here the resultant force acting on the body, and hence the acceleration depends on the slope of the track at each point, i.e., it varies from point to point. Thus the acceleration is variable. As such we cannot use the Newton’s equation of motion which holds for constant acceleration. Hence to solve the problem by other methods we shall have to determine the acceleration at each point and then applying integration. All this is avoided in the energy method. We have seen above that the speed at the bottom is independent of the shape of the surface. This conclusion would not hold if friction were present. This is so because the work of the friction force does depend on the path; the longer the path, the greater the work. Q.66. A body of mass 0.5 kg. starts from rest and slides vertically down a curved track which is in the shape of one quadrant of a circle of radius 1 met. At the bottom of the track the speed of the body is 3 m/s. What is the work done by the frictional force? Solution. Let m be the mass and R the radius of the circle. When the mass is at rest on the top of the track, its energy is entirely potential being equal to mg R. On reaching at
202
Mechanics
1 mv2 . Let the work done against 2 the frictional force be Wf. Then, by the conservation of total energy. We have the bottom the energy is entirely kinetic, being equal to
mgR =
1 mv2 + Wf 2
Wf = mgR −
1 mv2 2
= (0.5 × 9.8 × 1) –
1 × 0.5 × (3)2 2
= 2.65 Joules Q.67. A 0.5 kg block is released from rest at a point on a track which is one quadrant of a circle of radius 4m. It slides down the track and reaches its bottom with a velocity of 6 m/s. Then it further slides a distance of 9 m on a A level surface and stops. How much work is done against friction in sliding on the circular track and what is the 4m 4m coefficient of sliding friction on the horizontal surface? [Ans. 10.6 Joule, 0.20] B
C
Q.68. A body slides along a track with elevated 9m ends and horizontal flat central part. The flat part has Fig. 34 a length l = 3met. The curved portions of the track are frictionless. For the flat part the coefficient of kinetic friction is µk = 0.20. The body is released at point A which is at a height h = 1.5 meter above the flat part of the track. Where does the particle finally comes to rest. Solution. Suppose the body is released form rest at A where it has gravitational potential energy. It reaches B with kinetic energy which then carries it to C against friction. From C it rises up to D (say) where the kinetic energy in it at C is once again converted into potential energy. It then again descends to C, goes to B rises up to a point E and this process continue until finally it comes to rest some where on the horizontal part of the track.
Fig. 35
The initial (potential) energy of the body at the point of release A is mgh. Since the curved parts are frictionless energy is not spent up in moving the body on them. Energy is spent up in doing work against the friction only when the body moves on the horizontal part. The friction force against the direction of motion is fk = µN = µmg Suppose whole of the initial energy, mgh, is consumed in moving a distance d on the horizontal part, the work done being. W = fk × d = µ mgd
Work, Energy and Momentum
203
By the law of energy conservation, we write mgh = W = µmgd ∴
d =
h µ
Here h = 1.5 meter and µ = 0.20
1. 5 = 7. 5 meter 0. 2 Thus the body will cover a distance of 7.5 m on the horizontal part which is 3.0 meter long. This means that it will come to rest at a point F such that d =
BC + CB + BF = 7.5 3 + 3 + BF = 7.5 BF = 1.5 meter Thus the point F is the middle point of the horizontal part where the body would come to rest. Q.69. A 3000 kg automobile at rest at the top of an incline 30 m high and 300 m long is released and rolls down the hill. What is its speed at the bottom of the incline if the average retarding force due to friction is fk = 200 g ? Solution. Let m be the mass of the automobile, h the height of the incline, x the length and v the velocity at bottom. The potential energy of the automobile when at rest at the top of incline is mgh. The kinetic energy at the bottom is
1 mv2 , and the energy dissipated 2
against the friction is fkx. By conservation of total energy, we have mgh =
1 mv2 + fkx 2
v 2 = 2gh –
2x fk m
= (2 × 9.8 × 30) –
FG 2 × 300 × 200 × 9.8IJ H 3000 K
= 196 v = 14 m/s. Q.27. A block of mass m slides down at 30° inclined plane of length l and coefficient of friction µ. With what speed it will reach the bottom. If it further slides on a similar horizontal surface, how far will it go before coming to rest ?
204
Mechanics
Solution.
µN
N
l
θ mg
l sin θ
N µN
θ
mg
Fig. 36
The weight of the block, mg, acts vertically downwards. Let N be the normal reaction exerted by the plane of the block. Then
N = mg cos θ
The friction force opposing its sliding down will be fs = µN = µmg cos θ Let v be the velocity of the body on reaching the bottom of the plane. At the moment of start the vertical height of the block is l sinθ. Its potential energy is, therefore, mg (l sin θ).
1 mv2 to the block and in doing work against 2 the friction fs. Thus, by the conservation of total energy. We have This is used up in providing the kinetic energy
mgl sin θ = or or
1 1 mv2 + fs = mv2 + µ mgl cos θ 2 2
2gl sin θ – 2 µgl cos θ = v 2 v =
[2gl (sin θ − µ cos θ)].
1 mv2 . Suppose now it moves a 2 distance d on a horizontal surface before coming to rest. Thus the kinetic energy is used up in doing work against this friction. Now the friction force is µN = µmg and the work done is µmgd. Then
On reaching the bottom, the block has kinetic energy
1 mv2 = µmgd 2
or or
1 m [2 gl (sin θ − µ cos θ)] = µmgd 2 l (sin θ − µ cos θ) d = µ
Work, Energy and Momentum
205
Q.71. A 0.2 kg block is thrust up a plane inclined at an angle of 30° to the horizontal with an initial velocity of 5 m/sec. It goes up 2m along the plane and then slides back to the bottom. Calculate the force of friction and the velocity of the block with which it reaches the bottom of the plane. Solution. Let us first consider the upward motion. Let f be the force of friction. At the bottom where the block starts moving, the energy is wholly kinetic (K = 1/2 mv2). At the top where the block momentarily stops, the energy is wholly potential (U = mgh). The work done against the friction during this motion is f × x, where x is the distance travelled. By the general law of conservation of energy, we have K = U + fx or
h = 2 sin 3 0 °
3 0°
Fig. 37
1 mv2 = mgh + fx 2 Here, m = 0.2 kg, v = 5 m/s, x = 2 m, h = 2 sin 30° = 1. ∴
or
2m
1 × 0. 2 (5)2 = 0.2 × 9.8 × 1 + f × 2 2 2.5 = 1.96 + 2f
2. 5 − 1. 96 = 0.27 nt 2 Let us now consider the downward motion. Let v be the velocity of the block with which it reaches the bottom. Now the potential energy (U = mgh) of the block at the top is used f =
up in giving kinetic energy (K = 1/2 mv′2) to the block on its reaching the bottom and in doing work against the friction (f × x). Thus U = K + fx
1 mv′2 + fx 2 1 0.2 × 9.8 × 1 = × 0.2 × v'2 + 0.27 × 2 2 1.96 = 0.1 v'2 + 0.54 mgh =
or
1. 96 − 0. 54 = 14. 2 0.1
or
v' 2 =
or
v' 2 = 3.8 m/sec.
Q.72. A 12 kg block is pushed 20 m up the sloping surface of a plane inclined at an angle 37 to the horizontal by a constant force of 120 nt acting parallel to the plane. The coefficient of friction between the block and the plane is 0.25. Calculate (i) work done by F (ii) increase in potential energy of the block (iii) work done against the friction and (iv) increase in kinetic energy. What becomes the work done ? Solution. (i) The force F (= 120 nt) pushes to block through a distance x (= 20m). Hence the work done
206
Mechanics
W = Fx = 120 × 20 = 2400 Joule.
20
∆U = mgh
m F N
= 12 × 9.8 × 12 = 1411 Joule. (iii) The force of friction is µN, where µ is the coeff. of friction and N the normal force of reaction exerted by the plane on the block. But N = mg cos 37° (see Fig.). Thus the force of friction is µmg cos 37°. The work done against the force
37
µN
°
mg
h = 20 sin 3 7°
(ii) initially, the potential energy is zero, finally it become mg. Here m = 12 kg and h = 20 sin 37° = 20 × 0.6 = 12 m. Hence the increase in P.E.
3 7°
Fig. 38
W = force × displacement = µmg cos 37° × x = 0.25 × 12 × 9.8 × 0.8 × 20 = 470.4 joule This work is converted into heat. (iv) Initially, the kinetic energy is zero. Let ∆K be the change in kinetic energy. By the law of conservation of energy W = ∆U + ∆K + w ∆K = W – ∆U – w = 2400 – 1411 – 470 = 519 Joule Q.73. Figure 39 shows the vertical section of a frictionless surface. A block of mass 2 kg is released from position A. Compute its kinetic energy as it reaches positions B, C, D. (give gravitational field = 9.8 jm–1 kg–1). Solution. The (gravitational) force is conservative so that the mechanical energy is conserved. The loss in potential energy as the block comes down to B, C and D from A equals the corresponding gain in the kinetic energy ∴
K.E. at B = loss in P.E. between A and B (UA – UB) = mg (hA – hB) = 2 kg × 9.8 jm–1 kg–1 × (14 – 5) m = 176.4 Joule K.E. at C = UA – UC = mg (hA – hC) = 2 × 9.8 × 7.0 = 137.2 Joule K.E. at D = UA = mghA = 2.0 × 9.8 × 14 = 274.4 Joule
Work, Energy and Momentum
207
Fig. 39
Q.74. A small block of mass m slids along a frictionless track as shown in figure 40. If it starts from rest at P, what is the reaction exerted by the track on the block when it is at A, B and C. At what height from the bottom A of the loop should the block be released so that the force it exerts against the track at the top C is equal to its weight. P
C mg
5R NC O
NB
NA
A
B mg R
mg
Fig. 40
Solution. Let us first calculate the velocities at A, B and C. At P the entire energy is potential being equal to mg (5R) and at A it is entirely kinetic, being equal to
1 mvA2 . By 2
conservation of mechanical energy, we have mg (5R) = ∴
1 mv2A 2
v2A = 10gR
At B, the energy is partly kinetic
...(i)
1 mvB2 and partly potential mg(R). By energy conservation. 2
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Mechanics
mg(5R) = ∴
1 mvB2 + mg(R) 2
vB2 = 8gR.
...(ii)
Similarly, at C we have mg(5R) =
1 mvC2 + mg(2R) 2
vC2 = 6gR
...(iii)
Now, at any point on the track, the force acting on the block are its weight mg acting vertically downward and the reaction force N exerted by the track acting radially inward. The resultant radial force supplies the required centripetal force at that point.
mv2A R m = (10gR) R = 10mg + mg = 11mg.
NA – mg =
NA
[from eq. (i)]
For the point B, we have mv2B R m (8 gR) = 6mg = R = 8 mg
NB =
[from eq. (ii)]
Again for the point C, we write mv2c R m (6 gR) = 6 mg = R = 6mg – mg = 5mg.
NC + mg =
∴
NC
...(iv) [from eq. (iii)]
Now, let h be the height from which the block must be released so that NC equals mg. Then from eq. (iv), the velocity vc must be given by mg + mg = or
mvC2 R
vC2 = 2gR By energy conservation at C, we have
1 mvC2 + mg(2R) 2 1 = m(2gR) + mg(2R) 2 = 3mgR
mgh =
h = 3R
Work, Energy and Momentum
209
Q.75. A body is allowed to slide on an inclined frictionless track from rest position under earth’s gravity. The track ends in a circular loop of radius R. Show that the minimum height 5 R. h of the body, so that it may successfully complete the loop is given by h = 2 P
v
C N mg
h
O
R
B
A
Fig. 41
Solution. Let us first calculate the velocity v of the body which it acquires at the highest point C of the loop, when released from a height h. For this we use conservation of mechanical energy. At point P, the body is at rest so that its mechanical energy is entirely potential and is equal to mgh At point C it has a velocity v and a height 2R, so that it has kinetic energy equal 1 to mv2 as well as potential energy equal to mg(2R). By energy conservation the mechanical 2 energy at P must be the same as at C because the entire track is frictionless. That is,
1 mv2 + mg(2R) 2 1 mg(h – 2R) = mv2 2 v 2 = 2g(h – 2R) mgh =
or or
...(i)
Now, the forces acting on the body at C are its weight mg acting vertically downward and the reaction force N exerted by the track which is directed toward the centre O of the circle. In fact, at the point C both are directed toward O. Thus the resultant radial force at C is N + mg, and this supplies the required centripetal force mv2/R. That is mv2 R Since N cannot be negative, the minimum velocity of the body at C if it is to describe the circle must correspond to N = 0 so that
N + mg =
mv2min R = gR
mg = or or
v2min
vmin =
...(ii)
gR
If the velocity is less than g R , the downward pull of the weight will be larger than the required centripetal force and the body will lose contact from the loop. (The velocity can however be grater than g R ).
210
Mechanics
Now from eq. (i) v2min = 2g (hmin – 2R). Putting this value in eq. (ii) we get gR = 2g(hmin – 2R)
R 5 R + 2R = 2 2 Q.76. A small body of mass m slides without friction around the loop the loop apparatus, starting at a height 3R above the bottom of the loop, where R, is the radius of the loop. Compute its radial acceleration at the end of a horizontal diameter of the loop. From what minimum height above the bottom of the loop it should start so that it may loop the loop? or
hmin =
Fig. 42
LM Ans. 4 g, 5 R OP 2 Q N
Q.77. A body is moving on a vertical circular frictionless track. Calculate the minimum velocity which it should have at the lowest point of its path in order to go completely round the track. Solution. Let v1 be the velocity at the highest point of the path, and v2 the corresponding velocity at the lowest point. In order that the body does not leave the track at the highest point (where it is most likely to do so), the magnitude of V1 must be atleast such that the centripetal force mV12/R at that point utilises the entire weight mg of the body. That is mV12 = mg R
or
V12 = gR
Fig. 43
As the body slides down from the top to the bottom of the track, it loses potential energy by an amount mg(2R) but gains an equivalent amount of kinetic energy
1 m (V22 – V12) 2 = V12 + 4gR
mg(2R) = or
V22
1 m (V22 – V12). Thus 2
Work, Energy and Momentum
But
211
V12 = gR (proved above) V22 = gR + 4gR = 5gR
or
V2 =
( 5 gR )
Q.78. A particle of mass m = 0.2kg is moving inside a smooth vertical circle of radius R = 50 cm. If it is projected horizontally with velocity v0 = 4m/s from its lowest position find the angle θ at which it will lose contact with the circle. Solution. As the particle moves up the circular track, its velocity decreases. Let v be the velocity at the point P. In this rise the particle gains gravitational potential energy by an amount, mg (R + R sinθ), while it loses an equivalent amount of kinetic energy given by
1 m 2
(v02 – v2). By energy conservation, we have
1 m (v02 – v2) 2 = v02 – 2gR (1 + sin θ)
mg (R + R sinθ) = or
v2
...(i)
Let us now consider the forces acting on the particle at the point P. These are particle's weight mg acting vertically downward and the normal reaction N exerted by the track acting radially inward. The net force along the radius is N + mg sinθ which supplies the necessary centripetal force mv2/R. That is, N + mg sin θ = mv2/R. The particle will loose contact for that value of θ for which N = 0. Thus mg sin θ =
mv2 R
Putting the value of v2 from eq. (i) we get g sin θ = =
v02 − 2 gR (1 + sin θ) R 2 v0
R
v
– 2g – 2g sin θ m
2
or
3g sin θ =
v0 – 2g R
or
sin θ =
v02 2 − 3gR 3
Here v0 = 4 m/s and R = 0.5 m
g
s co
θ
P
θ mg
m g sin θ N R sin θ R
θ O
R
R
Fig. 44
(4 m/s)2 2 − sin θ = 3 ( 9. 8 m/s2 ) (0. 5 m) 3 = 1.09 – 0.67 = 0.42 or
θ = sin–1 (0.42) = 25°.
Q.79. A small mass m starts from rest and slides down the smooth surface of a solid sphere of radius R. Assume zero potential energy at the top. Find (a) the change in potential
212
Mechanics
energy of the mass with angle (b) the kinetic energy as a function of angle (c) the angle at which the mass flies off the sphere. If there is friction between the mass and the sphere, does the mass fly off at a greater or lesser angle ? Solution. (a) The potential energy of the mass m at A is zero. As it slides down to P, it descends vertically through. AC. Hence the loss in its potential energy is given by A A
N
C
m
P θ
θ
R
m
g
co
O
s
P
θ
θ m g sin θ mg
R B B
Fig. 45
U = –mg (AC) = –mg (AO – CO) = –mg (R – R cos θ) = –mg R (1 – cos θ) (b) This lost energy appears as gain in kinetic energy K, as the surface is smooth therefore K = + mgR (1 – cos θ) (c) The net radially inward force on the mass m at P is mg cos θ – N, where N is the normal reaction of the surface. This supplies the centripetal force, so that mg cos θ – N =
mv2 R
where v is the velocity at P the mass will fly off the surface at angle θ for which N is zero. i.e., mg cos θ =
mv2 R
1 mv2 , which is mgR (1 – cos θ), as shown above. Substituting 2 this in the last expression, we get The kinetic energy at P is
mg cos θ = 2mg (1 – cos θ) or
3mg cos θ = 2mg
or
cos θ =
or
2 3
θ = cos–1
FG 2 IJ . H 3K
Work, Energy and Momentum
213
In presence of fiction, the velocity at P will be smaller and hence the mass will fly off the surface at a large angle. Q.80. What will be the spring constant if it stretches 10 cm where it has a potential energy of 5600 joule ? Solution. The potential energy of a spring stretched through a distance is given by
1 Kx2 2 Hence the spring constant is given by U =
K =
2U x2
Here U = 5600 joule and x = 10 cm = 0.1 m. K =
2 × 5600 = 1.12 × 106 nt/m (0.1)2
Q.81. A spring-gun has a spring constant of 80 Nt/cm. The spring is compressed 12 cm by a ball of mass 15 gm. How much the potential energy of spring ? If the trigger is pulled, what will be the velocity of the ball be ? Ans. The spring (elastic) force is conservative and hence the mechanical energy of the system is conserved. Before the trigger is pulled the spring has an elastic potential energy of compression given by
1 2 Kx 2 where K is spring-constant and x is the distance of compression. U =
Here K = 80 nt/cm = 8 × 103 nt/m and x = 12 cm = 0.12 m ∴
U =
1 × (8 × 103) (0.12)2 = 57.6 joule 2
This energy, when the trigger is pulled is converted into the kinetic energy
1 mv2 of the 2
ball, where m is the mass and v the velocity of the ball. Thus
1 mv2 = 57.6 2 Here m = 15 gm = 0.015 kg ∴
v2 =
∴
v =
2 × 57. 6 2 × 57. 6 = = 7680 m 0. 015 7680 = 87.6 m/s
Q.82. A mass of one kg suspended by a spring of force constant 8 × 104 dyne/cm. Find the distance through which it should be pulled so that on releasing it passes through its equilibrium position with a velocity of 1.0 m/s. [Ans. 11.2 cm]
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Mechanics
Q.83. A block of mass 1 kg is forced against a horizontal spring of force constant K = 100 nt/m. Which is compressed by x1 = 0.2 m. When released, the block moves on a level surface a distance x2 = 1.0 m before coming to rest. Obtain the coefficient of friction µk between the block and the surface.
Fig. 46
1 Kx12 2 stored in the compressed spring. This energy is used up in doing work against the friction force µKN= µKmg, in moving the body a distance x2. Thus Solution. When the block is released, it receives the (elastic) potential energy
1 Kx12 = µKmg × x2 2 1 × 100 × (0.2)2 = µK × 1 × 9.8 × 1.0 2
or ∴
µK = 0.20
Q.84. A 2kg body moving on a level surface collides and compresses a horizontal spring of force constant k = 2 nt/m through 2 m. Compute the velocity of the body while colliding. (Between block and surface µK = 0.25).
Fig. 47
Solution. The kinetic energy of the block,
1 mv2 supplies the elastic potential energy, 2
1 2 Kx , to the spring and does work against the friction force µKN = µKmg, as the block moves 2 the distance x. Thus, by the general law of energy conservation, we have
1 1 mv2 = Kx2 + µKmgx (g = 9.8 m/s2) 2 2 1 1 × 2 × v2 = 2 × 2 × (2)2 + 0.25 × 2 × 9.8 × 2 2 v 2 = 4 + 9.8 = 13.8 v =
13. 8 v = 3.7 meter/sec.
Q.85. A 10 kg block slides from the top of a 30° inclined smooth plane and compresses a spring placed at the bottom of the plane through 2.0 meter before coming to rest. Calculate
Work, Energy and Momentum
215
the distance through which the block has slide before coming to rest and its speed on reaching the bottom. The spring can be compressed 1.0 meter by a force of 100 Newton. Ans. When a force F compresses a spring by x, we have F = kx where k is the force constant of the spring. S s sin 30 °
30 °
30°
Fig. 48
The given spring can be compressed by 1.0 meter by a force of 100 newton. That is 100 = K × 1 ∴
K = 100 Nt/m
Let s the distance through which the block has slide before coming to rest. The vertical distance through which it descends is s sin 30°. Now, at the moment of start, the block has gravitational potential energy, the kinetic energy being zero. At the moment the maximum compression of the spring occurs, again there is no kinetic energy. Hence the loss of gravitational potential energy of the block in sliding down, mg s sin 30°, equal to the gain of (elastic) 1 potential energy of the spring, Kx2 . That is 2 1 mg s sin 30° = Kx2 2 Here m = 10 kg, g = 9.8 m/s2 , K = 100 Nt/m and x = 2.0 m
s 1 = × 100 × 4 2 2 100 × 4 = 4.1 meter. s = 10 × 9. 8 This distance of 4.1 meter includes the distance of 2.0 meter travelled while compressing the spring before coming to rest. Thus the distance travelled on reaching the spring (but before compressing) is 2.1 meter. Up to this moment, the loss in potential energy of the block 1 2 is mg × 2.1 sin 30° which appears as kinetic energy of the block mv . Thus 2 1 2 mg × 2.1 sin 30° = mv 2 1 1 2 9.8 ×2.1 × = v 2 2 v = 9. 8 × 2.1 ∴
10 × 9.8 ×
v = 4.5 meter/sec.
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Mechanics
Q.86. A 20 kg block of mass, initially at rest is dropped from a height of 0.40 meter on to a spring whose force constant is 1960 Nt/meter. Find the maximum distance that the spring will be compressed. Ans. At the moment of release, the block has only gravitational potential energy. At the moment of maximum compression of the spring, the block loses some gravitational potential energy which is converted into the elastic potential energy of the spring. Let m be the mass of the block, and K the force constant of the spring. Let y be the distance through which the spring
Fig. 49
1 2 Ky . The total vertical fall 2 of the block is (h + y), so that the loss in its gravitational potential energy is mg(h + y). By the law of conservation of mechanical energy, we have is compressed. Then the elastic potential energy in the spring is
1 2 Ky = mg(h + y) 2 y2 –
or
∴
...(i)
2mg 2mgh y – = 0 K K
y =
LM MN
1 2mg ± 2 K
FG 2mg IJ HKK
2
+
8mgh K
OP PQ
...(ii)
The block of mass 20.0 kg and is dropped from a height of 0.40 meter. The force-constant K = 1960 newton/meter. On solving for y, we get y = 0.4 meter Q.87. A 20 kg body is released from rest so as to slide in between vertical rails and compress a vertical spring of force constant K = 1920 Nt/m, placed at a distance h = 1.0 meter from the starting position of the body. The rails offer a friction force of f = 36 Nt opposing the motion of the body. Find (i) the velocity v of the body just before striking with the spring (ii) the distance y through which the spring is compressed, and (iii) the distance h through which the body is rebounded off.
Fig. 50
Ans. (i) The body slides a distance h against the friction force f just before striking the spring. Hence, the loss of gravitational potential energy of the body equals the gain of kinetic energy plus the work done against the friction of f. That is, mgh = 20 × 9.8 ×1.0 =
1 mv2 + fh 2 1 × 20 × v2 + 36 × 1.0 2
Work, Energy and Momentum
217
196 − 36 = 16.0 10 v = 4.0 m/s
v2 = ∴
(ii) At the moment when maximum compression y of the spring occurs, there is no kinetic energy. Hence, the loss of gravitational potential energy of the body equals the gain of elastic potential energy of the spring plus the work done against the friction force f. Now the total fall of the body is (h + y). Thus
1 2 Ky + f (h + y) 2 1 20 × 9.8 × (1.0 + y) = × 1920 × y2 + 36 (1.0 + y) 2 160 (1.0 + y) = 960 y2 or 6y2 – y – 1 = 0 mg (h + y) =
1 ± 1 + 24 12
or
y =
or
y = 0.5 (– sign is inadmissible).
1 2 Ky rebounds the body to a height h'. 2 Therefore the loss of elastic energy equals the gain of gravitational energy of the body plus the work done against the friction force f. That is. (iii) The compressed spring with elastic energy
1 2 Ky = mgh' + fh' 2 1 × 1920 × (0.5)2 = 20 × 9.8 × h′ + 36h' 2 240 = 232 h' 240 = 1.03 m. 232 Q.88. A body of mass 4kg slides on a horizontal frictionless table with a speed of 2 m/ s. It is brought to rest in compressing a spring in its path. By how much is the spring compressed if the force constant of the spring is 64 kg/m. h' =
Ans. Kinetic energy of the body
1 1 mv2 = × 4 × 22 = 8 Joule. 2 2 Let the spring be compressed through a distance x so that the body loses its all kinetic energy and comes to rest. Obviously, the lost kinetic energy will be stored in the form of =
potential energy
1 2 Cx of the system. Thus, 2 1 2 1 Cx = mv2 = 8 Joule 2 2 16 16 1 = = x2 = C 64 4
whence, x = 0.5 meter.
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Mechanics
Q.89. A body of mass 2 kg is attached to a horizontal spring of force constant 8 N/m and then a constant force of 6 newtons is applied on the body along the length of the spring. Find the speed of the body when it has been displaced through 0.5 met. Now if the force is removed, how much farther the body will move before coming to rest? Ans. Work done by constant force 6 Newton in displacing the mass through 0.5 meter, W = 6 × 0.5 = 3 Joule. The potential energy of the system at this displacement
1 1 Cx2 = × 8 × (0.5)2 = 1 Joule 2 2 The work done W on the system increases the potential energy and kinetic energy of the system. If the speed acquired by the body is v m/sec, then U =
W =
1 1 Cx2 + mv2 2 2 1 × 2 × v2 2
3 = 1+ v =
2 = 1.4 m/sec.
At the displacement 0.5 meters, the total energy of the system is 3 joule. Now, if the constant force is removed, only 3 joule energy will remain with the system. Now if the body is further displaced through a distance α against the elastic force of the spring so that it comes to rest, the total energy will become in the form of potential energy i.e.,
1 8 C (0.5 + α)2 = 3 or (0.5 + α)2 = 3 2 2 0.5 + α =
or
3 / 2 = 0.866
α = 0.366 meter.
or
Q.90. A body of mass m is suspended from a spring. It comes to rest after a downword displacement x0. If a linear force is acting, then prove that (i) the spring (force constant) is mg/x0. (ii) the gravitational energy lost is mgx0. (iii) the elastic potential energy gained is
1 mgx0 . 2
What happened to the remaining energy ? Ans. In the equilibrium position, the downward force mg on the spring will be balanced by the restoring force in the spring. So that mg = –Restoring force = –(–Cx0) = Cx0 whence
C = mg/x0
...(i)
Loss in gravitational potential energy = mgx0 Elastic potential energy =
z
x0
0
Cx dx =
...(ii)
1 2 1 Cx0 = mgx0 2 2
...(iii)
Work, Energy and Momentum
219
1 1 mgx0 = mgx0 ...(iv) 2 2 This energy is first changed to kinetic energy and finally to heat, when initial oscillations stop. Remaining energy = mgx0 –
Q.91. A car carries a framework ABCD shown below, in which a 200 gm mass P is supported between two spring of force constant 5 N/m each. Side AB is kept horizontal and along the length of the car. And the pointer attached to P reads zero when the car is at rest. What will the pointer read when the car has: (i) uniform speed 20 m/s on a straight road. (ii) uniform speed 10 m/s on a circular road of radius 20 m (iii) uniform acceleration 0.5 m/s2 on a straight road. (iv) uniform acceleration –1.0 m/s2 on a straight road.
Fig. 51
Ans. (i) When the car is moving with constant speed on a straight road, then pointer will read zero. (ii) When the car is moving on a circular path, the force will be perpendicular to the spring and hence again the pointer will read zero. (iii) In the case when car has forward acceleration 0.5 m/s2 the mass P will experience a force F = ma = –0.2 × 0.5 = –0.1 newton. As the both spring exerts the force on the mass P, the force at the displacement x is F = – (C1 + C2)x i.e., 0.1 = (5 + 5)x or x =
0.1 = 1 cm = 10 mm 10
Hence, the pointer will read + 10 mm back to the zero position of the scale. (iv) In this case, or
F = 0.2 × 1 = –(5 + 5)x
0. 2 m = –2cm = – 20 mm 10 Hence the pointer will read –20mm forward to the zero of the scale. x = −
Q.92. A body of mass 1 kg, initially at rest is dropped from a height of 2 meters on to a vertical spring having force constant 490 N/m. Calculate the maximum compression. Ans. Loss in gravitational potential energy = Gain in potential energy by the spring i.e.,
mg(h + x) =
1 2 Cx 2
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Mechanics
Here, m = 1 kg, g = 9.8 m/s2, h = 2m, C = 490 N/m. ∴
1 × 9.8 (2 + x) =
490 2 x 2
100x2 = 4x + 8
or
25x2 – x – 2 = 0
or whence
x =
1±
1 + 200 18.177 = 50 50
= 0.3035 m (–ve value is inadmissible)
SELECTED PROBLEMS 1.
Explain clearly the work-energy theorem.
2.
A force of 5 Nt acts on a body of 10 kg mass initially at rest. Compute the work done by the force in the third second and also the instantaneous power exerted by the force at the end of zero second.
3.
Define conservative force (a) show that for a conservative force, ∇ × F = 0 .
→
→
(b) Central force is conservative. 4.
A body of mass 5 kg is released from a position of rest on a frictionless spherical surface. It then moves on a horizontal surface CD whose coefficient of kinetic friction is 0.2. An elastic spring with the force constant K = 900 Nt/m is placed at C. Find the maximum compression of the spring ; (g = 9.8 m/s2).
5.
Prove that the rate of change of K.E. of a body is equal to the power exerted by the force acting on it.
6.
An electric field is given by E = i (2 x + 3 y) + j ( 5 x + 4 y). Find the scalar point.
7.
What is a conservative force ? Show that for a conservative force, the work done around (a) closed path is zero. (b) Show that a conservative force can be expressed as negative gradient of potential energy. (c) Prove that curl of a conservative force is zero.
8.
(i) A light and heavy body have equal momentum, which one has greater kinetic energy ? (ii) A body of mass m is moved to a height h equal to the radius of earth. What is the increase in potential energy ? (iii) Two particles whose masses are in the ratio 1 : 4 have equal momentum. What is ratio of their kinetic energies ? (iv) A 2000 kg motor car was running at a speed of 10 m/s. It was brought to a stop suddenly by breaking. What is the amount of heat produced at the breakers ? (v) A ball is dropped from rest at a height of 12 m, if it loses 25% of its kinetic
Work, Energy and Momentum
221
energy on striking the ground, what is the height to which it bounces ? How do you account for the loss in K.E. ? 9.
A ball falls under gravity from a height of 10 m with an initial downward velocity v0. It collides with the ground and loses 50% of its energy in the collision and then rises back to same height. Find (i) initial velocity v0 (ii) the height to which the ball would rise after collision if the initial velocity was directed upward instead of downwards.
10.
A bullet of mass m moving with a horizontal velocity ‘v’ strikes a stationary block of mass M suspended by a string of length L.
222
Mechanics
5 LINEAR AND ANGULAR MOMENTUM 5.1 CONSERVATION OF LINEAR MOMENTUM →
If a force F is acting on a particle of mass m, then according to Newton’s second law of motion, we have →
F = →
→
→ dp d = (m v ) dt dt
→
Where p = m v is the linear momentum of the particle. Momentum is a vector quantity. If the external force acting on the particle is zero, then →
dp F = dt = 0 or →
→
→
p = m v = a constant
Thus, in absence of an external force, the linear momentum of a particle remains constant. The law of conservation of momentum for a system of two particles, which are interacting mutually, in absence of external forces is given by →
→
p1 + p2 = constant
or
→
→
m1 v1 + m2 v2 = constant
Now, let us consider a system of n particles whose masses are m1, m2 ........, mn. The system can be a rigid body in which the particles are in fixed positions with respect to one another, or it can be a collection of particles in which there may be all kinds of internal motion. Suppose that the particles of the system are interacting with each other and are also acted by external forces.
If
→
→
→
→
→
→
p1 = m1 v1 , p2 = m2 v2 , ..., pn = mn vn are the momenta of the
FG PIJ H K →
particles of masses m1, m2, ..., mn respectively, then the total momentum is the vector sum of the momenta of individual particles i.e., 222
of the system
Linear and Angular Momentum
223 →
→
→
→
→
P = p1 + p2 + p3 + ... + pn
→
→
→
→
→
p1 = m1 v1 + m2 v2 + m3 v3 + ... + mn vn
Differentiating it with respect to time t, we have −→
−→
−→
−→
−→
dP dp1 dp2 dp3 dpn = + + + ... + dt dt dt dt dt −→
dp → → → = F1 + F 2 + ... + F n dt
or →
→
→
Where F1 , F 2 , ... F n represent the forces acting on the particles of masses m1, m2, m3 ... mn respectively. These forces include external and internal forces both. But according to Newton’s third law, the internal forces exist in pairs of equal and opposite forces, they balance each other and so do not contribute any thing to the total force. Hence the right hand side in above →
equation represents the resultant force F ext only due to the external forces acting on all the particles of the system. The internal forces cannot change the total momentum of the system, because being equal and opposite they produce equal and opposite changes in the momentum. Hence, if we want to change the total momentum of the system of the particles, it is necessary to apply external force on that system. Then the sum of external forces is −→
→ → dp d → = ( p1 + p2 + ... + pn ) F ext = dt dt If the resultant external force is zero, then →
−→
→ dp = 0 or P = a constant dt Thus, if the resultant external force acting on a system of particles is zero, the total linear momentum of the system remains constant. This simple but quite general result is called the law of conservation of linear momentum for a system of particles. It is to be noted that the momenta of individual particles may change, but their sum i.e., the total linear momentum remains unaltered in the absence of external forces.
The law of conservation of momentum is fundamental and exact law of nature. No violation of it has ever been found and it has been thoroughly checked by all kinds of experiments.
5.2 CENTRE OF MASS Every physical system has associated with it a certain point whose motion characterizes that of the system as a whole. When the system moves under an external force then this point moves in the same way as a single particle would move under the same external force. This point is called the “center of mass” of the system. The motion of the system can be described in terms of the motion of its center of mass.
224
Mechanics
Let us consider a system of n particles of masses m1, m2, m3, ... mn with position vectors → → →
→
→
r1 , r2 , r3 , ..., rn relative to a fixed origin O. The position vector r cm of the center of mass of this system is defined by →
→
r cm
→
→
m r + m2 r2 ... + mn rn = 1 1 m1 + m2 + ... + mn n
→
∑m r
i i
=
i=1 n
∑m
=
1 M
i
n
→
∑m r
i i
...(i)
i=1
i=1
Fig. 1
F I GG = ∑ m JJ H K n
where M
i
is the total mass of the system. This is the expression for the position
i=1
vector of the center of mass of a system of particles. →
Qualitatively, r cm represents a geometric point located at the “average” position of the particle weighted in proportion to their masses. From the above definition, we draw two conclusions: (i)
If all the n particles have the same mass m, then →
r cm
=
m M
n →
∑
i=1
ri =
1 n
n →
∑r
i
i=1
That is, the center of mass coincides with the geometric center of the system. →
(ii)
If the origin O is at the center of mass ( r cm = 0), then n
→
∑m r
i i
i=1
= 0
Linear and Angular Momentum
225
That is, the sum of the moments of the masses of the system about the center of mass is zero.
5.3 CARTESIAN COMPONENTS OF THE CENTRE OF MASS →
→
The position vectors r cm and ri are related to their Cartesian components by →
r cm = xcm i + ycm j + zcm k →
+ jy + kz ri = ix i i i
and
→
Making these substitutions in Equation (i), the Cartesian components of r cm are given by xcm
1 = M
n
∑m x , i i
n
n
∑
∑
1 1 mi yi , zcm = mi zi , Mi=1 Mi=1
ycm =
i=1
where xi, yi, zi are the Cartesian co-ordinates of the ith particle
5.4 CENTRE OF MASS OF A SOLID BODY Let us now consider a body with continuous distribution of mass. If the surface may be sub divided into n small elements, the ith element being of mass ∆mi and located approximately at the point (xi, yi, zi). The Co-ordinates of the center of mass are then approximately given by n
∑
xcm =
n
∆mi xi
i=1 n
∑ ∆m
, ycm =
∑
i=1 n
i
i=1
z z
x dm
But ∴
and
z
dm = M (mass of the body).
dm
z z z
xcm =
1 x dm M
ycm =
1 y dm M
zcm =
1 z dm M
∑ ∆m
i
i=1
An n → ∞, the co-ordinates are defined precisely, by
xcm =
n
∆mi yi
∑ ∆m z
i i
, zcm =
i=1 n
∑ ∆m
i
i=1
226
Mechanics
The vector expression corresponding to these three scalar expressions is →
r cm
=
1 M
z
→
r dm
The center of mass of a “homogeneous body” (having a uniform distribution of mass) must coincide with the geometric center of the body. Thus if the homogeneous body has a point, a line, or a plane of symmetry, its center of mass must lie at its point, line or plane of symmetry. This fact can be easily understood. From symmetry, the first moment of mass
z
→
r dm of the “homogeneous” body must be zero with respect to its geometric center:
z
→
r dm = 0 →
r cm = 0
i.e., the center of mass coincides with the geometric center. From this we see that the center of mass of a body does not necessarily lie within the body. For example, in a homogeneous, ring the center of mass would be at its geometric center which does not lie within the material of the ring.
5.5 POSITION VECTOR OF THE CENTRE OF MASS Let us consider a system of n particles of masses m1, m2, m3, ... mn with position vectors → → →
→
→
r1 , r2 , r3 , ..., rn relative to a fixed origin. The position vector r cm of the center of mass of this system is defined by →
r cm
→
→
→
m r + m2 r2 + ... + mn rn = 1 1 m1 + m2 + ... + mn n
∑
=
→
mi ri
i=1 n
∑m
i
i=1 n
1 = M
F I GG = ∑ m JJ H K
→
∑m r
i i
i=1
n
where M
i
is the total mass of the system.
i=1
5.6 VELOCITY OF THE CENTRE OF MASS Let us write the expression for the position vector of a system of particles: →
r cm =
→ → → 1 (m1 r1 + m2 r2 + ... + mn rn ) M
Linear and Angular Momentum
227
Differentiating it with respect to time, we obtain
F GG H
→
→
→
→
d r cm = 1 m d r1 + m d r2 + ..... + m d rn n 1 2 M dt dt dt dt
I JJ K
→ → → 1 v cm = M (m1 v1 + m2 v2 + ... + mn vn )
→
or
→
v cm
=
1 M
n
→
∑m v
i i
i=1
→
→
where v cm is the velocity of the center of mass and vi the velocity of the ith particle. This is the required expression.
5.7 CENTER OF MASS FRAME OF REFERENCE It is a reference frame attached with the center of mass of a system of particles (or a body). It is also known as C-frame of reference. In this frame the velocity of the center of →
mass is zero by definition ( v cm = 0).
5.8 MOTION OF THE CENTER OF MASS OF A SYSTEM OF PARTICLES SUBJECT TO EXTERNAL FORCES Let us consider a system of n particles of masses m1, m2, m3, ... mn with position vectors → → →
→
→
→
→
r1 , r2 , r3 , ..., rn with respect to a fixed origin, and subjected to external forces F1 , F2 , ..., Fn exerted by the surroundings.
The position vector of the center of mass of the system is defined by →
r cm
→
→
→
m r + m2 r2 + ... + mn rn = 1 1 m1 + m2 + ... + mn
→ → → 1 r cm = M (m1 r1 + m2 r2 + ... + mn rn )
→
where M (= m1+ m2 + m3 + ... + mn) is the total mass of the system. Differentiating it with respect to time, we obtain
F GG H
→
→
→
→
dr dr d rn 1 d r cm m1 1 + m2 2 + ... + mn = M dt dt dt dt
or
→
v cm
=
→ → → 1 (m1 v1 + m2 v2 + ... + mn vn ) M
I JJ K
228
Mechanics →
→
Where v cm is the velocity of the center of mass and vi is the velocity of the particle m1, and so on. Differentiating again, we get
F GG H
→
→
→
→
dv dv dv 1 d v cm m1 1 + m2 2 + ... + mn n = M dt dt dt dt →
→
→
I JJ K
→
M acm = m1 a1 + m2 a2 + ... + mn a n
or
→
→
where a cm is the acceleration of the center of mass, a1 is the acceleration of the particle m1 and so on. →
Now, from Newton’s second law, the external force F1 acting on the first particle is given →
→
by F1 = m1 a1 and so on. Thus, →
→
→
→
→
→
M acm = F1 + F2 + ... + Fn or
M acm = F ext →
Where F ext is the sum of the external force acting on all the particles. Thus the product of the total mass of a system of particles and the acceleration of its center of mass is equal to the sum of the ‘external’ force acting on the particles. This means that the center of mass of a system of particles moves as if all the mass of the system were concentrated at it and all the ‘external’ force were applied at it. This result holds whether the system is a rigid body with particles in fixed position or a system of particles with internal motions.
5.9 LINEAR MOMENTUM IN CENTER OF MASS FRAME OF REFERENCE In the reference frame attached with the center of mass (C-frame) the velocity of the →
center of mass, v cm is zero by definition. Therefore, the total linear momentum of a system of particles is also zero: →
→
P = M v cm = 0
Whatever may be the velocities of the constituent particles. Thus the center-of mass coordinate system is also referred to as the center of-momentum coordinate system, to stress the fact that in such a system the total momentum is zero. The C-frame is important because many experiments that we perform in our laboratory or L-frame of reference can be more simply analyzed in the C-frame of reference (which →
moves with a velocity vcm relative to the L-frame).
5.10 SYSTEM OF VARIABLE MASS There are certain systems whose mass does not remain constant during their motion, but varies with time. A rocket is propelled by ejecting burnt fuel which causes the total mass
Linear and Angular Momentum
229
of the rocket to decrease as the rocket accelerates. A falling rain drop collapses with smaller drops which increase its mass. Let us see what form Newton’s law takes for such system of ‘variable mass’. →
Let a system of mass M be moving at a velocity v at any instant t in a particular reference frame. Let at a later instant t + δt, a part δM separated from the system be moving →
→
→
with a velocity u and the remaining system M-δM with new velocity v + δ v . If we still treat both parts as forming one and the same system, then for the finite time interval δt, we can write: →
→
F ext
→
→
→
→
→
P f − P i [(M − δM) ( v + δ v ) − δM. u ] − [M v ] = = δt δt →
→ → → = M δ v − v δM − δ v δM + u δM δt δt δt δt
M
→ v
δM
M - δM → u
At t
→ → v + δv
A t t + δt
Fig. 2
Now if δt → 0 (i.e., the mass of the system is continuously decreasing), then →
→
→ δ v = d v ; δM = − dM ; δ v = 0 dt dt δt δt →
→ → dM d v → dM → dM d = M v u M v u + − = ( ) − F ext dt dt dt dt dt This expresses Newton’s second law as applied to a body of variable mass. We note that →
so that
→ d (M v ) is not equal to the external force acting on the system (unless the ejected mass dt comes out with zero speed).
We can write the above equation as →
→ → → d v = F ext + ( u − v ) dM M dt dt →
→ → d v = F + v dM ext rel M dt dt
or →
Where v rel is the velocity of the ejected mass relative to the main body. The last term in the above equation is the rate of change of momentum of the system due to the mass leaving it. It can be taken as the reaction force exerted on the system by the leaving mass. Thus we can write →
M
→ dv = → F ext + F reaction dt
230
Mechanics
5.11 MOTION OF A ROCKET The rocket is the most interesting example of a system of variable mass. Its motion can be explained on the basis of Newton’s third law of motion and the momentum principle. It consists of a combustion chamber in which liquid or solid fuel is burnt. The heat of combustion raises the pressure very high inside the chamber. Therefore, hot gases (produced by combustion) are expelled from the tail of the rocket in the form of a jet with a very high exhaust velocity. Consequently, the rocket rushes in the forward direction. The rocket exerts an action force on the gas-jet in the backward direction, while the gasjet exerts a reaction force on the rocket in the forward direction. These are the internal forces in the (rocket + gas) system. In the absence of external forces, the total momentum of the system (rocket + gas) is constant. The gas-jet acquires momentum in the back ward direction and the rocket acquires an equal momentum in the forward direction. Let us now derive an expression for the final velocity of the rocket. Let M (a variable) →
be the mass of the (rocket + unburnt fuel) at an instant t and v its velocity in a fixed (laboratory) frame of reference. Suppose in a time-interval dt an amount of Mass dM is ejected from the rocket in the form of gas-jet. →
If u be the velocity of the gas-jet in the laboratory reference frame, then its velocity →
relative to the rocket v rel , would be given by →
→
→
v rel = u − v
→
v rel is known as “exhaust velocity”. Now, according to the Newton’s second law as applied to a system of variable mass, we have →
→ → d v = F + v dM ext rel M dt dt
Y → v0
→ v M
Th ru st → vR e s
d M /dt
x
O
Fig. 3 → → dM Where F ext is the ‘external’ force acting on the system and v rel in the reaction force dt exerted on the system by the leaving mass. In the case of a rocket, this term is called the →
‘thrust’ exerted on the rocket by the ejecting gas-jet. The external force F ext is the force of gravity on the rocket and the air resistance. →
To solve the above equation, let us assume that the exhaust velocity v rel is constant. Also, neglecting air resistance and the variation of the gravity with altitude, we may write →
→
F ext = M g , so that the last equation becomes:
Linear and Angular Momentum
231 →
→ → d v = M g + v rel dM M dt dt
Now, suppose the motion of the rocket is vertical. Then
→
→
v is directed upward and v rel
→
and g downward. The last equation may now be written as →
→ d v = − Mg − v dM rel M dt dt
dv = − gdt − vrel
or
dM M
Integrating from the beginning of the motion (t = 0), when the velocity is v0 and the mass of the rocket + fuel is M0, upto an arbitrary time t, we have
z
z t
v
dv = − g
M0
0
v0
z
M
dt − vrel
or
v – v0 = − gt − vrel log e
M M0
or
v – v0 = − gt + vrel log e
M0 M
or
v = v0 + vrel log e
dM M
M0 − gt M
If t is the time required for burning all the fuel, then M is the final mass and v is the maximum velocity attained by the rocket. If the force of gravity is ignored, then the above expression reduces to v = v0 + vrel log e
M0 M
Further, if the initial velocity v0 of the rocket is zero, then v = vrel log e
M0 M
Limitation of One-stage Rocket In the absence of gravity and air resistance, the ultimate velocity of the rocket at burn M out (when all the fuel has been used up) is vrel log e 0 . Thus it can be increased by increasing M M0 the exhaust velocity vrel and the ratio . However, the exhaust velocity cannot be more M than 2.5 km/sec with the conventional chemical fuels. Further, the fact that the rocket shell
FG IJ H K
232
Mechanics
must be strong enough to hold the fuel sets an upper limit on the ratio
FG M IJ also. HMK 0
In actual
practice, this ratio is not greater than 4. Therefore, the maximum velocity of the rocket at burn out is v = 2.5 loge 4 = 2.5 × 1.39 = 3.5 km/sec This velocity is much less than the escape velocity (11 km/sec) or the orbital velocity near the surface of the earth (8 km/sec). Thus a single-stage rocket is incapable to put space satellites in orbits or escape the earth’s gravitational field.
5.12 MULTI STAGE ROCKET To attain higher velocities, the rocket is designed in stages. For example, a two-stage rocket means that one rocket is placed on the top of another rocket. When the fuel of the first-stage (lower) is exhausted, its rocket casing is detached and drops off. The velocity attained so far becomes the initial velocity of the second stage which is now ignited . The removal of the surplus mass contained in the first stage considerably helps in attaining still higher velocity. Suppose that the initial mass of each of the first-and second stage rocket (plus fuel) is M0/2, and the mass of each rocket casing is M0/2. The total initial mass is thus M0. The velocity v′ attained when the first stage is detached is given by v′ = vrel log e
M = vrel log e 2 M0 /2
This is now the initial velocity for the second stage. The final velocity v attained, when the fuel of the second stage is exhausted, is given by v = v′ + vrel log e
M0 /2 M /2
v = vrel log e 2 + vrel log e Taking
M0 M
vrel = 2.5 km/sec and M0/M = 4, then we have v = 2.5 km/sec (loge 2 + loge 4) = 2.5 km/sec × loge 8 = 2.5 × 2.08 km/sec = 5.2 km/sec
Thus the final velocity attained by a two-stage rocket is greater than that attained by a single-stage rocket of the same weight and fuel supply. The velocity can be further increased by adding more stages. The rocket of equal stages is not generally the optimum construction. In fact the first stage should be made much larger than the second in order to obtain a high final speed.
NUMERICALS: SYSTEMS OF VARIABLE MASS: ROCKET Q. 1. Solve the last problem is M = 200 kg, m = 10 gm, n = 10/sec and vrel = 500 m/s.
Linear and Angular Momentum
233
Solution. The magnitude of the acceleration of the trolly at the instant the mass is M, a = vrel
is given by
mn M
500 × 10−2 × 10 = 0. 25 m/s2 200 The magnitude of the average thrust on the system is given by Freaction = vrel mn = 500 × 10–2 × 10 = 50 nt. Q. 2 (a). A man of mass m is standing on a trolly of mass M which is moving with a =
→
velocity v on frictionless horizontal rails. If the man starts running opposite to the direction →
of motion of the trolly and his velocity relative to the trolly is v rel just before he jumps off, find the change in velocity of the trolly. (b) Let us now assume that there are n men, each of mass, m, on the trolly. Should they all run and jumps off together or should they do so one by one in order to give a greater velocity to the trolly. Solution. (a) The initial mass of the (trolly + man) system is (M + m) which is moving →
with a velocity v in a fixed frame of reference. As the man jumps, he acquires a backward momentum and the trolly acquires forward momentum. The Newton’s second law, in the absence of external forces, gives →
→ d (M + m) t dV = V rel (M + m) dt dt →
→
→
→
( M + m) d V = V rel d (M + m)
...(i)
(M + m) ∆ V = V rel ∆ ( M + m)
Here ∆(M + m) = –m (the man of mass m jumps off). Therefore →
→
→
→
(M + m) ∆ V = − V rel m
∴ Change in velocity
∆ V = − V rel
m M+m
This equation shows that the speed of the trolly will increase by mvrel/(M + m) in its →
initial direction of motion (i.e., opposite to v rel ) (b) Let us now suppose that n men are standing on that trolly. If they all jumps off together, the change in velocity would be →
→
∆ V = − V rel
mn M + mn
...(ii)
234
Mechanics
But if they jump one by one, the mass of system will go on changing till the last man jumps. In this case we can write eqn. (i) as → → d ( mn + M) d V = V rel ( mn + M) →
z
M
→
∆ V = V rel
M + mn →
= V rel log e
d (M + mn) → = V rel log e (M + mn) ( M + mn)
M M + mn
→ M M + mn = − V rel log e M + mn M
FG H
→ = − V rel log e 1 + mn M
IJ K
→
mn ...(iii) M A comparison of eqns. (ii) and (iii) shows that the men would give a greater velocity to the trolly by running and jumping off one by one. Q. 3. Fine particles of sand are being dropped continuously from a fixed container on to a moving belt. Find out the force necessary to have the belt moving at a constant speed and show that the power supplied by this force is twice the rate of increase of the K.E. of the system. = − V rel
→
Solution. Suppose the belt is moving with a constant velocity V in a certain reference frame. The container is fixed in this reference frame. As sand particles are falling on the belt, the mass of the system (belt + material on it) is continuously increasing. Let M be the mass of the system at any instant, dM/dt the rate at which the material is falling on the belt. Then the Newton’s second law as applied to a system of varying mass, gives →
→ → d V = F + V dM ext rel M dt dt →
...(i) →
Where F ext is the external force acting on the system and V rel is the relative horizontal velocity of the falling mass relative to the belt. →
Here
d V = 0 (as → is constant) V dt →
→
V rel = − V and
dM is positive (the mass of the system is increasing with time). dt Therefore the equation (i) becomes →
→
0 = F ext − V
dM dt
Linear and Angular Momentum
235 →
→
F ext = V
dM dt
...(ii)
Y
X F → v
Z
Fig. 4
This is the external force acting on the system under which is moving at a constant rate. Alternative Method. We can also determine the force by applying momentum principle. Let M′ be the mass of the belt and M that of the material on the belt at any instant. The momentum of the system is →
→
P = (M ′ + M) V
Now the rate of change of momentum gives the force. Thus →
→
F =
→ dP d = (M′ + M) V dt dt →
= V
→ dM , (as V and M′ are constants) dt
dM is the rate at which the material is falling on the belt i.e., the rate of increase dt of mass of the system. where
→
Now, the power P supplied by the force F ext is the rate of doing work. In vector rotation, it is given by the dot product of the force and velocity. That is →
→
→
P = F ext . V →
= V =
dM → dM . V = V2 dt dt
d MV2 (3 V is constant) dt
= 2
d 1 dK ( MV2 ) = 2 , dt 2 dt
LM3 V. V = V OP Q N → →
2
236
Mechanics
FG H
IJ K
1 MV2 is the K.E. Thus the power supplied by the external force acting on the 2 system is twice the rate of increase of K.E. of ‘whole’ system. This is an example in which the mechanical energy is not conserved. Q. 4. A wagon filled with sand has a hole so that sand leaks through the bottom at a
where K =
constant rate −
→ dm = λ. A force F acts on the wagon in the direction of its motion. If its dt →
instantaneous velocity is V , write the equation of motion. →
Solution. Let m be the instantaneous mass and V the instantaneous velocity of the system (wagon + sand). The instantaneous momentum is →
→
P = mV →
By Newton’s law, the rate of change of momentum is the force F acting on the system. That is →
→
→ d V → dm dP d +V = ( m V) = m F = dt dt dt dt →
The rate of loss of mass is −
dm = λ, so that dt →
dV → − Vλ F = m dt →
If m0 be the initial mass, then m = m0 – λt, so that →
dV → . − Vλ F = (m0 − λt) dt →
This is the required equation. Q. 5. A rocket having initial mass of 240 kg ejects fuel at the rate of 6 kg/s with a velocity of 2 km/s vertically downward relative to itself. Calculate its velocity 25 seconds after start, taking initial velocity to be zero and neglecting gravity. Solution. Neglecting gravity, the velocity of a rocket at any time t is given by v = v0 + vrel log e
M0 . M
where M0 is the initial (at t = 0) mass of the rocket plus fuel and M is the mass remaining at time t. If the initial velocity v0 is zero, then v = vrel log e
M0 M
Linear and Angular Momentum
237
Here vrel = 2 km/s, M0 = 240 kg. As the fuel is ejected at the rate of 6 kg/sec, the mass consumed in 25 sec is 150 kg. Therefore, the mass remaining after 25 sec is M = 240 – 150 = 90 kg. Thus v = 2 log e
240 90
= 2 × 2. 3 log10 (2. 67) = 2 × 2.3 × 0.4265 = 1.96 km/sec. Q. 6. An empty rocket weighs 5000 kg and contains 40,000 kg of fuel. If the exhaust velocity of the fuel is 2.0 km/s, find the maximum velocity gained by the rocket. (loge10 = 2.3, log10 3 = .4771). Solution. Ignoring gravity effect, the velocity of a rocket at any time t is given by
M0 M When Mo is the initial (at t = 0) mass of the rocket plus fuel and M is the mass remaining at time t. The velocity v attains maximum value when all the fuel is burnt. M is then the mass of the empty rocket. Here, initial velocity v0 = 0, M0 = 5000 + 40,000 = 45,000 kg, M = 5000 kg and Vrel = 2 km/s. v = v0 + vrel loge
45000 5000 = 2 × loge(3)2 = 2 × 2 loge 3 = 2 × 2 × 2.3 × 0.4771 = 4.4 km/sec. Q. 7. From the nozzle of a rocket 100 kg of gases are exhausted per second with a velocity of 1000 m/s. What force (thrust) does the gas exert on the rocket.
∴
vmax = 2 × log e
→
Solution. The thrust (F reaction ) exerted by the escaping gas on the rocket is given by. →
→ F reaction = V rel dM dt
Here vrel = 1000 m/s and
dM = 100 kg / sec dt
→
F reaction = 1000 × 100 = 105 nt. Q. 8. The first and second stages of a two stage rocket have weights 100 kg and (10 kg and carry 800 kg and 90 kg of fuel supply. The velocity of ejected gases relative to the rocket is 1.5 km/s. Find the final velocity attained by the rocket (loge10 = 2.3). Solution. Ignoring gravity, the velocity of a rocket at any time t is given by v = v0 + vrel loge
M0 M
238
Mechanics
where v0 is the initial velocity, M0 the initial mass of the rocket plus fuel, and M the mass of the rocket plus unburnt fuel at time t. For the operation of the first stage. We have; v0 = 0, M0 = 100 + 10 + 800 + 90 = 1000 kg and M = 100 + 10 + 90 = 200 kg. ∴ v = 0 + 1.5 loge 5 This is the initial velocity for the second stage. Now, the first stage rocket of mass 100 kg drops off. Thus for the second stage, we have v 0 = 1.5 loge 5 km/s, Mo = 10 + 90 = 100 kg and M = 10 kg v = 1.5 loge 5 + 1.5 loge 10 = 1.5 loge 50 = 1.5 × 2.3 × log10 50 = 1.5 × 2.3 × 1.6990 = 5.86 km/s. If we would have a single stage rocket of the same total mass 110 kg carrying same amount of fuel of 890 kg, then for its operation we would have v 0 = 0, Mo = 110 + 890 = 1000 kg and M = 110 kg
1000 = 1. 5 log e 9.1 110 = 1.5 × 2.3 × log10 9.1 = 1.5 × 2.3 × 0.959 = 3.31 km/sec. Thus the final velocity attained by a multistage rocket is much greater than that attained by a single stage rocket of the same total mass and fuel supply. Q. 9. A 6000 kg rocket is set for vertical firing. If the gas exhaust speed is 1000 m/s, how much gas must be ejected each second to supply the thrust needed (a) to overcome the weight of the rocket, (b) to give the rocket an initial upward acceleration of 20 m/s2. Solution. (a) The equation for the vertical motion of the rocket is v = 0 + 1. 5 log e
∴
M
dM dv = − Mg − vrel dt dt
where M is the mass of the rocket + fuel at any instant t. The term M
...(i)
dv represents the dt
dM is the thrust. In order to dt just overcome the weight of the rocket, the thrust need not give any net upward force i.e.,
instantaneous net upward force acting on the rocket, and vrel
M
dv = 0. Then eqn. (i) becomes dt dM − Mg − vrel = 0 dt dM 6000 × 9. 8 −Mg = − 58. 8 kg /sec . = = − dt 1000 vrel Hence the gas should be ejected at the rate of 58.8 kg/sec. (b) If the rocket is to be given an initial upward acceleration of 20 m/s2, then plutting
dv = 20 in eqn. (i) we get dt
Linear and Angular Momentum
239
20M = − Mg − vrel
vrel
dM dt
dM = − Mg − 20M dt dM −6000 −M (9. 8 + 20) ( g + 20) = = dt 1000 vrel
= –178.8 kg/sec. Q. 10. A 8000 kg rocket is set for vertical firing. If the exhaust speed is 800 m/s, how much gas must be ejected/second to supply the thrust needed (i) to overcome the weight of the rocket, (ii) to give the rocket an initial upward acceleration of 3g. [Ans. (i) –98 kg/s, (ii) –392 kg/s] Q. 11. A rocket of mass 20 kg has 100 kg of fuel. The exhaust velocity of the fuel is 1.6 km/s. Calculate the minimum rate of consumption of fuel so that the rocket may rise from the ground. Also calculate the final vertical velocity gained by the rocket when the rate of consumption of the fuel is (i) 2.0 kg/s, (ii) 20 kg/s.
dM , supplied to dt the rocket will just overcome its initial weight Mog and raise it from the ground. Thus Solution. With the minimum rate of fuel consumption, the thrust, vrel
vrel
dM = M0g dt M0 g 200 × 9. 8 dM = = 1. 225 kg/sec. = vrel dt 1. 6 × 103
Now, the velocity attained by the rocket at any time t is
M0 − gt M where M is the mass of the rocket plus unburnt fuel at time t. If T is the time in which the entire fuel has burnt, then v is the final velocity attained by the rocket. Thus, if the initial velocity v0 = 0, we have v = v0 + vrel log e
vmax = vrel log e
M0 − gT M
180 M0 200 = 90 sec = log e = log e 10 = 2. 3 and T = 2 M 20 = (1.6 × 103 × 2.3) – (9.8 × 90) = 3680 – 882 = 2798 m/sec ∼ 2.8 km/s
(i) vrel = 1.6 km/s = 1.6 × 103 m/s, log e vmax
(ii) T =
180 = 9 sec 20 vmax = 3680 – 88.2 = 3591.8 m/sec ~ 3.6 km/sec.
240
Mechanics
Q. 12. A 103 kg rocket is set vertically on its launching pad. The propellant is expelled at the rate of 2 kg/s. Find the minimum velocity of the exhaust gases so that the rocket just begins to rise. Also find the rocket’s velocity 10 sec after ignition, assuming the minimum exhaust velocity.
dM , supplied to dt the rocket will just overcome its initial weight M0g and raise it from the ground. Thus Solution. With the minimum velocity of exhaust gases, the thrust, vrel
vrel
dM = M0g dt
103 × 9. 8 M0 g = dM / dt 2 = 4.9 × 103 m/s Now, the velocity attained by a rocket at any time t is given by vrel =
Mo − gt , M where M is the mass of the rocket plus unburnt fuel at time t. If the initial velocity vo is zero, then v = v0 + vrel log e
M0 − gt M Here vrel = 4.9 × 103 m/s, M0 = 103 kg, and t = 10 sec. As the fuel is expelled at the rate of 2 kg/sec, the mass consumed in 10 sec is 20 kg. Therefore the mass remaining is M = 1000 – 20 = 980 kg. Thus v = vrel log e
3 v = (4. 9 × 10 ) log e
1000 − (9. 8) (10) 980
b
g
= (4. 9 × 103 ) × 2. 3 log10 1000 − log10 980 − 98 = 4.9 × 103 × 2.3 × (3.0000 – 2.9912) – 98 = 99 – 98 = 1.0 m/s
6 COLLISIONS 6.1 COLLISION It is defined as the phenomenon of the change in the velocities of bodies during the very small time interval of their contact. In the event two bodies collide with one another they are deformed. The kinetic energy before collision of the system transforms completely or partially into potential energy of elastic deformation and into internal energy of the bodies. In collision process it is not at all assumed that the particles actually come in contact as contrary to every day life’s sense of collision. The significance of collision studies lies in the fact that they yield the information regarding the forces which act between particles.
Elastic Collision If the force of interaction between the colliding bodies are conservative, the kinetic energy remains conserved in the collision and the collision is said to be elastic. Collision between atomic, nuclear and fundamental particles are usually elastic. Collisions between ivory and glass balls are approximately elastic. Inelastic Collision: When the kinetic energy is changed in the collision, the collision is said to be inelastic (the momentum as well as the total energy is still conserved). Collisions between gross bodies are always inelastic to some extent. When two bodies stick together after collision, the collision is said to be completely inelastic. When a bullet hitting a target remains embedded in the target, the collision is completely inelastic.
6.2 ELASTIC COLLISION IN ONE DIMENSION Suppose two particles of following specification undergo elastic head-on collision. Let m1 = mass of particle 1, m2 = mass of particle 2, u1 = velocity of mass 1 before collision, u2 = velocity of mass 2 before collision, v1 = velocity of mass 1 after collision, v2 = velocity of mass 2 after collision. By the law of conservation of energy K.E. before collision = K.E. after collision ∴
1 1 1 1 m1u12 + m2u22 = m1v12 + m2v22 2 2 2 2 241
242
Mechanics
m1 (u12 − v12 ) = m2 (v22 − u22 )
or
...(i)
2
1
2
1
m1 m2 u1 u2 B e fo re co llision
m1 m2 v1 v2 A fter co llisio n
Fig. 1
From law of conservation of Momentum m1u1 + m2u2 = m1v1 + m2v2 or
m1 (u1 – v1) = m2 (v2 – u2)
...(ii)
Dividing (i) by (ii),
e m bu
j −vg
m1 u12 − v12 1
1
=
1
e bv
j −u g
m2 v22 − u22 m2
2
2
or
(u1 + v1) = (v2 + u2)
or
(u1 – u2) = – (v1 – v2)
....(iii)
Which suggests that relative velocity of the particles after elastic collision is equal and opposite to the relative velocity before collision. Putting Equation (iii), in Eq. (ii), we obtain m1(u1 – v1) = m2 (u1 + v1 – u2 – u2) or
u1(m1 – m2) + 2m2u2 = v1 (m1 + m2) v1 =
or
FG m Hm
2m2 − m2 u2 u1 + m1 + m2 + m2
FG m Hm
2m1 − m1 u2 + u1 m1 + m2 + m2
1 1
IJ K
FG H
IJ K
IJ K
FG H
IJ K
...(iv)
Similarly v2 =
2
1
...(v)
General Case: Newton experimentally showed that two smooth spherical balls, when approach, collide and separate, then they obey the rule. v1 – v2 = – e (u1 – u2) v1 − v2 = –e u1 − u2
That is, ratio of velocities of separation and approach are in the ratio e:1. Where ‘e’ has +ve value and is called coefficient of Restitution, which depends on (i) the material of the object (ii) shape and size of colliding objects. Its value lies between 0 and 1. For perfectly inelastic collision, e = 0 and for perfectly elastic collision, e = 1. Special cases: (1)
when m1 = m2 then the equation (iv) and (v)
Collision
243
v 1 = u2 and v2 = u1 i.e., colliding bodies, simply exchange velocities as a result of collision. (2)
when u2 = 0, i.e., second body is initially at rest, then
FG m − m IJ u Hm + m K FG 2m IJ u = Hm + m K
v1 =
and (3)
v2
1
2
1
2
...(vi)
1
1
1
...(vii)
1
2
When u2 = 0 and m2 >> m1 then from Equations (vi) and (vii) becomes v 1 = –u1 and v2 = 0 (nearly) Thus, when a lighter body collides with too heavy body at rest, the lighter body reverses the course of motion with no change in the rest state of heavier body.
(4)
When m2 << m1, in addition to u2 = 0, then from Equation (vi) and (vii) v 1 = u1 and v2 = 2v1 Thus, when a very heavy body collides with a lighter body at rest, then velocity of heavier body remains nearly unchanged and lighter body moves with nearly twice the velocity of heavy body.
Fraction of K.E. Transferred in Elastic Head on Collision Suppose a particle 1 collide with particle 2 which is initially at rest. After collision, particle 2 acquire velocity v2 and has K.E. = ½ m2 v22. Before collision, it has K.E. = 0. So kinetic energy transferred = ½ m2v22 . Initial total K.E. = ½ m1 u12 Thus, fractional K.E. transferred =
½ m2 v22 ½ m1 u12
Putting the value of v2 from equation (vii) v2 =
2m1 u1 m1 + m2
FG H
1 2m1 m2 u1 2 m1 + m2 We get, Fractional K.E. transferred = 1 m u2 2 1 1 If
IJ K
2
=
4 m1 m2 (m1 + m2 )2
m 1 = m2, Fractional K.E. transferred = 1.
Thus, when the mass ratio is unity, the whole of the kinetic energy of the moving ball is transferred to the ball initially at rest.
Inelastic Collision (Particle Stick After collision) (i) Description in Inertial frame: Let a particle of mass m2 be at rest at origin in an inertial frame. Suppose a particle of mass m1 moving with velocity u1 along x-direction
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collides with m2, both stick and then travel together with velocity v after collision in the same direction. From law of conservation of momentum m1u 1 = (m1 + m2)v final velocity of composite particle v, after collision m1 u1 v = ( m1 + m2 )
...(viii)
Ratio of K.E. after collision to K.E. before collision K.E. before collision Ki = ½ m1 u12 K.E. after collision Kf = ½ (m1 + m2)v2 ∴ ratio of final to initial K.E. of system
1 (m1 + m2 )v2 Kf 2 = 1 Ki m u2 2 1 1
Fig. 2
F GH
1 (m1 + m2 ) m1u1 . = 2 1 m 2 1 + m2 mu 2 1 1 Kf
or
Ki
=
I JK
2
using Equation (viii)
FG m IJ Hm + m K 1
1
2
It shows Kf < Ki which means during inelastic collision, there is a loss in K.E. of system. If a moving particle of mass m1 suffers inelastic collision with a stationary particle of mass m2, the loss in kinetic energy of system is given by δ.E = Since,
∴
v =
δ.E =
1 1 m1u12 − (m1 + m2 ) v2 2 2 m1 u1 (m1 + m2 ) 1 m1m2 u12 2 (m1 + m2 )
from Eq. (viii)
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245
Description in center of mass frame of reference (C-Frame): velocity of center of mass vcm is given by
or
vcm =
m1u1 + m2 . 0 m1 + m2
vcm =
m1u1 m1 + m2
Velocity of 2nd particle, before collision, in C-frame
u′2 = u2 − vcm = 0 − i.e.,
u′2 = −
m1u1 m1 + m2
m1u1 m1 + m2
Velocity of 1st particle, before collision, in C-frame
u′1 = u1 − vcm = u1 −
=
or
m1u1 m1 + m2
m1u1 + m2u1 − m1u1 m1 + m2
u′1 = +
m2u1 m1 + m2
Velocity of combined particle after collision in C-frame v′ = v − vcm =
m1u1 m1u1 − =0 m1 + m2 m1 + m2
v′ = 0 Thus combined particle is at rest in C-frame.
6.3 THE BALLISTIC PENDULUM It is a device for measuring the speeds of bullets. It consists of a large wooden block mass M hanging vertically by two cords. A bullet of mass m, moving with a speed v, strikes the block and remains embedded in it. If the collision time is very small compared to the time of swing of the pendulum, the supporting cords remain approximately vertical during the collision. Hence no external horizontal force acts on the system during the collision, and the horizontal momentum is conserved. Let v′ be the velocity of the block-bullet combination immediately after the collision. The initial momentum of the system is mv, and the momentum just after the collision is (m + M)v′, so that mv = (m + M)v′
...(i)
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Mechanics
1 (m + M)v′2 . The 2 block-bullet combination now swings up to a height y at which its kinetic energy is converted into gravitational potential energy. From the conservation of mechanical energy for this part of the motion, we obtain The kinetic energy of the system immediately after the collision is
1 (m + M) v′2 = (m + M) gy 2 or
v′ =
2 gy
Substituting this of v′ in Equation (i) and solving for v, we get v =
m+M m
2 gy
v
y
M
m
Fig. 3
By measuring m, M and y, we can compute the initial speed v of the bullet. The kinetic energy is not conserved in the collision. The ratio of the kinetic energy of block-bullet combination just after the collision to the initial kinetic energy of the bullet is 1 (m + M) v′2 m 2 = m+M 1 mv2 2 As m << M, a very small fraction of the original kinetic energy remains, most of the energy is converted into heat, etc.
6.4 COLLISION IN TWO DIMENSION (1) Description in Laboratory Frame: Suppose a particle of mass m1 collides with other particle of mass m2 which is at rest in the laboratory frame of reference. Suppose initial velocity of m1 is u1 in this frame and after collision it has velocity v1 in a direction that makes angle φ1 with the initial direction of travel. Here φ1 is the angle by which colliding particle is deflected and is called scattering angle. Let v2 be the velocity of m2, in direction φ2 with initial line. If direction of u1 is along x-axis and u1 and v1 are contained in x-y plane then there is no z-component of v2. According to principle of conservative of linear momentum: x-component m1u 1 = m1v1 cos φ1 + m2v2 cos φ2
Fig. 4
...(i)
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247
y-component 0 = m1v1 sin φ1 – m2v2 sin φ2
...(ii)
Since the collision is elastic, K.E. is also conserved
1 1 1 m1u12 = m1v12 + m2v22 ...(iii) 2 2 2 If we solve Equations (i), (ii) and (iii), we can know the three unknown quantities of interest. However, the velocity of center of mass in laboratory frame, vcm is given by ∴
(m1 + m2) vcm = m1 u1 + m2 . 0
m1u1 m1 + m2 (2) Description in C-frame: Centre of mass frame of reference moves with velocity vcm in lab frame i.e., center of mass remains at rest in c-frame. In this frame: vcm =
Initial velocity of m1 is,
u′ 1 = u1 – vcm
Initial velocity of m2 is,
u′ 2 = u2 – vcm = 0 – vcm
Final velocity of m1 is,
v′ 1 = v1 – vcm
Final velocity of m2 is,
v′ 2 = v2 – vcm
vcm is always zero in this frame, hence total momentum before and after collision is zero i.e. momenta of two particles are always equal and opposite as observed in C-frame. So the angle of scattering of both the particles should be same and, thus, they move along a linear path in opposite direction. Thus
m1 u′1 = m2u′2
and
m1v′1 = m2v′2
or
u′1 = v′1
Again K.E. conservation yields
1 1 m1u′12 = m1v′12 2 2 1 1 m2u′22 = m2v′22 2 2 u′2 = v′2
and or
Thus, in C-frame, the magnitude of velocities of the particles in an elastic collision do not alter.
6.5 VALUE OF THE SCATTERING ANGLE (i) In the center-of-mass frame of reference In this frame of reference, there are absolutely no limitations on the value of the scattering angle θ, so that it can have all possible values.
(ii) In the laboratory frame of reference In this frame of reference, there are some restriction on the value of the scattering angle, θ1, as will be clear from the following: we have, tan θ1 =
v1 sin θ1 v1 cos θ1
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Now, the y-component of the final velocity of the first particle being the same in the laboratory reference frame as well as the center-of-mass frame, we have v1 sin θ1 = v′1 sin θ. And, since the x-components of this velocity differ by V in the two frames of reference, we have v1 cos θ1 = v′1 cos θ + V. Substituting these values in the relation above, we, therefore, have tan θ1 =
v′1 sin θ = v′1 cos θ + V
sin θ cos θ +
As we known,
1
V =
FG H
IJ K F m IJ VG Hm + m K
V 1−
or
FG V IJ H v′ K
m1 m1 + m2 2
or
1
Whence
m1u1 m ( u′1 + V) = 1 + m1 m2 m1 + m2
=
m1u′1 m1V + m1 + m2 m1 + m2
=
m1u′1 m1 + m2
=
m1u′1 m1 + m2
2
V =
m1 m u′1 = 1 v′1 m2 m2
∴ u′1 = v′1
Substituting this value of V in Equation (i) above, we have tan θ1 =
sin θ cos θ +
FG m IJ Hm K 1 2
This shows that (a)
If m1 > m2, so that m1/m2 a little >1, the denominator can never be zero and hence tan θ1 can never be ∞. And therefore, θ1 must be less than 90°.
(b)
If m1 = m2, so that m1/m2 = 1, the denominator can be zero for cos θ = –1 and, therefore, tan θ1, can be ∞. Thus, in this case , θ1 can have any value up to the limiting value of 90°.
(c)
If m1< m2 , so that m1/m2 < 1, the value of tan θ1 can also be negative. In this case alone therefore, all values of θ1 can be possible.
It follows from case (a) that in an elastic collision, if a massive particle collides against a lighter one at rest, it can never bounce back along its original path. On the other hand, it follows from case (c) that if a lighter particle collides against a massive one at rest, it may well bounce back along its original path. And, it follows from case (b) that in an elastic collision between two particles of equal mass, one of which is initially at rest, the two particles move at right angles to each other after the collision.
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249
6.6 SCATTERING CROSS-SECTION We can consider three distinct stages in the entire scattering process. We show these three stages of the collision process in below figure. The first stage shown in figure (a) corresponds to a time long before the interaction of the colliding particles. At this stage each particle is effectively free, i.e., its energy is positive. As the particles approach each other (figure b), interaction forces much larger than any other force acting on them come into play. Finally, long after the interaction (figure C), the emerging particles are again free and move along straight lines with new velocities in new directions. The emerging particles may or may not be the same as the original particles. m1
v′1
m1 v1 m1
m2
v2
m2
v′2
m2 (a )
(b )
(c )
Fig. 5 Scattering of two particles
In a typical scattering experiment, a parallel beam of particles, also called projectiles, of given energy and momentum is incident upon a target (below figure). The particles interact with the target for a short time, which deflects or scatters them in various directions. Eventually these particles are detected at large distances from the target. The scattered particles may or may not have the same energies and momenta. An experimenter may be interested in knowing the velocities, linear momenta and energies of the particles before and after scattering. Then the changes brought about in these quantities can be determined.
Fig. 6 A typical scattering process
The probability of scattering in a given direction is found by determining the scattering cross-sections. Let us now define the scattering cross-section for a typical scattering process. Let us suppose that a uniform parallel beam of n particles, all of the same mass and energy, is incident upon a target containing N number of identical particles or scattering centers. Such scattering centers might, for example, be the positive nuclei of atoms in a thin metal foil which could be bombarded by α-particles. Let us assume that the particles in the beam do not interact with each other and the scattering centers in the target are sufficiently far apart. With these assumptions we can regard the incident particles and target particles
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Mechanics
to be sufficiently far apart. Then we can think of this scattering event as if at a given time only one projectile was being scattered by one target particle, without being affected by the presence of other particles. So, effectively at any instant we deal with a two-body collision process. For convenience, we choose the origin of the coordinate system at the position of the target and one of the axes, say z-axis, in the direction of the incidents beam. The direction of scattering is given by the angles (θ, φ) as shown in figure (a). The angle θ, called the angle of scattering, is the angle between the scattered and the incident directions. These two directions define the plane of scattering. The angle φ specifies the orientation of this plane with respect to some reference plane containing the z-axis. The shaded plane in figure is a reference plane. The probability of the scattering of a particle in a given direction (θ, φ) is measured in terms of the differential cross-section. x S catte rin g p la n e
dA D e tecto r r
S catte red P a rticle
θ In cide n t p article
O Targ et
φ In cide n t p article
Z
(a )
y
dΩ
(θ, φ) O ta rge t
z
(b )
Fig. 7
6.7 DIFFERENTIAL SCATTERING CROSS-SECTION Let F be the number of projectile incident per unit area per unit time on the target. F represents the incident flux. Let ∆n be the number of particles scattered into a small solid angle dΩ about the angle (θ, φ) in time ∆t. Then the number of scattered particles by a single target particle in time ∆t, must be proportional to the incident flux F, the duration ∆t and also the solid angle in which they are scattered, i.e., ∆n ∝ F (dΩ) ( ∆t ) The constant of proportionality is defined as the differential scattering cross-section and is denoted by the symbol
dσ . So that dΩ ∆n =
or
FG dσ IJ F (dΩ) (∆t) H dΩ K
∆n dσ = F ∆ t dΩ dΩ Thus, we can also express the differential scattering cross-section as the following ratio:
The number of particles scattered per unit time in a solid angle d Ω in the direction (θ, φ ) dσ = dΩ Incident flux i. e., the number of particles incident on the target per unit area per unit time
The differential scattering cross-section gives a probability. In fact, it is a measure of the probability that an incident particle will be scattered in solid angle dΩ in the direction
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251
(θ, φ). The dimension of
dσ dσ is area. This explains the use of the term “cross-section”. dΩ dΩ
is equal to the cross-sectional area of the incident beam that contains the number of particles scattered into the solid angle dΩ by a single target particle. The unit of
dσ is m2 sr–1. The dΩ
dσ depends only on the parameters of the incident particle, nature of the target and the dΩ
nature of the interaction between the two.
For the N scattering centers the number of particles scattered will be just N times the number scattered by a single scattering center. Thus for N scattering centers, the number of particles scattered is ∆n′ =
dσ NF dΩ ∆t dΩ
Above equation is valid only when the target scattering centers are far enough apart so that the same particle is not scattered by two of them.
6.8 TOTAL CROSS-SECTION Let us place the detector at all possible values of (θ, φ) and count the total number of scattered particles entering all the corresponding solid angles. Then we will get the total scattering cross-section. It is denoted by σ. It can also be calculated from the differential scattering cross-sections by integrating over all possible values of dΩ. Thus σ =
z FGH
IJ K
dσ dΩ dΩ
So the total scattering cross-section represents the number of particles scattered in all direction per unit flux of incident particles. It has the dimension of area. So its unit is m2. Now, we also define the solid angle subtended by an area to be dΩ = sinθ dθ dφ, where the limits of θ and φ are 0 to π and 0 to 2π, respectively. Using these relations we get,
z z FGH π 2π
σ =
0 0
IJ K
dσ sin θ. dθ. dφ dΩ
We can show that for the cases in which the force is central and its magnitude depends only on r,
dσ is independent of φ. We can integrate over φ so that dΩ
z FGH
2π
σ = 2π
0
IJ K
dσ sin θ . dθ dΩ
6.9 IMPACT PARAMETERS Let us suppose that the projectile does not make a head-on collision with the target. Instead, it travels along a path, which if continued in a straight line, would pass target at a distance b (as shown in figure 8). The distance b is known as the impact parameter. b is the perpendicular distance between the projectiles initial path and the target.
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Mechanics
(a) The impact parameter b
(b) The particle having impact parameters between b and b + db are scattered into angles between θcm and θcm + dθcm. Fig. 8
Let us now express the differential scattering cross-sections in terms of the impact parameter. We will study the scattering process in the c.m. frame of reference with θcm as the angle of scattering (figure b). Let us consider a circular ring having radii between b and b + db. The area of the ring is 2πbdb for infinitesimal values of db. If the incident flux is F then, The number of incident particles having an impact parameter between b and (b + db) = F(∆t) (2πbdb) Let us suppose that these particles are scattered into angles between θcm and θcm + dθcm. The particles with larger b will be scattered through smaller angles as shown in figure (c). This happens because larger b means lesser interaction, i.e., less scattering. For very large b, scattering will be minimal and the particles will go almost undeflected in a straight line. Now in the c.m frame of reference the number of particles scattered in the solid angle dΩ in time ∆t is given as ,n =
FG dσ IJ H dΩ K
F (dΩ)cm ∆t cm
dΩ θL b
O ta rge t
(c) The scattering angle decreases with increasing impact parameter. Fig. 8
This is the same as the number of incident particles in time ∆t having impact parameters between b and b + db; given by
FG dσ IJ H dΩ K F dσ IJ 2π.b.db = − GH dΩ K
F( dΩ)cm ∆t
F(∆t)2π.b.db =
cm
or
2π sin θcm dθcm cm
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253
Here we have assumed that
dσ is independent of φ. Taking into account all values of dΩ
φ in dΩ, we have dΩ = 2π sinθ.dθ. The negative sign expresses the fact that as b increases, θcm decreases, i.e., db and dθcm have opposite signs. From above Equation we get,
FG dσ IJ H dΩ K
= cm
b db sinθcm dθcm
We have not written the negative sign in above equation because
FG dσ IJ H dΩ K
has the cm
dimension of area and its magnitude has to be positive. So, if we know b as a function of scattering angle θcm, we can calculate the differential scattering cross-section using above equation.
6.10 RUTHERFORD SCATTERING The Rutherford scattering experiment was an important milestone in understanding the structure of the atom. Until the early twentieth century Thomson’s plum pudding model of the atom was believed to be valid. J.J. Thomson had proposed, in 1898, that atoms were uniform spheres of positively charged matter in which electrons were embedded. It was almost 13 years later that a definite experimental test of this model was made. Now, the most direct way to find out what is inside a plum pudding is to plunge a finger into it. A similar technique was used in the classic experiment performed in 1911, by Geiger and Marsden who were working with Lord Rutherford. They bombarded thin foils of various materials with α-particles (helium nuclei) and recorded the angular distribution of the scattered α-particles. It was found that most of the α-particles pass through the foil (i.e., scattering angle θ < 90°). However, about 1 in 6.17 × 106 alpha particles was scattered backward, i.e., deflected through an angle greater than 90°. This result was unexpected according to Thomson’s model. It was anticipated that the alpha particles would go right through the foil with only slight deflections. This follows from the Thomson model. If this model were correct, only week electric forces would be exerted on alpha particles passing through a thin metal foil. In such a case their initial momenta should be enough to make them go through with only slight deflections. It would indeed need strong forces to cause such considerable deflections in α-particles as were observed. In order to explain these results Rutherford proposed a nuclear model of the atom. Using this model he calculated the differential scattering cross-section. In doing so, he reasoned that the backward scattering could not be caused by electrons in the atom. The alpha particles are so much more massive than electrons that they would hardly be scattered by them. He assumed that the positive charge in the atom was concentrated in a very small volume, which he termed the nucleus, rather than being spread out over the volume of the atom. So the scattering of alpha particles was due to the atomic nucleus. Let us consider the scattering of a particle carrying charge q by the atomic nuclei having charge q′. For this scattering process Rutherford derived the relation between the impact parameter b and the angle of scattering θcm to be b =
θ r0 cot cm 2 2
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Mechanics
qq ′ , Ecm is the total mechanical energy of the projectile and the target 4 πε 0 Ecm in the c.m. system. Where r0 =
ε0 is known as the permittivity of free space. Its value is 8.8 × 10–12 C2N–1 m–2.
Fig. 9
The differential scattering cross-section in the c.m. system for Rutherford scattering is then given by
FG dσ IJ H dΩ K
= cm
b sinθcm
FG db IJ H dθ K cm
θcm 2 r0 . 1 cos ec2 θcm = 2 sin θcm 2 2 2 r0 cot
r02 cot
= 16 sin
or where
FG dσ IJ H dΩ K
θ cm 2
θ cm θ .cos cm 2 2
cosec2
θ cm 2
θ r02 cosec2 cm 16 2 cm qq′ r0 = 4 πε0Ecm =
This is the Rutherford scattering cross-section. For scattering of an α-particle by a nucleus of atomic number z, qq′ = (2e) (Ze) = 2Ze2 where e is the electronic charge. The Rutherford scattering cross-section is strongly dependent on both the energy of the incoming particle and the scattering angle. The number of particles scattered to increase as Z2 with increasing atomic number.
NUMERICALS Q.1. Assuming that all collisions are completely elastic, find the value of m for which both blocks move with the same velocity after m has collided once with M and once with the wall. (The wall has effectively infinite mass).
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255
Ans. Suppose after the collision, v 1 = speed of mass M towards right v 2 = speed of mass m towards left Hence, momentum conservation requires;
Fig. 10
momentum before collision = momentum after collision mu2 = Mv1 – mv2
...(i)
The mass m rebounds elastically from the wall and its speed is reversed after the collision with the wall. The mass m has the same speed as that of mass M after its collision with the wall. (v2 = v1 as given); so (i) is mu2 = (M – m) v1
...(ii)
The collision is elastic, so,
1 1 1 mu22 = Mv12 + mv12 2 2 2 or,
mu22 = (M + m) v12
...(iii)
Substituting the value of v1 from eqn. (ii) in eqn. (iii) we get, mu22 = or or
(M + m) (mu2 )2 (M − m)2
(M – m)2 = (M + m) (m) M2 + m2 – 2Mm = Mm + m2
or
M2 = 3Mm
or
M = 3m
or
m =
M 100 = = 33. 33 kg. 3 3
Q. 2. A ball moving with a speed of 8m/sec strikes an identical ball at rest such that after the collision the direction of each ball makes an angle of 30° with the original line of motion. Find the speeds of two balls after the collision. Is the K.E. conserved in this collision? Solution. Momentum before collision = m × 8 + m × 0 = 8 m
...(i)
where m is the mass of each ball. Suppose after collision their velocities are v1 and v2 respectively. Now the final momentum of the balls after the collision along the same line
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Mechanics
= mv1 cos 30° + mv2 cos 30° =
mv1 3 mv2 3 + 2 2
...(ii)
From the law of conservation of momentum 8 m =
mv1 3 mv2 3 + 2 2
8×2 = v1 + v2 3
...(iii) m
m
u = 8 m /sec
m
B a ll at rest
V1
30 30
V2
m
Fig. 11
The initial momentum of the balls along perpendicular direction is zero and final momentum of the balls along perpendicular direction = mv sin 30° – mv2 sin 30° = Then momentum conservation gives: 1 0 = ∴
FG m IJ (v H 2K
1
v1 – v2 = 0
or
8 8 m/s and v2 = m/s 3 3 Energy before collision = Energy after collision 1 1 1 1 mu12 + mu22 = mv12 + mv22 2 2 2 2 1 m (8)2 + 0 = 1 m 2 2
− v2 )
...(iv)
v1 =
∴
1
− v2 )
from eqns. (iii) and (iv), we get
Since,
FG m IJ (v H 2K
LMF 8 I F 8 I OP MNGH 3 JK + GH 3 JK PQ 2
2
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257
128 m 64 m = 2×3 2
or
L.H.S. ≠ R.H.S. Hence, energy is not conserved in this collision so, collision is inelastic. Since,
Q. 3. A plastic ball is dropped from a height of 1 metre and rebounds several times from the floor. If 1.3 seconds elapse from the moment it is dropped to the second impact with the floor. What is the coefficient of restitution. Ans. Suppose the plastic ball makes first strike on the floor with a velocity V, then ball rebounds with a velocity eV. Now the ball rises to a certain height where its velocity becomes zero and then it retraces its path. The ball strikes the floor with velocity eV making the second impact. After the second impact, it rebounds with a velocity e2V. Time taken by the ball before first impact is given by. V = gt1 (Initial velocity u = 0, Final velocity V, and acceleration = g) t1 =
V g
...(i)
Time interval between first and second impact t2 = 2 × time taken for the velocity to change from eV to O under gravity t2 = 2 × ∴
t1 + t2 =
eV g
...(ii)
V 2e V V + = (1 + 2e) g g g
...(iii)
According to question;
V (1 + 2 e) = 1.3 g Again,
...(iv)
V2 = 0 + 2g.1 V =
...(v)
2g
Using (v) in (iv)
2g (1 + 2e) = 1.3 g or or
2 / g (1 + 2e) = 1.3; or
(1 + 2e) = 1. 3 4. 9 = 2. 8777 1+2e = 2.8777
which gives
2 / 9. 8 (1 + 2e) = 1. 3
e = 0.94
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Mechanics
Q. 4. A meteor of M = 2 kg mass moving with velocity u = 12 km/s explodes above the atmosphere into two pieces which travel along the same path with 15 km/s and 11 km/s. Calculate the mass of each piece and energy set free by the explosion. Solution. Due to the absence of external forces; Linear momentum before explosion = After explosion 2 × 12 = 15m1 + 11m2
i.e., and
...(i)
m1 + m2 = 2
...(ii)
From (i) and (ii); m1 =
2 = 0.5 kg 4
m 2 = 2 – 0.5 = 1.5 kg Energy released ∆E =
FG 1 m V H2
2 1 1
+
IJ K
1 1 m2V22 − mu2 2 2
= 3 × 106 J Q. 5. A sand bag of mass 10 kg is suspended with a 3 meters long weightless string. A bullet of mass 200 gm is fired with speed 20 m/s into the bag and stays in bag. Calculate (i)
Speed acquired by the bag.
(ii)
Maximum displacement of bag.
(iii)
The energy converted to heat in collision.
Solution. (i) Collision is inelastic so the momentum is conserved not the K.E. if m = mass of bullet, M = mass of bag, u = velocity of bullet, V = velocity of bag after hit. then
mu = (m + M) V 0.2 × 20 = (0.2 + 10) V V =
0. 2 × 20 4 = = 0. 39 m/s 10. 2 10. 2
(ii) Bag shall oscillate like pendulum after it is hit. If x is the maximum displacement, Restoring force at maximum displacement; F = –(M + m) g sinθ i.e.,
F = − ( M + m) g
or
K =
x = − Kx l
10. 2 × 9. 8 3
...(i) ...(ii)
At maximum displacement, x, K.E. of system will be changed into potential energy
z x
Kx dx =
o
or,
x2 =
FG IJ H K
1 2 1 1 Kx = m V 2 = 10. 2 ( 0. 39)2 2 2 2
10. 2 × 0. 39 × 0. 39 10. 2 × 0. 39 × 0. 39 ×3 = K 10. 2 × 9. 8
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or,
(0. 39 × 0. 39 × 3)/(9. 8) = 0. 39 3 / 9. 8 = 0. 21
x =
max. displacement, x = 0.21 metre (iii) Energy converted into heat = =
1 1 mu2 − (m + M)V2 2 2 1 1 × 0. 2 × (20)2 − × 10. 2 × ( 0. 39)2 2 2
= 40 − 0. 8 = 38. 2 J Q. 6. A steel ball weighing 1 lb is fastened to a cord 27 inches long and is released when the cord is horizontal. At the bottom of its path, the ball strikes a block weighing 5.0 lb which is initially at rest on a frictionless surface as shown. The collision is elastic. Find the speed of the ball and the speed of the block just after collision. Solution. Let, m1, m2 = masses of the ball and block respectively u1 = velocity obtained by the ball after falling through 27′′, the velocity by which ball hits the block; V2 = velocity obtained by the block; V1 = velocity of ball after impact;
m1
2 7′′
Conservation of momentum gives. m1u1 + 0 = m1V1 + m2V2 K.E. obtained in 27”(= 9/4 ft) descent
9 1 m1u12 = m1gh = m × 32 × 4 2 u1 = 12 ft/sec.
or
m2
Fig. 12
Since collision is head on, we have V2 = Then from (i); or
2 u1 2 × 12 24 = = = 4 ft /sec. 1 + m2 / m1 1 + 5 / 1 6
1 × 12 = 1 × V1 + 5 × 4 V1 = – 8 ft/sec.
thus, ball recoils in opposite direction with a velocity = 8 ft/sec. Q. 7. A moving particle of mass m makes head on collision with a particle of mass 2m which is initially at rest. Show that the colliding particle looses (8/9)th of its energy after collision. Ans. Let initial velocity of colliding particle be ‘u’ and final velocity ‘ V ’, after collision. If V1 is velocity of the other particle after collision, from law of conservation of momentum. mu = mV + 2mV1
u− V 2 Again from law of conservation of energy
which gives,
V1 =
1 1 1 mu2 = mV2 + . 2m. V12 2 2 2
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Mechanics
or
2 u2 = V + 2
or
u 2 = V2 +
FG u − V IJ H 2 K
2
u2 V2 + − uV 2 2
3V2 = u2 + 2uV
or
3V2 – 2uV – u2 = 0
or
which gives final velocity of colliding particle, which is V =
2u ±
4 u2 + 12u2 6
= u or −
u 3
Since +ve value of V will not satisfy the law of conservation of momentum. ∴
V = −
u 3
Therefore, loss in K.E. of colliding particle ∆U =
=
FG IJ H K FG1 − 1 IJ = 8 initial K.E. H 9K 9
u 1 1 mu2 − m − 2 2 3
1 mu2 2
2
Q. 8. A bomb moving with velocity 40 i + 50 j − 25 k m/s exploded into pieces of mass ratio 1 : 4. The small piece goes out with velocity 200 i + 70 j + 15 k m/s. Deduce the velocity of larger piece after explosion. Ans. Force of explosion is the internal force acting on bomb. In absence of external forces, thus, total momentum of the system remains constant. ∴ Momentum before explosion = momentum after explosion i.e.,
5M (40 i + 50 j − 25 k ) = M (200 i + 70 j + 15 k ) + 4 MV2
or
4V2 = 200 i + 250 j − 125 k − 200 i − 70 j − 15 k
or
4V2 = 180 j − 140 k
or
V2 = 45 j − 35 k
Q. 9. A body at rest explodes and breaks up into 3 pieces. Two pieces having equal mass, fly off perpendicular to each other with the same speed of 30 m/sec. The 3rd piece has 3 times the mass of each of the other piece. Find the magnitude and direction of its velocity immediately after the explosion.
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261
Ans. Mass ratio is 1:1:3 in explosion. Since only internal force acts; so that momentum is conserved; Momentum before explosion = after explosion →
5M(O) = M × 30 i + M × 30 j + 3M V 3
i.e.,
→ 3 V3 = −30 i − 30 j
or
V3 = −10 i − 10 j = − (10 i + 10 j ) →
or or
|V3| =
102 + 102 = 10 2 m/s
The direction is given by
−10 = −1 ∴ θ = 90 + 45 = 135° 10 Q. 10. A U238 nucleus emits an α-particle and is converted into Th234. If the velocity of the α-particle be 1.4 × 107 m/s and the kinetic energy be 4.1 MeV, calculate the velocity of the recoil and the K.E. of the residual nucleus (Th234). tan θ =
or
Ans. Only internal force act; so total momentum is same before and after the fragmentation (or the nuclear reaction) i.e.,
momentum before reaction = after reaction
or
O × [MU] = MTh VTh + MαVα O = 234VTh + 4 × 1.4 × 107 m/s
or
VTh = –2.4 × 105 m/s
or So,
magnitude = 2.4 × 105 m/s and direction will be opposite to motion of α particle.
Kinetic Energy of Th234 nucleus:
Again;
LM N
1 2 MTh VTh (K.E.)Th 234 2. 4 × 105 2 = = 1 (K.E. )α 4 1. 4 × 107 Mα Vα2 2
OP Q
2
(K.E.)Th 144 × 10−4 = 171.19 × 10−4 = 58. 5 × 4.1 MeV 49 (K.E.)Th = 171.19 × 4.1 × 10–4 = .0704 K.E. of thorium nucleus = 0.0704 MeV Q. 11. A 5 kg object with a speed of 30 m/s strikes a steel plant at an angle 45° and rebounds at the same speed and same angle. What is the change (magnitude and direction) in the linear momentum of the object? Ans. Linear momentum before striking; is as below, →
P1 = mux i − muy j
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Mechanics
Linear momentum after striking is given as; → P2 = mux i − muy j
Change in linear momentum: →
→
→
∆ P = ( P2 − P1 ) = muxi − muy j − muxi + muy j →
∆ P = 2mux j But,
FG 1 IJ H 2K F 1 IJ = u sin 45° = 30 G H 2K
ux = u cos 45° = 30
uy →
∆P = 2 × 5 ×
So,
30 300 j= × 2 2
2 j 2
→
∆ P = 150 2 j
or →
So, magnitude, |∆ P| = 150 2 Kg m/s , in direction perpendicular to steel plate. Q. 12. A particle of mass M, moving with a velocity u, makes a head on collision with a particle of mass m initially at rest so that their final velocities V and v are along the same
2u . If the particle coalesce on 1 + m/M colliding, find the final common velocity and the loss in K.E. line. Assuming an elastic collision, prove that v =
Ans. (i) Law of conservation of momentum gives; Mu = MV + mv
...(i)
and law of conservation of energy gives;
1 1 1 Mu2 = MV2 + mv2 2 2 2 u 2 = V2 + (m / M) v2
...(ii)
Putting for V from (i), we obtain,
F Mu − mv IJ = G H M K F mI = u +G J v H MK
2
u2
+
m 2 v M
2
u2
2
2
− 2u
m m 2 v+ v M M
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263
2u
LMF m I + F m I OP MNGH M JK GH M JK PQ L m O or 2u = v M + 1P NM Q 2
2 m v = v M
v=
2u 1 + m/M
(ii) When particles stick together (inelastic collision); conservation of momentum gives Mu = (m + M)v v =
Loss of energy =
∴
M .u (m + M) 1 1 Mu2 − (M + m) v2 2 2
=
M2u2 1 1 Mu2 − ( M + m) 2 2 ( m + M)2
=
1 1 M2u2 Mu2 − 2 2 ( m + M)
=
1 Mu2 (m + M) − M2u2 1 Mmu2 + M2u2 − M2u2 = 2 (m + M) 2 m+M
Loss of energy, ∆E =
LM N
OP Q
LM N
OP Q
1 Mmu2 2 ( m + M)
Q. 13. A radioactive nucleus initially at rest, decays by emitting an electron and a neutrino at right angles to one another. The momentum of the electron is 1.2 × 10–22 kg-m/s and that of the neutrino is 6.4 × 10–23 kg m/s. Find the direction and magnitude of the recoil nucleus. If its mass is 5.8 × 10–26 kg, deduce K.E. of recoil. Ans. The nucleus is at rest, before decay. Let Pe = momentum of electron; Pn = momentum of neutrino; PN = momentum of recoil nucleus. Before decay; total momentum of nucleus is zero. In decay process, no external forces act (only internal one) so momentum remains constant before and after decay. x-component of momentum O = Pe – PN cos α PN cos α = Pe
...(i)
y-component of momentum
Fig. 13
O = Pn – PN sin α PN sin α = Pn (i) and (ii) give;
PN =
Pe2 + Pn2
...(ii)
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Mechanics
=
(1. 2 × 10−22 )2 + (6. 4 × 10−23 )2
=
1. 44 × 10−44 + 40. 96 × 10−46
So, magnitude of momentum of recoil nucleus =
1. 85 × 10−22 Kg-m/s
Its direction will be given by Pn 6. 4 × 10−23 0. 64 = = = 0. 53 Pe 1. 2 1. 2 × 10−22 α = tan–1 0.53 = 28° (nearly).
tan α =
From the given diagram, θ = 180 – α = 180 – 28 = 152° recoil nucleus will move making angle 152° from electron motion or 242° from neutrino’s path. K.E. of recoil nucleus; K.E. =
P2 1 MN VN2 = N 2 2 MN
( 1. 85 × 10−22 )2 1. 85 × 10−44 = = 2 × 5. 8 × 10−26 11. 6 × 10−26 =
18.5 1. 85 × 10−18 J = × 10−19 J 11.6 11. 6
=
18. 5 eV 11. 6
So, K.E. of recoil nucleus = 1.58 eV (nearly) Q. 14. A body of 3kg makes elastic collision with another body at rest and afterwards continues to move in the original direction but with one half of its original speed. What is the mass of the struck body? Ans. Let, mass of struck body = M Velocity of struck body = V Velocity of moving body which collides = v From law of conservation of momentum, Total momentum before collision = Total momentum after collision 3× v = 3×
v + MV 2
Law of conservation of energy gives
FG IJ H K
1 v 1 × 3v2 = ×3 2 2 2 9 2 v = MV2 4
2
+
or,
1 MV2 2
V=
3v 2M
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Substituting the value of V we get
FG IJ H K
9 2 3v v =M 4 2M
2
or
M = 1 kg.
Q. 15. The empty stages of a two stage, rocket separately weigh 6000 and 500 kg and contain 40000 and 3500 kg fuel respectively. If the exhaust velocity is 1.8 km/s. Find the final velocity. (loge 10 = 2.3; log10 2 = 0.30). Ans.
Total mass of rocket = (6000 + 500 + 40000 + 3500) kg = 50,000 kg
Mass at the end of I stage = 6000 + 3500 + 500 = 10,000 kg exhaust velocity = 1.8 km/s Thus, velocity at the end of I stage, V1; is V1 = V0 + v log e
FG M IJ = 0 + 1.8 log FG 50, 000 IJ H MK H 10, 000 K 0
e
10 = 1. 8 [log e 10 − log e 2] 2 = 1.8 × [2.3 – 2.3 × 0.3] = 1.8 × [2.3 – 0.69] = 1. 8 log e 5 = 1. 8 log e
= 1.8 × 1.61 = 2.898 V1 = 2.898 km/s, which gives initial velocity of second stage For II stage,
V0 = 2.9 (nearly)
M = 500 kg
M0 = 3500 + 500 = 4000 V2 = V0 + 1. 8 log e
FG 4000 IJ = 2.9 + 1.8 log 2 H 500 K
3
e
= 2.9 + 1.8 × 3 × 2.3 log10 2 = 2.9 + 1.8 × 3 × 2.3 × 0.3 = 2.9 + 3.726 = 6.626 km/sec ∴ Final velocity at the end of II stage = 6.626 km/sec. Q. 16. A 5000 kg rocket is set for vertical firing. If the exhaust velocity is 500 m/s, how much gas must be ejected per second to supply the thrust needed to (i)
Overcome the weight of rocket.
(ii)
Give rocket an initial upward acceleration 19.6 m/sec2
Ans. The net upward force on rocket at an instant is given by
dV dM = −V − Mg dt dt To just overcome the weight of rocket, net force is zero, i.e., F = M.
0 = −V or
dM − Mg, dt
dM Mg 5000 × 9. 8 = = 98 kg/sec = dt V 500
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Mechanics
i.e., rate of gas ejection should be 98 kg/sec. (ii) If ‘a’ is the upward acceleration of rocket, we have F = −V
dM − Mg = Ma dt
dM 1 1 . 5000 (9. 8 + 19. 6) = − M ( g + a) = − dt V 500 = –294 kg/sec.
or,
So, gas ejection should be at the rate of 294 kg/sec. Q. 17. A rocket starts vertically upwards with speed u0, show that its speed v at a height h is given by u02 − V 2 =
2 gh . h 1+ R
where R is the radius of earth, and g is acceleration due to gravity at earth’s surface. Hence deduce an expression for maximum height reached by a rocket fired with speed 90% of escape velocity. Ans. The velocity of rocket at height h is
z
h
2 V2 = u0 − 2 g ′ dh
...(i)
o
[As rocket rises up, h and g change] At earth surface
g =
GM R2
and g′ at height h is given by g′ =
GM GM g = 2 = (R + h)2 R (1 + h /R)2 (1 + h / R)2
putting the value of g′ in eqn. (i)
z
h
V2
=
u02
−2
0
g . dh (1 + h/R )2
LM R OP N 1 + h/R Q L R − R OP + 2g M N 1 + h/R Q L R − R − h OP = u + 2g M N 1 + h /R Q h
V2 = u02 − 2 g −
0
= u02
2 V2 = u0
2 0
−
2 gh 1 + h /R
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V2 – u02 = −
2 gh 1 + h/R
or u02 − V2 =
2 gh 1 + h/ r
Since the escape velocity = 2gR , the escape velocity in present case = 0 .9 2 gR . At highest point attained by rocket V = 0, hence the above equation gives 0 – 0.81 × 2gR =
0.81 =
2gh .R R+h h R+h
or 0.81R + 0.81h = h
h – 0.81h = 0.81 R h =
81 R = 4.26 R 19
Q. 18. The weight of an empty rocket is 5000 kg and it contains 40,000 kg fuel. If the exhaust velocity of the fuel is 2 km/s, find the maximum velocity attained by the rocket. (loge 10 = 2.3 and log10 3 = 0.4171). Ans. When gravity effect is neglected, the velocity of the rocket at any instant is given by; V1 = V0 + V log e
M0 M
where, M0 = initial mass; M = mass at any instant, V = exhaust velocity; V0 = initial velocity, V1 is velocity of the rocket at any instant. Maximum speed is obtained when all the fuel is exhausted. Now,
V0 = 0, m0 = 5000 + 40,000 = 45000 kg, M = 5000 Kg, V = 2 km/sec.
So,
Vmax. = 0 + 2 loge (45000/5000) = 2 loge 32 = 4 loge 3 = 4 × 2.3 × log10 3 = 4 × 2.3 × 0.4771 = 9.2 × 0.4771 km/s = 4.4 km/s
Q. 19. A rocket of mass 30 kg has 200 kg of fuel and the exhaust velocity of fuel is 1.6 km/sec. Calculate the minimum rate of consumption of fuel, so that the rocket may rise from the ground. Also calculate the final vertical velocity gained by the rocket when fuel consumption rate is (i) 2 kg/sec. (ii) 20 kg/sec. Ans. Rocket rises from the ground when fuel consumption occurs at minimum rate such that the thrust, balances the initial weight of rocket. ∴
Net force on rocket = 0
V Now,
dM = Mg dt M = 200 + 30 = 230 kg V = 1.6 km/s = 1.6 × 103 m/s
...(i)
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Mechanics
from (i);
dM Mg 230 × 9. 8 = = 1409 kg/s = dt V 1. 6 × 103
Final velocity,
V = V0 + V loge (M0/M) – gt
Case I:
V = 1.6 km/s = 1.6 × 103 m/s t =
given,
200 kg = 100 sec. 2 kg/sec
g = 9.8 m/s2 v0 = 0 M0 = 230, M = 30 Vmax = 1.6 × 103 × loge (230/30) – 9.8 × 100 = 1.6 × 103 × 2.3 log10 (23/3) – 980.0 = 2.284 km/s
Case II: All the quantities will be same as in case I: but,
t = ∴
200 kg = 10 sec 20 kg/s
Vmax = 3.16 km/s
Q. 20. A body moving in straight line suddenly explodes into two parts. If one of these is twice as heavy as the other one and the two parts move in opposite direction with same speed, show that mechanical energy released in explosion is at least 8 times the initial kinetic energy. Ans. Let, V1 and V2 be the speeds of smaller and larger parts. Then using law of conservation of momentum –mV1 + 2mV2 = 3mV But as given,
(V-speed of original body)
V2 = –V1 = V′ (say) –mV′ + 2mV′ = 3 mV V′ = 3 V
This means each part moves with thrice the initial speed of original body in opposite direction. So, from law of conservation of energy, the energy released in explosion. ∆U = =
LM N
1 1 1 (3m) V2 − m(3V)2 + × 2m (3V)2 2 2 2
OP Q
1 ( 3m) V2 1 − ( 3 + 6) = − 8 U 2
where U is initial K.E. This proves that released energy is 8 times the initial energy. Q. 21. A bullet of mass m moving with horizontal velocity v strikes a stationary block of mass M suspended by a string of length L. The bullet gets embeded in the block. What is the maximum angle made by the string after impact. Ans. Suppose block and bullet system moves horizontally with speed V, after bullet gets embeded. Then from law of conservation of momentum:
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mv = (M + m) V V =
mv M+m
The K.E. gained by block will be converted into P.E. when string reaches final deflected position. Hence if θ = is the maximum angle string makes after impact,
1 (M + m) V2 = (M + m) gh 2 1 2 V = g (L − L cosθ) 2 V =
2 gL (1 − cos θ) =
4 g L sin2 θ /2
Putting the value of V
mv = M+m sin
4 g L sin2 θ/2
mv θ = 2 2 ( M + m) − θ = 2 sin 1
gL
LM mV N 2 (M + m)
1/ gL
OP Q
Q. 22. The maximum and minimum distances of a comet from the sun are 1.4 × 1012 m and 7 × 1010 m. If its velocity nearest to the sun is 6 × 104 m/s, what is its velocity when it is in farthest position. Assume in both position that the comet is moving in circular path. Ans. From law of conservation of angular momentum; m1V1r1 = m2V2r2 (for circular orbit) here,
m 1 = m2 = m V1r1 = V2r2 V2 =
V1r1 6 × 104 × 7 × 1010 = r2 1. 4 × 1012
V2 = 3000 m/s Q. 23. A particle of mass m performs motion along a path which is given by equation →
r = i a cos wt + j b sin wt .
Calculate the angular momentum and torque about the origin. Ans.
∴
r = i a cos ωt + j b sin ωt
→
→
→
dr = ω ( i a sin ωt + j b cos ωt ) V = dt
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Mechanics
The angular momentum is given by →
→
→
→
→
J = r × P = r × mV = (i a cos ωt + j b sin ωt ) × m ( i a sin ωt + j b cos ωt ) ω = ab m k ω (cos2 ωt + sin2 ωt ) = m ω ab k
by definition, torque is, →
→
dJ τ = dt
∴
→
τ =
d (m ω abk ) dt
= 0 (as all quantities are constant) Q. 24. A particle of mass 20 gm moving in a circle of 4 c.m. radius with constant speed of 10 cm/s. What is its angular momentum (1) about the centre of the circle (2) a point on the axis of the circle and at 3 cm distance from its centre. Ans. (1)
J = mvr = 20 × 10 × 4 = 800 erg. sec.
indirection perpendicular to plane of circle. (2)
QP = distance of particle from point of reference =
So,
32 + 42 = 5 cm
J = mvr = 20 × 10 × 5 = 1000 erg. sec.
Its direction will also be perpendicular to plain containing the radius vector (QP) and →
instantaneous velocity; [Note: The direction of J change in this case in First case direction →
of J does not change]. Q. 25. A particle of mass 1/2 gm starts from the point (1, 0, 2) at a time t = 0. If force acting on it is 2 i + (1 + 6t ) j dynes. Calculate the torque and angular momentum about the point (2, 2, 4) at time t = 1 sec. Ans. Force
→
d2 r F = m 2 ; dt →
→
{
→
or
d2 r = 2 2 i + (1 + 6t ) j 2 dt
→
F d2 r = 2 m dt
}
Integration yields; →
{
}
dr = ...(1) 2 2ti + ( t + 6 t 2 / 2) j + K dt (K = constant of integration)
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as velocity,
V = 0, at t = 0; (1) then yields, K = 0 →
dr = 4 ti + (2t + 6t 2 ) j dt
So (1) is
...(2)
Further integration gives →
2 2 3 r = 2t i + ( t + 2t ) j + k ′ Given at t = 0; particle is at (1, 0, 2)
...(3)
→
ro = 1 i + 0 j + 2 k = i + 2 k = k′
→
r = 2. t2 i + ( t2 + 2t 3 ) j + i + 2k
thus, So, at t = 1;
→
r1 = 3 i + 3 j + 2 k
...(4) ...(5)
→
Distance R from given point of reference; (2 i + 2 j + 4 k) is ∴
→
R = 3 i + 3 j + 2 k − 2 i − 2 j − 4 k = i + j − 2 k
Thus at time t = 1, (refer eqns. (5), (2) and (1)) →
R = i + j − 2k →
V = 4 i + 8 j →
F = 2 i + 7 j So, torque;
→
→
→
OP PP PQ
...(6)
τ = R×F
=
i
j
k
1 2
1 7
−2 0
= i ( 0 + 14) − j ( 0 + 4) + k (7 − 2) →
τ = 14 i − 4 j + 5k →
→
and the angular momentum at t = 1, is = R × m V
i 1 1 = 2 4
j 1 8
k −2 0
1 i (16) − j ( 0 + 8) + k ( 8 − 4 ) 2 = 8 i − 4 j + 2k
=
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Mechanics
Q. 26. A particle of mass m moving in a circular orbit of radius r has angular momentum →
L about the centre. Calculate the K.E. of the particle and the centripetal force acting on it. Ans. Angular momentum of a particle moving in circular path L = mr2ω;
So,
rω =
or,
K.E. =
1 1 mv2 = m( rω )2 2 2
K.E. =
1 1 m 2 mr
FG IJ H K
2
=
1 mr
1 L2 2 mr2
FG IJ H K
mv2 m m L = .( rω)2 = . Centripetal force = r r r mr =
2
L2 mr3
Q. 27. A meter stick of mass M lies on a smooth horizontal table. A fall of mass m moving with velocity V collides elastically and normally with stick at a distance d as shown in figure. Find the velocity of ball after collision. Ans. The linear momentum, angular momentum as well as K.E. will be conserved in elastic collision. Let,
v′ = velocity of ball after collision, V = velocity of centre of mass of stick, ω = the angular velocity of stick after collision,
and
I =
M × l2 , the moment of inertia of stick about centre of mass. 12
From law of conservation of linear momentum mV = MV + mV′ therefore
V =
m ( V − V′ ) M
...(i)
Law of conservation of angular momentum gives mVd = mV′d + Iw
Mwl2 12 12 m (V − V ′ ) d ω = Ml2 Also law of conservation of K.E. yields m(V – V′)d =
1 1 2 1 2 2 1 mV2 = mV ′ + Iω + MV 2 2 2 2
...(ii)
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Mw2 l2 + MV2 12 Putting the values of V and ω in eqn. (iii) m ( V2 − V ′2 ) =
m ( V2 − V ′2 ) =
which gives
V + V′ =
...(iii)
12m2 d2 ( V − V ′ )2 l2 m2 ( V − V ′ )2 + M M l4 m ( V − V′ ) (12 d2 + l2 ) 2 Ml
m ( V − V ′ ) (12 d2 + l2 ) −V Ml2 Q. 28. Determine the impact parameters of 2 MeV, α-particle whose distance of closest approach to a gold nucleus (Z = 79) is 2 × 10 –11 cm.
V′ =
Ans.
Initial K.E. =
1 mu2 = [2 × 106 × 1. 6 × 10−12 ] ergs 2 = [3.2 × 10−6 ]
Angular momentum is conserved during scattering At G; Angular momentum of α++ particle about N = mαPu Angular momentum of
α++
particle about N: (when approach is closest) = mα.S.V
or
...(i) ...(ii)
mα Pu = mα.S.V V =
Energy conservation gives,
Pu S
1 Ze.(2e) 1 mu2P2 2 Ze2 1 mu2 = mV2 + = + 2 2 2 S2 S S
1 1 mu2P2 2 Ze2 mu2 − = 2 2 S2 S
F GH
1 P2 mu2 1 − 2 2 S
I JK
=
2 Ze2 S
1−
2 Ze2 2 P2 × = 2 s mu2 s
1−
P2 2 × 79 × (4 . 8 × 10−10 )2 1 = 0. 57 . = 2 − 11 S 2 × 10 3. 2 × 10−6
P2 = 1 – 0.57 = 0.43 S2 P = S 0.43 = 2 × 10−11 0. 43 P = 1.3 × 10–11 cm
...(iii)
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Mechanics
Q. 27. Calculate the angular momentum of 1 MeV neutron about a nucleus of impact parameter 10 – 10 cm. (m = 1.7 × 10–24 gm). Ans.
Initial K.E. =
1 mu2 = 1 × 106 eV 2
= 1.6 × 106 × 10–12 ergs = 1.6 × 10–6 ergs.
LM 2 × 1.6 × 10 OP N 1.7 × 10 Q −6
So,
u =
−24
1/2
cm/s
Neutron being a light neutral particle experiences no force and goes straight way (P = impact parameter). Its angular momentum. = m ×P× u −24 × 10−10 × = 1. 7 × 10
LM 2 × 1. 6 × 10 OP N 1.7 × 10 Q −6
1/ 2
−24
= 1.93 × 10–25 gm-cm/sec. Q. 30. A light weight man holds heavy dumb-bells in straightened arms when standing on a turn table. The turn table then start rotating at rate of 1 rev/min. The man then pulls the dumb-bells inside toward chest. If initially dumb-bells are 60 cms from his axis of rotation and are pulled to 10 cms from rotation axis, find the frequency of revolution of turn table. Neglect the angular momentum of man in comparison to that of dumb-bells. Ans. Initial angular momentum of system (neglecting the angular momentum of man) = 2Mω1 r12; where M is mass of dumb-bells, ω is initial rotational frequency, r1 is distance of dumb-bells from axis. Final angular momentum when arms are stretched; = 2Mω2 r22, where ω2 and r2 are changed frequency and distance due to arm stretching. Since no external torque is applied, angular momentum will be conserved. i.e.,
2Mω1 r12 = 2Mω2 r22 2πf1 r12 = 2πf2 r22
f1 r12 1 × ( 60)2 3600 = = 100 (10)2 r22 f2 = 36 rev/min. f2 =
Q. 31. Compute the orbital angular momentum and total energy of the electron in hydrogen atom, assuming the path to be circular. Ans. Hydrogen atom is one electron and one proton system. The required centripetal force for circular motion is provided by Coulombian attraction force; ∴
Ze.e e2 mV2 = 2 = r2 r r mV2r2 = e 2r
(mVr)2 = me 2r
...(i)
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putting the standard values, m = 9 × 10–28 gm; e = 4.8 × 10–10 esu; r = 0.5 × 10–8 cms: The angular momentum of revolving electron is J = mVr =
me2r
9 × 10−28 × (4. 8 × 10−10 )2 × (0. 5 × 10−8 )
J =
= 4. 8 × 10−28 × 3 0. 5 = 14. 4 × 10−28 = 1.02 × K.E. =
also
P.E. = − Total energy =
g
0. 5
cm2/sec.
e2 1 mV2 = 2 2r
from Eqn. (i)
∴
10–27
e2 r
e2 e2 e2 − ( 4. 8 × 1010 )2 − =− = 2r 2r 2 × 0. 5 × 10−8 r
= –23.05 × 10–12 ergs
−23. 05 × 10−12 = − 14. 4 eV = 1. 6 × 10−12 Q. 32. A nucleus of mass m emits a gamma ray photon of frequency νo. Show that the
FG H
loss of internal energy by the nucleus is not hνo but is hνo 1 + h
IJ K
νo . 2mc2
Ans. The initial momentum of nucleus is zero. Let its momentum after emission of photon be P. The momentum of photon =
h νo c
The conservation of linear momentum suggests, 0 = P+
h νo − hνo ; or P = c c
Then the internal energy which has converted into K.E. of nucleus; K.E. =
( − hνo )2 P2 = 2m 2mc2
Internal energy which has converted into radiant energy of photon = hνo So total loss in the internal energy of nucleus =
FG H
( hνo )2 hνo + hνo = hνo 1 + 2 2mc 2mc2
IJ K
276
Mechanics
Q. 33. Locate the centre of mass of a system of three particles of masses 1 kg, 2 kg and 3 kg placed at the corners of an equilateral triangle of 1 meter. Ans. Choosing the co-ordinate system as shown in figure: Position of m1 = 0, Position of m2 = 1 i
1 1 3 i + j sin 60 = i + j 2 2 2
Position of m3 =
→
→
→
m r1 + m2 r2 + m3 r3 R = m1 + m2 + m3 →
Then position of C.M.
1 1 × 0 + 2 × (1 i) + 3 × (i + 2 = 1+2+3 2 i + =
3 3 3 i+ j 2 2 6
F 7 i + GH 12
=
3 j )
3 j 4
I JK
Y (m 3 ) 3 kg
( ^j )
O
60 1 kg (m 1 )
2 kg (m 2 )
^ (i)
X
( k^ ) Z
So co-ordinates of C.M. is,
F7, GH 12
Fig. 14
I JK
3 , 0 meters. 4 →
Q. 34. A particle moves in a force field given by F = f ( r) r, where r is the unit vector along position vector of particle. Prove that the angular momentum of the particle is conserved. Ans. Torque acting on particle is →
→
→
→
→
→
τ = r × F = r × f ( r) r = f ( r) ( r × r), τ = 0
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Since external torque on the particle is zero, so according to theorem of conservation of angular momentum, its angular momentum shall be conserved. Q. 35. A stream of α-particles is bombarded on a mercury nucleus (z = 80) with velocity 1 × 107 m/s. If an α-particle is approaching the nucleus in head on direction. Calculate the distance of closest approach. Ans. Distance of closest approach (d) for α-particles is given, in head on approach, by d =
1 4 Ze2 4 π ∈o mVo2
9 × 109 × 4 × 80 × (1. 6 × 10−19 )2 = (6. 4 × 10−27 ) × (1 × 107 )2 = 1.15 × 10–13 m Q. 36. Express in term of angular momentum (J), the kinetic, potential and total energy of a satellite of mass m in a circular orbit of radius r. Ans. Angular momentum of the satellite in circular orbit is; (a) J = mVr if the K.E. be T; then T =
1 J2 mV2 = 2 2mr2
(b) Potential energy of satellite is: U = −
...(1)
GMm , where M = mass of the earth. But for r
a satellite moving in circular orbit we have;
GMm mV2 , = 2 r r then,
GMm = mV2 r
U = –mV2 and with J = mVr U = −
(c) total energy (E) = K.E. + P.E. =
then in view of Eqn. (2):
or
E = −
J2 mr2
...(2)
1 1 mV2 − mV2 = − mV2 2 2 J2 2mr2
Q. 37. “The ratio of maximum to minimum velocity of a planet in its circular orbit equals the inverse of the ratio of the radii of the orbits” – Establish the statement. Ans. Let V1 and V2 be the maximum and minimum velocity of the planet with mass m and r1 and r2 be the radii of the circular orbits. The principle of conservation of angular momentum suggests; mV1r1 = mV2r 2 or
r2 V1 V2 = r1 hence the statement.
278
Mechanics
Q. 38. Calculate the momentum of an electron accelerated by a potential of 100 volts. Ans. K.E. of electron,
T =
P2 1 1 mV2 = (mV)2 = 2 2m 2m
where P = mV = linear momentum of electron. But
T = eV; so we have eV =
P2 , or, P = 2me V 2m
here, m = 9 × 10–28 gm, e = 4.8 × 10–10 esu, V = 100 volts = P =
100 stat volts. 300
2 × 9 × 10−28 × 4. 8 × 10−10 × 100 / 300
= 5.37 × 10–19 gm cm/sec. Q. 39. A meteorite burns in the atmosphere before it reaches the earth surface. What happens to its momentum. Ans. Taking the earth, atmosphere and meteorite to form a closed system, the momentum lost by meteorite is taken up by the combustion products, molecules of the atmosphere (i.e., air) which interact with the meteorite and the earth. The momentum lost and gained remain same and so the total momentum of system is conserved. Q. 40. A 20 gm bullet passes through a plate of mass M1 = 1 kg and then comes to rest inside a second plate of mass M2 = 2.98 kg. It is found that the two plates initially at rest now move with equal velocities. Find the percentage loss in the initial velocity of the bullet when it is between M1 and M2. Neglect any loss of material of the plates due to action of bullet. Ans. Let V1 = initial velocity of bullet; V2 = bullet velocity after emergence from first plate M1. Also let m be the mass of the bullet and V be the velocity of plates after impact. Collision is inelastic, so linear momentum is conserved i.e., for first plate. mV1 = mV2 + M1V
...(i)
For second plate when bullet enters it with initial velocity V2, the conservation of momentum gives; mV2 = (m + M2)V from (i);
...(ii)
M1V = m(V1 – V2)
or
V =
m ( V1 − V2 ) M1
...(iii)
Using this value in eqn. (ii)
m mV2 = (m + M2 ) M ( V1 − V2 ) 1 M1V2 = (m + M2) (V1 – V2) = mV1 – mV2 + M2V1 – M2V2 or or
V2(m + M1 + M2) = (m + M2)V1 V2 =
(m + M2 ) V1 (0. 02 + 2. 98) V1 = = 0. 75 V1 (m + M1 + M2 ) (0. 02 + 1 + 2. 98)
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So loss in velocity after emergence from first plate = V1 – V2 = V1 – 0.75V1 = 0.25V1 hence % loss =
0. 25 V1 × 100 = 25% V1
Q. 41. A .5 kg block slides down from the point A as shown in figure on a horizontal track with an initial speed of 3 m/s towards weightless horizontal spring of length 1 m and force constant 2 Nt/meter. The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.2 respectively. If the distance AB and BD are 2m and 2.14 m respectively, find total distance through which block moves before it comes to rest completely. Ans. Speed of block when it comes at B is same as that at A because path AB is frictionless. Then K.E. of block at position B is =
1 1 9 Joules. mV2 = × 0. 5 × 32 = 2 2 4
Fig. 15
Suppose the block after travelling frictionful path BD, further travels distance x in which it comes to rest after compressing the spring. Total frictionful path traversed in this journey is (BD + x) i.e., (2.14 + x). Work done in this journey against force of friction is; W(BD + x) = µk.mg × (2.14 + x) J = 0.2 × 0.5 × 10 × (2.14 + x) J = (2.14 + x) J. Work done in compressing the spring by x;
1 2 1 Kx = 2 × x2 = x2 Joule 2 2 So total work done = (2.14 + x + x2) J then loss of K.E. of block = work done
1 mV2 − 0 = (2.14 + x2 + x) 2 1 × 0. 5 × 32 = 2.14 + x + x2 + x) 2 1 −1 ± 1 + 4 × .11 = 0.1 m 2 Now, the compressed spring forces back the block with a force. x2 + x – 0.11 = 0;
or;
x=
280
Mechanics
F = Kx = 2 × 0.1 = 0.2N But force of friction on the block = µs mg = 0.22 × 0.5 × 10 = 1.1 N It is evident that block will not move back as the pushing back force is smaller than the frictional force. Then total distance moved by the block. = AB + BD + x = 2 + 2.14 + 0.1 = 4.24 m Q. 42. Calculate the intrinsic angular momentum of (i) photon, (ii) electron and (iii) πmeson. Ans. (i) Photon: Ps = S (S + 1) .
So,
h , S = 1, h = 6.626 × 10–27 erg-sec. 2π 1 (1 + 1) ×
Ps =
6. 626 × 10−27 2 × 3.14
= 1.414 × 1.053 × 10–27 = 1.49 × 10–27 gm cm2/sec.
1 h ; here S = 2 2π
(ii) Electron: Again
Ps =
S (S + 1).
So,
Ps =
1 6. 62 × 10−27 (1 / 2 + 1) × 2 2 × 3.14
= (iii) So
1. 732 × 1. 054 × 10−27 = 9.127 × 10−28 gm cm2/s 2
π-meson Ps =
S (S + 1) ×
h ; here S = 0 2π
PS = 0
Q. 43. A cart of mass is M moving with a constant velocity v on a horizontal track. A monkey of mass m jumps from a tree on to cart just from above. Find the velocity of cart after the event. Ans. Considering cart and monkey as a single system, it is noted that there is no external horizontal force on the system. Supposing that cart is moving in the +x distance towards right, it is found, initial horizontal momentum of monkey = 0. Initial horizontal momentum of cart = Mv The final horizontal momentum of monkey–cart system if after the event system moves with velocity V. = (M + m) V The horizontal momentum of the system shall be conserved in the absence of external force, thus O + Mv = (M + m) V So;
V =
Mv M+ m
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SELECTED PROBLEMS 1.
State and explain the law of conservation of linear momentum, Explain the following (a)
A meteorite burns in the atmosphere before it reaches earth. What happens to its linear momentum ?
(b)
When a ball is thrown up, its momentum first decreases, then increases. Does this violate the principle of conservation of momentum.
(c)
Show that conservation of linear momentum is equivalent to Newton’s third law.
2.
What do you mean by “Centre of mass” of a system of particles/show that in absence of any external force the velocity of the centre of mass remains constant.
3.
Show that in head on elastic collision between two particles, the transfer of energy is maximum when their mass ratio is unity.
4.
Two bodies of mass M and m (at rest) collide and stick. Describe the collision in centre of mass frame. What is the ratio of kinetic energy transferred.
5.
Describe principle of rocket. Establish the relation for the motion of rocket in horizontal motion, V = V0 + log e
6.
7.
M0 , where the symbols have their usual meanings. M
(a)
What is multistage rocket. Discuss its motion when it is moving in a field fee space with frictional forces are present as also when it moves in a region where gravitational force is present.
(b)
What is multistage rocket? What is the advantage of a two stage rocket on a single stage rocket.
Show that the rocket speed is equal to the exhaust speed when the ratio Show that the rocket speed is twice the exhaust speed when
M0 = e2 . M
M0 = e2 . M
CENTRE OF MASS : CONSERVATION OF LINEAR MOMENTUM Q. 1. Show that centre of mass of two particles must lie on the line joining them, and the ratio of the distances of the two particles from the centre of mass is the inverse ratio of their masses. →
Solution. Let there be two particles of masses m1 and m2 whose position vectors are →
r1 and r2 with respect to a fixed origin O. Let c be the centre of mass whose position vectors →
→
with respect to m1 and m2 are d and d . 1 2 The position vector of C with respect to the origin O is defined by →
→
rcm
→
m r + m2 r2 = 1 1 m1 + m2
Now by the law of vector addition, we have →
→
→
r1 + d1 = rcm and
→
→
→
rcm + d2 = r2
282
Mechanics Y
→ m 1 d1
C d2
→ rc m
→ r1
m2
→ r2 O
X
Z
Fig. 16
These give →
→
→
→
d1 = rcm
→
→
→
m r + m2 r2 → m2 ( r2 − r1 ) − r1 = 1 1 − r1 = m1 + m2 m1 + m2 →
d2 = r2 − rcm
∴
→
→
→
→
→
→ → m r + m2 r2 m1 ( r2 − r1 ) = r2 − 1 1 = m1 + m2 m1 + m2
m2 → → d d1 = m 2 1 →
→
This shows that the vectors d and d are collinear i.e., the centre of mass O lies on the 1 2 line joining m1 and m2. Further, from the last equation, we have →
d1 →
d2
=
m2 m1
Hence the ratio of the distance of the particles from the centre of mass is the inverse ratio of their masses. This also shows that the position of the centre of mass is independent of the origin or the reference of frame chosen. Q. 2. The mass of the moon is about 0.013 times the mass of the earth and the distance from the centre of the moon to the centre of the earth is about 60 times the radius of the earth. Find the distance of the centre of the mass of the earth moon system from the centre of the earth. Take the radius of the earth 6400 km. Solution. Let d be the distance between earth and moon. Then if x be the distance of the centre of mass from earth, its distance from the moon will be d – x. The ratio of x to d – x will be equal to the inverse ratio of the masses of earth and moon. That is,
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283
x mass of moon = 0. 013 = d− x mass of earth x = (d – x) 0.013
or or
x (1 + 0.013) = 0.013 d
or
x =
0. 013 d = 0. 0128 d . 1 + 0. 013
Here
d = 60 × 6400 km
∴
x = 0.0128 × (60 × 6400 km) = 4928 km.
Q. 3. The distance between the centres of carbon and oxygen atoms in the carbon monoxide gas molecule is 1.130 × 10–10 meter. Locate the centre of mass of the molecule relative to the carbon atom. Solution. Let the molecule be along the x-axis, the carbon atom being at the origin. The centre of mass relative to the carbon atom is given by
m1 x1 + m2 x2 m1 + m2 where x1 and x2 are the distance relative to the carbon atom. xcm =
(12 × 0) + (16 × 1.130 Å) 12 + 16 = 0.6457 Å Along the line of symmetry.
xcm =
Q. 4. Locate the centre of mass of a system of three particles of masses 1.0 kg, 2.0 kg, and 3.0 kg placed at the corners of an equilateral triangle of 1 meter side. Solution. Let the triangle lie in the x – y plane with its corner m1 (1.0 kg) at the origin and side m1, m2 along the x-axis. Let (xcm, ycm) be the coordinates of the centre of mass C. By the definition of the centre of mass, we have Y
m 3 (3 gm )
Y (cm )
3
yc m
C
m2
m1
O (1 gm )
1
xc m
x (c m )
Fig. 17
(2 gm )
X
284
Mechanics
xcm =
m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3
=
(1. 0 kg) (0) + (2. 0 kg) (1 meter) + (3. 0 kg) (0. 5 meter) (1. 0 + 2. 0 + 3. 0) kg
=
7 meter 12
ycm =
m1 y1 + m2 y2 + m3 y3 m1 + m2 + m3
(1. 0 kg) (0) + (2.0 kg) (0) + (3.0 kg) =
=
(1. 0 + 2. 0 + 3. 0) kg
F GH
3 meter 2
I JK
3 meter 4
The co-ordinates of the centre of mass are
F 7 , 3 I meter . GH 12 4 JK
Q. 5. Two masses 6 and 2 units are at positions 6 i − 7 j and 2 i + 10 j − 8 k respectively. Deduce the position of their centre of mass. →
→
Solution. The position vectors of the masses m1 = 6 and m2 = 2 are r1 = 6 i − 7 j and
r2 = 2 i + 10 j − 8 k .
The position vector of the centre of mass is defined by →
→
m1 r1 + m2 r2 rcm = m + m 1 2
→
=
1 6 ( 6 i − 7 j ) + 2 (2 i + 10 j − 8 k ) 8
= 5 i − 2. 75 j − 2 k The coordinates of the centre of mass are (5, –2.75, –2). Q. 6. Calculate the position of the C.M. of a uniform semi-circular plane lamina. Ans. ACB is a semi-circular lamina of radius a. From symmetry we can conclude that the C.M. lies on OC, the perpendicular bisector of the base AB of the lamina. Let mass of the lamina be M. Let the C.M. be at a distance R from O on the straight line OC.
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|R| = OR =
z
a
1 M
...(1)
r dm
o
Here r is the radius of a semicircular element xy with centre at O and breadth ‘dr’ and mass ‘dm’. Area of entire lamina = ∴
Mass perunit area =
πa2 (semicircular) 2 2M M = 2 πa2 πa 2
...(2)
Area of semicircular element, xy = length × breadth dt = πr dr 2M 2M = 2 . rdr π a2 a
∴ Mass of xy element, dm = ( πr dr) Substituting in (1), we get,
1 OR = M =
2 a2
z
...(3)
L 2 M r drOP = 1 2 M rM Na Q M a LM r OP = 2a = 2a N 3 Q 3a 3
a
2
2
o
3
a
z a
r2 dr
o
3
2
o
C
dr
a
r
A x
0
B y
Fig. 18
The centre of mass of a plane semicircular lamina of radius ‘a’ will be at a height 2a/ 3 from the centre and lies on the perpendicular bisector of the base of the lamina. Q. 7. A system consists of masses 7, 4 and 10 gm located at (1, 5, –3), (2, 5, 7) (3, 3, –1) respectively. Find the position of its centre of mass. Solution. The coordinates of the centre of mass (xcm, ycm, zcm) are given by xcm =
m1 x1 + m2 x2 + m3 x3 7 (1) + 4 (2) + 10 (3) 45 = = 7 + 4 + 10 21 m1 + m2 + m3
ycm =
m1 y1 + m2 y2 + m3 y3 7 ( 5) + 4 (5) + 10 (3) 85 = = 7 + 4 + 10 21 m1 + m2 + m3
286
Mechanics
zcm =
m1 z1 + m2 z2 + m3 z3 7 ( −3) + 4 (7) + 10 ( −1) −3 = = 7 + 4 + 10 21 m1 + m2 + m3
The centre of mass is located at
FG 45 , 85 , −3 IJ . H 21 21 21 K
Q. 8. If the centre of mass of three particles of masses 2, 4 and 6 gm be at the point (1, 1, 1) then where should the fourth particle of mass 8 gm be placed so that the position of the centre of the new system is (3, 3, 3). Solution. Let (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) be the positions of the particles of masses 2, 4, and 6 gm respectively then xcm = =
m1 x1 + m2 x2 + m3 x3 2 x1 + 4 x2 + 6 x3 = 2+4+6 m1 + m2 + m3 1 ( x1 + 2 x2 + 3 x3 ) 6
It is given that (xcm, ycm, zcm) is (1, 1, 1). Thus 1 = or
1 ( x1 + 2 x2 + 3 x3 ) 6
x1 + 2x2 + 3x3 = 6
...(i)
Let (x4, y4, z4) be the position of the fourth particle of mass 8 gm, so that the new centre of mass is (3, 3, 3). Then we have 2 x1 + 4 x2 + 6 x3 + 8 x4 2+4+6+8 x1 + 2x2 + 3x3 + 4x4 = 30
3 =
...(ii)
Subtracting eqn. (i) from (ii), we get x4 = 6 In the same way we get y4 = 6, z4 = 6 Hence 8 gm mass should be placed at (6, 6, 6). → → →
Q. 9. Three particles of masses m1, m2, m3 are at positions r , r , r and are moving with 1 2 3 →
→
→
velocities V , V , V respectively (i) what is position vector of the centre of mass? (ii) what is 1 2 3 the velocity of the centre of mass ? Solution. (i) The position vector of the centre of mass is given by →
→
r cm
→
→
m r 1 + m2 r 2 + m3 r 3 = 1 m1 + m2 + m3 →
(ii) The velocity of the centre of mass is given by v cm
→
→
→
m v 1 + m2 v 2 + m3 v 3 = 1 m1 + m2 + m3
Q. 10. The velocity of two particles of masses m1 and m2 relative to an inertial observer
→
→
are v and v . Determine the velocity of the centre of mass relative to the observer and the 1 2 velocity of each particle relative of the centre of mass.
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287
Solution. The velocity of the centre of mass relative to the observer is given by →
→
m1 v1 + m2 v2 v cm = m1 + m2 The velocity of each particle relative to the centre of mass is, by Galilean transformation of velocities, given by →
→
→
→
v1 = v1 − v cm →
→
→
→
→
→
m v + m2 v2 = v1 − 1 1 m1 + m2 →
m (v − v2 ) = 2 1 m1 + m2 →
→
→
v2 = v2 − v cm
and
→
→
m v + m2 v2 = v2 − 1 1 m1 + m2 →
m (v − v1 ) = 1 2 m1 + m2 Thus, in c-frame, the two particles appear to be moving in opposite directions. Q. 11. Two bodies of masses 10 kg and 2 kg are moving with velocities 2i − 7 j + 3k and – 10i + 35 j − 3k meter/sec respectively. Find the velocity of the centre of mass. Solution. The velocity of the centre of mass in the laboratory frame of reference (Lframe) is defined by n
→ 1 mi vi v cm = m i=1
∑
→
in the case of two bodies only, we have →
v cm
LM N
OP Q
=
→ → 1 m1 v1 + m2 v2 m1 + m2
=
1 10 (2i − 7 j + 3k ) + 2( −10i + 35 j − 3k ) 10 + 2
1 20i − 70 j + 30k − 20i + 70 j − 6 k 12 = 2k meter/sec. Q. 12. In a system comprising two particles of masses 2.0 kg and 5.0 kg the positions of the particles at t = 0 are 4i + 3 j and 6i − 7 j + 7 k respectively, and the velocities are 10i − 6 k =
and 3i + 6 j respectively. Deduce the velocity of the centre of mass and the position of the centre of mass at t = 0 and t = 4 units. Solution.
→
v cm =
LM N
→ → 1 m1 v1 + m2 v2 m1 + m2
OP Q
288
Mechanics
=
1 2. 0 (10i − 6 k ) + 5. 0 (3i + 6 j ) 2. 0 + 5. 0
=
1 (20i − 12k + 15i + 30 j ) 7. 0
=
1 [35 i + 30 j − 12k ] 7. 0
The position vector of the centre of mass at t = 0. →
r cm
LM N
OP Q
=
→ → 1 m1 r1 + m2 r2 m1 + m2
=
1 2. 0 (4 i + 3 j) + 5. 0 (6i − 7 j + 7 k ) 2. 0 + 5. 0
=
1 8. 0i + 6. 0 j + 30i − 35 j + 35k 7. 0
=
1 38i − 29 j + 35k 7. 0
The displacement of the centre of mass in 4 sec is → 4 → ∆ S cm = 4 v = 7. 0 35i + 30 j − 12k ∴ position vector at t = 4 sec.
=
1 4 38i − 29 j + 35k + 35i + 30 j − 12k 7. 0 7. 0
=
1 178i + 91 j − 13k 7. 0
Q. 13. Two particles of masses 100 and 300 gm have position vectors 2i + 5 j + 13k and − 6i + 4 j − 2k and velocity ( 10i − 7 j − 3k ) and ( 7i − 9 j + 6 k ) cm/sec respectively. Deduce the instantaneous position of the centre of mass, and the velocity of the second particle in a frame of reference travelling with the centre of mass.
Solution. The position of the centre of mass in the laboratory frame (L-frame) is given by
LM N
→ → 1 r cm = m + m m1 r1 + m2 r2 1 2
→
OP Q
=
1 100 (2i + 5 j + 13k ) + 300 ( −6i + 4 j − 2k ) 100 + 300
=
1 ( −16i + 17 j + 7 k) cm . 4
290
Mechanics →
F = −6i + 8 j + 12i
Solution.
= 6i + 8 j newton We know that the product of the mass of the system M, and acceleration of its centre →
of mass a cm is equal to the resultant (external) force acting on the system. Thus →
→
a cm
=
F (6i + 8 j) newton = M (3 + 4 + 5) kg
=
1 2 i + j meter/sec2 2 3
its magnitude is
LMF 1 I + F 2 I OP = 0. 83 ms MNGH 2 JK GH 3 JK PQ 2
a =
2
−2
its direction with the x-axis is θ = tan−1
FG a IJ Ha K FG 2 / 3 IJ H 1 / 2K FG 4 IJ = 53°6' H 3K y x
−1 = tan
−1 = tan
Q. 17. Two particles P and Q of masses m1 = 0.10 kg and m2 = 0.30 kg respectively are initially at rest 1.0 meter apart they attract each other with a constant force of 1.0 × 10–2 nt. No external forces act on the system. Describe the motion of the centre of mass. At what distance from the original position of P do the particles collide? Solution. Initially the system is at rest so that its centre of mass is also at rest (zero →
velocity). If M be the total mass of the system and a cm the acceleration of the centre of mass, then →
→ M acm = F ext →
Since no external forces act on the system (F ext = 0) , we have →
a cm
→
= d Vcm = 0 dt
→ Vcm = constant
Hence the centre of mass remains at rest.
...(i)
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291
Let the particle m1 lie at the origin of co-ordinates, and the particle m2 at a distance r from it along the x-axis. The distance of the centre of mass of the system from the origin would be given by xcm =
(m1 × o) + (m2 × r) m2 r = m1 + m2 m1 + m2
...(ii)
Each particle is acted upon by a constant force F. Therefore the accelerations of these particles would be a1 =
F m1
and
a2 =
F m2
Suppose the particles collide after time t at a distance x from the origin. This means that in time t, the particle m1 travels a distance x and particle m2 travels a distance r – x. Thus, using the formula x =
1 2 at , we have 2 x =
1 2 a1t 2
r – x =
1 2 a2t 2
a2 m1 r−x (From above) = = a1 m2 x m1 r −1 = m2 x m1 m + m2 r +1= 1 = m2 m2 x x =
m2 r m1 + m2
Here m1 = 0.10 kg, m2 = 0.30 kg, r = 1.0 m x =
0. 30 (1. 0) = 0. 75 meter. 0.10 + 0. 30
Infact, the particle collides at the centre of mass. Q. 18. A bullet of mass m and velocity v passes through a pendulum bob of mass M and
v . The pendulum bob is at the end of a string length l. What is the 2 minimum value of v such that the pendulum bob will swing through a complete circle.
emerges with a velocity
Solution. In order that the pendulum bob may describes a complete circle, its velocity at the lowest point must be
5 gl . This must be the velocity after collision.
From the law of conservation of momentum.
292
Mechanics
mv = Mx +
mv 2
Mx = x =
mv where x is the velocity of the bob. 2
mv = 2M
5 gl
2M 5 gl. m Q. 19. A weightless string is passing over a light frictionless pulley of diameter 5 cm, and a tray of sand weiging 50 gm is tied at each of its two ends. The system is in equilibrium and at rest. Find its centre of mass. Now, 5 gm of sand is transferred from one tray to another but the system is not allowed to move. Find the new centre of mass. If the system is now set free to move, what would be acceleration of the centre of mass. v =
Solution. In the first case, by symmetry the centre of mass C will be at the mid point of the line joining the centres of the two trays. In the second case, let x be the distance of the first tray (45 gm) from some arbitrary origin on the left and on the same line as joining the centres of the two trays. The distance of the second tray (55 gm) from the same origin would be x + 5. Then the distance of the centre of mass C from the same origin
45 × x + 55 ( x + 5) = x + 2. 75 45 + 55
=
5 cm
5 cm
C
C
50 gm
50 gm
45 gm
(a )
45 gm (b )
Fig. 20
Thus the centre of mass lies at a distance of 2.75 cm from the lighter tray on the line joining the centres. When the system is released, the heavier tray goes down and the lighter one goes up. We know that the product of the total mass of the system and the acceleration of its centre of mass is equal to the vector sum of the external forces acting on all parts of the system. That is →
→
→
M a cm = F1 + F2 (45 + 55) acm = –45g + 55 g = 10 g acm =
10 g g = (down) 100 10
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Q. 20. A 10 kg boy standing in a 40 kg boat floating on water is 20 meter from the shore of the river. If he moves 8 meter on the boat toward the shore then how far is he now from the shore ? Assume no friction between boat and water. C x
C1 2 0m
C1
C2
C
(a )
C2
(b )
(X + 8 )m
d
Fig. 21
Solution. Let C1 be the centre of mass of the boy, C2 that of the boat and C that of the system (boy + boat), as shown in (a). Let x be the distance between C1 and C2. The distance of C1 (10 kg) from the shore is 20 meter and that of C2 (40 kg) is 20 + x meter. Therefore, the distance of the centre of mass C of the system is rcm = (10 × 20) + {40 × (20 + x)} 10 + 40 1000 + 40 x meter = 50 when the boy moves 8 meter towards the shore, the distance between C1 and C2 becomes (x + 8) meter. Let d be the new distance of the boy from the shore. The distance of centre of mass from the shore is given by
l
q
r′cm = (10 × d) + 40 × ( d + x + 8) 10 + 40 50 d + 40 x + 320 meter = 50 As no external forces are acting on the system, the velocity of the centre of mass C of the system remains constant. Initially the centre of mass was at rest, So it must continue to remain at rest i.e., its position will still remain unchanged with respect to shore. Thus rcm = r′cm 1000 + 40x = 50d + 40x + 320 d = 13.6 meter Q. 21. (a) What is the momentum of a 10000 kg truck whose velocity is 20 m/s? (b) What velocity must a 5000 kg-truck attain in order to have the same momentum (c) the same K.E. Solution. (a) The momentum of the truck of mass m, moving with velocity v is given by P = mv = 10000 × 20 = 2 × 105 kg m/s (b)
m 1v 1 = m 2v 2 10000 × 20 = 5000 × v2 v 2 = 40 m/s
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Mechanics
(c)
1 1 m1v12 = m2v22 2 2 10000 × (20)2 = (5000) v22 v 2 = 20 2 = 28. 2 m/s.
Q. 22. Two masses, one n times heavier than other, are dropped from the same height. How do their momenta compare just before they hit the ground. If these masses have same K.E. Then how do their momenta compare.
[Ans. n : 1,
n : 1]
Q. 23. A rifleman, who together with his rifle has a mass of 100 kg stands on a smooth surface and fires 10 shots horizontally. Each bullet has a mass of 10 gm and a muzzle velocity of 800 m/s (a) What velocity does the rifleman acquire at the end of 10 shots. (b) If the shots were fired in 10 sec. What was the average force exerted on him. (c) compare his K.E. with that of the 10 bullets. Solution. (a) Let m1 and m2 be the masses and v1 and v2 be the velocities of the rifleman and the bullet respectively after the first shot. The initial momentum of the system is zero, and so must be the final momentum. That is, m1v1 + m2v2 = 0 m2 v2 m1 Here m1 = 100 kg, m2 = 10 gm = 10–2 kg and v2 = 800 m/s
v1 = −
v1 = −
10−2 (800) = − 0. 08 m/s 100
∴ velocity acquired after 10 shots = 10 v1 = – 0.8 m/s The negative sign signifies that this velocity is opposite the velocity of the bullets. (b) The momentum acquired by rifleman is, P = mass × velocity acquired = 100 × 0.8 = 80 kg m/s This momentum is acquired in 10 sec. Hence the rate of change of momentum, which is equal to the average force exerted, is F =
dP 80 = = 8 nt. dt 10
(c) Kinetic energy of the rifleman, K1 = = Kinetic energy of 10 bullets, K2 =
1 m1 (10 v1 )2 2 1 × 100 × ( 0. 8)2 = 32 joule 2 1 m2v22 × 10 2
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= ∴
1 × 10−2 × (800)2 × 10 = 32000 Joule 2
K1 32 1 = = K2 32000 1000
Q. 24. A machine gun fires 50 gm bullets at a speed of 1000 m/s. The gunner holding the machine gun in his hands, can exert an average force of 180 nt against the gun. Determine the maximum number of bullets he can fire per minute. Solution. The initial momentum of the bullet is zero, while the momentum acquired on fire is P = mv = 50 × 10–3 × 1000 = 50 kg-m/s If n be the number of bullets fired per sec, then the rate of change of momentum is
dP = 50n kg-m/s2 dt But this must be equal to the average force (180 nt) exerted. Thus 50 n = 180 n = =
180 per sec 50 180 × 60 per minute = 216 per minute. 50
Q. 25. Find the average recoil force on a machine gun firing 120 shots per minute. The mass of each bullet is 10 gm and the muzzle velocity is 800 m/s. [Ans. 16 nt] Q. 26. A rocket of mass 20 kg with a capsule of mass 10 kg attached to it, is moving with a velocity of 25000 m/s. A compressed spring when allowed to re-expand, separates the capsule from the rocket which now moves with a relative velocity of 3000 m/s along the same line as the rocket. Calculate the new velocities of the rocket and the capsule. What happens to the kinetic energy of the system? Solution. Let mr and mc be the masses and vr and vc the respective velocities of the rocket and the capsule (after separation). As capsule is lighter, its velocity would be greater. Therefore vc – vr = 3000 m/s. Now, by the conservation of momentum, we have Initial momentum = Final momentum (mr + mc) × v = mrvr + mcvc 30 × 25000 = 20 × vr + 10 × (3000 + vr) 750000 = 30vr + 30000 vr = and
720000 = 24 , 000 m/sec 30
vc = 3000 + vr = 27,000 m/sec. But the rocket and the capsule move forward.
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Mechanics
Now, initial kinetic energy = =
1 (mr + mc ) v2 2 1 × 30 × (25000)2 2
= 9.38 × 109 joule Final kinetic energy = =
1 1 mrv2r + mcvc2 2 2 1 1 × 20 × (24000)2 + × 10 × (27000)2 2 2
= 9.40 × 109 joule This energy comes from the compressed spring. Q. 27. A bomb moving with velocity ( 40i + 50 j − 25 k ) m/s explodes into two pieces of mass ratio 1:4. The smaller piece goes out with velocity 200i + 70 j + 15 k m/s . Deduce the velocity of the larger piece after the explosion. Also, calculate the velocities of the piece in the centre-of-mass reference frame. Solution. The forces of explosion exerted by a part of the system (bomb) on other parts of the system are all internal forces which may the momenta of all the individual fragments but they cannot change the total momentum of the system. Before the explosion the bomb can be thought of as consisting of two pieces of respective masses M/5 and 4M/5, where M is the total mass of the bomb. Now using the given data, applying the conservation of linear momentum, we have Initial momentum = Final momentum
4M → M v (200i + 70 j + 15k ) + M ( 40 i + 50 j − 25k ) = 5 5 →
Thus the velocity v of the larger piece is → 5 ( 40i + 50 j − 25k ) − (200 j + 70 j + 15k ) v = 4 200i + 250 j − 125k − 200i − 70 j − 15k = 4 =
180 j − 140 k = 45 j − 35 k meter/sec 4
As there is no external force, the centre of mass moves after the explosion with same velocity 40i + 50 j − 25k as before the explosion. Hence the velocity of the smaller piece in the centre of mass reference frame is
(200 i + 70 j + 15 k ) − ( 40i + 50 j − 25 k ) = 160 i + 20 j + 40 k meter/sec.
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and that of the larger piece is ( 45 j − 35k ) − ( 40i + 50 j − 25k ) = −40i − 5 j − 10 k meter/sec.
Example 28. A U238 nucleus emits an α-particle and is converted into Th234. If the velocity of the α particle be 1.4 × 107 meter/sec and the kinetic energy be 4.1 MeV, calculate the velocity of the recoil and the kinetic energy of the remaining Th234 nucleus. Solution. Initially the U238 nuclues is at rest so that its momentum is zero. No external force acts upon it, the radio-active emission being purely an internal process. Hence, after emission, the sum of the momenta of the α particle and the remaining Th234 nucleus is also zero. That is →
→
ma v a + mth v th = 0 →
→
v th
The α-particle is
2He
4
− ma v a = mth
nucleus. Therefore
ma 4 = ; and v = 1.4 × 107 meter/sec. a mth 234
Therefore v th = −
4 × (1. 4 × 107 ) 234
= −2. 4 × 105 meter /sec Thus the Th234 moves with a velocity of 2.4 × 107 m/s opposite to the motion of the αparticle. The ratio of the kinetic energy of Th234 nucleus and α-particle would be equal to the inverse ratio of their masses. That is
ma K th = mth Ka ma 4 × Ka = × 4.1 = 0. 07 MeV mth 234 Q. 29. A nucleus of mass 190 amu emits an α-particle (mass 4 amu) with velocity ( 4i − 10 j ) 106 m/s in laboratory frame. Deduce the velocity of recoil of the residual nucleus, as also the fraction of total energy of disintegration shared by the recoil.
Kth =
Solution. We think of the system (α particle + residual nucleus) as initially bound and forming the original nucleus. The system then splits into two separate parts. The momentum of the system before splitting is zero. In the absence of external forces, the total (vector) momentum of the parts is also zero. Hence Initial momentum = Final momentum →
→
0 = ma v a + mn v n →
= 4 (4 i − 10 j ) 106 + 186 v n
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Mechanics
The velocity of the residual nucleus is therefore
4 6 v n = − 186 (4 i − 10 j ) 10 m/sec. The minus sign indicates that the residual nucleus recoils in a direction exactly opposite to the motion of the α-particle, so as to give a resultant vector momentum of zero. →
The ratio of the kinetic energies acquired by the recoiling nucleus and the α-particle is
But mnvn = –mava
1 mn vn2 kn (mnvn )2 mα 2 = = 1 kα (mava )2 mn mava2 2 ma 4 2 kn = = = mn 186 93 ka
The source of the K.E. of the products is the so called binding energy of the original nucleus. Q. 30. A shell of mass 2 kg and moving at a rate of 4 m/s suddenly explodes into two equal fragments. The fragments go in directions inclined with the original line of motion with equal velocities. If the explosion imparts 48 joules of translational K.E. to the fragments divided equally find the velocity and direction of each fragment. Solution. A moving shell has a small amount of K.E., but a large amount of hidden chemical energy, which is released when it explodes. Some of this energy is converted into heat, but a large part goes to the K.E. of the fragments of the explosion. As no external force acts on the shell (explosion is purely an internal affair), the vector sum of the momenta of the fragments is equal to the momentum of the unexploded shell. Further, the centre of mass of the fragments continue to move along the original line of motion. Since, here the two fragments are of equal mass, they move in directions equally inclined to the original line of motion. Now, by conservation of momentum, we have
Fig. 22
initial horizontal momentum = final horizontal momentum 2 × 4 = 1 × v cos θ + 1 × v cos θ v cos θ = 4
...(i)
1 × 2 × (4)2 = 16 joule. An amount of 48 joule is imparted 2 by explosion. Thus the total energy of the fragments is 64 joule i.e., each fragment has 32 joule of K.E. If v be the velocity of each fragment, then The initial K.E. of the, shell is
1 mv2 = 32 2
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1 × 1 × v2 = 32 2 v 2 = 64 v = 8 m/s substituting this value of v in equation (i), we get 8 cos θ = 4 cos θ =
1 2
θ = 60° Q. 31. A body of mass 8.0 kg moving at 2.0 m/s without any external force on it. An internal explosion suddenly breaks the body into equal parts and 16 joules of translational kinetic energy is imparted to the system by the explosion. Neither piece leaves the line of the original motion. Determine the speed and direction of each piece after the explosion. [Ans. One piece comes to rest the other moves ahead with a speed of 4.0 m/s]. Q. 32. A body at rest explodes and breaks up into three pieces. Two pieces having equal mass, fly off perpendicular to one another with the same speed of 30 m/s. The third piece has three times the mass of each of the other pieces. Find the magnitude and direction of its velocity immediately after the explosion. Solution. Let v be the velocity, and θ the direction as shown of the third piece. Before explosion, the momentum of the body is zero. Hence by conservation of momentum, the momentum of the system (all the three pieces) after explosion must be zero. Equating the momentum of the system along OA and OB to zero, we get and these give
m × 30 – 3m × v cos θ = 0
...(i)
m × 30 – 3m × v sin θ = 0
...(ii)
3 mv cos θ = 3 mv sin θ
B
cos θ = sin θ
or ∴ ∴
θ = 45° ∠AOX = ∠BOX = 180° – 45° = 135°
Putting the value of θ in eq. (i), we get 30 m = 3 mv cos 45° = ∴
m
m
0
F 1 IJ 3mv G H 2K
v = 10 2 meter/sec.
θ
A
v 3m
The third piece will go with a velocity of 10 2 meter/sec in a direction making an angle of 135° with either piece. Q. 33. A radioactive nucleus, initially at rest, decays by emitting an electron and a neutrino at right angles to each other. The momentum of the electron is 1.2 × 10–22 kg-m/sec, and that of the neutrino 6.4 × 10–23 kg-m/sec. Find the direction and magnitude of the momentum of the recoil nucleus. If its mass, is 5.8 × 10–26 kg, deduce its kinetic energy of recoil.
300
Mechanics →
→
→
Solution. Let Pe , Pn and PN be the momenta of the electron, neutrino and the recoiling nucleus respectively, with directions as shown. Before decay, the momentum of the nucleus is zero. As no external force is acting on the system, the vector sum of the momenta of e, n and N must also be zero. Pn n
θ O
e
Pe
N PN
Fig. 23
Equating the sum of the components of the various momenta along two chosen directions to zero, we have Pe – PN cos θ = 0 Pn – PN sin θ = 0 Pn =
Pe2 + Pn2
=
(1. 2 × 10−22 )2 + (6. 4 × 10−23 )2
=
(1. 2)2 + (0. 64)2 × 10−22
= 1.36 × 10–22 kg-m/sec Also,
tan θ =
Pn 0. 64 = = 0. 53 Pe 1. 2
θ = tan–1 (0.53) = 28° The direction of recoil is (180 – 28) = 152° from the electron track and 118° from the neutrino track. The kinetic energy of the recoiling nucleus is KN =
1 MN v2N 2
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=
PN2 (1. 36 × 10−22 )2 = 2mN 2 × (5. 8 × 10−26 )
= 0.16 × 10–18 joule. Now
1 eV = 1.6 × 10–19 joule KN =
0.16 × 10−18 = 1. 0 eV 1. 6 × 10−19
Q. 34. A cue striking a billiard ball at rest, exerts an average force of 50 nt over a time of 10 millisecond. The mass of the ball is 0.20 kg. Find the speed acquired by the ball. Solution. The impulse of the force exerted on the ball is F∆t = 50 nt (10 × 10–3 sec) = 0.50 nt-sec. This is equal to the change in momentum of the ball. The ball was initially at rest. It acquires a velocity v after the impact, the change in momentum is mv = 0.20 v. Thus 0.20 v = 0.50 ∴
v =
0. 50 = 2. 5 m/sec. 0. 20
Q. 35. A 1.0 kg ball falls vertically onto the floor with a speed of 25 m/sec. It rebounds with an initially speed of 10 m/sec. Find the coefficient of restitution. What impulse acts on the ball during contact. If the ball is in contact for 0.020 sec, what is the average force exerted on the floor. Solution. The coefficient of restitution e is defined as the ratio between the relative velocity of separation after the collision and the relative velocity of approach before the collision. Thus, in this case, e =
10 m/sec = 0. 5 20 m/sec
The initial velocity of the ball is v1 = 25 m/sec and the final velocity is v2 = –10 m/s. Thus m (v1 – v2) = 1.0 (25 – (–10) = 35 kg-m/sec This is also the impulse, F∆t, acting on the ball. Now,
F∆t = 35 nt-sec
35 35 nt - sec = = 1750 nt Ans. ∆t 0.020 sec Q. 36. The scale of a balance is adjusted to read zero. Particles falling from a height of 5 meters collide with a pan of the balance, the collisions are elastic. If each particle has a mass of 1/60 kg and collisions occur at the rate of 32 particles per second, what is the scale reading in kg? (g = 10 m/sec2). ∴
F =
Solution. The velocity of the particle before colliding in v =
2 gh = 2 × 10 × 5 = 10 m/sec.
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Mechanics
and so its momentum is
mv =
1 1 × 10 = kg m/sec. 60 6
The collision is elastic, i.e., the particle rebounds upward with the same speed. Hence the change in momentum due to a collision is 2mv =
1 kg m/sec 8
This is also momentum acquired by the pan in one collision. Since collision occur at the rate of 32 per second, the average force exerted on the pan. = rate of change of momentum =
Hence scale reading in kg =
1 × 32 = 4 kg m/sec2 8 4 kg m/sec2 = 0. 4 kg. Ans. 10 m/sec2
Q. 37. A 6.0 kg cart is moving on a smooth surface at a speed of 9.0 m/sec when a 12 kg package is dropped into it vertically. Describe the subsequent motion of the cart. Solution. The initial horizontal momentum of the cart is mv while that of the package is zero. After the package is dropped into the cart, the two move as a single mass. Thus the impact is inelastic, but the horizontal momentum is conserved. mv = (m + M)v′ where v′ is the velocity of the cart-package system. Putting the given values: 6.0 × 9.0 = (6.0 + 12)v′ or
v′ =
6. 0 × 9. 0 = 3. 0 m/sec 18
The speed of the cart slow down to 3.0 m/sec. Ans. Q. 38. A body of mass of 3.0 kg collides elastically with another body at rest and then continues to move in the original direction with one-half of its original speed. What is the mass of the struck body. Solution. Let m1 and m2 be the masses of the two bodies and v1i the initial velocity of m1. Let v1f and v2f be the velocities of m1 and m2 after the collision. By the conservation of momentum, we have m1 v1i = m1v1f + m2v2f or
3.0 v1i = 3. 0
or
v2f =
1 m2
FG v IJ + m v H2K FG 3 v IJ H2 K 1i
2 2f
1i
By the conservation of kinetic energy, we have
1 1 1 m1v12i = m1v12f + m2v22f 2 2 2
3 v1 f = vi /2
...(1)
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FG IJ H K
1 v 1 (3) v12i = (3. 0) 1i 2 2 2
or
2
+
1 m2v22f 2
3 2 3 1 v1i = v12i + m2v22f 2 8 2
or
Substituting the value of v2f from Eq. (1), we get
3 2 3 v1i − v12i = 2 8
FG 1 m IJ 1 FG 9 v IJ H2 K m H4 K 2
2 1i
2 2
or
9 2 9 2 v1i v1i = m2 8 8
or
m 2 = 1.0 kg
Ans.
Q. 39. A bullet of mass 10 gm moving horizontally with speed of 500 m/sec passes through a block of wood of mass 1 kg, initially at rest on frictionless surface. The bullet comes out of the block with a speed of 200 m/sec. Calculate the final speed of the block. Solution. Let m1 and m2 be the masses of the bullet and the block and v1i the initial velocity of the bullet. Let v1f and v2f be the velocities of the bullet and the block after the collision. By the conservation of momentum, we have m1v1i = m1v1f + m2v2f (0.01 kg) (500 m/sec) = (0.01 kg) (200 m/sec) + (1 kg) (v2f) ∴
v2f = 3 m/sec
Ans.
Q. 40. An open-topped freight car of mass 10,000 kg is coasting without friction along a level track. It is raining very hard, with the rain falling vertically downward. The car is originally empty and moving with a velocity of 1 m/sec. What is the velocity of the car after it has travelled to collect 1000 kg of rain water. Solution. The initial horizontal momentum of the car is (10,000 kg) (1 m/sec), while that of the rain water is zero (as it is falling vertically). Finally, the water sticks to the car. Now, the mass of the car plus water is 11,000 kg. If v′ be the velocity, then by the conservation of momentum, we have (10,000 kg) (1 m/sec) = (11,000 kg) v′
10, 000 × 1 = 0. 91 m/sec Ans. 11, 000 Q. 41. An empty freight car of mass m1 = 10,000 kg rolls at v1 = 2 m/sec on a level road and collides with a loaded car of mass m2 = 20,000 kg standing at rest. If the cars couple together find their speed v′ after the collision, and also the loss in kinetic energy. What should be the speed of the loaded car toward the empty car, in order that both shall be brought to rest by the collision. Solution. By the conservation of momentum, we have m1v 1 = (m1 + m2) v′ ∴
v′ =
so that
v′ =
FG m IJ v Hm + m K 1
1
1
2
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Mechanics
=
Loss in kinetic energy =
FG 10, 000 IJ (2) = 2 = 0.66 m/sec H 10, 000 + 20, 000 K 3 1 1 m1v12 − (m1 + m2 ) v′2 2 2
LM MN
FG H
1 2 2 = 2 10, 000 (2) − ( 30, 000) 3
=
IJ OP K PQ 2
40, 000 = 13, 333. 33 Joule . 3
Let v2 be the speed of the loaded car toward the empty car, in order that v′ = 0. Thus m1v1 – m2v2 = 0 or
v2 = =
m1 v2 m2
10, 000 ×2 20, 000
v 2 = 1 m/sec
Ans.
Q. 42. A slow-moving electron collides elastically with a hydrogen atom at rest. The initial and final motions are along the same straight line. What fraction of the electron’s kinetic energy is transferred to the hydrogen atom? The mass of hydrogen atom is 1840 times the mass of the electron. Solution. Let m1 be the mass of the electron and m2 that of the hydrogen atom. Let v1i and v1f be the initial and final velocities of the electron. Then the initial and final kinetic energies of the electron are ki =
1 1 m1v12i and kf = m1v12f respectively. The fractional decrease 2 2
in kinetic energy is ki − kf ki
=
v12i − v12f v12i
v12f
=1−
v12i
But, for such a collision, we have v1f =
so that
ki − kf ki
FG m Hm
1 1
= 1−
IJ K
− m2 v1i + m2
FG m Hm
1
1
=
− m2 + m2
IJ K
2
4 m1m2 4 (m2 /m1 ) = 2 2 (m1 + m2 ) m2 1+ m1
FG H
IJ K
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m2 = 1840 m1
Here
4 × 1840 ki − kf = 0. 00217 or 0.217% = (1841)2 ki
∴
Q. 43. A bullet of mass 5 gm is shot through a 1 kg wood block suspended on a string 2 m long. The centre of gravity of the block rises a distance of 0.50 cm. Find the speed of the bullet as it energies from the block if the initial speed is 300 m/sec. Solution. Let v be the velocity of the emerging bullet and v′ the velocity acquired by the block just after the collision. From the conservation of momentum, we have initial momentum of bullet = final momentum of bullet + momentum of block (0.005 kg) (300 m/sec) = (0.005 kg) v + (1 kg) v′ The kinetic energy
...(1)
1 mv′2 acquired by the block is converted into gravitational potential 2
energy mgy. so that v′ = =
2gy
[2 × 9. 8 m/s2 × (0. 50 × 10−2 m)] = 0. 3125 m/sec.
Substituting this value of v′ in Eq. (1), we get 1.5 = 0.005 v + 0.3125 or
v =
1. 5 − 0. 3125 0. 005
v = 237.5 m/sec.
Ans.
Q. 44. A moving particle of mass m collides head-on with a particle of mass 2m which is initially at rest. Show that the particle m will loose 8/9th part of its initial kinetic energy after the collision. Solution. Let v1i be the initial velocity of the mass m1 and v1f and v2f the velocities after the collision of mass m and 2m respectively. By conservation of momentum, we have mv1i = mv1 f + 2mv2 f or
v1i − v1 f = 2v2 f
...(1)
Again, by the conservation of kinetic energy, we have
1 1 1 mv12i = mv12f + (2m) v22f 2 2 2 or
v12i − v12f = 2v22f
or
( v1i − v1 f ) ( v1i + v1 f ) = 2v22f
Dividing Eq. (2) by Eq. (1), we get v1i + v1 f = v2 f
...(2)
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Mechanics
Substituting the value of v2 f in Eq. (1), we get v1i − v1 f = 2 ( v1i + v1 f )
1 v1 f = − v1i 3
or
...(3)
Hence, the initial and final kinetic energies of m are
and
∴
Ki =
1 mv12i 2
Kf =
1 1 1 2 mv12f = m v1i 2 2 9
FG H FG1 − 1 IJ H 9K
IJ K
1 mv2i ki − kf 2 1 Fraction lost, = 1 ki mv2i 2 1 =
8 Ans. 9
Q. 45. Two blocks of masses m1 = 2.0 kg and m2 = 5.0 kg are moving in the same direction along a frictionless surface with speeds 10 m/sec and 3.0 m/sec respectively, m2 being ahead of m1. An ideal spring with k = 1120 nt/m is attached to the backside of m2. Find the maximum compression of the spring when the blocks collide. v1
v2 k
m1
m2
Fig. 24
Solution. Let v′ be the speed of the combination after collision. By the conservation of momentum, we have
m1v1 + m2v2 = (m1 + m2 ) v′ or
(2.0) (10) + (5.0) (3.0) = (2.0 + 5.0) v′
(2. 0) (10) + (5. 0) (3. 0) = 5 m/sec (2. 0 + 5. 0) Now, by the conservation of mechanical energy, we have
∴
v′ =
1 1 1 1 (m1 + m2 ) v′2 + kx2 m1v12 + m2v22 = 2 2 2 2 200 + 45 = 175 + 1120 x2
or or ∴
x2 =
200 + 45 − 175 1 = 1120 16
x =
1 = 0. 25 meter Ans. 4
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Q. 46. A 5-kg object with a speed of 30 m/sec strikes a steel plate at an angle of 45° and rebounds at the same speed and same angle. What is the change in magnitude and direction of the linear momentum of the object. −→
Solution. The vector OA represents the momentum of the object before the collision, −→
−→
and the vector OB that after the collision. The vector AB represents the change in the →
momentum of the object ∆P
Fig. 25 −→
−→
−→
−→
As the magnitude of OA and OB are equal, the components of OA and OB along the plate are equal and in the same direction, while those perpendicular to the plate are equal and opposite. Thus the change in momentum is due only to the change in direction as the perpendicular components. Hence ∆P = OB sin 45° – (–OA sin 45°) = mv sin 45° – (–mv sin 45°) = 2mv sin 45° = 2 (5 kg) (30 m/sec) (sin 45°) = 212 kg m/sec. →
∆P is directed perpendicular and away from the plate. Q. 47. A ball of mass m is projected with speed v into the barrel of a spring-guest of mass M initially at rest on a frictionless surface. The mass m sticks in the barrel at the point of maximum compression of the spring. What fraction of the initial kinetic energy of the ball is stored in the spring? Neglect friction. Solution. Let v′ be the velocity of the system after the mass m sticks in the barrel. From the law of conservation of linear momentum, we have v
M
m
Fig. 26
mv = (m + M) v′
...(1)
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Mechanics
1 mv2 . It is converted into the elastic potential 2 1 2 1 (m + M) v′2 of the system. That is energy kx of the spring, and the kinetic energy 2 2 The initial kinetic energy of the ball is
1 1 1 mv2 = kx2 + ( m + M ) v′2 2 2 2 where k is the force-constant of the spring and x is its maximum compression. Dividing the last Equation by
1 mv2 throughout, we obtain 2 1 2 kx (m + M) v′2 1 = 2 + 1 mv2 mv2 2 1 2 kx (m + M) v′2 2 = 1− 1 mv2 mv2 2
or
Putting the value of
v′ 2 from Eq. (1) we get v2 1 2 kx 2 m2 (m + M) 1− . = 1 m ( m + M)2 mv2 2 = 1−
m M = (m + M) m + M
1 2 1 kx is the energy stored in the spring, while mv2 is the initial kinetic energy of the 2 2 1 2 1 kx / mv2 is fraction of the initial energy which is stored in the spring. This ball. Thus 2 2 M . fraction is m+M Q. 48. Two balls A and B, having different but unknown masses, collide, A is initially at
v and moves at right angles to its 2 original motion. Find the direction in which ball A moves after collision. Can you determine the speed of A from the given information?
rest while B has a speed v. After collision B has a speed
Solution. Let v′ be the speed of A after the collision by the conservation of momentum along the original direction of motion, we have
Collision
309 v 2
B
B
v
A
θ v′
Fig. 27
mBv = mAv′cos θ
...(1)
Again, by the conservation of momentum of right angles to the original motion, we have 0 = mB
mB
or
v − mA v′sin θ 2
v = mA v′ sin θ 2
...(2)
Dividing Eq. (2) by Eq. (1) we get tan θ = ∴
1 2
θ = tan–1 (0.5) = 27°
Thus the ball A moves at 27° + 90° = 117° from the final direction of B. We cannot determine v′ unless we know mA and mB or their ratio. Q. 49. A ball moving at a speed of 2.2 m/sec strikes an identical stationary ball. After collision one ball moves at 1.1 m/sec at 60° angle with the original line of motion. Find the velocity of the other ball. Solution. Let m be the mass of each ball, and v1i the initial velocity of the first ball. Let v1f and v2f be the final velocities of the balls after the collision in directions as shown in figure. By the conservation of momentum along the original line of motion, we have
Fig. 28
mv1f = mv1 f cos 60° + mv2 f cos θ or or
(3 cos 60° = 0. 5 )
2.2 = 1.1 (0.5) + v2f cos θ
v2 f cosθ = 2.2 – 0.55 = 1.65
...(1)
By conservation of momentum perpendicular to the original line of motion, we have 0 = mv1 f sin 60° − mv2 f sin θ
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Mechanics
0 = 1.1 (0.866) – v2 f sin θ
(3 sin 60° = 0. 866 )
v2 f sin θ = 1.1 (0. 866) = 0. 953
or
...(2)
Squaring and adding Eq. (1) and Eq. (2), we get v2f =
(1. 65)2 + (0. 953)2 = 1. 9 m / sec
Dividing Eq. (2) by Eq. (1), we get tan θ =
0. 953 = 0. 577 1. 65
θ = tan–1 (0.577) = 30°
or
Ans.
PROBLEM 1.
A ball of mass m and speed v strikes a wall perpendicularly and rebounds with undiminished speed. If the time of collision is t, what is the average force exerted by the ball on the wall.
2.
A 10 gm bullet moving horizontally at 100 m/sec strikes and is embedded in a 25-kg block placed on a rough surface. The block slides 3m before coming to rest. Find the frictional force retarding the motion of the block on the surface.
3.
Find the fractional decrease in the kinetic energy of neutron when it collides elastically headon with (1) lead nucleus (2) hydrogen nucleus, initially at rest. The ratio of nuclear mass to neutron mass is 206 for lead and 1 for hydrogen.
7 ROTATIONAL KINEMATICS 7.1 RIGID BODY If an external force applied to a body does not produce any displacement of the particles of the body relative to each other, then the body is called a rigid body. No real body is perfectly rigid. However, in solid bodies (leaving rubber, etc.) the relative displacement by the external force is so small that it can be neglected. Hence ordinarily, when we speak of a body, we mean a rigid body.
7.2 MOMENT OF A FORCE OR TORQUE When an external force acting on a body has a tendency to rotate the body about an axis, then the force is said to exert a ‘torque’ upon the body about that axis. The moment of a force, or the torque, about an axis of rotation is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the axis of rotation. In figure 1 shown a body, which is free to rotate about an axis passing through a point O and perpendicular to the plane of the paper. When a force F is applied on the body in the plane of the paper, the body rotates about this axis. If r be the perpendicular distance of the line of action of the force from the point O, then the moment of the force F, that is, the torque about the axis of rotation is
F
O r
L ine o f a ctio n
τ=F× r If the torque tends to rotate the body anticlockwise then it is taken as positive, if clockwise then negative. The unit of torque is ‘newton-metre’.
Fig. 1
7.3 ANGULAR ACCELERATION When a body rotates about a fixed axis, the rotation is called a ‘rotatory motion’ or ‘angular motion’ and the axis is called the ‘axis of rotation’. In rotatory motion, every particle 311
312
Mechanics
of the body moves in a circle and the centers of all these circles lie at the axis of rotation. The rotating blades of an electric fan and the motion of a top are examples of rotatory motion. If the angular velocity of a rotating body about an axis is changing with time then its motion is ‘accelerated rotatory motion’. The rate of change of angular velocity of a body about an axis is called the angular ‘acceleration’ of the body about that axis. It is denoted by α. If the angular velocity of a body about an axis changes from ω1 to ω2 in t second, then the angular acceleration of the body about that axis is α =
change in angular velocity ω2 − ω1 = t time-interval
The unit of angular acceleration is radian/sec2 and its dimensional formula is [T–2].
7.4 RELATION BETWEEN ANGULAR ACCELERATION AND LINEAR ACCELERATION In figure 2, a body suspended from a point O and free to rotate about an axis passing through O. When a torque is applied to the body, the body rotates with accelerated motion about the axis of rotation. Suppose, the body rotates through an angle θ in t second. It means that the angular displacement of the body in t sec is θ. Therefore, the angular velocity (=angular displacement/time-interval) of the body is given by ω =
θ t
Different particles of the body cover different distances in t second, hence their linear velocities are different. But all the particles rotate through the same angle θ in t sec, hence the angular velocity of each particle at any instant is same. Suppose a particle P of the body is at a distance r from the point O. As the body rotates, the particle P rotates through angle θ in t second and reaches P′. Thus the particle moves from P to P′ in t second. Therefore, its linear velocity is v =
PP′ rθ = t t
arc = radius × angle But ∴ or
A xis o f ro ta tio n
θ = ω t
O
v = r× ω ω =
v r
...(1)
Suppose, at any instant the angular velocity of the body is ω1 and after t second it becomes ω2. Then the angular acceleration of the body is α =
ω 2 − ω1 t
θ
r
P
P′
Fig. 2
Rotational Kinematics
313
If at the above instants the linear velocities of the particle P be v1 and v2 respectively, then according to equation (1), ω1 =
v1 v and ω2 = 2 . Therefore, r r
FG v IJ − FG v IJ H rK H rK = v 2
α =
1
2
t
− v1 rt
...(2)
v2 – v1 is the change in the linear velocity of the particle P in t sec. Therefore (v2 – v1)/t is the rate of change of linear velocity of the particle, that is, it is the linear acceleration a of the particle P. Substituting
α = or
v2 − v1 = a in equation (2), we get t
a r
a = r× α
Thus, the linear acceleration of a particle of a body is equal to the product of the angular acceleration of the body and the distance of the particle from the axis of rotation.
7.5 KINETIC ENERGY OF ROTATION Suppose a body is rotating about an axis with a uniform angular velocity ω. The angular velocity of all the particles of the body will be the same but their linear velocities will be different. Suppose a particle of the body whose mass m1 is at a distance r1 from the axis of rotation. Let v1 be the linear velocity of this particle. As the linear velocity of a particle is equal to the product of its angular velocity and its distance from the axis of rotation, we have v 1 = r1 ω The kinetic energy of this particle is
1 1 1 m1v12 = m1 ( r1ω )2 = m1r12 ω2 2 2 2 In the same way, if the masses of the other particles be m2, m3, ....... and their respective distances from the axis of rotation be r2, r3, ....... then their kinetic energies will be ½ m2 (r2)2 ω2, ½ m3 (r3)2 ω3, ....... respectively. The kinetic energy K of the whole body will be equal to the sum of the kinetic energies of all the particles: K =
1 1 1 m1r12ω2 + m2r22 ω2 + m3r32 ω2 + ....... 2 2 2
K =
1 1 Σ mr2 ω2 m1r12 + m2r22 + m3r32 + ...... ω2 = 2 2
e
j
e
j
But Σmr2 is the moment of inertia I of the body about the axis of rotation. Therefore K =
1 2 Iω 2
This is the formula for the kinetic energy of uniform rotation.
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Mechanics
7.6 ANGULAR MOMENTUM When a body moves along a straight line, then the product of the mass ‘m’ and the linear →
velocity v of the body is called the ‘linear momentum’ p of the body (p = mv). If a body is rotating about an axis then the sum of the moments of the linear momenta of all the particles about the given axis is called the ‘angular momentum’ of the body about that axis. It is →
denoted by j . Let a body is rotating about an axis with an angular velocity ω. All the particles of the body will have the same angular velocity, but their linear velocities will be different. Let a particle be at a distance r1 from the axis of rotation. The linear velocity of this particle is given by v 1 = r1 ω If the mass of the particle be m1, then its linear momentum = m1 v1 The moment of this momentum about the axis of rotation = momentum × distance = m1 v1 × r1 = m1 (r1 ω) × r1 = m1 r12 ω Similarly, if the masses of other particles be m2, m3, ........ and their respective distances from the axis of rotation be r2, r3, ........, then the moments of their linear momenta about the axis of rotation will be m2 r22 ω, m3 r32 ω, ........ respectively. The sum of the moments of linear momenta of all the particles, that is, the angular momentum of the body is given by J = m1 r12 ω + m2 r22 ω + m3 r32 ω + ........ J = (m1 r12 + m2 r22 + m3 r32) ω, = (Σmr2) ω
or
But Σmr2 is the moment of inertia I of the body about the axis of rotation. Hence the angular momentum of the body about the axis of rotation is J = I× ω The unit of angular momentum is ‘kg-meter2/sec’ or ‘joule-sec’. Its dimensional formula is [M L2 T–1].
7.7 RELATION BETWEEN TORQUE AND ANGULAR MOMENTUM Suppose a body is rotating about an axis under the action of a torque, τ and its angular acceleration is α. Then
τ = Iα
Where I is the moment of inertia of the body about the axis of rotation. Now, α = ∴
∆ω (rate of change of angular velocity) ∆t
τ = I
∆ω ∆t
...(i)
Rotational Kinematics
315
The angular momentum of the body about the axis of rotation is J = Iω The rate of change of angular momentum is given by
∆J ∆ω = I ∆t ∆t
[∴ I is constant]
Hence, from equation (i) we have
∆J = τ ∆t Thus, the rate of change of angular momentum of a body is equal to the external torque acting upon the body.
7.8 CONSERVATION OF ANGULAR MOMENTUM If no external torque is acting upon a body rotating about an axis, then the angular momentum of the body remains constant, that is, J = Iω = constant This is called the ‘law of conservation of angular momentum’. If I decreases, ω increases, and vice-versa.
7.9 TORQUE ACTING ON A PARTICLE A force is required to produce linear acceleration in a particle. In a similar way, a torque (or moment of force) is required to produce angular acceleration in a particle about some axis. Like force, the torque is a vector quantity. It is defined in the following way: Z → τ O
X
→ r
r sin θ
→ F P
θ
Y
Fig. 3 →
→
Let a force F acts on a particle P whose position vector is r with respect to the origin →
O of an inertial reference frame. Then the ‘torque’ τ acting on the particle about on the →
→
particle about the origin O is defined as the vector product of r and F . That is, →
τ
→
→
= r × F
Its scalar magnitude is τ = rF sin θ
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Mechanics →
→
where θ is the angle between r and F , and is along a direction perpendicular to the plane →
→
→
→
→
containing r and F . In the figure, r and F lie in the x-y plane and the torque τ is along the positive z axis. →
It is clear that torque is maximum when the force is applied at right angles to r (i.e., θ = 90°). This is why in opening or closing a heavy revolving door the force is applied (by hand) at right angles to the door at its outer edge. Besides this, the torque depends also on the position of the point relative to origin at which the force is applied a force applied at the →
origin 0 (i.e., r is zero) produces zero torque about 0. This is why we cannot open or close a door by applying force at the hinge.
7.10 ANGULAR MOMENTUM OF THE CENTER OF MASS OF A SYSTEM OF PARTICLES The total angular momentum of a system of particles about a fixed point can also be →
expressed in terms of the angular momentum of the center of mass about that point. Let r cm be the position vector of the center of mass c of the system with respect to the origin, and →
r ′ i the position vector of the particle i (mass mi) with respect to the center of mass. Then, by the rule of vector sum, we have →
→ → r i = r i′ + r cm
Z
X O → ri
→ rc m C
→ ri′
mi
Y
Fig. 4 →
→
Similarly, if v cm be the velocity of the center of mass and v ′i the velocity of the particle i with respect to the center of mass, then the velocity of the particle with respect to the origin is →
→
→
v i = v ′i + v cm
and the linear momentum with respect to the origin is
Rotational Kinematics
317 →
→
→
→
pi = mi v = mi v ′ i + mi vcm →
→
= p ′i + mi vcm →
where p ′ is the linear momentum with respect to the center of mass. Therefore, the total i angular momentum of the system about the origin is →
L =
n →
∑
→
r i × pi
i=1 →
or
L =
or
L =
→
F
∑ GH r ′ ∑ r′
→
IJ FG K H
→
→
→
L =
→
or
L =
→
→
Now, →
∑ p′
i
∑ r × p′ i
→
∑ r′
→
→
i
→
i
× p′ i + r cm .
→
→
cm
F
× mi v cm
∑ p ′ + ∑ GH r ′
× pi + r cm . →
→
∑r
. mi v cm +
i
→
∑ r′
IJ K
→
i
→
→
× p ′i + r cm ×
i
∑ p′ + ∑ r ′ or
→
+ r cm × p′i + mi v cm
i
→
→
→
IJ K
→
i
+ r cm × mi v cm
∑ p′ + ∑ ( r ′
+ r cm ) × mi v cm
i
→
→
i
→
i
→
→
i
= Lcm , angular momentum of the system about the center of mass, and
= 0 (we know that the total linear momentum of a system of particles about the center
of mass is zero). →
→
L = L cm +
→
∑
→
→
( mi ri ) × v cm
since
→
→
→
→
By the definition of the center of mass, r cm
∑ m r , where M is the total mass of the = i i
M
system. ∴
→
→
→
→
→
→
L = L cm + M r cm × v cm →
→
L = L cm + r cm × M v cm
or →
→
ri = ri ′ = rcm
→
But M v cm known to the total linear momentum P of the system. Thus
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Mechanics →
→
→
→
L = L cm + ri × P →
→
The term r cm × P is the angular momentum of the center of mass about the origin.
7.11 PRECESSION When a body rotates about a fixed axis, under the action of a torque, the axial component of the torque alone is effective in producing rotation, its component perpendicular to the axis being neutralized by an equal and opposite torque due to the constraining forces applied to the axis to maintain its original position, i.e., to keep it fixed. In the absence of any such constraining forces, obviously, the axis of rotation, and hence also the plane of rotation, of the body will turn from its initial position under the action of the component of the torque perpendicular to the axis, which we may denote by the symbol τ1. Since this torque τ1 (i.e., the axis of this torque) is perpendicular to the axis of rotation of the body, the magnitude of the angular velocity, and hence also that of the angular momentum of the body remains unaltered, with only its direction changed in exactly the same manner that in a linear motion, a constant force (viz, the centripetal force), acting on a body in a direction perpendicular to that of its velocity, simply changes its direction and not its magnitude. So that if the torque τ1 be a constant one, the rotation-axis and the plane of rotation of the rotating body continue to change their direction at a constant rate, with the axis of the torque always perpendicular to the axis of rotation of the body. This change in the direction of the rotation-axis, or in the direction of the plane of rotation, of the rotating body under the action of a constant torque perpendicular to the axis of rotation is called precession. The torque τ1 which brings this about is, therefore, called the precessional torque and the rate of rotation of the axis of rotation or the plane of rotation of the body is called the rate of precession, usually denoted by the symbol φ.
7.12 THE TOP (PRECESSION OF A TOP SPINNING IN EARTH’S GRAVITATIONAL FIELD) A symmetrical body rotating (or spinning) about an axis, one point of which is fixed, is called a ‘top’. Figure shows a top spinning with angular velocity ω about its own axis of symmetry, the fixed point O being at the origin of an inertial reference frame. Its angular momentum is →
represented by the vector L pointing upward along the axis of rotation (right-hand rule). The axis is inclined at an angle θ with the vertical. →
Let C be the center of mass of the top, with position vector r with respect to O. The →
weight of the top, m g , acting at C exerts a (external) torque about the fixed point O which is given by →
→
→
τ = r ×mg
Rotational Kinematics
319
Its scalar magnitude is rmg sin (180° – θ) = rmg sin θ
...(i) →
→
and its direction is perpendicular to the plane containing r and m g as shown in the figure. →
Thus the torque τ is perpendicular to L i.e., perpendicular to the axis of rotation of the top. →
→
The torque τ changes the angular momentum L of the top, the change taking place in the →
→
direction of the torque i.e., perpendicular to L . Let ∆ L be the change produced in a time ∆t. We know that the time-rate of change of angular momentum of a system is equal to the torque, that is →
∆L τ = ∆t
→
→
→
→
→
The angular momentum L + ∆ L, after a time ∆t, is the vector sum of L and ∆ L . These →
→
→
three vectors are shown clearly in figure. Since ∆ L is perpendicular to L (or paralled to τ ) →
→
and is very small, the new angular momentum vector L + ∆ L has the same magnitude as →
the initial angular momentum vector L , but a different direction. This means that the angular momentum remains constant in magnitude but varies in direction. As time goes on, →
→
the tip of the angular momentum vector L moves on a horizontal circle. But the vector L always lies along the axis of rotation of the top, the axis of rotation itself rotates about a vertical axis (z-axis) and sweeps out a cone whose vector is the fixed point O. This is the precessional motion of the top. →
The angle ∆φ through which the vector L turns in time ∆t, is given by ∆φ =
∆L L.sin θ
[ ∴ ∆L ≤ L]
∴ the angular velocity of precession, ωp is given by ωp =
But ∴
∆φ ∆L = ∆t L sin θ ( ∆t )
∆L = τ ∆t ωp =
τ L sin θ
...(ii)
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Mechanics z → L ω
c
1
θ 80°
r
θ
x
mg → τ y
Fig. 5
Putting the value of τ from equation (i), we get ωp =
mgr L
...(iii)
Thus the rate of precession is independent of the inclination of the axis of rotation with the vertical, and inversely proportional to the angular momentum of the spinning top. Smaller the angular momentum, larger the precessional velocity. As the spinning top slows down, say due to friction, its angular momentum L (= Iω) decreases and the precession correspondingly becomes faster. →
The precessional velocity would be represented by a vector ω p along the z-axis directed upward. Now, from Equation (ii), we have z
τ = ω p L sin θ →
∆p
→
We note in figure (6) that ω p and L are vectors
L
→
with θ as the angle between them, and τ is a vector →
→
→
→
→
→
τ = ωp × L
P
∆L
→ → L + ∆L θ
→
perpendicular to the plane formed by ω p and L . Thus
ω p L sin θ is the vector product of ω p and L . Hence the above expression can be written in the following vector form.
ωP → ω
x
O → τ y
Fig. 6 →
We obtain an important result from this. An angular momentum vector L of constant →
magnitude but rotating about a fixed axis with angular velocity vector ω p is always associated
Rotational Kinematics
321 →
→
→
with a torque τ which acts perpendicular to the plane formed by ω p and L and is directed according to the right-hand rule.
7.13 THE GYROSTAT A gyrostat, in actual practice, usually takes the form of a heavy circular disc, free to rotate at a high speed about an axle passing through its center of mass and perpendicular to its plane, i.e., about its axis of symmetry, and is so mounded that the disc, along with the axle can turn freely about any one of the three mutually perpendicular axes. Thus the disc or the gyrostat G is free to rotate about the axle, coincident with the axis of x and is carried by a circular ring R1 which is itself free to turn about the axis of y inside another ring R2, which is free to rotate about the axis of z, within a rigid frame work F. The three axes being obviously perpendicular to each other, intersect in the same point O and therefore, for any given position of F, the disc or the gyrostat G can turn freely about any one of them. Z Y
F G X′
X
R1 R2
Y′ Z′
Fig. 7
When the disc rotating fast then (i) any rotation of the frame work leaves the position and direction of the axis of rotation of the disc unaffected with respect to it and (ii) any torque applied to the axis of rotation of the disc results in its precession. Since the rate of precession is inversely proportional to the angular momentum Iω of the disc, the greater its M.I. and the higher its angular velocity about the axis of rotation, the smaller the precession of the latter. Obviously, therefore, a massive disc rotating at a high speed makes for a greater stability of its axis of rotation. Hence the use of gyrostats for steadying of motions and ensuring stability of direction. In view of the property of stability, the gyroscopes are used as stabilizers in ships, boats and aeroplanes. For examples, a gyroscope may be used to reduce the roll of a boat. For this, the gyroscope wheel, kept rotating at high speed by a motor, is mounted with its axis vertical in such a way that its axis may be tilted forward and backward but not sideways. When the
322
Mechanics
boat rolls, say to the right, a control-gyro closes contacts of a motor which tilts the axis forward, thereby producing a torque opposing the roll. On the other hand, if the roll is to the left, the motor tilts the axis of the gyroscope backward, again opposing the roll. The inherent stability of the gyroscope suggests its use as a compass, and a gyro-compass is preferable to the magnetic compass in many respects. It is so constructed that it always sets itself with its axis parallel to the axis of the earth. If it is in any other direction, there is a torque which causes a precession and brings it parallel to the axis of the earth. Another important application of the directional stability of a rapidly spinning (rotating) body is the rifling of the barrels of the rifles. Spiral grooves are cut in the barrels (the process is known as rifling) so that the bullet is forced to move along them before it emerges out in air, and thus acquires a rapid spin about an axis in the direction of motion. This spin motion prevents the deflection of the bullet from its path due to air and gravity effects, and causes only very little precession. Thus the uniformity of flight of the bulled is increased. The rolling of hoops and the riding of bicycles (which are statically unstable since both of them cannot remain in equilibrium when at rest) are possible because of the gyroscopic effect. This effect produces a movement of the plane of rotation, tending to counter balance the disturbing action of gravity. Many modern aircraft instruments such as automatic pilot, bomb sights, artificial horizon, turn and back indicators, etc. have been developed on gyroscope-controlled principles.
7.14 MAN WITH DUMB-BELL ON A ROTATING TABLE Let I0 be the total moment of inertia of the man (with outstretched arm), the dumb-bells, and the table about the axis of rotation and ω0 the initial angular speed. The angular momentum is I0ω0. As the friction is assumed to be zero, it exerts no torque about the vertical axis of rotation. The only external force is gravity acting through the center of mass, and that exerts no torque about the axis of rotation. Hence the angular momentum about the axis of rotation is conserved. → ω0
→ ω0
I0
ω
ω0 (a )
(b )
Fig. 8
When the man pulls in his arms figure (b), the moment of inertia of the system decreases to I. Therefore, in order that the angular momentum remains conserved, the angular speed must increases to ω, such that I0 ω0 = Iω
Rotational Kinematics
323
The table now rotates faster. The kinetic energy of the system, however, does not remain conserved as the man pulls in his arms. In fact, the kinetic energy increase. This increases comes from the work done by the internal forces of the system when the dumb-bells were pulled in. In other words, the kinetic energy increases at the expense of the chemical energy stored in the man’s body. The total energy is thus conserved.
NUMERICALS Q. 1. A top has a mass of 0.50 kg and moment of inertia of 5.0 × 10–4 kg-meter2. Its centre of mass is 4.0 cm from the pivot point. The top spins at a rate of 1800 rev/min with its axis making an angle of 30° with the vertical. Find the angular velocity of precession. If the rotation (spin) of the top is clockwise as one looks downward along the axis, is the precession clockwise or counter-clockwise ?
mgr mgr = L Iω Here m = 0.50 kg, g = 9.8 meter/sec2, r = 4.0 cm = 0.04 meter ωp =
Solution.
z
ω
→ L
180
c→ r θ
° θ
x
O mg → τ
y
Fig. 9
I = 5.0 × 10–4 kg-meter2 and ω = 2π × 30 rad/sec 0. 50 × 9. 8 × 0. 04 = 2. 08 rad/sec 5. 0 × 10−4 × (2 × 3.14 × 30) A consideration of the direction of the torque vector and angular momentum vector and the right-hand rule would show that the precession will be clockwise as seen from above.
∴
ωp =
Q. 2. A gyroscope wheel weighing 1.5 kg and having a radius of gyration 30 cm is spinning about its axle at a rate of 240 rev/min. The axle is held horizontal and is supported by a pivot at one end which is 10 cm from the centre of mass of the wheel. Calculate the angular velocity of the precession of the wheel. Ans. The gyroscope is shown in figure. The angular velocity of precession is given by
mgr mgr = L Iω where m is the mass of the wheel and L (= Iω) is the angular momentum about the axle. ωp =
324
Mechanics ω r
Fig. 10
Here m = 1.5 kg, r = 10 cm = 0.1 m, I = mk2 = (1.5 kg) (0.3 m)2 = 0.135 kg-m2 ω = 240 rev/min = 4 rev/sec = 4 × 2π rad/sec
and ∴
ωp =
1. 5 × 9. 8 × 0.1 = 0. 43 rad/sec 0.135 × 4 × 2 × 3.14
Q. 3. A torpedo boat with propeller making 4.5 rev/sec makes a complete turn in 84 second. The moment of inertia of the propeller is 94.5 kg-m2. Compute the processional torque on the propeller. Solution. ∴
ωp =
mgr Iω
torque mgr = Iω (ωp)
2π = 199. 66 nt - m. 84 Q. 4. Use the Bohr’s postulate of the quantisation of angular momentum to obtain (i) the possible orbits for the rotation of electron about the proton in the hydrogen atom, and (ii) the possible energy levels of a diatomic molecule behaving like a rigid rotator. = 94.5 × (4.5 × 2π) ×
Ans. (i) Orbits for the Electron Motion in Hydrogen Atom: In the hydrogen atom, an electron revolves in a circular orbit about the proton (nucleus). Then electron carries a negative charge e and the proton an equal positive charge. The electrical force of attraction between them is
e2 1 , where r is the radius of the 4 π ∈0 r2
orbit. This force supplies the necessary centripetal force to the electron which is
mv2 , where r
m is the mass and v linear velocity of the electron. Clearly e2 1 mv2 2 = 4 π ∈0 r r
or
r =
e2 1 4 π ∈0 mv2
...(1)
If I be the moment of inertia and ω the angular velocity of the electron about the proton, then the angular momentum of the electron would be given by
Rotational Kinematics
325
L = Iω = mr2
FG v IJ 3 I = mr H rK
2
and ω =
v r
Now, according to Bohr’s postulate of quantisation, the angular momentum of the electron, L, can only have values that are integral multiples of
h , where h is Planck’s constant. That 2π
is L =
nh 2π
or
mvr =
nh 2π
or
r =
where n is 1, 2, 3, ...
nh 2π mv
...(2)
Let us now eliminate v between Eq. (1) and (2). For this we divide the square of Eq. (1) by Eq. (2), when we obtain n2h2 / 4 π2 m2 v2 r2 = e2 / 4 π ∈0 mv2 r
or
r =
n2h2 ∈0 π m e2
Putting n = 1, 2, 3, .... we obtain the radii of the orbits allowed to the electron. (ii) Rotational Energy Levels for Diatomic Molecules: A diatomic molecule is a dumb-bell model, consisting of two atoms (point-masses) joined by a weightless rigid rod. It rotates about an axis passing through the centre of the joining rod (internuclear axis) and at right angle to it. If I be the moment of inertia of the molecule and ω the angular velocity of rotation, then its kinetic energy of rotation is given by K =
1 2 Iω 2
...(1)
The angular momentum of the molecule is given by L = Iω
...(2)
L2 2I
...(3)
The two equations give K =
Now, according to Bohr’s postulate of quantisation, the angular momentum L of the molecule can only have values which are integral multiple of L = J
h . That is, 2π
h where J = 0, 1, 2, ... 2π
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Mechanics
Substituting this value of L in Eq. (3) we have h2 8π2 I
2 K = J
Thus the kinetic energy also can only have a set of discrete values obtained by putting J = 0, 1, 2, 3,... in this expression. (In other words, the energy is also quantised). These values are
0,
h2 4 h2 9 h2 , , ... 8 π2 I 8 π2I 8 π2I
Figure represents an energy level diagram which indicates how the spacing between the energy levels varies as J increases. We can also obtain the possible angular velocities with which the molecule would rotate. J=6 J J J J J J
K
= = = = = =
5 4 3 2 1 0
Fig. 11
For this we put
L =
Jh in Eq. (2), we get 2π
Jh = Iω 2π
or
ω = J = 0,
h ; 2π I
J = 0, 1, 2, ....
h 2 h 3h , , ,... 2π I 2π I 2π I
Q. 5. The moon revolves about the earth so we always see the same face of the moon (a) How are the spin angular momentum and orbital angular momentum of the moon with respect to the earth related? (b) By how much its spin angular momentum must change so that we could see the entire moon’s surface during a month? Solution. (a) Let M be the mass and R the radius of the moon. Its spin angular momentum about its own axis is LS = IωS where I is the moment of inertia of the moon and ωS its angular velocity about its own axis. Treating it as a solid sphere, I =
2 M R2 , so that 5
LS =
2 M R2ωS 5
Rotational Kinematics
327
The moon not only spins about its axis, but also revolves in a orbit around the earth. The orbital angular momentum of the moon with respect to the earth is given by L0 = I0ω0 where I0 is the moment of inertia of the moon (now treated as a particle) about the earth and ω0 is its orbital velocity. If r be the distance between earth and moon, then I0 = Mr2, so that L0 = Mr2ω0
...(2)
Since we always see the same face of the moon, the moon makes one rotation about its axis in the same time as it makes one revolution around the earth. That is ωS = ω0 Then Eqs. (1) and (2) give
FG IJ H K
LS 2 R = L0 5 r
2
(b) LS must increase or decrease by one-half of its present value if we have to see the entire moon’s surface during a month. Q.6. The earth has a mass of 6 × 1024 kg and a radius of 6.4 × 106 meter. It spins on its own axis and also revolves in an orbit of radius 1.5 × 1011 meter around the sun. Compute its spin angular momentum about its own axis and its orbital angular momentum about the sun. Ans. The earth makes one rotation (turns through 2π radian) about its own axis in 24 hours. Treating it as a solid sphere, its spin angular momentum is given by LS = Iω where I is its moment of inertia about its own axis. Now I = =
2 MR2 5 2 (6 × 1024 kg) (6.4 × 106 m)2 5
= 98.3 × 1036 kg-m2 ωS =
and ∴
2 × 3.14 = 72. 6 × 10−6 rad / sec. 24 × 60 × 60
LS = (98.3 × 1036) × (72.6 × 10–6) = 7.13 × 1033 kg-m2/sec
The earth revolves around the sun once in an year. In order to find the angular momentum of the earth’s motion about the sun, we may treat the earth as a point mass. Then, its orbital angular momentum is L0 = I0ω0 where I0 is its moment of inertia about the sun and ω0 is its orbital velocity. If r be the distance between sun and earth, then I0 = Mr2 = (6 × 1024 kg) (1.5 × 1011 m)2
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Mechanics
= 13.5 × 1046 kg-m2 ω0 =
and ∴
2 × 3.14 = 1. 99 × 10−7 rad / sec 365 × 24 × 60 × 60
L 0 = (13.5 × 1046) × (1.99 × 10–7) = 2.7 × 1040 kg-m2/sec
Q. 7. A body of mass M = 30 kg is running with a velocity of 3.0 meter/sec on ground just tangentially to a merry go round which is at rest. The boy suddenly jumps on the merrygo-round. Calculate the angular velocity acquired by the system. The merry-go-round has a radius of r = 2.0 meter and a mass of m = 120 kg and its radius of gyration is 1.0 meter. Solution. The merry-go-round rotates about the central axis. The moment of inertia of the body about the axis of rotation is I1 = Mr2 = 30 × (2.0)2 = 120 kg-meter2 and that of the merry-go-round is I2 = mk2 = 120 × (1.0)2 = 120 kg-meter2 The body is running with an angular velocity about the axis of rotation which is given by ω1 =
FG v IJ = 3.0 = 1.5 rad/sec. H r K 2.0
while the merry-go-round is at rest (ω2 = 0). Let ω be the angular velocity of the system when the boy jumps on the merry-go-round. As no external torque is acting on the system (friction being ignored), we have initial angular momentum = final angular momentum I1ω1 + I2ω2 = (I1 + I2)ω (120 × 1.50) + (120 × 0) = (120 + 120)ω
120 × 1. 5 = 0. 75 rad/sec. Ans. 240 Q. 8. A meter stick lies on a frictionless horizontal table. It has a mass M and is free to move in any way on the table. A small body of mass m moving with speed v as shown in figure, collides elastically with the stick. What must be the value of m if it is to remain at rest after M the collision? ∴
ω =
Ans. There is no external force (or torque) acting upon the system, and the collision is elastic. Therefore, the linear momentum, the angular momentum about a point say, the center of mass C of the stick and the mechanical energy are all conserved. The initial linear momentum of the system is that of the body mv, angular momentum about C is 1 Iω = md2(v/d), and the kinetic energy is mv2 . Let V 2 be the linear velocity acquired by the stick after collision and ω the angular velocity about C. The body conservation of linear momentum, we have
C d m
v
Fig. 12
l
Rotational Kinematics
329
mv = MV
...(1)
By the conservation of angular momentum, we have
md2
FG v IJ H dK
=
MI2 ω 12
...(2)
where MI2/12 is the moment of inertia of the stick about C. By the conservation of mechanical energy, we have
FG H
IJ K
1 1 MI2 1 2 ω mv2 = MV + 2 2 12 2 From Eq. (1) V =
F GH
mv , from Eq. (2), ω = mvd 12 M MI2
...(3)
I JK
Substituting these values of V and ω in eq. (3), we get
F mv IJ + FG MI IJ FG mvd × 12 IJ = MG H M K H 12 K H MI K F 12 IJ m + md G 1 = M H MI K mF 12 d I 1+ = G J MH I K 2
2
mv 2
2
2
2
or
2
2
2
∴
m =
MI2 I2 + 12 d2
Q. 9. A point-mass is tied to one end of a cord whose other end passes through a vertical hollow tube, caught in one hand, and is caught in the other hand. The point-mass is being rotated in a horizontal circle of radius 2 meter with a speed of 4 meter/sec. The cord is then pulled down so that the radius of the circle reduces to 1 meter. Compute the new linear and angular velocities of the point-mass and compare the kinetic energies under the initial and final states.
v1 v2 r2
ω2
Ans. Let m be the point-mass v1 its linear velocity and ω1 the angular velocity of radius r1. Let v2 and ω2 be the linear and angular velocities respectively when the circle is shortened to a radius r2. The force on the point-mass due to the cord is radial, so that the torque about the centre of rotation is zero. Therefore the angular momentum must remain constant as the cord is shortened. Thus,
Fig. 13
ω1 r1
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Mechanics
initial angular momentum = final angular momentum I1ω1 = I2ω2
mr12
or or
r 1v 1 = r 2v 2 ∴
and
v1 2 v2 = mr2 r1 r2
v2 =
r1 v1 = 2 × 4 = 8 m/sec r2 1
ω2 =
8 v2 = 8 rad/sec = 1 r2
FG H FG H
v 1 2 mr22 2 I ω 2 2 r2 Final K.E. = = 2 1 Initial K.E. I1 ω21 mr2 v1 2 1 r1
IJ K IJ K
2
2
v22 (8)2 4 = 2 = (4)2 = 1 v1 we see that the kinetic energy increases. The increase comes from the work done in pulling in the cord against the centrifugal force. Q. 10. (a) The time-integral of a torque is called the ‘angular impulse’. Show that the angular impulse about a fixed axis is equal to the change in angular momentum about that axis. (b) Show that in order to make a billiard ball roll smoothly, without sliding on top or back spin, the cue must hit the ball at a height
7 R above the table, where R is the radius of the 5
ball. Ans. (a) Angular Impulse: When a large torque acts for a short time, the product of the torque and the time for which it acts is called ‘impulse’ of the ‘torque’ or ‘angular impulse’. We know that the time-rate of change of the total angular momentum of a system about an axis is equal to the resultant external torque acting on the system about that axis. That is τ = or
dL dt
τ dt = dL
If the torque acts for a short time ∆t, and the angular momentum of the system increases from L1 to L2 during this time, then
Rotational Kinematics
331
z
∆t
z
z
L2
τ dt =
dL = L2 − L1
L1
0
∆t
The quantity
τ dt is the ‘angular impulse’. Thus the impulse of a torque about an axis
0
is equal to the change in angular momentum about that axis. (b) Suppose the billiard-ball is initially at rest, and is hit by the cue at a height h above the table. Let the cue exert a force F for a short interval of the time ∆t. ω F
Cue
(h – R ) C
ωR
h R
Fig. 14
The moment of the force (torque) about the axis perpendicular to the paper through the centre C of the ball in F(h – R). If ω is the angular velocity of the ball after impact, then by consideration of angular momentum which increases from 0 to Iω, we have F(h – R) ∆t = Iω where I is the moment of inertia of the sphere.
2 MR2 5 2 MR2 ω ∴ F(h – R) ∆t = ...(1) 5 The original linear velocity of the ball is zero, and if there is to be no slipping at the point of contact with the table after impact, the velocity of the ball will be ωR. By consideration of linear momentum, we have But
I =
F∆t = MωR
...(2)
Dividing Eq. (1) by Eq. (2), we get
2 R 5 2 h = R+R 5
h –R = ∴ ∴
h =
7 R 5
Ans.
Q. 11. A block of mass M is moving with velocity v1 on a frictionless surface as shown. It passes over to a cylinder of radius R and moment of inertia I which has a fixed axis and is initially at rest. When it first makes contact with the cylinder, it slips on the cylinder, but
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Mechanics
the friction is large enough so that slipping ceases before it loses contact with the cylinder. Finally, it goes to the dotted position with velocity v2. Compute v2 in terms of v1, M, I and R. Ans. The initially angular momentum of M about C when it first makes contact with the cylinder, is L 1 = Moment of inertia about C × angular velocity 2 = (MR)
FG v IJ = Mv R H RK 1
1
The final angular momentum when it loses contact with the cylinder is L 2 = Mv2R The loss in angular momentum is ∆L = L1 – L2 = MR (v1 – v2) This must be equal to the angular impulse τ∆t on the cylinder about C. Thus MR (v1 – v2) = τ∆t
...(1)
Now, if α be the angular acceleration produced in the cylinder, then J = Iα = I
∆ω ∆t
...( 2)
M v1
v2
C R
Fig. 15
As the cylinder is initially at rest and finally moves with a velocity v2, we have ∆ω =
v2 R
τ =
I ∆t
So that from Eq. (2),
Putting this value in (1), we get MR (v1 – v2) = I or
v1 – v2 =
or
v2 =
FG v IJ H RK 2
FG v IJ H RK 2
FG I IJ v H MR K 2
v1 1+
I MR2
2
Ans.
Rotational Kinematics
333
Q. 12. As cockroach, mass m, runs round the rim of a disc of radius R with speed v counter clockwise. The disc has rotational inertia I about its axis and it rotates clockwise with angular speed ω0. If the cockroach stops, compute the angular speed of the disc and also the change in kinetic energy. Ans. As no external torque is acting on the system, the angular momentum is conserved. Now Initial angular momentum = Iω 0 − (mR)2
FG v IJ H RK
= Iω0 – mvR (the negative sign is used because the angular momentum of the insect of oppositely directed to that of the disc). Let ω be the final angular velocity of the system when the cockroach stops. Thus Final angular momentum = (I + mR2)ω By the conservation of angular momentum, we have Iω0 – mvR = (I + mR2)ω ∴
ω =
I ω 0 − mvR I + mR2
Though angular momentum is conserved, but the angular velocity has changed. Hence the energy is also changed. The change can be easily computed. Q. 13. If the earth suddenly contracts to half its radius, what would be the length of the day. Ans. Let R1 be the present radius of earth and T1 its period of revolution about its axis, which is the length of the day (T1 = 24 hours). Let R2 be the new value of radius and period. By conservation of angular momentum, we have I1ω1 = I2ω2
or
or
FG 2 MR IJ 2π = FG 2 MR IJ 2π H5 K T H5 K T 2 1
2 2
1
2
R22 (R1 / 2)2 hours) × = ( 24 R12 R21 = 6 hours Ans.
T2 = T1
Q. 14. A man stands at the center of a rotating platform holdings his arms extended horizontally with a weight in each hand, the platform making 1 revolution per second. The moment of inertia of the whole system is 6 kg-meter2. When the man pulls in his hands to his sides, the moment of inertia becomes one-third of its initial value. What is the new angular velocity of the platform. Has the kinetic energy been conserved, If not, explain it. Assume that there is no friction in the platform. Ans. If friction in the platform is neglected, no external torque acts about the vertical axis through the center of mass, and hence the angular momentum about the axis is constant. (the force of gravity acts through the center of mass and hence exerts on torque the axis of rotation). Thus
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Mechanics
I1ω1 = I2ω2 where I1 and ω1 are the initial and I2 and ω2 the final moment of inertia and angular velocity about the axis of rotation. I1 = 6 kg-m2, ω1 = 1 rev/sec = 2π rad/sec I2 = 2 kg-m2, ω2 = ? ∴
6 × 2π = 2 × ω2 ω2 = 6π rad/sec = 3 rev/sec
or Initial kinetic energy
1 1 I1 ω12 = × 6 × (2π)2 = 12 π2 joule 2 2
Final kinetic enegy
1 1 I2 ω22 = × 2 × (6 π)2 = 36 π2 joule 2 2
Clearly, the kinetic energy has not been conserved. It has increased. Increase in kinetic energy = 36 π2 – 12 π2 = 24 π2 = 24 × (3.14)2 = 236.6 Joule This increase in energy is supplied by the man who does work when he pulls in his hands to his sides. Q. 15. A uniform disc of mass M and radius R is rotating about its central axis which is horizontal with angular velocity w0 (a) write down its kinetic energy and angular momentum. (b) suddenly a piece of mass m breaks off the disc and rises vertically. To what height the piece would rise. (c) what are the final energy and momentum of the broken disc?
1 I ω20 2 1 1 1 MR2 ω20 = M R2 ω20 = 2 2 4
Ans. (a)Initial kinetic energy Ki
=
FG H
IJ K
1 MR2 ω0 2 (b) Thus moment of inertia of the mass m, which later on breaks off the disc is mR2. Initial angular momentum Li = I ω 0 =
1 (mR2 ) ω20 . On breaking off, the mass goes up and its 2 kinetic energy is converted into gravitational potential energy mgh, where h is the maximum height to which it goes. Thus, Its rotational energy is therefore
1 mR2 ω20 = mgh 2 R2 ω20 2g (c) Let us consider the disc to be made up of the piece of mass m and the remainder (broken) disc. Now, by the conservation of energy, we have
or
h =
Rotational Kinematics
335
ki = kf ( of broken disc) + k ′ f (of mass m)
1 1 MR2 ω20 = kf + m ( Rω 0 )2 4 2
or ∴
kf =
FG H
IJ K
1 1 M − m R2 ω20 2 2
Similarly, by the conservation of angular momentum, we have Li = Lf (of broken disc) + Lf (of mass m)
1 MR2 ω 0 = Lf + mR2 ω0 2
or ∴
Lf =
FG 1 M − mIJ R ω H2 K 2
0
Ans.
PROBLEM 1.
A wheel of moment of inertia 8 kg-m2 is momented on a shaft and revolving at a rate of 600 rev/min. When second wheel, initially at rest, is coupled to the same shaft, the combination revolves at a rate of 400 rev/min. What is the moment of inertia of the second wheel. How much energy is lost in the process ? (Neglect all bearing friction). [Ans. 4 kg-m2, 5258 joule]
2.
A certain mass tied to one of a light string of length 100 cm is making 4 rev/sec in a horizontal plane about the other end. How many rev/sec does it make if the string length is changed to 50 cm ? How much were kinetic energy does the revolving mass have if it weighs 0.5 kg ? From where does this additional energy come ? [Ans. 16, 48π2 joule]
3.
An ice-skater spinning at the rate 1 rev/sec changes posture in such a way that his moment of inertia is reduced to half its initial value. What becomes his speed of spin ?
4.
A turn table rotates about a fixed vertical axis, making one revolution in 10 sec. The moment of inertia of the turn table about the axis is 1200 kg-m2. An 80 kg man, initially standing at the center on the turn table, moves along a radius. What is the angular velocity of the turn table when the mass is 2 meter from the center.
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Mechanics
8 MOMENT OF INERTIA 8.1 MOMENT OF INERTIA The inability of a body to change by itself its state of rest or uniform motion along a straight line (Newton’s first law of motion) is an inherent property of matter and is called inertia. The greater the mass of the body, the greater the resistance offered by it to any change in its state of rest or linear motion. So that, mass is taken to be a measure of inertia for linear or translatory motion. In the same manner, a body free to rotate about an axis opposes any change in its state of rest or uniform rotation. In other words, it possess inertia for rotational motion, i.e., it opposes the torque or the moment of the couple applied to it to change its state of rotation. It must also be, therefore, of the nature of the moment of a couple, for only a couple can oppose another couple. Hence the name moment of inertial given to it. It will thus be seen at once that the moment of inertia (usually written as M.I.) about a given axis plays the same part in rotational motion about that axis as the mass of a body does in translational motion. In other words the moment of inertia, in rotational motion, is the analogue of mass in linear or translational motion. If a particle of mass m is at a distance r from an axis of rotation, its moment of inertia I about the axis is given by the product of the mass and the square of its distance from the axis. m1
Thus I = mr2
m2
r1
In an extended body, each particle of matter in the body add to the moment of inertia and amount mr2. Figure represents such a body rotating about an axis through O and perpendicular to the plane of the body. If m1, m2 , m3, ... be the masses of the particles composing the body and r1, r2, r3, ... their respective distances from the axis of rotation, the moment of inertia I of the body about that axis is given by
336
r2 O r3 r5
m3
r4 m4
m5
Fig. 1
Moment of Inertia
337
I = m1r12 + m2r22 + m3r32 + ... = Σ mr2
z
For a body having a continuous distribution of matter, we put I =
r2 dm
where dm is the mass of a infinitesimally small element of the body taken at a distance r from the axis of rotation. Hence the moment of inertia of a body about a given axis is the sum of the products of the masses of its particles by the squares of their respective distance from the axis of rotation. It thus depends not only on the mass of the body but also upon the manner in which the mass is distributed around the axis of rotation.
8.2 ROLE OF MOMENT OF INERTIA IN ROTATIONAL MOTION According to the Newton’s first law of motion, every body maintains its state of rest or uniform transnational motion unless some external force is applied on it to change that state. In other words, a force is necessary to change the motion of a body that is, to produce a linear acceleration in it. This property of bodies is called ‘inertia’. The greater is the mass of a body, the greater is the force required to produce a given linear acceleration in it. Thus the mass of a body is a measure of its inertia. Similarly, if the rotational motion of a boy about an axis is to be changed, that is, if an angular acceleration is to be produced, a torque (moment of force) about that axis must be applied. The body is, therefore, said to possess a ‘moment of inertia’ about that axis. The greater is the moment of inertia, the greater is the torque required to produce a given angular acceleration in the body. Thus, moment of inertia plays the same role in the rotational motion as mass plays in transnational motion.
8.3 RADIUS OF GYRATION The radius of gyration, K, of a body about its axis of rotation, which we have called the effective distance of its particles from the axis, is actually that distance from the axis at which if its center of mass (M) be supposed to be concentrated, its moment of inertia about the given axis would be the same as with its actual distribution of mass. If a change in the position or inclination of the axis of rotation of a body will bring about a change in the relative distances of its particles and hence in their effective distance or the radius of gyration of the body about the axis. And so will a change in the distribution of mass about the axis, e.g., the transference of a portion of matter (or mass) of the body from one part to another, despite the fact that the total mass of the body remains unaltered. Thus, whereas the mass of a body remains the same irrespective of the location or inclination of its axis of rotation, its radius of gyration about the axis depends upon (i) the position and inclination of the axis and (ii) the distribution of mass of the body about the axis. We usually put it as 2 2 I = Σ mr = MK
Where M is the total mass of the body (e.g. equal to m1 + m2 + m3 + ..... etc) and K, the effective distance of its particles from the axis, called its radius of gyration about the axis of rotation.
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Mechanics
8.4 ANALOGOUS PARAMETERS IN TRANSLATIONAL AND ROTATIONAL MOTION Translational Motion 1.
Mass (or translational inertia), m or M
2.
Distance covered (or linear displacement), S
3.
Linear velocity, v =
ds dt
4.
Linear velocity, a =
dv d2s = 2 dt dt
5.
Linear momentum, p = mv
6.
Force (or rate of change of linear momentum), F =
7.
Work done by force, W =
8.
Power, P = Fv
9.
Translational Kinetic Energy, ½ mv2
10.
z
dφ = ma dt
F dS = FS
Equations of translational motion: (i) v = u + at (ii) S = ut + ½ at2 (iii) v2 – u2 = 2as
Potential Motion 1.
Moment of Inertia, I = MK2 (or rotational inertia)
2.
Angular described, θ (radius or angular displacement)
3.
Angular velocity, ω =
4.
Angular acceleration, α =
5.
Angular momentum, J = Iω
6.
Torque or moment of couple (or rate of change of angular momentum), τ or dJ dω C= =I dt dt
7.
Work done by torque or couple, W =
8.
Power, P = τω or Cω
9.
Rotational kinetic energy =
10.
dθ (radian/sec) dt dω d2θ = dt dt2
1 2 Iω 2
Equation of rotational motion: (i)
ω 2 = ω1 + αt = ω1 +
FG dω IJ t H dt K
z
τ dθ or
z
cdθ or τθ or Cθ
Moment of Inertia
339
FG IJ H K
1 2 1 dω 2 t αt = ω1t + 2 2 dt dω 2 2 θ (iii) ω 2 − ω1 = 2αθ = 2 dt (ii) θ = ω1t +
8.5 GENERAL THEOREMS ON MOMENT OF INERTIA There are two important theorems on moment of inertia which, in some cases, enable the moment of inertia of a body to be determined about an axis, if its moment of inertia about some other axis be known. (i) The theorem of perpendicular axes: The moment of inertia of a uniform plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about any two mutually perpendicular axes in its plane intersecting on the first axis. This is the theorem of perpendicular axis for a laminar body. Thus, if Ix and Iy be the moments of inertia of a plane lamina, about the perpendicular axis OX and OY respectively, which lie in the plane of the lamina and intersect each other at O, its moment of inertia (I) about an axis passing through O and perpendicular to its plane is given by i.e.,
I = Ix + Iy
Y
For, considering a particle of mass m at P, distance r from O and x from y from the axes OY and OX respectively, we have
x
I = Σmr2, Ix = Σmy2 and Iy = Σmr2
r
Ix +Iy = Σmy2 + Σmx2
so that
m P y X
O
= Σm (y2 + x2) = Σmr2
[y2 + x2 = r2] Fig. 2
Ix + Iy = I
i.e.,
For a three-dimensional body: In this case, the theorem demands that the sum of the moments of inertia of a three-dimensional body about its three mutually perpendicular axes is equal to twice the summation Σmr2 about the origin. Suppose, we have a three-dimensional or a cubical body with OX, OY and OZ as its three mutually perpendicular axes, representing its length, breadth and height respectively. Consider a small element of mass ‘m’ of the body at a point P somewhere inside it. Drop a perpendicular PM from P on the x-y plane to meet it in M. Join OM and OP and from M draw MQ parallel to the x-axis and MN parallel to the y-axis. Also, from P draw PR parallel to OM. Then , clearly, the coordinates of the point P are x = ON = QM, y = OQ = MN and z = MP = OR. Z mP
R
r N
O
X
y Q Y
x
M
Fig. 3
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Mechanics
Since plane x – y is perpendicular to the z-axis, any straight line drawn on this plane is also perpendicular to it. So that, both OM and PR are perpendicular to the z-axis (PR being parallel to OM). ∠OMP is a right angle, [OM || PR and PM || OR.] We, therefore, have OM2 + MP2 = OP2, or OM2 + z2 = r2 But
...(i)
∴ OP = r
OM2 = QM2 + OQ2 = x2 + y2
∴ OQM is a right angle, being the angle between the axes of x and y. Substituting the value of OM2 in relation (i), we have x2 + y2 + z2 = r 2
...(ii)
Join PN and PQ. Then, PN and PQ are the respective normals to the axes of x and y. For ∠PMN being a right angle (the angle between the axes of y and z), we have PN2 = MN2 + PM2 = y2+ z2 and, therefore x2 + PN2 = x2 + y2 + z2 = r2 or ON2 + PN2 = r2 (x = ON). Angle PNO is thus a right angle and, therefore, PN perpendicular to the x-axis. Similarly, in the right-angled triangle ∆PMQ, we have PQ2 = MQ2 + PM2 = x2 + z2 Again, because x2 + y2 + z2 = r2, we have PQ2 + y2 = r2 or
∴ y = OQ and r = OP
PQ2 + OQ2 = OP2
Angle PQO is a right angle and, therefore, PQ perpendicular to the y-axis. Now, moment of inertia of the element at P about the z-axis = m × PR2 = m × OM2, because PR = OM is the perpendicular distance of the mass from the axis. ∴ moment of inertia of the whole body about the z-axis, i.e., Iz = Σm.OM2 = Σm (x2 + y2) Similarly, moment of inertia of the body about the y-axis, i.e., Iy = Σm.PQ2 = Σm (x2 + z2) because PQ is the perpendicular distance of mass m from this axis. And , moment of inertia of the body about the x-axis, i.e., Ix = Σm.PN2 = Σm (y2 + z2) because PN is the perpendicular distance of mass m from the x-axis, ∴ adding up the moments of inertia of the body about the three axes, we have Ix + Iy + Iz = Σm (y2 + z2) + Σm (x2 + z2) + Σm (x2 + y2) = 2Σm (x2 + y2 + z2) = 2Σm r2 C A The Theorem of parallel axes: The moment of inertia I of a body about any axis is equal to its moment of inertia Icm about a parallel axis through its center m r of mass, plus the product of the mass M of the body O P and the square of the perpendicular distance h between x the two axes. That is I = Icm + Mh2 This is the theorem of parallel axes. (a) For a plane laminar body: Let AB be the axis in the plane of the paper about which the moment
B
D
Fig. 4
Moment of Inertia
341
of inertia (I) of the plane lamina is to be determined and CD, an axis parallel to AB, through the center of mass O of the lamina, at a distance r from AB. Considering a small element of mass of the lamina at the point P, distance x from CD, we have moment of inertia of the element about the axis AB = m (x + r)2 and; therefore, moment of inertia of the whole lamina about the axis AB, i.e., I = Σm (x + r)2 I = Σmx2 + Σm2 + 2Σmxr
or
Σmx2 = Icm the moment of inertia of the lamina about the axis CD, through its centre of mass. So that, I = Icm + Σmr2 + 2Σmxr Now, Σmr2 = r2 Σm (r being constant) = Mr2, where M is the mass of the whole lamina, and Σmx = the sum of the moments of all the particles of the lamina about the axis CD, passing through its center of mass and, therefore, equal to zero, as we know, a body always balances about its c.m., showing that the algebraic sum of the moments of all its particles about the c.m. is zero, therefore, we have I = Icm + Mr2 i.e., the moment of inertia of the lamina about the axis AB = its moment of inertia about a parallel axis CD through its center of mass + mass of the lamina × (distance between he two axes)2 (b) For a three-dimensional body: Let AB be the axis about which the moment of inertia of a cubical or a three-dimensional body, is to be determined. Draw a parallel axes CD through the center of mass O of the body at a distance r from it. Imagine an element of the body, of mass m, at a point P outside the plane of the axes AB and CD and let PK ad PL be perpendicular drawn from P to AB and CD respectively and PT, the perpendicular dropped from P on to KL produced. PL = d, LK = r, LT = x and ∠PLK = θ.
Put
Then, if I be the moment of inertia of the body about the axis AB and Icm, its moment of inertia about the axis CD (through its center of mass O), we have
A
C r r
θ
K
T x L d
O B
D
P
Fig. 5
I = Σm.PK2 and Icm = Σm.PL2 = Σm d2 Now, from the geometry of the figure, we have PK2 = PL2 + LK2 – 2 PL.LK cos PLK = d2 + r2 – 2 dr cos θ.
...(1)
And, in the right-angled triangle PTL, we have cos PLT = LT/PL, where, ∠PLT = 180° –∠PLK = 180° – θ . So that cos (180° – θ) = x/d or –cos θ = x/d, where d cos θ = –x
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Mechanics
Substituting this value of d cos θ in the expression for PK2 above Eq. (1), we have PK2 = d2 + r2 + 2rx, and therefore, I = ΣmPK2 = Σm (d2 + r2 + 2rx) = Σmd2 + Σmr2 + 2r Σmx I = Icm + Mr2 + 2r Σmx
or
Σm = M, mass of the body and r is constant. clearly Σmx = 0, being the total moment about an axis passing through the center of mass of the body. We, therefore, have I = Icm + Mr2 the same result as for a plane laminar body.
8.6 CALCULATION OF MOMENT OF INERTIA In the case of a continuous homogeneous body of a definite geometrical shape, its moment of inertia is calculated by (i) first obtaining an expression for the moment of inertia of an infinitesimal element (of the same shape) of mass dm about the given axis, i.e., dm.r2, where r is th distance of the infinitesimal element from the axis and then (ii) integrating this expression over appropriate limits so as to cover the entire body, making full use of the theorems of perpendicular and parallel axes, where necessary.
8.7 MOMENT OF INERTIA OF A UNIFORM ROD (i) About an axis through its centre and perpendicular to its length: Let AB be a thin uniform rod, of length l and mass M, free to rotate about an axis YOY' passing through its center O and perpendicular to its length. Since the rod is uniform, its mass per unit length M/l. Y Considering a small element of the rod, of length dx at a distance x from the axis through O, we have mass of the element = (M/l).dx and therefore, its M.I. about the axis (YOY') through O
x dx O
A l 2
= (M/l).dx.x2 The moment of inertia (I) of the whole rod about the axis YOY' is thus given by the integral of the above expression between the limits x = – l/2 and x = + l/2 or by twice its integral between the limits x = 0 and x = l/2, i.e.,
z
l/2
I = 2
0
B l 2
Y′
Fig. 6
LM OP N Q
M 2 2M x3 x dx = l l 3
l 2 0
=
Ml2 2M l3 . = l 24 12
(ii) About an axis through one end of the rod and perpendicular to its length: The moment of inertia of the rod about the axis; passing through one end A of the rod, by integrating the expression for the M.I. of the element dx of the rod, between the limits x = 0 at A and x = l at B, i.e.,
z l
I =
0
LM OP N Q
M 2 M x3 x dx = l l 3
l
= 0
Ml2 3
Moment of Inertia
343 Y
l 2 dx O
A
B
x Y′
Fig. 7
8.8 MOMENT OF INERTIA OF A RECTANGULAR LAMINA (OR BAR): (i)
About an axis through its center and parallel to one side: Let ABCD be a rectangular lamina of length l, breadth b and mass M and let YOY' be the axis through its center O and parallel to the side AD or BC about which its moment of inertia is to be determined. Consider an element, or a small rectangular strip of the lamina, parallel to YY', and at a distance x from, the axis. The area of this strip or element = dx.b. And, since the mass per unit area of the lamina = M/(l × b), we have mass of the strip M M . dx. b = . dx, and, therefore, M.I. of the element about the axis or element = l×b l YOY' =
M dx. x2 l Y
dx
A
B
O
b
D
C x
l 2
l 2
Y′
Fig. 8
The moment of inertia (I) of the whole rectangular lamina is thus given by twice the integral of the above expression between the limits x = 0 and x = l/2.
z
l/2
2M M 2 x dx = I = 2 l l 0
LM OP N Q
2M x3 = l 3
l 2 0
=
z
l/2
x2dx
0
2M l3 Ml2 . = l 24 12
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Mechanics
This result show that if b small, the rectangular lamina becomes a rod of length l whose M.I. about the axis YOY’ passing through its center and perpendicular to its length would be Ml2/12. (ii) About One Side: In this case, since the axis coincides with AD or BC, we integrate the expression for the M.I. of the element of length dx and distance x from the axis, i.e., (M/l) dx.x2 between the limits x = 0 at AD and x = l at BC. So that M.I. of the lamina about side AD or BC is given by
z l
I =
0
LM OP N Q
M 2 M x3 x dx = l l 3
l
= 0
Ml3 3
(iii) About an axis passing through its center and perpendicular to its plane: This is obtained by an application of the principle of perpendicular axes to case (i), according to which, M.I. of the lamina about an axis through O and perpendicular to its plane = M.I. of the lamina about an axis through O parallel to b + M.I. of the lamina about an axis through O parallel to l, i.e., I =
Ml2 Mb2 M 2 + = (l + b2 ) 12 12 12
(iv) About an axis passing through the mid-point of one side and perpendicular to its plane: In this case the axis passes through the mid-point of side AD or BC, and perpendicular to the plane of the lamina, so that, it is parallel to the axis through O (the c.m. of the lamina) in case (iii). In accordance with the principle of parallel axes, therefore, the M.I. of the lamina about this axis is given by
F l + b I + MFG l IJ GH 12 JK H 2 K Fl + b +l I = MFl + b I = MG H 12 4 JK GH 3 12 JK 2
2
2
2
2
I = M
2
2
2
And if the axis passes through the mid-point of AB or DC, then
F l + b I + M FG b IJ GH 12 JK H 2 K Fl + b I = MF l + b I = MG H 12 JK GH 12 3 JK 2
2
2
2
2
I = M
2
2
(v) About an axis passing through one of its corners and perpendicular to its plane: Let the axis pass through the corner D of the lamina and it is perpendicular to the plane of the lamina, it is parallel to the axis through its center of mass O in case (iii). Again, by the principle of parallel axes, we have moment of inertia of the rectangular lamina about this axis through D is given by
F l + b I + Mr , GH 12 JK 2
I = M
where r is the distance between the two axes.
2
2
Moment of Inertia
345
F l I F bI l + b = G J +G J = H 2K H 2 K 4 Fl + b I + MFl + b I I = MG H 12 JK GH 4 JK F l + b + 3l + 3b I = MG JK 12 H Fl + b I I = MG H 3 JK 2
We have,
So that,
2
2
2
r2
2
2
2
2
2
2
2
2
2
2
8.9 MOMENT OF INERTIA OF A THIN CIRCULAR RING (OR A HOOP) (i) About an axis through its center ad perpendicular to its plane: Let the radius of the hoop or the thin circular ring be R and its mass M. Consider a particle of mass m of the hoop or the ring, its M.I. about an axis passing through the center O of the hoop or the ring and perpendicular to its plane = mR2 ∴ M.I. of the entire hoop or ring about this axis passing through its center and perpendicular to its plane, Y
I = ΣmR2 = MR2 Where åm = M, the mass of the hoop or ring. (ii) About its diameter: Due to symmetry, the M.I. of the hoop or the ring will be the same about one diameter as about another. Thus, if I be its M.I. about the diameter XOX', it will also be I about the diameter YOY' perpendicular to XOX'.
O
X′
X
Y′
Fig. 9
By the principle of perpendicular axes, therefore, the M.I. of the hoop or the ring about the axis through its center O and perpendicular to its plane is equal to the sum of its moments of inertia about the perpendicular axes XOX' and YOY' in its own plane and intersecting at O, i.e., I + I = MR2 or or
2I = MR2 I =
MR2 2
8.10 MOMENT OF INERTIA OF A CIRCULAR LAMINA OR DISC: (i) About an axis through its center and perpendicular to its plane: Let M be the mass of the disc and R its radius, so that its mass per unit area is equal to
M . πR2
Considering a ring of the disc, of width dx, and distant x from the axis passing through O and perpendicular to the plane of the disc, we have Area of the ring = circumference × width
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Mechanics
= 2π x dx and hence its Mass =
FG M IJ 2π. x. dx = 2M. xdx H πR K R 2
x
R
2
And, therefore, its M.I. about the perpendicular axis through O =
2M. x. dx 2 2M. x3 . dx x = R2 R2
Fig. 10
Since the whole disc may be supposed to be made up of such concentric rings of radii ranging from O to R, the M.I. of the whole disc about the axis through O and perpendicular to its plane, i.e., I , is obtained by integrating the above expression for the M.I. of the ring, between the limits x = 0 and x = R. Thus,
z
R
I =
0
=
LM OP N Q
2M 3 2M x4 = x dx R2 R2 4
R
0
2M R 4 MR2 = 2 2 R 4
(ii) About a diameter: Due to symmetry, the M.I. of the disc about one diameter is the same as about another. So that, if I be the M.I. of the disc about each of the perpendicular diameters XOX' and YOY', then by the principle of perpendicular axes, I + I = M.I. of the disc about an axis through O and perpendicular to its plane. i.e.
2I =
MR2 2
or
I =
MR2 4
Y O
X′
X
Y′
Fig. 11
8.11 MOMENT OF INERTIA OF AN ANGULAR RING OR DISC (i) An angular disc is just an ordinary disc, with a smaller coaxial disc removed from it leaving a concentric circular hole in it. If R and r be the outer and inner radii of the disc and M, its mass, then mass per unit area of the disc mass M = area π (R2 − r2 ) Now, the disc may be imagined to be made up of a number of circular rings, with their radii ranging from r to R. So that, considering one such ring of radius x and width dx, we have
=
dx r2 O
x
Face area of the ring = 2πxdx and ∴
its mass = 2πxdx [M/p (R2 – r2)] =
2M x dx ( R2 − r2 )
Fig. 12
R
Moment of Inertia
347
and hence its M.I. about the axis through O and perpendicular to its plane =
2Mx 2Mx3 2 dx . x = dx (R2 − r2 ) (R2 − r2 )
The M.I. of the whole annular disc i.e., I, is therefore, given by the integral of the above expression between the limits x = r and x = R
z
R
or
I =
r
I =
2Mx3 2M dx = 2 (R2 − r2 ) (R − r2 )
LM N
OP Q
z
R
x3 . dx =
r
LM OP N Q
x4 2M 2 2 (R − r ) 4
R
r
2M R 4 − r4 M (R2 + r2 ) = 2 4 2 (R − r ) 2
(ii) About a diameter: Due to symmetry, the M.I. of the annular disc about one diameter is the same as about another, say, I. Then, in accordance with the principle of perpendicular axes, the sum of its moments of inertia about two perpendicular diameters must be equal to its M.I. about the axis passing through its center (where the two diameters intersect) and perpendicular to its plane, i.e., I + I = M (R2 + r2)/2 2I = M
or
I =
(R2 + r2 ) 2
M (R2 + r2 ) 4
8.12 MOMENT OF INERTIA OF A SOLID CYLINDER (i) About its own axis of cylindrical symmetry: A solid cylinder is just a thick circular disc or a number of thin circular disc (all of the same radius) piled upon over the other, so that its axis of cylindrical symmetry is the same as the axis passing through the centre of the thick disc and perpendicular to its plane. ∴ M.I. of the solid cylinder about its axis of cylindrical symmetry, i.e., I = M.I. of the thick disc (or the pile of thin discs) of the same mass and radius about the axis through its center and perpendicular to its plane.
l 2 X′
R
2
Y
x
A
O
dx X
B l Y′
Fig. 13
MR case 8.10(i) 2 (ii) About the axis through its center and perpendicular to its axis of cylindrical symmetry: If R be the radius, l, the length and M, the mass of the solid cylinder, supposed to be uniform and of a homogeneous composition, We have its mass per unit length = M/l. I=
Now, imagining the cylinder to be made up of a number of discs each of radius R, placed adjacent to each other, and considering one such disc of thickness dx and at a distance x from the center O of the cylinder, we have mass of the disc = (M/l) dx and radius = R ∴ M.I. of the disc about its diameter AB =
M R2 dx. l 4
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Mechanics
and its M.I. about the parallel axis YOY', passing through the center O of the cylinder and perpendicular to its axis of cylindrical symmetry (or its length), in accordance with the principle of parallel axes, =
M R2 M dx dx. x2 + l 4 l
Hence, M.I. of the whole cylinder about this axis, i.e., I = twice the integral of the above expression between the limit x = 0 and x = l/2, i.e.,
z
l/2
I = 2
0
F M dx R GH l 4
2
2
I =
2
=
I JK
2M M dx. x2 = l l
LM OP N Q l O 2M L R l . + M P l N 4 2 8. 3 Q FR l I 2M F R l + =MG l GH 8 24 JK H4
2M R 2 x x 3 + = l 4 3
or
+
z FGH
l/2
0
I JK
R2 dx + x2 . dx 4
l /2
0
3
3
2
+
l2 12
I JK
8.13 MOMENT OF INERTIA OF A SOLID CONE (i) About its vertical axis: Let M be the mass of the solid cone, h its vertical height and R, the radius of its base. Volume of the cone = (1/3)π R2h and if ρ be the density of its material, its mass M M = or
1 2 πR hρ 3
A
X′ r
αx
3M ρ = πR2h
The cone may be imagined to consist of a number of disc of progressively decreasing radii, from R to O, piled up one over the other.
X
dx
h
O R
Fig. 14
Considering one such disc of thickness dx and at a distance x from the vertex (A) of the cone, we have radius of the disc r = x tan a, where a is the semi-vertical angle of the cone, and, therefore, its volume = πr2dx = πx2 tan2 α dx and its mass = πx2 tan2 α ρ dx Hence, M.I. of the disc about the vertical axis AO of the cone (i.e., an axis passing through its center and perpendicular to its plane) = mass × radius2/2 = πx2 tan2 αρ dx. r2/2 = πx2 tan2 αρ dx. x2 tan2 α/2 = (πρ tan4 a/2) x4 dx
Moment of Inertia
349
and M.I. of the entire cone about its vertical axis AO is given by
z
h
I =
0
πρ tan4 α 4 x . dx 2
πρ tan4 α = 2
=
z
h
x4 . dx =
0
πρ tan4 α 2
5
h
0
πρ tan4 α h5 πρR 4 h5 . . = 2 5 2h4 5
Substituting R/h for tan α or substituting the value of I =
LM x OP N5Q
ρ obtained above, we have
π 3 M R 4 h5 3MR2 . . = 10 πR2h 2h4 5
(ii) About an axis passing through the vertex and parallel to its base: Considering the disc at a distance x from the vertex of the cone, we have its M.I. about its diameter = mass × (radius)2/4 = πx2 tan2 αρ dx. r2/4 = π x2 tan2 αρ dx.x2 tan2 α/4 = (πρ tan4 α/4)x4 dx ∴ Its M.I. about the parallel axis XX', distant x from it. = (πρ tan4 α/4) x4 dx + π x2 tan2 αρ . x2 dx = (πρ tan4 α/4) x4 dx + π tan2 αρ.x4 dx Hence M.I. of the entire cone about the axis XX', parallel to its base is given by
z
h
I =
0
z z h
πρ tan4 α 4 x . dx + 4 0
πρ tan4 α = 4
z
πρ tan2 αx4 . dx
0
h
z
h
x4 . dx + πρ tan2α
0
LM OP N Q
πρ R4 x5 . = 4 h4 5
=
h
x4 dx
0
h
0
LM OP N Q
R2 x5 + πρ 2 5 h
h
0
πρR4 h5 πρR2 h5 . + . 5 h2 4 h4 5
or, substituting the value of ρ, we have M.I. of the cone about the axis XX', i.e.,
or
I =
π 3M R4 h5 3πM R2 h5 . . + . . 4 πR2h h4 5 πR2h h2 5
I =
3MR2 3Mh2 + 20 5
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Mechanics
8.14 MOMENT OF INERTIA OF A HOLLOW CYLINDER (i) About its axis of cylindrical symmetry: A hollow cylinder may be considered to be a thick annular disc or a combination of thin annular discs, each of the same external and internal radii, placed adjacent to each other, the axis of the cylinder (i.e., its axis of cylindrical symmetry) being the same as the axis passing through the centre of the thick annular disc (or the combination of thin annular discs) and perpendicular to its plane. The M.I. of the hollow cylinder about its own axis is, therefore, the same as that of a thick annular disc (or a combination of thin annular discs) of the same mass M and external and internal radii R ad r respectively about the axis passing through its center and perpendicular to its plane, i.e., I = M(R2 + r2)/2 (ii) About an axis passing through its centre and perpendicular to its own axis: If R and r be the external and internal radii respectively of the hollow cylinder, l, its length and M, its mass, we have mass per unit volume of the cylinder = M/π (R2 – r2) l. Imagining the hollow cylinder to be made up of a large number of annular discs, of external and internal radii R and r respectively, placed adjacent to each other, and considering one such disc at a distance x from the axis YOY' passing through the center O of the cylinder and perpendicular to its own axis, we have Surface area of the disc = p(R2 – r2), its volume (R2 – r2) dx and its mass 2 2 = π ( R − r ) dx .
M M. dx = l π (R2 − r2 ) l
So that, M.I. of the disc about its diameter (AB) = Mdx (R2 + r2)/4l And, its M.I. about the parallel axis YOY' distant x from it, in accordance with the principle of parallel axes,
F GH
I JK
M R2 + r2 M dx dx. x2 + l l 4
=
Hence, M.I. of entire hollow cylinder about the axis YOY' is equal to twice the integral of the above expression between the limits x = 0 and x = l/2 i.e.,
z
LM M (R + r ) dx + M x . dxOP I =2 l N 4l Q LM (R + r ) dx + x . dxOP 2M = l N 4 Q 2M L (R + r ) x x O = M 4 + 3 PQ = 2Ml LMN (R4 +× r2 )l + 8 l× 3 OPQ l N L R + r + l OP I =MM N 4 12 Q l/2
2
2
2
0
z
l/2
2
2
2
3
l/2
0
2
2
2
A
2
2
3
x
O l
2
0
2
l 2
Y
Y′
Fig. 15
B
dx
Moment of Inertia
351
8.15 MOMENT OF INERTIA OF A SPHERICAL SHELL (i) About its diameter: Let ACBD be the section through the center O, of a spherical shell of radius R and mass M, whose moment of inertia is to be determined about a diameter AB, its value being obviously the same about any other diameter. Surface area of the shell = 4π r2 and, therefore, mass per unit area of the shell = M/4πR2. Consider a thin slice of the shell, lying between two planes EF and GH, perpendicular to the diameter AB at distances x and x + dx respectively from its center O. This slice is, a ring of radius PE and width EG (and not PQ which is equal to dx, the distance between the two planes).
C
E G dθ
θ R A
P O x
Q
B dx
Area of the ring = circumference × width
D
= 2 πPE × EG and hence its mass
FH
Fig. 16
= 2 πPE × EG × M/4πR2 Join OE and OG and let angle COE be equal to θ and angle EOG = dθ. Then, PE = OE cos OEP = R cos θ, because OE = R and ∆OEP = alternate ∆COE = θ And OP = OE sin OEP, x = R sin θ and
∴ dx/dθ = R cos θ, OP = x and OE = R.
dx = R cos θ dθ = PE. dθ because R cos θ = PE and EG = OE dθ = R dθ ∴ mass of the ring = 2 π PE × R dθ × (M/4θR2) = Mdx/2R, PE dθ = dx Hence, M.I. of the ring about diameter AB of the shell (i.e., an axis passing through the center of the ring and perpendicular to its plane = mass × (radius)2 = (Mdx/2R) (R2 – x2) because PE2 = OE2 – OP2 = R2 – x2 M.I. of the whole spherical shell about the diameter (AB) is equal to twice the integral of this expression between the limits x = 0 and x = R, i.e.,
z
R
I = 2
0
=
M R
M (R2 − x2 ) dx 2R
z
R
(R2 − x2 ) dx
0
LM N
x3 M R2 x − = 3 R M =
OP Q
R
= 0
LM N
M R3 R3 − 3 R
OP Q
M 2 3 2 . R = MR2 R 3 3
8.16 MOMENT OF INERTIA OF A SOLID SPHERE (iv) About a diameter: Figure shows a section, through the center, of a solid sphere of radius R and mass M, whose moment of inertia is to be determined about a diameter AB, its value being the same about any other diameter. Since the value of sphere = 4πR3/3, its mass per unit volume (or density)
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Mechanics
M 3M = = 4 3 4 πR3 πR 3 A thin circular slice of the sphere at a distance x from its center O and of thickness dx, we have surface area of the slice (which is a disc of radius R2 − x2 ) = p (R2 – x2) and its volume = area × thickness = π (R2 – x2) dx and hence its mass = volume × density = π (R2 − x2 ) dx.
3M dx = 3M (R2 − x2 ) 3 4 R3 4 πR
M.I. of the whole sphere about its diameter (AB) is equal to twice the integral of the above expression between the limits x = 0 and x = R, i.e.,
z
R
I = 2
0
=
=
or
3M (R2 − x2 ) dx 8R3
2. 3 M 8R 3 3 M 4 R3
z z
R2 – x2 R A
R
=
B x
( R2 − x2 )2 dx
dx
0
Fig. 17
R
( R 4 − 2R2 x2 + x4 ) dx
0
LM N 3 M F GR 4 R H
OP Q 1 I + R J 5 K
3 x5 3M 4 2 x R R x − + 2 I = 3 5 4 R3
=
O
5
3
−
2 5 R 3
4
0
5
3 M 8R5 2 = MR2 5 4R3 15
(v) About a tangent: A tangent drawn to the sphere at any point will be parallel to one or the other diameter of it (i.e., an axis passing through its center or center of mass) and at a distance equal to the radius of the sphere, R, from it. By the principle of parallel axes. Moment of inertia of the sphere about a tangent, i.e., I = (2/5) MR2 + MR2 = (7/5) MR2
8.17 MOMENT OF INERTIAL OF A HOLLOW SPHERE OR A THICK SHELL (i) About its diameter: A hollow sphere is just a solid sphere from the inside of which a small concentric sphere has been removed. Since the M.I. about a given axis is a scalar quantity. The moment of inertia of the hollow sphere about a diameter= moment of inertia of the solid sphere minus moment of inertia of the smaller solid sphere removed from it, both about the same diameter.
Moment of Inertia
353
Let R and r be the external and internal radii of the hollow sphere, i.e., the radius of the bigger solid sphere and the smaller solid sphere (removed from it) respectively. Then, if r be the density of the material of the given hollow sphere, we have mass of the bigger sphere = mass of the smaller sphere =
4 πR3ρ and 3 4 3 πr ρ 3
so that mass of the hollow sphere, M = ρ =
4 π (R3 − r 3 )ρ, and therefore 3
3M 4 π (R3 − r3 )
Moment of inertia of the bigger and the smaller spheres about a given diameter are respectively
FG H
IJ K
FG H
IJ K
2 4 3 2 4 πR 3ρ R2 and πr ρ r2 5 3 5 3
Therefore, M.I. of the hollow sphere about the same diameter is given by I =
FG H
IJ K
FG H
IJ K
2 4 2 4 3 2 4 πR 3ρ R2 − πr ρ r2 = πρ ( R 5 − r5 ) 5 3 5 3 5 3
or, substituting the value of r obtained above, we have I =
F GH
2 4 3M 2 R 5 − r5 5 5 . π (R ) M r − = 5 3 4 π (R3 − r3 ) 5 R 3 − r3
I JK
(ii) About a tangent: A tangent to the sphere at any point being parallel to one diameter or the other (i.e., the axis passing through the center of mass) of the sphere and at a distance equal to its external radius R from it, we have, by the principle of parallel axes, M.I. of hollow sphere about a tangent, i.e., I =
F GH
I JK
2 R 5 − r5 M + MR2 5 R 3 − r3
8.18 M. I. OF A UNIFORM TRIANGULAR LAMINA M.I. about base of a triangular lamina: Let the mass of the triangular lamina ABC of mass M, the length of AN is equal to P. Divide the lamina by strips parallel to BC. Let PQ be one such strip of breadth δx at a distance x from A. Let AN be perpendicular to BC such that AN = P. Let BC = a. From similar ∆’s APQ and ABC, we have
A
x M
P
Q dx
P
x PQ xa = so that PQ = P a P M 2M = Now mass per unit area = 1 aP aP 2
B
N a
Fig. 18
C
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Mechanics
∴
Mass of the strip PQ =
∴
M.I. of the strip about BC =
∴ M.I. of the triangle about BC
2M 2M ax 2Mx . PQδx = . . δx = 2 δx aP aP P P 2Mx δx ( P − x)2 P2
z P
=
0
=
2M x 2M ( P − x)2 dx = 2 P P2
LM N 2M L P M P N 2
x4 2M P2 x2 2Px3 − + 2 3 4 P2 4
=
2
−
2P4 P4 + 3 4
OP Q
OP Q
z
P
x ( P2 − 2Px + x2 ) dx
0
P
0
2 = MP 6
8.19 KINETIC ENERGY OF ROTATION In pure rotation, in which the center of mass of the rotating body has zero linear velocity, e.g., the rotation of a body about an axis through its center of mass, and a rotation in rolling of a body along a plane or an inclined surface. (i) Kinetic energy of a body rotating about an axis through its center of mass: Consider a body of mass M, rotating with angular velocity ω about an axis AB, passing through its center of mass, O, so that the center of mass has zero linear velocity. It is thus a case of pure rotation.
A
ω
The body possesses kinetic energy in virtue of its motion of rotation which is, therefore, its kinetic energy of rotation. Let us obtain an expression for it. The body is made up of a large number of particles of masses m1, m2, m3 .…etc, at respective distances r1, r2, r3 ... etc. from the axis AB through O. Since their angular velocity is the same (ù), their linear velocities are respectively r1ω = v1, r2ω = v2 and r3ω = v3 ... etc. and hence their respective kinetic energies equal to ½ m1v12 = ½ m1r12ω2, ½ m2v22 = ½ m2r22ω2, ½ m3v32 = ½ m3r32ω2 ... etc. ∴ Total K.E. of all the particles = =
O
B
Fig. 19
1 1 1 m1r12ω2 + m2r22ω2 + m3r32ω2 + .... 2 2 2 1 2 ω (m1r12 + m2r22 + m3r32 + ...) 2
Moment of Inertia
355
=
1 2 ω 2
∑ mr
2
=
1 2 ω MK 2 2
where Σmr2 = MK2, with, M, as the mass of the body and K, its radius of gyration about the axis of rotation AB. Since MK2 = I, the moment of inertia of the body about the axis AB, we have kinetic energy of rotation of the body about the axis (AB) through its center of mass = ½ Iω2. (ii) Kinetic energy of a rotating body whose center of mass has also a linear velocity (a)
A body rolling along a plane surface: Let us consider a body, like a circular disc, a cylinder, a sphere etc. (i.e., a body with a circular symmetry), of mass M, radius R and with its centre of mass at O, rolling, without slipping, along a plane or a level surface, such that it rotate clockwise and moves along the x direction. Q
2v
O R P
X
Fig. 20
At any given instant, the point P, where the body touches the surface, is at rest, so that an axis through P, perpendicular to the plane of the paper is its instantaneous axis of rotation and the linear velocities of its various particles are perpendicular to the lines joining them with the point of contact P, their magnitudes being proportional to the lengths of these lines. Thus, if the linear velocity of the center of mass O (where PO = R) be v, that of the particle at Q (where PQ = 2R) is 2v. This shows that the particles have all the same angular velocity with respect to the point P or that the body is rotating about the fixed axis through P with an angular velocity ω, say, given by v/R, where v is the linear velocity of the centre of mass. The motion of the body is thus equivalent to one of pure rotation about the axis through P, with an angular velocity ω. The whole of the kinetic energy of the body is therefore, the same as its kinetic energy of rotation about this axis and hence equal to Ip ω2 where Ip is the M.I. of the body about the axis through P. If Icm be the moment of inertia of the body about a parallel axis through its center of mass, then by the principle of parallel axes, IP = Icm + MR2. So that, K.E. of the rolling body = ½ (Icm + MR2) ω2 = ½ Icmω2 + ½ MR2 ω2 = ½ Icm ω2 + ½ Mv2
...(1)
where v is the linear speed of its center of mass with respect to P. Now Icm = MK2, where K is the radius of gyration of the body about the axis through its center of mass, and ω = v/R. So that
F GH
I JK
v2 1 1 1 K2 +1 MK2 2 + Mv2 = Mv2 ...(2) 2 2 2 R R2 In equation (1) the first term gives its K.E. of pure rotation about the center of mass, i.e., its K.E. when it is simply rotating with angular velocity ù about the axis through its K.E. of the rolling body =
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Mechanics
center of mass, without executing any translational motion (i.e., with the linear velocity of its center of mass zero). And the second term gives its K.E. of pure translational, i.e., its K.E. when it is simply moving with linear velocity v without performing any rotational motion (i.e., with its angular velocity zero).
8.20 A BODY ROLLING DOWN AN INCLINED PLANE (ITS ACCELERATION ALONG THE PLANE) Let a body of circular symmetry (e.g., a disc, sphere, cylinder etc.) of mass M roll freely down a plane, inclined to the horizontal at an angle q and rough enough to prevent slipping and hence any work done by friction. If v be the linear velocity acquired by the body on covering a distance S along the plane, its vertical distance of descent = S sin θ. ∴ Potential energy lost by the body = Mg.S sin θ
...(1)
This is equal to the kinetic energy gained by the body, i.e., equal to its K.E. of rotation plus its K.E. of translation. Now, K.E. of rotation of the body = ½ Iω2 where ω is its angular velocity about a perpendicular axis through its center of mass, and its K.E. of translation = ½ Mv2, because its center of mass has a linear velocity v. ∴ Total K.E. gained by the body =
v2 1 2 1 1 1 Iω + Mv2 = MK 2 . 2 + M v2 2 2 2 2 R
Because I = MK2, where K is the radius of gyration of the body above the axis through its center of mass, and ω = v/R. ∴ Total K.E. gained by the body =
F GH
I JK
1 K2 +1 Mv2 2 R2
...(2)
From conservation of energy Eq. (1) are equal to Eq. (2)
F GH
I JK
1 K2 +1 Mv2 2 R2 v2
or
s
= MgS sin θ
(K 2 + R2 ) = 2g sin θS R2
v2 = 2
ω
s sin θ R
v
θ
R2 g.sin θ. S (K2 + R2 )
Fig. 21
Comparing this with the kinetic relation v2 = 2 a.s for a body starting from rest, we have acceleration of the body along the plane, i.e., a =
i.e., the acceleration is proportional to planes.
R2 g sin θ g.sin θ = 2 2 K2 (K + R ) 1+ 2 R R2 for a given angle of inclination (θ) of the (K + R2 ) 2
Moment of Inertia
357
Let us consider the case when the body is a solid sphere, cylinder, spherical shell and ring. M, K and R will be the mass, radius of gyration and the radius of the rolling section respectively in each case. (i)
Solid Sphere: The moment of inertia of a solid sphere about its diameter is given by 2 I = MK =
2 MR2 5
2 K2 2 = 5 R
or ∴ (ii)
a =
g sin θ 5 = g sin θ 2 7 1+ 5
Cylinder (or Disc): The moment of inertia of a cylinder or disc about the axis passing through its center and perpendicular to its plane (which is the axis of rotation) is given by 2 I = MK =
1 MR2 2
1 K2 = 2 R2
or
∴ (iii)
g sinθ 2 = g sin θ 3 a = 1+ 1 2 Spherical Shell: Its moment of inertia about the diameter (axis of rotation) is given by I = MK2 =
2 MR2 3
K2 2 = 3 R2
or ∴ (iv)
a =
g sinθ 3 = g sin θ 2 5 1+ 3
Ring: It moment of inertia about the axis passing through its center and perpendicular to its plane (axis of rotation) is given by I = MK2 = MR2
K2 = 1 R2
or ∴
a =
g sin θ 1 = g sin θ 1+1 2
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Mechanics
We see that the accelerators for a solid sphere, disc, shell and ring are in a decreasing order. Therefore, if all of them start rolling at the same instant, the sphere will reach down the plane first, then the disc, then the shell, and then the ring.
8.21 IDENTIFICATION OF HOLLOW AND SOLID SPHERE The given spheres are taken at the top of an inclined plane and allowed to start rolling at the same instant. The one which reaches the bottom of the plane earlier is the solid sphere. The other which reaches afterwards is ‘hollow’. Reason: The acceleration with which the sphere rolls is given by a =
g sin θ g sin θ = 2 K I2 1+ 2 1+ R MR2
∴ I = MK2
Now, the moment of inertia of a hollow sphere about the axis of rotation (diameter) is greater than that of a solid sphere of the same mass and radius, because in case of hollow sphere the material particle are, on the average, at a greater distance from the axis. This mean that the hollow sphere will roll down the plane with a smaller acceleration and hence reach the bottom after the solid sphere does so.
8.22 COMPOUND PENDULUM Any rigid body which is so mounted that it can swing in a vertical plane about a horizontal axis passing through it, is called a ‘physical’ or ‘compound’ pendulum. Figure represents the vertical section of an irregular rigid body pivoted at a point S in it. In the equilibrium position, the center of mass C of the body lies vertically below S. When the body is displaced to one side and then released, it oscillates in the vertical plane containing S and C about a horizontal axis perpendicular to this plane and passing through S. It is now acting as a compound pendulum. Motion of Compound Pendulum: Let m be the mass of the body and l the distance of C from S. Suppose at some instant during the oscillation, the body makes an angle θ with the vertical. At this instant the moment of its weight = mg (l sin θ). This is the restoring torque τ at this instant tending to bring the body to its equilibrium position. Thus τ = –mgl sin θ –ve sign is used since the torque acts in the direction of θ decreasing. If the angle θ is small, then sin θ = θ (radian). Therefore, τ = –mglθ
...(i)
d2θ the dt2 instantaneous angular acceleration, then the torque acting on the body at this instant must be given by n
If I be the moment of inertia of the body about the axis through S and
τ = I
d2θ dt2
From Eqs. (i) and (ii), we get I
d2θ = –mgl θ dt2
...(ii)
Moment of Inertia
359
mgl d2θ − θ = − ω2θ 2 = I dt
or
ω2 =
where
θ
mgl I
l
O
l sin θ
d2θ ∝ θ dt2
∴
k2 +l l
S
that is, the angular acceleration is proportional to the angular displacement. The motion is, therefore, simple harmonic and its time-period is given by mg
2π I = 2π T = ω mgl
Fig. 22
Let K be the radius of gyration of the pendulum about a plane of oscillation. Then its moment of inertia about this axis will be mK2. Therefore, the moment of inertia I about the axis through S will be I = mK2 + ml2
(Theorem of parallel axes) 2
mK 2 + ml2 T = 2π = 2π mgl
K +l l g
This is the expression for the time-period. Comparing it with the period of a simple pendulum.
L g We see that the period of a compound pendulum is the same as that of a simple T = 2π
pendulum of length L =
FK GH l
2
FK GH l
I JK
+ l . Thus
2
+l
I JK
is the length of the equivalent simple
pendulum. Let us suppose that the whole mass of the compound pendulum is concentrated at a point O on SC produced such that SO =
FK GH l
2
I JK
+ l . Thus it would be like a simple pendulum
of the same period. The point O which is at a distance equal to the length of the equivalent simple pendulum from S is called the ‘centre of oscillation’ corresponding to the center of suspension S. Interchangeability of the centres of Suspension and Oscillation: When the body is suspended from the centre of suspension S, its period of oscillation is given by
T = 2π where l is the distance of S from C.
K2 +l l g
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Mechanics
Let the body be now inverted and suspended from the center of oscillation O. The distance of O from C is K2/l. The new period of oscillation T′ is therefore obtained by substituting K2/l for l in the about expression i.e.,
K2 K2 + 2 l = 2π T′ = 2π K / l g
l+
K2 l g
Hence T = T' Thus the periods of oscillation about S and O are equal i.e., the center of suspension and oscillation are interchangeable.
8.23 FLY WHEELS It is a heavy wheel whose M.I. about its axis of rotation is large. Most of the mass is concentrated on rim (to increase its M.I.). These have wide application in stationary engines and various instruments of every day use. Determination of M.I. of Fly wheel: A cord is wound round the horizontal axle of fly wheel and a mass is attached to the free end of the cord. Releasing the mass, the fall of the mass is noted and number of turns in the noted time are counted when fly wheel comes to rest.
M Mg
Fig. 23
Let N is the number of revolutions of fly wheel till mass detaches from wheel, I is moment of inertia of fly wheel, v is velocity of the mass acquired, m is mass attached, h is distance descended, t is time taken to come to rest after mass is detached. n is the total revolutions till wheel comes to rest, after mass is detached. W is the work done/revolutions against friction. From Energy conservation gives: mgh = ½ mv2 + ½ Iω2 + nW where nW is the total work done against friction in n revolutions. or
2 mgh = mv2 + Iω2 + 2 nW
...(1)
Now work done in n revolutions = nW K.E. of rotation of wheel = ½ Iω2 i.e., or
nW = ½ Iω2 W =
1 Iω2 2 n
Substituting this value of W in Eq. (1) we get 2 mgh = mv2 + I ω2 + 2N. or or
1 Iω2 2 n
2mgh – mv2 = Iω2 (1+ N/n) I =
2mgh − mv2 2mgh − mr2ω 2 = N N ω2 1 + ω2 1 + n n
FG H
IJ K
FG H
IJ K
Moment of Inertia
361
2mgh − mr2 2 ω I = N 1+ n
FG H
or
IJ K
...(2)
Average angular velocity of wheel = (ω + 0)/2 = ω/2 If n rotation of wheel, angle described = 2ω n/t ∴ Average angular velocity = 2π n/t = ω/2 ω =
or
4πn t
Substituting this value of ù in Eq. (2) we get
2mght2 − mr2 2 2 m n 16 π = I = N N 1+ 1+ n n
FG IJ FG H K H F n IJ LM ght − r OP I = mG H n + N K N 8π n Q 2
or
LM ght IJ N 8π n K
2
2 2
− r2
OP Q
2
2 2
Hence, the value of I, the moment of inertia of the fly wheel about its axis of rotation, may be obtained.
NUMERICALS Q.1. The torque τ acting on a rotating body of moment of inertia I about the axis of rotation is given by
τ = a + bt + ct 2 , I where a, b, c are constants. Express angular displacement of the body (starting from rest at t = 0) as a function of τ. Solution. The torque τ and the angular acceleration a are related by
d2θ dt2 Therefore, from the given expression, we have τ = Iα = I
d2θ = a + bt + ct2 dt2
Integrating:
dθ t2 t3 = at + b + c + A dt 2 3
dθ = 0 so that constant A = 0. dt t2 t3 t4 + A′ Again integrating: θ = a +b +c 2 6 12 Again at τ = 0, θ = 0 so that A' = 0. At t = 0,
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Mechanics
∴
F GH
t2 bt ct2 a+ + 2 3 6
θ =
I JK
Ans.
Q.2. A flywheel of mass 10 kg and radius of gyration 50 cm is being acted on by a torque of 108 dyne-cm. Calculate the angular acceleration produced. Solution. In usual rotation : τ = Ia or
τ I 8 2 Here τ = 10 dyne-cm, I = MK = (10,000 gm) (50 cm)2 = 2.5 × 107 gm-cm2. α =
∴
108 = 4 rad/sec2 2. 5 × 107
α =
Ans.
Q.3. An automobile engine develops 100 kilowatt when rotating at a speed of 1800 rev/ min. What torque does it deliver ? Solution. The power developed by the torque τ exerted on a rotating body is given by P = τω ∴
P ω
τ =
Hee P = 100 kilowatt = 100,000 watt = 100,000 Joule/sec and ω = rad/sec.
1800 × 2π = 60π 60
100, 000 Joule/sec = 531 Joule Ans. 60 × 3.14 rad/sec Q.4. Show that the rotational inertial of a dumb-bell is twice as great about an axis through one end as it is about an axis through the centre. Compute these values if the dumbbell is a light rad of length one meter carrying 5 kg sphere, one at each end. ∴
τ =
Solution. Let M be the mass of each sphere, and l the length of the rod. Let us treat spheres as point masses. The moment of inertia of the system about the axis through l is
A
B
C
M
l 2
l 2
Fig. 24
IC = M = Here M = 5 kg and l = 1 meter
FG l IJ H 2K
1 2 Ml 2
2
+M
FG l IJ H 2K
2
M
Moment of Inertia
∴
363
IC = 2.5 kg-meter2
By the theorem of parallel axes, the moment of inertial of the system about an axis through A is
FG l IJ H 2K
2
1 2 1 2 Ml + Ml = Ml2 2 2 2 2 = 5 × (1) = 5.0 kg-meter .
IA = IC + (2M)
=
Here we see that IA is twice IC. Q.5. Deduce the moment of inertial of HCl molecule about an axis passing through its C.M. and perpendicular to the bond. Given : internuclear distance 1.3 Å, proton mass = 1.7 × 10–27 kg, atomic weight of chlorine = 35. Solution. The mass of H atom (proton) is 1.7 × 10–27 kg and that of Cl atom is 35 × (1.7 × 10–27) = 59.5 × 10–27 reduced mass of HCl molecule is therefore µ =
mH mCl (1. 7 × 10−27 ) (59. 5 × 10−27 ) = mH + mCl (1. 7 × 10−27 ) + ( 59. 5 × 10−27 )
= 1.65 × 10–27 kg. The internuclear distance is r = 1.3 Å = 1.3 × 10–10 meter. The moment of inertial about the centre of mass is therefore given by I = µr2 = (1.65 × 10–27 kg) (1.3 × 10–10 m)2 = 2.79 × 10–47 kg/m2. 0.9 the the the
Q.6. A uniform thin bar of mass 3 kg and length is bent to make an equilateral triangle. Calculate moment of inertia I about an axis passing through centre of mass and perpendicular to the plane of triangle.
Solution. Each side of the triangle, such as PQ has a mass m = 1 kg and a length l = 0.3 m. Its moment of inertia about a perpendicular axis through its middle point O is
P
30°
O C Q
R
Fig. 25
ml2 1 ×(0. 3)2 = 0. 0075 Kg-m2 = 12 12 The distance OC is d = PO tan 30° =
l 0. 3 × 0. 577 = 0. 0865 m. tan 30° = 2 2
The moment of inertial about C, by the theorem of parallel axes, is
ml2 + md2 = 0. 0075 + (1) (0. 0865)2 = 0. 015 kg-m2 12 By symmetry, the moment of inertial of each side about C will be same. Hence for the whole triangle I = 3 × 0.015 = 0.045 kg-m2. Ans.
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Mechanics
Q.7. The inter-nuclear distance between the two hydrogen atoms in a hydrogen molecule is 0.71 Å. Calculate its moment of inertia and the first two rotational energy levels. Mass of the proton is 1.6 × 10–24 gm and Planck’s constant is 6.6 × 10–27 erg-sec. Given that angular →
→
h2 with j = 0, 1, 2, ... . 4 π2 1 mH mH = mH Solution. The reduced mass of the H2 molecule is given by µ = mH + mH 2
momentum J is quantised by the rule |J|2 = j ( j + 1)
1 × (1. 6 × 10−24 gm) = 0. 8 × 10−24 gm. 2 The internuclear distance is r = 0.71 Å = 0.71 × 10–8 cm. Hence the moment of inertia of the molecule about an axis passing through the centre of mass and perpendicular to the line joining the two atoms is =
I = µr2 = (0.8 × 10–24 gm) (0.71 × 10–8 cm)2 The rotational energy of the system is E
FG = 1 Iω IJ , and angular momentum is J (=Iω). H 2 K 2
J2 2I
∴
E =
Here
J2 = j ( j + H)
h2 4π2
∴
E = j ( j + 1)
h2 8 π2I
For the first two levels j = 1, 2. ∴ But ∴
2h2 2 × ( 6. 6 × 10 −27 )2 = = 2. 8 × 10−14 erg. 8 π2I 8 × ( 3.14)2 4. 0 × 10−41 erg = 1 eV
E1 =
1.6 × 10–12
2. 8 × 10−24 = 1. 7 × 10−2 eV E1 = 1. 6 × 10−12 E2 =
6 h2 = 3E1 = 5.1 × 10−2 eV. 8 π2I
Q.8. Four particles each of mass m are symmetrically placed on the sim of a uniform disc of mass M and radius R. What is the moment of inertia of the system about an axis passing through one of the particles and perpendicular to the plane of the disc ? Solution. The arrangement is shown in figure 26. The moment of inertial of the disc about an axis through 1 2 its centre O and perpendicular to its plane is MR . Hence, 2 by the theorem of parallel axes, its M.I. about an axis through B and perpendicular to its plane
B m
2R R A
m
R
m C
O
m
Fig. 26
Moment of Inertia
365
=
1 3 MR2 + MR2 = MR2 2 2
Now, the M.I. of the particle at B about the same axis is zero, while that of the particle at A, C and D is m[ 2 R]2 , m[ 2 ]2 and m(2R)2 . Hence the M.I. of the entire system =
=
3 MR2 + m[ 2 R]2 + m [ 2 R]2 + m(2R)2 2
FG 3 M + 8mIJ R H2 K
2
Q.9. A circular hole of diameter R is cut from a disc of radius R, such that the circumference of the hole passes through the centre of the disc. Find the moment of inertia of the remaining disc about the straight line joining the centres of the disc and the hole in terms of the remaining mass. Solution. Let O be the centre of the disc and O' that of the hole. Let M be the mass of the remaining disc.
Area of the removed material disc = π
FG R IJ H 2K
Area of the removing disc = πR2 − ∴
Mass per unit area, σ =
R 2
=
1 2 πR 4 Fig. 27
1 2 3 2 πR = πR 4 4
M 3 2 πR 4
2 Mass of the original disc = πR σ =
Mass of the removed material =
O′
O
Area of the original disc = πR2
4 M 3
1 2 1 πR σ = M 4 3
Moment of inertial of the original disc about OO' (diameter) =
1 1 (mass) R2 = 4 4
FG 4 MIJ R H3 K
2
=
1 MR2 3
Moment of inertia of the removed material about OO' =
FG IJ H K
1 R (mass) 4 2
2
=
FG IJ FG R IJ H K H 2K
1 1 M 4 3
∴ Moment of inertia of the remaining disc about OO' =
1 1 15 MR2 − MR2 = MR2 . 3 48 48
2
=
1 MR2 48
P
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Mechanics
Q.10. A circular hole of diameter R is cut from a uniform disc of radius R and mass M such that the perimeter of the hole passes through the centre of the disc. Find the moment of inertia of the remaining disc about (i) an axis passing through the centre of the disc and perpendicular to the plane of the disc (ii) a tangent in the plane of the disc and touching the hole. Solution. (i) Refer to the above figure. M is now the mass of the original disc. The mass per unit area is σ = Mass of the removed material =
M πR2 1 2 1 πR σ = M 4 4
Moment of the inertia of the original disc about an axis through O and perpendicular to the plane. =
1 MR2 2
Moment of inertia of the removed material about an axis through O' and perpendicular to the plane
FG IJ FG IJ H KH K
2
1 1 1 R M MR2 = 2 4 32 2 By the theorem of parallel axes, the moment of inertia of the removed material about the parallel axis through O =
FG IJ U|V = 3 MR . H K |W 32
R| S| T
1 1 R MR2 + M× = 32 4 2
2
2
∴ moment of inertia of the remaining disc about an axis through O and perpendicular to the plane =
1 3 13 MR2 − MR2 = MR2 . 2 32 32
(ii) Moment of inertia of the original disc about a tangent in the plane of the disc and passing through P = and that of the removed material.
5 MR2 , 4
FG IJ FG IJ H KH K
5 1 R 4 4 2 ∴ moment of inertia of the remaining disc =
2
=
5 MR2 64
5 5 75 MR2 − MR2 = MR 2 4 64 64 Q.11. Find the M.I. of a rod 4 cm in diameter, 2m long and of mass of 8 kg (i) about an axis normal to its lengths and passing through its centre (ii) about an axis passing through one end and normal to its length (iii) about a longitudinal axis through the centre of the rod.
=
Moment of Inertia
367
Solution. (i) The M.I. about an axis normal to the rod and passing through its centre is given by
F l + R I = 8kg RS (2m) GH 12 4 JK T 12 2
I = M
2
2
+
(0. 02)2 4
UV = 2.67 kg-m W
2
(ii) By the theorem of parallel axes, the MI about an axis normal to the rod and passing through an end is given by
F l + R I + FG l IJ M GH 12 4 JK H 2 K F 2m IJ = 2.67 kg-m + (8 kg) G H2K 2
2
2
I = M
2
2
= 10. 67 kg-m2 .
(iii) The M.I. about the central axis of the rod is given by I =
1 1 MR2 = × (8 kg) × (0.02 m)2 = 0. 0016 kg-m2 . 2 2
Q.12. Four spheres each of radius ‘r’ and mass ‘m’ are placed with their centres on four corners of a square of side ‘l’. Calculate the moment of inertia of the arrangement about any (i) diagonal of the square (ii) any side of the square. Solution. (i) The moment of inertia of each of the spheres A and C about the diagonal AC of the square (an axis passing through their centres) is
2 2 mr . 5
Similarly, the moment of inertia of each of the spheres B and D about an axis through
2 2 mr . The distance between this axis and AC is l/ 2 , where 5 l is the side of the square. Therefore, the moment of inertial of each of the spheres B and its centre and parallel to AC is
D about AC is, by the theorem of parallel axes,
2 2 ml2 mr + . 5 2 B
A l/ 2 O
C
D
Fig. 28
Hence the moment of inertia of the all the four spheres about AC. = 2
LMF 2 mr I + F 2 mr MNGH 5 JK GH 5 2
2
+
ml2 2
I OP JK PQ
368
Mechanics
=
1 m (4 r2 + 4 r2 + 5l2 ) 5
=
m (8 r2 + 5l2 ) 5
(ii) The moment of inertia of all the spheres about a side of the square would be
m 2m (4 r2 + 5l2 ). (8 r2 + 10l2 ) = 5 5 Q.13. The flat surface of a hemisphere of radius R is cemented to one flat surface of a cylinder of radius R and length l and made of the same material. If the total mass be M, show that the moment of inertial of the combination about the axis of the cylinder is MR 2
FG l + 4 R IJ FG l + 2 R IJ . H 2 15 K H 3 K
Solution. Let m1 and m2 be masses of the cylinder and the hemisphere respectively and ρ the density of their material. The moment of inertia of the cylinder
1 m1R2 , and that of the hemisphere 2 2 2 about its own diameter is m2R . 5 about its own axis is
l
(The moment of inertia of a hemisphere about a diameter is same as that of a sphere of same mass and same radius). Therefore, the moment of inertial of the combination about this common axis is I = =
so that
Fig. 29
1 2 m1R2 + m2R2 2 5
FG H
LM1 lR N2
2
+
4 3 R 15
M = m1 + m2 = πR2lρ + πR2ρ = M
FG l + 2R IJ H 3K
Substituting this value of πR2ρ is eq. (i) we get I =
IJ K
1 2 2 2 ( πR2lρ)R2 + πR ρ R2 2 5 3
= ( πR2 )l
Now
R R
FG H
MR2 2R l+ 3
FG l + 4R IJ IJ H 2 15 K K
OP Q
...(i)
FG H
2R 2 3 πR ρ = πR2ρ l + 3 3
IJ K
Moment of Inertia
369
Q.14. What will be the radius of gyration for a solid sphere about a diameter whose radius is half meter ? Solution. Let K be the radius of gyration to be determined. The moment of inertia of the solid sphere about a diameter is given by I = MK2 =
∴
K =
2 MR2 5
FG 2 IJ R = FG 2 IJ H 5 K H 5K
× 0. 5 m = 0. 316 m.
Q.15. A circular metal disc of mass 4 kg and diameter 0.4 meter makes 10 revolution per second about its centre, the axis of rotation being normal to the plane of the disc (i) what is the moment of inertial about this axis ? (ii) What is the angular momentum about the same axis (iii) calculate the torque which will increase the angular momentum by 20% is 10 seconds. Solution. (i) The M.I. about the given axis is I =
1 1 MR2 = × 4 kg × (0.2 m)2 = 0. 08 kg - m2 2 2
(ii) The angular momentum about the same axis is L = Iω = (0.08 kg-m2) (2 × π × 10 rad/sec) = 5.02 kg-m2/sec. (iii) The torque is rate of change of angular momentum. τ =
dL dt
In 10 seconds, L increases by 20% i.e., dL = 5.0 × ∴
τ =
20 = 1. 0 100
1. 0 = 0.10 nt - m. 10
Q.16. A solid cylinder of mass 16 kg and radius 0.4 m is rotating about its axis. If its rotational speed increases in 10 second from 500 to 2500 revolution per minute, Calculate (a) its angular acceleration assuming to be constant (b) the torque which must be applied to the cylinder. Solution. (a) The moment of inertia of the cylinder about its axis is I =
1 1 MR2 = × 16 kg × (0.4)2 = 1. 28 kg - m2 2 2
The initial and final rotational speeds are ω1 = 2π × no. of rev. per sec = 2π × and
ω2 = 2π ×
500 50 π = rad/sec. 60 3 2500 250 π = rad/sec. 60 3
370
Mechanics
This increase takes place in 10 seconds. Therefore, the angular acceleration is
ω − ω1 = α = 2 ∆t
250 π 50 π − 3 3 = 200 π = 21 rad/sec2 . 10 30
(b) The torque to be applied isτ = Iα = (1.28 kg-m2) (21 rad/sec2) = 27 nt-m. Q.17. A flywheel in the form of a solid cylinder of mass 5000 kg and diameter 2 meters is rotating, making 120 revolutions per minute. Compute the kinetic energy and the angular impulse, if the flywheel is brought to rest is 2 seconds. Solution. The M.I. of the cylinder about its axis is I =
1 1 MR2 = ( 5000 kg) (1m)2 = 2500 kg-m2 . 2 2
If n is number of revolution per second, then the angular velocity is ω = 2πn = 2 × 3.14 ×
120 = 12. 56 rad/sec. 60
Therefore, the rotational kinetic energy is K =
1 2 1 Iω = × 2500 × (12. 56)2 = 1. 97 × 105 Joule. 2 2
The angular impulse is the time integral of torque, that is
z z t
τdt =
I
0
dω dt = I dt
z
ω2
dω = I (ω2 − ω1 )
ω1
provided I is constant. Thus the angular impulse for the fly-wheel coming to rest. = moment of inertial × change in angular velocity = 2500 × 12.56 = 3.14 × 104 kg-m2/sec2. Q.18. A thin plane disc of mass M and radius R, initially at rest, is set into rotation with angular velocity ω. Calculate the work done if the axis of rotation is (i) through its centre and perpendicular to its plane, (ii) through a point on its edge and perpendicular to its plane (iii) a diameter, (iv) a tangent parallel to a diameter.
1 2 Iω , is each case. 2
Solution. The work done appears as the kinetic energy of rotation, (i)
W =
= (ii)
1 2 Iω 2
FG H
W = Itω2 =
IJ K
1 1 1 MR2 ω2 = MR2ω2 2 2 4
FG H
IJ K
1 1 1 3 (I + MR2 )ω2 = MR2 + MR2 ω2 = MR2ω2 2 2 2 4
Moment of Inertia
(iii)
371
W =
= (iv)
W = =
1 I d ω2 2
FG H
IJ K
1 1 1 MR2 ω 2 = MR2ω 2 2 4 8
1 Itω2 2
FG H
IJ K
1 1 1 5 (Id + MR2 )ω 2 = MR2 + MR2 ω2 = MR2ω 2 . 2 2 4 8
Q.19. A circular disc of radius 0.1 m and mass 1.0 kg is rotating at the rate of 10 revolution a second about its axis. Find the work that must be done to increase the rate of revolution to 20 per second. Solution. The M.I. of the disc about its axis is I =
1 1 MR2 = (1. 0 kg) (0.1)2 = 5 × 10−3 kg - m2 . 2 2
Initial and final angular velocity are ω1 = 2π × number of revolution per sec = 2π × 10 rad/sec. ω2 = 2π × 20 rad/sec.
and
The work done is in increasing the rate of revolutions = final kinetic energy – initial kinetic energy =
1 2 1 2 Iω2 − Iω1 2 2
=
1 × (5 × 10−3 kg-m2 ) (40 π)2 − (20 π)2 / sec2 2
=
1 × 5 × 10−3 ×1200 × π2 = 29. 6 joule. 2
Q.20. Calculate the angular momentum of a body whose rotational kinetic energy is 10 joule, if the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is 8 gm-cm2. Solution. The angular momentum and rotational kinetic energy are given by L = Iω and K = ∴
L =
∴
L =
1 2 Iω 2
2KI Here K = 10 joule and I = 8 gm-cm2 = 8 × 10–7 kg-m2.
2 × 10 × 8 × 10−7 = 4 × 10−3 kg-m2/sec.
Q.21. Calculate the angular momentum and rotational kinetic energy of earth about its own axis. How long could this amount of energy supply 1 kilowatt power to each of the 3.5 × 109 persons on earth ? (mass of earth = 6.0 × 1024 kg, radius = 6.4 × 103 km).
372
Mechanics
Solution. Let us assume the earth to be a solid sphere. Its moment of inertia of about its axis is I =
2 2 MR2 = × 6. 0 × 1024 kg (6.4 × 106 meter)2 5 5
= 9. 8 × 1037 kg-meter2 The earth makes one revolution (traces an angle of 2π radian) is one day (24 × 60 × 60 sec.) Therefore its angular velocity ω =
2π = 7. 27 × 10−5 rad/sec. 24 × 60 × 60
Hence its angular momentum Iω = ( 9. 8 × 1037 ) × (7. 27 × 10−5 ) = 7.1 × 1033 kg-m2/sec.
1 1 2 −5 2 37 29 Iω = (9. 8 × 10 ) (7. 27 × 10 ) = 2. 6 × 10 Joule 2 2 The power supplied by this energy Its rotational energy
P =
=
energy 2. 6 × 1029 = Joule/sec (or watt) time t 2. 6 × 1029 kilowatt. 1000 t
We require 1 kilowatt power to each of the 3.5 × 109 persons, that is 3.5 × 109 kilowatt. ∴
or
2. 6 × 1029 = 3.5 × 109 1000 t
t =
=
2. 6 × 1029 sec 1000 × 3. 5 × 109 2. 6 × 1029 1000 × 3. 5 × 109 × ( 365 × 24 × 60 × 60)
= 2. 35 × 109 years. Q.22. A thin rod of length l and mass M is suspended freely at its end. It is pulled aside and swung about a horizontal axis passing through its lowest position with an angular speed ω. How high does its centre of mass rise above its lowest position ? Neglect friction. Solution. Let M be the mass of the rod. Suppose its centre of mass rises to a height h above the lowest position. The potential energy is then Mgh. As the rod passes its mean (lowest) position, the energy becomes entirely kinetic (rotational), neglected, then Mgh =
1 2 Iω 2
1 2 Iω . If air resistance is 2
Moment of Inertia
373
where I is the moment of inertia of the rod about the horizontal axis through its end. We know that I =
Thus
Mgh =
Ml2 3 1 Ml2 2 ω 2 3
(Iω )2 h = 6g
or
Q.23. A rod of mass 6.40 kg and length 1.20 meter carries a sphere of mass 1.06 kg at its each end. It is rotated with an initial angular velocity of 39.0 rev/sec in a horizontal plane about a vertical axle through its centre and fixed in ball-bearings. It is found that the system comes to rest in 32.0 sec on account of friction in bearings, (1 .06 kg ) calculate (i) the angular acceleration (ii) frictional torque (iii) work done by frictional torque and (iv) the number te r me 2 . of revolutions made before coming to rest. 1 C
Solution. The arrangement is shown in figure.
Ml2 , so that the 12 moment of inertia of the whole system about the vertical axis through the centre C is The moment of inertia of the rod is
(1 .06 kg )
Fig. 30
RS T
I = 1. 06 × (0. 60)2 + 1. 06 × (0. 60)2 + 6. 40 ×
(1. 20)2 12
= 1.53 kg-meter2. (i) The initial angular velocity is ω = 39.0 rev/sec. = 39.0 × 2π = 245 rad/sec. The final angular velocity is zero. Therefore the angular acceleration is
∆ω 0 − 254 = = − 7. 66 rad/sec2 ( retarding) ∆t 32. 0 τ = Iα = 1.53 × (–7.66) = –11.7 newton-meter
α = (ii) The frictional torque is
(iii) Work done by the frictional torque = initial kinetic energy of the system =
1 2 1 Iω = × 1. 53 × (245)2 = 4. 59 × 104 joule. 2 2
(iv) Let n be the number of revolution. Then Work = torque × angle traced before, coming to rest i.e., ∴
4.59 × 104 = 11.7 × 2πn n =
4. 59 × 104 = 624 rev. 11. 7 × 2 × 3.14
UV W
374
Mechanics
Q.24. A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the velocity of the other end when it hits the floor. Assuming that the end on the floor does not slip. Solution. Let M be the mass and l the length of the stick (l = 1 meter). When it is held
1 l from the floor. So that the potential energy 2 in the stick is (Mgl/2). On releasing, the stick falls i.e., it rotates about the end on the floor
vertically, its centre of mass is at a height
1 2 Iω , where I is the 2 moment of inertia of the rod about the lower end and ω the angular velocity when it hits the floor. Thus and the potential energy is converted into potential kinetic energy
But I =
Mg
l 1 2 = Iω 2 2
Mg
l 1 Ml2 2 ω = 2 2 3
Ml2 3
∴
ω =
or
3g l
If v be the linear velocity of the end hitting the floor, then v = lω =
3 gl
Now g = 9.8 meter/sec2 and l = 1 meter ∴
v =
3 × 9. 8 ×1 = 5. 42 meter/sec.
Q.25. A uniform disc of radius R = 0.5 m and mass M = 20 kg can rotate without friction around a fixed horizontal shaft through its centre. A light cord is wound around the rim of the disc and a steady downward pull T = 9.8 nt is exerted on the cord. Find the angular acceleration of the disc, and the tangential acceleration of a point on the rim. Also find the angular velocity after 2 second. Solution. The steady downward pull T exerts a torque τ about the central axis of rotation, which is given τ = TR
...(i) R
But we know that τ = Iα
...(ii)
where I is that moment of inertia of the disc and α its angular acceleration about the axis of rotation.
T
From eq. (i) and (ii), we get TR = Iα or
α =
TR I
Fig. 31
Moment of Inertia
Now, for the disc
375
I =
1 MR2 , so that 2
α =
2T MR
Given T = 9.8 nt, M = 20kg and R = 0.5 m. ∴
α =
2 × 9. 8 = 1. 96 rad/sec2 20 × 0. 5
The tangential acceleration of a point on the rim is given by a = Rα =0.5 × 1.96 = 0.98 m/sec2. If the disc started from rest, then its angular velocity after 2 sec is given by ω = αt = (1.96 rad/sec2) (2 sec) = 3.92 rad/sec. Q.26. Suppose we hang a body of mass m = 1.0 kg from the cord in the previous problem. Find the angular acceleration of the disc (M = 20 kg, R = 0.5m) and the tangential acceleration of a point on the rim of the disc. Also compute the tension in the cord. Solution. Let T be the tension in the cord. The forces on the downward suspended body m are its weight mg acting downward and the tension T in the cord acting upward. The net downward force on it is mg-T.
R
If this force produces a linear acceleration a in the suspended body (which is same as tangential acceleration of a point on the rim of the disc), then from Newton's second law, we have
T T
m
mg – T = ma or
T = m (g – a)
a mg
Let us now consider the disc. The tension T in the cord exerts a torque τ on the disc which is given by
Fig. 32
τ = TR But we know that
τ = Iα
where I is the moment of inertia of the disc and α its angular acceleration about the axis of rotation. From the last two equations, we get Τ R = Iα But T = m(g – a) = m(g – Rα), because we know that a = Rα. Thus, m(g – Rα) R = Iα α(I + mR2) = mgR
or or
But for the disc
α =
mgR I + mR2
I =
1 MR2 , so that 2
FG H
IJ K
mgR mg g 2m = = α = 1 1 M + 2m R MR2 + mR2 MR + mR 2 2
376
Mechanics
Given m = 1.0 kg, M = 20 kg and R = 0.5 meter α =
∴
FG 2 × 1.0 IJ 9.8 = 1.78 rad/sec H 20 + 2 × 1.0 K 0. 5
2
The tangential acceleration is a = Rα = 0.5 × 1.78 = 0.89 meter/sec2 We note that the acceleration are less for a suspended 1.0 kg body (equivalent to 9.8 nt weight) then they were for a steady 9.8 nt pull on the card (as in previous example). This is because the tension in the cord exerting the torque on the disc in now smaller (less than 9.8 nt). Infect, the tension in the string must be less than the weight of the suspended body if the body is to accelerate downward. Let us compute the tension T = m(g – a) = 1.0 (9.8 – 0.89) = 8.91 nt, which is less than 9.8 nt. it must be. Q.27. A uniform sphere, a spherical shell and a cylinder are released from rest at the top of an inclined plane. If the spherical shell reaches the bottom with speed 42 cm/sec. Calculate the speeds with which the sphere and the cylinder reach the bottom. Solution. A body of radius R and radius of gyration K rolls down a θ-inclined plane with an acceleration a given by α =
gsin θ K2 1+ 2 R
If the body is released from rest, and the length of the plane is S, the velocity of reaching the in given by v 2 = 2as =
For the spherical shall
2 gs sin θ K2 1+ 2 R
K2 2 = and v = 42 cm/sec. Thus. R2 3 (42)2 =
2 gs sin θ 6 = gs sin θ 2 5 1+ 3
....(i)
K2 2 . Thus 2 = 5 R
For the sphere
v2 =
2 gs sin θ 10 = gs sin θ 2 7 1+ 5
Dividing (ii) by (i), we get v2 10 5 × = ( 42)2 7 6
...(ii)
Moment of Inertia
377
or
v2 =
or
v2 =
10 5 × × 42 × 42 = 2100 7 6 2100 = 45. 8 cm/sec
2 For the cylinder K = 1 . Thus R2 2
v2 =
2 gs sin θ 4 = gs sin θ 1 3 1+ 2
...(iii)
Dividing (iii) by (i), we get v2 4 5 × 2 = ( 42) 3 6
v2 =
or or
4 5 × × 42 × 42 = 1960 3 6
v = 44.3 cm/sec.
Q.28. A disc of 10 cm diameter pivoted at its rim and made to swing like a compound pendulum. Find its time period. Solution. Let m be the mass and r the radius of the disc. When it swings like a compound pendulum, the time-period is T = 2π
I mgl
where I is the moment of inertial of the disc about the axis of suspension of l is the distance of the point of suspension (pivot) from its centre of mass. Here l = r. Now, the moment of
1 2 mr , and, 2 thus by the theorem of parallel axes, its moment of inertia about a parallel axis through the pivot is inertial of a disc about an axis perpendicular to its plane through its centre is
I =
∴
1 2 3 mr + mr2 = mr2 2 2
T = 2π
3 2 mr 2 = 2π mgr
3 r 2 g
The period is the same as that of a simple pendulum of length l = Now r = 10 cm and g = 980 cm/sec2.
∴
T = 2 × 3.14 ×
3 × 10 2 = 0. 78 sec. 980
3 r. 2
378
Mechanics
Q.29. A wire, bent into a circular ring of radius 40 cm hangs over a horizontal knife-edge. Assuming that there is no slipping on the knife-edge, find the periodic-time of oscillation and the frequency of the ring. Solution. Let m be the mass and r the radius of the ring. The ring is swinging like a compound pendulum with period T = 2π
I mgl
where I is the moment of inertia of the ring about the axis of suspension and l is the distance of the point of suspension (knife-edge) from the centre. Hence l = r. Now the moment of inertia of a ring about an axis perpendicular to its plane through its centre is mr2, and thus, by the theorem of parallel axes, its moment of inertia about a parallel axis through the knife edge is I = mr2 + mr2 = 2mr2 T = 2π
∴
2mr2 = 2π mgr
2r g
Now r = 40 cm and g = 980 cm/sec2 ∴ and
T = 2 × 3.14 × Frequency, n =
2 × 40 = 1. 8 sec 980
1 1 = = 0. 55 sec−1 . T 1. 8
Q.30. Show that a cylinder will slip on an inclined plane of inclination angle θ if the coefficient of static friction between plane and cylinder is less than
1 tan θ. 3
Solution. Let C be the centre of mass of the cylinder. The translational motion of, body can be obtained by assuming that all the external forces act at the centre of mass. Now, the external forces acting on C are the weight mg of the cylinder acting vertically downward, and the normal force N and the tangential (static frictional) force f exerted by the plane. Let a be the (linear) acceleration of the centre of mass. Resolving forces parallel and perpendicular to the plane, we have mg sin θ – f = ma and
...(i)
N – mg cos θ = 0
...(ii) a T1
R
T1
N
T2 T2 m2
gs m1 θ
i
nθ
a
m 1 g co s θ m2g
m 1g
Fig. 33
Moment of Inertia
379 N
C
f
R
mg
θ
sin
θ m g co s θ
mg
θ
Fig. 34
As the cylinder rolls down, it is acted on by a torque τ exerted by the frictional force f, such that τ = fR where R is the radius of the cylinder. But τ is equal to be product of the moment of inertia I of the cylinder about the axis of rotation and the angular acceleration α produced in it. That is τ = Iα The last two expression give fR = Iα = or
f =
Ia R
3α =
a R
Ia R2
...(iii)
Substituting then value of f in Eq. (i), we get mg sin θ = or
a = Now I =
Ia = ma R2 mg sin θ m + (I / R)
1 mR2 . Therefore 2 a =
mg sin θ 2 = sin θ m 3 m+ 2
2 gsin θ, which is 3 less than the acceleration of the centre of mass of a cylinder sliding down the plane (g sin θ). Thus the acceleration of the centre of mass of the rolling cylinder is Now, substituting the above values of a and I is Eq. (iii), we get
FG 1 mR IJ FG 2 g sin θIJ H2 K H3 K = 1 mg sin θ 2
f =
R2
3
This is the minimum force of static friction needed for rolling. If it is less, the cylinder would slip down the plane.
380
Mechanics
We known that, for just equilibrium f = µ sN = µ smg cos θ
1 mg sin θ = µ smg cos θ 3
or or
µs =
1 tan θ 3
This is the minimum value of µs needed for rolling. If µs is less than
1 tan θ, the cylinder 3
would slip. Q.31. A solid sphere of mass 0.5 kg and diameter 1 m rolls without slipping with a constant velocity of 5 m/sec along a smooth straight line. Calculate its total energy. Solution. When the sphere rolls about its diameter, it has translator y as well as rotatory motion. Hence its total kinetic energy is Ktotal = Ktrans + Krot = For sphere
and
∴
1 1 Mv2 + Iω2 2 2
I =
FG 2 IJ MR H 5K
ω =
v , where R is the radius of sphere R
Ktotal =
2
about diameter
LM N
1 1 2 Mv2 + MR2 2 2 5
OP LM v OP Q NR Q
2
=
1 1 Mv2 + Mv2 2 5
=
7 Mv2 10
=
7 × (0. 5 kg) (5 m / sec)2 10
= 8.75 joule. Q.32. A block of mass m = 5 kg slides down a surface inclined 37° to the horizontal. The coefficient of sliding friction is µk = 0.25. A cord attached to the block is wrapped around a flywheel on a fixed axis at O, as shown. The flywheel has a mass M = 20 kg, an outer radius R = 0.2 meter, and a radius of gyration with respect to the axis of rotation, k = 0.1 meter. Find the acceleration of the block down the plane and the tension in the cord. Solution. Let a be the acceleration down the plane. The force acting on the block are its weight mg, the normal force N and the tangential (frictional) force fk (opposite to the motion) exerted by the plane, and the tension T in the cord. mg can be resolved into a component mg sin θ along the plane, and a component mg cos θ perpendicular to the plane. The net force on the block down the plane is mg sin θ – T – fk, and that perpendicular to
Moment of Inertia
381
the plane is N – mg cos θ. The block is accelerated down the plane, while its acceleration perpendicular to the plane is zero. Therefore, by Newton’s second law, we have mg sin θ – T – fk = ma
...(i)
N – mg cos θ = 0
and
N = mg cos θ
or
...(ii)
Now fk = µkN = µkmg cos θ, so that Eq. (i) becomes mg sin θ – T – µk mg cos θ = ma
...(iii) T
N
R O
T
a θ
mg
si
nθ
m g co s θ mg
θ
Fig. 35
Let us now consider the flywheel. The tension in the cord events a torque τ = τR on it. If I be the moment of inertia of the flywheel about the axis of rotation and α the angular acceleration produced in it, then τ = Iα
FG IJ H K
τR = I a , R
or
where a is the linear acceleration. This gives T =
Ia R2
...(iv)
Substituting this value of τ in Eq. (iii), we get mg sin θ –
Ia − µ k mg cos θ = mg R2
FG H
mg (sin θ − µ k cos θ) = a m +
or
or
a =
I R2
IJ K
mg (sin θ − µ k cos θ) m + I / R2
Here m = 5 kg, sin θ = sin 37° = 0.6, µk = 0.25, cos θ = cos 37° = 0.8, I = MK2 = (20) (0.1)2 = 0.2 kg-m2 and R = 0.2 m ∴
a =
( 5) ( 9. 8) [ 0. 6 − ( 0. 25) ( 0. 8)] = 1. 96 m/sec2 5 + 0. 2 / ( 0. 2)2
Substituting the value of a in Eq. (iv), we get T =
Ia 0. 2 × 1. 96 = = 9. 8 nt. R2 (0. 2)2
382
Mechanics
Q.33. In an Atwood’s machine the pulley mounted in horizontal frictionless bearings has a radius R = 0.05 meter. The cord passing over the pulley carries a block of mass m1 = 0.50 kg at one end and a block of mass m2 = 0.46 kg at the other. When set free from rest, the heavier block is observed to fall a distance of 0.75 meter in 5 sec. Compute the moment of inertia I of the pulley. Solution. Let a be the linear acceleration of the system. The heavier block moves a distance 0.75 meter in 5 sec. Using
x = 0.75 =
∴
M R
1 2 at , we have 2 a
1 a(5)2 2
a m2 0 .46 kg
a = 0.06 meter/sec2
If v be the linear velocity at the end of 5 second, then v = at = 0.06 × 5 = 0.3 meter/sec. m1 0 .50 kg
The heavier block descends 0.75 meter, thus losing a potential energy of (0.50 × g × 0.75) joules, while the lighter one ascends the same distance and gains energy of (0.46 × g × 0.75) joule. Thus net loss is P.E. of the system.
Fig. 36
∆U = (0.50 × g × 0.75) – (0.46 × g × 0.75) = 0.294 Joule If there is no friction in the pulley, this must be equal to the gain in the kinetic energy of the blocks and the pulley which is ∆K =
=
1 1 (m1 + m2 ) v2 + Iω2 2 2
FG H
1 1 0.3 (0. 50 + 0. 46) (0. 3)2 + I 2 2 0.05
IJ K
2
= 0.0432 + 18 I Equating the loss in P.E. to the gain in K.E., we have 0.294 = 0.0432 + 18 I ∴
I =
0. 294 − 0. 0432 = 0. 0139 kg-meter2 . 18
Q.34. Suppose the disc (M = 20 kg, R = 0.5 m) in the problem 26 is initially at rest. As the body (m = 1.0 kg) is hanged from the cord, the disc begins to rotate. Calculate the work done by the applied torque on the disc in 2.0 second. Compute also the increase in rotational kinetic energy of the disc. Ans. The angular acceleration produced in the disc is α =
=
FG H
IJ K
mgR g 2m = M + 2m R I + mR2
3I =
1 MR2 2
RS 2 × 1.0 UV 9.8 = 1.78 rad/sec . T 20 + (2 × 1.0) W 0. 5 2
Moment of Inertia
383
The torque τ exerted on the disc is therefore τ = Iα = =
1 MR2α 2 1 (20 kg) (0.5 m)2 (1. 78 rad/sec2 ) 2
= 4.45 joule Let us now calculate the angle θ through which the disc has rotated in 2.0 second. We use θ = ω 0t +
1 2 αt . 2
Here ω0 = 0 (starts from rest), α = 1.78 rad/sec2 t = 2.0 sec ∴
θ =
1 (1. 78) (2. 0)2 = 3. 56 rad. 2
Hence, the work done by the torque on the disc is ω = τθ = (4.45 joule) (3.56 rad) = 15.8 joule. As there is no friction, this work is used up in giving rotational kinetic energy to the disc. Let us compute it Knet =
1 2 Iω 2
FG H
IJ K
=
1 1 MR2 (αt)2 2 2
=
1 (20) (0. 5)2 (1. 78 × 2. 0)2 4
= 15.8 joule. Q.35. Two mass m1 = 12 kg and m2 = 8 kg are attached to the ends of a cord which passes over the pulley of an Atwood’s machine. The mass of the pulley is M = 10 kg and its radius is R = 0.1 m. Calculate the tension in the cord and the acceleration a of the system. Solution. Since the pulley has a finite mass M, the two tension T1 and T2 are not equal. The net force on the larger mass m1 is m1g – T1 acting downward, and that on the smaller mass m2 is T2 – m2g acting upward. If a bet the (linear) acceleration of the system, we have by Newton’s second law and
m1g – T1 = m1a
...(i)
T2 – m2g = m2a
...(ii)
Let us now consider the pulley which may be assumed as a solid disc. The tension in the cord exert torques on the pulley, the net torque is (T1 – T2)R, which is clockwise.
M R
T2 a
T1
T2 m2
T1
m 2g
m1
a m 1g
Fig. 37
384
Mechanics
If α be the angular acceleration produced in the pulley, whose moment of inertia (I) than, we have torque = moment of inertia × ang. accel. i.e., But I =
(T1 – T2)R = Iα
a 1 MR2 and α = , so that R 2 T1 – T2 =
1 Ma 2
...(iii)
Adding (i) and (ii), we get
m1 g − T1 + T2 − m2 g = m1a + m2a (m1 − m2 ) g = (T1 − T2 ) + (m1 + m2 ) a
or
Putting the value of T1 – T2 from Eq. (iii), we have
(m1 − m2 ) g = ∴
a =
1 Ma + (m1 + m2 ) a 2 (m1 − m2 ) g 1 M 1 + m2 + 2
FG m H
IJ K
...(iv)
Given m1 = 12 kg, m2 = 8 kg and M = 10 kg ∴
a =
(12 − 8) × 9. 8 = 1. 57 m/sec2 (12 + 8 + 5)
Now from Eq. (i), we have T1 = m1 (g – a) = 12 (9.8 – 1.57) = 99 nt. Again, from Eq. (ii), we have T2 = m2 (g + a) = 8 (9.8 + 1.57) = 91 nt. Q.36. A sphere rolls up an inclined plan of inclination 30°. At the bottom of the incline the centre of mass of the sphere has a translational speed of 7m/sec. How for does the sphere travel up the plane. How long does it take to return to the bottom. Solution. We know the acceleration of a body rolling down an inclined plane is given by a =
For a sphere I = MK2 =
gsinθ K2 1+ 2 R
2 K2 2 = . His given MR2 (about a diameter), so that 5 R2 5
Moment of Inertia
385
5 sin θ 7 Lets be the distance travelled by the sphere up the plane, after which its velocity is reduced to zero. We apply the translational equation of motion a =
v 2 = v02 + 2as Here v = 0, v0 = 7 meter/sec, a = − ∴ or
5 5 5 g sin θ = − g sin 30° = − g. 7 7 14
0 = (7)2 –
s =
10 gs 14
7 × 7 ×14 = 7 meter 10 × 9. 6
After travelling the distance s up the plane, the sphere return. Let t be the time it takes in reaching the bottom. Thus, let us use s = v0t +
1 2 at 2
FG H
IJ K
We have
7 = 0 + 1 5 g t2 2 14
∴
t =
7 × 28 = 2. 0 sec. 5 × 9. 8
Q.37. A sphere/spherical shell rolling don on inclined plane without shipping, which is larger, the rotational kinetic energy of the translations kinetic energy find the ratio. Ans. In usual nattiness, the rotational end translational kinetic energies are given by Krot = and
Ktrans = Thus
1 1 v2 Icm ω2 = ?(MK2 ) 2 2 2 R 1 Mv2 2
K oat K2 = 2 K trans R
2 2 K2 K2 . Which for < thin spherical shell 2 = . Thus in each care 2 = 5 3 R R the translational energy in larger. For a solid sphere
Q.38. A ball starts for rest and rolls down a 30° inclined plane. How much time would it take to cover 7 meter? Solution. Let M be the mass and R the radius of the ball. Let v be the velocity of the centre of mass of the ball after it covers a distance of the plane. In doing so, it loses potential energy Mg (s sinθ) which appears as kinetic energy. Thus
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Mechanics
Mg (s sinθ) =
∴
v2
1 2 1 Iw + Mv2 2 2
FG H
s
IJ K
=
v2 1 1 2 + Mv2 MR 2 2 5 R2 2
=
7 Mv2 10
s sin θ
θ
10 gs sinθ = 7
Fig. 38
If a be the acceleration of the centre of men, then v 2 = 2as
10 gs sin θ = a2s 7
or ∴
a =
5 gsinθ 7
Here θ = 30° and g = 9.8 m/sec
FG IJ H K
5 1 (9. 8) = 3. 5 m/sec2 7 2 Let t be the time taken in covering 7 meter, than from the relation ∴
or ∴
a =
S =
1 2 at , we have 2
7 =
1 (3. 5)t2 2
t2 =
7 ×2 =4 3. 5
t = 2 sec
Q.39. A circular disc of man 50 gm and radius 10 cm rolls down a plan inclined at 30° to the horizontal. It rolls down form rest through 100 cm in 10 sec. Find the kinetic energy of the disc at the end of 10 sec and the moment of inertia of the disc about its areas. Solution. Let S be the distance on the inclined plane through which the disc rolls down. In doing so it comes down a vertical distance of S sin 30°, and thus loses potential energy equal to mg S sin θ which appears as kinetic energy K at the end of the journey. Thus K = Mg S sin 30° = (0.05 kg) (9.8 nt/kg) (cm)
FG 1 IJ H 2K
= 0.245 joule
The moment of inertia of the disc about its axis is given by I =
1 1 MR2 = (0.05 kg) (0.1m)2 = 2.5 × 10–4 kg–m2. 2 2
Moment of Inertia
387
Q.40. A small solid ball rolls without shipping along the track shown. The radius of the circular part of the track is r. If the ball starts from rest at a height of 6 r above the bottom, what are the horizontal and vertical force acting on it at the point Q. Solution. Let m be the mass and R radius of the ball. The ball on reaching the point Q dexends through a vertical distance of (6r – r) = 5, thus loosing gravitational potential energy, mg (5r). This appears as kinetic energy at Q.
6r
If v be the speed at Q, then the total kinetic energy is
2r
Q
Ktotal = Krot + Ktrans =
1 2 1 Iω + mv2 2 2
=
1 2 m R2 2 5
FG H
Fig. 39
IJ FG v2 IJ + 1 mv K HR K 2
2
2
=
7 mv2 10
Equating it to the loss in potential energy, we get
7 mv2 = mg(5r) 10 50 gr 7 The velocity v of the ball at the point, q is tangentially directed. The horizontal force F acting on the ball is the counterplot force directed toward the centre of the circular track. That is v2 =
or
F =
mv2 50mg = 7 v
The vertical force on the ball is its weight mg. Q.41. One and of a horizontal spring of force constant 6.0 nt/meter is tied to a fixed wall and the other end to a solid cylinder which can roll without shipping on the horizontal ground. The cylinder is pulled s distance of 0.1 meter and released. Calculate the kinetic energies of the cylinder when passing through equilibrium position. Solution. When the cylinder is pulled by 0.1 meter, the spring is stretched and acquires potential energy U, where
1 2 1 kx = × 6. 0 ×(0.1)2 = 0. 03 joule 2 2 This entire energy is converted into the kinetic energy of rotation and the kinetic energy of translation when the cylinder is passing through its equilibrium position. Thus U =
Krot + Ktrans = 0.03 joule Now, and
Krot = Ktrans =
FG H
IJ K
v2 1 2 1 1 1 w = MR 2 = Mv2 2 2 2 R2 4
1 Mv2 2
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Mechanics
∴ or
1 1 Mv2 + Mv2 = 0.03 4 2 Mv2 = 0.03 × Krot =
and
Ktrans =
4 × 0. 04 Joule 3
1 1 Mv2 = (0. 04) = 0. 01 Joule 4 4 1 1 Mv2 = (0. 04 ) = 0. 02 Joule. 2 2
Q.42. A 3.0 kg horizontal cylinder has threads wrapped round it near each end. The two threats are held vertical with their force ends tied to hooks on the ceiling. When the cylinder is released it falls under gravity and the threads are unwind. Find out the linear acceleration of the falling cylinder and the tension in the thread. Ans. The cylinder rotating under gravity has both a motion of translation as well as of rotation. Let M be the mass and I the moment of inertia of the cylinder about the axis of radiation. Let v be the linear velocity of its centre of mass and ω its angular velocity about the axis of rotation when it has fallen through a vertical distance h. Then
T
T h
v
ω
Fig. 40
Loss in potential energy = mgh gain in kinetic energy of translation =
1 Mv2 2
and gain in kinetic energy of rotation =
FG H
IJ K
v2 1 2 1 1 1 = Mv2 Iω = MR2 R2 4 2 2 2
where R in the radius of the cylinder. By the conservation of energy, we have Mgh = or
1 1 Mv2 + Mv2 2 4
4 gh 3 If a be the acceleration with which the cylinder falls, we have from the formula v2 =
v 2 = 2ah
4 gh = 2ah 3 2 g ...(i) 3 Let T be the tension in each thread acting vertically up ward. The weight Mg of the cylinder acts down ward. Then the resulting force acting downward on the falling cylinder in Mg – 2T. But this should be equal to Ma, where a is the accelerate of the falling cylinder. Then or
a =
Moment of Inertia
389
Mg – 2T = Ma or
1 M (g – a) 2 Substituting the value of a from Eq. (i) we have T =
1 1 2 M (g – g ) = Mg 6 2 3 Here M = 3.0 kg and g = 9.8 ht/kg. T =
1 × 3.0 × 9.8 = 4.9 nt 6 Q.43. A uniform circular slab 5.0 m and mass 100 kg rolls along a horizontal at a speed of 4 m/sec. How much work has to be done to stop it. T =
Solution. The total kinetic energy of a circular slab of mass M and radius R rolling without shipping is the sum of its kinetic energy of translation and that of ration, that is K = Ktrans + Krot =
1 1 Mv2 + Iω2 2 2
1 1 v2 Mv2 + (MR2 ) 2 2 2 R = Mv2 =
∴ For disc, I = MR2
Here M = 100 kg and v = 4m/sec ∴
K = 100 × (4)2 = 1600 Joule
By work-energy theorem the work need to stop it is W = K = 1600 joule Q.44. The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94 × kg-m2 about an axis perpendicular to the inter-nuclear axis and passing through the centre. The molecule in a gas has a mean speed of 500 m/sec, and its rotational kinetic energy is two-third of its translational kinetic energy. Find its average angular velocity. 10–46
Solution.
Ktrans = 1 Mv2 = 1 ( 5. 30 × 10−26 kg ) (500 m/sec)2 2 2 = 6.625 × 10–21 joule
1 2 1 Iω = (1.94 × 10–46 kg-m2) M2 2 2 = 0.97 × 10–46 M2 Joule
Krot =
But Krot =
or ∴
2 K (given). Therefore 3 trans 0.97 × 10–46 M2 = 2 (6.625 × 10–21) 3 2 × 6. 625 × 10−21 M2 = = 45.5 × 1024 3 × 0. 97 × 10−46 M = 6.75 × 1012 rad-sec.
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Mechanics
Q.45. A body of radius R and mass m is rolling horizontally without slipping with speed 3v2 , (a) what is the moment of inertia v. It then rolls up a hill to a maximum height h. If h = 4g of the body (b) what might the shape of the body be? Solution. (a) The total kinetic energy of a rolling body is Ktotal = Krot + Ktrans =
1 2 1 Iω + mv2 2 2
=
1 2 I 1 v2 1 +m I 2 + mv2 = v 2 R2 2 R 2
F GH
when it rolls up a hill to a maximum height h =
FG H
I JK
IJ K
3v2 . The entire kinetic energy is converted 49
into gravitational potential energy mgh. Thus
FG H
IJ K
1 2 I v +m 2 R2
F 3v I = mgh = mg G 4 g J H K 2
or
I 3 +m = m 2 R 2
or
I =
1 mR2 2
(b) The body may be solid circular cylinder or disc. Q.46. A small ball of mass m and radius r, starting from rest, rolls down on the inner surface of a large hemisphere of radius R. What is its kinetic energy when it reaches the bottom of the hemisphere? What forcing of it is rotational and what translatable. Solution. Initially the ball is at the top of the hemisphere. Its centre of mass is at vertical height (R – r) from the bottom of the hemisphere, so that it has gravitational potential energy mg (R – r). On reaching the bottom, this energy is totally converted into kinetic energy. Hence total kinetic energy
r R
= mg (R – r)
Fig. 41
We know that for a sphere of mass m and radius r rolling with linear velocity v and angular velocity w; rotational kinetic energy =
and
transactional kinetic energy = ∴
total energy =
FG H
IJ K
1 2 1 2 2 v2 1 Iω = mr = mv2 2 2 2 5 5 r 1 mv2 2 1 1 7 mv2 + mv2 = mv2 5 2 10
Moment of Inertia
391
1 mv2 2 5 = Now, fractional rotational energy = 7 7 mv2 10 1 mv2 5 2 = and fractional translational energy = 7 7 2 mv 10 Q.47. A uniform this bar of mass 4 kg and length 2 meter is bent to made a square. Calculate its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the square
LM Ans. N
1 kg − meter2 2
OP Q
Q.48. Three masses each 2 kg are placed at the vertices of an equilateral triangle of side 10 cm. Calculate the M.I. and radius of gyration of the system about an axis passing through (i) a centre (ii) mid point of one side and (iii) centre of man. In all cases the axis is perpendicular to the plane of the triangle. [Ans. (i) 0.04 kg-m2, 0.082 m, (ii) 0.025 kg-m2, 0.064 m, (iii) 0.02 kg-m2, 0.058 m] Q.49. Find the radius of gyration of a disc of mass 100 gm and radius 5 cm about an axis perpendicular to its plane and passing through its centre of gravity. [Ans. 3.5 cm.] Q.50. A square frame ABCD of side 40 cm and negligible mass, has four masses, 2, 5, 6 and 3 kg placed at the corners and a mass 10 kg at the centre X. It is be rotated about an axle passing through and perpendicular to the frame. Compute the torque needed to produce an angular acceleration of 20 per sec2. Q.51. A fly wheel of mass 10 kg and radius of gyration 20 cm is rotating with angler velocity 10 rotations per minute. Calculate its anguler momentum. [Ans. 0.42 kg-m2/sec.] Q.52. A solid cylindrical fly-wheel of 16 kg man and 20 cm radius is making 5.0 rev/sec. Find the kinetic and momentum. Q.53. A light cord is wrapped around the rim of a disc of mass M and radius R, which rotates without friction about a fixed horizontal axis. The free end of the cord is tied to a mass m, which is released from rest a distance h above the floor show that the mass m strikes the floor with a linear speed given by v =
[ 2 gh/{1 + ( M / 2m )] .
Q.54. A cord is wrapped around the rim of a fly wheel 0.5 m is radius and steady down word pull of 50 at is exerted on the cord. The wheel is mounted in fraction less bearings on a horizontal shaft through its centre. The moment of inertia of the wheat is 4 kg-m2. (a) compute the angular acceleration of the wheel (b) find the work done in unwinding 5 meter of cord (c) If a mass having a weight of 50 mt from the cord, compute the annular acceleration of the wheel. Why is this not the same as in part (a). [Ans. (a) 6.25 rad/sec2 (b) 250 Joule (c) 4.74 rad/sec2, it is smaller than that in part (a) because the pull on the wheel is now smaller.] Q.55. A block of mass m1 = 3 kg is put on a plane inclined at 0 = 30° to the horizontal and is attache by a cord parallel to the plane over a pulley at the top to a hanging block of
392
Mechanics
mass m2 = 9kg. The pulley has a mass m = 1kg and has a radius R = 0.1 meter. The coefficient of kinetic friction between block and plane is µk = 0.10. Find the acceleration a of the hanging block and the tension in the cord on each side of the pulley. Assume the pulley to be a uniform disc. [Ans. a = 5.7 meter/sec2, T2 = 37 nt. T1 = 34 nt.] Q.56. A thin hollow cylinder of 25 kg-wt, open at both ends (a) slides with a speed of 2 meter/sec (b) rolls with the same speed without slipping. Compare the kind energies of the cylinder in the two cases. Q.57. The moment of inertia of a reel of thread about its axis is MK2. If the loose end of the thread is held in the hand and the real is allowed to unroll itself while falling under gravity, show that it falls down with an acceleration of gR2 (R2 + K2) where R is the radius of the real. Q.58. A said cylinder (a) rolls, (b) slides from rest down an inclined plane. Neglet friction. Show that the ratio of the velocities of the centre of mass in the two cases, when the cylinder reaches the bottom of the incline is
3/ 2 .
9 MECHANICAL PROPERTIES OF MATTER 9.1 RIGID BODY A body is said to be rigid, if the distance between any two particles of which remains unaltered whatever the external forces applied to it and in whatever manner they may vary, so that it remains un-deformed, i.e., its size, shape and volume remain unaffected. In actual practice, however, we come across no such body, for all material bodies are found to get deformed to a greater or a smaller extent under suitably applied external forces, tending to recover their original state. (i.e., length, volume or shape) on the removal of these forces.
9.2 ELASTICITY A body remains in natural equilibrium under the internal molecular forces of attraction whose magnitude depends upon the spacing between the molecules. When external forces are applied on the body, new internal forces are developed which cause a change in the relative spacing between the molecules. Hence, the body changes in size, or shape or both and is said to be ‘deformed’. When the external forces are removed, the new internal forces bring the body to its normal state. The property of the body by virtue of which it recovers its original size and shape when the external forces are removed is called the ‘elasticity of the body.
9.3 PERFECTLY ELASTIC AND PERFECTLY PLASTIC If the deformation of a body under a given deforming force, at a given temperature, remains unchanged (i.e., neither increases nor decreases) by the prolonged application of that force and which completely regains its original state on the removal of that force, it is said to be perfectly elastic. On the other hand, if the body remains deformed and shows no tendency to recover its original condition on the removal of the deforming force, it is said to be perfectly plastic. The nearest approach to the former is a quartz fibre and to the latter, ordinary putty. All other bodies lie between these two extremes.
9.4 STRESS A body is equilibrium under the influence of its internal forces is in its natural state. But when external or deforming forces are applied to it, there is a relative displacement of its particles and this gives rise to internal forces of reaction tending to oppose and balance the deforming forces, until the elastic limit is reached and the body gets permanently deformed. The body is then said to be stressed or under stress. 393
394
Mechanics
If this opposing or recovering force be uniform, i.e., proportional to area, it is clearly a distributed force like fluid pressure and is measured in the same manner, as force per unit area, and termed as stress. Stress is equal in magnitude; though opposite in direction, to the applied or deforming force per unit area. Thus, if F be the deforming force applied uniformly over an area A, then stress = F/A. The stress is not uniformly distributed over a surface, let a force δF act normally on an elemental area δA of the surface, when stress at a point on the surface = δF/δA which in the limit δA → 0 is equal to dF/dA. If the deforming force were inclined to the surface, its components perpendicular and along the surface are respectively called normal and tangential (or shearing) stress. The stress is, however, always normal in the case of a change of length or volume and tangential in the case of a change of shape of a body.
9.5 STRAIN When a body suffers a change in its size, or shape, under the action of external forces, it is said to be ‘deformed’ and the corresponding measured per unit dimension i.e., per unit length, per unit volume or the angular deformations produced in it, is called strain and is referred to as linear, volume and shearing strain (or shear) in the three cases respectively.
9.6 HOOK’S LAW This is the fundamental law of elasticity and was given by Robert Hook in 1679 in the form ‘Ut tension sic vis’, i.e., ‘As the tension, so the strain’. It may given by: ‘Provided the strain is small, the stress is proportional to the strain. It follows, therefore, that if the strain be small, the ratio between stress and strain is a constant, called modulus of elasticity (a name given to it by Thomas Young) or Coefficient of elasticity (E). Thus:
Stress Strain The dimensions and units of the modulus or coefficient of elasticity are the same as those of stress or pressure. E =
9.7 ELASTIC LIMIT In the case of a solid, if the stress be gradually increased, the strain too increases with it in accordance with Hook’s law until a point is reached at which the linear relationship between the two just ceases and beyond which the strain increases much more rapidly than is warranted by the law. This value of the stress for which Hook’s law just ceases to be obeyed is called the elastic limit of the material of the body for the type of stress in question.
Behaviour of a Wire Under an Increasing Load Let us consider a wire suspended vertically from a rigid support. When a load is applied at its lower end, its length increases. The load divided by its area of cross-section gives the ‘stress’ while the increases in length divided by the initial length gives the ‘strain’. When the load is increased step by step, and a graph is drawn between the stress and the corresponding strain a curve is obtained. According to Hook’s law, the stress in a body is proportional to the corresponding strain produced, provided the strain is small. Now, up to the point A, the curve is a straight line,
Mechanical Properties of Matter
395
indicating that the length of the wire increases in proportion to the load applied. Further, it is found that the wire recovers its original length when the load is removed. The point A is called the ‘limit of proportionality’. C B
D
A S tress
S train
Fig. 1
When the load is further increased, the extension is greater than that allowed by proportionality. However, up to a small range beyond A the wire continues to return to its original length when the load is removed, but after that it refuses to do so and a permanent increase in length is produced. The stress at which the wire first refuses to return to its original length is called ‘elastic limit’. On further increasing the load, a stage is reached when the extension increases very rapidly for very slowly increasing load i.e., the material on the wire flows like a viscous liquid. This happens at point B, which is called the ‘yield point’. When the load is increased still further, a point C is reached when the stress takes a maximum possible value. This is called the ‘breaking stress’. The wire begins to thin down so that its cross-section does not remain uniform, and breaks with a release of stress. This point where the wire actually breaks is D.
9.8 YOUNG’S MODULUS When a body whose length is very large as compared to its breadth and thickness (a wire) is acted upon by two equal and opposite forces in the direction of its length, the length of the body is changed. The change in length per unit length of the body is called the longitudinal strain and the force applied per unit area of cross-section of the body is called the longitudinal stress. When strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is called the ‘Young’s Modulus’ of the material of the body. It is denoted by ‘Y’. Let there be a wire of length L and radius r. Its one end is clamped to a rigid support and a weight Mg is applied at the other end. Suppose within the elastic limit, its length is increased by l. Then Longitudinal stress = = Longitudinal strain =
L
Force (weight suspended) Area of cross - section
Mg πr2 Increase in length l = Original length L
l Mg
Fig. 2
396
Mechanics
∴ Young’s modulus of the material of the body is Y = =
Longitudinal stress Longitudinal strain M g/ πr2 MgL = l /L πr2l
‘Y’ can be determined only for solids and it is the characteristic of the material of a solid.
9.9 BULK MODULUS OF ELASTICITY When a uniform pressure (normal force) is applied all over the surface of a body, the volume of the body changes, but its shape remains unchanged. Such a strain may appear in all the three states of matter, solid, liquid and gas. The change in volume per unit volume of the body is called the ‘Volume strain’ and the normal force acting per unit area of the surface (pressure) is called the ‘normal stress’. When strain is small, the ratio of the normal stress to the volume strain is called the ‘bulk modulus’ of the material of the body. It is denoted by B. Let the initial volume of a body be V, which changes, by v when a pressure p is applied, when the pressure increases the volume decreases and vice-versa. Then Normal stress = p Volume strain =
v V
∴ Bulk modulus of the material of the body is B =
p pV = v /V v
Compressibility: The reciprocal of the Bulk modulus of the material of a body is called ‘Compressibility’ of that material.
9.10 MODULUS OF RIGIDITY (SHEAR MODULUS) When a body is acted upon by an external force tangential to a surface of the body, the opposite surface being kept fixed, it suffers a change in shape, its volume remaining unchanged. Then the body is said to be ‘sheared’. The ratio of the displacement of a layer in the direction of the tangential force and the distance of that layer from the fixed surface is called the ‘shearing strain’ and the tangential force acting per unit area of the surface is called the ‘shearing stress’. For small strain, the ratio of the shearing stress to the shearing strain is called the ‘modulus of rigidity’ of the ‘material of the body’. It is denoted by η. Let us consider a cube ABCDHIJK, whose lower surface is fixed. When a tangential force F is applied at its upper surface ABIH, then all the layers parallel to ABIH are displaced in the direction of the force. The displacement of a layer is proportional to its distance from the fixed surface. The cube thus takes the new form A′B′CDH′I′JK. The angle θ through which the line AD or BC, initially perpendicular to the fixed surface, is turned is called the ‘angle of shear’ or ‘shearing strain’. If θ is small, then
Mechanical Properties of Matter
397
θ = tan θ = =
BB′ BC
displacement of the upper surface distance of the upper surface from the fixed surface
If A be the area of the upper surface ABIH, then Shearing stress = F/A ∴ Modulus of rigidity of the material of the cube is H′
H
I
I′
F A′
A
B
B′ θ
θ
J
K Fixe d
D
C
Fig. 3
η =
FA F = θ Aθ
The unit of η is Newton/meter2 and its dimensional formula is [ML–1T–2]
9.11 POISSON’S RATIO When two equal and opposite forces are applied to a body along a certain direction, the body extends along that direction. At the same time, it also contracts along the perpendicular direction. The fractional change in the direction along which the forces have been applied is called the ‘longitudinal strain’, while the fractional change in a perpendicular direction is called the ‘lateral strain’. The ratio of the lateral strain to the longitudinal strain is called the ‘Poisson’s ratio’. It is a constant for the material of the body. In figure a wire of original length l and diameter D is subjected to equal and opposite forces F, F along its length. If the length increases to l + ∆l and the diameter decreases to D – ∆D, then F
∆l Longitudinal strain = l and
∆D D The Poisson’s ratio, σ, of the material of the wire is Lateral strain =
σ =
l + ∆l
D
D – ∆D
l
∆D/D ∆l/l F
9.12 POTENTIAL ENERGY IN STRETCHED WIRE
Fig. 4
When a wire is stretched, work is done against the inter-atomic forces. This work is stored in the wire in the form of elastic potential energy. Suppose on applying a force F on
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a wire of length L, the increase in length is l. Initially the internal force in the wire was zero, but when the length is increased by l, the internal force increases from zero to F (equal to the applied force). Thus for an increase l in length, the average internal force in the wire is
O+F F = 2 2 Hence the work done on the wire is W = average force × increase in length = ½ F × l This is stored as elastic potential energy U in the wire U = ½ F.l Let A be the area of cross-section of the wire. The last expression may be written as U = ½ (F/A) (l/L) LA = ½ stress × strain × volumes of the wire ∴ Elastic potential energy per unit volume of the wire is u = ½ stress × strain But ∴
Stress = Young’s modulus × strain u = ½ Young’s modulus × strain2
9.13 EQUIVALENCE OF A SHEAR TO A TENSILE AND A COMPRESSIVE STRAIN AT RIGHT ANGLES TO EACH OTHER AND EACH EQUAL TO HALF THE SHEAR Let ABCD be the section of a cube fixed at the lower surface. Let a tangential force F be applied to its upper surface, so that, it is sheared through a small angle θ and takes the shape A′B′CD. In this process the diagonal DB is extended to DB′ while AC is compressed to A′C. Let arcs BM and A′N be drawn with centres D and C, and radii DB and CA′ respectively. These may be taken as perpendiculars to DB′ and CA respectively, since θ is small. Again, since θ is small, ∠BB′M is very nearly equal to 45°.
A
A′
Fo rce
N
M
θ
θ
D
Let us consider the triangle BMB′.
C
Fig. 5
MB′ = BB′ cos BB′M = BB′ cos 45° = BB′/√2 Now, let us consider the triangle, DBC. DB = Extensional strain along DB = =
BC BC = = BC 2 cos DBC cos 45° Extension in DB Initial length DB′ − DB DB′ − DM = DB DB
= MB′/DB
B′
B
F
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399
Putting the value of MB′ and DB from above, we get
BB′ 2 = 1 BB′ Extensional strain = BC 2 2 BC =
1 BB′ 1 = tan θ 2 BC 2
When θ is small, we can put tan θ = θ Extensional strain = θ/2 Similarly, we can show that the compressional strain along AC is θ/2. Thus, a shear θ is equivalent to an extensional strain and a compressional strain at right angles to each other and each of value θ/2.
9.14 EQUIVALENCE OF COMPRESSION AND EQUAL PERPENDICULAR EXTENSION TO A SHEAR Let ABCD be the section of a cube of side l. Let it be compressed along the diagonal AC and extended along the perpendicular diagonal BD so that the new diagonals are A′C′ and B′D′. Let AA′ = CC′ = BB′ = DD′ = y, where y is small. Let
A
∠ ANA′ = ∠C′MC = ϕ
A′
B
B′
AO = AB Cos BAO = l cos 45° = l/√2 ∴ Similarly, ∴
But ∴
A′O = AO – AA′ = l/√2 – y
θ
θ
D D′
C C′
B′O = BO + BB′ = l/√2 + y A′B′ =
( A ′ O)2 + (B′ O)2
=
FG H
I F l − yJ + G K H 2
=
l2 + 2 y2
2
I l + yJ K 2
Fig. 6 2
2y2 << l2 A′B′ = l = AB
That is, the sides of the cube remain almost unchanged in length. Let us imagine that the deformed cube A′B′C′D′ has been rotated clockwise through the angle φ, and D′C′ has been made to coincide with DC. In this position A'D' makes with AD on anlge θ which is equal to 2φ. This position represents the case of shear where the plane DC is fixed and the planes above DC suffer parallel displacement. The angle of Shear is θ. Let A′E Fig. 7 be perpendicular to AD. Now θ = 2φ = 2 sin φ
A′ E A′ N Let us now consider the traingle AA′E = 2
(where φ is small, then sin φ = φ) ...(i)
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Mechanics
A′E = AA′ sin EAA′ = y sin 45° = y/√2
l 2 Putting these values of A′E and A′N in (i) we obtain Also
A′N =
θ = 2
y/ 2 2y =2 l /2 l
A′
N
B
O C′
compression α = initial length y AA ′ = = AO l/ 2
θ
...(ii)
Let α be the compressional strain along the diagonal AC. Then
=
B′ A
D M
C
φ
D′
Fig. 7
2y l
θ 2y = 2 l θ ∴ α = 2 Thus, the compressional strain along AC is one-half the angle of shear. Similarly the extensional strain along the other diagonal BD is one half the angle of shear. Thus, compression θ/2 and a perpendicular extension θ/2 are equivalent to a shear θ.
But from (ii)
9.15 EQUIVALENCE OF A SHEARING STRESS TO AN EQUAL EXTENSIONAL STRESS PLUS AN EQUAL AND PERPENDICULAR COMPRESSIONAL STRESS Let ABCDHIJK be a cube of side l having its base CDKJ fixed. Let a tangential force F act at its upper surface ABIH. The tangential (shearing) stress P is given by P =
tangential force F = 2 surface area l
This force must have a tendency of moving the cube as a whole. But actually the base of the cube is fixed. Hence an equal and opposite force of reaction F is developed in the base CDKJ. These two forces form a couple which tends to rotate the cube clockwise. Again, since the cube is not free to rotate, the base gives rise to an equal but anticlockwise couple by exerting forces F, F, on the faces BCJI and DAHK. Let us imagine a diagonal plane ACHJ, cutting the cube into two halves and consider the equilibrium of the upper half. The forces F, F acting on the faces. ABIH and BCJI give a resultant force F√2 parallel to DB acting normally over the plane ACJH. Since this upper half is in equilibrium, the resultant force F√2 must be balanced by an equal and opposite force. This force is provided by the lower half of the cube. Thus we have equal and opposite forces F√2, F√2 acting normally over the plane ACJH and tending to extend the material of the cube parallel to the diagonal DB. The area of the plane ACJH is
H
I F
A
B F
F K
J
F D C
Fig. 8
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401
AC × AH = l √2 × l = l2 √2 Therefore the extensional stress parallel to DB =
Force Area
=
F F 2 = 2 =P 2 l l 2
Similarly, if we cut the cube into two halves by an imaginary plane BDKI, we can prove that a compressional stress P is acting parallel to the diagonal AC. Thus a shearing stress P is equivalent to an extensional stress and a compressional stress at right angles to each other and each of value P.
9.16 RELATIONS CONNECTING THE ELASTIC CONSTANTS The elastic constant Y, K, n and the poission’s ratio σ, are all interconnected. Let us consider a cube whose sides each of length l, are parallel to the three coordinate axes x, y and z. Let a uniform normal stress P be applied over its surface. Then each side of the cube is under an extensional stress P. The stress P acting parallel to the x-axis will produce extension parallel to the x-axis, and compressions parallel to y and z axis. These will be as follows: Extension parallel to x-axis = longitudinal strain × initial length. =
stress × initial length , Young s modulus
=
P ×l Y
Compression parallel to y-axis = lateral strain × initial length = Poission’s ratio × longitudinal strain × initial length = σ
P ×l Y
Compression parallel to z-axis = σ
P ×l Y
y P
Similarly, the stress P acting parallel to y-axis produces. Extension parallel to y-axis =
P ×l Y
P ×l Y P and Compression parallel to z-axis = σ × l Y Compression parallel to x-axis = σ
P P
P
P
z
x
O
P
Fig. 9
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Mechanics
Also, the stress P acting parallel to z-axis produces Extension parallel to z-axis =
P ×l Y
Compression parallel to x-axis = σ
P ×l Y
P ×l Y Hence, net extension parallel to x-axis
and compression parallel to y-axes = σ
=
P P P P (1 − 2σ )l l−σ l−σ l= Y Y Y Y
Net extension parallel to y-axis = − σ
P P P P (1 − 2σ)l l+ l−σ l= Y Y Y Y
P P P P (1 − 2σ)l l+ l−σ l= Y Y Y Y P (1 − 2σ )l Thus each side of the cube increases from l to l + Y The new volume is therefore,
and net extension parallel to z-axis = − σ
LMl + P (1 − 2σ)lOP N Y Q Since the initial volume was
LM N
3
l3,
= l3 1 +
OP Q
3
LM N
= l3 1 +
3P (1 − 2σ ) Y
OP Q
approx
the change in volume
LM N
Hence,
P (1 − 2σ) Y
OP Q
= l3 1 + 3P (1 − 2σ ) − l3 = 3P (1 − 2σ ) l3 Y Y Change in volume Volume strain = Initial volume
3P (1 − 2σ )l3 3P Y (1 − 2σ) = = 3 Y l Now, considering the cube as a whole, its surface being under a uniform normal stress P undergoes the volume strain
3P (1 − 2σ). Its bulk modulus is therefore Y normal stress P Y = = volume strain 3P (1 − 2σ) 3(1 − 2σ) Y Y = 3K (1 – 2σ)
K =
...(i)
Let us now consider a cube of side l, upon which an extensional stress P parallel to x-axes, and a compressional stress P parallel to y-axes have been applied. There is no stress along the z-axes
Mechanical Properties of Matter
403
The extensional stress P produces
P ×l Y P Compression along y-axis = σ × l Y P and Compression along z-axis = σ × l Y The compressional stress P produces Extension along x-axis =
Compression along to y-axis =
and
P
x
Extension along to x-axis = σ
P ×l Y
Extension along to z-axis = σ
P ×l Y
and net compression along y-axis =
P
P
P ×l Y
Thus, net extension parallel to x-axis =
y
P z
Fig. 10
σP P P l+ l= (1 + σ ) l Y Y Y
σP P P l+ l= (1 + σ ) l Y Y Y
There are equal extension and compression along the z-axis. Hence there is no change along z-axis. Since, each side of the cube is of initial length l, the extensional and compressional strains along the x-and y-axes are each equal to
P (1 + σ)l P Y = (1 + σ) = l Y Now, simultaneous equal extensional and compressional strains at right angles to each other, are equivalent to a shear θ ∴ or or
P P (1 + σ ) + (1 + σ ) = θ Y Y 2P (1 + σ ) = θ Y
Y P = 2 (1 + σ) θ
Further, the extensional stress P parallel to the x-axis, and the compressional stress P parallel to the y-axis will be equivalent to a shearing stress P. Hence, the modulus of rigidity of the material of the cube is η =
Shearing stress P = θ Shear
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Mechanics
Putting the value of P/θ from above, we have η =
Y 2 (1 + σ)
Y = 2η (1 + σ)
or
...(ii)
Y = 1+ σ 2η
or
σ =
or
Y −1 2η
Let us put this value of σ in Eq. (i). Then
LM FG Y − 1IJ OP N H 2η K Q L YO 3KY = 3 K M3 − P = 9 K − η N ηQ
Y = 3K 1− 2
Y+
3KY = 9K η Y =
9ηK η + 3K
...(iii)
Equations (i) and (ii) give Y = 3K (1 – 2σ) = 2η (1 + σ) 3K – 2η = σ (6K + 2η) or
σ =
3 K − 2η 6K + 2η
...(iv)
Equations (i), (ii), (iii) and (iv) are the various relations among the elastic constants.
9.17 THEORETICAL LIMITING VALUES OF POISSON’S RATIO The relation between Poission’s ratio σ to bulk modulus K and modulus of rigidity η is 3K (1 – 2σ) = 2η (1 + σ) If σ is positive, right-hand side is positive. Therefore left-hand side must also be positive. This is so when 2σ < 1 or
σ < (1/2)
or
σ < 0.5
Now, if σ is negative, the left-hand-side is positive and hence right-hand side must also be positive. This is so when 1+ σ > 0 or
σ > –1
Mechanical Properties of Matter
405
Thus theoretically σ must lie between –1 and
1 2 1 . 2
σ is, however, never negative. Hence, in practice, σ lies between 0 and
9.18 POISSON’S RATIO FOR AN INCOMPRESSIBLE MATERIAL For a material whose volume remains unchanged by changing pressure, we have K= ∞ An infinite value of bulk modulus means that the body is incompressible. Putting
K = ∞ in the relation
G A
Y = 1 − 2σ, we get 3K
1 – 2σ = 0 σ = ½
or
W a te r
R
Determination of Poisson’s Ratio (σ) (i)
For rubber: About a meter long rubber tube R (like a cycle tube) is closely fitted with rubber bungs (smeared with glue) and metal caps (A and B) at its two ends. The upper cap A is clamped tightly in a wall bracket or a massive stand and a small weight placed in the scale pan suspended from a hook in the lower cap B, so that the tube hangs vertically with no bends or kinks in it.
B H
P W
Fig. 11
A glass tube G, about half a meter long and of radius r is just inserted into the rubber tube which is then filled with air free water until some of it rises in tube G. When conditions become steady, the positions of the water meniscus in G and a point or pin P carried by hook H are noted with the help of two separate travelling microscope. A suitable weight (W) is then placed in the scale pan so that the length of the rubber tube, as also its internal volume increases. Again, when conditions becomes steady, the positions of the water meniscus and the point or pin P are noted. Several such readings are taken by increasing the weight in equal steps of W each and the mean increase in length, say dL, of the tube for a stress P = W/A as also the mean fall dh in the level of the water meniscus in G obtained. If L be the original length of the rubber tube, longitudinal strain = dL/L and if Y be the Young’s modulus for the material of the tube, Y =
P dL L
And since the fall in the level of the water meniscus in G corresponds to increase in volume of the rubber tube R, we have increase in volume of the rubber tube = (πr2)dh = dV. And, therefore, Bulk modulus for the material of the tube is given by
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Mechanics
P , where P is the stress applied along the length of the tube, i.e., only in 3 ( dV / V) one direction (so that the increase in volume of the tube is one-third of what it would be with an equal stress (P) in all the three directions, and V, the initial volume of the tube (obtained by measuring the volume of the water filling it completely). K=
Y 3 (1 − 2σ) Substituting for K and Y, therefore, we have Now, we have the relation K =
dV dL (1 − 2σ ) = V L σ =
or (ii)
FG H
dV L 1 1− 2 V dL
IJ K
For Glass: We proceed here too in the same fashion as in case (i) for rubber. Since, however, the change in volume in this case is very much smaller, we use a capillary tube in place of tube G above.
9.19 TWISTING COUPLE ON A CYLINDER
(i) Case of a Solid Cylinder or Wire Let a solid cylinder (or wire) of length L and radius R be fixed at its upper end and let a couple be applied to its lower end in a plane perpendicular to its length (with its axis coinciding with that of the cylinder) such that it is twisted through an angle θ. This will naturally bring into play a resisting couple tending to oppose the twisting couple applied, the two balancing each other in the position of equilibrium. To obtain the value of this couple, let us imagine the cylinder to consist of a larger number of hollow, coaxial cylinder, one inside the other and consider one such cylinder of radius x and thickness dx as shown in figure (i). It is seen from figure (ii) that each radius of the base of the cylinder will turn through the same angle θ but the displacement (BB′) will be the maximum at the rim, progressively decreasing to zero at the center (O), indicating that the stress is not uniform all over.
dx
O x θ
2πx
A
O′
D
A φ
φ
B L
φ
L
B′
R
O (i)
x
θ ( ii )
B
Fig. 12
B′
B
B′
C ( iii)
C′
Mechanical Properties of Matter
407
Thus, a straight line AB, initially parallel to the axis OO′ of the cylinder will take up the position AB′ or the angle of shear (or shear) = ∠BAB′ = φ. This may be easily visualized if we imagine the hollow cylinder to be cut along AB and spread out when it will initially have a rectangular shape ABCD figure (iii) and will acquire the shape of parallelogram AB′C′D after it has been twisted, so that angle of shear = BAB′ = φ. Now BB′ = xθ = Lφ, whence, the shear φ = xθ/L and will obviously have the maximum value when x = R, i.e., at the outermost part of the cylinder and the least at the innermost. If η be the coefficient of rigidity of the material of the cylinder, we have η = or
Shearing stress Shear
Shearing stress = η × Shear Shearing stress = ηφ =
ηxθ L
FG η x θ IJ × face area of the cylinder H L K F η x θ IJ × 2π xdx = FG 2πηθ IJ x dx. = GH H L K L K
∴ Shearing force on face area of the hollow cylinder =
2
And the moment of this force about the axes OO′ of the cylinder =
FG 2πηθ IJ x H L K
2
dx . x =
z
R
∴ Twisting couple on the whole cylinder =
O
FG 2πηθ IJ x H L K
3
dx .
2πηθ 3 πηR4 x dx = θ L 2L
Or twisting couple per unit twist of the cylinder or wire, also called torsional rigidity of its material, is given by C =
πηR4 2L
(ii) Case of a Hollow Cylinder If the cylinder be a hollow one, of inner and outer radii R1 and R2 respectively, we have
z
R2
Twisting couple on the cylinder =
R1
2πηθ 3 πη 4 x dx = (R2 − R14 )θ 2L L
∴ Twisting couple per unit twist, C′ =
πη ( R24 − R14 ) . 2L
Now, if we consider two cylinders of the same material, of density ρ, and of the same mass M and length L, but one solid, of radius R and the other hollow of inner and outer radii R1 and R2 respectively, we have
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Mechanics
e
je
R22 + R12 R22 − R12 C′ R4 − R4 = 2 4 1 = C R R4
e
j
j
M = π R22 − R12 Lρ = πR2Lρ, we have
Since
C′ = C
or
eR
2 2
j
+ R12 R2 R
4
=
eR
2 2
+ R12 2
eR
2 2
j
− R12 = R2
j
R
Again, because
R22 − R12 = R2, We have R22 = R2 + R12
And therefore,
R22 + R12 = R2 + R21 + R12 = R2 + 2R12 i. e., R22 + R12 > R2
Clearly, therefore,
C′ > 1; C
C′ > C
or, the twisting couple per unit twist is greater for a hollow cylinder than for a solid one of the same material, mass and length. This explain at once the use of hollow shafts, in preference to solid ones, for transmitting large torques in a rotating machinery.
η) FOR THE MATERIAL 9.20 DETERMINATION OF THE COEFFICIENT OF RIGIDITY (η OF A WIRE 1. Statical Method (Barton’s Statical Method of Determining Modulus of Rigidity) The torsional rigidity for a rod of length l and radius r, which is fixed at the upper end and twisted at the lower end is, C =
πηr4 2l
Where η is coefficient of rigidity of the material. This expression has already given. The given wire or rod is hung vertically with its upper end rigidly clamped at A. To its lower end a brass cylinder C is attached. Two flexible threads are wound round the cylinder C such that they leave it tangentially at the opposite ends of a diameter and then pass over to two frictionless pulleys P1 and P2. The ends of these threads carry pans. A pointer is attached to the wire at a known distance from the upper end. When the wire is twisted, the pointer moves over a degree scale S. Theory: Suppose equal weights mg, mg are placed on the pans. So that the threads experience equal and opposite parallel forces mg, mg. These forces form a couple of moment mgD, where D is the diameter of the cylinder C. This is a twisting couple applied to the lower end of the wire. It causes each cross-section of the wire to twist through an angle proportional to the distance of the cross-section from the upper fixed end. In the equilibrium state, the twisting couple is equal and opposite to the elastic restoring couple set up in the wire. Let r be the radius of the wire and φ radian the twist in it at a distance l from the fixed end. The elastic restoring couple set up in the wire is
Mechanical Properties of Matter
409
πηr4 φ 2l
Cφ =
A
This, at equilibrium, is equal to the twisting couple mg D. That is
l
πηr4 φ = mgD 2l
S
2mgDl πr 4φ
η =
P
1 D Method: Equal weights are placed on the pans and the P2 twist is noted by reading both ends of the pointer on the scale mg C (to eliminate the error due to eccentricity of the wire with mg respect to the scale). The weights are then increased step by Fig. 13 step and the corresponding twists are noted. The procedure is repeated by decreasing the weights in the same steps. A graph is then plotted between the weights mg and the corresponding mean twists φ. This is a straight line. Its slope gives mg/φ. The length l is measured directly by a metre scale. To find the radius r, the diameter of the wire is measured by a screw gauge at a number of places in two mutually perpendicular direction and the mean value is calculated. The diameter D is measured by vernier callipers. η is calculated from the above expression.
mg
O
φ
Fig. 14
On the circular scale φ is read in degree. Hence its value is multiplied by π/180 to convert it into radian before substituting in the above expression.
2. Dynamical method (Maxwell’s vibrating needle method) In this method, the time period of torsional vibration of the suspended body is observed directly, its total mass kept the same throughout and yet its distribution ingeniously altered about the axis of suspension so as to bring about a known change in its moment of inertia about the axis and hence also a change in its time-period. The so called vibrating needle is a hollow tube of length a, into which can be fitted two hollow and two solid cylinders of equal length (a/4), one pair on either side, such that, placed end to end, they just completely fill the tube. The tube itself is suspended from a torsion head H by means of the wire under test, of length L and radius R and a small piece of mirror M attached to it to enable the torsional vibration of the tube to be observed by the telescope and scale method. The solid cylinders are first placed in the inner position, as shown in figure (i) and the time-period T1 of the loaded tube determined. The position of the solid and hollow cylinders are then interchanged (figure ii), and time-period T2 of the tube determined again.
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Mechanics
Since the total mass suspended from the wire remains the same, there is no question
πηR4 . 2L If I1 and I2 be the moments of inertia of the loaded tube about the wire as axis in the two cases respectively, we have of any possible change in its torsional rigidity (or restoring couple per unit twist) C =
I1 I and T2 = 2π 2 C C
T 1 = 2π
F 4 π I bI GH C JK 2
whence
T22 − T12 =
H
S
− I1
2
g
...(1)
H
H
M
M
S
H
H
S
( i)
H
S
( ii)
Fig. 15
To obtain the value of (I2 – I1), we note that the c.g. of either inner cylinder lies at a
a/4 a 3a / 4 3a , or and that of each outer cylinder at a distance or from the axis. 2 8 2 8 So that, if m1 and m2 be the masses of a hollow and a solid cylinder respectively, the change from position (i) to (ii) merely means the shifting of a mass (m2 – m1) from a distance a/8 to a distance 3.a/8 from the axis on either side. Therefore, in accordance with the principle of parallel axes, we have distance
LMF 3a I − F a I OP MNGH 8 JK GH 8 JK PQ 2
I2 = I1 + 2( m2 − m1 )
b
I2 – I1 = m2 − m1
or
2
g a4
2
Substituting this value of I2 – I1 in expression (1) above, we get
F 4 π I bm GH C JK 2
T22
−
T12
=
2
− m1
g a4
2
πηR4 we get 2L a2 4 π2 × 2L 2 2 m − m 2 1 T2 − T1 = 4 πηR4
Substituting the value of C =
b
which gives
η =
2πLa2 (m2 − m1 ) (T22 − T12 ) R4
g
Mechanical Properties of Matter
411
R must be measured most accurately, in view of its 4th power being involved in the expression for η.
9.21 TORSIONAL OSCILLATIONS Let a body, say a disc, be hung by a long and thin vertical wire whose upper end is rigidly clamped. Let the disc be given a slight rotation in the horizontal plane by applying a couple (by hand). The wire is twisted and an equal and opposite elastic restoring couple is set up in it, when the twisting couple is withdrawn, the restoring couple is unbalanced and produces an angular acceleration in the disc in a direction opposite to that of the twist. The disc therefore returns to its mean position, crosses it (due to inertia) and rotates further in the opposite direction. The wire is now twisted in the opposite direction. A restoring couple is again set up in it which arrests its motion and makes it return. The whole phenomenon is then repeated. Thus the disc oscillates in the horizontal plane about the wire as axis. These oscillations are called ‘torsional oscillations’ and the system is called a “torsion pendulum”. To find the periodic time of a torsion pendulum by considering a disc, when the disc is oscillating the twist at the lower end of the wire at an instant t is φ, the restoring couple at thus instant will be – C φ, where C is the restoring couple per unit twist. The minus sign is put as the couple is directed opposite to the twist φ. Now, it I be the moment of inertia of the disc about the axis of oscillation, the couple acting upon it at this instant must be I(d2φ/dt2), where (d2φ/dt2) is the angular acceleration at this instant. ∴
–Cφ = I
d2φ dt2
d2φ C 2 = − φ = −ω φ I dt2
or where ∴
ω2 =
C constant I
d2φ ∝ φ dt2
l r
Fig. 16
Thus the angular acceleration is proportional to the twist or the angular displacement. Hence the motion is simple harmonic, so that its period is given by T = or
2π ω
T = 2π
I C
9.22 DETERMINATION OF MODULUS OF RIGIDITY OF A TORSIONAL OSCILLATIONS The upper end of the given wire is clamped to a rigid support, and a disc is attached to its lower end. The disc is slightly rotated by hand and left. The system begins to perform
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Mechanics
torsional oscillations. The period of these oscillations is obtained by timing 50 (say) oscillations. The period is given by T = 2π
I C
...(i)
Where I is the moment of inertia of the disc about the axis of oscillation and C is the torsional constant of the wire. Now, a regular auxiliary body is placed centrally upon the disc and the period of the system is again obtained. Let it be T1. Then, T1 = 2π
I + I1 C
...(ii)
Where I1 is the moment of inertia of auxiliary body about the axis of oscillation. Squaring and subtracting equation (i) from equation (ii) we get T12 − T2 =
4 π2 I1 C
πηr4 , where l is the length, r is the radius and η is the modulus of rigidity of 2l the material of the wire, But C =
∴
T12 − T2 =
η =
or
4 π2 (2l) I1 πηr4
e
8 πI1l
j
...(iii) − T2 r4 The length l of the wire is measured by a meter scale. To obtain r, the diameter of the wire is measured with a screw gauge at several points of the wire in two perpendicular directions and the mean value is calculated. I1 is calculated from the mass and dimensions of the auxiliary body. Finally, η is calculated from the last expression.
T12
Defects: (a) The expression (iii) is obtained on the assumption that C remains constant. Actually it suffers a small change when the auxiliary body is placed upto the disc. (b) In calculating the moment of inertia of the auxiliary body from its mass and dimensions, it is supposed to be of uniform density throughout, which is not always the case.
9.23 BEAM A rod or a bar of a circular or rectangular cross section, with its length very much greater than its thickness (so that there are no shearing stresses over any section of it) is called a beam. Now, a beam may just rest on a support, like a knife-edge or have a small part of it firmly clamped or built into a wall at either end. In the former case, it is called a supported beam and in the latter, a built-in or on oncastre beam or usually, simply, a fixed beam. In a supported beam, obviously, the support can merely exert a force on the beam but in a fixed beam, it can also exert a couple on it. If the beam be fixed only at one end and loaded at the other, it is called cantilever.
Mechanical Properties of Matter
413
Longitudinal Filament: A rectangular beam may be supposed to be made up of a number of thin layers placed in contact and parallel to one another. Similarly, a cylindrical beam may be supposed to be made up of thin cylindrical layers placed in contact and coaxial to one another. Further, each layer may be considered as a collection of thin fibres lying parallel to the length of the beam. These fibres are called ‘longitudinal filaments’ of the beam. Neutral Surface: When a beam is clamped horizontally at one end and loaded at the other, it undergoes bending. The filament (or fibres) of outward side are lengthened and subjected to tension, while those of the inner side are shortened and compressed. In between these two portions, there is a layer or surface in which the filaments are neither elongated nor shortened. Such a surface is called neutral surface. Neutral axis: The line of intersection of the plane of bending with the neutral surface (both are perpendicular to each other) is called the neutral axis.
9.24 BENDING MOMENT When a horizontal beam is fixed at one end and loaded at the other, a bending is produced due to the moment of the load. In this position the beam remains in equilibrium, provided the limit of elasticity has not been exceeded. W A E N D F
B N′ C
W
Fig. 17
Let ABCD represent a section of the beam fixed rigidly in the wall at AD with the other end BC, loaded with a weight W, as shown in figure. Such a beam is called cantilever. Let us consider the equilibrium of a portion BCFE of the beam, cut by a transverse plane EF across it. A force W acts downwards at the end BC, hence an equal and opposite reaction force equal to it is acting vertically towards along EF. These two equal and opposite forces constitute a couple. This clockwise couple which bends the beam, is called bending couple and the moment of this couple is called the bending moment. But the beam is in equilibrium, therefore, there must also be an equal and opposite couple to balance it. The only other forces which act on EBCF are obviously the forces due to stretching and compression of the filaments of the beam passing across EF. The filaments in the upper portion of the beam are elongated and thus, they must be in tension, the portion of the filaments to the left of EF exerts a pulling force on the portion to the right of EF, similarly a portion of the filaments in the lower half of the beam, to the left of EF, exerts a pushing force on the portion of the right of EF, because the filaments below the neutral surface are shortened. The distribution of these forces at the section EF has been shown in above figure and their resultant is a couple which balances the bending couple.
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Mechanics
The moment of this balancing couple (due to tensile and compressive stresses) is called the moment of the resistance to bending. Since in the position of equilibrium it is equal and opposite to the bending couple, it is referred as bending moment of the beam.
Expression for bending moment Let us consider a small portion of the beam bounded by two transverse sections AB and CD close to each other. After bending as in 18(b), AC is elongated to A′C′ and BD is shortened to B′D′. Line EF represents the neutral surface, which is neither stretched nor shortened. Let the small portion, under consideration, be bent in the form of a circular arc, subtending an angle θ at the center of curvature O. Let R be the radius of curvature of the neutral surface E′F′. Consider a filament GH distant z from EF in the unbent position of the beam. After bending its position is shown by G′H′. Now G′H′ = (R + Z)θ Before bending
GH = EF E′F′ = EF (for neutral surface) GH = E′F′
But
E′F′ = Rθ A G
C Z
E
H
A′ G′ E′
F
B
C′ H′
Z
F′ B′
D′
D
θ
R
(a ) O (b )
Fig. 18
Change in the length of the filament = G′H′ – GH = (R + Z)θ – Rθ = Zθ
Strain =
Change in length G ′ H′ − GH Zθ Z = = = Initial length GH Rθ R
If PQRS represents a section of the beam at right angles to its length and the plane of bending. In figure 19 then, clearly, the forces acting on the filaments are perpendicular to this section. The line MN lies on the neutral surface. Let the breadth of the section be PQ = b and its depth, QR = d. The forces, producing elongations and contractions in filaments, act perpendicular to the upper and lower halves, PQMN and MNRS respectively, of the section PQRS and their directions being opposite to each other. Now, consider a small area δa of the section PQRS about a point A, distant Z from the neutral surface. The strain produced in a filament passing through this area will be Z/R, as shown above.
Mechanical Properties of Matter
415
Now Y = Stress/strain ∴
Stress = Y × strain
Hence, stress about the point A = Y × Z/R Where Y is Young’s modulus for the material of the beam, b Q
P A O δa Z
M
N
S
d
R
Fig. 19
Force on the area δa = Y (Z/R) δa Moment of this force about the line MN = Y.(z/R) . δa.z =
Y. δa. z2 R
But, the moments of the moments acting on both the upper and the lower halves of the section are in the same direction, therefore, the total amount of the moments acting on the filaments in the section PQRS is given by
∑
Y Y. δa. z2 = R R
∑ δa. z
2
=
YI R
Where I = Σδa.Z2 and is called the second moment (or geometrical moment of inertia) of the sectional area about MN and therefore, equal to aK2, where a is the whole area of the surface PQRS and K its radius of gyration about MN. This quantity YI/R is the restoring couple or the bending moment of the beam, being equal and opposite to the moment of bending couple due to the load in the position of equilibrium. Thus Bending moment =
YI R
The quantity YI is known as flexural rigidity. In case of a beam of rectangular cross-section of breadth b and depth d, the area of crosssection is bd and (radius of gyration)2 in equal to d2/12. Hence I = area of cross-section . d2/12 = bd.d2/12 = bd3/12 Similarly, in the case of cylindrical rod of radius r the area of cross-section is πr2 and the square of radius of gyration of the cross-section about an axis through the center of the section and perpendicular to the plane of bending is r2/4. Thus I = πr2.r2/4 = πr4/4 In case, the rod is hollow, having r1 and r2 as the internal and external radii, then I =
e
π 4 r2 − r14 4
j
416
Mechanics
9.25 CANTILEVER A horizontal beam, fixed at one end and loaded at the free end, is called a cantilever. Let a beam of length l be fixed at its one end A and loaded with a load W at B. Let the end B be deflected to the position B′ under the action of the load W. Suppose the axis of X lies horizontally in the direction of the unbent beam and the axis of Y, vertically downwards, consider a section of the beam, as at P, at a distance x from the end A. Then the moment of the bending couple due to the load W. = W (l – x) The cantilever is in equilibrium in the bent position, hence the bending couple must be balanced at the section under consideration by the restoring couple or bending moment. The bending moment =
YI R
Where Y is the young’s modulus of the material of the beam, I geometrical moment of inertia of the section and R the radius of curvature of the neutral axis at the section. YI = W (l – x) R
Thus,
...(i)
l ( l– x)
x A
B
y P
δ B′
W
Fig. 20
The radius of curvature R is given by the equation
|RS1 + FG dyIJ |UV |T H dxK |W
3 2 2
R =
d2 y dx2 Where y is the depression of the beam at a distance x from the fixed end. Since y is small,
FG dyIJ H dxK
2
can be neglected in comparison to l. Hence, we get
R =
1 d2 y / dx2
Substituting for R in equation (i) above, we have YI.
d2 y = W(l – x) dx2
Mechanical Properties of Matter
417
d2 y W (l − x) 2 = YI dx
or
...(ii)
This equation may now be solved to get the value of the depression of the free end i.e., at x = l. Integrating equation (ii), we get
F GH
I JK
x2 W dy lx − + C1 = YI 2 dx
...(iii)
Where C1 is constant of integration, which can be known from the conditions of the problem. One such condition is at x = 0,
dy = 0 dx Applying this condition in equation (iii), we have C1 = 0
F GH
I JK
x2 W dy lx − = ...(iv) and thus YI 2 dx It is to be pointed out that at P, the slope of the curve tan θ = dθ/dx, θ being the angle between x-axis and tangent at P of the bent portion APB′. Thus
F GH
dy x2 W lx − = tanθ = dx YI 2
Integrating once again y =
F GH
I JK
I JK
W lx2 x3 − + C2 YI 2 6
Where C2 is another constant of integration. But at x = 0 , y = 0, hence substituting these values in Eq. (v), we get C2 = 0 Thus, the expression for the depression y at a distance x from the fixed end is y =
F GH
x3 W lx2 − YI 2 6
At the free end x = l and let y = δ, then
F GH
W l 3 l3 − δ = YI 2 6 δ =
or
I JK
I JK
Wl3 3YI
For a beam of rectangular cross-section of breadth b and depth d, I = bd3/12 Therefore,
δ =
4 Wl3 Ybd3
...(v)
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Mechanics
For a beam of circular cross-section of radius r, I = πr4/4 δ =
4 W l3 3πr 4 Y
9.26 BEAM SUPPORTED AT ITS ENDS AND LOADED IN THE MIDDLE Consider a beam supported on two knifes edges K1 and K2 at a distance l apart in a horizontal plane. Let it be loaded at its middle point O. The reaction of each knife-edge is obviously equal to W/2 acting upwards. The beam bends as shown in the figure and the depression is maximum in the middle. The middle part of the beam is almost horizontal. Hence the beam may be considered as equivalent to two-inverted cantilever OA and OB. The length of each cantilever will be l/2, clamped at the end O and carrying an upward load W/2 at the other. Therefore, the depression of O below the knife edges in the elevation of the loaded end of such a cantilever from the lowest position O. Let us consider a vertical section at P, distant x from O and consider the part PB as shown in Fig. 22. The moment of the deflecting couple is =
W W . PB = 2 2
FG l − xIJ H2 K
In the equilibrium position this deflection couple must be balanced at the section under consideration by the restoring couple or bending moment, established inside the beam, i.e., W 2
FG l − xIJ H2 K
=
YI R
...(i)
If y is the elevation of the section P above x-axis, then radius of curvature of the neutral axis at this section is given by R =
1 d2 y dx2
W 2
W 2
l A
B O K1
K2 W
Fig. 21
Hence Eq. (i) is W 2
FG l − xIJ H2 K
= YI
d2 y dx2
Mechanical Properties of Matter
419
FG H
IJ K
W l d2 y − x 2 = 2YI 2 dx
or
Integrating this expression, we have
F GH
I JK
x2 W l dy x − + C1 = 2 2YI 2 dx Where C1 is the constant of integration. At O, x = 0 and dy/dx = 0, hence 0 = 0 + C1, or C1 = 0 ∴
F GH
x2 W l dy x − = 2YI 2 2 dx
I JK
W 2
Y
Integrating further, we have
W Fl x 2YI GH 2 2
2
y =
x I − +C 6 JK
B
3
P
2
δ
Where C2 is the constant of integration. Again, at 0 , x = 0 and y = 0 ∴ C2 = 0 ∴
y =
F GH
x3 W l x2 − 2YI 2 2 6
y
I JK
O
l 2
Now, at the free end, x = l/2 and y = δ (say). Therefore
F GH
l3 W l l2 − δ = 2YI 2 8 48 or
δ =
I JK
W 2
x
Fig. 22
Wl3 48 YI
If ‘b’ and ‘d’ be the breadth and thickness of beam respectively, then I = bd3/12 δ =
Wl3 4 Ybd3
...(ii)
If the cross-section of the beam be a circle of radius r, then δ =
Wl3 12 Yπr4
This is the same as the depression at the middle point O. If M be the mass with which the beam has been loaded in the middle, then Eq. (ii) is δ =
Wl3 Mgl3 = 3 4 Ybd 4 Ybd3
The use of the equation is made in the construction is girders.
420
Mechanics
The depression of a beam at the middle is directly proportional as the cube of its length and inversely as the width and cube of the depth or thickness. This, in order that the girder may not bend appreciably, its span, i.e., length is made small and its breadth and depth large. At the same time Young’s modulus for the material of the beam must be larger. When a girder is supported at its ends, its filaments above the neutral surface are compressed and those below the neutral surface (or in the lower half) are elongated. The compression or extension of the filaments is maximum at the upper or lower faces of the girder respectively; hence the stresses are maximum at these faces and they decrease, as we move towards the neutral surface from either side. Consequently, the upper and lower faces of the beam should be much stronger than its middle position. In other words, the middle positions may be made of much smaller breadth than the upper and the lower faces. This is done by manufacturing the girders with their section in the form of I so that breadth at the upper and lower faces may be sufficiently larger than that of the inner parts. This saves quite a good amount of the material without sacrificing the strength of the girder. Stiffness of a beam: The stiffness of a beam is defined as the ratio of maximum depression of the beam to its span.
9.27 DETERMINATION OF YOUNG’S MODULUS BY BENDING OF A BEAM In the last section it is shown that the depression of a beam of rectangular cross-section, supported at the ends and loaded in the middle, is given by
or
δ =
Mgl3 4 Ybd3
Y =
Mgl3 4 bd3δ Y
B a tte ry S
δ G
M
X
K2 K1
W
Fig. 23
Thus, to determine the young’s modulus of the material of a beam, the beam is supported horizontally and symmetrically on two knife edges, K1 and K2, fixed at a distance l apart, as shown in figure. A hanger, supported by a knife edge is then placed on the middle of the beam and the load is applied by placing the weight on it. The depression δ of the beam, so produced, is determined by a micrometer screw. At the moment, when micrometer screw just touches the beam, a current starts flowing in the galvanometer G, from the battery, causing to deflect the needle of the former and the reading is noted. A series of loads are applied and observations
Mechanical Properties of Matter
421
are taken for load increasing and decreasing. A graph is then plotted between the load of mass M as abscissa and the corresponding mean depression as the ordinate. The slope of the straight line, so obtained, gives the values δ/M. Now
Y =
gl3 M Mgl3 . = 4 bd3δ 4 bd3 δ
Thus, knowing the values of l, b and d, Young’s Modulus of the material of the beam may be determined.
9.28 DETERMINATION OF ELASTIC CONSTANTS BY SEARLE’S METHOD The elastic constants of the material of a thin short wire can be determined by this method. The apparatus consists of two exactly equal metal rods AB and CD of square or circular cross-section. They are connected together at their middle points by the specimen wire and suspended by silk threads from a rigid support such that the rods and wire lie in a horizontal plane. When the rods, so suspended, are brought near to each other through equal distances and then released, they execute torsional vibrations in a horizontal plane, due to the couple exerted by the specimen wire on the two rods. Young’s Modulus: In the Fig. 25 let θ be the angle through which each rod is turned from its normal position. Then the angle subtended by the bent wire at the centre of curvature O is 2θ and obviously l = R.2θ Where R is the radius of curvature of the bend wire. From the theory of bending of beam, moment of the internal couple or bending moment =
YI g R
Where Ig is the geometrical moment of inertia of the cross-section of the wire and Y the Young’s Modulus of the material of the wire. Moment of the couple = YIg 2θ/l For a wire of radius r, Ig = ∴
πr4 4 B
πr4 Y .θ Moment of the couple = 2l
Now this couple produces an angular acceleration d2θ/dt2 in each rod. If I is the moment of inertia of either rod AB or CD about the thread from which it is suspended, then the moment of the couple is Id2θ/dt2. Hence, the equation of motion is Id2θ πr4 Y .θ = 0 + 2l dt2
D
C
A
Fig. 24
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Mechanics
d2θ πr4 Y .θ = 0 + 2l I dt2
or
D
B R θ
A
C
O
Fig. 25
Therefore, the above equation represents a simple harmonic motion of period. T 1 = 2π
2lI 8 πl I whence Y = 2 4 4 πr Y T1 r
….(i)
Modulus of rigidity: Now the suspension threads are removed. One rod say CD, is clamped horizontally to a rigid support and the other is given a slight rotations so as to twist the specimen wire. When released, it executes torsional vibrations about the wire as axis, in a horizontal plane. Its time period T2 is determined. T2 = 2π
Then
∴
η =
I I.2l = 2π , C ηπr4
8 πl I T22 r4
C=
ηπr4 2l ….(ii)
Poisson’s Ratio: From Eqs. (i) and (ii), we have
Y T2 = 22 η T1 But Poisson’s ratio
σ =
2 Y − 1 ∴ σ = T2 – 1 2η 2T12
...(iii)
Thus, by substituting the values of various experimentally determined quantities in Eqs. (i), (ii) and (iii), we can find the values of elastic constants Y, η and σ. This method has the following advantages: 1.
This method requires only a small specimen of the material in the form of a wire.
2.
Young’s Modulus of the material of a wire is measured by measuring the periodic time and not by observing any deflection.
3.
Here, for determining the Poisson’s ratio, the radius of the wire is not required the measurement of the radius of the wire always introduces some error.
Mechanical Properties of Matter
423
NUMERICALS Q. 1. A copper rod of length 1 m and cross-section 3 cm2 is fastened end to end to a steel rod of length L and cross-section 1 cm2. The compound rod is subjected to equal and opposite pulls of magnitude 3kg-wt at its ends. (a) Find the length L of the steel rod if the elongations of the two rods are equal. (b) What is the stress in each rod ? (c) What is the strain in each rod ? Ycopper = 1 × 1012 dyne/cm2, Ysteel = 2 × 1012 dyne/cm2.
FG Y = stress IJ we have, H strain K
Solution. (a) By Hooke’s law
strain =
stress Y
In usual notations, l F/a = Y L
or
l =
FL Ya
The elongation, l, are equal and the pulls, F, are equal so that
Lcopper Ycopper × acopper
Lsteel Ysteel × asteel
=
Putting the given values, we get 12
(1 × 10
1m Lsteel = 2 2 12 (2 × 10 dyne / cm2 ) (1 cm2 ) dyne / cm ) (3 cm )
or
Lsteel =
2 m. 3
(b) The stress in the copper rod is F acopper
and in steel rod is
F asteel
=
3 × 9. 8 nt = 9. 8 × 104 nt / m2 3 × 10−4 m2
=
3 × 9. 8 nt = 29. 4 × 104 nt / m2 . 1 × 10−4 m2
(c) The strain in the steel rod is
29. 4 × 104 nt / m2 stress = 14. 7 × 10−7 = 2 × 1011 nt / m2 Ysteel and the strain in the copper rod is
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Mechanics
stress 9. 8 × 104 nt / m2 = = 9. 8 × 10−7 . Ycopper 1 × 1011 nt / m2 Q. 2. (a) Show that the potential energy per unit volume of a strained wire is
1 (stress 2
× strain). (b) Calculate the work done in stretching a uniform wire of cross-section 10–6 m2 and length 1.5 m through 4 × 10–3 m. The Young’s modulus is 2 × 1011 nt/m2. Solution. (a) Energy in the strained wire. Suppose a wire of length l and area of crosssection A is stretched through a distance x by applying a force F along its length. Then tensile stress =
F x and tensile strain = A l
Therefore the Young’s modulus of its material is Y =
stress F/A = x /l strain
Hence the force F required to stretch it through x is F =
YA x l
If the wire is now stretched a further small distance dx, the work done will be
YA x l Hence the total work done to stretch the wire from its original length to l + x (i.e., from x = 0 to x = x) will be dW = F dx =
z x
W =
0
2
=
LM x OP N2Q 1 L Yx OP LM x OP = ( A l) M 2 N l Q NlQ
YA YA x dx = l l
x 1 YA l 2
2
x
0
1 × volume × stress × strain. 2 1 Hence the work per unit volume is (stress × strain). This work is stored as the (potential) 2 strain energy in the wire. =
(b) For the given wire, we have l = 1.5 m, A = 10–6 m2, x = 4 × 10–3 m and Y = 2 × 1011 nt/m2 From this, volume of the wire = A l = 10–6 × 1.5 m3 tensile strain =
x 4 × 10−3 = l 1. 5
Mechanical Properties of Matter
425
F GH
−3 tensile stress = Y × strain = (2 × 1011 ) × 4 × 10 1. 5
and
I nt / m JK
2
Hence, the work done W =
=
1 × volume × stress × strain. 2
FG H
IJ FG K H
1 2 × 1011 × 4 × 10−3 4 × 10−3 (10−6 × 1. 5 m3 ) × nt / m2 × 2 1. 5 1. 5
IJ K
= 1.07 Joule. Q. 3. Young’s modulus Y for brass is 9 × 1010 nt/m2. Determine the energy stored in a brass rod of 2 cm2 cross-section, 0.01 meter in length if compressed by a force of 10 Newton. Solution. The work done in compressing the rod is given by W = Hence,
1 × volume × stress × strain 2
volume of rod = length × cross-section = 0.01 m × 2 × 10–4 m2 = 2 × 10–6 m3 stress = =
and
compression force cross - section 10 nt = 5 × 104 nt / m2 (2 × 10−4 m2 )
stress 5 × 104 = Y 9 × 1010 5 × 10−6 = 9 1 5 × (2 × 10−6 m3 ) × ( 5 × 104 nt /m2 ) × × 10−6 W = 2 9 = 2.78 × 10–3 Joule.
strain =
∴
FG H
IJ K
Q. 4. A steel wire of 2mm diameter is just stretched between two fixed points at a temperature of 20°C. Determine the tension when the temperature falls to 10°C coefficient of linear expansion of steel is 0.000011 per°C and Young’s Modulus for steel is 2.1 × 1011 newton/ meter2. Solution. Let l be the length of the wire and a its area of cross-section when the temperature falls, the wire contracts. The contraction (decrease in length) is dl = l α dθ where α is the coefficient of linear expansion and dθ is the fall in temperature. The strain in the wire is
dl = α dθ l
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Mechanics
If Y be the young’s modulus of the material of wire the stress is given by stress = young’s modulus × strain = Y α dθ The force (tension developed) in the wire is F = stress × area of cross-section = Y α dθ × a Substituting the given values: F = (2.1 × 1011 nt/m2) × (0.000011/°C) × (10°C) × 3.14 × (1 × 10–3 m2) = 72.5 nt. Q. 5. A material has Poisson’s ratio 0.20. If a uniform rod of it suffers longitudinal strain 4.0 × 10 – 3, deduce the percentage change in its volume. Solution.
Poissons ratio σ =
Here, and
lateral strain ( ∆r / r) longitudinal strain ( ∆l /l)
∆l = 4.0 × 10–3 l
∆r ∆l = 0. 20 × (4. 0 × 10−3 ) = 0. 80 × 10−3 = σ× r l Let V be the initial volume, and V + ∆V the final volume. Then V = πr2l V + ∆V = π (r – ∆r)2 (l + ∆l)
and
...(i) ...(ii)
substracting Eq. (i) from (ii) and neglecting smaller terms, we get ∆V = πr2∆l – 2πrl∆r ∴
∆V ∆l ∆r −2 = V l r
substituting the values:
∆V = 4.0 × 10–3 – 2(0.80 × 10–3) = 2.4 × 10–3 V ∴ percentage change in volume ∆V × 100 = (2.4 × 10–3) × 100 V = 0.24 Q. 6. Calculate Y for rubber if a rubber tube 0.4 meter long, whose external and internal diameters are 0.01 meter and 0.004 meter respectively, extends 0.0006 meter when stretched by a force of 5 kgm-wt. Solution. If r1 and r2 be the initial and external radii of the tube, then the area of crosssection of the tube is
Mechanical Properties of Matter
427
A = π ( r22 − r12 ) = 3.14 (0. 005 m)2 − (0. 002 m)2 = 3.14 × 0.000021 m2 = 6.6 × 10–5 m2 The stretching force is F = 5 kg-wt = 5 × 9.8 = 49 nt ∴
and
Stress =
F 49 = = 7. 4 × 105 nt /m2 A 6. 6 × 10−5
δl 0. 0006 = = 0. 0015 l 0. 4 m The youngs modulus of rubber is therefore Strain =
Y =
stress 7. 4 × 105 = = 4 . 9 × 108 nt/m2 strain 0. 0015
Q. 7. An iron wire of length 1 mts and radius 0.5 mm elongates by 0.32 mm when stretched by a force of 5 kg-wt, and twists through 0.4 radian when equal and opposite torques of 3 × 104 dyne-cm are applied at its ends. Calculate elastic constants for iron. Solution. Let a wire of length l and radius r be elongated by x with force F. By Hooke’s law, the young’s modulus of the material is given by Y = =
stress strain F / πr2 Fl = 2 x/l πr x
Here F = 5000 × 981 dyne, r = 0.5 mm = 0.05 cm, x = 0.32 mm = 0.032 cm l = 1 mts = 100 cm ∴
Y =
5000 × 981 × 100 = 19. 5 × 1011 dyne/cm2 3.14 × (0. 05)2 × (0. 032)
Now the torque necessary to produce a twist of φ radian in the wire is given by τ = ⇒
η =
π ηr4 φ 2l τ (2l) πr4 φ
Here τ = 3 × 104 dyne-cm, l = 100 cm, r = 0.05 cm and φ = 0.4 radian ∴
η =
3 × 104 × 200 3.14 × (0. 05)2 × 0. 4
= 7.64 × 1011 dyne/cm2.
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Mechanics
Now, the Poisson’s ratio σ, and Y and η are related as σ =
=
Y −1 2η 19. 5 × 1011 −1 2 × 7. 64 × 1011
= 0.276 Q. 8. Calculate Poisson’s ratio for silver: Given its Young’s modulus = 7.25 × 1010 nt/m2 and bulk modulus = 11 × 1010 nt/m2. Solution.
Y = 3K ( 1 – 2σ)
FG H
1 Y 1− 2 3K
σ =
or
IJ K
Putting the given values: σ =
FG H
1 7. 25 × 1010 1− 2 3 × 11 × 1010
IJ K
= 0.39 Q. 9. A wire of length 1 meter and diameter 2mm is clamped at one of its ends. Calculate the couple required to twist the other end by 45° (η = 5 × 1011 dyne/cm2). Solution. The couple required to twist the free end of a clamped wire of length l through φ radian is τ = For φ = 45° =
πηr4 φ 2l
π π2ηr4 radian, we have τ = 4 8l
Here l = 1 meter = 100 cm, r = 1.0 mm = 0.1 cm and η = 5 × 1011 dyne/cm2 ∴
τ =
( 3.14)2 × ( 5 × 1011 ) × ( 0.1)4 8 × 100
= 6.2 × 105 dyne-cm. Q. 10. A wire of radius 1 mm and length 2 meter is twisted through 90°. Calculate the angle of shear at the surface, at the axis of the wire and at a point mid-way. If the modulus of rigidity of material is 5.0 × 1011 dyne/cm2, what is the torsional couple ? Solution. Let l be the length of the wire, r the radius and φ the twist at the free end. Then the angle of the shear at a radial distance x from the axis of the wire is θ =
xφ l
At the axis of the wire,
x = 0, so that θ = 0.
At the surface of the wire,
x = r, so that
Mechanical Properties of Matter
429
θ =
rφ l
Here, r = 1mm = 0.1 cm, l = 200 cm and φ = 90°
0.1 × 90° = 0. 045° 200
∴
θ =
At a point mid-way,
x = r/2
∴
θ =
1 × 0. 045° = 0. 0225° 2
τ =
πηr4 φ 2l
The torsional couple is
φ = 90° = π/2 radian, we have
For
τ =
π2ηr4 4l
τ =
( 3.14)2 × ( 5. 0 × 1011 ) × ( 0.1)4 4 × 200
Substituting the given values:
= 6.2 × 105 dyne-cm Q. 11. Two wires A and B of the same material and equal lengths but of radii r and 2r are soldered coaxially. The free end of B is twisted by an angle φ. Find the ratio of the relative angles of the twist between the ends of the two wires, the twist at the junction, and the torsional rigidity of the complete wire. Solution. Let τ be the couple applied at the free end of B and φ′ the twist at the junction. The couple produce a relative twist φ′ between the ends A and a relative twist (φ − φ′) between the ends of B. Therefore, if η be modulus of rigidity of the material of the wires and l the length of each wire, we have
πηr4 φ′ 2l πη (2r)4 (φ − φ′ ) τ = 2l and comparing Eqs. (1) and (2) we get τ =
...(1) ...(2)
φ′ = 16 (φ – φ′) ⇒ Also, solving we get,
φ′ = 16 φ − φ′ φ′ =
16 φ 17
Now, substituting this value of φ′ in the expression τ =
πηr4 φ′ 2l
430
Mechanics
we get,
τ = =
πηr4 2l
FG 16 φIJ H 17 K
8 πηr4 φ 17l
∴ The torsional rigidity of the complete wire is c =
τ 8 πηr4 = φ 17l
Q. 12. The solid cylindrical rods of same material having lengths l and 2l and radii 2r and r respectively are soldered coaxially. The free end is twisted through 11°, the other end being clamped. Calculate the twist of the thicker rod. Solution. (1/3°). Q. 13. Two solid cylinders of the same material having equal lengths l and 2l and radii r and 2r are joined coaxially. Under a couple applied between the free ends the shorter cylinder shows a twist of 30°. Calculate the twist of the longer cylinder. Solution. Let τ be the couple, and let it produced a twist φ in the shorter cylinder and twist φ′ in the longer cylinder. Then,
πηr4 πη (2r)4 φ= φ 2l 2 (2l) φ 30° = = 3. 75° φ′ = 8 8 Q. 14. (a) Find an expression for the elastic (potential) energy stored (or work done) in a cylinder whose one end is fixed and the other end is twisted through an angle φ. τ =
(b) Find the work done in twisting a steel of radius 1 mm and length 0.25 mts through an angle of 45°. The modulus of rigidity of steel is 8 × 1010 nt/m2. Solution. (a) Let c be the couple per unit twist in the wire. Thus couple for a twist φ is cφ. The work done for additional small twist dφ is cφ dφ. Hence the work done for producing a total twist φ will be
z φ
W =
cφ dφ =
0
Now, ∴
c =
W =
1 2 cφ 2
πηr4 2l πηr4φ2 4l
This is also the elastic potential energy stored in strained wire. (b) For the given wire, we have, η = 8 × 1010 nt/m2, r = 1 mm = 10–3 m, l = 0.25 m and φ = 45° =
π radians 4
Mechanical Properties of Matter
∴
431
FG H
3.14 × (8 × 1010 ) × (10−3 )4 3.14 × W = 4 × 0. 25 4 = 0.155 Joule.
IJ K
2
Q. 15. A uniform tube of outer radius 5 units and inner radius 4 units is twisted by angle 20° under a couple c. If a solid cylindrical rod of same material and same length, but of radius 3 units, is placed under the same couple c what would be the angle of twist ? Solution. The torsional rigidities c and c′ of the solid and hollow cylinders are given by
and
c =
πηr4 2l
c =
πη ( r24 − r14 ) 2l
...(i)
where r → radius of solid cylinder r1 and r2 → external and internal radii of the hollow cylinder But
r 2 = 2r1 (given), so that c′ =
πη (15 r14 ) 2l
...(ii)
From eqs. (i) and (ii), we have r4 c = 15r4 c′ 1
...(iii)
The masses of two shafts are equal, so that
πr2lρ = π ( r22 − r12 ) lρ = π ( 3r12 ) lρ where ρ is the density of the material of the cylinders. This gives r 2 = 3r12 r =
3 r1
...(iv)
Making this substitution in eq. (iii) we get
3 r4 c = 15 r4 = 5 c′ 1 Now if φ be the twist produced at the free end of a solid cylinder under twisting torque τ, then τ =
πηr4 φ 2l
If θ be the (maximum) shearing strain at the surface, then θ =
rφ l
432
Mechanics
⇒
φ =
lθ r
τ =
πηr4 2l
Thus we have
FG lθ IJ = πηr θ H rK 2 3
...(v)
If θ′ be the maximum strain on the hollow cylinder for the same torque τ, then
FG lθ′ IJ Hr K πη (15 r ) F lθ′ I GH 2r JK 2l πη ( r24 − r14 ) 2l
τ =
2
4 1
=
1
πη (15 r13 θ ′ ) 4
=
...(vi)
Comparing (v) and (vi) we obtain, πη (15 r13 ) θ ′ πη r3θ = 2 4
θ 15 r13 = θ′ 2 r3
or But r = 3 r1 from eq. (iv) ∴
θ 15 r13 = θ′ 2 × 3 3 r13 5 5 = = 2 3 2 × 1. 732
=
1. 44 1
Q. 17. A cylinder of radius r and length l is recast into a pipe of same length. If the pipe possesses torsional rigidity which is 19 times greater than that of the original cylinder, calculate the inner radius of the pipe. Solution. Let c and c′ be the torsional rigidity of the solid and hollow cylinders respectively. Then,
πη ( r24 − r14 ) πηr4 and c′ = 2l 2l where r is the radius of solid cylinder and r1 and r2 the internal and external radii of the c =
hollow cylinder. Since
c′ = 19, we have c r24 − r14 = 19r4
Mechanical Properties of Matter
433
( r12 + r22 ) ( r22 − r12 ) = 19r4
or
...(i)
since the length and mass remain unchanged, we have π ( r22 − r12 ) lρ = πr2lρ r22 − r12 = r 2
or
...(ii)'
Making this substitution in eq. (i), we get r22 + r12 = 19r2
...(iii)
substracting (ii) from (iii) and solving, we get r 1 = 3r. Q. 18. A thin-walled circular tube of mean radius 10 cm and thickness 0.05 cm is melted up and recast into a solid rod of same length. Compare the torsional couples in the two cases. Solution. Internal radius of tube, r1 = 10 −
1 × 0. 05 = 9. 975 cm 2
1 × 0. 05 = 10. 025 cm 2 Let r be the radius of the recast solid rod of the same length, Since the mass remains unchanged, we have, and external radius of tube, r2 = 10 +
π ( r22 − r12 ) lρ = πr2 lρ
where ρ is the density of the material. Thus r 2 = r22 − r12 = (10. 025)2 − ( 9. 975)2 = 1. 000 r = 1.0 cm The torsional couple per unit twist for the hollow tube is rod is
πη ( r24 − r14 ) and that for solid 2l
πηr4 Therefore, their ratio is 2l (10. 025)4 − ( 9. 975)4 r24 − r14 = (1. 0)4 r4
= 200 : 1 Q. 19. A solid cylinder of radius 3 cm is melted and recart into 0 hollow cylinder of outer radius 5 cm, while length remain unchanged. By what ratio torsional rigidity change ? [Ans. It shall increase 4.5 times.] Q. 20. The restoring couple per unit twist in a solid cylinder of radius 5.0 cm is 1 × 108 dyne-cm. Find the restoring couple per unit twist in a hollow cylinder of the same material, mass and length but of internal radius 12.0 cm. Solution. The restoring couple per unit twist in a solid cylinder of length l and radius r is
434
Mechanics
πηr4 2l r = 5.0 cm and c = 1 × 108 dyne-cm c =
Here
1 × 108 c πη = 4 = ( 5. 0)4 r 2l If r1 and r2 be internal and external radii of the hollow cylinder of the same length, material and mass m, then
∴
m = π ( r22 − r12 ) lρ = πr2lρ
[ρ is density]
r22 − r12 = r 2
or
Here r1 = 12.0 cm and r = 5.0 cm ∴ ⇒
r 2 2 = r12 + r2 = ( 5. 0)2 + (12. 0)2 = (13. 0)2 r 2 = 13.0 cm
The restoring couple per unit twist for this hollow-cylinder is c′ =
πη ( r24 − r14 ) 2l
1 × 108 (13. 0)4 − (12. 0)4 (5. 0)4 = 1.25 × 109 dyne-cm. =
Q. 21. An inertia table is a uniform disc of mass M and radius r. Its period of oscillation is T0. Now a thin ring of mass 10 M and radius 0.8 r is placed symmetrically on the table. Deduce the new period of oscillation. How would the period further change if the length of suspension wire is reduced to one-third ? Solution. The period of oscillation of the inertia table is given by To = 2π
FG I IJ H cK
where I is the moment of inertia of the table about the axis of suspension and c the torsional contant of the suspension wire. Let I′ be the moment of inertia of the ring placed symmetrically on the table. The new period is T1 = 2π Now I =
I + I′ = c
I + I′ T0 I
1 Mr2 and I′ = (10 M) (0.8 r)2 = 6. 4 Mr2 2
F 0.5Mr + 6.4 Mr I ; T GH 0. 5Mr JK 2
∴
T1 =
2
0
2
=
FG 69 IJ T H 5K
0
Now let c′ be the torsional constant of the new wire. The period of oscillation becomes T2 = 2π I + I′ = c′
FG c IJ T H c′ K
1
Mechanical Properties of Matter
Now c =
435
1 πηr 4 πηr4 c and c′ = so that = 2l 2 ( l / 3) c′ 3
∴
T2 =
FG 1 IJ T H 3K
1
69 T 15 0
=
Q. 22. A steel wire of 1.0 mm radius is bent in the form of a circular arc of radius 50 cm. Calculate the bending moment and the maximum stress. (Y = 2.0 × 1012 dyne/cm2) Solution. The bending moment is
YI R
where I is the geometrical moment of inertia of the cross-section of the wire and R is the radius of circular arc For a wire of radius r, ∴
Bending moment,
I =
πr4 3.14 × (0.1 cm)4 and ρ = 50 cm = 4 4
(2. 0 × 1012 ) × 3.14 × ( 0.1)4 YI = 4 × 50 R
= 3.14 × 106 dyne-cm In bending of wire, the longitudinal stress on the cross-section of a filament at a distance Z from the neutral surface is
YZ R This is maximum when Z is maximum i.e., when Z = r ∴
Maximum stress =
Yr (2. 0 × 1012 dyne/cm2 ) × (0.1 cm) = R 50 cm
= 4.0 × 106 dyne/cm2 Q. 23. A cantilever of length 0.5 meter has a depression of 15 mm at its free end. Calculate the depression at a distance of 0.3 meter from the fixed end. Solution. The depression of the cantilever at a distance x from its fixed end is given by Y =
=
FG H
x3 W lx2 − yI 2 6
FG H
Wx2 x l− 2 yI 3
IJ K
IJ K ...(i)
and that at the free end (x = l) is δ =
Wl3 3YI
...(ii)
436
Mechanics
where l is the length of the cantilever, W is the load and Y is young’s modulus of the material. I is the geometrical moment of inertia. From eqs. (i) and (ii) we write y x2 / 2 (l − x / 3) = δ l3 / 3
or
y =
3 x2 ( l − x / 3) δ 2l 3
Here l = 0.5 meter, x = 0.3 meter and δ = 15 mm = 15 × 10–3 meter
FG H
∴
y =
IJ K
0. 3 × (15 × 10−3 ) 3 2 × (0. 5)3
3 × (0. 3)2 × 0. 5 −
= 6.48 × 10–3 meter Q. 24. One end of a 1.0 mts rod is clamped rigiditly. When a 0.5 kg load is suspended at the other end, it is depressed by 0.5 cm. Calculate the depression at 40 cm distance from the fixed end when a 5 kg load is suspended at (i) free end, (ii) 40 cm from the free end. Solution. The depression of the rod at a distance x from its fixed end is y =
Wx2 (l − x / 3) 2YI
...(i)
where W is the load suspended at the free end and l is the length of the rod. At the free end (x = l), the depression is δ =
Wl3 3YI
...(ii)
Here l = 100 cm, W = 0.5 kg-wt = 500 g dyne, δ = 0.5 cm 0.5 =
500 g × (100)3 3YI
...(iii)
(i) Now, when a 5 kg-wt is suspended at the free end (W = 500 g), the depressions at a distance of 40 cm (x = 40 cm) is, by eq. (i) given by
FG H
5000 g × (40)2 40 × 100 − y = 2YI 3 =
5000 g × (40)2 × 260/3 2YI
Dividing this eq. by eq. (iii), we get
y 5000 g × (40)2 × 260 = 0. 5 2 × 2000 g × (100)3 = 2.08
IJ K
Mechanical Properties of Matter
437
∴
y = 2.08 × 0.5 = 1.04 cm
(ii) When 5 kg-wt is suspended at 40 cm from the fixed end, then x = l = 40 cm Then by eq. (ii) we have, y =
5000 g × (40)3 3 YI
Dividing by eq. (iii) we get 5000 g × (40)3 y = 0. 64 cm = 500 g × (100)3 0. 5
∴
y = 0.64 × 0.5 = 0.32 cm.
Q. 25. The end of the given strip cantilever depresses 10 mm under a certain load. Calculate the depression under the same load for another cantilever of same material, 2 times in length, 2 times in width and 3 times in thickness (vertical). Solution. The depression δ at the end of a cantilever is δ =
Wl3 3YI
For a rectangular cantilever, I = bd3/12 ∴
δ =
4 Wl3 Ybd3
The depression for another cantilever of the same material (Y same) but length 2l, width 2d and thickness 3d would be δ′ = =
4 W (2l)3 Y (2b) (3 d)3 4 4 δ= × (10 mm) 27 27
= 1.48 mm Q. 26. The rectangular cross-section of a long rod (cantilever) has sides in the ratio 1:2. Calculate the ratio of depressions under the same load when (i) the smaller side is vertical, (ii) the longer side is vertical. Solution. The depression at the free end of a cantilever of rectangular cross-section under a load is δ =
4 Wl3 Ybd3
Where l is length, b is width and d is thickness of the cantilever rod. Let a and 2a be the sides (ratio 1:2) of the cross-section.
438
Mechanics
(i) when the smaller side is vertical, b = 2a, d = a ∴
δ1 =
4 Wl3 1 4 Wl3 = Y (2a) ( a)3 2 Ya4
(ii) when the longer side is vertical, b = a, d = 2a
4 Wl3 1 4 Wl3 = 3 8 Ya4 Y ( a) (2a)
∴
δ2 =
Thus,
δ1 1/2 4 = = 1/8 1 δ2
δ1 : δ2 = 4 : 1
or
Q. 27. If a cross-section of a beam of rectangular with breadth b and thickness d and if the depression of the free end for the given load are δ1 and δ2 respectively when b and d are vertical, show that
δ1 d 2 = 2. δ2 b
Q. 28. For the same cross-section areas, show that beam of a square cross-section is stiffer than one of circular cross-section of the same length and material, show also that for a given load the depressions are in the ratio 3:π . Solution. The depression δ at the loaded end of the beam whose other end is clamped horizontally is given by δ =
Wl3 3YI
Where W is load, l is length of beam, I is geometrical moment of inertia of its section. For a rectangular beam,
bd3 12 where b and d are breadth and depth of its section. If the beam has a square section b = d then I =
b4 12 Therefore, if δ1 be the depression for this beam, we have I =
δ1 =
Wl3
Fb I 3Y G J H 12 K 4
=
4 Wl3 Yb4
For a beam of circular cross-section of radius r, we have I =
πr4 4
Therefore, if δ2 be the depression for this beam, we have
...(i)
Mechanical Properties of Matter
439
δ2 =
Wl3
F πr I 3Y G H 4 JK
=
4
4 Wl3 3Yπr4
...(ii)
Dividing eq. (i) and (ii) we get,
δ1 3πr4 = δ2 b4 But the cross-section areas of two beam are equal ∴
b2 = πr2
∴
δ1 3π r 4 3 = = δ2 π ( πr2 )2
which is less than 1 ∴
δ1 < δ2.
Thus the depression at the loaded end of the beam of square cross-section is smaller than that of the beam of circular cross-section. Hence the former is stiffer than the later. Q. 29. Compare loads required to produce equal depressions for two beams of same material, length and weight when one has a square cross-section and the other has a circular cross-section. [Ans. π:3] Q. 30. A solid cylindrical rod of radius 6 mm bends by 8 mm under a certain load. Calculate the bending if the rod is replaced by a hollow cylindrical rod of the same material, mass and length but of internal radius 8 mm. Solution. If r1 and r2 be the internal and external radii of a same hollow cylinder of the same material (density ρ), same length l and same mass m as of a solid cylinder of radius r, then we have, 2 2 2 m = π ( r2 − r1 ) lρ = πr lρ
r22 − r12 = r 2
or
Here r = 6 mm, r1 = 8 mm so that r 2 2 = r2 + r12 = (6)2 + (8)2 = 100 ∴
r 2 = 10 mm
Let δ1 and δ2 be the depression for the solid and hollow cylinders. Then δ1 =
and
Wl3 = 3YI
Wl3
F πr I GH 4 JK 4
3Y Wl3
δ2 =
3Y
FG π (r H4
4 2
− r14 )
IJ K
440
Mechanics
∴
r4 δ2 = ( r22 + r12 ) ( r22 − r12 ) δ1
Here r = 6 mm, r1 = 8 mm and r2 = 10 mm δ2 = δ1
=
(6)4 (10)2 + (8)2 (10)2 − (8)2
n
sn
s
36 × 36 9 = 164 × 36 41
But
δ1 = 8 mm
∴
δ2 =
9 × 8 = 1. 76 mm. 41
This shows the longer stiffness of the hollow rod.
1 BC = CD = l. 2 If two equal masses m, m are suspended at A and D, the bar rises by height h in the middle. Deduce an expression for h in term of relevant quantities. Q. 31. A uniform bar ABCD is supported horizontally at B and C, here AB =
Solution. The bar ABCD is placed horizontally upon two knife-edges at B and C. On suspending weigths mg, mg at A and D, the bar bends. The bent position is shown in the figure. The reaction at each knife edge is also mg. The couple cutting at each knife edge is also mg. E
mg
mg h
B
C F
A
D mg
mg
O
Fig. 26
The couple acting at each end is T = mgl where l = AB = CD. The bar bend into a circular arc of radius R (say). Let us complete the circle. EF is the elevation h. Now by the property of circle, we have EF × FG = BF2 that is
h (2ρ – h) = (2l)2
[3 BF = 1/2 BC = 2l]
solving and neglecting smaller terms, we have h =
2l2 R
...(i)
Mechanical Properties of Matter
441
Now when the bar is bent, a restoring couple YI/R is developed where Y is Young’s modulus and I is geometrical moment of inertia of the cross-section. This is equal and opposite to the external couple τ. Thus,
YI YI = τ = mgl or R = R mgl putting this value in eq. (i), we obtain h =
2 mgl3 YI
This is required expression. Q. 32. A 10 cm wide and 0.2 mm thick metal steel is bent to form a cylinder of 10 cm length and 50 cm radius. If the young’s modulus of the metal is 1.5 × 1012 dyne/cm2, calculate (i) the stress and strain on the convex surface, (ii) bending moment. Solution. The cylindrical metal sheet is equivalent to a beam for which breadth b = 10 cm, thickness d = 0.02 cm and radius R = 50 cm. (i) When a beam is bent, the strain in a filament at a distance Z from the neutral surface is Z/R. For the convex surface, Z =
d = 0. 01 cm. 2 Z 0. 01 = = 2 × 10−4 R 50
∴
strain =
Now
stress = Y × strain = (1. 5 × 1012 ) × (2 × 10−4 ) = 3 × 108 dyne/cm2
(ii) The bending moment is
YI bd3 where I is geometrical moment of inertia which is R 12
bending moment =
Y R
F bd I GH 12 JK 3
1. 5 × 1012 10 × (0. 02)3 × 50 12 = 2 × 105 dyne-cm. =
Q. 33. A rectangular bar is 2 cm in breadth and 1 cm in depth and 100 cm in length is supported at its ends and a load of 2 kg is applied at its mid-point. Calculate the depression if the young’s modulus of the material of the bar is 20 × 1011 dyne/cm2. Solution. The depression at middle point of the beam supported horizontally on two-knife-edges distant l apart and loaded in the middle by a weight W is given by δ =
Wl3 48 YI
442
Mechanics
where I is geometrical moment of inertia of cross-section. For a rectangular beam of width b and thickness d, we have
∴
I =
bd3 12
δ =
Wl3 4 Ybd3
Here W = 2000 × 980 dyne, l = 100 cm, b = 2 cm, d = 1 cm and Y = 20 × 1011 dyne/cm2 ∴
δ =
(2000 × 980) (100)3 4 (20 × 1011 ) (2) (1)3
= 0.1225 cm. Q. 34. A bar 1 meter long, 5 mm square in sections, supported horizontally at its ends loaded at the middle is depressed 1.96 mm by a load of 100 gm. Calculate young’s modulus [Ans. 2.0 × 1012 dyne/cm2] for the material of the bar (g = 980 cm/sec2). Q. 35. A wire of length 2.0 meter and cross-section 1 × 10–6 m2 is stretched in horizontal position between two supports with force 200 N. If mid point of the wire is pulled through 5 mm, find change in the tension. (Y = 2.2 × 1011 N/m2). Solution. On being pulled, new length of wire = 2 (AC) = 2 12 + ( 5 × 10−3 )2
LM N
= 2 1+
OP Q
1 × 25 × 10−6 binomially 2
= 2 + 25 × 10–6 m. ∴ Change in length of wire AB, ∆l = 25 × 10–6 m If ∆F is extra force exerted in clamps, Y = or
2.2 × 1011 =
or
∆F =
∆F/A ∆l /l ∆F 2 × −6 10 25 × 10−6 2. 2 × 1011 × 25 × 10−6 × 10−6 2
= 2.75 Newton. Q. 36. Find the greatest length of steel wire that can hang vertically without breaking. Breaking stress for steel 7.8 × 109 dyne/cm2. Density of steel = 7.9 gm/sec Solution. Let the maximum length be ‘l’ and cross-sectional area of wire be A, then its weight, Mg = lA dg, ∴ stress =
Mg = l dg A
Mechanical Properties of Matter
For wire not to break,
443
ldg = Breaking stress
∴
l =
7. 8 × 109 = 1. 01 × 106 cm 7. 9 × 981
Q. 37. A light rod 2.0 meter long is suspended from ceiling horizontally by means of two vertical wires of equal length tied to ends. One is of steel having cross-section 10 – 3 m2 and other is of brass with cross-section 2 × 10 – 3 m2. Find the position along the rod at which a weight may be hung to produce (i) Equal stress in both wires, (ii) Equal strain in both wires. Given Ysteel = 2 × 1011 N/m2, Ybrass = 1011 N/m2 Solution. (I) If T1 and T2 are tensions in steel and brass wires. When weight W is hung at distance x from steel wire, for equal stress
S tee l
B ra ss
T1
T2
x
Fig. 27
T2 T1 −3 = 2 × 10−3 10 i.e.,
T1 1 = T2 2
...(i)
also for equilibrium of rod, T1 x = T2 (2 – x)
T1 2− x = T2 x
or From (i) and (ii), or
1 2− x = 2 x x = 1.33 m
(II) If equal strain in two wire is desired
T1 / 10−3 T / 2 × 10−3 = 2 Ysteel Ybrass
...(ii)
444
Mechanics
T1 T2 T1 =1 = i.e., 2 × 1011 2 × 1011 T2
or
∴ For equilibrium of rod if weight is suspended at distance y from steel wire T1 y = T2 (2 – y)
2− y y
or
1 =
or
y = 1.0 m.
Q. 38. By how much a rubber string of length 10 m increases in length under its own weight ? Density of rubber = 1.5 × 103 kg/m3 Y for rubber = 5 × 108 N/m2. Solution. If A is cross-sectional area, weight of rubber string (i.e., elongating) force, F = Aldg ∴
Normal stress =
F = lgd A
If ∆l is elongation in rubber string, this elongation will take places in only upper half length since the weight of string acts from C.G. of string. ∴
Tensile strain =
Hence,
Y =
= ∴
∆l =
∆l l/2 F/A ∆l l/2 ldg l l2dg . = ∆l 2 2 ∆l
l2dg 10 × 10 × 1. 5 × 103 × 9. 8 = 2Y 2 × 5 × 108
= 1.47 mm Q. 39. A gold wire 0.32 mm in diameter elongates by 1mm when subjected to force 330 gm-wt and twists through/radian when a couple 145 dyne × cm is applied. Find Poisson’s ratio for gold.
330 × 980 × L MgL = πr2l π × (0.016)2 × 0.1
Solution. We have
Y =
Also
2L.τ η = πr4θ =
RSas τ = πηr θ UV 2L W T 4
2L × 145 290 L = π (0.016)4 × 1 π ( 0. 016)4
Mechanical Properties of Matter
Using the relation,
445
Y = 2η (1 + σ), i.e., σ =
Y −1 2η
LMF 330 × 980L × π (0.016) I − 1OP MNGH π (0.016) × 0.1 2 × 290 L JK PQ 4
we get
σ =
2
= 1.428 – 1 = 0.428 Q. 40. A wire 1.0 m long and 0.5 mm in radius elongates 0.32 mm when 5 kg load is applied. It is twisted through 0.4 radian under a torque 3 × 10 – 3 Nm. Determine elastic coefficient for material of wire. Solution.
Y =
MgL πr2l
=
5 × 9. 81 × 1 π ( 0. 5 × 10 −3 )2 0. 32 × 10 −3
=
49. 05 × 109 = 2 × 1011 N/m2 3.14 × 0. 25 × 32
πη r4θ 2L
Also
τ =
∴
η =
Using relation,
Y = 2η (1 + σ) ⇒ σ =
Since
2 × 1011 − 1 = 0. 31 2 × 7. 6 × 1010 Y = 3K (1 – 2σ)
2 × 1 × 3 × 10−3 3.14 × ( 0. 5 × 10 −3 )4 × 0. 4 = 7.6 × 1010 N/m2
Y −1 2η
σ =
or
or
K = ∴
2 × 1011 = 1. 75 × 1011 N/m2 3 (1 − 2 × 0. 31)
K = 1.75 × 1011 N/m2.
Q. 41. One end of a wire 50 cm in length and 1 mm in radius is twisted through 45° at one end relative to the other. Calculate the angle of shear on the surface. Solution. If θ is the angle of twist at free end of a rod of length l and radius r, the angle of shear is given by φ =
rθ l
φ =
0.1 × 45° = 0. 09° 50
Putting the given values
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Mechanics
Q. 42. If a solid cylinder of length l and radius r is clamped at its middle section, what will be the couple acting on the clamp due to the cylinder, when the two ends are twisted around the axis through equal angles (i) In same direction. (ii) In opposite directions. Solution. For any twist of the couple acting on each half length of cylinder will be τ =
πη r4θ πη r4θ = 2 ( l / 2) l
(i) When the two ends are twisted in same direction, the couple in the two halves will have additive effect on the clamps and so total couple exerted at the clamp will be = 2τ =
2 π η r4 θ l
(ii) When the two ends of cylinder are twisted in opposite directions, the two will have neutralizing effect at the clamp and so net couple exerted at the clamp will be 0. Q. 43. Two cylinders of length L and 2L and radii 2R and R respectively are welded coaxially. One end being fixed, the other is twisted by 110°. Calculate the twist of the thicker rod. Solution. Let θ be the twist at welded point as shown in Fig. 28. Since in equilibrium condition, torque on the two cylinders will be same. L
2L 11 0° θ
Fig. 28
110 – θ = θ =
and
τ . 2L (for thinner cylinder) πη R4 τL (for thicker) πη (2R)4
Dividing we get
i.e.,
110 − θ = 32 θ θ = 3.33°.
Q. 44. A solid cylinder of radius 3 cm is melted and recast into hollow cylinder of radius 5 cm while length remains unchanged. By what ratio torsional rigidity changes in doing so? Solution. If r is inner radius of hollow cylinder but mass and length remaining same
π × (3)2 L ρ = π (52 − r2 ) L. ρ i.e.,
r 2 = 52 – 32 or r = 4 cm.
Now for solid cylinder, torsional rigidity, CS =
πη (3)4 πη = 81. 2l 2l
Mechanical Properties of Matter
447
and for hollow cylinder, torsional rigidity CH = ∴
πη ( 54 − 44 ) πη = 369 2l 2l
CH 369 41 = = CS 81 9
i.e., torsional rigidity of hollow cylinder will be
41 times that of solid. 9
Q. 45. A thin-walled tube of mean radius 10 cm and thickness 0.05 cm is melted to form a rod of same length. Compare torsional rigidity in the two cases. Solution. Since mean radius of tube is 10 cm and thickness of wall is 0.05 cm, the outer
0. 05 0. 05 = 10. 025 and inner radius r1 =10 − = 9. 475 cm. If tube is 2 2 melted to form a rod of radius r, we have
radius will be r2 =10 +
d
i
2 2 πr2ld = π r2 − r1 ld
i.e., or
r 2 = (10. 025)2 − ( 9. 975)2 r = 20 × 0.05 = 1 cm
If CH and CS are torsional rigidity
CH = CS
πη
Fr GH
4 2
− r14 2l
πη( r4 / 2l)
I JK
=
(10. 025)4 − (9. 975)4 (1)4
=
200 1
Q. 46. Two cylinderical steel rods have the same mass per unit length. One is solid while other which is hollow and has external radius twice the internal. Compare their torsional rigidity and show that maximum strains produced by equal twisting couple also bear the same ratio. Solution. Let rs be radius of solid rod and 2r, r be external and internal radii of hollow rod. Then mass per unit length being same, πrs2d = π (2r)2 − ( r)2 d i.e.,
rS2 = 3r2
The ratio of torsional rigidities is
πη rS4 / 2l r4 9 CS = S4 = = 4 4 15 15 r CH πη (2r) − ( r) 2l
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Mechanics
Also for equal torques
πη (2r)4 − ( r)4 πη rS4 θS = θH 2l 2l i.e.,
rS4 θS = 15 r 4 θH
rS4 9 θH = = 4 15 15 r θS
or
Since for any given cylinder φ α θ ∴
φH 9 = φS 15
Q. 47. Calculate the work done in twisting a steel wire 0.25 m long and 2 mm thick through 45° (η = 8 × 1010 N/m2). Solution. The work done in twisting the wire through of radian is given by
1 πη r4 1 2 2 θ2 W = cθ = 2 2l As given,
η = 8 × 1010 N/m2 r = 1 × 10–3 m, l = 0.25 m θ = π/4 radian
Substituting
1 3.14 × 8 × 1010 × (10−3 )4 × 2 2 × 0. 25 = 0.154 joule.
W =
FG 3.14 IJ H 4K
2
Q. 48. A steel bar of length 50 cm, breadth 2 cm and thickness 1 cm is bent into an arc of radius 2.0 meter. Determine, (i) Stress and strain at convex surface, (ii) Bending moment (Y = 2 × 1012 dyne/cm2). Solution. (i) The tensile strain at distance Z from neutral surface
Z 1 /2 = = 2. 5 × 10−3 R 200 stress = Y × strain =
and so
= 2 × 1012 × 2.5 × 10–3 = 5 × 109 dyne/cm2 (ii) Bending moment
=
YI g R
=
( Ybd3 / 12) 2 × 1012 × 2 × 13 = R 12 × 200
1 × 1010 = 1. 66 × 109 dyne × cm 6 Q. 49. A circular bar is supported horizontally on two horizontal knife edges 0.4 cm apart and projects by equal amounts beyond knife edges. The total length of bar is 0.6 meter and its radius is 0.4 cm. When a load 5 kg is suspended from each end, the centre rises by 0.5 cm. Find the value of Y for material of bar. =
Mechanical Properties of Matter
449
Solution. The situation is same as if beam of length, l = 0.4 cm supported horizontally is loaded in the middle by 10 kg, producing depression δ = 0.5 cm. Hence we must have δ = Putting the values,
we get
Mgl3 48 Y . I g
Mg = 10 × 9.8 N, l = 0.4 m Ig =
π (0. 4 × 10−2 )4 and δ = 0.5 × 10−2 m 4
Y =
10 × 9. 8 × (0. 4)3 × 4 = 13 × 1010 N/m2. 48 × π (0. 4 × 10−2 )4 × 0. 5 × 10−2
Q. 50. A rod AD consisting of three segments AB, BC and CD hangs vertically from a fixed support at A as shown in figure 29. Cross section of rod is uniform equal to 10 – 4 m2 and a weight 10 kg is hung from D. Calculate displacements of points B, C and D given, YAB = 2.5 × 1010 N/m2 YBC = 4 × 1010 N/m2 YCD = 1.0 × 1010 N/m2 Solution. The elongation of a wire given by I =
Mgl AY
Since the same load 10 kg will be effective on all the three section of rod, IAB
10 × 9. 8 × 0.1 = 3. 92 × 10−6 m = 10−4 × 2. 5 × 1010
IBC =
A 0 .1 B
10 × 9. 8 × 0. 2 = 4. 9 × 10−6 m 10−4 × 4 × 1010
0 .2 C
ICD
10 × 9. 8 × 0.15 = 14. 7 × 10−6 m = 10−4 × 1. 0 × 1010
Obviously the displacement of point B is = 3.92 × the displacement of point C is = 3.92 ×
10–6
10–6
m
0 .15 D
+ 4.9 × 10–6 m
= 8.82 × 10–6 m the displacement of point D is = 14.7 × 10–6 + 8.82 × 10–6 m
1 0 kg
Fig. 29
= 23.52 × 10–6 m Q. 51. Two different rod of length l1 and l2 but of same cross-section A are placed between two massive walls. The first rod has young’s modulus Y1 and coeff. of linear expansion α1. The corresponding values for second rod are Y2 and α2. If temperature of both rods is increased by T°, find the force with which rods act on each other. Solution. The thermal expansion in first rod due to increase in temp T° will be ∆l1 = l1 α1 T
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Mechanics
and increase in length of second rod ∆l2 = l2 α2 T ∴ Total expansion in length of composite rods = ∆l1 + ∆l2 = T (l1 α1 + l2α2) Since expansion cannot take place hence it is neutralized by elastic compressive forces, so we can write
Fl1 Fl2 T (l1 α1 + l2α2 ) = Y A + Y A 1 2 (where F is force exerted by one rod at other)
F×
i.e., or
LM l + l OP NY A Y AQ 1
2
1
2
= T l1 α1 + l2 α2
F =
A l1 α1 + l2 α2 T l1 / Y1 + l2 / Y2
Q. 52. Two rods of equal cross-section A, one of copper and other of steel, are joined to form a composite rod of length 2.0 m at 20°C. The length of copper rod is 0.5 m. When temp is raised to 120°C, the composite rod is increases to 2.002 m. If the composite rod is fixed between two rigid walls and not allowed to expand, the length of two components do not change. Find young’s modulus and coeff. of linear expansion for steel, given YC = 1.3 × 1013 N/m2 αC = 1.6 × 10 – 5/°C. Solution. The increase in length of Cu rod is (∆l)cm = lCu . α Cu . ∆t = 0.5 × 1.6 × 10–5 × (120 – 20) = 0.8 × 10–3 m If αS is expansion coeff. for steel (∆l)S = 1.5 × αS × (120 – 20) = 150 αS But total increase in length of the composite rod is 0.002 m ∴
150 αS + 8 × 10–4 = 20 × 10–4 αS = 0.8 × 10–5/°C
or
Since, stress in steel rod = Stress in Cu rod ∴ ∴
YS .αS ∆t = YCu . α Cu . ∆t YS = =
YCu .α Cu αS 1. 3 × 1013 × 1. 6 × 10−5 = 2. 6 × 1013 N/m2 −5 0. 8 × 10
Q. 53. Two rods of different materials have same length and cross-section and are joined to make a rod of length 2.0 m. The metal of one rod has young’s modulus 3 × 1010 N/m2 and coeff. of thermal expansion 10–5 /°C. The corresponding values for other metal are
Mechanical Properties of Matter
451
1010 N/m2 and 2 × 10–5/°C. How much pressure must be applied to prevent the expansion of rod when temperature is raised by 100°C ? Solution. The increase in length of two rods due to increase in temperature will be ∆l1 = l1α1∆t = 1 × 10−5 × 100 = 10−3 m ∆l2 = l2α2∆t = 1 × 2 × 10−5 × 100 = 2 × 10−3 m
Fl YA ∆l ∴F= YA l It means Tension developed in two rods would be
3
and
∆l =
F1 =
3 × 1010 × 10−3 A newton 1
F2 =
1010 × 2 × 10−3 A newton 1
Hence total pressure to be applied to prevent expansion must be F = F1 + F2 = 5 × 107 A Newton ∴
Force A = 5 × 107 N/m2.
Pressure =
Q. 54. A solid sphere of radius R made of a material of bulk modulus B is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of liquid. Find the fractional change in radius of the sphere, when a mass M is placed on the piston to compress the liquid. Solution. The volume of spherical body is v =
4 πR3 3
∴
dv = 4πR2 dR
Hence
dv dR = 3 v R
The Bulk Modulus to given by B = − v
...(i)
dP dv
∴
dv dP = v B
dv Mg Mg , so = v A.B A Substituting this value in (i) we get, Since dp =
dR Mg = R 3A.B Q. 55. A body of mass 3.14 kg is suspended from one end of a wire of length 10 meter. The radius of wire changes from 9.8 × 10–14 m at one end to 5 × 10 – 4 m at the other. Find
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Mechanics
the change in length of wire. What will be the change in length if ends are interchanged. α (Y = 2 × 10–11 N/m2) Solution. Consider an element of length dx at distance x from fixed end of wire from definition, the change in length of this element will be
Hence,
dl =
F. dx YA
A =
πr2
x
(∴ L = dx)
L
dx
= π (a + x tan θ)2 So,
dl =
Fdx Yπ ( α + x tan θ)2
The total change in length of wire is therefore;
Putting a + x tanθ = t, ∴
∆l =
z
dx =
dt tanθ
∆L =
= from fig. So,
L
a
z z
L
a b
a
b
Fig. 30
dx F . Yπ ( a + x tan θ)2
( a + L tan θ)
1−
F. t2 dt πY tan θ
F F = Yπ tan θ. t Y π tan θ
FG 1 − 1 IJ H a bK
tan θ =
b−a L
∆L =
F.L Yπ ab
∆L =
3.14 × 9. 8 × 10 = 10−3 m π (9. 8 × 10−4 ) (2 × 1011 ) (5 × 10−4 )
...(i)
Substituting the values given
If ends of wire are interchanged, values of a and b will replaced each other. Hence, from eq. (i), ∆l will remain the same. Q. 56. A gold wire, 0.32 mm in diameter, elongates by 1 mm when stretched by a force of 330 gm.wt and twist through 1 radian, when equal and opposite torques of 145 × 10–7 n.m are applied at its ends. Find the value of Poisson’s ratio for gold. Solution. When a wire of radius r and length L is elongated by l with a force F, the young’s modulus of its material is given by Y =
F L . πr2 l
Mechanical Properties of Matter
453
Here, F = mg = 0.33 × 9.81 newton, r = 1.6 × 10–4 m, l = 10–3 m ∴
Y =
0. 33 × 9. 81 × L n/m2 π × (1.6 × 10−4 )2 × 10−3
The torque applied to twist the wire by 1 radian, is given by −7 4 C = ηπ r /2L = 145 × 10 n. m
145 × 10−7 × 2L π × (1.6 × 10−4 )4 Y −1 Pisson’s ratio σ = 2η η =
where ∴
=
0. 330 × 9. 81 × L × π × (1.6 × 10−4 )4 −1 π × (1. 6 × 10−4 )2 × 10−3 × 2 × 145 × 2L
= 1.429 – 1 = 0.429 Q. 57. A circular rod of young’s modulus 2.04 × 106 kg. wt/sq cm. Poisson’s ratio 0.4, length 1 meter, and cross-section 0.95 sq.mm is stretched by a weight of 10 kg. Find the extension and dimension in cross-section. Solution. When a wire of length L and cross-section A is elongated through l by a force F, the young’s modulus of its material is given by
⇒
Y =
F×L A×l
l =
F×L Y×A
=
...(i)
10 × 9. 81 × 1 2. 04 × 10 × 9. 81 × 104 × 0. 95 × 10−6 6
= 5.16 × 10 –4 m Poisson’s ratio is given by
σ =
Lateral strain δr /r = Longitudinal strain l /L
where r is radius of wire and δr the diminuation produced in it ∴
∴
l ×r L cross-section area of wire A = πr2 = 0.95 × 10–6 sq.m δr = σ ×
r =
0. 95 × 10−6 / π
= 5.499 × 10–4 m Poisson’s ratio for material σ = 0.4. Substituting the values of various quantities in (ii), we get δr =
0. 4 × 5.16 × 10−4 × 5. 499 × 10−4 1
...(ii)
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Mechanics
⇒
δr = 11.35 × 10–8 m
A = πr2 ∴ δA = 2π r dr 3 where δA is the diminuation in cross-section A, ∴
δA = 2 × π × 5. 499 × 10−4 × 11. 35 × 10−8 = 3.92 × 10–10 sq/m.
Q. 58. A circular bar one meter long and 8 mm diameter is rigidly clamped at one end in a vertical position. A couple of magnitude 2.5 n.m is applied at the other end. As a result a mirror fixed at this end deflects a spot of light by 15 cm on a scale one meter away. Calculate the modulus of rigidity of the bar. Solution. When the mirrors turn through an angle θ, the reflected ray turns through an angle 2θ. As the spot of light is reflected through 15 cm on a scale 100 cm away, then tan 2θ = 0.15/1 ⇒
2θ = 0.15 if θ is small
∴
θ = 0.075 radian.
The mirror is fixed at the lower end of the bar, therefore its lower end is deflected by 0.075 radian. Now, to twist the bar by 0.075 radian, the couple required is Cθ =
ηπr4 × 0. 075 n/m 2l
Here l = 1 m, r = 4 × 10–3 m and Cθ = 2.5 n/m ∴
ηπ × ( 4 × 10−3 )4 × 0. 075 = 2.5 2×l
η =
or
2. 5 × 2 × 1 = 8. 29 × 1010 n/m2 . −3 4 3.14 × (4 ×10 ) × 0. 075
Q. 59. An iron wire of 1 mm radius 100 cm length is twisted through 18°. Where is shearing stress is maximum ? Find the maximum angle of shear. If modulus of rigidity of iron is 8 × 1010 n/m2, what is torsional couple? η =
Solution. ∴
shearing stress shearing strain
shearing stress = ηφ
Hence the shearing stress is maximum when φ is maximum. But φ is maximum at the surface of the wire, the reason is given below: The strain at distance x (radially) will be φ = xθ/l. It has maximum values, when x = r, the radius of the wire, i.e., φmax = rθ/l, at the surface of wire Here r = ∴
10–3,
θ = 18° and l = 1 m φmax = 10–3 × 18°/1 = 0.018°
Torsional couple Cθ =
ηπr4 2l
Mechanical Properties of Matter
455
=
π 8 × 1010 × 3.14 × (10−3 )4 = 18 × 2×1 180
= 3.8 × 10–2 n.m. Q. 60. Two cylinders A and B of radius r and 2r are soldered coaxially. The free end of A is clamped and the the free end of B is twisted by an angle θ. Calculate twist at the junction, taking the material of the cylinders to be the same and lengths equal. Solution. Let τ be the couple applied at the free end of B and θ′ the twist at the junction. The couple T produces a relative twist θ′ between the ends of the cylinder A and relative twist (θ − θ ′ ) between the ends of the cylinder B. Thus, if η be the modulus of rigidity of the cylinder and length l of each cylinder, we have τ =
πηr4θ ′ πη (2 r)4 (θ − θ ′ ) = 2l 2l
∴
θ′ = 16 (θ – θ′)
⇒
θ′ =
16 θ. 17
Q. 61. Two solid cylinders of same materials having length l, 2l and radii r, 2r respectively are joined coaxially. Under a couple applied between the free ends the shorter cylinder shows a twist of 30°. Calculate the twist of longer cylinder. Solution.
Torque, τ =
πηθ1 r 4 πη (2r4 ) θ = 2l 2 (2l)
θ 1 = 8θ
or ∴
θ =
θ1 30 = = 3. 75° 8 8
Q. 62. A thin walled cylinder tube of mean radius 10 cm and thickness 0.05 cm is melted up and recast into solid rod of the same length. Compare the torsional rigidities in two cases. Solution. Mean radius of the hollow tube = 10 × 10–2 m Its thickness = 0.05 × 10–2 m External radius r1 = 10 + 0.05/2 cm = 10.025 × 10–2 m Internal radius r2 = 10 – 0.05/2 cm = 9.975 × 10–2 m Now if r be the radius of recast solid rod having the same length and same mass as the hollow one, then
π ( r12 − r22 ) lρ = πr2lρ where l is length of rod and ρ is density of material ∴
r12 − r22 = r 2
r 2 = (10. 025 × 10−2 )2 − (9. 975 × 10−2 )2
or
= 0.05 × 20 × 10–4 = 10–4 ⇒
r = 10–2 m
456
Mechanics 4 4 Torsional rigidity of hollow rod C1 = ηπ ( r1 − r2 )/2l
Torsional rigidity of solid rod C2 = ηπr4 / 2l ∴
r14 − r24 (10. 025)2 + (9. 975)2 C1 = = C2 (1)2 r4
∴
C1 200 = C2 1
Q. 63. A body suspended symmetrically from the lower end of a wire, 100 cm long and 1.22 mm in diameter, oscillator about the wire as axis with a period of 1.25 sec. If the modulus of rigidity of the material of the wire 8.0 × 1010 n/m2. Calculate the M.I. of the body about the axis of rotation. Solution.
T = 2π I / C ∴ I =
CT2 4 π2
...(i)
Where I = moment of inertia of body about the axis of rotation and C = ηπr4 /2l Here,
η = 8 × 1010 n/m2, l = 1m, r = 0.61 × 10–2 m
∴
C =
8 × 1010 × 3.14 × (0. 61 × 10−2 )4 n. m 2 ×1
Time period T = 1.25 sec. Substituting the values of C and T in (i), we get, I =
8 × 1010 × 3.14 × (0. 61 × 10−4 )4 (1. 25)2 × 2 ×1 4 × (3.14)2
= 8.889 × 10–4 kg-m2. Q. 64. A wire of 4 mts long, 0.3 mm in diameter is stretched by a force of 8000 gm wt. If the extension in the length amounts to 1.5 mm, calculate the energy stored in the wire. Solution. The work done in stretching the wire through under a longitudinal force F is given by W = =
z
l
o
F. dl =
z
l
o
AYl dl L
AYl2 1 = Fl 2L 2
Here F = 8000 gm-wt = 8 kg wt = 8 × 9.8 n; l = 1.5 × 10–3 ∴
work done in the wire =
1 × 9. 8 × 8 × 1. 5 × 10−3 2
= 5.88 × 10–3 Joules. Q. 65. Find the amount of work done in twisting a steel wire of radius 1 mm and length
Mechanical Properties of Matter
457
Solution. The work done in twisting a wire from zero to θ angle is given by
Here
W =
z
C =
8 × 1010 × 3.14 × (10−3 )4 πη r4 = 2l 2 × 0. 25
θ
o
Cθ dθ =
1 2 Cθ 2
θ = 45 × π/180 = π/4 radian.
and ∴
Work done W =
=
1 2 Cθ 2
FG IJ H K
1 8 × 1010 × 3.14 × (10−3 )4 π × × 2 2 × 0. 25 4
2
= 0.1547 Joule. Q. 66. A steel wire of 1 mm radius is bent to form circle of 10 cm radius. What is the bending moment and the maximum stress, if Y = 2 × 1011 n-m–2. Solution.
Bending moment =
−3 4 4 I = πr /4 = 3.14 × (10 ) / 4
Here ∴
YI R
Bending moment =
stress =
2 × 1011 × 3.14 × (10−3 )4 = 1. 57 n-m 4 × 0. 1
YZ R
For maximum stress the value of Z should be maximum, i.e., Z = r ∴
Maximum stress =
Yr 2 × 1011 × (10−3 ) = = 2 × 109 n-m−2 R 0.1
Q. 67. A cantilever of length 50 cm is depressed by 15 mm at the loaded end. Calculate the depression at distance 30 cm from the fixed end. Solution. Depression
y =
Mgx2 (l − x/3) 2Yl
...(i)
where x = 0.5 m, Y = 0.015 m 0.015 =
Mg ( 0. 5)2 0. 5 0. 5 − 2 Yl 3
FG H
IJ K
y =
Mg ( 0. 3)2 0. 3 0. 5 − 2 Yl 3
FG H
IJ K
when x = 0.3 m, y = ?
Dividing eq. (ii) by eq. (i) we get,
...(ii)
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Mechanics
( 0. 3)2 1. 2 Y × = ( 0. 5)2 1. 0 0. 015
⇒
Y = 6. 48 × 10−3 m
Q. 68. Cantilever of length l and uniform cross-section shows a depression of 2 cm at the loaded end. What will be the depression at distances l/4, l/2 and 3l/4 from the fixed end? Solution.
FG H
IJ K
FG H
IJ = K
y =
Mg x2 x l− 2YI 3
0.02 =
Mg l2 l l− 2YI 3
when x = l, y = 0.02 m
When x = l/4,
When
when x = 3l/4,
y1
Mgl3 3 YI
FG IJ FG l − l IJ H K H 12 K
Mg l = 2YI 4
2
=
11 Mgl3 Mgl3 11 . = . 32YI 12 128 3YI
=
11 × 0. 02 = 1. 72 × 10−3 m 128
x = l/2, y2 =
FG IJ FG l − l IJ H K H 6K
Mg l 2YI 2
2
=
5 Mgl3 5 Mgl3 . = × 8 YI 6 16 3 YI
=
5 × 0. 02 = 6. 2 × 10−3 m 16
y3 = =
FG IJ FG l − l IJ H K H 4K
Mg 3l 2YI 4
2
Mgl3 27 Mgl3 81 . . = 2YI 64 3YI 128
81 × 0. 02 = 1. 27 × 10−2 m. 128 Q. 69. A 10 cm wide and 0.2 mm thick metal sheet is bent to form a cylinder of 10 cm length and 50 cm radius. If the young’s modulus of the metal is 1.5 × 1011 n-m–2, calculate =
(i) the stress and strain on the convex surface and, (ii) the bending moment. Solution. (i)
Strain =
Z 2 × 10−4 1 = × = 2 × 10−4 R 2 0. 5
Mechanical Properties of Matter
459
(3 for convex surface Z = half thickness of the sheet) stress = Y × strain = 1. 5 × 1011 × 2 × 10−4 = 3.0 × 107 n-m–2 (ii)
YI Y bd3 = R R 12 1. 5 × 1011 0.1 × (2 × 10−4 )3 × = 0. 5 12 = 2 × 10–2 n-m
Bending moment =
Q. 70. A light metal rod of length 60 cm and of radius 1 cm is clamped at one end loaded at the free end with 5.5 kg. Calculate the depression of free end assuming Y = 9 × 1011 dyne and g = 980 cm/sec2. Solution. When a beam of length l is fixed at one end and loaded by weight W at the free end, the depression is given by
Wl3 3YI For a circular cross-section of radius r, I = πr4/4 δ =
4 Wl3 3 Yπ r 4 Here W = 5.5 × 103 × 980 dynes, l = 60 cm, Y = 9 × 1011 dyne/cm2 and r = 1 cm
∴
δ =
Therefore,
δ =
4 × 5. 5 × 103 × 980 ×(60)3 3 × 9 × 1011 × 3.14 × (1)4 = 0.5495 cm.
Q. 71. A vertical rod of circular cross section of radius 1 cm is rigidily fixed in the earth. Its upper end is 3mts from the earth. A thick string which can stand a maximum tension of 2 kg is tied at the upper end of the rod and pulled horizontally. Find how much the rod will be deflected before the string snaps. (Y for steel = 2 × 1012 C.G.S. units, g = 1000 cm/sec2). Solution. The lower end of a vertical rod is fixed and the upper end is pulled by a horizontal force, hence this is the case of cantiliver.
B
The deflection of the upper end, is given by
Wl3 3YI W = 2 × 1000 × 1000 dynes/cm2, δ =
Here the horizontal force,
A
Fig. 31
length of the rod
l = 300 cm; value of Y = 2 × 1012 dyne/cm2;
radius of the rod
r = 1 cm
For circular cross-section ∴
πr4 π × (1)4 π = = 4 4 4 2 × 1000 × 1000 × ( 300)3 × 4 = 11. 46 cm δ = 3 × 2 × 1012 × 3.14 I =
W
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Mechanics
Q. 72. Compare loads required to produce equal depression for two beams, made of the same material and having the same length and weight with the only difference that while one has circular cross-section, the cross-section of the other is square. Solution. The depression δ at the middle point of a beam of length l of circular crosssection of radius r, is given by δ =
W1l3 12. Yπr4
...(i)
where W1 is the weight loaded. In the case of the rectangular cross-section beam the weight loaded by W2, then the depressions δ′, is given by δ′ =
W2l3 4 bd3Y
3 Two beams have same length and are made of same material But for square cross-section, b = d
∴
δ′ =
W2l3 4 b4 Y
...(ii)
The two beam have equal weights and made of same material, ∴ Their masses and hence volumes must be equal, i.e., πr2l = b2l or b2 = πr2 substituting the value of b2 in (ii), we get δ′ =
W2 l3 4π2r4 Y
...(iii)
But depression is same in both cases, i.e., δ = δ′, ∴ equating (i) and (iii) we get,
W1 l3 W2l3 4 = 12 Yπr 4π 2r4 Y W1 3 = which is the required ratio. W2 π
or
Q. 73. A cantilever of length l is loaded at a point x1, calculate the depression of the tip of the cantilever. If the same load is shifted to the tip, what is the depression of the point x1? Solution. The expression for the depression y at distance x from fixed end, is given by y =
Also at x,
F GH
W lx2 x3 − YI 2 6
F GH
I JK
dy x2 W = tanθ = bx − dx YI Z
I JK
Mechanical Properties of Matter
461
In the first case the cantilever is loaded at OP = x1. Hence the beam will not bend beyond x1 by the load W1 but the portion P′y will be target at P′ and the total depress δ = xy will be given by l
δ = xx ′ + P x′ y
P
= PP′ + P ′ x′ tanθ Hence PP ′ =
Wx13 , 3YI
h
P′x′ = l – x1 tan θ =
and
∴
δ = =
F GH
x
O
s
θ
Fig. 32
I JK
Wx12 W 2 x12 x1 − = YI 2 2YI Wx13 Wx12 + ( l − x1 ) 3YI 2YI
F GH
x3 W lx12 − 1 YI 2 6
FG H
Wx12 xl l− 2YI 3 This is the depression of tip of the cantiliver.
=
IJ K
I JK
In the 2nd case, the same load is shifted to the tip, then the depression at any point x1 is given by δ′ =
F GH
I JK
FG H
x3 Wx12 x W lx12 − 1 = − l− 1 YI YI 6 2YI 3
IJ K
We see that in two cases the depression is same. Q. 74. A lath of width 2 cm and thickness 3 mm supported horizontally on two knife edges 80 cm apart is loaded with the weight of 10 gm, hang its ends, which project 15 cm beyond the knife edges. If the centre of the lath is thereby elevated 2 mm, calculate young’s modulus for material. Solution. Let the beam be placed on two knife edges P and Q. It takes the shape as shown in fig., when loaded at both ends by 0.01 g newton. Therefore, according to statical laws, the reactions at each knife edge will be 0.01 g of newton. ∴ Couple acting at
YI P = 0.01 g × 0.15 n-m = R
0 .01 g
bd 12 0. 02 × (3 × 10−3 )3 = 4. 5 × 10−10 = 12 Consider the lath is bent into a circle of radius of curvature R, then from the geometry of the wide, I =
PO.OQ = (2R – OO′) OO′
O′
P 0 .4m
3
Here
0 .01 g Q
O 0 .4m
0 .01 g
0 .01 g R
R
C
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Mechanics
2 × 10–3 × 2R = 0.4 × 0.4[3 2R >> OO′] R = 40 m
or Whence
YI Y × 4. 5 × 10−11 = 0. 01 × 9. 81 × 0.15 = R 40
Now, ⇒
Y =
4. 0 × 0. 01 × 9. 81 × 0.15 = 1. 308 × 1010 n - m2 4. 5 × 10−10
Q. 75. A steel strip is clamped horizontally from one end. On applying a 500 gm load at the free end the bending in equilibrium state is 5 cm. Calculate (i) the potential energy of strip, (ii) the frequency of vibration, if the load is disturbed from equilibrium (neglect mass of strip itself). Solution. (i) The bending is 5 cm by the load 500 gm = 0.05 × 9.81 newton Hence force per unit bending or force constant is, C = ∴
Potential energy = =
(ii) Frequency
z
π
Cx dx =
0
0. 5 × 9. 81 = 98.1 n - m−1 0. 05
1 2 Cx 2
1 × 9. 81 × (0. 05)2 = 0.12 Joules. 2
n =
1 2π
C M
=
1 2π
9. 81 = 2. 23 sec−1 (nearly) 0. 5
Q. 76. A solid cylinder rod of radius 6 mm bends by 8 mm under a certain load. Deduce the bending when all other things remaining the same the beam is replaced by a hollow cylindrical one with external radius 10 mm and internal radius 8 mm. Solution. If δ and δ′ are the depressions of two beams, then, δ = For solid beam
Mgl3 Mgl3 I ; δ′ = δ and δ ′ = 3YI 3YI′ I′
I = πr4 / 4
For hollow beam
I′ =
π ( r24 − r14 ) 4
∴
δ′ =
64 × 8 × 10−3 = 1. 76 × 10−3 m. 10 − 84 4
10 FLUIDS 10.1 MOLECULAR FORCES There are two kinds of molecular forces: (i)
Force of adhesion or adhesive force and
(ii)
Force of Cohesion or Cohesive force.
The force of attraction between the molecules of different substances is called force of adhesion, while that between molecules of the same substance is called force of cohesion. The force of cohesion is entirely different from ordinary gravitational force and does not obey the ordinary inverse square law. Probably, this force varies inversely as the eight power of the distance between two molecules and therefore falls of rapidly with distance and becomes almost negligible at a distance of 10–7 cm. The maximum distance upon which the force of cohesion between two molecules can act is called the molecular range. It is of the order of 10–7 cm being different for different substances. The sphere, drawn with the molecule as center, having radius equal to molecular range, is called sphere of influence.
Free Surface of a Liquid Tends to Contract to a Smallest Possible Area The free surface of a liquid always behaves like a stretched elastic membrane and hence has a natural tendency to assume as small an area as possible. This peculiar property of a liquid can be illustrated by the following experiments: (a)
Let us tie loosely a closed loop of cotton thread between two points of a wire frame and dip it in a soap solution. A thin film of soap solutions is formed on the frame and the loop of thread is placed on the film in an irregular manner Fig (i). If now the film inside the loop is broken by a needle, the thread immediately assumes a circular form Fig. (ii). This shows that as soon as the film inside the loop is broken, the film outside the loop contracts, thus pulling the thread into a circle. We know that for a given perimeter the circle has the largest area. This means that the film outside the loop contracts to a smallest area.
(b)
A freely suspended drop of water, formed at the end of a tap, assumes spherical shape. We know that for a given volume, sphere has the least surface area. Therefore, to acquire minimum possible surface area, the drop takes the spherical shape.
(c)
If we place a greased needle on a piece of blotting paper and put the paper lightly on the surface of water, the blotting paper will soon sink to the bottom, but the 463
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Mechanics
needle is found to float on the surface although it is much heavier than water. On examining the surface of water below the needle, it is seen to be slightly depressed as if the liquid surface were acting as a stretched membrane, so that the weight of the needle is balanced by the inclined upward tension forces in the membrane. If one end of the needle is pressed to pierce a hole in the surface of water, it goes slantingly downward. N e ed le T
T
mg
Fig. 1
The above experiments clearly show that the surface of a liquid behaves as though it were covered with stretched elastic membrane, having a natural tendency to contract. But there is one important difference between the two that in an elastic membrane the tension increases with stretching or its surface area is increased, in accordance with Hook’s law, whereas the tension in the liquid surface is quite independent of stretching or of the area of its surface, unless the stretching force reduces the thickness of the film below 5 × 10–6 mm after which the tension diminishes rapidly.
(i)
( ii )
Fig. 2
10.2 DEFINITION OF SURFACE TENSION Let an imaginary line AB be drawn in any direction in a liquid surface. The surface on either side of this line exerts a pulling force on the surface on the other side. This force lies in the plane of the surface and is at right angles to the line AB. The magnitude of this force per unit length of AB is taken as measure of the surface tension of the liquid. Thus if F be the total force acting on either side of the line AB of length l, then the surface tension is given by
B A
Fig. 3
F l If l = 1 then T = F. Hence the surface tension of a liquid is defined as the force per unit length in the plane of the liquid surface, acting at right angles on either side of an imaginary line drawn in that surface. Its unit is ‘newton/meter’.
T =
The value of the surface tension of a liquid depends on the temperature of the liquid, as well as on the medium on the other side of the surface. It decreases with rise in temperature and becomes zero at the critical temperature.
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465
10.3 EXPLANATION OF SURFACE TENSION The tendency of a liquid surface to contract to a minimum possible area has suggested that it is in a state of tension, which we call as surface tension. This property of liquid surface has been explained on the basis of molecular theory given by Laplace. Consider three molecules A, B and C of a liquid, with their spheres of influence drawn around them. Since the sphere of influence of A lies completely inside the liquid, it is equally attracted in all directions by molecules within its sphere of influence. Hence there is no resultant cohesive force acting on it.
P R B
C
Q S
A
Fig. 4 In the case of molecule B, the sphere of influence lies partly outside the liquid. The number of molecules in the upper half, attracting the molecule B in the upward direction is less than the number of molecules in the lower half attracting it in the downward direction. Hence a resultant downward force is acting on it.
The sphere of influence of molecule C is exactly half outside and half inside the liquid. Hence this molecule is attracted in the downward direction with maximum force, perpendicular to the surface. Now, if we draw a plane RS parallel to free surface PQ of the liquid at a distance equal to the molecular range, hence the layer of the liquid between the planes PQ and RS is termed as the surface film. Clearly, what is true for molecules B and C is true for all other molecules in the surface film. Hence all the molecules in the surface film are pulled downwards due to the resultant cohesive force, the magnitude of which increases from surface RS to free surface PQ. Now, if a molecule from the interior of the liquid is brought up to the surface film, work has to be done against the downward cohesive force on it and its potential energy increase. Therefore the potential energy of the molecules in the surface film is greater than the potential energy of the molecules lying below it. A system is equilibrium has a tendency to acquire lowest possible potential energy. Thus to decrease the potential energy of the molecules in the surface film, the film tries to have minimum number of molecules in it. This can be done only by decreasing the surface area of the film, because its thickness is already fixed (being equal to molecular range). This is the reason, why the free surface of a liquid always tends to have minimum possible area.
10.4 SURFACE ENERGY When the surface area of a liquid is increased, the molecules from the interior rise to the surface. This requires work against the force of attraction of the molecules just below the surface. This work is stored in the form of potential energy in the newly formed surface. Besides this, there is cooling due to the increase in the surface area. Therefore, heat flows into the surface from the surroundings to keep its temperature constant and is added to its energy. Thus the molecules in the surface have some additional energy due to their position. This additional energy per unit area of the surface is called ‘surface energy’.
10.5 RELATION BETWEEN SURFACE TENSION AND WORK DONE IN INCREASING THE SURFACE AREA Let a liquid film be formed between a bent wire ABC and a straight wire PQ which can slide on the bent wire without friction. As the film surface tends to contract, the wire PQ
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Mechanics
moves upward. To keep PQ in equilibrium, a uniform force F (which includes the weight of the wire) has to be applied in the downward direction. It is found that the force F is directly proportional to the length l of the film in contact with the wire PQ. Since there are two free surfaces of the film, we have B
F ∝ 2l or
2 Tl
F = T × 2l
P
Where T is a constant called ‘surface tension’ of the liquid. Now, suppose the wire is moved downward through a small distance ∆x. This results in an increase in the surface area of the film. The work done by the force F (= Force × distance) is given by
Q
∆x A
C
F
Fig. 5
W = F × ∆x = T × 2l × ∆x But 2l × ∆x is the total increase in area of both the surfaces of the film. Let it be ∆A. Then W = T × ∆A or
T = W/∆A
If ∆A = 1, then T = W. Thus the work done in increasing the surface area by unity will be equal to the surface tension T. Hence, the surface tension of a liquid is equal to the work required to increase the surface area of the liquid film by unity at constant temperature. Therefore, surface tension may also be expressed in ‘joule/meter2’.
10.6 SHAPE OF LIQUID MENISCUS IN A GLASS TUBE When a liquid is brought in contact with a solid surface, the surface of the liquid becomes curved near the place of contact. The nature of the curvature (concave or convex) depends upon the relative magnitudes of the cohesive force between the liquid molecules and the adhesive force between the molecules of the liquid and those of the solid. In figure (i), water is shown to be in contact with the wall of a glass tube. Let us consider a molecule A on the water surface near the glass. This molecule is acted upon by two forces of attraction: (i)
The resultant adhesive force P, which acts on A due to the attraction of glass molecules near A. Its direction is perpendicular to the surface of the glass.
(ii)
The resultant cohesive force Q, which acts on A due to the attraction of neighboring water molecules. It acts towards the interior of water.
The adhesive force between water molecules and glass molecules is greater than the cohesive force between the molecule of water. Hence, the force P is greater than the force Q, their resultant R will be directed outward from water (figure i). In figure (ii), mercury is shown to be in contact with the wall of a glass tube. The cohesive force between the molecules of mercury is far greater than the adhesive force between the mercury molecules and the glass molecules. Hence, in this case, the force Q will be much greater than the force P and their resultant R will be directed towards the interior of mercury.
A
P
P W a te r
A M ercu ry R Q
R G la ss (i)
Fig. 6
G la ss ( ii )
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467
The resultant force R acts on all the molecules on the surface of water or mercury. For the molecules more and more away from the wall, the adhesive force P goes on decreasing while the cohesive force Q becomes more and more vertical. Consequently, the resultant R also becomes more and more vertical. In the middle of the surface, P becomes zero and Q becomes vertical. Hence the resultant R becomes exactly vertical (Figure iii & iv). If the surface of the liquid is in equilibrium, the resultant force acting on any molecule in the surface must be perpendicular to the surface. Hence the liquid surface sets itself perpendicular to the resultant force everywhere. This is why the water surface assumes a concave shape while the mercury surface assumes a convex shape in a glass tube. In either case the resultant force in the middle is vertical and the surface there is horizontal.
10.7 ANGLE OF CONTACT When the free surface of a liquid comes in contact of a solid, it becomes curved near the place of contact. The angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the point of contact is called the ‘angle of contact’ for that pair of solid and liquid. G la ss
G la ss
θ θ
W a te r M ercu ry
(i)
( ii )
W a te r
M ercu ry
( iii)
( iv )
Fig. 7
The angle of contact for those liquids which wet the solid is acute. It is zero for pure water and clean glass for ordinary water and glass it is about 8o. The liquid which do not wet the solid have obtuse angle of contact. For mercury and glass the angle of contact is 135o. The angle of contact for water and silver is 90o. Hence in a silver vessel the surface of water at the edges also remains horizontal. When a liquid is kept in container, a liquid molecule like A, touching the wall experiences three forces: (i)
Weight mg, acting downwards.
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Mechanics
(ii) (iii)
Cohesive force Q due to liquid molecules, acting inwards at 45°. Adhesive force P between liquid & solid molecule, acting horizontally outwards.
If cohesive force is less than adhesive force, net forces (P–Q cos 45°) and (mg + Q sin 45°) act on liquid molecule A. So that resultant force R is directed outwards as shown in figure (a). The meniscus then takes concave shape as in the case of water. If cohesive force is greater than the adhesive force, (Q cos 45° – P) and (mg + Q sin 45°) act as shown in figure (b) so that the resultant of these will be directed inwards giving convex shape to the meniscus as shown in figure (b). This is the case of mercury. (P – Q cos 4 5°) A P
4 5° Q
H
P
θ
Q
m g + Q sin 4 5 °
m g + Q sin 4 5 °
W a te r
M ercu ry
(a )
(b )
When a capillary is dipped a little in liquid the liquid rises or falls down a little. This action is called capillarity. Liquids having concave meniscus are found to rise and those like mercury which have convex meniscus are found to be depressed. This is because in concave meniscus, the vertical component T cos θ of surface tension force acts vertically upward per unit length, all along the circumference, which makes liquid to rise in capillary till it is balanced by the weight of liquid column in capillary. Thus for equilibrium, 2 2πr T cos θ = ( πr h) dg +
FG H
2 = πr h +
or
T =
Which gives the rise as, h =
FG H
IJ K
1 3 πr dg 3
θ T sin θ
T cos θ
10.8 CAPILLARY ACTION
T cos θ
Fig. 8
θ
θ
θ T sin θ
h
r
Fig. 9
r dg 3
IJ K
rdg r h+ . 2 cosθ 3
FG 2Tcos θ − r IJ H rdg 3 K
where d is the density of water. When meniscus is convex, the vertical component T cos θ acts vertically downwards and liquid is depressed inside capillary through depth given by above formula.
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469
10.9 RISING OF LIQUID IN A CAPILLARY TUBE OF INSUFFICIENT LENGTH If radius of capillary is negligibly small the rise of liquid is given by
2T cos θ ; neglecting r/3 rdg 2T cos θ . which gives h × r = dg The quantities on right hand side are constants for a liquid-solid contact and so h × r = h′ × r′ = ...... constant. It suggests that if capillary has sufficient length, liquid will rise up to the required height and radius of meniscus will be same as that of capillary. But if length of capillary is insufficient, liquid rises upto available height h and radius of meniscus is increased so as to keep product (h′ × r′) same as (h × r). In this condition liquid does not overflow. h =
r′′
r r
h ′< h
h
Fig. 10
10.10 EFFECTS ON SURFACE TENSION Surface tension is affected by the following factors: (i)
Effect of contamination: If the water surface has dust, grease or oil, the surface tension of water reduces.
(ii)
Effect of Solute: If the solute is very soluble then the surface tension of liquid increases, as by dissolving salt in water the surface tension increases. If the solute is less soluble then the surface tension decreases, as by pouring soap or phenol the surface tension of water decreases.
(iii)
Effect of Temperature: Rise of temperature, decreases inter molecular, force (called cohesive force) and so the surface tension in general, decreases as temperature is increased. It becomes zero at about 6o to 8o below critical temperature of liquids.
The most reliable relation showing variation of surface tension is the Etvos-Ramsey shield formula. 2
TpV 3 = To (θc − θ − d) where T = Surface tension at θoC, To = Surface tension at 0oC, θc = Critical temperature of liquid,
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Mechanics
V = Specific volume, d = Density of liquid. It is obvious that at temperature θ = θc – d, then the surface tension T will be zero.
10.11 IDEAL LIQUID An ideal liquid is one which has the following two properties: (i) Zero Compressibility: An ideal liquid is incompressible, that is, on pressing the liquid there is no change in its volume (or density). Most of the liquids may be considered approximately incompressible, because on pressing them the change in their volume is negligible. For example, on pressing water by one atmospheric pressure its volume changes only by a fraction of 0.000048. (ii) Zero Viscosity: An ideal liquid is non-viscous, that is, when there is a relative motion between different layers of the liquid then there is no tangential frictional force in between the layers. In actual practice, however, there is some viscosity in all liquids (and gases). It is less in gases, but larger in liquids.
10.12 STEADY OR STREAM LINE FLOW The flow of a fluid is said to be steady. Orderly, streamline or laminar, if the velocity at every point in the fluid remains constant (in magnitude as well as direction), the energy needed to drive the fluid being used up in overcoming the ‘Viscous drag’ between its layers. In other words, in a steady or streamline flow, each infinitesimally small volume element of the fluid, which we may call a particle of the fluid, follows exactly the same path and has exactly the same velocity as its predecessor, i.e., its velocity does not change with time. Thus, if a fluid flows along a path ABC, where A, B and C are points inside the fluid, each new particle arriving at A will always have the same velocity VA in a direction tangential to the curve ABC at A, and the pressure and density will always be the same there. Similarly, the particle arriving at B will always have the same velocity VB which may or may not be the same as VA. And the same is true for the point C. This line ABC along which the particles of the fluid move, one after the other, with their velocities constant at the various points on it and directed along the tangents to these points, is called a stream line.
C B A
vC
vB
vA
Fig. 11
The fluid-flow, however, remains steady or stream line if its velocity does not exceed a limiting value, called its criticial velocity, beyond which the flow loses all steadiness or orderlines and becomes zigzag and sinuous, acquiring what is called turbulence. Y
X
Fig. 12
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471
Stream-lined Tube If we imagine a tube made up of a large number of stream-lines, then this imaginary tube is called stream-lined tube. In figure 12 a stream-lined tube XY is shown. The boundary walls of the stream-lined tube are everywhere parallel to the direction of flow of the liquid. Therefore, the liquid can neither go out nor come in through the walls of the tube. The liquid that enters at one end X leaves at the other end Y, as if it were flowing through an actual tube.
10.13 EQUATION OF CONTINUITY OF FLOW The equation of continuity is a fundamental equation of fluid-flow and is a special case of the general physics law of conservation of matter. For an incompressible fluid it may be deduced as follows: Imagine the fluid to be flowing through a pipe AB, with a1 and a2 as its cross-sectional areas at sections A and B. Then, although the streamlines are all parallel to the axis of the pipe and perpendicular to its cross-section at every point, the velocity of flow is not necessarily the same at all sections of pipe. Let it be v1 at section A and v2 at section B. Consider an infinitely narrow tube of flow, shown dotted in the figure 13 of crosssectional areas da1 and da2 at sections A and B respectively. Then, if the fluid covers distances dS1 and dS2 in time δt at the two ends respectively and if ρ1 and ρ2 be the densities of the fluid at A and B, we have mass of fluid entering the tube of flow per unit time at section A = da1ds1P1/δt = da1v1P1 and mass of fluid leaving the tube of flow per unit time at section B = da2ds2P2/δt = da2v2P2 it being assumed that there is no appreciable change in the cross-sections of the tube at sections A and B over the small distances dS1 and dS2 respectively. ∴ Mass of fluid entering the whole section A of the pipe per unit time i.e.,
z
a1
mass rate of fluid-flow at A =
da1v1P1 = a1v1P1
0
and mass of fluid leaving the whole section B of the pipe per unit time;
z
a2
or mass rate of fluid-flow at B =
da2v2P2 = a2v2P2
0
Since the fluid is incompressible, ρ1 = ρ2 = ρ, say, and since there can be no flow across the sides of a tube of flow and there is no ‘source’ or ‘sink’ in between the sections A and B to add more fluid or to drain away some of it on its way from A to B, we have, from the law of conservation of matter, Mass of fluid entering section A per second = mass of fluid leaving section B per second. a 1 v 1 ρ = a 2 v2 ρ
or i.e.,
a1v1 = a2v2 B A
d a2
v1
v2
da1 a1
d S1
dS2
Fig. 13
a2
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Mechanics
This is called the equation of continuity and clearly means that the quantity av remains constant throughout the fluid-flow if it be steady or stream line. av is the volume of the fluid flowing across any cross-section of the pipe in unit time. It is therefore, called the volume rate of flow, or, simply, the rate of flow or the rate of discharge (or volume flux) and may be denoted by the letter V or Q, the latter being the more appropriate symbol, used in Engineering practice. Thus, the equation of continuity tells us that the volume rate of flow of an incompressible fluid (i.e., a liquid) remains constant throughout the flow, and so does it mass rate of flow (a vρ). Since a v = constant, it follows that v α 1/a i.e., the velocity of the fluid-flow at any section of the pipe is inversely proportional to the cross-section of the pipe at the section.
10.14 ENERGY OF THE FLUID A fluid in steady or streamline flow may posses any or all of the three types of energy, viz., (i) kinetic energy because of its inertia, (ii) potential energy because of its position relative to the earth’s surface and (iii) Pressure energy because of its pressure; for, if work be done on it against its pressure, it can do the same amount of work back for us and thus acquires energy. Let us obtain the value of each type of energy per unit mass and per unit volume of the fluid.
(i) Kinetic Energy If m be the mass of a fluid, of density ρ, flowing with velocity v, its kinetic energy is clearly ½ mv2. Kinetic energy per unit mass of the fluid = v2/2 and, since mass of unit volume of the fluid = its density ρ, we have Kinetic energy per unit volume of the fluid =
1 2 v ρ. 2
(ii) Potential energy If we have a mass m of the fluid at a height h above the earth’s surface, its potential energy = mgh and, therefore, potential energy per unit mass of the fluid = gh and potential energy per unit volume of the fluid = gh ρ = ρ gh
(iii) Pressure Energy Suppose we have a tube of uniform area of cross-section a, containing an incompressible non-viscous fluid, with its hydrostatic pressure equal to p. If we introduce an additional mass δm of the fluid into the tube against this pressure so gradually as not to impart any velocity, and hence any kinetic energy, to it and if the mass occupies a length δl of the tube, clearly, Work done on the mass = force × distance = (pa) × δl. Since the work done on mass δm of the fluid forms its pressure energy, we have pressure energy per unit mass of the fluid = ( pa) ×
δl δm
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473
δm = Volume × density = aδl × ρ So that, pressure energy per unit mass of the fluid = pa
δl p = aδlρ ρ
∴ Pressure energy per unit volume of the fluid =
p × ρ = p, its pressure ρ
10.15 BERNOULLI’S THEOREM This theorem states that the total energy of an incompressible, non-viscous fluid in steady flow remains constant throughout the flow. The theorem may be easily deduced as follows: Consider an infinitesimally small portion AB of a tube of flow of length dl, and in view of its small length, let its area of cross-section be assumed to be uniform and equal to da. If ρ be the density of the fluid (supposed to be incompressible); its mass in portion AB = m = dadlρ. Its weight mg = dadlρg, therefore, acts vertically downwards at its centre of gravity O, making an angle θ with the direction of flow, as shown. So that its component mg cos θ = dadlρg cos θ acts in the direction of flow, with its other component mg sin θ = dadlg sin θ acting perpendicularly to the wall of the tube, its effect being offset by the lateral thrust due to the adjoining tube of flow. da
0, v
p
mg
cos
dl
A
θ
O
B
θ m g sin θ
p + δp δl d l
m g = d ad l ρg
Fig. 14
δp dl, in the δl direction shown. And, therefore force acting on face A in the forward direction = pda and force If p be the pressure on face A of the tube, that on face B will be p +
FG p + δp dlIJ da. H δl K
acting on face B in the backward direction =
Hence, net force acting on mass dadlρ of the fluid in the tube, say
FG H
F = pda − p + or
F = −
IJ K
δp dl da + da. dl. ρg cos θ δl
δp ( dl) da + dadl. ρg cos θ δl
...(i)
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Mechanics
If v be the velocity of the fluid as it enters the tube at A, we have its acceleration = dv/dt, where dt is the time taken by the fluid to cover the length dl of the tube. So that, force acting on mass dadlρ of the fluid in the tube is also = mass × acceleration = da. dl. ρ ×
dv dt
...(ii)
Now, taking the most general case where the velocity, being a function of both distance and time, may change from point to point or from moment to moment, we have dv =
δv δv dl + dt and δl δt
dv δv dl δv δv δv + =v + = dt δl dt δt δl δt Substituting this value of dv/dt in expression (ii), above, we have
∴
FG H
F = dadlρ v
δv δv + δl δt
IJ K
δh , where h is the vertical height of tube AB from a choosen datum plane. δl Relation (i) above thus becomes
and cos θ = −
FG δv + δv IJ H δl δt K F δv + δv IJ ρ Gv H δl δt K
da. dl.ρ v
or
= −
δp δh dl. da − da. dl. ρ g δl δl
= −
δp δh − ρg δl δl
...(iii)
This is an important equation, applicable to both steady and unsteady fluid-flow and is referred to as Euler’s equation.
δv = 0 and the other partial derivatives all δt become total derivatives. So that, relation (iii) becomes In the case of steady or streamline flow, ρvdv + dp + ρgdh = 0 or
vdv + (dp/ρ) + gdh = 0 Integrating along the streamline, therefore, we have
1 2 v + 2
z
dp + gh = constant. ρ
This is called Bernoulli’s equation which, in the case of the fluid being incompressible, and, therefore, ρ, a constant, takes the familiar form
1 2 p v + + gh = Constant ρ 2
...(iv)
or, if we consider unit volume of the fluid, i.e., a mass ρ of the fluid (because mass of unit volume is the density ρ), we have on multiplying relation (i) by ρ,
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½ ρv2 + p + ρgh = Constant
...(v)
This is another form of Bernoulli’s equation in terms of pressure, every term on the left hand side has the dimensions of pressure. Now, the term (p + ρ gh) represents the pressure of the fluid even if it be at rest. It is, therefore, called static pressure of the fluid. And, the term ½ ρv2 represents the pressure of the fluid in virtue of its velocity v, and is, therefore, called dynamic pressure of the fluid. Thus we have, Static pressure + dynamic pressure = constant. Again, on dividing relation (iv) by g, we have p 1 2 v /g + + h = Constant 2 ρg
...(vi)
Which is yet another form of Bernoulli’s equation in terms of lengths or heads, as they are called. For, as can be easily seen, each term on the left hand side has the dimensions of length or a height and hence called a ‘head’. Thus ½ v2/g, is the velocity head, p/ρg, the pressure head, h, the gravitational head. So that, we have Velocity head + pressure head + gravitational head = total head = constant.
10.16 VELOCITY OF EFFLUX Let a vessel be filled with a liquid upto a height H and let there be an orifice at a depth h below the free surface of the liquid. The pressure at the free surface of the liquid and also at the orifice in atmosphere, and so there will be no effect of atmospheric pressure on the flow of liquid from the orifice. The liquid on the surface has no kinetic energy, but only potential energy, while the liquid coming out of the orifice has both the kinetic and potential energies. Applying Bernoulli’s theorem, we have total energy per unit mass of the liquid at the surface = K.E. + P.E.+ pressure energy = 0 + gh + 0 = gh and total energy per unit mass of the liquid at the orifice = ½ v2 + 0 + 0 = ½ v2 A h O
H B
h′
R
Fig. 15
Since total energy of the liquid must remain constant in steady flow, in accordance with Bernoulli’s theorem, we have ½ v2 = gh whence v2 = 2gh or
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Mechanics
velocity of efflux =
2gh
This formula was first established in 1644 by Torricelli and is called ‘Torricellip theorem’. If a body is dropped freely (u = 0) from a height h, then from the third equation of motion v2 = 2gh, we have v =
2gh
Clearly, the velocity of the liquid falling from a height h is 2gh . Hence the velocity of efflux of a liquid from an orifice is equal to that velocity which the liquid acquires in falling freely from the free surface of the liquid upto the orifice. No liquid being free from internal friction or viscosity, however, this ideal velocity is never attained in practice. The observed velocity is always less than v and is equal to Cv 2gh , where Cv is called the coefficient of velocity, its value lying between 0.95 to 0.99 in the case of water, depending upon the ‘head’ of water (i.e., depth of the orifice from the water surface) and the shape of the orifice. After emerging from the orifice the liquid adopts a parabolic path. If it takes t second in falling through a vertical distance (H-h) then according to equation S = ½ at2, we have H – h = ½ gt2
∴
t =
2(H − h) g
Since there is no acceleration in the horizontal direction, the horizontal velocity remains constant. The horizontal distance covered by the liquid is x = horizontal velocity × time = vt =
2 gh
2(H − h) g
= 2 h (H − h) This formula shows that whether the orifice in the vessel is at a depth h or at a depth (H-h) from the free surface of the liquid, the emerging liquid will fall at the same distance i.e., the range x of the liquid will remain the same. Now, h (H – h) will be maximum when h = H – h i.e., h = H/2. Hence the maximum range of the liquid is given by xmax = 2
FG HIJ × FG H − HIJ = H . H 2K H 2K
Therefore, when the orifice is exactly in the middle of the wall of the vessel, the stream of the liquid will fall at a maximum distance (equal to the height of the liquid in the vessel).
10.17 VELOCITY OF EFFLUX OF A GAS Since a gas issues out of an orifice in a reservoir under adiabatic conditions, we have, from Bernoulli’s equation for adiabatic flow of a gas,
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477
γ p 1 2 γ p0 1 2 v + v0 + 2 γ − 1 ρ0 γ −1ρ = 2
where v0, p0 and ρ0 are the velocity, pressure and density of the gas inside the reservoir and v, p and ρ their respective values as the gas issues out of the orifice in the reservoir, so that v is the velocity of efflux of the gas. Since, obviously,
v 0 = 0, we have
FG H 2γ F p G γ − 1H ρ
IJ K pI − J ρK
p0 p γ 1 2 − v = γ − 1 ρ0 ρ 2 0
v2 =
or
0
Because under adiabatic conditions,
v2
p0 p − γ , we have γ ρ0 ρ
LM F I OP MM GH JK PP N Q
p 2γ p0 1− = γ − 1 ρ0 ρ0
γ −1 γ
Whence, v the velocity of efflux of the gas, may be easily obtained.
10.18 VISCOSITY When a solid body slides over another solid body, a frictional force begins to act between them. This force opposes the relative motion of the bodies. Similarly, when a layer of a liquid slides over another layer of the same liquid, a frictional force acts between them which opposes the relative motion between the layers. This force is called internal frictional-force. M axim u m ve lo city c b a
A
B H o rizon ta l p la ne
Ze ro ve lo city
Fig. 16
Suppose a liquid is flowing in stream lined motion on a fixed horizontal surface AB. The layer of the liquid which is in contact with the surface is at rest, while the velocity of other layers increases with distance from the fixed surface. In the figure 16, the lengths of the arrows represent the increasing velocity of the layers. Thus there is a relative motion between adjacent layers of the liquid. Let us consider three parallel layers a, b and c. Their velocities are in the increasing order. The layer a tends to retard the layer b, while b tends to retard c. Thus each layer tends to decrease velocity of the layer above it. Similarly, each
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layer tends to increase the velocity to the layer below it. This means that in between any two layers of the liquid, internal tangential forces act which try to destroy the relative motion between the layers. These forces are called ‘viscous forces’. If the flow of the liquid is to be maintained, an external force must be applied to overcome the dragging viscous forces. In the absence of the external force, the viscous forces would soon bring the liquid to rest. The property of the liquid by virtue of which it opposes the relative motion between its adjacent layers is known as ‘viscosity’.
10.19 FLOW OF LIQUID IN A TUBE: CRITICAL VELOCITY When a liquid flows in a tube, the viscous forces oppose the flow of the liquid. Hence a pressure difference is applied between the ends of the tube which maintains the flow of the liquid. If all particles of the liquid passing through a particular point in the tube move along the same path, the flow of the liquid is called ‘stream-lined flow’. This occurs only when the velocity of flow of the liquid is below a certain limiting value called critical velocity’. When the velocity of flow exceeds the critical velocity, the flow is no longer stream lined but becomes turbulent. In this type of flow, the motion of the liquid becomes zig-zag and eddycurrents are developed in it. Reynold proved that the critical velocity for a liquid flowing in a tube is vc = kη/ρa, Where ρ is density and η is viscosity of the liquid, a is radius of the tube and k is Reynold number (whose value for a narrow tube and for water is about 1000). When the velocity of flow of the liquid is less than the critical velocity, then the flow of the liquid is controlled by the viscosity, the density having no effect on it. But when the velocity of flow is larger than the critical velocity, then the flow is mainly governed by the density, the effect of viscosity becoming less important. It is because of this reason that when a volcano erupts, then the lava coming out of it flows speedly inspite of being very thick (of large viscosity).
10.20 VELOCITY GRADIENT AND COEFFICIENT OF VELOCITY Suppose a liquid is flowing in stream-lined motion on a horizontal surface OX. The liquid layer in contact with the surface is at rest while the velocity of other layers increases with increasing distance from the surface OX. The highest layer flows with maximum velocity. Let us consider two parallel layers PQ and RS at distance Z and Z + ∆Z from OX. Let vx and vx + ∆vx be their velocities in the direction OX. Thus the change in velocity in a perpendicular distance ∆Z is ∆vx. That is, the rate of change of velocity with distance perpendicular to the direction of flow is
∆vx . This is called velocity gradient. ∆Z
Now let us consider a liquid layer of area A at a height Z above OX. The layer of the liquid immediately above it tends to accelerate it with a tangential viscous force F, while the layer immediately below it tends to retard it backward with the same tangential viscous force F. According to Newton, the viscous force F acting between two layers of a liquid flowing in stream-lined motion depends upon two factors: (i)
It is directly proportional to the contact-area A of the layers.
(ii)
It is directly proportional to the velocity-gradient
∆vx between the layers. ∆Z
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479 R
S → v x + ∆v x
∆z
A
F
Q → vx
P
F
z
z
O
X
O
(a )
X
(b )
Fig. 17
Combining both these laws, we have F α A
∆vx ∆Z
F = ±ηA
∆vx ∆Z
Where η is a constant called ‘coefficient of viscosity’ of the liquid.
∆vx = 1, then η = ± F. Thus, the coefficient of viscosity of a liquid is defined ∆Z as the viscous force per unit area of contact between two layers having a unit velocity gradient between them. If A = 1 and
In the above formula, ± sign indicate that the force F between two layers is a mutual interaction force.
10.21 POISEUILLE’S EQUATION FOR LIQUID-FLOW THROUGH A NARROW TUBE In deducing an expression for the rate of flow of a liquid through a narrow tube, poiseuille made the following assumptions: (i)
The liquid-flow is steady or streamline, with the streamlines parallel to the axis of the tube.
(ii)
Since there is no radial flow, the pressure, in accordance with Bernoulli’s theorem, is constant over any given cross-section of the tube.
(iii)
The liquid in contact with the wall of the tube is stationary.
All these assumption are found to be quite valid if the tube be narrow and the velocity of liquid-flow really small. Let a liquid of coefficient of viscosity η be flowing through a narrow horizontal tube of radius r and length l and when the condition become steady, let the velocity of flow at all points on an imaginary coaxial cylindrical shell of the liquid, of radius x, be v and, therefore the velocity gradient, dv/dx. Since the velocity of the liquid in contact with the walls of the tube is zero and goes on increasing as the axis is approached, where it is the maximum, it is clear that the liquid layer just inside the imaginary shell is moving faster, and the one just outside it, slower than it. So that, in accordance with Newton’s law of viscous flow, the backward dragging force on the imaginary liquid shell is given by F = ηA
dA dv = 2πxlη dx dx
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Where A is the surface area of the shell equal to 2πxl. And, if the pressure difference across the two ends of the tube be P, the force on the liquid shell, accelerating it forwards = P × πx2. Where πx2 is the area of cross-section of the shell. For the liquid-flow to be steady, therefore, we must have driving force equal to backward dragging force, i.e., P × πx2 = – 2 πxlη(dv/dx), the –ve sign of the dragging force indicating that it acts in a direction opposite to that of the driving force. We, therefore, have dv = –
v = –
Pπx2dx Pxdx =− , which, on integration, gives 2πxlη 2 ηl
P 2ηl
where C1 is a constant of integration.
z
xdx = −
P x2 . + C1 2ηl 2
Since at x = r, v = 0, we have 0 = – or
Pr2 + C1 4 ηl
Pr2 . 4ηl ∴ Velocity of flow at distance x from the axis of the tube, i.e., C1 =
v = –
Px2 Pr2 P + = ( r2 − x2 ) 4 ηl 4 ηl 4 ηl
Which, incidentally shows at once that the profile or the velocity-distribution curve of the advancing liquid is a parabola, the velocity increasing from zero at the walls of the tube to a maximum at its axis. Now, if we imagine another coaxial cylindrical shell of the liquid of radius. x + dx, enclosing the shell of radius x, the cross-sectional area between the two is clearly 2πxdx and, therefore, volume of the liquid flowing per second through this area is, say, dQ = 2πxvdx. Imagining the whole of the liquid inside the tube to consist of such coaxial cylindrical shell, the volume of the liquid flowing through all of them per second i.e., the rate of flow through the tube, as a whole, say, Q, is obtained by integrating the expression for dQ between the limits x = 0 and x = r. We thus have rate of liquid-flow through the arrow tube, i.e.,
z r
Q =
0
z r
2πxvdx =
LM N
2π x
0
πP x r x4 − = 4 2ηl 2 or
Q =
πPr4 8 ηl
or
η =
πPr4 . 8 Ql
2 2
OP Q
P ( r2 − x2 ) dx 4 ηl
r
= 0
LM N
OP Q
πP r4 r4 πP r4 − = . 4 2ηl 2 2ηl 4
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10.22 POISEUILLE’S METHOD FOR DETERMINING COEFFICIENT OF VISCOSITY OF A LIQUID The method is suitable only for liquids having a low viscosity, like water, consist in collecting in a weighed beaker the liquid flowing out in a trickle, in a given time t (in second), from a tall cylinder C through a horizontal capillary tube T of radius r and a known length l fitted near its bottom, the liquid head h being kept constant by means of an overflow tube O, as shown in figure. L iqu id O h T C
To sin k
Fig. 18
Dividing the mass of the liquid collected in the beaker by its density ρ and the time t, its rate of flow Q is obtained. Then, from the relation η =
πPr4 , where, P = hρg, the value 8 Ql
of η for the liquid can be easily calculated. There are two important sources of error in above experiments. (i)
The liquid-flow all along the length of the capillary flow-tube (T) is not uniform or streamline, as assumed. For, the motion of the liquid where it enters the flow-tube is clearly accelerated and does not become uniform or streamline until after covering a good length of it. This error is eliminated if the effective length of the flow-tube is taken to be (l + α) instead of l, where α is almost invariably found to be equal to 1.64 r. The correction (α) is thus quite independent of the length of the tube but depends upon its radius.
(ii)
Part of the pressure difference P between the two ends of the flow-tube is used up in imparting kinetic energy to the liquid flowing through the tube, so that, only the remainder, say P′, is really responsible for overcoming the viscous resistance of the liquid. The effective pressure difference for our purpose is P′, which should, therefore, be substituted for P in the relation for η.
Analogy between Liquid-flow and Current Flow The rate of flow of a liquid through a capillary tube is given by Poiseuillis equation Q = If we put
πPr4 8 ηl
8 ηl = R and call it effective viscous resistance, then πr4 Q = P/R
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Mechanics
Which is clearly a relation similar to I = V/R for flow of elastic current in accordance with ohm’s law. A liquid flow through a capillary tube is thus analogous to the flow of elastic current through a conductor, with the rate of liquid-flow, Q, corresponding to current or rate of flow of charge, I, the pressure difference P across the two ends of the capillary, to potential difference V across the two ends of the conductor and the effective viscous resistance,
8 ηl πr4
to electrical resistance R.
10.23 RATE OF LIQUID-FLOW THROUGH CAPILLARIES IN SERIES Let two capillaries A and B, of lengths l1 and l2 and radii r1 and r2 respectively be connected in series, as shown in figure 19 and let a liquid of coefficient of viscosity η flow through them is steady or stream line motion. Since liquids are incompressible, the same volume of liquid that passes through capillary A, in a given time, also passes through capillary B in the same time. The rate of liquid-flow through each capillary as also through the combination, as a whole, is the same (like the flow of current through conductors in series). Let it be Q. Let the pressure difference across the capillaries A and B be p1 and p2 and that across the combination, as a whole, P. So that P = p1 + p2
...(i)
If, therefore, R1 and R2 be the effective viscous resistances for capillaries A and B and R, for the combination, as a whole, we have Rate of liquid flow Q = Or
p p P = 1 = 2 R R1 R2
P = RQ, p1 = R1Q and p2 = R2Q A B
Fig. 19
Substituting these values in relation (i) above, we have RQ = R1Q + R2Q Or
R = R1 + R2
i.e., the total effective viscous resistance of the combination is equal to the sum of the effective viscous resistances for the individual capillaries. Therefore, rate of liquid-flow through the combination, i.e., Q =
Since ∴
R1 = Q =
P P = R R1 + R2 8 ηl1 8 ηl2 and R2 = 4 πr1 πr24 πP 8η
Fl GH r
1 4 1
+
l2 r24
I JK
−1
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10.24 RATE OF LIQUID-FLOW THROUGH CAPILLARIES IN PARALLEL Let A and B be two capillaries, of lengths l1 and l2 and radii r1 and r2 respectively lying in the same horizontal plane and in parallel with each other as shown in figure 20. Let a liquid of coefficient of viscosity η be flowing through them in steady or stream-line motion. The pressure difference across each capillary, as also across the combination as a whole will be the same, say, P. The rate of liquid-flow through each capillary will, however, be different. Let it be Q1 and Q2 through the two capillaries respectively. Then if Q be the rate of flow through the combination, we have Q = Q1 + Q2 Now,
Q = P/R, Q1 =P/R1 and Q2= P/R2.
where R is the total effective viscous resistance for the combination and R1 and R2, the effective viscous resistances for capillaries A and B respectively. Substituting these values in relation (i) we have P P P = R + R R 1 2 1 1 1 = R + R R 1 2
or
A l1 B l2
Fig. 20
Which corresponds to the law of electrical resistances in parallel. Thus, the reciprocal of the total effective viscous resistance for the combinations is equal to the sum of the reciprocals of the individual effective viscous resistances for the two capillaries. The same will, of course, be true for any number of them. Rate of liquid-flow through the combination is given by Q = or since R1 = ∴
FG H
P 1 1 =P + R R1 R2
8 ηl1 8 ηl2 , we have and R2 = 4 πr1 πr24
F GH
I JK
IJ K
F GH
I JK
4 4 4 4 Q = P πr1 + πr2 = πP r1 + r2 . l2 8 η l1 8 ηl1 8 ηl2
10.25 POISEUILLE’S FORMULA EXTENDED TO GASES According to Poiseuille’s formula, the volume of a liquid flowing per second through a capillary is given by Q =
πpr4 8 ηl
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Mechanics
where l and r are the length and radius of the capillary tube and p the difference of pressure across its ends. In deriving this formula it has been assumed that the volume per second crossing each section of the tube is constant. This is true for an incompressible substance and hence fairly true for a liquid. When a gas flows along a tube, however, the volume per second increases as the pressure decreases along the tube, and hence the above formula must be modified to take this into account. Let us imagine a short length dl of the tube for which the volume per second can be considered constant. Let p be the average pressure and dp the difference in pressure between its ends. The pressure gradient is –dp/dl, the minus sign indicating that the pressure diminishes as l increases. From Poiseuille’s formula, the volume of the gas flowing per second through this length is given by Q = –
πr4 dp . 8 η dl
Now, if p1 is the pressure at the inlet end and Q1 is the volume entering the tube per second then, from Boyle’s Law p1Q1 = pQ = − so that
p1Q1dl = −
πr4 dp p. dl 8η
πr4 p. dp 8η
Integrating this expression for the entire length of the tube, we get
z l
p1Q1
dl = −
0
πr4 8η
z
p2
p. dp
p1
Where p2 is the pressure at the outlet end of the tube. Hence p1Q1l = −
or
p1Q1 =
πr4 ( p22 − p12 ) 8η 2
πr4 ( p12 − p22 ) . 16 ηl
This is Poiseuille’s formula for the flow of a gas through a capillary tube.
10.26 DETERMINATION OF VISCOSITY OF A GAS One method for determining the viscosity of a gas is due to Rankine. The apparatus consists of a uniform tube ABDE bent as shown and provided with taps T1 and T2. A pellet of clean mercury is introduced into the tube. The tube is then joined to a capillary tube C by rubber tubes at A and E. The whole arrangement is then mounted on a board. Two fixed marks H and L are made on the tube such that the volume of the portion ABH = volume of the portion LDE = v (say). A quantity of the experimental gas is introduced into the system by opening the taps T1 and T2 and connecting them to the gas supply. The taps are then closed. Now, the board is held vertical causing mercury pellet to descent in BD under its own weight, thus forcing the
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gas below it through the capillary C. The time-interval t between the top surface of the pellet passing H and the bottom surface reaching L is measured. This is repeated by inverting the board several times. Theory: Let V be the total volume ABDE less the volume of the pellet. If the apparatus be placed horizontally, the pressure of the gas will be uniform throughout. Let it be P. Let the density of the gas at this pressure be ρP, where ρ is the density per unit pressure. The mass of the gas enclosed in this ρPV. With the tube vertical and the mercury pellet in position M, the pressure in ABH will be different than that in HDE. Let them be p and p1 respectively. The corresponding densities are ρp and ρp1 respectively. The volume ABH and HDE are v and (V – v) respectively. Since the mass of the gas is constant. We have ρPV = ρpv + ρp1 (V – v) or
PV = pv + p1 (V – v)
But p1 = p + (mg/a), where m is the mass of the pellet and a is the cross-sectional area of the tube ABDE. ∴ or
PV =
F mg IJ (V − v) pv + G p + H aK
p = P−
T1 B A
mg mg v + a a V C
mg v and p1 = P + a V At the end of the interval t, when the pellet reaches the position M', these pressures becomes
H
M
L
M′
E
D
T2
mg mg (V − v) + p' = P − V a a
Fig. 21
mg (V − v) . V a Now, with the mercury pellet at M, the mass of the gas below the pellet is ρp1(V – v), and with the pellet at M', the mass below it is ρp′1v. Hence the mass of the gas which has passed through the capillary during the interval t and
p'1 = P +
= ρp1(V – v) –ρp′1v
FG H
= ρ P+
IJ K
FG H
IJ K
mg v mg ( V − v) ( V − v) − ρ P + v a V a V
= ρP (V – 2v) Thus the average rate of flow is
ρP ( V − 2v) t
...(A)
Now, when the pellet is at M, the volume per second (Q1 say) entering the capillary is, from Poiseuille’s formula for the flow of a gas, given by 4 p1Q1 = πr
( p12 − p22 ) 16 ηl
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Mechanics
Thus the mass entering per second is p1Qρ =
πr4ρ 2 ( p1 − p22 ) 16 ηl
=
πr4ρ ( p1 − p2 ) ( p1 + p2 ) 16 ηl
=
2mg v πr4ρ mg mg 2P + − 16 ηl a a V a
FG H
IJ K
Similarly, when the pellet reaches at M′, the mass per second entering is
FG 2P + 2mg V − v − mg IJ . H a V a K
πr4ρ mg 16 ηl a
Hence the average rate of flow through the capillary is πr4ρ mg 2P 16 ηl a
Equating this value in that obtained in (A), we obtain πr4ρ mg ρR 2P = ( V − 2v) 16 ηl a t
πr4mgt 8 al( V − 2v) Where (V – 2v) is the volume between the two marks H and L less the volume of the pellet. Hence η for the gas can be calculated. η =
10.27 STOKE’S LAW OF VISCOUS FORCE When a solid body falls under gravity through a liquid (or gas), the layer of the liquid in contact with the body moves with the velocity of the body, while the liquid at a great distance from it is at rest. Thus, the body produces a relative motion between the layers of the liquid. This is opposed by the viscous force in the liquid. The viscous force increases with the velocity of the body and ultimately becomes equal to the force driving the body. The body then falls with a constant velocity, which is called the ‘terminal velocity’. Stokes showed that when a small sphere is moving slowly with a constant velocity through a perfectly homogeneous viscous fluid of infinite extension; the resistive force experienced by the sphere is F = 6πηrv Where r is the radius of the sphere, v its terminal velocity, and η the viscosity of the fluid. This is stoke’s law. The conditions for its validity are (i)
The sphere should be small in size,
(ii)
It should move slowly with a constant velocity,
(iii)
The fluid should be perfectly homogeneous and of infinite extension.
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10.28 STOKE’S FORMULA FOR THE TERMINAL VELOCITY OF A FALLING SPHERE Let us suppose that a small sphere of radius r and density ρ is falling freely from rest under gravity through a fluid of density σ and coefficient of viscosity η, when it acquires the terminal velocity v, the various forces acting upon it are (i)
downward force due to gravity = (4/3) πr3ρg
(ii)
upward thrust due to buoyancy = (4/3) πr3σg
(iii)
resistive viscous force = 6 πηrv
Thus the resultant downward driving force is (4/3)πr3 (ρ − σ)g. Since the sphere has attained a constant velocity, the resultant driving force must be equal to the resistive force. That is, (4/3) πr3 (ρ − σ)g = 6 πηrv or
v =
2 r2 (ρ − σ ) g η 9
This is the required formula.
10.29 VELOCITY OF RAIN DROPS Rain drops are formed by the condensation of water vapour on dust particles. When they fall under gravity, their motion is opposed by the viscous drag in air. As the rain drop falls with increasing velocity, the viscous drag increases and finally becomes equal to the effective force of gravity. The drop then falls with a constant terminal velocity v. Let r be the radius of the drop and ρ its density. Let σ be the density and η the viscosity of air. Now, the various force acting upon the drop are: (i)
its weight acting downward = (4/3) πr3ρg
(ii)
upthrust of the air = (4/3) πr3σg
(iii)
viscous drag = 6 πηrv
Since the drop is falling with a constant velocity, we must have 6 πηrv = (4/3) πr3 (ρ − σ)g
2 r2 (ρ − σ ) g η 9 2 v α r v =
or
The radii of the rain drops are very small. Hence their terminal velocities, being proportional to the square of radii, are also very small. Therefore, to an observer on earth, the drops appear as cloud particles floating in the sky. For the same reason, a larger dust particle (larger r) acquires a larger velocity v and hence falls faster than a smaller one in air.
NUMERICALS Q.1. Calculate the work done in blowing a soap bubble from a radius of 10 cm to 20cm. (S.T. = 25 × 10 –3N/m). Solution. Total surface area of two surfaces or soap bubble before blowing = 4πr12 + 4πr12
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⇒ 8rπ (0.1)2 m2 Similarly total surface area of bubble after blowing = 4πr22 + 4πr22 = 8π (0.2)2 m2 i.e., increase in surface area ∆A = 8π (0. 2)2 − (0.1)2 = 0.24π m2. The work done is therefore ∆W = T.∆A = 25 ×
10–3
× 0.24 π = 6π × 10–3 joules
Q.2. Calculate the work done in spraying a spherical mercury drop of 1mm radius into million identical droplets, the surface tension of mercury being 550 dyne/cm. Solution. Initially the surface area of mercury drop = 4π (1 × 10–1)2 = 0.4π cm2 When its sprayed into million droplets, let ‘r’ be the radius of one droplet then
4 4 3 πr = π (1 ×10−1 )3 3 3 (102r)3 = (1 × 10–1)3
106 ×
r = 10–3cm. Final surface area of million droplets = 106 × 4 π (10−3 )2 = 4 π cm2 Increase in surface area ∆A = 4π – 0.04 π = 3.96 π ∴ Work done against S.T. in spraying ∆W = T.∆A = 550 × 3.96 × 3.14 ⇒ 6839 ergs Q.3. Eight air bubble each of radius ‘a’ coalesce to form a bigger bubble of radius ‘R’. If P is atmospheric pressure, T the surface tension and ∈ =
T << 1, Show that ap
LM FG 2 IJ ∈OP N H 3K Q
R = 2a 1 +
Solution. Taking isothermal conditions, we must have for 8 air bubbles to coalesce PV + PV +..... = p'v' 8PV = p'v' Since ‘a’ is radius of original bubble p =
FG P + 2T IJ H aK
... (i)
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489
v =
4 πa3 3
Similarly ‘R’ is the radius of bigger bubble and we have for it p' =
FG P + 2T IJ H RK
v' =
4 πR3 . 3
Substitute these values in eqn. (i) gives
FG H
8 P+
2T a
IJ 4 πa K3
3
=
FG P + 2T IJ 4 πR H RK 3
3
8Pa3 + 16Ta2 = PR3 + 2TR2 Dividing both sides by 8Pa3, we get
1+ But
2T R3 TR2 + = Pa 8 a3 4 Pa3
T =∈ aP 1 + 2ε =
FG R IJ H 2a K
FG R IJ H 2a K
3
3
+
F GH
R2 ε 4 a2
= 1 + 2ε 1 −
R2 8 a2
I JK
L F R I OP R = 2a M1 + 2ε G 1 − MN H 8a JK PQ L 1 F R I + ....OP R = 2aM1 + . 2ε G 1 − MN 3 H 8 a JK PQ L 2 F R I + ....OP = 2a M1 + ε G 1 − MN 3 H 8 a JK PQ 1/ 3
2
2
2
2
2
2
Which concludes that
LM FG 2 IJ ε OP N H 3K Q
R = 2a 1 +
Q.4. A number of soap bubbles coalesce to form one spherical soap bubble under isothermal conditions. If ‘V’ represents change in volume ‘S’ the change in surface area, ‘P’ is atmospheric pressure and T is S.T. of soap solution, show that 3PV + 4ST = 0
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Mechanics
Solution. Let ‘n’ soap bubbles coalesce to form one soap bubble, under isothermal conditions. Then from Boyles law product PV for system will be constant. Hence
n(pv) = p'v'
Suppose ‘r’ is radius of small bubbles and ‘R’ be the radius of single bubble formed. p =
FG P + 4T IJ and v = H rK
p' =
FG P + 4T IJ and v' = H RK
Then Similarly for bigger bubble
4 3 πr 3
4 πR3 3
Putting these value
FG H
IJ 4 πr = FG P + 4T IJ 4 πR K3 H RK3 L 4 πr n − 4 πR OP 4 T (4 πR − n. 4 πr ) PM 3 Q = 3 N 3 F n . 4 πr − 4 πR I GH 3 3 JK = V, the change in volume n P+
4T r
3
3
3
3
2
or,
3
Now,
2
3
( 4 πr2n − 4 πR2 ) = S the change in surface area
and
PV = − which gives
4T S 3
3PV + 4ST = 0
Q.5. Two soap bubbles of radii ‘a’ and ‘b’ combine to form a single bubble of radius ‘c’. If external pressure is P. Show that surface tension S is given by
P( c3 − a3 − b3 ) . 4 ( a2 + b2 − c2 ) Solution. If bubbles coalesce under isothermal conditions, product of pressure and volume of enclosed air must remain same for the system. S =
i.e.,
or, or,
p1v1 + p2v2 = pv
FG P + 4S IJ 4 πa + FG P + 4S IJ 4 πb H aK 3 H bK3 3
3
=
FG P + 4S IJ 4 πc H cK3
3
P(a3 + b3 – c3) = 4S (c2 – b2 – a2) S =
P(a3 + b3 − c3 ) 4 ( c2 − b2 − a2 )
S =
P(c3 − a3 − b3 ) . 4 ( a2 + b2 − c2 )
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491
Q.6. Two air bubbles of radii 0.002 m and 0.004 m formed in same liquid of surface tension 0.07 N/m come together to form a double bubble. Find the radius of surface common to both the bubbles. Solution. The pressure inside air bubble formed in liquid of surface tension T is given by p =
FG P + 2T IJ H rK
Hence, pressure inside smaller bubble p1 =
FG P + 2T IJ H rK 1
and pressure inside bigger bubble p2 =
FG P + 2T IJ H rK 2
Since pressure inside smaller bubble will be greater, when the two bubbles come together, the common surface will have concave side towards smaller and convex side towards larger bubble. The difference of pressure on the two sides of common surface will be
FG P + 2T IJ − FG P + 2T IJ H rK H rK F 1 1I = 2T G − J Hr r K
p =
1
1
2
2
If R is the radius of curvature of common surface the excess pressure on its concave side must be p = So,
i.e.,
2T R
FG H
2T 1 1 = 2T − R r1 r2
IJ K
1 1 1 = − R r1 r2 R =
r1 r2 r2 − r1
Putting r1 = 0.002 m and r2 = 0.004 m we get R =
0. 002 × 0. 004 ⇒ 0. 004 m (0. 004 − 0. 002)
Ans.
Q.7. A plate of 100 cm2 is placed on upper surface of 2 mm thick oil having coeff. of viscosity η = 15.5 Poise. Calculate the horizontal force needed to move the plate with velocity 3 cm/sec. Solution. The viscous force due to viscosity which acts to oppose relative motion is
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Mechanics
dv dx η = 15.5 C.G.S. units A = 100 cm2 F = −ηA
As given and
velocity gradient
3 cm/sec dv = = 15 sec–1 0. 2 cm dx
Putting these values the viscous force is F = – 15.5 × 100 × 15 dynes = –23250 dynes Thus, the forward force needed to overcome the viscous force and move the plate will be F = 2.325 × 104 dynes. Q.8. A horizontal capillary of diameter 2 mm and length 20 cm is connected to a water tank and 0.2 cc. water flow per sec. Calculate the rate of flow, if another capillary of length 10 cm and diameter 1 mm is formed in series with the first capillary. Solution. The rate of flow of liquid through a capillary is given by V =
P πPr4 = 8η l 8 ηl . π r4
...(i)
If two capillaries are joined in series the combined viscous resistance will R =
=
LM 8ηl + 8ηl OP N πr πr Q l I 8η F l + J π GH r r K 1 4 1
2 4 2
1 4 1
2 4 2
Hence for same pressure difference the new rate of flow will be V =
8η F l π GH r
P
1 4 1
+
l2 r24
I JK
As given in first case V = 0.2 c.c., l1 = 20 cm, r1 = 1 mm = 0.1 cm 0.2 =
P 8η 20 π ( 0.1)4
FG H
IJ K
...(ii)
In second case, when another capillary is also used having. l2 = 10 cm and r2 = 0.5 mm = 0.05 cm V′ =
P 8η 20 10 + 4 π ( 0.1) ( 0. 05)4
FG H
IJ K
...(iii)
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493
Dividing (ii) and (iii)
0.2 = V′
=
V' =
LM 20 N (0.1)
10 (0. 05)4 20 (0.1)4
4
+
LM1 + 10 F 0.1 I MN 20 GH 0. 05 JK
4
OP Q
OP PQ = [1 + 8]
0. 2 = 0. 022 cc/sec 9
Q.9. A capillary of radius 0.2 mm and length 12cm is fixed horizontally at the bottom of a large reservoir filled upto 25 cm with alcohol (density = 0.8 gm/c.c., η = 0.006 poise). Find velocity of liquid flowing along axis of capillary. Solution. If pressure difference P is applied across a capillary of length l and radius r, the velocity of flow of layer lying at distance x from axis is given by V =
P 2 ( r − x2 ) 4ηl
Putting x = 0, We get the velocity of liquid flowing along axis of tube
Pr2 4 ηl P = hdg =25 × 0.8 × 9.8
Vaxis = As given
r = 0.02cm, l = 12cm and h = 0.006 Poise Substuting the values, Vaxis =
25 × 0.8 × 980 × (0.02)2 4 × 0. 006 ×12
=
25 × 8 × 98 × 4 × 10−5 4 × 6 × 12 × 10 −3
=
25 × 98 × 10−1 = 27. 2 cm/sec. 9
Q.10. Three capillaries of same length but radii in the ratio 3:4:5 are connected in series and liquid flows through them. If pressure across third capillary is 8.1 mm, deduce the pressure across the first capillary. Solution. Let l be the length of each capillary and radii 3r, 4r, 5r. Since they are in series rate of flow must be same through the three πp2 ( 4 r)4 πp3 ( 5r)4 πp1 (3r)4 = = 8 ηl 8 ηl 8 ηl
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Mechanics
p1(3)4 = p2 (4)4 = p3 (5)4 p1 =
8.1 × ( 5)4 8.1 × 625 = = 62.5 mm. 4 ( 3) 81
Q.11. Two tubes A and B of length 100 cm and 59 cm have radii 0.1 mm and 0.2 mm respectively. A liquid passing through both enters in tube A at pressure 80 cm of Hg and leaves tube B at 76 cm of Hg. Find the pressure at junction of two tubes. Solution. Let P be the pressure at the junction of tubes A and B. Then for tube A, p1 = (80 – P), l1 = 100 cm, r1 = 0.1 mm and for tube B, p1 = (P – 76), l2 = 59 cm, r2 = 0.2 mm. Since same quantity of liquid passes through both tube per sec, V =
πp1r14 πp r4 = 22 8 η l1 8 ηl2
(80 − P)(0.1)4 (P − 76) (0.2)4 = 100 59 80 − P P − 76 × 16 = 100 59 59 (80 – P) = 1600 (P – 76) 126320 = 1659 P P = 76.051 cm Q.12. A cylindrical vessel of radius 5 cm is completely filled with water and carries a horizontal capillary of length 20 cm and internal diameter 0.8 mm. What time it will take for vessel to become half empty ? (η = 0.01 poise) Solution. Let A be cross-sectional area of vessel. If height falls by ‘dh’ in vessel due to escape of liquid through capillary in time dt, rate of flow
−
πpr4 Adh = 8 ηl dt
−
π ( hdg) r4 Adh = 8 ηl dt
dt = −
8ηlA dh πdgr4 h
For vessel to become half empty, level will fall from h to given by t = −
=
8 ηlA πdgr4
z
h /2
h
dh h
8 ηlA log e 2 πdgr4
h . Hence total time will be 2
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495
As given A = π(5)2, l = 20 cm, r = 0.08 cm, η = 0.01 poise, d = 1 gm/cc, g = 980 cm/sec2 t =
8 × 0. 01 × 20 × π (5)2 = 99. 6 sec π × 1 × 980 × ( 0. 08)4
Q.13. A horizontal pipe of non-uniform bore has water flowing through it such that the velocity of flow is 40 cm/sec at a point where the pressure is 2 cm of mercury column. What is the pressure at a point where the velocity of flow is 60 cm/sec ? (Take g = 980 cm/sec2 and density of water = 1 gm/cc). Solution. Here p1 = 2 cm of mercury column = 2 × 13.6 × 980 dyne/cm2 ρ1 = ρ2 = ρ = 1 gm/cc, v1 = 40 cm/sec, h1 = h2 = h In accordance to Bernoulli’s theorem 1 p1 1 p + gh + v12 = 2 + gh + v22 ρ 2 ρ 2 Where p2 is the pressure at the second point
1 2 (v2 − v12 ) = 2
or, so that, or
FG p Hρ
1
−
IJ K
p2 = p1 − p2 ρ
1 ( 602 − 402 ) = 2 × 13.6 × 980 – p2 2
1000 = 26650 – p2 p2 = 26650 – 1000 = 25650 dyne/cm2 = 25650/13.6 × 980 = 1.925 cm of mercury column.
Q.14. Calculate the velocity of efflux of kerosene oil from a tank in which the pressure is 50 lb wt/square inch above the atmospheric pressure. The density of Kerosene is 48 lb/c.ft. Solution. We know the velocity of efflux v =
2gh , where h is the height of the liquid
surface from the axis of the orifice. Now, pressure due to kerosene at the level of the axis of the orifice = hρg poundals/ft2 = hρ lb wt/ft2. But this is given to be 50 lb wt/(inch)2 = 50 × 144 lb wt/ft2 hρ = 50 × 144 or, h = 50 × 144/ρ = 50 × 144/48 ft Hence,
velocity of efflux, v = =
2gh
2 × 32 × 50 × 144 × 48
= 97.97 or 98 ft/sec. Q.15. A water main of 20 cm diameter has pilot tube fixed into it and the pressure difference indicated by the gauge is 5 cm of water column. Calculate the rate of flow of water through the main. (g = 980 cm/sec2 and ρ for water = 1 gm/cc).
496
Mechanics
Solution. Radius (r) of the main = 20/2 = 10 cm and therefore its area of cross-section α = πr2 = π(10)2 = 100 π sq cm
1 2 Since loss of kinetic energy per unit mass on stoppage of flow = v = pressure energy 2 per unit mass = p/ρ = p/1 = p = 5 × 1 × 980 v 2 = 10 × 980 v =
9800 = 99. 0 cm/sec.
Rate of flow of water through the main = velocity of flow × area of cross-section of the main = 99 × 100 π = 31100 cc/sec. or
31.1 liters/sec.
Q.16. The diameters of a water main where a venturimeter is connected to it are 20 cm and 10 cm. What is the rate of water flow. If the water levels in the two piezometer tubes differ by 5 cm ? (g = 980 cm/sec2). Solution. The rate of flow of water through the main is given by the relation Q = a1v1 = a1a2
2 hg / ( a12 − a22 )
a1 = πr12 = π (20 /2)2 = 100 π sq cm, a2 = πr22 = π (12 /2)2 = 36 π sq cm and h = 5 cm, we have
Rate of flow of water through the main i.e., Q = 100 π × 36 π
= 3600 π
2 × 5 × 980 (100 π)2 − (36 π)2
9800 = 11920 cc/sec or 11.92 or 12 liters/sec. 8704
Q.17. Calculate the mass of water flowing in 10 minutes through a tube 0.1 cm in diameter, 40 cm long, if there is a constant pressure head of 20 cm of water. The coefficient of viscosity of water is 0.0089 C.G.S. units. Solution. From Poisueille’s Eqn. volume rate of water given by Q =
πPr4 π × 20 × 1 × 981 × ( 0. 05)4 = = 0.1353 cc/sec. 8 ηl 8 × 0. 0089 × 40
Volume of water flowing out in 10 minutes = 0.1353 × 10 × 60, say V = 81.18 cc mass of water flowing out in 10 minutes = V × P = 81.18 × 1 = 81.18 gm Ans. Q.18. A cylindrical vessel of radius 7 cm is filled with water to a height of 50 cm. It has a capillary tube 10 cm long, 0.2 mm radius protruding horizontally at its bottom. If the velocity of water is 0.01 C.G.S. units and g = 980 cm/sec2. Find the time in which the level will fall to a height of 25 cm.
Fluids
497
Solution. Let h be the height of the water column in the vessel at any given instant and dh, the fall in its height in a small interval of time dt. Then if A be the area of cross section of the vessel. We have rate of flow through the capillary tube, i.e., Q = –Adh/dt the –1 ve sign indicating that h decreases as t increases. But, as we know, the rate of flow of water through a capillary tube is given by poiseuille’s eqn., Q = πPr4/8ηl. Where P = hρg (ρ = 1 gm/cc) we have
−A
dh 8 ηlA dh πhgr4 or dt = − = 8 ηl dt πgr4 h
z z t
And ∴
h2
dt =
0
Or,
−
h1
8ηlA dh . , πgr4 h
h 8 ηlA log e 1 h2 πgr4
whence t =
putting the values of A = π × 72, r = 0.02 cm, h1 = 50 cm and h2 = 25 cm
We have
t =
8 × 0. 01 × 10 × π × 72 50 × 2. 3026 log10 25 π × 9. 8 × (0. 02)4
= 1.734 × 105sec = 48.16 hrs. The water level in the vessel will thus fall to 25 cm in 48.16 hrs. Q.19. Write down Poiseuille's formula for the rate of flow of a liquid through a capillary tube. From this show that if two capillaries of radii a1 and a2 having length l1 and l2 respectively are set in series the rate of flow Q* is given by
πP Q = 8η
Fl GH a
1 4 1
l + 24 a2
I JK
−1
Where P is the pressure across the arrangement and η , the coefficient of viscosity of the liquid. Solution. Let P1 be the pressure across the first and P2 across the second, capillary. So that, P = P1 + P2, and, therefore P2 = P – P1. Obviously, in accordance with the eqn. of continuity, the rate of flow through either capillary will be the same, say Q and therefore, from Poiseuille’s eqn., we have Q = πP1a14/8ηl1 = πP2 a24 /8ηl2 = π(P – P1) a24 /8ηl2,
P1a14 /l1 = Pa24 /l2 − P1a24 /l2
whence
or,
P1
Fa GH l
4 1
1
+
a24 l2
I JK
=
P1 =
Pa24 l2
e
Pa24 / l2 a14 / l1
j+e
a24 / l2
j
=
a14
e
Pl1 l2 / a24
j + e l /a j 1
4 1
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Mechanics
Fl GH a
Pl1 = 4 a1
2 4 2
l + 14 a1
I JK
−1
Substituting this value of P1 in expression Q = πP1 a14 /8ηl1, we have
πP Q = 8η
Fl GH a
1 4 1
+
l2 a24
I JK
−1
Q.20(a). Three capillaries of length 8L, 0.2L and 2L with their radii r, 0.2r and 0.5r respectively are connected in series. If the total pressure across the system in an experiment is p deduce the pressure across the shortest capillary. (b) Fig. 22 below shows two wide tubes P and Q connected by three capillaries A, B, C whose relative lengths and radii are indicated. If a pressure p is maintained across A, deduce (i) the ratio of liquid flowing through A and B (ii) the pressure across B and across C. A
( l, r) C
P
Q
B
( l/2, 2 r) ( l/2, r/2 )
Fig. 22
Solution. (a) Obviously the rate of flow of liquid across each capillary is the same. So that, if p1, p2 and p3 be the pressure across the three capillaries respectively, we have, in accordance with Poiseuille’s eqn., Q =
Whence
πp1r4 πp2 (0. 2r)4 πp ( 0. 5r)4 = 3 = 8 η (8L) 8 η (0. 2L) 8 η (2L)
p1 p2 p = 3 = 64 1000 256 p2 P2 × 64 8 32 p2 and p3 = × 256 = p2 = 1000 125 1000 125 8 p2 32 p2 165 + p2 + = p2 p = p1 + p2 + p3 = 125 125 125
p1 = Now
p2 = 125 p = 0. 7575 p . 165 4 (b) (i) Clearly, rate of flow liquid through capillary A is Q = πpr /8 ηl
and rate of liquid flow through B and C is, say Q′ = Whence
πp1 (2r)4 πp2 ( r /2)4 = 8 η( l /2) 8 η( l /2)
(2r)4 p1 = (r/2)4p2 or p2 = 256 p1
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499
Since p = p1 + p2, we have p = p1 + 256 p1 = 257 p1 ∴ Ratio of liquid flowing through A and B =
= (ii) Since
Q πpr4 /8 ηl = Q πp1 (2r)4 / 8 η(l /2)
257 πp1r4 /8 ηl = 257 : 32 32πp1r4 /8 ηl
p = 257 p1, we have
pressure across capillary B, i.e., p1 = p/257 and
256 p Ans. 257 Q.21. A gas bubble of diameter 2 cm rises steadily through a solution of density 1.75 gm/c.c. at the rate of 0.35 cm/sec. Calculate the coefficient of viscosity of the solution. (Neglect density of the gas).
pressure across capillary C, i.e., p2 = p – p1 = p – p/257 =
Solution. We know, the coefficient of viscosity of the solution is given by the relation
η =
2 9
FG ρ − σ IJ r g H v K 2
Here radius of the bubble r = 2/2 = 1 cm density of the solution σ = 1.75 gm/c.c., density of bubble ρ = 0 and velocity of the bubble v = – 0.35 cm/sec (because it is directed upward). We have
2 6 r2 g 2 1. 75 × 12 × 981 = × 9 v 9 0. 35 = 1.09 × 103 Poise.
η =
Q.22. A glass bulb of volume 500 c.c. has a capillary tube of length 40 cm and radius 0.020 cm leading from it. The bulb is filled with hydrogen at an initial pressure of 86 cm of mercury density 13.6 gm/c.c. and it is found that if the volume of the gas remaining in the vessel is kept constant, the pressure falls to 80 cm of mercury in 254 sec. If the height of the barometer is 76 cm and g = 981 cm/sec2, find the viscosity of hydrogen. Solution. This is obviously a straight application of Searle’s method for determining the coefficient of viscosity of a gas and we have, therefore η =
πr4Hρgt h 2H + h2 8lV × 2.3026 log10 1 . h2 2H + h1
FG H
IJ K
H = 76 cm, ρ = 13.6 gm/c.c., g = 981 cm/sec2, t = 25.4 sec, l = 40 cm, V = 500 c.c. h1 = 86 – 76 = 10 cm and h2 = 80 – 76 = 4 cm. We have coefficient of viscosity of hydrogen. η =
=
π (0. 02)4 × 76 × 13. 6 × 981 × 25. 4 10 152 + 4 8 × 40 × 500 × 2. 3026 log10 . 4 152 + 10
FG H
π ( 0. 02)4 × 76 × 13. 6 × 981 × 25. 4 8 × 40 × 500 × 2. 3026 × 0. 3815
= 9. 204 × 10−5 Poise.
IJ K
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Mechanics
Q.23. Two horizontal capillary tubes A and B are connected together in series so that a steady stream of fluid flows through them. A is 0.4 mm is internal radius and 256 cm long. B is 0.3 mm in internal radius and 40.5 cm long. The pressure of the fluid at the entrance is 3 inches of mercury above the atmosphere. At the exit end of B, it is atmospheric (30 inches of mercury). What is the pressure at the function of A and B if the fluid is (i) a liquid (ii) a gas ? Solution. Case (i): When the fluid is a liquid—Let the pressure at the junction of capillary tubes A and B be h inches of mercury, so that pressure across capillary A is say, P1 = (33 – h) inches of mercury, and pressure across capillary B is, say, P2 = (h – 30) inches of mercury. The two tubes being connected in series. The volume rate of flow of the liquid is the same through both and we have
Q =
πp1r14 πp r 4 p l = 2 2 , whence 1 = 1 8 ηl1 8 η l2 p2 l2
FG r IJ Hr K
4
2
1
Where l1, l2 and r1, r2 are the length and the radii of the two tubes respectively. Substituting the values of p1 and p2 from above, we have
FG IJ H K
33 − h 256 0. 3 = h − 30 40. 5 0. 4 or,
4
=
256 81 × =2 40. 5 256
33 – h = 2(h – 30) = 2h – 60, or 3h = 93, h = 31 inches i.e., the pressure at the junction of the two capillary tubes is 31 inches of mercury.
Case (ii): When the fluid is gas—The mass rate of flow of the gas through either capillary tube is the same and we thus have ηr14 ( 332 − h2 ) ηr24 ( h2 − 302 ) = 16 ηl1 16 η l2
Where h, as before, is the pressure in inches at the junction of the two tubes we have
332 − h2 = h2 − 302
FG r IJ Hr K 2
4
=2
1
332 – h2 = 2h2 – 2 × 302, or, h2 = (2 × 302 + 332/3) = 2889/3 = 963 h =
963 = 31. 03 inches
The pressure at the junction of the two capillary tubes is, in this case, equal to 31.03 inches of mercury. Q.24. A plate of area 100 cm2 is placed on the upper surface of caster oil 2 mm thick. Taking the coefficient of viscosity to be 15.5 Poise. Calculate the horizontal force necessary to move the plate with a velocity 3 cm/sec. Solution. The (horizontal) viscous force is, by Newton’s hypothesis, given by F = ηA
dv dy
Fluids
501
For the present case, we can put F = ηA
v y
η = 15.5 poise, A = 100 cm2, v = 3 cm/sec and y = 2 mm = 0.2 cm
3 = 2. 325 × 104 dynes. 0. 2 Q.35. Water is conveyed through a horizontal tube 8 cm in diameter and 4 km in length at a rate of 20 litre per second. Assuming only viscous resistance. Calculate the pressure required to maintain this flow. Viscosity of water may be taken as 0.01 cgs units. F = 15. 5 × 100 ×
Solution. The volume of water flowing per second through a capillary of length l and radius a under a pressure difference p is given by Q =
πpa4 8 ηl
the pressure difference required is, therefore p = Q
8ηl πa4
Q = 20 liter/sec = 20 × 103 cm3/sec, η = 0.01 cgs unit, l = 4 km = 4 × 105 and a = 4 cm 3 p = 20 × 10 ×
8 × 0. 01 × (4 × 105 ) 3.14 × (4)4
= 7.96 × 105 dynes/cm2 Ans. Q.26. Calculate the rate at which water flows through a capillary tube of length 0.5 meter with an internal diameter of 1 mm. Coefficient of viscosity is 1.3 × 10–3 kg/m-sec. The pressure head is 20 cm of water. Solution. By Poiseuille’s formula, the rate of water flow is given by Q =
πpa4 8 ηl
p = 20 cm of water = 0.20 m × (1 × 103 kg/m3) × 9.8 nt/kg = 1.96 × 103 nt/m2 a = 0.5 mm = 0.5 × 10–3 m, η = 1.3 × 10–3 kg/m-sec and l = 0.5 m
3.14 × (1. 96 × 103 nt / m2 ) × (0. 5 × 10−3 m)4 8 × (1. 3 × 10−3 kg/m - sec) × (0.5 m) = 7.4 × 10–8 m3/sec.
Q =
Q.27. A capillary tube, 1.0 mm in diameter and 20 cm in length is fitted horizontally to a vessel kept full of alcohol of density 0.8 gm/c.c. The depth of the centre of the capillary tube below the surface of alcohol is 30 cm. The viscosity of alcohol is 0.012 C.G.S. unit. Find the amount of alcohol that will flow in 5 minutes. Solution. The volume of liquid flowing per second through a capillary of length l and radius a under a pressure difference p is given by Q =
πpa4 8 ηl
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Mechanics
p = hρg = 30 × 0.8 × 980 dyne/cm2, a = 0.5 mm = 0.05 cm. η = 0.012 Poise and l = 20 cm
Q =
3.14 × 30 × 0. 8 × 980 × ( 0. 05)4 = 0. 24 cm2 /sec 8 × 0. 012 × 20
Volume flowing in 5 minutes = 0.24 cm3/sec × (5 × 60) sec = 72 cm3 The density of alcohol is 0.8 gm/cm3. Therefore, mass of alcohol flowing in 5 minutes = 72 cm3 × 0.8 gm/cm3 = 57.6 gm Q.28. Calculate the volume of water flowing in 10 minutes through a tube 0.1 cm in diameter and 40 cm long if there is a constant pressure head of 20 cm of water. The coefficient of viscosity is 0.0089 cgs units. Q.29. Find the coefficient of viscosity from the given data: volume of water coming out through the capillary in 10 minutes = 80.7 cc, head of water = 39.1 cm, length of capillary = 60.24 cm radius = 0.0514 cm, g= 980 cm/sec. Solution. By Poiseuille’s capillary flow formula η =
πpa4 8Ql
p = 39.1 cm of water = 39.1 × 1 × 980 dynes/cm2, l = 60.24 cm, a = 0.0514 cm Q =
80. 7 = 0.1345 cm3 /sec 600
η =
3.14 × (39.1 × 1 × 980) × (0. 0514 )4 8 × 0.1345 × 60. 24
b
g
= 0.01296 gm/cm-sec = 1.296 × 10–2 Poise. Q.30. Deduce the velocity profile for stream line flow of liquid through a capillary of circular cross-section. Deduce the fraction of liquid which flows through the section upto distance a/2 from the axis, where a is the radius of capillary. Solution. The velocity of liquid flow at a distance r from the axis of the capillary tube is given by v =
p ( a2 − r2 ) 8 ηl
Now, the volume of the liquid flowing per second through a thin cylindrical shell of radii r and r + dr is given by dQ = v(2πrdr) =
πp 2 ( a − r2 ) rdr 2ηl
the volume of the liquid Q, flowing per second through the section of the tube upto a distance a/2 from the axis is obtained by integrating this expression between the limits r = 0 and r = a/2
Fluids
Q' =
πp 2ηl
z
503 a /2
0
( a2r − r3 ) dr =
the volume Q flowing through the entire tube is ∴
7 πpa4 16 8 ηl
πpa4 8 ηl
Q′ 7 . = Q 16
Q.31. A liquid with coefficient of viscosity 0.50 Poise flows through a capillary tube having an internal diameter of 1 mm and length 25 cm due to pressure difference of 10 cm of mercury at its two ends. Find the volume of the liquid flowing out per minute and also its velocity in the capillary tube along its axis. Solution. The volume of liquid flowing per second through a capillary Q =
πpa4 8 ηl
p = 10 × 13.6 × 980 dyne/cm2, a = 0.5 mm = 0.05 cm, η = 0.50 Poise, l = 25 cm Q =
3.14 × 10 × 13. 6 × 980 ( 0. 05)4 = 0. 0261 cm3 /sec 8 × 0. 50 × 25
Volume flowing per minute = 0.0261 × 60 = 1.57 cm3/min. Now, the velocity of liquid flow at a distance r from the axis of the capillary tube is given by v =
p ( a2 − r2 ) 4 ηl
v =
pa2 10 × 13. 6 × 980 × ( 0. 05)2 = 4 ηl 4 × 0. 50 × 25
At the axis of the tube r = 0
= 6.7 cm/sec. Q.32. A capillary of radius 0.2 mm and length 12 cm is fixed horizontally at the bottom of a large reservoir filled upto 25 cm with alcohol of density 0.8 gm/c.c. and viscosity 0.006 Poise. Find the velocity of the liquid following along the axis of the capillary. Q.33. Water is flowing through a capillary tube 40 cm long and of 1 mm internal radius under a constant pressure head of 15 cm of water. Calculate the maximum velocity of water in the tube and verify that the flow is stream-lined. Given: for water viscosity = 0.0098 Poise. Reynold’s number = 1000 and g = 980 cm/sec2. Solution. The velocity of liquid flow at a distance r from the axis of the capillary tube (length l, radius a) under pressure p is given by v =
p ( a2 − r2 ) 4 ηl
It is maximum at the axis (where r = 0) thus vmax =
pa2 4 ηl
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Mechanics
p =15 cm of water = 15 × 1 × 980 dynes/cm2, a = 1mm = 0.1 cm, η = 0.0098 poise and l = 40 cm vmax =
15 × 1 × 980 × ( 0.1)2 = 93. 75 cm/sec 4 × 0. 0098 × 40
Now, the critical velocity of flow through the tube is Kη vc = ρa
Where K is Reynold’s number and ρ is density of water vc =
1000 × 0.0098 = 98 cm/sec 1 ×0.1
Thus vmax < vc. Hence the flow is streamlined. Q.34. 0.0314 c.c. of a liquid is flowing out per second through a capillary tube of 1 mm radius. Calculate the velocity of the liquid at a point (i) on the axis of the capillary, (ii) adjacent to the wall, (iii) at 0.5 mm distance from the axis. Solution. The velocity profile of a viscous liquid flowing in a capillary tube of length l and radius a is given by
P ( a2 − r2 ) 4ηl where v is the velocity of flow at a distance r from the axis of the tube. By Poiseuille’s formula, the volume of the liquid flowing per second is given by v =
πpa4 8 ηl Eliminating η between the above two formula: we get Q =
v =
2Q ( a2 − r2 ) ηa4
Q = 0.00314 cm3/sec and a = 0.1 cm v =
2 × 0. 0314 ( a2 − r2 ) = 200 ( a2 − r2 ) 3.14 × ( 0.1)4
(i) On the axis of the capillary, r = 0 v = 200a2 = 200 (0.1)2 = 2 cm/sec (ii) At the wall, r = a v = 200 (a2 ˆ– a2) = 0 (iii) At 0.05 cm from the axis, r = 0.05 cm v = 200 [(0.1)2 – (0.05)2] = 1.5 cm/sec Ans. ∴ Q.35. Water at 20 °c is flowing through a capillary tube of length 1 meter and of diameter 2 mm under the constant pressure head of 10 cm water. Calculate the velocity of water at a point (i) on the axis, (ii) adjacent to wall of the tube, (iii) at 0.5 mm distance from the axis. Coefficient of viscosity for water at 20°C = 0.01 poise. [Ans. [(i) 24.5 cm/sec, (ii) zero, (iii) 18.375 cm/sec]
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Q.36. Water flows in a streamline through a horizontal pipe of cross-sectional radius a from a vertical tank of cross-section. Find the time when the height of the level in the tank becomes half. Solution. Let h be the height of liquid in the tank above the horizontal tube at any instant t. The pressure difference across the ends of the tube at that instant is p = ρgh, Where ρ is the density of liquid. If the height of liquid falls through dh is a small time interval dt, then the volume of the liquid flowed through the tube in time dt is A(dh). Where A is the area of cross-section of the tank. Therefore the rate of flow of the liquid at the instant t is
A(dh) dt The minus sign is put because h decreases with increase in t. But from Poiseuille’s formula, the rate of flow −
= ∴
–
πpa4 π (ρgh) a4 = 8 ηl 8 ηl
π ( ρgh) a4 A ( dh) = 8 ηl dt
dt = –
8ηlA dh πρga4 h
The initial height of the liquid is H (say). Let t be the time required by the level to fall to H/2. Then, integrating the last expression between proper limits, we have
z
t
dt = −
0
t = −
=
8lηA πρga4
z
H/2
H
dh 8 lηA H/2 log e h H =− 4 h πρga
8lηA [log e H/2 − log e H] πρga4
8lηA 8lηA [log e H − log eH/2] = log e H 4 πρga πρa4
This is the required expression. Q.37. A cylindrical vessel of radius 0.05 meter has at its bottom a horizontal capillary tube of length 0.2 meter and internal radius 0.4 mm. The vessel is completely filled with water and then the water is allowed to flow out of the capillary. How long it will take for the water level to fall to half its initial height in the vessel. The viscosity of water is 0.001 mks unit (g = 9.8 mtr/sec2, density of water = 1 × 103 kg/m/r3, loge 2 = 0.693). Solution. The time required for the water level to fall to half its initial height is given by t = t =
8lηA log e 2 πρga4 8( 0. 2 m) ( 0. 001 kg /m - sec) { π × ( 0. 05 m)2 } × 0. 693 π (1 × 103 kg /m3 ) ( 9. 8 m /sec2 ) ( 0. 4 × 10−3 m)4
= 1.1 × 104 sec ≈ 3 hours.
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Mechanics
Q.38. A horizontal capillary tube of length 10 cm and internal radius 0.5 mm is attached to the bottom of a vessel of cross-section 20 cm2. The vessel is initially filled with water to a height of 20 cm above the capillary tube. Find the time taken by the vessel to empty onehalf of its contents coefficient of viscosity of water is 0.01 Poise. [Ans. 576.5 sec] Q.39. Prove that if the level of water in a reservoir falls from mark A to B in 120 sec when it flows through a capillary and in case of a liquid of specific gravity 0.9, the level falls from A to B in 100 sec, then the ratio of viscosity of the liquid with that of water is 3:4. Q.40. A horizontal capillary of diameter 2 mm and length 20 cm is connected to a tank of water and 0.2 c.c. of water flows out per second. What will be the rate of flow of water if another capillary of length 10 cm and diameter 1 mm is joined in series with the first capillary ? Solution. The rate of flow through the first capillary (length l1, radius a1) along under pressure p is Q =
πpa14 8 ηl1
πp Ql = 41 8η a1 Let Q' be the rate of flow when the second capillary (length l2, radius a2) is joined to the first in series. The pressure difference across the composite capillaries is same, p as was across the first capillary alone. Now for two capillaries in series, we have
F GH
πp l1 l + 24 Q' = 4 8 η a1 a2 =
Ql1 a14
Fl GH a
1 4 1
Q
=
1+
l2 a24
+
I JK
−1
l2 a24
I JK
Fa I GH l JK 4 1
1
Q = 0.2 cm3/sec, a1 = 0.1 cm, l1 = 20 cm, a2 = 0.05 cm and l3 = 10 cm ∴
Q' =
0. 2 0. 2 = = 0. 022 cm3 /sec. 4 1+8 10 × (0.1) 1+ (0. 05)4 × 20
Q.41. A tube of radius R and length L is connected in series with another of radius R/2 and length L/4. If the pressure across the two tubes taken together is p, deduce the pressure across the tubes separately. Solution. Let p1 and p2 be the pressure across the first and second tubes respectively. Then p = p1 + p2
...(i)
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507
the rate of flow of a liquid (viscosity η) throughout the system is the same. Thus
πp1R4 = 8 ηL
πp2 (R/2)4 8 η (L/4)
p1 = p2/4
...(ii)
Solving eqn. (i) and eqn. (ii), we obtain p1 =
p 4p and p2 = . 5 5
Q.42. Two capillaries A and B of lengths 16 cm and 4 cm have radii 0.2 cm and 0.1 cm respectively and are joined in series. In Poisuille’s experiment a pressure difference of 3 cm of water is employed across the ends of the composite tube. Find pressure difference across tube B. [Ans. 2.4 cm of water] Q.43. Two capillaries AB and BC of same length and joined end to end at B have radii r and 2r respectively. A is connected to a vessel of water giving a constant head of 0.1 meter and C is open to air. Calculate the pressure at B. Ans Let P be the pressure at B, then the pressure difference across AB is (0.17 – P) and that across BC is P relative to the atmosphere. Since the rate of liquid flow, Q, must be the same through both the capillaries we have Q =
π ( 0.17 − P) r4 πP(2r)4 = 8 ηl 8 ηl
(0.17 – P) = 16P P = 0.01 meter Q.44. Two capillaries A and B of diameters 0.4 mm and 0.3 mm and lengths 40 cm and 30 cm respectively are placed horizontally in series and water flows slowly through them from A to B. If the open end of B (water outlet) is at atmospheric pressure and the open end of A (water inlet) is at pressure 10 cm of water above the atmospheric pressure, calculate the pressure at the function of A and B relative to the atmospheric pressure.
LM Ans. 640 cm of waterOP N 91 Q
Q.45. Two capillaries of length 70 cm and 30 cm and radii 0.2 mm and 0.1 mm respectively are connected in series. If a liquid enters the first capillary at a pressure of 80 cm of Hg and leaves the second capillary at 70 cm of Hg, find the pressure at the junction of the two capillaries. [Hint. Let P be the pressure at the junction. Then the pressure difference across the first capillary is (80 – P) and that cross the second is (P – 70).] Q.46. Three capillaries of the same length and radii 2r, 3r, and 4r are connected in series and a liquid is flowing through them under streamlined conditions. If the pressure difference across the entire-system is 6.3 cm of mercury, deduce the pressure across the first tube. [Ans. 5 cm of mercury] Q.47. Let P1 and P2 be the pressure across the first and second capillaries. The rate of liquid flow is same in all of them. Thus,
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Mechanics
Q =
π P1 (3r)4 πP (4 r)4 π (8.1 mm)(5 r4 ) = 2 = 8 ηl 8 ηl 8 ηl
81P1 = 256 P2 = 625 (8.1 mm) P1 =
625 × 8.1 81
= 62.5 mm. Q.47. A cylindrical vessel of height h is kept maintained full of water. Three capillary tubes of same radius a and same length l are fitted horizontally in the wall of the cylinder at distances h/2, 3h/4 and h measured from the top. The water can flow through these tubes. Show that the length of a single outflow tube of the same radius which can replace the three tubes, when fitted at the bottom is 4/9. Solution. Let P1, P2, P3 be the pressure heads at the three tubes and Q1, Q2, Q3 the corresponding rates of flow. Then, by Poiseuille’s formula, we have Q1 =
πP1a4 π P2a4 πP3a4 , Q3 = , Q2 = 8 ηl 8 ηl 8 ηl
Q = Q1 + Q2 + Q3 =
πa4 ( P1 + P2 + P3 ) 8 ηl
...(i)
Let l be the length of the single tube fitted at the bottom (at pressure head P3) having rate of flow Q. Then Q =
πP3a4 8 ηl
...(ii)
Equating (i) and (ii)
P3 P + P2 + P3 = 1 l L h 3h , P2 = and P3 = h 2 4
Now
P1 =
Then
h/ 2 + 3h / 4 + h 9 /4 h h = = l l L
4 l. 9 Q.49. Three capillary tubes of the same radius r but of lengths l1, l2, l3 are fitted horizontally to the bottom of a long cylinder containing a liquid at a constant head which can flow through these tubes. Show that the length L of a single tube of the same radius r which can replace the three capillaries is given by L =
1 1 1 1 + + = l1 l2 l3 L Q.50. Calculate the maximum velocity of an oil drop radius 10 – 4 cm falling in air. (Density of oil = 0.9 gm/cc, viscosity of air = 1.8 × 10–4 gm. (cm-sec), g = 980 cm/sec2). Neglect air buoyancy.
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509
Solution. The drop (radius r1 density ρ) falling in air (density σ) acquires maximum (terminal) velocity v when the resultant driving (gravity) force equals the retarding viscous force i.e., when
4 3 πr (ρ − σ) g = 6 π η rv 3 Where η is the viscosity of air. If air buoyancy is ignored (σ = 0) then
4 3 πr ρg = 6 π η rv 3 v =
2 r2ρg 9 η
2 × (10−4 )2 × ( 0. 9) × 980 = 1.1 × 10–2 cm/sec. 9 × (1. 8 × 10−4 ) Q.51. Determine the radius of the drop of water falling through air, if the terminal velocity of the drop is 1.2 cm/sec. Assume the coefficient of viscosity for air = 1.8 × 10 – 4 Cgs units and the density of air = 1.21 × 10–3 gm/sec. [Ans. 0.001 cm/sec.]
v =
LMHint. MN
r=
9 ηθ 2( P − σ )g
OP PQ
Q.52. Two equal drops of water are falling through air with a steady velocity of 5 cm/sec. If the drops coalesce, what will be the new velocity? Solution. Let r be the radius of each drop. The (steady) terminal velocity v of a drop of radius r and density ρ, falling through air is given by
2 r2 (ρ − σ) g η 9 Where σ is the density and η the viscosity of air. Thus v =
v ∝ r2 = Kr2 Now, volume of each drop =
4 3 πr 3
∴Volume of the coalesced drop = 2 ×
4 3 4 πr = π [(2)1/ 3 r ]3 3 3
∴ Radius of the coalesced drop = 21/3 r. Hence the new terminal velocity of the coalesced drop. v1 = K{(2)1/3r}2 = (2)2/3 Kr2 = (2)2/3 v = (2)2/3 × 5 = 7.94 cm/sec. Q.53. A gas bubble of diameter 2.0 cm rises steadily through a solution of density 1.75 gm/cc at the rate of 0.35 cm/sec. Calculate the coefficient of viscosity of the solution. Neglect the density of the gas.
510
Mechanics
Solution. The weight of the bubble is negligible. The forces acting upon it are: (i) the
4 3 πr σg, where r is the radius of bubble and σ the density of solution, 3 and (ii) the viscous drag 6 πηrv . Since the bubble has attained a steady velocity, we have up thrust of solution,
6 πηrv =
η =
4 3 πr σg 3 2 r2σg 3 v
2 1 × 1. 75 × 980 × 9 0. 35 = 1.09 × 108 gm/cm-sec = 1.09 × 103 Poise.
η =
Q.54. An air bubble of 1 cm radius rises through a long column of liquid of density 1.47 × 103 kg-mtr3 with a steady velocity of 0.21 cm/sec. Find the viscosity of the liquid. Neglect the density of air (g = 9.8 m/sec2). [Ans. 1.52 × 102 kg/m-sec.] Q.55. With what terminal velocity will an air bubble 1.0 mm in diameter rise in a liquid of viscosity 150 centipoise and density 0.90 gm/cm3 ?
LMHint. v = 2 r σg , η = 150 centipoise = 1.5 Poise. 9 η N = 0.33 cm/sec.] 2
Q.56. A glass plate of length 10 cm, breadth 1.54 cm and thickness 0.20 cm weights 8.2gm in air. It is held vertically with the long side horizontal and the lower half under water. Find the apparent weight of the plate. Surface tension of water = 73 dynes/cm, g = 980 cm/sec2. Solution. Volume of the portion of the plate immersed in water is 10 ×
1 (1.54) × 0.2 = 1.54 cm3. 2
Therefore, if the density of water is taken as 1 then upthrust. = wt. of the water displaced = 1.54 × 1 × 980 = 1509.2 dynes Now, the total length of the plate in contact with the water surface is 2(10 + 0.2) = 20.4 cm. ∴ Downward pull upon the plate due to surface Tension = 20.4 × 73 = 1489.2 dynes ∴
resultant upthrust = 1509.2 – 1489.2 = 20.0 dynes =
20. 0 = 0. 0204 gm − ωt 980
apparent weight of the plate in water = weight of the plate in air – resultant upthrust ⇒
8.2 – 0.0204 = 8.1796 gm.
Q.57. A glass tube of circular cross – section is closed at one end. This end is weighted and the tube floats vertically in water, heavy end down. How far below the water surface is
Fluids
511
the end of the tube? Given radius of tube 0.14 cm, mass of weighted tube 0.2 gm, surface tension of water 73 dyne/cm and g = 980 cm/sec2. Solution. Let l be the length of the tube inside water. The forces acting on the tube are: (i) upthrust of water acting upward = πr2l × 1 × 980 =
22 (0.14)2l × 980 = 60.368 l dynes. 7
(ii) Weight of the system acting downward = mg = 0.2 × 980 = 196 dyne. (iii) Forces of surface tension acting downward = 2πrT = 2 ×
22 × 0.14 × 73 = 64.24 dyne. 7
Since the tube is in equilibrium, the upward force is balanced by the downward forces. That is 60.368 l = 196 + 64.24 = 260.24 l =
260. 24 = 4. 31 cm 60. 368
Q.58. Calculate the work done in blowing a soap bubble of 5 cm radius. The surface tension of soap solution is 25 dyne/cm. Solution. When a soap bubble of radius R is blown, the total surface are formed is = 2 × 4πR2 = 8 πR2 As the bubble has two surfaces. If T be the surface tension of soap solution, the work done in blowing is W = T × 8πR2 = 25 × 8 × 3.14 × (5)2 = 1.57 × 104 erg. Q.59. Calculate the work done against surface tension in blowing a soap bubble from a radius of 10 cm to 20 cm. If the surface tension of soap solution is 25 dynes/cm. Solution. Original area of both the surfaces of the bubble = 2 × 4π (10)2 = 800 π cm2 Final area of the surfaces of the bubble = 2 × 4π (20)2 = 3200 π cm2 Increase in surface area = 3200π – 800π = 2400π cm2 The work done against surface tension is W = S.T. × increase in area = 25 × 2400 × 3.14 = 1.88 × 105 erg. Q.60. A soap bubble of surface tension 25 dyne/cm has a radius of 5 cm. Find the work done in blowing the bubble to the radius of 10 cm. [Ans. 4.7 × 104 erg.]
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Mechanics
Q.61. Find the amount of work done in blowing a soap bubble of surface tension 30 dyne/cm from 10 cm radius to 15 cm radius. [Ans. 9.42 × 104 erg.] Q.62. Obtain an expression for the work necessary to break up a liquid drop of radius R into n equal smaller drops. Solution. When a liquid drop is broken into a number of smaller drops, the surface area increases. Hence work is to be done against the surface tension. Let us that compute the increase in surface area. Let r be the radius of each of the n smaller drops. Since the total volume of all the n smaller drops is the same as the volume of the original drop of radius R, we have n×
4 3 4 πr = πR3 3 3 r =
R (n)1 / 3
Thus, increase in surface area = n × 4πr2 – 4πR2 = 4π (nr2 – R2)
LM MN
= 4π n
R2 − R2 2/3 (n)
OP PQ
1/ 3 = 4πR2 (n) − 1 .
W = S.T. × increase in surface area 1/ 3 = T × 4πR2 (n) − 1 . 1/3 = 4πR2T ( n) − 1 .
Q.63. Calculate the work done in spraying a spherical drop of mercury of radius 1mm into a million identical droplets. The surface tension of mercury is 550 dynes/cm. Solution. The work done in spraying a liquid drop of radius R into n droplets is given by 1/ 3 W = 4πR2T (n) − 1 .
R = 1 mm = 0.1 cm, T = 550 dynes/cm and n = 106 W = 4 × 3.14 × (0.1)2 × 550{(106)1/3 – 1} = 4 × 3.14 × (0.1)2 × 550 × 99 = 6839 ergs. Q.64. Calculate the amount of energy needed to break a drop of water 1mm in radius into 106 droplets of equal size, taking surface tension of water = 72 dynes/cm. [Ans. 895 ergs.] Q.65. Find the increase in surface energy in breaking a water drop of radius 0.4 cm into 125 smaller drops. (S.T. of water = 72 dynes/cm) [Ans. 579 ergs.] Q.66. Calculate the amount of energy needed to break a drop of water 2 mm in diameter into 109 droplets of equal size. (S.T. of water is 73 × 10–3 nt/m).
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513
Solution. When a drop is sprayed into a number of droplets, the surface area and hence the surface energy increases. This increases in the surface energy is equal to the energy of the newly formed area. Now, the radius of the drop is R = 1 mm = 10–3 m. Let r be the radius of each of the droplets. The total volume of all the 109 droplets is the same as that of the original drop. Thus 109
109 ×
4 3 4 πr = π (10−3 )3 3 3 103r = 10–3 r = 10–6 m.
The increase in surface area = Surface area of 109 droplets – Surface area of the drop = 109 × 4πr2 – 4πR2 = 4π {109 × (10–6)2 – (10–3)2} = 4π (10–3 – 10–6) = 4π 10–3 m2. The work done W in spraying the drop is stored as the energy of the newly formed area and is equal to increase in area × S.T. W = 4 × 3.14 × 10–3 × 73 × 10–3 = 9.17 × 10–4 joule. Q.67. Calculate the loss of energy is 1000 drops of water each of diameter 2 mm coalesce to form one large drop. The S.T. of water is 72 dynes/cm. Solution. Let R cm be the radius of the large drop formed, when 1000 small drops, each of radius 1 mm coalesce. Since volume remains the same, we have
4 4 πR3 = 1000 × π (0.1)3 3 3 R = (1000)1/3 (0.1) = 1 cm. Now, the surface area of the 1000 small drops = 1000 × 4π (0.1)2 = 40 π cm2 and that of the large drop = 4π(1) = 4π cm2 Thus, decrease in surface area = 40π – 4π = 36 π cm2 Energy liberated = S.T. × decrease in surface area = 72 × 36 × 3.14 = 8139 erg. Q.68. Two drops of a liquid, each of radius r, coalesce to form a larger drop. Derive an expression for the rise in temperature. Solution. When two or more small drops of a liquid coalesce to form a single large drop, the surface area decreases, and hence the surface energy also decreases. The lost energy appears as heat, resulting in the rise in temperature of the drop. Let R be the radius of the larger drop formed. Its volume is the same as that of the smaller drops, each of radius r. Thus
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Mechanics
4 4 πR3 = 2 × πr3 3 3 R = (2)1/3 r. The decrease in surface area = surface area of 2 small drops – surface area of larger drop. = 2 × 4πr2 – 4πR2 = 4π (2r2 – R2) = 4π [2r2 – (2)2/3 r2]. = 8πr2 [1 – (2)–1/3] If W be the energy released, we have W = S.T. × decrease in surface area = T × 8πr2 [1–(2)–1/3]
...(i)
This energy is spent in increasing the temperature of the liquid Let t be the rise in temperature. If ρ be the density and s the specific heat of the liquid, the heat H absorbed by the drop is H = mass × Sp. heat × temp. rise =
4 4 πR3 ρ × s × t = π ( 2r3 )ρst . 3 3
...(ii)
Substituting for W form eq. (i) and for H from eq. (ii) in Joule’s relation W = JH. T × 8πr2 {1– (2)–1/3} = J ×
4 (2r3) ρst 3
3T −1 / 3 ]. t = rJρS [1 − (2) Q.69. Two water drops each of radius 0.00002 cm coalesce. What rise in temp. will result ? Surface tension of water is 74 dynes/cm and j = 4.2 × 107 erg./cal. [Ans. 0.055°C] [Hint. 1 – (2)–1/3 = 0.2063] Q.70. If a number of little droplets of water, each of radius r1 coalesce to form a single drop R, show that the rise in temp. will be given by
FG H
3S 1 1 − J r R
IJ K
where S is the surface tension of water and J the mechanical equivalent of heat. Solution. When a number of droplets coalesce to form a single drop the surface area decreases and the surface energy also decreases. The decrease in surface energy is equal to the energy of the decreased surface. Now suppose n droplets coalesce. Then Surface area of n droplets = n × 4πr2 Surface area of the combined drop = 4πR2 decrease in surface area = n × 4πr2 – 4πR2 = 4π (nr2 – R2)
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515
If W be the energy released, we have W = decrease in surface area × S.T. = 4π (nr2 – R2) × S
...(i)
This energy is spent in increasing the temp. of the water drop. Let t be the rise in temp. Then if both the specific heat and density of water are 1, the heat H absorbed by the drop is H = mass × sp.heat × t = =
4 πR3 × 1 × 1 × t 3
4 πR3t 3
...(ii)
Substituting for W form Eq. (i) and for H from eq. (ii) in the Joules relation W = JH, we have 4π (nr2 – R2) × S = J ×
F GH
4 πR3t 3
3S nr2 − R2 t = J R3
I JK =
F GH
3S nr2 1 − 3 J R R
I JK
...(i)
Since the total volume of n droplets is equal to that of the combined drop, we have
n×
4 3 4 πr = πR3 3 3 nr3 = R3 nr2 1 3 = r R
Putting this value in Eq. (i) we get t =
FG H
IJ K
3S 1 1 − . J r R
Q.71. Calculate the pressure inside a small air bubble of radius 0.01 mm situated at a depth of 20 cm below the free surface of a liquid of density 0.8 gm/cm3 and S.T. 75 dyne/cm taking the atmospheric pressure to be 76 cm of mercury. Solution. The excess pressure inside an air bubble of radius R is
2T R Where T is surface tension of liquid in which the bubble is formed. Here R = 0.01 mm = 0.001 cm and T = 75 dyne/cm. P =
2 × 75 = 0.15 × 106 dyne/cm2 0. 001 The pressure outside the bubble is atmospheric plus that due to 20 cm long column of liquid of density 0.8 gm/cm3 i.e., P =
(76 × 13.6 × 980) + (20 × 0.8 × 980) + (0.15 × 106) = 1.18 × 106 dyne/cm2.
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Mechanics
Q.72. Calculate the excess of pressure inside a soap bubble of radius 3 mm. Surface tension of soap solution is 20 × 10–3 nt/mtr. Also calculate the surface energy of the bubble. Solution. The excess pressure inside a soap bubble over that outside is
4T R Where T is the surface tension of soap solution and R the radius of the bubble P =
Here T = 20 × 10–3 nt/mtr and R = 3 mm = 3 × 10–3 mtr. P =
4 × (20 × 10−3 ) = 26.7 nt/mtr2 3 × 10−3
The soap bubble has two surfaces so its surface area is 2 × 4πR2 = 2 × 4 × 3.14 × (3 × 10–3)2 = 2.26 × 10–4 mtr.2 Now the surface energy of the bubble = S.T. × surface area = 20 × 10–3 nt/mtr × 2.26 × 10–4 mtr.2 = 4.52 × 10–6 joule. Q.73. The pressure of air in a soap bubble of 7 mm diameter is 8 mm of water above the atmospheric pressure. Calculate the surface tension of the soap solution (g = 980 cm/sec2). Solution. The excess pressure inside a soap bubble over that outside is
4T R Where T is the surface tension of soap solution and R is the radius of the bubble P =
PR 4 Here, the excess pressure inside the bubble over the outside atmospheric pressure is 8mm of water column. T =
P = (0.8) × 1 × 980 = 784 dyne/cm2
7 mm = 0.35 cm. 2 784 × 0. 35 T = = 68.6 dyne/cm. 4 Q.74. Two spherical soap bubbles. A and B of radii 3 cm and 5 cm coalesce so as to have a portion of their surface in common. Calculate the radius of curvature of this common surface. R =
Solution. Let r be radius of curvature of the common surface of the soap bubbles A and B. The excess pressure in A above atmospheric is 4T/3 while that in B is 4T/5. Thus there is an excess pressure on the A-side of the common surface above that on the B-side 4 T 4T − of . But this excess pressure must be 3 5 equal to 4T/r. Hence
r A 3 cm
B 5 cm
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517
4T 4 T 4T − = r 3 5 1 1 1 − = r 3 5 r = 7.5 cm. Q.75. A capillary tube of 0.8 mm bore stands vertically in a wide vessel containing a liquid of surface tension 30 dyne/cm. The liquid wets the tube and has a density of 0.8 gm/cm3. Calculate the rise of the liquid in the tube. What would be the rise if the capillary is inclined at 45° with the vertical? Solution. The surface tension T of a liquid is related to the rise h in a capillary tube of radius r by
FG H
T =
IJ K
r ρg 3 2 cos θ
r h+
Where ρ is the density of liquid. If the liquid wets the tube the angle of contact θ = 0. Further,
r can be neglected as compared to h. Then 3 rhρg 2 2T h = rρg
T =
0. 8 mm = 0.04 cm and ρ = 0.8 gm/cm2 2 2 × 30 h = = 1.9 cm 0. 04 × 0. 8 × 980 When the capillary is inclined at 45° with the vertical, the vertical height of liquid in it will still be h = 1.9 cm. The length of water column in the column would be T = 30 dyne/cm, r =
h = 1.9 × cos 45°
2 = 2.7 cm.
Q.76. Calculate the diameter of a capillary tube in which a liquid of specific gravity 0.85 and surface tension 35.0 dynes/cm rises 2.50 cm. [Ans. 0.067 cm.] Q.77. A liquid of specific gravity 1.5 is observed to rise 3.0 cm in a capillary tube of diameter 0.50 mm and the liquid wets the surface of the tube. Calculate the excess pressure inside a-spherical bubble of 1.0 cm diameter blown from the same liquid. Ans. The surface tension of the liquid is rhρg 0. 025 × 3 × 1. 5 × 980 = = 55 dyne/cm 2 2 Hence excess pressure inside a spherical bubble
T =
P =
4T 4 × 55 = = 440 dyne/cm2. R 0. 5
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Mechanics
Q.78. By how much will the surface of a liquid be depressed in a glass tube of radius 0.02 cm if the angle of contact of the liquid is 135° and its surface tension is 547 dyne/cm. Assume the density of the liquid to be 13.5 gm/cm3 and acceleration due to gravity 981 cm/sec2. Solution. The surface tension T of liquid is related to its elevation or depression h in a capillary tube of radius r by
FG H
T =
IJ K
r ρg 3 2 cos θ
r h+
Where ρ is the density of liquid and θ the angle of contact. The factor compared to h. Therefore, neglecting it T =
rhρg 2 cos θ
h =
2Tcosθ rρg
r = 0.02cm, θ = 135° so that Cos 135° = –
1 2
r is small 3
= –0.707
T = 547 dyne/cm, ρ = 13.5 gm/cm3, g = 981 cm/sec2. h =
2 × 547 × ( −0. 707) = –2.92 cm 0. 02 × 13. 5 × 981
The negative sign indicates that the liquid is depressed. Q.79. Water rises to a height of 10.0 cm in a certain capillary tube. In the same tube the level of mercury surface is depressed by 3.42 cm. Compare the surface tensions of water and mercury. Specific gravity of mercury is 13.6, the angle of contact for water is zero and that for mercury is 135°. Solution. The surface tension T of a liquid is related to its elevation or depression h in a capillary tube of radius r by T =
rhρg 2cos θ
where ρ is the density of the liquid and θ the angle of contact. Let Tω and Tm be the surface tensions of water and mercury respectively Then and
Tω =
r × 10. 0 × 1 × g = 5rg. 2 cos 0°
Tm =
r × ( −3. 42) × 13. 6 × g r × ( −3. 42) × 13. 6 × g = 2 cos135° 2 × ( −0. 707)
=
3. 42 × 13. 6 rg = 33rg. 2 × 0. 707
Tω 5 rg = = 0.15 Tm 33rg
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519
Q.80. Ripples are produced on the surface of liquid with the help of a tuning fork of frequency 60 and the wavelength observed is 0.5 cm. Calculate the surface tension of the liquid. Density of liquid is 1.0 gm/cm3 and g = 980 cm/sec. Solution. If n be frequency and λ the wavelength of ripples formed on a liquid of density ρ, then the surface tension of the liquid is given by T =
λ2ρ n2λ3ρ λ2ρg − = 2π 2π 4 π2
FG n λ − g IJ H 2π K 2
Substituting the given values, T =
LM N
(0. 5)2 × 1 980 (60)2 (0. 5) − 2 × 3.14 2 × 3.14
OP Q
= 0.04 × [1800 – 156] = 0.04 × 1644 = 65.8 dyne/cm. Q.81. Water is flowing through two horizontal pipes of different diameters which are connected together. In the first pipe the speed of water is 4m/s and the pressure is 2.0 × 104 N/m2. Calculate the speed and pressure of water in the second pipe. The diameter of the pipes are 3cm and 6cm respectively. Solution. If A is the area of cross-section of a pipe at a point and v is the velocity of flow of water at the point, then by the principle of continuity, we have Av = Constant A1v1 = A2v2 (πr12)v1 = (πr22)2 v2 v2
Fr I = G J Hr K 1
2
v1
2
Here r1 = 1.5 cm = 1.5 × 10–2 m, r2 = 3 cm = 3 × 10–2 m and v1 = 4 m/s
F 1. 5 × 10 I = G H 3 × 10 JK −2
v2
−2
2
× 4 = 1 m/s
By Bernoulli’s theorem:
P1 +
1 2 1 ρv1 = P2 + ρv22 2 2 P2 =
P1 +
1 ρ (v12 − v22 ) 2
For water ρ = 1 × 103 kg/m3, v1 = 4 m/s, v2 = 1 m/s and P1 = 2.0 × 104 N/m2
1 × 1 × 103 × (42 – 12) 2 = 2.0 × 104 + 0.75 × 104
P2 = 2.0 × 104 +
= 2.75 × 104 N/m2. Q.82. The pressure difference between two points along a horizontal pipe, through which water is flowing, is 1.4 cm of mercury. If, due to non-uniform cross-section, the speed of flow of water at the point of greater cross-section is 60 cm/sec, Calculate the speed at the other point.
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Mechanics
Solution. By Bernoulli’s theorem P1 +
1 1 ρv12 = P2 + ρv22 2 2
2 v22 – v12 = ρ ( P1 − P2 )
The speed of water will be greater at the place when the cross-section is smaller 2 2 v 2 2 = ρ ( P1 − P2 ) + v1
∴ Here
P1 – P2 = 1.4 cm of mercury = 1.4 × 10–2 × 13.6 × 103 × 9.8 = 1.866 × 103 N/m2,
P = 1 × 103 kg/m2, v1 = 60 cm/sec = 0.6 m/s v22 =
2 × 1.866 × 103 + (0.6)2 = 4.092 1 × 103
v 2 = 2 m/s (approx.). Q.83. Water flows into a horizontal pipe whose one end is closed with a valve and the reading of a pressure gauge attached to the pipe is 3 × 105 N/m2. This reading of the pressure gauge falls to 1 × 105 N/m2 when the valve is opened. Calculate the speed of water flowing into the pipe. Solution. According to Bernoulli’s theorem P1 +
1 2 1 ρv1 = P2 + ρv22 2 2
1 ρ(v22 − v12 ) = P1 – P2 2 Here v1 = 0 (the valve is initially closed and so the velocity of water is zero) ∴
2 v 2 2 = ρ (P1 – P2) =
2 × [3 × 105 – 1 × 105] = 2 × 2 × 102 = 400 1 × 103
v 2 = 20 m/s. Q.84. A horizontal tube has different cross-sectional areas at points A and B. The diameter of A is 4 cm and that of B is 2 cm. Two manometer limbs are attached at A and B. When a liquid of density 0.8 gm/cm3 flows through the tube, the pressure difference between the limbs of the manometer is 8 cm. Calculate the rate of flow of the liquid in the tube. Solution. The rate of liquid flow in non-uniform horizontal tube is given by Q = A1A2
2 gh A12
− A22
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A1 =
521
πr12
F 4I = π ×G J H 2K
2
= 4π
cm2,
A2 =
πr22
Q1 = 4π × π
F 2I = πG J H 2K
2
= π cm2, g = 980 cm/s2, h = 8cm.
2 × 980 × 8 (4 π)2 − ( π)2
2 × 980 × 8 15 = 4 × 3.14 × 32.3 = 406 cm3/s.
= 4π
Q.85. Water tank has a hole in its wall at a distance of 10 m below the free surface of water. The diameter of the hole is 2 mm. Compute the velocity of efflux of water from the hole and the rate of flow of water. Solution. The velocity of efflux of water is v =
2gh =
2 × 9. 8 × 10 = 14 m/s
The rate of flow of water is A × v = πr2 × v = 3.14 (1 × 10–3)2 × 14 = 4.4 × 10–5 m3/s. Q.86. The relative velocity between two layers of water is 8.0 cm/s. If the perpendicular distance between the layers is 0.1 cm, find the velocity gradient. Solution. Velocity gradient =
∆vx ∆Z
Relative velocity between layers, ∆vx = 8.0 cm/s and distance between the layers ∆Z = 0.1 cm.
8 = 80 per/sec. 0.1 Q.87. There is a 1 mm thick layer of glycerine between a flat plate of area 100 cm2 and a big plate. If the coefficient of viscosity of glycerine is 10 kg/m-s. Then how much force is required to move the plate with a velocity of 7 cm/s. ∴
Velocity Gradient =
Solution. To move the plate with a constant velocity, the necessary force will be equal to the viscous force F (say). Now F = ηA
∆vx ∆z
η = 1.0 Kg/m-s, A = 100 cm2 = 10–2 m2, ∆vx = 7 × 10–2 m/s and ∆z = 1 mm = 10–3 m
1. 0 × 10−2 × 7 × 10−2 = 0.7 N. 10−3 Q.88. A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 1.8 × 10–5 Kg/m-s. What will be the terminal velocity of the drop? (Density of water = 1.0 × 103 kg/m3 and g = 9.8 N/kg) Density of air can be neglected. F =
Ans. By Stokes’ law, the terminal velocity of a water drop of radius r is given by v =
2 r2 (ρ − σ ) g η 9
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Mechanics
Where ρ is the density of water, σ is the density of air and η the coefficient of viscosity of air. Here σ is negligible and r = 0.0015 mm = 1.5 × 10–3 mm = 1.5 × 10–6m. v =
2 (1. 5 × 10−6 )2 × (1. 0 × 103 ) × 9. 8 × 9 1. 8 × 10−5
= 2.72 × 10 – 4 m/s. Q.89. A metallic sphere of radius 1.0 × 10 – 3 m and density 1.0 × 104 kg/m3 enters a tank of water, after a free fall through a distance of h in the earth's gravitational field. If its velocity remains unchanged after entering water, determine the valve of h. Given coefficient of viscosity of water = 1.0 × 10–3 N-S/m2, g = 10 m/s2 and density of water = 1.0 × 103 kg/m3. Solution. The velocity attained by the sphere in falling freely from a height h is v =
2gh
...(i)
This is the terminal velocity of the sphere in water. Hence by Stokes’ law. we have
v =
2 r2 ( p − σ) g η 9
where r is the radius of the sphere, ρ is the density of the material of the sphere, σ (= 1.0 × 103 kg/m3) is the density of water and η is coefficient of viscosity of water ∴
v =
2 × (1. 0 × 10−3 )2 × (1. 0 × 10+4 − 1. 0 × 103 ) × 10 9 × 1. 0 × 10−3
= 20 m/s From Eq. (i) we have h =
v2 20 × 20 = = 20 m. 2g 2 × 10
Q.90. An air bubble of radius 1 cm rises up in a liquid column with terminal velocity of 0.21 cm/s. If the density of liquid be 1.47 × 103 kg/m3, then Calculate the coefficient of viscosity of the liquid. Density of air is negligible. (g = 9.8 m/s2) Solution. The weight of the air bubble is negligible two forces act upon it: (i) upthrust 4 3 πr σg, where r is the radius of the bubble and σ is the density of the liquid 3 and (ii) viscous force 6π η rv. Since the bubble has acquired terminal velocity, therefore
of the liquid
6π η rv =
4 3 πr σg 3
η =
2 r2σg 9 v
η =
2 (1 ×10−2 )2 × (1. 47 × 103 ) × 9. 8 9 0. 21 ×10−2
= 1.52 × 102 kg/m-s = 1.52 × 103 Poise.
11 HARMONIC MOTION 11.1 PERIODIC MOTION When a body repeats its motion continuously on a definite path in a definite interval of time then its motion is called ‘periodic motion’, and the interval of time is called ‘time-period’, T. The earth completes one rotation around the sun is 1 year. This motion of the earth is periodic motion whose time-period is 1 year. If a body is periodic motion moves along the same path to and fro about a definite point (equilibrium position), then the motion of the body is a ‘vibratory motion’ or ‘Oscillatory motion’. As the body goes to one side of its equilibrium position, comes back to that position, goes to the other side, and again returns to the same position; it is said to complete ‘one vibration’ or ‘one oscillation’. If figure, drawn the time-displacement curve of a body oscillating to and fro about its equilibrium position. The displacement y is measured from the equilibrium position and the time t is measured from the instant when the body passed through its equilibrium position. One oscillation is completed from the point O to the point A and the time T taken for one oscillation is the periodic-time of the body. T Y O
A t
Fig. 1
11.2 SIMPLE HARMONIC MOTION (S.H.M.); AS A PROJECTION OF UNIFORM CIRCULAR MOTION Simple harmonic motion is the simplest form of vibratory or oscillatory motion. Suppose a particle P is moving with uniform speed along the circumferences of a circle of radius a and center O. When the particle is at the point P, then the foot of the perpendicular drawn 523
524
Mechanics
from the particle on the diameter AA' of the circle is at the point N. When the particle was at B, the foot of the perpendicular was at the point O. When the particle moving along the circumference reaches the point A, the foot of the perpendicular moving along the diameter also reaches A. When the particle moving along the circumference reaches the point B', the foot moving along the diameter reaches from A to O. When the particle goes along the circumference from B' to A', the foot moves along the diameter from O to A', and when the particle moves from A' to B, the foot moves from A' to O. Thus the foot N moves in a straight line to and fro about the point O. This straight-line motion of N is simple harmonic motion. The motion of N from O to A, from A to A' and from A' to O is called 1 vibration. Thus, if a particle is moving with uniform speed along the circumference of a circle, then the straight-line motion of the foot of the perpendicular drawn from the particle on the diameter of the circle is called ‘simple harmonic motion’. The circle is called the ‘reference circle’ of the simple harmonic motion. This is the kinematical definition of simple harmonic motion. A N
θ
P
a
y
θ = ωt
B′
O
B
A′
Fig. 2
11.3 DISPLACEMENT EQUATION OF S.H.M. Suppose the particle P starts from the point B and rotates through an angle of θ radian in t second. If the angular velocity of the particle be ω, then ω = θ/t or θ = ω t. In t second the displacement of N from O is ON. If this displacement is written as y, then y = ON = OP sin NPO But OP = a and ∠NPO = ∠POB = θ = ω ∴
y = a sin ωt
This is the displacement equation of the simple harmonic motion. Amplitude: The maximum value of sin ωt is 1. Hence, by above equation, the maximum value of the displacement y will be a. This maximum displacement is called the ‘amplitude’ of motion. It is equal to the radius of the reference circle. Periodic Time: The time taken by N to complete one vibration is called the ‘periodic-time’ T of N. The time during which N completes one vibration, the particle P completes one round of the circle of reference, that is, it rotates through an angle of 2π radian. Thus the time of one rotation of P is 2π/ω. Therefore, the periodic-time of N is given by T=
2π ω
Hormonic Motion
525
Frequency: The number of vibrations completed by N in one second is called the ‘frequency’ (n) of N. The frequency is reciprocal of the periodic-time, n =
1 ω = T 2π
Phase: When a particle vibrates, its position and direction of motion vary with time. The phase of a vibrating particle at any instant expresses the position and direction of motion of the particle at that instant. A N y B′
α ωt φ
O
P P0 B
A′
Fig. 3
The phase is expressed either in terms of the angle θ or in terms of the periodic-time T. For example, in figure, when the particle P reaches the point A, then its phase is π/2 or T/4. If at any instant two vibrating particles are passing simultaneously through their equilibrium positions in the same direction, then at that instant they are in the same phase, and if they are passing in opposite directions then they are in opposite phase. Suppose the time is measured from the instant when the particle moving on the reference circle was at a point different from B, say at P0, where ∠PoOB = φ. Then the displacement y (=ON) of N from the equilibrium position O after a time t will be given by y = ON = OP sin NOP = OP sin POB y = a sin (ω ωt + φ) ωt is the angle through which the particle P has rotated in moving from Po to P in time t. Thus at the instant t the phase of N will be measured by the total angle (ωt + φ). φ is called the ‘initial phase’ or ‘epoch’ of N. The above equation is the general equation of simple harmonic motion.
11.4 CONDITIONS FOR LINEAR S.H.M. The following three conditions should be fulfilled for the linear S.H.M. of a particle: 1.
The motion of the particle should be in a straight line to and fro about a fixed point.
2.
The restoring force (or acceleration) acting on the particle should always be proportional to the displacement of the particle from that point.
3.
The force (or acceleration) should always be directed towards that point.
The S.H.M. is Defined on the Basis of These Conditions When a particle moves in a straight line to and fro about its equilibrium position in such a way that the force (or acceleration) acting upon it is always directly proportional to its
526
Mechanics
displacement and directed towards the equilibrium position, then the motion of the particle is called ‘simple harmonic motion’.
11.5 EQUATION OF MOTION OF A SIMPLE HARMONIC OSCILLATOR Let us consider a particle P of mass m executing S.H.M. about an equilibrium position O. By definition, the force under which particle is oscillating is proportional to the displacement of P from O and is directed toward O. Thus to the displacement of P from O at any instant t, the instantaneous force F acting upon it is given by F = – kx Where k is the proportionality factor, the minus sign shows that the force F is opposite to the displacement x. According to the Newton’s second law, the force acting on the particle is equal to the product of the mass and the acceleration. The instantaneous acceleration of the particle is d2 x . Hence dt2
–kx = m
or
d2 x dt2
d2 x = – (k/m) x dt2
Let us put the constant, k/m = ω2. Thus we have or
d2 x + ω2x = 0 dt2
...(i)
This is the differential equation of a simple harmonic oscillator. Let a solution to this equation be x = Ceαt Where C and α are arbitrary constant. On differentiating with respect to t, we get
dx = Ceαt.α dt and
d2 x = Ceαt.α2 dt2
Substituting these values of
d2 x and x in Eqn. (i), we get dt2
Ceαtα2 + ω2Ceαt = 0
or
Ceαt (α 2 + ω2 ) = 0
or
(α2 + ω 2 ) = 0
or
α = ± ( − ω2 ) = ± jω
Hormonic Motion
Where j =
and
527
−1. Thus, there are two possible solutions from Eqn. (i), which are x = Ce+ jωt
x = Ce− jωt Hence the general solution to Eqn. (i) will be the sum of these two solutions, i.e., x = C1e+ jωt + C2e− jωt Where C1 and C2 are arbitrary constants. This gives x = C1 (cos ωt + j sin ωt) + C2 (cos ωt – j sin ωt) = (C1 + C2) cos ωt + j (C1 – C2) sin ωt Let us make a change in arbitrary constants by putting C1 + C2 = a sin φ j(C1 – C2) = a cos φ
and
Where a and φ are now constants. This gives x = a sin φ cos ωt + a cos φ sin ωt x = a sin (ωt + φ)
or
This is, in fact, a solution of the equation of S.H.M. The values of the constants a and φ depends upon how the motion is started.
11.6 IMPORTANCE OF S.H.M. The study of simple harmonic motion is important for two reasons. (i)
There are a great many physical problem in mechanics, acoustic, optics, electricity and in atomic and molecular Physics, in which the force on a system is approximately proportional to its displacement from some equilibrium position. In such cases the resulting motion may be represented approximately be the simple harmonic model.
(ii)
Even the complicated periodic motion occurring in physical problem can be represented as a combination of a number of simple harmonic motion having frequencies which are multiple of that of the complicated motion. The vibrating strings and membranes, the vibrations of atoms in solid, the electrical and acoustical oscillations in a cavity can be treated in this manner.
11.7 ENERGY OF HARMONIC OSCILLATOR A harmonic oscillator possess two types of energy: (i)
Potential energy, which is due to its displacement from the mean position.
(ii)
Kinetic energy, which is due to its velocity.
Hence, at any instant the total energy of the oscillator will be the sum of these two energies. As the system is conservative (i.e., no dissipative or frictional forces are acting), the total mechanical energy E (=K + U) must be conserved. Let the displacement of the harmonic oscillator at any instant t be given by x = a sin (ωt + φ) Its velocity
v =
dx = aω cos ( ωt + φ) = ω a2 − x2 dt
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Mechanics
d2 x = − ω 2a sin ( ωt + φ ) = − ω2 x dt2
Acceleration ∴
Force = –m ω2x = –Cx, ω2 = C/m
If the oscillator is displaced through dx distance, then work done on the oscillator is dW = Cx dx If it is displaced from x = 0 to x = x, then work done
z x
W =
Cxdx =
0
1 2 Cx . 2
This work done on the oscillator becomes its potential energy U, i.e., U = ½ Cx2 Kinetic energy of the oscillator at the displacement x is K = ½ mv2 = ½ mω2(a2 – x2) = ½ C (a2 – x2) ∴ Total energy
E = U + K = ½ Cx2 + ½ C (a2 – x2) = ½ Ca2 = ½ mω2 a2 = ½ m (2π/T)2 a2 = 2m π2 a2 / T2 E = 2mπ π2a2n2
or
Thus, we see that the total mechanical energy is constant, as we expect and has the value ½ Ca2, i.e., the total energy is proportional to the square of the amplitude. Hence, if the system is once oscillated the motion will continue for indefinite period with t any decrease in amplitude, provided no damping (frictional) forces are acting on the system. E (To ta l E ne rgy) P.E . C u rve U = ½ C x2 E n erg y K .E . C u rve K = ½ C (a 2 – x 2 ) –a
O
+a
x
Fig. 4
If we plot a graph between the potential energy U and displacement x, we get a parabola having vertex at x = 0. The curve for the total energy E (constant) is the horizontal line. The particle cannot go beyond the points where this line intersects the potential energy curve because U can never be larger than E. These points are turning points of the motion and correspond to the maximum displacement. At these positions (i.e., x = ±a) the total energy of the oscillator is wholly potential (i.e., E = U = ½ Ca2) but the kinetic energy is zero and so the amplitude of motion is a = ± 2E /C. At the equilibrium position, the potential energy is zero, but the kinetic energy has the maximum values K = ½ mv2max = ½ Ca2 = E with 2E . While at other intermediate points the energy is partly m kinetic and partly potential but their sum (total energy) is always ½ Ca2.
maximum velocity vmax =
Hormonic Motion
529
11.8 AVERAGE VALUES OF KINETIC AND POTENTIAL ENERGIES The instantaneous displacement of particle of mass m executing S.H.M. under a force-constant k is x = a sin (ωt + φ) where ω2 = k/m The instantaneous kinetic energy is
FG IJ H K
1 dx K = m 2 dt
2
=
mω2 2 a cos2 (ωt + φ) 2
The time-average of the kinetic energy over a period T of the motion is
z
Kav =
1/2mω2 a2
kdt =
0
T
mω 3a2 = 4π mω 3a2 = 4π =
z
2π / ω
T
cos2 (ωt + φ) dt
0
z LM z MN z
, T = 2π/ω
2π /ω
2π / ω
1 + cos 2 (ωt + φ) dt
0
2π / ω
2π / ω
dt +
0
cos 2 (ωt + φ) dt
0
LM N
OP Q
OP PQ
1 mω 3a2 2π + 0 = mω 2 a2 4π 2 ω
K av = ½ ka2 The potential energy of the particle is U = ½ kx2 = ½ ka2 sin2 (ωt + φ) The time-average of the potential energy over a period T of the motion is
z
Uav =
=
(1/2) ka2
Udt =
0
T k ωa2 4π
sin2 (ωt + φ) dt
0
z LM z MN z
2π / ω
2π / ω
k ωa2 = 4π
=
z
2π / ω
T
1 − cos 2(ωt + φ) dt
0
2π / ω
2π / ω
dt −
0
LM N
k ωa2 2π −0 4π ω
Uav = ½ ka2
0
OP Q
cos 2 (ωt + φ) dt
OP PQ
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Mechanics
Thus the time-average kinetic energy is equal to the time-average potential energy and each is equal to ½ ka2, which is half the total energy. This is a characteristic property of the harmonic oscillator and is not true for anharmonic oscillators.
11.9 POSITION-AVERAGE OF KINETIC AND POTENTIAL ENERGIES Let us now find out the position-average of the kinetic and potential energies. The kinetic energy is K = ½ m (dx/dt)2 = ½ k (a2 – x2)
z
z
The position-average over the displacement from x = 0 to x = a is a
K =
a
1/2 k ( a2 − x2 ) dx
Kdx =
0
a
LM N
0
a
x3 1k 2 a x− = 2a 3 The potential energy is U = Its position-average is
OP Q
= 0
LM N
OP Q
1 k 3 a3 1 a − = ka2 2a 3 3
1 2 kx 2
z a
U =
a
z a
Udx
0
a
1/2 k =
0
a
x2dx =
LM OP N Q
1 k x3 2a 3
a
= 0
1 2 ka 6
Thus the position-average kinetic energy (1/3 ka2) is not equal to the position average potential energy (1/6 ka2).
11.10 FRACTIONS OF KINETIC AND POTENTIAL ENERGIES The equation for S.H.M. is x = a sin (ωt + φ) where ω2 = k/m So that u =
dx = ωa cos (ωt + φ) = ± dt
k 2 ( a − x2 ) m The total energy of the particle in S.H.M. is = ±
E = K + U
1 1 mv2 + kx2 2 2 1 1 = k (a2 – x2) + kx2 2 2 1 = ka2 2 =
FG k IJ (a H mK
2
− a2 sin2 (ωt + φ))
Hormonic Motion
531
Now the displacement is one-half the amplitude i.e., at x = a/2, we have
1 1 1 kx2 = k (a/2)2 = (1/8) ka2 = E 2 2 4 1 1 3 and (K)a/2 = k (a2 – x2) = k (a2 – a2/4) = (3/8) ka2 = E 2 2 4 Thus the potential energy is one-fourth and the kinetic-energy is three-fourth of the total energy. (U)a/2 =
11.11 MASS ATTACHED TO A HORIZONTAL SPRING Suppose a mass m attached to one end of a massless spring is free to move on a frictionless horizontal surface. Let x = 0 be the equilibrium position of the mass at which the spring is unscratched. When the mass is displaced toward right through a distance x, then the spring exerts an elastic restoring force F on the mass. By Hooke’s Law, F = – kx Where k is the force-constant of the spring. The negative sign indicates that F is directed towards left (opposite to displacement). If we displace the mass toward left, then F is directed toward right as in Figure (c). When the displaced mass is released, it oscillates to and fro about the equilibrium position. F (a)
m
x (b)
x=0 F=0 F
( c) x
Fig. 5
At any instant during oscillation the force F is equal to the mass m multiplied by the d2 x . That is instantaneous acceleration dt2 d2x dt2 Where ω2 = k/m. This is the differential equation of the simple harmonic motion of the mass. Its solution is
F = –kx = m
x = a sin (ωt + φ) Where ω and φ are constants. If t is replace by t + 2π/ω, the displacement x will be the same. Hence the period of the motion T =
m 2π = 2π k ω
532
Mechanics
Now, suppose the spring has a finite mass ms. If ms<< m, then the spring will stretch uniformly along its length, that is, the displacement of the different elements of the spring will be proportional to their respective distances from the fixed end. Let l be the length of the spring, then its mass per unit length is ms/l. Let ds be a small element of length at a distance S from the fixed end. Its mass is (ms/l) ds. When the displacement of the mass m is x, the displacement of the element ds will be (s/l)x and the velocity will be (s/l)(dx/dt). Hence the instantaneous energy of the element
FG H FG dxIJ H dt K
=
1 ms s dx ds 2 l l dt
=
ms 2l3
IJ K
2
2
s2ds
The instantaneous kinetic energy of the whole spring is ms 2l 3
FG dx IJ H dt K
z
2 l
s2ds =
0
1 ms 6
FG dx IJ H dt K
2
The kinetic energy of the system (mass + spring) is K =
FG H
2
+
1 ms 6
IJ FG IJ KH K
FG dxIJ H dt K
2
2
m 1 dx m+ s 2 3 dt This shows that the effective mass of the system is (m + ms/3). Hence the new time-period =
is
FG IJ H K
1 dx m 2 dt
T = 2π
m+
ms 3
k
11.12 MOTION OF A BODY SUSPENDED BY A VERTICAL SPRING Suppose a light spring, whose normal length is l, is hanging from a rigid support. When a body of mass m is suspended from its lower end, then due to the weight of the body, the length of the spring is extended say by xo. The spring exerts an elastic (restoring) upward force F on the mass m. By Hooke’s law, F = – kx0
F m mg
F′
...(i)
m
(b ) mg (c )
Fig. 6
–kx0 + mg = 0 x 0 = mg/k
x0 (a )
x
Where k is the force-constant of the spring. The other force acting on the mass is its weight, mg. Thus the total force acting on m is –kx0 + mg. Since the mass has acceleration, the total force on it is zero, that is, or
l
Hormonic Motion
533
This is the static equilibrium position of the mass m. Now, suppose during oscillation, the total extension in the spring at any instant is x0+x (figure c). The (upward) force exerted on m by the spring is then F' = –k (x0 + x) = –k (mg/k + x) = –mg – kx The other force acting on the mass m is its weight, +mg (downward). Therefore, the net force acting on the mass is (–mg – kx) + mg = –kx This, by Newton’s second law, must be m times the instantaneous acceleration
d2 x of dt2
the mass. That is –kx = m
or
d2 x dt2
d2 x = –(k/m) x = –ω2k dt2
where ω2 = k/m. This is the equation of the S.H.M. Thus the mass m executes S.H.M. about the position x0 = mg/k. The time period is T =
2π m = 2π ω k
...(ii)
we have k = mg/x0. Therefore
x0 ...(iii) g Eqn. (ii) and (iii) both can be used to determine the period of oscillations of themass-spring system. Eqn. (iii) has the advantage that we can find the period of oscillation simply by measuring the extension x0 of the spring without knowing the suspended mass and the forceconstant of the spring. T = 2π
Spring cut in two halves: For a given mass hung from the spring, the extension x0 is proportional to the unstretched length l of the spring. Therefore, for half length (l/2) of the spring the extension will be x0/2. Hence the new time-period will be T = 2π
( x0 / 2) = T/ 2 g
on the new frequency is √2 times the odd frequency.
11.13 MASS SUSPENDED BY A HEAVY SPRING Suppose a mass m is suspended from the end of a spring of mass ms. If ms << m, then the spring will stretch uniformly along its length. Let l be the length of the spring. Then its mass per unit length is ms/l. Let us consider an element of length ds at a distance S from the fixed end of the spring. The mass of this element is (ms/l) dS.
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Mechanics
Let x be the instantaneous displacement of the mass m from its position of rest. Since the displacement is proportional to the distance from the fixed end, the displacement of the element considered will be (S/l) x, and its instantaneous velocity is
S dx . Hence the l dt
instantaneous kinetic energy of the element.
FG H FG dxIJ H dt K
s dx 1 ms ds = l dt 2 l ms = 2l3
IJ K
2
2
s2ds
If the spring is uniform, the total kinetic energy of the suspended mass and the spring is
FG IJ H K
dx 1 K = m dt 2
=
FG IJ H K
FG IJ H K
2
ms dx 2l3 dt
+
FG IJ H K 1F IJ . G m + m3 IJK FGH dx 2H dt K 1 dx m 2 dt
2
+
ms dx 2l3 dt
z
2 l
s2ds
0
2
l3 3
2
=
s
Now, when the mass is at a distance x from its position of rest, the elastic potential energy of the system is
1 kx2 2 Where k is the force-constant of the spring. Therefore, the total energy of the system is U =
E = K+U=
FG H
m 1 m+ s 2 3
IJ FG dxIJ K H dt K
2
+
1 2 kx 2
Since E remains constant; dE/dt = 0. that is =
FG m + m IJ d x + kx H 3 K dt s
where
IJ FG IJ d x + 1 k(2x) dx = 0 K H K dt 2 dt 2
2
2
2
or
FG H
m dx 1 m+ s 2 dt 2 3
= 0
k d2 x . x = − ω2 x 2 = – m dt m+ s 3
FG H
ω2 =
IJ K
k
FG m + m IJ H 3K s
Hormonic Motion
535
Thus the motion is simple harmonic and its period is
2π T = = 2π ω
FG m + m IJ H 3K s
k
The frequency is n =
1 1 = T 2π
k
FG m + m IJ H 3K s
m . Thus when the mass of the spring k is taken into account, the period is increased or the frequency is decreased.
If the spring be massless, the period is T = 2π
11.14 OSCILLATIONS OF A FLOATING CYLINDER Let us consider a cylinder of mass ‘m’ and cross-sectional area A floating vertically in a liquid of density ρ. Let l be the immersed length of the cylinder. At equilibrium, the weight of the cylinder is equal to the upthrust of the displaced liquid (mg = Alρg). When the cylinder is slightly pressed down and released, it oscillates up and down. Let x be the instantaneous displacement of the cylinder. Then the mass of the further displaced liquid will be Axρ and the corresponding upthrust will be Axρg. This is the restoring force F on the cylinder, that is, F = –Axρg The force is directed to the displacement x. By Newton’s second law, the force F equals and 2
product of mass m and acceleration
d x of the cylinder. dt2
m
l
m x
Thus d2 x –Axρg = m dt2 d2 x Aρg x = − ω2 x = − m dt2
or
where ω2 = Aρg/m. This represents a S.H.M. of period. T =
m 2π = 2π ω Aρg
But m = Alρ ∴
T = 2π
l g
Fig. 7
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Mechanics
11.15 OSCILLATIONS OF A LIQUID IN A U-TUBE Suppose a U-tube of uniform cross-section A contains a liquid. Let ρ be density of the liquid and l be the total length of the liquid in the tube. Suppose, in the equilibrium position, the levels of the liquid are at B and C which are in the same horizontal plane. If the level at C is depressed downwards to a distance x, the level at B will rise to the same distance x. On releasing the liquid column oscillates up and down in the tube. Let x be the instantaneous displacement of the liquid level below C. Then the length of the extra column of the left hand limb will be 2x. The weight of this extra column is 2xAρg and this is the restoring force F acting on the liquid. That is F = –2x Aρg By Newton’s law, this equals the product of the mass m of the liquid and its acceleration 2
d x . Thus dt
A re a
–2xAρg = m
d2 x dt
x B
C x
2Aρg d2 x x = − ω2 x = − 2 m dt
or Where ω2 =
2Aρg . This represents a S.H.M. of period m T =
m 2π = 2π ω 2Aρg
l
Fig. 8
But m = lAρ T = 2π
g 2l
11.16 HELMHOLTZ RESONATOR It consists of a large spherical glass vessel having a small cylindrical neck A and a narrow stem B just opposite to A as shown in figure 9. When the neck A is directed towards a sound source, the air in the vessel vibrates with a natural frequency depending upon the volume of the vessel and the dimension of its neck. Let l be the length of the neck; A its area of cross-section and V the volume of the vessel. The mass of the air within the neck is m = lAρ Where ρ is the density of air. When sound waves enter the neck, the air in it acts like a piston, alternately compressing and rarifying the air inside the vessel. If the air within the neck moves inward through a small distance x then the air inside the vessel is compressed by an amount v is given by v = xA The resulting increase in pressure, ρ, given by Hooke’s law as
Hormonic Motion
537
K = or
ρ v/V
(Bulk modulus =
v V Therefore the (restoring) force on the air in the neck p = −K
F = pA = −
Volume stress ) Volume strain A
K vA KxA2 =− V V
(Since v = xA) But by Newton’s second law F = m
d2 x dt2
where m (= l Aρ) is the mass of the air in the neck. Hence
−
d2 x d2 x Kx A 2 = m 2 = lAρ 2 dt dt V
B
Fig. 9
d2 x KA = − x = − ω2 x dt2 lρV
or
where ω2 = KA/(lρV). This is the equation of S.H.M. whose frequency is given by ω 1 n = 2π = 2π
Now ∴
KA lρV
K = v, the velocity of sound in air. ρ n =
v 2π
A lV
This is the natural frequency of the air of the resonator. If a sound wave is incident on the neck A, then the air will, resonate only if natural frequency is equal to the frequency of the incident wave. To detect the resonance easily, the stem B is pressed into the ear.
11.17 COMPOSITION OF TWO SIMPLE HARMONIC MOTIONS OF EQUAL PERIODS IN A STRAIGHT LINE If two simple harmonic motions are acting on a particle simultaneously, it shall perform resultant vibration, which is obtained by compounding the component vibrations. There are two methods of compounding two S.H.Ms. the geometrical methods and the analytical method. We will discuss here only analytical method. Let two S.H.Ms. of equal periods (= 2 π/ω) but of different amplitudes and phases be simultaneously acted on a particle in the same straight line. Let the two component vibrations be represented by and
x 1 = a1 sin (ωt – φ1)
...(i)
x 2 = a2 sin (ωt – φ2)
...(ii)
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Mechanics
where a1 and a2 are the amplitudes and φ1 and φ2 the epochs or initial phases of the component vibrations. The phase difference between them is (φ1 – φ2). The resultant displacement from the principle of superposition, is given by x = x1 + x2 = a1 sin (ωt – φ1) + a2 sin (ωt – φ2) = a1[sin ωt cos φ1 – cos ωt sin φ1] + a2 [sin ω t cos φ2 – cos ωt sin φ2) = sin ωt [a1 cos φ1 + a2 cos φ2 ] – cos ωt [a1 sin φ1 + a2 sin φ2] Now, let and ∴
r cos θ = a1 cos φ1 + a2 cos φ2
...(iii)
r sin θ = a1 sin φ1 + a2 sin φ2
...(iv)
x = r [sin ωt cos θ – cos ωt sin θ] x = r sin (ωt – θ)
or
...(v)
where r is the amplitude and (ωt – θ) the phase of the resultant vibrations. The above relation shows that the resultant motion is simple harmonic of the same period (2π/ω) but of different amplitude r and epoch θ. Squaring Eqns. (iii) and (iv) and adding, we get, r 2 = (a1)2 + (a2)2 + 2 a1 a2 cos (φ1 – φ2)
...(vi)
Dividing Eqn. (iv) by (iii), we have tan θ =
a1 sin φ1 + a2 sin φ2 a1 cos φ1 + a2 cos φ2
...(vii)
Let us consider some special cases: (a)
If φ1 – φ2 = 0 or 2nπ, where n is an integer, the two vibrations are in same phase and the resultants amplitude has the maximum value, i.e., r =
a12 + a22 + 2a1a2 = a1 + a2
The negative root has been omitted, as the amplitude is essentially positive. (b)
If φ1 – φ2 = π or (2n + 1)π, where n is any integer, the two vibration are in the opposite phase and the resultant amplitude has the minimum values i.e., r =
a12 + a22 − 2a1a2 = a1 − a2
If in addition φ1 = φ2; the amplitude r = 0 i.e., the particle remains in rest position.
11.18 COMPOSITION OF TWO RECTANGULAR S.H.M. OF EQUAL PERIODS Lissajous Figures: When a particle is vibrating, simultaneously under two simple harmonic motions at right angles to each other, the resultant motion of the particle is called Lissajous figures. The nature of the motion, or the curve traced out, depends upon the amplitudes, frequencies and the phase difference of the two component motions. We shall consider below some simple cases of the composition of two rectangular vibrations. Let the two simple harmonic motions of equal periods be along the axis of X and axis of Y. If their displacement at any time t be x and y, then they can be represented by
Hormonic Motion
539
x = a sin (sin ωt + φ)
...(i)
y = b sin ωt
...(ii)
where a and b are the amplitudes of the component vibrations having each period 2p/w and the first vibration is ahead of the second by a phase angle φ, i.e., the phase difference between the two vibrations is φ. The resultant motion can be obtained by eliminating t from Eqs. (i) and (ii). Expanding Eq. (i), we get x/a = sin ωt cos φ + cos ωt sin φ But from Eq. (ii) sin ωt = y/b and cos ωt =
1−
y2 b2
x y y2 = cos φ + sin φ 1 − 2 a b b
FG x − y cosφIJ Ha b K
or
2
F GH
2 = sin φ 1 −
y2 b2
I JK
x2 y2 2 xy cosφ = sin2φ + − ab a2 b2
or
...(iii)
This expression represents the resultant motion, which is in general an ellipse inclined to the axes of co-ordinates. There are, however, a number of special cases: when φ = 0 , the relation (iii), becomes
(a)
x2 y2 2 xy + − = 0 ab a2 b2
FG x − y IJ H a bK F x yI ±G − J H a bK
2
or
= 0
= 0
This represent a pair of coincident straight line y = bx/a, passing through the origin and lying in the first and third quadrants (Fig. a) (b)
when φ = π/2, the expression (iii) is
x2 y2 + 2 = 1 2 a b This equation represents an ellipse, whose axes are coincident with the co-ordinate axes (Fig. b). Now, if a = b, the ellipse degenerates in the circle. x2 + y2 = a2
540
Mechanics
Hence, two simple harmonic motions at right angles to each other of equal amplitudes but with phase, differing by π/2, are equivalent to a uniform circular motion, the radius of the circle being equal to the amplitude of either simple harmonic motion. (c)
When φ = π, Eq. (iii) is x2 y2 2 xy + 2 + = 0 or 2 ab a b
FG x + y IJ H a bK
2
=0
bx , passing through the a origin and lying in the second and fourth quadrats as BOB' (Fig. c). which represents a pair of coincident straight lines y = −
For any other values of φ, except discusses above, the path will be an ellipse, inclined to the co-ordinate axes. 2a
Y
2a
Y
2b
X′
Y
2a
A
B
X
X′
O
O
2b X
X′
X O
Y′
Y′
Y′
(a ) φ = 0
( b ) φ = π/2
(c ) φ = 2π
Fig. 10
11.19 COMPOSITION OF TWO RECTANGULAR S.H.M.’S OF TIME PERIODS NEARLY EQUAL If the two component vibrations are exactly of equal periods, the elliptical paths, referred to above, remain perfectly steady. But, when the two time periods differ slightly from each other, there comes about a gradual but progressive change in the relative phase (φ) of the two vibrations and consequently the shape of the ellipse slowly undergoes a change. All these elliptic path lie within a rectangle of sides 2a and 2b. Now, starting with the phase in agreement, i.e., φ =0, the ellipse coincides with the diagonal
FG x − y IJ = 0 of the rectangle. H a bK
As φ increases from o to π/2, the ellipse opens out to the form
x2 y2 + = 1 passing a2 b2
through intermediate oblique positions. When φ increase from π/2 to π, the ellipse closes up again and finally coincides with the other diagonal
FG x + y IJ = 0 of the rectangle. H a bK
As the phase difference changes from π to 2π , the reverse process takes place, until the ellipse again coincides with the first diagonal. All these changes have been represented as follows:
Hormonic Motion
541
It should be noted that the frequency of the complete cycle is equal to the difference of the frequencies of the two component simple harmonic vibrations. P h ase differe nce
0
2π
π/4
π/2
π
3π/4
5π/4
3π/2
7π/4
P h ase differe nce
Fig. 11
COMPOSITION OF TWO RECTANGULAR S.H.M. OF PERIODS IN THE RATIO 1:2 Let the component vibrations be represented by and
x = a sin (2 ωt + φ)
...(i)
y = b sin ωt
...(ii)
where φ is the phase angle by which the first motion is ahead of the second. The resultant motion, which can be obtained by eliminating ωt from the equation (i) and (ii), in general will be a curve, having two loops for any value of phase difference and amplitudes. Phase difference 1:2 0
π/4
π/2
3 π/4
2π
7 π/4
3 π/2
5 π/4
π
P h ase d ifferen ce 1 :2 P h ase differe nce
Fig. 12
Here, we will consider only two special cases, given below: When φ = 0, we get
(a)
x = a sin 2 ωt = 2a sin ωt cos ωt x/a = 2 sin ωt cos ωt
or
x y2 2y 1− 2 = a b b
or
or
or
F GH
I JK
F GH
I JK
y2 4 y2 4 y2 y2 x2 1 − = − −1 = b2 b2 b2 b2 a2 x2 4 y2 + 2 a2 b
Fy GH b
2
2
I JK
−1
= 0
This equation represent the figure of ‘8’.
542
Mechanics
When φ = π/2, we have
(b)
x = a sin (2ωt + π/2) = a cos 2 ωt = a (1 – 2 sin2ωt), sin2ωt = y2/b2
x 2 y2 = 1− 2 a b x−a 2 y2 = 1− x/a = − 2 a b
or
∴ y2 = −
b2 ( x − a) 2a
This is the equation of a parabola with vertex (a, 0). If the time periods of the two rectangular vibrations depart slightly from the ratio 1:2, the form of the curve slowly varies as φ changes.
11.20 LISSAJOU’S FIGURES FROM TWO RECTANGULAR S.H.M. IN FREQUENCY RATIO 2:1 Let a particle be subjected to two mutually perpendicular simple harmonic vibrations along the x and y axes and having frequencies in the ratio 2:1. They may be represented by and
x = a sin (2 ωt + φ)
...(i)
y = b sin ωt
...(ii)
where a is the amplitude of the x-vibration whose angular frequencies is 2 ω, and b is the amplitude of the y-vibration whose frequency is ω. The phase difference is φ. The equation to the curve of resultant motion can be obtained by eliminating t from Eqs. (i) and (ii). Expanding Eq. (i), we get x/a = sin 2 ωt cos φ + cos 2 ωt sin φ = 2 sin ωt cos ωt cos φ + (1 – 2 sin2 ωt) sin φ But from Eq. (ii) sin ωt = y/b and cos ωt = 1 − sin2 ωt = 1 −
F GH
I JK
x y2 2y 2 y2 1 − 2 cos φ + 1 − 2 sin φ = a b b b
∴
F GH
I JK
F1 − y I cosφ GH b JK F 2 y I sin φ − 2x F1 − 2 y I sin φ = 4 y F1 − y I cos φ + G1 − J G b JK a GH b K b H H b JK x 2 y2 2y − 1 − 2 sin φ = a b b
or
squaring
y2 b2
x2 a2
2
2
2
2
2
2
2
2
2
2
2
2
2
4 y4 4 y2 2x 4 xy2 4 y2 4 y4 x2 2 2 2 2 sin sin sin sin sin cos cos2 φ + φ + φ − φ − φ + φ = φ − 2 4 2 2 2 4 a a b b ab b b 2x 4 y4 4 y2 4 xy2 x2 2 2 2 2 2 sin sin (sin cos ) (sin cos ) sin φ = 0 + φ − φ + φ + φ − φ + φ + a a2 b4 b2 ab2
Hormonic Motion
543
FG x − sin φIJ Ha K FG x − sin φIJ Ha K
or
2
+
4 y4 4 y2 4 yx − 2 + sin φ = 0 b4 b ab2
+
4 y2 b2
2
Fy GH b
2
2
−1+
x sin φ a
I JK
= 0
...(iii)
This is the general equation for a curve having two loops, which is the general path of the particle. Let us consider special cases: Y
When φ = 0, sin φ = 0 and the Eq. (iii) reduces to
(a)
Fy GH b
x2 4 y2 + 2 a2 b
2
2
I JK
−1
= 0 b
This represents a curve symmetrical about both the axis, like the figure of ‘8’ as shown in figure (a)
X′
a
when φ = π/2, sin φ = 1 and the Eq. (iii) reduces to
(b)
FG x − 1IJ + 4 y FG y − 1 + x IJ Ha K b Hb aK FG x − 1IJ + 4 y + 4 y FG x − 1IJ Ha K b b Ha K LMFG x − 1IJ + 2 y OP NH a K b Q 2
2
2
2
or
2
2
4
2
2
or
= 0 Y′
2
4
X
= 0
Fig. 13(a)
2
2
= 0
This represents two coincident parabolas symmetrical about the x-axis each being
y2 = −
b2 ( x − a), as shown in figure (b) 2a
(c)
when φ = π, sin φ = 0. Hence the path of the particle is again the figure of ‘8’.
(d)
when φ = 3π/2, sin φ = –1 and the Eq. (iii) reduces to
FG x + 1IJ + 4 y FG y − 1 − x IJ Ha K b Hb aK FG x + 1IJ + 4 y − 4 y FG x + 1IJ Ha K b b Ha K LMFG x + 1IJ − 2 y OP NH a K b Q 2
2
2
2
or
4
4
2
2
2
2
2
or
= 0
2
= 0
2
= 0
This again represents two coincident parabolas, each being y2 = b2/2a (x + a) which is the appropriate Lissajous figure (c).
544
Mechanics Y
Y
b
b
X′
X
O
X′
a
Y′
a
O
X
Y′
(b )
(c )
Fig. 13
11.21 USES OF LISSAJOUS FIGURES Lissajous figures are used in obtaining beautiful design for printing in cloth industry. They find useful applications in the acoustical measurement, e.g., determination of unknown frequency. If the frequencies of vibration of the two rectangular vibrations are not exactly equal in the ratio 2:1, the form of the Lissajous figure undergoes a gradual progressive change. The time, taken by the curve to undergo a complete cycle of changes, enables us to find the frequency of one vibration, if the frequency of the other is known. Let the frequency of one vibration be n and that of the other be slightly different n′, the frequencies being nearly in the ratio 1:1. If t be the time in which the cycle undergoes a complete change, then the frequency of the other will be
1 t If the frequencies are nearly in the ratio 2:1, then n' = n ±
1 t To find whether n′ is greater or less than n or 2n, a small piece of wax is attached to the unknown (e.g., the prong of the fork) and again the time of one cycle of changes is determined. If this time increases, the unknown frequency is higher and if it decreases, the unknown frequency is lower that n or 2n.
n' = 2n ±
One practical application of Lissajous figures is that they enable us to know by inspection the period ratio of its constituent vibrations. It is evident that if the horizontal and vertical lines are drawn on a Lissajous figure and if they cut it m and n times respectively, the required period ratio of the two vibration is m:n.
NUMERICALS Q.1. A particle is executing S.H.M. along the x-axis. Obtain expressions for its position velocity as function of time, assuming the initial conditions. t = 0, x = 0, v = v0. Solution. Let x be the displacement of the particle from its equilibrium position at any instant t. By definition of S.H.M., the instantaneous force acting upon the particle is given by F = –Kx
Hormonic Motion
545
where K is proportionality factor. But by Newton’s second law F =
md2 x dt2
where m is the mass of the particle and d2x/dt2 is the instantaneous acceleration. Thus, md2 x = –Kx dt2 d2 x = –ω2x dt2
where ω2 = K/m. Multiplying both sides of this eqn. by 2dx/dt. We have
FG H
2dx d2 x dx × 2 = − ω2 2x dt dt dt
IJ K
Integrating w.r.t. ‘t’ we get
FG dxIJ H dt K When x = 0, v =
2
= − ω2 x2 + c (constant) .
dx = v0 . Therefore dt c = v02
FG dxIJ H dt K
2
= v02 − ω2 x2
dx = dt dx v02 /ω2
− x2
sin−1
x v0 /ω
v02 − ω2 x2
= ωdt
Integrating, we get = ωt + φ (constant)
ωx v0 = sin (ωt + φ) or
ωx v0 = sin (ωt + φ) Again at t = 0, x = 0 ∴ φ = 0. Thus
ωx v0 = sin ωt
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Mechanics
x =
v0 sin ωt ω
Diff. w.r.t. t we get
dx v = ω 0 × cos ωt dt ω v = v0 cos ωt
Ans.
Q.2. If the displacement of a moving particle at any time is given by x = a cos ωt + b sin ωt. Show that the motion is simple harmonic. If a = 3, b = 4, ω = 2, find the period, maximum velocity and maximum acceleration. Solution. The displacement is given by x = a cos ωt + b sin ωt
...(i)
⇒
dx = –ωa sin ωt + ωb cos ωt dt
⇒
d2 x = –ω2acos ωt – ω2b sin ωt dt2
...(ii)
= –ω2(a cos ωt + b sin ωt) = –ω2x This acceleration
d2 x is proportional to displacement x and directed opp. to it. Hence dt2
the motion is simple harmonic of amplitude A =
a2 + b2 =
(32 + 42 ) = 5 cm .
The period of motion is T =
2π 2 × 3.14 rad = = 3.14 sec. ω 2 rad/sec
The maximum velocity is ωA = 2 × 5 = 10 cm/sec The maximum acceleration is ω2A = 22 × 5 = 20 cm/sec2 Ans. Q.3. A body on the end of spring oscillates with an amplitude of 5 cm at a frequency of 1 Hz. At t = 0 the body is at its equilibrium position x = 0. Write the equation describing the position of the body given the numerical values of A, ω and α in the form x = A cos (ωt + α). What will be its velocity at t = 8/3 sec. Solution. The required form of the motion is x = A cos (ωt + α) Here A = 5 cm and ω = 2 πn = 2 π × 1 = 2π rad/sec. Further at t = 0, x = 0 so that 0 = A cos α. or
α = π/2 Hence the equation of the motion is x = 5 cos (2πt + π/2) cm
Hormonic Motion
547
or
x = 5 cos
FG 2t + 1 IJ π cm H 2K
The instantaneous velocity of the body is
FG H
IJ K
dx 1 = −10 π sin 2t + π dt 2 At t = 8/3 sec we have
dx 35 π = −10 π sin dt 0
FG π IJ LMQ sin FG 6 π − π IJ = − sin π OP H 6K N H 6K 6Q π π 1 = 10 π sin LMQ sin = OP 6 N 6 2Q = −10 π sin 6 π −
= 5π cm. Ans. Q.4. A particle of mass 10 gm is placed in the potential field given by v = (50x2 + 100) erg/gm. Calculate the frequency of oscillation. Solution. The potential energy of the 10 gm particle is U = 10 (50x2 + 100) erg The force acting upon the particle is given by F = − But ⇒
F =
m×
dU = − 1000 x dyne dx
md2 x , by Newton’s law dt2
d2 x = –1000x dt2
⇒
d2 x −1000 x −1000 x = = − 100 x 2 = m 10 dt
⇒
d2 x = –ω2x dt2
Where ω2 = 100. Since the acceleration is proportional to the displacement, the particle is executing S.H.M. The frequency of oscillation is n = ω/2π = 10/2π = 1.58/sec or Hz. Q.5. Distinguish between two oscillation given by: y 1 =A cos ωt and y2 = A cos (ωt + π) Solution. The oscillation have an initial phase difference of π. At start (t = 0), the first particle is at one extreme end of its motion (y1 = A) while the second particle is at the opp. end of its motion (y2 = –A).
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Mechanics
Q.6. Write the equation of S.H.M. if (i) initial phase is zero (ii) initial phase is π/2. The amplitude is 5 cm and period 8 sec. Solution. The general solution of S.H.M. is x = a sin (ωt + φ) Here a = 5 cm, ω = 2π/T =
2 × 3.14 rad = 0. 785 rad / sec. 8 sec
(ii) When initial phase φ = 0, we have
FG H
x = 5 sin 0. 785t +
π 2
IJ K
= 5 cos (0.785 t) Q.7. A particle executes S.H.M. of period 31.4 sec and amplitude 5 cm. Calculate its maximum velocity and maximum acceleration. Solution. The eqn. of S.H.M. is x = a sin (ωt + φ) The velocity is u =
dx = ωa cos (ωt + φ) dt
The maximum value of cos (ωt + φ) is 1. Therefore umax = ωa Here ω = 2π/T =
2 × 3.14 = 0. 2 rad/sec and a = 5 cm. 31. 4 umax = 0.2 rad/sec × 5 cm = 1 cm/sec
The acceleration is given by f =
d2 x = ω2a sin (ωt + φ) dt2
The maximum value of sin (ωt + φ) is 1. Therefore (fmax) = ω2a = (0.2 rad/sec)2 × 5 cm = 0.2 cm/sec2 Q.8. Show that when a particle is moving in S.H.M. it’s velocity at a distance its amplitude from the central position is half its velocity in the central position. Solution. The displacement of particle in S.H.M. is x = a sin (ωt + φ) The velocity is u =
dx = ωa cos (ωt + φ) dt
= ωa 1 −
x2 = ω a2 − x2 a2
3/2 of
Hormonic Motion
At x =
549
3a , the velocity is 2 2 u = ω a2 − 3 a = 1 ωa, 4 2
which is half the velocity (ωa) in the central position (x = 0). Q.9. The equation of motion of a particle is x = 2 sin
FG πt + π IJ cm . Find the period and H 2 4K
the maximum velocity of the particle. Solution. The given eqn. is x = 2 sin
LM πt + π OP cm N 2 4Q
Comparing it with the general equation of S.H.M. x = a sin (ωt + φ) we have, a = 2 cm, ω = π/2 rad/sec. The period is
2π rad π /2 rad/sec
T = 2π/ω =
= 4 sec.
The maximum velocity of the particle is umax = ωa
π rad / sec × 2 cm 2 = 3.14 cm/sec. =
Q.10. The period of S.H.M. is 10 sec and amplitudes is 10–1m . Write its equation. What are the phase and displacement at a time 5 sec after a passage of the particle through its extreme positive elongation? What is the maximum velocity. Solution. The general equation of S.H.M. is x = a sin (ωt + ϕ ) Here a =10–1 m = 10 cm and ω = ∴
2π 2×π = = π / 5 rad/sec. T 10
x = 10 sin
FG πt + φIJ cm. H5 K
If the motion starts from of the extreme positive elongation, that is, x = 10 cm at t = 0 then we have 10 = 10 sin φ φ = π/2
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Mechanics
the equation of motion is x = 10 sin Now at t = 5, the displacement is x = 10 sin
FG πt H5
+
π 2
IJ K
FG π × 5 + π IJ H 5 2K 3π = − 10 cm. 2
= 10 sin the phase at t = 5 sec is 3π/2. The velocity is differentiating (i) given by u =
umax =
FG IJ H K
FG H
dx π πt π + = × 10 cos dt 5 5 2
IJ K
FG π IJ × 10 = 2π = 6.28 cm/sec. H 5K
Q.11. Two particle 1 and 2 start vibrating together in S.H.M. along the same straight line. If their periods are 40 and 60 sec resp. Find their phase difference (a) after 20 sec from start and (b) when one particle is at the end of its path while the other is at middle. Solution. The displacement of particle in S.H.M. about x = 0 is given by x = a sin
2π t T
(a) for particle 1 and 2, T = 40 sec respectively. Thus
and
x 1 = a sin
πt 20
x 2 = a sin
πt 30
at t = 20 sec, we have x 1 = a sin π and
x 2 = a sin
FG H
π 2π = a sin π − 3 3
IJ K
these two eqn. show that the phase diff between particle 1 and 2 is π/3 i.e., 60°. (b) Particle 1 is at the end of its path and particle 2 is at the middle i.e., x1 = +a and x2 = 0, thus at this instant, we can write +a = a sin ω1t 0 = a sin ω2t These give
ω1t = π/2 or 3π/2 and ω2t = 0
∴ phase diff. ω1t ~ ω2t = π/2 or 3π/2
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551
Q.12. A particle of mass 5 gm is executing S.H.M. has amplitude of 8 cm. If it makes 16 vibration/sec find maximum velocity and energy at mean position. Solution. The displacement of particle in S.H.M. is x = a sin (ωt + φ) and its velocity u =
dx = ω a cos (ωt + φ) dt
It is maximum when cos (ωt + ϕ ) = 1 i.e., umax = ωa = 2πna = 2 × 3.14 × 16 × 8 = 802.8 cm/sec. The energy at mean position is entirely kinetic and is given by E = Kmax = =
1 mu2max 2
1 × 5 (80. 8)2 = 1. 6 × 106 erg. 2
Q.13. A particle in S.H.M. has velocities u1 and u2 when its displacements from the mean position are x1 and x2 respectively. Calculate the period, amplitude and maximum speed of the particle. Solution. The speed of a particle in S.H.M. at a displacement x is given by u = ω a2 − x2 where ω is angular speed and a is amplitude Here
u1 = ω a2 − x12
...(i)
u2 = ω a2 − x22
...(ii)
u12 − u22 = ω2 ( x22 − x12 ) ω =
or
u12 − u22 x22 − x12
The period is therefore T = 2π / ω = 2π from eqn. (i) and (ii), we have.
u12 a2 − x12 = u22 a22 − x22 or
a2u12 − x22u12 = a2u22 − x12u22
x22 − x12 u12 − u22
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Mechanics
a =
x22u12 − x12 u22 u12 − u22
The maximum particle velocity in S.H.M. is umax = ωa =
u12 − u22 x22 − x12
=
x22 u12 − x12 u22 x22 − x12
×
x22u12 − x12u22 u12 − u22
Q.14. A particle is moving with S.H.M. along a straight line. If velocity when passing through the points 3 cm and 4 cm from the centre of its path is 16 cm/sec and 12 cm/sec respectively. Find the amplitude and period of motion and the maximum velocity of the particle. Solution. The displacement of a particle in S.H.M. is x = a sin (ωt + ϕ ) where a is amplitude. It’s velocity
dx = ωa cos (ωt + φ) dt
FG dxIJ H dt K Now when x = 3, when x = 4,
= ωa
1 − sin2 (ωt + φ)
= ωa
1−
x2 = ω a2
2
= ω2 (a2 – x2)
dx = 16 cm/sec dt
dx = 12 cm/sec dt
Substituting in eqn. (i) we get 256 = ω2 (a2 –9) 144 = ω2 (a2 – 16) Dividing we get
a2 − 9 16 = 2 a − 16 9 16 (a2 – 16) = 9 (a2 – 9) 7a2 = 156 – 81 = 175
a2 − x2
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553
a2 =
175 = 25 7
a = 5 cm. Substituting in Eqn. (iii) we get 144 = ω2 (25 – 16) ω2 =
144 9
ω = 12/3 = 4 ∴ The period is T =
2π 2 × 3.14 = = 1. 57 sec ω 4
Maximum velocity of the particle
FG dx IJ H dt K
max
= ω a.
Q.15. A particle executes linear S.H.M. about the point x = 0. At time t = 0, it has displacement x = 2 cm and zero velocity. If the frequency of motion is 0.25/sec find (i) period (ii) circular frequency (iii) Amplitude, (iv) maximum speed, (v) displacement and velocity at t = 3 sec. Calculate the time required for the particle to come half-way in toward the centre of motion from its initial position. Solution. (i) Frequency,
n = 0.25/sec
∴ period,
T =
(ii) Circular frequency
ω = 2π/T = 2π/4 = π/2 rad/sec.
(iii)
x = a sin (ωt + φ)
So that
u =
1 = 1/0.25 = 4 sec. n ...(i)
dx = ωa cos (ωt + φ) dt
...(ii)
At t = 0, x = 2cm and u = 0 (given). Therefore from eqn. (iii) we get a = + 2 cm
(Q sin π /2 = 1, sin 3π /2 = − 1)
(iv) From eqn. (ii) we have umax = ωa = π/2 × 2 = 3.14 cm/sec. (v) Putting t = 3 sec, ω = π/2, a = 2 cm and φ = π/2 in eqn. (i) and (ii) we get x = 2 sin 2π = 0 u = 3.14 cos 2π = 3.14 cm/sec. Let us now compute the time required for the particle to come halfway in toward the centre of motion from its initial Putting a = 2 cm, ω = π/2 and φ = π/2 in Eqn. (i) we get
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Mechanics
FG πt + π IJ = 2 cos πt H 2 2K 2
x = 2 sin At half way
x =
a = 1 cm. Hence 2 πt 2
1 = 2 cos cos
πt = 1/2 2
FG IJ H K
πt −1 1 = π/3 = cos 2 2 t =
2 sec 3
Q.16. The equaiton of motion of an oscillating body is x = 6 cos ( 3πt + π / 3 ) m. What is the period, frequency and phase constant of motion. Also find the displacement, velocity and acceleration at time t = 2 sec. Solution. The given eqn.
FG H
x = 6 cos 3πt +
π 3
IJ K
meters.
represents a S.H.M. Let us compare it with the general eqn. of S.H.M. in the cosine from which is x = a cos (ωt + φ) We see that for the given oscillating body a = 4 meter, ω = 2π and φ = π/3 period
T =
2π 2π 2 = = sec ω 3π 3
frequency,
n =
1 3 = vib / sec T 2
and phase constant
φ =
π rad = 60° 3
Now, the displacement is x = 6 cos Af t = 2 sec, x = 6 cos
FG 3πt + π IJ m H 3K
π = 6 cos 60° = 3 meter 3
FG H
IJ K
dx π = − 18 π sin 3πt + m / sec dt 3
The velocity is
u =
At t = 2 sec
u = −18 π sin 6 π +
FG H
π 3
IJ K
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555
= −18 π sin = −18 π ×
π = − 18 π × sin 60 3
3 = − 49 m/sec 2
The acceleration is f = At t = 2 sec
FG H
IJ K
π d2 x m/sec2 = − 54 π2 cos 3πt + 2 3 dt
FG H
f = − 54 π2 cos 6 π +
π 3
IJ K
1 = − 166 m/sec2 2 Q.17. A mass of 150 gm hangs from a vertical spring of elastic constant 2 N/m. Calculate the frequency of small vertical oscillations, assuming that the damping is small. 2 2 = − 54 π cos π / 3 = − 54 π ×
Solution. m = 250 gm = 0.25 kg and K = 2 N/m. n =
=
1 2π
K m
1 2 × 3.14
2 = 0. 45 sec−1 0. 25
Q.18. A spring is hung vertically and loaded with a mass of 400 gm and made to oscillate. Calculate (i) the time-period and frequency of oscillation. When the spring is loaded with 100 gm, it extends by 5 cm. Solution. If an applied force Fapp extends a spring by x, then by Hooks’s law Fapp = Kx where K is force constant of the spring. Here Fapp = 100 gm = 100 × 980 dynes and x = 5 cm. K =
Fapp x
=
100 × 980 = 1. 96 × 104 dynes/cm 5
Now, the period of a mass hung from a light spring is given by
m K 4 Here m = 400 gm and K = 1.96 × 10 dyne/cm T = 2π
The frequency is
T = 2π
400 1. 96 × 104
T = 2π
1 = 0. 9 sec 4a
n =
1 1 = = 1.1 sec−1 T 0. 9 sec
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Mechanics
Q.19. A mass M of 2000 gm hangs from a vertical spring. When a mass m of 600 gm is gently added to 2000 gm, the spring is further stretched by 4 cm. Now m is removed and M is set in oscillation. Calculate the period. Sol. The eqn. of S.H.M. of a body of mass m is T = 2π
m K
= 2 × 3.14 ×
2000 = 0. 73 sec. 147000
Q.20. A body of mass 0.1 kg hanging from a spring is oscillating with a period of 0.2 sec and an amplitude of 1 m. Find the maximum value of the force acting on the body and the value of force constant of the spring. Sol. The equation of S.H.M. of a body of mass m is x = a sin (ωt + φ) where a is the amplitude and ω =
...(i)
K the time period is m
T =
2π = 2π ω
K m
Here T = 0.2 sec K = 10π m The acceleration of the body is, differentiating (i) twice
ω =
f = Then
...(ii)
d2 x = − ω2 a sin (ωt + φ) dt2
fmax = ω2a
Here ω = 10π and a = 1 m (given) ∴ and
fmax = (10π)2 × 1 = 986 m/sec2 max. force = mass × max. acceleration = (0.1 kg) × 986 m/sec2 = 98.6 N
Now from eqn. (ii) we have
K = (10π)2 m K = 100 π2 m = 100 × (3.14)2 × 0.1 = 98.6 N/m. Q.21. The potential energy of a harmonic oscillator in its rest position is 5 joule and the avg. kinetic energy is 2 joule. What is the total energy? If the amplitude is 1 meter, what is the force constant? If the mass is 2 kg. what is the period?
Hormonic Motion
557
Solution. The potential energy in rest position is 5 joule the avg. Kinetic energy (2 joule) is half the energy of oscillation so that the energy of oscillation is 4 joule. Hence total energy. 5 + 4 = 9 Joule At the maximum displacement a, the gain in potential energy is
1 Ka2, where K is force 2
constant thus.
1 Ka2 = 4 Joule 2 Here a = 1 meter K = 8/a2 = 8 N/m The period of oscillation is T = 2π
m = 2π K
2kg = 3.14 sec. 8 N/m
Q.22. A simple harmonic oscillator of period 5 sec has 5 joule potential energy when its displacement is 2 cm. (Energy in the rest position may be taken as zero). Calculate (i) force constant, (ii) avg. kinetic energy if the amplitude is 4 cm, (iii) Kinetic and potential energy when the displacement is 1 cm, (iv) oscillation frequency if mass is reduced to 1/100th of it’s initial value.
1 2 Kx where 2 K is force-constant. Here at x = 2 cm = 0.02 m, the gain in potential energy is 5 joule. Thus, Solution. (i) The gain in Potential energy when the displacement is x is
1 K (0. 02)2 = 5 2 K = 2.5 × 104 N/m (ii) The avg. Kinetic energy is Kav =
1 K a2 2
Here a = 4 cm = 4 × 10–2 m Kav =
1 × (2. 5 × 104 N/m) × (4 × 10−2 m)2 4
= 10 Joule (iii) The kinetic energy when the displacement is x is K =
1 K ( a2 − x2 ) 2
Here a = 4 × 10–2 m and x = 1 × 10–2 m K =
1 × (2. 5 × 104 ) × (15 × 10−4 ) 2
= 18.75 Joule.
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Mechanics
The potential energy is U = =
1 Kx2 2 1 (2. 5 × 104 N/m) × (10−2 × 1 m)2 2
= 1.25 Joule. (iv) The frequency is inversely proportional to the square root of mass. The initial frequency is ni =
1 = 0. 2 /sec 5
If mi and mf be the initial and the final masses, the final frequency is given by
nf nc Here mf = ∴
=
mi mf
1 mi 100 mf
= 100 = 10 0.2 nf = 0.2 × 10 = 2/sec
Q.23. A body of mass 0.5 kg is projected with an energy 4 joules. If a restoring force of 2 N/m acts on it. What maximum displacement will be body acquire? What will be its periodic time? Solution. At the maximum displacement a (say), the energy would be entirely potential energy. That is
1 K a2 = 4 Joule. 2 Here
K = 2 nt/m a =
2×4 2
T = 2π
= 2π
= 2 meter.
m K 0.5 = π sec. 2
Q.24. The amplitude of harmonic oscillations of a point mass of 10 gm is 5 cm and the energy of oscillations is 3.1 × 10–5 joule. Write the eqn. of motion if the initial phase is 60° Solution. The energy of oscillation is given by 2 2 2 E = 2π mn a =
1 (2πn)2 ma2 2
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559
=
1 2 ω ma2 2
1 2E a m Joule = 310 erg, a = 5 cm, and m = 10 gm. ω =
Here E = 3.1 × 10–5
ω =
1 5
2 × 310 = 10
62 = 1. 57 rad/sec 5
Now, the Eqn. of motion is x = a sin (ωt + φ) Putting a = 5 cm, ω = 1.57 rad/sec, φ = 60 = π/3 rad, we get x = 5 sin
FG1. 57t + π IJ cm. H 3K
Q.25. A mass M is controlled by two massless springs between two rigit support as shown. If the mass is 50 gm and the force-constants of springs are 3000 and 2000 dyne/cm find (i) frequency of oscillations of the mass, (ii) energy of vibration of amplitude 0.4 cm, (iii) velocity of mass when passing through mean position. Solution. (i) When the mass is displaced through a distance x, the restoring forces developed in the spring are F1 = − K1 x and F2 = − K 2 x and act in the same direction. The total restoring force is
b
g
F = − K1 + K 2 x So that the force constant of the system is K = K1 + K 2 = 3000 + 2000 = 5000 dyne/cm Therefore, the frequency of oscillation is n = =
1 2π
K 1 = m 2π
5000 5 = 50 π
5 = 1. 6 /sec. 3.14
(ii) The energy of oscillation is the same as the maximum potential energy. Thus
E = K max =
1 Ka2 2
1 × 5000 ×(0. 4)2 = 400 erg. 2 (iii) The velocity, and hence the kinetic energy, is maximum when the mass passes through, its mean position. Again, the energy of oscillation is same as the maximum kinetic energy. E =
E = K max = umax =
1 m (umax )2 = 400 2
800 = m
800 = ± 4 cm/sec 50
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Q.26. A riffle bullet weighing 10 gm moving with a velocity of 16000 cm/sec strikes and embeds itself in a 790 gm block which rests on a horizontal frictionless surface and is attached to a spring of force constant 8 × 104 dyne/cm. Compute the amplitude of the resulting simple harmonic oscillation of the block. What happens to the K.E. of the bullet ? Solution. Let v' be the velocity acquired by the block M when the bullet m strikes it and comes to rest in it. By conservation of momentum, we have (M + m) v' = mv 800 + v' = 10 × 16000 ∴
v' =
10 × 16000 = 200 cm/sec . 800
The block is set in oscillation about its mean position. If a be the amplitude of motion, the eqn. of motion may be written as x = a sin ωt The velocity is
dx = ωa cos ωt dt
In the mean position, the velocity is a maximum and equal to ωa. This we have obtained as 200 cm/sec. Thus ω a = 200 a = If T be the period, then T =
200 ω
2π i.e., ω = 2π /T ω a =
200T 2π
M+m , where M + m is the mass of the block + bullet and K is the force K constant of the spring
But T = 2π
a = 200 Here
M+m K
M + m = 800 gm and K = 8 × 104 dyne/cm. a = 200 ×
800 = 20 cm. 8 × 104
The kinetic energy of the moving bullet was K = =
1 mv2 2 1 × 10 × (16000)2 = 128 × 107 erg. 2
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561
The maximum potential energy stored in the spring is U = =
1 2 Ka 2 1 × (8 × 104 ) × (20)2 = 1. 6 × 107 erg. 2
∴ The percentage of energy stored in the spring =
1. 6 × 107 × 100 = 1. 25 % 1. 28 × 107
Thus a very small fraction of the K.E. of a bullet is stored in the spring and the rest appears as heat. Q.27. A block rests on a horizontal surface at the rate of 2 oscillations. The coefficient of static friction µ between block and plane is 0.5. How large can the amplitude be without slipping block and surface. Solution. Let m be the mass of the block and a the largest possible amplitude without slipping between block and surface. Then the maximum horizontal force acting on the block must be equal to the frictional force the maximum horizontal force on the block is mω2a because ω2a is the maximum acceleration where ω is the angular frequency. The frictional force is µmg. Thus mω2a = µmg a =
µg ω2
But ω = 2 π n, where n is the frequency a =
=
µg 4 π2n2 0.5 × 980 = 3.1 cm 9 × (3.14)2 × 22
Q.28. A simple pendulum of length 2m and amplitude 3.03 meter has energy 0.10 joule. What will be its energy (i) if its amplitude is changed to 0.06 meter and length remains unchanged, (ii) amplitude remains 0.03 meter but length is changed to 1 meter. Solution. The energy of oscillation of amplitude a is E = In case of simple pendulum ω = E =
1 mω2 a2 2 2π = T
g l
mga2 2l
Originally l = 2 meter a = 0.03 meter, E = 0.10 joule. 0.10 =
mg × (0. 06)2 2×2
...(i)
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Mechanics
(i)
E =
mg × 0. 06 2×2
E =
0. 06 × 0.10 = 0. 40 joule 0. 03
E =
mg × (0. 03)2 2×1
...(ii)
From eqns. (i) and (ii)
(ii)
...(iii)
From eqns. (iii) and (i) we get E =
1 × 0.10 = 0. 20 joule 2
Q.29. One end of a horizontal spring of force constant 6.0 N/m is tied to a fixed wall and the other end to a solid cylinder which can roll without slipping on the horizontal ground. The cylinder is pulled a distance of 0.1 meter and released. Calculate the K.E. of the cylinder when passing through the equilibrium position. Solution. When the Cylinder is pulled 0.1 m, the spring is stretched and has a potential energy given by U =
1 1 Kx2 = × 6. 0 × ( 0.1)2 = 0. 03 joule 2 2
Whole of this energy is converted into the K.E. of rotation and the K.E. of translation when the cylinder is passing through the equilibrium position. That is Krot + Ktrans = 0.03 joule. Krot = Ktrans = ∴ or
FG H
IJ K
2 1 2 Iω = 1 × 1 mr2 v = 1 mv2 2 2 2 r2 4
1 mv2 2
1 1 mv2 + mv2 = 0.03 Joule 4 2 mv 2 = 0. 03 ×
4 = 0. 04 Joule 3
Krot =
1 1 mv2 = (0. 04) = 0. 01 Joule 4 4
Ktrans =
1 1 mv2 = (0. 04 ) = 0. 02 Joule . 2 2
Q.30. Two tunning forks produce Lissajous figures of the shape of st. line, ellipse and finally to the same straight line in 2 sec. If one of the tunning forks has a frequency of 100 vib/sec, find the possible frequency of the other.
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563
Solution. The Lissajous’ figures are straight line and ellipse. Hence the frequency of the two must nearly be in the ratio 1 : 1 As the cycle of figures is traced in 2 sec, the difference in frequencies must be
1 / sec . 2
Therefore, the other fork must have one of the following frequencies.
1 = 100.5 , 99.5 2 Q.31. The Lissajous’ figure in the case of two tunning forks is a parabola. Prove that the frequencies are in the ratio 1 : 2. 100 ±
Solution. Let us draw co-ordinate axes across the given parabola. Let one fork of frequencies n1 be vibrating along the X-axis and the other of frequency n2 be vibrating along the Y-axis. Then if the parabola cuts the Y-axis Py times and the X-axis Px, times
Py n1 = n2 Px Here Py = 2 and Px = 1. Therefore n1 2 = . n2 1
Q.32. The tunning forks of approximately equal frequencies produce Lissajous’ figures that go through a cycle of changes in 15 second. When one fork is loaded with wax, the cycle of changes of Lissajous’ figure takes 10 seconds. If the second turning fork has a frequency of 400, find the frequency of the first fork before and after loading. Solution. Let n1 and n2 be the frequencies of two forks. Before loading, the cycle of Lissajous’ figure takes 15 seconds. This shows that in 15 seconds one forks makes one complete vibration more (or less) than the other. Thus, Y
X′
X
Y′
Fig. 14
n1 × 15 ~ n2 × 15 = 1 n1 ~ n2 = 1/15 But
n2 = 400 n1 = 400 + 1/15 = 400.067 or 399.933
On loading the first fork with wax, the time of cycle decreases to 10 sec. Hence the frequency of first fork after loading n′ 1 = n2 ± 1/10
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Mechanics
= 400 + 0.1 = 400.1 or 399.9 But on loading, the frequency will decrease. Hence its frequency before loading 400.1 and after loading 399.9. Q.33. Two tuning forks A and B of nearly equal frequencies are employed to produce Lissajous’s figures. On slightly loading A, the cycle of change of figures slows down from 15 sec to 20 sec. If the frequency of B is 256, find the frequency of A before and after loading.
LM Ans. 256 + 1 , 256 + 1 OP 15 20 Q N
Q.34. A particle of mass 10 gm lies in a potential field V = 50x2 + 100 erg/gm. Deduce the frequency of oscillation. Solution. Potential energy of 10 gm. U = 10−3 ( 50 x2 + 100) ergs . Force = – dU/dx = –1000x, ∴ 10
d2 x = − 1000 x dt2
d2 x + 100 x = 0 dt2
or
This is the eqn. of S.H.M. whose frequency is n =
100 5 = = 1. 6 oscillations/sec. 2π π
Q.34. A particle moves in the potential energy field U = U0 − Px + Qx2 . Find the expression for the force. Calculate the force constant and time period. At what point does the force vanish ? Is this a point of stable equillibrium? Solution.
Force F =
− dU = + P − 2Qn dx
Evidently, this is a linear restoring force and the force constant is 2Q. Eqn. of motion is
m ∴
Periodic time = 2π m / 2Q .
The force vanishes, where ∴
d2 x d2 x 2Q P x x= P − 2 Q or + = 2 2 m m dt dt
dU = 0 i.e., P – 2Qx = 0 dx x = P/2Q
d2U = + 2Q , i. e., if Q is positive at x = P/2Q, there is the minima of potential dx2 energy curve and hence this point is in stable equilibrium. Now
Hormonic Motion
565
Q.35. A particle is moving with S.H.M. in a st. line, when the distance of the particle from the equilibrium position has the value x1 and x2 the corresponding values of velocity are u1 and u2. Show that the period is T =
Lx 2π M NU
2 2 2 1
− x12 − U22
OP Q
1/2
Solution. Let the equilibrium position of the particle be 0.8 if P, Q are the two positions of the particle at distances x1 and x2, from O respectively, then the velocity u1 of the particle at the position P is given by u1 = ω a2 − x12
...(i)
The velocity u2 at position Q is u2 = ω a2 − x22
...(ii)
Squaring (i) and (ii) and substracting we get u12 − u22 = ω 2 ( x22 − x12 )
u12 − u22 . x22 − x12
ω =
Hence,
time period T =
2π = 2π ω
x22 − x12 . u12 − u22
Q.36. A particle of mass 2 gm moves along the x–axis and is attracted towards origin by a force 8 × 10–3x newton. If it is initially at rest at x = 10 cm, find (i) then differential eqn. of motion, (ii) the position of particle at any time, (iii) the velocity of particle at any time, (iv) amplitude and frequency of vibration. Solution. (i) From
2 md2 x −3 −3 d x 10 x = − 8 × 10−3 x or 2 × 10 = –8 × dt2 dt2
d2 x + 4x = 0 dt2
or
This is the differential eqn. of motion. (ii) Let the solution of the equation be x = a sin (2t + φ) when t = 0
x = 0.1 m ∴ 0.1 = a sin φ
when t = 0 dx/dt = 0 ∴ 0 = 2a cos φ [dx/dt = 2a cos (2t + φ )] a = 0 or φ = π/2 but a ≠ 0, ∴ φ = π/2 Therefore 0.1 = a sin π/2 or a = 0.1 m. x = 0.1 sin (2t + π/2) = 0.1 cos 2t This equation gives the position of the particle at any time t.
566
Mechanics
(iii) Now, (iv)
v = dx/dt = – 0.2 sin 2t Amplitude = 0.1 m n = ω/2π = 2/2π =
Frequency
1 = 0.3 Hz. π
Q.37. A body having a mass of 4 gm executes S.H.M. The force acting on the body, when displacement is 8 cm, is 24 gm ωt. Find the period. If the maximum velocity is 500 cm/sec. Find the amplitude and maximum acceleration. (g = 9.8 m/sec.) Solution.
Mass of the body = 4 gm = 4 × 10–3 kg
Force acting on the body at 8 cm Displacement = mω2x = 4 × 10–3 × ω2 × 8 × 10–2 = 24 × 10–3 × 9.8 ω =
whence ∴
Period T =
6 × 9818 = 27.12 2π 2 × 3.14 = = 0. 2315 sec. ω 27.12
The maximum velocity of the body v = ωa = 0.5 m/sec. ∴
a =
0. 5 v = = 0.1844 m. ω 27.15
The maximum acceleration = ω2a = 735.75 × 0.1844 = 135.67 m/sec2. Q.38 A vertical U tube of uniform cross-section contains water upto height 30 cm. Show that if the water on one side is depressed and then released its motion up and down the two sides of the tube is simple harmonic and Calculate its time period. Solution. Let PQ be the initial level of water in the U tube upto a height h = 0.3 m. When the level of water in tube A is depressed through a distance y, it rises at the same time by the same amount in the tube B. Now the difference between the two levels P′ and Q′ = 2y.
A
B P
B
y
Q′ Q
P′
The weight of this column of length 2y will now act on the mass of the water in the tube, as a result it will oscillate up and down.
h
The weight of the column of length 2y = 2y × A × ρg newton. Where A is cross-section of the tube and ρ the density of the liquid.
Fig. 15
Or force acting on the mass of water = 2y Aρg newton Total mass of the water
...(i)
m = 2hAρ = 2 × 0.3 × Aρ kg
∴ Force acting on the mass of water = m(d2y/dt2) = 2hAρ(d2y/dt2) 2hAρ (d2y/dt2) = –2yAρg or
d2 y y + ×y = 0 dt2 h
...(ii)
Hormonic Motion
567
Hence the motion of water column is simple harmonic, whose time period is given by T = 2π
0. 30 h = 2π = 1.098 sec g 9. 81
Q.39. Show that the time period for the swing of a magnet in the earth’s field is given I where M is the magnetic moment of the magnet, I its moment of inertia MH about the axis of suspension and H the earth’s field.
by T = 2 π
Solution: Let a magnet Ns be suspended freely in the earth’s magnetic field H. Obviously two opposite forces each equal to mH will constitute a couple equal to mH × NS sin θ, where θ is the angle making the magnet with the direction of the earth’s magnetic field. But magnetic moment M = m × Ns Couple = –MH sin θ = –MHθ
(as θ is small)
(Since couple is opposite to θ increasing) H
The couple develops an angular acceleration d2θ/dt2 in the magnet. If I is the moment of inertia of the magnet about the suspension thread, then the couple can be depressed as I (d2θ/dt 2) Hence the equation of motion of the magnet is 2
2
d θ MH Id θ + MHθ = 0 or + θ = 0 I dt2 dt2
Therefore, the motion of the magnet is simple harmonic, whose period is t = 2π
mH
N
θ θ S
P
O mH
I MH
Fig. 16
Q.40. If the earth were a homogeneous sphere of radius R and a straight hole were bored in it through its centre. Show that a particle dropped into the hole will execute a S.H.M. and find out its time period. Solution. If the mass of the earth is M and R its radius then the acceleration due to gravity at the surface of the earthy is given by g = GM/R2 where G is gravitational constant. If the density of the earth is ρ, the M = ∴
g =
4 3 Rρ 3
4 Rρ 3
...(i)
If g' be the value at depth h below the surface of the earth, then g' =
4 (R − h). Gρ 3
...(ii)
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Mechanics
Dividing Eq. (ii) by Eq. (i) we have
R−h g' g (R − h) = or g′ = R R g If we measure the distance from the centre of the earth, putting R–h = x, acceleration of the particle
g d2 x = g′ = − x R dt2 Negative sign shows that the acceleration acts towards the centre of the earth i.e., opposite to x increasing. ∴
d2 x g + x = 0 R dt2
Thus, the particle, dropped into the hole, will execute S.H.M. Its periodic time is given by T = 2π
R g
Q.41. Two cities on the surface of the earth are joined by a straight smooth under ground tunnel of length 640 km. A body is released into the tunnel from one city. How much time will it take to reach the other city? Derive the formula used. G = 6.67 × 10–11 S.I. Units and ρ = 5.520 kg/m3. Calculate the velocity when the body would be nearest to the centre of the earth in its journey. Solution. Suppose that A and B are the two cities, which are connected by a straight tunnel of length AB = 640 km. Let O be the centre of the earth and R its Radius. OC is the perpendicular, drawn from the centre 0 on AB, where C is the midpoint of the tunnel AB. Let a body of mass m be released into the tunnel from its end A and let it be at P, distant x from c at any instant t. If M is the mass of the earth, then on its surface the weight (mg) of the body is given by mg = GMm/R2
A R P C
O
B
Fig. 17
The weight of the body at P, distant OP (= r) from the centre of the earth is mg' = GM′ m = r2
G
4 3 4 πr ρm G πR3ρmr GMm r 3 = 3 3 = . r2 R R2 R
LMQ m′ = 4 πr ρ and M = 4 πR ρOP 3 3 N Q 3
The direction of this force is PO, whose component along PC =
GMm r x GMm r GMm . . = x . cos OPC = 2 R2 R r R R R3
3
Hormonic Motion
569
Hence the eq. of motion of mass m is
m
d2x GM d2x GMm x + + 3 x = 0 = 0 or R3 R dt2 dt2
The equation represent a S.H.M. of period. T = 2π
R3 GM
i.e., this is the time in which the body will reach from A to B and again form B to A. Hence, time taken by the body to pass through the tunnel is
R3 T R = π G. 4 π R3ρ = = π 2 3 GM 3
=
velocity at c1(v) = ωa =
=
3π 4 ρ. G
3 × 3.14 = 2528 sec 4 × 5520 × 6. 62 × 10−11 GM AB . = R3 2 6. 67 × 10−11 ×
4 AB G. πρ . 3 2 4 640 × 3.14 × 5520 ×103 3 2
= 3.972 × 102 m/sec. Q.42. The total energy of a particle executing a S.H.M. of period 2π second is 10.24 × Joule. The displacement of the particle at π/4sec is 8 2 cm. Calculate the amplitude of the motion and the mass of the particle. 10 –4
Solution. The displacement of a particle executing S.H.M. is given by x = a sin ωt = a sin 2πt/T [Q ω = 2π/T] 8 2 × 10−2 = a sin
∴ ∴
a π 2π π = × = a sin 4 2 2π 4
Amplitude of vibration a = 0.16 m. If m be the mass of the particle, then its total energy is given by E =
Here
10. 24 × 10−4 × T 2 2mπ2a2 –4 or m = = 10.24 × 10 2π2 × a2 T2
T = 2π sec and a = 0.16 m m =
10. 24 × 10−4 × 2π × 2π = 0.18 kg 2 × π2 × 0.16 × 0.16
Q.43. If the potential energy of a harmonic oscillator in its resting position is 5 joules and the total energy is 9 Joules. When the amplitude is 1 m, what is the force constant? If its mass is 2 kg what is the period?
570
Mechanics
Solution. Potential energy in rest position = 5 joules. At the maximum displacement (equal to amplitude), the potential energy = 9 Joules. ∴
Gain in P.E. = 9 – 5 = 4 Joule. The gain in energy at the maximum displacement should be equal to
1 2 1 Cx max = C. 2 2
(1)2 = C/2 Joules. Thus
C/2 = 4 or C = 8 Joule/m. m 2 = 2π = 3.14 sec. C 8
Period T = 2π
Q.44. Using conservation of energy, show that the angular speed dθ/dt of a simple pendulum is given by
dθ = dt
LM 2 N ml
2
O { E − mgl(1 − cos θ ) }P Q
1 2
where E is total energy of oscillation. L and m are length and mass of the pendulum and θ is angular displacement from the vertical. S θ
Solution. The potential energy of the simple pendulum at any position P relative to the lowest position O of the bob is obviously given by
l
U = mg(l – l cos θ)
l sin θ
P
O
because the bob is displaced through a vertical height (l – l cos θ) The kinetic energy of the pendulum at the position
mg
P is
Fig. 18
K =
LMQ V (= rω ) = ldθ OP dt Q N
1 2 1 mv = m (ldθ / dt)2 2 2
1 2 ml (dθ/dt)2 + mgl(1 – cos θ) which should be a constant, 2 according to the law of conservation of energy ∴ Total energy E = K + U =
∴
dθ = dt
LM 2 N ml
2
O ( E − mgl (1 − cos θ) P Q
1 2
Q.45. A simple pendulum of length l and mass m is oscillating with a maximum angular displacement θ0 radians. Show that the tension in the string at the angular displacement θ is mg (3 cosθ – 2 cos θ0). Solution. Let the bob P of mass m be oscillating about the point S. At the angular displacement θ of the pendulum the weight mg can be resolved into two components mg cosθ and mg sin θ. The latter component provides the restoring force for S.H.M. and the former component reduces the Tension T of the string. So that the necessary centripetal force for the rotation of the pendulum in a circular path of radius l at the angular displacement θ is. T – mg cos θ = mv2/l where v is the vocally to the bob at P.
...(i)
Hormonic Motion
571 S θ
θ0
l
N
P0
P
N θ
θ θ
os θ mg c
s in mg
mg
Fig. 19
Total energy of the particle = mg ON (at Po relative to 0) = mgl(1–cos θ0) = mgl (1 – cos θ) + or
1 2 mv (at p) 2
2mg cos θ – 2mg cos θ0 = mv2/l
...(ii)
∴From Eq. (i) and Eq. (ii), we get T–mg cos θ = 2mg cos θ – 2mg cos θ0 T = 3mg cos θ – 2mg cos θ0 = mg (3cos θ – 2cos θ0) Q.46. Derive an expression for the small angle oscillations of a simple pendulum of length L comparable with the radius of earth R. What is the period of a pendulum of an infinite length on the surface of the earth?
S O
Solution. As shown in fig the restoring force on the bob is
md2x = – mg sin (θ + α ) dt2 where θ and α both are small. So that sin (θ + α ) = θ x x + α and θ = and α = therefore R L 2
LM N
OP Q
1 1 d x + g + x = 0 L R dt2
P x C
α
mg
O
Fig. 20
θ+α
572
Mechanics
Thus the motion of the pendulum is simple harmonic of Period.
2π
T =
g
When
FG 1 + 1 IJ H L RK
LR g(L + R)
= 2π
1 1 << or practically L is tending to infinite length L R T =
2π
FG 1 + 1 IJ g H ∞ RK
= 2π
R g
Q.47. A 1.6 kg wt extends a spring 8 cm from its unstretched position. The mass is replace by a body of 50 gm. The mass is pulled and then released. Find the period of oscillation. Solution.
F = –Cx or C = –F/x.
∴
C =
∴
1. 6 g 1. 6 × 9. 8 = = 196 n/m 0. 08 0. 08
m 50 × 10−3 = 2π c 196 = 0.11 sec.
Period T = 2π
Q.48. (a) Figure below shows a mass M resting on a smooth table between two firm support. A,B and controlled by two massless springs. If the mass M is 30 gm, and the force constants of the two springs are 2 and 1 n/m deduce (i) the frequency of small oscillations of M, (ii) the energy of oscillations for amplitude 0.5 cm.
A
C1
C2
B
M
Fig. 21
(b) If the two springs are connected as shown is fig. 21 find the period for oscillation. Solution. (a) If the mass M is displaced through a distance x towards left or right, the restoring forces in the two springs are F1 = –C1x and F2 = – C2x acting on the mass along the same direction. So that, the total restoring force on the mass M is F = –(C1 + C2)x. Thus the force constant of the system is C = C1 + C2 = 2 + 1 = 3 n/m. (i) Mass M = 30 gm = 0.03 kg ∴
1 C 1 3 5 = = 2π M 2π 0.03 π = 5/3.14 = 1.6 (nearly)
Frequency n =
(ii) Energy E = Max. P.E. =
1 2 cx (max) 2
=
1 × 3 × (0. 5 × 10−2 )2 = 3.75 × 10–5 J. 2
Hormonic Motion
573
(b) When the springs of force constant C1 and C2 are connected as shown in Fig. 21 the displacement x of mass M produces the same restoring force F in the two springs. If the two springs are extend through x1 and x2 then F = –Cx1 = –Cx2 and therefore x = x1 + x2 =
F =
LM N
−F F C + C2 − = −F 1 C1 C2 C1C2
OP Q
− C1C2 x C1 + C2
Thus the force constant of the system is C =
C1C2 2×1 2 = = C1 + C2 2 + 1 3
M 0.03 × 3 = 2π = 1.33 sec. C 2 Q.49. Show that the frequency of oscillation of a body of mass from M suspended from uniform spring of force constant C and mass m is given by ∴
Period T = 2π
1 C 2 π M + m/3 Solution. If l is the length of the spring, then mass per unit length = m/l. n =
Now, let us consider a length ds of the spring at a distance s measured along the length below the fixed upper end. The mass of this small element is m/l ds. If the velocity of the lowest point of the spring is v, then evidently the velocity of the length ds at a distance s from rigid support will be (s/l)v
FG H
IJ FG vs IJ KH lK
1 m ds 2 l ∴ Kinetic energy of the whole spring
∴Kinetic energy of the element=
z l
=
0
2
1 mv2 2 s ds 2 l3
mv2 1 mv2 l3 . = = 6 2 l3 3 Kinetic energy of the mass M at the lower end of the spring 1 Mv2 2 Total kinetic energy of the system =
=
1 mv2 v2 m Mv3 + = (M + ) 2 6 2 3
A ds B P M
Fig. 22
574
Mechanics
If x is the displacement of the mass from its mean position, then v = dx/dt and total K.E. = (M + m/3) (dx/dt)2 Potential energy of the spring
z x
=
Cxdx =
0
1 2 Cx 2
1 1 2 Cx = a constant (M + m/3) (dx/dt)2 + 2 2 Differenting it with respect to time and dividing by dx/dt, we have From the law of conservation of energy
FG M + m IJ d x H 3 K dt 2
2
+ Cx = 0 or
d2 x C + x = 0 2 ( M + m /3) dt
This is the equation of simple harmonic motion, whose frequency is n =
1 C 2π ( M + m/3)
Q.50. A solid sphere of mass 4 kg and radius 0.05 m is suspended from a wire. Find the period of oscillation, if the torque required to twist the wire is 4 × 10–3 newton–m/radiation. Solution. Moment of inertia of the Sphere about any diameter is I = ∴
2 2 MR2 = × 4 × 0. 05 × 0. 05 = 0.004 kg × m2 3 3
Period of oscillation T = 2π
I 0.004 = 2π C 4 × 10−3
= 2 × 3.14 = 6.25 sec. Q.51. A body is executing torsional oscillations about a wire of torsional rigidity C. If the total energy of oscillation is E and the moment of inertia of the body about the axis of rotation is I. Show that the angular velocity (dθ/dt) at the displacement θ is given by
dθ = dt
C I
2E −θ C
Solution. Let C be couple per unit twist. If the wire is twisted through an angle θ, then the restoring couple is –Cθ. Now if the wire is twisted to angle θ + dθ, then the amount of work done is dW = Cθ dθ Hence the total amount of work done in twistig the wire through an angle θ is
z θ
W =
0
Cθ dθ =
1 2 Cθ 2
At the angular displacement θ, this work done becomes the potential energy of the system i.e., U =
1 2 Cθ 2
Hormonic Motion
575
The instantaneous angular displacement θ of a body, executing torsional oscillation is given by θ = θ0 sin ( ω t + φ ), where ω =
C I
dθ = θ0 ω cos ( ω t + φ ) dt
So that
K.E. at the angular displacement is K =
FG IJ H K
1 dθ I 2 dt
2
=
LM N
=
1 2 2 θ2 Iθ 0 ω 1 − 2 2 θ0
=
1 2 ( θ0 − θ2 ) C 2
∴ Total energy
E = K + U =
∴ Angular amplitude
θ0 =
Now,
1 2 2 Iθ 0 ω cos2 (ωt + φ) 2
dθ θ2 = θ0 1 − 2 = dt θ0
OP Q=
e
1 2 2 Iω θ0 − θ2 2
j
1 1 1 C (θ20 − θ2 ) + Cθ2 = Cθ20 2 2 2
2E C C I
C I
θ20 − θ2 =
2E − θ2 C
Q.52. A sphere of moment of inertia 3355 gm.cm2 when suspended by a thin wire executes torsional oscillations of period 4.3 sec. Calculate the maximum angular velocity of the sphere and it’s average potential energy when the amplitude is π/2 radians. Solution. Let the equation for the angular displacement of the sphere be θ = θ0 sin ( ω t + φ ). ∴
Angular velocity
dθ = θ0 ω cos ( ω t + φ ) dt
Its maximum value = θ0 ω = 2πnθ0 Here ∴
θ 0 = π/2 radians and frequency n = Max. angular velocity = 2 × 3.14 ×
Average potential energy =
LM N
1 per sec. 4. 3
1 3.14 × = 2.3 rad/sec 4. 3 2
OP Q L 2π OP × LM π OP ×M N 4. 3 Q N 2 Q
1 2 2 1 Iω θ0 in place of mω2a2 4 2
1 × 3355 × 10−7 = 4 = 4.45 × 10–4 J.
2
2
12 WAVE MOTION 12.1 WAVE MOTION The concept of waves is very fundamental in nature and is one of the most important means of transferring of energy. Let us suppose a number of particles in an elastic medium (such as air) so that one of them cannot move without disturbing its neighboring particles. Thus any how if a particle is displaced then a disturbance will be produced in the medium which will be handed on from particle to particle till all of them have suffered greater or less displacements. Now, if the particle performs periodic motion, then other particles of the medium perform the same type of motion with the phase differing regularly from one particle to next. Such a disturbance is called wave motion. Thus, a wave motion may be regarded as a disturbance, which is caused by the particles of an elastic medium executing definite periodic vibrations about their mean positions, the phase varying particle to particle and which travels with a definite velocity, depending upon the density and the elasticity of the medium. If the disturbance or state of motion is continuously transmitted along the same direction, then it is called a progressive wave. It may be noted that in a wave motion there is no bodily transfer of the medium through which the wave propagates. What travels forward in the medium is a state of motion which is passed on from one particle to the next as the wave advances. Since during the propagation of the wave, the particles of the medium lying in its path are thrown into vibration in succession, obviously the waves pass on energy from one particle to next along their line of travel. Hence in a progressive wave motion there is always a transmission of energy along the direction of propagation of the wave. In stationary waves, the energy does not transfer from one place to another.
12.2 TRANSVERSE WAVE A transverse wave is one in which the direction of oscillation of the particles of the medium is at right angles to the direction of propagation of the wave. For example, the waves generated in ropes and stretched strings and the light (electro-magnetic) waves are transverse waves.
12.3 LONGITUDINAL WAVE A longitudinal wave is one in which the direction of oscillation of the particles of the medium is the same as the direction of propagation of the wave. For example, if a spring fixed 576
at one end is pulled at the other and then let free, a longitudinal wave is generated in it. Sound waves are also longitudinal waves.
12.4 GENERAL EQUATION OF WAVE MOTION Let a wave train move along the direction of x-axis with velocity v. If we assume the source of wave or any point of x-axis as the origin (x = 0), then at any instant t the displacement of a particle, situated at the origin, can be represented by the relation. y = f (t) Where f(t) is any function of time. The wave will reach a point P, distant x from O, in x/v second. Hence at any instant t the displacement y of the particle at P must be the same as the displacement at the origin x/v seconds earlier i.e., y = f (t – x/v)
...(1)
This is the equation for wave, travelling along the positive direction of x-axis with constant velocity v. The function f determines the shape and size of the wave. Similarly, the equation of a wave moving along the negative direction of x-axis, will be given by y = f (t + x/v)
.…(2)
Thus, the general equation of a wave, moving along x-axis, may be expressed as
FG H
y = f t−
IJ FG K H
x x + g t+ v v
IJ K
...(3)
Where f and g are any two functions. In Eq. (1) or (2) y is the function of time and distance both. Now partially differentiating Eq. (1), we get
FG H
x ∂y = f′ t − v ∂t
FG H
IJ K
x 1 ∂y = − f′ t − v v ∂x
FG H
IJ K
x ∂2 y f ′′ t − 2 = v ∂t 2 x 1 ∂ y f ′′ t − 2 = v2 v ∂x
FG H
.…(4a)
IJ K
...(4b) ...(5a)
IJ K
...(5b)
D isp la cem en t y
Y
p y X
O x
Fig. 1
578
Mechanics
where f ′ and f ′′ are the first and second differentials of f. From Eq. (4), we have
∂y ∂y = −v ∂x ∂t This
...(6)
∂y represents the particle velocity at the time t and distant x from the origin and ∂t
∂y is the slope of the y-x curve at the same instant and at the same point. Hence, for those ∂x waves, traveling along +ve direction of x-axis, we have particle velocity = –wave velocity × slope of y-x curve From Eq. (5) we have ∂2 y 1 ∂2 y = ∂x2 v2 ∂t 2
or
2 ∂2 y 2 ∂ y v = ∂t 2 ∂x2
…. (7)
This is called one dimensional differential equation of wave position. This is very important equation. The striking feature of this equation is that the coefficient of derivative on the right hand side represents the square of the wave velocity so that we need not to solve the equation to obtain the velocity of propagation of the wave.
12.5 EQUATION OF A PLANE PROGRESSIVE HARMONIC WAVE The simplest and most important type of wave is the simple harmonic wave in which the displacement of a particle is a simple harmonic or sinusoidal function of time and distance both. Such waves are produced by vibrating bodies, which execute simple harmonic motion. Let the source of the wave be situated at the origin 0 (x = 0) and let a regular train of plane waves travel in the positive direction of x-axis. Since the source of waves performing simple harmonic motion at any instant t the displacement of a particle at O will be given by yo = a sin ωt
...(1)
Where a is the amplitude of the vibration and ω/2π is the frequency. If the wave velocity is v, then the disturbance; produced at O, will reach at the point P distance x in x/v seconds. Hence the displacement of the particle P at any instant t will be given by y = a sin ω(t – x/v)
...(2)
This equation represents the displacement of a particle at any time t and is of the same form as Eq. (1) in previous section, and it is one of the solutions of general differential one dimensional wave equation. Thus it is the equation of a simple harmonic wave traveling with velocity v along the positive direction of x-axis. Similarly the equation y = a sin ω(t + x/v) will represent a wave traveling in the negative direction of x-axis with velocity v. The frequency of the wave is equal to the frequency of the oscillating body, hence the frequency of the wave n = ω\2π = 1/T, where T is the period. In one vibration (i.e., in period T) the wave moves through a distance equal to wavelength λ, hence v = λ/T = nλ.
Wave Motion
579
Now the various alternative forms of the equation of wave motion can be written as
FG x IJ H vK 2π F a sin G t − vx IJK = a sin 2π FGH Tt − vxT IJK T H F t xI a sin 2π G − J H T λK 2πv F a sin G t − vx IJK λ H
y = a sin 2πn t −
y =
y = y =
y = a sin
2π (vt − x) λ
Y
y
O
D istan ce x
P
X
Fig. 2
The quantity
2π 1 = 2πv = k is called the propagation constant, where v = is the wave λ λ
number. y = a sin (ωt – kx) If φ is the phase constant, then y = a sin (ωt – kx + φ)
Three-dimensional wave If a wave is moving in any direction (such a wave is called three dimensional wave), the wave equation can be written as 1 ∂2Ψ ∂2Ψ ∂2Ψ ∂2Ψ + + 2 2 2 = v2 ∂t 2 ∂x ∂y ∂z 1 ∂2Ψ = ∇Ψ v2 ∂t 2 2
One of the solutions of this equation is a three dimensional plane harmonic wave, in which the displacement (or disturbance) at time t and at a point r = xi + yj + zk is given by ψ = A sin (ωt – k1x – k2y – k3z + φ)
580
Mechanics →→
ψ = A sin (ωt − k x + φ )
or
is called the propagation vector. The direction of this vector is the + kk where k = i kx + jk y z direction of wave motion and its magnitude is 2π/λ, the propagation constant.
12.6 PRINCIPLE OF SUPERPOSITION A very important characteristic of all the waves is the principle of superposition. This principle states that when a number of waves are simultaneously propagated through a medium, the resultant physical disturbance (e.g., displacement) at any point is the sum of the disturbance due to separate waves, i.e., ψ = ψ1 + ψ2 + ψ 3 + ... Where ψ is the resultant disturbance at a point due to different waves, having ψ1, ψ2, ψ3 ... etc. disturbances separately at the same point. In case of a one dimensional wave, if y1, y2, ... etc. are the displacements at a point due to separate waves, then the resultant displacement due to all waves is y = y1 + y2 + y3 + ...
12.7 LONGITUDINAL WAVES IN RODS When longitudinal vibrations are produced in a solid rod, its every section vibrates to and fro about its mean position along the axis of the rod, provided that the rod has the sufficient length in comparison to its diameter. Let us suppose that longitudinal waves are traveling along an elastic solid rod. Now, consider two close planes A and B, perpendicular to the rod at distance x and x + δx respectively from some arbitrary origin. Suppose that at any instant A′ and B′ are the positions of the planes A and B respectively. Then, the displacement of x is y (= AA′) and at x + δx is y + δy (=BB′). Hence the initial length AB between the two planes = δx Change in this length = A′B′ – AB = (δx + δy) – δx = δy A′B′ = (x + δx) + (y + δy) – ( x + y) = δx + δy ∴
Longitudinal strain =
δy dy = δx dx
y + δy
y
A
A′
B
x x + δx
Fig. 3
B′
Wave Motion
581
The change in length AB of the rod occurs due to the difference of force acting on the two planes. The force acting on the plane A is given by Fa = Stress × Area of cross-section of the rod If α is the area of cross-section of the rod, the Young’s modulus of its material is given by Y =
Stress dy F dy or Stress = Y or a = Y Strain dx dx α
dy dx Similarly the force acting on the plane B is given by Fa = αY
∴
Fb = Fa + δ (Fa ) = Fa + = αY ∴
d (Fa ) δx dx
dy d2 y + αY 2 δx dx dx
Resultant force on AB = Fb − Fa = αY
d2 y δx dx2
If ρ is the density of material of the rod, then the mass of the element AB = ραδx. If the acceleration of this small element AB is d2y/dt2, then according to Newton’s second law of motion, we have
ραδx
or
d2 y d2 y α Y δ x = dt2 dx2 d2 y d2 y Y /ρ = dt2 dx2
This is the differential equation for the longitudinal waves in a rod. Comparing this with the differential equation of wave motion. 2 d2 y 2 d y , the velocity of the longitudinal waves in the solid rod is given by = v dt2 dx2
Y v 2 = ρ or
v=
Y ρ
Thus, the velocity of longitudinal waves in a rod depends upon the density and Young’s modulus of its material.
12.8 LONGITUDINAL WAVES IN GASES (PRESSURE VARIATIONS FOR PLANE WAVES) When a plane progressive longitudinal wave moves through a gas, the gaseous particles execute simple harmonic motion along the direction of propagation of the wave. The phases of these particles vary regularly. The distance between the particles so changes that at any instant particles are alternately crowded and spreaded out. Hence pressure varies from particle to particle inside the gas.
582
Mechanics
Let a progressive wave be moving along the positive direction of x-axis. Now, imagine cylindrical gas column of cross-sectional area α along x-axis. Let us consider in equilibrium position, two right planes A and B (being very close to each other) of the gas column, situated at x and x + δx distances respectively from the origin. At some later instant, when the wave propagates, the particles of the plane A are displaced by an amount y to the position A′ and those of the plane B are displaced by an amount y + δy to the position B′. As dy/dx is the rate of change of displacement with distance. ∴
y + δy = y +
dy δx dx
Initial volume of the gas column between the plane A and B = α(AB) = αδx
LM N
its final volume = α(A′B′) = α(δx + δy) = α δx + ∴
FG dy IJ δx H dx K F dyI α G J δx H dxK = dy
FG dyIJ δxOP H dxK Q
Change in volume = α
Thus, Volume Strain =
α δx
dx
If E is the Bulk modulus of the gas, then K =
Volume Stress decrease in pressure −p = = Volume Strain Volume Strain dy / dx
dy dx Which represents the variation of pressure in case of a longitudinal progressive wave travelling through a gas. Here the negative sign shows that an increase in volume is caused by a decrease in pressure. If there is a decrease in volume, p will be positive but negative sign will still be there. ∴
p = −K
If the equation of the plane waves is taken to be y = a sin then ∴
2π (vt − x) λ
dy 2πa 2π cos (vt − x) = − dx λ λ p = −K
2π 2π dy 2πa cos (vt − x) = po cos (vt − x) =K λ λ λ dx
where v is the wave velocity and ‘a’ the wave amplitude. The quantity po = 2πaK is called λ the pressure amplitude.
Wave Motion
583
If the rate of change of pressure is dp/dx, then the pressure difference across the small element AB is δp =
dp δx dx
FG dpIJ αδx H dx K d F IJ = − αδx K d y G −K dy = αδx dx H dx K dx
Hence the force on this element = δp × α =
2
2
If ρ is the density of the gas, then the mass of the element = αδxρ As the force on the element is in the direction B'A', hence the acceleration due to this force will be –d2y/dt2. Now according to Newton’s law of motion
−αδx ρ
d2 y d2 y −αδxK 2 2 = dt dx d2 y = dt2
FG K IJ d y H ρ K dx 2
2
This is the differential equation for the waves traveling in a gas. Comparing it with the standard differential equation of wave motion, we get v2 =
K or v = ρ
K ρ
Thus the velocity of longitudinal waves in a gas depends upon the elasticity and the density of the medium. The formula v =
K for the velocity of sound in a gas was first of all deduced by Newton ρ
and so it is called Newton’s formula. Newton assumed that during alternate compressions and rarefactions occurring when a sound wave travels through a gaseous medium, the temperature remains constant. Therefore he put the isothermal value of the elasticity K, which is equal to the pressure P of the gas. Thus v =
P ρ
12.9 WAVES IN A LINEAR BOUNDED MEDIUM If a wave travels along a fixed line in a medium, then we call such a medium as linear one. Transverse waves traveling on a stretched string longitudinal waves in a solid rod and sound waves traveling through a pipe are such type of examples. If a wave travels along a line, then it must satisfy the one dimensional wave equation, given by
d2 y 1 d2 y 2 = v2 dt2 dx when y is the longitudinal or transverse displacement.
...(1)
584
Mechanics
The solution of Eq. (1) must be some function of a linear combination of t and x. Let y = f (t + αx) be the solution, where α is a constant. So that on substitution in Eq. (1) we have
α 2 f ′′ ( k + αx) =
1 f ′′ (t + αx) v2
where a = ± 1/v and f may be any function. Therefore, the general solution of this equation will be given by
FG H
y = f t−
IJ FG K H
x x + g t+ v v
IJ K
...(2)
This means that the total displacement at a point is in general due to two waves, one of them is moving along the positive direction of x-axis and the other along the negative direction of x-axis. In the case of a linear medium of infinite length, the functions f and g depend only on the initial disturbance produced by the external cause. When the length of a medium is not infinite, it is called a linear bounded medium. In such a case, considerable restrictions are applied on the wave functions f and g. These restrictions depend on the boundary conditions. (a)
The boundary may be rigid: In such a case, the displacement of the particle at the boundary is zero at all the times i.e., particles remain permanently at rest. Such a boundary is formed by a fixed rigid body. If the medium tries to vibrate this body, then this rigid body exerts a large pressure on the medium. Thus, a boundary is said to be rigid when it does not yield under the pressure of sound waves, for example a rigid wall, closed end of a pipe etc.
(b)
The boundary may be absolutely Free: A boundary is said to be free when it has no inertia, for example the open end of an organ pipe. On such a boundary, the particles will have displacement and the change of pressure i.e.,
FG H
dy dy p = −K dx dx
IJ K
will be zero at all the times. In a linear bounded medium, generally a wave is incident on the boundary and on reflection we get a second wave. By superposition of these two similar waves, traveling in opposite directions, we get stationary waves. If the incident wave is a simple harmonic wave then Eq. (2) will have the following form: y = a sin
2π 2π (vt + x) + a′ sin (vt − x) λ λ
incident wave
...(3)
reflected wave
Where the incident wave is traveling along negative x-axis and the reflected wave is moving along positive x-axis. In general, a is the amplitude of incident wave and a' that of reflected wave. This equation gives the resultant displacement due to the two harmonic waves. Case 1. When the boundary of the linear medium is rigid: Let us consider a rigid boundary at x = 0, having linear medium on the positive side. As the displacement y = 0 at the point x = 0 hence from Eq. (3), we have
Wave Motion
585
0 = a sin
2πvt 2πvt + a′ sin λ λ
or
a′ = – a
This means that the harmonic wave, traveling along the +ve direction of x-axis, is reflected back with an equal negative amplitude, i.e., with a phase change of π. Thus, on reflection from a rigid surface, a reversal of phase occurs. Now, the total displacement at any point of the medium is given by y = a sin
2π 2π (vt + x) − a sin (vt − x) λ λ
y = 2a sin
2πx 2πvt cos λ λ
y = 2 a sin
FG 2πx IJ cos ωt HλK
But 2πv/λ = 2πn = ω ∴
...(4)
If the other end of the medium is not bounded, i.e., the medium extends to infinity, then frequency (n = ω/2π) can have any value. The amplitude at any point x is 2a sin 2π/λ, which is zero when sin
2πx = 0 λ
or
2πx = rπ λ
or
x =
rλ where r = 0, 1, 2, ... etc. 2
Such points at which the value of the amplitude is zero, are called nodes. First node exits at the rigid boundary and the other successing nodes are separated from each other by the distance λ/2. The maximum value of the amplitude is ±2a, when
sin or or
2πx = ±1 λ 2πx π = (2r + 1) λ 2 x =
rλ λ + where r = 0, 1, 2, ...etc. 2 4
Such point at which the value of the amplitude is maximum are called antinodes. First antinode occurs at a distance λ/4 from the rigid boundary and the adjacent antinodes are separated by a distance λ/2. This means that one antinodes is situated in between two nodes. Here, the resultant wave is stationary in which the net flow of energy is zero because equal amounts of energy are transmitted in opposite directions by the two component waves. Now, differentiating Eq. (4), with respect to x, we get
dy 4 aπ 2πx cos cos ωt = dx λ λ
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Mechanics
Therefore,
dy 4 aπ 2πx = −K cos cos ωt ...(5) dx λ λ Hence the variation of pressure is proportional to cos 2πx/λ. Its value is maximum, when p = −K
cos
2πx = ±1 λ
or
2πx = rπ λ
or
x =
rλ 2
where r = 0, 1, 2, ...etc.
Hence at the points where the value of the amplitude is zero, the variation of pressure is maximum. Similarly at the points of maximum amplitude, the variation of pressure is zero. If we differentiate Eq. (4) with respect to time t, we get the particle velocity, i.e.,
dy 2πx sin ωt = −2aω sin ...(6) dt λ Hence the velocity of any particle of the wave is proportional to sin 2πx/λ. This means that the particle velocity is zero at the nodes and maximum at the antinodes. For any (particular) value of x, we will discuss the variation of the displacement with respect to the time t. It is clear from Eq. (4), (5) and (6) that if cos ωt = ±1 or ωt = 0, then the value of displacement y and variation of pressure p will be maximum and the particle velocity will be zero, i.e., when ωt = mπ or 2π t/T = mπ ∴ or
t =
mT , where m = 0, 1, 2, ... 2
t = 0, T/2, T, 3T/3 ...
i.e., at these instants the value of y and p is maximum and that of velocity is minimum. These instants occur twice in every period. when
cos ωt = 0 or
sin ωt = ±1
ωt = (2p + 1) π/2
where p = 0, 1, 2, ...
or
t = (2p + 1) T/4
or
t = T/4, 3T/4, 5T/4, ...
Hence it is clear from Eqs. (4), (5) and (6) that at these instants the values of y and p are zero and that of dy/dt is maximum. Such instants occur twice in every period. Now, if the linear medium is also rigidly bounded at the other end, then in the case of the medium of length l, substituting y = 0 at x = l in Eq. (4), we get 0 = 2a sin
2πl cos ωt λ
As this relation holds at all values of t, ∴
2πl = rπ, where r is any integer. λ
Wave Motion
587
v rv 2l and frequency n = = r λ 2l Substituting this value in Eq. (4), we get ∴
λ =
rπx rπvt cos l l Thus, we see that when a linear medium has rigid boundaries at its both ends, only vibrations of certain discrete frequencies are possible (n = v/2l, v/l, 3v/2l, ...). When r = 1, the value of frequency (n = v/2l) is minimum and in such a case the mode of vibration of the medium is called fundamental. When r = 2, 3, ... etc. the medium can vibrate with other modes. The integral multiples (v/2l, v/l, 3v/2l, ...) of the fundamental frequency are called harmonics. y = 2a sin
Case 2. When the boundary of the linear medium is free: Let us now consider a free boundary at x = 0 in place of a rigid boundary. As the boundary is free, the pressure variation at x = 0 must be zero, i.e., at x = 0, dy/dx = 0. Differentiating Eq. (3) with respect to x, we get
dy 2πa 2π 2πa′ 2π cos (vt + x) − cos (vt − x) = dx λ λ λ λ when x = 0, dy/dx = o ∴
0 =
2πa 2πvt 2πa′ 2πvt cos − cos λ λ λ λ
This equation holds for all values of time t. ∴
a′ = a
This means that in this case the phase of the reflected wave does not change. Hence the displacement at any point of the medium is given by y = a sin
2π 2π ( vt + x) + a sin (vt − x) λ λ
incident wave or
y = 2a cos
2πx 2πvt sin λ λ
or
y = 2a cos
2πx sin ωt λ
∴ Strain
dy 4 πa 2πx sin sin ωt = − dx λ λ
reflected wave
...(7) ...(8)
dy 2πx cos ωt = 2a ω cos ...(9) dt λ From these equation it is evident that x = 0, displacement y and particle velocity dy/dt are maximum and variation in pressure p (p = –K dy/dx) is zero. Hence, in this case an antinodes is present at the free boundary. The other antinodes will be situated at x = λ/2, λ, 3λ/2,... . Here, the nodes will be at x = λ/4, 3λ/4, ... with the help of these equations, the changes with respect to time t of y, dy/dt and p can be described similar to the first case. and
588
Mechanics
Now, if another free boundary is present at x = 1 (e.g. in open end organ pipes, in a solid rod clamped in the middle point etc.), then substituting dy/dx = 0 at x = 1 in Eq. (8), we have 0 =
4 πa 2πl sin sin ωt λ λ
2πl = rπ, where r = 0, 1, 2, ... λ
or
λ =
v rv 2l and frequency n = = r λ 2l
n =
v v 3v , , , ... 2l l 2l
Thus, if both the boundaries of a linear bounded medium are free, only vibrations of certain discrete frequencies are possible. In the present case, the frequency of the fundamental is v/2l and the frequencies of harmonics are proportional to the natural numbers 1, 2, 3, 4, ... . Now, if there is a rigid boundary (as in the case of closed organ pipe whose one end is open and other is rigid) at x = l, we have there y = 0, i.e., 0 = 2a cos ∴ Where frequency or
2πl sin ωt λ
(from Eq. 7)
2πl π = (2r + 1) , where r = 0, 1, 2, 3, ... λ 2 n =
v v (2r + 1) = λ 4l
n =
v 3v 5v , , , ... 4l 4l 4l
Now, if we put r = 0, 1, 2, ...etc. the frequencies of the harmonics are proportional to the odd numbers 1, 3, 5, 7, ... . These are odd harmonics. Thus, we see that in linear bounded medium (i) when both the boundaries are free, all the harmonics odd and even are present and (ii) when one boundary is free and the other is rigid, only odd harmonic are present.
12.10 FLOW OF ENERGY IN STATIONARY WAVES In a linear bounded medium, the incident wave is always reflected back. In general, the displacement at a point of the medium is given by
LM N
OP Q
2π 2π ( vt − x) + a sin ( vt + x) + φ λ λ where first and second component waves are incident and reflected ones respectively. Now,
y = a sin
FG 2π vt + φ IJ cos FG 2πx + φ IJ H λ 2K H λ 2K 4 πa F 2πx + φ IJ sin FG 2πvt + φ IJ sin G − H λ 2K H λ 2K λ
y = 2a sin ∴
dy = dx
Wave Motion
∴
589
variations in pressure p = –K (dy/dx), where K is the elasticity of the medium
or
p = E
FG H
IJ K
FG H
4 πa 2π x φ 2πvt φ + sin + sin λ λ 2 λ 2
FG 2πvt + φ IJ ; where p = K 4πa sin FG 2πx + φ IJ H λ 2K H λ 2K λ dy 4 πav F 2πvt + φ IJ cos FG 2πx + φ IJ =− cos G U = H λ 2K H λ 2K dx λ F 2πvt + φ IJ U = u cos GH λ 2K 4 πav F 2πx + φ IJ u =− cos G H λ 2K λ = px sin
Particle velocity
or
IJ K
x
x
where
x
Now, at any point the energy is flowing across α area. Force on this area F = pα If in infinitely small time dt, the medium at that point is displaced through dy distance, then the work done is = pαdy = pα(dy/dt)dt = pUαdt
z T
∴ Work done or transferred energy in a period =
pUαdt
0
The rate of transfer of energy across unit area per second is called energy current ∴
1 Energy current = αT
=
1 T
z
z
T
pU α dt =
0
T
pxux sin
0
px ux = 2T
z
T
o
sin
1 T
z
T
p U dt
0
FG 2πvt + φ IJ cos FG 2πvt + φ IJ dt H λ 2K H λ 2K
FG 4 πvt + φIJ dt H λ K
LM FG IJ OP H KQ N LM− λ cos FG 4 πt + φIJ OP ∴ N 4 πv H T K Q FG − λ IJ cos b4 π + φ g − cos φ H 4 πv K T
px ux 4 πvt λ cos − +φ = 2T 4 πv λ =
=
px ux 2T px ux 2T
= 0
o
T
o
v 1 = λ T
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Mechanics
Thus, we see that in a linear bounded medium, the rate of transference of energy is zero.
12.11 CHARACTERISTICS OF STATIONARY WAVES The above theory tells us the following characteristics in stationary waves: (i)
All particles of the medium oscillate at the same time and with the same frequency.
(ii)
The amplitude of oscillation varies from zero at the nodes to a maximum at the antinodes.
(iii)
Consecutive nodes, as well as consecutive antinodes, are λ/2 apart. Nodes and antinodes occur alternately.
(iv)
The nodes remain permanently at rest but (in case of longitudinal stationary waves) suffer maximum pressure-variation and are alternately the center of compression and the center of rarefaction.
(v)
The antinodes have always maximum displacement and maximum velocity to other points but (in case of longitudinal stationary waves) do not suffer pressure-variation.
(vi)
All the points between any pair of nodes are in same phase, so that they all reach their maximum displacement at the same instant, but the phase suddenly reverses at each node.
(vii)
Twice in one period all the particles simultaneously pass through their mean positions. The direction of motion is reversed after each half-period.
(viii)
Twice in one period all the particles simultaneously have their maximum displacements and are momentarily at rest. The displacements are reversed after each half-period.
(ix)
There is no net flow of energy in any direction.
12.12 WAVE VELOCITY (OR PHASE VELOCITY) When a monochromatic wave (wave of a single frequency or wavelength) travels through a medium, its velocity of advancement in the medium is called ‘wave velocity’. The equation of a plane progressive harmonic wave, traveling along x-axis is given by y = a sin (ωt – kx) Where a is the amplitude, ω (=2πn) is the angular frequency and k (=2π/λ) is the propagation constant of the wave. For a wave, the ratio of ω and k is called the phase velocity or wave velocity i.e., vp =
ω k
Evidently, this is the velocity with which a plain wavefront, i.e., a plane of constant phase, given by (ωt – kx) = constant, propagates in the medium, because then Differentiating with respect to time, we get
ω−k
dx = 0 dt dx ω = dt k
Wave Motion
591
Which is the wave velocity v. Thus the wave velocity is the velocity with which planes of constant phase advance through the medium. Hence the wave velocity is also called as ‘phase velocity’. There may be cases when ω and k are not linearly proportional, but the ratio ω/k is certainly the speed of the propagation at which the phase of a single wave would move along.
12.13 GROUP VELOCITY In practice, we come across pulses rather than monochromatic waves. A pulse consists of a number of waves differing slightly from one another in frequency. A superposition of these waves is called a ‘wave packet’ or a ‘wave group’. When such a group travels in a dispersive medium, the phase velocities of its different components are different. The observed velocity is, however, the velocity with which the maximum amplitude of the group advances in the medium. This is called the ‘group velocity’. Thus the group velocity with which the energy in the group is transmitted. The individual waves travel ‘inside’ the group with their phase velocities. In figure (a) are shown two plane harmonic waves of equal amplitudes but slightly different frequencies traveling from left to right. The dotted curve represents the wave of lower frequency and is travelling faster. At a certain instant the two waves are in phase at the point P. Therefore, at this instant, the maximum amplitude of the group formed by them also lies at P (figure b): At a later instant the maximum will be build up a little to the left of P. That is the maximum will move to the left with time relative to the waves. As a result, the group velocity will be lower than the wave velocities. Expression for Group Velocity: Let us consider a wave group consisting of two components of equal amplitudes a but slightly different angular frequencies ω1 and ω2, and propagation constants k1 and k2. Their separate displacements are given by and
y 1 = a sin (ω1t – k1x)
...(1)
y 2 = a sin (ω2t – k2x)
...(2)
P
(a ) P
(b )
Fig. 4
592
Mechanics
Their superposition gives y = y1 + y2 y = a[sin (ω1t – k1x) + sin (ω2t – k2x)] Using the trigonometric relation,
sin A + sin B = 2 sin
A +B A−B cos 2 2
We get
LM (ω N L (ω 2a cos M N
OP LM Q N − k )x O L (ω sin M P 2 Q N
OP Q + k )xO PQ 2
y = 2a sin
1
+ ω2 ) t ( k1 + k2 ) x ( ω1 − ω2 ) t ( k1 − k2 ) x − cos − 2 2 2 2
y=
1
− ω 2 ) t ( k1 − 2
2
1
+ ω2 ) t ( k1 − 2
2
This represents a wave-system with a frequency (ω1 + ω2)/2 which is very close to the frequency of either component, but with an amplitude given by A = 2a cos
LM (ω N
1
− ω 2 ) t ( k1 − k2 ) x − 2 2
OP Q
Thus the amplitude of wave group is modulated both in space and time by a very-slowly varying envelope of frequency (ω1 – ω2)/2 and propagation constant (k1 – k2)/2; and has a maximum value of 2a. This envelope is represented by the dotted curve in Fig. (b). The velocity with which this envelope moves, that is, the velocity of the maximum amplitude of the group is given by vg =
ω1 − ω 2 ∆ω = ∆k k1 − k2
If a group contains a number of frequency-components in an infinitely small frequencyinterval, then the expression for the group velocity may be written as vg =
dω dk
12.14 RELATION BETWEEN GROUP VELOCITY AND WAVE VELOCITY Since ω = kv, where v is wave (phase) velocity, the group velocity is given by vg =
dω d dv = ( kv) = v + k dk dk dk
Now, k = 2π/λ, where λ is wavelength. Therefore vg = v +
vg = v +
2π dv 2π dv =v+ λ dk λ d 2π λ 1 λ
dv 1 d λ
FG IJ H K
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593
FG 1 IJ H λK
1 dλ . λ2 dv Therefore, vg = v − λ dλ This is the relation between group velocity vg and wave velocity v in a dispersive medium (A dispersive medium is one in which the wave velocity is frequency dependent. ω/k = constant). d
But
= −
12.15 NORMAL AND ANOMALOUS DISPERSION Usually dv/dλ is positive, so that the group velocity vg is smaller than the wave velocity vo. This is called ‘normal dispersion’. But ‘anomalous dispersion’ can arose when dv/dλ is negative, so that vg is greater than v. An electrical conductor is anomalously dispersive to electromagnetic waves.
12.16 NON-DISPERSIVE MEDIUM In a non-dispersive medium (v = ω/k = constant), we have dv/dλ = 0, so that vg = v that is, the two velocities are identical. For light waves in free space the group velocity is same as the wave velocity. Sound waves in gases are also non-dispersive. This is a fortunate circumstance. If sounds of different frequencies travelled at different speeds through the air it would result in chaos and aural anguish.
NUMERICALS Q.1. A simple harmonic wave train is travelling in the positive x direction with velocity 100 m/sec. The amplitude of the wave is 2 cm and frequency 100 Hz. Calculate (a) the displacement y, (b) the particle velocity and (c) the particle acceleration at x = 2m from the origin at t = 5 sec. Solution. The equation of a simple harmonic wave is given by
2π (vt − x) λ x = a sin 2πn t − (3 v = nλ) v where y is the displacement of a particle at a distance x from the arbitrary origin x = 0. y = a sin
FG H
IJ K
Here, a = 0.02 m, n = 100 Hz, v = 100 m/sec and x = 2 m ∴
y = 0.02 sin 2π × 100
FG 5 − 2 IJ H 100 K
= 0.02 sin (2π × 496) = 0 Particle velocity Here,
U =
FG H
sin2πn t −
x v
IJ K
FG H
∂y x = 2πn a cos 2πn t − ∂t v
= 0
or
FG H
cos 2πn t −
IJ K
IJ K
x =1 v
U = 2πna = 2 × 3.14 × 100 × 0.02
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Mechanics
= 12.56 m/sec. Particle acceleration
FG H
IJ K
∂2 y x = −4 π2n2 sin 2πn t − =0 2 ∂t v
Q.2. The equation of a progressive wave is
LM FG t − 0.5 xIJ + 2OP K Q N H 0.05
y = 8 sin 2 π
Find the amplitude, frequency, velocity, wavelength and wave number. Solution. The general equation of a harmonic wave is given by y = a sin where φ is a constant
LM N
y = 8 sin 2π × 0. 5
Here ∴
FG t − xIJ + 2OP H 0. 05 K Q
Amplitude a = 8 cm
2π = 2π × 0.5, λ
∴ ∴
LM 2π (vt − x) + φ OP Nλ Q
Wavelength λ = 2 cm velocity v = Frequency n = Wave number =
1 = 400 cm/sec. 0. 025 v 400 = = 200 Hz λ 2 1 1 = = 0. 5 cm . λ 2
Q.3. Which of the following are the solutions to the one dimensional wave equation (i) y = 2 sinx cos vt,
(ii) y = 5 sin 2x cos vt,
(iii) y = x2 – v2t2,
(iv) y = 2x – 5t.
Solution. One dimensional wave equation is
∂2 y ∂2 y = v2 2 2 ∂t ∂x (i) Differentiate (i) twice partially with respect to t and x separately i.e.,
∂2 y ∂2 y 2 and − v y =−y = ∂t2 ∂x2 ∴
∂2 y ∂2 y = v2 2 2 ∂t ∂x
Wave Motion
595
Hence (i) is the solution to the one dimensional wave equation. (ii)
∂2 y ∂2 y 2 and − v y = − 4y = ∂ t2 ∂t2
∴
v2 ∂2 y ∂2 y = 4 ∂x2 ∂t2
Hence (ii) is not the solution of one dimensional wave equation. In the same way do for (iii) and (iv) Eq. (iii) is not the solution and eq. (iv) is the solution of the wave equation. Q.4. Find the frequency 1 period and wave number for a light of wave length 6000 A.U. Solution. Velocity of light c = vλ where v is the frequency of light wave and λ the wavelength. ∴
v =
Period T =
c 3 × 108 = = 5 × 1014 Hz λ 6000 × 10−10
1 1 = = 2 × 10−15 sec. v 5 × 1014
1 1 = = 1. 7 × 106 m−1 Wave number V = λ 6000 × 10−10 Q.5. A train of sound waves is propagated along a wide pipe and it is reflected from an open end. If the amplitude of the waves is 0.002 cm, the frequency 1000 Hz and the wavelength 40 cm, find the amplitude of the vibration at a point 10 cm from the open end inside the pipe. Solution. The equations of incident wave and its reflected wave from open end are given by [Incident wave] y1 = a sin
2π (vt − x) λ
[Reflected wave] y2 = a sin
2π (vt + x) λ
By the principal of superposition, the resulting stationary wave is given by y = y1 + y2 = a sin = 2a sin 2π
2π 2π (vt − x) + a sin ( vt + x) λ λ
vt x x vt cos 2 π = 2a cos 2π sin 2π λ λ λ λ
The amplitude of the resulting wave is A = 2a cos 2π
x λ
Amplitude at x = 0.1 m
2π × 0.1 =0 0. 4 Q.6. A stationary wave produced in a certain rod is expressed as y = 0.01 sin (13x) sin (200t). A = 2 × 0.002 × 10–2 cos
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Mechanics
Calculate the maximum particle velocity and tensile stress at the point x = 4 cm if the young’s modulus of the material of the rod is 1011 n.m–2. Solution. Particle velocity
dy = 0.01 × 200 sin (13 x) cos 200t dt The maximum value of particle velocity occurs when cos (200t) = 1, i.e., maximum particle velocity = 0.01 × 200 sin (13x) = 2 sin (13x) But at x = 0.04 m,
13x = 13 × 0.04 = 0.52 radian = 0. 52 ×
180 degree = 30° 3.14
sin(13x) = sin30° =
∴
Maximum particle velocity = 2 × Tensile stress
1 2
1 = 1 ms−1 2
dy = 1011 × 0. 01 × 13 cos (13 x) sin (200t) dx
Therefore, at 0.04 m the maximum Tensile stress = 1011 × 0. 01 × 13 × 3 /2 = 11.26 ×
109
n/m2
3 cos 13 x = cos 30°
13 STATIONARY WAVES (WAVES IN A LINEAR BOUNDED MEDIUM) PROBLEMS Q.1. A string vibrates according to the equation y = 5 sin
πx cos 40 πt , 3
where x and y are in cm and t is second. Find out the amplitude and velocity of the two component waves superposition can give rise to this vibration. What is the distance between adjacent nodes? What is the velocity of a particle of the string at x = 1.5 cm at t = 9/8 second. Solution. The given equation
πx cos 40 πt 3 represents a stationary wave. It may be written as y = 5 sin
y =
FG H
IJ K
...(i)
FG H
πx πx 5 5 sin 40πt + − sin 40πt − 2 3 2 3
IJ K
The two sine terms represent two harmonic waves travelling in opposite directions. These two waves constitute the given stationary wave. The component waves may be written as y1 = and
5 2π sin (120t + x) 2 6
5 2π (120t − x) y 2 = − sin 2 6
comparing the component wave equations with the standard equation y = ± a sin amplitude, a =
2π (vt ± x), we see that λ
5 cm, wavelength λ = 6 cm, velocity v = 120 cm/sec. 2 597
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Mechanics
and frequency, n =
v = 20 sec−1 λ
The distance between adjacent node is
λ = 3 cm 2 Differentiating eq. (i) with respect to t, we get ∂y πx = −5 × 40 π sin sin 40 πt ∂t 3 At x = 1.5 cm and t =
9 sec, we have 8 ∂y π = −200 π sin sin 45 π ∂t 2 3 sin 45π = 0
= 0 Q. 2. The equation of a stationary wave is x = 6 cos
FG πx IJ sin 60 πt cm. H 3K
Find the (i) equations of waves constituting the stationary wave, (ii) wave length and frequency of these waves, (iii) distance between successive nodes.
LMHint. sin A − sin B = 2 sin A + B cos A − B OP 2 2 Q N LM Ans. (i) x = 3sinFG 60πt + πx IJ , x = 3sin FG 60 πt − πx IJ , (ii) 6 cm, 30 sec H H 3K 3K N 1
2
−1
, ( iii) 3 cm
OP Q
Q.3. A longitudinal stationary wave in a rod is expressed as y = 0.002 sin (0.15x) sin (4.8 × 104 t) Calculate the maximum particle velocity and maximum tensile stress at the point x = 3.5 cm. The young’s modulus of the material of the rod is 8.0 × 1011 dyne/cm2. Solution. The particle velocity at position x and time t is obtained by differentiating the given equation with respect to t. ∂y = 4. 8 × 104 ( 0. 002) sin ( 0.15 x) cos ( 4 . 8 × 104 t ) ∂t
The maximum velocity at x occurs when cos (4.8 × 104 t) = 1. Thus
FG ∂y IJ H ∂t K
= 4.8 × 104 (0.002) sin (0.15x) = 96 sin (0.15 x) max
At x = 3.5 cm, we have 0.15 x = 0.15 × 3.5 = 0.525 radian
= 0. 525 ×
180° = 30° 3.14
3 π rad = 180°
Stationary Waves (Waves in a Linear Bounded Medium)
FG ∂y IJ H ∂t K
∴
599
= 96 sin (30°) = 48 cm/sec. max
In longitudinal waves the tensile strain is given by modulus x tensile strain, that is, Y
∂y . So the tensile stress is young’s ∂x
∂y . Now, differentiating the given equation. With respect ∂x
to x, we get
∂y = 0.15 ( 0. 002) cos(0.15 x) sin(4. 8 × 104 t) ∂t ∴
tensile stress = Y
∂y = (8. 0 × 1011 dyne / cm2 ) (3 × 10−4 ) ∂x
cos ( 0.15x) sin ( 4 . 8 × 104 t )
The tensile stress at x is maximum when sin ( 4. 8 × 104 t ) = 1 Thus max. tensile stress = ( 8. 0 × 1011 dyne / cm2 ) ( 3 × 10−4 ) cos ( 0.15x) = 24 × 107 cos (0.15x) dyne/cm2 At
x = 3.5 cm, we have max. tensile stress = 24 × 107 cos 30° dyne/cm2 = 24 × 107 × 0.866 = 2.1 × 108 dyne/cm2 Ans.
Q.4. A stationary wave produced in a rod is expressed as y = 0.01 sin (0.13x) sin (200t) cm. Calculate the maximum particle velocity and tensile stress at the point x = 4 cm. The young modulus of the material of the rod is 1012 dyne/cm2. [Ans. 1.0 cm/sec, 1.1 × 109 dyne/cm2] Q.5. The refractive indices of a medium for two spectral lines of wave length 400Å and 5000Å are 1.540 and 1.530 respectively. Calculate the value of the group velocity. Velocity of light in vacuum is 3 × 1010 cm/sec. Solution. The group velocity vg is defined by vg =
∆ ω ω 2 − ω1 = ∆k k2 − k1
where ω is angular frequency and k is propagation constant. Now
ω = 2πn
where n is frequency, which is given by n =
c λ (vac)
...(i)
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Mechanics
where λvac is the wave length of light in vacuum. If λ be the wavelength in a medium of refractive index µ, then
and so
λ =
λ vac µ
n =
c µλ
ω = 2π c µλ Thus
k =
2π λ
from Eq. (i), we have
LM 1 Nµ λ F1 2π G Hλ
2πc vg =
=
2
1 µ1λ1 2 1 − λ1 2 −
IJ K
OP Q
c ( µ1λ1 − µ2λ 2 ) µ1µ 2 ( λ 1 − λ 2 )
Given: c = 3 × 1010 cm/sec, µ1 = 1.540, µ2 = 1.530, λ1 = 4000Å, λ2 = 5000Å ∴
vg =
3 × 108 (1. 540 × 4000 − 1. 530 × 5000) 1. 540 × 1. 530 ( 4000 − 5000)
= 1.897 × 1010 cm/sec Q.6. A wave packet of mean wavelength 3.6 × 10–5 cm is travelling with phase velocity 1.8 × 1010 cm/sec in a dispersive medium of carbon-di-sulphide. From the dispersion of the
dv is computed to be 3.8 × 1013 cm per sec per cm. Calculate the group velocity in dλ carbon-di-sulphide.
medium,
Solution. The group velocity vg is related to the phase velocity by vg = v − λ
dv dλ
Substituting the given values, we get vg = (1.8 × 1010 cm/sec) – (3.6 × 10–5 cm) × (3.8 × 1013/sec) = 1.8 × 1010 cm/sec – 13.68 × 108 cm/sec = 1.8 × 1010 cm/sec – 0.1368 × 1010 cm/sec = 1.66 × 1010 cm/sec
Stationary Waves (Waves in a Linear Bounded Medium)
601
Q.7. (i) Calculate the wave number for gamma rays of wavelength 1Å. (ii) The angular frequency is five times of the propagation constant. Calculate phase and group velocities. Solution. The wave number v (say) is defined by v = For the given gamma rays,
1 λ
λ = 1Å = 10–8 cm
1 1 = −8 = 108 per cm. Ans. λ 10 cm (ii) The angular frequency ω is 5 times the propagation constant that is v =
ω = 5. k ω is the phase velocity v. Thus k v = 5 cm/sec. Ans. If k is in per cm. Since here
ω is constant, that is k v = constant dv = 0 dλ
Now, the group velocity is given by vg = v − λ because
dv =v dλ
dv = 0 Thus dλ vg = 5 cm/sec.
Q.8. The intensity of a plane wave is reduced to half after travelling one meter in an absorbing medium. What will be its intensity after travelling three meters? What will happen if the medium is non-absorbing. Solution. Let I0 be the initial intensity of the wave and I the intensity remaining after the wave travels a distance x in an absorbing medium. Then we have I = I0 e–βx where β is a constant when x = 1 m, I =
1 I0 . Thus 2 1 = e–β 2
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Mechanics
β = loge 2 = 0.693
or
Let I′ be the intensity for x = 3m. Thus I′ = I 0 e –3β But 3β = 3 × 0.693 = 2.08 I′ = I0e–2.08 =
I0 e 2.08
we can see that e2.08 = 8 I′ =
1 I0 . 8
The given wave is a “plane wave”. Hence while travelling in an ‘non-absorbing’ medium the intensity will remain undiminished. Q.9. The intensity I0 of a plane wave is reduced by 30% after passing through 1 meter in an absorbing medium. What distance should it travel so that the intensity becomes approximately I0/2 ? Solution. For a plane wave, the variation of intensity with distance in an absorbing medium is given by an exponential law of the form I = I0 e–βx where I0 is the initial intensity, I is the intensity after travelling a distance x and β is a constant. when x = 1 meter, I = 0.7 I0. Thus 0.70 I0 = I0 e–β
1 = eβ 0. 70 β = loge (10/7) = loge (1.428) For I =
1 I0 we have 2 1 I0 = I0 e –βx 2 2 = eβx βx = loge 2 x =
=
log e 2 log e 2 = β log e 1. 428 log10 2 0. 3010 = = 2 meter. log10 1. 428 0.1549
Q.10. Which of the following are solution of the one-dimensional wave equation? (i) y = 4 sin x cos vt, (ii) y = 9 sin 2x cos vt, (iii) y = x2 – v2t2, (iv) y = 2x – 5t. (Lucknow, 1992)
Stationary Waves (Waves in a Linear Bounded Medium)
603
Solution. The general differential equations for a progressive wave y = y (vt ± x) are:
v∂y ∂y = − dx ∂t ∂2 y = ∂t2
and
...(1)
v2
∂2 y dx2
...(2)
where v is the constant wave-velocity. Let us examine the given equations: (i)
y = 4 sin x cos vt
...(i)
differentiating it twice partially with respect to t and x separately, we get
∂2 y = 4 sin x (–v2cos vt) = –v2y ∂t2 ∂2 y = ∂x2
and ∴
–4 sin x cos vt = –y
∂2 y ∂2 y = v2 2 2 ∂t ∂x
This is same as Eq. (2) above. Hence the given eq. (i) is a solution of wave equation. (ii)
y = 9 sin 2x cos vt
...(ii)
∂2 y = 9 sin 2 x ( − v2 cos vt ) = − v2 y ∂t2 ∂2 y = −36 sin 2 x cos vt = − 4 y ∂x2
and ∴
v2 ∂2 y ∂2 y . = 2 4 ∂x2 ∂t
This is different from Eq. (2). Therefore eq. (ii) is not a solution of wave equation y = x2 – v2t2
(iii)
...(iii)
∂ y = –2v2 ∂t2 ∂2 y = 2 ∂x2 2
∂2 y ∂2 y = −v2 2 2 ∂t ∂x Clearly, Eq. (iii) also is not a solution of wave equation (iv)
y = 2x – 5t
This equation can be differentiated only once ∂y = –5 ∂t
...(iv)
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Mechanics
∂y = 2 ∂x ∂y −5 ∂y = ∂t 2 ∂x
∴
5 . Hence the given Eq. (iv) is a solution of wave motion. 2
This is similar to Eq. (i) if v =
Q.11. What is meant by a propagation vector in a general plane harmonic wave? If the propagation constant of a wave is 132 per cm and its velocity is 300 m/sec. Which must be its wave number, wavelength and frequency? Solution. The equation of a plane harmonic wave moving along +x-axis, in terms of the propagation constant k is y = a sin (ωt − kx) A physical displacement y may not be associated with all waves. The common feature of all waves is all transmission of some sort of disturbance with a velocity characteristic of the medium. For example, y may correspond to pressure – change in sound waves or electric (or magnetic) field in electromagnetic waves. The disturbance may be represented by a general symbol ψ, so that we may write y = a sin (ωt − kx) The equation for a three-dimensional wave should be ψ = a sin ( ωt − kx x − ky y − kz z) → →
ψ = a sin (ωt − k . r )
or →
where the vector k ( = kxi + ky j + kz k) is called the ‘propagation vector.’ The direction of this vector is the direction of wave motion and its magnitude is 2π/λ, the propagation constant k. Now, propagation constant k = ∴
→
Wave number V =
Wave length λ = Frequency, n =
2π = 132 per cm. λ k 1 = = λ 2π
132 = 21 per cm. 22 2× 7
FG IJ H K
1 1 = = 0. 0476 cm. v 21 → v cm 21 = v V = 3000 = 6. 3 × 105 / sec. × λ sec cm
Q.12. Write down the equation of simple harmonic progressive waves having following characteristics: (i) amplitude 0.6 cm, period 0.5 sec, wavelength 36 cm, travelling along positive x-axis (ii) amplitude 0.003 cm, frequency 660 Hertz, velocity 330 m/sec travelling along negative x-axis. Assume the displacement y = 0 at x = 0 at t = 0.
Stationary Waves (Waves in a Linear Bounded Medium)
605
Solution. (i) The equation of a simple harmonic progressive wave along + x-axis, in terms of amplitude a, period T and wavelength λ is when y = 0 at x = 0 at t = 0 y = a sin 2π Putting
FG t − x IJ H T λK
a = 0.6 cm, T = 0.5 sec and λ = 36 cm, we get
FG H
y = 0. 6 sin 2π 2t −
IJ K
x cm. 36
(ii) Again, the equation of a simple harmonic wave travelling along x-axis, in terms of frequency n and velocity v is
FG H
y = a sinω t +
FG H
x v
IJ K
y = a sin 2πn t +
x v
IJ K
Putting a = 0.003 cm = 3 × 10–5 m, n = 660 sec–1 and v = 330 m/sec we get:
FG H
y = 3 × 10−5 sin 2π 660 t +
660 x 330
IJ K
−5 y = 3 × 10 sin 2π ( 660t + 2 x) meter. Ans.
or
Q.13. Two particles of a medium disturbed by the wave propagation are at x1 = 0 and x2 = 1 cm. The displacement of the particles can be given by the equations: y 1 = 2 sin 3πt y 2 = 2 sin
FG 3πt − π IJ H 8K
Determine the frequency, wavelength, wave-velocity and the displacement of the particle at t = 1 sec and x = 4 cm. Solution. The particle motion is simple harmonic (y = a sin ωt). This shows that a = 2 cm and ω = 3π. The frequency is ω 3π = = 1. 5 Hz 2π 2 π We see that two particles at a distance 1 cm apart have a phase difference of π/8. Therefore, the wavelength, i.e., the distance between two particles having a phase difference of 2π is
n =
λ =
8 × 2π = 16 cm π
The wave velocity is v = ηλ = 1.5 sec–1 × 16 cm = 24 cm/sec
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Mechanics
The wave equation y = a sin becomes At
2π (vt − x) λ
2π (24 t − x) cm 16 t = 1 sec and x = 4 cm, we have
y = 2 sin
y = 2 sin 2.5 π cm
FG H
= 2 sin 2π +
IJ K
π cm 2
π = 2 cm 2 Q.14. The equation of a wave is given by = 2 sin
y = 10 sin 2 π
FG t − x IJ , H 0.04 200 K
where y and x are in cm and t in second. Find the amplitude, wavelength, velocity and the frequency of the wave. Find the velocity of a point x = 400 cm at t = 0.08 sec. Solution. Let us compare the given equation with the standard form of the wave equation y = a sin2π
FG t − x IJ H T λK
we find amplitude, a = 10 cm, wave length, λ = 200 cm frequency, n = and velocity
1 1 = = 25 sec−1 T 0. 04
v = nλ = 25 × 200 = 5000 cm/sec
The particle velocity at a point x at time t is
FG H
IJ K
FG H
IJ K
FG H
t x 2π ∂y − cos 2π = 10 0. 04 0. 04 200 ∂t Putting x = 400 cm and t = 0.08 sec, we get
IJ K
2π ∂y cos 0 = 10 0. 04 ∂t = 500 π = 500 × 3.14 = 1570 cm/sec.
Q.15. The equation of a transverse wave travelling along a rope is given by y = 5 sin π (0.02x – 4.0 t) cm. Find the amplitude, frequency, velocity and wavelength. Calculate the maximum transverse velocity of any particle in the rope.
Stationary Waves (Waves in a Linear Bounded Medium)
607
y = 5 sin π (0.02 x – 4.0 t) cm
Solution.
FG H
= 5 sin 0. 02 π x − = 5 sin
IJ K
4. 0t cm 0. 02
2π ( x − 200t) cm 100
2π (200 t − x − π) cm 100 Comparing it with the standard wave equation = 5 sin
y = a sin
2π (vt − x) , λ
we get: amplitude, a = 5 cm, wavelength, λ = 100 cm velocity, v = 200 cm/sec, frequency, n=
v 200 = = 2 Hz. λ 100 The particle velocity is obtained by partially differentiating Eq. (i) that is
2πva 2π ∂y cos (vt − x) = λ λ ∂t The maximum value of the cos term is 1.
FG ∂y IJ H ∂t K
=
2πva λ
=
2 × 3.14 × 200 × 5 = 62. 8 cm/sec. 100
max
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Mechanics
14 DAMPED AND FORCED HARMONIC OSCILLATION 14.1 DAMPING FORCE The frictional force, acting on a body opposite to the direction of its motion, is called damping force. Such a force reduces the velocity and the kinetic energy of the moving body and so it is also called a retarding or dissipative force. These forces usually arise due to the viscosity or friction of the medium and are non-conservative in nature. When the velocities are not sufficiently high, the damping force is found to be proportional to the velocity (v) of the particle and may be expressed as F = –γv = –γ (dx/dt)
...(1)
Where γ is a positive constant, called the damping coefficient.
Y
If it is the only external force acting on the moving particle, then according to Newton’s second law, we have the equation of motion as
m or
1 .0
dv = – γv dt
dv γ + v = 0 dt m
–γv ...(2)
It is sometimes useful to define a constant τ = m/γ, called the relaxation time, so that the equation of motion becomes dv v + = 0 dt τ
e – t/τ
v/v 0 0 .5
τ
O t
Fig. 1
...(3)
This is a very important differential equation and may be written as
z v
or
v0
dv − dt = v τ
z t
−1 dv dt = τ v 0
608
...(4)
Damped and Forced Harmonic Oscillation
609
where v0 is the velocity at t = 0. On integrating Equation (4), we have logev – loge v0 = − log e
or So that
t τ
v t − v0 = τ
v = v0 e–t/τ
... (5)
Therefore, the velocity decreases exponentially with time and we say that the velocity has been damped with time constant τ. When t = τ, v = v0/e. Hence the relaxation time or time constant may be defined as the time in which the velocity becomes 1/e times the initial velocity. Sometimes, 1/τ or γ/m is put equal to 2k, where k is called damping constant. The instantaneous kinetic energy of the particle is t
K =
t
−2 −2 1 1 mv2 = mv02e τ = k0 e τ 2 2
...(6)
Thus, the kinetic energy also decreases exponentially with a relaxation time τ/2. Putting
v=
dx in Eq. (5) and then dt
z t
x =
−
t
v0e τ dt, where at t = 0, x = 0
0
∴
x =
L v τ M− e MN
−
0
t τ
OP PQ
t
F GH
= v0τ 1 − e 0
−
t τ
I JK
...(7)
The example, for representing eq. (1) is furnished by an ohmic resistance. If I is the current at any instant, then voltage drop across R is IR and the induced E.M.F. across inductance L is –LdI/dt. There is no external E.M.F., so that L
dI dI + RI = 0 RI = −L or L dt dt Which is exactly of the form (2) or (3) with x = L/R. The current will decrease in an undriven LR circuit as
I R
Fig. 2
I = I0 e–(R/L)t
14.2 DAMPED HARMONIC OSCILLATOR When the effect of frictional force on the oscillator is not considered then a pendulum or a tuning fork, if vibrated once, will oscillate indefinitely with a constant amplitude i.e., without any loss of energy. In actual practice, the amplitude of oscillation gradually decreases to zero as a result of frictional forces, arising due to the viscosity of the medium in which the oscillator is moving. The motion of the oscillator is said to be damped by friction and the vibrating system is called damped harmonic oscillator. If damping is to be taken into account, then a harmonic oscillator experiences.
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Mechanics
(i) a restoring force proportional to displacement (–x) and (ii) a damping force proportional to the velocity
FG dxIJ H dt K
but opposite to it
FG − γ dxIJ H dt K
dx is its velocity at this displacement. dt
where x is the displacement of the oscillating system and
Hence the equation of motion of the damped harmonic oscillator is m
dx d2 x −γ − Cx 2 = dt dt
d2 x dx +γ + Cx = 0 2 dt dt
or
m
or
d2 x dx + 2k + ω 20 x = 0 2 dt dt
...(1)
where 2k = γ/m = 1/τ and ω 0 = C/m is the natural angular frequency in absence of damping forces. This τ is the relaxation time and k is called damping constant. Equation (1) is the differential equation of damped harmonic oscillator and is applicable to any system for which the equation of motion have this form. Now, we have to solve equation (1) of damped harmonic motion. Let its possible solution be x = Aeat
...(2) 2
2
Substituting the values of x, dx/dt and d x/dt in Equation (1), we have
eα
2
j
eα
+ 2k α + ω20 eat = 0 or
∴
α = −k ±
2
j
+ 2k α + ω20 = 0
k2 − ω20
Hence the solutions of the equation (1) are
F − k+
x = A1eH
k 2 − ω 20
IK t
F − k−
x = A 2eH
and
IK
k2 − ω 20 t
Thus the most general solution of equation (1) is
F − k+
x = A1eH
F GH
k2 − ω 20
F
− kt A1eH x = e
i.e.
IK t
F −k−
+ A2eH
IK
k2 − ω 20 t
−
+ A2e
FH
k2 − ω 20
IK t IK
k2 − ω 20 t
I JK
...(3)
where A1 and A2 are constants, depending upon the initial conditions of motion. The quantity
k2 − ω20 is imaginary, real or zero, it depends on the relative values of
k and ω0. If k < ω0, then
k2 − ω20 is imaginary and it is called underdamped case. If k > ω0 or k = ω0, the situations are known as overdamped and critically damped respectively.
Damped and Forced Harmonic Oscillation
611
(i) Underdamped Case: If the damping is so low that If k < ω0, then =
e
j
− − ω 20 − k2 = i ω 20 − k2 = iω where i =
k2 − ω20
−1 and ω = ω20 − k2 , which is a real
quantity. Now, from Equation (3), we have
e
− kt A1eiωt + A2e− iωt x = e
j
x = e− kt A1 (cos ωt + i sin ωt ) + A 2 (cos ωt − isin ωt ) x = e− kt (A 1 + A 2 ) cos ωt + i( A1 − A 2 )sin ωt As x is a real quantity, (A1 + A2) and i(A1 – A2) must be real quantities. Clearly, A1 and A2 are complex quantities. If A1 + A2 = A0 sin φ and i (A1 – A2) = A0 cos φ , then x = A 0 e− kt sin(ωt + φ)
...(4)
This equation represents a damped harmonic motion. This motion is oscillatory or ballistic whose periodic time is given by T =
2π = ω
2π
...(5)
ω20 − k2
Thus, the effect of damping is to increase the periodic time. If k = 0, i.e.; no damping 2π then T = ω . But in actual cases, the effect of damping on periodic time is usually 0 negligible except few extreme cases.
The amplitude of the oscillatory motion is given by t − τ 2
A = A 0 e− kt = A 0 e Y
A = A 0 e – t/2 τ
D isp la cem en t
X Tim e
A = – A 0 e – t/2 τ
Fig. 3
612
Mechanics
Where A0 is the amplitude in the absence of damping. In the presence of damping, the amplitude decreases exponentially with time. In figure, time displacement curve is shown for damped harmonic motion. As the maximum value of sin (ωt + φ) is alternately +1 and –1, obviously the time displacement curve of the vibrating body lies entirely between the curves A = A0e–kt and A = –A0e–kt shown by the dotted lines.
14.3
LOGARITHMIC DECREMENT
The time interval between the successive maximum displacements (i.e., amplitudes) of left and right hand sides is T/2, hence if An and An+1 are the successive amplitudes, then
An = A n+1
∴
and An+1 = A 0 e
e− kt e
FG H
−k t +
FG H
−k t+
An = A0e–kt
T 2
IJ K
=e
k
T 2
T 2
IJ K
= d, a constant for motion
...(6)
Where d is called the decrement, indicating for the reduction in amplitude. Now
loge d =
kT =λ 2
...(7)
This quantity λ is called the logarithmic decrement and is equal to the ratio of the natural logarithmic of the ratio of two successive amplitudes of vibration. (ii) Overdamped Case: If the damping is so high k > ω0, then
k2 − ω20 , say β, is a real
quantity and then from Equation (3), We have
b g
− k+ β t
x = A1 e− ( k −β)t + A2 e
...(8)
As k > β, both quantities of right hand side decrease exponentially with time and the motion is non-oscillatory. Such a motion is called dead beat or aperiodic and its main application is in dead beat galvanometers. (iii) Critically Damped Case: If k = ω0, then from Equation (3), we have x =
bA
1
g
+ A 2 e− kt = Ce− kt , where C = A1 + A2
In this equation, there is only one constant, hence it does not provide us the solution of differential equation (1) of second order. Now, suppose
k2 − ω20 = h, which is a small quantity. Hence from Equation (3), we
have − kt A1eht + A2e− ht x = e
x = e− kt A1 (1 + ht + ...) + B(1 − ht + ...) Neglecting the small terms, containing h2 and higher powers of h, we get
b
g b
g
x = e− kt A1 + A 2 + h A1 − A2 t
Damped and Forced Harmonic Oscillation
613
x = e− kt ( P + Qt )
or
...(9)
Where P = A1 + A2 and Q = h(A1 – A2)
dx = e− kt Q + e− kt ( − k) (P + Qt) dt
Also,
If initially at t = 0 the displacement of the particle is x = x0 and there the velocity is v0, then x0 = P and v0 = Q – kP = Q – kx0 or Q = v0 + kx0 ∴
x =
x0 + ( v0 + kx0 ) t e− kt
...(10)
This equation represents that initially the displacement increases due to the factor x0 + ( v0 + kx0 ) t . But as time elapses, the exponential term becomes relatively more important and the displacement returns continuously from the maximum value to zero and the oscillatory motion does not just occur. Such a motion is called critically damped or just aperiodic. H e avy D am p in g C ritical D a m pin g
D isplacem e n t (x)
L igh t D am p ing
Tim e (t)
Fig. 4
14.4 POWER DISSIPATION IN DAMPED HARMONIC OSCILLATOR If a particle oscillates in a medium, then due to the viscosity of the medium damping forces act on the particle in a direction opposite to its movement. In this process work is done by the particle in overcoming the resistance forces. Consequently, the mechanical energy of the vibrating particle continuously decreases so that the amplitude of oscillation becomes less and less. Here, we want to find a relation for power dissipation (i.e., rate of dissipation of energy). At any instant t, the displacement of a damped harmonic oscillator is given by x = A 0 e− kt sin(ωt + φ) ∴ Velocity of the particle
dx = A 0 e− kt − k sin (ωt + φ) + ω cos(ωt + φ) dt
614
Mechanics
∴ Kinetic energy of vibration K=
1 mA20 e−2kt k2 sin2 (ωt + φ) + ω2 cos2 (ωt + φ) − 2k ω sin(ωt + φ) cos (ωt + φ) 2
Potential energy
1 2 1 2 2 U = 2 Cx = 2 m ω 0 x , U =
U =
z x
Cxdx
0
o
t
1 mA20 e−2kt ω20 sin2 (ωt + φ) 2
The average total energy for a period will be the sum of time average of kinetic and potential energy. If the amplitude of oscillation does not change much in one cycle of motion, then the factor e–2kt may be taken as constant. Now, we are left with the average of sin2 (ωt + φ), cos2 (ωt + φ) and 2 sin (ωt + φ) cos (ωt + φ), whose average values for a period are ½ , ½ and 0 respectively. Hence the average kinetic energy =
FG H
IJ K
1 1 1 1 mA 20 e−2kt k2 + ω2 . = mA 20 ω 2e−2kt 2 2 2 4
for damping is low so k2 << ω02 and average potential energy = ω ≈ ω0
1 mA20 e−2kt ω2 for low damping 2 t
∴
Average total energy E =
∴
Average power dissipation P = −
− 1 mA 20 e−2kt ω2 = E0 e−2kt or E0 e τ 2
dE E = mA20ω2ke−2kt = 2 kE or dt τ
This dissipation of energy is caused by the damping force γ
dx . dt
Hence, if we calculate the negative value of the average rate of doing work by this
FG dxIJ >, then we will get exactly the same expression for power dissipation, as H dt K 2
force, i.e., < γ
obtained above. This loss of energy generally appears in the form of heat in oscillating system.
14.5 QUALITY FACTOR Q To represent the efficiency of an oscillating system in a term known as quality factors Q, is widely used. The quality factor Q of an oscillating system is a measure of damping, or the rate of energy decay, of the system. It is defined as 2π times the ratio of the energy stored in the system to the energy lost per period. That is Q = 2π
Energy stored Energy loss per period
Q = 2π
E Eω 2π = , = ω T PT P
Damped and Forced Harmonic Oscillation
∴
615
Q = ωτ, P = E/τ
Evidently Q is a dimensionless quantity. In the case of low damping Q = ω0τ But
and
∴
ω0 =
C m
τ =
m , γ
mc γ
Q =
Thus high value of Q means that the damping of the oscillating system is low. This Q is a measure of the extent, to which oscillator is free from damping. For an undamped oscillator, γ = 0, so that Q is infinite. The energy of the oscillator (E = E0e
−
t τ)
decreases to e–1 of its initial value in the time
Q τ = oscillations. 2π 2π The Q of an excited atom is about 108, while for a violin string it is only about 103 due to a much larger damping in the string. τ and in this time the oscillator performs ω 0
14.6 FORCED (DRIVEN) HARMONIC OSCILLATOR If an external periodic force is applied on a damped harmonic oscillator, then the oscillating system is called driven or forced harmonic oscillator and its oscillations are called forced (or driven) vibrations. Let us consider system oscillating about an equilibrium position under and external periodic force. Let x be its displacement from the equilibrium position at an instant during the oscillation. Its instantaneous velocity is dx/dt. The forces acting upon the system at the instant are: (i) A restoring force proportional to the displacement but acting in the opposite direction, this may be written as –Cx, where C is the force constant. (ii) A frictional force proportional to the velocity but acting in the opposite direction. This may be written as −γ
dx dt
(iii) An external periodic force represented by F0 sin pt, where F0 is the maximum value of this force and p is its angular frequency. Thus the total force F acting upon system is F = − Cx − γ
dx + F0 sin pt dt
616
Mechanics
By Newton’s second law this must be equal to the product of the mass ‘m’ of the system and the instantaneous acceleration
m
d2 x . That is dt2
dx d2 x − Cx − γ + F0 sin pt 2 = dt dt
d2 x γ dx C F x = 0 sin pt + + 2 m dt m m dt
or
d2 x dx + 2k + ω20 x = f0 sin pt dt dt2
...(i)
C = 2πn0 = the natural angular frequency of the system in the m absence of damping and driven forces, and
where 2k = γ/m, ω0 =
f0 =
F0 m
The natural frequency of the oscillating system may be different to the frequency (p/2π) of the applied force. When a periodic force acts on a body, it delivers periodic impulses to the body so that the loss of energy in doing work against the dissipative forces is recovered. The result of this is that the body is thrown into continuous vibrations. In the initial stages the body tends to vibrate with its natural frequency whilst the impressed force tries to impose its own frequency upon it. But soon the free vibrations of the body die out and ultimately the body vibrates with a constant amplitude and with the same frequency as that of the impressed force. Such vibrations of constant amplitude, performed by a body under the influence of a impressed periodic force, with a frequency equal to that of the force, are known as forced vibrations and the oscillating system is itself called driven or forced harmonic oscillator. The impressed periodic force is called the driver and the body, executing forced oscillations, is called the driven. Now let us suppose that steady state solution of Equation (i) is x = A sin (pt – θ)
...(ii)
Because the amplitude (A) of the forced oscillations remains constant and the frequency of vibrations is equal to the (p/2π) of the force. Hence θ represents the phase difference between the force and the resultant displacement of the system. Now from Equation (ii), we have
dx = pA cos ( pt − θ) dt and
d2 x = –Ap2 sin (pt – θ) dt2
Substituting these values in Equation (i), we get
− p2 A sin( pt − θ) + 2kp A cos ( pt − θ) + ω20 A sin( pt − θ) = f0 sin ( pt − θ + θ)
Damped and Forced Harmonic Oscillation
617
A( ω 20 − p2 ) sin( pt − θ) + 2kp A cos ( pt − θ) = f0 cos θ sin ( pt − θ) + f0 sin θ cos ( pt − θ)
If this equation is to satisfied for all values of t, then the coefficients of sin (pt – θ) and cos (pt – θ) on the two sides must be equal. Equating them, we obtain.
e
A ω20 − p2 and
j
= f0 cos θ
...(iii)
2kpA = f0 sin θ
...(iv)
Squaring Eqs. (iii) and (iv) and on adding, we get ∴
A =
e
f0
...(v)
j
2
ω20 − p2 + 4 k2 p2
Dividing Equation (iii) by Equation (iv), we get tan θ =
2 kp 2kp or θ = tan−1 2 2 − p ω 0 − p2
...(v a)
ω20
Substituting these values in Equation (ii), we get x =
eω
f0 2 0
j
2
− p2 + 4 k2 p2
F GH
sin pt − tan−1
2kp − p2
ω20
I ...(vi) JK
Here, we will assume that the damping is low. Now we will consider the different cases, when ω0 >> p, ω0 = p and ω0<< p. For low driving frequency, i.e., when ω0 >> p, tan θ → 0 or θ → 0, that is, the driving force and the displacement are in the same phase. Hence the amplitude of the oscillation from Equation (v) is given by A =
F / m F0 f0 = 0 = 2 C/m C ω0
Thus, in this case the response (amplitude) does not depend on the mass of the oscillating system or damping, but it depends only on the force constant C. For very high driving frequency i.e., when ω0<< p, if the driving frequency is increased, the value of the amplitude A increases. F0 f0 A ≈ 2 ≈ p mp2
Which shows that the amplitude, which now depends upon the mass, continuously decreases as the driving frequency p is further increased.
14.7 AMPLITUDE RESONANCE Equation (vi) show that the amplitude of the forced oscillations depends upon ( ω20 − p2 ) which in turn depends upon the difference between the driving frequency p and the natural frequency ω0 of the oscillator. Smaller this difference, larger the amplitude.
618
Mechanics
To find the condition for maximum amplitude (Resonance) Equation (v) can be arranged as follows: A =
e
f0 ω20 − 2k2 − p2
j
2
...(vi)
+ 4 k2ω20 − 4 k4
Therefore, to have the maximum value of A, ω 20 − 2 k2 − p2 = 0
or
p =
ω20 − 2k2 = pr (say) f0
∴
Amax =
2k ω20 − k2
...(vii)
Hence for a certain value of driven frequency (pr) the amplitude of the oscillating system is maximum. This phenomenon in which the amplitude of the driven oscillator becomes maximum at a particular driven frequency, is called amplitude resonance and this frequency is known as the resonant frequency. For a forced harmonic oscillator, the resonant frequency is p/2π, where pr is the angular resonant frequency of that oscillator. The value of this angular frequency is less than ω0 or
FH
ω =
IK
ω20 − k2 , which is the frequency of the natural damped oscillations of the body. If the
damping is low, then ω0 = pr and the amplitude. f0 f0τ Amax = 2 ω or ω k 0 0
...(viii)
Hence in the case of zero damping the amplitude should be infinity. But it is not possible because the friction on the oscillating system is never zero. If the damping is low, then the ratio of the response at resonance to the response at zero driven frequency is given by =
f / 2 kω 0 ω 0 = = ω 0τ = Q 2k f / ω 20
Hence at resonance, this ratio is equal to the quality factor Q of the system. Thus, we see that the damping controls the response of resonance. When p = ω0 from Equation (va) and (v), we have tan θ = ∞ or θ = π/2 And A =
f0 fτ = 0 , which is less than the amplitude as given by Equation (viii). ω0 2k ω 0
In case of low damping, the resonant frequency is equal to the natural frequency of the oscillator and then the amplitude is maximum. For high driving frequency or when ω0 << p, tan θ = –2k/p tan θ → –θ or θ → p and amplitude A =
f0 nearly. p2
Damped and Forced Harmonic Oscillation
619
So on increasing the driving frequency (p), the response or amplitude decreases. Y L ow D am ping Ze ro D am p in g
A m plitu de
H ig h D am p in g
O
ω0
X
P
Fig. 5
The variation of the amplitude A with the frequency p of the periodic force is shown in figure. Initially, when the angular frequency p of the force is increased the amplitude A continuously increases and at a certain value (which is nearly equal to ω0 for low damping) of p, the value of the amplitude becomes maximum. This is the condition of amplitude resonance. Now, on further increasing p, the response decreases gradually. The x-coordinates corresponding to the peaks of these curves, as shown in figure, represent the resonant frequencies. When the damping is greater, the resonant frequency (pr) is less than ω0 but in case of low damping the resonant frequency of the force is nearly equal to the natural frequency (ω0) of the oscillator. It is also clear from these curves that for low damping the height of the peaks become greater and when the damping is zero (in the ideal case) the height of the peaks (i.e., the maximum amplitude) rises to infinity.
14.8 SHARPNESS OF RESONANCE At resonance, the amplitude of the oscillating system becomes maximum. It decreases from this maximum value with the change (decrease or increase) of the frequency of the impressed force. In above figure, for different values of damping curves have been drawn between the driven frequency and the amplitude. The term sharpness of resonance refers to the rate of fall in amplitude with the change of forcing frequency on each side of the resonant frequency. When the damping is low, the response (amplitude) falls of very rapidly on other side of resonant frequency and we say that the resonance is sharp. On the other hand, for high damping the response falls off very slowly on either side of resonant frequency and the resonance is said to be flat. Thus, the resonance is flat or sharp according to the damping for the oscillating system is large or small. Familar examples of a flat or sharp resonance are the resonance of an air column and a sonometer wire with a tuning fork respectively. Due to its large damping the air column responds with tuning fork over fairly wide range in the neighborhood of resonance. Thus in this case it is actually difficult to get an exact point of resonance and resonance is said to be flat. But in case of sonometer wire, the damping is small so that the resonance is sharp and hence to have the resonant points, the length of the wire is adjusted very precisely.
620
Mechanics
Half width of resonance Curve: If ph is the value of angular frequency when the amplitude falls to half the value at resonance, then the change in p is called the half width of the resonance curve i.e., Half width ∆p = |ph − pr| At resonance,
Amax =
f0 2k
ω20
2
−k
f0 2 2 4 k ω0
=
...(1)
− 4 k4
At the frequency ph the amplitude is Amax/2
A max = 2
∴
e
f0 ω20
− 2k − 2
ph2
j
2
...(2)
+
4 k2ω20
− 4k
4
From Eqs. (1) and (2), we have
eω
2 0
− 2 k2 − p2h
j
2
+ 4 k2ω20 − 4 k4 = 4 ( 4 k2ω 20 − 4 k4 ) p2r = ω 20 − 2k2
but
e
j e
2 2 4 2 2 4 ( p2r − ph2 )2 = 3 4 k ω 0 − 4 k = 3 4 k pr + 4 k
j
If damping is small, 4k2 is negligible and then we have ( p2r − p2h ) = ± 3 (2kpr ) p2h = p2r m
or
or
ph = ∴
3 (2kpr )
F p G1 m H
2k 3 pr
r
Half width ∆p = |ph − pr| =
IJ K
1 2
F GH
= pr 1 m
I JK
3k = pr m pr
3k
3k
14.9 VELOCITY RESONANCE The velocity of the driven oscillator is v =
dx = dt
e
f0 p ω20
− p
2
j
2
cos( pt − θ) + 4k p
2 2
= v0 sin( pt − θ + π / 2) where v0 =
e
f0 p ω20 − p2
j
2
= + 4 k2 p2
f0
FG p − ω IJ + 4 k H pK F 2kp I = tan G H ω − p JK 2
2
0
θ
−1
2 0
2
is the velocity amplitude and
Damped and Forced Harmonic Oscillation
621
We see that the velocity amplitude v0 varies with the driven frequency p. When p = 0, C m of the impressed force, the velocity has the maximum value and we call it velocity resonance.
v0 = 0 and when p = ω0, v0 attains the maximum value. Hence at the frequency p = ω 0 = At velocity resonance −1 θ = tan
∴
Velocity phase constant = − θ +
FG 2kp IJ = π H0K 2
π π π =− + =0 2 2 2
Y
L ow d am p in g v H ig h da m pin g X
ω0
O p
Fig. 6
This means that the velocity of the oscillator is in phase with the applied force. This is the most favourable situation for transfer of energy from the applied force to the oscillator, because the rate of work done on the oscillator by the impressed force is Fv, which is always positive for F and v in phase. When p < ω0, the velocity amplitude is smaller than at p = ω0.
14.10 POWER ABSORPTION When an oscillator executes vibration in presence of damping forces, it loses energy in doing work against these forces. This loss of energy is supplied by the periodic impressed force so as to continue the oscillations. In a small time interval dt the energy supplied by the driving force F(= F0 sin pt) will be equal to the work done on the oscillating system by this force, i.e., dE = F.dx = F (dx/dt) dt dE dx = F dt dt F = mf0 sin pt
or ∴
dx = dt
and
pf0 cos( pt − θ)
eω
2 0
− p2
j
2
+ 4 k2 p2
Hence at any time the energy absorbed by the oscillating system is
p=
FG H
dE dx = F dt dt
IJ K
=
e
mf0 sin pt ω20
− p
2
j
2
pf0 [cos ( pt − θ)]
+ 4k p
2 2
622
Mechanics
Hence the average power absorbed is
FG H
Pav = F
dx dt
IJ K
= av
e
mf02 p ω20
2
− p
j
2
[sin pt cos ( pt − θ)]av 2 2
+ 4k p
Now the average of sin pt cos (pt – θ) for one period T or 2π/ω
z z T
1 = sin pt cos ( pt − θ) dt T 0
T
=
1 1 1 {sin(2 pt − θ) − sin θ } dt = sin θ 2 T 2 0
=
Hence
Replacing
Pav =
eω
f02 p2 2 0
− p2
j
2
+ 4 k2 p2
1 2kp 2 (ω2 − p2 )2 + 4 k2 p2 0 1 2kp2 mf02 2 2 ω20 − p2 + 4 k2 p2
e
j
...(1)
by v02, we have
mv02 γv02 or 2τ 2 from equation (1), we see that when p = ω0, the absorbed average power is maximum and Pav = mkv02 =
1 1 mf02 2 or mf0 τ . Thus, the power absorbed is maximum at the frequency of 2 4 k velocity resonance and net at the frequency of amplitude resonance.
is equal to
When steady state is attained the average power dissipated should be the same as the average power supplied by the driving force. From equation (1), we see that the power drops to half its maximum value in the condition Y 1 .0
Pa v 0 .5 Fu ll w id th
ω0 p
Fig. 6
X
Damped and Forced Harmonic Oscillation
623
1 (Pav )Res = (Pav)half 2
power frequency
1 mf02 1 2kp2 = mf02 2 2 4k 2 ω20 − p2 + 4 k2 p2
e
8k2p2 =
or
eω
2 0
j
− p2
j
2
+ 4 k2 p2
ω20 − p2 = +2kp
or
2 pk 2k =± ≈ ±k ω ω0 + p 1+ 0 p
or
ω0 − p = ±
or
ω0 1 taking = 1 approx. p 2τ ∴ Half-width of the power verses frequency curve ∆p = Hence
ω 0 − p = k or
Q = ω0 τ =
1 2τ
ω0 Frequency at resonance = 2∆p Full width at half maximum power
Near resonance, when p = ω0, we see that Q = ω0τ, for low damping Q will be high. Hence Q measures the sharpness of resonance.
14.11 DRIVEN LCR CIRCUIT LCR circuit is an example of the driven harmonic oscillator in which the oscillations are sustained by an alternating electromotive force. R
L
C
EM F
Fig. 7
Let us consider a circuit containing the inductance L, resistance R and condensor C in series. An alternating E.M.F., E = E0 sin pt is applied to this circuit. If at any instant t, Q is the charge on the condenser C, then the potential difference across its plates is Q/C. If the current in the circuit at this instant is I, then the self-induced
dI . According to ohm’s law, the potential difference between dt the ends of the resistance is RI. Hence total Potential difference
E.M.F. in the inductance is −L
= RI +
dI Q = total E.M.F. = E − L dt C
624
Mechanics
i.e.,
L
dI Q + RI + = E dt C
Q d2Q R dQ E . + + = 0 sin pt 2 L dt LC L dt
or
∴ I = dQ/dt
...(1)
This equation is similar to the equation of driven harmonic oscillator only with the difference that here we have R/L in place of 2k, 1/LC for ω02 and E0/L for f0. Hence the steady state solution of the above equation is Q =
E0 / L
FG 1 − p IJ + FG pR IJ H LC K H L K 2
2
sin( pt − θ)
...(2)
2
θ = tan−1
where
pR / L 1 − p2 LC
...(3)
Differentiating Equation (2) with respect to t, we can find the current I = dQ/dt in the circuit. Practically, we require the knowledge of the current in the circuit. ∴
I =
or
I =
FG H
E0 p / L
FG pL − H
φ = θ−
where
IJ K
1 − pL LC E0
1 pC
+ R2
IJ K
π = tan−1 2
tan φ = − cot φ =
cos( pt − θ)
2
sin( pt − θ)
2
...(4)
+ R2
pL −
1 pc
R
...(5)
1 1 pL − pC LC = pR R L
p2 −
which represents that the current I lags in phase φ from the E.M.F. E. The quantity
FG H
R2 + pL −
1 pc
IJ K
2
is called the impedance of the circuit and is represented
by z. Its unit is ohm. Hence I = or
E0 sin ( pt − θ) Z
I = I0 sin (pt – θ)
...(6)
Damped and Forced Harmonic Oscillation
where I0 =
625
E0 is the peak value of the current. Z
The quantity (pL – 1/pC) is called the reactance of the circuit and is denoted by X. Therefore,
R2 + X2 =
Impedance Z = Now, the current amplitude I0 =
The expression for Z = R
2
E0 = Z
E0 2
R + X2
F 1 I + G pL − H pc JK
(Resistance)2 + ( Reactance)2 is smaller for the larger value of Z.
2
depends upon the frequency (p) of the applied
alternating E.M.F. We will consider the cases, when the driven frequency is increased. Case I: p << ω0 the term –1/pc in impedance is dominant and is very large and Z =
R
2
F 1 I + G pL − H pc JK
2
≈
1 pc
Therefore, the current amplitudes is (when p → 0) I0 =
E0 1 / pc
π 2 Thus, for low values of p, the current amplitude has a small value. On increasing the 1 driven frequency (p) gradually, the magnitude of reactance. pL − and hence the pc impedance Z decrease resulting in the increase of the current amplitude. φ =
and
FG H
Case II: When p = ω 0 =
IJ K
1 , the reactance LC X = pL −
1 = pC
L − LC
LC = C
L − C
L =0 C
and so the impedance Z is minimum and is equal to R. In consequence the current amplitude I0 = E0/Z = E0/R becomes maximum. This is the case of electrical resonance. If R = 0, then the current amplitude at resonance becomes infinite. The resonant frequency of the circuit is given by Pr =
ω p 1 = 0 = 2π 2 π 2π LC
626
Mechanics
In such a case φ = 0 and I = (E0/R) sin pt Thus at resonance the current and the applied E.M.F. are in the same phase. Case III: When p > ω0, the term LC in the expression for Z becomes dominant and very large and then I0 =
E0 pL
φ=
and
π 2
Y
π/2
R=0 H ig h R
L ow R
L ow R H ig h R O
I0
ω0 P
−π/2
R=0
ω0 p
Fig. 8
The variation of I0 with ω0 is shown in figure for different values of R. The smaller value or R, the sharper is the resonance. The variation of phase difference φ between the current and the E.M.F. is shown in figure. The current lags in phase for p < ω0 and exceeds in phase for p > ω0.
NUMERICALS Q.1. The differential equation of an oscillating system is d2 x dx + 2r + ω2x = 0 2 dt dt
If ω >> r, then calculate the time in which (i) amplitude becomes 1/e of its initial value, (ii) energy becomes 1/e of the initial value (iii) energy becomes 1/e4 of its initial value. Solution. The given differential equation with the condition ω>>r, is the equation of motion of an under-damped harmonic oscillator whose solution is x = ae–rt sin (ω′t + φ), where a and φ are arbitrary contants and ω′ = is ae–rt
(ω2 − r2 ). Thus the amplitude of oscillation
(i) The damped amplitude at any instant is ae–rt. Let a0 be the amplitude at t = 0 and a0/e at t. Then we have. a0 = a and
ao = ae− rt = a0e− rt e
Damped and Forced Harmonic Oscillation
627
from this, we have
e−1 = e− rt –1 = –rt t =
1 sec (mean life time) r
(ii) The instantaneous energy of the damped oscillations is given by E = E0e–2rt when the energy falls to E0/e, we have
E0 = E0e–2rt e e–1 = e−2rt –1 = –2rt t =
1 sec . (relaxation time) 2r
(iii) When the energy falls to E0/e4, we have
E0 = E0e–2rt e4 e–4 = e−2rt –4 = –2rt t =
2 sec . r
Q.2 . The equation of motion of a system is given by
d 2 y 2 dy + + ω2 y = 0 τ dt dt 2 Explain the significance of the various terms of the equation and show that τ has the dimensions of time. Solve the equation and discuss the motion of the system when ω2τ2 > 1. Solution. The given equation
d2 y 2 dy + + ω2 y = 0 2 τ dt dt
...(i)
represent the motion of damped harmonic oscillator along the y-axis let us write it as
d2 y −2 dy − ω2 y 2 = dt τ dt The term d2 y /dt2 is the instantaneous acceleration of the oscillator which is the sum of the acceleration due to damping force, ω2 y .
2 dy and the acceleration due to restoring force, τ dt
628
Mechanics
dy 2 dy is acceleration and is velocity, τ has the dimensions of time. Let us write dt τ dt the solution of equation (i) in the form Since
y = Ceαt
...(ii)
dy = Cαeαt dt d2 y = Cα2eαt dt2
Putting these values in equation (i), we get
Cα2eαt +
2 Cαeαt + ω2Ceαt = 0 τ α2 +
2 α + ω2 = 0 τ α =
FG 1 Hτ
−1 ± τ
2
− ω2
IJ K
Hence the general solution of equation (i) is
y =
when ω2τ2 > 1, the term
FG 1 Hτ
2
− ω2
FG 1 Hτ
2
where ω′ =
IJ K
− ω2
LM− 1 + F 1 M τ GH τ C1e N
2
− ω2
IJ OPt K PQ
LM− 1 − F 1 M τ GH τ + C2 eN
2
− ω2
IJ OPt K PQ
...(iii)
is imaginary. Then
IJ K
= j
FG ω H
2
−
1 τ2
IJ = jω ′ K
FG ω − 1 IJ . Now equation (iii) becomes H τK 2
FG − 1 + jω ′ IJ t K τ
y = C1eH
FG − 1 − jω ′ IJ t K τ
+ C2eH
e j nC bcos ω ′ t + j sin ω ′ t g
−t / τ C1e jω ′ t + C2 e− jω ′ t = e
= e− t / τ
1
b
+ C2 cos ω ′ t − j sin ω ′ t = e− t / τ
nbC
1
g
b
g
s
+ C2 cos ω ′ t + j C1 − C2 sin ω ′ t .
gs
Damped and Forced Harmonic Oscillation
629
Let us put C1 + C2 = a sin φ and j (C1 − C2 ) = a cos φ where a and φ are constants. Then we have y = e−t / τ ( a sin φ cos ω ′ t + a cos φ sin ω ′ t ) y = ae−t / τ sin (ω ′ t + φ) This is the solution of the given equation under the condition ω2τ2 > 1. This shows that the amplitude of motion, ae–t/τ decreases with time exponentially, and the period is T =
2π = ω′
FG ω H
2π 2
−
1 τ2
IJ K
The amplitude of motion at t = 0 is a and at t = τ is a/e. That is, τ is the time-interval in which the amplitude decreases to 1/e of its initial value. Thus here τ is the mean life-time of the (damped) oscillator. Q.3. A particle of mass 40 gm experiences only a damping force proportional to its velocity. If its velocity is decreased from 100 cm/sec to 10 cm/sec in 23.0 second, calculate (i) relaxation time, (ii) the damping force when its velocity is 50 cm/sec, (iii) the time in which its kinetic energy is reduced to one-tenth of its initial value, (iv) the total distance travelled if its initial velocity is 100 cm/sec. (Given loge10 = 2.30) Solution. Let v be the velocity of the particle at any instant t. Then the instantaneous damping force is –bv where b is a constant. If there is no other force acting, then by Newton’s law (force = mass × acceleration), we have
m
dv = –bv dt dv −b v = − 2rv, = m dt
where b/m = 2r. The expression may be written as
dv = –2rdt v Integrating
loge v = –2rt + A
(constant)
If at t = 0, we have v = v0, then A = loge v0. Then we have loge v = –2rt + loge v0
log e
v v0 = –2rt v –2rt v0 = e
...(i)
v = v0 e–2rt This is the expression for the (damped) velocity, which decreases exponentially with time.
630
Mechanics
Now, it is given that the velocity decreases from 100 cm/sec. to 10 cm/sec. in 23.0 10 1 v seconds. Hence putting v = 100 = 10 and t = 23.0 sec in equation (i), we have 0
1 = e–2r 10 10 = e2r
(23.0)
(23.0)
2r (23.0) = loge10 = 2.30 2r =
2. 30 1 = 23. 0 10
Equation (i) therefore becomes v = v0 e− t /10
...(ii)
(i) Let τ be the relaxation time (the time-interval in which the velocity reduces to 1/e 1 v of its initial value). Thus substituting v = e and t = τ in equation (ii), we get 0
1 = e−τ /10 e e–1 = e−τ /10
τ = 1 10 τ = 10 sec. (ii) The damping force is Fd = – bv = –2rmv. Here
2r =
1 sec−1 , m = 40 gm, v = 50 cm/sec 10
1 × 40 × 50 = − 200 dyne. 10 (iii) The instantaneous kinetic energy is Fd = −
K =
1 1 mv2 = mv02 e− t / 5 2 2
(by equation (ii))
If K = K0 at t = 0, then K = K 0 e− t /5 Let t′ be the time in which the kinetic energy is reduced to 1/10 of its initial value.
K 1 Putting K = 10 and t = t′ in the last expression, we get 0 1 = e− t ′ /5 10
Damped and Forced Harmonic Oscillation
631
t′ = loge10 = 2.30 5 t′ = 5 × 2.30 = 11.5 sec (iv) Putting v =
dx in equation (ii) and integrating, we get dt x = v0 e− t /10 ( −10) + C (constant)
At t = 0, we have x = 0, so that C = 10 v0. x = −10v0 e− t /10 + 10v0 = 10v0 (1 − e− t /10 )
∴
At t → ∞, e−t /10 → 0, and x → 10v0 Therefore, distance travelled if its initial velocity is 100 cm/sec is x = 10v0 = 10 × 100 = 1000 cm. Q.4. A 3 s gm particle is subjected to an elastic force of 48 dynes/cm and a frictional force of 12 dynes/(cm-sec–1). If it is displaced through 2 cm and then released, find whether the resulting motion is oscillatory or not. If so find its period. Solution. The equation of motion of a particle of mass m subjected to an elastic force and a frictional force is m
d2 x dx +b + Kx = 0 2 dt dt
where K is the elastic force per unit displacement and b the frictional force per unit velocity. Putting
b K = 2r and = ω2 we get m m d2 x dx + 2r + ω2 x = 0 dt dt
This represents an oscillatory motion provided r < ω, and the time period is T =
2π (ω2 − r2 )
Here m = 3 gm, K = 48 dynes/cm and b = 12 dynes/(cm-sec–1) ∴
r = ω =
and
b 12 = =2 2m 2 × 3 K = m
48 = 16 = 4 3
Thus r < ω hence the motion is oscillatory. The time period is T =
2π ω2 − r 2
=
2 × 3.14 2 × 3.14 = = 1. 81 sec 3. 464 (16 − 4)
632
Mechanics
Q.5. A body of mass 0.2 kg is hang from a spring of constant 80 nt/m. The body is subject to a resistive force given by bv, where v is the velocity in m/s. Calculate the value of undamped frequency, and the value of τ if the damped frequency is 3/ 2 of the undamped frequency. Solution. The undamped frequency of a mass m suspended by a spring of forceconstant K is given by n =
1 2π
FG K IJ H mK
1 2π
80 10 = = 3.18 sec−1 π 0. 2
Here m = 0.2 kg and K = 80 nt/m. ∴
n =
The damped frequency n′ (say) is given by n′ = where ω =
1 2π
(ω2 − r2 )
1 2π
FG K − r IJ Hm K
K /m and r = b/2 m.
∴
n′ =
2
But n′ = ( 3 /2) n. ∴
1 2π
FG K − r IJ Hm K 2
=
3 10 × π 2
K − r2 = 300 m Here ∴
K 80 = 400 = m 0. 2 400 – r2 = 300 r2 = 100 r = 10
Therefore, the relaxation time is τ =
1 1 = = 0. 05 sec . 2r 20
Q.6. If the relaxation time of a damped harmonic oscillator is 50 sec. Find the time in which (i) the amplitude falls to 1/e the initial value, (ii) energy of the system falls to 1/e times the initial value, (iii) energy falls to 1/e4 of the initial value. Solution. The equation of a damped harmonic oscillator is x = ae− rt sin(ω ′ t + φ)
Damped and Forced Harmonic Oscillation
633
where r is a damping constant. The relaxation time is
so that
τ =
1 = 50 sec 2r
r =
1 sec−1 100
(i) The damped amplitude at any instant is ae–rt. Let a0 be the amplitude at t = 0 and a0/e at t. Then we can write a0 = a
a0 = ae–rt = a0e–rt e
and From this, we have
e–1 = e–rt t =
1 = 100 sec. r
(ii) The energy (or velocity) falls to 1/e times the initial value in the relaxation time. Thus in this case t = τ = 50 sec. (iii) The energy of a damped harmonic oscillator is given by E = E0e–2rt where E0 is the energy at t = 0. Let t′ be the time during which the energy falls to E0/e4. Then we have
E0 = E0 e−2rt ′ e4 e–4 = e–2rt′ 4 = 2rt′ t′ =
2 = 2 × 100 = 200 sec. r
Q.7. The amplitude of an oscillator of frequency 200 cycles/sec falls to 1/10 of its initial value after 2000 cycles. Calculate (i) its relaxation time, (ii) its quality factor, (iii) time in which its energy falls to 1/10 of its initial value, (iv) damping constant. Solution. The oscillators having a frequency of 200 cycle/sec, completes 2000 cycles in 10 sec. The instantaneous amplitude of the (damped) oscillator is ae–rt where r is a damping constant. If a0 be its initial amplitude (at t = 0) and a0/10 after 10 sec, then we have a0 = a
a0 = ae–10r = a0e–10r 10 from this we have 10 = e10r
634
Mechanics
Taking logarithm loge 10 = 10r 10r = 2.3 log1010 = 2.3 r =
2. 3 = 0. 23 10
τ =
1 1 = = 2.174 sec. 2r 2 × 0. 23
(i) No, the relaxation time is
(ii) The quality factor is Q = ωτ = 2πnτ = 2 × 3.14 × 200 × 2.174 = 2730.5 (iii) The energy of the oscillator is E = E0 e−2rt Let t′ be the time after which the energy falls to E0/10. Then
E0 = E0 e−2rt ′ 10 10 = e2rt′ 2rt′ = loge 10 = 2.3 t′ =
2. 3 2. 3 = = 5. 0 sec. 2r 2 × 0. 23
(iii) The damping constant r = 0.23 (as determined above) Q.8. The quality factor Q of a tuning fork is 5.0 × 104. Calculate the time interval after which its energy becomes 1/10 of its initial value. The frequency of the fork is 300 sec–1 (loge10 = 2.3). Solution. The quality factor of a damped oscillator is given by Q = ωτ, where ω is angular frequency and τ is relaxation time. Therefore
Q ω Here Q = 5.0 × 104 and ω = 2πn = 600 π sec–1 τ =
5. 0 × 104 sec 600 π Now, the energy of the damped system is given by ∴
τ =
E = E0 e−2rt = E0 e− t / τ Let t′ be the time after which the energy becomes E/E0 =
1 and t = t′ in the last expression, we get 10
LMQ τ = 1 OP N 2r Q
1 of the initial value. Then putting 10
Damped and Forced Harmonic Oscillation
635
1 = e− t ′ / τ 10 10 = et′ / τ loge10 =
t′ τ
t′ = τ loge10 =
5. 0 × 104 × 2. 3 = 61 sec. 600 × 3.14
Q.9. Q of a sonometer wire is 2 ×103. On plucking, it executes 240 vibrations per second. Calculate the time in which the amplitude decreases to 1/e2 of the initial value. Solution. The quality factor is Q = ωτ 3
Here Q = 2 × 10 and ω = 2πn = 2 × 3.14 × 240 sec–1 τ =
Q 2 × 103 = = 1. 327 sec ω 2 × 3.14 × 240
Now, if a0 be the initial amplitude (at t = 0) and a0 /e2 the amplitude after time t, we have,
a0 = a0 e− rt e2 e–2 = e–rt 2 = rt t =
2 = 4τ r
LMτ = 1 OP N rQ
= 4 × 1.327 = 5.3 sec Q.10. The oscillator of a tuning fork of frequency 200 Cps die away of 1/e time their amplitude in 1 second. Show that the reduction in frequency due to air damping is exceeding small. Solution. The damped amplitude at any time t is ae–rt where a is a constant. If a0 be the amplitude at t = 0 and a0/e at time t, then we have
a0 = a0 e− rt e This gives
t =
1 r
Here t = 1 sec ∴
r = 1 sec–1
The angular frequency in presence of damping is
(ω2 − r2 ) =
ω2 − 1
636
Mechanics
ω = 2πn = 2 × 3.14 × 200 = 1256, so that ω2 = 1.6 × 106
Now ∴
damped frequency = ≅
(1. 6 × 106 − 1) ≅
(1. 6 × 106 )
ω2 ~ ω,
Which is almost same as damped frequency. Thus the effect of damping on frequency is exceedingly small. Q.11. A damped vibrating system starting from rest has an initial amplitude of 20 cm which reduces to 2 cm after 100 complete oscillations each of period 2.3 sec. Find the logarithmic decrement of the system. Solution. The logarithmic decrement λ is the logarithm of the ratio of two amplitude of oscillations which are separated by one period. Here the amplitude ratio of oscillations separated by 100 periods is
20 . Thus 2 λ = =
1 20 1 log e log e 10 = 100 2 100 1 × 2. 3 = 0.023. 100
Q.12. A damped vibrating system starting from rest reaches a first amplitude of 500 mm, which reduces to 50 mm after 100 oscillations, each of period 2.3 second. Find the damping constant, relaxation time correction for the first displacement for damping. Solution. The amplitude of damped vibrations is given by a = a0 e−Kt Time period T = 2.3 sec. First amplitude of the system will be at t = T/4 and the amplitude after 100 oscillations i.e, 20th amplitude will occur at (100 T + T/4) time, hence −K(100T + T/ 4) −KT /4 a1 = a0 e and a201 = a0e
∴
a201 50 e−100KT or = e−100 × 2.3K a1 = 500 10–1 = e–230K or loge10 = 230 K
∴ ∴ ∴
K = 2.3/230 = 0.01 sec–1 Relaxation time τ =
1 1 = = 50 sec 2K 2 × 0. 01
a1 = a0 e−KT /4 ∴ 500 = a0 e−0.001 (2.3 / 4) a0 = 500e0.023/4
or log10a0 = log10500 +
log10a0 = 2.699 + 0.0025 = 2.7015 hence
a0 = 502.9 mm.
0. 023 1 4 2. 3
Damped and Forced Harmonic Oscillation
637
Thus in absence of damping, the amplitude of motion would have been 502.9 mm. Q.13. The differential equation for a certain system is d2 x dx + 2K + ω 02 x = 0 2 dt dt
if ω0 >> K, find the time in which (i) amplitude falls to 1/e times the initial value, (ii) energy of the system falls to 1/e times the initial value, (iii) energy falls to 1/e4 times the initial value. Solution. (i) The given differential equation represent the equation of motion of damped harmonic, oscillator, whose amplitude of vibration is given by a = a0e–Kt when amplitude falls to 1/e times the initial value a a0e–Kt = a0/e or eKt = e' Kt = 1, (ii) Energy ∴ (iii)
∴ t=
1 sec K
Et = E0 e−2Kt = E0 / e e2Kt = e or E0–2Kt = E0/e4
2Kt = 1
or t = 1/2K sec.
or 2Kt = 4 or t =
2 sec. K
Q.14. The frequency of a tuning fork is 300 Hz if its quality factor Q is 5 × 104, find the time after which its energy becomes 1/10 of its initial value. Solution.
Energy E = E0 e−2Kt = E0 e− t / r
when energy becomes 1/10 of its initial value, E0/10 = E0 e− t / r or 10 = et/r
t = loge10 = 2.3, t = 2.3τ τ But
τ =
5 × 104 Q Q = = ω 2πn 2 × 3.14 × 300
∴
t =
2. 3 × 5 × 104 = 61 sec 2 × 3.14 × 300
Q.15. A box of 100 gm is attached to one end of a spring whose other end is fixed to a rigid support. When a mass of 900 gm is placed inside the box the system performs 4 vibration per second and the amplitude falls from 2 cm to 1 cm in 15 seconds. Calculate (i) the force constant, (ii) the relaxation time, (iii) Q of the system and (iv) Q when only 100 gm (empty box) is with the spring (loge10 = 2.3). Solution. (i) Frequency of vibration of the system n = 4 sec–1 ∴
ω0 = 2πn = 2 × 3.14 × 4 = 25 rad/sec
638
Mechanics
(if the damping is small) But
ω0 =
2 2 C / m or C = mw0 = 1 × 25 = 625 nm.
(ii) At any constant t, the amplitude of the system is given by a = a0e–Kt or 0.01 = 0.02 e–15K ∴ ∴ (iii)
loge(2 – 15 K) = 0, or K = Relaxation time τ =
log e 2 2. 3 × 0. 3 = = 0. 046 sec−1 15 15
1 1 = = 11 sec 2K 2 × 0. 046
Q = ω0τ = 25 × 11 = 275 Damping force = γdx/dt, where γ = 2 km = 2 × 0.046 × 1 = 0.092
(iv) when the box is empty, i.e., the mass is 100 gm the damping force γdx/dt and restoring force Cx remain unchanged. Now ω0 =
FG IJ H K
C 625 = m′ 0.1
1 2
= 79 rad. /sec
m′ 1 0. 1 τ = 2K = γ = 2 × 0. 046 × 1 = 1.1 sec
∴
Q = ω0τ = 79 × 1.1 = 87 (approx.)
Example 5. If the suspension of a galvanometer coil exerts a restoring couple 5 × 10–5 n-m/rad. the period is 6.28 sec and the amplitude decreases to 1/10 of its original value in 92 seconds, find the value of I, γ and Q factor. Solution.
Now, ∴
Period T =
2π ω20
− K
2
= 6. 28 or ω2o − K 2 = 1
...(i)
a 1 = e−92K or loge 10 = 92K . e−Kt , ∴ here ao = 10 K = 2.3/92 = 1/40
Substituting this value of K in relation (i), we have
ω20 − (1 / 40)2 = 1 or ω0 = 1 nearly Now,
ω2o = C/I, ∴ 1 = 5 × 10−5 / I, ∴ I = 5 × 10−5 kg × m2 1 γ γ =2× , ∴ γ = 2. 5 × 10−6 nm sec. = 2K or −5 40 I 5 × 10
∴
Q =
ω0 1 = = 20 2K 2 × 1 / 40
Q.17. A condenser of 2 microfarad capacity is discharged through a 1 ohm resistance and 2 henry inductance. Calculate the frequency and quality factor of LC circuit and find the time in which the amplitude of oscillations is reduced to 1%.
Damped and Forced Harmonic Oscillation
Solution.
639
n =
1 1 R2 − 2π LC 4L2
Here, C = 2 × 10–6 farad, L = 2 henry and R = 1 ohm
1 1 103 − = = 79. 5 Hz −6 16 2π × 2 2 × 2 × 10
n =
1 2π
Q =
Lω 2 × 2π 103 = × = 103 R 1 2π × 2
The resistance is low, so
R
The amplitude Q = θ e− 2L 0
Q a = a0e−Kt
As amplitude decreases to 1% of its initial value θ0, we have
θ0 = θ0 e− t / 4 or t1 / 4 = 100 100 Taking logarithm of both sides
t = log e 100 = 2. 3 × 2 or t = 18. 4 sec. 4 Q.18. In Case of a forced harmonic oscillator, the amplitude of vibrations increases from 0.02 mm at very low frequencies to a value 5 mm at the frequency 100 Hz. Find (i) Q factor of the system, (ii) damping constant K and relaxation time and (iii) half width of resonance curve. Solution. (i)
(ii)
A max ω0 = 2K A at zero or very low frequency = 5/0.02 = 250
Q = ω 0τ =
Resonant frequency = 100 Hz, ω0 = 2π × 100 τ =
K =
Q 250 = = 0. 4 sec ω 0 2π × 100
1 1 = = 1. 25 per sec. 2τ 2 × 0. 4
Half width ∆p =
3 × 1. 25 = 2. 2 rad/sec. Q.19. A damped vibrating system starting from rest reaches a first amplitude of 500 mm, which reduces to 50 mm after 100 oscillations, each of period 2.3 seconds. Find the damping constant and correction for the first displacement for damping. (iii)
Solution. The amplitude of damped vibrations as we know, is given by y = y0e–Kt
...(1)
Here, time period for 1 oscillation, T = 2.3 sec First amplitude of the system will be at T/4 and the amplitude after 100 oscillations i.e., 201th amplitude will occur at (100T + T/4) time, hence
640
Mechanics
a1 = a0e–KT/4 and a201 = a0e–K(100T + T/4) a201 = e–100KT a1
∴
50 = e−100 × 2.3K 100
or
10–1 = e–230K or
loge10 = 230 K
2. 3 = 0. 01 230
∴
K =
∴
a1 = a0e–KT/4 ∴ 500 = a0e–0.01(2.314)
0. 023 1 . 4 2. 3
a0 = 500e0.023/4 or log10a0 = log10500 + log10 = 2.699 + 0.0025 = 2.7015 whence,
a0 = 502.9 mm.
Thus in absence of the damping the amplitude of motion would have been 502.9 mm. Q.20. The time period of pendulum is 2 sec. If the angular retardation due to air is 0.04 times the angular velocity of the pendulum and the original amplitude is 1°. Calculate the amplitude after 10 oscillations. Solution. The amplitude of damped oscillations at any time is given by y = y0e–Kt
...(i)
where y0 is the initial amplitude and K, the damping constant and 2K, the frictional retardation per unit mass per unit velocity. y0 = 1°, 2K = .04 and
Here,
t = 2 sec.
∴ time required for 10° oscillations = 10 × 2 = 20 sec. Substituting these values in relation (i), we get y = 1 × e–.02 ∴
× 20
0. 4 = 00.1739 = 1 . 8261 2. 3 y = 0.67° = 0.67 × 60 = 40.2
log10y = 0 −
Q.21. The amplitude of forced vibrations is given by A =
f (ω02
If θ = 40, calculate the value of A/Amax Solution. Q = ω0τ = 50 and
or
− P ) + P2 / τ 2 when dP/ω0 = 0.99. 2 2
τ = 50/ω0 A =
f (ω20 − P2 )2 + P2 / τ2
=
F1 − P I GH ω JK 2
2 0
For small damping at resonance. Amax =
1 50 f = 2 ω20 × 1 / 50 ω0
f 2
+
1 P2 2500 ω20
Damped and Forced Harmonic Oscillation
When
641
P/ω0 = 0.99
A 1 A max = 50
f 1 (1 − 0. 99) + (0. 99)2 2500
= 0. 71
2
Q.22. If a constant torque of 8 × 10–5 n-m produces an angular displacement of 2° in a torsion pendulum whose moment of inertia is 10 – 3 kg × m2, Calculate the frequency of the periodic external torque which will produce resonance. Solution. Angular displacement = 2° = 2 × π/180 radian. Applied couple = 8 × 10–5 n-m Restoring couple for unit displacement =
8 × 10−5 × 180 n-m/rad . 2π
At resonance, frequency of the force = approx. natural frequency =
1 2π
C 1 = I 2π
8 × 10−5 × 180 = 0. 24 sec−1 . 2π × 10−3
Q.23. Calculate the average energy stored in a 20 gm mass attached to a spring and vibrating with an amplitude 1 cm in resonances with a periodic force whose frequency is 20 Hz. If the quality of the oscillator be 160, how much energy being dissipated per second. Solution. During resonance average energy stored = maximum potential energy
Now,
=
1 2 1 Cx0 = mω20 x02 2 2
=
1 × 0. 02 × (2 × 3.14 × 20)2 × ( 0. 01)2 = 0. 016 Joules 2
Q =
2π average energy stored energy dissipated per cycle
∴ Energy dissipated in 1/20 sec. (i.e., time for one vibration)
2π × average energy stored Q 2 × 3.14 × 0. 016 = 160 20 × 3.14 × 2 × 0. 016 = 0. 0125 Joule. ∴ Energy dissipated per second = 160 Q.24. A circuit contains a resistance of 4 ohms and an inductance of 0.68 henry and alternating effective e.m.f. of 500 volts at a frequency of 120 Hz applied to it. Find the value of the effective current in the circuit and power factor. =
Solution.
Impedence Z = =
R2 + P2 L2 =
R2 + (2πn)2 L2
42 + 4 (3.1416)2 × (120)2 × (0. 68)2
642
Mechanics
= ∴
Effective current = Power factor =
16 + 262600 = 512 ohms. Effective e.m. f . 504 = = 0. 98 amp. Z 512 R = Z
R R +P L 2
2 2
=
4 1 = 512 128
Q.25. An A.C. supply of frequency 10000 and 110 volts is connected across a circuit containing a resistance of 10 ohms, an inductance of 10 – 2 henry and a capacity of 1µF. Find the value of the current what must be the value of the capacity in order that the current may be maximum. Solution.
Impedance Z = =
=
R2 + ( PL − 1 / PC)2 R2 + (2πnL − 1 / 2πnC)2 2
10
F + G 2π × 10000 × 10 H
−3
1 − 2π × 10000 × 10−6
I JK
2
= 48 ohms approx. ∴
Current = e.m.f./Z = 110/48 = 2.293 amp.
Maximum current will flow when the total resistance is zero and resonance is produced. Hence the maximum current will be when PL – 1/PC = 0 i.e., when C =
1 1 1 = = LP2 L × 4 π2n2 10−3 × 4 × π2 × (104 )2
= 0.253 × 10–6 farad. Q.26. An A.C. potential of 1 × 105 sec–1 frequency and 1.0 volt amplitude is applied to a series LCR circuit, if R = 2 ohm, L = 0.50 mili henry, Calculate the value of C to secure resonance. Also calculate the rms value of current and peak potential difference across the condenser. Solution. Capacity, required to produce resonance, is given by
1 1 = 2 4 π n L 4 × (3.14 ) × (105 )2 × 0. 5 × 10−3 = 5 × 109 farad
C =
Irms =
2 2
Erms E0 1 = = amp. R R 2 2 2
[at resonance Z = R]
Peak potential diff. across the condenser = Peak current × reactance across condenser =
E0 1 1 1 × = × R PC 2 105 × 5 × 10−9
= 103 volt.
# RELATIVITY 15.1
BACK GROUND OF MICHELSON-MORLEY EXPERIMENT
To send a signal through free space from one point to another as fast as possible, a beam of light or some other electromagnetic radiation such as radiowave is used. No faster method of sending the signal has ever been discovered than light. This experimental fact suggests that the speed of light in free space denoted by C ( = 3.00 × 108 meters/sec) is an appropriate limiting reference speed to which other speeds, such as speeds of particles or mechanical waves, can be compared. The presently accepted value of speed of light is given by C = (2.997925 ± 0.000003) × 108 meters/sec. This one of the most fundamental importance due to the following facts: (1) The speed of light in free space is independent of the reference frame from which it is observed. (2) All electromagnetic radiations from longest radio waves to shortest wave travel with speed of light C independent of the frequency of radiation. (3) It is impossible to transmit any signal with a velocity greater than the speed of light, that is, C is an upper limit for the velocity with which a particle can move. (4) Maxwell’s equations of electromagnetic field in C.G.S. system involve this constant C, i.e., div D = 4πρ div B = 0 curl E = −
1 ∂B c ∂t
U| | ...(b) | | ...(c) V || | ...(d) | W ...(a)
...(1)
4π 1 ∂D j+ c c ∂t where D, B, j, E, H and ρ are electric displacement vector, magnetic induction vector, current density, electric field intensity, magnetic field intensity and charge density respectively. curl H =
(5) Lorentz force equation in C.G.S. system also involves the speed of light. 643
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F = qE +
q v×B c
...(2)
This is Lorentz force equation. F is the total force on a uniformly moving particle of charge q. This force is the sum of electrostatic and magnetic (Lorentz) forces. (6) Very important relations in relativity involve this constant as m=
m0 v2 1− 2 c
, showing variation of mass with velocity.
...(3)
and E = mc2, showing that mass and energy are equivalent and the factor connecting them is square of C. (7) The two important units in physics, i.e., electromagnetic units and electrostatic units of any electrical quantity are related to each other by this constant. For example, 1 e.m.u. of charge = c e.s.u. of charge 1 e.m.u. of capacity = c2 e.s.u. of capacity 1 e.m.u. of potential difference =
1 e.s.u. of potential difference. c
(8) The reciprocal of the fine structure constant also involves the speed of light given by
hc = 1370.4 2πe2 where h is the Planck’s constant and e the elementary charge. In the macroscopic world of our ordinary experience the speed u of moving objects or mechanical waves with respect to any observer is always less than c. For example, sound waves in air at room temperature move with speed 332 m/sec. So that u/c = 0.0000010. An artificial satellite circling the earth may move with speed 30000 kilometers/hour. So that there
u = 0. 000028. c
Such limiting examples of macroscopic world are the basis of formulation of our ideas about space and time in which Newton developed his system of mechanics. In the microscopic world it is readily possible to find particles whose speeds are quite close to that of light. An electron accelerated through a potential difference of 10 million volts has speed 0.9988 c. The validity of Newtonian mechanics can be extended from the u << 1 in which it was developed to this high speed region ordinary region of low speeds c u → 1 ; because experiments show that Newtonian mechanics does not predict the correct c answers when it is applied to such fast moving particles. In fact in principle there is no upper limit to the speed attainable by a particle in Newtonian mechanics, so that there should be no special role of the speed of light c at all. And yet if the energy of a electron of 10 MeV is increased four times (to 40 MeV), the experimental observation is that the speed is not
FG H
IJ K
FG H
IJ K
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doubled to 1.9976c, as one might expect from Newtonain mechanics relation
FG T = 1 mu IJ , H 2 K 2
but remains below c, it increases only from 0.9988 c to 0.9999 c, a change of 0.11 percent or if an electron of energy 10 MeV moves at right angles to a magnetic field of 2 weber/m2, the observed radius of curvature of its path is not 0.53 cm (as may be calculated from classical relation r =
meu ) but instead to 1.18 cm. Hence no matter how well Newtonian qB
mechanics may work at low speeds, but fails badly as
u → 1. i.e., when the speeds can not c
be neglected in comparison with the speed of light. In 1905 Albert Einstein published his special theory of relativity extending and generalising Newtonian mechanics as well. He correctly predicted the results of mechanical
u u = 0 to → 1. Newtonian mechanics c c was revealed to be an important special case of Einsteins theory. While developing the theory of relativity, Einstein critically examined the procedures used to measure length and time intervals. These procedures require the use of light signals and, in fact an assumption about the way light is propagated is one of the two fundamental assumptions upon which his theory is based. His theory ruled out the Newton’s concept of the space and time and resulted in a completely new view. The connection between mechanics and electromagnetism is not surprising because light is an electromagnetic phenomenon and plays a basic role in making the fundamental space and time measurements that underlie mechanics. However our low speed Newtonian environment is so much a part of our daily life that almost every one has some conceptual difficulty in understanding Einstein ideas of space and time when he first studies them. A careful analysis of basic assumptions of Einstein and of Newton makes it clear that the assumptions of Einstein are really much more reasonable than those of Newton.
experiments over the complete range of speeds from
We shall now develop the experimental basis for the ideas of Einsteins theory of relativity.
15.2 THE SPEED OF LIGHT RELATIVE TO EARTH At the beginning of the present century when the modern techniques were used to measure the speed of light in refined manner; it was questioned what was the reference system of co-ordinate frame with respect to which the speed of light was measured ? Was it relative to earth or stars of any other medium? It was to choose the medium for light as air for sound. We know that air is the medium through which sound travels. Therefore the speed of sound is measured with respect to air. We also know that velocity of sound relative to the earth can be calculated be imposing the velocity of air with respect to earth to the velocity of sound with respect to air according to Galilean transformations. If the velocity of air with respect to the earth is u in the direction opposite to that of sound, then the velocity of sound relative the earth is the velocity of sound relative to air plus velocity of air relative to the earth i.e., u. Then it was assumed as already described, that the preferred medium for light was ether which filled all space uniformly. This ether is perfectly transparent to light and the material bodies may pass through it without any resistance, and since the earth has
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orbital motion about 3 × 104 m/sec. It was, therefore supposed that acording to Galilean transformations the speed of light should depend upon the direction of motion of light with respect to the earth’s motion through the ether. Also since the ether was assumed to be at rest and it offers no resistance to material bodies moving through it, i.e., ether remains fixed in space, therefore, it was considered to be possible to find the absolute velocity of any body moving through it, but experiments were conducted to find the absolute velocity of the earth through ether by a number of scientists namely Fizeau, Michelson and Morely, Trouton and Noble and many others, but they could not get success in their attempts. Later Hertz proposed that ether was not at rest, but it was carried completely by bodies moving through it, that is, the velocity of ether is the same as that of the body moving through it. By this assumption the phenomenon of aberration could not be explained and hence the assumption was rejected.
15.3
MICHELSON MORLEY EXPERIMENT
According to Newtonian mechanics or classical mechanics there existed absolute space or luminiferous ether relative to which the measurements are taken. The ether is filled uniformly through all space and penetrates all matter and it was considered that light propagates through ether as the sound propagates through air and c is a constant independent of the direction of propagation of light. The ether was assumed to be at rest. The light passes through ether with fixed velocity 3 × 108 m/sec and the earth passes through this stationary ether with velocity 3 × 104 m/sec. It was attempted to determine the velocity of light relative to the earth, for the purpose of determination of velocity of the earth relative to medium, i.e., ether, through which the velocity of light was measured, was essential. For this purpose the best and most important experiment was first performed by A.A. Michelson in 1881 and then in 1887 in collaboration with E.W. Morley. Michelson-Morley Experiments laid the experimental foundations of theory of relativity. The principle of the experiments lies in noting the shift in fringes in the Michelson interferometer due to difference in time taken by light to travel along and opposite the direction of motion of the earth; for, the time taken by a beam of light to travel along the direction of motion of the earth is greater than that to travel distance opposite to the direction of motion of the earth. For his invention of the interferometer and many optical experiments, Michelson was awarded the Nobel Prize in Physics in 1907, the first American to be so honoured. The Michelson-Morley experimental arrangement consists of two excellent optically plane mirrors M1 and M2 highly silvered on their front surfaces to avoid multiple internal reflections and two plane glass plates P and Q of equal thickness and of same material. These plates are mounted vertically with an inclination of 45° to the interferometer arms as shown in figure. The plate P is semi-silvered glass plate so that a beam of light from a monochromatic extended source S is partly reflected and partly transmitted when it falls on the plate P. The reflected beam travels perpendicular to the direction of initial beam and falls normally on mirror M2 due to which it is reflected back to P. The transmitted portion travelling along the initial direction of the beam falls normally on mirror M1 at point A and is reflected back to P. The two rays returned to P superimpose to form interference pattern which is observed by telescope T. we see that the reflected ray passes two times through the glass plate P
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while the optical path of the transmitted beam lies wholly in air in the absence of plate Q. Due to this a path difference = 2(µ – 1)t is introduced in the reflected beam. Now if we place a plate Q of same material and of equal thickness parallel to plate P in the path of the transmitted beam, the extra optical path 2(µ – 1)t is compensated; due to this reason the plate Q is called compensating plate. If the apparatus is at rest in ether, the two waves, reflected and transmitted, would take equal time to return to P. But actually the whole apparatus is moving with earth. Consider the direction of motion of the earth in the direction of the initial beam of light. Due to motion of apparatus with earth, the positions of reflections of the mirrors and the paths of the rays are shown by dotted lines in figure. In this case the time taken by the two beams to return to P would not be same. This time difference may be calculated as follows: M2 B
B
S
A
P
A′
P′
O Q
M1
T Telesco p e
Fig. 1 Michelson-Morley Experiment
Let the two mirrors be at equal distances from the plate P, i.e., PA = PB = l (ray) and c and v be the velocities of light and apparatus (earth) respectively. Now the ray reflected from P strikes the mirror at B′ instead of B due to the motion of the earth. Total path traversed by reflected beam = PB′P′ = PB′ + B′P′ = 2PB′ since PB′ = B′P′ Also
PB′
2
= PO2 + OB′2
If t is the time taken by the beam to reach from P to mirror M2, then PB′ = ct and BB′ = vt ∴ Eq. (1) gives ∴
c2t2 = v2t2 + l2 since PB = OB′ t =
l 1
( c2 − v2 ) 2 If t1 is time taken by the reflected ray to travel total path, then
...(1)
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2
2l
t1 = 2t =
( c2 −
v U 2l R 1− V = S c T c W
1 v2 ) 2
−
1 2
2
=
RS T
v2 2l 1+ 2 c 2c
UV ...(2) W
The transmitted ray has velocity (c – v) relative to apparatus from P to B and (c + v) from A′ to P′; since from P to A the transmitted ray and apparatus are moving in the same direction while from A′ to P′ they are moving in opposite directions. ∴ Total time taken by transmitted beam to travel the total path, i.e., from P to A and then from A′ to P′ = t2 =
∴
l l + since PA = P′A′ = l c − v c + v′
t2
The time difference
∴
LM N
v2 2lc 2lc = 2 1− 2 = 2 2 c −v c c
∆t = t2 – t1 =
RS T
UV W
RS T
OP Q
−1
v2 v2 2l 2l 1+ 2 − 1+ 2 c c 2c c
The path difference = c∆t = c .
∴ The shift in terms of number of fringes =
RS T
v2 2l 1 + = c c2
UV = W
UV W
lv2 2l v2 . 2 = 3 c 2c c
lv2 lv2 = 2 c3 c
lv2 c2λ
In Michelson-Morley experiment PA and PB were not exactly equal to get straight fringes by proper inclination of the mirrors. In performing the experiment the transmitted beam travelled along the direction of motion of apparatus. Then the whole apparatus was turned through 90°. In this case the difference of path will be in opposite direction. The observed shift after 90° rotation = 2lv2/c2λ. If l = 10 meters, v = 3 × 104 meters/sec and c = 3 × 108 metres/sec and the visible light of wavelength λ ≈ 6 × 10–7 metres, the path difference =
2lv2 2 × 10 × (3 × 104 )2 1 = wavelength (in order). −7 = 2 8 2 cλ (3 × 10 ) × 6 × 10 3
In spite of all the necessary precautions, the experiment showed no shift. The experiment was performed by a number of scientists at the different times of year when the directions of earth’s orbital velocity were different, but no shift was observed. The negative results observed by the experiment suggest that the space or medium in which light propagates is not moving relative to earth. We may say that the existence of stationary medium carrying light (or absolute space) is disproved.
15.4 EXPLANATION OF NEGATIVE RESULTS: PRINCIPLE OF CONSTANCY OF SPEED OF LIGHT One way of explaining the negative results of Michelson-Morley experiment is to conclude simply that the measured speed of light is the same (i.e., c) for all directions in every inertial
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frame. This statement is called the principle of constancy of the speed of light and is one of the two fundamental postulates of Einstein’s special theory of relativity. This principle would lead to no shift in fringes in Michelson-Morley experiment: Since now the speed of light is c rather than |(c + v)| in any frame. Hence according to principle of constancy of speed of light no shift in fringes should be observed. Thus the principle of constancy of speed of light is capable of explaining negative results of Michelson-Morley experiment. The principle of constancy of light being incompatible with Galilean velocity transformations seemed to be too drastic philosophically at that time. If the observed speed of light did not depend on the motion of the observer, all inertial frames would be equivalent for propagation of light and there could be no experimental evidence to indicate the existence of an absolute ether frame.
15.5 THE RELATIVITY OF SIMULTANEITY: THE RELATIVISTIC CONCEPT OF SPACE AND TIME Michelson-Morley, Trouton Nobel and other carefully performed experiments could not observe the velocity of earth relative to ether. Lorentz and Fitzgerald tried to explain the negative results by introducing the hypothesis. It is necessary in every inertial frame to use a special time (Lorentz hypothesis) and special space co-ordinates (Fitzgerald hypothesis) which are different from the time and space co-ordinates in the absolute ether system. Lorentz and Fitzgerald, thus, suspected a new concept of space and time along with the concept of absolute ether frame. Einstein, in 1905, keeping in mind the negative results of Michelson-Morley and other experiments performed to determine the velocity of earth through ether and the scientific tradition not to postulate things that are by their nature unobservable took a bold step and stated that the ether does not exist and the motion relative to material bodies has physical significance while motion through ether in meaningless concept. In other words there is no absolute frame and all frames are equally suitable for description of motion, although one may be more convenient to use a specific frame in a specific case. He thus ruled out the absolute concept of space. To explain the following two contradictory statements: (1) According to classical mechanics the velocity of any motion has different values for observers moving relative to each other. (2) According to experimental observations the velocity of light is not affected by the motion the frame of reference. Einstein decided that the conflict between them must be due to an imperfection of the classical ideas about measuring space and time. He ruled out the concept of absolute time by attacking our pre-conceived ideas about simultaneity. The nature of his argument may be considered as follows: Let us see whether the statement “The two events x and y take place simultaneously without reference of any co-ordinate system” can have any meaning. Let us consider a light signal which travels in a straight line from one point x to another point y in a given inertial frame. If we imagine the time of emission t1 to be read on a clock placed at x and the time of arrival t2 to be read on a clock placed at y, the difference (t2 – t1) obtained in this way will give only the real time taken by the light to travel from x to y if the clocks at x and y are put right. This obviously requires that the hands of both clocks simultaneoutly are in the same position, i.e., the clocks placed at x and y are synchronized.
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Now how can we make sure that two events occurring in two different places are simultaneous ? Whether the two events simultaneous in one frame will also be simutaneous in any other frame ? Let us consider two events occurring at two points x and y which are fixed in inertial frame S. Since the velocity of light is c in all directions, the criterion for these events to be simultaneous relative to system S is obviously that the two light signals emitted from x and y at the moments when the events occur shall meet in the centre O of the line connecting x and y. A similar criterion for simultaneity is also true relative to system S′, having a constant velocity v relative to system S. Now let us consider that the two events are simultaneous relative to system S and also the line connecting x and y is parallel to the direction of velocity v of system S′ relative to system S. Then consider two points x′ and y′ in system S′ which, at the moment the events occur, coincide with the points x and y. Then simultaneously the centre O′ between x′ and y′ will coincide with O. Since now O′, just as x′ and y′ moves together with system S′ with a velocity v relative to system S, O′ will not coincide with O at the moment light signals from x and y meet in O. The light signals will thus not meet in O′ and hence, according to above mentioned criterion, the two events are not simultaneous relative to system S′. Thus simultaneity is a relative concept, not an absolute one. Hence the concept of simultaneity between two events in different space points has an exact meaning only in relation to a given inertial system. In other words each frame of reference has its own particular time. Therefore there is no meaning in declaring the time of an event, unless we refer the reference system to which the statement of time refers. On account of the absolute concept of simultaneity being ruled out, the absolute concept of space is automatically ruled out. To measure the length of an object means to locate its end points simultaneously. As simultaneity depends upon the frame of reference, the length measurements will also depend on the frame of reference. Thus the length i.e., the space is relative concept, not an absolute one. As a result of Einstein’s relativistic concepts of space and time it is possible to reconcile ourselves to the experimental fact that observers who are moving relative to each other measure the speed of light to be the same. Using these new concepts of space and time we have to replace Galilean transformation equations by a new type of transformation equations which will be based on the invariant character of the speed of light. The theory of relativity having new concepts of space and time is applicable not only to mechanical phenomenon, but also to all optical and electromagnetic phenomenon, and is divided into two parts. (1) Special or restricted theory of relativity. (2) General theory of relativity. The special theory of relativity deals with systems known as inertial system that is the systems, which move in uniform rectilinear motion relative to one another. According to this “All systems of co-ordinates are equally suitable for description of physical phenomenon”. If we extend this principle to accelerated system, i.e., the systems moving with accelerated velocity relative to one another, the theory of relativity is called “general theory of relativity”. The general theory of relativity is applicable to the laws of gravitation and explains it in a more refined manner than given by Newton.
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15.6 POSTULATES OF SPECIAL THEORY OF RELATIVITY (1) The fundamental laws of physics have the same form for all inertial systems, i.e., for reference systems at rest or moving with constant linear velocity relative to one another. (2) The velocity of light in vacuum is independent of the relative motion of the source and the observer. These are the two fundamental postulates used in the special theory of relativity. The first postulate is the extension of the conclusion drawn from Newtonian mechanics, since velocity is not absolute, but relative, which is a fact drawn from the failure of the experiments to determine the velocity of earth relative to ether. We know that the speed of light is not constant under Galilean transformations and the first postulate is the conclusion from Newtonian mechanics; thus second postulate is not true according to Galilean transformations. Actually this is true since the velocity of light calculated by any means is a constant. Thus the second postulate is very important and only this, postulate is responsible to differentiate the classical theory and Einstein’s theory of relativity. And according to Einstein the theory of relativity is applicable to laws of optics. Thus for the constancy of velocity of light we have to introduce the new transformation equations which fulfill the following requirements: 1. The speed of light c must have the same value in every inertial frame. 2. The transformations must be linear and for low speeds (i.e., v << c), they should approach the Galilean transformations.
Y′ S′
Y S
P
3. They should not be based on “absolute time” and “absolute space”.
The above requirements were fulfilled by H. A. Lorentz by introducing transformation equations relating the observations of position and time made by two observers in two different inertial frames and are known as “Lorentz Transformation Equations”.
Z
O′ → v
O
15.7 LORENTZ TRANSFORMATION EQUATIONS
X, X′
Z′
Fig. 2
Let S and S′ be two inertial frames of reference. S′ having uniform velocity v relative to S. Let two observers at O and O′, observe any event P from systems S and S′ respectively. For convenience let us consider that the x axes of two systems coincides permanently and the velocity is parallel to x-axis. The event P is a light signal and is produced when both t and t′ are zero and when origins of the two frames coincide. The event P is determined by co-ordinates (x, y, z, t) and (x′, y′, z′, t′) by observers O and O′ respectively. The light pulse produced at t = 0 will spread out as a growing sphere and the radius of wavefront produced in this way will grow with speed c, since (x, y, z, t) are co-ordinates of the event relative to observer in system S at rest therefore the equation of spherical surface whose radius grows at the speed c, is
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x2 + y2 + z2 = c2 t 2 x2 + y2 + z2 − c2t 2 = 0
or
...(1)
Similarly for observer O′ in system S′ having the co-ordinate of P as (x′, y′, z′, t′), the equation of spherical surface is x ′2 + y ′2 + z′2 = c2 t ′2 x ′2 + y′2 + z ′2 − c2 t ′2 = 0
or
...(2)
c is not primed since according to 1st postulate of special theory it is a constant for all inertial frames. As velocity of S′ is only along x-axis, thus from symmetry.
UV W
y = y′ and
z = z′
...(3)
Then from Eq. (1) and Eq. (2), we have
x2 − c2t2 = λ ( x ′2 − c2t ′2 ); λ being any, constant
...(4)
Now for the transformation equation relating to x and x′, let us put x′ = γ ( x − vt),
...(5)
γ being independent of x and t. The reasons for trying the above relation are (1) The transformations must reduce to Galilean transformation for low speed i.e., for speed v/c → 0. (2) The transformation must be linear and simplest. We see that above relation is sufficiently general to satisfy the required conditions. Since motion is relative, we may assume that S is moving relative to S′ with velocity –v along (+ve) direction of x-axis, therefore, x = γ ( x′ + vt ′ )
...(6)
γ is not primed done to first postulate. Now putting the value of x′ from (5) in (6), we have x = γ γ ( x − vt) + vt ′
x γ = γx − γ vt + vt ′
or
∴
∴
vt′ =
LM N
x x + γvt − γx = γ 2 + vt − x γ γ
LM N
t′ = γ t +
x x − 2 vγ v
OP Q
OP Q
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LM MN
FG H
1 x = γ t − v 1 − γ2
IJ OP K PQ
...(7)
Now substituting in Eq. (4) the value of x′ from Eq. (5) and for t′ from Eq. (7) we have x −ct 2
2 2
LM MN
R| S| T
= λ γ 2 ( x − vt )2 − c2γ 2 t −
F GH
1 x 1− 2 v γ
I U|V OP JK W| P Q 2
LMt − x F 1 − 1 I OP = 0 MN v GH γ JK PQ LM 2 xt F 1 I x F 1 I OP MNt − v GH 1 − γ JK + v GH 1 − γ JK PQ = 0 2
or x − c t − λγ ( x − 2vxt + v t ) + λc γ 2
2 2
2
2
2 2
2
2
or x2 − c2t2 − λγ 2 ( x2 − 2vxt + v2t2 ) + λc2 γ 2
2
2
2
2
2
2
2
...(8)
Since this equation is an identity, the coefficients of various powers of x and t must vanish separately. Equating coefficients of xt to zero, we have
|RS 2 FG 1 − 1 IJ |UV T| v H γ K W| 2c γ F 1I 2γ v − 1− G J v H γ K
λγ 2 . 2v + λc2 γ 2 −
2
or
2
= 0
2 2
2
= 0
(v2 − c2 ) γ 2 + c2 = 0
or
γ =
this given,
1
FG 1 − v IJ H cK
...(9)
2
2
Putting coefficients of t2 equal to zero in Eq. (8), we have
− c2 − λ γ 2 v2 + λc2 γ 2 = 0 λ ( v2 − c2 ) γ 2 + c2 = 0
...(10)
Substituting value of γ from Eq. (9), we get λ = 1 Further, from Eq. (5) we have x′ = γ ( x − vt)
...(11)
From Eq. (7) we have
LM MN
t′ = γ t −
FG H
1 x 1− 2 v γ
IJ OP K PQ
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LM x v OP from Eq. (9) N vcQ L vx OP γ Mt − N cQ 2
= γ t−
∴
t′ =
2
2
...(12)
Substituting value of γ from Eq. (9) in Eq. (11) and Eq. (12), we have
x − vt
LM1 − F v I OP N GH c JK Q
x′ =
2
...(13)
2
t′ =
t − vx / c2
FG 1 − v IJ H cK 2
2
thus combining Eq. (3) and (13) we have x′ =
and
t′ =
x − vt
, y′ = y, z′ = z
v2 1− 2 c
t − vx / c2 v2 1− 2 c
...(14)
These are called Lorentz transformation. If we exchange our frames of reference or consider the given space time co-ordinates of the event to be those observed in system S′ rather than in system S, the only change allowed by the relativity principle is the physical one of a change in relative velocity from v to –v, i.e., the system S moves relative to system S′ with velocity v along negative direction of x-axis where as system S′ moves relative to system S along positive direction of x-axis. when we solve equations (14) for x, y, z and t in terms of the primed co-ordinates x′, y′, z′ and t′ we obtain x =
x′ + vt ′
LM1 − v OP N cQ 2
2
y = y′ z = z′
...(15)
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t′ + and
vx′ c2
LM1 − v OP N cQ
t =
2
2
These transformation equations are identical in form with equation (11), except that, as required, v changes to –v and are called inverse Lorentz transformation equations. Example 1. Show that direction application of Lorentz transformations x 2 + y 2 + z2 − c2t 2 is invariant.
Solution. We have only to prove by the help of Lorentz transformations. x2 + y2 + z2 − c2t 2 = x ′2 + y ′2 + z′2 − c2t ′2
where (x, y, z, t) and (x′, y′, z′, t′) are the co-ordinate of the same event observed by two observers in systems S and S′ while S′ is moving with a velocity v relative to S. Let us consider the expression
x′2 + y′2 + z′2 − c2t ′2 =
( x − vt)2 + y2 + z2 − c2 v2 1− 2 c
FG t − vx IJ H cK
2
2
1−
v2 c2
Putting values of x′, y′, z′ and t′ from Lorentz transformations. ∴
x′2 + y′2 + z ′2 − c2t ′2 =
FG H
c2 c2 vx 2 2 2 ( ) . c2 t − 2 − + + − x vt y z 2 2 2 2 c −v c −v c
c2 c − v2
=
c2 c − v2
=
c2 c − v2
2
2
2
2
2 2
2
2
2
2
2 2
2
2
2
2 2
2 2
2 2
2
2
2
2
2 2 2
4 2
2
2
2
2
2
2
= x2 − c2 t 2 + y2 + z2 = x2 + y2 + z2 − c2 t 2
2
2
4
1 c2 ( x2 − c2 t2 ) − v2 ( x2 − c2 t2 ) + y2 + z2 c − v2 1 2 2 2 2 2 2 2 = 2 2 (c − v ) ( x − c t ) + y + z c −v =
2
LM FG IJ OP + y + z H K PQ MN LM x − 2vxt + v t − c F t − 2vxt + v x I OP + y + z GH c c JK PQ MN LM x − 2vxt + v t − c t + 2vxt − v x OP + y + z c Q N LM c x + c v t − c t − v x OP + y + z c Q N
c2 vx 2 2 = c2 − v2 ( x − vt ) − c t − c2
=
IJ K
2
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Thus we have proved that x ′2 + y ′2 + z′2 − c2t ′2 = x2 + y2 + z2 − c2t 2
That is, the expression x2 + y2 + z2 − c2t 2 is invariant under Lorentz transformations. Example 2. Show that for low values of v, Lorentz transformations approach to Galilean. Solution. We have Lorentz transformation equations
x − vt
x′ =
and
t′ =
v2 1− 2 c
, y′ = y, z ′ = z
t − vx / c2 v2 1− 2 c
...(1)
If v is small i.e., v < c so that v/c → 0, then we have from Eq. (1) x′ = x − vt , y′ = y, z′ = z, t ′ = t which are Galilean transformations.
15.8
LENGTH CONTRACTION
Consider two co-ordinate system S and S′, the latter moving with velocity v relative to former along (+ve) direction of x-axis. Let a rod of proper length (i.e., length at rest) lo be at rest in frame S′ along the x-axis. If x′1, and x′2 are the co-ordinates of abscissae of the ends of the rod, in S′ at the same time t′, then lo = x′2 − x′1
...(1)
since rod is at rest in frame S′. If l is the length of the rod in frame S and the abscissae of the ends of the rod in this system are x1 and x2, at the same time t, then l = x2 – x1
...(2)
According to Lorentz transformations
x′2 =
x′1 =
x2 − vt v2 1− 2 c x1 − vt v2 1− 2 c
Subtracting Eq. (4) from Eq. (3), we have
x′2 − x′1 =
x2 − x1 1−
v2 c2
...(3)
...(4)
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657
lo = +
or
l
;
v2 1− 2 c
F1 − v I GH c JK 2
∴
l = lo
...(5)
2
From Eq. (5) we see l < l0.
v2 :1 as measured by moving c2 observer with velocity v relative to rod along its length. We may also state that the length Thus the length of the rod is reduced in the ratio
v2 c2 in the direction of motion while the length at right angles to the direction of motion remains unchanged. This is also known as “Lorentz-Fitzgerald contraction.”
of the object moving with velocity v relative to observer is contracted by a factor
1−
Example 3. How, fast would a rocket have to go relative to an observer for its length to be contracted to 99% of its length at rest. Solution. According to length contraction l = lo Here l =
∴
99 lo i.e., 100
1−
v2 c2
l 99 = lo 100 99 = 100
v2 1 − 2 i.e., c
FG IJ H K
99 v2 = 1− 100 c2 v =
2
=
FG 99 IJ H 100K
2
=1−
v2 c2
(100)2 − (99)2 199 = (100)2 (100)2
199 c = 0.1416 C Ans. 100
Example 4. A vector in system S′ is represented by 8i + 6j. How can the vector be represented in system S while S′ is moving with velocity. 0.8ci with respect to S. i and j are unit vectors along the respective directions. Solution. S′ is moving relative to S with velocity 0.8c along x-axis, therefore the length of the vector will only change along x-axis while its length along y-axis will remain the same. We know
Here
l = lo 1 −
v2 c2
v 0. 8 c = 0. 8 = c c
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Mechanics
∴
lx = 8 1 − 0. 64 = 8 0. 36 = 8 × 0. 6 = 4. 8
Therefore in system S the vector may be represented by 4.8i + 6j.
v2 is the volume c2 viewed from a reference frame moving with uniform velocity v parallel to an edge of the cube. 3 Example 5. Show that if l03 is the rest volume of a cube, then l0 1 −
Solution. Suppose two systems S and S′, S′ moving with velocity v relative to S along +ve direction of x-axis. Now the volume of the cube in system S = lo3. Hence one edge of the cube = lo. Let one edge of the cube be along axis of x; then its length along x-axis as observed by an
v2 . The lengths along y and z-axis remain unchanged. Therefore c2 volume of the cube as observed from S'
observer in S′ , lx = l0
1−
= lx ly lz = lo
FG1 − v IJ × l × l H cK 2
o
2
o
= lo3 1 −
v2 . c2
Example 6. Calculate the percentage contraction of a rod moving with a velocity 0.8c in a direction inclined at 60° to its own length. Solution. Let lo be the length of the rod at rest. If 0.8c is the velocity of the rod along the axis of x, then we have lo = lo cos 60° i + lo sin 60° j where i and j are unit vectors along the x and y-axes respectively. As the contraction takes place in the direction of motion, the contraction will only take place along the axis of x, while the component of length along the axis of y will remain the same. If lx′ is the component of the length of the rod, when in motion, along the axis of x, then we have lx′ = lx
1−
v2 c2
= lo cos 60°
FG1 − v IJ H cK 2
2
= lo cos 60° 1 −
l (0. 8 c)2 = o × 0. 6 = 0.3 lo 2 c2
The component of the length of the moving rod along y-axis is given by
ly′ = lo sin 60° =
∴ The length of the moving rod =
lx2′ + ly2′ =
3 lo 2
(0. 3lo )2 +
FG 1 H2
3 lo
IJ K
2
= 0. 91 lo
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∴ decrease in length of the rod due to its motion = lo – 0.91 lo = 0.09 l0. Therefore the percentage contraction = 0. 09 lo ×
100 = 9% Ans. lo
15.9 TIME DILATION Consider two systems S and S′ , S′ moving with velocity v relative to S along positive direction of x-axis. Let the clock be situated at x in frame S and gives signals at intervals. ∆t = t2 − t1
...(1)
Then the interval observed by observer in system S′ will be ∆t′ = t ′2 − t ′1
...(2)
From Lorentz transformation we have
t′1 =
and
t′2 =
t1 − vx / c2 1 − v2 / c2 t2 − vx / c2 1 − v2 / c2
...(3)
...(4)
Substracting Eq. (3) from Eq. (4), we get
t′2 – t′1 =
∴ or
∆t′ =
t2 − t1 1 − v2 / c2 ∆t 1 − v2 / c2
∆t′ > ∆t.
Thus the time interval ∆t appears to be moving observer to the dilated or lengthened because the time interval in system S′ is greater than that in system S. We may state “A clock will be found to run more and more slowly if the relative velocity between the clock and the observer is increased more and more. The time interval measured by a clock at rest relative to observer is called the proper time interval.
15.10 AN EXPERIMENTAL VERIFICATION OF TIME DILATION To observe time dilation in the laboratory the following conditions must be satisfied. 1. The bodies must move with relativistic speed v. 2. The events occuring in bodies should be independent of v. 3. The time interval between events should be sufficiently short so that the events may occur within a reasonably short travel of bodies, otherwise the laboratory necessary will be inconveniently long.
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Mechanics
Some nuclear and elementary particles satisfy these three conditions, they undergo some transformation with a proper life time τ which is independent of their speed and they can be produced with large speeds so that their non-proper time τ′ (measured in laboratory frame) may differ appreciably from τ. The nuclear particles called π+ mesons are produced with speed 0.99c when high energy particle generated by a syncho-cyclotron strike a target. These mesons decay (break into µ mesons and neutrinos) in such a way that in every 1.8 × 10–8 sec half of them die out. The flux of π+ mesons was measured at two places separated by 30 meters. The laboratory time interval ∆t′ for travelling the distance was given by 30 m ≈ 10 × 10−8 sec 0.99 × 3 × 108 m/sec sec. If T is half life time of particles then after time
∆t = This is about 5.6 times of 1.8 × 10–8 t, the fractional flux N/N0 is given by
N N0 =
FG 1 IJ H 2K
t /T
=
FG 1 IJ H 2K
5.6
= 2 − 5.6 ≈ 2%
Hence the flux of π mesons should decrease to approx, 2% of the original flux in travelling 30 metres. But the actual flux at the second place was approx 60% of that at first place. This discrepancy is explained by calculating the proper time ∆t by the relation ∆t′ =
∆t 1−
or
v2 c2
∆t = ∆t ′ 1 −
v2 c2
− = 10 × 10 3 1 −
FG 0.99 c IJ H c K
2
= 1.4 × 10–8 sec.
This is 0.78 times of 1.8 × 10–8 sec. Hence in this much time the flux should fall to 2−0.78 ≈ 60% . This is exactly what is observed.
From this experiment it is clear that in the laboratory measurements time taken by π+ mesons for 30 m travel is 10 × 10–8 sec. while π+ mesons themselves measure the time as 1.4 × 10–8 sec. Thus a 7-fold time dilation has occurred in this case. Example 7. The half life of a particular particle as measured in the laboratory comes out to be 4.0 × 10–8 sec, when its speed is 0.8c and 3.0 × 10 – 8 sec, when its speed is 0.6c. Explain this. Solution. This can be explained on the basis of relativistic time dilation. The time interval in motion is given by ∆t′ =
∆t
FG 1 − v IJ H cK 2
2
where ∆t is the proper time interval.
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661
The proper half life of the given particle is
FG1 − v IJ H cK 2
∆t = ∆t ′
2
In first case given ∆t′ = 4.0 × 10–8 sec and v = 0.8 c. ∆t = 4 × 10−8
∴
R|1 − F 0.8 c I U| S| GH c JK V| W T 2
= 4 × 10−8 × 0. 6 = 2. 4 × 10−8 sec. As proper half life is independent of velocity, therefore half life of the particle when speed is 0.6c must be given by ∆t′ =
∆t 1−
=
v2 c2
=
2. 4 × 10−8 1−
FG 0. 6c IJ H c K
2
2. 4 × 10−8 = 3 × 10–8 sec 0. 8
which is actual observation. Thus the variation of half life of the given particle is due to relativistic time dilation. Example 8. A beam of particles of half life 2 × 10–6 sec. travels in the laboratory with speed 0.96 times the speed of light. How much distance does the beam travel before the flux fall to
1 times the initial flux. 2
Solution. The observed half life ∆t′ is given by ∆t′ =
∆t 1−
=
v2 c2
=
2 × 10−6
R|1 − F 0. 96 c I U| S| GH c JK V| W T 2
2 × 10−6 = 7.14 × 10–16 sec. 0. 28
According to definition of half life in this observed time the flux falls to
1 times the 2
initial value. Thus in this time the distance traversed by the beam is ∆t′ = 0.96 × 3 × 10 –8 × 7.14 × 10–6 = 2.1 × 103 metres. Example 9. A particle with a mean proper life of 1 micro-sec (µ-sec) moves through the laboratory at 2.7 × 108 m/sec.
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Mechanics
(i) what will be its life time as measured by observer in the laboratory. (ii) what will be the distance transversed by it before disintegrating. (iii) find the distance traversed without taking relativity into account. Solution. (i) By the result of time dilation, we have
∆t
∆t′ =
v2 c2
1− Given
∆t = 1 µ sec = 1 × 10–6 sec. v = 2.7 × 108 m/sec.
Therefore
∆t′ =
1 × 10−6
R| F 2.7 × 10 I U| S|1 − GH 3 × 10 JK V| W T 8
2
8
=
3 × 10−6
n3
2
− (2. 7)
2
s
= 2. 3 × 10−6 sec
= 2.3 µ sec. (ii) Distance transversed by the particle = v∆t′ = 2.7 × 108 × 2.3 × 10–6 metres = 620 metres (iii) Distance transversed without relativistic effects. v∆t = 2.7 × 108 × 1 × 10–6 = 270 meters. +
Example 10. The proper life of π mesons is 2.5 × 10–8 sec. If a beam of these mesons of velocity 0.8c is produced, calculate the distance, the beam can travel before the flux of the meoson beam is reduced to 1/e2 times the initial flux. Solution. According to time-dilation, we have ∆t′ =
∆t 1−
=
v2 c2
=
2. 5 × 10−8 sec
FG 1 − 0. 8c IJ H c K
2
2. 5 × 10−8 = 4.16 × 10–8 sec. 0. 6
If No is the initial flux and N is the flux after time τ, then we have N = No e− t / τ , τ being mean life Here
N =
1 No e2
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663
1 N o = N o e− t / τ e2
∴
t = 2τ = 2∆t′ (since here τ = ∆t′ = 4.16 × 10–8 sec).
or
2∆t′ = 2 × 4.16 × 10–8 sec = 8.32 × 10–8 sec.
∴
∴ The distance traversed by the beam before the flux of meson beam is reduced to
1 e2
times the initial flux. = 2∆t ′ × 0. 8 c = 8.32 × 10–3 × 0.8 × 3 × 108 = 19.96 meters. Example 11. At what speed should a clock be moved so that it may appear to lose 1 minute in each hour ? Solution. If the clock is to lose 1 minute, in one hour, then the clock must record 59 minutes for each hour recorded by clocks stationary with respect to the observer. If v is the required speed of the clock, then according to Einstein’s apparent retardation of clocks, we must have 59 = 60 1 −
v = c
or
1−
v2 c2
FG 59 IJ H 60 K
2
≈ 0.18
or or
v = 0.18 × c = 5.4 × 107 metres/sec Ans.
15.11 TRANSFORMATION AND ADDITION OF VELOCITIES Let there be two co-ordinate system S and S′ the latter moving with velocity v relative to former along the +ve direction of x-axis. Let a particle be moving with velocities u and u′ relative to frames S and S′ respectively; then u = i ux + j uy + k uz u′ = i u′ x + j u′ y + k u′ z
and
...(1)
where ux, uy and uz are components of u along the three axes, i, j and k being unit vectors along x, y and z-axis respectively.
u′ x , u′ y and u′ z are components of u′ along respective axes. From the definition of velocity, we have ux =
dx dy dz dx′ dy′ dz′ , u′ y = and u′ z = , uy = , uz = , u′ x = dt dt dt dt dt dt
...(2)
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Mechanics
Lorentz inverse transformations are given by x =
x′ + vt ′
, y = y′ , z = z′ and t =
v2 1− 2 c
t ′ + vx′ / c2 v2 1− 2 c
...(3)
Differentiating these transformation equations, we get dx =
dx′ + vdt′ v2 1− 2 c
, dy = dy′ , dz = dz′ and dt =
dt ′ + v dx′ / c2 v2 1− 2 c
...(4)
Now from these expressions, we have
dx ux = = dt
∴
ux =
dx′ +v dx′ + vdt ′ = dt′ v dx′ v dx′ 1+ 2 dt ′ + 2 c dt′ c
u′ x + v v 1 + 2 u′ x c
...(5)
dy′ v2 v2 1− 2 2 dy c c = dt′ uy = = v dx v dx ′ ′ dt 1+ 2 dt ′ + 2 c dt′ c dy′ 1 −
u′ y ∴
similarly
uy =
uz =
1+
1−
v2 c2
v u′ x c2
1 − v2 / c2 v 1 + 2 u′ x c
u′ z
...(6)
...(7)
Eqs. (5), (6) and (7) represent transformation of velocity components from one inertial system to another. If we consider the particle to be a photon moving with velocity c in frame S′ and the system S′ to be moving with velocity c relative to system S along +ve direction of x-axis, then from eq. (5), we have ux =
c+ c =c c2 1+ 2 c
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665
i.e., the speed of light is same for all inertial frames whatever their relative speed may be and no particle can obtain a speed greater than the speed of light. Example 12. Two particles, are travelling in opposite directions with speed 0.9c relative to the laboratory. What is their relative speed. Solution. To solve the problem consider the particle moving with velocity –0.9c to be at rest in system S. Then it may be considered that S′ is moving with velocity v = 0.9c relative to S, so that the particle in S′ has velocity u′ x = 0. 9 c. Then the velocity in S is given by ux =
u′ x + v 0. 9 c + 0. 9 c 1. 8 c = = 0. 995 c = v 1. 81 (0. 9 c)2 1 + 2 u′ x 1+ c c2
Example 13. An electron is moving with a speed of 0.85c in a direction opposite to that of a moving photon. Calculate the relative velocity of the electron with respect to photon. Solution. The speed of the photon = c The speed of the electron = 0.85c Let the electron be moving along –ve direction of x-axis while the photon along +ve direction of x-axis. Consider that the electron moving with velocity –0.85c is at rest in system S. Then it may be assumed that the system S′ (laboratory) is moving with velocity .85c relative to system S (electron). Then we have v = 0.85c, u′ = c; ∴
u =
u′ + v c + 0. 85c c + 0. 85 c = = =c vu c c ( . ) ( ) 0 85 1 + 0. 85 1+ 2 1+ c c2
Example 14. An particle has a velocity u′ = 3i + 4 j + 12k m/sec in a co-ordinate system moving with velocity 0.8c relative to laboratory along +ve direction of x-axis. Find u in the laboratory frame. Solution. Given or and
u′ = 3i + 4 j + 12k u′ x = 3, u′ y = 4, u′ z = 12 meter/sec. v = 0.8c
Therefore components of u are given by ux =
=
u′ x + v 3 + 0. 8 c 3 + 0. 8 c = = v (0. 8 c) (3) (0. 8) (3) 1+ 1 + 2 u′ x 1+ 2 c c c (3 + 0. 8 c) c ≈ 0. 8 c = 2.4 × 108 metres/sec. ( c + 2. 4) u′ y
uy =
1+
1−
v2 c2
v u′ x c2
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Mechanics
4 1− =
=
and
uz =
FG 0. 8c IJ H c K
=
(0. 8 c) (3) 1+ c2
(4 × 0. 6) c c + 2. 4
2. 4 c ≈ 2. 4 m/sec. c + 2. 4 1 − v2 / c2 v 1 + 2 u′ x c
u′ z
12 1 −
=
2
FG 0. 8 c IJ H c K
2
c + 2. 4
=
7. 2 c ≈ 7. 2 m/sec. c + 2. 4
Therefore u in the laboratory frame is given by
u = i ux + j uy + ku z = (2. 4 × 108 i + 2. 4 j + 7. 2 k ) m/sec. Example 15. A space-ship moving away from earth with velocity of 0.4c fires a rocket whose velocity relative to space-ship is 0.6c, away from the earth. What will be the velocity of the rocket as observed from the earth. Solution. We have
ux =
u′ x + v u′ v 1 + x2 c
Here ux is velocity of rocket relative to earth. v is the velocity of space-ship relative to earth. u′x is the velocity of rocket relative to space-ship. Here ∴
v = 0.4c, u′x = 0.6c ux =
0. 6 c + 0. 4 c (0. 6 c) (0. 4 c) 1+ c2
c c = 1 + 0. 24 1. 24 ux = 0.8 c Ans. =
15.12 RELATIVISTIC DOPPLER’S EFFECT Consider two systems S and S′, the latter moving with velocity v relative to former along positive x-axis.
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667
Let the transmitter and receiver be situated at origins O and O′ of frames S and S′ respectively. Let two light signals or pulses be transmitted at time t = 0 and t = T, T being true period of light pulses. Let ∆t′ be the interval between the reception of these pulses by receiver in S′. Obviously ∆t′ is the proper time interval T′ between these pulses (since observer or receiver is at rest in frame S′). Since observer continues to be at O′ all the time the distance ∆x′ covered by him in frame S′ during the reception of two pulses is zero. Y′
Y
Y ′′
S′ S X ′′
O ′′
X′
O′
X
O ∆x
Fig. 3
From inverse Lorentz transformation x =
x′ + vt ′ v2 c2
1− we have
∆x =
∆ x ′ + v∆ t ′ 1−
=
or
∆x =
v2 c2
v∆ t ′
since here ∆x′ = 0
v2 1− 2 c vT ′ 1−
since ∆t′ = T′
v2 c2
...(1)
This equation shows that the second pulse has to travel this much distance ∆x more than the first pulse in frame S along x-axis to be able to reach at origin O′ in moving frame S′. Also from inverse Lorentz transformations
t′ + t =
FG vx′ IJ Hc K
1−
2
v2 c2
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Mechanics
∆t ′ + ∆t =
we have
2
∆t ′
2
v2 1− 2 c
v 1− 2 c
since ∆ x′ = 0
T′
∆t =
or
FG v∆ x′ IJ H c K=
1−
...(2)
v2 c2
∆t′ = T′
since
This relation includes both the actual time period T of pulses and the time taken by the second pulse to cover the extra distance ∆ x in frame S. That is
∆x c Substituting values of ∆ x and ∆t from Eqs. (1) and (2) in Eq. (3), we get ∆t = T +
T′ 1−
2
FG ∆ x IJ H cK ...(3)
vT′
= T+
v c2
c 1−
v2 c2
on solving
FG H
T′ 1 − T =
v c
IJ K
v2 1− 2 c
= T′
FG 1 − v IJ H cK FG 1 + v IJ H cK
...(4)
If ν and ν′ be the actual and observed frequencies of light pulses respectively, we have ν =
1 1 and ν′ = T T′
∴ Equation (4) gives
1 1 = ν ν′
i.e.,
ν′ = ν
FG 1 − v IJ H cK FG 1 + v IJ H cK FG1 − v IJ H cK FG1 + v IJ H cK
...(5)
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669
This result is known as Doppler’s relativistic formula for light waves in vacuum. In terms of actual and apparent wavelengths λ and λ′ given by λ =
we get
c c and λ ′ = ν ν′
c c = λ′ λ
FG 1 − v IJ H cK FG 1 + v IJ H cK
λ′ = λ
FG 1 + v IJ H cK FG 1 − v IJ H cK
i.e.,
...(6)
In both relations (5) and (6), v is positive or negative according as the observer is receding form or approaching the source of light, so that during recession the apparent frequency of pulse ν′ decreases and its apparent wavelength λ′ increases and during approach the apparent frequency ν′ increases and the apparent wavelength λ′ decreases. The Doppler’s effect so far considered is the longitudinal Doppler’s effect since the observations are made along the direction of travel of light source. If the observations are made at right angles to the direction of travel of light source, the Doppler’s effect observed is known as transverse Doppler’s effect. This is found usually in the case of atoms. Unlike the classical theory, the theory of relativity predicts transverse Doppler’s effect. For transverse Doppler’s effect the second signal does not take on extra time since here y = y′ and z = z′ due to no relative motion along y and z axes. Therefore, from the results of time dilation, the real and observed periods are related by T′ =
T v2 1− 2 c
...(7)
If v and v′ are the actual and observed frequencies in S and S′ respectively, then T′ =
1 1 and T = ν′ ν
So that equation (7) then gives ν′ = ν 1 −
v2 c2
...(8)
This expression represents transverse Doppler’s Effect. If v << c, the longitudinal and transverse Doppler’s Effect take the form
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Mechanics
ν′ = ν 1 −
v c
(longitudinal Doppler’s Effect)
ν′ = ν (transverse Doppler’s Effect) Then according to the classical theory, there is no transverse Doppler Effect; but according to theory of relativity transverse Doppler’s change in frequency is given by Eq. (8).
15.13 CONFIRMATION OF DOPPLER’S EFFECT The longitudinal Doppler Effect was confirmed by H.E. Lves and G.R. Stilwell in 1941 spectroscopically using beams of hydrogen atoms in excited electronic states. Atomic hydrogen was produced by the result of breaking up accelerated molecular hydrogen ions using electric field. They observed the shift in the average wavelength of a particular spectral line of hydrogen atoms. The shift was observed by taking the average over forward and backward directions relative to the line of flight of atoms. The longitudinal Doppler’s Effect is expressed as
FG 1 + v IJ H cK FG 1 − v IJ H cK
λ′ = λ
...(9)
If v is +ve in forward direction, it is –ve in backward direction, thus if λ1 is wavelength observed in forward direction λ0 the wavelength emitted by atom at rest, then from Eq. (9)
λ1 = λ 0
FG vIJ H cK F vI 1− G J H cK 1+
If λ2 is wavelength observed in the backward direction, we have
λ2 = λ 0
v c v 1+ c 1−
Therefore the average observed wavelength is given by
λ1 + λ2 λ λ = = 0 2 2
=
λ0 1−
v2 c2
LM MM MN
v c + v 1− c 1+
v c v 1+ c 1−
OP PP PQ ...(10)
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671
The shift observed by them was 0.074 × 10–8 cm, while shift calculated from Eq. (10) taking value of
v = 0. 005 is 0.072 × 10–8 cms. c
Thus observed and calculated values very nearly agree, thereby verifying longitudinal Doppler’s Effect. Example 16. Calculate the wavelength shift in the relativistic Doppler Effect for the Hα (6563 Å) line emitted by a star receding from the earth with a relative velocity 0.1c. Is the classical (first order) result a good approximation? Solution. The new wavelength λ′ is given by
v 1 + 0.1 1 + v/ c = 0.1 = 6563 since c 1 − 0.1 1− v/c
λ′ = λ
1.1 = 7256 Å 0. 9
= 6563 ∴ The wavelength shift
λ′ – λ = 7256 – 6563 = 693Å
Now looking for the classical result, we have
FG H
1 + v/ c v = λ 1+ 1− v/c c
λ′ = λ
FG H
= λ 1+
v c
IJ K
IJ FG1 − v IJ K H cK 1 2
−
1 2
in classical limit
= λ (1 + 0.1) or
λ ′ − λ = 0.1 λ = 0.1 × 6563 = 656.3 Å Thus the classical result, in this case, differs with relativistic result by
FG 693 − 656.3 IJ × 100% H 693 K
= 5.3%
This percentage depends on the value of
FG v IJ . H cK
15.14 CONSERVATION OF MOMENTUM: VARIATION OF MASS WITH VELOCITY According to theory of velocity the laws of mechanics as well as those of electromagnetic, must remain unaltered when a transformation from one inertial frame to another frame is affected. According to principle of conservation of momentum. If the net force acting on the body is zero, its linear momentum must remain constant. F =
d (mv) = 0 or mv = constant. dt
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We shall see that if the laws of conservation of momentum and energy are to be retained under Lorentz transformations, the mass m must vary with velocity. Now we shall apply the laws of conservation of momentum and energy to simple hypothetical experiment first revised by Tolman. Consider two systems S and S′, the latter moving with velocity v relative to former along +ve direction of x-axis as shown in figure. Consider two exactly, similar elastic balls A and B placed in system S and S′ respectively. The two balls are thrown with equal and opposite velocities ± u by two observers situated on Y and Y′ axes of systems S and S′ respectively as shown in figure. The masses of the two balls are equal when measured in the same system. Now consider the balls A and B collide due to motion of S′ at the instant when the line of centres has the direction of Y-axis, so that the x-components of velocities do not change as a result of impact. Then according to law of conservation of energy, their total kinetic energy before impact and after impact must be equal. B
Y u
Y
S′
S
u
X
→ v
A
Fig. 4
Now before impact the components of velocity are given according to rule for the transformation of velocities. As observed from system S before impact. uax = 0, uay = u, uay = v and uby = − u 1 −
v2 c2
where uax is the velocity of ball A along x-axis uay is velocity of ball A along y-axis and so on. The total momentum of the two balls along y-axis before impact as observed from S is mauay − mbuby = mau − mbu
FG 1 − v IJ H cK 2
2
After impact each ball reverses its y-component of velocity; therefore the components of velocity of the two balls along y-axis after impact as observed from system S are given by uay = − u and uby = u
RS1 − v UV T cW 2
2
...(1)
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Therefore momentum of the two balls along y-axis after impact as observed from system S is given by
mauay + mbuby = − mau + mbu
FG1 − v IJ H cK 2 2
...(2)
According to principle of conservation of momentum, momentum before impact = moment after impact
mau − mbu 1 −
or
or
v2 v2 = − mau + mbu 1 − 2 2 c c
2mau = 2mbu 1 −
mb = ma
v2 c2
1 1−
v2 c2
According to this relation mass must vary with velocity. Now if u → 0, ma → mo which is known as rest mass and mb → m; the above equation takes the form m =
m0
...(3)
v2 1− 2 c
Thus the mass of a body increases with increase of velocity and m is known as apparent mass or momental mass. The momentum of a body of mass m moving with velocity v can be written as p = mv =
m0v v2 1− 2 c
...(4)
15.15 MASS-ENERGY RELATION Force is defined as rate of change of momentum, i.e., F =
d (mv) dt
...(1)
From the principle of conservation of energy, kinetic energy of a moving body is equal to work done by the force that imports the velocity to the body from rest. Work is defined as force multiplied by distance; therefore we have
z z
v=v
K.E. =
v
F ds =
v=0
0
d (mv) ds dt
674
Mechanics
z z v
=
0
z v
d ds dt = ( mv) dt dt
0
v
d ( mv) dt dt
v
=
vd ( mv) ,
0
According to theory of relativity
m0
m =
1−
z
v=v
∴
K.E. = T =
v=0
v2 c2
LM vd M MM N LM M vM MM N
z
m0v
0
z v
= m0
0
∴
K.E. T = m0 c2
2
1 v2 c2
1−
=
3 /2
2
2
2 2
3 /2
. vdv
2
v dv 2
v c2
H LM MM 1 MN 1 − vc 1−
IJ K
2 2
LM MM MN
2
2
2
2
z FG 0
OP P v /c dv + dvP FG 1 − v IJ P H c K PQ
FG 1 − v IJ + v H cK c FG 1 − v IJ H cK
v
= m0
v2 c2
1−
v
= m0
OP PP PQ
m0 v2 1− 2 c
3 /2
2 2
OP − 1P PP Q
OP .m P c PP Q 0
= m0 c2
LM MM 1 MM FG 1 − v IJ NH c K
2
= (m − m0 ) c2
3/ 2
OP PP PP Q
v
0
...(2)
Relativity
675
This is relativistic equation for kinetic energy. Thus the K.E. of a moving body is equal to gain in mass multiplied by c2. So m0c2 may be considered as rest energy of a body of rest mass m0. This rest energy may be considered as internal store of energy in the body. Then total energy E of a body is the sum of rest energy and kinetic energy of a body. ∴
E = rest energy + kinetic energy = m0c2 + (m – m0)c2 = m0c2 + mc2 – m0c2
∴
E = mc2
...(3)
which is well know Einstein mass-energy relation which states a universal equivalence between mass and energy. This principle has been confirmed a number of times in nuclear physics. Expanding 1st term in Eq. (2) by Binomial theorem, we have
RS T
2 T = m0 c 1 +
If v << c, then kinetic energy, T =
1 v2 3 v4 + + ... −1 2 c2 8 c4
UV W
...(3)
1 m0v2 as in classical mechanics. 2
Hence for non-relativistic cases we may use kinetic energy expression
1 mv2 but for 2
relativistic cases we must use the expression T = ( m − m0 ) c2 Example 17. The rest mass of an electron is 9.0 × 10–28 gm. What will be its mass if it were moving with
4 th the speed of light. 5
Solution. The mass of the electron if it were moving with speed v is given by m =
Here
v =
m0 v2 1− 2 c
, where m0 is the rest mass of the electron
4 c = 0. 8 c 5
m0 = 9 × 10–28 gm = 9 × 10–31 kg.
and ∴
m =
9 × 10−31 1−
=
FG 0. 8 c IJ H c K
2
9 × 10−31 9 × 10−31 = 1 − 0. 64 0. 36
676
Mechanics
=
9 × 10−31 = 15 × 10−31 0. 6
= 1.5 × 10–30 kg. Ans. Example 18. For what value of
b g
v = β will be relativistic mass of a particle exceed its c
rest mass by a given fraction f. Solution. We have f =
m − m0 m = −1 m0 m0
m = m0
But
∴
1 1− 1
f =
−1
v2 1− 2 c
l f (2 + f )q
v = c
Solving for
v2 c2
1+ f
Ans.
Example 19. A certain young lady decides on her twenty fifth birth day that it is time to slendrize. She weights 100 kg. She has heard that if she moves fast enough, she will appear thinner to her stationary friends. (i) How fast must she move to appear sleadrized by a factor of 50%. (ii) At this speed what will her mass appear to be to her stationary friends. (iii) If she maintains her speed until the day she calls her twenty ninth birth day, how old will her stationary friends claim she is according to their measurements? Solution. (i) According to Lorentz-Fitzgerald contraction l = lo
1−
v2 c2
Given that the dimensions of lady’s thickness are reduced by 50% i.e., l = ∴ or
50 lo 100
50 lo = lo 100 1−
1 v2 = 2 c2
1−
v2 c2 ...(1)
Relativity
∴
677
1−
∴
1 v2 3 v2 or = = 4 c2 4 c2 v =
3 c = 0. 866 c 2
(ii) The mass of a body moving with velocity v is given by m =
mo 1−
In this case
v2 c2
m o = 100 kg
1− ∴
1 v2 = from Eq. (1) 2 2 c m =
100 = 200 kg 1/2
(iii) According to time-dilation ∆t′ =
∆t 1−
∴
∆t′ =
2
v c2
; Here ∆t = 4 years,
1−
v2 1 = c2 2
4 = 8 years 1/2
∴ The lady will appear to be (25 + 8) = 33 years old. Example 20. Calculate the energy equivalent to 1 atomic mass unit in million electron volt. Given Avogadro Number = 6 × 1023/gm mol. Solution. We know
E = mc2
Here
m = 1 a.m.u. =
1 1 gm = kg 6 × 1026 6 × 1023
c = 3 × 108 m/sec. ∴
E =
=
F 1 GH 6 × 10
26
I JK
kg × (3 × 108 m / sec)2
9 × 1016 joules 6 × 1026
9 × 1016 ev = 6 × 1026 × 1. 6 × 10−19
= 937 × 106 ev = 937 Mev. Example 21. Calculate the fractional change in the mass of the hydrogen atom when it is ionized from the following data:
678
Mechanics
The binding energy of hydrogen atom = 13.58 eV The rest hydrogen atom = 1.00797 a.m.u. Solution. The change in rest mass ∆m0 =
∆E c2
=
13. 58 ev 13. 58 × 1. 6 × 10−19 Joules = 8 2 (3 × 10 m / sec) (3 × 108 m / sec)2
=
13. 58 × 1. 6 × 10−19 kg ( 3 × 108 )2
=
13.58 × 1.6 × 10−19 a.m.u. ( 3 × 108 )2 × 1. 66 × 10−27
= 1.46 × 10–8 a.m.u. So fractional change in the mass of the hydrogen atom,
∆mo 1. 46 × 10−8 = = 1.45 × 10–8 = 1.45 × 10–6% mo 1. 00797 Example 22. How much mass is lost when 1 kg. of water at 0°C turns to ice at 0°C. Solution. The heat energy lost by water, = (1 kg) × (80 kilo cal/kg) = 80 kilo cal = 80 × 4.2 × 103 joules ∴ Equivalent loss in mass ∆m =
∆E 80 × 4. 2 × 103 kg. = 3. 73 × 10−12 kg. Ans. 2 = 8 2 c (3 × 10 )
Example 23. A body whose specific heat is 0.2 kilo cal/kg°C is heated through 100°C. Find the percentage increase in its mass. Solution. The heat energy gained by body ∆E = ms ∆Q = m × (0.2) × 100 = 20 m kilocal = 20 m × 4.2 × 103 Joule ∴ Equivalent gain in mass,
∴
∆E c2 20 m × 4. 2 × 103 = (3 × 108 )2
∆m =
Percentage increase in mass =
=
∆m × 100 m 20 m × 4. 2 × 103 × 100 = 9. 3 × 10−11% ( 3 × 108 ) × m
Example 24. Find the velocity that an electron must be given so that its momentum is 10 times its rest mass times the speed of light. What is the energy at this speed.
Relativity
Solution. given by
679
The momentum of an electron of rest mass mo moving with velocity v is
p =
mov 1−
Given
mov v2 1− 2 c
v2 c2
= 10 moc
v v2 = 10 1 − 2 c c
or
FG H
(100 + 1)
or
IJ K
v2 v2 v2 100 1 − 2 = 100 − 100 2 = c c c2
Squaring it, we get
v2 = 100 c2 100 v2 = 101 c2
or
v = c
or ∴
100 = 0. 995 101
v = 0.995 c = 0.995 × 3 × 108 m/sec = 2.985 × 108 m/sec.
The mass of the electron at this speed is given by m =
m0 1−
v2 c2
where m = rest mass of electron = 9 × 10–31 kg ∴
mo =
9 × 10−31 9 × 10−31 = 100 1 1− 101 101
= 9 × 10−31 101 = 9 × 10−31 × 10. 04 = 90.36 × 10–31 kg = 9.036 × 10–30 kg ∴ Energy of the electron at speed 0.995 c is E = mc2 = (9.036 × 10–30 kg) (3 × 108 m/sec)2 = 8.13 × 10–13 Joules.
...(1)
680
Mechanics
Example 25. An electron and a positron practically at rest come together and annihilate each other, producing two photons of equal energy. Find the energy and equivalent mass of each photon. Solution. The rest mass of electron = 9 × 10–28 gm = 9 × 10–31 kg = rest mass of positron ∴
The energy of each photon = moc2 = 9 × 10–31 × (3 × 108)2 Joules = 81 × 10–15 Joules =
81 × 10−15 eV = 5 × 105 eV 1. 6 × 10−19
The equivalent mass of each photon = 9 × 10–3 kg. Example 26. Calculate the velocity of electrons accelerated by a potential of 1 millionvolts. Solution. We know, the kinetic energy T = ( m − m0 ) c2
=
Given
F GG GG H
1 1−
2
v c2
I J − 1J m c JJ K o
2
T = 1 mev = 106 eV = 106 × 1.6 × 10–19 Joules. mo = 9 × 10–28 gm = 9 × 10–31 kg.
∴
Solving we get
106 × 1.6 × 10–19 =
v =
LM MM MN
1
OP − 1P × 9 × 10 PP Q
−31
v2 1− 2 c
× (3 × 108 )2
2 2 2 2 c= × 3 × 108 m/sec 3 3
= 2.8 × 108 m/sec.
15.16 RELATION BETWEEN MOMENTUM AND ENERGY Relativistic momentum is written as p = mv =
mo v2 1− 2 c
v
Relativity
681
mo
E = mc2 =
and relativistic energy
1−
c2
2
v c2 2
mo mo2 c4 p2 = v2 E = 2 2 and v v 1− 2 1− 2 c c 2
Squaring
∴
E2 − p2c2 =
mo2 c2 2 mo2 c4 mo2 v2 c2 ( c − v2 ) − = v2 v2 v2 1− 2 1− 2 1− 2 c c c
FG H
mo2 c4 1 − =
1−
v2 c2
v2 c2
IJ K
E2 – p2 c2 = mo2c4.
or
We see that right hand side is invariant, therefore we may say that E2 – p2c2 is invariant under Lorentz transformations.
15.17 TRANSFORMATION OF MOMENTUM AND ENERGY Consider two systems S and S′, S′ moving with velocity v relative to S along +ve direction of x-axis. Now consider in system S a photon on the light sphere having momentum p and energy E. p = mc and E = mc2
We know
p2 = m2c2 and E2 = m2c4 = m2c2.c2
∴ or
E2 = p2c2
or
p2 =
E2 c2
...(1)
If px, py and pz are components of momentum in system S along the three axes, we have p2 = px2 + py2 + pz2
...(2)
Now from Eq. (1) becomes px2 + py2 + pz2 =
E2 c2
...(3)
If E′ and p′ are energy and momentum of photon as viewed from system S′, then we have
px′2 + py′2 + pz′2 =
E′2 c2
...(4)
682
Mechanics
The equation is similar to x2 + y2 + z2 = c2 t2
...(5)
Therefore energy and momentum transform in the same way as space and time. Here x has been replaced by px , y by py , z by pz and t by E/c2. Lorentz transformation for space and time are
x′ =
x − vt v2 1− 2 c
, y′ = y, z ′ = z and t ′ =
vx c2 v2 1− 2 c
t−
Therefore Lorentz transformations for momentum and energy are given by
vE c2 , p′ = p , p′ = p y y z z v2 1− 2 c
px − px′ =
and
or
...(6)
E vpx − 2 E′ c2 c = c2 v2 1− 2 c E′ =
E − vpx v2 1− 2 c
...(7)
Inverse Lorentz transformations for momentum and energy are
px =
and
E =
vE ′ c2 , p = p′ , p = p′ y y z z v2 1− 2 c
px′ +
E′ + vp′ x v2 1− 2 c
...(8)
Example 27. A proton has velocity 0.999c in the laboratory frame. Find the energy momentum as observed in a frame travelling in the same direction with a velocity 0.99c relative to laboratory (mass of the proton = 1.67 × 10–27 kg). Solution. Let the frame S′, in which momentum and energy are required, be moving relative to lab frame S along positive direction of x-axis. Also let the direction of motion of the proton in the lab frame be x-axis. If mo is the rest mass of the proton, the momentum of the proton in the lab. frame is,
Relativity
683
mo ux
p = px =
1−
u2x c2
mo (0. 999 c)
=
F 0.999 c IJ 1− G H c K
2
≈ 22. 4 mo c
The energy of the proton in lab. frame is E = mc2 =
mo c2 u2 1 − 2x c
≈ 22. 4 mo c2
Using transformation equations the energy and momentum of the proton in system S′ are
FG H
0. 99 c × 22. 4 mo c2 uE 22. 4 mo c − c2 c2 = 2 v2 0. 99 c 1− 2 − 1 c c
px − p′ x =
FG H
IJ K
IJ K
≈ 7. 09 × 0. 224 mo c = 1. 59 m0c = 1.59 × 1.67 × 10–27 × 3 × 108 = 7.966 × 10–19 kg.m/sec. and
E′ =
E − vpx 1−
v2 c2
=
22. 4 m0c2 − ( 0. 99 c) (22. 4 m0 c) 1 − ( 0. 99)2
= 7. 09 × 22. 4 × 0. 01 mo c2 = 1. 59 mo c2 = 1.59 × (1.67 × 10–27 kg) (3 × 108 m/sec)2 = 2.39 × 10–10 joules =
2. 39 × 1010 ev 1. 6 × 10−19
= 1.5 × 109 ev = 1.5 BeV Ans. Example 28. Write down the expression for the momentum of a photon. How much is the rest mass of a photon. Solution. The momentum of the photon of mass m is given by p = mc since the velocity of the photon is equal to c. The energy of the photon is given by
...(1)
684
Mechanics
E = mc2 = hv ∴
m =
hv c2
∴ Substituting the value of m in Eq. (1), we get p =
hv hv h .c= = λ c c2
...(2)
But the momentum is given by p =
mo v 1−
v2 c2
p 1− mo =
v2 c2
v
But for a photon v = c, therefore the rest mass of the photon is given by
p 1− mo =
v
c2 c2 = 0
Example 29. Calculate the momentum of a photon whose energy is 1.00 × 10–29 joules. Solution. The momentum of photon is p =
E 1. 00 × 10−19 = 3. 33 × 10−28 kg. m / sec = c 3 × 108
Ans.
APPENDICES APPENDIX A DEFINITION OF STANDARDS AND EQUIVALENTS Standard
Abbreviation
Equivalent
Meter
m
1,650,763.73 wavelengths in vacuo of the unperturbed transition 2p10—5d5 in Kr86
Kilogram
kg
mass of the international kilogram at Sevres, France
Second
sec
9,192,631,770 vibrations of the unperturbed hyperfine transition 4,0—3,0 of the fundamental state 2S1/2 in Cs132 a
Degree Kelvin
°K
defined in the thermodynamic scale by assigning 273.16 °K to the triple point of water
Unified atomic mass unit
amu
Mole
mol
Standard acceleration of free fall Normal atmospheric pressure Thermochemical calorie
1 the mass of an atom of the C12 nuclide 12 amount of substance containing the same number of atoms as 12 gm (exactly) of pure C12
gn
9.80665 meter/sec2
atm
101,325 nt/meter2
cal
4.1840 joules 0.001,000,028 meter3
Liter
l
Inch
in.
0.0254 meter
lb
0.453,592,37 kg
Pound (avdp.) a
There is no measurable difference between this and the previous standard of time, 1/31,556,925.9747 of the tropical year at 12h ET, 0 January 1900. For this reason and because even more accurate maser standards may soon be available, the Cs standard was adopted provisionally rather than “permanently”.
685
686
Mechanics
APPENDIX B FUNDAMENTAL AND DERIVED CONSTANTS Name
Symbol
Computational Value
Best Experimental Valueb
Avogadro constant Elementary charge Permeability constant Permittivity constant Speed of light
N0 e µ0 ∈0 c
6.02 1.60 1.26 8.85 3.00
× 1012/mole × 10– 19 coul × 10– 6 henry/meter × 10– 12 farad/meter × 108 metres/sec
6.02252 ± 0.00028 1.60210 ± 0.00007 4π × 10– 7 exactly 8.85418 ± 0.00002 2.997925 ± 0.000003
Electron rest mass Faraday constant Neutron rest mass Planck constant Proton rest mass
me F mn h mp
9.11 9.65 1.67 6.63 1.67
× 10– 31 kg × 104 coul/mole × 10– 27 kg × 10– 34 joule sec × 10– 27 kg
9.1091 ± 0.0004 9.64870 ± 0.00016 1.67482 ± 0.00008 6.6256 ± 0.0005 1.67252 ± 0.00008
Electron charge/mass ratio Electron Compton wavelength Fine structure constant Proton Compton wavelength
e/me λC α
1.76 × 1011 coul/kg 2.43 × 10– 12 meter 7.30 × 10– 2
1.758796 ± 0.000019 2.42621 ± 0.00006 7.29720 ± 0.00010
λ Cp
1.32 × 10– 15 meter
1.32140 ± 0.00004
– 15
Quantum/charge ratio
h/e
4.14 × 10
Bohr magneton Bohr radius Nuclear magneton Proton magnetic moment Rydberg constant
µB a0 µN µp R∞
9.27 5.29 5.05 1.41 1.10
× 10– 24 joule/teslaa × 10– 11 meter × 10– 27 joule/teslaa × 10– 26 joule/teslaa × 107/meter
9.2732 ± 0.0006 5.29167 ± 0.00007 5.0505 ± 0.0004 1.41049 ± 0.00013 1.0973731 ± 0.0000003
Boltzmann constant First radiation constant π 2 hc2 Second radiation constant hc/k Standard volume of ideal gas Universal gas constant
k c1 c2 — R
1.38 3.74 1.44 2.24 8.31
× 10– 23 joule/°K × 10– 16 watt/meter2 × 10– 2 meter °K × 10– 2 meter3/mole joule/°K mole
1.38054 ± 0.00018 3.7405 ± 0.0003 1.43879 ± 0.00019 2.24136 ± 0.00030 8.3143 ± 0.0012
Gravitational constant Stefan-Boltzmann constant Wien displacement constant
G σ b
6.67 × 10– 11 nt meter2/kg2 5.67 × 10– 8 watt/meter2 °K4 2.90 × 10– 3 meter °K
joule sec/coul
a
Tesla = weber/meter2.
b
Same units and power of ten as the computational value.
4.13556 ± 0.00012
6.670 ± 0.015 5.6697 ± 0.0029 2.8978 ± 0.0004
Appendices
687
APPENDIX C MISCELLANEOUS TERRESTRIAL DATA Standard atmosphere
1.013 × 105 nt/meter2 14.70 lb/in2 760.0 mm-Hg
Density of dry air at STPa
1.293 kg/meter3 2.458 × 10– 3 slug/ft3
Speed of sound in dry air at STP
331.4 meter/sec 1089 ft/sec 742.5 miles/hr
Acceleration of gravity, g (standard value)b
9.80665 meters/sec2 32.1740 ft/sec2
Solar constantc
1340 watts/m2 1.92 cal/cm2-min
Mean total solar radiation
3.92 × 1026 watts
Equatorial radius of earth
6.378 × 106 meters 3963 miles
Polar radius of earth
6.357 × 10 6 meters 3950 miles
Volume of earth
1.087 × 1021 meter3 3.838 × 1022 ft3
Radius of sphere having same volume
6.371 × 10 6 meters 3959 miles 2.090 × 107 ft
Mean density of earth
5522 kg/meter3
Mass of earth
5.983 × 1024 kg
Mean orbital speed of earth
29,770 meters/sec 18.50 miles/sec
Mean angular speed of rotation of earth
7.29 × 10– – 5
5
radians/sec
teslad
Earth’s magnetic field, B (at Washington, DC)
5.7 × 10
Earth’s magnetic dipole moment
6.4 × 1021 amp-m2
a
STP = standard temperature and pressure = 0°C and 1 atm.
b
This value, used for barometer corrections, legal weights, etc., was adopted by the International Committee on Weights and Measures in 1901. It approximates 45° latitude at sea level. c
The solar constant is the solar energy falling per unit time at normal incidence on unit area of the earth’s surface. d
Tesla ≡ weber/meter2.
688
APPENDIX D
THE SOLAR SYSTEMa Planet Mean diameter km Earth diameters Volume (earth volumes)
Mercury
Venus
Earth Earth
,
Mars
Jupiter
Saturn
Uranus, Uranus
Neptune
Pluto
5,000
12,400
12,742
6,870
139,760
115,100
51,000
50,000
12,700?
0.39
0.973
1.000
0.532
10.97
9.03
4.00
3.90
0.46
0.06
0.92
1.00
0.15
1,318
736
64
39
0.10
Mass (earth masses)
0.04
0.82
1.00
0.11
318.3
95.3
14.7
17.3
1.0?
Density (earth densities)
0.69
0.89
1.00
0.70
0.24
0.13
0.23
0.29
?
3.8
4.86
5.52
3.96
1.33
0.71
1.26
1.6
?
0.27
0.86
1.00
0.37
2.64
1.17
0.92
1.44
?
3.6
10.2
11.2
5.0
60
36
21
23
11?
58.6d
30d?
1d
1 d37m23s
9h55m
10h38m
10.7h
15.8h
?
Mean density gm/cm3 Surface gravity (earth’s) Velocity of escape, km/sec Length of day (earth days)
Mechanics
Appendices
Planet Planet
Mercury Mercury
Period, siderial days
87.97
224.70
—
Inclination of equator to orbit
Venus Venus
Earth , Earth
Mars Mars
Jupiter Jupiter
Saturn b Uranus Uranus, Saturn
Nepture Neptune
Pluto Pluto
365.26
686.98
4,332.59
10,759.20
30,685.93
60,187.64
90,885
0° ?
23°27′
25°12′
3°7′
26°45′
98.0°
29°
?
Oblateness
0.00
0.00
1/296
1/192
1/15.4
1/9.5
1/14
1/45
?
Atmosphere, main constituents
none
N2, CO2, A
N2, O2
N2, CO2, H2O
CH4, NH3
CH4, NH3
CH4, NH3
CH4, NH3
none
Maximum surface temperature, °K
700
700
350
320
153
138
110?
90?
80?
Distance from Sun, 106 km
58
108
149
228
778
1426
2869
4495
5900
The Sun
329,390 earth masses, mean density 1.42, mean diameter 1,390,600 km, surface gravity 28 (earth’s).
The Moon 0.01228 earth masses, mean density 3.36, mean diameter 3,476 km, surface gravity 0.17 (earth’s), distance from earth 38 × 104 km. a
Adapted from Payne-Gaposchkin and Handbook of Chemistry and Physics.
689
690
APPENDIX E
PERIODIC TABLE OF THE ELEMENTS Atomic weights are expressed in atomic mass units (amu), one atom of the isotope C12 being defined to have a mass of (exactly) 12 amu. For unstable elements the mass number of the most stable or best known isotope is given in brackets. Group → Period Series
I
II
III
IV
V
VI
1
1
1H 1. 00797
2
2
3 Li 6. 939
4 Be 9. 0122
5B 10. 811
6C 7N 12. 01115 14. 0067
8O 15. 9994
3
3
11 Na 22. 9898
12 Mg 24. 312
13 Al 26. 9815
14 Si 28. 086
15 P 30. 9738
4
19 K 39. 102
20 Ca 40. 08
21 Sc 44. 956
22 Ti 47. 90
5
29 Cu 63. 54
30 Zn 65. 37
31 Ga 69. 72
6
37 Rb 85. 47
38 Sr 87. 62
39 Y 88. 905
7
47 Ag 107. 870
48 Cd 112. 40
55 Cs 132. 905
56 Ba 137. 34
49 In 50 Sn 114. 82 118. 69 57 − 71 Lanthanide 72 Hf 178. 49 seriesa
9
79 Au 196. 967
80 Hg 200. 59
81 Tl 204. 37
10
87 Fr 223
88 Ra 226
89 Actinide seriesb
VII
VIII
9F 18. 9984
10 Ne 20.183
32. 064
17 Cl 35. 453
18 A 39. 948
23 V 50. 942
24 Cr 51. 996
25 Mn 54. 9380
32 Ge 72. 59
33 As 74. 9216
34 Se 78. 96
35 Br 79. 909
40 Zr 91. 22
41 Nb 92. 906
42 Mo 95. 94
43 Tc 99
51 Sb 121. 75
52 Te 127. 60
53 I 126. 9044
73 Ta 180. 948
74 W 183. 85
75 Re 186. 2
83 Bi 208. 980
84 Po 210
85 At 210
16 S
4
5
8 6
7
0
2 He 4. 0026
82 Pb 207.19
26 Fe
27 Co 28 Ni 55. 847 58. 9332 58. 71 36 Kr 83. 80 44 Ru 45 Rh 46 Pd 101. 07 102. 905 106. 4 54 Xe 131. 30 76 Os 77 Ir 78 Pt 190. 2 192. 2 195. 09 86 Rn 222
b
Actinide series:
89 Ac 90 Th 91 Pa 227 232. 038 231
92 U 93 Np 238. 04 237
94 Pu 242
95 Am 96 Cm 243 245
97 Bk 249
98 Cf 249
99 Es 100 Fm 101 Md 254 252 256
102 254
103 Lw 257
Mechanics
a Lanthanide 57 La 58 Ce 59 Pr 60 Nd 61 Pm 62 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu series : 138. 91 140.12 140. 907 144. 24 145 150. 35 151. 96 157. 25 158. 924 162. 50 164. 930 167. 26 168. 934 173. 04 174. 97
Appendices
APPENDIX F
THE PARTICLES OF PHYSICSa Family name
Particle name
Symbol
Mass
Spin
Strangenessb
Charge
Photon
γ (gamma
0
1
0
0
ray) Electron
Antiparticlec
No. of particles
Averaged lifetime, seconds
Typical decay products
Same
1
Infinite
—
2
Infinite
—
2
Infinite
—
2
2.212 × 10– 6
e– + ve + vµ
2
Infinite
—
3
2.55 × 10– 8
µ+ + v µ
1.9 × 10– 16
γ +γ
particle
Electron
e–
1
—
– e
e+
Electron’s
ve
0
—
0
Muon
µ–
206.77
—
–e
Muon’s
vµ
0(?)
—
0
π+
273.2
0
0
+e
π–
π0
264.2
0
0
0
π0
family
neutrino Muon
µ+
family
neutrino Mesons
Pion
Kaon
K
+
K0
966.6
0
+ 1
+e
974
0
+ 1
0
4
– 8
1.22 × 10
1.00 × 10– 10 and 6 × 10– 8
π+ + π0 π+ + π– (Cont...)
691
692
Family name
Particle name
Symbol
Mass
p+ (proton)
1836.12
n0 (neutron)
1838.65
Λ0
2182.8
Sigma
Σ+
2327.7
particle
Σ–
2340.5
Σ0
2332
Ξ–
2580
Ξ0
2570
Baryons Nucleon
Lambda
Spin
1 2 1 2 1 2
Strangenessb
Charge
Antiparticlec
No. of particles
Averaged lifetime, seconds
0
+e
p+
4
Infinite
0
0
n0
– 1
0
Λ0
– 1
+e
Σ+
– 1
– e
Σ−
– 1
0
Σ0
– 2
– e
Ξ−
– 2
0
Ξ0
2
Typical decay products
1013
p + e− + ve
2.51 × 10– 10
p + π–
8.1 ×10– 11
n + π+
1.6 × 10– 10
n + π–
about 10– 20
Λ0 + γ
1.3 × 10– 10
Λ0 + π–
about 10– 10
Λ0 + π0
particle
Xi particle
1 2 1 2 1 2 1 2 1 2
6
4
a
Adapted and modified from The World of Elementary Particles, by Kenneth W. Ford.
b
This is a “quantum number” whose assignment permits an understanding of the inter-relationships of the particles.
c
Antiparticles have the same mass and spin as the particles but their charges and strangeness numbers are opposite in sign.
d
The K0 meson has two different Lifetimes; all other particles have only one.
Mechanics
Appendices
693
APPENDIX G
SYMBOLS, DIMENSIONS, AND UNITS FOR PHYSICAL QUANTITIES All units and dimensions are in the mksq (rationalized) system. The primary units can be found by reading kilograms for M, meters for L, seconds for T, and coulombs for Q. The symbols are those used in the text. In practice, Q is defined in terms of M, L and T. However, the addition of Q to the traditional M, L, and T enables us to avoid the use of fractional exponents in dimensional considerations. The term ‘rationalized’ simply means that a factor 1/4π is separated out of Coulomb’s law in order to remove the factor 4π that would otherwise appear in many other formulas in electricity. Quantity
Symbol
Dimensions
Derived Units
Acceleration
a
LT – 2
meters/sec2
Angular acceleration
α
T
–2
radians/sec2
Angular displacement
θ
—
Angular frequency and speed
ω
T–1
Angular momentum
L
2
ML T
Angular velocity
ω
T–1
radians/sec
Area
A, S
L2
meter2
Displacement
r, d
L
meter
Energy, total
E
ML2T– 2
joule
K
2
–2
joule
2
–2
joule
kinetic potential Force
radian radians/sec
ML T
–1
U
ML T
F
MLT – 2
newton hertz ≡ cycles/sec
–1
Frequency
v
T
Gravitational field strength
g
LT – 2 2
kg-m2/sec
nt/kg
–2
Gravitational potential
V
LT
Length
l
L
Mass
m
M
Mass density
ρ
ML
Momentum
p
MLT – 1
kg-m/sec2
Period
T
T
second
joules/kg meter kilogram
–3
2
kg/m3
–3
Power
P
ML T
Pressure
p
ML– 1T – 2
watt nt/m2
Rotational inertia
I
ML2
kg-m2
Time
t
T
Torque
τ
ML T
Velocity
v
LT
–1
Volume
V
L3
meter3
Wavelength
λ
L
meter
second 2
–2
nt-m meters/sec
(Cont...)
694
Mechanics W
ML2T – 2
Entropy
S
2
ML T
Internal energy
U
ML2T – 2
joule
Heat
Q
2
ML T
joule
Work
joule
–2
joules/K°
–2
Temperature
T
—
Kelvin degree
Capacitance
C
M – 1L– 2T 2 Q2
farad
Charge
q
Q
coulomb
Conductivity
σ
M– 1L– 3TQ2
(ohm-meter)– 1
Current
i
T– 1Q
ampere
Current density
j
–2
L T
Electric dipole moment
p
LQ
–1
amp/meter2
Q
coul-meter
D
–2
M Q
coul/meter2
Electric polarization
P
–2
M Q
coul-meter2
Electric field strength
E
MLT – 2 Q –1
Electric flux
ΦE
ML T
Electric potential
V
ML2T – 2 Q – 1
Electromotive force
ε
ML T
Inductance
L
ML2 Q – 2
Magnetic dipole moment
µ
LT
Magnetic field strength
H
MT – 1 Q
amp-meter
magnetic flux
ΦB
ML2T – 1 Q– 1
weber = volt-sec
Electric displacement
Magnetic induction
B
3
–2
2
2
–2
–1
MT Q L
T
Q
–1
volt volts amp-meter2
tesla ≡ webers/meter2
–1
–1
volt-meter
henry
Q
–1
–1
Q
volts/meter
–1
Magnetization
M
Permeability
µ
Permittivity
∈
M L T Q
farads/meter
Resistance
R
ML2 T – 1 Q – 2
ohm
Resistivity
ρ
ML T
Voltage
V
ML2 T – 2 Q – 1
MLQ
Q
–1 – 3
3
amp/meter
–2
–1
henrys/meter 2
Q
2
–2
ohm-meter volt
Appendices
695
APPENDIX H
CONVERSION FACTORSa Conversion factors for common and not-so-common units may be read off directly from the tables below. For example, 1 degree = 2.778 × 10– 3 revolutions, so 16.7° = 16.7 × 2.778 × 10– 3 rev. The mks quantities are capitalized in each table. PLANE ANGLE ′
° 1 degree =
1
60
1.667 × 10–
1 minute = 1 second =
2.778 × 10
1 radian = 1 revolution =
″
2
– 4
rev – 2
3600
1.745 × 10
60
2.909 × 10–
1 1.667 × 10
radian
– 2
4
– 6
1
4.848 × 10 5
2.778 × 10–
3
4.630 × 10–
5
6
7.71 × 10
57.30
3438
2.063 × 10
1
0.1592
360
2.16 × 104
1.296 × 106
6.283
1
1 rev = 2π radians = 360°
–7
1° = 60′ = 3600″
SOLID ANGLE 1 sphere = 4 π steradians = 12.57 steradians a Adapted in part from G. Shortley and D. Williams, Elements of Physics, Prentice-Hall, Englewood Cliffs, New Jersey, Second Edition, 1955.
LENGTH cm 1 centimeter =
meter
1
10
–2
km 10
–5
in.
ft
mile
0.3937
3.281 × 10
1 Meter =
100
10– 3
1
39.37
–2
3.281
6.214 × 10– 6 6.214 × 10– 4
105
1 kilometer =
1000
1
3.937
3281
0.6214
8.333
1.578
× 104 1 inch =
2.540
2.540 × 10
1 foot =
30.48
–2
0.3048
2.540 × 10
3.048 × 10
1 statute mile =
1.609
1609
1.609
× 10
1 X-unit 10
– 13
1 micron = 10
10
meter
meter
– 6
meter – 9
1 millicron (mµ) = 10
× 10 12
–2
1
× 10– 5 1.894
–4
5
1 angstrom (A) = 10–
1
–5
× 10– 4 6.336 × 10
5280
1
4
1 light-year = 9.4600 × 1012 km
1 yard = 3 ft
1 parsec = 3.084 × 1013 km
1 rod = 16.5 ft
1 fathom = 6 ft
1 mil = 10– 3 in.
meter
1 nautical mile = 1852 meters = 1.1508 statute miles = 6076.10 ft.
696
Mechanics AREA 2
cm2
meter 1 square meter =
1
10
–4
1 square centimeter =
10
4
9.290 × 10–
2
929.0
1 square inch =
6.452 × 10–
4
6.452
5.067 × 10
– 10
1 barn = 10
– 3
1.076 × 10 1
6.944 × 10– – 6
5.067 × 10
1 square mile = 27,878,400 ft2 = 640 acres – 28
10.76
1
1 square foot = 1 circular mill =
ft2
3
– 9
5.454 × 10
in.2
circ mil
1550
1.974 × 109
0.1550
1.974 × 105
144
1.833 × 108 1.273 × 106
1 7.854 × 10
– 7
1
1 acre = 43,560 ft2
2
meter
VOLUME 3
Meter
cm3
1
1
106
1000
1 cubic meter = 1 cubic centimeter = 1 liter =
10
–6 3
– 2
1 cubic foot =
2.832 × 10
1 cubic inch =
1.639 × 10–
5
1.000 × 10
1000 2.832 × 10 16.39
6.102 × 104
3.531 × 10
– 5
3.531 × 10–
1 4
in.3
35.31 – 3
1
1.000 × 10–
ft3
28.32
6.102 × 10–
2
1
1.639 × 10–
2
2
61.02 1728
5.787 × 10–
4
1
1 U.S. fluid gallon = 4 U.S. fluid quarts = 8 U.S. pints = 128 U.S. fluid ounces = 231 in.3 1 British imperial gallon = the volume of 10 lb of water at 62 ° F = 277.42 in.3 1 liter = the volume of 1 kg of water at its maximum density = 1000.028 cm3 MASS Note: Those quantities to the right of and below the heavy lines are not mass units at all but are often used as such. When we write, for example, 1 kg “ = ” 2.205 lb this means that a kilogram is a mass that weighs 2.205 pounds. Clearly this “equivalence” is approximate (depending on the value of g) and is meaningful only for terrestrial measurements. Thus, care must be employed when using the factors in the shaded portion of the table. gm
kg
slug
amu
oz
lb
ton
1
0.001
6.852 × 10– 5
6.024 × 1023
3.527 × 10– 2
2.205 × 10– 3
1.102 × 10– 5
1 kilogram =
1000
1
6.852 × 10– 2
6.024 × 1026
35.27
2.205
1.102 × 10– 3
1 slug =
1.459 × 104
14.59
1
8.789 × 1027
514.8
32.17
1.609 × 10– 2
1 amu =
1.660 × 10– 24
1.660 × 10– 27
1.137 × 10– 28
1
5.855 × 10– 26
3.660 × 10– 27
1.829 × 10– 30
1 ounce (avoirdupois) =
28.35
2.835 × 10– 2
1.943 × 10 – 3
1.708 × 1025
1
6.250 × 10– 2
3.125 × 10– 5
1 pound (avoirdupois) =
453.6
0.4536
3.108 × 10– 2
2.732 × 1026
16
1
0.0005
1 ton =
9.072 × 10+ 5
907.2
62.16
5.465 × 1029
3.2 × 104
2000
1
1 gram =
Appendices
697 DENSITY
Note: Those quantities to the right or below the heavy line are weight densities and, as such, are dimensionally different from mass densities. Care must be used. (See note for mass table.) slug/ft3
kg/meter3
gm/cm3
lb/ft3
1
515.4
0.5154
32.17
1
0.001
1000
1
1 slug per ft3 =
lb/in.3 1.862 × 10–
2
3.613 × 10–
5
3.613 × 10–
2
1 kilogram per meter3 =
1.940 × 10–
1 gram per cm3 = 3
1 pound per ft = 3
1 pound per in. =
3
1.940 3.108 × 10
– 2
16.02
53.71
6.243 × 10– 62.43
1.602 × 10
2.768 × 10
4
2
– 2
– 4
1
27.68
5.787 × 10
1728
1
TIME yr 1 year =
day
1
365.2
2.738 × 10–
1 day = 1 hour = 1 minute = 1 SECOND =
hr
3
1.141 × 10
– 4
1.901 × 10
– 6
3.169 × 10
– 8
min
8.766 × 10
1
3
5.259 × 10
24
4.167 × 10
– 2
6.944 × 10
– 4
1.157 × 10
– 5
1.667 × 10 2.778 × 10
– 4
3.156 × 107
1440
8.640 × 104
60
3600
1 – 2
sec 5
1
60
1.667 × 10
– 2
1
SPEED ft/sec
km/hr
meter/sec
miles/hr
cm/sec
knot
1
1.097
0.3048
0.6818
30.48
0.5925
0.9113
1
0.2778
0.6214
27.78
0.5400
3.281
3.6
1
2.237
100
1.944
1.467
1.609
0.4470
1
44.70
0.8689
3.6 × 10– 2
0.01
1.852
0.5144
1 foot per second = 1 kilometer per hour = 1 meter per second = 1 mile per hour = 1 centimeter per second = 1 knot =
3.281 × 10– 1.688
2
1 knot = 1 nautical mile/hr
2.237 × 10–
2
1
1.151
1.944 × 10–
51.44
2
1
1 mile/min = 88 ft/sec = 60 miles/hr FORCE
Note: Those quantities to the right of and below the heavy lines are not force units at all but are often used as such, especially in chemistry. For instance, if we write 1 gram-force = 980.7 dynes, we mean that a gram-mass experiences a force of 980.7 dynes in the earth’s gravitational field. Thus, care must be employed when using the factors in the shaded portion of the table. dyne
nt
lb
pdl
gf
kgf
698
Mechanics dyne
1 dyne =
nt
lb
–5
1
10
2.248 × 10
5
1 newton =
10
1 pound =
4.448
–6
pdl
gf
7.233
1.020
× 10
–5
kgf 1.020
–3
× 10
× 10 – 6
1
0.2248
7.233
102.0
0.1020
4.448
1
32.17
453.6
0.4536
0.1383
3.108
1
14.10
1.410
5
× 10 1 poundal =
1.383 4
× 10 1 gram-force =
980.7
× 10 9.807
9.807
× 10– 2
2.205
–3
1 kilogram-force =
–2
7.093
–3
× 10
× 10
9.807
2.205
× 10
1
0.001
1000
1
–2
70.93
× 105 1 kgf = 9.80665 nt
1 lb = 32.17398 pdl
PRESSURE atm
dyne/cm2
inch of water
cm Hg
nt/ meter2
lb/in.2
lb/ft2
1
1.013 × 106
406.8
.76
1.013 × 105
14.70
2116
1 dyne per cm2 =
9.869 × 10– 7
1
4.015 × 10– 4
7.501 × 10– 5
0.1
1.450 × 10– 5
2.089 × 10– 3
1 inch of water at 4°Ca =
2.458 × 10– 3
2491
1
0.1868
249.1
3.613 × 10– 2
5.202
1 centimeter of 1.316 mercury at 0°Ca = × 10– 2
1.333 × 104
5.353
1
1333
0.1934
27.85
1 atmosphere =
1 newton per meter2 =
9.869 × 10– 6
10
4.015 × 10– 3
7.501 × 10– 4
1
1.450 × 10– 4
2.089 × 10– 2
1 pound per in.2 =
6.805 × 10– 2
6.895 × 104
27.68
5.171
6.895 × 103
1
144
1 pound per ft2 =
4.725 × 10– 4
478.8
0.1922
3.591 × 10– 2
47.88
6.944 × 10– 3
1
The electron volt (ev) is the kinetic energy an electron gains from being accelerated through the potential difference of one volt in an electric field. The Mev is the kinetic energy it gains from being accelerated through a million-volt potential difference. The last two items in this table are not properly energy units but are included for convenience. They arise from the relativistic mass-energy equivalence formula E = mc2 and represent the energy released if a kilogram or atomic mass unit (amu) is destroyed completely.
Appendices
ENERGY, WORK, HEAT
Again, care should be used when employing this table. Btu
erg
ft-lb
hp-hr
JOULES
cal
kw-hr
ev
Mev
kg
amu
1
1.055 × 1010
777.9
3.929 × 10– 4
1055
252.0
2.930 × 10– 4
6.585 × 1021
6.585 × 1015
1,174 × 10– 14
7.074 × 1012
1 erg =
9.481 × 10– 11
1
7.376 × 10– 8
3.725 × 10– 14
10– 7
2.389 × 10– 8
2.778 × 10– 14
6.242 × 1011
6.242 × 105
1.113 × 10– 24
670.5
1 foot-pound =
1.285 × 10– 3
1.356 × 107
1
5.051 × 10– 7
1.356
0.3239
3.766 × 10– 7
8.464 × 1018
8.464 × 1012
1.509 × 10– 17
9.092 × 109
1 horsepowerhour =
2545
2.685 × 1013
1.980 × 106
1
2.685 × 106
6.414 × 105
0.7457
1.676 × 1025
1.676 × 1019
2.988 × 10– 11
1.800 × 1016
1 Joule =
9.481 × 10– 4
107
0.7376
3.725 × 10– 7
1
0.2389
2.778 × 10– 7
6.242 × 1018
6.242 × 1012
1.113 × 10– 17
6.705 × 109
1 calorie =
3.968 × 10– 3
4.186 × 107
3.087
1.559 × 10– 6
4.186
1
1.163 × 10– 6
2.613 × 1019
2.613 × 1013
4.659 × 10– 17
2.807 × 1010
1 kilowatthour =
3413
3.6 × 1013
2.655 × 106
1.341
3.6 × 106
8.601 × 105
1
2.247 × 1025
2.270 × 1019
4.007 × 10– 11
2.414 × 1016
1.519 × 10– 22
1.602 × 10– 12
1.182 × 10– 19
5.967 × 10– 26
1.602 × 10– 19
3.827 × 10– 20
4.450 × 10– 26
1
10– 6
1.783 × 10– 36
1.074 × 10– 9
1 million 1.519 electron volts = × 10– 16
1.602 × 10– 6
1.182 × 10– 13
5.967 × 10– 20
1.602 × 10– 13
3.827 × 10– 14
4.450 × 10– 20
106
1
1.783 × 10– 30
1.074 × 10– 3
1
1 British thermal unit =
1 electron volt =
1 kilogram = 1 atomic mass unit =
8.521
8.987
6.629
3.348
8.987
2.147
2.497
5.610
5.610
× 1013
× 1023
× 1016
× 1010
× 1016
× 1016
× 1010
× 1035
× 1029
1.415 × 10– 13
1.492 × 10– 3
1.100 × 10– 10
5.558 × 10– 17
1.492 × 10– 10
3.564 × 10– 11
4.145 × 10– 17
9.31 × 108
931.0
1 m-kgf = 9.807 joules
1 watt-sec = 1 joule = 1 nt-m
6.025 × 1026
1.660 × 10– 27
1
1 cm-dyne = 1 erg 699
700
Mechanics POWER Btu/hr
1 British thermal unit per hour =
ft-lb/min ft-lb/sec
hp
cal/sec
kw
WATTS
1
12.97
0.2161
3.929 × 10– 4
7.000 × 10– 2
2.930 × 10– 4
0.2930
1 foot-pound per minute =
7.713 × 10– 2
1
1.667 × 10– 2
3.030 × 10– 5
5.399 × 10– 3
2.260 × 10– 5
2.260 × 10– 2
1 foot-pound per second =
4.628
60
1
1.818 × 10– 3
0.3239
1.356 × 10– 3
1.356
1 horsepower =
2545
3.3 × 104
550
1
178.2
0.7457
745.7
1 calorie per second =
14.29
1.852 × 102
3.087
5.613 × 10– 3
1
4.186 × 10– 3
4.186
1 kilowatt =
3413
4.425 × 104
737.6
1.341
238.9
1
1000
1 WATT =
3.413
44.25
0.7376
1.341 × 10– 3
0.2389
0.001
1
ELECTRIC CHARGE abcoul
amp-hr
coul
faraday
1
2.778 × 10– 3
10
1.036 × 10– 4
2.998 × 1010
– 2
3.730 × 10
1.079 × 1013
1.036 × 10– 5 1
2.998 × 109 2.893 × 1014
1 abcoulomb (1 emu) = 1 ampere-hour =
360
1
3600
1 coulomb = 1 faraday =
0.1 9652
2.778 × 10– 4 26.81
1 9.652 × 104
1 statcoulomb (1 esu) =
3.336 × 10– 11
statcoul
9.266 × 10– 14 3.336 × 10– 10 3.456 × 10– 15 – 19
1 electronic charge = 1.602 × 10
coulomb
ELECTRIC CURRENT
1 abampere (1 emu) = 1 ampere = 1 statampere (1 esu) =
abamp
amp
statamp
1
10
2.998 × 1010
0.1
1
2.998 × 109
3.336 × 10– 11
3.336 × 10– 10
1
ELECTRIC POTENTIAL, ELECTROMOTIVE FORCE
1 abvolt (1 emu) = 1 volt = 1 statvolt (1 esu) =
abv
volts
statv
1
10– 8
3.336 × 10– 11
10
1
3.336 × 10– 3
2.998 × 1010
299.8
1
8
1
Appendices
701 ELECTRIC RESISTANCE abohm
ohms
statohm
1
10– 9
1.113 × 10– 21
10
1
1.113 × 10– 12
8.987 × 1020
8.987 × 1011
1
1 abohm (1 emu) =
9
1 ohm = 1 statohm (1 esu) =
ELECTRIC RESISTIVITY abohm-cm µohm-cm
ohm-cm
statohm-cm
ohm-m
ohm-circ mil/ft
1
0.001
10– 9
1.113 × 10– 21
10– 11
6.015 × 10– 3
1000
1
10– 6
1.113 × 10– 18
10– 8
6.015
109
106
1
1.113 × 10– 12
0.01
6.015 × 106
8.987 × 1020
8.987 × 1017
8.987 × 1011
1
8.987 × 109
5.406 × 1018
1 ohm-meter =
1011
108
100
1.113 × 10– 10
1
6.015 × 108
1 ohm-circular mil per foot =
166.2
0.1662
1.662 × 10– 7
1.850 × 10– 19
1.662 × 10– 9
1
1 abohm-centimeter (1 emu) = 1 micro-ohmcentimeter = 1 ohm-centimeter = 1 statohm-centimeter (1 esu) =
CAPACITANCE abf 1 abfarad (1 emu) =
9
1
10
1 farad =
10– 9
1
1 microfarad =
10– 15
1 statfarad (1 esu) =
µf a
farads
10– 6
1.113 × 10
– 21
statf
15
10
8.987 × 1020
106
8.987 × 1011 8.987 × 105
1 – 12
1.113 × 10
– 6
1.113 × 10
1
a
This unit is frequently abbreviated as mf. INDUCTANCE
1 abhenry (1 emu) = 1 henry =
abhenry
henrys
µh
mh
stathenry
1
10– 9
0.001
10– 6
1.113 × 10– 21
109
1
106
1000
1.113 × 10– 12
1
0.001
1.113 × 10– 18
1 microhenry =
1000
1 millihenry =
106
–6
10
0.001 20
1 stathenry (1 esu) = 8.987 × 10
1000 11
8.987 × 10
1.113 × 10– 15
1 17
8.987 × 10
14
8.987 × 10
1
702
Mechanics MAGNETIC FLUX maxwell
kiloline
weber
1
0.001
10– 8
1000
1
10– 5
108
105
1
1 maxwell (1 line or 1 emu) = 1 kiloline = 1 weber =
1 esu = 299.8 webers MAGNETIC INDUCTION B
1 gauss (line per cm2) = 2
1 kiloline per in. = 1 weber per meter2 = 1 tesla = 1 milligauss =
gauss
kiloline/in.2
weber/ meter2 = tesla
1
6.452 × 10– 3
10– 4
1 gamma =
1.550 × 10
1.550 × 107
1
107
109
6.452 × 10– 6
10– 7
1
100
– 8
–9
0.01
1
1
104
64.52
0.001 10
10 5
1000 – 2
155.0
–5
milligauss
5
1.550 × 10
6.452 × 10
10
6
1 esu = 2.998 × 10 webers/meter
2
MAGNETOMOTIVE FORCE abamp-turn
amp-turn
gilbert
1
10
12.57
1
1.257
1 abampere-turn = 1 ampere-turn =
0.1 – 2
1 gilbert =
7.958 × 10
0.7958
1 – 11
1 pragilbert = 4π amp-turn
1 esu = 2.655 × 10
amp-turn
MAGNETIC FIELD STRENGTH, H abamp-turn/cm
ampturn/cm
amp-turn/in.
ampturn/ meter
oersted
1
10
25.40
1000
12.57
1
2.540
100
1.257
0.3937
1
39.37
0.4947
0.01
2.540 × 10– 2
1
1.257 × 10– 2
0.7958
2.021
79.58
1
1 abampere-turn per centimeter = 1 ampere-turn per centimeter =
0.1
1 ampere-turn per inch = 1 ampere-turn per meter = 1 oersted =
– 2
3.937 × 10 0.001
– 2
7.958 × 10 1 oersted = 1 gilbert
– 9
1 esu = 2.655 × 10
1 praoersted = 4π amp-turn/meter
amp-turn meter
Appendices
703
APPENDIX I
MATHEMATICAL SYMBOLS AND THE GREEK ALPHABET Mathematical signs and symbols = equals ≅ equals approximately ≠ is not equal to
≡ is identical to, is defined as > is greater than (>> is much greater than) < is less than (<< is much less than) ≥ is more than or equal to (or, is no less than) ≤ is less than or equal to (or, is no more than)
± plus or minus
ee.g.,
j
4 =±2
∝ is proportional to (e.g., Hooke’s law : F ∝ x, or F = –kx) Σ the sum of x the average value of x
The Greek alphabet Alpha
Α
α
Nu
Ν
ν
Beta
Β
β
Xi
Ξ
ξ
Gamma
Γ
γ
Omicron
Ο
ο
Delta
∆
δ
Pi
Π
π
Epsilon
Ε
ε
Rho
Ρ
ρ
Zeta
Ζ
ζ
Sigma
Σ
σ
Eta
Η
η
Tau
Τ
τ
Theta
θ
θ, ϑ
Upsilon
ϒ
υ
Iota
Ι
ι
Phi
Φ
φ, ϕ
Kappa
Κ
κ
Chi
Χ
χ
Lambda
Λ
λ
Psi
Ψ
ψ
Mu
Μ
µ
Omega
Ω
ω
704
Mechanics
APPENDIX J
MATHEMATICAL FORMULAS
Quadratic formula If
ax2 + bx + c = 0, then x =
− b ± b2 − 4 ac . 2a
Trigonometric functions of angle θ y sin θ = r tan θ =
y x
r sec θ = x
Y
x cos θ = r cot θ =
x
x y
r θ 0
r csc θ = y
Fig. App. I
Pythogorean theorem x2 + y2 = r2 Trigonometric identities sin2 θ + cos2 θ = 1 sec2 θ – tan2 θ = 1
csc 2 θ – cot2 θ = 1
sin (α ± β) = sin α cos β ± cos α sin β cos (α ± β) = cos α cos β + – sin α sin β tan (α ± β) =
tan α ± tan β 1 m tan α tan β
sin 2 θ = 2 sin θ cos θ cos 2 θ = cos2 θ – sin2 θ = 2 cos2 θ – 1 = 1 – 2 sin2 θ sin θ =
eiq + e − iq eiq − e − iq cos q = 2i 2
e±iθ = cos θ ± i sin θ
Taylor’s series f(x0 + x) = f(x0) + f'(x0)x + f "(x0)
x2 x2 + f "'( x0 ) + .... 2! 3!
Series expansions (these expansions converge for – 1 < x < 1, 1 = 1 – x + x2 – x3 + .... 1+ x 1+ x = 1+
x x2 x3 − + − .... 2 8 16
y X
Appendices
705 1 1+ x
= 1−
x 3 x2 5 x3 + − + .... 2 8 16
ex = 1 + x +
R| sin x || x in radiansS cos x || ||T tan x
x2 x3 + + .... 2 6
(– ∞ < x < ∞)
= x−
x3 x5 + − .... 6 120
(– ∞ < x < ∞)
= 1−
x2 x4 + − .... 2 24
(– ∞ < x < ∞)
= x+
x3 2 x5 + + .... (– π/2 < x < π/2) 3 15
n (x + y)n = x +
n n−1 n( n − 1) n − 2 2 x y+ x y + .... 1! 2!
(x2 > y2)
Derivatives and indefinite integrals In what follows, the letters u and v stand for any functions of x, and a and m are constants. To each of the integrals should be added an arbitrary constant of integration. A Short Table of Integrals by Peirce and Foster (Ginn and Co.) gives a more extensive tabulation. 1.
dx = 1 dx
1.
2.
du d ( au) = a dx dx
2.
3.
d du dv + ( u + v) = dx dx dx
3.
4.
d m x = mxm – 1 dx
4.
5.
d 1 ln x = x dx
5.
6.
dv du d +v ( uv) = u dx dx dx
6.
7.
d x e = ex dx
7.
8.
d sinh x = cosh x dx
8.
9.
d cosh x = sinh x dx
9.
10.
d 1 arctan x = dx 1 + x2
10.
z z z z z z z z z z
dx = x
z
au dx = a u dx (u + v )dx =
xm dx =
z z u dx +
xm + 1 (m ≠ – 1) m+1
dx = ln x x u
dv dx = uv − dx
z
v
x e x dx = e
cosh x dx = sinh x sinh x dx = cosh x dx 1 + x2
v dx
= arctan x
du dx dx
706
Mechanics
1
11.
d arcsin x = dx
12.
1 d arc sec x = dx x x2 − 1
12.
13.
d cos x = – sin x dx
13.
14.
d sin x = cos x dx
14.
15.
d tan x = sec2 x dx
15.
16.
d cot x = – csc2 x dx
16.
17.
d sec x = tan x sec x dx
17.
18.
d csc x = – cot x csc x dx
18.
11.
2
1− x
z z z z z z z z
dx 1 − x2 dx x x2 − 1
= arcsin x
= arcsec x
sin x dx = – cos x cos x dx = sin x tan x dx = ln sec x cot x dx = ln sin x sec x dx = ln sec x + tan x csc x dx = ln csc x − cot x
Vector products Let i, j, k be unit vectors in the x, y, z directions. Then i . i = j . j = k . k = 1,
i . j = j . k = k . i = 0,
i × i = j × j = k × k = 0, i × j = k,
j × k = i,
k × i = j.
Any vector a with components ax, ay, az, along the x, y, z axes can be written a = axi + ay j + azk. Let a, b, c be arbitrary vectors with magnitudes a, b, c. Then a × (b + c) = a × b + a × c (sa) × b = a × (sb) = s(a × b)
(s a scalar).
Let θ be the smaller of the two angles between a and b. Then a . b = b . a = axbx + ayby + azbz = ab cos θ
i a × b = – b × a = ax bx
j ay by
k az bz
= (aybz – byaz)i + (azbx – bzax)j + (axby – bxay)k a×b
= ab sin θ
a . (b × c) = b . (c × a) = c . (a × b) a × (b × c) = (a . c)b – (a . b)c
Appendices
707
APPENDIX K
VALUES OF TRIGONOMETRIC FUNCTIONS TRIGONOMETRIC FUNCTIONS Radians
Degrees
Sines
Cosines
Tangents
Cotangents
.0000
0
.0000
1.0000
.0000
∞
90
1.5708
.0175
1
.0175
.9998
.0175
57.29
89
1.5533
.0349
2
.0349
.9994
.0349
28.64
88
1.5359
.0524
3
.0523
.9986
.0524
19.08
87
1.5184
.0698
4
.0698
.9976
.0699
14.30
86
1.5010
.0873
5
.0872
.9962
.0875
11.430
85
1.4835
.1047
6
.1045
.9945
.1051
9.514
84
1.4661
.1222
7
.1219
.9925
.1228
8.144
83
1.4486
.1396
8
.1392
.9903
.1405
7.115
82
1.4312
.1571
9
.1564
.9877
.1584
6.314
81
1.4137
.1745
10
.1736
.9848
.1763
5.671
80
1.3963
.1920
11
.1908
.9816
.1944
5.145
79
1.3788
.2094
12
.2079
.9781
.2126
4.705
78
1.3614
.2269
13
.2250
.9744
.2309
4.332
77
1.3439
.2443
14
.2419
.9703
.2493
4.011
76
1.3285
.2618
15
.2588
.9659
.2679
3.732
75
1.3090
.2793
16
.2756
.9613
.2867
3.487
74
1.2915
.2967
17
.2924
.9563
.3057
3.271
73
1.2741
.3142
18
.3090
.9511
.3249
3.078
72
1.2866
.3316
19
.3256
.9455
.3443
2.904
71
1.2392
.3491
20
.3420
.9397
.3640
2.748
70
1.2217
.3665
21
.3584
.9336
.3839
2.605
69
1.2043
.3840
22
.3746
.9272
.4040
2.475
68
1.1868
.4014
23
.3907
.9205
.4245
2.356
67
1.1694
.4189
24
.4067
.9135
.4452
2.246
66
1.1519
.4363
25
.4226
.9063
.4663
2.144
65
1.1345
.4538
26
.4384
.8988
.4877
2.050
64
1.1170
.4712
27
.4540
.8910
.5095
1.963
63
1.0996
.4887
28
.4695
.8829
.5317
1.881
62
1.0821
.5061
29
.4848
.8746
.5543
1.804
61
1.0647
.5236
30
.5000
.8660
.5774
1.732
60
1.0472
.5411
31
.5150
.8572
.6009
1.664
59
1.0297
.5585
32
.5299
.8480
.6249
1.600
58
1.0123
Cosines
Sines
Cotangents
Tangents
Degrees
Radians (Cont...)
708
Mechanics Radians
Degrees
Sines
Cosines
Tangents
Cotangents
.5760
33
.5446
.8387
.6494
1.540
57
0.9948
.5934
34
.5592
.8290
.6745
1.483
56
0.9774
.6109
35
.5736
.8192
.7002
1.428
55
0.9599
.6283
36
.5878
.8090
.7265
1.376
54
0.9425
.6458
37
.6018
.7986
.7536
1.327
53
0.9250
.6632
38
.6157
.7880
.7813
1.280
52
0.9076
.6807
39
.6293
.7771
.8098
1.235
51
0.8901
.6981
40
.6428
.7660
.8391
1.192
50
0.8727
.7156
41
.6561
.7547
.8693
1.150
49
0.8552
.7330
42
.6691
.7431
.9004
1.111
48
0.8378
.7505
43
.6820
.7314
.9325
1.072
47
0.8203
.7679
44
.6947
.7193
.9657
1.036
46
0.8029
.7854
45
.7071
.7071
1.0000
1.000
45
0.7854
Cosines
Sines
Cotangents
Tangents
Degrees
Radians
Appendices
709
APPENDIX L
NOBEL PRIZE WINNERS IN PHYSICSa 1901 1902 1903
Wilhelm Konrad Rontgen
1845-1923
German
Discovery of X-rays.
Hendrik Antoon Lorentz
1853-1928
Dutch
Influence of magnetism on the
Pieter Zeeman
1865-1943
Dutch
phenomena of atomic radiation.
Henri Becquerel
1852-1908
French
Discovery of natural radioactivity
Pierre Curie
1850-1906
French
and of the radioactive elements
Marie Curie
1867-1934
French
radium and polonium.
1904
Baron Rayleigh
1842-1919
English
Discovery of argon.
1905
Philipp Lenard
1862-1947
German
Research in cathode rays.
1906
Sir Joseph John Thomson
1856-1940
English
Conduction of electricity through gases.
1907
Albert A. Michelson
1852-1931
U.S.
Invention of interferometer and spectroscopic and metrological investigations.
1908
Gabriel Lippmann
1845-1921
French
Photographic reproduction of colours.
1909
Guglielmo Marconi Karl Ferdinand Braun
1874-1937 1850-1918
Italian German
Development of wireless telegraphy.
1910
Johannes Diderik van der Waals
1837-1923
Dutch
Equations of state of gases and fluids.
1911
Wilhelm Wien
1864-1928
German
Laws of heat radiation.
1912
Nils Gustaf Dalen
1869-1937
Swedish
Automatic coastal lighting.
1913
Heike Kamerlingh-Onnes
1853-1926
Dutch
Properties of matter at low temperatures; production of liquid helium. Diffraction of X-rays in crystals.
1914
Max von Laue
1879-1960
German
1915
Sir William Henry Bragg Sir William Lawrence Bragg
1862-1942 1890-
English Study of crystal structure by English— means of X-rays. his son
1916
(No award)
1917
Charles Glover Barkla
1877-1944
English
Discovery of the characteristic X-rays of elements.
1918
Max Planck
1858-1947
German
Discovery of the elemental quantum.
1919
Johannes Stark
1874-1957
German
Discovery of the Doppler effect in canal rays and the splitting of spectral lines in the electric field.
1920
Charles Edouard Guillaume 1861-1938
Swiss
Discovery of the anomalies of nickel-steel alloys.
1921
Albert Einstein
German
Discovery of the law of the photo-electric effect.
1879-1955
710
Mechanics
1922
Niels Bohr
1885–1963
Danish
Study of structure and radiations of atoms.
1923
Robert Andrews Millikan
1868-1953
U.S.
Work on elementary electric charge and the photoelectric effect.
1924
Manne Siegbahn
1886
Swedish
Discoveries in the area of X-ray spectra.
1925
James Franck
1882–1964
German
Laws governing collision between
Gustav Hertz
1887
German
electron and atom.
1926
Jean Perrin
1870–1942
French
Discovery of the equilibrium of sedimentation.
1927
Arthur H. Compton
1892–1962
U.S.
Discovery of the scattering of X-rays by charged particles.
Charles T.R. Wilson
1869–1959
English
Invention of the cloud chamber, a device to make visible the paths of charged particles.
1928
Sir Owen Williams Richardson
1879–1959
English
Discovery of the law known by his name (the dependency of the emission of electrons on temperature).
1929
Louis-Victor de Broglie
1892
French
Wave nature of electrons.
1930
Sir Chandrasekhara Raman 1888
Indian
Work on the scattering of light and discovery of the effect known by his name.
1931
(No award)
1932
Werner Heisenberg
German
Creation of quantum mechanics.
1933 1934
1901
Paul Adrien Maurice Dirac
1902
English
Discovery of new fertile forms of
Erwin Schrodinger
1887–1961
Austrian
the atomic theory.
(No award)
1935
James Chadwick
1891
English
Discovery of the neutron.
1936
Victor Hess
1883–1964
Austrian
Discovery of cosmic radiation.
Carl David Anderson
1905
U.S.
Discovery of the positron.
1937
Clinton Joseph Davisson
1881–1958
U.S.
Discovery of diffraction of elec-
George P. Thomson
1892
English
trons by crystals.
1938
Enrico Fermi
1901–1954
Italian
Artificial radioactive elements from neutron irradiation.
1939
E.O. Lawrence
1901–1958
U.S.
Invention of the cyclotron.
1942
(No awards)
1943
Otto Stern
1888
U.S.g
Work with molecular beams and magnetic moment of proton.
1944
Isidor Isaac Rabi
1898
U.S.
Nuclear magnetic resonance.
1945
Wolfgang Pauli
1900–1958
Austrian
Discovery of quantum exclusion principle.
1946
Percy Williams Bridgman
1882–1961
U.S.
High-pressure physics.
Appendices
711
1947
Sir Edward Appleton
1892
English
Upper atmosphere physics and discovery of Appleton layer.
1948
Patrick Maynard Stuart Blackett
1897
English
Discoveries in cosmic radiation and nuclear physics.
1949
Hideki Yukawa
1907
Japanese Prediction of existence of meson.
1950
Cecil Frank Powell
1903
English
Photographic method of studying nuclear processes; discoveries about mesons.
1951
Sir John Douglas Cockeroft 1897
English
Transmutation of atomic nuclei
Ernes Thomas Sinton Walton 1903
Irish
by artificially accelerated atomic particles.
Felix Bloch
1905
U.S.
Measure of magnetic fields in
Edward Mills Purcell
1912
U.S.
atomic nuclei.
1953
Frits Zernike
1888
Dutch
Invention of phase contrast microscopy.
1954
Max Born
1882
English h
Work in quantum mechanics and statistical interpretation of wavefunction.
Walther Bothe
1891–1957
German
Analysis of cosmic radiations using the coincidence method.
Willis E. Lamb, Jr.
1913
U.S.
Fine structure of hydrogen.
Polykarp Kusch
1911
U.S.
Magnetic moment of electron.
1952
1955 1956
1957
1958
1959
John Bardeen
1908
U.S.
Invention and development of
Walter H. Brattain
1902
U.S.
transistor.
William B. Shockley
1910
U.S. c
Chen Ning Yang
1922
Chinesed
Non-conservation of parity and
Tsung Dao Lee
1926
Chinesed
work in elementary particle theory.
Pavel A. Cerenkov
1904
Russian
Discovery and interpretation of
Ilya M. Frank
1908
Russian
Cerenkov effect of radiation by
Igor Y. Tamm
1895
Russian
fast charged particles in matter.
Owen Chamberlain
1920
U.S.
Discovery of the antiproton.
e
Emilio Gino Segre
1905
U.S.
1960
Donald A. Glaser
1926
U.S.
Invention of bubble chamber.
1961
Robert L. Hofstadter
1915
U.S.
Electromagnetic structure of nucleons from high-energy electron scattering.
Rudolf L. Mossbauer
1929
German
Discovery of recoilless resonance absorption of gamma rays in nuclei.
1962
Len D. Landau
1908
Russian
Theory of condensed matter; phenomena of superfluidity and superconductivity.
1963
Eugene B. Wigner
1902
U.S. f
Contributions to theoretical atomic and nuclear physics.
712
Mechanics
1964
1965
Maria Goeppert-Mayer
1906
U.S.g
Shell model theory and magic
J.H.D. Jensen
1907
German
numbers for the atomic nucleus.
C.H. Townes
1915
U.S.
Invention of the maser and theory
Nikolai Basov
1922
Russian
of coherent atomic radiation.
Aleksandr Prokhorov
1916
Russian
Richard Feynman
1918
U.S.
Development of quantum electro-
Julian Schwinger
1918
U.S.
dynamics.
Shin-Ichiro Tomonaga
1906
Japanese
a
See Nobel : The Man and His Prizes, by Schuck et al., Elsevier, N.Y.
b
Born in Germany; naturalized British citizen.
c
Born in England; naturalized U.S. citizen.
d
Both have permanent U.S. resident status.
e
Born in Italy; naturalized U.S. citizen.
f
Born in Hungary; naturalized U.S. citizen.
g
Born in Germany; naturalized U.S. citizen.
Appendices
713
APPENDIX M
THE GAUSSIAN SYSTEM OF UNITS Much of the literature of physics is written, and continues to be written, in the Gaussian system of units. In electromagnetism many equations have slightly different forms depending on whether it is intended, as in this book, that mks variables be used or that Gaussian variables be used. Equations in this book can be cast in Gaussian form by replacing the symbols listed below under “rationalized mks” by those listed under “Gaussian.” For example, Eq. 37–26, B = µ0 (H + M)
B = c
becomes or
FG 4 π IJ . FG c H + cM IJ H c K H 4π K 2
B = H + 4πM
in Gaussian form. Symbols used in this book that are not listed below remain unchanged. The quantity c is the speed of light. Quantity
Rationalized mks
Gaussian
Permittivity constant
∈0
1/4π
Permeability constant
µ0
4π/c2
Electric displacement
D
D/4π
Magnetic induction
B
B/c
Magnetic flux
ΦB
ΦB/c
Magnetic field strength
H
cH/4π
Magnetization
M
cM
Magnetic dipole moment
µ
cµ
In addition to casting the equations in the proper form it is of course necessary to use a consistent set of units in those equations. Below we list some equivalent quantities in mks and Gaussian units. This table can be used to transform units from one system to the other. Quantity
Symbol
Mks system
Gaussian system
Length
l
1 meter
102 cm
Mass
m
1 kg
103 gm
Time
t
1 sec
1 sec
Force
F
1 newton
105 dynes
W, E
1 joule
107 ergs
Power
P
1 watt
107 ergs/sec
Charge
q
1 coulomb
3 × 10 9 statcoul
Current
i
1 ampere
3 × 109 statamp
Electric field strength
E
1 volt/meter
Electric potential
V
1 volt
Electric polarization
P
1 coul/meter2
Work or Energy
1 × 10− 4 statvolt / cm 3 1 statvolt 300 3 × 105 statcoul/cm2
714
Mechanics Electric displacement
D
1 coul/meter2
Resistance
R
1 ohm
Capacitance
C
1 farad
1 × 10− 11 sec cm− 1 9 9 × 1011 cm
ΦB
1 weber
108 maxwells
1 tesla ≡ 1 weber/
104 gauss
Magnetic flux Magnetic induction
B
meter Magnetic field strength
H
12π × 105 statvolt/cm
2
1 amp-turn/meter
Magnetization
M
1 weber/meter
Inductance
I
1 henry
2
4π × 10– 4 oersted 1/4π × 10– 3 gauss 1 × 10− 11 9
All factors of 3 in the above table, apart from exponents, should be replaced by (2.997925 ± 0.000003) for accurate work; this arises from the numerical value of the speed of light. For example the mks unit of capacitance (= 1 farad) is actually 8.98758 × 1011 cm rather than 9 (= 32) × 1011 cm as listed above. This example also shows that not only units but also the dimensions of physical quantities may differ between the two systems. In the mks system (see Appendix F) the dimensions of capacitance are M– 1L– 2T2Q2; in the Gaussian system they are simply L, the Gaussian standard unit of capacitance being 1 cm. The student should consult Classical Electromagnetism, p. 611, by J.D. Jackson (John Wiley and Sons, 1962) for a fuller treatment of units and dimensions.
