Analytical Dynamics

ANALYTICAL DYNAMICS

DYNAMICS

ANALYTICAL

Haim Haim Baruh Baruh Rutgers

University

WCB ~WCB

McGraw-Hill ti.tfili McGraw-Hili Bun Ridge, Boston Burr Ridge, IL IL Dubuque, Dubuque, IA IA Madison, Madison, WI WI New New York York San San Francisco Francisco St. St. Louis Louis Bangkok Bogota Caracas Lisbon London Madrid Mexico City City Milan Sydney Taipei Mexico Milan New New Delhi Delhi Seoul Singapore Sydney Taipei Toronto

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WCB/McGraw-Hifl WeB/McGraw-Hill A Division Corn paths A Division of of The McGruw'J-IIU McGmw·HiU Companies

ANALYTICAL DYNAMICS International Editions Editions 1999 1999

This book Exclusive rights by McGraw-Hill Book Co — _ Singapore, for manufacture and export. export. This cannot country to to which consigned by cannot be be re-exported re-exported from from the the cmmtry which itit is consigned by McGraw-Hill. McGraw-Hill. Copyright © 1999 Copyright © 1999 by by The McGraw-Hill McGraw-Hill Companies, Companies, Inc. All rights reserved. Except as permitted under the United Act of of 1976, under United States States Copyright Copyright Act 1976, no part part ofofthis thispublication publication may may be bereproduced reproduced or or distributed in any system, without any form or or by by any any means, means, or or stored stored in a data data base base ororretrieval retrieval system, without the the prior written permission of the the publisher.

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Barth, Baruh, Haim. Haim. Analytical dynamics I/ Hairn Haim Baruh. Barth. cm. p. Includes bibliographical references and index. index.

ISBN 0-07-365977-0 I. Title. 1. Dynamics. 1. QA845.B38 1999 531'.I 531 '.11-dc21 1—dc2l www.mhhe.com

When ordering this title, When ordering title, use use ISBN ISBN 0-07-116094-9 0.. 07-116094-9 Printed in Singapore Singapore

98-34940

Dedicated in em ory of of my my dear in other, Ester Barn/i, in in memory mother, Baruh, rry rr~

CONTENTS

PREFACE

Chapter 11 Chapter

xiii Xlll

BASIC PRINCIPLES 1.1 1.2 1.3

1

Introduction 11 Systems of Units 11 Review of of Vector Analysis 1.3.1 1.3.2

4 Rectilinear (Cartesian) Coordinates Curvilinear Coordinates 8

5

Newtonian Particle Mechanics 26 Degrees of Freedom and Constraints 33 Impulse and Momentum 36 Work and Energy 43 WorkandEnergy 1.7.1 Gravitational Potential Energy 44 1.7.2 Potential Energy of Springs 46 1.7.3 Elastic Strain Energy 48 1.7.4 Work-Energy Relations 48 1.8 Equilibrium and Stability 54 1.9 Free Response of Linear Systems 58 1.10 Response 1.10 Responseto to Harmonic Harmonic Excitation Excitation 62 1.11 Forced Response of Linear Systems 68 1.12 First Integrals 71 1.13 Numerical 1.13 NumericalIntegration Integrationof ofEquations Equations of of Motion Motion References 77 Homework Exercises 77

1.4 1.5 1.6 1.7

Chapter 2 RELATIVE Chapter RELATIVE MOTION

87

2.1 Introduction 87 2.2 Moving MovingCoordinate Coordinate Frames Frames 88 2.3 Representation of of Vectors Vectors 91 2.4 Transformation of Coordinates, Finite Rotations 97 2.5 Infinitesimal Rotations, Angular Angular Velocity Velocity 107 RateofofChange ChangeofofaaVector, Vector, Angular Angular 2.6 Rate Acceleration 117 2.7 Relative Velocity and Acceleration 124 2.8 Observations from a Moving Frame 133 References 144 Homework Exercises 144

vii

75

viii viii

CoNnwIs CONTENTS

Chapter Chapter3 3DYNAMICS DYNAMICS OF A SYSTEM OF PARTICLES

153

Introduction 153 Equations of Motion 153 Linear and Angular Momentum 157 Work and Energy 161 Impact of Particles 164 Variable Mass Mass and and Mass Mass Flow flow Systems 170 Variable Concepts from Orbital Mechanics: The Two Body Problem 173 TheNatureofthe 3.8 The Nature of the Orbit Orbit 177 3.9 Orbital Elements 187 PlaneKinetics Kineticsof ofRigid Rigid Bodies Bodies 193 3.10 Plane 3.11 Instant Centers and Rolling 200 3.12 Energy Energyand andMomentum Momentum 203 References 206 Homework Exercises 207 3.1 3.2 3.3 3.4 3.5 3.6 3.7

Chapter 4 ANALYTICAL ANALYTICAL MECHANICS: MECHANICS: BASIC CONCEPTS

4.1 4.2 4.3

215

Introduction 215 Generalized Coordinates Constraints 219 4.3.1 4.3.2

216

Holonomic Constraints 220 Nonholonomic Constraints 223

4.4 4.5 4.6

Virtual Displacements and and Virtual Virtual Work Work Generalized Forces 237 Principle of Virtual Work for Static Equilibrium 241 4.7 D'Alembert's Principle 245 4.8 Hamilton's Principles 249 4.9 Lagrange's Equations 253 4.10 Lagrange's Lagrange'sEquations Equationsfor for Constrained Constrained Systems 260 References 265 Homework Exercises 265

Chapter 5 ANALYTICAL Chapter ANALYTICAL MECHANICS: ADDITIONAL ADDITIONAL TOPICS 5.1

5.2 5.3 5.4

273

Introduction 273 Natural and Nonnatural Systems, Equilibrium 273 Small Motions about Equilibrium 279 Rayleigh's Dissipation Function 286

230

CONTENTS

Eigenvalue Problem for Linearized Systems 289 5.6 Orthogonality and Normalization Nonnalization 293 5.7 Modal Equations of Motion and Response 296 5.8 Generalized Momentum, First Integrals 300 5.9 Routh's Method Method for for Ignorable Ignorable Coordinates Coordinates 302 5.10 Impulsive 5.10 Impulsive Motion Motion 304 5.11 Hamilton's Equations 308 5.12 Computational ComputationalConsiderations: Considerations: Alternate Descriptions of the Motion Equations 311 5.13 Additional AdditionalDifferential DifferentialVariational Variational Principles 314 References 317 Homework Exercises 317

5.5

Chapt.r 6 RIGID RIGID BODY GEOMETRY Chapter

323

6.1 Introduction

323 6.2 Centerof ofMass Mass 323 6.2 Center

6.3 Mass 6.3 MassMoments Moments of of Inertia Inertia 326 6.4 Transformation Transformation Properties Properties 334 Translation of Coordinates 335 Rotation of Coordinate Axes 337 6.5 Principal Moments of Inertia 343 6.6 Inertia Ellipsoid 347 References 351 Homework Exercises 351

6.4.1 6.4.2

Chap'.r RIGID BODY KINEMATICS KINEMATICS Chapter 7 RIGID 7.1 7.2

355

Introduction 355 Basic Kinematics of of Rigid Bodies Bodies

355

Pure Translation 356 7.2.2 Pure Rotation 356 7.2.3 Combined Translation and Rotation Euler's and and Chasles's Chasles's Theorems Theorems 360 7.2.1

7.3 7.4 7.5

Relation between Direction Cosines and Angular Velocity 365 Euler Angles 367 7.5.1 7.5.2 7.5.3

7.6 7.7

359

Euler Angle Sequences 368 Angular Velocity and Acceleration Axisymrnetric Axisymmetric Bodies 372

369

Angular Velocities Velocities as Quasi-Velocities Angular (Generalized Speeds) 378 Euler Parameters 380 7.7.1

to the Euler Relating the Direction Cosines to Paramaters 382

Ix

x

CONTENTS CONTENTS

7.7.2 7.7.2 Relating the the Euler Euler Parameters Parameters to to Angular Angular Velocities Velocities 384 384 7.7.3 7.7.3 Relating Relating the the Euler Euler Parameters Parameters to to the the Direction Direction Cosines Cosines 386 386 7.7.4 7.7.4 Relating Relating the the Euler Euler Parameters Parameters to to the the Euler Euler Angles Angles and and Vice Vice Versa Versa 387 387 7.7.5 7.7.5 Other Other Considerations Considerations 388 388

7.8 7.8

Constrained Motion Motion of of Rigid Bodies, Interconnections 391 391 7.8.1 7.8.2 7.8.3

7.9

Rotational Contact, Joints Joints 391 391 Translational Motion and Sliding Contact Contact 395 395 Combined Sliding and Rotation 396 Rolling 400 References 412 Homework Exercises 412

Chapter 8 RIGID RIGID BODY DYNAMICS: BASIC BASIC CONCEPTS 8.1

8.2 8.3

421

Introduction 421 and Angular AngularMomentum Momentum 422 Linear and 422 Transformation Properties of Angular Momentum Transfonnation Translation of Coordinates 425 Rotation of Coordinates 426 Resultant Force and Moment 429 General Equations of Motion 431 Rotation about a Fixed Axis 443 8.3.1 8.3.2

8.4 8.5 8.6 8.7 Equations of Motion in State Form Fonn 445 8.8 Impulse-Momentum Relationships 448 8.9 Energy and Work 451 8.10 Lagrange's 8.10 Lagrange'sEquations Equations for for Rigid Rigid Bodies 456 Work and and Generalized Forces 456 Virtual Work Generalized Coordinates 458 the Euler Euler Lagrange's Equations in Terms Terms of the Angles 462 the Euler 8.10.4 Lagrange's Equations in Terms of the Parameters 463 8.10.5 Lagrange's Equations and Constraints 464 8.10.5 8.11 D'Alembert's D'Alembert's Principle 8.11 Principle for for Rigid Bodies 474 8.12 Impact 8.12 Impact of of Rigid Rigid Bodies 477 477 References 481 Homework Exercises 481 8.10.1 8.10.2 8.10.3

Chapter 9 DYNAMICS OF RIGID BODIES: Chapter DYNAMICS OF BODmS: ADVANCED ADVANCED CONCEPTS 9.1 9.1 9.2 9.2

489 489

Introduction 489 Introduction Modified Euler's Euler's Equations 490 Modified 490

425

Corinwrs CONTENTS

Moment Moment Equations about about an an Arbitrary Arbitrary Point 498 503 9.4 Classification of the Moment Equations 503 9.5 Quasi-Coordinates and and Quasi-Velocities Quasi-Velocities 9.5 Speeds) 504 (Generalized Speeds) Generalized Speeds Speeds and and Constraints Constraints 510 510 9.6 Generalized 9.7 Gibbs-Appell Equations Equations 517 517 9.8 Kane's Kane's Equations Equations 524 524 9.8 9.9 The Fundamental Equations and Constraints 531 Constraints Relationshipsbetween betweenthe theFundamental Fundamental 9.10 Relationships Lagrange's Equations Equations 532 Equations and Lagrange's 9.11 9.11 Impulse-Momentum Impulse-MomentumRelationships Relationships for for Generalized Speeds 539 References 542 Homework Exercises 542 9.3 9.3

OFRIGID RIGIDBODY Bony Chapter QUALITATIVE ANALYSIS ANALYSIS OF Chapt.r I100 QUALITATIVE MOTION

547

10.1 Introduction 547 10.2 Torque-Free 10.2 Torque-Free Motion Motion of of Inertially Symmetric Bodies 548 10.3 General 10.3 GeneralCase Caseof ofTorque-Free Torque-Free Motion Motion 556 10.4 Polhodes 557 10.4 Polhodes 10.5' Motionof ofaaSpinning Spinning Top Top 563 10.5 Motion 10.6 Rolling Disk 573 10.6 Rolling 10.7 The Gyroscope Gyroscope 580 10.7 The 10.7.1 Free Gyroscope 580 10.7.2 The Gyrocompass 582 10.7.3 Single-Axis Gyroscope 585 References 587 Homework Exercises 587

Chapter I1 1

OF LIGHTLY FLEXIBLE DYNAMICS OF FLEXffiLE BODIES 591 11.1 11.2 11.2

Introduction 591 Kinetic and Potential Energy: Small or No Rigid Body Motion 592 Kinematics and Geometry 592 Kinetic and Potential Energy for for Kinetic 597 Beams 597 Beams Kinetic and Potential Energy for for 11.2.3 Kinetic 11.2.3 Torsion 599 599 Torsion 601 Operator Notation Notation 601 11.2.4 Operator 11.2.4 Equations of Motion 604 604 Equations 604 Extended Hamilton's Hamilton's Principle Principle 604 11.3.1 11.3.1 Extended 11.2.1 11.2.1 11.2.2

11.3 11.3

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xli xII

CONTENTS

Boundary Conditions 606 Simplification 608 Operator Notation 611 Eigensolution and Response 616 11.3.2 11.3.3 11.3.4

11.4 11.5

Approximate Solutions: The Assumed Modes Method 626 Approximation Techniques and Trial Functions 626 11.5.2 Method of Assumed Modes 628 sues 632 11.5.3 Convergence Convergence Is Issues 11.5.1

11.6 11.7

Kinematics of Combined Elastic and Large Angle Rigid Body Motion 637 Dynamics of Combined Elastic and Large Angle Rigid Body Motion 645

Equations of Motion in Hybrid Form 646 11.7.2 Equations of Motions in Discretized Form 649 Fonn 11.8 Analysis of the Equations of of Motion 651 11.9 Case 3, Rotating Shafts 659 11.10 Hybrid 11.10 Hybrid Problems Problems 663 References 664 Homework Problems 664 11.7.1

Appendix A A ABRWF BRIEF HISTORY OF OF DYNAMICS DYNAMICS

669

A.1 Introduction 669 A.l A.2 Historical HistoricalEvolution Evolution of of Dynamics Dynamics 669 A.3 Biographies BiographiesofofKey KeyContributors Contributors to to Dynamics 678 References 685

Appendix

B CONCEPTS B CONCEPTS FROM FROM THE CALCULUS OF VARIATIONS

687

B.1 B.l Introduction 687 StationaryValues Values of ofaa Function Function 687 B.2 Stationary B.3 Stationary Values of a Definite Integral 693 B.4 The TheVariational Variational Notation Notation 696 B.5 Application Applicationofofthe theVariational Variational Notation Notation to Dynamics Problems 699 References 700 Homework Exercises 700

Appendix

C COMMON COMMON INERTIA PROPERTIES INDEX

707

701

PREFACE

ApPROACH APPROACH

Two occurrences in in the the second second part part of the the 20th 20th century have radically changed the Two occurrences nature of the field of dynamics. dynamics. The first first is the increased need to model and analyze complex, as satellites, satellites, robot robot complex, multibodied rnultibodied and often elastic-bodied structures, such as manipulators, and vehicles. The second second is the the proliferation proliferation of of the the digital digital computer, computer, which has led to the development of numerical techniques to derive the describing equations of a system, integrate the the equations equations of of motion, motion, and and obtain obtain the the response. response. This new computational capability has encouraged scientists and engineers to model and numerically analyze complex dynamical systems which in the past either could not be analyzed, or were analyzed using gross gross simplifications. simplifications. techniques to to model model a dynamical system The prospect of using computational computational techniques has also led dynamicists to reconsider existing methods of obtaining equations of motion. motion. When evaluated in terms of systematic application, ease of implementation by computers, and computational effort, some of the traditional approaches lose part of their appeal. appeal. For example, example, to to obtain obtain Lagrange's Lagrange's equations, equations, one one isis traditionally traditionally taught first to generate a scalar function called the Lagrangian and then to to perform a series of differentiations. This approach approach is computationally computationally inefficient. inefficient. Moreover, Moreover, certain terms in the differentiation of the kinetic energy cancel each other, resulting in wasted manipulations. As a result of the reevaluation reevaluation of the the methods methods used used in in dynamics, dynamics, new approaches have been proposed and certain certain older older approaches approaches that that were were not not commonly commonly used used in the past have been brought back into the limelight. What has followed in the literature is a series of papers and books containing claims by proponents of certain methods, each extolling the virtues of of one approach over the other without a fair and balanced analysis. This, This, at least in the opinion of this author, has not led to a healthy envianalysis. author, has ronment and ideas. It is now possible for basic graduate level and. fruitful fruitful exchange of ideas. courses in dynamics to be taught at different schools with entirely different different subject subject material. These developments of recent recent years years have have inspired inspired me to compile These developments of compile my lecture lecture textbook. Realizing that much of the research done lately in the field of notes into a textbook. dynamics subjectively, II have tried to present in this book book dynamics has been reported very subjectively, a fair and and balanced balanced description description of of dynamics dynamics problems problems and and formulations, fonnulations, from from the the classical methods to the newer techniques used in today's today's multibody multibody environments. environments. I have emphasized the need to know both the classical methods as well as the newer newer techniques and have shown that these approaches are really complementary. Having the knowledge and experience to to look look at a problem in a number of of ways ways not not only only facilitates the solution but also provides a better perspective. For For example, the the book book discusses which lead lead to to fewer singularities singularities in the solution solution and discusses Euler parameters, parameters, which which lend themselves to more efficient computer implementation.

xlii xiII

xlv

PREFACE

The focus of this book book is is primarily primarily the the kinematics kinematics and derivation of the the describdescribing equations of dynamics. We also consider the qualitative qualitative analysis analysis of of the response. We discuss discuss a special case of quantitative analysis, namely the response We response of motion motion linearized about equilibrium. We discuss discuss means to analyze analyze the kinematics and to describe the equations of momahon. force and moment balances, as well as analytical methods. In most tion. We study force dynamics problems, the resulting equations equations of of motion motion are are nonlinear nonlinear and and lengthy, lengthy, so so that closed-form solutions solutions are are generally not available. We discuss analytical solusolutions, motion integrals and basic stability concepts. We We make use of of integrals of of the motion, which are derived quantities that give qualitative qualitative information about the system without having to to solve for the exact exact solution. solution. For linearized systems, we discuss We outline outline concondiscuss the the closed-form closed-form response. response. We cepts from vibration theory and eigenvector eigenvector expansions. expansions. This This also also is is done for continuous systems, in the last chapter of We discuss discuss the the importance importance of of numerical numerical of the book. We solutions.

I

CONTENTS

book is organized into eleven eleven chapters chapters and three appendixes. organized into appendixes. The first eight The book chapters are intended for an introductory level graduate or advanced undergraduate course. of aa second, second, more more adadcourse. The later chapters chapters of the book can be used as part part of vanced graduate level course. The book book follows a classical approach, in which which one one first deals with particle mechanics and then extends the concepts into rigid bodies. first The Lagrange's equations equations are initially discussed for a system of particles and plane motion of rigid bodies. We follow in this book this school of thought for a number We follow number of of reasons. reasons. First, First, from a variety of backgrounds. Many times, students have graduate students come from not considered dynamics dynamics since the sophomore sophomore dynamics, dynamics, or the freshman physics physics course. Also, Also, this this organization presents aa more more natural course. organization presents natural flow flow of the concepts used in dynamics. Nevertheless, Nevertheless, the book is suitable suitable also also for for instructors instructors who prefer prefer to to teach three-dimensional rigid body dynamics before introducing analytical methods. Following is a description of the chapters: dynamics and see their appliIn Chapter 1 we study fundamental concepts of dynamics cations to particle mechanics problems. problems. We We discuss discuss Newton's laws and energy and motion and basic ideas from stability momentum principles. We look at integrals of motion theory. an intheory. The chapter outlines the response of linearized systems, which forms an theory. This This chapter chapter should should be covered in in detail if the course troduction to vibration theory. one. Less Less time should should be be spent on on itit for for a graduate graduate course course or or if if is an undergraduate one. the students taking the course are familiar with the basic ideas. Chapter 2 discusses relative relative motion. Coordinate frames, rotation sequences, angular velocities, and angular accelerations are introduced. The The significance significance of of taking time derivatives in different coordinate systems is -emphasized. emphasized. We derive the relative motion equations and consider motion with respect to the rotating earth. Chapter 3 is a chapter on systems of particles and plane kinetics of of rigid bodies. It is primarily included for pedagogical considerations. I recommend its use for an undergraduate course; for for aa graduate graduate level level course, undergraduate course; course, it should serve as independent independent

CONTENTS CONTENTS

reading. A number of sections in this chapter are devoted to an introduction to cereading. lestial mechanics mechanics problems, problems, namely the two-body problem. The sections lestial sections on on plane kinetics of rigid bodies basically review the sophomore level material. This review is included here mainly because the approaches in the next chapter are described in terms of particles and plane motion of rigid bodies. The subject of classical analytical analytical mechanics mechanics is discussed in Chapters 4 and is discussed in Chapters introduces the the basic basic concepts, concepts, covering covering generalized generalized coordinates, coordinates, 5. Chapter 4 introduces constraints, and and degrees degrees of of freedom. We derive derive the principle freedom. We principle of of virtual virtual work, work, constraints, D'Alembert's D' Alembert's principle, principle, and and Hamilton's principle principle and then we develop Lagrange's equations. Analytical mechanics makes makes use of the calculus of variations, a subject equations. Analytical covered separately separately in in Appendix Appendix B. B. While Chapters Chapters 4 and 5 are written so that one covered does not not absolutely absolutely need need to to learn learn the calculus of variations variations as a separate subject, it does has been the experience of this author that some initial exposure of students to the has calculus of variations is very helpful. Chapter 5 revisits the concept of equilibrium and outlines the distinction distinction between natural natural and nonnatural tween nonnatural systems. systems. We We derive the linearized linearized equations equations about about equilibrium. The The response response of linearized linearized systems systems is analyzed, equilibrium. analyzed, which in in essence is vibration theory for multidegree of freedom systems. Generalized momenta and motion integrals are considered. In Chapter 6, we discuss the internal properties of a rigid body. In a departure from traditional traditional approaches, approaches, we we discuss discuss moments from moments of inertia independent of the kinetic energy and angular momentum. Chapter 7 is devoted to a detailed detailed analysis analysis of the the kinematics kinematics of a rigid body, body, where we learn of methods of quantifying the angular angular velocity velocity vector. vector. We present a discussion of Euler angles and Euler parameters. We then discuss constraints acting on the motion motion and quantify these constraints and the resulting kinematic relations. Chapter 8 explores of rigid rigid bodies. bodies. Chapter explores basic basic ideas associated associated with the kinetics kinetics of We first of force force and and moment moment balances. balances. We We express the first begin with the application application of equations motion in in terms terms of both the Euler Euler angles angles and and angular angular velocities. velocities. We We equations of motion discuss in terms of generdiscuss the relative relative merits merits of deriving the equations equations of motion motion in alized as well alized coordinates coordinates as well as in in terms terms of ofangular angularvelocity velocity components. components. We We anaphysical ininlyze impulse-momentum and work-energy principles. We discuss the physical terpretation of Lagrange's equations and and integrals integrals of of the motion motion associated with the Lagrangian. Chapter 9 introduces more advanced concepts in the analysis of rigid rigid body body moanalysis of tion. We We analyze analyze the the modified modified Euler's Euler's equations and then consider the moment equations about an arbitrary point. We discuss quasi-velocities, also known as generalized speeds, and their applications. applications. We We demonstrate demonstrate that such such coordinates coordinates are desirable when dealing with nonholonomic nonholonomic systems. We demonstrate the equivalence of the Gibbs-Appell and Kane's equations equations and and discuss momentum balances in terms of the generalized speeds. speeds. Many of the analytical methods described in this chapter chapter could could have have been been introintroduced in are equally equally in Chapters Chapters 44 or or 5. 5. However, the power of these methods, which are applicable to both particles and rigid bodies, is better appreciated when we consider applications to complex rigid body body problems. problems. Chapter of rigid body motion, and in particular Chapter 10 10 covers the qualitative qualitative analysis analysis of gyroscopic effects. The chapter initially goes into a qualitative study study of of torque-free torque-free

xv

xvi xvI

PREFACE

the response response between between axisymmetric and arbitrary bodmotion and the differences in the We then discuss interesting classical applications of gyroscopic motion, such as ies. We a spinning top, a rolling disk, and gyroscopes. 11 investigates the subject of dynamics of of lightly lightly flexible flexible bodies. bodies. ReReChapter 11 cent problems in dynamics have demonstrated the importance of including the elasticity of a body in the describing equations. The emphasis is the analysis of of bodies that undergo undergo combined rigid and elastic motion, motion, typical examples examples being being robot robot mamanipulators and spacecraft with appendages. We derive the classical boundary value nipulators problem and examine a shortcoming in the traditional traditional formulation. formulation. The combined of a large-angle rigid and elastic motions are modeled in terms tenns of of the superposition superposition of primary motion, as the motion of a moving reference frame, and a secondary motion, the motion of the body as observed from the moving reference frame. Appendix A is a historical survey of of dynamics dynamics and and aa synopsis synopsis of of the work of of the many people who contributed contributed to to this this field. field. Appendix B presents an introduction to to the calculus of variations. variations. It is recomAppendix part of of this this appendix appendix be be studied studied before before Chapter Chapter 4. 4. However, However, mended that at least part Chapter 4 is written such that a brief brief introduction to virtual displacements should be to understand basic concepts from analytical mechanics. sufficient to the mass mass moments moments of of inertia inertia of of common 'common shapes. shapes. Appendix C gives the

PEDAGOGICAL PEDAGOGICAL TOOLS contains several several examples and homework problems. It It has been my exThe book contains homework problems. perience see many many examples. examples. II perience that students understand understand a subject best when they see encourage anyone anyone teaching teaching dynamics, dynamics, either at the undergraduate encourage undergraduate or the gradutite graduate level, to use as many examples as as possible. possible. Another pedagogical tool emphasized in the book is computational techniques. into details details of numerical integration, integration, we discuss the numerical While we do not go into motion. Many of the examples and homework problems integration of equations of motion. in the book can be assigned as computer projects. I encourage every student to keep advances in in scientific scientific software, software, such such as as symbolic symbolic manipulators, manipulators, bebepace with new advances cause, as discussed earlier, the availability of computational tools tools has changed the nature of dynamics.

SUPPLEMENT The book is supplemented by an Instructor's Solutions Manual which includes detailed solutions to all of of the problems in the book.

ACKNOWLEDGMENTS

Writing a textbook is a lengthy and difficult task and II would would like like to to acknowledge acknowledge the support of many people. First, First, I am richly indebted to all the many people. the teachers teachers II have have had in my life, life, from from my my elementary elementary school school teacher in Istanbul, Mrs. Mrs. Cahide sen, ~en,

ACKNOWLEDGMENTS

my advisor in graduate school, Dr. Meirovitch, who who has always been a to my Dr. Leonard Meirovitch, advice and inspiration. I am very grateful to to my my employer, employer, Rutgers Rutgers Universource of advice sity, and and especially especially to to the leadership of the Mechanical and Aerospace Engineering sity, Department, the current chair, Professor Abdelfattah Zebib, and the previous chair, Professor Samuel Samuel Temkin, Temkin, for for giving giving their their faculty faculty the the time time and resources to conProfessor conduct scholarly work. I thank Verica Radisavljevic emphatically, who read the entire manuscript, helped helped with with figures and made very valuable suggestions. Professor Professor Leon Bahar gave valuable suggestions on variational mechanics. Many graduate students students who used earlier versions of the manuscript as their course notes helped correct erwho rors and improve the text. Many thanks are are due due to to McGraw-Hill, McGraw-Hill, and and especially especially to to Senior Editor Editor Debra Debra Riegert, Riegert, who who gave gave me the opportunity to publish with them. Senior opportunity to Victoria St. outstanding copy competent staff of Victoria St. Ambrogio Ambrogio was was an an outstanding copy editor editor and and the the competent McGraw-Hill and and Publication Publication Services Services facilitated facilitated the the production production of ofthe the book. book. I am grateful to the reviewers of the manuscript, Professors Maurice Adams of Reserve University, University, Vincent Vincent K. Choo of New Mexico State University, Case Western Reserve William Clausen of Ohio State University, John E. E. Prussing William Clausen University, John Prussing of the University of Illinois at Urbana-Champaign, Urbana-Champaign, and and Thomas Thomas W. W. Strganac of of Texas A&M University; all provided very meaningful comments and and suggestions. suggestions. I thank my beloved wife Rahel Rabel for all her her support support and and encouragement, encouragement, as well as for reading parts of the manuscript and improving the text. I am forever thankful to my dear parents, Ester and and Yasef Yasef Baruh, who worked tirelessly so their children could get a good good education and also be good people. people. They taught me, by teaching teaching and by example, what the most important things in life really are. I am very grateful to my dear uncle, Selim Selim Baruh, Baruh, who who has always always been been aa source source of of inspiration inspiration and and who made it possible for me to attend graduate school. school. I dedicate this this book book to the memory of my mother, who passed away a year before the publication of this this book. book. Ester Ester Baruh Baruhwas was aa symbol symbol of ofhonesty, honesty, integrity, integrity, decency, decency, publication of kindness, kindness, compassion, compassion, and and common common sense. She was unfailingly unfailingly devoted devoted to her family and to her community. May her memory be a blessing. While both the author author and and the the publisher publisher have have made made every every effort effort to to produce produce an error-free error-free book, book, there there may nevertheless nevertheless be errors. errors. I would would be be most most appreciaappreciative if if aa reader reader who who spots spots an an error error or or has has aa comment comment about about the the book book would would bring bring it to to my my attention. attention. My My current current e-mail e-mail addresses addresses are [email protected] [email protected] or or [email protected]. To inform inform the the reader reader of any discovered errors I am estabdiscovered errors [email protected]. To lishing a page on my professional professional Web Web site site (www-srac.nitgers.edul---'baruh). (www-srac.rutgers.eduJ--baruh). Just Just go to my Web site and you will find the link. If my web address at any time has changed, try www.rutgers.edu www.rutgers.edu and and find find me me from from there. there.

xvII xvii

chapter

BASIC PRINCIPLES

I •1

INTRODUCTION

This chapter discusses basic principles and concepts used in the field of dynamics. It describes these principles and concepts within the context of particles.weFirst, look at the systems of units commonly used in dynamics. Then, we analyze the kinematics of a particle and outline the coordinate systems used to describe the motion. We distinguish between coordinate systems that deal with components of the motion in fixed directions and coordinate systems based on the properties of the path followed by the particle. The kinematic analysis is followed by the kinetics of a particle, and Newton's laws are given. We move on to analyze the concept of force and discuss the integration of the equations of motion. The distinction is drawn between the qualitative and quantitative, quantitative analysis analysis of of the the motion. motion. Integration Integration of of the the equations equations of of motion motion lead to energy and momentum momentum expressions expressions and and in in some cases to other integrals of the motion, which are useful in analyzing the nature of the motion. There is an introduction to the concepts of equilibrium and stability, and the closed- Ibrm integration of linearized equations of' motion. This subject Ibrrns lbrrns the basis of vibration analysis. analysis. This chapter is a collection of the fundamental principles that one uses in obtaining the equations of motion and in analyzing these equations of motion. The develare built on these principles. The reader opments in subsequent chapters in this book arc is encouraged to understand all the concepts discussed in this chapter thoroughly before continuing with the rest of the text.

1.2

SYSTEMS OF UNITS

While anyone can develop a set of units to describe the evolution of a dynamical system, in classical mechanics two basic sets are widely used: Système International 1

2

CHAPTIR I • BASIC PRiNCIPLES

(SI). or metric, and and U.S. U.S. Customary. Customary. The The primary difference between these systems is that the SI system is universal and absolute, and the U.S. system is local (or gravitational), that is, valid on earth. Today, most countries in the world have adopted the SI system as their standard. To describe the evolution of a body. three fundamental quantities are needed. needed. In all commonly used systems there is agreement on two: length and time, whose dimensions are denoted by L and T, respectively. The U.S. and SI systems diflèr on the nature of the third quantity. The SI system uses mass (M). defined as the amount of matter (absolute) contained in a body. By contrast, the U.S. system usesforce (F). The justification for this system is that the weight of a body. as the frrce with which the body is pulled toward the center of mass of the coordinate system (in our case, earth). is easier to visualize. Weight is a relative quantity. It changes depending on the amount of gravitational attraction. Corresponding to each fundamental quantity there is a base unit1 that describes standardized amounts of a fundamental quantity. Evolution of the base units has been in many cases based on convenience (e.g.. jhot), but in some cases a rational explanation is not available. The base units are usually abbreviated by symbols. The fundamental quantities and corresponding base units in the U.S. and Si systems are as follows Ibilows (the dimensions and symbols are denoted in the parentheses):

SI Units

Mass M: Length (L

Time (T):

):

U.S. Cus tomarv Units

kilogram (kg.)

Force (F):

pound (Ib)

meter (iii) (m)

Length (L):

loot (It)

second (s)

Time (T):

second (sec or s)

In the SI system. force is a derived quantity. The base unit of force (F = ML/T2) is denoted by a newton (N). where 1 N = kg• rn/s2. In the U.S. system, mass is a derived quantity, and the unit of mass (M = FT2/L) is denoted by a slug, with I slug = I lb•sec2/ft. When using the SI system many people commonly, and at the same time erroneously, refer to mass as a unit of weight. This is due to the perpetuation of the original definition of kilogram as a unit of weight. The kilogram was formally redefined as a unit of mass in the year 159. The selection of base units in a given system of units is not absolute. In many 1

cases one switches to a different system of' units to describe the motion. For example. the speed of an automobile is usually described in kilometers per hour (or miles per hour). and the speed of a ship is described by knots (I knot = nautical mile/hr. where 1 nautical mile = 1.852 km). To describe rotational displacements one may use degrees (°) or radians Going around a full circle takes 360 degrees or 2ir radians radians and is referred to as a revolution. A revolution. A radian is a dimensionless unit. The angle of I radian = 57.2958° is depicted in Fig. 1.1. 1

I

Some refer to the fundamental quantities as base units and to the base units as dimensions.

1 .2

FIgure 1.1

OF UNITS

One radian

An interesting source of' debatc has been the need to reexamine and to recal-

ibrate the standards used in defining the base units. Many precision calculations require formal definitions for several quantities. For example. one meter was originally defined as one 10-millionth of the distance from one of the poles to the equator. The current formal definition of a meter is the distance that light travels in one 299,792,458th of a second. One inch is delined as being exactly equal to 2.54 cm. The current formal definition of' a second is the time it takes for 9, 192,631 ,770 vibrations to occur in cesiuni atoms excited by microwave. A second was originally defined as 1/60 of' a minute, which is 1/60 of an hour, which is 1/24 of a day. A kilogram is defined as the mass of water contained in a volume of one liter and at a temperature of 4° Celsius. A block with a mass of 1 kg is maintained in a vault in Paris as the standard kilogram. At the time of writing of' this text, there was ongoing debate on revising the formal definition of the kilogram. To relate the mass of an object to its weight. we make use of the gravitational constant. The gravitational constant is denoted by g and it has the units of acceleration. The general form for this constant will be given in Section 1 .4. On earth at sea level the value of g is approximated approximated as as g = 9.81 rn/s2 org = 32.2 f't/sec2. On earth an object of mass 1 kg weighs 9.81 N. and an object of weight 1 lb has a mass of 1/32.2 slugs. It is interesting that the weight of a medium-sized apple is about one newton. We have so far treated the three Fundamental quantities as independent of each other. This is a correct assumption as long as relativistic effects are ignored (when the speeds involved are much less than the speed of light). As the speeds involved approach the speed of' light, mass, length. and time become interrelated. This is the foundation of relativistic mechanics and it will not he pursued in this text. It is always important to check that the dimensions of the quantities being manipulated match. Equations of motion and equilibrium equations must have all of' their terms with the same dimension. Checking dimensional homogeneity is a good way of spotting errors. Analogous to the preceding discussion regarding standards, one must also he

careful in the accuracy of the solution and rounding off numbers when solving a problem. Many engineering problems require an accuracy of about one part in a thousand or better. Accordingly, in most problems in this text we will retain four

3

4

CHAPTIR I • B1tsic PRINCIPLES

sometimes five significant digits in the solution (e.g., 0.7655. 8.975. 12.34 or 0.76549, 8.9762. 12.345). Following this line of thought. we note that the standard assumptions for the gravitational constant given above are not extremely accurate. In celestial mechanics problems, which we will discuss in Chapter 3, one needs to go to niuch higher levels of accuracy. and

Example 1.1

Suppose a new coordinate system is developed such that the density of water and the acceleration of gravity arc both of unit magnitude. If the pound is taken as the unit of force, how

do the units of length and time in the new system compare with the U.S. Customary systenl1?

Solution Denote the base units in the old system by lb. ft. and see and the units in the new system by lb. tf, and lb. ands*. s*.We We are are given given that

Acceleration of gravity g = 32.2 ft/sec2 =

I

[a]

l)ensity of water y = Specific wtlgravity = (62.8 lb/ft3 )/(32.2 ftlsec2 = 1 lh/ft44/(1 1.950 lb sec2/ft4 =

I

Ibi

lb

solved for the two relations are two equations that can he solved These two unknowns ft and From the first relation. From ft Substituting this into the second relation, we obtain I

1.950 lb sec2/ft4 =

1

lb

ft

[ci

which can be solved to give 1 sec/st = 11 11.31. .3 1. Substitution of this result into the first relation yields II ft/ft = 3.973. 3.973.

1.3

REviEw OF VECTOR ANALYSIS

In this section we review the sets ol coordinates and associated unit vectors commonly used in dynamics. As with systems of units, one can develop a specific set of vectors to describe the orientation of a body. The study of dynamics uses two basic types of coordinates: rectilinear and curvilinear. Rectilinear coordinates describe the components of the motion in fIxed directions. Curvilinear coordinate systems incorporate the properties of the path that the particle follows. In a coordinate system, one defines a set of three principal directions. The coordinate axes and associated unit vectors are directed along the principal directions. X When a set of of unit unit vectors vectors e1, e1,e21 e2, e3 obeys the cross product rule e1 X = = lifi,j.kareinorder(/ = 1,j = 2,k = 3,ori = 2,j 3,k = 1, on = 3,j = l.,k = = —lifi,j,karenotinorder(i = 1,j 1,) = 3,k = 2,

ori=2,j=

are repeated. repeated, the set is referred to as a mutually orthogonal triad. The coordinate system system that that uses uses these unit vectors is called a right-handed coordinate system.

1.3

1.3.1

REVIEW OF VECTOR ANALYSIS

RECTILINEAR (CARTESIAN) COORDINATES

In this coordinate system, the origin and the principal directions remain fixed. These directions are usually referred to as x, v, and z axes or X, Y, and Z axes, with associated unit vectors i, j, and k or I, J, and K, respectively. These unit vectors are time invariant. Because the unit vectors are independent of the path followed, this coordinate system is called extrinsic. Consider Fig. 1.2, where a particle P is moving along a curve. The displacement vector r(t) that describes the location of point P is

r(t) =

xi+vj+zk

11.3.11

The distance r from the origin 0 to point P then becomes =

•

r(i) =

by r(t + st), st),the thevelocity velocity of of the the

Denoting the position of the particle at time t + particle is defined by

v(t) =

dr(t) di

= r(I) =

r(t +

Jim

11.3.2]

y2 + z2 + y2

...J'x2

—

r(t)

[1.3.3] 11.3.31

Eq. [1.3.1] into Eq. 11.3.31 yields Introduction of Eq. v( I) =

ii + vj + ±k =

I+

+ v.. k

[1.3.4] 11.3.41

Because the unit vectors are time invariant, their derivatives vanish, in a similar fashion, one can obtain expressions for the acceleration

a(ñ= i'(i)= r(i)=

[1.3.5] 11.3.51

=

This set of coordinates is useful when the components of the motion can be analyzed separately from each other. Projectile motion problems provide a classical example. V

I

..

Path

'I .•

I

I

p

I

t

0

t

FIgure 1.2

x

5

6

CHAPTER CHAPTER 1 1 ••BAsic B.&sicPRINCIPLES PRINCIPLES

Let us investigate a few interesting cases lbr motion in one direction, also known as rectilinear motion. Taking the displacement variable as x, we can express the acceleration in the most general case as a = a(x, .,i, t). If the acceleration is constant,

a=

then we can obtain the velocity and displacement by direct integration as v(t) = v0 + ct, x(t) = x0 + v01 + c12/2, with x0 and denoting the initial displacement and initial velocity. Direct integration can again be used if the acceleration is only a function of time, a = f(r). If the acceleration is a function of the displacement only, a = f(x), one uses the transformation a = dv/dt = f(x). Multiplying both sides by dx, we obtain c,

f(x)dx =

dv

[1.3.6]

= vdv

where each side is a function of x or v only. It follows that each side can be integrated

separately from the other. If the acceleration is a function of velocity alone, a = we write a = dv/di = f(v). f(v). Carrying Carrying the the dt dt term term to to the the right right side and the f(v) term to the left, we obtain

dv

f(v)

[1.3.7]

= dr

where we have again separated the variables so that each side of the above equation can be integrated independently of the other. The case when a = f(x, k) .k)isisleft left as as aa homework homework exercise. exercise. Any Any other other combination lends itself to more complicated problems. The situation gets worse when there is two- or three-dimensional motion, and the acceleration in one direction is related to motion in more than one direction, such as = f(x, y). One has to then go into coordinate systems that take advantage of the properties of the path fbllowed by the particle, as we will see next.

Example 1 .2

I

A projectile is thrown from an inclined surface with an initial velocity v0 and an angle of 0 = 30° from the incline, as shown in Fig. 1.3. The plane of the incline makes an angle of = 15° with the horizontal. Find the time elapsed before the particle falls to the ground and the distance traveled along the incline.

'0

x

Encline

Figure 1.3

1.3

REVIEW OF VECTOR ANALYSIS

Solution The equations of motion in the horizontal and vertical directions are independent of each other and have the form

=0 =

lal

= —mg

which can be integrated directly to yield the velocity and displacement relations

— z(t) ==V:Ot V:Ot— z(t)

— gi — gi

= where are

x(t) =

= constant

=

ibi Ibi

5gt2

of the the initial initial velocity. velocity. From Fig. 1 .3. the initial velocities and v:o are the components of' = VØCOS(O VØCOS(O

= V0COS = V0COS 150

—

V() ==V()Sifl(fI — V()

sin 15° 15°

id

Denote the time it takes fc)r the projectile to touch the ground by t1 and the coordinates From Fig. 1.4 (not drawn to scale) we can relate of that point by xj = x(tj) and Zj = xj by Zj and x1 —

[dl

(f) Xj tan tan(f) = Xj

and z1 z1 from from Eq. [hi, we obtain Substituting the values for for x1 and tan th

Ic]

tan tan 4')t, = 0

[11 if]

=

—

which leads to the following equation for +

—

Solving for

for the final time, we obtain V.0 + 1

vosin 15° + v0cosl5°tan 15°

4v0 sin 15°

=

g

1

g

.v

—- XI

Figure 1.4

igi Igi

7

B

CHAPTER

1

• BAsIc PRINCIPLES

The final distance traveled is calculated by substituting the value for d], and we have .1

)

sin 150 cos 15 0

V , 0tj = = V

-,

1.3.2

(f) =

—

g

[hi

g

g

z(11) = — Xf

into Eqs. [hi and

tan 150 =

g

[Ii

CURVILINEAR COORDINATES

Properties of the path followed by a particle turn out to be extremely useful quantities for understanding the nature of the motion. This observation has led to the development of curvilinear coordinate systems. There are several such coordinate systems that one can develop. We will begin with a coordinate system that is entirely based

on the properties ol' the path. We will then study two commonly used coordinate systems that make use of the properties of the path as well as the position of the particle. Curvilinear coordinate systems are referred to as intrinsic, as these coordinate systems depend on the path followed by the particle.

Normal.Tang.ntlal Coordinates

This coordinate system is attached to the

particle whose motion is considered (see Fig. 1 .5). The distance traversed along the path is designated by s, measured from a reference position. This coordinate system is primarily used to describe the nature of the path followed by the particle.2 The variables used are referred to as path variables. While it may appear to be more more desirable desirable from from aa mathematical mathematical perspective to just look at the distance between the initial and final positions. considering the path

AS

t

r( .v) v)

r(s + As) r(.s

Figure 1.5

2ln this 21n thisregard, regard, the the normal-tangential normal-tangential coordinates do not constitute a coordinate system that can describe displacement, ment, velocity, velocity, or or acceleration. Rather, they are a means of obtaining information about the path followed by the particle. The other curvilinear coordinates that we will study in this section can, in essence, be derived from the path variables.

___

I .3

REVIEW OF VEcTOR ANALYSIS

followed and the total distance traversed along the path is very important. Someone

raveling from one location to another. say by driving, usually selects a route that minimizes the distance traveled along the road (the path). We define two principal directions to describe the motion. The first is along the angent to the curve and along the direction of the motion. This direction is called the tangential direction. We denote the unit vector associated with this direction as er.. er..Consider ConsiderFig. Fig.1 1.5.5and andthe theposition positionofofthe theparticle particleafter afteritithas has traveled traveled distances distances of s and s + The associated position vectors, measured from a fixed location, are denoted as r(s) and r(s + as), respectively. Define by the the difference difference between between r(s) r(s) and r(s + and thus = r(s +

—

r(s)

11.3.81 11.3.81

From Fig. 1.5, as

becomes small and the same same length length and and become have the 'arallel to each other. Further, 'arallel becomes aligned with the tangential direction. We define the unit vector in the tangential direction as

e1 == lim e, lim

=

As '0

dr ds

[1.3.91

The unit vector e1 changes direction as the particle moves. We can obtain the elocity of the particle by differentiating the displacement vector with respect to rnne. Using the chain rule for differentiation. v(t) =

dr

=

di

dr ds ds di

11.3.10]

Now using the definition of of e, e, from from Eq. Eq. [1.3.9] and noting that the speed v is the [1.3.9] and the rate rate change of the distance traveled along the path, v ds/dt. we obtain v(t)

—

11.3.11]

The second principal direction is defined as being normal to the curve and di-

rected toward the center of curvature of the path, as shown in Fig. 1.6. This direction defined as the normal direction. The associated associated unit unit vector vector is denoted by The center of' of curvature curvature of of aa path path associated associated with with a certain point on the path lies along a line perpendicular to the path at that point. An infinitesimal arc in the vicinity of that point can be viewed as a circular path, with the center of curvature as the center the circle. The radius of the circle is called the radius of curvature. Because the normal to the curve is perpendicular to the tangential direction, the two unit vectors are orthogonal. that is, • = 0. Differentiation of Eq. [1.3.111 with respect to time gives the acceleration of the particle a(t) as

a(t) =

= i'e, + vet

11.3.12]

obtain the derivative of e1 we displace the particle by an infinitesimal distance ds. and refer to the unit vectors associated with the new location by e,(s + ds), as shown in Fig. 1.6. The center of curvature associated with all points oti the arc is the same. The arc length can be expressed as 1s = pd4: d4 isisthe the infinitesimal infinitesimal angle To

9

__

10

CHAPTIR 1 • B&sic B&sic PRINCIPLES

er(s)

+d.s) + ds)

I

+

Center of curvature

'1

Figure 1.6

Figure 1.7

traversed as the particle moves by a distance dx. ds. Defining the vector connecting et(s) and + ds) by de1. we can can write write de1 de, = e1(s + ds) — e1(s). From Fig. 1.7, the angle between e1(s between e,(s + ds) and e,(s) is very small, so that fde,J de,J

C/S

sin dcble,(s)I

— =— p

11.3.131

or de,

—

dx

[1.3.14] 11.3.14]

p

The radius of curvature is a measure of how much the curve bends. For motion along a straight line, the curve does not bend and the radius of curvature has the value of infinity. For plane motion, using the coordinates x and y such that the curve is described by); = y(x). the expression for the radius of curvature can be shown to be

I

d2v/dx2 (1 + (dv/dx)2)312 I

P

11.3. 151 11.3.151

The absolute value sign in this equation is necessary because we defined the radius of curvature as a positive quantity. Considering the sign convention that we adopted above, we have de1 de,

ds

[1.3.161

p

Using the chain rule, we obtain the time derivative of e, e1 as as

de, de, il1s

e,,

ds di

p

—=

[1.3.17J

1.3

REVIEW OF VECTOR ANALYSIS

Introduction of this this relation relation into into Eq. Eq. 111.3.12] 1.3.12] yields

a(t) = ij(t) =

11.3.181

+ p

The first term on the right in this equation is the component of the acceleration due to a change in speed, referred to as tangential acceleration (a,). The second term is the contribution contribution due due to a change in direction, referred to as the normal acceleration or cent ripetal acceleration (a,1). The acceleration expression can be written as

11.3.191

a(i) = a,e, + a,1e,7

with

=V

a,

U-

11 .3.20a,bl

p

It is a common misconception to treat an object moving along a curved path with constant speed as having no acceleration. Note that the normal component of the acceleration is always directed toward the center of curvature. The normal and tangential directions define the plane of the motion for that particular instant. This instantaneous plane of' motion is called the osculating plane tafter the Italian word osculari, which means to kiss). The orientation of the osculating plane changes as the particle moves. This is referred to as the twisting of the o.vculatingplane. osculating plane.The Theunit unit vector vector e/). e/). referred referred to to as the unit vector in the binorinal direction and defined as e1, = e1 X is perpendicular to the osculating plane. Consider, for the sake of illustration, plane motion as shown in Fig. I .8. The direction of e1, and hence the osculating plane alternates as the curvature of the path switches from convex to concave. The unit vector in the binomial direction is used to obtain the derivative ol' the unit vector in the normal direction. While e1, is not needed to describe the velocity or acceleration, it is used to describe the evolution of the plane of motion. The twisting of the osculating plane is a path parameter. e1)

(into pagc)

a

e1

e,, )

I

,

1

Figure 1.8

11

12

PRINCIPLES CHAPTER 11• •BASIC BASICPRINCIPLES CHAPTER

To analyze the movement of the plane of motion, we investigate the derivatives and e1, with respect to s. To this end, we take the derivative of e() using the of definition of' the cross product as de,, c/s

=

dc,,

—XXe,1 e,1 +— dx ds

e, X

11.3.21]

e,j/p, so that the second term in the right side of Using Eq. 11.3.16]. de,/ds the above equation vanishes. From the first term on the right side we conclude that the rate of' change of along the path has no tangential component. To show that de1,/ds has no component in the binorrnal direction either, we differentiate the dot product • eb = 1. with the obvious result (I

— (e1,

deh

• e1,)

[1.3.221

= 2eb• —— = 0

ds It It follows that deh/ds can only have a component in the normal direction. This can be explained physically the same way we explained Eqs. 11.3.13] through [1.3.16]. We define a quantity denoted by T, which is called the torsion of the cun'e, ds

as de1,

1

[1.3.231

— —---e,, T

The torsion of the curve is a measure of how much the plane of motion twists, or how the osculating plane changes direction. For plane motion, while the curve may change from convex to concave, the binortnal direction does not change and hence r is equal to infinity. We next obtain the the derivative derivative of' of with respect to s. From the above discussion we know that de,1/ds cannot have a component in the normal direction. We proceed to write this as

ci,,

[1.3.24] 11.3.24]

=

and c2 are coethcients to be determined. We make use of the property that the unit vectors are orthogonal to each other and write c1

•D =

=0

11.3.25]

0

Differentiation of these equations with respect to s and using Eqs. [1.3.161 and [1.3.231 yields de,1 de,3 dc,, (/Cj, =0 +eh' = ——+e,,• dx T dx (is ds (is

e,•

de (IS (IX

+e1•

de,1

1

= —+e,•

—

()

c/s (IS

ds

Taking the the dot dot product productof' of Eq. Eq. 11.3.241 11.3.24] with e, and 11.3.26], we obtain

Cl =

1

1)

=

I —

T

=0

(1.3.26]

and comparing with Eqs.

[1.3.271

1 .3

REVIEW OF VECTOR ANALYSIS

We are now in a position to write the rate of change of the unit vectors in the normal-tangential coordinate system as I de, = ds p

I

=

i/s

+

——e,

p

de,,

I

—eh

r

1

=

dx

——e,,

11.3.28a,b,cI

Eqs. Eqs.

[I .3.28a,b,c] are known as Frenet's formulas. They enable us to describe how the curve and the associated unit vectors change. Note that up to now, we derived the unit vectors and their derivatives as a function of s, the distance traversed along the path. But we did not find an expression for s itself. To accomplish this, we need to express position in terms of a path variable. Denoting this variable by a, we can write the position vector as

r = r(a) = x(a)i

11.3.291 11.3.29]

+ y(a)j + z(a)k

Using the definition ole1 in Eq. [1.3.291. we obtain

dr

dr i/a

dx ds

dcx ds

(dx.

=(

+

dy.

dz

dcx

dcx

\da j dx

kj

[1.3.301

We then take the dot product of e with itself: /

e,•e, =

=

1

I

\ ds

2

+1 \dcx

j \da) I

fdz\ +(

fdv

I

I

\dcx)

[1.3.311

which can be solved to yield 1,

ds =

iidx \

+

I

fdy

\

+

I

i idz

\

dcx

\da) I

I

[1.3.321

Integrating this expression, we obtain

s

U iidxi =

dcx

/

/

)+

dcx

)+

dcx

dcx

)

[1.3.331

where a0 is the initial value of the path parameter. When the path parameter is time, we get the familiar expression of displacement being the integral of the velocity over time. We next express the radius of curvature, torsion, and the unit vectors in terms of the path variables. We first write the unit vector in the tangential direction and the derivative of the position vector as =

dr = ds

dr dcx

dcx ds

=

r'

r = se1

—

s s'

[1.3.34a,b]

Primes denote differentiation with respect to the path variable a. Difièrentiation of Eq. [1 .3.34bJ with respect to the path parameter a, and using Eq. [1 .3.16], yields =

s"e1

+

,de1

ds

dsda

=

s"e1

+

p

11.3.351

13

14

CHAPTER 1 S BASIC PRINCIPLES

As one would expect. if the path variable is selected selected as time. Eqs. Eqs. [1 [1 .3.34b] .3.34b] and [1.3.351 [1 .3.35j yield yield the expressions for velocity and acceleration. To express the unit vector in the the normal normal direction in terms of the path variables, we multiply Eq. [1.3.35] by s' and subtract from it Eq. [1 [1.3.34b1 .3.34b1 multiplied multiplied by by s", with the result by

(r"s' — r's") (r"1s" [1.3.361 (s) It is customary to express e,, in terms of the first derivative of s. Noting from Eqs. [1 .3.34h] and [1.3.351 that r" r' = s's" and substituting this relationship in the e,1 = e,1

above equation, we obtain (s

(r"(s')2

—

r'(r" • r'))

[1.3.371

To find the unit vector in the binormal direction, we use the definition

= x and substitute the values for e1 and e,1 e, X e,, from from Eqs. Eqs. [1 [1 .3.34a] .3.34a] and and [1.3.36], [1.3.36], with with the result e/) =

X s

—

(S )-

r's") =

—

(s )

.,

(r' X r")

[1.3.381

Finding the radius of curvature and the torsion requires manipulation of the above equations. The radius of curvature is found by making use of Eq. Eq. 11.3.371 11.3.37] and the property that the norm of e,1 is 1. To find the torsion r we invoke Eq. [1 .3.28c] and carry out the algebra. The results arc —

p

= =

1

(s')3

•• r")s'2 r")s —

—

(r' • r")2

I

T

IF _pX i")] r")] (f)(,Ir [r"ii ••(r (r' X —

(1)6

11 .3.39a,b]

The derivations are left as an exercise. Note that to find p one needs the second

derivative of r with respect to the path variable, whereas to find the torsion r the third derivative of r is required. This can he explained by the following observation: Let time be the path variable. Given the velocity and acceleration of a particle at a point in time, one can determine what the plane of the motion is and the radius of curvature, but not how the plane of motion is twisting. Finally, in terms of the derivatives with respect to s. we can write d2r dr =e1 = —e,, [1.3.401 [) ds (IS

dr

1

We now compare the two coordinate systems that we have seen so far. With rectilinear coordinates, one describes velocity and acceleration as the rate of change of absolute distance from an origin. With normal-tangential coordinates, one describes velocity and acceleration using properties of the path that the particle follows. The distance from an origin does not come into the picture. The two descriptions can be used together to give a beflei- understanding of the nature of the motion.

Example

1.3

A particle moves on a path on the xv plane defined by the curve v = 3x2. where x varies with the relation x = sin a. Find the radius of curvature of the path and the unit vectors in the normal and tangential directions when a = ir/6.

1.3

11

REVIEW OF OF VECTOR ANALYSIS

Solution The position vector can he written as

r=

Ia] lul

ai + 3 sin2 aj

sin

so that the unit vector in the tangential direction is

dri

dr

1

=— —[cosai [cos ai + sin a cos au e, = + 66sinacosajl = — — = dcxs ds SI where

s' = ds/da. Noting that the magnitude of e, is s.c

1

cos2aa ± 36

=

I

[bi Ibi

we write

a cos- a

[ci

find the radius of curvature we use Eq. [1.3.39aJ. which requires expressions for r' and r". These derivatives are To

ar 1

r =

da

.

.

H

ar

I

.

r = —--,- = — sinai +

= cos ai + 6 sin a cos aj

dcr-

.

a — sin

S

a)j [dl

Evaluating these expressions at a = ii-/6. we obtain =

= =

+ 3j)

+ 3j

[ci Ic]

which, when substituted into Eq. [1 .3.39a1 yields = =

_!_

p When

a=

• r")s'

(s)-

(r' r")2 =

—

0.

[1] 11]

1897

ir/6, the unit vector in the tangential direction has the value =

+

,L

/7 5

\

2

j

2

0.31621 + 0.9487j + 3j) 3j) ==0.31621

= 10 I0

which yields

To find the unit vector in the normal direction we need to use Eq. =

(s)

—

[g]

r'(r" er'))

= —0.94871 + O.3162j

1k]

Note that we determined all the path parameters using their given formulas. We could Note have solved this problem by assuming that the parameter a is time and by using the expressions for the normal and tangential accelerations.

The motion of a particle is described in Cartesian coordinates as .v(t) = 2t 2 + 4t .v(t)

v(t) =

0.

It + cos t

I

z(t) =

Find the radius of curvature and the torsion of the path at time t =

3t

[a] Ia]

0.

Solution From Eq. 11.3.20bj, the radius of curvature is related to the speed of the particle and its normal

acceleration as

Ibi

Example

1.4

5

16

CHAPTER 1 • B.ksic PRINCIPLES

and the components of the acceleration in the normal-tangential and Cartesian coordinates are related by +

=

+

=

+

[c] Ic]

We We can obtain the components of the velocity and acceleration by differentiating Eq. [a]: [a]:

= 4, 41 + 4

= 0.3t2

= 4

=

—

sin!

[dj Id]

cost

—

The square of the speed of' of the particle at t = 0 is +

v2(0) =

[42

+ ±2(0)]

+0+

=

321

1.] 1.1

25

The tangential acceleration can be found by differentiating the expression for the speed (1 (7

At

t

d.... d...

= — 1V2(t) 'v2(t) + v2(t) +

=

cit

1

—

—

.

1

+

[f] If]

+

= 0, we have

4(4)+0(—1)+3(0) 4(4) + 0( — 1) + 3(0) =

+ 02 + 32}1/2

{42

16

Igi [gJ

=

Substituting the numerical values into Eq. [c]. [c]. we obtain

+

=

+

=

+

42

+ 12 + 0

[hI

+

which can be solved for p to yield =

P5

252

/

- (16/5)2

III

9.615

=

Next, we find the radius of curvature using Eq. [1 .3.39a]. Noting that the path variable variable

is time, at t = 0 we have 41 + 1(0) = 41 + 3k

=

41 —

j

+

=

+ ±2(0) ±2(0) =

5

flJ [fi

Substituting these these values values into into Eq. Eq. 11.3.39a1 1l.3.39a1 we obtain I

I

—

p

125

17

2 — —

62 1162

13 1.

[kI 1k]

125

which gives the same result as the one obtained in Eq. [ii. To find the torsion torsion of' of the curve we make use of Eq. [1 .3.39b1. We note that the third derivative of the position vector is required. From Eq. [dl [dl we obtain

r" so so

that that at attt =

0,

r(t) =

= r(t) = (0.6 + sin t)j

[I]

0.6j. Introducing this and the expressions for r' and r" into Eq.

[1.3.39b1 we obtain 1

=

96152 {(4i

—

j). j). [(41 [(41 ++ 3k) X 0.6j1} 0.6j1} =

0.04260

Im]

1 .3

REVIEW OF VECTOR ANALYSIS

that the torsion of the curve at -r = 0 is T = 1/0.0426() = 23.47. Comparing with the .alue of .ilue of the the radius radius of of curvature, curvature, we we see that the curve is bending more than it is twisting at instant.

The above analysis could also be carried out using a vector approach from the beginWe can write the velocity as v = 4i + 3k, from which we obtain thc the unit vector in the direction as

4i+3k 41+3k

v

In]

=

The component of the acceleration in the tangential and normal directions can then he as a1

a,, = a — (a•e,)e,I =

= a•e1

For the problem at hand. a =

41

—

101

—

j, so that substituting it into the above equation yields

he results that we obtained earlier. This latter approach is in many cases more efficient in btaining the path parameters.

Cylindrical Coordinates

This set of coordinates is particularly useful if the

particle article is moving along a curved path, the position of the particle is of interest, and one component of the motion can be separated from the other two. Fig. 1.9 describes this coordinate system. The inertial coordinate system xv: is chosen such that the component of the motion that can be separated from the other two is defined as the z direction. We take the path of the particle and project it onto the xv plane. The parameters describing the motion are: 1.

2.

The height z4 The absolute distance from the origin of the coordinate system to the projection of the path of the particle on the xv plane, denoted by R or r.

a..

Original path

'I

Project iOfl Onto xv plane

x p.

Figure 1.9

1

7

18

CHAPTER

3.

1•

BAsic PRINCIPLES

On the xv plane. the amount of counterclockwise rotation from a starting line (usually selected as the .v axis) to reach the line joining the origin and projection of the particle on the xv plane. Denoted by 6 and measured measured in in radians. radians.

The selection of the origin of the coordinate system is very important when using cylindrical coordinates, as R and 0 change by changing the location of the origin. The first principal direction is fixed, and it is the thc z direction, with associated unit vector

k. I describe the component of' the motion in the xv plane, we define two perpendicular directions. The radial direclion is defined as outward from the origin of the coordinate system to the projection of the particle on the x xv plane. plane. The The associated associated unit vector is referred to as er. The transverse direction is perpendicular to the raEssentially, the radial and dial direction and it is denoted by (9, with unit vector transverse directions are obtained by rotating the x and v axes counterclockwise by and k form a mutually orthogonal triad, 0 about the z axis. The unit vectors with e,. X e0 = k. For plane motion, these coordinates are referred to as polar coordinates. The unit vectors can be expressed in terms of the unit vectors in Cartesian coordinates as e,. = cos(9i + sinOj

e9 =

—

sinOi + cos0j

11.3.411

The position of a particle is expressed in cylindrical coordinates as = xi + yj + zk [1.3.42] r(t) = Re,. + zk = Rcos0i + RsinOj + To obtain the velocity we differentiate this equation, to get To = is(t) = Rer + + ±k 11.3.431

of the which requires the derivative of' the unit unit vector vector in the radial direction. To calculate this derivative, consider consider the the particle particle at ataa point P', as shown in Fig. Fig. 1. 1.10. 10. The coordiDenoting the unit vectors associated with nate system has rotated by an angle of we relate them to erW) and e0(0) by the new position as er(0 + and e0(O +

er((9 +

=

e0(8 +

= —e,.W) —e,W) sin L\0 +

sin cos L\0

e,. ((1 + e,.((1+M

V

e9(O)

er(E1)

.1

Figure 1.10

[1.3.44] [1.3.441

1.3

Using a small angles assumption of sin approaches zero, we obtain as •

urn

•

Jim

cos

[er(O +

—

er(O)I

[e6(0 +

—

eo(O)1

=

19

REVIEW OF VECTOR ANAlYSIS

and taking the limit

I

der dO

de0 dO

=

—er

[1.3.45] 11.3.451

Using the chain rule of differentiation and Eq. [1.3.44], we obtain = der dO = 0e0

.

•

dO

=

dee, dO dO

di

=

Oe,.

[1.3.46] 11.3.461

which, when substituted in the expression for the velocity, yield +

v(i) =

+ ±k

[1.3.47] 11.3.471

The first term on the right side of this expression corresponds to a change in the radial direction and the second term to a change in angle. The third term is the component of the velocity in the z direction. In a similar fashion we can find the expression for acceleration. Differentiation of Eq. [1.3.471 yields

a(t) =

+

=

+

ROe6

+ ROe6 ROe11 + ROe6 ++ ROe11

[1.3.481

Substituting in the values fbr the derivatives of the unit vectors and combining terms. we obtain

a(t) a(t) = =

(R

—

R02)e,.

+ (RO + 2R0)e0 +

[1.3.491

We can attribute a physical meaning to the acceleration terms. The first term. R. describes the rate of the change of the component of the velocity in the radial direction. The second term, R02, is the centripetal acceleration. It is always in the negative radial direction as R is a positive quantity. The term RO describes the acceleration due to the change in the angle 0. The last term. 2R0. is known as the Coriolis acceleration, named after the French military engineer Gustave Coriolis. who first explained the significance and existence of this component of the acceleration. it results from two effects: The first effect is due to the differentiation of the er term is due to in Re,., and and itit is is associated with a change in direction. The second effect is in Rer, and it is associated with a change in the differentiation of R in the expression ROeS. and magnitude.

A car travels with constant speed v on a spiraling path along a mountain. The shape of the mountain is approximated as a paraboloid with base radius a and height h. as shown in Fig. mountain 1.11. It takes the car exactly six full turns around the mountain to reach the top. Find the velocity of the car as a function of its radial distance R from the center of the niountain.

Example 1 .5

20

CHAPTIR II • BASIC PRINCIPLES

I:

-L

V

Figur. 1.11 Solution The height reached by the car is related to the radial distance by the relation

h—:=

[a]

The height is also related to the angle traversed by

Ib] One can express the height and the angle traversed as a function of the radial distance R as 12'iT (7

a

Ii

= 127r(l 121r(1

—

[cI [ci a—

Differentiating Eqs. [ci and introducing into the expression for velocity, we obtain

0=

=

—24w

RR

[dl [di

a

a— —

and Re,-++ ROe() Rt9e() + zk v == Re,zk == Re,. Re,. —24ir —24w

R2R

a-

—2/i

Rk

,k=ue,

a-

[.1 1.1

We are given that tile the speed is constant, so that we take lake the magnitude of the the velocity velocity as =

R4

1 +5761T2—

=

Defining the variable G as 4 I

[g]

we can write R = v/G. Equation [fj gives the relation for the rate of change of the radial distance as a function of the radial distance itself. In a similar fashion, we can come up with an expression thr the magnitude of' the acceleration.

1.3

REvIEw OF VECTOR ANALYSIS

1• 1'

.1'

x

Figure 1.12

Spherical Coordinates

When neither one of the components of the motion is from the other and a path-related coordinate system is needed, spherical coordinates coordinates are are suitable. suitable. The The configuration configuration of this coordinate system is shown in Fig. 1.12. The parameters describing the path are the absolute distance from the origin of the coordinate system to the particle, denoted by R, and two angles 0 and referred to as the polar and azi,nuthal angles, respectively. Note that the parameter R used here is different from the parameter R used in cylindrical coordinates. The principal directions are referred to as the radial, polai; and azimuthal. The radial direction is defined as outward from the origin to the particle. The corresponding unit vector is denoted by eR. The other directions depend on the polar angle angIe 0 and the azimuthal angle The azimuthal angle is the angle between r and the z axis, and the polar angle 0 is the angle between the x axis and the projection of r the xy plane. Note the similarity between the polar angle 0 and the transverse angle when using cylindrical coordinates. It follows that the polar direction, with the unit vector e6. is tangent to to the the circle circle obtained obtained by by travel-sing traversing the component of r(t) in the xv plane. The unit vector associated with the azimuthal direction is denoted by and it is tangent to the circle on the x'z plane. which is obtained by rotating the xz plane about the z axis by an angle of 0. The three unit vectors are mutually orthogonal, with the relation

exXem =

11.3.501

We express the coordinates of a particle as

x = Rsin4cos0

vv = Rsindsin0

z=

[1.3.511

and the displacement vector has the form

r= ReR

[1.3.521

To obtain the velocity, we differentiate the above equation.

v(t) = r =

ReR

+

R

(9eR dO

dek

+ R94

dt

[1.3.531

21

___

22

CHAPTER 1 • BAsic PRINCIPLES

indicating that the derivatives of eR must he found with respect to both angles. To accomplish this, we will obtain the derivatives of eR using their expressions in Cartesian coordinates. From Eqs. [1.3.511 and [1 .3.52], we can write

+ sin4sinOj sin4sinOj + cos4k

eR =

yields

Differentiation of this equation with respect to 0 and thcos cosOj Oj = sin sin Oi + sin sin th

= — sin deR -

= cos

U

COS

+

.

•

S U

sin Oj Oj — sin

COS

11.3.55]

sin Oi + cos Oj)

—

/k =

[1.3.541

4(cos Oi + sin Oj) — sin U

cos

•

From Fig. 1.12, we can express the unit vectors in the polar and azimuthal directions as

e0 =

—

11.3.561

+ sinOj) — sin

=

sin0i +

=

Note that when the azimuthal angle

9Q0,

spherical and cylindrical coordi-

nates coincide. Indeed, we have

= 900) =

= 900) =

= 900) =

e,

—k

(1.3.57J 11.3.57J

Eqs. [1.3.56] into 11.3.55] we obtain Introducing Eqs. rieR -

dO

deR

.

= sin

c14)

= ed, ed,

11.3.581 11.3.581

as timederivative derivative of of CR as toobtain obtainthe thetime used to The above relations can he used

dek sin = 00 sin di

[1.3.59]

+

As aa result, one expresses the velocity as

v(t) =

ROsin4)e0 + R4e4, ReR + ROsin4)e0 ReR

(1.3.60] 11.3.60]

acceleration, we we need need to to generate generate the time derivatives of the unit To obtain the acceleration, To vectors in the poiar polar and azimuthal directions. Using a procedure similar to the above, we obtain cit

= —o sin 4)eR 4)eR

=

—4)eR

—

0 cos

+

[1.3.61] [1.3.611

Using this equation and manipulating the algebra, the expression for the accel-

eration has the form

a(t) =

= (R

—

R62 sin2 4'

—

R4i2)eR + (RO sin

+ (R4' + 2Rth — R02

+ 2R0 sin 4) + 2R04' cos

[1.3.62]

1.3

23

REvww OF VECTOR ANALYSIS

In Eq. [1.3.62], the terms having the squares of the first derivatives correspond to ne accelerations, and the mixed first derivative terms correspond to the

Cciolis accelerations. accelerations. The The second second derivatives indicate accelerations in the radial and in the polar and azimuthal angles. The length of the spherical pendulum shown in Fig. 1.13 varies by the relation L = 2 + iTt iTt m. m. The The pendulum pendulum spins spins with the constant constant rate rateof of00 = 2 radls. We are given that the .mgle /3 is related related to to the the length length of the pendulum and 0 by L26 sin2 /3 =

Iai IaI

8 1W/S

Assuming that that /3 —TI/2 /3 is always in the range —TI!2 the pendulum at t = 0. 0. s.

iT/2, find the velocity of the tip ii-/2,

/3

1

attach a set of spherical coordinates to the pendulum. Note that according to our definition the polar and azimuthal angles, // and and are related related by by/3 /3 = — th. We observe that = sin/. Fig. 1.14 shows the coordinate system in side view using the z and radial ixes. Considering the range of /3. we can obtain the expression for sin sin [3 [3 from trom Eq. [aj as We

I_i.=

sinf3=/ sinf3 = I

L20 V V L20

Ibi [bi

A-.

2 + Sin iT!

To find /3. we differentiate this equation. with the result cos/3/ =

2ircos TI! —

/3=

(2 +

1I

—

sin2

—2ircosrn 171)2 (2 ++ sin (2 sin171)2

[ci

the velocity velocity is is given given hy Eq. 11.3.60]. We identify the individual values The expression for forthe as

R

= I. =

[dJ

ITCOS7T! ITCOS ITt

I.. —

(f)

V

H3

L

V

\

(e8 inside paper)

F

0 /

I

I

Figure 1.13

eth

Figure 1.14

eR

Example

1.6

24

•

CHAPTIR I CHAPTIR

PRINCIPLES

At! = O.ls.wehave R=

2

+sin(O.l'ir) + sin(O.l'ir)— = 2.309 2.309rn rn — = 0.8662 + sin(0.1,r) 2 + .

lr)'

I

.

= -

sin4 = sinf3[e]

R = ircos(ftlir) 2.988 'ii ircos(O.lir) = 2.9S8iii =

(2 +

—

= 2.243radJs

which. when substituted in Eq. [1.3.60]. yields

v(t) =

=

ReR + RA sin 4)e() +

+

+

5.

1 79eth mis

III

Equation Eal is an angular niomenturn conservation relation. This relationship arises hecause the speed of the pendulum can be written as + R2(02 R2(02 sin2

±

Ig]

The speed u is independent of 0. Many systems in dynamics have this property.

Mixed Descriptions

As discussed earlier, in certain cases it may not be sufficient to use a single set of coordinates to solve a problem: one may also need to take advantage of another coordinate system. The different coordinate systems provide different types of information about the motion, and the added intbrmation arising from the use of more than one coordinate system facilitates our understanding of the nature of that motion. Usually, one exploits the relationships between the two coordinate systems by first writing the unit vectors of the coordinate systems in terms of the motion variables. Then, one considers the relationship between the unit vectors in the different coordinate systems. One question that arises is whether one can use any set of parameters to define a coordinate system. The answer to this question is positive, provided that three parameters m P2e p3 p3 can he found such that there exists a unique transformation between the components of the motion in rectilinear coordinates x. v, and z (or any other proper coordinate system) and and For example. for cylindrical coordinates, = R, P2 = 0. and transformations from .v. .v. v. and z to R, = z. The transfbrmations 6. and z are given in Eq. [1.3.42]. Considering these equations and Fig. 1.9. we can relate R. 0. and z to .v, v. and z as

i'.

R=

+ v2

00

=

=z

11.3.631

where the range of interest interest in in the the inverse inversetangent tangentisis(—17-/2, (—ii-12, 7r/2). ir/2). A proper selection

of the coordinate frame is crucial to the understanding of the problem and to the solution.

There are several other parabolic and hyperbolic coordinate systems that facilitate derivation of the governing equations, as well as the solution, for a specific problem. Examples can be found not only from motion analysis but from other problems [ems such as heat transfer. To develop a coordinate system with orthogonality properties among its components, additional requirements need to he introduced. These requirements are described in the text by Ginsberg listed listed in in the the reference reference section section at the end of' the chapter.

1.3

25

REVIEW OF VECTOR ANALYSIS

The reader is always encouraged to explore the possibility of using more than one coordinate system when tackling a dynamics problem. One note of caution is in order, though. When selecting coordinate systems and the variables associated with them, be careful to avoid the ambiguities that can result from an improper selection of the variables. A good way to avoid this problem is to make sure that the transformation from one set of variables to another is indeed unique.

In Fig. 1.15. the pin attached to the circle 01:6-inch of 6-inch radius is sliding in the slot with the constant speed of speed of vv = 22 in/sec. Find the values of 0 and 0 at the instant when = 900.

Solution will make use of polar as well as as normal-tangential normal-tangential coordinates. The The geometry geometry of' of the system is shown in Fig 1.16. The expressions for Rand 0 arc

We

,J62 + 92 = 10.82 in R R = ,J62

0 = tan'

[a]

= 33.69°

The relationship between the two sets of unit vectors can he written as

e, =

e,, e,,

cos Oer — sin

=

— cos —

—

[hi

sin OeR

where sinO 0.5547, cosO = 0.8321. Writing the expression for velocity in the two cooriinate systems,

V = tie1 = kCr +

Ic] [c]

R(9e(,

so that to find R and (1 we take the inner product of v with er and v • er = V • e(,

k =

ye1 • e,.

= R0 =

thus

= v cos 0 = 2(0.8321) = 1.664 in/sec

tie1 • eA

=

—v —

sin 0 Sin

=

109 in/sec

[d] Id]

= —0.1025 — 0.1025 rad/sec =

[e] le]

2(0.5547) =

—

— 1.

with the conclusion R

= 1.664 1.664 in/sec

=

— 1.

109 = —

11

()

Cr

F

p e,,

6" 911

A

FIgure 1.15

91P

FIgure 1.16

0

Example

1.7

26

CHAPTER 1 •

PRlNcIPI2is

To find the acceleration. we make use of the fact that the speed is constant, So that the acceleration has the form

a=

e,,

p

=

1? — (( 1?

with the radius of curvature being p —

6

R02 )er )er + (RO

Ifi

+ 2 Ro Ro

in. Taking the inner product of a and e() we obtain

= Rq9 = Rq9 + 2R6

= —(—cos0) p

Igi

Solving for 0 we obtain (1

= —

1.4

(v2cos0 ± + 2R0) ± p+

(40.8321 ÷÷ 6— (40.8321 6—2' 2• l.664•0. l.664•0. 1025)

R

10.82

= —0.01974 radlsec

[hI Ihi

NEWTONIAN PARTICLE MECHANICS

Newtonian mechanics is based upon the three laws of Newton, and the counterpart of Newton's second law for rotational motion. It is valid for reference systems at rest or moving with uniform velocity with respect to each other. Such reference systems approximationto toan an inertial reference frame is to are known as inertial. The best best approximation consider the distant stars lixed. A less accurate approximation is to consider the sun fixed. in the vicinity of the earth, if the velocities involved are much less than the Or ifif the the distances distances traversed traversed are are much smaller than the radius of the escape velocity,3 or earth, and if the duration of the motion is not long, the earth is used as an inertial reference frame. Isaac Newton developed his three laws while he was working on the motion of rigid bodies. While the laws govern the motion of the center of mass of a body, they do not account for the rotational motion. The rotational equations relate the moments applied to a body to the change in angular momentum: these were developed by Euler around 1750. (Newton published his three laws in his Principia in 1687, but he had developed them about 20 years before. while he was at Cambridge University.) Here, we will consider Newton's equations within the context of particles. To this end, we first describe what we mean by a particle. A particle is defined defined as as aa body bodywith withflO flO physical physical dimensions, thmenszons, implying that the entire mass of the. the particle particle isis concentrated concentrated at at one one point. point. This, This. of of course, course, is is an an idealidealization and it is a valid assumption only when the physical dimensions of the body are small compared to its weight. Neglecting the physical dimensions implies that the rotational motion of the body is ignored. Consider a particle of mass in and let F(t) denote the sum of all forces acting on the particle. Define the linear niomeniwn of the particle by p(t) = Fnv(t). The 3The escape velocity of a spacecraft is the speed that the spacecraft must attain in in order order to to escape escape From the

gravitational attraction of the celestiol body it is is orbiting and to assume a parabolic orbit.

1.4

ma

F F-,

27

NEWTONIAN PARTICLE MECHANICS

=

I.,' I,,'

Figure 1.17 explanation of linear momentum is the tendency of the body to continue to The mass of the body is its resistance to translation. Newton's laws can be as follows. tbllows. First law: If no forces act on a particle, the particle retains its linear momentum. If the particle is at rest it remains at rest, and if' it is moving, it moves with linear momentum. A particle of constant mass moves along a straight line vU) = constant. constant velocity. In other words. F = 0 Second law: The rate of change of the linear momentum of a particle is equal the sum of' all forces acting on it.

dp(i) di

—

d(rnv(t)) cit

—

[1.4.1]

F

For a particle whose mass remains constant, the above relation reduces to

11.4.21

tna(t) = F

Newton's first law is a special case of his second law. Fig. 17 illustrates this law. We will treat variable mass systems in Chapter 3. exert forces forces upon upon each other, these forces are particles exert Third law: When When two two particles equal in magnitude and opposite in direction, and they lie along the line joining the two particles. the force exerted on particle i by particle we arrive at the Denoting by = —F11. Consider. for example, Fig. 1.18. a dog pulling a cart. conclusion that The free-body diagrams of the dog and cart are given in Fig. 1.19. The friction the dog's legs and the ground provides the forward thrust. The dog and cart exert forces on each other that are equal in magnitude and opposite in direc1

.

tion.

ing

pp1 t1V2 LV,,

Figure 1.18

Figure 1.19

tN1

N4

28

B,&sic PRINCIPLES CHAPTIR 1 O B.&sic

p.

In

F

Figure 1.20 A very important application of Newton's third law is in celestial mechanics. The forces that two bodies exert on each other, as shown in Fig. 1.20. are governed by Newton's law of gravitation as 1)12

F12

=

=

-

[1.4.31

which G is the universal constant of gravitation, and r is the distance between the particles. The value of G is G = 6.673(10 ") rn3/kg• s2 in SI units, and G = in

3.439(10-8) ft4/lb

' sec4 in U.S. Customary units. Because G is such a small quantity,

for any two small bodies the gravitational attraction is extremely small. The gravitational force becomes significant when at least one of' the bodies involved is very in the vicinity of a celestial body. Equation large. such as in an analysis of motion in [.1.4.31 canbe beexpressed expressed in in vector form as [.1 .4.3 1 can

= where

—F21

=

Gm1!n7

[1.4.41

r is the position vector between the centers of mass of the two bodies. For

motion near the earth, using the values of mass of the earth as me = 5.976(1024) kg = 6,378 km. we assume that the distance of the body from the and mean radius as surface of the earth is negligible compared to the radius. Defining the gravitational we obtain the force of gravity as F = rn2g, where the mean constant as g = value for the gravitational constant is 124 [1 .4.5] m/s2 mIs2 = 9. )}2 [6. 3Th( 1

Using the above equation to calculate the gravitational constant g is not accurate. Equation .4.51 is is based based on treating the earth as a particle (or as a rigid uniform Equation [.1 [1.4.51

sphere), and it ignores centrifugal effects due to the rotation of the earth. Furthermore. gravitational effects due to the sun and moon also affect the value of g. The radius of the earth is not constant14 and its density is not uniform, resulting in different values of g at different points on the earth. For dynamics problems that do not require a substantial amount of precision. average values of g = 9.81 (or 9.807) 32.2 (or 32.17) ft/sec2 are used at sea level. m/s2 or g rn/s2 4The actual shape of the earth is an oblate spheroid. The earth looks more like an apple, with the radius larger around the equator, smaller at the poles, and with the poles slightly pressed in.

I4

NEWTONIAN PARricLE MECHANICS

A more accurate approximation is the 1980 International Gravity Formula. It into consideration the rotation of issumes that the earth is a rigid ellipsoid and takes into the earth. The approximation for g is given as a function of the latitude as

g=

9.780327[1 + 0.005279 sin2 A + O.OO()023 sin4 A] mis2

[1.4.61

A = 90 — çb is the latitude angle. with 4, being the azimuthal angle defined hen we considered spherical coordinates in Fig. 1 12. Even this approximation is accurate, because it fails to lake take into consideration both the dip in the exactly accurate. shape at the poles and the nonrigidity of the earth. At a latitude of 45° and at level, thecommonly used value of g to 5-digit accuracy is g = = 9.8066 rn/s2 or .

We introduced above the concept of a force without rigorously defining what it :s. It turns out that we know of the existence ol' of forces and we know their effects, we cannot ngorously define what a force is. Here is a commonly used detIniion: son: A force is the effect of one body on anothe!j

We also note that no method exists to directly measure a t'orce. We have developed theories on how forces af'Ièct aflèct systems systems and and have have validated validated those those theories theories by the forces forces in in the the form form of deformations or accelerations. In measuring the effects of' of the essence, in dynamics we use cause and effect relationships. For example. people mechanical bathroom bathroom scale scale read read an an output outputof' of the spring spring eighing themselves on a mechanical Jeflection caused by their weight. multiplied by the spring constant. We can categorize the forces acting on bodies into two two general general types: types: (1) (I) Contact forces, such as friction, impact. impact, spring, dashpot, and so forth. These forces are to a point or an area on the body. (2) Field forces, such as gravitational and electromagnetic. These forces are applied to the body uniformly. A special class of force that is of importance in dynamics is friction. Friction torces are developed when a body in contact with another moving or fixed body has a tendency to slide (slip) over that other body. The contact force between the two bodies is the normal force, hence a reaction forcc. The amount of slippage deon the material properties and surface characteristics (rough. smooth, etc.) the contacting bodies. The study of friction is a very complex subject, and an representation of' of friction friction forces forces is is difficult. difficult. Often, we use the model known as dry friction or coulomb friction to describe friction. This model approximates the friction force as a coefficient of friction multiplied by the normal force. The is an approximate quantity whose value deof friction, denoted by pends on the material properties of the contacting bodies, the relative speed of the contacting bodies, and to a lesser degree. the temperature. Many engineering have tables of the coefficient of friction for a variety ol' of materials and aces.

Another factor that affects the value of the coefficient of friction is whether the ,int ofofcontact contactisismoving movingorornot. not.IfIfthere thereisisslip slip between between the the point of of' contact contact and the

29

30

CHAPTER

1•

B.kslc PRINcwUs

surface. one uses the coefficient of kinetic friction. Mathematically.

/ikN

F1 =

11.4.71 [1.4.71

where F1 is the friction force. N is the normal force, and is the coefhcient of kinetic friction. If there is no slip between the contacting bodies the friction force is a quantity less than or equal to the static coefficient of friction multiplied by the normal force, and it can be expressed as F,

11.4.81 [1.4.81

in which is the static coefficient of friction. In general, (When no other information is given, one assumes Sliding begins when the friction force = reaches and after that point is used to describe the amount of friction. The friction force acting on a body always opposes the impending motion. For moving bodies the friction force opposes the velocity velocity o.f of the the contacting point relative to the point of contact. Fig. 1 .21 illustrates this concept. The friction force can be represented in represented invector vectorform form as F1

=

or

——

F1

vt

= —,lkNef

11.4.91

For rectilinear motion one can write F1. F1

=

[1.4.101

in which sign(v) =

1

when v > 0

sign(v) =

—1

when v < 0

[1.4.11] 11.4.11]

The The friction friction force is a nonlinear function. When solving problems involving

friction one must be careful in determining the direction the friction force should be acting and whether there is slipping or not. In many cases the direction of the friction force may not be obvious. To determine whether there is slipping, one can begin by assuming that there is slipping (or that there is no slipping) and then check the validity of this assumption. For example. if one assumes no slipping, one can calculate

V

F

I.' I..

p

F,

I. I-f 1V

FIgure 1.21 FIgure

.Iv N

i

31

NEWTONIAN PART!CI.E MECHANICS

magnitude ofthe friction force and compare it with the maximum magnitude the friction force can attain. When solving a dynamics problem one should: the

1.

Isolate the bodies involved.

2.

Select a coordinate system and positive directions. directions, and draw free-body dia-

3.

grams. Write the fbrce balances. Use the kinematics of the problem to eliminate redundant variables.

4.

If the objective is an instantaneous analysis. the accelerations and reaction forces the kinematics is used furare calculated. If one desires to solve for the response. the ther to eliminate the reaction forces. The resulting equation(s) are differential equa-

tions in terms of' the motion motion variables variables only. only. Such Such equations are called equations of the motion.

When considering the response. one first needs to decide whether to obtain a qualitative or a quantitative solution of the equations of motion. A quantitative solution implies actual solution of the differential equations of motion and it depends on the nature of the differential equation of motion as well well as as the the ft)rm form of the forcing F. The tremendous evolution in the field of' ol' differential equations in the past few centuries has produced several methods of' solution. Several approximate methods have of solution. also been developed that can he implemented on digital computers. A qualitative solution gives information about the nature of the response. or it gives the response at specific points in time and space. without having to solve fbr the response explicitly. Such qualitative analysis includes I. 2. 3.

4.

Impulse-momentum relationships Work-energy relationships Other motion integrals Equilibrium and stability

We will discuss these approaches in the remaining sections of this chapter.

A collar of mass in slides in a circular track of radius R. as shown in Fig. 1 .22. The coefficient given an an initial initial velocity velocityvO. The collar is given of friction between the collar and the track is vO. Find the distance traveled by the collar when it comes to a rest.

Solution We can use either normal-tangential or cylindrical coordinates to solve this problem. The free-body diagram of the collar is given in Fig. 1 .23. Summing forces perpendicular to the plane of the motion yields =

N1

lal

perpendicular to to the the plane plane of of the the motion. motion. where N1 is the component component of of the the normal normal force perpendicular Summing forces in the normal and tangential directions, we obtain

I

Example

1.8

32

CHAPTER 1 • BASIC PRLNCH'LES

g V

ing

I

N1

2'

Figure 1.23

Figure 1.22

[bI Ibi

=N2

= ni

dv F1 = ma, = in— dd t

[c1

where N2 is the component in the normal direction and N is the total normal force N =

\Nl

/

=

+

n-g-I +

1fl2 V4

[dl Idi

Introducing Eq. [dl into Eq. [ci and eliminating the mass term, we obtain dv

'4

(it

R2

lel

which represents the equation of motion. It can be solved by moving the di tenri to the right and the radical to the left. The problem. however, asks us to find the distance traveled, so that we seek to convert Eq. [el to one in terms of the displacement, as Liv

dt

=

dv d.c dv ds

ds di

dv

= (I— =

+

ds dx

V4

R-

If I

or

vdv R2 g2 + R2 g2

Ig] [9]

= =

Equation Igi can be integrated from the initial velocity —

s.s

=

I

n( kg) —

In

+ +

to the final velocity 0 to yield - g-

+

[h] IhI

1.5

33

CONSTRAINTS

DEGREES OF

which can he solved for the distance traveled by the collar as R

2p.

R2 g2 +

In

[1] III

Rg

I

L

The distance traveled is inversely proportional to the coefficient of friction

mass-spring system shown shown in in Fig. Fig. 11.24 .24 consists of a block of mass rn attached to aa wall wall The mass-spring with a spring of constant k. The coefficient of friction between the block and the surface it A force F(1) acts on the block. Find the equation of motion. slides on is

Solution We draw the free-body diagrams in Figs. 1.25 and 1.26. Summing forces in the vertical direction yields

N=nig

[a] [ul

To find the friction force we multiply the normal force with the friction coefficient, To tind the direction of the friction force. we note that the block moves back and forth, so that the friction force is in a different direction depending on the velocity of the block. We have that

> 0,

F=

± <0, when i•

F1 =

when

—,aN —pEN

[b]

=

[ci Ic]

=

that the equation of motion can be written considering the two regimes as when ,i• >

when

0,

<0,

ping

Id] Ed]

F + ,uing

I.]

mx + kx = F

,ni ++ kx =

—

here we note that both equations of motion are linear. The two equations of motion [dl and le] can he combined into a single nonlinear equation by

'lix + kx = F

1.5

—

[f]

/1fllgSigfl(X) ji.nigsign(x)

DEGREES OF FREEDOM AND CONSTRAINTS

In the beginning of this chapter we studied coordinate systems commonly used in to specify the position of a -ivnamics. .ivnam i Cs.Three Threephysical physical coordinates coordinates were required to

tug

mx Ør- .1 A

I

k

F F

F

kx

kx

F

pN LI

Figure 1.24

N

Figure 1.25

N

Figure 1.26

Example 1 .9 1 .9

34

CHAPTIR I •

BASIC PRINCIPLES

particle. In the previous section we studied Newton's second law, which related the resultant acceleration of a particle to the applied force. Equation [1.4.2] is a vector relationship, and it can be separated into three scalar components. In many circumstances the motion of a particle is constrained to move in a subset of the three-dimensional configuration space. Examples of this include a vehicle In these cases, it is not necessary to use moving along a track or on a fixed surface. In all three coordinates associated with a reference frame to describe the motion; depending on the constraints, one or two coordinates are sufficient. For example, if the particle shown in Fig. 11.27 .27 is is constrained constrained to niove along a surface whose mathematical description is [1.5.11 t) = 0 f(x, v, f(x,v,z,t) then two coordinates are sufficient to describe the motion. Selecting them as x and v,

the z coordinate can be ascertained by solving Eq. [1.5. 1]. A constraint of the form above is known as a configuration constrain!. A more general form of a constraint

is when the constraint is expressed in the form of differentials, such as

a1dx + or,

+

+ a0di =

0

[1.5.2] [1.5.2]

dividing Eq. [1.5.2] by di. +

which the a0. the coefficients coefficients a0. in which

+

+ a0 =

0

[1.5.31 [1.5.31

will see see and z. We will of x, x, y, y. and are functions functions of as., and a— are

a more complete analysis of' constraints in Chapter 4. A constraint equation is a geometric or kinematic representation of the restrictions on the motion of a body. What causes the body to execute such restricted mo-

tion is the constraint force associated with the constraint. When the constraint can constraint force is althe associated constraint [1 .5.11, .5.1 J, the represented as as aa surface, surface, as as in in Eq. [1 be represented ways normal to the surface. While we will prove this formally in Chapter 4, we can explain it physically here by noting that the particle velocity is always tangent to the surface. Hence, the particle cannot have a motion normal to the surface. This is caused by the constraint force.

r) = ()U J'(x, v. . C, z, r)

Figure 1.27

1.5

DECREES OF F'REEDOM

35

CONSTRAINTS

define by degree of of freedom freedom the the minimum number of independent coordinecessary to describe the configuration of a system. Each constraint applied to stem reduces the number of degrees of freedom by one. For example. the orientaof a vehicle moving up a spiral in Example 1.5 can be described using cylindrical The radius is a function of the angle traversed, which constitutes one The rise of the spiral is also described in terms of the angle, which conanother constraint. Hence, only one of the coordinates coordinates K. R. &. or z is sufficient r• the motion. The number of degrees of freedom can be calculated using relationship We

No. of degrees of freedom

No. of coordinates

—

No. of constraints

[1.5.41

Newton's second law for a particle yields three scalar equations. The number

equations of motion is determined by the number of constraints. If one (or two) a particle there will be two (or one) equations of motion and the equations will represent reactions. The reaction equations can usually be :&ntified easily. We will see a more complete analysis of' constraints in Chapter 4. system shown in Fig. 1.28 consists of a vehicle undergoing rectilinear motion. A penduis attached to the vehicle. Find the equations of motion.

This is a two degree of freedom problem. if unrestricted, the vehicic has one degree of freeand the mass has three. The wire restricts the motion of the pendulum. Denoting the .isplacements of the pendulum in the horizontal and and vertical directions by x1' xj' and Vp. directions by Vp. we can them as = x + L sin 0

=

—

L cos 0

La]

constitutes two constraint equations. Hence, the combined system has two degrees of

The free-body diagrams of the vehicle and pendulum are given in Fig. 1 .29. For the summing forces along the horizontal we obtain

+F

= Tsin0

[bJ Ibi

For the pendulum. we have the force balances in the horizontal and vertical directions as

m11 = —Tsin0

or

rnjp = T cos 0 — mg rnip

m(i ÷ L6cosO — or

LU2

sin0) = —TsinO sinO)

[ci

[dl [di

sin 6 + L02 cos 0) = T COS COS 00 — — ing Mg

.I. T

I

F

F V

'vi

N1

:o L

T

(a)

hi

FIgure 1.28

FIgure 1.29

mx (b)

I

Example 1I .1 .1 00

36

PRINCIPLES CHAPTER 11••BAsic CHAPTER BAsicPRINCIPLES

Eqs. [a], [cJ, and [d] are three equations that contain the constraint force (in this case the tension in the pendulum) explicitly. We can eliminate T from the above equations and obtain two equations of motion, the same number of equations as the degrees of freedom. This can be accomplished in a number of ways. One way is as follows: First, we introduce Eq. [ci into Eq [b}. which yields

(M + ?n)l ?n)i + inLO cos 0

—

mLO2

sin 0 = F

1.1

Then, we multiply Eq. Eel by cosO and Eq. [d] by sin 0 and add the resulting expressions. Doing so yields mL0 + ml cos 0 + mg sin 0

0

IfI

which can be put into a more familiar form by multiplying it with L. so that

+ mLl cos 6 + mgL sin 0 =

[g]

0

Equations [eJ and [gi are the two equations of motion in terms of the two independent

variables .v variables x and and 6.

1.6

IMPULSE AND MOMENTUM

Newton's second law states that the rate of change of the linear momentum of a or particle is equal to the applied lhrce, or F(i)

If we multiply the above equation by di

[1.6.11

= and

integrate from an initial time t1 to final

time t2, t2, we obtain 1pU2)

((2

j

F(t) di

=

dp = p(t2) J

—

p(11) = mv(12) mv(12) p(11)

—

invOi) invO1)

[1 .6.21

1

the left is called the impulse, which is equal to the change in linear linear momentum? The unit of linear momentum is mass times velocity, ML/T. In the U.S. system one commonly uses lb• sec and in the SI system, N • s. When there is more than one particle, the linear momentum of the entire system

Eq. [1.6.2] is the impulse-momentum Eq.

is obtained by adding up the linear momentum of each particle that comprises the of masses masses m1 and rn2. rn2. the linear system. For example. for a system of two particles particles of momentum is obtained by simply adding the linear momenta of each particle. particle, p = lfliVi + m2v). An interesting special case is when F(t) = 0, or over the interval (ti. 12) the integral of F(i) is zero. It follows that in such cases the initial and final values of I

5Note that this definition is different than the common use of the word

1.6

IMPULSE AND MOMENTUM

the linear momentum are the same, p(t2) = p(t1). which denotes the principle of p(t1). which conservation of linear conseri'alion The principle states that if the net effect of forces acting on a particle is zero over a time period, then the linear momentum of the particle has the same values at the beginning and end of the interval. The principle ol conservation of linear momentum is generally of more use when more than one particle is involved. Note that linear momentum of a system can be conserved in a certain direction of the motion only and not be conserved in the other directions. Defining Defining aa unit unit vector vector ee along alongthe thedirection directionthe thelinear linearmomentum momentumisisCOflconserved and taking the dot product of Eq. [1.6.21 with e. we obtain II

dp•e

F(r)•edt =

it1

.

= p(t2)•e—p01)•e

[1.6.3]

p(i

An interesting application of the impulse-momentum theorem is when the duration of the applied force is very short. A large Force applied through a very short time period is defined as an impulsive (orce. Denoting by by the duration of the impulse, and taking the limit as approaches to zero. we have tI

F(r)dt F(z'1) F(r)dt==F(z'1)

urn

[1.6.4] [1.6.41

where is the impulsive force and has the dimension of force X time. Theoretically. Theoretically, the amplitude of the impulsive force approaches infinity. To mathematically describe

an impulsive force we make use of the Dirac delta function. Consider Fig. 1.30. The Dirac delta function at point t a is denoted by by 6(t — a) and is defined as6

(S(t—a)=O

a

-x —

I

a)dt

[1.6.5]

I

J—x

I

I I (1

FIgure 1.30

Dirac delta function

We are using a different notation than the traditional delta to differentiate between the Dirac delta, the Kronecker and the variation Functions.

37'

38

CHAPTER

1 • BAsic

PRINCIPLES

— a) at II — a approaches infinity. Therefore, this function is defined The value gets not by what its amplitude is but by what its integral is. When a function J'(t) gets multiplied by the Dirac delta function and integrated, the result is cc

—

a) dt

J

=

f(a)6(t

- a) di =

f(a)

J

6(t

—

a) di

= 11.6.61

so thatmultiplying f(t) by the Dirac delta function at point a and integrating yields the value of f(t) at rt = a. This is similar to taking a snapshot of the function f(t) at 1 = a. can be expressed as a continuous (in time) The impulsive force in Eq. function as 1

F(i) =

[1.6.7] 11.6.7

(1 —

The Dirac delta function is not a discontinuous function. Rather, it can be shown to arise from a limiting process of a Continuous function, and it obeys the laws of dif5(t — a) exists. Denoting the derivaferentiation. This implies that the derivative of 5(t — a). we multiply it with the function j'(1) and tive of the Dirac delta function by integrate by parts, parts. which yields the relation

f(t)6(t

—

a)dt =

—

a)Ix,x —

J

I

f(r)6(r f(t)6(t

—

—J(a) a)di = —f(a)

11.6.8] 11.6.81

The interested reader is referred to the text by Greenberg for more details. The effect of an impulsive force is a sudden change in velocity, with no apparent change in position. To demonstrate this, consider motion in one direction and write the equation of motion in the general form [1.6.9] 11.6.91 g(x(t).iii)) i(i)) = F(t) ,nx(t) + g(x(t). the sum sum ol ol all forces acting on the particle that are a function in which g(x(i). g(x(i). i(t)) isisthe Examples of such forces include gravity, springs, and dashpots. and of x(t) and When the force F(t) is impulsive and it is applied at I = 0. multiplying Eq. [.1 .6.91 0. we obtain 0 to I = E. and taking the limit as by dt, integrating from 1 urn

I1

{mi(i) + g(x(t).

di =

urn

F(t) di dt

[1.6.101 11.6.101

The first term on the left side becomes urn Urn

I

lirn mI(t)dt = tim

di' dt

di =

lim{,n.*(E)

=

—

[1.6.11] 11.6.11]

The second Using Eq. 11 .6.41, the right side of Eq. [1 .6.101 becomes equal to term on the left side of Eq. 11.6.101 vanishes when integrated over the time period This is because finite time is required for a displacement to develop. Recall that

is a force of finite magnitude. Hence, the limit of its integral over the g(x(t), duration of' of the the impulse impulse is zero. It follows that immediately after the impulse we have = x(0)

= F(0)

11.6.12]

1.6

IMPULSE AND MOMENTUM

yV

r

F:)

V (1) (I)

0

x

Figure 1.31 In real situations. one does not have actual impulsive forces, but forces of large applied over very short periods of time. Hence, the consideration of a as impulsive is an approximation. The validity of this approximation should he checked, especially when friction forces are involved. Let us next consider the motion of constrained bodies subjected to impulsive fDrces. As an example, consider the system in Fig. 1.28 and the tension in the penduWhen the force F applied to the mass is impulsive, the tension T also becomes Lmpulsive. Elnpulsive. while the weight of the pendulum does not. If a body or series of bodis constrained in some fashion, the impulsive force applied to one body will be forces. to the other interconnected bodies by means of impulsive reaction Ibrces. Next, we define angular momentum. Consider Fig. 1.31 and denote the position •:f the particle as measured from point 0 by r and its absolute velocity by v. The (denoted by H0). which is also referred to as the momentum about point 00 (denoted .ngular momentum .ngular of of the linear mnonienturn vector about point 0, is defined as

H0=rXmv

11.6.13]

r is the vector connecting the reference point 0 to the particle. Angular mois a vector perpendicular to both the linear momentum vector and the posiAngular momentum momentum about 0 vector from the reference point 0 to the particle. Angular an be physically explained as the tendency of a particle to rotate about point 0. If r and p are in the same direction, there is no tendency to rotate about 0. The dimension of angular momentum is ML2/T or FLT. In the SI system one in the the U.S. U.S. system. system.lb•ft•sec. lb•ftsec. or kg•m2/s, and in uses While linear momentum is an absolute quantity. one calculates angular inornenabout a specific point, an indication that it is a relative quantity. The choice point is important when solving dynamics problems. A basic guideline is to noose a point that is fixed, or to select 0 such that the angular momentum term has physical significance or is simplified. Now, differentiate Eq. [1.6.13], which yields ,,iv) dH0 —= d(r X ,iiv) di dt

=rxrna=rXF

11.6.14]

39

40

CHAPflR 'I

•

PRINCIPLES

The quantity on the right side of this equation is recognized as the moment of the X F, so that resultant of the external forces about point 0, M0 = r X dH() =

di

[1.6.15]

Mo

The rate of change of angular momentum of' a particle about a point is equal to the

applied moment about that point. The above equation is of considerably more use when dealing with systems of particles. There is debate among scientists on whether Eq. [1 .6. 1 5.1 is a derived relationin support support of argument in stated principle. like Newton's second Jaw. The argument ship or a stated in the stated principle viewpoint is based on the idea that shear forces are neglected in the derivation, and that every physical body has a nonzero volume. What makes the derivation above possible is the assumption that a particle has no physical dirnensions.

Similar to the linear niomenturn niomentum case, we can integrate Eq. [1.6.15] over time, theorem as yielding the angular unpulse—moineiztum theorem 1-H0(:2) H0u2)

M0(t)dt =

dH() = H0(t2)

—

H0(t1)

[1.6.161

JH0(t1

The term on the left is defined as the angular impulse. As with linear momentum, The

if the integral over time of the moment about a point is zero. angular momentum is conserved about that point. Also, because Eq. 11.6.161 is a vector relationship, angular momentum may be conserved in a particular direction while not conserved in another direction. If the applied moment is impulsive, that is, its duration is infinitesimally short, the angular impulse-momentum relationship for zero initial conditions becomes =

11.6.17]

An impulsive angular moment can be generated by a very large large torque torque applied applied over a very small interval, or by an impulsive applied through a moment arm. For panicle mechanics problems. the commonly used coordinates for angular momentum problems are cylindrical coordinates. The central force problem, as demonstrated in Example 1.1 is a classic case. Remember that when we discussed cylindrical coordinates, we emphasized the importance of attentively selecting the origin of the coordinate frame. The same argument is valid when selecting the point about which to calculate the angular momentum.

Example 1

.1 1

One of the earliest industrial applications of the principle of conservation of angular niomenturn is the centrifugal governor. James WaR turn Watt used the centrifugal governor to control flow in

steam engines. Fig. I .32 shows a typical governor. As the speed of the governor changes. the arms of the governor move. Because the only external force acting on the governor is along the shaft, the angular momentum about a point along the axis of the shaft is conserved.

The links are of length L = 0.2 m and assumed to be massless. The balls are 0.6 kg each. The governor is originally rotating at 50 rpm and the arms make an angle of 6 = 300 with the vertical. What is the value of 6 when the governor's speed becomes 75 rpm?

1.6

41

IMPULSE AND MOMENTUM

C')

0.6 kg

.4 .1

1•

Figur. 1.32

C

as the point 0 as the moment moment center. The angular momentum of the balls about 0 is along to the fixed plane of rotation, so it can be expressed in scalar form, The magnitude angular momentum is

= 2)r

I

= 2nir2w

X

lal

w is the angular velocity and r is the horizontal distance from the shaft to the balls. = L sin 6. We can thcn write write the the conservation conservation of of mornen momentum turn relationship relationship as SJfl = [w 1w sin2

1w sin2 Iw

--

[b] [hi

can be solved for 6(t7) as =

sin

(t)(t1 ) 0(t, 0(t,)() )()

[C]

Substituting the given values we obtain

0(t2) =

0.

'so

= 0.1667

Id]

1667) = 24.100. 24. 100. Notice Notice that a small change in the angle 0 leads t .i. substantial change in the angular velocity. The same phenomenon occurs when figure bring their arms inward to increase their rotational speed. ±.at ±.at 6(t2) 6(t2) = sin

FORCE FIELD PROBLEMS A very interesting example of the conserof angular momentum is in central force field problems. Consider a particle moving a path. as shown in Fig. 1.33. It is acted upon by a force F that is always directed toward i 0 (hence. the name central force). Selecting point 0 as the origin of the coordinate 'me. and using cylindrical coordinates, we express the applied force as

F=Fe,.

[a]

that force F causes about 0 is X F M0 = r X F = re,. re,. X X Fe,. Fe,. = 0

11'I IbI

Example

1.12

42

CHAPTER 'I

•

BAsic PRINCIPLES

H7

0

6?

r

Figure 1.33

which is an obvious result: as the line of force passes through point 0 it does not cause a mo1.16.16] that that the the angular momentum about 0 is conserved, ment about 0. It follows from Eq. [[1.16.16] and H0 = constant. and Because the angular momentum is constant in both magnitude and direction. the particle can only move in a plane perpendicular to the angular momentum. We select the z direction as the direction of the angular momentum. Using Eq. [1.6.131, we obtain nh[rer ++ rOe0 rOe0l1 = inr26k = constant Hc) = r XX lflV = rep X ,n[rer

Ic]

We can express momentum per unit mass, Ii, as express the the angular angularmomentum

r20=h

Ed]

where the value of Ii depends on the initial conditions. wherc Equation [d] can also be derived by directly integrating Newton's second law expressed in polar coordinates. Indeed, writing Newton's second law in the radial and transverse components as — 111(1: —

,.02) = Fr

m(rO + 2i'O) = F0 = 0

Eel

and considering the derivative of'

'It

=

rO + 2r10 2riO = r(rO + 2i'O)

111I

is constant when the force always lies along the line connecting We conclude that the particle and point 0. A typical example of central force problems is in orbital mechanics, as the gravitational attraction between two bodies is along the line joining the centers of mass of the two bodies. In the absence of other external disturbances, the motion of one body with respect to the other (that is, the orbit) lies on a fixed plane.

1.7

1.7

WORK AN!) ENERGY

WORK AND ENERGY

In the previous section. we integrated Newton's second law over time to obtain momentum relationships. Here, we integrate it over the spatial variable. To this end, we a particle whose position is described by the vector r and which is acted upon by a force F. as shown in Fig. 1 .34. We define by dW the incremental work that the force does Ofl the particle as the particle moves by an incremental distance

dras (1W =

Fdr F'dr

= ,na•dr

=

[1.7.11

Recalling that and dr. dr. Recalling that v = dr/dr and a = dv/dt, the angle between FF and and multiplying and dividing the right side of the above equation by dt, we obtain

where

dW =

is

F•dr =

dv dr m—•—dt di

di

dv = in di

1

•vdt = inv•dv = —rnd(v•v) 2

of the particle as

We next define by T the kinetic T =

[1.7.21 11.7.21

1

in v • v =

1

[1.7.31

mv

speed. The The incremental incremental change change in kinetic energy is dT = v is the speed. where v = and, hence, hence. dT is rnd(v • v)/2 = my • dv. The kinetic energy is an absolute quantity and. write Eq. Eq. [1.7.21 [1.7.2] as a perfect differential. We then write

[1.7.4]

(1W = dT

The work done on the system by the. force F is denoted by W1 by integrating the motion from point I to point 2, hence

Jr1

it is

1T2

F'dr=

W1_.2=

and

[1.7.5]

dT= T2—T1 - 11

hich gives the woi*-energv theorem. =

T1 +

[1.7.61 11.7.61

T2

t

dr

0

Figur. 1.34

43

44

PRINCIPLES CHAPTER 1 • B2tsic PRINCIPLES

differential and when it is We distinguish between cases when dW is a perfect dif'ferential differential. we cannot express it as the differential of not. When cIW is not a perfect differential, a function. but merely as an infinitesimal element. The necessary condition for dW to be a perfect differential is that the force depends on the position vector alone, F = anddW dWisisnot nota perfect differential). F(r) (although there are cases when F = F(r) and One can take advantage of cases when some of the forces acting on a body lead /brces. Examples to perfect differentials. Such forces are known as conservative /brces. of conservative forces include spring forces, gravitational forces, and certain electromagnetic lorces. forces. The incremental work can be expressed as the derivative of a potential function as [1.7.71 F(r)dr = —dV(r) dW = F(r)•dr where the potential function V is an explicit function of r only. Because dW is a per-

fect difiBrential. differential. its its integral is independent of the path followed and it is dependent only on the end points of the integration. Over a closed path the value of the integral is i.ero, zero, or

F(r)'dr = F(r)•dr

0

[1.7.8]

position FR integrating from a reference reference position 11.7.71 can can be be evaluated by integrating Moreover, Eq. 11.7.71 (or datum) to the location of the particle to yield • dr F(r)'dr F(r)

V(r) = —

J

[1 [1.7.91 .7.91

rR

Physically, the potential energy. Physically, The potential function V(r) is also known as the

potential energy is explained as the potential of a body to do work, or the stored energy.

Note that while kinetic energy is an absolute quantity, potential energy is relative: its value depends on the reference point about which it is measured. Because the interest is in increments of potential energy, selection of the reference point does not make any difference. One selects the datum to either simplify computations or to give the problem at hand a better physical interpretation. The dimension of' of work, kinetic energy, and potential energy is force times disorML2/T2. In In the the SI SI system system one one commonly uses the units N•m. Ajoule tance, FL. orML2/T2. In the U.S. Customary system in. In (J) is defined as a unit of energy as J = N • in. Do not not con confuse fuse the unit of energy with the unit of the commonly used unit is ft ft • lb. Do a moment, which has the same dimension as energy. In the U.S. system, the unit of a moment is usually denoted by lb • ft. Also, joule is never used as a unit of moment in the SI system. We now consider three types of lhrces that lead to potential energy. 1

1

.7.1

1

GRAVITATIONAL POTENTIAL ENERGY

Ii .4.3] as The gravitational attraction force between two bodies was given in Eq. Il F = Gm1m2/r2. where G is the universal gravitational constant, ni1 and are the

1.7

WORK AND ENERGY

of the two bodies, and r is the distance between the two bodies. Consider m1 be a celestial body. The incremental work dW done by in1 on m.2 can be expressed a perfect differential. so that the potential energy can be written as 1r

V(r) = — F(r)dr =—J V(r)= —II F(r)dr= — JrR

Gin1 1112 2

dr = dr=

Gin in.) 1

r I

JrR

(1.7.101

The reference value the potential energy is measured from is commonly chosen as distant stars, so that rR = which leads to the expression for the gravitational energy:

V(r) =

Gm11n2

[1.7.11]

— 1•

Around the surface of the earth, we can calculate the expression for gravitational i:tential energy in a number of ways. One way is to write the gravitational force as as derived in Section 1.4. Denoting by z the height of the particle from the of the earth, we obtain for the potential energy

[1.7. 121 11.7.12]

V(z)=—I —m2gdz = Jo

Another way is to substitute for in1 the mass of the earth me and express r as =

re

+ zt where

is the mean radius of the earth and z is the distance of the particle

the surface of the earth (altitude). Note that r is being measured from the center :i the :1 the earth, earth, as as shown shown in in Fig. Fig. 1.35. 1.35. Equation Equation [1.7.111 [1.7.111 can can then then be be written as 1

V(r) = V(s) = —

1+

re

by considering that near the surface of the earth be approximated as

- GIfleifl2 re a

11.7.13]

—

z

is much smaller than

[1.7.141

There are two terms in the above expression; one is a constant, and the other function of z. The constant term, Gin can be eliminated, and doing so fl11)

1•

FIgure 1.35

45

46

CHAPTIR 1 S BASIC PRINCIPIIFS

moves the datum to the surface of the earth. Considering the definition of the gravitathe potential energy expression becomes tional constant from Sec. 1.4, g = Eq. [1.7.12].

1.7.2

POTENTIAL ENERGY OF SPRINGS

We will be concerned here with two types of discrete springs, axial and torsional. The subject of continuous springs, which exert their forces over an area, will be covered in Chapter 11. Consider an axial spring that is compressed (or stretched) by a distance ..v,asasshown shown in Fig. 1.36, and the associated free-body diagram. The simplest model x, resisting force Such a spring exerts a resisting spring.7 Such for an an axial axial spring is a linear axial spring.7 approximated as linearly linearly proportional proportional to to its its deflection, deflection, F = — kx, where k is the approximated as spring constant. The dimension of the spring constant is force/distance. It is easy to see that the spring force is conservatives and the potential energy becomes V(x)

=

—kxdx =

[1.7.1 51

—J

When dealing with spring deflections. deflections, the datum point for the potential energy has to be taken as the undeformed position of the spring. The potential energy of a spring is the same when the spring is stretched or compressed by the same amount. Torsional springs resist rotational motion and hence exert a moment on the body they act on. on, as Fig. 1.37 shows. When the spring deforms by an angle 0, the moment exerted by the spring has the fonii MU)) = MU))

11.7.16]

—kTO

k

0 V

Lion

F::: —k,v —k,v F=

LTndctomed position posit ion

FIgure 1.36

Axial spring

Figure 1.37

Torsional spring

7We prefer the terminology axial so as to not cause confusion between an axial spring and a linearly varying spring Force.

1.7

Wou

AND ENERGY

k1 is is the the torsional torsional spring spring constant, constant, having the dimension of moment = X distance. The spatial variable is the angle 0. and the potential energy can be written as V(E1)

=

—

I

M(O)dO =

—

Jo

=

I

11.7.171

Jo

As in the axial spring case. the datum point for a torsional spring must be selected the undeformed position. Equation [1.7.161 and F = — kx. representing the resistive monients and forces ctf torsional and axial springs. are linear approximations to the actual spring force

moment. The linearity assumptions are valid when the spring deflections are a certain range. This range, referred to as the linear range. is a property of material the spring is made of, the size of the spring, and the range of operation. The range of operation is a function of' the applied forces. More accurate models of springs have nonlinear expressions that describe the of the springs. Two such models are sqftening sqflening and stijfrning springs. 1.38 shows the spring force as a function of the spring deflection. A stiffening spring is one whose resistive force increases more than that of a linear spring the applied load increases. A common model of a stiffening axial spring is F(x) = —kx

—

[1.7.181

k1x3

k1 a positive constant. By contrast, the resistive force of a softening spring as the spring deflection increases. The spring force here is commonly cxressed as

F(x) = — —kx + k2x3

[1.7.191

= which k2 is positive. This model is accurate only until the resistive force F(x) zero at a point x 0; after that, either the spring provides no resistance or breaks.

-F Stiffening Sti ifenirig spring .k. k

Linear range

—U..

.'x

Softening

spring

Figure 1.38

47

48

CHAPTIR 1 • BASIC PRINCIPLES

1.7.3

ELASTIC STRAIN STRAIN ENERGY

This form of potential energy is associated with the elasticity of a deformable body. We will study this subject in more detail in Chapter 11. For now, we write the strain energy for a simple case and draw analogies with discrete springs. Consider the axial deformation of an element, such as of the rod as shown in Fig. 1.39. Using standard assumptions from linear elasticity theory, the only component of strain is in the axial direction. The axial strain at any point is defined as the change of length per unit length. The strain in the x direction is denoted by For small deformations, it can be approximated as = iIu(x)/dx, where u(x) is the axial elongation of the rod at point x. The associated axial stress is denoted by and, in the linear range, it is related to the strain by Hooke's law, = with E denoting the modulus of elasticity (Young's modulus). The modulus of elasticity is a measure of the stiffness of the material used. It is analogous to the spring constant k. We next consider the potential energy associated with the deformation, called

the strain energy. Consider a differential element of the rod shown in Fig. 1.40. In the linear range, the strain energy has the form

[1.7.20]

V

=

= J

J

The factor of one half comes from the fact that the stress is integrated over the strain

to get the strain energy per unit volume. This is entirely analogous to integrating the spring force over the displacement for a discrete spring.

1.7.4

WORK-ENERGY RELATIONS

We next relate conservative Forces and potential energy. Because Eq. [1 .7.8J involves a line integral, we can invoke Stokes's theorem.8

VxF=O

[1.7.21]

where V is the del operator. This term is zero only if F can be expressed as the gradient of a function. It is easy to show that this function is the negative of the potential energy. We then have

11.7221 11.7.221

F(r) = —VV(r)

x

Figure

•

II

utv)

1.39

+ (Li

FIgure 1.40

where A A is is any any vector, denotes a line integral, and x A) • ds, where °f A dr Stokes's theorem converts a line integral into a surface integral.

denotes a surface integral.

1.7 In

WORK ANDENERGY

the Cartesian coordinate system. the del operator has the form a

a

a

[1.7.231

Ôz

+ FJ F%j +

that, expressing the force vector as F =

one relates the compo-

of the force to the potential energy as

=

(IV —

=

ax

(9V —

F: =

-

cIv

(9V —

dz

[1.7.241

When using cylindrical coordinates the del operator has the form C)

V=

dR

l9 er+—-—-eo+—k 1(9

(9

Rd6

(9:

[1.7.251

the components of the force are related to the potential energy by Fr =

av

Fe

—

=

I(9V

F- =

——

9V —

a:

11.7.261

Now consider that some of the forces acting on a particle are conservative and are are not. Forces that arc not conservative are refened to as nonconservative. the force vector as F = F(. + Fne, the notation being obvious. It follows that incremental work done by the force also can be divided into two parts: that is, dW =

[1.7.271

dV +

=

+

= dr is the work done by the nonconservative total work by integrating the overall work as frl dW (Fe. + F.)•dr = V1 — V2 + = J r1 = J

b-iere

We obtain

[1.7.28] [1.728]

J

the work done by the nonconservative forces is --2

Ffl(. S

= Jr,

dr

11 .7.29]

[1.7.28J and and1.11 .7.29] .7.29] into into Eq. [1.7.6] yields Substituting Eqs. [1.7.281 1

The The

+

=

11.7.301

T2

+

V2

rota! energy of the system is defined as E =

7'

+ V. One writes the energy

T1 +

W,1e1 •2

as

=

E1 +

E2

[1.7.311 11.7.311

The total energy of a dynamical system under the influence of nonconservative

changes as the system moves. If all the forces acting on the body that do b:rk are conservative. Conservative.Eq. Eq.[1 [1.7.31] .7.31] indicates that its total energy remains the same. is known as the principle otconservalion of ofenergy. energy. and and itit explains explains the name force. This principle can he written as E1

=

E2

[1.7.321

49

50

CHAPTER 11 •• B.tsic Btsic PRINCIPLES CHAPTER

convenient to express work as an integral over time. We divide and multiply the integrand in the expression for work by di, so that It is sometimes

et)

F'dr

W1_2 =

=

dr F• —dt = di

I

F'vdi

11.7.331 [1.7.331

which t1 in which t1and and12 12 denote denote initial initial and and final times, respectively. The integrand on the

and it is denoted by P.

right side of the above equation is defined as

P=F'v

[1.7.341

Power is basically the measure of how fast a force can do work. The dimension of power is work/time. In the SI system, the unit watt (W) is defined to represent power as 1 W I J/s = N • rn/s. The U.S. system represents power using either ft• lb/mm or horsepower (Hp). where 1 Hp = 550 ft • lb/sec. We have 1

dW =P ut

Pdt

Wi-.2 =

11.7.351

.11

Some texts defIne define work as the integral of power over time. The above equation

can also be viewed as the integral of the expression

di

=

=

11.7.36]

over time. It indicates that the rate of change of the total energy (rate of work done) is equal to the power of the nonconservative forces acting on the body.

A special category of forces comprises the forces that do flO work. From Eq. 11.7.1]. (1.7.1]. for for aa nonzero force force to to not not do do any any work, work, either either dr dr = 0, 0, or or F is perpendicular to dr. Included in this category are normal forces, other reaction forces perpendicular to the direction of motion (perpendicular to the tangential direction). and forces applied to points that have zero velocity. Obviously, if a force is applied to a stationary point, there is no work done. A very important force that does no work is the friction force in a rigid body rolling without slip.

Example 1 .1 3

I

Consider the 30 iii long incline of 15° shown in Fig. 1.41. Along the incline there are two

paths. one a straight line and the other a semicircle. A mass of 0.5 kg is released at the bottom with an initial speed of' of 15 15 rn/s. rn/s. lirst lirst along along the straight path and next along the circular path = 0. 0. 1. What will The coefficients of friction between the mass and the incline are = be the speed of the mass as it reaches the top of the incline following the straight path. and then, following the circular path?

Solution This problem can be solved using the work-energy theorem. From the free-body diagram on

Fig. 1.42, the forces acting on the mass are gravity, contributing to the potential energy: the normal force N, which does no work; and the friction force F, a nonconservative force. From the force balance along the incline we obtain

N=

mgcos 150 15° = O.5(9.807)(O.9660) = 4.736 N

[al

1.7

AND ENERGY

4g

F

mg

I115°' 15°'

Figure 1.42

Figure 1.41 that the friction force is

F =

= 041

mgcos 15° =

0.

1(0.5)(9.807)(0.9660)

0.4736 N

Ibi

that the friction force is in the direction opposing the motion. We assume that friction only at the bottom of the paths and that the walls of the path are friction less. Hence, the of the friction force is the same for both paths. even though its direction is different thebottom bottom of the incline. IT the two paths. We take take as as aa datum datum point pointfor forthe thepotential potentialenergy energythe way, V1 way, V1 = 0. The initial kinetic energy is =

!flV

=

)2 15)2 (0. 5)( 15

= 56.25 N' in

Ic]

Let us first consider the motion of the mass along the straight path. Assuming that the reaches the top. the potential energy at the top of the incline is 38.07N'm N'm 15° = 38.07

V2 =

[dl

work done by the nonconservative force can be expressed as = —F(30) —F(30) = —0.4736(30) = —14.21 N•rn

W1

the

[.1

work-energy theorem. we obtain

T2 =

T1

+

V1

+

W1

—

V2

N• m 14.21 —38.07 3.970N•m +0— — 38.07 = 3.970 0 — 14.21 = 56.25 +

if If I

calculate the velocity as =

/2

V2

In

= 3.985 3.985 rn/s

[g]

Considering motion along the semicircular path. V2 remains the same, but the nonconsere work changes. as the path is longer. To reach the top. the path length that is traversed is of a semicircle of radius 15 rn, so that the nonconservative work becomes W1

= —F('zr)( —F('zr)( 15) = —0.4736(3. 142)( 15) = —22.32 N • m

1k]

Comparing Eqs. Ed. [dl, [dl. and [hi. we conclude that the mass does not reach the top of the as the value of kinetic energy at that point would be negative. We then ask how far travels along the incline before it comes to a stop. To find this value, it is preferable as shown in in Fig, Fig, 1,43, 1 ,43,The Themass masshas hastraveled traveled aa distance distance of with the angle

51

52

BASICPRINCIPLES PRINCIPLES CHAPTIR11 ••BAsic CHAPTIR

R( 1I — cos 4)) sin 150. It follows that with R being the radius of 15 m. The height of the mass is R( the potential energy and nonconservative work become V.2 = mgR(1

—

cos4))sin 150 = 0.5(9.807)(15)(0.2588) = 19.04

—

19.O4cos4)

= —FR4) = —0.4736(15)4) = —7.1044)

[11 [II

write the work-energy theorem rest, we write Noting that T2 = ()when the mass comes to a rest. as

T1 +

—

= 56.25

—

19.O4cos4) 19.04 + 19.O4cos4) — 19.04 7.1044)— 7.1044)

= 37.21

—

19.O4cos4) = 0 7.1044) + 19.O4cos4)

Ill

The solution of Eq. Iii can be obtained numerically, and ii can be shown to he

Iki [ki

4) = 2.756rad = 157.9°

rest, slides back or not. To examine One can next ask whether the mass, after coming to a rest. line tangent tangent to to the the path pathat. at 4) = 157.90. this issue, we need to perform a force balance along a line Fig. 1.43 shows the configuration. We define the plane of the incline as the xz plane. with perpendicular LO to the theincline. incline. The The normal force and the magnitude of friction the v direction perpendicular force remain the same as before. However, we are now summing forces along the tangential direction, that is, the direction of impending motion. We write the gravity force as =

[I] [II

— ing sin I 5°i — ing cos I 5°j

tangential direction is The mass will move if the component of the gravity force along the tangential larger than the friction force. From Fig. 1 .43 we express the unit vector i in terms of normaltangential coordinates as

i == —— sin in which i.11i = 180°

—

4)

=

—

[ml

cos i/ie,3

10.The Thecomponent component of of gravity gravity along along the 22. 10. = mgsinl5°sin22.l° = 0.4773 N

path becomes

I'll

Comparing Eq. In] with Eq. [hi we conclude that the mass moves hack. as the component of gravity in the tangential direction is greater than the friction force.

1

1/!

t

1?

Figure 1.43

1.7

Consider the system in Fig. 1.28. The pendulum is released from rest from the position 0 =

Find a relation for the velocity of the base as a function of 0.

for

F =

0.

Solution treat the two masses as one system. The only ftrces external to the system are gravity and the normal force acting on the cart. We conclude that (1) there are no nonconservative forces that do work, so energy is conserved, and (2) there are no external forces in the horizontal direction, so the linear momentum in the horizontal direction is conserved. Using as datum the horizontal position of the pendulum, we can obtain the total energy by considering the initial condition as We

E(6

[a]

no initial initial motion motionand andVV = = 0. as there is no

where where i'

that the total energy

0. it

= is E = 0 = constant. We next take an arbitrary position otthe pendulum and write the kinetic

and potential energies T =

=

I

+

I

[b]

V = —mgLcos0

and

+

+ Lsin9) Lsin8)

=

k ++ LOcosO. .i. LOcosO.

= LOsinO

=

Ic]

Substituting Eqs. [h] and [c] into Eq. [al we obtain for the total energy

E =

0

=

+

+

+

—

[dl

We next look at the conservation conservation of linear momentum in the horizontal direction. The aitial linear momentum is zero. The linear momentum for any value of 0 is MX p = Mi

=

±

Mi

LU in(i ++ LU + in(i

cos

(1) = 0

Eel [01

can be rewritten as m)i == —mLO —mLO cos U (M + m).i (i4r

0=— We We eliminate 0

from

I

——(M +

2'

+ ,n)i in)I

Ill III IgI

iiiL COS inL COS 0

Eq. [d] by substituting Eq. [gi into Eq. [d] and simplifying, which

.

?fl)X +

1

(M + in )2 -)

2m

cos-0

—

53

WORK AND ENERGY

iflgLcOSf9 = 0

[hI [hj

When cos 0 = 0, the above relation cannot he used. However, from the initial conditions The cos 00 = 0 then i = 0 as well. Also, note that Eq. [h] is in terms of Eq. [fl. when cos of the velocity of the cart can be determined from from Eq. Eq. [1 111. 1.

Example 1 .1 4

54

CHAPTIR 1 • B1tsic PRINCIPLES

1.8

I

EQUILIBRIUM AND STABILITY

Equilibrium is an essential and very useful concept in dynamics. For a system of particles or interconnected bodies, static equilibrium is defined as the state when all the particles and bodies comprising the system are at rest. Both the velocity v(t) and the acceleration a(t) of each body arc zero at equilibrium. When discussing the equilibrium of a system. two important questions come to mind: (I) How does one calculate the equilibrium position? (2) What happens to the system if it is displaced from equilibrium? To calculate the equilibrium position one one can use Newton's second law. Setting the acceleration to zero results in the equilibrium relation F = 0

11.8.11

where F is the sum of all the external forces. An alternate procedure that gives more

of a qualitative insight is to look at the energy. Consider a particle and the case when the kinetic energy is only quadratic in terms of the velocity of the particle.9 particleY We take the differential fbrrn form of' the work-energy principle and remove from it all velocity- and time-dependent terms (including the kinetic energy as well as all time-dependent parts of the nonconservative force). From Eq. (1.7.281 we have

dW =

(Fe.

+ F,ZL.) F,ZL) 'dr = —dv —dV +

= 0

[1.8.2]

The right side of this equation is zero because at equilibrium the particle is not rnovmoving; hence, no work is done on it by the external forces. The equilibrium equation then becomes

dV = dWne

[1.8.31

When the system is conservative, the equilibrium condition is defined as

dV = 0

11.8.41

which can be interpreted as the potential energy having a stationary value at equi-

librium. For a single degree of freedom system. Eq. [1.8.31 [1.8.3] can be solved easily, and it is the preferred approach for systems consisting of interconnected components. In Chapter 4 we will derive the general form of Eq. [1.8.31 for fbr multiple degree of freedom systems. Now consider the second question: What happens to the particle when it is disturbed from equilibrium'? Three scenarios are possible:

The particle returns to the equilibrium position and stays there. In this case, the equilibrium position is called stable, or asymptotically asvmpwticailv stable. 1. I.

I

°The relevance of this requirement will be discussed in Chapter 5.

1.8

STABILITY

The particle hovers

around equilibrium without staying at one point, but it does -etum to or get away from the equilibrium position. The equilibrium position is to as critica fIr stable, merely stable, or neutrally stable. The particle moves away from the equilibrium point and it never returns there. position is called unstable.

.

flere are several several approaches that enable one to examine behavior in the

vicinity equilibrium position. The simplest is to linearize the equations of motion about position. A theorem from stability theory states that if the linearized of motion in the neighborhood of equilibrium reveal sigtitjicant significant behavior decaying or growing in amplitude), then the general motion (in the large) by this significant behavior. a particle of mass and write the equation of motion as F(x(t), .i:(t)) i(t)) = F(x(t), = /(x(i), x(r)) [1.8.5] !fl

Al equilibrium, all the velocity and acceleration terms are zero; therefore, the condition is obtained by f(Xe,O) = 0

z=

[1.8.6]

denotes the equilibrium position. Depending on the nature of function .1. f. be more than one equilibrium position. position. We now define a local variable — Xe. and expand each term in the equation of motion about the equilibrium x= = 0, 0, = 0. The term 1(r) simply simply becomes We expand side of Eq. [1 .8.5] in in a Taylor Taylor series series and retain the linear terms:

f'(x,

0) 0) +

+

[1.8.7]

'. -Ic

The first term on the right side of Eq. 11.8.71 is zero by defInition of the the equilib.equilib-

-dation elation in Eq. [1.8.6]. Defining by equilibrium position, position.

and

Y2

the partial derivatives evaluated

[1.8.8]

dx

the local variable s. we can express f(x. •[(x, f(x,

—

as

y2è(t)

[1.8.9]

:e-sults in the linearized equation of motion for small motions (local behavior) equilibrium. + y2è(t) +

= 00

[1.8.10]

\te that all coefficients in the above equation equation are constant. Consider a timesolution of E(t) = Eelkt, where A denotes the time dependency and E is Substitution of this into Eq. [1.8. I OJ. yields -'

At

=0

[1.8.11]

55

56

CHAPTIR I • B.AsIc PRINcIPLLs PRINcIPu.s

4

Unstable

Unstable

Stable

Stable

Figure 1.44 EeAt cannot be zero for a nontrivial solution, one must have Because Eelkt

A2+y2A+71 =0 A+y2A+71

[1.8.121 11.8.121

This is known as the characteristic equation. whose roots are Y2

±

—

2

11.8.13]

The behavior in the neighborhood of equilibrium is dictated by the roots of the characteristic equation. If the roots A are both real negative or complex with negative real parts, decays exponentially and the equilibrium position is stable. If any one of the roots has a positive real part, then E(t) grows exponentially. The equilibrium position is unstable. If the A roots are purely imaginary, then E(t) oscillates with constant amplitude. The linearized equations in this case do not represent significant behavior and they are not conclusive. One has to conduct additional analyses to determine the nature of' the motion. Such analyses include qualitative approaches such as energy theorems. theorems, the the Liapunov Liapunov method, method, or quantitative analyses such as as numerical integration. More advanced concepts from stability theory are beyond the scope of this text. We will discuss one important stability theorem here, referred to as the potential energy theorem. theorem. The theorem states that fèr conservative ftr conservative svstemn1v, systems, ifif(lie (liepotential potential

has a mninimnum mninimnum in in the the equilibrium equilibrium position. then then the the equilibrium equilibrium position is critically stable. Otherwise, is unstable. Otherwise, itit is

The theorem is illustrated conceptually in Fig. 1.44.

Example

1.15

Find the equilibrium position for the two links attached to a spring in Fig. 1.45. The spring is not stretched when the links are horizontal.

Solution have a conservative, single degree of freedom system. The displacement of every component (both links and the spring) can he expressed in terms of the angle 0. Denoting the spring deflection by x, we can write We

—

2L(1

—

cosO)

[a]

1.8

V

EQUILIBRIUM AND STABILITY

k

0 0

.v

in. L.

L

Figure 1.45 To solve for the equilibrium position by a vector approach. we need to draw a free-body diagram for each link and invoke the equilibrium relations. This is quite cumbersome, as we need to model and include in our calculations the reaction forces at the joints. It is preferable to seek a scalar solution by means of the potential energy function. The free-body diagram of the system is shown in Fig. 1.46. There are three external forces that do work acting on the system. Two of these are the force of gravity on the rods: we denote them by

F1 =

F2

[ci

= —mgj

and they act at the midpoints of the beams. The third force is the spring force and it is expressed in the form F3 =

—hi = —kL(l —cosO)i

Idi Id]

We write the potential energy as

V=

sin

—

o) =

cosO)j2

—

Eel [.1

—

We obtain the equilibrium position by differentiating the potential energy and setting it tü zero; thus. 0

(IV

=

dO

= 4kL2(1

0) 0) sin 0

— COS COS

—

mgL cos 0

If I

Upon rearranging, this gives sin 8

—cosO)

[gJ

= 4kL2 = 4kL

to solve for 0. Solution of Eq. [gl can he obtained numerically.

2kL(1 —cosO)

x

•1

Figure 1.46

ing

57

•

58

CHAPTER

Example

uponby byananexcitation excitation F(F(x) .v ) = —x A particle ofmass of massinin == I is acted upon — x + x2/4. Find the equitheir stability. libriuni positions and

I .1 6

1

BAsic BAsic PRINCII'LES PRINCII'LES

Solution siniilar to a mass-spring system with a nonlinear spring constant. We can find This problem is similar the equilibrium position by setting F(x) = 0. with the result

f(x) [Cv) =

m

.v = ——x(l

— —

4j = 00

lal Ia]

4 .Ve .Ve = 4

or

We have two equilibrium We

positions. To tind the linearized equations, we expand expand f(x) f(x) = 0 for all all positions. positions. as as there there are are no noterms terms inin .1.f that about equilibrium. about equilibrium. From From Eq. We functions of of i. Differentiating Eq. [aJ [aJ with with respect respect to loxx we obtain f'(x) = — + x/2. We arc functions Differentiating Eq. then evaluate y, for each of the equilibrium positions to yield 1

0, For For .Ve = 0, so

y=

I

4, For For .Ve = 4.

yi

— =—

I

[b]

that the linearized equations of motion about equilibrium become For .Ve == 0. 0. For Xe For

=

4,

ë(t)

+

= 0

[ci

—

= 0

[dl [di

Equation tel represents a simple sinusoid, so that the linearized equations do not exhibit Equation

significant behavior. One can further analyze this case using the potential energy theorem. The potential energy has a minimum at that point, point, as as from from Eqs. Eqs. [1 [1.7.241 .7.241 and [1 .8.81 the term describes the second derivative of the potential energy. Hence, the equilibrium position Y describes Yi is critically stable. By contrast, the equilibrium position .ç = 44 is unstable, as it has an increasing exponential solution. solution. The characteristic equation for this case is

A2—I=0

[.1

= ±± 1. indicating one real positive root. One can also physically which has the solution AA = explain this result: result: If If —f(.v —f Cv) is regarded as the spring force, once the variable x passes the equilibrium point x,. = 4. the spring force becomes negative, thus offering negative resistance. Of course, course, in in an anactual actualmass-spring mass-springsystem. system.OflCC once the the point point = 4 is crossed there would no longer he a spring. One can verify verify the the result resultfor br x, = 4 by invoking the energy theorem. The second derivative for the potential energy is negative. negative, indicating a local maximum and an unstable equilibrium position.

1.9

FREE

OF LINEAR SYSTEMS

In previous sections we obtained the equations equations of motion and equilibrium, and then linearized the equation of motion about equilibrium. We now consider the explicit response of a linear (or linearized, linearii.ed. constant coefficient, single degree of freedom system. in this section. we shall analyze the free response. that is. when when there arc no external forces. The linearized equation of motion of a system with one degree of

1.9

FREE RESPONSE OF LINEAR SYSTEMS

and = notation is given in Eq. [1.8.101. Introducing the notation = and the variable x to describe the motion in the vicinity of equilibrium, we

Eq. 1 l.8.lOJ as

1(1) + quantities w,1 and

= 0

+

are kfloWfl as

[1.9.11

natural frequency and dwnpingfactoi: respec-

stiffness versus versus mass in a The natural frequency is a measure ofthe amount of stif'fness gem. It is related to the potential energy. The damping factor is a measure of the dissipation in the system. Energy dissipation is caused by internal friction, as by dissipative forces, such as forces generated by a dashpot. When = 0 motion is referred to as undamped and when > 0. the motion is called damped. To observe the physical significance of these expressions. we begin by considthe mass-spring-dashpot system in Fig. 1.47. Consider a linear model for the = — kx(t) and the and dashpot, so that the spring force is described by force is in the form Fd = —c±(1), with c referred to as the viscous dwnpcoefficient. This coefficient indicates the strength of the dashpot. This way of dissipative force acting on a body is convenient, as it leads to equations motion that are linear. The external force is denoted by F. Summing forces, we

= —ci(t)

—

ci(t) ++ kx(t) kx(t) = F ml(t) + c.i(t)

kx(t) + F

11.9.2]

= = c12\/km. We analyze the natural and In In the above equation. uency as in and k are varied. As the spring constant is increased, we have a stiffer and the natural frequency becomes larger. As the mass increases the system heavier, so the natural frequency decreases. We first consider systems with no damping, = 0. The equation of motion reto

1(t) +

= 0

solve this equation. we introduce the general solution x(t) =

11.9.3] and collect

SO

(A2

=

+

0

[1.9.4]

For there to be a nontrivial solution. XeAI cannot he zero, from which we con-

that aide that

= 0 ,k2+o2—0 A2 +

[1.9.5]

is recognized as the characteristic equation. The roots of the characteristic are A1,2 = ±iw,,, where i2 = — 1. Complex roots indicate an oscillatory

// 'H

() ///////////////////////////////////////

Figure 1.47 Figure 147

59

60

CHAPTER 11 •• BAsic CHAPTER B!tslcPRINCIPLES PRINCIPLES

system. The response can he expressed as

x(t) = and X2 are complex constants of integration. Because x(t) is real and and are complex conjugates. X1 and X2 must be complex conjugates of each other as well. Introducing the real valued constants amplitude A and phase angle 4) such that X1

X1 = and

=

11.9.71

substituting in Eq. 1 .9.6j, we obtain

x(t) =

A cos(w,,t

—

4))

[1.9.8]

The constants A and 4) are determined from the initial conditions. Given initial conditions of x(O) = x0 and i(O) = it is easy to show that

A=

[1.9.9] 11.9.91

= tan1

) and that the response can also be expressed as

+

.v(I) =

V()

11.9. 10] 11.9.10]

wfl

The nature of the motion is harmonic, repeating itself in cycles. Using the relation A cos(w,,l

—

4)) = A cos(2ii- + w,,,

—

[1.9.111

the amplitude of the motion attains the same value after a time of period T so that

2ir Wi'

[1.9. 121

The value of T is known as the period ojovciliation, ojovciliatwn; it is usually measured in in seconds, and it describes the length of a cycle. The units of natural frequency are

in radians/second. Another quantity commonly used to describe harmonic motion is the frequency, defined as I. 'C

[1.9.13]

The unit of frequency is cycles per second (cps) or hertz (Hz). A frequency of 1 Hz = 2'rr 2'rr radls. Fig. 1 .48 shows a plot of x(t). An interesting property of the nat-

ural frequency of a system is that it is a quantity dependent on the parameters of a system and not on the initial conditions. We next consider the case when the damping is not zero. Introducing x(t) = to Eq. ( 1.9. lj Ij and using the same line of thought. we obtain the characteristic equation A2 + 2(w,,A + = (.) 11.9.141 [1.9.141

1.9

FREE RESPONSE

61

LINEAR

x (1)

A

£

.1

—7

t

t

1.48

Figure

Undamped response

1

.49

Damped response

roots are

±

A1,2 =

The nature of the motion depends the following five cases:

1)

—

11.9.151

the values of the damping coefficient. We

1. the roots are real, negative, and distinct. The motion is in the I'orm When decaying exponential and it is not periodic. Such a system is called overdamped. response has the general form + A2e

x(t) =

:

[1.9.161

-

A1 and A2 being real quantities whose values depend on the initial conditions, above equation can also he expressed in terms of hyperbolic sines and cosines. When 0 <4 < I. the roots are complex conjugates with negative real parts in form

±

=

=

—

[1.9.171 11.9.171

— is is the damped naturalfrequefl('V. This quantity basically = the frequency of oscillation of the damped system = e"e1', so that To obtain the response, we note the identity

Wd

x(t) =

=e

+

11.9.181

•1— •+-

F' flowing the approach used for the undamped system. one can express this equation

x(t) = x('t)

cos(w(/! —

(1))

11.9.191

The motion is in the form of a decaying sinusoidal, with an exponential decay as shown in Fig. 1.49. Such a system is known as underdamped. One can that, in terms of the initial displacement x0 and initial velocity v0, the response the form

62

CHAPTER 1 . BASIC PRINCIPLES

x(t) =

+

+

SlflWdt

/

[1.9.201

Note that, similar to the natural frequency, the damping factor is also not dependent

on the initial conditions, but it is a function of the system parameters. 1 represents the border between underdamped and overdamped 3. The systems. It is called critically damped. The roots of the characteristic equation are A2 = real, negative, and equal to each other, A1 = A2 The motion is in the form of a decaying exponential. The response has the form

x(t) = (A1 x(t) (A1 +

[1.9.211

in which both A1 and A2 are real.

case that we saw above. Here, Wd = cofl. and Eq. [1.9.201 reduces to the undamped response of Eq. [1.9.10]. 5. When < 0, the roots of the characteristic equation Oave positive real parts, and they may or may not be complex. A positive real root implies an exponentially growing solution and instability, as we saw in the previous section. Such a system is sometimes called negative/v damped. 4.

Example 1 .1 .1 7 1

I

The

Consider Fig. 1.47. The block weighs 20 lb and the spring is of constant k = 5 lb/in. Damping

negligible and F = 0. The block is released from rest with an initial displacement of x0 = in. Find its position and velocity after four seconds. is

3

Solution The equation of motion is

rni(r) + kx(t) kx(t) == 00

[.1

with rn = 201g = 20/32.2 = 0.6211 slugs = 0.6211 lb• sec2/ft = 0.05176 lb sec2/in. The

natural frequency is

=

=

176

= 9.829 md/sec

[b]

From Eq. [1.9.101 we can write the response to an initial displacement as

x(t) = xocosa,,t

i(z) =

[ci

so that at t = 4 sec we have 3cos(9.829X4) = x(4) x(4) = 3cos(9.829X4)

—0.1382 in

.i14) = —3(9.829)sin(9.829X4) = —29.46in/sec

Id] [di

1.10

RESPONSE TO HARMONIC EXCITATION

In this section we consider single degree of freedom systems subjected to harmonic excitation. Rather than studying the general case of response to arbitrary excitation

I.I0

RESPONSE

TO HARMONIC ExcITATIoN

then considering the special case of harmonic excitation, we prefer to study excitation independently. The reason for this is twofold: First, excitations -:1 a harmonic nature play a very important role in engineering. For example, most have rotating parts, which generate harmonic forces on their supports or bodies they are in contact with. Second, one can study response to harmonic excitation using the steady state motion approach, where one is primarily concerned the motion amplitudes and phase angles, rather than with initial conditions. Consider a dynamical system, whose linearized equations equations of of motion motion have the th-st and

:-:rrn

1(t) +

+

=

[(I)

[1.10.1]

f(r) is the external excitation. For harmonic excitation with a single driving frequency w, w, f(i) f(i) can can be be written written in in the the general compact complex form

f(t) ==

[1.10.2]

One of the advantages in writing the excitation in this form is that the excitation

irnplitude A has the same units as x(t). For example. example, for translational mechanical stems f(t)has hasthe stems f(t) theunits unitsof offorce/mass force/massand and A A has has units units of' of' displacement. displacement. Also, be= COS wt + i sin wi, one can solve for the response for a cosine or a sine simultaneously. The response to a complex excitation will also be corn*x. so that the real part of the solution will be the response to a cosine excitation, the imaginary part, response to a sine excitation. We are interested in the steady state motion. At steady state, all effects due transient sources, such as initial conditions and impulsive forces, have died out, the response is only due to the external harmonic excitation. Hence, we assume the response is of the same nature as the excitation and consider a steady state in the form x(t) = Introducing this solution and Eq. 11.10.21 into [1.10.1] and collecting the coefficients, we obtain (—w2 +

= =

+

[1.10.3]

can be rearranged and solved for X(iw) as X(iw) = X(kv) —

+

[1.10.4]

The amplitude X(iw) can be characterized by the ratio of the excitation amplimultiplied by a nondimensional ratio. This ratio dictates the amplitude of the so we define it as G(iw)

=

I —

+

[1.10.5]

G(iw) is called the frequency response. Recalling that a complex number — ib can be expressed as where M = + b2 and = tan 1(b/a), we express the frequency response as

bbere

G(iw) =

[1.10.6]

63

64

CHAPTER CHAPTER 'I1 •• BASIC BASICPRINCIPLES PRINCIPLES

where G(iw)I =

([1

—

i/i

(w/w,)2]2 (w/w,)2]2 +

=

tan I

- (w/w,7)2 —

[1.10.71 The quantity quantity G(iw)I is is referred referred to as the niagnificationfiwtoi; magnification facto,; and and the the angle i/i

is the phase angle. not to be confused with the phase angle fr defined in the previous section. The steady state response can then he written as =

[1.10.81

The value of the magnification factor is affected by the amount amount 01 01 damping, as as well as by the ratio of the driving frequency to the natural frequency. A plot versus is given in Fig. 1 .50 for various values of the damping factor. The magnification factor becomes smaller with increasing amounts of damping. as one would expect. For a given amount of damping. the magnification factor becomes larger as approaches unity. For an undamped system when = I the magnification factor becomes infInity. This phenomenon. called resonance, manifests itself as very high amplitude vibrations when the driving frequency becomes very close to the natural frequency. Dynamical systems subjected to harmonic excitation are designed such that resonant frequencies are avoided, or encountered as briefly as possible. Furthermore, 4

3.5

3

2.5

I.5 l.5

0.5

0 0

0.5

I

1.5

a) la),1 a)

Figure 1.50

Magnification factor

2

2.5

3

1 •

10

65

RESPONSE TO TO HARMONiC HARMONiCExC11A'rIoN ExcltvrloN

systems have a certain amount of damping to reduce vibration amplitudes. A designed system can experience dangerous levels of vibration, which can physical damage, discomfort to occupants, and reduction in precision. A plot of the phase angle i/i versus is given in Fig. 1 .5 1. The phase angle always less than 900 when < 1 and greater than 900 when > 1. The resonance point point becomes becomesthe thedefining definingfactor factorfor forthe thephase phaseangle, angle, as all curves go the point = IT and = I. < 1. the response has the same sign as the excitation. In this case When response is referred referred to to as as being beingin inpha.ve phase with the excitation. By contrast, when > 1. the response has the opposite sign as the excitation, indicating that ii it is of phase with the excitation. An interesting special case to analyze is that of undamped vibration. Here, set= 0 we can write the steady state response as A

xç(t) = 1

= I and f(t) =

When

x(t) =

11.10.91

- (w/w,1)2 cos

the response can he shown to be

A

[1.IO.10J

indicating that the amplitude of the response increases with time in a linear amenvelope. Fig. 1.52 shows the response. An infinite amplitude is never 0.1

1800

1500

120°

900

60°

300

00

0

0.5

1.5

1

(0

Figur. 1.51

Phase angle

2•2

2.5

3

66

CHAPTIR 1 • BASIC PRINCIPLES

.4

L 0

FIgure 1.52

I

Resonance

reached in reality. As the amplitude increases, either the mathematical analysis becomes invalid and nonlinear and other effects begin to dominate the motion, or the amplitudes become high enough to damage the system. Also, all physical systems have some energy dissipation mechanism, in the form of viscous damping or in some other form, which helps reduce vibration amplitudes.

Example

1.18

A very interesting application of response to harmonic excitation is the motion of a vehicle over a wavy terrain. Consider such a vehicle, modeled as a single degree of freedom system, traveling with constant speed v over a road that has a sinusoidal shape. as shown in Fig. 1 .53. Derive the magnification factor.

Solution The road surface can he modeled as (x) =

"IT'.

Ysin(L

[a]

)

(x)

k

.1•

L

Figure 1.53

1

. I1 0 0

RESPONSE TO HARMONIC HARMONIC ExcIT.kTION

here Y and L denote the amplitude and length of' of tile the wave. Because the vehicle is traveling constant speed we can express its location on the x axis as x(t) = ut. ut. Substituting this this the above equation, we obtain

= Ysin

f2iwt\ L

= }'}'sinwr sinwr

[bi

w = 2irv/L. The problem now is reduced to one of a mass-spring-damper system base is undergoing a prescribed harmonic motion. The speed of the vehicle dictates frequency of the excitation. Consider the free-body diagram of the vehicle. The deflection of the spring is S(t) = — v(i). Writing Newton's second law, we obtain — —k8(t)

—

c6(t)

[ci

—

btich leads to the equation of motion as bnich + c±(t) + kz(t) = —tug —tug +

+ kvO)

Idi

We divide this equation by in and make use of the expressions for for natural natural frequency' frequency' and the factor. Further, we we can can put put v(t) v(t) and and i(t) into into exponential exponential form as as v(I) y(I) = Y lm(euut), •'r) = and write the equation of motion as +

+

+ = —g —g +

+

[e] le]

The first term on the right side has the effect of lowering the mass by a height of = It is the static deformation of the mass due to its own weight. Measuring z(t) from static

Nuilibriurn, we have two terms for the excitation, so that the response also consists Nuilibriuni, consists of' of two s

z(!) zU)

H(iw) =

[f]

—

+

1)

[gI [g]

the frequency response of this system. The magnification factor (also known as transinis:thilit for this problem) is the magnitude of H(iw). Recalling from complex algebra that :thilitv a + ib)(c + J(c + id)J, we have = (a + ib)I X J('c =

+ (2Z'w/w,,)2 G(iw)I

[hi

Fig. 1.54 plots the transmissibility for various values of the damping factor. As expected. the driving frequency gets close to the natural frequency. the transmissibility becomes very i_rge. Peak values occur at frequencies slightly lower than the natural frequencies. It is interto note that as gets larger—that is, as the vehicle goes faster—the transmissibility becomes less than 1 and approaches zero. Also, this ratio goes to zero faster when the damping becomes smaller. Anyone can observe these results when driving a vehicle. As a vehicle goes faster, the effects of uneven terrain and of potholes arc felt less by the passengers. (It should be cautioned when a vehicle goes faster over potholes. the damping damping forces frrces exerted by the dashpots also larger: hence, such driving wears out the shock absorbers.)

67

68

CHAPTER 1 • BASIC PRINCIPLES

4

3.5

3

2.5

IH(iw)I

2

1.5

0.5

0 0.5

0

'S '-.

2

1.5

1

(010),,

Figur. 1.54

Transmissability

1.11

FORCED RESPONSE OF LINEAR SYSTEMS

We will obtain the general response of a linear system by making use of its response to impulsive loading. Consider the system in Eq. [1.10.1] subjected to an impulsive excitation .f = F/rn at time t = 0

1(t) +

+

=

f5(t

—

0)

11.1 1.11

which f(t) is the impulsive force per unit mass. In Section 1.6 we learned that an impulsive force applied to a system results in a sudden change in velocity. For a system originally at rest. from Eq. [1.6. [1.6.12], 12], the the position position and and velocity velocity immediately immediately after the impulse are in

x(0)

=

f

[1.11.2]

One can view the response of the system after the application of the impulse as the free response with the position and velocity right after the impulse as initial conditions. From Eq. [1.9.201. [1.9.20]. for for zero zero initial displacement, the response has the form

x(i) =

(lid

e

sin

[1.11.3]

1.1 1.111

FORCED RESPONSE OF LINEAR

f( 1) f(a)

aI

I

Figure 1.55 generalize this result to the case when the amplitude of the impulsive force unity by calling it impulse response g(t) as We

g(t) =

11 .1 1 .4]

W(jt

next represent an arbitrary force as a summation ol' impulses. Consider a f(t), as shown in Fig. 1.55. At any point i a, the impulse due to the force over a time period of has the form f(r — a) = — Considering that is very small, we can treat as impulsive and, from the above equation, c'e the response to an impulse applied at t = a the form We

=

f(t

the subscript in mpulse at tt = a, and u(t

To

—

a)g(I

a) =

—

f(a)g(t

a)

—

11.1 1.5]

signifies that it is the part of the response due to the —

a) is the. unit step ('unction, defined as

u(i

—

a) =

u(t

—

a) = 0

I

when I a when t < a

find the response to the entire excitation, we sum Eq. [1.11.5] over all the As becomes smaller, the summation is replaced by integration and

get I'

x(1) =

'

I

f(a)g(t

—

[1.11.6]

a)da

Jo

E4uation [1.11 .61 isis known known as as the the convolution convolution integral. By a proper change of van[1.11.61 tie-s (say, I — a = T, a = t — T. da = —d'r), one can show that

x(t) =

I

Jo

f(t

—

T)g(T)dT =

I

f(r)g(t

—

T)dT

[1.11.71

Jo

When solving for the response. one can use either form of Eq. [1.11.71. One bases

-

selection of which form to use on the ease with which the convolution integral be evaluated. For example. if the excitation is constant, use of the form with — T) is preferable.

69

70

CHAPTER 1 •• BASIC BASICPRINCIPLES PRINCIPLES

Note that the response obtained from the convolution integral is the response of Note

the system to zero initial conditions. To find the total response. we superpose this with the response to initial conditions conditions only. only. given given in in Eq. Eq. [1 [1.9.20], .9.20], with the result .v(i) =

/

(x0

COS

V()

+

+

,,

I

.

Sifl Sin co1jt jj + +

/

I I

(Oil JO

— T)e T)e ft (t —

-

sin wilr dr

[1.11.81

For an undamped system, system. the general response reduces to V()

x(t) == .V0 COS w,,t + x(t)

.

+

Sin

1

1 I

.1(1 — T) 1(1 —

(IT

Jo

[1.1 1.91 [1.11.91

stated earlier, there are other ways of solving Eq. [1.10.1]. One way is the homogeneous plus particular solution approach (obtain the homogeneous solution, the particular solution, add them UI) and then impose the initial conditions), and the other is the Laplace transform solution. Actually, the Laplace transform solution is equivalent to the convolution integral. Note that with this approach we obtain the solution as the sum of two quantities. the response of an otherwise free system with the given initial conditions plus the response to the excitation for zero initial conditions. As

Example 1 .19

An interesting intcresting special case is the response of systems subjected to Coulomb type frictional forces, such as the mass-spring system in Fig. 1.24 in Example 1.9. While the friction force

is nonlinear. the equation ol motion can he split into two linear equations. depending on the value of the velocity. Ignoring the external force F. Eqs. Idi and [c] in Example 1.9 can be written as when .i > 0, when 0,

+

when .iwhen .i <

= —p.g —p.g

[al

=

[b] (b]

Consider as initial conditions an initial displacement x, and zero initial velocity. We first need to determine whether motion will take place or not. This depends depends on whether the spring force is is larger larger than the friction force. force. At At the the point point when when the velocity is zero. zero, there will be subsequent motion if

or

> p.p. mg

I

x > p. g

[cJ

Assuming Assuming that at the onset of motion, the spring force is larger than the friction force. force,

the velocity flrst encountered will bc negative. Therefore. we use Eq. Ihi as the equation of motion. From Eq. I. 1. 11 .91 we obtain

x(t) = .V() cos w,,t w,,t +

I

w,1T dr sin sin w,1T

(0,1 _() 1

1

= .V0COSW,11 + —p.g———(1 A()COSW,,T + —p.g———(1

—

cosw,,!) = cosw,,!)

/ + (x0 (xo

i,o\ —

)cosw,,t

/

[dl [dI

The velocity becomes zero at a half cycle. t = T/2 iT/tv,1. The displacement at this point T/2 = iriw,,. is

,

(7r

=—

(.0,,

/ =—

2p.g

lel

1 .1 2

FIRsT INTEGRALs

x(1)

2pg .V()

pg t

Figure 1.56

Response to Coulomb friction

We now switch to Eq. [aJ [aJ as as the the equation of motion, as long long as Eq. [ci holds. The response

second half cycle can be shown to he

x(t) =

—

,

/ +(xo

—

\

r= T =

IT

2ir

wfl

WI,

If'

the velocity is again zero, at which point the displacement has the

x

'27T\ (

—--—

= .V() —

/ It turns out that at every half cycle the amplitude of vibration is reduced by

Igi

the response subject to Coulomb Coulomb Friction friction is in the form of' a decaying curve, with the envelope in the form of two straight lines of slope A typical response curve in Fig. 1.56. The motion stops when the velocity is zero and the spring force is less the friction force. The problem with dealing with Coulomb friction is that the solution has to be obtained half cycle individually, making the analysis cumbersome. Often. engineers replace Coulomb friction model with an equivalcnt viscous damping model and use linear equaWhile this is a gross simplification, if one considers the uncertainties in determining fr frction force and the damping factor, the simplification does not look as bad. It is difficult the damping properties of a system accurately.

1.12

FIRST INTEGRALS

sections we learned about quantities that were derived by integrating second law. For the linear and angular momenta. Newton's second law integrated over time, and for energy. the integration was carried over the disvariable. We also studied the circumstances under which energy and mowere conserved. When applicable, the principles of conservation of momentum and energy give information about the motion without solving for the response explicThis is is a desirable feature, especially for systems systems with complicated equations motion and when one needs to know the nature of the motion but not the cxresponse. When they are conserved we refer to energy, linear momentum, and

71

72

CHAPTER 1 •

PRINClvLis PRlNclvLis

angular momentum momentum asfirsi integrals, or integrals ofthe motion. The The terminology terminology is due to the equations of motion being integrated once to arrive at the first integrals. First integrals involve expressions in which the highest order derivative is one less than the highest order derivative in the equations of motion. Energy and momentum are not the only first integrals that can be found for a dynamical system. There exist other first integrals such as the Jacobi integral and generalized momenta associated with ignorable coordinates. In most cases, integrals of the motion have physical explanations, even though one can generate a first integral that may not have an obvious physical interpretation. First integrals also come in handy when integrating the equations of motion numerically. as they can be used to check the accuracy of the numerical integration. One of the first steps when analyzing a dynamical system should be to search for the existence of first integrals in order to ascertain the qualitative description of the system behavior. First integrals are also used for the geometric analysis of motion. A typical example of this is phase plane analysis. where one plots the dependent variable versus its time derivative. We will not go into this subject in detail but will outline some of the uses of the phase portrait when dealing with conservative systems. Consider. fbr example, a particle of mass in moving in one direction and being Consider, acted upon upon by by a force F(x). From From Newton's Newton's second secondlaw, law, we have have ini(t) ini(t) = F(x). Energy is conserved. One can arrive at this conclusion by integrating the equation of motion over x, which yields —

where dGLv)/dx dG(.v)/dx = 2F(x)/m.

G(x) = C = constant

[1.12.1] 11.12.1]

C is basically a measure of the total energy. and it

depends on the initial conditions. We can rewrite Eq. [1.12.11 [1.12.1 Jto to show the relationship between x and as

=

—

G(x)

11.12.2]

From Eq. [1.12.21. the phase portrait of ii versus versus x has a number of properties: It

is symmetric about the x axis. Also, the phase portrait is continuous with continuous derivatives, so that there are no sharp corners in the phase portrait. For conservative systems. each curve of the phase portrait corresponds to a specific energy level. One

can plot the phase portraits for different energy levels and analyze the motion. It can be shown that phase portrait curves below (or above) the .v axis corresponding to different energy levels never cross each other.'0 If the phase portraits did cross each other, one would not know the energy level associated with the point where the curves met. The phase portrait is also a useful tool for systems that do not admit integrals of' the motion. In such systems. the phase portrait will not be symmetric about about the the .vx axis. Phase axis4 Phase plane plane analysis is one of' the cornerstones of nonlinear stability theory; it is widely used in mechanical and other systems (electrical, chemical, etc.) as well as in control theory. '°Except a curve that crosses the x axis, which ends up intersecting its symmetric counterpart. Such a curve defines critical points and regions on the phase space, separating stable and unstable regions of the phase space. Ii called a sepcratrix.

is

1.12

versus versus x, the the phase phase portrait portrait

for a particle of unity mass subjected to the forces forces (1) (1)

—xx ++ .v3/9, =— .v3/9,and and(2) (2) F(x) F(x) = —x — x — x3/9. the

section. we we can can write write Eq. Eq. 11 . I 2.1] for case (!)as analysis in the previous section.

x., + .r

x4

[a]

= E1 = constant

—

constant, of course, can be evaluated using the initial conditions x(O) and 1(0) '.alues of displacement and velocity at any other time when they are known. For case integral of the motion becomes + x2 ±

=

Ibi

= constant

Before we plot the phase trajectories, we investigate the equilibrium positions and their

case (1), the equilibrium points can he shown to he x = 0 and x = ±3. A wtlitv analysis and plot of' the potential energy indicates that the equilibrium point ..v = 0 stable and the two equilibrium points x = ±3 arc unstable. To see this, we note For

f(x) —x + f(x) == —x

x3/9, so that its derivative is

f'(x) = —l —l

[ci

+

At the equilibrium position x = 0. 0, = 1, I, and for x = ±3. is only one equilibrium point. Xe = 0. 0. and andiiii isisstable. stable.

y, =

—2.

For case (2),

The phase portraits are shown shown in in Figs. Figs. 1.57 plotted versus 1 .57and and1.58. 1.58. where where velocity velocity is plotted for various values of the energy constants E1 and and E2. E2. Li It should should he he noted noted from 1.7 that that the the two cases considered in this example are representative of softening and springs, respectively. The quantity F(x) can he treated as the spring force acting it the the body. body. In In case (I) the spring force it force is is zero or negative in the range —3 .v.v 3, and when Ixi > 3. 3. This implies that as the when Ixi the deflection deflection of' of the spring gcts bigger, the spring smaller. When Whenxx > 33 (or .v < —3). the spring force becomes less than zero, leading gets smaller. unstable equilibrium position. By contrast, contrast. case (2) represents a spring force that keeps growing as x gels larger. larger. Conthere is no equilibrium point other than x = 0. As .v moves further away from the force pulling it toward equilibrium becomes beconìes larger.

r

3

—3

1.57

Case (1)

FIgure

1.58

73

FIRST INTEGRALS

Case (2)

I

Example 1.20 1 .20

74

CHAPTER

1 • BAsic

For case (1) when disturbed from unstable equilibrium (in this case for Ixf < 3). the particle leaves the vicinity of the unstable equilibrium point and moves toward the stable equilibrium point. This is typical of systems that have stable and unstable equilibrium positions.

Example 1.21

A particle of mass in. shown in Fig. 1.59. is acted upon by a force expressed in polar coordinates as F = k/r2er. where r is the distance from the origin to the particle. Find the integrals of the motion.

Solution This problem is similar to the central force problem discussed in Example 1.12. The angular

momentum about the origin 0 is conserved, so that ii. is an integral of the motion. To show that this indeed is the case, we write Newton's second law in poiar coordinates as

F=

ma

k

—jer

r-

= ,n(i

—

r02 )er ± m(r() + 2i0)eo rO')er

[a]

We resolve Eq. [a] into its radial and transverse components as —

k

r02) =

,n(rO +

[hi

r-

[c]

= 0

We first manipulate Eq. [ci and note that d

(1.2(1)

dt

=

+ 2ri0 = r(rO + 216) =

IdI

0

which implies that r26 is constant. Hence, one integral of the motion is

[eJ I.]

= h = constant

momentum of of the the particle particle This first integral has a physical significance. as it is the angular momentum about the origin of the coordinate system '9

k

FIgure 1.59

I3

1 ..1 1

= r X my =

NtmIERIcAI. INTEGRATION OF EQUATIONS OF MOTiON

re1 X X

in( ier ++ ?n(ier

)

= mr mr26k 6k

If]

of the motion can be directly identified by noting that the applied force F does a moment around 0. Next. consider Eq. [bJ and eliminate the 0 term from it using \Vriting Eq. [e] as t9 \Vnting h/r2. and introducing it into Eq. rh]. we obtain

F(r) F(ii

mr

+

k —,

1•-

Igi

equation describes a conservative system. with F(r) denoting the the conservative The energy is conserved, and it has the form 1

2

1

+— — 2

mn/i2 k mh2 , + — — = E = constant

r

[h]

the above expression as another integral of the motion. For general central force F= Eq. + Eq. [h] becomes + V(r) = E, where -r

V(r) =

f(r)dr

— -

*

III

r0

orbital mechanics problems it is convenient to introduce the variable = hr and the equations of motion in terms of' u. We will study this subject in more detail in

r3.

1.13

NUMERICAL INTEGRATION OF EQUATIONS OF MOTiON

sections, we saw closed-form approaches for obtaining the response of a sections. degree of freedom system described by linear, constant coefficient equations. equations are almost always approximations to more complex equations, often £Ec equilibrium. equilibrium. Even £Ec Eventhen, then,ififthe theexternal externalexcitation excitationisisaacomplicated complicated function, function, a closed-form solution of the convolution integral may become prohibitive. when no analytical tool is available to obtain the response, numerical inteof the equations of motion by means of a digital computer is often a viable

There are several types of computational methods for integrating the equations of a system and getting its response. The different types of methods are useu-eating different types of equations. We will not go into the various approaches but we will discuss the general principles behind numerical integration. Dynamical systems are governed by differential equations, where the excitation response are usually continuous functions of time. Digital computers are deto deal with discrete phenomena. Hence, integration of equations of motion z digital computer needs to be discretized and carried out on an incremental basis. basic idea is as follows: Consider the equation of motion 1(1) = Iz(x(t), We We select a time

.1

+ f(r) f(t)

11.13.1]

increment, say and initial time, say t0. The time increment also called the sampling period. We feed the computer the initial conditions and velocity at = t0, as well as the value of the t0, x(t0). x(t0). and

75

CHAPTER CHAPTER 11 •

76

PRINCIPLES

of the and invoke the numerical integration routine. The output of and external force. numerical integration routine will he approximations to the position and vclocit.y at This is considered as one step ofthe integration. + and i(!() + +& + I = We then go to the second step of the integration. Values generated for the disare fed and i(t() + and placement and velocity at the end of the first step. x(a'() + + into the computer as initial conditions. together with the external force position and and velocity at the end of the second The output is approximations to the position The The process continues until the final time or other and i(to + step. .v(l() + selected condition is reached. Two obvious questions involve the selection of the numerical integration apand the selection of the time increment & One selects the numerical intewell as as of of f(l). f(i). One selects gration approach based on the nature of h(x(t). i(t)) k(t)) as well the sampling period such that the results of the numerical integration are accurate. A very small value of increases the computational efiort quite a hit, while a large value of leads to inaccuracies. Often. the nature of the equations of motion gives vibrating system like the one example. in a vibrating For example. of & & For guidelines for the selection of one considered in Sections 1 .9—1 .11, the sampling period should he less than T/6. one sixth of the period of vibration.

integration methods require that the It turns out that almost all numencal numencal integration describing equations be cast into what is known as stale jorni. In state form. the system is represented by fIrst-order differential equations. The left side of the equations contain first-order derivatives of the variables and the right side ot the equations do not have any time derivatives. For example, to write the vibratnew variables. 0. 11 11 in in state state form, we introduce two new ing system of Eq. [1 I 0. [1.13.2] 22(1) = .v(1) 11.13.21 = x(t) as the two equations and write the equation of motion in state = Z2(!) . I

+ f(t)

[1.13.3]

For multidegree of freedom systems we have a set and can write the state equations as (t). 27(1) 27(1) . . . :,,,(l)TT. and [21(t),

of variables {z(1)} =

Z2(l) =

.

—

.

[1.13.4] = [A(jz(l)})1 [AUz(l)})1 + [B({z(t)})]4f(1)} constant coefficient systems are in which {f(i)} is the excitation vector. When linear, constant expressed in state form, the equations have the form {±(i)} = [A}{z(t)}

[1.13.5] 11.13.5]

+ [B}{f(t)}

IBI are constant coefficient matrices. where IAI and IBI

Example 1.22

I

13.3]. A]and and [Ri [Rimatrices matrices for for the the system system described described by by Eqs. Eqs. [1. [1. 13.3). Identify the ff A]

Solution so that have one one input. f(fl. so We have

Thus.[Al [Alisis a 2 X is a scalar. Thus. is order 2. We have vector of order matrix 1, hence aa vector matrix of order 2 x 1. =

2(t)

[A]

=

1

1 i

[B] =

10 L1

2

[B] is aa matrix, and [BI

=

(1) f(!)

[a]

EXERCISES

REFERENCES •tzt berg. •.tzt berg. J. J. H. H. Ath'aneed nainis. New NewYork: York: Harper Harper & Row, 1988. Dvna,nis. D H. Classical classical Mechanics. Mechanics. 2nd 2nd ed. Reading, Rcading, MA: Addison-Wesley. 1980. 1980. Cliffs. NJ: Prentice-Hall, 1978. M. D. Foundations f Applied Mathematics. Englewood Cliffs, D. T. Principles of Dynamics. 2nd ed. Englewood Cliffs. NJ: Prentice-Hall. 1988. C. The Variational Principles of Mechanics. New York: Dover Publications. 1 L. Methods of Analytical Dynamics. New York: McGraw-Hill. 1970.

HOMEWORK EXERCISES

1.2 Given the values for g = 32.174 ft/sec2 and that a block of mass I kg has a weight of' weight of 2.2046 Ib, Ib, determine how many newtons newtons equals equals aaforce force ol' 1 lb.

1.3 1

Show that the radius of curvature of a projectile's trajectory reaches a minimum at the top of the trajectory. Consider normal and tangential coordinates and show that the radius of curvature and torsion are related to the path variable s by Eqs. [1.3.391. Then, consider twodimensional motion (x, v) and derive the expression for the radius of curvature given in Eq. [1.3.151.

A pin is constrained to move move in in aa guide guide slot slot whose whose curve curve isis defined definedby byvv = —

+ x, as shown in Fig. 1 .60. The pin is being pushed by a motor such that it V

y(x) = —x+x

Figure 1.60

77

78

I • B.kslc BAsic PRINCIPLES

has an acceleration = + 0.2i. The initial conditions are at t = 0, x(0) = —0.6. = 0. Find the acceleration of the particle when t = 0.5 s, and find the radius of curvature at that instant. 1

5. A particle moves along a surface defined by z = 2xv, such that x = 3 sina and y v = 3 cos a, in which a is a parameter. Find the unit vectors in the nonnal. normal. tangential. and binomial directions, as well as the radius of curvature and torsion of the ol' the curve when a = ir/4. 6. At 6. At aa certain certain instant, instant, the velocity and acceleration of a particle are defined by v = 3i + 4j — 6k rn/s and a = —21 + 3k rn/s2. Find the radius of curvature and change of speed of the particle at that instant.

7. Consider Example 1.3 and find the distance traveled by the particle on the curve between the points a = 0 and a = rr/6. 8. A 8. A vehicle vehicle modeled as a particle of mass in is moving up a spiraled road of constant radius R, as shown in Fig. I .61. It takes the vehicle five full circles to reach the top. which is at a height Ii from the bottom. a. Express the position. velocity, and acceleration of the particle using cylindrical coordinates. h. Obtain the relationships between the unit vectors in the cylindrical coordinates and the normal-tangential coordinates. What is the radius of

curvature'?

9. Using a spherical coordinate system. express R, R. 0, and 4) in terms of and :. In other words, find the counterpart of Eq. 11 .3.63] in spherical coordinates. 10. A particle is traveling along an elliptic path, described by the equation x2/a2 + rib- = I. with a b. Show—using path variables—that an approximate solution for the perimeter of the ellipse is + b2), b2). and that the solution becomes exact when a = 1'. Hint: I-Iint. You You will will need to use elliptic integrals and their tables to arrive at your result. 11. Equations I 1.3.321 and [1 .3.33] relate the distance traversed along the path in terms of the rectilinear coordinates x. v. and z as well as the path parameter a. Derive the equivalent expressions for ds in terms of spherical and cylindrical coordinates.

ii

Figure 1.61

HOMEWORK EXERCISES

20cm/s

A

3 in/sec

O.6m 0.6m

im

)

V

V

B

B

x

Figure 1.62

.vV

Figure Figure 1.63

A rod is attached to two sliders moving in guide bars by universal joints as shown in Fig. 1.62. Find the velocity of slider B when slider A is at a height of 25 in. and is sliding down with a speed of 3 in/s. A rod is attached to two sliders moving in guide bars by universal joints as shown in Fig. 1 .63. Find the velocity of slider B when slider A is at a height of 60 cm and it is sliding up with a speed of 20 cm/s.

1.4 Using spherical coordinates, find the equations of motion of the pendulum in

Fig. 1.13.

Find the equations of motion for the double pendulum in Fig. 1.64. Consider the vehicle moving up a spiraled road of problem 1.8. There is a coefticient of friction of M between the road and the vehicle. Derive the equation of motion. First, use all three cylindrical coordinates as motion variables and generate three constrained equations. Then, using the traversed angle as the motion variable, reduce the equations of motion to one. Treat the vehicle as a bead sliding around a helical wire. A force pushes the vehicle up.

nil

I8,

L-,

Figure 1.64

79

80

CHAPTER 1 . BASIC PRINCIPLES

g

rn

4-

'U

tx 4-—

4

p

Figur. 1.65 17.

Figure 1.66

Figure 1.67

A vehicle modeled as a particle of mass m is moving with constant speed v along a curved road with radius of curvature p. as shown in Fig. 1.65. The coefficient

of friction between the road surface and the vehicle is p,. In order to reduce slipping, the road is banked by an angle 0. Find a relationship between p. and 0 that will prevent the vehicle from slipping when the brakes are applied and the vehicle slows down with deceleration a. 18. A bead of mass m slides without friction on a wire shaped as the curve z = x2/4, as shown in Fig. 1.66. Find the equation of motion for the bead.

19. Find the equation of motion for the mass sliding with friction over a disk of radius R, as shown in Fig. 1.67. A spring connects the mass with the top of the disk. The spring is unstretched when 0 = 0. 20. A particle of niass in is being acted upon by a Ibrce F(x, t) = Find its response using initial conditions x(0) = x0, i(0) = v0. Consider two cases: (a) < vO > and (b) 21. Find the equation of motion of the system in Fig. 1.68. The two sliders have mass

m and the link is massless. Friction affects only the slider moving horizontally.

22. A steel ball of mass 0.2 kg is released into the frictionless inner surface of a cone with an apex angle of 15°, at a distance of 65 cm from the apex. as shown

g3 .'fl

Massless

A

B

in

Figure 1.68

=0.1

Flgur. 1.69

HOMEWORK ExERcisEs

g

/

p

FIgure 1.70

in Fig 1.69. What should he the initial speed of the ball, so that the ball does not fall to the bottom of the cone and its elevation remains the same? Use spherical coordinates. 23. The wedge (15°) in Fig. 1.70 of mass M rests on a rough platfbrni with coefficient of friction A mass m1 is suspended by a string and is attached by a pulley to another mass in2 which slides without friction on the wedge. a. Solve for the accelerations of m1. m2, and the tension in the string when is sufficient to keep the wedge From moving. b. Find the smallest value of for which the wedge remains at rest.

SECTION 1.5

24. Determine the number of degrees of freedom for the systems shown in Figs. 1.71 and 1.72.

/ 8

k

FIgure 1.71

FIgure 1.72

81

82

CHAPTER 1 • BASK PRINCIPLES

I'

Figure 1.73 SECTION 1.6

25. A fugitive who weighs 180 lb is running on top of a train. The train car weighs 10,000 lb and is moving at a speed of 5 mph, as shown in Fig. 1.73. The fugitive's goal is to jump onto the next car, which has the same mass and speed and which has just separated from the car behind it. As the fugitive jumps, he resembles a projectile which has lefi lell the ground with an angle of 30° and a speed of 20 mph relative to the train. What is the speed of the second car after the fugitive jumps on it and he comes to a stop on the train? P is applied 26. Consider the double pendulum in Fig. 1 .64. An impulsive to in2 in the horizontal direction. Find and 02 immediately after the impulse.

SECTION 1.7

The forces acting on a particle, expressed in the Cartesian coordinate system, + 4. — = 2v + — 3xv2 3x2 + 1,1. F-. = 0. What is the are = 2x + 3x2y 3x2y— potential energy V? arbitrary motions of the 28. Find the period of oscillation of a simple pendulum pendulum. 29. A particle slides over a circular cylinder of radius I?, as shown in Fig. 1.74. The angle that the line connecting the center of the cylinder with the particle makes with the vertical is denoted by 0. The particle is slightly tilted with speed 27.

V

S..

P x

V

(a)

Figure 1.74

EXERCISES

in the x direction at the top of' of the the cylinder. cylinder. Find Find the value of the angle 0 when

the particle and cylinder lose contact. Consider the particle in the previous problem. The particle is released from the top with an initial velocity of V0 = + Vv0J.and and p. p. = 0. 1. Derive the equations of motion, and discuss the difficulties in solving for for the the value value ol' of 0 when the particle loses contact with the cylinder.

SECTION 1.8

£ Find the equilibrium position for the system in Fig. 1 .72. Find the equilibrium position(s) of a particle of mass in = 1. which is acted upon by the forces: b. b. F(x,iT) = + x + + O.051.2 F(x,i) =O.6i+x—0.1x3 = O.6.i+x—0.1x3 A particle of mass in is at the tip of a massless rod of length L and is being used an inverted pendulum attached to two springs, as shown in Fig. 1.75. Find the positions for the mass and check for their stability. Assume small notions and that the springs always deflect horizontally. Assess the stability of the equilibrium points associated with the mass in Prob1.32.

1.9 Obtain the equation of motion motion ('or lbr the the system shown in Fig. 1.76. The rod is rnassless. Then. assume small motions and linearize. Obtain the equation of motion and calculate the natural frequency of the two.77. Assume Assume that that the the pulleys pulleys are smooth and system shown in Fig. 11.77. mas sic s S.

Consider Exan1ple 1.17 and that a dashpot of constant c = 0.3 lb • sec/in actmass-spring system. system. Find the displacement and velocity after two :ng on the mass-spring seconds.

k

In

k

/

g

I— L/3

L/3

L/3 Tn

Figur. 1.75

Figur. 1.76

83

84

CHAPTIR i • BASIC PRINCIPLES

k1

FIgure 1.77

I Figure 178

SEcTIoN 1.10 38.

A machine of weight 50 lb is mounted on a 110 lb table (vibration isolator), which is supported by three springs of constant 500 lb/in each, as shown in Fig. 1 .78. The rotor inside the machine rotates with a speed of 1000 rpm. and it

generates a force that varies harmonically between —30 and 30 lbs. Find the amplitude of the response, assuming that the table has no rotational motion. What is the force transmitted to the support? springs and four 39. A machine with rotating components is to be placed on

dashpots. The machine weighs 200 lb and it operates at 600 rpm. Design the spring and dashpot constants such that only 50 percent of the shaking force of = 0.3. the unit is transmitted to the supporting structure, 40. A mass-spring-damper system has a mass of 10 kg and spring constant of 500 N/rn. The viscous damping coefficient is not known and is to be determined experimentally from the frequency response. When the system is excited by a frequency w = 14.5 rad/s, the magnification factor is observed to be 0.3, and when = 11 radls. the magnification factor is 0.65. Find the damping coefficient that will give values for the magnification factor closest to the two measurements. 41. In Fig. 1 .28. the mass M moves according to the relation .v(t) = x0 cos wi. Find

the equation of motion for the pendulum. assuming that 0 remains small at all times. 42. Consider the small motions of the double pendulum in Fig. 1 .64. Let be specified and find the resonance condition for 02. Then let 02 be specified and find the resonance condition for SECTION 1.11 43.

The step response of a system is the response to the excitation f 0) = lu(t) in Eq.

.10.11, with zero initial conditions. Find the step response lix an undamped

EXERCISES

and show that the derivative of the step response is the impulse response

An undamped mass-spring system (ml + kx = F) is subjected to the excitation Fit) = tt4t). with zero initial conditions. Find the response. An undamped mass-spring system is subjected to the excitation F(t) = F0 (0 to) and no excitation aftert > The initial displacement is xO = —F0/3k, ith no initial velocity. Find the response. An An

undamped undamped mass-spring system is subjected to the excitation F(t)

sin 2oi,1t. with zero initial conditions. Derive an expression for the response. Compare this expression with the frequency response in Section 1.10 and cornnent on the difference.

1.12 A particle of mass mass in in isis being being acted acted on by a force expressed in polar coordinates F = kOir where k is a constant. Find the integrals of motion of' this system. A particle of mass m is being acted on by a force expressed in spherical coordi-

as F =

kReR, where k is a constant. Find the integrals of motion of this kRek,

Find the integrals of the motion for the pendulum shown in Fig. 1.13.

85

chapter

RELATIVE MOTION

2.1

INTRODUCTION

motion is observed from an inertial frame, the expressions for velocity have simple forms. Often it is necessary or advantageous to obmotion from a moving reference frame rather than an inertial frame. One rind examples of this from, among other cases, machine dynamics, vehicle and motion relative to the rotating earth. In machine dynamics. one to relate the motion of one component to the other. Measurement of motion a moving vehicle or platform is a common necessity. And, in an expanded iespect to a rotating coordinate all motion measured on the earth is with respect

•*

the earth's rotation is a realistic assumption in a number of problems. over short distances and over short time intervals can be accurately analyzed considering the earth's rotation. In a number of cases, though. the effect of a noninertial reference frame must be considered. For example. for computaassociated with weather patterns and ocean dynamics. and just about any type over long time periods, neglecting to consider rotation of the earth gives results. In this chapter we consider the motion of reference frames with respect to each and establish relative motion equations. A major difference between an inertial noninertial reference frame arises when we calculate the derivatives of vecWe distinguish between local derivatives, that is. derivatives calculated from reference frames. and global (or total) derivatives, which are calculated with to inertial reference frames. We discuss the differences in form between the lational velocity vector and angular velocity vector and emphasize that angular is a defined vector, rather than the derivative of another vector. We consider involving the rotating earth. Within this context, the reader is introduced to a perturbation technique to obtain approximate solutions to complex problems.

87

88

CHAPTER 2 . RELATIVE MOTION

2.2

MovING COORDINATE FRAMES

Consider two coordinate frames: XYZ. a fixed frame with origin at 0. and .vvz, .vvz, a moving moving frame with origin at B. as shown in Fig. 2.1. We define the unit vectors along the fixed axes X. Y, and Z by 1. 1, J. and K and along the moving axes axes x, y, and z by i, i. j, and k. respectively. We are primarily interested in the case when the moving xyz frame rotates. Consider the motion of a point P. One can observe the displacement of P from the inertial frame, using the vector rp. rp, or from the relative frame, by the vector rp/B, From vector calculus we write

rp =

r8 + rp18

12.2.11

For the sake of discussion, assume that points 0 and B coincide, and drop the

subscript P. One can express r = rp as r = X1 + YJ + ZK or r = xi + vj + zk. To investigate the relationships between the velocities as observed from the different frames, we differentiate r with respect to time. In terms of the inertial frame components, because the unit vectors I. J. and K are fixed in direction, we have (j (/

[2.2.2]

= —r = XI + YJ + ZK (It

and in terms of the moving frame

v=

12.2.31

The first three three terms terms on on the theright rightininEq. Eq.1.2.2.31 describe the velocity velocity as as observed from the relative frame. The last three terms describe the rate of change of the unit vectors and, hence, the contribution due to the motion of the relative frame itself. The normal-tangential, cylindrical, and spherical coordinates that we studied in Chapter 1 are in essence rotating coordinate systems. We identify two types of terms: the local derivative terms, taken in the relative frame, and an added set of terms that depend on the motion of the relative frame. The local derivatives together with the added terms constitute the global derivative terms, measured in the inertial frame. From this comes the simple but important conclusion that derivatives of a vector are different quantities quantities when when taken taken in diffrrent frames. One One must must clearly clearly specify specify which which reference reference frame frame the the derivative derivative is is taken in. V

x B

y

'V

z

Figure 2.1

Vs

ii, Is

I

2.2

89

COORDINATE FRAMF,S

Using a similar approach. we obtain the expression for the acceleration of point

a=

di

12.2.41

= -

The first first three threeterms terms onon thethe right right in this in this equation equation describe describe the acceleration the acceleration as The

as

from the relative frame. They constitute the local second derivative. The three terms describe the acceleration of the relative frame. The last three terms because there is motion with respect to a moving reference frame. These terms from two sources: the differentiation in time of the unit vectors in the expression and the differentiation of the displacement variables in the expression — vj + — vj + zk. These are known as the Coriolis terms. We obtained Eqs. 12.2.31 and [2.2.4 1 by a straightforward differentiation of the expressions. We followed this procedure in Chapter 1 when obtaining of unit vectors associated with the coordinate systems we were considg.

One may ask whether there is a more general way to obtain derivatives of Indeed, there is, as we will see later on in this chapter.

When selecting a moving coordinate system. one must establish the relationship the inertial and the moving coordinate systems. A simple illustration of a transformation is given in the example that follows.

Water is flowing out of the garden sprinkler in Fig. 2.2 with the constant speed of 2 rn/s. The rpm. As As it. it exits the sprinarm rotates counterclockwise at the constant rate of 20/ir rpm. the water makes an angle of 15° with thc horizontal. Find the velocity and acceleration particle of water as it leaves the sprinkler arm. and the velocity 0.05 seconds later.

—-

have two convenient locations to define the origin of the relative axes: to place point B at pivot, or to place point B at the tip of the sprinkler. Let us locate point B at the pivot, so 0. We define the inertial coordinate frame with the Z direction along the rB = 0. yR = 0. The orientation of the relative axes is selected such that the Z and z axes coincide that the projection of the sprinkler on the XV plane is along the .v axis. Defining the angle the X and xx axes axes by by 6. 6. the the.vyz xyz frame is obtained by a counterclockwise rotation about Z axis by 6. as shown in Fig 2.3. We have

I=

cos

01 + sin e9J

0J — j = cos OJ

sin 01

k= K

[a] Ia]

2rnIs y

y

O.3rn

B. kier O.4m

Figure

2.2

X

11

Figure

2.3

I

Example

2.1

90

CHAPTER 2• RELATiVE RELA'fIVE MOTION

of change of' of the unit vectors Differentiation of Eq. [a] with respect to time yields the rates ot'change i, j. and k as dr

= —OsinOI + OcosOJ =

=

Oj

=

di'

—

—01

di'

= 0

/2 =

= —Oi—02j

=

= Oj—021

OsinOJ =

0

[b] indicating that the component of the motion in the :-direction is not affected by the motion of

the relative frame. The rotation of the moving frame is described by 2 20 rev 1mm 2iTrad = —radls (1 = —-—--— 3 60s rev 7T mm

[ci

associated with with aa water particle inside the sprinkwrite the the position position vector vector associated It is useful to write ler as [dl Ed] = xi + :k = r1 r1 =

that the coordinate v and all its derivatives are zero. Using this information, Eqs. and the the vector derivatives in Eqs. [hi, [2.2.4]. and + x0j + + .11 = + Vp =

so

ap =

021)+ x(Oj— 02i)+

11 + = 11

[.] I.] [f] [fJ

As the waler particle is about to leave the sprinkler, we have from Fig. 2.2. 0.4 m x = 0.4rn

1.932 mIs = 2cosIS° 2cos 15° = 1.932m1s

0.3 m z = 0.3m

0.5176 mIs = 2sin 15° 15° = 0.5176

I=

=0

[91 [9]

obtain If]. we [cJ and Substituting the the above above values into into Eqs. Eqs. [cJ and If]. we obtain Substituting

ap =

[hi

mIs 1.9321 0.2667j ++ 0.5176k mIs 1.9321 ++ 0.2667j

Vp

4\

f

= —0.1778i —0.17781 + 2.576j

—

rn/S2

[II

direction. the the velocity velocityand and acceleration acceleration can Because the xvz axes arc continuously changing direction. he expressed more meaningfully using their components in the vertical and in the horizontal plane. We can hence write 176111/S If/S = 0.5 0.5 176

Vhoriz

=

=

1.9322 + 1.9322

0.26672 = 1.950 1.950 if/S

[j] Ii]

it is is gravity. As As soon as the water particle leaves the nozzle, the only force that acts on it It is is more more convenient Also, we are no longer viewing motion froni a set of rotating coordinates. It now to look at the horizontal and vertical components of the velocity, ignoring air resistances the horizontal component of the velocity does not change. The vertical component changes due to gravity, which we can express as a = —9.807K rn/s2. After 0.05 seconds. the vertical component of the velocity becomes mIs — 0.05(9.807) = 0.02725 m/s 176 — 0.5176 = 0.5

that the the water particle is about to start going down. indicating that

[k] [ki

2.3

2.3

REPRESENTATION OF VECTORS

REPRESENTATION OF VECTORS

this section we describe different ways of representing vectors. Consider a reccoordinate system with unit vectors e1, and e3 and two vectors r and u as

r=

r1e1 + rie7 + r3e3

U = u1e1 + u2e2 + u3e3

12.3.1]

will refer a set of vectors described this way as geometric vectors or spatial The The dot dotand andCrOSS cross products of these vectors yield the results r • u = r1 UUi i +

r

U

= (r2u3

—

r1 u3)j + (r1u, (Ii U2— — r2uJ r2uJ)k )k r3u2)i + (nut — r1u3)j

also express the vectors r and u in column vector

[2.3.21 12.3.2]

as

r1

{r}

=

{u}

r7

142

(2.3.3] 12.3.31

r3

column vectors {r} and {u} are also referred to as algebraic vectors. Using this we can express the dot product of two geometric vectors in colunm vecformat by

{r}'{u}

r• u

12.3.4]

T denotes the matrix transpose. To express the cross product r X u using umn vectors, we introduce the skew-symmetric matrix [F] associated with the r, and write =

0

13

r,

ri

0

—r1

—r2

r1

0

12.3.51 [2.3.51

that

r Xu

[F]{u} =

(2.3.6]

?3U1 — ?3U1 —

?1U2

—

r2u1

the geometric vector operation operation of of rr xx U u and the column vector operation X r, we can also write —u X :( {F1{u} are equivalent. equivalent. Note Note that that because becauserrXXuu = —u = In dynamics, one frequently encounters the vector product r X (r X U), which is to describe centripetal acceleration. The expression is commonly shortened to r r X u, with the understanding that the cross product between r and u is performed Using the notation introduced above,

r X (r X U)

[F][Fj{u} [F][F1{u}

12.3.71

we note that the matrix multiplications in [?][Fj{u} [F]fjj{u} can he performed in any order.

91

92

CHAPTER 2 • RELATWE MoiloN

The relations derived above are only valid when all vectors involved are expressed in the same coordinate system. If the vectors r and u are not represented in the same coordinate system, Eqs. [2.3.4] and [2.3.61 no longer hold. For example, if we have two coordinate systems with unit vectors e1 e2e3 and e',e and two spatial vectors r and u in the form

r=

r1e1

u=

+ r2e2 + r3e3

u1e

+

+

[2.3.81

their dot product becomes

33

r•U = > r

and the matrix [ci has the form

in

id =

e2•e'1

e1'e

e1'e

e2•e.

e2•e

[2.3.10]

e3'e;

The expression [c]{u} can be viewed as the column vector representation of u using {p.}i the e1e2e3 triad. Conversely, the expression {r}T [cj can be viewed as the transpose of triad. In the next section, the column vector representation of r in the we will formally define the entries of [Cl as direction cosines. It should be noted that sets of geometric vectors can also be represented in colurnn vector format. The elements of the column vector will be geometric vectors. For instance, the triad e1 e2e3 can be written as e1

{e} {e}

=

12.3.111

e2 e3

This form is also useful when taking dot or cross products of unit vectors. For exam1, 2, 3) are scalars, can be + g3e3, where gj(i + ple, the expression g = g3]T, g3]T. written using the column vector format as g = {e}T{g} in which {g} = The column vector notation is not restricted to describing geometric vectors. This formulation is commonly used to represent variables in a Euclidean space. Next, consider differentiation of scalars and vectors with respect to other vectors. This procedure can be conveniently illustrated using column vectors. Consider ... of dimension ii. where the elthe scalar S and a vector {q} = ements ... . ., q,, are independent of each other. The derivative of S with respect to {q} is defined as the n-dimensional row vector dS/d{q}, whose elements have the form

dS = d{q}

(9q2

... aq,,j

I

[2.3.12]

{q}isis the the column column vector repalso written written as S{q}. When {q} In compact form, dS/d{q} is also

resentation of a geometric vector in rectilinear coordinates, the above operation becomes similar to the gradient operation. We can write VS = [dS/d{q}IT.

2.3

REPRESENTATION OF VECTORS

In a similar fashion, we obtain the derivative of a column vector with respect to

... unjJT, 112 2nother. Given the the column column vector vector {uJ {uJoforderrn, oforderrn,where where{u} {u} = ±ae derivative of with respect to a vector {q} of order ii is obtained by differentievery element of {u} with respect to every element of {q}. The end result is the referred to to as as the the Jacobian, Jacobian.and anddenoted denotedby by[{U}{q}i [{U}{q}iororbybyLII.J], having n matrix, referred form dq2 d{u}

d{q} cRq}

(lIb

/2117

=

1{U}{q}] 1{U}{q}J

(/117

[2.3.131

L"qI

(qi

cIq2

now investigate some of the special forms of the scalar S and its derivative b!th respect to a vector. Consider, !br example. the n-dimensional column vectors = = [U1 v2 ... u,j" and {q} and define the scalar S as We

S=

r

[2.3.141

{v}7'{q} = {q}1{v} =

. ., ii) are the = 1. 2, .., of' {q}. Taking the derivative of S with respect to {q}, we obtain

vk(k vk(k =

1,

2. . . ., ii) are the elements of {v} and 2,

.

.

d({v}T{q})

d{qf fl

11

(1

...

= k=I

k=1

(-1q2

dVk vfl+>qk.dUk k—i k.1

For the special case when the elements of {v} are not functions of qk. vj = 1.2,..., n) we obtain

dS = /2 ({v}T{q}) = {v} T d{q} d{q}

t

•t

vj(qk).

[2.3.161

1D]isisaasquare square matrix matrix of of order where IPI {D]{h} where If we write the vector {v} as {v} = {DJ{h} = {q}'[D}{h} {q}"[D}{h} and {h} is an n-dimensional vector, so that S = {h}are area af'unction f'unctionof of {q}. {q}. and{h} the case where where none none of of the theelements elementsofofIDI IDIand

d({h}'[Dl'{q}) d{q}

d{q}

—1

that if [DI is not a function of {h} the derivative of S with respect to {h} is

k

dS

s

In

1 ({h}'[DJ'{q}) d{h}

d({q}1[D]{h}) a'

= {q} T IDI

[2.3.181

mechanics, one commonly encounters scalar quantities that are quadratic in of the motion variables, such as kinetic and potential energy. Define S as

93

94

CHAPTER22• •REI1:V1'IvE CHAPTER REI1:V1IvE MO'I'LON MO'I'LON

S = {q}'[Dl{q}. The derivative of S with respect to {q} in this case has the form

'is

{q}7'[DJ + {q}"[DJ1'

d{q}

[2.3.19] 12.3.19]

and when the matrix D is symmetric, this becomes (1 ({q}T [D]{q})

d{q}

=

[2.3.20]

Consider the symmetric form above and the case when the matrix ID] has elements that are functions of {q}. We frequently encounter this situation in Lagrangian mechanics. The derivative of a matrix with respect to a vector is a tensor of' order three. We can avoid dealing with tensors of order three by taking the derivative of the product fDJ{q}, so that we now have d ({q} ({q} TfDJ{q}) d{q}

+ {q}1(h1D1{C1})

=

[2.3.21] 12.3.21]

d{q}

Next, let us obtain the derivatives of functions of several variables with respect to time. Consider Consider the the scalaiscalar x, which is a function of nn variables x, which .....,., q,1 and time t. so that that xx = x(q1, x(q1, t). The derivative of x with respect to time is ., t). obtained using the chain rule as

dx = di

P1

.,

dx

(9x

[2.3.22]

We can express this relationship in column vector format. indeed, introducing the column vector {q} = [qi q2 we write Eq. [2.3.221 as

dx

dt =

dx (IX

.

dx

(9x

+ —

=

[2.3.23]

where we recall that the derivative of a scalar scalar with respect to a column vector is aa

row vector. Extending this to the case when the time derivative of a column vector {r} JT is sought, where {r} = [r1 r2 where r1 q t), (.1 (.1 = 1, 2, . in) we obtain .

. ,

=

[2.3.24]

= I'qI{q}

+{

+ f in which [{r}{cj}J = Irq] is recognized to be a matrix of order X in x n. n. When When {r} is the column vector representation of' a geometric vector, in = 3 and [rqJ becomes a 3 X iiii matrix with [rq [rqljk Jjk = 1,2, 2. (j = 1, k = 1,2, 1, 2, ..., ., n). To To visualize this better we express the vector {r} in terms of a set of unit vectors as .

.

ruT

{r} = [r1

r=

r1e1

+ r2ei r2ei + r3e3 r3e3

12.3.25]

so that

r=>T ar. a

k=-1

c9r

dl

3

or

3

1

j1

j=i Lk=I

+ + a,

jei

__

2.3

95

OF VECTORS

To evaluate the second derivative of {r}. we use the chain rule again. We wish to taking the partial derivative of a matrix with respect to a vector, so we make of Eq. 12.3.211 12.3.2 lJ and write =

+

+

,*r at

+ {

_1 I

at2

[2.3.271

second term on the right side of this equation can also be expressed as {4}

[rj{q}

=

z which the elements of the matrices [rj(J(J}(j = —

1,

{c•/}1•['3qq]{c;'}] ['3qq]{c;'}]'

[2.3.28] [2.3.281

2, 3) are

[2.3.29]

.1 / ==1,2,3: 1,2,3: i,k =

It is important to remember that the differentiation operation is conducted in same reference frame as the vector r is measured in. In the above equations we an inertial frame.

the two coordinate systems XYZ and xv:. with unit vectors UK and ijk. The cosystem is obtained by rotating the XYZ system first by an angle of 300 counterc counterc locksystem about the Z axis and then rotating the resulting intermediate intermediate x'v'z' x'v'z' coordinate coordinate system clockwise about the axis. Fig. 2.4 shows the rotation sequence. Given the vectors the matrix matrix id. 2k, find a • b and the a = 3! + 4J + 6K and b = 21 + j ++ 2k,

first represent represent the the unit unit vectors vectors ijk ijk in in terms terms of of UK. UK. From From Fig. 2.4. we can write the unit of the intermediate x' v'z' axes as = cos 300! + sin sin 30°J

j'

cos 30°J = — sin 30°! + cos

k' = K

[a]

xvz coordinate coordinate system system is is related related to the intermediate axes by a clockwise rotation of The xv: .L5 about the v' axis. so that—using the short notation for the sine (s) and cosine (C) of an of an

Y

•v

.v

\ 30° \\ 30°

450

I'

Z, —.

'.

Figure 2.4

Example

2.2

20

96

CHAPTER 20 RELATIVE RELATIVE MorioN MortoN CHAPTER

angle—the unit vectors in the coordinate systems systems arc arc related by =

30°J) +s45°K= +s45°K= 45°(c 30°I 30°I ++ss 30°J) c45°i'+s45°k' = c 45°(c

j k =

—s

3001 + c 30°J = —s30°1+c30°J =j' = jl ==—s

45°k' = 451' ++ cc 45°k'

—s

45°(c 300! 300! + ss 30°J)

1

—

c

'

+

—

1')

"jJ+

1+

/.,

Ibi [c] [ci

J

45°K =

+

—

[d] [dJ Introducing these expressions to the vector b. we obtain

b=2i

= +j + 2k 2k =

+

—

Yj± VK)_ ji + VK)_ ii,,;

4% +

4

=

2

I

—

+

.J

/

+

2(_

4

2K

+ 2

1.] 1.1

We can now find the dot product of a and b as We

a•b =

If]

= 18.94

3

LU

)+ Let us now generate the direction cosine matrix and write

I•i I'i

[cj [cj =

J•i K•i

J.j

'V — —

K•k K'k

K•j —

=

I'k J•k == J'k

I•j

1/2

—

3/2

—

c45°c30° 3Ø0 s cc45°s30° s45°

—s45°c 300 30° —s45°s 45° s 30° —s c45°

—s30° cc30° 30° 0

[g]

,6/4 2/2

0

To find the dot product using column vectors we use Eq. [2.3.9], which yields To —1/2 — 1/2

= {i,}T[c]{h} = [3 4 61

2

—

—

= 18.94

I

0

[hi 1k]

2

which, of course is the same as the result in Eq. Ff1.

Example

2.3

I

Figure 2.5 shows shows a door opened at an angle 0. On the door at point C is an ant that starts Figure 2.5 crawling upwards in a straight line. Its path makes an angle which is fixed. with the bottom of the door. Determine the position of the ant, and obtain the velocity of the ant using Eqs. [2.3.231 through [2.3.261. [2.3.261

Solution to We attach the inertial coordinates XYZ to the door frame and the moving moving coordinates coordinates .vyz xyz to We

the door. as shown in Fig. Fig. 2.6. 2.6. The The .vyz xyz axes are obtained by a clockwise rotation of XYZ XYZ by by an angle of about the Z axis. The position of the ant is

r=

—Lk + hi + s(—

+

= (h

—

(—L + .scosd)i + (—L

[a] Ia]

coordinates by by We relate the inertial and moving coordinates We

k= K

I=

cos(—O)I + sin(—O)J sin(—O)J = cosO!

—

sinOJ

(bJ Ib]

2.4

ThANSFORMATION OF COORDINATES, FINITE RomiioNs

V

z.:

V

4

x x

L

L

1)

I

C

C.

Figure 2.6

FIgure 2.5 when introduced when introduced into Eq. [a]. yields

r=

(Ii — — .scos4))cos6l .scos4))cos6l

—

(Ii

id [ci

+ (—L (—L +

—

51T• To To find find the arc()()and andssso sothat that{q} the motion motion arc that can describe the Two variables that Two {q} = [0 51T, column vector vector form. We then then find find of the ant. we write the the position position vector vector in in column we write

will he a 3 X 2 matrix, with its columns containing derivatives of {r} with respect to 0 We thus have s. We {r}

=

(h

—

—(h

—

scosj4cosO

Fri [rql

—(Ii

—

= —(h

—

scosç&)sinO (9 cos 4)) cos U s COS S 4)) COS

-1- ssinçb —L -1ssinçb —L

—

COSØCOSO

costhsinO sind

0

[di Idi

vector. we can write the position vector. that there is no explicit time dependence in the the position

as

{v}

=

= [

—(Ii

—

—(h

—

scosçb)sinU scos4))sin(9 scos4))cost9 0

scos 4)) —(Ii — .ccos 4)) sin (10 cos

cos

cos4)sinO cos4)sin6 sin4)

S

—

Os 4) sin Os

1.1

sin

While one can obtain the above relation by direct differentiation of Eq. [c], the use of [2.3.24J lends itself to efficient computer implementation and is preferred for more cornproblems.

2.4

ROTATIONS TRANSFORMATION OF COORDINATES, FINITE ROTATIONS

Cnsider two coordinate frames, x1 X7X3 with unit vectors e1. e2, and e3; and e. Without loss of generality we select the frames such unit vectors

97

9B

CHAPTIR 2• RELATIVE MOTION X2

x'2

xI

013

x-4

Figure 2.7 that their origins coincide, as shown in Fig. 2.7. The position vector to a point P can

be expressed as r = x1e1 + x2e2 + x3e3 =

xe + .4e +

12.4.1]

in which in which

= r•e1

X3 = r•e3

x2 = r•e2

= r•e

= r•e

= r•e1

[2.4.2] 12.4.2]

These terms can be generalized to resolve the vector r into its components along a coordinate system with unit vectors e1 e2e3 as (r'e1)e1 + (r'e2)e2 (r•e2)e2 + (r•e3)e3 rr == (r'e1)e1

[2.4.31

in a similar fashion, we can express unit vectors in the two coordinate systems in terms of each other. For example, we express the primed unit vectors as

e ==

(e

•e1)e1 +

•e2)e2 +

•e3)e1

= (e •e1)e1 + (e •e2)e7 + (e; •e3)e1

e=

(e •

e1

)e1 + (e!

•

+

e3)e3

[2.4.41

Let us now examine the nature of the dot product terms in Eqs. [2.4.41. Take. Evaluating that expression. we obtain e2 = e2 for example, ••

e7 =

e2 •

=

1e

I

cos 021 021 == cos cos 021 je2j cos

[2.4.51

where 02! is the angle between the x2 and xI axes (Fig. 2.7). The dot product between two unit vectors is equal to the cosine of the angle between them. We define cosine a quantity called direction cosine between the two axes x by by the thecosine of the angle between the x1 and and axes, and denote it by = cos = • (i, f = 1, 2, 3). The angles that the coordinate axes make with the axes of another coordinate system are called direction angles. Considering the preceding relations,

2.4

TRANSFORMATION OF TRANSFORMATION OF COORDINATES. FINITE ROTATIONS

can write for x =

=

+ x2e2 x2e2++ x1e3)•[(e xie3)•[(e •e1)e1 •e1)e1 + (e •e2)e7 + (e

(x1e1

= x1e x1e •e1 •e1 + x7eç x7eç •e2 •e2 +

•e3 = x1c11 •e3 x1c11 + x2c21 x2c21 + x3c31

Similar expressions are derived for

=

and

[2.4.6J [2.4.61

DefIne the column vectors {r} and

[2.4.7] 12.4.71

=

[x1

the directio,z cosine matrix fc] as Ic] =

C?j C21

C1)

Cl3

C27 C22

C23

C3j C31

[2.4.8] [2.4.81

C33

leads to the relationship between {r'} and fr}

as

= [cJT{r} Next, we express express .v1, x1,

[2.4.91

x3 in terms of

Following the same procedure

we obtain

clix, + CJ2.V-, CJ2.V, ++ C13X3

_vI =

= C7JX + +

= C31X1 = C31X1

+ C73X1 C23X1 C12.Xii

+

12.4.101

can be written as {r}

= IcI{r'} 1c1{r'}

[2.4.111

= [cJ1'{r'}. [cJ1{r'}. Comparing with Eq. llJ. we conclude that the direction cosine matrix is a unitary (also called ormatrix, that is, its inverse is equal to its transpose, or

Equation 12.4.91 can be inverted to yield

Ic]

'

=

[CIT

IcJIcJT = [1]

[2.4.12a,b] 12.4.12a,b]

re [11 is the identity matrix.

Observe from the preceding equations that the unit vectors can also he expressed of each other using the direction cosine matrix. Defining the column vectors e1

e

t

easy

to show that {e'}

[cJ'{e}

{e}

= [cJ{e'}

12.4.141

Be aware that the definition of direction cosine we arc using here is not univeraccepted. Some texts instead define the direction cosine as c,1 = • e. e1.

We defined one direction cosine cosine for for each each angle angle between between the theith ithand andjth jth coordinate for a total of nine. The question arises as to how many of the direction cosines

99

100

RELATIVE MOTION CHAPT1R 2 S S RELATIVE CHAPT1R

are independent of each other. Equation [2.4.1 2h1 represents six independent equations that relate the direction cosines (six because of the symmetry of' the equations). reducing the number of independent direction cosines Cj,. and hence independent to three) It follows that. at most. three parameters are necessary to repangles resent the transformation from any given configuration of coordinate axes to another one.

The important question is how to select these parameters. One possibility is to use three rotation angles. In this case, any two successive rotations need to be about nonparallel axes. Otherwise the rotation angles will not be distinguishable. Another possibility is to use a single rotation about a particular axis. We will make use of this case in Chapter 7. Let us consider three rotation angles and analyze how one can accomplish rotations of coordinate systems. and explore means of expressing rotations of coordinate systems and the rates of change of' these rotations. To this end, we identify two approaches: a body-fixed rotation sequence and a space-fixed rotation sequence. To carry out a body-fixed rotation sequences begin with an initial frame and rotate it about one of its axes. Make the next rotation about one of the axes of the rotated coordinate system, which leads to a third coordinate system. Then rotate this third coordinate system about one of its axes to obtain the final frame. This rotation sequence can he visualized by imagining a box attached to the moving reference frame. Each rotation is performed along onc of' the edges of the box. The position of the box with respect to the final rotated coordinate frame is the same as its position with respect to the. initial frame. Consider an initial frame x1x2x3 as shown in Fig. 2.8, and rotate it by an angle of about the x1 axis. Denoting the resulting frame by Yi = x1

Y2

+ X3SIflOI +X3SIflOI

=

V3 = —x2sin01 —X2S1flO1 + x3cos01

12.4.15] y3

CO

y1

x"

0

V.'-

.11.

—

Yi

Figure 2.8

A 1 rotation

These vectors are orthonormal vectors, and they represent the direction angles of the axes of the transformed coordinates. Equation [2. 14.1 2b] represents the six possible dot products among these vectors. 1One can demonstrate this by writing [ci as three column vectors

{a7}

2.4

TRANSFORMATION OF COORDINATES, FINITE ROTATIONS

m matrix form. {y} =

[2.4.16]

1R1 J{x}

r which xI {Y}

=

{x} = x,

Y2

[R1] =

V

0 0

cos01 —sin01

sin cos01

[R1j is referred to [R1J

as the rolalion matrix. We recognize that [R1 is the transof the direction cosine matrix between the two coordinate systems, systems, [c1] [c1] = RJT. The above transformation is also known as a I rotation, denoting the axis which the rotation takes place. Take the Y1Y2.v3 axesand androtate rotatethem themby by an an angle angle 02 about the Y1Y2Y3 axes axis (Fig. 2.9). This type of rotation is called a 3 rotation. Denoting the resulting frame by Zi can show that 1

= [R,]{v}

12.4.18]

which cos

C'

I-,

—

/ -7

[R2] =

COSO2

0

0

1

0

[2.4.19]

rotation matrix [R2 is the transpose of the direction cosine matrix between {z} {v}. Finally, rotate the Zi Z2Z3 axes by 03 about the Z2 axis (a 2 rotation) to obtain axes (Fig. 2.10). Similar to the previous rotations, we have J

[2.4.20]

{x'} = 1R31{c}

Y3. '-3

Plane of

and

0, Plane of YIY2 and

Vi

Zl

Figure 2.9

Zi, X.,

Z1

A 3 rotation

Figure 2.10

x Xj

A 2 rotation

101

1 02

CHAPTER 2 S RELATIVE MOTION

in which cos63 0 sin 03

=

{x'} =

0 1

—sinO3 0

0

[2.4.21]

COS 03

axes from the x1x2x3 x1x2x3 axes, axes, we wecombine combine Eqs. Eqs. 12.4.161 To obtain the through [2.4.21], which yields {..v'}

[2.4.22]

[R]{x}

= [R11[R7}{Ri]{x} [R11[R7]{Ri]{x} =

in which [RI is the rotation matrix between the unprirned and primed coordinates.

It is clear that [RI =

where [c] =

flc2 I[c3J.The Thetransformation transformation we we have just performed is referred referredto to as as aa 1-3-2 1-3-2 rotation rotation sequence. describing the order of the axes about which the coordinate systems are transformed. We can continue to perform more rotations of the kind above. and in some cases it may he convenient to do so. However, performing more rotations than three introduces a redundancy. As an illustration, consider again the same rotation sequence. Given the direction cosine matrix between the initial and rotated frames, and what This is the rotation sequence is, one can uniquely determine angles 62, and 62, because Eq. 12.4.1 2b 2b11 describes describes three three independent independent equations which can be solved If we have a fourth rotation, with an angle 64. for the three unknowns 62. but now we have fbur unknowns with no we still have three independent unique solution. All rotation matrices [R11 (i = 1, 2, 3) have determinants equal to 1, that is, det[R11 = 1. From linear algebra. det(1R3J1R2 hR1 I) detIR3 detlR,J detiR, detER1 detIR1 J = detiRl = det(IR3JIR2JIRI detIRI J) = detlR3I I

J

J

1

[2.4.23]

that the combined Iran from {x} to {x'} is carried out by a matrix whose determinant is equal to I. This implies that the direction cosine matrix between {x} and {x'} has a determinant of unity. The determinant of a general orthogonal matrix is ± 1, 1, so that we have in effect shown that for an orthonormal matrix matrix to to represent represent aa direction cosine matrix, its determinant must he equal to unity. One can show independently of the preceding argument that the direction cosine matrix between any two right-handed coordinate systems has a determinant equal to 1. The second approach mentioned for describing iotation rotation transformations between coordinate systems systems isis by bymeans meansof ofthe the.s'pace-fLved space-fLved rotation rotation sequence, where the rotation transformations are carried out about the initial axes. Consider a set of initial coordinates x1 x2x3. We We first rotate this frame about about the the x1 axis by an angle Oj to obtain the V1 axes and call call this this rotation rotationmatrix matrix[R1 [R1]. 1. Then, we rotate the Y2Y3 coordinates about the 12 axis by an angle 02 to obtain ZI coordinates. We denote this transformation matrix by [R21. In a similar fashion, we perform the third transformation about transformation about the the.v3 by 63 to obtain the final coordinate system .Vi axis by so

Denoting the rotation matrix by [R3], one can show that the final coordinates are

related to the original coordinates by

= I R]{.v} = I

R, IfIf R3 I R1 R1 If If R, I

J{x} J{x}

[2.4.24]

2.4 x3

OF COORDINATES,

103

ROTATIONS

Rotation by 90° about

Rotation by 900 about x1

I

z3..

I,. ' :' i. -i-l

xI

(c)

(h)

(a)

/

Rotation

Rotation by 90° about X3. x3.

.t, .1

by

about Yi

V..,

Yi

< 33

(d)

Figure 2.1 1

(e)

(f)

Finite rotations do not commute

Looking at Eq. 12.4.241, we see that the final transformation matrix is in reverse

rder compared with the transformation matrix for body-fixed transformations. In general. one uses body-fixed transformations to relate one coordinate system i another. It is usually more convenient and meaningful to visualize the motion and i express angular velocities and accelerations in terms of a set of axes attached to body. Nevertheless, space-fixed rotations provide an alternate description, and help one to visualize the rotation angles. We are interested in expressing transformations from one coordinate system to the preceding analysis that the order in which rotations are performed makes a difference in the orientation of the transformed rdinate system. One can verify this visually, by just taking a hook and rotating t'or sequences. The The concept concept isis illustrated illustratedin inFig. Fig.2.2.11 11 for about two axes in different sequences. rotation sequence. One can illustrate this concept using a space-fixed i -'cation sequence as well. Therefore. it is not possible to represent rotations of coorsystems by finite angles as vector operations, because the commutativity rule not hold.

We analyze the minimum amount of information needed to determine the direction cosine We have uniquely. We begin with a set of axes x1x2x3 and transform it into direction cosines and six independent equations resulting from Eq. [2.4. 1 2h1. SO that of the direction cosines have to be specified. Consider first the case where the following

Example

2.4

104

MotioN

CHAPTER 2 S

x

Figure 2.12 information is given:

Angle between the

axes is 60°

Xi and

Angle between the x1 and x4 axes is 45°

Angle between the x1 and

axes is 60°

We are given the task of finding the direction cosines. We need to first check to see if the information given is consistent, or. in other words. geometrically compatible. Given the angle between two axes, the locus of points that are compatible defines a right circular cone with the apex angle as the angle between the two axes, as shown in Fig. 2.12. In this example. tile axis .v defines a right circular cone about with apex angle 60°. Similarly, the axis defines a cone about x1 with apex angle It follows that the maximum angle between any two lines on the cones is 105°, making it possible for the angle between the and axes to be 90°. In a similar fashion, it is possible to have a 90° angle between the x and axes and the x.1 and axes. Therefore. the information given is compatible with a right-handed coordinate system. (As an illustration of a geometrically incompatible case. suppose we were given the problem above, except that the angle between the x1 x1 and and x axes is 30°. It follows that the maximum angle that one can have between the .v and axes is 75°. making it impossible to have a right angle between and x.) Returning tO the original problem, once we determine that the information we have is consistent. we proceed with finding the direction cosines. From the above relations ('jj = 0.5 0.5 = 0.707 0.5 (-13 = 0.5 Ia] Equation [2.4.12bJ written in terms of +

+

=I

C12C22 Cit (-'21 + CI2C22 Cit (-'21

results in the six equations

+ C-, C2 +

+

C13C23

C21C31 ('31 +

= 0

= II =

C-11 + + ((-i,

+

+ C12C32 + C13C33

+ c23e33 = 0

=1

=0

Ib]

The values in Eq. [aJ satisfy the first of Eqs. [h] [h] uniquely, uniquely, so that that we are reduced to live equations for the six unknown direction cosines. Hence, the direction cosines cannot be solved for. The physical explanation of this is that only the .v1 axis is uniquely specified with

2.4

105

TRANSFORMATION OF' COORDINATES, FiNITE FLNITE ROTATIONS ROTATIONS

to the frame. The x2 and axes are not specified at all. A coordinate system by any amount of' rotation about the x1 axes will satisfy Eq. [bJ. The conclusion is that, to define the direction cosines, one needs consistent and compatiinformation about the direction angles of at least two different axes in any frame. Consider the following information:

Angle between the X3 and

axes is 15°

Angle between the .v2 X2 and Angle between the Xi and

axes is 35° axes is 125°

information results in the direction cosines c33 == 0.9659 0.9659

c22 c22

= 0.8192 0.8192

= —0.5736

[ci

A quick quick examination examinationof ofc12 c12 and and c22 indicates that that c32 c32 = 0. c22 indicates 0. so that the axis axis lies on plane generated by the x1 and and •v2 fromx1x1X2X3 •v2axes. axes. The The transformation transformation from is X2X3 to x is rotation about the the x3 x3 axis by in the intermediate coordinate system v1y2y3. The second rotation the axis by an angle of 15°. What we do not know at this point is whether this rotation is clockwise or counterclockwise. Fig. 2. 13 illustrates the rotations. It fbllows m this case, we need one more piece of information to uniquely determine the transformed transformed This additional information can be in the form of a sketch.

way to visualize body-fixed rotation transformations is to attach an imaginary box to the 'qdinatc frame and observe what happens to the box as the coordinate system is rotated. 'qdinate Cci.iider the box in Fig. 2.14. Rotate the box about line OA clockwise by 300, then about line by 105°. What are the coordinates of' point D in the initial frame after rotations?

I

2.5

relative frame is attached to the box. We denote the initial frame by XYZ. The intermeframe after rotation about OA (the X axis) is denoted by x'v'z', x'y'z', as shown in Fig. 2.15. rinal configuration xvz is obtained by a counterclockwise rotation about OB (the axis) ,'. 05°, as shown in Fig. 2.16.

.13.

p,.

/ l,1_

3

A

x,

y

I)

// V. .1

7

2.13

FIgure 2.14

Example

x

2•

106

RELATIVE MOTION CHAPTER 2• RELATIVE CHAPTER

a

A

x. .v

B

V1

D

C

xv

Y

/

z'

D

B V

Figur. 2.15

FIgure 2.16

Using rotation transformation equations, we write the relations between the coordinate frames as

.v'

y' z'

.ix y

=

z

I

=

0 0

—

cos(105°) cos( 105°) 0 sin( 105°) sin(105°)

0 cos( — 300 ) sin(—30°)

0 1

o0

—

X

C)

300 )

Y

cos(—30°)

Z

sin(

—

sin( 105°)

x'

0 cos( 105°)

p7

=

0

0

1

0

V

A. —0.2588 0

0

0.9659

0

z'

x

x = [R1j

x'

0

1

—0.2588

Ia]

z

z

—0.9659

Y

z'

xl. = [R21 y' yl

[bJ Ibi .

so that the relation between between the original and the rotated frames is

x

-vV

yV

= [R21[R11

x = LRI

Y

z Z

z

[ci

V

z Z

in which

[Ri = [Ri

[R2 [R21[R,1 11 R, 1 = =

0.4830 0. 866() —0.1294 —0. 1294

—0.2588 0

—

0.9659

—0.X365 —0. 5(X)0 —0.5000

[dl

._() —0.2241

is the final transformation niatrix. Equation Ic] is valid when relating the initial, initial, as well as the final, orientations of points on the box as the box is moved. moved Denoting these initial values by the subscript i and the final coordinates by the suhscriptf, we can write

[X,i [X,1

HI

= [R] [R]

A1

x.f

yf

Yj = IR] Yi Zf

[e, I] I] le,

LzJ On the other hand. because the moving frame is attached to the box, the coordinates of a point on the box before rotations in the initial frame are the same as the coordinates of that I

2.5

107

INFINITESIMAL ROTATIONS, ANGULAR ANGUL..%R

xint in the rotated frame after rotations. We therefore have Xi

.Vj•

=

y1

Igi Zi Z,

ZI. Zi

the inverse of Eq. [f] into Eq. [g], we obtain

xi

x.i X

=

Vj

xi. X, =

LRJT

[hi

Y,

To To find the coordinates of point D.

we denote the initial and final positions of point D as and D1. The initial coordinates of point D are (0. 2, 1): ii: thus, thus, its its final final coordinates coordinates are

:'

0.9659

0

=

ERJT

YD1 YDI

= IR]T 2

Zflf

2.5

1

=

1.603

[I] II]

—1.224

INFINITESIMAL ROTATIONS, ANGULAR VELOCITY

the previous section we saw that consecutive rotations of coordinate frames by ±nite angles do not lend themselves to representation as vectors. Hence, one does have a vector to differentiate in order to represent rotation rates. rates. To To express express rates rotations, we begin by analyzing infinitesimal rotations. Consider, for example, a :-2-3 body-fixed rotation sequence with rotation angles of 0 02, and 03 about about and about Z3). The final transfbrmation matrix can can be shown to •

[R] = —cO,s03 [R]= —cO,s03

c01s03 ++ sO1sO,c03 c01c03 — S01S02S03 cOjcO3—s0150,s03

502

S01C02

—

cO,s02c03

+ c01s02s03 c01c02

12.5.11

consider that all the rotation angles 0,, 02, and are very small, and them with and We also assume that these small rotations ake place during a short time period of Invoking the small angle assumption of cos 1. and I. and neglecting neglecting secondsecond- and higher-order terms in Now,

rotation matrix becomes 1

[R] =

1

12.5.21

It should be stressed that and are not the differentials of finite expressions but differential quantities themselves. This observation is critical to •inderstanding the definition of angular velocity. Examining Eqs. [2.5.11 and [2.5.2] closely, it becomes clear that no matter what the order of transformation is, closely. :R] the same R] in Eq. [2.5.2] has the sameform, form,indicating indicating that that infinitesimal infinitesimal rotations rotations are

108

MortoN

CHAPTER 2•

commutative. The iOnof ofthe thetransfi)rrned transfi)rrned coordinate coordinate system system does does not depend The onentat onentation

on the sequence of the infinitesimal rotations. Let us explore ways to express infinitesimal rotations, and their rates, as vectors. We write the relation between the coordinates of a point in terms of the initial and transformed coordinates as {x'} = [RJ{x} We

{x}

express the rotation matrix as [Ri =

[11

[2.5.3]

=

in which [I] is the identity

—

matrix, and 0

[2.5.4]

0 0

= —[SB]. It follows that [RIT =

is a skew-symmetric matrix, that is,

[lj

+

We now obtain a relationship between the initial arid final coordinates of a point as the refei-ence reference frame frame isis transformed. transformed. Denoting Denoting quantities quantities pertinent to the initial and

final positions by the subscripts i =

= [R1{x,}

{x'1}

=

12.5.5]

Next, we define define by by{&v {&v'}'}the thechange change in in the the coordinates by

=

=

— —

— —

{4} == [R1T{4} [R1T{x;} — — {4}

IR]T = Now, dropping the subscript i, and using the relation IR]T

[2.5.6]

[Ii + [SO], [I] [SO], we can

express

{x'} = (LII +

—

{x'} =

[2.5.7] [2.5.71

Note that we could have derived the equivalent of Eq. [2.5.7J in terms of the

initial coordinates x1x2x3. Indeed, defining the change in coordinates as = {xj} — {x1} and substituting into Eqs. [2.5.5], one obtains {&v} = Eq. Eq. f2.5.7j can he viewed as the column vector representation of the relation

=

[2.5.8]

Xr

where

=

+

+

[2.5.9]

an infinitesimal rotation vector.2 The concept is illustrated in Fig. 2.17. The rotation takes place about an axis passing through the vector The rotation is by an amount which is the magnitude of Note that the boldface in the above equation. indicating that the quantity is a vector, is over the entire expression Aft and not just over the 0. This signifies that is not the infinitesimal value of a vector but a defined quantity consisting of a collection of infinitesimal rotations. is

2We defined by Eq. [2.5.9] the infinitesimal rotation vector without rigorously proving that it indeed is a vector. The proof requires thai certain transformation properties be satisfied. It can be found in the text by Meirovitch.

2.5

INFINITESIMAL ROTATIONS, ANGULAR VELOCITY

Rotation axis

Figure 2.17 during which the rotations by the time increment Let us divide Eq. approaches zero. The left side leads to a simple place, and take the limit as term, written urn

a

Lit

dr =—=r di

[2.5.101 12.5.101

To evaluate the right side. we take Eq. [2.5.91, divide it by Lit, and take the limit approaches zero. We define the resulting expression as the angular velocity of frame with respect to the initial frame and write it as (0

U) =

= hm

[2.5.11J

lirn

+

+ w3e

i = 1,2,3

[2.5.12]

12.5.131

components of the angular velocity, also referred to as the instantaneous velocities of the rotating frame. We have shied away from writing the right side of Eq. [2.5. II] as a derivative. should be emphasized is that angular velocity is a defined quantity and that the derivative of any vectoi: For this reason, the angular velocity vector is to as nonholonomic, a term that is associated with expressions that cannot as derivatives of other terms. A nonholonomic expression cannot be to another expression. The way one arrives at the angular velocity vector is different from the derivation of the expression for translational velocity, rate of change of any defined vector.

109

.uu

CHAPTIR

2•

RELATIVE MOTION

In view of this discussion, one can write the rate of change of the position vector as (I

toXr in which

is

—{x } = [thJ{.i'} di

[2.5.14]

the matrix representation of the angular velocity vector to, to. -W3

0

[2.5.15)

0

0

WI

It should be reiterated that r is a vector whose components are fixed in the relative frame x Equation [2.5. 141 is is illustrated illustrated in Fig. 2. 18. The The change in r is [2.5.141 2.18. due to a change in direction, and hence, r is orthogonal to r. It is also orthogonal to the angular velocity vector, as r can he visualized as rotating about the axis generated by the angular velocity vector. By definition of the cross product, r is perpendicular to the plane generated by the vectors w and r. NIow that we have defined angular velocity as a vector, we can obtain the angular velocity of a reference frame by adding up the angular velocities associated with the rotations that lead to the orientation of the reference frame. Equation [2.5.14] is valid not only fora vector describing the velocity ofa point. hut for any vector u whose components are constant relative to the moving frame. We have

u=toXu

[2.5.16]

An example is when u is a unit vector. For the unit vectors considered in this section. we have

= to X = to

=

Xee ==

to x = to

—

+ w1e — —&1e!,

+

Plane generated by

Figure 2.18

[2.5.17J

2.5 that

R0TvnoNs, ANGUIIAR VELOCITY

the magnitude of the time derivative of a unit vector. e1 (i =

1,

2, 3), is

equal to unity. The preceding definition of angular velocity is not the only way angular velocity be defined. In the following. we present a more abstract defInition. Consider a

reference frame which is rotating with respect to a fixed reference frame. angular velocity of the rotating frame is defined as the vector c& which. which, when into any vector fixed in the moving frame. gives the rate of change of that viewed viewed from from the inertial frame. The angular velocity of the relative frame w quantity that makes the the relationship relationship 12.5.161 f2.5. 161 hold. hold. Using the unit vectors of the moving reference frame, which in this section we taken as , e, and e, and their rates of change, one can define the angular vector as to =

.1

e)e

+ (è

•

+

e)e e;)e

[2.5.181

definition definition can can he he verified verified by by analyzing analyzing the the expressions expressions ftr the rates of change unit vectors from Eqs. While this definition is more abstract than

way we arrived at Eq. 12], it is mathematically more sound, and it can be more easily in mathematical operations that involve angular velocity. In 7 we will see yet another definition of angular velocity. Note that in this section so far, we have defined angular velocity in a number of *I"IS. discussed what it is physically, and derived expressions for rotating reference What we have not done is to come up with a general way to quantify angular as a function of rotational parameters. We will analyze the quantification for the general case in Chapter 7. within the context of rigid bodies. Now let us discuss a special case of angular velocity. Previously, we defined anr velocity as a vector with certain properties and stated that it is not the derivaof any quantity but rather it is a defined one. There is an exception to this. When velocity is along a fixed direction, then angular velocity is called simple velocity, and it becomes the time derivative of the rotation angle about the direction. if we denote the unit vector along this fixed direction by, say, J, we can express te angular velocity by to wJ, and can write v as an exact differential in the form dO cit

12.5.191

0 is the angular displacement about the fixed axis. The commonly studied cases of plane motion and rotation about a fixed axis involve simple angular Let us next consider more than one relative reference frame. We begin with a frame X YZ and rotate it by an angle Oi 01 about the X axis to obtain the x The angular velocity of the x'v'z' frame with respect to the inertial frame is as simple angular velocity. Denoting it by to1, we can write = 011 =

12.5.201 [2.5.201

then rotate the x'y'z' frame about the z' axis by 02 and obtain the xz frame. The velocity of the xvz frame with respect to the x'v'z' frame, which we will thus is also "simple" when this second rotation is considered by itself, thus by

1 11

1 12

2•

CHAPTIR 2• RELATIVE RELATIVE MoTioN

w2=0-,k

12.5.21] 12.5.211

The angular velocity of the rotated frame xv: can be written by adding the angular velocity terms associated with the two rotations as

0) = Wj Wj +

+ 07k

=

12.5.221

Let us express the angular velocity in terms of the different reference frames. First, consider the final relative frame. We find 1' by reading the first column of Eq. [2.4.191 as I Weintroduce = cos 621 — sin 0j. 0j. We introducethis thisinto intoEq. Eq. [2.5.221 [2.5.221 and and obtain 0.) =

+ (02 (02 = 0, 0, cos62i

—

sin61j + 02k

[2.5.23] 12.5.231

We observe that o cannot he expressed as the derivative ol' another vector. Hence, it cannot he classified as simple angular velocity, although both to1 and o2 are simple angular velocities when considered individually. This can he explained by noting that while is about a set of fixed axes. is is actually with respect to a set of rotating axes. The situation does not change when we express the angular velocity vector using the unit vectors of the inertial frame. Indeed, if we use, use. from Eq. I171. the relation k = k' = — sin6,J + cosO1K and substitute it into Eq. (2.5.22), we obtain =

+0)7 +0)7 ==

[2.5.24]

We hence conclude that for a sequence of rotations about nonparallel axes, the combined angular velocity will not be "simple."

Example

2.6

A momentum whcel is a useful classroom tool to demonstrate angular momentum conservation. It is basically like a bicycle wheel with a thickened rim and handles along the spin axis. The angular momentum principles are illustrated by asking a student to hold the wheel and spin it. and then to move the wheel around or stand on a platform that is free to rotate, as shown in Fig. 2.19. At a given instant, the momentum wheel is spinning counterclockwise (viewed from the right) with angular velocity of 3 radls, and the student holding the wheel is leaning left with

/ Figure 2.19

2.5

INFINITESIMAL R0TArloNs, R0TArloNs,A.NGULAR ANGULARVELOcIlY VELocrlv

1

V

wI)eel/sludent

V

WstudentJpiactornl

-I

Figure 2.20 velocity of 0.2 rad/s and making an angle of 15° with the vertical. At the same the platform is rotating with a clockwise angular velocity of 0.5 rad/s. Find the total velocity of' the wheel.

•

e 2.20 illustrates the reference frames involved. We attach the frame x'v'z' to the platThe inertial Z axis is in the vertical, and it is aligned with the z' axis. The orientation wheel can be obtained by rotating the x'v'z' axes by an angle of 15° counterclockwise the x' axis. Referring to this coordinate system by xv:. the the momentum momentum wheel's wheel's spin is v direction. We write the angular velocity as U) =

+

+

= — 0.5k' + 0.21' + 3j rad/s

Ia]

instant shown, the unit vectors in the .vz and x'v'z' coordinate frames are related by xz and 1

1

j = 0 k

0

0

0

Il is

150 cos 15° — sin sill 150 15°

sin 15° cos 150

,j1

Ibi

k'

inverse of Eq. [b], we have k' = sin 15°j + cos I 5°k = 0.2588j + 0.9659k and so we can express the angular velocity in terms of a set of coordinates attached to the wheel as

the

I=

L

w == —0.5(0.2588j —0.5(0.2588j + 0.9659k) + 0.21 + 3j = 0.2i 0.2i ++ 2.S706j 2.S706j — — 0.4S3Ok radls It] Note that to express w in terms of the unit vectors associated with an inertial reference XYZ requires that the exact relationship between the coordinates xvz .vz and XYZ he In this problem, we conducted an instantaneous analysis and did not specify the iner\YZ axes, except for the vertical direction. Expressing w in terms of a moving coordinate is more meaningful.

robot arm makes an angle of' 400 with the rotating shaft. which oscillates about the y' with the relation 6(/) = cos 2t rad. rad, as shown in Figs. 2.21 through 2.23. The shaft has velocity of = 0.5 rad/s. At the tip of the arm, there is another shaft. A disk is

Example

2.7

13

1 14

CHAPTER 2• RELATIVE Mo'rIoN

z

0

0

vcrtica 1

400

0 1

/

I

/

/

/

/

B

(03

B

V

x. x' x,

/

(03

/

/

/

yf, y

Figur. 2.21

Figure 2.22

Figure 2.23 = 7 rad/s about this shaft. Find the angular velocity oi

spinning counterclockwise with thediskati = 3s.

Solution As shown in Fig. 2.22. the Z axis is the vertical, and the X'Y'Z axes are attached to the shaft

The x'v'z' axes are obtained by rotating the X'Y'Z' axes about the X' axis. The v' axis along the robot arm. We use a rotation about the v' axis by 0 to go Irom the x'v'z' axes to the xvz axes of the second shaft, about which the disk turns. We have = cos0i'

—

j =j'

sin Ok'

k = sin 01' + cos0k'

[aJ

The angular velocity' velocity of the disk can be written as (.0 =

+ (.02 + (03

[bJ

in which

= 0.5K = 0.5(sin40°k'

—

cos40°j') = —0.3830j' + 0.32 14k' rad/s

IT

=

= _iOsinôj = iOsinôj = 0.08778j rad/s

(.03 = 7i (.03 7i radls

[ci

Air = 3s,0(3) = 0.l5O8rad1sothat 0.lSO8rad,sothai k' = —sin — sin 01 01 ++ cos cos0k Ok = —0. —0.15021 15021 ++ 0. 0.9887k 9887k

[dJ

and we can express the total angular velocity in terms of the xvz coordinates as

w=

0.3214(—0..1502i + 0.9887k) — 0.3830j + 0.08778j + 71

= 6.9521

—

0.2952j + 0.3178k rad/s

[eJ

Note that in this example. as well as in the previous one, we did not express the angular velocity in terms of a set of coordinates attached to the rotating body. Rather, we used the xz xvz coordinates attached to the spin axis of the disk. This is commonly done when analyzing axisymmetric bodies, as we will see in Chapters 7 and 8.

2.5

INFINITESIMAL ROTATIONS, ANGULAR VELOCITY

the derivatives of the unit vectors in the cylindrical. spherical, and normal-tangential Example using Eq. Eq. [2.5.16] and compare with the results in Chapter 1. 2.8 begin by examining the angular parameters associated with the coordinate systems. First cylindrical coordinates, as shown in Fig. 1.9. The z direction is fixed and the coorframe rotates in the xv plane with angular speed 0, so that = (1k. The unit vectors expressed as the mutually orthogonal triad Cr, Cr, e0. e0.and andk. k.For For the the time time derivatives derivatives we then = Ok

X

e, =

= Ak

=

X

k(1kxk=()

[a]

k = Ak x k = 0

—Ac,.

For For spherical coordinates (Fig. 1.12). the mutually orthogonal unit vectors arc eN. ecu,

Denoting the unit vectors in the inertial frame by i. j. k and those in the rotated frame j'. k', k', from 12 we can write from Fig. Fig. I1.12 .j'. .

=

1'

[b]

=

ecu,

Fig. I . 12, there are two angular components, (1 about the z axis and Fig. 1.12,

about the polar

The combined body-fixed rotation matrix [R] is is

cd [RI =

0

sj

cO

0

0

1

—sO

cf

0

0

cbsO

0

sO cO 0

0=

cO

—sO —SO

Ic]

0

s4.s0

1

Eq. [2.4.91. we obtain for the unit vectors coscfcos(1i + e4, = coscfcosOi =

—

sin4)k

SiflOi+COSAJ

[dl

+

+

CR =

are the same as Eqs. 11.3.541 11.3541 and and [1.3.561. [I .3.561. The The angular angular velocity vector is a superposiof the two angular velocities. so that

[.J [.1 the unit vector in the z direction as k = cos — sin decu,, we arrive at the angular of the coordinate frame in terms of the unit vectors in spherical coordinates as = 99cos4e/? cos

(i

—

sin

[f]

+ fre0

to the derivative expressions eN =

—

= (0 cos

= (0 (0 COS

t9 — A

—

sin

+

=

+

=

Osifl thecu, + thee ()sin

+ 0

cos

= —o cos

—

— (1 0

are the same as Eqs. 11.3.591 [1.3.591 through [1.3.61]. For the normal and tangential coordinates, the unit vectors are e1. showed in Chapter 1 that e

V

—e,,

[g]

sin be/?

e,

X e,,.

[hi [hJ

11 5

1 16

CHAPTER

2• RELATIVE MOTION h WI,

en I

=0 ii

Figure 2.24 where p is the radius of curvature. We can write the angular velocity vector in its general

form as

= w,e, +

[I]

+ w1,e1,

where we have yet to determine w,, cofl. and Wh, the components of the angular velocity in

the tangential. normal, and binormal directions. Fig. 2.24 shows the angular velocities. We obtain the time derivative of e, using Eq. as w,,e,1 + = (w1e, + w,,e,1

X e1

=

+

(L)flel)

U]

Comparing Eqs. [hJ and Liii Lii' we conclude that Ii

1k]

p

The above relations can be explained physically and by inspecting Fig. 2.24 more closely. Because the hinormal direction is perpendicular to the osculating plane. the component of the

angular velocity in the binormal direction is the speed divided by the radius of curvature. or the rate at which the path path bends. bends. That That w,1 = 0 can be deduced from the same argument. Because at any point the motion of the particle can he considered as going along a circular path whose center is the center of curvature, there is no rotation in the normal direction.

We next consider the time derivatives of the normal and hinormal unit vectors, and write + w/,eJ)) w/,eJ)) XX e,1 e,1 =

= =

[I] II]

—

[ml Im]

+ w,,e,,) X e1, =

Recalling Recalling the delinition of the torsion of the curve as Idebf = (IsIT, (IsIT, and how the torsion is

linked to the twisting of the osculating plane, we obtain the component of the angular velocity in the tangential direction as V

Wi= —

T

V

In]

= ——e, T

As the torsion i gets larger. the plane of the curve twists less. Equation [1] indicates that the

rate of change of the unit vector in the normal direction depends on the way the curve bends as well as on the amount by which it twists, an expected result. The angular velocity of the reference frame can thus be written as =

/l 11

1

\

+ WhC/, = vv —e, —e, + —e1, I p J I

I

I.]

2.6

2.6 C

RvrE OF OFChANGE ChANGEOF OF,k VECTOR, RvrE VECTOR,ANGULAR ANGULARACCELERATION ACCELERATION

RATE OF CHANGE OF A VECTOR, ANGULAR a vector u observed from a moving coordinate system xvz. The coordinate is rotating with angular velocity w. The vector u is expressed as

u=

+

u can be found by differentiating Eq. [2.6. 1] as

U = uJ + UyJ + 14-k u-k +

+ uJ ++ u-k

12.6.2] 12.6.21

first three terms on the right side of this equation denote the change in u as by an observer on the moving frame. Hence, the differentiation is carried out

e the moving frame. We denote this local derivative term by

fdu\ \dt JreI II

UyJ ++ u-k u-k + UvJ

=

[2.6.31

next three terms on the right side of Eq. [2.6.21 denote the change in u due to the of the coordinate system. Considering Eq. [2.5.161, we can express them as

[2.6.4] 12.6.41 to the relation

du fdu\ U= + to XU wX =I dt \dt \dtJrei JreI I

12.6.5] 12.6.51

relation is known as the transport theorem. relation theorem. In In column vector format we can

it as

d —{u} = dl

[<u} + [oltU}

12.6.61

notation the transport theorem is written as

di cit

=

di

)rei + wx()

[2.6.7]

The physical interpretation of the transport theorem is that the rate of change of i is a different quantity when viewed from different reference frames. When with moving reference frames, one must he careful that the differentiation iteration is carried out in the proper reference frame. A natural application of the transport theorem is the calculation of the derivative the angular velocity, known as the angular acceleration. The angular acceleration coordinate frame. denoted by a, is defined as

a=

a'

12.6.81

'te that the time derivative is being taken here in the inertial reference frame. We b-rite the angular velocity and acceleration in terms of the unit vectors of the relative

1 17 1

1 18

CHAPTIR 2S RFL\TIVE MolloN

frame as

to = w i + w

a=

+w k

a i+ a

+ a -k

[2.6.91

Differentiating the angular velocity, we obtain

a

= w11 + W1j + (0-k + to (0 X (A) w == th1i + Wvj +

[2.6.10]

If we write the angular velocity in terms of the unit vectors of the inertial frame as to =

+ wyJ + w7K, the angular acceleration has the form

a

=

=

(It

+ th1J + th7K

[2.6.11]

in both both Eqs. 12.6. 10] and [2.6. 111, the components of the angular acceleration are the rates of change of' the angular velocity, = thj (i = .v. .v, v, i = X. Y. v. Z, or I Y, Z).

We draw the following important conclusion: If the ai angular ig u lar 'elocilv corn poi i en Is of a ni components oi 'ing coordmate coordinate fra iize are expressed moving franze in ternis inertial coordinates or in terms terms of of the the coo coorthnates rthnates of 0/'(lie (lie nwving nwving frame. frame. the comnponents components oft/ic of theangular angular acceleration acceleration can be be obtained obtainedby byaasimple simple diffirenti— di:tftrenti— ation of the angular velocity components. When the angular velocity of the reference frame frame is expressed in terms terms of the unit vectors of another frame that is not attached to the relative frame and is rotating with angular velocity, say, say. where to, then the expression for the angular acceleration has the forni = Note that in this case,

aj a,

+

[2.6.12]

X (0

th1.

An interesting application of the transport theorem is in systems involving more than one reference frame. Consider the disk in Fig. 2.25. As the disk spins, the axis about which it rotates also turns. We attach a moving reference frame to the axis and find the angular acceleration of the wheel. Denote the angular velocity of the disk with respect to its axis by to4 and the angular velocity of the axis as An observer sitting on the axis of the disk sees it rotating with angular velocity to2. The total angular velocity of the wheel is to (Uj + (02. (02.

(03

V V

-t

FIgure

2.25

2.6

1 19

RvrE OF cHANCE OF A VECTOR, ANGULAR

differentiate the To the angular angular yeyeTo find the angular acceleration of the wheel, we diftBrentiate expression

a=

11

d

cit

di

dci di

—w1 + —W2 = —(W1 + (02) = —w1

[2.6.131

The first term on the right side of Eq. [2.6.13] is the angular acceleration of the ius about which the wheel is rotating. This term can he obtained by straightforward To obtain the angular velocity velocity of' of the second frame, we invoke the transport transport thetheUsing Eq. [2.6.5]. we obtain

d —(02 = —(07 di

+

)<

+ (01 X>< (02

=

12.6.141

1

We get the total angular acceleration by adding Eqs. [2.6.13] and [2.6.141: thus

a=

(01

+

[2.6.151

+ (Oj X (0'

Let us contrast the difference between Eqs. [2.6.15] and [2.6.101. In Eq. [2.6.10] have a straightforward form for the angular acceleration, because the angular yeof the frame (the disk) was expressed in terms of the reference frame attached t the disk only. The derivations that led to Eq. [2.6.15] are based on an intermediate attached to the disk axis, rotating rotating with with tot. To illustrate the point further, consider the coordinate frame transformation in the section with an inertial frame XYZ, an intermediate frame x'v'z' obtained about X, and the final relative frame xvz obtained by a rotation 02 a rotation 01 about z'. From Eq. [2.5.22]. the angular velocities of the two frames are

=

u which I = has

ft =

1 1'

611

[2.6.161

= 02k'

— sin Oj so that using Eq. [2.6.151. the angular accelera= cosO2i COS 021 —

the form (01

+W7reI+(O1 +W2reI+(01 XW2 XWi = 01i'+Oik+01i'X67k

= 01(cosO2i

—

sin 02j) + 02k + 0102(cos02i

=

—

0162 sin 02)i

cosO7 cos02

—

—

sin

62j) x k

sin 02 + 0102cos02)j + 02k

12.6.171

let us obtain the angular acceleration by direct differentiation of Eq. [2.5.231, write

a =(O1cosO2—0162sin62)i—(01sin62+0102cos02)j+02k [2.6.181 of course, is the same answer as Eq. [2.6.17]. The transport theorem is most often the preferred approach for obtaining derivaespecially for complex problems. and it is more adaptable to implementation Jigital computers. When analyzing the relative motion of bodies and especially when studying

rotation problems. one may need to transform velocities and

120

CHAPTER 2• RELATIVE MOTION

1)3

P,Q b

a a1

a

Figure 2.26 2.26 Figure

accelerations from one reference frame to another several times. The resulting ex-

pressions can become quite complicated. To avoid confusion, we are going to adopt the following notation for dealing with multiple reiBrence reference frames. We will denote reference frames by capital letters, such as A. B, C. and so on. In general. the A frame will he inertial and the B frame will be relative, and when rigid bodies are involved, this frame will be attached to the body. The coordinate axes in these frames will he defined with lowercase letters corresponding to the frames. For example. for the the A A frame. frame, the the axes axes will willbe bedenoted denotedby by a1, a2, a,, and a3, and the corresponding unit vectors by a1 a2, and a3. The origins of the relative frames will usually be denoted by the sanie capital letter used to denote the reference frame. The origin of the inertial frame will usually be denoted by 0. The frames A and B are illustrated in Fig. 2.26. The angular velocity and angular acceleration of one frame with respect to anand AaB. other, say of frame B with respect to frame A. will be denoted by respectively. When describing translational velocities and accelerations, as well as the difTerentiation operation. the frame in which the differentiation is performed will be denoted by a superscript on the left side of these vectors. The differentiation operwhere u is a vector. Expressions for the velocity ation can he be written as AA*U, of point P. where the position vector is r = rp, have the form I?

4Vp =

=

[2.6.191

[2.65]. Using Using the notation introduced above Consider the transport theorem. Eq. [2.6.5].

and using the A and B frames as the inertial and relative frames, we can write Eq. 12.6.51 as

Ad —u= dt

dt di

12.6.201

It is clear that the transport theorem can be used to relate derivatives in any two

reference frames. We will denote the expression on the left side of Eq. [2.6.20] as the rate change of the vector in frame A. the first expression on the right as the frame B, and the second term on the right side as the rate of change change of of the the vector vector in infraine

2.6

RATE OF CHANGE OF A VECTOR, ANGULAR ACCELERATION

term. In operator form we can write the transport theorem as

4d

X() —()+w X() —() = •—o= di di

[2.6.21]

B

4

Equations [2.6.20] and [2.6.21] describe the transport theorem between any two frames with no need to have or to identify an inertial frame. Considering we defined angular velocity previously, the notation introduced here is more For example, the transport theorem between the B and A frames becomes

B_()

=

di =

A( di

)

+

X (

[2.6.221

)

when compared with Eq. 12.6.2 1 I. leads to the expected conclusion that

When there are a number of intermediate frames of reference between the A and one can express the angular velocity of frame B with 3 frames. say A2, A-i, ..., to the A frame as +

=

+

...

[2.6.23] 12.6.23]

+

The second derivative of a vector u in the A and B frames using this notation is +

(Bthu). Note that in both expressions. all the derivative terms are con-

in the same frame. Following this argument one can write the acceleration 4

d (A a'

— —

a1

B

ap

— B —

d

(B d \

[2.6.24] [2.6.241

In dynamics, one frequently encounters the need to take the derivative of an

in one frame that has been derived by differentiation in another frame. n 5uch cases we again invoke the transport theorem. For example. given a vector by taking the derivative of a vector u in the B frame, we use the transport rem and obtain 1

d —

"B

\

d

B

—u I = di \ di ,/

d "B d \ "B a' —u I — I

(

di \ dt j

B

+'

X

d \ —u 1

\\ \ dt /

I

12.6.251

can show that A

(B

dt\

dt )

di

12.6.261

more than one derivative is taken in thfferent refi?rence frames. changing the thjjerentiaiion gives different results. results. This applies not only to differendifferen;"der of the th:ffrrentiaiion ion with respect to time, but to differentiation with respect to other variables as Consider, for example, a vector u that is a Function of the variables We denote the time derivative of u in frame A by =

(lU

.

dt

[2.6.27] [2.6.271

121

CHAPTER 2 .

MOTION

When considering the second derivatives of u. one can show that 4(9 dq1 dq,

-

riqi

dq1

I

(9qj

\

j

14(9u\

ô

B

1q1 dq1

\

i,j=l,2,...,n = l,2,...,ii i,j

[2.6.281

,/

Also, be aware that a vector u may be a function of a variable qj in one reference

frame and not in another. Consider next the derivative of Eq. [2.6.23]. We differentiate each term individually to find the angular acceleration. The first term is measured from the inertial frame. thus its derivative is straightforward. To obtain the derivatives of the subsequent terms we use the above relation. For example. for two intermediate frames. we have 4

a" == 1?

=

=

B"

(1

dt'

)

/

A2

—

+

(In d di

+

dt 4ah12

1df4

=

A

8) +

(A

A (01%

2x

A

+ 4wA2 X

+

[2.6.291

The expression for the angu[ar acceleration for the general case of several frames

is lefi as an exercise. We end this section with an important note, it is crucial that one be able to distinguish between the reference reference Frame frame in which a derivative is taken and the coordinates of' of the the reference reference frame frame in in which which the the differentiated differentiated vector is resolved. Usually. one expresses a vector to he he difTerentiated differentiated in in aa particular particular reference frame in terms of the unit vectors of the frame. However, exceptions to this general procedure do exist.

I

Consider the robot arm in Example 2.7 and find find the angular acceleration disk, given acceleration of the the disk,

that w and w3

are

both constant.

Solution From Example 2.7. the angular velocity of the disk is written as (0 = =

w1 +

+

1.1

iii which iii which = 0.5K = —0.04827i

—

= 0.08778j rad/s

0.3830j + 0.3178k rad/s (03 03 = 71

rad/s

We use the transport theorem to get the angular acceleration, which gives

a= = W3rel + = W3rel

+

=0

(1)2rcI + + (1)2rcI

X

X (1)3 0)2)XX(1)3 (1)3 (1)3== (wi (WI+ +(02)

Ic]

so that the angular acceleration becomes

a=

+ a2 +

= W2rcj + (Oj X

+

±

X (03

[dl [dJ

____

2.6

We now evaluate the individual terms in the above equation. Given that at! = 3s = X

—

cos 2tj rad/s2 =

= (—0.04827i

+

cos

—

123

VECTOR, ANGULAR ACCELERATION

RATE OF CHANGE OF

cos 21

ôj radls2 = —0.6033j radls2

0.3830j + 0.3 178k) X (71 +

—

+ 2.677k rad/s2

= —0.027891 +

lel

7i = —0.6145k radls2 (02 >< (03 = 0.08778j X 7i Adding the individual terms. we obtain the total acceleration as = —0.027891 + 1.622j + 2.063k rad/s2

t

LU Lf]

the angular acceleration of the disk shown in Fig. 2.27, which is spinning at the constant of 60/ir rpm. The disk is attached to a collar, which is rotating at the rate of' 3/7r rpm. the rotation rate increasing by 0.6/ir rpm/mm. A rod connects the disk to the collar and :; pinned to the collar. It makes an angle of 300 with the vertical, which is increasing at the rate of 1 $/ir °/sec. Express the angular acceleration in terms of a reference frame to the collar.

attach an x'v'z' coordinate system to the collar, with the Z = z' direction denoting the vertical. The v coordinate attached to the arm is obtained by rotating the x'y'z' axes so that j = sin 30°j' — cos 30°k'. We the x' axis counterclockwise by an angle of the total angular velocity of the disk as +

= 0)collar + =

(0collar =

lar

)i' radls =

K rpm =

—1 Is =

IT 18// IT 18

IT

0.

1k' rad/s

1•, radls = u. ii rad/s rad/s 0.11'

IT

60.

60 = —(sin 30°j' = —j rpm iT

Z, I'

—

cos30°k')

2IT 21T

7

1'.,•

radls =

4..

I

rpm

'I

Figure 2.27

lal

(0tjisk/rod

j'

—

3k rad/s

[bJ

Example

2.10

124

2•

RELATIVE CHAPTER 2• RELATIVE CHAPTER

MortoN MorloN

To find the angular accelerations, we differentiate each of the angular velocity terms sepTo arately. Since the angular velocity of the collar is measured from a fixed frame. its derivative is obtained through straightforward differentiation as acollar = tCollar

=

rpmlmin)K

=

= 3.333(l04)k' 3.333(104)k' radls2

[ci

The angular velocity of the rod is measured from a Frame that is rotating with the angular velocity of the collar, collar. so we can express the angular acceleration as x 0. Ii' = 0.0 lj' rad/s2 [dl Wrod/collar = = 0 + 0. 1k' X a rod/collar = rel + t0collar X Wrod/collar rd The angular velocity of the disk is relative to the rod: thus we can write its angular acceleration

as adick/roil = cldick/roiI

rd

± ((0coIlar + Wrod/coijar) X

= 0O+(0.lk'+O.li')X(j'— — + (0.1k' + 0.11') X Ii' + = —0. —0. Ii'

/3

+ 0.1k' rad/s2 +

[.1

Adding Eqs. [c]—[ej.we weobtain obtainthe thetotal totalangular angular acceleration acceleration of of the the disk disk as Eqs. [c]—[ej. =

—0.

rad/s2 Ii' ++ 0.0. 11 832j' 832j' + 0. 0. 1003k' rad/s2

[fi

Note that, as discussed before, this selection of the coordinate axes makes it much easier to visualize the motion than would a reference frame attached to the disk.

2.7

RELATIVE VELOCITY AND ACCELERATION

Consider the Consider thetwo tworeference reference frames ft-ames shown shown in Fig. in 2.26. Fig. 2.26. The position The position of pointofPpoint is P is expressed as = r8 +

12.7.11

rp111

The vectors rp and rB are measured from the inertial frame, and r1111 is measured from the relative frame. To obtain the velocity, we differentiate Eq. [2.7. 1] for

[2.7.21

+

Vp =

We We find the expression for VP/B by means of the transport theorem as Vpj8

X r,,8 r,,8 +w X

= rP/R =

12.7.31

The difference in derivatives is because rp and r8 are measured from the inertial frame, while rp18 is measured from the rotating frame. Introducing Eq. 12.7.31 into Eq. [2.7.21, we obtain the relative velocity expression. written Vp/s = Vp = yR yR + Vp/s

VJ?

+

+ W X rp,a

[2.7.4] 12.7.4]

The first term on the right side of this equation, VB. is known as the base velocity: it denotes the absolute velocity of the origin of the moving frame. The second term. velocity, as it denotes the velocity of point P as viewed is known as the relatit'e i'eiocitv, by an observer attached to the relative frame. The third term. w X is called the

2.7

RELATIVE

AN!) ACCELERATION AN!)

velocity; it describes the change in the position vector rp18 as the relative rotates. Equation [2.7.41 can also be written in terms of a point Q. which is coincident point P but is not moving with respect to the rotating frame. We write

[2.7.51

VQ + Vp/Q

Vp

VQ is the absolute velocity of point Q, and Vp/Q is the relative velocity of P respect to Q, having the forms +

VQ =

X rp111

Vp/Q

12.7.6a,bJ

Vp/Bd

motion of of two analyzing the relative motion two points points both both fixed fixed on the same reference the relative velocity term VP/Q vanishes. P = Q represents represents the same point. we use Eq. [2.7.6a] to relate the velocities. To find the acceleration of point P. we differentiate Eq. [2.7.4] once more, with result + ap = aB aB +

d di

a'

+ —((A) —((A) X rp/B) di'

[2.7.71

of the left side of Eq. 12.7.71 and the first term on the right side is

a di dt

—Vp = ap = —Vp

, r,r ,rp

di-

a —V8 = a8 = —V8 , dt dtdi-

12.7.81

of the second and third terms requires that we invoke the transport for each of these terms, with the result d

=

(A) X + + (A)

a'

12.7.91

di

Eqs. [2.7.81—[2.7.91 into Eq. [2.7.7] and combining terms, we obtain

ap == a8a8++a aXXrp18 ap rp18++wwXX(w (w X X rp111) rp111) + apIBR. + '2w 2w X

V/)//Irt

12.7.10]

The term a X rp18 is due to the angular acceleration of the rotating frame, while

(w X rp18) is the centripetal acceleration ol point P. For the general case of ±ree-dimensional motion w X (w X rp111) lies on the plane generated by the angular wand Fp18. For the special case of plane motion, the centripetal acceleration the form

w X (w (w X X F/i/B) F/i/B) = —ü2rp,8

12.7.1 1]

The fourth term ap18 = is the acceleration of point P as measured by observer located on the moving frame. The fifth term. 2w X is known as as i

coriolis acceleraijon. It is due to two effects: a directional change in and (I) X a change in magnitude of' rp/B. Both Both terms terms contributing to the Coriolis of rp/B. effect arise because there is translational motion with respect to a relative frame.

1 25 25 1

126

CHAPTIR 2• RELATIVE Mo'I'IoN

The direction of the Coriolis acceleration is perpendicular to the plane generated by w and (Vp/B)rej, so that it always results in a change in direction from (Vp/B)reI. (Vp/B)rel.

as shown in Fig. 2.28. Even as Even in in cases when the magnitude of this acceleration

is

small, because it always causes a change in direction the Coriolis acceleration must be considered in the analysis of' several systems. Equation [2.7.101 can also he expressed in terms of a point Q coincident with point P but not moving with respect to the coordinate frame as

ap =aQ+ap/Q ap = aQ + ap/Q

12.7.121

in which which aQ = aB + a. a11 + aQ a Xx rp/fi rp/fi ++wX w X (w Xx rp/fi)

+ 2w 2w X X Vp/fl1

a,)/Q a,),Q

[2.7.1 3a,bJ The term aQ is the absolute acceleration of point Q. The term ap,1Q is the acceleration erat ion of of point point P due to its motion with respect to the reference frame. Unlike Vp/Q. it it contains two terms. The difference is the Coriolis acceleration. When there is no motion with respect to the moving frame, such as with the motion Of two points fixed on a rigid body. the relative motion equations reduce to Eq. [2.7. 1 3a1, and one replaces Q with P

ap — =

+ aa XX

+ w XX (w (w X X rp18)

[2.7.14]

Now Now let let us write the relative velocity velocity and and acceleration expressions using the notation introduced in the previous section. For the relative velocity expression from Eq. [2.7.4j we write AVp =

Av8+Bvp+AwBXBrp + 8Vp + AwB X

[2.7.15]

and, for the relative acceleration from from Eq. [2.7.101, [2.7.101, we write

xX rp +

4ap = '4a8 +

X rp) +

>(

11ap

+

12.7.16] 12.7.161

X 8Vp X

with

'tap == /A

(I

('1Vp) Vp)

-i-c'

=

d (4 /4 d

4

)

a8 a8 = =

Wxv

V

Plane of W.V

FIgure 2.28

A

(1%

B) = =

V8) V

(A d

A

.4

at

\

—r111

di ,i [2.ff.17J [2.7.17]

2.7

127

RELATIVE VELOCITY AND ACCELERATION

The most effective way of dealing with problems where more than one reference

frame is involved is to systematically break the problem into several parts and to the terms for each part individually. We next discuss two very important issues associated with the kinematics of relmotion: how to select the origin of the relative frame(s), and how to select the xientation of the relative frame(s). There is no clear-cut answer to these questions. One guideline is to select, if possible, the origin and orientation of the relative frame that it minimizes the number of expressions in the relative motion equations. ConFig. 2.26, if B is selected so that it coincides with P (B = Q), then rp1 BB = 0. if B is selected such that it coincides with the origin of the coordinate system 0, then = 0 and = 0. Another guideline is to select the relative frames so that the of relative frames is minimized and the angles that have to be calculated are un pie. The way to learn how to select reference frames is by gaining experience solving problems. When there is more than one reference system involved, the relative motion can become lengthy and complicated. One way to avoid confusion is to a tabulation approach when obtaining the components of the relative motion expression.

platform in Fig. 2.29 is rotating with a constant angular velocity of w = 0.2 rad/s. Plyon the platform is a tube oscillating according to the on the relationship relationship 0(1) = sin 2t rad. A particle particle of of mass mass m m slides slides without friction inside the tube. tube. The The particle particle is is attached attached to to the the of the tube by a spring of constant k and dashpot of constant c. Find the velocity of the at t = 3.6 s, at which point it is given that = 40 cm and " = —30 cm/s. Also find expression for its acceleration.

I

Example

2.11

an X'Y'Z' coordinate system moving with the platform and an xyz coordinate sysattached to the tube, as shown in Fig. 2.30. The Z' = Z axis is the vertical. When 0 = 0, v and Y' Y' axes axes coincide. coincide. The coordinate axes are related to each other by Y

x=

v

= Y'cosI9 + Z'sinO

2z =

[a]

+ Z'cosO

We have

= wK + 0!' =

Oi

+ +&sin6j sin 6j +wcosOk + w cos Ok

=

[bj

VJ

zzg V

V

YI

Figure 2.29

Figure 2.30

2•

1 28

CHAPTER 2 • RELATIVE RELATIVE MOTION MOTION

coordinate frames coincide. the relative velocity cxprcssion beBecause the origins of the coordinate Because comes

r = vj + (fli +

V=

(OS1flOJ --

wcosOk) X vj

[ci [cJ

=

Au ==

= 0.4156rad.sinO 0.4l56rad.sinO = 0.4037,cos6 0.4037,cosA = 0.9149,0(3.6) = 0.6371 0.637 1 rad/s. Substituting in these values, we obtain for the velocity at t = 3.6 s —0.073191 V = —0.073191 V

0.3j 0. 3j

—

+ 0.2548k ni/s

[di [dl

The relative acceleration expression for this problem is

[.1 in which arci = arci = vj vj

If]

=

obtained the Note that we obtained the angular acceleration expression by direct differentiation of the velocity expression in Eq. [hj, angular velocity [h], rather than using the transport theorem. H was possible to do this because in Eq. Ihi the components of w were expressed in terms ofthe coordinates

of' the relative frame. We perform the cross products. writing

xr=

(01 + (01

wOcOj w0cOj — — w0sOk)X wOsOk) X

= wOvs6i w0vs6i + Ovk

oX oX (o. (o.XXF) r) = (01 + weOk) (—wvcOi ++Ovk) (01 +wsOj + wcOk) X (—wvcOi Ovk) = w0vsOi w6vsOi +

=

ws

—

w2vc2O)j +

w c Ok) X

= —2"w c 01 +

[gJ

Introducing these expressions into the relative acceleration, we obtain j = (2wOvsO (2wOvsO — —

Example 2.1 2

I

— 02v + (i —

—

w2vc2O)j + (2i'O ++ w2vsOc& w2vsOc& + Ov)k

[hi

An airplane, shown in Fig. 2.31, is moving with a speed of 420 mph. A flight attendant who weighs 120 lb is standing 15 ft from the center of mass. To avoid a turbulent region, the pilot

initiates an emergency maneuver. The aircraft begins to pitch upward at the constant rate of 0.1 rad/sec. and it begins to pursue a curved trajectory toward the left of the pilot with a radius of curvature of 30,000 ft. The speed of the center of mass of the airplane does not change with these maneuvers. Find the forces exerted on the flight attendant's feet tèet if the attendant wishes to move forward with a speed of 2 ftlsec.

Solution We consider two reference frames. The first frame is associated with the curved trajectory and has an angular velocity velocity of of 420(88/60)1(30. 420(88/60)/(30. (X)0) = 0.02053 radls. The second frame is attached to the airplane. and its angular velocity is the pitch rate of 0.1 radls.3 At the instant considered, the two reference frames coincide. We can write the angular velocity and angular order to simplify this problem, we do not consider any roll of the aircraft. In general, when an airplane makes a turn the pilot rolls the aircraft so that the resultant acceleration vector due to the turn and due to gravity lies as much as possible along the local vertical direction, through the spinal cords of the passengers. This way, passengers experience less discomfort. 31n

2.7

RELATIVE VELOCITY RELATIVE Vn.ocirv AND ACCELERATION

as

=

0.

lj

—

a=

0.02053k rad/sec

—0.02053k

X 0. lj =

0.0020531

rad/sec2 [a]

We write the relative acceleration relation as

a, =

a8

+aX

+ w X Co X

+ 2(0 X

+

Lb]

each term. The acceleration of the center of mass is due to the change in curvature -an be written as (60 mph = 88 ft/sec) I

=

a8 = a,, =

[ci Ic]

= —12.65j ft/sec2

terminology. 12.65/32. 17 = 0.3932g. The angular acceleration term is very terminology. 12.65/32.17 and can be ignored. The centripetal acceleration term becomes = (0. lj

•wX

—

0.02053k) x (0. lj — 0.02053k) X 15i = —0.15631 —0. 15631 ft/sec2

Ed]

term, (a,,fi)mI, is zero, because we are assuming a constant speed tbr the flight inside the aircraft. The last term is the Coriolis term, which has the form = 2(—0.02053k + 0. lj) lj) xx 21 2i = —0.08212j

2w X

—

0.4k ft/sec2

[.1 Eel

As As can be seen from Eqs. through Eel, the dominant term in the acceleration is acceleration due to the change in curvature. The Curious term has the smallest nude, and and itit is is also also in in the the direction direction of of the the gravitational gravitational attraction. attraction. The The total total acceleraaccelera-

If] [fJ

2.73j — 0.4k ftisec2 —0.15631 — 12.73j—0.4kftisec2 a1) ap = —0.15631— 1

Let us now draw a free-body diagram of the flight attendant (Fig. 2.32), treating the Let a point mass. The forces that the airplane exerts on the flight attendant are transby the normal force A' and the friction forces F, and Using Newton's second law,

map =

m(—0. 15631

—

//,/ p

I

/

2.73j

—

0.4k) = F,1

+ FJ

+ (mg

—

[g]

N)k lb

/

/ mg V

F'

N

V

15' V

Figure 2.31

Figure 2.32

x

129

130

CHAPTIR 2• RELATIVE MOTION

Solving for the unknowns

N= F1.

ing + 0.4m = 120 + 0.4

= —0.1563tii =

170

= 121.5 lb

3217 = —12.73,ii =

[h]

—47.49 lb

Note that the coefficient of friction between the attendant's shoes and the airplane floor must be large enough to permit the resultant of and to be less than the normal force times the coefficient of friction. Even then, because of the large forces acting on the attendant, it is very difficult for the attendant to keep standing or to walk. The Coriolis acceleration makes the flight attendant feel heavier, and for someone further away from the center of mass of the airplane the centrifugal force becomes much largcr. Even for passengers who are sitting down. any change in the curvature of the path of the airplane causes aa substantial down. substantial amount of discomfort. It is for all these reasons that pilots navigate aircraft such that the path followed by the center of mass of the airplane is as close to a straight path as possible and any angular velocity is very small.

Example

2.13

Consider the robot in Fig. 2.33 mounted on a rotating shaft. The robot arm is attached to the shaft with a pin joint (in robotics terminology, a ret'olute joini). With a motion similar to that of an automobile antenna, a second arm can extend from the outer end of the first (in robotics terminology, a prismatic joint). Given that the shaft angle 0(i) and first arm angle 4(i) vary with the relationships O(t = 0.2t rad. 4(t) = 'ir/4 (1 + sin in) rad and that the second arm rad, th(t) is extending with the relation r = 3t cm, find the angular velocity and angular acceleration of the robot arm as well as the velocity and acceleration of the tip.

Solution solve this problem using two approaches. In the first approach we use two relative frames, OflC attached to the shaft and rotating with 0, the other attached to the robot arm and rotating with with respect to the shaft. In the second approach we use a single relative frame attached to the robot arm. We

First Approach

The two relative coordinate frames are denoted by H and B, as depicted in Fig. 2.34. The inertial frame is denoted by A. The H frame is attached to the rotating shaft.

06 /Z2 •1•

B

P tending arm

Ii

1)3

(b

(1)

J) P

FIgure 2.33

FIgure 2.34

b2

2.7

ACCELERATION

the B frame is taken such that the the axis is along the extending arm and attached to the the first arm. The angular velocities velocities of the reference frames can can be be written written as

=

=

HwB = thh1 =

rad/s

cos

in h1 rad/s

lal

Considering first the B frame, frame. we can express the the position position and and relative relative velocity' velocity of P as rp18 = O.03tb2 m rp18

Ibi

= 0.03b2 ni/s

the unit vectors of the B and H frames are related by b1

b- == sin

= h1

—

b3 b3

cos

=

COS cos

[ci

+ sin

relative velocity expression for the H and B frames is Hv

=

V8

+ + "0) 0)

= O.03b2 O.03b2 mis

= 0.6 d)b3 d)b3 radls radls H0)1I

X rf)/,9 =

[dl

X rp/B +

I.]

X 0.O3th, 0.O3th, = 0.03k/b3 rn/s rn/s

thb1

'1Vp = (0.6 (0.6 +

[f] En

+ 0.03b2 rn/s

Now transfer the velocity of point P to the inertial frame. The relative velocity equation een the A and H frames is '4Vp

A0)!! 4VH + = '4VH

Igi [gi

+

X

.n

+ sin

Noting that that a3 a3 = — cos

=

—o sin di(O.6

Second Approach

[hi

X (0.6 + 0.03t)b2

X rp111 =

we obtain for the velocity of point P

+ O.03t)b1 + 0.03b2 + (0.6 +

rn/s

[ii

The angular velocity of the single frame is

=

+

= O(i)h3 +

=

—

+

[Ii [II

the relative velocity expression is 1Vp

+

=

X rp18 + 8Vp

[kJ

nnd 0) X X

X rB/fl =

r118 =

—

—

(9cos4bi + (-1 sin + 8Vp

x X 0.6h2 = —0.66 sin thb, +0.64)b sin thb,

X 0.03tb2 = —0.O3rOsin4b1 + 0.03ufb3 = (103b2 0.03b2 rn/s [I]

when added up. yields Eq. [i]. Note that if we attach the relative frame to point H, = 0. Or, if the relative frame is attached to the tip of the protruding arm, then = 0.

131

132

CHAPTIR 2•

RELATIVE MoTioN

A ccc leration To find the acce'eration of point P. let us use the single coordinate frame approach. The angular acceleration between the A and B frames has the form AaR =

+

[Ml

><

in which cKh

=

x

= Oh3 x

=

th1 =

Liii

= 0.05zr2cosirth2

So that we find AaB =

—

Tr'

[oJ

nih2 cos lTtll2 nih, + 0.05ir2 cos

sin

7T1h1

—

The expression for the acceleration then becomes = 4afi + The term

X "w8 X

+

X

[p]

XB Vp + 11ap

+

is constant in the B frame. We find the absolute acceleration of

= 0. as

point B using

Aa == Aa

x

so that the first three terms on the right in Eq. Aa

+

x

x

+

[q]

x 4w8 x r8111

+

can

be expressed as Er] [r]

xX

xX

x rP/H + X

>( rP/R =

Evaluating the individual terms, we have

X rI'/H rp/H =

sin

nib1 +

x (0.6 ++ O.03t)b2 0.03t)b2

4b2 ++COS C05 —

COS = (0.6 + 0.031) 0.03t) [—0.05 1T COS

x

)< r,P,J/

=

—

Sin

+ 0 sin thb3) X

cos

lTtb3] rn/s2

—

0 cos

4b2 + 0 sin

X (0.6 + 0.03t)b2

= (0.6 + 0.031)[ —0d X 24w'1 X

= 2(4b1

— U

cos

+

+ sin2 4.)b2 4)b2 + 02 sin2

—

cos

0 sin sin 4b3) 4b3) xx 0.03b2 =

—

—

02 cos

sin

0.060 sin çbb1mls2

[sJ

so so

that the acceleration of the tip of the robot becomes

= —[(0.6 + 0.03t)(0.05ir2 cos —

[(0.6 +

+ [_(o.o +

+

+

cos

+ 0.060 sin

sin2

+

b3m/s2

['I

2.8

2.8

OBsERvATIoNs FROM A MoviNc

OBSERVATIONS FROM A MOVING

many cases. either by choice or by necessity. one observes motion from a movreference frame and manipulates the equations of motion within that reference ::ame. Typical examples are motion with respect to the earth and measurement taken a moving platform. When one cannot measure absolute velocities and accelone has to take relative measurements and to use the relative velocity and equations to find the actual accelerations. Similarly, when analyzing or the equations of motion, one has to use the variables associated with the relative motion.

We first apply the relative motion equations to Newton's second law, which is for inertial frames. Given the inertial and relative coordinates from the precedsection, we write the equations of motion for P as + F = map = + a X rp18 + w X w X r118 + 2w X

[2.8.11 bhere F is the sum of all external forces acting on the particle. We can rewrite the equation as

map,81 = F + F*

[2.8.21

F* = — m(a8 + a X rp/B + (.*) X (0 X rp/B +2w X

which is the resultant all forces that need to be considered due to the motion of the moving reference ±ame. In general. it is preferable to write the equations of motion by by placing placing on on the the side every term that includes variables associated with the relative frame. Doing we obtain for Newton's second law

+aX

+ 0) X W X

+ 2w

X

= F

—

[2.8.3]

A word of caution is in order. Often, we analyze motion with respect to a moving

by assuming that the motion characteristics of the moving frame are known not affect the wit/i respect to the relative frame does not that the motion of the characteristics of ofthat that frame. frame. For For example. example. for for aa car car traveling traveling on on the the earths earth, can safely assume that the motion of the car does not affect the rotation of the While this assumption is valid where the mass of the body to which the relative is attached is much larger than the mass of the body whose motion is analyzed, assumption begins to lose its validity as the bodies involved become comparable mass. Be cautious when assuming that the motion observed from a relative frame not change the characteristics of that frame. One of the most common applications of analyzing motion from a moving frame motion with respect to the rotating earth. As stated earlier, motion over short disor with small velocities and involving short time periods can he analyzed relely accurately without considering the motion of the earth. Otherwise, the earth's needs to be included in calculations. Consider a particle near the surface of the earth, as shown in Fig. 2.35, and attach moving frame B to the surface of the earth using an xvz coordinate system. The ±rection is the vertical, the x direction is toward the north, and the y direction is

133

134

CHAPTIR 2• RELATIVE MOTION

North pole

x North) .t (North) .t

UcosA P

z

(Vertical)

c1sinXx

0

Figure 2.35

Figure 2.36

toward the west. We assume that the earth is rotating about its own axis with constant angular velocity Fig. 2.36 shows the coordinate system from the side view. To calculate the spin rate of the earth, we note that it takes the earth about 365.25 days to orbit the sun, and that the earth rotates about its own axis at the rate of one revolution per day.4 Both rotations are counterclockwise, which leads to

=

+ 365.25)

= 7.2921

X

rad/s

[2.8.41

that, considering Fig. 2.36. one can describe the angular velocity of the earth in vector form as so

fl

=

Ak + cos Ai)

[2.8.5]

where A is the latitude. We ignore the angular acceleration of the earth and set

a=

0.

This assumption and the assumption that the rotation rate of the earth are constant are not exactly true. The earth's rotation about itself is not along a fixed axis. The axis about which the earth rotates exhibits a small wobbling motion with a period of 433 days. primarily because the earth is not totally rigid and not totally spherical. The rate of the earth's rotation is not constant; it is slowing down at an extremely low rate. In addition, we ignore the inclination between the equatorial plane (the plane generated by the equator) and the ecliptic plane (the plane generated by the orbit of the earth around the sun). We also ignore any subsequent relative motion of the sun with respect, to the fixed stars. 4The measured orbital period of the earth earth is is 365.256 365.256 360 360 55 days. days. The The difference difference in in the the third thirddecimal dedmal place place from from the commonly used value of 365.25 days leads to a difference of one day about every 150 years. The calendar was adopted to compensate For this difference. In this system, in a 400-year period, three years normally should be leap years are not considered as leap years. These years are selected at the beginning c,

centuries. For example, the year 2000 C.E. is a leap year, while the years 2100, 2200, 2300 will not be considered as leap years. The year 2400 will be a leap year.

2.8

OBSERVATIONS FROM A MOVING FR1UiW

Using these assumptions, we calculate acceleration of a point on the surface of

the earth (origin of' of the relative frame) as

aB = where

[2.8.6]

X (111 X

nil = rek, rek. denoting the vector from the center of the earth to the surface of'

the earth. Using Newton's second law, the absolute acceleration of' point P is written as

= F+

12.8.71

here F denotes the sum of all other external forces acting on the particle. Introducing Eqs. 12.8.61 and [2.8.71 into Eq. [2.8.31, we obtain

the equations of motion in terms of the relative coordinates as +

+

X

X rp/B) = F +

X

—

mfl X (li X 12.8.81

X As discussed in Chapter 1. we include the centrifugal force X r11) in the gravitational force and define the term Fg as the augmented gravitational and approximate it as

=

—

,nfl

X

X r8)

[2.8.91

—ingk

This simplification is possible because the radius of the earth is almost a constant, x making the term cos2 Ak + sin Acos Al) nearly constant X rB) = .n magnitude. The component along the vertical (z direction) is used to augment the gravitational force. The component in the x direction (North) is usually ignored. The maximum value of the centripetal acceleration is = 3.39 cm/s2. which is at the equator. The effect of the centripetal acceleration is to make the earth more squat. Howbecause the earth is not made up of fluid only, the flattening occurs mostly around the equator. Around the equator, the combined effects of flattening and cenacceleration make the value of g about 0.53% less than its value at the poles. One can then write the equations of motion of a particle in the vicinity of the earth as

+

X

VP/BrCI

+

X

X

= F

—

rngk

12.8.10]

Writing the equations of motion in terms of' relative quantities is mathematically equivalent to applying a different force than the external forces. The equations of can be written as

Feti Feti +

[2.8.1 11 [2.8.111

here the effective force Feft denotes the force felt on the surface of the earth. It has form FefT = F

—

X X rp18) rp18)— 2mw X mw X (w X

12.8.121

135

1 36

CIIAPTIR

2

•

RELATIVE MoTIoN

After all the simplifications have been made with regard to the nature of the angular velocity of the earth and with regard to gravity. the difference between the actual and effective effective forces forces consists consists of ofaa centi-ijiigalftrce, centi-ijiigalftrce, — —mw mw X (w X rp18). and the coriolisfbrce, X The centrifugal force is very small: it is a function of the square of' the earth's rotational rate, and rp18 is usually very small compared to the radius of the earth. The Coriolis force has a distinctively more pronounced effect. The magnitude of the Coriolis force is dependent on both the velocity of the particle and on the latitude. Considering Eq. 12.8.5], for for a particle moving in the east-west direction, maximum values of the Coriolis force are observed at the North and South Poles. The maximum value of the Coriolis fhrce per unit mass is 1.5 X 104v, where v is the speed with respect to the earth. While the magnitude of the Coriolis is very small, its direction is a/wa vs perpendicular to the velocity; thus, this force causes a change in direction. II' there are no significant fbrces acting perpendicular to the velocity, the effect of the Coriolis force builds over time. The Coriolis force for the Northern Hemisphere is depicted in Fig. 2.37. It causes a particle to veer to the right. Note that this analysis ignores the vertical component of' of the Coriolis force, as this component is much less than the effect of gravity. The Coriolis effect is used to account for several kinds of natural and physical phenomena. Pertinent to weather analysis, analysis. the motion of air masses in the atmosphere is affected by the Coriolis force. For example, in the Northern Hemisphere. the spin of the air masses in cyclones and hurricanes is counterclockwise counterclockwise as shown in Fig. 2.38. A hurricane occurs when a low low pressure pressure center center attracts attractsair airpanic panicles inward les inward with large speeds. In the Southern Hemisphere, the spin spin of a cyclone is clockwise. Rossby waves, which include the Coriolis effect, are widely used to predict wave motion in the oceans as well as for weather analysis. The motion of projectiles such as missiles is affected substantially by Coriolis forces, And the Coriolis effect iniluences the whirl of water as it goes down the sink.

z

(Height)

plane

Low pressure (West)

—2(0

Northern Hemisphere X (North)

Figure 2.37

Coriolis force

Figure 2.38

Hurricane formatiore

2.8

OBSERVAI1ONS

FROM A MoviNc

1

next consider the motion of a particle of mass in acted upon by an external + xi + yj + zk and Writing the position vector as rp111 F = F,i + rn-oducing these terms into Eq. [2.8.101. and separating into components in the x, and z directions, we we obtain We

x

., •, sin A — ft(x sin- A — z sin Acos A cos A) v sin •

.,

.

.

A —

—

+

+

sin A

—

A) — 112v

= —

[2.8.1 [2.8.1 3a1 3a1

=

[2.8.1 3bJ

in

A + xxsinAcosA) sin A cos A) = —g +

A+ -

in

[2.8.13cJ [2.8.1 3cJ

We can perform a qualitative analysis to investigate the magnitudes of the cornof the motion with respect to the rotating earth. Consider, for example. the of free motion, = = F. = 0 and a particle thrown upward. with miconditions in the z direction only. Because is a small quantity. terms of order namely the Coriolis terms. terms, will dominate terms of order We thus ignore the terms. In 3c 1. the terms. InEq. Eq.[2.8.1 [2.8.13cJ, the gravitational gravitational acceleration acceleration g will will dominate dominate response. We conclude that the component of the motion motion in in the the zz direction is of

n*r 1. From Eq. 12.8.1 3hJ, the response in the v direction will be influenced by the in the x and z directions. That is,

0(v) =

+

=

+

[2.8.14J

Investigation of' Eq. {2.8.13aJ shows that the component of the motion motion in in the

: ±rection is influenced by the motion in the v direction and can he considered as of order 0(x) = This implies that motion in the x direction will be smaller than the motion in the v direction. Hence. Hence, from Eq. [2.8.141, motion nfl the v direction will will be be dominated dominated by by the the motion motion in in the the zz direction and it will be of' 11. It follows that the motion in the .v direction will he of order Because v axis axis denotes denotes the the west west and and the the x direction denotes the north. for a freely moving the Coriolis effect will he much more significant in the east-west direction, much less along the north-south direction. The results of the above dimensional analysis will be different when there are forces in the x. y, v, or or zz directions directions and and for for initial initial conditions.

Example

the equation of motion of the mass sliding in the tube in Example 2.11.

2.14 be

make use of Eq. [2.8.3]. We introduce the acceleration expression from Eq. [hI of 2.11 into 2.11 [2.8.3]. which yields —

2o.OvcO)i + ,nN — 02v 2vOvcO)i

—

w2vc2O)j +

m(2v0 +

± Ov)kJ = F

[a] find the forces acting on the mass, mass, we draw a free-body diagram (Fig. 2.39). The acting on the particle can be classified into three groups:

To

37

138

CHAPTER

2• RELATIVE MoTIoN z

V

C' C

F..

Figure 2.39 nigcosOk

1.

Gravity: —mgK =

2.

Normal reaction threes that the tube exerts on the particle: ELi + Spring and dashpot: —(kv —(kv ++

3.

—

The total force thus becomes F = F,i + (—nigsinO

—

kv

—

cv)j ++

—

Ibi

ingcosO)k

The reaction forces arise fronì the fact that the i)article is constrained to move inside the tube. Introducing Eq. [hj into Eq. [a] and separating into the components along the x. y, and z directions, we find the force balances to be In x direction:

m(2wOv sin

In y direction:

—

—

—

2wOvcos 0)

=

w2 cos2O) = —mgsinO

rn(2"O ++ w2vsinOcosA m(2"0 w2vsinOcosA + Ov) =

In z direction:

id

F..

—

—

kv

mgcos6

—

Ed] Id]

[0J

Of the three force balances. Eq. [d] represents the equation of' motion and Eqs. [ci and [ci give expressions for the reaction threes [c] and that is, the constraint forces. Note that we have a single degree of freedom, because the motion of the platform and tube are defined as known quantities. The only variable is the motion of the particle inside the tube, described by y. In essence, Eqs. [cl—Fe] represent an equation of motion and two constraint equations. As stated earlier, it is customary to write the equation of motion by placing all of the dependent variables on the left side of the equation. Doing so, we rewrite Eq. [dl as +

Example

2.15

+ (k

—

—

mw2 cos2 A)v

= —mg sin 0

MOTION ANALYSIS USING PERTURBATION THEORY

Equations [2.8.131

are a set of nonlinear equations. which cannot he integrated analytically. 11' quadratic terms in the angular velocity are neglected, these equations become linear. Their qualitative analysis requires solution of an eigenvalue problem. and their integration requires simultaneous integration of three equations. The integrals of the motion for this set of equations do not give much additional insight. One can conduct a numerical integration of the equations of motion. hut doing so gives answers for the particular set of forcing and initial conditions. It turns out that there is yet another analytical approach to analyzing the equations of motion. Because fi is a very small quantity. Eqs. [2.8.131 can be analyzed by means of a perturbation approach by treating as the perturbation parameter.

2.8

OBSERVATIONS PROM PROM A A MOVING FRAME FRAME

Here, we carry out what is known as a

siraighzfirward expansion expansion and and demonstrate demonstrate peranalysis.5 We assume that the solutions in x. v. and z can be expressed in a perturseries in terms of the perturbation parameter as

x(t) =

x0

2

+ Ii x1 +

...

x2 +

v( t)

+

zu +

z(t)

=

vo

+

+ fl Yi +

+

lal

...

the subscripts indicate the order of the solution (zeroth. first, second, .. .). Given the conditions x(O), k(O). k(O). v(O), v(O), z(O). ±(O). we consider them as the initial conditions with the zeroth-order problem. All initial conditions associated with the higher:rder solutions are zero. Substituting Eqs. [a] into Eqs. [2.8.13] and neglecting terms of order than 1).-. we obtain

...

+ —

90

112(x0 sin2 A 1l2(x0

+

+

—

+

-i-

+...)

zosinAcosA

...

+ ...) = +

+

II?

sinA rn

+

-

•,. ...

+ +

+

+ + ...)

= —g+

F—b

"1

Ibi

Collecting terms of like orders in IL we obtain a set of linear differential equations in form

)rdcr ci° X() =

=

!fl

F —h

x0(O) x(O) x0(O) == x(O)

x0(O) = x(O)

v(O) yo(O) = v(O) yo(O)

=

F11

=

—g

+

F-

zo(O) = z(O)

Z()(O) = ±(O) Zo(O)

id

Il

Ij

=

=

—

=

X2

= 2 sin

A

sin A sin A

Z2 = Z2

[dl

A"0

—2

—

sin

±i cos A) +

A

Yo

± zocos2A — x0sinAcosA

lel

-The field of perturbation analysis is very broad and several perturbation techniques exist. An in-depth treatment t the subject sublect is beyond the scope of this text. The straightforward expansion we use here is the simplest form of expansions.

139

140

CHAPTIR 2• RELATIVE Mo'I'lON depend on the external force, and Zo depends on both the external and We observe that first- and second-order equations indicate which terms dominate the force and gravity. The firstequations are motion. The order ci equations are due to the Coriolis effect, and the order due to both Coriolis and centrifugal effects.

Let us select the case of projectile motion. and consider a projectile launched in the Hemisphere with speed v toward the west and at an angle of 0 with the vertical. We have the following initial conditions:

x(O) =

i(0) =

v(O) = z(O) = 0

±(0) = vcos0

= vsinO

0,

[(1

We assume that the external force consists of gravity only, neglecting the effects of wind resistance. The zeroth-order problem yields

x0(0) =

=

0

=

= u sin(O)t

U sin 6

[gJ

vcos(O)t — — gt212 gt212 = vcos(O)t

= vcosO

zO(0) = 0

—g

0

= 0

0

Recalling Recalling that the higher-order solutions all have zero initial conditions, substitution of the zeroth-order solution into the first-order equations yields = 2sin = —2(.i0sinA —2(i0sinA

—

A

sin

A

2cosA(vcosO z0cosA) = 2cosMvcosO

—

cosMvcos(0)12 v1(t) = cosA(vcos(0)t2

go

z,(t)

= —2vcosAsin8

=

sin(O)t2

—

—

—vcosAsin(0)t2

[hJ [h]

[hi, we obtain the first-order approximation to the solution as Combining Eqs. [g] and [h],

x(t) =

/

vsin(O)t + v(t) = vsin(O)l

= vcos(6)t vcos(6)t

cos A vcos(O)t2

—

—

L(13 —

civcosAsin(6)r

III

denoting the time at which the projectile with If denoting t ti-. ti-.with This above solution is valid forO setting z(t) z(t) == 00 in falls to the ground. The first-order approximation approximation to to t1 t1 can be found by setting the above equation. which gives 1

00 =

= ucos(0)tj

g12. gil. —

.

g

vcosü

—

+ civcosAsinO)ti' = o

—

If'

which yields the final time as —

t

.1

VCOSO

[kJ

— g/2 + civcosAsinO

Note that in the absence of the Coriolis effect, the. final time becomes the well-known result of = 2v cos 01g. The Coriolis effect, for this particular set of initial conditions. reduces the time the projectile stays in the air. The Coriolis deflection in the x direction can he found by introducing the expression for the fInal final time to the response. with the result

/

=

VL

= UvsinAsin0I \g/2 + ciucosAsin0 .

iii

2.8

141

OBSERVATIONS FROM A MOVING FRAME

expected. the magnitude of x(t1) is dictated by the latitude angle A. In the Northern Hemiwhere A is greater than zero, x(t1) is positive, indicating a Coriolis deflection to the is less By contrast. in the Southern Hemisphere. where A is negative, x(1j) is less than zero.

a deflection to the left. These results agree with the discussion on the Coriolis xflection in the previous section.

PENDULUM Another interesting application of motion with respect r the rotating earth is the FoucauR Foucault pendulum (Fig. 2.40). This pendulum consists of a large of a sphere, suspended from a very high ceiling by a mass, usually in the wire.5 As the pendulum swings, its swing plane rotates slowly in the clockwise direction. Thi This rotating motion may seem hard to explain, because the motion of a pendulum is expected r along a fixed plane or a fixed elliptical path. (The initial conditions dictate whether the motion is along a plane or an ellipse.) We see the swing plane of the pendulum rotate because of the rotation of the earth and —- resulting Coriolis acceleration. To an observer in an inertial frame, the pendulum executes motion along a fixed plane or a fixed elliptical path. to and are due to From the free-body diagram in Fig. 2.41, the external forces Tx., tension ininthe r.ension thewire. wire.Denoting Denotingthe thetension tensionininthe thewire wirebybyT,T,and anditsitscomponents components byby T.1., Tx., we have =

F:= T.= T:= LZZT F:.

Fy=

=

[aJ

x. v. y. and z denote the coordinates of the pendulum. We next make use of the great I. We hence treat the amplitude of the of the pendulum and assume that (L — z)/L 0. In essence, we are 0. in the z direction as negligible, and write zIL 0. ± that the pendulum is moving on the xv plane only. In addition, we assume that

f 1.

7,7.-

1_

V

.1•

V

Figure 2.40

Figure 2.41

States,there thereisisaaFoucault Foucaultpendulum pendulumininthe theUnited UnitedNations Nationsbuilding buildingininNew NewYork YorkCity, City, at at the the American American reUnited States, rr re of Natural History in Washington, D.C., at the Franklin Institute in Philadelphia, and at the Museum of Industry in Chicago.

Example

2.16

142

RELATIVE MortoN CHAPTIR 2 . RELATIVE

the sphere. sphere. of the the wire is very thin, its mass is negligible compared to the mass of and that air resistance is negligible. Using these assumptions. introducing Eqs. [al into Eqs. because

[2.8.131. 131. and and neglecting terms quadratic in

I — 2fl

=

—

'V

niL

we

7'

=

1-

V

—

niL

T

= T

2thcosA ++ gg =

[hi 'C' [ci

,fl

[hi and ignoring terms cos A). introducing it into Eq. [hI RewritingEq. Eq.[C] [C]asas TT = in(g + Rewriting in x and y, we obtain the linearized equations of motion as quadratic in

v=

V + 2 si n V+211.sinAi+

= 00

+

— 2f1 sin 1—

+

0

[di 'dl

Equations Idi represent a gyroscopic system where the rate of rotation is governed sin A. We note that sin A is the component of the angular velocity of the earth in the vertical direction. To analyze the nature of the response, we assume a solution in the form

x(t) = Xe"

[.1

v(t) =

of where X and and Y Y are are amplitudes, introducing Eqs. [e] into Eqs. [dl and collecting powers of

c", we obtain g

1

sin Al

11

sin,'

—w

2

g

rx1

[11

= [oj

14

In order for Eq. [fj to hold, the determinant of the coefficient matrix must vanish. Setting the determinant equal to zero leads to the characteristic equation (

=

—

[gi

U

Solving the characteristic equation for w. we obtain four values as +

U) =

[hJ

v(t) isis harmonic harmonic with with frequency frequency w. w. The The quantity quantity inside the radical in The response x(t) and and v(t) Eq. [h] is always positive, it follows that all values of w are real and very close to each other. After carrying out the algebra! algebra, the response can be expressed in the general form C1cos(w11 cos(w11 + x(t) x(t) = C1

—C1sin(w11 Sjfl(W1! + vU) vU) = —C1

+ C2 cos(w't +

[I] w

+ C2 C- SIfl(W21 sin(w2t + '/,,)

(1 = 1,1, 2) depend on the initial conditions, and where C1 where C1 and 4.', (i

+

(01

+

U),

[LI

—

are the frequencies. Because the angular velocity of the earth is much smaller compared gIL. gIL dominates the roots of the characteristic equation. Indeed, approximating w1 and series expansion. we get using a

(01=

/

+

sin2 A +

+ 1)'- sin A —

A

sin

A

'lid

-

[kJ

_

2.8

t

FROM

143 1 43

A MovIN(;

the rotation of the earth and set fl = 0. then then o = (Li), and and it becomes clear Eqs. [ii that the motion of the pendulum is an ellipse or a straight line. We next examine the effects of the rotating earth. We We concluded concluded earlier earlier that that the two and w2 are very close to each other. The type of motion of a system when two of are very close to each other is the classical case of the beat plienoinenoii. We the expressions

=

1

LiI2sin2A\ LiI2sinA\

II ++

+ W2) W2)

—11 i —

1

= —(Wi (')2) = —(Wi — — (1)2)

I

j

2g

2

sin SIflAA

II]

frequencv and w,, the average frequency w, the the beat beat frequency. frequency. The average frequency is the of the pendulum as observed from an inertial reference frame (of order 1) plus a change (of order due to the rotation of the earth. The heat frequency is of Il. the same as the coefficient of the middle terms in Eqs. [dJ and the component of the velocity velocity of the earth in the z direction. One can express w and w2 in terms of the and beat frequencies as is

ii

Wi = Wi

=

+ Wh Wh

Imi

—

Without loss of generality, we consider the case where the local motion of' the pendulum swing plane. For this. we specify the initial conditions on the velocity as

3

= UU

= 0

at t =

tt == 00

at

En] [nJ

only one of x or v has to he nonzero. We select the swing plane as the so that the remaining initial conditions are 0

x = C,

[ol

v = 00 V

these initial conditions into Eqs. [1] [I] and solving, we obtain

=

.

Cw2

—

=

Wb)

W7) (Wi (Wi + W7)

-

=

Cw1

+ w17) + w17)

(w1 (w1 + to2)

Eqs. [11and andmu, Irni,the theconstants constants[p], [p],as aswell wellas asthe thetrigonometric trigonometric identities identities cos(a cos(a ++ h) = Eqs. [11 h — sin a sin b. b. sin(a sin(a ++ h) h) = sin a cos b + COS COS aasin sinb,b,we wecan canexpress express Eqs. Eqs. [I] [i] as —

x(t) =

wi,)

cosw1t +

C(COSWatCOSWbt C(COSWatCOSWbI +

v(t) =

— —

(1

.

sinw1t +

2w1,,

—

To

— toj,) C(w —wj,) C(w,

to"

C(cos to111 to111 sin sin w17t —

Wi7

to U)

+ W/)) 2w(,

Cosw2t Cos w2! =

to1,!) sm sin tot tot sin sinto1,!)

C(w a + C(w1 +Wh) Wh)

Sin 0)21 = Sin

sin Wat WatCOS cos wj,t) toj,!)

[qJ Iqi

get a feel for the motion, motion. we note that is an order of magnitude smaller than Wa. toa' and the terms with with the the coefficient coefficientto h/Wafrom fromthe theright right side side of of the the above above equations. This leads to an interesting explanation of the motion to a first-order approximation:

motion of the FoucauR Foucault pendulum can be explained as an amplitude modulated swing both xx and and yy where where the the amplitude of the swing varies with the rate Wh sin A. for both tob =

____

144

CHAPTIR 2 • RELA'IIVE

(Li1,

V

x

Swing plane of pendulum

FIgure 2.42

Rotation of Foucault's pendulum

Furthermore, if we take the ratio of v(t) over x(t), we obtain y(t) x(t)

— tan(w1,t)

=

—

tan(fl sin At)

In

which indicates that the plane of the pendulum rotates at the angular rate = sin A, as depicted in Fig. 2.42. Because of the negative sign, the direction of rotation in the Northern Hemisphere is clockwise and opposite to the rotation of the earth, a conclusion we can visualize easily. One intuitively expects the pendulum to rotate in the opposite direction of the rotation of the earth. While the wrms that were ignored in Eq. [rJ change this interpretation slightly, the main result is the same. At the North Pole, as A becomes 900, 90°, the Coriolis effect is the most pronounced. At the equator, the Coriolis effect disappears.

REFERENCES Ginsberg. J. H. Ath'anced Engineering Dyna,nic.v. Dynamics. New New York: York: Harper Harper & & Row, Row, 1988. 1988. Goldstein. H. Classical Mec/,anic.s. Mechanics. 2nd 2nd ed. ed. Reading. Reading. MA: MA: Addison-Wesley. Addison-Wesley. 1980. 1980. Greenwood. D. T. Principles of Dynamics. 2nd ed. Englewood Cliffs. Cliffs, NJ: Prentice-Hall, 1988. Kane. T.. and D. Levinson. Dvnwnics. Theory Theory and Applications. Applications. New York: McGraw-Hill, 1985. NewYork: York:McGraw— McGraw—Hill. I 970. Meirovitch, Mci rovitch, L. L. Methods oJflnalvneal A nalvtieal Dynamics. New Hill. 1970. Elements of Vibration Anal Eletnenis A,za!vsis. vsis. 2nd ed. New York: McGraw-Hill, 1986.

Horwwoiu ExERcIsEs SECTION 2.2 I.

A coordinate system XYZ is transformed into a coordinate system xyz by the following series of transformations: transformations: Fit-st. First, aa counterclockwise counterclockwise rotation of about the Y axis, resulting in the .v'v'z' coordinate system, and then a clockwise

HOMEWORK EXERCISES

of 300 about the x' axis, resulting in the xvz coordinate system. Find the coordinates in the inertial frame of a vector which originally is r = 21 + 3J moves with the transformed coordinate system. Then, find the components a vector p = —31 + 4k in terms of the inertial system and the components of the vector v = —4j + 1.2K in both the XYZ and ..vyz coordinate systems. .

A coordinate system X YZ is transfbrmed into an xyz coordinate system, first by

about the Z axis, which gives the x'v'z' coordinates, and then h a rotation 02 about the y' axis. Express the unit vectors i, j, and k and their in terms of the unit vectors of the XYZ coordinates. rotation of

2.3 Given the column vector {q} =

the matrix ID) as

q2 q3 5

1

0

1

4

1

0

1

6

—1

—1

0

—l —1

3

[D1=

—1

0

and the quadratic form S = evaluate S. Then, calculate dS/d{q} and show that the value you obtain is identical to what would be obtained using Eq. .3.201.

The motion of the double pendulum in Fig. 2.43 is described by the angles 02. Find expressions for the position vector r associated with the motion of the tip of the pendulum, as well as the velocity v and acceleration a of' the tip. Eqs. [2.3.231—[2.3.281, and verify Eq.

the matrix [D] is

the column vector {q} =

[D1=

4q2 6

[

I

V A' x

'1

Figure 2.43

145

146

CHAPTER 2• RELATIVE MOTION

{q}T[D]{q}, evaluate S. Then calculate dS/d{q} and and the quadratic form S show that the value you obtain is identical to what would be obtained using Eq.

12.3.21].

SECTION 2.4 6.

Show (independently of the arguments in this chapter) that the determinant of the direction cosine matrix [ci between two coordinate systems is always equal to 1.

7. Two coordinate systems XYZ and xyz are related to each other as shown in Fig. 2.44. Find the direction cosine matrix between the two coordinate systems. 8. The direction cosine matrix resulting from a 3-1-3 body fixed transformation (U about x3. 02 about and about Z3) is given below. Find the values of the rotation angles 02, and

=[

0.8839 —0.4330

—0.3062 —0.2500

0.3536

9. The rectangular box in Fig. 2.14 is first rotated clockwise by 15° about the line OA and then by 45° counterclockwise about line OB. Find the coordinates of point C after this rotation sequence. l(.). The rectangular box in Fig. 2.14 is rotated counterclockwise by 45° about a line passing through points A and B (viewed from B). Find the coordinates of point C after this rotation sequence. 11. The rectangular box shown in Fig. 2.45 is rotated counterclockwise by an angle of 30° about the axis passing through points 0 and B. Find the coordinates of point A after this rotation in terms of the inertial coordinates.

y V

Y

/

/

2.5

900

/

x

/

z

z

Figure 2.44

3

Figure 2.45

x

HOMEWORK EXERCISES

Consider the double-link robot mounted on a rotating base shown in Fig. 2.46, and 04 measured from the vertical. Find the position of the ith the angles = 300, 01 inertial coordinates coordinateswhen when/ / = ip of the robot arm in terms of inertial 01 = 600, = — 15°. The XYZ coordinates are inertial. AND 2.6 2.6 2.5 AND

sander shown in Fig. 2.47 has a sanding disk that spins spins at at the the constant constant of 150() rpm. The arms AD and DB, which are used to position the sander, angles of 4) 4) and and i/i with the vertical. At the instant shown, their values angles of 6 = 900 qi 60°. and they are moving with the constant angular speeds 6 = 0.2 rad/s and i,Ii = —0.3 rad/s. Find the angular velocity and angular of the sanding disk in terms of inertial coordinates. the airplane in Fig. 2.31. The airplane is moving with speed of 600 in a curved trajectory p = 3000 ft. At the same time, the airplane is pitchupwards at the rate of 0.1 rad/s (constant). The propeller is spinning with constant counterclockwise angular rate of 4000 rpm. Find the total angular of the propeller. = 15 radls rotates with angular velocity 02 = disk shown in Fig. = 1.2 radls2 about a rotating shaft. The shaft is angular acceleration and it rotates with angular velocity w = 4 rad/s and angular accelera= —3 rad/s2. At this instant, the center of the disk coincides with its position. Find the angular acceleration of the disk. Find a general expression for the angular acceleration AaB, given the relation 3wB = + A2WA + ... + single gimbal gyroscope (inner gimbal not moving), such as the one shown in 2.49, is used to measure the angular motion of vehicles. The spin rate of the i

;

Z

z

Z

0.1 m Y

m

I)

I

0.6 rn

-S

V

x

/) 0.5 m P

—z

X, v

2.46

Figure 2.47

147

148

CHAPTER 2S

RELATIVE MOTION

-v 'p

150

V

Figure 2.48 rotor is denoted by ib while the gimbal makes an angle of 0 with the platform. The angular velocities of the platform are wX, Wy. and Find expressions for the angular velocity and angular acceleration of the rotor using a set of relative coordinates attached to the outer gimbal.

18. Find the angular acceleration of the momentum wheel in Example 2.6, given that all angular velocities are constant. 19. Consider Problem 2 and solve for the angular velocity and angular acceleration of the moving frame using the transport theorem. 20. The flywheel of the gyroscope shown in Fig. 2.50 has a constant angular of w3 = 5000 rpm about its axis. The outer gimbal has an angular velocity of = 3 radls, which is decreasing at the rate of rad/s2. The inner gimbal is at a position such that the angle between the outer gimbal axis and flywheel axis is 0 = 75°, with U = 0 = 3 rad/s2. Find the angular acceleration of the flywheel.

Wy

Y

V

/ k

z z

x4

x

FIgure 2.49

FIgure 2.50

/

I 1

EXERUSES

2.7 A bead of mass m is free to slide on a hoop of radius R (Fig. 2.51). The hoop spins

with a constant angular velocity the bead,

about the vertical axis. Find the acceleration

Consider Example 2.1 and using the relative velocity and acceleration relations the velocity and acceleration of a water particle as it exits the rotating sprinConsider Example 2.3. This time, the ant has started moving from point C along

path with point D as the center of the circle with constant speed v. Find the acceleration of the ant. spring pendulum is attached to a rotating shaft by an arm of distance d. as in Fig. 2.52. At the instant shown, the shaft is rotating with the constant velocity u = 0.4 radls, 0 = 30 0 0 = 0.3 rad/s, 0 = 2 rad/s-. Ihe of the pendulum is L = 1.3 m and it is getting shorter at the constant rate : f 0.1 mIs. Given also is that I = 0.8 rn. Find the acceleration of the tip of the .i

the bent shaft in problem 17. Now, the shaft is bent such that its deforcan be expressed by the relation sin('nx/2) m, as shown in = 0. 18 sin('rrx/2) 2.53. The disk is located at x = 0.9 iii. in. Find the velocity and acceleration -f point B. Then derive an expression for the acceleration of an arbitrary point the edge of the disk. Assume the angular velocity of U) is constant and the is of radius R. i

hunter shown in Fig. 2.54 is aiming at a moving target with her rifle. The is moving the rifle upwards with with aa constant constant angular angular velocity velocity of' of 6 1r.'sec and pulls the trigger when 0 = 105°. The bullet, which weighs 1/8 lb. a constant speed of 900 mph relative to the rifle. Find the velocity and acceleration of the bullet. A

z d

,fl

'0

Figure 2.51

Figure 2.52

149

150

CHAPTER 2•

MonoN

ii'

(/;

(0,

z

Q1 -v

-

0.9 m

2m

FIgure 2.53

Y

Figure 2.54

27. Consider the previous problem and find the average force exerted by the rifle on the hunter as the hunter pulls the trigger, assuming that it takes the bullet 0.01 seconds to reach its speed of 900 mph.

28. Consider the hunter in Problem 26 and find the velocity and acceleration of the at bullet if the hunter is also turning at the constant angular speed of = 15°/s 15°/sat the instant the bullet is about to leave the rifle. 29. Consider the two-link robot mounted on a rotating base in Fig. 2.46. Given thai that the base and arm angles vary with the relationships = 0.1 sin t rad, Oi(t) = 0.3trad,O,(t) = —0. it2 rad. find the angular velocity and angular acceleratioc of the robot arms and the velocity of the tip at I = 2 seconds. 30. A disk of radius R shown in Fig. 2.55 spins at the constant rate of w2 = 0 abou an axle held by a fork-ended horizontal rod that rotates itself at the rate w i = th. An ant is walking toward the center of the disk with constant speed v with to the disk. Find the acceleration of the ant as a function of the angle 0 and whet the ant is at the edge of the disk (at point P). 31. An airplane. shown in Fig. 2.56. is flying at a speed o1 500 km/h. The airplane has a constant pitch rate (wi.) of 0.05 radls and a constant roll rate (wy) of 0.0 radls with no yaw, = 0. The trailing edge flaps are being extended to give

the airplane more lift at the constant rate of 0.04 rn/s. Find the velocity acceleration of point C on the flap. 4;

P

FIgure 2.55

HoMEwORK EXERCISES

0.6 ff1

C

Flap

I

20 rn

4

p V

Figure 2.56 An airplane, shown in Fig. 2.56, is executing a circular turn with a radius of

4000 m, while flying at a speed of 600 km/h. The airplane has a constant pitch = 0.01 radls. The trailing = 0.04 rad/s and a constant roll rate of rate of edge flaps are being extended to give the airplane more lift at the constant rate of point C On the flap. 0.05 mIs. Find the velocity and acceleration of'

2.8 Find the equation of motion of the head in Problem 21 using Newton's second law.

Figure 2.57 shows a bead of mass in sliding without friction over a thin wire shaped in the form of a parabola governed by the equation z = x2/2. The thin wire is rotating about the z axis with the constant angular velocity ft There is a spring acting on the bead of constant k that deforms only in the vertical direction. the equation of motion of the bead using Newton's second law.

x

FIgure 2.57

151

152

CHAPTER 2• RELATIVE MOTION

V

V

z (Vertical)

6() mph

x

0 (Westi

(oz

(a)

Figure 2.58

(b)

x(North)

Figure 2.59 35. A satellite in space has angular velocities w.v, 0v, and about the x, y, and z axes, which are attached to the satellite at point 0, as shown in Fig 2.58. On the xv plane, making an angle of 300 with the x axis and going through 0, is a tube, inside which a particle slides without friction. The particle is attached by a spring and a dashpot to each end of the tube. The spring is unstretched when the particle is at point 0. Find the equation of motion of the particle, using Newton's second law. 36. Consider a 90-ft-long bowling alley in Rio de Janeiro, Brazil. A bowler releases the ball with a constant velocity of 20 ft/sec, aimed directly at the pocket. What is the Coriolis deflection. and in which direction is it? 37. Integrate the perturbation expressions to find the deflection of a stone thrown in the air vertically with speed v as it falls to the ground. Then compare this result with the exact solution.

chapter

DYNAMICS OF A SYSTEM OF PARTICLES

3.1

INTRoDUcTIoN

chapter serves three purposes. First, it extends the developments of Chapter 1 to with more than one particle. Second, it prepares the reader for the analysis jr bodies, addressing the most basic form of rigid body motion, that is. plane of a rigid body. Third, it introduces basic concepts in orbital mechanics. Analysis of systems of particles and rigid bodies is greatly simplified when the of center of mass is introduced. Newton's laws of' translational motion and law of rotational motion for a single particle can he expressed in the same

for a system of particles as well as br rigid bodies in terms ol' the center of This chapter is written such that one can skip it and go directly into analytical This or into rigid body dynamics. This chapter actually belongs with Chapter L it is an extension of the basic concepts studied in that chapter. For pedagogical the developments here are presented separately. The sections in this chapter on plane kinetics are intended to serve as a review 1 this this special case of rigid body motion. This review is essential, especially for who have not studied the plane kinetics of rigid bodies for a long time. Plane is relevant to Chapter 4, where we carry out the developments developments in in analytical analytical for particle motion or rigid body motion on a plane.

3.2

EQUATIONS OF MOTION section. we extend the developments in Newtonian particle mechanics to syssection,

consisting of several particles. Consider a system of N particles, as shown in

153

154

CHAPTER 3

•

SYSTEM OF PARTICLES

DYNAMICS OF

. ,flI In1

r8

.

B

. •.

,fljv

Figur. 3.1 . ., N) and is located by the displace(i = 1. 2, .. Fig. 3.1. Each particle is of mass ment vector r. The motion of the individual particles is not necessarily independent of each other, as there may be forces that relate the behavior of some of the particles to each other. the total mass by m the We next introduce the concept of center of mass. mass. Denoting Denoting by of the system .

N ?fl = in

[3.2.11

r,li ,,li i—i

Denoting the center of mass by the point G and locating it by the vector rG, it

is

defined as 1

rG =

[3.2.21

—>Fn1r1 —>Fn1r, Pfl I—I

We can express the position of the ith particle relative to the center of mass as

r =rG+pi r1 =rG+pi

[3.2.31

in which is the vector connecting the center of mass with the ith particle. Intro13.2.21, we we obtain obtain [3.2.3] into intoEq. Eq. 13.2.21, ducing Eq. [3.2.3] I

+

rG

= rG rG +

—

[3.2.4]

which leads to the conclusion N

>}fliPi

= 0

[3.2.5]

i=I

One can differentiate the above equations to find expressions for fbr the velocity and and aG, a(;, respectively. acceleration of the center of mass, which we will denote by

3.2

EQUATIONS OF MOTION

Differentiating Eq. 13.2.2] with respect to time, we obtain N

-i-'ss>.'/flivi

vG=rG=

in /

—j

[3.2.6] 13.2.6]

1—I

leads to the relation for the relative velocity N

=0

[3.2.7] 13.2.71

From this we find the linear momentum p of a system of particles to have the 'V N

N

=

p = m1v, p, = m1v1

(i =

1, 2,

.

.

)

=

-J

111V(; I11V(;

[3.2.8]

., N) is the linear momentum of the ith particle. . ,

In a similar fashion, we find the acceleration of the center of mass as N

aG —r(;= aG =r(;=

1

in

[3.2.9] -

eading to the relation for the relative acceleration terms

[3.2.10]

= 0 i—i

Next, we apply Newton's second law to a system of' particles. We separate the force acting on the ith particle into two parts:' (1) Ibrces acting on the ith /th parfrom outside the system of particles. referred to as the external or impressed and denoted by F1 (i = 1, 2, ., and (2) forces exerted on ni1 m1 by the other within the system, referred to as internal or constiwn! constiwn! forces, forces, and and denoted F1. Newton s second law for each particle is .

=

F1

.

I = l,2,...,N

+ F;

[3.2.11] [3.2.111

The N equations above are usually not independent ol each other, so that it is not to analyze the motion of each particle individually. The number of degrees -ff freedom. denoted by ii, is in general smaller than the number of coordinates (for the here 3N). The reduction is due to the action of' one particle on the other and from the restrictions in the motion of the particles that they cause. These actions constrain motion of the particles within the system. The internal forces F; (I = l,21....iV) ze the forces associated with the constraints. Considering the system as a whole, we sum the force balances for all of the thus N

N

= i—I

+ F;) 1—I

We will see this separation also n Chapter 4, when we study the principle of virtual work.

[3.2.12] [3.2.121

155

156

CHAPTER 3• DYNAMICS OF A

OF PARTICLES

Substitution of Eq. [3.2.3] to the left side of this equation yields N

N

N

=

+ P1) =

i=I

i—I I—

I

1

d

=

i=I j

I

[3.2.13]

=

I

Now evaluating the right side of Eq. [3.2.12], because F; (i =

1,

2,

.

. .. N) rep-

resents the forces that one particle exerts on another, their sum over all particles must be zero, by virtue of Newton's third law. Defining by F the sum of all external forces over all particles. N

N

[3.2.14]

F =

we obtain Newton's second law for a system of particles as ci

maG= maG= -1--p=F

[3.2.15] [3.2.151

Another way of writing Newton's second law for a system of particles is to cornhine Eqs. Eqs. 13.2.131 13.2.131and and[3.2.15] [3.2.15] to to yield yield

[3.2.16]

>",n1a1 = F

Depending on the need and on the problem, one can use Eqs. [3.2.11 J, [3.2.15], or [3.2.1 6J to describe the force ba]ance of a system of particles.

Example 3.1

The Iwo masses ni1 and in2 are connected by a massless rod, and they are acted upon by a

shown in Fig. 3.2. Write second law using Eqs. [3.2.11], [3.2.15], and [3.2.16]. and evaluate if these equations qualify as equations of motion.

force 1',

as

Solution We first separate the two masses and draw free-body diagrams, shown in Fig. 3.3. We denote

the displacements of the masses by x1 and (i = Using Eqs. [3.2.11] we obtain the four equations =

i?liX2 =

flfl In

1,

and

2).

=

Fy

+P

are internal reaction forces.

[.1

rn2Y2 =

in terms of the reaction forces.

L

?fl

F71,

6

ii (a)

Figure 3.2

Figure 3.3

(b)

.1

(C)

Free body diagrams

3.3

1 57

LINEAR AN!) ANGULAR

use Eq. [3.2.15], we first need to locate the center center of ofrnass mass G. ii is easy to show that the distance from to the center of mass is L1 = rn2L/(rn1 + m2). Considering the system is a whole, the only external force is P. Denoting the displacements of the center of mass by and Newton's second law becomes To

=0

(m1 + m2

(fli, +

)YG

IbI

= P

Finally, we add the first two of Eqs. [a] and the last two, to obtain Newton's second law the form of Eq. Eq. [3.2.16] as [3.2.16] as ,n111 + m2x2 = 0

,n1V, +

= P

Id

Let us now investigate the nature of these equations. Eqs. [a] have four equations and Eqs. [b] and [ci have two. To find if any of these equations qualify as equations of of Eqs. we calculate the number of degrees of freedom. which can he shown to be 3. Hence, we three differential equations, void of internal forces, to describe the system. We conclude Eqs. [b] represent two of the equations of motion, in terms of XG associated with XG and and translation of the center of mass. Similarly, Eqs. [ci represent the same, but in terms of x, (i = 1, 2). We still need a third equation. (1 To find the equations of motion, we need to find expressions for and in Eq. [a] .irxi write one of x1, x2, or P2 in terms of the other three variables. The procedure is at best, especially if one realizes that this problem is ideally suited for treatment with coordinates and angular momentum balances. We discuss this approach in the next ,ection.

3.3

LINEAR

ANGULAR MOMENTUM

The linear momentum of a system of particles is defined in Eq. [3.2.81 and, as in Chapter 1, it is an absolute absolute quantity. quantity. We can integrate the equaof motion for of particles for aa system system of particles over over time time to to obtain obtain impulse-momentum Indeed, following the approach in Section 1.6, we can integrate Eqs. Eqs. [3.2.111. 13.2.151, 13.2.151, and and 13.2.161 13.2.161 over a time period (11,12) to obtain the linear relationships, writing relationships. rf h m1v1(t1

)+

(F1(1) ++F(t))di (F1(1) F(t))di

i=

1. 2, 1.

. . . ,

N

J tl

N

N

+

(

ill

i=1

F(i)di F(t)dt =

+

?flVG(12)

[3.3.Ia,

b, ci

J tI

i

In order to solve Ibr the individual velocities of each particle, one has to integrate of Eqs. [3.3. Ib]. One may not he able to eliminate the F(t) terms directly: hence, flumber of constraint forces may have to be solved for. number

When the sum of all external forces acting on a system of particles is zero or integral over a time period is zero, the linear momentum of the system remains

1 58

CHAPTIR3 •3DYNAMICS • DYNAMICS CHAPTIR OFOF AA SYSTEM SYSTEMOF OFPARTICLES PARTICLES

unchanged, which is the statement of the principle of consen'ation of linear momentum for a system of particles. The principle can be written as

p(t1) = p(t2)

[3.3.2J 13.3.21

or N

mVG(tI) = mvG(t,) These

N

'>

=

>

[3.3.31

i=1 i==1

equations are vector relationships. If the sum of forces or their integral

over a time period is zero along a certain direction, the linear momentum is conserved only along that particular direction. For example. if linear momentum is along a certain direction, and we express the unit vector in that direction by e, the conservation of linear momentum equations become N

= rnVG(t2) e

PflV(;(11) • e

N

mv1(t1)

>

'e

13.3.41 = > rn1v,(t2) • ee [3.3.4J 1=1 i=I

1=1

Impulsive forces are treated the same way as in Chapter 1. We next define the angular momentum, or moment of the linear momentum, of a particle in1 about a point B by

=

H81

r81

X ,n1v1 ,n,v1 =

r81 X P1

1

1,2,..., N

[3.3.5j [3.3.51

where r81 is the vector connecting point B and the ith particle (Fig. 3.1). Unlike linear momentum, which is an absolute quantity, angular momentum is relative: its depends on the point about which it is calculated. The total angular momentum of particles about point B is denoted by H8 and is expressed as N

H8 =

> H8 = > > r81 r8 i=I

[3.3.6J

X

1=1

We next relate the angular momentum of a system of particles to the center of mass motion. From Fig. 3.1, we write r81 in terms of the center of mass as r8, = rc,B + p, (i = 1, 2, ..., rc/B .., N). For the ith particle, we write the angular momentum expressions as .

H81 = (rG,B

+

[3.3.7J [3.3.71

X mf(vG X ml(vG + iii)

We sum the individual angular momenta about point B and obtain the angular momentum of the system of particles about point B as N

H8 =

N

H81

=

B

+ Pi)

X

+ p,)

[3.3.8J

1=1

Considering the definition of the center of mass, this equation reduces to N

H8 = rG/B >( X fliVc; fliVc; + >' >( m1ii1 rG/B X 13.3.9) 13.3.91 1=1 i=I where the first term on the right is associated with the motion of the center of mass. and the second is due to motion with respect to the center of mass. The second tent is also referred to as the apparent angular momentum. Differentiation of Eq. [3.3.9

33 3.3

LINEAR AND ANGULAR MOMENTUM

:'elds N

V(;/8 X !flVG + HB = rG/B X 'flaG + VG/8

=

rG,B X IflaG ±

V Gill

X mipj +

X

f3.3. 10]

P X

X FflVG +

The moment about point B of all the forces acting on the ith particle is defined

MB, The

X r8, X

=

i = 1, 1, 2, 2, .. ., N

r81 X ni1a1

[3.3.1 13.3.1 11

total of all of the moments acting on the system of particles can be obtained by the individual moments, which yields N

M8 =

> rm

N

X F = >r111 r11, X in,a, ma1 =

+

>

X

j1

+ j,) [3.3.12] + [3.3.121

Luvoking Luvokingthe thedefinition definition of of the center of mass, Eq. [3.3.12] reduces to N

MB = rG/B X IflaG

+> +>

Pi X Pa

13.3.13]

The second term on the right side of Eq. f 3.3.101 can be written as VG/B X

IVVG = ?flV(; X = (VG — Va) Va) XXIVVG (VG —

[3.3.14] [3.3.141

that introducing Eqs. [3.3.131 and [3.3.14] into Eq. 13.3.10], we obtain the angular balance fbr a system of particles as

[3.3.15] [3.3.151

H8 = M8 + /flVG X V8

Equation [3.3.15] is a general relation describing the angular momentum balabout a point B. whether whether B B is is fixed fixed or moving. Under certain circumstances.

depending on the choice of point B, the equation can be further simplified: When the point B is selected selected as the center of mass. mass. BB = G, then G, then we have

[3.3.16]

H(; = MG

:

vanishes

When the point B is fixed in an inertial coordinate coordinate frame, frame,then thenVB VB = 0, the second in Eq. [3.3.15] vanishes, and we get

[3.3.17] 13.3.171

= M8

When VG/B VG,B is parallel parallel to to VG VG (11w (fhr any anyreason), reason), the the cross cross product product in Eq. [3.3.151 This mathematical possibility is not of any physical significance.

Integration of the moment balance over time yields the angular impulserelationships. For each particle m, and and considering considering a fixed point B about the center of mass (denoting such a point by D), HD,(tI) + JI'

MD,(t)dt = HD,(t7) MDI(t)dt

i — 1,2, . . ., N

[3.3.18] [3.3.181

1

59

160

CHAPTER 3 • DYNAMICS OF A SYSTEM OF PARTICLES

Evaluation of this equation equation fi)r fi)r each each mass massm1 m, (1 = 1, 2. . .. N) requires that the internal forces acting Ofl the particles be calculated, reducing its usefulness. When we consider the entire system and either a fixed point B B or the center of mass G. summation of Eq. [3.3. 181 over all particles yields .

(•

HD(tl)

+

r,

13.3.191

= HD(t2) J

If the applied moment about the fixed point B or center of mass is zero, or the integral of

over the interval (t1, t2) vanishes, we have

[3.3.20]

=

I

the principle principle of ofconservation conservation I which is the cles.

Example

3.2

inoinentuin for a system of parti-

Example 3.1. The system is initially at rest with 0 = 300 when ills hit by the impulsive force P. Find the velocities of the two masses immediately after the impulsive

Consider

force acts.

Solution and VG. Henceq Eqs. [hi in Example It is more convenient to use center of mass coordinates 3.1 become two of the equations of motion. To find the third equation, we make use of the angle 0. We write the displacements of the two masses in terms of the coordinates of the center of mass as

cos00 — L L1 cos = = XG + L2 COS 0

Yi =

VG —

L1

sinG0 sin

V2 = YG YG + L2 sin 0 V2

[a]

where L1

m1L 1171 + !fl !fl 22

L

+ 7Th fl!1 + ill

—

[hI

so that the velocities of the masses can he written as

+ L1OsinA)i +

V1 =

=

—

L20 sin 0)1 +

—

L1OcosO)j

+ L20 cos0)j

[ci

The position vectors from the center of mass to the individual masses are

r1 = so so

—L1 — L1

cosOi COS Oi — — L1

r2 = L2cos0i L2 cos 01 + + L2 L2sin0j sin Oj

sinOj sin Oj

Id]

that the angular momentum about G becomes (after (after aa few few manipulations) manipulations) =

r1

X in V1 +

>K >K ?n7V2 v2 = (m L LI

++mm22L )O k )Ok

[.]

The moment generated about the center of mass is simply = — PL1 cos Ok, so we can say that the impulsive moment due to the impulsive force is = — PL1 cos 0. From the linear impulse-momentum theorem applied to Eqs. [hi of Example 3.11 we obtain immediately after the impulse •

XG=O

.

YG

P in1 + in,

If]

3.4

toy

WoRK

The angular impulse-momentum prob\ern

() =

PLcosO

—

——

,n1L1 +

IgI Ig'

+ + ,,iiL;)

2(in1

—

Substituting Eqs. [f] and tgJ into Eqs. [ci yields the velocities of the individual masses after the impulse. thus

/

—

p

+

—

1L2

+

+ 315L1L2___ +

—

-1

3PL

+—

V1

1/b 1'?!i + + 1/b

1k]

the considerable considerable advantage of using the center of mass. If we This example illustrates the have to first solve for 1.we we would would have in Example Example 3.3.1. binted to solve this problem using Eqs. [al [al in ftrces and substitute into the impulse—momentum relations, which is impulsive reaction frrces procedure.

3.4

WORK AND ENERGY

ftrces acting acting on the ith particle is prom Chapter I the incremental work done by all ft'rces i = 1, 2, .. ., N) and has the form by .

(1W1

.

l,2,...,N 2.

= (Fr +

1

,

.

.

.

,

13.4.11

From the work-energy theorem, we write for the ith particle dW1 =

13.4.21

dT1

bhere T is the kinetic energy associated with the ith particle =

T T,

I1

[3.4.3]

• Vi 'Vi

To express the work done by all forces on the system of particles, write dr1 in of the center of mass as dr1 = drG + dp,

13.4.41 13.4.4]

3.4. 1] 1] and separating the total ftrce into its ztroducing this expression into into Eq. Eq. 13.4. nternal and external components. we obtain I

dW, = (F, dW1

+ F)'(dr(] F)•(dr(; ++ dp1) dp,)

13.4.51

We find the total work done by all forces acting on the system by slimming the

miividual incremental work for each particle and integrating over the displacement

each particle. Denoting by r1 and r,2 the initial and final locations of the itli ith the total work is N

wI__.2

=

=

-r,

(F, + (F1

+ dg1)

[3.4.6]

162

CHAPTER

3

•• DYNAMICS DYNAMICS OF OF A A SYSTEM SYSTEM OF PARTICLES

Using Eq. [3.2.14]. we can simplify the expression for work to N

wI__.2

=

F drG +

J

(F1

[3.4.7) [34.7)

+ F) dp1 +

. r1 J

describes the work done due to The first term on the right in this equation describes includes the contribution due second term term includes The second motion of the center of mass. mass. The

the motion with respect to the center of mass. Note that this second term F (i == 1,1, 2, . ., N), indicating that the constraint forces between the particles contribute to the work done. Although these internal forces cancel each other we are summing forces to get the equations of motion. they do not necessarily vanish when the work is calculated. As we did for a single particle, we can write the expression for work as an integral over time. Multiplying and dividing each term in Eq. [3.4.7] by di. we obtain .

'V

W

= ill ill

F• VGdt +

(F1

134S1 [3.4.SJ

+ F) F).p1dt

and 12 denote the times at which the particles are at positions r11 and

where

respectively. We next evaluate the expression for kinetic energy by summing the

expressionv1 v = T, isis defined defined in in Eq. Eq. [3.4.3]. Substituting the expression in which T1

VG

+

in

above equation, we obtain N

+ Pi)'(VG j)j)(VG +

T

rnj(Vc;VG +

=

+pj•pj) +

i=1

=

V(j) +

term, Ttran Ttran = The first term,

'>

=

13.4.101

+ Trot

•VG/2. VG/2. is due to the translation of the center of mass. due to the motion of the individual particles with respect w

FflVG

is the second term, the center of mass. The potential energy of a system of particles does not lend itself to a formulation, and we use the expressions in Chapter 1. In general. the potential is due to elastic forces between particles, such as interconnecting springs, and due w gravity. The gravitational potential energy for a system of particles can be in terms of the potential energy of each particle.

V1 =

i=

1,

2,... . ., 2,

N

1J [3.4.1 [3.4.111

is the the height height of each particle from a common datum. One can show in which h1 is the potential energy of a system of particles can also be written in terms of the

3.4

163

WORK AND ENERGY

:enter of mass as N

V

13.4.121

=

=

of particles can be expressed in terms of The work-energy theorem for a system particle as = 1i2 + Vi2

T11 + V,1 +

I = l,2,...,N

[3.4.13]

in terms of the system of particles as

[3.4.14] 13.4.141 the work done by all nonconservative forces and the subdenote the initial and final stages of the motion. When all the forces acting system of particles are conservative4 the total energy of a system of particles is and we write ., denotes

E = T + V = constant

[3.4.I5J

of mass 0.05 kg is shot with speed 400 m/s at a target on wheels of mass 3 kg. shown a. 3.4, 3.4, which which has has a plate a plate in itinthat it that is is attached attachedtotoaaspring springof of constant constant kk = 80.000 N/rn. et is initially at rest and it can slide without friction. Find Find the the maximum maximum compression compression after the bullet hits the target.

a

takes place as follows. The bullet hits the target and begins to compress the spring. compression of the spring is reached when the target and bullet are moving speed. Because there are no forces external to the bullet-target system, linear is conserved along tile the horizontal. At the point of maximum compression Pflj,Vj, = (,n1, (in1, + in1)u ni,)u

[a]

subscripts b and t correspond to the bullet and target, respectively, and v is the corn'.elocity at the instant of maximum deflection of the spring. Substituting in the values, the

v=

illj,Uj, in1,

+ in,

-.fflD

Figure 3.4

=

0.05 x 4(X) = 6.557 ni/s 3.05

[b]

I

Example

3.3

164

CHAPTER 3 • DYNAMiCS OF A SYSTEM OF PARTICLES

There is no loss of energy, as the energy transfer between the bullet and the target involves the spring and the bullet does not get lodged in the target. We thus invoke the conservation of energy between the initial state and the maximum deformation of the spring as

T,=T2+V-'

Ic]

in which

T1 T1 so

T, =

=

+ rn1)v2 rn1)v2

V2 =

[dl

that the deflection of the spring is found as in1,

—

(mb + in1 ) v2

1.1

Substituting in the values, we obtain 1(0.05 x

=

3.5

(3.05 X 6.5572) = 0.3 136 m 80,000 —

If]

IMPACT OF PARTICLES

There are many ways two bodies come in contact with each other. The contact may take place at an isolated point, over a line, or over a surface. The contact between two bodies generates a constraint or reaction force on both bodies. From Newton's third law, the constraint force on each body is equal in magnitude and opposite in direction. In this section, we consider the special case of collision of two particles. that of' of impact. impact. Chapter Chapter 88 conducts conducts aa general general study of impact of rigid bodies. In order to have a better understanding of impact, we initially assume that each particle is a solid sphere. We consider two spheres of masses in1 and in2 that have velocities (of their centers of mass) of v1 and v2 before impact and u1 and 112 after impact, as shown in Fig. 3.5. The changes in velocity, as they occur in a very short period ol' of time, time, must be caused by impulsive forces, from which we conclude that impact generates an impulsive reaction (or impulsive constraint force) F on each of the two spheres. This impulsive constraint lies along the line connecting the center points of the spheres and the point of contact; it is illustrated in Fig. 3.6. The line joining the two centers is commonly referred to as the line qt impact and its direction the tiorinal direction. The normal direction here should not be confused with the normal direction associated with normal and tangential coordinates. Because the impulsive constraint I'orce force is is along along the the center center of each sphere. there is flO no change change in rn the linear momentum about the centers of mass of' of both spheres. The impulsive fbrce E can be expressed as

F = PH Fn where n is the unit vector along the line of impact.

[3.5.1]

3.5

/

v

-

Before

---

After

.-_,fl1

— —

IMPACT OF PARTICLES

——

_3

,nI

-

—

-

U

Ia)

Figure 3.5

(b)

Colliding particles

The linear impulse-momentum relation for the two masses can be written as —

We

+ F' F' = ?fl2Ui

lfliUi

F =

[3.5.21

add the two equations above to obtain the conservation of linear momentum for the system of the colliding particles as FfliVi + Ifl2V2 = 1fl1U1 +

[3.5.3]

[3.5.31 can also be written directly by using the impulse-momentum refor a system of particles. Eq. [3.3.lh]. Noting that during impact (here impulse external to the system of the two spheres, the linear momentum

u the system of two particles is conserved. Equation [3.5.3] indicates that the of mass of the system of two spheres does not change position during the Because the impulse impulse is only in the normal direction, the components of Because velocities of both masses in the plane perpendicular to the normal direction do change, either. We separate the components of the velocities along the line of (i (i = 1, 2). in which u, = u,n + as v = un + the line of impact. From Eqs. [3.5.2] and [3.5.3], we v and u for the velocities perpendicular to the line of impact VjJ)

=

=

[3.5.41

along the line of impact, the linear momentum expression becomes

+

=

rn1u1 + 1fl2U2

[3.5.5]

u2. To Equation [3.5.51 is a relation in terms terms of of two two unknowns, unknowns, u1 u1 and and u2. To solve solve

i-r the two unknowns, we need another relation. This relation is derived from Poiss hypothesis. Poisson's hypothesis is based on the assumption that the contacting are not exactly rigid, and it states that the impact takes place in Iwo stages. the bodies compress each other the first stage, called the period of the relative velocity between the colliding particles becomes zero along the

Figure 3.6

1

65

166

CHAPTER 3 • DYNAMICS OF A

OF PARTICLES

?fl1

0---

Approach

Restitution

--———a————.A

A

Fr

Fr

(a)

(c)

Compression

Separation Uj

F (b)

FIgure 37

U1

(d)

The stages of impact

line of impact. In the next stage, called the period of restitution, the bodies regain their original shapes, as shown in Fig. 3.7. The ratio of the strength of the two impulses is denoted by the coefficient qf restitution e. We separate the impulsive force

F into two parts associated with the compression and restitution periods penods as during the compression stage Fr during the restitution stage

13.5.61

Because the entire impact is in one direction, the absolute values of the strengths can

be expressed as P. and Pr. The coefficient of restitution is defined as

e=

[3.5.7]

which leads to the relations

=

F

A

Fr =

J+e

Fe

[3.5.8] 13.5.81

in which which! P = Pc + Pr is is the the total total strength strength of tlie the impact. To To find the velocities of the colliding bodies along the line of impact, we integrate the equations of motion for the two stages of impact. Denoting the velocity at the end of the compression period by the linear momentum balances along the line of impact become Mass 1

Compression: ,nlvJ —

=

Mass 2

Compression: in2v2 +

=

Restitution:

vi..

Restitution:

in1

—

P??1 Fr = P??1

+ Fr = !fl2U2

13.5.9] Introducing Eqs. 13.5.81 to these equations and eliminating the total force F, we obtain —

U,

=

—

)

[3.5.10]

which is the commonly seen relation. The interpretation of this equation is that the coefficient of restitution represents the ratio of the relative velocity after impact to the

relative velocity before impact. The relation [3.5.10] was first observed by Newton.

3.5

IMPAcT OF 01'

1

generalized Newton's result to bodies of any shape. We will make use of Ei>. EL>. [3.5.81 [3.5.81 in Chapter 8. when dealing with the impact ofrigid bodies. Solving Eqs. 13.5.5 1 and 13.5. 101 simultaneously yields the results I

e)nz7v7] —— = —I("i I("i — — eln2)v1 eln2)v1 + (( 1 + e)nz7v7]

UI

Ti! I

—L(n12

112

[3.5.1 11

('in1 )Vi )Vi + ( 1 + e)ni,v1 I ('nil

I

?11

,n = ,n + is the total mass niass of the colliding particles. that is The coefficient of restitution e is a quantity in the range 0 e on the material properties of the colliding particles. as well as on the speed of the colliding masses. The coefficient of restitution decreases in 1

tihie as the relative relative speed speed of of impact impact gets getshigher. higher.The Thespecial specialcase caseofofee = 1 is known perfectly elastic impact. In this case. the strength of the impact is the same in the and restitution stages. and there is no energy loss. The case of e = 0 is to as plastic in/pact. Setting e = 0 in Eq. [3.5. 1 1 ]. we obtain 1

1

1=

1.1 1 = 1.1

I

.

lfl VV I + + lfl lfl 2 V2 )) — ( lfl 1?1 in I

1.! 1.! 22

in

( Ffl

L'2 + 111

)

I 3 . 5 . 1 21

to the cOnclusion that in this case = 112. 1.12. After impact. the colliding parhave the same velocity along the line of impact. We next examine the energy loss associated with impact. The energy loss ocbecause the strength of the impact diminishes in the restitution phase. The lost gets translèrred to the colliding bodies through internal vibrations as well as increase. i show that that the the energy energy loss loss is is due due to to the the change change in in the the relative relative velocities One Can can show ii the colliding particles. For perfectly elastic impact there is no energy loss. On the hand, when there there is is plastic plastic impact. impact.ee = 0, all of the kinetic energy associated the relative motion of the colliding masses is lost. Be aware that the above derivation of impact relations represents a gross simof what actually happens when two bodies collide. First, we assume that collision takes place in a very short period of time. This is possible if the speeds

with the impact are large. Then, we implicitly assume that there is no damage due to impact. For this to hold, the speeds involved should not be In realistic impact situations, situations. the impact takes place over a finite time period. small. As we discussed in Chapter 1. UrIC OIiC should always check the validity of assumpions used assumptions used during during impact. impact. The The coefficient coefficient of restitution itself is an approxtion, as it is determined experimentally. As discussed earlier, its value depends variety of factors. Another assumption whose whose validity validity comes comes into into question question isisregarding regardingthe thecorn compop0of the motion orthogonal to the line of impact. especially when frictional forces ;nvolved. For example, consider a ball thrown from a certain height with a horlior-

velocity, as shown in Fig. 3.8. As the ball hail collides with the ground. it has a horizontal as well as a vertical velocity. The line of impact is perpendicular i.' the ground. The impact results in an impulsive normal force. The impulsive norforce leads to an impulsive friction force. In this section. we assume that the friction force is zero. This assumption assumNion may not he valid in all cases.

67

168

CHAPTER 3 • DYNAMICS OF

A

SYSTEM OF PARTICLES

/ nix

Af

F

A

N

FIgure 3.8

Example

3.4

I

A sphere of mass in and radius R is dropped from a height Ii onto another sphere of mass 2m and radius II .5R. .5R. which which is is at at the bottom of an inextensible cord, as shown in in Fig. Fig. 3.9. 3.9. Find Find the the velocities of the spheres immediately after impact. given that the coefficient of restitution is (? =0.9

Solution Figure 3.10 shows the geometry of the impact and the free body diagram. Denoting the spheres by A and B. the angle 4 with which impact takes place is =

sin'

= sin '(0.4) = 23.58°

(a]

The line of impact joins the centers of the spheres. and the components of velocities before befbre and after impact are —O.4u VA: —vsin VA: == —V sin 'f' = —O.4u

=

u

cos

= 0.91 65v

UB,,

=0

(bI

where the subscripts ii and i denote the components along the line of impact and perpendicular to it. After impact, the velocity of B is in the horizontal direction.

/ "Ag

A

A

I,,, "p.

"I

I

I

I I'

A

'I I

B fl

(a)

FIgure 3.9

FIgure 3.10

fling (b)

3.5

OF PARTICLES IMPACT OF

Consider the free-body diagrams of spheres A and B. Of the two forces acting on A.

t

on B, B, two two are are impulsive. impulsive, P, and of the three forces acting on one force is impulsive, one t Hence, Hence, there there is no conservation of the impulsive tension in the string, denoted by f. momentum for the two spheres. We need to write the impulse-momentum relationships for each sphere. We denote the components of the velocity after impact impact by by UA UA and and Ufi. Ufi. Because Because Ufi Ufi is it can be expressed in terms of its components as = U11 cos

4) =

id Ic]

sin 4) = 0.4u8

=

0.91 65 U11

write the total impulsive force as

En

—

D cos 4)n +

I

Id] [dl

sin 4)t 4)t I sin

we have have directions. we the linear momentum balances in the n and t directions. A

A, n direction

—

r direction

mv41 =

t cos 4) = 2mvp,, ++ I 2mvp,, I sin sin 4) = ± ft

B. n direction B, t direction B.

Eel [01

P =

—

[fI Ff1

FflUAI

2rnu87,

= 2mufi sin 4)

2?flUBI 2?flUB,

=

COS 4)

[g] igI [h] 1k]

of the velocities along the line of impact are related by 118n

=

—

—

=

III [II

evA,1

P. and 1. Equation [fj If I directly. Multiplying Eq. [gj by sin sin 4) 4) and and Eq. Eq. [hJ [hJ by by cos cos 4) 4) and and adding adding the the two two

We then solve equations [e]—[ij for forthe theunknowns. unknowns. U11. U11.

we obtain

=

2rnu,1

Ill

sin4)

when introduced introduccd into Eq. Ic] [ci yields 2mu8 FflVAn —

.

SIfl4)

[kI 1k]

=

Equations [kJ and [i] can he solved together for U11 as v.,%,1(l + e) v.4,,(l e)

148= sin 4) D = 2/ sin 4) + sin 4) =

D

vcos4)(I + e) e) I) 1)

[II Ill

5.4. so that the velocity of sphere B immediately after impact

3 = v

can be obtained using the work-energy relationship of En] ml

v= V

we make use of Eq. [ii

To find

Lt. = Us,,

—

= u8sin4)

—

evcos4) = 0.3225(O.4)v

—

0.9(O.9165)v = —O.6959v

Eel

169

1 70

CHAPTIR 3 S DYNAMICS OF

OF PARTICLES

A

Hence, the velocity of sphere A immediately after the impact is u,1

3.6

=

=

+

+ 0.69592 = O.8027v

VARIABLE MASS AND MAss

SYSTEMS

Two interesting applications of impulse-momentum principles are variable mass mass flow systems. In such systems, either the mass of the bodies involved or mass flows in and it flows flows out out of' of a body at a certain rate. A typical of variable mass systems is the rocket problem, depicted in Fig. 3.11. A

gains speed not only because of the thrust generated but also because the results in loss of mass. in mass flow systems, air enters the system in a certain threction with a certain speed and comes out in a different direction with a speed. To analyze variable mass systems, we use the linear impulse-momentum rehtionship in its general form

rt, p(t1)

+ j ti

F(t)dt = p02)

[3.6.11

Consider a single particle of mass m(t), acted upon by an external force F(t),

shown in in Fig. Fig. 3. 3. 11. Il. The The initial mass of the particle at t = is m(11), and its velocity is v(t1). The initial linear momentum is = rn(t1 )v(t1). The of the particle at time t2 is denoted by v(t2) and its mass by !n(t2). The between the mass at and the mass at t2 can be written as m(11 ) + in(t2). from the figure that is defined as a negative quantity and that we are treating the body and the lost mass as one system. The linear momentum of the system i

t=

12 12 is

p(t2) = tn(12)v(!2)

—

v(11)

rn(11)

f

t=tl (a)

Figur. 3.1 1

(h)

[3.6.2J

3.6

VARIABLE MASS AND

FLOW SYSTEMS

is the exit velocity of the mass that has left the particle. We can write the momentum balance as Ve

I1_

,n(11

)v(tj) + I

F(t)dt

lfl(t2)v(t2)

—

=

)V(t2)

—

=

)V(12) +

JtI

Vrej(1) = v(t)

—

13.6.31

denoting the relative velocity of the exiting mass. To solve the above equation. we need to know how the change in mass takes —

and the exit velocity of the mass leaving the system. This requirement ol about the characteristics of the separation is analogous to the need to the coefficient of restitution of colliding bodies. We need to have information rn how the separation takes place. Now, assume that the time interval is small and inthe notation = — and approximate the integral in the above equation This approximation is valid as long as F is not impulsive. Introducing the notation = v(12) — v(/1) v(/,) into Eq. [3.6.3] and dropping the subscripts 2, we obtain I

= Dividing both sides of this equation by gives

+

and taking the limit as

[3.6.41 approaches

+ th(t)Vrci(t) = F(t)

13.6.51 This is the general equation of motion for a variable mass system. Note that this ation is different than F(t) = dlm(1)v(n]/dr. This is because the body which is mass and the lost mass are considered together as one system. Next, consider mass flow systems. We restrict our analysis to steady mass flow such that the rate of mass flowing into the system is the same as the rate

mass flowing out. Consider, for example. the air-blowing machine shown in 3.12. Air enters the container through a duct of cross-sectional area A, with v1 and density p,. It leaves through through aa duct duct of of cross-sectional cross-sectional area areaA2 Ai with v2 and density P2. We denote the mass flow rate by in'. in', recognizing that in'

A1

V

Figure 3.12

1

71

172

CHAPTER

3

.

DYNAMICS OF OF .k k

OF PARTICLES

is not the rate of change of mass, but it is the rate of' mass flow in and out of the system. The mass flow rate can he expressed as in' = A1p1v1 = A2p2v2

13.6.61

The external forces acting on the system are the reaction and support forces that hold the container in its place. These forces counteract the change in linear and angular momentum due to the flow of mass. We denote the resultant of all forces by F and the resultant of' all moments through a fixed point 0 by M0 (or about the center of mass). The change in lineat- momentum is due to the change in speed direction of the mass flow. Denote the amount of mass that flows in and out of the such that system in a time period by in in'

13.6.11

=

The change in linear momentum during

= &flV,

can he expressed as

[3.6.83

—

Dividing the above equation by taking the limit as approaches zero, and ing the change in linear momentum to the resultant force yields

F = p = in'(v2

—

[3.6.93

v1)

relate the resultant moment about a fixed point 0 (or center of mass) tc the change in angular momentum in a similar fashion. Denoting by and t2 the vectors from 0 to the centers of the entry and exit ducts, the sum of moments 0 becomes We

= H0

Example

I

X V2 V2 in(e2 X

—

X v1)

[3.6.101

Find an expression for the speed of' the rocket fired vertically shown in Fig. 3.13.

Solution We assume that the motion takes place along a straight line and drop the vector notation.

recognize the term as the thrust. The external force F acting on the rocket is due to three sources: the pressure differential between the exit nozzle and air, friction due air resistance, and gravity. The force due to the pressure differential can be expressed as

F=peA

[CI

where Pe is the pressure difference at the exit nozzle and A is is the the cross-sectional cross-sectional area area of of the

exit nozzle. It is customary to assume that the rate of change of mass is constant and therefore the mass of the rocket can he written as

m(t) =

—

bt

[bJ

is the initial mass and b is the mass loss rate. It follows that thVrei = force of gravity is simply Introducing this and Eqs. [aJ and [hj into obtain, for a rocket moving vertically, where

JJ Figure 3.1 3

,n(t)v(t) + m(1)g = The term F1 is referred to as the static 1/u-us!.

=

+ bVrci

hVrci

.

The

we

[cJ

1 73

CONCEPTS FROM ORBITAL MECHANICS: rilE Two BODY

3.17

\Ve We next next calculate the velocity of the rocket when all the fuel is expended (burnout). (burnout). the mass of the the propellant propellant by bym1,. m1,. and and using using the the constant mass loss rate iii = b. it

the rocket a time of = in,,/b to usc up all its fuel. We write Eq. [cJ in terms of iii(I) iivide by in(r). in(r). which yields

dv(t) = peA + +g m(t) di di m(t) ,n(t)

Idi

both sides by di and and noting from Eq. Lhl that di = —din/b we obtain \ din

(11'=— (IV=—

I

/

in in

tel

1,

differential form. form. SO SO that Eq. [cJ can be in differential sides in in the the above equation are expressed in to yield I/

+

V = V0 — I

L()

is

\

iii in

b ,ij

ifl() m0

)ln

+

ill — lfl() fll()

b

If I

the above equation mU) in the the initial velocity. Substituting the expression for iii(i)

V

= V( —

p,,A

I

h

\

)In )ln

ifl() — ?fl()

hi —

gt

[gi

into this equation. = velocity is found by substituting the final velocity the fInal However, [gJ may look misleading at first. because of the two negative terms. However. is always less than unity and its natural logarithm is a negative quantity, thus giving the are derived by assuming its upward velocity. It should also he noted that Eqs. [f I and the gravitational constant remains the same during ascent of the rocket. A more accurate for the velocity should include the change in the gravitational attraction as a result altitude, as well as the air resistance.

3.7

CONCEPTS FROM ORBITAL MECHANICS:

Two BODY PROBLEM problem in mechanics is that of the motion of two celestial bodies moving the gravitational attraction of each other. The class of problems is referred to The two bodies are depicted in Fig. 3.14. Typical examthe two body pmblein. pmblein. The include the earth—moon, the earth—satellite, and the sun—planet pairs. For our we assume that the only force acting on each body is the gravitational atdue to the other. We ignore the gravitational force exerted by other bodies. Fc example, if we are considering the earth—satellite problem. we ignore the effects the moon and the sun. This assumption is valid only as long as hc distance hethe satellite and earth is much smaller than the distance between the earth and or any other planet. or the sun. The only force acting on the particles is the gravitational attraction in the

F

= Gm1m7

[3.1.11

_______

174

CHAPTER 3 • DYNAMICS OF A SYsTEM OF PARTICLES

r

in

r,

r1

0

FIgure 3.14 where r = r2 — r1 is the vector connecting the two masses. Newton's second for each particle is iii1r1 = F =

m2r2 =

—F

Gm1 m7

= —-

r3 r3

r [3.7.2a,bJ [3.7.2abJ

To manipulate the equations of motion, consider the center of mass of the particle system. From Eq. [3.2.2] we have that FF1 r1 + ,,z2r2 = (in1 (in1 + + 1fl2)FG ifl.2)FG

[3.73J [3.73J

Adding Eqs. 13.7.2] and differentiating Eq. [3.7.3 with respect to time, we conclude that 1

mIr1 + nilr1

=

+ 1fl2)rG = 0

13.7.41

We can see that the center of mass of the system of two particles does not have an acceleration, an expected result because we originally assumed that there are forces acting externally on the system of two particles. Equation [3.7.4] hints that one may simplify the governing equations. by considering the motion of the two masses relative to each other. Actually, we are more interested in this relative motion of the masses than the absolute motion of each Let us express the equations of motion in terms of the vector r connecting the two bodies. Dividing Eqs. [3.7.2a] and [3.7.2b] by ru1 ru1 and and in2, in2,respectively, respectively, and subtracting Eq. [3.7.2a] from [3.7.2b], we obtain

=r=— Gm r-

i

G (in G(m

Gm2 G 1112

r-

r-

If we introduce the gravitational parameter relative motion as the differential equation

++ ni,)

= G(,n1 +

r+4r=o r+4r=o

we we

[3.7.5J can express the

[3.7.61

3.7

ORBITAL MECHANICS: THE Two BODY BoDy

CONCEPTS

1

The question may he asked as to what information was given up when we re-

the two equations of iliotion motion into a single equation. The answer is: the location

md the motion of the center of mass of the two body system. Using Eq. 13.7.61 one one cannot analyze the motion of the center of mass. En many orbital memeproblems. one is not interested in the location of the center of mass but in relative motions of the two masses. Also, in several celestial mechanics applicathe mass of one body is much smaller than the mass of the other, and one can ignore the mass of the smaller body when calculating the center of mass. The of the larger body is assumed to be the center of mass of the two body system. made this assumption when when he he stated stated his his laws laws of' of planetary motion. Equation [3.7.6] represents represents aa central central force force problem. problem. as discussed in Chapter 1. conveniently analyzed by polar coordinates. Attaching a set of polar coordinates center of mass and separating the motion into the radial and transverse cornwe write the radial and transverse components of the equation of motion

P

=—

—

rO

[3.7.7a,bI

+ 2H) = 0

2 showed that the angular momentum associated with a central ,.20 = h is constant. That is, the angular momentum is conserved and that r20 the center of mass is conserved. As we will see later. this is the mathematical of Kepler's second law. Note that the angular momentum considered here the apparent angular momentum. Because the acting force is conservathe energy of this system is also conserved. Indeed, introducing the expression = h/r2 into Eq. [3.7.7a] we obtain Example 1

. I

=

13.7.81

I..

can he integrated to yield the energy integral +

Ih—

—— — == E E ==Constatil Coristaill

2r

r

or

— = E = constant constant r

—v — 2

[3.7.9]

=

in

+ h2/r2 h/r2 isisrecognized recognizedas asthe thesquare squareof ofthe the total total velocity. velocity. This expresof the energy integral is for the relative motion with respect to the center of However, because the center of mass executes motion with constant velocity. energy associated with the center of mass motion can be absorbed in the constant the right side of Eq. [3.7.9 1. (This is entirely analogous to Eq. [3.4. 10]. which the kinetic energy for a system of particles.) It should also be noted that the and momentum momentuni expressions are per unit mass. We can calculate the energy integral from the expressions for the kinetic and energies. The kinetic energy has the form

T=

I

in1 lii)

—

2 in

i

m2 + m2

v•

V

=

I

1?Zj ,m

—

2 in + im i

., (r +

•

-,

)

=

I —

in1 inim

2 in +

+

—5-)

[3.7.10]

75

176

CHAPTER

3

.

DYNAMICS OF A SYSTEM OF PARTICLES

The potential energy is obtained by integrating Eq. [3.7.11. The gravitational force is rer, so that dr = c/re,. + rdOe0. We select the datum in the radial direction and r so that position for the potential energy as the distant stars, r() r0 = r

F • dr = F•dr V = V = — —1I = —J — J

Gm1 in,

r •'dr dr

r

= J

G,n1 in2 , —dr =

r-

=

Gin1 in, —

r [3.7.111

in2) yields Adding Eqs. [3.7.101 and [3.7.111, [3.7.111, and and dividing dividing by by m1m2/(m1 + in2)

Eq. [3.7.91.

As Chapter 1 mentions, many celestial mechanics problems require higher accuracy than most engineering problems. Many times one manipulates very large numbers and obtains a very small number as a result. For this reason. it may be necessary to use more than four significant figures. Here are pertinent celestial data: Universal gravitational constant: 6.668462(10

Example

3.6

h1)

s2 m3fkg s2

Mass of earth: 5.977414(1024) kg.

Average radius of earth: 6,378.1 km

For earth, p. =

Mass of sun: 1 .987323(10

n13/s2,

30) kg

A particle is thrown from the surface of the earth with a very large initial velocity, as shown in Fig. 3.15. Calculate the maximum height reached by the particle.

Solution Because the initial velocity is very high. we will dispense with standard projectile motion equations, and we will use concepts from celestial mechanics. We denote the initial comActually, at the onset onset of of motion motion the the .v x and z axes are and ponents of the velocity by basically the radial and transverse directions. When the particle reaches its maximum height. the component of the velocity in the vertical direction is zero. The velocity now is purely in the transverse direction. We denote its vertical velocity by w. We have conservation of energy as well as conservation of angular momentum about the center of the earth. Denoting the maximum altitude reached by L. we describe the conservation of angular momentuni momentum by =

7n(R

V

Figure 3.15

+ L)w

[a]

_________________

___________

3.8

THE NATURE OF THE ORifiT

R is the radius of the earth. This yields an expression for the velocity at maximum by =

14'

Ru,

Ibi [bi

R+L

The components of the kinetic and potential energy are =

T2 =

)

1

I

\2

/ R

I

= =

+

R + 1. )

(

,,i

?11

V1 =

—

V2 = V2

—

R

Gni,

[ci [cJ

117

R+L

The work-energy theorem gives I

,

Gm,n Gme'n

,

,n( v; + tv:)

= =

R

—

1

2"

/

R

I

R+L)

Gnl!fl

Idi Id]

R+L

represents a nonlinear equation for L. To obtain an approximate, hut still still meaningful. we note that L is much less than R. Defining by e = LIR. we linearize the terms L as

/ I

R I

(R ± L)

= =

/

I

\-

I

\I +e) e) (i I

I

—2e) (1— 2e)

R-1L R+L

-

e)

lel

roducing Eqs. [e} into Eq. Eq. [dj and solving for e we obtain If I 2 the maximum height reached is the

L=

[g]

It It is interesting interesting to to note notethat thatififv1. is set set to to zero, zero, [hat [hat is, is, if if the particle is launched vertically, vertically. height it reaches becomes

I

,

•,

2G

v: 2g

[hj [hi

is the same equation obtained from projectile motion analysis. The value for the height Eq. Eq. [gj is higher than than the the value value for for LL in in Eq. Eq. [hi, [hi, an an expected result. When we consider the total motion of the body. we see that it has some sort of orbital although the particle cannot complete an orbit hut falls to the earth. This situation is encountered in the launching of spacecraft. when the booster rockets malfunction fail to place the spacecraft into a sufficiently high orbit.

NATURE OF THE ORBIT 3.8 TwTwNATURE 3.8 now wish to determine what kind of motion the masses execute with respect to other. We will accomplish this by solving the equations of motion. Eqs. [3.7.7],

177

3

1 78

•

DYNAMICS OF .4 SYSTEM 0F PARTICLES

and by making use of the energy and momentum integials. Consider the equation of motion in the radial direction. &.j. [3.7.7a1. This equation is nonlinear. It turns out that Eq. [3.7.7] can be put in a simpler form if we introduce introduce the the transformation

=-

IL

[3.8.1]

F.

and replace differentiation with respect to time with differentiation with respect to the transverse variable 8. Noting Noting that the rate of change of the transverse angle can we express the time derivative of r as be written as 0 = h/r2. we

dr dO didO dO

di

dr /1 Ii

=

[3.8.2]

do

with respect to 0 Taking the derivative of u with

dii

=

1

dO

dO

so

1 dr

d

[3.8.3]

—--i---r— dO

,

that (11-

di

ii d r

=

h/

,

dii \

111.1

I = -h---do, dO dO,i I

r—dO

[3.8.41

In a similar fashion. we write the second derivative of,• with respect to time as

IP

d /du

= —h— di

Ii d2udO u dO d2u , d2 u d2 d2 u Ii = —hir = —h = —h —h dO- di do- rdO—

dO

[3.8.51

The remaining terms in Eq. [3.7.7aJ can be expressed in terms of u as 0 =

h

= hu

= /alC

13.8.61

which, when introduced together with Eq. 13.8.51 in Eq. [3.7.7a1. yield d2 u + I +

[3.8.7]

ii =

This relationship is in the form of a second-order differential equation with con-

slant coefficients. It is considerably simpler to solve than Eq. [3.7.7a]. The solution of Eq. [3.8.7J can be written as u(O) =

+ cr cos(O cos(0

[3.8.8]

—

and 0, depend on the initial conditions. This solution is used to find r(O). We are interested in the value of the magnitude of the radial distance r as a function of the transverse angle 6. as well as r as a function of time. The energy integral in Eq. [3.7.9] can be expressed in terms of u as where

/

I

—Ii 2

\dOJ

(

I

+ —hic 2

—

=E

[3.8.9]

__

3.8

Ti-w

OF THE ORBIT

This expression can also he obtained by integrating Eq. [3.8.7] over 0. Recall

the energy expressions here have units of energy per unit mass. We explore the between the energy E and the amplitude a. Introducing Eq. [3.8.8] into [3.8.9] and carrying out the algebra. we obtain sin(O

+

—

+ a cosW — 6i)]

= 11a2/12 2L

—

h-j

+ a cos(O — Of)]

—

[3.8.10]

=E E

we can rewrite as

a=

1+ /7

Introducing the variable

e=

2E/i 2E/i2

[3.8.11]

Iithe

above equation can be expressed

a=

[3.8.121

substituting Eq. [3.8.12] into Eq. [3.8.8]. we write the solution for u. u(f1) u( 6) =

follows that the solution

r(6) = r(9)

/2-

[I + cos( 6

—

0)j)J

13.8.13]

is /72

1

u(0)

1

13.8.141

— —

1+

14] represents represents the equation of a conic section, with the value of Equation [3.8. [3.8.14] referred to as the eccentricity of the section. determining its shape, with the vals s = I being the critical values. If < I the conic section is a :iosed one, in the form of an ellipse, as the values thr r(0) remain finite. The conic becomes a circle when = 0. When I, r(O) I, r(O) assumes assumesinfinite infinitevalues, values, in a parabola or hyperbola. The term is called the semilatus rectum; it de±.nes the size of the conic section. The semilatus rectum is a function of the angular The initial condition is usually selected as zero. Using Eq. [3.8.121. we can write the energy as 1

1 e

S

E=

1

-

—

[3.8.15]

2/i-

) can make use of the nondimensional ratio E = = C- — 1. We next consider the different types of conic sections. When < I.I. the the tratraetory is an ellipse, with the center of mass as a focus of the ellipse, as shown in 3.16, 3.16, and the motion is periodic with a closed orbit. An ellipse is defined by its o foci, its Selnima/or selnima/or axis axis aa and and semiminor semiminor axis b. The point Ofl on the ellipse closest a focus is called pericentei; pericentei: and the point farthest away, apoceniei: The pericenter md apocenter are known as the apsides (plural of apsis) of the ellipse. An apsis is as the point where dr/dO = 0. This definition of apsis is valid for any conic

we

*

_)

179

180

CHAPTER CHAPTER 33 IIDYNAMICS DYNAMICS OF .k SYSTEM OF PARTICLES

F.

Flight path angle

Ui

Apocenter

Pericenter

a

FIgure 3.1 FIgure 3.166

Elliptic orbit

section and is not limited to ellipses. Considering the motion of the body on the orbit.

the angle y between the transverse and tangential directions is referred to as the theflighr flighi path angle. This angle is easier easier to to measure measurethan than other other orbital orbital quantities quantities and, and, hence. hence. is useful in determining the nature of an orbiL orbi1

When dealing with two masses of comparable magnitude. magnitude, the focus of the ellipse is the center of mass of the two body system; the ellipse describes the trajectory of the distance between the two masses. For sun—planet or earth—satellite type problems, because of the very large difference in the mass ratios, one can assume that the larger mass is located on a focus. The situation is difiBrent when the earth—moon pair

is considered. The mass ratio between the earth and moon is 81.30, and the mean distance between the two bodies is km. The center of mass of the earth— moon system lies 4670 km from the center of the earth, roughly 2/3 the mean radius of the earth. It has become customary to use different names for the apsides when motion about the earth and sun are considered. Table 3.1 gives the names used for the apsides and the names of commonly used coordinate systems when the earth and sun are assumed to be on a focus.

Denoting the distances from the focus to the pericenter and apocenter by arid respectively, and without loss of generality selecting the initial condition as O — = 0, so that 0 is measured from the pericenter, we obtain h2

Table 3.1

i

h2

1+6

/L

16

[3.8.1 6a,bJ

Nomenclature for celestial motion problems

Types of Orbit

Apsis Near

Apsis Far Away

Generic ellipse

Pericenter

Apocenter

Earth as focus

Perigee

Apogee

Geocentric-equatorial or perifocal

Sun as focus

Perihelion

Aphelion

Heliocentric-ecliptic

Coordinate System

3.8 From

NATURE OF 11W ORRI'I'

these equations. and noting from simple geometry that a = for the semimajor axis a

=

i we consider Eqs.

1/12

1

J+ + j J

11,21

1

2/1L

J

h[ = —1

1

ILL

1

+ r11)/2. we

[3.8.17]

J

15],we we can can express express the energy as 13.8. 15], [3.7.91 [3.7.91 and 13.8.

E =

—

[3.8.18]

=

that the semimajor axis is a measure of the energy in the system._In a similar — 62. Also. one can show that the semirninor axis is of length b = Eqs. [3.8.1 [3.8.16] 6] and and [3.8.171. we obtain

= a(1 We

= a(1 a(1 + e)

—

13.8.19]

can then then write the equation for the ellipse as a(1 — a(1 —s) s) I1

[3.8.201

+

At this point, we consider Kepler's laws of' planetary motion. The astronomer Kepler. after studying the observations of Galileo and Tyco Brahe and the ol planets, developed the following laws for the motion of planets (his calof ulations were based on the motion of Mars) in the solar system. Law 1: Each planet revolves in an elliptic orbit about the sun, with the sun at one focus of the ellipse.

Las' 2: Equal areas are swept per unit time by the line joining the sun to the Lai' planet.

Law 3: The squares of the periods of the planets are proportional to the cubes semimajor of the semi major axes of the ellipses. The first two laws of Kepler were published in 1609 and the third in 1619. These predate Newton's Jaws. (first published in 1687) by over 70 years. Kepler did know about Newton s laws of motion when he formulated his laws. Newtons

laws of motion and his law of gravitation are based in part on Kepler's laws. It mteresting that we traditionally study Newton's laws Iirst and then Kepler's laws. laws. more natural order but in reverse historical order. Kepler's first law becomes correct when the mass of the smaller second body planet in sun—planet. moon in earth—moon, satellite in earth—satellite) is nefrom fL. fL. To To demonstrate demonstrate Kepler's second law, consider Fig. 3. 1 7 and the area swept by the position vector r as

Area A

= f 1A = J

= =

j

=

j

write an expression

13.8.211

area! rate—the area swept per unit time—is found by differentiating the area respect to time. Noting from earlier that r20 = 1, = constant, we obtain the

1

81

1 82

CHAPflR 3 • DYNAMiCS DYNAMicS OF A

OF

r-dO

r + dr

Figure 3.17 areal rate A

=

/7

= constant

[3.8.221

which is proof of Kepler's second law. The period of the orbit can be obtained from Kepler's second law by noting that the period T is equal to the area of the elliptic orbit divided by the areal rate, = ITab h/2

A

A

[3.8.23) 52

where irab is the area of the ellipse. Introducing b = tion and noting from Eq. [3.8.17] that /1

=

— E2

1—

into the above equa-

[3.8.241

we obtain — S2

I 1

= 27r I—

[3.8.253

—

This is the mathematical statement of Kepler's third law. The special case of 0 corresponds to a circular orbit. Circular orbits became very important in the second half of the 20th century, as satellites were placed into circular orbits around the earth. A circular orbit in which the relative position of the satellite with respect to the earth does not change is called geosynchronous. As of this writing there were several communications satellites in geosynchronous orbi! about the equator. In a circular orbit, the body has constant velocity, which can easily be obtained from the energy expression. From Eq. [3.8.15] for a circular orbit, E = —1i2/2h. We verify this by writing the energy expression as 21,2

2

2

so that we obtain the velocity in circular orbit, —

13.8.261

as

13.8.271

We next consider the case when the eccentricity is greater than or equal to 1, resulting in an open orbit. When e = 1 the orbit is parabolic, and when > I the orbit

3.8

THE NATURE OF THE ORBIT

hyperbolic. Parabolic orbits rarely exist, as they describe a limiting case between an open and closed orbit. Fig. 3. 18 depicts a parabolic orbit. From Eqs. (3.8.16], as approaches 1. the sernimajor axis becomes longer. When = 1. the orbit becomes an open one. The orbit equation, setting E = 1 in Eq. [3.8. [3.8.14J, 14J, becomes

13.8.281

1 +cosO

The expression for energy for a parabolic orbit gives insight into the nature of orbit. Setting s = I in Eq. [3.8. [3.8.15] 15] yields, yields, for the energy, E 0, so that we can E

[3.8.291

0

2

The velocity required to achieve a parabolic orbit and escape the gravitational

attraction of the large mass is called the escape From the above equation, a spacecraft to transfer from an elliptic to a parabolic orbit it must have the escape '.elocity, denoted by as Ve

=

[3.8.301

minimum value of the escape velocity is at the perigee. Many orbital maneuvers that involve a change of orbits are carried out at the perigee. It is interesting to note The

happens as the satellite keeps moving in a parabolic orbit. As r approaches from Eq. [3.8.301 the velocity approaches zero. This implies that it' a body achieve a true parabolic orbit and there arc no other disturbances acting on it, t body would eventually come to a rest. Physically this never happens. When the eccentricity is greater than 1. the resulting orbit is hyperbolic. Space-raft launched on interplanetary missions arc given hyperbolic orbits. Fig. 3.19 a hyperbolic orbit. From Eq. Eq. 13.8. 13.8.15]. 15]. the the energy for a hyperbolic orbit is

P

Di rectrix

Parabolic orbit

Figuro 319

Hyperbolic orbit

183

I DYNAMICS CHAPTER 3 CHAPTER 3 I DYNAMICS OF A

1 84

O' PARTICLES O' PARTICLES

Table 3.2

Types of orbits and associated eccentricities

Type of Orbit

Eccentricity

Circular

Energy Ratio

0

Elliptic

—1

0 <

Parabolic

—1

1

< E' < 0

I

Hyperbolic

0

E' >

1

0

positive, and the energy expression can be written as

-

E

>0

13.8.31]

implying implying that as r approaches infinity, the speed of the body is not zero, hut approaches a finite value. This value is called the hyperbolic excess speed. Table 3.2 summarizes the relation between orbits and eccentricity. T\vo essential problems in space mechanics are 1.

2.

To place a satellite in a desired orbit and to change the path of the satellite from one orbit to another. To determine the orbit of a satellite or a planet from measurements of its location and motion.

Analysis of both subjects is very lengthy and beyond the scope of this text. For more details on these subjects. the reader is referred to texts on orbital mechanics, in the following examples we illustrate two simple approaches.

Example

3.7

I

At the At the burnout of a rocket, rocket, the the following followingdata dataarc arc givcn:r givcn:r = 65(X) km, v = 9750 rn/s, rn/s, flight path angle y = 16.4°. Has the rocket achieved an earth orbit or will it crash into the earth?

Solution The orbit of the rocket is shown in Fig. 3.20. We need to find the properties of the orbit

and determine if the distance to the perigee is less than the radius of the earth. The angular momentum and energy per unit mass are Ii

= rvcos y = (6.5 =

E=

V

— —

2

f.L

X

106)(9.75

= (9.75 x 10)

r

16.4°) = 6.0797 X 1010 m2/s

X

—

3.986 X

6.5x10

2

= —1.3792 —1.3792 X

m2/s2

[a] [bJ

We next calculate. the eccentricity. Wc have 2Eh2

p

x

2(— 1.3792 x

(3.986x 10i4)2

=

= —0.64172 0.64172

Ic] Id

so that from Eq. [3.8.1 5} so

=

—

0.64172 = 0.59857

[dJ [dl

38

185

THE NATLTRE OF THE ORBIT

1• 1'

Burnout 01

Impact

Figure 3.20

from Eq. [3.8.181. [3.8.18]. the the semimajor axis is =

=

—3.986 X

2(—l.3972 xx JØ6) 2(—l.3972 Jo6)

2E

= 14.450x

106111

I.] Fe]

iM from Eq. [3.8.19bj, the distance to perigee is

rp =

a(

I

14.45 X 106(1

—

—

0.59857) = 5.8008 X 106m

If] I:fJ

unfortunately, unfortunately, is is smaller than the radius of the earth. Thc The rocket crashes into the earth !ts way hack.

UOHMANN TRANSFER A Ho/unann transfrr is a common way of moving a spacein a circular orbit to another circular orbit that lies on the samc plane. The transfer DIves applying two impulses to the spacecraft. Denote the radii of the first and second orby r1 and r2, as shown in Fig 3.2 1. The first fIrst impulse changes the initial orbit to an elliptic called the iransfc'r or/ni, orbit, whose semimajor axis is of length (r1 + r2)/2. Consider the case when r2 > r1. The location of' the first impulse becomes the perigee of transfer orbit (apogee when r2
[a] [al

— —

2r1

Immediately after the impulse is applied to move the spacecraft into the transfer orbit, = = r1 and 2a = r1 + so that denoting the velocity at perigee by v,,. the energy becomes I.-'

—

= —

p —

2

——

r1 + '2

FbI [bi

Example

3.8

186

CHAPTER 3 • DYNAMICS Oh A SYSTEM OF PARTICLES

L\

Final orbit (r2)

r,

Iran stèr orbit

Initial orbit (r1)

FIgure 3.21 which can be solved for

Hohmann transfer

as

- /2,u V—I V r1 so

r1+r2

+12

[cJ Ic]

that the impulse needed at the first step is =

—

=

in1

—

2r2

r1 + r2

[dl

The second impulse is applied at the apogee of the transfer orbit. The velocity at apogee before the impulse can be found from the angular momentum conservation

[.1 1.]

= where

denotes the velocity at apogee and has the form 2r2

r1

= —Vp —Vp = —

r1 +r,

r2

r1 +r2 ?•I +r2

Ifi

Because the desired final orbit is circular, the velocity in orbit should be V2

so

=

1g1 Lgj

that the second impulse is of magnitude =

—

[hi

Va)

=

Consider the issue of what happens when the impulses are not applied at the proper places in the orbit. For the first impulse. the location of the impulse defines the perigee

3.9

ORBITAL ELEMENTS

orbit. This This location the transfer orbit. location is not critical. The critical impulse is the second. If the impulse is not applied at apogee. the resulting orbit will not be circular.

3.9

ORBITAL ELEMENTS

in the previous section, we showed that the general solution of the orbital problem ofthe the motion, motion, angular angular momentum and a conic section. Based Ofl the two integrals of axis a and eccentricity Given energy. we developed two quantities. :nitial conditions, one can determine the nature of the orbit by calculating a and s. The orbital problem as given in Eq. [3.7.61 requires six initial conditions or constants :f integration, :•f integration,as asititisisaasecond-order second-order differential differential equation equation in in three three dimensions. dimensions. Given a set of initial conditions, OflC can solve Eq. [3.7.61 and obtain a solution. This solution, while being a mathematical answer. (toes (toes not not describe describe the the properties properties the orbit. It is desirable to express the position and velocity of the body in orbit in erms of parameters that lend themselves to a physical explanation. It turns out that in addition to a and that orient the orbit in space and can find four parameters in the body on the orbit. Of these four parameters. two describe the orientation of' the :cbital plane. one describes the orientation of' the orbit on the orbital plane. and the body on on the orbit and introduces time. The six parameters are one locates the body one to as the orbital parameters or orbital elements. We begin by describing the position of' of the the body body on the orbit. Our primary inis in closed orbits. The angle 6, referred to as the true anomaly, is measured one of the foci of the ellipse. It is the angle between the line joining the focus the body and the semirnajor semimajor axis, relative to the pericenter. A more convenient aav of measuring the orientation of the particle is to introduce the variable eccentric denoted by and shown in Fig. 3.22. We draw a circle of radius a with the same center as the ellipse. The eccentric anomaly is measured positive counterfrom the semimajor axis. It is the angle between the semirnajor axis and line that connects point P' and the center of the ellipse, in which point P' is the ztersection between the circle and the vertical line perpendicular to the semimajor This line goes through point P. with P denoting the position of the mass.

V

P1

P

-V

Orbit

FIgure 3.22

187

188

3•

DYNAMICS OF A SYSTEM OF PARTICLES CHAPTER 3• DYNAMiCS CHAPTER

Denoting the coordinates of P by x and v, from Fig. 3.22 we can write

= r sin 0 v v =

x = aa cos

=

a cos

+ r cos (9

Recalling the equation of the ellipse x2/a2 + v2/b2 =

13.9.1 a,b,cJ

and that b = and introducing Eq. [3.9.1 a] into the ellipse equation. we obtain

rsin0 = rsinO

a

1. 1,

—

e2.

[3.9.2]

—

Combining Eqs. [3.9. Ic] and [3.9.21 gives

=

r2 cos2 + r2 cos200 = a( 1 sin2 U +

f 3.9.31 3.9.31

cos

—

which is a simpler equation to deal with than Eq. 13.8.201. In a similar fashion, we relate the true and eccentric anomalies. Introducing Eq. [3.9.3] into Eq. [3.9.IcI, we obtain —

—

cosU =

r

(1

1

[3.9.41

—

Using Eq. [3.9.4] with the trigonometric identity tan2(6/2) = we write

(1

—

cos (fl. cos (9)1(1 + cos OL

13.9.51 1

from which we conclude that /(9\ tan(— 2

1E

=

[3.9.6]

tan(—) 2

We now introduce time into the formulation. Differentiating the two expressions for r in Eqs. [3.9.3] and 13.8.20] [3.8.20] and and using the expression for the momentum integral h = r20, we obtain

11 =

r =

aE(1 (1

—

a(l

+

—

a(l —

E2)

2

smU =

2

r

(1

[3.9.7a,bJ 13.9.7a,bI

si n Si

13.9.8]

—

introduce it into Eq. [3.9.7bJ, and equate the result with Eq. [3.9.7a]. We express as dVdt and collect all the terms involving on one side, with the result (1

—

=

/1

a-

[3.9.91

_€2

angular velocity, i'elocitv, in in which which r is the period of' the Define by n = 27r/T 27r/T the mean angular orbit given by Eq. (3.8.25J, T = 2ira3"2/p)'2. Recalling that h = — E2), we I

3.9

ORBITAL ORBITAL ELENIENTS

;an express the right side of Eq. [3.9.9J as

di

17

._____ =

a

,idt

13.9.101 [3.9.101

—

It remains to integrate Eq. [3.9.91. To this end, we select the initial conditions

:: coincide with with the the position position of the body at the pericenter. At that location = 0, the time is selected as defined as the time q/ of pericenter pericenter passage. passage. We then

ft (1 —

=

JO

fldt

[3.9.11]

i.:th the result n(t

— ET)

=

—

[3.9.121 13.9.121

This is known as Kepler ss equatioii. equation. We define the mean anomaly anomaly .14. .14. by by

= n(t

—

13.9.13]

that Kepler's equation becomes

At At

=

—

[3.9.14] 13.9.14]

The mean anomaly basically describes the angle that would have been described the

position of the particle if the motion was uniform, as in a circular orbit

mean angular velocity n. Using Kepler's equation, one can calculate the mean At if the eccentric anomaly and the eccentricity are given. Often, one and can measure At more easily than the other parameters. so one needs and This requires the solution of Eq. [3.9.141, which usually is calculate numerically. We have defined a constant of integration that permits us to find the exact locaof a mass on the ellipse. We have yet to locate the ellipse on the orbital plane. the plane of the orbit as the x'v' plane. the xv x axes, axes, which which are are along along the and sernirninor axes of the ellipse, are obtained by a counterclockwise about the z' axis. The angle of rotation is denoted by w (not to be confused angular velocity) and is called argument oft/ic pericente!: Next, we orient the orbital plane with respect to an inertial frame. We need to Next, both the origin and orientation of the inertial frame. There are several choices, on the bodies being analyzed:

planetary motion, the sun is selected as the origin and the reference frame :alled heliocentric-ecliptic. The coordinate axes, denoted by XYZ. are selected that the XY plane coincides with the plane of the orbit of the earth around the (the ecliptic plane), shown in Fig. 3.23. The XY plane is also referred to as the plane, and the positive Z axis is referred to as the noii/i polar axis. The I is selected as being toward the vernal equinox, also called the first point of On the first day of autumn, the line joining the centers of the sun and earth nts toward the vernal equinox. -

For

1

89

1 90

CHAPTER 3 •

OF A SYSTEM (iF PARTICLES

z First clay

First day of spring of' spring

of summer

Y

First day of autumn

x

First day of winter

(Seasons are for Northern Hemisphere)

Verna' equinox direction

Figure 3.23

Heliocentric-ecliptic coordinate system

Because the XYZ plane is assumed to be inertial, the vernal equinox is assumed

to be fixed as well, even though it is actually precessing very slowly. The causes of moon, asaswell wellas asthe thefact fact that that the earth the precession are the disturbing effects of the fllOOfl. is not a perfect sphere. both of which add terms to the equations ol motion describing the motion of the earth with respect to the sun. The period of precession can be shown to be, and has been observed as. about 26.000 years. When the vernal equinox was first defined, it was pointing toward the constellation Aries. At the writing of this text, it was pointing toward Pisces.

For satellite motion, a geocentric-equalorial or a perifocal coordinate system is often used. We describe here the geocentric-equatorial system. The origin of the 2.

coordinate system is selected as the center of the earth. The Z axis is selected towan± the North Pole. The XY plane is the equatorial plane. and the X direction is toward the vernal equinox.

We are ready to orient the orbit. It turns out that this orientation is accomplished by a 3-1-3 coordinate transformation. Consider an inertial frame as shown pushed is about about the the Z axis. The rotation angle is in Fig. 3.24 and a sphere. The first rotation is The ascending node called the longitude of the ascending node and is denoted by is defined as the point at which the orbital plane crosses the fundamental plane with a northerly velocity, that is. moving in the positive Z direction. The descending ,iode is delined as the point at which the orbital plane crosses the fundamental plane. moving in the negative Z direction. The line joining the ascending and descending nodes nodes. Denoting the rotated coordinates by X'Y'Z'. the line of is called the the line line of of nodes. is nodes is the X' axis, and it describes the intersection of the orbital plane and the fundarnental plane. Historically, the term line of nodes originated from this coordinate transformation sequence. is The second rotation is about the K' axis. The rotation angle. angle. denoted denoted by by /.i. is called the orbital inclination. The resulting coordinate axes are denoted by x'v'z'.

3.9

ORBITAL ELEMENTS

z

Orbit plane

Y

Fundamental plane

x Vernal equinox

direction

Figure 3.24 the x'v'

Line of flodeS

Orbital parameters

describing the orbital plane. The angle i hence describes the between the fundamental plane and the orbital plane (or the angle bethe Z and z' axes). The angular momentum is in the z' direction. Finally, the : axes arc rotated about the z' axis by the argument the pericente': denoted to orient the orbit on the orbital plane. The remaining remaining three three orbital orbital parameters parametersui-c are the semimalor axis a. eccentricity passage In many cases, and especially in problems involving orbits the sun, the mean anomaly is used instead of The transformation from the three displacement and three velocity coordinates the body on the orbit to the orbital parameters is a unique transformation. That ziven all three displacement and velocity components at a certain instant, one detennine the orbital parameters, and vice versa. For the ideal two body probprobwhere there are no disturbing effects of other bodies, the orbital plane as well the size and orientation of the orbit are all constants. However, when there are functions, such as gravitational forces of other bodies and atmospheric the right side of' Eq. [3.7.6] is no longer zero. In addition, the equations in this section assume that the bodies involved are perfect and homogespheres. Any deviation in geometry from a homogeneous sphere results in gravitational terms. Furthermore, the rotational motion of the bodies inalso must be considered. The analysis of these effects is beyond the scope of text.

plane

191

.

192

CHAPTER 3

Example

Consider Example 3.7 and calculate the time elapsed alter burnout until the rocket crashes.

3.9

DYNAMICS OF A

OF PARTICLES

Solution We will make use of the eccentric anomaly to solve this problem. First, we find the transverse angle 0 the rocket makes at burnout. Denoting this angle by 0, we use Eq. [3.8.20] to find it •

as

cos0, =

a( I —

14450(1 14450(1

—

— —

0.598572)

—

6500

6500(0.59857)

= 0.71266

[.1 [a]

from which we get we get

01 = cos'(0.7 1266) = 0.7775 1 rad = 44.5480

[bJ

From Eq. [3.9.6] we have the eccentric anomaly at burnout burnout as

- /i-€

—

\/ 1.59857 tan (22.274°) = 0.50 112(0.40962) = 0.20527

[ci Ed so that

['I

= 2tan '(0.20527) = 0.40492 rad = 23.2000 We now make use of Kepler's equation, Eq. [3.9.12]. Sening the initial time as using Eq. [3.8.24J. [3.8.24J. we can solve for time as —

1= and /—

v/i

-

/ (1.445 x 'V

1014 3.986 xx 1014 3.986

=

= /—

n

=

X

—

sin

[.1

= 2.7513 X i03 s. Denoting by

11

the time

at burnout and using Eq. [C]. we obtain =

13 xx = 2.75 13

— 0.59857 sin(0.40492)) —

465.3 seconds

[1]

We next find the above parameters at impact, and denote them with the subscript 2. find the angle 02 we make use of Eq. [a], with r replaced by R = 6378. 1 km cosf97 =

a( II—— a(

e2)

—

R

14450(1 — 0.598572)

—

6378.

6378.1(0.59857)

1

= 0.75821

Noting that the rocket crashes in the fourth quadrant. we can find the angle

0, =COS =COS

+

= 5.4226rad =

3

[ii

[11

from

10.69°

[bJ [hI

The corresponding eccentric anomaly is found from

-

= 2tan 2tan'

1—e

f(h

= 2tan'((0.501 12)(—0.45900)) = —0.45216rad = —25.910°

Ii]

3.10 by

PLANE KINETICS

RIGID BODIES

to the the value we obtain adding 27r to = —0.45216 —0.45216 ++

+ 360° =

= 5.83 5.83 lOrad = —25.9

334•Ø90

[jJ

Introducing the mean anomaly to Kepler's equation, we obtain for the time

t) = t.) -

/

= 2.7513 x

—

-

—

O.59857sin5.8310)

[ki

= 16.762scconds 16.762 scconds

Subtracting t, from t2. we obtain the time the rocket stays in orbit before it crashes as —

3.10

it ==16,762 16.762 I

—

4.527 hr 465 ss = II 6,297 6,297 ss == 4.527

III

PLANE KINETICS OF RIGID BoDiEs

section we analyze the kinetics of rigid bodies that move on a plane. In this section and the ones following it constitute a review of sophomoredynamics. This review is primarily aimed at refamiliarizing the reader with basic concepts. as the developments in the following chapter are described terms of particles or plane motion of rigid bodies. A detailed analysis of the tinematics and kinetics ol' of three-dimensional rigid body motion is carried out in this

Thapters 7 and We denote the inertial frame by XYZ and consider the XY plane as the plane of zs'tion. The moving reference frame .vvz. obtained by rotating the XYZ frame about ze Z axis by 0. is attached to the body. The angular velocity and angular acceleration one component each:

a=aK=OK a=aK=6K

[3.10.1J [3.10.11

begin with defining the center of mass. Consider a system of N particles, in Fig. 3.1. The center of mass is denoted denoted by by the the point point GGand anddefi defined tied by [3.2.21. A rigid body can be considered as a collection of particles in which the of particles approaches infinity and in which the distance between the mdimasses remains the same. As N approaches infInity, each particle is treated as i mass clement, ni, dm, and the summation in Eq. 13.2.2] is replaced integration over the body. We then define the center of mass G as We

=

!fl ifl

rd,n

13.10.21

bodV

r is the vector from the origin 0 to differential element din and in = I

dm

13. 1 0.3] 13.10.31

mass of the body. Considering Figs. 3. 1 and 3.25. 3.25, for a system of' particles write r1 = r(-; and I'or r(-;++ p, p, and I'or a rigid body. body. rr = rG + p. Introducing Introducing this this term in

fle

193

194

OF PARTICLES

CHAPTER 3 • DYNAMICS ø'

z.Z

(IF

Y

0 A.

Figure 3.25 Eq. [3.10.2], we obtain 1

rG=—

1

rdm =

—

fli

(r(;

I

=

+

.?hody

1

pdin

+—

In

[3.10.41 13.10.41

- body -thod y

leading to the conclusion that

pdm =

0

[3.10.5]

Ibody Ihody

about This equation indicates that the weighted average of the displacement vector about the center of mass is zero. The center of mass is a very important quantity, as its use simplifies the analysis of rigid bodies considerably. To see this, let us write the rigid body equivalent of the combined equations of motion. Considering the differential element and its equation of motion

adm = (IF adin

[3.10.61

where dF is the total force acting on the differential element. We write for the entire body.

dF=F

adnz =

13.10.7]

Jhody

ibodY

in which F is the resultant of all forces. This resultant contains contributions only from the external forces, as the internal forces cancel each other. Introducing Eq. [3.10.4] into Eq. 13.10.71 gives the translational equations of motion of a rigid body. that is. Newton's second law for a rigid body, as

[3.10.81

lnaG = F

The derivation of the translational equations of motion above is valid for the general motion of a rigid body. For plane motion, we define the moment of all forces acting on the body about the center of mass as N

M(; =

x F1

= .body

pXdF=M(;K

[3.10.9]

3.10

PlANE KINETICS PlANE KINETICS OF OF RIGrn BODIES

MG is the resultant moment about the center of mass. Next, introduce 13.10.61 to the above equation. Before doing that, we express the velocity and of the differential element as

v= V(;+wXp

a=aG+axp—cop

[3.10.101

proceed, so that

MG

= -fh ody

pXdF=j

pxadm body

=

(ar; + a X p — wp)dm p X (a(; body

[3.10.111 first term inside the brackets vanishes because of the definition of the center of and the third term vanishes because of the cross product. To evaluate the middle consider that the cross product a X p is on the plane of motion, perpendicular p. and it has the magnitude Hence. the cross product between p and a X p is to the plane of motion, parallel to a. and with magnitude ap2. Thus. [3.10,111 can be written as

The

MG =

p X (a X X p) p)dm din Jhody

=

a,)2 dm K = I( body

pdmK [3.10.121

a Jhody

The integral on the right side of this equation is not dependent on time or any variable, so that it can he evaluated independently of the angular acWe define it as the mass moment of inertia of the body about the center mass, and denote it by 1c;. so that =

p2 c/ni

[3.10.131

J body

The mass moment of inertia is a property of the body itself: it is a measure of

the mass of the body is distributed about an axis passing through the center of s and perpendicular to the plane of motion. We discuss ways of calculating the moment of inertia in Chapter 6. Figure 3.26 gives the centroidal mass moment nertia for two common shapes. Appendix C gives a more comprehensive list. Combining Eqs. [3. 10. 1 2] and [3. 10.13], we obtain the rotational equation of 1(;a = MG MG

Thin disk:

= ink2

FIgure 3.26

Slender rod: J(; =

13.10.141

195

196

DYNAMICS OF .k SYSTEM OF PARTICLES CHAPTER 3 SSDYNAMICS

This equation, together with the two equations in Eq. [3.10.8]. [3.10.8], gives the three equations of motion for the plane motion of a rigid body. The rotational equation of motion can also be represented in terms of the angular momentum. Indeed, we define the angular momentum about the center of mass as

HG=(

() X

=

Jbody

+ (0 (0 XX p)din p)din =

X

X X (to X

Jhody

- I body

[3.10.151 13.10.151

considering the the integral integral in in Eq. Eq. [3.10.121 [3.10.12] and the definition of the For plane motion, considering mass moment of inertia, we can write

HG=I

[3.10.16)

(In, K (In, K = l(7wK

J body

and the rotational equation of motion can he written as

[3.10.17J

HG HG = MG MG

or

just as the translational equations of' niotion can he written as

p=

[3.10.18J [3.10.181

F

in which p = mVG is the linear momentum of the body. It should be noted that while in vector vector ftwni form holds foT we derived Eq. 13.10.171 for plane motion, this relationship in the general three-dimensional rotational motion of a body. As discussed in there is is debate debate about about whether whether Eq. Eq. 13. 13. 10. 10.171 171 is a derived relationship, or a stated II. there law of motion.

The equations of motion can be illustrated by means of' equivalent and resultant force diagrams. as shown in Fig. 3.27. The suni of the external forces is equal to the resultant, which is equal to the rates of change of the linear and momentum. Up to now, we considered the angular momentum and sum of moments

the center of mass. It turns out that under certain circumstances, it becomes more convenient to write the rotational equations of motion about another point. For ample. such a case arises when motion about a fixed point is considered. To analyze the moment equation about an arbitrary point, consider the third part of Fig. 3.27 and sum moments about an arbitrary point D. We have Mo

= IGa +

or

+ r(;/fl r(;/fl XX?flaG !naG

ML) =

F1 F'

F,

External input input

Figure 3.27 FIgure

Resultant

Accelerations

[3. 10.191 [3.10.191

3.10

197

PLANE KINETICS OF RIGH) BoDws

Ct). (X

I

/

/

I)

'I

Figure 3.28

Rotation about a fixed point fixed point

which d is the perpendicular distance from D to the the acceleration acceleration vector for the of mass G. A special application of the above equation is for rotation about a fixed point. Consider Fig. 3.28. where the rigid body rotates about point point 0. 0. It follows that the ol the center of mass can he expressed in terms of the distance between 0 and and G. which we denote by b. as

aG = aG

).

[3.10.201

bw2e,, +

The component of in the normal direction does not affect the moment about Hence, Hence, we can write the sum of moments about 0 as Mo

.

+ ar;, =

is

iflba = (I(; (J(; + inh)a inb)a = IGa + ?flfra

I0cr

[3.10.211

the mass moment of' of inertia inertia about about point point 0. 0. and we have written the parallel axis described in detail in Chapter 6. This theorem essentially relates the mass of inertia of a body about the center of mass G and another point D by IL)

= 'G 'G + mci2

[3.10.221

d is the distance between the two points. Another way of describing the inertia of a body is by means of the the radius radius qf of gyration, denoted by K. K. such that 1flK2. The radius of gyration is often used when dealing with complex bodies, it is usually listed as a property of a body.

When solving plane kinetics problems. one should select the moment center that the number of reactions to he solved and the total number of calculations a minimum. The procedure in solving plane kinetics problems is the same the procedure outlined in Chapter 1 for fir particle particle motion. motion. The The only only difference is that must invoke the rotational equations of motion.

t:± OA is of length 0.4 iii and has a mass of 3 kg. while rod AB is of mass 12 kg and length The two rods arc released from rest from the confIguration configuration shown in Fig. 3.29. 3.29. Find Find the the accelerations accelerations of both rods at this instant. There is no friction at point B.

Example

3.10

198

CIIAPTUR 3

•

DYNAMICS OF A SYSTEM OF PARTICLES

/

O.4rn

I

3kg

A

I

0.8 m

im

B

Figure 3.29 Solution first isolate the two bodies and draw free-body diagrams. shown in Fig. 3.30. Note tha1 triangle DBA is a 3-4-5 triangle. The next task is to write the force and moment balances. For rod OA it is convenient to write the sum of moments about point 0 as (counterclockwise positive) We

[a]

M0 == 10a = —0.2(3g) +

and so The force balance equations merely give relations for the reactions for they are not of much use. unless one wishes to calculate those reactions. Introducing the + = mL2/3 = 3(Ø•42)/3 = 0.16 kgm2 = expressions = 1(3 + rn( Eq.

=

—

[b]

O.6g

O.ôm

I—

1)

A

Im

G •1'

0.8 ni

12g

A'

ov

A

0

B

A

IN (a)

Figure 3.30

/

(b)

/

3.10

PLANE PLANE KINETICS KINETICS OF' OF' Ricm Ricm BODIES

rod AB. there are three unknown unknown reactions. reactions. A1, A1. A1. and N. Considering that the of forces in the horizontal vertical directions and the sum of moments about the of mass involve all three unknowns, we prefer writing the moment balance about D. a point about which which N N and and A1 A do donot notexert exertaamoment. moment. To To do do this, this, we we need an for the acceleration of point G. We will obtain the acceleration of point G by writing a relative acceleration relation between points A and B and then a relation B and G. Consider points A and B first. Because motion is just being initiated, the angular velociFor

.:tf OA and AB are 1)0th zero. Also, point B moves in the horizontal plane. We hence. write

of points A and B as O.4a1j = aAj = 0.4a1j

aA

a,1 =

id

u± the relative relative acceleration acceleration relation becomes becomes u± =

+ a2 XX

X (0.61 (0.ôi ++ 0.8j) =

£1AJ = a8i +

(a,1

+ 0.6a2j

—

[dJ [d]

we conclude that

a8 =

= 0.4cr = 0.6cr, i

[eJ

0.8x2

Next. write the relative acceleration relation between points B and G aG

a8 + a7 XX rGIB =

X (0.31 (0.3i ++ 0.4j 0.4j) = 0.4cr2i + 0.3cr2j

+

ifi Ifi

We are now in a position to write the moment equation about point I). Because a1) has in both x and y directions, we write Eq. [3. 10.21] in vector form as

x maG maG = (—0.3 XX 12 12 gg — — 0.6 = lGa2k + rG/n X

,

is the mass moment of inertia of rod AB. AB, having the value of 12 x 12/12 = = 0.3i — 0.4j m. Substituting in these values to the above equation. .irn2 and

la2k + (0.3i

—

2(0.4a21 + ().3cr2j) 0.3cr2j) = (—3.6g 0.4j) X I12(0.4a21

—

[hi

reduces to

+ 1.92)a, =

(1 +

4cr2

= —3.6g

Equations [h] and [ij are two equations with the unknowns f Eq. [e] we have cr1

[ii

—

a2, and A.,. A,. From From the first

l.5a2

[II

Eq. [bI can he written as 0. 24cr2 = ().4A 0.24cr2 0.4A

—

0. 6g 0.6g

[ki

Eq. [k] by 1 .5 and adding it to Eq. [ii. we obtain for a2 —4.5g

4.5 X x 9.807 9.807

= 4.36 =

4.36

asa1 =

= 10.12 rad/s

15.l8rad/s. 1.5cx2 1.5cx2 = 15.l8rad/s.

[II

199

200

CHAPTIR 3

3. 1 1

•

DYNAMICS OF A

OF PARIlcIEs

INSTANT CENTERS AND ROLLING

An important concept associated with the plane motion of rigid bodies is that of instant center. At any instant oF the motion. there exists an axis perpendicular to zern Pelocirc, t'elocirc, such that the plane of' motion, called the instantaneous axis ot zero can he viewed as rotating about that axis at that instant. The intersection of this zero velocity, or insrar: centerqf qfzero and the plane of motion is called the instantaneous center instar instant center center of a rigid body is located by visual inspectioa ce,itei: In general. the instant ce,itei: In generaL To establish the location of the instant center. one needs to know the velocities two points on the body. If the velocities are not in the same direction, one draws lines, beginning at the points at which the velocities are known and perpendicular the velocities. Their intersection is the instant center. Figure 3.3 1 illustrates. If velocities of the two points are in the same direction, one again draws two lines: joining the points at which the velocities are known and the other joining the of the velocity vectors (drawn to scale). Their intersection gives the instant as shown in Fig. 3.32. In Chapter 7. we prove the existence of the instant center fr plane motion. It should be noted that while the instant center has zero velocity, its location 10.21 acceleration isis not not zero. zero.Hence, Hence,Eq. Eq.[3. [3.10.21.. every time instant is different, and its acceleration the moment equation about a fixed point, cannot he written about an instant center. over another body or over An interesting case of motion is that of a body rolling rolling over fixed surface. For rolling to take place between two bodies, a continuous sequence points on one of the bodies must be in continuous contact contact with with aa continuous continuous seq p/acc'. the coacactiug coacactiug bodcocicinuous ccin(acC cc otpoin(s on (lie other Oody. ies must have smooth Contours. with nojunips. Rolling can occur in a variety of ways. as Chapter 7 will show in detail. In this The surface surface can be section we consider roll of a circular body over a fixed surface. The planar or curved. Consider Fig. 3.33. We denote by w = —wk the angular velocity Let n denote the unit vector The contact point is denoted by C, with velocity of perpendicular to the plane of contact. The kinematic relation describing rolling is

V(fl V('fl

[3.11.11

= 0 V

'V

(1)

'I

I

'I

C'

A

FIgure

3.31

FIgure

3.32

Figure

3.33

3.1 '1

1

I

201

INSTANT CENTERS AN!) ROLLING

if the contact point has a finite velocity along the plane of contact, the motion to as roil wit/i slip. If the contact point has zero zero velocity, velocity, the the motion motion is is to as roll wit/iou! without 511/). sli/). Whether Whetherslipping slippingexists exists or or not not depends depends on the forces on the rolling bodies, as well as the friction between the rolling bodies. Maththe no slip condition is represented as

"C =

0

[3.1 1.2J

Because the velocity of the contact point is zero. the contact point can be treated

an instant center. center. However, the acceleration of the contact point is not zero, an instant there is slipping or not. The contact point approaches the plane of contact, has contact, and then it moves away. The constraint associated with the contact is to the contacting points only during the instant of contact. Consider rolling without slipping and a point P on the body. The velocity of a P on the body can be expressed as Vp = VG+(*) X rp/c

13.11.3]

We write the above equation for the point of contact C. with the result VC

O=v(;+wXrc7G

[3.11.41

we can use to express the velocity of the center of mass as

v(;=o)XrG/c=

[3.11.5]

This equation can be physically explained by noting that the point of contact C is the center.

£ sphere of mass in and radius r rolls without slip inside a circular curved surface with radius ?. as shown in Fig. 3.34. Obtain the equation of motion as a function of 0.

0 I'

I---

/

R

C

FIgure 3.34

Example

3.11

202

CHAPTU3 3• DYNAMJCS • CHAPTU OF OF

.4

OF PARTICLES

0

FIgure 3.35 Solution The free-body diagram of the sphere is given in Fig. 3.35. We write the force balance

normal and tangential coordinates, thus = mg sin

= N

—

—

F1

rngcost9 =

EM

P

where p is is the the radius radius of curvature for point G G and and F1 F1 is is the the friction friction force. The center curvature is the center 0 of the curved surface and the radius of curvature is constant has the value p ,. The moment balance about the center of mass is R — r. positive)

IGa = Fjr

[4

Because we assume assume rolling without slipping, the speed and and tangential acceleration of

center of the sphere are

=

wherc w is the angular speed of the sphere and where

[4

=

Ut)

a

ci. ci. Solving lbr for F1 F1 in in Eq. Eq. [a] and

the expression for fbr aGt aG1 in Eq. [dJ. we obtain = mg sin fIfI

—

inra

which. when introduced into Eq. [cJ. yields

(JG engr sin U )a = ingrsinO (JG + mr4)a

To obtain the equation of motion. we need to express the angular acceleration of sphere in terms 01 0. We make use of the property that the center of mass of the sphere also he considered as moving about the center of curvature of the surface, so that we can = rw = —(R

—

r)O

[4

3.1 2

203

ENFR;Y ANI) ENER;Y

we obtain

[hi we write the equation of motion as introducing introducing Eq. [hi into Eq. Eq. 1l1'l. fI. we

(I(, + ;nr2)(R

—

+ ingi2 sine)

")O

Iii

0

Note that whenever one assumes roll without slip, one must check the validity of assumption. This can he done by calculating the magnitude of the friction force comparing it with the maximum value of the friction force. force. It' If this maximum is one in muu St St redefine rede Ii nethe theproblem problemasasaaroll rollwith with slip slip problem. problem.

3.12

ENERGY AND MOMENTUM

kinetic energy of a rigid body is defined as kinetic T =

If

v•vdm

[3.12.11

J bodV

introduce the expression of of the velocity velocity in terms of the center of mass to this we obtain

(vG+oXp)'(vG+coXp)dnl

T =

[3.12.21

J

body

J

bod v

first term on the right side of this equation gives mv(; v(;/2. v(7/2. The The second vanishes due to the defInition of' of the the center center of mass. To evaluate the third term. note that the cross product of and p is a vector of' of magnitude magnitude pw. so that The

X p)dni X p).(w p)•(w X (w X J body

=

j

p2w2

(1,11

13.12.3]

body

the kinetic energy ol' a rigid body undergoing plane motion can be written as 71

=

mVGV(; +

[3.12.4]

the rigid body is rotating about a fixed poirn point C. its kinetic energy becomes

T == T

I

Icw

[3.12.51

204

CHAPTER 3 •• DYNAMIcs D\'NAMIcs OF OF A A

OF PARTICLES

The kinetic energy can be written in this form about the instant center, as the center has zero velocity. When the body is rotating about a point C that is moving. Eq. [3.12.5] is not valid. The gravitational potential energy for a rigid body is V

=

13.12.61

in which is the perpendicular distance between between the the center center of of mass mass and and the the datum line. All the other potential energy expressions are the same as we have seen them before hetbre for for particles. The work-energy theorem discussed in Chapter 1 is valid for afl types of bodies. An interesting property of bodies undergoing rolling without slipping is that the friction force does no work. This can be easily shown using the definition of won as a time integral. Because the the friction friction force force isis always always applied appliedto toaapoint pointwith withzeitb velocity, the the power er of the force force PP = F1 is zero. Hence, the work done, which is the integral of power over time, is zero. We obtain the impulse-momentum relations the same way we did for particles. by integrating the equations of motion over time. Doing so between two time and 12 gives gives

+J

F

di =

mV(;(t2)

JGW(tI) + JMG ('1

13.12.71

[3.12.8J 13.12.81

The integrals The are known integrals in in Eqs. Eqs. [3.12.71 and known as the impulse and and [3.12.81 are the

angular impulse, impulse. respectively. A very interesting application of the impulseimpulse-

momentum relationships is in cases where the integral of the sum of moments or or the the

sum of forces vanishes. In such cases, we have conservation vi linear momentum or COflSerwzIiOfl of angular Coflsert'aliofl angular momentum. momentum.

Example 3.1 2

A solid uniform sphere of mass ni m and radius R is placed on top of a fixed sphere of the

radius, and and it is slightly tipped (Fig. 3.36). Find the value of the angle begins as aa function function of the coefficient of friction frictionia.. coefficient of M•

at which sliding

Solution The displaced position of the sphere and its free-body diagram are depicted in Fig. 3.37. We

denote by t9 the angle made by the line joining the centers of the spheres and the vertical. B? we denote the angular displacement of the top sphere. The friction force F is less than for no slipping, where N is the normal force. When there is slipping, F = MN. The speed of the center of' the sphere is given by

= 2R0

I.] [a]

=

(bJ LW

and for no slip

3.12

ENERGY AND MOMENTUM

3

F

t

?1

Figure 3.37

Figure 3.36

for the no-slip case the two angles are relatcd to each other by

Ic] id to consider no slip, and noting from Appendix C that the centroidal mass moment

of a sphere is 'G =

2mR2/5.

we can write the kinetic energy about the point of

contact

18

[di

=

= =

T=

= 'G + mR2 is the mass moment of inertia about the point of contact. Considering

;enter of the lower sphere as the datum. the potential energy has the form [0J

V = 2mgRcosO The sum of forces in the normal direction is mgcos6

F =

—

N =

IflV(,

—

m(2R0)2

2R

f)

= 2mRO2

If I

The sum of the kinetic and potential energies is constant. Therefore. 14

that when 0 =

0.

[g]

+ 2mgRcosO = E

0 is also equal to zero, we can evaluate E as

[hi 1k]

E = 2mgR

Eqs. [gi and (hJ, the relation between U and 0 can be expressed as

=

—

cos0)

[11

Eq. [ii into Eq. [fI we obtain an expression for the normal force N in terms of 0

V

=

mug

cos 0 —

2mn RU2

= ing cos 0 — 2niR

5g —

cos 0)

= ing

7( l7cosO

—

10

[II

205

206

CHAPTIR 3 • DYNAMICS OF

SYsTEM OF OF PARTICLES

Summing moments about the center of mass of the sphere we obtain

FR =

=

which relates the friction force F to /.

We

ma, =

[k]

mR2çli mR2çb

sum forces in the tangential direction, thus —F +

[II

Noting that a1 = 2R0 and introducing Eq. 1k] into Eq. [1] we obtain 4

'p

2mR() = 2mR()

..

..

0A =

[J

'Si?

we write write an expression for the friction force in terms of 6 as Combining Eqs. [k] and [ml we

F =

2

.4

mnRth

=

••

inRO =

4

2

sin A = 7mngsinO sinO 7mg sin 0

The instant slipping begins. F = pIN. Hence. considering Eqs. [II relation for U at the instant slip begins. =

—

10)

and

1.1 In], we obtain ml,

[o]

This can be solved for 0 given a value of p.. It should he noted that the solution is independent of the mass and radius of the sphere. we can solve Eq. 101 using For very small values values of of p. p. ..wc usingasmall a smallangles anglesassumption assumption of of sinU sin 6 1. which yields the result 0 = 7 p.12. For larger values of the coefficient of friction. cos U one can use a numerical approach. For example. example, when p. = 0.5, 0 can be found to he 6 = 41.5°. The case of having a rough contact between the spheres, that is. when the coefficient to] by p. and setting p. = of friction is p. = is of interest. Dividing both sides of Eq. 1°] we obtain

—

10) = 0

10/17. or 0 = 53.97°. We observe from Eqs. [jJ and [o] that which has the the answer answer cosO cos0 = 10/17, Eq. basically implies that the the normal normal force force isis zero. zero. Therefore, Therefore,atatU0 = 53.97°. the top sphere loses contact with the lower sphere arid and goes into a free-fall mode.

REFERENCES Rate. R.B.. I). 1)1). I). Mueller, and i.E. White. Fumidainentals Fuiidainentals o/Astrodvnanizcs. o/Astrodvnanncs. New York: Dover Publications, 1971. Greenwood. D.T. Greenwood. D.T. Principles Principlesof ofDynamics. Dynamics. 2nd ed. Englewood Cliffs, Cliffs. NJ: Prentice-Hall, 1988. New York: York: John John Wiley. Wiley. 1976. Kaplan. M.H. Modern .Spacecra/i Spacc'cra/i 1)vnwniL's !)vna,nics Control. New Meirovitch, 14. L. Introduction IntroductiontotoDvnwnies Dvnwnirs and and Control. Control. New New York: York: John John Wiley, Wiley, 1985. Methods oJAnalvtu a! a, Dynamics. New York: McGraw—I-li II, 1 970.

207

HOMEWORK EXERCISES

EXERCISES

3.2

:

The blocks shown in Fig. 3.38 are released from rest. Find the acceleration of each block. The pulleys are massless. A chain of length 0.7 m is at rest over two slopes, as shown in Fig. 3.39. First, assume there is no friction between the chain and slopes. If the chain is released from rest, which end of it will rise, and what will its speed be when it reaches the top? Then, assume that there is friction and calculate the minimum amount of friction necessary to prevent the chain from moving. The pulley system in Fig. 3.40 is released from rest with the spring unstretched. Find the acceleration of each block. The pulleys and the cord are massless and frictionless.

7 I

0'

'S. F

/ Figure 3.39

Figure 3.38

k

Figure 3.40

208

CHAPTER 3 • DYNAMICS DYNAMICS OF OF A A SYSTEM OF OF

PARTJCLFS

SECTION 3.3 4.

The three masses in Fig. 3.41 are connected by links of negligible mass. A force of magnitude F is applied as shown. Find the acceleration of the center of mask as well as the acceleration of the individual masses.

SECTION 3.4 5.

Find the direction of sliding in Problem 2 using energy methods for the (case no friction).

SECTION 3.5 6.

7.

A ball is released from a height Ii from a plane. The coefficient of between the ball and is e. Show that the time it takes for the ball t = + e)/(l — e). Two spheres. made of the same material, the lower with radius 2b and of radius b, b. are dropped from a height Ii. as shown in Fig. 3.42. Assuming centers of the spheres lie on a vertical line and a]] all collisions between the spheres. and the ground are elastic, find the maximum height the upper sphere reaches after impact. Assume that the lower sphere collides with the ground first and then collides with the second sphere.

t

8.

Four small spheres of equal size and weight are aligned in a straight line and spaced equally (L). as shown in Fig. 3.43. Sphere A is given an initial velocir v along the line. The coefficient of restitution e = 0.8 0.8 is the same for all spheres. Find the velocities of the spheres after 5 collisions have taken plac& and the position of sphere A.

lii F

L

L

450

450

11

FF1

Figure 3.41

Figure 3.42

209

EXERCISES ExERcisEs

0.4 iii 0.4 iii )C k e

I-.

in

-.... Shall

-_.--_.--/

Lu e A

C

B

L

I

FIgure 3.43

I

1)

— I.51T1

L—H

FIgure 3.44

Figure 3.45

Figure 3.44 shows five pendulums of equal mass and length arranged so that takes the pendulum at someone takes they touch each other when they arc at rest. If someone pendulum at the very right will the very left. swings it, and lets it go, only swing out. If one takes N pendulums and swings them. N pendulums swing out. Explain why. Also, explain why. if Iwo two of the pendulums are glued together and then released, the above phenomenon is not observed. In a pool game. a player must hit the cue ball and bounce it from a wall (e = 0.9),

into the pocket. as that the cue ball hits the 8 ball head on and the 8 ball falls into illustrated in Fig. 3.45. Find the direction in which the player must hit the cue ball and location of cue ball on the table. A mass in is attached to a massless rod. The mass is at rest in the position shown in Fig. 3.46 when it is hit by another body of mass 2iii 2i,i and velocity v (e = 0.8). Find the angular velocity of the rod immediately after impact.

SECIION 3.6

L Using the results of Example 3.5, calculate the altitude reached by a vertically tired rocket at burnout. Half of the rockets weight is its propellant. Assume that = constant. constant. = 0. A rocket is fired vertically such that dm/dt and Vrd are both constant. Derive expressions for the velocity and position as functions of time given that initial value of the acceleration is zero. A bucket of mass 1.5 kg is filled with 3 kg of water and is being pulled up a .1 well. The tension 1' in the rope is kept constant at T = 3() N. The bucket has a hole in it which lets the water leak out at a steady rate, and the bucket becomes empty in 38 seconds. Find the speed of the bucket at this instant. and cross-sectional area A is moving with speed v0, If A spacecraft of' mass as it encounters a dust cloud, as shown in Fig. 3.47. The dust cloud has densit p. As the spacecraft moves inside the cloud, dust begins to stick on its cross section. Derive an expression for the speed of the spacecraft as a function of time. Assume the dust offers no resistance to the spacecraft. 1

L

0i

I

iii

2ni

FIgure 3.46

210

CHAPTIR 3

•

DYNAMICS OF A SYSTEM OF PARTICLES

Oball I

dust cloud

V

44

(i,)

0

•

. S

.S

S

S

S U

S U

S

•

S

U S

S

S

S U

U S

S U

•

S

•

•S

•S

•

•

•S

•

I.

S

.

I. •

S U

•S

•S

•S

U

•

U

S

U

S

•S

•

•

.•

.•

•

S

•

S

S

S

S

S

•

S

U

S

S

S

S

S

S

S S

p

g

•

(b)

FIgure 3.48

Figure 3.47

pong ball alloat by blowing on it, as shown a 16. A performer wants to keep a ping poflg Fig. 3.48. Given is that when he blows on the ball, the opening of his mouth ol area A. the speed is v. and the density of the air is p. Find the largest mass the ball that he can keep afloat. 17. The cart in Fig. 3.49 is of mass 1250 kg and is moved by the action of a jet. The water jet is of radius 0.08 m, and it squirts water with the speed of 2C' mis. Find the speed of the cart 3 seconds after aller the water jet is turned on. Ttt cart has a coefficient of friction of = 0.08. Hint: First develop an fc)r the fhrce acting on the cart as a function of the flow rate.

SECTION 3.7

Calculate the gravitational potential energy of a small mass inside a from sphere of mass and radius R (Fig. 3.50). The distance of' the mass from Consider the the sphere sphere as consisting Hint. Consider center of the sphere is Ii, with Ii < R. Hint. an infInite number of' thin spherical shells and calculate the potential energy each shell. 19. A tube of length 500 km is dug inside the earth from one city to another. shown in Fig. 3.51 (not drawn to scale). The inside of the tube is frictionless. A mass is released from rest from one end of the tube. Find the maximum the mass attains in the tube.

18.

500 km

300 /1

Figure 3.49

Figure 3.50

Figure 3.51

EXERCISES

kLTION 3.8

ill

Calculate the period of the earth's rotation around the sun using Kepler's third and Eq. [3.8.251 and compare with the value given in Chapter 2. Calculate the altitude of a satellite in geosynchronous orbit (geo) about the equaThen, calculate how much time it takes for an electrical signal to travel beween the equator and the satellite. Determine the minimum number of satellites n geo required to make it possible for any two people on earth to have a teleconversation via satellite. A satellite is launched into earth orbit. Burnout is at the perigee at an altitude of :24 in, at which point the satellite is parallel to the earth's surface and has .ii speed of v 90(X) mIs. Find the properties of the orbit. satellite of mass 800 kg is to be placed at circular orbit around the earth of 4(X) km. Find the energy required. as well as the period of the orbit. A spacecraft is in elliptic orbit with = 0.5 and = 2R, where R is the radius :f the earth. It is desired to change this orbit into a circular one with radius Find the impulse required to accomplish this maneuver. Then, calculate the of' the resulting orbit if the impulse is applied incorrectly at 6 = 185°. Calculate the speed and flight path angle a rocket must have at burnout at altitude km so that it can achieve an orbit with = 5400 km and = 0.7. A meteorite is at a circular orbit around the earth with an altitude of 3500 The meteorite collides with another another meteorite, and in doing so loses 3 of its kinetic energy. Find the properties of the orbit after this colisjOfl.

3.9 satellite is in orbit and its position and velocity at a certain instant are measured be r = 30001 + 2000J + 6500K km and v = 5222! — 4000J + 6000K rn/s

the origin is at a focus. Find the orbital parameters associated with this :rbit. Tbit.

A satellite is in an elliptic earth orbit with = 0.6 and and = 8R, where R is the radius of' the earth. Calculate how long it takes for the satellite to go from = 150 to 0 = 135°. Compare your answer with the period of the orbit.

3.10 Consider the mechanism shown in Fig. 3.52, consisting of a rod of mass rn and length L and disk of mass rn and radius R = L/4. The mechanism is held in place with the the pin pin joint joint at at 00 and and the the string. string. Suddenly, Suddenly, the string breaks. Find the angular acceleration of the rod at that instant if The disk is welded to the rod. h. The disk is attached to the rod with a pin joint.

21 2

CHAPTER 3 • DYNAMICS

OF

OF

&

/

B 111,

171

0

,,z. L

0

ni, L

300

A

150

--

Figure 3.53

Figure 3.52

30. The two bars of equal length and mass are connected by pin joints and they B. as shown in Fig. 3.53. Find the are supported by a string at point B, acceleration of each bar when the string breaks. 31. Figure 3.54 shows the schematic of a car. Explain why the body of the car counterclockwise when the car is accelerated. 32. The disk in Fig. 3.55 rotates with constant angular velocity w = 5 rad/s. pin attached to it moves in the slotted bar OA. Bar OA has a mass of 10 kg centroidal mass moment of inertia of 1.4 kg m. Determine magnitude of force exerted by the pin on rod OA when rod OA makes an angle of 10° with tir = 0.15 horizontal. The coefficient of friction between the pin and the slot is •

•

.

.

S

SECTION 3.11 33.

A bar of mass ni and length L (Fig. 3.56) is connected to a disk of mass 2iii radius R = L/2. The assembly is released from rest with t9 = 300. Given radius R friction between the disk and the surface is sufficient to prevent slipping, the angular acceleration acceleration of of the the disk disk at at this thisinstant1 instant.

34. Consider Fig. 3.53. Let the string be broken and the following parameters = 0. 1 m/s downwards. Find the instant center and anguir ply: LL = 0.6 m. ply: velocity of the link AB. 35. The arm AOC rotates with angular velocity 3 radls ccw, with point 0 stationary. as shown in Fig. 3.57. Gears A. B. and C are of the same radius R. Use centers to find the angular velocity of' gear B if (a) gear I) is fixed, and (b) gez = 2.5 radls. D is not fixed but is rotating clockwise with '4

c Iii—.. Iii—.. U..) U..) 0-

G

l.2m

Rear wheel drive

Figure 3.54 Figure 354

Figure 3.55

213

HONWWORK EXERCISES

D

ni, L

ml.

C

(1)

m,L

A

3.56

Figure 3.57

Figure 3.58

Consider Example 3. 11. Find the minimum value of the coeffIcient of friction that will prevent slipping as a function of 0.

3.12 Consider the bar attached to a disk in Problem 35. which is released from rest U = 300. Find the velocity of the center of the disk when the rod becomes horizontal.

rods of equal length and mass are connected by a pin joint and they are at rest. as shown in Fig. 3.58. An impulsive force F is applied at point A perpendicular to the line AB. Find the angular velocities of the rods immediately after the impulse. Consider Example 3. 12, and assume that friction is sufficient to prevent slip at all times. If the spheres involved were of different diameter, would the angle at which the spheres lose contact be different than in Example 3.12? h. Consider now that the cylinder is rolling over a cylinder. How would the final results change if (ci) the sphere and cylinder have the same radius and (b) the sphere and cylinder have different radii? Two

B

chapter

ANALYTICAL MECHANICS: BASIC CONCEPTS

4.1 a

INTRODUCTION

chapter and Chapter 5 introduce analytical techniques for describing the motion namical systems. The dynamical system is considered as a whole and scalar

such as energy and work are used. Constraint forces threes and moments are differently than in Newtonian mechanics. Constraint forces that do no work appear appear in the formulation, and they are accounted for by appropriately appropriately sethe variables used to describe the motion. Sometimes, one may need to find the magnitudes of the constraint lhrccs. This can he accomplished by calculating magnitudes of' the constraint k)rccs after the problem is solved, or by leaving in the thesystem system formulation formulation by means of Lagrange multipliers. The aches described in this chapter are analytical approaches and they are based principles of variational calculus. Appendix B provides aa more more detailed look principles. Generalized coordinates, which do not necessarily have to .* sical sical coordinates, are used used as motion variables. This makes the analytical apmore flexible than the Newtonian. as the Newtonian approach is implemented physical coordinates.

We derive the analytical equations of motion in this chapter for particles for plane motion of rigid bodies, though these equations arc valid for threerigid body motion and deformahie bodies as well. Chapter will with with D'Alembert's principle and Lagrange's equations for the general three-

a

motion of rigid bodies. One question often asked is whether it is more convenient convenient to to use use aa Newtonian Newtonian or an analytical one when obtaining the equations 01' ol' motion. motion. There There is no set to this question, with the possible exception of dynamical systems consisting interconnected components. When the number of' coordinates needed to

215

216

B&siu CoNcEPi's CoNcEPis

CHAPTUR 4 • ANAJXTICAL ANAJJYFICAL

than the the number number of ofcornponents, components, it is usually describe the system is much less than

able to use analytical analytical techniques. techniques. When Whenamplitudes amplitudesof ofreaction reactionthrees threesare aresoughL soughLLt is usually better to use a Newtonian analysis. Looking at the problem from both t± is Newtonian and analytical points of view gives one more insight and a better Newtonian standing. Analytical techniques use scalar scalai functions like work and energy in the formslation, rather than vector quantities. While this approach makes a lot of sense. experiences of dynamicists in recent years have shown that vector approaches bined with analytical techniques are more desirable when modeling complex tems. One advantage of a vector approach is that it can be implemented on a computer more readily.

4.2

GENERALIZED COORDINATES

A system of' N particles requires 3N physical coordinates to specify the systems sition. Consider an inertial coordinate system and let the the vector vector r1 = r1(x,, the mapping of the ith particle particle in in this this coordinate coordinate system.1 system.' We We express as (Fig. =

Xj1

l,2....,N i = 1,2,...

+ Vij + z,k

[4.2.11 [4.2.11

The 3N coordinates required to represent the system span a 3N-dimensional

which is called the the configuration space spaceof' of'dimension dimensionix ix = 3N. In many cases. we will soon see, it is more advantageous to use a different set of variables than physical coordinates to describe the motion. This approach is analogous to thai using different coordinate systems that we saw in Chapter I. We introduce a set variables q,,, related to the physical coordinates by vi

= =

= =

=

•.

. ...' Un)

We will refer to a set of variables that can completely describe the position We a dynamical system as generalized coordinates. The space spanned by the genera— ized coordinates is the configuration space. As an illustration, consider the spherica pendulum in Fig. 4.2, whose length can change. The motion of the pendulum can described by the Cartesian coordinates x. v. and z. or by and q3. where = L describes the length of the pendulum. and = 6 and q3 = describe the a noninertial coordinate coordinate system system is is used, used, one one has to include the variables describing the motion of the

frame in the set of coordinates that describe the motion, unless the characteristics of the reference frame are as known.

4.2

"ii

.

GENERALIZED COORDLNATES C00IWLNArEs

Pfl1

V

—

V

yi

Figure 4.2

A system of N particles

A spherical pendulum whose Iengfh changes

The choice of L, 0. and 4) as generalized coordinates is equivalent to spherical coordinates. The two sets of coordinates are related by

= ql COS

y=

q3 = L cos 0 sin 4)

z=

— qi —

cos q3 =

—

sin

Sin q3 = L sin 0 sin 4)

L cos

If the length of the pendulum is constant, = L = constant, we do not need it as a variable; = 0 and qi = 4) arc sufficient. If we use the coordinates .. and z to describe the motion, we have to relate relate them them employing the constraini

+v+

=

14.2.41

= constant

relations, such as the one in this equation. indicate that the generalized are related to each other, and that the system can be analyzed by a smaller of coordinates. The double link in Fig. 4.3. where the lengths of the rods are requires at least two generalized coordinates to describe the configuration two rods. One can conveniently select them as the angles 01 and 02.

F

P lU,, '-'2 — '-I

FIgure 4.3

A double link

217

218

CONCEPTS

MECHANiCS: ANALYTICAl1 MECHANICS: CHAPTER 4 • ANALYIICAL

We hence need to distinguish between sets of generalized coordinates when each coordinate is independent of the others and where these variables are not pendent.2 In general. if a system of N particles has in constraint equations acting it. we can describe the system uniquely by p independent generalized (k = 1, 2, .. p), where .

.

p=

3N

—

in = n

—

m

in which p is called the number of degrees of freedom of the system. The term minimum number of independent can he defined as the minirnuni gree 0/ freedom can Sets of generalized coordinates where necessary to describe a system uniquely. Sets coordinate is not independent of the others are called constrained generalized dinates or depeiideiit generalized coordinates. The number of degrees of freedom a characteristic of the dynamical system and is independent of the coordinates motion. While WhileOflC one can can select select the number and types of to describe the motion. in more more than than one one way. way. p = ii — in is coordinates and associated constraints in The rate of change of a generalized coordin ate with respect to time is called and is denoted denoted by by Ck(t) Ck(t) (k (k = 1, 2, ., n). The generalized space spanned by the generalized coordinates and generalized velocities is called . .

SW te SW teS/)aCC. space.

For the pendulum in Fig. 4.2 we generated two sets of generalized

We could select other sets of generalized coordinates as well. For example.. we and .v. However, this would introduce and L. select the generalized coordinates coordinates as as L. ambiguity into the description of the pendulum. as x has the same value when angle 0 is positive or negative. Such coordinates are known as ambiguous ized coordinates. Another example of ambiguous generalized coordinates would and Vp Vp of of the the endpoint endpoint P of the double link in Fig. 4.3. to use the coordinates .Vp and can easily show that a given coordinate ol' the endpoint can be reached by two f'erent configurations of the links, the two being mirror images about the line points 0 and P, as shown in Fig. 4.4. First configuration conhguration

0

P Figuration Second con Second con Figuration

Figure 4.4

this regard, the definition of generalized coordinate here is slightly different than the traditional older texts, which often restrict the term's meaning to only an independent set. 21n

4.3

CONSTRAINTS

draw two conclusions from the above. First, the generalized coordinates, they are independent or not, do not constitute a unique set. This actually is a advantage, as it gives a lot of flexibility. Second, one must exercise care selecting generalized coordinates, especially independent generalized coordito avoid redundancies and ambiguities. A poor choice of generalized coordimake the problem formulation and solution unnecessarily difficult. The discussion here with regards to generalized coordinates is similar to the of coordinate systems in Chapter 1. When we go from Cartesian to cylinor spherical coordinates, all we are doing is going from one set of generalized to another. We choose the coordinate system so that it simplifies the forWe

—iiOfl.

43 b

.t

CONSTRAINTS

section we analyze constraints that act on dynamical systems. We describe in terms of physical as well as generalized coordinates. The interest in equality constraints. in dynamical systems, constraints are usually encountered as a result of contact two (or more) bodies. Constraints restrict the motion of the bodies on which act. Associated with a constraint are a constraint equation and a constraint The constraint equation describes the geometry and/or kinematics of' the conThe constraint force is the contact force, also called the reaction. (Constraint can also be written when the motion is viewed from a moving reference and there is no contact. The relative motion equation becomes the constraint

Consider Fig. 4.5 and a particle moving Ofl a smooth surface whose shape is by

f(x, v. z, t) = f(x,v,z,t)

0

[4.3.1]

has continuous second derivatives in a!! all its variables. The motion of the parthe surface can he viewed as the motion of an otherwise free particle subto the constraint of' moving on that particular surface. Hence, (x, y, z, I) =

a constraint equation. equation. The The constraint constraint equation equation14.3.1] 14.3.1] is referred to • i configuration constraint. For a system described in terms of n generalized f(x.

z,

t) =

0

V

Figure 4.5

A particle movin9 on a smooth surface

219

220

MECHANICS: CHAPTER 4• CHAPTER 4• ANALYTICAL ANALYTICALMECHANICS:

CONCEPTS

coordinates, we can express a confIguration configuration constraint as

f(q1, L/2,

•

•

1)

•

= 0

The differential of the constraintf (in terms of physical and generalized is

df = df

=

dq1

di

(V

dx

+

+

+

+

c9qi

= 0

dl

jbrin. % The expressions [4.3.3] are said to be constraint relations in constraint con straint in Pfaffian form is one that is expressed in the form of differentials.) viding these equations by di. we write the constraint equations in velocity form called velocity constraints or motion constraints) as Of df = ----X .

dx

di

Of

dJ'

---- = at

.

Of

.

+ -;---V

Of. Of dl. + ---Z + — = 0 (9

dz d:

c1V

c9f

;

. .

qi + . q2 + dq2

...

+

[4.3.44

di

Of 9f

.

q,1

Of + ---- = 0

di

written in terms of physical coors The general form of a velocity constraint can be written nates as

.i +

14.3.a

+ a2± ++ a0 a0 = 0

and, in terms of a system with n generalized coordinates subjected to in con I?

>ajkqk a3kqk++ ajo a30 = = where

a0, and

In 1,2,..., 2, . ., in / = 1,

0

and aJ() af() (j =

.

1,

. ., rn: 2, . .., 2.

k=

1,

2,

.

.

., n) are

= a/k(ql, . .. of the generalized coordinates and time. for example. Note that once the constraints are imposed to a set of independent generalized cxvdinates, these coordinates are no longer independent. A constraint that can he expressed as both a configuration constraint as well * in velocity form is called holonomic. Constraints that do not have this property .

called non/wionomic. In other woi-ds, nonholonomic constraints cannot he as configuration constraints.

4.3.1

HoLoNoMic CONSTRAINTS

An unconstrained dynamical system or one subjected to a holonomic constraint is not an explicit function of time, for example. f1(q1. . .. q,,) = 0, is called t scieronomic system. It' the holonomic constraint is an explicit function of time. system is called rheonoinic. Throughout this text we will deal mostly with nomic systems, as they constitute the majority of situations encountered in ing applications. .

4.3

CONSTRAINTS

Consider the single particle discussed above and the case when the holonomic

[is not an explicit function of' time. That is, the plane defined by the from Eq. Eq. 14.3.4a1 14.3.4a1 yields Elimination of of the the of/ôt term from ;nstraint is fixed. Elimination

(If = (I I

C'

dx

d/.

.

14.3.71

£1V

Denote the position and velocity of the particle by r(i) = x(I)i + v(i)j + z(t)k The gradient of the constraint is + v(t) = 1(t) = 1(t)i +

vf= Vf=

14.3.81

dv (V

4'.

Taking the dot product between the gradient of the constraint and velocity v(f) '-I

.—

df.

Vf•v= —---x+------v —---x+-----v ('1.1 ()X

(iv

[4.3.9J [4.3.9]

when compared with Eq. (4.3.4a1. yields 0

14.3.101

b:th the expected result that the particle velocity is always tangent to the surface.3 same relation can he derived for generalized coordinates. same Given the holonomic constraint of a particle moving on a surface. the question arises as to what keeps the particle on the surface. The answer is a constraint normal to the surface. as shown in Fig. 4.6. To every constraint relation correa constraint force. Considering a single particle and denoting the constraint by F'. one can express it as

F'=F'n

14.3.111 [4.3.111

n is a unit vector representing the direction perpendicular to the surface. usureferred to as the normal direction. (This direction is similar to the normal direcin normal-tangential coordinates, hut here it can he taken as in either direction to the surface.) Since F' is perpendicular to the surface, it must be perto the velocity. It follows from Eq. [4.3.9] that the unit vector n. which is 11

FIgure 46

Constraint Force For a holonomic constraint

the derivation in Chapter 1 when analyzing path variables that the particle velocity is always tangent to

221

222

CHAPTER 4• ANALYTICAL. MECHANICS:

CONCEPTS

normal to the surface, should be parallel to

iviH

One can define n as (9/'. t9/'.

(9J

ely dy

(Z

—I + + —I —2—J —J +—k (IX

=±

)2

I.

[4.3.121

1/2

•,

((9

Since the constraint force is expressed as .

F' =

+

14.3.13]

+

when we compare Eqs. [4.3.12 I and 14.3.131 we conclude that the components the constraint force must be proportional to the partial derivatives of the or —

\\ —

—

(iif (°f

[4.3.14)

\\ —

k\ay)

1y9x)

Now. consider the work done by the constraint force as the particle dr. Denoting this incremental work by cIW and considering from position r to r + 1r. Eqs. 14.3.111 14.3.11] and 14.3.12], we obtain

dW

F'•dr

=

F'n•dr

=

[4.3.15J

= 0

This relation indicates that the work c/one by aa holonomic constrain! fo,re which done by which independent of time in any possible displacement is zero. Such constraints are referred to as work/ess icork/ess constraints. This result is to be expected, because the constraiit force is always perpendicular to the velocity. Note that, while the total work done by the constraint forces that are of time is zero, the individual constraint forces are doing work themselves. This won is in the ibrm of transferring energy from one component of the system to the other The sum of the transferred energies is zero. To visualize this, consider the double link in Fig. 4.3, whose free-body diagram is given in Fig. 4.7. If the first link is given initial motion, the second link will begin moving, and vice versa. The motion of second link is initiated by the constraint forces acting at point B. ()

0 B' J

I-,' I-.'

A

nz1g

(a)

Figure 4.7

(b)

Free-body diagram of double link

4.3

CONSTRAINTS

Considering Fig. 4.7, reaction forces, such as the forces at the pin at 0 and at B, are holonornic constraint forces. Normal lirces are also holonomic constraint

forces. However. friction forces are not constraint forces, even though their magniaide is directly dependent on a constraint force. Nevertheless, for static problems one :an treat friction as a reaction force, because in such cases friction prevents motion. Next consider a holonomic constraint that is an explicit function of time. For the considered earlier. this implies that the surface is moVing moving and and the the constraint in the the form form .1' 1' = j•(.t. v, z, t). Using Eqs. 14.3.4a1 and [4.3.9] we obtain .

c?t

[4.3.161

implies that Vf.dr

Itfollows workdW. 0. It follows thatthe that the incremental incremental work dW. which now not a perfect differential as time is explicitly involved, is not not zero. The incremental ork has the form dW =

F'•dr F'dr

=

Fn•dr

=

14.3.171 14.3.171

When the holonomic constraint is time dependent, the work performed by the

:.:rresponding constraint force is not zero. The path followed by the particle can no be described by the path variables associated with the surface. The vector n the normal to the surface, but it is not the normal to the path followed by particle.

4.3.2

CONSTRAINTS

When the constraint is nonholonomic. it can only he expressed in the form of Eqs. [4.3.5] or [4.3.6]. as an integrating factor does not exist to permit exprestheform formof ofEqs. Eqs.[4.3.11 [4.3.11 oror[4.3.21. s-n ininthe [4.3.21.Consequently. Consequently. none none of of the preceding we obtained regarding the work done by the constraint force are valid for constraints. The constraint force associated with a nonholonornic cannot be expressed as a force normal to a surface, as the nonholonomic does not define a surface. One can go into the space spanned by q,(I) and ; 1) (i = 1, 2, ... ., ii) and define a surface there, but this does not give any physical .

or significant results. Hence, there is flO general expression for the constraint when the constraint is nonholonomic. A common example of a nonholonomic constraint is the rolling without slipping a body with no sharp corners or edges. such as a disk or a sphere.

In general. constraint equations in terms of relative velocities and especially involving angular velocities that are not turn out to be nonholonomic.

j

the discussion of angular velocity in Chapter 2. When a reference frame is deby successive rotations about nonparallel axes, the resulting angular velocity he described as the derivative of a vector. Other examples of nonholonomic systems are from vehicle dynamics. Included this category are the motions of ships. missiles, airplanes, airplanes. automobiles, wheelshopping carts, and sleds. Figure 4.8 is a simplified illustration of such undergoing plane motion, such as a sled. Vehicles usually have a plane

223

224

CHAPTIR 4. ANALYTICAL MECHANICS: B.4sIc CONCEPTS

/ E

y

"C '.4

Y

/

x

F1)

Figure 4.8

Generic model of a vehicle

of symmetry, and they are propelled in a way that the guiding forces act primarily along the symmetry plane. with a very small component of the force used to change direction. A steering mechanism usually accomplishes the change in direction. One then makes the assumption that there is a point along the plane of denoted by A. point A is is always always along the plane of symmetry. A. such such that that the the velocity velocityof ofpointA The location of this point depends on the vehicle and the types of forces that point A from having a velocity component perpendicular to the plane of In a tricycle or automobile, automobile. the point A is in the middle between the rear wheels. hydrodynamic forces determine the location of A. a boat, the hydrodynaniic Consider the vehicle in Fig. 4.8. The configuration of this system can be dethe body body makes makes scribed by the coordinates andby by the the angle angle the coordinates of of point point Aq Aq X.4 X.4 and and Y,%,and with the inertial X axis, denoted by 0. The nonholonomic constraint is write ititas as with the translational translational velocity velocityof ofpointA. point A. Denoting Denoting this this velocity by VA, we write VA =

A141

+ Y4J

[4.3.11)

The constraint is written as

v4.j where j =

cos 9J

—

sin VA

= 0

[4.3.19) 14.3.19)

01. Introducing Eq. [4.3.191 into Eq. [4.3.18], we obtain OS OJ) Si fl01 01 + C C OS OJ) = (XA 1 + V4 J) • (— sill

= —XAsinO + YAcosO = 0

14.3.20) 14.3.201

This equation can conveniently be expressed as tanO

-0

[4.3.21) 14.3.211

It is clear that this constraint is nonholonomic. The associated constraint perpendicular to to the lire is basically the resistance of point A to have any motion perpendicular

4.3

CoNSTRAINTS

if flotiofl. In an automobile, for example. this force would be the friction force hethe rear tires and the road in the direction perpendicular to the velocity the tires. A very strong wind in the lateral direction. collision with another vehicle. a turn with high speed would violate this constraint. In general. the constraint force associated with a nonholonornic constraint perwork. A special case when this is not valid is rolling without slipping, where frictionforce isis applied te friction applied to to aa point point with with zero zero velocity. For roll without slip, friction a constraint lorce. as it reduces the number of degrees of freedom. next look into determining whether a constraint is holonomic or not. In genwhether it is or is not can be ascertained by visual inspection. Mathematically.

r

for a constraint in Pfaffian or velocity form to be integrable to configurafarm. the constraint relation must satisfy differentiability conditions. The con— must represent an exact differential. Consider Eq. [4.3.6]. If the jth constraint is holonornic, one should he able to write it as . ., q,1, 1) = 0. the differential of' fj and, for the most general case. dividing it by an intefactor . ., q,1) we obtain Eq. [4.3.3bj. Comparing Eq. [4.3.4bJ with 3.6j. we obtain for the general case of a holonomic constraint .

.

J

= g1a/k ma/k

k = 1. 2, .. ., ii

= g,a10 di = gj"jo

[4.3.22]

F.r a constraint given by Eq. [4.3.61 to be holonomic. there must be a function ., rn) where the partial an integrating factor gj(qj, . ., q,,) (j = 1. 2, .., = 1,2, .. ., in) satisfy Eq. [4.3.22J. To check this, we evaluate ol' of Indeed, considering considering an an index indexi•. r. we we obtain derivatives of Indeed, obtain .

.

e'92f' e'92f

=

(92f. (92/' ..

(Icir dcjr

(gjajk)

.

= =

and

(gjajr) [4.3.23]

'

(92/.

d21,

and

(?(/rdf (?(/,.df =

riq;.dt

=

(

k,r = 1,...,n:j = l,2,...,in l,2,...,m

[4.3.24]

prom Eqs. [4.3.23) and [4.3.24] if an integrating factor g1 exists such that

satisfy the relations (9

(k/,. (k/r

=

£1 -

(1

(gjajr) (g1a11.)

-

(h/r

. . ., ii: j = in = I,2,....ni 1,2 = 1.l,....n:j

=

CI. ci! di

[4.3.25]

5e constraint is holonomic. holonornic. The The problem problem with with using using the the above above procedure procedure is is that that not he easy to find the integrating factor. especially for systems having more degrees of freedom. thatis. is. A constraint of the form q,1. t) 0. or akc'k + ao akc'k ao 0.0.that .

.

. .

quality constraint, is nonholonomic because it cannot he reduced to a form . .. .. q,1. q,1. i) = 0. Such constraints require a different treatment than .

.

—ai.nts. We also encounter constraints that are valid in

225

226

CHAPTER 4

•.

BASIC CONCEPTS CONCEPTS AN.u4YT!c.ti ANtLYTIctI MECHANIC'S: MECHANICS: BASIC

or during certain intervals of the motion. Such constraints can also be classified inequality constraints. They can be found in problems involving contact. Consider now a system that originally has n degrees of freedom and is subjectei± to in holonomic constraints. Introduction of in constraints reduces the degrees of freedom by in to p = — m. resulting in a set of in excess, or surplus, coordinates.. It is possible, at least mathematically, to eliminate the surplus coordinates from the formulation, formulation. which results in an unconstrained system of order ii — in. of this, unconstrained systems are referred to as holonomic. By contrast, a nonholonomic constraint constrains only the generalized velocities, without affecting the generalized coordinates. In such systems there are ii independent generalized coordinates and n — in independent generalized velocities. 11

Example

4.1

I

shown1a tube, whose shape is given by the equation—v x2, = as shown A bead is sliding in a tube. Fig. 4.9. Find the direction of the normal to the tube. 1

Solution One can solve this problem in a number of ways. We lirst consider the problem from a physica in the the tube. tube. the equation defining the shape of the the bead bead isis sliding sliding in standpoint. Because the

becomes the constraint equation. and it has the form +

f( .v,) = V — I

=0

[.1

Taking the partial derivatives off, we obtain LW

clx

that, using Eq. [4.3.8]. [4.3.8], the gradient off has the form Vf = 2xi +j. From Eq. unit vector in the normal direction (chosen, (chosen. for convenience, positive outward) becomes so

2x1+j

(4

v 1 + 4x2 1

As expected, because the constraint is not an explicit function of time, neither is the of the constraint force. The constraint is, of coursc. coursc, holonomic. To solve this problem geometrically, we define the angle 0 between the horizontal the tangent to the curve. The tangent of 6(x) describes the slope of the tube, and

tanO =

dx.r d

= —2x —2x

V

fl

t

y= i—x i—x y=

x

FIgure 4.9

Bead sliding inside a tube

14

43

227

CONSTRAINTS

We find the sine and cosine of 0 by

SjflO: sin() =

2x

cosO=

-

[0]

—

± 4x2

I ++ 4x2

-.J I

we can express the unit vector describing the normal as

n=

sin th

—

cos (Jj

2xi+j \.l + 4x2

=-

If] [U

1

We can also determine the normal direction directly from the geometry. using the approach in Chapter 1. 1, without going into any constraint equations. Denoting the path variable x. we write the position vector as x.

r = vi + (1

—

x)j r)j

[gi

and the expressions for the slope and the unit vector in the tangential direction become

r'=

dr

e1=

$

(IX

I— 2xj

S= F

Is.

\/1 + 4x2

\/

4

I

+ 4X

1

[hi

Use of of Eq. Eq. [1.3.36] [1.3.36] yields n. When the path parameters associated with the motion of a body specified. in essence a constraint has been imposed imposed on on an an otherwise otherwise free free body. body.

A block of of mass mass in in isis attached to a cord of original length L and is rotating about a thin hub, ,is shown in Fig. 4.10. Friction is negligible. Find the constraint constraint fbrce fbrce ifif (a) (a) the the cord is not wrapping around the huh, and (h) the cord is wrapping around the hub.

Solution When the cord is not wrapping around the hub. the constraint is holonomic and independent

time. The constraint equation basically describes that the length of the cord is constant, and has the form f (( x.

(a)

Figure 4.10

) == )

±

—

2

[al

= 0

(I))

Mass rotating around a thin hub (a) Cord is not wrapping around hub (b) Cord is wrapping around hub

Example

4.2

228

CHAPTU

4•

CONCEPTS

MECHANICs: ANAlYTICAL MECHANICS:

Once motion is initiated, the mass keeps rotating with the same speed and the energy of the particle does not change. The constraint force is the tension in the rope. and it does no work. b. The situation is quite different when the rope wraps around the huh. Assuming that the radius is very smalL the tension in the rope is directed toward point 0. Summing moments about 0. we obtain

Ib] [b]

M0 = 0

angular momentum momentumabout about 0 is conserved. In essence, we have a central force that the angular problem. Let us use polar coordinates r and Consider that the length of the rope. denoted by r.. reduces continuously by the relation so

r= r=L—r00

[c]

where r0 is the radius of the hub. In one revolution of the mass. the rope shortens by given by by H0 H0 = mr2O. Because the angular momenThe angular momentum about 0 is given

turn is conserved. [41

= constant = Ii

where we note that the constant Ii is always greater than zero, h > 0, and that Ii is a function of the initial condition. Differentiating the relation between r and 0. we write =

1.1

= —-f-

r(10 —.

'(1

and substituting the above relation into Eq. [dl. we obtain

''

=

= 11

or r2

C<0

0 ==— —rji r,,/i == C = — 1(110

[g.J

Now, let us find the response r(t). We can rewrite Eq. [gJ as where C is constant. Now.

dr =

[hi 111

C di

which, when integrated, gives [II

where D is a constant of integration, determined from the initial conditions. We note that a: D. so that D = L3/3. Considering that the length from Eq. Eq. [iJ [iJ r313 r313 = D. t = 0. r = L, and from

x2 + v2, we can write Eq. [i] as

the rope is related to x and v by

f(x,v,t)=

(

v2 +

1 3

'Ct—()

[fl UI

constraintrheonomic constraintThe constraint is a time-dependent holonomic constraints that is. a rheonomic The constraint force, which is the tension in the rope, does perform work. To show that the constraint force does indeed perform work, we consider the configuration vector r and its derivative

r = rer

rOes r = ier /er ++rOes

[k)

229

4.3 (the tension tension in in the the rope) rope) can can be be expressed expressed as F' = The constraint force (the

—

Fer. so that the

dot product between the constraint force and the particle velocity becomes

F•r =

Ill [fi

—Fi

not changing, changing,i i = 0, and is not is not zero. Note that for the case when the length of the rope is ordec wdec to \s ieco. Mso note done b'j fc'r fc'r rU), the initial angular velocity must be

Siven a system with generalized coordinates (3qi sinq2 sinq2 + +

+

and the constraint equation

and

+ 2q2)dq2 2q2)dq2 = =

+

Example

I

4.3

0

J

whether the constraint is holonomic or not.

Solution q2). such that

The constraint equation is holonomic if there exists an integrating factor holds, or

(if

= 3,gqi

..

sin

+ x'i;- +

c?f c?t

+ 2gq2

cos aq2 = gql

[01 [01

q:) = q. then

observe that if

= 3qfsinq2 ±

+ 2qIc/2

=

+

Ibi

C1q2

ntegrating the two expressions, we obtain

JJ == qisinq2 sinq2 +

+

+ h1(q2) ++ C1 C1

ff ==

sinq2 + sinq2 +

+ h2(q1) h2(q1) + C2 C2 id id

respecand are functions that appear as a result of the integration over C2 are are constants. Comparing the two integrated terms, we conclude that and C1 and C2 C2. The constraint, constants are are related related by by C1 C1 == C2. and that the constants = 00 and and h1(q2) and h1(q2) = and has the the form form is holonomic and

h1

and

/12

,f(qI.q2) ,f(q,, ff2) = qjsmq2 +

+

+C C=

0

IdI [dl

C is is a constant. constant. For For this this problem the integrating factor factor was was found by visual inspection. guidelines for for finding the integrating factor. 2eneral. there are no set guidelines

The tip of ofthe the double-link double-link mechanism mechanism in in Fig. 4.11 is constrained to lie on the inclined plane. The tip the constraint equation and express express itit in velocity form. the

Solution coordinates. generalized coordinates. This is a single degree of freedom system. We use 0, and 62 as gencralized Hence. we need need one one constraint constraint equation. equation. We We can can simplify simplify the the formulation formulation by by expressing the Hence. we

of the tip along the incline by the variable s. To derive the constraint equation we the position vector of the tip in two ways: using the links and using the incline. Using links, the position vector has the form + cosO, + rp = (L, cosO,

+ (L,

sinO2)j + 1i2 1i2 sinO2)j

[a] Ia]

I

Example

4.4

4*

230

ANALYTICAL MECHANiCS: BAsic BAsIC CONCEVrS CONCEVrS

V

Li

0

t.

3L1/2

Figure 4.1 1 and using the incline, it has the form

3L1. 1

2

+ S COS +

+ .v,vsini/;j sin i/ij

Ib]

We equate the above two expressions and separate components in the x and v directions, thus

L1cosO1 L1 cosO1 ++L, L2 cosO2 =

3L1

+ +scosçli scosçli

Ic] [c]

a-

L1 sin61 sin6, +

ssini/i

Ed]

To obtain the constraint equation, we eliminate s by multiplying Eq. Id by sin i/i and and Eq. Eq. Idi by —cos i/s and adding the two equations. Dividing the result by L1 sin we obtain cosO1 +

L2

— —

L1

-

L,tani/i

tani/i

sinO2

—2

101 I.]

which is recognized as the holonomic constraint equation. To express this constraint in veyelocity form, lhrm, we differentiate Eq. je] with respect to time, with the result

/

.

(sin

4.4

+

-

\.

tan i/i )

+

/ (sin0 ++ .

L1

=0

Ian i/i J -

Ifi

VIRTUAL DISPLACEMENTS AND VIRTUAL WORK

At this point, we introduce the variational notation. The variational notation is ideally suited for dynamics problems because it makes the formulation concise, and it has a meaningful physical interpretation. When applied to dynamical systems, the variations of displacements are known as virtual displacements, denoted by

etc. In terms of generalized coordinates, the virtual displacements etc. The variations of the velocities are denoted have the form . ., (k = 1, 2, ... n) ft)r generalized for physical coordinates and by 61,

6x, 6v.

.

..

velocities.

4.4 4,4

VIRTUAl. DISPIJACFMENTS DISPLACEMENTS AND ANT)

WORK

Virtual displacements have the following properties • •

They are infinitesimal displacements. They are consistent with the system constraints, but are arbitrary otherwise. The variation of displacenients (or velocities, etc.) is obtained by holdiiig holding time fixed: fixed: thercfhre, virtual displacements can be considered as occurring instantaneously, and time is not involved in their applications.

Dealing with virtual displacements is like imagining the system in a different p0that is physically realizable, while freezing time. It is as if a different set of forces were applied and, as a result, the system moved to another location by one of the admissible paths it can follow. The rules for calculating virtual displacements are intimately related to the rules differentiation. For the position vector r = .v(t)i .vWi + v(t)j + z(t)k, or r = r(q1, t), the variation of r becomes (Sr

= 6x1 6x1+6yj+&k + Svj + &k

or

(9r

6r =

dq1

+

or

+... +

6q,1 6q,,

[4.4.1 o,bJ Figure 4.12 depicts the concept of a variation (for the coordinate v). When expressing the motion r = xi + + zk in which x, v. and z are all functions of the generalized coordinates, the variation of r has the form (Sr

(C

+ ------kj6qk

=

/

[4.4.21

As As discussed in Appendix B, we distinguish between dependent and independent variables. For dynamical systems, time is the independent variable. The coCOc'rdinates x, v, z, z. as well . .., q,1 are functions of time and are refened to as the dependent variables. The term dependent is used here to denote explicit depeiiJence of the generalized coordinates on time, tune, rather than on each other. It follows that can interchange the time differentiation and the variation operators. That is, = = (k = 1, 1, 2, . ., n). The variation of a position vector can he obtained in two different ways. One way by obtaining an analytical expression for the position vector and taking its variation .

V

" + or v(t)

I

Figure 4.12

Variation of y

231

232

CHAPTER 4 S ANALYTICAL MECHANICS: MECHANICS:BASIC BAsic CoNCEPTs CoNcEvrs

by differentiating with respect to the generalized coordinates. Basically this is the use of Eq. (4.4. 1 bJ: it is known known as as the the analytical analytical approach. approach. This This approach may lead to lengthy expressions for certain complex problems. When r is expressed in terms of the coordinates of a moving reference frame, one must also take the variation of the unit vectors of the moving reference frame. The exception to this is when the motion of the relative frame is prespecified as a known quantity and is not treated as as a motion variable. In the second way. known as the kinematical appmach, one explores similarities between velocities and virtual displacements. When taking the variation of an expression, the independent variable is not varied. We use this property, as time is the independent variable. The time derivative of r is ..

r=

r=xi+vj+zk+— lit •1.

.

.

or

.

q,+

(9r

.

.

(9r

q,1+— lit

dq2 -

[4.4.3a,bJ 14.4.3a,bI Elimination of Elimination of the partial derivative of r with respect to time, elimination elimination of of all expressions explicit in time. time, and replacement of by 6x. by 6v, by in Eq. [4.4.3a] and of (k = 1, 2,.... n) with in Eq. [4.4.3b1 yields the variation

of r. This implies that if the expression for the velocity is known, the associated virtual displacement can he obtained directly from it. This approach of calculating virtual displacements from velocities is especially useful when the velocity of a point can be found using an instant center or a relative velocity expression, such as

[4.4.4] [4.4.41

= V4 + w X r814 + The variation of the displacement of point B is

=

cSr4

+

X rB/A + 6rnrei

[4.4.5] 14.4.51

where we note that the where

term is left intact and that represents the variation of an infinitesimal rotation. Also, keeping in line with the developments in Chapter 2, we extend the boldface to the entire term 80 to denote that 60 is a variation of a rotation and that it is not obtained by differentiating a vector. Consider Eq. 14.4.3b1 and the derivative of r with respect to Of all the terms in Eq. [4.4.3h] only one survives and we obtain the important relationship (1'

k= l,2,...,,i

-. =

14.4.61

so that the variation of r can be expressed as (/r

Sr =

14.4.71

Note that Eq. [4.4.7] is in essence the mathematical representation of the kinematical method of calculating virtual displacements. Next, consider the holonomic constraint y, z, t) = 0 and obtain its variation, which has the fbi-rn

6f =

dx

+

dv

+ -

liz

z

= 0

[4.4.81

4.4

233

VIRTUAL DISPLACEMENTS AND VIRTUAL WORK

time is held fixed while 1' is varied. (Sf has the same form form whether the 1' is the constraint is time dependent or not. When a constraint is given in velocity form by Eqs. [4.3.5] and [4.3.6J. in terms of physical coordinates the virtual displacements Because

satisfy

+ and,

[4.4.9]

+ a-(SZ = 0

-I-

in terms ol generalized coordinates and the jth consfl-aint, they satisfy (Sfj

=

+ aj2(Sq2 +

j = 1.2,

+ ajn(Sq,j

. ., in

. .

[4.4.101

Let us next consider the work done by a force over a virtual displacement. Cona body acted upon by a force F and the virtual displacement associated with the point at which the force F is applied. We define the work done by the force over the virtual displacement (Sr as the virtual work or variation variation of of work and denote it by Hence (SW

= F'(Sr

[4.4.111

We will examine the virtual work associated with a general force in the next For now, let us consider the holonomic constraintf(x, v. z, t) = 0 and the virtual work. Recall that whether the constraint is time dependent or not immaterial. From Eqs. [4.3.1 l]—[4.3.14]. the constraint force F' has the form

k)=

14.4.12]

define by (SW' the work performed performed by by the the Constraint constraint force in any virtual disi 'irma! vi 'ork 'oi* due to constraint fèrces, ftrces, as 6 W' = F' (Sr

=

+

[4.4.131

+

Using Eqs. [4.4.11 and [4.4.81 we conclude that (SW'

=

F'•6r = çffvf.6r =

= 0

[4.4.141

that the work performed peiformed by a holonoinic constraint anyvirtual virtual displace— displace— constrainttorce torceininany l.s zero.

A disk of radius R rolls rolls without without slipping slipping on on a rod of' of length L pivoted at one end, as as shown shown in in 4.13. Denoting the pivot pivot angle angle by by 0 and and the the angular angular displacement displacement of of the the disk disk by find virtual displacement of the center of the disk.

S.lvtlon 'Ae will solve this problem using both a kinematical and an analytical approach. We begin iith the kinematical approach. We select an inertial frame XYZ and a relative frame xyz, that the xv: axes arc obtained by rotating the XYZ frame by an angle 0 counterclockwise the Z axis. The velocity of point G can be written as =

X Vfi + +(1.) (0 X

r(;/8 r(;/8 ++ "ret

I

Example

4.5

234

ANALYTICAl. MECHANICS: BASIC CONCEPTS CHAPTER 4 • ANALYTICAL

4)

I',, p111

-t

R41,

Y V

x

Figure 4.13

w in which v11 = 0. w

Disk rolling over bar

Ok, and = Ok,

rG/B = rG = (L

—

+ Rj

Lb, ci

=

Substituting the above values into Eq. [a} we obtain VG

=Okx

1(L — R(tI)i

+ 0)1 0)1 ++ (LO (UI — R40)j

+ Rjl Ru — R41 =

Idi

Thus, we write the variation of rG as =

1.1

Now we will find the variation of rG analytically. The position vector rG is given in Eq. IhI. There are two ways to obtain its variation. In the first. we express rQ in terms of the inertial coordinate frame and then differentiate. In the second, we take the variation of Eq. [bJ directly. which requires the variation of the unit unit vectors vectors Ii and j of the moving frame. The relation between the unit vectors of the inertial and relative frames is

Iji Iii l.1 I—.,

k=I(

Li]

sinO cosO

COSO

—sinO

I

J

If]

Introducing this into Eq. [b}, we obtain

= [(L

—

—

Rsin 011 + I(L

—

R4)sinO + RcosOJJ

Ig]

The virtual displacement then becomes =

—

Rçb)sint96O Rçb)sin(hSO

+ [(L [(L— — Rç1)cos(flSO

— —

Rcos0&,b Rcos0&b

Rsin(h34

—

—

RcosO&)]1 RsinO6OlJ

[hi

To convert the virtual displacement in terms of the relative frame, we introduce the relationships I = cosOI cosOi

gives Eq. into Eq. fhJ. which give.s Eq. [ej. [ej.

—

sinAj

= SiflOl + COSOJ

II]

4.4

VIRTUAL DISPLACEMENTS

235

WORK

Next, take the variation of Eq. [h] directly. with the result

II] [I] From

Eq. [fj. the variations of the unit vectors have the form

(Si

=

—

sin 0 60 1 + + cos cos(1 0 60 J = 8Oj

6j =

— cos

0 60 1

sin 0

—

.J

= —(SOi —601 1k]

Equations [ki can also be obtained directly from the rates of change of the unit vectors.

indeed, recalling that the angular velocity is ô = 6k. the derivatives of the unit vectors are

di —=oXi=Oj di

dj

—=toxj=—Oi —=wxj=—Oi dt

[II III

from which the variations can he calculated easily. Introducing Eqs. [k] into Eq. UI, we obtain

have thus obtained the \'ariation variation of of three different ways. It is clear that the number three different -•f manipulations is the least when we obtain the variation of rG from the velocity expressions. ot'rG We

Consider the two-link mechanism in Fig. 4.3. A force F is acting at point P. Find the virtual expression for each link and demonstrate that Eq. [4.4.14] holds. The free-body diagrams of the link arc shown in Fig. 4.7. For the first link, the forces that The :4Jntrihute to the virtual work are the reactions at point B and the force ot gravity at the mass center G1. The forces can be expressed in vector tbrm form as

F8 =

+ B,j

[a]

—m1gj FG, FG, = —m1gj

The associated displacement vectors are = LisO1i—LicO1j L1 sO1i — L1 S

S

I

rG1 =

sO1i

.

I

[b]

—

that the virtual displacements become

= We

L1

c01 601i 60,i ++ L1 sO1

=

+

661j

[c] Ic]

thus find the virtual work for the first link as = F8•6r11

—

in1gj•6r(;]

c01 (SO +

sO,

—

migL1 mjgL1

[dl

For For the second link, the virtual work is due to the reactions at point B, gravity acting through the center of mass of the link G2. and the external force F at the tip P. The forces at B are equal and opposite of F8. The other forces can he expressed by

F =

+

FG. =

—m2gj

[0] [e]

with with associated associated displacements

rp =

(L1s01

r(;, = (LisAJ (Li sO1 +

+L2c02)1j CO1 + c01 +

[II Ill

I

Example

4.6

236

4• ANALY11CAI. ANALYTICAl MECHANICS: MECHANICS: BAsic Btsic

whose variations are

+L2s02602)j + L2s02602)j

c01601 + L,c02602)i + (L1 (L1c01601 = (L1 (L1c01601 + =(L ôrG, =

L

2

C

+

02602)11 + (L1

Igi

We now find the virtual work for the second link as tn2gj = —F8 • 6r8 + F• 6rp — m2gj

1k]

•

The virtual work of the entire system is found by adding Eqs. [dl and [hi. with the result CSW = 6 Wl1flkI + (SW + 66 W[j(1k2 = =

F' 6rp

—

Ii'

— ?fl ?fl1gj gj ni2gj • 6rG2 —

The only terms that contribute to the virtual work are those associated with the external forces. The contribution of the holonornic constraint tirces (in this case the reaction at point B) to the virtual work is zero. [ii in terms of the generalized coordinates as Taking the dot products. we write

6W = 6W — img

=

i

602) +

+

c01

+ L1s026Th)

sO1

sO, sO1 66, +

SO1

— c01 + F%.L1 sO1 —

,mgLi ,mgL, I

+ F1L2c 02 +

—

sOi) 60,

—

Iii Ii]

m2gL2 sO7 SO?

Consider next the problem of having not a pinned joint at point 8. hut a joint that permits we will will see see in more detail as we sliding motion, such as the collar shown in Fig. 4.14. Such ajoint. ajoint. as in Chapter 7. is called aprisina!ic aprisinasic join!. The free-body diagram is illustrated in Fig. 4.15. The friction force at the sliding joint must be considered, considered, and and the the forces forces that that the. the two bodies exert on each other are split into two parts: a normal force N and a friction force F1. Introducing the unit vectors e1 and e2 along and perpendicular to the link, we express the normal and

g

F

rn1. L1 rn1. L,

F e1

N q

L1

N

F1

.1' .1

(a)

FIgure 4.14

Prismatic joint

Figure 4.15

(b)

Free-body diagrams

4.5

GENERALIZED FORCES

friction forces on the two rods as F\; — F\ —

[k]

F1 = F1e2 F1

Ne1

Note also that because of sliding, the points 11 on the first link and on the collar do not have the same velocity. Denoting these points by and B2 and introducing a generalized coordinate q to describe the sliding of link 2, the position vectors for B1 and B2 become

Ill III

rfi, = r8+qe2

= rB

lorm The virtual work expression has the form = (F,v +

6

F,)•

—

,n1gj ••

[ml

F' 6r1 — in2gj • = —(F.v + F ) • 6r8, + F• so that the virtual work for the entire system is

6W =

= —F18q±

+

F•6rp

—

—

lnlgj•6rG1

[n] In]

The contribution of the friction force to the virtual work is clear. Note that in order to determine the magnitude of the friction force, we need to have the normal force, which is absent from the above expression. This. basically, is the typical problem encountered when formulating problems involving friction. Also note that the position vectors for the center of mass and for the tip of the second rod change when the sliding joint is introduced to the problem.

4.5

GENERALIZED FORCES

Consider the system of particles in Fig. 4. 16. The jth particle particle exerts exerts aa force force of of F,1 . . ., N). The resultant of all forces acting on the ith on the ith ith particle particle (i, (i, // = 1. 2, .. particle is denoted by R1 and has the form N

=

F1

+F=

F1

ii== 1.2,....N

F11 + Y F,1

•j=I j=I

In I PH

F

in3 .

•

FIgure 4.16

[4.5.1]

237

238

CHAPTER 4• ANAINTICAI. ANAI.YTICAI. MECHANICS:

B&SIc CONCEVFS CONCEPTS

where F1 denotes the sum of all external (impressed, applied) forces exerted on the ith particle and F; is the sum of all internal forces (constraint or reaction forces that one particle exerts on the other). The virtual work each particle is defined as 6W, 6W1

[4.5.2] 14.5.2]

= R1'6r1

One obtains the virtual work work ft)r ft)r the the entire system by summing over the individual particles N

N

(Swi=

6W 6W'

14.5.31 [4.5.31

Substituting Eq. [4.5.1] [4.5.1] into into Eq. Eq. [4.5.3L [4.5.31. N

N

=

(SW

[4.5.4] [4.5.41

1 -I

i—I

i—I

We showed showed in in Eq. Eq. [4.4.141 [4.4.141 that the total work performed by the constraint forces in any virtual displacement is zero. It follows that the second term on the right side of the above equation vanishes because N

= 0

[4.5.5]

1=1

and the expression expression for for the virtual work becomes N

[4.5.6]

6W = i—I

It is of interest to examine the virtual work in terms of generalized coordinates. We express the displacement of each particle in terms of a set of n generalized coordinatcsqk(k dinates (k = 1,2,...,n)asr, 1. 2, . ., ii) as r, = rj(q1. (i = l,2,...,N).Thevarjation of' r is .

.

N

=>:

6r,

.

.

ar,

[4.5.7]

Substitution of Eq. [4.5.7J into the expression for virtual work yields

=y =N' N

3W =

>'F1.6r1

fl

cSqk

1

k=1

1=1

7N

n

9r1

=y 'V\ji .7

2

6qk

[4.5.8]

We define the term inside the fences in the above equation as generalized forces and write N

=

>'F1 i—I

•

k== E,2,...,n k

[4.5.9] 14.5.91

4.5

GENERALIZED FORCFS

where Qk is the generalized force associated associated with with the kth generalized coordinate.

We can then express the virtual work as '1

[4.5.10]

6W = k=I

The relation between a generalized coordinate and a generalized force is analogous to the relation between a physical coordinate and the force applied in the direction of that coordinate. Also, the dimensional relation between generalized COordinates and generalized forces is worth noting. The product of Qk and has the same units umts as the vanation of energy. For example, ii the generalized coordinate describes a displacement, the generalized force has the units of force. If the generalized coordinate describes a rotation, the generalized force becomes a moment. Recalling from the previous section that the variations are often calculated with more ease by velocity relations. relations, we make use of Eq. 14.4.6] di•

=

k

14.5.111

'1k

to express the generalized forces as Qk =

"> F•

_ri

[4.5.12]

=

I—I

;—

Another way of calculating generalized forces is based on the nature of the applied forces. For a conservative system. because is a perfect differential, the virtual work can be written as the variation of the negative of the potential energy. or 'V N

= —6V

[4.5.13]

1=1

in n which V is the potential function, or the potential energy. The variation of the potential energy in terms of physical coordinates is (9V

/ c?V

6V=>(.

rIZ, cIz,

1—I

14.5.141

system. x,. When there are no constraints acting on the system, and Zi (i = 1, 2, . ., N) are independent. It follows that and 3z, are arbitrary, arbitrary, and and using Eqs. [4.5.13 1 and [4.5.14], we obtain . .

(9V

(.1k (.Ij

=

dV —

aVF =—F(Iv

=— —

[4.5.151

In terms of independent generalized coordinates, and when all the applied forces are conservative, the virtual work expression can he written as

6V=—6W=

-

6qk

[4.5.16]

239

240

CoNcEvrs

CHAPTER 4• ANALYTICAl. MEchANics:

Comparing Eqs. 114.5.16] and [4.5.10]. and considering the independence of the variations ot' the generalized coordinates, we conclude that the generalized forces are related to the potential energy by (9V

[4.5.171

= —.

d qk

In the presence of both conservative and nonconservative forces, the virtual work and generalized forces can be written as

6W = —6V

[4.5.181

6W?l(. N

+> —=— rlqk

Qk = Qk' Qk' +

14.5.19]

where the notation is obvious. When they are constant, nonconservative forces can

be treated as conservative. In summary. one can use a number of ways to calculate generalized forces: 1.

Write Eq. [4.5.61 and after the virtual work is calculated collect coefficients of (k = 1, 2,. ., n). and use use Eq. [4.5.121. Calculate (or Take advantage of the potential energy and use Eq. 14.5.171 14.5.17] for fbr the conservative .

2.

3.

forces.

The reader is encouraged to use and compare all three approaches.

Example

I

Find the generalized forces for the mechanism in Fig. 4.3 (Example 4.6).

Solution Wc will calculate the generalized forces in a number of ways. First, we take the expression for virtual work from Eq. Iii in Example 4.6. thus

The generalized coordinates are

c01 +

=

+ SO

and

sO1

—

in2gL1

—

Ia]

— CO2 + F1L2 sO2 —

that we can identify the generalized forces as

= Q Q =

c01 ++

—

+

Q2 =

in2gL1 sO1 — sO2

sO1

s02

—

[bJ

For the first link we have one external force,

Next, consider each force and

gravity. and •

= —in1gj

=

L1 —i--

sU1i

—

[ci Ic]

4.6 so

PRiNCIPLE OF VIRTIAL WORK FOR STATIC EQIJ1LIBRIUM PRINCIPLE

that cir(J -

dO,

L1 L;

L1

eO1i+ cO1i+ —sO1j

I

2

2

(-10)

= 00

[dl

There are two external forces acting on the second link, written

F=

+

rp = (L, sO1 + L2sTh)i rp

—

(L, (L1

c01 +

± L2 L'2 = (L1 sO, sO1 s02)i

—

(L1

cO, + c01

FG: =

c02)J

lel

that

= LicO1i+LisO1j L,c011+L,sO,j = L1c01i+L1s01j L,cO,i+L,sO,j Applying Ecj.

= L2c62i+ L2c021+ L2s02j L2s02j

(/02 (102

Ifi

= —

121 we obtain 121 we

=

+ F•

=

sO1

Q2 =

FG'

-

d02 4102

+

=

ôr,

+ F(;2•

(irG, (IrG,

sO, + sO1

—

F•

+ FG,•

+

+

c01 +

sO1

(9r(7,

(/02 (102

[g] [91

are the same as Eq. [bj. Finally, we make use of the potential energy to calculate the portion of the generalized associated with the the gravitational gravitational Taking point 0 as the datum, we write the energy as as

V=

I

CA, —m2gLicO1 — m2gL, c01

—

cO2

[hi IhI

the generalized forces due to the conservative forces lorces become =— — -.

—in2gL,sO,

=

(101

=— —

(IV

= ——m2gL2 S 02 =— — tfl2gL2 sO2 2

III Ill

It is easy to see that the use of potential energy simplifies the calculation of the generalzed forces.

4.6

PRINCIPLE OF OF VIRTUAL WORK WORK FOR FORSTATIC STATIC EQUILIBRIUM EQUILIBRIUM

Let us now consider static equilibrium. equilibrium. For For a dynamical system, static equilibrium described as the state where all components of the system are at rest, with zero

241

242

CHAPTER CHAPTER 44

ANALYTICAL MECHAMCS: BASEC CONCEPTS

velocity and zero acceleration. To find the equilibrium position. one can write the equilibrium equations using Newton's second law and solve these equations. The disadvantage of doing so is that if the motions of any two components arc related to each other with a Constraint relation, then the associated constraint forces must be calculated in the process. This may become tedious for systems with several interconnected components. Encouraged by the results of the previous section, we seek a different solution to the equilibrium problem that does not require one to solve for the constraint equations. At equilibrium, the resultant lorce lowe on each component of a system must he zero. Hence. we have 1, 2, .. .,N). O(i 0(1 It follows fromEq. from Eq. 1,2,..., N).Itfollows that since sinceevery every resultant R1 = 0. 0, the virtual work must vanish as well and we must have 6 W 0. Introducing this into Eq. [4.5.61 gives N

6W =

= 0

[4.6.11

i= i=I I

above equation. fIrst first formulated by Johann Bernoulli, Bernoulli. is known as the principle qf of virtual work fbr static equilibrium. It basically states that, at static equilibrium, the work performed by the external, impressed forces through virtual displacements compatible with the system constraints is zero. It can easily he cxtended to rigid bodies if we consider r, r to to be be the the displacement displacement of the point on the The

body to which the force is applied. Let us consider the principle of virtual work in terms of generalized forces. It follows from Eq. [4.5.101 that at equilibrium 11

6W =

= 0

[4.6.21

k— II

When the system is represented in terms of independent generalized coordinates,

because the generalized coordinates are independent of each other, their variations also are independent. Therefore, for Eq. [4.6.2] to hold, each of the coefficients of that is. Qk. must vanish individually. We write N

-

Qk—,?F,. Qk = ,> F.

-

-.

N

-'

dr1

F. = ,, F1•,. —,, -

.

=0 —0

k = 1, 2. ..., ii k—l,2,..,n

14.6.3]

In the presence of conservative forces we can take advantage of the potential energy

and write

—

(?V -

+

= 0

[4.6.4]

The above results can also be interpreted as follows: Because independent general-

ized coordinates represent the independent motion of each degree of freedom, their corresponding generalized forces must vanish at equilibrium. As in the previous section. section, one can follow two approaches when solving static equilibrium problems using the principle of virtual work:

______

______

4.6 I. 2.

FOR FOR

OF VIRTUAL

243

STATIc EQUILIBRIUM

One can work with physical coordinates and use Eq. 14.6.11. One can select a set of generalized coordinates, calculate the associated generor 14.6.4]. ahzed forces, forces, and and use useEq. Eq.14.6.31 or

In the second approach. Eq. [4.6.41 is usually recommended over Eq. 14.6.3 1 in the presence of conservative forces, as it makes use of the potential energy. On the N; k = 1,1,2,..., 2, . ., ii) in 2, (i = 1,1,2,..., or ., N; other hand. computation of Eq. [4.6.3] can be done in a systematic fashion and tabulated. thereby mechanizing the derivation of the equilibrium equations. Next, consider the principle of virtual work in terms of constrained generalized coordinates. To this end. write the constraint equations in Pfaffian form as . .

> k— A:—

2

dc/k

+ a30 af 0 cIt dt =

= L 1, 2, J =

0

.

..

. .

,

[4.6.5]

in

II

Now write the variation of the generalized coordinates as Pt

[4.6.6]

= 0 k— I

We add this relation to the principle of virtual work via the Lagrange multipliers

A7 (j =

1,

2, .., 2,. ... in), resulting in the expression for the augmented virtual work as .

/ nn

OW = OW

—YA,(>ajk6qk)=

— Qk6qk Qk6qk —

I

> j.=I

/

ii

0

\A=I

/ ,1

[4.6.7] Rearranging this equation as Fl

=

/

\

> AJUIk Qk — Qk —> AJUJk )8qk = 00

k—1\

./—i

14.6.8] 14.6.81

J

ind by selecting the Lagrange multipliers such that the coefficients of :ndividually, we write the equilibrium equations as

vanish

'H 'U

k = l,2,...,n

Qk = ">A/afk in

[4.6.9] 14.6.9]

equation. [4.6.4] to to this equation. the presence of conservative forces, we introduce Eq. [4.6.4]

which leads to

=

aV

+

>

[4.6. 1 0]

Find the equilibrium position of the two links in Fig. 4. 17. The springs are unstretched when Find both

rods arc horizontal. Both springs deflect only vertically.

1

Example

4.8

____

244

4.

CHAPTER

ANALYTICAL MECHANICS: BAsic CONCEPTS

in1. L1

B

''2

L2

C

Figure 4.17 Solution it is conservative. it is preferBecause this problem involves two interconnected bodies and it able to use potential energy to find the equilibrium position. Noting that the spring deflections and taking as the datum position the horizontal position L1 sin 1- L2 sin are L1 sin Lu and L1 of both of' both links, the potential energy is

V=

sO1 + sO1

sO1)2

s02)+

sO1 +

—

sO1 ++ L2s02)2

[al The equilibrium positions are found from (IV

=0

(902

=0

Ib] Ibi

and, taking the partial derivatives derivatives of' of V. we obtain TV'

I

—

—

c01 + m1gL1 cL)1

I

C 02

—

2

c01

+ k2( k,( L1 L1 s

+

c03 id [ci c01 ±± k2(L1 c01 k2(L1sO1 sO1 ++ L2s02)L1 c01

—

s 02 )L2 C 02

and cosO2 are common to the first into Eqs. Ib]. Because We introduce Eqs. respectively, we eliminate them from Eqs. [c] and obtain and second of Eqs. I

+k2L1L2sLI2 =

(k1

k2L1L2 sO1 + k2L1L2

+rn2gL1

Id] Idi

=

andC0502 cos02from fromthe theformulation, formulation, we we arc arc concluding that COSO1 and Note that by eliminating COSOI 0 represent equilibrium positions themselves. This basically is the both links can he vertical, or one can. If vertical position of the links. At equilibrium either both then the equilibrium position for link 2 is found by solving the second of vertical, then link 1 is is vertical,

cos

0

Eqs. {d]. and vice versa. To find the equilibrium positions where neither link is vertical, we solve Eqs. [dl simultaneously. To this end. we express Eqs. [d] in matrix form by 1

(k1 + k2L1L2

+ in2gL1

k2L1L2 [sin Oi] 01] slnO2 k2L;

[eJ Eel 2

in2gL2

4.7 which can he written written as as [KJ{q} [KJ{q} =

D'ALEMBERT'S PRINCIPLE

and whose solution is {q} =

The solution

be shown to be

[fJ

+ in,) m')

=

+ ,m) + 2k2L2m2

sinO7 S1fl07 =

An interesting case arises when k1 is set to zero, or when there is no spring attached to the middle link. In this case, case. dct[K] = 0, which implies that one cannot solve for thr the equilibrium position by inverting (cj. The double link can assume an infinite nuniber of equilibrium positions.

4.7

D'ALEMBERT'S PRINCIPLE

D'Alemhert's principle extends the principle of virtual work from the static to the dynamic case. Consider the system of N particles discussed in the previous sections. If the system is not at rest, we can write Newton's second law for the ith particle as R1

=

m,a1

i =

=

1,2,...,N 1,2....,N

[4.7.1]

p1 = my, is the linear momentum of the ith particle and R1 is the resultant of all forces acting on the ith particle. As in the static case, we split the resultant R1 into the sum of the externally applied and constraint forces as where

R1

=

[4.7.2]

F;

introducing Eq. 1 4.7.2] into Eq. 14.7.1]. we obtain

F ++ F; F,

— —

p1

= 0

14.7.31

This equation is known as the dynamic equilibrium relation, where the negnegative of the rate of change of linear momentum. — = — is treated as a force, referred to as as the the inertia inenia firce, that that provides providesequilibrium. equilibrium. We We can now treat the dynamic system as if it is a static system and invoke the principle of virtual work. Equation [4.7.31 is sometimes referred to as D'Alemhert's principle. We proceed with the dot product of Eq. [4.7.31 and the variation in the displacement, and write

(F, ++ F (F1

—

• t5r1

= 0

14.7.4] 14.7.41

Summing over all the particles gives N

+ F; —

= 0

14.7.51

245

246

CHAPTER 4• CHAPTER 4• ANALYTICAl. ANAlYTICAl. MECHANICS: BASIC CONcU'fs

Recalling from Section 4.4 that work done by the constraint fbrces over virtual displacements is zero, or N

[4.7.6] 14.7.61

>'F;.6r1 = 0 i-H

Eq. [4.7.6] [4.7.6] from from Eq. Eq. 14.7.51. we arrive at and subtracting Eq. liv

=

—

[4.7.71

0

i—I

c principle. We of D 'Alemberi, 'Alemberi. or D 'Alemberi s' This we call the observe immediately that the principle of virtual work, given in Eq. 14.6.11, becomes a special case of D'Alembert's principle. D'Alemhert's principle is a fundamental principle that provides a complete mulation of all of the problems of mechanics. Hamilton's principle and Lagrange's equations are all derived derived from from D'Alernhert's principle. principle, as will be shown in the next sections. D' Ale Alembert's over aa Newtonian Newtonian approach approach sect ions. The The advantage advantage of using D mhert s principle over is that constraint forces and interacting forces between particles are eliminated from the formulation. This advantage becomes more pronounced pronounced for tbr systems with several several degrees of freedom. We next next extend extend D' Alembert's principle principle to to rigid rigidbodies. bodies. We We consider here plane plane We motion only (the general three-dimensional case will he derived in Chapter 8). We N approaches collection of of particles, particles, so so that that in in Eq. [4.7.7). [4.7.7 N treat aa rigid body as as a collection rigid body infinity. Define the angular velocity of the rigid body as w = 0. Also, we express the position, velocity, and acceleration in terms of the center of mass motion as

r1 =rG+pi =rG+pi a1

where

—

Ok.

= a(-; +

x

—

w2pj

ii == 1,

2,

... ., N

14.7.81

.

variation of of r1 r1 can be written as so that the variation = 6k. so 6r1 = 6rG

[4.7.9]

+ 60k 60k X x p,

D'Alemberf = 60 k. Introducing Eqs. L4.7.81 and [4.7.9] into D'Alemberf:

and we recognize that

principle, we obtain N —

X p1 p1 +

m,aG —

Pn1wp,)'(6r(; + 60k X p3 p3 =

0

[4.7.101

i--H

Now, consider that the number of particles approaches infinity. The summation din. p, and dF, respecis replaced by integration, and p. and F, are replaced by din, p,. tively. Evaluating the individual terms and using the definitions of center of mass and mass moment of inertia, we obtain

dF• 6r(; 6r(1 = F• 6r(; J

din(Ok I

j X

dF• (60k x p) =

MG 60

d?naG

orG =

?fla(;

OrG

J

p)•(OOk X p) = 060J p2drn = 'G680

[4.7.11] (4.7.11]

4.7

247

D'ALEMBERT'S PRINCIPLE

where in is the total mass, mass, F is the resultant of all forces, 'G is the centroidal centroidal mass moment of inertia, and MG is the sum of moments about the center of mass. All other remaining terms in Eq. [4.7.101 are zero. It follows that D'Alemhcrt's principle lbr a rigid rigid body body in plane motion is

(F

mae;) (S r(; + (M(; ma(,)•(Sr(; (M(;

—

—

[4.7.121

I(; 0) (SO = 0 I(;O)d!O

For a system system of of N N rigid rigid bodies bodies in in plane plane motion. motion, D'Alembert's principle becomes N

mlaG1)

—

6rG + (M(;

—

0

(So, J IGOi) 6(), IGOf)

[4.7.131

I

where where the subscript Ii now denotes the ith rigid body.

Up until the second half of the 20th century. the property of D'Alcmhert's principle being a vector relationship was usually viewed as a disadvantage, and D'Alembcrt's principle was primarily considered as a tool to obtain Hamilton's and Lagrange's equations. Equations [4.7.71 or 14.7.131 were rarely used in the form given here. The need to deal with complex multibody problems and the availability of digital computers has led scientists and engineers to take another at D'Alemhert's D'Alcmhert's principle as a primary method of solution. For example, if we :ntroduce 1 bJinto into Eq. Eq. [4.7.71, [4.7.71, we obtain :ntroduce Eq. Eq.[4.4. [4.4.lbJ fl

—

injaj).(V

=

\k=

y

rN

=0

—

k=1

[4.7. 14] [4.7.14]

When we have a set of independent generalized coordinates, the coefficients of must vanish independently, with the result N A"

-

>_(F1

dr1 —

=

0

k = 1,2, . . .. ., n .

14.7.15]

Extending this to the case of N rigid bodies in plane motion, we obtain —

inIaG)•

+ (MG

—

= 0

14.7.16]

Equations 14.7.151 and (4.7.16] represent direct use of D'Alemhcrt's principle derive equations of motion.

cinsider a bead of mass in free to slide on a ring (hoop) of radius R. as shown in Fig. 4.18. The ring is rotating with the constant angular velocity 1. Find the equation of motion using Y Alernhert's principle.

S.Iuflon 3ecause we are dealing with a single particle, we drop the subscript in Eq. [4.7.71 and write .:as (F

—

ma)•6r =

0

[a]

Example

4.9

248

CONCEPTS MECHANICS:BASIC BAsic CONCEPTS CHAPT1R 4 • ANAIXTICAE. ANALYTICAl. MECHANICS:

ci.

g

171

mg

N1

FIgure 4.18

Figure 4.19

Bead on a

Free-body

rotating ring

diagram

The free-body diagram is given in Fig. 4.19. The b1b2b3 axes are attached to the hoop. The generalized coordinate is selected as 0. We first derive an expression for the acceleration. The moving frame is attached to the ring. The position vector is

r=

[bJ

—

so its variation is = RcosOSOb2 + RsinO6Ob3

Ic]

Because the motion of the relative frame, that is. of the hoop, is treated as a known quantity. its variation is zero. Hence, it is possible to calculate the variation of r in the relative frame. To see this better, write the velocity of the bead as V = Vrej +

to x r

=

+ RU sin

+

= ROcosOb2 RU cos 0b2 + RU sin

—

x (RsinOb2

—

Rcos6b3)

[dl

RU sin sinOb1 0b1

Since

is a constant and it is not the derivative of a motion variable, it cannot he expressed in terms of a variation. variation. Consequently. Consequently,the thethird thirdterm term on on the the right right side side of of Eq. Eq. [dl does not not contribute to the virtual displacement. Because the angular velocity is constant, the expression for the acceleration has the form

toxxto to xx rr ++ 2to x a = arei arei ++to

= R6 cos

+ Ub.3 Ub3 x fib fib3XX(R (Rsin sin Oh2 Oh2 —

a=

+ RO sin 0b3 —

RU2

sin (lb2 + RU2 cos 0hz

R cos Oh3) + 2flb3 2flb3 xX(RU (RUCOS cos 0b2 0b2 ++ RU sin 0b3) 6b3)

—2ROIIcos6b1 + (—Rsin0(02 +

+ ROcosO)b2 + (RU2 cosO + ROsinO)b3

Ic]

The.only The only force force acting acting on on the the system system which is not a constraint force is gravity, and it has the form F = Substituting Eqs. [c] and Eel into the generalized principle of D'Alembert yields (F

—

6r = [—mgb3 + m(Rsin 0(62 + U2) —

—

RO

cos O)b2 0)b2

in( RU2 cos 0 + RO RO sin sin O)b3 O)b3 — — 2mROU cos 0 b1]

+ RsinOôOb3) =

0

If]

4.8

HAMILTON'S PRINCIPLES

product and and setting setting the the coefficient coefficientof of61-1 61-1 equal equal to to zero, zero, we we obtain the After evaluating the dot product equation of motion as + 0+

—

cI2cosO =

[gI

1)

Let us compare the procedure we used in this example with a Newtonian ap-

proach. From the free-body diagram, there are two normal (reaction) forces. N1 and N2. After applying Newton's second law, we get three equations and we need to eliminate the reactions. It is obvious that using D'Alembert's principle is simpler. The difference becomes more pronounced where there are several degrees of freedom.

4.8

HAMILTON'S PRINCIPLES

From D'Alembert's principle we develop the scalar variational principles that provide a complete formulation of the problems of' mechanics. These principles were stated for the most general case of motion by Sir William Rowan Hamilton. Consider a system of N particles and D' Alembert's principle N

[4.8.11

= 0

—

the virtual work of all the impressed forces. To denote by 6W = manipulate the first term in the above equation. consider the expression We

=

+

ij•61j

i == 1

l,2,...,N

[4.8.2]

The second term on the right in Eq. 14.8.2] can he recognized as

=

[4.8.31

2

The kinetic energy of the ith particle is

[4.8.4]

=

that the variation of the kinetic energy of the ith particle becomes =

[4.8.51

=

md we can express Eq. (4.8.21 as in1

•

6r1) =

• 6r1 6r +

• 61',

=

•

6r1

+

[4.8.61

249

250

CHAPTIR 44 CHAPTIR

S

MECHANICS:

The variation in the total kinetic energy of the system is j 'p

(ST

[4.8.71

6T1 =

= i—I

principle as we express express D'Alembert's principle Using Eq. [4.8.6]. we ii N

N

= —6T = 00=

—

(1

m,—(r1•6r1) > ni1—(r1

i—I 1—1

(It

1—I

—

(SW

[4.8.81

potential energies so that we have an expression for the variation of the kinetic and potential A" N

/

6T+öW 67' + 6W

•6r) • (5r1)

di' dt

i—I i— I

[4.8.91

Next, we integrate the right side of Eq. [4.8.9] over two points in time, say. t., thus —1'

'V

— ,,,

+ 6W)dI = (6T+6W)dI (6T -'I'I

.i: Ir I

=

I I

J

and

(At

i—I

"

N

It,

I'V N

I

—

•6r) = '> ,n11'1 •6r1 ,n1d(i'1 •Sr1)

i—i

[4.8.10] [4.8.101 I,''I

The term in1i1 is recognized as the partial derivative of T, with respect to

so that

we may write N

= =

•

-.

'I

i— I i—I

.

N

[4.8.11]

or Iti

yields

which, when introduced back into Eq.

(1 1' .p\,

(5T ((ST -

+ 6W)dt

—

Il

i—I

-7,

—

('s'

=00 =

[4.8.12]

t

Hanijiton v principle (or This equation is known as Hamilton

varying action. One

can put this principle into more general form, by expressing it in terms of generalized we obtain coordinates alone. Introducing Eq. 14.4.7 I into Eq.

i=1

dr1

•6r

= =

1-'

t2

.'\ (lIe dr

'I

(1

14.8.13] 14.8.131

A

k=I

'I

—

'I

write Hamilton's Hamilton's principle principleof olvarying varying Introducing Eq. Introducing Eq.14.&131 (4.8.13J into Eq. (4.8.121 we write action as — (1 -(a

F,

— (6T + 6W)dt —

It

t-.

-

lIT

(Sq'

= =00

[4.8.14]

k

Note that the derivation above does not put any any restrictions restrictions on on the the time time instances instances 11 11 Note and

i'2•

4.8

HAMILTON'S PRINUPI1ES PRINCIPlES

A special case of Hamilton's principle of varying action is obtained when we

consider the variation of r as time is held lIxed. We reexamine Fig. 4. 1 2, which is analogous to Fig. B 1 in Appendix B. The varied path can take any value within the set ofadmissihlc displacements ofthe system. and it coincides with the true path at the end points. It follows that the variation ofthe displacement 6r and ofthe generalized have values of zero at I = and! = I,, provided r is specified at and Of interest is the case when r1 (i = 1, 2. . .. N) are specified. which eliminates the integrated term in Hamilton's principle of varying action, resulting in .

r ti I

it,

(5T+6W)dr =

[4.8.151

0

This equation is known as the exiended Hainilion Ham//iou S principle. Writing the virtual work as 6 W = 6 — 6 V. one can express the extended Hamilton's principle also

1•

(S7'

—

6V +

= 0

14.8.161

J

Even though we derived it here for a system of particles, the extended Hamil-

ton's principle is valid both for particles and for rigid or elastic bodies. It is. again, fundamental principle of mechanics from which the motion of all bodies can be Jescrihed. In this sense, the extended Hamilton's principle is not exactly a derived principle. Rather, it is more like a law of nature. in the same way that Newton's sec:ind law is a law of nature. Further, only scalar quantities like work and energy are needed. No acceleration terms need to be calculated to invoke this principle. Introduce the Lagrangian L such that L = T — V. For conservative systems, 51V = —6V, and we can write

6Ldt =

I

Jit,

0

[4.8.171

and Eq. [4.8. 171 71 isis relèrred relèrred to to as Hamilton 'S principle. This principle was first stated 's principle. 1

by Lagrange and originally called Principle qf k'asr least action. When the system is hubriomic, one can interchange the integration and variation operations. which yields rionlic,

Ldt =—

6

0

[4.8.181

(I

Hamilton's principle for a holonumic system basically states that among all the

paths that a system can take, the actual path filbowed fillowed renders renders the the definite definite integral = ((2 L dt stationary. This integral is also known as the action integral. The implementation of the extended Hamilton's principle for finding the equanons ions of motion requires the evaluation of' the variations of' the kinetic and potential energies. The procedure can become tedious, primarily because of the large number c'f integrations by parts that one must perform to relate the variations of generalized velocities to the variations of the generalized coordinates. A simpler and more

251

CHAPTIR 4 •

B&sic CONCEPTS MECHANICS: BASIC CONCEPTS

general procedure for deriving the equations of motion motion for for systems systems with with aa finite finite numnumber of degrees of freedom is by means of Lagrange's equations. as we will see in the next section. The direct use ol' the extended Hamilton's principle is effective when deriving the equations of of' motion motion of' of' def'orrnable defbrrnable bodies, such as for the vibrations of beams.

plates, and shells. In such problems, the extended Hamilton's principle yields the equations of motion in the form of partial differential equations with accompanying boundary conditions. We will investigate the dynamics of deformable bodies in Chapter 11. Hamilton's principle is also used in transformation theory and in optimal control theory. One may wonder why we list two major principles in this section that encompass

nonconservative forces when the first, Hamilton's law of varying action, is lengthier and has the appearance of being redundant when compared with the extended principle. The difference between the two principles is in how they treat the time instances r1 and r2. II' we view and 12 as arbitrary time instances, we obtain the extended Hamilton's principle from Hamilton's law of varying action and the two principles become

the same. But ii' we view t1 as a point at which we know the values of the generalized coordinates, then we can make use of Eq. [4.8.14] to find the values of the generalized coordinates at time t2. To do this we do not need to derive any equations of motion, just the variation of the Lagrangian and the virtual work. This approach comes in handy in numerical integration, as 12 can be taken as Ii + & in in which which is a small time increment.

I

Obtain the equation of motion of the bead problem in Example 4.9 using the extended Harnil-

toil's principle.

Solution To

find the kinetic energy, we need the velocity of the bead. From Example 4.9 we have

r=

[a]

Rcos(Th3

—

v = —RflsinOb1 +

[b]

+

The kinetic energy is

T=

+

+ (ROcosO)2 + (ROsinO)21 (ROsinO)21 =

=

fh2è2 Ic]

Using the position of the head at the bottom of the ring (0 = 0) as the the datum, the potential

energy becomes

V = ,ngR(l so

—

[dl

cosO)

that the Lagrangian has the form L = T

—

V

inR—

=

2

,

mR2

•

+

2

—

tngR(1

—

cosO)

[e]

4.9 The

E()UATIONS

variation of the Lagrangian is = dO

—

= mR2

+

[f]

+ mR2O6O

sinO I?

dO

The second term in this equation is in terms of 60. To invoke the extended Hamilton To accomplish this, we integrate we have to express all the tcrms in terms of' second term by parts and write second (12

(t2

•

4960d1 =

I

I

.

•

O—(60)dt = 060 (11 di

it1

—

Il

A60d'

I

Eg]

J:1

integrated term on the right side of Eq. [g] vanishes by virtue of the definition the variation operation. (The values of the variation at the beginning and end of the path zero.) The second term. when used with Eq. III and the Extended Hamilton's Principle. :dds The

/ I

—fliRO + mR2 (n2 shiOcosO

sinO

—

I?

)

[h]

60 di = 0

/

In order for the equality to hold, the integrand must vanish at all times. Because &1 is In must he identically zero. Thus wc

irbitrary. for the integrand to he zero the coefficient of -ecognize as the equation of motion + sinO

III vi

= o

—

Afler obtaining the kinetic and poLet us review the operations we carried out. After

energies and taking the partial derivatives, we perfirrned an integration by arts on the term 6fl. We could have done the integration by parts on the general 60 rather than the correspondiiw specific term in this problem, 060. The question then arises as to whether, manipulating the extended Hamilton's princi:le. one can perform the integrations by part in advance and develop a general ft)rm :Ie. ft)rm the equations of motion. This is the question we will explore in the next section.

4.9

LAGRANGE'S EQUATIONS

From Hamilton's principle, we derive Lagrange's equations. which present themas a convenient way of deriving the equations of motion. motion. The extended Hamilprinciple can be expressed as ft2

f12

(6 T

—

6

V+

6

di

=

—

6L J

'I

dt

dr =

6

+

14.9.11

0

J

The Lagrangian L can be written in terms of generalized coordinates ii) as L = L(qi, L(q1, velocities c'k (k = 1. 2, . •, q,,, .. ., .

.

,

..

.

•.

and I).

253

254

4S 4•

CHAPTIR

CONCEPTS BtsicCONCEPTS MECHANICS: Btsic ANALYTICAL MECHANICS:

The variation of L is

(SL =

+

[4.9.2] [4.9.21

clqk

using Eqs. [4.5.10] and 14.5.191. the variation of the nonconservative work is written in terms of the generalized forces as and,

'I

[4.9.3]

=

Making use of the property that the variation and differentiation (with regard to time) operations can be interchanged, we integrate by parts the second term in Eq. 14.9.211and and obtain obtain Eq. 14.9.2 (12

t'L 1" t'L I

/iiL d —(&/k)dt = (9qk dt

=

.

.

..

t2

(9L .

dC/k

6qk

—

/_9L \ /_9L J6qkdt — (

.

J

.

14.9.41

(k =

.. .n) at the beginning and the end of the time intervals. By the definition of the variation, the varied path = vanishes at the end points, thus = 0 for all values of k. ConsiderThe integrated term requires evaluation of

1.2.

.

ing this, and introducing Eqs. Eqs. [4.9.21—14.9.4] [4.9.2]—[4.9.4] into into the the extended extended Hamilton's principle. we obtain

J(6T

—

6V + 6Wne)di

=

J

>'

dt \dqk /

+ Qkrl(.]

= 0

[4.9.5] [4.9.51

the integral over time to vanish at all times, the integrand must be identically equal to zero, which can be expressed as For

H

dt dqkj

+ Qkiic]

=0

[4.9.6]

It should be noted that this equation can be directly obtained from D'Alembert's principle, without using Hamilton's principle. Because of this. Eq. [4.9.6 1 is some-

times referred to as Lagrange of D 'AIe,nbert 'AIe,nbert principle. qfD Consider now a set of independent generalized coordinates, it follows that the only way Eq. 14.9.6] can be equal to zero is if the coefficients of vanish individually for all values of the index k. Setting the coefficients equal to zero, we obtain Lagrange 's equations of motion

d/9L\ —I I— .

.

.-

=

k

l,2...,n 1,2...,n

[4.9.7] 14.9.7]

dt \dqk / clqk Equation [4.9.71 the most most general general form form of of Lagrange's Lagrange's equations. equations. They They can can also also [4.9.71 isis the he expressed in terms of the kinetic and potential energies. Noting that the potential energy is not a function of the generalized velocities (except for electromagnetic

4.9 systems), we wewrite write Eq. systems), Eq. 14.9.71 14.9.71

EQUATIONS

as

nv

d ci /

dqk

=

k= 1,2,...,,n 1,2,...,n

Qk,, Qk,,•

[4.9.81

This form of Lagrange's equations is preferred by many. as it reduces the possibility of making a sign error when evaluating the partial derivatives. It also is similar to the format Lagrange first presented these equations in 1788. Under certain circumstances it is more convenient to write Lagrange's equations in terms of the kinetic energy alone, alone. in the form of

(if

ci

-

dr

—

[4.9.9J [4.9.9]

=QL

where the values of Qk contain contributions from the conservative as well as nonconservative forces. The principle of virtual work given by Eq. [4.6.4], is a special case of Lagrange's equations. In the static case, the fIrst two terms in Eq. [4.9.8] vanish. For a holonomic conservative system. one can use Eq. [4.8.15] directly in conjunction with the Euler-Lagrange equation in Appendix B to derive Lagrange's equa-

tions. The order of variation and integration can be exchanged, and one seeks the stationary values of the integral I = L di, leading to

d

—

[4.9.1OJ [4.9.10]

0

Lagrange's equations can conveniently be expressed in colunm vector format4 Introducing the n-dimensional generalized coordinate and generalized force vectors {q}

=

q2

T -

{Qm:}

= [Qne [QJfl(.

•

r'h •

I

[4.9.111

we can write Lagrange's equations as

c? {

q}

14.9.121

Let us now compare the steps involved in obtaining the equations of motion using Lagrange's equations and using the Newtonian approach. When using Newton's second law, we 1.

2. 3.

4. 5.

Isolate the different bodies involved. Select a coordinate system and draw free-body diagrams. Relate the sum of forces and sum of moments to the translational and angular accelerations. Use kinematics to express the accelerations in terms of translational and angular parameters. Eliminate the constraint and reaction forces and derive the equations of motion.

255

256

CHAPTIR 4 •

CONCEPTS MECHANICS: B%sIc B%slc CONCEPTS

When When using the Lagrangian approach. we 1

.

2. 3.

4. 5.

Determine the number of degrees of freedom and select a set of independent

generalized coordinates. The free-body diagram is a useful tool for this. Use the kinematical relations to find the velocities and virtual displacements involved. Identify the forces that arc conservative and those that arc not. Write the kinetic and potential energies. as well as the virtual work. Apply Lagrange's equations.

There are two distinct differences between the two approaches. The first difference is in the order of the steps involved: In the Newtonian approach, one first writes the force and moment balances for all bodies separately and then uses kinematical relations and the constraint forces to reduce the number of equations. In the considersthe theconstraints constraintsand and kinematics kinematics of of the the problem problem Lagrangian approach. OflCconsiders first. Then, Then. the equations of motion are written, one Ibr each degree of freedom. The bulk of the work involved in Lagrangian mechanics is to find a proper set of generalized coordinates and to express the kinematics. Once this is done, the rest is straightforward. approach uses velocities and scalar The second difference is that the quantities. whereas the Newtonian approach uses accelerations and vector quantities. Dealing with velocities involves considerably less algebra than dealing with accelerations. equations should be It may appear, from the above discussion, that preferable to the Newtonian approach at all times; hut this is not so. By eliminating

the constraint forces from the firmulation. the Lagrangian approach does not calculate the amplitudes of these forces. While this may be acceptable for classroom examples. it certainly is not in many real-life applications, where one must know acting Ofl a body. Furtherthe amplitudes of the reaction and other contact more, for certain geometries a Newtonian approach is more suitable. The best way to determine which approach is most suited to one's needs is by gaining experience looking at at aa problem problem from from both both aa in solving mechanics problems. In many cases. looking Lagrangian and Newtonian point of view increases the physical insight and makes it easier to understand the characteristics of the system. We should add here that the historical development of analytical mechanics did not follow fbllow the sequence in which it is presented in this chapter. Lagrange's equations were derived before the extended Hamilton's principle, and they were derived fbr conservative systems only. It was Hamilton, horn born after Lagrange. who put together the developments in variational mechanics and Lagrange's equations to develop a general scalar principle from which all the equations of motion can he derived.

Example

4.1

1

Fig. 4.20. 4.20. find find the the equations equationsof ofmotion motionusing using Lagranges equations. Assume For the system in Fig. and that that the the force force FF is always applied that the spring and dashpot deflect only horizontally and horizontally.

4.9

257

L.A(;R..tN;E's EQUATIONS L.A(;R..tNGE's

/ k

kx

\ / (1

cv'

£1

(.

,fl

F

Figure 4.20

FIgure 4.21

Free body diagram

Solution This is a two degree of freedom system. and we select the generalized coordinates as the displacement of the mass x and rotation of the bar The free-body diagram of the entire cystem is shown in Fig. 4.21. The kinetic energy of the cart is = The kinetic energy of the bar is due to the translation and rotation and can he expressed as I

+

=

1

+ v;)

La]

??tIr/l 2. and where 'G is the mass moment of inertia about the center of niass. 'G = ??tIr/1 are the velocities of the center of mass of the bar. found as It

tjy

dl

ci

=

—'.v+

dt

cli

d

= —VG =

dr

sinO) =

•

.V +

and

b•

5OcosO

\

d(b

=

I,

Ibi

The total kinetic energy is

T=

+

I

+ +

+ +

+ in)x4 +

=

+

b•

+

Ic]

-:

6

The potential energy is due to the deflection of the spring and the vertical movement of the center of mass of the bar. written V = 2kx2 kx2

=

—

2

Id]

cosO

The virtual work of the nonconservative forces is due to the external force F and the dashpot, so =

—

6XG =

=

±

cos 0

F6(x + —

CX oX — cxbx —

—

+

ebO COS 0 Sx

+

—

cbk cos

—

Ic]

258

4• ANALYTICAL MECHANiCS: BAsic CoNcErrs

from which we recognize the generalized forces as —

cx

1

—

Fbcos0

—ehcosO

—

cb2O cos2 0

If]

—

Taking the appropriate derivatives. we obtain (IT

(IT

4-ynb0cos() ynbOcos() = (M + in):i: ÷

1

= —inhxcosO + 2

dO

(IT dx

=0

(if (IT

= kx

=

(IV

I1

[gi

—

Substituting the above values into Lagrange's equations equations we we obtain obtain the the equations equations of' of molion as flon

(M + iii)I + 1

3

Example

4.1 2

sinO c++ SiflO ++CX

—

nifrO +

+

+

—chOcosO + kx = F —ch0cosO

_(..b20 eQs2 eQs2 ü +

1

ynxbsinO = FhcosO ynxbsinO

[hi

Figure 4.22 shows a collar of mass in sliding outside a long, slender rod of mass M and length There is a force F acting at the L. The coefficient of friction between the rod and collar is tip of the rod. Find the equations of motion.

Solution We will solve this problem as a two degree of freedom unconstrained system. Polar coordinates arc suitable as generalized coordinates. The free-body diagrams are given in Fig. 4.23. There arc four external forces: two gravity forces, which we will account for in the potential energy, the friction force. and the force at the tip. 4,

() ox

\

0

r

rn

\

N

/

L \

mg

M

G

r

F Mg

Figure 4.22

Collar slidslid. ing on a rod ing on

Figure 4.23

F Free-body diagram

4.9

EQUATIONS

The position and velocity of the collar are

v= = ie,. ie, +

r= re,.

[cii

The virtual work associated with the two external forces can be written as

[bi

6W = in which Fcosq;er + F F = Fcosq;er

+ = ?rer 6rer +

= —F,sign(1)er —F,sign(i)er

F1

r6Oe()

6rp= L60e0 [ci

so that the virtual work becomes

6W =

=

= FL sin

Qr = — Fjsign (i) and

as

the generalized forces due to the nonconser-

'ative forces. The kinetic energy is +

+

+ Tcoiiar Tc(,IIar =

=

[el

and the potential energy is

V=

—in1çrcos()

—

L Mg-5- cos0

[f]

Application of Lagrang&s equations yields the equations of motion as In,: ml: —

+

— —

mg cos (9 =

—

F1

sign (r) (i)

6 = FLsinqi FLsinqi

+ 2rnriO +

[gi [hi

The friction force is related to the normal force N between the collar and rod by F, However, at this point we do not know what the normal force is. To find the normal force, need to go to a Newtonian Newtonian analysis. Reconsidering the free-body diagram and summing farces along the transverse direction. we obtain = rn( rn( r() + 21t9) 21(9) = N

—

tiig sin 0

[II [Ii

from which we obtain the magnitude of the normal force as m(rO +

iV

+ gsinO)

III

We can eliminate the normal force from the equations of motion by introducing Eq. [ii :nto Eq. [gJ. Note that the friction force is always a positive quantity, as it is proportional to the magnitude of the normal force. Thc expression involving N in Eq. U] can lead to both and negative values. Therefore, we express the friction force as F1

=

IN INI I

= /L rO ++ 2i0 + gg sin sin 61 01 ILInIrO

and and use Eq. [kJ in the equations of motion.

[ki

259

260

CHAPTIR 4 •

MECHANICS: BASIC CONCEVFS

The preceding example illustrates the problems that one encounters when dealing with problems involving friction. As stated earlier. friction is not a constraint force, but its magnitude depends on a constraint force. If we select a set of unconstrained generalized coordinates to describe the motion, as we did in this example. we cannot obtain the magnitudes of the friction fhrce without an additional Newtonian analysis. In the next section. we will see an analytical approach that calculates magnitudes of constraint forces.

4.10

lAGRANGE 'S EQUATIONS FOR CONSTRAINED SYSTEMS

The formulation of' Lagrange's equations in the previous section was for unconstrained systems and for constrained systems where the generalized coordinates are selected such that all constraints are accounted for and the surplus coordinates eliminated. This approach is not feasible under a number of circumstances: I.

2.

3.

When the constraints are nonholonornic. Because noriholonomic constraints involve velocity expressions that cannot be integrated to displacement expressions, one cannot find a set of unconstrained generalized coordinates. When the constraints are holonornic and one cannot eliminate the surplus coordinates easily, for one of the following reasons: a. The constraint equation is complicated. b. Finding the transformations that lead to unconstrained equations makes the equations of motion very complicated. Some of the forces acting on the system are functions of constraint forces. c. c. Some When the constraints are holonomic but one does not want to eliminate the surplus coordinates from the formulation, usually because of the need to know the amplitudes of ot the reaction forces.

Consider a system originally of nn degrees of of freedom, to which rn rn constraints constraints are applied. For the most general case, we express the constraints in velocity fOrm as '1

j

+ ajo = 0

= 1, 2, ... .,..

14.1 0.1] 0.1] 14.1

0

[4.10.2]

•

k=1

whose variation is fl

>a/k6qk k=1

Multiplying Eq. [4. 10.2] by the Lagrange multipliers Aj (j = 1, 2, ., m) and introducing these constraints to the extended Hamilton's principle, we obtain .

-I, I

(t1

6L di +

I

-

m ,'fl

.

fl

—

When the constraints are holonomic, the coordinates

dt =

14.10.31

0

....

no longer constitute a set of independent generalized coordinates. They are now constrained q,,

4. 1 0

L&GRANGE 'S EQUATIONS FOR CONSTRAINED

generalized coordinates. When the constraints are noriholonomic. only the gener-

alized velocities are constrained. while the generalized coordinates are still mdcpendent. In both cases, the variations of the generalized coordinates are constrained. Following the same procedure as when deriving Lagrange's equations for the unconstrained case, we take the appropriate partial derivatives and perform the integrations by parts to obtain d

= 0 [4.10.41

As in the static case, we select the Lagrange multipliers A1 such that the coefficients (k = 1, 2, ..., ii) vanish, which leads to a modified form of Lagrange's

equations. written

d d —

f9L \ .

dL

1—

di \cIqk/ I

•

-

+ '>

=

k

Qk11 Qkn

=

1,

2,

. .

., fl ..

14.10.51 [4.10.51

ftrces. They have the same units as the where A/aJk are the generalized constrain! firces.

generalized forces (which do not necessarily have the units of force). In column vector notation, Eq. [4.10.5] is expressed as — —

(u

+ {A}Tla] =

—

di

/ matrix of of order order in x ii in which [al is a matrix

[4.10.61

whose entries are afk and {A} is a column

vector of order ni that contains the Lagrange multipliers. After obtaining the equations of motion, one has two courses of action Ihr finding a solution. The first is to eliminate the Lagrange multipliers from the equations of motion and obtain a set of ii — in unconstrained equations. One accomplishes this by algebraic manipulation of the equations of motion. Many times, such an approach results in complicated expressions. The second course of action is to take the ii equations of motion in Eq. [4.10.5] and the m constraint relations in Eq. [4.10.11 and then to solve them together for the The resulting n + in equa.., .. q,1, A,, A2. ..., ± m = p + 2m unknowns q,. nons are not a set of differential equations, as there is no derivative of the Lagrange equations. multipliers involved. Such equations are known as Their analysis requires a different treatment than that for differential equations. When the constraints are holonomic and expressed in the configuration form :4.3.21, one can add them to the extended Hamilton's principle by .

. . . .,

fli fli

6Ldt+ 6 L di + .11 -'I J

and

....

6 W,k dt 6W,ZL.dt— — J ti

_tI

q,,)

= 0

[4.10.71

j—1

obtain the contribution of the constraint by replacing afk in Eq. [4.10.51 with Or, we can add them directly to the Lagrangian as Pfl ,fl

£ = T

—

V

q2.

—

i—i

.

.

.. qn) q,,)

[4.10.81

261

262

ANALYTICAL CHAPTIR 44 S CHAPTIR S ANALYlICAL

BASIC CoNcErn's

When the objective is to obtain the amplitude of' of a constraint force. an analytical approach that can be used is the constraint relaxation method. This method is mathematically equivalent to the Lagrange multiplier approach. However. it is more intuitive and it is particularly useful when dealing with holonomic constraints expressed in configuration form. Following is a description of the method. We relax the constraint from the formulation and represent represent the the effects effects of' of the constraint by a constraint force. Then, we write the Lagrangian and virtual work. The constraint force enters the formulation via the virtual work. We invoke Lagrange's equations and obtain the equations of motion. We next impose the constraint, which enables us to calculate the magnitude of the constraint force.

Example

4.133 4.1

I

12, in which a collar of mass in is sliding on a rod of mass Al and length Consider Example 4. 12. collar is is p.. p.. Obtain Obtain the the equations equations of of motion motion the rod and collar cocflicient of friction between the L. The coefficient using constrained generalized coordinates and find the value of the normal force N.

Solution describe this system in terms of constrained generalized coordinates, consider the rod and A. the collar separately. We express the motion of the collar using polar coordinates. i and A. rod, we we introduce 4), The introduce another angle. 4). as in Example 4.12. To express the motion of the rod, constraint equation is

To

[a) form as in Example 4. 1 2. We write them here The kinetic and potential energy has the same fbrrn

in terms of the constrained generalized coordinates as

T=

I

± ±

L

+ r2&) + r0)

I

V = —mgrcosO —mgr cos A — — Mg3- cos cos4) 4)

[bi [bI

The normal force N acts in the transverse direction and it contributes to the virtual work.

Considering that the velocity of the collar in the transverse direction is V() = the virtual work expression as (SW =

We

+ Nr(5A — &b)

+

r0e11.

we write

[ci

obtain the the Lagrange's equations as

For r —

in? — ?n1

For

mi2O +

A

For4)

nirA2 ,nr(92

—

ingcos A = —F, sign(/) mgcosA sign(i) +

sin A = Nr Sin

+ ±

= —Nr +

Ed]

I.] [e] If]

These These equations have to be solved together with Eq. [aJ. Equation Ed] is the same as Eq. [gJ in Example 4.12. and ii we add Eqs. Ic] and and [fi [fi and and use the constraint equation [al we eliminate the normal force and obtain Eq. [hi in Example 4.12. From Eq. (ci, we find the

normal force as ,nrA + 2,ni() 2,n1() + N N = inrA

(9 mg mx cos cos19

[g] Ig]

which which is the same value obtained in Example Example 4.12. 4.12. The The friction friction force force is is given given in in Eq. Eq. Iki Iki of' of Example 4.12.

4.10

263

LAGRANGE 'S EQUATIONS FOR CONSTRAINED SYSTEMS

The difference between the approach here and the approach in Example 4.12 is that here we calculated the normal force directly from the Lagrang&s equations. while in Example 4.12

we conducted a force balance in addition to the Lagrange's equations.

Consider the vehicle vehicle in in Fig. 4.8. Given that the velocity of point A is along the line of symConsider the Gravity acts acts perpendicular to the plane inetry of the vehicle, derive the equations of motion. Gravity of motion.

Solution We denote the coordinates of' of the the center center of mass by X and Y and select the generalized coordinates as X. Y, and 0. The kinetic energy of the vehicle is

T=

+ >2) + m(X2 m(X +

Lal lal

in is the mass and 1c; is the centroidal mass moment of inertia. There is no potential We can lind expression involves the two forces F(' and energy. and the virtual work expression the virtual work conveniently by calculating the velocities of points C and D. Defining a coordinate system xv attached to the vehicle, we write the velocities of points G and A as

(Xcos0 + YsinO)i Ysin0)i ++ (—Xsin0 (—Xsin0 + Ycos0)j

= XI + YJ

v4 = vA

+ Ok x —Li = (Xcos6 + Ysin0)i + (—XsinO (—XsinO ++ YcosO Ycos0 — L0)j

Ibi [ci

The constraint is defined as = —Xsinf9 + Ycos0 — LO = 0 f=v4'j= —Xsin19+Ycos0—LO=O f =

[dl

thus the velocity of A reduces to

V4 V4

= (Xcos6 (Xcos6 + Ysin0)i

Eel

and the variation of the constraint becomes

[fI IfI

= sin 0 6X — cos 0 6 Y + L 60 = 0

Hence, the velocities of C and D become = V/) V/) The

Ysin6— + Ok X hj = (Xcos0 + Ysin6 —

V4++Ok 19kxx(—hj) (—hj) == (XcosO (XcosO = V4

f

4-

[gJ EgI

/zO)i

[hi

Iz(1)i sin0 + Iz(1)i

F,, = F1,i. F,,i. so the virtual work expression becomes external forces are F(-' = F(•i. F/) = =

(Fe-

6r0 + + A6/ A6 + 1D FD•6r0 ens 06 X + (F(' ++ F,,) sin 06 Y + + FD) FD)cosO6X F,,)sinO6Y

+A(6XsinO—6YcosO+L60)

—

PC )Ii 50

Lii Iii

The physical interpretation of the Lagrange multiplier is that it is the resultant of forcesthat thatkeep keepVA VAalong alongthe thexxaxis, axis,and andititacts actsin inaa direction direction perpendicular perpendicular to this all forces axis. Introducing Eqs. [a] and Iii into Lagrange's equations. we obtain the equations of motion as

I

Example

4.1 4

264

CHAPTER 4• ANALYTICAL MECHANICS: BASIC CONCEPTS

niX = (Pc + FD)cos0 + Asin0 mY =

sin 0

+

[k]

COS 0

[I]

)h+LA

(FD —

'GO

— 1k

Ii]

Note that while deriving the equations of motion we did not introduce Eq. [dl directly into

the expression for the kinetic energy, thus eliminating one of the generalized coordinates from the outset. Had we done done so, we would would have eliminated the contribution due to the variation of that coordinate and ended up with an incorrect representation. This procedure is crucial to the treatment of nonholonomic constraints.

Even if we eliminated one of the generalized velocities and the Lagrange multiplier, writing the equations of motion in terms of Y and 0 (or X and 0) would not give the most meaningful description of the motion. A quantity critical to the understanding of the motion is the speed of point A. if the equations of motion can be expressed in terms of that speed. speed, one gets a clearer picture of the nature of the motion. One can introduce VA to Eqs. [j]—[l1 with a substitution. subs ti tut ion. Indeed, if we multiply Eq. IJJ by cos 0 and Eq. [k] by sin 0 and add the two we obtain ?n(XCOSO + Ysin0) = ?n(XcOSO

F(-

+ F,,

Em]

Recalling from Eq. [ej that v,1 = (X cos 0 + Y sin 0), differentiating this expression we obtai ii

VA = Xcos0

+ Ysin0 + 0(—Xsinf9 + YcosO)

In]

Introducing Eq. [dJ to Eq. [n] ml we can write X cos (3 +

sin (3

=

—

lo]

L02

SOthat so thatEq. Eq.[ml [nil can can be be written written as —

L62) = Fc + FD

which is recognized as the force balance along the x direction. We next find an expression for the Lagrange multiplier A and introduce it to Eq. [e}. [j] with sin (3 and Eq. [kI by — cos 0 and add the two, with the To this end, we multiply Eq. U] result mn(Xsin(3 m( X sin 0 — — Y Ycos0) cos 0) =

Iqi

A

We can introduce introducethis thisrelationship relationshiptotoEq. Eq.111, Ill, but a more meaningful expression can he generated if we consider Eq. [d] and differentiate it X sin 0

—

Y

cos 0 + O(X cos 0 + Y sin 0) =

—

LO

Er] [r]

Considering Eq. [qj. [qj, we express the Lagrange multiplier as A A

=

— —mvAO —

mU)

[s]

Introducing this this equation equation into intoEq. Eq.[11 [II we obtain (It- + mL2 )O +

= (PD

—

)h

which we recognize as the moment balance about point A. Equations independent equations of motion of the vehicle.

[I] and [t] are the two

EXFRCISFS

REFERENCES lnitial-%kzlue Problems in Brenan. K. E., S. L. Campbell. and L. Petzold. Numerical Solution of l,iitial-%kzlue Diffrrential-A igebraic

New York: Elsevier Elsevier North North Holland, Holland,11989. 989.

Ginsberg. J. J. H. Advanced Engineering Dynamics. New York: Harper & Row. 1988. Ginsberg.

Goldstein. H. Classical Mee/zanics. Goldstein, Meelianics. 2nd 2nd ed. ed. Reading, Reading, MA: MA: Addison-Wesley. Addison-Wesley. 1980. 1980. Greenwood, Dwzamics.2nd 2nd ed. ed. Englewood Englewood Cliffs. Cliffs. NJ: NJ: Prentice-Hall. Prentice-Hall. Greenwood. D. T. Principles (?/ Dynamics.

Haug. E. J. Intermediate Dvnan?ics. Haug, Drnan?ics. Englewood Cliffs. NJ: Prentice-Hall. 1992. Linczos. C. The Variational Principles of Mechanics. 4th ed. New York: Dover Publications. 1 970. o/ Analwical Analytical Dynamn ics.New New York: York: McGraw—Hill. McGraw —Hill. 11970. 970. Meiro vi tch, L. Methods of Meirovitch, Dynamics.

HOMEWORK EXERCISES SECTION 4.3 1.

The four-bar linkage in Fig. 4.24 is a single degree of freedom system. Show that this is so by separating the mechanism into its three components and by writing the constraint equations that relate the configurations of the links. A bead slides up a spiral of constant radius R and height /i, as shown in Fig. 4.25. It takes the bead six full turns to reach the top. Express the characteristics of the path of the bead as a constraint relation.

smooth paraboloid paraboloid of of revolution revolution described describedby byzz = r2/b. A particle slides inside a smooth as shown in Fig. 4.26. Using cylindrical coordinates, find an expression for the constraint force on the particle. &

The radar tracking of a moving vehicle by another moving vehicle is a common

problem. Consider the two vehicles A and B in Fig. 4.27. The orientation of vehicle A must always be toward toward vehicle vehicle B. B. Express the constraint relation between the velocities and distance between the two vehicles and determine whether this is a holonomic constraint or not.

I ?fl

.

I?3

I Figure 4.24

Four-bar

linkage

FIgure 4.25

265

266

CHAPTER 4•ANALYTICAL CHAPTER 4• ANALYTICAL MECHANICS: MECHANICS:BAsic BAsic CONCEPTS CoNcEvrs

B I

V

/, /

/

, I/

£

'I

I

I I

I

/

I

YR .11 1

R2/h A

9

7

I

VA

I

A V

Figure 4.26 5.

Figure 4.27 Consider the double pendulum in Fig. 4.3. It is desired to have the velocity of

the tip of the pendulum point toward the pinned end 0. Express this condition as a constraint and determine whether the constraint is holonomic or not. SECTION 4.4 6.

7.

Consider a particle moving along a path and the description of motion by path variables. Express the force keeping the particle moving along the path in terms of its components in the tangential, normal, and binormal directions and evaluate the virtual work expression. Identify which of these forces are constraint forces and verify Eq. [4.4.141. Find the virtual displacement of point P in Fig. 4.28. The mass is suspended from an arm which is attached to a rotating column. The pendulum swings in the plane generated by the column and arm. z

—L1

) 1

P J71 '77

FIgure 4.28

HOMEWORK EXERCISES

B

I.,

F

o

FIgure 4.29 8.

Slider-crank mechanism

FIgure 430

Express the virtual displacement of the slider in the slider-crank mechanism shown in Fig. 4.29 using (a) the relative velocity relations, and (b) the analytical expressions.

9. A uniform solid cylinder of radius R rolls without slip on a horizontal plane and an identical cylinder rolls without slip on ii it (Fig. 4.30). Find the virtual displacements of the centers of the cylinders. 10. Consider Fig. 4.2 and the case when the cord is getting pulled down by an external force, such that the length of' the cord varies by L(t) = Find the virtual displacement of the mass. SECTION 4.5

Find the generalized force associated with the system in Fig. 4.29. 12. The spherical pendulum of mass in shown in Fig. 4.2 has its length being reduced by a force F. according to the relationship L(t) = L() — hi, where L0 is the initial length and b is a constant. Calculate the generalized forces using spherical coordinates as generalized coordinates. 13. Consider Fig. 4.13. and calculate the associated generalized forces. The disk is of mass in and the rod is of mass 2in. There is a moment M acting on the rod at the pin joint. 11.

SECTION 4.6

For the two links attached to a spring as shown in Fig. 4.31, [md the equilibrium position. The spring is not stretched when the rods are horizontal. 15. Find the equilibrium position of the rod of' mass ni and length L sliding in the guide bars shown in Fig. 4.32. The spring is not stretched when the rod is vertical. The sliders are massless and the contact between the horizontal slider and the guide bar involves friction with coefficient 14.

16.

Find the equilibrium position for the system shown in Fig. 4.33, with the middle mass equal to zero. Assume that the displacements are small, and that the springs

267

268

CHAPTIR 4

ANALYTICAL MECHANICS: BASIC CONCEVTS ANALYTICAl.

g

L. in

I'

o

k

0 in, I.

ii,

F??, I.

,i,

L 1

B

p

()

Figure 4.32

Figure 4.31

deflect only in the vertical direction. Lise Use as generalized coordinates the translation of the center of the rod and the rotation of the rod. Then, use the deflections of the springs at A and B as generalized coordinates and obtain the equilibrium configuration. Compare the results.

17. Consider the two systems in Figs. 4.3 and 4.14 and set up the equations to find the magnitude and direction of the force F necessary to keep the systems at equilibrium at = 30°. = 15°. Find Find the equilibrium position of the system in Fig. 4.34. 19. Find the equilibrium position of the pulley system in Fig. 1 .77 (Problem I .36). Use the constrained coordinate approach. I

/

x k

/ --------.

----------

1.

I k

1

Figure 4.33

Figure 4.34

269 SECTION 4.7

D'Alembert's principle. the equation equation of motion of the rod in Fig. 4.32 using D'Alemhert's 20. Find Find the using D'Alembert's 21. Find the equations of motion of the pulley system in Fig. 3.40 using principle. The pulleys are massless.

SECTION 4.8

in Fig. Fig. 4.34 4.34 by by Hamilton's the system system mechanism in Find the equation of motion of the 22. 22. Find principle. ilton's principle. Hamilton's 23. Find the equation of motion of the rod in Fig. 4.32 by Ham

SECTION 4.9

24. Find the equations of motion of the system in Fig. 4.35 using Lagrange's tion S. tions.

25. Use Lagrange's equations to derive the equations of motion for the Foucault's pendulum in Chapter 2. 26. Figure 4.36 depicts a simplified illustration of a spacecraft to which a robot arm nioves by a moment T exerted is attached at the center of mass. The robot arm moves to it at the pin joint by a motor on the spacecraft. Considering only plane motion for both the spacecraft and the robot arm, derive the equations of motion by Separating the two masses, writing lbrce and moment balances. and a. Separating eliminating constraints. b. Using Lagrange's equations. Compare the complexity in both cases. block semicircular block positioned over over a semicircular is positioned length LL is 27. A block ol mass in and length (Fig. 4.37). It is given that friction is sufficient to prevent slipping. Derive ovcr the semicircular block. the equation of the rod as it rocks over

V

/ g

y

F1\

-t

a/2

(•1

x 0

Figure 4.35

Figure 4.36

270

CHAPTER 4• ANALYTICAL MECHANICS: BASIC CONCEPTS

(;

L

Figure 438

Figure 4.37

radius R rolls without slipping on a wedge of mass m1 (Fig. 4.38). The wedge is moving under the influence of the force F with no friction. Obtain the equations of motion. 29. Find the equation of motion of the rod in Fig. 4.32 using Lagrange's equations. 30. Find the equations of motion of the pulley system in Fig. 3.40 using Lagrange's equations. 31. Find the equation of motion of the bead in Fig. 4.39 sliding without friction in the parabolic tube described by z = x2/4, while the tube is rotating about the z axis with constant angular velocity 32. Find the equations of motion of the system in Fig. 4.33, using Lagrange's equations. Assume small motions and that the springs and dashpots deflect only ver-

28. A cylinder of mass m2 and

tically.

33. A particle of mass in is constrained to slide without friction down a channel as shown in attached to a cone spinning with constant angular velocity Fig. 4.40. Derive the following: a. A differential equation of motion describing the motion of P using Newton's second law. b. The equation of motion using Lagrange's equations. 1)3

R

-,

xS.

4 h

/

Figure 4.39

Figure 4.40

_____ _______

HorwwoRK EXFRCISES HOIIEWORK EXFRCISES

??Ii. 1..

fl?4. 1.4 1.3 m$.

,

(:14

1.5 i

.

L x

05

Figure 4.41 34. Consider the pendulum in Fig. of mass in, L1 = L. L. = 2L. The pendulum turn swings swings in in the the plane plane generated generated by by the the column column and and arm. arm. Derive Derive the the equations equations of motion. using the pendulum angle 0 and rotation angle 4 of the shaft as generalized coordinates. The combined mass moment of inertia of the column and arm about the Z axis is 1. 35. Consider the particle in Problem 3. Find the equations of motion.

SECTION 4.10

Consider the four-bar linkage mechanism in Fig. 4.24 and derive the equations of motion using constrained generalized coordinates. Consider each link separately. A moment M acts on the first link. 37. Consider the platform robot in Fig. 4.41. Derive the equations of motion using constrained coordinates for the following cases: (a) each link is considered sepaand (h) links I and 2 and links 4 and 5 are each considered as one system 36.

each.

3S. Given the two-link system in Fig. 4.3. derive the equations of motion using the constrained coordinate formulation and using as generalized coordinates 02, Xp, and Vp. Calculate the kinetic energy of the second link using the motion of its center of mass, expressed in terms of' 02. Xp. and Vp.

39. Consider the vehicle in Fig. 4.8. We are given the constraint that the velocity of the tip of the vehicle, that is. point E. is along the x direction. Derive the equations of motion using constrained generalized coordinates. 40. Consider the pendulum in Fig. 4.28. The angular velocity of the column is kept by a motor that generates a moment about the Z axis. Find the equation of' motion of the pendulum and using the constraint relaxation method find the moment necessary to maintain the constant angular velocity of the column. 1. Find the equation of' motion of the rod in Fig. 4.32 4.32 using using the the constraint constraint relaxation relaxation method. constant at 4) =

271

chapter

ANALYTICAL MECHANICS:

ADDITIONAL Tonics

5.1

INTRODUCTION

This chapter continues with the analytical techniques discussed in the previous chap:er and introduces additional concepts. We first discuss natural and nonnatural sys-

:ems, and revisit the concept of equilibrium. We consider small motions around equilibrium and derive the linearized equations of motion about equilibrium. A new of describing damped systems is introduced. We analyze the response of tinearized systems. which in essence is multidegree of freedom vibrations. We look at coordinates and generalized momenta and discuss integrals of the motion. We develop Routh's method to eliminate cyclic coordinates. Then we revisit the idea of Irom a Lagrangian perspective. motion from We continue with Hamilton's canonical equations. These constitute an alternaive to Lagrange's equations and have the advantage of being first-order equations. The chapter ends with a look at computational issues involving the derivation and :ntegration of the equations of motion and with the introduction of additional differvariational principles, namely Jourdain's and Gauss's. These additional vanprinciples make it easier to analyze nonholonomic systems. The concepts in this chapter enhance the understanding of the principles studied :n Chapter :zi Chapter as well as providing additional insight and perspective. rfhe reader is iiways encouraged to analyze a problem using more than one perspective.

5.2

NATURAL AND NONNATURAL SYSTEMS, EQuILIBR1urI

of frecdom freedom Consider the kinetic energy of a system of N particles that has ii degrees of N

T T =

[5.2.11

273

________

274

CHAPTER 5 • ANALYTICAL ANALYTICALMEChANICS: MEChANICS:ADDITIONAL ADDITIONAL Pr

For the most general case, the position vector of particle i can be expressed as

= r,

•

•.

•.

15.2.21 15.2.2]

1) q, t)

We have not excluded the possibility that displacements (i = 1, 2. 2, .. ., N) have an explicit dependence on time. This may seem strange at first, as the generalized coordinates should be sufficient to describe the evolution of the system completely. Viewing r1 also as explicit functions of time implies that the system has a prescribed motion or that part of the motion is treated as known. From Newton's third law, the motion of each body is related to the motion of every other body that it interacts with: thus. in absolute reality such prescribed motion does not exist. However, in several cases reasonable approximations can be made. For example, when we analyze the motion of a body on the surface of the earth, the motion of that body both is influenced by the motion of ui the earth. The motion of the earth is not affected by the motion of the body. and it is treated as known. The time derivative of' r, is .

1

vi= dt V1

dr1

=

i=I

(?t

2.

,

. .

.

.

.

[5.2.31 15.2.31

or ô'r,

or dr lit

.

15.2.41

Introducing Eq. [5.2.4J into Eq. [5.2.11. [5.2.11, we obtain iV

dir1

T =

-

Ix' in1 Lx' =—> =__> I

2ii=1

tlk+ tIk+

ii

/__ri

s-=

1

I

(9qk

.•

.

(?qs

+2

(1r1\l

+

(II) dr1

-,

I

In dr1 I/fl

\ •.

(if)]

dr1

dr1

•

k—i

.

or1

dr1

CII

(1/

0)41k

15.2.51

Denoting by N

akc = ak5

13k

flu nil

i—I

(?f1

k, s = 1.2,.. k,s 1. 2,

. .

iqk

N

=

lir,

) in. lir1 lir,

((is

•.

ii

11

1

lir1

lir1

2

lit

lit

T= — ) in,

lii

.

[5.2.6] [5.2.61

where and T are functions of the generalized coordinates and time, one can express the kinetic energy as 7'

+To = 12+7! +To

15.2.71

in which

= i I>TI = ,,

'I

ri

k=i s—i k=1

kc '1k '1k qs

T1

= >1[3kqk k—I

[5.2.8]

5.2

275

NATURAL AND

Examining the nature of the terms that contribute to the kinetic energy. we noice that T2 is quadratic in the generalized velocities. T1 is linear in the generalized and it is usually related to Coriolis Codolis effects. T0 has 110 no generalized velocity and and is usually related to centrifugal effects. Both T1 and exist because of exist the explicit time dependence of r1 due to aa component of the motion being being treated treated as The kinetic energy can he expressed in matrix form as +

T=

+r

:n which :ri which {c} = is the generalized generalized velocity velocity vector. vector.{/3} = referred to as the gvmscopic vectol: and the matrix •

(M]=

a11 a21

a12 a,2 a22

•

•

15.2.91 •

••

••

P,1]'

a1,1

•

15.2.101

referred to as the iizass referred masS matrix matrix or or the the inertia inertia incitrix. matrix. One can show that fM] is and positive definite. When T1 = T0 = 0. the system is called a natural S stein, stein, and and when when T1 T1 or T0 not zero the system is called called noimatural. noimatural. The name name nonnatural is associated the fact that in a nonnatural system. a component of the motion is treated and/or as known; hence, the system being observed is artificial and not a natural :ne. One should always keep in mind the procedures used and the assumptions made when viewing a system as nonnatural and evaluate whether the assumptions are —ealistic or not.

Tile Foucault's pendulum studied in Chapter 2 represents a nonnatural system, The system. we treat the motion of the earth as known and unaffected by the motion of the So is the bead on a hoop example considered in Chapter 4, where we that the motion of the bead does not affect the rotation of' the hoop. As an illustration of what constitutes a nonnatural system, system. consider a particle :1 mass in sliding along an incline a with speed q. as shown in Fig. 5.1. The in(of mass M) is moving in the horizontal direction with speed The total

/

V

Figure 5.1

Mass sliding on a moving wedge

276

CHAPTER 5 • ANALYTICAL MECHANICS: ADDITIONAl. Torics

velocity of the particle is V= so

15.2.111

qSiflLrj

+ (JCOS(Ji

the kinetic energy becomes

7= 7,

I

=

1

.

V

+

1

.-,

= =

I

.

1

..

..'

+ 2qwcos cT a ± it'J +

.

-)

.

.

+ £/COS(T) i/C0so) ++ (qsino)

,) -) I

+

I1

1

[5.2.121

Whether this system is natural or not depends on how the velocity of the in-

to be known a priori, then dined surface is treated. If we consider the velocity to there is only one generalized coordinate. q. and the system is nonnatural. The term has no gencos 0 becomes linear iii in the generalized velocities. in addition. eralized velocity terms. On the other hand. if we treat w as a variable. the system has two degrees of freedom and every term in the kinetic energy is quadratic in the generalized velocities. The system becomes a natural system. Nonnatural systems can he categorized into two distinct distinct groups: groups: (1) (I) when both T1 and T1, are nonzero. and (2) when only Jo is nonzero. The second case can basically he treated as an otherwise natural system with an equivalent kinetic energy T2 and equivalent potential energy U = V — T0. referred to as modified potential energy or dynamic potential. The Lagrangian for such a system can be written as

L=

T

—

V

=

T2 + To

V=

12

—

Li

15.2.13]

Recall that T2 is quadratic in generalized velocities. The case when T1

0 usually

describes more complex problems: its response characteristics are quite different than when T1 = 0. We now now reexamine the equilibrium of a dynamical system. We considered static We equilibrium in Chapters 1 and 4 and defined it as the state at which velocities and accelerations of all parts of the system are zero. All of the components of the system are at rest. Of course, because all physical velocities and accelerations are zero, so are the generalized velocities and accelerations. For a nonnatural system equilibrium is defined as the the slate slate where where all all generalized wiLed velocities and accelerations are zero. The velocities and accelerations of certain parts of the system are not zero at equilibrium. For example. for a nonnatural system where there is motion with respect to a moving frame, at equilibrium, only the motion viewed from the moving frame is at rest.

Consider an ii degree of freedom system with generalized coordinates •, q,,. At equilibrium, all generalized velocities are zero, and all forces and c12, moments acting on the system are not functions of times that is. (time. . .

it follows that Lagrange's equations become —

rIL = Qk11

k

= 1.2

ii

15.2.141

and T, are functions of the generalized velociFor a nonnatural system. because T2 arid ties. they vanish at equilibrium, which results in the equilibrium condition ties,

5.2 —

—

277

AND NONNATURAII NONNATURAI. SYsTEMs, EQuILIBRIuM NATURAL NATuRAL AND

V) = iiU • = rlqk

= T0 = For a natural system T = 7'2' and the equilibrium position is given by

15.2.1 51 reduces to Eq. [4.6.4].

0. so Eq.

(1V

15.2.161

=

When all forces acting on the system are derivable from a potential. the equilib-

riurn position for a nonnatural system corresponds to the point where the dynamic has a stationary value, that is.

15.2.171

= 0 qk

For = 0. which is the relation obFor natural systems this relation reduces to in Chapter 4 when deriving deriving the the principle principle of of virtual virtual work work For for static equilibrium.

We return to the bead problem considered in Chapter 4. However, here we treat the rotation of ±e hoop as another degree of freedom. Let us write the rotation rate f1 as th. The Lagrangian.

we derived in Example 4.10. can now he written as

L= T

V

—

mRcb sin 0 + mRçb

=

;nR202

—

ingR(

I

—

cos

[a] lal

I is the mass moment of inertia of the hoop about the axis of rotation. With the term we have included the kinetic energy associated with the rotation of the hoop. The hoop-bead system can now be treated as a natural system. Observe that cb is absent from the the Lagrangian. Lagrangian. This puts the problem at hand into a special category. which we will evaluate the partial derivatives of the Lagrangian. writing iialyze later on in this chapter. We evaluate iiL dO

=

= 0

mR2th2 sin (1 cos 0

(11

= mR2O

sin 0

—

.

.

= 1(f) 1(f) + ?flR(f)Sifl ?flR(f)S1fl00 = (1 + mR2 sin2 0)f)

IbI Ib] Id

ntroducing Eqs. [bi [bi into into Lagrange's equations we obtain the equation of motion for 0 as +

—

o

= 0. thus

for 4.). if). we note that To evaluate evaluate the equation o1 motion niotion for

diL d)L

d

=

•

,.

Idi Id]

= 0o

cos

.

(It

-

= 0

[.1

constant

If] IfI

then have an integral of the motion as then ÔL (1ç11 (1(f)

=

.

1(f)

+.

sin 0 =

Example 5.1

278

CHAPTER

5

ANALYTICAl. MECHANICS ADDITIONAL

rfol,I(s

Equations [dJ and [f] need to be integrated simultaneously to obtain the response.

= constan: ever, we can simplify this problem if we take advantage of the fact that th Denoting this constant by C. where C depends on the initial conditions, we can express 4) C ?flR2 = I + ?flR2

sin•2

The constant C is an integral of the motion. Introducing Eq. IgI into Eq. [dl, we rewrite the equation of motion fbr for 0 as

/

0 + sin0

—

R

C .,

I

I

\1+mR2sin'0) + mR2sin'0)

[hI

= 0

can be he solved by itself, and the value ot 4 at any point in time can Equation ascertained by substituting the value of ii into Eq. [g]. We can treat the angular velocity of the hoop as a constant in two ways. The first is ir: the presence of a motor that keeps the angular velocity constant. In this case, we have: nonnatural system. In the second way. we assume that 4) remains constant on its own. us analyze the accuracy of this assumption. The mass moment of inertia of the hoop is I = MR2/2. where M is the mass of the hoop. Introducing the mass ratio p. = m/M, Eq. [gJ car: be rewritten as

._2 •_2 —

C

ui

MR2 1 + 2p. sin2 0 2p. sin2

When p. is small, one can expand Eq. [hi in a Taylor series

—2p.sin20)

LI]

variable 4) 4 is is depends on the mass ratic Eq. UI It is clear from Eq. UI that how much of aa variable When p. is very small, one can assume that 4) is constant and treat the problem as nonnatura with a single degree of freedom. The initial conditions on 0 also affect the accuracy of this

Let us next analyze the equilibrium positions for this problem. When we consider the rotation of the hoop as a variable, the equilibrium positions are 4)

[k]

sin fi = 0

= constant constant

ir, that that is, the bead is either in the top or at the That is. the hoop is not rotating, and 09 = 0 or 'n-, bottom of the hoop. the dynamic When we consider the motion of the hoop as a known quantity. say potential U is

U=

V

—

T0

= mgR( 1

—

cosO)

—

[II

12 sin2 0

Taking the variation of U and setting it to zero, we obtain the position describing equilibrium as 6U

=

60 = (mgR sin 0 —

leading to the equilibrium equation

,iiRfl sin 0 cos 0)60 =

0

[ml

5.3

,ngRsinO(l ,ngRsinO(1

SMALL MOTIONS ABOUT EQUILIBRIUM

In]

= 0

—

g

/

'ote that for this problem. setting the partial derivative of V with respect to 0 equal to zero rind the equilibrium position position would would give give incorrect incorrect results. results.Solving Solvingfor br the equilibrium powe obtain

sinO =

0

eosO

lol

= Ru2

.eading to the equilibrium angles =

=

=

Let us next analyze the moment that is necessary to maintain a constant angular velocity

o :s

this end, we make use of the constraint relaxation method described in Section 4. 10 consider that a monlent moment T is acting on the hoop. The associated virtual work expression so that thatconsidering consideringEq. Eq.[lIthe [lItheequation equation of of motion motion of of the the hoop hoop becomes becomes = T 8b, SO

To

cit

Imposing the Imposng the constraint constraint that angular velocity becomes

=

T T =

5.3

[qJ

= T

= constant, the moment required to maintain the 2mR2QOsinOcosO

In

SMALL MOTIONS MoTioNs ABOUT ABOUTEQUILIBRIUM EQUILIBRIUM

As discussed in Chapter 1. the behavior of a system in the neighborhood of equilibrium is of utniost interest. Here, we extend the developments of Chapter 1 to multiiegree of freedom systems. We derive the equations of motion, find the equilibrium = 0 (r = 1. 2, .. ii). and linearize by a Taylor denoted by qr(' and series expansion.' As we saw in Chapter 1, 1. if the linearized equations exhibit sigbehavior, then the nature of the motion in the neighborhood of equilibrium governed by this significant behavior. Otherwise, one must perform higher-level stability analyses. Another physically intuitive way of analyzing analyi.ing motion in the neighborhood of equilibrium is to expand the kinetic and potential energies in the neighborhood of and express equilibrium. Let us consider a system with flO nonconservative the potential energy V (or modified potential energy U) in a Taylor series scrics expansion .. ci,,), about the equilibrium position. Noting that V = V(q1, qn), and denoting the the Taylor series expansion of V has the equilibrium positions by . ., .

.

.

.

.

Note that we are switching to the index r for the generalized coordinates, to avoid confusion with the symbol k that denotes stiffness.

279

280

CHAPTER 5

I

ANALYTICAL MECHANICS: MECHANIC'S: AI)I)ITJONAL

Topics

form fl

V(q1,

. .. .

Y

+

•

(( (

(qr

)

—

qre)

—

fl

1

2

where ye

/ d

Fl

;/

(c — qre)(qs )(qs

;z

—

) +f

h.o.t.

15.3.11

is the value of the potential energy at equilibrium. Without loss of generality, we select the datum position for the potential energy being zero at equilibrium, so that V(l = 0. We measure the generalized coordinates from equilibrium, so that = 0 (r = 1, 2, . ., ii). Recalling that for a system at static equilibrium all the first derivatives of V vanish, .

.

dV

0

(2q,. Oq,.

so

.

15.3.21 [5.3.21

that Eq. [5.3.11 reduces to fl

V(q1, V(q1.

•.

•.

.. ., qn) =

F?

>'

krcq,.q.s. + h.o.t.

15.3.3]

in which k rs =

)

r, r,

s = 1.2, 1. 2, .. ...,., ii

15.3.4] 15.3.41

are referred to as stiffiwss coe//icienl.v. coe//icienls. This

name is used in analogy with the potential energy of a spring. For nonnatural systems. at equilibrium '3U/ôq,. vanishes and the stiffness coethcients have the form Jr

— —

[5.3.51

Making use of the generalized coordinate vector {q} •• = fqj q,,]7' can write the quadratic approximation to the potential energy in matrix form as •

V

one

[5.3.6] 15.3.6]

in which [K is known as the stiffness matrix. The stiffness matrix is symmetric. FurJ

thermore, if for a conservative system [K] is positive definite, the potential has a minimum at the equilibrium confIguration. This can be concluded by comparing the potential energy V with the developments in Appendix B regarding the minimization of a function. The Hessian matrix in Appendix B becomes the stiffness matrix when the function whose stationary values are sought is the potential A positive semidefinite stiffness matrix is usually an indication that the system possesses rigid body motion. It should he be reiterated that Eq. [5.3.6] is valid when the generalized coordinates assume small values, and that it is for small motions about equilibrium only.

5.3

Moiioxs ABOUT EQUILIBRIUM

A theorem from stability theory states that for a natural conservative system. if

the potential energy has a local minimum at equilibrium. then the equilibrium posiuon is stable, implying that if the system is disturbed from its equilibrium position it either returns to the equilibrium position or oscillates around it. The instability theorem stales that if the potential energy does not have a local minimum at equilibrium. the equilibrium position is unstable. We conclude that the equilibrium position of a natural conservative system is stable if the stiffness matrix associated with that equilibrium position is positive definite. For nonnatural conservative systems, the corresponding stability theorem is different. When T1 = 0. 0, the system is treated as an otherwise natural system with kinetic energy T2 and potential energy U. However, the situation changes when T1 0. The theorem states that when U V — T0 is a local minimum at equilibrium, the equilibrium position is stable. But when U does not have a local minimum at equilibrium, the equilibrium equilibrium position position is is not not necessarily necessarily un unstable. stable. This concept can explained physically by noting that gyroscopic effects usually increase the stability properties. Table 5. 1 the stability theorems, We next investigate linearization ol the kinetic energy. For natural systems. the. kinetic energy is

[5.3.7]

T = T, =

we note that it is already in quadratic form. It follows that, for small motions .ibout equilibrium, the kinetic energy can he expressed as

[5.3.8]

T =

the subscript e denotes denotes that the inertia matrix is evaluated at at the the equilibrium equilibrium We conclude that, small motions of a natural system about equilibrium, the Lagrangian can be written as

!{q}T[K]{q}

L=

Table 5.1

[5.3.9]

Summary of stability theorems for conservative dynamical systems

Is Dynam ic Potential U Minimum? Or. Is 1K] Positive Definite? Yes

Natural systems

No

Stable

Unstable

T1 = 0

Stable

Unstable

0

Stable

No COflCIUS!Ofl

Nonnatural systems T1

281

282

MECHANICS:ADDITIONAl. ADDITIONAl. Toiics CHAPTER 5 • ANALYTICAL ANALYTICAL MECHANICS: Toiics

Lagrange's equations aregiven givenininEq. Eq.14.9.12 14.9.12 1. 1. Using Using the equations in in column column vector vector format formatare properties in Chapter 2 of the derivative of a scalar with respect to a column vector. we have d = ( di \d{q}j

(IL

=

15.3.101

d{q}

introducing Eqs. [5.3. 10] into Eq. [4.9. [4.9.12] 12] we we obtain obtain the linearized equations of mo-

tion in matrix form as 15.3.1 1]

+ [K1{q} =

that after calculating the kinetic and potential energies about equilibrium, one can use them directly to derive the equations of motion. Next, consider nonnatural systems. The kinetic energy is T = T2 + T1 + The To term is absorbed into the modified potential energy U. and it enters the stiffness matrix via Eq. [5.3.51. The T2 term is treated the same way the entire kinetic energy is treated in a natural system. It follows that. for the case where T1 = 0. Eq. [5.3. 11] is still the linearized equation of motion, with the entries of IKI obtained by using Eq. [5.3.5] [5.3.5] and and [MC] obtained from T2. = When T1 = f3,.q,. is not zero. the equations of motion have an is aa Function Function of added term. Noting that the gyroscopic vector [3,, 1' is = ($1 the generalited coordinates, we write the Taylor series expansion of T1 as so

\

..

Ii

T1

=

T1 ({(;'e})

1?

-.

r1

+

;'

('/r

ji

—

/ -'fl-/I (dl -'

q,e) q,e) +

I

I

yiq,.

r—

c/i.

F,

11

F1

(ci, — — q,ce) + — qie)(qc —

2

/

£/rqs

r= II s— s— I Fl 11

) le

—

+yy r

..

..

17 P7

)e (qr — —

1 S = II

+ h.o.t.

IC

15.3.12] Of' all the ternis on the right Of all side of of this this equation. equation. only two survive: the third right side term and the last. The tIrst. second, and fourth fourth terms terms vanish vanish because {O} at second, and = {O} equilibrium, and the fifth fIfth term vanishes because T1 is not quadratic in the generalized velocities. Introducing the notation as the value value of of f3r f3r at at equilibrium equilibrium and andBc,. = a,Bc/rlq,. evaluated at equilibrium, and noting that

=

ti2T1

/3

dq, dq,.

=

r, s =

Bçr

1

,

2,

.

.

.

,

ii

15.3.13]

we obtain the approximation to T1 in the neighborhood of equilibrium as fl

F,

T

-.

I

r—I r--I

P F

,-

Ft II

r

15.3.141

s=Ir=I

or. in column vector format T1

+

15.3.151

5.3

We now introduce T1 into the Lagrangian and write

the entries of [B] are

L=

283

SMALL MOTIONS MorloNs ABOUT ABOUT

I

÷

-

+

[5.3.16]

riking the appropriate derivatives, we obtain =

1M1,

I

B

=

- [K]{q} + [5.3.171

leads to the equations of' motion as

+ ([B] ([B] — — [B]'){cfl [B]' ){cfl ++ [Kj{q} =

15.3.181 15.3.18]

It is of interest to note the coefficient of the generalized velocity vector

A

mirix subtracted from its transpose results in a skew-symmetric matrix (a null maif the original matrix is symmetric). Denoting [B] — [B]' by [G], where (GJ linearized equations equations of' of motion for a we write the linearized matrix, we called the gvroscopic matrix, system as

[5.3.191

+ IKJ{q} IK]{q} =

[M11{i/} +

For relative motion problems when the motion of the relative frame is treated as

T1 leads to the Coriolis effect. For linear or linearized systems. the Coriolis is manifested in a skew-symmetric matrix. Note that in Example 2.16. the pendulum. the linearized equations of motion in column vector format a skew-symmetric matrix as the coefficient of the velocity vector. The same •=ic be said about the projectile motion example, Example 2.15. The first step associated with understanding the nature of motion in the neighof equilibrium is to linearize the equations of motion about equilibrium and imply significant significant behavior. behavior. IfIf there there isisflO no significant —- if the linearized equations imply then higher-order analyses need to be conducted, including the stability summarized in Table 5. 1.

a

the stability of the equilibrium poSitionS of of the the bead bead problem problem by analyzing the motion neighborhood of equilibrium.

I

Err this problem. T1 = 0 and the second derivative of U is

(IL (1L

cosOU — = engR cos —

that T2 = mR2O/2. and has the form

=

0 6

— —

R S1fl 6 -ftcosO ft COS 0 ++ —cksino

R

-,

•

-.

the the

47n1?2E 7flRE 4-

[al Ia]

linearized equation of motion about equilib-

[bJ

= 0

Substituting the values of U at equilibrium, we obtain

= ()() For 00 = For

k=

c10 dO-

/ = mc R (1

R

\

g

/

— —

[C]

Example

5.2

284

CHAPTER 5

=

•

ANALYTICAL MECHANICS: ANALYTICAL MECHANICS:AI)Ln'r!oN..tL AI)1rr!oN.tL Tonics Tonics

= _rngR(l +

k=

77.

Ed]

-' clU

=

R

g

2-

R

-

=

•,

2

-

I

+—

2

—

R

g

)2]

[.1

For the equilibrium point to be stable, k must he positive. For the position

corresponds to the bead lying on the bottom of the hoop. to he stable

I

—

0, which

RW/g must be

positive, or

[f]

which corresponds to the head being at the The second equilibrium position, 0e = top, is unstable. as the value of d2 U/dO2 is always negative. This result is easily verified from a physical point of view, as the bead will always fall from the highest point on the hoop. From Eq. [e], the third equilibrium position is stable when

[g] The motion of_the bead can he explained as follows. When the angular velocity of the hoop the only stable equilibrium position is the bottom of the hoop, Oe = 0. The is less than

point. As the angular velocity of the hoop becomes bead oscillates_around that and the bead greater than ..Jg/R. the stable equilibrium position becomes Oe = COST begins to oscillate about Os.. The equilibrium point Or = 0 is no longer a stable equilibrium is the critical angular velocity that dictates which = point. The angular velocity equilibrium point is stable. Note that until the critical angular velocity is reached the the location location Of As the the angular angularvelocity velocityincreases increasesover over%Jg/R. the of librium point is still still °e = 0. 0. As the stable equilibrium point moves up. The highest stable equilibrium position approaches = IT/2forveryhighvaluesoffl. for very high values of fl.

Example

5.3

I

Consider the three mass-spring-damper system shown in Fig. 5.2, and obtain the equations of

motion directly from the energy expressions.

Solution We will solve for the equations of motion of this linear system by expressing the kinetic and

potential energies in matrix form. The generalized generalized coordinates coordinates q,.

H '/i

H 'I:

and q3 represent the

H- 'h F

FIgure 5.2

Free-body diagrams

5.3

285

MOTIONS ABOUT SMALL MOTIONS SMALL

k1(q2—q1)

q2) F 1-

'/1

Figure 5.3

diagrams

.zisplacementsof ofthe themasses. masses.From FromFig. Fig. 5.3 5.3 the the kinetic kinetic and and potential potential energies have the form

T=

+

V=

+

Introducing the column vector energies in matrix form as

)2

+

—

[a]

C12)2} 12)2}

can write the kinetic and potential q: q3V we can

=

T=

—q

Ibi

V=

[M}fr'}

the mass niass and stiffness matrices have the form 0

1711

[M1=

0 0

0 0

[Ki =

—k1

0

0

—k1

k1

+ k2

k1

Ic]

"2

0

k2

Note that [Ki is positive positive sernideilnite. sernideilnite.The Thedashpots dashpotscxcii cxciiforces forcesofofmagnitude magnitudeu1 (c': (': to the masses. The virtual work is given by — =

—

ql)8q2 +

—

c2(q3 — £/2)6L/3 £/2)6(/3 +

—

—

—

(/2)(Sc/2 + (/2)(Sq2

[dl

6W = Qi

+ QQ-

Eel

tSq3

±

which

Qi =

—

Q2 Q2 =

— — q' ) +

C1

'/2) + F '/2) [f I

=

— '/2)

that the generalized force vector has the form Cl C1

=

—

)

[gi

(/2) + C2(q3 — q2) — (/2) '/2) —c2(q3 — '/2) + F 1 —

[

Using Eq. [5.3.1 [5.3.11]. 1]. the the equations equations of of motion become m1411

+ k1c11 kicj1 — — k1q2 + c'i(/i 0 — cIq2 c1q2 = 0 c'i(/i —

k2)q2— ni2ij2 + (k1 — kiq1 ni2q2 (k1 ++k7)q2

—

kq3

—

1fl3q3 +

+ (c1 (c1 + C2)( k2(/2 + C2(13 '2'13

—

C2(/2

— —

ciq1 Cl 'Ii

= F

—

= 0

[hi

In the next section we will learn about a simpler way to handle the dissipative forces by a dashpot.

286

S ANALY'1'1CAI4 ANALYTiCAL MECHANICS: ADDITIONAL Toi'ic:s CHAPTER 5 S

5.4

RAYLEIGH'S DISSIPATION FUNCTION

An iniportant important class of nonconservative Forces is the class of Forces that dissipate energy. A common way of approximating energy-dissipating forces is by friction. Two models of' friction are widely used, even though both are crude simplifications of a complex phenomenon: The first is dry or Coulomb friction. where the friction force opposing the impending motion has the magnitude of the friction coefficient times the normal force, and it opposes the velocity. As As we we saw saw in in Chapter Chapter I,I, F, =

=

[5.4.1) [5.4.11

where

is the friction force, N is the normal force, v is the relative sliding velocity between the contacting points, and ,u is the coefficient of friction. In general. ,u /1 is approximated by two values, the static coefficient of friction and the dynamic is

ion. Equation Equation [5.4. lJ represents a non linear relationship. coefficient of frict friction. The second approximation is viscous damping. where the damping force is modeled as opposing the velocity and proportional to it. The special case of linear proportionality is commonly used. In this case, the friction force is assumed to be in the form F, = —cv

15.4.21 [5.4.21

in which c is the viscous damping coefficient. This approximation approximation is is used used especially especially when modeling light amounts ol of damping. because it is a linear approximation and easier to deal with mathematically. Its applications include modeling of shock absorbers (dashpots). In analytical mechanics, a convenient way of treating VISCOUS ViSCOUS damping forces is by the use of Rayleigh 's dissipation function. Consider a single particle moving in one direction, such as a mass-spring-dashpot system. In terms of the generalized coordinate x, the viscous damping force is given by Pd = —ciT. —ci. Rayleigh's dissipation function is defined as

=

I

[5.4.3)

and the generalized force Q is obtained by LI

Q =—

15.4.41

the similarity between the expression for potential energy of springs (kx2/2) and Rayleigh's dissipation function. We extend this definition to multiple Note

dashpots by -—

=

— v1_1)

[5.4.5] 15.4.5]

where = 1, 2, Nd) are the viscous damping coefficients and and denote the components of the velocities of the points connected to the ith dashpot. .

.. .. ,

i

5.4

's

FUNCTIoN

The Rayleigh's dissipation function can be written in terms terms oF of the generalized

coordinates as = =

1

1> >

[5.4.61

k—I r—1

the coefficients dkr depend on the viscous damping coefficients. The contribution of viscous damping forces to Lagrange's equations can be obtained from Rayleigh's dissipation function by where

It -

.

k

(1krqr

=

1.2,...,,,

15.4.7]

One can then express Lagrange's equations in the presence of viscous viscous friction friction Forces forces as

d a!

\ (

.

+

J—

k

=

,i J

=

1, 2,

.

.

,, ii

15.4.81

where where Qk,l(. Qk,l(. no longer include contributions from viscous damping forces. Note that, •.. using the column vector we can write Rayleigh's dissi-

pation function as —

= =

[5.4.9]

{q}

in which [DJ is the damping matrix matrix with with entries entries dkr. dkr. It can be shown that the damping

matrix is symmetric and positive semidefinite. Another class of' dissipative forces is the circulatory circu/ator\' fbrces. Circulatory forces Forces in power transmission devices, pipes, and as constraint damping in structures undergoing rotational motion. For systems subjected to circulatory forces, the dissiration function has the general form = =

+ {flTIHI{q} I{q}

15.4.101 [5.4.101

[H] is the circulatory matrix. [Hi To analyze the effect of viscous damping and circulatory forces for small motions the neighborhood of equilibrium, we note that is quadratic in the generalized coordinates and velocities. Considering the developments in Section 5.3, a Taylor series expansion of in Eq. [5.4. 10] up to quadratic terms has the form +

[5.4.11]

is the circulatory matrix, both evaluated positive semidefinite seniidefinite and and [H(l] [He] at the equilibrium equilibrium position. position. [Dr] is symmetric arid and positive is skew symmetric. It follows that the equations of motion in the neighborhood of equilibrium can be expressed as where IDe is the damping matrix and

+ (IDel ++ [G1){c'} [G1){c'}++ ([K] ([K] + [Hej)4q} = {Q}

15.4.121

287

288

CHAPTER 5 . ANALYTICAL MECHANICS: ADDITIONAl. Tonics CHAPTER

Finally, we examine the units of Rayleigh's dissipation Function Consider a = ci2/2. The single dashpot, with The damping dampingforce furceis is F F= —ci. It follows that the unit of Rayleigh's dissipation function is k)rce >< velocity, or power.

Example

5.4

Consider Example 5.3 and obtain the equations of motion using Rayleigh's dissipation function.

Solution Rayleigh's dissipation function is given by

-

=

—

qi) +

I

[a]

—

which can he expressed in the matrix form =

[bJ

I D I '1

where the damping matrix is 0

Cl

Cl (i (• + Ci

[D1=[ — Cl

— C,

C)

[ci

— C., C)

It follows that the equations equations of motion are given by

[dJ

+ [K]{cj} = {Qne}

L114i{4} +

where and [K1 [Ki are are defined defined in in Example Example 5.3. 5.3. For For this this example. example. because because all nonconservative krces are kwces areaccounted accountedfor forin inthe the Rayleigh's Rayleigh's dissipation dissipation function, the generalized forces forces VeCtOr is is .{Q,J {Q,J = [00 FIT. A brief examination of Eq. [dj and Eq. [iJ in Example 5.3 indicates that they arc equivalent.

Example

5.5

s stem in Consider the system in Example Example 4. 4. 11. 11. Obtain Rayleigh's dissipation function, linearize itit around equilibrium, and and obtain obtain the the [DC]j matrix.

Solution From Fig. 4.2 1 Rayleigh's dissipation function can be written as =

= =

I

I,

1.

CON 0

•{-

+

if.., ± hxfl.. COSAA ±±

= =

(.

.i.2

COS

4

/r0 cos2 cos2 ()

The equilibrium points can he he obtained by visual inspection as = 0, 0, consider the position = 0. In the neighborhood of equilibrium becomes b*O + bi:O

==

=

4

==

and, considering that the generalized coordinate vector can be written as (q} = [x dampi damping matrix matrix becomes

[a]

I

0,

ir. We

FbI Fbi (flT, the the

b

I

1/)el =

/)

-2

/)2

4-

Ic]

5.5

5.5

PROBLEM FOR LINEARIZED SYSTEMS

EIGENVALUE PROBLEM FOR LiNEARIZED SYSTEMS

In the previous Sections sections we saw the linearized equations of motion for dynamical Here. we obtain the solution of ol the linearized equations. analogous to the treatment of a single degree of freedom system. The developments in Sections 1.9 through 1. 11 and the developments of this and the next two sections constitute the basis of linear vibration theory. The equations of motion of a ]inearited undamped. nongyroscopic system are in Eq. [5.3. lii. iii. To obtain the response. we first consider the free vibration case, that is, = {O}. Dropping subscript e from [Me]. the equations of motion have the form IM]{cY(1)} + [Kj{qW} = {O} IM]{ci(1)}

[5.5.1J [5.5.1]

[M] and [Ki are constant coefficient matrices of order n X n. As in Chap1. we I. we are are interested interested in in analyzing analyzing stable stable motion about equilibrium, so we consider

where

cases when the stiffness matrix I K 1 is positive definite or positive semidefinite. Such

are basically vibratory systems. The case of a positive sernidefinite seniidefinite [K] is encountered in systems admitting rigid body motion. The above equations represent n second-order coupled differential equations. To analyze the response. we make the synchronous motioii assumption and express the in the fonii

[5.5.21

{q(t)} =

{u} is an n-dimensional n-dimensional vector and and A A is is aa scalar. scalar. The synchronous motion assumption is based on the observation that the response of every component of the Jvnamical system has the same time variation. The amplitudes are different, but the nature of the motion is the same. Hence, A describes the time variation and {u} Jescribes the different amplitudes. We introduce Eq. 15.5.2 I into 15.5.11 and collect terms, which gives (A2[M] +

For a nontrivial solution, nly if

{u}eAt cannot cannot

=

15.5.31

be zero. It follows that Eq. [5.5.31 vanishes

(A2[M] + [KJ)'(u} = {O}

[5.5.41

Equation [5.5.4] converts the problem to a set of n homogenous algebraic equa-

tions lbr the parameters (u} and A2. The task at hand is to find the values of A2 and corresponding values of {u} for which Eq. [5.5.41 has a nontrivial solution. This problem is known as the ('igenvalue problem. For the set of iiii equations equations [5.5.4] to have a nontrivial solution, the determinant the CocUiCient CoeffiCient matrix must vanish, that is A2[MI + [Ku = 0 det(A2IMI + [K1) = A2[M1

15.5.51

This is known as the characteristic equation. completely analogous to the deand definitions in Sec. 1 .9. It represents an iith order polynomial in A2,

289

290

CHAPTIR 5

•

ANALYTICAL MECHANICS: ADDITIONAL TOPICS

known as the characterislic polynomial. The characteristic polynomial has ii roots. denoted by (r = 2, eigenvaiues' or characteristic values.,i). and known as eigenvaiue.s' It follows that for each root of the characteristic equation, there is a vector 1 ,

{Ur} such

. . . ,

that + IKI){Ur} IKI){Ur}

{O}

r = 1,2,..., 1, 2, iin . . . ,

[5.5.6]

is satisfied, in which {ur} is the solution of the eigenvalue problem corresponding to The tenri {Ur} is called the eigenvector eigenvector belonging belonging to to the theFl/i rth eigen value, or the nh eig en vecto 1..

Let us examine the nature of the eigenvalues and eigenvectors. We know from

the previous section that [M] is positive definite, and we are considering cases where [KI is positive definite or positive semide finite. A theorem from linear algebra states that the eigenvalues and eigenvectors of a symmetric matrix are real. Further, if the matrix is positive definite, all eigenvalues > 0. Also, the eigen values of the general eigenvalue problem A[A]{x} =

[B]{x}

15.5.71

where [A] is positive definite and [B] is positive semidefinite, are real and nonnegative. It follows that the eigenvalues of Eq. [5.5.61, = 1, 2, ..., n), are all real and nonpositive. Hence ±Ar are all pure imaginary. This is to be expected, because we are dealing with a critically stable system. Introducing

r = l,2,...,n

=

[5.5.8] 15.5.81

where Wr is a positive quantity called the nh natural frequency, the response can be represented as a summation of the individual frequencies multiplied by the eigenvectors {q(t)} =

+ + YrC_IWnl) YrC")

>

[5.5.9]

and depend on the initial conditions. Because {q(t)} and {u,.} are realvalued quantities, following an argument similar to the one in Sec. 1.9 we conclude that Xr and Yr Yr are are complex complex conjugates conjugates of each other, so that the free response can be written as where

'I

{q(t)} =

> "{u,.}A,. COS(Wrl —

4r)

[5.5.10]

I

with Ar and 4,. being the amplitude and the phase angle corresponding to the rth natural frequency, and their values depend on the initial conditions. Beibre analyzing the amplitudes and phase angles, let us discuss the physical interpretation of the eigenvalue problem. The eigen value problem has as its solution n natural frequencies (for systems admitting rigid body motion, one or more of the natural frequencies is usually zero) and corresponding eigenvectors. The ii natural

frequencies describe the n unique ways (or modes) the system can vibrate. For a

I

2The word eigenva!ue comes from the German word eigen, meaning characteristic.

5.5

291

E1GFNVALIJE PROBLEM FOR FOR LINEARIZED LINEARIZED SY.STEMS SYSTEMS

degree of freedom system we had one One natural natural frequency. frequency. and for an n degree :f freedom system there are ii n natural frequencies. In each mode of the motion, the free response is harmonic (except for a rigid body rnode with zero frequency). The eigenvector corresponding to a particular mode represents the amplitude ratios ol' the generalized coordinates. That is, it describes the shape of the motion when the system vibrates with that that particular mode. The eigenvectors are also referred to as modal vectors or natural modes. Note that, from Eq. [5.5.6]. the amplitudes of the modal vectors are not unique. A modal vector multiplied by a nonzero constant is still a modal vector. It is useful to adopt a procedure to specify the magnitudes of the modal vectors, and we will do exactly that in the next section. via a procedure called normalization. section,

Obtain the natural frequencies and cigenvalues associated with the system in Example 5.3, Obtain :or the values ni1 = rn2 rn2 = = rn, k1 = k1 = k, c1 = C2 = 0. From Eq. [5.5.4]. the Eq. [5.5.4]. the eigenvaluc eigen value problem problem is

100 0 0 0 ±± k 001

/

0

1

(A'[u14] ± [K]){u} (A'[M] = ( A'rn

k

1I

0

—l

1

2

— —1

00—1 —l

I

—

1

1

iJ

)

Dividing the above equation by k. and introducing the quantity quantity yy = = the characteristic equation becomes

/

1

y 00 deti y 0

\

0 1

0

0 0

1

—

2

— 11

0

I

0\\

y—1 'y—l

0

—11 —

— I

)

1//

—l1 —

= det det

1

I

I

— 2 y—

0()

1

{u}

—

[a] Ia]

=

A2in/k A2 in/k

= w2rn/k,

0 1

= 0

[bJ

y—l

Evaluating the determinant gives gives a third-order polynomial in y in the the form form —

I)(y l)(y

—

2)(y

—

1—

—

1) = I) = (y (y

—

l)(y2

—

3y + + 2—2) 2—2) = — l)(y2 1)(y2 (y — = (y

—

3y) = 00 3y) [ci

whose roots are (listed for convenience in ascending order. which is the common way ol ranking them) = 0

Y2

=

I

Y3

= 33

Id]

IL IL should should be be noted noted that the characteristic equation for most systems is much harder to than the example here. Recalling the definition of y above, the natural frequencies can written as w,. w,. = sjyrk/m (r = 1, 2, 3). with the result U)

=0

2= =

U) 2

—

in

w

=

/3k in

Eel

One of the eigenvalues is zero, indicating a rigid body mode. We can verify' verify the presence a rigid body mode visually, by noting that the two ends of the mass-spring system arc are free. Hence. if the system is given an initial velocity, it will keep translating, translating. until opposing forces are applied or friction takes its toll. Next, we calculate the eigcnvectors. For each mode of the motion there exists a unique

eigenvector. To solve for the cigenvectors. we make use of Eq. [5.5.61, which can he

Example

5.6

292

MECHANICS:ADDITIONAl. ADDITIONAl. Tonics CHAPTIRS 5•. AN,tlxTrlc&L MECHANICS: CHAPTIR

written as 0

1

I

—

0

2

Yr'

1

r = 1, 2, 3

{ur} = {O}

1

(fJ

The coefficient matrix in the above equation is singular, so that the three equations obtamed from Eq. ff1 are not independent. Equation [Ii gives only two independent which raises the question of how to obtain the three elements of {Ur}. It turns out that only

two of the elements of {Ur} can be found, and the third is expressed as a ratio. = 0 we obtain for fori•r == 1. Writing {Ur} as =[111r = [lilr ld2r —l

0

1

1

—2

1

1

—1

0

u11

[,J

={O} 113' 1131

Taking the first and third equations as the independent ones, we obtain U11 1421 = 0 Uii ++1421

1121

1431 — 1431

== 0

[hI

which leads to the conclusion =

1111

=

[IJ

1431

This result can be confirmed by substituting in the second equation in Eq. fg}. the first cigenvector can he expressed as =

a,isis an au amplitude amplitude ratio. Let us examine this first eigcnvcctor. Recall that the where a of the eigenvectors denote the amplitudes of the generalized coordinates, the displacements the three masses in this problem. For the rigid body mode, all of the elements of the tor are the same. indicating that all three masses move with tile same amplitude. It follows the springs between the masses arc not stretched, and the entire mass-spring system is aa single rigid body with no moving parts. as one piece. as if it is = I and For the second mode we have o

0o

i

o

= {0}

142,

1

1

[kJ

0

1

and we take the first two equations from !rorn above 0 1122 = 0 1122 =

U12 — 1422 + 1137 =

0

thus, the second eigenvector can be written as I

=

For the third mode we have

= 2 I

o

3 1

0

and 0

I

I

tb3

I

2

1133

1

=

[.1

5.6

0

AND NORMALIZATION

——

—1

—,

1—

0 —t

-

±

q

'1:

Figure 5.4

Cl 3

Plot oF the eigenvectors

the first and second equations +

= 0

[o]

+ 21113 21113 = 0

that the third eigenvector becomes {u3} = = a

The three cigenvectors are plotted in Fig. 5.4. Wc see that all of the elements of the first have the same sign, elements of the second cigenvector have one sign change. and the third eigenvector there are two sign changes. This is a common feature of modal vectors. elgenvalues are arranged in can he generalized to a system of order as: When the eigenvalues — I zero crossings. ending order, the the rth rth eigenvector eigenvector has hasr —

5.6

ORTHOGONALITY AND NORMALIZATION

The natural modes possess an important property called called orthogonalitv. orthogonalitv. Consider the and two of its its solutions. solutions. say say the therth rthand and.vth sth modes problem in Eq. with r s and Wç. A theorem from linear algebra states that {Ur}T[Ml{Uc} {Ur}'[Ml{tic}

= 0

= 0

r, S S == I

I

,

2,

.

.

.. ,

ii,

r

s, w,.

(5.6.11 This relation is true for any two nonequal natural frequencies. In the case of' repeated eigenvalues (w,. = w.v. r the associated modal vectors are not unique.

Any linear combination of two such modal vectors is a modal vector itself. However, is possible to select the modal vectors associated with repeating eigenvalues such that Eq. [5.6. 1] holds. In Chapter 6, we will see an interesting application of repeated eigenvalues.

293

294

CHAPTER 5 . ANALYTICAL MECHANICS: ADDITIONAL Topics

r=

the product {tdr}T[Mi{lIr} > 0. and the magnitude of the product depends on the magnitude of the modal vector. Because these amplitudes are arbiIrary, a common way to deal with them is to normalize them. One way of normalizing the eigenveclors eigenvectors is to set the above product to unity When

s,

{Ur}T[M1{iir} {ldr}T[MI{iir}

[5.6.21

1

This equation is one way of normalizing the modal vectors. There are other approaches, such as {u1}'{u,.} = 1. In this text we use Eq. [5.6.21. Using this equation.

and Eq. [5.5.61, one can show that

[5.6.31

{u,.}'[KI{u,.}

Equations [5.6. 1 J—[5.6.3J J—[5.6.3J can can be be combined combined to togive givethe theso-called so-calledortho,wr,ncziitv ortho,wr,naiitv

relations

{t'r}'[K {Ur}'[K I{i'.c} I{1'.c} = = Wr.c

=

15.6.41

where s,., is the Kronecker delta, delta. defined as

=

1

when

r=

=

0

when

r

15.6.5]

Sç

A corollary of the result above is that the modal vectors constitute an indepen-

dent set. This implies that any vector of order n can be expressed as a linear combination of the cigenvectors multiplied by appropriate coefficients. Hence, given an n-dimensional vector vector {z}. {z}. one one can can expand expand it as n-dimensional {z}

[5.6.6] 15.6.6]

=

where ak are the coefficients of the expansion. Note the similarity oF Eqs. 15.6.61 and 15.5.101. The 15.5.101. The question questionbecomes becomesthat thatofofgiven given{z}1 determining ak (k = 1. 2, ..., n).

For this, we make use of orthogonality. and lell-multiply both sides of the above equation by {Ur}T[M] r = 1,1. 2, n) with the result . . .

,

{ji,}T[M]{_} = {ji,.}T[M]{_}

= =

=

r = 1,1,

2.

..,n .,,i

. .

[5.6.7]

We conclude that any vector {.} of order n can he expressed in terms of the modal

vectors fl

ar{u,} '>' a,.{u,} {z} = '>' r=

= ar ('p =

{Ur}TIM1{Z}

[5.6.8]

1

This equation is known as the the expansion expansion theorem. theorem.AAvery veryimpOrtallt ifliportatit application of the expansion theorem is in obtaining the response of a vibrating system, as we will discuss in the next section.

Example

5.7

Consider Examples 5.3 and 5.6. and demonstrate that the orthogonality relations hold. Normalize the eigcnvectors. Then. take take the vector = [I 2 51T and expand it in terms of

the e.igenvectors. eigenvectors.

5.6

ORTHOGONALITY AND NORMALIZATION

S.Iution where [11 Ill isisthe the identity identity The mass matrix for this problem probleni can be written as [Mi = ,n[1'J, m[11. where matrix. Hence, the orthogonality relations in Eq. [5.6.1] reduce to

[a' Ia]

rv= 1,2,3 ,.v=

Iii{Ur}T{Uc}

{ Ur}' [M1{U,c} [M1{Uc}

Evaluating the possible combinations, we get {ui}TFM1{u2}

1) = 0 ma1a,(1 + 0— 1)

= ma1a3(l ma1a3(1

{U2}T [M]{u3} =

I —

+ I) =

0

—2+ 1) =0 [bi IbI

0

Next, we normalize the cigcnvectors according to Eq. 15.6.21 For the first mode we have {14}T[Mri{IA}

= =1

= ?nal(l ?nal(l + + II + 1) =

[ci Ed

I

which can be solved for a1 as

a1 =

÷1' —\/ —\/

[d]

3k?? 3,ii

The choice of the plus and minus sign is up to the analyst. Let us select the positive value. For the second and third modes we have =

{u2}"1M1{u2} =

{143}1[J%'fl{1d3}

I 1

[.1

6,naf = 1

from which we obtain

/1 \/ 2m

ôrn

we select the positive values for a2 and [I Let us now consider the vector {b}

[M]{b} = [MJ{b}

so that

in ,n

=

is expanded is

/1(1 /—(I

\ 3,fl 2iii

xX 1 ++ II X X 22 ++ II X x 5) 1

oI,n ojrn 0l

(1 xx II ++ Ox (I Ox 2— 2— 11 )< )< 5) 5) =

1

= rn{u3}T{b} =

(i13 03 ==

Eg] [gJ

2, 3) 3) are are tèund tèund from from Eq. Eq. [5.6.81. [5.6.8]. We then have = 1, 1, 2,

(1

}T{b} = = {ui}1'IM]{b} {u,}1'IM]{b} = m{Ui}T{b}

=

and expand it as

a3{U3} a2{u2} + a3{u3} a1 {Ui }++a2{u2} = (li{Ui}

{b}

the coefficients

51T

2

6in

(IX l—2x2+ 1

1k]

as —

/iii

'/3 ru 811 81'

—., I

r I

11—21

Lii

I

ii I

I

[-1] I

I

1

—

_1

3— I

Ii]

295

296

CHAPTER

5.7

5

.•

ANAI1Y1ICAL MECHANICS: ANAI1YTICAL MECHANICS:ADDITIONAL ADIHIIONAL Topics Topics

MoDAl. EQUATIONS OF MOTION AND RESPONSE

Consider the undamped vibration problem. repeated here as

[5.7.1]

= {Q(i)}

[M1{c/(r)} +

We will make use of the developments of the previous sections to find the response. We first solve the eigenvalue problem and fInd the natural frequencies and the modal vectors. We then apply the expansion theorem to the generalized coordinates and and expand expand {q(i)} {q(i)} as subject to the initial conditions {q(O)} and

fl

[5.7.2J

{q(n} = '> TJk(1){U/} TJk(1){U/.} k—I

are known as modal coordinates or principal principal cuorthnates. cuorthnates. Note that because {qW} is a function of time, so are the coefficients of its expansion in terms of the modal vectors. In essence, the principal coordinates (I), i1z(t), ij,1(t) constitute another set of generalized coordinates. Next, we introduce Eq. [5.7.2] into Eq. [5.7. ii. Obviously, we wish to make use of the orthogonality properties. properties. so so we we left-multiply left-multiply both bothsides sidesof ofEq. Eq.[5.7. [5.7.1] 1] by 1, 2, n), with the result {{ tir}' tir}' (r = 1. where

.

i

.

.

.

.

. ,

=

+ + {u,.}nfKI

{ur}11M1> {u,.}11M1>

[5.7.3]

which can be rewritten as t1

11

= .{I,,.}T{Q(t)}

+

r=

1,

2, 2.

.

.

.

,

n

k—I

[5.7.4] Invoking the orthonormality relations in Eq. [5.6.41. [5,6.41. we see that of' all the terms on the left side of Eq. [5.7.41. [5.7.41. two two terms terms survive, survive, those those corresponding correspondingto to r = k. We define the quantity llzodli modal force where

[5.7.5]

= {Ur}1{Q(t)} {Ur}1{Q(t)} SO

that the describing equations become

[5.7.6]

=

+

have converted the equations of motion from a set of ii coupled secondorder differential equations to F? ii independent independentequations equations of of motion. motion. This conversion serves a number of purposes. It permits the user to view the equations of motion as a collection of single oscillators. It also makes it much simpler to obtain the solution. Indeed, rather than solving a set ol' of iiii coupled coupled differential equations, one solves ii independent equations. For each modal coordinate, from Sec. I 11. the response has the form of the convolution sum We

.

=

cos Wrt +

w,.t + WI.

I(

Jo

—

dff U) Sifl WrU (1ff

[5.7.7]

5.7

MoDAL EQUATIONS OF' MOTION AND RESPONSE

the initial conditions. They can be obtained from the initial where 71r(O) and iIr(O) are the by means of the expansion conditions on the generalized coordinates {q(O)} and theorem as

[5.7.8:1

= {Ur}1

[M]{q(O)} {u,.}'[M]{q(O)} iJr(O) = {u,.}

the response of each modal coordinate is obtained. Eq. 15.7.2 1 is invoked and the response of the principal coordinates is found. In the presence of a rigid body mode with zero frequency (denoted by R), the modal equation of motion is Once

'17R(!) 17R(I)

15.7.91

= NR(!)

=

subject to the initial conditions

= {uR}T[M1{4,(O)}.

The response can be shown to be -T

7j,?(O)t + qR(t) = l)R(O) + iJ,?(O)1 ?7R(t)

[5.7.101

I I

Jo .0

For damped systems. the cigensolution associated with the undamped part of the system does not usually lead to a decoupled set of equations. L.et us add a [DJ{c'(i)}

of Eq. [5.7. 1], thus term to tile the left side ol'

+ [K]{q(r)} [K]{q(t)} = {Q(r)}

[M1{qu)} +

[5.7.1 1]

{t.ir}T This This yields yields Introduce Eq. [5.7.2] into it and left-multiply by {t.ir}T. .

+ + {Ur}T[D] {UrY[D]

+

[K]

= {Ur}7{Q(t)}

15.7.121 can be reduced to 'I

dkrllk(t) + dkrhlk(t)

+ k=

= Xr(t)

[5.7.131 15.7.131

I

dk, = {uk}TLDI{u,.} {u k}T ID 1{u,.}

15.7.141 15.7.1 41

equations of' of motion are no longer independent hut are As a result, the modal equations through the damping terms. There are cases when the damping matrix has pmportional modal equations. One such case is proportional special form that decouples the modal special damping. where the damping matrix is a linear combination of the mass and stiffness matrices in the form

[I)] = a1[M] + a2[K} so

that the values of'

dkr =

15.7.1 51 15.7.151

beconie

(a1[M] + a2[K]){ui,} a2lKI){uir} = (ai +

[5.7.161

Actually, proportional damping is more of a mathematical convenience and not a very realistic model. When the amount of damping in a system is small, a simplifying

297

298

CHAPTER 5

.•

ANALYTICAL MECHANICS: ANALYTICAL MECHANICS: AJ)LHIIONM.Topics Topics

assumption is to ignore the values of dk,. when k

r. This assumption decouples

the modal equations. One can justify such an assumption by noting that the damping matrix ID1 is itself a gross approximation and that the damping is not known accurately. The decoupled modal equations can then he written in the form

+ 0$7)r(t) = Xr(t)

+

[5.7.1 71 7] [5.7.1

where = drr (r = 1, 2, . .., ii). The response for each mode is obtained by Eq. [1.11.8]. Gyroscopic and circulatory systems require a different analysis. One has to express the equations of motion in state form and then solve the associated cigenvalue problem. This same analysis can be carried out for damped systems as well. It turns out that for undamped gyroscopic systems. there is a very elegant solution. The interested reader is referred to the texts by Meirovitch.

Example

5.8

I

Consider the undamped mass-spring system in the previous examples. A uniform force F is applied to the third mass. The initial conditions are {q(O)} = L[I 0 — I]". 1]", = {0}. Find

the response.

Solution = [0 0 F]'. and using Eq. [5.7.51. [5.7.5]. the modal

The generalized force vector is written as

forces become

= {u}

=

T

=

{ Q}

\

1il 3m

[111]

=

3m

0

ii / \2m \2in

=

{

H

F

'I•l

U

0 0

0o

11

/

ij 0 o = \2m / 1]

—

1

—

0

= =

u

Q}=

— 2 11 —2 11 o

\ / 6m [1

11

/ 6,,, F

=

F

Ia] lal

The initial conditions are = .{ui}'[M]{q(O)} =

=

=

\

/1 1

11

-

/1

, L[ L[ 1

0

1

— 11 11

—

= /I

\6m \6,n

L[1 L[ 1

—2

0

0

in

—l —I

0

0 0

'ii

0

0

()

in

0

0

0

0

in

—l

in

=

in

0 0

in

1]

0 0

0 0

0

1

0 0 in

1

0= —1

0

(bJ

5.7

Mor'1i. EQUATIONS OF MOTION AND RESPONSE

[ci

7:12(0) ==1J3(O) = 71(0) 1J3(O) = 0

should both be zero. This simply looking looking at at {q(O)} {q(O)} that that iji(O) and can ascertain by simply second cigenvector. so the initial condition is recognized as a constant times the second it should have no contribution to the first and third modes. Hence. we have for the first mode, = = V 3m

Idi

=0

=

response is

i (Ft2

/

Th(t) —

Eel [.1

)

For the second mode we have =

+

112 (

0i 0i =

772(0) = 0

,i L

Ef 1I Ef

response can be shown to be

/i

772(1) = —

=

Id Co CoSS W in Id 2 in

=

upon introduction of the value of

(1

2m k/ni k/rn

in L co s

=

=

3k/ni,

F/

—

COSW2I

[gJ

(02

can he CUfl hewritten written as

—

In a similar fashion, noting that

I (

\%W2

/k\ F / t) —

/

do

Sifl

2,,, 2,ii

/1 F —

112(1) = 112(1)

\'

F

cos 2

Ihi

—r

—

111

)

/

the response of the third mode becomes

Iii

—

The total system response is then written as + {u2}ij2(t) {u2}rj2(t) + +

= 1

=1

br" ôni 6rn

I

±

0

1_f

( lv

—

\

F — LOS LOS (1 — A

t)) + t))

1-1

F

-COS

mt))

—1 —l

[ii Let us conduct a dimensional analysis of the terms in the response. The coefficient of the coefficients coefilcients of the the first eigenvector has units of force x tirne2/mass = length. For thc second eigenvector. the component due to the initial condition has units of' length. and the due to forcing has units of force/spring constant. But the spring constant has units of force/length. SO that the component due to forcing has units of length. also. The same can terms in the above equation be said about the contribution of the third mode. so that all of the ternis have the unit of' length. As always, such dimensional analysis is a good way of spotting errors.

299

300

ME('t-IANICS: A1)DfI'LONAL ADDITiONAL Topics CHAPTER 5 S ANAlYTICAl1 ME('t-IANICS:

5.8

GENERALIZED MOMENTUM, FIRST INTEGRALS

As discussed in Chapter 1 first integrals are expressions that involve derivatives up to ()I1C order less than the highest derivative in the equations of motion. They come in handy when analyzing the behavior of a dynamical system qualitatively, instead of solving explicitly for kr the theresponse. response. We We investigate investigate here here first first integrals integrals associated with associated Lagrangian mechanics. Lagrangian mechanics.We Wefirst firstdefine defineby byirk irkthe the4generalized generalized with the kth generalized coordinate as

aL

k k =

l,2...,,i l,2....,ii

15.8.11 [5.8.11

The relationship between the generalized coordinates and the generalized mo-

menta is very similar to the relationship between a translational coordinate and linear momentum or between a rotation angle and angular momentum. Because the poten-

tial energy does not contain any terms in the generalized velocities (except for in electromagnetic systems). one can express the generalized momentum as aL = ... = ..

[5.8.2]

(Iqk

Consider now a system where the lth generalized coordinate qj does not appear

in the Lagrangian. Such a coordinate is referred to as crc/ic or ignorable. The name cyclic is due to the fact that such coordinates are encountered mostly in rotational motion. It follows that in the /th equation of motion, IILMq1 = 0, and Lagrange's equations become d /oL\ /ôL\ I= — di \aqi / I

.

[5.8.3] 15.8.3]

=

When the generalized coordinate is cyclic, the rate of change of the generalized When momentum is equal to the generalized force. One can integrate Eq. 1 5.8.3] over time to obtain the generalized momentum. In the special case when the generalized force associated with the ignorable coordinate is zero, we obtain from Eq. 15.8.31 that —

(aL\ I

.

\rlqj ,Ij di \rlqj .

I

=

d

0 = —(ir1) CIT

—

= constant

[5.8.4] 15.8.4]

When a generalized coordinate is absent from the Lagrangian, and the external

excitation is not a function of that generalized coordinate, the associated generalized momentum is conserved. This is an integral of the motion. One can then raise the question as to whether it is possible to take advantage of the cyclic coordinates and simplify the equations of motion, The answer to this is positive. If a system has n degrees of freedom and I cyclic coordinates, the n equations of motion can be reduced to ii — / equations of motion that can be solved independently of the cyclic coordinates, plus / first integrals associated with the cyclic coordinates. We actually carried out such a procedure in Example 5. 1. One way to identify and separate ignorable coordinates is described in Section 5.9.

5.8

GENERALIZED MOMENTUM, Fiisi LNTECRALS

We encounter an interesting integral of the motion when the Lagrangian is not explicit function of time and when no nonconservative forces act on the system. In case, considering case, considering that L = L(q1, the time derivative •. .. ., 1( the Lagrangian becomes3 •

IlLI

dL dt

(9Ld.. dLd..

fl

.

15.8.5]

k=1 (kik

we rewrite Lagrange's equations in the absence of nonconservative forces as k

= 1,2,...,

[5.8.61

fl

) into Eq. [5.8.5] we obtain

1

I

,I

cIL

\

..

fl F,

dt =

fl

d (aL

= >A dt

[5.8.7]

Jr

d di'

n

k—I

ilL. IlL. .qk _L)=O Il

[5.8.8]

k

And, integrating this equation, we obtain the Jacobi integral h, defined as n fl

h=>T

—

(IL (1

>

k1

—

L = constant

[5.8.9]

k=1

k-TI

The Jacobi integral is yet another integral of the motion. But Ii can be expressed

a simpler form: We first write the generalized momenta in column vector format T , and using Eqs. [5.8.2] and [5.2.91. [5.2.91, we have = p

{ir}=(. {ir}=L. \rI{q} \d{q}

\T

= [M]{q} +

15.8.101

)

?2

-I

=

=

[M]{4} +

= 2T2 + T1 T,

15.8.11] 15.8.111

k=1

this into Eq. [5.8.91. we obtain

I=

—

L = 2T2 + T1

—(T2+T1+T0--V) = = T2—T0+V= T2+U

k=I

15.8.121 For a natural system T0 = 0 and T2 = T. so the Jacobi integral becomes

the similarity between this approach and the procedure in Appendix B to find first integrals.

301

302

Tonics

ANALYTICAL MECI1AN1CS MEcilANics: CHAPTER 5 • ANALYTICAl.

[5.8.131

constant 1' + V == constant = T2 + V = 1'

Ii

a natural system. the Jacobi integral is the system energy. We next demonstrate denionstrate that the generalized momenta are indeed a set of independent variables that are derivable from the generalized velocities, and vice versa. The relation between the generalized momenta and generalized velocities is given in Eq. 15.8.101. 15.8.10]. Inverting Inverting it. it. we we can can express the generalized velocities in terms of the generalized momenta as Thus.

=

[5.8.14]

—

For Eq. [5.8.14] to hold, [MI must he nonsingular. It was stated earlier that

is positive definite, so it is guaranteed to be nonsingular. Equation [5.8.14J leads to the conclusion that the generalized momenta and generalized velocities are related to each other by a linear one-to-one relationship and that they can be used interchangeably in the problem k)rmulation. formulation. We had earlier expressed the Lagrangian in terms of the generalized coordinates and generalized velocities as

L = L(qj, q'.

. . .

' c/n' '11' £/2.

Based on the discussions above, we can now L = L(q1.

cJ2.

[5.8.151

t)

also represent the Lagrangian as

7T1' 7T2, 7T1' 7T2,

•

ci,,,

..

.

.

.•

.•

.

..

[5.8.1 6] [5.8.16]

lTn. 1)

It follows that one can use any combination of generalized velocities and generalized momenta. (We cannot use Trk ITk and qk(k = 1, 2, . ., ii) together, though. as they would constitute a redundant set.) .

Example

5.9

Consider the head problem problem again. The Jacobi integral is given by Eq. [5.8.12] so that it has the form I,

=

+

•i .

= 2

+

I I

—

Cos

0)

mR2 —

-—

2

,

(1 = constant sur (1

[a]

which obviously obviously is different different than than the sum of the kinetic and potential energies of the head. head. kinetic and signifying that that br is not the sum of the kinetic nonnatural systems systems the the energy integral is signifying br nonnatural potential energies.

5.9

FORIGNORABLE IGNORABLE COORDINATES COORDINATES ROUTH 's METHODFOR 's METHOD

If a system has ii degrees of freedom and / of those coordinates are cyclic. Routh's method permits one to reduce the ii equations of motion to ii — / equations that can equations he solved separately from the cyclic coordinates. The reduced set of ii — 1 equations is usually easier to solve than the full set of ii equations. The remaining / equations associated with the cyclic coordinates are expressed as first integrals. Consider a scieronomic conservative system that has ii degrees of freedom, with 1

I1 of' of'them themcyclic. cyclic.Order Order the the generalized generalized coordinates coordinates such that the first n — 1 coordinates

5.9

ROUTH 'S METHOD FOR IGNORABIJE COORDINATES

.. ., /1—1) ii—!) are not cyclic, cyclic. and the next (k = iin—!+ — I + 1. 1, nn —1 + 2. .., n) are. The I, ii — / + 2, . ., ii) associated with are. The generalized generalized momenta momenta 'irk Irk (k (k = nn — / + 1,

(k (k =

1,2, 1, 2,

.

.

.

.

,

. .

the cyclic coordinates are constant. We will write the Lagrangian in terms of the Leneralized coordinates and the generalized momenta. This is perfectly acceptable because, as demonstrated in the previous section, generalized momenta are independent variables. For a natural system with no cyclic coordinates the Lagrangian can be expressed in terms of the generalized momenta as

[5.9.1]

L=

the presence of I cyclic coordinates, we write the Lagrangian Lagrangian in in terms terms of' of the generalized velocities l'or for the coordinates that are iw! 1w! cyclic cyclic and in terms of the generalized momenta fbr the coordinates that are cyclic. For the cyclic coordinates the coordinates are absent from the Lagrangian In

L = L(qi. L(q1. c/i,

q2'

. .

.. .. .. ,.

c',,j,

'ir,,_1 ir,,_1

.

.

.

'ir,,) 'fl,,)

,

[5.9.2]

and

k = 1,2,

= constant We

next introduce the Roui/iian. denoted by

.

..,

[5.9.3] [5.9.31

1

as

I

> >.i

15.9.41

k=I

Comparing the Lagrangian and Routhian, we observe for the coordinates that

are not cyclic (9

=

-

k= l,2,...,n—1 I,2,...,n—1

= ,• ,.

.

dq1,

(-Iq/%

[5.9.51

and for the coordinates that are cyclic and

=

.

aL

• •

k

—

'lTn_/÷k 7Tn_l±k

k

d'TT,,

1.2,. 1,2,..

k

-

.

•, / .

,

[5.9.6] 15.9.61

f+A

The last term on the right of this equation can be written using the chain law for ,IiftBrentiation as -I+k =

(I(J -

.

-l

k

•

n—i+k

i

= =

15.9.71

-

k

which, when introduced into Eq. [5.9.6] yields (

k = 1,2.

=

.•

.•

.•

[5.9.8] 15.9.8]

'1/ ,

One can substitute Eqs. [5.9.5 J into Lagrange's equations and write them as

d \ I— di \dqk I —

I

.

I

= 0

k =

1,

2,

.

.

.

,

ii

—

/

[5.9.9] 15.9.91

303

5 5 •

304

ANAI1Y'I1CAI1 MECHANICS: ANAI1Y'flCAI1 MECHANICS: ADDITiONAL Topics

Equations [5.9.9] and [5.9.8] represent two sets of equations, of orders n — 1 and One first lirst solves Eqs. respectively, which can be solved separately from each other. One 1, respectively, [5.9.9] for the generalized coordinates and velocities ., . •• associated with the noncyclic coordinates. The results are then substituted into Eqs. 15.9.8] to solve for the generalized velocities associated with the cyclic coordinates. are obtained from the initial The values for the constants ir,, -1+ 1' ., 2' .., conditions. The motivation behind defining the Routhian comes from Hamiltonian mechanics, as we will will see see in in Sec. Sec.5.5.11. II. .

.

Example 5.1 0

1

.

Return again to the bead problem. As in Example 5.1, we treat the rotation of the hoop as the Lagrangian becomes as another degree of freedom. Writing the rotation rate

L = T—V ==

[.1

is the mass moment of inertia of the hoop. where I is

We observe that is absent from the Lagrangian. which makes it an ignorable coordinate. The generalized momentum associated with j is constant. The partial derivatives of the Lagrangian are

=

sin 00 COS COS 0 — nzgR sin 6 = ,nR2th2 ,nR2th Sin

ilL .

-

=

,. = (I+mRsinO)d = 14+rnR'4sin0 = 14+rnR'4sin0 ••

.

mR2O

Ib]

= constant constant

[ci [c]

c?1

is the can seek a physical explanation of the generalized momentum. Indeed, component of the angular momentum of the hoop-head system along the axis of the hoop. We derived the equations of motion in Example 5. 5.1. 1. To To obtain obtain the the equation equation of motion using Routh's method, we calculate the Routhian as One

= L — —

=

6+ sin2 6

+

rngR(l

the Routhian can be expressed without

Using the expression for = 31 =

—

ingR(l

= —rnR2(1 — 2

5.10

—

—

cosO)

—

cosO) — — -

—

cosO)

—

/nr(1

[dl

by

—

'

2(1 + mR2 sin2 6)

1.1

IMPULSIVE MOTION

acted upon upon by a The linear impulse-momentum relationship fbr a particle of mass in acted force F(t) is obtained by integrating Newton's second law over time, for

5.10 (ti ft's

mv(!1) mv(!1) +

IMPULSIVE MOTION IMPULSIVE MOTION

mv(r2) F(i) di == mv(r2)

15.10.11

J

We saw in Chapter 1 that if a true impulsive force were applied, the particle velocity would change immediately after the impulse but the position would not change. Displacements require finite time to develop. We now consider Lagrange's equations and impulsive external excitation. Similar to the approach in Chapter 1, we integrate the equations of motion. In general, integration of Lagrange's equations over time does not yield important results. But the situation is different when the applied forces are impulsive. We rewrite Lagrange's equations in the form of Eq. 14.9.81 as d —

di (it

aT

9T

—-.---—+ —.--—+ clqk

.

ac/k

l,2,...,,i

=

k

=

-

15.10.2] 15.10.21

Integrating the the above equato. Integrating force is is applied applied at at time time t = to. Assume an impulsive force non from t0 to to t0 + and taking the limit as approaches zero. we obtain for each term flo+E

lim urn

I

•OJ(()

I faT \ d lilT faT \)dt )dr = lirn hiii d( —1 .

.

..

I

cit \aqk dt \aqk /

.

\dqk /

= limIiTk(tQ + E)

= (to +

f()

(9T (IT

urn

+

C

dt

€Of() lim urn

15.10.3] 15.10.31

I

urn

U

_4O J10

7Tk(1o)l 7Tk(1O)l

dt = 00

I

(qk

oJ,()

k == 1,2...., l,2,...,n ii

=

I

(2 V

—

=

[5. 11 0.41 [5.

[5.10.5]

have fiEquations 15. 15.10.41 10.41 are are both both equal to zero because IIT/iiqk and aite magnitudes. We refer to as the generalized impulse. The only forces that contribute to the generalized impulse are impulsive forces, as the contribution of all 'ther forces disappears as the duration of the impulse becomes infinitesimally small. Chapter 1 presented this same argument. It follows that

=

kk

= = 1,2,...,iz 1,2. ...,n

[5.10.6] [5.10.61

to the the generalized impulse, impulse, a that the change in generalized nzoinenrunz is equal to result totally analogous to the impulse-momentum theorem for particles. The virtual expression in terms of impulsive forces becomes A

6W fl(. flC

N

=)F =) F —i

1?IC Inc

=

• r5r.

'

1

N /n N

1>' >'

n

A

I

i=1'

;'jiiJ

= \_

\/ F• N

A

/

j

()I'• -

jiiij ("1k

\

•

=

1

dC/k

)

16,,'ik

Y'

[5.10.7] 15.10.7]

305

306

CHAPTIR 5

•

ADDITIONAL Toflics

ANAISTICAI.

so that the generalized impulse has the form N

A

or1

= ='>

[5.10.8]

-

rlqk

J{ij} + Equation [5.10.61 can be expressed in matrix form. Noting that {ir} = [M J{ij} vector do not change during and assuming that the inertia matrix and the the impulse. we can write Eq. 15.10.6] as

[5.10.91

=

is the generalized impulse vector. Hence. given the impulsive ftrces, one can calculate the generalized impulses. and by inverting Eq. [5. 10.9] one calculates the generalized velocities immediately after the impulse as in which {Q} =

[5.10.10]

= [M] '{Q}

Note that if an impulsive force is applied to a body subject to constraints, the

constraint forces, such as reaction lbrces, Ibrces, also become impulsive. One should exercise caution in determining those krces that are impulsive and those that are not. Also, as discussed in Chapter 1. always keep in mind the fact that Eq. 15.10.6] is an approximation. as it assumes the ideal situation of the impulse taking place in zero reality, iiripulses impulses take place over finite time, so that there is some change of time. In reality. position as well.

Example 5.1 11 5.1

massless collar in Fig. 5.5 is free to slide over a guide bar. Attached to the collar with a pin joint is a rod of mass 'ii and length L. A ball of mass M is attached to the tip of the rod. The system is at rest when an impulsive force E is applied to the mass at the tip of the rod in the horizontal direction. Find the velocity of the collar and angular velocity of the rod ininiediately after the impulsive force is applied. immediately

The

This system has two degrees oU freedom. We select the generalized coordinates as the translation of the collar and the rotation angle 0. The configuration of the system disturbed from equilibrium is shown in Fig. 5.6, together with the external forces. The kinetic energy is due to the kinetic energy of the ball and the kinetic energy of the rod

I N

,n.L A

F A

M

Figure 5.5

a-F 1_F

Figure 5.6

5.10 5. 0

T T =

1

1

1

+

+

IMPULSIVE MOTION

(a]

+

+

arc the arc and directions. ball in the x and y directions. are the velocities ot the hail and velocities of the center of mass of the rod, and 'G 'G is is the the centroidal mass moment of inertia of the. rod, the rod, 1G 1G == mL2/l mL2/12.2.From FromFig. Fig. 5.6, 5.6, we we can can express the displacements of the center of mass of the rod and of the ball as where

= x+

L.

YG

L

=

= .v.v+LsinO -F L sin 0

YM = —Lcos0 YM

Differentiating the above terms, we obtain the velocities in the x and

=X i• +

L -0

directions as

= LOsinO

.i- ++ LOcosO

sin 0

Ib]

Ic]

Substituting Eqs. [ci into Eq. [al we obtain T = T

M

=

+ LOcos

2] + ( LO sin o) +

+ L202 +

cos 01 +

in m

2j

[(i. +

H

+

cos

)

+

+

[di

cos 6]

+

The change in potential energy during the impulse is zero, as we assume that the position of the system does not change during the impulse. Using Eq. 15.8.21, the generalized momenta have the form

.1

.

—rnLOcos0 = = Mx + MLOcosO + mx ++ —rnLOcosO 2 .

cix

!mL2O + PVIL2O ML!ccos0 + + LnL26 + +MLicos0 = tvlL2O 3

!rnLicos6

=

2

1.1

The virtual work is

6W = 6W

=

F6(x + Lsin0)

=

F6x+FLcosO(S0

En

= PLcosO

[g] [gi

so that the generalized impulses arc

0. Combining Combining Eqs. rod is is vertical verticaland and00 == 0. At the point of application of the impulse. the rod as :eI and [gj. we obtain two equations for the two unknowns and 00 as

(M + in)x + (M

ML+

1

+

+ (ML2 + +

=

= EL

[hi

immed lately after after the the impulse. impulse. we obtain Solving Eqs. [hi for the values of i and 6 immediately

-2F V= 4M + in

0=

(4M + rn)L rn)L

[g]

Note that the velocity of the collar right after the impulse is in the opposite direction of the impulse.

307

308

CHAPTIR 5 • ANALYTICAL MEcII.&NIcs: MECHANICS:

5.11

TOPICS

HANIILTON 'S EQUATIONS

Hamilton's equations have the property that they directly yield 2ii first-order equalions of motion that are in state form. tions Ibrni. They also find find applications applications in in the the stability analysis of dynamical systems, in transformation theory, and in control systems. Consider the generalized momenta. {'7T} = + It follows that Lagrange's equations can be written in terms of the generalized momenta as

[5.11.11

{'fr} + {g} =

where {g}T = state form as

It follows that OflC can express Lagrange's equations in

=

(M[

—

{ir}=

{f3})

—{g}

[5.1 1.2a,b]

+ {Q,1j

The equations in the above form can be derived from a scalar function called the Hamiltonian, defined in a way similar to the Jacobi integral and the Routhian

(a

= =

—

L

=>

—

[5.11.3]

L =

Considering Eq. [5.8.121 we can write :i(

= T7+U

[5.11.4]

As {c'}can canbe beexpressed expressedin in terms terms of of {ir}. {ir}. the variation of {q} and {7r} {ir} as as the the independent independent variables, thus = ii { q} q}

If we take the variation ol

{6q} +

can be written using

[5.11.5]

(1{IT}

using its definition in Eq. [5.11.31. [5.11.3 we obtain

+ = {57r}'{q} {57r}'{q} +

—

[5.11.61

Because Because the the generalized momenta

are defined as {ir}1' = ÔL/O{q}. ÔL/O{q}.the thesecond second and third terms on the right side of this expression cancel each each other. other. Equating Equating Eqs. Eqs. to the the remaining remaining terms termsinin Eq. Eq. [5. [5. 11 11 .6]. .6]. we we write [5.11.51 to aR. d{IT}

(1{q}

(9{q}

[5.1 1.7a,b] 1 .7a,b]

= HgV

Now introducing Eq. [5. 11 .7b] into Eq. 15. 11 11 .2b], .2b], we weobtain obtainHamilton Hamilton c cationlea! equations written written as as =

[5.11.8]

+ (9{q} (9{q}

or, in scalar form, as as I

,•

h.p,

.

.

.

,

1?

[5.11.9]

___

S. I I

309

EQUATIONS

constitute 2ii first-order equations with all time derivatives tieing on on the the left left side. Hence. the equations of motion are in state tbrm. 1 .3] is as folThe rationale behind defining the Hamiltonian Haniiltonian as in Eq. Lagrange's equations are a set of second-order differential equations, with the = 1, 2, 2,. ..,., n). derivative terms arising from the expression (1 One then ponders the possibility of having an augmented form of the Lagrangian. = 0. Consequently. second-order derivatives are such that by eliminated. The Hamiltonian is defined in a similar way to the Routhian, but considall the generalized coordinates. Another rationale comes from the definition of Yet another motivation comes Jacobi integral, which uses the expression Jacobi from Legendre's dual transformation. to which Lagrangian mechanics and generilized velocities and momenta are an application. It should be noted that Hamilton much before Routh and that it was Routh who was inspired from the Hamiltoian to develop the Routhian. It can be shown easily that when all forces acting on the system can be derived ii), we have = 0 (k = 1, 2. .. ., n), from a potential, so that Equations (5. 1 1

.

.

inC

.

riL ilL

=

.

15.11.101

(It

system, if the Hamiltonian does not depend on For a holonomic conservative system. = 0. and the Hamiltonian becomes explicitly, its time derivative is zero. Jacobi integral IL an integral of the motion. One advantage of Hamilton's equais that they make it easier to identify integrals ol' of the motion. Hamilton's equaare also useful in the stability analysis of dynamical systems. However, it may be SO simple to obtain Hamilton's equations. especially when a large number :r degrees ot of freedom are involved. Also, from an algebraic standpoint, Hamilton's equations arc usually lengthier than Lagrange's equations. limiting their usefulness low-order systems. Hamilton's equations are viewed more as a tool to extract information from the system and to identify integrals of the motion. f.:rmation

pendulum varies varies because the length length of of the the pendulum 5.7 describes a spring pendulum. where the acted upon by two springs of' constant k/2 each and a dashpot of constant c. Find the motion using using Hamilton's canonical equations. of motion

0 g k

fl? '1?

Figure 5.7

A swing pendulum

I

Example

5.1 2

310

ADDITIONAL MECHANICS: ADDITIONAl. ANALYTICAL MECHANICS: CHAPTER 5 . ANMNTICAL

Solution denote the We denote We use polar coordinates to describe the motion, as described in Fig. 5.8. We unstretched length of the spring as a. The velocity of the mass is

[.1

v = i-e,. +

leading to the kinetic energy expression

kr+ci-

.• vV =

T =

(hi [hI

+ r262) r262)

nig

The potential energy is due to the stretch of the spring and due to gravity, thus

Figure 5.8 Free-body

V =

diagram

—

a)2

—

[ci id

ingrcos0

The dashpot gives rise to the Rayleigh's dissipation function term 11

=

(dJ (dl

We next find the generalized momenta terms using Eq. L5.8.21. which yields IT,. TT, =

(?r or

(IL

= nil.

=

=

o'O

I.'

= c?()

express i-i and and 0 as From Eqs. Eel we can express IT,.

r=— !fl

IT0

0=

U?

and. using these expressions. we write the Harnilionian in terms of the generalized coordinates and generalized momenta as

1(= 7Trr±7r,jOL 7Trr±7r,jOL 7T

2w

,lflrL

2 inr-

+ 4k(r

—

a)2 — — ingrcosO ingrCosO

191

The Rayleigh's dissipation function leads to the generalized forces associated with the nonconservative forces acting on the system as Q r,ie = —

= — ci- =

—

--

in

Qonc = — —

r

.

=0

[hJ

Using the Hamilton's canonical equations in conjunction with Eqs. [gJ and {h]. we obtain the equations of motion 17r

in 'ir,.

=— —

dr

+ Qrnc =

0=

1

—

k(r

(1

IT')

(I'7T(1

mr2

—

a) + mg COS 0—

(C

lTr

,

= =

+ —

=— — ingr ingr sin 0

Ill 'Ii

5.1 2

€bMPUTATIONAL CONSIDERATIONS : ALTERNATE DESCRIPTIONS OF 'L'IIE MOTION EQUATIONS

5. 1 2 COMPUTATIONAL

ALTERNATE

DESCRIPTIONS OF THE MOTiON EQUATIONS With the advent of digital computers and powerful software. it has become possible derive the equations of motion. as well as to solve them. using computational :echniques. We discuss here issues associated with the derivation and integration of equations of motion from a computational standpoint. We begin with derivation of the equations of motion. The classical derivation of the equations of motion in ternis of Lagrange's equa-

cons is elegant, and it gives a lot of physical insight. As the nuniber number of degrees freedom become larger. however, using Lagrange's equations directly to derive the equations of motion becomeS increasingly diffIcult and time consuming. This is for hand calculations as well as for derivations performed by a computer. using mbolic manipulation software. To make matters worse, an experienced user of La= equations will notice that some of the terms in the evaluation of ., ii) cancel terms in :. 2, resulting in excess, wasted manipulations. . .

An alternative way to derive the equations of motion is to use D'Alembert's cmnciple directly(Eq. (Eq.14.7.161). [4.7. l6J). Fhe advantages of using D Alembert cinciple directly Alembert Ss principle are that the kinetic energy does hot not need to be calculated, and, because the entire process vectors, the derivation process can he mechanized and accomplished more efficiently by by digital digital computers. The disadvantage is that acceleration acceleration terms terms must must calculated, which requires more efibri than the velocity terms used in Lagrange's

Consider a system of N bodies having ii degrees of freedom undergoing plane We express the position of the center of mass of the ith body in column vector {r1} (i 1, 2, {rj(q1. q2. {r1(q1. N) where q,1, t)} is a column vector of xder 3. Similarly, the angles (i = 1, 2, .. ., N) can he expressed in terms of the Leneralized coordinates as = 1).The The time time derivatives derivatives of and . .. q,1, 1). . . . ,

.

.

,

.

tare

I hr1 1

=

: 'Fl. 'Fl

..

1 h)r1

+

+

.

1

= [rjqBq}

T

15.12.11

= {Oiq} {q} +

1

= [h101/ôq1 ... a01/hiq,,]. aO1Mq,,]. The expression for is the vector representation of Eq. [5.2.4]. Note that is a matrix of dimension

From From the above equation. equation. we write the variations of

{Sr} {Sr1}

=

and Oj

as

{Ojq}T{ôq}

I

[5.1 2.21 [5.12.21

a similar fashion, we obtain the translational and angular acceleration of the öthbody body as In

= dr2 =

= [r [riq}(c/} ]44} + +

d{q}

01 h/

.

q

()t{q}

+

.

+

01

h/r1 1 1 dr1

+

+ at

II

h/

..

Ii dl—

.

{q}

+

LI ..., N

= 1,2,

hit2

311

312

CHAPTER 5 • ANALY1JCAL MECHANICS:

ol =

Topics

+

where

= dOi/d(/k can he written as

(k1 S

+

+ 1, 2, 1,

.

iv

.

.

.

/2).

The generalized work expressior

N

+

=

+

=

6t91)

is.i 2.31 [5.12.31

= {Q}'{6i}

[5.12.4J 15.12.41

We We next combine all these relations. format is

principle in column

N

Y

—

+ ('Gi0i

—

MGi)

= 0

[5.12.5j [5.12.51

Introducing Eqs. Eqs. [5.12.2j—[5.12.4] [5.12.2j—[5.12.4]into into this equation, equation, and considering the case of constrained systems, we obtain N

(tniiriqii

+ IG,{Oiq}

-

+

+

}([FlqJ{q} +

+

+

= {Q +

[5.12.6J [5.12.61 in in which [ails [al is the the constraint constraint matrix matrix of order in X ii. n. The above above equation, equation, of of course. course. needs to be solved simultaneously with the constraint equation, Eq. [4.10.11. [4.10. Ij. It clear that calculation of the matrix and vector vector is crucial to obtaining the equations of motion. Considering the above derivation, one may question why we study Lagrange's equations in the first place. if' the equations of motion can be more derived using Eq. [5.12.61. There are several answers answers to to this this question. question.Lagrang&s Lagranges equations are based on D'Alembert's and Hamilton's principles, and one can derive the equations of motion using variational principles and fioni scalar functions. Doing so gives a very powerful result, putting in better perspective perspective the the relation relation betwe.en between the equations of motion and energy. It points to an order in dynamical systems. It gives a better understanding of the nature of the motion, what the integrals of the motion and it makes it easier to establish these integrals. Next, we consider numerical integration of the equations of motion. As discussed in Chapter 1. the equations of motion have to be cast in state form, which implies that they have to he first-order differential equations. In each equation there must he a single derivative term, on the left side. Neither Lagrange's equations the direct application of D'Alembert's principle yields equations of motion in stare fbi-in.On tbrin. Onthe the other other hand. hand. Hamilton's Hamiiton's equations are in state fbrm. One way to convert Lagrange's equations into state lorni is to introduce the variables (k = 1, 2, .., ii). Then, = Then, realizing that the equations of of motion motion can can he written as [MJ{i/} + {g({q}. {g({q}. {'Th} {'i}fl = {Q}

15.12.71

S.1 2

MOTiON EQUATIONS EQuATIONS thMPUTATIONAL CONSIDERATK)NS: O)MPUTATIONAL CONSIDERATK)NS: ALTERNATE DESCRIPTIONS OF THE MOTiON

write the equations as IM]{±} + {g({q}, IMI{±}

[5.1 2.81

= {Q}

The 2ii first-order equations are obtained as the inverse of Eq. [5.12.81, which n of the equations, together with the relationship between {z} and {q}, which

the other n equations. They can be expressed as

= —[Mf

1{g}

= {z}

+

15.12.91

Recall that EM] is a function of the generalized coordinates only, so its calculation

md inversion do not involve velocity expressions. Another way to convert the equations of motion into state form is to define {z}

[5.1 2.1 01

= [T]{c'} + {y}

and {y} is a vector of order n. [T] [T] is a nonsingular matrix of order n X n and introduce this transformation into the equations of motion. When [TI = [11 and {z} becomes the = {0} we get Eqs. [5.12.911. When LTI = [M] and {y} = momentum vector and we obtain Hamilton's equations. In general, the of the problem. of constrained constrained systems. We next discuss integration of the equations of motion of' We repeat Eqs. [4.10.11 and and [4.1O.5j [4.10.5] here as on

a10 = >: ajkqk + aJ()

/ == 1,1, 2,

0

1

. . . ,

in

k=I I (1

I(YL.

—(

..

\

?fl

SL

=

k

AJQJk

=

l,2,...,;i [5.12.111 1,2,...,;i [5.12.111

can be written in matrix form as IA]{4} ++ {b} {b} = {0}

[M]{j} + [A1'{A} + {g} =

{Q,11}

15.12.12a,bI 15.12.12a,bl

which [A] is a matrix of order in X n, whose elements are afk, as defined in Section 10. It is preferable to combine the two expressions in Eqs. [5.12.121. The highest-

derivative in Eq. [5.12.12a1 is of order one, while Eq. [5.12.12b] has time of order two. Differentiating Eq. (5A2.12a] with respect to time yields [Ai{j} =

—

15.12.131

{b}

This can now be combined with Eq. [5.12.1 2bj to give one matrix —mm as — as

[MI [A]

t the

[A]T

{4;}

[01

{A}

—

—

-IAIW}

{g} {g}. - {b}

equation of order

[5 12 14 ]

of qj qj (i = 1, 2, ..., n) are moved all the terms not involving involving second second derivatives derivatives of left side of Eq. [5.12.1 4j. The coefficient matrix on the left side of the above is symmetric, but it is no longer positive definite. However, it can be shown

i be nonsingular. Equation 15.12.14] represents a set of hybrid equations. corn-

differential and algebraic equations. They are called differential-algebraic

313

314

ANAIX1ICAL MECHANICS: MECHANICS: A[)DUI0NAE. A[)DFII0NAE. TOPICS TOPICS CHAPTER S • ANALYTICAL

equations. Several new methods have been developed in recent years to solve equations. The interested reader is referred to the texts by Brenan et al., and by

5.13

ADI)ITIONAL DIFFERENTIAL VARIATIONAL PRINCIPLES

Previous developments in this chapter and in Chapter 4 are based on variational principles. namely D Alembert s and Hamilton S principles and the special case of problems. the principle of virtual work. These principles are based on the variatiot

of the displacement coordinates while keeping time tinic fixed. In this section we arialyze the variational principles that one can derive using variations of velocities a

system of N particles

N —

F,)

•

[5.13.1)

6r, = 0

1=1 /=1

the dependent dependent variable, variable, which which where we note that the variation is taken by keeping the where ., 1, 2, position vector vector r1(t) r1(t) (i(i = 1, is the time 1, constant. We now look at the position later. Given the value of the position vector associated an increment of time by means of i one can obtain its value at I = t + by the /th particle at time t, Taylor series approximation as .

rU + Lit) = r,(t) +

+

+ r(t)

2

!+

.

... [5.13.21 .

.

.

The variation of r,(t + it). using time fixed at t. becomes

6r,(t +

! + .. ...

6 r1(l)

+

= 6r1(1) +

.

[5.13.3)

Substitution of this equation into Eq. 15.13.11 yields

>

(m,r

—

F1)•• F,)

I'

.

+ 6r1(t)

+

2

-1- 6

r(t)

3

!+

... = . . .

)I

0

1

15.13.4) 0 gets one back to Eq. [5. 13.1]. Now, consider a different variation: one one obtained obtained by by setting setting 6r1(t) = 0. We will denote this variation by the subscript 1. so that we write 61r, = 0(1 O(i = 1,1, 2. .., N). Using and takvariation and keeping in mind that 6r1(i) = 0. dividing Eq. 15.13.4] by approaches zero, we obtain ing the limit as Obviously, taking the limit as

.

\

I',

—

F,) • F1)

= 0

(61r, (61r = 0)

[5.13.51

i=l

Equation [5.13.5] we note that the evaluation took place at I = I + N'. Jourdain s's' variatio,ial variatio,ial principle. Note that 6, 61, (i = 1.1. 2. 61', known as Jourdam which are are used used in in Hamilton's Hamilton's principle principle and Lagrange's equations. equations, are because 6i,, which because obtained by taking the variation of 1', i', holding holding only only time time fixed, fixed, with with no no restrictiot on r,. where

. . . .

5.1 3

31 5

VARIATIONAL PRiNCIPLES VARIATIONAL PRiNCIPLES

ADDITIONAL ADDITIoNAL

variational principle principle can can be be used used to to deal deal with systems subjected to Jourdains variational [5. 13.5J 13.5J leads to the relation constraints. One can show that that Eq. Eq. [5. a I()'L a I()'L —I

11

,>

di \aqk .

\

I I— 1—

•1 (IL

-

.

61'Jk

—

15.13.61

= 0

,/

are chosen as indepenFor a holonomic system that is unconstrained and the variations ofthe gencralized coordinates are also independent and the above yields Lagrange's equations. in the presence of nonholonornic constraints are no longer independent, but one can introduce the the constraint into the formulausing Eq. [5. 13.6]. Hence, Hence. Lagrange's equations can he considered as a special of Eq. [5.13.6]. [5.13.6].

fashion,consider consideraa second second In aa similar In similar fashion,

variation, denoted denoted by the subscript 2, variation, iuring which both r1 = 0. 621'i = 0. 0. Divide Eq. r1 and i'1 I', are are held held fixed. fixed. such such that approaches zero to obtain and take the limit limit as as by N

(S2r, =

—F1) •• 62r1 = 0 —F1)

= 0)

[5.13.7] 15.13.71

principleo/o/least leastcon— con— principleororGauss's Gauss'sprinciple This is known as Gauss's van ational ationalprinciple

This principle can he shown to lead to a least squares interpretation of the of the quantity Z as N

Z == '1

2m,

(mr1

—

F)

•

—

F,)

15.13.81

takingthe thevariation variation of of Z Z we obtain Eq. (5.13.71. The variations :deed. taking variations of ofF1 (i = that of of all acforcing is a known quantity. We conclude that ., N) are zero .. zero as the the forcing : 2. that are compatible with the constraint equations. the actual accelerations the quantity Z aa minimum and, hence, constitute the solution. In practice, practice. one uses the variational principle that yields the equations of mowith the most ease. For unconstrained unconstrained holonomic holonomic systems. systems. there is usually no to use the the Gauss or Jourdain variations. The advantage of using Jourdain's or to are nonnonvariational principle becomes becomes more more pronounced for systems that are constrained. Because nonholonornic nonholonomic constraints can only be written velocity form. one can impose these constraints using the Gauss and Jourdain vanwithout manipulating the generalized coordinates. We will discuss this issue 9. in Chapter 9. It is interesting to note that D'Alemhert's principle is dated to the year 1743, and s pnnciple was stated in 1829. while Jourdamn s principle—which establishes link between the other two principles—was stated in 1909. .

.

.

It

c:nsider the vehicic in Example 4. 14 and apply Jourdain's variational principle to it. The

vehicle is shown in Fig. 4.8. The nonholononiic constraint is that the velocity of point A only he in the x direction. The constraint equation is written in terms of the generalized

I

Example

5.13

316

CHAPTER 5 • ANALYTICAL MECHANICS: MECHANICS: ADDITIONAL. ADDrFIONAL Topics

coordinates X. K and 8 as —XsinO —XsniO + YcosH

—

LO

=

[a] Ia]

C) 0

We will use Eq. 15. 13.61. Th this end, end, we write the kinetic energy as 1

f

-

=

.,•,

I

., •,

+ Y) ++

1

[b] Ib]

and the generalized forces as

= (Fç + FD) FD) sin sin00

+ F1)) F1)) COS Cos 8

Qx

=

—

F(' )/i F( )/l

Ic]

so that Eq. [5. 13.6J has the form (niX

—

+

61X

+(mY—(Fc + FJ))sjnO)61Y +(JGO —(F/)

—

Fc)h)610

=00

[d] Id]

The Jourdain variation of the constraint has the form

+ cos

sinO

—

1.1 [.1

0

We can manipulate this expression and express the variation of one of the variables in terms of the other. Expressing the Jourdain variation of 0 in terms of the Jourdain variations of X and I is futile, because in this problem X and Y are not sufficient to describe the motion completeR. Hence, we eliminate the variation of either X or Y. Let us select Y, so we can write COSO

= sinO

Y

+ 1.

If]

We multiply Eq. [dl by cos 0 and introduce Eq. [fj to it, which yields (inX

+ F/)) F/)) cos cos 0) cos cos 00 66 X X ++ (in (mY Y

—

+

+ PD) sin 0)( sin 0

—

=

—

—

X + L 6i 0)

0

IgI

Because the Jourdain variations of X and 0 arc independent, their coefficients must vanish independently, which yields

,nXcos() + = (F((F(- +

+

= F( + FD

sinO ± (Ff)

—

[hi

These equations still contain derivatives of three variables. To simplify, consider incorporating the velocity olA as a motion variable. The velocity of point A can be written as V4 =

= (Xcos0 + Ysin0)i

[I] [1]

The acceleration of A then becomes

a4 = a4 =

Differentiating

=

+ totoXXV/% V/% = VAi VAi +

II]

and using Eq. [a], we obtain

+ ?sin6 — XOsinO + YOcosO = Xcos0 + ?sinO + L02

[kJ 1k]

which, when introduced into into the the first first of of Eqs. Eqs. [hi yields in(v4 — L02) = F(' + FL) m(v4

[IJ III

HOMEWORK EXERCISES

In

a similar fashion, fashion. one can show that the second of Eqs. [hJ can he written as +

(JG

—

[ml

Fc)Ii

(). Upon that Eqs. [1] [1] and and [ml [ml constitute constitutethe thetwo twoequations equationsofofmotion motionininterms termsot'ot'VA and (). as the force balance in the x direction, and Eq. ;k)ser examination, one recognizes Eq. rn] as the moment balance about point A. These equations are, of course. the same as the of motion obtained in Example 4.14.

REFERENCES K. E.. S. L. Campbell. and L. Petzold. Numerical Solution

Initidil-Value Problems Initidil-Value Problems iii

Differential-Algebraic Equations. I)ifferential-Algebraic hquatwns. New York: Elsevier. 1989. Dynamics. New York: Harper & Row. 1988. •3!nsberg. Dvnwnics. •3msberg. J. H. H. Advanced MA: Addison Addison Wesley. Wesley. 1980. Reading. MA: H. Classical Mechanics. 2nd ed. Reading. 1988. I). T. 2nd ed. ed. Englewood Cliffs. Cliffi. NJ: Prentice-Hall. 1988. 1). T. Principles Principles ofDvnamic.v. of Dynamics. 2nd 1992. E. J. J. Intermediate Dynamics. Englewood Englewood Cliffs, Cliffs, N.J: N.J: Prentice-hall. Prentice-Hall. 1992. E. Publications. 1970. ed.New NewYork: York: I)ovcr I)ovcr Publications. Principles of of Medzanics. Medumics. 4th ed. C. The The Variational Prineiples McGraw—Hill. Hill. 1970. 970. oj/tnalvtical York:McGraw— nalyticalDvnami Dynami's.s.New NewYork: L. Methods oj/t and Thchniques Thchniques of of Vibrations. Vibrations. Eriglewood Cliffs. NJ: Prentice—Hall. Principles and Prentice—HaIl, 1997. 1997. Principles 1

HOMEWORK EXERCISES SECTION 5.2

length L is pinned to the edge of a disk of mass M and A link of mass in and and length radius R. as shown in Fig. 5.9. A servomotor keeps the angular velocity ol' the

Ffl. I.

Figure 5.9

31 7

318

CHAPTER 5 • ANALYTICAL ANALYTICAl. MECHANiCS: MECHAMCS: AIM)1TIONAL Tonics

(,b (1

\ \

r 'H

\ U''

:o

2

____,

————I————

'H

in 2

Figure 5.10

FIgure 5.11

Figure 5.12

disk constant at a value of [1. Find the equilibrium position for 0 as a function ol the angular velocity of the disk. of 2. Find the equilibrium equations for the rotating double pendulum in Fig. 5.10. The angular velocity of the shaft is constant. 3. Consider the rotating spring pendulum in Fig. 5.11 and obtain the equilibrium positions when th = is a constant and c = 0. 4. Find the equilibrium position of the bead shown in Fig. 5.12 for f 1. = constant constani

SECTION 5.3

Consider the equations of' of motion motion for for the the double double pendulum in Fig. 4.3. Linearize them in the neighborhood of the equilibrium position(s). 6. Consider Example 4.11 and linearize the equations of motion about the equilibriurn point Xe = 00. O( = 0. 7. Consider Problem 3 and evaluate tile stability of the equilibrium points. Given the bead problem considered earlier, plot the dynamic potential U as a = 1.5g/R. function of 0 for the following conditions: (a) = 0.5g/R. (b) Verify that the stability relations discussed in Sec. 5.3 hold. 9. A particle of mass in moves on a smooth surface described by the relation z = + v2 — xv. with the z axis being the vertical. Derive the equations of motion and obtain the linearized equations about equilibrium. 5.

SECTION 5.4

10. For the mass-spring system shown in Fig. 5.13. calculate the expressions for the kinetic and potential energies. as well as the Rayleigh's dissipation function. Then. write the equations of motion in matrix form fbrm directly from the energy terms, without using Lagrange's equations or Newton's second law.

HOMEWORK EXERCISES

FIgure 5.14

Figure 5.13 SECTION 5.5

1. A thin square metal plate, of mass in and sides L, hangs from its corners attached to two springs, each of constant k, as shown in Fig. 5.14. Find the equations of motion for small oscillations, as well as the natural frequencies. The horizontal motion of the center of mass is negligible. Find the equations of motion, natural frequencies, and eigenvectors of the system shown in Fig. 5.13 for k1 = k7 = k, k3 = 2k, in1 = in, in, = 2iii and no damping. 3. Find the linearized equations of motion, natural frequencies. and eigenvectors of the system shown in Fig. 5. 15. The system consists of two links of the same length and mass, attached by a spring. Assume small motions and consider only the horizontal deformation of the spring.

Consider Problem 5 and obtain the natural frequencies and modal vectors for fl?1 lflm,L1 = Li = L. 5. Consider Problem 6 and obtain the natural frequencies and modal vectors for M = 2iii and k = mg/b.

6. Fout- identical masses are connected to four identical springs and constrained to move in a circle, as as shown shown in Fig. 5.16. 5.16. The springs are unstretched when the masses are all equidistant. How many rigid body modes does this system have? Find the natural frequencies.

(1 1??

g

k

Figure

5

In 17?

Figure 5 16

319

320

CHAPTIR 5 • ANALYTICAL MECHANiCS: AIMMTIONAL TOPICS

SEc1II0N 5.6

Check for for the the orthogonality orthogonality of the natural modes in Problem 12. Check 18. Check for the orthogonality of the natural modes in Problem 14. 17. 17.

SECTION 5.7 19. 19.

Consider Problem 13 and the case when the spring is weak, that is, the potential energy of the spring is much less than the potential energy due to gravity Approximate the two natural frequencies and plot the response to zero initial conditions, 20. Obtain the response of the system in Problem 12 to a unit impulse on mass 1 Initial conditions are zero. 21. Consider the double pendulum in Fig 4.3. Set ni1 = rn2 = m. L1 = L2 L and find the response to the initial conditions Oi(0) = 30°, 02(0) = 00. Calculate the maximum value of' the amplitude of 02(l).

SECTION 5.8 22. Identify Identifythe theintegrals integrals of of the the motion motion for for the the pendulum pendulum in Pro blem 3 for (a) 4) is 4) iS variable, and (b) 4) constant = Take c = 0 in both cases. 4) is a constant

23. Identify the integrals of the motion for the Foucault's pendulum. 24. Find the integrals of the motion for the rotating double pendulum in Fig. 5.10.

SECTION

5.9

25. Consider 25. Consider the rotating spring pendulum in Problem 3 and obtain the Routhian.

SECTION 5.10 26. Obtain the velocity of the collar shown in Fig. 5.17 immediately after an impulsive force of P is applied applied to the bottom of the disk of radius R = L/6. At At the instant the impulse is applied. 6 = 0, and the disk is not rotating.

27. Two rods of equal length and mass are connected by a pin joint, and they are at rest, as shown in Fig. 5.18. An impulsive force is applied at point B and perpendicular to the line AB. Find the angular velocities of the rods after the impulse. Hint: Include the coordinates of point point A in the generalized coordinates.

28. The double pendulum in Fig. 4.3 is at rest when an impulsive force P is is applied applied in the horizontal direction to the bottom of the second rod. rod. Find Find the the ensuing angular velocities.

EXERCISES

in. L

0

4

2m

flz, 1.

g

nz,L

B 3 in A

F

Figure 5.17

Figure 5.18

Figure 5.19

SECTION 5.11 29.

Derive the equation of motion of the bead on a hoop that we have considered earher using Hamilton's equations. Consider the rotation of the hoop 4 as a variable.

SECTION 5.12

Consider the system in Example 4.11 and derive the equations of motion using the direct application of D'Alembert's principle by Eq. 12.61. 31. Solve Problem 30 using D'Alembert's principle, hut without explicitly calculating the [riqi matrices. Compare the ease with which the solution is obtained with 12.6] and by using Lagrange's equations. the solution obtained by using Eq. 30.

SECTION 5.13

Solve Problem 4.41 using Jourdain's variational principle. 33. Find the equation of motion of the rod in Fig. 5.19 using Jourdain's variational principle. Friction is negligible and the rod maintains contact with the wall. 32.

321

chapter

RIGID

6.1

Bony

GEOMETRY

INTRODUCTION

in this chapter we consider the geometrical properties of rigid bodies. The term idealization, because because all bodies defbrm by a ceris in reality a mathematical idealization, is amount under the application of loads. If the deformation is small compared to energy dissipation due to elastic effects is and energy the overall dimensions of the body. and the rigid body assumption can safely he used. Let us distinguish between particles and rigid bodies. A particle was defined no physical physicaldimensions. dimensions. This This definition, definition, of course. is also an a body with flO of the body arc dimensions. If If the the dimensions dimensions of as all bodies have physical dimensions.

it.becomes becomes possible possible to neglect the much smaller than the path followed by the body. it dimensions of' of the the body. body. One One does does not not consider consider any rotational motion. and translational degrees of' freedom are arc suffIcient sufficient to to describe the motion. Whether of freedom Can use use this approximation depends on the conditions under which the motion Can of' the the bodies involved. place, the loading, as well as thc the material properties of' A rigid body is defined as a body with physical dimensions where the distances the particles that constitute the body remain unchanged. One needs to conthus. six degrees of freedom. three transla,;der the rotational motion of a rigid and three rotational, arc required to completely describe its motion. In addition. :ne needs to develop quantities that give information regarding the distribution of mass along the body. Just as the mass of a body represents its resistance to translamotion. the distribution of the mass about a certain axis represents the body's to rotational motion about that axis.

6.2

CENTER OFMASS

Consider a system of N particles viewed from a fixed reference frame with origin denotes the mass of the ith particle particle (i(i == 1.1. ',7. as shown in 1. The term in Fig. Fig. 6. 6.1. Chapter 3. 3. the the center of mass of a system of In Chapter N) and r, its distance from 0. 0. In

323

324

CHAPTER

RIGID BODY GEOMETRY

V

'V 'S ?71

• Ffl,

(1?71

//

•G• 0

'V V

¾

Figure 61

FIgure 6.2

A system of particles

Differential element for

a rigid body

particles was denoted by the point G. and its location was defined as N

rG =

In in

[6.2.11

ni1r1

— ,> I

I—I

in which in denotes the total mass N

16.2.21

liii

=

A rigid body can be considered as a collection of particles in which the number of particles approaches infinity and in which the distances between the individual each particle can he treated as a masses remain the same. As N mass element, as shown in Fig. 6.2, diii, and the summations in the above equations are replaced by integrations over the body. We then define the location c4 the center of' mass G as I

rG=— rn

rd,n -

16.2.31

body

where r is the vector from the origin to the differential element din, and

in=J

[6.2.4J 16.2.41

drn

hod

is the mass of the rigid body. Considering Figs. 6.1 and 6.2. for a system of we can we canwrite writer1r = r(; ++ P1. P1. and and for a rigid body r rG + p. Introducing this term into Eq. [6.2.31. we obtain

rdin = 11

I

r(;=— body

in JJ body

(r(; + P) drn = r(; +

I

in —

I body

pdm

[6.2.5]

leading to the conclusion that

pdm=O Jbody

[6.2.61

6.2

325

CENTEROFMASS

This equation indicates that the weighted average of the displacement vector

the center of mass is zero. Considering concepts from statistics, one can reto the definition of the center of mass as the first moment of the mass distribuNaturally. Eq. [6.2.6] is identical to its counterpart for a system of particles. [3.2.51.

The center of mass is a very important quantity, as its use simplifies the analof bodies considerably. One has to perform the integrations in Eqs. [6.2.31 and in order to find the center of mass. These integrations are in general triple Many times, the body that one is dealing with consists of parts that are .Lniform in density and that have shapes whose masses and centers of mass are known

or found more easily. Under such circumstances, one can calculate the :enter of mass of' of the the composite composite body body by by using using the center of masses of the individual If the body is separated into N components, and the mass and center of mass Tithe each component is denoted by m1 and r(1, (1 = 1, 2, . ., N) the center of mass can advance

. .

found from N

rG =

16.2.71

—

in

i—I

When dealing with several components. we will also use the notation of denoting center of mass by an overbar. The velocity and acceleration of the center of mass be obtained by differentiating the expression for position and integrating it over body. Using the expression for rG in Eq. [6.2.31 we obtain VG

If

— I m Jhocly

rdm

I

i'dm

16.2.81

T!e velocity and acceleration of a point can be expressed in terms of the center of ass as

v = VG+P

a = aG+p

16.2.91

ntroducing these expressions to Eqs. [6.2.81 and performing the integrations gives

pdni =

0

pdm = 0

16.2.101

body

6.3 shows a decorative pencil holder. niadc of uniform material of thickness 0.4 cm. Find the center of mass.

first select the coordinate frame. For convenience, we attach it to point 0. The pencil can be analyzed as consisting of three parts: the base. the hollow cylinder that holds pencils, and the circular hole. As the entire pencil holder is niade of uniform material, we with the density and base our calculations on the volume. For each component, we determine the volume and location of center of mass. We thus have:

I

Example

6.1

326

RIGID BODY GEOMETRY

CHAPTER

6cm

hole (radius 2 cm)

/

6

9cm

0 i

cm

cm

'4

Flgur. 6.3

=9i+6j+0.2kcm

Volume V1 = 18(12)(O.4) = 86.4cm3 For the cylinder: Volume V2 = 77.(32 — 2.62)(8) = 56.30 cm3

For the base:

For the circular hole: Volume V3 =

5.027cm3

r2 = 9i+6j+4.4kcrn r3

151+8j+0.2kcm 151+8j+0.2kcm

Combining all these values, we obtain V1 + V2 — V = Vi

V3

= 86.4 + 56.30 — 5.027 = 137.7 cm3

Es]

so that the center of mass is located at

rG=

86.4(91 + 6j + 0.2k) + 56.30(91 + 6j + 4.4k)

= 8.7801 + 5.926j +

6.3

—

5.027(151

+ 8j + 0.2k)

137.7 1.9 17k

cm

[bJ

MAss MOMENTS OF INERTIA

While the center of mass provides valuable information and simplifies the analysis of translational motion. it gives no measure of the way the mass is distributed on body. The mass of a body describes the amount of matter contained in the body and the resistance of the body to translational motion. We know of the need to develop a quantity that describes the resistance of a body to rotation. Such a quantity is dependent on how the mass is distributed. As the center of mass is located using the first moment of the mass distribution, we consider the second moment of the mass distribution.

6.3

MAss

OF INERTIA

V

'I 'a

'I 'a

'

drn

x

I -v

FIgure 6.4 We select a coordinate system xyz fixed to a point on the body and describe the configuration of a differential mass element by the vector r = xi + yj + zk. also written as the column vector {r} = [x y z111. Such a selection of the coordinate frame is possible because the body is assumed to be rigid. We are primarily concerned with two types of quantities: the distribution of the mass with respect to a :ertain axis: and the distribution of the mass with respect to a certain plane. Consider the x axis first. From Fig. 6.4, the perpendicular distance of a differential element dm from the x axis is Rr = + z2. We define the mass moment of inertia about

thexaxisas lxx

= Jbody

= Jbody

(y_ + z)dm

[6.3.11

in a similar fashion, the mass moments of inertia about they and z axes are defined as

=1body =1-

= Jbody

'zz =1body

= Jbody

+ z2)dm.

[6.3.2]

+V v )dm )dm

One quick observation is that the mass moment of inertia of a body about a :ertain axis becomes larger as the axis is selected further away from the body. This is !n indication that mass moments of inertia will be useful in describing the rotational motion of a body. Considering the distribution of the mass with respect to the xv, xz, and vz yz planes, introduce the three products of inertia Ixy = I

i body

x din

1yz

= Jbody

yzdrn

Ixz=1

xzdtn [6.3.31

Jbody

It is clear that = and so forth. In general, the products of' inertia do contribute too much to the physical description of the mass distribution, unless there are certain symmetry properties with respect to the coordinate axes. If any two coordinate axes form a plane of symmetry for the body, then the products of inertia

327

328

CHAPTER 6• RIGID

BODY

x

Flgur. 6.5

A body symmetric about the yz plane

associated with coordinates normal to that plane vanish. That is,

If there is symmetry with respect to the yz plane, then if there is symmetry with respect to the xz plane, then If there is symmetry with respect to the xv plane. then

=

= =

= 0. = 0. = 0.

The concept is illustrated in Fig. 6.5. Assume there is symmetry with respect to the yz plane. Then, for each point defined by the coordinates (x, y, z) there corresponds vanish because and the point (—x, y, z). It follows that the products of inertia both terms require integration over x, and the contribution of the differential mass on the left side of the yz plane is countered by the mass on the right side. If a body is almost symmetric about a certain plane. then the corresponding product of inertia sign of will be a small quantity. However, the converse is not necessarily true. The sign a product of inertia does not usually give much insight, either. If a rigid body is symmetric about an axis then it must have symmetry about at least two planes. Thus for a body that has an axis of symmetry, all products of inertia vanish when one of the coordinate axes is along the symmetry axis. It should be noted that a body need not have planes or axes of symmetry for the products of inertia to vanish. A proper orientation of the xyz axes leads to the same result, as we will see later. The moments and products of inertia form the so called inertia matrix, denoted

by [Jjanddefinedas 1,1: .1

[11

=

_lxy Iyy

[6.3.4]

—!xzz

Note that the inertia matrix is symmetric. Furthermore, it is positive definite. To demonstrate the positive definiteness of [I], one can use Sylvester's criterion, which

positive definite, all the diagonal states that for a symmetric matrix to be positive and all principal minor determinants must be positive. The diagonal elements of [1] are the principal moments of inertia and they are all positive quantities. each obtained by integration of a positive integrand. The first principal minor determinant

6.3

I,

MASS

OF INERTiA

inequaliiv, show that it is greater than zero. consider the statesthat states that for for any any two two functions f(x) and g(x) continuous over an interval — —

.

To

1

.1)

_I)

—

Lf(x)P Lf(.v)P Ia

( 11

[6.3.5]

(I..v .1(2

Theequality The equality hoidsonly holds only when when f(x) = y1g(x). where 'y is aconstant. Considering

In integral over the rigid body. and the products of inertia. we can write

I

body

I

Jhodv

(12 +

(\12

:2)din :

+ :2)(IF,?

16.3.61

body

<

which shows that the first principal minordeterminant is greater than zero. In a similar fashion. one can show that the determinant of [I] > 0. En some _ases. such as when dealing with slender rods or thin plates. the components of the matrix about certain axes will be very sinai! compared with the inertia compoabout other axes. Then. one can assume that the very small inertia components ire negligible. For example. a slender rod is assumed to have zero moment of inertia inertia Ut an an axis axis along along the rod. The positive definiteness of the inertia matrix is to be expected. because it dethe distribution of the mass about a certain set of axes. axes. The The expression expression 'from from analogous to mass moments of inertia is variance. Two rigid bodies may the same inertia matrix and not the same shape. Such bodies are called equimothat

If the coordinate axes are selected such that the products of inertia vanish, the .-xrdinate -xrdinate axes are referred to as principal axes and the corresponding mass moof inertia are called principal moments of inertia. For an axisymmetric body axis of symmetry is a principal axis. The units of' mass moments of inertia are mass X length squared. The moments :f inertia of of'aa body are usually written as the mass of the body multiplied by a length parameter and an appropriate appropriate constant. constant. For For example, example, the the mass massmoment momcn of ..nertia of a disk about the symmetry axis is written as ,nR2/2. where in is the mass md R is the radius of' the disk. The mass, center of mass, and inertia matrix of a rigid body specify what are the internal properties of the body completely. For an elastic body, one needs know measures of' the resistance of' the body to deformation, in addition to the properties. We will study these in Chapter 1 I. We next calculate the entries of the inertia matrix. This is dependent on the of the differential element to he integrated. We have two choices:

Select the differential element as a differential mass clement and perform a the differential difTerenrial clement clement is is din dm = integration. For Cartesian coordinates xvz, xv:, the bodies with p being the density. Polar Polar coordinates coordinates conic conic in in handy handy For bodies circular cross sections. The differential element is dm = prdrdO dz.

329

330

CHAPTER 6• RiGiD

tRY

BODY

4

V

V

V

dx

V I

I

(1V

x

Figure 6.6 6.6 Figure

Figur. 6.7

Differential element as a rod or a thin plate

Differential element as a thin plate

One or two of' the sides of the differential element has a finite size. Select the differential element as a thin rod or as a thin plate. as shown in Figs. 6.6 and The differential element then has a finite mass moment of inertia about one or c axes. The expression for mass moment of inertia is written about such an axis (say, about the z axis) 2.

[6.3.71

Izz

in which

= J is the mass moment of inertia of the differential element about the:

axis.

The advantage of using such elements is that the number of integrations is reduced. When the differential element is a rod, one can perform a double integration. and when the differential element is a plate. one can evaluate the mass moment c'f inertia by a single integral. Consider a thin flat plate of thickness t as shown in Fig. 6.8. We take a axis of inertia inertia about about the the zz axis mass moment moment of element in the form din = pt dx dv. The mass becomes )

pij v2)din = p1 (x + v")d,ii

(x + v) dx dv =

[6.3$) [6.3.$J

area

body

is recognized as the area polar moment of inertia about the z axis. suniing that the thickness is small, the monient of' inertia about the x axis can where

approximated as lxx

= Jbody

(y2 +

=

ptj area

+ 12)dxdv

pt pt f

f

. area

v' dx d v

[6.3.9) [6.3.91

is the area moment of inertia of the plate about the. x axis. In a in which From area moments of inertia, we know = fashion, one can show that and [6.3.91, for a thin flat plate oriented as ix. Considering Eqs. + = + Fig. 6.8, Fig. 6.8,1..- =

6.3

MASS

LNERTIA

V

iliii

FIgure 6.8

Differential element on a thin plate

Now, consider the differential element in Fig. 6.7. specifically a slice of the section with an infinitesimal thickness, say dz. Denoting by x*v*z* the set of axes that are attached to its center of mass, the differential element has a using + One then finds moment of inertia of = Izz

=

=

J

[6.3.10]

+

is thus obtained by a single integral. If there is sonic some The mass moment of inertia of symmetry in the body. one can select the coordinate axes to exploit this symThis helps determine such that = and have = well.

we have to use the parallel axis theorem, which relates and To calculate the moments of inertia of a body about two axes that are parallel to each other with :ne of the axes centroidal. The theorem was stated in Chapter 3. and its general form derived in the Section 6.4. Here, we slate it between the x and x* axes, which are by a distance of z, z, as shown in Fig. 6.7. lithe :z axis is the symmetry axis, a centroidal axis, can be expressed as

[6.3.1 11 [6.3.111

+ Z2 diii z2 din

=

element, For a circular plate as the differential clement, = —

+ z2dm z2 dm

=

and

[6.3.12]

when integrated, when integrated, gives the mass moment of inertia about the x axis (and y iiis)

as It. .t..

=

=

I--

•

(III? + + jj z din .,,2

[6.3.131

The above approach is most appealing when the.re there is is aa symmetry symmetry axis, axis, as as the the rtducts of inertia vanish. For bodies where there is no symmetry axis, then ..1::/2, and and itit must must be be evaluated evaluated independently. Further, as the products of inertia do vanish in such cases, the above approach is of little use.

An interesting property of a mass moment of inertia is that it is an additive uantity. Given the task of finding the mass moment of inertia of a body (about an

331

332

CHAPTIR 6• RIGID

BODY GEOMETRY

axis or a plane), one can accomplish it by taking advantage of the fact that an integral from one point to another can be expressed as the sum of two or more integrals, appropriately separating the domain of the integration. The procedure is analogous to obtaining the center of mass of a composite body. The body is separated into more than one part. The moment of inertia of the entire body is calculated as the sum of the moments of inertia of the individual parts. To see this, take a body, separate it into N parts (body1, body2, ..., bodyN). and consider any of the inertia expressions We then have, say, for N

'XX ix

.i•

j

= Jbody

(y2 (y2 +

d rn

—

N

+ z2)din z2)din =

> 1.body, f

16.3.141

> i—

I

in which is the mass moment of inertia of the ith body about the x axis. For bodies that can be broken down into simple parts whose individual inertia properties are known or easily calculated, one can obtain the moments of inertia of each of the parts and add them together to find the moment of inertia of the entire body. This process is facilitated by use of the parallel axis theorem. Mass moments of inertia of bodies with common shapes are given in Appendix C. A useful quantity when dealing with mass moments of inertia is the radius gyration The radius of gyration about the v. and z axes is defined as in 'Ti

-.

—

'1

ISV Ivy

Jzz

Yin

in

[6.3.15)

and it represents the distance that an equivalent particle of mass in would have to placed from an axis so that it would have the same mass moment of inertia as the entire body. Hence, the mass moments of inertia are expressed as = in The radius of gyration is frequently used in engineering practice when the inertia properties of commonly used components such as 1-beams and brackets.. One can define the radius of gyration for the products of inertia as well, thus

lix / '.l•

= ¶1/ V

V

V..

K%- =

=

in In

in

Ixz IXz

in

[6.3.16)

with the understanding that

Example

6.2

Find the mass moments of inertia of the paraboloid shown in Fig. 6.9 about the x and v axes The paraboloid is obtained by rotating the parabola xIL = (y/R)2 about aboutthe the.x.x v R)R) axis. ax is.

(0

Solution The differential element for the paraboloid is taken as a thin disk of thickness dx. At a of x (0 x L). the radius of the disk is r = The mass and mass moment inertia of the differential clement are din = pA

=

=

f)ITR2

-)

dx = p-rrr dx =

=

L

x dx

(SI

=

[bJ

6.3

333

MAss MOMENTS OF INERTIA

(fr

V

x

FIgure 6.9 To find the mass moment of inertia

/

we integrate Eq. [hJ over x. which yields

.vdx vdx

2L2 J0

—

=

[ci

6

The mass of the paraboloid is found by integrating Eq. [a] over x. thus rn

= J

im

'

=

L

L

xdx= x dx =

pirR2L 2

[d] Ed]

thtroduction of Eq. [dl into Eq. [Cl yields the mass moment of inertia about the x axis as 1

=

Eel

To find the moment of inertia about the axis, attach an axis to the center of the element. The y axis is separated by a distance x from the axis, so axis, so that using [6.3.13] gives =

+

v2

If]

dm

f

Eq. [a], the second term on the right becomes r

xLdrn

pITR2

L

xdx =

p7rR2L3 4

=

Igi

the mass moment of inertia about the y axis (and (and z a,is) becomes 'V',

=

121

—niL' = —mR + —fl?L' 6

2

[hi

Because of the axial symmetry. aU products of inertia are zero.

Pind the mass moments of inertia and products of inertia of the right triangular prism of density p shown in Fig. 6.10.

S.Iutlen We first find the mass of the prism as in = pV. where the volume V = abc/2. so that the of the prism is

pabc 2

lal Ia]

Example

6.3

334

CHAPTER 6. RIGID BODY GEOMETRY

V

b

a

x

C

Figure 6.10 moments of inertia, we need to calculate the expressions x2 din, J To find the the moments I z2 diii. Using a rectangular differential element, we write

y2

din = pdzdydx

[bJ

We now evaluate x2 din, J v2 din, and z2 dm. Note that we perform the integrations in z2 dm. order over the z, y, and x axes. I J

J

h—tx

a

Jx2dm

jf

=

I Jo

o

ch

y2dm =

I

I

Jo .0

-

3J0(

pz2dzdydx

.00j

=

py2dzdydx = —

cc cc.

b——x)dx= a

Jo\

3

pa3 he pa3hc

d— 2

a

—

ma2

—

6

pa/3 c

[cJ

mb2

12

/

=

J0

J

—

a

io

Jo JO

ra

z2dm

px2dzdvdx = pcf x 2(b_" ) dx

cc

JO

r

a

pabc3

rnc2

6

3

[.J

The mass moments of inertia then become

=m

!xx = Note the similarity between 1a(I) (I) ja

Ixy=I =

b 0

a

'c

h

pxz dz dv dx =

Jo J a

JO Jo JO

Izz

a

J

a

PC-

J

J

I

I pyz dz dy dx —

/

b ——x Cl

pcf (b-

.r

Jo

If'

=

/ b\2 pa2 b2 x(b— —xI dx =

PC

=

Jo 0

= I

ly

pxvdzdvdx p xv d z d y dx =

I

JoJo Jo Jo

(a2

We next find the products of inertia

1e

I

I

and

I

icr

a a //

4Jo\

I

)2

mab

24

12

pa2 hc2

mac

12

6

pabc 1

12

=

mbc 6

[I] [bJ

En

Note again the similarity between

6.4

TRANSFORMATION PROPERTIES

Given the inertia properties of a body about a point and a certain orientation of the set of axes, it is desirable to relate these properties to the inertia matrix obtained

6.4

PROPERTIES

or aa diFferent diFferent onentation onclitationof of the coordinate axes. This permits •ibout difierent point point or •ibout aadifierent :'ne U)tind tind the the mass mass moments moments of of inertia inertia once once and to then use the transformation :'ne U)

equations to find the moments of inertia about other points and about other sets of axes, without having to perform the necessary integrations again. Also. moments of inertia read from tables are given in terms of a specific set of axes. We will first how the inertia properties change as the coordinate system is translated and then the coordinate system is rotated. Recall that in the previous seCtion section there there were were the no restrictions on the location of the origin and on the orientation onentation of the coordinate frame about which the inertia properties were calculated. First, note that the inertia matrix can be expressed in column vector format and terms of the position vector {r} as 111

:n

1

I—

(1.2111 — {,.}{,.}T )

{r}{,.}T)

=

=

[6.4.11

J

which r2 = {,.}T{,.} = r • r. The inertia matrix can also be expressed as

[6.4.21

1/1

ahere [?1 [?1 is is the the skew-symmetric skew-symmetric matrix formed from the elements of the column ;ector{r} = [x v :]'. We will use the above definition of the inertia matrix when :]'. We discuss dynamics of rigid bodies. Yet another way of expressing the elements of the inertia matrix is as follows.

so that the unit vectors associated with the coordinate frame frame by by e1, e1. e2, e2, arid and The ijth element of' of the the inertia matrix can he expressed as r = r1e1 + r2e, +

x

=

c

X e1)dm X

1. 1,

[6.4.31

/ = 1,2,3

derive Eq. Eq. [6.4.11 [6.4.11 from from Eq. Eq. 16.4.31 16.4.31 by using the vector identity (a X b)• One can derive. r. b = e,, e, d = e1. d)b• C) C) and by setting a = c = r, x d) = (a• c)(b• d) — (a • d)b

6.4.1

TRANSLATION OF COORDINATES

the origin of the coordinate coordinate system system by by 00 and and con consider sider a different coordinate .v"v"z", which is parallel parallel to to the the xvz xvz system system and andhas hasits itsorigin origin at at point point 13 x = ..v" + v = v" + coordinates(drn1, d\, coordinates d\, as shown in Fig. IL That + + d-k. The moment of inertia about the .v" .v" axis r1110 = .i.. C or r1110 C = z" + found from =

(y"2 (V'2 ++ :"2)dm I

f (v2 +

= f

_2)

=

—

d-)]din d\)2 ± (z (z ——d-)]din

I

+ d? ) din —

(/111 ± J

— 2di/rn— j z din

j

." dlfl Y i/in

[6.4.41

The second The third third and fburth fourth terms arc basically the first moments +d?). The reduces to + 0 = these :f the mass mass distribution. distribution. If the axes xvz are selected as centroidal The first term on the right side of this expression is recognized as

335

336

CHAPTER 6• RIGID BODY

/ / 'p

y

V

B

/

-

C

-

--

/d_

-

/1

-

'p

Figure 6.1 1

Translation of coordinates

axis theorem, which states that terms vanish. We then obtain the parallel cuis

16.4.51 16,4.51

= 'Gxx + m(d; +

with respect respect to the x where 'G.V.V denotesthat thatthe themoment moment of of inertia is calculated with where 'GV.V denotes poin also indicates indicates that the poin mass. Equation Equation 16.4.5 16.4.5 1 also of mass. center of passing through the center center of of mass body are the smallest is the center about which the moments of inertia of a body Note that we have modified the notation for the mass moment of inertia by indicating the point about which it is calculated by the first subscript.

We calculate the remaining moments and products of inertia associated with x "y"z" axes in a similar fashion. For example, for the product of inertia term 'Br' we have — (x — = X")"d?fl X'Y'd?fl = (x = J = J

dd rn

d

J

J

[6.4.6) [6.4.61

+ rnd.VdV rndVdV

= We

dd in

can express the parallel axis theorem in matrix form as + il? ci?

—

—

+

[1(3] ++ in [/81 = [1(5]

—d%ldz

[6.4.71

+ d;

Using column vector notation, denoting by {r"} the position vector associated {r} — {d}. in which {d} = with a point with the primed axes, such that {r"} — {r} we obtain ['BJJ ['B

,,

r ,, }{r }

I

=

j

= J

+ d2

—

d in £1

)

2{r}T{d})[11

- c{,.H,.}T

+ {dHd}T

—

2{r}{d}T)] dm

[6.4$)

6.4

TRANSFORMATION PROPERTIES

Considering the centroidal axes. j {r} din vanishes, and we end up with I

+ (/2 [11 — — {rHr}T

= + = [JG] [JG] +

1] —

—

{d}{d}T] dm

.{dHd}T)difl

16.4.91

where d = {d}"{d}. {d}'{d}.

The above relation is significant in that it permits a simple calculation of the moments of inertia about any set of axes parallel to the centroidal axes, as long as the inertia matrix about the centroidal axes is known. For bodies with complex shapes. one splits the body into smaller parts whose centroidal moments of inertia can be inertia of the calculated easily or looked up in a handbook. The total moment moment of of inertia body is found by using the parallel axis theorem and by adding up the individual moments of inertia. Appendix C gives the mass moments of inertia of bodies with commonly found shapes.

6.4.2

ROTATION OF COORDINATE AXES

set of of axes axes .v'v'c'. obtained obtained by by applying a set of rotations to the Consider now a set riginal coordinate system xvz. The origins of the two coordinate systems coincide. The position vector of a point is {r'}. From Chapter 2, using the direction cosine [c] or transformation matrix [RI. [c'] [Rb one can relate the vectors {r} and {r'} by the -elationship {,J} =

or

[6.4.10]

{r'} = [R1{r}

The distance of the differential element from the origin is the same in both co-

,.2 = = {r}T{r} == ,.2 :rdinate frames, :cdinate frames. r'2 = [c'} Consider now a element whose location is described by {r'}. The inertia matrix about the nmed axes is defined as

[/'l

•

11 — (r'2 Ill —

Substituting Eq. [6.4.10] [6.4.1011 into II ][C], ][c], we obtain

[I'I ["I

lid

—

dm

= J

— [11 — (r2[fl (r2

drn

16.4.111

this expression. and noting that ,.2[II r2[l] =

(?[l]

[(.]T{,.}{r}/[(I1)c!,,: [(.]T{,.}{r}T[(Il) [cIT dm = [CIT

—

{,.Hr}T) dm[cj

— J -

[6.4.12] 16.4.121

Recognizing the term in the middle as f I]. we obtain the result

I"] = [cjT[JJ[cl

or

[/'l

[R1[fl[RIT [R1[/1[RIT

16.4.13a,b]

that for a rotational transformation, it is not necessary to begin with a set of axes. Any set of axes will do. In general, when calculating mass D:'ments of inertia, inertia, select select the the .vvz xv: axes axes such such that that the the calculation calculation of the moments inertia is simpler (e.g., using symmetry axes or symmetry planes) and then Note

337

338

CHAPTER 6• RIGID BODY GEOMETRY

a coordinate transformation to obtain the moments of inertia about the desired axes axes. A very important coordinate transformation is one that yields a diagonal inertia matrix. The question then arises as to whether and how a transformation matrix [RI inertia matrix matrix is is diagonal. diagonal. We will show in such that that the resulting inertia can he found such Section 6.5 that the transformation matrix can he found by solving an eigenvalue problem. When the coordinate axes must he both translated and rotated, the order of these operations does not affect the linal result. result, use

Example

6.4

the square square cross cross section to which a concentrated mass ,iz/4 rod of mass in with the Given the rod as shown in Fig. 6.12. find the mass moment of inertia about a set of axes C. as attached at point C.

x"v"z" which are parallel to the sides of the rod and which pass through point A.

Solution There are two approaches to solve this problem. The first is to find the center of mass the rod combined with the concentrated mass and to then use the parallel axis theorem. The second is to treat the rod and the concentrated mass separately, find their mass moments inertia about the x"v"z" axes. and then add the two moments of inertia. We demonstrate latter approach first as it is usually simpler, and is so for this example, because one neeth use the parallel axis theorem only once for each component. Using the second approach, we find the inertia matrices associated with the bar and mass separately and then use the parallel axis theorem. For the bar, we attach the xv: the center center of mass of the bar denote the to the center of mass, as shown in Fig. 6.12. We denote

U

V

y,p

77777

'p I,

V

V

1

-*

a-—-

Figure 6.12

6.4

)

=

Gi,ar.

TRANSFORMATION PROPERTIES

From Appendix C, the inertia matrix is diagonal with its elements Fm(a2 + L2) L2)

= [1o] = [la]

ma2

in(a2 + L2) in(a2 L2)

6

12

12

lal

Points A and 0 are related by

a.

= that the the

L. L

Ibi

+ 3-J — —

x"v"z" axes axes are are apart apart from the xvz axes by

=

=

a/2,

L/2,

d- =

—a/2.

Eq. [6.4.7] the combined inertia matrix becomes + L2 [1.4harl

=

o 2a2 o

0

0 14a2

rn

+ 4L'

o

—3aL

—

I

3aL 6cr 3aL

3a2

-

—3aL

a- + L2]

3aL

3a2

Ffl

0

8cr 8cr

'2L

r3a2+3L2

-1

3cr

3aL 3a2 + 3L2 I

3a2

Ic] IC]

3aL 4a2 + 4L2 I

For the point mass, the inertia matrix about the center of mass is a null matrix. The of point C in the x"v"z" axes are rc,n = —aJ2i — Lj + ak, so that —a

d-=a

2

[dJ Id]

yields the inertia matrix

of Eq.

I

—aL 2

2

I

5cr 4

in

=— 4

a

—

aL

2

[eJ lel

aL a—

4

+L2

Addition of the the two two inertia inertia matrices yields t'A] =

['Ahar] +

[IAmaJ =

+ 28L2 —l8aL 18a2

—l8aL 47cr 24aL

18a-

24aL 19a2 + 28L2

Ifi

In the other approach. we first find the center of mass of the combined system, the asso:..1aied inertia matrix, and then the inertia matrix about point A. To find the center of mass we the xvz frame. The properties of the bar and point mass can now be written as 1flhar =

ifl

Fhar

=00

=

L.

=

+

a

Igi

we have denoted the position of the centers of mass of the rod and point mass by an The center of mass of the combined system. denoted by G. has the coordinates

rG=

?flbarFbar + Tllhar + Tflhar

—

L.

+ ma/8k 5rn/4

=

+

a

[hi 1k]

339

340

CHAPTU 6• Ricin BODY GEOMETRY Next, we need to find the inertia matrix about the center of mass by using the parallel axis theorem on each component. The vectors from the centers of mass of the individual components to the center of mass are For the rod:

ro,G =

For the point mass: so 0,

L.

a —

L.

=—

/ L.

a

+

—

+

a

2L.

=—

2a

+

[I]

that for the bar we have = L/lO, —a/lO. and for the point mass = 0, 0, = = —2L/5, = 2a/5. Introducing these values into the parallel axis theorem, we obtain [I'Gl =

[U

+

in which

a2+L2 0

=

0 0

0 2a2 0

0

a2+L2

L2+a2

0

0

0 0

a2

aL

aL

L2

+

Iki 1k]

and

L2+a2

(±)

[1Grnuj = [0] +

0

0

()

[11

aL

0

L-

To find the moment of inertia about point A, we calculate the distances between points

GandA

as

rAIG

— rG,o

=

a.

i+

L.

a —

k

f —

L, +

a

k) =

a.

11 +

3L. 3u jj — —

so that so

a

=-

=

3L

3a

=

—s--

[a]

—

The inertia matrix about point A becomes

=

± 11(71

4

9(L + a-)

—3aL

25

10

10

—3aL

61a2

9aL

10

1(X)

25

9aL 10

3a

a-

9L

I

25

Adding Eq. [oj to Eqs. [kJ and [I] gives Eq. [F]. Note that this approach is much lengthier than when we dealt with the individual components directly. On the other hand, it gives us

the location of the center of mass as well as the centroidal inertia matrix of the composite body.

Exomplo

6.5

I

Given the mass moments of inertia of a body, find the resulting moments and products of inertia when the coordinate system is rotated about the z axis by an angle of 0.

6.4

PR0I'ERTIEs

approach this problem in in two two ways. The first is to find the direction cosine matrix the two coordinate coordinate systems systems and andto tothen thenUSC use Eq. Eq.[6.4.131. [6.4.131. The The second second is to take adthe of the simplicity of the coordinatc transformation transtbrmation and to obtain the moments and of inertia by direct substitution. We will show the second approach first. ol axes .vyc with the theprimed primed axes axes obtained obtained by by rotating Consider the two sets ol .vy: and x'v'z'. x'v'z'. with angle H 9 about the axes by an angle two sets of axes are the zz axis. The two arc related by .v'

cos 0 + + vsinll V Sln II = xxcosO

.v sin 0 V == ——xsin0 V

cos H + vycosH

=

[a]

z

Substitution of Eqs. [a] into the moment of inertia expressions about the primed axes + +

(v

=

I

-,

:)dni ==

..

+ (x sin 60 +

[(—.vsinO sinO++vcosO) vcosO) ++ z]dni zjdni [(—x

v cos 0

..

[b]

sin 0Hcos cos66 ++ z) z) din 2.vv sin 2xv

—

J

_2

cos2 0. which. when substituted into Eq. [hi. yields

as z2 sin2 0 +

= sin2

J(x2 1(x2 + z2)dm + cos2 H f(v2 ± -2)(Inz

U

—

xvdm

2sin0cos0 2sin0cosO J

=

Sifl 0 ± Sin

COS00 — c0s — 21, sin Ucos 0

Ic]

Eq. [cJ further, we make use of the trigonometric identities

siir 00 = S1fl

— COS

20

cos

-

0=

1 ++ COS Cos 20

[dl

we get qibstituting Eqs. [dl [di into Eq. [ci we =

/

±

y"

cos26 cos20

—

sin 26)

1.1

are the familiar relations from the transformation of coordinates. Tn a similar way. we rotation of of the the axes axes isis about aboutthe thezzaxis, axis,the themass mass moment moment olinertia olinertia Because the rotation the z' axis is the same as the moment of inertia about the z axis. we obtain The products of inertia are found the same way. For 0)(—x sin sin00++vvcos cos 0) din (x COS U + v sin H)(—x

dm f .v'v' = =

=

=

[(v2

—

0 COS 6— — sin2 0)1 sin H COS 00 ++ .vv(cos2 xv(cos2 H 0)] dIn dm

sin 20

+

The remaining products of inertia. 1%...f =

and +

can

En

he shown to become =

+

[gJ

and by using can also solve this problem by generating the tile direction cosine matrix [cj and [6.4.13a]. The direction cosine matrix (cJ has the form We

341

_____—

342

CHAPTIR 6• RIGID BODY GEOMETRY

—sinO cosO 0

cosO

[c] = [R]" =

sinO 0

0 0

[II)

1

[6.4.1 3a], yield the inertia terms derived above. which, when substituted into are both zero. Given the entries of the inertia and Suppose now that we are told that matrix associated with the transformed system. we can calculate the rotation angle that will equal to zero and solving for 0. Setting Eq. [f 1 equal yield the principal axes by setting to zero yields tan2O

111

—

ly' — —

which is the same result obtained from a Mohr's circle analysis (of stress or area moments el inertia) for a two-dimensional system.

Example

6.6

I, cc = 3. 1, a a = = 2, b = 1, Consider the right triangular prism in Example 6.3. Given that m = 1, find the elements of the inertia matrix associated with a set of coordinates obtained by rotating find the .v' x' axis axis is parallel with the incline. the xvz axes about the z axis such that the

Solution The configuration is shown in Fig. 6.13. The rotation is clockwise with rotation angle 0 = tan '(b/a) = tan '(1/2) = 26.57°. From Example 6.3. the inertia matrix has the form —ab/2

111=

—ac

—ab/2

—ac

a-+2c + 2c2

—1w —bc

a2

=

a+b a2+b22

—bc

19

—1

—6

—1

22

—3

— —6

—3

5

1.]

The rotation matrix between the xyz and x'v'z' coordinates is 0.8944 [RJ = 0.4472 [RJ 0A472 0

—0.4472 0.8944 0. 8944 0

0 0

[hI

1

From Eq. [6.4.13bj, the inertia matrix associated with the transformed coordinates is [R][fl[RIT = [I'] = [R][fl[RIT [1']

20.40

—1.800

—4.025

—1.8(X)

20.60 —5.367

—5.367

1

—

6

—4.025

5.000

V

V

h

9

a=2b

x

Figure 6.1 3

Side view of prism

[ci

6.5

PRINCIPAL

OF

As can be seen

is the same as I-,. by virtue of the flict that the coordinate transfor::iation is a rotation about the z axis. 1-lowever. the products products of of inertia inertia involving involving the thez.z axis 'v7) do change. because of the change in the .v and y coordinates.

6.5

PRINCIPAL MOMENTS OF INERTIA

will see in Chapter 8, it is desirable to work with an inertia matrix that is diagonal. The equations for rotational motion become substantially simpler. When the coordinate axes x, y, and z are chosen such that the products of inertia vanish. the x, V. and z axes are called the principal axes, and the moments of inertia y, and md are called the principal moments qf of inertia. There are two ways of finding the principal moments of inertia. The first is by visual inspection. For example, if the coordinate axes are selected such that they lie a plane of symmetry. then at least two of the products of inertia vanish. If one of As we

the coordinate axes is an axis of symmetry. symmetry, all products of' inertia vanish. The second method of finding principal moments of inertia is by solving the eigenvalue problem associated with the inertia matrix. Since II] is symmetric and definite, all of its eigenvalues are real and positive. Its eigenvectors are real. as well.

Examining Eq. [6.4.13a1 and considering the case where [11 is a diagonal mathx, we observe that the transformation equation for the inertia matrix represents in orthogonal transformation. Because the direction cosine matrix is orthogonal. Eq. [6.4.1 3a} can be written as

["I

felT1! fC] T11 JfcJ

[cJ'[c] = [1]

[6.5.11

These equations are essentially the mathematical description of the eigen value

coblem associated with the matrix [11. [1]. The diagonal matrix [1'] contains the eigenin its diagonal elements, and [c] [cj is the normalized eigenvector matrix with the as the columns of [c]. To verify this, consider the eigenvalue problem with [I]. [I],

[6.5.21

[I]4u} = A{u}

A denotes the cigenvalues and {u} the eigenvectors. Equation 16.5.21 can also expressed as ([1] — A[ I J){u} = {O}. A solution to the above equation exists if — A[l]) is singular, hence det([/1

—

[6.5.3] 16.5.3]

A[1J) = 0

This is called the characteristic characteristic equation, equation, and and itit is is a third-order polynomial in A.

S.lution of' the characteristic equation yields three eigenvalues A1 (i = 1, 2, 3) and eigenvectors eigenvectors{u1}. {u1}.The The eigenvectors eigenvectors are are orthogonal orthogonal to each other, such that = 0

{u1}TlIl{u1} = 0

i

j: i, j = 1, 2, 3

16.5.4] 16.5.41

.rnd. as we saw in Chapter 5. they can be normalized to yield {u1}T{u1} = 1 (i = :. 2. 3). It follows that = A1. A,. We then form the eigenvector matrix = [{u1} {u2} {u3}1. The eigenvector matrix is an orthogonal matrix, such that

343

344

Boiv GEOMETRY GEOMETRY CHAPTER 66 • R1GIU Boiw CHAPTER

[U]'LUJ = I I. The orthogonality relations with respect to the inertia matrix can written in compact form as I

ILJIrI/IIU1 = [A] IUInI/IIUl [A] = [I']

IuITIL/1 =

[11

16.5.51

is a diagonal matrix containing the cigenvalues. which which are are the the principal principal moments of inertia. Eq. 16.5.5 I can be inverted to yield in which

[UIIAflL/JT

[6.5.6]

I/l

comparingEq. Eq.16.5.51 16.5.51and andEq. Eq.[6.5. [6.5.lJ We conclude by comparing 1 1that thatthe the direction direction cosine We

matrix [ci that leads to the principal axes is the same as the eigenvector niatrix elgenvector matrix [Li obtained by solving the eigenvalue problem associated with the inertia matrix. The = eigenvalues of [I] are the principal moments of inertia. The eigenvectors 1, 2. 3) are the direction cosines associated with the principal directions. Care must he taken when relating the eigenvector matrix to the direction cosine [cj matrix fri that leads to the principal axes. In Chaptei Chapter we learned that for [c] describe a proper rotation. its determinant must be equal to I The detenmnant of the normalized eigcnvector normalized eigcnvector matrix matrixisis±±1 with the choice of plus or minus depending so with is an eigenvector associated with is on the analyst. This is because if For [U] to represent a proper transformation matrix. the eigenvectors should be normalized such that det [Ui = I. is the the ordering ordering of of the the eigenvalues eigen valuesand andeigcn eigcnvectors. vectors.Lr.i In Another issue to consider is general, in eigenvalue problems associated with real symmetric matrices, one writes the cigenvalues in ascending (or descending) order and constructs the cigenvector matrix accordingly. This is primarily done for convenience, as we did in the vibration problems in Chapter 5. In vibration problems. the lower frequencies arc usually of more interest than the higher frequencies. When dealing with inertia matrices there is no such iieed. need. One can select the order of eigenveclors entirely arbitrarily. One guideline for selection is to look at the magnitudes of' the diagonal elements of .

1

and then to order the cigen values such that they are similar to the order of the diagonal elements of [I]. When the products of inertia are small quantities. such an approacti usually leads to the smallest rotation angles between the original coordinates and the principal axes.

An interesting special case in eigenvalue analysis is that of systems possessing repeated eigenvalues. In mathematics, mathematics. a matrix that has repeated is referred to as degenerate. Inertia matrices having repeated eigenvalues have an elegant physical interpretation in dynamics. = A1. with Consider a symmetric matrix that has two repeated eigenvalues Because the eigenvalues eigenvalues are are repeated.. repeated. and corresponding cigenvectors Any linear combination and there is a certain amount of arbitrariness in {u1} + these two cigenvectors cigenvectorsisisalso alsoan aneigen eigenvector vectorbelonging belongingtotoA1. A1.that thatis.is,,Bi {u1} and P2. {u,}. P2. There There isisno nounique uniqueeigenvector eigenvector{u1} or {u1}. is an cigenvector for any

usually selects these two eigenvectors so that {u1}'{u1} = 0. Consider next a rigid body that has two of its principal moments of inertia mosz Themosz and12. 12. Such a body is referred to as inertia//v symmetric. The same, say Ii and encounteredisisin inaxisvmnmnetric axisvmnmnetric bodies. bodies. such situation is common case when such a situation is encountered common Mathematically. inertial symmetry as disks, circular rods. or an American football. Mathematically,

6.5

345

PRINCIPAL MOMENTS OF INERTIA

exist for bodies that have no physical symmetry properties. Using a set of axes b1 b2b3 attached to the body. the symmetry axis is the b3 axis. The b1b2 plane is refened to as the principal plane. If we take a rotation of the coordinate axes about axes also also constitute constituteaaset seto1 of we see that the rotated coordinate axes the symmetry axis principal axes. There is no unique way of defining the principal axes, the same as with the eigenvcctors associated with repeated eigenvalues. We shall see in Chapter 8 that bodies possessing inertial symmetry have better stability properties than arbitrary bodies. Bodies that spin are usually designed to be axisymmetric, with the spin axis and the symmetry axis coinciding. can

Example

Given a rigid body with the inertia matrix 400

0

111=

—125

6.7

—125

0

350 0

kg•rn2 0 100

the principal moments of inertia and the transformation (or direction cosine) matrix that •iiagonalizes (.1].

S.Iution = 0, in this problem. leading to the = 0 and of the products of inertia are zero. onclusion that we are probably dealing with a body which has symmetry about the xz plane. is defined defined in in Eq. in Fig. Fig. 6.14. 6.14.The The eigenvalue eigenvalue problem problem is Eq. 16.5.31 16.5.31 as as is depicted in (111 —

lal

A[l]){u} = {0}

requires that that the the determinant determinantdet(1i] det(1i]— —

Af 11) = 0, or A[

400—A

0

—125

0

350—A

0

—125

0

100—A

det

=0

[b]

to the characteristic equation 400— A)(35()

—

A)(l00

—

A)

—

(350— A)1252

= (350— A)[(400

—

A)(lOO

—

A)

—

15,625] = 0

Ed

solution of which yields the cigenvalues. which are the principal moments of inertia, as kg•m2 445.26 kg•m2 11 = 445.26 A: == 11

A2

= '2 = 350 kg•m2

A3 =

13

= 54.74 kg• kg. m2

Id]

V

FIgure 6.14

A vehicle that is symmetric about the xz axes

346

CHAPTER

6 • RIGID Boiv

GEOMEIRY GEOMETRY

Note that we ordered the eigenvalues such that they follow the order of the diagona elements of [I], which in this case happens to be in descending order. To find the correin conjunction with the eigenvalues. Using the sponding eigenvectors. we use Eq. u31]T. we have for the first eigenvalue A1 = 445.26kg•m2 u31]T,wehaveforthefirsteigenvalueAi 445.26 kg. m2 {u1} = [u11 [u11 U21 0 0 —

1

35()

—125

0

25

1 00

—

[.J

= {0}

0

— A1

A

which leads to the simultaneous equations —45.26u11

—

125u31

= 0

U21 = 0

—

[9

= 0

—

As discussed in Chapter 5. only two of the equations are independent, so that the multiplicative constant. constant. Eqs. Ff1 yield can be determined to to aa multiplicative =

where

dill

0

_0•36211T

Iii

is a normalization parameter. parameter. We We normalize normalize the the eigenvectors eigenvectorssuch suchthat that{u1 {u1}'{u1 } } =

= d?(l d?(1 + 0.36212) =

[hI

±0.9403. When selecting the sign of d1(i = 1, 2, 3) we must so that the value of d1 into consideration that the cigenvector matrix [UI = Li{ui} {u3}1 must have a LI{ui} {U2} {u3}1

minantequalto 1. Actually, one can The second eigenvalue is 350 kg• kg. m2, m2, unchanged unchanged from from 122. 122. Actually, ± [0 The second second eigenvector eigenvector is is {u2} {u2} = ±[0 this from the symmetry about the xz plane. The 01'. This 01". This choice choice implies implies that that the the y axis is not The logical choice choice isis {u2} {u2} = [0 [0 101T (flT would have resulted in a 180° = [0 — — after the rotation, whereas selecting The v axis is a principal axis. It follows that the coordinate transformation that leads to principal axes is a rotation about the v axis, also known as a 2 rotation. principal Using the same same procedure procedureas asthe theone onefor forthe thefirst firsteigenvalue. eigenvalue.the thethird thirdnormalized normalizedeigec— vector is found to he 1

1

{U3}

±O.9403[0.3621 ± 0.9403 [0.3621

0

1]' = ±10.3404 ±10. 3404

0

094031T 0.94031

in

We We

construct the eigenvector matrix keeping in mind that its determinant has to equal to 1. Considering what a rotation about the vy axis looks like, we write the matrix as 0

0.3404

0

1

0

—0.3404

0

0.9403

0.9403 [U] = [ci = [Ri' = [Ui

EU

Comparing [U] with the transformation matrix for a 2 rotation, the cosine of the angle is 0.9403 and its sine is 0.3404. The rotation is counterclockwise and the rotation turns out to he 19.90°. 19.90°. The The rotation rotationisisshown shownininFig. Fig.6.6.15. 15. the sequence sequence of of rotations that leads to the principa Had we selected {u2} as [0 — I 01'. the coordinates would not he the single clockwise rotation about the y axis, but a more compi— cated one. Also, when the eigenvector matrix [U] is fully populated there is no set way finding the transformation to obtain the principal coordinates. This further demonstrates tha there is no unique way of transforming one coordinate system into another.

6.6

INERTIA ELLIPSOID

l9.90z

-V -V

19.9v xl

Principal axes

Figure 6.15

6.6

Principal axes of the vehicle

INERTIA ELI4IPs0ID

The eigenvalue problem associated with the inertia matrix is given in Eq. [6.5.21. {u}T gives Left-multiplying it by {u}" {u}T[1]{u} = A{u}T{u} 16.6.11 If we use a coordinate system xvz and the expansion {u} = [x can be expressed as

+

+ 1--z2 — —

= A(x2 +

—

—

Eq. 1.6.6.11

v

\?22

+ z2) z2)

16.6.2] [6.6.21

This equation represents the intersection of two closed quadratic surfaces, the

an ellipsoid and the right a sphere. This is because the expressions on either side of the equation are positive definite. The ellipsoid on the left side is called the inertia ellipsoid. The inertia ellipsoid is fixed to the body and is dependent on the point about which the inertia matrix is considered. Let us next normalize the vector {u} such that {14}T{u}

x2 + x2

+

=

[6.6.3]

Now we have reduced our points of' interest to the locus of all points that define a

sphere of unit radius. As we vary the parameter A. the shape of the ellipsoid does change, but its size varies. For each value of A. the ellipsoid and sphere intersect it different points. Without loss of' generality, ordering the principal moments of 13. one can show that the values of A are bounded by the nertia such that 'i principal moments of inertia = It

Amax =

13

16.6.41

In this this case. case. the the longest The smallest ellipsoid of inertia is obtained when A = 11. In

axis of the ellipsoid has a length of unity and lies tangent to the sphere. and the entire ellipsoid lies inside the unit sphere. The largest ellipsoid is obtained when = 13, when the shortest axis of the ellipsoid is of length unity and is tangent to the unit sphere, with all of the sphere lying inside the ellipsoid. When A = '2. the intermediate axis of the ellipsoid becomes tangent to the sphere. with the smallest axis of the ellipsoid lying entirely inside the sphere and the largest axis extending

347

348

CHAPTER

6•

RUIID Bony GEOMETRY

A

nun

Unit circle

Figure 6.16

Inertia ellipsoid in two dimensions

outside the sphere. Fig. 6.16 illustrates this concept in two dimensions. We refer to

the longest. shortest, and intermediate axes of the ellipsoid as the major axes. The major axes of' the ellipsoid correspond to the eigenvectors of the inertia matrix. One can ascertain this from the discussion above, as well as by writing Eq. 16.6.21 when the coordinate axes are selected as the principal axes. Denoting the coordinates associated with the principal axes as u1. and u3, the inertia ellipsoid equation reduces to

U ++/2 11111 '2 I

+

= AA

13

16.6.5] 16.6.51

The sphere equation is simply + + = I. Now examine the relationship between the lengths of' the axes of' of the the ellipsoid, ellipsoid, which which are the principal axes, and the principal moments of inertia. At the tip of the ellipsoid along each principal direction, only one of has a value of unity, and the rest are zero. This u2 or confIrms that the axes of the ellipsoid are the principal axes. The normalization in Eq. 16.6.3] and the associated inertia ellipsoid equation are not unique. unique. The normalization we used gives a good physical explanation of the inertia properties. but it has the disadvantage of of dealing dealing with ellipsoids of different sizes. An alternate normalization is to use ,k{u}'{u} = A(x A(x++)?2 + z2) z2) =

16.6.61

which leads to the ellipsoid equation written +

+

—

—

21,-xz 2/,-xz

—

21'vzYZ 21'yzYZ

=

I

16.6.71

or. in terms of' of principal coordinates 11u1 +

I

16.6.81

In this case, as shown in Fig. 6.17, the size of the ellipsoid is fixed but the sphere

with which the ellipsoid intersects is no longer of unit radius, but of radius The length of the axes of this ellipsoid is equal to (i 1, 2, 3), so that the axis about which the moment of inertia is the largest will have the smallest axis in the inertia ellipsoid, and vice versa. versa. Equation Equation 16.6.7] 16.6.7] is is commonly used when relating the inertia ellipsoid and the kinetic energy of' a rigid body.

6.6

349

INERTIA ELLipsoiD

Largest_sphere 'mm

Smallest sphere

R=

FIgure 6.17

Inertia ellipsoid of fixed size

the inertia ellipsoid for the rectangular prism shown in Fig. 6. 1 8 about the center mass, and about point A.

S.Ivf Ion the xvz

axes

attached to the center of mass. the inertia matrix is diagonal with ,fl

)

(.'Gi = —diag(b2 + c2

b) a- ++ b2)

12

Point A and the center of mass are separated by ±e inertia matrix about point A becomes

['Al =

/n

ICa2

{JGI +

Ha c

0 + (.2 0

(1 C

0 -I

m

lal

(1

=

irT)

a

+ h4.C

0

12

3ac

(I —

= O,d- = —e/2. So (hat

3ac

0 I 4a + 0

0

4a + b-

V

I,

V

b

'I

x

A

C

a

Figure 6.18

I

[b]

Example

6.8

6• RIGID

350

BoDY BODY

coordinates. an x"v"z" coordinate system attached to point A and parallel to the .vvz coordinates.. Using an

the inertia ellipsoid equation can he written as V"2 + 14 V"2

1

s.,.

V

+ 14.t1:hI

.v"z" .v":" =

—

[ci [cJ

1

The inertia ellipsoid about point G is relatively simple to draw. The ellipsoid has .v, v, v, and and z axcs. axcs. as as these these axes axes are are principal axes. The largest axis of the major axes about the .v, ellipsoid corresponds to the smallest moment of inertia. The inertia ellipsoid associated with point A is harder to draw. as two of the axes of the ellipsoid no longer longer lie lie parallel parallel to to the the.v", x". v". or z" axes. Only one of its axes is parallel to the v" axis. Looking at Eq. [hJ and comparing it with the results of Example 6.7, we conclude thai

xvz axes axes about about the the v axis. The rotatio the principal axes are obtained by a rotation of the xvz. angle can he found by solving the associated cigenvalue problem. Note. also that the inertia matrix is recognized as one obtained by rotating a set of principal axes about the component of ofthe theinertia inertia ellipsoid ellipsoid is not going to yield any significant v axis. the v coiTipoflent Because of this. and because visualization in two dimensions is simpler. we plot and comtbr the value v = 0. Doing so. pare the inertia ellipsoids associated with points G and A for and assigning the parameters rn = 1.1, a = 1,I, h = 2, c = 3. we obtain the following ellipsoid equations:

For point point G G (with with v = + 5.2) =

Ed]

I

For point A (with v" = 0): !

(40 VH +

—

1

8.v":") = 8v":"

1.1

1

Figures 6.19 and 6.20 show the ellipsoids associated with Eqs. [dl and [el.In Fig. 6.2( = the points in which the ellipsoid intersect the .v and : axes are at x = ± and z = ±\/12/5 = 1.549. To find where the ellipsoid associated with Fig. 6.20 intersecb the principal axes, we can carry out a graphical analysis and visually find these points. 0r we solve the eigenvalue problem associated with the inertia matrix and determine the location of the principal axes. Solving the eigenvalue problem gives the cigenvalues as and 1I .283. .283. Note Note again again that that we we have have ordered the eigenvectors similar to the diagonal elements

1.5

Principal Pri tici pal axes

—1

1

—1.5

Figure 6.19

Inertia ellipsoid for [IG]

FIgure 6.20

Inertia ellipsoid for [/4

HOMEWORK EXERCISES

of l'Al• The normalized eigen vector matrix is written [Ui — [RIT —

—

10.95 JO 10.9510

—0.3092

[0.3092

0.9510

Recalling that we are dealing with a rotation about the y axis, we conclude that the rota1(0.9510) = from the x'y'z' axes to the principal axes is a clockwise rotation of 0 = IS.0l°. One can verify the rotation angle and the lengths of the major axes of the ellipse from IS.Ol°. Fig. 6.20. A clockwise rotation of 7 1.99° also leads to a set of principal axes.

REFERENCES 1988. Row, 1988. New York: York: Harper & Row, Engineering Dvnwnics. New Ginsberg. J. H. Cliffs, NJ: Prentice-Hall, 1988. ed. Englewood Englewood Cliffs. Gieenwood, D. 1. Principles Principles of oJDvnamws. Dynamics. 2nd ed. Theoryand andApplications. Applications.New NewYork: York: McGraw—Hill, McGraw—Hill, 1985. 1985. Kane. T. T. R.. R.. and and D. D. A. A. Levinson. Levinson.Dvnamic1c: Dynamics: Theory Kane. Meirovitch, L. Methods of Analytical Dynamics. Ncw York: McGraw-Hill. McGraw-Hill, 1970.

HOMEWORK EXERCISES SECTIoN 6.2 I.

Find the center of mass of the body shown in Fig. 6.21.

SECTION 6.3 2.

Find the elements of the inertia matrix associated with the figure shown in Fig. 6.22 about point 0 and using the xvz axes.

V

X

Hole Hole diamctcr = (1

5(1

Figure 6.21

0

a

351

352

CHAPTUR 6. RIG II) BODY GEOMEtRY

V

V

y=x 4

I

4

6 z

I

I

FIgure 6.22

Figure 6.23 3.

4. 5.

6.

Calculate the mass moment of inertia about the x, v and z axes of the body of revolution generated by rotating about the y axis the area between the curves y = x2/4andy = xqwhereO x 2,0 v 1, as shown inFig. 6.23. Find by direct integration the inertia matrix associated with the paraboloid it Example 6.2. Use a differential element in the form dm = pr dr dO dx. Calculate the mass moment of inertia of a torus of mass ni, mean radius R. core radius of the inner cross section a about the generating axis (Fig. 6.24). Find the mass moment of inertia of the body shown in Fig. 6.25 about the x. and z axes.

SECTION 6.4 7. 8.

Consider the composite shape in Fig. 6.21, without the hole. Determine the inertia matrix about point 0. Determine I and of the square plate shown in Fig. 6.26 with three holes cut in it. V

R(x) =

R0—

bR0 x

a

4R0

J

FIgure 6.24

FIgure 6.25

353

HOMEWORK EXERCISES

V

V

I

x

d2

a

p

—x

.1•

/ x,I

Li

Figure 6.27

FIgure 6.26 9.

Consider the rectangular box in Fig. 6.27 and determine its inertia matrix about

point 0. point 0. using usingthe thex'y'z' x''z' axes. The x' axis goes through points 0 and D. The x'v'z' axesare areobtained obtainedby byrotating rotating the the .vvz .vvz axes axes about the v axis by 01. and then x'v'z'axes axis by rotating the resulting frame about the z" axis by 62. 10. Consider the right circular cone in Fig. 6.28. Find the elements of the inertia matrix about point 0 and a coordinate system x'v'z' obtained by rotating the xyz coordinates about the z axis, so that the x' axis goes through point B. Let RIL = 0.2041. 11. Calculate the inertia matrix of the body in Fig. 6.29 about point 0. using the .v. v, and z axes. The body is composed of thin slender rods.

V

?fl, L

'I

B

in. L

0 x

in.'. I-.

"I!. "I.,!.

A.

FIgure 6.28

FIgure 6.29

354

CHAPTER

61

RIGID

Boiv

xl

4

V

2R V

4

R

t

-4

FIgure 6.30

Figure 6.31 Using the x'v'z' axes, fInd the mass moments of inertia of the triangular prism in Fig. 6.30 about 0. The x' axis goes through points 0 and B and it is obtained by a counterclockwise rotation about the z axis. 13. Find the inertia matrix of' the composite body shown in Figure 6.31 about point 0 and using the .vvz axes. The body consists of a thin triangular plate attached 12.

to a solid shaft. SECTIONS 6.5 AND 14.

6.6

Find the principal moments of' of inertia inertia of' of' aa body body whose inertia matrix is given below for a = 7. Then, suggest a set of rotations that will transfbrm the inertial into the principal axes. Next., find the principal moments of inertia and axes into axes transformation matrix for a = 0 and compare the results. 450 [11

0 100

0 500

100 a

a

550

the principal principalmoments moments of' of inertia of of the box in Fig. 6.27 about 15. Find the about point point 0. Sketch the principal principal axes. axes. the principal principal moments 16. Find the moments of of inertia inertia of of the the shape shape in in Problem 7. 17. Find 17. Find the the principal moments of inertia of the composite body shown shown in in FigFigure 6.31 6.31 about aboutpoint pointC). 0. Sketch Sketch the the inertia inertia ellipsoid. ellipsoid. 18. Find the principal moments of inertia of the body in Fig. 6.30. about point 18.

Sketch the inertia ellipsoid.

chapter

RIGID BODY KINEMATICS

7.1

iNTRODUCTION

In this chapter we consider kinematical relations that describe the motion of rigid The analysis follows the developments of reference frames and relative mofl-on equations in Chapter 2. To extend these concepts to rigid bodies, we attach movng reference frames to the bodies and express the angular motion components using Using ng the moving frames. We also quantify the angular velocity by means of Euler angles.

well as Euler parameters, the latter being quantities that eliminate the singulariassociated with the Euler angles. We then discuss commonly encountered conwhen rigid bodies are interconnected or when they move against each other. The chapter ends with a discussion of rolling. The field of kinematics can be viewed as consisting of two major components: inalysis and synthesis. The focus of this chapter is kinematic analysis. Kinematic is usually needed when designing interconnected bodies and mechanisms. is mostly a specialty field and is beyond the scope of this text.

7.2

BASIC BASIC KINEMATICS KINEMATICS OF OF RIGifi RIGifi BODIES

We can classify the general motion of a rigid body into three categories:

• • •

Pure translation Pure rotation

Combined translation and rotation

To describe the kinematics, we make use of the relative motion equations dein Chapter 2. The velocity of a point P. whose motion is observed from

355

356

CHAPTER 7 S RIGID B0D\' KINEMATICS

a rotating coordinate system with origin at B. is =

VH + (0 X rp18 + VrcI

+

17.2.11

is the velocity ofthe ongin ofthe relèrcnce frame, w is the angular of the reference frame. and and VreI VreI iS the the velocity velocity of point P as observed from the moving reference frame. The expression for the acceleration of P is where

a/) = aD a/) aD +

X rp/B + (J)

><

(o. X rp,B) +

X Vrel Vrel +

[7.2.21

with the terms having their obvious meaning. These two equations have a significant signilicant application Ihr rigid bodies. The moving reference frame can be attached to a point on the body and it moves with the body the or the In this configuration. the relative axes are referred to as the bod -fixed axes, or as the the center center of of mass body axes. The origin of' the reference frame is usually selected as or. if it exists, the center at rotation. of the the body body are are then the angula The angular velocity and angular acceleration of' velocity and angular acceleration, respectively, of the reference frame and VreI anc arel become the velocity and acceleration of point P with respect to the body. If P = 0. = 0 and a point fixed on the rigid body. then In the majority of dynamics problems involving three-dimensional motion, one attaches the relative axes to the body. We will later see a special case of employing relative frame riot not attached to the body. The study of axisymmetric bodies makes use of this special case.

7.2.1

TRANSLATION

in this case, the rigid body moves with no angular velocity and no angular acc&eration, that is. o = 0. = 0. Every point on the body has the same velocity and acceleration, so that three translational parameters are sufficient to describe the motion. For the most general case we have three degrees of freedom. 1: should be noted that a rigid body is capable of moving along a curved trajectory without any rigid body rotation. An example of this is landing of an aircraft.

7.2.2

PURE ROTATION

Here, the motion of the rigid body is described using rotational parameters alone The velocity and acceleration acceleration of' of any point on the body can he expressed in of the angular velocity, angular acceleration, and the distance of the point from rotation center. We separate this type of motion into two categories:

•

Rotation about a fixed axis.

•

Rotation about a fixed point.

Rotation about a fixed axis is a special case of rotation about a fixed point. For twc:dimensional motion, the above two categories coincide.

Rotation About a Fixed Axis

Consider a body rotating about a fixed axis k

The direction of the angular velocity vector w is along the fixed axis, as shown LI

7.2

BASIC KINEMATICS OF RIGiD BODIES

Fig. 7.1. According to the definition in Chapter 2, the angular velocity is "simple." Denoting by e1, the unit vector along the fixed axis. we can write (*3 = = (0

a=

we;,

17.2.31

= we1, = ae1,

The unit vector e1, is similar to the binormal vector we considered in Chapter

1

in conjunction with normal and tangential coordinates, the difference here being that the direction of is fixed. Consider a point P on the body and express its position in tei-ms of its components along the h axis and the plane perpendicular to the h axis. We write rp as

[7.2.41

rp = h + b in

which which hh =

he1,

and b =

where b is the perpendicular distance from point is the associated unit vector. We recognize that is

P to the axis of rotation and the normal direction. When P is fixed on the body, its velocity is Vp

(he,1 X rp = we1, X (h + b) = we1, X (he,, X

=

—

be,,)

= wbe, wbe [7.2.51 [7.2.51

The magnitude of the velocity is bw, leading to the conclusion that the velocity

of a point on the body is dependent only on the perpendicular distance between that and the axis of rotation. Rotation about a fixed axis is a single degree of freedom 'roblem. Eq. [7.2.51 [7.2.5] and write the acceleration of point P as We differentiate Eq.

X rp) = a X rp + w X (w X rp)

ap =

[7.2.6]

which is recognized as the sum of the tangential plus normal components. We have

[7.2.7] [7.2.71

ap = a1+a,, where

a,,=oX(wXrp)

[7.2.8]

the acceleration is at = ab, and of tangential component component of the The magnitude of the tangential = w2b. The components of the acceleration can be the normal component it is I

Vp

P

P I,

Figure 7.2 (1)

FIgure 7.1

Rotation about a fixed axis

Normal and tangential coord i..

'1

h is out of paper

notes

357

358

CHAPTER 7•

RIGID BODY KINEMATICS

expressed as

= bae,

Rotation About a Fixed Point

[7.2.9J

= bwe,1

a,1 =

In this case, the angular velocity vector w

does not lie on a fixed axis. The rate of change of the angular velocity depends on change in direction as well as a change in magnitude. A body rotating about a point has three degrees of freedom, and the angular velocity is usually a combinatioE of two or more rotation components. Consider the cylinder in Fig. 7.3. The shaft rotating about the fixed Z axis with angular velocity w The cylinder is spinning with angular velocity W3 about an axis h. which lies on on the the yz plane and makes angle of 13 with the rod. The angular velocity of the cylinder can be expressed as (0 = (01 +

= w1K +

+ (03 = w1K + f3i f3i + w3h /31

[7.2.1 OJ

+ w3(cos/3k — sin/3j)

In general. w3 w3 is is known known as as the thespin spinrate rate and and w1 w1 as as the the precession precession rate. The angk (3is referred to as the nutation angle. The rate of change of /3 is called the rate. We will formally quantify these rotational parameters in the next two

Even if the spin rate and precession rate are constant—that is, the of the angular velocity are constant in magnitude—the angular acceleration of the body is not zero, because the direction of the angular velocity vector is changing indeed, applying the transport theorem we obtain

a =w =

w1

+.02+w3+(o1 X(o2+w3)+w2Xo3

[7.2.11)

The last two terms in this equation are also known as gyroscopic effects. Figure 7.4 The terms terms ój + describe the change in the magnithis effect effectfor foro2 o2 = 0. The shows this describes the change in directioa. tude of the angular velocity vector, and (Oi X

(0

Z.z

0)1

Z1z

/1

(03

0 V

(01X(03

x x

Figure 7.3

Rotation about a Fixed point

FIgure 7.4

Angular velocities and gyroscopic effect

7.2

BAsic KINEMATICS OF RIG1I) BoDws

The line specifying the direction of the angular velocity vector w is known as

the instantaneous axis of acceleration, or the instantaneous axis of rotation. mtation. The unit vector along this axis is defined as 0)

[7.2.121

U)

which w = is the magnitude of the angular velocity. The rigid body can he viewed as rotating about the axis defined by w(t) at a particular instant. The trajectory of the instantaneous axis of rotation defines the body and space

cones. Body and space cones are helpful in visualizing the motion of rigid bodies. When the trajectory of the axis of rotation is viewed from the rotating body (the bodyaxes), the cone that is generated by the angular velocity vector is the body cone. When the trajectory of the axis of rotation is viewed from an inertial frame, the space cone is generated. The body and space cones are always in contact with each other. The line of contact is the angular velocity vector w. which is also referred to as the &neratrix. Figure shows body and space cones for an arbitrary body. We. will ;tudy body and space cones for axisymmetric bodies in Chapter 10.

7.2.3

COMBINED TRANSLATION AND ROTATION

A body undergoing combined translation and rotation requires both translational and angular parameters to describe its motion. The unrestricted three-dimensional moof a rigid body is a six degree of freedom problem. Combined translation and rotation of an arbitrary rigid body is too general to be in broad terms. Chapter 3 presents the special case of plane motion. An nteresting concept associated with plane motion is the existence of an instant center. Once the instant center is located, the velocity of a point P on the body can be found from the relation Vp =

wXq

[7.2.131

q = rpic. and C is the instant center. Space cone

0

Figure 7.5

Body and space cones for an arbitrary body

359

360

CHAPTER 7• RIGIt)

BODY

The situation is quite different for three-dimensional motion. One can prove existence of an instantaneous axis of rotation, as we will do in Section 7.3. ever, barring a few exceptions. a center of instantaneous velocity does not exist three-dimensional motion. In three dimensions, Eq. [7.2. 131 has an infinite Eq.[7.2. [7.2.131 131 can can of solutions. Using Using the the column column vector vector notation notationfrom fromChapter Chapter2, Eq. written as

[7.2.141

{Vp} = [thJ{q}

[7.2.13J has matrix [th I is a singular matrix of rank 2, another proof that Eq. [7.2.131 The matrix infinite in finite number number of solutions.

7.3

EIJLER'S AN]) CHASLES'S CHASLES'S THEOREMS

A rigid body has three rotational degrees of freedom. From this, and from the opments of Chapter 2, we deduce that three rotations (about nonparallel axes) be suffIcient to describe the most general transformation from one coordinate to another. According to Euler's theorem, one can view a transformation from coordinate system into another as a single rotation about a certain axis. Euler's theorem states that:

The most general displacement of a rigid body with one point fixed can be described as a single rotation about some axis through that fixed point, called the axiS axis of of rotation or principal line.

Proof of' Euler's theorem can be carrier The rotation angle is denoted by out by an eigenvalue analysis. Consider a coordinate system A with with coordinate coordinate rotations, resulting in the frame B with the a sequence of rotations. b1b2b3.For b1b2b3. For convenience. convenience, we we use use the the same same origin origin for for both both coordinate coordinate systems. systems. denote the unit vector along the principal line by n, n. and consider an arbitrary r. The coordinate systems and the principal line are sketched in Fig. 7.6. Note thx a3 1)3

1,-, 71

a

Figure 7.6

Rotation about principal line

7.3

EULER'S AND

THEOREMS

to Euler's theorem the coordinates oft/u' principal line remain unchanged the A and B frames. The angles (i = 1, 2, 3) that n with the (i = 1, 2, 3) axes arc the same as the angles n makes with the b,

We can express the vector r in terms of the components of the two coordinate as

r=

Aria, + = Anal + Aria7

8r3b3 = Br = 8r1b1 + 8r,b2 Br2b2 ++ 8r3b3

[7.3.1] 17.3.11

which the subscript on the left of r identifIes the coordinate axes in which the vector ..s resolved. The usual convention is that a vector is resolved along the coordinates with the reference frame from which it is being viewed. We will use the in Eq. 17.3.11 only when it is necessary to make a distinction between the reference frame from which a vector is viewed and the unit vectors used to resolve it. In column vector format we have {Br}

[7.3.21 17.3.21

= [c]'{Ar}

r} = r1 8r2 and {{ A r} = A r1 A r2 A 1i I T• II n a si i II ar ar fashion, fashion, we 812 B 13 a-an express the unit vector along the principal line as {

n=

I

An = 1tniai 1tniai

+

I

n=

+

8n =

11n1b1

+ 8tz2b7 + 8n3b3 8n3b3

[7.3.3] in column vector format, as

{Bn} = [c1T{An} {Bn}

17.3.41

The expressions Ani and 8n1(i = 1, 1, 2. 2. 3) 3) are are the the direction direction cosines of the principal n • a. Rn, = n b, so they are equal to each other. Hence, A'1i = = n A'1i

=

{Bn} = {Bn}

[clT{/%n}

= {c}

17.3.5]

1T where the vector {c} = [c1 c2 is recognized as the direction cosine vector aswith the principal line. Substituting Eq. [7.3.5] [7.3.5J into Eq. [7.3.4] we can write

{c}

= [c]1{c} Eel1 {c}

[7.3.6]

Mathematically, Mathematically, Eq. [7.3.6] [7.3.6] can can be be explained explained as as{c} {c} = {8n} = (.4n} (4n} being the of [ciT associated with the eigenvalue A = 1. To prove Euler's theorem need to establish that [c] has an eigen value equal to I. Consider the eigenvalue problem associated with [c]1 [RI. where [R] [RI is the matrix that relates the b1h2b3 axes to the ala2a3 axes. axes. We can write the left and right eigenvalue problems as [R]{x} = A{x}

IRIT{v} = A{y}

17.3.7] 17.3.71

Solution of the eigenvalue theorem yields three cigenvalucs A1. A2, and Cor-esponding right eigenvectors {x1}. {x2}, and {x3} {x3} and and left tell eigen vectors and }, }' We know from Chapter 2 that the determinant of [R} = [ciT is equal to one. From a theorem in linear algebra the product of the eigen eigenvalues values of of aa matrix matrix is is equal equal the determinant, so that A1 = I. 1.

361

362

CHAPTIR

7

•

RIGII

KINEMAI'ICS KINEfS1AI'ICS

The relations between the eigenvalues and right and left eigenvcctors of [RI be written as =

[R]'{v1} =

i == i

I ,

2, 3

17.3.8a,hJ

Now. consider the eigenvalue problem associated with the inverse of [Ri. It easy to show that the eigenvalues of the inverse of' a matrix arc the the eigenvalues eigenvalues of the original original matrix. matrix. while the eigenvectors the originverses of ofthe ofthe eigenvectors of ofthe ma! matrix and its inverse are the same. We can therefore write the right fbrrnulation as [R] '{.v,} =

[7.3.91 17.3.91

One can arrive at Eq. [7.3.9J [7.3.9J by by left-multiplying left-multiplying Eq. Eq. [7.3.8a1 [7.3.8a1 by by [R1

.

But

inverse of f R] is equal to its transpose. SO that Eqs. [7.3.8h] and [7.3.91 define same eigenvalue problem. problem. Hence, Hence, for for aa unitary unitary matrix, matrix, the theleft. left and right coincide. {x1} = (1 = 1, 2, 3). Furthermore, we have (i 1

A

Because deti deti Ri Ri =

17.3.101

and the product of the three cigenvalues is unity, there two possible cases: (a) all eigenvalues eigenvalues are are i-cal real with with values 1, 1. 1 or 1, —1, —1. (1) one eigenvalue is real and unity. and the other two are complex conjugates moduli equal to I. The first case when all eigenvalues are real does not make physically, as it would indicate at least two repeated roots and hence an axes about which a rotation can be performed. This is impossible, as one can fInd an axis that will not yield the desired rotation. We are left with the second sihility. Hence, [R] (or FelT) has one real eigenvalue equal to 1. The eigenvector gives the direction cosines of the principal line. Note that the tor must he using the relation {c}T{c} = 1. Also, if {c} is the associated with the cigenvalue A1 = I. so is —{c}. One can take the unit along the principal line along either of the two directions of the principal line. difference will he the direction of the rotation angle We next discuss determination of the rotation angle 1. Consider an coordinate system d1d2d3, where the d1 axis is aligned with the principal line axes are selected arbitrarily. We can write the relationship between the £/2 and d1d2d3 axes and the A frame as I

{a} = Ic*1{d}

17.3.111

where Fc*1 is the associated direction cosine matrix. Using the relationship the A and B frames, we can write {b} = IcJT{a} =

(7.3.12]

Let Let us now perform the rotation by the angle about the d1 axis, which leads to the coordinate system It follows that the relation between the a1a2a3 and d1d2d3 axes has to be the same as the relationship between the h1h2b3 and did,a:

7.3

363

EULER'S AND CHASLES'S

Both coordinate axes were transformed in the same way. We can express this relationship as axes.

{h}

=

17.3.13] 17.3.131

J{d'}

However, the relationship between {d} and {d'} is

{d'} = [R']{d}

17.3.14] 17.3.141

where R'] is a rotation matrix in the form I

0

1

cosF

[R'I = 0 0

0

[7.3.15]

sinF

—

Combining Eqs. [7.3.11 ]—[7.3. I 3j, we can write

Ic*lT[clTIcl = [R'l

17.3.161

This equation represents a similarity transformation between between Id' Id' and and II R' R' I; thus

the traces (sum of the diagonal elements) of' these two matrices are the same. We can now write + C22 +C33 C33 = [R']1, + [R'122 [R'h2 ++ [R']33 [R'133 = C22 +

I

+

[7.3.17]

which can be solved for 'ii) to yield the rotation angle.

it is interesting to note that in Chapter 2 we considered a three-parameter rotation to describe a general transformation from one set of coordinates to the other. whereas in this section we used Euler's theorem and accomplished the transformation using four parameters: the three direction cosines associated with the principal line, and the the rotation rotationangle angleci). ci). ItIt turns turns out out that that one one can take advantage of the fourparameter description used here to simplify the kinematical equations. equations, as we will in Section 7.7. that, we will discuss the relationship between the angular velocity vector and the direction cosine matrix, and we will parametrize the angular velocity vector in terms ot of the transformation angles. The above developments were for a body with one point fixed. If we consider an unrestrained rigid body, we can extend the results of Euler's theorem into what is known as Cliasles's theorem, which states: The most general displacement. displacement of of aa rigid rigid body body is is equivalent equivalent to to aa translation translation of of some point in the body. plus a rotation about an axis through that point.

Depending on the point selected, the instantaneous axis axis of' of rotation will be different.

The coordinate coordinate system systema1a2a3 a1a2a3 isis transfonned into the b1 b2h3 system by first rotating by an = 300 about the Q3 axis. and then by rotating the resulting angle jj = frame about orientation of of the the principal principal line, as well as the rotation the a; axis by 0 = 45°. Find the orientation angle.

I

Example

7.1

364

CHAPTER 7O RIGID BODY KINEMATICS

Solution The rotation matrix associated with the transformation given is

[R2][R] [Ri = [R21[R,]

I.]

in which 0.707 11 0.707

0

0

1

[R2] =

0.707 0.707 1

[R1] =

0 0.7071

0

1

0.8660

—0.7071

—0.5

0

0.5 0.5

0

0.8660

0

0

1

[bJ

so that

0.6124 [RJ =

—0.5000

0.6124

0.3536 0.8660 0.3536

—0.7071

Ic)

0 0.7071 0.707 I

The eigenvector associated with the eigenvalue A = = 1I can be found manually or numerically. This eigenvector can be shown to be x = [—0.2195

0.8192

0.5299] T

[dl

The elements of the cigenvector are the direction cosines of the principal line. The angles the principal line makes with the inertial coordinates are then = cos '(—0.2195) = 102.7° 102.7°

= To To find the rotation angle 1, we make use

= COS1

4-

02 =

35.00° 58.000

(0.5299)

[.1

of Eq. [7.3.17J. thus

/?22 + R33

[fi

= cos'(0.5928) = 53.65°

— 1)

coordinates, principal line, and rotation angle are shown in Figs.

The and 7.8.

Plane of a

02.7°

(j)

a

580

a

FIgure 77

Figure 7.8

a1

7.4

7.4

ANt) ANGULAR ANGULAR VELocITY [)IRECIION CosINES CosiNEs ANt) RELATION BETWEEN [)IRECTION

RELATION BETWEEN DIRECTION

AND ANGULAR

VELOCITY z this section. we explore the relationship between the direction cosine matrix and angular velocity vector. Consider two coordinate frames A and B and associated axes a1a2a3 and h1b2b3. The origin of the two frames coincide. The relvelocity expression for a point whose position is given by r is Av

[7.4.11

+ 4w'3 X r

=

Using the notation in Eq. [7.3.11 of denoting the reference frame in which a

is resolved by a subscript on the left, we can relate the position and velocity ectors by

{8r} = jA

{Ar} = [c]{Br} ,1TjA

—

t-Ui

jA

1

—

1AUYI —

1AVI

(clT{Rv}

r

17.4.2a,bI

ijA

[7.4.3a,b,c,d]

=

(ci isisthe (C] thedirection direction cosine cosine matrix matrix between the A and B frames. relationship between We next find a i-elationship between [cj and the angular velocity of the B frame To this end, we first express 4to13 along the cornrespect to the A frame. of the A frame as

[7.4.4] [7.4.41

+

1a, + = Awla, using the matrix 0

AW3

— —

AW2 — AWl — 1

.• .•

0

can write the transport theorem as =

[7.4.6] [7.4.61

+

In In a similar fashion. we write the velocity of a point in terms of the B frame as 4

4-8 1{Br} B i{Br} {BV} + = {BV}

17.4.7]

Left-multiplying Eq. [7.4.6] by [cf and using Eq. (7.4.71 relates the angular velocity matrices expressed in the A and B frames as {A&B}

=

17.4.81

ion, transformation, One should compare this equation, which represents a similarity transforinat Eq. [6.4.1 3aJ. We next obtain a relationship between the direction cosine matrix md the angular velocity vector. To this end. we consider Eq. (7.4.2bj. The left side the equation is in terms of a vector resolved in the A frame, and the right side is in of a vector viewed in the B frame. Therefore. when we take the time derivative •:f Eq. •:'f Eq. [7.4.2b] [7.4.2b1we wehave have to to differentiate differentiate the the left side in the A frame and the right side rn the B frame. Doing so yields =

+ [ij{8r}

17.4.9]

365

366

7

CHAPTER

.

RIGID Bony

We convert Eq. [7.4.91 to one expressed in terms of the components of the A frank

by introducing Eqs. [7.4.2b1 and [7.4.3d1 into it, which yields =

17.4.101

+ ki[c]'{11r}

Comparing Eqs. [7.4.61 [7.4.6] and [7.4.10]. [7.4.101, we we conclude that = =

In terms of the rotation matrix [RI = In

[7.4.111 17.4.111

ft]T, kiT, we can express

as

17.4.121

EkITLRI

and [7.4.12], we obtain the angular velocity matrix express& Using Eqs. in terms of its components in the B frame as

[111

=

=

k.JTk]

[R][RIT

[7.4.1 17.4.1 3J

It is usually more convenient to write the components of the angular velocity [7.4.13] more extensively that terms of the body coordinates, so that we will use Eq. [7.4.131 Eq. [7.4.111.

Example

7.2

Consider two frames A and B, where the coordinates b,b2b3 are obtained by rotating using Eqs. [7.4.11] Find about and then by an angle about by an angle 17.4.131. 17.4. 131.

Solution Denoting the transformation matrices by [R1 I and [R2]. the combined transformation matrn is [RI = [R2j[R1j. The direction cosine matrices are

[c1.] =

cO, c01

—sO1 —sOs

0

sO1

c01

0

0

0

C02

[e2] =

0 —S02

1

0 1

0

502 0

[c] = [clj[c2] [clj[c2j

[.1

CO cO

Taking the derivative of [cJ with respect to time yields Fe] =

[bJ

1[i'21 +

Let us first use Eq. [7.4.111 and evaluate we obtain Ic2ITLc1 Ic2 I'jci

[hi by [ciT Right-multiplying Eq. [b]

= kllclT = [ciI[C:lIc2jT[ctlT + = kj[c]T

=

(cJ [cJ

is what what one term is has an an interesting interesting interpretation. interpretation. The The second second term of Eq. Eq. Id Id has The right side of transformation. The first term for the angular velocity matrix when there is only one transformation. secondtransformation, transformation, [e2l[c2]T, with aa proper coorthe angular velocity expression for the second rotation angle. first rotation dinate transformation by the first When the angular velocity is obtained in terms of the B frame we have

= [CJT[cJ = [C21'[C21 [c,]T[c1]T[C11[c21 [C21'[C21 + [c2]T[c1]T[è1][c2]

[dJ

which has a similar explanation as the one for Eq. Eq. [ci. tel.

It remains to calculate the derivative matrices and use them in Eqs. [c] and [dl. The derivatives of [c1J [c1j and [c2] are

7.5

—01s01

—01c01

01c01

—&1s01

VI] VI] =

0

V2] =

0

0

—0c02 —02C02

0

02c02 0 02S0) —02S02

0 0 0

—02s02 —02S02

EULER ANGLES

1.1

Consider writing the angular velocity vector in terms of its components along the a1a2a3 axes.

We first evaluate

and related terms. Using Eqs. [a] and tel, we have =

c61

sO1

0

—sO1

c01

0

0

0

—01s01 01c01

—01c01 —01s01

I

0

0

0 0 0

0

—01

0 0

= 0

0

[f]

0

0

hich represents the angular velocity matrix associated with a I rotation. As expected, we find [i.11[c1]T that [i.1][c1]T Icilr[ci]. Performing Performing the the same matrix multiplications with 02, we obtain [c21[c21 k211c21

0002

=

=

0

0

—02

0

Igi IgI

0 0

is the angular velocity matrix for a 2 rotation. Carrying out the algebra we obtain in terms of the A frame ][è2J[c2 17'[c1 1T + = [è][c]' = [ci [c1 11è2][c2

1T

0

—01

02C0I

0 01

•00

02S 01

—02C01

—02s61

0

=

[hJ [hi

We We confirm this result using the formula for the angular velocity as

= 0)a3 +

[ii [I]

= 0!a3 + 02(cos01a2 — sin01a1)

that

AWl = —O2sinO1

/%W2

=

[ii Ii]

=

O2cosO1

These match the entries of the angular velocity matrix in Eq. In terms These Eq. [hI. thl. In we terms of of the B frame we

[cIT[c 1"[ii l'[i11[c21 1['21 = = [c21"[i21 + [clT[c

=

0

—01c02

01c02

0

02

0 In

a similar fashion, we write the angular velocity in terms of the coordinates of the B

frame as Awfi =

SWL

+ +0,a', =

= —01 02 —Oi sin 02

91(—sin02b1 sin 02b1 + +cosO,b3)+02b2 + 02b2

BW2 = 02

11W3

=

cos02

[II [I]

[ml Fm]

confirms our earlier results in terms of frame B, when we compare this with Eq. [ki.

7.5

EULER ANGLES

We saw earlier that, at most, three successive rotations about nonparallel axes are sufficient to define a rotation transformation from one coordinate system to an.::ther. The transformation can he expressed in many ways and is not unique. In this

367

368

CHAPTIR 7 • RIGm BoDy KINEMATICS

section, we quantify the different choices for carrying out the transtbrmation from one coordinate frame to the other.

7.5. 1

EULER ANGLE SEQUENCES

Let us review what we have done in the previous chapters. In Chapter 1 , we studiei curvilinear coordinates and discussed the rate of change of the unit vectors. This cussion was in terms of specific coordinate systems, without a general formulation. In Chapter 2, we saw that three transformations about nonparallel axes are to describe the most general transformation from one coordinate system to another We outlined two ways of accomplishing these rotations: body-fixed and rotations. When the coordinate transformation angles are infinitesimal, the rotations can be viewed as vector operations, from which we defIned the angular vector. We obtained expressions for the rates of change of unit vectors. Most of the analysis in Chapter 2 was instantaneous. In this section. we use body-fixed rotations and select the nonparallel axes which the rotations are carried out as the coordinate axes of the rotated frames. The three angles used for transforming one coordinate set into another are referred to as Euler angles. In generating three sets of rotations to transform one set of coordinates into another, say, say, a1a2a3 a1a2a3 to toh1b2h1, h1b2h1, there there are twelve twelve choices in which no two adjacent rotation indices are the same. We begin with the a1a2a3 frame and rotate it about frame. Here we have three choices. The next rotatior of the axes to get the

a.;, or a; axes, excluding the axis that coincides with the a;, is about one of the previous rotation. That is. if the first rotation is about the a2 axis. the second rotation should not be about the a; axis. Hence, we have two possible rotations for each previous rotation. We follow the same procedure for the third transformation, resulting in two possible transformations for each previous rotation. As a result, we have 3X X 2 = 12 possible combinations: x 22 X

1-2-1, 1-2-3, 1-3-1, 1-3-2, 2-1-2. 2-1-2, 2-1-3, 2-3-1. 2-3-2, 3-1-2, 3-1-3. 3-2-1, 3-2-3

The twelve sets are called Euler angle sequences. They can be classified inte two main categories, each showing similar characteristics. The first category is when the first and third indices are the same (e.g.. 3-2-3, 1-2-1) and the second category consists of rotations where the first and third indices are different (e.g.. 1-2-3. One selects the sequence depending on the application, such that the sequence used gives a better physical visualization visualization and, and, as as we we will will see see later lateron onin inthis thissection, section,lea&s leads to fewer singularities. One of the most commonly used Euler angle sequences is the 3-1-3 sequence. often used to describe rotating rigid bodies. In recent years the 3-2-3 transformation has been used more widely for such problems. We learned learned in in Chapter Chapter 33 that that the the 3-1 3-I the orbit orbit of of aa body body in in coordinate transformation has traditionally been used to locate the space.

For purely historical reasons. reasons, we begin our discussion of Euler angles with a 3-1-3 transformation. (Readers are urged to familiarize themselves with other trans-

75

EULER ANGLES

sequences, especially the 3-2-3 and the 3-2-1. Properties of the 3-1-3, 32-3, and 3-2-1 3-2-I sequences sequences are are summarized summarized in Table 7.1 at the end of' this chapter.) In a 3-1-3 transformation, first, the axes a3 by an angle to yield the aç aa aa axes. axes. Then, Then, aa rotation rotationisisperformed performedabout aboutthe the a axis axis by by 0, the axes. The axis is also known as the line of nodes (see Chap3 for the source of this name). This axis describes the intersection of the a a2 and planes. Finally, the axes are rotated by about the a;' axis, which results in the b1b2b3 axes. 0, and are known as the For a 3-1-3 transformation, the rotation angles precession, nutation. and spin angles, respectively. For a 3-2-3 transformation, the angles are taken as (precession). 0 (nutation). and c/ (spin). To obtain the combined transformation from the A frame to the B frame, we use matrix notation and successively apply the transformations. For a 3-1-3 sequence result is I

c4s40

{a'} =

—s4

c4

0

0

0

0 cO

0

—sO

1

0 {a}

{a"} =

1

ecu

= 0

sçls

0

c4/J

0 {a"}

0

sO {a'} cO

[7.5.1]

01

C.mbining the three transformations, as shown in Fig. 7.9, we obtain {b}

[7.5.2]

=

i here

s0si/i —c4s4/J—s4cOcI// s(t)sO

7.5.2

17.5.31 —cfrs0

ANGULAR VELOCITY AND ACCELERATION

We next obtain the angular velocity vector, which has a component due to each roas

= AwN + A'wA" + A"wB =

+ 0a + i/ia'

[7.5.4]

We wish to express the angular velocity in terms of the coordinates of the trans-

frame. From Eq. [7.5.3] we obtain

a=

+

+ cOb3

= cqib1

a;' = b3

—

[7.5.5]

The inverse ol' [RI [R] is its transpose, {a} = [RJT{h}, so an easy way to obtain a3 is we read thethird thirdcolumn column of of IRI. IRJ. To To obtain obtain read down the the third third row row of of [Rf' [Rf ororthe the fIrst first row of [R]T (or the first column of IR1) while setting 4 = 0. Introducing

Eqs. [7.5.5] into Eq. [7.5.4J, [7.5.4J, we obtain

=

O(ci/b1 + soccu;b2 soccuib2 + cOb3) + 0(ci/b1

=

+ Ocqi)b1 Ocqi)b1 ++(4sOcq; (4sOcq;—— Osq;)b2 +

+

—

+

17.5.61

369

370

CHAPTER

7

RiGID BODY KINEMATICS

/,,

)

7 Plane by a1 a2

8 4,

a, b1 a1 4)

Plane defined by b1

FIgure 7.9

Line of nodes 3-1-3 Euler angle sequence

which can be written in the matrix form as

js Os t/i +

AwI11

=

t/i

4)s Oc i/i — Os i/i 4)c 0 + i/i

=

s Oc

0

4)

0

0

17.5.71

0

in which is the ith component of AwIJ expressed in terms of the body-fixed Equation [7.5.7] can also be written as {w} = [B1{0}, where the notation is A few observations about the matrix [B] that relates the angular velocities to the rates of the Euler angles are in order. First of all, [B] is not orthogonal. This is because and a;' axes, which do a3. the rotations 4), 0, and i/i are performed about the a3, 0, and i/i are independern form an orthogonal set, even though the rotation angles 4), 0. of each other. We also observe that the sines or cosines of the precession angle 4) are absent in Eq. [7.5.7]. Second, the matrix [B] becomes singular when 0 is equal to zero or to a mulupie of ii-. This can be explained by noting that when sin () = 0, the rotation reduces to a 3-3 sequence, and that the 4) rotation cannot be distinguished from the i/i

tation. Because [B] can become singular at times, the rates of change of the angles cannot always be obtained from the components of the angular which causes problems when integrating the equations of motion. To visualize singularity, we invert Eq. [7.5.7], to yield ct// 0

H I';

sO

sO

cl/i st/i

—si/I —si//

0

[7.5.$J

0

ct/i

Aw1B I

tO

10

which tO = tan 0. Equation [7.5.8] is referred to as the kinematic tial equations relating the angular velocities to the Euler angles. In scalar fortz.. in

7.5

EULER ANCLES

Eq. 17.5.81 is

=

(w1 sine/i (w1

.

.

1

=

.

sinO

+

0 =

-

—

-

I

(—w1 cosOsnu/i cosOsiiu/i (—w1

+ w3 w3

—

17.5.91

where we have adopted the compact notation w, 1, 2, 1, 2, 3). 3). Existence Existence of the singularity at 0 = 0 and at multiples of 'ir is obvious. Moreover. Eqs. [7.5.9] are highly highly nonlinear, thus reducing their suitability for manipulation in rigid body dyequations, especially for numerical calculations. The singularity associated a particular Euler angle sequence can he overcome by switching to another Euler er angle sequence in the neighborhood of the singularity, but this makes the analysis analysis Each of the twelve Euler angle sequences have singularities at certain values of the second angle. One of the objectives when selecting an Eulcr angle sequence is avoid singularities as much as possible. For example, in aircraft or other vehicle ivnamics problems, usually a 3-2-1 sequence is used. The Euler angle sequences no index is repeated (3-2- I, 1I -3-2. -3-2. etc.) all have a singularity when the second angle, 0, has the value 0 = ±ir/2. For the most common flight operations, all of these angles remain small quantities. The text by Junkins and Turner contains a of the transformation matrices and singular values for all twelve Euler angle sequences. We next analyze the angular acceleration, extending the results in Chapter 2 to to bodies. If we express the components of the angular velocity in terms of an frame or using a set of body-fixed coordinates, then the angular acceleration a simple form. Mathematically,

AB a

=

B —('(*))+'w XW di

1t

=

B —((0) dt

[7.5.10]

Given the angular velocity components along the body axes in Eq. 17.5.7]. it follows the components of the angular acceleration along the body axes are, for a 3-1-3 sform atio II.

=

= =

= =

—

=

+

—

—

—

17.5.111 [7.5.111

In general, evaluating the components of the angular acceleration along inertial

yields complicated expressions and does not provide much insight. We hence :'rltinue to to see see that that expressing expressing the angular velocity by its components components along body leads to a more compact form and and a more meaningful understanding. This ohseris totally in line with the results in Chapter 6. that most of the time, expressing

371

372

CHAPTER

7• RIGID

BODY KINEMATICS

the inertia properties of a body using a set of axes attached to the body is simpler and more meaningful.

7.5.3

AXISYMMETRIC BODIES

We next consider a case where it is more desirable to view the motion using a set of relative axes that do not coincide with the body axes. Such a case arises when dealing with axisymmetric bodies. A suitable choice for analyzing this type of motion to use a reference frame that does not contain the spin angle. That is. if we are using. transfonnation. the relative frame is obtained after the 3say, a 3-2-3 Euler angle transformation, transformation. The motivation behind selecting a reference frame different than body frame stems from the fact that in many problems involving axial symmetr'. the actual value of the spin angle is not of too much importance. while the spin rate center of of mass mass or orc4 is. One is more interested in the velocity and acceleration of the center another point along the symmetry axis, as well as the angular velocity and acceleration, than one is in the motion of sonic specific point on the body. The precession and nutation rotations completely describe the the orientation orientation ol' of the symmetry axis. being 1a2a3 being Consider a 3-1-3 transformation. transformation. Using Usingthe thenotation notationabove aboveofofaata2a3 inertial axes, the coordinate axes associated with the new reference frame relèrence frame axes, and we can find the angular velocity of the reference the the 3-1 rotation in Eq. [7.5.4] as +

=

+

= çb(sinOa' gb(sinOa' +

17.5.121

We will, from now on. refer to this relative reference frame as the and use the associated unit vectors The concept is illustrated in Fig. 7.10 for a spinning top. For axisyrnmetnc is equivalent equivalent to viewing the general shape of a disk bodies, viewing the F frame is top. without following the motion of a specific point on the body. We can write the angular velocity of the body as = = + [7.5.131 denote the associated coordinate axes by

LI

i-i

.1 .1. C) () F

I

IF

F

Li'

a-, a1

e a a

Figure 7.1 0

The F frame

7.5

373

EULER ANGLES

= = and is the spin, Where appropriate, the above flotation flotation will will be be shortened in o,r,. tof. and where the subscripts b,f, and s stand body, frame, and spin, respectively. We then write here

17.5.141 For a 3-1-3 transformation, we can write the angular velocities of the body and

f the F frame in terms of the components of the F frame as

to = to,, = 4)(sin to = (Of

Of1

+ cos cos6a') 6a') +

+ çbsinOf +

+ cos = 4(sin 0a' + cos

+

+

+ 6a7 == Of1 Of1 +

sin Of2 + 44 cos cosOf3 Of3

(0 ==

[7.5.1 5] 51

(Os

The expression in Eq. 17.5.6].

for angular velocity above is noticeably simpler than its counterpart

Next, consider consider the expression for the angular acceleration. When using the F frame, the the angular angular acceleration of of the the body body can can be found using the transport theorem. Noting Noting the form form of of w,, from fromEq. Eq.[7.5.1 [7.5.14J. 4J. we we have have

a1, =

d

d

X (Oç + (Of (Of X

+ (Ot Wj, = Wi )< (01, cit

cit

r

AaB

di

=

x X

+

di

+ 4wF X

[7.5.16]

Expanding Eq. [7.5.161 in terms of the Euler angles using the 3-1-3 set, we obtain

d (I = (it cit

X + (0 X

= 0f 0f1 ++ (4s0 (4s0 +

+

+ (Of1 + çbsOf + =

—

+

X i/ifs

+(4s0+çbOcO—Th/I)f2+(4c0—40s0+I//)f3 +(4s0+çbOcO—Th/i)f2+(cfcO—40s0+I//)f3

[7.5.17] is considerably simpler than its counterpart when the spin angle is included, Eq. [7.5.11]. The above result can he verified by setting = 0 in Eq. 17.5.111, while retaining retaining i/i. i/i. Note Note that that if one wishes to deal with with the the actual actual value valueof ofthe thespin spinangle anglecli, can still use Eq. [7.5.1 [7.5.171. 7J. When the manipulations are complete one can switch the the coordinate system that includes ±

the angular velocity expressions for aircraft dynamics problems, using a 3-2-1 transirrnation.

;oordinate system traditionally used with aircraft is shown in Fig. 7.11. We will denote coordinates by XYZ and the body-fixed ones by xvz. The z axis is the local vertical. angle transtorrnat,on,s are:

I

Example

7.3

374

CHAPTER 7•

R1G11) R1G11)BODY BODYKINEMATICS KINEMATICS

(/)

xvvI x xv,,

}/H

Y

Zn

zIz,

Figure 7.11

3-2-1 Euler angle sequence

a.

A rotation about the Z axis. by angle i/i (the heading angle), leading to the X' Y'Z' dinate system.

b.

A rotation about the V' axis, by angle U (the attitude angle), leading to the coordinate system.

c.

A rotation rotation about byby angle aboutthe theX"X"axis, axis, angle 4 (the bank angle). leading to the xyz system.

Using this conflguration. the xz plane denotes the plane of symmetry of the aircraft. shown in Fig. 7.12. 7.12 The rotation matrices are ci/i FR1] =

—si// 0

si/i ci/i cu,

0

0

01

[R2]=

0 sO

o

—sO

1

0

0

0

I

[R3j

=0 0

c4

s4

-sç1

cc/p

V

(0.r Roll

w..: Yaw

Pitch V

FIgure 7.12

0

Body axes For an aircraft. The xz plane is the plane of symmetry.

[sJ [.1

7.5

375

EULER ANGLES

that the relation between the inertial and body-fixed coordinates is written as x

x

K

=

v

= [Ri

Y

Ibi

Y

7

'-4

md the combined transformation matrix [R} is si/icO

[R] = —sqic4) —sqic4)++cqisosd cqisosd +

—sO

cqic4 + si/isOs4)

cOs4)

—ct/is4) + sifisOcth

cOc4)

[ci Ic]

The angular velocity vector is written as

[d] Using the same approach as in the previous section, we obtain K and J' in terms of the vectors associated with the body axes by reading the the third third and and second second columns columns of of [R], [Rj, respectively, respectively, and and setting setting sji = 0 when reading the second column. which yields =

+

+

+ 4)i +

+ s4cOj + c4)cOk) + 6(cthj — s4)k)

=

+Oc4))j + (t/ic4)cO

—

Os4))k

+ 4)1

lel

The body angular velocities w.v, and W: are commonly referred to as rn/I, pile/i. and rates, respectively, as shown in Fig. 7.12. We relate these angles to the heading, attitude, md bank angles i/i. 0. and 4) using Eq. [e] as

= s4)cO s4)cO

0 c4)

0

c4c0

—sth

0

—sO

1

[f] En

6

When 0 = rr/2, the the first and third columns of the coefficient matrix become similar. similar, so a singularity is reached. For civilian aircraft this is not a big problem. as 0 rarely exceeds for most flight operations.

The spinning symmetric top in Fig. 7.13 has the following rotational parameters: precession 4) = 0.3 tad/s and increasing with the rate of 0.05 radls2, nutation angle 0 = 300, zero rate 0 = 0, 0 = 0, and constant spin rate of = 5 rad/s. Assuming that the bottom of the top does not move, find the velocity and acceleration of the center of mass of the as well as its angular velocity and angular acceleration.

Solution use the 3-1-3 Euler angle transformation sequence and distinguish between the body frame and the relative frame, as we are interested in the velocity and acceleration of a point ing on the symmetry axis. We can write the angular velocity of the body frame and the relative frame in terms of the unit vectors of the F frame as -elative We

WI, = 4(sinOf2 +

WI =

+ Of, + + Of1 Of, =

4)(sin Of2 + cos

=

= 0. 0. 0.

+ 5.2598f3 rad/s

1Sf2 + 0.25981'3 rad/s

= Sf3 Sf3 rad/s radls

[a] [b] Ic]

I

Example

7.4

376

CHAPTIR 7• RIGID BODY KINEMATICS a3

f3

4;

12

6 0

a,

0

6

J'i' aj

Figure 7.13 We

can find the angular acceleration by using Eq. [7.5.17] as

d

+ (Of X

Idl

(U.S

46c0 + (4.s0 + 40c0

= (0 +

+ (4.c0

—

—

c/6s0 + t/i)f3 c/0s0

Substituting in the values, we obtain

= 50.30.5f1 50.30.5f1 + 0.05 0.5f2 +

O.75f1 ++ 0.02Sf2 + 0.04330f3 raclls2 O.75f1

I.'

We find the velocity of the center of mass as Of2 + cos Of1) + X rc,c rc,c = [4(sin [4(sin Of2

VG

=

—

Of1

+

X Lf1

L0f2 = O.15Lf1

(fJ

To find the acceleration of the center of mass, one can use a number of approaches. One perform aa straightstraigh;approach is to express the velocity in terms of the inertial coordinates and perform forward differentiation. This is lengthy, and it requires that the results then be converted back. to the body axes. We prefer to use the F frame and transport theorem, which gives A

aa = =

Fd

AF (U Xv0 sin Of1 +

cos Of1 — L0f2 +

= (L4 sin 0 + Ld0 cos 0 +

X + Of1] Of1] X

+

—

L0f2

cos O)f1

— (L42 (L42sin20 sin2 0 + L02)f1 L4)2sinOcosO)f2 sin 0 cos O)f2 — + (—LO + L42

[Ii ui

Substituting in the appropriate values, we obtain

aG = 0.025Lf1 + 0.03897Lf2

—

0.0225Lf3

[hI

A third approach to solve for the acceleration of the the center center of of mass mass is is to to use use the the formth tbrmth for the rotation about a fixed point, which is X a0 ac=ahXrG,o+wbXwbXrG,o=abXrG,o+cohXvG ab X rG/o + Wh X t0b X rub = ab X rG/o +

III

7.5

377

EULER ANCLES

equation values values for for the the angular angular velocity velocity and and acceleration acceleration parameters parameters Substituting in the above equation

X Lf3 X Lf3 + (0. 1Sf2 + + 0.7890Lf2

= (O.75f1 + O.025f2 +

= —O.75Lf2 + 0.025Lf1

—

= O.O25Lf1 + 0.03897Lf2

—

X X 0. 0.1

1

Iii

0.O225Lf3

of course. is the same as the result in Eq. [hJ.

A gyropendulum, consisting of a disk of radius R attached to a shaft of length L. rotates with a spin rate of about the shaft. The shaft is pivoted to another vertical shall which itself rotates coeflicients of with the the rate rate d. The pivot angle is 0, as shown in Fig. 7.14. Find the inertia coefilcients the disk about point B. the angular velocity, and the angular acceleration.

us defIne the inertial frame XYZ with the Z-axis along the precessing shaft. a second preccssing shaft, shaft, rotating rotatingwith withb.b.and andaathird thirdframe framexyz xz obtained frame x'v'z' attached to the preccssing rotating the shaft about the y' axis by the nutation angle 0. In essence, we are using a 3-2 The resulting frame is the F frame. The mass moments of inertia of the disk about its center of mass niass are Let

=

=

=

[a]

ith all products of inertia zero. About the center of rotation B, the moments of inertia can be using the parallel axis theorem, which yields 'Bzz

= =

mR2

IRyr =

7?!

+ L2)

R2

Noting that the unit vector K = k' can be expressed in the xvz frame as k' = :os Ok. the angular velocity of the disk and the frame can be written as

M

yf. y y'.

e u/i

Figure 7.14

.v

[b] —

sin Oi +

Example

7.5

378

CHAPTER 7 • RIGII RIGI1 Bony Bony

= to1

+ i/;k i/;k = 4)k' + 0j +

+

= dK ++ Oj' O,j' =

+ Oj Oj =

—

= —4isinOi + Oj Oj +

sin Oi Oi ++ Oj Oj +

Ic)

+

Ed] Idl

cos

w3=qik

[.1

The angular acceleration of the disk is simply + + WI WI XXw, w, =

a1, =

+

+

= (—thsinO

7.6

—

+ 6j +

+

+

—

X (t/;k) X

+

+ (6 (6 ++ (bi/JsinO)j (i/i + (bi/IsinO)j ++ (i/i

thOcosO +

—

4OsinO)k

[11

ANGULAR VELOCITLES AS QuAsi-VELocITIEs (GENERALIZED SPEEDS)

From Chapter 4. a generalized velocity is obtained by diflèrentiating the corresponding generalized coordinate with respect to time. The three Euler angles that the angular velocity are in essence generalized coordinates, and their time tives are the generalized velocities. However, the components of' the angular velocth cannot be classified as generalized velocities, as there is no corresponding ized coordinate. A similar statement is true for a nonholonornic constraint, cannot be expressed as the derivative of a function. The questions can then be posed as to how one can classify angular and whether there are other cases in dynamics where one deals with such We introduce here a set of variables called quasi-velocities or generalized We define these as linear combinations of the generalized velocities, but they therz.selves are not necessarily derivatives of any coordinates and thus cannot always integrated to generalized coordinates. Quasi-velocities are not perfect differenti2k a fact from which their name is derived. We will use both terms, generalized and quasi- velocities, interchangeably. The introduction of generalized speeds increases the choice of parameters can use to describe motion. Consider a holonomic system having n degrees of freedom, with generalized coordinates .... and generalized velocities We define a set of' quasi-velocities or generalized speeds iii. u2, u2, ...., ... as tI

k = =

Uk = '>

1,2,..., 1,2,...,

n

[7.6.IJ

j=I = q,1, t), Zk (k, / = 1,2, . ., Zk = q,,, 1) (k,/ are functions of the generalized coordinates and time. In column vector form we can write Eq. [7.6.11 as where

.

.

.

,

•

{u}

{Z} = IYJW} + {Z}

.

. ,

.

[7.6.2)

In order for the set of generalized speeds to completely describe a system.

must be nonsingular. We invert Eq. [7.6.2J to express the generalized velocities iz z

7.6

ANCULAR VEI.ociiw.s .&s

379

(GENERALIZED SPEEDS)

terms of the generalized speeds as terms

= I.YJ I.YJ '{u}

—

[7.6.3J [7.6.31

[YJ '{Z} [Y]

The angular velocities of a body can be classified as quasi-velocities. The relabetween the body angular velocities and the Euler angles, Eq. [7.5.7J, defines defines a

set of generalized speeds.' By contrast, the angular velocities of the F frame not constitute a set of quasi-velocities, as the relation between them and the Euler ingles is described by a matrix that is singular at all times. Representation of angular velocity as a set of generalized speeds is not the only application of these quantities. Generalized speeds are very useful when one deals with dynamical systems subjected to nonholonomic constraints and in cases when the use of generalized velocities makes the problem formulation cumbersome. We saw an example to this in Examples 4.14 and 5.13. where switching to the velocity :f point A as a motion variable simplifies the equations. Consider the posit position ion vector vector r.r. In In terms terms of the generalized coordinates, one can express r as r = r( q,1, 1). Differentiating this expression with respect to ime. we obtain . . . ,

r=

dr di

(9r

=

aq1

9r.

+

+

°?q2

ar. dq,1

9r

q,1

+ —

17.6.41

at

From Eq. [7.6.4], we can write the rate of change of r in terms of the quasi-

as

v= = v=r=

I cr

F?

•

q•,

which vk (k = 1,2,

. .

fl

'S

— [Yl/kZk)+ [Y17k'Zk) + >([Ylikuk >'([Y1 7k' Uk — .

k=1

., n)

j=I -.

=

>V

+ V1 V' Uk +

[7.6.51 17.6.51

k=I

and v' are called partial velocities and have the

I }'] yj

=

dl

k

v' =

Zk

J

+ — 9t

[7.6.61

Throughout this text, we will denote the index associated with partial velocities a superscript. In a similar fashion, we can express the angular velocities assowith a body or systems of bodies in terms of the partial angular velocities = as fl

(*)

+

[7.6.71

the angular velocities associated with the F frame in terms of the quasi-coordinates with the body frame, and show that they do not constitute a set of generalized Then find the partial angular velocities associated with

Eq. [7.5.7J has has aa singularity singularity when when the the second second Euler Euler angle angle reaches reaches aa certain certain value. value. One One can can avoid avoid this this singularity singularity

switching to a different set of Euler angles at that instant. Hence, the transformation between angular veloci'es ønd Euler is considered a valid one-to-one fransformation between the generalized velocities and the speeds.

Example

7.6

380

RIG[I) BODY KINEMATICS

CHAPTER

Solution will use a 3-1-3 Euler angle set. From Eqs. [7.5.15]. the angular velocities of the body and frame are

We

=

=

Of1

+ 4) sin Of2 + (4) cos 6 +

=

Of1

Of1 Of2 ++ 4) cos 6f1 + 4) sin 6f2

[a]

We define the generalized speeds as

Ui Ui = Wi =

U2 =

0

=

= W3 W3 =

COSO

± i/I

[b]

The angular velocities associated with the F frame then can be related to the generalized speeds as

wji=O=uI

w,2=4)SIflO=u2 w,2=4)sInO=u2

-

tanO

[ci

frame do do not not constitute As U3 U3isisnot notpresent presentininthe thedescription, description,the the angular velocities of the F frame As a complete set of generalized speeds. For the body angular velocity, using the generalized speeds defined in Eq. [hi we have =

'Ywkuk ++w' d =

[d] [dl

+w1f3 +&hf2±w3f3

so that the partial angular velocities are recognized as

o2=f2

w'=O

u2 = Note that had we selected our generalized speeds as u1 = 0. U2

[.1 u3

= iii, the partial

angular velocities would have the form

w2=sinOf2+cosOf3

7.7

to3=f3

td=O

If] In

EULER PARAMETERS

We studied Euler angles in the previous sections to relate the angular velocities to the rotation angles 0, i/i and their rates. As we saw in Section 7.5, these equations are highly nonlinear and have singularities, which makes it difficult to integrate them and to do any analytical as well as numerical work with them.

To alleviate these difficulties it is preferable to work with another set, called Euler para meters. These These parameters parameters increase increase the number of variables one deals with parameters. from three to four, but they eliminate the nonlinearities and many of the numerical problems. Because a set of three variables (4., 0. ijs) is being expressed in terms of four variables, the use of Euler parameters introduces a redundancy. This implies that there is no unique way of expressing the Euler parameters. and one can easily come up with

different sets of parameters. Other commonly used quantities include the CayleyKlein parameters. Rodrigues parameters. and quaternions. All these quantities are. essentially, different forms of the Euler parameters. A comparison of these quantities can be found in the text by Junkins and Turner. The mathematics of the Euler parameters was first introduced by Hamilton in 1843. The vector formulation of Euler parameters. which is used with quaternions, was developed by Oliver Heaviside.

ii

EULER PARAMETERS

a1

fl 03

I

a1

FIgure 7.15

Principal line

Euler parameters are inspired from Euler's theorem, which states that any roof a rigid body about a point can be accomplished by a single rotation by an angle (principal angle) about a line fixed in the body and passing through the cener of rotation (principal line). Euler parameters are in essence a characterization of this rotation in terms of the principal angle and direction cosines of the principal line. Let us describe the principal line by its direction cosines c1, c2, C3, ('3, as shown

m Fig 7.15. Denoting by coordinate axes, we have

02. 03 the. angles the principal line makes with the

=

= cosO,

[7.7.11

C3 =

The Euler parameters are defined as the four parameters e0. e1. e2, and e1 =

e() = COS

.fq:, .fq;,

e2

•fct

=

e3 such

that

= c3 sin

[7.7.2] We use the half angle so as to eliminate ambiguities associated with the rotation angle. We observe from Eq. [7.7.2] that =

[7.7.3]

1

We will relate the Euler parameters and their rates of change to the Euler angles. angular velocities, and direction cosines, and vice versa. That is, we will discuss the relationships Cjj

= f(eo. e1, e2,

i,

= f(eo,el,e2,e3) f(e01e1,e2,e3,w1,w2,w3) èJ = f(e0,e1,e2,e3,w1,w2,w3)

j

= 1,2,3

j = 1.2.3 1.2,3 / = 1,2,3,4

[7.7.41

We denote the unit vector along the principal line by n. Section 7.3 discussed that the direction cosine matrix [ci has an eigenvalue A = 1 and that c1, c2, and c3 are elements of the associated eigenvector. From the eigenvalue problem [c]T{c} = A{c} we

set A = I and find {c} = [c1 ('2 ('1T We now explore each of the relations in Eqs. [7.7.41 individually.

[7.7.5]

381

382

CHAP1IR

7.7.1

7. RIGID BODY

KINEMATICS

RELATING THE DIRECTION COSINES TO THE EULER PARAMETERS

Consider two coordinate framesA (ala2a3) and B (b1 b7b3), where b1b2b3 is obtained by rotating a1 a2a3 by an angle of (I) about the unit vector n. We use the notation in Section 7.3 and express the unit vector n in terms of its components in the coordinate frames as AnIal + 4n2a2 + An3a3 = 8n1b1 + Bfl2b2 +

[7.7.6J

= (i = 1, 2, 3). We = cos cos = relate the unit vectors in the A and B frames and the components of n in the A and B frames as considering Fig. Fig. 7. 7.15, 15, Naturally, considering

b1

a1

b2 = [c]'

a2 a3

b3

8n1

[7.7.13

= [cJ' flfl3

Let us consider the vectors a1 and b1 and think of an inverted cone with and with the unit vector n as the axis of the cone, as shown shown in in Fig. Fig. 7.16 angle Because frame B is obtained by rotating frame A around n, this rotation defines a circle of radius sin on a plane perpendicular to the vector n. The angle betweet a1 and n is the same as the angle between b1 and n, and the projections of a1 and b on n are identical. Mathematically. [7.7.$J n•a1 = n•b1 = cos01 We denote by Q the point at which the vector n intersects the circle defined by aand b1. The vector connecting points 0 and Q is defined as x1, and it can be expressed as x1 = cos 0 in, as shown in Fig. 7.17. We introduce the vectors Yi and z1 such thz Yi is along the line connecting point Q to the tip of a1, and z1 is perpendicular

such that its tip coincides with the end of b1, as shown in Fig. 7.18. We can write

b1=x1+y1+z1 bi=Xi+yi+Zt

[7.7.9J [7.7.9J

fl

P.

ni

Q

Yo

sinø1

P

b1

cosO1 xi

a'

0

Figure 7.16

0

Figur. 7.1 7

Side view

77

EuI1ER PARAMETERS

Yo

Q Yi

P

Figure 7.18

Top view

The objective is to express b1 in terms of the components of the A frame. To this, we consider Fig. 7.18 and the circular arc generated by rotating a1 b1. Introducing the notation Yo Yo = sin 01, we can write

Yi = p0 =

lYi I I

= Yo COS (1) — xjf = lbi —xii

sin (1) = fz1 = v0 sin zi=fzjl= -xd xiI

[7.7.10] 17.7.101

—

from which we have

[7.7.11] [7.7.111 — cos61 —x1)cos(1) x1)cos(1) = Yi = (ai — The vector vector z1 z1 is is perpendicular to both n and Yi. Denoting the unit vector along by

we can express it as

nXa1 nXy1 e-= nxy1f = (nXaj (nXaiI

[7.7.12] 17.7.121 as sin(I). we express z1 as

a1 = sin Oi = V(). and Izil = sin fl XX a1 However, since In

nXa1

z1 = fz1lefz1Ie- = z1

sin

=

17.7.13] 17.7.131

X a1

and [7.7.13 [7.7.13 we have Eqs. 17.7.9]. 17.7.9].[7.7. [7.7.111, III, and Now combining Eqs.

b1 b1 =

X1 X1

c1n + + Yt + z1 z1 = c1n Yt +

— c1 — c1

coscFn +

X a1 X

17.7.141

—cos 11 —cos (1)1) = 2 Using formulas cos cos = 2 Using the half-angle formulas and Eq. Eq. [7.7.6]. [7.7.6]. we can can write write the above equation as 2 md sin 1

b1

+

c2a2 + a1 + c2a2

= (2c1

•(I) .(I)

+

—

l)ai (7.7.15] 17.7.151

a1 + c3a3) XX a1

Introducing the definition of the Euler parameters given in Eq. [7.7.2] into Eq. [7.7. [7.7.151, 151, we we obtain obtain the the relation between the vector b1 of the rotated axes and the unit vectors of the original coordinate frame. in terms of the Euler parameters as b1

=

I+

— I

(2e1e7 + + (2e1e7

+

—

2e0e2)a3

[7.7.16]

383

384

CHAPTER

RIcID BoDy KINEMATICS

One can repeat the procedure used above to relate the vectors b2 and b3 to a. a2, and a3 by the Euler parameters. The results can be expressed in matrix form as b1

e1e2

a1

=

—

1) a2

a2 ++ 22

e1e2

a3

e1 e-4 e-4

0

e1e3

a1

e7e3

a7 a7 — — 2eo 2eo

e2

0

—e1

a2

a

e-)

17.7.17] Introducing the column vector {e} = [ej

we can write Eq. [7.7.17] as

e2

a1

=

b.

—

1

)[1] + 2{eHe}'

—

2e0[ë]) a2 2e0[eI)

[7.7.1SJ 17.7.1S1

a3

b3

The first two terms on the right side of Eq. [7.7.181 [7.7.18] are symmetric matrices. while the third term is a skew-symmetric matrix. A comparison of Eq. [7.7.18] anc Eq. [7.7.7] indicates that the direction cosine matrix between the coordinate frames A and B. in terms of the Euler parameters, has the form

-

[cIT = [R] =

+

=

—

1)[l1 + 2{eHe}T 1)[l]

- 2e0[ë])

—

—

2(e1e3 + e0e2) e0e2)

+

—

2(e2e3

—

2

—

e0e1)

+ eoej) 2(e2el+eoeJ) 2(e2el + — —

[7.7.191 17.7.191

The elements of the direction cosine matrix are quadratic expressions in term.s

of the Euler parameters. When using Euler angles. the direction cosine matrix [ci. given in Eq. 17.5.31. [7.5.31. contains contains trigonometric trigonometric and other nonlinear terms, and it to singularities. There are no singularities in [cJ when it is expressed in terms of Euler parameters. For any possible orientation of two coordinate frames there exists. and it is possible to find, associated Euler parameters. By contrast, every Euler angle transformation sequence has a nonlinearity at which point one cannot determine values of the Euler angles.

7.7.2

RELATING THE EULER PARAMETERS TO ANGULAR VEL0cn1Es

In Section 7.4 we derived the relationship between the angular velocities of the coordinate frame and the direction cosines in Eq. [7.4. 13], 13]. which we repeat here as

=

[7.7.201 17.7.201

Equation [7.7.191 gives the direction cosine matrix in terms of the Euler parameters. We use those results to obtain the angular velocities in terms of the Euler parameters. For example. let us calculate the component of the angular velocity in the h1 direction. w From Eq. 1.7.4.13.1 v is the the 3.2 3.2element elementofof iR] and andititcan canbe beobtained obtained multiplying the third row of [cIT with with the second column of [è]. Mathematically, I

= >.Cj3Cj2 >Cj3Cj2 i—I

17.7.211

7.7

385

EULER

Substituting the values for cjj from Eq. [7.7.191 we obtain cU from

= 4(e1e3 4(e1e3 ++ e0e2)(e1è2 ++ e2è1 — — eoèi — — e0è1 + 4(e7e3 + 4(e7e3 — — eOe1)(e0è() + e7è2 eOe1)(e0è0 — e1è1 + +

— —

2

ef — eI —

e2

.

+

e3è0) — —

e3è3)

+ e-4e2 + e3è2 ++ eoej eoej ±±

Using the identity of Eq. [7.7.3] and its derivative, that is, eoèO++ e1è1 e1è1 ++ e7è7 e7è2 + + e3è3 e3è3 = 0, Eq. [7.7.22] reduces to eoèO

= =

EA&BI

=

2[—e1è0

+ e0è1 + e3è7

+

—

e2è31

[7.7.221 [7.7.221 +

+

=

1,

[7.7.23]

Similar expressions can be developed for (02 and w3. Introducing the vectors e3]T, we e3]T, v2 w1 W2 = [0 w1 and {e} = [e0 e1 e1 we can express them in matrix form as 0

e0

=

(02

2

(03

e2

e3

è0

e0 e0

e3 e3

—e2

e1

—ei

e0

ei

e2

e,

—e1

e0

è3

—ei

e2

[7.7.24] [7.7.24]

3r, 3r, in in compact compact form

2[E*J{è} = 2[E*J{ê}

[7.7.25]

-there the notation is obvious. One can show that the coefficient matrix [E*1 is or[E*1_i = IE*]T. Using this thonormal, that is. its inverse is equal to its transpose, [E*1_I

property one can write {e}

and

0.5[E*]_I{w*} = 0.5[E*lT{w*}

[7.7.26] 17.7.26]

can rearrange this as 0 e2

= 0.5

—WI —w1

(02

—W2 W2

0

(03

W3

0

(03

e3

eO

—w2 WI

e1

0

e3

e2

[7.7.27] 17.7.27]

The The equations relating the Euler parameters to their derivatives are linear, and

they contain no singularities, making them easier to deal with both analytically and If Eq. [7.7.271 [7.7.27] is is compared compared with with Eqs. Eqs. [7.5.9], the derivatives of the Euler angles, the resulting simplification becomes clear. We also observe that the coefficient matrix in Eq. [7.7.27] is skew symmetric. Noting that the first row of Eq. [7.7.24] is essentially 0 = 0, we can generate a new matrix matrixIE.I IE.I of of order order 33 xx 44 by by eliminating eliminating the the first first row of I 1. writing e0

[E] =

—e2

e3

—e3 e7

—e2 e1

—e1

17.7.28] 17.7.281

e0

Thus, Eq. [7.7.25] can be expressed as {w} = 2[EJ{è}

17.7.29]

386

7•

CHAPTIR 7• RIGID RIGIDBODY BODY KINEMATICS KINEMATICS

The matrix [Eli has a few interesting properties. One can show that 2[EJ[E]T [&] = 2[E1[EIT

[E][E1T = [1] [EIIE1T

However, note that [EIT[E1

=

{O}

17.7.30a,b,cJ

[fl. Indeed, one can show that IEITIE]

[1J

—

{e}{e}T

17.7.311

The relationships discussed so far have been been for for the the Euler Euler parameters that describe the transformation from one coordinate system to another. We next explore the relations among the Euler parameters when there is more than one transformation We begin with a frame A and transform it to arrive at frame B. We denote the of Euler parameters that describe this transformation by {e}. Let us assume that the transformation from A to B is carried out in two steps: first a transformation from A to frame A' with associated Euler parameters {e'}, and then a second transformation from A' to frame B, with associated Euler parameters {e"}. We wish to relate {e} {

e'} and {e"}.

Denoting by J the the counterpart of the IE*i matrix in Eq. [7.7.24] for the Euler parameters {e'}. we can show that the transformed Euler parameters obey the relationship {e} = [El*]u{ehl}

17.7.321 This relation can be derived by taking the coordinate transformation matrices between the A, A', and B frames.

7.7.3

RELATING TILE EULER PARAMETERS TO ThE DIRECTION

This process is in essence the reverse of the process in in Sec. Sec. 7.7.1. 7.7.1. As As itit is is an an inverinversion, the task can be accomplished in a number of ways. One way is to make use of the property of the direction cosine matrix and to note that the direction cosines the principal line are elements of the eigenvector associated with the eigenvalue 1 The procedure to follow is this: Obtain the eigenvector matrix corresponding to the eigenvalue 1, and calculate the rotation angle 1 from from Eq. Eq. [7.3.171. [7.3.171. Another and more straightforward way is to use Eqs. [7.7.2] and [7.7.19], which yield C21 — C12

= 4eoei =

C13 —

= 4e0e2 4e0e2 =

C32

C11

+

C72

— C23 = 4e0e1

+ c33 =

—

=

ef

—

cos

sin

sine

=

= 4c1

cos

—

cPcI? sin

=

= 4 cos2

2c1

—

I

=

—

1

17.7.3 31

These equations can be manipulated in a number of ways to yield expressions for and The goal in this algebraic exercise is to come up with relations

7.7

EULER PARAMETERS

that do not have singularity problems. For example, we can obtain an expression for sin directly from the first of Eq. [7.7.33] in the form

sinct = sirict

C21 —

Cj7

17.7.34]

2c

which which has a singularity singularitywhen whenC3 C3 = 0. On the other hand, multiplying the first three of Eqs. [7.7.33] with c3, c3, c2, c2, and and c1, c1, respectively, respectively, and adding them yields [C3(C21 — C12) + C2(C13

C31) + C1(C3, Ci (C3,

—

—

c,3)] =

+

+

= 2sincl

[7.7.351 which exhibits no singularities. Another suitable choice is to use the half-angle formula cos = 2 cos2(C1/2) — with the last of Eq. [7.7.33J, which yields 1

C11

+ C22 C22++ c33 c33 = 4cos2

—

=

+

[7.7.361

C33 C33 — — 1)

17.7.37] 17.7.371

I

1

This is identical to Eq. [7.3. 17] and leads to

cos:F =

+ C22 C2) +

A commonly used set of expressions, provided e0 e0 is is not equal to or close to zero, is derived by manipulating the diagonal elements of Eq. [7.7.19] as = C11 = C11

+

+

4e5 = —c11 +

C33

+

=

I

— C33 +

I

— c22 — c22 — c33

+I

= —c11 —c11— — C22 C22 + C33 + 1

17.7.381

Usually, one first finds from Eqs. [7.7.38] the Euler parameter with the largest

magnitude. This larger quantity is used together with the off-diagonal elements of' the direction cosine matrix in Eq. [7.7.19] to solve for the Euler parameters. For example, if e0 e0 isis the largest largest in in magnitude, magnitude, one uses uses the the value value of of e0 e0 from the first of' [7.7.38] and finds the remaining Euler parameters using Eqs. 17.7.33] as e0 = e0

1

/

2

+ c22 C33 + + C22 ++C33

— C13 C13 —

4e0

7.7.4

-

I

e1 e1

=

e3 = e3

C32—c23 C32—C23

4e0 C,1 — — C12 C12

4e0

[7.7.39] 17.7.391

RELATING THE EULER PARAMETERS TO THE EULER ANGLES AND VICE VERSA

This task can be accomplished by generating the direction cosine matrix for a specific Euler angle transformation, and using the relations given in Eqs. [7.7.39] for relating the Euler parameters to the direction cosines. The results, of course, vary from one of transformation to another. We dispense with the algebra and give the results

387

388

CHAPTIR

7S

RIGID BODY KINEMATICS

for a 3-1-3 Euler angle rotation sequence with the transformation angles

0, i/i.

which are

e0 = cos

4+fs

cos

(1

e1

= cos

e3

= sin

sin —

-

= sin

sin

2

.4+i/i 2

0 COS

17.7.40]

2

Results for the 3-2-3 and 3-2-1 sequences are given in Table 7.1 at the end of

the chapter. Once the Euler parameters are found, one can use Eq. [7.7,19] [7,7,19] to to calcalculate the direction cosine matrix. The reverse problem determination of the Euler angles from the Euler parameters—is more complicated. complicated, lt it can be accomplished in a number of ways. In one way, we calculate the Euler parameters and the direction cosine matrix in terms of the Euler parameters. Then, we make use of the algebraic relations in the direction cosine matrix in terms of the Euler angles. This procedure is usually tedious. Another approach for finding the Euler angles from the the Euler Euler parameters parameters is is b) direct use of Eq. [7.7.40] or its counterpart for other sequences. For a 3-1-3 sequence we manipulate the Euler parameters as e0

=

2

tan e1

- 'I'

[7.7.41]

2

After taking the inverse tangents of the above terms and manipulating, we obtain -

—

\eoJ

—

\eiJ

\eoJ

\ei/

determine the exact values of these angles, that is, the quadrants they lie in, we can use Eqs. [7.7.401, or we can compare the results with the entries of the direction cosine matrix. The angle 0 can be found by noting that To

2

e1

so

+ e22 = sin

[7.7.43] 17.7.431

2sin'

[7.7.44]

that 0 =

The same approach is used for other Euler angle sequences.

7.7.5

OTHER CONSIDERATIONS OTHER CONSIDERATIONS

Yet another advantage of using Euler parameters arises when conducting numerical computations. Because Because the the Euler Euler parameters parametersare arerelated relatedtotoeach eachother otherby byEq. Eq.[7.7.3 [7.7.3]. 1. Eq. [7.7.31 can be used as a constant of the kinematic equations of motion, and it can be computed at each step of the numerical integration to check for accuracy and numerical stability. Other constants associated with this analysis are the diagonal elements of Eq. [7.7.201, which should be zero, by virtue of the skew symmetry of the (i = 1, 2, 3). angular velocity matrix. The values of the diagonal elements,

7.7 can

EULER PARAMETERS

be verified using =

()

k =

1.

[7.7.451

2, 3

i—I

These identities can also he checked to verify the accuracy of the numerical computation.

Disadvantages associated with using Euler parameters include the need to use four parameters and four differential equations instead of three for the case of Euler angles. However, the linearity of the resulting equations and lack of singularities far outweigh the complexities introduced by the additional differential equation. Another disadvantage is that Euler parameters do not lend themselves to a simple physical interpretation as Euler angles do. But here, too, one can always calculate the values of the Euler angles from the Euler parameters at every instant of the motion and seek a physical interpretation or visualization. We conclude the discussion of Euler parameters with this important generalization and an additional definition of angular velocity. Assume we have a vector r and we are rotating r by an angle about an axis that passes through the origin. Let ii be the unit vector in the direction of the axis of rotation and r' be the rotated vector. We can write the relationship between r and r' by substituting r for a1 and r' for b1 in Eq. [7.7.14], which yields [7.7.46] + sin (I)n X r r)n + cos r' = (1 — cos This recognize that c1 = cos in Eq. [7.7.14] is equivalent to n• equation is a very useful general relation for determining the tr ectory of a vector that is being rotated. A special application of Eq. [7.7.461 is for small rotations over time intervals for finding the rate at which r changes. Indeed, as (1) approaches where we

zero, we have I

r'

sin ci) —'

—

r

—b

dr

[7.7.46], yields which, when substituted into Eq. [7.7.461,

[7.7.47]

dr = d1(n X r)

that particular by dt yields the angular velocity expression

Because the infinitesimal rotation is taken about a single axis at

instant, we recognize that division of in the form

di

to = wn = —n

=w

di

[7.7.48a,bJ

Equation [7.7.48b1 is yet another definition of the angular velocity vector. Dividing Eq. [7.7.47] by di and using Eq. 17.7.48b1 we obtain

dr — = to w xr dt

[7.7.491

We have hence showed another way at arriving at the angular velocity of a referframe and the relation describing how the angular velocity of a reference frame iffects the rate of change of a vector fixed in that reference frame.

389

__________

390

CHAPT1R 7.

Example

Consider the transformation sequence in Example 7.1. Obtain the associated Euler parameters. eters. and substitute them them into into Eq. Eq. [7.7.19] [7.7.19] to calculate the values of [Ri to toconfirm confirm your

7.7

BODY KINEMA11('S

results.

Solution From Example 7. 1. the transformation matrix is

[RI =

0.6124

0.3536

—0.7071

—0.500()

0.8660

0.6124

0.3536

0 0.7071

The rotation angle was calculated as line were found to he

[al

= 53.64° and the direction cosines of the principal Cl = ci 0.8192 192 = 0.8

= —0.2195

C3 = 0.5299

[bJ

The sine and cosine of halt' the principal angle arc sin

= 0.4512

= 0.8924

cos

[ci

The Euler paranieters for this problem become

e0 =

= 0.8924

cos

=

= c1 c1 sin

(I)

= 0.8192(0.4512) = 0.3696

= —0.2195(0.45 —0.2195(0.4512) 12) = —0.09904

=

= 0.5299(0.4512)

0.2391

We next calculate the entries of the transformation matrix. For example, for R1 and we have, from Eq. [7.7.19]. +

=

R11

=

—

+

0.5) = 2(0.89242 + 0.099042

= 2(0.3696)(0.2391)

—

—

0.5) = 0.6124

2(0.8924)(0.09904) = 0

[.1

One can show that the rest of of the the entries entries of of [RJ [RJ can be found using Eq. [7.7.19] as as well.

Now we will calculate the Euler parameters without using the eigensolution of [RI. To this end, it is preferable to find the largest of the Euler parameters and use calculations that involve division by that parameter. We can accomplish this by looking at the sines and cosines of half the principal angle. From Example 7.1 or using Eq. [7.7.35J we determine that 1 = 53.64 In Eq. [ci we determined the sine and cosine of the half half angle. angle. From that information, conclude that e0 = cos((1/2) = 0.8924 0.8924 is the largest of the Euler parameters. We find the remaining Euler parameters using Eqs. [7.7.39] as

e1 =

C37 C32 —

=

4eo e2

R73 —

4e()

4e(,

=

C21 — Cl2

=

— 0.3536) W—

4 X 0.8924

0.6124 0.6 124 + 0.707 0.7071 1

— ('13 —

=

-- =

x 0.8924 0.8924 4 X

= —0.09906

= (.).3696

0.3536 + 0.5 = 0.2391 4 X 0.8924

En

Note that there is a slight difference in the fiurth fiurth significant significant figure figure of of some some of of the the results

in Eqs. [dj and Eqs. If], due to roundoff error.

7.8

7.8

CONSTRAiNED MOTION OF RIGID BODIES, INTERCONNEC'flONS

CONSTRAINED MOTION OF RIGID BoDIES, INTERCONNECTIONS

Constrained motion takes place because of contact between one or more points on c'ne body with a point or a surface on another body. Such contact between two bodies take place in a number ways: Motion restriction or transmission through sliding and interconnections. Impact. Rolling.

In this section, we analyze constrained motion from a physical perspective and the interconnections that obtain such motion. Chapter 8 will analyze impact of bodies, as it cannot be treated using kinematics alone. And Section 7.9 will consider rolling. The number of degrees of freedom of a system is equal to the number of coordinates minus the number of constraints. Constraints generated by contact of bodies reduce the number of degrees of freedom. To describe the general three-dimensional motion of a rigid body. one needs six degrees of freedom. Plane motion is a three degree of freedom problem. A set of two or more bodies in contact with each other for the purpose of transmission of motion from one of the bodies to another is called a mechanism. Compoof a mechanism are generally referred to as links.

7.8.1

ROTATIONAL CONTACT, JOINTS

A joint basically prevents the bodies it connects to translate with respect to each ether, while permitting some type of rotational motion between the two bodies. The common joint is the pin joint, also known as a revo/ute joint, shown in Fig. 19. The contact is over the external surface of the pin. The angular velocity of the second body, as viewed from the first body, is denoted by w2 and is along the axis of the pin. We express as (02

W211

/ fl

Axis of rotation

Figur. 7.19

A revolute loint

[7.8.11

391

392

CHAPTII 7. RIGID BoDY

KINEMATICS

where n is the unit vector along the axis of the pin. The constraint relation can

expressed as

[7.8.2J

(R)2Xfl =0

which represents two rotational constraint equations.2 Because the joint does not permit translational motion at all, the total number of constraints imposed is five.. Hence, a rigid body connected to a system of rigid bodies by a revolute joint increases

the degrees of freedom of the resulting system by one. Denoting the angular velocity of the first body by w1, we can express the total angular velocity of the second body as the sum of the two angular velocities, thus

(0 =

+

[7.8.3J

= (01 + (.02fl

Differentiation of Eq. [7.8.3] yields the angular acceleration expression as = ó1 ó1 +th2n+o1 XW2fl

[7.8.41 [7.8.41

For multilink bodies in plane motion, one can use Gruehier equation to determine terrnine the number of degrees of freedom, freedom. The formula states that the number of degrees of freedom p is related to the number of active links and number of joints b? s s'

p = 3k—2f where k is the number of active links and is the number of joints. For example. it Fig. 4.41, 4.41, there there are are five fiveactive activelinks linksand andsix sixjoints, joints,leading leadingice the linkage shown in Fig. the conclusion that there are three degrees of freedom. Gruebler's equation can extended to different types of joints as well. The interested reader is referred to on kinematics or mechanism theory. It should be noted that Gruebler's equation does links. One can find geometries where not take into consideration the lengths of the links. Gruebler's equation states that there are zero degrees of freedom and the mechanism can move. Another commonly used joint is a hail ha!! and and socket socket joint, joint, also also known known as as aa globular globular or spherical pail: The joint is shown in Fig. 7.20. Ball and socket joints do not place a restriction on the angular velocity of the second body as viewed from the first. angular velocity W2 can have components in any direction. Such joints impose constraints that prevent translational motion at the joint.

Figure 7.20

A ball and socket joint

2We deduce that Eq. [7.8.2] mathematically mathematically represents represents two two constraint constraint equations equations by byconsidering consideringSection Section772 scalar = TO), which represents three scatar Equation [7.8.2] can be written in column vector notation as is a singular matrix with rank 2, only two ol the equations are independent. Because

78

Q)NSTRAINED MoTIoN OF RIGID BODIES, INTERCONNECTIONS

An important use of joints is in power transmission. In machine dynamics it is often necessary to transfer the rotational motion of a shaft onto another shaft rotating in a different direction. A common mechanism used to accomplish this is a univer-

sal joint. There are several kinds of universal joints available, such as the Cardan, Rzeppa, Weiss, and Devos joints. All these joints accomplish the task of power transmission in different ways; each is useful lhr specific engineering applications. The Cardan joint shown in Fig. 7.21 is one of the common and low-cost universal joints. The joint consists of two shafts attached to flrks, known as yokes, linked to each other by a cross. The two shafts are called the input and output shafts. The relative position of the cross with respect to the yokes can change, as the cross can rotate inside each of the yokes, making it possible to align the two shafts in almost any direction. direction, This gives the universal joint its tremendous versatility and its name, it can operate in transmission problems where the shafts connected to the yokes are not aligned, vibrate, or change orientation. A universal joint is a single degree of freedom mechanism, because the angular velocity of one of the shafts determines the angular velocity of the other. To explore the relationship between the two angular velocities we split the universal joint into two parts, A and B, as shown in Fig. 7.22. We select the coordinate axes such that both shafts lie on the xy plane. The input shaft, shaft B, is aligned with the x direction, and its angular velocity is —Oi. —61, with 0 measured from the z axis. We rotate the xy axes about the z direction by an angle of to get the x'v' axes and place the output shaft A along the negative x' axis. The angular velocity of the output shaft is —4,1', with 4, measured from the z axis as well. We denote the unit vectors along the sides of the cross as a and b, respectively, and express them as b = sinflj + cos0k sinflj+cos0k

a=

sin 4,j' 4)j' + cos 4)k

—

4, cos sin 4, cos /3j /3j ++ cos cos 4)k 4)k 17.8.51 4) sin /31 + sin 4)

It follows that the two unit vectors must be perpendicular to each other. The dot product between the two gives the constraint relation between between the the angles angles 00 and and 4,. 4). b

0 x

Input shaft a

Output shaft

FIgure 7.21

A Cardan joint

393

394

CHAPTER

7• RICED

BODY KINEMATICS

4-

V

V

I

x a

(h)

(a)

Figur. 7.22 We have a•b

= sin 0 sin

cos /3 + cos 6 cos

=

0

17.$.'1

It is of interest to compare the angular velocities of the input and output shafts We express 4) in terms of 0 and 0. We rewrite Eq. [7.8.6] with the 4) and 0 terms opposite sides of the equation as tan4)

—

sin4)

—

—

cOSc/

cos0 sin0cos/3

[7.8.71

which, when differentiated, yields 0

= COS- 4) COS-

sin- 0 cos /3 sin j3

[7.8.11

To eliminate eliminate 4) 4) from from this equation we make use useof ofthe theidentity identitycos2 cos2 4) 4) = 11(1 — Substituting the value of tan 4) from Eq. Eq. [7.8.7] into this identity and usin& it with Eq. [7.8.8], we obtain the relation between the angular velocities as 4)

=

cosf3

0 .1 1 ——sin6sin/3 sin)sin/3

[7.8.9)

angular velocities of the input and output shafts are not linearly related t: each other, except when /3 = 0. in which case the joints are aligned. The Cardat joint is not what is called a constant velocity joint, making it impractical to use it applications such as the steering of front wheel drive automobiles, as it would lead slipping tires and large stresses on power transmission components. In the case, when /3 = 900. the two shafts are perpendicular and no motion can be transmitted. This is known as gimbal lock. However, if two Cardan joints are used to joit two parallel shafts, the angular velocities of the two parallel shafts will be the it should also be noted that there are several constant velocity joints available such The

7.5

395

O)NSTRIIINED MOTION OF RIGID BODIES, CONSTRAINED

Many of these joints provide a near constant velocity ones discussed earlier. Many re'ationship between between the the input input and and output outputshafts shaftsfor br a specific range of operation. operation. the

7.8.2

ThANSLATIONAL MOTION AND

CONTACT

Here one part of a body slides over another body. The simplest form of sliding is to have a point, a line, or a surface on a body sliding on a surface. This surface is the contact plane. The associated constraint is the same as the very lirsi constraint relations we studied in Chapter 4. The velocity of the contact point does not have a component normal to the plane of contact, and the constraint can be written as

Vp'fl =

17.8.101

0

where n is the unit vector in the normal direction at the the point point of of contact. contact. If If more than one line on the rigid body slides on a surface, then the angular velocity of the body only have a component perpendicular to the plane. so that w is parallel to n. The resulting two constraints can be described by

wXn=O

17.8.111

A special case of sliding on a surface is sliding in a guide or in a slot (Fig. 7.23).

The joint formed by the sliding components is referred to as a prismatic joint. In this case translational motion is possible in only one direction, and there are two constraints on the translational motion. Denoting the unit vector in the direction of the guide by h, we can write the constraint as

[7.8.121 where we note that Eq. [7.8.121 represents two constraint relations. Further, no rota-

tion is permitted, so a prismatic joint represents a total of five constraint equations. A prismatic joint for translational motion is basically the equivalent of a revolute joint for rotational motion. In another form of sliding contact, a collar slides on a bar while turning around it. The simplest case here is when the collar is not attached to another body. as shown rn Fig. 7.24. 7.24. Such aa connection is referred to as a slide,: Denoting the unit vector along the rod by h, we note that the collar can only have a velocity in the direction translational motion is Eq. so that the constraint associated with the translational of h, h. Vp = 7.8.12]. :7.8.12].

I,

I' P

FIgure 7.23 FIgure 7.23

(i)

A prismatic joint

Figure 7.24

VI,

A slider

396

CHAPTER 7• RLGm BODY BODY KINEMATICS KINEMAncs

In addition, the collar can only have an angular velocity about the axis of the rod, = Wcouarh, so that the constraint associated with the rotational motion is

=0

[7.8.131

There are a total of four constraints restricting the motion, so the slider has two degrees of freedom: one translational and one rotational.

7.8.3

COMBINED SLIDING AND ROTATION

In this case, a body is attached to a slider by means of a pin, tbrk, or universal joinL

as depicted in Fig. 7.25. These two elements in contact (rod and slider) are also referred to as a cylindrical pail: The collar is sliding on and turning around a guide bar. Consider the coordinate frame XVZ where the x axis is aligned with the collar. = The coordinate axes The angular velocity of the collar becomes It follows .vv'z' are obtained by rotating the xyz axes about the x axis by an angle .vv'z' that Wc(,jlaf = The xv' plane defines the plane of the collar to which the pin joint is attached. The body attached to the collar can have an angular velocity in the direction

perpendicular to the plane of the collar, in the z' direction. We can then write WcolJarI' + + WcolJarI'

[7.8.141 Thus, the angular velocity of the rod cannot have a component in the y' v' direction. The constraint for lbr the total angular velocity can then be written as 'I [7.8.15J =0 The total number of constraints on the rod becomes three, two for the translational motion and one for the rotational. The angular acceleration of the rod can be obtained by differentiating Eq. [7.8.141, W10d —

Grod

+ Ok

Wrod/coijark WcollarI + WrodJcoljarl(

I,

acollarl arodJcoljark ++ WcoIIarl' acollarl ++ arodJcoljark WcoIIarl' X (Vrocljcoilark' =

+ Ok'

[7.8.161 yg

0=

x, x' '0coltar =

FIgure 7.25

A cylindrical pair

7.8

397

CONSTRAINED MOTION OF RIGID BODIES, INTERCONNECTIONS

When the collar is attached to the rod by a ball and socket joint, there there is is no restriction on the angular motion of' the rod with respect to the collar. The rod then has four degrees of freedom: one translational and three rotational. As an illustration of the constraint relations for combined sliding and rotation, first consider plane motion and the the rod AB where where points points A A and B rest on collars that slide on bars perpendicular to each other, as shown in Fig. 7.26. Each collar introduces one constraint on the motion, resulting in a one degree of freedom system. The two constraint equations are

VBJ

0

17.8.171

These constraint relations can be expressed in different forms as well. Also, the rotational constraint relations associated with the pin did not not come into the picture, because the rotational motion of the rod is restricted to be perpendicular to the plane of motion. We next extend the discussion to the three-dimensional case, where we consider a rod AB in which points A and B are attached to collars and the collars slide on perpendicular guide bars (Fig. 7.27). However, in this case the guide bars are not on the same plane. Let us analyze the attachments of the points to the collars. If both points A and B are attached to the collars with pin joints, we end up with six constraint equations, three tbr dons, for each pin joint. It follows that the resulting system has zero degrees of freedom. No motion of the rod AB is possible. When one of the joints is a ball and socket joint, say, the joint at A, the number of constraints is reduced to five, two translational constraints for joint A and three for joint B. The system has one degree of freedom. If the motion of any point on rod AB is specified, one can determine the characteristics of the motion of the rod, as well as the collars. In order to be able to do this we have to generate an expression for the angular velocity of the rod. It is preferable to work with collar B, as we will make use of the properties of' the angular velocity of the rod with respect to the collar. From V

A

V

1'

b A

V

B

0

B

x a

z

Figure

7.26

Guide bars on same plane

FIgure

7.27

Guide bars not on same plane

398

7•

CHAPTUR CHAPTUR 7• RIGID Bony KINEMATICS

Eq. [7.8.141 [7.8.141 we can write

=

WcoIIar

[7.8.1 8]

+

of' the the collar collar B is The angular velocity of' t0col}ar t0collar

[7.8.191

=

The angular velocity of the rod as viewed from the pin joint is perpendicular to

the plane defined by rod A B and the guide bar. The unit vector perpendicular to the plane of symmetry oF collar B can be obtained by taking the cross product of any two vectors along rod AB and the guide bar. For example, taking the vectors as rB/n and r810. we obtain the unit vector, denoted by e, as ±

e

r4/fl X rB/o

[7.8.20]

X

The sign of the unit vector can be chosen by convenience. In the xy'z' coordinate

system (Fig. 7.27) the unit vector in the z' direction. k'. can be found using the relation

k' =

IX

[7.8.21]

•

I' X

so that

[7.8.22]

=

Equation [7.8.18] can also be analyzed from an Euler angle point of view. Us-

ing a 1-3-1 coordinate transformation (similar to 3-1-3 but with x and z changing places, and we are starting with coordinates xvz and going into x'y'z'), the angular velocity of the collar becomes the precession with the angle 4 and nutation with the angle 0, such that U = Because of the constraints imposed on the rod, the precession and nutation rates are related to each other. There is no spin. When both the joints are ball and socket joints, the system has two degrees of freedom, with the spin of the rod about its own axis constituting the second degree of freedom.

Example

7.8

rod AB in Fig. 7.27 is constrained to move between the horizontal and vertical guides. Given that the velocity of point point A A is is 22 rn/s rn/s downward. downward. and and aa = 6 m. m. bb = 3 m. c = 4 m, find the velocity of point B and the angular velocity of the rod.

The

Solution We

write the relative velocity expression for points A and B as

= V4 V4 +

[a] I.]

X rB/A

where VA =

—

2j rn/s

V8 =

rn/s

rB/A = 6i

—

3j + 4k m

[bi

7.8

CONSTRAINED MOTION OF RICH) BODIES, INTERCONNECTIONS

With the unit vector i describing the direction along the collar B. we use Eq. [7.8.2 1] to With find k', thus

k'=

IXFA/B iXrA/B

3k+4j

=

.•

IIXrAJBI

[ci

5

so that the angular velocity of the rod can be expressed as =

3k-i-4j

WcojIarI + WcojIarl

+ (L)rod/collark (t)roiJ/collark

[di Idi

s

Comparing the elements of k' and Fig. 7.27 we conclude that f has has aa negative negative value (clockwise rotation) for this problem. So does the angular velocity of the collar. We introduce Eqs. {bJ and [dJ into Eq. [al, which yields = —2j —2j +

+ 4j)] 4j)]XX(61 (61— — 3j + 4k)

+

—2j + Wcoiiar(3k

—

4j) + + Wrod/coIIar Wrod/coIIar

+ 91 91 — — 24k + 161) 161)

[.1

Equating the components of Eq. Ic] in the x, V. and z directions, we have a set of three equations and three unknowns WCOjIar. in the form WCOjIar. vrocjlcol!ar, vrocjlcollar,and and i components

0=

j components k components

If]

= —2 2

0 =

Ff1

[g]

+

—

24 3WcolIar

[hi Ihi

—

Solving Eqs. [f]—[h], we obtain Wrod/coflar = Wrod/coflar

=

radls

—

rad/s

=

1

mIs

[ii [i]

Note that we have three equations for the three unknowns. even though the system has a single degree of freedom. Next, consider finding the angular velocity of the collar A. To accomplish this, we write the angular velocity of the rod in terms of the angular velocity of the collar B as (R)rod = (R)rod

+ + Wrod/cc,IIarfl

+ 4j) =

(—81

+ 4j + 3k) rad/s

[I] fi]

The angular velocity can also he expressed by noting that the angular velocity of collar A is in the y direction + Wrod = WcolIarA WcolIarA +

Wcoflafl4j + +

[8j + (4 —

2SWcolIar4)J 2SWcoiiarn)J

+ 3k] rad/s

[kI 1k]

Use of the relative velocity relation will not lend any more information, so we need to look at the nature of the angular velocity. Previously, we used a 1-3-1 Euler angle transformation to quantify the angular velocity and said that there is no spin associated with an axis along the length of the rod. When visualizing the angular motion with respect to collar A or collar B, we can then quantify this as the constraint and (*)rod/collarB not having a component

399

400

CHAPTU

7• RIGID BoDY KmwMA'ncs

along the axis of rod AB. We can then write • rB/A

= 0

Wrod/collarB • rB/A

=

0

[1]

The second of these equations is automatically satisfied. The first equation can be solved for the angular velocity of the collar. Carrying out the calculations, we obtain [—81 + (4

—

25WcolIarA)J + +

3kJ .. [61 3kJ

—

3j

+ 4kj = = =

7.9

16

—48

—

12

+

+ 12 =

0

[.j

rad/s

ROLLING

An interesting case of constrained motion of rigid bodies is rolling. A body can roll over a fixed surface or over another body. For rolling to take place between twc' bodies, a continuous sequence of points on one of the bodies must be in continuous contact with a continuous sequence of points on the other body. A necessary condition for the continuity is that the contacting bodies must have smooth contours, so that the radii of curvature exist at each contact point on both bodies.3 The contact that takes place is point contact (for thin disks and spheres) or line contact (for and thick disks, disks. such as rigid wheels). The contour of a body is considered smooth enough to permit rolling if the peaks and dips on its surface are exceedingly small with respect to the overall dimensions of the body. The motion of bodies with sharp edges or with rough surfaces cannot be classified as rolling. Rolling can occur in a variety of ways. as depicted in Figs. 7.28 and 7.29. The most common form is for a body to roll over a fixed surface; the surface can be plaroll together. together. Examples nar or curved. A second form of rolling is for two bodies bodies to to roll of this are gear mechanisms and a sphere rolling over another sphere. The contacting points define the plane of contact, which is tangent to both contacting bodies. The rolling constraint is defined as no relative velocity of the contacting points perpendicular to the plane of contact. At the point of contact, the radii of curvature of Plane of contact

P (1)

Y

x

FIgure 7.28

I

Normal n

/ Rolling of a body over a surface

FIgure 7.29

3Note that For a straight line the radius of curvature exists and it is at infinity.

Rolling of two bodies

7.9

ROLLING

the contours of the rolling bodies are along the same line. This last statement statement is is aa the geometric constraint that determines whether one body can roll over another. Consider two bodies rolling over one another, as shown in Fig. 7.29. The conV(',. respectively. tacting points are denoted by by C1 C1 and and C2, C2, with with velocities velocities of V(' and V(',. Let II denote the unit vector perpendicular to the plane of contact. The relative yelocity of one contacting point with respect to the other is along the plane of contact. The rolling constraint can then be expressed as

[7.9.11

= 0

(VC,

A rolling body is a five degree degree of offreedom freedom system. Ifthe If thecontacting contacting points points have have the same velocity, the motion is referred to as roll without slip. If the contacting points have different velocities, the motion is referred to as roll with slip. Whether slipping exists or not depends on the forces acting on the rolling bodies, as well as on the amount of friction between the rolling bodies. The no-slip condition can be expressed as

[7.9.2]

=

A body rolling without slipping has three degrees of freedom. An interesting point to note about rolling contact is that the accelerations of the contacting points are different, whether there is slipping or not. The contacting points approach each other, they have contact, and then they separate and move away from each other. The constraint force is applied to the contacting points only for one instant. Also, the accelerations of the contacting points usually have components along the contact plane. Hence, there is no direct constraint relation associated with the acceleration of rolling bodies. This can be remedied by taking into consideration that the shapes of the bodies are fixed and that the radius of curvature exists for all contacting points. which permits one to write the acceleration expressions for each body. Let us first consider plane motion of an axisymmetric body—such as a disk or cylinder —of radius R that rolls on a fixed flat surface as shown in Fig. 7.28. We use as generalized coordinates coordinates the the position position of of the the center center oF of mass, X and Y. and the rotation angle 6. The angular velocity of the body is w = —vK, where cv = 0. We can express the velocity of a point P on the body as Vp =

V0

+ W X rp/G rp,G = XI +

YJ

+

X X

re/c

[7.9.31 [79.31

where rp/0 is is the the position position vector vector from from the the center center of of mass mass to to P. P. The The path path followed followed by by shows a point point on on the the circumference circumferenceof ofthe therolling rollingbody bodyisiscalled called aacc cloid. Fig. 7.30 shows

a cycloidal path. To find the velocity of the contact point C. we write Eq. [7.9.31 in terms of C. The no-slip condition can be expressed as V( = 0. Introducing this into Eq. [7.9.3], we obtain VC VC = 0

V( V( ++(.0 XXrc-7G rc-7G

[7.9.4] [7.9.4J

which we can conveniently write as which V0

(.0 X FG/C = —wK

X RJ = RwI = R0I

[7.9.5]

so that that X = Rw, Y = 0. and so roll without slip for plane motion is a single degree so of freedom problem. Because there is no velocity of the point of contact C. the body

401

402

CHAPTER 70 Ricrn Boiw KINEMATICS y

9= 3ir

2R

x 0 —

2irR

FIgure 7.30

Cycloidal path

can be visualized as rotating about C at that instant. For roll without slip the point of contact is the instant center. The acceleration of a point P on the rolling body can be obtained by differentiating Eq. [7.9.3],

[7.9.6)

ap =aG+a.Xrp/(;+cz)X(wXrpfG)

The acceleration of the center of mass can be expressed as aG = d(RwI)/dt RaI. The angular acceleration is a = —iK = —aK. The acceleration of a point on the rolling body is =

+ (—aK X

—

17.9.71

O.)2rpfG U)2rpfG

If we select point P as the contact point C rpfG rp/G = —RJ and we obtain

[7.9.S) [7.9$)

—Ri) — w2(—RJ) = Rw2J = Ral ++ (—thK XX —Ri)

Now consider roll without slip on a fixed, concave-upward planar curve as shown in Fig. 7.31. We use a set of' normal and tangential coordinates. It follows that we can write the velocity of the center of mass as

[7.9.9) [7.9.91

= Rwe,

=

The acceleration of the center of mass thus becomes .,

a6

V2

p

=

+

Rw p

-)

[7.9.101

Center of ol curvature p ?1

a) t

/ Figure 7.31

Rolling inside a concave surface

7.9

ROLLING

Substituting this expression into the relative velocity expression gives the acceleration anon of any point point on the body. For the special case of the instant center, we we have and rc/G = — = a(; + a X FCIG —

=

+

R2w2

W2FCIG

+ (—aK X —Re,1)

—

w2(—Re,1)

P

=

1+

Roi2

R\ —

(

17.9.11] 17.9.111 If the curve on which the disk is rolling is convex, the direction of the unit vector in the normal direction is reversed and FC/G Rep. The acceleration of the instant center C becomes

Rw(—l +

ac

R\

17.9.12] 17.9.121

—

PJ

Next consider rolling in three dimensions. We make use of of aa 3-. 3-1-3 1-3 transformatransforma-

tion sequence to describe the rolling of a flat axisymmetric body, such as a thin disk non or a coin. We begin with an inertial set of axes XYZ, with the initial position of the coin as lying flat on the XY plane, as shown in Fig. 7.32. Figures 7.33 and 7.34 depict the disk after the first two Euler angle transformations have been carried out. The generalized coordinates used to describe the general motion of the disk are the coordinates of the center of mass X, Y, Z, and the three Euler angles 6, and cli. We use the F frame as the relative frame. From Eq. [7.5.3], the axes xyz and XYZ and unit vectors ijk and IJK are related by =

j = —s4'cOI+cc/cOJ+sOK k = s4s61—c4s6J+cOK

17.9.13] 17.9.131

It should also be noted that the contact point between the disk and the surface is along the y axis. The angular velocity of the disk is

(0=

[7.9.14] z,z, (/) I

F

F

Y

x x,

Figure 7.32

Initial position and precession

403

404

CHAPTER 7• RIGID BODY

z V

z V

---

R

RcosO i

/ -0----

x

Figure 7.33

RsinO

Figure 7.34

N u tati on

so that = 0. reference frame is

=

sin 6, and

=

th cos

Side view

6 + i/i. The angular velocity of the

= Oi + 4sinOj + (/)cos6k (/)cosOk = WI +

+

tan6

[7.9.15]

We thus have

=0

W.1.x =

U)1 v

=

=

WI,, =

tan6

= 4cosO 4cosO

[7.9.16)

First, consider roll with slip. When the disk is rolling and slipping, the point of contact C has a nonzero velocity along the roll surface. Let us write the velocity of point C using the relative velocity expression between C and the center of mass, and express the result in terms of the inertial coordinates. Thus VG

+

= = (X

—

'1C =

X

rc,G

R0s(/)sO + R(4c0 +

+ (Y +

I c4isO +

+

+ (Z

R0c9)K

[7.9.1

The Constraint associated with roll with slip states that the velocity of point C has no component in the vertical, that is, Vç K = = 0. Setting the components of in the Z direction to zero, we obtain

Z = ROcosO

[7.9.18)

This Constraint is is recognized as being the time derivative of Z = Rsinf) [7.9.19] so that the constraint associated with roll with slip is holonomic. The constraint a physical explanation: it represents the height of the center of mass from the roll surface.

7.9

ROLLING

next consider roll without slip, where the velocity of the contact point C is = 0 in terms of the components zero. Hence, there are three constraints. Writing along the inertial axes X and Y. we obtain from Eq. [7.9.17] We

- R(4c0 +

X=

Y = —R0c4s6

—

R(4c6 + i/i)s4

[7.9.20]

equations are not perfect differentials and cannot be integrated to a form similar to Eq. [7.9.191. Hence, constraints associated with the no-slip condition are nonholonomic. For rolling without slipping, the velocity of the center of mass can be expressed in terms of the inertial coordinates as

These

XI+YJ+ZK

VG =

= (R0s4s0

R(4c0 + + i/i)s4)J + ROcOK

+

—

We can also express VG

—

[7.9.21]

by using the relative velocity expression with VC = 0, which

yields

=

VG =o)bxrGfc

[7.9.22] [7.9.221

velocity components components along the F frame In terms of the angular velocity V6 =

—

[7.9.23]

Rw2i

Consider the acceleration of the center of the disk. One can calculate aG in a

11. This apnumber of ways. The most cumbersome is to differentiate Eq. [7.9.2 1}. proach does not yield any insight into the nature of the problem, except for the component of the motion in the Z direction. A more convenient approach is to use the F frame and to apply the transport theorem to Eq. [7.9.231, which yields A

1A _I' —VG-r A a(;— (0

VG

Substituting the values for VG and wj into Eq. [7.9.24], we obtain aG =

=

—

•

+

+

+ wJ

+

tanO

/

+ RI—vT

—

k)

X

jj + tanOj

—

[7.9.251

+

In terms of the Euler angles, we write the expression for the acceleration by substituting Eq. [7.9.14] and its derivatives into Eq. [7.9.25J, which yields

aG = R(—4c0 + 24)Ose aG

—

+ R(O + 4s0(4c0 +

+

+ qi)

—

[7.926]

The advantages of using a reference frame not attached to the body are obvious. of the the spin spin angle angle i/i in Attaching the relative frame to the body would necessitate use use of X + X to use the relation aG = all calculations. One would not be able to (cOt, X rG/c) to find the acceleration of the center of mass. Point C has a nonzero

405

___\ 406

CHAPTER 7• RIGm BODY KINEMATICS

acceleration. Another reason for not using a body-fixed frame is that the position of the contact point between the disk and the surface changes with respect to a bodyfixed frame as the disk rolls. One would have to take into consideration the motion of the point of contact. By contrast, viewed from the F frame, the point of contact always lies on the XY plane and 'VC K = 0.

Now consider the acceleration of the contact point C. We first calculate the angular acceleration of the body using the transport theorem. Noting that = — to1

=

tan O)k. we write

—

=

ci

= =

[7.9.27]

+ (Oj X to5 i + th

'

(

+

+

+

+

1.

JI+ tan 6,,

/

klxlw-— tan6J tan 6j

tans jj

:WVW z +

tan

6)

The relative acceleration equation between C and G is

ac =

+

a1,

X rC/G rc/G +

Substituting in the values of rc/G =

[7.9.281

X (tot, X X rC/G) rc/G)

and the values for aG, Wb, tob, and from Eqs. [7.9.26], [7.9.141 [7.9. 14j and [7.9.271, respectively, and carrying out the algebra. —Rj

we obtain

ac =

Example

7.9

R (U)

Yw z + R (U) YU)

-

Z

)k

[7.9.29]

—

The cone shown in Fig. 7.35 is of height L and apex angle It rolls on a smooth surface without slipping. The center of the base has a constant speed v. Find the angular velocity and angular acceleration of the cone.

Solution find the angular velocity of the cone we make use of instant centers. Figure 7.36 shows the geometry. The h1h2h3 axes move with the line of contact between the cone and the horizonta' surface. They are similar to the F frame. The vertical distance between point B and the surface

To

11,

WI

C')

C)

/11

3

/11

Figur. 7.35

A

Figure 7.36

0

7.9

ROI1UNG ROI1uNG

L sing. Because of the no-slip condition, A is an instant center and the velocity of B is equal to the angular velocity times the height. The total angular velocity of the cone is then is

[aJ We express the angular velocity in vector form as to = —v/(L sin $)h3. To see this. we note that Vfi = uh1 and rB/A = L sin /3h2 and rB/A. = o X rB/A. To find the angular acceleration, we visualize the the motion motion of' of the cone as the superposition of two angular velocities. The first angular velocity w1 is that of a cone which is only rotating in the horizontal plane. The second angular velocity. (02. (02, is the angular velocity of the cone about its thethe h1h1 /12113 about its symmetry symmetryaxis. axis.We Wenote notethat that /12/13coordinates coordinatesare arerotating rotatingwith withangular angular velocity velocity We thus have (01

Ibi

= w1h2

From Fig. 7.36 we find to1 as to1 as to

Ic]

h2 h7

—

This can he explained by considering the velocity velocity of' of point B and that point B is rotating about the h4 axis. The distance between B and the /12 /12 axis is is LL cos cos The total angular velocity is

[dl

(0 = (0

where W2 is the spin velocity of the cone. As the direction of this spin is along the the symmetry symmetry axis of the cone, W2 can he expressed as

= w2f3 = w2 sin

+

[0] Eel

cos f3h3

We then use Eqs. [c] to [ci to relate the angular velocities as We

to =

U —

to, = = Wi + w2

.

Lsinf3

V

[f]

(02 sin sin /3h2 /3h2++ w2 cos h2 ++ w,

Lcos/3

can he solved for w, (0 W

V

cos

and R = L tan f3. f3. we we can can express the angular velocity as

sin

—

v

V

—

.

[gI [gi

Lsinf3

h3= h3 =—f2— -42— L

..

Lsinf3

v

v

L

R

fi=—f2——f3

1k]

The angular acceleration acceleration of the cone can be expressed more easily using the H frame, frame. thus

a=

=

to X (02

V_--h2 V_--h7

Lcosf3

X—

-

V

Lcosf3sirif3 Lcosf3sinf3

(sin f3h2 (sin f3h2++cosf3h) cosf3h)

1

=

This problem can also he viewed from a Euler angle point of view, with rate, constant nutation angle rate of' of w,. and spin rate

[II denoting the

407

408

CHAPTER 7• RIGID Bony

Example

A disk rolls over a horizontal surface while it rotates about a shaft, which is attached by a revolute joint to a bar with an arm, as shown in Fig. 7.37. The shaft is rotating with speed Find the angular velocity and angular acceleration acceleration of of the the disk. disk.

7.10

Solution relative frame frame fi The reference frames arc shown in Fig. 7.38. We obtain the relative by rotating the reference frame attached to the shaft (rotating with = about the fixed a3 axis) an angle of 2700 — f3 about the d2 axis. We write

f2 =d2

=

The angle

= — cos

—

I.'

sin /3d.3

which is lixed, is related to the height of the arm by Ii

[bi Ibi

+ Rcos/3 = Lsin/3 +Rcos/3

The angular velocity of the disk can be written as the sum of the angular velocities of the shaft and of the disk with respect to the shaft. or (I) (i)

[ci id

W1 + (02

in which =

= w1cosf3f1

[dl Edl

(02 = —w2f3

are the respective angular velocities. Note that (02 is not known, and it will be found by using the rolling constraint. To do this, we obtain an expression for the velocity for the contact poin: as

=

[.1

+ (I) X rc•,u = 0

where =— — (L + L co s /3 )w if2

[fi En

=— —Rf1 Rf1

so that we have

Lcos/3)wif2 + = —(L + Lcos/3)w1f2 = (—Lw1 (—Lwi

—

cosf3f1

—

(o1 sinf3 sinf3++o)2)f3) XX—RI'1 —RI'1

Lcos/3w1 + Rsinf3wi + Ro2)f2 = 0

Eul Eli

(01 (Oi

L ci'

L

B

fi fl ,1 I,

,1

C

Figure 7.37

Figure 7.38

79

409

ROLLING

Solving the above, we obtain w2 in terms of w1 as (02 (02

(L + Lcos/3

= Wj

—

Rsin1B)

[hi

R

We recognize the term in the brackets above to be the horizontal distance between 0 and C. Introducing the ratio r = LIR. we can now write the angular velocity of the disk as

w= =

w 1d3

w2f3 =

—

w1 cosf3f1

—

w cos /3f

— (V

1(r + rcosf3

—

sin/3 + sin /3)f3

[ii

w1r(1 +

To find the angular acceleration we differentiate Eq. [ij with the result

a=

x X

+ (0

= w1d1

+ (wi cos/3f1 — Wi sin /3f1) X

—

U]

Note that ó2 i2 isis obtained obtained by by direct direct differentiation of E.q. [hi. Substituting in all all the the values values and carrying out the algebra we obtain

a

cos $f1 + o4(r + r cos /3 — sin /3) cos /3f2

—

r(l + cos

1k]

We cannot obtain the expression for the angular acceleration by direct differentiation of Eq. [ij because f, is not attached to the disk.

Figures 7.39 and 7.40 show the top and side views of a tricycle. Do a kinematic modeling on

Example

the tricycle and determine the number of degrees of freedom, assuming that all wheels roll without slip and the tricycle does not tip or roll.

7.11

Solution We select the inertial coordinates XYZ such that the tricycle moves on the XV plane, and we attach an xvz coordinate system to the tricycle at point G. The xyz axes are obtained by rotating the XYZ coordinates by an angle of i/i about the Z axis, axis. as shown in Fig. 7.41. The velocity of point G and the angular velocity of the main body are

VG=XI+YJ and considering that I = cos the xyz coordinate system as

—

sin

= (Xcosçli (Xcosçli +

J=

[a] cos

+ sin

+

D2

we write the velocity of G in

[bI

—

V.

T

I

.'

FIgure 739

lop view of tricycle

410

H) BODY KINEMATICS CHAPTER7• CHAPTER 7•RIG RIGH)

'I

-

G

Cl

FIgure 7.40

Y

x

x

C3

V

FIgure 7.41

Side view

We next calculate the velocity of point D1, which is the point to which the right rear wheel is connected. connected. Noting Noting that that r/)1/(; r/),/(; = —L1j — hk + L31, we obtain X (—L,j — — uk + L3i) = = uk X

X

[ci

+

and

['I

=VG+flI xrD1/G xrD,/G =VG+fll

The point of contact between back wheel on the right and the ground is denoted by C The angular velocity of that wheel is expressed as =

and, noting that and.

= — R1 k,

=

—

[.1

—

the velocity of C1 becomes

+ WI X =

+

+ Ysini/i + L1i/;)i Lii/;)i + (Ycos&Ls

—

Because of the roll without slip assumption. the velocity of C1 is zero, tantamount to two constraint equations Kcosçli +

+

= 0

Ycosi/i

+

—

—

[fi

R101)j

R101

=

=0

[g,bJ

and we note that both constraints arc nonholonomic. We can obtain the expression for velocity of the point of contact for the back wheel on the left, Va..,. by substituting —L3 for and 02 for in Eq. [f]. Since that wheel also rolls without slipping, we set Vç2 to zero, whict yields one additional constraint, namely the component of vc, in the direction —

—

—

R102

=00 =

[II

The component of the velocity of c2 in in the x direction is the same as its counterpart C1. It is interesting to note the velocity of point A, which is obtained by setting L3 = 0 Eq. [dl as V4 = vc +

X r,4 r4fG =

X co s

+ Ysin

+

+ (Ycosçli s — — X Si fl j' )j

[Li [fi

7.9

ROLLING

However, from Eq. [gI, the x component of VA is zero, leading to the conclusion that

Iki

=

The velocity of point A is always only in the direction, that is, in the direction of the body of the tricycle. This is the constraint mentioned in Chapter 4, when discussing vehicle dynamics problems. By virtue of the roll without slip approximation. we end up with this constraint.

Next. consider the steering mechanism. The turning of the wheel is denoted by d and the rotation rotation of of the thewheel wheelby by03. 03.We Weattach attachaacoordinate coordinate system x''z' to the wheel (obtained by rotating rotating the the XYZ XYZ axes axes about the z axis by an angle 4)). The velocity of point E, about which

the front wheel turns, is obtained by VE =

VB

+ a2

X rE/a rE/a =

VG

fl)XXrp18 rp18 + fl1 X rrnG + fl)

III Ill

in which

U2 =

= L4 sin

rg/G = L2j

(if, + 4))k

L4 cos

—

=

cos

—

sin Sifl

[ml [mI Substitution of all these values into Eq. fI} yields — (L2 + = (Xcosif' + Ysini/i —

cos i/i + (Y (Y cos

—

X sin i/i

—

—

sin 4,(i/i 4)(i/i ± th))j L4 sin f3 sill

In]

To find the velocity of the contact point between the front wheel and the ground, we first develop an expression for the angular velocity of the front wheel. The wheel is at an angle of 6 with the body of the tricycle, so that the angular velocity of the wheel becomes (2)3 =

Noting that

fl7 fl2

—

03i' 031

—Oisin4)j 03 COS — O3cos4)i + 4))k 4))k— — 03 sin 4)j

b)k — 03F = = 0/i + 4,)k—

[.1

we have = —R3k. —R3k. we (2)3 (2)3

rC3/E == ((u' (0/' + 4))k XX rc3IE

—

03

cos

—

sin 4)j) 4)j) X X (— (h sin (— R3)k

= R3O3sin4)i—R3Oicos4)j R3O3sin4)i — R1O3cos4)j we obtain for the velocity of C3

Combining Eqs. [nJ and (03 X X rc;IE = VE + (.03

— (L2 + = [Xcosqi [Xcosi/i + Ysini/; Ysini/;—

+

—

Xsini/i

—

R3O3cos4)

—

— 144

L4sin/3cos4)4)+

sin4.'O/' ++ 4))IJ sin4.1(l/F 4')IJ

[qi

The roll without slip condition states that the velocity of C3 is zero, which yields two more constraints that are also nonholonomic, which we write

Ysint/i— — + Ysintli

+ L4sinj3cos4))i/i

Ycosi/i Ycosi/i — — XSIn Xsin

iii

—

—

—

R3O3sin4) = 0 L4sinf3cos44) + R3O3sin4)

sin4)0/i sin 4)(çIi++ 4)) 4)) = 0

[rj In [.1

Let us now calculate the number of degrees of freedom for this system. We introduced 63.We Wegenerated generatedfive five constraints, constraints, and 6. generalized coordinates: X, Y. Y, c/i, c/i. Oi, 02, 02, Eqs. [gj, [hJ, [hj, [iJ, [r], and [Si. Hence, the tricycle is a two degree of freedom system.

411

412

CHAPTER 7•

RIGID BODY KINEMATICS

REFEREN CES Ginsberg, J. H. Advanced Engineering Dynamics. New York: Harper & Row, 1988. 2nd ed. ed. Reading. Reading. MA: Addison-Wesley, 1981. Goldstein, Goldstein. H. Classical Classical Mechanics. 2nd Spacecraft Rotational Maneuvers. Amsterdam—New Yori Junkins. J. L.. and J. D. Turner. Optimal Spacecraft Junkins, Elsevier, 1986. 1986. Spacecrqft I)vnamics. New York: McGraw-Hill. McGraw-Hill. 1983. 1983. Kane. T. R.. R.. P. P. W. Likins. Likins. and D. A. A. Levinson. Spacecrqfl York: McGraw-Hill. McGraw-Hill. 1970. 1970. Meirovitch, L. Methods oJAnalvtical oJAnalvtica! Dynamics. Dynamics. New New York: Meirovitch, Orthwein. W. Machine Componeni Design. St. Paul. MN: West Publishing, 1990. Orthwein, Sandor. G. N.. and A. G. Erdrnan. Advaneed Mechanism Design. Vol. 2. Englewood Cliffs, Cliffs. NJ: PrenticeHall, 1984. York: McGraw-Hill. 1969. of Mechanisms. Mechanisms. 2nd ed. New York: Shigley. J. E. Kinematic Analysis or Whittaker. E. T. Analytical D namics qf Particles and Rigid Bodies. Cambridge: Cambridge Universir' Whittaker. Press. 1933.

I

HOMEWORK EXERCISES SECTION 7.2 1.

2.

The bar in Fig. 7.42 is always in contact with the corner D and the curved path Find the angular velocity and angular acceleration of the bar as a function of the angle 6. For the system shown in Fig. 7.43, find the instantaneous center of the rod AB and the angular velocity of the rod.

0 5 rad/s 4

L

13cm

a

—

5cm

—

B .1

L

12cm

Figure 7.42

Figure 7.43 Figure 7.43

413

HOMEWORK EXERCISFS

SEcTIoN 3.

7.3

750 about A coordinate frame B is obtained by rotating frame A by an angle of a line joining the origin and the point (0.5, —0.7,0.4). Find the direction cosine matrix [c] between the two frames. Then, calculate the direction cosines of the principal line from [cJ and compare your result with the given information.

SECTION 7.4 4.

A reference frame A is transformed into a reference frame B using a transformation consisting of two rotations. First a rotation about the a1 axis. by an angle = 0.3t, and then a transformation about the resulting a; axis by an angle 0(1) = 0.6 sin 'irt/4. Using Eq. [7.4.1 3J find the angular velocity of the frame B with reference to frame A at t = 3 seconds.

SECTION 7.5 5.

6.

7.

Solve the previous problem using Euler angles. A disk of radius R is attached to a rod of' length L2, and it rotates about it with angular velocity w (Fig. 7.44). The rod pivots about an arm ot' length L1 with angle 0. This arm is attached to a vertical shaft that is rotating with the constant angular velocity of' ft Find the angular velocity and angular acceleration of the disk, and the velocity and acceleration of' the center of the disk. Figure 7.45 depicts a spacecraft undergoing general motion. There is a point mass s'iding inside a frictionless tube along the b2 axis. Using a 3-1-3 Euler an0. i/i) and denoting the position, velocity% and acceleration gle transformation of the point mass with respect to the symmetry axis of the spacecraft by r, v, and derive expressions for the absolute velocity and acceleration of a, the mass. First write these expressions in terms of angular velocity parameters 2. 3). 3). Then Then express them in terms of the Euler angles. 1, 2. (i (i = 1, top

1)3

\\

/

1)1

R

((I)

Figure 7.44

Figure

7.45

VICW

414

CHAPTER 7S RIGiD BODY

p

z

x

V

y

FIgure 7.46 & Figure 7.46 depicts a spinning top on a cart. The cart is moving along a curved track on the X Y plane, the radius of curvature of the apex is p. and the constant speed of the apex is v. Find the velocity and acceleration of the center of mass of the top using a 3-2-3 transformation (vi, 0. 4)), with the inertial axes as shown.

SECTION 7.6 9.

Consider a 3-2-1 transformation and express the partial angular velocities associated with the angular velocity vector, using the components of the angular velocity along the body axes as the generalized speeds.

SEcTIoN 7.7 A coordinate frame B is obtained by rotating frame A by an angle of 75° about a line joining the origin and the point (0.5, —0.7, 0.4). Find the Euler parameters associated with this rotation. From the Euler parameters, calculate the transforination matrix [RI between the two coordinate frames. 11. The orientation of a rigid body is described by a 3-2-3 Euler angle rotation, with the rotation angles i/i, 0. and 4). At a certain instant, the values of the Euler angles and their rates of change are = 45°. i/i = 0.2 rad/s, 0 = 30°, 0 = 0. 1 radls. 4) = 30°. 4) = 1.0 rad/s. Find the values of the Euler parameters and their derivatives for this instant. 12. Find the counterpart of Eq. [7.7.401 for a 3-2-1 transformation and for a 3-2-3 transformation. 13. Solve Problem 2.10 using Eq. [7.7.46]. 14. Solve Problem 11 using Eq. [7.7.46]. Problem 2. 2.11 10.

HOMEWORK EXERCISES

15. Rodrigues parameters are defined as p, = (i = 1, 2, 3). The Rodrigues vector is defined as p = n where n is the unit vector associated with the principal line. Derive an expression for the direction cosine matrix Ic] in terms of the Rodriguez parameters. 16. Show using the definition of the Rodrigues vector from Problem 15 that the vectors r and r' in Eq. [7.7.46] are related by r' = r — (r + r') X p. SECTION 7.8

Consider the Cardan joint in Fig. 7.21. Plot Eq. [7.8.9] for different values of /3 and identify the range in which the Cardan joint can be assumed as a constant velocity joint (specified here as less than 3 percent variation in angular velocity). 18. The collars A and B are attached by a rod of length 30 cm, as shown in Fig. 7.47. The joint at A is a ball and socket joint and at B a pin joint. Collar A moves in the Z direction, while the guide bar for collar B is on the XY plane. At the point shown collar A is at a height of 24 cm and collar B is moving with a speed of 10 cm/s. Find the angular velocity of the rod. 19. The collars A and B are attached by a rod of length 30 in. The joint at A is a pin joint and at B a ball and socket joint, as shown in Fig. 7.48. The motion of collar B is restricted to the XY plane. At the instant shown. the rod is at a height of 25 in and collar A is moving down with the speed of 10 in/sec. Find the velocity of collar B and the angular velocity of the rod. 20. The two guide bars are on the same plane, as shown in Fig. 7.49. A rod of length 20 in. is attached to two collars, one at each end, by pin joints. Find the velocity of collar B if collar A has a speed of 3 in./sec at the position shown. 21. A sketch of a centrifugal governor is shown in Fig. 7.50. As the angular speed increases, point D moves upward, forcing the point C to move up as well. 400, point D is moving up with speed 0.2 rn/s. and the We are given that 0 17.

z A

10cm/s

I

/

20cm

-I

z

20 cm y B

TI

10 10 in/sec

A

25 in

10 x

Figure

y

7.47

Figure 7.48

4

415

416

CHAPTER 7. RIGH) BODY

3 in/sec 3m

300

5. A

E

04m 0.4m D

B

C

Figure 7.50

Figure 7.49

governor is rotating with 11 = 200 rpm. Find the angular velocity of the arm DE and the velocity of the collar E with respect to the shaft AC.

SEcTIoN 7.9 22. A cylinder of radius R rolls inside a parabolic surface defined by the relation y = kx2, as shown in Fig. 7.51. Find the critical value of k that permits the cylinder to roll continuously inside the parabola. 23. Consider the disk in Fig. 7.33. Show that the velocity of the contact point, as always lies on the XY plane. viewed in the F frame, that is, 24. Consider the gyropendulum of Fig. 7.14. Here, the disk is rotating without slip+ R)/2, as shown in Fig. 7.52. Given ping inside a cylinder of radius D = a constant precession rate of 1.8 radls, find the corresponding spin rate, as well as the angular velocity and angular acceleration of the disk.

V

y=kx2 x

Figure 7.51

D

Figure 7.52

EXERCISES

r=Ji

4R

3

FIgure 7.53 Figure 25. Consider the disk in Example 7.10. Now, the shaft arm is moving upward with respect to the horizontal surface with the relationship h = constant. This causes the point of contact between the disk and the surface to slip in the d1 direction, while the rolling motion is still without slipping. Given that w1 = constant, find the angular velocity and angular acceleration of the disk for R = L/4. 26. The disk in Fig. 7.33 is rolling without slipping. Consider that R = 8 in, and that when the nutation angIe angle 0 = 750 and is constant, the precession rate is observed to be 0.4 rad/s and the angular velocity of the disk about its symmetry axis is 15 radls. Find the angular velocity of the disk and the velocity of the center of the disk. 27. The small cone in Fig. 7.53 rolls on the large cone (both cones are right-angled) and makes a trip every four seconds. Find the angular velocity and acceleration of the small cone.

28. Consider the unicycle in Fig. 7.54. The wheel wheel is of mass m m and radius R and is approximated as a disk, while the rider is modeled as a slender rod of mass 3m and length 3R. Find the velocity and acceleration of the center of mass of the rider in terms of the Euler angles i/i, 0, and 4) and a 3-2-3 transformation. Assume roll without slip and that the rider is upright. 29. Consider the axle in Fig. 7.55 to which two wheels of identical size are attached. The wheels roll about the axle without slipping. Perform a kinematic analysis

JR V

T

2R

C

/

Side vicw

Figure FIgure 7.54 7.54

I C

Front view

x

Figure Figure 7.55 7.55

417

418

CHAPUR 7. Ricm BODY

I

KINEMATICS

I

j

Top view

Side view of rear wheel B

A

x

C

Figure 7.56 B

I,

Figure 7.57 and determine the number of degrees of freedom. How would you select quasivelocities for this problem?

30. Consider the bicycle in Fig. 7.56. Considering roll without slip of both wheels, determine how many degrees of freedom the bicycle has, and write an expression for the velocity of point B. Identify whether the constraints are holonomic or not. Begin with the frame and the coordinates ofpointA. The angle () is with respect to the XY plane and it represents the tilting of the bicycle. 31. The wheel in Fig. 7.57 rolls without slipping, with its center having the constant speed v. Find the velocity and acceleration of the collar as a function of 0.

Table 7.1 Summary of common Euler angle transformations

3-1-3 (Historically Significant)

to b1 b2b3, or XYZ to xyz)

3-2-3 (NASA Standard Aerospace)

i/i (heading). (heading). 80 (attitude). 1.' (bank)

3-2-1 (NASA Standard Airplane)

Name of Sequence (spin)

t/i (precession). ii; (precession). 0 (nutation).

si/i

0

0 (precession). 0 8 (nutation). i/i t/i (spin) d

ct/i ct/I

Sequence of Angles

0

Vehicle motion, attitude dynamics

st/i

Spinning or rolling bodies

ct/i

Spinning or rolling bodies, orbital parameters

0

0

—sO

ct/I ct/i

0

0

—st/i

c8 cO

I

[R1J

—sO

()

1

0

0

0

1

0

ct/i

cO c8 0

1

Application

'4'

[R1]

[Rfl]

0

cO

0

0

1

s4,

s0 s8 0

[R3J.1—— 0 [R3

cOs4,

0-s4,c4,

(w2sOs4,

±

0 0 o

w

+ w1 w1

+ Oc4,)b2 Oc4,)b,

—sO —so

sd'cO si/is0s4,

ct/1c4)

cOc4,

0

cosO cost)

0 ==W2C4, — WjS(f) 8 W2C4,—

•.

1

• st/isOcth

—sO

cth cO

ct/jell cilicO —st/ic4, ++ ct/isOsth ct/ististh —st/ic4, sWsd + ct//sOc4) sijisd ()

cOs4,

-i-I-

tirI2 :tirl2

0

0—

+ (Ia2 + Oa; ÷

.

—s4, I

4, =

+ tl'cOcth -- 8s4,)b;

/i)b1 (—t/;sO + 4,)b1 = (—l/;sO

tba1 tba3

c/

cO sd.'

0

0 cO cth

c4,

01

—s4' —

s8 sO

0

w1: pitch. w2: roll. w3: Wi: yaw

c4 0 1

cf 0

0

0 sO

[R1] =

0

—

—sOc4,

sa/icOc4, st/icOc4,

ciliscb ci/iscb

st/is4i ct/ic6c0 — st/;s4i cçl'cOcd

(—w1c4

w= w

cOcth

cO

sOsd' SOS'1'

s4,

—stlicOs4, -r cijicth —St/JCOS4, cij'cth

—sOc4, —s0c4,

—cçi!ic0s4 —ci/icOs.b — — st/ic4, s111c4)

s6s4,

cO cth 0

si/isO st/isO

cO

:tir

ct/isO c/isO

-

(1 = 0. .

= tba3 + 0a2 ++ thb3 thb1

i

W2C(j)

W2sd')

Os4,)b1 ++ (t/isOs(/) ( islls4, + Oc4,)b2 t/,s9c4, ++ Os4,)b1 = ((-t/is8c4,

i/I i/i

=

(w1cOc4, — (w1cOc4, —

(:1 = = W1S4) W1S4) 0

.

+ (t/ic0 + d)b3

4) = =

() 0

w3: spin rate

-

0

cO

cO

st/i

01

0 0

--sO ciii ct/i

0 O

W3

Osi/i)b' Osi/1b2

-.-

—si' 0

w: spin w3: spin rate

[R1J = [R11

= 0 0

=

— s4,costb s4icOstb c4ct/; —

ct/i

0 0

-

cO

sOst/i c4,c8si/' s(Isa/i skt/i sdct/i — — c4,c0si/' s(ici/i ---s4,st/;++c4,ct/icO c4,ctjic() sOci/i

sOsi!'

—si/i

I

—cths()

s0ci/i sOct/i

0 0.

—

sal; W2C1/i) st/i + W2c1/i)

(Jct/i)b1 (Jct/i)b1 ±

i

o

(—ai1c0si/i (—oi1c8si/i —

— WiSt/i WiSt/J 6 0 ==W1Ci// —

.

.

0=

cO

—cast/i — S4>cOCtII —cOst/i s4cOct/'

)s0

(çf)s(?sij;

-s- 0a1 + = /ia3 -s=-.

Cl)

.

sinfl sinfI

I

(cticO + + (d'cO

—

1

Names of Angular Velocities

Transformation • Matrices

Combined Transformation

[RI = [R3J[R2][R11 FBI Matrix [B] Matrix

Singularity at AngularVelOClt% Angular Velocity in Body-Fixed Reference Frame

Derivatives of Euler Angles in Terms of Angular Velocities i/i

Table 7.1 (Continued)

Transformation Matrix for F Frame

Angular Velocities in F Frame (for Inertially Symmetric Bodies)

Derivatives of Euler Angles in Terms of Angular Velocities Expressed in F L' 12 frame Acceleration of Body in Body

Frame

Angular Acceleration of Body in F Frame Euler Parameters in

TermsofEuler Angles

sthsO

of ++

Of

—

ccôc()

s4) sO

0

•

cO

—

4)cOf3

—cths6

i

-. sinO

4)-Of

= 4)a3 + fia1

=

.

O=w1

= 4)sOsdi 4)s0sd;

—

Osçli

+ 4)OcOci/i

•

—

04)ct/i

— Oç&s4) d)i/JsOc4) + Oct/i — d)i/JsOcl/I

+

3-1-3 (Historically Significant)

a

— —

+

4- th 4)Os() 4th = 4)cO -- 4)Os()

=

(/)uJctI - Oi/i]f2 + (ths0 4 (/)uJcO

C

4,

2 —

2

çb-'-4) 2

0

S2

0

— 4)OsO i- [4)c€1 [4)c€1 — 4)Os() +

e() =

(.

S

e2s =

Name of Sequence

-

-

(Ids4)

.

Oth]f1

—

I/iOcOc4)

0

0

04)c4)

[ticO [qco ±

cO

0

3-2-1 (NASA Standard Airplane)

—

— OC(/) 0c4) —

i/;Oc()

- i/iOsO i/iOs() + th th —

—

iisdsO If2

+

- t/iOsO 1i0s() —

4, 2

a2 =

=

—

i/ithcOs4) i/ubc0s4)

4)

i/i

(1

0

0

4,

di

4)04)

e1 =c-4)

=

sOs4)

—

()sth

i/i

—

Odcd £)thc(b

Oths4)

0

4, 4)

4)

4, 4)

—

4) 4)

4)

0

0 -s-sic

4)

Not of any significant use

—

— i/iHs0c4) 4)c0c4) — I/ICOC4)

+ b4)cOcth b4)cOcth-1- Octh

-I-

-- i/iOcO 4)Oc() + =— — 4)st) --

Not of any significant use

Not of any significant use

3-2-3 (NASA Standard Aerospace) CI/J Cl/i

s,'sO stt'sO + 4)f (/if

.

Not of any significant use

1.'

ciJic& —

cl/IsO

I/ia3 +

4)f3

sm 0

(Vj

Of2

-t- Of2 -4)sOf1 -t-l/JsOf1

-

siJicO

=

.

4)=

çlicO

bths0c4)

-4- iji(IcOsth iji(IcOsth iJisOs4) 4)s0s4) -4-

+ l/I4)s8sth 4)4)s0sth + Os4) Os4)

= — f;sOccl fis0c(1'

wi cosO sinO

to, ==—&sOf1 —i7&sOf1

a

—

s

-tt

[—4)sO = [—l/isLi

a2 ==

cxi,

+ [0

e1

0

e2=c 4)-4)0 2 b+d'

chapter

RIGID BODY DYNAMICS: BASIC CONCEPTS

8.1

INTRODUCTION

This chapter presents basic concepts associated with the kinetics of rigid bodies. We develop the linear and angular momentum expressions. The Newtonian and Eulerian laws of motion arc introduced as force and moment balances. We discuss analytical methods of obtaining equations of motion, such as Lagrange's equations and the direct application of D'Alembert's principle. We analyze qualitative and quantitative integration of the equations of motion, together with motion integrals. The developments in this chapter make extensive use of the geometric and kinematic properties developed in the previous two chapters. When describing the motion of a particle three coordinates are used, one for each direction of' the motion. motion. For For analyzing analyzing the the motion of a rigid body one needs to of the studied using both inertial use angular variables. The kinematics of' a rigid body is studied coordinates and relative coordinates. As we saw in the previous chapter. in most cases using a relative frame attached to the body gives better insight; this continues kinetics of rigid bodies. The use of body-fixed coordinates to be the case for the kinetics is attractive, as the moments of inertia of the body with respect respcct to a set of bodyfixed coordinates remains constant, and it becomes a simple task to differentiate the angular velocity. Thus, we derive several forms of the equations of motion, analogous to the concept of using different generalized coordinates. We extend the impulsemomentum and work-energy relationships to three-dimensional motion. This leads to a further study of impact problems. first introduced in Chapter 3.

421

422

CHAPTER 8• RIGID CHAPTER RicmBODY BODY

8.2

BASIC CONCEPTS BASIC CoNcEvrs

LINEAR AND ANGULAR MOMENTUM

Consider a rigid body, such as the one in Fig. 8.1, with center of' of mass G. Also consider an arbitrary point B, located on or off the body. The linear momentum of the body, denoted by p. is defined as

p=jr

vdm== vdm

dr

—c/rn

I

Jbody

body

di

[8.2.11

The position vector of the differential element is

18.2.21 where p is the vector connecting the center of mass with the differential element. Differentiating the above expression, we obtain the velocity of the differential element as

dr

V

—

di

=VG+WXP VG+WXP

[8.2.31

where w is the angular velocity of the body. Substituting Eq. [8.2.3] into Eq. [8.2.11. we obtain

(vG+wXp)dm body

pdm = rnvG+WX I pdm Jbody Jbody

VGJ

I

body

[8.2.41

Recalling the definition of center of mass from Chapter 6, we write

pdm =

0

[8.2.5]

Ibody

so that introducing Eq. [8.2.51 into Eq. [8.2.4], we obtain the linear momentum as

p=

mv(;

18.2.61 [8.2.61

As expected, the linear momentum of a rigid body is equal to its mass multiplied by the velocity of its center of mass. Hence, the translational motion of a rigid body

r

0 (fixed)

Flgur. 8.1

8.2

LINEAR AND ANGULAR MOMENTUM

be treated the same way as the motion of a particle having the same mass as the rigid body and where the entire mass of the body is concentrated at the center of

can

mass. The angular momentum, or moment of linear momentum, of the body about the

center of mass, denoted by H0, is defined as p

HG = I

x

[8.2.7] [8.2.71

v drn

Jbody

If we introduce Eq. [8.2.3] into this equation and make use of Eq. [8.2.5], [8.2 .5 ii, we obtain HG = f

Jbody

p X (VG + W X p) dm =

p

X (wX p) p) drn

18.2.8] 18.2.81

Jbody

We express the vectors p and w in terms of their conWonents in the body-fixed coordinate system xyz as

= W(i wi +

p = xi ÷ yj + zk zk

+

[8.2.9] [8.2.91

(p.o)p

18.2.10]

Using the vector identity

p X (w X p) = (p•p)w

—

and expressing H0 in terms of its components as

[8.211]

HG =

and carrying out the algebra, we obtain [(y2 [(y2 +

= I

—

—

XZWd drn

+ z2)wv

—

—

dm

+

—

—

dm dni

ibody

= I

Jbody

=

1(x2

18.2.121 18.2.12]

Recalling the definitions for mass moments of inertia derived in Chapter 6, Eq. 18.2.121 becomes

= IGxxWx v = 'G v

v—

H(;z = — IGzzWz lGzzWz or,

—

—

—

'Gx

x —

I;xzO.)x IGxzO.)x

—

lGxzWz

'G

z

lGvzWy lGvzWv

[8.2. 18.2.11 31

in column vector format,

[8.2.14] 18.2.141

{H0} {HG} = {!G1{w} where {HG}

= [HGx

I1Gv

{w}

and {!GIJ [IG] isisthe theinertia inertiamatrix matrixdefined defined in in Eq. Eq. 16.3.41. 16.3.41.

=

18.2.1 5]

423

424

CUAPTIR

RIGH) RIGID BO1)Y DYNAMiCS: BASIC CONCEPTS

The expression for the angular momentum about the center of mass can also be derived conveniently in column vector format. Writing Eq. [8.2.8] as HG =

I

Jbody

p X (w X p) dm =

—p X (p X (0) drn

18.2.161

Jbody

and using Eqs. [2.3.7], we can express Eq. [8.2.16] in column vector format as {HG}

18.2.17] 18.2.171

—1i3][,51{w}dm —I,3][,51{w}dm

= I

Jbody

Noting that

is a skew-symmetric matrix and using Eq. [6.4.2], we obtain [IG] =

[8.2.1 8] 18.2.18]

drn

I

J body

which leads to Eq. [8.2.14]. The expression for angular momentum is simplified considerably when the coordinate axes used coincide with the principal axes or, for axisymmetric bodies, the F frame. The products of inertia vanish, and the angular momentum expression about the center of mass becomes

18.2.191

HG = JGxxWx1 + lGvvWvJ + !GzzWzk

or, using b1b2b3 as the coordinate frame, or.

=

11w1b1

18.2.201

+ 12w2b2 + 13w3b3

where the body-fixed axes .xyz .xyz or b1 b2b3 are aligned with the principal axes. A special case of motion is rotation about a fixed axis. For example, for rotation about the z axis, w = wk, so axis, w H(; =

Exampi. 8.1

I

IGxzW1

+ IGzzwk IGzzwk

—

18.2.211

Find the angular momentum about the centroid of the rolling cone shown in Fig. 7.35 using

a set of centroidal axes that go through the symmetry axis of the cone.

Solution This cone is the one used in Example 7.9. Noting from Fig. 7.36 that h3 = cos f3f3 — sin f3f: and R == LLtan tan/3, /3,we wecan can express express the the angular angular velocity as (0

u

ucos/3

v

Lsin/3h3 = Lf2

v

u —

=

—

The centroidal moments of inertia of a right circular cone are 13 = + L2). The angular momentum then becomes

H( =

12w2f2

+

=

+

—

(a] = '2 =

[hi

It is clear that the angular momentum and angular velocity are in different directions.

8.3

TRANSFORMATION PROPERTIES OF ANGULAR MOMENTUM

8.3 TRANSFORMATION PROPERTIES OF ANGULAR MoMENTUM The transformation properties associated with angular momentum for a rigid body are very similar to the transformation properties associated with mass moments of inertia. We will see both the translational and rotational properties in sequence.

8.3.1

TRANSLATION OF

Consider angular momentum expressions about the center of mass and about an arbitrary point B. From Fig. 8.1 we express the distance from B to the differential element as s, where

s=rG,B+p

18.3.11

Introducing Eq. [8.3.1] into the definition of the angular momentum about B, we obtain =

s Jbody

X v dm = I

p) X (rG,B (rG,B + p)

+ to X p) drn

[8.3.21

Jbody

which, when expanded. yields

= I

[8.3.31 (r(;/nXv(;+pXvG+rG,RX(toXp)+pX(wXp))dm 18.3.31

ihody

The second and third terms on the right side of this equation vanish due to the definition of the center of mass. The last term is recognized as the angular momentum about the center of mass. The angular momentum about point B can then he written as

H8 = H(; + rnr(;18 X V(;

18.3.4] 18.34]

If the body is rotating about a fixed point C on the body, one can express the which, when introduced into velocity of the center of mass as VG = to X Eq. [8.3.3], yields

Hc = I

(rG,c+p)X(WXrG,c+toXp)drn (rGIc+p)X(WXrG,c+toXp)drn

Jbody

= I

X (r(;/c (r(;/c ++ p))dm (rc;/c + p) X (to X

18.3.51

Jbody

The column vector representation of Eq. [8.3.5] is recognized as the inertia matrix about point C times the angular velocity vector, or

{Hc} [IC]{w} {Hc} = [Iç']{w}

[8.3.61

Note that if point C is not attached to the body, Eq. [8.3.6] does not hold (except

when point C can be viewed as an extension of the body) and one should use Eq. [8.3.4] to calculate the angular momentum.

425

426

CHAPTIR B•

8.3.2

RIGID BODY DYNAMICS: Btsic CONCEPTS

ROTATION OF COORDINATES

While the translation theorem for angular momentum is valid only when one is considering the center of mass, the rotation relations are valid for angular momentum calculated about any point. Finding mass moments of inertia by rotation of axes can be carried out without any restriction on the origin of the coordinate system. We use a point B to which a set of coordinates is attached. We are given the angular momen turn of the rigid body about B, and using this set of coordinates we are asked to find the components of the angular momentum vector about a rotated set of coordinates. This is essentially the same question asked in Chapter 2: Given a vector in a certain coordinate system. what are the components of that vector in a transformed coordinate system? Given the angular momentum vector in the xyz coordinates as

+

+

H8 =

18.3.71

we wish to compute its components in the x'y'z' frame, obtained by rotating the xyz frame. We express the angular momentum in the x'v'z' frame as

+

=

[8.3.81

+

Given the direction cosine matrix between the two frames as [cl = can write in column vector format that

[R1T, we

[8.3.91

= [c]'{H,?} I c]'{H,?} = in which —

iz.j ILl

1118! — — L"Bx'

ii

LI

''By'

LI

1T

the components components of the the angular angular momentum momentum in the coordinate system x'v'z'. If' the the point point B B coincides coincides with with the the center center of of mass mass or or if' if' the the point point B B is is fixed fixed in in If' rotation, we have an interesting result. Denoting, Denoting. as we did in Chapter 3, such a point by D, and considering that {H0} = I in]{w}. the expression for the angular momentum in the x'v'z' frame should have the form are

18.3.1 11

= where {w'} =

18.3.121

cv

are the components of the angular velocity vector expressed in the coordinates and is the inertia matrix in terms of the primed coordinates. We proceed to show that this is indeed the case, One can relate the the angular angular velocities velocities in in the the two two frames as {cv'} = [clT{w}

{w}

= [c]{w'}

[8.3.13) [8.3.131

Introducing Eq. [8.3.13] into Eq. [8.3.9], we obtain [c1T[lv]{w} = {H;)} = [cIT{HD} = [clT[lv]{w}

18.3.141

8.3

427

TRANSFORMA11ON PROPERTIES OF ANGULAR MOMENTUM TRANSFORMA1'ION

[8.3. 14j 4j with Eq. [8.3.11] we recognize that Comparing Eq. [8.3.1

[8.3.1 5]

= [c]7[IDI[c}

which is the same same relation relation obtained obtained in in Eq. Eq.[6.4. [6.4.13] 131 when when discussing transformation of coordinates. Considering the complicated nawre of expressions associated with moments of inertia and angular momentum, the advantages of using the principal axes of a body as reference axes become more obvious. When finding the angular momentum of a body about a set of axes not coinciding with the principal axes, it is often more to then use Eq. desirable to find the angular momentum about the principal principal axes axes and and to [8.3.9] to find the angular momentum about the desired axes.

Find the angular momentum of the disk in Fig. 2.21 about point 0. The distance from 0 to B is L. from B to C is L/3 and the disk is of radius R.

Solution We will use Eq. [8.3.4] and the F frame to calculate the angular momentum. The reference frame is shown in Figs. 2.22—2.23. From Example 2.7 the angular velocity has the form

w=w,K+6j'+w31 w=w,K+6j'+w3i

[a]

where K = sin 40°k'

—

= cos Ok

cos4O°j'

—

j =j

Ib] Ibi

sin4O° cosOk sin40°

id

sin t9i 91

so that the angular velocity has the form

w=

(w3

sin40° sin 0)i + (0 (0

—

—

cos40°)j +

w, w1

w1

The angular momentum about the center of mass (denoted by C in Fig. 2.21.) is = HG =

= rnR2/2, about the center of mass as in which

Hc

= —

=

w1

[di Id]

+ IGvvWvJ I(;VvWVj +

=

Hence, we write the angular momentum

?nR2/4.

sin 400 sin 0)1 + (0

w

—

cos Ok]

[.1

next calculate the velocity of C. We first find the velocity of point B as We next VB =

w,K X Lj' = w,(sin40°k'

From Fig. 2.23 we have i' = VB VB

—

cos40°j)

x U'

= —Lw, sin 40°i'

[II [11

cos 01 + sin sin Ok. Ok. SO SO that that

sin 40° cos 01 = —Lw1 — Lw1 sin40°cos0i

—

sin4O°sin0k 40° sin Ok Lw1 sin

[g] Igi

We find the velocity of' the center of the disk as

v( =

O.)j X + O.)j

[hi

I

Example

8.2

428

CHAPTU

S•

R1G11)

BODY DYNAMICS: BASIC

in which rc,8 = L/3i, so that [—w1 sin 400 sin

=

X

01 + (0 — —

cos

w

sin 400 400 cos

[I]

cos 40°)k

Hence, the velocity of C becomes

=

—

Lw1 sin sin

400

cos 01 +

Lw1 sin Lwi sin 40° 40° cos Oj

1(cos 400 (e — w 1(cos

—

—

3

sin 400 sin 0))k

III We now calculate the second term in Eq. [8.3.4]. We note that rub FG/0 FG/0 X VG

L"

= Lj + L/3i so that

sin 40° sin 0 — — cos 40°))i

—

(o — w

+

— cos 40°))j (è — — w 1(3 sin 40° sin sin 00—

400))j + +

L2w1 sin 40° cos

Ok Iki

Multiplying Eq. 1k] by in and adding it to Eq. [ci gives the angular momentum of the disk about point 0. Note that Eq. [kj does not have any terms involving w3. w3, the spin of the disk. It is cleM that using Eq. 18.3.6] would lead to an incorrect result.

Exampi.

8.3

I

Find the angular momentum associated with the rolling cone in Fig. 7.35 about the pivot point

0, using the set of h21z3. of coordinates coordinates h1 h1h21z3.

Solution We We can solve this problem in a number of ways. We can calculate the angular momentum directly from Eq. 18.3.9] using the direction cosine matrix. We can find by direct integration the inertia elements and use Eq. 18.3.15]. Also, because the cone is rotating about point 0. we can find the mass moment of inertia and angular momentum components about 0 using the parallel axis theorem. We begin with the third approach. First, we use the frame ft and express the angular momentum about point 0 using these axes. Given the centroidal moments of inertia in Example 8.1, and that the center of mass lies on the away axis and is a distance of 3L/4 away from 0, Jo = = Jo

2) m(4R2 + II 6L 6L 2)

+m m )' =

1 '02 ==

101

13 103 = 13

=

rnR2

[a]

with with all the products of inertia zero. We proceed to find the inertia matrix about the h1h2h3 axes. Expressing the unit vectors as

h1 =

f1

cos j3f2 + sin h2 = cosj3f2

h3 =

—

sin /3f2 ++ cos cosj3f3 J3f3

Ibi

the direction cosine matrix has the form 1

Icc = I

0 0

0

0 cos f3

sinf3

—

sin f3

cos/3

[ci

8.4

RESULTANT FORCE AND MOMENT

the inertia inertia matrix in terms of the H frame as Using Eq. [8.3.! 51 we obtain the

[dl Idi

[u Jo] = [c17[,'Jo][c]

Note that we are using the notation introduced in Chapter 7 to denote the coordinate axes in which quantities of interest interest are are resolved resolved by by aa subscript subscript on on the the left. left Substituting in the values for for [C] [C] and and carrying carrying out out the the calculations, calculations, we obtain IH'OI =

0

0

101

0

101 COS2

0

1o1)sin(3cos(3 (/03 — 101)sin(3cos(3

(3 ++ '03

sin2

(103

13

—

(3 cos (3 101) sin (3 101)

Eel [.1

103cos213 + Io3cos2hB

The angular velocity vector can be expressed in the h1h2h3 frame as

(.0=—

V .

Lsinf3

[Ii

h3

The angular momentum can now he written as H0 = =

—(103

—

Ioi)

V COS V —

(103 cos2 (3 + 101 sin2

L sin

13h3

[g] Igi

Let us compare this result with the direct use of Eq. [8.3.9]. which yields

0

o

cos(3

0 sin(3

/3

cos (.3

0

1

—

0 0

0

0 0

0

0 vIL —v cos f3/L sin $

0 —

—Io1vsinf3/L v

!0i)vcosf3/L —

(3

[hi

which, of course, is the same answer as Eq. [g]. We note that the use of Eq. [8.3.9] is simpler, but that the use of Eq. [8.3.151 [8.3.15] gives gives more insight. This is because one sees what the inertia matrix associated with the transformed coordinates consists of.

8.4

RESULTANT FORCE AND

In many cases, a body is subjected to a multitude of forces and moments, and it becomes convenient to express the combined effects of these external excitations. These combined effects are called resultant resultant Jhrce and and resultant resultant moment. Consider a system of N particles, in which an external force and internal force act on each particle. The resultant of all forces is defined by F and given as N

N

F = i=I

+ F;) =

>Fj >F1 i=1

[8.4.1] [8.4.11

429

430

CHAPTER 8• RIGID

BODY DYNAMICS: BASIC CONCEPTS

FN F1

M1

MM

Flgur. 8.2

Applied forces and moments

The internal forces cancel each other. For a rigid body, we use the above relationship for N discrete forces. If there are distributed forces acting on the body, we consider a differential mass element with an external force dF acting on on it. it. The The resultant force is then defined as the sum of all external forces as N

F=>"Fj+ 1=1

dF

[8.4.21 18.4.21

body

Consider now a body acted upon by N forces F, (i = 1, 2, . . ., N) through points P, and M* external moments M (i = 1,1, 2, .. ., M*), as shown in Fig. 8.2. The resultant moment about point B is defined as Mt

N

N

F, + >'rp.,n XX F, MB= >'rp.,n

x Fj +

=

M*

18.4.31

1=1

in which

M*

18.4.41 i=1

is the sum of all external moments acting on the body. Note that because we are dealing with bodies that we assume to be rigid, the location of the individual individual momoments is immaterial when generating Eq. [8.4.3]. For flexible flexible bodies, bodies, this this is is not not the case; the location of the applied forces as well as moments becomes important. The combined effects of all external excitations are usually expressed as a resultant force applied through a certain point and the resultant moment about that point. A common choice is the center of mass, as shown in Fig. 8.3. In this case, we write Eq. [8.4.3 1 in terms of the center of mass as N

x F + M*

MG=

[8.4.51

i=1

When the resultant is expressed using a point other than the center of mass, say. B, the resultant moment becomes M8 = MG+rG,BXF

The concept is illustrated in Fig. 8.4.

18.4.61

8.5

EQUATIONS OF MOTION GENERAL EQUATIONS

F

Figure 8.3

Figure 8.4

8.5

Resultant force and moment

Resultant force and moment about B

OF MOTION

GENERAL

In this section we consider a Newtonian approach and write the laws of motion for translation as well as rotation. Consider the translational equations first. For a system of N particles, the force balance for each particle has the form = where

F

i = 1,2,..., N

18.5.11

F; is F is the the

is the mass of the ith particle and

the individual equations of motion, and

a

take the limit as N We replace replace rn1by bydrn. drn. aa by by a1 a1 and and the the summation with integration. The left side of Eq. [8.5.11 then becomes N

urn urn

>

=

I Jhody

adm = pnaG =

d

18.5.21 [8.5.21

The right side of Eq. [8.5.11 becomes Eq. 18.4.1]. Thus, for a rigid body, Newton's second law is expressed as

d —p=maG 7-p=maG di

=F

18.5.31

From a historical perspective, Newton developed his laws for the motion of rigid bodies, even though we first study them within the context of particles. Defining the inerlia force acting on the body as —rnaG, Newton's Newton's second second law can be described as the inertia force being equal and opposite to the applied tbrce. The law governing rotational motion was formally stated by Euler in 1775, together with Eq. [8.5.311. in its form above. The law states that the rate of change of the angular momentum about the center of mass of a rigid body is equal to the sum of all applied moments about the center of mass, or

s-HG = MG

[8.5.41

431

432

CHAPTER 8. RIGID BODY DYNAIvUCS: BASIC CONCEPTS

Equations [8.5.3] and [8.5.4] provide a complete description of the governing equations of a body. They constitute the basis of Newtonian mechanics, also known as the Newton —Euler rinulation. Equation [8.5.41 can also be derived from Newton's second law using a differof N N particles, taking the ential element or particle formulation. Indeed, for a system system of of mass mass distance from from the the center center of cross product of Eq. [8.5.11 with p,, where p, is the distance to the ith particle, we obtain = pxF1

= l,2,...,N l,2,...,N i=

[8.5.51

where the internal forces are not included in the formulation because they drop out. The term on the left side of this equation can be written as =

rn,

dv1

di

+

X m1v,

=

d

X rn1v1) rn,v1) =

d

18.5.61

is the angular momentum about the center of mass associated with the ith particle. If we sum Eq. [8.5.61 over N, we obtain the angular momentum of all the particles. For a rigid body, if we take the limit as N approaches infinity and add to it the sum of all applied moments, we obtain Eq. [8.5.4]. As discussed in Chapter 1, there is debate in the literature on whether Eq. [8.5.41 is a derived law or a stated law. One can argue that Eq. 18.5.41 is a stated principle, that is a fundamental law of mechanics, because of two major points: The internal forces of a rigid body are unknown, and for a deformable body the moment balance is a fundamental equation. as it leads to the symmetry of the stress tensor. (For a compelling argument in support of this issue, see pp. 259—271 in the text by where

Truesdel 1.)

The resultant of the applied forces and moments can be viewed in terms of the changes in the linear and angular momentum, as shown in Fig. 8.5. Hence, Figs. 8.2. 8.3, 8.4, and 8.5 are equivalent. We next discuss two issues associated with writing the equations of motion:

• •

Whether one can write the rotational equations of motion about points other than the center of mass. What types of variables and reference frames one can use to express the equations of motion.

The translational equations of motion are not about a particular point. Recall that linear momentum is an absolute quantity. To obtain the rotational equations of motion about a point other than the center of mass, consider the angular momentum

Figur. 8.5

Resultant changes in linear and angular momentum

8.5

GENERAL EQUATIONS OF MoTIoN

about about an arbitrary point B, given by Eq. [8.3.41. Recall that B is an arbitrary point

and not necessarily attached to the body. Differentiating Eq. 18.3.41 we obtain

d

=

d

+

d

X VG)

d

!nrG,R X aG + m(VG/B) X = —HG + !nrG/R

18.5.71

(Ii (/1

The sum of moments about B is given by Eq. 18.4.6], which we can write as

M8

X FF = MG ++ rG,B rG,B x

X + mr(;/fl X

18.5.81

The last term on the right of Eq. [8.5.71 can also be expressed as m(V(;/B) X V(; = tfl(VG

Vjj) X VG

—

PflVG X

18.5.91

so that introducing Eq. [8.5.81 [8.5.81 into into Eq. Eq. [8.5.7] [8.5.7] and considering Eq. [8.5.9] we obtain

d

= M8 +

18.5.101

X V8

Equations Equations [8.5.8J [8.5.8J and and[8.5.10.1 [8.5.10] are the most general forms of the moment balance equations. When point B is attached to the body, one can show that rG,B) X v8 = M8 + rn(co X rG/B)

18.5.111 (8.5.111

When the moment balance is written about a point C (on or off the body) that is fixed

in translation, we have =

18.5.121

We move on to the second issue: selection of the types of coordinates to express the equations of motion. For three-dimensional motion, using inertial parameters does not give much physical insight. For example, if we use an inertial coordinate system XYZ with unit vectors uK and denote by X, Y. and Z the coordinates of the center of mass, the force balances become rnA = F1 rnX=F'I

mY=FJ = F•J

rnZ=F'K = F'K

(8.5.13] 18.5.131

Unless the coordinates X, Y, and Z can be analyzed independently of each other, Eqs. [8.5.13) [8.5.131 are are not convenient convenient to to use use by by themselves. But consider writing these equations using a moving coordinate frame, which we will denote by xyz. Selecting the moving frame as a body-fixed frame, we write the components of the velocity, angular velocity, and force as V(; = (0 (0 =

WvJ + + WvJ

ui +

+

F =

+

[8.5.14] 18.5.141

+

To obtain the rates of change of the linear and angular velocities, we use the transport theorem. For the angular acceleration we have, as shown in Chapter 7, =

+ w7k + o X o =

+

+

[8.5.151

433

434

CHAPflR 8•

RIGID BODY DYNAMICS: BASIC CONCEPTS

a set of body-fixed coordinates. Differenso that tiating the expression for the velocity of the center of mass, we obtain

d VG = di =

.

+ VyJ + v-k u.k + w X VG

+

+

+

—

—

VzWx)J

+ [8.5.16] 18.5.16] — introducing this expression into Newton's second law, we obtain for the translational equations +

+

—

=

+

—

=

+

—

=

[8.5.17]

In column vector form, the translational equations are written as

[8.5.18] 18.5.18]

= {F} or,

using the notation for multiple reference frames, as

+

X V(;

[8.5.19]

F

We obtain the rotational equations in a similar fashion. Consider Eq. [8.5.41. We

express the angular momentum vector as

[8.5.20]

HGVj + HGZk HG = HG (1 + HGVJ

are defined in Section 8.2. We write the resultant moment about the center of mass as in which

Hc;y, and

[8.5.21) 18.5.211

M(; = Mc;xi + MGYj + MGZk

Using the transport theorem, the rate of change of the angular momentum vector

becomes

[8.5.22] 18.5.221

HGrel+(OXHc;

The gyroscopic moment is defined as w X HG. Hence, the rotational equations

can be viewed as the gyroscopic moment plus the relative change in angular momentum being equal to the applied moment. In column vector form we have

18.5.23]

{AGrel} + = {AGrel}

Substitution of Eq. [8.5.22] into Eq. [8.5.41 yields the rotational equations in

terms of the body-fixed angular velocities as —

Jyvav

—

—

1yz(0!z

—

—

—

(t)xwy)

— —

+

+ +

— —

—

—

—

=

—

—

—

=

—

—

—

—

MG1

= MG18.5.241 18.5.24)

8.5

GENERAL EQUATIONS OF MOTION

In column vector format we have

[IGI{a} + [&][JGI{o-} = {MG} [IGI{a} and,

18.5.251

in the notation for multiple reference frames, we have —HG

di

+ Al? w X HG

— —

MG

18.5.261

If there is a fixed point C on the body about which the rigid body rotates, the rotational equations of motion have the form t!c]{cr} tlc]{a} ++ [&l[lCl{w} [&l[lCl{w} =

18.5.271

The rotational equations are quite complicated even though there are no transla-

tional velocity terms in them. The complicated terms arise because both the angular momentum and the angular velocity change directions. We can achieve significant simplification of the rotational equations of motion if we select the body axes as principal axes. In that case, all products of inertia vanish, and we have 11a1

13a3

—

= M1

(12 —

—(/3

—

li)wjcoi =

—

—

12)W1W2

= M3 M3

[8.5.28] [8.5.281

where the indices denote the principal directions. Equations [8.5.28] are known as

Euler equations of motion, stated originally by Euler in 1750. 1750. Except Except for for certain certain special cases, such a4s some vehicle dynamics problems, Eqs. 18.5.28] are preferred over Eqs. [8.5.24]. Note also the order of the mass moments of inertia in this equation, which lends itself to an easy way of remembering these equations. Euler's equations are for the general three-dimensional motion of a body. In many cases, part of the motion of the body is constrained; thus, the body has less than three rotational degrees of freedom. In such cases, Euler's equations yield the equation(s) of motion as well as expressions for the reactions. The equations of motion expressed in terms of body-fixed coordinates give a lot of insight. Among their common uses is when conducting an instantaneous analysis, such as finding angular accelerations, finding the gyroscopic moment, or calculating the reactions at supports. Euler's equations can be integrated qualitatively to come up with integrals of the motion as well as for conducting a stability analysis. However, one cannot integrate these equations by themselves to find the orientation of the body at a given instant. The above derivation of Euler's equations was based on using a set of rotating axes attached to the body. As discussed in Chapter 7, when dealing with the motion of an axisymmetric body, it is more convenient to make use of the F frame. The F frame is suitable for axisymmetric bodies, because the inertia matrix matrix is constant in the F frame. One can use the F frame in conjunction with Euler's equations in two ways: One can perform the differentiation in the F frame, or one can express the equations of

435

436

CHAPTU a. RiGID RiGID Bony Bony DYNAMICS: DYNAMICS:BASIC BASICCoNCEPTS CoNCEPTS

motion using the components of the F frame. When performing the differentiation in the F frame, similar to Eq. [8.5.20] the translational equations of motion become

Pd

+

X rnVC = F

or

m{IvGrcl} + m1th11{vG} = {F}

18.5.291

The rotational equations of motion become

b.dH +

or

X HG = MG

{HG} =

+ 11w1 II'G I{wh}

= {MG}

[8.5.30] Equations [8.5.301 are known as the Euler equations. We will study these equations in more detail in Chapter 9. When resolving the equations of' motion along the components of the F frame, one no longer has a1 = (i = 1, 2, 3), and one must calculate {a} using the transport theorem. We write {IHG} =

[IG]{fa} + [f&j[JGJ{Fw} =

[8.5.31]

in which {Fa} =

Exaispi. 8.4

+ [FWfj{FW} =

+

[8.5.321

A slender bar of mass m and length L is attached to a vertical shaft, which is rotating with constant angular speed CL as shown in Fig. 8.6. Find the equation of motion for the bar.

The only variable in this problem is the angle that the bar makes with the shaft, so that the three Euler's equations should yield one equation of motion and two other equations for the reactions. Because the bar is rotating about the fixed point B, it is preferable to write the

B

L

Figur. 8.6

85

GENERAL EQUATIONS OF MOTION

b2

M1

F3

G

b3

FIgure 8.7 Euler's equations about the center of rotation. This way, we avoid calculation of the reaction forces at B. The free-body diagram of the bar is shown in Fig. 8.7. We express the moment about

Bas =

M1

b1 + M2b2 + rG//? X X mg(cos (lb1 — — sin Ob,) 8b2)

[aJ lal

Noting that rG,s = 1J2b1, we obtain — M8 = M1 b1 M2bL2 — b1 ++M2bL2

mgL sin (lb3

[bJ

The angular velocity and angular acceleration have have the form form

=

w1b1

a ==a1b, a1b,

+ o2b2 + w3b3 w3b1 =

cos (lb1 +

+ a2b2 + a3b3 = flOsinOb1 +

sin (lb2 + (lb3

[ci

+ Oh3

Id]

The mass moments of inertia about B are L—

1213 We

I•J

now invoke Euler's equations, which yield

Forh1

/1WlsinO = M,

For h2

I

cos 0 +

Igi —

mgL

[hi liii

Equation Equation [hi is the equation of motion, while Eqs. If] and Igi give expressions for the

reactions. We simplify Eq. [h] to

i—fi2sinOcosO+

=

0

'Ii Ill

437

438

CHAPTER 8•

RIGID BODY DYNAMICS: BASIC CONCEPTS

The expressions for the moment reactions become

II] LI]

Al2 = 2l2UOcosO

Al1

The reaction forces at B can be obtained from a force balance.

Example

8.5

Consider the gyropendulum in Example 7.5, repeated here as Fig. 8.8. Derive the equations of motion, assuming there is a moment M acting on the vertical shaft and that both shafts are massless.

Solution We will solve this problem in two ways: First, we will use the modified Euler equations. Second, we will resolve Euler's equations along the coordinates of the F frame. We split the gyropendulum into two parts: the vertical shaft and the disk and smaller shaft. The free-body diagram of the disk is shown shown in in Fig. Fig. 8.9. 8.9. Using Using the the relative relative axes axes xyz. xyz. are components of the reaction moments at point B in the x and z directions. The M, and and reaction thrces arc Fr. Because the assembly can be considered as rotating about point B, we can write the equations of motion about B. Observe that the xyz coordinate system describes the F frame. and it constitutes a set of principal axes, so that thai all products of inertia vanish. We have K = —

Ok. and cos0k. sin 01 + cos

/ Rvv =

'R.. ==

=4K+Oj'+

'liii thu

+ L2)

=

(a' Ibi

= —4sinOi + Oj + (qi + 4cosO)k

Oj' = —4sin0i+Oj+4cos0k

(ci

x

x

M

Fr

0 yf,

z y

yV

•1'

K_I. t.. K_I,

rng 'I, 7 S

Figure 8.8

z

Figure 8.9

8.5

Solution by Modified Euler Equations

GENERAL EQUATIONS OF MoTioN

The angular momentum about B has

the form

H8 —

+ L2)Oi +

—rn

+ 4cosO)k

Idi Id]

We next differentiate the expression for the angular momentum. The first term is the time derivative of HB in the F frame and it is obtained by simple differentiation of Eq. [dj as =

+

+

sin0 + 4)OcosO)i

+

+

+

—

4)6 sin sin 0)k 4)6

[.1 1.1

The second term in the derivative is

x

= w1XH8 = (—4) (—4) sin 01 + (3j + 4)cos0k)

+

61 +

+

>

+

+ 4)cos0)O 4)cos0)0 — m(_R2 +

=

+

sin0 — + 4cosO)4) sinO

+ 4)coso)k] I

If'

+

The external moment vector is M8 =

+ Lk X ?ngK =

+

- mgL sin Oj +

Igi [gi

Combining Eqs. [e]. [1], and and [g] [g] and separating the x. y, and z components, we obtain

+

iI

jj

+

+

+

k

—

+

[h]

2rnL24)Ocos0 = 2tnL24)Ocos0

m(

sin

)

0 cos 0 =

+ çbcos() çbcos0 — 4)0 sin 0)

—

rngL sin 0111

[I] [Li

We We have three degrees of freedom; thus, each one of Eqs. [h], [ii, and [ii is an equation

of motion.

Solution by Euler Equations Resolved Around F Frame

To obtain the solution this way, we first need to obtain the expression for the angular acceleration. Given and of in Eqs. [bj, we obtain the angular acceleration as (Yb

=

+ (Oj X

(*)brel ++ (Oj X = (*)brel

0)i + Oj + (i/i + 4) — (f)0 4)0 sin 0)k = —(4) —(4 sin th 0 + 4)0 cos 0)j 4 cos 00 — i/ik (—4)sin sin01 01++Oj Oj++4)cos0k) 4)cos0k) XX i/ik + (—4)

= (—4) sin 6

—

4)0 cos

0 + 04r)i 04r)i + (0 (0 +

sin 0)j +

cos 00 — 0)k 1k] Iki + 4) cos — 4)0 sin 0)k

439

440

8• RIGID BODY DYNAMICS: BASIC CONCEPTS

CHAPTER 8• CHAPTER

x

M

z

Figure 8.10 Substituting the values for the angular velocities velocities in in Eq. Eq. [bJ [bj and angular acceleration in Eq. [k] into the Euler's equations, we obtain

nO + 4UcosO

rn

R2

+

+

sin 0) + rn

+

—

sin

—

R2

+ 4)cosO) =

—

Eli EU

0

0) =

[U] Eu]

which can be shown to be the same as Eqs. [hi, [iJ, [ii, and and UI. [Ji.

Analysis of the Equations

rh], [ii, [jj, and UI U] are in The equations of motion in Eqs. [hI, in terms of the the internal internal reaction moments M does not not

appear. We relate the external moment M to the internal moments by conducting a moment balance for the vertical shaft about the Z axis, as shown in Fig. 8.10. Doing so, we obtain —

M +

— M., M.( sin cos (1(1 — cos

6=

0

1.1 1.1

We still stillneed needanother anotherrelationship relationshipfor for these moments. This This added these moments. added relationship be relationship can can be

obtained by separating the rod and the disk and analyzing the forces and moments these these two two members members exert on each other. We will not pursue this, but rather, consider the case when there is a motor between the rod and disk, which makes the disk maintain a constant angular velocity with respect to the shaft. Such a motor is called a servomowr. In the presence of a servomotor that keeps ç& constant so that it is no longer a variable. the system is reduced to two degrees of freedom. freedom. From From Eq. Eq. U] we get an expression for as =

which, when substituted into Eq. [o], gives an expression for M as

—M 4 M1cosO -M + —Al = sinO = sinO sin() 2 1

ffrcos2 o

.

sinO

_4)Ocoso)

['I IqI

8.5

GENERAL EQUATIONS

441

MoTioN

Let us next calculate the value of the external torque needed in order to keep the precession 4) rate constant. Introducing Eq. [qJ into Eq. [h] and setting 4) to zero, we obtain M =— In this case we have

Thu

cos 0 sin 0

sin 0 + 2mL24)O cos 0 sin 0 —

[rJ In

a single degree of freedom system, and Eq. [ii is the equation of

motion.

ROTATIONAL MOTION MOTION QUALITATIVE STABILITYANALYSIS ANALYSISOF OF ROTATIONAL QUALITATIVE STABILITY

Conb2b3 are are attached attached at the center sider a rigid body to which a set of body-fixed principal axes b1 b2b3 of mass. The body has an initial motion in the form of a rotation about one of the axes, say, b1. We We have wi b1. = (03 = 0. No external moments act, so that from Eq. [8.5.281 wo WO, wi = WO, remains constant. To investigate whether this motion is stable or not, a small moment is applied to the body, so that the angular velocities after the application of the moment become

= (0() ++ where

1.1 [.1

(03 =

=

(i == 1,1, 2, 3) are small quantities. We wish to determine the evolution of these per-

turbed angular velocities in time. Introducing them into Euler's equations and noting that no further moments are applied, we obtain = 0

+ è1) —(12 — — ('2 — 1363

— (11

—

1282 '282

12)(WO

—

(13

—

li)(ioo + Ii)(wo

Ibi IbI

= 0

+

=0

Because the change in the angular velocities is small, we linearize the above equations is constant, by eliminating quadratic and higher-order terms in (1 = 1, 2, 3). Noting that we get

[ci

0 — /282 + (1 —

= 0

[di Ed]

1363 ++ (12 1363 (/2 — Ii — 1,

= 00

Eel [.1

remains constant. To understand the behavior of the remaining two angular velocities, we conduct an eigenvalue analysis. Introducing the expansions Equation [c] states that

e2(t) = E2eM

E3(t) =

If]

E3eM

into Eqs. [d] and [e] we obtain into 12A

— —

13)WO

(/2 (12 — —

E2

At e =

0 0

Igi [gi

The solution of Eq. Fg] requires that the determinant of the coefficient matrix be zero, which yields the characteristic equation 12!3A2

—

(12

—

—

= 0

Ihi IhI

ExampI.

8.6

442

CHAPTER 8 CHAPTER 8

•

BASIC CoNcErrs

RIcED Boor

The sølution is

II)

A = A

and 13 < , both , or if 12 < 13 > solutions are are possible. possible. IfIf 12 12 > and 13 Two types of solutions are no no nonconservative there are imaginary. Because there roots of the characteristic equation arc pure imaginary. analysis indicates that rotation critically stable. A higher-level stability analysis lhrces, the system is critically of inertia axes representing representing rotation rotation about about the the minimum minimum and maximum moments of about the axes if is the largest or smallest moment of inertia, rotation about the are indeed stable. That is. it b1 axis is stable. is that is, if axis b1 b1 is 13 > 13 < or ifif 12 12 < and /3 On the other hand, if '2 > and 13

the intermediate moment of inertia axis, the roots of the characteristic equation are real, one positive and one negative. Hence, rotation about the intermediate axis of inertia is unstable. One or both of the angular velocities grow exponentially to a level such that they can no longer he considered as perturbations. A set of equations other than the linearized equations be used after that point. must be Next, consider the special case when two of the mass moments of inertia are the same. the symmetry symmetiy is the We have two possibilities. In the first, b1 is body. We such as in an axisymmetric body. critical stability. A imaginary, so there is critical From Eq. Eq. lii all A are pure imaginary. all roots A From axis and and 12 12 = shows that that perturbed perturbed angular angular motion which shows higher-level stability analysis can he conducted which of axisymmetric bodies spinning about the symmetry axis is indeed stable. Note that, unlike in the case of three distinct moments of inertia, the magnitudes of the moments of inertia did not affect this result. not The second case is descriptive of an axisymmetric body rotating about a transverse axis. leads to to zero is.ififb1 That is. b1 nY the symmetry axis, the rotation is about b2 or h3. This system leads the tumbling motion and transof the he motion motion can can be be explained explained as as aa continuation continuation of eigenvalues. II he to about the symfer of motion between the two transverse axes, with no rotation transferred to metry axis. We can summarize the results as follows. For general bodies, rotation about the axes with the minimum and maximum moments of inertia is stable. Rotation about the intermediate axis motion will will turn turn itself itself into angular motion with components of inertia is unstable. Any initial motion along every axis, resulting in some sort of wobbly motion. This phenomenon can readily observed by taking a book or other object with three distinct principal moments of inertia and spinning it about its three principal axes, as illustrated in Fig. 8.11.

1)-i

W2 (stable)

(unstable)

Flgur. 8.11

86

RotATioN ABOUT k FIXED Axis

The instability phenomenon is not encountered in axisymmetric bodies. Another way

ofexplaining this added stability due to axisymmeiry (or. for the general case. case, inertial syrnmetry) is to recognize that for such a body, every principal axis represents a minimum or maximum moment maximum moment of of inertia. inertia. In In enginecring enginecring applications, applications,one onetakes takesadvantage advantageofofthis thissEastability feature and designs rotating components. such as engine parts, as axisymmetric. In Chapter 10, we will study stability issues associated with rotating bodies in more detail. It turns out that in real-life applications. rotation is stable only around the axis of maximum moment of inertia. This is due to energy loss and transfer due to the flexibility of the body, and energy loss as a result of friction. In space mechanics, these issues were brought into the limelight after the Explorer satellite was launched in 1958, and nuuuional instabilities were observed. Ideally, the shape of a space vehicle would be like a disk. Because of practical considerations, satellites launched into space are spherical or slender cylindrical bodies.

8.6

ROTATION ABOUT A FIXED Axis

An important special case is rotation about a fixed axis. Representation of' such motion requires requires only only one one angular angular velocity velocity component. component. making making the the angular angular velocity velocity simple.

Two types of rotation about a lixed axis are of interest. One is plane motion. Chapter 3 discusses the kinetics of plane motion. In the second type. one is primarily interested in bodies that spin in a fixed direction, hut the axis of rotation is not a principal axis. In several situations a body spins about an axis that is fixed, for example a shaft used in power transmission, or the wheels of a vehicle. if the rotating body is not symrnetiic about the axis of rotation, the angular velocity of the body fluctuates or dynamic reactions are generated at the supports. Both of these are undesirable. Such behavior increases the loads on the supports, causing unnecessary vibration, noise, and possible damage. We select the axis of rotation as one of the coordinate axes, say, z, and note that this axis is not necessarily necessarily a principal axis. That is, the products of inertia do not not vanish. The components of the angular velocity vector are =

w

18.6.11

Substituting this into the rotational equations of motion, Eqs. [8.5.241, we obtain

M =

[8.6.31

in this problem there is only one degree of freedom, so that Eqs. [8.6.2J represent relations for the reactions, while Eq. 18.6.31 [8.6.31 is the equation of motion. It is interest-

ing to note that the reaction moments M1. and are influenced by the square of the angular velocity: this means that the problem of the axis of rotation not being a principal axis becomes more critical in high-speed situations. In many applications, such as the balancing of a wheel or a shaft, weights are removed or added to a rotating body. If only static balancing is being done, one aligns

443

444

CHAPTER 8• RiciD BODY

DYNAMICS: BASIC CONCEPTS

the center of mass and axis of rotation, but this does not eliminate the products of inertia. Dynamic balancing—such that the products of inertia become zero is a more difficult task.

Example

8.7

I

A disk of mass m and radius R is attached to a rotating shaft, as shown in Fig. 8.12. Due to a manufacturing defcct. the symmetry axis of the disk is not aligned with the shaft, but makes an angle of y. The shaft rotates with the constant angular velocity ft Find the reactions at the bearings.

Solution use a set of inertial coordinates XYZ. with the Z axis along the shaft, and a set of xyz axes attached to the shaft. shaft, so that the z axis is along the shaft. The x'y'z' axes are principal coordinates for the disk, and they arc obtained by rotating the xyz axes clockwise by an angle contiguration of y about the x axis. Fig. 8.13 shows the con tiguration and and free-body free-body diagram diagram of the system. The centroidal mass moments of inertia are = = rnR2/4, L,..1 = mR2/2, with Wc We

all products of inertia zero. From Eqs. [8.6.2] and 18.6.31, because .fI is constant we only need to find and it is easy to show that = 0, as the yz plane is a symmetry plane. To find we use the definition of a product of inertia and express the and z coordinates as

y = y'cos-y + z'siny

z = —y'sin'y + z'cos.y

1.1

so that that I,,,I'.- becomes becomes

= J

yzdm =

= sin'ycosy(L..1

+ z'siny)(—y'siny + z'cosy)drn —

sinycosy

=

Ibi

£1

V

Y

B A

lz L

I

Figure 8.12 y

a'

Y

U I

G

F4

mg

x

Figure 8.13

8.7 Introducing the values of I.,. and disk as

EQUATIONS

OF MolloN iN STATE FORM

into Eqs. [8.6.2] yields the moments acting on the

=

= 0

[ci [cJ

Consequently, Consequently, the disk exerts a moment moment — —

on the shaft. Consider now the reactions at the bearings. To this end, we will sum forces and moments for the shaft. To this end we express —M1 in terms of its components along the inertial coordinates XYZ as cos 0

=

sin 0

=

[di [dl

where () where 0 =

is the rotation angle between the XYZ and xyz frames. We are assuming that the center of the mass of the disk does not move. move. Hence, Hence, summing summing forces forces in in the X and Y directions we obtain F,tx = — FRX

= mg

FAY +

1.1

Summing moments about G for the shaft gives

Mx =

—

FAY4

(FAX

—

[(I

Introducing Eq. [c] into Eq. [dJ and solving solving Eqs. Eqs. [ci [ci and and [f [f 11 for for the the reaction reaction forces, forces, we [dl and obtain

FBX FBX

FAY FAY

lmR2

= —FAX

=

1

4

+

L

lmR2 L

lmR2

1

= 2mg —

4

L

flsinycosysinIkt 2

sinycosycosclt SIfl yycos SIfl cos yy cos cos

Ig]

The misalignment between the shaft and disk gives rise to bearing forces. These forces are related to the amount of misalignment and to the square of the angular velocity, and they are cyclic loads. For high-speed machinery, such misalignment can be detrimental, causing damage to the bearings as well as to the rotating parts.

8.7

EQUATIONS OF MOTION IN STATE

the previous two sections we derived the equations of motion. In order to extract more information from these equations, and to understand how the motion evolves, the equations of motion must be integrated. This integration can be carried out qua!itatively or quantitatively. The qualitative integration leads to impulse-momentum and work-energy relationships, which we cover in Sections 8.8 and 8.9, as well as integrals of the motion. Here, we look into the quantitative integration of the equations In

of motion.

445

446

CHAPTER

RIGID BODY DYNAMICS: BASIC CONCEPTS

Integration of rotational equations of motion is usually difficult to carry out by hand, except for a few special cases. Computational techniques require that the equations of motion be expressed in state form. Consider the rotational equations of motion independently from the translational equations for the time being, which is a valid assumption as long as the moments acting on the body are not functions of the translational velocities. We need to integrate Euler's equations, Eqs. [8.5.281, together with a set of equations that relate (i = 1,2. 3) to inertial quantities. We saw two the body-fixed angular velocities sets of equations that accomplish this in Chapter 7: the Euler angles and the Euler veparameters. The relationship between the Euler angles and body-fixed angular yefor a 3-1-3 transformation and at the table locities is given in Eqs. at the end of Chapter 7 for the 3-2-3 and 3-2-1 transformations. Let the body-fixed axes be the principal axes. One can then combine Eqs. [8.5.281 and 17.5.91 together. for a set of six first-order differential equations in terms of the variables 4), 0, 4), w1. W2,

as

•

M1

= ('2—13) I

'I

'I

•

(02=

7

= (Ii

—

—(01(03+ 12) '2)

(01(02+

'3

M2

'2 M3

'3

Siflt/J + w2C051/J)

= e =

—

= —

sinO

Sin i/I

(02 cos cos0 cos t// + w2 (w cos 0 sin t/;

+ &)3 1)3

[8.7.11

where t0 is conditions w(to) w(to) (i (i = 1,2,3), 4)(t0), 0(t0), and initial conditions subject to initial the rates of the Euler angles are given at t = t0. the initial time. If, instead of using Eqs. [7.5.71. If the external can be found from cb(Io), 0(t0), and

excitations M1 (i (i = 1, 2, 3) are not explicit functions of the Euler angles, then the excitations first three of the above equations, the Euler equations, can be integrated separately from the second three, the kinematic differential equations. One problem associated with the above approach is that Euler angles have discontinuities at certain values of the second rotation angle. For example, a 3-1-3 or 3-2-3 transformation is discontinuous at 0 = 0 or 0 = ± ir. The 3-2-1 transforrnation has a singularity when the second transformation angle is ±ir/2. In the neighborhood of a singularity, one can switch to a different set of Euler angles, but the process is tedious. By contrast, the Euler parameters discussed in Section 7.7 easily lend themin Eqs. Eqs. [7.5.9J. selves to numerical computation. Rather than using the three relations in

8.7

EQUATIONS OF MOTION IN STATE FORM

use the four relations in Eqs. [7.7.27] and have a set of seven first-order differential equations in the form we

(/2—/3)

.

WI = .

'I

.

(02

eo =

(01(03

=

+

M2

.

=

12)

=

1

1

—

— w3e3)

+

e1 = I

.

1

e2

7

11

(03

,

M1

+

I

=

w1e3) — cv3e1 + w)e3)

+

—

w2e3)

— w1e2)[8.7.21

The initial conditions for for the the Euler Euler parameters, parameters,e1(t0) e1(t0) (j (j = 0, 1, 2, 3) can be

found from the initial values of the Euler angles using the relationships given in Eq. [7.7.40]. The initial conditions from the angular velocities can be input either as quantities,or. or.ifif4)(io), given quantities, and are given instead, they can be converted to the angular velocities by using Eq. 17.5.71.

Note that even though the Euler parameters are related to each other by the Note

equation {e}1'{e}

=

+

+

+

=1

[8.7.3]

not need to solve Eqs. [8.7.21 together with Eq. 18.7.3]. Any accurate numerical solution that begins with an accurate description of (j = 0, 1, 2, 3) should lead to the correct solution. In fact, Eq. 18.7.31 can be used to check the accuracy of the numerical solution. In this regard, the Rodrigues parameters (see Problems 7.15 and 7.16) are useful, as they result in six differential equations. Another set of constains that can be used to check accuracy are the entries of the matrix staifls calculated by Eq. [7.7.30b]. Because this matrix is skew symmetric, its diagonal elements should be zero.

one does

Consider next the translational equations. When the equations of motion are written in terms of the center of mass, the translational and rotational equations are independent of each other, unless coupling exists as a result of the applied external forces. Consider a 3-1-3 transformation and the translational equations of motion as six first-order equations. As variables, we will use the body-fixed translational velocities v1, v2, and v3 and the coordinates of the center of mass in the inertial frame, which we will denote as A1, A2, and A3. The translational equations are given in Eq. 18.5.18]. To relate the rates of change of the inertial coordinates to the body-fixed velocities, we note that the relation between the velocity of the center of mass in inertial coordinates and body-fixed coordinates is {AVG}

= [R]"{RVG}

[8.7.41 18.7.41

[U1 v2 v3]T and [R] is the transformain which {Av(;} = [A1 [Ai A2 A3]T and {BVG} = [u1

tion matrix, given in Eq. [7.5.31 [7.5.3] for for a 3-1-3 transformation. Introducing Eq. [7.5.3] into Eq. [8.7.4] and using Eq. 18.5.181, we obtain the translational equations of

447

448

CHAP1IR B. Ri;rn BODY

DYNAMICS: BASIC CONCEPTS

niotion in state form as El V3W2 + V2W3 + —

=

V1

Ui ill

= A1

.

V7 = —VIW3 + V3WI +

—1)2W1 + V1W2 —1)2W1 + V1W2 +

F,

-

m

F3

= (cos 4) cos if; — sin 4) cos 6 sin i/i)ui + (— cos 4) sin i/i — sin sin4) 4)cos cos 1) 1) cos + sin = (sin 4)cos i/i + cos 4)cos 0 sin i/i)vi

+ (— sin4)sini/; sin4)sini/; ++ cos4)cosqicoso)v2 — — cos4)sin6v3

0 sin

A3

+ sin (3 cos

18.7.51

+ COS 0V3

Equations [8.7.1] [8.7.11 (or (or [8.7.21) [8.7.21) can can be be integrated integrated independently of Eqs. f8.7.5J as long as the the applied applied moments momentsM1 (I (I = 1, 2, 3) are not functions of or A1(i(i == 1, 2, 3).

Also, even though the latter three of Eqs. I18.7.51 8.7.51 are complicated expressions, they do not have the singularity problems that plague Eqs. [8.7.11. [8.7.1]. The 12 equations [8.7.1] and [8.7.51 (or the 13 equations [8.7.2] and 18.7.51) comprise the complete equations of an unrestrained rigid body in state form. In the presence of constraints, one must either obtain the equations of motion in terms of independent coordinates or include the constraint equations in the system description.

8.8

IMPULSE-MOMENTUM RELATIONSHWS

In this section. we integrate the translational and rotational equations qualitatively with respect to time. Integration Integration of' of' Eqs. Eqs. [8.5.31 [8.5.3] and [8.5.4] over two points in time t1 and 12 yields f12

d

—pdr = di f2 d I

—HG di = iii iii

I

ill

I

Fdt

p(r2)

—

it1

MG dt

p(t1) =

it1 ill

Fdz

f2 HG(12)

—

HG(1I) =

I

ill ill

M0 dt

18.8.11 [8.8.11

[8.8.21

which constitute the linear and the angular impulse-momentum relationships for a rigid body, respectively. The angular impulse-momentum equation, Eq. [8.8.2], can

be written in the same form forni when the rigid body is rotating about a fixed point. The primary uses of Eqs. [8.8.1] and 18.8.21 are for cases when the excitation is a function of time, in the presence of impulsive forces and moments, or when momentum is conserved. The drawbacks associated with working with these relations are that they are vector relations and their manipulation often involves calculation of the reactions and constraint forces as well as integration of these fbrces over time. Also, when the forces and moments acting on the body are functions of time, Eqs. [8.5.3J and [8.5.4] cannot be integrated by themselves.

8.8

449

IMPULSE-MOMENTUM RELATIONSHIPS

When the applied forces and moments are impulsive, the linear and angular When impulse-momentum relations are extremely useful. One must be careful when dealing with bodies subjected to impulsive loads, as some of' the reactions become impulsive as well. A special case of impulsive tbrcing is that of collisions, which we discuss in Section 8.12. When the applied moment about the center of mass is zero, or its integral over time during a certain period of interest is zero, the angular momentum in the beginning of the period is the same as the angular momentum at the end of the period. The angular momentum is conserved. Similarly. if the applied forces or their integral over a period of oftirne time is zero, the linear momentum is conserved. In some cases, cases, linear and angular momentum may be conserved about a certain direction, so that HG(t1) is is used. Denotonly a component of the relations p(t2) = p(ti) or HG(t2) HG(t1) ing the unit vector along which the linear momentum is conserved by e and the unit vector along which the angular momentum is conserved by f, we can write for linear momentum

p(12)•e = p(t1)•e

[8.8.3:1

and for angular momentum conservation

H(;(t1)'f H(;(t2)'f = H(,(t1)'f

18.8.41

The impulse and momentum conservation equations are very useful when more than one body is involved and motion is transferred from one body to another. A typical example is contact or loss of contact—between two bodies. In the absence the system of external forces, the only forces acting on the bodies are internal of two bodies. Hence, the linear momentum of the combined system and its angular momentum about the center of mass are preserved.

The axisymmetric spacecraft shown in Fig. 8.14 consists of a main body and a booster. The mass and centroidal radii of gyration of the main body and booster arc given in the table. The xvz coordinates denote the F frame and point C is the center of mass of the combined system. Just before burnout, the spacecraft has acquired a velocity of 2500 rn/s along the symmetry axis and an angular velocity about the symmetry axis of 0.1 rad/s. At burnout, the craft releases its booster. Immediately after the release, the center of mass of the booster has a velocity of 2000 rn/s along the x direction and 2 rn/s in the v direction. The angular

V

Booster

Main body

Ct)

4.

FIgure 8.14

Example

8.8

________-—

450

DYNAMICS: BASIC BASIC CoNcErrs CoNcEi'm CHAPTER 8• RIGID BODY DYNAMIcs:

= 0.9 radls. Find the = —0.4 rad/s, and = 0.5 rad/s. velocities of the booster are angular velocity of the main body and the velocity of its center of mass at this instant.

Mass

Location of C enter of Mass(on x axis)

1600kg

3.5m from A

0.35m

2 m

0.5 m

0.4 m

0.6 m

Part Main body

400 kg

Booster

from

A

Solution internal to the original original system, which is the All forces and rnonients monients during separation separation are are internal main body and booster together. Hence, during separation, the linear momentum and angular

momentum about the center of mass of the spacecraft are conserved. Rather than calculate the centroidal mass moments of inertia of the system. we prefer to work with Eq. [8.3.4], which relates angular momentum about two points. We denote quantities pertaining to the main body. booster. and system by subscripts M. B, and S. respectively. First we calculate the center of mass of the system. Using point A as the origin, we have in11 in11 0.5 0.5

[.1

3.5 = (m8 +

+

with the result 0.5(4(X)) + 3.5(1600) =2.9m = 2.9 m = 200()

r,ç=

[bJ

Let us first consider the linear momentum. Before separation the linear momentum is 200()i +2j. p = (rn8 + mM)25(X)i = 5(106)1 kg. rn/s. After separation we are given that so that the linear momentum after separation is + 2j) +

FflM(vMX1

+

UMYJ

p = 5(106)1 kg. mis VMZk) = p + vMZk)

id

Solving for the components of the velocity of the main body. we obtain 5(106)

V

=

—

mM

1J1B(VB1) rnB(vBX)

=—

5(106)

—

400(2000)

—0.5 rn/s —0.5

1600

VM z

0

[dl

Next we consider the angular momentum. We write the angular momentum of each body about the center of mass of the system. point C. as

Hfi + mBrn/c

X v11

HCM

X mBrM,c X = HM + mBrM,c

1.1 [.1

and H,,, denote the angular momenta of the booster and main body about their own center of mass, and rB/c = —2.41 m. m, rMfc rM/c = 0.61 in are vectors from the center of mass of the combined system to the centers of the mass of the booster and main body, respectively. Before separation. both components have a translational velocity in the x direction; thus, the where

8.9

ENERGY ENERGY AND WORK

cross products in the above equation vanish and we have + = Hc8 + HeM HeM = ('lxx + = = (400(0.42) + 1600(0.352)0. 1)1 = 26i kg • m2/s

[fJ

After separation, the angular momentum for the booster and main body are

+ IRywflJ JR w8J +

=

2.4i) XX(20001 + 1flB( 1flB( —2.41) (20001++ 2j) 2j)

HcM = (IMXXWMXI (IMXXWMXI + JMyyWMyJ JMyyWMyJ +

+ in,w(0.61)

X

(2625i

—

0.5j)

Igi

where the mass moment of inertias components are

'lxx = IflIK 'lxx

= 400(0.42) = 64 kg .

=

= 400(O.6) = 144 kg•

'Mxx =

= 1600(0.352) = 196 kg m2

'Myy =

= 1600(22) = 6400 kg• m2

[hi

= 0.9 radls. Hence, the angular momenta

and WIA- = 0.5 rad/s, Wjh = —0.4 radls, and

after separation are 321 32i

HCR HCM =

—

kg• m2/s 57.6j + (144(0.9) (144(0.9) — — 1920)k kg. +

+

—

480)k kg. m2/s

[ii Ii]

Summing the two angular momenta, equating to the angular momentum before separanon, and solving for the unknown angular velocities, we obtain

= 26—32

= —0.0306 1 radls

1920 + 480— 129.6 1920+480— 129.6

8.9

= 0.3548 rad/s

=

= 0.009 rad/s

[I]

ENERGY AND AND WORK WORK

In this section we examine the kinetic and potential energy for rigid bodies and develop expressions for the work done by forces and moments that act on rigid bodies. In essence, we extend the developments of Sec. 1.7 to rigid bodies. The kinetic energy of a rigid body is defined as T =

v•vdm -2 Jbody I

18.9.11 [8.9.11

where v is the absolute velocity of the differential element. Considering Fig. 8.1 we can express the velocity of the differential element in terms of the velocity of an arbitrary point B on the body as

v=v8+wXs

18.9.21

451

452

CHAPTER 8• RIGID BODY DYNAMiCS: BAsic CONCEPTS

Introducing this into Eq. 18.9.11 and performing perlbrming the dot product gives

T=

2

•

I

din +

s)' (0) (0) X X s) s) dm + VB • to (w X s)•

I

J

s dm

X i body

18.9.31 The first term on the right side in this equation involves a translation, while the

second term involves a rotation. The last term is a combination of translational and rotational terms, and it. involves the first moment of the mass distribution. This term becomes zero if one of the following holds: 1.

2. 3.

4.

B, the origin of the body-fixed coordinates, is v8 = 0, implying that the point B. fixed, so that the rigid body is rotating about point B. to = 0, implying that the body has no rotational motion. B coincides with the center of mass G. If the origin of the coordinate system is selected as the center of mass. then the integral of s dm vanishes over the body. If the vectors v8. to. and s are such that the dot or cross product vanishes.

The fourth case is mathematically possible. but not practical. Hence, for all practical purposes, the kinetic energy expression is simplified if the body is rotating about a fixed point, if it is not rotating at all, or if it is expressed in terms of the center of mass.

Let us consider the kinetic energy in terms of the center of mass motion. The first term in Eq. Eq. 18.9.3] gives gives the the translational translational part part of' of the kinetic energy, denoted by [ran,

as I

Ttran =

V(; VG

[8.9.41

The second term in Eq. [8.9.3] gives the rotational component of' the kinetic energy,

denoted by

To evaluate it we make use of the vector relationship

(aXb)•c = a•(bXc) so that letting a = to, b = p. and c = to X p. we have Trot =

11 I

w•pX(toXp)drn w'pX(toXp)drn

18.9.51

Jbody

Recalling from from Eq. 18.2.8 1 the definition of angular momentum, we can express the

rotational kinetic energy as 1

Trot =

18.9.61 [8.9.61

This equation can also be obtained using the column vector representation of

Eq. [8.9.5]. Indeed, noting that

(toXp).(wXp)=(pXw)•(pxw) (toXp).(wXp)=(pXw)•(pXw)

18.9.71

8.9

ENERGY AND WORK

whose column vector representation is

=

[fljl [fl]l [fillw}

18.9.8] 18.9.81

and recalling Eq. 16.4.2] we obtain the column vector representation of Eq. [8.9.7] as Trot Trot =

!

J 2hody

{w}'[15]'[j51{w}dm =

= !{W}T{HG}

2

2

[8.9.9] [89.91

Let Let us summarize the kinetic energy energy expressions. expressions. For For the the general general case of com-

bined translational and rotational motion we have have TT = Ttran Trot, in in which Ttran ++ Trot, Ttran

Trot = w HG =

=

=

18.9.10] 18.9.101

if the body is rotating about a fixed point C, we can write the kinetic energy as 7' and

=

Trot

=

=

[8.9.11]

if the body is only translating, with no rotational motion, we write T =

Ttran

=

I

Trot Trot = 0

V

[8.9.12] 18.9.12]

for this case, all points on the body have the same velocity v = VG. For plane motion, the kinetic energy can also be written about the instantaneous center of zero velocity. However, as the body moves the location of the instant center changes. Hence, the inertia matrix can also change. which creates difficulties if one needs to manipulate the angular momentum. Taking the center of mass as the origin of the body-fixed reference frame, denoting {VG} = and considering the inertia matrix and angular velocity vector, we get as,

V%I

=

+

+ v?)

Trot =

=

+

+

—

—

—

[8.9.13] 18.9.13]

If the coordinate axes are selected as the principal axes, the expression for the rotational kinetic energy further simplifies to Trot

in which the indices indices

=

+

+

[8.9.14]

IG(i = 1,2,3) denote the principal moments of inertia about the center of mass, and = 1,2,3) are the components of the angular velocity.

453

454

8.

DYNAM1CS RIGID BODY DYNAMiCS: CHAPTER 8 • RIGID CHAPTER

BAsic CoNcEvrs

along the principal axes. Observe that Eq. [8.9.101 is another proof that the inertia matrix is positive definite. Kinetic energy is an absolute quantity; both the translational and rotational kinetic energy are always greater than or equal to zero. The only way lèr {W}1[IG 1{w} to he greater than zero for all nonzero values of the angular to he he aa positive definite matrix. velocity {w} is for ['G I to Next, we write the rotational kinetic energy in terms of the Euler angles. Consider a 3-1-3 transformation. We recall that the angular velocity expression for a 3-1-3 31 -3transformation transformation is I

18.9.151 The rotational kinetic energy then becomes Trot

+

+

+

[8.9.16]

[B]' f/Gl [BIJ{fI} [/Gl [B]{O}

[8.9.1 7] 71

+

—

This relationship can he expressed in column vector form as [I(;1{w} fIG 1{w} =

Trot =

denote the time derivatives of the 0 in which {w} {w} = IBI{O} where {O} in Euler angles. The above form is not commonly used, because the corresponding IBIT[I(;][BI, is is time dependent. inertia matrix, IBIT[I(;][BI, = '2' the expression for rotational kiFor inertially symmetric bodies, say, netic energy simplifies to Trot = l[J(02

sin2 0) +

+

11/)2]

[8.9.18]

describe the work done on a rigid body, consider Section 8.4 and assume that M* moments M, (i = 1,2,..., points r1, r,, and M.* N forces F (i = 1,2, . ., N) at points on the body. Because the body is assumed to be rigid, the location of the are acting oil applied moments is immaterial. The incremental work performed by the forces and moments is defined as To

. .

+ M* •dO

dW =

[8.9.191

in which M* is the sum of' all external moments acting on the body, and we have treated the incremental angular displacement as a vector quantity by considering dO as a vector. It is more convenient to divide the above equation by dr and work with power. Doing so we obtain

P=

dW

dt

N

Fj.vj+M*.w

[8.9.20] [8.9.201

where v, is the velocity of the point to which the force F1 is being applied. The expression for power can conveniently be expressed in terms of the motion of the center of mass. To this end, we write the velocity v, in terms of the center of

8.9

ENERGY

WORK

mass as

I = 1,2,. 1, 2, . ., N

ti) X V, = V(; V, V(; + ti) and

introduce it into >. >,

• v1, v1, which which yields

N

N

18.9.2 1]

. .

N

+ w X p1) = F•VG

=

Pi

xE

18.9.221

Introducing this expression into Eq. 18.9.201 and using, from Eq. 18.4.5], the expression for lbr the resultant moment about the center of mass we write the expression for power as

P=F•VG+MG•W

[8.9.231

The work done is the integral of power over time ft2

Pdt

W

=

J "I

[8.9.24] 18.9.241

(F•vG+MG•o)dt

=

J

We can also write the expression for power in terms of the resultant of the forces

and moments about an arbitrary point B as

18.9.25] The general work-energy relation is 72 — T1

=

TIran2 + Tm17

=

—

—

[8.9.26] [8.9.261

W1

As we saw in Chapter 1, some of the forces and moments acting on a body may be conservative, conservative, that that is, is, derivable derivablefrom fromaapotential potentialfunction. function.ItItusually usuallyisismore moreCOflconvenient venient to work with the potential energy associated with such forces and moments. Recalling the potential energy energy V1 V, such such that that the infinitesimal work done by conservative forces and moments can be expressed as dW(. = —dV. we write •2

=

V1

—

[8.9.27] [8.9.271

V2

The reader is referred to Chapter 1 for further details. One can then separate separate the the work work expression into the potential energy and work of the nonconservative forces as

W1_, = —V2 + V1 + Wfl(.1.2 and

18.9.281 18.9.28]

write the work-energy theorem as + Trot + V1 + I

=

1tran2

+ lro(2 +

V2

18.9.291

The expressions for potential energy for rigid bodies do not lend themselves to any special form. The gravitational attraction between two bodies is a body force that is applied uniformly to every point on the body. Except for certain celestial mechanics or spacecraft dynamics problems. the gravitational attraction between two bodies is negligible compared to the gravitational attraction of the earth, and the force of gravity is conveniently represented as a single force acting through the center of mass. The gravitational potential energy becomes nig times the distance from a datum to the center of mass. For nonnatural systems, the energy integral is = T2 — To + V (note that here the subscripts denote parts of the kinetic energy), so that in conservation problems one should make use of the property that is constant.

455

8•

456

CHAPTER

Example

Find the kinetic and potential energies of the rotating slender bar in Example 8.4 and identify

8.9

the

RIGID

DYNAMiCS: BASIC CONCEPTS

integrals of the motion.

Solution The angular velocity of the bar is

w2b3 = —flcosOb1 = w1b1 + w2b, + w2b3 —QcosOb1 +

+ 0b3

The mass moments of inertia about point B are It

/2 = 12

0

13

[bJ Ibi

=

so that the kinetic energy is

T=

= hJ(fl2sjfl2O + sin2 0

=

Ic]

sin2 0 +

Using point B as datum, the potential energy has the form —

mgL

cos 0

Ed]

We observe that this is a nonnatural system, and that the external forces and moments do not do any work, as they arc all reaction moments. Hence, an integral of the motion is the Jacobi integral. =

T2

— To T0

+V=

— —

sin2 0)

If we are given the value of 0 when 0 =

cos 0 = constant

—

1.1

and arc asked to find the value of 0 when

0 = 02. 02. we can solve for 02 using —

8.10

ftSiflOL)

—

=

—

Il2 sin2 02)

—

[f]

LAGRANGE'S EQUATIONS FOR RIGID BODiES

The extended Hamilton's principle and Lagrange's equations discussed in Chapter 4 have the same form whether they are written for particles or rigid bodies. in this section, we analyze analyze the generalized forces, quantities suitable to be used as generalized coordinates, coordinates, and the nature of the Lagrange's equations for rigid body motion. We We also develop develop a physical interpretation of generalized momentum expressions.

8.10.1

VIRTUAL WORK AND GENERALIZED FORCES

One can obtain expressions for the virtual displacements by calculating the velocities and can replace time derivatives with the variational notation. To calculate the virtual work, we can take the expression for incremental work or for power, and make the

8.10

LAGRANGE 'S EQUATIONS EQUATIONS FOR FOR RJGw RJGw BODIES BODIES

appropriate substitution. For example, one way of expressing the incremental work done on a rigid body is Eq. [8.9.19], repeated here as N

M*.dO dW= >'Fj.dr1 ++M*.dO

18.10. 11

i—I

The virtual work has the form N

>'F1.8r1 >'Fj.8r1

18.10.2]

i—I

We express the virtual work in terms of the generalized coordinates. For a system with n independent generalized coordinates, the variation of 6r1 can be written as n

6r1 =

i

>1

=

18.10.31 [8.10.31

k

k=1

The problems of dealing with and its derivatives were discussed earlier. We also saw in Chapter 4 that if a vector r is a function of n generalized coordinates (k = 1, 2, ..., .. ., n) and time 1. we we can write Introducing this = substitution into the variation expressions, we obtain

6r1=

=

8qk k—I

k=1

[8.10.41 18.10.41

cIqk

Substituting this equation into Eq. [8.10.2] gives

n/N

a

6W =

=

aw

di'1

>1 k--I k--i \i=I i=I n(N

18. 10.5] 18.10.5]

so that so thatthe thegeneralized generalized forces forces havehave the form the form N

Qk=> Qk=> F1•

(20)

81'l

k = 1,2,

. . .

[8.10.6]

The expression fbr for the incremental work or power can be expressed in terms of the resultant of all lbrces forces F acting through the center of mass and the resultant moment MG about the center of mass as Eq. [8.9.23]. The associated virtual work expression has the form =

18.10.71

and the associated generalized forces become

Qk=F' dVG +MG• Qk=F .

18.10.81 [8.10.81

When the resultant resultant of' of forces and moments is expressed as the resultant force F about a point B and resultant moment MR about B, as in Eq. [8.9.251, the virtual

457

458

CHAPTIR S•

RIGID BODY DYNAMICS: BASIC CONCEPTS

work can be expressed in terms of the motion of point B, and the generalized forces become

Qk =

t-1(j)

,.

[8.10.9]

,.

Depending on the problem at hand, one can use any one of Eqs. 18.10.61. [8.10.8], or [8.10.91 to calculate the generalized forces. For a system of N rigid bodies, Eq. 18.10.8] can be extended as

Ni Qk

= >

dVG. .

+ MG1

.

do1

18.10.10]

.

denotesthe theresultant resultant of of all all external external forces forces acting on the ith body, and which F denotes in which is the resultant moment about the center of mass of the ith body.

8.10.2

GENERALIZED COORDINATES

A set of generalized coordinates suitable to describe the translational motion of a rigid body are the displacements of its center of mass. However, the associated eralized velocities do not give too much insight: it is more suitable to use the components of the velocity in a moving coordinate frame, such as a set of body-fixed axes or the F frame. In essence, velocity and angular velocity components in a body frame are quasi-velocities (generalized speeds). Lagrange's equations. on the other hand. deal with generalized coordinates and generalized velocities. Using quasi-velocities is not feasible with the traditional form ot' Lagrange's equations. associated Consider the Euler angles of precession (d), (d). nutation (s), and spin with a 3-1-3 Euler angle transformation. For the translational motion, we use the inertial coordinates A1, A2, and A3 of the center of mass. Recall that the angular velocity expression for a 3-1-3 transformation is =

+

+

[8.10.11] 18.10.11]

+

+

—

The kinetic energy is T =

+

=

+

+ Ocqi)2 + +

Al ++

+

+ —

+

+

+

[8.10.121 18.10.121

Note that the precession angle is absent from T, so that can become a cyclic coordinate if it is not present in the potential energy and in the virtual work. Differentiation of the kinetic energy with respect to the Euler angles and their derivatives is cumbersome. An alternate way of dealing with the derivatives of the kinetic energy is to retain the angular velocity expression in the formulation as long as possible, as we will demonstrate. In the next chapter, we will see a modification to the

8.10

LAGRANGE'S EQUATIONS FOR RIGID BODIES

Lagrange's equations that makes it possible to derive derive the the equations equations of' motion in motion in terms of the angular velocities. terms velocities. Consider the physical Consider physical interpretation interpretation of the generalized generalized momenta terms assowith the Euler ciated with angles.The Therotational rotationalkinetic kineticenergy energyisiswritten writtenas asTrot Euler angles. Trot = w' Note that because we are are considering considering an unconstrained unconstrained rigid body the the translational kinetic energy energy has has no terms involving the derivatives of the Euler angles. We first first analyze the the generalized generalized momentum associated with the precession precession 18.10.121 with respect to 4) angle 4). 4). Differentiating Differentiating Eq. 18.10.121 4) yields

= aT = • •

18.10.131

.

(:14) (14)

= 11(disosqi + (Ici/i)s(sfi (Ici/i)s(sfi++12(4)sOci/i 12(4)sOci/i— —

Osi/J)sOcL/J

i/t)cO + 13(4)C6 13(4)c0 ++ çl')cO

Recalling the definition of the rotational kinetic energy. we can write 9w (9w

I1

= =— —

.

2c24)

=

(9w .

HG

•HG + (t.) •HG+w•

(9

..

1

—{w} II!GI{W} —{w} [JGI{W} =

.

= 11w1

.

+ 12W2 /2W2

(9(tJ .

+

[l(;}{w}

[8.10.14] [8.10.141

..

(94)

(94)

(94)

.

(14)

(94) (9w1 (9W1

(1W'

also as

(94)

velocity vector vector with with respect respect to to the the rates rates of of change change The derivatives of the angular velocity from Eq. [7.5.4] as of the Euler angles angles can he obtained from (9w

=a3

(9w

(9w

.

(90

(94)

=b

.

[8.10.15] [8.10.151

(1k/I

In order to generalize Eq. [8.10.151 to any Euler angle sequence, we denote by the the unit vector about whose direction the Euler angle transformation with 4) is

conducted. Considering our coordinate system obtained by a 3-1-3 transformation, and e11, as the unit vectors about which the 0 and = a3. Similarly, we define rotations are performed. For the 3-1-3 system under consideration

=

(9w •

=

e() =

(9w •

.

= a1 a1 =

—

=

sin

(96

(94)

(9w •

=

b3

(9k/I

18.10.161 velocities associated with are in essence the partial velocities are + a3 = sin 0 sin the rates of the Euler angles. Also, from Eq. [7.5.5] we have a3 the rates +cos 0b3. lightof ofEqs. Eqs.f8.10. 18.10.131—18.10.151, 131—18.10.15], can can be be expressed expressed sinO6 cos sin 0b3, which, which,ininlight

e(/), e0, e0, and The unit vectors e(/),

as (9W1

(9W,

(94)

(14)

.b1+

dw1/ô4) (i = Physically, dw1/ô4)

.b2+

:b3

[8.10.17] [8.10.171

cI4) (94)

2, 3) is equivalent to the direction cosine, or. the cosine of the angle between the vectors vectors b1 b1 and 1,

459

___

460

Ricw Bony BonyDYNAMICS: DYNAMICS: B.&stc CONCEPTS CONCEPTS CHAPTU 88•. RIGiD CHAPTU

Considering Eqs. [8.10.161 and Eq. [8.10.14]. [8.10.14], one can express the generalized momenta associated with the Euler angles as =

= ep•H; ep H;

= eØ•HG e0 • HG

HG

[8.10.181

Thus, the generalized associated wit/i the Euler angles are the components which the Euler angle rolattons of the the angular ;nomentwn along tile directions about which

have been pertorined. The rates ol change of the unit vectors de0

=

dt

=

di

can be shown to be

and (9w

de,,,

(90

dt

=

(9w

[8.10.19J

(911/

Equations [8.10. [8.10. 119J 9J can be obtained by differentiation of Eqs. [8.10.16] and making the proper substitutions. We next examine the generalized forces associated with the Euler angles. The virtual work expression can he written as

= are

and

where

=

+

[8.10.20] [8.10.201

+

the generalized forces. We can write the rotational equa-

tions of motion as il d

(91'

-

d

=

— —

d

=

(96

Qo

= Qp

—

[8.10.21]

Now let us demonstrate that Lagrange's equations for an unconstrained rigid

body are the angular momentum balances in the directions about which the Euler angle transformations are made. and are the components of applied exterWe begin by showing that nal moment about the center of mass G (or center of rotation C) in the directions of the Euler angle rotations. Assuming N forces are applied to the body, we write the resultant moment as N

[8.10.221 1=1

where

(i =

1,

2.

.

.

., N) are the position vectors of the points at which the forces

are applied. Let us define the components of the external moment about the through which the Euler angles arc transformed as =

eçj,

M(;

D1)

=

•

MG

=

•

M(; MG

[8.10.231

Take, for example. a 3-1-3 transformation and the spin angle i/i. Considering Eq.

[8.10.22], we can write the the component component D11, D11, as as N

Dli, = edl•MG ed,•MG

XF1)

= i—i

[8.10.24] [8.10.241

8.10

LACRAN(;E's Eou.vrloNs FOR RIGID BoDws

b.)we we can write Recalling the vector identity a• (b X C) == cc • (a XXb.) N

=

/ F1 • i=1

(e,1,

18.10.251

X

-J

We next evaluate the expression e11, X p. Recalling the definition of the generalized force Qk as N

k = 1,2,...,ii

18.10.261

To this end, we write X and we explore the relationship between terms of the body coordinates and in column vector form as

[8.10.271

{BPi} = 1R3(l/J)11R2(O)IIRI(441{Apj} {BPI}

so that differentiation of {BPI} with respect to (9{J?pI} (9{J?p1}

in

yields

= a[R3011)] a[R3011)j

(ct)I{API}

[8.10.28]

(?lIJ

It is It is easy easy to show that

[8.10.29] [8.10.291

= we obtain

and then, introducing Eq. [8.10.291 into Eq.

th/i

=

{.4pi} {.4pi} =

[B. 10.30] [8.10.30]

N

[8.10.31]

whose vector counterpart is

=C1pXpi

,

(?

Equation [8.10.311 is a very useful relationship. Introducing it into Eq. [8. 10.25 1. is indeed indeed the thesame sameasasD111. Di,,. and and itit has the we conclude that

-"F• i=l N

(1

=

= e,1,

M(;

[8.10.321

-a

i—i

[8.10.321 is is applicable applicable to the other In a similar fashion, one can show that Eq. [8.10.32] which Euler angle transformation sequence is Euler angles as well, regardless of which used, so that the generalized generalized Forces forces associated with the Euler angles are indeed the components of the applied moment in the direction of the Euler angle rotation, and N

QI, QI,

=

>Fj. 1=1

e() 'M(;

Q4) = Q4)

= e(/)•M6

[8.10.331

461

462

CHAPTER 8

8. 1 0.3

BASICCONCEPTS BASIC CoNcErrs

RIGID BODY

ANGLES OF TIlE EULER ANGLES

lAGRANGE'S EQUATIONS

Consider the dot product between the rate of change of the angular momentum and the axes about which the Euler angle transformations are made. the unit vectors for instance, instance, HG'es, e0, which which can be expressed as Take, for = Hc;'e()+HG'èo—HG'é() Hc;'e()+HG'èo—HG'è() d d = = di di .

18.10.341

Considering Eq. [8.10.191 as well as the definitions of the rotational kinetic energy and angular momentum. we obtain T

HG

=

= {

=

=

18.10.35]

Introducing Eq. Eq. [8.10.35] [8.10.35] into Eq. [8.10.34]. repeating the procedure for the Introducing other Euler angles, and considering Eqs. [8.10.32] and [8.10.33], we conclude that Lagrange's equations for an unconstrained rigid body can be recognized as the components of the moment balance along the directions about which the Euler angle rotations are performed: ci

—

cli Li

d —

di

—

—

=

or

HG •e4, = MG •eth

=

or

HG •

=

or

HG e1p = MG ep

(:IT

e0

= M(;

•

[8.10.361

If the motion of the body is constrained in some fashion, Lagrange's equations for the rotational motion no longer have this form. Rather, one can only write them in their traditional form. Also, for rotation about a fixed point, Lagrange's equations are the moment balances about the center of rotation. Unlike Euler's equations, Lagrange's equations are angular momentum balances about a set of nonorthogonal axes. The decision as to which form of the rotational equations to use depends on what type of explanation is sought. In general, if all three Euler angles need to be used and they are independent of each other and the general motion of the body is analyzed, it is easier to deal with Euler's equations, as they are in terms of the angular velocities and lead to simpler expressions. The advantage of Lagrange's equations over Euler's equations becomes more apparent when one studies the motion of interconnected bodies. With Euler's equations, one must separate each coniponent of the body and write the Euler equations for each body. Then, one eliminates the reaction forces by combining the individual equations. When using Lagrange's equations, one can circumvent part of the labor and avoid lengthy expressions by writing the kinetic energy in terms of the angular velocity components and then using the partial derivatives of the angular with respect to the Euler angles.

___ 8.10

8.10.4

LAGRANGE'S EQUATIONS FOR RIGifi BoDiEs

lAGRANGE'S EQUATIONS IN TERMS OF EULER PARAMETERS

Another choice for generalized coordinates is the Euler parameters. In this case, we deal with a set of constrained generalized coordinates. Recall Eqs. [7.7.291 and [7.7.30c], repeated here as {w}

= 2[EJ{è}

IEI{e} = {0}

[8.10.37a,b] 18.1O.37a,bJ

The rotational kinetic energy and virtual work have the form The

Trot =

[8.10.38] [8.10.381

= {M(;}1'{60}

We We are not considering the translational kinetic energy, as it is not a function of the Euler parameters. Using Eq. 18.1 0.37a], we can write

[8.10.39]

{60} =

Introduction of this and Eq. [8.lO.37a] [8. l0.37a] into Eqs. [8.10.381 yields the kinetic energy and virtual work in terms of the Euler parameters as

Trot = 2{e}TIEITIIGIIE]{e}

[8.10.40] [8.10.401

=

{e}T{e} = 1, this exBecause Because the Euler Euler parameters parametersare arerelated relatedtotoeach eachother otherbyby{e}T{e} pression needs to he used as the constraint relation. We proceed with taking the partial derivatives of the kinetic energy as

(9{e} (1

di'

)

=

= 4{j,}T[E1T1J 4{e}T[E1TIJIIEI IIEI + 4{è}T[EJTIJG 4{è}T[E1TIJGIIE] lIE] + 4{è}T [E]1

[8.10.411 can show, by examining the elements of [El, [EL that that [É]{è} [É]{è} = {0}, so the second term on the right side of this equation vanishes. To obtain the derivative of the kinetic One

energy with respect to {e}. we note from Eq. [8. 1 0.37hJ that

[8.10.42] [8.10.421

[E1{è} = —IEI{e} —1E1{e} and

then. introducing Eq. [8.10.42] into dTrot

(i{e}

=

ci

and differentiating. we obtain

= 4{e}T 4{e}T iEiT iEiT 1/GlEE [E]' [El' [1;1 [I(;l [EHe} = [/Gl[E1 1

c9{e}

[8.1 0.43] 0.43]

Using Eq. [8.10.42] we Using Eq. [8.10.42] we can also show that that the the last last term term on the right side of Eq. [$.10.41] Eq. 10.411 can be written as

4{è}TIE1TI/GIIEI 4{è}TIE1TI/G IIEI = —4{e}T[Ej7[IG][Ej —4{e}T[EP'[IGJ[Ej

[8.10.44] 18.10.441

The modified virtual work, in the presence of the constraint, has the form

= 2{MG}TIEI{6e} 2{MG}TIEJ{6e} + 2A{e}T{6e}

[8.10.45]

463

464

CHAPTER

8

RiGID BODY DYNAMICS: B.?tsic CONCEPTS

Combining Eqs. [8.10.41], t$.1O.431, and and [8.10.45], and taking the transpose

01 the resulting expression. expression, we we obtain obtain Lagrang&s equations in terms of the Euler parameters as = O.5[EIT{M(;} + O5,k{e}

—

[8.10.461

subject to the constraint {e}T{e} = I.

8. 1 0.5

LAGRANGE'S EQUATIONS AND CONSTRAINTS

The Lagrangian treatment of dynamical systems subjected to constraints was discussed in Section 4. 1 0. The treatment of three-dimensional motion is essentially same. One basically has to decide how to handle the constraints. When the constraint is holonomic. holonomic, one has a choice: generate a set of independent generalized coordinates that take into account the constraint, or deal with constrained generalized coordinates. When the constraint is nonholonornic. one use constrained generalized coordinates. When using constrained generalized coordinates, one has two options:

Introduce the kinematics of the constraint into the problem by of Lagrange multipliers. When the constraint is expressed as a constraint, it augments the Lagrangian. When the constraint is expressed iz

a.

velocity form, it augments the virtual work. Both cases lead to a set of equations t: terms of the Lagrange multipliers. After obtaining the equations. one can seek t eliminate the Lagrange multipliers and generate a set of independent equations motion.

Introduce the constraint forces into the formulation. One does this by relaxing the constraints and accounting for the effect of the constraints by means of the constraint forces. These forces enter the equations of' motion via the virtual work. the equations of motion are obtained, the kinematics of the constraints are intrcduced. and an expression is developed for the constraint force. Mathematically. this approach is equivalent to expressing the constraint in velocity form. The force is essentially the Lagrange multiplier. While mathematically not any from the others, this approach gives better physical insight. It is particularly usefui when dealing with holonomic constraints and in the presence presence of of friction forces. also is a suitable approach to calculate the magnitudes magnitudes of of constraint constraint forces forces that that do do r: b.

work.

Example 8.1 0

Derive the equations of motion for for the the spinning spinning top top shown shown in inFig. Fig.7. 7.13 13 using a 3-1-3 angle transformation, and analyze the motion integrals. The ccntroidal moments of inertia

I.

from the the bottom bottom of of the the top. top. '1. 12 = 1. and 13. The center of mass is at a height of L from that the point of contact between the top and the ground is stationary.

Solution The motion of the top can be viewed as rotating about the fixed point 0. The kinetic energ) T =

Trot

=

1, I

(101

+ + 1Q7W lmw-,

+

103(1)3

[.1

8.10

LAGRANGE's EQUATIONS FOR RIGID BODIES

where the components of the mass moment of inertia about point 0 are + mL2

=

+ niL2

'02 =

In3 =

[bJ [bI

/3

Introducing the values for the angular velocities for a 3-1-3 transformation from Eq.

[8.9.15]. we obtain for the kinetic energy (4)2

T

+ 0+ sin2 o

+

cos A + +

)2)

[cJ

The potential energy is

V = ingLcosO

Idi

The only other forces acting Ofl the spinning top arc those at the point of contact. They do no work, as they are being applied to a fixed point. It follows that the virtual work expression is zero.

Before obtaining the equations of motion, let us first analyzc the integrals of the motion. The virtual work is zero, the Lagrangian is not an explicit function of time, and all terms in the kinetic energy are quadratic in the generalized coordinates. It follows that the first integral of the motion is the Jacobi integraL integral, which for this case is the total energy

T + V = constant

1.1

Examining Eqs. [c] and [dl closer, we note that the Lagrangian does not have any and dependency. We conclude that the generalized momenta associated with these two coordinates are constant, which gives us two more integrals of the motion in the form =

aT

Slfl UU + = '01 / Slfl

=

A + i/i) cos cos U

=

+ i/i)

A = constant U

constant

[I]

[gi

first integrals are the components of the angular momentum along the a3 and directions. We will discuss their physical interpretation further in Chapter 10. Equation [g] is recognized as = constant. The first two equations of motion are then obtained by differentiating the generalized momenta associated with 4) and '1'. and they have the ftrm These

=

0

IhI

= 0

by invoking invoking Lagrange's equations. We We find the equation of motion associated with A 6 by have + ,nL2)O ,nL2)O = (11 (11 +

8T

=

+ inL2)4)2sinOcosO inL2)4)2sinOcosO — —

+

m9

= —rngLsinO

[I] [U

leading to the equation of motion + inL2)O inL2)O

—

+ ,nL2 — J3)42 Sin sin 0 COS cos 00 +

sin 0

—

ingL sin 0 = 00

[ji [ii

465

466 Example

8.11

CHAPTII B. RIGID BODY DYNAMICS: BASIC C0NCEVIS

I

slender bar of mass m and length L, shown in Fig. 8.15, is attached to the arm of a shaft which is rotating with a constant speed of ft The length of the arm is a. Find the The

equation of motion for the bar and the equilibrium position(s). Then, analyze the stability of the equilibrium.

Solution We attach a h1 b2b3 coordinate system to the rod and write the angular velocity of the rod as

the summation of the rate of change of (1 and the angular velocity of the shaft as

I.] 1.1

+0b1

=

The mass moments of inertia about the center of mass are

13 13 =

12

[bJ

To To evaluate the kinetic energy, we need to find the

velocity of the center of mass. Using

the relative velocity relation

[ci

VG =VR+WXrG/B where rG,B =

v11 =

L

Ed]

we obtain X = —fLab3 +(—fLcosOb1 + fLsinOb, ++ 6b3) 6b3) X

= =

—

+a)b3

2

L

[.1

We write the kinetic energy as

T=

+

+

+

a

G

/

Figur. 8.15

If)

8.10

LAGRANGE'S EQUATIONS FOR FOR R1G11) R1G11) BODIES

and, and, substituting the appropriate terms, we obtain

T=

+

mL2Ô2 + +

sin2 0

Note that we could not have written the kinetic Note

+

a)

[gi [91

energy as purely rotational about B because

point B is not fixed. The potential energy is

[hi

V=

There are no nonconservative forces that do work. Application of Lagrange's equations

yields the equation of motion as —

cosO

—

+ 2L

sinO = sin6

Iii

0

Finding the equilibrium position requires that we solve Eq. [ii with 0 = —

0, for

[ii

SiflO = 0

+

Let us assume small motions and approximate sin 6 by 0 and cos 9 by 1. Solving Eq. [ii we we obtain for the equilibrium position 2

3

=

.g

2L

[ki

a:

2L

These equations also give a range of validity for the small angle assumption. namely that should he less than 3g/2L. Comparing this result with the head problem in Chapters 4 and 5. we observe that 6 = 0 is not an equilibrium point as long as the arm length a is not zero. Next, we analyze the equilibrium point. From the above equation. as the rotational speed Il gets larger, becomes larger. This can be explained by noting that the rod lifts up higher as the shaft spins faster. Considering the stability of the equilibrium point, Eqs. [gi and [hi has tndicate that the system is nonnatural, with T1 = 0. The dynamic potential U = V — the form L

U=

1

cosO

,

.,

1

•

—

1fLsinO ,fLsinO

—

\

+ a)

[Ii Ill

The first derivative of U gives the equilibrium equation (Eq. Ui Lii divided divided by 3). The second The derivative of U is =

—

For small values of

—

sin2

6) +

Imi Im]

if we introduce the small angle assumption to this equation. we

."ihtai n

=

L +

+

aOe)

=

En] liii

(—

thus. the derivative of' of U U is is largcr largcr than than zero zero for for all all times when the small angle assumption is

[ni is greater than From Eq. [ku, the first term on the right side side of' of Eq. [n] From

467

468

CHAPflR 8•

RI(;ID BODY

BASIC CONCEPTS

valid. We, of' course. intuitively expected this to happen. Indeed, the stability property is valid also for large values of the equilibrium position. When a = 0. we get that 0 = 0 is an equilibrium position for the bar as well, and that for small values of the rotation speed = 0 represents a stable equilibrium position. To understand why 0 = 0 is not an equilibrium position when there is an arm. it is helpful to visualize the equilibrium position as a point where the centrifugal force and the gravitational force create moments that balance each other out. The instant an arm is placed on the shaft and the rod is suspended from that arm, the moment generated by the centrifugal force becomes much larger, and the equilibrium position shifts up.

Example

8.12

A spinning top of mass and centroidal moments of inertia inertial1 and and 13, where is measured about the symmetry axis. is moving on a cart of mass 1112 which which is constrained to move hor-

izontally and in one direction, as shown in Fig 8.16. Find the equations of motion using a Lagrangian approach.

Solution This problem is a four degree of freedom. holonornic problem. We select the translational

coordinate thr the cart and the three Euler angles as the generalized coordinates. We write the kinetic and potential energies of the cart and spinning top separately. We select a reference frame a1a2a3 such that the cart always moves in the a2 direction and use a 3-1-3 Euler angle transformation 0, i/i) to orient the top. We take advantage of the symmetry and use the F frame. The angular velocity of the top can he expressed as

to =

=

+ (0712 +

Of1

+ (4)cosO +

+

1.1

Next, let us write the kinetic and potential energies of the cart and top. For the cart we have

Ibi

= 0

Tcan =

'1' 'V .12

a2

F

/ El O

a

Figure 8.16

8.10 and

LAGRANGE'S EQUATIONS FOR RIGID BODIES

for the top we have fl1 V( •• Vc; ?fl Vc; +

=

HG =

?fl1 flfl VG •• VG VG

[ci

+

(w? +

+

where the velocity of the center of mass of the top is

= Ya2 ±± to X rG/(' = )'a2 + (w1f1 + =

+

X Lf1

[dl Ed]

+w2Lf1

Ya2

andW3 W3 toto preserve preserve the the compact nature of and We We prefer to use (for notational purposes) w,. frame by by taking taking the the second column of [RI the equations. We can express a2 in terms of the the F frame as and setting = 0 in Eq. [7.5.3] as a2 a2

= sin sin

+ cos cos4cos 4,cos Of2

le]

—

velocity expression expression for for the the center of mass of the top. written which results in the velocity (Ysin VG VG = (Ysin

+ w7L)f1 + (Ycos

0

L)f2 — Ycos 4.sin

—

[11

Substituting the above equation in Eq. [ci. carrying out the algebra. and combining with the

kinetic energy of the cart yields I

T = 3(11 +

-,

iniLj(w1

+

+ m1L2)(& +

= +

I

,.

w;) +

+

I

± ,m)Y + m1LY(Wis4 +

s2 6) S2 6) +

+

—

w1 w

cjcO)

+ m2)Y2 ?fl2)Y2

[gJ

—

provides the coupling between the translational and notice that the last term in Eq. [gJ [gJ provides rotational motions. The external force that does work is F, which is along the a2 direction, The potential energy and virtual work and the moment M. which is along the spin axis, We

are

V = m1gLcos6

(SW =

F6Y+M(cos084+5i/i)

[hJ [hi

with respect implement Lagrange's equations. we need to take the partial derivatives with respect to the generalized generalized coordinates and generalized velocities. Once we write the angular velocities to the in terms of the Euler angles and start taking partial velocities, the equations become complex and lengthy. In addition, it becomes increasingly difficult to attribute a physical meaning to the the equations of motion. We thus keep the angular velocities in the formulation as long as to possible. We evaluate the partial derivatives of the angular velocities with respect to the Euler [a] we we have. have angles and their rates of change. Using Eq. [a] To

=0

=

•

•

=sin0

(9w, .

=0

dW.i .

=0

(1()

(9(03

— cosH =

(9W3 .

(90

=0

= 00

= 00

i

(1(03 (1W3

-----

= II =

(1W2

rIO

dW3

=

[I]

469

470

CHAPTER 8• RIGID BODY DYNAMiCS: BASIC CoNcEPTS

With respect to the precession angle, we have

aT

()W1

+

=

+

=

m1L2)w2sin0 + + m1L2)w2sinO

aw1 law2 cos4)cosO sin4)— m1LY I+ ,n1LY

ôW2

.

w-)-———-.-—

(94)

(94)

C94)

+ rn1LYsin4)sinO rn1LYsin4)sin0

aT — = ,n1L Y(wi COS 4) + w sin 4) cos 0) .

— =

[U

0

c?çb

w, = 0, the equation of motion associated with the precession so that, noting the relation w angle becomes = McO + rn1Lrs4)sO — w1w2cO) + ,n1L)(w2sO + w1w2cO)

(ii + ,nIL)(w2sO +

nutation angle. we have With respect to the nutation ?fl1L2)(Wl ('I ++ FflLL2)(W1 = (Ji

2

+W1 (96 oe

+ 130)3 +

iiw3

+ ,n1LY

.

I

.•sin4)— .-sin4)—

-

cos4)costI

(90

(10

(90

CIwi

— ,n1LYcos4)cosO ,n1LYcos4)coS0 ,pj1L2)w1 — ('I + ,n1L2)wi

(if

=

,n1LY(4)sincbCoS0 + w1 cos 4) sin 0) + !3w3(—4)sin6) + ,n1LY(4)sin4)COSO

+

(90

[II 111

(9V

— = —in1gLsin0 —in1gLsinO (90

Noting that

=

nutation rate as sin 0. we obtain the equation of motion for the 1

,p,1LYcoS4)cos0 (li (!i + miL2)thi — ,n1LYcos4)cos0

—

('I ++ (Ii

tan0 tanO

+

—

in1gLsin0 = 0 in1gLsinO

[—1

For the spin rate.

—=0

/3(1)3

=0

I.1 [.1

(91/F

so that the equation of motion has the form

[.1

hw3 = M

and [ki. we observe that Eq. to] is embedded in Eq. [k]. This is precession angle 4) is nothing but he expected, because the equation of motion for the moment balance about the precession axis (a3). and the equation for the spin angle is moment balance about the b3 axis. The angle between the two axes is 0. We multiply Eq. [ce: by cos 0. subtract it from Eq. [kJ. and divide the result by sin 0, which gives Comparing Eqs.

+

-

+

WIW2

tanO

)+

=0

—

Finally, with regards to the translational coordinate Y. we note that (i(l)1/dY = 1.2, 3. thus dT aT

+ m2)Y + ,n1L(w2s4) — WI c4)cO) = (in1 +m2)Y+rnjL(w2S(W1C4)CO) =(,n1

(if

=0

=0

0 (i

[qJ

8.1 0

471

EQUATIONS FOR Ricm

and the equation of motion becomes (in1 + ?fl2)Y + (rn1

—

i1c4c0) +

+

+

= F

I']

= F

Iii

which we can rearrange as

+ ?fl2)Y ?fl2)Y +

.

—

+ ,n1L ,n1L

4

tanO

+

I)

An interesting observation can be made with regards to the equations of motion. The or (03. final expressions do not contain a mixed product product of of Y Y and and one one of of w . (03. This This is to be terms, expected. for two reasons. First, we discussed in Chapter 4 that many of the are generalized coordinates, cancel. Second. mixed velocity terms in the equations where

of motion are indicative of Coriolis effects and of motion with respect to a rotating frame. While we have mixed angular velocity terms in the equations of motion, there are no mixed velocity terms involving Y, because Y is not a rotational coordinate. One can use this property when checking the accuracy of the equations of motion. Let us now consider the spinning top without the cart. We should be able to derive its terms from from the equations of motion. Indeed, equations of motion by eliminating all the rn2Y terms eliminating the Y V terms from Eqs. [m}, [ol. and Ip] and rearranging, rearranging. we obtain + ni1

— w2

+

+ ni1 ni1 L2)

tanO

— in1

—

rn1L 2 ') + w1 w1 (Ii (Ii ++ rn1L

tan6

—

gL sin 0 =

()

= 00

= M

It]

which can also be derived from Eqs. (8.5.30] or [8.5.31]. as we did in Example 8.10.

The above example illustrates some of the difficulties encountered when using Lagrange's equations in conjunction with the general three-dimensional motion of rigid bodies. In this particular case, we alleviated some of the complexities in a roundabout way, by retaining (i = 1, 2, 3) in the equations of motion and taking their partial derivatives with respect to the Euler angles. If, on the other hand, we wanted to apply Eulefs equations to this problem directly, we would have to separate the top and the cart into two separate systems. We would then write the individual equations of motion, which contain the reaction terms. After that, we would need to eliminate the reactions and obtain the equations of' motion for the system. The question arises as to whether there are direct methods that permit the use of the body angular velocities in the equations of motion and also at the same time facilitate the solution for interconnected bodies. The answer to this question is discussed in Chapter 9. Derive the equations of motion for a disk of mass in and radius R that is rolling without slipping on a flat surface. Use Lagrange's equations.

Solution discussed in in Section Section 7.9. 7.9. Consider Consider Figures 7.32—7.34. The kinematics of the rolling disk is is discussed

and use as generalized coordinates the translation of the center of mass X. Y. and Z. and the

Example 8.1 3

472

S•

CHAPTIR S• RIGID CHAPTIR RIGID

BASIC CONCEPTS

BODY

z V

x

FIgure 8.17 0, and i/i in conjunction with a 3-1-3 sequence. The free-body diagram of the Euler angles 0. and are friction forces. is the normal force and disk is shown in Fig. 8.17. The force We obtained in Section 7.9 that the motion is governed by one holonomic and two nonholonornic constraints of the form

Z = RsinO X = ROsin4)sinO X Y

=

[.1 [b]

+

—

+

—

id Ic]

To find the equations of motion using Lagrange's equations, we write the expressions for To

the kinetic and potential energies. For a uniform thin disk we have /3

'I = 12

=

(dl

We write the kinetic energy as T =

+

= 'rn(k2 +

= _

+

+

+

+ R202cos20)+

sin2 o +

+

+

cos U + ip)2]

+ !3(çbcosO +

Note that in Eq. Eel we did not substitute for the values of X and Y from Eqs. [bi and [cJ. but we substituted the value of Z. This is because the constraints [b] and [C] are nonholonomic

arid, while and, while the expression for the kinetic energy would be correct if this substitution were made, the changes in the translational generalized coordinates would not be accounted for when the derivatives of the kinetic energy are taken. After deriving the equations of one can perform the substitution. If Eqs. [b] and and [c] [c] are are introduced introduced into into the the kinetic kinetic energy energy arid and Lagrange's equations are invoked, we get an incorrect result. The potential energy is

V = mgRsin6

If] ifi

8.10

LAGRANGE 'S EQUATIONS FOR RIGID BODIES

The virtual work expression is due to the nonholonomic constraints [hi and [c], and it is expressed as

= A1[SX

+ R(cosO&/ +

—

+

+

+

+

Igi

The generalized forces are obtained from the virtual work as

Qx =

Qy =

A1

=

A2

+

= RcosdA1 Rcos&t, + RsinqtA2

+

[hi

Applying Lagrange's equations for a constrained system, we obtain five differential equations

+ I1çbc2fI +

nix = A1 ,nX A1

[ii

mY =

[Ii

A2

13011/SO + 211 thOsOcO — 1301//SO

—

21;çbOcOs6

= + mR2c26)6

—

inR2O2sOcO

—

+ RsthcOA2

+ 1/i )4)sO++ingRcO ingRcO

—

= —Rs4)sOA1 —Rs4sOA1 + Rcq')sOA2 Rcd)sOA2 !3(çbCO

+ 1/I) = Rc4)A1 +

—

[ki [kJ [II III

[ml

The five equations above and the two constraints Fbi arid [ci can he used together to solve for the seven unknowns: the five generalized coordinates and the two Lagrange multipliers. One can reduce the number of equations by first introducing Eqs. [ii and Iii into Eqs. [ki, [1]. and [rn]. This eliminates the Lagrange multipliers from the formulation. One can realize a further reduction by differentiating Eqs. [b] and [cJ. and substituting the results into Eqs. 1k], il], and [m], which are now in terms of and Y. P. As As aa result, result, we we end end up up with with three equations of motion. in terms of the Euler angles. The procedure is tedious; we outline here the procedure for the the i/i i/i equation only. Introduction of Eqs. [i] and and UI [ii into Eq. [ml yields 13((J) cos 0

—

i/s) d'6 sin (1 + 1/i) d'O

= ,nRX cos

ml [ni

+ ,nR? sin d-) d.)

Differentiation of Eqs. fbi and [ci gives

K=

ROs4)sO

P =

—

—

ROcd'sO

—

lol

+ R02s4)cO + R4)2sthc0 +

R(dic0 + 1//)cth 1/;)cth +

R(ç&cO + 1/s)s4) + R(4)cO + 2R4)Os4)sO — —

—

Rd2cbcO

—

R4n/ic4)Ipi Ipi

Introduction of Eqs. [0] and [pJ into Eq. ml and carrying out the algebra yields (13 ++ mR2)(dcos0 (13 mR2)(dcos 0

—

4)0

sin 0 ++ i/i)

—

sin 0 = 0 mR2frOsinO

[qJ IqI

The other two equations of motion arc obtained in the same way, and they have the form 114)2sOcO + mgkcO = 0 + 'I') — 11c/2s0c0

(Ii + mR2)O + (13 + 114)sO +

211 4)OcO —

130(4)cO

+ 1/i) = 0

In

[sI [SI

473

474

CHAPTER 8• 8. RIGm BODY

BAsic CONCEPTS CONCEPTS

This example further illustrates the difficulties associated with obtaining the equations of motion of a system when all three Euler angles are present as generalized coordinates. Given the simplicity of the unconstrained equations. one whether there is a more suitable way of obtaining the equations of motion. Also, this example is more of a special case, where elimination of the Lagrange multipliers and redundant coordinates leads to a simpler set of' of' equations. equations. More Moreoftefl ofter than not, one ends up with considerably more complicated equations after eliminating the Lagrange multipliers from the formulation.

Example 8.1 4

I

Consider Example 8.4 and calculate the moment acting on the shaft so that the shaft rotates with a constant constant angular angular velocity. velocity.

Solution Solution will use the constraint relaxation method. We treat the rotation of the shaft as a variahk the shaft. shaft.From From Example Example 8.9, th. and consider aa moment moment M acting on the 8.9,the the rotational rotational energy has has the the form form We We

Trot =

,nL2(d? sin2 0 +

+

[.1

where I is the mass moment of inertia of' the shaft about the vertical. The virtual work is to themoment moment M M to the

[bJ 4ii does does and 4ii and

not contribute to the potential energy. invoking Lagrange's equations we obtaiT the equation of motion associated with 4' as = Al

next invoke the constraint that

We

=

[ci

is constant, so that the expression for M

becomes

M M =

Ed]

Let us compare this answer with Eq. in Example 8.4, where we obtain obtain the moments. Considering Fig. 8.7, we recognize that

M = M,sinO which we confirm upon comparing Eq. [dJ with Eq.

8.1 8.1

1

[.1 Example 8.4. in Example 8.4.

D'ALEMBERT'S PRiNCIPLE FOR RIGID BODIES

In Section 5.13 we discussed certain shortcomings associated with Lagrange's tions and saw a way of writing the equations of motion for a system of particles rectly from D'Alembert's principle. In this section, we do the same for rigid bodies

8.11

PRINCIPLE FOR RIGED Bonws

For a system of N particles. particles, D'Alemhert's principle is written as N

N

m1a, •

>

=

[8.11.1]

•

1=I

1=1

which F1 denotes all the forces external to the ith particle. For a rigid body, we replace the summation by an integration. integration, m1 in1 by din, din. F1 by dF, and we drop the subscript i, with the result in

a•6rd,n J body

dF'6r

= I

[8.11.2]

J body

Writing from Fig. Fig. 8. 8.11 the position vector as r = rG + p. we expand a and 6r in terms of the center of mass as

a=aG+axp+(ox(toxp)

[8.11.3]

Note that, as before, the variation of the rotation is denoted by 60. indicating that this is not a derived quantity. but rather a defined one. introduction of' Eqs. [8.11.31 side of of Eq. Eq. [8. [8.11.2] to the lefI side 11.2] yields

-

body

a' 6rdin = fI a•

+

X pp ++ o (0 X X (w (w X

+ 60 X p)dm

Jhody

=

(

Jbodv

X X

(o X X p)).(60

X

[8.11.4] [8.11.41

p))drn

All other terms drop out due to the defInition ol the center of mass. The first term on the right side of the above equation is recognized as maG • 8rG. To evaluate the second and third terms, we make use of column vector fbrmulation to write

X p)•(60 X p) =

[fiJ{a}

[8.11.5]

and realize from Eq. [8.2. 18] that [j5Jdin = [iG] and that EIG]{a} {dHG/dl}rel. Manipulation of the last term on the the right right side sideof ofEq. Eq.[8.11 [8.l1.4J .4J is more complicated. After a number of manipulations one can show that

(wXcoXp)'(6Oxp)= (wXtoXp).(6Oxp)= 60•[wX(px(üxp))J 60'[wX(px(üxp))J

[8.11.6] [8.11.61

where we recognize the expression that leads to the angular momentum. p X (to X p). It It follows that the second and third terms in Eq. 18.11 .4] reduce to

J body

(aXp).(60Xp)+(coX(wxp)).(6OXp) (aXp).(60Xp)+(coX(wxp)).(6ftXp)

=60.

din

[8.11.7]

Introducing the expression for 6r into the right side of Eq. [8.11.2] and using

the results from Section 8.4. we obtain

dF(6r(;+6OXp)dJ?7 =

= ibod y

Ihody

18.11.81

475

476

CHAPTER 8•

DYNAMIcs: BASiC BASiC CONCEPTS CONCEPTS RIGID BODY DYNAMICS:

in which F is the resultant force and M(; is the resultant moment about the center of mass. mass. Substituting Substituting Eqs. Eqs.t8.11.71 t8.ll.71 and and [8.11.8] into Eq. [8.11.2], we obtain D'Alembert's principle for a rigid body as

18.11.9]

= F'6rG F•6rG + MG•80

+ HG

When there is more than one body in the system. D'Alembert's principle becomes N

N

•6r(;+ +HG1 HG1 >(mac;, >(mac;, •6r(;

18.11.101

+ MG1 MG1

=

where all the internal forces and moments that one body exerts on another cancel. Actually. Eq. [8.11.10] is the most general definition of D'Alernbert's principle. For a system of particles, all one does is to eliminate the rotational terms from Eq. [8.11.101.

We next consider writing the equations of motion directly from D'Alemberts principle. To this end, we express the variation of the centers of mass and of the rotations in terms of independent generalized coordinates and velocities as PS

6rG1

kI

8qk=

c/wi

.6qk 18.11.111 .8qk

qk

k=I

We introduce introduce Eqs. Eqs. [8. [8. 11 11 . .111 11J into intoEq. Eq.[8.11.10]. [8.11.10]. with with the result

+HG.üqk +HG'

= 0

.

[8.11.12]

We mak.e use of the property that the variations of independent generalized co12] to hold, the coefficients of ordinates are independent themselves. For Eq. [8.11 121 the variations of the. the generalized generalized coordinates coordinates must vanish individually. We hence obtain the equations of motion, as the coefficients of the variations of the generalized coordinates, as ..

N

(9qk

(1(0,

+

=

'

Qk

l,2,...,n [8.11.13] k k = 1,2,...,n

in which the generalized forces are given by Qk =

/))

(F• c?VG. ,

-

-I-M(7•

$

)

(9qkJ

[8.11,141

Note the similarity of the above equations with Lagrange's equations derived in the previous section. The right right sides sides are are identical. identical. On On the the left leftsides, sides,in inLagrang&s Lagranges equations we have derivatives of the kinetic energy, and and in in Eq. Eq. [8.11 [8.11.13] .13] we we have have vector products. Writing the rotational equations of motion in this form is usually more convenient than taking derivatives of the kinetic energy.

8.12

IMPACT

OF RIGiD Rmm Bonws Bonws

The disadvantage of Eq. [8.11.13] is that the acceleration and rate of change of

the angular momentum need to be calculated, which involves many more operations than computation of the kinetic energy. In a sense, Eq. [8. 11 13] is a compromise between Lagrange's equations and Euler's equations, combining aspects of both. In Chapter 9, we study additional ways of writing the equations of motion that are based on D'Alembert's principle. .

8.12

IMPACT OF RIGID BODIES

An interesting application of the impulse-momentum relations is when two moving rigid bodies or a moving body and a stationary object collide. The duration of the collision is very important when analyzing the properties of the ensuing motion. When the collision takes place in an extremely short period of time, it can be considered as impulsive motion. The impulsive reaction forces that are generated form an impulsive constraint at the location of the collision. The velocities and angular velocities before and after the collision are related to each other by linear and angular momentum principles, as well as by Poisson 's hypothesis. A collision takes place along the line of impact, which is perpendicular to the surface tangent to both colliding bodies. This surface is defined the same way the rolling surface is defined in Chapter 7. For particles, smooth spheres. or disks. disks, the line of impact joins the centers of mass of the colliding bodies. For collisions of rigid bodies, the line of impact does not necessarily go through the centers of mass of the colliding bodies. Rather, the line of impact connects the impact point and centers of curvature of the contours of the colliding bodies at the point of impact, as shown in Fig 8.18. Also, one of the impacting bodies may not have a radius of curvature at the point of impact, because the impacting point is a

S '—'2

Corn mon

tangent

•G1

Line of impact

Figur. 8.18

477

478

CHAPTER 8 •. RIGID BODY DYNAMICS: BAsic CONCEVFS

0 -

—

(-

/

—

(00

mi t ial

position C

Figure 8.19 initial sharp edge. An example is given in Fig. 8.19. The rod on the left is given an initial motion by releasing it from an angle. The line of impact is perpendicular to the rod on the right. If, at the point of contact, the contours of both colliding bodies do not have a radius of curvature, one then has to make a reasonable assumption regarding the line of impact.

As a result of the property that the line of impact does not lie along the line joining the centers of mass of the colliding bodies, Poisson's hypothesis, which we studied in Chapter 3, does not always reduce to Eq. [3.5.101, repeated here as = —e&1

[8.12.1]

is the difference in the components of the velocities of the colliding points along the line of impact, and the subscripts 1 and 2 denote before and after the impact, respectively. Also referred to as Newton 's experimental law,1 Eq. [8.12.1], which can be regarded as a kinematic relationship. holds for the component of the velocity along the line of impact when the contours in which e is the coefficient of restitution and

of the colliding bodies are smooth at the point of impact and there is no friction involved. Recall that Poisson's hypothesis states that the impulse takes place in two stages: a compression stage, where the bodies compress each other until the relative velocity of the two bodies along the line of impact is zero, and a restitution stage. where the colliding points split from each other and the bodies regain their original shapes. Inherent in this hypothesis is the consideration that the colliding bodies have some elasticity in them to permit compression and restitution. The coefficient of restitution e indicates the strengths of the two stages of impact. It varies between 0 and 1, its value depending on the material properties of the colliding bodies as well as the circumstances under which contact takes place. When 1, the collision is known as perfectly elastic impact and there is no energy loss. e The strengths of the impact during the compression and restitution are the same. A

value of e between 0 and 1 indicates that the strength of the impact has lessened in the restitution phase. This reduction in the strength is due to the elasticity and Firstformulated formulatedthe thevelocity velocityrelations relationsFor Forimpact. impact. Poisson Poissonlater lateron ongeneralized generalized what what happens happens was Newton Newton who whoFirst during impact to what is known as Poisson's hypothesis. It

8. 1 2

479

IMPAcT OF RICH) BODIES

internal energy dissipation. When e = 0, there is plastic impact arid no restitution.

The colliding bodies do not separate from each other immediately after impact. The shapes of the bodies also affect the coefficient of restitution. When the collision takes place over a sharp edge. the value of e becomes smaller than its value for impact over a smooth surface. It should be noted that the coefficient of restitution represents a gross simplification of what happens during impact. Another issue that needs needs consideration is the material damage that may ensue as a result of impact. Since the implicit assumption is that there is no damage to the colliding bodies as a result of impact, we are inherently considering low-velocity impact. This itself challenges the validity of the assumption that impact takes place during an extremely small period of time. Yet another issue is the presence of frictional lhrces during impact. Recalling that friction forces are obtained from normal forces through the multiplication of coefficients of friction and that the coefficients of friction themselves are largely simplifications, the issue of dealing with frictional impulsive lbrces becomes even more complicated. Take care to see if the assumptions used are correct and whether dealing with a frictional impulsive force is realistic.

Consider a cube resting on a smooth subjected to an impulsive force as shown in Fig. 8.20. Find the velocity of the center of the mass of the cube and its angular velocity after impact as a function of the coefficient coefficient of' of restitution e.

Solution draw in Figs. 8.21—8.22 the free-body diagrams of the cube, during the compression and restitution, respectively. The impact of the cube with the smooth surface takes place at point C. We conclude this by noting that point C can move horizontally but it cannot move vertically because F causes a clockwise rotation of the cube. We denote the impulsive reaction during compression by k. We denote the components of the velocity of the center of mass and angular velocity at the end of the compression period by and after the collision is over, by vi-, VVL.. and and w, respectively. After compression. compression, the linear impulse-momentum theorem yields in the x and v directions We

F

[a]

=

2h

-'

A

2b

F

A

F

Y

fG

x

I

B B

C

C A

R

Figure 8.20 Figure 8.20

Figure 8.21

Compression

Example

8.1 5

_____

480

CHAPTIR 8S RIGID BODY DYNAMICS: BASIC CoNcEvrs 2h

—

x

C

B

A

Figur. 8.22

Restitution

The angular impulse-momentum relationship about the center of mass yields

[hi

= h(R h(R—— F)

where 'G where 'G

(2b)2)/12 2mb2/3 is the centroidal mass moment of inertia. = ,n((2h)2 ,n((2h)2 + (2b)2)/12 [a] and [b] constitute three equations that need to be solved for the four unknowns

we.,

and

t-.

-

The fourth equation comes from the kinematics and from the realization that poth

C has no vertical velocity at the end of the compression period. To this end, we write the

relative velocity equation between the center of mass and point C as +

=

X r(,u =

-I-(un. (un. ++ w.b)j w.b)j

(vu. +

[cJ [ci

-f-

= V).I + and rc/G = hi — bj. Noting that the vertical component of vc is zert. and we obtain for the component component of' of the motion in the direction where

=

= V( +

V(

[di [dl

0

Solving Eqs. [al through [dl simultaneously. we obtain

F =— in

Pb2 = ye — nib2 'G + mb'-

-

Pb

A

we= - 1G ++ nib2 nib2

mEb2 mPb2

IG+,flh2

[.1

Next, we write the impulse-momentum relationships for the restitution phase. From

free-body diagram, the velocity in the x direction direction does does not not change. change. In In the the yy direction direction we we have

+ eR = Introducing the values for

Ff1 Eq

and k into this equation, we obtain Pb2 -

,(l+e) ,(1+e)

[I]

14:7++mb— 'G mb—

Similarly, for the angular impulse momentum about the center of mass we have

[hi [hI

IGW( + ekb = whose solution is !GW

(—IG + emnb2)F/, emnb2)F/, = ee — — — if!)2 'G + mb2

2\ mPh3 + mb2 rnb2

The sign of the angular velocity after impact depends on the value of the of restitution. For perfectly elastic impact. the angular velocity is positive,

[II

HOMEWORK EXERCISES

our sign convention. Because there is no energy loss, loss, itit is is as as ifif the the cube cube bounces bounces back. back. However, as the cube is resting horizontally, a counterclockwise rotation is impossible and a second impact is immediately generated by an impulsive force at point B. The cube cube rocks hack and forth as it translates forward. As the coefficient of restitution becomes smaller, the resistance of the cube to clockwise rotation becomes smaller. When e = 2/3. the angular velocity w becomes zero and the cube just acquires a translational velocity with no rotation. When e is less than 2/3. the cube acquires a clockwise angular velocity. As the coefficient of restitution becomes qwres becomes even smaller. smaller, the angular velocity becomes large enough to tip the cube over. Note that in order to solve this problem using Eq. Eq. [8.1 [8.1 2.1 2.1 J, J' we would have to assume assume that the cube is originally resting an infinitesimal distance above the surface: then we would use the kinematic relations as the infinitesimal distance goes to iero. by

REFERENCES Etkin, B. !)vnamics of of Flight: Flight: Stability Stability a,z1 a,zd ('onirol. ('omml. 2nd 2nd ed. ed.New NewYork: York: John John Wiley, Wiley, 1982. Ginsberg, J. H. Advanced Dynamics. New York: Harper & Row, 1988. Greenwood. D. D. T. T. Principles ofif Dynamics. 1988. Dynamics. 2nd ed. ed. Englcwood Englcwood Cliffs, Cliffs, N.J.: Prentice-Hall, 1988. Amsterdam-New York: SpacecraftRotational Rotational Junkins. J. L.. and J.J. I). Junkins. I).Turner. Turner. Optimal Optimal Spacecraft Elsevier. 1986. Kilmister, C. W.. and J. E. Reeve. Rational Mechanics. New York: American Elscvicr, 1 966. Kilmister. Mci rovitch. L. Methods of Analytical Dynamics. Dynamics. New York: McGraw—Hill. McGraw—Hill. 1 970. Truesdcll. C. Essays in the History o/ Mechanics. New York: Springer-Verlag, 1968.

HOMEWORK EXERCISES SECTIONS 8.2 AND 8.3 1.

Consider the slender bar in Fig. 8.15. Find the angular momentum of the bar

2.

about point B. Replace the slender bar in the previous problem by a rectangular plate. as shown in Fig. 8.23. Find the angular momentum about point B.

(/ f

1)

I

C.

FIgure 8.23

481

482

CHAPTIR 8• RiGID BODY DYNAMICS: BASIC CONCEPTS

B L12

112

D

w

Figure 8.25

FIgure 8.24

3. Figure 8.17 shows a disk of' radius R rolling without slipping. Using a 3-1-3 Euler angle transformation, tInd its linear momentum and angular momentum about the center of mass and about the contact point. 4. The vehicle in Fig. 8.24 has an inertia matrix as given below. If for the instant rad/s,os,. os,. = —0.04 = 0. 1 rad/s, considered the angular velocities of the vehicle are rad/s. and w-. = —0. 15 rad/s, find the angular momentum of the vehicle exrad/s. pressed in terms of principal axes.

I'GI =

100 0 50

0 300 0

50 kg•m2 0 250

5. The light shall of a spherical pendulum is bent as shown in Fig. 8.25. Find the angular momentum of the pendulum about B. 6. Find the angular momentum of the disk in Fig. 2.47 about point A. SECTION 8.5

7. Find the equation of motion and magnitudes of the reaction forces for the spinfling pendulum in Fig. 8.15 using Euler's equations. 8. Consider Example 8.5. Find the equation of motion when the rod connecting the shaft to the disk has a mass of ,n/2. Both 4) and i/i are kept constant by servo motors.

9. Consider the spinning top in Fig. 8.16. The cart is not moving, and at the instant shown 0 = 15°. 0 = 0, 0 = 0, and 4) = 0.2 radls and is constant. Find the spin rate necessary to maintain this condition. 10. Consider the rolling cone in Fig. 7.35. Pivot 0 is lIxed. and RIL = 0.2. Given that friction is sufficient to prevent slipping, find the necessary condition on the

EXERCISES

V

L

B

ci

R N

L—-'

FIgure 8.26

Figure 8.27

angular velocity of the cone that will prevent the cone from tipping.

Model

the normal force acting on the cone as a single force acting from a distance d relate the angular velocity to d. from 0 find find d and and relate from 11. 11. The disk of mass rn and radius R in Fig. 8.26 is being held by a light bar. The disk with constant constantangular angularvelocity velocity ft The other end of the bar is connected rotates with to a joint joint atatB. B.The Thexvz xz axes are attached to the bar. Find the external moment necessary to have the bar rotate and the reaction moments when the joint permits axis, and (c) the z axis. motion about (a) the x axis. (h) the y axis, 12. Consider the spinning top in Fig. 7.13. Find the equations of motion using Eqs. [8.5.31] and [8.5.32]. 13. The disk in Fig. 8.27 rotates about the light rod BG and it rolls on a horizontal surface. The shaft to which point B is attached rotates with the constant angular velocity fl. ft Find Find the the reactions reactions at at B B and and the normal force at the point of contact C. 14. Find the equation of motion of the plate in Fig. 8.23 for a = 0, b = d. Calculate the reactions at B. 15. Consider the rolling disk in Example 8.13. Show that when the nutation rate is zero, the center of the disk traverses a circular path. Relate the radius of this path to the spin rate. 16. Consider the unicycle in Fig. 7.54. Assume roll without slip and find the normal and friction forces at the point of contact. SECTION 8.6

The triangular plate plate in in Fig Fig 8.28 8.28 isis attached attachedto toaamassless masslessrod. rod.'The 'Th rod spins with Find the reactions at the supports and the moment constant angular velocity velocity. that needs to be exerted on the shaft to maintain this angular angular velocity. 18. The rectangular plate shown in Fig 8.29 is attached to a massless rod. The rod A thereactions reactions at at the the supports supports A spins with with constant constantangular angularvelocity velocity ft Find Find the and B. 17.

483

484

CHAPUR

80

RIGID BODY DYNAMICS: BASIC CONCEPTS

M

A

B

B

A

—L/2

I--L/4H

a

/

4a

IOu—

Figure 8.29

Figure 8.28

A

M

-lOa

Figure 8.30 19.

The rectangular prism of mass m is attached to a shaft that goes through points B and C (Fig. 8.30). Find the reactions at the supports and the moment necessary to exert on the shaft to keep the angular velocity constant.

SECTION 8.8 20. The rectangular plate of mass 5 kg in Fig. 8.31 is resting on the xz plane on one

of its edges, which is attached to a spherical joint. Suddenly, the plate is hit by an impulsive force of magnitude 300 N • s in the negative y direction. Find the angular velocities of the plate and the velocity of the center of mass after this impulse is applied. 21. The cylinder in Fig. 8.32 of length L and radius R initially has an angular veloc= 0.4 radls. The cylinder breaks into two equal parts at point C, and ity of it is observed immediately after the break that the difference in velocities at C = 0.2 j mis, and the cylinder on the left has angular velocities of IS VC, — iS = —0.1 radls. Find the angular velocity of the cylinder on = 0.2 rad/s, the right. 22. The slab from Space Odyssey 2001, shown in Fig. 8.33, is tumbling freely space with Vi-, = 0. The slab is of mass 10 kg and has dimensions a = b = 0.2 m, and c = 0.5 m. Suddenly, a rock of mass 0.5 kg traveling it. RigL speed 12 rn/s in the negative y direction hits the slab and gets lodged in it. 0.2 = 0.1 rad/s, before impact. the slab has angular velocities of 1

485

HOMEWORK

g -V

V

V

1

x B

10cm

/

L/2

Figure 8.31

/

L/2

Figure 8.32 V

'1,

.t

CL). CL)..

Figure 8.33 W: = 0.4 radls. Find the velocity of the center of mass of the slab and its angular velocity immediately after impact. and

SECTION 8.9 23.

The double pendulum in Fig. 8.34 is swinging in the local xz plane. which is rotating with the constant angular velocity of Write the kinetic and potential energies and the equilibrium equations, and identify the integral(s) of the motion.

24. Consider Fig. 8. 15. with a = 0.3 m, L = 0.6 m, mass in = 2 kg. The shaft is rotating with constant angular velocity = 2 rad/s. Initially, the pin joint at B is locked in the position 0 = 900. The lock breaks and the rod begins to swing. Calculate the angular velocity of the rod when 0 = 60°. 25. Consider the previous problem, with a = 0, and in the absence of a motor that maintains a constant angular velocity of the shaft. Given the same initial conditions as above, find the angular velocity of the shall and of the rod when 6 = 60°. Hint: What are the integrals of the motion?

486

CHAPTER BS RIGID BODY DYNAMICS: BASIC CONCEPTS

a.

x

flis L1

Figure 8.34 SECTION 8.10 26. Derive the equations of motion for a spinning top using a 3-2-3 Euler angle transformation and by using Lagrange's equations. Compare your answer with the results of Example 8.10. 27. Find the equation of motion of the gyropendulum in Example 8.5 using Lagrange's equations. grange's 28. Find the equation of motion of the spinning plate in Fig. 8.23 using Lagrange's equations. equations. = = 29. The spacecraft in Fig. 8.35 has the following centroidal inertias: 5mK2, 13 = PflK2, where K is the radius of gyration. Along the b2 axis there

m/l0 moves. The point mass is a frictionless tube. inside which a point mass rn/l0

77,

FIgure 8.35

HOMEWORK EXERCISES

connected to the ends of the tube by a spring k and a dashpot e. An external torque M = Mb3 is applied. Derive the equations of motion using a 3-1-3 Euler angle sequence. ignore the translation of the spacecraft. 30. The shaft in Fig. 8.23 rotates with constant angular velocity Find the moment that must he exerted on the shaft to keep its angular velocity constant. Do this by treating the rotation of the shall as a constrained generalized coordinate. Let is

a=0.

31. Consider Example 8.11 and, using constrained generalized coordinates, find the moment on the shaft needed to keep constant. SECTION 8.11 32.

Consider Example 8.10 and derive the equations of motion by using Eq. [8.11.13].

33. Find the equation of motion of the plate in Fig. 8.23 using Eq. [8.11.1 3j.

SECTION 812 Consider the collision of the two rods in Fig. 8.19. Assume that the rod on the left is released from some initial position and calculate the angular velocities of the rods after impact as a function of the coefficient of restitution. The rod on the left is of mass in and length L, and the rod on the right of mass 4m and length 2L. The distance between 0 and 0' is L/2. 35. A beam of mass in = 2 kg and length L = 0.6 in hits a table as it is falling down (Fig. 8.36). At the time of impact the beam is horizontal, with its center of mass having a speed of 0.5 mIs and a clockwise, angular velocity of 0.2 radls. Find the velocity of the center of mass and angular velocity immediately after impact for e = 0 and e = 1. 36. The plate in Fig. 8.37 has a mass of 3 kg and width a = 20 cm. It is dropped

34.

from a height and hits a stationary table at corner P. At the time of impact the plate is horizontal and it has a vertical velocity of 0.5 mIs, with no

G I

I

L

// // // / Figure 8.36

FIgure 8.37

487

488

CHAPTER

a. RIGID

BODY DYNAMiCS:

CONCEPTS

L

B

L

P

/

Figure 8.38 angular velocity. Find the velocity of point P and the angular velocity of the plate after impact for e = 0.5. 37. The double pendulum shown in Fig. 8.38 impacts the horizontal surface at the 300, 02 02 = 45°, = —0.2 radls, and 02 = —0.3 rad/s. Find = 30°. instant when the angular velocities immediately after impact, assuming that there is no friction between point P and the surface, and e = 0.8. 38. Consider the slab in Fig. 8.33. except that the impact between the rock and slab has a coefficient of restitution of 0.9 and that the slab has no initial angular velocity. Find the velocity of the center of mass of the slab and its angular velocities immediately after impact.

39. The coin of weight 0.5 oz and radius 0.4 in, such as the one in Fig. 8.17, is dropped onto the ground. At the point of contact, the angular velocities of the w = 5 radls, the center of mass has a vertical coin are speed of 9 in/sec. and (1 = 60°. Assuming perfectly elastic impact, calculate the angular velocities of the coin immediately after impact.

chapter

DYNAMICS OF RIGm BODIES: ADVANCED CONCEPTS

9.1

INTRODUCTION

This chapter presents additional methods for deriving the equations of motion of rigid bodies. These methods are particularly suitable for bodies subjected to nonholonomic constraints, and bodies that are interconnected. These additional equations are based on two concepts: First, the rotational equations of motion represent the moment balance of a body. The angular momentum of a body can can be written in many ways, about the center of mass or about an arbitrary point. When differentiating the angular momentum, one can use a set of coordinates attached to the body, a set of coordinates that take advantage of symmetry of the body (if any), or a totally different set. Further, when expressing the moment balance, one can resolve the equations using the body frame or another frame. Each one of these choices leads to a different way of writing the equations of motion.

Second, analytical methods for obtaining the equations of motion, such as in Lagrange's equations, are derived based on D'Alemhert's and Hamilton's principles. As discussed in Chapter 5, there are other variational principles that one can consider. and they lead to additional forms of the equations of motion.

The chapter begins with the modified Euler's equations, discussed briefly in Chapter 8 and particularly useful for inertially symmetric bodies, It moves on to a discussion of moment equations about an arbitrary point. There is an introduction to the equations of motion in terms of quasi-velocities (generalized speeds). The Gibbs-Appell and Kane's equations are derived and compared. We demonstrate that the two are indeed equivalent.

489

490

CHAPTER 9 S

RIGID BODIES: AD\tNCED CONUEVI'S CONCEVI'S

When studying the equations of motion niotion in terms of quasi-velocities, one

question the reasoning behind introducing such coordinates in association with rigid bodies. After all. one can define such quantities associated with a system of partides or any other system. These variables and the Forms of' the equations that make use of them could have been been introduced introduced in in Chapter Chapter 4. 4. The The reason reasonfor forinEroducing introducing them here is that generalized speeds are the most useful for nonholononiic or interconnected systems and that such properties are, for the most part. associated with three-dimensional rigid body motion.

9.2

MODIFIED EULER'S EQUATIONS

The modified Euler's equations are a variant of Eule(s Eulers equations that are applicable to inertially symmetric bodies. They are based on writing the equations of rotational motion with respect to a reference frame other than one fixed to the body. Consider the rotational motion equations about the center of mass = {MG}

19.2.1]

If we view the angular momentum from a relBrence frame N, we can write the above equation as A

{Hc; } =

{ HG } + [

4

\' N

H(; } = {MG 1{HG} {MG}}

19.2.21

where where the A frame is inertial. When selecting frame N. the objective is to find a reference frame that will make the calculations simpler. Particularly appealing is a

reference frame in which the elements of the inertia matrix remain time invariant. A reference frame that satisfies this criterion is one that is attached to the body. The rotational equations of motion can then he written as Eq. [$.5.251. or [Iellth} ++ [thJIl(;J{w} [Iellth} [thJll(;J{w} = {M6} {MG}

[9.2.3]

For inertially symmetric bodies, when one uses the F frame, the elements of the

inertia matrix are time invariant. As discussed in Chapter 8, one can use the F frame with the the moment moment balance balance equations in two ways: in Conjunction conjunction with ways: One One can perform

the differentiation in the F frame, or one can express Eulers equations using the components of ol' the the FF frame. frame. In In this this section, section, we we discuss discuss the the first first approach approach in in more more detail. Consider an inertially symmetric body. with body axes b1b2b3. b1b2b3. where where 13 /3 is the symmetry axis. It tollows that = and the body axes are principal axes. Further. any rotation of coordinates about b3 yields another set of principal axes. We introduce the F frame to the formulation. Chapter 7 introduced the notations of and to denote the angular velocities of the body, frame, and spin, respectively. Mathematically, if the body-fIxed body-fixed frame is obtained from, say, a 3-1-3

transformation, the F frame is obtained after the 3-1 3-I transformation. This type of transformation completely describes the orientation of the symmetry axis. For a 3-1-3 transformation, the angular velocities are

9.2

= w,f, + w2f, + w3f3 w3f3 =

=

+ wf2f2 wf2f2 ++ wpf3 wpf3 =

= =

Of1

+ frsin frsin Of2 (If2 ++ (4)cosO (4cosO + Of1

=

=

M0DW1ED EULER'S EQUATIONS

+ 4)sin Of2 Of2 + 4)cost9f3 Of3

[9.2.4] [9.2.41

= W/ w1 + w

+

We then have We

= wf1 = 0

4)SiflO = (Op = 4)Siflf9

0)2 (02

=

0)3 (03

= 4cosO (/)COSO + i/i

[9.2.5]

= w2/tanO

4)COSO

For a 3-2-3 transformation, with the sequence i/i, 0, and 4). the angular velocities of the body and frame become

= —i/i sin Of, üf, ++ Of2 Of2 + (i/i cos 0 + = —i/i sin

[9.2.6]

Of1 Of2 + Of, ++ Of2

and we have

= wji = WI

= (Of WJ'7 = 00 = —W1/tanO —W1/tan()

=

= i/;cos6 + 4)

[9.2.7]

We can write the rotational equations of motion using the angular velocity of the F frame as

Ad{H}

+

= =

[9.2.8]

{MG}

where we note that all the column vectors are resolved in the F frame, and

[9.2.9]

=

The equations of motion become

+

= {M(;}

or

[lc;]{th}rei + 1th, IIIGI{W} II/GI{W}

{MG}

[9.2.10] Equations [9.2.10] are known as the modified Euler 's equations. Denoting the Equations and carrying out the matrix algebra, we can inertiamatrix inertia matrixbyby[ZIG] I'G] = diag( express them as —

= M, M,

— — w1(13w3 — —

3) == M,

11w1 + W2(13W3 W2(130J3

130)3 = M3 13(03

[9.2.11]

Also, for rotation about a fixed point C on the body. the modified Euler's equations can be written about C as well as the center of mass. Depending on the Euler angle transformation sequence used, the modified Euler's equations look slightly different when we substitute the value of (Of wf 33 into intO them. them. Substituting Substituting for for wp from from Eqs. [9.2.51 and [9.2.7). [9.2.7), Table 9.1 gives the modified Euler's equations thr the 3-1-3 and 3-2-3 transformations.

491

492

CHAPTER 9 S DYNAMICS OF RIGID BODIES: ADVANCED CONCEPTS

TabI. 9.1

Modified Euler equations for 3-1-3 and 3-2-3 transformations

3-1-3 Transformation Sequence I 0)1 +

2

110)2 \ /10)2

(130)1

tan

\ —

(110)1

tan 6) tand

3-2-3 Transformation Sequence

= M1 M,

IIó, I 1ón

= M2

12W2 — Ufl

± w,2 (110)3 +

tan6) \

tan

= 0) 0) =

M1

M2

I;(0

Table 9.2

Kinematic differential equations for the modified Euler's equations

3-1-3 Transformation Sequence • •

cj)

3-2-3 Transformation Sequence

(L = ————

. .

sin6

() =

()

•

tan0

+

(1) th

W1

=—

=

•.

sin (9 (02

(tfl tan (1

(0;

To write the associated kinematic differential equations, we solve Eqs. 19.2.51 and [9.2.71 k)r for (i = 1, 2, 3). Table 9.2 gives the result. Be aware that the equations of motion obtained by the modified Euler's equations are not a restricted form of what one would obtain using the original Euler's equations. Once one has solved Eqs. [9.2.11] together with the kinematic differential equations. one has the complete time history of 'PU). An interesting observation is that the singularity at sin () = 0 is still present. Furthermore. the modified Euler's equations and corresponding kinematic differential equations are coupled through the tan £1 term. This coupling diminishes the usefulness of the equations in numerical analysis. The primary uses of the modified Euler's equations are to derive the equations of motion and to aid in a qualitative analysis. The modified Euler's equations are for the rotational component of the motion. We next investigate the translational equations of motion in terms of a set of axes other than the body-fixed axes. Recall that the translational equations ol motion are given in terms of the body axes by Eq. [8.5. 19] as —{mvG} +

di

= {F}

[9.2.12]

The massofofthe the body does The mass body does not change under aa coordinate coordinate transformation, transformation, so so that Eq. [9.2.12] can be written about any reference frame. Considering the F frame, we have

di

+ ['[' Cs) (V FF ]{rnuc;} ]{rnU(;} = {F}

[9.2.131

Equations 19.2.101 and 19.2.131 can be useful not only to simplify the equations of motion of an inertially symmetric body but also to to solve solve nonholonomic nonholonomic problems. problems.

9.2

493

MODIFIED EULER'S EQUATIONS

We can write the above equations in terms of the kinetic energy. Expressing the kinetic energy as =

T

I

[J J{w} +

19.2.141

we can write )T \T

= =

{mvG} = {p}

(aT [IC; II

)

)T

}

19.2.151

where the notation is obvious. The equations of motion are )T

d

\7

+

d.

[&fI(a{v}) ={F} \i

d

di

6u,bl 19.2.1 6u,bI

= {MG}

+

Example 9.1

Consider the spinning top in Fig. 7.13. Write the rotational equations of motion about 0.

Solution The modified Euler's equations are particularly useful, as we have an axisymmetric body and

the external moments are not a function of the spin angle. The angular velocities are

wi=O

=

sinO

(03 = 4C0S0 + cli

Wf3=

tanO

= 4cosO

[a] Eel

The free-body diagram of the top is given in Fig. 9.1. The external moments about 0 are due to gravity and an applied moment about the axis, so that

M

a3 .3

41;

I:

F

fl

Figure 9.1

[bJ Ibi

494

CHAPTER 9• DYNAMICS OF RIGID BODIES: ADVANCED CONCEPTS

Introducing Eqs. [a] and [b] into the modified Euler's equations, written about the fixed point 0, we obtain sinO

+

+

101 Jo,

—

+

I03(IP +

= 00

= M

[c) [c] Ed]

[.J [.1

These equations are, of course, identical to those obtained in Example 8.10. Using the modified Euler's equations is much simpler than using Lagrange's equations.

Further, they are much simpler than Euler's equations. Had we used the regular Euler's equations, we would have obtained two equations that are combinations of Eqs. [ci and [dl multiplied by sin

Example

9.2

and cos i/i. cli.and andEq. Eq. [e] [e] would would be be unchanged.

A dual spin satellite consists of a main body with an attached spinning disk (rotor). Such

a satellite is also called a gyrostat) gyrostat. The spinning disk can be placed outside or inside the main body. and it is there primarily for stability and maneuvering purposes. Derive the equations of motion for the dual spin satellite shown in Fig. 9.2. The disk is located at the center of mass of the main body. The spin axis of the disk is aligned with the principal axis b1 of the main body. The disk is spinning with an angular velocity of with respect to the main body, and it is powered by a motor that imparts a moment M along the b: axis.

Figure 9.2

A dual spin satellite

'More sophisticated spacecraft hove three rotors on nonparallel axes that permit rotational maneuvers in three dimensions.

9.2

MODWIED EULER'S EQUATIONS

h1

b1

M

M3

Cb3

Figure

9.3

Free-body diagrams

Solution We separate the disk and satellite into two bodies and draw the free-body diagrams, as shown in Fig. 9.3. A motor exerts a moment M on the disk to keep it rotating about the spin axis. and the two bodies exert moments M2 and M; M; on each other about the transverse axes. The equations of motion for the main body obtained by using Euler's equations have the standard form of 11(01

/2(02

—

—

13(1)3 —

—M +

—

(13 — /1)WLW3

=

(/i

=

—

—M2

[al

N1

+ N2

[bJ

+ N3

[cJ [ci

where '1. '2, and /3 are the centroidal moments of inertia of the main body. WI. (1)1.W2, W2, and

are

the corresponding body angular velocities, and N1, N2, and N3 are the excitations external to the satellite. To derive the equations of motion of the rotor, we view its motion from the main body. Hence, the angular velocities of the rotor in the h2 and b3 directions are the same as that of the main body. These two factors make the modified Euler's equations ideally ideally suited suitedfor tir analyzing the motion with the b, b2b3 axes as the F frame for the rotor. The angular velocity vector for the rotor is written as [WI + — [Wi =

and

(1)3J

7-

[d] Ed]

the angular velocity of the reference frame is {wj} = = ILo1 Iwi

w2 w,

w31

7-

1.1

Denoting the centroidal mass moments of inertia of the rotor by J1, J2 = Euler's equations for the rotor become

+1k) J2th2 J2W3

0

±

(1)3

J1(w1 J1(WL +

—(03

0 (Oi

—WI

0

J2W2

J3.

the modified

M

=

If :i

M3

495

496

CHAPTER 9• DYNAMICS OF RICID Bonws: ADVANCED CONCEPTS

which reduce to

=M

Jl((i)l + JI((i)I

IsJ [b]

./2(02 + (J1 — J2)wiw3 + .11.Uw3 = M2

+ (J- —

M3

—

El:'

Equations Ia] and [gi are the equations of motion w1 and fl. ft To To find find the of motion for w2 and we need to eliminate and M3. To this end, we add Eq. [bJ Eq. [hI and Eq. [cJ to Eq. Iii, with the result +

(12 +

+ .11 —

(13 + J7)W3 J7)W3 + (12 (12 +

13 —

J2)w1w3 + J1)w1w3

—

—

—

=

ifi

= N3

[kJ

The four equations of motion are Eqs. [a]. [a]. [gJ. [gJ. [jJ, [jJ, and and [k]. One can substitute for

[gI from Eq. [al to make all the equations of motion have a single angular [g] term, thereby putting them in state form. Doing so yields Eq.

JI

=

Il)

ED

'I

Also, one can add Eqs. [a] and [gJ to get an equation for w1 as

(Ii + J1)i1 +

J111 —(12

—

13)W2W3

= N1 N1

[—I

When a servomotor is used to keep [1. constant. constant, Eqs. U]- [k]. [k], and [m] [mj become become the the describin& equations of the satellite.

Example

9.3

Consider the rolling (without slipping) disk of mass m and radius R shown in Fig. 9.4. will niake use of the modified Euler's equations and a 3-1-3 transformation to derive equations of motion. Note that in Example 8. 13, we derived the equations of motion for system using 1.dagrange Lagrange1ssequations equations for for constrained constrained coordinates.

(1

.1•; .1.

Fl .11 Ii

FIgure 9.4

____

9.2

MoDIFIED EULER'S EQUATIONS

Solution will write both the translational and rotational equations of motion and then use the rolling constraint to eliminate three of the variables. This elimination of the nonholonornic nonholonomic constraint 13 is possible because we are dealing with the angular velocities of the body. From Eq. the translational equations of motion have the form We

V1

+ + W2V3 W2V3

F,

Wf3V2 =

[a]

fli

Ibi FbI

—gsinO V2 + Wj3U1 — W1V3 (01V3 == —f — g sin o rn 7, r3

1)3 + WIV2 — — W2V1 W2V, =

where v1

in

—

[c] Ic]

gcosO

denote the velocities of the center of mass of the disk along the coordinate and 1)3 1)3 denote

axes of the F frame. The modified Euler's equations about the center of mass are I1WI + JIWi

—

I1cii2 — 111:ii2

w1(13W3

Idi

!IWf3) = —F3R —

1lWf3) 11(1)f3)

lel le]

= 0

[f]

13W3 = F1R

The rolling constraint can be expressed as ruic = t0b X ruic = w1Rf3

—

= (o1f1 (v1f1 + w2f2 +

u2f2 + v,f, + u,f2

X Rf2 Rf

Igi

w3Rf,

leading to the three constraint equations

1)2=0

1k]

Introduction of Eqs. [h] into into Eqs. Eqs. [a] [a] and and [ci [ci yields yields expressions expressionsfor forF1 F, and F3 as F, = ?n(—th3R

+ w,2R)

= in(thiR + w2w3R) + rngcosO

[I] [ii

Substituting these values values of of F1 F1 and and F3 F3 into into Eqs. Eqs. [dl [dl and and[1 [1]1 and and using using the the relation relation W[3 = (02/ tan 0, we obtain the equations of motion as I

(13 + + mR)w, ++(13 (I, + mR)w1

—

LanA

= —rngR cos A

[j] Li]

Iiw,w'- = 0

1k]

rnR2wiw2 = 0 (13 + inR2)th3 — rnR2w,w2

[I]

— 13W$W3 + 11W2 /1W2 —

tanO

If desired, one can write the equations of motion in terms of the Euler angles by substituting the values values of of cvi.

w3 into W2. and w3

— j/j) — + j/j)

(1, + rnR2)6 rnR2)6 + (13 + —

di (13

the above equations, which yields

i/i) + liO(4)cos0 + i/i)

sinOcosO = = 0

Imi In]

-

+

— ,nR2O4 inR2O4sinO sin 0 = 0 (4) cos () + 4,) i/i) —

10] 1°]

497

498

CHAPTER 9• 9• DyNAMICS DyNAMIcS OF RIGID BODIES: ADVANCEI) CoNcEvrs

9.3

MOMENT EQUATIONS ABOUT AN ARBITRARY POINT

The Euler's equations and the modified Euler's equations that we have studied so far can be written about the center of mass. or, ii the body is rotating about a fixed point. about that point. In this section, we examine the moment balance equations about an arhitraiy point. Consider a body whose motion is being viewed from a reference frame. We make no restriction on the nature of the reference frame; it can be either fixed to the body or not fixed to the body. Furthermore, the origin of the coordinate frame. which we denote by B. can move with respect to the body. As discussed in Chapter 8. the rotational equations of motion about an arbitrary point can be written in several forms. We use the form given in Eq. [8.5.101 [8.5.10] and write M8 = H8 + !flVB X VG

19.3.11 19.3.1]

where where M8 is the sum of all external moments acting on point B. H8 is the angular

momentum about point B. and VA is the velocity of point B. We wish to express Eq. 1 9.3.1 J in column vector format. We use the angular yelocity of the body and the velocity of point B. both expressed in terms of the reference frame attached to B. as the motion variables. The angular velocity of the reference frame is denoted by It follows that the angular momentum about B can be related to the kinetic energy by / c?T . \d{W}J

{H8} = We

(

[9.3.2J [9.3.2]

)

also express the velocity of the center of mass of the body as {VG}

=(

-

[9.3.31

)

\d{VG} /

introducing Eq. [9.3.2] and [9.3.3] into Eq. [9.3.11 we write the rotational equation of motion of a body about an arbitrary point B as d di \d((t)) / +

}

= {M8}

+

j

[9.3.4]

or (I / (17

—1

dt

\f J

/

+

\T J

\d{W} /

/I

+

/ uT \i \a{VG} /

= {M8}

19.3.5]

The translational equation of motion is not dependent on the location of point B. It can be expressed in terms of wN as d

rn—{v(;} + (It

J{vG} I{VG} = {F}

19.3.61

or

d /

\'

+ —I.. j itt \o1{V(;}J

t?T f c?T

\7

\o'{VG}J

= {F}

19.3.71

9.3

499

EQUATIONS ABOUT AN ARBITRARY POINT

Equations [9.3.4] and [9.3.6J are known as as the Boltzmann-Harnel equations.

Compare Eqs. 19.2. 16bJ and 19.3.4]. The added term in the moment equation is due

to the point B not being the center of mass mass or center of rotation. Also, it should be reemphasized that the rate variables in the Boltzmann-Hamel equations are the components of the velocity of point B and not the components of the velocity of the center of mass, even though to write these equations in compact form we use the partial derivatives of the kinetic energy with respect to the velocity of the center of mass. The moment equations about an arbitrary point make no restrictions on whether the body should be inertially symmetric or not. However, their usefulness diminishes if the inertia matrix is time varying. We thus identify the tbl]owing tbllowing major uses: a. b.

If there is no inertial symmetry, one should select the reference frame such that it is attached to the body. This implies that WN = If there is inertial symmetry. one can select the reference frame as the F frame. However, note that the moment center is now a point lying on the reference frame and not necessarily on the body and that the moment center can move with respect to the body.

Selection of the moment center is an important issue. In general, one selects the point B such that reaction forces do not create a moment around it, or as a point whose velocity is described by a simple expression. In this regard, the Boltzmann-Hamel equations conie come in handy for considering the motion of multiple connected bodies. Also, as with the Euler's or modified Euler's equations, nonholonomic constraints can be dealt with effectively.

A spinning top of mass and centroidal moments of inertia and 13. where 13 13 is about the symmetry axis, is moving on a cart of mass which is constrained to move horizontally and in

one direction. Find the equations of motion using the Boltzmann-Hamel equations.

Solution This is the same problem considered in Example 8.12. We have four degrees of freedom, and

we will use the velocity of the cart (Y = 14) and the angular velocities of the top observed from a 3-1-3 Euler angle transformation as the rate variables. Refer to Example 8.12 for details. The free-body diagram of the top is shown in Fig. 9.1 and of the cart in Fig. 9.5. Note that

we arc conveniently expressing the reaction forces at 0 in terms of both the inertial and F frames. 1'

F

a1

Figur. 9.5

Free-body diagram of cart

I

Example

9.4

500

CHAPTIR 9.

DYNAMiCS OF RIGm BODIES: ADVANCED CONCEPTS

The kinetic energy of the system has the form 1

T =

+

1

2 rn1L)(w1 rn1L)(w1 ++z.i;) z;) +

+

I

2

+

+ ,niLu(W2 pniLu(w2 sin4

w1

—

[a] Taking the partial derivatives, we obtain

= (ii ÔT

— + rn1L2)w1 rn1L2)w1 —

?n1L2)w2 + (it (it + ?n1L)w2

[b] Ib]

=

2

The angular velocity of the reference frame is (Oj I (Oj

= wi WI =

(Uf2 =

6

£02

=

(013 =

(I)COSO

tan0

[cJ Ic]

The time derivatives of Eqs. Ibi are

'I

'I

dt dt

)

+ rn1L2)th, —

=

d = dt dw2, dw2,

(11

in1

+ ,n1Lu

Là

+

\

The term

cosctsin0

W

+ ,n1Lu SiflO sin6

+

d

+ ,n1Luw1

tanO

=

(dl

/3(03

becomes r

o

tO

•'r 1(11 /

—

I

,ntL2)wi —— in1Lu [(Il + ,n1L2)w, + m1L2)w2 + rn1Lusçl'

I

0

Ii

13(03

[—W2

0

Wj WI

[ —(Ii +

]

— m1Lu tO

tO tO

.,

(ii (ii ± m1L)

—

tO

m1Lu

+

I

I

W2 tO

I

I

j

+ w1 wi s4)

I

[.1

— I3WiW3 13W1W3

The velocity of the point of contact C is

= ua2 = usçbf1 + uc4cOf2

If EnI

—

and. from Example 8.12. the velocity of the center of mass is

vu = vG

(14

sin

+ wL)f1 ++ (ii (ii cos cos 0 —

w1 L)f'2 L)f'2

—

cos u cos

cj

sin 0

[g]

and the angular expressed in in terms terms of of uu and Note that the velocity of the center of mass is expressed

velocities. The velocity term in the Boltzinann-Hamel equations yield

—uw1 Wi.)j cosdsin6 nil

= m1 L

uo2 cos cos sin 0 — uw1 sin / — uw4 cos c/ COS 0 —

[hi

9.3

MOMENT EQUATIONS ABOUT

501

ARBITRARY

The moment about point C can be expressed as

=

—m1ga3 + —mjga3 + Mf

X

[II

where rG/c = Lf3 and a3 = sin 0 f2 + cos 00 f1, so so that that in column vector format tbrmat we have in1gL sin 0 M

We write the rotational components of the moment equations by combining Eqs. Idi. Eel. [hi, and [jJ, which yields ?fl1L2)(W1

(Ii +

—

/fl1Li4COS4)COSO + /3(D2W3 — — ni1gLsinO ni1gL sinO = 0

tan0j

+

+

tanO

+

=0

—

13(03 = M

IkI

[II

[ml

Next, we obtain the translational equation of motion using Eq. [9.3.6]. We note that there is only one degree of freedom involved. Y, which is in the a2 direction. To solve for the equations of motion, we can use the relative frame F and deal with all three equations. or we can express [thN], and d{UG}/dt in the a1a2a3 coordinate system and use only the second row of Eq. (9.3.6). Another option is to write the sum of the forces in the a2 direction for the combined system. Here, we opt for the first approach (primarily tbr for illustrative purposes). Looking at the free-body diagram of the cart (Fig. 9.5), we can write the. force balance in the a2 direction as

F

—

P2

from which we solve for the reaction P2 as P2 = F the form

F =

P1a, P1a1

= (F

liii

= ,n'U —

m2à. The forces forces acting actingOfl on the top have

+ P2a2 + P3a3 /nliL)cosçbcos0f7 + (F — ?n2iL)cos4cos0f7

—

+ P1 cos thf1 + (— P1 sin

—

(F

—

sin O)f2 O)f2 + (P1 sin ff sin 0 + Pi cos 0 + P3 sin

[o]

Introducing the expression for the force into Eq. [9.3.61 [9.3.6J and carrying out the algebra yields the equations = (F—rn2ü)s4+P1c4

in1

—

math 1L

m1 rn1àc4s0 àc4'sO ++ m1

+

rn1L +

= (F

= (F

—

— P1

—

?n,i4)c(/s0

—

s4c0 + P3s0

P, P,sq's0 sq's0

—

P3c0

[qJ Iqi In 1.1

502

CHAPTIR

9•

RIG11) BODIES: ADVANCED CONCEPTS OF RIG11)

We next eliminate P1 and P3. Multiplying Eq. [pj by cos and adding them. we obtain (in1 ('n1 +

+

+w

+

+ in1L

—

Eq. 1q1 by

= F [a]

[m}, and [Si [si with with their counterparts in Example 8.12. we Comparing Eqs. [k], that they are the same. Note that the actual values of the magnitudes of P1. P2, and P3 were not calculated in this problem. due to selecting the moment center as point C. not

Example

9.5

I

Derive the equations of motion of the rolling disk considered in Example 9.3 using Boltzmann-Harnel equations. Boltzmann-Hamel

Solution The disk configuration is is shown shown in Fig. 9.4. For roll without slip, the velocity of the point is zero, which constitutes the nonholonomic nonholonoinic constraint. We use the 3-1-3 Euler angit transformation and F frame. The kinetic energy can be written as transforniation = T T =

[.]

+

+

=

the components where the components of the mass moments of inertia about the contact point and velocity are '('2 = 'I 'C2 =

+ mR2 ,nR2

=

=

= bsin()

= 0

13

+ mR2 ,nR2

[bJ

= thcosO + i/i

Note that we have automatically incorporated the nonholonornic velocity constraint the kinetic energy. We proceed with solving the problem as if it were a three degree of freedor unconstrained problem. Because of roll without slip, the velocity of the contact point is zero. However, recalling the argument with regards to the components of the velocity of the moment center, we ne& to have the velocity of a point on the reference frame that coincides with the contact Denoting that point by C' and noting that C' is always in contact with the rolling surface. write the relative velocity expression between the the center center of' of mass and point C' in the We have =

+ (0

[ci

X rGICP

Using the values A

V(; =

—

Rw f1 + Rw Rw1 f3 f1 + i

A

wI. =

w -f + ifi + w

(i) 1f1

a),

tan6

rc;,c =

Rf2 = Rf

[dl and

substitution of these values into Eq. Id yields the velocity of the moment center as Av

=

This result may seem surprising at first. Imagine a coin rolling without slipping.

look at the bottom point of the coin from a distance distance we we think think that that point point is is moving. moving. In In

[.1

______

9.4

UASSWICA11ON OF TIW MOMENT EQuATIoNs

our eyes are fooled by the change in location of the contact point between the coin and the

surface. What we do when looking at the motion of the coin is to follow the motion of the F frame. We do not visualize the spin of the coin unless there are special markings on the coin that draw attention. The Boltzmann-Hamel equations are written about point C' as

(Ii (!i

1. + mR—)w1 mR2)th1

(1).)

A —

+

l1W7 (13

C02

0

+ mR 0 Li

tan 19 tanf9

mR2 )w, (ii ++ inR2)w, -

—

0

—WI

WI

0

0

11w2

0 — —

0

+ mnR2)w3 inR2)w3

(13

—ingRcos6

—Rw3 0)2

0 Rw1

W3 +

=

0

IfI

0

Expansion of Eq. [f I gives the equations of motion as I

2

(ii mR)w, ++ (13 (13++rnR2)w2w3 rnR — ('I ++mR2)th, — 11w2 110)2

—

13W1W3 130)10)3 +

'1w,

—mgRcosU - == —mgRcosU

tanO

IIW;(Lh - == 00 tanO

rnR2)ô..3 — — ,nR2w1w2 = 0 (13 + rnR2)i3 (13 +

[g]

we note that they are the same as Eqs. [j]—[l] in Example 9.3. Using the BoltzmannHamel equations gave the equations of motion much more rapidly than using the modified Euler's or Lagrange's equations. By selecting the moment center as coincident with the point of contact, we were able to eliminate the contributions of the forces acting there. Had we used a body-fixed frame to write the equations equations of of motion, motion, the the point point of' of contact at a particular instant would he at different locations on the body throughout the motion. One would then derive an expression for the velocity of point C and incorporate that into the Boltzmann-Hamel equations. The end result would be more complicated expressions. This situation is common in rolling problems. and

9.4

CLASSIFICATION OF THE MOMENT EQUATIONS

In Chapter 8 and in the earlier sections of this chapter we saw several ways ways of forsaw several mulating the moment equations. In this section we outline the process of selecting the best form for a given problem. The principal choices are with respect to: a. b. c.

Selecting the point about which these equations are written. Selecting the reference frame in which the differentiation is performed. Selecting the coordinate axes to resolve these equations.

For arbitrary bodies one usually differentiates the moment equations and resolves them in a reference frame attached to the body. Otherwise, the inertia matrix will be time varying—in all probability, any advantages realized from using a

503

504

CHAPTER 9• DYNAMICS OF RICH) BODIES: ADVANCED CONCEPTS

Table 9.3

Classification of the moment equations

Type of Body

Moment Equations about

Differentiat ion Performed in

Equations Resolved in

Arbitrary

Center of mass or center of rotation

Body frame

Body frame

Euler's [8.5.281

Arbitrary

Arbitrary

Time-varying inertia matrix

Name of Equations, Comments

[9.2.2]

An arbitrary Point point

menially symmetric

Center ot' of mass or center of rotation An arbitrary point

Body frame

Body frame

Arbitrary

Arbitrary

Body Frame frame

F frame

[8.5.31]. [8.5.32] Eulers [8.5.3111. [8.5.32]

F frame Body frame

F frame F frame

F frame

F frame

Modified Eulers [9.2.3] Boltzmann-Hamel Angular acc. term Boltzmann-Hamel Boltzmann-Harnel

Boltzmann-Hamel [9.3.4]. [9.3.6] Time-varying inertia matrix [9.3.4]. [9.3.6]

reference frame different than one attached to the body will be counteracted by the time-varying inertia matrix. The choice ol of a different frame becomes attractive for bodies possessing inertial synimetry. in which case one can use the F frame to differentiate or resolve the moment balance equations. One should be careful, though. to correctly calculate the velocity of the contact point, especially if it is located along the F frame. Table 9.3 summarizes the different possibilities. There are a few other possible cases, but those do not have common applications. When selecting the approach for writing the equations of motion. take inte consideration a variety of factors, including the acting forces, the additional terms you need to calculate, and the complexity of the resulting equations. Of course, you should also compare these different approaches with the analytical techniques to derive the equations of motion. Experience with a variety of problems is the best to develop guidelines for selecting the method to derive the equations of motion.

9.5

QUASI-COORDINATES AND QUASI-VELOCITIES

(GENERALIZED SPEEDS) In the previous sections, we saw additional ways of writing the equations of motion. What makes these equations so useful is that they are first-order equations and the angular velocity components along the body axes are used directly in the equations of motion. We also saw in Chapter 8 that in certain cases it is more convenient to retain the angular velocity components in the formulation. Angular velocities are not generalized velocities and they are not perfect differentials. Recall Examples 4.14 and 5.13, which dealt with a nonholonomic constraint in conjunction with a vehicle dynamics problem. We saw that switching to a rate

9.5

AND QUASI-VELOCITIES (GENERALIZED SPEEDS)

variable different than the generalized velocities simplified the equations of motion

considerably. The question arises as to whether one can define other such quantities and make use ofthern ofthem in deriving the equations of motion. As we saw in Chapter 7. it is possible to define defineaaset setofofvariables variablescalled called quasi-velocities or generalized ss2 These quasi-velocities or generalized speeds.2 These are quantities that are linear combinations of the generalized velocities but which cannot necessarily be integrated to the generalized coordinates. We define quasi-coordinates as a set of coordinates whose derivatives are mean-

ingful quantities (quasi-velocities), but they themselves do not necessarily have a physical meaning. We will denote the quasi-coordinates by q. . ., and the quasi-velocities by u1, quasi-velocities by u1, u2, u2, .. u,. Using this definition, generalized velocities become a subset of the quasi-velocities. An alternate definition for quasi-velocities is variables that cannot always be expressed as derivatives of displacement coordi.

. .

nates.

Let us first consider a holonomic holonomic system system having having n degrees of freedom, with n independent generalized coordinates . ., qn and generalized velocities We defined earlier a set of independent generalized speeds u1, . ., u1, u2, u2,.. , q21 .

.

. .

.

fl

k =

=

1,2,...,n

19.5.11

r=l functions of the generalized coordinates and time. In column vectorwe form can write Eq. [9.5.1] as {u}

= [Y]{4'} + {z}

[9.5.21

order for the set of generalized speeds to completely describe the system under consideration, it must must be be defined defined such such that that [Y1 [Yl is nonsingular. For example, the relation between the body angular velocities and the Euler angles, Eq. [7.5.7], defines a valid set of quasi-velocities.3 By contrast, as we saw in Example 7.6, the angular velocities of the F frame do not constitute aa set set of of quasi-velocities, quasi-velocities, as the relation between them and the Euler angles is defined by a matrix that is singular at In

all times. Inversion of Eq. [9.5.2] yields = [YIH({u} in which [W] =

I

Y]' and {X} =

—

{Z}) = [W1{u} + {X}

[9.5.3]

—[Y] '{Z}. The above equations are referred

equations ororkinematic to as kinematic equation1s kinematic differential differentialequations equations for the generalized speeds. 2Each of these these names has been used used in in conjunction conlunction with with one one of of two two distinct distinct ways ways to to derive derive equations equations of of motion, motion, the Gibbs-Appell equations and Kane's equations. Because the two methods are are equivalent, equivalent, we we use use the the terms terms interchangeably. 3Equation [7.5.7] has a singularity when the second Euler angle reaches a certain value. This singularity may be avoided by switching switching to to aadifferent Hence, the the transformation transformation different set set of of Euler Euler angles, angles, or by using Euler parameters. Hence, between angular velocities and Euler angles is considered here as a valid transformation between the 9eneralized 9enerahzed velocities and the quasi-velocities.

505

506

CHAPTIR

9•

DYNAMICS OF RIGID BODIES: ADVANCED CONCEPTS

Consider next a position vector r. For an n degree of freedom system r can be r has the form expressed as r = ar 9r [9.5.41 v= r = qk + --di

or, in column vector form,

= frqI{q}

[9.5.5) 19.5.51

+ dt

whose elements are

in which [rq] is a matrix of order 3 X

c9{r}

[9.5.6]

1.2,...,ii = L2,...,n

[9.5.7) 19.5.71

c9{r}

[!qI [!qI =

ôq2

Further, in Chapter 4 it was derived that 9r

r9v

kk

cIqk

so that calculation of the matrix [rq] can be accomplished using the position vector so r or the velocity v. In the latter case, we have iI{v}

[rq] = [rqiJ

(9{v} -

.

.

[9.5.SJ 19.5.S1

.

This is a very important result, as it permits calculation of [rq] when only the

velocity vector is available, such as in cases involving angular velocities. Introducing Eq. [9.5.3] into Eq. [9.5.5], we obtain {v}

We

+

+

=

[9.5.9) [9.5.91

define the following quantities: Partial velocity matrix: [vi']

I

W], a matrix of order 3 X n

Time partial velocity vector: {v') = [rq]{X} ±

a vector of order 3

Thus, the velocity can be expressed as {v}

{v'} = [v'9{u} + {vt}

[9.5.10) [9.5.101

In algebraic vector format, we can express this relation as n

v=

+v

[9.5.11) [9.5.111

is the the time partial velocity of v.4 It in which vk is the kth partial velocity of v and v' is

thilows that with a properly defined set of generalized speeds, one can express the 4Note that this notation is slightly different than the notation used by Kane. We denote the indices of the partial velocities by superscripts, whereas Kane denotes these indices by subscripts.

___

9.5

AND QUASi-VELOCITIES (GENERALIZED SPEEDS)

velocity of a point in terms of the generalized speeds and the partial velocities of the system.

Observe that partial velocities can also he expressed as k

Vt

19.5.121 [9.5.121

at

This equation can be interpreted as follows: The kth partial velocity denotes the

direction of the velocity along the direction affected by the kth generalized speed.

The reader should compare this interpretation with the interpretation in Chapunit vectors about which Euler angle transformations are e6, and ter 8 of taken.

One can also define partial velocities for angular velocities. The procedure is the same, where one first expresses the angular velocities in terms of the generalized velocities, and then relates the generalized velocities to the generalized speeds. with the result = (1) (1) =

'I

+

[9.5.13] 19.5.13]

&

k—I

in which wk

and & are partial angular velocities. Also, we have so far considered inertial reference frames and the associated partial velocities. It turns out that partial velocities can be calculated for velocities that are measured from relative frames. We can then write the above expressions in the more general form =

fl

=

+ Avt

Av

\" 11

'4tU8

=

k=i

+

k—i

denote reference frames. Let us now consider the virtual work. For a system of N particles, the virtual work is B

N

N

F, 8r 8r F, •

=

> '{Fj}T{5rj}

19.515]

i=I 1=1

In terms of the generalized coordinates, the expression for the virtual work is written In as

6W=

= {Q}"{6q} {Q}T{6q}

[9.5.161

where the. generalized fbrccs are found from N

Qk = Qk i—I

9r1

N

=

>'Fj. i.:=I

19.5.171

507

508

CHAPTER 99 •• DYNAMICS OF RIGID BODIES: A1)VANCED CoNcEvrs CHAPTER

or, in column vector format. N

{Q} =

>I,uI{I}

[951$j

is a column vector of order 3 containing the components of F.

where

We next express the generalized forces in terms of the generalized Left-multiplying the above equation by [W]' = [y]_T and defining by {U} [W1T{Q} = (LI1

=

U,...U,,J'.weohtain {U}

I

wJT{Q} = wIT{Q}

Introducing the definition of the partial velocity matrix to the equation. we have N

=

{Li}

\'[v'J'{F,}

[9.5.201

i—I

or.

in algebraic vector form, N

=

Uk

Y

19.5.211

F

where is the kth partial velocity associated with the point to which is applied. applied We refer to Uk as the the generalized generalized fi)rce fi)rce associated assoeiaied wit/i with the kt/i generalized and to {U} as the generalized force vector associated associatedwith with generalized generalized speeds. 1: follows that the virtual work can he written in terms of quasi-coordinates as

= {U}{6q'}

=

[9.5.221

Note that we are using the variations of the quasi-coordinates for derivation pwposes only. When the generalized speeds are selected as the generalized velocities = (k = 1, 2. 2........ii)ii)the theassociated associatedgeneralized generalizedforces forcesUk Uk become become the the same the generalized forces QA. Next, we express the generalized forces associated with with aa rigid rigid body. body. a rigid body subjected subjected to to N N forces forces F1 (i (i = 1,1. 2, ... .. N) and and M* torques torques M, M, (i(i = 1, 2, .. ., M"). The generalized force for the quasi-velocities takes the form .

.

.

.

M

V

=

• F1

i=I

+

Y

k

•

=

1, 1.

2,

.. .. ii .

.

19.5.231

i=I

can express the generalized forces in terms of the resultant resultant of of all all the the forces forces and moments. Writing the resultant of all applied forces and moments as force F passing through a point B and a moment M11 about B. where We

N

F = YF1 i=I 1=1

M,1

=

>:(Mj

+

r

X X F1)

[9.5.24]

i—i

which r1 is the position vector from point B to the point where F1 is applied, we can express Eq. [9.5.23 1 as in

9.5

509

QUASI-COORDINATES AND QUAsJ-VEIocrrws (GENERALIZED SPEEDS)

=

Uk Uk

•

F+

•

[9.5.251

M3

When using the center of mass to express the resultant force and moment, we write •F +

Uk =

•

[9.5.261

MG

The decision about which of Eqs. [9.5.23], [9.5.25], and [9.5.261 to use depends

on the number and nature of the forces and moments. For a system of N rigid bodies. we extend Eq. [9.5.261 to N

N

Uk

=

+>

>'

19.5.271

denote the in which G1 denotes the center of mass of the ith body, and F1 and resultant of all forces external to the ith body and the resultant moment about the center of mass of' the ith body due to these external forces.

Consider the spinning top on a cart in Example 9.4 and fInd the partial velocities associated with the cart, the center of mass of the top, and the angular velocity of the top. Then, calculate the associated generalized forces.

Solution The free-body diagram of the top and cart are given in Figs. 9.1 and 9.5. The velocities of the cart and the center of the top, top. and the angular velocities of the top arc +

= Ya, =

—

= (Y sin 44 ++ w2L )f1 + (Y cos = w1f1 + w2f2 + w3f3 =

Of1

lal

YCOSISIflOf3

cos U — wi WI L)f2

—

Y cos

Ibi

sin Of3

[ci Ic]

frsinOf, ++ (bcosO —— + frsinOf2

We define the generalized speeds as Uj

= Wj Wj

=

=

= (02

[dl

Y

so that the velocity and angular velocity expressions can be written as

[el

+ cos4cosf, — cos4sinOf3)

Vç = u4a2 =

If] if]

u4(sindif1 + cos4cos8f2 — vG = —Lu1f2 + Lu2f1 + u4(Sind)f1

[gJ Ig]

u2f2 + u3f3 u3f3 = u1f1 u1f1 + u2f2

from from which we identify the partial velocities as

=

=

= ft

=

=

a2

= (sin 4f1 + cos4.cos Of2

—

= (sin çbf, — çbf, + cos cos 4cos Of2 Of2 —

w2=f2

Of3)

1k]

Ii]

w4=O

Ill II]

The forces external to the system that do work are gravity and the applied force F. The

cart exerts a moment moment M M about about the the f direction direction to to the the top top through a motor, and there is a counter

Example

9.6

510

9•

CHAPTER CHAPTER 9• DYNAMICS DYNAMICS

RIGID BODiES: BODiES: ADVANCF4D ADVANCF4DCoNcEvrs CoNcEvrs

moment of equal magnitude applied to the cart by the top. To find the generalized forces we use Eqs. 19.5.23] for each force and moment. We identify the external inputs as =

F( =

= — ing( sin 0 f2 ++ cos

—

Fa2 = F( sin th f1 + COS

cos 0

—

acting through G

acting through C. C, and

cos ci.) sin 0

M=Mf3

Iki

The generalized forces arc calculated as =

k

= 1.2,3.4 1.2,3,4

[I] [II

Evaluating the individual expressions, we find LI, = LI,

+

= =

•

±

•-

M + Fa2

+

•

.0

Mt'3 •f1

ingL sinO

M

+ COS Lf1 + Pa2 • 0 + Mf3 •• f2 COS Of:4) f2 = 0 Of:4)'' Lf1

=

9.6

•

±

—ing(sin t9 = —ing(sin

(14

+ w' ±

•

= —mga3•0+ —mca3•0+

±

Fa'•O+ Fa''O+

M

M = —inga3a.2 + Pa2 + 'a2 +Mf3'O + Mf '0 = F —lnga3 Fa2'a2

=

Em]

GENERALIZED SPEEDS AND CONSTRAINTS

Consider a system that originally has ii ii degrees of freedom subjected to in equality constraints. We write the constraints in velocity form as + a01 = 0

/1 = 1,2...., in

19.6.1]

wherethecoeflIcientsafk(j = 1.2, 1,2, n)andao1(j = 1,2, ,m: k = 1,2, ., n)andao,(j arc functions of ol the generalized coordinates and of time. When the constraints are nonholonomic. one cannot readily eliminate them from the formulation. SO so we we exex. .

. .

.

.

.

.

,

plore the use of generalized speeds. We write Eq. 19.6. 1] in matrix form lbrm as

IaBfl +

{b} = {O}

19.6.21

into Eq. Eq. 19.6.2], we express the constraint in terms of the Introducing Eq. [9.5.3] into quasi-velocities as {b} [al[WJ{u} + [a]{X} + {b}

[A]{u} + {B} =

19.6.31

and we note that the matrix [A] = [al[WJ is ol order in X X ii. ii.The The generalized generalized velocities and the generalized speeds are no longer independent. We now refer to {u} as the constrained generali:ed constrained generali:ed speeds. Because we we have have pp = ii — — in degrees of freedom, it is of interest to generate a set of p unconstrained (independent) generalized speeds. We order the generalized speeds such that the first p are coordinates from from are coordinates

which the independent generalized speeds arc constructed and collect them in a vector {u1,} = [u1 114 The remaining generalized speeds, called surplus 112 ..

..

..

9.6

generalized speeds, are expressed as {uS} = tition the generalized speeds into {u}

= [{uf)}7

GENERALIZED SPEEDS AND CONSTRAINTS

...

u1

I

uniT. We hence par-

{u5}T]T {us}T]T

19.6.41

We define a set of independent generalized speeds by The vector containing the p independent generalized speeds

••• Up]7. = ["i by is related to

= [T] '{u,}

[9.6.5]

=

{Ifl

[T] is a nonsingular transtbrmation matrix of' order p X p. Without loss of in which [TI and {u1,} coincide, so that [TI = [1]. We next generality, we assume here that partition the constraint matrix [Al as =

[9.6.6]

[A5]]

Introduction of of this this partition is of' of' order m X p and [As] is of order in X in. Introduction where [Ar] is into Eq. [9.6.3] gives

[9.6.71

[A1,]{u,,} [A1,]{u,,} + IA5 1{us} + {B} = {O}

Solving for {uS}, we obtain {us} =

—

[Aç]

—

I

[Asf '{B}

=

+ {B'}

19.6.8]

in which

[A'] [A'l =

{B'} {B'}

=

[9.6.91

which is the requirement when For Eq. [9.6.8] [9.6.8] to hold, lAs] must he nonsingular. which

partitioning the generalized speeds into independent and surplus sets. Next, we write the generalized velocities in terms of the independent and surplus generalized speeds. The relation between the generalized velocities and the generalized speeds, Eq. [9.5.31. can be partitioned as {X} [Wf,j{uf)} + I Wsl{us} {X} = [Wf,]{uf)} = {W]{u} [W]{u} + {X} Wsl{us} + {X}

19.6.1 0] 19.6.1

X p and X in, respectively. and n X in. respectively. are partitions partitions of I Wi of orders nn X where [We] and [W5 1 are the above above from Eq. [9.6.8] into [W5]]. Introducing Introducing the the expression expression from into the [Wa]]. [W] = [Wi equation. we we obtain 1

+ {B}) + {X} = [14']{ui} +

=

19.6.111

in which

[W] = [*1

= [Wv] + [Ws][A'] [Ws][A']

—

19.6.12]

next extend the concept of independent generalized speeds to the partial velocities. We rewrite Eq. [9.5.10] as We

{v}

= [V]{u} +

[9.6.1 31 [9.6.1

where [V] is of order 3 x ii and is the partial velocity matrix associated with {v}. We and partition [V] fV] in terms of the independent and surplus parts as [V] = write {v}

=

{v'} + [Vsl{us} ++ {v'}

10.6.141

51

1

______

512

CHAPTER 9• DYNAMICS OF RIGID BOIMFS: ADVANCED CONCEPTS

Introducingthe theexpression expressionfor for{US} from Eq. [9.6.8 1 to this equation and carrying out Introducing

the algebra gives

19.6.15]

+ [V5]4B'} + {v'}

=

{v}

The partial velocity matrix associated with the independent generalized speeds

can be expressed as

[9.6.16]

[c'l = [Vp] + [VsI[A'l

Note the similarity in form between Eqs. [9.6.121 and (9.6.16]. To express the partial

velocities in algebraic vector form we note that contains the first p partial velocities, and [VsJ[A'l the remaining in. Hence, the partial velocities associated with the independent partial velocities can be expressed as = vk+>TAI,kvi+P

k

=

ft

l,2,...,p l,2....,p

=

j=I 19.6.171 where A,k and

are the entries entries of of [A [A 'I'I and and {B'}. respectively, respectively, which which were defined in

Eq. [9.6.9]. When the transformation matrix [T] [Tj is not the identity matrix, we have [A'] =

[W] =

+ IW5I[A'i IW5I[A'I

19.6.18] [9.6.19] 19.6.19]

= in which Tkr are the entries of [TI.

We repeat the procedure for independent partial angular velocities. The results turn out to to be the same as Eqs. [9.6.17], with the v terms replaced by w: =

In

+

=

j=I .1=I

&+

[9.6.20] [9.6.201 j=I

We now calculate the generalized forces associated with the independent gener-

alized speeds. To this end, we make use of the quasi-coordinates. From Eq. 19.6. 19.6.111] 1] we write the virtual displacements as 6{q} = [W]6{4'}, [W]6{4'}, where 6{4'} are the virtual where 6{4'} displacements associated with the independent quasi-coordinates. As before, we are using this expression only for derivation purposes. Introducing the virtual displacements into the virtual work, we can write 6W = {Q}T6{q} = {Q}TI

= {U}T6{4'}

[9.6.21]

where {U}T

= {Q}T[W1 {Q}'[Wl =

+

19.622] 19.6.22]

a vector of order p in the form {U}' {U}" = [UI [UI U2 U2 ... U,,, 1. Toexpress express the the gener1. To alized forces associated with the independent generalized speeds, we partition the generalized forces as is

{U}T {U}T

= =

[Ws11 = [WsI1

[9.6.23]

9.6

513

SPEEDSAND ANDCONSTRAINTS C0NsmAINTs GENF.RALIZEI) GENER&uzEI SPEEDS

Introducing Eq. 19.6.231 into Eq. 19.6.221 gives {U}T =

[9.6.24]

+

=

be expressed in scalar form as which can be 'H

k

Uk

=

l,2,...,p

[9.6.251

j—1 f—I

We have eliminated the surplus generalized speeds from the formulation and arrived at a set of' independent generalized speeds, partial velocities, and generalized forces. This is a significant advantage of using generalized speeds rather than generalized velocities. Recall that with Lagrangian mechanics. nonholonomic constraints and associated Lagrange multipliers cannot be eliminated from the formulation before the equations of motion are derived. Only after deriving the equations of motion in constrained form can one use a series of substitutions to eliminate the Lagrange multipliers and surplus coordinates.

One can arrive at a set of independent generalized speeds. partial velocities, and generalized forces in two ways. One is by observing the kinematics of the sysspeeds. The other is tem and directly identifying a set of independent generalized speeds. by beginning with a set of constrained generalized speeds and then applying the constraint equations derived above. The latter approach is most often a complicated of independent independent generalized speeds one is usually better off generating a set of directly.

Under certain circumstances, such as in problems involving Coulomb friction, it may be more desirable to deal with constrained quasi-velocities than with unconstrained ones. In such cases, one deliberately leaves a constrained coordinate in the system formulation. This procedure is similar to the relaxation of constraints using Lagrange's equations.

Consider the rolling disk problem. First consider the case when the disk is rolling and slipping.

obtain the partial velocities associated with the center of mass and the partial angular velocities. Then, impose the rolling constraint and obtain the corresponding partial velocities and partial angular velocities. and

Solution frame. We We select the inertial frame angle transformation transformation and the F frame. We use the 3-1-3 Euler angle such that the a1a2 plane is the rolling surface. and the a3 direction is the vertical. We invoke the constraint that the disk is always in contact with the surface, so that for roll with slip we have five degrees of freedom. We select the quasi-velocities as UI

=Wj =0

= (U3 (U3 =

142

+

W2

u4 U4 = X

=

Y

lol

where X and Y Y denote the coordinates of the center of mass in the a1 and a2 directions, re-

spectively. Noting that the height of the center of mass is given by the holonomic constraint relation Z = R sin 0. the component of the velocity of the center of mass in the a3 direction mass and the angular velocity of the k cos 0 = Ru1 cos 0. The velocity of the center of niass

Example

9.7

514

CHAPTER 9• DYNAMICS OF RIGID BODIES; ADVANCED CONCEPTS

body can he written as VG

= Xa1 + Ya2 + Za3 Za3 = u4a1 ++ u5a2 u5a2 ++ Ru, cos 0

= u1f1 u1f1 +

1k]

+ u3f1

Ic] [ci

The partial velocities and partial angular velocities can now he written as

= RcosOa1

=

=

0

=

a1

a2

Id] find the partial velocities associated with the constrained system, we first calculate the velocity of the contact point C and set it equal to zero, thus To

=

VG + (a)

=

(u4

—

X

rc,G = u4a1 +

Ru, s4,sO Ru,sçbs6

+ Ru,cOa3 + (u1f1 + u2f2 +

+ Ru3c4,)a, +

X (—Rf2)

+ Ru1c4s0 + Ruis4,)a2 =

1.1

0

As expected. components of the motion in the a3 direction canceled each other. We select the independent generalized speeds as = u1. 142 = u2, u2, and = so that the surplus coordinates are = [u4 u5]T. We are retaining the angular velocities in the

formulation. Comparing Eq. [el with Eqs. [9.6.7] and [9.6.8], we have 1

[A S1

—o 0

0

[A 1 — —R 1

i1

—

[A'I — = — —[A S F'[A

I

SIfl (/) sin 0 sn 4,

—

0 0

cos 4, sin 6

= = —[A [Ar]I — — —

COS 4,

{B} {B} —

sin 4,

Sin — sin4,sin0 _R[ —R

cos4,sinO

—

0 0

0

0

cos4, sin4, ]

Ifi

The matrix [As] is nonsingular: nonsingular; thus, our choice of independent generalized speeds is a

valid one. Next, we introduce the constraints into the expression for the velocity of the center of mass. Introducing the expressions for u4 and u5 from the right side of Eq. [ci into Eq. [hi and expressing the result in terms of the F frame, we obtain VG

= (Ru, sin 4, sin 0— Ru1 cos 4,)a1 —

=

(Ru1

cos4,sinO +

Ru3

sin 4,)a2 +

Ru1

cos6a3

—Ru3f1 +

Igi [gi

which which is the well-known result for the velocity of the center of mass. The partial velocities

associated with the velocity of the center of mass in terms of the independent generalized speeds are = Rf3

= 0

=

—Rf1

1k]

The independent partial angular velocities remain the same as before and have the form

r=f2

w =fi

w =f3 w =f

[II

We We

next calculate the partial velocities associated with the independent generalized speeds using Eqs. [9.6.17]. We find =

+

+

= —

=

+

+

R cos

6 a3 + R sin 4, sin 0 a1 — — R cos 4, sin 0 a2 = Rf3

5_

2..LA' '

"2257G —

R cos 4, a1 — = 0— 0 — Rcos4,a1 R Sin 4, a2 = — Rsin4,a2

— Rf1 —Rf1

11] 111

9.6

Consider the rolling disk. First analyze the case when the disk is rolling and slipping and obtain the generalized forces. Then impose the rolling constraint and obtain the associated generalized forces.

Solution will use the expressions generated for the partial velocities in Example 9.7. For roll with slip, there are five degrees of freedom. We select the quasi-velocities as in the previous example as We

=

112 = (02 = 4.)SiflO

0

+

U3 = (03 =

U4 = X

i/i

115

=

lal

and can write the velocity velocity of ofthe thecenter center of of mass mass and the the angular angular velocity velocity vector as as

=

u4a1

+ u5ai + Ru, Ru1 cos cosOa3 0 a3

w=

u1f1

+

+ u3f3

Ibi

The associated partial velocities are I

=

= R cos 0 a3

= 00

4

= a1

VG

VQ

= a2 a2

o4=0

[cJ

The forces forces acting acting on on the disk arc the force of gravity.

= —mga3. and three reaction f1 ++ F2f2 F2f2 + forces at the point of contact, F1 f, whcrc P, and F, = P1 a1 + P2a2 + P3a3 = F, are their components in the A and F frames. We can use two approaches to find the generalized forces associated with the quasi-coordinates. One approach is to treat each force separately and use Eq. [9.5.23]. which requires the partial velocities associated with the individual points at which the forces are applied. For the problem under consideration, consideration. this requires calculation of the velocity of point C and of the center of mass G. As there are no applied torques. there is no need for partial angular velocities. The second approach is to take the resultant and use Eq. [9.5.26]. Here, one needs to calculate the resultant moment, the partial velocities associated with the center of mass. and the partial angular velocities. To demonstrate the merits of the two approaches, we will use both.

In the resultant approach. we first calculate the resultant force, F, which has the lorm F =

+ Fç =

P1a1 P,a1

+ P2a2 +

rnX)a3

—

— mg sin (1)f2 + (F3 (F2 — = F1f1 F1f1 ± (F2

—

[d]

ing cos

The resultant moment about the center of mass is

MG = rcIG X

=

—

Rf- X (F1 f1 f1 ++ F,f2 + F3f3) F3f3) = —

Rf1 ++ P1 F,

1.1

Note that we are Note are conveniently conveniently using using unit vectors vectors from from both both the the A A and F frames to simplify the

expressions. The associated generalized forces become = (P3—mg)RcosO—F1R (P3—mg)RcosO—F1R

U1 = U,

U2 U3

=

=

U4 =

U5 =

51 5

GENERALIZED SPEEDS AND AND CONSTRAINTS

= 0 =0 = F,R F1R

= P, P1

'F + w5 'MG =

P2

Ifi

Example

9.8

516

CHAPTER

9• DYNAMICS OF RIGID BODWS: ADVANCED CONCEPTS

When treating each force separately. we first find the velocity of the contact point C. which is

vG+wXrcIG

Vç

u4a1 + uia2 ++ Ru1 Ru1 cosfla3 cosfla3 ++(u1f1 (u1f1++u,f2 u,f2++u3f3) u3f3)XX(—Rf2) (—Rf2)

= U4à1 + uca2 u5a2 ± Ru1 cosOa3 + Ru3f1 Ru3f1 — —

[gJ

so that the partial velocities associated with the velocity of point C have the form vi-.

=

—

=

Rf3 + R cos 0a3 Rf1

=

0

=

Rf1

a1

=

a2

Ihi

We proceed to find the generalized forces. Noting that there are no applied torques, we we write the generalized forces as

k k

Fg +

Uk

=

II]

1, 2, . . ., 5 1,

Carrying out the dot products. we obtain

'• RcosO a3 + (P1a1 + P2a2 + P3a3)•(-Rf1 P3a3)'(-Rf3 + RcosO a3)

U1 =

= —mgRcosO

—

F3R + P3RcosO

U2=o U3 =

—rnga3

'0 + (F1f1 + F2f2 .0 F2f2 ++ F3f3)• F3f3)' Rf1 = F1R

+(P1a1 +P2a2+P3a3)•a, +P2a2+P3a3)'a1 U4 = —rnga3'a, --rnga3'a1 +(P,a1 U5 =

+ (P1a1 (P1a1 ++ P2a2 P2a2++P3a3) P3a3)• 'a2 a2 =

—rnga3 • a2

P2

IV

which is identical to Eqs. [f]. Now we impose the no-slip constraint and obtain the generalized forces associated with the independent generalized speeds. Here, we use the resultant approach. From Example 9.7. the independent partial velocities are

= Rf3

-3

= 0

=

-1 = f, f1

—Rf1

=

f2

=

f3

[ki

so that the generalized forces become U1

=

'-'I

vG

'F + w •MG •F 'MG =

(F3

—

rngcosO)R

=

F3R = —rngRcosO

= 0

U2 = 02 U3

—

• F+ 'F +& & 'MG 'MG =

—F1 R + F1 —F, F, R = 0

[I] [LI

As expected, expected, the reaction Iorces at the contact point do not appear in the generalized forces. Let us now calculate the generalized forces associated with the independent generalized speeds using the constraint equation. Eq. [9.6.25]. From Example 9.7 we have A'

—

— —

—

sin 4, sin 0

cos4,sin0

0 0

cos 4, cos4 sinçb

thus, we can write Eq. [9.6.25] as U, = U, U1 + A,U4 ++A',, A',, U5 U5 = U, + U4Rsin4,sinO U4Rsin4sinO — U5Rcos4,sino UsRcos4sinO

= —mgR —mgRcosO cos 0 — — F3R + P3R P3RcosO P1Rsin4sinO cos 0 ++P1R sin 4, sin 0

—

P2R cos 4, sin 0 P2RcosthsinO

9.7 02 = U, U3

=

=

+ A,U4 + A,Uc =

U2

=

0

=

LI3

—

U4Rcoscb

+ F1 R F1

+

-

P1 R COS fr —

Noting that F1 = P1 P1 cos

+ P2 sin

we observe that that all all the the P, P and F Eq.

=

1,

GIBBS-APPEIJEI EQUATIONS

—

P2R sin

En]

and F3 = P1 sin sin 0 — P2 cos / sin 00 + P3 cos 0. 2, 3expressions cancel from Eqs. ml. and we obtain

mu.

9.7

GIBBS-APPELL EQUATIONS5

The Gibbs-Appell method makes use of a scalar function in terms of accelerations to derive the equations of motion. analogous to the concept of using kinetic energy in Lagrange's equations. We describe the method first for a system of particles and then for rigid bodies. Consider a system of N particles that has n degrees of freedom. We will initially consider independent quasi-velocities. The force balance for each particle can be written as m1a, =

ji = l,2,...,N

F1 +F

[9.7.1]

order to simplify the derivation and to establish that the Gibbs-Appell and Kane's equations are indeed identical. we will carry out the derivation of the GibbsAppell equations using partial velocities. This way of deriving the Gibbs-Appell equations is not the traditional derivation. We select a set of n independent quasi-velocities Ut' ..., u,1 and write the In

velocity and acceleration of each particle. The velocity of each particle can be written as fl

—

I

"\

i

•

V, y1

•

k=1

The acceleration of each particle can be written as a1

=

+

+

>

[9.7.3]

Define the Gibbs -A ppell function,denoted denoted by by S, S, as as having having the form N

S

=

• a,

[9.7.4]

The Gibbs-Appell function is also referred to as the energy of acceleration or

the Gibbs function. Let us next differentiate S with respect to the time derivative of the kth quasi-velocity, which yields 5lhere is debate on whether these equations should be called the Gibbs-Appell or the Appell equations. The name Gibbs-Appell is attributed to Pars.

51 7

5 'I 8

9

• DYNAMICS OF RIGm BoDws: ADVANCED CONCFYFS

. rI as

1

=> ,>,n1a1

.

(9Uk

Noting from Eq. [9.7.3] that

(Ia1 •

(lUk

.

1=1

and

do

[9.7.5]

• .

not contain any derivatives of the general-

ized speeds, we can write 9a1 .

=

/Iv,

= vik

k =

1,2,...,n

19.7.6]

SO that. substituting this result result into into Eq. Eq. 19.7.51 19.7.5 1 and definition of ofthe the generand using the definition alized forces associated with the quasi-velocities, we obtain =

,

(hLIk

,n1a1

=

= Uk

19.7.7]

i=1 1=1

where where the contributions of the reaction forces have canceled out. Equation

can be rewritten as

= = (I U

Uk

k =

1,2,...,iz

19.7.81

Equation [9.7.8] represents the Gibbs-Appell ppellequations equations or Appell Appell equations in

terms of independent quasi-velocities. They arc n first-order equations. When combined with the kinematic differential equations relating the generalized velocities to the generalized speeds. they form a set of 2rz first-order fIrst-order equations that describe the evolution of the dynamical system completely. To write the Gibbs-AppeIl Gibbs-Appell equations, equations. we first determine the number of degrees of freedom of the system and select an an appropriate appropriate set set of' of' independent independent generalized coordinates and quasi-velocities. We proceed with the kinematics of the problem, developing expressions for the velocity and for the acceleration of each particle. From the velocity terms, we identify the partial velocities. From the acceleration terms, we generate the function S. Because we will be taking partial derivatives of S with respect to the rates of change of the generalized speeds, we can ignore any term in S that does not contain derivatives of generalized speeds. After obtaining S. we take the partial derivatives. We obtain the generalized forces Uk after calculating the forces acting on each particle. We then invoke Eq. [9.7.8J. The term ÔS/dàk can can also be calculated by using the second term on the right side in Eq. [9.7.7]. without the need to calculate S explicitly. Doing so, as we will see in the next section, is tantamount to using Kane's equations. Let us compare the Gibbs-Appell Gibbs-AppelI equations and Lagrange's equations. In Lagrange's equations. first the kinetic energy is calculated, which is a function of the velocities. Lagrange's equations can handle holonornic constraints as—at least in theory one can find a set of independent generalized coordinates that take into account the constraints. In the presence of nonholonomic constraints, one cannot eliminate the constraints beft)re beibre deriving the equations of motion. But the Gibbs-Appell Gibbs-Appell method uses an acceleration function. Hence, for systems acted upon by nonholonomic constraints, it becomes possible to find a set of independent quasi-velocities that are compatible with the constraints. In such systems. systems, one ends tip with n—rn = p independent equations and n kinematic differential equations, for a total of n + p first-order equations.

9.7

GIIiBS-APPELL EQUATIONS

Lagrange's equations can be shown to be a special case of the Gibbs-AppeII Gibbs-Appell

equations. Selecting the quasi-velocities to be the same as the generalized velocities. ii), we observe that Uk qk (k = 1, 2, . = Uk, and by manipulating S one can show that .

(9S .

.

. ,

=

dS ..

=—

—

k =

1,2,...,ii

(911k

19.7.91

It is for these reasons that the function S is also referred to as the function. fttnctioii. One might wonder at this point why we bothered to learn about Lagrange's equathe Gibbs-Appell equations represent the more general case and can handle tjOflSq holonomic as well as nonholonomic nonholonomic constraints, constraints, as as well as leading to equations of motion that can be put into state form with greater ease. There are several answers. First of all, for holonornic systems, the Gibbs-Appell equations have no advantage over Lagrange's equations1 The Gibbs-AppeII Gibbs-AppeIl equations are more cumbersome; they they require the calculation of acceleration terms. terms, as opposed to the velocity terms needed for Lagrange's equations. From a physical perspective, velocities are easier to visualize than accelerations,

so that dealing with velocities gives a greater physical insight. Following this line of argument. the kinetic and potential energy are quantities that are much easier to visualize than the function S. which looks much more abstract. Dealing with energy expressions helps one differentiate between natural and nonnatural systems easier. easier, and it simplifies the solution of equilibrium problems. problems. Finally, Finally, integrals integrals of' of the motion are much more easily derived from the kinetic and potential energy than they are from the fundamental fundamental f'unction function S. The definition of S is motivated by Gauss's principle for virtual accelerations, By which is obtained by considering a special class of variation (see Section S contrast, Lagrange's equations are based on D'Alemhert's principle and its extension to scalar functions, the extended Hamilton's principle. Another shortcoming of the Gibbs-AppeIl Gibbs-Appell equations is when deriving the equations of motion associated with spatially continuous systems. As we will see in Chapter 1 1, for deformable bodies the extended Hamilton's principle leads to both the equations of' of motion motion and and boundary boundary conditions. conditions. By contrast, contrast. the Gibbs-Appell Gibbs-AppeIl equations can be used only after the flexible motion is discretized. We are ready to extend the Gibbs-Appell equations to rigid bodies. The general form of the equations of' motion is still Eq. [9.7.8]. so so we develop the expression for the fundamental function. Intuitively, we expect it to contain contain expressions expressions associated associated with the acceleration of the center center of' of mass and with the rate of change of angular angular momentum. From Fig. 8.1, the acceleration of a differential mass element on the body is written in terms of' the center of' mass motion as I

X (to X p) a = aG aG + a X p + to X

[9.7.101

where to and and aa are arethe theangular angularvelocity velocity and and acceleration acceleration of of the body. respectively.

and p is the position vector from the center of mass to the differential element. It follows that ( p dm = 0. The Gibbs-Appell function can be written as an integral

519

___

520

CHAPTER 9O DYNAMICS OF RIGID RIGID BODIES: BODIES: ADVANCED ADVANCED CONCEPTh CONCEFTh

over the entire body as S

=

if a'adrn a•adrn

[9.7.11] 19.7.1 1]

We introduce Eq. [9.7.101 into Eq. [9.7.111 and integrate integrate over the body. All of

the 5 p dm terms vanish and terms that do not contain any accelerations or angular accelerations can be ignored, as they do not contribute to ÔS/9ük. ÔS/9ilk. The expression for

S reduces to

X p)•(a p).(a X p)drn

+

S =

+J(a

X

X(w X p))dm

[9.7.12] It is convenient to express S in terms of the angular momentum. Here, we follow an approach similar to the one in Section 8.11. The second term on the right side of Eq. [9.7.121 can be expressed as

(a X p)•(a

X

p) >

[9.7.13]

[,5]T[151 din and we realize that J [,5]T[fjj

['GI and that that I!G1{a} I!Gl{a} = {ArG}rel. Manipulation of the last term on the right side of Eq. [9.7.12] is more complicated. After a number of manipulations one can show that X p) • (w X (w X p)) = a X (p X (w X p))) a •(w (w X

(a

[9.7.14]

where we recognize the expression that leads to the angular momentum term, p X

(w x p). Carrying out the integration, we can write the Gibbs-AppelJ function for a rigid body as

a • (co X HG)

S = i/naG

[9.7.15]

The last two terms in this equation indicate that S is related to the rate of change of the angular momentum. In column vector format we have + 1{a}T[l {a}T[l ]{a} + {a}T[&1[IGI{w}

S =

1

[9.7.16]

When When taking the partial derivatives of S with respect to the derivatives of the generalized speeds. a simplification takes place when when we we recall recall Eq. Eq. 1.9.7.61. 19.7.61. Indeed.

for the acceleration of the center of mass, one can write

=

dVQ

auk

1<

VG

k =

1,2,...,n

[9.7.17]

and. using the same argument for the angular acceleration vector, write and,

(9Uk

(9Uk

=w

[9.7.18] 19.7.18]

Taking the partial derivative of S with respect to ilk and introducing the above two equations into these partial derivatives, we obtain

9.7

\i

(9S

=

\T

+ dUk

= v(.ma(; v(ma(; +

+

)

(A)k

.(

'Grel

521

GIBBS-APPELL EQuA'rloNs

+

)

19.7.191 [9.7.191

+

X H(;) =

the partial derivatives of S can be calculated without evalthe generalized forces tbrccs are given uating S itself'. itself. Considering Considering that the expressions by Eq. [9.5.26]. we can write the Gihhs-Appcll equations for a rigid body as so that, as with particles, so

F+

•HG =

•Ina(7 IflaG +

k

= 1,2,..., 1,2,..., iiii

19.7.20]

For a system consisting of several bodies, the Gihhs-Appell equations can he written as N

N

>

+ (0, (0, • HG1) =

•

i=1

>

+

•

k=

M

i—i

19.7.21] 19.7.211 When the body is rotating about a fixed point C. the Gibbs-Appell function can

be written about that point. The Gibhs-Appcll function takes the form S

I

[9.7.22] 19.7.221

On the issue of the selection of the quasi-velocities: A good choice of these should simplify the equations of motion and, if possible. give the generalized speeds a physical interpretation. If there are nonholonomic quantities, such as constraints, angular velocities, or body-fixed velocity components, the selection should reflect those quantities.

Derive the translational and rotational equations of motion for a rigid body, using the GibbsAppell equations.

Solution We will use the column vector notation. We select the quasi-velocities as the three angular velocities w w2, and W3 of the body and thrce components of the velocity of the center of mass along body-fixed axes, denoted by v1, v2, and Ui. Hence. Uk =

143—k = Vk

k

[a] lal

1,2,3

To obtain the rates of change of the quasi-velocities, we note that '{VG} = [Uj

t'2

V3]

T_r —

1144

= [w1

U5

W2

(2)31'

=

[u1

U2

[b] Ibi and that

=

d

+ [&1{v(;} [&1{v(;} = 1i'i

+ [th1{v(;} [th1{v(;} =

d = —{w} {a} = [at

di

a2

alT = [uI

+

V2

ü31'

[ci Id]

I

Example

9.9

522

CHAPTER

9•

DYNAMICS OF RIGID BODIES: ADVANCED CONCEPTS

The partial velocities of the center of mass and the partial angular velocities have the form

= = {w'}

{O}

k

[1

0

OJT OIJT

= [0

1

[1

0

01T

{w2} = [0

1

= 1,2,3

k=

{w'} = {0}

0]T

= [0

0

01T

= [0 [0

0

jjT

4, 5, 5, 6

[.1

write the Gibbs-AppeIl Gibbs-Appell function in column vector format. For a set of body-fixed coordinates, We

S= S

+ {a}T[th1[JG]{w}

+

If]

Consider a resultant force vector {F} = [F1 F2 F3J1 F1]1 and a resultant moment vector {MG} = [M1 {MG} [M1 M2 M317 M311 about the center of mass. We calculate the generalized forces, using

Eq. [9.5.26], as + {wk}T{MG}

Uk =

[g]

Evaluating the expressions, we obtain k = 1,2, 3

=

[hI

Consider the rotational equations of motion first. They correspond to the first three Gibbs-

Appell equations associated with UI, U2, and {u'} = [14i [141

U2

u3JT = {w} u3]T

Introduce the notation

{u"} = [u4 [u4

u5

u6]T = [v1

u2

1)31T

=

{v}

[I]

It follows that {ü'} = [u1

{14"}

=

U2

Ü6JT

=

=

[j]

obtain the left side of the Gibbs-AppelI Gibbs-Appell equations, we differentiate S with respect to the derivatives of the first three generalized speeds. From Eq. [d] this corresponds to differentiating S with respect to {a}. Doing so, we obtain To

(es)'

= [JGI{a} [JGI{a} + [&J[JGI{w} [&J[JGI{w}

1k]

From Eq. [hI, the right side of the Gibbs-Appell equations associated with the first three

and last three generalized speeds have the form [U1

U2

1j31T 1j31T

= [M1

M2

[U4

U5

U6]T

=

F2

[F1

M3JT =

{M}

F3JT = {F}

[II

Combining Eq. [k] and the first of Eqs. III yields the rotational equations of motion as [Ial{cx} + [iiJ[JGJ{.o} = {M}

[ml

We next consider the translational equations. Differentiating S with respect to the generalized speeds associated with translation, we obtain cI.S

= m aG

Td{aG}

= tn aG a(

T

[nJ [nJ

__-

9.7 because

is

523

GEBBS-APPELL EQUATIONS

equal to the identity matrix. matrix. Combining Combining Eq. In] and the second of Eqs.

yields

lol lo]

= {F}

+

which are the translational motion equations.

Consider the vehicle in Example 4.14 and derive the equations of motion using the GibbsAppell Appell equations. it is given that the velocity velocity of of point point A A is is always always along along the the x-axis. x-axis.

Solution The vehicle configuration is shown in Fig. 4.8. The orientation of the vehicle can be described

by three generalized coordinates, X. Y, and 6, with X and Y denoting the coordinates of the center of mass. (Another reasonable choice for the generalized coordinates would be the coordinates of point A: YA, and 19.) The nonholonomic constraint indicates that the velocity of point A is always in the direction of' the vehicle, or

[a] [al We select the generalized speeds as u1 that VA = u1i. w u2k. VA and u2 = 6. so that In order to write the generalized forces, we need to calculate the partial velocities associated with points of application of the forces, CandD. To derive the fundamental equations without calculating S explicitly, we also need the partial velocities of G. The velocities of G. C. and

Dare V6

=

uk Xx Li + Ct)XX r(3/A== u,i u,i + u-k Li = u1i +

v

=

+ w X rc,

Vf)

= %4 =

to XX rl)IA = u1i + 0.) uii ++142k u2k x —hj —Iij = (Ui +

= u1 I + uu 2k 2k xx hj = (u1

—

Ii u2 ) I

[b] [bi

The partial velocities are

= Lj• =

= i

I

•

I

V(-. = V(1

to w

•

I

2 V(-. = —hi

•

I

co=k

=0

[ci

The acceleration of' of the center of mass is

a0 = uii + Lu2w Lu2w Xj Xj == (ii, Uii + LñJ ++U10) XI + (Ii,

—

[dl Ed]

+ (Lu2 +

The angular momentum is = 1(3i42k. in which 1(3isisthe themass mass moment moment of inertia about the center of mass. The rate of change of angular momentum is simply HG = IGu2k. Because this is a plane motion problem. the angular velocity and angular momentum are in the same direction; direction: thus, the Gibbs function reduces to S =

•a(3 + •a0

=

(m(üi

Taking partial derivatives of S we obtain

us

=

+

•

= V6

• ma(3

—

I.

+ to w ••

Lid)2 + tn(Lü2 +

.

= m(u1 — —

+

[.1

-,

= ,HL(LU2 ,nL(Lu2 + U1 142) + 1(;u2 Ui 112) l(;u2 = (JG (JG +

+ inLu1 inLu1 u2 u2[11 [II

I

Example 9.1 0

524

CHAPTER 9• DYNAMICS OF RIGID Bonws: ADVANCED CONCEPTS

The external forces are equations become

= Fri. F/) =

ED1.

so that the right sides of the fundamental

U1

U2 =

=

+

—

Fc)

Ig] [gi

Equating Eqs. [f] and Fg] gives the equations of motion as rn ( a1

—

L

=

[h]

+ Ff)

h(FJ) mL2)ü2 )ü2 ++inL ,nLu1u2 U1 U2 == h( (IG + mL2

—

II] III

the equations equations of of motion. motion. Equation Fh] is the One can assign a physical interpretation to the force balance in the x direction. Equation [i} represents the moment balance about A. Let us compare the effort to obtain the equations of motion with Example 4.14. where Lagrange's equations are used and with Example 5.13. where Jourdain's variational principle is used. In both these examples we begin with a set of constrained coordinates and impose the constraint. By contrast, when using the Gibhs-Appcll equations, we begin with a set of independent variables, it is clear that we obtained the equations of motion more easily using Gibbs-Appell equations. the Gibbs-AppeIl The kinematic differential equations are obtained by expressing the rates of change of the generalized coordinates coordinates K. X. Y. Y, and and (1 0 in in terms terms of of the the generalized generalized speeds. To this end, we write the velocity of the center of mass as cos 0 + YsinO)i Y sin 0)i + (—XsinO (—X sin 0 + YcosO)j Y cos 0)j (XcosO = X1 + YJ YJ = (X

[H

and equating this equation to the expression of VG in terms of the generalized speeds in Eq. Fbi we obtain

XcosO + YsinO Xcos0 Ysin0 =

UI

—

XsinO + Ycos6 = Lu2 Xsin0

[k] 1k]

Solving these equations for X and Y. we reach — Lu2 sin 0 X = Ui COS COS 0 —

Y

=

Ui sin 0 + Lu2 cos

0

[I]

[ml

The kinematic differential equation for 6 is simply

O=U2

[n] In]

The five five equations equations [hJ, [hJ, Fil, Fil, [11, [11, [ml, [ml, and and FnJ FnJ constitute constitute the the complete complete set set of of differential differential equations that describe the motion of this system.

On the selection of the quasi-velocities for Example 9.10: We selected u1 as the velocity of AA in view of the constraint. Had we selected one of the quasi-velocities velocity as X or Y, the resulting equations of motion would he quite complicated. Also, components of the velocity along a body-fixed frame provide physical insight.

9.8

EQUATIONS

Kane's equations can be derived in a variety of ways. One can take a system of particles and use the equation of motion of each particle, or one can begin with

9.8

KANE'S EQUATIONS

D'Alernbert's principle and use the transformation from the generalized velocities to the quasi-velocities. In order to establish the equivalence between the Gibbs-Appell and Kane's equations, we begin with the first approach.

Consider a system of N particles that has n degrees of freedom. The equation of motion for each particle can be written as Eq. [9.7.11. [9.7.1]. We select an appropriate set of independent generalized coordinates and quasi-velocities. Associated with each particle and each quasi-velocity there is a partial velocity vector (k = 1, 2, ...,., n; i = 1. 2, . ., N). To obtain Kane's equations, one takes the dot product of each of the N equations of' motion with the partial velocities and adds the resulting equations, which yields .

.

N

=

k

= 1,2,

...,n

19.8.11

The right side of this equation is recognized as the generalized force Uk. We define by F7 = the inertia Jhrce associated with the ith mass, and the sum of the inertia forces multiplied by the partial velocities as the generalized inertia force, is defined as6 N

N

=

[9.8.2]

= 1=1

Introducing this expression into Eq. [9.8.11 yields (J

k = 1,2,...,ti

=

[9.8.3]

which is known as Kane v equations. To establish the equivalence between the Kane's equations and the Gihbs-Appell equations. we focus on Eqs. [9.7.71 [9.7.71 and [9.8.21, from which N

-

dS

=

[9.8.4] [9.8.41

It is clear that the Gibbs-Appell and Kane's equations are identical.

can be found in a way similar to finding the generalized forces. Indeed, we define a resultant inertia force For For rigid bodies, the generalized inertia forces

=

acting through the center of mass, and a resultant inertia torque M* = —dHG/dt, similar to the negative of the gyroscopic moment. The generalized inertia force has the form — lnaG,

=

•

—

•

=

•

+(O• o• HG +

[9.8.5]

Compare this equation with Eq. [9.7.191. For interconnected bodies, bodies, one uses the resultant acceleration and change in angular momentum for each body and sums, as we we did for particles, to obtain Eq. [9.7.211. Even though we performed near-identical manipulations to obtain the GibbsAppell and Kane's equations, the two are inspired from very different viewpoints.

I

6lhis definition is different than Kane's definition of generalized inertki forte.

525

526

CHAPTER 9. DYNAMICS OF Ricm Bonws:

ADVANCED CONCEPTS

The motivation behind the development of the Gibbs-Appell equations was to de-

velop a fundamental function, in a sense similar to the extended Hamilton's principle but one that can handle nonholonomic constraints. These equations can be derived without using partial velocities. On the other hand, Kane's equations are based on Kane's initial work with D'Alembert's principle, and on Lagrange's fonn of D' Alembert's principle. D'Alembert's As discussed before, Kane's equations can be derived from D'Alembert's principle is as follows: We invoke D'Alembert's principle for rigid bodies, derived in Section 8.11. For a system of N bodies, the principle is written in terms of the generalized velocities as N

c9VG. •

+HG•

.

k

1=

=

l,2,...,n

19.8.61

I

in which the generalized forces are given by

Qk =

-'

+MGI•

[9.8.7] [9.8.71

.'1 ôqkJ

We introduce to these two equations the relationships between the generalized

velocities and the generalized speeds. Making use of column vector notation and the partial velocity matrix, we can express the partial derivatives of VG1 w1 with and w1 VG1 and respect to the generalized velocities as 19{v(;1 } } 9{u} 9{u} — s9{vG1 } — 19{v(;,

c9{4}

-

-

(9{u}

— —

—

-

d{u}

(9{vG1

[Wi —'

d{wi} [W]'

-

q

—

-

—

.1 .j

[

[W] 1

I

[9 8 8]

so that Eq. [9.8.6] can be expressed as +

= {Q}T

[9.8.9]

= {Q}T[WI {Q}T[Wi = {U}T

[9.8.10] [9.8.101

We right-multiply both sides of this equation by [WI, for

+

The right side of this equation is recognized as the generalized force vector as-

sociated with generalized speeds. The left side is recognized as the column vector representation of the generalized inertia forces. Hence, Kane's equations can be derived directly from D'Alembert's principle for rigid bodies, by a simple change of variables from the generalized velocities to the generalized speeds. Derivation of Kane's equations can also be accomplished without the use of any variational principle, such as D'Alembert's or Hamilton's or of variational calculus. Nor do they require the development of scalar functions such as energy and S. In this regard, they are different than the others that we have studied.

9.8

527

KANE'S EQUATIONS

Another advantage of Kane's equations is that the approach. especially the use

of partial velocities, is desirable for interconnected and large-order systems. This is because the procedure of obtaining the partial velocities and the generalized forces can be mechanized and made suitable for computer implementation. Rather than adding to the controversy on differences and similarities between the Gibbs-Appell and Kane's equations, we will refer to the equations of motion derived in this and the previous section as the fi€ndamental equations motion. We will differentiate the forms by referring to the Kane ofthe qf the fundamental fundamental equations of motion and the Gibbs -A -A ppell ppell .tbrrn jbrm of the fundamental equations of nZOIIOn. Another interesting interpretation of the fundamental equations is as follows. Consider that we have one body, whose translational and rotational equations of motion are

ma(;—F

0

HG—MG = 0

19.8.11]

The fundamental equations are obtained by taking the dot product of the first of Eq. [9.8.11] with and of' the second Eq. [9.8.111 [9.8.11] with and summing the two expressions. This results in the equation of motion for the kth generalized speed. In essence. the kth fundamental equation is simply the sum of the components of'

the force and moment balance along the directions of the partial velocities. This interpretation should be compared with the interpretation of Lagrange's equations in terms of the Euler angles.

Obtain the equations of motion of aa disk disk rolling rolling without without slipping slipping using the fundamental equations.

Solution We discussed this problem in Examplcs 9.3 and 9.5. We begin with a set of independent generalized speeds eralized speeds Uk = 1, 2. (k = 1, 2. 3) and and use use the the FF frame. frame. First, First, we we calculate calculate the the acceleration acceleration of the the center of mass. The velocity of the center of' mass is given by

Ia]

= —Ru3f1 +

so that the acceleration becomes

au = =

d

+

X V(; V

+ Ru1 u2)f1

(—

f —

+

\ -

+ (Ru1 + Ru2u3 )f3

[bi Ibi

At At this point, we can either invoke the Gihhs-Appell or the Kane's forms of the fundamental equations. Let us use the Gihhs-Appell form and write the Gibbs function

S S

=

+

[ci

+

We will take partial derivatives of each term with respect to the quasi-velocities. We note that

ôaGIRf C,

(1142

63Rf

Ed]

Example 9.1 1

528

B0DWs: ADVANCED ADVANCED CONCEPTS CHAPTER 9S 9• DYNAMIICS DYNAMIICs OF RIGID BoDws:

For the first term of S (9

rn{aQ}T{aG}

=

k = 1,2,3

dUk

[e]

and carrying out the algebra we obtain UI + U2U3 9{ü}

=mR

T

0 £43

If I

— UIU2

= 111. The derivatives of the second and third terms in S yield Euler's equations expressed in the F frame, Eqs. [8.5.31 Theseequations equations are are the the same same as as the the modified modified Euler's equations. as given in [8.5.3 lJ.1.These the left side of Table 9.1. Adding these equations to Eq. [fi, and noting that the generalized forces are given by Eq. [11 of Example 9.8. we arrive at the equations of motion as To differentiate the second and third terms of S, we note that

—

(ii ±

'U22 I

tanO 11U2+

+ (13 + mR2)u2u3 = —rngRcosO !1u1u2 11u1u2

= 0

tanO

(13 + ,nR2)ü1

—

[ii

rnR2u1u2 rnR2u1 U2 = 0

Egi

Use of the Gibbs-AppelI form of the fundamental equations led to a physical interpretation. The equations of motion are basically the modified Euler's equations with an additional moment term. This additional expression, given in Eq. [f], is in essence the contribution of the moment generated about the center of mass by the forces acting on the contact point.

We have now solved the problem of the rolling disk four times. We first studied

this problem in Chapter 8, using Lagrange's equations with constraints. Next we used the modified Euler's equations, where we discussed both a force and moment balance. Then we used the Boltzmann-Hamel equations and summed moments about the contact point. Finally, in Example 9.11, we used the fundamental equations. The reader is urged to compare all these methods of solution.

Example

9.12

Obtain the equations of motion for the spinning top on a cart using Kane's form of the funda-

mental equations.

Solution We saw this problem in Examples 8.12 and 9.4. The forces and moments that do work on the combined system are

F1 =

—mga3

applied at G

F2

= Fa2

atC

applied at C [01

9.8

KANE'S EQUATIONS

Note that the Note that the top exerts exerts an equal and opposite moment moment on the cart. The other forces that the cart and top exert on on each other do not enter the formulation, as they are internal to the system. s y stern.

We need to calculate the following partial velocities and partial angular velocities:

to evaluate U and to find the component ofUk (4 due due to to F1 F, component of to evaluate to evaluate U and to find the component of Uk due to M

to evaluate

and to find the component component of of(iA (4. due to F2

[bJ Ibi

All these partial velocities were calculated in Example 9.6. We summarize the results as Generalized speeds:

=

Angular velocity:

=

Partial angular velocities:

w1f1

+ w2f2 +

Wi = f1 f1

Partial velocities for C:

(

=

=

Velocity of G:

=

= W2

=

a2

=

—

=

Generalized forces:

Ic]

= u,f1 u1f1 + u212 + u3f3 =

f2

[dl [di

1.]

0

f3

= 0

+ cos4cosOf2

=

Ifi

—

Lf2u1 + Lf1 u2 Lf2u,

+ (sin df1 df, ++ cos cos / cos Of2 Partial velocities for G:

114== Y 144

=

= Lf, Lf1

= a7 a = (sin

+ cos

U1 = mgL Sill U, Sin 0

—

cos d

[gi

)U4

= 0

U2 = 0

[hi Ihi

sin

cos Of7 — cos

U3 = M

U4 = F

Ii]

To derive the equations of motion, we need to obtain the accelerations of G and C, and the rate of change of angular momentum. The acceleration of C is simply =

II]

We find the acceleration of G using the relation

ac = in which FGKFGK- =

a=

/1.

+ w X (w X rG/c)

Iki

and

+ (01 X

(u1 — = (UI

aX

/

= ((i)1!'1 ++ w2f, w2f, ++(J)3f1) (J)3f1)++(W1f'1 (W1f'1++w2f, w,f, +

ii' /. 111111 + + U2U3 Ifi (U2 + Ifi ++ (u2 tanO tanO J

\ /

U;

.

— U1 U3 )f2 )f U1U3

- f3 )X

tanO

/

/ \

—

Jf1

tanOj

[II II]

+

As mentioned before, this differentiation is is necessary because the angular velocity is being expressed in the F frame and not a body frame. Carrying out the necessary steps. the acceleration of point G becomes =

sin

+ Là2 + Là +

+ (— a4 cos

sin 0

+ —

L

—

L

COS

)f3

0

—

Là1 ±

±

Imi

529

____

__

530

CHAPTIR 9• CHAPTIR 9• DYNAMICS DYNAMICSOF OFRIGID RIGIDBODIES: BODIES:ADV&NCED ADV&NCEDCONCEPTS CoNCErrS

As expected. all u3 = (1)3 terms drop out of the acceleration expression. Taking the dot products between the accelerations and the associated partial velocities yields

=0

=0

v(-•a( = v(-.•a(

= 00

L 2 u1 — Lu4 cos cos4)cosO — 4.cosO —

In]

.

-,

+ Lu4 sin b + = Lu2 Lu2+Lu4sln(b+

= Lf1 •

•

tanO

tan0 tan6

-

= 0 (sin •• aG = (sin

+ cos cos 4)cos cos Of2 — cos4)sin sin Of1) Of1) 'à( — cos

= 144 144 ++ Lw, Lu2 —

(/) +

Lu1 u2 sin cfr 4)

tanO

Làa1 cos 4)cos cos 00 ++ L uf cos

+

sin

i

L u,2 cos cos 4)

lol

sinO

We We

change of of the the angular angular momentum. next find the rate of change momentum.We Wederived derived a general expression for this earlier in this chapter. when studying the modified Euler's equations. Using Table 9.1, for a 3-1-3 transformation the rate of change of angular momentum of the top is

/

II(')2

+W2 HG = [l,w +W2 HG=

(13W3

—

tanOj I

Iii]

fz +

The contribution from angular momentum of the cart is zero, as the cart has zero angular velocity. Taking the dot product between HG and the partial angular velocities from Eqs. [e], we obtain

= =

fI.HG

11 112

11111 1i111 + U2

—

(1) 'HG = f2'HG = 11i12_U1(13U3 'I — UI (13 U3 (JL)3HG = (JL)3'HG =

f•HG f3•HG

—

tan6 11u2) U2 tail tanO

=0

13U1

Iqi

The generalized inertia forces U are calculated as =

Combining Eqs. [n}. equations of motion as

•

m2a( +

+

•

0 + in1

+ ni1 L2)à1 — — I T*

U-,

= U2

0 + in1

''

' 3,4 3,4

Er]

and [q] with the generalized forces in Eq. [ii. [i]. we obtain the ,

= UI

k=

S

—Lu4 c4)cO— in1 Lis4 Là4 Cc

4)c 0

—

+ + Lü4s4)+

141U2 hi 142 + (Ii + miL2)(u2 +

tO

)

I

.,

+ u(13u3

to ) +

(/t + L2u1 112

to

in L 114 s 4) — —

U)

—

11U2

tO

)=

ingL sO

+ 13U2U3 Iiu2u3 — — ingL sO = 0

+ Jul2 — + 11112 —

Ut (13113

I! U-' tO

I3UIU3 UI == 0

Is]

)=o It]

9.9

531

THE FIJNI)AMENTAL EQUATIONS AND

[uJ [uI U4 =

U4 7.

(m1

.

,n1Lu1Lhstf ,il1Lu1Lhstf

+ —-—-—-—--

9.9

+ ,iz1Lü, s4

+ lfli)U4 lfli)U4 —— fl?1LU1 tn1Là1

[vi lvi

-,

+nhiLuIcd)s6+

sO

THE FUNDAMENTAL EQUATIONS AND CONSTRAINTS

As we saw in the previous sections. the fundamental equations have the advantage that by a judicious choice of the generalized speeds one can account for nonholonomic constraints in the formulation and obtain equations of motion in terms of independent generalized speeds. In certain cases, however, it is desirable or necessary to leave the problem formulation in terms of constrained constrained coordinates. coordinates. Consider a system with n degrees of freedom described in terms of n generalized coordinates (k = 1, 2, .., ., n) and n generalized speeds Uk 14k (k = 1, 2, .. ., ii). We now apply in constraints to the system of the form .

.

+ {B} =

[9.9.1] [9.9.11

{O}

where [A] and {B} are defined as in Section 9.6. From Eq. [9.5.221. [9.5.221. the expression

for virtual work in terms of the quasi-coordinates is

=

6W =

19.9.21

As we did in Lagrangian mechanics, mechanics, we we take take the the variation variationof ofthe thecon constrai strairit lit in in Eq. Eq. [9.9.1] and left-multiply it by the Lagrange multipliers {A}T = {A1 A2 ... ••• Am}. ikm}. with the result with {A}T[AJ{6q?} {A}T[A]{6q?}

=

19.9.31

0

We then add the above expression to the virtual work and create an augmented virtual work function

= {U}T{6q1} + {A}7[A]{6q'} {A}7[A]{6q'} =

[9.9.4]

in which {U}

= {U} + [A}T{A}

[9.9.5]

a set of generalized forces associated with the constrained generalized coordinates. Individually, the elements of {U} have the form is

ill

Uk =

Uk +> AJkAI

k = 1,2,

..., n

19.9.61

j=1 This expression for the generalized forces is then used in the the Gibbs-A Gibhs-AppelI ppell equations.

with the result

__ 532

CHAPTER 9• DYNAMICS OF RICH) BODIES: ADVANCED CONCEPTS

as

=

(fIUk

Uk

[9.9.7] 19.9.71

+> A +> A •kAJ kAJ j=I

or m

= Uk +

[9.9.8] [9.9.81

AJkAJ

j=1

equations [9.9.8], [9.9.81, the the in Fflconstraint constraintequations equations[9.9.1], [9.9.1],and andthen then kinematic kinematic Then equations Then differential equations [9.5.3] are used to solve for tbr the 2ii + in variables . .. ., q,,, U1,

A1, k2,

. . . ,

•

•

•

'

RELATIONSHIPS 9.10RELATIONSHIPS 9.10

BETWEEN THE FUNDAMENTAL

EQUATIONS AND LAGRANGE'S EQUATIONS

The relationship between Lagrange's equations and the fundamental equations is given in Eq. [9.7.9], for when the system is unconstrained and the generalized speeds are selected as the generalized velocities. In this case the generalized forces Qk and Uk (k = 1, 2, .. ., n) coincide, If the applied forces are conservative, one can take advantage of the potential energy to calculate the generalized forces Uk

=

Qk

=

k =

—

1,2,...,n

19.10.11 [9.10.11

Consider now the case when the generalized speeds and generalized velocities are related by Eq. [9.5.3] as

= [W]{u}+{X}

19.10.21 [9.10.21

We make use of Eq. [9.5.191, which relates the generalized forces by We

[9.10.31

{U} = [W]1{Q}

and and

separate the generalized forces into those that can be derived from a potential and those that cannot, as

19.10.4]

+

{U} =

It follows that

=

=

'av _[W]T 'dv

\T

(I

[9.10.5] 19.10.51

or, in or, in scalar scalarform, fl

= UL.k UL.k =

k

=

l,2,...,n

We write write the Lagrange's equations in matrix form as We

ddT dt

dv d{q}

d{q}

—

tiC tic

19.10.61

9.10

BETWEEN THE IkTNI)AMENTAL EQUATIONS AN!) LAGRANGE'S EQUATIONS

in which

contains the contribution of all external forces not derivable from a

potential. Right-multiplying the above equation by [WI and using Eq. 19.10.3], we obtain

(doT (doT

-

or

c2V\ J[W1

(1{q} O{q}

= =

=

a{q} j

{U,l(S}T

[9.10.8]

which can be expressed as ci OT

(IT (IT

av\ av

+

[9.10.9] [9.10.91

J

= where U,1(.2 ... Ufl(.,,j is the generalized U,:(.2 force vector. If one does not not wish to make use of the potential energy tbrmulation, the equation becomes (Cc

I

-

q,v

/)

[9.10.10] 19.10.10]

Uk

k

In the presence of constraints, the relationship between the generalized forces is given by Eq. [9.6.22] as

{U} = = IWIT{Q} [WJT{Q}

19.io.iij 19.10.111

Let us right-multiply Eq. [9.10.71 by the LW] [Wj matrix. matrix. From

Eq. [[9.6.12] 9.6.12] we have

[WI = [W,,J + [Ws][A'j [WsJ[A'j

[9.10.12]

in which {W1,] and [WsI are the partitions of [WI. [W]. Hence, we obtain fir the equations

of motion

nfd aT nfd aT

dV av\(

OT

+

.s

In

Wck+'> +

= UnCk UflCk

WS,p.f j

i—I

k

=

19.10.131 A special case of Eq. [9.10.6] arises when the generalized speeds are selected as the same as the independent generalized velocities, k = 1, 2, ..., p. In this Uk = case, case. coincide with Qk, and the first p X p partition partition of the [W] matrix matrix becomes an identity matrix, or I'Vck

=

s,k= l,2,...,p s,k= l,2,...,p

&vk

19.10.14]

and, as a result, we write Eq. [9.10.8] [ 9.10.8] as

(d OT OT I—

.

OT (?V (9V —--.---+ +

nH 1

+

Ifl 17i

IA/

I

VP,

A

s—I

j=I

p +11•1 j k

)

[9.10.15] = tzck 19.10.15]

Equation 19.10.15] [9.10.15] in essence represents the removal Equation removal of the Lagrange multimultiplier from Lagrange's equations. This procedure, procedure, while relatively straightforward in derivation, often often involves involves several complicated steps for actual problems and is not commonly carried out. Equations [9.10.101 and [9.10.15] are known as the Passarello-j-juston equations.

533

534

CHAPTER DYNAMICS OF RIGID BODIES: ADVANCED CONCEPIS CHAPTER 9O 9• DYNAMICS

Note that we could have derived Eq. [9.10.15] in Section 4.10, when we discussed Lagrange's equations for constrained systems. Instead, we left it for here, so that it could be done for both Lagrange's as well as the fundamental equations. We next consider Lagrange's equations in terms of quasi-velocities. By substituting the quasi-velocities into the kinetic energy, we write the kinetic energy as

T=

[9.10.16] 19.10.16]

T({q}, {u}) {u})

where the overbar denotes that a change of variables has occurred. We also recall

the relationships {u}

= [Y]{4} + {Z}

{c;,} = =

[W]{u} + {X} {X}

19.10.1 7a,bJ

Using Eq. [9.10.1 7aJ, we can express the derivative of the Lagrangian with respect to the generalized velocities in terms of the generalized speeds as

aT

a{u}

19.10.181 [9.10.181

a{U}[Yl

which, upon differentiation with respect to time, becomes

dôf do7

ci(9T dt

dtô{u} dtô{u}

[Y]+

19.10.19] 19.10.191

[Y] [Y]

a{u} 9{u}

Noting thatthe thegeneralized generalized speeds Noting that speeds are are functions functions of of both both the the generalized generalized coordinates and generalized velocities, the derivative of the kinetic energy with respect to the generalized coordinates becomes

aT

c9T c?{U}

ô{q}

- a{q}

19.10.20] 19.10.201

+ ?{u}d{q}

In this equation, the first term is the explicit derivative, and the second term uses the chain rule to find the contribution from the generalized speeds. Introducing Eqs. [9.10.19] and [9.10.20] into Lagrange's equations, we obtain

daT — [Y]+ dta{u} +

(iii a{u}

7.

(1{q}

19.10.21]

We right-multiply this equation by [Wi and make use of the relationship for the generalized forces {U} = [W]T{Q}. with the result

— —

dta{u}

+

aT

[Wj = {U}T

a{u}

[9.10.22] 19.10.22]

in which Id c){U}

—

=

—

[9.10.23]

Equation [9.10.22] is Lagrange's equations for quasi-coordinates. Their primary difference from the traditional Lagrange's equations is the calculation of the coefficient matrix One can make use of the several different ways to express [9fl. One way is to use the preceding definition directly. Once is calculated, if it is not in terms of generalized speeds, it can be expressed in terms of them by

91 0

BETWEEN THE FUNDAMENTAL EQUATIONS

LAGRANGE'S EQUATIONS

simple substitutions. The same procedure can be tbllowed when evaluating the time

derivative of Another way to evaluate [2fl is as as follows. follows. can can be written entirely in terms of the generalized coordinates and generalized speeds. To demonstrate this, differentiate an element of say. as 7?

=> =

y

0

{}[W1{u}+ d{}{X}

=

=

i,j ==

1,2,...,iz

19.10.241 and and

note that this operation is performed for each component of [Y]. We expand as ô{u}

a{q}

=

a{u}

(9{u}

ôqj

19.10.251

where each of the column vectors can be expressed as

a{u}

+ {Z}) {Z})

=

= aIYJ [WJ{u}+ ..

=

ô{X}

(9[Y]

+

.

{q}+

d{Z}

k

-

= 1,2,...,n 19.10.261

In the presence of forces that can be derived from a potential, one can express the contribution of these forces using the potential energy V. In this case, Lagrange's equations for quasi-coordinates have the form + di d{u}

+ (-2{u}

—

(9{q}

[WI =

[9.10.27] 19.10.271

For constrained systems, once a set of independent generalized speeds are se-

lected, Eq. [9. 10.231 can still be used, noting that both [Yl [Y] and [W] are rectangular matrices of orders p X x ii and nn X X p, p, respectively. respectively. A A Lagrange Lagrange multiplier matrix enters the formulation in in the the same way that it does in the traditional Lagrange's equations. Let us compare Lagrange's equations for quasi-coordinates with the fundamental equations. Both sets of equations are fundamental, and they can handle nonholonomic systems. Other forms of the equations of motion, such as Euler's equations. the Boltzmann-Hamel equations, and the traditional form of Lagrange's equations, can be derived from them. The right sides of both equations are identical. The advantage of Eq. [9.10.221 is that it does not involve calculation of acceleration terms or of rate of change of angular momenta. Its disadvantage is the effort required in the calculation of The fundamental equations, especially in the Kane's form, lend themselves to efficient calculation of partial velocities, making it more convenient to derive the equations of motion in many cases. On the the other other hand, hand, Eq. Eq. [9.10.221 [9.10.221 retains the Lagrangian formulation and has more physical insight. It utilizes the kinetic and potential energies and eases the generation of motion integrals. One can consider Lagrange's equations for quasi-coordinates as a set of equations lions derived from the extended Hamilton's principle, which is an integral principie. By contrast, the fundamental equations are not based on an integral principle. ple.

535

536

RIGID BODIES: BODIES:ADVANCED CONCEPTS CHAPTER 9• DYNAMICS OF RIGID

Another advantage of the Lagrangian formulation will become evident in Chapter 11, where we study the dynamics of deformable bodies. Similar to the debate about the difference between the Gibbs-Appell and Kane's forms of the fundamental equations. there is debate about whether the fundamental equations or Lagrange's equations for quasi-coordinates are better and easier to use, or more fundamental themselves.

Example 9.1 3

I

Again consider the spinning top on cart problem. Oht4in the equations of motion using Lagrang&s equations for quasi-coordinates.

Solution Because the right sides of Lagrange's equations and the fundamental equations are the same.

we will only calculate the left sides of the equation of motion. The kinetic energy for this problem is calculated in Example 8.12 and it has the form (using a 3-1-3 transformation and F frame) T =

I

+ w;) + + ,niL)(w1 +

I

1

I

+

± + ,n2)Y ,n2)Y +

—Wj cd)cO) — Wj cd)cO)

(a] We

select the generalized coordinates and speeds as =

so

q2 q2 =

=

0

q4

II

u1 =

Y

= 1,2.3

Ib]

= Y

that the kinetic energy is written as ÷ in1Lu4(u2sqi — Ui Ui cql cq2) (c]

+

+

+

+

+

T =

c0+ we write the relationship between the generalized speeds and generalized velocities as Noting that the angular velocity vector has the form w = 0 sq2

U1

143

cq2

144

0

1

0 0 0

0 0

0

1

0 0

0

1

+

s Of2 +

'?'

[dl q4

and the inverse relationship as Il Ii

0

q2

I

'13

0

'14

The coefficient matrices in Eqs.

sc/2 Sc112 () —

/ 0

0

0

U

0

0

112

1

0

0o

1.1

U4 1

and [e] are recognized as I Yj and [Wi, with

{Z}

= {0}. The derivatives of the kinetic energy with respect to the generalized speeds is

d111

=

± m1 L2)u1 L2 )u1

— —

Lu4 ccqi ni1Lu4 nh qi Ccq2 q2 (9T

= 13113 (9114

=

,n1L2)u2 L2)u2 + in1Lu4 Lu4 sqi sqi + ni1

in1L(u2 sqi — = (in1 (in1 + ,n2)u4 + ?n1L(u2

u1 U1

cq2) cqi cq2)

[fj

__

9.

and

537

EQUATIONS

EQUATIONS AND BETWEEN THE THE FUNDAMENTAl EQUATIONS RELATIONSHIPS BETWEEN

the explicit derivatives with respect to the generalized coordinates are = m1 m1 Lu4(u2 c

+ uI tiI sqi

(IT c c/2)

=

-

(TIC/i

=0

dq3

We next calculate the

(IT cIq4 1q4

m1

Lu4u1 c q,

s

=0

Igi IgI

matrix. The The tinie tinie derivative derivativeof of[YJ (YJmultiplied multipliedby byIW1 Wi is

o

0

o

0

0

0

ulcq2

0

0

0

0

0

0

1

[Y}[W1 [Yj[W1 — Ui S

Sq2

o

—

0

0

0

0

0

0

i1

0

0

—u

01

0

0

1

tq2

0

o

0 t q2

0

0

0

0

[hi 1k] 0

0

00

To evaluate the second term in Eq. [9.10.23] we note that FYi contains contributions only from the second generalized coordinate, so that 9{u}

= {0}

9{u}

1, 3,4 k = 1,3,4

III

8q2

is similar to the time derivative

Also, evaluation of the derivative of FYI with respect to but without the U1 terms, so that we obtain

d{u}

dq2

o

0

0

o

Cq2 cq2

0

0

0

0

0

0

—s

0

0 0

1

0

0

0

0

0

0

0

1

0

0

—1

01

0

0

—

tq2

0000

0

0

0

(Y

0

0

u2 U2

0

0

U3

1q2

0

00U4 OOU4

0

Iii Ii] Thus, combining all generalized coordinates, we obtain 0

fl

0 0

0 142

q2 U2

0

We can now write the

0

0

0

0

0

0

Sq2

0

0 U2

0

0

0

0

0

0

[ki

q2

0

0

0

0

0

0

1

—

tq2 0

1

0

0

1

0

0

0

0

0

0

0

matrix as 0

0

U2

[Y][W1 [2T1 = [Y][W] [2T1 =

— —

.,

[WI

tq2

tq2

U2

—u1

0

0

00 0

0

[I] 0

0

00

538

CHAPTER 9• DYNAMICS OF RiGiD BODIES: ADVANCED CONCEPTS

Next, we evaluate the individual expressions in Lagrange's equations. The time derivaare tives of the

di dU1

= (Ii + rn1L2)ü1

doT di

dOT di du4

0U2 —

cq2 +

—

in1 LU2U4

tq2

dOT

—

di ()143 £1!

—

sq2

= 13t43 13143

u1cqlcq2 + ulcqlcq2

= (in1 + 1112)144 + ,n1L

sq2

c qi

ni1 ti?1

(/i in1 Lu4 sq1 sq1 + iP?1L)142 ++in1Lii4 (ii ++ iii1L)ü2 —

+ m1Lu1u4

u2eqj —

+

+

sq2

Ui U2SCIJ

tq2

[ml

)

The transpose of second term on the left side of Lagrange's equations becomes —

tq2 1q2 T

UI

0

0

—Ui

tq2

(Ii ++ni1L2)u1 m1L2)u1 — — in1Lu4cq1 in1Lu4cqi cq2 in1Lu4sq1 sq1 + in1L2)u2 iniL2)u2 + m1Lu4 13U3

o

0

0

_0

0

0

0

(ni1 + in2)U4 + in1 m1 L(u2 sq1 —

ni1Lu2u4 in1Lu•2u4

—(ii —(ii +

L2)'tq2

+ (ii (Ii +

+ 13147113 13117113

tq2

t '12

+

C qi Cc

U1 U1 C

114 sq1 sqi In1 Lu1 114

tq2

— /3111113 13141143

[RI

0 0

and the transpose of third term is 0 )T

—[W]T

1

o

——

0

in1Lu4(u2cq1++ u1sq1cq2) in,Lu4(u2cq, u1sq1cq2) in1Lu4u1 Lu4uicq1sq2 sq2 ml

0

1

0

0 0

0

0

1

(p!.

Sq2

0

0 1q2 tq2

1fli Lu4u1cq1sq2 Lu4u1cqisq2 tn1Lu4(u2cq1 -1- uisq,cq7) uisq,cq2) sq2

[01

0 L

0

The potential energy and virtual work are

V = rn1gLcosO rn,gLcosO = in1gLcosq2

= Mt5q +

9.1 1

so

IMPULSE-MOMENTUM RELATIONSHIPS FOR GENERALIZED SPEEDS

that the generalized force vector becomes

0M

[m1gLsinq2

{U}

p1T

Iq] Iqi

n],[o]. [o].and and[41 [qi leads to the equations of motion, for the fourth time. Combining Eqs. [rn], In], The reader should compare these methods of solution and judge, at least for this problem, the merits and disadvantages of each method. Note that, as encountered in many cases when using

the traditional Lagrange's equations, certain terms in Eqs. [m], [n], and [o] canceled each other. This is one of the disadvantages of using Lagrange's equations: in many cases they lead to wasted manipulations. However, considering the proliferation of symbolic manipulation software, this disadvantage loses its importance.

9.1

1

IMPULSE-MOMENTUM RELATIONSHIPS FOR GENERALIZED SPEEDS

The impulse and momentum relationships lbr systems described by generalized speeds are very similar to those obtained using generalized velocities. In Section 2,

..., n) associated with the

1,2,...,ii 1,2,...,n

[9.11.11

5.8, we defined the generalized momentum 'irk (k = generalized coordinate as

= In

k =

.

1,

terms of independent generalized speeds. generalized momenta are defined sim-

ilarly as

[9.11.2]

= (9Uk

For system of of N N particles particles the generalized momenta can be shown to be For aa system N

[9.11.3]

= >m1vj 1=1

that = relate the two generalized momenta by noting from Eq. = [WI. Hence, writing Eq. [9.11.1] in column vector [W]{u} + {X}, so that we can invoke the chain rule of differentiation and have = format as We

aT =

=

a{ç'} -

=

a{u}

[9.11.4] [9.11.41

so that

[WTT{lr} {'Tr'} = [WTT{ir} {ir'}

19.11.5]

The generalized impulse associated with a generalized coordinate was defined

in Section 5.10 as the integral of the generalized force over time, or f

Qk

=

Qkdt J

k = 1,2,...,n

[9.11.61

539

540

9•

CHAPTER 9• DYNAMICS CHAPTER DYNAMICS OF OF RIGID BODIES: ADVANCED CONCEPTS

For generalized forces associated with generalized speeds we define the generalized impulse as -

[9.11.7]

Ukdt Jr1

From Eq. [9.5.191 we relate the generalized impulses associated with generalized coordinates to those associated with generalized speeds by

19.11.8]

{U} =

For generalized coordinates, we saw in Sec. 5.10 that for a true impulsive For generalized force—that is. is, for a very large force applied over an infinitesimal time instant is equal the difference in the generalized momenta between any two time instances is

to the generalized impulse. or

r"

lirn urn

[9.11.9] 19.11.9]

Qkdt =

I

To see the corresponding relationship for generalized speeds. we write this equation in column vector form as if1

lirn lilT)

[9.11.10]

{Q}dt =

I

Jt1 Jt,

[9. 11 .8], we Left-multiplying both sides by I WIT and considering Eqs. [9.11.51 and [9.11.8], obtain ft1

urn

[9.11.111 19.11.111

{U}dt =

I

—.oJtI

indicating that when acted upon an impulsive force, the generalized momenta associated with generalized speeds behave in the same way as the generalized momenta associated with generalized coordinates. Hence, when nonholonornic variables such as angular velocities are involved, it is more convenient to use Eq. [9.11.11] over Eq. [9.11.10].

Example

9.14

Consider the spinning top on a cart once more. It is given that at a certain instant the top is spinning while the cart is moving. An impulsive force is applied to the cart in the a: direction. Find the change in the velocity of the cart and in the angular velocities of the top immediately after the impulse.

Solution It is more convenient to use generalized speeds in this problem. From Example 9.13, the

kinetic energy is

T=

+

(U2 Sin 4, — + in Lu4 (U2 Sin 4) —

+

+ COS 4) 4, C(.)S 6)

+

+

[a] Ia]

9.1 1 so so that that the generalized

IMPULSE-MOMENTUM RELATIONSHIPS IMPULSE-MOMENTUM RELATIONSHIPS FOR GENERALIZED SPEEDS

momenta associated with the generalized speeds have the form

(Ii +

= du1 = aT = aT

=

L2)u1

+

—

m1 m1 Lu4 Lu4

cos 4)cos 6

+

=

=

+ ifl2)114 + in1 + 4) — in1 L(u2 L(u2 sin 4) —

(1114

u1

[bI

cos 4)cos 0)

From Eq. [mj in Example 9.6, the generalized forces have the form U1

=

m 1gL Sin sin 0

= 0

U2

(/3 =

M

U4 = F

Ic] [C]

Of the applied forces, only FF is impulsive, so that only it contributes to the generalized forces during the impulse, with the result

01=0

02=0 02=0

Uo 0=o U4=E U4=E

Ed]

equate Eqs. [b] and Eqs. [b] and [d] to solve for the generalized speeds after the impulse. This gives four equations for the four unknowns (k = 1, 2, 3.4). We note that u3 u3 only only appears in one of the equations, so that it can be solved for independently of the others, with the result = 0. This indicates that the spin of the top is not affected by the impulse, and w3 is unchanged. To solve for the remaining unknowns, we denote the changes in the generalized speeds by and write the balance equations as We

—

I, + ,n1L2

0

—in1LcosgbcosO

0

1, + m1L2

in1Lsin4)

in1

L cos 4) COS 0

in1 L sin 4)

0

L1u1

0

1.] [.1

+ m2 m2

From the first and second of these equations we have

m,Lcos4)cosO miLcos4)cosO

m1Lsin4)

+ in1 L2

=

III Ifl

+ ,n1

which, when introduced into the third equation yield

' '

—

) mTL- cos 2' cos- 0 inTL+ in L2 L2

•'' 2' L- sin 4)

in ——

11 + in1 L2

+ in1 in1 + + im rm

A

=F

—

Igi

The change in the horizontal speed of the cart is dependent on the angles 4) and 6, 0, so the

position of the top influences the velocities and angular velocities after the impulse.

541

542

BODiES: ADVANCED ADVANCED CONCEPTS CONCEPTS CHAPTER 9• CHAPTER 9• Th'NArsIICS Th'NAMICS OF' RICID BOIMES:

REFERENCES Ginsbcrg. J. H. Athancc'd Athanced Engineering Engineering Dynamics. New York: Harper & Row. 1988. 2nd ed. Reading. MA: Addison-Wesley. 1981. Goldstein, H. Cla.ssiraf Classical Mechanics. 2nd Greenwood. 1). T. Classical Dynamics. Englewood Cliffs. NJ: Prenticc-HalI, 1977. Kane. T. R., and D. A. Levinson. Dynamics: Theory and Applications. New York: McGraw-Hill. 1983. Meirovitch. L. Methods o/Analvt:cal Dynamics. New York: McGraw-Hill. 1970. Pars, L. A. A Treatise of Analytical I)vnanzics. I)vnamics. New York: Wiley, 1965.

HOMEWORK EXERCISES SECTION 9.2 1.

2.

Using the modified Euler's equations. derive the equations of motion of a disk rolling without slipping on a surface that is like a wedge, as shown in Fig. 9.6. Using the modified Euler's equations. derive the equations of motion for the disk in Fig. 8.26. The xyz axes also constitute principal axes for the fork, which has a mass ,n/2 and can be modeled as a slender rod.

SECTION 9.3 3.

4.

5.

Solve Problem I using the Boltzmann-Harnel equations. Obtain the equations of motion of the dual spin spacecraft shown in Fig. 9.7. The centers of mass of the rotor and the main body are not at the same point, but along the b3 axis, which is a principal axis. Use the Boltzmann-Hamel equations and consider oniy the rotational motion of the main body. The shaft connecting the main body to the rotor is light. Find the equation of motion of' the gyropendulurn in Example 8.5 using the Boltzmann-Hamel equations. Assume that both and i./i çb are constant.

U?

ti

b1

Figure 9.6

FIgure 9.7

ExERcisEs

vertical

z

b2

Y

3L

Figure 9.9

FIgure 9.8 SECTIONS

9.5

AND

9.6

Consider the satellite in Fig. 9.8 with the solar panels. The panels are operated by motors that exert moments N1 and N2 about the b1 axis, and moments M1, speeds, and write M2, and M3 act on the main body. Select a set of generalized speeds. the partial velocities. Consider only rotational motion of the satellite. inertia forces forces for the pre7. Generate the generalized forces and the generalized inertia and 13 about ceding problem. The main body has mass moments of inertia 11, the b1 b2b3 axes, which are principal axes, and the solar panels can be treated as plates. constraint is is that the tip 8. Consider the vehicle in Fig. 4.8 and consider that the constraint of the vehicle moves in the x axis only. Select a set of dependent dependent generalized speeds, invoke the constraint, and arrive at a set of independent generalized speeds. 9. Consider the tricycle in Figs. 7.39—7.40. Select as independent generalized partial velocities associspeeds the speed of point A and 4), and express the partial ated with the centers of masses of all components (main body. wheels, and handlebars) as well as all the partial angular velocities. 10. For the preceding problem, can you think of a better choice of generalized speeds? Compare with the results of selecting the orientation and speed of the front wheel as the variables. 11. Reconsider problem 10. This time the tricycle is on an incline as shown in Fig. 9.9, 9.9. Calculate the generalized forces. Assume that the handlebar is masswheels iii/l0 each, and less, the main body of the tricycle has mass in., the rear wheels the front wheel ,n/5. 12. Consider the rod to which two wheels are attached in Fig. 7.55. The rod is of mass m and the wheels are each of mass 2m. Consider the following sets of generalized speeds: (a) the angular velocities 01 and 02 of the wheels, (b) the speed of the center of the rod, and the turning rate of the rod. Consider that the system is on an incline, like the one in Fig. 9.9. Find the generalized forces for both sets of quasi-velocities. 13. Consider Example 9.7 and express the translation of the disk and the constraint equations by beginning with the contact point of the disk.

6.

543

544

CHAPTER

9•

DYNAMICS OF RIG11) BODIES: ADVANCE[ CONCEPTS

(03

WI (1),

I)

Figure 9.10 SECTIONS 9.7 AND

9.8

14. Generate the Gibbs-Appell function for Problem 4. 15. Generate the Gibbs-Appell function for Problem 6. 6. 16. Solve Problem 2 using the Kane's form of the fundamental equations. 17. Solve Problem 5 using the Kane's form of the fundamental equations. 18. Derive Hamilton's equations from the fundamental equations. 19. Solve the dual spin spacecraft problem in Fig. 9.7 using the fundamental equations. Consider rotational motion only. 20. Figure 9. 10 shows a spacecraft that has three rotors of identical size and shape. used primarily for attitude maneuvers. Each rotor is driven by its own motor and is at a distance L from the center of mass. Derive the equations of motion using the fundamental equations. 21. Consider the spacecrafi in Fig. 8.35. Obtain the equations of motion using the fundamental equations. 22. Consider the rod with the two wheels attached in Fig. 7.55. Derive the equations of motion, motion. using the two sets of quasi-velocities in Problem 13. SECTION 9.10

23. Consider the rotational motion of an axisymmetric rigid body. Show that the modified Euler's equations can be derived from Lagrange's equations for quasicoordinates.

24. Solve for the equations of motion of the the vehicle in Example 9.10 using Lagrange's equations for quasi-coordinates. 25. Solve Problem 2 using Lagrange's equations for quasi-coordinates. 26. Solve Problem 4 using Lagrange's equations for quasi-coordinates.

HOMEWORK EXERCISFS

4-

Ic A

F x

Figure 9.11

SECTION

9.11

Consider the tricycle in Figs. 7.39—40. The tricycle is at rest on a flat surface, with 4 = 15°. An impulsive force P is applied at point A along the V axis. Find the the resulting angular velocities of the wheels, and the velocity of G. Treat the wheels as disks and the main body of the tricycle as a plate. and ignore the weight of the front handle and connecting rods between the main body and wheels. 28. Consider the disk in Fig. 9.4, rolling without slipping with a constant nutation direction at direction angle of' angle of 450• An impulsive force is applied to the disk in the point A. Calculate the resulting precession, nutation, and spin rates immediately after the impulse, assuming that the disk is still rolling without slipping after the impulsive force is applied1 29. The ellipsoid of revolution (about thex axis) in Fig. 9.11 is at rest on a horizontal surface. It is struck by an impulsive force in the z direction and applied through center of' of mass and a/2, y = b/2. Find the velocity of the center the coordinates x the angular velocity of' the ellipsoid, assuming that the ellipsoid rolls without slipping on the horizontal surface.

27.

545

chapter

QUALITATIVE ANALYSIS OF RIGID BODY MOTION

10.1

INTRODUCTION

In this chapter, we analyze the motion of rigid bodies bodies from a qualitative perspective. We study the nature of the motion and make use of motion integrals. The main interest is in gyroscopic motion and on the effects of the gyroscopic moment. Gyroscopic motion is commonly described as motion in which the angular momentum and angular velocity vectors have different different directions directions and and one of the components of the angular velocity is much larger than the other two. This type of motion is commonly encountered in bodies that spin and the direction of the spin axis is not stationary. The gyroscopic moment becomes large and this enables tops, disks, and gyroscopes to continue their motion and prevents them from falling down. In essence, gyroscopic motion is a battle between gravitational moment and gyroscopic moment. A specific spin rate is required to overcome the effects of gravity.

We begin the analysis in this chapter with the simplest case of moment-free motion. We study axisymmetric as well as nonsymmetric bodies. We focus on bodies under the action of forces and the rolling disk and spinning top are studied, as well as gyroscopes. While the qualitative analysis of gyroscopic motion and of spinning bodies is a very useful and important tool, it becomes even more interesting to supplement the qualitative analysis with quantitative results. Numerical integration of the equations of motion was discussed in Chapter 1. The reader is urged to take advantage of this. By numerically integrating the equations of motion one can quantitatively observe many of the interesting results that are presented in this chapter.

This chapter does not have many examples. In essence, each section can he considered as an example analyzing a specific situation.

547

548

CHAPTER 10

10.2

MolloN

AIJTATIVE ANALYSIS ALITATIVE ANALYSIS01 øi RIGW RIGw

TORQUE-FREE MOTION OF INERTIALLY SYMMETRIC BODIES

We begin with the motion of a rigid body that is not subjected to any external momoments. Since the translational equations are straightforward. we will only analyze the rotational equations. Selecting the principal axes as the body coordinates and in the absence of external moments. Euler's equations. Eqs. [8.5.281, simplify to —

—

I3)w2w3

12w2

—

(13 —

11)w1w3

13(03

—

—

M1

=

= 0

=0

J,)w)w2 = M3 = 0

[10.2.1]

next consider the special case where there is inertial symmetry in the structure and that two of the principal moments of inertia are the same, say = 12. '2. This assumption not only simplifies the equations of motion but increases the stabilit' margin as well. (Example 8.6 analyzed this increased stability.) From a mathematical standpoint, an arbitrarily shaped body with two equal principal moments of inertia will behave in the same way as an axisymmetric body. However, one does not frequently encounter this situation. Many times, air flow or resistance by another fluid around a body has a substantial effect on the nature of the motion; the shape of the body then becomes a significant factor and there is a difference in the response of axisymmetric and inertially symmetric bodies. Several bodies that undergo rotational motion are specifically designed as axisymmetric. Examples of this include rockets and satellites, machinery parts, frisbees, and balls such as American footballs and rugby balls. Soccer and ping pong balls have all three principal moments of inertia as the same. in Eqs. [10.2.1] yields Setting 11 = We

—

—

13)w2w3

= 0

—

(13

—

JI)w,w3 = 0

= 0 [1 0.2.2a,b,cJ 13(1)3 13W3

Equation I 10.2.2cJ indicates that w3 w3 is is constant, constant,SO SOitit isis an integral of the motion.

Introducing the rotation constant

[10.2.31 Eqs. Eqs. II !0.2.2a] !0.2.2aJ and and [10.2.2b] [10.2.2b] can he expressed as

+

0

—

=0

I10.2.4a,bI

10.2.4] 10.2.4] describe describe a gyroscopic gyroscopic system. with with

being the parameter describing the gyroscopic properties. They can be written in column vector format as Equations

I

(01

=—

0 0

[10.2.5]

10.2

TORQUE-FREE MOTION OF INER'riALIx SYMMETRIC

Boows

where we notice that the coefficient matrix is skew symmetric. A skew-symmetric coefficient matrix is usually the sign of gyroscopic behavior.' The rotation constant describes the rate at which the gyroscopic motion unfolds. Multiplying Eq. [l0.2.4a] by and Eq. [10.2.4bJ by w2 and adding the two equations leads to

Id +W2W, = ——(w,2

= 0

2dt

[10.2.61

from which it is realized that wf + + w 12 = the magnitude of the projection of the angular velocity vector on the b1b2 plane, as shown in Fig. 10.1. Hence, we have a second integral of the motion. The two first integrals can be combined to yield a third integral of the motion, thus +

+

= constant = wi2 =

[10.2.7]

indicating that the magnitude of the angular velocity vector is also constant. Because the body is torque free, both the angular momentum about the center of mass and the rotational kinetic energy are conserved, pointing to two other integrals of the motion as HG =

!iw 1b + I, w2b2 +

in which b, (i =

+

= 1iW1 +

2Trot =

= constant

13w3b3

= constant

[10.2.8]

2, 3) denote the unit vectors along the principal axes. We use the integrals of the motion to explain the motion of the body. Defining the projection of the angular momentum vector onto the h1b2 plane by H,2, we observe that the 1,

magnitude of H12 is 1-112

=

1H121

=

+

=

11(012

[10.2.9] 110.2.91

so that the magnitude of H,2 is also constant. The above equations imply that the projections of HG and w onto the h1h2 plane lie along the same line and hence,

(01

/I I

I

—

I,I I, H— b1,

Figure 10.1 1The reoder is urged to compare Eq. [10.2.5] with the linearized equations For the Foucoult's pendulum, Example 2.16, Eq. [e].

549

550

CHAPTER 1O• QUALITATIVE ANALYSIS OF RIGID BODY MOTION

b

H(;

h1, u-i

Figure 10.2 as described in Fig. 10.1, the vectors vectors b3. as described b3. HG, and w must lie on the same plane. such that that H12 Defining the axis b12 b12 such H12 = H12b12, where b12 is the unit vector in the the defined by the h3 and plane is defined and b12 b12 axes. axes. Fig. Fig. 10.2 depicts the b12 direction, direction, this plane b12 plane.

The motion can be interpreted as the rotation of this plane. The rotation is about the angular momentum vector HG, as both the magnitude and direction of' this vector are constant. constant. The rotation of the angular velocity vector can also be described as the movement of two imaginary cones, called the and space cones, on top of each other. The body cone is fixed to the body. It is generated by the rotation of the angular velocity vector about the b3 axis (the spin and symmetry axis). The space cone is fixed in the inertial space and is generated by the rotation of the angular velocity vector about the angular momentum vector. In Chapter 7. we saw that the body and space cones can take arbitrary shapes. Because of the inertial symmetry, the cones generated here are right circular cones. Figures 10.3 and 10.4 show the body and space cones, as well as the orientation of H(; and w along the b3 and h12 directions. depending on the angles a and /3. We select the inertial frame such that the angular momentum vector (which is constant in both magnitude and direction) coincides with the a3 direction. We thus have

HG =

Space cone

110.2.101

"3

a

H(;

/

Body cone

Body cone

Figure 10.3

2

Flat body

Figure 10.4

Slender body

1 0.2

TORQUE-FREE MOTiON MOTiON OF OF INERTIALLY INERTIALLY SYMMETRIC BODiES

where 11G is the magnitude of the angular momentum vector. From Fig. 10.2 the angles a and f3 are constants, and they can he expressed as tan a

=

H12

H3

=

wp - = constant tan /3 = W3

= constant 13(1)3

[10.2.1 1]

in which H12 and H3 are components of the angular momentum along the h12 and in b3 axes, respectively. The relationship between a and /3 depends on the magnitude of the mass moments of inertia and 13. We have two possible cases: 1.

If <13, then a
2.

If

> /3, then a > /3. This case corresponds to the motion of a slender body,

such as a rod or a football.

When viewed from the body-fixed frame, w generates the body cone, which has an apex angle of /3 around h3. When viewed from the inertial frame, generates the space cone with an apex angle. of /3 — a flat body with /3, the body and space cone rolls inside the body cone, and for a slender body, cones lie outside of each other, as shown in Figs. 10.3—10.4. The space cone is easier to visualize, because it shows the direction of the angular velocity vector from an inertial frame. The visualization is also simpler for slender bodies.

Consider, for example. a football being thrown. Under ideal circumstances, the ball would just have a spin along its longitudinal axis, implying an angular velocity vector for the ball that has constant direction in addition to constant magnitude.2 However, if there is a wobble associated with the throw, the angular velocity now has two components: the wobble and the spin. Consequently, the angular velocity vector no longer has a constant direction, but a direction that rotates itself.

the component of the We next address the issue of the rate with which cv angular velocity vector along the b1 b2 plane, rotates. This is, of course, the same as the rate with which the plane defined by b3, H12, and w12 rotates, or how fast a body cone is swept. Considering Eqs. [10.2.41, we intuitively expect this rate to be the rotation constant To verify this, consider Fig. 10.1 and the = W3(13 = w3(13 — — angle y, which is the angle cv 12 makes with the h1 axis. The rotation rate is then j'. Simple geometry indicates that sin->' sin-y

cosy

=

(07

110.2.121

Differentiation of both sides of the above equation with respect to time yields —. sin y(—j, sin->') sin y) cosy(j'cosy) cos cos y) —.

cos2y

cv1w2 cv 1w2 — — w2w1

[10.2.131

2The motion of the football is actually a much more complex phenomenon, due to the aerodynamics and gravily moment.

551

552

CHAPThR lOS QUALITATIVE ANALYSIS OF RIGID BODY MOTION

Introducing Eqs. [10.2.41 into Eq. [10.2.1.3]. we obtain a—'

=

-)

.

-

+ wT)

[10.2.141

I

COST

We observe from Fig. 10.1 that cos2 y

=

which leads to the expected

+

result that j' = The total motion of the body, as stated earlier, is some kind of wobbling, unless

motion is initiated with a perfect spin about one of the principal axes. The amount of wobble, as observed from the value for Ii. depends on how the principal moments of inertia are different from each other. However, the wobble remains constant. the apex angles of the body and space cones are the same. An example of wobbling motion is the rotation of the earth. The earth is not a perfect sphere—it is an oblate spheroid. If we attach a set of body axes to the earth and select the /23 axis as the polar axis, we have (from measurements that have been taken) Il

[10.2.151

= 0.0033

rev/day, we get fl = should sweep sweepone one body body cone cone in in 1/0.0033 = rev/day. This implies that the earth should earth as as angular velocity velocity of the earth taking the the angular and, taking and,

I

300

days.

Astronomers have observed that the polar axis of the Earth indeed has some variation in latitude. The period of the phenomenon called called variation wobbling motion, a phenomenon wobble has been measured as 433 days and the radius swept by the poles in one body cone as about 4 meters. The difference from the predicted period of 300 days is attributed to two reasons. The earth is not rigid and not entirely torque free. Because it is an ohiate spheroid and not a perfect sphere, the gravitational attraction of the sun and moon creates a gravity gradient torque. Also, the lack component of of the the rotation rotation tc' to of rigidity causes some energy transfer from one component another.

before, we consider a Let us introduce the Euler angles to the formulation. As before. (nutation). and t/i (spin). The 3-1-3 sequence with the rotations of çb (precession), 6 (nutation), translbrmation matrix (R] is such that 11

a1

b2

= [R] a2

[10.2.16]

£11

as

given in Eq. [7.5.3J. From this we have a3

sill 00 cos i/ib2 + cos 6b3 = sin 0 sin i/ib1 + sin

110.2.17]

body-fixed coordinates coordinates as We express the angular momentum in terms of the body-fixed H(; = HGaI = HG(Sjfl 0 sin

=

11w1b1

+ 11W2b2 +

COS + sin sin00COS

+ cos 0b3)

[10.2.18]

___

10.2

TORQUE-FREE MOTION OF INERTIALLY SYMMETRIC BODIES

from which we obtain 11w, =

= HGsinOsint/J

13(1)3

= H(;cosO

110.2.191 Notice from the last of these equations that the nutation angle 0 is is constant and

hence the nutation rate is zero. Indeed, we have

cos6 =

H(;

110.2.201

= constant

In addition, we realize that the angle between the a3 and b3 axes is 0. so that In = a. To obtain expressions for the precession and spin rates, we make use of

0

the relationship between the body-fixed angular velocities and the Euler angles, Eq. [7.5.7]

cqi =

SOCçII

(03

0

CO

0 0

4)

110.2.211

0

li/i

In view of the nutation rate being zero, these reduce to

w1 w1

=4)s0sçls

w2=4)sOcç(i

w3=4)cO4)+1/; 110.2.221 w3=4)cO4)+i/; 110.2.22]

Comparing Eqs. [10.2.19] and [10.2.221, [10.2.221, we conclude that [10.219] and HG

••

=

= constant

•

c's

=

.

—4)c 0

+

cos 0 13

=

fH(; fH(;

\13

H(; \\ H(;

)cos 0

—

=

(1)3(1 (03(1

= constant constant

— 11

110.2.23] that the precession and spin rates are constant as well. Combining Eqs. [10.2.23], we arrive at the relation between the precession and spin rates as so

11

(Il

—

13)cos0'i'

[10.2.24] 110.2.24]

This equation states that for torque-free motion of a body with two equal moments of inertia, the precession rate 4) and spin rate c's are proportional to each other.

The nutation rate is zero. For example, if a football is thrown with a wobble, the magnitude and angle of the wobble should remain as it is and not change.3 Furthermore, the direction of the wobble remains the same. From Eq. [10.2.24] and Figs. 10.3 and 10.4 we observe that again two cases are possible, depending on the magnitudes of the moments of inertia: 1.

I

When /1 > 13, 13, such such as in a slender body, the precession and spin have the same sense. This type of motion is called direct precession. The space cone rolls

31n reality, the wobble changes due to aerodynamic and gravitational effects.

553

554

CHAPTER

2.

10

•• QUALITATIVE ANALYSiSOF OFR1(,ll) RICH)BODY BODY Molios QUALITATIVE ANALYSiS MotioN

outside the body cone as shown in Fig. 10.4. The wobble is in the same direction as the spin. When < 13. such as in a flat body. the precession and spin have opposite senses. The motion is called retrograde precession. The space cone rolls inside the body cone as shown in Fig. 10.3.

Direct precession is relatively easy to observe, as for a slender body the ratio of the largest to smallest moment of inertia can take on any value. Making 11/13 a large quantity. such as 4 or 5. one can easily view direct precession. On the other hand, because in flat bodies the ratios of the moments of inertia about the principal axes are closer to each other than in slender bodies, it is usually more difficult to visually observe retrograde precession. For a circular cylinder /3 ,nR2/2 and 1] = rnR2/4+ mL2/12. mL2/1 2. The The maximum maximum value value of of the the inertia inertia ratio is for a thin disk, 13/lJ = 2. As the disk thickness increases, the two moments of inertia become closer. We now relate the values of the body and space cone angles to the Euler angles. As stated above, a, the angle between the angular momentum vector and b3 is the same as 6, so a = 0. From Eqs. [10.2.111 and [10.2.191 we can write a relationship for the angle /3

as

tanf3=

U) j' U))

HG/It sin0 = HG/Il cos 0

—

—

110.2.251

which verifies that whether the angle /3 is larger than a depends on the shape of the body.

We now relate the Euler angles and body angular velocities to each other. Considering Eqs. [10.2.101. 110.2.23], and [10.2.25], one can show that S1fl4 /3 + sill2

=

çb=w

cos2 /3

0=tan -i

cos2 j3

= ('—

tan$ /

[10.2.26] 110.2.261 thus, given the total angular velocity w and the angle /3. one can determine the angular momentum, the precession and spin rates, and the nutation angle. Torque-free motion of axisymmetric bodies lends itself to an interesting Euler parameter description. Because precession and spin rates and the nutation angle are constants. one can find closed-form expressions for the Euler parameters in terms of the initial conditions. The Euler parameters are expressed in terms of the Euler angles angles in in Eq1 Eq1 [7.7.40]. Hence, Hence, we obtain for a 3-1-3 transformation o0

'9

= COS

()COS

e2 = sin

t)SIfl

\.

e1

0

= sin

0

t)cos5.

110.2.271

10.2

An American football (axially symmetric body) with a mass moment of inertia ratio of 1, 413isis thrown thrown with with a spin about its axis of symmetry. A player tips the football, and in doing 413

in a transverse direction, where s is a small number. so exerts an angular impulse of The change in the linear momentum is assumed to be negligible. Obtain the precession and spin rates after the impulse is applied.

Solution Without loss of generality, we assume that the angular impulse is applied about the b1 axis.

The angular momentum before the impulse is

'a] [a]

HG(before) = Jd'l.b3

After the impulse, using the angular impulse-momentum theorem. the angular momentum becomes

[bi

+

HG(after) = so so that the angular velocity after the impulse is

w=

[c] [ci

+

we have have aa case case of direct now construct the body and space cones. Because > /3, we precession; the body cone lies outside the space cone, a is the angle between b3 and HG, and and is the angle between b3 and w. Considering Fig. 10.4 we can write We We

tana== tana

el1fl eI1fl

tan/3== tanf3

=

=

[di Ed]

so that the body and space cones depend on the magnitude of the impulse. From Eqs. [10.2.23] 0.2.23] we have the precession and spin rates 13 13\ / 1/I = (.1)3(1 1/l=W31l—J

HG

It)

Ii

and that the nutation angle 0 = a = tan

Eel

From Eq. [b]. we obtain

l

+ 16s2

HG

ii:i [I]

which then leads to the precession and spin rates after the impulse as HG

= /

555

TORQUE-FREE MOTION OF INERTIALLY SYMMETRIC BODIES

i3n.fF+ i3n,/F+lf)F 4' 41

cLJl + l6E2 4

3

=

4

EgI

1k] [hi

independent of the magnitude of the angular impulse. The nutation angle and the precession rate depend on the strength of the impulse. For very small values of the precession rate can be approximated as Jr is interesting to note that the spin rate

is

+

indicating that the minimum value of the precession rare after the impulse is

[I] 'II

Example 10.1 1 0.1

_556

CHAPTER 1O• QUALiTATIVE ANAixsIs OF 0F RiGifi RiGifi BODY BODY MoTIoN

From Eqs. [gJ and [h] we obtain two additional results: First, the precession and spin

rates depend on the ratios of the mass moments of inertia. Secondly, the spin rate after the angular impulse is always less than the spin rate before the impulse. known intuitively w players and spectators all around.

10.3

GENERAL CASE OF TORQUE-FREE MOTION

We next analyze the torque-free motion of an arbitrary rigid body whose principal moments of inertia are different from each other. Using a 3-1-3 Euler angle transformation, we write the rotational kinetic energy as +

Trc)t =

+

+

+

—

[1O.3.1J is absent from the Lagrangian, so that it is a cyclic coordinate. Therefore, in the absence of moments about the axis, which is the case considered here, u-4, is an integral of the motion and has the form where we note that

dçb

= = constant

+

+

+

—

+ ifi)cO

[10.3.21

the Unlike the axially symmetric bodies studied in the previous section, component of the angular momentum along the spin axis, is not constant. As a result.

the precession, nutation. and spin rates are no longer constant, either. This makes it difficult to visualize the rotational motion of arbitrary rigid bodies. Because the motion is torque free, the angular momentum about the center of mass is conserved, and it provides three additional integrals of the motion. These integrals can be used to obtain relations between the precession. nutation, and spin rates. Combining Eqs. [10.2. 191 and [10.2.211 gives 11w

sin 0 sin t/' + 6 cos f) = HG sin 0 sin çlí

=

sin sin 66 cos ç&

12w2 13W3

— (1 6

= 13(4'cosO !3(4'cosO +

=

sin i/i) = HG HG Sjfl 6 cos cli

[10.3.31

HGCOSO

which can be solved for the precession. nutation, and spin rates. The result is

=

+

HG

= (H(;

1

13

— II , —

\1I

.

tli +

HG ,

cos

' i./i

0 =

/

cos 6

—

cos 0

13

110.3.41

__ 10.4

POLHODES

These represent three first-order differential equations. They are expressions for

the Euler angles in terms of' of the the integrals integrals of of the the motion. Because they are in state form, they can be integrated once the initial conditions are specified. The precession rate is always positive, indicating that even though the precession rate of a body may change during motion, the direction of precession does not change. The same cannot be said of the nutation and spin rates. These rates can oscillate, depending on the initial conditions. Note that we observed this same relationship when we conducted a stability analysis in Example 8.6 with regards to the stability of rotational motion about different principal axes. The nutation rate is dependent dependent on on the the ratios ratios of of the the two two moments momentsof ofinertia inertia 'i and 12. For the same initial conditions the nutation rate will be different depending on whether is larger larger than than12. 12. On On the the other other hand, hand, the the spin equation is the same for general as well as axially symmetric bodies. Considering the special case of axisymmetric bodies with '2 = Ii, Eqs. [10.3.41 reduce to =

0

= 0

fi

11

(HG

= \13 (

—

Iij)cosO

110.3.5]

which are the same as Eqs. 110.2.23]. The nutation angle is constant for axisymmetric bodies, constituting one of the biggest diftérences between the torque-free motions of general and inertially symmetric bodies. Observe from this and the previous sections that in order to understand the nature

of the motion qualitatively, one needs to analyze the motion using variables from both the inertial and body-fixed frames. Each of these reference frames provides a different kind of insight into the problem.

10.4

POLHODES

An interesting qualitative interpretation of the torque-free rotational motion of arbitrary rigid bodies is due to Poinsot's construction. The approach makes use of the energy and angular momentum integrals, as well as the inertia ellipsoid. Consider a set of body-fixed coordinates. These coordinates have a general configuration and they do not have to be principal coordinates. The rotational kinetic energy and angular momentum about the center of mass are

= We express

'{w}T[lj{w}

{HG} = [l(;]{w}

[10.4.1]

the angular velocity vector using the unit vector along which it is directed

as

[10.4.2] [10.4.21 is the unit vector along the instantaneous axis of rotation. The rotational =

where

or

{w}

=

kinetic energy can be expressed as Trot

'{w}T[!]jw} = !w2{e

}rI1cI{e

}

=

[10.4.3]

557

558

CHAPTIR 1 0 • QUALI'IATIVE ANALYsIs OF RIGII BODY MOTiON }T }isisthe themass mass moment moment ofinertia of inertia aboutthe about theinstantaneous instantaneous axis {e}T[JI{e} where where 1 [/G of rotation. Note that 1 is not constant. and its value changes as the angular velocity changes.

We next compare the expression for the kinetic energy with with the the inertia inertia ellipchipsoid relations defined in Chapter 6. We defined the quadratic expression {u}T[IGXu}, We magnitude was was 1, 1, {u}"{u} {u}"{u} = I. We [u1 U2 u3jT was a vector whose magnitude where {u} saw that the expression =

[10.4.4] [10.4.41

11

quadratic surface surface (Fig. (Fig. 6.16). 6.16). The value of A ellipsoid, a closed quadratic defined the inertia ellipsoid, depends on the orientation of the vector {u}. When {u} coincides with the unit vector along a principal axis, A becomes the principal moment of inertia corresponding to

the principal axis. The maximum and minimum moments of inertia are the major and minor axes of the ellipsoid of inertia. is analogous to {u} in In the expression for kinetic energy in Eq. [10.4.31. Eq. [10.4.4]. Siniilarly. the term A is analogous to Trot. It follows that the expression

for kinetic energy can be described in terms of an ellipsoid of inertia. For a given level of kinetic energy. the angular velocity will he smallest about the axis with the largest moment of inertia. It is uset'ul to express the inertia ellipsoid in terms of normalized quantities. To rT = p,,J/X• Substituting these expresor this end, we define a vector p = sions Ibr for the kinetic energy yields Trot

= =

lw2{e}T[lI{e} w2{e }T[J 1{e} =

[10.4.5]

1

from which we conclude that {p}'T

[IGl{p} == 1 {p}' [IGl{p}

11

1

0.4.6]

From now on we will use Eq. 1 10.4.6] to describe the inertia ellipsoid associated

with the kinetic energy. Note that the inertia ellipsoid is fixed on the body. This ellipsoid is illustrated in Fig. 6.17. We next explore the relationship between the angular momentum about the center of mass and the inertia ellipsoid. We define the intersection between the inertia ellipsoid and the angular velocity vector by P. We denote the unit normal vector to the inertia ellipsoid at point P by n, as shown in Fig. 10.5. In Chapter 4. we saw that surface defIned defIned by by f(x, f(x,y, y,z) z) = 0. we can find the unit vector normal to that given a surface surface by means of the del operation and by then normalizing the magnitude of the = R2. the normal foundvector. tiund vector. For For example, example, given given the equation of a circle of x2 + afh9vj = 2xi + 2yj. to the circle is found by Vf = V(x2 + V22 — R2) = 9fh9x1 + afh'9vj Extending this case to the inertia ellipsoid and the column vector representation. we can express the normal to the ellipsoid as K{n}' = V({p}T[laj{p}

= d({p}T[1GJ{p} cJ({p}T[IGI{p} —

1)

-

1)

= 2{p}T[IcJ

[10.4.7]

104

POt HODES

S

S

S

I

3.

S S

Invariable plane

S

S

S

(0

n

IS

a S

I S

S S

S S S

S

Figure 10.5

I2

S 5

S.

Inertia ellipsoid and the invariable plane

where K is a normalization constant. However, the vectors {w} and {p} are parallel, We can then write and {HG} = K'{n} = {JI(;}

[10.4.8]

where K1 is yet another constant, showing that the normal to the inertia inertia ellipsoid ellipsoid at point P iS is a/wa a/wa vs parallel 10 the angular momen turn vecroi: vectoi Let us analyze the physical interpretation of this result. The vector {p} lies on the instantaneous axis of rotation, so that the rotational motion can be viewed as the rolling (without slipping) of the inertia ellipsoid, with the normal to the ellipsoid at point P always perpendicular to a plane whose orientation is fixed. This plane is called the invariable /)lafle. While this plane can translate, it cannot rotate. As discussed before, [lie the point of tangency (contact) between the inertia ellipsoid and the invariable plane is on the instantaneous axis of rotation. As the angular velocity changes. so does the orientation of the inertia ellipsoid and of the point of contact with the invariable plane. The loci of point P. that is. is, of the tangent points on the ellipsoid as it rolls on the invariable plane, are called poihodes. and the loci of the tangent points on the invariable plane are called lierpoihodes. The poihodes polhodes are closed curves because the kinetic energy is constant. It is of interest to look at the component of p along the fixed direction of the angular momentum. From Fig. 10.5. denoting this distance by PG we can write

= p.n

[10.4.9]

where n is the unit vector along the direction of the angular momentum. n = HG/HG. Noting that the vector p can be written as p = and that 2Trot = J*w2

559

560

CHAPUR 10• CHAPTRR 10•

BO[)Y MOTION MOTION Ricm BODY ANALYSIS OF RICH)

we obtain

P(; =

2T

I I

/J*

I

\

w

I=

/

/2TrI)t

=

H(; \/2Tro(

HG

= constant

[10.4.10]

so that the perpendicular distance from the center of mass to the invariable plane is con constant. stant. Equation Equation [10.4. [10.4. 10] 10] is is yet yet another another integral of the motion and, as other integrals of' of the the motion, motion, itit depends depends on on the the initial initial conditions with which the body is released. Up to now, the orientation of the body-fixed coordinate axes was not specified. Without loss of generality, and in order to simplify the derivations, we now consider the body axes to be the principal axes. The inertia matrix is now diagonal, and 110.4.1 1] 110.4.11]

= {w}'[IG][Jc]{w} {WV[IG][Ic]{w} = i—I

and, using the relation {w} = w -,

I

=

(i =

=

or 3

""

1,

2, 3). we can write

I

110.4.121

i=1

which can he rearranged as

+

+

=

=

= constant

110.4.13]

This equation defines yet another closed quadratic surface, that is, another ellipsoid of inertia that is attached attached to to the the body. body. Comparing ComparingEqs. Eqs.[10.4.101 [10.4.101and and[10.4. [10.4.13]. 131, we conclude that (I

= =

1

-

110.4.141

rut

We next relate the motion ol the ellipsoid defined in Eq. [10.4.131 to the inertia ellipsoid defined in Eq. [10.4.6]. The two ellipsoids are in contact with each other at point P. leading to the conclusion that as the body rotates the intersection of the two ellipsoids describes the poihodes. polhodes. We can construct the poihode curves by solving Eqs. [10.4.61 [10.4.6] and and [10.4.13] [10.4.13] simultaneously. simultaneously. Note that if we consider Eq. [10.4.6] as describing a closed surface and Eq. 110.4.131 as a constraint relation, the result becomes a set of curves on the closed surface. We gel different curves for different starting points. One can simplify the construction and interpretation of the poihode curves by looking at their projections along the principal axes. To this end, we define the variable L)

=

=

[10.4.15] [10.4.151

10.4

POLHODES

whose value depends on the initial conditions. It follows from the above discussion D and that when D = 11, (i 1, 2, 3) this corresponds to an that 'mm angular motion about the axis axis b1. b,. Without Without loss of generality, we select our body axes < 13. This implies that the inertia ellipsoid is longest in the b1b2b3 such that < b1 direction. We next find the projections of the poihode curves on the b1 b2. b1 h3. and b2h3 planes. Recall that the poihode equations are +

Ii t1 + "P1

Ijpf +

I

+

= D

[10.4.161

Multiplying the first of Eqs. [10.4.1611 by 13 and subtracting the second from it yields the projections of the curves on the b1 h4 plane as

+ '2(13 12(13

—

[10.4.171

'2)1)2 = 13 — — D

In a similar fashion, we obtain the poihode polhode equations on the h2b3 and h1h3 planes as 2 2 — i i r i b2b3 plane: 110.4.181 — '1 'I — — '2)P2 + 13(11 —

b1b3 plane:

'3(12

+

—

11(12 —

110.4.191

— D 12 — = 12

All the coefficients of Eqs. [10.4.17] and 110.4.181 have the same sign, indicating that these two equations represent ellipses for all values of D. By contrast. contrast, the coefficients on the left side of Eq. [10.4.19] have differing signs, thus describing hyperbolas. Furthermore, the sign of 12 — D depends on the problem. The three projections arc given in Figs. 10.6. The special case of D = '2 has an interesting 12 has interpretation. The projection on the b1b3 plane is in the form of straight lines that separate the hyperbolas. In analogy from stability theory. these lines are called separatrices (plural of separatrix), separating different regions of the motion that that have have different properties. From the information above, we can construct the poihode curves in three di< < /3, mensions. For a rigid body with /3. the curves are given in Fig. 10.7. Note that when D < '2, the poihode curves are closed around the b1 axis, and when 1)

h1

D=t

4 1

D
N' N'

h1 4

I..

11

(D> -

D>12

(D<1,) (a)

Figure 1 0.6

'j

(1.')

Polhode prolections. (a) b1 b7 and b2b3 b2b3 planes, planes, (b) (b) b1 b3 plane.

1

561

562

lo.

ANALYSIS OFRIGID RIGID BODY BODY MOTION MOTION ANALYSIS OF

CHAPTER 10 CHAPTER

n I,

D>I,

1

N'!1 Nj1

Invariable plane

N

Figure 10.8

Figure 10.7

the poihode poihode curves curves are closed around the b3 axis. The separatrices, which D > 12i 121 the 13, the give the limiting cases. As D approaches or 13, are constructed for D = poihode curves become smaller in size. The locus of the line joining the center of polhode curves defines the body cones when viewed from a body-fixed mass and the poihode reference frame. However, unlike the symmetric body considered in Section 10.2. the cones have no symmetry. We can analyze the motion of the body by examining the poihode curves. Con-

sider the case where we give a large angular velocity in one direction and much smaller angular velocities in the other directions. According to the stability results predoniinantly about the largest or smallest in Example 8.6. initial motion that is predoniinantly axes of inertia will be stable, and initial motion about the intermediate axis of inertia (h2 in our case) will be unstable. In this latter case, angular velocities about the other axes will grow with time. and all three angular velocities will become compapolhode curves. The angular rable in magnitude. We can demonstrate this using the poihode polhode velocity of the body is along the line connecting the center of mass to the poihode or h3 axes, the value of D curves. When the initial angular velocity is along the is slightly larger than or slightly smaller than 13. respectively. In both cases, the polhode curves are small. Hence, the motion retains its characteristics. By contrast. poihode and the if the bulk of the initial angular velocity is along the b2 axis, then D polhode curves are close to the separatrices. Hence. the poihode curves are large. poihode which means that the magnitude and direction of the angular velocity varies a lot. This, in essence, is the explanation for the wobbly motion and also why it is difficult to visualize the motion of' an arbitrary body. Now consider the special case of' axisymmetric bodies. Without loss of generality, we set = '2 and < 13. The inertia and momentum ellipsoids are described by +

+

+

I

+

= I)

[10.4.201

so that both ellipsoids are symmetric about the b3 axis. It follows that the intersection of the ellipsoids defines circles about the 13 axis, with the radius of the circles determined by the value of D. indeed, introducing '1 = /4 into Eq. 10.4.171 )

= 117 717

1

— 13)

110.4.211

10.5

MOTiON OF A SpiNNIrSG MoTioN SPINNING Toi

this case D varies between Ii Ji and 13 and D = '2 = no longer represents a separatrix. The body cone generated by the locus of the line joining the center of mass and the poihode curve is axisyinmetric. To show that these body cones are arc indeed the same as the cones discussed in Section 10.2, consider Fig. 10.8. which shows the ellipsoid for a given value of D. This figure can also he considered as the projection of the ellipsoid to the plane generated by the the intersection intersection of of h3 h3 and a3. To show that the angles ij and (from the definition of body cone) are the same, we need to demonstrate that Eq. 110.2.251 holds. or In

110.4.221 110.4.22]

I1tan-q = I4tant9 I4tanO

The proof is left as an exercise. The projections of the poihode curves onto the b2h3 b2h3 or or b1b3planes planesrepresent represent straight straight lines lines and and they they do do not not have have a special physical interpretation. One can show that the cone described by the points G. P. and C is the space cone.

10.5

MOTION OF A SPINNING

In the previous sections we considered moment-free bodies. bodies. We We 110W flOW switch our attention to bodies subjected to external moments. Au An interesting case is the motion of' a spinning top. such as the one shown in Fig. 10.9. A top is basically described as a body that possesses inertial symmetry and that terminates at a sharp point along the synMnetry axis. symirletry axis. This This point point is called an apex or t'er!ex. Tops are designed to he axisyinmetric. as such a construction increases stability and minimizes the friction due to the air mass around the top. The motion of the top can be viewed as the balancing of the gravitational moment about the apex by the gyroscopic moment. The spin of the top gives it its stability. We assume that the apex is in continuous contact with a plane in a way that it has iio translation. (The translational motion of the apex. commonly referred to as drifi. is due to the initial translational motion given to the top as well as to the unevenness and roughness of the plane on which the top moves.) The equations of motion for a spinning top were derived in Example 8.10 using a 3-1-3 Euler angle transformation. Here, we will analyze the integrals of the motion (14

(1

I

/

/

1

I

I

Figure 10.9

/ I

/

I

,

563

564

CHAPTER

10•

ANALYSIS OF RiCH) BODY MOTION

and qualitatively assess the behavior of the top. The top being an axisymmetric body,

we can write the kinetic energy of the top as T =

Trot

2

=

[10.5.1]

+ 1iw5 +

where the components of the mass moment of inertia are about point 0. Introducing the values for the angular velocities for a 3-1-3 transformation, we obtain for the kinetic energy +

= T T =

+

+

[10.5.2]

The potential energy is

[10.5.3] [10.5.31

V = ingL COS 0

The only other forces acting on the spinning top are those at the point of contact. They do no work, as they are being applied to a fixed point. Note that the moment about the apex 0 is due to the weight of the top. The integrals of motion are the total energy and generalized momenta associated with precession and spin. The latter two integrals are recognized by noting that neither nor it/i are present in the Lagrangian. From Example 8.10, these integrals of the motion are =

(9

0+

=

= =

+ i/i) =

=

= constant

+

[10.5.4] 110.5.4]

= constant

These integrals of the motion represent the components of the angular mornenaxes. The equations of motion associated with the precesturn along the a3 and equations. Using Using SiOfl and spin can he obtained by differentiating the above two equations. Lagrange's equations. the equation of motion for 0 is —

—

sin 6 cos 0 +

sin 0

—

mgL sin 0 = 0

[10.5.5]

From Eqs. [10.5.41 one can solve for the precession and spin rates as — IT11, COS

Ii sin2 sin2 0

0

cos- 0

1

'I,

sin

3

cos0

[10.5.6] [10.5.61

and substitute into Eq. [10.5.5]. Once the nutation angle is solved tbr. the precession and spin rates can be calculated using Eqs. 110.5.61 and the initial conditions. Another way of deriving the equation of motion for 6 is to make use of Routh's method for ignorable coordinates, as and are such coordinates. Defining the Routhian as

= L

—

1T4(t'

—

ITipLil

[10.5.71 110.5.7]

___-

I 0.5 and

introducing the expressions for •,

1 I

= —ho-

and i/i into it. we obtain , 2 IT11,

(

•

•1

—

COS

0 )—

—

2/,sinO

2

MOTION OF A SPINNING TOP

,,

—

110.5.8]

2/3

The Routhian can now be treated as the Lagrangian of a single degree degree of' of freedom natural system with kinetic and potential energies T' and V' of the form I

,

T = T'

I

V VI 2

(7r(/)

=

ir,11cosO)2

, 211sine9 ,,

+

.

213 2h

+ mgLcosO

[10.5.9]

Obviously. the energy integral is E = T' + V'. To analyze the energy integral in more detail, we introduce the constant quantities

a ==

/

2 — (E

=

Ii\ 11\

2ingL

a =

Ii

b 11 Ii

11

=

IT,h

[10.5.10]

and the variable U = cos 0. Note that a and b have the units of angular velocity and that all of the above quantities are calculated from the initial conditions. Also, u is a nondimensional quantity describing the elevation of a point on the symmetry axis that is at a distance of' of unity unity from the apex. The energy integral can thus be written as a cubic function = f(u) = = (a (a

II —

—

—

(a

—

[10.5.1 110.5.1

hi.i)2

11

will study the characteristics of the motion by qualitatively analyzing the iUflCtiOfl f(u). Other methods of analysis include numerical integration of Eq. [10.5. 11] or of Eq. [10.5.5], as as well as a separation of variables. The latter leads to an elliptic integral integral For for time in terms of u. A very interesting special case is when i€ = 0. as it corresponds to zero nutation rate. The values of u that lead to a zero nutation rate range can he obtained by solving f(u) = 0. To this end, we consider the characteristics of f(u). Because u is defined 1(u) as u = as = cos 0. we are interested in the roots roots of of f(u) (ii) in in the range range — — I u 1. Further, we are interested in the values of ii that are larger than zero. as a negative value for u implies that the top is spinning below the platform it is on. We make the following observations regarding the roots of f(u): We

I. 2.

Both f( 1) and f( — I) are less than zero. As u becomes larger, f(u) > 0. since

>

0.

Hence, [(U) has a root for u > I and [(U) < 0 for u =

This implies that in the iange range of — u there either are no roots, or there are two roots. We discount the possibility that f'(u) f(u) has no roots in this range, as this would imply that there are no positive values lhr the nutation angle. We thus consider the case of two roots for f(u) in the range — u I. Fig. 10.10 shows a typical plot of f(u). The two roots, denoted by u1 and and Ui, m, are are in ingeneral general both positive or both negative. Let us consider the case when they are arc both positive, corresponding to an upright top, and order them without any loss of generality as u2. This way, u2 corresponds to a smaller nutation angle. The physical range I

I

I

I

—

1.

565

566

CHAPTER 10

.

QUALITATIVE ANALYSIS OF RIGID Bony MOTION

f(u)

U

()

1.)

Figure 10.10 of the motion of the top is then between u1 and and U2. U2. The The motion can be visualized as the change in the nutation angle from 02 corresponding to 62 = COS '(U2). corresponding to the highest elevation of the top. to Oi = cos 1(u1). 1(u corresponding to the lowest corresponding elevation. The motion of a point on the syinmetiy symmetiy axis of the top at a distance of unity from the apex can be viewed as tracing a curve on the unit sphere, as in Fig. 10.11. The circles on the sphere that correspond to the locus of all points with 0 = cos 1(u1) and are called the houmling hounding circles, and the axis basically moves moves between these circles. The precession rate, from Eq. (1O.5.6j (10.5.61 can be shown to be •

—

a—bi,

110.5.12]

1—u2

so that the precession rate depends on the value of

a — hu.

Let us introduce the

quantity u0. based on the initial conditions, and defined as a

l'0 iT11,

'I

Oo4)o

h

+ i11i0)

a3

Bounding circles

Figure 1 0.1 1

Bounding circles

+COSO()

110.5.13]

10.5

MOTION OF A SPINNING Tot'

Ii

a1

cos

II

'S (h)

(a)

Figure 10.12

(a) Unidirectional and (b) looping precession

For the general case of motion. "o COS Oo• Oo. It turns out that characteristics of the motion of the top depend on the relationship of' u() to U1 u1 and and u2. To find uo one begins with the initial conditions and the property that and = 0. The generalized momenta 'n-v,. and a and b are calculated using Eqs. [10.5.101 and [10.5.41. The results are substituted into Eq. [10.5.131 to find 110. We identify the following different types of motion based Ofl the initial conditions:

Unidirectional Precession

In this case, 140>> u2 U2 (or (or

1.11)

< u1). u1). This case can

also he described as having the same precession direction at the bounding circles, so that a > bu2 (01- a < bu1). The precession rate is not zero at both hounding circles. This kind of motion can be initiated by releasing a top from its highest elevation point U = with zero nutation rate and a precession rate high enough to make U0 u0 greater greater than 112. After the top is released, it begins to fall as a result of the action of gravity. In the process, the top attains precessional motion. At u = U1, when the maximum kinetic energy is reached, the nutation rate changes sign and the top begins to rise until u = U,. and the process repeats itself. The resulting motion is a periodic motion of the nutation angle, like a sinusoid, as as shown shown in in Fig. Fig. 10.1 lO.12a. 2a.

< uo < u2, and and the direction of precession is different at the bounding circles (Fig. 10.1 2b). At the bounding circles, the nutation rate becomes zero. At the top hounding circle, 0 = 02, the precession rate 62, the is negative; when 0 = 61, the the precession rate is positive. When uu becomes becomes equal to toU() u0 the precession rate becomes zero, which explains the looping nature of the motion of the symmetry axis. Such motion can he initiated by releasing the top at 6 = 62. with a small precession precession rate rate and andzero zeronutation nutationrate ratesosoasastotosatisfy satisfy U1
Looping Precession

Here,

u1

is reached, reached. the precession again changes sign. Fig. 10.13 gives a typical plot = of the precession rate 43. In essence, the precession rate oscillates between its bounds U

at

0 =

567

____-

568

CHAPTER io.

ANALYSIS OF RIGID BODY MOTION

a1

COS

Figure 10.14

Figure 10.13 FIgure

Cuspidal (or Cuspidial) Motion

This is the limiting case between the two bu2. From Eq. [10.5.12] we so that a = bu2. u2. so precession.Here, Here.U() U() = u2. other types of' precession. and is greater than zero at all observe that the precession precession rate rate becomes becomeszero zeroatatuu = u2 and other times. This type of motion can be initiated by releasing releasing the the top top at at U ii = 142, with or nutational nutational niOtiOfl. motion. The Themotion motion characteristics characteristics are similar to no initial precessional or

those in unidirectional precession. with the difference being a zero precession rate begins to with zero precession. As the top begins is released released atatuu = u2 with at uu == 112. The top is energy precess. its symmetry axis begins to fall, resulting in an increase of kinetic energy the top begins to rise, and when it u1. the in potential potential energy. energy.At Atuu = u1. and reduction in the precession rate becomes zero. The symmetry axis traces cusps. the reaches = as shown in Fig. 10.14. One cannot have cuspidal motion if the top is released with because the tendency of the top is to fall. u1, because U0 = u1, U0 To obtain more insight into the nature of the motion of tops. we analyze Eq. = (u) = [10.5. 111further. further.Recall Recallthat thatall allofofthe theabove aboveresults resultswere wereobtained obtainedfbr for f(u) [10.5.111 0. which corresponds to zero nutation rate as an initial condition. Because cuspidal motion represents a limiting case. consider an initial condition corresponding to this and zero initial precession. type of motion. Hence, we have an initial spin rate of with an initial nutation angle Of and no initial nutation rate. In this special case, U(, It follows from Eqs. [10.5.41 cos and that thatHO HO = COS are related by Eq. Eq. [10.5.13] [10.5.13] and and w3 is is constant. We also have as initial conditions = that = /1W3. /iWi. and w3 = bu0 a=

or

U() = U()

[10.5.14] 110.5.14]

U2 U2

substitutedinto intoEq. Eq.[10.5.111 [10.5.111and and .t(u)set set to to zero. zero. yields yields a = /3u0. which, when substituted Introduction of this into Eq. [10.5.111 yields

f(u) J(u) ==

(uo

—

u)[/3(1 u)[/3(l

—

u2)

—

+

—

u)]

[10.5.15]

The remaining equation to solve

is 112 denoted that byby 1.12. 112 = that one oneroot rootofoff(u), f(u),denoted 1.12. is tbr the other roots can be written as so

U4 —

b2(uo

—

1

= 0

[10.5.16]

where )

=

-k--

2/3

=

2

411m.gL 411in.gL

=

3/0 4J1mgL 4IimgL

[10.5.17]

10.5 From From a physical perspective,

MOTION OF OF

SPINNING

denotes the ratio of kinetic

energy to the gravitational potential energy, multiplied by the ratio of the mass moments of inertia. The kinetic energy is associated with the gyroscopic moment and the potential energy is due to gravity. The other two roots are =

—

+I

—

=

+

+1

—

1

[10.5.181

Of these roots. is greater than one and without physical significance._To show this, recall that the magnitude of is smaller than 1. so that — 2 U() + I>

+1=

By the same argument. one can can show showthat thatJu1 argument, one u1 I < 1. If the top is given a fast spin as as an initial condition, then and hi can he approximated using a two-term Taylor series expansion as —

—

1

1

—

1

110.5.19] 110.5.191

which gives an indication of how the lower bounding circle varies as a function of Because the difference between u1 and UØ is not large, one can introduce the perturbation parameter = — iiu to to Eq. [jO.5. 11] and obtain =

—

— —

[10.5.20]

—

Neglecting terms terms in e that are of higher order than 2, and noting that b2

fast spin, this equation can he simplified as =

—

—

u0 for

[10.5.21]

h282

This equation can be solved by converting it into an elliptic integral by taking the

square roots of both sides and separating variables. A more physically meaningful solution solution can be obtained by differentiating it with respect to time and dividing the resulting equation by which yields 1

+

[10.5.22] [10.5.221

=

subject to the initial conditions

= 0. è(O) = 0. This basically is the equation of a simple sinusoid subjected to a constant excitation. The solution can be shown to be —

= =

2b-

° (I

— COS

bt) == bt)

I

—

0(1

— COS cos

br) bt)

110.5.23] 11 0.5.23]

Differentiating the above equation and recalling the definition of 8. we obtain the rate of change of u as U

=

—0

sin 0

h(1 —un) h(l

=

sin bi

[10.5.24] 110.5.24]

and, because the value of the angle fI fI does does not change_too much during the motion, its

sine can be approximated as as sin sin 00 =

—

\/l

—

Hence, an approximate

569

570

10

.

expression lbr 0

ANALYSiSOF OFRIGiD RIGID BODY BODY MOTION MOTION ANALYsIs

can

be written as

sin hi hi = sin =

0(t)

h sin

110.5.25] 110.5.251

sinbt

In a similar fashion, we approximate the precession rate. Introducing Eqs. using the the approximation approximation (1 (1 —— 110.5.23] into Eq. Eq. [10.5.121 [10.5.121 and using obtain for the precession rate (1

—

U2)

(1

—

we

110.5.261

cosb/) cos hi)

The motion can be interpreted as follows. For cuspidal motion the precession rate oscillates between th = 0 and (1 = h/2& so that its average value is h

4av =

=

= 2b

[10.5.27] 110.5.27]

t

13qJ0

The frenutation rate oscillates like a sinusoidal of amplitude b sin quency of oscillation is the same for both nutation and precession. At the point of zero nutation. sin bt = 0 and cos bi = ± 1. so that the precession rate is either zero (which corresponds to the highest elevation of the top) or at its maximum value (which corresponds to the lowest elevation). The

Steady Precession

As the initial spin rate of the top is increased, the amplitudes of the precession and nutation become smaller, while the frequency of oscillation b becomes larger. As a result, U1 and U2 become very close to each other, the nutation

and the oscillatory part of the precession become smaller and more difficult to observe visually, which gives the appearance that the top is precessing uniformly with precession. prece.vsion. nutation angle. This case is known as no variation in the nutation To analyze steady precession in more detail, we set 0 = 0 in the equation of motion for the nutation angle. We rearrange Eq. [10.5.51 as a quadratic f'unction in terms of as 13w3

—

ingL = 0 cosO

+

COSO

We We solve this equation for 1) (1)1,2

to

[10.5.28] 110.5.28]

yield ±

= 2/1cosO 2l1cosO

41i mgL ingL COS 0

2/1cosO

[10.5.29]

these solutions must he real, which implies that in order to have steady precession. these ln equation must must be bepositive. positive. From From this this we we obtain the critical the radical in in the the above equation the velocityrequirement requirementfor forsteady steadyprecession precession as as angular velocity -,

>

4I1mgLcosfl 13

[10.5.30]

I 0.5 Equation

MoiioN

571

OF A SPINNING

0.5.301 describes the minimum angular velocity that must be irnparted in the 1)3 direction for steady precession. For large values values of ofthe the spin, one can )2 treat the term mgL cos as small and approximate the precession rates [

I

in Eq. [10.5.291 as .

=

.

t

/3(03

=

COSO '1 COSO

b COS()

110.5.31]

The two two roots rootscorrespond correspond to fast and slow precession. The smaller root.

cor-

responds to the slow precession rate: it is the same as the average precession rate motion in in Eq. Eq. [1[1 0.5.27J. The larger root, derived for cuspidal motion root. corresponds to fast precession: it is usually not attainable, attainable, because of the very high spin rate required to achieve this motion. Actually, there is an interesting interpretation of the fast precession rate. Expressing it as =

+

=

110.5.321

we can relate the precession and spin rates as as

= —

13)COSEI 13)cosEi

[10.5.33]

This equation is the same as Eq. [10.2.241. which we derived for torque-free motion. The spin in this case is so high high that the moment exerted exerted on on the the top top by by gravity gravitybecomes becomes insignificant. A special case of steady precession is when the nutation angle 0 is zero, which implies that the top is rotating in the upright position. To an observer the top appears motionless (the axis of the top and F frame are not moving) unless there are distinguishing marks on it. This type of motion is known as the sleeping top. The precession and spin rates are added linearly. w3 = + i/i. and from Eq. [10.5.301 the minimum value of the angular velocity required becomes -,

Wcr

= =

2

13

[10.5.341

To have the sleeping top motion, the top must he released in an exactly upright position with a minimum rotation rate given in the above equation. This initial condition, of course, is hard to achieve. Also, there always is some amount of friction at the apex. which slows the top. Most tops, when released almost vertically and with a high spin, begin their motion as steady precession and with a very small nutation angle. As the effects of friction begin to build up over time, the the spin rate decreases decreasesbelow below the the critical critical speed and nutalional nutational motion is observed. As As the spin rate decreases decreasesfurther, further, the the nutation nutation angle gets bigger, angle gets bigger, until until the the lop lop hits the ground. Table 10. 1 summarizes the types of motion that can be achieved depending on the initial conditions. Here, we denote by the initial angle with which the top is released (note that U() = COS O() for this special case), and recognize from Eq. [10.5.17] that the the ratio ratio of the the initial initial angular velocity = to

572

CHAPTER 10 • QUAI1rNrIvF: QUALITATIVF: ANAIxsIs OFRIGID RIGID BODY BODY MOTION ANAIxsIs OF

Table 10.1

Summary of top motions

Initial Precession Rate

Initial Spin Rate

initial Condition U(,

0

1

0

Ugj

0

<1 1.
U() == Uj Uj ==112 112

Ut1 < 1.

0 0

U1) >

or

I/Jo

Type of Motion

W3

Sleeping top

WI

Steady precession

Cuspidal motion

= U()

< ii,

Unidirectional precession Unidirectional precession

u1 < u0 <

Looping precession

can be expressed as (w3

—

cr

—

I i3qi0

—

(1

4l1mgL

(1O.5.35J

Be aware that the values of u1 and u2 are dependent on the initial precession and spin rates, as well as on the nutation angle. When deciding on the type of initial condition to impart to the top, one has to conduct an analysis relating u1 and to the initial values of the precession nutation and spin. Up to now we hardly considered the conditions under which a top will no longer spin and fall to the ground. Confining our analysis to tops on horizontal platforms without elevated pivots, the instant the side of the top makes contact with the platfbrm is when the nutation angle i9 plus the half angle of the top becomes equal to 90 degrees (Fig. 10.15). Once this value of the nutation angle is found, the correspond-

ing u1 is calculated and the remaining conditions are then determined. A simple special case is cuspidal motion and setting 0 = 900, so that = 0. Setting u1 = 0 in Eq. [10.5.18] and solving for U() we obtain U()

=

cosO()

=

=

2I1mgL

a3

6

Figure 10.15

[10.5.36J

10.6

ROLLING DISK

which can be solved for the initial spin rate as

= =

[10.5.37] [10.5.371

cos

Equation 110.5.37] gives some interesting results. If the top is released from = 1. I. the minimum speed required to prevent the top the upright position, cos Any from falling down is = Any initial spin between and results in a precessional motion. Also, as the initial nutation angle increases, the initial angular velocity required to keep the top spinning increases, an expected result.

10.6

ROLLING DISK

In Section Section 7.9 we explained explained the kinematics of a rolling disk, and in Chapters 8 and 9 we derived the equations of motion for the case of no slip, using a variety of approaches. Here, we qualitatively analyze the equations of motion for the case of no slip and investigate the range of validity of the no-slip assumption. As in the spinning top discussion, the general motion of a disk can be described as the contest between the moment due to gravity and the gyroscopic moment. Intuitively, we know that a disk rolled with a higher initial speed will fall down later than a disk rolled with a lower initial speed. The disk configuration is shown in Fig. 10. 16. Because of the no-slip assumption, the friction force at the contact point does no work, so there are no forces that are nonconservative. We hence conclude that energy is conserved. With regards to the generalized momenta, recall the kinetic energy expression from Eq. [e] in Example 8.13 as

T=

+

+ R202 cos2

sin2 0 +

+

+

+ 110.6.11 110.6.11

z

V

S.

F

1

I

p

1

I

/I

F

I

/

1

I

.

I

I

0

I

I

I

I 1•

Figure 10.16

I

/I

/

573

574

CHAPTIR 1 0 CHAPTIR

•

Mo'uoN

ANALYSiS OF RIGID

Substituting the values of the velocity components of the center of mass X and Y from Eqs. [b] and Ic] in Example 8.13 X = RO sin

Y=

—

RO

sin 0

cos

COS 0 +

—

[10.6.21

cos ff sin sin00 — — R(4 cos 0 + t/i) t/i) sin

we obtain for the kinetic energy

T =

+

((ii + mR2)0 + (13 + ,nR2)(ØcosO +

sin2 o)

[10.6.31

Even Even though the kinetic energy does not have contributions from 4 and t/i, the and t/i,

generalized momenta associated with these generalized coordinates are not constant. This is because we are dealing with a constrained system, and Lagrange multipliers are present in the formulation. The equations of motion were derived to be (13 +

+

—

110.6.41

= 0

—

+ ,nR2)O + (13 (13 + ,nR2)d)sin0(bcos0 +

sinOcosO + ingRcos0 0 + 21140cosf1 21140cosf1

—

qi) = 130(4cosO + qi) 130('4cosO

0

[10651 [10.6.61

()

These These

are considerably more complicated than the equations for the spinning top. Consequently, it is more difficult to conduct a qualitative analysis. One special case is steady motion, with a constant nutation angle ().To analyze the characteristics of this motion. we remove all derivatives of 0 from Eqs. 11 0.6.4J—[ 10.6.6]. with the result

(/3 i/i) = (13++ ,nR)(thcost9 iiRi(thcost9 ++ i/i)

(13 ++ (13

mR2)c/sin6(thcosf) + 'I')

—

I142sinOcosO

I1cjsinO =

[10.6.7J [10.6.7]

0

+ mgRcos0 =

0

110.6.81

[10.6.9J 110.6.9]

0

From Eq. [10.6.91 we conclude that çb is zero, corresponding to a constant precession rate 4. This is to be expected. as zero nutation results in steady precession in

axisymrnetnc bodies. Introducing this to Eq. 110.6.7], [10.6.7], we conclude that the spin rate is also constant. We calculated the velocity of the center of mass for the general case in Section 7.9 as v(-; = V(;

—

Roi-i =

+

+ ROk

[10.6.10]

For steady precession 0 = 0. 0, and because the precession and spin rates are constant,

one concludes that the velocity of the center of mass has constant amplitude. The center of the disk ft)llows a circular path. The radius of this path. denoted by p, is basically the speed of the center of mass v divided by the precession rate V

[10.6.111

10.6

RoI,L4ING DIsK ROILiNG

in which

v=

[10.6.12] 110.6.121

—R(4)cosO + t/i) t/i)

is the Constant speed of the

Combining Eqs. Eqs. [10.6. [10.6.11] center of mass. Combining 11] and [10.6. 12]. we can express the spin rate in terms of' the precession rate as

=

[1o.6.13J [10.6.13]

—4)

The motion can be described in terms of two parameters, the precession and spin

rates, or in terms of v and one of' p or 0. We explore the relationship between v and these parameters by introducing Eqs. [10.6.111 and [10.6.12] into Eq. 110.6.8]. with the result —(13

v

/

Iv\

v

.

+ tnR-)— sinO— R p

(—

—

j

sin6cosO +

= 0

110.6.141

which can be rearranged and solved for v2 as -,

rngR2p2 Cot COt 0

= (13

+

[10.6.15J [10.6.15]

+ I1RcosO

For a thin disk. the mass moments of inertia can be written as /3 = mi<-, mK-, = FnK2/2, FnK2/2, where where K is the radius of gyration. so that the above equation can be expressed as I?2 p2 cot 6 2gR2p2 2g

[10.6.16] 110.6.161

2(K2(K- + R-)p ++ KRCOSO KRCOSO For a uniformly flat disk,

= R2/2. Hence, given an initial speed of v and ini-

tial nutation angle 6, with no nutation rate, one can determine the radius of curvature for steady precession from Eq. [10.6. 11]. Of course, this relationship holds as long as the no-slip assumption is not violated. A special case of Eq. [10.6. 161 is when the disk is released in the upright position. The minimum velocity for this case can be obtained by making appropriate substitutions in Eq. [10.6.16) (0 approaches 90 degrees and p approaches ac). ac), Another way is to conduct a perturbation analysis. We expand the nutation angle and precession and spin rates about their nominal values by 0

=

in which

+ cOi

4)

= 4)o = &

±

=

+

110.6.17] 110.6.171

is a small parameter. This is tantamount to assuming that the disk, rolling with steady precession, is acted upon by an impulsive force and as a result the precession, nutation, and spin values have slightly changed. Introducing these perturbed values into the equations of motion, Eqs. [10.6.41—1 [I0.6.41—1 10.6.6]. and ignoring terms that are quadratic or higher order in one gets a set of three linear equations in terms of Oi. and i. The eigenvalues of these equations are then used to ascertain the conditions under which steady precession is possible. .

575

576

CHAPTER

1

O• QUALITATIVE ANALYSIS OF' RIGID BODY MOTION

For the special case of upright release there is no precession, 4o = = 'rr/2. We have nutation angle is cosO

sin 0

4)

1

0 =

=

'I' =

0.

and the

+

110.6.181 Introducing these values into the equations of motion and neglecting higher-order terms, the linearized equations can be shown to be

110.6.191

(13 + mR2)dii = 0

+ rnR2)61 + (13 +

—

mgROj = 0

110.6.211

= 0

—

[10.6.201

is zero, showing. as expected. the constancy Equation 110.6.191 indicates that of the spin rate. Integrating Eq. [10.6.2 lj, we obtain = (ti (/;j =

[10.6.22] 110.6.221

13tf100

which, when introduced into Eq. [10.6.201, yields + mR2)Oi +

((13 + —

= 0

[10.6.231

In order for the above equation to have a nondivergent solution, the coefficient initial spin rate must satisfy the condition of B must must be be positive. positive. For this, the initial •,

ml1gR (13 + mR2)!3 mR2)!3 (13

gR 2(K2 + R2)

110.6.241

larger than the value given above A disk rolled with an initial angular velocity will continue rolling upright. For a disk to roll in an exactly upright position it has to be released exactly upright. Otherwise, the disk will attain some precessional motion. as evidenced by Eq. [10.6.161. In reality, because of friction and air resistance a disk rolled upright initially will eventually begin to tilt and precess. As this happens, the center of the disk no longer follows a circular path. but a path like a spiral. Ultimately either the no-slip condition is violated or the angular velocity of the disk becomes too small to sustain rolling. To analyze the no-slip assumption in more detail, it is useful to look at the motion using the X'Y'Z' frame. frame, which is obtained after the first Euler angle rotation. Fig. 10. 17 shows the free-body diagram for this case. The normal thrce is in the Z direction and the friction forces are shown along the X' and Y' axes. We see that if the disk slips, the slip can take place in two forms: One type of slip is along the X' direction, which is the line of nodes and basically denotes the tangent to the path of the point C'. Point C' is on the F frame and coincides with the contact point at all times. Recall from Eq. [f I in Example 9.5 that the velocity of C' is calculated as —Ri/if1 = —Ri/if1

110.6.25]

106

ROLLING DISK

z.z, 'II a..

Y,

.-—

' C

of C' - N AA

Figure 10.17 Hence, for slip that occurs along the line of nodes, the disk basically spins in place. We refer to this type of slip as spill spin slip. An example is someone moving a car from rest by depressing the accelerator fully or trying to move a car on an icy surface. If the friction between the car and road is not sufficient to prevent slip, the tires spin in place. Intuitively, we expect that in order for spin slip to occur, a large force or moment must he applied to the disk. The second type ol slip is along the Y' axis, and is very similar to the sliding of a body on a surface. We refer to this type of slip as sliding slip, and expect intuitively that for a freely rolling disk. if slipping occurS it will be of this type. To examine whether slipping occurs or not, we write the translational equations of motion of a disk along the X' Y'Z' axes and equate them to the external forces = F = F11' + FJ' + (N — rng)K 110.6.261 in which N is the normal force and the expression for a(; is given in Eq. [7.9.26]. The relationship between the unit vectors in the F frame (xvz) and the X'Y'Z' frame is given by

j

= cos OJ' + sin sin OK' OK'

k =

—

sin

or +

cos OK'

[10.6.271

SO, introducing Eqs. Eqs. [lO.6.27J [l0.6.27J into Eq. 17.9.26J, the components of Eq. 10.6.261 become F1

=

+

=

+ i/i)— i/i)— 02Jc()

=

N

—

=

+ i/i)

+ —

—

iiiR[O + ç&sO(4c0 + ifi)]sO

—

ing =

(1!

+ i/i)

—

02Js0 (12Js0

+ [0 +

+ t/i)JcO

d

= inR—(OcO)

[10.6.281

(It

The no-slip condition is given by +

,aN

110.6.291

577

578

ANAlYSIS OF RIGID BoDy MoiioN

CHAPTER 1 0

is the coefficient of friction. After fInding the values of F1 and F2, one where should check if Eq. I 0.6.29 J holds. If analyzing the motion of' the disk by numerical simulation. one should check Eq. [10.6.29] at every time step. 1

Let us examine the physical significance o1 Eqs. [10.6.28]. The Z component depends only on the change in nutation. This is logical. as only the change of elevation of the disk should affect the value of the normal force. The first term for the angular velocity in the z direction. In the abdescribes the rate rate of ofchange change of sence of large torques applied to the disk, this change will be small. The second term for F1 is the product of the precession rate multiplied by the nutation rate, so that if the nutation rate is small, this term is small also. We conclude. that unless large torques are app1ied along the z axis. F1 is usually a small quantity. corroborating the earlier intuitive comments regarding spin slip. By contrast, the first term for F2 has the precession rate multiplied by the spin rate, a much larger term than the precession rate multiplied by the nutation rate. Hence. Hence, in the absence of large torques about the z axis, the friction force in the Y' direction uses up most of the available friction force.

An interesting application of these results is for steady motion. In this case, the nutation angle and precession and spin rates are constant. Using Eqs. 0.6.1 1] and reduce to [10.6.12], Eqs.

F1 =

0

N

0

—

F2 =

V

rn—v

p

=

p

[10.6.301

Hence, there is flO no spin spin slip slip in in this this case case and and the friction force is required to only prevent sliding slip. The normal force is equal to the gravitational force. This is to be expected. because as the elevation of the disk does not change, the normal Ihrce force remains the same. The last term in Eq. 110.6.301 is the same expression as the centripetal acceleration for constant speed. Indeed, examining Fig. 10.17 and considering steady precession we see that the X' axis corresponds to the tangential direction and the Y' axis to the normal direction. The binormal direction is the vertical. The above results can be visualized by taking a coin and rolling it with different initial values of 0 and different initial speeds of the center of the coin. Above a certain

elevation of the coin, if the initial speed is large enough. the coin will roll. Below a certain elevation or if the initial speed is not large enough, the coin will slip and, depending on the initial speed. will exhibit a rolling motion with a very small radius of curvature or just fall flat down.

Examplo 1 0.2

I

released with with aa speed speed of of its its center centerof' of mass mass of' of 5 m/s and at a nutation A disk of radius 1 2 cm is released angle = I 5° and zero nutation rate. Determine the radius of curvature tracked by the center of the disk arid the minimum value of the coefficient of friction required to prevent slipping.

Solution assume the disk to he perfectly thin, so that the radius of gyration is related to the radius by K = R2/2. and Eq. F 10.6.16] becomes We

=

4gp2cotO 6p + RcosO

10.6

ROLLING DISK

which can be written as a quadratic expression in terms of the radius of curvature as

4gcotOp2

6v2p

—

Rv2

cos6 =

[bI [bJ

0

Introducing the given information, we obtain 4g cot 6 = 146.4 mJs

6v

,

—

= bO rn1s

,

'1

Rv cos (3 = 2.898

m/s I

[ci

We solve the quadratic equation for p. The result is a positive and negative root. We discard the negative root, as it has no physical significance. The positive root is p = 1.044 m. precession, the friction force is F1 = 0. so that the total friction Because we have steady precession. force is F2. Also. the normal force is equal to the force of gravity. Introducing the values for u and p into the force balance. we obtain for the limiting case of friction = We

=

[dl [d]

=

can solve this equation for the friction coefficient coefficient j.t j.t to prevent slipping v v-

25

= pg = 1.044 x 9.807

= 2.442

[.1

which indeed is a very high friction coefficient. This is to be expected. because the initial

angle was very low. To try a more realistic case, we set nutation angic 4g cot (3 = 22.65 m/s 4gcot(3 rn/s2

6v2 = ISo I So m2/s2 rn2/s2

Ru2

= 60°, which gives cos (3 = cos6

1.5 rn3/s2

[fI If]

and and

solve solve for p. with the result p = 6.633 m. rn. It follows that the minimum coefficient coefficient of' of friction needed to sustain the no-slip condition is 25

pg

6.633 x 9.807

[gi [g]

An examination of Eqs. [c] and [Ii shows that for this choice of disk radius (12 cm. indicating a small disk) and for other problems when Rip is small, the Ru2 cos (3 term will always he much smaller than the other terms. For such a case only, we can drop the last term in Eq. [hi and approximate the radius of curvature curvature as 3v2

tan 6 2g

[hi

so that the friction coefficient necessary to prevent slippage becomes 2 cot cot(3 6 pc'?

3

Iii

For example. example. when when U0 = 15°, Eq. [IJ [iJ gives the approximate value for = 2.49. and when 0 = 600. = 0.38, which match the results above very closely. This indicates that

for a small disk, the friction coefficient required to sustain no slip is dependent primarily on the nutation angle with which the disk is released. For a larger disk. the speed becomes more important as the factor to sustain the no-slip condition. If we compare the roll of a coin to the free roll of an automobile tire (not attached to the automobile), we have two important distinctions. The radius of the coin is much smaller and the coefficient of friction between the tire and any platform is higher than that of the coin and the platform. Hence, given some initial motion, the tire keeps rolling for much longer distances, a fact known to people who have chased a tire or watched one roll off a racing car. The higher coefficient of friction and the size of the tire keeps the tire rolling more.

579

580

CHAPTIR 1 0 0 QUALITATIVE QLJMATATLVEANAlYSIS ANAlYSISOF OFRicin RicinBODY BoDy MOTION

1

0.7 THE GYROSCOPE

Gyroscopes have traditionally been integral parts of vehicles as navigation systems and as devices that provide guidance by measuring directions and angular velocities. A gyroscope basically consists of a rotor spinning rapidly about its symmetry axis. The symmetry axis is allowed to rotate through a system of gimbals. When there are two gimbals. the outer gimbal permits precessional motion and the inner gimbal, to which the symmetry axis is attached. permits the nutational motion. In essence, the F' frame is attached to the inner gimbal. The location of the rotor on the symmetry axis, axis. the number, weight. and inertia properties of the gimbals, gimbals, as well as as weights. springs, and dashpots attached to the gimbals determine the nature of the motion. As inertial guidance systems for vehicles. gyroscopes provide an an inertial inertial referreference frame that moves with the vehicle. A gyrocompass is designed such that its precession rate is the same as that of the earth. so that it always points in the same direction. Gyroscopes that measure angular velocities are usually designed as singleaxis gyroscopes. Gyroscopes used for navigation are manufactured to very high precision standards. The analysis here can be considered as a gross simplification. From a historical standpoint, towards the end of the 20th century the importance of the classical gyroscopes—those with rotors and girnbals—somewhat diminished. This is because of the development of' of satellite-based satellite-based navigational navigational systems. as well as the manufacture of certain types of gyroscopes out of piezoelectric materials, with fewer moving parts.

In this section we study the free gyroscope. gyroscope, the gyrocompass, and the singleaxis gyroscope. Except where noted, we ignore the mass and inertia properties of the gimbals. For a more complete analysis of gyroscopes. the reader is encouraged to consult the references at the end of this chapter.

10.7.1

FREE GYROSCOPE

A free gyroscope is illustrated in Fig. 10. 18. with L denoting the distance of the centhe rotor rotor from from the the center of the inner gimbal. When L = 0, the gyroscope is ter of' of the referred to as a balanced gyroscope. Such gyroscopes are used for inertial navigation systems. In this case, the rotor can rotate without any precession and nutation. The symmetry axis of the rotor is fixed in rotation. which provides the navigational requirement of a translating reference frame. The orientation of the vehicle is incasured with respect to this reference frame. This measurement is also used to drive servomotors that maintain a platform inside the vehicle in a fixed orientation with respect to the earth. Accelerometers on the platform measure the translational motion of the platform with respect to the earth. It is obvious that the design and operation of such a device requires tremendous precision. When L 0, and the center of the rotor rotor is is above above the the center center of' of the gyroscope. gyroscope, the gyroscope behaves like a spinning top. Unless the initial conditions are specified as those of a sleeping top, top. any motion of the gyroscope involves precession and nutation. When the center of the rotor is below the center of the gyroscope, gyroscope, we have a gyropendulum. such as the one considered in Example 8.5.

107

Tiw

GYROSCOPE

03

.1.,

()

/

Figure 10.18 An interesting case is a gyroscope in steady precession when a disturbance is encountered. To analyze this case, we use the Routhian in Eq. [10.5.8} and derive the equation of motion in terms of the generalized momenta. momenta. The result can be shown

tobe 116 +

ab(1 + cos-O)

—

h-)cosO (a- + h—)cosO —

sin3 sin31) 0

mgL sinO = 0

110.7. 1]

where a and b are defined in Eq. 110.5. l0J. Let us now take the the case when the angle 6 is disturbed from its steady value and analyze the ensuing motion. We can accomplish this in two ways. One is to take the expression br V in Eq. [10.5.91, differentiate it twice, and substitute the value of 0 for steady precession. This approach is based on the developments in Chapter 5, for small motions about equilibrium. A second and simpler way is to replace 0 in Eq. 110.7.11 with + where c is a small parameter and is the nutation angle at steady state. We expand the transcendental functions in Eq. [10.7.11 and retain terms linear in with the result 5jfl cosO(3 = cos + e) cos — E sin sinO sin 1) = sin(Oç + Sifl0s COS Os +

110.7.2] introduce Eqs. I into introduce Eqs.[10.7.2 [10.7.21 into Eq. [10.7.1] and recognize that the terms terms not involving describe the steady precession condition, so that they vanish. After some manipulation. we obtain the linear equation We

+

= 0

110.7.31

581

582

QUALITATIVE ANALYSIS CHAPTER 1 0 • QLALITATIVE

RIGm BODY MonoN

where b2

+

sin2os

4ingL —

110.7.4]

IS the value of the precession rate for slow steady precession. given in which in Eq.{1O.5.3 {1O.5.31]. As the the system system is is conservative, from the potential energy theorem. in Eq. 1 1. As

the ensuing motion will he stable if fl2 > 0. We show that this is indeed so, considering from Eq. [1 [10.5.30J. 0.5.30J, the the critical critical value of the angular velocity for which we steady precession is possible. Considering the definition of b as b = l0.5.30J as rewrite Eq. 1 10.5.30J Ir-— fr—

4mgL cos

>0

110.7.5]

[l0.7.5J with withEq. Eq. [10.7.4]. [10.7.4]. once once motion motion is initiated with steady Comparing Eq. [10.7.51

will be larger than zero. We conclude that a free gyroscope subjected precession. to a small impulsive moment will oscillate about the steady precession position. In reality, as the friction and air resistance take effect, the small oscillation eventually dies out and the motion reverts to that of steady precession. We can actually visualize this by taking a spinning top and putting it into motion with steady precession. Tipping the top slightly will make the symmetry axis oscillate a bit, with the oscillation dying out. In this case, because the apex of the top is not attached, there is energy loss associated with the translation of the apex. This energy loss changes the nutation angle, as well as damping out the oscillation of the symmetry axis.

10.7.2

GYROCOMPASS

The gyrocompass is a device designed to determine the direction of true north (or south). Unlike a free gyroscope, whose symmetry axis is fixed in space. the gyrocompass is designed to always point to the north. This is accomplished by adding a counterweight under the inner gimbal, which results in precessional motion that can be adjusted to match the precession of the earth. Fig. 10.19 shows a schematic of the gyrocompass. The rotor is placed at the middle of the inner gimbal and a pendulous mass tn is attached to the bottom of the inner gimbal. We use a coordinate system e1 e2e3 attached to the earth, where e1 is in the north, e2 is toward the west, and e3 is the local vertical. This system is the same as the xyz coordinate system attached to the earth that we used in Chapter 2. system is The angular velocity of this coordinate systeni =

+ sin Ae3)

[10.7.6]

= 7.292( 10-i) radls. As before, the inner gimbal—the F frame—is obtained by a 3-1 transformation, so that the angular velocity of the rotor with respect to the earth is

where A is the latitude and We is the angular velocity of' the earth,

Of1 ++ 44sin6f2 sin 6f2 + = Of1

cos 1-) +

110.7.7]

10.7

Tiw

GYROSCOPE

e3 (vertical)

I.'

ct;

/ Figure 10.19 Using Table 7.1 at the end of Chapter 7, we find the unit vectors in the E and F frames are related by

e1 =

cos 4)f1

—

e3 =

Sifl 4) cos 012 + sin 4) sin

sin Of2 + cos

110.7.8]

so we can write the angular velocity of the rotor as

48

=

4F o + EwB

+ w2f2 +

=

[10.7.9]

where

= W3

=

We COS

A COS 4) + 0

LV)

=

We COS

A sin 4) COS 0 + We Sjfl A sin U + (/) sin 0

COS A sin 4) sin 0 + We Sjfl A COS 0 + 4) COS 0 + I/I

[10.7.10] [10.7.101

Because the precession and nutation rates are slow, we ignore the kinetic energy

of the pendulous mass, so that the kinetic energy of the system is only due to the rotor. Further, we ignore terms quadratic in We, as the angular velocity of the earth is very small. Recalling that our objective is to see if there is a set of precession and nutation angles for which the rotor axis and spin rate are constant, we consider the precession and nutation rates to be zero. Ignoring also the gimbal inertias, the kinetic energy becomes

T=

+

+

+

+ sin AcosO)]

110.7.111

583

584

CHAPTER

10 •

QUALITATIVE ANALYSIS OF RIGID BODY MOTION

The potential energy is due to the pendulous mass as

V = —mgLsin6 next invoke Lagrange's equations for 4) nutation, and spin rates are constant. We obtain We

for4).

= 00

for 6,

= 0

and

110.7.121 6, noting that the precession,

= 0 A sin 4)cos 0

—

sin A srn sin 0) + mgL cos 6 = 0

[10.7.131 the spin angle, we note that the associated generalized momentum. constant, which we can express as For

+

0 + sin A cos 0) = constant

A sin 4) sin

is

110.7.14]

are interested in the values of 4) and 6 that make Eqs. [10.7.13] valid. For the first equation to hold, either cos 4) = 0 or sin 6 = 0. The case when sin 6 = 0 is not meaningful, as it corresponds to the case where the inner gimbal has not moved. 7r/2. Let us take the We then consider that cos 4) = 0. This is possible fbr 4) = ± 7r/2. case when 4) = IT/2. so that sin 4) = 1. We introduce this value into the second of Eqs. [10.7. 13] and solve for 6. with the result We

cotO =

sin A .

cos A + mgL

[10.7.15] [10.7.151

so that if the gyrocompass is released from rest with the initial nutation angle given in Eq. [10.7.151, it will acquire a precession rate that is equal to the component of the angular velocity of the earth along the local horizontal. We can simplify the expression for the nutation angle by noting that ingL is much larger than /3 1//We, and that the mgL term dominates the above equation. Further, cot 0 is very small, and 6 is very close to 'n12. Introducing the expression 6 =

—

[10.7.16]

where c is small, we can simplify Eq. [10.7.151 to 13,1/lW1) -

sin A A

mg!1

110.7.171

One can show that upon the application of a small impulsive moment, the gyrocompass retains its stability. The translational motion motion of of the the gyrocompass gyrocompass is neglected from the above analysis. It turns out that translational motion of the gyrocompass, especially motion along a curved trajectory. reduces the accuracy of the gyrocompass even more. For this reason, reason. gyrocompasses are more widely used in slowly moving vehicles, such as ships.

10.7

10.7.3

THE GYROSCOPE

SINGLE-AXIS GYROSCOPE

A single-axis gyroscope is a useful tool for measuring the angular velocity of a vehicle in a particular direction. This gyroscope consists of a platform. which is attached to the vehicle, a single gimbal. and a rotor attached to the gimbal, as shown in Fig. 10.20. The symmetry axis of the rotor is normal to the axis of the gimbal. Furthermore, a torsional spring of' constant k and torsional dashpot of constant c are attached to the gimbal, such that they resist the motion of the gimbal. We denote by XYZ the reference frame attached to the vehicle, with the F frame (xvz or attached to the gimbal. rotating about the X axis by an angle 0. The angular velocities of' the vehicle, gimbal, and rotor are

+ wyJ + wzK

Wvehicle = (Of (Of

+ t91

=

= (wx + 6)1 + (Wy COS 0 +

+ /ik =

w =

Sin O)j + (—Wy Sin 0 +

+ 0)1 + (wyCOSO++w7sin0)j wzsin0)j + (—Wy (—Wy Sin sin 0B +

coO +

[10.7.181

We use a Lagrangian approach to analyze the system. To this end, we note that the angular velocities of the vehicle are given quantities and not generalized coordinates. Hence, we have a two degree of freedom system. Ignoring the gimbal inertia and the effects due to the translational motion of ol the vehicle, the kinetic energy has the form

7= 71

=

+

[10.7.191

+ WZCOSO +

V

0))

X.x

z

(Ok. WV

Figure 10.20

x

585

586

10•

CHAPTER I o. QUAliTATIVE QUALITATIVEANALYSIS ANALYSIS OF RIGrn Bony MOTION

The potential energy and Rayleigh's dissipation Function have the form = =

V=

[10.7.20] 110.7.20]

from which we notice notice that that i/i i/i isis aa cyclic cyclic coordinate coordinate and and that that the the associated generalized

momentum iS is constant

= 13(—wysinO + w7cosO +

= constant

[10.7.21]

Invoking Lagrange's equations, we obtain the equation of motion for 0 as kO — + cO ++kO — 11(wy cos 0 +

sin 00 + wz cos 0) sin

Sin sin

[10.7.22] 110.7.221

+ ir,f,(w}'cosO + WZ sin0) =

Assuming that 0 is small and that the spin of the rotor is much larger than the

angular velocities of the platform, we can linearize the above equation to + cO + kO

—

[10.7.23]

Ijth,y

which represents the equation of motion of a damped oscillator. For steady motion

of the vehicle in the X direction, or when thx is either small or it takes place over a short period of time, its effect damps out. The transient effects also die out over time, and the steady-state solution of Eq. [10.7.231 becomes

[10.7.24]

0=

Thus, the gimbal of the rotor is tilted by an angle proportional to the angular yeve-

locity of the platform platform in in the the Y I direction, direction, that that is, is, perpendicular perpendicular to the plane of the platform. By placing three such gyroscopes on each of the XY, XZ, and YZ planes, one can measure the angular velocities of the vehicle about the X, Y, and Z axes.

Returning to the gyroscope in Fig. 10.20. 10.20, if the spring is removed, then one can obtain a value of 0 by integrating the angular velocity. Indeed, using either a convolution integral or a Laplace transform solution, the response of 0 can be shown tO he t9(i) 0(1)

(I

wy((r) wy(1)

1

—

exp

—C(1 —c(1 — —

.00

'1

(r)

do-

[10.7.25]

where we note that wy(t) does not need to he constant. If the the damping damping ratio ratio isis highe high. the second term on the right side of the above equation vanishes rapidly and we have 0(1)

=

'TTçfrt C

.0o

wy(r)d'r wyUr)d'r

[10.7.261 110.7.261

so that the tilt of the gimbal is proportional to the integral of the angular velocity Wy over time. Such a gyroscope is called an integrating gyroscope.

HOMEWORK

REFERENCES Arnold, R. N.. and M. Maunder. Gv,vdvnwnics and Its Engineering Aj;plieaiunis. New York: Academic Press, 1961. Ginsberg. J. H. Advanced Engineering 1)vnarnics. New York: Harper & Row. Row, 1988. Goldstein, H. Classical Mechanics. 2nd cd. Reading. MA: Addison-Wesley. 1981. Greenwood. D. T. Classical Dynamics. Englewood Cliffs. NJ: Prentice-Hall, 1977. Junkins. J. L.. and .1. 1). Turner. Optimal Spaceerqft Rotational Maneuvers. Amsterdam-Ncw York: Elsevier, 1986. Elsevier. Mcirovitch, L. L. lvi Method.s ethod.sof of Analytical Analytical I)vnamics. I)vnwnics. New York: McGraw-Hill. McGraw-Hill. 1970. 1970.

HOMEWORK EXERCISES SECTION 10.2 1. I.

A football (mass 400 g, radii of gyration 4.5 cm. 6 cm) is thrown and at the rad/s, instant considered has an angular velocity about the longitudinal axis of 4 rad/s, and its center of mass has a speed of 100 km/hr. A player tips the football, and begins to wobble with a nutation angle of observers can see that the 15°. Assuming that the tipping of the ball can be treated as an impulse, find (a) the angular velocity and velocity of the center of' mass, (b) the angle

precession as well as spin rates, (c) if the impulse took 0.05 seconds, the magnitude of the applied force by the tip. in Problem 8.39 sketch the body and space cones before and after impact. An axisymmetric spacecraft, with '1//3 = 2.5 has a precession rate of 0.2 rad/s

and

2. 3.

and spin rate ol'0.3 rad/s. It is desired to eliminate the precession by firing rockets attached to the satellite. The rockets exert impulsive moments on the spacecraft. Find the minimum minimum value value of the angular impulse that the rockets need to exert. What is the spin rate afterwards?

SECTION 10.3 4.


SECTION 10.4 5.

Show that Eq. [10.4.22] holds.

SECTION 10.5 SECTION 103 6.

Write a computer program to simulate the motion of a top. Then, supply the program with the initial conditions derived in Section 10.5 and verify the results

587

588

CHAPTER 1 0 •

OF R!Gw R!GwBony Bony MolloN

in Section 10.5. When dealing with the sleeping top. the initial condition of 0 should be taken as a very small number. 7. Use the modilied Euler's equations and derive the relationship for for steady precession. 8. Given a top with the fbllowing properties in = 250 g. # = 4 cm. 1(3 = 2 cm (both about center of mass). L = 6 cm. select the initial precession, nutation, and spin rates such that the top will have the following initial conditions: a. Unidirectional precession = 0.6. 112 = 0.8, 0.8. = 0.9). h. Looping precession (u1 = 0.6, u2 = 0.8. 0.8. U() = 0.75). c. Cuspidal motion (u1 = 0.6. U2 = 0.8. UJ = 0.8). 9. Consider a spinning symmetric top. The apex of the top is on a rough surface. with coefficient of friction Assuming that the normal force and friction forces are the only source for the reaction forces, derive a relationship for the minimum value of ,u necessary to prevent the apex from slipping. Consider cuspidal moi

tion.

SECTION 10.6

Consider a disk of weight 0.5 lb and radius R = 4 in. The disk is released with an initial speed of its center of 2 ftlsec, a zero nutation rate. and a finite nutation angle. Obtain the coefficient of friction required to prevent the disk from slipping as a function of the nutation angle. 11. Consider the rolling disk, rolling without slipping with a constant nutation angle of 750• An impulsive impulsive force force isis applied appliedtotothe thedisk diskininthe the h direction direction at point A. Calculate the maximum value of the impulsive force, as a function of the precession and spin rates, as well as the coefficient of friction that will prevent the disk from slipping.

10.

SECTION 10.7 12.

Consider a free gyroscope and that a servomolor is used to keep the spin rate = = constant. constant. Derive the equations of motion, and identify the integrals integrals of' of the motion. Then, given that the initial conditions are specified as steady precession, for small linearize the equation for 0 for small disturbances, disturbances, and and comment comment on whether instability can result.

13. Consider the free gyroscope in Fig. 10.18. Now. consider that the mass moments of inertia of the gimbals arc at-cnot notnegligible. negligible. The The mass mass moments moments of inertia of the inner gimbal about the F frame are f1. J2, and J3. and the mass moment of inertia of the outer gimbal about the inertial a3 axis is J. Derive the Lagrangian, and the integrals of the motion. Use the motion integrals to obtain a single equation for 0. 14. Consider the gyrocompass in Fig. 10. 19 and derive the equations of motion in the IT presence of gimbal inertia. Then, linearize the resulting equations about 0 = and show that Eq. 110.7.17] holds. .

.

.

.

HOMEWORK EXERCISES

15. Consider the single-axis gyroscope in Fig. 10.20 and derive the equations considering the gimbal inertia. Discuss the effects of the gimbal inertia on the measurement of the angular velocity. 16. Consider the single-axis gyroscope in Fig. 10.20 and that a servomotor is used to keep the spin rate i/i = = constant. Derive the equation of motion, linearize it, and discuss the performance of the resulting angular angular velocity velocity measurement system. 17. Consider the gyropendulum in Example 8.5 and find find the condition for steady precession. The angular velocity of the shaft is 4,. Evaluate the stability of' steady precession.

chapter

DYNAMICS OF LIGHTLY FLEXIBLE BODIES

1 1 •1

Considering a body as rigid is an approximation whose validity needs to be checked at all times. Under many circumstances elastic effects have to be included in the mathematical model. Dynamical systems consisting of both rigid and elastic components have widespread applications. Examples include rotating shafts, spacecraft with flexible solar panels and antennae, and robots with both rigid and elastic links. Interactions and energy transfer between the rigid and elastic motions of a system is of utmost importance. We saw a significant example of this in 1958 in the field of spacecraft dynamics. when the Explorer satellite was launched. The vibration of the antennae resulted in energy translèr between the rigid and elastic motions of the satellite and led to nutational instabilities. In this chapter we analyze dynamical systems that undergo large-angle rigid body motion as well as a small amount of elastic motion. The subject of motion of deformable bodies is commonly treated in depth in vibration hooks and there are several excellent texts available. We do not go into such an analysis in detail here and assume that the reader is familiar with the introductory theory of vibration of continuous systems. We use near linear elasticity theory and assume that the elastic deformation is small compared to the overall dimensions of the body. When the rigid body component of the motion is small, one can use linear theory and the principle of superposition for the entire motion. Small rigid body rotations generally correspond motions, linear superposition of the rigid and to 20 degrees or less. For larger-angle motions. elastic motions loses its accuracy and interaction effects between the rigid and elastic motions become significant. There are a number of ways to derive the equations of motion of a deformable body. One approach is Newtonian, and it is based on force and moment balances. It

591

592

CHAPTEP 1 1 • DYNAMICS DYNAMICS 01' 01' LI(;IITIX LI(;IrrIx FLEXIBLE FLEXIBLE BODIES

uses the geometry of the problem and stresses as a starting point. Another approach is analytical. and it is based on strain energy. Yet another analytical approach is the integral formulation. We use the strain energy approach in this chapter. because it is a natural extension of the analytical approaches we developed earlier. We will see a very elegant application of the extended Hamilton's principle, where both the equations of motion and the boundary conditions are derived simultaneously.

11.2

KINETic AND POTENTIAL SMALL OR No RIGm BODY MOTION

We first consider the case where the rigid body motion is small, so that it is possible to linearly superpose the rigid and elastic motions. The equations of motion can be derived the same way as when the motion is due to the elastic effects only. We first study the kinematics of the structure, and then develop expressions for the kinetic and potential energies.

11.2.1

AND GEOMETRY

Consider an axially long structure, such as a straight beam of length L. We use use an inertial .vvz coordinate system. where the x axis is along the axis of the beam, going through the centroid of the cross section. section, as shown in Fig. 11.1. This axis is refelTed to as the heani axis. As a result of deformation, the beam axis moves and point A on the beam axis moves to point as shown in Fig. Fig. 11 11.2. .2. The position of A* can be expressed as

r(x, t)t) = (x (x + u(x, t))i + v(x, t)j + r(x,

t)k

111.2.11

where the u(x, 1). I). v(x, 1). and w(x, w(x, 1) denote the components of' the deformation in v(x, 1). the x, x. v, and z directions. directions, respectively. The distance along the beam axis to point Ax is denoted by s. similar to the definition in Chapter 1 of the distance traversed along a curve.

V

4.

4,

A*

V

'I p

.v (beam

t

$

I' 3,

il,, p1" LI

a V

Figure 11.1

Figure 11.2

Deformation of beam axis

1 11.2 .2

'_*

—

z—

M

KLNETIC AND AND POTENTIAL ENERGY: SMALL SMALL OR OR No No RiGm RiGm BODY BODY MOTION MOTION

- *

•

4.

V -

*

pze

* _v V

I*

**

x

x

yV

Figure 111.3 Figure 1 .3

*

Figure 1 1.4

In addition to the deformation of of the the beam beam axis. axis, the the plane plane shown shownin inFig. Fig.111.1 1 rorotates, and a differential element around the beam axis assumes an orientation t), O%(x, O%(x, t), t), shown in Fig. 11.3.' We will quantify this rotation by the angles and O-(x, 1). We hence have six variables at each point x to describe the deformation of the beam, the three displacements and the three rotations. Fig. 11 .3 shows the free-body diagrani of the cross section. where there are three internal forces and three internal diagram moments. shown along their components in the x z coordinate system. These are moments, referred to as . 1

Mi..,

Pr.: Axial force Bending moments

P...: Shear forces P..•: Twisting moment

We now introduce additional assumptions that simplify the problem: All deforto the the length length of of the the beam, beam, all all angles angles6.1, mations u. u, v, and w are small compared to and 0-. are are small, small, and all shear deformation effects can be ignored. The first two assumptions are justified by limiting the analysis to beams that deform very little. The assumption of no shear deformation effects is justified analytically by taking a full-blown model of a beam and analyzing the effects of shear deformation. For slender beams, it can be shown that the shear deformation effects become negligible compared to the bending and torsion. To examine the effects of these assumptions, consider an infinitesimal element of the beam that has an undeformed length of dx, whose deformed length becomes d1v (Fig.11.4'). 11.4).The The coordinate coordinate axes ds (Fig. and at the two ends of the differential element are obtained obtained by by the the rotations rotations dO1, dOs, and dO:. Because these angles are very small, the rotation sequence to go from the x z to the x z coordinates is immatenal. Consider now the projections of the differential element onto the the xx*v* ii .6. One can relate the rotation v and x z planes, shown in Figs. 11 .5 and lI angles and Pz' the radii of' curvature of the projections of' and dOs. to ds by the curve on the xv and xz planes, planes. by p

1

= dO-

ds

—=— 1

p.. Pz

.

ds

[11.2.21

'We are using starred coordinates xy'z to denote the coordinates associated with the rotated plane, as we use primes in this chapter to denote spatial derivatives.

593

594

1 11••DYNAMICS OF LIGHrLv FLEXIBLE BODIES

¼.

V

Beam axis

Beam axis

.v*

A*

A4

—y*

FIgure1115 FIgure 1.5

Figure 111.6 Figure 1.6

From Chapter 1 the radii of curvature along the xv and xz planes are given by 1

Li'' U''

p' p.'

(I ++ v'2)312 (1

H'''

I

(I + (1

111.2.3] 111.2.31

in which the prime denotes partial differentiation with respect to the spatial variable x, that is, v'(x, t) = 1)/dx. Because of the small angles and deflections, we approximate theexpressions expressions1/pv lIP' and approximate the and11PZ w", 1"Pz in the above equation by v" and w". respectively, and we approximate differentiation with respect to S with differentiation with respect to x in Eq. 111.2.21. Equating the different expressions for the radii of curvature and integrating both sides over overx. .v.we we obtain obtain =—

c?W(X, -

dx

t)

= —w(x,i)

dv(x, t)t) oz = dx

= v(x,t) 111.2.41 111.2.4]

Hence, the six variables used For the description of the deformation are reduced

to four: u, v, w. and The relations from the force and moment balances corresponding to to Eqs. Eqs. [11 [II .2.4J .2.4J indicate that shear lhrce is the derivative of the bending moment. We next assume that the torsional motion of the beam can be separated from the axial and bending motions. Axial and bending motions are coupled to torsional motions only through higher-order terms. Having separated torsion from the rest of the motion, we now consider the axial stretch and bending in the v and z directions. Thus, the only component of the stress and strain that we consider are and Effects that will be included in the formulation are the shortening of the projection of the deformed member, which results from the curvature of the beam, and rotatory inertia, which contributes to the kinetic energy as a result of rotational motion of the cross-section due to the curvature. A beam modeled this way is referred to as a Rayleigh beam. Table 11 .1 summarizes the commonly used beam models and the assumptions they make. The validity of the assumptions we use depends not only on the physical dimensions 01 of the croSs cross section section of the beam, hut but also on the boundary conditions and the loading. The significance of the shortening of the projection actually depends flot not so much on the amplitude of the elastic motion, but on the presence of large axial loads or rapid rigid body motions. Rotational motion can be viewed as generating a centrifugal force, which can be treated as an axial load. The rotatory

1 1.2

Table 11.1

ENER(;Y: SMALL OR No RIGID BoDY MOTION

KINETIC AND

Comparison of beam models

Effects Included

Rotatory Inertia

Bending Strain

Shear Strain

Shortening of Projection

Euler-Bernoulli

No

Yes

No

Yes or no

Rayleigh

Yes

Yes

No

Yes or no

Timoshenko

Yes

Yes

Yes

No

Shear

No

No

Yes

No

Model

inertia is usually significant when there is rapid rotation about the x axis. We also ignore temperature effects. To illustrate the shortening of the projection. consider a deformed beam. Fig. 11 .7 shows the projection of the deformation of the beam axis on the xy plane. The slope of the deformation along the xv and and xz planes planes is given by Eq. [11.2.41. It follows that the slope of the beam axis at any point x is O(x, t)t) = O(x,

11 1.2.5] 1.2.5] 11

t) + w'2(x, t)

Because shear effects are ignored we neglect any other rotational deformation of the

beam. We denote by s(x, t) = x + e(x, t) = Distance traversed along the beam axis from = 0 to the deformed position A* x x = How much the beam axis has stretched e(x, 1) == How 1) 1) = xx + u(x, 1) = Projection of' the beam axis position onto the x axis

Consider now a differential element of the beam, whose undeformed length is dx and whose deformed length is ds. Fig. 11.8 shows the projection of this differential element on the xy plane. The projection of ds onto the x axis is and it can be expressed as )2

= ds COS 0 = ds

)2

[11.2.6]

+1

\c2x

Beam axis

V

dx

X

'"'id"-

x

I

dx

dç

FIgure 1 1.7

Figure 11.8

1

595

596

CHAPTIR 1 1 •

OF LIGHTLY FLEXIBLE BODWS

The difference between dE and dx, — dx = die. is referred as the shortening of the projection. The nomenclature can be explained by noting that if there is no stretch. du will be a negative quantity. Because the deformations are small, we approximate cos 0 in the above equation by a Taylor series expansion, so that dE

1.c

(i

—

= ds

—

+

I

111.2.7] 111.2.71

ds

Integrating Eq. [II [II .2.7], .2.7], we we obtain obtain t) = x + u(x, t)

fdv\2

s(x, t)

—

1\'rI(T -

.0 J

+

80 )

)

dcr

[11 [11.2.81 .2.81

We introduce the definitions of s and into this equation. Also, because the stretch e(x, t) is small we replace the upper limit in the integral. s = x + e, by x. Hence, we have an expression relating the deformation in the x direction to the stretch and the curvature in the v and z directions as u(x, I)

e(x, 1)

fdv

I

— —

\d0j (

2 -o

dcr

+( \d(TJ

J

[1

1.2.9]

This relation can also be obtained by considering the nonlinear strain displace-

ment relationships. The first and second-order terms in the expression for the strain in in the x direction are are =

au a .

cix

+

I

/ dv

/a

-, -,

+--— i 2 \dxJ \dx J \dx \dxJ/ —

i

i

/ aw

+higher-ordcrterms

, \dxJ \dx ,/ ,

i

111.2.10] 111.2.101

Assuming that the axial defoi-mation deformation isis much much smaller smaller than than the the transverse deforma-

tions, we ignore the as well as any cubic or higher-order terms, Integration of the remaining terms in Eq. 111.2.10] over x leads to to Eq. Eq. [11.2.91. [11.2.91. We We also also observe from Eqs. [11.2.81—111.2.10] [l1.2.81—[1l.2.lOJ that

[11.2.11] [11.2.111

= e'(x,t)

indicating the well-known relation between the axial strain along the beam axis and

the stretch.

To find the components of the strain at any point along the beam, consider .9, drawn drawn for the case of deformation in the v direction only. We now invoke Fig. 11 11.9, the assumption that plane sections remain plane after deformation. This permits one to express the deformation of a point on the beam in terms of the distance of that point from the beam axis. One can show that the beam axis coincides with the neutral axis. The neutral axis is defined as the axis at which there is no stretch or contraction when the beam is subjected to a pure bending load. Be aware that for a curved beam, the neutral axis and the centroidal axis do not coincide. For straight beams, the stretch of a point away from the beam axis is given by

s(x,

y,

t) = s(x, s(x, t)

—

vv'(x, I)

or

e(x,

1) == e(x, e(x, t)

v, 1) v,

—

yv'(x, t)

[1 1.2.121

1 1.2

RIGrn OR No RIGrn OR No

AND POTENTIAL ENERGY: KINETIC AND

MOTIoN MoTIoN

V

Neutral axis ,vx, )') ,v(x, )') v '(x)

0. (.x) I x Vt

.r(x) s(x)

FIgure 11.9 It follows that for the general case where there is transverse deformation in both the y and z directions, the expression for strain becomes V, z,

[1 1.2.131 [11.2.131

vv"(x, 1) 1) zw"(x, t)t) — — vv"(x, t) = e'(x, t) + zii"(x,

law as The stress is obtained by Hooke's law

[11.2.141

o_.I..x = =

with with all of the remaining components of the stress being zero. E is the modulus of elasticity, a material property. 1 1.2.2 11.2.2

BEAMS KINETIC AND AND POTENTIAL POTENTIAL ENERGY FOR BEAMS

We next calculate the kinetic and potential energies for a beam. Differentiating Eq. 111 .2.1]. the velocity of a point on the beam axis becomes -

ih(x, t) t) = ii(x.

t)i + t'(x, t)j +

1.2.151 111.2.151 11

r)k t)k

considerFigs. Figs.11 11 .5 .5 and To find the angular velocity of of aa differential differential element, element,consider To 11 .6. We saw earlier that, as a result of the elastic deformation, the cross section rotates. The angular velocity of the cross section is due to this rotation. The rotation 1) about the y axis. Because angles are 0 = v'(x, t) about the z axis and 0 = — we assume the rotation angles to he small, the order of the rotation is not significant. We then approximate the angular velocity of the differential element as

1.2.16] 111.2.16] 11

i)'(x, t)k —W(x, t)j + 1'(x,

(,)(x. I)

We also ignore the effects of the angular velocity of the differential element on the We velocity of a certain point on the differential element. Hence, the translational velocity of every point on the cross section is approximated as being the same. Definwe can write the translational kinetic ing the mass per unit length at .v as energy as =

•'•

1

—

I

p,(x)is(x, I)•r(x, t)dx t)dx p1(x)i(x, I)•r(x,

—Jo

/

=

'-.0

1) + ,u.(x)(1i2(x. t) + i'2(x, t)

\

r))dx t) )dx /

[112.171

597

598

CHAPTIR 1 1 S DYNAMICS OF OF L1;HTL%' L1;HTLv FLEXIBLE BODIES

To find the rotational kinetic energy we note that the angular velocity has cornponents along the v and z directions only and consider the mass moments of inertia of the differential elements = SL(x)dx. SL(x)dx, = = dx. denoting denoting mass mass moments moments of of inertia inertia per per unit unit length. length. The The rotational rotational kinetic energy has the form i

I1

=

'

1.

{w} fI

.''

1

= =

I

I

•,., i., w+

+

—

\

dx

111.2.181

general. the contribution of the rotatory inertia to the kinetic energy is very small and is ignored. The potential energy is due to the elastic deformation. Noting that, for a beam where the shear deformation is ignored and the beam is not subjected to any torsional loading, all of the stress components except vanish, the potential energy can he shown to he In

2

V

= =

zw"(x, 1) 1)— 1) + zw"(x, — yv"(x, vv"(x. 1)) d(Vol)

d(Vol) = J

J

111.2.191 [11.2.191

can write the differential volume element as d(Vol) = dA dx, where 1A = is a differential area element oF the cross section. Because the beam axis is also the neutral axis, the following area integrals hold We

dA = A ( x) - .4

v ddAA = = . .4

v2dA = I-(.v) I-(x)

j

(

ill

z2dA =

J

: dA f

= 0

vzdA =

111.2.201

where A(x) A(.v)isisthe thecross-sectional cross-sectional area area of of the the beam beam at at point point x. and are the area moments of inertia about the v and z axes, respectively, and is the area product of inertia. If the v and zz axes are selected such that they are principal principal axes,2 axes,2 the product of inertia vanishes, = 0. The reader should not confuse area product of inertia and mass product of inertias. Expansion of Expansion of Eq. Eq. [11.2.191 [11.2.191and anduse useofofEqs. Eqs.111.2.20.1 II l.2.20j yields yields the the following following terms for the potential energy:

.v)[e'(x. x)[e'(x. ()]2 +

V=

EI-(x)[v"(x. t)12 t)J2 + EI-(x)[v"(x,

[11.2.21J [11.2.211

—

The term EA(x) is known as the axial sliffizess, and the

and ELT(x) terms denote the bending stiffness. sti/fliess. When the v and z axes are selected as principal axes. = 0 and one can show through a stress analysis that the internal bending moments and the internal axial forces are related to the deformations by t) = t)

1)

t) = EI-(x)v"(x. EI-(x)v"(x. t) t)

P(x.t) = EA(x)e'(x) I

2See Chapter 6 For cases on how

vanishes due to symmetry.

[11.2.221

1 1 .2

OR No No RICH) Ricm Bony Bony MOTION MoTIoN

KINETIC AN!) POTENTIAl. ENERGY:

The virtual work is due to the external forces and moments acting on the beam.

One can write the virtual work as 1L

3W =

t)(Su

t)3v + Pz(X,

+

+

—

111.2.231 II 1.2.231 in which /)x, in and p.. denote the distributed external forcing (force/length) in the x, y, v, and zz directions, directions, respectively, and and denote the distributed moments (moment/length) about the v and z axes. 1

1.2.3

KINETIC AND POTENTIAL ENERGY FOR TORSION

Next, we consider torsion. In linear elasticity theor , the axial and transverse motions are not coupled to torsion. Hence, we analyze torsional deformation by itself. t). As Fig. 11.10 shows a slender bar subjected to a distributed torsional load a result, the cross section at a distance x.vfrom fromaafixed fixedend endtwists twistsby byan anangle angleO1.(x, t) about the neutral axis. The internal moment at pointx is denoted by Using the semi-inverse method of St. Venant. Venant, we assume that the only stress and strain com(Tv-, ponents are and The components of the stress can conveniently be studied using the stress function formulation. We will not pursue this analysis here. but concentrate on expressions for the strain. Neglecting nonlinear terms, the components of strain have the form =

+

ilU -

=

CIX (IX

+

dii ciZ

and the stress and strain components are related by Hooke's law, law. in which G is the shear modulus, related to the modulus of GE

[11.2.241 = = elasticity E via the

v ratio v by G =

2(1 2(1

+

111.2.251 ii')

A major difference between torsion and bending is that in torsion, the shape

of the cross section affects the nature of the deformation. If the cross section is not circular, warping occurs, where plane sections no longer remain plane hut warp out of their original planes. As hypothesized by St. Venant. Venant, fbr small detbrmations the projection of the warped cross section onto the yz plane coincides with the original shape of the cross section. Fig. Fig. 11 11.11 .11 shows shows an arbitrary point B, at a distance r from the neutral axis and at an angle from the v axis that deforms to point B'. The distance between B' and the neutral axis is still r. Approximating the arclength between B and B' as a straight line we have

w=

11 1.2.261 = We We further assume that the deformation in the x direction is proportional to the rate of change of (x. (x. t) with respect to x. Introducing the angle of twist per unit length 1)/ax, we write t)/ax, 1) = /) V

=

=

u(,v, t)t) = i/i(x, i/i(x, t)J'(v, t)f(v, z)

[1 1.2.271

599

600

Bol)us FLEXIBLE BoDws

OF DYNAMICS OF CHAPTER1111• •DYN.&MICS CHAPTER

A

Neutral axis V

-----------

IL

Figure 11.11

FIgure 11.10

pointx. The The warping warping function is a property in which f(v, funclion at at pointx. f(v, z) z) is is the the warping warpingfunc!ion of the cross section. One can show that the warping function satisfies the relation V2f = 0. in which V2 is the two-dimensional Laplace operator. Calculation of the warping Function lunction is, is, in in general, general, aa complicated complicated task, and beyond thc the scope of this text. Introducing Eqs. [11.2.26] and [11.2.271 to Eqs. [11.2.241. we obtain for the strain Exy =

= uIi( uii(

(9/

+

[11.2.28]

We can then write the potential energy as (a-

+

6x y

d(Vol) d ( Vol) =

:6 )2

(c?J

+1

I

+

2

+v -

dv dz dx

[11.2.29]

+v

[11.2.30]

define the the iorsioii iorsioiiconstant constant J(x) as We define )2

J(x) =

dvdz

and can write the potential energy as

v=

t)]2 dx

[11.2.311

The term GJ(x) is also called the torsional stiffness. It can be shown through a stress analysis that the torsion constant is related to the internal twisting moment by

J(x)

[11.2.32]

= Gui

For a circular cross section there is no warping. so that f(v. f(v, z) vanishes, and considering Eqs. [11.2.301. the expression br J(x) becomes

+ :)dydz ==

J(x) = 11

and J(x) is recognized as the the polar polar area area moment

+ 1-(x)

[11.2.33] [11.2.331

___

1 1.2

KINETIC AND Ik)TENTIAL Ik)TENTIAI. ENERGY: ENERGY: SM1tI4I. SM1tI4I. OR ORNo NoR!Gm RIGm BODY BODYMOTION MoTIoN

calculate the kinetic energy we consider Fig. 11.11 and the velocity of point B, which we can write as To

y, z, t) =

[11.2.341

t)

The differential mass element can be expressed expressed as as din din =

pdvdzdx, in which p is

the density, that is, mass per unit volume. The kinetic energy can then be written as

T=

=

=

+

2

[1 1.2.35] 1.2.351 We defined earlier earlier by by

&1(x) the the mass mass moment moment of of inertia inertia per unit length about the x axis. The explicit expression for has the form P(Y2 + z2)dydz

=

J

= pJ(x). We cannow nowwrite write We can

that for circular rods with constant density the kinetic energy expression as so

T = T =

[1 1.2.361

"'S "'•

1 —

111.2.371

J

The virtual work is

6W =

J

t)60k. dx

[1 1.2.381

Jo

To analyze the general case of coupled axial, transverse, and torsional motion one needs to consider higher-order terms in the bending and shear stresses. The interested reader is referred to the text by Rivello.

1

1.2.4

OPERATOR NOTATION

The kinetic energy and quadratic parts of the potential energy can conveniently be expressed in operator notation. To this end. end, we introduce the displacement vector in which is the spatial domain. For the beam considered above, {°U.} = 01T and [Li [LI V W = x. Recall that the rotational deformation of the beam about the y and zz axes was was expressed by the slopes w'(x, I) and v'(x, t). Hence, we have four degrees of freedom. Note that what we refer to as a degree of freedom here is different from a degree of freedom for a rigid body. Here, the degree of freedom is for each point on the body. perhaps more properly described by the term continuous degrees of freedom. The derivatives of with respect to the spatial variables have the form =

ax

=

[11.2.39]

601

602

CHAPTIR 1 1 • DYNAMICS OF Licirny FLEXIBLE Bonws

Here, we introduce the inner product notation between two vectors

and

as

111.2.40] El 1.2.40]

>

= J and we can express the kinetic energy as T =

111.2.41] 111.2.411

<{4t'},

<{°it}' [M1{°tL}> <{°it}.

and {M1 [M1 I have the form in which the matrices [MI and 0 0

0

[Mi

0 o

—

—

0

o 0

0 0 o

0 0 0 o 0

[M1 1

0

0

0

0

&-(x)

0 o 0

0 0 (1 1.2.42] 111.2.42]

0

The [M1] matrix contains the rotational inertia terms. To describe the potential

energy, we note from Eq. 111 .2.19] that the potential energy is more conveniently expressed in terms of the stretch e, as Eq. [11 .2. .2.19] 19] is is in in quadratic quadratic form. form. We We introduce and the matrices matrices [S01. [S01.[S1 [S11.1. and and IS2]. We can the vector = Ie v w then write the quadratic part of' the potential energy as

V=

<{.W},

+

+

[S1

{S21{W"}> [S21{W"}>

111.2.431 111.2.43] For the beam considered,

{S2]

—

o

o

0

EI-(x)

0 —E11-(x)

— 00 o

0

0

0 0 0 0

[S0] [S01 = [01

EA(x) —

0

—

0 0

0 0 0 0

0 0 0 0

0 0 0

GJ(x) [1 11 1.2.441 1.2.441

The matrix [S0] has nonzero entries in the presence of springs attached to the The body. Also, in the presence of axial forces that can be expressed as part of the potenenergy,[S1 [S1I Ihas hasadditional additional terms. terms. Often, Often, the the notation for potential energy in Eq. tial energy, liii .2.43] is shortened to the energy inner product

V V = =

[11.2.45] 111.2.451

The energy inner product is a quadratic term. Gravitational potential energy is

not a quadratic function, thus it is not included in the energy inner product, but as an additional term.

Example 1 1 .1

I

Write the kinetic and potential energies and the virtual work For the beam shown in Fig. 11.12. which is of uniform density and isis allowed allowed to to deflect deflect in in the the zz direction direction only. only. Ignore Ignore rotatory rotatory inertia.

1 1.2

MoTION ORNo NoRicw Ricii BODI' MOTION SMALL OR KINETIC AND IkTENTIAL ENERGY: SMALL

4.

F

"4

k

C C •0 '

I?

L

Figure 11.12 Solution taken as upward. From Eq. [II .2. 17] The positive z direction for the deflection of w(x, t) is taken the kinetic energy has the form L

t)dx ++

T(t) =

1)

The potential energy is due to the elasticity, the concentrated springs, and the gravitational force. It has the form 1

1)12

+

t))dx

+

± k2[w(L, t)]2)

±

ibi Ibi

in which g is the gravitational constant. To calculate the mass density and area moments of length. The expressions for inertia, we recall that the mass density is the mass per unit length. and

1%.(x)

for a circu Ear cross-secton are

=

,n(x) -

L

= pA(x) =

pirrix)

1(x)

= =

14 (x) =

[ci id

R(1 density. The The radius radius isis given givenby byr(x) r(x) = R(1 where r is the radius and p is the density. '•I

= pITR (i

so that

.v

1(v) ==

2L

—

1TR4(1

—

2L)

Id]

The virtual work is due to the external force F and torque rr. We can write it as cSW

'C' Ic]

= —F&v(a) + TôW'(d)

the disThis expression is obtained by substituting into Eq. [11 .2.231 the expressions for the tributed force and moment =

—

a)

in which (S denotes the Dirac delta function.

t) t) = r6(x

—

Ifi

603

604

01'LI6HTLY LI6HTLY FLEXIBLE FLEXIBLE BOI)ws CHAPTIR 1 1 • I)YN.&MICS I)YN.&MICS OF

1

1 .3

EQUATIONS OF MOTION

In this section. we use the expressions for the kinetic and potential energy developed in the previous section and invoke the extended Hamilton's principle. The process yields the equations ol motion. as well as the boundary condititons. 1

1.3.1

EXTENDED HAMILTON'S PRINCIPLE

Consider the axial and transverse deformations only only and and select select the the vv and and zz axes axes as Consider principal axes. Here we have two choices of variables to use: u, v. and or e, u, and w. Let us use u, u. v. and w. From From Chapter Chapter 4. 4. the the extended extended Hamilton's principle states -t2

6Wdt =

— V(t)Jdt + IT(l) — IT(l)

6J II

[11.3.1]

0

. "1 I''

and12 12are areany anytwo twotime timeinstances. instances. The The expressions expressions for for the the kinetic kinetic and powhere t1 t1 and tential energies are given in Eqs. [11.2.171 and [11.2.211. We ignore the contribution of the rotatory inertia. Taking the variation, we obtain +

+

EA(x)e'

—

—

+

\c9U

EI-(x)v"(5v" +

—

+ p.6W p.6W ++ ,n-6v'

+ nV

+

dxdt =

—

dw

0

[11.3.2] [11.3.21

and 6w 6w only, only, write the the extended extended1—lamilton's Hamilton's principle principlein interms terms of of 6u. 6v, and we eliminate the variations of the derivatives of it, v, and w by integration by parts. Recall that the order of differentiation and variation can he exchanged when the differentiation is with respect to an independent variable, in this case the time t and To

the spatial variable .v. The procedure yields integrals as well as integrated terms. We observe that three types of terms result. Examples of' each type are 1L

.0

in -6 -6 vvddx .v =

m..

.0

= iii:6v •

=

d

dx

(6 v) dx •LL

L

—

—(m-)6vdx

—

(11k

((2

[11.3.31

LI

= =

(rn-)6vdx

LI

• (•)

I

dx

0 o

di

ji(x)ii6u

112 1'2

I, — '1 ti

I

,u(v)ii6udt

J11 Ji,

— 1'

p.(x)ii6udt 11L(x)ji(5udt

=— •

111.3.4]

1 1.3

•L

OF MOTION

L —

w" dx = )w "6w" )w"6 Jo

6w

dX

•

=

1L

I.

dx-

\

w' clx cix

) ,j

d2

+

18w

—

jJo

—

\

dx

111.3.51

Jo

'I

The integrated term in Eq. [11.3.31 vanishes because m.(x) is a moment per

unit length. The integrated term in Eq. [11.3.41 vanishes due to the definition of the variation. Recall that the varied paths for u. v, and w are selected such that they vanish at times times t1 t1 and t2. By contrast, there are two integrations by parts in Eq. [11.3.5], both involving spatial differentiation, and none of the integrated terms disappear automatically. Performing all the integrations and rearranging yields

r,

rh- ri. /

(1

(

j j0 +

/..L(X)V

+

—

dxdX

—

dX

/

IEJ'\

(IV

(IX

(lx

+

dw d w

P(x,

---—

I

+

) ]]

( (I

—

—m- 6v dx (5'X d

+ p- +

dxdt

2

/

dv

+

11

ii (V dv

+

(X)

C)

d

I

) + ç.

2

/

2

1(2

ill —

I

-

/ 49jj

+ ---- P(r. t)---1)---- +

dww

+

+ !EA(X)

(9X dx

(1

I

1

\

\—

P(x,t)---- 6v dx

—

cIt

dff '+ —I(E/ ,+ —I(E/ d.t

•W V

)

P(x,i)

—

(;V2

\

L

jlo

d9x1J

di =

0

[11.3.61

in which we have used the notation r

P(x, 1) = EA(x)

U

dx

1

+ —EA(x) 2

I /'dU (1 V I

.

(I x

-ri

1

I

+I (

w

I

dx)

= EA(x)e EA(x)e'(x, 1) (x, 1)

I

)J

[11.3.71

the internal internal force in the x direction. By virtue of the arNote that P(x, t) denotes denotes the bitrariness of the variations of u. U. vv and and w, w. in order for for the the integrals integrals to he equal to zero. the integrands and integrated terms must vanish, and they must vanish individually. The three integrands in [11.3.61 lead to the equations of motion, and the integrated terms lead to the boundary expressions. We have, for the axial deformation + —

+

=

c? -V

I..

(114

1

/ dv

i. r + I(Iw

+ !EA(x) L—I +L EA(x)-—+—EA(x) dx 2 \d.vJ \dxJ \dx)

)

L

8u

o 0

=0

111.3.8]

605

606

11•

FLEXIBLE Boows

I)YNAMICS OF OF

For the transverse deformation in the

,u(x)U + J.L(X)U

direction ( d

(1 V (1'

d—

dx

v

——-

-

d.vdx-

— —

(9V

(?V2

dx

EI..(x)

—

(I.V'

=pv—

t-lx nIx

(9x

(1L'

(9

(1 (9

P(x,t)---P( .1, t) --—

1??

=0 =

o

dv

P(x,

dxx C)

/.

1)---—

(IX dx

=00 =

111.3.91 11 1.3.91

0

For the transverse deformation in the : direction -, ,LL ii' ,LL (( vx)) •.

+

.

-T1

,

()X

Eli. ( x)

(9

P(x, t)

(IX dx

nVrIX

ii— 'pt' 1

+ fll%, fll%. = pdx dx

L

=00 Ow L = O14'

dx—

(9

I

(-)

L )

—

r9.V -

P(x, t) P(x,t)

dx

111.3.101

= 0 0

In a similar we obtain the equation of motion and boundary expressions for the torsional vibration. The results can be shown to be (x (x )

11.3.2

—

(IX

[G J .v)

dx

=

11! L

(?x ] (2.1

= 0

(11.3.11] 111.3.111

BOUNDARY

The boundary conditions are ascertained from the boundary expressions by examinexarnining the geometry at at the the boundaries. boundaries. Fig. Fig.11. II . 13 13 shows commonly encountered bound-

ary conditions. In each boundary expression there are two terms. One term is a variation of a deformation (or variation of a slope), and the other term is an internal force (or moment). By examining the geometry. one determines which one of the terms is zero, and identifies the boundary condition. This procedure is basically an application of a fundamental principle from structural mechanics:

At any point on the body, if the displacement (slope) is known, then the internal force (moment) is not known

at that point; and if the loading force (moment) is known, then the displacement (slope) is not known at that point, unfIt th0 problem is solved.

This principle is applicable at the boundaries as well as in the interior. Confining the

analysis to the boundaries, if the beani beam is restrained to move in a certain direction at a boundary. then the associated boundary condition is geometric, as it is determined from the geometry of the deformation. For example, if the beam is fixed at x = 0.

11.3 1 1.3

EQUATIONS OF MOTION

Fixed end

Pinned end

Figure 11.13b

Figure 1 1.13a I,.

Guided end

Free end

Figure 1 1.13d Figure

Figure 1 1.13c Springs at

boundary

Figure 11.13e the boundary conditions become

u(O,t) = 0 v'(O, 1) = ()

v(O,t) = 0 w'(O, 1)

=

0

w(0,t) =

t) =

0 0

111.3.121

Din chief chief type Such boundary boundary conditions are known as essential, or geometric, or or Din Such

boundary conditions, or boundary conditions of the first kind. When the boundary conditions are essential, it follows that the force and moment balance terms, which are unknown. Furthermore, 6w'. and '5v, Fw, 6v'. 6v', 6w'. are the coefficients of 5u. '5v, the magnitudes of these internal forces and moments do not vanish: if a point is restrained to move in a certain direction, there must he a force or moment at that point preventing the structure from moving. The magnitudes of the internal forces and moments can only he determined after the system differential equations are solved, as stated in the fundamental principle above, We will refer to these internal reactions (the coefficients of' 6u, 6v, 6w, as complementary boundary conditions (CBC). The reason for 6v'. and that. when constructing approximate solutions by series exdefining these terms is that, pansions, trial functions can he selected that violate the force and moment balances at the boundaries. Unless the CBC arc considered, the approximate solution may incorrectly set the internal forces and moments to zero at boundaries and interfaces. The net effect is use of an incomplete set of trial functions and slow or lack ol convergence. Table 11.2 gives the geometric boundary conditions and associated CBCs. When the boundaries are unrestrained, the variations of the deformations and their slopes do not vanish. It follows that the coefficients of the variations must he

zero. For example. ii' the beam is free to move in the v direction at x = L, the

607

608

TabI.

CHAPTER 1 1 • DYNAMICS OF LIGHTLY FLEXIBLE BoI)IEs

1 1 .2

boundary conditions and complementary boundary conditions

Geometric Boundary Condition (Known)

Type of Boundary

Geometric Boundary Condition

CBC (Unknown Quantity)

CBC

Transverse deformation (v) V == o

Fixed end

(El- (9

=0

Pinned end

=0

)—

(1.V

El

—

1\

= ()

Gwded end

—

Axial deformation (ii)

ii =

Fixed end

EM

0

+ +

Torsional deformation (Ok) Torsional (0) =

Fixed end

GJ .v )

associated boundary associated boundary conditions become = .V X

1.

ci .V

P(x,i) P(x,i)

dx —

J

dx

= 0

[11.3.13a,b]

x—L

These boundary conditions are known as dynamic or natural. Equation 1 .3.1 3aj shows a type also called the Neumann type. or a boundary conthtion of the second 11 .3. 113b kind. Equation I 11.3. 3b1 I isisalso alsoknown knownas asaa boundwy boundwy ('Ofl(IitiOIl ('Ofl(IitiOIl oJ the third kind, as it involves not only the shear force hut also the transverse due to the axial motion. Such boundary conditions are also encountered in the presence of springs at the boundaries. Natural boundary conditions indicate that the and moment balances are zero at the boundaries. This is intuitively expected. because, if the structure is free to translate (rotate) at a boundary. there should be no force (moment) at that point to restrain it from moving.

1

1.3.3

SIMPLWICA'IIoN

Equations [11.3.81—111.3.101 can be simplified in many cases by ignoring the deformation in the axial direction, setting the axial stretch e(x, t) and all its derivatives to

zero. The simplification is valid when the axial internal fbrce P(x. t) is not large or not a function of ume time (or it is a slowly varying function of time). Typical examples include the helicopter blade problem and the buckling of columns. Note that once the axial deformation is ignored. u(x, 1) is no longer an independent variable. From

1 1.3

EQuATIoNs OF MonoN

Eq. [11.2.9]. setting e(x, t) = 0 we obtain 1

u(x,t) =

(ivy

LV

——

/ \2 1(1W \

+1

[11.3.141

do

I

\ckrJ

2 J ()

The time derivative of u(x, 1) is also usually ignored. The force P(x) now becomes the internal internal axial axial force force that that balances balancesthe theexternal externalaxial axialforce forceP.v(X). A simple simple force balance in the axial direction (or a reexamination of Eq. [11.3.81) indicates that (1

— P( x) = dx

— p1. (

111.3.15] 111.3.151

x)

which basically states that the external force (per unit length) is balanced by the derivative of the internal force. The equations of motion for the transverse deflection become +

p(x)%i'

a2

82v

clv

ax2 dx2 x)

—

[1 1.3.16a1

—

aw (9w

a

a2

+

= P'

(IX

a —Hi— (IX

= Pz +

ax dx

Ô d

[1 11 1.3.16b] 1.3.16b1

When one ignores the deformations due to axial elasticity from the beginning, the equations of motion can be derived more simply. One can treat the effect of the internal axial force P(x) in two ways: as part of the virtual work or as added potential energy. Consider the virtual work expression due to the external axial load (x) 1L

6W=I

111.3.171 111.3.17]

(x)6i.i dx

Jo

Introduce Eq.

1 .3.15] to this equation and integrate by parts to obtain

rt d 1P(x)6udx = —P(x)6u —J

L.

0

rL

+I

fL

P(x)6u'dx =

dx

Jo

[11.3.18]

The integrated term disappears because at a fixed end u vanishes, and at a free end the internal force P is zero. Differentiating Eq. [11.3.14], we get the expression U (x, (x, 1) 1) as

(dv \ lay 1—1+1 I+11(914' \dxJ

u(x,t) =

-

111.3.19]

so the virtual work expression becomes p)

6W=

1

\_

1+1 \(9x) \dx)

1

(1X=—

+

1x

.0 [1 [11.3.201 1.3.201

609

610

CI4APTIR

1

1•

DYNAMICS OF Li;wnx Li;irrix FLEXIBLE BODIFAS

The effect of an axial force can also be expressed as a potential energy contrihution due to the internal axial force P(x) acting through the shortening of the projection. Recall that that P(x) is is taken taken as as positive positive along the x direction and the shortening of the projection dx — d4 = —du is positive along the negative x direction. We hence write the additional potential energy as P(x)(dx

=

—

[1 1.3.211 [11.3.211

dE)

0

introducing the expression for dii from Eq.[11 .3.19] into the above equation yields Vadditional

=

I

P(X)

+

[1 1 .3.221

clx

J

Introduction of Eq. [11.3.221 {lI.3.22J or orEq. Eq.[11.3.201 [11.3.201 into into the the extended extended Hamilton's prin-

ciple yields Eqs. [11.3.16]. When the axial loading is not large or is not present. the P(x) term is eliminated and one arrives at the well-known Eu/er-Bernoulli equations. If P(x) is tensile, then according to Eq. [11.3.22] the additional potential energy

is a positive quantity. Considering that the potential energy is strain energy. representing the stiffness or resistance to deformation, a tensile force can be considered as one that adds to the stiffness. This increase in stiffness has a stabilizing effect, a phenomenon readily observable, such as in the helicopter blade. By contrast, a compressive axial force reduces the potential energy, thereby making the component less stiff. A classic example of this is the buckling of columns. Columns become unstable and collapse when the applied compressive load becomes too high. The above derivations are in terms of an internal axial force along the x axis. Sometimes it is more convenient to describe the internal axial force along the beam axis (x* direction). Consider an external force which generates an internal force F(x) along the beam axis, axis. as depicted in Fig. 11.14 for a two-dimensional system. We resolve the internal force into its components along the .v and v axes as

F(x)cosA-i + F(x)sinOj F(x) = E(x)cosA-i

F(x)i ++ F(x)v'(x, F(x)v'(x, i)j

[11.3.231

The contribution of the component in the x direction is the same as in Eq. [11 .3.161,

with F replacing P. The contribution of ol' the the component component in in the the v direction direction can be obtained by noting noting from from Eq. Eq. [11.3.15] [11.3.15] that thatF(..v) F(x) isis related related to to the the distributed distributed external load by Pt(X) = —F'(x). Hence, the component in the v direction can be expressed * V

F(x)

C)

Figure 11.14

V

1 1.3 as

FIQUATIONS OF MOTION

an additional distributed external force as

[11.3.241

= —F'(x)v'(x) From Eq. 111.3. 16aJ, the total contribution of F(x) becomes £9

dx

(IV

F(x)-—— + F(x)---+

= F(x)

dx

(1V

— F (x)v'(x) = F(x) + F'(x)v F'(x)v (x) (x)—

.,

(1x (1X

[11.3.25] In a similar fashion, we obtain the contribution of F(.v) in the

z

direction. Conse-

quently, the equations of motion, Eqs. [11.3.161. have the form /.L(X)V +

(7 V

a

- F(

I

-7

1

(9-.

+

)

-)

-,

(.1—w

•1

dx—

1 1.3.4

= Pv f)

—

•1

£9

[

—

,

rix—

1(x)

(

—

— in fl1—

d —Tn V

(9.

0x

[1

-

OPERATOR NOTATION

It is convenient to express the equations of motion of a flexible body in terms of mass and stiffness operators. especially when the equations are linearized. Let us consider The an elastic foundation under the beam, modeled as a distributed spring added potential energy due to the spring is

1j 1

t)dx

Vspring

111.3.271

J

and the effect on the equation of' motion in they direction is an additional

t)

term on the left side of Eq. [1 L3.16a]. is the spatial domain. For in which Recall the displacement vector = x. x. Equations the simplified beam model considered above {01L} = [v wIT and [11.3.161 (with the added term due to the elastic foundation) can then be expressed in operator form as

[11.3.28] [1 1.3.28]

= {F}

+

is the matrix stiffness operator and .M is the matrix mass operator in the

in which form

( —

[

dx—

ax cLv

P(v)

(7 .

+

0

C1X

d

d

0

—

0

—

0

a

— P(x)—

[11.3.29] [1 1.3.29]

61

1

612

CHAPTER 1111 • DYNAMIcs CHAPTER DYNAMICS

LIGHTLY FLEXIBLE

The and {F} is the forcing vector in the form {F} = Ipx, — p- + operators are diagonal because the v and z axes are selected as principal axes. The equations of motion in the z and y directions are uncoupled. unless P(x) is a function

of both v and w.

We observe from the above equation that the mass operator is nothing but the mass matrix in Eq. [11.2.42]. and the stiffness operator can be expressed in terms of the [Se] (i = 0, 1, 2) matrices as =

,1 (i •

clx-

/ d (ItS2]. I

-

()

I

/

/

(1

111.3.301

+ [S0] [S0] (is1 , )— --—(1S11-—-

dx

dx \

clx-

This expression can be derived from the kinetic and potential energies. And the boundary conditions conditions can can also also be be expressed expressedin interms termsof ofthe thematrices matrices[.S1j (i = 0, 1, 2). and .M have have the the property property of being selfFor conservative systems, the operators and that adjoini. which is defined as follows. Consider two functions {(k)(9,)} and satisfy all boundary conditions and are as many times differentiable as the highestorder derivative in the stiffness operator. Such functions are called comparison junc/uncself-adjoint if the following relationships tions. The operators and At are called seif-adjoint hold:

<

> = < {F(9)},

>

111.3.31] 1.3.31] 11

<

>=<

>

11 1.3.32] 11 1.3.32]

Also, for two comparison functions the following relationship holds: Also,

[1 1.3.33] [1 1.3.33]

>=

<

Another important group of functions is admissible junctions. These satisfy only

the geometric boundary conditions and are half as many times differentiable as the highest-order derivative in the stiffness operator. The relationship relationship [11.3.3 [11.3.31] 1] cannot cannot be written for admissible functions. However. Eq. [11.3.321 holds. Also, one can write the energy inner product for two admissible functions =

Example 1 11 .2 .2 1

{e(9;)}jJ

[1 1.3.34]

Consider the beam in Example 11.1: derive the equation of motion and identify the boundary

conditions.

Solution We will take the kinetic and potential energies from Example 11 .1 as well as the virtual work, obtain their variations, and perform the necessary integrations by parts. We can write the kinetic energy as

T=

x, x,

0

Ia]

in which the mass operator jtt has the form .11.

+ M5(x — Ii)

[bJ

1 1.3 13

EQUATIONS

0' MOTION

The stiffness terms can also he expressed as S2

=

S1

=

So = k16(x

—

—

h) + k2S(x k2S(x— — L)

[ci

Note that the contribution due to the weight of' the beam is riot in quadratic form. It is more convenient to treat the weight as part of the external forces, in the virtual work. The potential energy is L

I

V = —1J

+ k16(x — — b)w2(x, b)%v2(x, t)

t)12

I

,1

+

Id] [d]

c)w ix, k7S(x— (x,I)t)++k26(x — L)w2(x, L.)w2(x, t)} dx

—

We can express the virtual work as L

x

d)6w' d)6ii"

a )8 u' + r6( .v .v — —

)g6 wl dx dx

1.1 I.1 = The variation of the kinetic energy is straightforward, as all integrated terms drop out due to the definition of the variation. In the virtual work expression, the only term we must integrate by parts is due to the torque. Integrating by parts gives (5 W

—

I

—

1L

—

-L TO ( x TO(

IJ

Jo)

—

d ))6 d 6 w'dx dx =

rS '(x '( x

—

—

d )cSw )6w d dxx

[1 1I If

—

Taking the variation of the potential energy and performing the necessary integrations by

parts. the individual terms become •L

/

dx

(El (1 .k

- (J

(1— 1's' (1— W

dx

—

-

6

k1 88(x k1

-

) w2 (

x, r) dx

J

S (x

—

b)w8 w dx b)wFi

1)

1

[1JL '2

k•1•6 .v x —c)

x, 1) 1) d

—

in

.il =

k78(x k7'8(x

— C C ))

w

6 w' d ..v .v = I — .0

.fo

L)w2(x,

t)dx] = t)dv]

—

(91

t)] = k2 w( L. t )6 w( L)

(5

[gI

Note that we treated the spring at the end differently than the springs in the middle. This is because the variation of the displacement at the end x = L enters the boundary terms. Performing ün in Eq. Eq. Eel, Eel, we obtain r

6W=I

L

—F6(x

it) Jo

—

a) a) — —

( a

dx

[hi

—

We next invoke the extended Hamilton's principle. For the equation of motion, which is the sum of all the integrands. we obtain + M6(x

=

+ k16(x

+

—

—[T6(x—d)]—/.Lg --

b)w

—

[ki6(x

O<x
—

[II

613

614

CHAPTER 1 1 • DYNAMICS OF LIGHTLY FLEXIBLE BODIES

and the boundary terms become r

'E'lv

Ow —

I

At the end x =

0.

+

¶

(I.1

(?X

J

[ii Ii'

w( L. t )S w( Id) = 0

()

the beam is fixed, so that the boundary conditions arc geometric and

they arc

w(0,t) w(0, t) =

0

%i"(O, t)

[k]

= 0

fixed. hence their variations are At the end x = L. the displacement and slope are not fixed, not zero. The coefficients of and eu" must vanish. The boundary conditions are recognized as

.ITL

=0

[I, m]

= kw(L, I)

(?,V

/

r — — L L

These arc natural boundary conditions, with Eq. (mj being a boundary condition of the third kind. This boundary condition represents the force balance at x = L. where the spring force counteracts the shear force.

Example

Given the uniform rotating rotating blade in Fig. 11 find the find the equations equations of motion.

11.3

.

15

spinning with the constant angular velocity

Solution solving this this problem. One way is to consider that the rotation of There are several ways of solving

the blade causes a centrifugal force. Another is to consider a relative frame to which the base of the blade is attached. Using the first way. the external forces acting on a differential element at-c shown in Fig. 11. 16. Note that the .v axis is the original undeformed axis. and the .v*axis x* axisisisalong along thc thc elastic elastic curve. curve. The The external external forces forces acting acting on on the the differential differential element are dx and the centritugal force p.1dx dx = ,..tft-x dx. The total the force ol gravity dx = centrifugal force acting at any point x can he obtained by integrating the centrifugal force on

V

ci £LV-

*

V

.v

p L

Figure 11.15

pg dx

Figure 11.16

___

1 1.3

EQUATIONS

OF MoTIoN

element (from the free end to point x)

the

= J.L

P(x)

= —

[a]

—

J JL

Introducing Eq. [a] and the force of gravity into Eq. [11 .3.1 .3.16a], 6a], we obtain obtain the equation of motion a

ci4V

.,

[bi

We next derive the equation of motion for the rotating blade using a rotating reference frame approach. We attach a reference frame to the undeformed position of the beam. with the z-axis along the vertical. The position of a point can be expressed as u(x, 1)11 + v(x, t)j r(.v,t)t) = [x ++ u(x, r(x,

[ci

The .vvz frame is rotating with constant angular velocity point can he expressed as

so that the velocity of a

iix, t)t) = ii..v.

t)i +

X X

t)j +

r(x, t) ii(x, t)i + t) = ii(x,

t)j

+ u(x, r)]k [dl

—

expression leading to the kinetic energy energy expression

T =

1

—

'

(x.

t)•

JU

=

1)

t) +

t) + Q2[x + u(x, t)]2)dx

1) +

= 1) +

t))dx

t) +

+

[.1

Because angular velocity of the blade is large, the term 1) is much smaller than Because the angular and we ignore it. The term 1) is quadratic in u and can any term containing containing IV, and be ignored as it is much smaller than the 1) term. ignoring the axial elasticity, we write u(x. u(x, 1) as

u(x.t) =

'a

I

——

I

2 Jo

\2

\d(TJ

dr

[f]

1), which, in light of the above It follows that the kinetic energy has one added tenm tenm 2IVxu(x, 2Wxu(x, t).

equation, has the form

j/

1-i

Tadditic,nal = Tadditic,nal =

— —

-

J0

xJ

L) V

) dcr do dx

igi Igi

Note that the term in Eq. [ci does not affect the equations of motion, as its derivatives with respect to the variables u and v are zero. zero, By applying some integration trickery. Eq. {g] can he expressed as =

-

=

__/hfl-J 4

f

(av)2

(L (L

JL

dodx =

—xidx —xidx

—

2

)2 )2 1L

o do dx

Ihi

615

616

Licirnx FLEXIBLE FLEXIBLE BoDw.s BoDws CHAPTER 1 1 • DYNAMICS OF Licirnx

The potential energy is 1L (L

v=3J

/

() 0

—

Ii'

dx \dx.., )) dx

EL(..v) (

Integrating by parts gives the variation of the additional kinetic energy the form (L2

=—

= =

—

-

1

dx

v2) L

—

\(?X

/

L

+ +

10

/ -i?[(L2 (L [

- x)(-\(9.lJ —

ijj

i

and at at x = L. the coefficient boundary terms drop out because because at x = 0. v(0, t) = 0, and The boundary Eq. [a] [a] we we can express Eq. Iii as L2 — x2 vanishes. Invoking the definition of P(x) from Eq. (I.

= =

I

(31

y

[k]

cix J Jo

Yet another another way of obtaining the equations of motion is to use Eq. [11.3.221 and treat the Yet

axial force P(x) in conjunction with the shortening of the projection. In this case, we have the additional term in the potential energy. (.1 v

1

Vaddttional =

210

,

)2

P( x)

dx = dx

-

2

J

(L22 (L

/ \ , f(')Ll

—

.t)L---—

I

\cIx) \dXJ

III

d.v

in Eq. [1]. [1]. which is to be expected.

IhiIhiisisthe Taclditjoiiat in Eq. thenegative negative of Taclditjoiiat in Eq.

11.4

I

EIGENSOLUTION AND RESPONSE

In this section, we summarize the solution of the eigenvalue problem associated with linear, seif-adjoint continuous systems and obtain a general form for the response by modal analysis. Consider the simply supported beam in Fig. 11.17 and assume that and transverse deformation in the z direction are negligible. the axial and that the axial firce P(x) is zero. The equation of motion becomes /L(X)ii /.L(X)V +

dv

C)

ax -

O<x
[11.4.1] 111.4.1]

The boundary conditions are due to the deflection and moment balance being equal

V

EL(x). /1(X) -V

.v

L

Flgur. 11.17

_____ ______

1 1.4

EIGENSOLUTION AND RESPONSE

to zero at both ends. thus

v(x,t) =

atx = 0,x =

= 0

0

L

111.4.21

We will seek a solution by modal analysis. The procedure that we will use is very

similar to the eigensolution of a multidegree of freedom vibrating system in Section 5.5. The reader is urged to compare every step here with the procedure in Section 5.5. We first consider the homogeneous problem (no external excitation and no explicitly time-dependent boundary conditions) and use separation of variables to express the deformation v(x. t) as v(x, 1) =

111.4.31

where is the amplitude function and denotes the time dependence. Introducing Eq. [II [11.4.3] .4.3] into into Eq. Eq. [11.4. [11.4.1] 1] and setting = 0. = 0, we obtain +

(

(p. ci

=0

dx-

[11.4.4]

the boundary boundary expressions expressions in in Eq. Eq.[11 [11.4.2]. Introducing Eq. [II .4.3] to the .4.2]. we obtain the boundary conditions that has to satisfy as

-'ii 0 dx—

= 0

atx

0,x == LL

111.4.5]

In order for Eq. [11.4.4] to have a nontrivial solution, we must have

=0

111.4.61

which is which is aavariable variablecoefficient, coefficient, ordinary ordinary' differential differential equation equation of order of order four. four. Equation Equation [11.4.6]. subject to the boundary conditions [11 .4.5], constitutes the boundary value problem, where one needs to find the values of 1k that lead to nontrivial solutions of Eq. [11.4.6]. The solutions A are called eigenvalues, and the corresponding functions 4(x) are called the eigenfunciions. To demonstrate the solution procedure. we further simplify the problem by assuming that the beam has a uniform cross-section, setting = = constant, = El = constant. Expecting A to be pure imaginary, we write Eq. 111.4.6] as

where

f34

=

—

A2

= 0

111.4.7] [11.4.71

,i/El. The ,u/El. The genera] genera] solution of ot Eq. [11.4.71 [11.4.7] is

çb(x) =

c1

sin/3x + C2C0S/3x + C3 sinhf3x + c4coshf3x

111.4.8]

where the coefficients c1(i = 1, 2, 3, 4 are found by substituting in the boundary conditions. Note that because the right side of Eq. [11.4.7) is zero. f(x) can only be determined to within a multiplicative constant.

617

618

OF LIGHTLY FLExIBLE BoI)lEs OF

CHAPTER 1 1 S

First, take take 4'( çb(x) and evaluate evaluate at x = 0. This gives c2 + c4 = 0. Then, take 4"(x) x) and 0. This gives —c2 + c4 = 0. from which we conclude that and evaluate evaluate itit at at xx = 0. = c4 = 0. We repeat the same procedure for the boundary x = L. Doing this, we end up with c3 = 0. and the relation that remains is

[11.4.91

c1Sifl/3L = 0 c1sifl/3L

For Eq. [11.4.91 to have a nontrivial solution we must have

111.4.101

sin/'3L = 0

The characteristic equation has an inwhich is known as the characteristic equation. The finite number of solutions. For the problem at hand, the solutions arc )31L = rir(r = all the eigcnvalues A,. are pure imaginary. 1,2,...). 1, 2, . .). Considering the definition of From Eq. [11.4.31 we conclude that the motion is oscillatory. Defining the natural trequencv Wr trequency Wr = —iAr(r = — 1), we can write .

Ar = i(rir)2

El

(r'7T)2 Wr = (rn-)2

,iL4

El

r = 1,2,... 1,2,... 111.4.111

= C4 = 0. the eigenfunction corresponding to each eigenvalue and (or natural frequency) is f,.(x). and Since c-, c-, =

C3

th(i)

= Ar

f3rX = Ar

Sifl

r,7rx r,7rv

L

1.4.12] 11 111.4.121

where A,. is an arbitrary constant. The solution to the cigenvalue problem yields the shape of the eigenfunctions uniquely, but not their amplitudes. It is convenient to normalize the eigenfunctions. A commonly used normalization scheme is with respect to the mass distribution, thus =

1

111.4.131

Jo

Using Eq. [11.4.131, one can show that for the pinned-pinned uniform heani, Ar = .j2/,uL. The first three eigenfunctions are plotted in Fig. 11.18. We note

r

Figure 11.18

First three eigenfunctions of a simply-supported uniform beam

1 1.4

E!GENS0LUTION AND RESPONSE

that except for the first, the cigenfunctions cross the x axis and the number of zero

crossings increases with the mode number. The points of zero crossing are called nodes. It can be shown that the rth mode mode hasr — I nodes. Another very interesting and useful property of the eigenfunctions is that they are orthogonal to each other. The orthogonality results from the self-ad jointness of the mass and stiffness operators. For the beam problem at hand, the orthogonality relations can be expressed as -

L

= 0

.0

when ,

s,

r, s =

1,

2,

L

h(x)

J

dx-

LIx-

/

dx =

when r

0

[11.4.14]

s

Considering the normalization scheme in Eq. [11.4.13], one can combine Eqs. [11 .4. .4. 1133 and 111 .4. 1 4] to obtain the ort/ionor,na/ relations orthonor,na/ il-v itv relations J

=

f

o

1.1.

Ia

dx =

dx

dx— clx—

[11.4.1 5a,bJ

where

was defined earlier as the Kronecker delta. The eigenfunctions constitute a complete set. That is, any comparison function Y(x) can be expanded as a uniformly convergent series of the eigenfunctions as Y(x) = The coefficients

111.4.161

ar can be found by multiplying Y(x) by

integrating along the domain and invoking the orthogonality relation 111.4.1 5aJ. We then have X

( X)

.0

(.v) Y ( i) clx dx =

= Jo

=

( X)

Y

r

(

x) dx

L

a,. dr

I

dx =

Jo

Y

= = aç

[1 [11.4.17] 1.4.17]

so that we can write -L a,.

= (') 0

Equations[11 Equations [11.4.1 .4.16J 6J and I

I

11

[11.4.18]

.4. 1 8] describe the expansion theorem. Note the sim-

ilarity of the expansion theorem to Fourier series expansions or other orthogonal series expansions. One very important application of the expansion theorem is in derivation of' of the modal equations of motion. Here, the function to be expanded is

619

620

FLEXIBLE Bonws Bonws CHAPTER 1 1 • DYNAMICS OF LIGHTLY FLEXIBLE

the

displacement profile v(x,

1)

v(x, I) =

[1 1.4.191 [11.4.191

>

time-dependent terms. We refer to as the modal coordinates. Introducing this expansion into the equation of motion. Eq. [11.4.1 multiplying with integrating along the domain, and invoking the orthogonality conditions, we obtain The coefficients of the eigenfunctions.

I

.,,

d-

,u(v)v + ,u.(v)v

-

0

are

.,

(1

-,

I

dx-

I)

j— J—

-——-m_(x, 1)

—

dx

rL

92

=

+ r

dx -

,

J

dx .0 —

>

+ '>

[11.4.201

—

r=I

r=I Where

-1.

(I) = is

.0

.x, I) — x) p (( .x, ((x) —

d -———

dx

in in (x. (x.I)I)

dx

[1 1 .4.2 1 1

called the inodal/orce. Hence, we obtain the modal equations of motion as +

=

s = 1,2,

.

.

.

[1 1.4.221

The initial conditions associated with each mode can be found by applying the

expansion theorem to the initial conditions v(x, 0) as x

x

V(X, V(X, 0)

>' r=I

[11.4.23J [1 1.4.23J

i'(x, 0) = r=1

Multiplying the above equations by integrating along the domain, domain. and invoking the orthogonality conditions, one obtains the modal initial conditions 0) dx

= .0

=

J

Jo

/.L(x)çbç(x)1:(X, 0) dx /.L(x)çbç(x)i'(x,

111.4.241

Once the modal equations [11.4.221 [II .4.221 are solved, the response can be obtained by substitution of the modal responses into Eq. [11.4.19]. In a real situation, one can integrate and sum only a finite number of modal equations. This raises issues about

the number of modal equations of motion to retain in a specific problem and the participation ol each mode to the motion. We begin analyzing this issue qualitatively. by making use of the asymptotic behavior of the eigensolution.

I I .4

F'JGENSOLUTION AM) RESPONSE

An extension of a theorem from Courant and Hubert states that as the index

r

the asymptotic 4)r(X) are governed by —'

behavior of the natural frequencies çbr(x)

'-'-ar

r=

— f(Drx)

Wr and

1, 2,

.

.

.

eigenfunctions

[1 1.4.25]

in which 2p is the highest-order spatial derivative associated with the elastic motion.

The constants C and D depend Ofl Oflthe themass mass and and stiffness properties. properties.When When pp = I. such as in axial or torsional deformation. the natural frequencies increase with arithmetic progression, and [(Drx) = Ar sin(Drx) + B,. cos(Drx). When p = 2, such as in the beam equation. the natural frequencies increase in geometric progression. and f(Drx) = A,. A,. sin(Drx) sin(Drx) + B,. cos(Drx) + A. sinh(Drx) + cosh(Drx). In the simply supported (pinned-pinned) beam we discussed, p = 2 and and w,. w,. = (nT)2 all of the natural frequencies increase with geometric progression. When studying beam vibrations, retaining the first four or five modes is usually sufficient for an accurate mathematical model, except for cases when the initial conditions on certain modes are large and the external forces and moments excite higher modes more than the others. By contrast, when studying axial and torsional vibrations, because the asymptotic behavior of the rth mode is governed by r, the amplitude of' of each mode falls off much slower. Hence, one must retain a larger number of modes For an accurate representation of the motion. A similar statement can be made about plates and shells, where the modes are much more closely spaced than in beams. The solution of Eq. [11.4.22] subject to the initial conditions [11.4.241 [11.4.24] can be accomplished in many ways. as discussed in Section 5.7. Equation [5.7.71 gives the general form of the response. The preceding developments can be generalized by making use of the operator notation. Using the equations of motion in the form of Eq. [11.3.28]. we assume a solution in the form of

t)} =

[11.4.261

is the amplitude vector and denotes the time dependency. Introducing this into the homogeneous part of' of Eq. Eq. [11.3.281 [11.3.281 and using the same argument as before. we obtain the eigenvalue problem where

1' 1.4.27] solution yields an infinite number of eigenvalues A,.(r = 1, 2,...) and cone+

The

=

{O}

sponding vector eigenfunctions When the mass and stiffness operators are self-adjoint. the eigenfunctions can be normalized as seif-adjoint.

< <

)} > = 66 r.s

) }'

>=

i; S = 1,2,

. . .

[1 1.4.281

The inner product notation between two vectors was defined in Eq. [11 .2.401.

One can then invoke the expansion theorem and expand any comparison function

as

= '5"

11

1.4.29]

621

622

CHAPTIR 1 1 •

FI,Ex!BI,E Bonws

DYNAMICS OF LIGHTIX

and the coefficients Ir can be shown to he

ar = <

>

{I1(c.Yi)},

(1 1.4.30] [1

To find the response. we expand the deformation

t)} as

=

t)}

111.4.31]

Introducing this expression into the equations of motion, left-multiplying with and integrating OVCF the domain yields the modal equations [11.4.22]. with the modal excitation

and and initial conditions

=< ijç(O)

Example

11.4

being

>

t)},

< <

T)ç(O)

and

0),

>

0),

>

(1I.4.32J 111.4.321

Find the the eigensolution eigensolution of the uniform pinned-free beam shown in Fig. 11.19.

Solution We consider Eq. [I 1 .4.11 as the equation of' motion. The boundary conditions are that the displacement and moment vanish at the pinned end. and that the moment and shear vanish at the free end; thus v(O.t)t) = 00 v(O.

atx = Oandx Oandx = L L

= 0 t) ox3

= 0

at .i.v == 1..

1°]

Introducing Eq. Introducing Eq. [11 [11.4.3] .4.3] to Eq. Eq. [11.4. [11.4.1]. 1]. we obtain the differential cigenvalue problem

A2p4)1) + El A2p4)(i)

dx4

=

(b] (hI

0

and associated boundary conditions (/)(O)

Using the variable

=0

= 0

= 0

= 0

(c] Ic]

= —A2 /2/El. we we write write Eq. Eq. [hi [hi as —

[34th(.v) = () [34th(x) 0

V

0

z.

pE! L

Figure 11.19

(d] Id]

1 1.4

E!GENSOI4UTION AND RESPONSE E!GENSOI4UTION

which has the general solution =

To find the constants 0 yields

Ct

(i =

sin 1,

Iii +

[.1

cos ,Bx + C3 sinh /3x + C4 cosh

2, 3, 4) we invoke the boundary conditions. At x

=

0,

=

+ C4 C4 = o = C2 C2 +

Ifi

Differentiating Eq. [e] twice gives = f32(_ci sinf3x

Atx ==

—

C2COSf3X + +

c3sinhf3x ++c14coshf3x) c14coshf3x)

[gJ Igi

Owe have

[hI [h]

+ (:4) c4) = 0

=

A comparison of Eqs. [fJ and [hi indicates that C2 C4 = = 0. The The boundary boundaryconditions conditions C2 = C4 at .v x = L yield

p2(ç sin f3L ++ sinf3L

sinh,BL) = 0 sinh/3L)

III

cos/3L +

= 0

II]

=

From Eq. [iJ we obtain c1 = yields the characteristic equation

c3

sinh sinh f3L/sin f3L. which, when introduced into into Eq. Eq. [ii [ii

[k] 1k]

=

Unlike the pinned-pinned uniform beam. beam, solution of the characteristic equation here cannot be obtained in closed form, and one has to resort to numerical techniques to solve Eq. [kj. The first five roots of the characteristic equation can be found to he 131L = 0

= 3.9266

= 10.210

= 7.0686

13.352

The first root is zero, which implies that the first natural frequency ol' the pinned-free beam is zero. This mode is known as the rigid body mode. To find the corresponding eigenfunction, we take Eq. [d] and set = 0, leading to the equation = 0. which has the solution i

(x) == (x)

A

A .v .v ++AA2X 2X++A3 .v3 +A .v3

[II Ill

3

Introducing the boundary conditions. we obtain ç!c(0) = 00

The only term that survives is A

We

A2 A2 = 00

= 00

A3 A3

=0

[mi [ml

conclude that the rigid body mode has the ft)rm

[nI [nJ where A,is_the amplitude. Using the normalization procedure in Eq. [11 .4.131, one obtains

= We observe that, because the mass moment of inertia of a rigid uniform beam about the pinned end is = mL2/3 = ,uL3/3, the orthogonality constant can he expresA1

sed as3

A1=

L to

[0] lo]

3Equation [oJ is valid for pinned-free beams with nonuniform cross sections as well. To verify, introduce Eq. [n] into 3Equation Eq. [11 .4.1 3] and invoke the definition of the moss moss moment moment of of inertia, inertia, — /L(x)x2 dx. /.L(x)x2

023

624

11S

FLEXIBLE BODIES LICHTLY FLEXIBLE DYNAMICS OF LICHTLY

Let us investigate the rigid body eigenfunction in more detail. With the rigid body mode, the beam does not deform elastically. hut moves at an angle with respect to the original unde1t itit had had no no elasticity. formed axis. That is, the first mode represents the motion of the beam as if elastic restoring Because in the rigid body mode there usually is no elasticity, elasticity, there there is no elastic force.4 Hence, one can physically explain that there is no natural frequency' associated with this mode, or that the natural frequency is zero. The remaining modes describe the elastic deformation of the beam. From Eqs. Eel, [ii, motion have have the the form we can show that the eigenfunctions associated with the elastic motion and

r=

sin [3rL ,8rX ++ SjIl [3rL sinh f3rX) = Ar(SIflh I3rL I3rL Sjfl PrX

2, 3,

(11.4.13J. 13J. The The total total where the the amplitudes amplitudesA1. A1.can can he he found found by by introducing introducing Eq. Eq. [pJ [pJ into into Eq. Eq. (11.4. of the the rigid rigid and and elastic motions. The motion of the pinned-free beam is a linear superposition of' expansion of the motion can be written as -ç

v(X,

r

t) ==

1) 1) ±

=

1)

4)1(.v)7J1(t)

[qi

+ r=2

I

Note that in the same way the rigid body mode contains no contributions from the elastic

motion, the elastic motion has no contributions from the rigid motion. We can show this by invoking the orthogonality orthogonal ityproperties propertiesof of the the modes. modes. Multiplying Multiplying the the rigid and elastic motions and integrating over the domain, we obtain i_

— 1.

I

Jo

/LVngiii(X.

dx

/.LA1x7)i(t) dx = t) dx = .0 r=2 :.

= = ;•=2 r=2

=

(3

=0

In

The first three eigenfunctions are plotted in Fig. 11 .20. in order for the linear superposition of the rigid and elastic motions to be possible. the rigid body motion must be small. In general. rigid body angles less than 20 degrees can be analyzed using this approach.

(p1(x)

L

Figure 11.20

First three eigenfunc-

tions of a pinned free beam

I

is possible to Find cases where a rigid body mode will have a nonzero natural frequency.

1 1.4

Consider the pinned-free uniform link in Example 11.4 and obtain the modal equations of r(t) is applied at the pinned end, as shown in Fig. 11.19. motion considering that a moment rU) Then, analyze the contribution of each mode to the response.

Solution The external excitation can he expressed as

I) =

[al

in-(x, 1) = in-(x, 1)

0

Substituting Eqs. lal into Eq. 111.4.211 and using integration by parts yields L

=

(

(

—

I

[r(t)8(x)]dX = -ET(t)8(X)]dX

(IX (1.1

Jo

—

L

r = 1,1, 2,

=

—T(!)

[b]

Jo

For the first mode, the rigid body mode. the excitation can be written as =

çW1(O)T(t)

/3

= A1r(t) A1r(t) == 1—

[ci

which. when substituted into the modal equations of motion, gives the equation of motion for the first mode as =

/3 it)

[di

/

\'

The rigid body modal modal coordinate. coordinate. Tfl Tfl(I). (1).can canhe herelated relatedto tothe the rigid rigid body body angle angle as as follows. From From Eq. [qi [qi in Example 11.4 we have Vrigiti(X.l) gb1(.x)771(t) Vrigiti(X.1)==gb1(.x)771(t)

Eel

Since we consider the angle angle (1 (Itotohe hesmall, sniall,we wecan can write write for for any any point point x t) =

=

q1(t)

.1

[fi

Substituting this into Eq. [dl. and noting that the mass moment of inertia about the pin joint is given by = we obtain for the rigid body mode

Ig] Igi

r(t) JoO(t) = T(t)

which is recognized as the rotational equation of motion of a uniform rigid pinned-free link of length L. From Eq. [11 .4.221. the modal equations associated with ihe elastic coordinates have the form +

r = 2,3,...

=

[h] [hi

Let us next investigate the mode participation. Consider zero initial conditions. If we introduce the excitation from Eq. [hi into the response equation [5.7.7] we obtain ij,(r) = ij,(t)

— I T(t —

WI. Jo Jo

625

E1GENSOI.UTION AND RFSPONSE

u)sin WrTdcT

Ill Eli

Example

11.5

626

CHAPTIR

1

1•

DYNAMICS OF 141G11T1Y I4IGIITLY FLEXIBLE FLEXIBLEBODIES BoDIEs

we estiniate we estiniare the orders of magnitudes of

From &is.

= 0(r)

the terms in this equation as

=

Li]

and the order of the integral depending on the nature of T(t). Let us denote the order of the follows that the modal coordinates have orders of magnitude of integral in Eq. Ii] by O(1. kIt follows O(llr(t)) = For the case when r =

1

o(!)o(1)

the integral has the form

/1 \

1

T(t — (T)SlflWrUdU (T)SlIlWrUdU == -—-(1 1

-

1k]

U

(np.

—

COSWrt) = 01

I

/

II]

so that the response is of order =

o(4)

[ml

As the mode number increases, the amplitudes of the individual modes become smaller by the cube of the mode number. The second mode has an amplitude of about 1/8 the first mode, the third mode 1/27, and the fourth mode. 1/64. Hence, the participation of' the higher modes rapidly falls, which is typical for beams.

1

1.5

APPROXIMATE SoLuTioNs:

THE

MODES METHOD

The pinned-pinned uniform beam considered in Section 11.4 and the pinned-free beam in Example 11.4 lend themselves to closed-form solutions for the eigenfunctions. These two examples are more of' an exception than the general case. In most cases, and even for simple geometries, it is not possible possible to to find find aa closed-forni closed-fonii solution solution to the eigen eigenvalue value problem. pro blem.Members Memberswith withnonuniform nonuniform cross cross sections. sections. interconnected bodies, and almost any structure with a complex geometry fall in this category. When

a closed-form solution cannot be found, one resorts to approximate techniques to solve for the eigenvalues. All approximations to Continuous systems are in essence discretization procedures. The eigenvalue problem associated with Eq. 111.4.6] is in differential form. which implies that the solution is of infinite order. By discrelizing the eigenvalue problem. one converts the differential eigenvalue problem into an algebraic one that is of finite order. Algebraic eigenvalue problems are generally easier to solve than differential eigcn value problems. as they permit the use use of of matrix matrix algebra. algebra. 1

1.5.1

APPROXIMATION TECHNIQUES AND TRIAL FUNCTIONS

Approximation techniques can be classified into two broad categories: lumping and series expansions. When using the lumping approach. one lumps parts of the system into discrete elements. The stiffness parameters become the connections between the

1 1 .5

APPROXIMATE SOLUTIONS: THE ASSUMED

627

Mon:s

lumped masses, and they are modeled as springs. This approach is not analytical. It

is crude but easy to use and leads to discretized models of' relatively low order. The approach was more popular befire the advent of fast digital computers that made it possible to ftwmulate and solve discretized problems of very large order. When using series expansions. one expands the deformation in a finite series of known functions, multiplied by undetenuined coefficients such as n

v(x,

i')

">

i/ir(x)qr(t) t/ir(x)qr(t)

[1 1.5.1]

ii) are the known trialfunctions, qr(t) are their ampli= 1, 2, . tudes, and ii is the order of the expansion. The coefficients qr(t) are in essence a set of generalized coordinates. The trial functions must be from a complete set and must satisfy certain conditions, depending on the discretization method. In general. the trial functions can he classified into two groups: where

.

. ,

Comparison functions. As discussed earlier, these functions satisfy all the boundary conditions and are as many times differentiable as the highest-order spatial derivative in the equations of motion. In order to be consistent with the force and moment balances, the comparison functions should not violate the complementary boundary conditions. 2. Acfinjssjbiefiuicijons. Athnjssjbiefiuicijons. These These functions functions satisfy satisfy only only the geometric boundary conditions and are half as many times dif'ferentiable differentiable as the highest-order spatial derivafive in the equations of motion. In order to be a consistent set, they should not violate the complementary boundary conditions, and they should not prevent the natural boundary conditions from being satisfied. 1.

We. introduce two new categones of' trial functions: consistent admissible func((OilS ((Oilsand andconsistent consistent compa compa risen risen functions. functions. The The properties properties of these and traditional admissible and comparison functions are summarized in Table 11 .3. As before, the highest-order spatial derivative is denoted by 2p.

Table 1 1 .3

Classification of trial functions

Properties Trial Function

Complete Set

Differentiability

Boundary Conditions to Be Satisfied

Complementary Boundary Conditions

Admissible functions

Yes

p times

All geometric boundary conditions

No requirement

Comparison functions

Yes

2/) times

All boundary conditions

No requirement

Consistent admissible functions

Yes

p times

All geometric boundary conditions do not prevent dynamic boundary conditions from being satisfied

Do not violate

Consistent Consistent Comparison comparison functions

Yes

2p times

All boundary conditions

Do not violate

628

OF LIGhTLY Licirnx FlExIBLE FLEXiBLEBODiES BODiES CHAPTER 1 1 S DYNAMICS OF

The added requirements regarding consistent trial functions are for physical completeness. Trial functions that violate the CBC and do not make it possible to satisfy the natural boundary conditions lead to slow convergence. It is usually more desirable to deal with admissible functions than comparison functions. as ii. easier to to generate generate aa set set of admissible functions. Whether one caii it isis easier use admissible functions depends on the discretization method. Methods that permit use of admissible functions are more widely used. Having selected a set of trial functions. one performs an operation to the series expansion in Eq. [ 1 1 .5. 1 J, which leads to an algebraic eigenvalue problem. The nature of the eigenvalue problem depends on the type of expansion and discretization. Commonly used discretization procedures are a.

Those based on variational principles, such as the Rayleigh-Ritz. assumed modes, and finite-element methods.

h.

Those based on weighted residuals, such as the Galerkin and colocation methods.

When the discretization method is based on variational principles, the series expansion is plugged into the variational principle. Stationary values of the variational principle lead to an eigenvalue problem. For example. with the RayleighRitz method, one seeks stationary values of the Rayleigh quotient. For the method of assumed modes, the variational principle used is the extended Hamilton's principle.

When using a weighted residual approach, one seeks minimization of the difference between the actual solution and the series expansion in Eq. 11 1.5.1]. The difference, referred to as the residual, is minimized by the use of weighting functions.

Another consideration when selecting the trial functions is the domain of the trial functions. We distinguish between global trial functions. defined over the entire body. and local trial functions, defined over a portion of the domain. The tInite element method basically makes use of local trial functions, a very suitable approach for problems with complex geometry. In this chapter. we consider global trial t'uncfunctions.

1

1 .5.2

METHOD OF ASSUMED MODES

We outline the assumed modes method, as it is convenient to use and is applicable to nonlinear systems. as will be the case when we consider combined large-angle rigid and elastic motion. The trial functions qir(x) can be admissible functions. We first take the expansion in Eq. Eq. (11 (11.5.11 .5.11 and and substitute it into the expressions for the kinetic and potential energies. The method requires use of the energy inner product expression for the potential energy. Otherwise, one cannot use admissible functions as the expansion functions. We demonstrate the method with an illustration. For a beam that deforms only in the y direction and is subjected to an internal axial load P(x) and where the axial

1 1.5

APPROXIMATE SOLUTIONS: TILE ASSUMED MODES METHOD

elasticity is ignored, the kinetic and potential energies can be written as L

.v, V t) dx

Jid IL ( .1• )

,

1L V

=

{EI2

[11.5.2]

dx

+

Substituting the expansion [11.5.11 into these equations yields

T=-J a-Jo 1

fl U

fl U

,u(x) "> 1u(x)

s=1

F. — I

Ii

fl

=

(IX

51 -L

/

r=l s—I i•=1 s-I

dx

,i(x)1/i,.(

t)

d

—

-

U

?1

I,

=

(x)qr(t)

r-I r

r

2

y/

]

U

(x )

(

I,

t

is-.'

[r—i iiU

(x)q5(t)

I[,is=i.

I>

+

=

V z

1.

fl

—

1

t)

dx ] } dx

•L

(x) +

[11.5.3]

ii ( .v ) 1/Jr(( xx ) i//s i//s( (x)x)dx dx

[11.5.41

t)q.r(t)

A

0

1 s= s=1 1

I"

introducing the notation 1'

=

k,.3.

=

I

Jo

r, s =

+

I

Jo

1,

2,

. . . ,

[11.5.51

where mr.s are the mass coefficients and the kinetic and potential energies as I

(1)

rs

are the stiffness sti fness coefficients. we express

V

[11.5.61

=

r— 5= 1 — 1 s

r= 1I s— I

or. in matrix form

T=

[M1{q( [M1{4'( t)}

V=

[11.5.7]

1,2,... n) in which {q(t)} is a column vector containing the coordinates qr(t) (r = l,2,...n) and [MI and [KI are the mass and in the form {q(r)} {q(t)} = Lqi(t) q2(t) stiffness matrices. The mass matrix is always symmetric and positive definite. For the system described by Eq. [11.5.21. the stiffness matrix is symmetric. The sign-definiteness of of K K II depends depends on the internal axial axial force P( x) as well as on the .

.• .•

629

630

CHAPTIR 1 1 • DYNAMICS OF LIGHTIX FLEXIBLE BODIES

boundary conditions. For example. for a beam admitting rigid body motions, [K] is positive semidefinite. If P(x) is tensile, as in a helicopter blade, [K] is positive definite. When P(x) is compressive, as in columns, the sign of [K] depends on the value of P(x). t). the virtual work can he expressed In the presence of an external force in terms of the expansion [11.5.1] as L

1.

6W =

t)6v(x, i)dx JO

ii

dx

t)6

= .0 f

1.—I

fl

=

>6q,.

111.5.81

I

Jo

Introducing the notation

Qr(t) Q10) =

r=

dx

1,

2,

.

.

. ,

ii

[1 1.5.91

Jo

and substituting it into [11 .5.8]. we can express the virtual work as fl

>1 > Qr6qr = {Q}T6{q}

6W

11 1.5.101

...

We recognize Qr(t) to he the generalized force associated with the generalized coordinate qr(t). We have now expressed the kinetic and potential energies in terms of ii coordinates, thus approximating the system as an n-dimensional discrete one. We then use these discretized forms of the kinetic and potential energies together with the extended Hamilton's principle. Because the problem is now in terms of a finite set of generalized coordinates, one can obtain the equations of motion directly by using Lagrange's equations. Next, let us discuss the solution when the mass and stiffness coefficients are independent of time, which is the case for the linear problem considered here. From Section 5.3. we can directly write the equations of motion of the approximate system as where {Q(i)} =

(')

{Q(t)} [M){q(t)} + [K]{q(t)} [K]{q(t)} = {Q(')}

111.5.111 11 1.5.111

weuse use the the same same approach approach as in Sections 5.5— To find the response of Eq. [11.5.11 1.we

5.7. We first solve the eigenvalue problem. Introducing {q(t)} = {a}eAt

[1 1 .5.1 2]

where {a} is a time-invariant amplitude vector, and eAt denotes the time dependency.

into Eq. [11.5.11] yields (A2{M] +

[11.5.131

=

which leads to the characteristic equation

det(A2[M]+[K1) det(A2[M] + [K]) =

0

[11.5.14]

11.5

APPROXIMATE SOLUTIONS: rfFIF ASSUMED MODES METHOD

The roots of the The the characteristic equation arc the the eigenvalues of the discretized

system. As discussed in Chapter 5, when the stiffness matrix is positive definite, all the the eigenvalues eigenvalues A,. A,(r(r = I. 2, are pure imaginary. n) are imaginary. We We introduce introduce the notation = — jthr, where c2,. are the the natural natural Frequencies frequencies of of' the the approximate system. Associated with each eigenvaluc A,.. there is is a corresponding cigenvector {a,.} such that .

I .

.

.

,

+ IKD{a,.} [K]){ar} =

=

{U}

1,2,...,,,

[11.5.15]

The eigenvalues eigenvalues are approximations to the tile first ii eigenvalues of the actual sys-

tem. The eigenvectors can he used to approximate the first n cigenfunctions by

&(v)

T = = {a,.}

[1 1.5.161 [11.5.161

where 4,.(x) are arc the approximate eigentunctions elgentunctions and

is a column vector con-

taining the trial functions in the lbrm = [1/Jj (X)t/12(X) The eigenvectors eigenvectors {ar} {ar} can can he he shown shown to to he he orthogonal orthogonal with with respect to the mass and stiffness matrices. They can be normalized to yield .

{a,.}T1

=

K

.

.

[11.5.17] 11 1.5.1 7]

=

The approximate eigenluncuons approximate eigenlunct ions are orthogonal with respect to the original mass

distribution and stiffness, stillness, expressed in terms of the energy inner product as •11

L

) dx =

j

Jo

)(x ))dx

(EIx)'(x)'(x) +

J ()

=

r,s =

rs

1,2,

.

.

.

[11.5.18] 111.5.18]

, n ,n

To prove prove this, we introduce introduce Eq. Eq. IIII 1I .5. 16] into into Eqs. Eqs. [11 [II .5.181, .5.181, which yields •JL

dx =

[11.5.191

)}

Li

I 4)

(E1=( x

x)

x )))dx

x )b.(

I.

=

)dx{ac} 11 1.5.201

+ J

A closer examination of' of the term inside the square brackets indicates that x

)}T

•v

=

1L

+ .0

)dx

—

[K]

111.5.21] 111.5.21]

I

Thus. Eqs. [11.5. I 9]—[ 11.5.20] become identical to Eq. [11 .5. 17]. Note that the orthogonality with respect to stiffness is demonstrated using the energy inner product

631

632

CHAPTER 1 1 • DYNAMICS OF LIGH'rIx FLEXIBLE Bonws

only. only. To have orthogonality with respect to the stiffness operator, one must use cornparison functions as trial functions.

1

1 .5.3

CONVERGENCE ISSUES

It can he shown that the approximate eigenvalues are higher in magnitude than the actual eigenvalues. This is because using a finite series of trial functions is mathematically equivalent to placing constraints on a system. An exact representation of a flexible system requires expansion by an infinite series. Constraints make a system stiffer and raise its natural frequencies. This property is common to all discretization vanational principles. Increasing procedures that are based on series expansions and variational the order of approximation in essence relaxes some of the constraints. Hence, the eigenvalues become lower in value. As the model order is increased, the eigenval— discretization is ues approach their actual values from above. By contrast. when the discretization in the form of lumping. the approximate eigenvalues usually approach their actual values from below. The property that the approximate eigenvalues approach the actual eigenvalues from above can he demonstrated by the inclusion principle, which can be stated as follows: Let ó-Jr (r = 1,2, .. ,n) denote the natural frequencies associated with an nth-order approximation. Now, consider an approximation of order ii + i, where the first n trial functions are the same as before. and the next terni in the set of admissible function. functions is the (n + 1)th trial I'unction. F? ++ 11), )q Denoting the eigenvalues of the order n + system by r (p = 1,2, .. F? the inclusion principle states that . . .

1

WI

(02

... •..

. ,

[11.5.22]

Proof of the inclusion principle is based on the minimax theorem: it can be found in

advanced vibration texts. We next consider the issue of convergence. Convergence of the approximate solution to the actual solution depends on the type and number of trial functions used, as well as on the mass and stiffness properties and the loading. For example, if the system has discrete (also called concentrated, to distinguish from continuous) springs acting on it. these concentrated springs cause a discontinuity in the shear profile. This discontinuity becomes hard to approximate by continuous trial functions. Also, if the trial functions are selected such that the complementary boundary conditions are violated or it is not possible to satisfy the natural boundary conditions, convergence becomes slower. The sensitivity of the trial functions to various parameters in them are also important. For example, dealing with hyperbolic sines and cosines often presents problems. as these functions are very sensitive to their arguments. In general, if one uses an nth order model, one can assume the first iz/2 eigenpolynomials are are used as trial values to be accurate. The exception is when simple polynomials functions, where usually fewer than ,i/2 cigenvalues are estimated accurately. Also, functions. if the trial functions arc selected as functions resembling the cigenfunctions. convergence is much faster. For example. to find the eigensolution of a beam with a nonuniform cross section. a good choice of admissible functions is the eigenfunctions of a

633

API'ROXIMATE SOLUTIONS: Tiir ASSuMED Moiws METHOn

1 1 .5

uniform beam with the uniform the same same boundary boundary conditions. conditions. Convergence issues associated with trial function expansions of solutions constitute a very broad subject. one that is beyond the the scope scope oF of this text. The assumed modes method can he described in terms of the general operator notation. We expand the elastic deformation as

[1 1.5.23] [1

=

ot admissible functions. The entries of in which in ,n) are a set of )} (r = 2, the mass and stiffness matrices and the generalized forces become I ,

.

.

.

=

k,.ç

=

r,

=

s = 1,1, 2.

.. .

.

1n ,n

[11.5.24] [11.5.241

Lagrange's equations equations in which {F} is the external excitation vector. One then invokes Lagrange's in and obtains the equations of motion, which have the form of' of Eq. Eq. [11.5.11 [11.5.11 when when IM IM I and [K] arc constant coefficient matrices. I

.3. which which is is modeled modeled as a uniform fixedConsider the rotating rotating blade problem in Example I 1I .3. Consider the Compare the the natural natural frequencies frequencies free beam of length L. rotating with angular velocity velocity IL Compare of the beam for the rotation speeds speedsU/(EI/pL4) U/(EI/1iL4) = 0, 0. 1, I, and ID, and verify that the asymplotie behavior of the eigenvalues is governed governed by by Eq. Eq. [11 [11 .4.251. .4.25].

Solution We use the method method of aSsumed aSSumed modes. modes. The The trial trial functions functions must must satisfy satisfy the the geometric geometric boundary

conditions of 0

r == 1.2, 1.2,

0

. .

.

ii

lal Ia]

The trial functions must also not violate the CBC, such that at least one of the trial functions andone onemust musthave haveaanonvanish— nonvanish— must have a nonvanishing nonvanishing second secondderivative derivativeatatxx==0 0and lIJr(X) imist tunctions should not prevent the natural boundary ing third derivative at .v = 0. Also, the trial functions conditions to he satislied at the free end. A suitable set of consistent admissible functions is the cigensolution ola uniform, fixed— free. nonrotating beam. Another choice is polynomials in the form = I, 2. .. ..

V)

=

n

[bi

I

We saw in Example 11 .3 that the effects of the rotation can he modeled in a number of let us us consider consider itit as as an an added addedpotential potentialenergy. energy.From FromEq. Eq.[a] [a]of ofExample Example11 Ii .3, the ways. Here. let

internal axial force has the form

P(v) P(x)

=

/2ft(L2 /2ft(L

—

.v2

[ci

Example 1 11.6 .6

634

DYNAMICS OFLIGHJi,y LIGHrIxFLF:xIBLF: FLF:XIBLE BODIEs CHAPTER 11 11••DYNAMICS OF CHAPTER

Table 1 1 .4

Natural frequencies of helicopter blade as a function of the rotor speed

Ratio of

R

AR -10

Mode no.

0

(11

3.516

1

1

3.68

3.533

1

10

AR.0

4.91

1.399

2

22.03

22.05

22.18

23.46

1.065

3

('1.72 61.72

61.73

61.86

('3.15 63.15

1.024

128.4

128.4

4

1.010

129.7

The entries of the mass and stiffness matrices matilces are found using Eqs. f II .5.41 .5.41 and and 111.5.5]. to have the form 1

ci dx

1

= L±

(r ++ 1 )r(s + I1)s )v

L

1

-

=

1) J

1-ix Hx

d.v+

ç

.v

+

r + iI xs

L

+

P(x)xxdx P(x).vxdx ()

El )rs = —• = —•ft + I )(s + II )rs + + r s—i s—

L

(1. )(s ++ 1) 1) / ft ++ II )(s

1

2

I1

I1

rfs+ r f s ±I

I

—

r+s+3 r+s± 3)

Ed]

and introduce the ratio Let us divide both the mass and stiffness stiffness coefficients coefficients by by 1iL 1iL and Let p.L4/El. As R is increased, the blade rotates faster. We can then write the mass and R= stiffness coefficients as

r+s+3 =

El

—(r± —(r ÷

Letting El/pd14 Letting El/pd14 =

Il)(s+ )(s + 1)

is

r+s—1 r+s— 1

RI

+ —I

I

2\r+s+1 I

I

—

r+s+3j I

Eel

we solve the elgenvalue problem for various values of R. We use a first four eigen values are given in Table 11 .4. together discretized model of order n = 6. The fIrst with the ratio of the eigenvalues for R = 10 and R = 0. We observe that as R increases and the rotation speed gets bigger. all of the eigenvalues become larger. As expected from the asYmptotic analysis. the increase in the natural frequencies is more pronounced in the lower modes.

Example 1

1 .7

1.

an illustration of the importance ol' satistying the complementary conipkmcntary boundary conditions and As an associated convergence issues. we consider a uniforni bar of length L in axial vibration, fixed

spring of of constant constant kk at the other end (x = L). as shown at one end (x = Oh and attached to aa spring in Fig. 11 .2 1. The kinetic and potential energies have the kwni T = T

t)dx

=

dx + EA[u'(x. t)12 t)12d.v

I)

[a]

1 1 .5

MODES METHOD METHoD ASSLMEL) MODES SOIimONS: TIlE TIlE ASSLMEL) APPROXIMATE SOLUTIONS APPROXIMAI'F

L

k

,

Figure 11.21 For For /_L =

1

.

simplicity we select unit length and unit niass and stiffness distributions. L = boundary boundary conditions can he shown to be EA =

I

1 .

1) 14(0, 1)

= = 00

IbI

EAu'(L,1) 1) ++ ku(L. ku(L. t)t) = 0 EAu'(L,

The boundary condition at .v = 0 is geometric. and the associated complementary condition isis that that EAu'W, EAu'W,t)t) isis unspecified. unspecified. At x = L. we have a boundary condiboundary condition tion of the third kind. To find the cigensolution we consider three sets of admissible functions ,ft1,.(x), l/;2,.(X), i/i37.(x). cacti each from a complete set:

Set I:I: lIJir(X)

[ci

X

(2r —1 (2r

Set 2:

= sin

Set 3:

= (sin /3,.L

—

Idi — Sifl Sin h /3r-V) fir-V) ± (cos sinhh f3,L)(sinf3,x sin f3,L )(sin 13,X—

lel

h f3rX) + COSh/3,L)(COSI3rX cos — cos COSI1,Br.V) 13,.x —

The mass and stiffness matrices have the entries L

=

)i/115(x) clx )i/115(x) dx —

J

/

a' a.

krv

=

EAqI,. .v .v )'/'r x) dx + ki/Jir( L

(L) ( L)

ij = .2,3 1

[f]

J0

violate the CBC or prevent the natural set is simple polynomials. which do not violate The boundary condition from being satisfied. Hence. they qualify as a set of consistent admissible functions. The second set is the eigen functions of a fixed-free bar in axial motion. This set is the exact eigensolution when there is no spring. i.e., k = 0. In the presence ot a spring, this set of functions cannot satisfy the natural boundary condition at .v = L. as all trial functions

have vanishing Iirst derivatives at x = L. The third set. Eq. Id. is the cigenfunctions of f3rL are the solutions of the associated characteristic a fixed-free beam in bending. where /3rL areZCFO at equation. For this at .v .v = 0. thus violating the set, both boththe thedisplacement displacementand andslope slopeare this set, do not qualify sets do complementary boundary boundary condition conditionat at xx = 0. Hence, the second and third sets complementary functions. as consistent admissible functions. We compare the accuracy of the solution obtained by using all three sets for varying orders of' of approximation approximation and and for for different different values of the spring constant k. Table 11 .5 Comfunctions. and for varying pares the first three cigen values for the different sets of admissible functions, values of the spring constant, using approximation orders of five live and seven. Even though simple polynomials constitute a relatively poor choice of admissible functions, the difference in accuracy and rates of convergence between simple polynomials and the other two sets is oh v iO io us.

Analyzing the eigcnvalues cigenvalues we make the following observations: Analyzing

As expected. in all cases the eigenvalucs become smaller as the order of approximation increases and the inclusion principle holds. 1.

635

636

CHAPTER 1 1 S DYNAMICS OF LI(;IrrLv FLEXIBLE BODIES

Table 1 1 .5

Comparison of eigenvalues

k=100

n=5

n=7

n=5

11=7

k=1000

n=5

,z=7

Set I: Polynomials as consistent admissible functions L570S 1.570S

1.5708

3.1105

3.1105

3.1385

3.1385

4.7132

4.7124

6.2227

6.2211

6.2787

6.2769

7.8550

9.9508

9.3479

10.091

9.4154

12.174

Set 2: Eigenfunctions of fixed-free

w,

uniform bar as trial functions

1.5708

1.5708

3.2417

3.2027

3.2717

3.2348

4.7124

4.7124

6.4938

6.4091

6.5515

6.4726

7.8340

7.854()

9.7727

9.6239

9.8524

9.7157

Set 3: Eigenfunctions of fIxed-free fixed-free uniform beam as trial functions Wi

1.6221

1.6187

3.2090

3.2043

3.2387

3.2341

4.8963

4.8816

6.4390

6.4322

6.4965

6.4903

8.1518

8.0818

9.5568

9.5163

9.6339

9.5978

Wheti When kk = 0, which implies iniplies a free end at x = L, the eigenfunctions of the fixed-free bar give the exact natural frequencies. This is because the eigenfunctions of the fixed-free bar become the closed-forni eigensolution for this case. Comparing polynomials and set 3, 2.

,1 = 7. polynomials exactly match the first two natural frequencies. with the third being off by about 0.0 1 percent. However, using the eigenfunctions of the fixed-free beam gives errors in all eigenvalues. with the error in the first natural frequency being about 2.5 percent. (t)i

Interestingly. the error in the third natural frequency is also about 2.5 percent off the actual natural frequency. This is an indication of problems encountered when the CBC are violated. because one expects the lower natural natural frequencies frequencies to tobe be more moreaccurate accurate than than the the higher higherones. ones. 3.

For nonzero values o1 ol' k. polynomials again give the best results, results. except for (03 for n =

5.

This is expected. because the polynomials are a set of consistent admissible functions and the other two sets are not. The natural frequencies again have an error of 2.5 to 3.0 percent for sets 2 and 3. The lower frequencies are not estimated more accurately than the higher frequencies, clear indications of the problems encountered when the CBC are violated and the natural boundary condition cannot he satisfied. 4. For nonzero values of k, while the accuracy obtained from sets 2 and 3 is comparable for w1. eigenfunctions eigentunctions of the fixed-free beam give better results for w3. For z.2. set 2 generally gives more accurate results. 5. Also. for nonzero values of k. the eigen values obtained from sets 2 and 3 never get very close to the actual values. For example. for k = 100. the solution for w1 does not improve at all for set I as we go from ,i = 5 to n = 7. This is because the solution has almost converged. However. for set 2 thcre is quite a change as the order of the approximation is increased. For set 3. there seems to he convergence, but not to the correct natural frequencies. 6. When polynomials are used as trial functions, one obtains very accurate results for the lower modes hut the higher modes are inaccurate. Usually, for an an approximation approximation of order ii, ii. fewer than n/2 modes give accurate results. This can be observed by looking at for n = 5.

I I1.6 1 .6

ELASTICAND AND LARGEANGLE ANGLERicio RicioBoDy BoDyMOTION MOTION ELASTIC

Finally, one can observe ohsen'e the the asymptotic behavior of the eigenvalues by looking at the L so SO that we k. Consider the results for polynomials and ,i = 7. results for different values of k. thc fIrst can treat the first three natural t'rcquencies t'icquencies as exact. For k = 100 and k = 1000. the increases by 33 percent and natural frequency is doubled than w for k = 0. while (03 increases by about 20 percent. As in Example I I .6. addition of an extra source of stiffness affects the lower modes much more than the higher modes. 7.

1

1 .6

KiNEMATICS OF COMBINED ELASTIC AND LARGE

ANGLE RIGm BODY MoTioN An important important case case in in vibrations vibrationsisisthe thecombined combinedelastic elasticand andlarge—angle large-angle rigid rigid motion example. the. heani in Fig. 11 .22. Analyzing the position of a body. Consider, of a point compared with an initial. undel'ormed position, the rigid body and elastic motions can flO no longer longerhe helinearly linearly superposed. superposed. as as the the large-angle large-angle motion changes the projection of the beam axis onto the undefornied axis substantially. When the elasticity of the body is small, it is convenient to view the elastic motion from a set of reference axes, selected such that the motion observed from these axes is small and linear vibration theory can he applied. When the elastic tion becomes large compared with the dimensions of the body. one needs to consider theories other than those presented in this chapter. Consider Figs. 1 .22—11 .24. describing beams undergoing both elastic and I

large-angle rigid body motion. The two ends of the beam are denoted by c and d. These beams represent commonly encountered situations when dealing with the combined elastic and large-angle large-angle rigid rigid body body motion. motion. Figure Figure11 II .22. depicting pinned-l'rec beam, is representative of a robotic arm: Figure 11.23 describes a a pinned-l'rcc free-free beam. as in a satellite with antennae: antennae; and Fig. 11.24 is representative of an intermediate link of a mechanical system. We will observe the motion of these beams from a relative frame such as the one shown in Fig. 11 .25. The frame has an origin 0 and a certain orientation, and is called the shadow franie fralile or the tracking frame. We will refer to the motion of this frame as the f)r!marv motion, and the motion observed from the frame as the secondary and Fnotion. The reason for describing the motion in two parts parts is is to to select select origin origin and

De )rrned

C. C

Original C.

Figure 1 1.22

pinned-free beam Deformed pinned-free Deformed beam

Figure 11.23

Deformed freefree beam

637

638

CHAPT!I 1 1 •

DYNAMICS OF LIGHTLY FLEXIBLE BODIES

d

C.

Figure 1 1.24

Deformed intermediate link

V

Inertial frame

Figure 1 1.25 orientation of the tracking frame such that one can treat the secondary motion using the near linear theory discussed earlier in this chapter. With this in mind, we write the displacement of a point on the beam as

r(x, t) = r0 + (x + u(x. 1))i + v(x. t)j +

ilk

111.6.11

in which which r0 r0++xixidenotes denotes tuetue primary primary motion motion and and u(x, 1)1 u(x,+r)iv(x, + v(x, t)j +t)j w(x, + w(x, t)k denotes t)k denotes the secondary motion. We note that u(x, t), V(X, t)e and w(x, 1) are not necessarily the same as their counterparts in the previous sections of this chapter and that the primary and secondary motions do not necessarily correspond to the rigid and elastic motions. The u(x, t), v(x, t). or w(x, t) depend on the choice of the tracking reference frame and they may contain components from the rigid body motion. The angular velocity of the reference frame is written in terms of its components along the reference frame as =

+ Wj

+ co-k

111.6.21

The velocity of point .v becomes .v, t)

üi + vj + ii'k + w to X [(x + u )i + vj + wk = i'o + üi = r0 r(-) + (u (u— W-L W-L + + (1 + w-x + co-u — w)j + + — — [11.6.31

We will discuss the following issues associated with the modeling of lightly flexible bodies undergoing combined rigid and elastic motions: 1.

2. 3.

Selection ol' the origin and orientation of the tracking reference frame. The type of trial functions to use to expand the secondary motion. Analysis of the equations of motion.

1 1 .6

KINEMAtICS OF

ELASTI(; AND LARGE ANGIE Ri;in Boiv MOTION

can expect the nature of the reference reference frame frame to dictate the nature of the equations of motion. The primary criterion is to enable one to use the near linear One

theory and to describe the secondary iliotion motion in a series expansion. As we will see in the next section. in many cases it is necessary to to keep keep the the u(.v. u(x. 1) in the formulation. because for fast rigid body motions the centriFugal ftrce is quite large. The expansions for v( x. I). t)., and e(x, t) have the form ,I

F:

v(x.

=

1)

W(X, 1)

'>Yr(X)qns ,.(l)

,fl

('(V. II)) ('(V.

[1 1.6.41

>T 1

in which which111r(X), 1/J,.(x),)',.(X), y,.(x), and and f3,(x) f3,(x) are sets of suitable admissible Functions and qr(t) are generalized coordinates. with ii. #( . and in denoting the order of the expansion.

It follows from Eq. [I [II1.2.61 .2.61 that u(x. 1) can be expressed in terms of the generalized coordinates as u( x, t) = e( u(x, v. 1) e(v. 1)

1

—— 2

. 2 \2 +I ri(T) du) \drJ \daJ I

-

I

in

c/a =

2

/3r( .i

t)

r— F.—. I I

I

I

11 I

I

There are two primary choices for the selection of the relative frame:

relative relative franu' franu' sue/i sue/ithat thatits its motion met ionisisknowmi knowmi a priori. This choice is relevant when one has a good idea as to what the rigid component ol of the motion is going to look like. An example is the maneuver of robots or spacecraft with little flexibility. A set of inputs is applied to move the body along a desired trajectory. For a variety' of reasons, the actual trajectory of the arm will differ from the desired one. The secondary motion makes up for this difference. With this model, the motion is measured from a tracking frame whose niotion motion is known.

a.

To select the the

Note that in this case both r() and 0 are treated as known quantities. so that when deriving the equations of motion their variation is zero. Hence, in this formulation the primary motion is independent of the secondary motion. The boundary conditions on the secondary motion motion remain remain the the same as in tile the initial description of the beam. h. To To SeleCt select the relative relative frwne frame sue/i sue/ii/mat its location localion (mini (111(1 motiOn motiondepend dependomi cii the orientation of the structure. In this formulation one takes into consideration the curorientation rent position and orientation of the body. Hence. the primary motion is not selected f/it'

independently and it is related to tile the secondary motion. The quantities describing the primary motion are variables, because their values depend on the location of the body. This implies that we may be adding redundant degrees of freedom. Such redundancy necessitates imposition of constraints on motion. TileThe 1111— on the thesecondary secondary motion. imposition of ofconstraints constraintscan canchange changethe theboundary boundaryCOfl(litionS conditions of ofthe thesecondary secondarymotion. motion.

639

640

CHAPTER 1111 • DYNAMI(S CHAPTER DYNAMI('S ot ot LIGHiLI LIGHTLYFLEXIBLE FLEXIBLE BODIES BODIES

One can view One can view the the geometric geometric boundary conditions of a body as configuration constraints imposed on on an otherwise free-free member. From this perspective. one can view the constraints imposed on the secondary motion as changing the boundary conditions. However. one should keep in mind that the imposition of constraints on the secondary iiiotion motion is at a mathematical level. so that the existing internal forces and moments nioments at the boundaries. hence the CBC. do not change. One must be cautious in describing the secondary motion such that the newly generated geometric boundary conditions are satisfied and the already existing CBC are not violated. As a consequence oF of using using eonstrained constrained coordinates. coordinates. the the resulting resulting equations equations of

motion will be in the form ol constrained equations. One way to circumvent dealing with a constrained set of equations is to select the generalized coordinates that describe the secondary rnotioii motion in a way such that the imposed constraints are automatically satisfied. We will adopt this approach. In essence, by selecting the trial functions as consistent admissible functions for the secondary motion one can satisfy the COnstraints automatically.

There are several ways ofdelining ofdeliniiig the primary motion. Selection olthe relative frame is, in general, complicated for the case of general three-dimensional motion. Furtherniore. fbi. for three-diniensional three-di mensionalproblems. problems, physical physical representation representation of the choice ofthe relative frame becomes difficult. as well as any attempt to analyze analyze the nature of attemptto the resulting equations of' motion. On the other hand. hand. when when the the angular angular velocity velocity has has certain properties, some interesting descriptions of the motion result. Here, we will analyze cases where the angular velocity of the reference frame is in in one direction only. and the secondary motion is on on a plane. We consider the the undeformed beam axis to he the x direction, direction, one one of of w(x, w(x, i) i) or v(x, t) to be zero, zero, and and the the angular angularvelocity velocity to be along along OflC OflC o1 of the .v, v, or or zz directions. It turns out that there are three three distinct distinct types of motion motion possible depending on the directions of the the secondary motion and angular velocity.

Example 11 .3 showed the rotating blade problem. where the angular angular velocity velocity was in the v direction and the secondary secondary motion motion was on the xv xv plane. plane. with with w(x, w(x,t)t) = 0. When the angular velocity is in the : direction and v(x, t) is set to zero we get the same type of behavior. behavior. We We refer to such problems as Case 1. 1. In In this type of problem. there are no constraints imposed on the secondary motion. The angular angularvelocity velocity of' of the reference frame frame is is along along the the plane plane of of the the reference the secondary secondary motion: thus thus the the additional additional rotational rotational variable variable due due to to the rotation of the tracking frame is an added degree of freedom and and not not a redundant one. For Case Case 11 problems. problems. the the angular angular velocity velocity of the has aa stiffening tracking frame has effect on stiffening effect onthe the secondary secondary motion, as we saw in Example II 11.3. .3.This Thiscase caseisisthe themost moststraightforward straightforward of of the three types of motion. When the angular When angular velocity velocity is in the .v direction only. only. the the motion motion is is similar to that that of a rotating shaft. We will analyze this case. referred to as Case 3. referred to 3. in in Section Section 11 .9. together together with with the effects of of the the rotat rotation ion on on the the secondary secondary motion. motion. As in 1. in Case Case 1, selection of the tracking frame does not impose a constraint on the secondary motion. for the same reasons. By far the most interesting case is Case 2. 2. when the the secondary motion is in the and the the angular angular velocity is in the zz direction (or xv plane and (or secondary secondary motion in the xz xz and angular angular velocity velocity in plane and in the the v direction. direction).The Theangular angular velocity velocity is perpendicular

1 1.6

KINEMATICS OF COMBINED

Table 1 1 .6

Types of large-angle rigid motion

of Motion Case

1

AND LARGE ANGLE RiGID BODY MOTION

(rotating blade. stiffcning effect)

Plane of Secondary Motion

Angular Velocity Direction

.vv xv

v z

Case 2 (robotic arm. spacccrat't) ('ase

.vv xv —

z

'' )

Case 3 (rotating shall. whirling)

v:

to the 1)lafle 1)Iafle ofthe ofthe secondary secondarymotion. motion.Hence, Hence,selection selectiono(the o(thereference referenceframe frameimpOSeS imposes a constraint on the secondary motion. Table 1 1 .6 outlines the three cases. For the case of angular velocity in all three directions. general conclusions cannot he drawn, because of the mixed terms that enter the formulation. The exception is when the angular velocities are then, one can i-ely on the results from this then. chapter to describe the stiffening and softening effects. in the remainder of this section we consider Case 2 and outline three approaches

to select the origin and orientation onentation of the reference frame, together with the constraints they impose on the secondary motion. Without loss of generality, in all descriptions we consider the secondary motion to he in the xv plane and the angular velocity in the

z:

direction.

The Rigid Body Constraint

In this approach. one selects the moving coor-

dinate frame such that there is no rigid body component in the secondary motion. That is. all of the rigid component of the motion is described by the angle 0 and by r0. Figures 11 .26 to 11 show this constraint for the three beam configurations considered earlier. Let us consider the selection of the origin of' the reiCrence reference frame. For the pinnedfree beam. the origin 0 is selected as the pin joint. For the free-free and interconnected beams the origin is selected as the center of mass, SO that (1

r(x. t)dm = mr()

[1 1.6.61

—tC —

where i/ni = ,tt(x)dx. This leads to the constraint fot' the center of mass LI

p.(x)[(x + u(x, t)i + v(.v, i),j]dx = 0

111.6.71

—C

Note the component of this equation in the v direction indicates orthogonality

with the translational rigid rigid body body mode. mode. The The component component in inthe the.v x direction is basically the lateral motion. This motion is usually very small, as it is the integral of u(x, 1) over the beam. For the intermediate link in Fig. 11 .28. one can select the origin at one of' the joints. hut this complicates the physical interpretation of' the motion.

641

__ 642

CHAPTER CHAPTER 11 11 •• DYNAMiCS DYNAMiCS OF' oi'

FLEXIBLE BODIES

(I t.

0

Figure

1

0 link

Figure Figure 11.28

Rigid body constraint

To determine tile orientation of the tracking frame, a number of approaches can be followed. One choice is to have the components of the motions in the x and y directions be orthogonal to each other. This leads to the principal axes constraint ,

d

p

)I

x + ii ( x, II )) jj V( (X,x, I) (1 (1.V.V = = 0

[11.6.8] [11.6.81

J

Another Anotherchoice choiceisistoto select orientation thatt)v(x, t) contains select thethe orientation suchsuch that v(x, contains no rigid body components. That is. v(.v, t) is orthogonal to the rotational rigid body mode. For both a pinned-free and a free-free beam, the rotational rigid body eigenfunction is (note that x is measured from point 0) th1(x) =

[11.6.91 Jo

The constraint can then he expressed as • (I

p(x)41(x)v(.v, t)dx = 0

[11.6.10] [11.6.101

C

For the interconnected beam beam in in Fig. Fig. 11 11.28 .28 ii is much harder to distinguish between the rigid and elastic motions, as both ends of the beam are not free, but one can use the above equation as the constraint.

1 1 .6

AND LARGE ANGIE RIGID BODY MOTION

KINEMATICS OF

the orientatloil orientation constraint constraintisiskIlOWfl known as asthe the ,nodifieI ,nodifieI Tisserand Tisserand Another choice Ibr the (0flS1r(liflI,and anditithas hasthe the Form .

(I

i)jdx =

ji(.x)[.v + u(x. 1)11 x

111.6.11]

0

For the types of motion of interest here. the deformation u(x, t) is usually small. of' ii(.v. 1) and v(x, t) and their derivatives can he ignored. Using so that that any anypn)dUCt product of .6. 1 1 1 become equivalent to Eq. .6.8and andI 1I 11 1.6. this assumption. assumption. both boththe theconstraints constraintsI..I I .6.8 I

[1 1.6.10]. We will use Eq. [1 1.6. 10] as the rigid body constraint from now on. Let us next investigate the boundary coiiditions on the secondary motion. For

aller the pinned-free and free-free beams the boundary conditions do not change alter the primary frame is introduced. introduced. Hence. Hence. to to expand expand iiCi, iiCi, t)t) and andv(.v. v(x. I). one can use For functions used in the analysis of pinned-Free or free-free free-free bcams. For pinned-free or any set set Of Oftrial trialtUflCtiOflS the pinned-free and free-free beams. use of the rigid body mode constraint has the natural choice for the shadow frame. appearance of being a natural For the intermediate link, the situation is different, introduction of the reference frame changes the boundary conditions of the secondary motion at c and c/to basically free ends. ends. The The iiiteiiial intenial Force firce atataafree freeend end isis iero. iero. However. However. For the actual beam. at that point there is an internal force due to the joint. So one ends up with a situation where the displacement and slope are unspecified at the boundary. boundary, while the internal forces and moments are not zero. lorces A variant of the rigid body mode constraint is to not use the mass density term in tile the constraint equations. Rather, one has v(x, v(..v,

i)dx =

.0

/ 0

I

Jo

(1

—

\

— )v(x. )v(x,

Lj

t)dx =

0

[11.6.12]

This constraint makes it easier to generate a set of trial functions. The rigid body constraint has the advantage that it provides a physical explanation for Fortile theselect selection jOIlof ofthe the reference reference frame. frame. Also, for the pinned-free and free-free into the beams, it separates the rigid and elastic components of the overall motion into primary and and secondary secondarymotions. motions.On Onthe theother otherhand, hand,illinananactual actualnlCasUl'CIlleflt measurement or or concontrol application, where one needs to have measurements of the origin and orientation with this this constraint. constraint. Because the referof the tile reference frame. iit is difficult to deal with ot the body. the location and orientation of the reference ence axes are not attached to the frame cannot be measured directly and they must be calculated from other measurements.

The Zero Slope Constraint

reference Frame frame is attached to a The origin of the reference

point on the body. and the orientation orientation of' of the reference frame is selected such such that the secondary motion has zero slope at the origin, as shown in Figs. 11 .29 to 11 .3 1. In general. if there is a hub on the body the hub location is selected as the origin, as it is easier to attach a sensor or actuator there for measurement, navigation, or control purposes. For the interconnected beam. the reference frame is usually placed at one constraint on the secondary motion is expressed as of the joints. v'(O,t) = 0

[11.0.131

643

644

CHAPTER 1 1 • DYNAMICS DYNAMICS OF' OF LIGHTLY FLEXIBLE BODIES

(1

-v

'I £1

V -v

Pinned-free

0

U

C.

Figure 1 1.29

Free- free

Figur. 1 1.30 x

V

(1

0.c

Intermediate link

Figure Figure 111.31 1 .31

Zero slope constraint

The The

resulting geometric boundary conditions on the secondary motion at the origin become V(O,t)

= 0

v'(O,i)

0

111.6.141

in essence giving the secondary motion fixed-free end conditions. This choice of the reference frame changes the boundary conditions on the secondary motion of all three types of beams considered in this section. For the beams in Figs. 11.30 and 11.31, if the origin 0 is placed Ofl a point in the interior of the beam, one can view the beam as consisting of two beams, one from 0 to one end c. and the other from 0 to the other end d. As in the rigid body mode constraint, the fixed-free end conditions specify that the internal force is zero at end d. For the interconnected beam, there is the problem of having the boundary conditions of a free end and the CBC of a pinned end when modeling the secondary motion.

The Zero Tip Deformation Constraint

The orientation of the reference

frame is obtained by drawing a straight line between the two ends of the beam, as shown in Figs. Figs. 11 II .32 .32 to to 11 11 .34. .34. The origin is taken as one of the ends. As a result, the constraints on the secondary motion become

v(0,t) =

0

v(L,t) =

0

[11.6.151

where L is the original length of the beam. These constraint equations are the geometric boundary conditions associated with the secondary motion, those of a

1 1 .7

E1.&sTIc

DYNAMICS OF OF

645

LARGE ANGLE Ru;JD Ru;ID BODY Mo'noN

V S

(1

ci

Free-free

Pinned-free

O,c O.c

Figure 1 1.32

Figure 11.33

V

E)

Intermediate link

Figure 11.34

Zero tip deformation constraint

pinned-pinned beam. Note that the distance between the two ends is not necessarily

equal to L. due to the shortening of the projection. Intuitively, this choice of the reference frame seems suitable for the intermediate link. Fig. 11 .34. With this choice of the tracking reference frame, frame. the boundary conditions as well as complementary boundary conditions at ends c and d are changed for the pinnedfree and free-free beams. At the free end, the internal force and internal moment are zero. Once the constraint is imposed, the free end becomes a pinned end. However. because tile the internal force and moment are restriction on the types are zero, zero,there thereisis110 no restriction of functions one can use to expand the secondary motion other than the geometric boundary conditions. Hence, the imposition of the zero tip deformation constraint does not represent an inconsistency from a physical standpoint.

11.7

OF COMBINED ELASTIC AND LARGE

ANGLE RIGm BODY MOTION We will consider elastic deformation along the beam axis and along the v direction and use an analytical approach to derive the equations of motion. We will write the kinetic and potential energies. as well as the virtual work, and then invoke the extended Hamilton's principle. We first introduce the displacements u(x, /) and v(x. 1) in their general form to the kinetic and potential energies. As a result. the equations of motion will have the form of partial differential equations for u(x, t) and u(x. u(x, t). and ordinary differential equations ftr and OW. OW. Then, Then.we wediscretize discretizethe the(115— (us— placement. ii(.v, placement. the ecluations equations of motion in discretized form. ii(.v, ) and v(x. /) and derive the Consider a differential element of the heani, beam, and Case 2 discussed in the previous section. The angular velocity of the differential element can be expressed as

646

CHAPTER 11 • DYNAMICS DYNAMICS OF OF LI(;IITIx LI(;IITIX FLEXIBLEBonw.s Bonw.s

+ ô'(x, t)]k. Hence. the kinetic and potential energies are -i'

1

I 'iran iran + 'rot.

11

=

t) dx +

.v, 1).

i

(Jv + i"(x, t)12 dv J

J

[1 1.7.IaJ V = =

(EI-(.v)v"2(x. (EI:(x)v"2(x,

t)+

)e'2(x, t))dx P(x)v'2(x, I) 1) + EA(x )e'2(x,

j

[1 1.7.lbJ

we have included in the potential energy an axial force in the x direction. The virtual work has the form

where

,t)6r + ,t)•6r

=

t)i ±Pv(X. t)j denotes the distributed external force. The distributed moment acts through the total rotation of point point .v. x. which which is is 0(1) 0(1) ++ v'(x, v'(x, t). Note that when the reference frame is selected a priori. 0(1) becomes a known quantity. instead of a variable. in which f(x. t) =

1

1.7.1

EQUATIONS OF MOTION

HYBRID

Derivation of the equations of motion in the form of partial differential equations is quite cumbersome, even for the case considered here. For this task, we simplify

the model further by ignoring the stretch associated with the axial deformation, e(x, t) 0. Further, we consider a pinned-free beam, so that r0 = 0. We also ignore the rotational kinetic energy Trot, which is due to the rotatory inertia. As an external input, we consider a concentrated moment at the pin joint of magnitude r. The kinetic energy then has the form

1rL T =

t) — Ov(x, t)]2

J

1) + Ox + Ou(x. i)J2)dx t)J2)dx

+

Jo 1.

+

=

—

2Oiiu +

+ 201'x 201'x + 201'i, 201'i, ++ 2&ux)d.v

± Ô2x2 +

[1 1.7.31

We simplify the expression further by noting that u(x. 1) is much smaller than v(x, t). Any term quadratic in U or involving as well as terms involving products of u t, and

v, are ignored. We are then left with 1

+ J

+

+ 20i.ix)dx 20ux)dx

[11.7.41

_

1 1.7 11.7

lARGE

OF COMBINED OF

BODY MOTION RICID BODY

When the mass density is constant the last term in the kinetic energy can be written

in terms terms of' of v. using Eq. [hJ in Example 11.3, as -L

i

= 20 u x dx =

-

(iv \-

1.

i I

.-)

I

x

—

- 1)

(

J ()

0

-J

)

L

= =

• I

4

\2

/

— —

/1(L

jj

(1(7(/ X

•il(--—-

)I

\(1.V \(l.V J

Jo

[11.7.51

d.v

The potential energy has the same form as in Eq. 111 .7. I bJ. b]. The virtual work is

[11.7.61

OW = TMO+V'(O)I

one variable—the body angle 0—is s—is added added to to the the system system description. description. one cot-responding constraint equation needs to he specified. For the time being. describe the constraint in the genera] general lorm Because

-L

[11.7.7J [11.7.71

((X)V(X, c(x)v(x, t)d.v t)dx = 0

C C 0

where c(x) is the constraint operator. We continue with the case of' constant mass

density. We add Eq. [1 I .7.7] to the Lagrangian by using the Lagrange multiplier mu]tip]ier A. A. which yields L = T — V + AC. AC. and we invoke the extended Hamilton's principle.

Taking the variation of the Lagrangian with respect to the two variables 6(t) and v(x. t) and their derivatives, we write aL

6L =

.

60 +

.

(IL (IV

dO

6V 6V +

+ XV XV +

= .0 —

•

I

dv

6v + 6V

--0601 2

—

1L

dV' (IV'

()xl6Odx

.7

dx /x

— —

j .0

J

/

L

E1(x) o ()

/L(v /L(v++6x)6i'dx 6x)6i'dx

.0

"(IV

— —

H

/10VbV(IX + /10VbV(IX

+

I

.

— —

IlL + dv"

\dX-

&"

P(x) —

J

.

o 0

\dX /

6v' (Ix

111.7.81

The applied applied load load can he expressed expressed as as m;.x) m.(x) = r6(x). r6(xh and the virtual work can be expressed as '1.

=

+

I

Jo

—

I

T6(x)6v'dx = T6(x)6v'dX

T(S(X)bOdX

—

.0

Jo

r6(x)6vdx [11.7.91 111.7.91

Perlbrming the integrations by parts in the variation of the Lagrangian and col-

lecting coefficients of 60 and 6v, we obtain two equations of' motion. For the primary of motion. motion we have + {

+H—

—

x2)v12] t.2)Li'2] + 20 20

!(L2

—

= T6(X)

[11.7.101

647

648

CHAPTER CHAPTER 1 1 • DThAM1CS DThAMICS OF OFLIc,IITIX Lic,iiuy FLEXIBLE FLEXIBLEBODIES BODIES

and for the secondary motion p. 1.

+

Id

—

Li —

(9X

E1-(x)

—

d_dX dX

1.

+

-,

12)] (?X

(V — -—— P(x) -, dx dx (9

,

.,

riX— dx—

=

(x)

—

Ac(x)

111.7.11]

When the mass density is not constant. the above equations have a more com-

plicated form. A few observations about these equations of motion are in order. The equation for the primary motion can be expressed as as ., I p.— x0 /L— x0 ++xv xv ++ v() v0 ++—O(L —O(L — xiv 2 C

.

=

v20 + 2xuOJ 2xuOJ =

+

di

,-)

111.7.12]

To understand this equation better, it is useful to look at the expressions for the linear and angular momentum. The linear momentum is expressed as -I. -L

p=

.0

p.(x)iix, t)dx = p(x)iix,

—

.0

Ov)i +

U) ++ Ox Ox + Ou)jjdx

[11.7.13]

The angular momentum about 0 is -L

I) I) XX i'(x, i'(x, t)dx

H0 = ..0

-L

=

.0 -o

[(ii

X ••

(.1 ))[([(xx + + u ) I ++ p. (.1

—

Ov)i

vj]

+ (i' + Ox + Ou)j] dx

I.

p.(x)[(x+u)(i+Ox+Ou)—v(ü—vO)lkdx

= -

0

I

Jo

p.(x)(xi' + x20 + 2xuO + v20)kdx v26)kdx

111.7.14] 1.7.14] 11

in which we ignored terms quadratic in u, ü and andproducts productsofof1.1 ii and v. Comparing Eqs. [11.7.12] represents the rate of change [I 1.7.121 and [11.7.141. we observe that Eq. [11.7.121 ol' angular momentum per unit length about 0. This is expected, because we were considering a pinned-free beam whose equation of motion for the primary motion

is a moment balance. One can show that Ibr for the free-free beam. the translational equations of motion represent force balances. When we use the approach of selecting 0 and its derivatives a priori, then 0 and its derivatives become known quantities and they have no variations. Equation [I 1 .7.111 remains the sole equation ot' motion, as there is no longer an equation of motion for the primary motion.

1 1 .7

11.7.2

DYNAMICS OF QMBINED ELASTIC ELASTIC ANt) ANt) LARGE LARGE ANGLE ANGLERK;ID RK;ID BoDy MOTION MOTION

MOTION IN DISCRETIZED FORM

EQUATIONS

The equations of motion derived above are in the form of hybrid equations. a combination of' of ordinary ordinary and and partial partial differential differential equations. equations. One way of dealing with these equations is to discretize discretize Eqs. Eqs. [11.7. [11.7. 101 101and and111.7. 111.7.111. 111.To Tothis thisend, end, one one can can make make use of several discretization methods. An alternative is to begin with a discretized description ol the secondary motion and introduce this description to the extended Hamilton's principle. We illustrate this approach next and make an an assumed assumed modes expansion of' (lie secondary motion of order n as fl

= v(x, t) = v(.x.

[11.7.15]

i/i,.(x)qr(t)

where are admissible functions and q,.(!) are are generalized coordinates. For the pinned-free beam under consideration, introducing Eq. [1 1 .7. 15] to the kinetic energy energy and and potential potential energy energyininEqs. Eqs.111.7.41 III .7.4 Iand andIII III.7. .7.11bJ, bJ, we we obtain obtain -L

ii

t)dx =

2 1)

A2

2J0

= =

> r—

fL

fl

I

s—I

U

/.L. ( .v

1) dv = i)d..v

( .v,

[M 1{c'}

"> ,.=

I

,c= I '1

2J0

—I

'-

H

I

-V

2.0

I

(L

=

-

I

[Ii 1 { q }

2

L

P(x)v'2(x, P(x)v ix, t)dx t) dx J

I

•)

2

--

jr 1',.

-J J

111.7.161

=

s-I

and k,.çare aredefined definedin in Eqs. Eqs. [ii [ii .5.4J .5.4J and and Ill Ill .5.5J. .5.5J. and -I.

r '-L

.v/L( v .V/L(.V —

ilrsqrqsO hrsqrq.vO

"2 E1(x)v E1( x )I/2( (x. 1)dx 1) dx +

=

=

•1

)20'uxdx == — )2Ouxdx 1• — v—I 1• — I .v

I

iii which

=

t)dx = (9

dx .v) (IX

h,.r =

(A

xp1(.v) X1tL(.V)

LILT

[1 1.7.17o,b]

Jo

()

When the mass density is constant Eq. [

1L IirS

=

2 — JO

1

11

.7. 1 7hj becomes

[11.7.18]

f

— —

From Eqs. Eqs. [11 [11.7.61 .7.61 and and [11.7.91. [11 .7.91.the thevirtual virtual work work is is expressed expressed as 'I

ô14' = T[(5() + (SV(O)} =

+

[11.7.19]

649

650

CHAPTIR

11•

BoviEs

DYNAMICS OF LIGIrrI1v

and the variation of the constraint becomes C =

r

A>

Cr(5qr

Cr =

C(X)1,lJr(x)x)dx C(X)1,lJr( dx

j

[1 1 .7.201

r=.1

From now on. we consider that the admissible functions have been selected such that Cr Cr = 0 (r = 1,2, . . ,n). Next, we invoke Lagrange's equations. For the coordiOfs nate 0. the integral ofs the term in the kinetic energy gives Carrying out the algebra. we obtain the equation of motion for the primary motion as .

.

.

Ii

1!

a,4,.

+

\

H

+ 26

+ 00 >

ii

— hr.s

)qrq3

H

r=I r=I

-

11 1.7.21] 11 1.7.211

T

— I

or in column vector form as {a}T{q} + {q}T(IMI be + {a}

—

[/i]){q}O + [/iJ){q}O

—

IhJ){q}O =

r [11.7.221 111.7.221

This equation can also be written as

(lao + {a}TW} +

—

[hJ){q}0) =

111.7.231

T

and the term in the big brackets is recognized as the spatially discretized form of the

angular momentum about point 0. For the secondary motion, we follow a similar procedure and obtain the equations of tiOns of motion motion as + +

>

+

— —

lflrs) +

=

r = 1,2, ..

.

[11.7.24] or.

in column vector format as {a}Ô

+ [M]{j} [M]{/} ±± (([h] (([h]

—

[MI)02 + IKI){q} = T{t/J'(O)} I

111.7.25]

Both the primary and secondary motion equations are nonlinear. The nonlinearity anda source of coupling between the two motions is through the [Mi — [h] term and 62. Had we not included the shortening of the projection to the formula-

tion, the [hJ matrix would be absent from the equations of motion and the coupling would be influenced only by [M]. For Case 2 problems. it turns out that that [/'l, [/'l, II MJ, MJ, and Ihl — IMI are all positive definite (see article by Smith and Baruh). so that neglecting the shortening of the projection changes the sign of the coupling and leads to an erroneous mathematical model. The contribution of [h]0 in Eq. 111 II 1 .7.25] is known as the centrifugal stiffening 102 is known as the centriflistiffeniizg and and the contribution contributionofofIM IM102 centriñigal softening. For Case 2 problems. the centrifugal stiffening is always larger than the centrifugal softening. When the primary motion has three rotational components. centrifugal stiffening does not always dominate the centrifugal softening.

I 1I .8 .8

ANALYSIS OF' OF THE THE EQCAUONS EQUATIONS oil OF Mo'I'IoN Mo'I'IoN

When the angular velocity of the tracking frame is slow. one can ignore the term — [hl){q}6 trom ([hi — [M1)02 the secondary and trorii the primary

motion. As a result, one obtains a set of linearized equations that describe the overall motion in fliOtion in the the Foriii Fonii = r + + [M]i/ + [KF{q} = T{l/I'(O)}

[1 1.7.261

These two equations can be combined into a single matrix equation of order ii + I

[m*]{(} + where

matrices 1

{F'} = ri (0) rI j are of order iiii ++ I and have the form

= 10 10 qi

q,,

. . .

and I (il

1/fl I [inj

if1 1fl1

I

fl?, i i

Un an

.

I

1.7.27] 1.7.271

t/1,(0)1T,

. .

and the

1

a1 (11

11 11

=

/1122

ii

.

.

.

. .

. .

.

in1,,

.

.

.

im,,

.

.

.

in,1??

.

[k* Ij =

0

0

0

.

.

.

0

0 0

k11

k12

.

.

.

ky,,

..

.. .

.

. .

k2 i k1

()

i

k,,2

k,1,7

[1 [1 1.7.281

with Ffl,-, Ffl,-,and and (r,s = 1. 2. (r.s ii) denoting the entries of [M} and [K]. It should also be noted from Eq. 111.7.251 mo111.7.251 that the location of the external mo.

..

.

ment affects the equations of motion for the secondary motion, but not that of the primary motion. corroborating the statements in Chapter 8 that the location of a concentrated moment does not affect the rigid body motion but that it affects the elastic motion.

1

1 .8

ANALYSiS OF THE EQUATIONS OF MOTION

In this section we examine the nature of' the equations of motion, motion. the individual terms in the equations of motion for the various constraints discussed in Section 11 .6. and suitable trial functions to be used as admissible functions. We primarily consider the pinned-free beam from the previous section. so that r0 = 0. We also discuss issues associated with the free—free and interconnected beams.

The Rigid Body Constraint

In this this approach approach one selects the body axes such

that there is no rigid body component component in in v(.v. v(.v. t). t). All All of' of the rigid component of the motion is described by the body angle 0. The constraint equation is given by Eq. [111.6. [1 .6.10]. 10]. from from which which the the constraint constraint function function c(x) is recognized as

cix) =

111.8.11

fL(X)X

We We now investigate investigate trial trial functions functions /i,.(x) (r = = 1,2. .ii) .iz) to to use when expanding the secondary motion. We are looking for trial functions that automatically satisfy this constraint. One suitable suitable set set Consists consists of the eigenfunctions of a pinned-free beam with the same mass and stiffness characteristics except for the rigid body . .

.

65 11 65

652

CHAPTIR 11 • DYNA\IICSOF OFLicirnv LicirnvFI,KxIBLE BoDws BODIES

eigenfunction. We then have = &. 1(x) 1(x)

111.8.2]

which, of course, satisfies the relationship -I..

X).V(br+i (.\) (lx = 0

= ()

[1 1.8.31

()

by virtue of the orthogonality of the cigenfunctions to the rigid body mode

(x) = (x) x. Another suitable set oI'trial oltrial functions functions is is polynomials polynomials orthogonalized using a GramSchmidt process. Such polynomials can he normalized to satisfy satisfy Eq. Eq. [11 [11 .8.31. .8.3]. When using the cigenfunctions of the pinned-free beam When beam as trial l'unctions. functions, we obtain the following values for the coefficients discussed in the previous section =

kr.v krs

Sr,s Sr.s

Lir = 0

=

C1 = 0 Cr

[1 1.8.41

where w,w,-are are the the natural natural frequencies frequencies associated associated with the elastic motion of a pinnedfree beam. When When orthogonalized orthogonalized polynomials polynomials are used, the entries of are used, are no

longer natural frequencies of the corresponding beam. Also, [Ki is fully populated. as opposed to being diagonal. When the eigenfunct ions of of the pinned-free beam arc eigenfunctions are used as admissible functions the equation of motion for the Pnrnary primary motion becomes

\'

fl

F,

10O(t) + (9

Fl

F,

+ 20

—

>

1

>(cS

= T(t)

—

11

1.8.51

r=Ls—1

As discussed earlier, this equation can he expressed as an angular momentum

balance

(ho -f- {q}'([M J

=

—

T(1)

[11.8.6] [11.8.61

where the terms inside the square brackets

Jo +

[M I

Jo

I

—

(hi

11

1.8.7] 1 .8.7]

recognized as the mass moment of inertia of the deformed beam about the z: axis. axis. To see this, we consider the definition of the mass moment of inertia about the : axis, are

Jo

=

f r• rdiii

((x + ii)i + vjj [(x + u)i + v,jj din ((x

= -

= =

Ignoring the

1

1

2xu ++ v2)din v)dm (x + u + 2xu

11 1.8.81 1.8.8]

term, introducing the the discretization discretization[11 L. 11.7. .7. 151 15 I and and the definition of

ii. gives Eq. [1 1.8.7j. In essence. the elastic deformation changes the magnitude of the mass moment of inertia. This change is dependent on the sign of As — discussed earlier, — [M] [MI is positive definite fbr the type of' of motion motion considered here, so that while these terms increase the stillness in the secondary motion equations. they reduce the mass moment of inertia. When {q}T( MJ — [hi ){q} ){q} becomes becomes very I

1 'I .8

ANALYSIS OF THE

OF MOTION

large. one may reach a very small or even negative value for the inertia iflertia term in Eq.

[I I I11.8.7], .8.7],which whichisisnot notpossible possible physically. physically. This. of course. takes place when either olic or both of 0 and the the secondary secondary motion amplitudes are high. These parameters basically define the accuracy range for the assumption used in viewing the motion as a primary plus a secondary motion. For the secondaiy motion fllOtioflwe we obtain obtain (Jr(t) + (lr(t)

+ k,.ç Iqc(l) =

.

i

(0)

,• = 1,2, ,•

.

.

.

[1 1.8.9] [11.8.9]

,ii

An interesting property when using the rigid body constraint is that the equation of motion ftr the primary motion contains no acceleration terms from the secondary motion, and vice versa. The inertia matrix for the combined system is diagonal. thus = diag(10. 1. 1, 1..... 1...., 1)

[11.8.101 111.8.101

This feature makes it easier to integrate or manipulate the equations of motion nu-

merically. In addition, the rigid body constraint gives one a better physical interpretation of the terms involved. While these features are very desirable for analysis and simulation of the system behavior, the rigid body constraint loses its appeal when one needs to take measurements or conduct experiments. This is because of the additional effort needed in locating the relative frame from actual measurements. For the free-free beam, one can select as trial functions the eigenfunctions of that beam, or orthogonalized polynomials, as in the pinned-free case. For the interconnected beam, the situation is again different. Here, Here. if' if one one selects selects the the eigenfunctions of a free-free beam as the admissible functions, the CBC at the ends will be violated, because these trial functions describe a system with zero shear force, and there are nonzero shear furces at the ends of interconnected links. Hence, pinned-free or freefree eigenfunctions do not constitute a set of consistent admissible functions for the interconnected beam. Orthogonalized polynomials do not have this problem. However, the problem they do have is regarding convergence and numerical accuracy of' the solution. Polynomials of order higher than seven or eight often lead to singularity. Also. as discussed in Section 11 .5. a larger number of terms are required from polynomials than from other types of trial functions, for the same desired accuracy levels.

The Zero Slope Constraint

This constraint selects the rigid body angIe 0 such that the slope of the secondary motion is zero at the pinned end. or v'(O. v'(O. t) = 0

[1 1] [1 1.8.1 1.8.111

The constraint function c(x) is expressed as

[11.8.12]

c(.v) = c(v)

Substituting the constraint constraint in in Eq. Eq.[11.8. [11.8.11 liJ Jinto intoEq. Eq.[11 [11 .7.20] .7.20] we obtain -I -L

•11.

= .0

C(.V)1/I,.(x) dx C(.V)1/1r(X) dx

=

dx .0

— —— —

= 00

,

= 1,2, 1.2,

.

.

.

[11.8.131

653

654

.

CHAPTIR

11•

OF

LIGHiLY FLEXIBLE Bovws

The trial functions functions(/Jr(X) /i,.(x) that are used in the expansion ofthe secondary motion

are selected such that they have zero slopes at x = 0. It follows that the essential boundary conditions these trial functions should satisfy are [1 1.8.141

= 0

ifi,(O) = — 0

There are several choices when selecting the trial functions. One reason why

there are more choices here than tbr the rigid body mode constraint is that the trial functions need not satisfy' any orthogonality relations. Two functions that immediately come to mind are the eigenfunctions of a fixed-free beam and simple poiypoly.VrI nomials, where nomials. Using the trial functions that satisfy Eqs. [11 .8. 14] x) = •VrI together with the assumed modes method, we obtain the equation of motion for the primary motion of the pinned-free beam as II

toO /00 ++

and

\

Ii

11 Ft

fl

F?

+ > (trill (lrij,. +

N

r=I

r—1,v—1 r—1 s—I

—

hr.v)qrqv + 2f1 20 hr.v)qrq.v

Il

('flrs Y > (lflr5

—

hr.y)ijrqs = T hrs)ijrqs

r=-1

[11.8.15j

for the secondary motion as I,

11 Ft

F?

a,.0 + a,.O

+

+

—

s-=1

If the trial functions t/Jr(X)

= ri/i.(O)r = 1. 2,

. . . ,

ii

[11.8.16] 111.8.161

selected as the eigenfunctions of a fixed-free beam with the same cross section and stiffness properties as the beam at hand, then and = (r.s = 1,2. ,n). However, the a,. terms, which lead to = a nondiagonal inertia matrix, do not vanish, regardless of the choice of trial functions. The discussion in the preceding subsection with regards to the dominance of the centrifugal stiffening over the softening terms is valid when the zero slope constraint is used. However, one can show that the magnitudes of the hr,s — mrs terms are much larger when the zero slope constraint is used, as compared with the rigid body constraint. This leads to a less accurate mathematical model, whose range of applicability for 0 is much smaller than the model obtained using the rigid body are

. . .

For a free-free beam, the formulation and the types of trial functions one can use are the same as for the pinned-free beam. The origin can be selected as a point in the interior of the beam, or as one of ot the ends. When the interior is selected one needs to use two sets of functions, from 0 to the ends. For an interconnected beam, one can select the origin as any point on the beam, including the end points. However, the eigenfunctions of the fixed-free beam do not

constitute a set of consistent trial functions, as they violate the CBC at the interconnected ends. The same situation is encountered when using eigenfunctions of a pinned-free beam with the rigid body constraint. The primary advantage of this constraint is that it makes it very easy to measure the location and orientation of the reference frame.

I

5See 1. Yu, master's thesgs, Rutgers University, 1995.

1 1 .8

The Zero Tip Deformation Constraint

ANAIXsIS OF Ti-IF EQUATiONS OF' MortoN

This constraint leads to the following

geometric boundary boundary conditions conditions for br the the admissible admissible functions:

= 00

çlir(L) i./i,.(L)

ri = 1,1,

= 0

2,

. . .

,n

[1 1.8.171 111.8.171

where L is the undeformed length of the beam. The equations of' motion for the pri-

mary and secondary motions have the same as Eqs. [1 1 .8. 151 and {11.8.16J. Again. ii' the eigenfunctions of a pinned-pinned beam with the same measurements and mass and and stiffness properties ofthe beam considered are used as trial ftinctions, then mrs = and n). Also, a,. a, 0, (r,s = 1,2, 0, regardless = regardless of the choice oftrial functions. Another set ofsuitable trial functions are simple sinusoidals. which become the actual eigenfunctions if the beam is uniform. Sinusoidals are easicr to deal with than hyperbolic sines and cosines or polynomials. Further, they have better convergence characteristics and very few singularity problems. Another advantage of the hero tip deformation constraint is that the trial functions do not violate internal internal force force and and moment balances for all of the pinned-free, free-free. free-free, and interconnected beams. This is especially important for the interconnected beam, as in the other two constraints we discussed, the only set of trial functions that did not violate lbrce and moment balances at the boundaries were polynomials. Also, from an implementation perspective, the location and orientation of the tracking reference frame can he measured directly by taking measurements at the joints. . . .

Comparison Comparison of of the the Constraints Constraints and and Convergence Convergence Issues

When model-

ing Case 2 problems. the type of constraint to select and the types of trial functions to use depend on a variety of factors. The rigid body constraint appears more logical. especially for pinned-free and free-free beams, as it permits use of the exact eigensolution of the beam as trial functions. The boundary conditions of the secondary motion arc the same as the original beam. The discretized equations of motion are likely to have fewer terms. making it easier to numerically integrate these equations. Plus a,. = 0 (r = 1,1, 2, .... ii), another desirable feature for numerical integration. On the other hand, the choice choice of of trial trial Functions functions is limited. And, there is the problem of locating the origin and measuring the orientation angle 0 when seeking measurements.

With the zero slope constraint, the discretized equations are more complicated. but there are more and easier choices for selecting the trial functions. The trial functions do not need to satisfy any orthogonality conditions. A disadvantage is that the assumption of viewing the motion as as a superposition of rigid and elastic motions loses its accuracy much faster than the rigid body mode constraint. In addition, polynomials have much slower convergence properties than sines or hyperbolic sines. On the other hand, the origin and orientation of' the reference frame can be easily measured. The zero tip deformation constraint emerges as an alternative somewhere in the middle. For pinned-free and free-free beams it does not have the nice properties of the rigid body mode constraint, but it does not have the convergence problems associated with the zero slope constraint. For the interconnected beam, it definitely

655

656

CHAPTER 11 11••DYNAMICS CHAPTER DYNAMICS OF

is

Li;iiiix FLEXIBLE FLEXIBLE

the better alternative. as a set of consistent admissible functions is very easy to

find. summarizes the the different differenttypes typesofof trial trial functions discussed in this Table 1 1.7 suiiimarizes

section and whether they qualify as consistent admissible Functions or not. When we refer to eigenf'unctions of pinned-Free, fixed-free, or free-free beams, we mean eigenfunctions of' of aa beani beani with with the the same same dimensions. dimensions. material properties; properties: and crosssection as the beam being considered. The accuracy of the eigenfrequencies of the linearized problem is one indicator of the accuracy to be expected of the overall motion. hut it is not the only indicator. Another indicator of accuracy is the numerical complexities associated with the manipulation of trial functions. Especially when one deals with hyperbolic sines and cosines, as well as with polynomials. the trial Functions are difficult diffIcult to handle numerically. Hyperbolic sines and cosines arc extremely sensitive to the coethcients coethcicnts involved. For instance, if the model parameters are not known accurately. the trial functions constructed using these erroneous values will not be a complete set and, lunctions = and even hence, they will give incorrect results. The function the higher modes. This can remay not be small, small. especially if is small. model substantially. substantially. With polynomials, one duce the accuracy of the mathematical mathematical model should not use too many trial functions, as this leads to singularity problems. Also

Table 1 1 .7

Comparison of certain trial functions

'l'ype of Beam

Type of Constraint

Pinned-free Consistent Admissible Function

Free-free Consistent Admissible Function

Interconnected Consistent Admissible Function

Diagonal

Rigid body Eigcnfunctions of Eigcntunclions pinned-free beam

Yes

Yes

No

Yes

Yes

Yes

Yes

No

Yes

Yes

Yes

No

Yes

No

No

Yes

Yes

No

Yes

No

Figenfunctions of fixed-Iree uniform beam fixed-tree

Yes

Yes

No

No

No

Simple polynomials

Yes

No

Yes

No

No

Simple sinusoids

Yes

Yes

Yes

No

No

Eigent'unctions of Eigenfunctions pinned-pinned beam

Yes

No

Yes

Yes

No

a, =

Elgenfunctions of

free-free beam Orthogonalized polynomials

Zero slope Eigenfunctions of m Ii x ed - free bca iii

Zero tip deformation

0

ANAIxsIs OF THE EQUATIONS OF Morios ANAIxsls

1 II 1 .8

with polynomials. oniy vcry few of the approximate eigenvalues are close to their actual values. involves a comparison of Another interesting comparison ofthe different magnitude of ofthc the strain. While the magnitude the secondary secondary motion motion is different for the different constraints used. the strains are comparable. as they are derivatives of the secondary motion. We summarize the issues to be considered when selecting the model to describe the secondary motion and the trial functions: I.

The objective in the mathematical modeling. Is it analysis. measurement, or navigation and and control'? control?

2.

3.

4. 5.

Are the trial functions a set of consistent admissible functions? 0! Are the coefficient matrices diagonal and are a1 = 0? How sensitive are the trial functions to numerical as well as other types of errors? What are the convergence characteristics of the trial functions and how many trial functions arc needed to have a certain number of accurate modes?

A pinned-Free uniform link is made of steel and has the following dimensions properties: cross psi. unit weight = 0.28 lb/in3. Find the 12 ft. E = section 3/8 3/8 x 66 in. section in. length eigcnsolution of the beam for small motions, and calculate the /lrc tcrms and conipare them with the natural frequencies using the rigid body mode constraint.

Solution The normalized cigenfunctions of a pinned-free beam are used as trial functions. These func.4. For tions were calculated calculated in in Example Example11 11.4. For the the linear linear elastic elasticmodel, model,nir., nir, = 5rs, and krc krc = of orthonormality orthonormality of the trial functions. by virtue virtue of where w,. are the natural frequencies, frequencies. by These diagonal elements, elements, along along with with the the centrifugal centrifugal stiffening stiffeningcoeflicients coeflicientsh,.., h,., calculated us-

ing Eq. [11.7.18]. are given in Table 11.8. Further, comparing the As one can can see. see. the thehrs. hrs.ternis ternisare aremuch muchlarger largerthan thanihe ihe111rr terms. Further. — in,., terms. so that these hesubstantially substantiallylarger largerthan thanthe the/1g.!, h,.., — elements kr, WC lind them to he terms have an effect on the secondary motion only when the primary motion is quite rapid. However, their effect on the primary motion is niore more important (see article by Baruh and Tadikonda).

Table 11 1 .8

Comparison of coefficients in equations of motion

h,.ç

rls 1

2

1

6.397

2

1.861

3

(1366

4

0.121

1.861 1

—

7.90

3

4

0.121

—0.366

6.195

6.195

35.Y9

1.48()

12.64

—

1

1.480

2.64

60.67

1

1

1

1

inn.

k,,.

.000

250.3()

.000 .(.)00 .000

.000

2629.1 1

1.445

Example 1 1.8

658

Example 1

1 .9

CHAPTER 1 1 • DYNAMICS OF Licwnx FLEXIBLE BoDIEs

I

Consider the pinned-free beam in the previous example and compare the cigenvalues using the three approaches.

Solution The linearized equations are given by Eq. 111.7.27]. Expressing the configuration vector as .{q*(tfl = {z}eM, the associated eigenvalue problem has the form

+ Fktl){z} We

[al

{O}

use the following trial functions in each problem:

Rigid body constraint: (a) Eigenfunctions of a pinned-free uniform beam with the same dimensions as the beam in the previous example. (b) Orthogonalized polynomials. Zero slope constraint: (a) Eigenfunctions of a fixed-free uniform beam of the same geometry as in the previous example. (h) Simple polynomials. x' (r = 1, 2, .

Zero tip deformation constraint: Simple sinusoids, sin rirx/L. where L =

12

.

ft. the

beam length.

All the trial functions qualify as consistent admissible functions. The polynomial trial functions give the same results for both constraints they are used with, even though the [mx] and [k11 matrices will be different for the different constraints used. The nonzero natural frequencies are given in in Table Table 11 .9. are given the As expected. the pinned-free eigenfunctions eigenfunctions give givethe thebest best results, results, because because they are the As exact solution for the linearized problem. The results obtained using the fixed-free eigenfunctions arc very close to the actual values, except for the last natural frequency. The frequencies obtained using simple sinusoids are slightly less accurate than the fixed-free eigenfunction results, except for the last frequency. when they are much better than the fixed-free results. Polynomials give relatively good results only for the first four frequencies. after which their accuracy diminishes substantially. Note that this problem can also he viewed as an example for Section 11.5, where the accuracy of different trial functions is compared for the linear vibration of a uniform pinnedfree beam. The rigid body mode constraint results are basically the exact solution.

Table 1111 .9

Comparison of nonzero natural frequencies For pinned-free beam

Mode

Rigid Body

Zero Slope

No.

Constraint

Constraint

Zero Tip Deformation Constraint

Poly nomials (for Both Constraints)

2

15.827

15.827

15.827

15.827

3

51.288

51.289

51.293

51.292

4

107.01

107.02

107.05

107.15

5

182.99

183.07

183.21

194.72

6

279.24

279.72

280.08

325.45

7

395.74

398.32

398.82

1030.6

8

532.51

895.21

549.96

1943.4

1 1.9

11.9

3, ROTATiNG ROTATiNG SHAFTS SHAFTS CASE 3,

CASE 3, ROTATING SHAFTS

When the angular velocity of the reference frame is along the x axis. the motion is similar to that of a rotating shall. Because in such problems the cross section is circular, we will consider both v(x, 1) and w(x, 1) 1) in the formulation. Consider a rotating shaft, shown shown in in Fig. Fig. 11 11.35. .35. The reference frame is attached to one end of the shaft and the x axis is along the neutral axis. The velocity of a point and on the center line of the shafi can be obtained by setting r0, to zero in Eq. [11.6.31 and using the notation which yields =

+ wk+wi + Wk+WrI

i'(x,t) =

X

[(x+ u)i+ vj+wki [11.9.11

=

Assuming that ü is negligible compared to the other terms, the velocity expression further reduces to r(x. 1) r(x,

—

—

(11.9.2] (11.9.21

thv)j 9w)j + (ii' + Ov)k

The expression for velocity gives the indication that we are dealing with a gyroscopic system. The kinetic energy associated with the translational motion of the shaft then becomes t) ••ihI(x, t) i'(x, 1

çL

+

2J()

+

t) dx + w2) +

—

(11.9.3]

dx

To find the rotational kinetic energy. we need to calculate the angular velocity of a differential element on the shaft. Consider the formulation in Section 1 1.2. The difference here is that the xvz frame is rotating with 0 while in Section 11.2 11.2 the V

9

x I

FIgure 1 1.35

A rotatin9 flexible shaft

659

660

CHAPTIR 1 1 •

DYNAMICS OF LIGHTLY FLEXIBLE Bonws

xyz frame was fixed. Here, the rotations arc by v'(x, v'(x. t) about the z axis and then by —w'(x, 1) about the rotated V axis (or first by —w' and then by u'). This is basically a 1-3-2 (or 1-2-3) Euler angle transformation. Making a small-angles assumption and recalling from Fig. 11.3 that is the rotated frame, that of the beam axis, we approximate the angular velocity of the differential element as

=

+ i'k — Wj* =

Oi

sV'j

cv'swk) +

—

—

(0 +

+

— —

[11.9.4]

+ (—Ow' (—Ow' +

+ (—Ov' (—Ov' —

Wj*

view of the fact that 0 is often much larger than the secondary motion, the angular velocity can he simplified to In

=

—

Ovl,j*

[11.9.5] [1 1.9.5]

—

differential element has mass moments of inertia of = = = Assuming Assuming that the the xvz axes are selected as principal axes, so are and the rotational kinetic energy can be approxiThe

mated as =

+

I

dx

+

11 11 1.9.61 1.9.61

Often, Trot Trot is is small small enough enough to to be be ignored. ignored, or, as in Eq. 111.9.61. only very few

terms are retained in its description. Trot usually becomes significant in the presence of disks attached to the shaft. and it is calculated only for the disks. The potential energy has the form 1L

v =

[1 1.9.7]

+

Invoking Invoking the extended Hamilton's principle, we obtain the equations of motion.

Note that when evaluating the variation of a term like W, one uses two integrations by parts. parts, one with respect to time and the other with respect to the spatial variable. Omitting the details of the integration by parts. parts, the equations of motion can he shown to have the form +

+ 201;'v') +

+ .. —

+

p.(x)Ow ,u.(x)Ow

+

+ w2) +

L

.

—

+

Op(x)Ow —

+

+

—

ji(x)62w + o

x

+ )

—

F, J

+ +

L

wü) Wi))

= 0

,, ,, F,,, )

= 0 = 0

[11.9.81

The equations of motion associated with the secondary motion contain Coriolis

as well as centrifugal terms. The centrifugal terms arise from the rotatory inertia and from the centripetal acceleration. They have a softening effect, while the Coriolis terms contribute to added stiffness.

1 1.9

661

CASE 3, ROTATING

Obviously, the equations of motion are quite complicated, so that one usu-

ally seeks an approximate solution. We demonstrate the use of the assumed modes method in Example 11.10.

Consider the uniform circular shaft of length L in Fig. 11.36. On the shaft there is a rigid disk

of mass M and moments of inertia I and J, with I being about the shaft axis. located at a distance a from end 0. Derive the equations of motion using the assumed modes method, by assuming that the shaft speed is kept constant by a servomotor at kinetic energy energy of of the the shaft. shaft

=

11

and by ignoring the

Solution The secondary motion has the same boundary conditions as that of a pinned-pinned beam in w(x, t) in and w(x, both transverse directions. We will use a one term expansion for both v(x, t) and

the form

i!/(x)q1 (t) v(x. t) = i!i(x)q1

[aJ

w(x. t)t) = w(x. = 'f'( 'f'(

= sin sin irx/L is the first eigenfunction of a pinned-pinned uniform beam. There justification t). The The justification is no problem with using the same trial function for both v(x. t) and and w(x, w(x, t). in which

for ignoring the kinetic energy of the shaft is based on the fact that the mass moment of inertia of the disk is usually much larger than that of the shaft. The rotational kinetic energy has the

Trui

[bi

1) + w'2(a, t)1)

+

=

From Eq. [ii .9.31 the translational kinetic energy has the form

=

M

t) +

i)ii(a, 1)]) — w(a. i)ii(a. t) —

1)] + '2U[v(a,

1) + 112Lv2(a. 1) +

[c] with the potential energy defined by Eq.

I .9.7].

We next introduce the assumed modes expansion expansion of of v(x. v(x. 1) 1) and and w(x, w(x, t)t) into Eqs. [hi, Ic] and [11.9.7], with the results Trijt = = Ttra,, Twa,,

+

= =

+

=

±

=

+

±

+

.1-1-

+

V=

I

—

+

+

1

q;) =

I, 1

a-

$..— aa — —

Figure 1 1.36

1

q,/1

)]

Example

11.10

662

CHAPTER 1 1 • DYNAMICS OF L1;IITI.v L1;IITIx

Boiiws

i/i(a). Z = = i/i(a). (U (U isIsthe thefirst firstnatural natural frequency frequency of of aa pinned-pinned pinned-pinned bar with the same geometry and material niaterial properties. and Ig] is a skew-symiiietric niatrix matrix in the form which )r'• in which Y

IeJ

=:

II g I

•

j

The constant constant B B relates relatesthe thefirst firsttmrmalivecl norinalivcd cigen cigen tuntions tuntions(I)(.v (I)(x to ) by 8/1(.v. for 8/1(.v. for a uniform uniforni beam B = The ternis terms in Eqs. can he combined as )

T-

+

If]

V=

[G1{qJ

)= )

ill which [M21 = MY

)

2MY[gj IGI = 2MY{gj

?

=-

igi [gi

(JZ2 + M M Y-)

(K1 = [

Ii

Because all coefficient matrices are time invariant, we make use of the developments in Section 5.3 and directly write the equations of motion as

+ ([K I

+ 11

Roth

2

—

I

[hJ

= {O}

I

the terms that that have have iF iF in in them them(JZ2 (JZ and and MY2) MY2) contribute contribute to the

effect. and it can he visualized easily. The contribution

The centrifugal centrifugal component componentA1)P2 MY2 is from

due to the rotatory inertia. JZ-. may seem puzzling at first, hut can he explained by considcring that rotatory inertia adds to the kinetic energy that has to he balanced by the potential energy. The kinetic energy becomes higher when one includes the rotatory inertia term, so so that in essence this corresponds to a lower relative stiffness. Be aware that Eq. [hj represents a gyroscopic system system SO so that that the the reduction reduction in in stiffness stiffness due to increasing does not describe the entire dynamics. When a = L/2. that is, when the disk is in the middle of the shaft. the rotatory inertia terms have no contribution for the tile order order of of expansion expansion considered, considered, because because i/i'(L/2) i/i'(L/2) = 0. in In this case, to capture the rotatory inertia contribution and to have a more accurate model one should use more than one trial function in the expansion of of'V( v( x. x, I) and 1). Let us next introduce the the inertia inertia of' of the shaft to the problem. In this tills case. we can safely assume that the contribution of the shaft to TF(,t will he negligible and the contribution to is given by Eq. [11.9.3]. Introducing the assumed mode expansion in Eqs. [al to Eq. [11.9.3J, we obtain =

+

+

+

±

— —

q241

[I] [ii

in which in -L

=

x) Lix J0

Iii

This expression is then added to the total energy. It follows that the equation of motion

retains the same form as Eq. [hJ. The elements of the coefficient matrices change as we replace MY2 with MY2 MY2 ++ iW in [M2]. [Gjand andFM0 I. As before, when seeking a two or higher-order [M2]. [Gj term expansion of the secondary motion, the additional will he more complicated. A very important phenomenon in the motion of shafts is that of whirling. At certain rotational speeds the shaft undergoes large amplitude vibrations. Such vibrations are often

11.10

HYBRID PROBLEMS

detrimental to the structural integrity of shafts. An eigenvalue analysis o1 the equations of

motion. Eq. [hi. is a useful tool to calculate shaft speeds under which whirling occurs.

1 1

. 1 0 HYBRID PROBLEMS

In previous sections we saw derivation of the equations of motion and analysis of these equations for continuous beams. The angular velocity was always in one direction. More complex models flexible systems undergoing rigid body motion include angular velocities iii more than one direction and bodies comprised of rigid as well as elastic components. In such a system the kinetic energy will have contributioiis from both the rigid and elastic parts. However. the potential energy will he butions due to the elastic part alone. This makes it difficult for a single continuous function to approximate the entire body. Consider, Consider, as as an example. the the model model of a satellite satellite with a huh of radius radius R and two antennae, as shown shown in in Fig. Fig. 1 .37. A suitable placement for the tracking reference frame is to attach its origin origin to the center of the huh and to align it along the undeformed position of tile the antennae. The position of a point on the antenna can be I

WFittCil as written

Antenna I: r1(x1. t) = [R + x1 + u1Lv1. u1(x1. t)]i + v1(x1. t)j + w1(x1, ok + U2(X2, 1)11 + V2(.V7. V2(.V7.i)ji)j++ W2(X, W2(X, t)k [1 1.10.11 Antenna 2: r2(.v2, r2(x2, 1) i) = [R + with .v1 x1 and and .12 .12 denoting the spatial variables. The secondary motion of the antennas

are then obtained using separate expansions for each antenna. One can expand and w, WI (i

1, 2) using the zero slope constraint and use fixed-free eigen functions or simple polynomials polynomials as ti-ia! trial functions. functions.One Onecan can also also invoke invoke the the other other constraints constraints we have discussed, but these constraints do not have as good a physical interpretation as the zero slope constraint for this problem. For example. if the zero tip deformation constraint is used one draws a line from one end of the antenna to the other end. If one wishes to use a single set of admissible functions to describe the motion

(from the tip of r1 to the tip of r2). the functions used should he able to model the zero elastic deformation in the huh. This may he a viable approach if the hub is small compared to the rest of the structure, but not so if the hub dimensions are anywhere close to the sii.e size of the antennae. For intercwrnected bodies, such as the one shown in Fig. 11 .38. the zero tip deformation constraint appears more suitable than the other two, because of the attractiveness of' using simple sinusoids as consistent admissible functions. The deformations on the individual links can he expressed as Link I Link 2: :

r1 (.v1

.

x1 + t) = f x1 t) = r1(L1,

(x1 ,, 0)

w1 (x1 + vi v1 (x1 (x1 t)k1 t)k1 (x1 t)j + w1

w2(x2, i)k2 + 1x2 + U-,(.Vi, t)j2 + w2(x2, m(.V2, t)112 t)1i2 + v2(x2, v2(x2, t)j2

111.10.2] where x1 x1 and and .v2 are the spatial variables and L1 iS is the length of the first link. Note that there is a separate set of unit vectors and orientation angles and for each link.

663

664

LII6HTLY FLEXII3LE BODIES CHAPTER 1 1 S DYNAMICS OF LII6HTLY

y

II

Xl

V., V .1•

V

Figure 11.37

Figuro 1 1.38

REFERENCES and Control Control of Flexible Robot Dynamics and Robot Manipulators." Baruh. H.. and Baruh, and S. S. S. S. K. K. Tadikonda. Tadikonda."Issues Issues in in the Dynamics Manipulators." 12. no. !)vnamics 12. no. 5. 5, 1989. 1989. pp. (;iizdance, ('mitral, and and !)vnamics Jouri:al of Journal at Guidance, pp. 659—71.

Courant, R., and I). Hubert. Methods Courant.

lnterscience, 1953. Phvsi('s. 1 . New York: Wiley lnterscience. Phvsu's. 1

.

7'/u'orv and wid Applications. Applications. Methods: 7'lu'orv Gottlich, D., D.. and S. A. Orszag. Nunu'rieal Numerical Analysis oj .Specira/ .Speciral Methods: J.\V. Arrowsmith. 1977. Society tbr Industrial and Applied Mathematics, .!.\V.

Likins, Laskin. R. A., P. W. Lik ins, and and R. R. W. W. Longman. Longman. "Dynamical "Dynamical Equations Equations of a Free-Free Beam Subject 507—27. Sciences331.1.1983. 1983.pp. to Large Overall AstronauticalSciences Overall Motions," Motions," .Journal .Journaloft/ic oft/icAstronautical pp.507—27. Eleme,its of of Vibration Meirovitch. L. Eleme,its

2nd ed. ed. 2nd

New New York: McGraw-Hill. 1986.

Computational Methods in Structural Dvnaniics. The Netherlands: Sijthoff Noordhoff. I Rivello. R. M. Theory and

of Flight Structures. New York: McGraw-1-Iill. 1969.

Smith, C.. C.. and Dominance of of Stiffening Effects in Rotating Flexible Beams." Journal Smith, and H. H. Baruh. Baruh. "The Dominance 1991, pp. no. 5. 1991, Guidance. Control, and Dynamics 14. no. of Guidance. of pp. 1072—74.

S. S. K.. and H. Baruh. "Gibbs Phenomenon in Structural Mechanics." Mechanics." AIAA AIAA Journal Journal 29. 29. no.9. 1991.pp. 1488—97. Multilink Manipulators1" Manipulators," master's thesis. of Flexible Multilink thesis, Rutgers University. Yu. 1.. Yu. 1.. bModeling and Control of

Tadikonda, Tadikonda.

May 1995.

EXERCISES

SEcI'loN 11.2

2.

Write the kinetic and potential energies. as well as the virtual work, for the beam is constant, shown in Fig. 11.39. The cross section varies as follows: follows: bb is constant, while h(x) = — .v/2L). Find for the beam in Example 11 1.

3.

Find the kinetic and potential energies for the bar undergoing torsion in Fig.

1.

.

11.40.

HOMEWORK EXERCISES

V

V

Ii I

-.—

cross section £1

L

—

Figure 11.39

V

-— ———--.--——— 7

Ed

_—_

---- -

Figure 1 1.40 SECTION 11.3 4.

Find the equation of motion and boundary conditions of the tapered bar in axial vibration shown in Fig. 11.41. The cross section varies as follows: b is constant, while /i(x) = h0(l — x/2L).

5.

Find the equation of motion and boundary conditions of the beam shown in Fig. 11.39.

6. 7.

Find the equation of motion and boundary conditions of the bar undergoing torsion in Fig. 11.40. Find the eigensolution and first five eigenvalues of a free-free uniform beam of length L. stiffness El, and mass density j.L.

/ t ho

1)

Ii

y

b

Figure 11.41

665

666

CHAPTER 1 1 • DYNAMICS OF LIGHTLY FLEXIBLE Bonws

V

•

.1

/7/

/ / //// 7/7 /// /7/////// L

I

I

FIgure 11.42

8.

9.

Find the eigensolution for the free-free uniform beam resting on an elastic foundation, shown in Fig. 11.42. The elastic foundation is modeled as a continuous spring of constant = O.4E1/L4. Comment on the rigid body mode. Derive an expression for the response of a pinned-pinned uniform beam to an impulsive force I) = FS(t)6(x — a). in which a denotes the point of application of the force. Analyze the mode participation, participation. and draw conclusions arbitrary values of a. Then, let a = L/2 and reevaluate your results.

SECTION 11.5

10. Using the assumed modes method, find the approximate eigensolution of the tapered fixed-free beam shown in Fig. 11.39. Ignore the dashpot. let = 0. a = L/2, k = 2E1/L3. Use approximation orders of 3.5. and and use as trial functions the eigenfunctions of a fixed free beam. Verify numerically that the approximate eigenfunctions obtained from the discretizalion discretization are indeed orthogonal with respect to the mass distribution. 11. For a pinned-pinned uniform steel beam of length L = 2 ui m and circular cross section with radius R = L/10, analyze the contribution of the rotatory inertia. Consider deformation in both the y and z directions. and obtain the eigensolution with and without the rotatory inertia terms. Then compare the two sets of' eigenvalues. Use as trial functions the eigenfunctions of a pinned-pinned beam and three terms in the expansion. 12. Given a uniform column subjected to an axial load P as shown in Fig. 11.43. find the load P that makes the first eigenvalue zero using the assumed modes method and three trial functions. Then, compare this value for P with the critical critical load required for buckling, = 7r-EI/L-. Plot the first natural frequency versus P. Note: Finding P requires some trial and error calculations.

p.El

-L

Figure 1 1.43

/

667

EXERCISES

/ H

ILL. El

—liii

I

1

1

1

0.75 in

1.45

13. Using the assumed modes method, find the eigensolution of the beam shown in Fig. 11.44. Use simple polynomials as trial functions and an approximation order of 5. Compare Compare the the results results for for the the following followingvalues valuesof ofk:k:0.0.1 1 EI/L3. EI/L3. 0.5 E!/L3, 3 E1/L3. 14. Consider the shall undergoing torsion in Fig. 1 .45. The shall manufactured of steel (G = 79 GPa. p = 7860 kg/rn3) by connecting two shafts oF of different diameter. Calculate the eigensolution of the shaft and list the first three frequencies. Use as trial functions the cigensolution of a uniform fixed-Free fixed-free shaft. 1

SECTIONS 11.7 AND 11.8

Consider Example 11.8 and calculate the values of /lrs when the zero slope constraint is used and as trial functions simple polynomials (x2. .) are used. Normalize the trial functions by setting in1, = I (i 1. 2, .. 4). and compare hjj with the the results results in in Example Example11 I I.8. .8. 16. Write a computer program to model the beam in Example Example 11 .8. Use the rigid body constraint and the trial functions in Example 11 .8. Simulate the response for a period of 15 seconds for an external moment moment of otM(i) M(i) = 1.2 l.2sin2Ot sin 20t Niii Nm at the pin joint. Do the simulation for n = 1. ii = 2. and n = 3. Plot the values for 0(t) and qr(t) for each case. Compare the results as the number of terms that describe the secondary motion are varied. 15.

.

.

.

.

SECTION 11.9 17.

Derive the equations of motion of the rotating shaft considered in Example 11

.

10. 10. This This time time use use two two trial trial functions functions each to expand v(x. I) and and %t'tv, %t'tv,

t). Cal-

culate the elements of the coefficient matrices. Include gravitational potential energy in your formulation. 18. Consider the previous problem and a shaft made of steel ol' length L = 2 m and radius 3 cm. The rigid disk is of mass 20 kg and radius 10 cm. Conduct a parametric study and plot the first two eigcnvalues of the approximate model as a function of the driving frequency IL in the U 2000 rpm. of 0

I

668

CHAPTER

1

1 ••DYNAMICS OF' 1 OF

FI.F:xlrn.E BoDIES FI.F:xlBI.E

SECTION 11.10

Write the the kinetic kinetic and and potential potential energies for the double-link robot arm shown in 9. Write Considerthat links are are flexible flexibleonly onlyininthe theplane planeof'ofmotion (XY). that both links motion (X Y). Fig. I I .38. Consider 20. Find the equations of motion for the robot arm shown in Fig. II .38. Assume that link is rigid and link 2 is flexible. Use one trial function to describe the secondary motion of link 2. Justify your choice for the reference frame. 11 .46 describes a uniform robotic arm of length L attached to a pin joint. 21. Figure Figure 11 directions. Find the equation of mumoThe beam has elasticity in both transverse transverse directions. Discuss your your choice choice of trial lion tion of the beam, using the assumed modes method. Discuss f'unctions for functions for each each transverse transverse direction. Neglect axial stretch. 22. Find the equations of motion for the robot arm in Problem 20 by considering that both arms are flexible. Use one trial l'unction function for each arm and justify the motion constraints that you are using for each arm. 23. Consider the model of a satellite in Fig. II .37. The satellite consists of a hub of radius R. R. and mass moment of inertia J. To the hub are attached two mass M. radius antennae, each of' of length L = 3R. mass density p. = M/5L. and stiffness El. arc along the same line. Write the In the undeformed position the two antennae are kinetic and potential energies of the huh. 24. Using the assumed modes method, derive the equations of motion for the hub in the previous problem. Which constraint should you use? Note: Each antenna needs to he described as a separate body. 25. Consider the equations of motion of the satellite in Problem 23. Linearize these (L3MR2. Use equations and solve the resulting eigenvalue problem. Let I = O.3MR2. two trial functions for each antenna. Plot the eigenfunctions. 26. Consider the satellite model in Fig. 11.47. The satellite has the same dimensions as the satellite considered in Problem 23. but now there are four identical antennae. whose undelormed positions are perpendicular to each other. Obtain the equations of motion using the assumed mode. method. and one trial function per antenna. 27. Derive the linearized equations of motion for the satellite in the previous problern and solve for the cigensolution. Use one trial function per antenna. lem I19.

I

V-i

y _.

L -V -t.

I

Initial

A4

Figure 1 1.46

Figure 1 1.47 Figure

.r

APPENDIX

A BRIEF HISTORY OF DYNAMICS A. 1

INTRoDucTIoN

This appendix presents a historical overview of the field of analytical mechanics. with emphasis on the history of rigid body dynamics. We also discuss the lives and achievements of the people who made key contributions. We present the accomplishments acconiplishinents of each individual. and explorc the interactions that took place among these brilliant minds. One ofthe greatest challenges when writing a historical document is that of perspective. Many of tile the scientists who contributed to this field also made significant contributions in other fields. For example. when studying the life of Isaac Newton. dynamicists tend to think of Newton's Ncwton's laws of motion as his greatest contribution, while mathematicians view Newton's greatest contribution as his development of the calculus. There are several publications about the history of mechanics, each one emphasizing different accomplishments. Even two major books on the subject (Truesdell (Truesdell and and Dugas) Dugas) at at times times disagree disagreeon onthe thelevel levelof' ofcontribution. contribution. Although a well-balanced history of dynamics may be hard to attain, the process is exextremely fascinating and rewarding. In this author's view, we get a much deeper understanding of' mechanics when when we know the historical background. It is my of mechanics my recommendation to every student of mechanics (or any other branch of science, mathematics, or engineering) to learn its history and benefit from it. The interested reader should consult not only this appendix and the references cited. but also other texts on the history of science and engineering. The sections that follow give a historical survey of the evolution of dynamics and the life stories of key contributors. contributors. The The technical technical community community isismuch muchindebted indebtedtotothese thesepeople.. people. We We should also acknowledge those who thought to study the history of science and technology. In mechanics. Pierre Duhem Duhern (1X61—1916) (1X61—1916)deserves deservesthe themost most credit: he he is is regarded regarded by by many many as the founder of' of the the field field of of history history of of science. Another individual who contributed to this subject significantly is Ernst Mach

A.2

HisToRIcAL EvoImloN OF DYNAMICS

That the motion of moving bodies has intrigued people throughout history should not come as a surprise. People have have always always be.en been interested not only in moving things from one place place to another, hut also in the motions of animals, clouds, and celestial bodies such as the earth, moon, moon, sun. sun, and stars. For moving and lifting objects. we have used two mechanisms since earliest times: levers and pulleys. Expectedly, some of the first significant contributions to mechanics were for levers and pulleys, as well as for weaponry. And the ancient study of celestial mechanics was carried out for scientific as well as religious purposes. The seminal contribution to mechanics is attributed to the Greek scientist Aristotle (384— 322 B.cF.). He is credited with the first written work, Problems of Mechanics. Aristotle derived

669

670

APPENDIX A • A

BRIEF

OF

lever rotates. rotates. the velocities of the when a lever the lever. the lengths of the arms of the weights at the ends of the lever will be proportional to the as Aristotle stated the principle that Interesting to note is that his starting point was theyare areinversely inverselyproportional proportional to the velocities. "forces balance each other 11' if they ofaa lever lever involves involves Because the principle Aristotle stated involves velocities. and balance of equilibrium and small deviations from equilibriulil. equilibriuni. Aristotle is also credited as having introconcept ofa ofa virtual displacement. Aristotlc's other research. which includes general duced the concept n-lotionanalysis. motion analysis.celestial celestial mechanics. mechanics. and the shape of the earth. is rwt as well posed as his must he a force motion. there must believed that that wherever wherever there there is a niotion. on levers. Aristotle believed results on on celestial celestial mechanics mechanics was priand he tried ried to to relate relate velocity velocity to force. His work Ofl causing thought that. that the only natural motion was niarily intuitive and based on his ohservations. He thought circular and that the sun and planets moved around the earth in circular paths. Aristotle's slightly by Ptolemy. hut it was hot not challenged for theory on celestial celestial mechanics iiiechanicswas wasretmned retined slightly over 1500 years. until the time otCopernicus and Galileo. Similarly'. his work on levers unrefined for a very long period of time. The next significant contribution to mechanics is attributed to Archimedes cqui), ), another Greek scientist. Archimedes built on Aristotle's work. refined Aristotles equllaws were published in his treatise. On libriuni librium laws. and proposed some oF his own.

law of equilibrium for a lever. He stated stated that that the law the

tiUf tiUf

UnlikeAristotle. Aristotle. who who at— o/ Planes. Planes. Unlike ofGravitv o/ i/u' Ce,,ier.c oJGravi!v Pla,it's ()F ()F on i/u' Equilibrium of Pla,it's

variety of static and dynamic problcnis. problems. Archimedes concentrated tempted to solve a primarily on bodies at rest. The lame ot Archimedes is primarily due to his contributions for principle). He also made contrihustatic fluid mechanics and floating bodies tions to geometry and the measurement of circles and of the number not usually flOt usually ii.) (365—---300iic iic1:.) Although the the famous Egyptian mathematician mathematicianEuclid Euclid(365—---300 niechanics, many Arab authors refer to him in thcir credited with a major contribution to mechanics, writings as having had sonic some contribution. particularly with regards to the equilibrium of levers. However, there seems to he no historical evidence that Euclids work on mechanics was known to Aristotle or Archimedes. We should note that Euclidean geometry is ideally contributions to to mathematics mathematics suited for the study of Newtonian mechanics: thus. thus, Euclids contributions paved the the way way For kr the theadvances advancesin in Newtonian Newtonian mechanics. PappUS (second century century cc L.) L.) and and Pappus In Egypt. other mathematicians. mathematicians. Hero Hero of' of Alexandria (second (fourth century CE.). made contributions to mechanics. Hero did work on pulleys and levers and considered levers of nonuniform cross sections and density. He knew about Aristotle's results and introduced the concept of a moment, albeit implicitly. Pappus is more known for his contributions to mathematics hut lie be explored equilibrium problems on inclined planes. as well as the center of gravity. Between the time of Aristotle and Galileo there were minor contributions to mechanics. which we briefly discuss here. In the first millennium millennium of. of the the common conimon era, era, primary contributions came from Alexandria and the Arab Peninsula. As mentioned. Hero of Alexandria contributed through his discussions of machines, machines. like levers. pulleys, and screws. Thahit ihn Kurrah (836—901 ). also of Alexandria. did work on geometry and on levers. There is uncertainty whether the hook Liher Cliaristonis is the Latin translation of his book or was written by the second-century mathematician Charistion. In the 1 3th century. the equilibrium of levers and of bodies on inclined surfaces was the focus of attention. The German scientist Jordanus (1225—1260) developed a theory where bodies would be heavier and lighter depending on the type of motion they were undergoing. accelerations. One One of' of the Hence, he realized the concept of inertia, hut withoul considering accelerations. reasons mechanics did not c','olve as fast as many other subjects (mathematics. for example) is that the study focused primarily on position and velocity. This is understandable, as it is visualii.e accelerations. accelerations. much easier to visualize displacement and velocity than to visualiie

A.2

HISTORICAL HISTORICAL

EvoLurloN

OF

The 14th and 15th centuries were the precursors to the scientific revolution. During this

period, three major concepts were analyzed more extensively, and certain Aristotelian concepts began to be challenged:

shape FflOtiOfl ojthe Ffl()tiOfl ()jthe cart/i. cart/i. Even though the correct shape of the earth was known to many Greek and Egyptian astronomers and physicists. the flat or cylindrical shape theory persisted fèr thr centuries. perhaps because of the influence of the church and the widely accepted Ptolcmaic theory. which placed the earth at the center of the universe. Albert of Saxony ( 1 3 I (i—I 390) theorized that the earth was round in the north—south direction. 2. The concept of This related the tendency of a body to continue moving. Impenis was a concept close to linear momentum, but it was not defined as such. Impetus was a combination of momentum and a hit of energy. and it made way for the development of momentum and energy as concepts that indicate indicate why why illotion continues. continues. Inipetus contradicts the Aristotelian theory. The person credited with first discussing impetus is Buridan (— I 300— — 1357) of France. Albert of Saxony and his French contemporary Oresmc( I( I323—1387) contemporary Oresmc 323—1387) are also credited with development of impetus as as a property of motion. 3. The use (?taCceIeratio,l (?tacceIeratio,l (iS a variable to describe inotion. Albert of Saxony and Oresrne are credited also with contributions in this area. Truesdell indicates that scholars at Merton College in Oxford. England. between the years 1132X—1350 32X— I 350considered considered accelerations accelerations to describe motion. However. it was only in the I 5th and 1 6th centuries that acceleration began to be applied more widely. I

.

The

—.-

The Italian scientists Parma. Nicolas ofCues, Leonardo Da Vinci (1452—1519), and Dominic De Soto (1494—1560) are credited with blending these concepts and furthering new new thoughts in mechanics. Da Vinci is the most famous, though primarily for his paintings and contributions to geometry. Leonardo did work on inclined planes, resolution of forces (his results were incorrect), free-falling objects (again, with incorrect results, even though he consuits sidered accelerations), shape of the earth, energy and momentum, and weaponry. De Soto is credited for work on the effects of gravity. The discussions initiated by the Italian school continued in other European institutions, especially Italian and French universities. At this point, let us consider a brief history of the evolution of orbital theory, as this evolution was instrumental in bringing about the scientific revolution and the developments in dynamics. Copernicus was the first to propose a heliocentric system for the motion of planets. He theorized a system where the planets and the earth were in circular orbit around the sun. (One should not he overly critical of his error iii in theorizing that the orbits were circular. The orbits of the earth and Mars around the sun have very small eccentricities, of 0.0 167 and 0.093. respectively.) His theory, refined during the years 1513—1543, was widely spurned, as it went against the teachings of the church that the earth was the center of the universe. However, scientists began to notice, notice, among them theni Galileo, Tycho Brahe of Denmark and later on Prague, and Kepler. Astronomical observations soon began to show that the Copernican theory was on the right track hut not very accurate. In 16(X). Kepler (1571—1630) became assistant to Brahe (1546—1601), (1546—1601). who was making accurate observations of the planets. But Tycho Brahe's thinking was deeply influenced by his religious beliefs and he (lid not accept the Copernican theory. After Brahe died in 1601, Kepler continued his work. The level of accuracy of Tycho Brahe's and Kepler's observations is very impressive, especially Kepler's in his resolve to explain the smallest discrepancies in the motion of planets. Kepler showed that a planet moves around the sun in an elliptical orbit that has the sun in one of its two foci. He also showed that a line joining the planet to the sun sweeps out equal areas in equal times as the planet describes

671

672

APPENDIX A • A BRIu II1S'IOR\ II1S'I'OR\OF OF I)vNA\iics ()ft)it. Both its ()ft)it. Both these laws were first formulated for the motion of Mars around the sun. sum and

publishcd in IOO9. JODY.Keplcrs Keplcrsthird thirdlaw lawappeared appearedinin1619. 1619.ItItisisironic ironicthat thatwhile while Kepkr's third law was widcl'y accepted from the beginning. his first two Jaws laws were were not not warmly warmly received received for of Newton. 75 years. years. ItIt Newton. as well as of Hooke and Halley, that finally (he confirmed KepIerss laws. Newton and Hooke both take credit for the inverse square law that describes gravitational attraction. As a result of discussions and exchange of information between 1-lalky. l-laIIev. Hooke. Hooke. Wren. Wren. and and Newton. Newton. both both Hooke Hooke and and Newton Newton showed showed that that the the inverse inverse square law of gravitational attraction leads to elliptic orbits. These developments led to the publication of the Principia in I 687. Returning to the history of dynamics. one 16th-century scientist to make an important contrihution \Vas Stevin Stevin (( 1548—1620). 1548—1620). who who lived lived in in Belgium Bclgiuni and and Holland. Holland. Stevin Stevin refined refined the the theory of statics and considered equilibrium of masses on inclined planes. and of pulleys. He used coiicepts concepts of and. the principle of virtual work. His work was followed by the contributions ofGalileo. who confirmed Stevin's results and showed equilibrium directly froiii f'roiii virtual virtual work. Galileo Galileo (( LS64—1642 recogniied that. contrary to Aristotelian theI 642 ) recognized ory. applied force is not necessary to maintain applied O!V. with no change in velocity. This observation is (ialileos (ialileos/au' lawofi/lertia, ofiiiertia,and andit it observation is lays laysthe thefoundation foundationofofwhat whatconstitutes constitutes an an iner— iner— tial retcrciice hal reference (jalilco also deduced froni horn his measurenients of pendulums that inertial ITLtSSan(l an(l gra vi tat i(JllaIniass niassarc arcidentical. identical. Galileo worked ofl the projectile probleni problem and obtained accurate results. primarily behehe used acceleration in analyzing motion. Interestingly. Interestingly. correct results on the projectile P1()tlIeIllcame projectile camewithout withoutknowledge knowledgeof ofNewton's Newton's second second law. law. Galileo Galileo is one of the first iirst discuss that in the free-fall problem. displacement is proportional to the square of time. Toricelli ( I 608— tinie. 608— II (47 (47). ). was a student—more a disciple—of Galileo, put Ofl projectile 011 projectile motion iflt() iflt() systematic 'ysteiflatic torm. Of course. Toricelli is most famous for his work on pressure and the barometer. Galileo is also credited with an accurate analysis of tides, which included the effects of both the sun and the moon. I 'he more (1 dissemination di sscminat ionofinforimition of inforimition and and theories 'I'he more rapi rapid theories from from one institution of learning to the next next had had begun. begun. Works ol scientists and astronomers astronomers were being translated, translated. and the influence influence of of the church was beginning to decline. These OCCUITCflCCS occurrences were were important importantcatalysts catalysts in the rapid developments in dynamics that began with the second half of the 17th century. One scientist who made no significant contributions contributions himself. himself. hut hut Who who was responsible for initiating discussion and correspondence and spreading spreading inf'oniiation. information, is Father Marinne Mersenne (1588—IM$). Mersenne lived most of his litè in Paris. He was in continuous comcornwith his his French Frenchcolleagues colleaguesDescartes Descartes( (l596—1650. munication with 1596—1650. Pascal Pascal(1623—1662), (1623—1662), and and Roherval ( 1(02—1675). as well as with Galileo in Italy and Huygens (1629—1695) in Holland. Mersenne defended Descartes' and Galileo's work to others. Contributions around the middle of the 1 7th—century also came from Roberval, who is credited with the parallelogram law of forces: Huygens. who invented the pendulum clock and made it possible to measure time more accurately: accurately; and Descartes. Descartes, whose contributions in algebra and geometry are Far more significant. Huygens also worked on the geometry of curves, oscillations, and centers of oscillations. Huygens is credited with the first published treatise of relative motion. Descartes contributed the concepts of work and linear momentum: however, his free-fall theory was incorrect, as he (like many before him) assumed that velocity was proportional to distance traversed. Another major (levelopment in the middle of the 17th century was the increased interest in impact of bodies. The problem attracted the interest of the Royal Society in London, which promoted research on the subject. The fact that bodies with different levels of elasticity behave differently in a collision was known quite early, and and aa distinction distinction was was made made between so/I impact and hard unpaci. precursors to the of restitution. It is interesting to note I

'

A.2

HISTORICAl. EvoLuTioN oi•' OF'

that the initiative of the Royal Society is concurrent with Newton's discoveries—but before Newton publishcd his discoveries. Actually. Huygens Huygens in in 1700 1700 used used Newton's Newton's fIrst first law law as

one of his hypotheses to describe the equations governing impact. Other contributors to the developments in impact impact include includethe theBritish Britishscientists scientistsWallis Wallis( 1(16—I developments in 703) and ( 1(16—1703) andWren Wren 1632— 1632— 723). and Mariottc ( of France. We next conic to isaac Newton. Newton ( 1642—1727) began his contributions by laying the fbundation thundation for differential and integral calculus. several years before its independent disI

covery by Leihniz ( 1 646— 1I 77 16) of Germany. Newton considered differentiation as the basic operation of calculus and thus was able to develop solution methods that unified many sep-

arate techniques such as finding areas, tangents. the lengths of curves. and the maxima and minima of functions. Newton's laws of motion are dated to 1I 666. at which time he had already lbrmulatcd formulated their early versions. Newton made use of the concepts of momentum and forCe as the fundamental quantities. He delined delined quantities quantities as as mass. mass.impressed impressedforce force((vis vis impressa, and and inertia force vista, or innate force of matter). He showed that the period of a pendulum is independent of its mass. Newton also made signiticant significant contributions to celestial mechanics. The Copernican theory and Kepler's laws had been proposed lbr the motion tilotion of planets. Newton proposed that the niotion of the iiioon motion moon around the earth was governed by the same laws that described the moLion of planets. Newton's laws of motion and the universal gravitational law were refined in part due to his communication (often unpleasant—see biography at the end of this appendix) with his fellow British scientists Hooke ( 1635—1702 and Halley ( 1656—1742). Newton published the PIii/osnp/ziae PIii/osnp/ziae ?laturalis ,iaturalis prineiffla prineifna matheinanca, mathematica, or or Principia, in 1 687, which included his laws of motion and his contributions to celestial mechanics. Newton also discovered the law giving the centrifugal force on a body moving uniformly in a circular path. However. he did not have a correct understanding of the mechanics of circular motion. In addition. Newton studied the precession of the vernal equinox and the shape of the earth. Newton's laws provide a firm basis for describing the motion of particles, and his law of universal gravitation unities unitIes the works of Copernicus. Kepler. and Galileo. It should he noted that Newton stated his laws of motion for rigid bodies hut did not con sider rotational motion. This may be attributed to his incomplete understanding of circular motion. What we know as Newtonian analysis. which is based on force and moment balances. balances, continued with Euler. Euler (1707—1783) built on Newton's work, and he developed descriptive equations for systems of particles in 1740. He then proposed the rigid body rotational equations of motion in 1750. while he was working on the motion of ships. In that same paper, Euler restated Newton's second law as a fundamental law of motion applicable to all bodies. In dynamics, Euler is remembered primarily for his contributions to the calculus of of varivariations and rigid body dynamics. dynamics, as well as for his book hook TIu'oria motus ?notus corporum solidorum solidorum ,ceu .ceu rigidorum (Theory of tile the Motions Motions of Rigid Bodies), Bodies), published published in 1765. He is considered

to be the first to use the components of the angular velocity as kinematic variables. Euler's equations use a set of body-fixed coordinates, a novel novel concept concept for ftr its time. They also involve mass moments of inertia and principal axes. all of which were developed by Euler. Euler is also credited with considering angular momentum momentum as as aa fundamental fundamental quanquantity. By the early 1730s Euler. Daniel Bernoulli. and Johann Bernoulli had recognized that Newton's laws of motion could not explain rigid body motion, and they began to look for other quantities to describe rigid body rotations. F-Iuygens had already developed the correct equations for a pendulum by this time. Euler and Daniel Bernoulli began to look into rigid body motion and Euler continued the work after Daniel Bernoulli gave up on this sub•ject. ject. While While developing developing the the rigid rigid body body equations equations of motion. motion, Euler began to consider angular

673

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APPINDUX A • A BRIEF HIS'I'ORY HISTORY OF DYNAMICS

momentum as a fundamental quantity. His results were also influenced by his work on deformable bodies. where the angular lilomentuni leads to the synimetry of the stress tensor. Finally. in 1775. Euler formally stated F = p and M = H to be the two fundamental laws of mechanics. These two equations are known as Euler 's laws ofinechanics. and they have gi rise tO I.the heterm tenii Ne%t'toFl Ne%t'toFl lertonitulation. lerfonitulation. The work of Newton and Euler conipletes completes the analysis of rigid bodies and particles by fcrce and moment force nioment balances. At the same time as the Newtonian mechanics (or Newton-Euler formulation) was being refined. the analytical approach to mechanics was being developed. Actually. 1)0th both Newton Newtonand andEuler Eulercontributed contributedtotoanalytical analyticalniechanics. niechanics.Newton Newton br the the develdevelopmcnt of calculus. and Euler for the calculus of variations and the principle of least action. Lcihniz ( 1646—1716) took a different approach to describe motion: in contrast to Newton's approach tOfl's approachof ofusing usingchange change in in linear linear momentum. momentum, Lcibniz Lcibniz used energy (he defined a quantity called vis i'is viva, which is equal to twice the kinetic cnergy. He stated that motion could he described by equaling equating the change in kinetic energy to the work done by the acting forces, hence the work-energy theorem. Leihniz is also credited for inventing calculus independently of Newton's findings, as well as for his work on impact oF bodies. The next major development in analytical mechanics came from Johann Bernoulli D'Alenihert ( 17 17—1783). Making use ofNewtons and Leihniz's Leihiiiz's contrihu( 1667—1748) and D'Alemhert tions to calculus and variational calculus. in 1 7 1 7. Johann Bernoulli formally stated the principle of virtual work as a general principle. from which all problems of statics could be solved. Many consider him to he the most important contributor to this subject. DAlembert added to this by considering moving bodies and the inertia force. He viewed the inertia force as a force that produces equilibrium. By including the inertia force in the formulation. D'Alcmbert showed that the principle of virtual work can be used to analyze systems that are in motion. This statement is dated to 1743. Actually. there is debate on the scope of what D'Alemhert actually stated. In mechanics. two equations are referred to as DAlembert's principle. as discussed in Chapter Chaptcr4. 4. D'Alcmbert D'Alcnibert also also worked on impact problems. as well as the living living force force concept and the center of oscillation. The concept of a differential equation of motion was put forward by and Euler in I1 743. 743, in in two two separate separate articles articles that that each published that year. Variational calculus and the principle of virtual work continued to attract interest. Johann II and Daniel Bernoulli made contributions to the subject. and Euler worked maximum and minimum problems. These developments eventually led to the formulation of the principle of least action. Lagrange is usually credited with formally stating this principle. hut the principle was initially conceived by Fermat (I 6() I — 1665 665 of ofFrance, France,who who isis credited credited with with stating the fIrst minimum principle that is not trivial, and Mauperitus who lived in France and Switzerland. Mauperitus theorized that there is a fundamental quantity called action, and that its minimization described motion. However, lie he was not able to establish exactly what action was. As a result, his contributions were not appreciated by many. The next major contribution came from Lagrange (1736—1813). In addition to his contributions to celestial mechanics and mathematics, Lagrange is credited for formulating the principle of least action and for taking advantage of generalized coordinates. This made it possible to extend the treatment of dynamical systems by different coordinate systems and different variables, and gave one the power to select the set of variables best suited to solve a problem. Lagrange's equations were introduced in the Méclianique Analitique in I 788. In the introduction, Lagrange states that one needs only kinetic and potential energy and a set of generaliied coordinates to solve for the equations equations of of [notion. motion. In their original original derivation, derivation, these equations were written for conservative systems only. Lagrange is also credited with formally stating the generalized principle of D'Alernhert in the form that we study it today. Lagrange was influenced by both Euler's and D'Alemhert's work. What we call Lagrange's equations are also referred to as the Euler-Lagrange equations (see Appendix B). —

1

A.2

HlsToRic1u. EVOLUTION OF

Another of' of Lagrange's Lagrange's contributions to mechanics was in the treatment of systems subjected to constraints. 1-Ic introduced introduced the the Lagrange multipliers. which we use to formulate. and

solve constrained systems. He also considered equilibrium problems. developing the theory of small motions motions around equilibriuni. equilibrium. and stability theorems based on potential energy. The contributions of Lagrange put the field of analytical mechanics into a structured lbrni. The next significant contributions to the field came from from Hamilton Hamilton(1(1805—1865). 805— 1 865). along with his developments in relativistic mechanics. In between the two. a number of notables made made minor contributions contributions to mechanics—minor in the sense that, although significant. they were not nearly as signifIcant significant as Lagrange's or Uamilton's contributions. All these scientists made more significant contributions to other branches of science. Wc discuss their contributions to to mechanics next. next. Coulomb ( I1736—1806) 736—1 806) developed developed the niodel ofdry friction in 1 78 781. L as the winning essay of a competition sponsored by the French French Academy Academy of' of Sciences. Coulomb's priniary contrihutions to science were on electricity and magnetism. Carnot ( 1753—1823) developed the impulse-momentum theorem. He also worked on impacts. His work was similar to that of Lagrange. Carnot was certainly familiar with D'Alernhert's work. How much he was aware of Lagrange's work when he published his essays in 1 783 and I 803 is not accurately known. Laplace ( I 749—1 827), who was a contemporary of Lagrange. contributed to celestial mechanics and the analysis of the the solar system. He discovered the invariability of planetary mean motions. These results appear in his Traité di, Mécanique Céleste published in five tive voluines over 26 years (1799—1825). Laplace also introduced introduced the potential function and Laplace coefficients. Fourier coefficients. Fourier ( 1(768— 1 830) 1 768— 1 830) workedwith withLagrange Lagrange and Laplace on mathematics, with worked emphasis on analysis. He succeeded Lagrange in his position at the École Polytechnique. His contributions to mechanics include work on virtual displacements and constrained motion. Gauss ( I 777—1 777—1855) 855) developed the principle of least constraint in I 829. which is a precursor to the Gihhs-AppelI Gihhs-Appell equations. He regarded Lagrange's work as brilliant. hut did not consider the principle of virtual work as very intuitive and thus worked on developing a more intuitive principle. Gauss's principle does not involve time integration and gives the true minimum rather than a stationary value. However, the principle makes use of accelerations, which are harder to deal with than velocities. Poisson (1781—1840). (1781—1840), who was a student of Lagrange and Laplace. considered generalized momentum (without giving giving it that name) and generated the Poisson brackets. His work was included in the second edition edition of of the theMéchanic1iw Méchanique Analitique, which was published in 1 811. Poisson also contributed to the theory of impact. by developing the Poisson's hypothesis. hypothesis, and to the theory of elasticity, elasticity. by by developing the Poisson

ratio. In the 18th 18th and and 19th 19th centuries, centuries, while while the the advances advances were being made in dynamics, a nurnnumber berof of scientists scientists (primarily (primarily French scientists at the Ecole Polythechnique) refined refined the the concepts in kinematics, rotations and relative motion: motion: Clairaut Clairaut (1713—1 (1713—1 765) obtained a more accurate but still incorrect incorrect form for relative acceleration. He also worked on celestial mechanics prob-

lems, such as analyzing the motions of the moon moon and of Halley's Halley's comet. comet. He studied the shape of the the earth and and verified verified Newton's result result that the earth is an oblate spheroid. Coriolis (1792— 1843) 1843) analyzed analyzed the dynamics of motion motion viewed from from a moving reference frame and developed the concept of Coriolis force. It is interesting to note that while his is regarded as a contribution to kinematics. Coriolis always considered considered the dynamics of rclative relative motion, motion, hence the name Coriolis force. Foucauli (1819—1868) conducted experiments on the effects of the rotation of the earth on bodies. He was inspired by Reich's experimental work on free-fall in mine shafts (1833). One of his earlier experiments was with a sphere of 5 kg suspended from a steel wire of length 2 m. He later used a gyroscope attached to the pendulum and demonstrated that the symmetry axis of the gyroscope did not change orientation as the pendulum moved. He related the rotation of the swing plane of the pendulum to the latitude of the location of the

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FLISIORYOF OFDYNAMICS DYNAMICS BRIEF FLISIORY APPENDIX A • A BRIEF

whose primary work was in geometry and conic sections. experiment. Chasles theorem. an extension of Euler's theorem. is known for The next most important important C(.)fltflhUtiOflS COfltflhUtiOflS to mechanics came From Hamilton and the Ger804—185 185 ). Hamiltonisiscredited creditedwith with unifying unifying the field of analytical 1 ).1 Hamilton man scientist Jacobi ( I 804— mechanics. He developed the most general form of the principle of least action (called in this text the extended Harnilton's principle). He showed that the fundaniental principles of mechanics and optics are very similar. Haiiiilton Hamilton also succeeded in developing the Hamiltonian. from which the equations of niotion can he written as tirst-ordcr differential equations that Jacobidevel— devel— equations. equations.Jacobi v Jacobi called these equations are in state form. Jacobi oped the transformation theory of canonical equations and showed that these equations have applications in many other fields. He gave a new formulation to the principle of least action. eliminating the time integral. He is also credited for the Jacobean. which becomes the most Hamilton's other signilIcant contribution to mechanics general form of the energy integral. integral. Haniilion's includes quaternions. principles that that govern govern the motion of contributions consist ol' the fundatlxbntal principles Hamilton's contributions and bodies. The contributions that came after Hamilton deal with the analysis of Lagrange's and spccial cases. cases. We We outline these developments next. equations. as well as certain special equations. By the time of Hamilton. communication among scientists had become more frequent and information was being disseminated more rapidly. This led to substantial collaboration. in many cases. makes it hard to credit a certain development to a particular individual. Hanillionian methods was the examination One of the offshoots of the Lagrangian and Hamiltonian of the types of variables that we can use as generalized coordinates. Advances in this subhalt of of the the 19th 19th century century and and the the tuist first Few years of the 20th lect came mostly in the second halt' ge,wralized eoordi,:ates eoordi,:ates was introduced by Thomson ( 1824—1907) century. The terminology terniinology ge,zeralized 867 text on Natural Philosophy. Routh 1—1901 ) ofScotland. Scotland.inintheir their11867 text Treatise on — 90 1 ) of and Tail. ( I Germany. and Thomson Thonison recognized the importance Hc!mhoIt7. (( 1 82 1— 1 894 ) of Gcrniany. ( 1 83 1 — I 907 )., ). HcInThoIt7. cliniinated from the of cyclic variables. i.e.. ignorable coordinates, and the way they could he eliminated formulation. Such variables are mostly encountered in rotating systems. Kelvin. Thomson. and Tait are also credited with work on gyroscopic systems. Helmholtz analyzed energy conservation. Hertz (1857—I (1857—1894) 894) looked at how mechanics can he described as a whole using 19 19) introduced Lord Rayleigh Rayleigh (1 (I the four concepts of space. mass. force, and motion. Lord the the Rayleigh's quotient, a way of treating dissipative effects in Lagrangian mechanics. Nonholonomic systems began to attract more interest toward the end of the 19th century. (1877—1954) 877—1954) are recThe German mechanician inechanician Heun (1859—1929) and his student Hamel (I nonholonomic constraints ognized as the initial contributors. 855—I 930) considered nonholonomie contributors. Appell Appell (1 (1855—I and the types of approaches one can use when dealing with them. He used quasi-velocities equations, called with nonholonomic systems. He laid the loundattons for the Gihhs-Appell equations. by many the Appell equations. Jourdain (1879—1921) developed Jourdain's variational principle, which closes the gap between D'Alemhcrt's principle and Gauss's principle of least constraint. 8() years after Gauss's principle was stated. Gauss's principle was extended for inequality constraints in 1879 by Gibbs. An interpretation of the various theories was written by Duhem in 1903. in (1849—I 849—I 925) 925) and and Cay Cay Icy Icy (182 (182 II—i —1895) 895) did work hook Evolution tie hi Ia Mécanique. Klein (I his book on description of rotations. Cayley's work based on Hamilton's research on quaternions. More recent contributors to mechanics, such as Lanczos (1893—1974) and Pars (1896—1985). did work on both classical mechanics as well as relativistic mechanics. Hanii!tonian methods has been the qualitaAnother extension of the Lagrangian and Hamiltonian tive analysis ol of dynamics problems. especially regarding stability theory. Lie (1842—1899) introduced the group theory to canonical transformations. He also considered infinitesimal transformations. 1

1

A.2

HISTORICAL EVOLUTION OF OF DYNAMIcs

Significant contributions to the analysis of motion and stability theory came from

Poincaré (1854—1912), who looked into stability issues. integral invariants. perturbation theory. differential geometry. and topology. The index of Poincaré and analysis of the three-body problem by perturbation techniques are some of his contributions to mechanics. Table A. I gives the dates of major contributions to the field of analytical mechanics. Be aware that many of the dates are approximate, and therc are differences of opinion regarding the contributor and levels of contribution. The dates used here are mostly the ones given by Papastavridis and Dugas. Also, Also. we list contributions up to 1910. 1910, as after that date there was substantially more cooperation among dynamicists. dynarnicists.

Table A. 1

Key dates in the history of dynamics

Contributor(s)

Contribution

1513

Copernicus

Copernican theory introduced

1609

Kepler

1638

Galileo

1687

Newton

1717

Johann Bernoulli

1740

Euler

1743

D'Alembert

1750

Euler

1760

Lagrange

First two of Kepler's laws Particle motion in one dimension, projectile motion Principia published, laws of motion and law of gravitational attraction Principle of virtual work Systems of particles Inertia force. principle of D'Alemhert Rigid body rotational equations Principle of least action

1765

Euler

Theory of the motions of rigid bodies published

1775

Euler

Declares force and moment balances as the two fundamental equations of mechanics

1780

Lagrange

1788

Lagrange

1811

Lagrange

Lagrange's equations Méchanique Analitique published Lagrangian derivation of rigid body Euler's equations. 2nd edition of Méchanique Analitique

1829

Gauss

Principle of least constraint

1830

Hamilton

Canonical equations of motion. Hamilton's principle

1840

Jacobi

Transformation theory

1860

Kelvin, Thomson, Tait

Gyroscopic systems

1870

Routh, Routh. Hel mholtz

Cyclic (ignorable) coordinates, stability

1879

Gibbs

Heun. Hamel. Appell

Gauss's principle for inequality constraints Nonholonomic systems, quasi-coordinates, velocity constraints. Gibbs-AppeIl Gibbs-Appell equations

Jourdain

Jourdain's variational principle

Year

1870—1910

1909

677

678

APPENDIX AA •• A APPENDIX

A3

hISTORY HISTORY

Bloi:;it&PFIIEs OF KEY CONTRIBUTORS TO DYNAMICS

Thk section presents brief life lile histories of key contributors to mechanics. Albert of Saxouv Saxoiiv (13 (13 16—1 16—139W. 39W.Few Fewdetails detailsarc areknown known about about the the life life ol ol Albert Albert of Saxonv. except that he studied in the Sorhonne and was rector of the University of Paris. He became rector of' of the the Liniversity University of Vienna in 1365 and Bishop of Halberstadt from I 366 until his death. Albert did work on the concept of impetus. and began to use acceleration as a kinematic i.juantitytoto describe motion. motion. He is credited with work on locating the center of gravity. He also i.juaiititv did work on pro projectile jeLlilemotion, motion,with withincorrect incorrectresults. results. Albert Albert is is credited credited more more for dissemination of sc scienti ienti tic (ic in lormation formationthan thanFor tor developing developingnew newtheories theorieshihimself. nisel 1. Appell. Paul Emile (1855-1930). Appell was born horn in Strashourg. France. In 1885 he was appointed to the Chair Chairof of Mechanics Mechanics at at the the Sorbonne. Sorbonne. He He served served as as rector of the University of Paris from I 92() to 1925. based on on projective projective geometry. continuing the work of Appell's first paper (1876) was based C'hasles.He ('basics. 1-kilieti thencontinued continuedresearch researchon onalgebraic algebraic functions, functions, differential differential equations. and corna good friend of ol' Poincaré. Poincaré. His His paper paper •'Notice "Notice sur lea travaux sciena ple.\ tilique." titique." Acta Mathei,,atica 45 (1925). lists his 140 works in analysis. 30 works in geometry. 87 works in mechanics. and many textbooks, addresses, addresses, and and lectures lectures on on the the history history of' of matheniatics and Ofl on mathematical mathematical education. education. His His primary primary contributions contributions in mechanics were the treatment of non nuiihuionomic holonomic systems and the development development of of the the Appel AppellI equations. Appell took high government positions in addition to scientific posts. which exposed him to war, espionage, espionage. and political controversy, lie was involved with the Dreyfus affair. and served on the commission that exonerated Dreyfus. Dreyfus. During During World World War War I.I. Appell Appell founded founded Secours National. National,aasemi—othcial senii-othcial organi/atiun. of the the Secours organhiatiun.which whichgave gavehelp help to to civilian civilian victims of' the Ibr the French Association to the League war. After Alter the war, he served as secretary-general secretary-general for Nations. I)espite his niany conu'ibutLons I)espitc contributions to toscience. science.politics. politics. and and government, government, Appell Appell is not nearly known as his peers. This is mostly because he was a problem solver, not a developer of' as It may he he aa lilting conclusion conclusion to to his his life life story story that the importance olAppell's theories. It

1)1 UI.

to he appreciated 50 years after his death. e()fltFihlIti()flS began to contributions began Aristotle (384—322 (384—322 FfC FfC p.), Born in northern Greece. Aristotle made contributions to a

broad range of fields. including niechanics, mechanics, logic, anatomy. and nature. in In 367 B.C Aristotle came as aa student student to to Plato's Plato'sacademy academyinin.Athens. Athens.where wherehe hesoon soonbecame becamea ateacher. teacher.InIn335 33513C.h.. he lormed he formed his own school. school, the Lyceum in Athens. In 343 ii c.i-, c.i-. he became tutor to Alexander the Great. He died a year after Alexander the Great died. One of (he (he most remarkable of' Aristotle's attributes was his tremendous capacity to observe phenomena and then deduce laws that describe those phenomena. His contributions to mechanics primarily ai'e are intuitive intuitive deductions deductions from from his his observation observation of ol niotion. motion. Archimedes (287—212 ii i•.j. Archimedes lived almost his entire life in Syracuse, Syracuse, Sicily. i•..). Archiniedes with a short stay in Egypt. His primary contributions were in the areas ot geometry. statics, and Floatingbodies. floating bodies.1ills us contributions contributions to to geometry geometry are viewed as a precursor to integral calculus. He did calculations on the area of a circle and the volume ol' a sphere. He narrowed down the value of r to between 3 and 3 (3. (3.1408 1408 and and 3.1429). 3.1429). In In mechanics, Archimedes worked on the laws of equilibrium for levers and centers of' of gravity. He used his results in centers of gravity to study buoyancy and developed the famous Archimedes' principle. While he was in Egypt. he developed a machine called Archimedes' screw, which is used as a pump even to this day. '

Bernoulli family. This was a large Swiss family who for three generations dominated the field of' of mathemafics. matheniatics. Several members of the family occupied the chair in

A.3

BI0GRAPHWs OF KEY CONTRIBUTORS TO DYNAMICS

mathematics in the University of Basel. The first major contributors. Johann—also known

as Jean—( 1667—1748) and Jacob—also known as Jacques—(1654--1705) were the children of Nicolaus (1623—1708). Their other brother Nicolaus (1662—1716) was not as prominent, prominent. but his son Nicolaus I (1687—1759) was. Jacob Bernoulli studied the center of oscillation problem, which became a precursor to principle. Johann took his brother Jacob's chair at Ba.sel after Jacob died in 1705. Based on New-

ton's and Leibnizs work. he formulated the principle of virtual work and put the field of calculus of variations on a solid ftundation. He was the first to recognize that the principie of virtual work was a general principle from which static equilibrium problems could be solved.

Johann Bernoulli had three Sons who become prominent mathematicians: Johann II (

1 771 1 0— 1 1 790). 1 0— Daniel 1 700—1 1782), 782),who whoworked worked closely closely with Daniel ( 1( 700— with his his brother Johann II. and Nico-

laus H (1695—I 726), whose contributions appear to be mainly in mathematics and who died very young. Johann II took his father's chair when his father died in I 748. He was 38 years old at ()ld at the time. Johann 11 and Daniel did a lot ofcoilaboration, with Johann's work primarily concentrating on heat and light. and Daniel's on hydrodynamics. Johann 11 initially studied law, and he pursued mathematics independently of his father. He won the Prize of the Paris Academy four times by himself. Johann II is credited for extending extending the works of Newton and Leibniz into a broader spectrum of problems. Johann Bernoulli TI introduced the brachistochrone problem. Daniel is credited with Bernoulli's Bernoulli's principle. principle. In In 1725., 1725, Daniel Daniel and his brother Nicolaus were invited to work at the St. Petersburg Academy of Sciences. There he collaborated with

Euler (who came in as his student). One of the areas of collaboration was ship dynamics, based on which Euler developed the rigid body equations of motion. Daniel also had a wider range of interests, such as anatomy, statistics, life insurance, and kinetic theory of gases. Johann H had three sons, Johann 111 (1744—1807), Daniel H (1751—1834), and Jacob 11(1759— 1789). Of these three children, only Johann III became prominent in mathematics (probability theory. decimals, and theory of equations). Copernicus, Nicolaus (1473—1543). Copernicus was born and died in Poland. He studied in Krakow (then the capital of Poland) and also in Bologna, Padua, and Ferrara. Italy, and he graduated in 1503 with a degree in canon law. Shortly after, he returned to Poland and settled in Frauenberg (Frombok). In addition to his religious duties, he practiced medicine and wrote articles on monetary reform. While doing all that, he began to work on the motion of the earth and planets. He was the first to challenge the views of Aristotle and Ptolemy and to formally state that the sun is the center of the solar system. In his theory, theory. the earth and the other planets revolve around the sun in circular orbits. He further theorized that the earth rotates about its own axis. Copernicus first wrote his theory. called Commentoriolus, and known to this date as the Copernican theory, in 1513. His theory was widely rebuffed, as it went against what was accepted as fact for well over 1000 years. He was spurned by many as a result of his writings. For 30 years he worked on refining his theory and his hook hook On On tile the Revolutions of the I-leaven/v Spheres (in Latin) was published in 1543. a few days before he died. A couple of generations after his death. Kepler and Galileo came to his defense (Galileo was under house

arrest for eight years for doing this) and their use of' the Copernican theory was the basis for the development of the laws of planetary motion. Many historians regard the Copernican theory as the beginning of the scientific revolution. Coriolis, Gustave Gaspard de (1792—1843). Coriolis was horn and died in Paris, and lived there for most of his life. He studied mechanics and engineering mathematics mathematics at the École Polytechnique. Paris, then taught mathematics there. He also worked as a military engineer. His major contribution is what is known as the Coriolis force and the laws of motion

679

680

APPENDIX A • A BRIEF HISTORY

OF

in a rotating reference frame. He showed that to write write the the equations equations ot' of motion in a rotating frame of reference. the Coriolis force has to he included. Coriolis also did work on solid mechanics and on collisions, the latter based on collisions of billiards. In I 835 he published E/fets dii Jeti de Billiard. his Théorie Mathérnarique des F/fets Coulomb, Charles Augustin de ( 1736—180ô). While this French physicist and engineer is primarily known for his contributions to static electricity and magnetism magnetisni (a Coulomb is a unit of electrical charge), Coulomb also contributed to mechanics. Guided by the interest of the French Academy of Sciences in friction and ropes. as well as their military applications. Coulomb began experimental work on friction. extending the results of Amontons. 1-us awardwinning essay "Theory ofSimple Machines" discusses the relation between forces (pressure) exerted by contacting bodies and friction, and develops Coulomh's Coulomb's law (of proportionality) to describe friction. To this day. dry friction is referred to as Coulomb friction. Coulomb also worked on torsional vibrations and inclined surfaces. Alembert lived all his Ilife ife in in Paris. Paris. He D'Alembert,Jean JeanLeLe Rond DAlemhert D'Alembert, Rond ( 1 (71717—1783). 1 7— 1 783 ). D turned down many lucrative offers. such as an offer from Frederick II to go to Prussia as president of the Berlin Academy. and an invitation from Catherine 11 of Russia to tutor her son.

In 1741 In 1 74 1 at the age of 24. D' Alenihert was admitted to the Paris Academy of Science. where he worked for the the rest rest of of his his life. life. He He was was aa contemporary contemporaryand andfriend friendof ofVoLtaire. Voltaire. His primary work was the study ofdifferential equations and their applications. as well as theory of functions. derivatives. and limits. In mechanics. he is best known tbr the D'Alemherts principle. which appeared in his Traité tie Dvnamique ( 1 742 ). He is credited with developing the concept of an inertia force ( — ma) which makes it possible to apply the principle of virtual work to systems in motion. A year later he applied his results to the equilibrium and motion of fluids.

D'Alembert's other major work is on the concept of equations of motion as differential equations. He first published an article on this in 1743. His more famous article, entitled Difjérenriel. was published in 1754. Euler, Leonhard (1707—1783). Born in Basci, Basci. Euler joined the University of Basel as a theology student, following in the footsteps of his father. There, he began to like mathematics and, at the insistence of Johann Bernoulli II, II. he became Bernoulli's student. After graduating, Euler went to St. Petersburg Academy of Sciences in 1727. He began working with and living home of Daniel Bernoulli. Tn 1741, 1741. Euler joincd the Berlin Academy of Science at the in the home invitation of Frederick the Great. He stayed there for 25 years, returning to St. Petersburg in 1766. Euler lost his vision in one eye at age 31 and became blind in 1765. hut he nevertheless continued his research for many years. contriEuler's hook Euler's hook Theory Theoryofofthe theMotions Motionsoff Rigid Bodies was published in 1765. His contributions also include logarithms, the Euler number. function theory. and the function notation. He is credited. too. with contributions to the solutions of differential equations, geometry. and topology. In addition, Euler worked on the mathematical theory of music, as well as on dethrmable bodies. Euler is regarded by many as the most prolific writer ol mathematics of all time. Some view him as the great architect of mechanics, at a level higher than Newton and Lagrange. who traditionally receive more credit for their contributions than Euler did for his. Euler's complete works include over 800 books and papers. which continued to be published by the St. Petersburg Academy for So years after his death. Despite his tremendous achievements. Euler remained a very modest person. always praising other people's work and refraining from criticism. In an incident involving himself. Lagrange, and Mauperitus (.1698—1759). Euler went to great lengths and fought hard to defend Mauperitus from charges of plagiarism and to get Mauperitus the credit due for his work on the principle of least action. action. His His conduct conduct should should serve scn'e as a guide and example to all.

A.3

B1o;Rw!iws OF KEY KEY CONTRIBUTORS CoNTRIBuTORS TO

Galilel, Galileo ( 1 564—1642). Galileo was horn and lived in Italy. His father was a profèssional musician. Galileo was a professor of' mathematics in Pisa and Padua. where he also taught astronomy. Tn addition to his contributions in astronomy and natural philosophy. Galileo contributed to the analysis of free-falling objects and also to the theory of vibrations through his research on the simple pendulum strings. lie is known as tile the father of the experimental method. He is credited with developments in hydrostatics. While teaching astronomy at Padua. Galileo began to develop an interest in the Copernican theory and Kepler's observalions. In 1609. after seeing a new spyglass in Venice. Galileo made his own telescope. which was was the thefliOSt fllOStpowerful powerfulofits ofitstime. time. He He is is reported to have seen mountains Ofl the moon. and the satellites of Jupiter, and to have discerned the nature of the Milky Way. He received an open letter from Kepler in 1610. praising him him for his measurements and endorsing his discoveries. In I1613. 6 1 3.while while he was Mathematician and Philosopher to the Grand Duke of Tuscany. Galileo in his observations discovered that Venus orbits orbits the the Sun. Sun. His His rejection rejection ofthe ofthe theory theory that the earth was the center of the universe and adoption of the Copernican theory earned him many enemies. in 1616 1 6 1 he 6 hewas waswarned warnedby by the the Italian government government not to defend the Copernican theory. He did not heed and his Dialogue Concerning the Th'o Greatest World World was published in Florence in 1632. Galileo was summoned to Rome, found to be suspected ofhcresy. and condemned to house arrest. for life, at his villa. Under house arrest. he continued working with with his his followers frllowers and wrote a book on the strength ofmatcrials. which was smuggled out of Italy and published in the Netherlands in 1638. He died tinder house arrest Gauss, Carl Carl Friedrich Friedrich((II777— 777—1855). 1 855 ). Born Born in the Duchy of Brunswick, Gauss showed himself to be a genius genius at a very early age. Gauss entered the academy Brunswick Collegium Collegiuni Carolinum in I 792; he left. returned. and obtained his degree in I 799. I-Ic received tinancial suppo1.t from suppol.t from the the Duke Duke of of Brunswick. Brunswick. He published the book Disqui.s'itiones Disqui.sitiones Arithinetieae in 180 1 His second second hook, Theoria corporuin coelectiiun coelestiiun in sectionihus seciwnihus conicis conicis Soleiii wnbientium, was published in 1 809, a major two-volume treatise on the motion of celestial ambientiu,n, bodies. In 1822, 1 822, he won the Copenhagen University Prize. Gauss worked in a wide variety of fields in both mathematics and physics. including number theory. analysis, differential geometry. geodesy. magnetism. magnetism, astronomy. astronomy, and optics. His primary contribution to mechanics is the principle of least constraint. Some of Gauss's principal contributions include Bode's law, the binomial theorem, theorem. the arithmetic-geometric mean, prime numbers, construction of the I 7-gon by ruler and and compass, least squares approximation. orbital measurements, Gaussian curvature, and potential theory. Gauss's life was full of personal tragedies. Soon after his first marriage, the Duke of Brunswick (lied, leaving him without financial support. He was forced to move to Gottingen. A year later his wife died after childbirth, soon to he by the death of the baby. He remarried and then his second wife died a few years later. Gibbs, Josiah Willard (1839—1903). Gibbs was horn, raised, worked, and died in Connecticut. He was the son of a professor of literature at Yale. He received his doctorate in engineering from Yale in 1863, the first engineering doctorate given in the United States. He studied until 1869 in Paris, Paris. Berlin, and Heidelberg. and became professor of mathematical nlathematical physics at Yale ill in 1 871. Gibbs's major publications began in 1873. on thermodynamics and equilibrium. His subsequent work included electromagnetic theory of light, statistical mechanics. astrodynarnics, and vector analysis. En mechanics he is is recognized for his work on inequality constraints and contributions toward the Gihhs-AppcIl equations. His contributions were appreciated more iii Europe than they were in in ill the United States. Gibbs was known as a true gentleman, kind and unassuming in his manners. He lived a quiet and dignified life. Hamilton, SirWilliam WilliamRowan Hamilton, Sir Rowan(1805—1865). (1805—1865). Born in Dublin. Ireland. Hamilton gave evidence of genius as a yOung child. By age five he knew four languages. 1-Ic was very .

681

682

APPENDIX A • A

HISTORY OF DYNAMICS ThSThRY

interested in mathematics and. by the time he was 15, was studying the works ofNewton and céleste. Laplace. At age 17, Hamilton Hamilton found found an an error error in in Laplace's Laplace'sMéC/zanique Méc/zanique céleste. even Hamilton entered Trinity College in Dublin at the age of I 8. Tn I 827. at age though he was an undergraduate. he was appointed astronomer royal at Dunsink Observatory and professor of astronomy at Trinity College. He had not applied for the post. for which there were applicants. but rather. was invited to accept it. In 1 833. Hamilton published a study of' vccwrs as ordered pairs. as well his first work of Three Three Bodies by my Characteristic Function. He used algebra on dynamics. The Problem ut in treating dynamics in On a General Method in Dynamics Iivna,nics in 1834. This work. in essence. In 1843, he began to formulate the describes his most important contribution to theory of quaternions (Euler parameters). He devoted the rest of his life to quaternions. and wrote two books on the subject. hut neither hook was well received. The second book was published posthumously. with the final chapter incomplete. Hamilton was the first foreign member elected to the U.S. National Academy of Sciences. The latter years of his life were unhappy ones, and he suffered from alcoholism. Jourdain, Philip Edward Bertrand (1879—1921 This British scientist lived a short life that was full of medical problems. He was already severely handicapped when he began his studies at Cambridge in 1898. His undergraduate years were very difficult. He did poorly in his degree. However. he was awarded the Allen studentship for research in 1904. In mechanics. Jourdain is credited with Jourdain's variational principle ( 1 909). which closes the gap between D'Alernhert's principle and Gauss1s principle. His most famous paper ofMaihematical Physics Physics ((1 on this 011 this subject is On some Points in the i/u' Foundatkn: ofMaihematical Jourdain also worked in mathematical logic and set theory. He wrote a number of articles of TraizsTra,zson this subject between I 906 and 1 9 1 3 under under the the title ot' the Theory of finite Nu,nher.s'. Nu,nher.s. In In 1913 1913 he he proposed proposed the the card paradox. This was a card which on one side sentence on the other side of this card is TRUE." On the other side of the card the said: sentence on the other side of this card is FALSE." sentence read: Johannes (1571—1630). Kepler. horn in Germany, is known primarily for formulating the three mathematical laws of planetary motion. He is also credited for work on mathcniatics. primarily on logarithms, as well as optics. mathcmatics, Kepler attended the University of Tuhingen and took a teaching position in mathematics at Graz. At Tuhingen he was exposed to both the Ptolcrnaic Ptolcmaic and Copernican theories, theories, and and he came to accept the Copernican theory. In 1596. 1596. Kepler Kepler published published his his Mvsteriwn Mvsteriwn CosmoCosnwgraph icum, which graphicum, which argues argues for for the the truth truth of the Copernican theory. In 1600. Kepler went to Prague, as assistant to Tycho Brahe. Brahe died in 1601, hut Kepler went on to use his observations to calculate planetary orbits. The level of accuracy of his measurements nieasurements is astounding. His first two laws were published in Astronoinia Nova in 61 9.Kepler's Kepler'sother otherbooks books are are Epitome Epitome 1609. His third law appeared in Hunnonice mundi in 11619. 621 1). ). and andRudolp/zine Rudoiphine Tables Tables (1 627). based on Tycho Avtronomiae Astronomiae Copernicanae Copernicanae(1(1661188 to 11 62 Brahe's observations and Kepler's laws. Italy, Lagrange's interest in mathItaly. Lagrange, Joseph Lagrange, Joseph Louis Louis(173&—1813). (173&— 1813). Born in ematics began at a very early age when he read a book by Halley. Lagrange served as professor of geometry at the Royal Artillery School in Turin from 1755 to 1766 and helped to found the Royal Academy of Science there in 1757. In 1764 he was awarded his first prize of many. when the Paris Academy awarded a prize for his essay on the libation of the moon. When Euler left the Berlin Academy of Science, at his request, Lagrange succeeded him as director of mathematics (1766). Development of what we know as Lagrange's equations was accomplished during this period. These developments were made possible by the advances that Lagrange introduced to the calculus of variations. He also contributed to hydrodynamics. .

A3

BIOGRAPHIES OF KEI CONTRIBUTORS TO DYNAMICS

In 1787 Lagrange left Berlin Berlin to become become a member of the the Paris Paris Academy of Science. where he remained remained for forthe the rest of his career. ofhis career. He He published his Méchanique MklzaniqueAna/jtjqj,e Analitique in I

which includes Lagrang&s equations. This work summarized all the work done in the field of mechanics since the time of Newton: it is notable for its use of the theory of differential equations. In it, Lagrange transformed mechanics into a branch of mathematical analysis. He survived the French revolution. During the 1790s he worked on the metric system and advocated a decimal base. I-Ic also taught at the École Polytechnique, which lie he helped to found. In I 797 he published the first theory ol" of functions of a real variable. Napoleon named him to the Legion ofHonor and Count ofthe Empire in in 1808. I 808. In In 11 811 81 1he hepublished published the the second edition edition ofMéchanique Méchanique Aiialiiique. Analilique. Lagrange also excelled in all fields of ofanalysis analysis and number theory. as well as in celestial mechanics. Hamilton referred to Lagrange as the "Shakespeare of in appreciation of the mathematical depth and elegance of Lagrang&s contributions. contributions. Lagrange also worked on the history of science.

Newton, Sir Isaac ( I1 642— 642— 1 727 ). Regarded by many as the most important contributor to the field of' of dynamics. Isaac Newton was born in Lincoinshire to a family of farmers. In

his teenage years his academic performance was very weak. His mother removed him from grammar school, and at the urging of an uncle, in 1661 Newton entered Trinity College. Cambridge, to study law. At Cambridge. Newton was attracted to philosophy. hut but also to algebra and analytical geometry. He was very impressed irnprcssed with the Copernican theory. The plague in 1665—which closed Cambridge University for two years—gave him the opportunity to go home to to Lin-. Un-. coinshire and work on his own. He made niade several significant signilicant contributions in mathematics. niathematics. optics, physics. and astronomy during that time. His tirst first contribution was in calculus and unification of the previously scattered developments in this subject. His De Methoclis Serieruni ci Huxwnum Huxwnwn was was written in 1671 (he failed to get it published and it did not appear until 1736). After returning to Cambridge in 1669. he was appointed to the Lucasian chair. He was 27 years old at the time, and his genius was well known, Newton held this chair until 1687. Newton was elected a fellow of the Royal Society in 1672. Also in 1672 he published his first scientific paper on light and color in the Philosophical Transactions Transactions of the Royal Royal Society. Society. While well received in general, his paper was objected to by Hooke and Huygens. This objection resulted in further strained relations among them.' them. He published his Opticks in 1704, a year after the death of' of Hooke. Newtons laws of mechanics were published after debates with Hooke and Halley. In 1684, Halley asked Newton what orbit a body would follow under an inverse square force, and Newton replied that it would be an ellipse and that he had solved this problem five years earlier. Halley, being very impressed with this answer. urged Newton to publish his law of planetary motion, which he did in 1687 in the third third part part of of his his Principia. Newton suffered a nervous breakdown breakdown in in 11 693. 693. after which he lost interest in science. He retired from research to take up a government position in the Royal Mint in London. In 1703 he was elected president of the Royal Society and was re-elected each year until his death, death. He was knighted in 1708. The falling apple episode seems to he a myth. as it is reported that this incident occurred in 1666. whereas by 1666 Newton had already thrmulated his laws of motion.

Like

many before and after him, Newton did not take kindly to scientific criticism. In 1679, Hooke wrote to Newton proposing a central force that governs the motion of planets. Newton's response to Hooke discussed on inverse square law, but it also had an error, which Newton had to admit grudgingly. As a result, both Hooke and Newton worked independently on the law of planetary motion, and both claimed credit For the inverse square law.

683

684

APPENDIX A • A BRIEF HISTORY OF DYNA\IICS

Poincaré, Jules Henri (

was an applied mathematician who is credited with several advances in algebraic topology. stability theory, and the theory of analytic functions. Poincaré studied in the Ecole Polytechnique and at the Ecoic des Mines. In I 88 1 he was appointed to a chair of mathematical physics at the Sorhonne. He studied optics. electricity. telegraphy. elasticity. thermodynamics. potential theory. quantum theory, theory. theory ofrelativity. and cosmology. He also studied the three-body problem. and theories of light and electromagnetic waves. He is acknowledged as a codiscoverer. with Einstein and Lorentz, of the special theory of relativity. His His major works include Analysis situs ( 1895), an early systematic treatment of topolI 854— 1 9 1 2 ). Poincaré

ogy, Les Methods izourelle de Ia inéchanique celesie celesie (( II 892—1 892—I 899). and LeçonS de celevte (1 celeste 1-Ic also wrote many popular scientific articles. In addition to his work on manifolds and what is known in dynamics as the index ofPoincare, ojPoincare, Poincaré was the first to

consider the possibility of chaos in a deterministic system. Little interest was shown in this work until the modern study of chaotic dynamics began in 1963. Poincaré was critical of the way mechanics was taught by the British. He thought the British treated mechanics as an experimental science, science. whereas in continental Europe mechanics was taught as a deductive science. Poisson, Shnéon Denis ( 1781—1840). Originally forced to study medicine. Poisson began to study mathematics in I 798 at the École Polytechnique. His teachers were Laplace and Lagrange. He also made contact with Legendre, Legendre. as a result of a memoir on finite diftèrences, written when he was 18 years old. Poisson taught at École Polytechnique from 1802 until from 1802 1 X08 when he became an astronomer at Bureau des Longitudes. In II809 809he hewas was appointed appointed to to the chair of pure mathematics in the Faculté des Sciences. Poissons most important works were in mathematics, mechanics. astronomy. electricity, and magnetism. En mathematics. he advanced the theory ofdefinite integrals and Fourier Fourier series, series, as well as Poisson's integral. Poisson's equation and Poisson's brackets. His contribution to probability theory is best remembered by the Poisson distribution. The Poisson distribution is widely used to model several phenomena (such as the famous gas station attendant problem: arrival of cars is modeled by the Poisson distribution while service time is modeled using the exponential distribution). In electricity, he is known for the Poisson's constant. Poisson was a pro prolific hue author: he wrote over 300 articles. His Traité c/c de inecanique, 'necanique, published in 1811 and again in 1833. was considered the standard text on mechanics for many years. Claudius(85—165 Ptolemy, Claudius (85—165 c.L.).Ptolemy Ptolemy was was born born in in Egypt Egypt and and lived most of his life in Alexandria, where he died. He developed the geocentric theory. with the motion of the planets and the sun as epicycles (combination of circles). His astronomical observations and his theory on the motion and location of the planets were considered valid for over a millennium. until they were disputed by Copernicus. Routh, EdwardJohn Routh, Edward John(1831—1907). (1831—1907). Born Bornin in Canada. Canada. Routh Routh went to England in 1842. He entered Peterhouse College at the same time as MaxwelL Maxwell. but Maxwell transferred to Trinity College. Much of his life was spent in competition with Maxwell. In 1854. the Smith Prize was divided between Routh and Maxwell (the first time the prize had been awarded jointly). Routh published fuimous flimous advanced advanced treatises treatises which which became became standard applied mathematics of Rigid texts. such as A Treatise Treatise on on Dvnatnic.c Dvnatnics of Rig it!Bodies Bodies (1 (1 860). 860). A Treatise on Analytic Statistics (1891). (1891 ). and A Treatise Treatise on on Dynamics Dynamics u/a Particle (1898). Routh was elected to the Royal Society in 1872. His contributions to mechanics include advances to stability theory, gyroscopic systems. cyclic coordinates, and Routh's method fèr

ignorable coordinates. In 1877. 1877, Routh was awarded the Adams Prize for work Ofl dynamic stability. I-Ic He was was revered as an outstanding teacher and coach. He died in Canibridge. Cambridge.

REFERENCES

REFERENCES Dugas. René. René. A A Hi.vzorv History of ofMechanics. Mechanics. New New York: York: Dover Publications, 1988. Lanczos. C. The Variational Principles qi u/Mechanics. Mechanics. 4th 4th ed. ed. New New York: Dover Publications. 1970. O'Connor. John J.. and Edmund F. Robertson. History ofMathematicsArchive. (World Wide Web site) URL: http://www-groups.dcs.st-andrews.ac.uk/---history/index.html Papastavridis. J. J. G. G. "A Panoramic Panoramic Overview of the Principles and Equations of Motion of Advanced 239—65. Engineering Dynamics.' Dynamics.' Applied AppliedMechanics MechanicsReviews Reviews51. 51.no.4 no.4(1998). (1998).pp. pp.239—65. Hixtor of ofMechanics. Mechanics. New New York: York: Springer-Verlag. Springer-Verlag. 1960. Truesdell. C. Essays in the History

685

APPENDIX

CONCEPTS FROM THE CALCULUS

OF VARIATIONS B. 1

INTRODUCTION

In dynamics one frequently uses concepts from the calculus of variations. Variational principies such as principle and Hamilton's principle are considered as foundations of analytical mechanics. One of the earliest problems in the calculus was the determination of the minimum and maximum values of a function. The first major developments in this subject came in the second half of the 17th century. Newton. in 167 1671, 1, and and Leihnitz. Leihnitz. in in 1684. 1684. made made the first original contributions, by noting that the rate of change of' of a continuous function must be zero for a minimum or maximum. Newton also looked into the problem of minimization of a functional. Johann Bernoulli and his son Daniel extended the works of Newton and Leihnitz into a broader spectrum of problems. Johann Bernoulli introduced the brachistochrone problem. Finally. Euler. who was a student of Johann's sons Johann H and Daniel Bernoulli. built on the Bernoulli family's work and developed the calculus calculus of' of variations to its current form. Toanalytical mechanics mechanics but but also in several day the calculus of variations is used not only in in analytical and engineering, engineering, such such as as Sturm-Liouville. Sturm-Liouville probother problems from mathematics, physics. and lems and optimal control theory. as well as in generating methods—such as the Rayleigh-Ritz and finite element methods—for finding approximate solutions to boundary value problems. An introduction to the variation of a function is also given in Chapter 4, which develops variational principles used in dynamics. One can follow the developments of the text by just reading Chapter 4. without consulting this appendix. Here, the reader will find a source of additional details and more in-depth developments.

B.2

STATIONARY VALUES OF A FUNCTION

an interval interval where x is the variable and Consider a function of a single over an single variable variable f(x) f(x) over extremwn, aa local minimum or max= (a, h) is the interval. The function f(x) has an extremwn, such that /'(x) flx) •f(f) oror imum, in the interior of this interval if there exists a point for all values of x. For 1(x) to have have an an extremum in the interior of the interf(x) to f(x) val, f'(x) = dfCv)/dx has to vanish at x = or it has to not exist. Otherwise. the minimum and maximum values of f(.v) f(.v) will will be be at at the the boundaries, boundaries,atatf(a) f(a) or or atat f(b). represents a local minimum or One determines whether the extremum point .v = > 0. represents a 0. f represents maximum by examining the the second second derivative derivative of of f(x). f(x). If 0. no local minimum: if f"(f) < 0, 0. x4 .v4represents representsaalocal localmaximum. maximum.When When •f"(f) = 0.

f.

f

687

688

APPENDIX B • CONCEPTS FROM THE CAI.cul.us CAI.cul.us OF OF

conclusion can be drawn. A simple example of this is to compare compare the the functions functions .f(x) /(x) — and f(x) = The first derivative of a function detennines determines the the slope. slope. and and itit isis aa necessary necessary condition condition for the existence of a niinimum oi niaximum. The second derivative describes the shape of the function. and it constitutes a sufficient condition for for the the extremurn to be a local maximum or minimum. When its second derivative is positive. positive. f(x) f(x) isis convex convex upwards. upwards. as in a bowl. When the second derivative is negative, the curve is convex downwards (also called concave). as in an inverted bowl. A point where the shape of the curve changes from convex to concave or vice versa is called an an inflection inflection point. point. At At an an inflection inflectionpoint. point.f"(.v) f"(x) = 0. Next, consider a function of n independent independent variables, variables, ,f(x,. f(x1.X2 in which X2 x1, are the variables. We denote by X(x1, X(x1. x2 a a point point in the interval 1, 2, . ., ii). Let Let •f(X) f(X) he continuously differentiable in all its vari(ak bk, k = 1, ables. The necessary condition for f(X) j(X) to have a local nhininiuni minimum or a maximum in the interior of i' is for all of its first derivatives with respect to x1. x1, .v2, to vanish. That is, the point X = (xc. xç. .v4 is a local minimum or maximum if . .

c/f

=0

k=l,2,...,n

at

IB.1J

The point

where all the fIrst derivatives off with respect to vanish, is referred to as a .stationarv .stationari value value of the function f(X). One One arrives arrivesatatEq. Eq.lB. lB. 11 11 also also by by considering considering the differential of •f(xi, .f(xi, which has the form •. .v,1). which .

•

dl =

oX1

dx1 d x1

+

(if dX2

dx2 +

•.. ++

dx

I B. 2]

Because the variables .vk xk (k (k = 1,1,2, 2. .... .. ii) are are independent, independent, their differentials (k = 1,2, 1, 2, .. ., ii) are also independent. independent. The The differential differential of off vanishes at a stationary point, dJ. df = 0. For df to vanish, all of the coefficients of dxA (k = 1. 2. . , n) must vanish independently. .

.

.

,

and hence the result in Eq. lB. Ii. To determine whether the stationary value represents a minimum or maximum, one needs to investigate the second derivatives of ,f(X). f(X). Define Define the the Hessian Hessian matrix matrix IHJ by

(I X (:1

-

[H] =

(1 .V (1 .t'2

(/ (IX

Ii

.1,1

(9-f

dx2d.V1

(12/

[B.3J IB.31 (v-f (9-f

(/Xp1 (.VJ

whose whose entries are = (92f/(9x1(9x1(i, / = 1. 2, ..., nii) evaluated at X = X'. Note that [Hi is square and symmetric. A theorem from the calculus of variations states that if the Hessian matrix is positive (negative) definite at the stationary point X. X, then the stationary point is a local minimum (maximum). For a symmetric matrix to be positive detinite, detinite. it must satisfy Sylvester's criterion, which states that the following must hold:'

For a symmetric matrix to be negative definite, all the diagonal elements must be negative, and the sign of the principal minor determinants must alternate. 1

B.2

I

.

2.

STATIONARY VALUES VALUES OF OFAAFUNCTION FuNcTIoN

All diagonal elements must he greater than 0. and All principal minor determinants must be greatcr than 0.

If the determinant of (11J = 0. no conclusion can he drawn. If the determinant of [HI is not not zero and [Hi is indefinite. the stationary point is neither a maximuni nor nilnimum and is for a minimum referred to as a saddle point. The Hessian matrix is is also also denoted denoted by by [HJ [HI = V2f.

Consider now the case when the variables related to each other by the equality constraints .

.

.

..

/ = 1,1,

= 0

. ,

.

x1,

.v,,

2,

are not independent and are

. . . .,

[L4J

in

where in is the number of constraints. To lind the stationary values of f(x1, x2 subject to the the constraints constraintsc1(x1, c1(x1,.v2, .v2...., ..., a number of approaches can be adopted. One approach is based on rewriting the constraint equations such that in of the variables are expressed in of the the remaining remainingiiii— terms of — in in variables. Theoretically, at least. it is possible to do so. Then the function is described by n — in indcpcndcnt independent variables and the above developments apply. For instance, say is the variable we wish to eliminate from the formulation. We seek to express x1 in terms of the other n — I variables as xi = xi

.vi, .v2 .v .v2

I'

/—

+

II. 51

,,) ,i)

1

and and to substitute substitute Eq. Eq.FB.5] FB.5] into into f(X). If this can can be be done done for all the constraint relations, one one effectively express express.f(X f(X in in terms termsof ofiiii—— in independent variables and use the procedure can effectively outlined earlier to find the stationary values.

Many times, it is difficult to find a relation of the the form form {B.5j {B.5j for for a!! all the constraint relations.

Even if such relations are found, they may complicate the expression expression for for f(X) f(X) immensely. immensely. Also, certain constraints arc described in the differential form a / dx1 +

(1/2

dx2 d +

•

± a',, a dx11 dx11 == 0

IB.61

and it is not possible to to integrate integrateEq. Eq.{B.61 {B.61to tothe the form form of of Eq. FB.41. For these cases, there is another procedure for finding the stationary values of f(X). This method makes use of Lagrange multipliers. We construct an augmented function. denoted by d(X) = and .V2 ,...,x,,) ,,) and Lagrangian. which has the same value as f(X). referred to as the Lagrangian, f(X). hut hut has has the the form form

"' "s' In

(x1, x2 ., .x,,) = f(x1,

+

.V2, A1c,(.v1. .v2.

. .

. ,

x,,)

[B.7] IB.71

1=1

where

A, (/ = 1,2,

. .

.. ,,

in) are called the Lagrange multipliers. Consider the differential

of the augmented function

To

this end, we first take the differentials of the constraint rela-

t 1011s i oils

d d cc1 =

de2 =

dc1 .

a' x + dx i

1

()XI

- (/.l'I d.ii +

cIXj

d.ki d.Vi

(/,12 + (1.12

de'

— dx2 —

±

•

',

•+

'.

.+

di:,, = 0 a'

-

— dx,1== 0 —

(1,V2

(9.VI

dc,,, dc,,, =

cICj

+

(ICm

(I.V7

± ... +

dx,, = 0

IB.81

Multiplying Multiplying each one of Eqs. [B.8] by the Lagrange multipliers and adding these equations to Eqs. IB.21. one obtains d4, the differential ol' the augmented function, function. as

689

690

APPENDIX B I CONCEPTS FROM THE

= df + =

OF VARIATIONS

A1 dc1 =

+ XI

+

(9f

+

+

+

(C1

dx,,

A1 A1 +

- A-

-

(IX,,

-A+

-

9C,,1

+"•

(IX'

+

9c

+

cIx1

±

= A2 A2 +

A1 A + i

(1c1

1

c1x I

+

-

"

dx1

(1X1

-

A,,,

d.v + —

=

A,,,

IB.91

0

When the constraint is in the form of Eq. [B.6]. we multiply each constraint equation

by A1 and add it to df to obtain d4. In essence, essence,we wereplace replaceÔCI/ô.Vk iIc//ôxk by in Eq. [B.9j. To solve Eq. [B.9] one selects the Lagrange multipliers .., in) such that each of if = 1.1. 2,2. ..., 1/ .

the expressions in the square brackets, that is. the coefficients of (k = 1, 2,2. ..,., n), vanish individually. This leads to a total of,i equations in Eq. [B.91 and in equations in Eq. {B.41 or Eq. [B.6]. which can be solved for the ii ++ ni ni unknowns x1, .v' A1, A,, . ., .

.

.

. .

,

The determination of whether the stationary values of constrained systems constitute minima or maxima is a more complicated subject. This determination requires use of developments from nonlinear optimization theory and Kuhn-Tucker conditions. We will not go into this subject in detail, hut briefly mention the theorem from Avricl (see the text by Pike), which establishes a set of sufficient conditions to establish minima or maxima. After finding the stationary values of a constrained function using the approach outlined above, one constructs the matrices IH,d (p = in tn + I. m + 2, n), such that . ,

S

P

—

[c 1

[01

The matrices [H'] and and [c'l [c'l are are defined defined as = =

(

i./

j. f //

—

= 2. . ., p = 1.I.2,....p

,=

r

r = 1,1,2,...,m:s= 2, in: s =

.

.

.

.

,

l,2,...,p 2, . ., p 1,

.

[B.1 11 where all the derivatives are evaluated at the stationary points. Avriel's theorem states that I)'" If (— 1)'"

> 0 for all p = in + 1, in + 2,

.

.

., ii. the

stationary point is a local

minimum. If (— 1)"

> 0 for all p =

in + I, in — 2,

. . .

,

ii. the stationary point is a local

111 ax i iii u iii. maximum.

It• neitherone oneof of the the above above conditions conditions holds, no It neither no conclusions conclusions can be drawn. The interested reader is referred to texts on optimization theory for further details on equality constraints, as well as frw inequality constraints.

Example

Find the stationary values of the function f(x, v) = •v2/2 .v3/24 + .vv + .v + I in the interval v2/2 + .v3/24

B.1

Solution The stationary value off is given by df = 0. 0. Taking the differential off we obtain

df

= =

dx +

dv = 0

Ia]

_____

1.2 which holds only if both we obtain

= 0 and (1.V

dx

dv

691

S1'ATIoNARY VALUES VALUES OF OF A A FUNcTIoN FuNcTIoN

= 0. SettinQ these first derivatives equal to zero -

'=v+x=O dv

8

Ib]

Substituting the second of the above equations into the first we obtain

[ci [c] the solution of which yields two stationary values:

+

= 4 — 2\/2. Both stationary values are inside the interval given above. It follows from the second of Eqs. [hi that = = —4— and = —4 + To check whether the stationary points are local minima, maxima, or neither, we analyze the Hessian. The second derivatives derivatives of,f off have the form (1

—

4

d2J d2f

v

d

—i

— —

—4

and

so that the Hessian becomes so

= =

(x,

= (4 (4 +

—4— 2\/2)

= [1 + V2f = [HJ = V2f

= (4—

—4 +

[Hi = V2f

=

1]

[1

[.1

—

For For the stationary point

Hessian is positive definite. Hence, the stationary is a local minimum. The Hessian for the stationary point point (xr, is neither is positive nor negative definite. It is indefinite. It follows that this second stationary point is neither a minimum nor a maximum, but a saddle point. To check if constitutes a global minimum in the interval, one must evaluate the values of f at the boundaries of the interval as well. the

Find the stationary values of the function f(x, y. v.z) z) = c(v c(x, VV z) = vV— — = 0.

+ y2 + z2 z2 subject subject to the constraint y2 +

Solution We can solve this problem either by introducing the constraint into f(x, y, z) and and expressingf ing f in in terms terms of of two two variables, or by using the Lagrange multipliers. For illustrative purposes, we select the latter approach. We form the augmented function y, f(x, y,y, z) + y, z) = f(x, Ac(.v, Ac(x, v, v, z) z) and and take take partial partial derivatives derivatives of

which yields

=3x2—2Ax

= = 2y+A

[aJ

= 2z

-

Selecting A such that all of the first derivatives of

with respect to x, v, and z vanish, vanish, and considering the constraint equation, we obtain the four equations —

2Ax = 0

2v + A = 0

2z =

0

y — x2

= 0

[b]

to he solved for for the the unknowns unknowns x, x, v, v,z.z. and and the Lagrange multiplier A. The third equation yields = .v2, .v2,and and from from the the second second equation, A = —2y. Combining z = 0. From the last equation.

Example B.2

692

APPENDIX B

•

FROM THE CALCULUS OF VARIATIONS

these equations we obtain A =

2..v2,which. which. 2..v2,

—

3x

—

when substituted into the first equation gives

2Ax =

+ 4x3

= 00

Ic]

3.12(1 F = 0. 0. This equation has three solutions, Equation [ci can he factored to yield 3v2( — 3/4. Combining with Fqs. I hi. we identify the stationary and xx = — a double root at .v = 0 and and X to he points A1 and 1

—a

= 0

= 0z =

3=—

X

0

—

=()

=

Id]

To calculate whether the stationary points constitute local minima or maxima, we generandtwo twovalues valuesforp. forp. p= p =2. 3. in = 1. I. and matrices. For this problem we have ii = 3. in ate the matrices to check. The tB'] matrix is diagonal with the entries There are two *

H11 1111

_()-4) — —

1

d .v dx

H,=

= 6.v—2A 6.v—2A

— —

=2

c?V—

=

(1—4) .•

(1: (1::

=2

le]

All other entries of [H4 I are zero. Substituting the value of the Lagrange multiplier into H1'1, we obtain

=6x—2A=ôx—2(—2x2)=6.v+4x = 6x — 2A = ôx — 2(—2x2) = Ox + .v

If If]I

l-4)r the the second second stationary stationary point = 0. Nor Nor the the first first stationary stationary point .v' = 0, so that = —2.25. Consider the first stationary point. For the case when p = 2 = —0.75 —0.75 and 01

= This results in the

k'i

2]

Lo

—2.v

= [_2x1 =

[(j)] 1)

=1

Ig] Igi

matrix

1H21 =

0 0 0

o

o

21

[hJ

10

whose whose determinant is zero. The first stationary point is neither a minimum nor a maximum.

but a saddle point. We dispense with the calculation ol tbr this stationary point and for consider the next stationary point. For the fH2j matrix has the form 0

—2.25

[H2] =

1.5

0

I

1.5

0

Ii]

and (— II )"' )"' detIjI2J detIjI2] = 2.25.

To determine

we form the [H!1 [H!i and and [c'] matrices matrices as 0

00 20

0

0

—2.25

[Fr] =

• I.5

Ic] =

Iii 0

2

resulting in —2.25

[H31 = [H31

0 o 1.5

0

0

1.5

201

020 1

0

0

[kJ [ki

B.3

DEFINITE INTEGRAL INTEGRAL .A.A DEFINrI'E STATIONARY VAJJUF.So'OF

shown that (— 1)" detf H3] = 4.50. As both quantities are larger than zero, the it can he shown

second equilibrium position constitutes a local minimum.

B.3

STATIONARY VALUES OF A DEFINITE INTEGRAL

A large number of problems problems in in mechanics and in mathematics can be posed as the determination of the stationary values of' of aa definite definite integral. A simple definition of the problem is as follows: Find the function v(x) which is continuously differentiable in the interval (a, bb) which yields the stationary values of the integral h

1=

IB.121 —

a

v(x) (x) satisfies certain boundary conditions. For (x) to he the solution. any function other than v(x) should not lead to a stationary value. We introduce a continuous function rj(x). which is arbitrary. together with I. When as shown in Fig. B. 1. the constant and we construct the varied function v(x) + must he consistent with the constraints. The values of dealing with constrained systems at the boundaries depend on the boundary conditions of v(x). Consider initially the 1/u specified. Such Such boundary boundary conditions are known as case where where y(a) y(a) and and v(b) r(b) are are specified. conditions of the first kind. In this case, the varied function boundary conditions or vanishes at the boundaries. ij(a) = r1(b) = 0. If v(x) has fixed values at certain points then the varied function cannot he arbitrary at those points. Introducing the varied function v(x) + into the integral whose stationary values are sought. we obtain

this equation. equation.I Iisisreferred referred to to as as thethe .ttUl(11011(11 .tunctional and In this

'1'

E(.v, I'(.v,

= —

V

±

[B.131 IB.131

y'' +

(I

with primes denoting differentiation with respect to x. For the integral Ito have a stationary value, it must satisfy L/ I t/

[B.141

= 00

when

= ()() we we obtain

Differentiating Eq. [B. 131 using the chain rule and evaluating at

dl

1'

de

-

(IF

+-

(1( 0

t9('

a

(' +

+ E'rJ')

oF

=1

,9v

Ja \'9Y

/

a

Yb

I

a

=

(B.151

v(x) V

) dx

?J(x)

Figure B.I

693

694

APPENDIX B • CONCEPTS

FROM 11W

OF VARIATIONS

We intein-ate integrate the second term on the right hand side of the above equation by parts to obtain2 /'

)F

dv =

Ja

h

F (IV

1

,71 —

Ja

"(IF \ dx __( dv

IB.16]

The objective of this integration by parts is 1.0 have as a coefficient of all the terms in the integrand. introducing Eq. [B. 161 into Ft1. [B. 15] we obtain b

J

i/F ?F

d / CiCiFF

ôF oF q( x dx + ,ij ---— — Ci Ov clx (Iv —— — ( —,)

"1'

= 0

[B.171 lB. 1 71

The integral and boundary ternis in this equation vanish individually. Because ij(x) is arbitrary in the interval (a1 b), in order for the integral to he equal to zero the integrand must vanish at every point in the interval, so that OF

a<x
=0

C;,',

[B.181

Equation [B. 18] is known as Euler's equation or the Euler-Lagrange equation. In addition. each boundary term in Eq. IB.171 is zero by itself', leading to OF (9F

=0

dv

iiF OF

[B. 1 9a,b]

Equation [B. 1 81 and the end conditions constitute the necessary condition For the establishment of the stationary value of the definite integral. Previously we considered geometric boundary conditions, so that ij vanishes vanishes at at x = a and x = b. As a result. Eqs. [B.19a1 and [B.19h1 are automatically satisfied. Consider now the case when when v(x) isisnot not specified specified at one or or both both of of xx = a and .v = b. Hence does not vanish at those boundaries and has arbitrary values, in addition to the interior. To To drive drive the the boundary boundaryterms termsininEqs. Eqs.[B. [B.19a] 1 9a] and [B. I 9b1 to zero, the coefficient of [B.19b1

ij(x) must vanish, or

(if

[1 v(a) is not specified. then

(V'

=0

iiF0 i/F0

If v(b) is not specified. then

(Iv' (IV'

at

x=a

at

.v =

The conditions above are the boundary conditions. Note that these boundary conditions

are related to the function. They are called natural boundary conditions. or boundary conditions of the second kind. In mechanics they are also referred to as dynamic boundary conditions. Sometimes. the boundary conditions have the form g(v(a), v'(a)) = 0. h(v(b), v'(b)) = 0. Such boundary conditions are known as boundary conditions ?f qf the third kind. An interesting special case is encountered when the function F is not an explicit function of x, that is. F = F(v, v') only. We can take the derivative of F with respect to x and obtain

(iF \ dv dF = (OF dd .1 .i

.v Ci vv )) dd .v

+

(OF '' dv'

—

0 v') d dx.t

(OF dv (iiF (OF \ d2v '\ x2 ) dx2 )d +

2Note that we are assuming that F is twice differentiable with respect cases when the second derivatives of F hove discontinuities.

[B.20J

yf• x, y, and and y'. See Seetext textby byHildebrand Hildebrand For

B3 B.3

YtLUES OF VALUES OF

695

DEFINiTE LNTEGRAI.

From Eq. [B. 181 we have (IF

= d — dx

[B21]

/

Introducing Eq. [B.21 I into Eq. [B.20J we obtain

di (ii

ci // f9 FF\ \ddvv --I dx \c9y' dx dx

dx

ff9 / F

\d

d '1 dv ) + \f9v' ) dx2 = (1 ci ,ff9F = = 00 dx \cIv \t9v' /

or

/

v

I

-.

\Iy') [B.22J

•

leading to the result F

[B.231 [B 231

= constant constant

v'

—

j so so that that when F is not an

then the the expression expression F — v'(iiFThv') represents explicit function of.v. ofx. then a firs! integral. The subject sublect of first integrals For dynamical systems is discussed in Section 1.12.

The above relations describe a very useful property which lets one recognize tile the existence of first integrals. In Chapters 4 and 5 we make use of the above properties and observe the relation between the function F and the Lagrangian.

'2

For the functional F(x, v, v. v') given by F = associated fIrst integral.

v'2

—

4v2

—

6v4.

find Euler's equation and the

Solution Taking the partial derivatives derivatives of of I-', F, we obtain

= —8v —8v

dv

—

oI.•

24v3 24v

ci (f9F ci f9F\

=2v = 2v

(I

dx \\dv' )

.

.•.'

= 2v

,I

[a]

which, when introduced into Eulc(s equation. equation, gives —

8v

—

—

24r3 = 0 = 2v" + $v + 24v3

To find the first integral we note that .v To V v'

—

-

so so

[bi

is absent from F. From Eq. Ia] we write = 2v/2

[ci

)=

that the first integral becomes

F

—

•v'

(

= v' ),/ =

—

4v2 4r

—

6v4 6v

1

—

2v'2 = constant 2v'

[dl

which can be rewritten as + 4v2 4v ++ 6v4 ôv4 = constant 1

v -

I.] 1.]

We can draw an analogy between this example and tile the stiffening spring considered in Chapter 1. Indeed, if we think of x as time. v as the spring deflection, then F represents the difference between twice twice the the kinetic kinetic and and potential potential energies. energies.and andf9F/f9v 9FMv represents represents the the spring spring force. The integral of the motion is. of course. course, the total energy.

Example B.3

696

APPENDIX

B.4

B•

IRE CALCULUS OF CONcFVIs FROM [HE

THE VARIATiONAL NOTATION

Ffh

variational notation streamlines the formulation of variational mechanics problems. and ii demonstrates the difference between differential calculus and variational calculus more clearly. Consider the function F(x. v. v') and stationary values of its integral. In the previous section. we defined the varied function as v(x ± Let he a small quantity and define and denote it by so that the variation of v.v) as the change

IB.241

= E7)(X)

The change in F due to the variation in v, v. for a fixed value of .v. can be expressed as = F(x, v +

v' +

—

V. v') F(x, v.

[B.25] IB.251

which can he expanded in a Taylor series

=

+

a

(IF dv

so

that =

where ofF. of F.

IB.261

is defined as the terms in the above equation linear

The variation of F. denoted by in

+ (higher-order (higher-order terms in

(IF = -bv+ dv dv

(IF

dv'

(V

denotes the variation of v', 6v' =

IB.271

5F is is also also referred referred to as the first

is different. different. The The above relationship is similar to the exact differential differential of of F. F. hut itit is exact differential of F has the form

dF

= cix

ix ++

dv

iiv

± oF

dv'

IB.28:l

it has one extra term due to the derivative of ot F with respect to .v. The difference between the exact differential and variation of F arises from the way the variation operation treats the variables .v variables .vand andv.v.For Forthe thefunction function F. F. .i.i isis the the independent independent variable and v(x v(x) is is the dependent variable, dependent in the sense that it is a function of x. and and that that itit is is not not constrained constrained in we hold hold the the independent independent variable fixed and any form. When we obtain the variation of F. we and

vary the dependent variable v(x) and its derivative v'(x) to get r(x) + (x) and and v'(x) + is held The variation is from one curve (or path) to another while x is The held fixed. fixed. so that there is is no no variation variationininx.x.Hence, Hence,6x 6x = 0. 0. By contrast, of of a function describes contrast,the thedifferential differential a function describes the change in that Function along a particular curve or path). and the independent variable varies along x varies along that that path path as as well. well. This This point point highlights highlights the the difference difference between between differential differential calculus and variational calculus; The variation of a function includes changes of only the dependent variables, whereas the differential of a function considers changes of both the dependent and independent variables. The rules associated with mathematical operations on variation are the same as the rules of differentiation. The variation operation and the differentiation operation commute when the differentiation is with respect to an independent variable. We have. fhr example =

=

= =

[B.291 IB.291

B.4

697

THE THE VARIATIONAL VARIATIONAL NOTATION NOTATION

The The

variation operation is applicable to scalar functions as well as to vectors. Also, when we refer to the variation of a function f(xi, X2, where all the variables Xk ,,), ,,), where (k = 1, 2, .. ., n) are independent of each other, the variation is exactly analogous to differ.

.

entiation and has the form

6f

= (if 5 = ii.f

+ +

+ ... ++ (It

i9x1

[B.30J

One should compare Eq. [B.301 with Eq. [B.2]. We next derive the Euler-Lagrange equation by using the variation of the integral in Eq. [B. 12]. Taking its variation, we obtain = 1bb 61 = 6 dx v')dx dx = 0 6v + F(x. v, i") IB.3 11 EB.311 dV = J f Jj 1

Integration of the second term in the integrand by parts yields

=

-

ci d

dx \ d vv ,ij

Svdx+ 6vdx+

9V v

= 0

IB.32]

Noting that the integral and the boundary expressions vanish individually, we obtain

i"IoF — I

d

dx

dv

L1

Svdx = 6vdx

I

0

iIF

Sv 6v =0 (IV'

(B.33o,bJ (B33o,bJ

(4

)J

By virtue of the arbitrariness of6v, Eq. [B.33a] leads to the Euler equation in Eq. [B.18]. Equation [B.33bJ leads to the boundary conditions. The variational formulation can also be used to find the stationary values of an integral subject to constraints. The general formulation of the constrained problem will not he derived here. We consider this problem in the study of analytical mechanics in Chapter 4. within the context of applications.

A very important application of variational calculus is the Sturm-Liouville problem, which

can bc derived by seeking the values for v(x) which render stationary values of the ratio of integrals 1, 1, and and12, '2' defined as two integrals as

[p(x)v 1)

j

-)

q(x)vi.x)J dx

—

[a] Ia]

Ia r(x)v r(x)v (x)dx 2

2

where y is the ratio of the integrals, p(x), p(x). q(x). and r(x) are known functions of x, and a and b are the boundaries. Moreover, p(x) and r(x) are positive. Note that x is the independent variable. Invoking the product rule, we obtain for the variation of y 12611 —11612

Ibi [bi

= 0

I'., I..,

Because 12

0.

we can multiply the above equation with it, which yields I

12=0

Id [ci

The variations of the individual integrals are I

1)

61,

= 2

[p(x)y' 6v' [p(x)v'

1

.1

a

—

q(x)v6v] dx

r(x)y6vdx

612

=

a

Id]

I

Example

BA

698

FROM TIlE APPENDIX B B • CONCEPTS APPENDIX • CONCEPTSFROM

OF

We integrate the first of Eq. [dJ by parts and obtain d

=2

— .

) —

clx

'I

q(x)v 6vd.v + f)(v)v' f)(x)v' Sv

'p

lel

Li

-

Introduction of the second of Eqs. 1(1] and tel into Eq. Id and division by 2 yields I

dv

—

dx

p(x)—--

dx

-

q(x)v

h

Svdx + p(x)v' Srdx

—

= 0

j

a

111 (II

evaluated over overthe theinterval interval (a, a, bh and and the the integrated term terni is evaluated evaluated at the boundaries. By definition of the variation of v. the two expressions are independent of each other and. for Eq. Ff1 to hold, they must both vanish individually. For the integral to vanish for all possible values of 6v, 6v. the integrand must be zero. leading to the Euler-Lagrange equation Eq. ff1 ff1 is The integral in Eq. is

ci

dx

a < xx <: <:bb

yr(.vx )) vv = 0 q(x)) vV ++ yr( [j'( .v ) V 11 ± q(x [j'(

[gi

The second term in Eq. [f 1 contains the boundary expressions. expressions, from from which which the the boundary boundary vtv) isis specified specified at .v = a. a, say conditions arc ascertained. Consider the boundary x = a. If v(.v)

then the variation of vr at at that that point point is is zero. zero. This type of boundary condition is then fi,st kind. If, on the other C(flldiliofl, or or aa boundary boundary(i)flthtiOll conditioii of if ific first an essential essential boundary boundaryC(flldiliofl, does not vanish vanish at that point. It teldoes not a, its variation specified at hand. v(x) hand. at xx == a, v(x) is not specified lows that, to render the boundary term zero, the coefficient of Sv, p(x)v'. must vanish. The

v(a) =

Va,

relation

p(x)V(x) =

0

at

.v x= a

or

p(a)y'(a) =

0

[hI [hi

becomes becomes

the boundary condition. This type of boundary condition is referred to as a nalural boundary Ll)flditwfl, condition, or oraaboundary boundary condition condition of of the the second second kind. The Euler—Lagrange Euler—Lagrange equation and the boundary conditions constitute the so—called boundary ralue problem. Examples of Sturm-Liouville problems from mechanics include the transverse vibration of strings, axial vibration of bars, and torsional vibration of rods. Table B. I gives the p(x). q(x). and r(x) for these problems. corresponding quantities for

Toble B. 1

Sturm-liouville Problems

Type of Problem

y(x)

p(x)

String vibrations

Transverse deformation v(v) v(x) Axial deformation u(x)

Tension

Axial vibration of bars Torsional vibration of rods

Torsional deformation

0(x)

—

T(x)

Stiffness —EA(x) Torsional stiffness —GJ(x)

q(x)

r(x)

0

Mass/unit length p..(x )

0

Mass/unit length

0

)

per unit length .'? x )

In all three cases the expression for p(x) is a negative quantity. Derivation of the equations of motion of deformahie bodies is considered in Chapter 11. Note that even though the form of the equations looks similar, the transverse vibration of beams is not a Sturm-Liouville two in four, as as opposed opposed to to two in Sturm-Liouville Sturm-Liouville problem. The highest order spatial derivative is four,

_/ B.5

PRORLEMS VARIATIONAL NOT:\TION NOT:\TI0N TO DYNAMICS PRORLEMS OF THE VARIATIONAL APPLICATION OF

problems. However, the way the equations of motion are derived is the same. For the simin more ple vibration considered considered here. here.qqx) x) = 0. This expression becomes ple cases of vibration

complex vibration problems.

B5

APPLICATION OF TIlE VARIATIONAL NOTATION

TO DYNAMICS PROBLEMS forniuThe variational notation is is ideally suited for dynamics problems because itit makes the formulation concise and it has a meaningful physical interpretation. Consider. for example, the motion of a particle. When the motion is described in terms of Cartesian coordinates, we use the variables .v. .v. v. and z to describe position and the variable 1) describes describes the position. It :(t), 1) tto describe the time dependence. The vector r(x(i). v(t), y(t), :(t), follows that .v. v, and z are the dependent variables and t is the independent variable. There is 110 no variation variationofofthe theposition positionvector vector with with respect respect to time. and

6r =

dx

[B.341

-.--&

+

The variations of the spatial coordinates are known as virlual displacements. Because there is no variation in time, the virtual displacements displacements can can he he considered considered as as occulTing occurring instantaneously. This leads to a very interesting physical interpretation. A virtual displacement can be thought of as imagining the dynamical system in a different position while holding time fixed. The arbitrary position is consistent with the system constraints and it is within the admissible paths the particle can follow. This concept is illustrated in Fig. B.2 for the x coordinate. The variation of the position vector is intimately related to the velocity. Indeed, the velocity of the particle considered above can be written as v(x,

.

1)

dr (/1'

iir.

ar.

(1! (I!

oX

(1\' (1V

ar.

—% + +--—-Z --—-z + — —==---—X —x -I- -——v = — cit

IB.351

expresThis brings about the possibility of calculating a virtual displacement from the expression for velocity, instead of taking derivatives of of the the position position vector. vector. We We demonstrate denionsuate this in Chapter 4.

:::: :::;:

ox

11

1

Figure B.2

699

700

APPENDIX

CONCEPTS FROM THE

Cucui.us OF VARIArloNs

REFE REN CES Englewood Cliffs. Cliffs. NJ: Prentice-Hall. 1978. Greenberg. M. I). Ioii,idations a/Applied a/AppliedMcitheinarics. Englewood I 978. Englewood Cliffs. Hildebrand. F. B. Methods a/Applied Mathematics. Englewood Cliffs. NJ: NJ: Prentice-Hall, Prentice-HaIl, I1965. 965. 970. McLrovitch. L. Methods Dynamics. New York: York: McGraw—Hill. McGraw—Hill, 11970. Pike. R. W. Opzimizwion fir Engineering Systems. New York: Van Nostrand Reinhold, 1970.

HOMEWORK EXERCISES SEcTIoN B.2 1.

2. 3.

on the the curve curve xv xv = that arc the closest to the origin. Find the points on Find the points on the ellipsoid 4x2 — xv = that are the closest to t•v + 4v2 — — v: + 3z2 = the origin. Note: This problem problem can can he he posed posed as as an an eigen eigenvalue value problem. + + 2x1.v2 Find the stationary values of the function f(x) = 2x1x2 + 6x1 + 2.v2 2x2 subject subject = 4. Check for minima and maxima. to the constraint constraint 2x1 2.ii — I

1

SEc1'IoNs B.3 AND B.4 4.

Given two points on the xv plane that one must minimize the integral

Vi)) and and (x2. (x2. v2). v2). make make use of Eq. [1.3.331 to show Vi

Vi

1=

/

+ v'2

di

w liii With

(. t1 .v( .ti) ) =

.V •v

= Y2

1.2) Y( .12)

-

5. 6.

iii order to find the arc of mininwni in minimum length length that that passes passes through through two two given given end end points. points. Show that, given given no no restrictions restrictions on on the thepath pathfrom from(x1, (x1,V1 v1 )) tO to (X2, the the solution is a straight line. Obtain the Euler-Lagrange equation associated with with the the functional functional F(x, F(x, v, v, v') xv'2 — v') == .vv'2 yy' ++•v. .v. in a similar way to Problem 4, show that the minimal surface of revolution about the .v axis passing through the the two two points points (x1. (x1. Y1) Yl) and (X2, y2) V2) can he found by minimizing the Ii ntegral

1=

-f-

v' dx

with

v(x1 ) = v(x1

v1.

(x2) =

.v2

'I

Then, find the resulting Euler's equation. and the solution. 2+

7.

Find the first integral associated associated with with the the function function F(.v. F(x. v. v') =

8.

Consider the problem of axial deformation ot a circular rod of length L and uniform density p. fixed at x = 0 and free at .v = L. as shown in Fig. B.3. The radius of the cross section of the rod R(x) R(x) varies varies with with the the relation relation R(x) R(x) = Rc)( Ro( I — x/3L). Find Euler's I

equation and boundary conditions. 'I

2R11/3

L

Figure B.3

—

2v4.

APPEND

Ix

COMMON INERTIA PROPERTIES Note: xz

always denote centroidal coordinates.

L12

Volume = irR2L

= I.\x Ix.\.

= 'V Iy.v =

I

I -F1-

.1

=

— —

Lljml?2 +

S.

Circular Cylinder

Volume

—

TrRL

= L

x

Thin Disk

=

V

L

= 4inR

very small

R

)

V

V. S.

Volume

= Area A

Slender Rod

F.' F,'

very small)

701

=

= AL = =

mL2

=

L 1

in!12

702

APPENDIX c.

COMMON J.NERTL4 J.NERTL4PRoPERTIES PR0PEIrnES

Volume =

Sernicyl inder

I

V

=

inR +

36ir2 1

1

4

12

Ivy = —inR + _,fl12 97T2—32

= V.

1

1

=

4k

,nR

+ i!;2 ,nL2

3ir

=

inl? a-

Rectangular Prism

Volume = abc

V

=

+ C2) C2)

I

=

+

I..

=

1

•1-

1

1x' 1'

= —n,(b2 I

Ii-

+ 4c) + 4c2)

= Thin Plate (One of a, b. or c Say (•) C)

Volume = abc abc

'I

lxx = 1

—

V

12

m(a2 + /)2) -v

a

Sphere V

Volume = 1 %t

¼.

.t.

=

= 1zz ==

APPENDIX C

703

PROPERTIES

Hemisphere

Volume = = VS VS =

VX

I-.. I-..

=

:1

m

mR2

=

=

mR2

.1

Right Circular Cone V V

Volume = y

= 3L/4

I

=

+

m R2

=

-V

in L

Volume = 1-laif Cone V

yr V

I.t'

3

=k-

3L/4

=

+

(3

I.-

10 1

InRL

4.

= I.'.z I.'.z = 0

:i.-

k/jr

=

=

+

=

Eli ipsoid

4

Volume = —7rabc

V

=

+ c2) c2)

= --m(a2 + (.2) -l

I.

+ b2) -,

.2

Surface defined by

a

+ —

I

704

APPENDIX C• C0NITIoN LNERTIA PROPERTIES

Right Triangular Prism

I

Volume = —abc

V

a /3

1

Ix.v 1xx =

1

1

lzz

mc

=

+ b2)

— 'xv = — 'xv

= 'xz = 'vz 'xz 'Vt =

<<1)) a. CC <<1)) Triangular Plate (C (C << <
1

Volume = Volume

1a/3

1

Ivy 1

1.-

a

1)13

.1

1

Volume = I

—abc

=

+ c2) c2)

V

+ c2) + ()

IyIvI = =

Rectangular Tetrahedron

I..,., I,, =

1,n(a2 +b2 + b2) 1

Note: Centroidal coordinates x vz not shown. Coordinates of G:

(a

3 -nz(b +c2)

I IIvy = =

in a2 + c2) c2)

1:: =

+ b2)

I

=

(

1

— 1

j6mac Ix: = — — Ivz =

b

________

705

INERTIA PROPERTIES

APPENDIX C

1-laif Torus 1-laif' v

—,

irF?

Volume = rra2R —

=

Ixv

11 +

'3,

.v

=

1y'_v'

2

—mr = ,nR ,nR ++ —mr --

a

R

R

Hall Circular Rod Halt

/0

Volume 7TRA Volume == nRA small)

/1

)R

= V

,nR

4)R2

=

Cross sCction A

1

1T

I

= =

Conical Shell V,1 VI,

= =

=

I

1

niL2

+

V

VI

I.... = I... I

2L/3

1

= IyI Iyl v' =

I

= I vI'yIf "Ivy"

L2

+

= =

?flR

1

+

'-

a.

Circular Cylindrical Shell V

1... = ink2

=

S.

I 1.

= =

Ivy

= yi1R2 =

4.

+

1

2

706

APPINDIX C.

INERTIA PROPERTiES

Half Cylindrical Shell

4\

V

—

7TJ

= rnR rnR2

=

1

=

+

1vx

2R

=

.1•

Spherical Shell V

= jvy =

=

Xx

Hemispherical Shell

Ix.i. Iyy = lxx

= = =

I

V y

1v)vf 1."yf

=

!flR'

mR2

1

INDEX

INDEX

A Acceleration centripetal. 11. 11, 19. 125 Coriolis. 19. 23. 125 normal. 11 plane motion. 1 25 relative. 124 tangential. 11 Action integral, 251 of Saxony, Saxony, 67 1 678 Albert of Angle attitude. 374 azimuthal, 21 bank. 374 heading, 374 phase. 65 polar, 21 Angle of twist. 599 Angular acceleration, 11 7 apparent. 158 axisymmetric bodies. 372 conservation, 40. 449 in terms of Euler angles. 371 Angulur momentun particles. 39. 158 rigid bodies. 96, 423. 652 transformation properties, 425 Angular velocity critical, 570 definition of, 109, definitionof, 109, 111,389 of' differential element. 597 matrix. 110. matrix. 110,365 365 mean, mean. 188 partial. 379. 379, 507 as quasi-velocity, 378 simple. 111,357 simple, in terms ol' Euler angles. 369 in terms of' Euler parameters, 385 vector. 109, 389 Anomaly eccentric, 187 eccentric. mean, 189 true, 187 Apex, 563

Apocenter. 179 Apogee. 180 Appell. P.E.. 517, 517. 676. 678 Appell equations, 518 Archimedes. 670. 678 Areal rate. 181 Argument of pericenter. 1 89 Aristotle. 669. 678 Assumed modes method. method, 628, 649 Asymptotic behavior, 621 Axial force, 593 Axis of rotation. rotation, 359, 359. 360 Axis of of' symmetry. 328, 372, 565 Axisymmetric body. body, 341. 372. 499, 548 499.

B Baruh, H.. 650 Beam axis. 592 Rayleigh. 594 theory, 592 Beat phenomenon, 143 Bending moment, 593 Bending stiffness, 598 Bernoulli. D.. D., 679. 687 Bernoulli, Johann. 242, 674, 679 Bernoulli, Johann. II. 679. 687 Body-fixed axes, 356. 370 axes. coordinates. 371. 435 rotation. 100, 105 Boltzmann-Hamel equations, 499 Boundary condition complementary, 607 dynamic. 608 essential. 607 of the first kind. kind, 607. 693 geometric. 607, 693 natural, 608. 694 of the second kind. 608, 694 of the third kind. kind, 608, 694 Boundary value problem. 617

709

710

IN1)EX

Bounding circles, 567 Brahe. T., 671

C Center of' curvature. 9

Center of mass, 154, 154. 324 composite bodies. 325 Centrifugal softening, 650 Centrifugal stiffening. 650 Characteristic equation. equation, 56, 59, 289. 289, 343. 618, 630 Characteristic value, 290 Chasles's theorem. 363 Coefficient of restitution, 166, 193, 478 Complete set, 619 Compression period, 165. 478 Cone body, 359. 550 space. 359. 550 space, Configuration space. 217 Conic section, 179 Constraint, 217 configuration, 219 equation. 219 equation, force. 219 holonomic, 220, 404 modified Tisserand, 643 nonholonomic, 220, 405 number of, 35 in Pfaffian form, 220 rigid body, 641. 651 velocity. 220 workiess, 223 zero slope, 643, 653 zero tip deformation, 643. 655 Constraint relaxation method, 262, 464 Convolution integral, 69 Coordinates, body-fixed, 356 Coordinates, Cartesian, 5 constrained, 218 curvilinear, 8 curvilinear., cyclic. 300

cylindrical, 17 extrinsic, 5 generalized, 216 ignorable, 300, 300. 303 independent. 218 intrinsic. 8 normal-tangential, 6 number of, 35 polar, 18 rectilinear. 5 right-handed system. 4 spherical. 21 Copernicus, N., 671, 679 Coriolis. G. G.. 675, 679 Coriolis, Coriolis acceleration, 19, 23, 125

Coriolis force, 136 Coulomb, C. A., 675. 680 Coulomb friction. friction, 29 Courant. R., 621 Courant, Critical angular velocity. 570 Curvilinear coordinates, 8 Cuspidal motion, 568 Cycloid, 402 Cylindrical pair. 396

D D'Alembert. J. L. R.. 680

D'Alembert's principle, 246 direct application of, 247. 311 Lagrange's form, 254 rigid bodies, 247, 474. 474, 526 Da Vinci, L., 671 Damped linearized systems, 287 Damped natural frequency, 61 Damping coefficient, 59 factor, 59 matrix. 287 matrix, proportional, 297 Datum, 46, 46. 47 Degree of freedom, 34, 218, 601 Dependent variable, 231

INI)EX

Derivative in different reference frame, 121 global, 88 locaL local, 88. 117 Differential-algebraic equations. equations, 313

Dirac delta function. 37 Direction Di rect ion

azimuthal. 21 hinormal. 11 normal. 9 polar. 21 21 radial. radial. 18,21 tangential. 9 transverse, 18 18 Direction angle, 98 Direction cosine, 98. 343 in terms of Euler parameters, parameters. 386 Dissipation function, 286 Distance traversed, 8. 595 Dry friction, 29 Dual spin satellite. satellite, 494 Dugas, A., 669 Dugas. Duhem. P.. 669 Dynamic balancing, 444 Dynamic potential. 276

E Earth mass ol, 1 76 as a moving reference frame. frame, 133. 582 Eccentricity. 179 Eigenfunction, 617 Eigenfunction. Eigenvalue, 290, 343, 361, 617, 617. 631 631 repeated. 344 Eigenvector. 290, 343. 361, 63 1 Ellipse apsis, 179 semi major axis, 179 semiminor axis, 179 serniminor Energy principle of conservation of, 49 total, 49, 453 total.

Energy inner product. 602 Energy of acceleration, 517 Equation of motion, 31 damped systems, 255 differential-algebraic, 261 Euler's, 437 Gihhs-Appell, 518 Gihhs-AppeIl. Kane's, 525 Lagrange's, 255 linearized, 55, 282, 287 in matrix Ihrm. lhrm. 287 rigid body. 431 in state form, 76, 31 447 Equilibrium critically stable, 55 dynamic. 245 nonnatural systems. systems, 276 small motions about, 279 stable. 54 static, 54, 241 static. unstable, 55 Escape velocity. velocity, 26. 26, 183 Euclid. 670 Euler. L., L, 26. 360, Euler. 360. 673. 680. 687 Euler angles for aircraft problems. 373. 41 9 angular acceleration. acceleration, 371 angular velocity. 369 properties of, 419 sequence of. 368, 460 Euler parameters, 38 1, 463 in terms of direction cosines, 387

Euler's equations, 435, 437 modified, 436, 490 Euler-Bernoulli equation, 610 Euler-Lagrange equation. equation, 255 Expansion, 138. 138, 441. 575, 58 I Expansion theorem. theorem, 293, 619 Extremum, 687

F F frame, 372 First integral. 72. 301, 695

71

1

712

INDEX

First variation, variation. 696 Flat body, 554 Flight path angle, 180 Force

central. 41, 175 centrifugal, 136 circulatory, 287 conservative, 44 con straint, 136, 155 constraint, Coriolis, 136 definition of. of, 29 effective, 135 external. 155 friction, 30 generalized, 239, 508 impulsive. 37, 164, 305 inertia, 245, 245. 431 internal, 155 nonconservative, 49 normal, 30 Foucault, 675 Foucault's pendulum, 141, 283 Free body diagram, diagram, 31, 31 255 Frenet's formulas, 13 Frequency average, 143 beat. 143 natural. 59, 290 response, 63 Friction coefficient of, 29 Coulomb. 29 dry, 29 treatment in analytical mechanics,

260,444 Function admissible, 612, 627 comparison, 612, 619, 627 consistent admissible, 627 consistent comparison, 627 Functional. 693 Fundamental equations, 527 Gibbs.-AppeIl form, 527 Gibbs.-Appell Kane's form, 527 Fundamental function, 518

G Galileo, G., 672. 681

Gauss, C.F., 675, 681 Gauss's variational principle, 315 Generalized coordinate, 216 constrained, 218, 261 ignorable, 300 independent, 218 Generalized force, 239, 457 in assumed modes method, 630 inertia, 525 for partial velocity, 507, 507. 513 Generalized impulse for generalized coordinates, 305 for generalized speeds. 539 Generalized inertia force, 525 Generalized momentum, 300. 459 Generalized speed, 378, 504 constrained. 504 independent, 504, 513 surplus. 504, 513 Generalized velocity. velocity, 232 Generatrix. 359 Geosynchronous orbit. 182 Gibbs, J.W., 517. 517, 681 Gibbs function, function. 517 Gibbs-Appell equations. 518 Gibbs-Appell function, 517 for a rigid body, 520 Gimbal lock, 394 Ginsberg. J.H., 24 Gradient operation, 92 Gravitational constant. 3, 28 universal, 28, 173 Guide bars, 397 Gyrocompass, 582 Gyropendulum, 377. 438 Gyroscope balanced, 581 free, 581 integrating, 586 single-axis, 585 Gyroscopic effect, 358 Gyroscopic matrix, 283

INDEX

Gyroscopic moment. 434 Gyroscopic motion, 547 Gyroscopic system. 142 Gyroscopic vector, 275 Gyrostat. 494

H Hamilton. W.R., W.R.. 249. 249, 681 Hamilton's equations. Hamilton's 251 ellJaIioIls. 251 Hamilton ss principle toi Continuous for continuous systems. 604 extended.25 extended. 25 11

of varying action. 250 Hamiltonian, 308 Hamiltonian. Herpolhode. 559 Hessian matrix, 688 Hubert. D.. D.. 62 62 1I Hohmann transfer. 1 85 Holononiic constraint. 220, 220. 405 system. system, 220. 226 Hooke's law. law. 48, 48 597. 599 Hybrid systems. 663 Hyperbolic excess speed. speed. 1I

I ignorable coordinates, coordinates. 300 liiipact impact elastic. 167 167 line ot. of. 164. 164, 477 particle. 164 plastic, I 67 plastic. rigid body. body, 477 Impetus. 67 6711 Impetus. Impulse. 36 Impulse response. 69 IImpulse—momentum mpu ise—niomentum theorem aiuzu Jar.40, 40. 159. 159. 204, 204, 449 aiuzular.

linear. 36. 36, 157. 204. 449 Impulsive force. force. 37 Inclusion principle. 632

Inertia ellipsoid, ellipsoid. 347. 558 normal tO, normal to. 559 Inertia matrix. 275, 275, 328 Inertia properties axisymmetric bodies. bodies, 344 parallel axis theorem. 336 rotation of coordinates, 337 translation ot coordiiiates. coordinates. 335 Inertia torque. 525 Inertial symmetry. 372. 548 Inflection point. 688 Inner product, product. 602 Instantaneous center zero velocity. 20()

Integral of the motion. 72, 301, 695 Internal prOperties. 329 International Gravity Formula. 29 Invariable plane. plane, 559

J Jacobi integral. 301. 309 Joint hail and hail and socket. 392 Cardan, 393 393 constant velocity. 394 prismatic. 130, 395 revolute, 130. 391 universal. 393 Jordanus. 670 Jourdain. P.E.B.. P.E.B., 678. 682 variational principle. principle. 3314 Jourdain's variational 14 Junkins, J.L.. 37 1

K Kane's equations, 525 Kepler, J.. 181, 181. 671. 682 Kepler's laws of planetary motion, motion, 181

Kepler's equation. equation. I 89 Kinematic differential equations. 370, 446, 505

713

714

INDEx

Kinematics infinitesimal rotation, 107 joints, 391 plane motion rotation, 195 rigid body, 355 rolling, 200, 400 Kinetic energy beams, 597 linearized form, 281 particles, 43, 162 rigid bodies, 203, 451 rotational, 162, 452 torsion, 599 translational, 162, 453, 453. 597 two body problem, 175 Kronecker delta, 293

L Lagrange, J. L., 255. 674, 683 Lagrange multipliers, multipliers. 260, 464, 689 Lagrange's equations, 255 for constrained systems, 260, 464 for quasi-coordinates, 534 for rigid bodies, 456 in terms of Euler angles, 462 in terms of Euler parameters, 463

Lagrangian. 251. 303 Line of impact, 164, 477 Line of' nodes. 190, 369, 577 Linear momentum conservation of, 37 particles, 26, 155 rigid bodies, 422 Linear range. 47 Linearization damped systems, 287 about equilibrium, 283 of free gyroscope, 581 nonnatural systems, 283 of rolling motion, 576 Taylor series expansion, 55, 278 Longitude of ascending node. 191 Longitude of descending node, 191

M Mach, E., 669 Magnification factor, 64 Mass, 2 Mass flow rate, 172 Matrix circulatory, 287 identity, 99 inertia, 275

orthogonal, 343 positive-definite, 280, 290, 328, 688 rotation, 101 stiffness, 280 symmetric, 280, 328, 688 unitary, 99 Meirovitch, L., 108, 298 Mixed descriptions, 24 Modal analysis, 296, 617 Modal coordinates, 296 Modal equations, 296 Modal force, 296. 620 Modal vector, 291 Modified Euler equations, 436, 490 Modulus of elasticity. 48, 597 Moment equations arbitrary bodies, 434 about an arbitrary point, 498 axisymmetric bodies, 436, 490 classification, 503 Moment of inertia area, 330 area polar, 330 mass, 195, 327, 423, 598, 652 principal, 329, 343 Momentum angular, 39. 196, 423 generalized, 300. 459 linear, 26, 36, 196, 422 Momentum wheel, 112 Motion primary, 637 relative to earth, 134 secondary. 637

INDEX

N Natural frequency, 59, 59. 290 Natural mode. mode, 29 1 Newton. 1., 1.. 26, 166. 478, 673, 683

law of gravitation of. 28 laws of, 27, 434 Newton-Euler formulation, 432 Node. 619 Nonholonomic, 109 angular velocity as. 109 constraint. 220 system. 223 North polar axis, 189 Nlutation. 369

0 Orbit circular. 179 e]liptic, 179 hyperbolic. 183 parabolic, 183 Orbital inclination, 190 Orbital parameters. 187 Orthogonality, 293, 631 Orthonorma]ity, 293 Orthonormality, Osculating plane, 1 1

P Parallel axis theorem, 197. 336 Partial velocity. 506 matrix. 506 Particle, definition of. of, 26 Particle assumption, 26 Passarello-Huston equations, 533

Path variables. variables, 8 Pericenter, 179 Perigee. 180 Perturbation, 138, 441. 575

Pfaffian form, 220 Phase portrait. 72 Pitch rate, 375 Plane motion, 193 Plane of contact, 400 Plane of symmetry, 328 Poincari, J.H.. 677, 684 Poiiisot's construction. 557 Poinsot's Poisson, S.D.. 165. 478, 675, 684 Poisson's hypothesis. 165, 478 Poissons ratio, 599 Polhode, 559 Position vector cylindrical coordinates. 18 rectilinear coordinates, 5S spherical coordinates. 21 Potential energy, 44. 239 beams, 598 elastic. 48, 598, 600 gravitational. 44, 163 gravitational, modified, 276 springs, 46 56, 281 theorem, 56. torsion, 600 Power, 50. 454 Precession, 369 direct. 553 looping, 567 retrograde, 554 retrograde. steady, 569. 574 unidirectional, 567 Principal axes. 343, 348 axisymmetric bodies, 343, axisyrnmetric 348 Pri ricipal coordinates, coordinates. 296 Principal line. 360 Principal plane. 345 Principia, 26 Principle of least action, 251 Principle of virtual work, 243 Product of inertia area, 598 I

•

—

Ptolemy. Ptolemy, C.. 684

Pure rotation, 356

715

716

INDEX

Q Quasi-coordinate, 505

variation of, 508 Quasi-velocity. 378, 505 constrained, 378. 505

R Radian. I Radian, Radius of curvature, 9. 400. 593 Radius of gyration. gyration, 197. 197, 332 Rayleigh s dissipation function,

with slip, 200, 401 without slip, 201, 401, 573 Rotating shaft, 659 Rotation, 102 of earth, 133 finite, 103 about a fixed axis, 356. 443 about a fixed point, 358, 563 infinitesimal, 107 about principal line. line, 360 Rotation sequence, 107 Routh. E.J.. 684 Routhian, 303

286

Reference frame for axisymmetric bodies, 372 body-fixed, 356 fixed, 88 fixed to earth. earth, 133 133 inertial, 26 moving, 88

relative, 133.630 Relative acceleration, 125 Relative velocity, velocity. 124 Resonance, 64 Response damped systems, systems. 61, 298 free. 58, 298 general, 68 to harmonic excitation. 65 in phase, 65 out of phase, 65 Restitution period, 166, 478 Resultant, 196, 429 Rigid body assumption, 323, 591 591 equations of motion. 431 kinematics. 355 Rigid body mode, 291, 624 Rivello, R.M., 601 Rivello. Rodriques parameters. 415 Roll rate, 375 Rolling of inertia ellipsoid. 559 kinematics, 400

S Saddle point, 689

Sampling period. 75 Semilatus rectum, 179 Semimajor axis. 179 Semiminor axis. 179 Separatrix. 72. 561 Servomotor, 440 Shadow frame, 637 Shear force, 593 Shear modulus, 599 Shortening of projection, 595 Sleeping top. top, 571 Slender body, body. 553 Slider, 395 Sliding in rolling, 577 Smith, C., 650 Space-fixed rotation, 100 Spherical pair, 392 Spherical pendulum. 23 Spin. 369 Spin slip in rolling, 577 Spring axial, 46 axial. linear range, 47 softening, 47 stiffening, 47 torsional, 47 Spring constant, 46 St. Venant's hypothesis. 599

INDEX

Stability asymptotic. 55 asymptotic, critical. 55 free motion. 441. 562 neutral, 55 theorem. 56, 281 State form. form, 76. 308. 3 I 2 Stationary value of constrained systems. 690 of a definite integral. 693 of a function, 688 Steady fliolion, motion, 574 Steady precession. 574, 582, 584 Stiffness axial. 598 bending. 598 torsional, 600 Stiffness coefficients, 280 Strain axial, 48 shear, 599

Torsion of curve, 12 Torsional stiffness. stiffness, 600 Tracking frame, 637 Transfer orbit. orbit, 1 85 Transmissibility, 67 Transport theorem. 117, 121 Trial function, 628 Truesde]l, C.T.. 432. Truesde]], 432, 669 Turner. J., 371 Twisting moment, 593 Two body problem. 173

U Unit step Functioii. Function. 69 Unit vector, 4 cylindrical coordinates. 1 8 rectilinear coordinates, 4 Units S1,2 SI.,2 U.S. Customary, 2

S tress

axial, 48 shear. 599 Sturni-Liouvilic problem, 697 Sylvester's criterion. 328, 328. 688 Synchronous motion, 289 System conservative. 56 holonomic. 220, 226 natural, 275 nonnatural, 275 rheononiic, rheononiic. 220 scleronomic, 220 scleronomic. undamped, 59 undamped. underdaniped, 61 underdamped,

T Tadikonda, S.S.K., 654 Taylor series expansion, 55, 279

Time of pericenter passage. 189 rroI.que4ree Forquc-tree motion, motion. 548. 556

V Variable mass. 170 Variation of latitude, 552 Variation of quasi-coordinates, 508 Variational notation, 230, 696 Vector algebraic, 91 column, 91 column. geometric, 91 gyroscopic, 275 spatial, 91 Vehicle dynamics. 224 Velocity base, 124

partial. 379. 459, 506 relative. 1124 relative, 24 transport. 125 Vernal equinox. 189 Vertex, 563

717

iie

INDEX

Virtual displacements, 230, 699 analytical approach, 232 kinematical approach, 232 Virtual work, 233 constraint force, 233 independent quasi-velocities. quasi-velocities, 512

w

Work

done by a conservative force, 48 done by a constraint force, 223 done by a nonconservative force, 49 incremental, 43 virtual. 230. 233, 457 Work-energy theorem, 43, 163, 455

Y

Warping function, 598

Weighted residual, 628 Whirling, 662 Wobbling motion, 552

Yaw rate, 375 Yu. L., 654

E-1

Analytical Dynalllics Errata Sheet Please Note: My word processor does not print boldface Greek characters. So, to denote a boldface Greek character, I underline it in this errata sheet.

Location

Error and its Correction

17 54 91 100 135 155 187 203 213 233 234 235 238 240 245

Eq. [b] First line before Eq. [1.8.2] Eq. [2.3.2] Footnote Eq. [2.8.12] Eq [3.2.7] Fig. 3.22 Eq. [1] Problem 37 Before Ex. 4.5 Fig. 4.13 Eq. [d] Eq. [4.5.7] Eq. [c] Before Sec. 4.7

Middle line units should be 1 Ib/ft*3 T = 0 should read t = 0 Last sentence of paragraph should read From Eq. [1.7.27] we have Unit vectors should be el, e2, e3, not i, j, k Eq. [2.14.12b] should be Eq. [2.4.12b] w should be 0 (also on next page in text and in Fig. 2.37)

258

1st term in Eq. [f]

262 263 265 284 288 295 308 309 312 314 320

betw Eqs. [b]&[c] Eq. [h] Fig. 4.24 Fig 5.2 Eq. [c] Eq. [e] Eq. [5.7.11 b] Fig. 5.7 eq. [5.12.6] After [5.13.1] Problem 19

320 321 321 336 340 340 351 354 362 379 380 410 412

Problem 20 Problem 29 problem 32 Eq. [6.4.5] Eq. [j] Eq. [1] Problem 1 Problem 16 Eq. [7.3.10] [Eq. 7.6.6] Eq. [b] Fig. 7.40 Problem 1

412

Problem 2

414

Problem 9

Page

4

L;f

1 miP = 0 should read i Line closest to x axis it should read aE Last term on left side should read mg(R - r) sinO Should read: Consider the bar attached to a disk in Problem 33 ... remove th word holonomic from last line R4> is mislabeled. Length of short line is RrP First term on right side should be BxLICOl00l Index of summation should be k and not i left side of second expression should be r G1 (boldface for r) Last sentence should read: The only equilibrium solution is if cos 01 = cos O2 = 0, which is the upright position of the linkage. Should read Qx = F - cx - ~cbe cos 0 . . v() = rOe() should be v() = rO the sin 0 term should be Y sin 0 Grounded link is of length L 4 Caption should be Mass-spring-damper system 33 element of [D] should be C2, not 0 Middle terms in the two expressions should be 2mQ~ and 2mQ~ Last term should be {g}T Caption should read A spring pendulum Right parenthesis missing on top line.... 8{q} ([riq] {q} + 2 [riq]{ q}) + ... Say keeping the independent variable, Last sentence should read: Approximate the two natural frequencies and plot the free response. Add to end of problem the following: The initial angular velocities are zero. Should read: Derive the equations of motion ... Should read: Solve problem 4.39 using ... last term should be m( d~ + d;) Right paren missing after IGba.r Second term on right should be [IG1nM J Length of body along x axis should be b Add: for the value b = 4a Should read i = ~i In second term, minus sign missin.g before ~ummation signs Third term should be U3 = W3 = rP cos 0 + 7/J radius of front wheel is R 3 Last sentence should read: Find the velocity of the tip of the bar as a function of the angle O. problem should read: For the system shown in Fig. 7.43, find the instantaneous center of the rod AB and its angular velocity. last part of sentence should read: using the derivatives of the Euler angles

tt

1

E-2

415 418 419 436 440

Problem 21 Problem 31 [R] for 3-1-3 Eq. [8.5.31] Eq. [0]

444 445 475 483 486 498 503 507 512 526 535

Fig. 8.13 Eq. [f] Eq.[8.11.8] Problem 16 Fig. 8.35 Eq. [9.3.3] Eq. [f] Eq. [9.5.12] Eq. 9.6.19 over Eq. [9.8.8] under Eq. [9.10.27]

543 543 544 544 553 564 576 585 585 583 586 586 586 586 587 587 597 598 598 598 599 603

Problem 6 Problem 8 Problem 19 Problem 25 Eq/ [10.2.22] Eq. [10.5.1] above Eq. [10.6.25] 4th line Fig. 10.20 Fig. 10.19 Eq. [10.7.23] Eq. [10.7.24] Eq. [10.7.25] Eq. [10.7.26] Problem 1 Problem 3 Eq. [11.2.13] Eq. 11.2.19] Eq. [11.2.21] Eq. [11.2.22] Eq. [11.2.27] Eq. [b]

604 605 611 615 625 627

Eq. [11.3.2] Eq. [b] under [11.3.27] Eq. [e] Eq. [b] Table 11.3

633

Eq.[l1.5.24]

as the generalized speeds. In this problem B is constant. In this problem, the length of the rod is 2.5 R. 21 element should be -ups'lj; - stPcBc'lj; All the left subscript f should be subscript F should read -M - M z cosB + M x sinB = 0 Consequently, all the signs in Eqs. [q] and [r] are reversed. Direction of mg should be along negative Y axis. M x My should be replaced with -Mx and - My There should be po dm term in the middle expression (the one with the integral sign) In this problem B = 0 Axis pointing down should be bl Left side should read {mvG} and not {vG} In second line, 32 element of fist matrix should be -W3 + t~~ () Second term (v t = ... )is incorrect, ignore it. Tkr should be Trk In next to last line of paragraph Wi should be boldface WI Paragraph should read: For constrained systems, a Lagrange multiplier vector enters the formulation the same way that it does in the traditional Lagrange's equations. Alternatively, one can select a set of independent generalized speeds. In this case, [Y] should be written such that Eq [9.10.18] is satisfied. Last line should read: the partial velocities of points D and P. Add to problem: Calculate the partial velocities of G. Last sentence should read: Consider rotational motion of the main body only. Solve Problem 3 using .... not 2 . problem . Third term should be W3 = tPcB + 'lj; No i after the one half, ~ (llwi ... In line above Eq.[10.6.25], it should say Recall from Eq. [e], not Eq. [f] Note that the inner gimbal is stationary. This figure should be the same as Fig. 2.49 (p. 148). 12 should read - 12 Minus sign missing from 1st term on right side Wy ... Minus sign missing from 1st term on right side Minus sign missing from 1st term on right side Minus sign missing from 1st term on right side Length of football should be 25 cm, and player tips the football at the front Precession rate should be 0.25 rad/s, not 0.2 rad/s. +zw ll (x, t) should be +zw ll (x, t) Same mistake as above -2Elyz should be +2Elyz Last term should be EA(x)e'(x, t) Left side should be u(x, y, z, t) Second line should have added term Mgw(h, t) due to potential energy of ring Also, in first line the 1/2 should be before Ely .. It should not cover the f.l(x)gw(x, t) term Third term should be wOw and not wOw First integrand in Fourth line should begin with a minus sign term on right on line immediately below Eq. [11.3.27] should read ky(x)v(x, t) Add dx to first term on top line The integrand should be tP~(x)o(x)dx with the delta double underlined 4th column 3rd term (corresponding to Consistent admissible functions) should read: All geometric boundary conditions. Do not prevent dynamic ... In krs term the triangular brackets should be replaced by square brackets, as in [11.2.45]

1r:

E-3

641 648 649 649 656 662 665 666 667 668 668 674 690

Eq. [11.6.7] eq. [11. 7.12] Above Eq 11.7.15 Eq. [11.7.1 7b] Table 11.7 Below Eq. [e] Fig. 11.41 Problem 10 Problem 18 Problem 23 Problem 25 and also p. 680 Eq. [Bll]

Integrand is fl(X) [(x + u(x, t))i + v(x, t)j]dx On right side, o(x) should not be boldface Two lines above, replace "the extended Hamilton's principle" with "Lagrange's equations" First integral is missing a dx 3rd column (yes, yes, no, yes, yes, no, yes, no) should be all Yes. ¢J(x) should be ¢Jl(X) Length of bar is not given. It should be L. 3rd line should read a = L/2, k = 2EIo / L 3 , where 10 = bh 3 /12. Radius of shaft should be 2 em, not 3 em. Also, a = 0.8m. Last word of problem should be satellite, not hub. In second line, J = 0.3M R2, not I = 0.3M R2 Mauperitus should read Maupertuis. Second term should be c~r = aacX ro

698 698

Eq. [e] Table B.l

First term in integrand should be (p(x) ~;) Last column, last element should read: Mass moment of inertia per unit length I(x) Poincari should be Poincare.

715

-lx

Last revised April 6, 2005 If you find any other errors please bring to my attention ([email protected]). Thank you