Intermediate Dynamics for Engineers: Newton-Euler and Lagrangian Mechanics [2 ed.] 1108494218, 9781108494212

Table of contents :
Contents
Preface
Part I A Single Particle
1 Kinematics of a Particle
1.1 Introduction
1.2 Reference Frames
1.3 Kinematics of a Particle
1.4 Cartesian, Cylindrical Polar, and Spherical Polar Coordinate Systems
1.5 Curvilinear Coordinates
1.6 Examples of Curvilinear Coordinate Systems
1.7 Representations of Particle Kinematics
1.8 Kinetic Energy and Coordinate Singularities
1.9 Constraints
1.10 Classification of Constraints
1.11 Closing Comments
1.12 Exercises
2 Kinetics of a Particle
2.1 Introduction
2.2 The Balance Law for a Single Particle
2.3 Work and Power
2.4 Conservative Forces
2.5 Examples of Conservative Forces
2.6 Constraint Forces
2.7 Conservations
2.8 Dynamics of a Particle in a Gravitational Field
2.9 Dynamics of a Particle on a Spinning Cone
2.10 A Shocking Constraint
2.11 A Simple Model for a Roller Coaster
2.12 Closing Comments
2.13 Exercises
3 Lagrange’s Equations of Motion for a Single Particle
3.1 Introduction
3.2 Lagrange’s Equations of Motion
3.3 Equations of Motion for an Unconstrained Particle
3.4 Lagrange’s Equations in the Presence of Constraints
3.5 A Particle in Motion on a Smooth Surface of Revolution
3.6 A Particle in Motion on a Sphere
3.7 Some Elements of Geometry and Particle Kinematics
3.8 The Geometry of Lagrange’s Equations of Motion
3.9 Lagrange’s Equations of Motion for a Particle in the Presence of Friction
3.10 A Particle in Motion on a Helix
3.11 A Particle in Motion on a Moving Curve
3.12 Closing Comments
3.13 Exercises
Part II A System of Particles
4 Lagrange’s Equations of Motion for a System of Particles
4.1 Introduction
4.2 A System of N Particles
4.3 Coordinates
4.4 Constraints and Constraint Forces
4.5 Conservative Forces and Potential Energies
4.6 Lagrange’s Equations of Motion
4.7 Construction and Use of a Single Representative Particle
4.8 Kinetic Energy, Mass Matrix, and Coordinate Singularities
4.9 The Lagrangian
4.10 A Constrained System of Particles
4.11 A Canonical Form of Lagrange’s Equations
4.12 Alternative Principles of Mechanics
4.12.1 Principle of Virtual Work and D’Alembert’s Principle
4.12.2 Gauss’ Principle of Least Constraint
4.12.3 Hamilton’s Principle
4.13 Closing Comments
4.14 Exercises
5 Dynamics of Systems of Particles
5.1 Introduction
5.2 Harmonic Oscillators
5.3 A Dumbbell Satellite
5.4 A Pendulum and a Cart
5.5 Two Particles Tethered by an Inextensible String
5.6 Closing Comments
5.7 Exercises
Part III A Single Rigid Body
6 Rotations and their Representations
6.1 Introduction
6.2 The Simplest Rotation
6.3 Proper Orthogonal Tensors
6.4 Derivatives of a Proper Orthogonal Tensor
6.5 Euler’s Representation of a Rotation Tensor
6.6 Euler’s Theorem: Rotation Tensors and Proper Orthogonal Tensors
6.7 Relative Angular Velocity Vectors
6.8 Euler Angles
6.8.1 3–2–1 Euler Angles
6.8.2 3–1–3 Euler Angles
6.8.3 The Other Sets of Euler Angles
6.8.4 Application to Joint Coordinate Systems
6.8.5 Comments on Products of Rotations
6.9 Further Representations of a Rotation Tensor
6.10 Rotations, Quotient Spaces, and Projective Spaces
6.11 Derivatives of Scalar Functions of Rotation Tensors
6.12 Exercises
7 Kinematics of Rigid Bodies
7.1 Introduction
7.2 The Motion of a Rigid Body
7.3 The Angular Velocity and Angular Acceleration Vectors
7.4 A Corotational Basis
7.5 Three Distinct Axes of Rotation
7.6 The Center of Mass and Linear Momentum
7.7 Angular Momenta
7.8 Euler Tensors and Inertia Tensors
7.9 Angular Momentum and an Inertia Tensor
7.10 Kinetic Energy
7.11 Attitudes of Constant Angular Velocities
7.12 Closing Comments
7.13 Exercises
8 Constraints on and Potential Energies for a Rigid Body
8.1 Introduction
8.2 Forces and Moments Acting on a Rigid Body
8.3 Examples of Constrained Rigid Bodies
8.4 Constraints and Lagrange’s Prescription
8.5 Integrability Criteria
8.6 Potential Energies, Conservative Forces, and Conservative Moments
8.7 Closing Comments
8.8 Exercises
9 Kinetics of a Rigid Body
9.1 Introduction
9.2 Balance Laws for a Rigid Body
9.3 Work and Energy Conservation
9.4 Additional Forms of the Balance of Angular Momentum
9.5 Moment-Free Motion of a Rigid Body
9.6 The Baseball and the Football
9.7 Motion of a Rigid Body with a Fixed Point
9.8 Motions of Rolling Spheres and Sliding Spheres
9.9 Chaplygin’s Sphere
9.10 Closing Comments
9.11 Exercises
10 Lagrange’s Equations of Motion for a Single Rigid Body
10.1 Introduction
10.2 The Lagrange Top
10.3 Configuration Manifold of an Unconstrained Rigid Body
10.4 Lagrange’s Equations of Motion: A First Form
10.4.1 Proof of Lagrange’s Equations
10.4.2 The Four Identities
10.5 A Satellite Problem
10.6 Lagrange’s Equations of Motion: A Second Form
10.6.1 Summary
10.7 Lagrange’s Equations of Motion: Approach II
10.8 Rolling Disks and Sliding Disks
10.9 Lagrange and Poisson Tops
10.10 Closing Comments
10.11 Exercises
Part IV Systems of Particles and Rigid Bodies
11 The Dynamics of Systems of Particles and Rigid Bodies
11.1 Introduction
11.2 Preliminaries
11.3 A Planar Double Pendulum
11.4 A Particle on a Rotating Circular Hoop
11.5 Constraints
11.6 A Canonical Function
11.7 Integrability Criteria
11.8 Constraint Forces and Constraint Moments
11.9 Potential Energies, Conservative Forces, and Conservative Moments
11.10 Lagrange’s Equations of Motion
11.11 Two Pin-Jointed Rigid Bodies
11.12 A Simple Model for a Spherical Robot
11.13 A Semicircular Cylinder Rolling on a Cart
11.14 A Single-Axis Rate Gyroscope
11.15 Orthogonality of Generalized Forces and Gimbal Lock
11.16 Closing Comments
11.17 Exercises
Appendix A Background on Tensors
A.1 Introduction
A.2 Preliminaries: Bases, Alternators, and Kronecker Deltas
A.3 The Tensor Product of Two Vectors
A.4 Second-Order Tensors
A.5 A Representation Theorem for Second-Order Tensors
A.6 Functions of Second-Order Tensors
A.7 Third-Order Tensors
A.8 Special Types of Second-Order Tensors
A.9 Derivatives of Tensors
A.10 Exercises
References
Index

Citation preview

Intermediate Dynamics for Engineers

Suitable for both senior-level and first-year graduate courses, this fully revised edition provides a unique and systematic treatment of engineering dynamics that covers Newton–Euler and Lagrangian approaches. New to this edition are: two completely revised chapters on the constraints on, and potential energies for, rigid bodies, and the dynamics of systems of particles and rigid bodies; clearer discussion on coordinate singularities and their relation to mass matrices and configuration manifolds; additional discussion of contravariant basis vectors and dual Euler basis vectors, as well as related works in robotics; improved coverage of navigation equations; inclusion of a 350-page solutions manual for instructors, available online; a fully updated reference list. Numerous structured examples, discussion of various applications, and exercises covering a wide range of topics are included throughout, and source code for exercises, and simulations of systems, are available online at https://rotations.berkeley.edu. is a Professor of Mechanical Engineering at the University of California, Berkeley and a Fellow of the American Society of Mechanical Engineers. He is the author of several textbooks, has coauthored over 100 papers in archival journals, and is the recipient of multiple teaching awards including a Distinguished Teaching Award from the University of California, Berkeley. Oliver

M.

O’Reilly

Intermediate Dynamics for Engineers

Newton–Euler and Lagrangian Mechanics O L I V E R M . O ’ R E I L LY University of California, Berkeley

University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India 79 Anson Road, #06–04/06, Singapore 079906 Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning, and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781108494212 DOI: 10.1017/9781108644297 c ±

Oliver O’Reilly 2020

This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2008 Second edition 2020 Printed in the United Kingdom by TJ International Ltd, Padstow Cornwall A catalogue record for this publication is available from the British Library. Library of Congress Cataloging-in-Publication Data Names: O’Reilly, Oliver M., author. Title: Intermediate dynamics for engineers : Newton-Euler and Lagrangian mechanics / Oliver M. O’Reilly. Description: Second edition. | New York : Cambridge University Press, [2020] | Includes bibliographical references and index. Identifiers: LCCN 2019031993 (print) | LCCN 2019031994 (ebook) | ISBN 9781108494212 (hardback) | ISBN 9781108644297 (epub) Subjects: LCSH: Dynamics–Textbooks. Classification: LCC TA352 .O745 2019 (print) | LCC TA352 (ebook) | DDC 620.1/04–dc23 LC record available at https://lccn.loc.gov/2019031993 LC ebook record available at https://lccn.loc.gov/2019031994 ISBN 978-1-108-49421-2 Hardback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

This book is dedicated to my adventurous daughter Anna. She’s photographed above with the author at Dún Aonghasa in 2017.

Contents

Preface Part I

1

A Single Particle

Kinematics of a Particle

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 2

Kinetics of a Particle

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 3

Introduction Reference Frames Kinematics of a Particle Cartesian, Cylindrical Polar, and Spherical Polar Coordinate Systems Curvilinear Coordinates Examples of Curvilinear Coordinate Systems Representations of Particle Kinematics Kinetic Energy and Coordinate Singularities Constraints Classification of Constraints Closing Comments Exercises

Introduction The Balance Law for a Single Particle Work and Power Conservative Forces Examples of Conservative Forces Constraint Forces Conservations Dynamics of a Particle in a Gravitational Field Dynamics of a Particle on a Spinning Cone A Shocking Constraint A Simple Model for a Roller Coaster Closing Comments Exercises

Lagrange’s Equations of Motion for a Single Particle

3.1

Introduction

page xiii 1 3 3 3 5 6 9 15 19 21 22 28 35 35 42 42 42 44 45 46 48 54 58 65 69 71 76 76 81 81

viii

Contents

3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 Part II

4

A System of Particles

Lagrange’s Equations of Motion for a System of Particles

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12

4.13 4.14 5

Lagrange’s Equations of Motion Equations of Motion for an Unconstrained Particle Lagrange’s Equations in the Presence of Constraints A Particle in Motion on a Smooth Surface of Revolution A Particle in Motion on a Sphere Some Elements of Geometry and Particle Kinematics The Geometry of Lagrange’s Equations of Motion Lagrange’s Equations of Motion for a Particle in the Presence of Friction A Particle in Motion on a Helix A Particle in Motion on a Moving Curve Closing Comments Exercises

Introduction A System of N Particles Coordinates Constraints and Constraint Forces Conservative Forces and Potential Energies Lagrange’s Equations of Motion Construction and Use of a Single Representative Particle Kinetic Energy, Mass Matrix, and Coordinate Singularities The Lagrangian A Constrained System of Particles A Canonical Form of Lagrange’s Equations Alternative Principles of Mechanics 4.12.1 Principle of Virtual Work and D’Alembert’s Principle 4.12.2 Gauss’ Principle of Least Constraint 4.12.3 Hamilton’s Principle Closing Comments Exercises

Dynamics of Systems of Particles

5.1 5.2 5.3 5.4 5.5 5.6 5.7

Introduction Harmonic Oscillators A Dumbbell Satellite A Pendulum and a Cart Two Particles Tethered by an Inextensible String Closing Comments Exercises

83 85 87 90 94 97 103 107 109 114 118 119 135 137 137 137 139 141 144 145 147 153 154 155 158 164 164 165 166 167 167 170 170 170 176 179 183 187 189

Contents

Part III

6

A Single Rigid Body

Rotations and their Representations

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

6.9 6.10 6.11 6.12 7

Kinematics of Rigid Bodies

7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 8

Introduction The Simplest Rotation Proper Orthogonal Tensors Derivatives of a Proper Orthogonal Tensor Euler’s Representation of a Rotation Tensor Euler’s Theorem: Rotation Tensors and Proper Orthogonal Tensors Relative Angular Velocity Vectors Euler Angles 6.8.1 3–2–1 Euler Angles 6.8.2 3–1–3 Euler Angles 6.8.3 The Other Sets of Euler Angles 6.8.4 Application to Joint Coordinate Systems 6.8.5 Comments on Products of Rotations Further Representations of a Rotation Tensor Rotations, Quotient Spaces, and Projective Spaces Derivatives of Scalar Functions of Rotation Tensors Exercises

Introduction The Motion of a Rigid Body The Angular Velocity and Angular Acceleration Vectors A Corotational Basis Three Distinct Axes of Rotation The Center of Mass and Linear Momentum Angular Momenta Euler Tensors and Inertia Tensors Angular Momentum and an Inertia Tensor Kinetic Energy Attitudes of Constant Angular Velocities Closing Comments Exercises

Constraints on and Potential Energies for a Rigid Body

8.1 8.2 8.3 8.4 8.5 8.6

Introduction Forces and Moments Acting on a Rigid Body Examples of Constrained Rigid Bodies Constraints and Lagrange’s Prescription Integrability Criteria Potential Energies, Conservative Forces, and Conservative Moments

ix

201 203 203 204 206 209 211 217 219 224 227 232 234 236 238 240 243 246 249 257 257 257 261 262 265 267 269 270 276 277 279 288 288 299 299 299 301 307 310 313

x

Contents

8.7 8.8 9

Kinetics of a Rigid Body

9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 10

10.5 10.6 10.7 10.8 10.9 10.10 10.11

11

Introduction Balance Laws for a Rigid Body Work and Energy Conservation Additional Forms of the Balance of Angular Momentum Moment-Free Motion of a Rigid Body The Baseball and the Football Motion of a Rigid Body with a Fixed Point Motions of Rolling Spheres and Sliding Spheres Chaplygin’s Sphere Closing Comments Exercises

Lagrange’s Equations of Motion for a Single Rigid Body

10.1 10.2 10.3 10.4

Part IV

Closing Comments Exercises

Introduction The Lagrange Top Configuration Manifold of an Unconstrained Rigid Body Lagrange’s Equations of Motion: A First Form 10.4.1 Proof of Lagrange’s Equations 10.4.2 The Four Identities A Satellite Problem Lagrange’s Equations of Motion: A Second Form 10.6.1 Summary Lagrange’s Equations of Motion: Approach II Rolling Disks and Sliding Disks Lagrange and Poisson Tops Closing Comments Exercises

Systems of Particles and Rigid Bodies

The Dynamics of Systems of Particles and Rigid Bodies

11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9

Introduction Preliminaries A Planar Double Pendulum A Particle on a Rotating Circular Hoop Constraints A Canonical Function Integrability Criteria Constraint Forces and Constraint Moments Potential Energies, Conservative Forces, and Conservative Moments

319 320 329 329 329 331 333 336 343 347 352 355 363 365 379 379 380 383 386 387 388 390 393 401 402 403 411 416 417 429 431 431 432 433 436 439 443 445 447 450

Contents

11.10 11.11 11.12 11.13 11.14 11.15 11.16 11.17 Appendix A

Lagrange’s Equations of Motion Two Pin-Jointed Rigid Bodies A Simple Model for a Spherical Robot A Semicircular Cylinder Rolling on a Cart A Single-Axis Rate Gyroscope Orthogonality of Generalized Forces and Gimbal Lock Closing Comments Exercises

Background on Tensors

A.1 A.2 A.3 A.4 A.5 A.6 A.7 A.8 A.9 A.10

Introduction Preliminaries: Bases, Alternators, and Kronecker Deltas The Tensor Product of Two Vectors Second-Order Tensors A Representation Theorem for Second-Order Tensors Functions of Second-Order Tensors Third-Order Tensors Special Types of Second-Order Tensors Derivatives of Tensors Exercises

References Index

xi

453 455 456 461 470 473 480 480 488 488 488 489 490 490 493 496 498 499 500 503 522

Preface

Preface to First Edition

The writing of this book started over a decade ago when I was first given the assignment of teaching two courses on rigid-body dynamics. One of these courses featured Lagrange’s equations of motion, and the other featured the Newton–Euler equations. I had long struggled to resolve these two approaches to formulating the equations of motion of mechanical systems. Luckily at this time one of my colleagues, Jim Casey, was examining the elegant works [274, 276, 277] of Synge and his coworkers on this topic. There, he found a partial resolution to the equivalence of the Lagrangian and Newton–Euler approaches. He then went further and showed how the governing equations for a rigid body formulated by use of both approaches were equivalent [37, 38]. Shades of this result could be seen in an earlier work by Greenwood [105], but Casey’s work established the equivalence in an unequivocal fashion. As is evident from this book, I subsequently adapted and expanded on Casey’s treatment in my courses. My treatment of dynamics presented in this book is also heavily influenced by the texts of Papastavridis [226] and Rosenberg [245]. It has also benefited from my graduate studies in dynamical systems at Cornell in the late 1980s. There, under the guidance of Philip Holmes, Frank Moon, Richard Rand, and Andy Ruina, I was shown how the equations governing the motion of (often simple) mechanical systems featuring particles and rigid bodies could display surprisingly rich behavior. There are several manners in which this book differs from a traditional text on engineering dynamics. First, I demonstrate explicitly how the equations of motion obtained by using Lagrange’s equations and the Newton–Euler equations are equivalent. To achieve this, my discussion of geometry and curvilinear coordinates is far more detailed than is normally found in textbooks at this level. The second difference is that I use tensors extensively when discussing the rotation of a rigid body. Here, I am following related developments in continuum mechanics, and I believe that this enables a far clearer derivation of many of the fundamental results in the kinematics of rigid bodies. I have distributed as many examples as feasible throughout this book and have attempted to cite up-to-date references to them and related systems as far as possible. However, I have not approached the exhaustive treatments by Papastravridis [226] nor its classical counterpart by Routh [247, 248]. I hope that sufficient citations to these and several other wonderful texts on dynamics have been placed throughout the text so that the interested reader has ample opportunity to explore this rewarding subject.

xiv

Preface

Using this Text

This book has been written so that it provides sufficient material for two full-length semester courses in intermediate engineering dynamics. As such it contains two tracks (which overlap in places). For the first course, in which a Newton–Euler approach is used, the following chapters can be covered: Chapter 1. Kinematics of a Particle (Sections 1.5 and 1.6 can be omitted). Chapter 2. Kinetics of a Particle. Appendix on Tensors. Chapter 6. Rotations and their Representations. Chapter 7. Kinematics of Rigid Bodies. Chapter 8. Constraints on and Potential Energies for a Rigid Body. Chapter 9. Kinetics of a Rigid Body. Chapter 11. The Dynamics of Systems of Particles and Rigid Bodies. The second course, in which a Lagrangian approach is used, could be based on the following chapters: Chapter 1. Kinematics of a Particle. Chapter 2. Kinetics of a Particle. Chapter 3. Lagrange’s Equations of Motion for a Single Particle. Chapter 4. Lagrange’s Equations of Motion for a System of Particles. Chapter 5. Dynamics of Systems of Particles. Appendix on Tensors. Chapter 6. Rotations and their Representations (with particular emphasis on Section 6.8). Chapter 7. Kinematics of Rigid Bodies. Chapter 8. Constraints on and Potential Energies for a Rigid Body. Chapter 9. Kinetics of a Rigid Body. Chapter 10. Lagrange’s Equations of Motion for a Single Rigid Body. Chapter 11. The Dynamics of Systems of Particles and Rigid Bodies. In discussing rotations for the second course, time constraints permit a detailed discussion of only the Euler angle parameterization of a rotation tensor from Chapter 6 and a brief mention of the examples on rigid-body dynamics discussed in Chapter 9. Most of the exercises at the end of each chapter are highly structured and are intended as a self-study aid. As I don’t intend to publish or distribute a solutions’ manual, I have tailored the problems to provide answers that can be validated. Some of the exercises feature numerical simulations that can be performed with M ATLAB or M ATHEMATICA. Completing these exercises is invaluable, both in terms of comprehending why obtaining a set of differential equations for a system is important and for visualizing the behavior of the system predicted by the model. I also strongly recommend semester projects for the students during which they can delve into a specific problem, such as the dynamics of a wobblestone, the flight of a Frisbee, or the reorientation of a dual-spin satellite, in considerable detail. In my courses, these projects feature simulations and animations

Preface

xv

and are usually performed by students working in pairs who start working together after 7 weeks of a 15-week semester.

Image Credit

The portrait of William R. Hamilton in Figure 4.7 in Section 4.12 is from the Royal Irish Academy in Dublin, Ireland. I am grateful to Pauric Dempsey, the Head of Communications & Public Affairs of this institution, for providing the image.

Acknowledgments

This book is based on my class notes and exercises for two courses on dynamics, ME170, E NGINEERING M ECHANICS III and ME175, I NTERMEDIATE DYNAMICS , which I have taught at the Department of Mechanical Engineering at the University of California at Berkeley over the past decade. Some of the aims of these courses are to give senior and first-year graduate students in Mechanical Engineering requisite skills in the area of dynamics of rigid bodies. The book is also intended to be a sequel to my book Engineering Dynamics: A Primer, which was published by Springer-Verlag in 2001. I have been blessed with the insights and questions of many remarkable students and the help of several dedicated teaching assistants. Space precludes mention of all of these students and assistants, but it is nice to have the opportunity to acknowledge some of them here: Joshua P. Coaplen, Nur Adila Faruk Senan, David Gulick, Moneer Helu, Eva Kanso, Patch Kessler, Nathan Kinkaid, Henry Lopez, David Moody, Tom Nordenholz, Jeun Jye Ong, Todd Lauderdale, Sebastién Payen, Brian Spears, Philip J. Stephanou, Meng How Tan, Peter C. Varadi, and Stéphane Verguet. I am also grateful to Chet Vignes for his careful reading of an earlier draft of the book. Many other scholars helped me with specific aspects of and topics in this book. Figure 9.2 was composed by Patch Kessler. Henry Lopez (B.E. 2006) helped me with the roller coaster model and simulations of its equations of motion. Professor Chris Hall of Virginia Tech pointed out reference [156] on Lagrange’s solution of a satellite dynamics problem. Professor Richard Montgomery of the University of California at Santa Cruz discussed the remarkable figure-eight solutions to the three-body problem with me, Professor Glen Niebur of the University of Notre Dame provided valuable references on Codman’s paradox, Professor Harold Soodak of the City College of New York provided valuable comments on the tippe top, and Professors Donald Greenwood and John Papastavridis carefully read a penultimate draft of this book and generously provided many constructive comments and corrections for which I am most grateful. Most of this book was written during the past 10 years at the University of California at Berkeley. The remarkable library of this institution has been an invaluable resource in my quest to distill over 300 years of work on the subject matter in this book. I am most grateful to the library staff for their assistance and the taxpayers for their support of the University of California.

xvi

Preface

Throughout this book, several references to my own research on rigid-body dynamics can be found. In addition to the students mentioned earlier, I have had the good fortune to work with Jim Casey and Arun Srinivasa on several aspects of the equations of motion for rigid bodies. The numerous citations to their works are a reflection of my gratitude to them. This book would not have been published without the help and encouragement of Peter Gordon at Cambridge University Press, and would contain far more errors were it not for the editorial help (and wit) of Victoria Danahy. Despite the assistance of several able proofreaders, it is unavoidable that some typographical and technical errors have crept into this book and they are my unpleasant responsibility alone. If you find some on your journey through these pages, I would be pleased if you could bring them to my attention. Oliver M. O’Reilly [email protected] Berkeley December 2007

Preface to Second Edition

In the decade since the first edition of this book was published, I’ve had the opportunity to teach a course on I NTERMEDIATE DYNAMICS from the first edition ten times. Teaching the course from the book has allowed me to identify areas where improvements and corrections were needed. The improvements include additional examples and exercises, a discussion of geometry and geodesics, a more coherent discussion of constraint forces and constraint moments in rigid-body dynamics, clearer treatments of dual Euler basis vectors and coordinate singularities, and a discussion of the application of Euler angles to joint coordinate systems in biomechanics. The book also has a companion online resource: https://rotations.berkeley.edu/ This resource contains M ATLAB code for simulations and animations of the motions of rigid bodies. The website was constructed and maintained in partnership with Alyssa Novelia and Professor Daniel Kawano (Rose-Hulman Institute of Technology). Many of the additions to the book can be attributed to questions and constructive comments of undergraduate and graduate students who have taken my class on I NTER MEDIATE DYNAMICS at the University of California at Berkeley (UCB). I have also had the great fortune to collaborate with students and faculty on research topics that eventually made their way into the second edition of this book. These collaborators include: 1. Evan Hemingway (M.S. 2016) with whom I collaborated on establishing new perspectives on coordinate singularities and gimbal lock in mechanical systems [123].

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2. Professors Brian Feely and Jeffrey Lotz (University of California at San Francisco), Dr. Julia Nichols, and Dr. Mark Sena with whom I researched joint coordinate systems and applications of the dual Euler basis to anatomical joints [204, 217]. 3. Dr. Alyssa Novelia (M.S. 2015, Ph.D. 2018) with whom I collaborated on geodesics of SO(3) and motions of the eye [206, 207]. 4. My long-term collaborator, Professor Arun Srinivasa (Texas A&M University), on constraint forces and constraint moments [220]. 5. Drs. Nur Adila Faruk Senan (M.S. 2008, Ph.D. 2011) and Melodie Metzger with whom I worked on Cartesian stiffness matrices and quaternions [78, 185]. Material from the papers I have just cited all appear in this edition. I would like to take this opportunity to thank my collaborators and former students for their contributions and Daniel Kawano for taking the time and care to meticulously proofread an earlier draft of the book. The topic of gimbal lock is discussed in Section 11.15 and it is a great pleasure to acknowledge a helpful discussion with Professors Jeremy Kasdin and Rob Stengel (Princeton University) on the Apollo program and gimbal lock. I am also grateful to Steven Elliot at Cambridge University Press for inviting me to write a second edition and for coordinating an extensive set of reviews. These reviews served to improve the book and make it more accessible to a wider audience. I am pleased that students at many universities world wide, including UCB, will have free access to digital versions of the book from the publisher’s website. Although I hope that there are no typographical errors in this edition, my experiences have shown that such errors are inevitable. So please feel free to let me know by email at [email protected] should you find such errors or have questions about the book. An updated errata will be posted on my UCB homepage. All of the author royalties from the sales of this book will be donated to causes that support students at the University of California at Berkeley.

Image Credits

The portrait of William R. Hamilton in Figure 4.7 in Section 4.12 is from the Royal Irish Academy in Dublin, Ireland. I am grateful to Pauric Dempsey, the Head of Communications & Public Affairs of this institution, for providing the image. The images of Leonhard Euler shown in Figure 6.6, of Siméon Denis Poisson in Figures 7.19 and 10.7, and of Joseph-Louis Lagrange in Figure 10.1 are reproduced under license from Getty Images. Oliver M. O’Reilly Berkeley

Part I

A Single Particle

1

Kinematics of a Particle

1.1

Introduction

One of the main goals of this book is to enable the reader to consider a mechanical system, model it as a system of particles and rigid bodies, and then interpret the results of the model. For this to happen, the reader needs to be equipped with an array of tools and techniques, the cornerstone of which is to be able to precisely formulate the kinematics of a particle. Without this foundation firmly in place, the conclusions from the model either do not hold up or lack conviction. Much of the material presented in this chapter will be used repeatedly throughout the book. We start the chapter with a discussion of coordinate systems for a particle moving in a three-dimensional space. This naturally leads us to a discussion of curvilinear coordinate systems. These systems encompass all of the familiar coordinate systems and the material presented is useful in many other contexts. At the conclusion of our discussion of coordinate systems and their application to particle mechanics, you should be able to establish expressions for gradient and acceleration vectors in any coordinate system. The other major topics of this chapter pertain to constraints on the motions of particles. In earlier dynamics courses, these topics were intimately related to judicious choices of coordinate systems to solve particle problems. For such problems, a constraint is usually imposed on the position vector of a particle. Here, we also discuss time-varying constraints on the velocity vector of the particle. Along with curvilinear coordinates, the topic of constraints is one most readers will not have seen before, and for many it will hopefully constitute an interesting thread that winds its way through this book.

1.2

Reference Frames

To describe the kinematics of particles and rigid bodies, we presume the existence of a space with a set of three mutually perpendicular axes that meet at a common point P. The set of axes and the point P constitute a reference frame. In Newtonian mechanics, we also assume the existence of an inertial reference frame. In this frame, the point P moves at a constant speed. Depending on the application, it is often convenient to idealize the inertial reference frame. For example, for ballistics problems, the Earth’s rotation and the translation of

4

Kinematics of a Particle

its center are ignored and one assumes that a point, say E, on the Earth’s surface can be considered as fixed. The point E, along with three orthonormal vectors that are fixed to it (and the Earth), are then taken to approximate an inertial reference frame. This approximate inertial reference frame, however, is insufficient if we wish to explain the behavior of Foucault’s famous pendulum experiment. In this experiment from 1851, Léon Foucault (1819–1868) ingeniously demonstrated the rotation of the Earth by using the motion of a pendulum.1 To explain this experiment, it is sufficient to assume the existence of an inertial frame whose point P is at the fixed center of the rotating Earth and whose axes do not rotate with the Earth. As another example, when the motion of the Earth about the Sun is explained, it is standard to assume that the center S of the Sun is fixed and to choose P to be coincident with S. The point S is then used to construct an inertial reference frame. Other applications in celestial mechanics might need to consider the location of the point P for the inertial reference frame as the center of mass of the solar system with the three fixed mutually perpendicular axes defined by use of certain fixed stars [106]. For the purposes of this text, we assume the existence of a fixed point O and a set of three mutually perpendicular axes that meet at this point (see Figure 1.1). The set of axes is chosen to be the basis vectors for a Cartesian coordinate system. Clearly, the axes and the point O are an inertial reference frame. The space that this reference frame occupies is a three-dimensional space. Vectors can be defined in this space and an inner Path of the particle

A(t) m

v (t)

r(t)

r(t +

Δt )

O

The path of a particle moving in E 3. The position vector r, velocity vector v, and areal velocity vector A of this particle at time t and the position vector of the particle at time t + ±t are shown.

Figure 1.1

1 Discussions of his experiment and their interpretation can be found in [82, 180, 276]. Among his other

contributions, Foucault is also credited with introducing the term “gyroscope” in [85] (cf. [285]).

1.3

Kinematics of a Particle

5

product for these vectors is easy to construct with the dot product. As such, we refer to this space as a three-dimensional Euclidean space and we denote it by E 3.

1.3

Kinematics of a Particle

Suppose a single particle of mass m is in motion in E3 . The position vector of the particle relative to a fixed origin O is denoted by r (see Figure 1.1). In mechanics, this vector is usually considered to be a function of time t: r = r(t). The velocity v and acceleration a vectors of the particle are defined as the respective first and second time derivatives of the position vector: v=

dr , dt

a=

dv dt

=

d 2r . dt2

It is crucial to note that, because r is measured relative to a fixed origin, v and a are the absolute velocity and acceleration vectors. By definition, the velocity vector can be calculated from the following limit: v(t) = lim

±t → 0

r (t + ±t ) − r(t) . ±t

We also use an overdot to denote the time derivative: v = r˙ and a = r¨ . Supplementary to the aforementioned kinematical quantities, we have the linear momentum G of the particle: G = mv. Further, the angular momentum HO of the particle relative to O is HO

=

r × mv.

As we now show, this vector is related to the areal velocity vector A. As used in celestial mechanics, the magnitude of the areal velocity vector is the rate at which the position vector r of the particle sweeps out an area about the fixed point O (see, e.g., [199]). To establish an expression for this vector, we consider the position vector of the particle at times t and t + ±t. Then, the area of the parallelogram defined by these vectors is ±r(t) × r (t + ±t)± (see Figure 1.1). This is twice the area swept out by the particle during the interval ±t. Taking the limit of the vector r(t)×2r±(tt+±t) as ±t → 0 and using the fact that r(t) × r(t) = 0, we arrive at an expression for the areal velocity vector A (t) as follows: r(t) × r (t + ± t) 2±t ± ² r (t + ±t ) 1 r(t) × lim 2 ± t→ 0 ±t ± ² 1 r (t + ±t ) − r (t ) r(t) × lim . ± t→ 0 2 ±t

A (t ) = lim

±t→ 0

=

=

6

Kinematics of a Particle

That is,

1 r × v. (1.1) 2 The vector A plays an important role in several mechanics problems in which either the angular momentum HO is constant or a component of HO is constant. Several other examples of its use are discussed in the exercises at the end of this chapter. Finally, we recall the definition of the kinetic energy T of the particle: A=

1 mv · v. 2 The definitions of the kinematical quantities that have been introduced are independent of the coordinate system that is used for E3 . In solving most problems, it is crucial to have expressions for momenta and energies in terms of the chosen coordinate system. It is to this issue that we now turn. T

1.4

=

Cartesian, Cylindrical Polar, and Spherical Polar Coordinate Systems

Depending on the problem of interest, there are several suitable coordinate systems for E3 . The most commonly used systems are Cartesian coordinates {x = x1 , y = x2 , z = x3}, cylindrical polar coordinates {r, θ , z}, and spherical polar coordinates {R, φ , θ }. All of these coordinate systems can be considered as specific examples of a curvilinear coordinate system {q1 , q2 , q3 } for E3 , which we will discuss later on in this chapter. Cartesian Coordinate System

For the Cartesian coordinate system, a set of right-handed orthonormal vectors are defined: {E1, E2, E3}. Given any vector b in E3 , this vector has the representation b=

3 ³

bi Ei .

i=1

For the position vector r, we also have r=

3 ³

xi Ei ,

i=1

where {x1 , x2, x3} are the Cartesian coordinates of the particle. Because Ei are fixed in ˙ = 0. both magnitude and direction, their time derivatives are zero: E i Cylindrical Polar Coordinates

A cylindrical polar coordinate system {r, θ , z} can be defined by a Cartesian coordinate system as follows: r

´

=

x21 + x22 ,

θ =

tan

−1

±

x2 x1

²

,

z = x3,

1.4

where θ

∈

7

Cartesian, Cylindrical Polar, and Spherical Polar Coordinate Systems

[0, 2π ) and r x1

≥

0. Provided r

r cos(θ ),

=

²=

x2

0, we can invert these relations to show that =

r sin(θ ),

x3

=

z.

In other words, given (x1, x2 , x3 ), a unique (r, θ , z) exists provided (x1 , x2) Otherwise, when r = 0, the coordinate θ is not uniquely defined. Given a position vector r, we can write

²=

(0, 0).

r = x1E 1 + x2 E2 + x3 E3 =

r(cos( θ )E1 + sin(θ )E2) + zE3

=

rer + zE3 ,

where, as shown in Figure 1.2, er = cos(θ )E1 + sin(θ )E2 . It is convenient to define the set of unit vectors {er , eθ , ez }: er

=

cos(θ )E1 + sin(θ )E2 ,

eθ

cos(θ )E 2 − sin(θ )E1 ,

=

ez

=

E3.

We also notice that ˙er = θ˙eθ , whereas e˙ θ = −θ˙ er . We should also verify that {er , eθ , ez } is a right-handed orthonormal basis for E3 .2

Spherical Polar Coordinates

A spherical polar coordinate system {R, φ , θ } can be defined by a Cartesian coordinate system as follows: R

´ =

x21 + x22 + x23 ,

θ =

tan−1

±

x2 x1

⎛´

²

,

φ=

tan−1 ⎝

x21 + x22 x3

⎞ ⎠,

E3 r z

O

E2

eθ

r

E1

θ

er Figure 1.2

Cylindrical polar coordinates r, θ , and z. µ

¶

(

2 A basis p , p , p is right-handed if p · p 1 2 3 3 1

)

p2 > 0 and is orthonormal if the magnitude of each of the vectors pi is 1 and they are mutually perpendicular: p1 · p2 = 0, p2 · p3 = 0, and p1 · p3 = 0. ×

8

Kinematics of a Particle

where R ≥ 0, θ relations to find x1

=

∈

[0, 2π ), and φ

∈

R cos(θ ) sin(φ ),

(0, π ). Provided

x2

=

φ ²=

0 or π , we can invert these

R sin(θ ) sin(φ ),

x3 = R cos( φ).

Given a position vector r, we can now write r = x1 E1 + x2 E2 + x3E3 =

R sin(φ )(cos(θ )E1 + sin(θ )E2 ) + R cos( φ)E 3

=

ReR ,

where, as shown in Figure 1.3, eR = sin(φ ) cos(θ )E1 + sin(φ) sin( θ )E2 + cos(φ )E3. This vector is a unit vector that points along r. For future purposes, it is convenient to define the right-handed orthonormal set of vectors {eR , eφ , eθ }. Referring to Figure 1.4, we first rotate E1 and E2 through an angle

eR

r

E3

eφ R

φ

O

z

eθ

E2 r θ

E1

er Figure 1.3

The spherical polar coordinates R, φ, and θ . (a)

eθ

(b)

E2

E3 er

θ

θ

E1

eR

φ

φ

er eφ

The rotations used to define the basis vectors er , e θ , e R, and eφ : (a), a rotation about E3 through an angle θ ; and (b), a rotation about eθ through an angle φ.

Figure 1.4

1.5

θ φ

Curvilinear Coordinates

9

about E3. This rotation defines er and eθ . We then rotate er and E3 through an angle about eθ to define eR and eφ : ⎡

⎤

⎡

⎡

⎤

⎡

cos(θ ) er ⎣ eθ ⎦ = ⎣ − sin(θ ) 0 E3

⎤⎡

⎤

⎤⎡

⎤

E1 sin(θ ) 0 ⎦ ⎣ E2 ⎦ , cos(θ ) 0 E3 0 1

eR sin(φ) 0 ⎣ eφ ⎦ = ⎣ cos(φ ) 0 eθ 0 1

cos(φ ) er − sin(φ ) ⎦ ⎣ eθ ⎦ . 0 E3

After performing a matrix multiplication, we find that ⎤

⎡

⎡

cos(θ ) sin(φ ) sin(θ ) sin( φ) eR ⎣ eφ ⎦ = ⎣ cos(θ ) cos(φ ) sin(θ ) cos(φ ) − sin(θ ) cos( θ ) eθ

(1.2)

⎤

⎤⎡

E1 cos(φ ) ⎦ ⎣ E2 ⎦ . − sin(φ ) E3 0

An alternative perspective on the sets of rotations we have just discussed can be found in Exercise A.7 on page 500 and Exercise A.8 on page 501. As with the cylindrical polar coordinate system, the basis vectors we defined for the spherical polar coordinate system vary with the coordinates. Indeed, assuming that θ and φ are functions of time, a series of long calculations using (1.2) reveals that ⎡

⎤

⎡

e˙ R ⎣ e˙ φ ⎦ = ⎣ e˙ θ

⎤⎡

⎤

eR ˙ ˙ ⎦ ⎣ −φ 0 θ cos(φ ) eφ ⎦ . −θ˙ sin(φ ) −θ˙ cos(φ ) 0 eθ 0

˙ φ

θ˙ sin(φ)

(1.3)

These relations have an interesting form: notice that the matrix in (1.3) is skewsymmetric. We shall see numerous examples of this later on when we discuss rotations and their time derivatives. Our later discussion will enable us to easily verify (1.3).

1.5

Curvilinear Coordinates

The preceding examples of coordinate systems can be considered as specific examples of curvilinear coordinate systems. The development of the vector calculus associated with a curvilinear coordinate system will be the focal point of this section of the book. Curvilinear coordinate systems have featured prominently in all areas of mechanics, and the material presented here has a wide range of applications. Most of our discussion is based on classic works and can be found in various textbooks on tensor calculus. Of these books, the one closest in spirit (and notation) to our treatment here is [265]; [182, 268] are also recommended. Consider a curvilinear coordinate system {q1, q2 , q3} that is defined by the functions q1

= q ˆ ( x1 , x2 , x3) ,

1

q2

= q ˆ ( x1 , x2 , x3) ,

q3

= q ˆ ( x1 , x2 , x3) .

2

3

(1.4)

10

Kinematics of a Particle

q

3

coordinate curve

q

a1

2

coordinate curve

a3

S

a2

q

1

coordinate surface

O

An example of a q1 coordinate surface S . At a point on this surface, a1 is normal to the surface, and a2 and a3 are tangent to the surface. The q1 coordinate surface S is foliated by curves of constant q2 and q3.

Figure 1.5

We assume that the functions qˆ i are locally invertible: ·

¸

·

¸

·

¸

x1 = xˆ 1 q1 , q2, q3 , x2 = xˆ 2 q1 , q2, q3 , x3 = xˆ 3 q1 , q2, q3 .

(1.5)

This invertibility implies that, given the curvilinear coordinates of any point in E3 , there is a unique set of Cartesian coordinates for this point and vice versa. Usually, the invertibility breaks down at several points in E3 . For instance, the cylindrical polar and spherical polar coordinate θ is not uniquely defined when x21 + x22 = 0. This set of points corresponds to the x3 axis. Assuming invertibility, and fixing the value of one of the curvilinear coordinates, q1 say, to equal q10 , we can determine the values of x1 , x2 , and x3 such that the equation q10

1

= q ˆ ( x1,

x2, x3 )

is satisfied. The union of all the points represented by these Cartesian coordinates defines a surface that is known as the q1 coordinate surface (cf. Figure 1.5). If we move on this surface we find that the coordinates q2 and q3 will vary. Indeed, the curves on the q1 coordinate surface that we find by varying q2 while keeping q3 fixed are known as q2 coordinate curves. More generally, the surface corresponding to a constant value of a coordinate qj is known as a qj coordinate surface. Similarly, the curve we obtain by varying the coordinate qk while fixing the remaining two curvilinear coordinates is known as a qk coordinate curve.

1.5

Curvilinear Coordinates

11

Covariant Basis Vectors

Again assuming invertibility, we can express the position vector r of any point as a function of the curvilinear coordinates: r=

3 ³

ˆxi

·

¸

q1, q2 , q3 Ei .

i =1

It is also convenient to define the covariant basis vectors a1 , a2 , and a3: ai

∂r ∂ qi

=

=

3 ³ ∂x ˆk k=1

Ek .

∂ qi

Mathematically, when we take the derivative with respect to q2 we fix q1 and q3 ; consequently, a2 points in the direction of increasing q2. As a result, a2 is tangent to a q2 coordinate curve. In general, ai is tangent to a qi coordinate curve. The matrix [aik ], where aik

=

ai · ak ,

is known as the metric tensor. You should notice that we can express the relationship between the covariant basis vectors and the Cartesian basis vectors in a matrix form: ⎡

⎤

⎡

∂ xˆ1

∂ q1

a1 ⎢ ⎣ a2 ⎦ = ⎢ ⎢ ⎣ a3

∂ xˆ1

∂ q2 ∂ xˆ1

∂ q3

¹

∂ xˆ 2

∂ ˆx3

∂ q1

∂ q1

∂ ˆx3

∂ xˆ 2

∂ q2

∂ q2

∂ xˆ 2

∂ ˆx3

∂ q3

∂ q3

º»

⎤

⎡ ⎤ E1 ⎥ ⎥⎣ E2 ⎦ . ⎥ ⎦

E3

¼

=A

It is a good exercise to write out the matrix A in the preceding equation for various examples of curvilinear coordinate systems, for instance, cylindrical polar coordinates.

Contravariant Basis Vectors

Curvilinear coordinate systems also have a second set of associated basis vectors: 1 2 3 { a , a , a }. These vectors are known as the contravariant basis vectors and they are defined as the gradients of the curvilinear coordinates: ak

= ∇q

k

.

(1.6)

Using the representation for the gradient in Cartesian coordinates, we can express the definition (1.6) as follows: a

1

=

3 1 ³ ∂q ˆ i =1

∂ xi

Ei ,

2

a

=

3 2 ³ ∂q ˆ i=1

∂ xi

Ei ,

a

3

=

3 3 ³ ∂q ˆ i=1

∂ xi

Ei .

12

Kinematics of a Particle

Using a matrix representation, the definition of the contravariant basis vectors can be expressed as ⎡

⎡

⎤

a1

⎣ a2 ⎦ =

a3

⎢ ⎢ ⎣ ¹

1 ∂q ˆ

1 ∂q ˆ

1 ∂q ˆ

∂ x1 2 ∂q ˆ

∂x2 2 ∂q ˆ

∂ x3 2 ∂q ˆ

∂ x1

∂x2

∂ x3

∂ x1 3 ∂q ˆ

∂x2 3 ∂q ˆ

∂ x3 3 ∂q ˆ

º»

⎤

⎡ ⎤ E1 ⎥ ⎥ ⎣ E2 ⎦ . ⎦

E3

¼

=B

Geometrically, ai is normal to a qi coordinate surface. However, as in the case of the covariant basis vectors, the contravariant basis vectors are not necessarily unit vectors, nor do they½ form an orthonormal basis for E3 . We leave it as an exercise to show that ¾ the matrix aik , where aik

=

ai · ak ,

is the inverse of the metric tensor [aik ].

Representations of the Gradient of a Function

We can use the definitions of the contravariant basis vectors to define a useful ( ) representation for the gradient of a function f = f q1, q2 , q3 : ∇f =

∂f ∂f ∂f 1 a + 2 a 2 + 3 a 3. 1 ∂q ∂q ∂q

(1.7)

This result will allow us to quickly write down the representation of the gradient for spherical polar and cylindrical polar coordinate systems.

Relationships Between the Covariant Basis Vectors and the Contravariant Basis Vectors

It is interesting to examine the relationships between the covariant and contravariant basis vectors. Using the chain rule of calculus, we can show that3 ai · ak

i

= δk ,

where δki is the Kronecker delta. As discussed in the Appendix, δ ik = 1 if i = k and 0 otherwise. The orthogonality result can be established in a straightforward manner using the chain rule: ai · ak

i

= ∇q ·

=

∂r ∂ qk

⎛ ⎞ ± 3 i ³ ∂ qˆ ∂ ˆx1 ⎝ Ej ⎠ · E1 j=1

∂ xj

∂ qk

+

∂x ˆ2 ∂x ˆ3 E2 + k E3 ∂ qk ∂q

3 Using the matrices A and B defined previously, the nine identities ai · a k equation BA T = I where I is the identity matrix.

²

i are equivalent to the matrix

= δk

1.5

=

3 ³ 3 ³ ∂ qˆ i ∂ ˆxs j=1 s =1

∂ xj ∂ qk

13

Curvilinear Coordinates

Ej · Es

3 ³ 3 i ³ ∂ qˆ ∂ ˆxs s

δj ∂ xj ∂ qk j=1 s =1 i i i ∂q ˆ ∂x ∂ qˆ ∂ x ∂q ˆ ∂x ˆ1 ˆ2 ˆ3 = + + k k ∂ x1 ∂ q ∂ x2 ∂ q ∂ x3 ∂ qk ∂ qi =

=

∂ qk i = δk ,

which is the result we sought to establish. Given ak , the relations ai · ak = δ ki can be considered as nine scalar equations to determine ai . Following [103, p. 20], the solutions to these equations can be expressed in the compact form4 a1 =

a2 × a3 , a 1 · ( a 2 × a 3)

a2

a3 × a1 , a2 · (a3 × a1 )

=

Similarly, given ak , the relations ai · ak to determine ai :5 a1

=

a2 × a3 ( ), a1 · a2 × a3

i

= δk

a1 × a2 . a 3 · ( a 1 × a 2)

(1.8)

can be considered as nine scalar equations

a3 × a1 ( ), a1 · a2 × a3

a2 =

a3 =

a3

=

a1 × a2 ( ). a1 · a2 × a3

(1.9)

Vector identities can be used to show that −1

(a1 · (a2 × a 3 ))

·

=

¸

a1 · a2 × a3 .

We leave establishing this result as an exercise.

Covariant and Contravariant Components

As {a1, a2 , a3} and {a1, a2 , a3} form bases for E3 , any vector b can be described as linear combinations of either sets of vectors: b=

3 ³ i=1

bi ai

=

3 ³

bk ak .

k=1

4 To see how these solutions were established, consider a1 . As a1 ⊥ a and a1 ⊥ a , we conclude that 2 3 a1 = α a2 × a3 where α still needs to be determined. However, a1 · a1 = 1, and so we conclude that α (a1 · (a 2 × a 3)) = 1. The representation (1.8)1 for a1 now follows. 5 These relations are particularly useful in situations where it is nontrivial to invert the relations

·

¸

qˆ k (x 1 , x2 , x3 ) to determine xˆ j q1 , q2 , q3 .

14

Kinematics of a Particle

The components bi are known as the contravariant components and the components bk are known as the covariant components: ¿

b · ai b·a

i

3 ³

=

=

k =1 ¿ 3 ³

À

bk a

k

·

3 ³

ai

=

À k

b ak

·

ai

=

k =1

k =1 3 ³ k =1

bk δ ik

=

bi ,

bk δ ki

=

bi .

It is very important to note that bk ² = b · ak in general, because ai · ak is not necessarily equal to δki . Additional relationships between the covariant bk and contravariant bj components of a vector are presented in Exercises 1.8 and 1.9 and Figure 1.8 below. The trivial case in which xi = qi deserves particular mention. For this case, ∑3 i r = k=1 xi Ei . Consequently, a i = Ei . In addition, a = Ei , and the covariant and contravariant basis vectors are equal. Some Comments on Derivatives

(

)

Several partial derivatives of functions ² q1, q2 , q3, q˙ 1 , q˙ 2, q˙ 3 , t play a prominent role in this book. When taking the partial derivative of this function with respect to q2, say, we assume that t, q1, q3 , and q˙ k are constant. A related remark holds for the partial derivatives with respect to the velocities q˙ j and time t. That is, ∂q

k

∂ qj

and ∂q

∂q ˙

k

= δj ,

k

0,

=

∂ q˙ j

k

∂ q˙

k

k

= δj

∂q ˙j

∂t

0,

=

∂ qj

=

0,

(1.10)

∂t = ∂ q˙ j

0.

(1.11)

∂ qj

,

In all these equations, j and k range from 1 to 3. You may have noticed that (1.10)1 was used in our calculations of ai . It is easy to be confused about the distinction between the derivative dtd and the derivative ∂∂t . The former derivative assumes that qi and q˙ k are functions of time, whereas the latter assumes that they are constant: ˙ = ²

d² dt

=

3 ³ ∂ ² dqi i=1

∂ qi

dt

+

3 ³ ∂ ² d 2 qk k=1

∂q ˙k

dt2

+

∂² ∂t

.

For example, consider the function ·

²=

q1 + q˙ 3

¸2

+

10t.

Then, ∂² = ∂ q1

1,

∂² ∂ q˙ 3

=

2q˙ 3,

∂² ∂t

=

˙ ²= It should be clear from this example that ²

10,

∂² ∂t

.

1 ˙ = q ² ˙ +

2q˙ 3q¨ 3 + 10.

1.6

1.6

15

Examples of Curvilinear Coordinate Systems

Examples of Curvilinear Coordinate Systems

Following our discussion of curvilinear coordinates, we now revisit the three coordinate systems discussed at the start of this chapter and establish expressions for the covariant and contravariant basis vectors associated with these systems. We then turn to several examples of coordinate systems where the covariant basis vectors are not orthogonal. The reader is also referred to Exercises 1.5, 1.6, and 1.7, where three additional coordinate systems are discussed.

Cartesian, Cylindrical Polar, and Spherical Polar Coordinate Systems Revisited

The three examples of curvilinear coordinate systems we have discussed previously have the unusual property that their coordinate surfaces and coordinate curves intersect at right angles. As a result, the associated covariant and contravariant basis vectors for these systems are orthogonal. We leave computing the vectors as an exercise and summarize our results. ( ) For the Cartesian coordinate system q1 = x, q2 = y, q3 = z : r = xE1 + yE2 + zE3 , a1

=

a

1

=

a2 = a2

E1,

=

E2 ,

a3

=

a 3 = E 3.

The coordinate surfaces for this coordinate system are planes and the coordinate curves are straight lines. ( ) The cylindrical polar coordinate system q1 = r, q2 = θ , q3 = z is one of the simplest examples where the magnitudes of ai are not necessarily unity: r = r cos (θ ) E1 + r sin (θ ) E2 + zE3 , 1 a1 = a1 = er , a2 = reθ , a2 = eθ , a3 = a3 = E3 . r The respective coordinate surfaces for this system are cylinders, half planes, and horizontal planes. Further, the coordinate curves are rays, circles, and straight lines, respectively. With the help of the representation (1.7), we note that the gradient of a function f = f (r, θ , z) is ∇f =

∂f ∂r

er +

1 ∂f eθ r ∂θ

+

(

∂f ∂z

E 3.

Finally, for the spherical polar coordinate system q1 = R, q2

= φ,

q3

= θ

)

:

r = R sin (φ) cos (θ ) E1 + R sin (φ) sin (θ ) E2 + R cos (φ) E3, a1 = a1

=

eR,

a2

a3 = R sin (φ) eθ ,

=

Reφ , a3

=

a2

=

1 eφ , R

1 eθ . R sin (φ)

The coordinate surfaces and coordinate curves for the spherical polar coordinate system are shown in Figure 1.6. We leave it as an exercise to describe the geometry of the

16

Kinematics of a Particle

(a)

(b) φ

φ

c.c.

(c)

c.c. R

R

c.c.

c.c.

θ

c.c.

c.c.

θ

O

O

R

c.s.

θ

c.s.

φ

O

c.s.

Coordinate surfaces (c.s.) for the spherical polar · curves (c.c.) and coordinate ¸ coordinate system q1 = R, q2 = φ, q3 = θ : (a) an R coordinate surface is foliated by curves of longitude (φ coordinate curves) and circles (θ coordinate curves); (b) a θ coordinate surface is a half plane which is foliated by semicircles (φ coordinate curves) and rays (R coordinate curves); and (c) a φ coordinate surface is a cone which is foliated by circles (θ coordinate curves) and rays (R coordinate curves).

Figure 1.6

surfaces and curves. Again, with the help of the representation (1.7), we note that the gradient of a function f = f (R, φ , θ ) is ∇f =

∂f ∂R

eR +

1 ∂f eφ R ∂φ

+

1 ∂f eθ . R sin (φ) ∂ θ

This representation is very useful when establishing representations for contravariant basis vectors for the parabolic coordinate system in Exercise 1.5.

An Example of a Nonorthogonal Coordinate System

It is instructive to consider a simple example of a coordinate system having nonorthogonal basis vectors: q1

=

x,

q2

=

q3

y − α x,

=

z.

(1.12)

Here, α is a constant, and x = x1, y = x2, and z = x3 are Cartesian coordinates. Representative images of the q1 and q2 coordinate curves are shown in Figure 1.7. The relations (1.12) are invertible provided α is finite: x = q1 ,

y = q2 + αq1 ,

z = q 3.

The covariant basis vectors can be found from the following representation for the position vector: ·

¸

r = q1E1 + q2 + αq1 E2 + q3E3 . That is, a1

=

∂r = ∂ q1

E 1 + αE2 ,

a2

=

∂r = ∂ q2

E2 ,

a3

=

∂r = ∂ q3

E3.

1.6

a2

y

a

2

17

Examples of Curvilinear Coordinate Systems

a2

a1

= E2

a1

a2

a1 = E1

a1 x

q

q

2

1

coordinate curve

coordinate curve

Representative examples of the q1 = x and q2 = y − αx coordinate curves for coordinate system (1.12) on a q3 = z coordinate surface. For the results shown in this figure, α > 0.

Figure 1.7

The easiest method to compute the contravariant basis vectors is to use the definition ak = ∇ qk : a1

= ∇q

1

=

a2 = ∇ q2

E 1,

= −α E1 +

a3

E2 ,

3

= ∇q

=

E3 .

We leave it as an exercise to verify that ai · ak = δik . Representative examples of the basis vectors are shown in Figure 1.7. It should be clear from this figure that a1 · a2 ² = 0 and a1 · a2 ² = 0. Thus, the covariant and contravariant bases are nonorthogonal. An example of the covariant bi and the contravariant bk components of a vector b are shown in Figure 1.8. It should be noted that because a1 and a2 are not unit vectors, the Á Á 2 Á projection of b onto these vectors is not b1 and b , respectively: ±a2 ± = a1Á = 1 and Á Á √ 2 1 + α 2. ± a1 ± = Áa Á =

A Second Example of a Nonorthogonal Coordinate System

A more complex nonorthogonal coordinate system is the following example that could be used to study the motion of a particle on a parabolic surface or the optical behavior of a parabolic mirror: q1

=

q2

y,

=

x − β y2 ,

q3

=

z.

(1.13)

Here, x = x1 , y = x2 , and z = x3 are Cartesian coordinates and β is a constant which has dimensions of m−1. We set β = 1 in the sequel. Representative projections of the coordinate surfaces for q1 and q2 are shown in Figures 1.9 and 1.10. For this coordinate system, it is straightforward to invert (1.13) to see that x = q2 + ( )2 1 q and y = q1 . Thus, ±

r= q

2

· +

q

1

¸2²

E1 + q1 E2 + q3E3 .

Kinematics of a Particle

2

b

(a)

b · a1

a2

a1

a2

2

a

2b

/

2

a

=

·b

1

=

b

b

)2 a · b( =

18

a1 (b 1

(b

=

·a

) 1)

/

a1

(b) b

b

2

1

a1

a

a2

b

a2

a 2 = E2 2

b2

a1 = E1 + αE2

= E2 − αE1

a1 = E1

b1

a1

Components of a vector b for the coordinate system (1.12): (a) covariant and contravariant components of a vector b and their relation to the perpendicular projections of a vector on the basis vectors; (b) decomposition of a vector b into its covariant and contravariant representations. For the example shown in this figure, b · E 3 = 0.

Figure 1.8

y q

1

2 q

q

1

1

1

2

= 2

2

= 4

= 1

=

−1

−4

6

= −3

Projections of the q1 (1.13) on the x–y plane.

Figure 1.9

= −4

q

x

q q

2

q

= 3

−4 =

y and q2

=

x − y2 coordinate surfaces for coordinate system

1.7

19

Representations of Particle Kinematics

a1 y

q

1

>

a1

0

q

1

c.c.

a2 x

a2

a

1

a1 q

1




0,

a22 = a2 · a2

a11a22 − a12 a12 det (M) = m

3

>

0,

= ± a1 ×

a2 ±2

a33 >

0,

2 (a1 · (a2 × a 3 )) >

0.

=

a3 · a3 > 0,

Thus, because the components of M are formed by the inner products of a set of vectors, it suffices to check that (1.16) holds or, equivalently, the determinant of M is positive in order to verify that M is positive definite. We have just established that {a1 , a2 , a3 } forming a basis for E3 is equivalent to the mass matrix M being positive definite. Positive definiteness of M is also equivalent to stating that T is a positive definite function of the velocities q˙ k . In the sequel, we shall use the kinetic energy of the system to characterize coordinate singularities.7 Specifically, if det (M) > 0, then {a1 , a2, a3 } spans E 3 and the coordinate system is free of singularities. In contrast, if det (M) = 0 at a specific point, then we say that the coordinate system µ 1 2 3¶ q , q , q has a singularity at that point. Examples

For example, for a Cartesian coordinate system det (M) = m and M is positive definite. This system is devoid of singularities. In contrast, for the cylindrical polar coordinate system det (M) = mr2 and this system has singularities when r = 0 (or the entire x3 = 0 axis). As det (M) = mR4 sin2 (φ), the same conclusion can be established for the spherical polar coordinate system. To see why coordinate singularities present difficulties, consider a particle moving on a plane as shown in Figure 1.12. If we use Cartesian coordinates to describe the motion of the particle, then x˙ = v0 cos (θ0 ) and y˙ = v0 sin (θ0 ). However, if we use cylindrical polar coordinates to describe the motion of the particle, then r˙ switches from −v0 to v0 as the particle passes through the origin and θ switches from θ0 + π to θ0 as the particle passes through the origin. In other words, θ˙ is not defined at this instant and neither is r¨.

1.9

Constraints

A constraint is a kinematical restriction on the motion of the particle. Constraints are introduced in problems involving a particle in three manners: either as simplifying assumptions, prescribed motions, or because of rigid connections. The constraints on 7 Our treatment of this topic is based on the recent paper [123].

1.9

Constraints

23

E2 ˙ = v0 , θ = θ0

r θ0 θ0

+π

E1

O

˙ = −v0 , θ = θ0 + π

r

The path of a particle moving in a straight line at a constant speed v0. At the instant that the particle passes through O, θ˙ and ¨r are undefined.

Figure 1.12

the motion of a particle dictate, to a large extent, the coordinate system used to solve the problem of determining the motion of the particle. In this section, we examine the simplest class of constraints on the motion of a particle. Later, these constraints will be classified as integrable.

Classic Examples

Consider the four mechanical systems shown in Figure 1.13. The first system is known as the spherical pendulum. Here, a particle of mass m is attached by a rigid rod of length ³0 to a fixed point O. The constraint on the motion of the particle in this system can be written in the equivalent forms r · eR

= ³0 ,

± r± = ³0.

By differentiating either of these representations, we see that the velocity vector satisfies the relation v · eR = 0. The second system we consider is the planar pendulum. Again, the particle is attached by a rigid rod of length ³0 to a fixed point O, but it is also assumed to move on a vertical plane. The constraints on the motion of the particle can be written as follows: r · er

= ³0,

r · E3

=

0.

After differentiating these equations with respect to time, we observe that the velocity vector of the particle has a component only in the eθ direction: v · er = 0 and v · E3 = 0. The third system involves a particle moving on a horizontal surface that is moving with a velocity vector f˙(t)E3 . The constraint on the motion of the particle is r · E3

=

f (t).

24

Kinematics of a Particle

(a)

(b)

E3

E2

m

0

E2

O

O

E1

0

E1 m

(c)

˙E 3

f

Moving plane

E3

(d) m

E3

E2 O

E1

m

α

Spinning conical surface

O

Four mechanical systems featuring constraints on the motion of a particle: (a) the spherical pendulum; (b) the planar pendulum; (c) a particle in motion on a plane that is being given a vertical displacement f (t); and (d) a particle moving on a spinning cone.

Figure 1.13

The final system of interest consists of a particle moving on a spinning cone. The constraint on the motion of the particle can most easily be described with the help of a spherical polar coordinate system: φ + α(t) −

π

2

=

0.

For all four systems, notice that we have selected a coordinate system in which the constraint(s) on the motion of the particle is(are) easily described.

A Particle Moving on a Surface

Turning to the more general case, consider a particle constrained to move on a surface. With the help of a single smooth function ´ (r, t ), we assume that the constraint can be described in a standard (canonical) form: ´ (r, t) =

0.

At each instant in time, this equation can be interpreted as a single condition on the three independent Cartesian coordinates of the particle. Thus, the condition ´ = 0 defines a two-dimensional surface (see Figure 1.14). The unit normal vector n to this surface is parallel to the gradient ∇ ´ = ∂∂´r (see Figure 1.14). Observe that we have introduced an alternative notation for ∇ ´ that is often

1.9

25

Constraints

∇Ψ

m

r Surface Ψ = 0

O

Figure 1.14

constraint.

A particle moving on a surface ´

=

0. The particle in this case is subject to a single

more illuminating in particle mechanics. Depending on the coordinate system used, the vector ∇´ has numerous representations: ∇´ =

∂´ ∂r

=

3 ³ ∂´ i =1

=

=

Ei

3 ³ ∂´ i a i =1

=

∂ xi

∂ qi

1 ∂´ ∂´ eθ + E3 ∂r r ∂θ ∂z 1 ∂´ 1 ∂´ ∂´ eR + eφ + eθ . ∂R R ∂φ R sin(φ ) ∂ θ

∂´

er +

(1.17)

You should notice how simple the expression for the gradient is in curvilinear coordinates.8 A simple differentiation of the function ´ helps to provide the restriction it imposes on the velocity vector: ∂´

˙ = ´

∂r

·

v+

∂´ ∂t

.

˙ = 0. Consequently, the However, if r satisfies the constraint, then ´(r, t) = 0 and ´ constraint ´ = 0 implies that the velocity vector satisfies the restriction

∂´ ∂r 8 With the help of the representation v

·v+

∂´ ∂t

=q ˙1 a 1 + q ˙ 2 a2 + q ˙ 3 a3 ,

·

˙ q1 , q2 , q3 (or, equivalently, (1.7)) by first noting that ´

·

¸

∑

=

¸

0. we can establish the representation (1.17) 2 v. Taking the time derivative of ´ , we

= ∇´ ·

3 ∂´ i ˙ ˙ q1 , q2 , q 3 = find that ´ i=1 ∂q i q˙ . Substituting for v and comparing both expressions for ´ , we arrive at the representation (1.17)2 .

26

Kinematics of a Particle

That is, the component of the velocity vector of the particle that is normal to the surface is prescribed by the constraint. This result will be important in our discussion of the mechanical power of the constraint forces.

A Particle Moving on a Curve

We now consider the more complex case of a particle moving on a curve. A curve can be defined by the intersection of two surfaces. Using the previous developments, we consider the condition that the particle moves on the curve to be equivalent to two (simultaneous) constraints: ´1 (r, t) =

0,

´2 (r, t) =

0.

This situation is shown in Figure 1.15. The normal vectors to the two surfaces at a point of their intersection are assumed not to be parallel: ∇ ´1 × ∇ ´2 ² = 0. That is, the two constraints ´1 = 0 and ´2 = 0 are assumed to be independent. Once ´1 and ´2 are given, then expressions for the two normal vectors to the curve can readily be established. We also note that deriving the restrictions these constraints impose on the velocity vector follows from the corresponding results for a single constraint: ∂ ´1 ∂r

·v+

∂ ´1 ∂t

=

∂ ´2

0,

∂r

·v+

∂ ´2 ∂t

=

0.

(1.18)

If the curve is fixed, then ´1 and ´2 are not explicit functions of time. In this case, (1.18) can be used to show the expected result that v is tangent to the fixed curve formed by the intersection of the two surfaces.

Surface Ψ2 = 0

∇ Ψ1

t

m

∇Ψ2 r Surface Ψ1 = 0

O

A particle subject to two constraints. The particle is free to move on the curve formed by the intersection of the surfaces ´1 = 0 and ´2 = 0. The vector t is the unit tangent vector to this curve. Figure 1.15

1.9

Constraints

27

A Particle Whose Motion is Prescribed

The case in which the motion is prescribed can be interpreted as a particle lying at the intersection of three known surfaces. In other words, the particle is subject to three constraints: ´1(r, t) =

0,

´2(r, t) =

0,

´3 (r, t) =

0.

We assume that these constraints are independent and thus their normal vectors at their intersection point form a basis for E3: ∇ ´ 3 · ( ∇ ´1 × ∇ ´ 2) ² =

0.

The three conditions ´i = 0 can also be interpreted as three equations for the three components of r. The two primary situations in which a particle is subject to three constraints arise when either the motion of the particle is completely controlled or the particle is subject to static friction and is therefore in a state of rest relative to a curve or surface. Coordinates and Constraints

The constraints we have considered on the motion of the particle have been described in terms of surfaces that the motion of the particle is restricted to. These surfaces can be described in terms of the coordinate system used for E3 . The description is greatly facilitated by a judicious choice of coordinates. For instance, if a particle is constrained to move on a fixed plane, then we can always choose the origin O and the Cartesian coordinates such that the constraint is easily described by the equation x3 = constant. Similarly, if a particle is constrained to move on a sphere, then spherical polar coordinates are an obvious choice. The more complex the surfaces that the particle is constrained to move on, the more difficult it becomes to choose an appropriate coordinate system. Help is at hand: the surfaces ´ (r) = 0 of interest in this book can be described in an appropriate curvilinear coordinate system by a simple equation, q3 = constant. Furthermore, a moving surface 3 ´ (r, t) = 0 can, in principle, be described by the equation q = f (t), where f is a function of time t. For example, suppose a particle is moving on a sphere whose radius is a known function R0 (t). Then the constraint that the particle moves on the sphere is simply described by R = R 0(t). Here, we are choosing the spherical polar coordinate system to be our coordinate system. The Classic Examples Revisited

Returning to the four mechanical systems shown in Figure 1.13, you should convince yourself that the constraint(s) on the motions of the particle in these systems are individually of the form ´ = 0. Specifically, for the spherical pendulum: ´ =

r · eR − ³0

=

R − ³0 ,

∇´ =

eR .

28

Kinematics of a Particle

That is, we may imagine the particle in a spherical pendulum as moving on a sphere. For the planar pendulum, we have ´1 =

r · er − ³0 = r − ³ 0,

´2 =

r · E3

=

z,

∇ ´1 = ∇ ´2 =

er ,

E3.

In this case, the particle can be visualized as moving on the intersection of a cylinder of radius ³0 and a horizontal plane. This intersection defines a circle. If the rod’s length ³0 changes with time, then the circle’s radius also changes. For the particle moving on the horizontal surface: ´ =

r · E3 − f (t) = x3 − f (t),

∇´ =

E3.

Notice that in this example ´ = ´ (r, t). For the final system, the particle moving on a spinning cone, the constraint on the motion of the particle can be represented by ´ = 0, where 1 eφ . 2 R If it were possible to vary the angle α = α(t), then the function example, suppose α = α0 + A sin(ωt), then π

´ (r) = φ + α −

´ (r, t) = φ −

π

2

+ α0 +

,

∇´ =

A sin(ωt),

∇´ =

´ = ´ (r, t).

For

1 eφ . R

˙ = 0, but ∂ ´ = Aω cos(ω t). The spinning motion of the cone You should verify that ´ ∂t has no effect on ´, but it must be considered in any formulation of the friction forces acting on the particle. Further, in the event that the particle is stuck to the cone, then the particle will be subject to three constraints. This situation is discussed in Section 2.9.

1.10

Classification of Constraints

All of the constraints discussed so far can be written individually in the form ´(r, t) =

0.

Thus they are often known as positional constraints. We now define a further type of constraint: π =

0,

(1.19)

where π =

f · v + e,

and f = f(r, t) and e = e(r, t). The constraint π = 0 does not restrict the position of the particle – it only restricts its velocity vector. For a given instant in time and a given location of the particle, the constraint π = 0 determines the velocity of the particle in the direction of f. However, the components of v that are perpendicular to f are not restricted. Consequently, the constraint π = 0 is often known as a velocity constraint.

1.10

As we demonstrated earlier, every constraint of the form differentiated to yield a restriction on the velocity vector: ∂´ ∂r

·v+

∂´ ∂t

29

Classification of Constraints

=

´(r, t)

=

0 can be

0.

This restriction is of the form (1.19). Thus every positional constraint ´(r, t) = 0 provides a velocity constraint f · v + e = 0. However, the converse is not true. A constraint π = 0 that can be integrated to yield a constraint of the form ´ (r, t) = 0 is said to be an integrable (or holonomic) constraint. More precisely, given a constraint π = 0, if we can find an integrating factor k = k (r, t) and a function ´(r, t) such that 9 k ( f · v + e) = ∇ ´ · v +

∂´ ∂t

,

then the constraint π = 0 is said to be integrable. Otherwise, the constraint π = 0 is said to be nonintegrable (or nonholonomic). In the sequel, a velocity constraint π = 0 is said to be nonintegrable unless we can prove otherwise. The terminology for constraints dates to Heinrich Hertz [126] (1857–1894). As noted by Lanczos [163], integrable constraints were further classified by Ludwig Boltzmann (1844–1906) as rheonomic when ´ = ´(r, t) and scleronomic when ´ = ´(r) (i.e., when ´ is not an explicit function of time t). The distinction between integrable and nonintegrable constraints becomes particularly important when rigid bodies are concerned. However, for pedagogical purposes, it is desirable to introduce them when discussing single particles. Shortly, we shall discuss the forces needed to enforce the constraints: such forces are known as constraint forces. To explore the differences between integrable and nonintegrable constraints, it is best to first consider some examples. Following such an exploration, we shall discuss known criteria to determine whether or not a set of constraints is integrable. Three Examples

As a first example, we suppose that the particle is subject to the constraints xy − c = 0,

z = 0.

That is, the particle is constrained to move on a hyperbola in the x–y plane (see Figure 1.16). Two points A and B are also shown in this figure, and it is important to notice that it is not possible for the particle to move between A and B without violating the constraint xy − c = 0. The constraints xy − c = 0 and z = 0 imply the velocity constraints (xE2 + yE1 ) · v =

0,

E3 · v = 0.

(1.20)

These conditions imply that v has no component normal to the hyperbola xy = c. Constraints (1.20) are both clearly of the form f · v + e = 0, where e = 0 and f = xE2 + yE1 , and e = 0 and f = E3 , respectively. By construction, constraints (1.20) are both integrable. 9 Further background on integrating factors can be found in most texts on differential equations or

differential forms (see, e.g., [81, 84, 152]). It is well known that integrating factors are not unique.

30

Kinematics of a Particle

y

= x2

B

x

= x1

A

The motion of a particle subject to the constraints xy = c and z = 0 where c is a positive constant. The arrows shown on the hyperbolae indicate the possible directions of motion of the particle.

Figure 1.16

y

= x2 B

x

= x1

A

The motion of a particle subject to the constraints ˙yx = 0 and z = 0. The arrows indicate the possible directions of motion of the particle. Observe that there are an infinite number of ways to move from A to B without violating the constraints.

Figure 1.17

As a second example, let us examine the following constraints: (xE2) ·

v = 0,

z

=

0.

(1.21)

The motions of the particle that satisfy these constraints are shown in Figure 1.17. Notice that it is possible to move between any two points A and B on the x–y plane without

1.10

Classification of Constraints

31

violating the constraint ˙yx = 0. The restriction this constraint places is that it restricts how one can go from any A to any B. This is in marked contrast to the constraint xy − c = 0. By multiplying y˙ x = 0 by x1 , for example, we see that, away from the y axis, this constraint is integrable.10 Considering the possible motions shown in Figure 1.17, it is not surprising to note that we cannot find a smooth function ´ to conclude that the constraint ˙yx = 0 is integrable throughout the entirety of E3 . Instead, we classify the constraint y˙ x = 0 as a piecewise-integrable constraint.11 We shall discuss further unusual aspects of this constraint in Section 2.10. Our third example is the simplest possible nonintegrable constraint on the motion of a particle.12 The constraint is (−zE1 + E2 ) · v =

0.

(1.22)

That is, y˙ − zx˙ = 0. To demonstrate the type of restrictions y˙ − zx˙ = 0 imposes, we choose two points A and B and use the x coordinate to parameterize a path between them. Choosing df y = f (x), z= , dx where f (x) is any sufficiently smooth function, we observe that the constraint −zx˙ + y˙ = 0 is satisfied. In order that the particle be able to move between any two points A and B, f (x) is subject to the restrictions yA

=

f (xA ) ,

zA

=

df (xA ) , dx

yB

=

f (xB ) ,

zB

=

df ( xB ) , dx

where rA = xA E1 + yA E2 + zA E3 and rB = xB E1 + yB E2 + zB E3 . Graphically constructing a function f (x) that meets these restrictions is not difficult, and some examples are presented in Figure 1.18. A specific example of f (x) is discussed by Pars [227], and a slightly modified version of his example is presented here: ±

f (x) = (3 (yB − yA ) − (xB

−

xA) (zB

+ zA ))

x − xA xB − xA ±

²2

x − xA − (2 (yB − yA ) − ( xB − xA ) (zB + zA )) xB − xA ± ² 2 2 2 π (x − xA ) + c ( x − xA ) (xB − x) + d sin xB − xA ± ² xB − x + zA ( x − x A ) + yA , xB − xA

²3

10 Here, 1 is an example of the integrating factor k ( r, t ) mentioned earlier in the definition of a x

(1.23)

nonintegrable constraint. The fact that this factor is not unique can be appreciated by noting that xc , where c is any nonzero constant, is also an integrating factor. 11 With the notable exception of [226], this classification is not typically mentioned in the textbooks on analytical mechanics. Further discussion of piecewise-integrable constraints can be found in the interesting paper by Ruina [249]. As discussed in his paper, constraints of this type also arise in many locomotive systems such as passive walking machines that involve impacting mechanical systems. 12 A proof of this statement can be found in [84, Section 163]. Our discussion of constraint (1.22) is based on the treatments presented by Goursat [100] and Pars [227].

32

Kinematics of a Particle

f v

E3 E2 A O

E1

B

Three possible motions between two given points A and B of a particle subject to the constraint ˙y − z x˙ = 0. The arrows indicate the directions of motion of the particle and the vector f = − zE1 + E 2. The motions presented in this figure were constructed with the assistance of (1.23).

Figure 1.18

where c and d are arbitrary constants. It is left as an exercise for the reader to verify that an infinite number of paths between A and B are possible without violating the constraint y˙ − zx˙ = 0. Shortly, we shall verify that this constraint is indeed nonintegrable.

Integrability Criteria

Suppose a constraint π = 0 is imposed on the motion of the particle. As mentioned earlier, this constraint is integrable if we can find a function ´ (r, t ) and an integrating factor k such that ˙ = ´

k (f · v + e) .

(1.24)

Otherwise, the constraint f · v + e = 0 is nonintegrable. It is desirable to know if a constraint is integrable, because we can then, in principle, find a coordinate system µ 1 2 3¶ q , q , q such that f · v + e = 0 is equivalent to the constraint q˙ 3 + e = 0. The latter constraint in turn is equivalent to the constraint q3 = g, where g˙ = −e. Using this coordinate system, the dynamics of the particle is easier to analyze. With this in mind, several classic integrability criteria are presented for single and multiple constraints.13

A Single Scleronomic Constraint

The first criterion we examine pertains to constraints of the form f·v = 0, where f is µ ¶ 1 2 3 not an explicit function of time. Using a coordinate system q , q , q , we can write the constraint π = 0 in the form 13 For additional discussion and illustrative examples from mechanics, the texts of Papastavridis [226] and

Rosenberg [245] are recommended. Our historical remarks on these criteria are based on the paper by Hawkins [120].

1.10

33

Classification of Constraints

f1 q˙ 1 + f 2q˙ 2 + f3 q˙ 3

=

0.

A necessary and sufficient condition for f · v = 0 to be integrable is that14 Ic for all possible choices of qi . Here, ±

Ic

= f1

∂ f3

∂ q2

−

²

∂ f2

±

+ f2

∂ q3

=

∂ f1

∂ q3

0

(1.25)

−

∂ f3

²

∂ q1

± + f3

∂ f2

∂ q1

−

∂ f1

²

∂ q2

.

It is convenient to recall at this point the expression for the curl of a vector field P in Cartesian coordinates: ¿

curl(P) = ± =

3 ³ ∂ i=1 ∂ P3 ∂ x2

∂ xi −

À

Ei ∂ P2

×

P

²

±

E1 +

∂ x3

∂ P1 ∂ x3

−

∂ P3

²

∂ x1

±

E2 +

∂ P2 ∂ x1

−

∂ P1 ∂ x2

²

E 3, (1.26)

where Pi = P · Ei . With the help of this expression, it is easy to see that criterion (1.25) can also be expressed in the compact form f · (curl (f)) = 0. We refer to (1.25) as Jacobi’s criterion after its discoverer, Carl G. J. Jacobi (1804–1851). Satisfaction of (1.25) does not tell us what ´ (r) or k(r) are; it indicates only that these functions exist. Further, this criterion is local – it does not tell us if these two functions are the same for each point in space. For instance, although the constraints xy˙ + yx˙ = 0 and x˙y = 0 trivially satisfy integrability criterion (1.25), only the former has a continuously defined ´ (r). The function ´ (r) for the latter constraint can be defined only in a piecewise manner (see Figure 1.17). If we use (1.25) to examine the constraint y˙ − zx˙ = 0, then we find that 15 Ic

= −z ( 0 −

0) + 1 (−1 − 0) + 0 (0 − 0)

= −1.

As Ic

= −1 ² =

0, the constraint y˙ − zx˙ = 0 is nonintegrable.

A Single Rheonomic Constraint

It is clearly of interest to present the generalization of Jacobi’s criterion to rheonomic constraints: f · v + e = 0. The result is very similar in form to that for a scleronomic constraint, but it is more tedious to evaluate. To proceed, we express the constraint π = 0 in the form f1 q˙ 1 + f2 q˙ 2 + f 3q˙ 3 + f4

=

0,

14 Proofs of this result can be found in [84, Section 151], [99, Section 442], and [226]. A proof using

differential forms can be found in the book by Flanders [81], who refers to this result as Frobenius’ integration theorem. 15 That is, we choose q1 = x, q2 = y, and q3 = z.

34

Kinematics of a Particle

and define the variables U1 Clearly, f4

=

IJKL

=

q1 ,

U2

=

q 2,

U3

=

q3 ,

e. We next form the functions ±

= fJ

∂ fK ∂ fL − ∂ UK ∂UL

²

±

+ fK

∂ fL ∂ fJ − ∂ UL ∂UJ

U4

²

± + fL

=

t.

∂ fJ ∂ fK − ∂UJ ∂ UK

²

.

Here, the integer indices J, K, and L range from 1 to 4. A necessary and sufficient condition for the constraint π = 0 to be integrable is that the following four equations hold for all q1 , q2, q3, and t: IJKL

=

0,

for all J, K, L ∈ {1, 2, 3, 4 }, L ² = J ² = K, K

²=

L.

(1.27)

For a proof of this theorem, the reader is referred to [84, Section 161] or [81].

Systems of Constraints

When particles are subject to several constraints, their independence needs to be examined. For the case of two constraints, we first express the constraints as follows: f1 · v + e1

=

0,

f2 · v + e2

=

0.

If f1 × f2 ² = 0, then the constraints are said to be independent. For integrable constraints, this is equivalent to the condition ∇ ´1 × ∇ ´2 ² = 0. That is, the normal vectors to surfaces ´1 = 0 and ´2 = 0 are not parallel. The case of three constraints is similar. We first express the constraints as f1 · v + e1

=

0,

f2 · v + e2

=

0,

f3 · v + e3

=

0.

(1.28)

Then, the condition for their independence is that f1 · (f2 × f3 ) ² = 0. If the constraints are integrable, then this condition is equivalent to ∇´1 · (∇´2 × ∇ ´3) ² = 0. Geometrically, this means that the normal vectors at the point of intersection of the surfaces ´1 = 0, ´2 = 0, and ´3 = 0 form a basis. The presence of more than one constraint can also imply that a system of constraints that are individually nonintegrable can become integrable. The most well-known instance occurs when two scleronomic constraints are imposed on a particle [226]: f1 · v = 0, f2 · v = 0,

(1.29)

where the functions f1 and f2 are functions of r, and f1 × f2 ² = 0. In this case, the system of constraints is integrable. The proof of this result, which is presented on page 313, uses a criterion that is due to Ferdinand G. Frobenius (1849–1917), which we

1.12

Exercises

35

postpone discussion of until Chapter 8. This criterion can also be used to show that if the three constraints (1.28) are imposed on a particle, then the system of constraints is integrable. Consequently, the motion of the particle is prescribed. Other instances of multiple constraints on the motion of a particle are discussed in the exercises at the end of this chapter.

1.11

Closing Comments

In this chapter, we have assembled many of the kinematical concepts and tools needed to solve problems in particle dynamics. For most readers, the novel aspects of the chapter will have been the discussion of curvilinear coordinates and kinematical constraints. These two topics are intimately related and will feature prominently in the forthcoming chapters.

1.12

Exercises

Exercise 1.1: Consider a particle whose motion is described in Cartesian coordinates as r(t) = cE2 + 10tE1 , where c is a constant. Determine the areal velocity vector A of the particle, and show that the magnitude of this vector corresponds to the rate at which the particle sweeps out a particular area. Does the particle sweep out equal areas during equal periods of time? In your solution, you should also consider the case in which c = 0. Exercise 1.2: Consider a particle whose motion is described in cylindrical polar coordinates as r(t)

=

10er ,

θ (t) = ωt,

where ω ² = 0. Determine the areal velocity vector A of the particle. Under which conditions does the particle sweep out equal areas during equal periods of time? Exercise 1.3: Recall that the cylindrical polar coordinates {r, θ , z} are defined in Cartesian coordinates {x = x1, y = x2, z = x3 } by the relations r

´

=

x21 + x22 ,

θ =

tan

−1

±

x2 x1

²

,

z = x3.

Show that the covariant basis vectors associated with the curvilinear coordinate system, q1 = r, q2 = θ , and q3 = z, are a1

=

er ,

a2

=

reθ ,

a3

=

E3 .

=

E3.

In addition, show that the contravariant basis vectors are a1

=

er ,

a2 =

1 eθ , r

a3

36

Kinematics of a Particle

It is a good exercise to convince yourself with an illustration that a2 is tangent to a θ coordinate curve, whereas a2 is normal to a θ coordinate surface. Finally, for this coordinate system, show that T

¸ m· 2 r˙ + r2θ˙ 2 + z˙2 . 2

=

Exercise 1.4: Recall that the spherical polar coordinates {R, φ , θ } are defined in Cartesian coordinates {x = x1, y = x2, z = x3 } by the relations R= θ =

´

x21 + x22 + x23,

tan

−1

±

x2 x1

⎛´ φ =

tan−1 ⎝

²

, x21 + x22 x3

⎞ ⎠.

Show that the covariant basis vectors associated with the curvilinear coordinate system, q1 = R, q2 = φ , and q3 = θ , are a1

=

eR ,

a2

Reφ ,

=

a3

=

R sin(φ)eθ .

In addition, show that the contravariant basis vectors are a 1 = eR ,

a2

=

1 eφ , R

a3

=

1 eθ . R sin(φ)

Exercise 1.5: In the parabolic coordinate system, the coordinates {u, v, θ } can be defined in Cartesian coordinates {x = x1 , y = x2, z = x3} by the relations Â

u = ³ x3 +

´

Â

v=³ θ =

−x 3 + −1

tan

±

x2 x1

x23 + (x21 + x22),

´

²

x23 + (x12 + x22 ),

.

In addition, the inverse relations can be defined: x1

=

uv cos( θ ),

x2

=

uv sin(θ ),

x3

=

1 2 (u − v2). 2

(a) In the r–x3 half ´ plane (i.e., a θ coordinate surface) where r is the cylindrical polar coordinate r = x21 + x22, draw several representative examples of the projections of the u and v coordinate surfaces. You should give a sufficient number of examples to convince yourself that u, v, and θ can be used as a coordinate system. (b) In the x1 –x2 –x3 space, draw a u coordinate surface. Illustrate how the v and θ coordinate curves foliate this surface.

1.12

Exercises

37

(c) Show that the covariant basis vectors for the parabolic coordinate system are a1

=

a2 = a3

=

∂r ∂u ∂r ∂v ∂r ∂θ

=

ver + uE3,

=

uer − vE3,

=

uveθ .

Illuminate your results from (a) and (b) by drawing representative examples of these vectors. (d) Show that the contravariant basis vectors for the parabolic coordinate system are16 a1

= ∇u =

a2

= ∇v =

a3

= ∇θ =

1

a 1, u 2 + v2 1 a 2, u 2 + v2 1 eθ . uv

Again, illuminate your results from (a), (b), and (c) by drawing representative examples of these vectors. (e) Where are the singularities of the parabolic coordinate system? Verify that, at these singularities, the contravariant basis vectors are not defined. (f) For a particle of mass m that is moving in E3, establish expressions for the kinetic energy T and linear momentum G in terms of {u, v, θ } and their time derivatives. Exercise 1.6: Consider the classic problem of determining the motion of a particle on a circular helix (see Figure 1.19). In terms of the cylindrical polar coordinates {r, θ , z}, the equation of the helix is Bead of mass m

g

E3 E2 O

E1 Figure 1.19

A particle moving on a circular helix.

16 One method to establish these results is to use the representation (1.17) for the gradient of a function 3 ´ ( r, θ , z) .

38

Kinematics of a Particle

r = R0 ,

z = α R 0θ ,

where R 0 and α are constants. Here, we use another curvilinear coordinate system to define the motion of a particle: q1

q2

= θ,

=

q3

r,

= ν =

z − α rθ .

A q3 coordinate surface is known as a right helicoid. (a) Show that the covariant basis vectors associated with this coordinate system are a1

=

r (eθ

+ α E3 ) ,

a2

=

er + αθ E 3,

a 3 = E 3.

Verify that the covariant basis vectors are not orthonormal. (b) Show that the contravariant basis vectors associated with this coordinate system are17 1 a1 = eθ , a 2 = er , a3 = E3 − αθ er − αeθ . r Verify that the contravariant basis vectors are not orthonormal. (c) Show that the kinetic energy of the particle has the representation T

=

¸ m· (1 + α2 θ 2 )˙r2 + (1 + α 2)r 2θ˙2 + ν˙ 2 2 · ¸ m + 2ν˙ ˙rαθ + 2ν˙ θ˙ αr + 2r˙θ˙α 2r θ . 2

Exercise 1.7: Consider the following curvilinear coordinate system: q1

=

x1 sec(α ),

q2

=

x2 − x1 tan(α ),

q3

=

x3,

where α is a constant. This coordinate system is one of the simplest instances of (a nonorthogonal coordinate system. Referring to (1.5), calculate the functions ) xˆ k q1 , q2, q3 . Draw the coordinate curves and surfaces for the curvilinear coordinate system and then show that the covariant and contravariant basis vectors are a1

=

cos(α)E 1 + sin(α)E2 ,

a2

=

E2 ,

a3

=

E3 ,

(1.30)

and a1 = sec(α)E 1,

a2

= −

tan(α )E1 + E2 ,

a3

=

E3 ,

(1.31)

respectively. Illustrate these vectors on the coordinate curves and surfaces you previously drew. For which values of α is {a1 , a2, a3 } not a basis? Exercise 1.8: Given a vector b = 10E1 + 5E2 + 6E3, 17 One approach to determine the contravariant basis vectors is to use the definition ak

k along with

= ∇q

the representation (1.17) 3 for the gradient of a function ´ (r, θ , z) . Alternatively, given the results of (a), the relations (1.8) can be used.

1.12

39

Exercises

calculate its covariant components bi and contravariant components bi when the covariant and contravariant basis vectors are defined by (1.30) and (1.31), respectively. In addition, verify that b=

3 ±

bi ai

=

i=1

Furthermore, show that b ³ a1, a2 , a3 not a basis?

²

∑3

i=1 b

±=

3 ±

bi ai .

i=1

i i

a and b

∑3

i=1 b i ai .

±=

For which values of

α

is

Exercise 1.9: The purpose of this exercise is to illustrate how the covariant and contravariant components of a vector are related. To start, we define the following scalars by using the covariant and contravariant basis vectors: aik

=

aik (qr ) = ai · ak ,

You should notice that aik range from 1 to 3.

=

aki and aik

aik

=

=

aik (qr ) = ai · ak .

aki . The indices i, k, r, and s in this problem

(a) For any vector b, show that the covariant and contravariant components are related: bi

3 ± =

aik bk ,

bi

k= 1

3 ± =

aik bk .

k=1

In other words, the covariant components are linear combinations of the contravariant components and vice versa. Using a matrix notation, these results can be expressed as ⎡

⎤

⎡

⎡

⎤

⎡

⎤⎡

⎤

⎤⎡

⎤

a11 b1 ⎣ b2 ⎦ = ⎣ a21 a31 b3

a12 a22 a32

b1 a13 ⎦ ⎣ b2 ⎦ , a23 b3 a33

b1 a11 2 ⎣ b ⎦ = ⎣ a21 b3 a31

a12 a22 a32

a13 b1 a23 ⎦ ⎣ b2 ⎦ . a33 b3

(b) By choosing b = ar and as , and using the symmetries of akm and ars , show that 3 ± k =1

aik akj

=

3 ±

aki akj

k =1

=

3 ±

aki ajk

k =1

j

= δi .

It might be helpful to realize that, by using matrices, one of the preceding results has the following representation: ⎡

a11 ⎣ a21 a31

a12 a22 a32

⎤⎡

a13 a11 a23 ⎦ ⎣ a21 a33 a31

a12 a22 a32

⎤

a13 a23 ⎦ a33

⎡

1 ⎣ = 0 0

0 1 0

⎤

0 0 ⎦. 1

40

Kinematics of a Particle

½

¾

(c) Compute the expressions for the matrices [aik ] and aik for the simple example of a nonorthogonal coordinate system (1.12). Verify that the relations shown in Figure 1.8(a) for the projections of a vector onto the covariant and contravariant basis vectors are correct. Exercise 1.10: With the help of integrability criteria (1.25) and (1.27), show that only one of the following constraints is integrable: π1 =

0,

π1 =

π2 = π3 =

0,

0,

xx˙ + yy˙ + e(t),

π2 = zy ˙ + x, ˙ π3 =

cos(z)y˙ − sin(z) ˙x.

In addition, show that the integrable constraint corresponds to a particle moving on a cylinder whose radius varies with time. Exercise 1.11: With the help of integrability criterion (1.27), show that one of the following constraints is nonintegrable: π1 =

0,

π1 = zx ˙+y ˙ + e(t),

π2 =

0,

π2 = z˙ .

If both constraints are imposed on the particle simultaneously, then show that the system of constraints is integrable. Give a geometric description of the line that the particle is constrained to move on. Exercise 1.12: As shown in Figure 1.20, a particle of mass m is free to move on a rough catenary curve that is defined by the pair of equations ±

y = A cosh

x − x0 ³

²

+

y0 ,

z = 0,

where A, ³, y0 , and x0 are constants. To establish the equations of motion for the particle, the following curvilinear coordinate system is defined for E3 : q

1

=

x,

q

2

±

= η =

y − A cosh

x − x0 ³

²

,

q3

=

z.

E2 m

O

E1

g

Catenary

Schematic of a particle of mass m which is moving on a curve which has the shape of a catenary in E 3 under the influence of a gravitational force −mgE2.

Figure 1.20

1.12

41

Exercises

(a) What are the covariant basis vectors ak for this coordinate system? You will find ( ) it helpful here, and in the sequel, to use the abbreviations f = A cosh x−³x0 and ( ) ½ ¾ fx = A³ sinh x−³x0 . Compute the matrix aij . (b) Show that the contravariant basis vectors for this coordinate system are a

1

=

E1 ,

a ½

2

=

¾

E2 −

A ³

±

sinh

x − x0 ³

²

a3

E1 ,

=

E3 . ½

¾

Compute the matrix aik and show that it is the inverse of the matrix aij that you computed in (a). What are the unit normal vectors to the catenary? (c) Show that the kinetic energy of the particle that is in motion on the catenary has the representation ±

T˜ =

m A2 1 + 2 sinh 2 2 ³

± 1 q

−

³

x0

²² ·

q˙ 1

¸2

.

(d) Consider a particle in motion on a plane curve y = f (x). Using the results from (a)– (c), establish expressions for the velocity vector and kinetic energy of the particle in motion on the curve and the pairs of unit normal vectors to the plane curve.

2

Kinetics of a Particle

2.1

Introduction

The balance law F = ma for a single particle plays a central role in this chapter. This law is used to examine models for several physical systems ranging from planetary motion to a model for a roller coaster. Our discussion of the behavior of these systems predicted by the models relies heavily on numerical integration of the equations of motion provided by F = ma, and it is presumed that the reader is familiar with the numerical integration of ordinary differential equations. Two of the most important types of forces present in many applications are conservative forces and constraint forces. For the former, the gravitational force between two particles is the prototypical example, whereas the most common constraint force in particle mechanics is the normal force. It is crucial to be able to properly formulate and represent conservative and constraint forces, and we will spend a considerable amount of time discussing them in this chapter. In contrast to most texts in dynamics, here we consider friction forces to be types of constraint forces. For most applications, exact (or analytical) solutions are not available and recourse to numerical methods is often the only course of action. In validating these solutions, any conservations that might be present are crucial. To this end, conservations of momentum and energy are discussed at length and we also show (with the help of two examples) how angular momentum conservation can often be exploited. The examples discussed in this chapter are far from exhaustive. Although several other examples are included in the exercises, they too do not come close to encompassing the vast array of solved problems in the mechanics of a single particle. To this end, it is recommended that the interested reader consult the classical texts by Routh [248] and Whittaker [306], and the more recent texts by Baruh [20], Moon [193], and Sheck [254].

2.2

The Balance Law for a Single Particle

Consider a single particle of mass m that is moving in E3 . As usual, the position vector of the particle relative to a fixed origin O is denoted by r. The balance law for this particle is known as the balance of linear momentum, Newton’s second law, or Euler’s first law. The integral (or impulse–momentum) form of this law is

2.2

The Balance Law for a Single Particle

G(t) − G (t0 ) =

± t t0

F(τ )dτ ,

43

(2.1)

where F is the resultant force acting on the particle and G = mv is the linear momentum. Notice that this form of the balance law does not assume that v is differentiable with respect to time t. As a result, (2.1) is valid in impact problems, among others. An example of the application of (2.1) is discussed in Section 2.10. If we assume that G is differentiable with respect to time, then we can differentiate both sides of the integral form of the balance of linear momentum to obtain the local form: F

˙ = G.

Assuming that the mass of the particle is constant, we can write F

mr¨ .

=

(2.2)

This law represents three (scalar) equations that relate F and the rate of change of linear momentum of the particle. We refer to F = ma as the balance of linear momentum.1 Our emphasis in this chapter is on the principle F = ma, but it is misleading to believe that this is the sole accepted principle in dynamics. Indeed, since the creation of this principle by Newton over 300 years ago, several alternative (and often equivalent) principles of dynamics have been proposed, but we postpone discussion of these alternative principles until Section 4.12. It is convenient to write the balance law as a set of first-order ordinary differential equations: v = r˙ ,

F = mv˙ .

In the absence of constraints, these represent six scalar (differential) equations for the six unknowns r(t) and v(t). To solve these equations, six initial conditions r(t0 ) and v(t0) must be specified. Alternatively, if the problem is formulated as a boundary-value problem, then a combination of six initial and final conditions on r(t) and v(t) must be prescribed. If we express F = ma using a Cartesian coordinate system, then we find the following three equations: mx¨ 1 = F · E1, mx¨ 2 = F · E2, mx¨ 3 = F · E3. In contrast, if a cylindrical polar coordinate system is used, we have ²

m r¨ − r θ˙ 2 (

³

m r θ¨ + 2r˙θ˙

)

=

F · er ,

=

F · eθ ,

1 Discussions of the historical threads from Newton’s Principia [203] that lead to Euler’s explicit

formulation of (2.2) in [67] can be found in several works by Clifford A. Truesdell (1919–2000); see, for example, [287, 288].

44

Kinetics of a Particle

m¨z = F · E 3.

(2.3)

Finally, if we use a spherical polar coordinate system, we find that ²

²

(

m R¨ − R φ˙ 2 − R sin2 (φ)θ˙ 2

m Rφ¨ + 2R˙ φ˙ − R sin(φ ) cos(φ)θ˙ 2

³ =

F · eR ,

=

F · eφ ,

=

F · eθ .

³ )

m R sin(φ)θ¨ + 2R˙ θ˙ sin(φ) + 2Rθ˙ φ˙ cos( φ)

(2.4)

Notice that these equations are different projections of F = ma onto a set of basis vectors for E3. Establishing the component representations of F = ma for various coordinate systems can be a laborious task. However, Lagrange’s equations of motion allow us to do this in a very easy manner. We will examine these equations in Section 3.2.

2.3

Work and Power

The mechanical power P of the force P acting on a particle of mass m is defined as

P = P · v. Clearly, if P is perpendicular to v, then the power of the force is zero. As shown in Figure 2.1, consider a motion of the particle between two points: A and B. We suppose that at time t = tA the particle is at A: r(tA ) = rA . Similarly, when t = tB , the particle is at B: r(tB ) = rB . During the interval of time that the particle moves from A to B, we suppose that a force P, among others, acts on the particle. The work WAB performed by P during this time interval is defined to be the integral, with respect to time, of the mechanical power: WAB

± tB =

tA

P · vdt.

P v

A

B

r

Path of particle

rA rB

O

Figure 2.1

A force P acting on a particle as it moves from A to B.

2.4

Conservative Forces

45

Notice that this is a line integral, and we are using t to parameterize the path of the particle. Depending on the choice of coordinate system, the integral in this expression has several equivalent representations. As an example, suppose that a force P = Peθ acts on a particle, and the motion of the particle is r(t) = ±eαt (cos( ωt)E1 + sin(ωt)E 1), where P, ±, α, and ω = θ˙ are constant. A straightforward calculation shows that the power of this force is P · v = ωP±eαt , and the work performed by the force is WAB

± tB

=

tA

αt

ωP± e

dt =

ωP± α

(

eαtB

−

)

eαtA ,

where, in evaluating the integral, we have assumed that α ± = 0.

2.4

Conservative Forces

A force P acting on a particle is said to be conservative if the work done by P during any motion of the particle is independent of the path of the particle. When a result from vector calculus is used, the path independence implies that P is the gradient of a scalar function U = U (r): P

= −∇ U.

The function U is known as the potential energy associated with the force P, and the minus sign in the equation relating P to the gradient of U is a historical convention. Various representations of the gradient can be found in (1.17). ˙ It is important to notice that, if P is conservative, then its mechanical power is −U. To see this, we simply examine U˙ and use the definition of a conservative force: ˙ = − −U

∂U ∂r

· v = −(−P) · v =

P · v.

This result holds for all motions of the particle and is very useful when we wish to establish expressions for the rate of change of the total energy E of a particle. To check if a given force P is conservative, one approach is to find a potential function U such that ˙ P · v = −U

holds for all motions of the particle. This approach reduces to solving a set of coupled partial differential equations for U. For example, if cylindrical coordinates are used, one needs to solve the following three partial differential equations for U (r, θ , z): Pr

= −

∂U ∂r

,

Pθ

= −

1 ∂U , r ∂θ

Pz = −

∂U ∂z

,

(2.5)

where P = P r er + Pθ eθ + Pz E3 . You might notice that the solution to (2.5) will yield a potential energy U(r, θ , z) modulo an additive constant. This constant is usually set by the condition that U = 0 when the coordinates have a certain set of values.

46

Kinetics of a Particle

Another approach to ascertain if a given force P is conservative is to examine its curl. The idea here is based on the identity curl (grad(V)) = 0, where V = V(r) is any scalar function of r. Clearly, if the given force P is conservative, then curl(P) = 0.2 Consequently, if curl (P ) = 0, then the Cartesian components of P must satisfy the following conditions: ∂ P3 ∂ x2

=

∂ P2 ∂ x3

∂ P1

,

∂ x3

=

∂ P3 ∂ x1

,

∂ P2 ∂ x1

=

∂ P1 ∂ x2

,

where P i = P · E i . We leave it as an exercise to show that the force P previously is not conservative.

2.5

=

Peθ discussed

Examples of Conservative Forces

The three main types of conservative forces in engineering dynamics are constant forces, spring forces, and gravitational force fields. Constant Forces

All constant forces are conservative. To see this, let C denote a constant force and let Uc = −C · r. Now, ∇ Uc = −C and, consequently, Uc is the potential energy associated with C. The most common examples of constant forces are the gravitational forces −mgE2 and −mgE 3, and their associated potentials are mgE2 · r and mgE3 · r, respectively. Spring Forces

Consider the spring shown in Figure 2.2. One end of the spring is attached to a fixed point A, and the other end is attached to a particle of mass m. When the spring is unstretched, it has length ±0 . Clearly, the stretched length of the spring is ²r − rA ², and the extension/compression of the spring is ² = ² r − rA ² − ±0 .

The potential energy Us associated with the spring is Us = f (²), where f is a function of the change in length of the spring. Evaluating the gradient of Us , we find the spring force F s : Fs = −

∂ Us ∂r

= −

∂f ∂²

r − rA . ²r − rA ²

To establish this result, we used the identity ∂² ∂r

=

∂ ∂r

(² r − rA ² − ±0) =

r − rA . ² r − rA ²

2 An expression for the curl of a vector field was presented earlier in (1.26).

2.5

47

Examples of Conservative Forces

m

Spring

r E3 E2 E1

O

rA A

Figure 2.2

A spring whose ends are attached to a particle and a fixed point A.

This identity is left as an exercise to establish.3 The most common spring in engineering dynamics is a linear spring. For this spring: Us

=

K 2

(²r − rA² − ±0)

2

,

Fs

= −K ( ²r −

rA² − ±0)

r − rA . ² r − rA ²

In words, the potential energy of a linear spring is a quadratic function of its change in length. Examples of nonlinear springs include those in which f is a more complicated function of ² . For instance, f ( ²) = A ²2 + B ²4 , where A > 0 and B are constants. Such a spring is known as hardening if B softening if B < 0.

>

0 and

Newton’s Gravitational Force

Dating to Newton in the late 1600s, the force exerted by a body of mass M on another body of mass m has been modeled as Fn

=

where g=−

mg,

GM ² r²3

r.

The body of mass M is assumed to be located at the origin O, and G is the universal gravitation constant. In this model for the gravitational force, both bodies are modeled 3 To help you with this, it is convenient to first establish that

to showing that the gradient of ²r² is eR .

∂

√

x·x

∂x

x = √

x·x

x

= ² ². x

This result is equivalent

48

Kinetics of a Particle

as mass particles. Later on, in Sections 4.5 and 8.6, we shall examine generalizations of this force field to systems of particles and rigid bodies, respectively. Because the magnitude of Fn depends on the distance squared between the two bodies, Newton’s force field is often known as the inverse-square law. The force Fn is conservative, and its potential energy is GMm . Un = − ²r² It should be transparent from the expressions for Un and F n that they have rather simple representations when a spherical polar coordinate system is used. Newton’s gravitational force field is attractive: it tends to pull m toward M. Thus, we have the interesting question of what keeps the two bodies from colliding. The answer, as you know from other courses, is the change of momentum of m. It is this delicate balance that enables m to steadily orbit M in a circular orbit.

2.6

Constraint Forces

A constraint force Fc is a force that ensures that a constraint is enforced. Examples of these forces include reaction forces, normal forces, and tension forces in inextensible strings. Given a constraint ³ (r, t) = 0 on the motion of the particle, there is no universal prescription for the associated constraint force. Choosing the correct prescription depends on the physical situation that the constraint represents. However, when we turn to solving for the motion of the particle by using F = ma, we see that we need to solve the six equations 1 r˙ = v, v˙ = F, m subject to the following restrictions on r and v: ³ (r, t ) =

0,

∂³ ∂r

·

v+

∂³ ∂t

=

0.

To close this system of equations, an additional unknown is introduced in the form of the constraint force Fc . The prescription of Fc must be such that F = ma and ³ (r, t ) can be used to determine F c and r(t). There are no unique prescriptions for constraint forces. The prescription most commonly used, which dates to Joseph-Louis Lagrange (1736–1813), is referred to in this book as the Lagrange prescription. As shown by O’Reilly and Srinivasa [218], this prescription is necessary and sufficient to ensure that the motion of the particle satisfies the constraint. However, freedom is available to include arbitrary forces, including frictional forces, that are in the direction of motion of the particle. This is the reason why prescriptions for constraint forces other than Lagrange’s are valid. A Single Constraint

Consider the case of a particle subject to a single constraint: ³ (r, t) =

0.

2.6

Constraint Forces

49

∇Ψ Surface Ψ = 0

n s2

m

r

s1

O

A particle moving on a surface ³ = 0. The vectors s1 and s 2 are unit tangent vectors to this surface at the point of contact of the particle and the surface.

Figure 2.3

Referring to Figure 2.3, we recall that the unit normal vector n to this surface is n=

∇³ ²∇ ³ ²

.

Knowing n, we can construct a unit tangent vector s1 to the surface. In addition, by defining another unit tangent vector s2 = n × s1, we have constructed a right-handed orthonormal basis {s1 , s2, n} for E3 . This is not in general a constant set of vectors; rather, it changes as we move from point to point along the surface. The final ingredient we need is to denote the velocity vector of the point of the surface (which the particle is in contact with) by vs . With this background in mind, we now consider two prescriptions for the constraint force Fc . The first prescription is known as the Lagrange prescription: Fc

= λ ∇³ ,

where we must determine λ = λ(t) by using F = ma. In other words, λ is an undetermined Lagrange multiplier. As discussed by Casey [37], the constraint force Fc is normal to the surface ³ = 0 and is identical to a virtual work prescription that is sometimes known as the Lagrange principle or the Lagrange–D’Alembert principle. On physical grounds, Lagrange’s prescription is justified if the surface on which the particle is constrained to move is smooth. In the event that the surface is rough, an alternative prescription, which is due to Charles Augustin Coulomb (1736–1806), can be used.4 We refer to this prescription as Coulomb’s prescription for Fc : Fc

= λ∇ ³ +

Ff ,

4 This prescription is often known as Amontons–Coulomb friction, after Guillaume Amontons (1663–1705).

50

Kinetics of a Particle

N = λ∇ Ψ

vrel

Ff

n

Fc = N + Ff

s2 s1 The constraint force F c acting on a particle moving on a rough surface. The velocity vector vrel = v − vs is the velocity vector of the particle relative to the point of its contact with the surface.

Figure 2.4

where the normal force N

and the friction force is v − vs F f = −µ d ²λ∇ ³ ² . ²v − v s ²

= λ∇ ³

Here, µd is known as the coefficient of dynamic friction. Notice that the tangential components of Fc are governed by the behavior of vrel = vrel1 s1 + vrel 2 s2 and oppose the motion of the particle relative to the surface (cf. Figure 2.4). The velocity ²v − vs ² is sometimes known as the slip speed. The mechanical power of the constraint force Fc is Fc · v = λ∇ ³ · v + Ff · v = −λ

∂³ ∂t

+

Ff · v,

where we used the identity ˙ = ∇³ · ³

v+

∂³ ∂t

=

0.

For the Lagrange prescription, Ff = 0, we can now see that, if the surface that the particle is moving on is fixed (i.e. ³ = ³(r)), then Fc does no work. Otherwise, this constraint force is expected to do work because its normal component ensures that part of the velocity vector of the particle is vs . For the Coulomb prescription, except when vs = 0, it is not possible to predict if work is done on the particle by the constraint force. As a first example, consider a particle moving on a rough sphere of radius ±0 whose center is fixed at the origin O. For this surface, the constraint is ³(r) = 0, where ³ (r) = r · eR − ±0. Consequently, ∇ ³ = eR . In addition, v − vs = ±0 φ˙ eφ + ±0 sin(φ)θ˙ eθ . In conclusion: φ˙ eφ + sin(φ )θ˙ eθ . F c = λeR − µ d |λ| ´ ˙ 2 + sin2(φ )θ˙ 2 φ

2.6

Constraint Forces

51

In this expression, |λ| is the magnitude of the normal force exerted by the sphere on the particle. If we now consider the spherical pendulum, then Fc is given by the Lagrange prescription: Fc = λeR . In the spherical pendulum, −λ is the tension in the rod connecting the particle to the fixed point O. Two Constraints

When a particle is subject to two constraints, ³1(r, t) = 0 and ³2 (r, t) = 0, then it can be considered as constrained to move on a curve. The curve in question is formed by the instantaneous intersection of the surfaces defined by the constraints. At each point on the curve there is a unit tangent vector t. We can define this vector by first observing that ∇ ³1 and ∇ ³2 are both normal to the surfaces that the curve lies on (cf. Figure 1.15). Consequently, t

=

∇ ³ 1 × ∇ ³2 ²∇ ³ 1 × ∇ ³2 ²

.

For each instant in time, the point of the curve that is in contact with the particle has a velocity. We denote this velocity by vc . The velocity vector of the particle relative to the curve is vrel = v − vc = vt. We now turn to prescriptions for the constraint force. The first prescription is the Lagrange prescription: Fc

= λ1∇ ³ 1 + λ2 ∇ ³2 ,

where λ1 = λ1(t) and λ2 = λ 2(t) are both determined by use of F = ma. As in the case of a single constraint, this prescription is valid when the curve that the particle moves on is smooth, and it provides a constraint force that is normal to the curve. For the rough case, we use Coulomb’s prescription: Fc

= λ1 ∇ ³ 1 + λ2 ∇ ³2 +

where the friction force is Ff

= −µ d ² λ1 ∇ ³ 1 + λ2 ∇ ³2 ²

Ff ,

v − vc . ²v − vc ²

The friction force opposes the motion of the particle relative to the curve and the normal force N = λ 1∇ ³1 + λ2 ∇³2 . The mechanical power of the constraint force Fc for this case is Fc · v = λ1 ∇ ³1 · v + λ2 ∇ ³2 · v + Ff · v = −λ1

∂ ³1 ∂t

− λ2

∂ ³2 ∂t

+

Ff · v,

where we used the identities ˙ 1 = ∇ ³1 · ³

v+

∂ ³1 ∂t

=

0,

˙ 2 = ∇ ³2 · v + ³

∂ ³2 ∂t

=

0.

For the Lagrange prescription, Ff = 0, and we can now see that, if the curve that the particle is moving on is fixed (i.e., ³1 = ³1(r) and ³2 = ³2 (r)), then Fc does no work.

52

Kinetics of a Particle

Otherwise, this constraint force is expected to do work because its normal components force part of the velocity vector of the particle to be vc . As in the case of a single constraint, for the Coulomb prescription, except when vc = 0, it is not possible to predict if work is done on the particle by this force. We now consider some examples. Recall that the planar pendulum consists of a particle of mass m that is attached by a rod of length ±0 to a fixed point O. The particle is also constrained to move on a vertical plane. In short: 0,

³1 =

³2 =

³1 =

0,

³2 =

With a little work, we find that ∇ ³1 = er and Lagrange’s prescription is appropriate: Fc

r · er − ±0 , r · E3 .

∇³2 =

E3. For this mechanical system,

= λ1 er + λ2 E3.

For this system, λ1 er can be interpreted as the tension force in the rod and λ2E 3 can be interpreted as the normal force exerted by the plane on the particle. If we let ± 0 = ± 0(t), the prescription for the constraint force will not change. A system which is related to the planar pendulum can be imagined as a particle that moves on a rough circular track which has a radius ± 0 = ± 0(t). The particle is subject to the same constraints as it is in the planar pendulum; however, Lagrange’s prescription is not valid. Instead, we now have Fc

= λ1 er + λ2E 3 − µd

where we used the fact that v − vc

´

˙ 2 2 θ λ1 + λ 2 µ µ eθ , µθ˙ µ

= ±0 θ˙ eθ .

Three Constraints

The reader may have noticed that our expressions for the constraint force when we employed Coulomb’s prescription were not valid when the particle was stationary relative to the surface or curve that it was constrained to move on. This is because we view this case as corresponding to the motion of the particle subject to three constraints: ³i (r, t ) = 0, i = 1, 2, 3. As mentioned earlier, when a particle is subject to three constraints, the three equations ³i (r, t) = 0 can in principle be solved to determine the motion r(t) of the particle. We denote the resulting solution by rˆ (t) (i.e., r = rˆ (t)). In other words, the motion is completely prescribed. In this case, the sole purpose of F = ma is to determine the constraint force Fc . For the case in which the particle is subject to three constraints, the Lagrange prescription and a prescription based on static Coulomb friction are equivalent. This equivalence holds in spite of the distinct physical situations these prescriptions pertain to. To examine the equivalence, let us first use Lagrange’s prescription: Fc

= λ1 ∇ ³ 1 + λ 2∇ ³ 2 + λ3 ∇ ³ 3 .

2.6

Constraint Forces

53

Here, λ 1, λ2 , and λ 3 are functions of time. Because the three constraints are tacitly assumed to be independent, {∇ ³1 , ∇ ³2, ∇ ³3} forms a basis for E3 . Consequently, Lagrange’s prescription provides a vector Fc with three independent components. Coulomb’s static friction prescription for a particle that is not moving relative to the curve or surface on which it lies is Fc

=

N + Ff ,

where the magnitude of Ff is restricted by the static friction criterion: ¶ ¶ ¶F ¶ f

≤ µ s ² N² ,

where µs is the coefficient of static friction. Again, the Coulomb prescription provides a vector Fc with three independent components. In other words, both prescriptions state that Fc consists of three independent unknown functions of time. If we now assume that the resultant force F has the decomposition F = Fc + Fa , where Fa are the nonconstraint forces, then we see how Fc is determined from F = ma: Fc

= −Fa +

ma = −Fa + m¨ˆr.

The solution for F c will be the same regardless of whether one uses Lagrange’s prescription or Coulomb’s prescription.

Nonintegrable Constraints

Our discussion of constraint forces has focused entirely on the case of integrable constraints. If a nonintegrable constraint f ·v+ e = 0 is imposed on the particle, we need to discuss a prescription for the associated constraint force. To this end, we adopt a conservative approach and use the prescription Fc

= λf.

The main reason for adopting this prescription is as follows. In the event that the nonintegrable constraint turns out to be integrable, then the prescription we employ will agree with Lagrange’s prescription we discussed earlier. As a further example, suppose the motion of the particle is subject to two constraints, one of which is integrable: ³ (r, t) =

0,

f · v + e = 0.

Using Lagrange’s prescription, we find that the constraint force acting on the particle is Fc

= λ1 ∇ ³ + λ2 f.

Suppose that the applied force acting on the particle is F a; then the equations governing the motion of the particle are ³ (r, t) =

0,

f · v + e = 0,

54

Kinetics of a Particle

1 (Fa + λ 1∇ ³ + λ 2f) . m This set of equations constitutes eight equations for the eight unknowns: and v. r˙ = v,

2.7

v˙ =

λ1 , λ2 ,

r,

Conservations

For a given particle and system of forces acting on the particle, a kinematical quantity is said to be conserved if it is constant during the motion of the particle. The conserved quantities are often known as integrals of motion. The solutions to many problems in particle mechanics are based on the observation that either a momentum or an energy (or both) is conserved. At this stage in the development of the field, most of these conservations are obvious and are deduced by inspection. However, for future purposes it is useful to understand the conditions for such conservations. We shall consider numerous examples of these conservations later on. Conservation of Linear Momentum

The linear momentum G of a particle is defined as G = mv. Recalling the integral form of the balance of linear momentum G(t) − G (t0 ) =

± t t0

F(τ )dτ , ·

we see that G(t) is conserved during an interval of time (t0 , t) if tt F(τ )dτ = 0. The 0 simplest case of this conservation arises when F(τ ) = 0. Another form of this conservation pertains to a component of G in the direction of a d given vector b(t) being conserved. That is, dt (G · b) = 0. For this to happen: ˙ ·b+ G·b ˙ = F · b+ G·b ˙ = 0. G ˙· b = G

In words, if F · b + G · b˙ = 0, then G · b is conserved. Examples of conservation of linear momentum arise in many problems. For example, consider a particle under the influence of a gravitational force F = −mgE3 . For this problem, the E 1 and E2 components of G are conserved. Another example is to consider a particle impacting a smooth vertical wall. Then the components of G in the two tangential directions are conserved. For these two examples, the vector b is constant. Conservation of Angular Momentum

The angular momentum of a particle relative to a fixed point O is HO = r × G. To establish how HO changes during the motion of a particle, a simple calculation is needed: ˙O H

=

˙ = v × mv + r × F v×G+r ×G

=

r × F.

2.7

Conservations

55

It is important to note that we used F = ma during this calculation. The final result is known as the angular momentum theorem for a particle: ˙ H O

=

r × F.

In words, the rate of change of angular momentum is equal to the moment of the resultant force. Conservation of angular momentum usually arises in two forms. First, the entire vector is conserved and second, a component, say c(t), is conserved. For the first case, we see from the angular momentum theorem that HO is conserved if F is parallel to r. Problems in which this arises are known as central force problems. Dating to Newton, they occupy an important place in the history of dynamics. When the angular momentum theorem is used, it is easy to see that the second form of conservation, HO · c is constant, arises when r × F · c + HO · c˙ = 0. Conservation of Energy

As a prelude to discussing the conservation of energy, we first need to discuss the work– energy theorem. This theorem is a result that is established using F = ma and relates the time rate of change of kinetic energy to the mechanical power of F: T˙

=

F · v.

This theorem is the basis for establishing conservation of energy results for a single particle. The proof of the work–energy theorem is very straightforward. First, recall that T = 1 mv · v. Differentiating T , we find 2 T˙ =

d dt

¸

¹

1 mv · v 2

1 ˙ ) = mv ˙ · v. (mv˙ · v + mv · v 2

=

However, we know that mv˙ = F, and so substituting for mv˙ , we find that T˙ = F · v, as required. To examine situations in which the total energy of a particle is conserved, we first divide the forces acting on the particle into the sum of a resultant conservative force P = − ∂∂Ur and a nonconservative force Pncon : F = P + Pncon . Here, U is the sum of the potential energies of the conservative forces acting on the particle. From the work– energy theorem, we find T˙

=

F ·v

P · v + P ncon · v ∂U · v + P ncon · v = − ∂r ˙ + Pncon · v. = −U =

Defining the total energy E of the particle by E

=

T + U,

56

Kinetics of a Particle

we see that E˙

=

P ncon · v.

This result states that if, during a motion of the particle, the nonconservative forces do no work, then the total energy of the particle is conserved. To examine whether energy is conserved, it usually suffices to check whether P ncon · v = 0. To see this, let us consider the example of the spherical pendulum whose length ±0 = ±0 (t). For this particle: P

= −mgE3 ,

Pncon

= λeR .

Consequently, E

=

T + mgE3 · r

and P ncon · v = λeR · v = ±˙0λ . As a result, when the length of the pendulum is constant, ±˙0 = 0, then E is conserved. In contrast, if ±˙0 ± = 0, then the constraint force λeR does work by giving the particle a velocity in the eR direction. Conservation of an Energy-Like Quantity: The Jacobi Integral

On certain occasions it is possible to find a conserved quantity for a problem that is not the total energy or a component of a momentum. We now examine a class of particle problems where this is the case and show two methods to establish conservation of an energy-like quantity. To proceed, consider the system shown in Figure 2.5. A particle m

E2 Spinning smooth rail

E1

O

Ω Figure 2.5 Schematic of a particle of mass m that is attached to a fixed point O by an elastic spring. A vertical gravitational force −mgE3 acts on the particle and the particle is free to move on a smooth rail that is being rotated about the vertical axis with a constant speed ´ = ´0 .

2.7

57

Conservations

is free to move along a smooth rail or track that is being spun about the E3 axis with a constant angular speed. The motion of the particle can be described using a cylindrical polar coordinate system: r = rer ,

²

³

a = r¨ − r´20 er + 2˙r´0eθ .

v = r˙er + r ´0 eθ ,

The total force acting on the particle consists of a spring force gravitational force −mgE3 , and a normal force N

=

Nθ eθ

+

−K ( r − ± 0 ) e r ,

a

Nz E3.

Using a balance of linear momentum, the equation of motion for the particle,5 mr¨ − mr´20 + K (r − ± 0) = 0,

(2.6)

and the normal force, N = 2m˙r´0eθ

+

mgE3 ,

can be determined. For this problem, the total energy E is not conserved by the solutions to (2.6) because N performs work: E˙

=

N·v

2mrr˙´20 d ² 2 2³ = mr ´0 . dt

=

However, we can use the result for E˙ that we found from the work–energy theorem to propose a conserved energy-like quantity. Observe from the expression for E˙ that ³ d ² E − mr2 ´20 dt

=

0.

This result implies that the energy-like quantity V

=

E − mr2 ´20

=

³ K m² 2 2 2 2 ˙r − r ´ + (r − ±0 ) 2 2

is conserved during the motion of the particle. In other words, the trajectories in the phase portrait of (2.6) lie on the level sets of the function V = V (r, ˙r). The conservation of V can also be established using the identity vdv = a(r)dr, where v = r˙ and a(r) = mr´20 − K (r − ± 0). The conservation of an energy-like quantity V is also present in related systems such as the bead on a hoop shown later in Figure 3.14 and the particle that is free to move in a groove milled into a plate that is being rotated about a vertical axis shown later in Figure 3.19. As discussed by Greenwood [104, Section 2-3], the energy-like quantity V is sometimes known as the Jacobi integral.6 5 A classification of the solutions to (2.6) can be found in Exercise 3.12. 6 Greenwood’s text contains a complementary discussion of this conservation from a perspective that

exploits Lagrange’s equations of motion.

58

Kinetics of a Particle

2.8

Dynamics of a Particle in a Gravitational Field

The problem of a body of mass m orbiting a body of mass M is one of the centerpieces in Isaac Newton’s Principia.7 Over 100 years later, it was also discussed in wonderful detail in Lagrange’s famous text Mécanique Analytique.8 Newton was partially motivated to study this problem because of Johannes Kepler’s (1571–1630) famous three laws of motion for the (then known) planets in the solar system: I. The planets move in elliptical paths with the Sun at one of the foci. II. The vector connecting the Sun to the planet sweeps out equal areas in equal times. III. If a denotes the semimajor axis of the elliptical orbit and T denotes the period of the orbit then, for any two planets,

a31 T2 = 12 . 3 a2 T2

Concise discussions of Kepler’s laws can be found in [199, 234, 252, 296]. Some of these authors note that the laws were based on astronomical data taken with the naked eye. In our analysis we assume that the body of mass M is fixed. As can be seen in Exercise 4.6 (Chapter 4), this restriction is easily removed and the results presented can readily be applied to deduce the motions of m and M. Many of the results presented feature terminology associated with ellipses and, for convenience, a summary of several of the terms associated with an ellipse, such as its eccentricity e and axes a and b, is presented in Figure 2.6. The area swept out by the particle can be determined by integrating 1 the areal vector (1.1): A = 2m HO . y

a

m b

r θ

O A

a

ˆ

h
0

Schematic of a particle of mass m moving about a fixed ´ point O in an elliptical orbit. 2

One of the foci of the ellipse is at O, and the eccentricity e = 1 − b2 of the ellipse is less than a 1, where a and b are the lengths of the semimajor and semiminor axes of the ellipse, respectively. Point A is known as the apocenter, and P is known as the pericenter. 7 See Section III of Book 1 of [203]. 8 See Section VII of Part II of [159].

2.8

Dynamics of a Particle in a Gravitational Field

59

The third law is remarkable when we note from [252] that the semimajor axis a and orbital period T for the planet Mercury are 0.387 astronomical units (AU) and 0.241 years, the Earth’s are 1 AU and 1 year, Jupiter’s are 5.203 AU and 11.862 years, and Mars’ are 1.524 AU and 1.881 years, respectively. We note that 0.3873 0.2412

=

1.00,

5.2033 11.8622

1.00,

=

1.5243 1.8812

=

1.00.

All of these results are in agreement with Kepler’s third law. Here, we start by setting up the coordinates for this problem and establishing the equations of motion. Our analysis of the equations of motion then exploits conservation of angular momentum to show that the motion must be planar. Following this, we reduce the equations of motion to a single second-order differential equation that we nondimensionalize and integrate numerically. 9 An alternative approach, which is used in most textbooks, will also be discussed. These analyses enable us to classify all five possible types of trajectories of the particle.

Kinematics

We pick as the origin O the fixed particle of mass M. Then the position vector of the particle of mass m is r, and it is convenient to pick cylindrical polar coordinates for this position vector: r = rer + zE3. Representations for the velocity and acceleration vectors in terms of cylindrical polar coordinates were established earlier and we do not rewrite them here.

Equations of Motion

The equations of motion for the particle can be obtained from F due to Newton’s gravitational force: Fn = −

GMm 3

² r²

=

ma, where F is solely

r.

You should recall that this force is conservative and its potential energy is denoted by Un . Using (2.3), we can write out the component forms of F n = ma. The result will be three differential equations: ²

m r¨ − rθ˙ 2 (

³

GMmr

= −² √

)

m rθ¨ + 2˙rθ˙

r2 + z2

=

³3 ,

0,

9 The reduction procedure we use is equivalent to the so-called Routhian or Lagrangian reduction procedure

that is used to incorporate momentum conservation in a variety of mechanical systems ranging from the problem at hand to Lagrange and Poisson tops. For further details on this procedure, the reader is referred to [90], [144, Chapter 2], and [180].

60

Kinetics of a Particle

GMmz m¨z = − ²√ ³3 . r2 + z2

(2.7)

For a given set of six initial conditions, 10 these equations provide r(t), θ (t), and z(t), and hence can be used to predict the position of the particle of mass m.

Conservations

The solutions of differential equations (2.7) conserve two important kinematical quantities. First, they conserve the total energy E = T +U, where U = Un . Second, the angular momentum HO of the particle is conserved. It is left to the reader to demonstrate these results by using the work–energy and angular momentum theorems. The conservation of angular momentum implies that r (t ) × v (t ) = r (t0) × v (t0) . Now the initial position r (t0 ) and velocity v (t 0) vectors define a plane, in general, and thus the motion of the particle remains on this plane (which is known as the orbital plane). We also observe that the normal to this plane is parallel to HO . If we allow ˆ ourselves the freedom to choose E 3, then we can pick this vector such that HO = hE 3, 2 where hˆ = mr θ˙. Consequently, r and v will have components only in the E1 and E2 directions: z(t) = 0 and z˙(t) = 0. We henceforth exploit the fact that angular momentum is conserved and the motion is planar. Because HO and the areal velocity vector are synonymous, angular momentum conservation for this problem is often known as the “integrals of area.”11 We have tacitly ignored the case in which r (t0 ) ² v (t0 ). In this case, HO is zero and must remain so. Consequently, the motion of the particle is a straight line. For convenience, we can choose this line to lie on the E1 − E2 plane. It can be shown that the motion of the particle will eventually lead to a collision with the particle of mass M at the origin. Thus, without an initial angular momentum, a collision for this system would be unavoidable.

Determining the Motion of the Particle

Because the angular momentum is conserved, we choose E3 such that z(t) = 0 and z˙(t) = 0 for the particle. That is, the direction of HO is E3. As a result, the equations of motion reduce to ²

³

GMm , r2 ( ) m rθ¨ + 2˙rθ˙ = 0.

m r¨ − r θ˙ 2

= −

10 The six initial conditions needed are r ( t ) , θ (t ) , z (t ) , r˙ ( t ) , θ˙ (t ) , and z˙ ( t ) . 0 0 0 0 0 0 11 See, for example, [199, Section 86].

(2.8)

2.8

61

Dynamics of a Particle in a Gravitational Field

Now the second of these equations can be expressed as d ² 2 ³ mr θ˙ dt

=

0.

(2.9)

This equation is equivalent to the conservation of HO · E3. Using this conservation, we can eliminate θ˙ from (2.8) and arrive at a single governing differential equation: m¨r −

hˆ 2 mr3

= −

GMm . r2

(2.10)

Here, hˆ is determined from the initial position and velocity of the particle: hˆ = HO · E3

= ( mr ( t0 ) ×

v (t0 )) · E3 .

For a particle with no angular momentum, we see from (2.9) that θ˙ = 0 and hence r¨ = − GM . It is not difficult to see that this equation implies that r(t) → 0 as t increases. r2 This event corresponds to the collision we discussed earlier. ˆ and then integrate (2.10) to determine Given r (t0 ) and v (t0 ), we can determine h, 12 r(t). We can then compute the coordinate θ (t) by integrating θ˙ =

hˆ . mr 2

(2.11)

Finally, after finding expressions for x(t) = r(t) cos (θ (t)) and y(t) = r(t) sin (θ (t)), the orbit of the particle can be constructed. The easiest solution of (2.10) to compute is the one for which r is constant: r(t) = r0 and ˙r(t) = 0. In this case, (2.10) is satisfied provided r0

=

hˆ 2 . GMm2

We also show, using (2.11), that θ˙ is constant: θ˙ (t) = ωK =

hˆ mr20

(2.12) º

=

GM r30

.

(2.13)

The frequency ωK is known as the Kepler frequency. Physically, the particle is moving in a circular orbit of radius r0 at constant speed r0ωK about the fixed body of mass M. In numerically integrating (2.10), the time scale of the integration is very long and it is convenient to nondimensionalize the equations of motion. To do this, we choose the dimensionless variable w = rr0 and time τ = ωK t. Now using identities of the form

r˙

dr

dτ dr

dr

= dt = dt dτ = ωK dτ ,

we can simplify (2.10) and (2.11) to d 2w dτ 2

=

1 w3

−

1 , w2

dθ dτ

=

1 . w2

(2.14)

12 The differential equation (2.10) has an analytical solution that can be expressed in terms of Jacobi’s

elliptic functions. However, this is beyond our scope here, and the reader is referred to Whittaker [306] for details on how such an integration can be performed.

62

Kinetics of a Particle

2

h

h 1

p c

dw dτ

e

e

e

−1 −2

1

w

4

The phase portrait of (2.14)1. The trajectories labeled e and h correspond to elliptical and hyperbolic orbits of the particle, the point c corresponds to a circular orbit, and the trajectory labeled p corresponds to the parabolic orbit. The arrows in this figure correspond to the directions of increasing τ .

Figure 2.7

Notice that we have reduced the problem of determining the motion of the particle to the integration of two differential equations. Differential equation (2.14)1 is a second-order differential equation for w(τ ). As opposed to exhaustive displays of w(τ ) for several sets of initial conditions, a qualitative method of representing the solutions of (2.14)1 is to construct what is known as 13 a phase portrait. In this portrait, dw dτ is plotted as a function of w(τ ). Equilibria of the second-order differential equation correspond to points in the phase portrait where dw d2 w dw d τ = 0 and w( τ ) is a constant. To find such points, we set dτ = 0 and dτ 2 = 0 in the governing second-order differential equation for w(τ ) and solve for the resulting constant values of w(τ ). Later examples in this chapter will examine differential equations with multiple equilibria. Returning to the problem at hand, the²phase³portrait of (2.14) 1 is shown in Figure 2.7. There we see an equilibrium point at w, ddwτ = (1, 0 ) that corresponds to a circular orbit of the particle. The closed orbits enclosing this point represent elliptical orbits of the form shown in Figure 2.8.14 The remaining orbits shown in this figure correspond to hyperbolic orbits of the particle. For these orbits, the particle circles around the equilibrium once and never returns. The interesting case in which the particle describes a parabolic orbit is also shown in this figure. The trajectory on the phase portrait corresponding to this orbit separates the elliptical and hyperbolic trajectories.15 13 Phase portraits are a standard method for the graphical representation of solutions to ordinary differential

equations (see, e.g., [12, 26, 108, 307]).

14 A second integration involving d θ = 1 is needed to construct these orbits, and is left as an exercise. dτ w2 15 In the parlance of dynamical systems theory, the homoclinic orbit that passes through the point

²

³

²

w, ddwτ = (0.5, 0 ) connects the fixed point at w, ddwτ of the particle.

³

= ( ∞, 0 ),

and corresponds to the parabolic orbit

2.8

y

(a)

=

63

Dynamics of a Particle in a Gravitational Field

x2

y

(b)

= x2

r θ x

O

=

x1

y

(c)

x

O

= x2

= x1

ˆh < 0

r θ

θp x

O

= x1

ˆh < 0 (d) y

=

(e) y

x2

=

x2

r

ˆ h > 0

r

x θp

=

x1

θp

O

x

O

=

x1

ˆh > 0

Schematic of the four types of orbits of a particle of mass m moving about a fixed point O: (a) line; (b) circular orbit (e = 0); (c) elliptical orbit (0 < e < 1); (d) parabolic orbit (e = 1); and (e) hyperbolic orbit (e > 1). For each of the orbits (c)–(e), distinct values of θp are considered. Figure 2.8

The Orbital Motions

An alternative approach to the one outlined in the previous section is followed in most textbooks on dynamics. This approach involves solving for the motion of the particle as a function of θ rather than of time t. For this approach, we first use the chain rule and (2.8)2 in the form (2.11) to show that hˆ d r˙ = − m dθ

¸ ¹

1 r

,

hˆ 2 d2 r¨ = − 2 2 2 m r dθ

¸ ¹

1 . r

64

Kinetics of a Particle

Using the second of these results to rewrite (2.8)1 , we find the differential equation d2 dθ 2

¸ ¹

1 r

+

1 r

1 , r0

=

(2.15)

where r0 was defined earlier (see (2.12)). Equation (2.15) is a linear ordinary differential equation for has the exact solution r where e and

θp

=

(

r (θ ) = r0 1 + e cos

(

θ − θp

)) −1

1 r

,

as a function of θ that (2.16)

are constants (which are determined from the initial conditions for the 2

position and velocity of the particle). We can then integrate mr ˆ(θ ) dθ = dt to determine h an analytical expression for θ (t). This is left as an exercise. Solution (2.16) represents a conic section and from the theory of conic sections, it is known that, when e = 0, the orbit (r(θ )) is circular; when 0 < e < 1, the orbit is elliptical; when e = 1, the orbit is parabolic; and when e > 1, the orbit is hyperbolic. Referring to Figure 2.6, for the hˆ 2 elliptical orbit, it is easy to see that b = GMm 2. To solve for e and θp, let us assume that r (t0 ) and v (t0 ) are given. Then we can ˆ To compute θp , we first calculate ˙r by using (2.16) determine HO and specify E3 and h. and the chain rule: ¸ ¹ ( ) GMm ˙r(t) = − e sin θ (t) − θp . (2.17) hˆ We also compute the value E0 of total energy of the particle: E0 =

GMm m v (t0 ) · v (t0 ) − . 2 ² r ( t0) ²

(2.18)

Now as the total energy is conserved, the value of this kinematical quantity when θ = θp is also equal to E0 . With some manipulations of (2.16) and (2.17), it can be shown that E0

=

G2 M 2 m3 ² 2hˆ 2

³

e2 − 1 .

(2.19)

We now have a method of determining e and θp from a given set of initial conditions. The procedure is to compute E0 by use of (2.18) and then use (2.19) to compute e ≥ 0. Once e is known, then (2.17) can be used to compute θp . With these values, r(θ ) is specified and θ (t) can be calculated. Depending on the value of e, the orbits will be one of four types: a circle, an ellipse, a parabola, and a hyperbola (see Figure 2.8). For completeness, we could also have nondimensionalized (2.15): d 2u dθ 2

+

u = 1,

(2.20)

where u = w1 = rr0 . The phase portrait of this equation is shown in Figure 2.9. In contrast to the earlier phase portrait, here the trajectory corresponding to the parabolic orbit is easily distinguished. However, as in the previous case, to gain a physical interpretation of the trajectories shown in Figure 2.9, it is necessary to reconstruct a position vector of the particle corresponding to the particular orbit.

2.9

2

h

h

h

h

1 c

du dθ

65

Dynamics of a Particle on a Spinning Cone

e

e

p

−1 −2

1

u

4

The phase portrait of (2.20). The trajectories labeled e and h correspond to elliptical and hyperbolic orbits of the particle, the point c corresponds to a circular orbit, and the trajectory labeled p corresponds to the parabolic orbit. In this figure, the arrow indicates the direction of increasing θ . For the parabolic trajectory, this angle ranges from −π → π .

Figure 2.9

Comments

For this problem, the rare event occurs that a complete classification of the motions of the particle is possible. For many of the problems that are discussed later on in this book, this classification has not been performed and indeed may not be possible. As a result, our previous discussion will be a benchmark. We have not exhausted the literature on this problem, and discussions of related problems involving escape velocities and transfer orbits can be found in several textbooks (see, e.g., [20]). Generalizations of the problem also abound, and we shall discuss two of them at later stages in this book. Before we leave the problem for now, we wish to show that the elliptical (and circular) orbits are in agreement with Kepler’s laws. We first note that satisfaction of the first law is trivial, and the second law is a consequence of angular momentum conservation. To see that the third law is satisfied, we need to compute the period T of the particle executing an elliptical orbit. We leave it as an exercise to show 3 that T = √2π a 2 , and consequently the third law is satisfied. GM

2.9

Dynamics of a Particle on a Spinning Cone

As shown in Figure 2.10, a particle of mass m moves on the surface of a cone. It is attached to the fixed apex O of the cone by a linear spring of stiffness K and unstretched length ±0. We assume that the surface of the cone is rough and that the cone is spinning about its axis of symmetry with an angular speed ´0 . Our goal is to establish the equations of motion for the particle and discuss some aspects of its dynamics.

66

Kinetics of a Particle

Ω0

(a)

g

(b)

m

m

z

Rough cone α

r

O

O

(a) Schematic of a particle of mass m which is attached to a fixed point O by an elastic spring. A vertical gravitational force −mgE3 acts on the particle and the particle is free to move on the rough inner surface of a cone which is rotating about the vertical z axis with a constant speed ´0 . (b) Representative motion of the particle on a smooth cone.

Figure 2.10

Coordinates, Constraints, and Velocities

As discussed in Section 1.9, when the particle is moving on the surface of the cone, it is subject to a single constraint ³ = 0. This constraint can conveniently be expressed by use of a spherical polar coordinate system: π

³ = φ +α −

2

.

For future reference, we note that the gradient of ³ is ∇³ = R1 eφ . Because the cone is rotating with a speed ´0 , the velocity of the particle relative to the cone is vrel

˙ R+ = Re

R cos(α)

(

θ˙ − ´0

)

eθ .

We defer discussion of the case where the particle is stuck to the cone. Forces

The particle is under the influence of a gravitational force −mgE 3 and a spring force Fs

= −K ( R − ±0) eR ,

where K is the stiffness of the spring and ±0 is its unstretched length. Assuming that the particle is moving relative to the surface of the cone, we find that the constraint force F c acting on the particle has the representation Fc

=

N + Ff ,

where the normal force N is parallel to ∇ ³ : N

=

λ

R

eφ ,

Ff

= −µ d ² N²

Thus the total force on the particle is F = Fc

+

vrel . ²v rel ²

Fs − mgE3.

2.9

67

Dynamics of a Particle on a Spinning Cone

The Equations of Motion

To obtain the equations of motion, we express F = ma in spherical polar coordinates (see (2.4)) and impose the constraint ³ = 0 to find ²

m R¨ − R cos2(α )θ˙ 2

³

= −K (R − ±0 ) +

(

m R cos(α)θ¨ + 2R˙ θ˙ cos(α) 2

−mR sin(α ) cos(α )θ˙

=

λ

)

+

R

F f · eR − mg sin(α ),

=

Ff · eθ ,

mg cos(α ).

(2.21)

The first two of these equations are ordinary differential equations for R and θ , and the third equation can be solved for λ (and hence the normal force) as a function of the motion of the particle. To integrate these equations, it is desirable to nondimensionalize them. We use ±0 as ´ a measure of length and

±0

g

as a measure of time: » τ =

t

g ±0

d dτ

¸

¸

w2 cos2 (α)

dθ dτ

R ±0

. ´

d τ dR

= dt = dt dτ =

¹2

dθ = w cos (α ) dτ 2

w= dR

With the help of identities of the type R˙ as d2 w dτ 2

,

− ω

2

g dR dτ ,

±0

we can rewrite (2.21) dw

(w − 1) − sin(α ) − µ k n dτ , u

¹ = −µk n ( w cos(α ) )

w

²

dθ − ω0 dτ

u

³

.

(2.22)

In (2.22), the constants ω and ω0 and dimensionless normal force n and speed u are ω

º

u=

¸

w2

dθ dτ

2

=

¹2 − ω0

¸ +

º

K±0 , mg dw dτ

¹2

ω0 = ´

,

n=

±0

g

² N²

mg

,

=

cos(α) +

h2 tan(α). w

We can also show that the dimensionless versions of total energy E and angular momentum HO · E3 are E mg±0

=

1 2

¼¸

dw dτ

¹2

¸ +

h=

dθ dτ

¹2

½

w2 cos2(α )

HO · E3 ´

m gL30

=

+

w 2 cos2( α)

ω2

2

2 (w − 1) + w sin(α),

dθ . dτ

Notice that, by nondimensionalizing the equations of motion, we have reduced the number of parameters by two.

68

Kinetics of a Particle

The Static Friction Case

When the particle is stuck to the cone, its velocity vector is v = R 0´0 cos(α )eθ . In addition, the particle is subject to three constraints and the friction and normal forces constitute three undetermined forces that enforce these constraints: Fc

=

λ

R0

eφ +

λ1

R0 cos(α)

To determine λ , λ 1, and λ2 , we examine F that Ff

² =

=

eθ

+ λ2eR .

ma. With some manipulations, we conclude ³

˙ e , K (R0 − ± 0) + mg sin(α ) − mR 0 cos2 (α)´20 eR + mR 0 cos(α )´ 0 θ

²

³

N = − mg cos(α ) + mR 0´20 sin(α) cos(α) eφ . Such a state of the particle is sustained provided sufficient friction is¶ present, and to ¶ check this sufficiency we need to examine the static friction criterion ¶Ff ¶ ≤ µ s ²N². If this criterion holds, then a particle that is stuck on the surface of the cone will remain stuck on the surface. Otherwise it slips, and the initial direction in which it slips is parallel to Ff .16

The Smooth Cone

When the particle is moving on a smooth cone, we can simplify (2.22) considerably. Indeed, as in the earlier particle problem, we can exploit the conservation of angular momentum to write a single equation for w: d 2w dτ 2

=

h2 w3 cos2 (α)

− ω

2

(w − 1) − sin(α ).

Integrating (2.23), we can find w( τ ). Another integration using w2 cos 2(α ) ddτθ ²

provides θ (τ ). The equilibrium point at w, of h2 w30 cos 2(α)

−ω

2

dθ dτ

³

= ( w0 , 0),

(2.23) =

h

where w0 is the solution

(w0 − 1) − sin(α ) = 0,

corresponds to a circular orbit of the particle that has radius r = ±0 w0 cos( α). Some of the other trajectories of w in the w − ddwτ plane are shown in Figure 2.11. In this figure, we have also constructed possible trajectories of the particle corresponding to w(τ ). Unlike the problem of the particle subject to F n that we discussed in Section 2.8, here the classifications of the trajectories for the particle on the cone defy a simple classification. 16 The initial slip direction must be specified in order that the initial motion of the particle slipping on the cone can be determined. The prescription of the initial slip direction allows one to specify ²vvrel ² even rel though vrel = 0.

2.10

A Shocking Constraint

69

x2

x2

x1

x1

20

1 dR 0

dτ

1

8

R 0

x2

−20

x1

x2

x1

The phase portrait of (2.23) and the corresponding planar projections of the trajectories of the particle. For this figure, we assumed that α = 20◦, h = 5, and ω 2 = 10. Consequently, the equilibrium point corresponding to a circular trajectory of the particle has the coordinates (w0, 0) = (1.58896, 0).

Figure 2.11

2.10

A Shocking Constraint

We now return to the constraint y˙ x = 0, discussed earlier (see (1.21)). Our interest is to determine the equations of motion of a particle that is subject to this constraint and that is also under the influence of an applied force Fa = P 1E1 + P 2E2 . First, we assume that the constraint force which enforces y˙ x = 0 has the standard prescription Fc

= λxE2 ,

(2.24)

where λ is a Lagrange multiplier. We note that, when x = 0, Fc = 0. From a balance of linear momentum, we find that the equations of motion for the particle are xy˙ = 0, m¨x = P1 , my¨ = P2 + λ x, m¨z = 0.

(2.25)

The equation for motion in the z direction is trivial to integrate and interpret and, for convenience, we henceforth ignore this direction and assume that the motion is planar. When some modest restrictions are imposed on P1 and P2 , governing equations (2.25)1,2,3 have exact solutions that are easy to establish provided the motion is rectilinear: y(t)

=

y0,

x(t) = x0 + x˙ 0 t +

± t± 0

τ

0

P1 dudτ , m

λ=−

P2 x(t)

(2.26)

70

Kinetics of a Particle

(a)

y

(b)

= x2

y

Ic

=

x2

2

x

=

x1

B

A

B

x

= x1

A

Ic

1

Two possible motions of a particle subject to a constraint yx ˙ = 0. In (a) the particle moves from A to B and there is no impulse Ic when x = 0, whereas in (b) the particle experiences two instances where v is not continuous. Figure 2.12

and x = 0,

y(t) = y0 + y˙ 0 t +

± t± 0

0

τ

P2 dudτ . m

(2.27)

Here x0 = x(0), x˙ 0 = x˙(0), y0 = y(0), and ˙y0 = y˙ (0) are initial conditions. From these solutions to the equations of motion, we observe that λ is not defined when x = 0. The Shock

Referring to Figure 1.17, we recall that this constraint has the unusual feature that it can be decomposed into two piecewise integrable constraints: y˙ = 0 when x ± = 0,

and x = 0.

At the points where the particle makes a transition from one of these integrable constraints to the other, its velocity vector v can be discontinuous and therefore its acceleration vector may not be defined. At such a transition, the prescription (2.24) for F c does not hold and instead we can calculate only the impulse Ic that is due to this force by using (2.1). Supposing that the transition occurs at time t = T , we will have ¸

Ic

=

lim

σ →0

mv (T + σ ) − mv (T − σ ) −

± T +σ T −σ

¹

Fadτ

.

For example, to achieve a motion that goes from point A to point B in Figure 2.12(b), the particle needs to perform a motion for which it will possess a discontinuous velocity vector in at least two locations. That is, the particle will experience a shock. Impulses Ic1 and Ic2 shown in the figure enable these shocks. This is in contrast to the situation shown in Figure 2.12(a), where there is no discontinuity in the motion of the particle. That is, the shock is absent and consequently Ic = 0. If the constraint cannot supply the impulse Ic , then the particle is effectively subject to a single holonomic constraint. This constraint is either y = y0 or x = 0, depending on the initial position and velocity of the particle. It is left as an exercise for the reader to imagine a rigid wall placed to the left of the y axis in Figure 1.17 as a method of realizing the constraint y˙ x = 0.

2.11

2.11

71

A Simple Model for a Roller Coaster

A Simple Model for a Roller Coaster

Imagine being in a cart at the top of a roller coaster. If there is no friction, then the slightest nudge will set the cart in motion. The presence of Coulomb friction with stick– slip changes this scenario. It will eventually bring the cart to a halt, and it may bring the cart to a halt near the top of the roller coaster. Indeed, if there is sufficient static friction, then the cart can come to a halt at any location on the track, and the chief quantity that governs how fast the halting occurs in this extreme case will be the dynamic friction coefficient. Here, a very simple model is presented for the roller coaster that captures its stick–slip behavior.17 First, we establish a differential equation governing the motion of the roller coaster, and then we use numerical integrations to investigate the dynamics of the roller coaster. The Equations of Motion

One model for the dynamics of a cart on a roller coaster is to model the cart as a particle of mass m that is moving on a fixed plane curve: y = f (x1), z = 0. That is, the particle is subject to two constraints, ³1 = 0 and ³2 = 0, where ³1 =

y − f ( x1) ,

³2 =

z.

These constraints will be enforced by Fc = N + Ff . It is a standard exercise to calculate the unit tangent et , the unit normal en, and the binormal eb vectors to this curve [215]: et

=

´

1

1 + f ³2

(

)

E1 + f E 2 , ³

( )

en

=

sgn f ³³ ( ) E2 − f ³ E1 , ´ 1 + f ³2

eb

=

et

× en,

where the prime denotes the derivative with respect to x1 . These three vectors constitute the Frenet triad, and in calculating this triad we assume that x˙ 1 > 0. A normal force

E2

A

y

E1

x1

B

Figure 2.13

Schematic of a particle on a cosinusoidal path.

17 The work presented on this model was performed in collaboration with Henry Lopez [172].

C

72

Kinetics of a Particle

N = Nen + λ2 E3, a friction force Ff et , and a vertical gravitational force −mgE2 act on the cart (see Figure 2.13). Now, for a particle moving on a curve with velocity v = vet , the acceleration vector of the particle is a = v˙ et + κ v2en , where κ is the curvature of the space curve: κ =

¸´

µ µ µf ³³ µ

1+f

³2

¹3 .

Taking the et and en components of F = ma, we can easily calculate the equations governing the motion of the cart and the normal force: m

²²

1+f

³2

³

x¨ 1 + f f

³³ ³

x˙ 21

³

´ = −mgf − µd ³

1 + f ³ 2 ² N²

x˙ 1 |˙ x1|

,

(2.28)

where N= ´ =

´

²

1 1+f mg

³2

1+f

³2

(

³

µ µ

)

sgn f ³³ mg + µf ³³ µ mx˙21 en,

en,

when f ³³

0.

=

if f ³³

±=

0, (2.29)

These equations apply when the cart is moving and the friction is dynamic. In the event that the cart is stationary, static friction acts and, provided the static friction criterion is satisfied: µ µ µ µ µ mgf ³ µ µ´ µ µ µ µ 1 + f ³2 µ

≤ µ s ²N² ,

(2.30)

the cart remains stationary. With the help of (2.29), (2.30) can be expressed in the simple form µ µ µf ³ µ ≤ µ . s

(2.31)

This equation can be viewed as the basis for the classical experiment to measure the coefficient of static friction: we place a block on an inclined plane and slowly increase the angle of inclination until slipping occurs. The tangent of the angle of inclination is equal to µ s . We shall shortly use (2.31) to establish a continuum of points at which the roller coaster can remain in a state of rest. We now choose a cosinusoidal track ¸

f (x1) = A cos

π x1 ±0

¹

.

(2.32)

2.11

A Simple Model for a Roller Coaster

73

The phase portrait of (2.28)–(2.30) when friction is absent. The points on the x = ±x 1 0 axis labeled by · correspond to equilibria of the cart. For the results shown in this figure, A = 0.25 and µ s = µ d = 0.0. ±

Figure 2.14

0

In addition, we employ the following nondimensionalizations: x=

x1 ±0

»

,

τ =

g ±0

t.

Of course, other tracks are possible, and the reader is referred to Shaw and Haddow [258], where other interesting choices of f (x) can be found. A further interesting choice would be Euler’s spiral (clothoid) that features in “loop-the-loop” roller coasters.

States of Rest

For a cart on a smooth roller coaster, the motion of the cart will be perpetual, and a portion of its phase portrait is shown in Figure 2.14. We note the presence of an equi) ( librium at x = 0, ddxτ = 0 . This point corresponds to the cart’s being stationary at the top of the roller coaster. Referring( to Figure 2.15(a),) we refer to equilibria of this type as saddles. The two equilibria at x = ´1, ddxτ = 0 represent a stationary cart at the bottom of one of the valleys of the roller coaster. Examining the phase portrait in Figure 2.15(c), we easily see why equilibria of this type are known as centers. The equilibria at ( ) x = (−2, 0, 2), ddxτ = 0 correspond to a stationary cart at one of the crests of the roller coaster. When stick–slip friction is present, the phase portrait changes dramatically (see Figure 2.16). First, all of the saddles have split, and between their two split halves we have what we call a sticking region (see Figure 2.15(b)). Depending on the value of x1, this region contains either x˙ 1 > 0 or x˙ 1 < 0. If the cart’s state enters this region, then the cart will stop. That is, the cart will come to rest near a crest of the roller coaster. Similarly, the equilibria of the smooth roller coaster at the floor of its valleys have now transformed from discrete points to regions surrounding these points (see Figure 2.15(d)).

74

Kinetics of a Particle

dx dτ

(a)

(b)

0.2

0. 2

−0 2

x

0. 2

.

−0 2

0. 2

s

.

x

−0 2

−0 2

.

.

dx dτ

(c)

dx dτ

(d)

0.2

dx dτ

0.2

0. 8

1.2

x

0 .8

−0 2

1. 2

s

x

−0 2

.

.

Expanded views of the phase portraits in the neighborhood of equilibria of (2.28) with f specified by (2.32). For (a) and (c), µ d = 0 and the roller coaster is smooth, whereas for (b) and (d), µd = 0.1. For the latter cases, the sticking regions s are shown for the case in which µs = 0.3.

Figure 2.15

dx dτ

1. 5

−2

2

x

−1 5 .

The phase portrait of (2.28)–(2.30). Although it is not evident from the figure, the discrete equilibria of the frictionless case shown in Figure 2.14 are now replaced with families of equilibria that correspond to possible resting (sticking) states for the cart. For the results shown in this figure, ±A = 0.25 and µ d = 0.1.

Figure 2.16

0

2.11

A Simple Model for a Roller Coaster

75

f

µs

= 0 .3

0.2

−2

2

s

s

s

s

−2 s

2

s

s

x

= x01

x

= x01

s

s

Figure 2.17 A graphical method to compute the possible sticking regions s of the cart on a smooth roller coaster when µ = 0.3. The method is based on examining (2.31) for the choice (2.32). µ µ µµ Aπ s ² π x ³µµ ³µ µ That is, f = µ ± sin ± 1 µ . Examples of the sticking regions can be seen in Figure 2.15.

0

0

These regions are also sticking regions, and if the cart’s state enters this region, then the cart will stop. The size of the sticking region is easy to compute by use of (2.31) and a graphical method is shown in Figure 2.17. As µ s gets larger, the size of the sticking region (or sticking states) surrounding the equilibria when µd = 0 grows, and eventually any point (x, 0) on the ddxτ axis will become an equilibrium, and thus a state of rest for the roller coaster.18 This phenomenon is also easy to explain physically. We note for completeness that, for the present choice of f (x), if µs ≥

πA ±0

,

then it is possible to stick at any location on the roller coaster. This can also be inferred from the graphical technique shown in Figure 2.17. 18 For the phase portrait shown in Figure 2.15(b), the sticking region s is the interval [−0.124755, 0.124755]

and for the situation shown in Figure 2.15(d), the sticking region s is the interval [1. − 0.124755, 1.124755].

76

Kinetics of a Particle

2.12

Closing Comments

A vast amount of material has been covered in this chapter, starting with descriptions of various forces, discussions of the balance laws, and analyses of various applications. The analyses we employed invariably featured the numerical integration of an ordinary differential equation and an interpretation of its solutions. Developing physical interpretations of the results provided by the model is one of the most rewarding aspects of dynamics; however, it can also be the most time consuming. In many of the chapters to follow, several more examples of such interpretations are presented and you are strongly encouraged to take the time to do this when completing the exercises in this book or performing your own research.

2.13

Exercises

Exercise 2.1: Consider the following force fields: P = x1 E1 + x3E2 , P = x2 E1 + x1E2 , P = x1 x2 E1 , P = −±0 sin(θ )E 1 + ±0 cos(θ )E2, where ±0 is a constant. Determine which of the force fields are nonconservative and which are conservative. For the conservative force fields, what are the associated potential energies? Exercise 2.2: Consider a particle of mass m that is moving in E 3. Suppose the only forces acting on the particle are conservative. Starting from the work–energy theorem, prove that the total energy E of the particle is conserved. Suppose, during a motion for which the initial conditions r0 and v0 are known, the position r (t1) at some later time t1 is known. Argue that the conservation of energy can be used to determine the speed ²v² of the particle. Provide three distinct physical examples where this result can be applied. Exercise 2.3: In contrast to Exercise 2.2, consider a particle that is moving on a smooth fixed surface. The constraint force acting on the particle is prescribed by use of Lagrange’s prescription, and the applied forces acting on the particle are conservative. Prove that E is again conserved. In addition, show that the speed of the particle can be determined at a known position r(t1 ) if the initial position and velocity vectors are known. Finally, provide three distinct physical examples where this result can be applied. Exercise 2.4: A particle is free to move on a smooth horizontal surface x3 = 0. At the same time, a rough plane propels the particle in the E1 direction. That is, the constraints on the motion of the particle are ³1 = 0 and ³2 = 0, where

2.13

³1 = ³1 (r) = x3,

Exercises

77

³2 = ³2 (r, t) = x1 − f (t).

Give a prescription for the constraint force acting on the particle. Exercise 2.5: Suppose a particle of mass m is in motion and has a position vector r and a velocity vector v. (a) Show that the areal velocity vector A is conserved if the resultant force F acting on the particle is a central force.19 (b) Show that conservation of angular momentum HO is synonymous with conservation of the areal velocity vector. (c) Suppose a particle is moving on a horizontal table under the action of a spring force, a normal force, and a vertical gravitational force −mgE 3. The coordinate system is chosen so that the table corresponds to the surface x3 = 0. One end of the spring is attached to the fixed origin O and the other end of the spring is attached to the particle. The spring has a stiffness K and unstretched length ±0 . What can you say about the area swept out by the particle in a given period of time? (d) Derive the equations of motion for the particle in (c). Using the nondimensionalizations »

τ =

K t, m

x=

r ±0

,

and conservation of angular momentum, show that the motion of the particle can be found by integrating the following differential equations: d2 x dτ 2

−

2 β = − (x − 1) , x3

dθ dτ

=

β

x2

,

(2.33)

where β =

h

±2 0

√

Km

and h is a constant that depends on the initial conditions of the motion. For a selection of values of β (e.g., β = −20, −2, −1, 0, 1, 2, and 20), construct the phase portraits of (2.33) 1. For a selection of the orbits on each of these phase portraits, construct images of the motion of the particle.20 (e) Verify that the areal velocity and the dimensionless total energy e=

E K ±20

=

1 2

¼¸

∂x ∂τ

¹2 +

x

2

¸

∂θ ∂τ

¹2

½ + (x −

1)

2

are conserved for the motions of the particle you found in (d). 19 A force P is said to be central if P is parallel to r. 20 Some of your results will be qualitatively similar to those presented in Section 2.9 for a particle moving

on a smooth cone.

78

Kinetics of a Particle

Exercise 2.6: A particle of mass m is in motion about a fixed planet of mass M. The external force acting on the body is assumed to be a conservative force P. The potential energy UP associated with this force is a function of || r|| , where r is the position vector of the particle relative to the fixed center O of the planet. (a) Prove that r is parallel to P. (b) Show that the angular momentum HO of the particle is conserved and that this conservation implies that the motion of the particle is planar. This plane, which is known as the orbital plane, also contains O. Show that the particle sweeps out equal areas on its orbital plane in equal times. (c) Write out the equations of motion of the particle using a spherical polar coordinate system. (d) Using the conservation of HO , show that the equations of (c) can be simplified to ²

m r¨ − rθ˙ 2

³ = −

mr 2θ˙

=

∂ UP ∂r

,

h,

(2.34)

where h is a constant. (e) Show that the solutions to (2.34) conserve the total energy E of the particle. Exercise 2.7: A particle of mass m is free to move on the inner surface of a rough sphere of constant radius R0 . The center of the sphere is located at the origin O, and the particle is attached to a fixed point A, whose position vector is aE1 + bE2 , by a linear spring of unstretched length ±0 and stiffness K. A vertical gravitational force −mgE 3 also acts on the particle. (a) Using a spherical polar coordinate system r = R0 eR , derive expressions for the acceleration vector a and angular momentum HO of the particle. (b) What is the velocity vector of the particle relative to a point on the surface of the sphere? (c) Give a prescription for the constraint force Fc acting on the particle. (d) If the particle is moving relative to the surface, show that the equations governing the motion of the particle are mR0(φ¨ − sin(φ ) cos(φ) θ˙ 2)

=

mg sin(φ) − K (²x² − ±0) −µd ² N²

d ² 2 2 ˙³ 1 mR0 sin (φ )θ R 0 sin(φ ) dt

˙ φ ˙ 2 + sin φ

= −K ( ²x ² − ±0 ) −µd ² N²

where x = R 0eR − aE1 − bE2 .

´

´

x · eφ ²x ²

2

(φ)θ˙2

,

x · eθ ²x ² sin(φ )θ˙

˙2 + φ

sin

2

(φ)θ˙2

,

2.13

79

Exercises

(e) Show that the normal force exerted by the surface on the particle is ¸

¹

N = mgE 3 · eR + K (²x² − ±0 )

x · eR eR − mR0 (φ˙ 2 + sin2 (φ)θ˙ 2)eR . ²x ²

(f) For the case in which the particle is not moving relative to the surface, show that Fc

=

mgE3 + K (²x² − ±0 )

x . ²x ²

What is the static friction criterion for this case? (g) Show that the total energy of the particle decreases with time if the particle moves relative to the surface. (h) If the spring is removed and the surface is assumed to be smooth, prove that the angular momentum HO · E 3 is conserved. Using this conservation, show that the dimensionless equations governing the motion of the particle simplify to 21 dθ dτ

=

h sin2(φ )

¸

In these equations, h =

1 mR20

´

d2 φ dτ 2

, R0 g

=

h2

¹

cos( φ)

+

sin3 (φ)

H0 · E3 and τ

´ =

sin(φ).

g R0 t.

(i) Suppose that the sphere is smooth. Using the fact that the total energy E of the particle is conserved, show that the criterion for the particle to remain on the outer surface of the sphere is mg (3 cos( φ) − 2 cos (φ0 )) − mR0

²

˙ 2 + sin2 (φ ) θ˙ 2 φ 0 0 0

³

>

0,

where φ0 is the value of the initial φ coordinate of the particle and θ˙0 and φ˙ 0 are the initial velocities. Suppose the particle is placed on top of the sphere. If the particle √ is given an initial speed v0 > gR0 , show that it will immediately lose contact with the sphere. Exercise 2.8: Consider a particle of mass m whose motion is subject to the following constraints: (xE3 + E2 ) · v =

0,

(E2 ) ·

v + e(t)

=

0.

(2.35)

(a) Show that one of the constraints is integrable whereas the other is nonintegrable. In addition, for the integrable constraint, specify the function ³(r, t) = 0. (b) Suppose that, in addition to the constraint force Fc

= µ 1 ( xE3 +

E2) + µ2 E2,

a gravitational force −mgE3 acts on the particle. With the help of the balance of linear momentum F = ma, specify the equations governing the motion of the particle and the constraint forces. 21 These differential equations are also the equations of motion for a spherical pendulum. The normal force

can be identified as the tension force in the cable connecting the particle to the fixed point O.

80

Kinetics of a Particle

(c) With the help of the work–energy theorem T˙ = F · v, prove that the total energy of the particle is not conserved. Give a physical interpretation for this lack of conservation. (d) Using the results from (b), determine the motion of the particle and the constraint force Fc .

3

Lagrange’s Equations of Motion for a Single Particle

3.1

Introduction

The balance of linear momentum F = ma for a particle can be traced to Newton in 1686. Euler subsequently expressed this balance law as a set of three ordinary differential equations. The equations that Euler presented in Berlin in 1750 [67, Section XXII] read as follows: 2Mddx = Pdt 2,

2Mddy = Qdt2,

2Mddz = Rdt2 ,

where P, Q, and R are the Cartesian components of F. In the centuries that followed, alternative principles of mechanics were proposed. Some of them, such as the principle of least action, also yielded equations of motion that were equivalent to those obtained with F = ma. Others did not, and the equivalence of, and interrelationships between, the principles of mechanics remain one of the central issues for any student of dynamics. At the end of the eighteenth century, a formulation of the equations of motion for a single particle appeared in a famous text by Lagrange [159].1 Among their attractive features, Lagrange’s equations of motion could easily accommodate integrable constraints, and they (remarkably) have the same canonical form both for single particles and systems of particles, as well as for systems of rigid bodies. In this chapter, Lagrange’s equations of motion for a single particle are discussed and several forms of these equations are established. For example (see (3.2)) d dt

±

∂L ∂q ˙i

²

−

∂L ∂ qi

=

Fncon · ai .

Many of the forms presented can be used with dynamic Coulomb friction and nonconservative forces. One of the most important features of our discussion is the emphasis on the equivalence of Lagrange’s equations of motion to F = ma. Although this equivalence is not discussed sufficiently in most textbooks, it can be found in several texts, such as those of Synge and Griffith [276] and Whittaker [306]. A recent paper by Casey [37] explores this equivalence in a transparent manner and we follow many aspects of his exposition in this chapter. Casey’s work will also appear when we discuss systems of particles and rigid bodies. 1 Four editions of Lagrange’s great work Mécanique Analytique appeared in the years 1789, 1811, 1853, and

1888. The last two of these editions were posthumous. An English translation of the second edition was recently published [160].

82

Lagrange’s Equations of Motion for a Single Particle

λ1

er

λ2

E3

E2

O

g

θ

mg

E1

E1 m

A planar pendulum where ± = ± (t) and the free-body diagram of the particle of mass m. Varying the length ± can be achieved either by a telescoping rod or a cabling system.

Figure 3.1

Two Approaches

Two approaches to Lagrange’s equations will be presented. The first, Approach I, starts with Lagrange’s equations for an unconstrained particle moving in E3 and then imposes the constraints and introduces the constraint forces. The resulting equations can then be used to determine the motion of the particle and the constraint forces. The second, which we call Approach II, is applicable when the constraints are integrable, each constraint is represented using a single coordinate, and the constraint forces are prescribed using Lagrange’s prescription. In this case, we use the constrained kinetic energy to calculate Lagrange’s equations of motion for the generalized coordinates. As an example, consider the planar pendulum shown in Figure 3.1. The length of the pendulum is assumed to be variable: ± = ±(t). We use cylindrical polar coordinates to describe the motion of the particle. The constraints on the particle and the associated constraint forces are ²1 =

r − ±(t),

²2 =

z,

Fc

= λ1 er + λ 2E3 .

Clearly, λ1 er is a tension force and λ2 E3 is a normal force. The equation of motion for the particle can be obtained from the constrained kinetic energy T˜ : m ³ 2 ˙ 2 ˙2 ´ T˜ = ± θ +± . 2

M

The configuration manifold is equivalent (or homeomorphic) to a circle S1. The equations of motion are found from a single equation: d dt

µ

˜ ∂T

∂ θ˙

¶

2 = m± θ˙

µ

−

∂ T˜ ∂θ

¶

=

0

= (F =

mgE1 + λ1 er + λ2 E3) · ±eθ .

Evaluating the derivative d /dt, the equation of motion for the generalized coordinate θ is found: ˙ θ˙ m±2 θ¨ + 2m±±

= −mg ± sin (θ ) .

When ± is constant, this differential equation simplifies to the classic case of the simple pendulum.

3.2

3.2

Lagrange’s Equations of Motion

83

Lagrange’s Equations of Motion

There are several approaches to deriving Lagrange’s equations of motion that appear in the literature. Among them, a variational principle known as Hamilton’s principle (or the principle of least action) is arguably the most popular, whereas an approach based on D’Alembert’s principle was used by Lagrange [159]. Lagrange’s original developments were in the context of mechanical systems subject to holonomic constraints, and his equations were subsequently extended to systems with nonholonomic constraints by Edward J. Routh (1831–1907) (see [246, Section 24, Chapter IV]) and Aurel Voss (1845–1931) [297].2 Here, an approach is used that is rooted in differential geometry and is contained in some texts on this subject (see, e.g., [277]). It probably migrated from there to the classic text by Synge and Griffith [276] and has recently been revived by Casey [37].

Two Identities

We assume that a curvilinear coordinate system has been chosen for vector v consequently has the representation v=

3 ·

E3 . The velocity

q˙ i ai.

i=1

In addition, the kinetic energy has the representations 3

3

T

=

m ·· m ai · ak q˙ i q˙ k . v·v= 2 2 i=1 k=1

(

)

It is crucial to notice that T = T q1 , q2 , q3 , q˙ 1 , q˙ 2 , q˙ 3 . We now consider in succession the partial derivatives of T with respect to the coordinates and their velocities. We wish to establish the following identities: ∂T = ∂ qi

∂T = ∂ q˙ i

mv · a˙ i,

mv · ai .

These two elegant results form the basis for Lagrange’s equations of motion. First, we start with the derivative of T with respect to a coordinate: ∂ ∂T = ∂ qi ∂ qi =

³m

mv ·

2

±

v·v

∂v ∂ qi

´

²

.

2 For further details on the historical development of Lagrange’s equations, see [224, 226]. Equations of

motion (3.9)2,3 are examples of what could be referred to as the Routh–Voss equations of motion.

84

Lagrange’s Equations of Motion for a Single Particle

∂T

To proceed with the goal of concluding that ∑3

k=1 q˙

ka

k

and that

k ∂q ˙ ∂ qi

=

µ

mv ·

=

mv ·

3 · k=1

mv ·

µ 3 · µ

k=1 3 ·

=

mv ·

=

mv · a˙ i .

∂ ak q˙ k i ∂q

2 ∂ r q˙ = ∂ qi ∂ qk

q˙ k

k=1

∂ ai

=

¶

k

¶

3 · k =1

The remaining steps use the fact that ak µ

mv · a˙ i , we first note that v

=

=

0. Consequently, ∂T = ∂ qi

∂T = ∂ qi

∂ qi

3 · k=1

∂r

∂ qk

∂ ak q˙ k i ∂q

¶

.

:

2 ∂ r q˙ = ∂ qk ∂ qi k

3 · k =1

∂ q˙ k k ∂q

±

∂r = ai ∂ qi

²¶

∂ qk

We achieve the last step by noting that f˙

) ( f q1, q2 , q3 .

=

∑3

∂f k ˙ k=1 ∂ qk q

for any function f

=

The next result, which is far easier to establish, involves the partial derivative of T with respect to a velocity. The reason this result is easier to establish is because the basis vectors ai do not depend on q˙ k . Getting on with the proof, we have ∂T = ∂q ˙i

µ

mv ·

3 · ∂q ˙k

k =1 µ 3 ·

=

mv ·

=

mv · ai .

k =1

∂q ˙i

¶

ak ¶

k δi a k

This completes the proofs of both identities. A Covariant Form of Lagrange’s Equations

It is crucial to note that Lagrange’s equations are equivalent to F = ma. The form of Lagrange’s equations of motion discussed here is derived from this balance law by taking its covariant components (i.e., dotting it with ai ). To start, we consider d dt

±

∂T ∂q ˙i

²

−

∂T ∂ qi

d (mv · a i ) − mv · a ˙i dt = ma · a i + mv · a ˙ i − mv · a ˙i

=

=

ma · ai

=

F · ai .

3.3

Equations of Motion for an Unconstrained Particle

85

In conclusion, we have a covariant form of Lagrange’s equations of motion: d dt

±

²

∂T

−

∂q ˙i

∂T ∂ qi

=

F · ai .

(3.1)

We can appreciate some of the beauty of this equation by using it to establish component forms of F = ma for various curvilinear coordinate systems.

The Lagrangian

Another form of Lagrange’s equations arises when we decompose the force F into its conservative and nonconservative parts: F = −∇ U + F ncon , where the potential energy U

=

U(q1, q2 , q3). As

∇U =

3 · ∂U k a, k=1

∂ qk

∂U

=

∂q ˙k

0,

we find that Lagrange’s equations can be rewritten in the form d dt

±

∂T ∂U − ∂q ˙i ∂ q˙ i

Introducing the Lagrangian L equations: d dt

=

±

²

±

−

∂T ∂U − ∂ qi ∂ qi

²

=

Fncon · ai .

T − U, we find an alternative form of Lagrange’s

∂L ∂q ˙i

² −

∂L = ∂ qi

Fncon · ai .

(3.2)

If there are no nonconservative forces acting on the particle, then the right-hand side of these equations vanishes. In addition, to calculate the equations of motion, a minimal amount of vector calculus is required – it is sufficient to calculate v and U.

3.3

Equations of Motion for an Unconstrained Particle

To illustrate the ease of computing equations of motion using Lagrange’s equations, we consider the case where the curvilinear coordinates chosen are the spherical polar coordinates: q1 = R, q2 = φ , and q3 = θ . For these coordinates: a1

=

eR ,

and T

=

a2

=

Reφ ,

a3

=

R sin(φ)eθ ,

´ m³ 2 R˙ + R2 sin2(φ )θ˙ 2 + R 2φ˙ 2 . 2

Notice that T does not depend on θ . If we evaluate the mass matrix m [aik ] associated with T , then we find that the determinant of this matrix is mR4 sin2 (φ). Thus, this system of coordinates for E3 has singularities along the entire x3 axis.

86

Lagrange’s Equations of Motion for a Single Particle

We obtain Lagrange’s equations of motion for the spherical polar coordinate system by first calculating the six partial derivatives of T: ∂T = ∂R

mR sin2(φ )θ˙ 2 + mR φ˙ 2 , ∂T

∂T = ∂θ

˙ ∂φ

=

∂T = ∂φ

mR 2 sin(φ ) cos(φ) θ˙ 2, ∂T

0,

˙ ∂R

∂T

mR2 φ˙ ,

=

∂ θ˙

=

˙ mR,

mR 2 sin2(φ )θ˙ .

Using these results, we find the covariant form of Lagrange’s equations: ±

²

²

±

d ∂T ∂T 2 2 ˙2 ˙ = mR sin (φ )θ˙ + mRφ = F · eR, = mR − dt ∂ R˙ ∂R ² ± ± ² ∂T d ∂T 2˙ 2 2 ˙ = mR φ − = mR sin(φ ) cos(φ)θ = F · Reφ , dt ∂ φ˙ ∂φ ² ± ± ² ∂T d ∂T 2 2 = mR sin (φ )θ˙ − = 0 = F · R sin(φ )eθ . dt ∂ θ˙ ∂θ (3.3) Clearly, these equations were far easier to calculate than an alternative approach that involves differentiating r = ReR twice with respect to t. d to find a canonical form of the We can expand the equations (3.3) by evaluating dt equations: ⎡

m 0 ⎣ 0 mR2 0 0 where

⎡

⎤⎡

⎤

⎡

⎤

⎤

⎡

0 R¨ F · eR C1 ⎦ ⎣ φ¨ ⎦ + ⎣ C2 ⎦ = ⎣ ⎦, 0 F · Reφ 2 2 ¨ mR sin (φ) F · R sin (φ) eθ C3 θ ⎤

C1 ⎣ C2 ⎦ C3

⎡

⎤

mR sin2 (φ) θ˙ 2 2 2 ⎦. ˙φ ˙ = ⎣ −mR sin(φ) cos(φ )θ˙ + 2mRR 2 2 ˙ ˙ ˙ 2mR sin(φ ) cos(φ)θ φ + 2mR sin (φ) R˙ θ ˙ −mRφ

2

(3.4)

−

Observe that the matrix on the left-hand side of (3.4) is none other than the mass matrix M = m [aik ]. As discussed in Exercise 3.2, the terms C1 , C2 , and C3 are quadratic in the velocities and can be attributed to the dependency of the mass matrix on the coordinates. That is, if we were to write F = ma using Cartesian coordinates, then the terms C1, C2 , and C3 would vanish. Let us now suppose that the only force acting on the particle is due to gravity: F

= −mgE 3,

U

=

mgE 3 · r = mgR cos(φ ).

For this case, the Lagrangian L is L = T −U ´ m³ 2 = R˙ + R2 sin2( φ)θ˙2 + R2φ˙ 2 − mgR cos(φ ). 2 We can calculate Lagrange’s equations of motion using L: d dt

±

∂L ∂ q˙ k

²

−

∂L = ∂ qk

0,

3.4

Lagrange’s Equations in the Presence of Constraints

87

or by substituting for F in (3.3). It is left as an exercise to show that both approaches are equivalent.

3.4

Lagrange’s Equations in the Presence of Constraints

The previous discussion of Lagrange’s equations did not address situations in which constraints on the motion of the particle were present. It is to this matter that we now turn our attention. With integrable constraints, whose constraint forces are prescribed by use of Lagrange’s prescription, the beauty and power of Lagrange’s equations are manifested. In this case, it is possible to choose the curvilinear coordinates qi such that the equations of motion decouple into two sets. The first set describes the unconstrained motion of the particle, and the second set yields the constraint forces as functions of the unconstrained motion. There are two approaches to obtaining Lagrange’s equations. We refer to them throughout these sections as Approach I and Approach II. For the novice, we highly recommend the first approach. As in the previous section, our exposition follows that of Casey [37]. Preliminaries

We assume that the particle is subject to a position constraint ²(r, t) =

0

and a velocity constraint f · v + e = 0. Further, we assume that the curvilinear coordinates are chosen such that the integrable constraint ² = 0 can be expressed in a simple fashion with a single coordinate: ² (r, t) =

q3 − d(t),

∇² =

a3 ,

and that the constraint forces are prescribed by use of Lagrange’s prescription: Fc

= λ 1a

3

+ λ2

µ 3 ·

¶

fi a

i

.

i=1

Notice that fi = f · ai . At this stage, we do not presume that the velocity constraint f · v + e = 0 is integrable. However, we assume that f is not parallel to a3 so that the constraints ² = 0 and f · v + e = 0 are independent. Suppose that there is an applied force Fa acting on the particle. This applied force can be decomposed into conservative and nonconservative parts: Fa

= −∇ U +

Fancon .

88

Lagrange’s Equations of Motion for a Single Particle

The resultant force acting on the particle is µ

F = λ 1a

3

+ λ2

3 ·

¶

fi a

i

− ∇U +

Fancon .

i=1

The total nonconservative force acting on the particle is Fncon The kinetic energy of the particle is T

=

=

Fc + Fancon .

3 3 m · · ³ 1 2 3´ i k m v·v = aik q , q , q q˙ q˙ , 2 2 i= 1 k =1

where aik = ai · ak . Imposing the integrable constraint on T , we find the constrained kinetic energy: T˜

˜ +T ˜ +T ˜ , = T 2 1 0

(3.5)

where 2

2

T˜ 2

=

m ·· a˜ ik q˙ i q˙ k , 2

T˜ 1 = m

i=1 k=1

2 ·

˙ a˜ i3q˙ i d,

T˜ 0

i=1

=

m a˜ 33 d˙ 2. 2

(3.6)

In these expressions, the components of the constrained metric tensor and the mass matrix are a˜ ik

= a ˜ ik (q

1

2

1

2

, q , t) = aik (q , q , q

3

¸

=

d(t)),

M

=

m

a˜ 11 a˜ 12

a˜ 12 a˜ 22

¹

.

Notice that we use a tilde ( ˜ ) to denote imposition of the integrable constraint(s) and that the subscripts on T˜ 2 , T˜ 1, and T˜ 0 refer to the powers of q˙ i . The coordinate singularities associated with using q1 and q2 as coordinates for the surface that the particle is moving on can be determined from the mass matrix M associº » ated with T˜ 2 . Specifically, paralleling the discussion in Section 1.8, in order for a˜ 1 , a˜ 2 ¼

¼2

to span E2 it is necessary and sufficient that ¼a˜ 1 × a˜ 2 ¼ > 0. The latter condition is equivalent to det (M ) > 0. Thus, if det (M) = 0 at a point on the surface, then the º » coordinate system q1 , q2 has a singularity at that point. A direct calculation shows that3 ½ ½ ˜ ∂T = 1½ 3 1, ∂q ˙ ∂q ˙ ˙ q =d,q ˙ 3 =d ½ ˜ ∂T ∂T ½ = , ½ ∂ q1 q3 =d,q ∂ q1 ˙ ˙ 3 =d ½ ˜ ∂T ½ ∂T ±= 3½ 3 3 = 0, ∂q ˙ 3 ∂q ˙ ∂T

q

˙ =d,q ˙ =d

½ ½ ˜ ∂T = 2½ 3 2, ∂q ˙ ∂q ˙ ˙ q =d,q ˙3 =d ½ ˜ ∂T ½ ∂T = , ½ ∂ q2 q3 =d,q ∂ q2 ˙ ˙3 =d ½ ˜ ∂T ½ ∂T = 0. ±= ½ 3 ∂ q3 3 ∂ q3 ∂T

q

(3.7)

˙ = d,˙q =d

In these relations, the partial derivative of T is evaluated prior to imposing the constraint q3 = d(t). These relations imply that we can use T˜ to obtain the first two Lagrange’s ½ ½

10t 2. Then ∂∂gt ½ = 2(10)(5) = 100. In words, we evaluate the derivative of g with respect t=5 to t and then substitute t = 5 in the resulting function.

3 Suppose g

=

3.4

Lagrange’s Equations in the Presence of Constraints

89

equations of motion, but not the third. Results that are identical in form to (3.7) pertain ˜ to the partial derivatives of L and L˜ and U and U. Notice that we did not impose the constraint f · v + e = 0 on the kinetic energy T and the Lagrangian L. It is possible to do this, but the result is of no use to us here. Approach I

In the first approach, which we refer to as Approach I, we evaluate the partial derivatives in Lagrange’s equations of motion (3.2) in the absence of any constraints: d dt

±

∂L

²

−

∂q ˙k

∂L ∂ qk

=

Fancon · ak + (F c

=

0) · ak .

Explicitly, these equations are µ

3

· d m ai1 q˙ i dt µ

i =1 3

· d m ai2 q˙ i dt µ

i =1 3

· d m ai3 q˙ i dt i =1

¶ −

3 · 3 · m ∂ air i r q˙ q˙ i=1 r =1

¶ −

3 3 · · m ∂ air i r q˙ q˙ 2 i=1 r =1

¶ −

2

∂ q1

2

∂q

3 3 · · m ∂ air i r i=1 r =1

2

∂ q3

q˙ q˙

+

∂U = ∂ q1

F ancon · a1 ,

+

∂U = ∂ q2

F ancon · a2 ,

+

∂U ∂ q3

=

F ancon · a3 .

Notice that we have not introduced the constraint forces on the right-hand side of these equations. That is, in the preceding equations F = −∇ U + Fancon . We now impose the integrable constraint q3 = d(t) and introduce the constraint f · v + e = 0 and the constraint forces. The resulting equations govern the motion of the particle and the constraint forces:

µ

3 d · mai1q˙ i dt

µ

i=1

3

· d m ai2q˙ i dt µ

i=1 3

· d m ai3q˙ i dt i=1

q3

=

q˙ 3

˙ = d,

d,

f1 q˙ 1 + f2 q˙ 2 + f 3d˙ + e = 0,

¶ −

3 · 3 · m ∂ air i r q˙ q˙ i=1 r =1

¶ −

3 · 3 · m ∂ air i r i=1 r =1

¶ −

2

∂ q1

2

∂ q2

q˙ q˙

3 · 3 · m ∂ air i r q˙ q˙ i=1 r =1

2

∂ q3

+

∂U ∂ q1

= λ2 f1 +

Fancon · a1 ,

+

∂U = λ2 f2 + Fancon · a2 , ∂ q2

+

∂U = λ1 + λ 2f3 + ∂ q3

Fancon · a3. (3.8)

We have refrained from ornamenting U, fi , ai , and aik with a tilde in the last four of these equations.

90

Lagrange’s Equations of Motion for a Single Particle

It is crucial to notice that if the constraint f · v + e = 0 were absent, then (3.8) would reduce to two sets of equations. The first of these sets, (3.8)4,5 , would yield differential equations for the unconstrained motion q1 (t) and q2 (t) of the particle, whereas the second set, (3.8)6, would provide the constraint force Fc = λ1a3 acting on the particle. Approach II

In the second approach, which we refer to as Approach II, we work directly with L˜ ˜ Here, as T˜ − U. ³

=

´

L˜ = L˜ q1 , q2 , q˙ 1 , q˙ 2 , t , the partial derivatives of L˜ with respect to q3 and q˙ 3 are zero. Consequently, using (3.2) and (3.7), we find only two Lagrange’s equations: d dt d dt

µ µ

˜ ∂L

∂q ˙1 ˜ ∂L ∂q ˙2

¶

−

¶ −

˜ ∂L

=

Fc · a˜ 1 + Fancon · a˜ 1,

˜ ∂L = ∂ q2

Fc · a˜ 2 + Fancon · a˜ 2.

∂ q1

Introducing the expression for the constraint force Fc and the additional constraint f · v + e = 0, we can determine the equations governing λ 2, q1 (t), and q2(t): d dt d dt

µ µ

f 1q˙ 1 + f2 q˙ 2 + f 3d˙ + e = 0, ˜ ∂L

∂q ˙1 ˜ ∂L ∂q ˙2

¶

−

¶ −

˜ ∂L

= λ 2f1 +

Fancon · a˜ 1 ,

˜ ∂L = λ 2f2 + ∂ q2

Fancon · a˜ 2 .

∂ q1

(3.9)

Notice that λ1 is absent from these equations. In addition, if the constraint f · v + e = 0 were absent, then the differential equations provided by Approach II are all that are needed to determine q1 (t) and q2(t). Equations (3.9)2,3 are examples of the Routh–Voss equations of motion.

3.5

A Particle in Motion on a Smooth Surface of Revolution

The first set of examples that we consider involve a particle of mass m that is in motion on a smooth surface of revolution. Such surfaces can be generated by rotating a curve about the z axis as shown in Figure 3.2. For the surfaces discussed here, we assume that the curve can be described uniquely by specifying z = f (r). In some cases, for instance when the surface of revolution is a cylinder, this is not possible and instead we can specify r = h(z). We leave the developments for the case r = h(z) as an exercise.4 Our 4 See Exercise 3.10.

3.5

91

A Particle in Motion on a Smooth Surface of Revolution

m

a3 a3

g z

a1 a1

E3

r

E2 E1

z

θ

η

= f (r )

= 0 coordinate surface

Schematic of a particle of mass m which is moving on a smooth surface of revolution in E3 under the influence of a gravitational force −mgE3 . The inset image shows an r coordinate curve formed by the intersection of the η = 0 coordinate surface with a θ coordinate surface.

Figure 3.2

goal is to show how to obtain the equations of motion for the particle using Approach II. We later apply our results to a particle on a spinning cone that is subject to spring and gravitational forces. Additional examples of this type of particle problem are discussed in Exercises 3.1, 3.6, and 3.9. The latter exercise includes the case where the surface is rough and Approach I must be used to obtain the equations of motion. To establish the equations of motion for the particle, we use cylindrical polar coordinates to define the surface of revolution and define the following curvilinear coordinate system for E3: q1

=

q2

r,

q3

= θ,

It is imperative to note that the constraint ² of revolution has the simple representation

=

= η =

z − f (r).

0 for the particle moving on the surface

² = η.

(3.10)

With respect to the coordinates r, θ , and η, the position vector r has the representation r = r cos (θ ) E1 + r sin (θ ) E2 + (η + f (r)) E3. Using this representation, we compute the covariant basis vectors: a1

=

∂r ∂r

=

er + f ² E3 ,

a2

=

∂r ∂θ

=

reθ ,

a3 =

∂r ∂η

=

E3 . 2

df and f ²² = ddr2f . We Here, and in the sequel, we employ the shorthand notations f ² = dr also assume that f ² is bounded. To compute the contravariant basis vectors, we use the representation (1.17)3 of the gradient operator in cylindrical polar coordinates:

a1

= ∇r =

er ,

a2

= ∇θ =

1 eθ , r

a3

= ∇η =

E3 − f ² er .

92

Lagrange’s Equations of Motion for a Single Particle

We leave it as an exercise to verify that ai · ak = δik . Some examples of the covariant and contravariant basis vectors are shown in Figure 3.2. For the problem at hand, a˜ i = ai and a˜ i = ai , so we will often drop the tildes ornamenting these vectors in the sequel. Because the surface of interest is smooth, we can use Lagrange’s prescription for the constraint force. There is a single constraint on the particle that is expressed in terms of a single coordinate: η = 0. This constraint implies that v · a3

0.

=

Thus, Fc

3

= λa

.

That is, Fc = N, the normal force acting on the particle. For completeness we note that if the surface of revolution was spinning about E3 with a speed ³, then (

vrel

= r ˙a 1 + θ˙ − ³

)

a2

(3.11)

and any dynamic friction force on the particle (if there were such a force) would oppose this velocity vector. For the problem at hand, in addition to the normal force, the other force acting on the particle is gravitational: F

3

= −mgE3 + λ a

.

(3.12)

It is important to note here that F c · a1 = 0 and F c · a2 = 0. We are now in a position to use Approach II to determine the equations of motion for the system. The constrained velocity vector and kinetic energy of the particle are ´ ) m ³( v˜ = r˙ a1 + θ˙ a2 , T˜ = 1 + f ² f ² ˙r2 + r2 θ˙ 2 . 2 ¾

¿

As the surface is fixed, T˜ = T˜ 2 . If we evaluate the 2 × 2 mass matrix m a˜ ik) associated ( with T˜ , then we find that the determinant of this matrix is mr2 1 + f ² f ² . Thus, the system of coordinates for the surface of revolution has singularities along the entire x3 axis. The equations of motion for the particle can be found from the pair of Lagrange’s equations: d dt

µ

∂ T˜ ∂ r˙

d dt

µ

¶

−

∂ T˜

¶

˜ ∂T

∂r −

∂ θ˙

=

(

F · a˜ 1

˜ ∂T

∂θ

=

=

(

F · a˜ 2

)

er + f ² E3 , =

)

reθ .

Evaluating the partial derivatives and expanding terms, we find the pair of differential equations for r and θ : ¸

(

m 1 + f ²f ² 0

)

0 mr2

¹¸

r¨ θ¨

¹

¸

+

mf ² f ²² r˙ 2 − mrθ˙2 2mrr˙θ˙

¹

¸

=

−mgf

0

²

¹

.

The second of the equations of motion is equivalent to the conservation of angular ( ) momentum HO · E3 of the particle (i.e., d/dt mr2 θ˙ = 0). Because Fc does no work

3.5

A Particle in Motion on a Smooth Surface of Revolution

Ω(t )

(a)

g

(b)

93

m

m

z

Smooth cone α

r

O

O

Figure 3.3 (a) Schematic of a particle of mass m which is attached to a fixed point O by an elastic spring. A vertical gravitational force −mgE3 acts on the particle and the particle is free to move on the inner surface of a smooth cone which is rotating about the vertical z axis with a speed ³ = ³ (t). (b) Representative motion of the particle on the cone.

(i.e., Fc · v = 0) and the gravitational force is conservative, it is straightforward to show that the solutions to the equations of motion conserve the total energy T˜ + mgf .

A Particle Moving on a Smooth Cone

The results we have just presented can be applied with a minimum amount of modification to determine the equations of motion of the particle moving on the smooth inner surface of the cone shown in Figure 3.3(a). Here, the surface of revolution is z = f (r) = r tan (α) , where α is a constant angle. The constraint force acting on the particle is Fc = λa3 and the remaining forces acting on the particle are conservative. Because there is no friction between the particle and the surface of the cone, the spinning of the cone has no effect on the motion of the particle. We invite the reader to compare the derivation of the equations of motion for this problem to a related derivation in Section 2.9 of Chapter 2 where spherical polar coordinates were used. To establish the equations of motion for this problem, we will use the Lagrangian. To compute L˜ we first note that ³r³ =

À

r2 + z2

À

=

r 1 + tan2 (α) = r sec (α) .

Thus, the constrained potential energy of the particle is K 2 ´ + mgr tan (α) , 2 of the spring is U˜

where the extension ´

=

´ =

r sec (α) − ±0 .

˜ of the particle is The Lagrangian L˜ = T˜ − U

L˜ =

´ m³ 2 K sec (α) r˙ 2 + r2 θ˙ 2 − ´2 − mgr tan (α) . 2 2

94

Lagrange’s Equations of Motion for a Single Particle

Because Fc · a˜ 1 form:

=

0 and Fc · a˜ 2 d dt

µ

˜ ∂L

¶ −

∂ ˙r

0, Lagrange’s equations of motion have a simple

=

˜ ∂L

∂r

=

d dt

0,

µ

˜ ∂L

¶

˜ ∂L

−

∂ θ˙

∂θ

=

0.

(3.13)

Evaluating the partial derivatives and expanding terms, we find the pair of differential equations for r and θ : m sec 2 (α) r¨ − mrθ˙2 = −mg tan (α) − K ´ sec (α) , d ³ 2 ´ mr θ˙ = 0. dt We can rearrange these equations into a canonical form: ¸

m sec2 (α) 0

0 mr2

¹¸

r¨

¹

θ¨

¸

+

−mrθ˙ 2

¹

2mr˙rθ˙

¸

=

−mg tan (α) −

K ´ sec (α)

¹

0

.

This pair of ordinary differential equations can be integrated to determine the motion of the particle. A representative solution for such a simulation is shown in Figure 3.3(b). The solutions to the equations of motion conserve the total energy T˜ + U˜ and HO · E3.5

3.6

A Particle in Motion on a Sphere

To further illustrate and contrast Approaches I and II, we now consider the example of a particle moving on a smooth sphere whose radius R is a known function of time: R = d(t). The particle is subject to a conservative force −mgE3 and a nonconservative force DReθ , where D is a constant. Later on, we shall impose an additional constraint on the motion of the particle. Referring to Figure 3.4, it is convenient to use a spherical polar coordinate system: q1

q2

= θ,

q3

= φ,

=

R.

Using this coordinate system, we can write the integrable constraint R = d(t) in the form ² =

0,

² (r, t) =

R − d(t).

As the sphere is smooth and a3 = eR , we can use Lagrange’s prescription: Fc

= λa

3

= λeR .

The kinetic and potential energies of a particle in the chosen coordinate system are ´ m³ 2 R˙ + R2 sin2 (φ)θ˙ 2 + R2 φ˙ 2 , U = mgR cos(φ ). T= 2 The constrained kinetic and potential energies are ´ m³ 2 ˜ = mgd cos( φ). d˙ + d2 sin2 (φ)θ˙ 2 + d2φ˙ 2 , T˜ = U (3.14) 2 5 Additional results pertaining to this particle problem, including time traces of H and T ˜ O

found in [215, Chapter 6].

˜ + U,

can be

3.6

95

A Particle in Motion on a Sphere

a˜1 = d sin (φ) eθ m

˜a2 =

eφ

d

Figure 3.4 A particle in motion on the surface of a sphere whose radius changes with time: R = d(t). Some examples of q1 = θ and q2 = φ coordinate curves on the sphere are shown.

Finally, the covariant basis vectors are a1

=

R sin(φ)eθ ,

a2

=

Reφ ,

a3

=

eR .

Their constrained counterparts a˜ i are easily inferred from these expressions. If we evaluate the 2 × 2 mass matrix associated with ´ m³ 2 2 d sin (φ)θ˙ 2 + d2φ˙ 2 , T˜2 = 2 then we find that the determinant of this matrix is md4 sin2 (φ ). Thus, the system of coordinates for the sphere has singularities at the north and south poles. First, we use Approach II to obtain the equations governing θ (t) and φ(t). There are two equations: d dt

d dt

µ

˜ ∂L

¶

−

∂ θ˙

µ

˜ ∂L ˙ ∂φ

¶ −

˜ ∂L

∂θ

˜ ∂L

∂φ

=

Fncon · a˜ 1

=

Fc · a˜ 1 + Ddeθ · a˜ 1

=

Dd2 sin(φ),

=

Fncon · a˜ 2

=

Fc · a˜ 2 + Ddeθ · a˜ 2

=

0.

Evaluating the partial derivatives of the constrained Lagrangian, we find that these equations become ´ d ³ 2 2 md sin (φ )θ˙ dt

=

Dd2 sin(φ),

d ³ 2 ´ md φ˙ − md2 sin(φ ) cos(φ)θ˙2 − mgd sin(φ) = 0. dt

(3.15)

96

Lagrange’s Equations of Motion for a Single Particle

Notice that the constraint force λ eR is absent from these equations. Alternatively, using Approach I, we start with the unconstrained Lagrangian L and establish three equations of motion (cf. (3.3)): d dt

±

∂L ∂ θ˙

=

d dt

2

mR sin

±

∂L ∂ φ˙

d dt

±

=

2

²

(φ)θ˙

mR2 φ˙

± −

²

∂L ∂θ

² =

0

Fancon · R sin(φ )eθ , ± ² ∂L 2 2 − = mR sin(φ ) cos(φ)θ˙ + mgR sin(φ )

=

∂φ

Fancon · Reφ , ² ∂L ∂L 2 2 2 ˙ = mR − = mR sin (φ )θ˙ + mRφ˙ − mg cos( φ) ˙ ∂R ∂R = Fancon · eR . ²

=

±

Next, we impose the integrable constraint and introduce the constraint force Fc to find the equations of motion: ´ d ³ 2 2 md sin (φ )θ˙ dt

=

Dd2 sin(φ),

d ³ 2 ˙´ md φ − md2 sin(φ ) cos(φ)θ˙ 2 − mgd sin(φ) = 0, dt ´ d ( ˙) ³ md − md sin2 (φ)θ˙ 2 + mdφ˙ 2 − mg cos(φ ) = λ . dt

(3.16)

Notice that the first two of these equations are identical to (3.15), whereas the third equation is an equation for the constraint force Fc . We could now introduce an additional constraint: f1 θ˙ + f2φ˙ + f3R˙ + e = 0. Using Lagrange’s prescription, we find that the total constraint force on the particle is ±

Fc

= λ eR + λ2

f1 eθ R sin(φ )

+

f2 eφ R

² + f3 eR

.

To obtain the equations of motion for the case where the additional constraint is active, we need to introduce the constraint force associated with the constraint f · v + e = 0 on the right-hand side of (3.16) and also append the constraint to the resulting equations: f1θ˙ + f2 φ˙ + f3 d˙ + e = 0,

´ d ³ 2 2 md sin (φ )θ˙ dt

=

Dd2 sin(φ) + λ2 f1 ,

d ³ 2 ´ md φ˙ − md2 sin(φ ) cos(φ)θ˙ 2 − mgd sin(φ) dt

´ d ( ˙) ³ md − md sin2 (φ)θ˙2 + mdφ˙ 2 − mg cos( φ) dt

= λ2f 2,

= λ + λ2 f3 .

3.7

Some Elements of Geometry and Particle Kinematics

97

It is left as an exercise to show what additional simplifications to these equations arise if the constraint f · v + e = 0 has the form f 1 = 0, f2 = 1, f3 = 0, and e = 0. In this case, the velocity constraint is integrable and the particle moves on a circle of radius d sin(φ0 ).

3.7

Some Elements of Geometry and Particle Kinematics

As a prelude to our discussion of Lagrange’s equations and their geometrical significance, some material from differential geometry is required. Our treatment is limited to the ingredients we shall shortly need and, as such, it cannot do justice to this wonderful subject. Mercifully, there are several excellent texts that can be recommended to remedy this: [61, 197, 210, 268]. Reading Chapter 1 of Lanczos [163] for a related discussion on kinetic energy and geometry and the recent paper by Lützen [174] for a historical overview of the interaction between geometry and dynamics in the nineteenth century is highly recommended. Here, we are interested in surfaces and curves that are in E3 . We assume that these entities are smooth. That is, they are without edges and sharp corners, and we call them manifolds. In the case of a curve, a single coordinate is needed to locally parameterize the points P on this manifold, and so it is considered to be a one-dimensional manifold. For a surface, two coordinates are needed to locally parameterize the points on the surface and so the surface is considered to be a two-dimensional manifold. In an obvious generalization, subsets of E3 such as solid spheres and solid ellipsoids are considered to be three-dimensional manifolds. Previously, in Section 1.5, curvilinear coordinates were introduced. For a given surface (or curve), we used these coordinates both to label points on the manifold and to define the manifold. For example, for a sphere of radius R0 , the spherical polar coordinates φ and θ label points on the sphere and the coordinate R can be used to define the sphere: R = R0 . Similarly, for a circle, the cylindrical polar coordinate θ can be used to label points on the circle, and the coordinates r and z can be used to define the circle. The curvilinear coordinate system we use to label points on the manifold is known as a chart. For some manifolds, such as a plane, a straight line, and a circle, a single chart suffices to enable the labeling of each point on the manifold. For surfaces such as spheres, for which a set of spherical polar coordinates will not be defined at the poles, at least two charts are needed. With terminology borrowed from cartography, the set of all charts for a manifold is known as an atlas. Tangent Spaces

At each point P of a manifold M we define a tangent space, and we denote this space by TP M . If the manifold is n-dimensional, then T PM is also n-dimensional. For example, the tangent space TP C is a line for the curve shown in Figure 3.5(a), and the tangent space TP S is a plane for the sphere S shown in Figure 3.5(b). Continuing with the

98

Lagrange’s Equations of Motion for a Single Particle

(b)

(a) TP

C

TP

S

P

C

P

S Two examples of manifolds and the tangent spaces to a point P on them: (a) a curve C ; (b) a sphere S .

Figure 3.5

P1

V P2

vrel

S

A curve V connecting two points on a sphere S . The velocity vector vrel of a particle moving on this curve would lie in the tangent plane TPS at each point P of the curve. Figure 3.6

sphere as an example, if we fix a point P on the sphere, this is equivalent to fixing the polar coordinates φ = φ0 and θ = θ0 . The vectors eφ

=

eφ (φ0 , θ0 ) = sin (φ0 ) (cos (θ0 ) E1 + sin (θ 0) E2) + cos (φ 0) E3, eθ

=

eθ

(θ0 ) = − sin (θ 0) E 1 + cos (θ0 ) E2

S

form a basis for the tangent space TP at P, and any tangent vector to the sphere at this point can be expressed in terms of these vectors. Related remarks apply at a point on a curve, but now only a single vector is needed to span the tangent space. As a final 3 example, for a particle that is free to move in = E , the dimension of T P is 3.

M

M

Distance Traveled

V

Returning to the example of a sphere, we choose two points P1 and P 2 on the sphere of radius R and consider a path between them (see Figure 3.6). We wish to measure the distance one would travel along . To do this, we first parameterize the curve with a parameter u, where u = uα at Pα . The curve can then be described uniquely by the

V

3.7

99

Some Elements of Geometry and Particle Kinematics

functions θ (u) and φ(u).6 To determine the distance method would be to evaluate the following integral: µs =

Á u2 Â u1

R2 sin2 (φ)

dθ dθ du du

µs

traveled along the curve, one

R2

dφ dφ du. du du

+

(3.17)

Referring to (3.5), (3.6), and (3.14), we can express the integrand on the right-hand side of this equation in terms of the kinetic energy of a particle moving on the sphere: µs =

Á u2 u1

Á t 2 =

t1

Á t 2 =

t1

Ã

Ã

2T˜ 2 m

±

²

dt du du

2T˜ 2 dt m ³v rel ³ dt.

For the second integral, we changed variables and parameterized the path using t rather than u. Thus, the distance traveled that is measured using (3.17) agrees with the interpretation of integrating the speed that one is traveling along the path. Turning to the example of a circle of radius r0, the reader is invited to show that a measure of distance corresponding to (3.17) can be established by using the single polar Ä u2 ½½ dθ ½½ coordinate θ : µ s = r0 u1 ½ du ½ du. For a particle moving in space, a measure of distance can be defined in a standard manner using Cartesian coordinates: µs =

Å Á u Æ 3 2 Æ· dx dx k k Ç du. u1

k =1

du du

It is easy to see that this expression can be rewritten as µs = Configuration Manifold and Degrees of Freedom

From the previous discussion, for an n-dimensional manifold similar to that provided by (3.17) can be defined:7 µs =

Ä t È 2T 2 t1

m

dt.

M, a measure of distance

Å Á u Æ n n 2 Æ· · ∂ qi ∂ q k Ç du. a˜ ik u1

i=1 k=1

∂u ∂u

Parameterizing the path by using time t instead of a variable u, we find that µs =

Å Á t Æ n n 2 Æ· · Ç a˜ ik q˙ i q˙ k dt. t1

(3.18)

i=1 k=1

6 For example, if the curve were a segment of the equator, then θ (u) = θ (u ) + (u2 −u1 ) and φ (u) = π . 1 2πR 2 7 The index n here is either 1, 2, or 3. In later chapters, we shall see that n can range from 1 to 3N for a

system of N particles and from 1 to 6 for a single rigid body.

100

Lagrange’s Equations of Motion for a Single Particle

Following the advocacy of Hertz [126] for a mechanical system with n degrees of freedom, we can conveniently imagine a (representative) particle of mass m moving on a manifold . The manifold is known as the configuration manifold and the n coordinates q1 , . . . , qn are known as generalized coordinates. We emphasize that the number of generalized coordinates is equal to the number of degrees of freedom of the mechanical system. In addition, the number of generalized coordinates only depends on the number of integrable constraints imposed on the system and is unaffected by the number of nonintegrable constraints. For a single particle, the configuration manifold is typically the space, surface, or curve that the particle is constrained to move on. To describe these manifolds, we use notation from topology for certain manifolds:

M

S1 is a circle with unit radius, T 2 is a two-torus, and S2 is a unit two-dimensional sphere in a three-dimensional space (or two-sphere).

M

We also say that the configuration manifold of the particle is homeomorphic to a specific manifold. Thus, the configuration manifold of a simple pendulum of length ± 0 and the configuration manifold of a pendulum of unit length are both homeomorphic 1 ∼ to the unit circle S1.8 We denote this equivalence as follows: = S . We take this opportunity to list some examples:

M

M

M

3 ∼ Unconstrained particle: = E = E × E × E. 2 ∼ Particle constrained to move on a plane: = E = E × E. 2 ∼ Particle constrained to move on a sphere: = S . 1 ∼ Particle constrained to move on a circle: = S . 1 ∼ Particle constrained to move on a cylinder: = S × E. 2 ∼ Particle constrained to move on a rotating circular wire: = T

M M M M

M

=

S1 × S1 .9

The last example listed above is one of the few examples that we are aware of where the configuration manifold of a particle does not correspond to the surface that the single particle is moving on. In listing the configuration manifolds, we have used the symbol × for the topological product of two manifolds. ∑ The velocity of the particle relative to is simply vrel = nk=1 q˙ k a˜ k . Thus, (3.18) can be expressed as the time integral of a speed:

M

µs =

Á t2 t1

³ v rel ³ dt.

8 A homeomorphism between two spaces is a continuous function that has a continuous inverse. The word

homeomorphic derives from a combination of the Greek words for similar (or same) and shape (or form). Homeomorphisms can stretch, compress, shear, twist, and distort surfaces and spaces. Thus, a sphere of radius R1 is homeomorphic to a sphere of unit radius and a rectangle is homeomorphic to a square (and vice versa). 9 This example is discussed in Exercise 3.7(e) and Section 11.4. The symbol T 2 denotes a two-torus and the generalized coordinates for this torus are φe ∈ [0, 2 π ] and θ ∈ [0, 2π ].

3.7

101

Some Elements of Geometry and Particle Kinematics

M

We see from this expression that, in defining µs, we have also defined a measure of the magnitude of a vector vrel ∈ T P . Such a measure is known as a metric, and a manifold that is equipped with a metric is known as a Riemannian manifold. The terminology here pays tribute to Georg F. B. Riemann (1826–1866) and his remarkable work [241].

Geodesics and Line Elements

A geodesic is a curve of shortest distance between two points. The curve lies entirely on the manifold and the distance in question is measured along the curve. For instance, the shortest distance between two points on a plane or in space is a straight line, on a sphere it is a great circle, and on a cylinder it is a circle, line, or helix depending on the relative location of the two points. It can be shown that a necessary condition for the ( ) curve between two points to be a geodesic is that the path q1(t), . . . , qn (t) traveled by the particle of mass m satisfies Lagrange’s equations of motion: 10 d dt

µ

˜ ∂T ∂q ˙k

¶

˜ ∂T = − ∂ qk

0,

where T˜ = T˜ 2 =

n · n ·

a˜ ik q˙ i q˙ k .

(3.19)

i=1 k=1

The easiest example (to demonstrate this result is to imagine a particle moving on a ) m 2 2 ˜ plane. Then T = 2 ˙x + y˙ and Lagrange’s equations of motion (3.19) imply that x¨ = 0 and y¨ = 0. Thus, the path traveled by the particle is a straight line at constant speed. For a particle moving on a cylinder of radius r0, Lagrange’s equations of motion (3.19) imply that the particle moves with constant z˙ and θ˙ , a result which validates our earlier remarks. We leave it as an exercise to verify that, after integrating (3.19) for a particle moving on a sphere, one would find that the particle traces out a great circle.11 Because the distance measure (3.18) is clearly intimately related to the kinetic energy of a particle, following Synge [274]: ds =

Å Æ n n Æ· · Ç a˜ ik q˙ i q˙ k dt

(3.20)

i=1 k=1

¾

M

¿

is known as the kinematical line element. We can also use the mass matrix m a˜ ik to identify singularities in the coordinate systems used to parameterize . Thus, we say ¾ ¿ that a coordinate singularity is present when the determinant of a˜ ik is 0. The use of the kinematical line element ds as a measure of distance has a long history, with important contributions by Jacobi [139] and Ricci and Levi-Civita [239]. Other choices of ds are available (see, e.g., [162, 163, 197, 274]) and some of them feature prominently in Einstein’s theory of relativity. The freedom of selection is similar to the 10 One proof of this result uses the Euler–Lagrange necessary condition for a curve ³ ´ Ät À ˜ q (t) = q 1 (t ) , . . . , qn (t ) that minimizes t 1 Tdt for all paths connecting q ( t0 ) and q (t1 ) . We refer the 0

reader to [163] or any textbook on the calculus of variations for a proof.

11 One of the easiest ways to show this result is to first select the two points on the sphere of interest and

then choose the spherical polar coordinate system so that both points have the coordinate φ the spherical polar coordinate system is chosen so that points of interest lie on the equator.

=

π

2 . That is,

102

Lagrange’s Equations of Motion for a Single Particle

notion that different measures of distance, such as meters and feet, are possible. The curves of shortest distance between two points that minimize one measure of distance are not necessarily the same as those which minimize another measure of distance. We now turn to an example to demonstrate this observation. An alternative line element to ds in classical mechanics is the action line element ds A. If we imagine a particle that is free to move on a (stationary) configuration manifold under the action of conservative forces only, then the total energy E of the particle is conserved and the motion of the particle satisfies Lagrange’s equations of motion:

M

d dt

µ

∂ T˜ ∂ q˙ k

¶

−

˜ ∂T

∂ qk

= −

˜ ∂U

∂ qk

.

As discussed in [163, 277], Jacobi showed that the solutions qi (t) minimize the distance between points on provided the distance is measured using the action line element 12 ds A:

M

ds A

=

Å Æ n n Æ· · Ç (E0

−

U ) a˜ ik q˙ i q˙ k dt,

i=1 k=1

where E0 is the constant value of E and is determined by the initial conditions. You might notice that the SI units of sA are kg1/ 2 m2 s−1 , whereas the units of s are m. By way of a simple illustration that the geodesics measured using ds and dsA can be distinct, the reader is referred to Figure 3.7. This figure illustrates that for curves in space where the potential energy U = mgy, the curves of shortest distance between two points with respect to the distance measure ds A are parabolas which depend on the selected value of E0 . By way of contrast, the curve of shortest distance between two points in space with respect to the distance measure ds is a straight line.

A Representative Particle

M

We remarked earlier on a (representative) particle of mass m moving on . Clearly, for a single particle, such a construction can easily be achieved. Indeed, for a single particle the configuration manifold typically corresponds to the physical surface or curve that the particle moves on. However, for a system of particles or rigid bodies subject to constraints, this is not the case, and the construction of a single representative particle is nontrivial. Indeed, for a system of particles, such a construction was explicitly recorded recently by Casey [37].13 Subsequently, he extended this construction to a single rigid body [38] and a system of rigid bodies [40]. The construction of the representative particle and the associated configuration manifold enables new perspectives on coordinate singularities and the phenomenon of gimbal lock (see [123]). 12 This result is sometimes known as the principle of least action [182, Section 5 of Chapter XVII] or

Jacobi’s principle of least action [163, Section 6 of Chapter V].

13 We shall examine his construction in Section 4.7.

3.8

103

The Geometry of Lagrange’s Equations of Motion

y

x

E0

inc.

A

B

È

(

)

Geodesics with respect to the line element ds A = (E0 − mgy) dx2 + dy2 that connect two points A and B in space where the potential energy function is U = mgy. While the energy value E 0 for each of the geodesics is different, each of the paths À is a parabola. The corresponding unique geodesic with respect to the line element ds = dx2 + dy2 is a straight line that is shown as a dashed line in the figure. Figure 3.7

3.8

The Geometry of Lagrange’s Equations of Motion

Some readers will have gained the perspective that Lagrange’s equations of motion obtained by use of Approach II are projections of F = ma onto the covariant basis vectors for the unconstrained coordinates. That is, we are projecting F = ma onto the basis for TP . For those who have not yet found this perspective, let us recall the example of the particle moving on the sphere of radius R = d(t). There, we obtained the two Lagrange’s equations for θ and φ by taking the d sin(φ)eθ and deφ components of F = ma. These two vectors, d sin(φ )eθ and deφ , form a basis for the tangent space T P to a point P of the sphere . Furthermore, because the constraint force λ eR associated with the integrable

M

S

S

104

Lagrange’s Equations of Motion for a Single Particle

constraint is perpendicular to the sphere, this force did not appear in the two Lagrange’s equations. An important feature of nonintegrable constraints is that the constraint force associated with these constraints is not decoupled from the equations governing the unconstrained motion. This deficiency in Lagrange’s equations of motion can be removed by use of alternative forms of F = ma that are suited to nonintegrably constrained systems, but we do not approach this vast subject here. We now delve a little more deeply into the geometry inherent in Lagrange’s equations of motion. Our discussion is based on [37, 163, 274].

A Particle Subject to a Single Integrable Constraint

First, let us consider the case in which the particle is subject to a single integrable constraint ² (r, t) = 0: ² (r, t) =

q3 − d(t),

where d(t) is a known function. When considered in relation to E3 , the constraint 3 ² = 0 represents a moving two-dimensional surface: in this case, a q coordinate surface. As mentioned earlier, this surface is known as the configuration manifold (see Figure 3.8). The velocity of the particle relative to this surface has the representation

M

vrel

1

2

= q ˙ a ˜1 + q ˙ a ˜2 .

The coordinates q1 and q2 in this case are known as the generalized coordinates and, as mentioned previously, the number of these coordinates is the number of degrees of freedom of the particle. Thus an unconstrained particle has three degrees of freedom, whereas a particle constrained to move on the surface has only two. q

1

coordinate curve

q

a3

2

coordinate curve

a1

M

a2

q

3

= d (t) coordinate surface

O

M

M

The configuration manifold of a particle moving on a surface. Here, the coordinates q1 and q2 are known as the generalized coordinates and is the q3 = d(t) coordinate surface.

Figure 3.8

3.8

The Geometry of Lagrange’s Equations of Motion

105

M

M M

Recall that at each point P of , {a1, a2 } evaluated at P is a basis for the tangent plane TP to at P. We can also define a relative kinetic energy T rel = T˜ 2 : T rel

=

m vrel · vrel 2

=

2 · 2 · m i=1 k=1

M

2

a˜ i · a˜ k q˙ i q˙ k .

Consider a particle moving on . To calculate the distance traveled by the particle on the surface in a given interval t 1 − t 0 of time t, we integrate the magnitude of its velocity vrel with respect to time: s(t 1) − s(t0 ) = =

Á t1

√

t0

vrel · vrel dt

Á t  1 2Trel

m

t0

dt.

We can differentiate this result with respect to t to find the kinematical line element ds: Â

ds =

2Trel dt m

=

Å Æ 2 2 Æ· · Ç a˜

i k i · a˜ k dq dq .

(3.21)

i=1 k=1

As emphasized previously in Section 3.7, notice that the measure of distance is defined by the kinetic energy Trel . It is left as an exercise to show that the integral of (3.21) is none other than (3.17). With regard to Lagrange’s equations of motion, suppose that the constraint forces associated with the integrable constraint are prescribed by use of Lagrange’s prescription: Fc = λa˜ 3 . Then Fc · a˜ 1 = Fc · a˜ 2 = 0, and the constraint force does not appear in the first two Lagrange’s equations. The forces Q1

=

F · a˜ 1,

Q2

=

F · a˜ 2

are known as the generalized forces, and the expressions we use for them are equivalent to those used in other texts on dynamics (e.g., [20, 106]). In addition, Approach II can be used to obtain the differential equations governing q1(t) and q2 (t): d dt d dt

M

µ µ

˜ ∂L ∂q ˙1 ˜ ∂L ∂q ˙2

¶ −

˜ ∂L = ∂ q1

Fancon · a˜ 1 ,

−

˜ ∂L = ∂ q2

Fancon · a˜ 2 .

¶

(3.22)

Imposing a nonintegrable constraint on the motion of the particle will not change . Furthermore, this constraint will, in general, introduce constraint forces into (3.22). These forces will destroy the decoupling that Lagrange’s equations achieve for integrable constraints.

106

Lagrange’s Equations of Motion for a Single Particle

A Particle Subject to Two Integrable Constraints

We now turn to the case in which a particle is subject to two integrable constraints 0 and ²2 (r, t) = 0:

²1(r, t) =

²1(r, t) =

q3 − d3 (t),

²2 (r, t) =

q2 − d2 (t).

Notice that we have chosen the curvilinear coordinates so that the constraints are easily represented. At each instant of time, the intersection of the two surfaces ²1 = 0 and ²2 = 0 in E3 defines a curve – in this case a q1 coordinate curve (see Figure 3.9). The configuration manifold in this case corresponds to the q1 coordinate 1 curve. The coordinate q is the generalized coordinate for this single degree of freedom system. We can easily represent the velocity vector vrel of the particle relative to :

M

M

vrel

1

= q ˙ a ˜1 .

M

M

It should be clear that a˜ 1 is tangent to . Indeed, a1 evaluated at P is a basis vector for the one-dimensional space TP . In addition, we can associate with vrel a relative kinetic energy: m m T rel = vrel · vrel = a˜ 1 · a˜ 1q˙ 1q˙ 1 . 2 2 Paralleling previous developments (see (3.20)), the kinematical line element for Â

ds =

2T rel dt = m

È

q

q

3

=

3

d

M is

a˜ 1 · a˜ 1 dq1 dq1.

2

= d 2 (t ) coordinate surface

(t) coordinate surface

m

a˜ 1 r

O

A particle moving on a curve. In this figure, a˜ 1 is tangent to the q1 coordinate curve corresponding to q2 = d2(t) and q3 = d3 (t). That is, the q1 coordinate curve is the configuration manifold of the particle. The vectors a˜ 2 and a˜ 3, which are not shown, are normal to .

Figure 3.9

M

M

3.9

Lagrange’s Equations of Motion for a Particle in the Presence of Friction

107

With regard to Lagrange’s equations of motion, if the constraint forces are prescribed by use of Lagrange’s prescription, F c = λ1 a˜ 3 +λ1 a˜ 2, then we can easily find the differential equation governing q1(t) by using Approach II: d dt (

µ

˜ ∂L ∂q ˙1

¶

−

˜ ∂L = ∂ q1

F ancon · a˜ 1 .

)

Here, L˜ = L˜ q1, q˙ 1 , t and F · a˜ 1 is the generalized force for this problem.

3.9

Lagrange’s Equations of Motion for a Particle in the Presence of Friction

In traditional treatments of Lagrange’s equations of motion, Approach II is used. As a result, the constraint forces acting on the particle that enforce the integrable constraints cannot be computed. However, for a particle in motion in the presence of friction, the dynamic friction force depends on the normal force, and so the normal force must be computed in order to determine the motion of the particle. Approach I and a hybrid approach discussed below are valid in these cases. We now present overviews on using Lagrange’s equations for a particle moving on a rough surface and a rough curve. These overviews are intended to complement our discussion in Section 3.8. The corresponding developments for the cases where the particle is subject to static friction forces are easily inferred and are left as exercises.14

A Particle in Motion on a Rough Surface

Consider a particle moving on a rough surface. The surface is defined by the constraint ² (r, t) = 0: ²(r, t) =

q3 − d(t),

where d(t) is a known function. The constraint force acting on the particle is given by Coulomb’s prescription: Fc

=

where N = λa˜ 3,

Ff

= − μ d ³ N³

N + Ff , vrel , ³v rel ³

vrel

1

2

= q ˙ a ˜1 + q ˙ a ˜ 2.

If the coordinate system is such that F f · a˜ 3 = 0, then Lagrange’s equations give a decoupled set where the first two equations can be used to determine the equations of motion and the third equation gives λ . An example of this instance is a particle moving on a rough sphere of radius d(t) where a set of spherical polar coordinates is used to determine the motion of the particle: a˜ 1 = d(t) sin (φ) eθ , a˜ 2 = d(t)eφ , and a˜ 3 = eR . The equations of motion for a particle moving on such a sphere can easily be inferred from our discussion in Section 3.6 and are left to the reader. 14 See Exercise 3.4 for instance.

108

Lagrange’s Equations of Motion for a Single Particle

When the coordinate system is such that F f · a˜ 3 ± = 0, it is best not to use the third of Lagrange’s equations of motion to compute the normal force. Instead, a hybrid approach can be used where Lagrange’s equations of motion found using Approach II provide the two equations of motion and, because Ff · a˜ 3 = 0, a contravariant component of F = ma gives the normal force: ³

´

˜ ˜ ∂T d ∂T − = dt ³ ∂ q˙ 1 ´ ∂ q1 ˜ ˜ ∂T d ∂T dt ∂ q˙ 2 − ∂ q2 = 3

F · a˜

=

⎫

F · a˜ 1 ⎬ F · a˜ 2 ⎭

ma · a˜ 3

equations of motion,

−→

−→

equation for λ.

An example of a coordinate system where Ff · a˜ 3 ± = 0 is the coordinate system (r, θ , η) used to parameterize the motion of a particle moving on a surface of revolution in Section 3.5. Other relevant examples are discussed in Exercises 3.5 and 3.8.

A Particle in Motion on a Rough Curve

We next consider a particle in motion on a rough curve. The particle is subject to a pair of integrable constraints ²1 (r, t) = 0 and ²2(r, t) = 0: q3 − d3 (t),

²1(r, t) =

²2 (r, t) =

q2 − d2 (t),

where d1 and d2 are known functions of time. The constraint force acting on the particle is given by Coulomb’s prescription: Fc

=

N + Ff ,

where N = λ 1a˜ 2 + λ2 a˜ 3 ,

Ff

= −μd ³N³

vrel , ³ v rel³

vrel

1

= q ˙ a ˜1 .

Now suppose the coordinates chosen for this problem are such that Ff · a˜ 2 ± = 0 and/or Ff · a˜ 3 ± = 0. Then, for this problem, it is easier to use the hybrid approach discussed previously to determine the equations of motion of the particle and the normal force acting on the particle rather than using Approach I. In the interests of brevity, we refrain from discussing using Approach I for this case and instead refer the reader to the discussion of a particle moving on a rough helical space curve in Section 3.10. The hybrid approach for this case uses Approach II to determine the equation of motion and two components of F = ma to determine the normal force: d dt

³

˜ ∂T

´

∂q ˙1

F · a˜ 2 F · a˜ 3

−

˜ ∂T

= ∂ q1

ma · a˜ 2 3 = ma · a ˜

=

F · a˜ 1

É

−→

Ê

−→

equation of motion,

equations for λ1 and λ2.

Additional examples of the application (and benefits) of this hybrid approach can be found in Section 3.10 and Exercise 3.11.

3.10

A Particle in Motion on a Helix

Increasing s &

109

θ

eb

et

en

eb

en et

A right-handed helix and its associated Frenet triad {et , en , eb}. Here, et is the unit tangent vector, en is the unit principal normal vector, and eb = e t × en is the binormal vector.

Figure 3.10

3.10

A Particle in Motion on a Helix

As an illustrative example, we turn our attention to establishing results for a particle that is in motion on a helix (see Figure 3.10). The helix can be either rough or smooth, and a variety of applied forces are considered. This example is interesting for several reasons. First, it is a prototypical problem to illustrate how the Serret–Frenet formulae and the Frenet triad {et , en , eb} help to determine the motion of a particle on a space curve. Second, we can use this example to illustrate a nonorthogonal curvilinear coordinate system.15

Curvilinear Coordinates, Basis Vectors, and Other Kinematics

A helix is defined by the intersection of two surfaces: a cylinder r = r0 and a helicoid z = cθ , where c and r0 are constants. To conveniently define these surfaces, we define a curvilinear coordinate system: 1

q

= θ =

3

q

−1

tan

= η =

±

x2 x1

z − α rθ

²

=

,

q

x3 − α

2

=

r=

È

x21 + x22,

È

x21 + x22 tan−1

15 This is a coordinate system in which ai are not necessarily parallel to a . i

±

x2 x1

²

.

110

Lagrange’s Equations of Motion for a Single Particle

It is appropriate to notice that r = x1 E1 + x2E2 + x3 E3 =

r cos( θ )E1 + r sin(θ )E2 + (η + αrθ ) E3 .

Our labeling of the coordinates minimizes subsequent manipulations. You should note that the curvilinear coordinate system is not defined when r = 0. That is, it has the same singularities as the cylindrical and spherical polar coordinate systems. The coordinates θ , r, and η can be used to define bases for E3 : a1

=

∂r ∂θ

=

reθ

+ α rE3 ,

a2

=

∂r

=

∂r

er + αθ E3,

a3 =

∂r ∂η

=

E3 .

For completeness, we note that a1 · (a2 × a3 ) = −r. As anticipated, the covariant basis fails to be a basis along the x3 axis. Using the representation of the gradient in cylindrical polar coordinates, we find that the contravariant basis vectors are 1 a 1 = ∇ θ = eθ , a 2 = ∇ r = er , a3 = ∇ η = E3 − αθ er − αeθ . r j

You should notice that ai · aj = δi , as expected. Further, neither the covariant basis nor the contravariant basis is orthogonal. You may recall that the Frenet triad for the helix of radius r0 is (from [215]) et

= √

1 1 + α2

(eθ + αE3 ) ,

e n = −e r ,

eb

= √

1 1 + α2

(E3 − α eθ ) .

We have tacitly assumed that θ˙ > 0 when writing these expressions. Furthermore, the torsion τ , curvature κ , and arc-length parameter s of the helix are τ =

α

r0 (1 + α2 )

,

κ=

1 r0 (1 + α2 )

,

À

s = r0 1 + α 2 (θ

− θ 0) −

s0 .

These results also apply to a helix for which α and r0 are functions of time. You should verify that a1 is parallel to et and that a2 and a3 are in the plane formed by en and eb . ∑ For a particle moving freely in E3 , we have the general representation v = 3i =1 q˙ i ai . From this result, we can immediately write v = θ˙ (reθ

+ α rE3 ) + r˙ (er + αθ E3 ) + η ˙ E3.

Furthermore, the kinetic energy of the particle is ´ m³ 2 2 2 2 ˙r + r θ˙ + (η T= ˙ + α r˙ θ + α rθ˙) . 2 When the particle is in motion on the helix, it is subject to two constraints ²2 = 0: ²1(r, t) =

q 2 − r0 ,

²2(r, t) =

q 3.

²1 =

0 and

3.10

A Particle in Motion on a Helix

111

In preparation for writing expressions for the constraint forces acting on a particle moving on the helix, you should calculate the gradient of these two functions. You might also notice that θ is the generalized coordinate for a particle moving on the helix that we will be using.

Forces

We assume that an applied force Fa acts on the particle. In addition, we assume that the friction is of the Coulomb type. Consequently, if the particle is moving relative to the helix: F = F a + λ1 a˜ 2 + λ 2a˜ 3 + Ff , where Ff

=

¼ ¼ 2 −μd ¼λ1 a ˜

¼ ˙ θa ˜1 ¼. + λ 2a ˜ ¼ ¼ ¼θ˙ a˜ 1¼ 3¼

In contrast, if the particle is not moving relative to the helix (i.e., θ is constant), then F = F a + λˆ 1 a˜ 2 + λˆ 2a˜ 3 + λˆ 3 a˜ 1. We ornament the λs with a hat to distinguish them from λ1 and λ2 that were used previously to define the components of the normal force N. The friction force in this case is subject to the static friction criterion ¼ ¼ ¼F f ¼

where

≤ μ s ³ N³ ,

µ

Ff

=

N

a˜ 1 (λˆ 1a˜ 2 + λˆ 2 a˜ 3 + λˆ 3a˜ 1 ) · ¼ ¼ ¼a˜ 1 ¼

2 3 1 ˆ 1a ˆ 2a ˆ 3a = λ ˜ +λ ˜ +λ ˜ −

¶

a˜ 1

¼ ¼, ¼a˜ 1 ¼

Ff .

Balance of Linear Momentum and Lagrange’s Equations

For an unconstrained particle moving in E3 , we have the three Lagrange’s equations d dt

±

∂T ∂q ˙i

²

−

∂T = ∂ qi

F · ai .

For the present coordinate system {θ , r, η}, these equations read d dt

±

∂T ∂ θ˙

²

2 = mr θ˙ + mα r(η ˙ + α r˙θ + α rθ˙ )

± −

d dt

±

∂T ∂ r˙

∂T ∂θ

² =

mα ˙r(η˙ + αr˙θ

+ α rθ˙ )

²

= mr ˙ + mαθ (η ˙ + α r˙θ + α rθ˙ )

=

F · (a1

=

reθ

+ α rE3 ),

112

Lagrange’s Equations of Motion for a Single Particle

± −

∂T ∂r

2 = mrθ˙ +

d dt

±

∂T ∂η ˙

²

mα θ˙ (η˙ + αr˙θ + α rθ˙ )

=

²

m(η˙ + αr˙θ

=

+ α rθ˙ )

F · (a2 ±

−

∂T ∂η

er + αθ E3 ),

=

²

0

=

=

F · (a3 = E3) . (3.23)

As with the equations of motion (3.4) for an unconstrained particle expressed in spherical polar coordinates, (3.23) can be expanded and expressed in a canonical form that highlights the mass matrix m [aik ], but we leave this as an exercise.16

Equations of Motion for the Particle on the Helix

We obtain the equations of motion for the particle on the helix from the preceding equations by substituting for the resultant force and imposing the constraints. With some algebra, for the case in which the particle is moving relative to the helix, we find three equations: ¼¼ ¼ ˙ ¼ ´ θ d ³ ¼ ¼ m(1 + α2 )r20 θ˙ = F a · a˜ 1 − μ d ¼λ1 a˜ 2 + λ2 a˜ 3 ¼ ¼ a˜ 1 ¼ , dt | θ˙ |

´ d ³ 2 mα r0θ θ˙ − m(1 + α2 )r0θ˙ 2 = F a · a˜ 2 + λ1 dt −μ d

¼ ¼ 2 ¼λ 1a˜

) d ( mαr0 θ˙ = F a · a˜ 3 + λ2 dt −μ d

¼ ¼ 2 ¼λ 1a˜

¼ ¼

˙ ˜1 · a ˜2 3 ¼ θ¼a ¼

+ λ2 a ˜

¼θ˙a˜ 1¼

,

¼

˙ ˜1 · a ˜3 3 ¼ θ¼a ¼ , ¼ ¼θ˙a˜ ¼ 1

+ λ2 a ˜

(3.24)

where a˜ 1

=

r0 e θ

a˜ 1 =

+ α r0 E3 ,

1 eθ , r0

a˜ 2

=

a˜ 2 = er ,

er + αθ E 3, a˜ 3

=

a˜ 3

=

E3 ,

E3 − αθ er − αeθ .

It is interesting to note that, as anticipated, a˜ 1 ³ et and a˜ 2 and a˜ 3 span the normal plane to the helix. The normal plane is also spanned by the unit normal en and unit binormal eb vectors. Equations (3.24) provide a differential equation (3.24)1 for the unconstrained motion of the particle and two equations for the unknowns λ 1 and λ2 . However, because a˜ 1 · a˜ 2 ± = 0 and a˜ 1 · a˜ 3 ± = 0, solving for λ 1 and λ2 from (3.24) is nontrivial. The hybrid approach ˙ discussed in Section 3.9 is ideally suited for this case. To proceed, we calculate v: a = v˙ = θ¨ a˜ 1 − θ˙ 2r0 er . 16 The canonical form is also discussed in Exercise 3.2.

3.10

Next, we compute F · a˜ 2 find that ¸

1

=

ma · a˜ 2 and F · a˜ 3 ¹¸

−αθ

(

1 + α2 1 + θ 2

−αθ

)

λ1

113

A Particle in Motion on a Helix

=

ma · a˜ 3 . Omitting details, we quickly Ë

¹ =

λ2

F a · a˜ 2 2 mr0αθ θ˙ − F a · a˜ 3 2 −mr0 θ˙ −

Ì

.

Inverting the matrix in this equation and performing some rearranging gives the final expressions for λ 1 and λ2 in terms of the motion of the particle and the applied forces: ¸

λ1 λ2

¹

=

¸

1

(

2 (ma − Fa ) · αθ E3 − α θ eθ + (ma −

1 + α2

(

)

1 + α 2 er

) ¹

Fa) · (E3 − αeθ )

.

These expressions for λ1 and λ2 can be used in (3.24)1 to establish a differential equation for θ that can be integrated numerically. For the case in which the motion of the particle is specified (i.e., the particle is not moving relative to the helix), we find, from F = ma, three equations for the three unknowns: ˆ = −F · a λ 3 a ˜ 1,

ˆ = −F · a λ 2 a ˜2,

ˆ = −F · a λ 1 a ˜3.

It remains to invoke the static friction criterion to examine when these three equations are valid, but this is left as an easy exercise. The Particle on a Smooth Helix

When the particle is moving on a smooth helix, the friction force is absent: F and, because Fc · a˜ 1

=

=

Fa + λ1a˜ 2 + λ 2a˜ 3,

0, Lagrange’s equations of motion decouple: ´ d ³ m(1 + α2 )r20 θ˙ = F a · a˜ 1, dt

´ d ³ 2 mα r0θ θ˙ − m(1 + α2 )r0θ˙ 2 = F a · a˜ 2 + λ1 , dt ) d ( mαr0 θ˙ = F a · a˜ 3 + λ2 . dt ∼ For this case, the configuration manifold = E. From Lagrange’s equations of motion, we can immediately identify the equation of motion for the particle:

M

m(1 + α2 )r20 θ¨

=

Fa · (r0eθ

+ α r0E3 ) .

The two remaining Lagrange’s equations provide the constraint force: Fc

2

˜ + λ 2a ˜ = λ1 a

³

=

mα2 r0θ¨θ

3

−

´

(

)

mr0 θ˙ 2 − F a · a˜ 2 a˜ 2 + mαr0 θ¨ − Fa · a˜ 3 a˜ 3 .

Once θ as a function of time has been calculated from the ordinary differential equation, then Fc , which is the normal force acting on the particle, as a function of time can be determined.

114

Lagrange’s Equations of Motion for a Single Particle

To illustrate the previous equations, consider the case in which the applied force is gravitational, F a = −mgE3 . Then, from the preceding equations: m(1 + α2 )r20 θ¨

= −mg α r0.

(3.25)

Subject to the initial conditions θ (t 0) = θ0 and θ˙ (t0) = ω0, this equation has the solution θ (t) = θ0 + ω0 (t − t0 ) −

gα ) (t − t0 )2 . 2r0 1 + α2 (

(3.26)

Using this result, we find that the constraint force is ±

Fc

=

mgαθ (t) 1 + α2

²

−

mr0 θ˙ 2(t) a˜ 2 +

mg (E3 − α(θ er + eθ )) , 1 + α2

where θ (t) is given by (3.26).

Some Observations

Suppose one is interested in determining only the differential equation governing the unconstrained motion of the particle moving on a smooth helix. In other words, the constraint forces are of no concern. One can obtain this differential equation by imposing the constraints on the expression for T : ´ m³ T˜ = 1 + α2 r02θ˙2 . 2 Furthermore, d dt

µ

˜ ∂T

∂ θ˙

¶

−

˜ ∂T

∂θ

=

F · a˜ 1

=

F · (r0 eθ

=

Fa · (r0 eθ

+ α r0E3 ) + α r0 E3)

.

A quick calculation shows that the resulting differential equation is identical to that obtained previously in (3.25). Because the constraint force Fc does no work (Fc · v = 0) and the gravitational force is conservative, the solution θ (t) to (3.25) conserves the total energy T˜ + mgαr0 θ . Clearly, Lagrange’s equations calculated with Approach II (i.e., with T˜ ) have their advantages, but they cannot accommodate dynamic friction forces. It is, however, the standard approach to Lagrange’s equations in the literature and textbooks. You should ˜ ˜ ˜ ˜ note that ∂∂T˙r = ∂∂ ηT˙ = ∂∂Tr = ∂∂ Tη = 0. Consequently, we cannot recover the other two Lagrange’s equations once we have imposed the constraints.

3.11

A Particle in Motion on a Moving Curve

As a second example of a particle moving on a curve, we consider the particle of mass m that is attached to a fixed point O by a linear spring of stiffness K and unstretched

3.11

A Particle in Motion on a Moving Curve

A O

E2

115

cos(ωt)

0

a1

a2

m g

E1

Plane curve y = f (x) + A cos(ωt) A

cos(ωt)

Figure 3.11 Schematic of a particle of mass m which is attached to a fixed point O by a spring and is free to move on a plane track. The ends of the track are subject to a vertical displacement A cos(ωt).

length ±0 shown in Figure 3.11. The curve is subject to a vertical oscillation A cos( ωt) and the particle is subject to a vertical gravitational force −mgE2 . We seek to determine the equations of motion of the particle. Curvilinear Coordinates, Basis Vectors, and Other Kinematics

The following coordinate system is used to describe the motion of the particle: q1

=

x,

q2

q3 = z.

y − f (x),

= η =

(3.27)

You may recall that this system is similar to the coordinate system that was used in Exercise 1.12. Thus, r = xE1 + (η + f (x)) E2 + zE3 . Using this coordinate system, when the particle is constrained to move on the curve it is subject to a pair of integrable constraints: ²1 (r, t ) = 0 and ²2 (r) = 0. Here, ²1 = η − A cos (ω t) ,

²2 =

z.

Observe that the first of these constraints is rheonomic (time-dependent) and the second is scleronomic. The covariant basis vectors associated with this coordinate system are easily computed using the above expression for r: a1

=

∂r ∂x

=

E1 + f ²E 2,

a2 =

Here, we use the familiar abbreviation f ² ⎡

∂r ∂η

df

= dx .

1 + f ²f ² [aik ] = ⎣ f² 0

=

E2 ,

a3

=

∂r ∂z

=

E3.

For completeness, we note that f² 1 0

⎤

0 0 ⎦. 1

116

Lagrange’s Equations of Motion for a Single Particle

The contravariant basis vectors for this system can be computed using the representation for the gradient in Cartesian coordinates: a1

= ∇q

1

= ∇x =

a2

E1 ,

= ∇η =

E2 − f ² E1 ,

a3

= ∇q

3

= ∇z =

E 3.

For this coordinate system and pair of constraints, a˜ k = ak and a˜ j = aj so we drop the tildes in the sequel. We note that ∇ ²1 = a2 and ∇ ²2 = a3. The velocity vector v and kinetic energy T for a particle that is unconstrained are

T

=

m ³( 2

v = ˙xa1 + η˙ a2 + z˙a3, )

´

1 + f ² f ² x˙ 2 + 2f ² x˙ η˙ + η˙ 2 + z˙2 .

When the particle is subject to the pair of constraints, the expressions for v and T simplify to

T˜

=

v˜ = x˙ a1 − Aω sin (ωt ) a2,

m ³( 2

)

´

1 + f ² f ² x˙ 2 − 2f ² x˙ Aω sin (ωt ) + A2 ω2 sin2 (ωt) .

In preparation for establishing an expression for the friction force, we note that the velocity vector of the particle relative to the curve is vrel

= x ˙ a1 .

The dynamic friction force opposes this velocity.

Forces

We assume that an applied force Fa acts on the particle: Fa

= −mgE2 +

Fs ,

where the spring force F s has the representation Fs

= −K ´

r , ³ r³

´ = ³ r³ − ±0 .

(3.28)

The constrained potential energy function for the particle is U˜

=

mg (f (x) + A cos (ωt)) −

K 2

±È

x2 + (f (x) + A cos (ωt)) 2 − ±0

²2

.

(3.29)

In addition to the conservative forces, we assume that the friction is of the Coulomb type. Consequently, if the particle is moving relative to the curve, then F = F a + λ1 a2 + λ2a3 + Ff , where Ff

= −μ d

¼ ¼ ¼ 2 3 ¼ v rel ¼λ1 a + λ2 a ¼ , ³˙ xa1 ³

¼ ¼ È ¼ 2 3¼ ¼λ1 a + λ2 a ¼ = λ21 (1 + f ²f ² ) + λ22 .

3.11

A Particle in Motion on a Moving Curve

117

In contrast, if the particle is not moving relative to the curve (i.e., x is constant), then F = F a + λˆ 1 a2 + λˆ 2a3 + λˆ 3 a1 =

(

F a + λˆ 1 E2 − f ² E1

)

ˆ E + λ ˆ E . +λ 2 3 3 1

The hats ornamenting the λs are used to distinguish them from λ1 and λ 2 that we used ¼ ¼ ¼ earlier. The friction force in this case is subject to the static friction criterion Ff ¼ ≤ μs ³ N³, where ±

Ff

=

a1 (λˆ 1a + λˆ 2 a + λˆ 3 a ) · ³a 1³ 2

3

1

²

a1 , ³ a 1³

N = λˆ 1 a2 + λˆ 2 a3 + λˆ 3a1 − Ff .

Balance of Linear Momentum and Lagrange’s Equations

We now examine the equations of motion for the particle. If the curve that the particle is in motion on is rough, then the three components of F = ma must be computed. As the developments will be similar to those in Section 3.10, we leave establishing these equations as an exercise. Because of our careful selection of the coordinates x, η, and z for the particle on a smooth curve, Approach II can be employed. In addition, as the ˜ to applied forces on the particle are conservative, we can use the Lagrangian L˜ = T˜ − U establish the equations of motion: d dt

µ

˜ ∂L

¶

˜ ∂L

−

∂x ˙

∂x

=

0.

The equation of motion of the particle is (

)

m 1 + f ² f ² x¨ + m f ² f ²² ˙x2 where f ²²

=

d2 f dx2

²

³

g − Aω 2 cos (ωt )

´ −

K´

d´ , dx

and È ´ =

d´ dx

= −m f

=

È

x2 + (f (x) + A cos (ωt)) 2 − ±0 , 1

)

(

x2 + (f (x) + A cos (ωt)) 2

x + (f (x) + A cos (ωt)) f ² .

Because the normal force acting on the particle does work, F c · v˜ =

³

2 3 λ1 a + λ2a

´

· ( ˙xa 1 −

Aω sin (ωt) a2)

= −A ω sin (ω t) λ 1,

we cannot expect the total energy T˜

˜ +U

of the particle to be conserved.

118

Lagrange’s Equations of Motion for a Single Particle

3.12

Closing Comments

In this chapter, several forms of Lagrange’s equations of motion for a particle were presented. The most fundamental of these forms is (see (3.1)) d dt

±

∂T ∂ q˙ i

² −

∂T = ∂ qi

F · ai .

In Exercise 3.2, we establish two other forms of these equations by expanding the partial derivatives with respect to the coordinates and their velocities. These two forms are a covariant form (3.33) and a contravariant form (3.34). If we decompose the forces acting on the particle into conservative and nonconservative forces, then we can transform (3.1) to (3.2): d dt

±

∂L ∂q ˙i

²

−

∂L = ∂ qi

Fncon · ai .

Now suppose that an integrable constraint is imposed on the particle, that this constraint can be written as q3 − f (t) = 0, and that the constraint force associated with this constraint is F c = λa3 . In this case, Lagrange’s equations of motion can be used to readily provide a set of differential equations for the generalized coordinates q1 and q2 : d dt

µ

˜ ∂L

q

∂ ˙α

¶

−

˜ ∂L

∂ qα

=

Fncon · a˜ α ,

α=

1, 2.

(3.30)

These equations feature the constrained Lagrangian L˜ that we obtain from L by imposing the integrable constraint q3 = f (t) and, most importantly, do not feature λ. That is, equations of motion (3.30) are reactionless. This case, in which all the constraints are integrable and the constraint forces are prescribed by use of Lagrange’s prescription, is an example of a mechanical system subject to “ideal constraints.” We also discussed the situation in which nonintegrable constraints were imposed on the system and outlined how the equations of motion could be obtained in these circumstances. The imposition of nonintegrable constraints will not affect the number of generalized coordinates, the configuration manifold, or the kinematical line element. The summary just presented will be identical for systems of particles, rigid bodies, and systems of particles and rigid bodies. The only major differences are that the calculation of the kinetic energy becomes significantly more complicated for these systems and that the right-hand sides of Lagrange’s equations feature several forces and moments. Despite these differences, the decoupling of the equations of motion into a set of reactionless equations governing the generalized coordinates will hold if Lagrange’s prescription for the constraint forces is used. This is one of the most remarkable features of Lagrange’s equations for systems subject to integrable constraints.

3.13

v

Exercises

119

coordinate curve

z

a1

a2 a1

O

r

a1

a2

a2

u

coordinate curve

The u and v coordinate surfaces on a θ coordinate surface (i.e., a half plane) for a parabolic coordinate system À {u, v, θ }. Observe that a1 and a2 are perpendicular to each other and have the same magnitude u2 + v2 .

Figure 3.12

3.13

Exercises

Exercise 3.1: Recall from Exercise 1.5 that, for a parabolic coordinate system {u, v, θ }: a1

=

∂r ∂u

=

ver + uE3 ,

a3

=

∂r ∂θ

a2

=

=

∂r ∂v

=

uer − vE 3,

uveθ ,

and a1

=

1

a1 , u2 + v2

a2

=

1

a 2, u 2 + v2

Representative u and v coordinate curves on a Figure 3.12.

θ

a3 =

1 eθ . uv

coordinate surface are shown in

120

Lagrange’s Equations of Motion for a Single Particle

(a) Consider a particle of mass m that is acted on by a force F and is free to move in E3 . Show that the equations of motion of the particle are

´ d ³ m(u2 + v2)u˙ − m( u˙ 2 + v˙ 2 )u − mv2uθ˙ 2 = F · a1, dt ´ d ³ m(u2 + v2 )v˙ − m( u˙ 2 + v˙ 2 )v − mu2vθ˙ 2 = F · a2 , dt d ³ 2 2 ´ mu v θ˙ = F · a3 . dt (b) Next, we are interested in a particle that is moving on the inner surface of the smooth surface of revolution

c 2 = −z +

À

z2 + r2,

where c is a constant. A vertical gravitational force −mgE 3 acts on the particle. Using the results of (a), derive the equations governing the unconstrained motion of the particle and show that the normal force acting on the particle is ³

´

N = − mu˙ 2c + mu2 cθ˙ 2 + mgc a2. Show that the two second-order differential equations governing the generalized coordinates can be written as a single second-order differential equation: ³

³

´´

h2 = −mgu, (3.31) mu3 c2 where h is a constant. (This constant is none other than HO · E3 , which is an integral of motion.) Noting that the units of u and c are m1/2 , what is a dimensionless form of the equation of motion (3.31)? (c) Show that the solutions of (3.31) conserve the energy m u2 + c2

u¨ + mu˙ 2u −

´ m³ 2 h2 u + c2 u˙ 2 + 2 2mu2c2 How can this expression for E be established?

E

=

+

´ mg ³ 2 u − c2 . 2

Exercise 3.2: For many mechanical systems, a canonical form of Lagrange’s equations can be established that is suited to numerical integrations. Here, we establish one such form (see (3.34)).17 This problem is adapted from the texts of McConnell [182] and Synge and Schild [277]. We take this opportunity to note that (3.34) can be found in an early paper by Ricci and Levi-Civita [239]. We start by recalling the covariant component forms of Lagrange’s equations of motion for a particle that is in motion under the influence of a resultant external force ∑ ∑ F = 3i =1 Fi ai = 3i=1 F i ai : 18 d dt

±

∂T ∂q ˙k

²

−

∂T ∂ qk

=

F · ak ,

17 As will become evident from the developments of later chapters, a related form can be established for any

mechanical system that features scleronomic integrable constraints and constraint forces and moments that are prescribed by use of Lagrange’s prescription. 18 The indices i, j, k, r, and s range from 1 to 3.

3.13

Exercises

121

where T

=

3

3

m ·· aik q˙ i q˙ k 2

T(qr , q˙ s ) =

(3.32)

i=1 k =1

and aik

=

aik (qr ) = ai · ak ,

aik

=

aik (qr ) = ai · ak .

You should notice that aik = aki and aik = aki . Here, we wish to show that Lagrange’s equations can be written in two other equivalent forms. The first one is the covariant form: m

3 ·

aki q¨

i

+

m

3 · 3 ·

[si, k]q˙ i q˙ s

i=1

˙ · = G

where a Christoffel symbol of the first kind is defined by

=

±

1 2

[si, k] = The symmetry aik

ak

=

Fk ,

(3.33)

i=1 s=1

∂ aki ∂ ask ∂ a si + − s i ∂q ∂q ∂ qk

²

.

aki implies that [si, k] = [is, k].

The second form of Lagrange’s equations is known as the contravariant form: mq¨

k

+

m

3 · 3 · i=1 s=1

k i s ˙ · ak = ¶si q ˙ q ˙ = G

Fk,

(3.34)

where a Christoffel symbol of the second kind is defined by k ¶ij =

3 ·

akr [ij, r]

r=1 3

=

1 · kr a 2 r =1

±

∂ air ∂ a rj ∂ a ij + − j i ∂q ∂q ∂ qr

²

.

As with the Christoffel symbols of the first kind, the symmetry aik

=

aki implies that

k k ¶rs = ¶sr .

The contravariant form (3.34) of Lagrange’s equations is used in numerical simulations of mechanical systems.19 To compute these equations it is imperative that [aik ] is invertible. In other words, the coordinates used to describe the motion of the particle are singularity free for the locations of the particle that are of interest. 19 Most numerical integration packages assume that the differential equations to be integrated are of the form ˙x = f(x, t). By defining the set of variables (states) x1 = q1 , . . . , x3 = q 3, x4 = q˙ 1 , . . . , x 6 = q˙ 3 , the

contravariant form (3.34) of Lagrange’s equations can easily be placed in the form x˙ = f( x, t).

122

Lagrange’s Equations of Motion for a Single Particle

(a) Show that the Christoffel symbols for this case are identical to the connection coefficients: [si, k]

k k ¶ij = γ ij ,

= γsi,k ,

(3.35)

where each of the 27 connection coefficients γsi,k and γsik are defined by the relations γ si,k =

∂ as · ak , ∂ qi

∂ as ∂ ak k k γ si = ·a = − · ∂ qi ∂ qi

as.

(3.36)

In your solution you will also observe that the connection coefficients have a symmetry k k γsi = γ is

γsi,k = γ is,k ,

that can be traced to the following commutativity result: 2 ∂ as ∂ r ∂ ai = = . i s i ∂q ∂q ∂q ∂ qs

(3.37)

(b) Starting from Lagrange’s equations d dt

±

∂T ∂q ˙k

² −

∂T = ∂ qk

F · ak ,

derive the following representation for the covariant component form: 20 m

3 ·

aki q¨ i + m

3 3 · ·

[si, k]q˙ i q˙ s

˙ · = G

ak

=

Fk .

˙ · = G

ak

=

Fk ,

i=1 s =1

i=1

(c) Starting from Lagrange’s equations in the form m

3 ·

aki q¨

i

+

i=1

m

3 · 3 ·

[si, k]q˙ i q˙ s

s =1 i=1

derive the following representation for the contravariant component form:21 mq¨ k

+

m

3 · 3 · s=1 i =1

k i s ˙ · ak = ¶si q ˙ q ˙ = G

Fk .

(d) For which coordinate system do the Christoffel symbols vanish? 20 Hint: Expand the partial derivatives of T using the representation (3.32). Then, take the appropriate time

derivative and reorganize the resulting equation by using the aforementioned symmetries. You may need to relabel certain indices to obtain the desired results. 21 Hint: Multiply the covariant form by ask and sum over k. After some rearranging and relabeling of the indices, you should get the final desired result. Notice that the covariant component and contravariant component forms of these equations can be viewed as linear combinations of each other.

3.13

Exercises

123

Exercise 3.3: Recall that for spherical polar coordinates, {R, φ, θ }, the covariant basis vectors are a1

=

eR ,

a2

=

Reφ ,

a3

=

R sin(φ)eθ ,

1 eφ , R

a3

=

1 eθ . R sin(φ)

and the contravariant basis vectors are a1

=

a2

eR ,

=

Furthermore, the linear momentum and kinetic energy of a particle of mass m are ´ m³ 2 ˙ 1 + mφ ˙ a 2 + mθ˙ a 3, R˙ + R2 φ˙ 2 + R2 sin2(φ )θ˙ 2 . G = mRa T= 2

(a) For a particle of mass m that is in motion in E3 under the influence of a resultant force F, establish the three covariant components of Lagrange’s equations of motion. In your solution, avoid explicitly calculating the 27 Christoffel symbols of the first kind. (b) For a particle of mass m that is in motion in E3 under the influence of a resultant force F, establish the three contravariant components of Lagrange’s equations of motion. In your solution, avoid explicitly calculating the 27 Christoffel symbols of the second kind. Exercise 3.4: Consider a particle that is in motion on a rough surface. A curvilinear coordinate system is chosen such that the surface can be described by the equation q3

=

d(t),

where d(t) is a known function of time t. (a) Suppose that the particle is moving on the rough surface. (i) Argue that vrel = q˙ 1 a˜ 1 + q˙ 2a˜ 2. What is the kinematical line element ds for the surface? (ii) Give a prescription for the constraint force acting on the particle. (b) Suppose that the particle is stationary on the rough surface. In this case, two equivalent prescriptions for the constraint force are Fc

=

N + Ff

3 · =

i λia ,

i=1

where the tildes are dropped for convenience. (i) Show that ⎡ ⎣

λ1 λ2 λ3

⎤

⎡

a11 ⎦ = ⎣ a12 a13

a12 a22 a23

⎤⎡

⎤

F 1f 0 ⎥ ⎢ 0 ⎦ ⎣ F 2f ⎦ , 1 N

where N and F 1f , Ff2 uniquely define the normal force N and friction force Ff , respectively, and aik = ai · ak with i, k = 1, 2, 3. (ii) For which coordinate systems do Ff1 = λ1 , F2f = λ2 , and N = λ 3? Give an example to illustrate your answer.

124

Lagrange’s Equations of Motion for a Single Particle

(c) Suppose that a spring force and a gravitational force also act on the particle. Prove that the total energy of the particle is not conserved, even when the friction force is static. Exercise 3.5: Consider a particle of mass m that is in motion on a helicoid. In terms of cylindrical polar coordinates {r, θ , z}, the equation of the right helicoid is z = βθ , where β is a constant. A gravitational force −mgE3 acts on the particle. (a) Consider the following curvilinear coordinate system for E3 : q1 = θ ,

q2

=

q3 =

r,

ν =

z − βθ .

Show that a1

=

and that a1

=

reθ

+ β E 3,

1 eθ , r

a2

a2

=

er ,

=

er , a3

a3

=

=

E3 −

E3

β

r

eθ .

(b) Consider a particle moving on the smooth helicoid. The generalized coordinates for the particle are θ and r. (i) What is the constraint on the motion of the particle, and what is a prescription for the constraint force Fc enforcing this constraint? (ii) What is the kinematical line element ds for the helicoid? (iii) Show that the equations governing the unconstrained motion of the particle are ´ ´ d ³ ³2 m r + β 2 θ˙ = −mgβ, dt d 2 (mr˙ ) − mrθ˙ = 0. dt

(3.38)

(iv) Prove that the angular momentum HO · E3 is not conserved. (c) Suppose the constraint r θ˙ + h(t) = 0 is imposed on the particle. Establish a second-order differential equation for r(t), a differential equation for θ (t), and an equation for the constraint force enforcing the constraint. Indicate how you would solve these equations to determine the motion of the particle and the constraint forces acting on it. Exercise 3.6: Consider a particle of mass m that is free to move on the smooth inner surface of a hemisphere of radius R0 (cf. Figure 3.13). The particle is under the influence of a gravitational force −mgE3 . (a) Using a spherical polar coordinate system, what is the constraint on the motion of the particle? Give a prescription for the constraint force acting on the particle.

3.13

Exercises

125

E3 g

E2

O

r

E1

m

Figure 3.13

Schematic of a particle of mass m moving on the inside of a hemisphere of radius R0 .

(b) Using Lagrange’s equations, establish the equations of motion for the particle and an expression for the constraint force. (c) Prove that the total energy E and the angular momentum HO · E3 of the particle are conserved. (d) Show that the normal force acting on the particle can be expressed as a function of the position of the particle and its initial energy E 0: ²

±

2E 0 − R0

N=

+

3mg cos (φ) eR .

(e) Numerically integrate the equations of motion of the particle and show that there are instances for which it will always remain on the surface of the hemisphere. Exercise 3.7: As shown in Figure 3.14, consider a bead of mass m that is free to move on a smooth circular wire of radius R0 . The wire has a constant angular velocity ³0 E3 and whirls about the configuration shown in the figure. The particle is also under the influence of a gravitational force −mgE3. This is a classic problem that is discussed in several textbooks (see, e.g., [104]). 22 (a) Using a spherical polar coordinate system, what are the two constraints on the motion of the particle? Give a prescription for the constraint force Fc acting on the 1 ∼ particle. Observe that the configuration manifold = S and that the generalized coordinate φ = φe ∈ [0, 2π ]. (b) Using Lagrange’s equations, establish the equation of motion for the particle:

M

± ¨ = φ

2 ³0 cos (φ) +

g R0

²

sin (φ) .

(3.39)

22 A more comprehensive model, based on the recent paper [123], that incorporates the inertia of the rigid

hoop is discussed in Section 11.4.

126

Lagrange’s Equations of Motion for a Single Particle

Ω(t )

E3

g

E2

O

r

E1

m

Schematic of a bead of mass m moving on a smooth circular wire that is being whirled about the vertical at a speed ³(t) = ³0.

Figure 3.14

After nondimensionalizing (3.39), numerically integrate the resulting differential equation and construct its phase portrait for values of 0.5, 1.0, and 1.5 of the g parameter R ³2 . 0

0

(c) Recall that an equilibrium point x = x0 of the differential equation x¨ that ˙x = 0 and f (x0 ) = 0. Show that (3.39) has three equilibria: µ

φ0 =

0,

φ0 = π ,

φ0 =

−1

cos

−

g R 0³20

=

f (x) is such

¶

.

Give physical interpretations for these equilibria and show that the third one is possible if, and only if, ³20 is sufficiently large. How do these results correlate with your phase portraits? (d) Starting from the work–energy theorem T˙ = F · v, prove that the total energy of the particle is not conserved: E˙ = Nθ R0 ³0 sin (φ) , where Nθ is the eθ component of the normal force acting on the particle. (e) Show that the energy-like function23 ´ mg m³ 2 ˙ − ³2 sin2 (φ) + φ cos (φ) V= 0 2 R0 is conserved by the solutions to (3.39). 23 This function can be established using the work–energy theorem E ˙

= N · v after substituting for N. Equivalently, it can be determined from (3.39) using the identity vdv = a(φ )d φ, where v = φ˙ . The reader is also referred to our earlier discussion on page 57.

3.13

y

m q

2

Exercises

127

= −4

2

−4

6

x

−4 Figure 3.15

Schematic of a particle moving on a parabola in the x–y plane.

(f) Suppose that the wire is free to rotate about E3 so now the particle has two degrees of freedom. The generalized coordinates for the system are θ ∈ [0, 2π ] and φ = φe ∈ [0, 2π ] and the configuration manifold is homeomorphic to a two-torus T 2 . Establish Lagrange’s equations of motion for the particle. In contrast to the particle shown in Figure 3.14, the solutions to these equations of motion conserve the total energy E of the particle and the angular momentum component HO · E3.

M

Exercise 3.8: Consider a particle of mass m moving in E3 . If the coordinate system (1.13) is used to describe its kinematics, then establish expressions for the velocity vector v and the kinetic energy T of the particle. (a) Suppose a particle is constrained to move on a rough parabolic surface described by the equation x − y2

= −4.

Give a prescription for the constraint force Fc acting on the particle, and establish the equations of motion for the particle. 24 (b) As illustrated in Figure 3.15, suppose a particle is constrained to move on the smooth parabola x − y2 = −4,

z = 0.

Give a prescription for the constraint force Fc acting on the particle, and establish the equation of motion for the particle. Exercise 3.9: Consider the following curvilinear coordinate system, which is suited to examining the dynamics of a particle on a surface of revolution x3 = z = f (r) such as the one shown in Figure 3.16: 24 As discussed in Section 3.9, for the coordinate system used in this problem, F f

·a ˜ 3 ± = 0. Consequently, you will find it easier to use the first two Lagrange’s equations to compute the equations of motion and the equation F · a˜ 3 = ma · a˜ 3 to compute λ, where N = λa˜ 3 . Observe that we have labeled the coordinates (1.13) so that q1 = y, q2 = z, and q3 = x − β y2 .

128

Lagrange’s Equations of Motion for a Single Particle

E3 E2

g

r

E1 m

z z

η

= f (r )

= 0 coordinate surface

Schematic of a particle of mass m which is moving on a smooth surface of revolution in E3 under the influence of a gravitational force −mgE 3.

Figure 3.16

q1

=

r=

È

q2 = θ

x21 + x22,

± =

arc tan

x2 x1

²

q3

,

= ν =

z − f (r).

(a) Show that the covariant basis vectors are a1

=

er + f ² E3 ,

a2

=

reθ ,

a3

=

E3 ,

where the prime denotes the derivative with respect to r: f ² = df /dr. Note that this basis is not orthogonal. (b) Show that the contravariant basis vectors for this coordinate system are a1

=

er ,

a2

=

1 eθ , r

a3

=

E3 − f ² er .

(c) Compute the matrices [aik ] and [ars ] and establish an expression for the kinetic energy T of a particle of mass m using the {r, θ , ν } coordinate system. (d) Draw a representative r coordinate surface and show how it is foliated by ν and θ coordinate curves. Show the vectors a2, a3, and a1 in this figure. (e) Draw a representative θ coordinate surface and show how it is foliated by ν and r coordinate curves. Show the vectors a1, a3, a1 , a2 , and a3 in this figure. (f) Verify that ∂∂ θT˙ = HO · E3 , where HO = r × mv is the angular momentum of the particle relative to a fixed point O. (g) Consider a particle of mass m moving on the smooth surface of revolution ν = 0. If a gravitational force −mgE3 acts on the particle, show that the equations of motion of the particle can be found simply from the following pair of functions:25 T˜ =

´ ) m ³( 1 + f ² f ² ˙r2 + r2 θ˙ 2 , 2

U˜

=

mgf .

Show that one of these equations is equivalent to conservation of a component of the angular momentum of the particle. 25 Observe that T˜ can be used to define the kinematical line element ds for the surface of revolution.

3.13

Exercises

129

a2 θ

coordinate surface

a3

z

a2

r

a3 The q2 = z and q3 = η coordinate curves on a θ coordinate surface. Representative examples of the covariant basis vectors a2 and a3 are also shown.

Figure 3.17

(h) Show that the following energy-like function V is conserved by the equations of motion you found in (g): ) h2 m( 1 + f ² f ² ˙r2 + O 2 + mgf , 2 2mr where hO is the value of the conserved angular momentum component of the particle. (i) Show that the normal force N acting on the particle is

V

=

³

´

N = mg + mf ²² ˙r2 + mf ² r¨ a3 . Exercise 3.10: Consider the following curvilinear coordinate system, which is suited to examining the dynamics of a particle on a surface of revolution r = h(z), such as a cylinder: 1

q

±

= θ

x2 = arc tan x1

²

q2

,

=

z,

q3

= η =

r − h(z).

Representative examples of the η and z coordinate curves for this system are shown in Figure 3.17. We invite the reader to contrast this coordinate system to the one used in Exercise 3.9. (a) Show that the covariant basis vectors are a1

= (η +

h ) eθ ,

a2

=

E3 + h² er ,

a3

=

er ,

where the prime denotes the derivative with respect to z: h² = dh/dz. Note that this basis is not orthogonal. (b) Show that the contravariant basis vectors for this coordinate system are a1

=

1 η +h

eθ ,

a2

=

E3 ,

a3

=

er − h² E3.

130

Lagrange’s Equations of Motion for a Single Particle

(c) Compute the matrices [aik ] and [ars ] and establish an expression for the kinetic energy T of a particle of mass m using the {θ , z, η } coordinate system. (d) Draw a representative z coordinate surface and show how it is foliated by η and θ coordinate curves. Show the vectors a1 , a3, and a2 in this figure. (e) Draw a representative η coordinate surface and show how it is foliated by θ and z coordinate curves. Show the vectors a1 , a2, a1, a2 , and a3 in this figure. (f) Verify that ∂∂ θT˙ = HO · E3 , where HO = r × mv is the angular momentum of the particle relative to a fixed point O. (g) Consider a particle of mass m moving on the smooth surface of revolution η = 0. If a gravitational force −mgE3 acts on the particle, show that the equations of motion of the particle can be found simply from the following pair of functions: 26 T˜

=

´ ) m ³( 1 + h² h² ˙z2 + h2 θ˙ 2 , 2

U˜

=

mgz.

Show that one of these equations is equivalent to conservation of a component of the angular momentum of the particle. (h) Simplify the equations that you found in (g) to the case of a particle moving on a cylinder of radius r0 . If gravity is ignored, show that the path of the particle is either a circle, a straight vertical line, or a circular helix. Exercise 3.11: A candidate plane curve for a roller coaster design is a curve known as the “cubique d’Agnesi:”27 y = f (x) =

8a3 , x2 + 4a2

z

=

0,

where the parameter a in the expression for f (x) is a constant. We seek to establish the equations of motion for the particle. When friction is absent, Lagrange’s equations of motion provide the differential equation of motion for the particle in a direct manner. However, when dynamic friction is present, we will find that, for the coordinate system we have chosen, solving for N is nontrivial. 28 The following coordinate system and its associated covariant basis vectors are used to analyze the motion of a cart on the roller coaster: q1

=

x,

q2

= ν =

q3 = z.

y − f (x),

In the sequel, we use the shorthand notations f ²

df

= dx

and f ²²

=

d 2f . dx2

(a) Show that the covariant basis vectors for this coordinate system have the representations a1

=

E1 + f ² E2 ,

a2

=

E2 ,

a3

=

E3 .

26 Observe that T˜ can be used to define the kinematical line element ds for the surface of revolution. 27 Maria Gaetana Agnesi (1718–1799) was an Italian mathematician and a faculty member at the University

of Bologna.

28 An alternative approach for this problem would be to use the Serret–Frenet triad and project F

= ma onto eb }. The e n and e b = ´ E3 components would provide the normal force N and the et component would provide the differential equation of motion (3.40). We exploit this observation in (e) below.

{ et , en ,

3.13

131

Exercises

E2 g m

E1 Plane curve y = f (x),

z

= 0

Schematic of a particle of mass m which is free to move on a plane space curve. A vertical gravitational force −mgE2 acts on the particle.

Figure 3.18

Compute the matrix [aik ] associated with this coordinate system. Illustrate a1 at three locations (one with x < 0, one with x = 0, and one with x > 0) on an x coordinate curve. (b) Show that the contravariant basis vectors for this coordinate system have the following representations: a1

=

a2

E1 ,

=

a3

E2 − f ² E1 ,

=

E3 .

Illustrate a2 at three locations (one with x < 0, one with x = 0, and one with x > 0) on an x coordinate curve. (c) Referring to Figure 3.18, consider a particle of mass m moving on a rough track which can be modeled using the coordinate system as an x coordinate curve. Give a prescription for, and a physical interpretation of, the constraint force enforcing the constraints on the motion of the particle. In this instance, it is helpful to note that vrel

= x ˙ a 1.

You should also observe that the friction force Ff is not orthogonal to a2 : Ff · a2 (d) Show that the motion of the particle is governed by the differential equation (

)

m 1 + f ² f ² x¨ + mf ² f ²² x˙ 2

= −mf

²

g − μ k ³ N³

À

1 + f ²f ²

x˙ . |˙ x|

±=

0.

(3.40)

Using the other pair of Lagrange’s equations and the help of (3.40), show that µ

N

=

=

mg + mf x˙ ²²

2

+

mf x¨ + μk ³N³ ²

³ ´ 1 ²² 2 mg + mf x ˙ a 2. 1 + f ²f ²

À

|˙ x|

f ² x˙ 1 + f ²f ²

¶

a2

Clearly, solving for the normal force is nontrivial when friction is present. For the coordinate system used in this problem, this difficulty can be traced to the fact that Ff · a2 ± = 0. (e) Show that the acceleration vector of the particle in motion on the x coordinate curve is a = ¨xa1 + x˙ 2f ²² E2 . Show that by computing the equations F · a2

=

ma · a2 ,

F · a3

=

ma · a3,

132

Lagrange’s Equations of Motion for a Single Particle

m

Spinning plate

E2 E1

O

Ω

Schematic of a particle of mass m that is attached to a fixed point O by an elastic spring. A vertical gravitational force −mgE3 acts on the particle and the particle is free to move in a smooth groove on a plate that is rotating about the vertical axis with a nonconstant speed ³ = ³ (t).

Figure 3.19

an expression for the normal force can readily be found: N=

³ ´ 1 ²² 2 mg + mf x ˙ a2 . 1 + f ²f ²

Provide a geometric reason as to why the approach taken for solving for N in (e) is significantly easier than the approach used in (d).29 Exercise 3.12: As shown in Figure 3.19, a particle of mass m is attached to a fixed point O by a linearly elastic spring. The spring has a stiffness K and an unstretched length ±0 . The particle is free to move on a smooth groove on a disk that rotates about the vertical axis with a speed ³(t). A vertical gravitational force −mgE 3 acts on the particle. Using a cylindrical polar coordinate system, the two constraints on the motion of the particle can be expressed as ²1 =

0,

where ²1 = θ − θ ( 0) −

²2 =

Á t 0

0,

³ (τ ) d τ ,

The constrained potential energy of the particle is ˜ U

=

K 2 (r − ±0) . 2

29 See Section 3.9 for assistance with this problem.

²2 =

z.

3.13

133

Exercises

(a) Give prescriptions for the constraint forces enforcing the two constraints ²1 and ²2 = 0. (b) Show that the motion of the particle is governed by the differential equation mr¨ − mr³2

= −K (r − ±0 ) .

=

0

(3.41)

(c) Show that the motion of the particle does not conserve the total energy E, ´ K m³2 r˙ + r 2³2 + (r − ±0 )2 , 2 2 because the nonconservative forces acting on the particle perform work when ³ ± = 0. (d) Show that the energy-like function30

E=

´ m³ 2 K 2 2 2 (r − ±0 ) ˙r − r ³ + 2 2 is conserved by the solutions to (3.41) È provided ³ is constant.

V

=

t and position u = (e) By nondimensionalizing time τ = K m of motion (3.41) can be written in the form

r

±0

, show that the equation

u²² + (u − 1) − ω2 u = 0,

(3.42) m³2 ω2 = . K

where the prime denotes the derivative with respect to τ and (i) What (if any) are the equilibria of (3.42) and what energy-like function e do the solutions to (3.42) conserve? (ii) Numerically integrate (3.42) and construct its phase portraits when ω = 0, ω2 = 0.5, and any value of ω2 > 1. Your phase portraits should be restricted to the half plane u > 0. (iii) With the help of the results from (ii), show that the solutions to the equations of motion can be categorized as (1) impact-free motions where the particle either escapes to infinity or oscillates about the equilibrium point and (2) motions where an impact occurs when u µ 0 and (assuming an elastic collision) the particle either escapes to infinity or oscillates about the equilibrium point. (iv) Suppose that the collisions that happen when u µ 0 were inelastic with a coefficient of restitution 0 < ´ < 1. Argue that the particle will either escape to infinity if ω > 1 or eventually oscillate about the equilibrium, albeit with the stretched length of the spring becoming 0 periodically.

30 This function can be established using the work–energy theorem E ˙

= N · v after substituting for N. Equivalently, it can be determined from (3.41) using the identity vdv = a(r)dr, where v = r˙ . The reader is also referred to our earlier discussion on page 57.

Part II

A System of Particles

4

Lagrange’s Equations of Motion for a System of Particles

4.1

Introduction

In this chapter, we establish Lagrange’s equations for a system of particles by starting with the balances of linear momentum for each of the particles. Our derivation is based on the results presented by Synge and Griffith [276, Chapter 15].1 We supplement their work with a discussion of constraints and potential energies. To examine the geometry inherent in Lagrange’s equations of motion for the system of particles, we also discuss the explicit construction of a representative single particle by Casey [37]. In his construction, the particle of mass m moves on the configuration manifold and has the same kinetic energy as the system of particles it represents. The work presented in this chapter emphasizes the equivalence of Lagrange’s equations of motion for a system of particles and the balances of linear momenta. For completeness, a brief discussion of the principle of virtual work, D’Alembert’s principle, Gauss’ principle of least constraint, and Hamilton’s principle are also presented in Section 4.12. The chapter closes with a discussion of a canonical form of Lagrange’s equations of motion in which time-independent integrable (i.e., scleronomic) constraints are present. For many specific problems, one can obtain Lagrange’s equations by merely calculating the kinetic and potential energies of the system. This approach is used in most dynamics textbooks, and neither the construction of a single particle nor the components of force vectors are mentioned.2 For many cases – that are not possible to treat using the approach adopted in most dynamics textbooks – we find that the use of Synge and Griffith’s representation of Lagrange’s equations of motion allows us to increase the range of application of Lagrange’s equations tremendously. For instance, as will be shown in some of the examples in Chapter 5, dynamic Coulomb friction can be accommodated.

M

4.2

A System of N Particles

Here, we are interested in establishing the equations of motion for a system of N particles. The first step in this development is to discuss the individual elements in the system of particles. 1 Related derivations for systems of particles can be found in several texts. For example, Greenwood [105,

Section 6-6] and Whittaker [306, Section 21].

2 These texts use either the principle of virtual velocities or Hamilton’s principle (also known as the

principle of least action) to derive Lagrange’s equations.

138

Lagrange’s Equations of Motion for a System of Particles

Fi E3 mi

ri E2

O

E1 3 Figure 4.1 A single particle of mass mi in E . The position vector of this particle is denoted by ri and the resultant external force acting on the particle is denoted by F i.

We consider a system of N particles, each of which is in motion in three-dimensional Euclidean space E 3. For the particle of mass mi (see Figure 4.1), the position vector is ri

=

3 ± j j=1

xi Ej .

We also recall that the kinetic energy of the particle is Ti

=

1 mi v i · v i . 2

It is also convenient to recall that the linear momentum of the particle of mass mi is Gi = mi r˙ i = mi vi . The resultant force acting on the particle of mass mi has the representation Fi

=

3 ± j j= 1

Fi Ej .

The balance of linear momentum for the particle of mass mi is Fi

=

mi v˙ i .

As a consequence of the balance of linear momentum for the particle, we have the angular momentum theorem ˙ H Oi

=

ri × F i ,

where HO i = ri × mi vi is the angular momentum of the particle relative to the fixed point O. A second consequence of the linear momentum balance is the work–energy theorem T˙ i for the particle of mass mi .

=

F i · vi

4.3

Coordinates

139

The Kinetic Energy

For the system of particles, the combined (total) kinetic energy T is the sum of the kinetic energies: T

=

T1 + · · · + T N .

With the help of the work–energy theorem for each particle, we can determine the corresponding result for the system of particles: T˙

=

F1 · v1 + · · · + FN · vN .

(4.1)

This result is used to establish energy conservation (or lack thereof) in a system of particles. The Center of Mass

For the system of particles, we can define a center of mass C. This point, which lies in E3 , has the position vector r¯ , where r¯ =

1 m1 + · · · + mN

( m1 r 1 + · · · +

mN rN ) .

It is easy to show from this result that the linear momentum of a system of particles has the representation G = (m1 + · · · + mN ) r˙¯ . Later on, when we examine a particle of mass m moving in E3N , it is important not to confuse the representative particle with C.

4.3

Coordinates

In many problems, separate sets of Cartesian coordinates for each position vector ri are not a convenient choice. Indeed, for many systems of two particles, we use one coordinate system to describe r1 and another to describe the relative position vector r2 − r1. For instance, for the system shown in Figure 4.2, we might use Cartesian coordinates for r1 and spherical polar coordinates for r2 − r1: r1 = x E1 + y E2 + z E3 , In this equation, R 2 = eR2

=

±0

r2 = x E1 + y E 2 + z E3 + R2 eR2 .

(4.2)

is the length of the rod connecting the particles, and

sin(φ2 ) (cos(θ2 ) E1 + sin(θ2 ) E2 ) + cos( φ2 ) E3,

with related definitions for eφ2 and eθ2 . Notice that eR 2 points from m1 to m2 . The coordinate system most suited to a given system of particles will, as in the case of a single particle, depend on the constraints and applied forces. In general, for the system of N particles we will use a coordinate system denoted by ² 1 ³ q , . . . , q3N . As in the case of a single particle, we assume that we can determine the

140

Lagrange’s Equations of Motion for a System of Particles

m1

E3 m2

g

E2

O

E1 Figure 4.2

A particle of mass m1 attached by a rigid rod of length ± 0 to a particle of mass m2 .

unique Cartesian coordinates for a system of particles once q1 , . . . , q3N are known, and vice versa: qK j

xi

= q ˆ

K j

= x ˆ i

´

´

x11, x21 , x31 , . . . , x1N , x2N , x3N

q1 , . . . , q3N

µ

µ

(K

=

1, . . . , 3N),

(i = 1, . . . , N and j = 1, 2, 3).

To calculate Lagrange’s equations of motion, we need to calculate the following 3N 2 vectors: ∂ ri ∂ qK

(K

=

1, . . . , 3N and i = 1, . . . , N).

These vectors play the role of the basis vectors ak that we used with the single particle earlier. We shall use the mass matrix M associated with the kinetic energy of the system of particles to determine the presence and absence of singularities in the coordinate ² ³ 1 3N system q , . . . , q used to describe the motion of the system of particles. As an example, let us return to (4.2). In these equations, we gave examples of the coordinates for a two-particle system: q1

=

x = x11,

q2

=

y = x21 ,

q3

=

z = x31,

q4

=

R 2,

=

0

q5

q 6 = φ2 .

= θ 2,

For this coordinate system, and with the help of (4.2): ∂ r1

∂q j

=

Ej ,

∂ r1

∂ q( j+3)

and ∂ r2 = ∂ qj

E j,

∂ r2 = eR2 , ∂ q4

∂ r2

∂ q5

=

R2 sin(φ2) eθ 2 ,

∂ r2

∂ q6

=

R2 eφ2 ,

where j = 1, 2, 3 in these two sets of equations. It is left to the student to realize how coordinate system (4.2) can also be used to describe the kinematics of the particle system shown in Figure 4.3.

4.4

Constraints and Constraint Forces

141

m1

E3 Linear spring

E2

O

m2

E1 A particle of mass m1 attached by a linear spring of stiffness K and unstretched length to a particle of mass m2 .

Figure 4.3

±0

4.4

Constraints and Constraint Forces

For a single particle, one of the key notions we encountered earlier was kinematic constraints. We now turn to examining this topic for a system of particles. As we shall see, the extension to this case is not as difficult as it first may seem. Let us start with a physical system and use it to establish expressions for constraint forces. We will later show that these constraint forces are compatible with Lagrange’s prescription. Suppose we have two particles connected by a rigid rod of length ±0 (see Figure 4.2). Then, the constraint imposed by the rod on their motion is ±r1 − r2± − ±0 = 0. To enforce this constraint, the particles will be subject to equal and opposite constraint forces: F c1

= −µ t,

F c2

= µ t,

(4.3)

where µ is the tension force in the rod and the unit vector t points from m1 to m2 : t=

r2 − r1 . r1±

± r2 −

If we describe r2 − r1 = ± 0eR2 , which we did earlier, then t = eR2 . Writing the constraint as ² =

0,

where ² = ± r2 − r1 ± − ±0 ,

then as3 ∂² ∂ r1

= −

r2 − r1 , r1 ±

± r2 −

∂² ∂ r2

=

r2 − r1 , r1±

±r2 −

3 For a function ² (r , . . . , r , t ), the partial derivative ∂ ² /∂ r (for example) assumes that the position 1 N 3 3

vector r 3 of the particle of mass m3 is varied while the position vectors of the remaining N − 1 particles and time are held fixed. In the sequel, several examples of these partial derivatives will be discussed.

142

Lagrange’s Equations of Motion for a System of Particles

m3

e2

E2

e1

E1

C

E1

O

e1

φ

m1

Q

Vehicle

m2

A very simple model for a braking three-wheeled vehicle. The distributed mass of the vehicle is modeled as three mass particles of masses m1 , m2, and m3 . The particles of mass m1 and m2 model wheels that are rolling without slipping, and the particle of mass m3 models a slipping wheel.

Figure 4.4

we observe that (4.3) can be expressed as Fc 1

= µ

∂² ∂ r1

,

F c2

= µ

∂² ∂ r2

.

(4.4)

This result will help motivate Lagrange’s prescription for the constraint forces acting on a system of particles. As a second example, consider the three-particle system shown in Figure 4.4. This model is a prototypical model for a braking vehicle and is used to explain the vehicle instability that often occurs when the front wheels lock during braking. 4 We assume that the motion of this cart is planar. As a result, the motion of the center of mass C of the system is planar: r¯ = xE 1 + yE2 . Further, we need to supplement x and y only by an angle φ to determine the position of any point on the cart. That is, the system is subject to six integrable constraints and has three degrees of freedom. Thus, for this system, we make the following choices for the nine coordinates q1, . . . , q9: q1

=

x,

q2

=

y,

q3

= φ,

q3+k

=

r2 · Ek ,

q6+k

=

r3 · Ek .

Here, k = 1, 2, 3 and r¯ = xE1 + yE2 . The unit vector e1 = cos(φ)E1 + sin(φ )E2 in Figure 4.4 is perpendicular to the line connecting m1 and m2 , and e2 is tangent to this line. The constraint that the velocity of the particle of mass m1 is always normal to e2 can be written in several equivalent forms. For example, e2 · v1 = 0. This constraint is nonintegrable. By inspection, the force that enforces this constraint is parallel to e2: Fc1

= µ e2.

(4.5)

This force ensures that m1 moves only in the e1 direction. 4 For further details on this matter, and references to the vast literature on this prototypical nonintegrably

constrained system, see [27, 222, 249].

4.4

Constraints and Constraint Forces

143

Lagrange’s Prescription

We now have sufficient motivation for Lagrange’s prescription for the constraint forces acting on a system of particles. We suppose that the system of particles is subject to an integrable constraint: ² =

0,

(4.6)

where ² = ²(r1 , r2, . . . , rN , t).

Constraint (4.6) can be differentiated to yield a constraint of the form N ±

fi · vi

+

e = 0,

i =1

where fi

=

e=

∂² ∂ ri ∂² ∂t

, .

When evaluating ∂∂ r²1 , for example, we emphasize that r2, . . . , rN are fixed. Lagrange’s prescription for the constraint force F ci acting on the particle of mass mi is Fci

= µfi = µ

∂² ∂ ri

.

Result (4.4) discussed earlier is an example of this prescription. As with a single particle in the presence of dynamic friction, Lagrange’s prescription is not universally applicable. Indeed, we shall encounter some examples later on when this prescription cannot be used because of the presence of friction.5 Clearly Lagrange’s prescription can also be applied to nonintegrable constraints. Such constraints have the functional form f1 · v1 + · · · + fN · vN

+

e = 0,

(4.7)

where fi = fi (r1 , . . . , rN , t ) and e = e (r1 , . . . , rN , t ). Lagrange’s prescription for such a constraint is F ci

= µ fi

(i

=

1, . . . , N).

This prescription is equivalent to one we discussed earlier for the example of a nonintegrably constrained system of particles (see (4.5)). It should be obvious how Lagrange’s prescription can be extended to systems of (integrable and nonintegrable) constraints on the system of particles. 5 See, for example, Section 8.3 (Chapter 8).

144

Lagrange’s Equations of Motion for a System of Particles

The Power of the Constraint Forces

The constraints we consider can all be written in the form (4.7). Further, if the constraint forces are consistent with Lagrange’s prescription, then the power expended by these forces is easily calculated:

P=

N ±

F ci · vi

= µ

i=1

N ±

fi · vi

= −µe.

i =1

Thus, when e = 0, the constraint forces do no work. For an integrable constraint this occurs when ² is not an explicit function of time: ² = ² (r1 , . . . , rN ). In a previous example, shown in Figure 4.2, we note that ² depended on r1 and r2 . If, in addition, ±0 were a function of time, then ² = ² (r1, r2 , t ).

Newton’s Third Law

For the constraint ²

= ±r1 −

r2 ± − ±0 mentioned earlier, Lagrange’s prescription yields

Fc1 Fc2

r1 − r2 , ∂ r1 ± r1 − r2 ± ∂² r2 − r1 . = µ = µ ∂ r2 ± r1 − r2 ± = µ

∂²

= µ

Notice that these constraint forces point along the rod – which is what we would expect from physical grounds. Furthermore, Fc2 = −F c1 , which is none other than Newton’s third law.6

4.5

Conservative Forces and Potential Energies

Conservative forces in the dynamics of systems of particles occur commonly. For instance, they arise when two particles are connected by a spring (see Figure 4.3) or when each particle is attracted to a central body by a gravitational force field. In this section, we discuss how to prescribe the conservative forces Fcon1 , . . . , FconN acting on their respective particle of a system of particles. The system is assumed to have a potential energy U

=

U (r1, r2 , . . . , rN ) .

(4.8)

Notice that we are presuming that this is the most general form of the potential energy of conservative forces in a system of particles. General form (4.8) encompasses the inverse gravitational law between two particles of mass m1 and m2: Gm1m2 , Un = − ±r2 − r1± 6 For more details on this interesting result, see [205, 219] and Section 11.8 of this book.

4.6

145

Lagrange’s Equations of Motion

and the potential energy of a spring force between two particles of mass m1 and m2 : Us

K 2

=

2 (±r2 − r1 ± − ±0 ) .

Here, G is the universal gravitational constant and K is the spring constant. To calculate the conservative forces, we equate the time derivative of U to the negative power of the conservative forces. After some rearranging, we find that N ±

¶

∂U

Fconi +

·

∂ ri

i=1

· vi =

0.

(4.9)

Assuming that, for each i, Fconi + ∂∂ rUi is independent of v1, . . . , vN and that (4.9) is true for all possible v1, . . . , vN , we conclude that Fconi

= −

∂U

(i = 1, . . . , N).

∂ ri

(4.10)

This is the prescription for the conservative forces. For the spring force and gravitational forces associated with Un and Us that we defined earlier, prescription (4.10) yields F con1

= −

r1 − r2 , 2 ±r − r ± ±r2 − r1± 2 1 Gm1 m2

Fcon2

= −

r2 − r1 2 ±r − r ± ± r2 − r1± 2 1 Gm1m2

and F con1 F con2

r1 − r2 , r1± r2 − r1 , = −K ( ±r2 − r1 ± − ±0 ) ±r2 − r1±

= −K ( ±r2 −

r1 ± − ±0 )

±r2 −

respectively. You might again notice that each of these pairs of conservative forces obey Newton’s third law identically. This interesting result was first pointed out by Lanczos [163] and Noll [205].

4.6

Lagrange’s Equations of Motion

For a system of unconstrained particles, the equations of motion consist of N balances of linear momenta: mi r¨ i

=

Fi

(i = 1, . . . , N).

We now show that these equations are equivalent to Lagrange’s equations of motion: d dt

¶

∂T ∂q ˙K

·

−

∂T = ³K ∂ qK

where ³K =

N ± i=1

Fi ·

(K =

∂ ri

∂ qK

.

1, . . . , 3N ) ,

(4.11)

146

Lagrange’s Equations of Motion for a System of Particles

As there are no constraints on the system, all of the coordinates q1, . . . , q3N are generalized coordinates and ³K are the generalized forces. Our results here are adapted from the work of Synge and Griffith. Their derivation is presented in [276, Section 15.1]. Shortly, an alternative derivation of (4.11) that is due to Casey [37] will be discussed.7 Lagrange’s equations presume that each individual particle’s position vector is expressed as a function of the coordinates q1, . . . , q3N : ri

´

=

ri q1 , . . . , q3N

µ

(i = 1, . . . , N).

With these coordinates, we can compute vi and construct T .

Derivation of Lagrange’s Equations of Motion

Before proceeding with the derivation of Lagrange’s equations of motion, we establish an important result: ∂ ri ∂ vi = . (4.12) K ∂q ˙ ∂ qK This identity is often (fondly) referred to as “canceling the dots.” To verify the identity, we use the chain rule to calculate vi : vi

= r ˙i =

3N ± J =1

q˙ J

∂ ri . ∂ qJ

Taking the partial derivative of both sides of this expression with respect to q˙ K then yields (4.12). We are now in a position to show that ∂T = ∂ q˙ K

N ± i=1

¶

mi v i ·

∂ ri ∂ qK

·

∂T = ∂ qK

,

First, ∂T ∂ = K ∂q ˙ ∂q ˙K

N ± =

¸ N ± mi i=1

mi vi ·

i=1 =

N ± i=1

mi vi ·

2

N ±

mi v i ·

i=1

d dt

¶

∂ ri ∂ qK

·

.

(4.13)

¹

vi · vi

∂ vi ∂q ˙K ∂ ri

∂ qK

.

Notice that we used (4.12) in the final stages of this calculation. The other result follows similarly and so we skip some of the intermediate stages: 7 See (4.20) in Section 4.7.

4.7

∂T = ∂ qK

=

=

147

Construction and Use of a Single Representative Particle

N ±

mi vi

·

i=1 N ± i=1 N ±

∂ vi ∂ qK

¸

mi vi mi vi

i=1

·

·

¸ 3N ±

∂ ∂ qK

d dt

¶

J =1

∂ ri

∂ ri q˙ J J ∂q

¹

3N ±

=

J=1

·

∂ qK

∂ q˙ J J ∂q

¶

∂ ri ∂ qK

·¹

.

In the last stages of this calculation, we used the identities ∂ 2ri ∂ 2ri = , ∂ qK ∂ qJ ∂ qJ ∂ qK

˙f =

q˙ J

J =1

)

(

3N ±

∂f , ∂ qJ

where f = f q1, . . . , q3N . To derive Lagrange’s equations of motion, we now follow a familiar series of steps with the help of (4.13): d dt

¶

∂T ∂q ˙K

·

∂T − = ∂ qK

=

¸

N d ± ∂ ri mi v i · K dt ∂q i=1

N ±

¶

mi v˙ i

·

i=1 =

N ±

Fi ·

i=1

∂ ri ∂ qK

¹

N ± −

·

i =1

mi vi

·

d dt

¶

∂ ri ∂ qK

·

∂ ri

∂ qK

= ³K ,

(4.14)

where the force ³ K is ³K =

N ± i=1

Fi ·

∂ ri . ∂ qK

We have now shown how Lagrange’s equations of motion for a system of particles can be established by using the balances of linear momenta for each of the particles.

4.7

Construction and Use of a Single Representative Particle

An alternative derivation of Lagrange’s equations was presented recently by Casey [37]. In this work, he constructs a single representative particle moving in a 3N-dimensional space subject to a force Φ (cf. Figure 4.5). With the help of this construction, a derivation of Lagrange’s equations of motion follows easily and additional insight can be gained into the geometry of Lagrange’s equations of motion for a system of particles subject to integrable constraints. We note that omitting the material in this section should not greatly impact the reader’s ability to comprehend the material in subsequent sections of this book.

148

Lagrange’s Equations of Motion for a System of Particles

Φ F1

m

m1

r1

equivalent to

O

FN

rN

r O

Fi

ri

v

mi mN

Schematic of a system of N particles moving in E3 and the equivalent representative particle of mass m moving in E3N .

Figure 4.5

In this section, we follow Casey [37] and construct a single particle of mass m that is moving in E3N . The position vector of this particle is r. The kinetic energy T of this particle is the same as the kinetic energy of the system of particles that it represents, and the force on the particle is such that T˙

=

Φ · r˙ ,

Φ = mr¨ .

In Casey [37], m is chosen (without loss of generality) to be the sum of the masses: m = m1 + · · · + mN . Here, we assume only that m > 0. As noted earlier in Section 3.7, the interested reader is also referred to Lanczos [163, Chapter 1] for a related discussion of the geometry of a mechanical system. The Configuration Space

The space E3N , which is known as the configuration space, is equipped with a Cartesian coordinate system. Consequently, for any vector b: b=

3N ±

bK e¯ K

N ± 3 ± =

K=1

b3i+j −3e¯ 3i+j−3 .

i =1 j =1

Here, {¯e1 , . . . , ¯e3N } is a fixed orthonormal basis for of basis vectors:

E3N . We also define two other sets

º

e3i+j−3 e3i+j−3 Here, i

=

mi e¯ 3i+j−3 , m º m = e¯ 3i+j−3 . mi =

1, . . . , N and j = 1, 2, 3. Notice that eK · eJ

J

= δK .

The particle has a position vector r and a force vector Φ. Both of these vectors have representations similar to that for b.

4.7

Construction and Use of a Single Representative Particle

149

Prescription for the Position Vector r

As mentioned earlier, the position vector r is defined by the criterion that the kinetic energy of the particle of mass m is identical to the kinetic energy of the system of particles: N

T

1 1± mv · v = mi vi · vi . 2 2

=

i =1

Substituting v = r˙ =

N ± 3 ±

r˙3i+j− 3e¯ 3i+j−3 ,

vi

= r ˙i =

3 ± j j=1

i=1 j=1

x˙ i Ej

and equating kinetic energies, we find one solution for r: r=

3 N ± ± (

ri · Ej

)

º

i=1 j=1

mi e¯ 3i+j−3 m

=

3 N ± ± (

)

ri · Ej e3i+j−3 .

(4.15)

i=1 j=1

It is a good exercise to show that the other solutions for r can be shown to be equivalent to (4.15) modulo a translation of the origin and a relabeling of axes of E3N . Notice how the mass ratios mmi are subsumed into the basis vectors e3i+j−3 in (4.15). Prescription for the Force Vector

Φ

The force vector Φ is prescribed by the requirement that mv˙ = Φ. Substituting (4.15) and using the balance of linear momentum for each particle, one is led to the prescription Φ=

3 N ± ± (

Fi · Ej

i =1 j =1

)

º

m e¯ 3i+j−3 mi

=

Again it is interesting to note how the mass ratios e3i+j−3 in (4.16). Next, it is often convenient to note that

3 N ± ± (

)

F i · Ej e3i+j− 3.

(4.16)

i=1 j=1 mi m

are subsumed into the basis vectors

Φ · e3i+j −3 = F i · Ej ,

which follows from (4.16). It is also an easy exercise to show that the work–energy theorems for the individual particles, T˙ i = Fi · vi (no sum on i), give a work–energy theorem for the particle of mass m: T˙

=

Φ · v.

This theorem can be used to establish energy conservation results for the system of particles.

150

Lagrange’s Equations of Motion for a System of Particles

Curvilinear Coordinates

We now use the curvilinear coordinates q1 , . . . , q3N to define basis vectors for E3N : aK

=

∂ ∂r = K ∂q ∂ qK

⎛ N ± 3 ´ ± ⎝ ri · E j i=1 j=1

and aJ

= ∇ (q

J

j

= x ˆi

)=

N ± 3 ±

¸

∂q ˆ

J

⎞

e3i+j−3 ⎠

¹

e3i+j −3.

j ∂x i

i=1 j=1

µ

It can be shown that aK · aJ = δ JK .8 For most problems, it is not necessary to explicitly calculate the contravariant basis vectors aJ .

Equivalences

For future reference, it is important to note that Φ · aK

N ±

=

Fi

·

i=1

∂ ri . ∂ qK

(4.17) p

To establish this result a series of identities and the definition ˆxs Φ · aK

Φ·

=

⎞ ⎛ N ± 3 ± ( ) 3i+j −3 ∂ ⎠· ⎝ F i · Ej e

∂ qK

∂ qK

i=1 j=1 N ± 3 ± N ± 3 ±

(

)

N ± 3 ± N ± 3 ± (

)

=

F i · Ej

i=1 j=1 s =1 p=1

F i · Ej

i=1 j=1 s =1 p=1 =

N ± 3 ± (

F i · Ej

i=1 j=1 N ± =

Fi

·

i=1 N ± =

i=1

rs · Ep are invoked:

∂r

=

=

=

Fi

·

⎛ ∂ ∂ qK ∂ ri

∂ qK

⎝

)

¸

j ∂x ˆi

⎛ ⎝

N ± 3 ± (

⎞ )

rs · Ep e3s+p−3 ⎠

s=1 p=1

¶

p· ∂x ˆs e3i+j−3 · e3s +p−3 ∂ qK

¶

p· ∂x ˆs i j δsδp ∂ qK

¹

∂ qK

3 ± j =1

⎞ j

xˆ i Ej ⎠

.

We conclude that the desired identity has been established. 8 The proof closely follows a related proof in Section 1.5 and is left as an exercise for the reader.

4.7

Construction and Use of a Single Representative Particle

151

Identity (4.17) is similar to several other results pertaining to the system of particles and the single particle of mass m. For instance, consider a function ¯ ( r1, . . . , rN , t ) . ´ = ´ ( r, t) = ´

With some minor manipulations, we find several representations for the derivative of ´ with respect to a curvilinear coordinate: ∂´ ∂r ∂´ = · = ∂ qK ∂ r ∂ qK

N ± ¯ ∂´ ∂ ri

i=1

·

∂ ri

∂ qK

.

(4.18)

These results can be used to establish equivalences between conservative forces and constraint forces acting on a system of particles and those acting on the single particle of mass m.

Constraints and Constraint Forces

For the single particle, we can express the constraint ² = 0, where (as in (4.6)) ² = ² (r1 , r2, . . . , rN , t) ,

as a constraint on the motion of the single particle of mass m: ¯ (r, t). ² =²

This implies that the motion of the single particle is subject to the constraint and, by use of Lagrange’s prescription, a constraint force Φc , respectively: f · v + e = 0,

Φc

= µ f,

where f=

∂² ∂r

,

e=

∂² ∂t

.

M

Furthermore, the integrable constraint implies that the particle moves on a configuration manifold that is homeomorphic to a (3N −1)-dimensional subset of the configuration space E3N . It is important to note that the prescriptions for Φc and F ci are consistent with each other. Indeed, the equivalence of the constraint forces F ci acting on the system of particles and the constraint force Φc acting on the single particle of mass m can be inferred using (4.17) and (4.18).

Conservative Forces and Potential Energies

For a single particle, we can express the potential energy U as a function of the motion of the single particle of mass m: ¯ U (r1 , r2 , . . . , rN ) = U(r).

152

Lagrange’s Equations of Motion for a System of Particles

˙¯ and equating it to −Φcon · v, we find that 9 Calculating U

Φcon = −

¯ ∂U

∂r

.

(4.19)

Again, it is important to note that the prescriptions of Φcon and Fconi (see (4.10)) are consistent with each other. As with constraint forces, the equivalence can be inferred with the help of (4.17) and (4.18).

Lagrange’s Equations: An Unconstrained System of Particles

We are now in a position to establish Lagrange’s equations of motion for the single particle of mass m. Because ⎧ ⎨ m1v˙ 1 ⎩

=

⎫

F1 ⎬

...

mN v˙ N

=

FN

is equivalent to mv˙ = Φ,

⎭

these equations will be none other than Lagrange’s equations for the system of N particles. First, for the single particle of mass m, it is easy to see that v=

3N ±

q˙ K aK .

K =1

Consequently, we have the two intermediate results ∂T = ∂ q˙ K

∂T = ∂ qK

mv · aK ,

mv · a˙ K .

With the assistance of these results we find that d dt

¶

∂T ∂ q˙ K

·

−

∂T ∂ qK

d (mv · a K ) − mv · a ˙K dt d (mv) · a K = dt = Φ · aK .

=

We have just established Lagrange’s equations for a system of particles: d dt

¶

∂T ∂q ˙K

These equations give the motion r motion of the system of particles. ´

·

=

−

∂T = ∂ qK

Φ · aK .

(4.20)

r(t) of the particle of mass m and, hence, the

9 Essentially, we are solving the equation Φ con +

¯ ∂U ∂r

µ

v = 0 for all possible motions of the system. For this to hold, the terms in parentheses, assuming they are independent of v, must vanish, and we arrive at (4.19). ·

4.8

Kinetic Energy, Mass Matrix, and Coordinate Singularities

153

We notice that, because of (4.17), Lagrange’s equations (4.20) are identical to those we established earlier. In particular: ³K =

N ±

Fi

·

i=1

∂ ri = ∂ qK

Φ · aK .

For many problems it is more convenient to use F1 , . . . , FN instead of Φ. However, the construction of the representative particle and (4.20) can provide additional insights into the geometry of Lagrange’s equations for a system of particles. For instance, the construction of the representative particle involves the definition of a covariant basis ² ³ 3N a1, . . . , a3N for E . An M-dimensional subset, a˜ 1, . . . , a˜ M for E3N , of these basis vectors forms a basis for the tangent space to a point on the M-dimensional configuration manifold . As we shall discuss shortly, the curvilinear coordinate system ² 1 ³ ² ³ q , . . . , qM has a singularity when a˜ 1 , . . . , a˜ M are not linearly independent. As with a single particle, the loss of linear independence can be determined by examining the » ¼ M × M-dimensional mass matrix m a˜ KJ associated with T˜ 2 = Trel .10

M

4.8

Kinetic Energy, Mass Matrix, and Coordinate Singularities

We consider a system of N particles and suppose that a set of curvilinear coordinates ³ q1, . . . , q3N has been chosen to describe the motion of the system of particles. An expression for the kinetic energy of the particle can be established: ²

3N 3N

N

T

=

1± mi vi · vi 2

=

1 ±± mJK q˙ J q˙ K , 2

(4.21)

J =1 K=1

i=1

where mJK are the components of an 3N × 3N mass matrix M. In principle, we can also construct the representative particle of mass m and compute the covariant basis vectors {a1, . . . , a3N }. The coordinate system is said to have a singularity when {a1, . . . , a3N } fails to be a basis for E3N . By construction, the kinetic energy of the representative particle is equal to the kinetic energy of the system of particles. Thus, as v=

3N ±

q˙ J aJ ,

T

=

J =1

m v · v, 2

we conclude that the components of the mass matrix are related to the inner products of the covariant basis vectors: ⎡

m11 ⎢ .. M=⎣ . m13N

...

..

.

...

⎤

m13N ⎥ .. ⎦ . m3N3N

10 The kinetic T will be defined later in (4.25). rel

⎡

a11 ⎢ . .. = m⎣ a13N

...

..

.

...

⎤

a13N ⎥ .. ⎦. . a3N3N

154

Lagrange’s Equations of Motion for a System of Particles

To examine if M is positive definite, we consider a set of scalars x1 , . . . , x3N , and compute the following quadratic form: 3N ± 3N ±

xI (mIK ) xK

=

m

I=1 K=1

3N ± 3N ±

xI a I · xK a K

I=1 K=1

=

m

¸ 3N ±

¹ ¸

xI a I

·

I =1 =

3N ±

¹

xK aK

K=1

mx · x,

(4.22)

∑3N

where we have defined the vector x = I =1 xI aI . As {a1, . . . , a3N } is a basis for E3N , x = 0 if, and only if, x1 = . . . = x3N = 0. After recalling the definition of a positive definite matrix, 11 we conclude that the mass matrix M is positive definite if, and only if, {a1, . . . , a3N } is a basis for E3N . In contrast, if M is not positive definite at a given point, then {a1 , . . . , a3N } is not a basis for E 3N and the ² ³ coordinate system q1 , . . . , q3N has a singularity at the same point. Our conclusions about the relationship between singularities of the coordinate system and positive definiteness of M are identical to those we found for a single particle in Section 1.8. Often the easiest way to check if an M × M matrix A is positive definite is to check that the following conditions are satisfied: a11 ¶½

det

a11 a12

a12 a22

>

0,

a22 > 0, ⎛⎡

¾· >

0,

a33 a11

det ⎝⎣ a12

a13

>

0,

a12 a22 a23

...,

⎤⎞

aMM

a13 a23 ⎦⎠ > 0, a33

>

0, ...

,

det (A) > 0.

Thus, for an M × M matrix, a set of 2M − 1 conditions must be satisfied.

4.9

The Lagrangian

In many mechanical systems, the sole forces that act on the system are conservative. In this case, we can use the potential energy to define a Lagrangian for the system of particles: L = T − U. Using the Lagrangian, and with the help of the identity12 ∂U = ∂ qK

N ± ∂U i=1

∂ ri

·

∂ ri

∂ qK

,

it is easy to see that Lagrange’s equations for a system of particles are d dt

¶

∂L ∂q ˙K

·

−

∂L = ∂ qK

QK .

(4.23)

11 That is, a matrix A is positive definite if y · Ay > 0 for all y ² = 0 and y · Ay = 0 if, and only if, y = 0. 12 Readers familiar with Section 4.7 will have seen this identity previously as (4.18).

4.10

A Constrained System of Particles

155

Here, QK are the nonconservative forces acting on the system of particles. The nonconservative forces have the equivalent representations QK

= ³K +

=

N ±

F nconi

i=1

=

∂U ∂ qK ·

∂ ri ∂ qK

Φncon · aK .

Here, F nconi

=

Fi +

∂U ∂ ri

,

Φncon

=

Φ+

¯ ∂U

∂r

.

Notice that if QK = 0 – as it is in many celestial mechanics problems – one can write the equations of motion for a system of particles without ever having to calculate an acceleration vector. When constraints are present, a form of Lagrange’s equations similar to (4.23) can also be obtained, provided the constraints are integrable and the constraint forces are prescribed by use of Lagrange’s prescription. In this case, the equations are identical to (4.23), with the kinetic and potential energies being replaced with their constrained counterparts.13

4.10

A Constrained System of Particles

Consider a system comprising particles that are subject to one integrable and one nonintegrable constraint. We choose the curvilinear coordinates to express these constraints as q3N N ±

−

h(t) = 0,

fi · vi + e = 0.

i=1

The pair of constraints are equivalent to the following constraints on the motion of the single (representative) particle: q3N − h(t) = 0,

f · v + e = 0.

Thus, the generalized coordinates for this system are q1, . . . , q3N . These coordinates parameterize the (3N − 1)-dimensional configuration manifold . This manifold lies in the 3N-dimensional configuration space that was discussed in Section 4.7 in conjunction with Casey’s construction of the representative particle. A representative example is sketched in Figure 4.6.

M

13 That is, in the notation used previously, T is replaced with T˜ and U is replaced with U. ˜ As a result, L is

replaced with L˜ = T˜

˜ − U.

156

Lagrange’s Equations of Motion for a System of Particles

Φc

M v

m

r O

MM

The representative particle moving on a configuration manifold . For the example shown in this figure, the configuration manifold is fixed and Φc is normal to .

Figure 4.6

Assuming that the constraint forces associated with these constraints are compatible with Lagrange’s prescription, the forces acting on the system of particles have the decompositions Fi

= −

∂U

+ μ1

∂ ri

3N ∂q ∂ ri

+ μ2 fi +

Pi,

where Pi is the resultant of the nonconservative and nonconstraint forces acting on the particle of mass mi . It is important to note that ³K =

N ±

Fi

·

i=1 = −

∂ ri

∂ qK

∂U 3N + μ1 δ K + μ2 fK + µK , ∂ qK

(4.24)

where µK =

N ±

Pi

i=1

·

∂ ri , ∂ qK

fK

N ±

fi

·

∂ ri . ∂ qK

K=1 ³K a

K

.

=

i=1

From these results, it is easy to construct Φ =

∑3N

Approach I

For the first approach, we expand the partial derivatives of T and U and then impose the integrable constraint q3N = h(t) on the resulting equations. It can be seen that the resulting Lagrange’s equations of motion decouple into two sets: ½

and

½

d dt

¶

d dt

∂T ∂q ˙ 3N

¶

∂T ∂q ˙S

·

·

−

∂T ∂ qS

= −

∂U ∂ qS

¾

+ μ2 fS + µ S

˙ q3N =h(t),q˙3N =h(t)

¾

∂T ∂U − = − + μ 1 + μ2 f3N + µ3N , 3N ∂q ∂ q 3N ˙ q3N =h(t),q˙ 3N =h(t)

4.10

A Constrained System of Particles

where S = 1, . . . , 3N − 1, and we emphasize that the constraint q3N after the partial derivatives of T and U have been calculated.

=

157

r(t) is imposed

Approach II

For comparison, we now use the second approach. In this case we first impose the integrable constraint on the kinetic and potential energies, in addition to the basis vectors and constraints. For instance: T˜

˜ (q1 , . . . , = T =

q3N−1 , q˙ 1 , . . . , q˙ 3N−1 , t)

T (q1 , . . . , q3N−1 , q3N

=

h(t), q˙ 1 , . . . , q˙ 3N −1, q˙ 3N

˙ = h(t)).

M

As in the case of a single particle, T˜ can be used to construct a line element for the configuration manifold . We note that, because ˜ ∂T = ∂ q 3N

˜ ∂T = ∂ q˙ 3N

0,

0,

we are unable to obtain an expression for μ 1. In other words, because we have eliminated the coordinate associated with the integrable constraint, we can obtain only 3N − 1 Lagrange’s equations: d dt

¸

˜ ∂T ∂ q˙ S

¹

−

˜ ˜ ∂U ∂T ˜ . = − + μ2 ˜fS + µ S S ∂q ∂ qS

Further, no information on the constraint force enforcing the integrable constraint is obtained using this approach.

Geometric Considerations

Once the integrable constraint has been imposed, the single particle of mass m is constrained to move on a (3N − 1)-dimensional submanifold of the configuration space E3N . As before, this submanifold is known as the configuration manifold . A measure of the distance traveled by the single particle on this manifold can be found by use of T˜ . Indeed, this energy can be decomposed as

M

T˜

˜0 + T ˜1 + T ˜2 , = T

where

Trel

T˜ 0

=

m a˜ 3N3N h˙ 2 , 2

T˜ 1

=

m

˜2 = = T

m 2

3N −1 ±

˙ a˜ K3N q˙ K h,

K =1 3N −1 3N −1 ± ± K=1 J =1

a˜ KJ q˙ K q˙ J .

(4.25)

158

Lagrange’s Equations of Motion for a System of Particles

In the event that the basis vectors aK and aJ associated with the single particle of mass m have been computed, we also have the decomposition aJK

=

aK · aJ

(J, K =

1, . . . , 3N ) .

The mass m in (4.25) is usually chosen to be m1 + · · · + mN , although other selections such as m = 1 are equally admissible. Paralleling our work with a single particle in Section 3.7, we can use T rel to establish an expression for the mass matrix of the system of particles: ⎡

M

=

a˜ 11 m⎢ . ⎣ .. 2 a˜ 1M

⎤

a˜ 1M .. ⎥ . ⎦, a˜ MM

...

..

.

...

where M = 3N −1. Following from our earlier results in Section 4.8, if M is positive def³ ² ³ ² inite, then a˜ 1 , . . . , a˜ 3N−1 span E 3N −1 and the generalized coordinates q1, . . . , q3N −1 for are free of singularities. Thus, by simply examining the mass matrix, potential coordinate singularities can be determined without explicitly computing the basis vec² ³ tors a˜ 1, . . . , a˜ 3N−1 . As we shall see from the examples in the next chapter, singularities are often present in the coordinate systems used for systems of particles. The kinematical line element ds for is

M

M

¸º

ds =

2Trel m

¹

dt.

This quantity may also be written as ds =

¿ À3N−1 3N −1 À± ± Á a˜

K J KJ dq dq .

K=1 J= 1

M

As was the case previously, the imposition of a nonintegrable constraint will not change or ds. Generalizing the developments of this section to multiple position and velocity constraints follows in a straightforward manner. We limit our discussion of these cases to the illustrative examples and exercises in Chapter 5.

4.11

A Canonical Form of Lagrange’s Equations

We now consider a system of particles in which the constraints are integrable and time independent and the constraint forces are prescribed by use of Lagrange’s prescription. 14 In this case, it suffices to examine those equations associated with the generalized coordinates in order to find the equations of motion. In what follows, we establish two alternative forms of Lagrange’s equations of motion (see (4.29) and (4.31) below). 14 That is, the constraints are ideal.

4.11

A Canonical Form of Lagrange’s Equations

159

Preliminaries

We are interested in a system of N particles subject to a set of C (scleronomic) integrable constraints, where15 ²1 = ²1 (r1, . . . , rN ) , . . . , ²C = ²C (r1 , . . . , rN ) .

(4.26)

We assume that the coordinates q1 , . . . , q3N are chosen such that the generalized coordinates for the system are q1, . . . , qM , where M = 3N − C. That is, the constraints M +1 ²1 = 0, . . . , ²C = 0 are expressed in terms of the coordinates q , . . . , q3N . The constrained kinetic energy for the system has the representation T˜

=

M M m ±± a˜ KJ q˙ J q˙ K , 2 K =1 J=1

where we have chosen m = m1 + · · · + mN . We note that a˜ KJ = a˜ JK and assume that » ¼ this matrix is invertible. The inverse is the matrix a˜ JK . For example, if M = 3, then ⎡

a˜ 11 ⎣ a˜ 12 a˜ 13

a˜ 12 a˜ 22 a˜ 23

⎤⎡

a˜ 13 a˜ 11 a˜ 23 ⎦ ⎣ a˜ 12 a˜ 13 a˜ 33

a˜ 12 a˜ 22 a˜ 23

⎤

a˜ 13 a˜ 23 ⎦ a˜ 33

⎡

1 = ⎣ 0 0

0 1 0

⎤

0 0 ⎦. 1

We also define the Christoffel symbols of the first kind as [SJ, K]

=

1 2

¶

∂ a SJ ∂ a KS ∂ aKJ − + J S ∂q ∂ qK ∂q

·

(J, K, S =

1, . . . , 3N ) .

There are (3N)3 of these symbols, but many of them are not distinct: [SJ, K] = [JS, K]. The Christoffel symbols of the second kind are K ´IJ =

3N ±

aKR [IJ, R]

(I, J, K =

1, . . . , 3N ) .

R=1

To gain further insight into these symbols, it is very useful to examine the single particle of mass m.

The Representative Particle

Casey’s construction of the representative particle is very useful for exploring the Christoffel symbols. Using the particle and its configuration space, we have a set of covariant aK and contravariant aJ basis vectors for E3N . These vectors can be used to define aJK , aJK , and the connection coefficients: aJK

=

γSI,K =

aJ · aK ,

∂ aS

∂ qI

·

aK ,

aJK

=

aJ · aK ,

∂ aK J J γ KS = ·a . S ∂q

15 The generalization of our developments to instances in which the constraints are rheonomic can be found

elsewhere (e.g., [96]).

160

Lagrange’s Equations of Motion for a System of Particles

As in our earlier work on page 122, it can be shown that the connection coefficients and Christoffel symbols are simply related: [SI, K ] = γSI,K

∂ aK J J J ´KS = γ KS = ·a . ∂ qS

∂ aS · aK , ∂ qI

=

Here, all the indices range from 1 to 3N. Using the symmetries aJK = aKJ and aJK = J aKJ , the symmetries [SI, K] = [IS, K] and ´JKS = ´SK should be transparent. We also observe that the Christoffel symbols are none other than the covariant and contravariant components of ∂∂aqKS .

Derivatives of the Kinetic Energy

Preparatory to establishing the covariant and contravariant forms, we first record the derivatives of the kinetic energy. What follows is based entirely on the constrained kinetic energy, and we now drop the tildes ornamenting the various kinematical quantities. First, we note that ∂ ∂T = R ∂q ˙ ∂q ˙R

¸

M M m ±± aKJ q˙ J q˙ K 2 K =1 J=1

⎛

=

¹

J δR

M M ⎜ J✓ m ±± ⎜ ∂ q˙ ✼ aKJ ⎜ ✓R q˙ K ⎝ ∂ q˙ 2 ✓ K=1 J =1

=

m

M ±

✓✼

K ˙ J ∂q ∂q ˙R

+q ˙

✓ ✓

K δR

⎞ ⎟ ⎟ ⎟ ⎠

aRK q˙ K .

K =1

To arrive at this result, we used the symmetries aRK independent of q˙ S . Similarly, ∂T ∂ = ∂ qR ∂ qR

=

¸

=

aKR and the fact that aRK are

M M m ±± aKJ q˙ J q˙ K 2

¹

K=1 J =1

M M m ± ± ∂ aKJ J K q˙ q˙ .

2

K =1 J=1

∂ qR

∑

K We next differentiate m M ˙ with respect to time. With some rearranging and K=1 a RK q 16 relabeling of the indices, we find 16 Although

holds:

∂ aRK

∑M

∂q S

K=1

²=

∑M

∂ aRS

, because we are summing over the indices S and K, the following identity always

∂q K ∂ aRK

S=1

∂ qS

q˙ S q˙ K

=

∑M

K =1

∑M

S =1

∂ aRS ∂ qK

q˙ S q˙ K .

4.11

d dt

¶

∂T ∂q ˙R

· =

=

=

m m m

M ± K=1 M ± K=1 M ±

161

A Canonical Form of Lagrange’s Equations

aRK q¨ K + m aRK q¨ K + aRK q¨ K +

K=1

m 2 m 2

M ± M ± ∂ aRK S K q˙ q˙ ∂ qS

K=1 S=1 M M ± ±

∂ a RK S K q˙ q˙ + ∂ qS

K=1 S=1 M ± M ¶ ± K=1 S=1

M M m ± ± ∂ aIR I J q˙ q˙ 2 ∂ qJ

∂ aRK ∂ a SR + S ∂q ∂ qK

J =1 I=1

·

q˙ S q˙ K .

Summarizing, the derivatives of T have the representations ∂T = ∂ qR

d dt

¶

∂T ∂q ˙R

M M m ± ± ∂ aKS S K q˙ q˙ , 2 ∂ qR

· =

m

K=1 S=1 M ±

aRK q¨ K

+

M M ¶ m ± ± ∂ aRK 2 ∂ qS

+

K =1 S=1

K =1

∂ aSR ∂ qK

·

q˙ S q˙ K .

(4.27)

A Covariant Form of Lagrange’s Equations of Motion

A covariant form of Lagrange’s equations of motion was previously established: d dt

¶

∂T ∂q ˙R

·

−

∂T = ³R ∂ qR

(R =

1, . . . , M) .

(4.28)

Here, ³1, . . . , ³M are the generalized forces. We now expand the derivatives of T to establish another form of this equation. With the help of (4.27) and the definition of the Christoffel symbol of the first kind, it is straightforward to show that Lagrange’s equations of motion can be written in the form m

M ± K=1

aRK q¨ K

+

m

M ± M ±

[SK, R] q˙ K q˙ S

= ³R

(R =

1, . . . , M) .

(4.29)

K =1 S=1

As mentioned in the exercises at the end of Chapter 3, this form of Lagrange’s equations appears in several texts on differential geometry17 and, in the case of a single particle, was an exercise at the end of Chapter 3. From (4.29), it should be apparent that, if we know aJK , then we can immediately write the left-hand side of Lagrange’s equations of motion. Further, for the system at hand, the kinematical line element ds is ds =

¿ À M M À± ± Á a

˙ Rq ˙ K dt. RK q

R=1 K=1

Knowledge of ds (which also implies knowledge of aJK ) enables us to write the left-hand side of Lagrange’s equations. 17 See, for example, [182] or [277].

162

Lagrange’s Equations of Motion for a System of Particles

A Contravariant Form of Lagrange’s Equations of Motion

If we multiply both sides of the covariant form of Lagrange’s equations by the inverse of aRK , then we will find the contravariant form of Lagrange’s equations of motion. Because we have imposed the integrable constraints and are interested only in the first M equations of motion, we can find only the M contravariant equations from the M covariant equations provided:18 ⎡

a11 ⎢ . ⎣ .. a1M

···

..

.

···

⎤⎡

a11 a1M .. ⎥ ⎢ .. . ⎦⎣ . aMM a1M

···

..

.

···

⎤

⎡

1 a1M .. ⎥ = ⎢ .. . ⎦ ⎣ . MM 0 a

⎤

0 .. ⎥ . . ⎦ 1

···

..

.

···

(4.30)

That is, we require several aKJ s to be zero: ⎡ ⎢ ⎢ ⎢ ⎣

a1(M+1) a2(M+1) .. . aM(M+1)

a1(M+2) a2(M+2) .. . aM(M+2)

··· ···

..

.

···

a1(3N) a2(3N) .. . aM(3N)

⎤

⎡

⎥ ⎥ ⎥= ⎦

⎢ ⎢ ⎢ ⎣

0 0 0 0 .. .. . . 0 0

··· ···

..

.

···

0 0 .. . 0

⎤ ⎥ ⎥ ⎥. ⎦

When (4.30) holds, the right-hand side of Lagrange’s equations of motion transforms from the covariant components of Φ to the contravariant components: J ³ =

M ±

aJR ³R .

R=1

In summary, the contravariant form of Lagrange’s equations of motion is mq¨

J

+

m

M ± M ± K =1 S=1

J K S J ´KS q˙ q ˙ = ³

(J =

1, . . . , M) .

(4.31)

This form of Lagrange’s equations is very useful in numerical simulations because it provides explicit expressions for q¨ 1 , . . . , q¨ M .

Examples

A simple example with which to explore the two forms of Lagrange’s equations of motion (4.29) and (4.31) is to consider a single particle moving in E 3 under the action of a force F. We describe the motion of this particle using a spherical polar coordinate system: q1 = R, q2 = φ , and q3 = θ .19 That is, M = 3. We consider our discussion here to be a solution to Exercise 3.3. It should by now be trivial to establish that T

=

µ m´ 2 R˙ + R2 sin2(φ )θ˙ 2 + R 2φ˙ 2 . 2

18 In terms of the representative particle, these restrictions are equivalent to the orthogonality conditions

aJ · aK = 0 for all J = 1, . . . , M and all K = M + 1, . . . , 3N. 19 The covariant form of these equations was also presented previously (cf. (3.4)).

4.11

163

A Canonical Form of Lagrange’s Equations

From the expression for T , we can immediately deduce that ⎡

⎤

⎡

⎤

⎡

a11 ⎣ a12 a13

a12 a22 a23

a13 1 ⎦ ⎣ = a23 0 a33 0

0 R2 0

a11 ⎣ a12 a13

a12 a22 a23

1 a13 23 ⎦ ⎣ 0 = a 0 a33

0 R−2 0

⎡

⎤

0 ⎦, 0 2 2 R sin (φ)

⎤

0 ⎦. 0 −2 −2 R sin (φ)

We readily deduce the covariant form of Lagrange’s equations of motion from dtd ∂T = ∂ qi

´

∂T ∂q ˙i

µ −

F i by expanding the time derivative: ⎡

1 m⎣ 0 0

0 R2 0

⎡

⎤

⎡

⎡

⎤

⎡

⎤

⎤⎡

⎤

⎡

⎤

0 R¨ C1 ⎦ ⎣ φ¨ ⎦ + ⎣ C2 ⎦ 0 R 2 sin2 (φ ) θ¨ C3

where

⎡

)

(

⎤

F1 = ⎣ F2 ⎦ , F3 ⎤

C1 ⎣ C2 ⎦ C3

sin2(φ )θ˙ 2 + φ˙ 2 ⎦, ˙ φ˙ = ⎣ −mR 2 sin(φ) cos(φ )θ˙ 2 + 2mRR 2 2 2mR sin (φ )R˙ θ˙ + 2mR sin(φ ) cos(φ)φ˙ θ˙

F1 ⎣ F2 ⎦ F3

F · eR ⎦, ⎣ F · Reφ = F · R sin(φ)eθ

F1 ⎣ F2 ⎦ F3

a11 ⎢ ⎥ 12 = ⎣ ⎦=⎣ a 1 F · R sin(φ) eθ a13

−mR

(4.32)

⎤

⎡

⎡

⎤

F · eR F · R1 eφ

⎡

a12 a22 a23

⎤⎡

⎤

a13 F1 23 ⎦ ⎣ F2 ⎦ . a 33 F3 a

From (4.32), we can read off the Christoffel symbols of the first kind. For example, [22, 1] = −R, [33, 1] = −R sin2(φ ), [12, 2] = R, and [13, 3] = R sin2 (φ). We can determine the contravariant form of (4.32) by multiplying both sides of (4.32) » ¼ by the inverse of aij : ⎡

m⎣

R¨ ¨ φ

θ¨

⎤

⎡

(

)

⎤

sin2 (φ )θ˙ 2 + φ˙ 2 ⎦ + ⎣ −m sin(φ) cos(φ )θ˙ 2 + 2m R ˙ ˙ ⎦ R φ 2m ˙ ˙ ˙ ˙ R R θ + 2m cot(φ )φθ −mR

⎡

⎤

F1 = ⎣ F2 ⎦ . F3

(4.33)

Again, we can read off the Christoffel symbols of the second kind from the right-hand side of these equations. For example, ´122 = −R, ´233 = − sin(φ ) cos(φ), and ´323 = cot(φ). The results for a particle that is free to move in E3 can readily be modified to give the equations of motion for a spherical pendulum such as the one shown in Figure 1.13(a). First, we assume that the particle of mass m is attached to a fixed point O by a rod of length ±0 and negligible mass and inertia. The particle is subject to a constraint ² = 0, where ² = R − ±0 . The constraint force associated with this constraint is prescribed

164

Lagrange’s Equations of Motion for a System of Particles

using Lagrange’s prescription Fc of the equations of motion is ½

m

2 ±0

0

0 2 2 ±0 sin (φ )

¾½

¨ φ

¾

+

θ¨

= μ eR .

½

It follows from (4.32) that the covariant form ¾

2

sin(φ ) cos(φ )θ˙2 2 2m±0 sin(φ ) cos(φ)φ˙ θ˙

−m±0

½ =

F a · ±0 eφ Fa · ±0 sin(φ)eθ

¾

.

The contravariant form of the equations of motion is also readily obtained using (4.33): ½

m

φ¨

¾

θ¨

½ +

2 −m sin(φ) cos(φ )θ˙

Â

¾

2m cot(φ)φ˙ θ˙

=

Fa · Fa ·

1

±0

1

Ã

eφ e

±0 sin( φ ) θ

.

As anticipated, the constraint force Fc is absent from these equations. We leave the substitution Fa = −mgE3 to the reader.

4.12

Alternative Principles of Mechanics

Newton’s tremendous contributions to mechanics in the seventeenth century still left many unanswered questions. Among these questions were the formulation of equations of motion for rigid bodies and deformable media. To this end, several principles of mechanics were postulated in the subsequent centuries: among them, Jean Bernoulli’s principle of virtual work from 1717, D’Alembert’s principle from 1743 [57], Gauss’ principle of least constraint in 1829 [91], and Hamilton’s principle in 1835 [116].20 The purpose of this section is to briefly outline how some of these principles are related to the balances of linear momenta used to establish the equations of motion for a system of particles. 4.12.1

Principle of Virtual Work and D’Alembert’s Principle

We first consider the principle of virtual work and D’Alembert’s principle applied to a system of N particles. These principles are the basis for treatments of Lagrange’s equations of motion in many texts. 21 We assume that the particles are subject to a single constraint: f1 · v1 + · · · + fN · vN

+

e = 0.

(4.34)

The principle of virtual work and D’Alembert’s principle collectively state that the motion of the system of particles is such that the following equation is satisfied: (

Fa1

−

m1r¨ 1

)

·

(

d1 + · · · + FaN

−

mN r¨ N

)

·

dN

=

0,

(4.35)

for all possible choices of the vectors d 1, . . . , d N that satisfy the condition f1 · d1 + · · · + fN · d N

=

0.

(4.36)

20 Further background on these (and several other principles) can be found in [62, 278, 287, 288]. 21 See, for example, Baruh [20, Section 4.9], Greenwood [104, Section 2.1], or Synge [275, Section 46]. As

noted by many authors, the principle of virtual work pertains to static problems, and its extension to dynamics requires the application of D’Alembert’s principle; hence the combination of these principles in this section.

4.12

Alternative Principles of Mechanics

165

Notice that the function e present in (4.34) is notably absent from (4.36). The vectors dK are known as virtual displacements and are usually denoted by δ rK . The virtual work performed by the applied force FaK is defined as FaK · d K ; thus, (4.35), states that the combined virtual work of the applied forces FaK and the inertial forces −mK r¨ K is zero. Our aim is to obtain the equations of motion of the system of particles from (4.35). To this end, we can introduce a Lagrange multiplier, which we denote by the scalar function μ = μ(t), to accommodate the single constraint on the vectors dK : 22 N ± ( K=1

FaK

−

mK ¨rN

)

·

dK

+μ

N ±

fK · dK

=

0.

(4.37)

K=1

As a consequence of the Lagrange multiplier μ, the vectors dK can be varied independently. For (4.37) to hold for all such displacements, it is necessary and sufficient that mK ¨rK

=

FaK

+ μfK

(K =

1, . . . , N ) .

(4.38)

However, (4.38) are none other than the balances of linear momenta for a system of particles subject to constraint (4.34), for which the constraint forces are prescribed by use of Lagrange’s prescription: F cK

= μ fK

(K =

1, . . . , N ) .

(4.39)

We can now easily proceed to establish Lagrange’s equations of motion for the system of particles. Thus, the principle of virtual work and D’Alembert’s principle combined with Lagrange’s prescription for the constraint forces collectively lead to the balance laws for a constrained system of particles. When constraint (4.34) is integrable, we can interpret (4.36) as a normality condition for the vectors dK . Further, it is not too difficult to see that constraint force prescription (4.39) implies that the constraint force Φc is normal to the configuration manifold in this case.

M

4.12.2

Gauss’ Principle of Least Constraint

Gauss’ principle of least constraint was published in 1829 [91]. It is a remarkable interpretation of the role played by constraint forces in a mechanical system. Restricting our attention to particles, suppose that we have a system of N particles that are subject to constraint (4.34) and suppose that the constraint forces are prescribed by using Lagrange’s prescription (i.e., (4.39) holds). Then, in any motion of the system that satisfies the constraints, the constraint forces FcK = μfK are the least needed to ensure that the constraint is satisfied. That is, Lagrange’s prescription is in a sense optimal! Since its introduction in 1829, the principle of least constraint has played a key role in several seminal developments in mechanics (see, e.g., [126]). A lucid discussion of the principle can be found in [294], and its extension to cases for which Lagrange’s prescription does not hold can be found in [218]. 22 When constraint (4.34) is integrable, this is precisely the approach taken by Lagrange in [159, 160] (cf.

Dugas [62, Section 4, Chapter 11]).

166

Lagrange’s Equations of Motion for a System of Particles

4.12.3

Hamilton’s Principle

The remaining principle of interest was discovered by William R. Hamilton (1805– 1865) and first published in [116] for a system of unconstrained particles subject solely to conservative forces. For ease of exposition, we restrict our discussion to a single ( ) particle of mass m and suppose that three coordinates q = q1, q2 , q3 are used to parameterize its position vector. Then Hamilton’s principle states that the motion of the system between a given initial configuration q (t0 ) and a given final configuration q (t1 ) is such that it extremizes the action integral:23 I

Ä t1 =

Ldt,

(4.40)

( )

)

(

t0

where the Lagrangian L = T qi , q˙ i − U qi and t0 and t 1 are fixed instances of time. As illustrated in Figure 4.7, there are an infinite number of paths q(t) that can connect two possible configurations, and so finding the one that extremizes I appears to be a daunting task. However, it was known long before (4.40) appeared in print in 1835 that the necessary conditions for the components of q (t ) to extremize I are that q (t) satisfy the following differential equations: d dt

¶

∂L ∂ q˙ k

·

−

∂L ∂ qk

=

0

(k =

1, 2, 3 ) .

(4.41)

In fact, in the context of extremizing I, (4.41) are known as the Euler–Lagrange equations.24 Thus, Hamilton’s principle implies that the motion of the system satisfies (4.41). At this stage in the book, it should be transparent that (4.41) are equivalent to F = ma q ( t 1) q

q (t 0 )

t0

t

t1

Some of the possible paths q connecting two configurations, q (t0) and q (t1), of a particle. The actual path is the one that satisfies Å t the differential equations F = ma. Hamilton’s principle states that this path extremizes I = t01 Ldt (see (4.40)). The portrait of Hamilton is reproduced with the kind permission of the Royal Irish Academy ³ c RIA in Dublin, Ireland.

Figure 4.7

23 To extremize is to minimize or maximize. 24 Problems featuring the extremization of I can be solved by use of the calculus of variations. In the middle

of the eighteenth century, Euler and Lagrange played seminal roles in the development of this calculus [97]. It was on the subject of this calculus that a 19-year-old Lagrange first wrote to Euler in 1755.

4.14

Exercises

167

for a particle. In conclusion, for an unconstrained particle, Hamilton’s principle is equivalent to the balance of linear momentum. Thus, the motion of the system provided by the solution to F = ma extremizes I! Subsequent extensions to Hamilton’s principle were made by several authors and these include the case where the system of particles was subject to integrable constraints. Jacobi [138] in particular used this principle to great effect when examining the motion of particles on smooth surfaces and the shortest distance between two points on a surface. Although Hamilton’s principle cannot readily be applied to systems of particles subject to nonintegrable constraints, it has helped form several pillars of modern physics.

4.13

Closing Comments

In this chapter, a derivation of Lagrange’s equations of motion was presented for a system of particles, and it was shown how they can be modified when constraints are introduced. Our developments emphasized that these equations are equivalent to the Newtonian balance laws of linear momenta for each particle. This important feature ∑ ∂ ri enables us to confidently calculate the forces ³K = N i=1 Fi · ∂ qK that appear on the right-hand side of Lagrange’s equations of motion: d dt

¶

∂T ∂q ˙K

·

−

∂T = ³K ∂ qK

(R =

1, . . . , 3N ) .

We are also able to introduce integrable and nonintegrable constraints and the constraint forces associated with them. On a deeper level, if we wish to consider the system of particles as a single particle moving on a configuration manifold that is embedded in a 3N-dimensional Euclidean space, then Casey’s construction of the representative particle enables us to do this in a straightforward manner. In the next chapter, we consider examples in which Lagrange’s equations of motion for several systems of particles are established and discussed.

M

4.14

Exercises

M

Exercise 4.1: What are the configuration manifolds , generalized coordinates, and kinematical line elements ds of the following systems? (i) (ii) (iii) (iv) (v)

a particle attached to a fixed point by a spring; a particle attached to a fixed point by a rod of length ±(t); a harmonic oscillator consisting of a single particle; a planar double pendulum; a spherical double pendulum.

Exercise 4.2: Under which circumstances do constraint forces perform no work?

168

Lagrange’s Equations of Motion for a System of Particles

Exercise 4.3: It is typical to assume that the kinetic energy of a system of particles is a positive definite function of the velocities q˙ K . As T

=

3N ± 3N ± m I =1 J=1

2

aIJ q˙ I q˙ J ,

this assumption is equivalent to stating that the mass matrix formed by the components m 25 2 a IJ is positive definite. Using the spherical pendulum as an example, show that certain representations of T are not always positive definite. Specifically, if one uses spherical polar coordinates, then the representation for T becomes indefinite at the singularities of this coordinate system. Exercise 4.4: In Casey’s construction of the single particle, what are the distinctions between the bases {eK }, {eK }, and {¯eK }? For a given system of two particles, how does one construct these bases for E6 ? Exercise 4.5: Consider the following function that depends on the motion of two particles: V

˜ (r , r , t) = = V 2 1

V (±r2 − r1 ± , t ) = Vˆ (x, t ) ,

where x = ±r2 − r1 ± ,

x = r2 − r1 .

This function is representative of a potential energy function and an integrable constraint. (a) For any vector a, establish the following result: d ±a± dt

=

a ·a ˙. ± a±

One way to show this result is to represent a using a Cartesian basis and then differentiate ±a±. (b) Show that V˙

=

˜ ∂V

∂ r1

· v1 +

˜ ∂V

· v2 +

∂ r2

˜ ∂V

∂t

,

where ˜ ∂V

∂ r1

= −

˜ ∂V

∂ r2

= −

ˆ ∂V

x . ∂ x ± x±

You should recall that the partial derivative ∂∂rV1 assumes that r1 is varied while r2 and t are held fixed. As a result, the easiest method to establish the required results is to express x, r1 , and r2 using three distinct sets of Cartesian coordinates: ˜

r1

=

x1E1 + y1E 2 + z1 E3,

r2

=

x2E1 + y2E2 + z2 E3,

x = xE1 + yE2 + zE3 .

25 Recall that a matrix C is positive definite if, for all nonzero x, xT Cx > 0 and xT Cx

=

0 only when x = 0.

4.14

(c) Consider a system of two particles subject to a constraint ²

169

Exercises

=

0, where

² = ² (± r2 − r1 ± , t ) .

(4.42)

Using Lagrange’s prescription, show that Fc1 = −Fc 2 . Notice that this is Newton’s third law of motion. Give two examples of a physical constraint for which ² has the form (4.42) and for which the constraint forces that enforce this constraint are equal and opposite. (d) Consider a system of two particles subject to a conservative force that has a potential energy function U

=

U (±r2 − r1±) .

(4.43)

Using the identity, ˙ = −F con · v 1 − F con · v 2, U 1 2

show that Fcon1 = −Fcon2 . Notice that this (again) is Newton’s third law of motion. Give two instances in which U has the form (4.43) and demonstrate that the conservative forces are equal and opposite. Exercise 4.6: The two-body problem consists of a system of two particles, one of mass m1 = M and the other of mass m2 = m. The sole forces on the system are the result of a . Newtonian gravitational force field that has a potential energy function Un = − ±rGMm −r ± 2

1

(a) For this system, show that the linear momentum of the system is conserved and that the center of mass C moves at a constant speed in a straight line. In addition, show that m M (r2 − r1 ) , r1 − r¯ = − (r2 − r1 ) . r2 − r¯ = m+M m+M Here, r¯ is the position vector of the center of mass. (b) Show that the angular momentum of the system of particles relative to C is conserved. (c) Show that the total energy of the system of particles is conserved. (d) Show that the differential equations governing the motions of m1 and m2 can be written in the following forms:26 (

)

M r¨ 1 − ¨¯r (

)

m r¨ 2 − ¨¯r

¶ = −GMm

¶ = −GMm

m M +m M M +m

·2 ·2

r1 − r¯ ±r1 − r ¯±

r2 − r¯

3

3 ±r2 − r ¯±

, .

(4.44)

(e) Argue that the results of Section 2.8 can be applied to (4.44) to determine the orbital motions of the particles about their center of mass C.

26 In celestial mechanics, expressing the equations of motion in this manner is equivalent to using what is

known as a barycentric coordinate system (see, e.g., [296]), that is, a coordinate system whose origin is at the center of mass.

5

Dynamics of Systems of Particles

5.1

Introduction

In this chapter, several systems of particles are examined. We pay particular attention to how the equations of motion for these systems are established by use of Lagrange’s equations. The classic examples we discuss range from simple harmonic oscillators to dumbbell satellites and pendula. Our goals are to illuminate the developments of the previous chapter and to present representative examples. Examples that are closely related to the ones we discuss can be found in many dynamics texts. Most of these texts use alternative formulations of Lagrange’s equations of motion that do not readily accommodate nonconservative forces. Here, because we have already established an equivalence between Lagrange’s equations of motion and the balances of linear momenta, we are easily able to incorporate nonconservative forces such as dynamic Coulomb friction. This chapter closes with a brief discussion of some recent works on the dynamics of systems of particles.

5.2

Harmonic Oscillators

We first consider simple examples involving a system of two particles. The system shown in Figure 5.1 is the first of several related systems that we discuss in this section. Referring to the figure, we see that a particle of mass m1 is connected by a spring of stiffness K1 and unstretched length ±01 to a fixed support. It is also connected by a spring of stiffness K2 and unstretched length ±02 to a particle of mass m2. Both particles are constrained to move in the E1 direction: r1 = r2 =

( (

±01 + x1

)

E1 + y1 E2 + z1 E3,

±01 + ±02 + x2

)

E1 + y2 E2 + z2E3 .

Notice that x1 and x2 measure the displacement of the particles from the unstretched spring states. This parameterization will lead to simple expressions for the spring forces and their potential energies.

5.2

Harmonic Oscillators

m1

m2

171

E2 g

O

E1

Smooth surface Figure 5.1 The system of two particles moving on a smooth horizontal surface. The particles are connected by linear springs.

Coordinates

We organize the six coordinates as follows: q1

=

q 2 = x2 ,

x1,

q3

=

q4 = y2 ,

y1,

q5

=

q6

z1 ,

=

z2 .

For the systems of interest, the particles are constrained to move on a line. Thus the system is subject to four constraints: 0,

²1 =

²2 =

0,

²3 =

0,

²4 =

0

where ² 1 = y1 ,

² 2 = y2,

² 3 = z1 ,

∂ ²2

∂ ²3

² 4 = z2 .

It should be noted that ∂ ²1 ∂ r1

=

E2 ,

=

0,

∂ r1

=

0,

∂ r1

=

E3,

=

0,

∂ ²4 ∂ r1

=

0

and ∂ ²1 ∂ r2

∂ ²2 ∂ r2

=

∂ ²3

E2,

∂ r2

∂ ²4 ∂ r2

=

E 3.

These expressions will be used later to prescribe constraint forces.

Kinetic and Potential Energies

It is easy to see that the kinetic and potential energies of the system are T

=

U

=

² m ± ² m1 ± 2 2 x˙ 1 + y˙ 21 + z˙21 + x˙22 + y˙ 22 + ˙z22 , 2 2 K1 2 K 2 2 x + (x2 − x1) + m1 gy1 + m2 gy2 . 2 1 2

(5.1)

Imposing the four integrable constraints on these energies, we can determine their constrained counterparts: T˜ =

m1 2 m2 2 x˙ + x˙ , 2 1 2 2

˜ U

=

1 1 K1 x21 + K2 (x2 − x1)2 . 2 2

172

Dynamics of Systems of Particles

Constraint Forces

If the surface that the particles are moving on is smooth, then Lagrange’s prescription can be invoked: Fc1

= µ1 E2 + µ3 E3 ,

F c2

= µ 2E2 + µ 4E 3.

Here, µ 1, µ2 , µ3 , and µ 4 are normal force components: N1 = µ 1E2 + µ3E 3 and N2 = µ 2E2 + µ4 E3. If the surface is rough, then these prescriptions must be altered to include the friction forces: x˙ 1 E1, Fc1 = µ1 E2 + µ 3E3 − µ d ||µ1 E2 + µ 3E3 || |˙ x1 | x˙ 2 E1. Fc2 = µ2 E2 + µ 4E3 − µ d ||µ2 E2 + µ 4E3 || |˙ x2 | The total force acting on each particle is composed of the constraint force and the conservative forces that are due to gravity and the springs. The Representative Particle

It is easy to construct the representative particle of mass m for this system. First, the position vector of this particle, which is moving in E6 , is defined with the help of (4.15): (

)

(

)

r = x1 + ±01 e1 + y1 e2 + z1e3 + x2 + ±01 + ±02 e4 + y2 e5 + z2 e6. Using this expression, we easily calculate the six covariant vectors aJ and the six contravariant vectors aJ : a1

=

e1,

a2 = e4 ,

a3

=

e2,

a4

=

e5 ,

a5

e3 ,

a6

=

e6

a1

=

e1,

a2

a3

=

e2 ,

a4

=

e5,

a 5 = e3 ,

a6

=

e6.

=

and =

e4 ,

If the kinetic energy T = m2 ˙r · r˙ were calculated, it would be identical to (5.1). The constraints on the motion of the single particle have the representations ²1 =

r · e2,

²2 =

r · e5 ,

²3 =

r · e3 ,

²4 =

r · e6.

Using these representations, we easily see that the constraint force Φc that we would prescribe using Lagrange’s prescription is Φc

2

5

3

6

= µ1 e + µ2 e + µ3 e + µ4 e

.

(5.2)

When friction is present, this prescription is inadequate and components of the frictional forces are required. The force vector Φ can be determined using the prescription (4.16): ³

Φ=

´

x˙ 1 −K1 ( x1) + K2 ( x2 − x1 ) − µ d|| µ1 E2 + µ3 E3 || e1 |˙ x1| m1g) e2 + µ3 e3 + (µ2 − m2 g) e5 + µ4 e6 ´ x˙ 2 e4. −K2 ( x2 − x1 ) − µd || µ2 E2 + µ4 E3 || |˙ x2|

+ (µ 1 −

³

+

5.2

173

Harmonic Oscillators

Notice that the constraint force contributions to this vector are consistent with our prescription for Φc in (5.2) when µ d = 0. We have now completely specified all the ingredients needed to establish Lagrange’s equations of motion by using the representative particle of mass m.

The Generalized Coordinates and Configuration Manifold

M

There are four integrable constraints on the system, so the generalized coordinates are 2 ∼ x1 and x2. The configuration manifold = E and the kinematical line element is µ

ds =

³

m1 m1 + m2

dx 1 dt

´2

m2

+

³

m1 + m2

dx 2 dt

´2

dt.

Notice that this line element is easily related ¶ to the standard measure of distance traveled ±

along a curve (x(τ ), y(τ )) on a plane:

dx dτ

²2

M

±

+

² dy 2 dτ . dτ

Imposing a nonintegrable

constraint on the system will neither affect nor ds. It is straightforward to show from the expression for T˜ that the mass matrix is invertible and hence the coordinates x1 and x2 for are free of singularities.

M

Equations of Motion for the Oscillator

We first consider the case shown in Figure 5.1 in which friction is absent. For this system, the constraint forces are prescribed using Lagrange’s prescription and the other ˜ ˜ forces acting on the system are conservative. Thus, ³1 = − ∂∂qU1 and ³2 = − ∂∂qU2 . In conclusion, the equations of motion can be obtained by use of the following form of Lagrange’s equations of motion: d dt

·

˜ ∂T

q

∂ ˙α

¸

−

˜ ∂T

∂q

α

= −

˜ ∂U

∂ qα

,

(5.3)

where α = 1, 2. With a small amount of work, we find the equations of motion: m1x¨ 1

= −K1x1 −

K2 (x1 − x2) ,

m2x¨ 2

= −K2 ( x2 −

x1 ) .

These equations are classic and are easily seen to be equivalent to F1 · E 1 and F2 · E1 = m2r¨ 2 · E1, respectively.

=

m1r¨ 1 · E1

The Influences of Dynamic Friction and an Applied Force

We now consider the same system of particles, but assume that dynamic Coulomb friction is present and a forcing P = P cos(ω t)E1 is applied to the particle of mass m2 (see Figure 5.2). We have already determined all of the necessary ingredients to determine the equations of motion for this system. The major difference is that Lagrange’s equations of motion in the form (5.3) are inadequate. It is necessary to use the following form of these equations:

174

Dynamics of Systems of Particles

E2 g

m1

E1

m2

P

O

Rough surface Figure 5.2

A system of two particles moving on a rough horizontal surface.

d dt

³

∂T

´

∂ q˙ K

−

∂T ∂ qK

= ³K .

(5.4)

Here, ³K =

F1 ·

∂ r2 ∂ r1 + F2 · , K ∂q ∂ qK

or equivalently, if we use the single representative particle, ³K = Φ · aK . Clearly we need to append P = P cos( ωt)E1 to F2 and P cos(ω t)e4 to Φ. From (5.4) and the expressions we have established for T and the forces, we find the following six equations: m1 ¨x1

= −K1 x1 −

K2 (x1 − x2 ) − µ dm1 g

m2 ¨x2

= −K2 ( x2 −

x1) − µd m2g

0 = N1y − m1g,

x˙ 2 |˙ x2|

+

x˙ 1 , |˙ x1|

P cos(ω t),

0 = N2y − m2g, 0 = N1z , 0 = N2z . The last four equations yield the constraint forces: N1y N2z = µ4 . Thus, N1 = m1 gE2 and N2 = m2 gE2.

(5.5) = µ1 ,

N1z

= µ 3,

N2y

= µ2 ,

and

Stick–Slip Oscillations

One of the most interesting features of the oscillator arises when either or both of the velocities of the particles vanish. In this case, the number of integrable constraints increases and the constraint forces must be altered. The resulting oscillations are often termed stick–slip, behavior we encountered earlier in our model for the roller coaster (cf. Section 2.11). Because the number of differential equations governing the motion is different for these two types of friction, numerical simulations of the equations of motion can be challenging.1 1 For further details on numerical schemes for stick–slip oscillators, see [33, 66, 149]. Interesting examples

of such oscillators and their application to various fields, including brake squeal, can be found in [31, 148].

5.2

Harmonic Oscillators

175

For instance, suppose m1 is instantaneously at rest (sticks), then the additional constraint x1 = x10 , where x10 is a constant, needs to be imposed. The constraint force Fc1 is now F c1

= µ 1E2 + µ 3E 3 + µ5 E1,

where µ 5E1 is the static friction force. The friction force is subject to the static friction criterion ¹

|µ 5| ≤ µ s

2 2 µ1 + µ 3,

where µs is the coefficient of static friction. Instead of (5.5), the equations of motion for the oscillator in this case are µ5 =

m2 x¨ 2 where

(

K1x10 + K2 x10

= −K 2(x2 −

º ºK1x1

0 +

−

)

x2 ,

x10 ) − µd m2g (

x˙ 2 |˙ x2|

+

P cos(ω t),

(5.6)

)º

K2 x10 − x2 º ≤ µs m1 g.

Thus, if there is sufficient friction to match the spring forces, then m1 will remain stationary. Otherwise, this mass particle will tend to slip in the direction of the resultant spring force and (5.5) is then used to determine the motion. Imposing an Integrable Constraint

We now consider the introduction of an integrable constraint in the system considered in Section 5.2. The constraint is x˙1 x2 − x˙ 2 = 0. We can also express this constraint as (x2E1 ) · v 1 + (−E1) · v 2 =

0,

or, for the representative particle of mass m, as ±

x2 e1 − e4

²

·v =

0.

It is left as an exercise to show that the constraint is integrable. Using Lagrange’s prescription, we find that the additional constraint forces on the particles are µ 5x2 E1 and −µ 5E1 . For the representative particle of mass m, Φc needs to ) ( be augmented by µ 5 x2e1 − e4 . Starting from Lagrange’s equations of motion in the form (5.4), we quickly arrive at the equations of motion: m1x¨ 1

= −K1 x1 −

K2(x1 − x2 ) − µ d m1g

m2x¨ 2

= −K2 (x2 −

x1 ) − µd m2g

x˙ 2 |˙ x2 |

x˙ 1 |˙ x1|

− µ 5.

+ µ5 x2,

176

Dynamics of Systems of Particles

m1

r1

C

¯r

E3

r2 O

E2

m2

E1 A particle of mass m1 attached by a linear spring of stiffness K and unstretched length to a particle of mass m2 . Each particle is attracted to the fixed point O by a Newtonian gravitational force field.

Figure 5.3

±0

The last four equations are identical to those recorded in (5.5). The two equations of motion are supplemented by the constraint x˙ 1x2 − x˙ 2 = 0. 5.3

A Dumbbell Satellite

In the 1960s, several simple models for deformable satellites orbiting a planet of mass M appeared. Here we consider one of these models and discuss some features of the dynamics predicted by it. In particular, we will notice a coupling between the motion of the center of mass and the rotation of the satellite. This coupling, which is elegantly explained in Beletskii’s text on satellite dynamics [22], is induced by the gravitational forces acting on the satellite. The model we discuss here lumps the mass distribution of the satellite into two mass particles at the extremities of a spring of stiffness K and unstretched length ± 0 (see Figure 5.3). The gravitational force exerted on the satellite by the planet of mass M that it is orbiting is modeled as a Newtonian gravitational force field exerted on each of the particles. In what follows, we discuss how the equations of motion for this model can be determined. Coordinates

Our first task is to choose coordinates for the position vectors r1 and r2 of the mass particles. One reasonable choice would be to pick Cartesian coordinates for both particles; another would be to choose Cartesian coordinates for r1 and spherical polar coordinates for r2 − r1. A third alternative would be to pick Cartesian coordinates for the position vector r¯ of the center of mass C of the system and a set of spherical polar coordinates for r2 − r¯ and r1 − r. ¯ We elect the third choice here. Denoting r2 − r1 by ReR , where R = ±r2 − r1 ±, eR

=

sin(φ) (cos( θ )E1 + sin(θ )E 2) + cos( φ)E 3.

5.3

A Dumbbell Satellite

Some elementary algebra shows that ³

r1

= r ¯−

´

m2

³

ReR,

m1 + m2

r2

= r ¯+

177

´

m1 m1 + m2

ReR .

With our choice of coordinates, ¯r = xE 1 + yE2 + zE3. For future reference, we label our coordinates q1

=

q2 = y,

x,

q3 = z,

q4

=

q5

R,

q6

= φ,

= θ.

(5.7)

We also record the following 12 partial derivatives: ∂ r1

=

∂x ∂ r1 ∂R

³ = −

∂ r2 ∂x

∂R

³ =

=

∂y

³ =

= −

´

m2 m1 + m2

∂ r2

E2 , ∂ r2

eR ,

³

Reφ ,

R sin(φ )eθ

m1 + m2 =

E3 ,

=

´

m2

∂ r2

m1 + m2 ∂θ

∂φ

´

m1

∂z

∂ r1

³ = −

∂ r1

E2 ,

eR ,

E1 ,

∂ r2

=

´

m1 + m2 ∂θ

∂ r2

∂y

m2 ∂ r1

and

∂ r1

E1 ,

∂φ

m1 m1 + m2

∂z

³ =

=

E3 , ´

m1 m1 + m2

Reφ ,

´

R sin(φ )eθ .

Because the forces acting on the system of particles are conservative, we can determine Lagrange’s equations of motion without calculating these vectors. Our motivation for calculating them here is that they will illuminate the combination of equations that constitute Lagrange’s equations of motion for this system and this choice of coordinates. Kinetic and Potential Energies

The kinetic energy of the system is the sum of the kinetic energies of the particles: m2 m1 v1 · v1 + v2 · v2 T= 2 2 m1 m2 m1 + m2 v¯ · v¯ + (v1 − v ¯ ) · ( v1 − v ¯) + (v2 − v ¯ ) · ( v2 − v ¯) = 2 2 2 ± ² ± ² m1 + m2 2 m1 m2 = x˙ + y˙ 2 + ˙z2 + R˙ 2 + R 2φ˙ 2 + R2 sin2 (φ) θ˙ 2 . 2 2(m1 + m2) The potential energy of the system is due to Newtonian gravitation and the spring potential energy: U

= −»

» »r¯ −

GMm1

±

m2 m1 +m 2

²

» » ReR »

−

GMm2

» » ² ± » » m »r¯ + m +1m ReR » 1 2

+

K 2 (R − ±0 ) . 2

178

Dynamics of Systems of Particles

We could also have included the Newtonian gravitational force that m1 exerts on m2 and vice versa, but this will not add significantly to the discussion. Lagrange’s Equations of Motion

For this system of particles, there are no constraints and the applied forces are conservative. As a result, we can use Lagrange’s equations of motion in the form d dt

³

∂L ∂q ˙S

´

−

∂L ∂ qS

=

0

(S =

1, . . . , 6) .

Equipped with the expressions for T and U recorded in the previous subsection, we can derive the equations in a straightforward manner; the details are not presented here. Because the sole forces acting on this system are conservative, the solutions to the equations of motion should preserve the total energy E = T + U. In addition, the angular momentum of this system relative to O, HO , should also be conserved. The former momentum has the following representation: HO

r 1 × m1 v 1 + r 2 × m2 v 2 m1 m2 2 ( = r ¯ × mv ¯ + R φ˙ eθ m1 + m2

=

− θ˙ sin(φ)eφ

)

.

The conservation of HO implies that the linear speeds x˙ , y˙ , and z˙ of the center of mass are coupled to the angular speeds θ˙ and φ˙ of the satellite. We shall observe this coupling later in rigid-body models for satellites. The Generalized Coordinates and Configuration Manifold

M

There are no constraints on the system, so the generalized coordinates are x, y, z, R, 6 ∼ θ , and φ . The configuration manifold = E , and the kinematical line element is ¹ 2T dt. m 1 +m2

ds = It is straightforward to show from the expression for T that the mass matrix for this system of particles is not always invertible. This calculation will also reveal that the coordinate system we are using for has singularities along a line where sin (φ) = 0.

M

Comments on the Equations of Motion

The equations of motion found by use of Lagrange’s equations are equivalent to those that would be obtained directly from F1 = m1r¨ 1 and F 2 = m2 r¨ 2. Indeed, if we return to the derivation of Lagrange’s equations in Section 4.6, then it is easy to see that these equations are linear combinations of F1 = m1r¨ 1 and F2 = m2 r¨ 2. For instance, Lagrange’s equation of motion for R: d dt

³

∂L ˙ ∂R

´

−

∂L ∂R

=

0

is equivalent to (F1 =

m1 r¨ 1) ·

∂ r1 ∂R

+ (F 2 =

m2 r¨ 2) ·

∂ r2 ∂R

=

0.

5.4

179

A Pendulum and a Cart

If we change our selection for the coordinates used for the system, this changes the precise linear combinations of the components of F1 = m1 ¨r1 and F2 = m2r¨ 2 that constitute Lagrange’s equations of motion.

Related Systems

We can modify the model presented in this section in a variety of manners. First, the spring can be replaced with a rigid bar. For this model, E and HO are still conserved. It is interesting to observe that, even for this simple system, the gravitational force field is equivalent to a force acting at the center of mass C and a moment relative to this point.

5.4

A Pendulum and a Cart

Consider the system of two particles shown in Figure 5.4. A particle of mass m1 is free to move on a smooth horizontal rail and is connected to a particle of mass m2 by a spring of stiffness K and unstretched length ± 0. The motion of both particles is assumed to be planar. We seek to determine the equations of motion for this system.

Coordinates and Constraints

As usual, the first task is to choose coordinates for the position vectors r1 and r2 of the mass particles. In anticipation of imposing the constraints, we choose Cartesian coordinates for r1 and cylindrical polar coordinates for r2 − r1 : Smooth horizontal rail

E2 g

m1

O

E1 Linear spring θ

r

m2

A system of two particles connected by a linear spring of stiffness K and unstretched length ±0 . The particle of mass m1 is free to move on a smooth horizontal rail and the second particle moves on the x–y plane.

Figure 5.4

180

Dynamics of Systems of Particles

r1

=

xE1 + yE2 + z1 E3,

r2 − r1 = rer + z2E3 , r2

=

xE1 + yE2 + rer + (z1 + z2 ) E3 .

The angle θ is measured from the E1 direction and is counterclockwise positive. That is, when m2 is directly below m1, θ = 32π . We label the coordinates as follows: q1

=

q2 = r,

x,

q3

q4

= θ,

=

y,

q5

=

q 6 = z2

z1 ,

= ( r2 −

r1) · E3 . (5.8)

The constraints on this system of particles are all integrable: 0,

²1 =

²2 =

0,

²3 =

0

where y = r1 · E2,

²1 =

²2 = z1 = r1 ·

E3 ,

²3 = z2 = (r2 − r1 ) · E3 .

At this stage, it is prudent to compute the following gradients: ∂ ²1 ∂ r1 ∂ ²2 ∂ r1 ∂ ²3 ∂ r1

=

E 2,

=

E 3,

= −E3,

∂ ²1 ∂ r2 ∂ ²2 ∂ r2 ∂ ²3 ∂ r2

=

0,

=

0,

=

E3.

We could also have used Cartesian coordinates to parameterize r2 − r1 , and this choice would be a good exercise to pursue. Kinetic and Potential Energies

The potential energy of the system has contributions from gravity and the spring potential energy: U

=

m1 gy + m2gy + m2 gr sin(θ ) +

K 2

2 (||r2 − r1|| − ±0 )

.

Imposing the integrable constraints, we find that K 2 (r − ±0 ) . 2 The kinetic energy of the system is also easy to calculate: U˜

T

=

=

m2 gr sin(θ ) +

² ² m1 + m2 ± 2 m2 ± 2 x˙ + y˙ 2 + ˙z21 + r˙ + r2θ˙ 2 + z˙22 2( 2 ) + m2 x ˙ ˙r cos( θ ) + y ˙ r˙ sin(θ ) − xr ˙ θ˙ sin(θ ) + y ˙ r θ˙ cos( θ ) + m2z˙1 ˙z2.

This expression simplifies when we impose the constraints: T˜

=

² ( ) m1 + m2 2 m2 ± 2 2 2 x˙ + ˙r + r θ˙ + m2 x ˙ ˙r cos(θ ) − xr ˙ θ˙ sin(θ ) . 2 2

(5.9)

5.4

181

A Pendulum and a Cart

Generalized Coordinates and Constraints

M

The system has three generalized coordinates: x, r, and θ . The configuration manifold for this system is homeomorphic to a three-dimensional manifold in E6 that is parameterized by these coordinates. Because r ranges from 0 to ∞ and θ ranges from 0 to 2π , these two coordinates parameterize E2 in its entirety. Further, as x ranges from −∞ to 3 ∼ ∞ , we see that this coordinate parameterizes E. We conclude that = E . ˜ We can find the kinematical line element for the configuration manifold by using T:

M

µ

ds =

2T˜ m1 + m2

dt,

where we have chosen m = m1 + m2. Examining the expression for T˜ , we find that the mass matrix fails to be invertible when r = 0. Thus, the coordinate system we are using for has a singularity.

M

Constraint Forces and Resultant Forces

The constraint forces on the system can be computed by use of Lagrange’s prescription: Fc1 Fc2

3 ¼ =

=

i =1 3 ¼

µi

µi

i =1

∂ ²i ∂ r1 ∂ ²i ∂ r2

= µ1 E2 + µ 2E3 − µ3 E3,

= µ3 E3 .

Notice that the Lagrange multipliers µi are equivalent to the normal forces: N1 µ1 E2 + (µ2 − µ3 ) E3 and N2 = µ3 E3 . For completeness, the total resultant forces F1 and F2 acting on the particles are F1

=

F2

= −K ( r − ±0) er −

=

K (r − ±0) er + (µ1 − m1g) E2 + (µ2 − µ3 ) E3 , m2 gE2 + µ3 E3 ,

respectively. We shall use these expressions later on to determine the force vector Φ.

Lagrange’s Equations of Motion

We can find the equations of motion for this system by using any of the variety of forms for Lagrange’s equations of motion. Arguably, the easiest approach is to use the form (4.23) involving the Lagrangian: d dt

³

∂L ∂q ˙S

´ −

∂L = ∂ qS

QS

=

2 ¼ i=1

Fnconi

·

∂ ri ∂ qS

(S =

1, . . . , 6) .

(5.10)

However, as the constraint forces are compatible with Lagrange’s prescription, for the first three of these equations the right-hand side will be zero: Q1 = 0, Q2 = 0, and Q3 = 0.

182

Dynamics of Systems of Particles

To find the equations of motion, it suffices to examine the first three of Lagrange’s equations of motion: d dt

·

˜ ∂L ∂q ˙J

¸

−

˜ ∂L = ∂ qJ

0

(J =

1, . . . , 3) .

˜ Evaluating the partial derivatives of L ˜ = T ˜ −U ˜ and then calculating Here, L˜ = T˜ − U. d , we find with some rearrangement that the equations of motion can be expressed in dt the form

⎡

m1 + m2 ⎣ m2 cos(θ ) −m2 r sin(θ )

m2 cos(θ ) m2 0

−m2 r

⎤⎡

⎤

⎡

⎤

sin(θ ) x ¨ f1 ⎦ ⎣ ¨r ⎦ = − ⎣ f2 ⎦ , 0 m2 r 2 θ¨ f3

(5.11)

where f1 , f 2, and f3 are quadratic functions of the velocities: f 1 = −2m2r˙ θ˙ sin(θ ) − m2 rθ˙2 cos( θ ), f 2 = −m2 rθ˙2 + m2g sin(θ ) + K (r − ±0 ) , f 3 = 2m2 r˙rθ˙ + m2 gr cos(θ ).

(5.12)

The canonical form (5.11) of the equations of motion is easy to implement numerically. As emphasized previously, the canonical form is relevant to the equations of motion of many mechanical systems with time-independent (scleronomic) integrable constraints. You should also notice that the mass matrix on the left-hand side of (5.11) can be obtained by inspection from T˜ . Solving for the Constraint Forces

To determine the constraint forces Fc1 and Fc2 , we first solve for µi by using three of Lagrange’s equations of motion: d dt

³

∂L ∂q ˙J

´

−

∂L ∂ qJ

=

QJ

(J =

4, . . . , 6) .

We leave the intermediate steps as an exercise: Fc1

= (m1 +

Fc2

=

0.

m2) gE2 +

) d ( m2r˙ sin(θ ) + m2r θ˙ cos( θ ) E2, dt

It is easy to observe from these expressions the expected result that F c1 · v1+F c2 · v2

=

0.

Conservations

The solutions to equations of motion (5.11) conserve two kinematical quantities. The first is the total energy E of the system. To see this conservation, it suffices to note that none of the constraint forces do work and all of the remaining forces acting on the system are conservative. Thus, the work–energy theorem easily leads to the conclusion that ˜ The second conservation can be deduced from (5.11)1 . That E˙ = 0, where E = T˜ + U. is, the linear momentum of the system in the horizontal direction, G · E1 , is conserved.

5.5

Two Particles Tethered by an Inextensible String

183

The Single Representative Particle

We computed Lagrange’s equations of motion for the system without explicitly calculating the position vector r of the single particle of mass m moving in E6 . If we were to use this particle to compute (5.11), then we would first need to use (4.15) to define r = xe1 + ye2 + z1 e3 + (x + r cos( θ )) e4 + (y + r sin(θ )) e5 + (z1 + z2 ) e6. Using this expression, we can easily calculate the six covariant vectors aJ and the six contravariant vectors aJ . For example, a1 = e1 + e4 and a6 = e6. As expected, the kinetic energy T = m2 r˙ · r˙ will be identical to (5.9). The force vector Φ can be determined by use of the prescription (4.16): Φ = K (r − ±0) cos(θ )e1 + (µ1 − m1 g + K (r − ± 0) sin(θ )) e2 + (µ2 − µ 3) + ( − m2 g −

e3 − K (r − ±0) cos( θ )e4

K (r − ± 0) sin(θ )) e5 + µ3 e6 .

We used the expressions for F 1 and F2 recorded earlier to compute this vector. If we compute Φ · aS and compare the results with those we obtained by using F1 · ∂∂ qrS1 + F2 · ∂

r2

∂ qS

, then the two sets of expressions for ³S should be identical.

Remarks

The system discussed here is also a good candidate to explore the covariant form (4.29) and the contravariant form (4.31) of Lagrange’s equations of motion½that¾ we discussed earlier. Indeed, by using (5.11), we can readily compute the matrix aij and the Christoffel symbols of the first kind.

5.5

Two Particles Tethered by an Inextensible String

As shown in Figure 5.5, a particle of mass m1 is connected by an inextensible string of length ± 0, which passes through a smooth eyelet at O to a particle of mass m2. The particle of mass m1 moves on a rough horizontal plane, and the particle of mass m2 is free to move in space. The goal of the following analysis is to establish the equations of motion for the system of particles and then discuss certain conserved quantities associated with their solutions. To make the problem tractable, we assume that the string remains taut and that the particle of mass m2 does not collide with the underside of the horizontal plane.

The Coordinates and Other Kinematical Quantities

To describe the kinematics of this system of particles, a cylindrical polar coordinate system {r1 , θ1, z1} is used to describe the motion of the particle of mass m1 and a spherical

184

Dynamics of Systems of Particles

Rough horizontal plane

E3 g m1

E2

O

E1

m2

Inextensible string

A system of two particles connected by an inextensible string of length ±0 .

Figure 5.5

polar coordinate system {R 2, φ2 , θ2 } is used to describe the motion of the particle of mass m2 : r1

=

r1er 1 + z1 E3,

r2

R2 eR 2 .

=

We define six coordinates as follows: q1

=

q2 = θ1 ,

r1 ,

q3

q4

= φ2 ,

q5

= θ 2,

=

x,

q6

=

z1 .

(5.13)

Notice that we have introduced a new coordinate x: R2

= ±0 − r1 +

x.

Using the cylindrical polar coordinate and spherical polar coordinate systems, we have r1

=

r1er1

+ z1 E3,

r2

= (±0 +

x − r1 ) eR2 .

Hence, ∂ r1 ∂ r1

=

∂ r1

er 1 ,

∂ θ1 ∂ r2 ∂ r1

∂ r2 ∂ θ2

=

r1eθ1 ,

= −eR2 ,

= (±0 +

∂ r1 ∂ φ2 ∂ r2 ∂ θ1

=

∂ r1

0,

=

0,

∂ θ2 ∂ r2 ∂ φ2

x − r1) sin(φ2 )eθ2 ,

=

∂ r1

0,

∂x

= (±0 +

∂ r2 ∂x

=

=

0,

∂ r1 ∂ z1

=

E3 ,

x − r1) eφ2 ,

eR2 ,

∂ r2 ∂ z1

=

0.

The Potential and Kinetic Energies

The potential energy of the system is entirely due to gravity: U

=

m1 gE 3 · r1 + m2 gE3 · r2

=

m1 gz1 + m2g (±0 + x − r1) cos (φ 2) .

To calculate the kinetic energy of the system, we first need expressions for the velocity vectors. The expression for v1 is easy to find: v1

= r ˙1 er 1 +

r1θ˙1eθ1

+z ˙1E 3.

5.5

Two Particles Tethered by an Inextensible String

185

To calculate v2 , we recall the expression for this vector in spherical polar coordinates and then substitute for R2 and R˙ 2 : v2

= (x ˙ − r˙1) eR2 + (±0 +

x − r1 ) sin (φ2 ) θ˙2 eθ2 + (±0 + x − r1) φ˙2 eφ2 .

The kinetic energy T of the system of particles is T

m2 m1 v1 · v1 + v2 · v2 2 ± 2 ² m1 2 2 2 2 ˙r1 + r1 θ˙1 + ˙z1 = 2 ± ² m2 2 2 2 2 2 ˙2 ˙ − r˙1) + (±0 + x − r1) sin (φ2 ) θ˙2 + (±0 + x − r1) φ + (x 2 . 2 =

Notice that T is a function of q1, . . . , q6 and their time derivatives.

The Constraints and Constraint Forces

The constraints on the motion of the system are twofold. First, the particles are connected by an inextensible string of length ±0 and second, the motion of m1 is planar. In terms of the coordinates q1 , . . . , q6 , the two constraints are x = 0,

z1

=

0.

The constraint forces associated with these constraints correspond to the tension force in the string and the friction and normal forces on m1 : Fc 1

= µ1 er1 + µ2 E3 − µ d ||µ2 E3||

Fc 2

= µ1 eR2 ,

vrel , ± v rel ±

where vrel = r˙1er 1 + r1 θ˙1eθ1 . The constraint forces associated with the inextensible string can also be prescribed by use of Lagrange’s prescription. Let ²1 = ± r2± + ±r1± − ± 0,

and observe that ∂ ²1 ∂ r1

=

r1 , ± r1±

∂ ²1 ∂ r2

=

r2 . ± r2±

The inextensibility constraint is ²1 = 0 and, using Lagrange’s prescription, we find F c1

= µ1

F c2

= µ1

r1 ± r1 ±

r2 ± r2 ±

= µ1 er 1 , = µ1 eR2 .

Notice that we imposed the constraint z1 = 0 to simplify the expression for Fc1 . The constraint forces associated with the constraint ²2 = 0, where ²2 = z1, are not prescribed by Lagrange’s prescription because it features a dynamic friction force.

186

Dynamics of Systems of Particles

The Equations of Motion

Lagrange’s equations will provide four differential equations for the generalized coordinates and two equations for µ1 and µ2 . For ease of exposition, first the differential equations are presented: 2 (m1 + m2 ) ¨r1 − m1 r1 θ˙1

±

+

²

vrel · er 1 ±v rel ± + m2 g cos (φ2 ) , ² d ± vrel · eθ 1 , m1r12θ˙1 = − µ d r1||µ 2E 3|| dt ± vrel ±

m2 (±0 − r1) sin2 (φ2 ) θ˙22 + φ˙ 22

= −µd ||µ 2E 3||

² d ± m2 (±0 − r1)2 φ˙ 2 dt 2 2 − m2 (±0 − r1 ) θ˙2 sin (φ2 ) cos (φ 2) = m2 g (±0 − r1) sin (φ2 ) , ² d ± m2 (±0 − r1)2 sin2 (φ2 ) θ˙2 = 0. dt

In addition, the two equations for µ1 and µ 2 are µ1 =

±

(5.14)

²

m2g cos (φ 2) − m2 r¨1 − m2 (±0 − r1 ) sin2 (φ2 ) θ˙22 + φ˙22 , µ2 =

m1 g.

That is, the normal force on m1 is N1 = m1 gE3 and the tension forces µ1 eR2 and µ 1er1 in the segments of the string can be computed using the expression given above for µ1 . To find the preceding equations, we started with Lagrange’s equations: d dt

³

∂T ∂q ˙K

´

−

∂T = ∂ qK

F1 ·

∂ r1 ∂ r2 + F2 · K ∂q ∂ qK

(K =

1, . . . , 6) .

We then substituted for T , F1, and F 2. After the partial derivatives of T were calculated, we imposed the two constraints and performed some rearranging. Some of the details of these calculations are subsequently provided: vrel · er 1 + m2 g cos (φ2 ) , ± v rel ± vrel · eθ1 , F1 · r1eθ1 + F2 · 0 = −µd r1|| µ2 E3|| ± vrel ± F 1 · 0 + F 2 · (±0 − r1 ) eφ2 = m2g (±0 − r1 ) sin (φ 2) , F1 · er 1 − F2 · eR2

F 1 · 0 + F2 · (±0 − r1 ) sin (φ 2) eθ2 F1 · 0 + F2 · eR2

= −µd ||µ 2E3 ||

=

0,

= µ1 −

m2 g cos (φ2 ) ,

F1 · E3 + F2 · 0 = µ2 − m1 g.

The Lack of Energy Conservation

To prove that the total energy of the system of particles is not conserved, we start with the work–energy theorem T˙ = F 1 · v1 + F2 · v2 and substitute for the applied and constraint forces:

5.6

T˙

=

(

−m1 gE 3 +

Fc1

)

187

Closing Comments

(

· v 1 + −m2 gE3 +

Fc2

)

· v 2.

Making use of the potential energy function U, we can express this equation as ( ) vrel E˙ = µ1 er 1 · v1 + eR 2 · v2 + µ2 E3 · v1 − µd ||µ 2E 3|| · v1 , ± vrel ± where the total energy E = T µ2 = m1 g, we surmise that E˙

= µ1

However, er 1 · v1 + eR 2 · v2 conclusion:

+

U. Thus, as v1

(

er1 · v1 + eR2 · v2

˙ 2, = r˙1 + R

)

=

vrel and is normal to E3 , and

− µd m1 g||v 1|| .

and this sum is zero because r1

+

R2

= ±0 .

In

E˙ = −µ d m1g ±v1± . Notice that E˙

≤

0 as expected because of the friction force.

Conservations of Angular Momenta

If friction is absent, then we find from (5.14)2,4 that HO 1 · E3 = m1r21 θ˙1 and HO 2 · E3 = m2 (±0 − r1 )2 sin2 (φ 2) θ˙2 are conserved. That is, the angular momentum of each particle relative to O in the E 3 direction is conserved. Configuration Manifold and its Geometry

The configuration manifold for this system is homeomorphic to a four-dimensional subspace of E6 parameterized by r ∈ (0, ±0), θ1 ∈ [0, 2π ), θ2 ∈ [0, 2π ), and φ2 ∈ (0, π ). The kinematical line element ds for this manifold is given by µ

ds =

2T˜ 2 dt, m1 + m2

where we find T˜ 2 from T by imposing the constraints and collecting all those terms that are quadratic in the generalized velocities: ² ² m1 ± 2 m2 ± 2 T˜ 2 = r˙1 + r21 θ˙12 + r˙1 + (±0 − r1)2 sin2 (φ2 ) θ˙22 + (±0 − r1 )2 φ˙ 22 . 2 2 To visualize the configuration manifold, one would give a two-dimensional picture of a disk of radius ±0 with the coordinates r1 cos (θ1 ) and r1 sin (θ1 ). This would be supplemented by a three-dimensional image of a unit two-sphere parameterized by φ2 and θ2. We leave it as an exercise to examine the mass matrix for this system and show that singularities in the coordinate system for are present.

M

5.6

Closing Comments

Problems involving systems of particles have played a key role in the development of dynamics. Specifically, mention is made here of a model for the celestial system of the

188

Dynamics of Systems of Particles

Sun, Earth, and Moon, known as the three-body problem. In this problem, the three bodies are modeled as particles subject to the mutual interaction that is due to a Newtonian gravitational force field. That is, the potential energy for the system is (cf. (4.9)) Un

= −

Gm1 m2 ±r2 − r1 ±

−

Gm3m1 ±r1 − r3±

−

Gm2 m3 , ± r3 − r2±

(5.15)

where r1 , r2 , and r3 are the position vectors of the particles of masses m1 , m2, and m3 , respectively. Famous exact solutions to special cases of the three-body problem range from the equilateral triangle solution by Lagrange [155] in 1772, 2 to the figure-eight solution that was only recently found numerically by Moore [194, 195] and proven to exist by Chenciner and Montgomery [49, 192] (see Figure 5.6). 3 Apart from its paucity of exact solutions, the three-body problem is also well known because of the profound analysis of this system by Henri Poincaré (1854–1912) in the late 1880s (see [5, 18, 59]). His analysis is considered to be the first description of chaos in mathematical models for physical systems and formed one of the cornerstones for the field of chaos in dynamical systems that achieved popular attention some 100 years later in the late 1980s. The three-body and two-body problems are special cases of the more general n-body problem. In celestial mechanics, the n-body problem is synonymous with models for our solar system and has attracted some of the most celebrated scientists in history. It was also the problem that led Hamilton to discover his famous equations of motion in [115] and his variational principle in [116]. Unfortunately, we do not have the opportunity to m1

(b)

(a) m1

m3

m2

(c)

m3

m3

m2

m2 m1

Representative orbits from the three-body problem: (a) and (b) examples of Lagrange’s equilateral triangle solutions; (c) the figure-eight solution. For the solutions shown in this figure, m1 = m2 = m3 .

Figure 5.6

2 A discussion of this famous solution can be found in many texts on celestial mechanics, for

example [127, 199, 296].

3 The interested reader is referred to the online article by Casselman [41], where simulations of several

three-body problems can be found.

5.7

Exercises

189

explore the three-body and n-body problems in any detail here; the interested reader is referred to the previously cited texts. The problems we have mentioned do not feature constraints on the motions of the particles, and they are often formulated without using Lagrange’s equations of motion. However, problems involving particles connected by rigid links often feature in simple models for artificial satellites orbiting a celestial body and in various pendulum systems. For these models, Lagrange’s equations of motion are ideally suited to the task of establishing a set of governing ordinary differential equations that are free from constraint forces. In the following exercises, problems for systems of particles of this type are emphasized.

5.7

Exercises

Exercise 5.1: Consider the systems of particles discussed in Section 5.2. Suppose a time-dependent force P(t)E1 acted on the particle m2 .4 What are the equations of motion for each of these systems? Exercise 5.2: Again, consider the systems of particles discussed in Section 5.2. Suppose, in addition to the springs, there are viscous dashpots in these systems. 5 What then are the equations of motion? Exercise 5.3: Here, we are interested in establishing a particular representation for the equations governing the motion of two unconstrained particles. In a subsequent exercise, one can impose constraints to yield the equations of motion of a pendulum system. Consider the system of particles shown in Figure 5.7. The particles are free to move in E3 under the influences of resultant external forces F1 and F 2, respectively. To establish the equations of motion for the single particle, we use a cylindrical polar coordinate system {r1, θ1 , z1 } for the particle of mass m1. For the second particle, it is m2

m1

E3 g

O

E1 Figure 5.7

A system of two particles.

4 This force is not conservative. 5 The forces from these dashpots are not conservative.

E2

190

Dynamics of Systems of Particles

eθ

E2 er

1

E2

eθ

er

2

1

θ1

eθ

2

eθ

1

er

2

θ2 θ1

2

θ21 θ2

E1

θ21

E1

er

1

Figure 5.8 Transformations of unit vectors associated with the cylindrical polar coordinate systems used in the exercises. The angle θ21 = θ2 − θ1 .

convenient to describe its motion with the assistance of the relative position vector r21 = r2 − r1 . We describe this vector by using a spherical polar coordinate system {R2, φ2, θ2 }: r1

=

r1er 1 + z1 E3,

r2

=

r1 + R2 eR 2 .

In the sequel, the coordinates are organized as follows: q1

q2

r1 ,

=

q3 = z1 ,

= θ1 ,

q4

=

q5

R2 ,

= φ2,

q6

= θ2 .

(a-1) Establish the following results: ∂ r1 ∂ r1

=

∂ r2 ∂ r1

=

∂ r1 ∂ R2

∂ r2

=

∂ R2

∂ r1

er 1 ,

eR2 ,

∂ θ1 =

=

∂ r2 ∂ θ1

∂ r1

0,

∂ φ2

∂ r2 ∂ φ2

=

=

=

∂ r1

r1 eθ1 , ∂ r1

0,

∂ θ2

∂ r2

R 2eφ 2 ,

∂ θ2

=

∂ z1 =

=

∂ r2 ∂ z1

=

E3,

0,

R2 sin (φ2 ) eθ2 .

For partial assistance with the basis vectors in these expressions, the reader is referred to Figure 5.8. (a-2) As an alternative to (a-1), show that the position vector of the single representative particle is r = r1 cos(θ1)e1 + r1 sin(θ1)e2 + z1 e3 + ( r1 +

cos(θ1) + R2 sin(φ2 ) cos(θ2 )) e4

(r1 sin(θ1 ) + R 2 sin(φ2 ) sin(θ2 ))e5 + (z1 + R2 cos( φ2 ))e6.

(i) Using r and the curvilinear coordinate system it induces on E6: q1

=

r1 ,

q2 = θ1 ,

q3

=

z1 ,

q4

=

show that the six covariant basis vectors aJ a1 a2

=

=

∂r ∂ θ1

∂r ∂ r1

=

= −r1

R 2,

=

∂r

∂ qJ

q5

= φ2,

q6 = θ2 ,

are

cos( θ1 )e1 + sin(θ1 )e2 + cos(θ1 )e4 + sin(θ1 )e5 ,

sin(θ1 )e1 + r1 cos(θ1)e2 − r1 sin(θ1)e4 + r1 cos( θ1 )e5, a3

=

∂r ∂ z1

=

e3 + e6,

5.7

a4 a5

=

=

∂r ∂ φ2

∂r

=

∂ R2 =

a6 =

191

Exercises

sin(φ2 ) cos(θ2 )e4 + sin(φ2) sin( θ2 )e5 + cos(φ2 )e6 ,

R2 cos(φ2 ) cos(θ2 )e4 + R2 cos( φ2 ) sin(θ2 )e5 − R2 sin(φ2)e6, ∂r ∂ θ2

= −R2

sin(φ2) sin( θ2 )e4 + R2 sin(φ2) cos(θ2 )e5.

(ii) Show that the six contravariant basis vectors have the following representations: a1 a2

=

cos( θ1 )e1 + sin(θ1 )e2,

= −

sin(θ1) 1 cos(θ1 ) 2 e + e , r1 r1 a3 = e3 ,

a4

±

=

a5 =

²

sin(φ2) cos(θ2 )(e4 − e1) + sin(θ2 )(e5 − e2)

+

cos( φ2 )(e6 − e3),

² sin(φ2 ) 6 cos( φ2 ) ± cos(θ2)(e4 − e1 ) + sin(θ2)(e5 − e2 ) − (e − e3 ), R2 R2

a6

= −

cos( θ2 ) sin(θ2) (e4 − e1 ) + (e5 − e2). R2 sin(φ2 ) R 2 sin(φ2 )

(b) Show that the kinetic energy T of the system of particles (or, equivalently, the representative particle of mass m) has the representation T

=

² m1 + m2 ± 2 r˙1 + r21θ˙12 + z˙21 2 ² m2 ± 2 R˙ 2 + R 22φ˙ 22 + R22 sin2 (φ2)θ˙22 + 2 ( ) ˙ 2z˙1 + φ ˙ 2 ˙r1R2 cos(θ 21 ) + φ ˙2 θ˙1r1 R2 sin(θ 21 ) + m2 cos(φ2 ) R (

)

−

m2 sin(φ2) R2 φ˙2 ˙z1 − r1 R2θ˙1 θ˙2 cos( θ21) − ˙r1 R˙ 2 cos( θ21 )

−

m2 sin(φ2)

(

˙ 2θ˙1 − r1 R

where we have used the abbreviation θ21 energy follows from the definition T=

)

sin(θ21 ) + r˙1 θ˙2R2 sin(θ21) , = θ2 − θ 1.

This expression for the kinetic

m1 m2 m v·v= v1 · v1 + v 2 · v 2. 2 2 2

(c) If the forces acting on the particles are F 1 = −m1 gE3 and F2 = − m2gE3 , then what is the potential energy U of the system? If you are computing the representative particle, determine the force Φ corresponding to this pair of forces. (d) What are the six Lagrange’s equations for the system of particles (or, equivalently, the representative particle of mass m)?6 6 You should refrain from expanding the time derivative d /dt in your solution.

192

Dynamics of Systems of Particles

E3 g

E2

O

E1 m1

m2

Figure 5.9

A planar pendulum coupled to a spherical pendulum.

Exercise 5.4: As shown in Figure 5.9, two particles of mass m1 and m2 are connected by a rigid massless rod of length ±2 . The rod is connected to m1 by a ball-and-socket joint. In addition, the particle of mass m1 is connected by a rigid massless rod of length ±1 to a fixed point O. The connection between the rod and the point O is through a pin joint and is such that the motion of m1 lies on the plane spanned by E1 and E2 . (a) What are the three constraints on the motion of the system of particles (or, equivalently, on the single representative particle)? (b) Using Lagrange’s prescription, what are the constraint forces on the particles (or, equivalently, the constraint force Φc acting on the single representative particle)? (c) Starting from the final results of Exercise 5.3, establish Lagrange’s equations of motion for the pendulum system. In your solution, clearly distinguish the equations governing the motion of the particle and the equations giving the constraint forces. (d) Let us now establish some of the equations of (c) by using an equivalent approach. ˜ In Impose the constraints on T to determine the constrained kinetic energy T. ˜ Verify that the following addition, determine the constrained potential energy U. 7 equations correspond to those you obtained from (c): d dt d dt d dt

·

˜ ∂T

∂ θ˙

1

·

˜ ∂T ˙2 ∂φ

·

˜ ∂T

∂ θ˙2

¸ −

¸ −

¸ −

˜ ∂T

∂ θ1 ˜ ∂T

∂ φ2 ˜ ∂T

∂ θ2

=

Φ · a2

= −

=

Φ · a5

= −

=

Φ · a6

= −

˜ ∂U

∂ θ1 ˜ ∂U

∂ φ2 ˜ ∂U

∂ θ2

, , .

(e) Suppose a nonintegrable constraint is imposed on the pendulum system discussed in (d): f1 · v1 + f2 · v2 + e = 0. 7 It is crucial to note that

˜ ∂T ∂ r˙1

=

∂ T˜

˙ ∂R 2

˜ ˜ ˜ ˜ = ∂∂˙zT = ∂∂rT = ∂∂RT = ∂∂zT = 0. 1 1 2 1

5.7

Exercises

193

m1

m2

E3

E2

O

E1 Figure 5.10

m3

Schematic of a model for a satellite orbiting a fixed body of mass m3 .

Show that this constraint can be expressed as f · v + e = 0. In addition, what are the equations governing the motion of the nonintegrably constrained system? Illustrate your solution with a nonintegrable constraint of your choice. Exercise 5.5: As shown in Figure 5.10, a model for an artificial satellite consists of two particles of mass m1 and m2 connected by a rigid massless rod of length ±0 . A third particle of mass m3 is assumed to be stationary at the fixed point O. In addition to the constraint force in the rod, the system is subject to conservative forces whose potential energy function is given by (5.15). (a) What are the four constraints on the motion of the system of particles? (b) Using Lagrange’s prescription, what are the constraint forces acting on the particles of mass m1 and m2? You should also, if possible, verify that the components of these forces are physically realistic. (c) Using a set of Cartesian coordinates to describe the location of the center of mass C of the satellite of mass m1 + m2 and a set of spherical polar coordinates to parameterize the position of m2 relative to C, establish an expression for the kinetic energy of the system. (d) Establish the equations of motion for the system. (e) Show that the solutions to the equations of motion for the system conserve the total energy of the system and the angular momentum of the system relative to O. (f) Show that it is possible for C to execute a steady circular motion about O. What are the possible orientations of the rigid massless rod of length ±0 during such motions? Exercise 5.6: As shown in Figure 5.11, a particle of mass m1 is connected by a linear spring of stiffness K1 and unstretched length ±0 to a fixed point O. A second particle of mass m2 is attached by a rod of length ±2 to the particle of mass m1 with a pin joint. For this system, which is a variation on the classical system of a planar double pendulum,

194

Dynamics of Systems of Particles

E2 g

E1

O

Linear spring

Rigid massless rod m1

m2

Figure 5.11

A system of two particles connected by a rigid massless rod of length ± 2.

we assume that the motions of m1 and m2 are constrained to move on the plane spanned by E1 and E 2. To describe the kinematics of this system, a cylindrical polar coordinate system { r1, θ1 , z1 } is used to parameterize the motion of the particle of mass m1 and another cylindrical polar coordinate system {r2 , θ2 , z2} is used to parameterize the motion of the particle of mass m2 relative to m1 : r1

=

r1er 1 + z1 E3 ,

r2

=

r1 + r2er 2 + z2 E3 .

We define six coordinates as follows: q1 =

θ1 ,

q2

= θ2 ,

q3

=

r1 ,

q 4 = r2 ,

q5

=

z1 ,

q6

=

z2 .

(5.16)

(a) With the help of (5.16), what are the 12 vectors ∂∂qrK1 and ∂∂qrK2 ? Here, K = 1, . . . , 6. (b) What are the three constraints on the motion of the system of particles? Argue that the constraint forces F c1 and Fc 2 acting on the individual particles have the prescriptions F c1

= µ1 er2 + µ2 E3 ,

F c2

= −µ 1er2 + µ 3E3 .

(5.17)

Compute the following six components: ³cK =

F c1 ·

∂ r1 ∂ r2 + Fc 2 · . K ∂q ∂ qK

Comment on the values of the first three components. (c) In terms of the coordinates q1, . . . , q3 and their time derivatives, what are the kinetic energy T˜ and potential energy U˜ of the constrained system of particles? For partial assistance computing T˜ , the reader is referred to Figure 5.8. (d) What are Lagrange’s equations of motion for the generalized coordinates of this system of particles?

5.7

195

Exercises

(e) Suppose a nonintegrable constraint8 r1 θ˙1 + ±2θ˙2

=

0

(5.18)

is imposed on the system of particles. After expressing this constraint in the form f1 · v1 + f2 · v2 = 0, argue that F c1

= µ 1er2 + µ 2E3 + µ4

(

)

eθ1 − eθ2 ,

Fc2

= −µ 1er 2 + µ 3E3 + µ4 eθ 2 .

With the help of your results from (d), determine the equations of motion for the system of particles. (f) Starting from the work–energy theorem T˙ = F1 · v1 + F2 · v2 , show that the total energy E is conserved. (g) Suppose that the spring is replaced with a rigid rod of length ±1 and the nonintegrable constraint (5.18) is removed. In this case, which is the classic planar double pendulum, show that the equations governing the motion of the system are (1 + α)

d 2θ 1 dτ 2

+ αβ

cos (θ2 − θ1 )

d 2θ 2 dτ 2

³ − αβ

dθ2 dτ

´2

= − ( 1 + α)

2

d θ2 dτ 2

+

1 β

cos (θ2 − θ1 )

2

d θ1 dτ 2

+

1

³

β

= −

1 β

dθ1 dτ

sin (θ2 − θ1 )

cos (θ 1) ,

´2

sin (θ 2 − θ1)

cos (θ 2) .

(5.19)

In writing (5.19), we have used the following dimensionless parameters and time variable: ¶ ±2 g m2 , β = , τ = t. α= m1 ±1 ±1

M

M

The configuration manifold for this system is homeomorphic to a torus. What is the kinematical line element for ? (h) Numerically integrate (5.19) for a variety of initial conditions and illustrate your solutions on the configuration manifold for the planar double pendulum. You should verify that your solutions conserve the total energy of the system. Exercise 5.7: The swinging Atwood’s machine (SAM) is a variation of Atwood’s machine, a system consisting of two point masses which are connected by an inextensible string and supported by a pair of massless pulleys. In a regular Atwood’s machine, both masses can only move up and down. However, in the SAM, one of the masses is allowed to swing on a plane. In this problem, the equations of motion for this system and some of their features are examined. This problem is adapted from [290] and the reader is referred to this paper for further details on the dynamics of the system. As shown in Figure 5.12, let m1 be the swinging mass and m2 be the vertically oscillating mass. A string of length ± 0 connects the two masses and the distance between the 8 Referring to the discussion of constraint (1.22) in Chapter 1, this constraint is arguably the simplest

nonintegrable constraint that we can impose on this system.

196

Dynamics of Systems of Particles

1 O

g

E2

r

E1

x

θ

m1 m2

P

(t)E 1

A system of two particles known as the swinging Atwood’s machine. The particle m1 is allowed to swing freely on a plane, while m2 is free to move in a smooth vertical guide under the action of an applied force P(t)E1 . You should assume that the pulleys have negligible dimensions. Figure 5.12

two pulleys is given by ±1 . Gravitational forces m1 gE 1 and m2gE1 act on the respective particles. In addition, an applied force P(t)E1 acts on m2 . Using a cylindrical polar coordinate system, the positions of the particles can be expressed as r1

=

rer + ±1 E2 + z1 E3 ,

r2

=

xE1 + yE2 + z2 E3.

In writing these vectors, the dimensions of the pulleys are assumed to be negligible. To describe the dynamics of the machine, we choose the following coordinate system: q1 q3

=

r + x + ±1 − ±0 ,

=

q2

r, q4

=

= θ,

y,

q5

=

z1 ,

q6

=

z2 .

(a) Compute the 12 vectors ∂∂q˙vKi for i = 1, 2 and K = 1, . . . , 6. (b) What are the four constraints on the system of particles? Show that one of these constraints can be expressed as v1 · er + v2 · E1

=

0.

What is Lagrange’s prescription for the constraint forces Fc1 and Fc2 enforcing the four constraints? Give a physical interpretation for the constraint forces this prescription provides. (c) Compute Fc1 ·

∂ v1 ∂ v2 + F c2 · K ∂q ˙ ∂q ˙K

(K =

Two of the resulting summations should be zero.

1, . . . , 6) .

5.7

197

Exercises

˜ (d) Show that the constrained kinetic energy T˜ and the constrained potential energy U for the system have the following representations:

T˜

1 1 2 2 2 (m1 + m2 ) ˙r + m1 r θ˙ , 2 2 ˜ = m2 gr − m1gr cos (θ ) . U

=

(e) Show that the equations of motion of the system of particles can be expressed in the form ¿

( m1 + m2 )

0

0 m1 r 2

À¿

r ¨ θ¨

À

¿

+

2 −m1rθ˙

À

2m1rθ˙ ˙r

¿

=

− m2 g +

m1g cos (θ ) − P −m1 gr sin (θ )

À

.

(f) Suppose P(t) = P0 is a constant. Give an expression for the total energy E of the system and prove this energy is conserved during a motion of the system. If P(t)E1 were not a constant force, then explain why the total energy of the system is not conserved. (g) Suppose P(t) = P 0 is constant. Consider an equilibrium state of the system: r = r0 and θ = θ0. Show that an equilibrium state is possible only when either P0 + m2 g = m1 g or P0 = − (m1 + m2 ) g. (h) Using the mass matrix from the equations of motion (or, equivalently, T˜ ), show that 2 the coordinate system for = E has a singularity when r = 0.

M

Exercise 5.8: A simple model for the suspension system of a pendulum of mass m2 assumes that the pendulum is connected by a pin joint to a mass m1 which is free to move in a rough vertical guide (cf. Figure 5.13). The mass m1 is suspended from a fixed point O by a linear spring of stiffness K and unstretched length ± 0. The particles are under the influence of the respective gravitational forces m1 gE1 and m2 gE 1, and the motion of each particle lies in the vertical plane. The coefficient of dynamic Coulomb friction between the guide and m1 is denoted by µ d . A Cartesian coordinate system {x, y, z1} is chosen to parameterize the motion of m1 and a cylindrical polar coordinate system {r, θ , z2 − z1} is chosen to parameterize r2 − r1: r1

= ( x + ± 0) E1 +

yE 2 + z1E3 ,

(a) Compute the 12 vectors q6 = z2 .

∂ ri

∂ qK

where q1

r2

= ( x + ± 0) E1 +

yE 2 + z2E3 + rer .

=

x, q2

y, q4

= θ,

q3

=

=

z1, q5

=

r, and

(b) What are the four constraints on the motion of the system of particles? Give prescriptions for the constraint forces Fc1 and Fc 2 acting on the respective particles. For full credit, provide a clear expression for the friction force acting on m1 and give physical interpretations for your prescriptions of F c1 and F c2 . (c) Show that the kinetic energy of the system of particles has the representation T

=

² ² m1 + m2 ± 2 m1 2 m2 ± 2 x˙ + y˙ 2 + z˙1 + r˙ + r2 θ˙ 2 + z˙22 2( 2 2 ) + m2 ˙rx ˙ cos (θ ) + ˙ry ˙ sin (θ ) − rθ˙ x ˙ sin (θ ) + rθ˙y ˙ cos (θ ) .

198

Dynamics of Systems of Particles

Linear spring K ,

0

O

E2

Rough vertical track

E1

m1

g

Massless rod of length θ

m2

A system of two particles. The particle of mass m1 is free to move on a rough vertical guide while the particle of mass m2 is suspended by a rod of length ± underneath. The mass of the rod is negligible and the motion of the system is planar.

Figure 5.13

(d) Compute the following four summations: Fncon 1 ·

∂ r2 ∂ r1 + Fncon 2 · k ∂q ∂ qk

(k =

1, 2, 3, 4 ) ,

where F nconα is the nonconservative force acting on the particle of mass mα, where α = 1, 2. ( ) (e) Compute the constrained Lagrangian L˜ = L˜ x, θ , x˙ , θ˙ for the system of particles. (f) If friction is ignored, show that the equations of motion of the system can be expressed in the form ¿

( m1 + m2 )

−m2 ± sin(θ )

−m2± sin(θ )

m2±2

À¿

x¨ θ¨

À

¿

=

À

m2±θ˙ 2 cos (θ ) − Kx + (m1 + m2 ) g . −m2 g± sin (θ )

(g) Suppose friction cannot be ignored. Establish the equations governing the motion of the system of particles. (h) What are the generalized coordinates, configuration manifold , and kinematical line element for the system of particles? (i) Show that the time rate of change of the total energy for the system of particles only depends on the product of the coefficient of friction µd , the normal force on m1, and the speed |˙x| of m1. ˜ show that (j) Using the mass matrix from the equations of motion (or, equivalently, T), 1 the coordinate system for = E × S is free of singularities.

M

M

Exercise 5.9: This problem is adapted from Whittaker [306, Section 156] and the introduction to [40]. Consider a system of N particles and, following Jacobi [138, Lecture 4], define the following function:

5.7

199

Exercises

N

J=

1¼ mk ±rk ±2 . 2 k=1

The quantity 2J is often known as the moment of inertia of the system of particles, and can also be considered as a measure of the distance squared from the origin of the configuration space E3N for the representative particle discussed in Section 4.7. (a) Assuming that the center of mass C of the system is stationary and located at the origin, show that J has the equivalent representation J

=

N N 1 ¼ ¼ mk mj » »rk 4 M

−

»2

rj » ,

(5.20)

k=1 j=1

where M = m1 + · · · + mN . (b) As in (a), assuming that the center of mass C of the system is stationary, show that the kinetic energy of the system of particles has the representation T

=

N N 1 ¼ ¼ mk mj » »vk 4 M

−

»2

vj » .

(5.21)

k=1 j=1

(c) Now suppose that the system of particles is in motion subject to a conservative Newtonian force field Un = −

N N 1¼ ¼ 2

j=1 k =1, k²= j

Gmk mj ». − rj »

» »rk

(5.22)

The presence of 21 in the expression for Un should be noted: it is needed to ensure that the summations on k and j yield the correct expression for Un. With the help of (5.20)–(5.22) and a balance of linear momentum for each particle, establish Jacobi’s equation J¨ = 2T + Un.

(5.23)

This equation is also known as the Lagrange–Jacobi equation (see, e.g., [296]).

Part III

A Single Rigid Body

6

Rotations and their Representations

6.1

Introduction

One of the key features of the rigid-body dynamics problems that we will examine shortly is the presence of a variable axis of rotation. This is one of the reasons for the richness of phenomena in rigid-body dynamics. It is also a reason why the subject of rotations can be intimidating. To quote the mechanician Louis Poinsot (1777–1859): “. . . if we have to consider the motion of a body of sensible shape, it must be allowed that the idea which we form of it is very obscure” [229]. In this chapter, several representations of rotations are discussed that will enable us to establish both a clear picture of rigid-body motions and straightforward proofs of several major results. To this end, numerous results on two key kinematical quantities for rigid bodies – rotation tensors and their associated angular velocity vectors – are discussed in considerable detail. The subject of rotations in rigid-body dynamics has a wonderful history, a wide range of interesting results, and an impressive list of contributors. Here, however, space limits the presentation to only the handful of results that are most relevant to our purposes. From a historical perspective, much of what is presented was established by Leonhard Euler (1707–1783) in his great works on rigid-body dynamics that started to appear in the 1750s. The foundations Euler established were built upon by such notables as Cayley, Gauss, Fick, Gibbs, Hamilton, Helmholtz, and Rodrigues in the nineteenth century. Despite the wealth of their results, the subject of rotations remains an active area of discovery (and some unfortunate reinvention) to this day. Our exposition makes extensive use of tensors, and the background required for these quantities is provided in the Appendix. Following related developments in continuum mechanics, tensors are an invaluable tool for providing concise explanations of many important results. They are not universally used in dynamics texts, and the interested reader is invited to compare our treatment with those in the textbooks cited in the References. We start this chapter with a discussion of a rotation in which the axis of rotation is fixed, and this example is used to introduce several of the key concepts in this chapter. The example also serves to illuminate many of the generalizations that are required for characterizing more elaborate rotations. To this end, our exposition then turns to proper orthogonal tensors in Section 6.3, and many of their interesting features are shown there.

204

Rotations and their Representations

Then, Euler’s representation of a rotation tensor is presented and we examine his important theorem that any proper orthogonal tensor is a rotation tensor. This result has many consequences for rigid-body dynamics. One of the intriguing aspects of rotation tensors is the vast range of representations. In this chapter, the Euler angle representation is emphasized, but sufficient material is present for the interested reader to examine many of the other representations (such as the Euler–Rodrigues symmetric parameters, often synonymous with quaternions). Our principal reference for this chapter is the authoritative review by Shuster [263]. This material is supplemented with the elegant relative angular velocity vector proposed by Casey and Lam [39], the dual Euler basis [214], and a proof of Euler’s theorem by Guo [110]. In the interests of presenting a unified treatment, our notation differs from these references in several places, but the differences are easily deciphered once the material in this chapter has been mastered.

6.2

The Simplest Rotation

To motivate many of our later developments, we start with the simplest case of a rotation about a fixed axis p 3 through an angle θ = θ (t). This example should be familiar to you from many different venues. To describe this rotation, we consider the action of this ² ± rotation on the set of orthonormal right-handed basis vectors p 1, p2 , p 3 . As shown in Figure 6.1, we suppose that these vectors are transformed to the set {t1 , t2 , t3 } by the rotation. Using a matrix notation, we can represent the transformation from the set of basis ² ± vectors p1 , p 2, p3 to {t1, t2, t3} as ⎡

⎤

⎡

⎤

p1 t1 ⎣ t2 ⎦ = B ⎣ p ⎦ , 2 p3 t3

(6.1)

p2 t1 t2 θ

θ

p1 θ

p3 = t3 Figure 6.1

The transformations of p1 and p2 produced by a rotation about p3 through an angle θ .

6.2

where

⎡

cos(θ ) sin(θ ) ⎣ B= − sin(θ ) cos(θ ) 0 0

205

The Simplest Rotation

⎤

0 0 ⎦. 1

It is easy to see that the matrix B in this equation has a determinant of +1 and that its inverse is its transpose: B−1 = BT . That is, the matrix B is proper orthogonal. By differentiating (6.1) with respect to time, we find that ⎡

⎡

⎤

cos(θ ) ⎣ ˙t2 ⎦ = θ˙ ⎣ − cos(θ ) − sin(θ ) ˙t3 0 0 Using the result B−1 ⎡

˙t1

⎤

⎡

˙t1

=

− sin(θ )

⎤

BT , we can easily replace p i in this equation with ti : ⎤⎡ ⎤⎡

cos(θ ) ⎣ ˙t2 ⎦ = θ˙ ⎣ − cos(θ ) − sin(θ ) ˙t3 0 0

cos( θ ) 0 0 ⎦ ⎣ sin(θ ) 0 0

− sin(θ )

⎡

0 1 = θ˙ ⎣ −1 0 0 0 ³

⎤⎡

0 p1 0 ⎦ ⎣ p2 ⎦ . 0 p3

´µ

⎤⎡

cos(θ ) 0

⎤

⎤

t1 0 0 ⎦ ⎣ t2 ⎦ 1 t3

− sin(θ )

0 t1 0 ⎦ ⎣ t2 ⎦ . 0 t3

(6.2)

¶

˙ T BB

Notice that the above equation is equivalent to the familiar results ˙t1 = θ˙ t2 and ˙t2 = ˙ T is a skew-symmetric matrix. A vector −θ˙ t1 . It should also be clear from (6.2) that BB θ˙p 3 = θ˙ t3 can be introduced with the useful property ˙tk = θ˙p × tk 3

(k =

1, 2, 3 ) .

(6.3)

˙ T. You should notice how the vector θ˙ p3 can be inferred from the components of BB It is convenient to use a tensor notation to describe the rotation we have been discussing. In particular, we can write (6.1) in the form1

tk

=

Rpk

where R is the tensor R

=

(

cos(θ ) p1 ⊗ p1 + p2 ⊗ p2

)

(k =

−

1, 2, 3) ,

(6.4)

(

sin(θ ) p 1 ⊗ p 2 − p2 ⊗ p1

)

+

p 3 ⊗ p 3.

It is left as an exercise to verify that this representation of the rotation is equivalent to (6.1). Indeed, because2 I − p 3 ⊗ p 3 = p 1 ⊗ p 1 + p 2 ⊗ p2 ,

εp 3

= −skwt

(

p3

)

=

p1 ⊗ p2 − p 2 ⊗ p 1,

we can express the tensor R entirely in terms of the axis of rotation p 3 and the angle of rotation θ : (

R = cos( θ ) I − p 3 ⊗ p 3

)

+

(

sin(θ ) skwt p3

)

+

p3 ⊗ p3 .

1 Background on tensor notation can be found in the Appendix. 2 The definition of the alternating tensor ε is given in Section A.7 (see, in particular, (A.11)).

(6.5)

206

Rotations and their Representations

Here, skwt (a) defines a second-order skew-symmetric tensor such that skwt (a) b = a × b for any vector b. As we shall see later, the representation (6.5) naturally leads to the general form of a tensor that represents a rotation about an arbitrary axis through an arbitrary angle of rotation. It is left as another exercise to show that RT = R−1 and that det (R) = 1. In order to unambiguously define a rotation using a matrix, it is also necessary ± ² to define one of the bases p1 , p2 , p 3 or {t1, t2, t 3}. By contrast, when a rotation is described using a tensor (such as (6.5)), both the matrix components and the bases are automatically defined. This is one of the chief advantages of using tensors to define rotations and, as we shall see later, inertia tensors. Differentiating (6.4), we find that 0 ±±✒

˙t

˙ ˙k k = Rpk + Rp T ˙ Rt = R k

³ ´µ¶

=

(

=p k

)

˙ T t . RR k

Some straightforward computations are used to show that (

˙ = −θ˙ sin(θ ) p ⊗ p + p ⊗ p R 1 1 2 2

˙ T RR

(

= θ˙ −p1 ⊗

)

)

− θ˙ cos( θ )

(

)

p1 ⊗ p2 − p 2 ⊗ p 1 ,

p2 + p 2 ⊗ p 1 .

With the help of these results, we conclude that ˙tk =

(

)

˙ T t RR k

= θ˙p 3 × tk

(k =

1, 2, 3 ) .

As expected, this result is in agreement with (6.3). You might already have noticed that ˙ T. θ˙p 3 = θ˙ t3 is the axial vector of RR We have demonstrated how several results for a familiar rotation can be expressed using a tensor notation. This notation will prove to be very useful when we examine more complex rotations. To this end, we need to answer several questions. The first among them is this: what is the representation for a rotation about an arbitrary axis? Once this is known, the question of whether its time derivative leads to a vector such as θ˙ p3 then presents itself. The answers to these questions were first formulated by Euler in the 1750s. However, numerous alternative representations for his solutions have appeared since then, and we shall exploit several of these representations in the remainder of this chapter.

6.3

Proper Orthogonal Tensors

To proceed with our treatment of rotations, we will appear to regress somewhat by discussing proper orthogonal tensors. This discussion will lay the foundations for the possibility of three-parameter representations of rotations and subsequently the

6.3

Proper Orthogonal Tensors

207

existence of angular velocity vectors. It is also a starting point for several investigations on experimental measurements of rotations. We recall that a proper orthogonal second-order tensor R is a tensor that has a unit determinant and whose inverse is its transpose: RT R

=

RRT

=

I,

det(R)

=

1.

(6.6)

The first of these equations implies that there are six restrictions on the nine components of R. Consequently, only three components of R are independent. In other words, any proper orthogonal tensor can be parameterized using three independent parameters. The tensors of interest here are second-order tensors, any one of which has the representation R=

3 3 · · i=1 k=1

R ik pi

⊗ pk .

Let us now consider the transformation induced by R on the basis vectors p1 , p 2, and p3 . We define t1 t2 t3

=

=

=

3 ·

Rp1

=

Rp2

=

Rp3

=

i=1 3 · i=1 3 · i=1

Ri1p i , Ri2p i , Ri3p i .

You should notice that by using the vectors ti , R has the representation R = t1 ⊗ p 1 + t2 ⊗ p 2 + t3 ⊗ p3 . We now wish to show that {t1 , t2 , t3 } is a right-handed orthonormal basis. First, let us verify the orthonormality:3 ti · tk

=

Rp i · Rpk

=

p i · pk

=

RT Rpi · p k

= δik .

Hence the vectors ti are orthonormal. To establish right-handedness, we use the definition of the determinant (see (A.6)): [t1 , t2 , t3 ] = [Rp 1, Rp 2, Rp3 ] =

det(R)[p1 , p 2, p 3]

=

(1) (1) = 1.

Hence, {t1 , t2 , t3 } is a right-handed orthonormal basis. 4 3 Observe that we are using the identity a · ( Ab ) = AT a · ( b), which holds for any second-order tensor A

and any pair of vectors a and b.

4 It is left as an exercise to verify that this result holds for the example of a simple rotation presented in (6.1).

208

Rotations and their Representations

A proper orthogonal tensor also has a rather unusual representation. To arrive at it, we embark on a series of manipulations:5 ¸

R = RRRT

=

R

3 · 3 · i=1 k=1

3 · 3 · =

(

R ik R pi

⊗ pk

¹

Rik pi ⊗ pk RT ) T

R

i=1 k=1 =

3 3 · · i=1 k=1 3 3 · ·

=

(

R ik Rpi R ik ti

⊗ Rp k

)

⊗ tk .

i=1 k=1

In summary, we have the following representations for R: R

=

3 3 · · i =1 k =1

Rik p i ⊗ pk

=

3 3 · ·

Rik ti ⊗ t k

=

3 ·

i =1 k=1

i=1

ti ⊗ pi .

Notice that the components of R for the first two representations are identical, and the handedness of {p 1, p2 , p3 } is transferred without change by R to {t1 , t2 , t3 }. We observed earlier that the components Rik of R are equal to tk · p i : ⎡

R11 R = ⎣ R21 R31

R 12 R 22 R 32

⎤

That is,

⎡

ti

=

⎡

t1 · p 1 R 13 R 23 ⎦ = ⎣ t1 · p 2 t1 · p 3 R 33 ⎤

t1 ⎣ t2 ⎦ t3

Rp i ,

t2 · p 1 t2 · p 2 t2 · p 3 ⎡

⎤

t3 · p1 t3 · p2 ⎦ . t3 · p3

(6.7)

⎤

p1 T = R ⎣ p2 ⎦ . p3 ±

²

If we know the components of ti in terms of the basis p1 , p2 , p 3 , then the matrix R and the tensor R can be determined. As the product tk · p i is equal to the cosine of the angle between tk and p i , each component Rik is often referred to as a direction cosine. Consequently, the matrix R = [Rik ] is often known as the direction cosine matrix. Clearly, the nine angles whose cosines are tk · p i are not all independent, for if they were then [Rik ] would have nine independent components and this would contradict the requirement RT R = I. Indeed, as we shall see shortly, it is possible to arrive at three independent angles to parameterize R, but these angles are not all easily related to the angles between p i and tk . 5 Observe that we are using the identity bA

vector b.

=

AT b, which holds for any second-order tensor A and any

6.4

6.4

209

Derivatives of a Proper Orthogonal Tensor

Derivatives of a Proper Orthogonal Tensor

Here we consider a proper orthogonal tensor R that is a function of time: R Consider the derivative of RRT : d ( T) ˙ T RR = RR dt However, RRT Hence,

=

I and I˙

=

+

=

R(t).

T

˙ . RR

0, so the right-hand side of the preceding equation is zero. ˙ T RR

˙T = − = − RR

(

˙ T RR

)T

.

˙ T is a skew-symmetric second-order tensor. We define In other words, RR

ΩR

˙ T, = RR

in part because this tensor appears in numerous places later on. The tensor ΩR is known as the angular velocity tensor (of R). ˙ T allows us to define the angular velocity vector ω : The skew-symmetry of RR R ωR

=

(

˙ T ax RR

)

= −

1 º T» ˙ ε RR . 2

˙ T )a for all vectors a.6 The most common As mentioned in the Appendix, ωR × a = (RR example of the calculation of an axial vector arises when we consider the motion of a rigid body rotating about the E3 direction. In this case, we will see later that the skew-symmetric tensor

ΩR

= ± (E2 ⊗

E1 − E1 ⊗ E2) .

Consequently, we can compute that ax (ΩR ) =

±E 3.

It is useful to verify that (± (E2 ⊗

E1 − E1 ⊗ E2)) a = ±E3 × a

for all vectors a. ˙ is a skew-symmetric tensor and In a similar manner, we can also show that RT R define an angular velocity tensor Ω0R and another angular velocity vector ω0R : Ω0R

=

˙ RT R,

ω0 R

=

(

)

˙ . ax RT R

(6.8)

At a later stage, you should be able to show that Rω0R = ω and RΩ0R RT = ΩR . A key to establishing one of these results is the identity, which holds for all orthogonal Q, (

ax QBQT

)

=

det (Q) Q ax (B) .

Notice how this identity simplifies when Q is a proper orthogonal tensor. 6 Several examples showing the calculation of the axial vector ω of a given skew-symmetric tensor Ω are R R

presented in Section A.7 of the Appendix.

210

Rotations and their Representations

Corotational Derivatives

We recall the following representation for a proper orthogonal tensor R: R=

3 ·

ti

⊗ pi .

i=1

Now suppose that p˙ i ΩR

˙ = RR

T

=

=

0. Then,

¸ 3 ·

˙ti ⊗

i=1 =

3 ·

pi

+

3 · i=1

¹

0

✼i ti ⊗ ✓ p˙✓

R

T

=

¸ 3 ·

¹ ˙ti ⊗

i=1

pi RT

˙ti ⊗ ti .

i=1

If we now consider ΩR tk , we find a familiar result: ˙ti =

Ω R ti

=

ωR × ti .

It is left as an exercise to show the less familiar result ˙ti =

R(ω0R × pi ).

It is important to note that if ti are defined by use of a proper orthogonal tensor R and a fixed basis p i , then their time derivatives can be expressed in terms of the angular velocity vector of R and the basis vectors ti . Given any second-order tensor A and any vector a, we have the following representations: a=

3 ·

ai ti ,

A=

3 3 · ·

Aik ti

⊗ tk .

i=1 k=1

i =1

If we assume that a is a function of time, then a˙ =

3 · i=1 3 ·

=

a˙ i ti + ai ˙ti a˙ i ti + ai (ωR

× ti )

i=1 =

3 ·

a˙ i ti + ωR × a.

i=1

Similarly, if we assume that A is a function of time, then ˙ = A

3 · 3 ·

A˙ ik ti

⊗ tk +

i =1 k =1 3 · 3 · =

=

i =1 k =1 3 · 3 · i =1 k =1

3 · 3 ·

A ik ˙ti ⊗ tk +

i=1 k=1

A˙

ik ti ⊗ tk +

3 · 3 ·

⊗ tk +

Aik ti

⊗ ˙tk

i =1 k =1

A ik (ΩR ti ) ⊗ tk

i=1 k=1

A˙ ik ti

3 · 3 ·

ΩR A − AΩR .

3 · 3 · +

i=1 k=1

A ik ti ⊗ (ΩR tk )

6.5

◦

Euler’s Representation of a Rotation Tensor

211

◦

The derivatives A and a are known as the corotational derivatives (with respect to R) of A and a, respectively. They are the respective derivatives of A and a if the vectors t i were held constant: ◦

A=

3 3 · ·

A˙ ik ti ⊗ tk ,

◦

a=

3 ·

a˙ i ti .

i=1

i=1 k=1 ◦

Using the earlier expression for a, we observe that ◦

a˙ = a + ωR × a. Similarly, for any second-order tensor A, we have ◦

˙ = A + Ω A − AΩ . A R R

The terms involving the angular velocity vectors and tensors in these expressions are the result of the orthonormal vectors ti changing with time. We shall subsequently use several distinct corotational derivatives, and we wish to do so without introducing a laborious notation. To this end, it will be stated explicitly which proper orthogonal tensor the corotational derivative pertains to in situations where any confusion is possible.

6.5

Euler’s Representation of a Rotation Tensor

Leonhard Euler defined a rotation by using an angle of rotation φ and an axis of rotation r.7 Using notation introduced over a century later by Gibbs,8 Euler’s representation for a tensor that produces this rotation can be written as R

=

L(φ , r) = cos(φ )(I − r ⊗ r) + sin(φ) skwt (r) + r ⊗ r,

(6.9)

where r is a unit vector and φ is a counterclockwise angle of rotation. We refer to (6.9) as Euler’s representation of a rotation tensor and use the function L to prescribe the rotation tensor associated with an axis and an angle of rotation. The three independent parameters of R are the angle of rotation and the two independent components of the unit vector r. The reason r is known as the axis of rotation lies in the fact that it is invariant under the action of R: Rr = r. We shall examine shortly the role of φ. It is interesting to examine some of the features of representation (6.9). To this end, we define an orthonormal basis {p1 , p 2, p 3} with p3 = r. Using this basis, we have I = p 1 ⊗ p 1 + p2 ⊗ p2 + p 3 ⊗ p 3, r ⊗ r = p 3 ⊗ p 3, I − r ⊗ r = p 1 ⊗ p 1 + p2 ⊗ p2 , 7 This representation can be seen in Section 49 of one of Euler’s great papers on rigid-body dynamics from

1775 [72]. There, he provides expressions for the components of the tensor R in terms of an angle of rotation φ and the direction cosines p, q, r of the axis of rotation. In our notation, r = pp1 + qp2 + rp3 . 8 See [309, Section 129].

212

Rotations and their Representations

skwt (r) = p 2 ⊗ p 1 − p1 ⊗ p2 . Consequently, we can write R = L(φ , r = p 3)

cos(φ )(p1 ⊗ p1 + p 2 ⊗ p 2)

=

+ sin(φ )(p2 ⊗

Using the identity (a ⊗ b)T RT

=

L

(

φ, r =

=

p3

p1 − p 1 ⊗ p 2) + p3 ⊗ p3 .

b ⊗ a, we find

)

=

cos( φ)(p 1 ⊗ p 1 + p2 ⊗ p2 ) +

sin(φ )(p 1 ⊗ p 2 − p2 ⊗ p1 ) + p 3 ⊗ p 3.

It is now easy to check that RRT is the identity tensor, and the details are left to the reader. Next, we examine the determinant of R: ⎡

cos( φ) ⎣ det(R) = det sin(φ) 0 =

− sin(φ)

cos(φ ) 0

⎤

0 0 ⎦ 1

1.

In conclusion, RT R = I and det(R) = 1. Thus, R is a proper orthogonal tensor. We shall examine the converse of this statement shortly. Composition of Rotations

Suppose we have a rotation through an angle φ1 about an axis r1 and we follow this by a rotation through an angle φ2 about an axis r2; then the tensor Rc

=

L (φ2 , r2 ) L (φ 1, r1)

represents the composite rotation. It is left as an exercise to show that Rc is a proper orthogonal tensor. As can be shown later by Euler’s theorem, this implies that Rc is a rotation tensor. Thus, the composition of two rotations is also a rotation. As tensor multiplication is not commutative, the order in which we consider the composition is important. In general, L (φ 2, r2) L (φ1 , r1 ) ± = L (φ1 , r1) L (φ 2, r2 ) . The exceptions arise when r1 ² r2 or one of the angles of rotation is 0. The fact that rotations are sequence dependent will play a major role later on in the definition of Euler angle sequences. Euler’s Formula

We now examine the action of R on a vector a. As shown in Figure 6.2, the part of a that is parallel to r is unaltered by the transformation, whereas the part of a that is perpendicular to r is rotated through an angle φ counterclockwise about r. To see this, it is convenient to decompose a: a = a⊥

+

a² ,

6.5

Euler’s Representation of a Rotation Tensor

213

Ra⊥ Ra = Ra⊥ + a

φ

a a⊥

a

r

Figure 6.2

The transformation of a vector a by the rotation tensor R = L (φ , r).

where a⊥

=

a − (a · r)r = (I − r ⊗ r)a,

a²

=

(a · r)r = (r ⊗ r)a.

It is helpful to observe that r × a² = 0,

r × a = r × a⊥ .

With the assistance of the decomposition a = a⊥ + a², we now compute that Ra = L(φ, r)a = ( cos(φ )(I −

r ⊗ r) + sin(φ) skwt (r) + r ⊗ r) a

=

cos( φ)(I − r ⊗ r)a + sin(φ) skwt (r) a + (r ⊗ r)a

=

cos( φ)(I − r ⊗ r)a + sin(φ)r × a + (r · a)r

=

cos( φ)a⊥ + sin(φ )r × a⊥ + a².

(6.10)

Noting that cos( φ)a⊥ + sin(φ )r × a⊥ is a rotation of a⊥ about r, we obtain the desired conclusion. The final expression for Ra in (6.10) is known as Euler’s formula.

Remarks on Euler’s Representation

Euler’s representation (6.9) is unusual in several respects. First, you should notice that L(φ , r) = L(−φ , −r). This implies that there are two equivalent representations for the same rotation tensor. Second, as R is a rotation tensor, R−1 = RT , and this leads to two useful representations for the inverse of R:

214

Rotations and their Representations

R−1

=

RT

=

L(−φ , r) = L(φ , −r).

In words, the inverse can be calculated by either reversing the angle of rotation or inverting the axis of rotation. Another peculiarity is that L(φ = 0, r) = I holds for all vectors r. Finally, we note that Euler’s representation is used to define other representations of rotation tensors in this book.

Calculating the Axis and Angle of Rotation

Given a rotation tensor R, it is a standard exercise to calculate the axis of rotation r and the angle of rotation φ associated with this tensor. First, one is normally presented with the matrix components of R with respect to a basis, say {p 1, p2 , p 3}: R=

3 3 · · i=1 k=1

R ik pi

⊗ pk .

If we compare this representation with (6.9), we find that Rik

=

=

(

(cos(φ )(I −

cos(φ ) (δik

r ⊗ r) + sin(φ) skwt (r) + r ⊗ r) p k

− ri rk ) + ri rk −

3 ·

)

·

pi

sin(φ )²jikrj ,

j =1

where r=

3 · i=1

ri p i .

Expanding the expressions for the components of R, we find the matrix representation (cf. (6.7)) ⎡

R 11 ⎣ R 21 R 31

R 12 R 22 R 32

⎤

R13 R23 ⎦ R33

⎡

1 − r21 = cos (φ) ⎣ −r1r2 − r1 r3 ⎡

r21 + ⎣ r1 r2 r1 r3

⎡

r1 r2 r22 r2 r3

0 ⎣ + sin (φ) r3 −r2

− r1 r2

1 − r22 − r2 r3

⎤

r1 r3 r2 r3 ⎦ r23 − r3

0 r1

− r1 r3

− r2 r3

⎤ ⎦

1 − r32

⎤

r2 −r1 ⎦ . 0

(6.11)

Notice that this matrix can be expressed as the sum of a symmetric matrix and a skewsymmetric matrix. The skew-symmetric part is the only part that changes when we transpose the tensor. Suppose± that the matrix R on the left-hand side of (6.11) is given and either one of ² the bases p1 , p2 , p 3 or {t1, t2, t3} is prescribed. Our goal is to determine φ and rk associated with the rotation. To find the angle of rotation, we calculate the trace of R:

6.5

cos(φ ) =

215

Euler’s Representation of a Rotation Tensor

1 1 (tr (R) − 1) = (R11 + R22 + R 33 − 1 ) . 2 2

(6.12)

If φ = 0, then R = I and r can be any unit vector. If φ = π , then R = RT = L (π , r) = L (π , −r). Assuming that r1 ± = 0 (i.e., R 11 ± = − 1), we compute the components of r from (6.11): ⎡

⎤

⎡

¼

⎤

1+R11 r1 2 ⎥ ⎢ ⎥ ⎣ r2 ⎦ = ⎢ √ R12 ⎣ 2(1+R11 ) ⎦ . R13 r3 √ 2(1+R11 )

If r1 = 0, then it is straightforward to consider a row of Rik other than the first row in order to solve for r2 and r3. If φ ± = 0 or π then, after examining the skew-symmetric part of R, we find that ⎡

⎤

⎡

⎤

r1 R23 − R32 1 ⎣ r2 ⎦ = − ⎣ R31 − R13 ⎦ . 2 sin(φ ) r3 R12 − R21

(6.13)

An alternative method to compute the axis r and angle φ is to compute the eigenvector, which we denote by r1 , corresponding to the unit eigenvalue of R and, as before, the angle φ using (6.12). This yields four possible pairings: (r1 , φ), (r1 , −φ), (−r1, φ), and (−r1 , −φ). Substituting each of these pairings into the representation (6.11) enables us to eliminate two of the four pairings. The eliminated pairings correspond to the axis and angle for the rotation RT . ∑ It is interesting to notice that, if R = 3i=1 ti ⊗ pi , then r has the same components with respect to both sets of basis vectors: r=

3 · i =1

ri pi

=

3 ·

ri ti .

(6.14)

i=1

The proof of this result, which we leave as an exercise, is based on the observation that R has the same components with respect to the bases pi ⊗ p k and ti ⊗ tk . That is, R = ∑3 ∑3 ∑3 ∑3 i=1 k= 1 Rik p i ⊗ pk = i=1 k =1 Rik ti ⊗ tk . The identity (6.14) finds application in ophthalmology where rigid-body models for the human eye are employed (see Figure 6.3). In this field of application, the axis of rotation is subject to Listing’s law [122, p. 12].9 As a consequence of (6.14), this constraint has two equivalent representations: r · p1

=

0

or

r · t1

=

0.

(6.15)

Studies of the dynamics of a rigid-body model for the eye subject to this constraint can be found in [94, 95, 207] and references therein. 9 Johann Benedict Listing (1808–1882) was a student of Gauss. He is credited with inventing the term

“topology” in 1847.

216

Rotations and their Representations

(a)

(b)

p3

p3

t1 = Rp1

= 0

t3 = Rp3 ϕ

p2

O

O

L

F

p1

p1

p2

θ

r

t2 = Rp2

Reference configuration (a) and present configuration (b) of a rigid body modeling the eye. The gaze direction is assumed to be parallel to t 1 and the rotational motion of the eye is ∑ characterized by a rotation tensor R = 3i=1 t i ⊗ pi . Following Listing’s law, the axis of rotation r lies on the intersection of the p2 –p3 (Listing’s) plane L and the t 2–t 3 (focal) plane. As a result, r = r 2p2 + r 3p3 = r2t 2 + r 3t 3 and the angle ³ = 0. In ophthalmology, the reference configuration is known as the primary position and the present configuration is known as a secondary position. Additional details on the angles ϕ and θ in the figure are discussed in Section 7.11 (see (7.33)). This figure is adapted from [207].

Figure 6.3

An Example

As an example, suppose that the components of a rotation tensor R are10 ⎡

R11 ⎣ R21 R31

R12 R22 R32

⎤

⎡

R13 0.835959 R23 ⎦ = ⎣ 0.271321 0.47703 R33

−0.283542

−0.469869

0.957764 −0.0478627

−0.0952472

⎤ ⎦.

0.877583

We can compute the angle of rotation φ of this tensor by using (6.12): cos(φ ) =

1 (0.835959 + 0.877583 + 0.957764 − 1) . 2

That is, φ = 33.3161 ◦. We can calculate the axis of rotation r by using (6.13): r = 0.043135p1 − 0.861981p 2 + 0.505103p 3 =

0.043135t1 − 0.861981t2 + 0.505103t3 .

In writing this result, we are emphasizing that the components of r in the basis ² ± p 1, p2 , p3 and {t 1, t2 , t3 } are identical (cf. (6.14)). The Associated Angular Velocity Vector

Given Euler’s representation (6.9), we next assume that R = R(t). This implies, in general, that φ = φ (t) and r = r(t). We now seek to establish representations for ωR . 10 As we shall see later when we discuss Euler’s theorem in Section 6.6, it is prudent to check that R is a rotation tensor by verifying that the matrix R is a rotation matrix: RT R = I and det ( R) = 1.

6.6

Euler’s Theorem: Rotation Tensors and Proper Orthogonal Tensors

217

As a preliminary result, we note that, because r is a unit vector, r · r˙ = 0. In addition, ε˙ = 0 and I˙ = 0. Now, starting from (6.9),11 R = L(φ , r) = cos( φ)(I − r ⊗ r) − sin(φ)( εr) + r ⊗ r, we differentiate to find ˙ = −φ ˙ sin(φ )(I − r ⊗ r) − φ ˙ cos(φ ) ( ε r) + (1 − cos(φ )) (r R ˙ ⊗ r+ r⊗ r ˙ ) + sin(φ) skwt ( r ˙) .

To proceed further, we define a right-handed orthonormal basis {t1, t2, t3} such that t3 r at a given instant in time. Then, at the same instant in time: − skwt ( r) =

=

εr = (t1 ⊗ t2 − t2 ⊗ t1 ), ˙r =

at1 + bt2 ,

r × r˙ = at2 − bt1 , −skwt ( r ˙) =

ε˙r = a(t2 ⊗ t3 − t3 ⊗ t2) + b(t3 ⊗ t1 − t 1 ⊗ t3 ).

(6.16)

In these expressions, a and b are the scalar components of r˙. Using (6.16) along with some manipulations, we find that ˙ T RR

˙ (t1 ⊗ t2 − t2 ⊗ t1 ) + = −φ +

(a(1 − cos( φ)) + b sin(φ ))(t1 ⊗ t3 − t3 ⊗ t1 )

(−b(1 − cos(φ)) + a sin(φ )) (t3 ⊗ t2 − t2 ⊗ t3)

˙ εt3 − = −φ

(a(1 − cos(φ )) + b sin(φ ))εt2 − (−b(1 − cos(φ )) + a sin(φ ))εt1.

With the assistance of (6.16), we can now write the desired final result: ΩR

˙ T = −φ ˙ εr − = RR ˙ = φ

(1 − cos( φ))ε(r × r) ˙ − sin(φ)ε r ˙

skwt (r) + (1 − cos(φ )) skwt (r × r˙ ) + sin(φ) skwt (r˙ ) .

The associated angular velocity vector is ωR

˙r + = φ

sin(φ)r˙ + (1 − cos(φ ))r × r. ˙

(6.17)

If r is constant, then the expression for the angular velocity vector simplifies considerably. It is interesting to note that a constant ωR does not necessarily imply a constant r. However, as discussed in [207, 216], apart from rigid-body models for the eye, it is ± ² ∑ usually possible to choose a fixed basis p 1, p2 , p 3 where R = 3k =1 tk ⊗ pk so that a constant angular velocity implies a constant φ˙ and a constant r.

6.6

Euler’s Theorem: Rotation Tensors and Proper Orthogonal Tensors

A tensor Q is proper orthogonal if, and only if, QT Q = I,

det (Q) = 1.

From our discussion of rotation tensors, it follows that a rotation tensor is a proper orthogonal tensor. However, is the converse true? In other words, is every proper 11 Recall that skwt (r )

= −εr.

218

Rotations and their Representations

orthogonal tensor a rotation tensor? The affirmative answer to this question is known as Euler’s theorem. Our approach to presenting the proof of Euler’s theorem is adapted from a wonderful paper by Guo [110]. One of the key results needed is to recall that we can define two of the invariants, IA = tr(A) and IIIA = det(A), of any second-order tensor A by using the scalar triple product [Aa, b, c] + [a, Ab, c] + [a, b, Ac] [Aa, Ab, Ac]

= IA [a, b, =

c],

IIIA [a, b, c],

where a, b, and c are any three vectors (see Section A.6). For a proper orthogonal tensor, the fact that det (Q) = 1 implies that this tensor has an eigenvalue λ = 1. There are several ways to see this, but first consider QT (Q − I) = I − QT . Taking the determinant of both sides, and using the fact that det (QT ) = 1, one finds that det (Q − I) = (−1)3 det (Q − I) . It follows that det (Q − I) = 0, and thus Q has an eigenvalue of 1. As Q has a unit eigenvalue, the tensor Q has an eigenvector u such that Qu

=

u.

When this equation is multiplied by QT , it follows that u = Qu = QT u. We conclude that u is a unit eigenvector of both Q and QT . Now consider a vector v ⊥ u. Some manipulations show that Qv · Qu

=

Qv · u,

T

Qv · Qu = Q Qv · u = v · u = 0. Consequently, v⊥u

if, and only if,

Qv ⊥ u.

We henceforth assume that u and v have unit magnitudes and define a vector w such that [u, v, w] = u · (v × w) = 1: w=

u×v ² u × v²

=

u × v.

Let us now calculate IQ : IQ

=

[Qu, v, w] + [u, Qv, w] + [u, v, Qw]

=

[u, v, w] + [u, Qv, w] + [u, v, Qw]

6.7

Relative Angular Velocity Vectors

=

1 + v · Qv + w · Qw

=

1 + v · Qv + (u × v) · (Qu × Qv)

=

1 + v · Qv + (u · u ) (v · Qv ) − (u · Qv) (v · u)

=

1 + 2v · Qv.

219

We define an angle ν such that cos(ν ) = v · Qv. It is important to notice that this angle is an invariant of Q and that Qv = cos(ν )v + sin(ν )w =

cos(ν )v + sin(ν ) (u × v) .

(6.18)

As mentioned previously, any vector a can be decomposed into two components: a² and a⊥ , where a⊥ · u = 0 and a = a² + a⊥. Now, Qa²

=

a ²,

and, from (6.18), Qa ⊥

=

cos(ν )a⊥ + sin(ν ) (u × a⊥ ) .

It follows that Qa = (a · u) u + cos(ν )(I − u ⊗ u)a + sin(ν )u × a. Comparing this equation with (6.10), we conclude that Q has the form (6.9) of a rotation tensor. Further, as the preceding equation holds for any proper orthogonal Q, we conclude that every proper orthogonal tensor is a rotation tensor and henceforth use these terms interchangeably. The result that every proper orthogonal tensor is a rotation tensor is credited to Euler and dates to 1775 [71]: Every proper orthogonal tensor is a rotation tensor. We shall revisit this result shortly and see why it is also known as E ULER ’S THEOREM ON THE MOTION OF A RIGID BODY . To verify that a tensor A is a rotation tensor, we invoke Euler’s theorem and simply show that A is proper orthogonal: AT A = I and det (A) = 1.

6.7

Relative Angular Velocity Vectors

Consider two rotation tensors R1 = R1(t) and R2 = R2 (t). It is straightforward to show that the product R = R2 R1 of these two tensors is also a rotation tensor: det (R) = 1 and RT R = I. We now wish to calculate the angular velocity tensor ΩR and vector ωR . To perform these calculations, we use the work of Casey and Lam [39], who defined a very useful and intuitive relative angular velocity vector.

220

Rotations and their Representations

ψ

t

XA

t

1 2

1 1

= 2 t2

θ

¯

X

p 3 = 1 t3

t

2 3

A pair(of rigid ) bodies connected by a pin joint at A. The rotation tensor of the( first ) body is R1 = L ψ , p3 and the rotation tensor of the second body is R = L (θ , 1t 2) L ψ , p3 .

Figure 6.4

To provide additional motivation, consider the pair of rigid bodies shown in Figure 6.4. Here, a shaft is rotating about a fixed vertical axis p3 = 1t3 with an angular speed ˙ . The shaft is attached to a rigid body by a pin joint at XA . The pin joint allows the ψ rotation of the second body relative to the first body through an angle θ and an axis 1 t2 = 2t2 . Thus, the rotation tensor of the first and second rigid bodies is, respectively: R1

=

L

(

ψ , p3 = 1 t3

)

,

R = L (θ , 1 t2) L

(

ψ , p3 = 1 t3

)

.

Using the relative angular velocity vector, we will easily be able to show that the angular velocity vectors of the respective bodies are ω1

˙p , = ψ 3

ω2 = θ˙ 2t2 + ω1 .

That is, the angular velocity vector of the second body relative to the first body is simply θ˙2 t2, even though the axis 2t2 of rotation is not fixed. To discuss the relative angular velocity vector, it is convenient to define three sets of right-handed orthonormal vectors: { 1t1 , 1 t2, 1t3 } ,

±

{ 2t1 , 2 t2, 2 t3 } ,

²

p1 , p 2, p3 .

In this section, we assume that p˙ i = 0. For a given R1 and R2 , two of these sets can be defined by use of the representations R1

3 · =

i=1

1 ti ⊗ pi ,

R2

3 · =

i =1

Notice that R=

3 · i=1

2 ti ⊗

pi .

In words, R transforms the vector pi into the vector 2ti .

2ti ⊗ 1 ti .

6.7

Relative Angular Velocity Vectors

221

Following Casey and Lam [39], let us consider the following relative angular velocity tensor: ˆ Ω R2

=

ΩR

−

ΩR 1 .

˙ = R ˙ R +R R ˙ Using the definition of the angular velocity tensors and the fact that R 2 1 2 1, we find that ˆ Ω R2

=

˙ T −R ˙ RT = RR 1 1

ΩR − ΩR1

˙ 2 R1 RT RT + = R 1 2 ˙ RT + = R 2 2 ˙ RT + = R 2 2

½

˙ 1 RT RT R2 R 1 2

R2Ω R1 RT2

˙ +R Ω R 2 2 R1

=

−

R2Ω R1 RT2 + +

˙ 1 RT −R 1

ΩR1 ΩTR1 ¾

ΩTR1 R2 RT2 .

However, ◦

R2

˙ + = R 2

R2 ΩR1

+

ΩTR 1 R2,

◦

where the corotational derivative R2 is defined to be ◦

R2

=

3 3 · · i=1 k=1

R˙ 2ik 1 ti ⊗ 1 tk ,

◦

where R2ik = ((R2 )1t k ) · 1 ti . In words, R2 is the derivative of the tensor R2 assuming ˆ that 1ti are constant. In conclusion, the relative angular velocity tensor Ω R 2 is ˆR Ω 2

ΩR − ΩR1

=

◦

=

R2RT2 .

If we denote the axis of rotation of R2 by r2 and its angle of rotation by φ2 , then we can parallel the derivation of (6.17) to find that the relative angular velocity vector has the representation ω ˆ R2

=

ωR

−

ωR1

˙ r + = φ 2 2

◦

In this equation,

¿

ω ˆ R2

and the corotational derivative of r2 ◦

r2

◦

sin (φ 2) r2 + (1 − cos (φ 2)) r2 × r2.

=

=

ax

∑3

◦

R2RT2

(6.19)

À

i=1 ri (1 ti )

,

(6.20)

is

= r ˙1 ( 1t1 ) + r˙2 ( 1t2 ) + r˙3 ( 1 t3) .

Formula (6.19) will prove to be exceedingly useful when calculating the angular velocity vector associated with various representations of a rotation tensor. In particular, for the Euler angle representation, we decompose R into the product of three rotation tensors: R = R3 R2R1 . We will then invoke (6.19) twice to get a representation for the angular velocity vector corresponding to R: ωR = ω ˆ R3 + ω ˆ R2 + ωR 1 . The manner in which we perform such calculations is similar to the example we are now about to present.

222

Rotations and their Representations

An Example

To illustrate the convenience of relative angular velocity result (6.19), let us consider an example. Suppose we have two rotations R1 and R, where R1

=

cos(ψ )(I − E3 ⊗ E3 ) + sin(ψ ) skwt (E3) + E3 ⊗ E 3,

R = R2 R1 . Here, the relative rotation tensor R2 is chosen to correspond to a rotation through an angle θ about t2 = cos( ψ )E2 − sin(ψ )E1 : R2

=

cos(θ )(I − t2 ⊗ t2) + sin(θ ) skwt (t2) + t2 ⊗ t2 .

The example can be motivated in several manners. First, it pertains directly to the pair of rigid bodies shown in Figure 6.4. Second, we met an example of such a compound rotation earlier when we defined the basis vectors eR , eθ , and eφ associated with the spherical polar coordinate system by rotating the vectors E3 , E 2, and E1 (cf. Figure 1.4 and Exercise A.8). We shall also see several examples of this compound rotation when discussing the 3–2–3 and 3–2–1 sets of Euler angles. The tensor R1 defines a transformation consisting of a rotation through an angle ψ about E3. This rotation transforms Ei to ti , where t1

=

cos(ψ )E1 + sin(ψ )E1 ,

t2

= −

sin(ψ )E1 + cos(ψ )E2,

t3

=

E3 .

Referring to Figure 6.5, R consists of the rotation R1 followed by a rotation through an angle θ about t2.12 We do not record explicit expressions for the components of R and instead refer the interested reader to Section 6.8.5.13 To calculate ωR1 we can appeal to (6.17) to find that ωR1

˙ E3 + = ψ

˙3 sin(ψ )E˙ 3 + (1 − cos(ψ ))E 3 × E

˙E . = ψ 3

(a)

t2

(b)

t3

E2 t1

ψ

ψ

E1

R 2t 3 θ

θ

t1 R2 t1

The rotations R1 and R2: (a) a rotation about E3 through an angle ψ ; (b) a rotation about t 2 through an angle θ .

Figure 6.5

12 In the notation of the previous section, p is replaced with E and t is replaced with t . i i i 1 i 13 Referring to (6.44), the components of R can be obtained by setting R = I (or, equivalently, φ 3

=

0).

6.7

223

Relative Angular Velocity Vectors

To calculate ωR we cannot appeal directly to (6.17) because we do not know the axis and angle of rotation of R.14 Instead, we use the relative angular velocity vector. To do this, we first need to write R2 with respect to an appropriate basis. As R1

= t1 ⊗

E1 + t2 ⊗ E2 + t3 ⊗ E3,

the appropriate basis is ti ⊗ tk : R2

=

cos(θ )(t 3 ⊗ t3 + t1 ⊗ t1) − sin(θ ) (t 3 ⊗ t1 − t1 ⊗ t3) + t2 ⊗ t2 .

Calculating the corotational rate of this tensor, we take the time derivative of R2, all the while keeping t i fixed: ◦

tk

=

0,

◦

R2 = −θ˙ sin(θ )(t3 ⊗ t3 + t1 ⊗ t1) − θ˙ cos(θ ) (t3 ⊗ t1 − t1 ⊗ t3 ) .

Hence, ˆ Ω R2

◦

=

R2 RT2

= θ˙ ( t1 ⊗ t3 − t 3 ⊗ t1 )

and, with the help of (6.20), we conclude that ω ˆ R2

= θ˙ t2 .

(6.21)

As an alternative method of calculating (6.21), we can use (6.19) directly. Thus we replace r2 and φ2 in (6.19) with t2 and θ , respectively. We then appeal to the fact that ◦ t 2 = 0. In summary: ω ˆ R2

= θ˙ t2 +

◦

◦

sin(θ ) t2 + (1 − cos(θ )) t2 × t2

= θ˙ t2.

This alternative method is clearly equivalent to, but more attractive than, the method that ◦

involved calculating R2. Combining the expressions for ωR1 and ω ˆ R 2 , we arrive at an expression for the angular velocity vector for R: ωR

˙E . = θ˙ t2 + ψ 3

The intuitive nature of this result is often surprising and always appealing. It is easy to use (6.17) to see that ωR2 ± = ωˆ R2 . In particular, ◦

t2 ˙t2 = −ψ ˙ t1,

ωR2

=

0,

ω ˆ R2

˙ = θ˙ t2 − ψ

= θ˙ t2,

sin (θ ) t1 + ψ˙ ( 1 − cos (θ )) t3.

The second of these results can be used quickly to verify that ωR

±=

ωR2

+

ωR1 .

14 The interested reader might wish to use the famed Rodrigues formula (6.58) that is discussed in the

exercises to compute these quantities and then one could use (6.17) to compute ωR .

224

Rotations and their Representations

p3 = g1 t3

= g3

t2

φ

t2 = g2 φ θ ψ

p2 ψ

p1

θ

t1

φ

t1

t1

Schematic of the 3–2–3 set of Euler angles and the individual rotations these angles represent. In this figure, the three Euler angles are denoted by ψ = γ 1, θ = γ 2, and φ = γ 3 , respectively, and the rotation tensor R that they parameterize transforms pi to t i . The image on the left-hand side is a portrait of Leonhard Euler.

Figure 6.6

6.8

Euler Angles

The most popular representation of a rotation tensor is based on the use of three Euler angles. As discussed in [50, 308], this representation dates to works by Euler [70, 73] that he first presented in 1751.15 In these papers, he shows how three angles can be used to parameterize a rotation and he also establishes expressions for the corotational components of the angular velocity vector. One interpretation of the Euler angles involves a decomposition of the rotation tensor into a product of three fairly simple rotations: R

` = R

½ γ

1

, γ 2, γ 3

¾ =

L

½ γ

3

¾

, g3 L

½ γ

2

¾

, g2 L

½ γ

1

¾

, g1 .

(6.22)

Here, {γ i } are the Euler angles and the set of unit vectors {gi } is known as the Euler basis. The function L (ϑ , b ) is defined by use of the Euler representation: L (ϑ , b ) = cos(ϑ )(I − b ⊗ b) + sin(ϑ ) skwt (b ) + b ⊗ b, where b is a unit vector and ϑ is the counterclockwise angle of rotation. In general, g3 is a function of γ 2 and γ 1 and g2 is a function of γ 1. As we shall see shortly, there are 12 possible choices of the Euler angles. For example, Figure 6.6 illustrates these angles 15 For discussions of these papers, and several other interesting historical facts on the development of

representations for rotations, see Blanc’s introduction to parts of Euler’s collected works in [75, 76], Cheng and Gupta [50], and Wilson [308]. Although [73] dates to the eighteenth century, it was first published posthumously in 1862.

6.8

225

Euler Angles

for a set of 3–2–3 Euler angles. Because there are three Euler angles, the parameterization of a rotation tensor by use of these angles is an example of a three-parameter representation. If we assume that g1 is constant, then the angular velocity vector associated with the Euler angle representation can be established by use of the relative angular velocity vector. In this case, there are two relative angular velocity vectors (cf. (6.19)). For the first rotation, the angular velocity vector is γ˙ 1g1 (cf. (6.17)). The angular velocity of the ( ) second rotation relative to the rotation L (γ 1 , g1 ) is (γ˙ 2g2, and the angular velocity of ) the third rotation relative to the rotation L γ 2, g2 L γ 1, g1 is γ˙ 3g3.16 Combining the two relative angular velocity vectors with γ˙ 1g1 , we conclude that ωR

= γ˙

3

g3 + γ˙ 2g2 + γ˙ 1 g1 .

(6.23)

If the rotation tensor R transforms the vectors pi into the set ti , then it is possible to express the Euler basis in terms of either set of vectors. Alternative approaches to establishing (6.23) can be found in textbooks. In one such approach, all three Euler angles are considered to be infinitesimal. A good example of this approach can be found in Section 2.9 of Lurie [173]. Another approach, which can be found in [145, 248] and dates to Euler, uses spherical geometry. Finally, a third (lengthy) approach involves differentiating (6.22) directly and then computing ΩR and its axial vector. For the Euler angles to effectively parameterize all rotations, we need to assume that we can find γ k and γ˙ k such that, for any given ωR , (6.23) holds. For this to happen, it is necessary and sufficient that the vectors g1 , g2, and g3 span E 3: º

g 1, g 2 , g 3

»

=

(

g1 · g2 × g3

)

±=

0.

(6.24)

When the Euler basis vectors are not linearly independent, we say that the Euler angles have a singularity. As shown in 1964 by Stuelpnagel [272], any three-parameter representation of a rotation will unavoidably have singularities and such singularities are present in any of the 12 sets of Euler angles that we discuss. 17 We shall find that this singularity occurs for certain values of γ 2, and we shall restrict the range of the angle 2 γ to avoid these singularities. We can view γ 1 and γ 2 as spherical polar coordinates for g3 . In this manner, we allow γ 1 ∈ [0, 2π ) but restrict γ 2 ∈ [−π/2, π/2] or γ 2 ∈ [0, π ]. The restrictions placed ) ( on the second angle enable γ 1, γ 2 to provide a unique set of coordinates for the unit vector g3.18 The third angle is allowed to range from 0 to 2 π . In some of the examples that are discussed later in this book, such as the sliding cylinder shown in Figure 8.3, the 16 The calculation of these angular velocity vectors is similar to the example discussed in Section 6.7. 17 The Euler angle singularities should not be confused with a phenomenon known as “gimbal lock.”

Mebius et al. [183] make a clear delineation between the Euler angle singularity and the gimbal lock when they state “We remark that while the gimbal lock is a mechanical phenomenon, the Euler angle singularities are of a purely mathematical nature and have nothing to do with flight mechanics per se.” We also refer the reader to Baruh [20, Chapter 7, Section 8] and the paper [123]. 18 Stated in another manner, γ 1 ∈ [0, 2 π ) and γ 2 ∈ [0, π ] or ∈ [− π/ 2, π/2] provide a one-to-one covering on the unit sphere S2 . The reasoning is identical to geodesy, where longitude ranges from 0◦ to 360◦ but latitude only ranges from − 90◦ to 90◦ .

226

Rotations and their Representations

third angle will be constrained to be 0. In such cases, the range of the second angle will be extended to 0 to 2π in order to capture all possible orientations of the rigid body. For future purposes, it is also convenient to define the dual Euler basis {gj }: gj · gi

j

= δi ,

j

where δ i is the Kronecker delta. Clearly, ωR · g i

= γ˙

i

.

(6.25)

The dual Euler basis vectors date to the late 1990s and are analogous to the contravariant basis vectors. We follow [214, 219] and use the dual Euler basis later to represent constraints on the motion of rigid bodies and to represent conservative moments. They can also be used to represent moments in a joint coordinate system in orthopaedic biomechanics [58, 63, 204, 217]. A related basis, where a screw theory representation for the motion of a rigid body is employed, was introduced independently by Kumar and coworkers [131, 298]. To determine the dual Euler basis, we consider the nine scalar equations gj · gi = j δ i . Paralleling an earlier result (1.8) for the contravariant basis vectors and taking into account that g2 ⊥ g1 and g2 ⊥ g3 , the solutions to these nine equations are as follows: g1

=

g2 × g3 (

g1 · g2 × g3

g2

),

=

g3 =

g 2,

g1 × g2 (

g3 · g1 × g2

).

(6.26)

Another method to compute the dual Euler basis vectors is to first express gi in terms of a right-handed basis, say {ti }. Then, to determine g2 say, we write g2

=

at1 + bt2 + ct3,

and solve the three equations g1 · g2

=

0,

g2 · g2

=

1,

g3 · g2

=

0

for the three unknowns a, b, c. When the Euler angles have singularities, one will find that the two dual Euler basis vectors g1 and g3 cannot be defined. Paralleling (3.36), we can define connection coefficients for the Euler and dual Euler basis vectors: νsi,k =

∂ gs · gk , ∂γ i

k ∂ gs ∂g k k νsi = ·g = − · gs. ∂γ i ∂γ i

(6.27)

In contrast to the connection coefficients γsi,k and γsik , the connection coefficients νsi,k and νsik do not always possess the symmetry (s, i) → (i, s). The lack of symmetry can be traced to the fact that, because gi are not partial derivatives of a vector, we no longer have commutativity of the partial derivatives: 19 19 These identities can be verified for all possible sets of Euler angles using the relations (6.33) and (6.38).

They were first noted by Metzger et al. [185].

6.8

∂ g1 = ∂γ 2

0 ±=

∂ g2 = ∂γ 1

227

Euler Angles

g 1 × g 2.

(6.28)

These results should be contrasted to (3.37). We now turn to examining two different choices of the Euler angles: the 3–2–1 set and the 3–1–3 set. Both of these sets are popular in different communities. For instance, the aircraft and vehicle dynamics community favors the 3–2–1 Euler angles to parameterize heading–attitude–banking and yaw–pitch–roll behavior, respectively, whereas problems involving spinning rigid bodies in dynamics often lend themselves to the 3–1–3 set. We shall subsequently discuss examples of both sets. Exercise 6.2 at the end of this chapter involves a comprehensive investigation into the set of 3–2–3 Euler angles. It is highly recommended that you complete this exercise after reading this section of the book.

6.8.1

3–2–1 Euler Angles

To elaborate further on the Euler angles, we now consider the 3–2–1 set of Euler angles (see Figures 6.7 and 6.8). These are arguably among the most popular sets of Euler angles.20 In several communities, they are known as examples of the Tait and/or Bryan angles (after Peter G. Tait (1831–1901) and George H. Bryan (1864–1928)), Fick coordinates (after Adolf E. Fick (1829–1901)), or the Euler–Cardan angles (after Euler and Girolamo Cardano (1501–1576)). First, suppose that the rotation tensor has the representation R=

i=1

²

±

3 ·

ti

⊗ pi ,

where p 1, p2 , p3 is a fixed Cartesian basis. The first rotation is about p 3 through an angle ψ . This rotation transforms pi to t³i . The second rotation is about the t³2 axis through an angle θ . This rotation transforms t³i to t³³i . The third and last rotation is through an angle φ about the axis t³³1 = t1 . Thus, R

` γ 1 = ψ , γ 2 = θ , γ 3 = φ) = = R(

p2

L (φ , t1) L

t1

t2

t1

t1 ψ

p1

θ , t2 ³

)

L

(

ψ , p3

)

.

t3

t3

t3 θ

ψ

(

t2 φ

θ

p3

φ

t2

Figure 6.7 The transformations of various basis vectors induced by the individual angles in a set of 3–2–1 Euler angles. In vehicle (aircraft) dynamics, if t 3 points downwards and t 1 points forward along the longitudinal axis of the vehicle, then ψ is known as the yaw (heading) angle, θ is known as the pitch (attitude) angle, and φ is known as the roll (bank) angle.

20 For example, they are used in Greenwood’s text [105, Section 7-13], Rao’s text [237], and numerous texts

on ophthalmology, vehicle dynamics, and aircraft dynamics.

228

Rotations and their Representations

g1 = p3 ψ

g2

= t2 =

t2

p2

p1 g1

g1

g2

θ

θ

= g2

g3 t2

θ

g3

g2 = t2 φ

t3

g3 = t1

±

²

±

²

1 2 3 Graphical representation of the ± Euler g1², g2, g3 and dual Euler g , g , g basis vectors and their relationships with the p1 , p2 , p3 and { t1 , t 2, t 3} bases vectors, where ψ –θ –φ are a set of 3–2–1 Euler angles. In this figure, θ > 0.

Figure 6.8

Here, ti

=

L (φ , t1 ) t³³i ,

t³³i

=

L

(

θ , t2 = t2 ³³

³

)

t³i ,

t³i

=

L

(

ψ , t3 = ³

)

p3 pi .

It is not difficult to express the various basis vectors as linear combinations of each other: ⎡

⎤

⎡

⎤

⎡

⎤

⎡

t³1 ⎣ t³ ⎦ 2 t³3

cos( ψ ) sin(ψ ) = ⎣ − sin(ψ ) cos(ψ ) 0 0

t³³1 ⎣ t³³ ⎦ 2 t³³3

cos( θ ) 0 ⎣ = 0 1 sin(θ ) 0

t1 ⎣ t2 ⎦ t3

1 ⎣ = 0 0

⎡

⎡

⎤⎡

⎤

p1 0 0 ⎦ ⎣ p2 ⎦ , 1 p3

− sin(θ )

⎤⎡

⎤

⎤⎡

⎤

t³1 ⎦ ⎣ t³ ⎦ , 0 2 cos(θ ) t³3

0 0 t³³1 cos( φ) sin(φ ) ⎦ ⎣ t³³2 ⎦ . − sin(φ ) cos( φ) t³³3

(6.29)

The inverses of these relationships are easy to obtain once you realize that each of the three matrices in (6.29) is orthogonal. As the inverse of an orthogonal matrix is its transpose, we quickly arrive at the sought-after results:

6.8

⎡

⎤

⎡

⎡

⎤

⎡

⎡

⎤

⎡

p1 cos(ψ ) ⎣ p ⎦ = ⎣ sin(ψ ) 2 p3 0

⎤⎡

229

⎤

sin(ψ ) 0 t³1 cos( ψ ) 0 ⎦ ⎣ t³2 ⎦ , 0 1 t³3

−

t³1 cos(θ ) 0 ⎣ t³ ⎦ = ⎣ 0 1 2 t³3 − sin(θ ) 0

1 t³³1 ⎣ t³³ ⎦ = ⎣ 0 2 0 t³³3

Euler Angles

0 cos(φ) sin(φ)

⎤⎡

⎤

⎤⎡

⎤

sin(θ ) t³³1 ⎦ ⎣ t³³ ⎦ , 0 2 cos( θ ) t³³3 t1 0 − sin(φ ) ⎦ ⎣ t2 ⎦ . cos(φ ) t3

(6.30)

Relationships (6.29) and (6.30) can be combined to express ti in terms of p k and vice versa. Later, they will also be used to obtain representations for the Euler basis in terms of pk and ti . By using (6.29), we can find a representation for the components Rij = tj · pi = ( ) ( ) Rp j · pi = Rtj · ti . The components are easily computed using a matrix representation: ⎡

R11 ⎣ R21 R31

R12 R22 R32

⎤

⎡

R13 cos( ψ ) R23 ⎦ = ⎣ sin(ψ ) R33 0 ⎡

1 ×⎣ 0 0

− sin(ψ )

cos(ψ ) 0

0 cos(φ) sin(φ)

⎤⎡

0 cos(θ ) 0 0 1 0 ⎦⎣ 1 − sin(θ ) 0 ⎤

0 − sin(φ ) ⎦ . cos(φ )

⎤

sin(θ ) ⎦ 0 cos(θ ) (6.31)

Representation (6.31) is the transpose of what one might naively expect. However, recalling that Rik = p i · tk will hopefully resolve this initial surprise. Indeed, it is helpful to note that ⎡

⎤

p1 ⎣ p ⎦ 2 p3

⎡

R11 = ⎣ R21 R31 ³

R 12 R 22 R 32 ´µ

⎤⎡

⎤

R 13 t1 R 23 ⎦ ⎣ t2 ⎦ , R 33 t3 ¶

=R

⎡

⎤

t1 ⎣ t2 ⎦ t3

⎡

R 11 = ⎣ R 12 R 13 ³

R 21 R 22 R 23 ´µ

⎤⎡

⎤

p1 R31 R32 ⎦ ⎣ p2 ⎦ . R33 p3 ¶

=RT

(

)

Additional details on how the tensor product L (φ , t1) L θ , t³2 L results expressed in (6.31) are discussed in Section 6.8.5.

The Euler Basis

(

ψ , p3

)

is related to the

As mentioned earlier, by examining the individual rotations (cf. (6.29)), we can show that the Euler basis vectors have the representations

230

Rotations and their Representations

⎡

⎤

⎡

⎤

g1 ⎣ g ⎦ 2 g3

⎡

⎤

⎡

⎤

⎡

p3 ⎣ = t³2 ⎦ = ⎣ t1

⎡

sin(θ ) sin(φ) cos(θ ) 0 cos( φ) 1 0

⎤⎡

⎤

⎤⎡

⎤

cos(φ ) cos(θ ) t1 ⎦ ⎣ − sin(φ) t2 ⎦ . 0 t3 (6.32) Alternatively, we can also express the Euler basis in terms of the basis vectors ± ² p 1, p2 , p3 : −

g1 p3 0 ⎣ g ⎦ = ⎣ t³ ⎦ = ⎣ − sin(ψ ) 2 2 g3 t1 cos(θ ) cos(ψ )

0 cos(ψ ) cos(θ ) sin( ψ )

1 p1 ⎦ ⎣ 0 p2 ⎦ . − sin(θ ) p3 (6.33) This representation is useful for establishing the components ωR · pk .

The Dual Euler Basis

Having found the Euler basis vectors, we can now compute the dual Euler basis vectors gk . The easiest method to determine these vectors is to use the solution (6.26) to the equations gi · gk = δik that define the dual Euler basis vectors. Omitting details, with the help of (6.32) we compute that ⎤

⎡

⎤⎡

⎡

0 g1 ⎣ g2 ⎦ = ⎣ 0 1 g3

⎤

t1 sin(φ) sec(θ ) cos(φ ) sec(θ ) ⎦ ⎣ cos(φ ) − sin(φ ) t2 ⎦ . sin(φ ) tan(θ ) cos( φ) tan(θ ) t3

(6.34)

A similar calculation using (6.33) in place of (6.32) will lead to the following representations: ⎡

⎤

⎡

⎤⎡

⎤

g1 cos( ψ ) tan(θ ) sin(ψ ) tan(θ ) 1 p1 ⎣ g2 ⎦ = ⎣ ⎦ ⎣ − sin(ψ ) cos(ψ ) 0 p2 ⎦ . 3 g cos(ψ ) sec(θ ) sin(ψ ) sec(θ ) 0 p3

(6.35)

Expressions for the dual Euler basis vectors in terms of t³k are easily inferred from (6.35) and are shown in Figures 6.8 and 6.9(a). For completeness, we note that, when θ ± = ´ π2 , one can also express the Euler basis vectors in terms of the dual Euler basis vectors: ⎡

⎤

⎡

g1 ⎣ g ⎦= ⎣ 2 g3

1 0 0 1 − sin(θ ) 0

− sin(θ )

0 1

⎤⎡

⎤

g1 ⎦ ⎣ g2 ⎦ . g3

(6.36)

The simplicity of this relationship (related versions of which hold for the other 11 sets of Euler angles) is surprising.

Singularities

If we examine (6.32), we can compute that (

g1 · g2 × g3

)

=

cos (θ )

and conclude that the Euler basis fails to be a basis for E3 when θ = ´ π2 . In particular, when θ = ´ π2 , g1 = p3 = ´g3 . To accommodate the aforementioned singularity, it is

6.8

(a)

(b)

g1 = p3 g

t3

g1 = p3

1

t3

g 2 = t2

θ

g2 = t2

θ

p2 ψ

t1

p1

231

Euler Angles

p2 g3

ψ

p1

θ

t1

φ

g3 = t1 (a) The dual Euler basis vectors gk for the 3–2–1 set of Euler angles: g1 ² t³³3 , g2 = t ³2, and g3 ² t ³1. (b) The Euler angles ψ and θ serve as coordinates for the Euler basis vector g3 = t 1 in a manner that is similar to the role that spherical polar coordinates play in parameterizing eR and eφ .

Figure 6.9

º

»

necessary to place restrictions on the second Euler angle: θ ∈ − π2 , π2 . The other two angles are free to range from 0 to 2π . One of the easiest ways to visualize the existence of the singularity at θ = ´ π2 is to consider ψ and θ + π2 to be spherical polar coordinates for t1 (see Figure 6.9(b)). When ( π) π θ = − , t1 points towards the north (south) pole of the sphere, ψ is not uniquely 2 2 defined, and the spherical polar coordinate system has a singularity.

Angular Velocity Vectors

The angular velocity vector ωR associated with the 3–2–1 Euler angles has several representations: ωR

=

ax

(

˙ T RR

)

3 · =

γ˙

i

gi

i=1 ˙ t1 + θ˙ t³ + ψ ˙p = φ 3 2 ˙ = (−ψ +

sin(θ ) + φ˙ )t1 + (ψ˙ sin(φ) cos(θ ) + θ˙ cos(φ ))t2

(ψ˙ cos(φ ) cos(θ ) − θ˙ sin(φ ))t3.

To arrive at the primitive representation ωR = φ˙ t1 + θ˙t³2 + ψ˙ p3 , we needed to compute two relative angular velocity vectors. We calculated the first of these, θ˙ t³2, assuming that t³i were fixed. The explicit details of this calculation are easily inferred from our earlier example in Section 6.7. The final stage of the calculation is to compute the expression ˙ t1 for the second relative angular velocity vector by using the relative rotation tensor φ ( ) L φ, t³³1 assuming that t³³i were fixed. It is also interesting to note that the angular velocity vector ω0R has the representations ω0R

=

(

˙ ax RT R

)

=

RT ωR

T T ³ ˙ RT p = φ˙ R t1 + θ˙ R t2 + ψ 3

232

Rotations and their Representations

(−ψ˙ sin(θ ) + φ˙ )p 1 + (ψ˙ sin(φ) cos(θ ) + θ˙ cos(φ ))p2

=

+

( ψ˙ cos( φ) cos(θ ) − θ˙ sin(φ))p 3.

In establishing this result, we used the fact that RT ti

6.8.2

=

pi .

3–1–3 Euler Angles

We can parallel the developments of the previous section for another popular set of Euler angles: the 3–1–3 Euler angles (see Figure 6.10). These are the set of Euler angles that Lagrange used,21 and they are also used in Arnol’d [11], Landau and Lifshitz [164], and Thomson [284], among many others. For motions of a spinning top, the Euler angles are identified with precession, nutation, and spin, respectively. A closely related set of Euler angles, the 3–2–3 set, is discussed in Exercise 6.2 at the end of this chapter. Paralleling the developments for the 3–2–1 Euler angles: R

=

L

(

φ , t3 = t3

³³

)

where {p3 , t³1 , t3 } is the Euler basis and t³i

=

L

(

ψ , p3

)

pi ,

t³³i

=

L

L (

(

θ , t1 ³

θ , t1 ³

)

)

L

(

ψ , p3

t³i ,

ti

)

=

,

L

(

φ , t3

³³

)

t³³i .

Harking back to many of the celestial mechanics applications for this set of Euler angles, the line passing through the origin that is parallel to t³1 is often known as the line of nodes [284]. The angular velocity vector has the representations ωR

=

(

˙ T ax RR

)

=

3 · i=1

γ˙

i

gi

˙ t + θ˙t³ + ψ ˙p . = φ 3 3 1

(6.37)

We are using the same notation for the three Euler angles as we did for the 3–2–1 set. However, it should be clear that θ and φ represent different angles of rotation for these two sets of Euler angles.

p2

p3

t2

t3

t1 ψ

t2 θ

ψ

p1

t2

t2

t1 φ

θ

t2

φ

t1

The transformations of various basis vectors induced by the individual angles in a set of 3–1–3 Euler angles. In a spinning top, if t3 points along the longitudinal axis of the top, then ψ describes the precession, θ describes the nutation, and φ describes the spinning of the top. An example of such a top can be seen in Figure 10.1 (Chapter 10).

Figure 6.10

21 See [156] and Section IX of Part II of [159]. As discussed in [124], Lagrange’s φ, ω , and ψ correspond to

our φ, θ , and ψ , respectively.

6.8

233

Euler Angles

Euler Basis

It is not difficult to show that the Euler basis {gi } has the representations ⎡

⎤

g1 ⎣ g ⎦ 2 g3

⎤

⎡

⎡

⎤⎡

⎤

p3 sin(φ) sin( θ ) cos( φ) sin( θ ) cos( θ ) t1 ³ ⎣ ⎦ ⎣ ⎦ ⎣ = cos(φ ) − sin(φ ) 0 t2 ⎦ = t1 t3 0 0 1 t3 ⎡

0 ⎣ = cos(ψ ) sin(θ ) sin( ψ )

0 sin(ψ ) − sin(θ ) cos(ψ )

⎤⎡

⎤

1 p1 ⎦ ⎣ 0 p2 ⎦ . cos(θ ) p3 (6.38)

With these results and (6.37), two other representations for ωR can be obtained, but this is left as an exercise.

Singularities

As with all sets of Euler angles, the 3–1–3 Euler angles are subject to restrictions. For the 3–1–3 set, we can find the restrictions by examining when the Euler basis fails to be a basis: (

g1 · g2 × g3

)

= − sin (θ ) .

It is easy to see from this expression that the Euler basis fails to be a basis when g3 = t3 = ´p 3. This conclusion can be confirmed by inspecting Figure 6.11(a). As a result, restrictions are placed on the second angle: φ ∈

[0, 2π ),

θ ∈

[0, π ],

ψ ∈

[0, 2π ).

The fact that the restriction needs to be placed on the second Euler angle is consistent with corresponding results for the other 11 sets of Euler angles.

Dual Euler Basis

The easiest method to determine the dual Euler basis vectors is to use the solution (6.26) to the equations gi · gk = δ ik that define these vectors. We find that the dual Euler basis ± ² gi has the representations (a)

(b)

g1 = p3

g3 = t3

g1 = p3

g3 = t3

θ

θ

φ

t2

φ

p2 ψ

p1

g1

g 2 = t1

p2 3

g

ψ

p1

g2 = g2

= t1

(a) The Euler basis and (b) the dual Euler basis vectors for the 3–1–3 set of Euler angles. For this set of Euler angles, g1 ² t³2 , g2 = t ³1, and g3 ² t ³³2 .

Figure 6.11

234

Rotations and their Representations

⎡

⎤

⎡

g1 sin(φ)cosec(θ ) ⎣ g2 ⎦ = ⎣ cos(φ ) g3 − sin(φ ) cot(θ ) and

⎡

⎤

⎡

g1 − sin(ψ )cot(θ ) ⎣ g2 ⎦ = ⎣ cos(ψ ) g3 sin(ψ )cosec(θ )

⎤⎡

⎤

cos(φ )cosec(θ ) 0 t1 ⎦ ⎣ − sin(φ ) 0 t2 ⎦ − cos( φ) cot(θ ) 1 t3 ⎤⎡

(6.39)

⎤

cos(ψ )cot(θ ) 1 p1 sin(ψ ) 0 ⎦ ⎣ p2 ⎦ . − cos(ψ )cosec(θ ) 0 p3

The dual Euler basis vectors are shown in Figure 6.11(b). It is important to observe that they are not defined when this set of Euler angles has its singularities at θ = 0 and θ = π. Using (6.39), we can show that ωR · g1 = ψ˙ ,

ωR · g 2

= θ˙ ,

ωR · g3

˙. = φ

We can also establish the following results: ⎡

⎤

⎡

1 0 g1 ⎣ g ⎦=⎣ 0 1 2 cos( θ ) 0 g3

⎤⎡

⎤

g1 cos(θ ) ⎦ ⎣ g2 ⎦ . 0 g3 1

(6.40)

It is a valuable exercise to compare the results just presented for the 3–1–3 Euler angles with those presented earlier for the 3–2–1 set. One issue that will arise in this comparison is the different ranges that the second Euler angle θ possesses for the two sets. 6.8.3

The Other Sets of Euler Angles

For the Euler basis, one has three choices for g1 and, because g1 ± = g2 , two choices for g2. Finally, there are two choices of g3 . Consequently, there are 2 × 2 × 3 = 12 choices of the vectors for the Euler basis. The easiest method to see which set of Euler angles is being used is to specify the angular velocity vector. Here, expressions are given for each ∑ of the 12 sets of Euler angles for a rotation tensor R = 3i=1 ti ⊗ p i : 1–2–3 Set:

ωR

˙ p + θ˙t ³ + φ ˙t , = ψ 3 1 2

3–2–3 Set:

ωR

˙ p + θ˙t ³ + φ ˙ t3 , = ψ 3 2

1–2–1 Set:

ωR

˙ p + θ˙t ³ + φ ˙t , = ψ 1 1 2

1–3–1 Set:

ωR

˙ p + θ˙t ³ + φ ˙ t1 , = ψ 1 3

1–3–2 Set:

ωR

˙ p + θ˙t ³ + φ ˙ t2 , = ψ 1 3

2–3–1 Set:

ωR

˙ p + θ˙t ³ + φ ˙t , = ψ 1 2 3

2–3–2 Set:

ωR

˙ p + θ˙t ³ + φ ˙ t2 , = ψ 2 3

2–1–2 Set:

ωR

˙ p + θ˙t ³ + φ ˙ t2 , = ψ 2 1

2–1–3 Set:

ωR

˙ p + θ˙t ³ + φ ˙t , = ψ 3 2 1

3–1–3 Set:

ωR

˙ p + θ˙t ³ + φ ˙ t3 , = ψ 3 1

2–3–1 Set:

ωR

˙ p + θ˙t ³ + φ ˙ t1 , = ψ 2 3

3–1–2 Set:

ωR

˙ p + θ˙t ³ + φ ˙t . = ψ 2 3 1

6.8

Euler Angles

235

The 1–2–1, 1–3–1, 2–3–2, 2–1–2, 3–1–3, and 3–2–3 sets of Euler angles are known as the symmetric sets, whereas the other six sets are known as asymmetric sets. For all sets of Euler angles, a singularity is unavoidably present for two values of the second angle θ . At these values, g1 = ´ g3 and the Euler basis fails to be a basis for E3 . For the symmetric sets, one of the singularities coincides with the situation where Q = I. To avoid Euler angle singularities, it is often necessary to use two different sets of Euler angles and to switch from one set to the other as a singularity is approached.22 Euler’s original work [73] on Euler angles pertained to the symmetric sets and Lagrange can be credited for popularizing their use in the dynamics of rigid bodies (see Section IX of Part II of [159]). Starting in the nineteenth century and prompted by research in optometry and vehicle dynamics, the asymmetric sets were developed. The physician and physiologist Fick, in work dating to 1854 that is not well known in the mechanics community, used a 3–2–1 set of Euler angles to examine the kinematics of the eye having a gaze direction t1 (see Figure 6.3(b)).23 The 3–2–1 set of angles, sometimes known as “Fick coordinates,” also appear in the classic work by Helmholtz [121, p. 207] from 1865. Later, in his highly influential treatise [122, pp. 43–44 and 73–75], Helmholtz (1821–1894) uses both a 2–3–1 set of Euler angles and a 1–3–1 set of Euler angles. The 2–3–1 set of Euler angles is sometimes known as “Helmholtz coordinates” (cf. [257]). In mechanics, the asymmetric sets are also known as the Cardan angles, Tait angles, or Bryan angles. Tait’s original discussion (of what we would refer to as 1–2–3 Euler angles) can be seen in Section 12 of his 1868 paper [279]. In his seminal text [34] on aircraft stability that was published in 1911, Bryan introduced what we would refer to as a 2–3–1 set of Euler angles (see Figure 6.12). 24

z

φ G

φ θ

φ

θ x

Reproduction of Figure 3 in Bryan’s seminal text [34] on the stability of aircraft. The angles θ and φ in this figure are the pitch (or attitude) and roll (or bank) angles, respectively, of the aircraft. The point G is the center of mass of the aircraft, and x and z label the corotational bases for the aircraft.

Figure 6.12

22 An example of an analysis where two sets of Euler angles are needed can be found in the paper by

Ashbaugh et al. [13] on the motion of a tumbling tennis racket.

23 For perspectives on Fick’s work and additional references to the kinematics of the eye,

see [119, 161, 200, 257].

24 Bryan’s work is remarkable given the fact that the Wright brothers’ first successful flight was in 1903.

236

Rotations and their Representations

g3

p2

= t3

p3

γ

g3

OP

p1

g1

β D

α

C

g1 = p1

g2 = g2

t3

OD

t2

t1

±

²

Schematic of the right knee joint showing the proximal p1, p2, p3 and distal {t 1, t 2, t3 } bases which corotate with the femur and tibia, respectively. The Euler and dual Euler basis vectors associated with the rotation of this joint when it is parameterized using a set of 1–2–3 Euler angles (1 = α, 2 = β , 3 = γ as in Grood and Suntay [107]). The condyles C and D are also shown. This figure is adapted from [204, 217].

Figure 6.13

6.8.4

Application to Joint Coordinate Systems

Analyzing the kinematics of anatomical joints in orthopaedic biomechanics provides an interesting area of application of the Euler angles. As a concrete example, consider the knee joint between the femur and the tibia shown in Figure 6.13. The rotation tensor R of interest describes the rotation of the tibia relative to the femur. To parameterize the rotation, we follow the seminal paper of Grood and Suntay [107] and use a set of 1–2–3 Euler angles. The vector p 1 is fixed to the femur and the vector t3 is fixed to the tibia. Both vectors are determined using prescribed bony landmarks [107]. A third vector is prescribed as being parallel to t3 × p 1. The prescription of these three vectors is equivalent to prescribing the Euler basis vectors: g1

=

p1 ,

g2

=

t3 × p1

Á Át3

×

Á,

p1 Á

g3

= t3.

6.8

Euler Angles

237

The resulting 1–2–3 set of Euler angles has clinical interpretations: is known as flexion/extension of the knee, β is known as adduction/abduction of the knee, and γ is known as internal/external rotation of the knee. α

An example of clinical data from [204] showing the variation in these angles for the right knee during a drop vertical jump is shown in Figure 6.14. The associated dual Euler basis vectors are g1

=

g2

sec 2 (β) g1 − sec (β) tan (β) g3 ,

=

g 2,

3

g

= −

sec (β) tan (β) g1 + sec2 (β) g3 .

Unlike g1 and g3 , g1 and g3 have no apparent clinical interpretations. The moment M transmitted by the knee joint from the femur to the tibia is known as the knee abduction moment.25 This moment has several representations: (a)

I

II

III

IV

V

Percentage of stance

100

(b) β

20 )◦( elgnA

γ

α

−120

0

(a) Schematic of the five stages of the drop vertical jump task. (b) Variation in the right knee joint angles during a drop vertical jump task. The stance phase starts at initial contact, Stage II, and ends at take-off, Stage IV. This figure is adapted from [204]. Figure 6.14

25 More precisely, the knee joint transmits a force and moment to the tibia. This force–moment system is

equivalent to a moment M and a force F which acts at the center of mass of the tibia. For additional details on joint forces and moments, the reader is referred to [243] and to Section 11.8.

238

Rotations and their Representations

M

=

M 1p 1 + M 2p 2 + M 3p 3

=

m1 t1 + m2t 2 + m3t3

proximal basis distal basis

1 2 3 = M E g 1 + ME g 2 + M E g 3 1 2 3 = M E1 g + ME 2 g + M E3 g

Euler basis dual Euler basis.

The moment components in a joint coordinate system that are typically reported in the literature are ME1 , ME2 , and ME3 . These components can be computed by projecting M onto the unit vectors gi. It is interesting to note that ME1 , ME2 , and ME3 can be calculated without explicit knowledge of the dual Euler basis vectors.26 The developments we have discussed here for the joint coordinate system can be applied to other joints such as the shoulder, hand, wrist, and elbow. The relatively recent standardization of the axes used to describe these joints can be found in [16, 312]. 6.8.5

Comments on Products of Rotations

For the 3–2–1 Euler angles, the components Rik of the rotation tensor R were obtained in an indirect manner and presented in (6.31). We now take this opportunity to show the ( ) ( ) precise equivalence of our earlier work to the product R = L (φ , t1 ) L θ , t³2 L ψ , p3 . Our discussion of the products can readily be translated to any of the other 11 sets of Euler angles and to the compound rotations discussed in Section 6.7. ) ) ( ( At first glance, computing L (φ , t1 ) L θ , t ³2 L ψ , p3 may appear to be tedious because each of the rotation tensors must be converted to one of the bases p i ⊗ pk or(ti ⊗ tk) prior to computing the product. However, because )g2 = t³2 is related to p 2 by ( ) ( L ψ , p 3 and g3 = t³³1 is related to p1 by L θ , t³2 L ψ , p 3 , some helpful simplifications are possible. To explore the simplifications, we first establish the following useful identity for any rotation tensor Q: L (ν , Qr) = QL (ν , r) QT for any proper orthogonal Q.

(6.41)

The identity (6.41) can be proven using some straightforward manipulations: L (ν , Qr) = cos (ν) (I − (Qr) ⊗ (Qr)) + sin (ν) skwt (Qr) + (Qr) ⊗ (Qr) =

(

cos (ν) QQT + ( Qr) ⊗

(

− ( Qr) ⊗

) T

(

rQT

))

+

det (Q) Q sin (ν) skwt (r) QT

rQ

=

Q ((cos (ν) (I − r ⊗ r)) + sin (ν) skwt (r) + r ⊗ r) QT

=

QL (ν , r) QT ,

which was to be established. Next, we consider the following three rotation tensors and apply (6.41) to establish some additional representations for the rotation tensors R2 and R1: R1

=

3 3 · · i=1 k=1

R 1ik pi

⊗ pk =

L

(

ψ , p3

)

,

26 Comparisons of the different components of a moment vector obtained from clinical data can be found

in [58, 204].

6.8

R2

3 · 3 · =

i=1 k=1

R2ik t³i

⊗ tk = ³

=

=

L

(

)

θ , t2 = ³

R1 L

(

¸

R1

θ , p2

R3

=

i=1 k=1

R3ik t ³³i ⊗ t ³³k

(

θ , t1

)

L

=

R2 R1 L

=

R2 R1

³³

)

R1

3 · 3 ·

=

θ , R1 p 2

) T

i= 1 k =1 3 · 3 ·

(

L

239

Euler Angles

=

¹

R2ik pi (

L

(

⊗ pk

φ, R2R1 p1

φ, p1

RT1 , )

) T T

R1 R2

¸ 3 · 3 · i=1 k=1

¹

R3ik pi ⊗ pk RT1 RT2 .

(6.42)

Observe that representations for the tensors have been established so that the basis pi ⊗ pk can be used when computing the product of the three rotations. Expressions for the components R 1rs , R2rs , and R3rs are presented in (6.44) below. ) ) ( ( The product L (φ , t 1) L θ , t³2 L ψ , p 3 can be expressed as the product of three tensors as follows: R

=

R3 R2 R1

¸

¸ 3 · 3 ·

R2 R1

=

¸ =

R2 R1

i=1 k=1 3 · 3 · i=1 k=1

¹

R 3ik pi

⊗ pk

¹

RT1 RT2 L

(

θ , t2 ³

)

L

(

ψ , p3

¹

R3ik p i ⊗ pk I.

We next use the representation (6.42)2 for R2: R3R2 R1 = R1

=

R1 ¸

=

¸ 3 · 3 · r =1 s =1 ¸ 3 3 ·· r =1 s =1

¹

R 2rs pr ⊗ ps RT1 R1 R 2rs pr ⊗ ps

R1mn pm ⊗ p n m=1 n=1 3 3 · 3 · 3 · ·

i=1 k=1

¹¸ 3 3 ··

¹¸

3 · 3 ·

¸ 3 · 3 ·

i=1 k=1

R 3ik pi

3 · 3 · 3 · r= 1 s=1 k=1

Thus, if R

=

R3 R2 R1

=

i =1 k =1

⊗ pk

¹ ⊗ pk

R2rs R3sk pr ⊗ pk

i=1 k =1 m=1 n=1

3 3 · ·

R 3ik pi

¹

R 1im R2mn R3nk p i ⊗ pk .

=

¹

Rik p i ⊗ pk ,

)

240

Rotations and their Representations

then Rik

3 · 3 · =

R1im R2mn R 3nk .

(6.43)

m=1 n=1

Expressing (6.43) as a matrix product: ⎡

R 11 R = ⎣ R 21 R 31 where

⎡

R 12 R 22 R 32 ⎤

⎤

R13 R23 ⎦ R33

=

R1 R2 R3

⎡

R111 R1 = ⎣ R121 R131

R112 R122 R132

R113 cos( ψ ) R123 ⎦ = ⎣ sin(ψ ) R133 0

R211 R2 = ⎣ R221 R231

R212 R222 R232

R213 cos(θ ) 0 R223 ⎦ = ⎣ 0 1 R233 − sin(θ ) 0

R 311 ⎣ R3 = R 321 R 331

R 312 R 322 R 332

R 313 1 ⎦ ⎣ R 323 0 = R 333 0

⎡

⎡

⎤

⎡

⎤

⎡

⎤

− sin(ψ )

0 0 ⎦, 1

cos(ψ ) 0

0 cos(φ ) sin(φ)

⎤

sin(θ ) ⎦, 0 cos(θ ) ⎤

0 − sin(φ ) ⎦ , cos(φ )

(6.44)

we observe that (6.43) is precisely equivalent to our earlier representation (6.31) for the components of R.

6.9

Further Representations of a Rotation Tensor

Apart from Euler’s representation and the Euler angle representation, there are several other representations of a rotation tensor. Most of them are discussed in Shuster’s review article [263], and we now discuss two of them. These are representations due to Olinde Rodrigues [244] in 1840 and the Euler parameter representation. The Rodrigues Vector

The Rodrigues representation is based on the vector ¿ À

λ = tan

φ

2

r.

This vector is sometimes called the Gibbs vector. Clearly, the Rodrigues vector λ is not a unit vector. Indeed, when φ = 0, λ = 0, and when φ = π , λ is undefined. Consequently, if φ varies through π , then we cannot use the Rodrigues representation subsequently discussed. With the assistance of the identities sin(φ) =

2 tan( φ2 )

1 + tan2 ( 2φ )

,

cos( φ)

=

1 − tan2 ( 2φ )

1 + tan2 ( φ2 )

,

6.9

Further Representations of a Rotation Tensor

241

you should be able to verify that cos(φ ) =

1−λ·λ , 1+λ·λ

sin(φ) =

2λ · r . 1+ λ·λ

Substituting for r and φ in (6.9), we find the Rodrigues representation: R

˜ λ) = = R(

1 ((1 − λ · λ)I + 2 λ ⊗ λ + 2 skwt (λ)) . 1+λ·λ

The angular velocity vector associated with this representation is ωR

=

( ) 2 λ˙ − λ˙ × λ . 1+λ·λ

This vector can be calculated by substituting directly into the earlier result associated with Euler representation (6.17). The Euler–Rodrigues Symmetric Parameters

One of the most popular four-parameter representations uses the four Euler–Rodrigues symmetric parameters e0 and e.27 These parameters are often known as the Euler parameters or unit quaternions and have a rich history. 28 The four parameters for a rotation can be defined using the angle φ and axis r of the rotation: ¿

e0 = cos

φ

2

À

¿ À

,

e = sin

φ

2

r.

As a consequence of their definition, the parameters are subject to what is known as the Euler parameter constraint:29 e20 + e · e = 1. We also note that λ=

e . e0

It is possible to express r and φ in terms of the parameters e0 and e, but this is left as an exercise. By substituting for r and φ in Euler representation (6.9), we find the Euler–Rodrigues symmetric parameter representation: ½

¾

2 ¯ R = R(e 0 , e) = e0 − e · e I + 2e ⊗ e + 2e0 skwt (e) . 27 The four parameters, e and the three components of e, are often considered to be the four components of 0 a unit quaternion e = e 0 + e1 i + e 2 j + e3 k, where ei are the components of e relative to a right-handed

orthonormal basis, and i, j, and k are basis vectors for the quaternion. Consequently, Euler–Rodrigues symmetric parameters are sometimes referred to as (unit) quaternions. It is also common to denote the quaternion as either e = ( e0 , e) or e = e0 + e. 28 Discussions of Hamilton’s discovery of quaternions [117] on October 16, 1843 and their relation to works by Gauss and Rodrigues can be found in Altmann [3, 4], Gray [102], and Pujol [235]. Hamilton’s considerable body of work on quaternions can be appreciated by examining his treatise [118]. 29 The relaxation of this constraint is discussed in O’Reilly and Varadi [223] who show, among other matters, how it can be visualized by using Hoberman’s sphere. This topic is also intimately related to Gauss’ mutation of space [92].

242

Rotations and their Representations

∑

Suppose R = 3i=1 ti ⊗ p i , then it is easy to show that e · tk = e · pk . It is straightforward to write out expressions for the components of R ik = tk · pi :30 ⎡

R11 ⎣ R21 R31

⎤

R12 R22 R32

⎡

R13 1 ½ ¾ R23 ⎦ = e20 − e21 − e22 − e23 ⎣ 0 0 R33 ⎡

2e21 + ⎣ 2e1e2 2e1e3 ⎡

0 ⎣ + 2e0e3 −2e0e2

2e1 e2 2e22 2e2 e3

0 1 0

⎤

2e1e3 2e2e3 ⎦ 2e23

−2e0e3

0 2e0e1

⎤

0 0 ⎦ 1

⎤

2e0 e2 −2e0 e1 ⎦ . 0

(6.45)

Clearly, the rotation tensor is a quadratic function of the four parameters e0 , e1 , e2 , and e3. As a result, this representation has several computational advantages over other representations. You may also notice how easy it is to establish an expression for the components of RT from (6.45). The angular velocity vectors associated with this representation are ωR ω0 R

=

=

2 (e0e˙ − e˙ 0e + e × e˙ ) ,

T

R ωR

=

2 (e0e˙ − e˙ 0e − e × e˙ ) .

Again, we calculated ωR by substituting directly into the earlier result associated with the representation (6.17) for ωR . Notice that both angular velocity vectors are relatively simple functions of the Euler–Rodrigues symmetric parameters and their derivatives. Indeed, the angular acceleration vector has a remarkably simple representation that is easy to establish by evaluating ω ˙ R: αR

= ω ˙R =

2 (e0 ¨e − e¨ 0e + e × e¨ ) .

In Exercise 6.4, further results pertaining to the Euler–Rodrigues parameters for the composition of two rotation tensors are presented. These results, which date to Olinde Rodrigues (1794–1851) in 1840, are remarkably elegant. The Euler–Rodrigues symmetric parameters find application in numerous fields, including computer graphics and animation, ophthalmology, robotics, and satellite attitude determination. As advocated by Shoemake [261] in 1985, in contrast to simply linearly interpolating Euler angles, these parameters provide a smooth method of interpolating between two distinct orientations. As discussed in [206] and in Section 7.11, the interpolation procedure is equivalent to using constant angular velocity motions and can be related to geodesics on the group of rotations (known as SO(3)).31 The use of the parameters to study the kinematics of the human eye can be traced to a wonderful paper by Westheimer [305]. His paper laid the foundation for the recent work by Tweed and 30 The corresponding matrix representation for Euler’s representation was established earlier; see (6.11). 31 Geodesics are curves of shortest distance on manifolds. Common examples include straight lines on a flat

plane, great circles on spheres, and circles, helices, and straight lines on cylinders.

6.10

243

Rotations, Quotient Spaces, and Projective Spaces

coworker [291–293] characterizing experimental results on the saccadic motions of the eye. In the area of attitude determination, one considers measurements of two sets of vectors that are related by an unknown rotation tensor R and a translation d. The estimation of R and d can be rendered as a least-squares estimation problem that is known as the orthogonal Procrustes problem [130, 256] and in the satellite dynamics community as the Wahba problem (after Grace Wahba [299]). That is, given the N ≥ 2 measurements a1 , . . . , aN and b 1, . . . , b N , which are related by a rotation R and a translation d, determine the optimal R and d such that the following function is minimized: W

=

N 1· αK ² bK 2

−

RaK − d ²2 ,

(6.46)

K =1

where αK is a scalar weight for the Kth measurement. When d = 0, Davenport showed that, by parameterizing R in (6.46) by e0 and e, it is possible to find a very elegant solution to the Wahba problem.32 The corresponding elegant solution to the Procrustes problem (i.e., when d ± = 0) was established later by Horn [130]. 33

Summary of the Angular Velocity Vectors

It is useful to summarize the representations we have discussed for the angular velocity vector corresponding to a rotation through an angle φ about an axis r: ωR

=

3 ·

γ˙

i=1 ˙r + = φ

i

gi

sin(φ)r˙ + (1 − cos(φ ))r × r˙

2 (e0 e˙ − e˙ 0 e + e × e˙ ) ( ) 2 = λ˙ − λ˙ × λ . 1+λ·λ

=

(6.47)

With some minor manipulations of these equations, one can also obtain expressions for r˙ , e˙ , and λ˙ .

6.10

Rotations, Quotient Spaces, and Projective Spaces

The set of all possible rotations from E3 to E3 forms a group known as SO(3). The identity I is a member of this group and the inverse L (φ , −r) of any element L (φ , r) is also a member of the group. The group SO(3) is homeomorphic to the real projective space RP 3 where a homeomorphism is a continuous function from one space to another 32 Discussions of Davenport’s solution and those of others, along with extensions to Wahba’s original

formulation, can be found in [178, 179, 262, 264].

33 Additional references to other solutions to this problem can be found in [60, 65, 267].

244

Rotations and their Representations

that has a continuous inverse.34 To discuss the space RP3 further, we need to discuss the notion of quotient spaces and through a series of examples build a picture of what RP3 constitutes. To help attain this goal, we make extensive use of the discussion of real projective spaces in the recent textbook by Tu [289, Section 7.6]. However, our discussion is primarily descriptive and we refer the reader to [7, 51, 289] for technical details. We begin by considering a line of length ´ shown in Figure 6.15(a) and identify (or glue) the endpoints of the line to each other. As a result, the line becomes a circle. We can deform the circle so that it evolves into a unit circle S1. Thus, S1 can be considered as the quotient space of the line. As a second example, consider the rectangle shown in Figure 6.15(b). By identifying (i.e., gluing) two edges of the rectangle, a cylinder can be formed. If we then identify the ends of the cylinder, the resulting space can be deformed into a torus T 2. Thus, T 2 is the quotient space of a cylinder, which in turn is the quotient space of a rectangle. Because a curve is homeomorphic to a line and a square is homeomorphic to a rectangle, we can replace “rectangle” by “square” and “line” by “curve” in the above discussion without any additional modification.

(a)

2θ

P1

P2

P1

(c)

( b)

= P2 S

◦

1

◦ θ

◦ T

S

1

2

(a) The quotient space of a line whose endpoints P 1 and P2 are identified is homeomorphic to a unit circle S1. (b) The quotient space of a rectangle is homeomorphic to a cylinder and the quotient space of the cylinder is homeomorphic to a torus. (c) The real projective line RP1 is homeomorphic to a unit circle S1 . Pairs of antipodal points are highlighted in each of the images of these spaces.

Figure 6.15

34 Recall that homeomorphisms can stretch, compress, shear, twist, and distort surfaces and spaces. Thus, a

circle of radius R 1 is homeomorphic to a circle of unit radius and a rectangle is homeomorphic to a square (and vice versa).

6.10

The Real Projective Line

245

Rotations, Quotient Spaces, and Projective Spaces

RP1

We next consider a circle and parameterize the circle using an angle θ ∈ [0, 2π ) as shown in Figure 6.15(c). Observe that 2θ ranges from 0 to 4π as we traverse the circle in a counterclockwise fashion. The real projective line RP1 is obtained by identifying antipodal points on the circle and several pairs of these points are highlighted in the aforementioned figure. The resulting quotient space of the circle is a half circle whose endpoints are identified. The quotient space of the latter space is homeomorphic to the unit circle S1. Thus, RP1 is homeomorphic to S1. Now the set of all rotations about a fixed axis, say E3 , defines a group and each element of this group can be identified with a rotation tensor R = L (θ , E 3). We can say that RP1 is homeomorphic to this group of rotations.

The Real Projective Plane

R

RP2

Next, we consider a set of rotations where the axis r is restricted to lie on a plane so that r = r2 E2 + r3 E3.35 We can show that this set of rotations is homeomorphic to the real projective plane RP2 in two manners. First, we suppose that the rotations are parameterized by a set of Euler–Rodrigues symmetric parameters (or unit quaternions) 2 2 (q0 , q ), where q = q2 E2 + q 3 E3 and q2 0 + q 2 + q 3 = 1. Referring to Figure 6.16(a), the unit quaternion lies on the surface of a unit two-sphere S2 . However, the rotation corresponding to (q0 , q) is identical to the rotation corresponding to (−q0 , −q). The quotient space of the unit sphere with antipodal points identified is homeomorphic to the real projective plane RP 2. The second manner is to consider a flat disk D of radius π centered at the origin (cf. Figure 6.16(b)). Apart from points on the outer rim of the disk, we can uniquely identify a rotation R = L (φ , r) ∈ with a point of the disk whose position vector relative to the center of D is a rotation vector φ r. The points on the outer rim of the disk correspond to rotations through 180 ◦ and the rotation vector π r corresponds to the same rotation as the point on the outer rim whose position vector relative to the center of the ball is −π r. It follows that the quotient space that is homeomorphic to the

R

(a) ( q 0, q 2 , q 3)

(−q0 , −q 2, −q3 )

(c)

(b)

S

E3

r

r E2

2

P1

φ

φ

P2

(a) The unit two-sphere S2 parameterized by a set of unit quaternions with q1 = 0 and illustrating a pair of antipodal points. (b) The solid disk D of radius π showing a rotation vector φ r where r = r 2p 2 + r 3p 3 and four pairs of antipodal points. (c) The solid ball B of radius π showing the rotation vector φ r and three pairs of highlighted antipodal points. Figure 6.16

35 We shall see several examples of such rotations in Section 7.11.

246

Rotations and their Representations

real projective plane RP2 corresponds to D with the antipodal points on the boundary of D identified. It is well known that the space RP2 can be embedded in E4 but not in E3. One method of visualizing RP2 is to first exploit the manner in which RP 2 is related to the unit two-sphere S2 with antipodal points identified. Then, one can use a construction due to Steiner that maps S2 to a self-intersecting two-dimensional surface known as Steiner’s surface in E3 . This construction will be used in Section 7.11, where additional details can be found. The Real Projective Space

RP3

The real projective space RP3 can be visualized by considering a solid ball B of radius π in E3 (see Figure 6.16(c)). We can identify a rotation R = L (φ , r) with a point in B whose position vector relative to the center of the ball is a rotation vector φ r. The point, say P1, whose position vector relative to the center of the ball is π r corresponds to the same rotation as the point P2 whose position vector relative to the center of the ball is −π r. That is, the point P1 can be identified with the antipodal point P2. We can now define RP3 as the ball of radius π where the antipodal points are identified with each other. Every rotation R = L (φ , r) in SO(3) can be identified as an element of RP3 , and the group of rotations SO(3) is said to be homeomorphic to RP 3. An alternative construction of the quotient space that is homeomorphic to RP3 starts with the unit three-sphere S3 in E 4. The set of Euler–Rodrigues symmetric parameters (or unit quaternions) can be used to parameterize points on S3. Similar to the construction shown in Figure 6.16(a), we next identify antipodal points (i.e., (q0, q1 , q2, q3 ) is identified with (−q0 , −q1 , −q2, −q3 )) to form a quotient space of S3 . The resulting quotient space is homeomorphic to RP3 . This concludes our discussion of projective spaces. In the sequel, we shall see that the configuration manifold of a rigid body rotating about a fixed point is homeomorphic 3 ∼ . If the axis of rotation is constrained to be orthogonal to a given RP to RP3 : = 2 ∼ vector, say E1 , then = RP . Finally, if the rigid body is constrained to perform a 1 ∼ fixed-axis rotation about a fixed point, then = S .

M

6.11

M M

M

Derivatives of Scalar Functions of Rotation Tensors

Consider a function U = U(R). We wish to calculate the time derivative of this function. One of the complications is that R has several representations, and for each of them a different representation of the derivative will be found. It will subsequently be revealed that a simple expression for the derivative of U can be found in terms of a vector uR and ) ( the dual Euler basis when U = Uˆ γ 1, γ 2, γ 3 : uR

=

3 · ˆ ∂U i i=1

∂γ i

g.

(6.48)

This result can be used to establish an elegant representation for conservative moments.

6.11

247

Derivatives of Scalar Functions of Rotation Tensors

To start, we use the following representation for R: R=

3 · 3 · i=1 k=1

where p˙ k

=

R ik pi

(

⊗ pk ,

)

0. Then, as U = U(R) = U Rik , pj , we find by using the chain rule that U˙

3 · 3 · ∂U

=

i=1 k=1

We can express this result using the trace operator ¿

˙ U

tr

=

R˙ ik .

∂ R ik

∂U ∂R

˙ R

T

À

,

where ∂U ∂R ˙ and R ˙ Noting that ΩR R = R

T

=

=

3 · 3 · ∂U i= 1 k =1

pi ⊗ pk .

∂ Rik

RT Ω TR , we can then write ¿

˙ = U

tr

∂U ∂R

(

R

T

ΩTR

À

.

)

However, as ΩR is skew-symmetric, tr AΩTR = 0 for all A ˙ the skew-symmetric part of ∂∂ UR RT contributes to U: ¿

U˙ =

tr (

∂U ∂R

R

T

ΩTR

=

AT . Consequently, only

À

)

tr UR Ω TR ,

=

where we have introduced the skew-symmetric operator UR

=

1 2

¸

∂U ∂R

R

T

¿ −

R

∂U ∂R

ÀT ¹

.

As UR is a skew-symmetric tensor, we can calculate a vector that corresponds to twice the axial vector of UR :36 uR

=

skwt (UR ) .

(6.49)

˙ The existence of this vector allows us to establish the following representations for U:

¿

˙ U

=

tr (

∂U ∂R

˙ R

T

=

tr UR ΩTR

=

uR · ωR .

À

)

(6.50)

36 The reason for the absence of the factor 1 in (6.49) can be inferred from the identity (A.12), which is 2

discussed in the exercises at the end of the Appendix.

248

Rotations and their Representations

Because we have established numerous representations for ωR , we next invoke the final ˙ to determine representations for u R . form of the representation for U ( ) ˆ γk . As an example, suppose U is parameterized by use of Euler angles: U = U Then, invoking (6.50)3: ˙ = U

3 · ˆ ∂U k =1

∂γ k

γ˙

k

3 ·

uR ·

=

j =1

γ˙

j

gj .

With the help of the dual Euler basis, we conclude that uR

=

3 · ˆ ∂U gi . i i=1

∂γ

This is the simplest, and most useful, representation that we know of for u R . It first appeared in print in [219]. We now assume that R is parameterized by use of one of the other three methods men˙ using identity (6.50)2 and representations (6.47), tioned previously. By evaluating U, and following the procedure that led to (6.48), we can find three other representations for uR . With details omitted, a summary of the representations is now presented: uR

=

3 · ˆ ∂U gi ∂γ i

i=1 =

˜ ∂U

∂φ

¿

r+

¿

1 cot 2

¿ À

À

φ

(I − r ⊗ r) − εr

2

¯ ∂U

=

1 (e0I − εe) 2 ∂e

=

˘ ∂U 1 (I + λ ⊗ λ − ελ) , 2 ∂λ

¯ ∂U

−

∂ e0

À

˜ ∂U

∂r

e (6.51)

¯ 0, e) = U( ˆ γ 1, γ 2, γ 3) = U( ˜ φ , r) = U( ˘ λ ). Representation (6.51) 4 was where U(R) = U(e first established by Simmonds [266]. We also note that a representation that is closely related to (6.51)2 is discussed in Antman [6]. Several of the partial derivatives in (6.51) need to be evaluated carefully. For example, ˜ because r is a unit vector, the derivative ∂∂Ur must be evaluated on the surface r · r = 1.

U Related remarks pertain to ∂∂Ue , ∂∂eU0 , and ∂∂ R . In other words, these are tangential or surface derivatives. One method of evaluating them is to parameterize R by the Euler angles, and then transform from the parameters of interest to the Euler angles. Indeed, (6.51), the chain rule, and the identity εgi = − ∂∂γRi RT can be used to show that ¯

¯ ∂U

∂ e0

=

∂U ∂R

¯

uR · 2e,

= − ( ε uR )

¯ ∂U

∂e

R,

=

2(e0I + εe)u R ,

˜ ∂U

∂φ

=

u R · r,

6.12

˜ ∂U

∂r

¿ =

sin(φ)(I − r ⊗ r) + 2 sin2

¿ À φ

249

Exercises

À

εr uR .

2

As discussed by Faruk Senan and O’Reilly [78], O’Reilly [214], and Simmonds [266], results (6.51) can be used to establish moment potentials associated with conservative moments. We shall use (6.51) 1 extensively.

6.12

Exercises

Exercise 6.1: Consider the tensor R

=

cos(θ )E1 ⊗ E1 + sin(θ )E 2 ⊗ E1 + cos( θ )E2 ⊗ E2 − sin(θ )E1 ⊗ E2 + E3 ⊗ E3.

(a) Show that R also has the representations R = er ⊗ E1 + eθ =

⊗

E2 + E3 ⊗ E3

cos( θ )er ⊗ er + sin(θ )eθ

⊗ er +

cos(θ )eθ

⊗ eθ −

sin(θ )er ⊗ eθ

+

E 3 ⊗ E3 .

(b) Show that ΩR

= θ˙ (E2 ⊗

E1 − E1 ⊗ E2)

= θ˙ (eθ ⊗

er − er ⊗ eθ ) ,

ωR = θ˙ E3 .

Exercise 6.2: Recall that three Euler angles can be used to parameterize a rotation tensor R. In this exercise, we consider the 3–2–3 set of Euler angles: R=L

(

φ , t3 = t3

³³

)

L

(

θ , t2 ³

)

L (ψ , E3 ) .

This set of Euler angles is used in several texts, for example, [96, Section 4.2]; [145, 247, 248, 306], 37 and illustrated in Figures 6.6 and 6.17. (a) Establish relationships between (i) Ei and t ³i , (ii) t³i and t³³i , and (iii) t³³i and ti .

E2

t1

t2

t1

t1 ψ

t3 θ

ψ

E1

t2

t2

t1 φ

θ

E3

φ

t1

The transformations of various basis vectors induced by the individual angles in a set of 3–2–3 Euler angles.

Figure 6.17

37 Whittaker’s notation is similar to ours, except his ψ corresponds to our φ and vice versa. If you are

comparing his expression for the components of ωR with ours, you will see that there is a typographical error in his expression for ω2 in [306, Section 16]. Routh’s notation is similar to ours, but his coordinate axes are left-handed.

250

Rotations and their Representations

(b) Explain why the second angle of rotation θ is restricted to lie between 0 and π . (c) For this set of Euler angles, show that the Euler basis has the representations ⎡

⎤

⎡

g1 ⎣ g ⎦= ⎣ 2 g3

−

⎡

⎤⎡

sin(θ ) cos(φ) sin(φ) 0

0 − sin(ψ ) = ⎣ cos( ψ ) sin(θ )

⎤

sin(φ ) sin(θ ) cos(θ ) t1 ⎦ ⎣ t2 ⎦ cos(φ ) 0 0 1 t3 0 cos(ψ ) sin(ψ ) sin(θ )

⎤⎡

⎤

E1 1 ⎦ ⎣ E2 ⎦ . 0 E3 cos(θ )

In addition, show that the Euler angles are such that ⎡

⎤

t1 ⎣ t2 ⎦ t3

⎡

⎤⎡

cos( φ) sin(φ) = ⎣ − sin(φ) cos( φ) 0 0

0 cos( θ ) 0 0 ⎦⎣ 0 1 1 sin(θ ) 0 ⎤⎡

⎡

⎤

⎤

− sin(θ )

⎦ 0 cos(θ )

E1 cos(ψ ) sin(ψ ) 0 × ⎣ − sin(ψ ) cos( ψ ) 0 ⎦ ⎣ E2 ⎦ . 0 0 1 E3

(6.52)

(d) Using (6.52), derive expressions for the components Rik of R. These components have a variety of representations: Rik

= (Rtk ) · ti = (REk ) · Ei = tk ·

Ei .

Alternatively, apply the results developed in Section 6.8.5 to R = L (φ , t3) L

(

θ , t1 ³

)

L (ψ , E3 )

and show how the components of R ik can be determined using the product of three matrices. (e) Recall that every rotation tensor R has an axis of rotation r and an angle of rotation. With the help of the results presented in Section 6.5, select four different values of the set (φ , θ , ψ ) and determine the corresponding axis of rotation and angle of rotation. Give physical interpretations for the four sets of values of the Euler angles that you have selected. (f) For this set of Euler angles, show that the dual Euler basis has the representation ⎡

⎤

g1 ⎣ g2 ⎦ g3

⎡ =

⎣

− cos(φ )cosec(θ )

sin(φ) cos(φ ) cot(θ )

⎤⎡

⎤

t1 sin(φ )cosec(θ ) 0 cos(φ ) 0 ⎦ ⎣ t2 ⎦ . t3 − sin(φ) cot(θ ) 1

With the help of these results, verify the following expressions for the dual Euler ± ² basis in terms of the bases {E1, E2, E3} and g1, g2 , g3 : ⎡

⎤

⎡

⎤⎡

⎤

E1 g1 − cos(ψ )cot(θ ) − sin(ψ )cot(θ ) 1 ⎦ ⎣ ⎣ g2 ⎦ = ⎣ E2 ⎦ − sin(ψ ) cos(ψ ) 0 3 E3 cos( ψ )cosec(θ ) sin(ψ )cosec(θ ) 0 g

6.12

⎡

cosec2 (θ ) 0 ⎣ = 0 1 −cot( θ )cosec(θ ) 0

−cot( θ )cosec(θ )

0 cosec2 (θ )

251

Exercises

⎤⎡

⎤

g1 ⎦⎣ g ⎦ . 2 g3

(g) For this set of angles, show that the angular velocity vector has the representation ωR

˙ t 3 + θ˙ t³ + ψ ˙ E3 . = φ 2

(6.53)

You will need to use two distinct corotational derivatives to find this representation. The first of these fixes t³i , whereas the second fixes t³³i . (h) For this set of angles, show that the components of ωR relative to the basis {E 1, E2, E3} have the representations ˙ sin (θ ) cos (ψ ) − θ˙ sin (ψ ) , ±1 = φ ˙ sin (θ ) sin (ψ ) + θ˙ cos (ψ ) , ±2 = φ ˙ cos (θ ) + ψ ˙, ±3 = φ

where ±i = ωR · Ei . These representations can be found in [248, Section 257]. (i) Suppose that ωk (t) = ωR · tk are known functions. Show that ⎡

˙ ψ

⎤

⎣ θ˙ ⎦

⎡

=

⎣

− cos(φ )cosec(θ )

sin(φ ) cos( φ) cot(θ )

˙ φ

⎤⎡

sin(φ )cosec(θ ) 0 cos(φ ) 0 ⎦⎣ − sin(φ) cot(θ ) 1

ω1 ω2

⎤

⎦.

(6.54)

ω3

(j) This problem involves the numerical integration of (6.54). Given ω1 (t) =

0.2 sin(0.5t),

ω2(t) =

0.2 sin(0.05t),

ω3 (t) =

10ω1(t),

and initial values for the Euler angles of your choice, determine φ (t), θ (t), and ψ (t). How can these results be used to determine tk (t)? Exercise 6.3: Recall that a rotation tensor R representing a counterclockwise rotation about an axis p through an angle ν has the representation R = L(ν , p) = cos( ν )(I − p ⊗ p) + sin(ν ) skwt (p) + p ⊗ p, and its associated angular velocity vector has the representation ωR

= ν ˙p +

sin(ν )p˙ + (1 − cos( ν ))p × p. ˙

Consider two rotation tensors R1 = L(θ , E3 ),

R = L(φ , e1 )L(θ , E 3),

where e1

=

cos( θ )E1 + sin(θ )E2 ,

e2

=

cos( θ )E2 − sin(θ )E 1,

e3 = E3.

(a) Show that R1 has the representation R1

=

e1 ⊗ E1 + e2 ⊗ E2 + e3 ⊗ E3 .

(b) Give an example of a system of two rigid bodies in which the rotation tensor of one body is R1 and the rotation tensor of the second body is R.

252

Rotations and their Representations

(c) Given that the relative rotation tensor R2 = RRT1 , show that ωR2

= φ˙ e1 +

sin(φ)θ˙ e2 + (1 − cos(φ ))θ˙ e3.

(d) Explain why ωR2

±= ω ˆ R2 =

ωR − ωR 1 .

Exercise 6.4: The parameterization of the rotation tensor by use of Euler parameters (unit quaternions or symmetric Euler–Rodrigues parameters) has the beautiful consequence that the formula for the composition of two rotations is very elegant. Indeed, the same results for two tensors described by Euler angles are very unwieldy. In this problem, we use Euler parameters to explore some results pertaining to rotation tensors. Consider two rotation tensors A and B: A = (e20 − e · e)I + 2e ⊗ e − 2e0 (εe), B = (f02 − f · f)I + 2f ⊗ f − 2f0(εf).

(6.55)

Here, {e0, e} and {f0 , f} represent two sets of Euler parameters: ¿ À

e0

=

cos

φ

2

¿ À

,

e = sin

φ

2

¿ À

a,

f 0 = cos

θ

2

,

f = sin

¿ À θ

2

b,

(6.56)

where a is the axis of rotation of A, b is the axis of rotation of B, φ is the angle of rotation for A, and θ is the angle of rotation for B. That is, the tensor A corresponds to a counterclockwise rotation of φ about a. (a) Recall the representation for a rotation tensor A in terms of the angle of rotation φ and the axis of rotation a: A(φ, a) = cos(φ)(I − a ⊗ a) + sin(φ) skwt (a) + a ⊗ a. Verify that A has the representation (6.55)1 . ∑ (b) Letting e = 3i=1 ei Ei , what are A ik = (AEk ) · Ei ? (c) In 1840, Rodrigues [244] established a remarkable representation for the Euler parameters associated with a compound rotation: C

=

BA = (g20 − g · g)I + 2g ⊗ g + 2g0 skwt (g) ,

(6.57)

where g0

=

e0 f0 − e · f,

g = e0 f + f 0e + f × e.

(6.58)

The earliest English commentary on Rodrigues’ result was by Cayley [42] in 1845.38 In terms of e0, f0 , f, and e, what are the Euler parameters of the rotation tensors AB and BT AT ? (d) Using the results of (c), show that the compositions of rotations are not, in general, commutative (i.e., AB ± = BA). 38 You may notice that (6.58) is an expression for the product of two quaternions e

= e 0 + e and f = f 0 + f that Hamilton discovered [117] on October 16, 1843. The product can also be found in one of Gauss’ posthumous notebooks [92]. The easiest proof of (6.57) that we know of uses quaternion algebra. The result can also be established by tedious algebraic manipulations and is not presented here.

6.12

ex

North

E3

Figure 6.18

ex

−e z

ey

East

ey

E2

λ

E1

253

Exercises

ez

θ

Down

The angles of longitude

θ

and latitude λ .

(e) Recall that the angular velocity vector associated with A has the representation ωA

=

2 (e0 e˙ − e˙ 0e + e × e˙ ) ,

)

(

˙ T . where ωA = ax AA (i) What are ωB and ωC ? (ii) Give an explanation for the following result:

ωC

±=

ωB + ωA .

(f) If e0 = f0 = √1 , a = E3 , and b = E2 , what does the rotation tensor C represent? 2 Illustrate your solution by showing how C transforms the basis {E1 , E2 , E3 }. Exercise 6.5: The latitude (λ ) and longitude (θ ) of a point on the Earth’s surface are illustrated schematically in Figure 6.18. In navigation systems, one uses these angles to define the downward direction ez , the northerly direction ex , and the easterly direction ey : ⎡

⎤

⎡

ex ⎣ ey ⎦ = ⎣ ez

⎤⎡

⎤

E1 cos( λ) ⎦ ⎣ E2 ⎦ . − sin(θ ) cos(θ ) 0 E3 − cos( θ ) cos(λ) − sin(θ ) cos(λ) − sin(λ) − cos( θ ) sin(λ )

− sin(θ ) sin(λ )

(6.59)

Here, the triad {E1 , E2 , E3 } is a set of fixed right-handed Cartesian basis vectors. (a) Suppose that R = ex ⊗ E1 + ey ⊗ E2 + ez ⊗ E3 . Verify that this rotation tensor can be composed of two rotation tensors R1 and R2 , where R1 corresponds to a rotation about E 3 through an angle θ and R2 corresponds to a rotation about ey through an angle − π2 − λ . (b) Given a vector x = xx ex + xy ey + xz ez

=

X1 E1 + X2 E2 + X3 E3 ,

show that xx

=

3 · i=1

Ri1 Xi ,

xy

=

3 · i =1

Ri2 Xi ,

xz =

3 · i=1

R i3X i ,

254

Rotations and their Representations

where R

3 · 3 · =

Rik E i ⊗ E k .

i =1 k =1

How are the nine components R ik related to the matrix in (6.59)? Exercise 6.6: Consider a pair of rotation tensors A and B: A=

3 ·

ti ⊗ Ei ,

B

3 · =

i=1

ei ⊗ ti .

i=1

Here, {E1 , E2 , E3 } is a fixed, right-handed orthonormal basis for E3 . (a) Show that ◦

˙ = B + Ω B − BΩ , B A A ◦

where B is the corotational derivative of B assuming that ti are fixed, and ΩA

˙ T. = AA

(b) Consider the rotation tensor C = BA. Using the results of (a), show that ◦

T ˆ Ω B = BB , ˆ where the relative angular velocity tensor Ω B and the angular velocity tensor ΩC are, respectively, ˆ Ω B = ΩC − ΩA,

ΩC

˙ T. = CC

ˆ ˆ Why is Ω B skew-symmetric, and what does this imply for the product Ω Bb, where b is any vector? (c) Consider the following example of a pair of rotation tensors:

A = cos(ψ )(I − E3 ⊗ E3 ) + sin(ψ ) skwt (E3) + E3 ⊗ E 3, B = cos(θ )(I − t1 ⊗ t1 ) + sin(θ ) skwt (t1) + t1 ⊗ t1 , where t1

=

cos(ψ )E 1 + sin(ψ )E2.

◦

˙ and B? (i) What are A (ii) Using (6.17), what are ωA and ω ˆ B? (iii) With the help of (6.17), verify that

ωB = θ˙ t1 − ψ˙ (cos( θ )E3 − sin(θ )t2 ) + ψ˙ E3 .

Here, ωB is the angular velocity vector associated with B.

6.12

Exercise 6.7: Show that the rotation tensor L rotation tensor L

½

2π 3 ,p

¾

(π

)

2

, E3 L

(π 2

Exercises

255

)

, E1 is equivalent to a

, where p=

1

√

3

(E1 + E2 + E3 ) .

This result is very useful in analyzing material symmetry groups of crystals. Exercise 6.8: Show that39 L

½π

¾

, E2 L

½π

¾

, E3 L

½ −

π

, E1

¾ =

L

½π

¾

, E3 .

2 2 2 2 This result has an interpretation that is sometimes used to show that successive rotations about three perpendicular axes can be reproduced by a single rotation about one of the axes. It is also the source of one explanation for a phenomenon in biomechanics that is known as Codman’s paradox [53].40 Exercise 6.9: Examine Euler [70, Figure 2 and Sections 1, 24–30], where he introduces ∑ the Euler angles (p, q, r) to parameterize a rotation tensor R = 3k =1 tk ⊗ p k . By specifying the basis vectors p i appropriately, show that a set of 1–3–1 Euler angles is being used, where ψ =

p,

θ =

q,

φ = π − r.

You may wish to examine his expressions for the components R = t 1 · ωR , P = t2 · ωR , and Q = t3 · ωR on [70, p. 205] to help with this. By relabeling the unit vectors p i , demonstrate that Euler’s results can also be used to prescribe a set of 3–2–3 Euler angles. Related results, but with a different notation, can be found in [73]. You can find, among other matters, Euler’s discussion of the components of the Euler and inertia tensors in [70] and [73]. Exercise 6.10: If p1 , p 2, and p 3 are any right-handed set of orthonormal basis vectors, then establish the Rodrigues–Hamilton theorem L

(

π , p3

)

L

(

π , p2

)

L

(

)

π , p1 =

I.

(6.60)

This result is discussed in Whittaker [306, Section 3] and he credits it to Rodrigues [244] and Hamilton [118]. Whittaker presents a proof that involves purely geometric arguments, and it is a good exercise to compare your proof with his. The Rodrigues– Hamilton theorem will be used later to show that a constant moment is not conservative. Exercise 6.11: Suppose a set of 3–2–1 Euler angles is being used to parameterize a rotation tensor R. If we denote the values of these angles by α1 , α2 , and α3 , respectively, 39 For the rotation tensors discussed in this problem, you will need to compute εE . If you need assistance k

with this, please see (A.11) in the Appendix.

40 As quoted in Politti et al. [232], Codman’s paradox occurs if you first place your right arm hanging down

along your side with your thumb pointing forward and your fingers pointing toward the ground. Now elevate your arm horizontally so that your fingers point to the right, then rotate your arm in the horizontal plane so that your fingers now point forward, and finally rotate your arm downward so that your fingers eventually point toward the ground. After these three rotations, you will notice that your thumb points to the left. That is, your arm has rotated by 90 ◦ . The fact that you get this rotation without having performed a rotation about the longitudinal axis of your arm is known as Codman’s paradox.

256

Rotations and their Representations

what are the corresponding values of the 3–2–1 Euler angles for the inverse of this rotation? Exercise 6.12: In Kane et al. [142, Section 1.7], two distinct sets of Euler angles are defined. The first corresponds to “body-angles” and the second corresponds to “spaceangles.” Thus they define 24 sets of Euler angles. Verify that the 3–2–1 Euler angles discussed in Section 6.8.1 are equivalent to the “body-three: 3–2–1” angles in [142]. In addition, show how a set of “space-three: 3–2–1” angles can be used to parameterize R. Denoting the “space-three: 3–2–1” angles by β1, β2, and β3 , respectively, in your solution to this problem you will find the representation ⎡

R11 ⎣ R21 R31

R12 R22 R32

⎤

⎡

1 R13 ⎣ ⎦ 0 R23 = 0 R33 ⎡

0 cos (β3) sin (β 3)

cos (β2 ) ×⎣ 0 − sin (β2 ) ⎡

cos (β 1) × ⎣ sin (β1 ) 0

⎤

0 − sin (β3 ) ⎦ cos (β3 ) 0 1 0

⎤

sin (β2 ) ⎦ 0 cos (β 2)

− sin (β1)

cos (β1 ) 0

⎤

0 0 ⎦. 1

This result should be compared with the corresponding representation (6.31) for the “body-three: 3–2–1” angles. For assistance with this problem, the discussion of “bodyfixed” and “space-fixed” rotations in Ginsberg [96, Section 3.2] and a related discussion in Pars [227, Section 7.14] might be helpful.

7

Kinematics of Rigid Bodies

7.1

Introduction

This chapter contains results on the three-dimensional kinematics of rigid bodies. We discuss several useful classic representations for the velocity and acceleration vectors of any material point of a rigid body. We also discuss the angular velocity vector ω, the linear momentum G, angular momenta H, HO , and HA , and kinetic energy T of rigid bodies and the inertias that accompany them. The chapter concludes with a discussion of the rotational motions of rigid bodies with constant angular velocities. The origin of most of the material in this chapter can be traced to Euler’s seminal work on rigid-body dynamics in the 1750s. Since that time, his theory has been used to develop models for a wide range of mechanical systems and various treatments of his work have appeared. Recently, a tensor-based notation has been used by Beatty [21], Casey [36, 38], Fox [87], Greenwood [106], and Gurtin [111]. The tensor (or, as it is also known, dyadic) notation was pioneered by Gibbs [309] over a century ago. In this book, we use a tensorial treatment as it leads to transparent developments, particularly with regard to inertias and angular velocities.

7.2

The Motion of a Rigid Body

To discuss the kinematics of rigid bodies, it is convenient to follow some developments in continuum mechanics and define the reference and present configuration of a rigid body. First, a body B is considered to be a collection of material points (mass particles or particles). We denote a material point of B by X. The position of the material point X, relative to a fixed origin, at time t is denoted by x (see Figure 7.1). The present (or current) configuration κt of the body is a smooth, one-to-one, onto function that has a continuous inverse. It maps material points X of B to points in three-dimensional Euclidean space: x = κ t (X). As the location x of the particle X changes with time, this function depends on time, hence the subscript t. It is important to note that κt defines the state of the body at time t. We also define a fixed reference configuration κ0 of the body. This configuration is defined by the invertible function X = κ0 (X). Using the invertibility of this function, we can use the position vector X of a material point X in the reference configuration to uniquely define the material point of interest. Later, we will use the reference

258

Kinematics of Rigid Bodies

κ0 ¯

X

Y

X

Π

π

x ¯ X

X

¯

X

Y

x¯

X

Y

κt

y O

Figure 7.1

The reference κ0 and present κ t configurations of a body B .

configuration to determine many of the properties of a body, such as its mass m and inertia tensor J0. Using the reference configuration, we can define the motion of the body as a function of X and t: x = χ (X, t). Notice that the motion of a material point of B depends both on the instant of time t and the material point X of interest.

Euler’s Theorem

For rigid bodies, the function χ (X, t) can be simplified dramatically. First, for rigid bodies the distance between any two mass particles, say X1 and X2 , remains constant for all motions. Mathematically, this is equivalent to saying that ± x1 −

x2 ± = ±X1 − X2 ± .

(7.1)

Second, the motion of the rigid body preserves orientations. In 1775, Euler [71, 72] showed that the motion of a body that satisfies (7.1) is such that x 1 − x 2 = Q ( X1 − X2 ) ,

(7.2)

where Q is a rotation tensor. As you may recall, this rotation tensor has an associated axis and angle of rotation. We can use (7.2) to tell if the rotation of a rigid body is nontrivial. That is, if Q ² = I, we can pick two points X 1 and X2 of the rigid body and examine how the relative position vector x1 − x2 varies as a function of time. If, for all choices of X1 and X2 , the relative position vector is unaltered in direction, then Q = I, otherwise the rigid body is rotating. This is, perhaps, the most useful interpretation of Q: it is the transformation that takes vectors between two points in the body and transforms them into their present state. If we assume that one point of the body is fixed, then we can simplify (7.2) by choosing the fixed point to be the origin: x(t) = Q(t) X.

(7.3)

7.2

The Motion of a Rigid Body

259

We can then infer E ULER ’S THEOREM ON THE MOTION OF A RIGID BODY : Every motion of a rigid body about a fixed point is a rotation about an axis through the fixed point. The axis here is the axis of rotation of Q(t). Because the motion of the body in question is from the configuration κ 0 to κt , this axis depends on the choice of reference configuration. We can arrive at an alternative, and more common, interpretation of Euler’s theorem that does not involve the reference configuration κ0 . To do this, we again consider the motion of the body with a fixed point during the interval t ∈ [t 0, t]. We find from (7.3) that x (t ) = Q (t) QT (t0) x (t 0) . Thus the motion of the body at the end of the time interval is characterized by the rotation tensor Q (t ) QT (t0 ).1 Invoking Euler’s theorem, the axis of rotation of Q (t ) QT (t0 ) is the axis of rotation for the motion of the rigid body. It is emphasized that, during the motion in question, the body’s present configuration changes from κt0 to κt .

Representations for the General Motion

A general motion of a rigid body is one in which the body may not have a fixed point. In this case, it is easy to argue that a uniform translation of the reference configuration can be imposed on the rigid body so that an arbitrary one of its material points XP is placed at its location xP (t) in the present configuration. The rigid body is then rotated about X P so as to occupy its present configuration κt : x(t) − xP (t)

=

Q(t) (X − XP ) .

That is, x = Q(t) X + d(t),

(7.4)

where d(t) = xP (t) − Q(t) XP . In words, (7.4) states that the most general motion of a rigid body is a translation and a rotation.2 It is one of the general representations of the rigid-body motion that we will often use in our subsequent discussions. To discuss an alternative to (7.4) that does not involve the reference configuration, we consider the motion of the body during a time interval [t0 , t] (see Figure 7.2). With the help of (7.4), we find that x (t0 ) = Q (t0) X + d (t0 ) ,

x (t ) = Q (t ) X + d (t ) .

1 In general, the tensor Q ( t ) Q T (t ) has an axis of rotation and an angle of rotation that differ from those 0

associated with Q ( t) .

2 It should not come as a surprise that this result was also known to Euler. See his remarks in the

introductory sections to [68, 70].

260

Kinematics of Rigid Bodies

Q (t) QT (t0)

κ t0

X

X

x (t0 )

Q (t0 ) X

x (t)

κt

O

X

Q (t )

κ0

Schematic of the configurations κt0 and κt of a rigid body. The reference configuration κ0 and illustrations of the roles played by several rotation tensors are also shown.

Figure 7.2

Combining these equations, we arrive at an alternative representation of (7.4): x (t ) = Q (t) QT (t0 ) x (t0) + z (t) ,

(7.5)

where z (t ) = d (t ) − QT (t0 ) d (t0 ).3 Result (7.5) is a convenient departure point to discuss a third alternative representation of rigid-body motion. This representation of rigid-body motion is synonymous with a famous theorem, credited to Michel Chasles (1793–1880), on this topic (see Section 5 of [306] and references therein). Here, the motion of a rigid body is decomposed into a screw motion (see Figure 7.3). That is, the motion is considered to be a rotation through an angle φ about an axis s(t) followed by a translation σ (t)s(t) along that axis. The screw axis s(t) is the axis of rotation of Q (t) QT (t 0), and φ(t) is this tensor’s angle of rotation. The translational component σ s(t) and the location ρ of the intercept of the screw axis can in principle be determined from the three components of z(t) by use of three scalar equations: (

)

I − Q(t)QT (t0 ) ρ(t) + σ (t)s(t) = z(t).

(

(7.6)

)

However, I − Q(t)QT (t0) is noninvertible,4 and so ρ(t) is not uniquely defined. As a result, several choices of ρ can be found in the literature (see, for example, the two choices shown in Figure 7.3). Choosing ρ (t) to be normal to s leads to the following solutions for σ (t) and ρ(t): 3 You may notice that we can choose the reference configuration κ to be identical to κ . Then, Q ( t ) t0 0 0

=

and representations (7.4) and (7.5) will be identical. 4 There are many ways to see this result. The first is to refer the reader to the derivation of (6.18). Alternatively,(one can use the Euler representation (6.9) for a rotation tensor and observe that s is in the ) null space of I − Q(t)Q T (t 0 ) .

I,

7.3

261

The Angular Velocity and Angular Acceleration Vectors

()

φ t

s(t)

X

κt

x (t ) ρ2

Q (t) QT (t0 )

O

ρ1

x (t 0 ) X

κt0 Schematic of the configurations κt0 and κt of a rigid body showing the screw axis s(t) and angle of rotation φ (t). Here, the rigid body is translated along the screw axis by an amount σ s and rotated about s(t) through an angle φ. Two possible choices of ρ (t) are also shown. For the first, ρ (t) = ρ 1 = ρ1 E1 + ρ 2E2 is chosen to be the intercept of the screw axis with the E1 − E2 plane and for the second, ρ(t) = ρ2 , where ρ 2 is the vector from the origin that intersects the screw axis at a right angle: ρ2 · s = 0.

Figure 7.3

ρ(t) =

1 2

±

±

z⊥ (t) + cot

φ(t)

2

²

²

s(t) × z(t) ,

σ (t) =

s(t) · z(t),

(7.7)

where z⊥ (t) = z(t) − (z(t) · s(t)) s(t). It is left as an exercise for the reader to verify (7.7). You should notice that a special case of the solution occurs when Q(t)QT (t0) = I and ρ is indeterminate.

7.3

The Angular Velocity and Angular Acceleration Vectors

The motion of a rigid body is defined by (7.4) or, equivalently, by (7.5). Because Q is a rotation tensor, we can define an angular velocity tensor Ω and an angular velocity vector ω:5 5 As Q ˙ T (t ) 0

0, ω is also the angular velocity vector associated with the rotation tensor Q(t)Q T ( t0 ) . Thus, representations (7.4) and (7.5) have the same angular velocity vectors. =

262

Kinematics of Rigid Bodies

)

(

˙ T . ω = ax QQ

˙ T, Ω = QQ

The vector ω is the angular velocity vector of the rigid body, and Ω is the angular velocity tensor of the rigid body. As we know, the rotation tensor Q can be represented in a variety of manners, for instance, Euler angles or the Euler representation, and so too can its angular velocity vector. However, here it is convenient to omit explicit mention of these representations. By differentiating the angular velocity vector, we find the angular acceleration vector of the rigid body: α=ω ˙.

You should notice that ( ) 1 1 T T ¨ ˙Q ˙ ] = − ε [Ω ˙ ] = ax Ω ˙ , α = − ε[QQ + Q 2 2 where we used the fact that ε is a constant. That is, ε˙ = 0, where 0 represents the third-order zero tensor. We can use result (7.2) to determine the relative velocity and acceleration vectors of any two points X1 and X2 of the rigid body:

v1 − v2

= x ˙1 − x ˙2 ˙ ( X1 − = Q

X2 )

˙ T Q (X − = QQ 1

X2 )

=

ΩQ (X1 − X2 )

=

ω × (x1 − x2 ) .

A further differentiation and some manipulations give the relative acceleration vectors: a1 − a2

= v ˙1 − v ˙2 = ω ˙ × (x 1 − =

x2) + ω × (x˙ 1 − x˙ 2 )

α × (x1 − x2) + ω × (v1 − v2 ) .

The final forms of the relative velocity and acceleration vectors are expressed as functions of t, x1, and x2 . They can also be expressed as functions of t, X1, and X2.

7.4

A Corotational Basis

It is convenient, when discussing the dynamics of rigid bodies, to introduce another basis 6 { e1 , e2 , e3} , which is known as a corotational basis. Here, we define such a basis and point out some of the ways it is used. Our discussion of the corotational basis follows Casey [36], with some minor changes. As is well known in the kinematics of rigid bodies, knowledge of the position vectors of three material points suffices to determine the motion of the rigid body. Indeed, 6 This basis is commonly referred to as a body-fixed frame or an embedded frame.

7.4

A Corotational Basis

263

e2 e3

E3 X3

E1

X2 X3 X1

X2

E2

e1

X1

κt

κ0 O

Figure 7.4

The corotational basis {e1 , e 2, e3} and the fixed Cartesian basis {E1 , E2 , E3}.

this is the premise for many navigation and motion capture schemes and is the motivation for our construction of a corotational basis. Referring to Figure 7.4, we start by picking three material points X1 , X2 , and X3 of the body. These points are chosen such that the orthonormal vectors E1 and E 2 point from X3 toward X1 and X2 , respectively: E1

±

X1 − X3 ,

E2

±

X2 − X3 .

We then complete the (fixed) right-handed Cartesian basis by defining E3

=

E1 × E2.

Now consider the present locations of the three material points. Because Q preserves lengths and orientations, the two vectors x1 − x3 , x2 − x3 will retain their relative orientation. As a result, using (7.2), we define two orthonormal members of a corotational basis by choosing them to point from x3 toward x1 and x2 , respectively: e1

±

x1 − x3,

e2

±

x2 − x3 .

We then define e3 = e1 × e2. As mentioned earlier, the basis {e1, e2, e3 } is known as the corotational basis. It should be apparent that Q = e1 ⊗ E 1 + e2 ⊗ E2 + e3 ⊗ E3

3 ³ 3 ³ =

i =1 k =1

Qik Ei

⊗ Ek =

3 ³ 3 ³

Qik ei ⊗ ek .

i =1 k=1

This result follows from (7.2) and our previous discussions on representations of rotation tensors. In motion capture, knowledge of the location of the markers (i.e., points such

264

Kinematics of Rigid Bodies

as X1 , X 2, and X3 in Figure 7.4) and a fixed basis {E1, E2, E3} enables one to compute the components of Q using the identities ⎡

Q=

Q11 ⎣ Q21 Q31

Q12 Q22 Q32

⎤

Q13 Q23 ⎦ Q33

⎡

e1 · E1 = ⎣ e1 · E2 e1 · E3

e2 · E1 e2 · E2 e2 · E3

⎤

e3 · E1 e3 · E2 ⎦ . e3 · E3

(7.8)

You may recall that we made a similar comment in the previous chapter (p. 208). In motion capture, the basis {E 1, E 2, E 3} is often known as the laboratory-fixed frame and the position vectors x1 , x2 , and x3 are determined with the help of reflective markers placed on X1, X2 , and X 3. Because the corotational basis moves with the body, we can use our previous results for relative velocities and accelerations to show that e˙ i

=

ω × ei ,

e¨ i

=

α × ei + ω × ( ω × ei ) ,

where i = 1, 2, 3. Additionally, as the corotational basis is a basis for E 3, for any vector r we have the representation r=

3 ³

ri e i .

i=1

When the components ri are constant, then the vector is known as a corotational vector. The most trivial examples of corotational vectors are e1, e2, and e3 . The time derivative r˙ of r has the representations v = r˙

3 ³

=

˙ri ei +

i=1 =

3 ³

ri e˙ i

i=1

◦

r + ω × r,

◦

where r is the corotational derivative (with respect to Q) of the vector r. A related expression can be obtained for ¨r: a = v˙

3 ³

=

=

i =1 3 ³

ri ei + ¨

2

3 ³

r˙i e˙ i +

ri e¨ i

i=1

i=1

ri ei + ¨

3 ³

◦

2ω × r + ω × (ω × r) + α × r.

i =1 ◦

The presence of the Coriolis acceleration 2ω × r in the expression for a arises because we have chosen to express r in a basis that is not fixed. You should also observe that, ∑ ◦◦ if r is a corotational vector, then r = 3i=1 ¨ri ei = 0 and the Coriolis acceleration also vanishes.

7.5

7.5

Three Distinct Axes of Rotation

265

Three Distinct Axes of Rotation

It is possible to define three distinct axes of rotation for a rigid body. These axes are commonly used in mechanics and navigation, and to discuss them it is convenient to recall the representations of rigid-body motion (7.4) and (7.5). The rotation tensor Q(t) associated with the former has an axis of rotation q(t) and an angle of rotation θ (t), and the screw axis s(t) and angle φ(t) are the axis and angle of rotation, respectively, of Q (t) QT (t0). A third axis of rotation, which is known as the instantaneous axis of rotation i, can also be defined. This axis is a unit vector parallel to the angular velocity vector ω:7 i(t)

=

ω ± ω±

.

Except in the simple case in which ω is constant, i does not have an associated angle of rotation. The terminology “instantaneous axis” can be appreciated from the observation that ˙r = ω × r for any corotational vector r. Thus, if ω is constant then, as shown in Figure 7.5, r(t) will appear to rotate about i. Our definition of the instantaneous axis of rotation is identical to that used in classic works on rigid-body dynamics, for example, Poinsot [229] and Poisson [231, Sections 405 and 406]. It is not universally adapted. For instance, in the literature on kinematics

ω r

r⊥

q

r

Figure 7.5 An example of r(t) for a motion of a rigid body for which ω is constant and r is a corotational vector. In this figure, r⊥ · ω = 0 and r⊥ + r± = r. The evolution of a possible axis of rotation q(t) for Q(t) is also shown.

7 You may wish to recall that the angular velocity vectors associated with Q (t ) and Q ( t ) Q T ( t ) are 0

identical.

266

Kinematics of Rigid Bodies

of anatomical joints, the terminology instantaneous axis of rotation often refers to s and not i.8 In general, the axes q, s, and i are not identical. However, (6.17) can be used to relate the axes and two angles of rotation: ω = ±ω± i ˙s + = φ

sin(φ)s˙ + (1 − cos( φ))s × s˙

= θ˙ q +

sin(θ )q˙ + (1 − cos(θ ))q × q. ˙

(7.9)

In addition, the Rodrigues formula (6.58) can be used to relate q, s, φ, and θ , but this is left as an exercise. In the course of examining the rotation tensors from various problems in rigid-body dynamics, you can numerically compute s, q, and i. There you will easily find examples in which these axes are distinct. It is, however, also of interest to consider examples in which some of these axes are equal. Two such examples are now presented. First, suppose that the body’s rotation tensor describes a steady rotation about E3 through an angle ν , where ν˙ = ν˙0 is constant: Q(t) = L (ν˙ 0t, E3 ). In this case, it is easy to compute that s(t) = i(t) = q(t) = E3 and ω = ν˙0E 3. Now consider an example in which Q describes a rotation about a time-varying axis of rotation at constant speed:9 Q(t)

=

2q(t) ⊗ q(t) − I,

(7.10)

where the axis of rotation of Q(t) is q(t) = cos

´ν µ

2

E1 + sin

´ν µ

2

E2,

ν (t) = ν˙ 0 (t − t0 ) + ν0 ,

and ν˙0 and ν0 are constants. You should notice, with the possible help of (6.17), that ω = ν˙ 0 E3, and that the angle of rotation θ for this rotation tensor is π . A standard calculation also reveals that Q(t)QT (t0 ) = L (−ν (t) + ν0, e3 ) . That is, the rotation tensor Q(t)QT (t0 ) corresponds to the familiar rotation about E3 through an angle ν (t) − ν0. It now follows that s ² = q. Indeed, (7.10) is the simplest example of a rotation where ω is constant but the axis q is not parallel to ω that we know of. The temporal behavior of ei (t) = QEi , the axes of rotation, and the angles of rotation θ (t) and ν (t) are shown in Figure 7.6. We shall return to this example later in Section 7.11, where it is classified as a Type II motion. 8 The interested reader is referred to [310, 311] for further discussion on this matter. 9 This example is adapted from [216]. Other examples of rotations with constant angular velocity vectors

but distinct axes s and q can be found in [206, 216] and in Section 7.11.

7.6

267

The Center of Mass and Linear Momentum

i = s = E3 2π

( ) − ν0

ν t

e1 e2

E1

E2

q

π

( )

θ t

0

e3

t T

0

1

Figure 7.6 Plots of the loci of the extremities of the corotational vectors ei (t) and the axes of rotation s, q, and i for the rotation tensor (7.10). A plot of ν (t) is also shown in this figure where T = 2ν˙π .

0

7.6

The Center of Mass and Linear Momentum

It is convenient (and traditional) to define a special point, which we refer to as the center ¯ We then use the velocity vector v of mass X. ¯ of this point in the present configuration to define a very useful expression for the linear momentum G of the rigid body.

The Center of Mass

The position vectors of the center of mass of the body in its reference and present configurations are defined by ¶ ¯ = X

¶

R0 Xρ 0dV , R0 ρ0 dV

x¯ = ¶R

¶

xρ dv

R ρ dv

,

(7.11)

where ρ 0 = ρ0 (X) and ρ = ρ (x, t) are the mass densities per unit volume of the body in κ0 and κt , respectively. The regions 0 and denote the regions of E3 occupied by the body in κ0 and κt , respectively. If a body is homogeneous, then ρ 0 is a constant that is independent of X. The principle of mass conservation states that the mass of the body is conserved. That is,

R

R

dm = ρ0 dV or, equivalently,

= ρ dv,

·

m=

R0

· ρ0 dV =

R

ρ dv.

It follows immediately from (7.11) that · ¯ = mX

R0

·

Xρ 0dV,

mx¯ =

R

xρ dv.

268

Kinematics of Rigid Bodies

We can also find from these results that ·

0=

R0

·

¯ ρ dV, (X − X) 0

0=

R

(x − x) ¯ ρ dv.

These identities play key roles in establishing expressions for momenta and energies of a rigid body. For rigid bodies, the center of mass behaves as if it were a material point, which we ¯ To see this we follow [36] and consider denote by X. ·

mx¯ =

R

xρ dv

· =

R

(QX + d ) ρ dv

· =

(QX + d ) ρ0 dV

R±· 0

²

=

Q

=

¯ + md. mQX

R0

Xρ0 dV

±· +

R0

² ρ0 dV

d

Consequently, ¯ + d. x¯ = QX

(7.12)

Recalling (7.4), this implies that the center of mass of the rigid body behaves as if it were a material point of the rigid body. For many bodies, such as a rigid homogeneous sphere, the center of mass corresponds to the geometric center of the sphere, whereas for others, including a rigid circular ring, it does not correspond to a material point. It is convenient in many treatments of rigid-body dynamics to define a reference frame consisting of the center of mass X¯ and the corotational basis {e1 , e2, e3 }. Such a frame is known as the corotational reference frame. In the treatment presented in this book, we often express the relative position vectors of particles x − x¯ in this frame.

Linear Momentum

By definition, the linear momentum G of a rigid body is ·

G=

R

vρ dv.

That is, the linear momentum of a rigid body is the sum of the linear momenta of its constituents. Using the center of mass, we can establish an alternative expression for G with the help of the definitions of m and x: ¯ G = mv¯ , where v¯ = x˙¯ is the velocity vector of the center of mass. You may recall that a related result holds for a (finite) system of particles.

7.7

269

Angular Momenta

Relative Position Vectors

For a material point X of a rigid body, it is convenient to define the relative position vectors π and Π: π

=

x − x, ¯

¯ Π = X − X.

Representative examples of these vectors are displayed in Figure 7.1. With the assistance of (7.2) and (7.12), we see that π

=

QΠ.

(7.13)

Using the corotational basis, we also easily see that π · ei

=

Π · Ei .

This implies that the relative position vectors have the representations Π

=

3 ³

±i E i,

π

=

3 ³

± i ei .

Furthermore, the corotational derivative (with respect to Q) of π is zero: π˙ That is, π is a corotational vector.

7.7

(7.14)

i=1

i =1

=

ω × π.

Angular Momenta

Angular momenta of a rigid body are its most important distinctive feature when compared with a particle. In particular, the angular momentum relative to two points, the center of mass X¯ and a fixed point O, is of considerable importance. For convenience, we assume that the fixed point O is also the origin (see Figure 7.1). ¯ H, By definition, the angular momenta of a rigid body relative to its center of mass X, a fixed point O, HO , and a point A, HA , are ·

H=

·R

HO

=

·R

HA

=

R

(x − x) ¯ × v ρ dv, x × vρ dv, (x − xA) × vρ dv.

The position vectors in these expressions are relative to the fixed point O, and xA is the position vector of the point A. The point A is not necessarily a material point on the rigid body. You should notice that the velocity vector in these expressions is the absolute velocity vector of the material point X. The aforementioned angular momenta are related by simple and important formulae. To find one of these formulae, we perform some manipulations on HO :

270

Kinematics of Rigid Bodies

·

HO

=

· =

=

=

R

x × vρ dv

(π + x) ¯ × vρ dv

·R

R

·

π × vρ dv + ·

H + x¯ ×

R

R

x¯ × vρ dv

vρ dv.

That is, HO

=

H + x¯ × G.

(7.15)

This relation states that the angular momentum of a rigid body relative to a fixed point O is the sum of the angular momentum of the rigid body about its center of mass and the angular momentum of its center of mass relative to O. Paralleling the establishment of (7.15), it can also be shown that HA

=

H + (x¯ − xA ) × G,

HO

HA + xA × G.

=

(7.16)

These results have obvious similarities to (7.15).

7.8

Euler Tensors and Inertia Tensors

To use the balance laws for a rigid body, it is convenient to consider some further developments of the angular momentum H. These developments are aided considerably by use of the Euler tensors E0 and E and the inertia tensors J0 and J. Euler Tensors

We next define the Euler tensors (relative to the center of mass of the rigid body): ·

E0 =

R0

·

Π ⊗ Πρ0 dV,

E=

R

π ⊗ πρ dv.

You should notice that E and E0 are symmetric. It is also possible to show that they are both positive definite.10 Using mass conservation, (7.13), and the identity (Aa) ⊗ (Bb) = A(a ⊗ b)BT , we easily see that E

=

QE0QT .

Furthermore, (Eei ) · ek = (E 0Ei ) ·

Ek .

10 Recall that a (symmetric) tensor A is said to be positive definite if x · ( Ax)

only if, x = 0.

≥

0 and is equal to 0 if, and

7.8

Euler Tensors and Inertia Tensors

271

This implies that E has the representation E=

3 ³ 3 ³

Eik ei ⊗ ek ,

i=1 k =1

where Eik = (E0 Ek ) · Ei are the constant components of E0 . In other words, although E is a function of time, its components, relative to the corotational basis, are constant. Furthermore, these constant components are identical to the components of the constant Euler tensor E0 .

Inertia Tensors

The inertia tensors J0 and J can be defined by use of the Euler tensors: J0

=

tr(E0 )I − E0,

J = tr(E)I − E.

Using the definitions of the Euler tensors and the identity tr(a ⊗ b) these definitions as ·

J0 =

R0

=

a· b, we can restate

·

((Π ·

Π)I − Π ⊗ Π) ρ0 dV,

J=

R

((π ·

π)I − π ⊗ π) ρ dv.

(7.17)

It is easy to see that J = QJ 0QT ,

JT0

=

JT

J 0,

=

J.

The first of these results implies that J=

3 ³ 3 ³

Jik ei

⊗ ek ,

i=1 k =1

where Jki = (J0E i) · Ek are the constant components of J0.11 The symmetry of the inertia tensors also implies that Jki = Jik .

Additional Relationships

Another set of interesting results follows by inverting the relationships between the Euler and inertia tensors. To perform the inversion, we use the definition of the inertia tensor in terms of the Euler tensor and the fact that tr(I) = 3: E0

=

1 tr(J0 )I − J0, 2

E=

1 tr(J)I − J. 2

(7.18)

These results are useful for obtaining the Euler tensors from tabulations of the inertia tensor that are found in numerous undergraduate textbooks on dynamics. 11 Not surprisingly, this is similar to the situation we encountered earlier with the Euler tensors E and E . 0

272

Kinematics of Rigid Bodies

It is a good exercise to substitute (7.14) into (7.17) to see that the components of J0 relative to the basis {E1 , E2 , E3 } are volume integrals involving quadratic powers of the components of Π. For example: ·

J011

= ( J0 E1 ) · E1 =

J012

= ( J0 E2 ) · E1 = −

J013

= ( J0 E3 ) · E1 = −

´

R·0

y 2 + z2

µ

ρ0dV,

R · 0

xy ρ0 dV,

R0

xzρ0 dV,

(7.19)

where x = π · e1

=

Π · E1,

y = π · e2

=

Π · E2 ,

z

=

π · e3 = Π · E3 .

A similar exercise with the components of E0 shows that ·

x2 ρ0 dV,

E 011

= ( E0E 1) ·

E1

=

R0 ·

E 012

= ( E0E 2) ·

E1

=

R · 0

xyρ0 dV,

E 013

= ( E0E 3) ·

E1

=

R0

xzρ 0dV.

It is interesting to note the simple relationship between the off-diagonal components of E0 and J0 . Again, one can then use tables of inertias found in textbooks to determine the components of J0 and J. Both inertia tensors J0 and J are symmetric. It can also be shown that they are positive definite. This allows us to choose {E1 , E2 , E3 } such that these vectors are the eigenvectors of J0 and, consequently, J0

= λ1 E1 ⊗ E1 + λ2E 2 ⊗ E2 + λ 3E3 ⊗

Here, λi are known as the principal moments of inertia. As J we also have

E3 .

=

QJ 0QT and ei

=

QEi ,

J = λ1 e1 ⊗ e1 + λ2 e2 ⊗ e2 + λ3 e3 ⊗ e3. It is common to refer to ei as the principal axes of the rigid body. Now because of the definition of the components J0ik of the inertia tensor, the principal moments need to satisfy certain inequalities: λ1 + λ 2 > λ 3,

λ2 + λ3 > λ1 ,

λ3 + λ1 > λ 2.

(7.20)

These inequalities are easy to establish using the definitions (7.19) and are useful when selecting representative examples of inertias. For instance, (7.20) imply that a rigid body with λ1 = 1, λ 2 = 2, and λ3 = 3 is not physically realizable. It can also be shown that the eigenvectors of J0 are the eigenvectors of E0 . Consequently, if we choose {E1, E2 , E3 } to be the eigenvectors of J0 , then E0

=

e1 E1 ⊗ E1 + e2 E2 ⊗ E2 + e3E3 ⊗ E3 .

7.8

Euler Tensors and Inertia Tensors

273

m3 m1

1 2

m2

¯

X

m2

3

E3 m1 m3

E2

O

E1 The reference configuration of a dumbell satellite consisting of six masses joined to a center by three massless, rigid rods of lengths 2²1 , 2² 2, and 2² 3.

Figure 7.7

By use of the identity (7.18)1 , the constants ei can be related to the principal moments of inertia: 1 1 e2 = (λ1 + λ3 − λ 2) , e1 = (λ2 + λ3 − λ 1) , 2 2 e3 As E = QE0 QT and ei

=

=

1 (λ1 + λ2 − λ 3) . 2

(7.21)

QEi , we also have

E = e1 e1 ⊗ e1 + e2 e2 ⊗ e2 + e3e3 ⊗ e3 . Notice that the principal axis ej corresponding to the maximum value of λi corresponds to the minimum value of ei . Some Examples

In many studies of satellites, it is convenient to model the satellite as a set of connected dumbbells. For such systems, the integrals in the definition of the Euler and inertia tensors degenerate into summations over a discrete number of particles. An example of such an arrangement is shown in Figure 7.7. For the body shown in this figure, it is easy to calculate E0: E0 =

3 ³ k= 1

Consequently, the inertia tensor is J0

´

=

2mk ²2k Ek

µ

⊗ Ek .

´

µ

2 m2 ²22 + m3 ²23 E 1 ⊗ E1 + 2 m1²21 + m3²23 E2 ⊗ E2 ´

+

µ

2 m1 ²21 + m2 ²22 E3 ⊗ E3 .

It is left as an exercise to write expressions for E and J.

274

Kinematics of Rigid Bodies

The simplest inertia and Euler tensors arise when the body is a homogeneous sphere of radius R0 or a homogeneous cube whose sides have length ²: 2mR20 I, 5 mR20 I, E = E0 = 5

m² 2 I, 6 m²2 E = E0 = I, 12

J = J0 =

J = J0

=

respectively. For these bodies, any three mutually perpendicular unit vectors are principal axes. The next class of bodies is for those with an axis of symmetry. For instance, a homogeneous circular rod of length ² and radius R0 has the following respective Euler and moment of inertia tensors: mR20 m²2 E1 ⊗ E1 + (I − E 1 ⊗ E1 ) , 12 4 ¸ ¹ mR 20 mR 20 m² 2 E1 ⊗ E1 + + (I − E1 ⊗ E1 ) , 2 4 12

E0 J0 =

=

(7.22)

where E1 is the axis of symmetry of the circular rod in its reference configuration. Most bodies, however, do not have an axis of symmetry. Consider the homogeneous ellipsoid shown in Figure 7.8. The equation for the lateral surface of the ellipsoid is x2 a2

+

y2 b2

+

z2 c2

=

1.

The inertia tensor of the ellipsoid is J0

=

µ µ µ m´ 2 m´ 2 m´ 2 b + c2 E 1 ⊗ E1 + a + c2 E2 ⊗ E 2 + a + b2 E3 ⊗ E3. 5 5 5

It is left as an exercise to write the Euler tensor E0 for the ellipsoid.12 Ellipsoid

E3

¯

X

E1

Figure 7.8

An ellipsoid of mass m.

12 For assistance, see (7.21).

E2

7.8

Euler Tensors and Inertia Tensors

275

As a final example, we consider the output from a computer-aided design program. These programs output a matrix of inertias for a given rigid body with respect to a set of axes. An example of such a symmetric positive definite matrix is ⎡

J

=

J11

⎣ J12

J13

J12 J22 J23

⎡

⎤

J13 ⎢ ⎢ J23 ⎦ = ⎢ ⎣ J33

27 8 º −

º

−

5 º8

3 32

3 32

5 8

⎥

3 ⎥. ⎥ 32 ⎦ 27 8

9 º4

3 32

⎤

We denote the axes by a set of fixed, right-handed basis vectors {B1 , B2 , B 3}. Thus, the inertia tensors of the rigid body are J0

=

3 3 ³ ³

Jik Bi

⊗ Bk ,

J=

3 3 ³ ³

Jik b i ⊗ bk ,

i=1 k=1

i=1 k=1

where bi = QBi are corotational basis vectors. To determine the principal axes and principal values of J, we compute the eigenvalues λk and eigenvectors lk of the matrix J . That is, Jlk = λk lk : λ1 =

4,

λ2 =

3,

λ3 =

2,

l1

1 »

= ³√

2 ½ 8¾ l2 = ³ 3 1 »

l3

= ³√

8

1

0 1 º

−1

1

√

6

¼T

2 3

,

1 −1

¿T ¼T

, .

Observe that the inequalities (7.20) required for J to represent the moment of inertia matrix of a physically realizable rigid body are satisfied. In addition, the ³ signs in the expressions for lk are intended to emphasize that if lk is an eigenvector, then so too is −lk . We next use the eigenvectors lk to construct a right-handed set of principal axes for J0 : E1

1

= √

2

½ ¸

(B1 + B3 ) ,

E3

E2

1 ´

= √

8

=

−B 1 −

8 3

√

½

−B1 +

¹

2 B2 + B 3 , 3

µ

6B2 + B3 .

The expressions for E1 and E2 follow from the + signed versions of l1 and l2 , while the − sign for l3 that is used to construct E3 is selected to ensure that (E1 × E2 ) · E3 = 1. The corresponding principal axes for J are ei = QEi . Thus, we find the following representations of the inertia tensors: ½

27 3 9 J0 = (B1 ⊗ B1 + B3 ⊗ B3 ) + B2 ⊗ B2 − (B1 ⊗ B 2 + B2 ⊗ B 1) 8 4 32 ½ 5 3 + (B1 ⊗ B3 + B3 ⊗ B1 ) + (B2 ⊗ B3 + B3 ⊗ B2 ) 8 32 = 4E1 ⊗ E 1 + 3E 2 ⊗ E2 + 2E3 ⊗ E3

276

Kinematics of Rigid Bodies

and

½

9 3 27 J= (b1 ⊗ b 1 + b 3 ⊗ b 3) + b2 ⊗ b2 − (b1 ⊗ b2 + b2 ⊗ b 1 ) 8 4 32 ½ 5 3 + (b1 ⊗ b3 + b 3 ⊗ b 1) + (b 2 ⊗ b 3 + b3 ⊗ b2 ) 8 32 = 4e1 ⊗ e1 + 3e2 ⊗ e2 + 2e3 ⊗ e3. We leave it as an exercise to compute the corresponding results for the Euler tensors E0 and E with the help of the relations (7.18). For all of the preceding examples, we have exploited the property that any body has three principal axes. Writing the inertia and Euler tensors with respect to these axes provides their simplest possible representations.

7.9

Angular Momentum and an Inertia Tensor

The result we now wish to establish is that H = Jω. This is arguably one of the most important results in rigid-body dynamics. In particular, with the assistance of J0, for a particular rigid body it allows us to write down a tractable expression for H. We now reconsider the angular momentum H: ·

H=

(x − x¯ ) × vρ dv

R

· =

R

π × vρ dv

R

π × (v¯ + ω × π)ρ dv

R

π × v¯ρ dv +

· =

· =

·

R

π × (ω × π)ρ dv.

However, because X¯ is the center of mass and the velocity vector v¯ of X¯ is independent of the region of integration, we can take v¯ outside the integral: ·

H= · =

=

·

R

π × v¯ ρ dv +

R

πρ dv × v¯ +

=

·

0 × v¯ + ·

R

R

R

π × (ω × π)ρ dv

π × (ω × π)ρ dv

¶

R πρ dv

·

R

π × (ω × π)ρ dv

π × (ω × π )ρ dv.

Notice that we also used the identity calculation. Summarizing, we have H=

R

·

π × (ω × π)ρ dv

=

0 in the next-to-last step of this

· =

R

((π · π) ω − (π · ω)π) ρ dv.

(7.23)

7.10

Kinetic Energy

277

In writing this equation, we used the identity a × (b × c) = (a · c)b − (a · b)c = ((a · c) I − c ⊗ a) b. With the help of the definition of the inertia tensor, J, it should now be apparent that H = Jω. As mentioned earlier, this is one of the most important results in the kinematics of rigid bodies. You should notice that H = Jω implies that there is a linear transformation between angular velocity and angular momentum. Furthermore, unless ω is an eigenvector of J, H and ω will not be parallel. The triple product identity that was invoked to express the vector H in terms of the tensor J is an example of the use of the following pair of useful identities: 2 ( a × b ) · ( a × b ) = ( a · a ) (b · b ) − ( a · b ) =

a · (Ba) ,

b × (a × b ) = (b · b) a − (a · b ) b = Ba,

(7.24)

where a and b are any pair of vectors and the tensor B is B = (b · b ) I − b ⊗ b. Observe that the tensor B is symmetric.

7.10

Kinetic Energy

The kinetic energy of a rigid body has a very convenient representation. This representation, which was first established by a contemporary of Euler, Johann S. Koenig (1712–1757), is known as the Koenig decomposition: T=

1 1 mv¯ · v¯ + (Jω) · ω. 2 2

(7.25)

Here, a derivation of this result is given. The kinetic energy T of a rigid body is defined to be T

=

1 2

·

R

v · vρ dv.

We can simplify this expression for the energy by expressing the velocity vector v as v = v¯ + ω × π. Substituting this expression into T , we have ·

T

=

=

1 v · vρ dv 2 R · · · 1 1 v¯ · v¯ ρ dv + (ω × π) · v¯ ρ dv + (ω × π) · ( ω × π)ρ dv. 2 R 2 R R

278

Kinematics of Rigid Bodies

However, with the help of (7.24)1 , we have the following identities: ·

R

·

R

v¯ · v¯ ρ dv

= v ¯ ·v ¯

±

(ω × π) · v¯ ρ dv

=

(ω × π) · (ω × π )ρ dv

=

R

·

·

ρ dv =

R±·

ω× ·

mv¯ · v, ¯ ²²

R

πρ dv

·v ¯ =

(ω × 0) · v¯ = 0,

( ω · ω)(π · π) − (ω · π)2ρ dv

R ±±·

=

ω·

=

ω · (Jω) .

²

R

²

(π · π)I − π ⊗ πρ dv ω

Substituting these results into the previous expression for T , we find the desired result: 1 1 mv¯ · v¯ + (Jω) · ω. 2 2 This result is known as the Koenig decomposition of the kinetic energy of a rigid body. In words, the kinetic energy of a rigid body is equal to the sum of the kinetic energy of its center of mass and the rotational kinetic energy T rot = 21 ω · (Jω) of the rigid body. T

=

Comments on the Rotational Kinetic Energy

The previous development of the Koenig decomposition showed that 2T rot = ω · Jω. This representation is used in the vast majority of works on rigid-body dynamics. An equivalent, complementary representation using J0 can also be found,13 and we now discuss this representation. Consider the angular velocity vector ω0 :

If ω =

∑3

i=1 ωi ei ,

ω0 = Q T ω.

then it follows that ω0

3 ³ =

ωi Ei .

i=1 ˙ In addition, it can also be shown that ω0 is the axial vector of QT Q:

ω0

=

(

˙ ax QT Q

)

=

(

)

ax QT ΩQ .

It is a good exercise to see what the representation of ω0 is when the 3–2–1 or 3–1–3 Euler angles are used. We now use the relationship between J and J0 to see that 2T rot

=

ω · (Jω)

=

Qω0 · (JQω0 )

13 See, for example, [38] and [180, Chapter 15].

7.11

279

Attitudes of Constant Angular Velocities

(

=

ω0 · QT JQ ω0

=

ω0 · (J0 ω0) .

)

In summary, T rot =

1 1 ω · Jω = ω0 · J0ω0 . 2 2

The advantage of the representation involving ω0 is that J0 is a constant. Thus, when taking derivatives of Trot with respect to the parameters used for Q, we only need to consider the derivatives of ω0 .

7.11

Attitudes of Constant Angular Velocities

Constant angular velocity motions occupy a special place in studies on the motion of a rigid body and are present in most applications of the theory of a rigid body, including torque-free motion of a rigid body and rolling rigid bodies. The classic case where the axis of rotation r of Q and the angular velocity vector are parallel (i.e., ω = θ˙r) is the most familiar case. However, there are two other types of rotational motions where ω is (a)

(b)

E3

i = ωω

e1

e3 E3

e2 E2

e2 E1

e1

E2

r

e1 E3

E1

(c)

e3

(a0, a)

Q

(q 0 (t ), q (t )) Θ

e1 (b 0, b)

e1

E2

e1 E1

Three realizations of constant angular velocity rotations Q = L (θ , r) from rigid-body dynamics, computer graphics, and optometry. (a) Conic surfaces swept out by the moving basis vectors ei for a constant angular velocity motion. The trajectories (or spherical indicatrices) of ei are circular arcs centered about i. (b) Examples of spherical indicatrices of the motions of the gaze direction e1 of the eye which satisfy Listing’s law (r · E1 = r · e1 = 0). The point Q, through which all the indicatrices pass, is known as the occipital point and the inset image shows the eye and its reference configuration (or primary position). (c) A great circle on the unit three-sphere S3 and a schematic of spherical linear quaternion extrapolation (Slerp) between two unit quaternions (a0 , a) and (b0, b) lying on S3 (cf. (7.35)).

Figure 7.9

280

Kinematics of Rigid Bodies

constant but r is not parallel to ω. Further, the latter rotations appear in many applications. We take this opportunity to emphasize that, in our discussion in this section, we use r to denote the axis of rotation of Q and q to denote the vector associated with a unit quaternion q = q0 + q. As shown in Figure 7.9(a), motions with constant angular velocities are characterized by motions of a body-fixed axis about the angular velocity vector. In optometry, they are central to Helmholtz’s studies on the motion of the eye and his celebrated characterization of the saccadic motion of the gaze direction in [121, 122] (see Figure 7.9(b)). Recent commentary that the saccades are related to geodesics (curves of shortest distance) of SO(3) can be found in the optometry literature (see, e.g., [125, p. 3239]; [291, p. 106 ]). More precise analyses on these geodesics can be found in works by Ghosh and coworkers [94, 95, 233]. In the field of computer graphics, these geodesics manifest as great circles on the unit sphere S3 in four dimensions and play a central role in Shoemake’s Slerp algorithm [261] for interpolating between two rotations that are parameterized by unit quaternions (see Figure 7.9(c)). In the interests of completeness, the first instance we found where geodesics were equated with constant angular velocity motions is in the text by Synge and Schild [277, Section 5.5], published in 1949. We now proceed to show how rotational motions of a rigid body with constant angular velocity can be classified into three distinct types: Type I. Motions where the axis of rotation r is parallel to ω and θ˙ is a nonzero constant: q ± i and q˙ × q = 0 (cf. Figure 7.10). Type II. Motions where θ = π and the axis of rotation r is perpendicular to ω. That is, q0 = 0 and q · ω = 0 (cf. Figure 7.11). The periods of r(t), q0(t), and q(t) are twice the period Tp of the corotational basis vectors. Type III. Motions where θ = θ (t) but the axis of rotation r is neither normal to nor parallel to ω. That is, q0 ²= 0 and q · ω ²= 0 (cf. Figure 7.12). The periods of θ (t) and r(t) are twice the period Tp of the corotational basis vectors (cf. Figure 7.9(a)).

Here, q0 and q are the scalar and vector components of a unit quaternion (Euler– Rodrigues symmetric parameters), θ is the angle of rotation, and r is the axis of rotation: (a)

(b)

(c)

e3

1

1 Q21

e2 = i = r e1

q0

−1 −1

1 Q32

1 −1

Q13

e3 = i = r

−1 −1

1 q2

1 −1

e2

q3

e1

A representative pair of Type I motions and their representations. (a) The motions appear as straight lines on Steiner’s (or Roman) surface, (b) the projection of the great circles on S3 onto E3 appears as circles of unit radii, and (c) the corresponding spherical indicatrices of the instantaneous axis of rotation i, axis of rotation r, and the vectors ei . This figure is adapted from [206].

Figure 7.10

7.11

(a)

(b)

(c)

r

1

1

e3

e1 Q21

−1 −1

281

Attitudes of Constant Angular Velocities

q0

−1 −1

1

Q32

1

−1

Q13

e2

1 q2

1 −1

i

q3

A representative Type II motion and its representations. (a) The motion appears as a straight line on Steiner’s (or Roman) surface, (b) the projection of the great circle on S3 onto E3 appears as a circle of unit radius, and (c) the corresponding spherical indicatrices of the instantaneous axis of rotation i, axis of rotation r, and the vectors e i . This figure is adapted from [206].

Figure 7.11

(a)

e3

(c)

(b)

Q21

1 Q32

1 −1

Q13

e1 e3

q0

−1 −1

e2

r

1

1

−1 −1

e2 e1

1 q2

1 −1

q3

r

A representative pair of Type III motions and their representations. (a) The motions appear as circles on Steiner’s (or Roman) surface, (b) the projection of the great circles on S3 onto E3 appears as circles with unit radii, and (c) the corresponding spherical indicatrices of the axis of rotation r and the vectors ei . For these motions ei (t) rotate at constant speed about ω (cf. Figure 7.9(a)). This figure is adapted from [206].

Figure 7.12

± ²

q0 = cos

θ

2

± ²

,

q = sin

θ

2

r.

As q20 + q · q = 1, the unit quaternions are said to lie on a unit three-sphere S3 . This sphere lies in the four-dimensional space E4. Type I motions are prototypical constant angular velocity motions and, given the freedom to choose the reference basis {E1, E2, E3} or, equivalently, Q (t 0), constant angular velocity motions can always be restricted to this type. Unfortunately, in many application areas, including ophthalmology, the reference basis for Q is prescribed and Type II and Type III motions must be considered. Our forthcoming exposition is based on the works of Novelia and O’Reilly [206, 207] and O’Reilly and Payen [216], in addition to the aforementioned works. Given a rotation R = L (θ , r), we recall that the angular velocity vector has the representations

282

Kinematics of Rigid Bodies

ω = ω1 e 1 + ω2 e 2 + ω1 e 3 = θ˙ r + =

sin (θ ) r˙ + (1 − cos (θ )) r × r˙

2q0q˙ − 2q˙ 0 q + 2q × q. ˙

The corresponding representations for α = ω ˙ are α = ω˙ 1 e1 + ω ˙ 2e2 + ω ˙ 1 e3 = θ¨r + θ˙ (1 + =

cos (θ )) r˙ + sin (θ ) r¨ + θ˙ sin (θ ) r × r˙ + (1 − cos (θ )) r × r¨

2q0 q¨ − 2q¨ 0 q + 2q × q¨ .

(7.26)

From the first representation for α , we observe that if ω is constant, then the components ωi are also necessarily constant. We also note that ω determines ˙ek , e ˙ k = ω × ek , and that the direction of ω defines the instantaneous axis of rotation i: i=

ω . ± ω±

From the second representation for ω, we conclude that the axes r and i are parallel only in instances where r˙ = 0.

The Differential Equations for r and

θ

The vector r is a unit vector. By differentiating the identity r · r identities can be found: r · r˙ = 0,

=

1, the following

r˙ · r˙ = −r · r¨ = (r × r˙ ) · (r × r˙ ) .

(7.27)

Consequently, r is perpendicular to r˙ . The vectors r, r˙ , and r × r˙ can be used to form a basis for E3 provided ˙r = 0. We now examine the components of α with respect to this basis. With the help of the representation (7.26)2, for α: α · r = θ¨

−

sin (θ ) r˙ · r˙ ,

α · r˙ = θ˙ (1 +

cos (θ )) r˙ · r˙ + sin (θ ) r¨ · r˙ − (1 − cos (θ )) (r × r˙ ) · r, ¨

α · (r × r˙ ) = θ˙ sin (θ ) r˙ · r˙ + (1 − cos (θ )) r˙ · r¨ + sin (θ ) r¨ · (r × r˙ ) .

(7.28)

We seek solutions θ (t) and r(t) to these equations where α = 0. The three equations governing θ and r can be found from (7.28) by setting α = 0: θ¨ =

sin (θ ) r˙ · r, ˙

0 = θ˙ ( 1 + cos (θ )) ˙r · r˙ + sin (θ ) r¨ · r˙ − (1 − cos (θ )) (r × r˙ ) · r, ¨ 0 = θ˙ sin (θ ) r˙ · r˙ + (1 − cos (θ )) r˙ · r¨ + sin (θ ) ¨r · (r × r˙ ) .

(7.29)

Setting θ = 0, in these equations we find that (7.29)1 implies that θ¨ = 0. In addition, (7.29)2 implies that θ˙ r˙ · r˙ = 0 and (7.29) 3 is identically satisfied. We conclude that if θ (t) = 0 at any instant t, then r ˙ = 0 and r is constant. This is the case of a Type I motion. For motions of this type, ω is parallel to r.

7.11

283

Attitudes of Constant Angular Velocities

To discuss the other types of constant angular velocity motions, we consider the case where θ (t) ² = 0. In this case, the differential equations (7.28) simplify to θ¨ =

sin (θ ) r˙ · r˙ ,

(r × r˙ ) · r ¨ =

r˙ · r¨ = −

0,

θ˙ sin (θ )

1 − cos (θ )

r˙ · r˙ .

(7.30)

The second of these equations implies that r¨ lies in the plane formed by r and r˙ . We next exploit (7.27) and choose an orthonormal set of vectors {E1 , E2 , E 3} so that r and r˙ lie in the plane formed by E2 and E3 : r = cos (ϕ) E 2 + sin (ϕ) E3 ,

r˙ = ϕ˙ cos (ϕ) E3 − ϕ˙ sin (ϕ) E 2,

r × r˙ = ϕ˙ E1.

We will need to determine the function ϕ = ϕ (t) in the sequel. After substituting for r and its derivatives in the differential equations (7.30)1,3 , we find the following pair of differential equations for ϕ and θ : 2 θ¨ = ϕ˙

±

sin (θ ) ,

ϕ¨ = −ϕ˙ θ˙

²

sin (θ ) . 1 − cos (θ )

(7.31)

Analytical expressions for the solutions to (7.31) can be found in [216, Section 4]. Substituting for r and r˙ in the representation for ω, we find that ω = θ˙r + sin (θ ) r˙ + (1 − cos (θ )) r × r˙ = θ˙r +

sin (θ ) r˙ + ϕ˙ (1 − cos (θ )) E1 .

To interpret the solutions to (7.31), we note that they have two integrals: ω · E1

E

=

= ³ = ϕ ˙ (1 −

1 2 ± ω± 2

=

cos (θ )) ,

2 1 ˙2 ³ θ + . 2 (1 − cos (θ ))

That is, the solutions to (7.31) conserve E and ³: E˙ = 0 and ³˙ = 0. We denote the value of these conserved quantities with a subscript 0 in the sequel. As a result, we can reduce (7.31) to a single second-order ordinary differential equation θ¨ =

2 ³ 0 sin (θ )

2 (1 − cos (θ ))

.

(7.32)

The phase portrait of (7.32) is shown in Figure 7.13 and representative solutions to (7.31) are shown in Figure 7.14.

The Three Types of Rotations

We first consider the case where θ (t ) = 0 for some instant t = t1 . It follows that Q (t1 ) = I. The axis of rotation r in this case is constant and we choose E1 = r. Thus, Q=L

(

θ (t) = θ0 + θ˙0 t, E 1

)

,

ω = θ˙0 E1 ,

³0 =

0,

E0 =

1 ˙2 θ , 2 0

284

Kinematics of Rigid Bodies

5 .0 dθ dt

0

−5 0 .

0

2π

θ

The phase portrait θ –θ˙ of the differential equation (7.32) for a constant value of 1. The phase portraits for other values of ³0 ²= 0 are qualitatively similar and can be ( ) obtained by rescaling the θ˙ axis. The equilibrium point at θ = π , θ˙ = 0 corresponds to a Type II motion.

Figure 7.13

³2 0 =

(a) 2π

(b) ν

2π ν

ϕ θ

π

ϕ θ

π

ν

0

0

ν

0

2

t/Tp

(c)

0

2

t/Tp

(d) 2π

2π

ν

ν

ϕ

π

ϕ

π

θ

θ ν

0

ν

0

0

2

t/Tp

0

2

t/Tp

Figure 7.14 Time traces of the representative solutions θ (t) = f θ (t) and ϕ (t) = f ϕ (t) to (7.31). By way of comparison, ν (t) = ±ω± t is also shown. For the results shown in this figure, ³0 = 0.5, ϕ(0) = 0, θ (0) = π , and θ˙ (0) = θ˙0 is varied: (a) θ˙0 = 0, (b) θ˙0 = 0.2, (c) θ˙0 = 1.0, and (d) θ˙0 = 5.0. This figure is adapted from [216].

where θ˙0 is a constant and quaternions is simply ±

q0

=

cos

θ (t)

2

θ0 = θ (t =

0). The representation of this motion using

²

±

,

q = sin

θ (t)

2

²

±

E1

=

sin

θ (t)

2

²

e1.

Thus, the components of the quaternion describe a great circle on a unit three-sphere S3. The corotational vectors e2 = QE2 and e3 = QE3 describe great circles on a unit two-sphere S2 and they traverse these circles at a constant angular rate and a period T p:

7.11

Attitudes of Constant Angular Velocities

Tp =

2π ± ω±

2π

= √

2E0

285

.

The period of the motions of q0 and q is T p/2. Representative examples of the components Qik (t), ek (t) and the quaternion representation for two other Type I motions where we choose E2 = r and E3 = r can be seen in Figure 7.10. ( ) The differential equation (7.32) has a single equilibrium: θ , θ˙ = (π , 0). For this case, which we refer to as a Type II motion, we find from (7.31) that ϕ˙ is a constant which we define as ϕ˙0 (cf. Figure 7.14(a)). To characterize these rotations, we choose E1 to be parallel to ω. Thus, the axis of rotation, rotation tensor Q, unit quaternion, rotation tensor, and angular velocity vector for this case have the representations r(t) = cos (ϕ˙ 0 t + ϕ0 ) E2 + sin (ϕ˙ 0t + ϕ0 ) E3 , Q(t) = 2r(t) ⊗ r(t) − I, q0(t) = 0,

q(t) = r(t), ω

=

Q(t) = 2r(t) ⊗ r(t) − I,

2ϕ˙ 0E 1 = −2ϕ˙0 e1,

where ϕ0 = ϕ (t = 0) and QE1 = −e1 . Whence, the axis of rotation rotates at a constant speed ϕ˙0 about E 1. However, the corotational basis vectors e2 and e3 rotate about E1 at twice this rate: e2 (t)

=

cos (2 (ϕ˙ 0t + ϕ0)) E2 + sin (2 (ϕ˙ 0t + ϕ0 )) E3,

e3 (t)

=

cos (2 (ϕ˙ 0t + ϕ0)) E3 − sin (2 (ϕ˙ 0t + ϕ0 )) E2.

Thus, the period of r(t) is Tp /2 and the periods of e2 (t) and e3 (t) are Tp . Representative examples of the trajectories (spherical indicatrices) of ek (t) and r(t) are shown in Figure 7.11. Type III motions correspond to periodic solutions (θ (t), ϕ(t) ) of (7.31). We denote these solutions by θ (t) = f θ (t) and ϕ (t) = fϕ (t) and a sample of these solutions are shown in Figure 7.14(b)–(d).14 We observe from this figure that the functions f θ (t) and fϕ (t) have period 2Tp . The axis of rotation r(t) lies on( a plane,) rotates at a nonconstant rate ϕ˙ = ³0/ ( 1 − cos (θ )), and is periodic: r (t ) = r t + 2T p . To determine the rotation tensor corresponding to a Type III motion we substitute for r(t) in L (θ , r) and expand terms. After much rearranging, we find that ⎡

⎤

⎡

⎤

e1 E1 ⎣ e2 ⎦ = AT BA ⎣ E2 ⎦ , e3 E3 where the rotation matrices A and B are ⎡

1 A=⎣ 0 0 ⎡

0

0

cos (ϕ (t)) − sin (ϕ (t))

cos (θ (t)) ⎣ 0 B= sin (θ (t))

0 1 0

(7.33)

⎤

sin (ϕ (t)) ⎦ , cos (ϕ (t)) − sin (θ (t))

0

⎤

⎦.

cos (θ (t))

14 As remarked earlier, the explicit expressions for these solutions can be found in [216, Section 4].

286

Kinematics of Rigid Bodies

∑

The matrix Q formed by the components of Q = 3i=1 Qik Ei ⊗ E k is related to A and B by the identity Q = AT BT A. Representative examples of the trajectories (spherical indicatrices) of ek (t) are shown in Figures 7.9(a) and 7.12, and a graphical representation of the angles can be seen in Figure 6.3. In contrast to a Type II motion, e2 and e3 no longer lie in the plane of r(t).15 However, ek (t) describe closed circular paths centered ( ) around the constant angular velocity vector with period T p: ek t + T p = ek (t).

Steiner’s Surface

Steiner’s surface, sometimes known as a Roman surface, dates to the Swiss mathematician Jakob Steiner (1796–1863) who discovered this surface while in Rome in 1844. The surface is formed by a mapping of RP2 into a self-intersecting two-dimensional surface.16 To construct the surface, consider the set of points x, y, and z lying on a unit two-sphere S2 in E3 and map these points using the following function (from Apéry [7]): F (x, y, z) = (xs

=

2yz, ys

=

2zx, zs

=

2xy) .

The point (x, y, z) on the sphere is transformed to a point (xs , ys , zs ) on Steiner’s surface. For us, we identify q0 with x, q2 with y, and q3 with z. Thus, F (q0, q2 , q3) = (2q2 q3 , 2q0 q3, 2q0q2) = ( Q32 =

e2 · E3 , Q21 = e1 · E2 , Q13 = e3 · E1 ) .

We used the representation (6.45) of a rotation tensor to establish relations between Qik and the products of the quaternion components. Thus, if we were to consider the set of all possible Q21, Q13, and Q32 = Q23 associated with a rotation matrix Q with q1 = 0, then they would form a two-dimensional surface known as Steiner’s surface. The mapping F defines a self-intersecting mapping of the real projective space RP2 into E3 . We refer the reader to Section 6.10, where RP2 and its relation to rotations is discussed. We now plot the components Q21 = e1 · E 2, Q13 = e3 · E 1, and Q32 = e2 · E3 for the three types of constant angular velocity motions and find the results shown in Figures 7.10(a)–7.12(a). It is interesting to note that the Type I and Type II motions both appear as straight lines, whereas the Type III motions appear as circles on this surface. We leave it as an exercise to show how the angles θ and ϕ used earlier to describe rotations with constant angular velocities can be used as curvilinear coordinates for the surface. 15 Recall that we chose {E , E , E } so that r(t) would lie in the plane formed by E and E . This choice 1 2 3 2 3 was made without loss of generality and other choices are possible. For instance, one could choose E1 and E2 to define the plane that r(t) describes. 16 For additional details on the topology of this surface, the reader is referred to [88, Figure 3, Chapter 5].

The chapter is titled “Shadows from Higher Dimensions” and contains beautiful illustrations of, and insightful comments on, Steiner’s surface.

7.11

Attitudes of Constant Angular Velocities

287

Quaternions and Slerp

The representation of the Type III motions in terms of the components of a unit quaternion is the simplest known. With a significant amount of algebra using the solutions fθ (t) and fϕ (t), we find that 17 qK (t)

=

q˙ K (0)

qK (0) cos (ς0 t) +

ς0

where 2 ς0 =

E0 2

sin (ς0 t)

=

(K =

0, 1, 2, 3) ,

(7.34)

1 ω · ω. 4

The representations for qK (t) are also valid for the Type I and Type II motions and the periods of qK (t) are twice those of the corotational basis vectors. In addition, the components of the unit quaternion describing the rotation associated with a constant angular velocity motion trace out great circles on the unit three-sphere S3. By choosing the basis vectors {E1 , E2 , E3 } appropriately, one of the components of q can be set to zero and the great circles can be visualized on the unit two-sphere S2 (see Figures 7.10– 7.12 for examples where q · E1 = 0). Spherical linear quaternion extrapolation (Slerp) was proposed by Shoemake [261] in 1985 as a means of interpolating between two rotations. His scheme is preferable to interpolation schemes that use Euler angles, and extensions to his algorithm are used in graphics and animation to this day. We now examine how this interpolation scheme is related to the constant angular velocity motions (7.34) we have just described. Referring to Figure 7.9(c), suppose we are given two rotations characterized respectively by the unit quaternions (a0 , a) and (b0, b ). As shown in [206], the Slerp algorithm generates a rotation between these two orientations that can be described using (7.34). To see this, we suppose that the interpolation starts at time t = 0 and terminates at time t = t1: (q0 (0), q(0)) = (a0 , a) and (q0 (t1 ) , q (t1 )) = (b0 , b ). The angle ´ between the two quaternions is given by the equation cos (´) = a0b0 + a1 b1 + a2 b2 + a3 b3. We now examine the solution (7.34) and use it to describe the sought-after interpolation. After some algebraic manipulations, we find that qK (t) =

1 sin (ς0 t1 )

{a K

sin (ς0 (t1 − t)) + bK sin (ς0 (t ))}

(K =

0, 1, 2, 3) , (7.35)

where the time t1 of the interpolation and the angular rate ς0 are specified by the relation ς0t 1 = ´.

The formula (7.35) is none other than the Slerp algorithm. In conclusion, this algorithm is based on constant angular velocity motions. The simplicity of (7.35) often obfuscates the possibility that r and ω may not be parallel. 17 An alternative derivation of these results can be found in [206].

288

Kinematics of Rigid Bodies

7.12

Closing Comments

We have now assembled most of the kinematical quantities required for characterizing the motion of a rigid body. The precise representations for these quantities that are used generally depend on the problem at hand. For instance, we shall later use a set of Euler angles for a spinning-top problem that differs from a set used for a problem featuring a satellite. Another example arises when we examine the dynamics of a rolling sphere. There, we will choose to use a fixed-basis representation of ∑ its angular velocity vector, ω = 3i =1 µi E i , whereas we will use the Euler basis for the satellite problem. These choices are guided by experience and are often not intuitively obvious. It is hoped that the examples in the exercises and chapters ahead will help you gain this needed experience. You might also have noticed that we have yet to discuss constraints on the motions of rigid bodies. The next chapter is devoted to this topic.

7.13

Exercises

Exercise 7.1: Recall that the rotation tensor Q of a rigid body has the representations Q=

3 ³

ei

3 3 ³ ³

⊗ Ei =

Qik ei ⊗ ek

=

Qik Ei

⊗ Ek.

i=1 k=1

i =1 k =1

i=1

3 3 ³ ³

We recall that the corotational rates of a vector a and a tensor A relative to the rotation tensor Q are defined as ◦

◦

a = a˙ − ω × a,

˙ − ΩA + AΩ. A=A

(

)

˙ T and ω = ax QQ ˙ T . In these expressions, Ω = QQ

(a) If ai

=

a · ei and A ik

= ( Aek ) · ei , ◦

a=

3 ³

then show that 3 ³ 3 ³

◦

a˙ i ei ,

A

i=1

=

A˙ ik ei

⊗ ek .

i=1 k=1

Give physical interpretations of these results. (b) A vector z is said to be corotational if z = QZ, where Z is constant. Give examples of such vectors from rigid-body dynamics and show that the corotational rate relative to Q of a corotational vector is 0. (c) Construct a definition of a corotational tensor and give two prominent examples of such tensors from rigid-body dynamics. (d) Establish the following identities: ◦

ω ˙ = ω,

◦

˙ = QT QQ. Q

7.13

g3

Exercises

289

Robotic arm Axle

Payload

θ φ

g2

XP

Drive shaft

ψ

E3 O

E2

E1 Schematic of a robot consisting of a robotic arm, drive shaft, and axle. The motors used to actuate the robot are not shown.

Figure 7.15

(e) Suppose a rigid body is in motion and the corotational rate of ω is 0. What is the angular acceleration of the rigid body and what is the rotation tensor of the rigid body?18 Exercise 7.2: As shown in Figure 7.15, a robotic arm can be used to move a payload. The robot consists of a drive shaft that rotates about the E3 axis through an angle ψ , an axle A that rotates about g2 through an angle θ relative to the drive shaft, and an arm that rotates about g3 through an angle φ relative to axle A. The payload is rigidly attached to the robotic arm. In this question, the drive shaft, the axle A, the robotic arm, and the payload are assumed to be rigid. (a) What are the angular velocity vectors of the drive shaft, the axle A, the robotic arm, and the payload? (b) Which set of Euler angles is being used to parameterize the rotation tensor Q of the payload? (c) If the position vector r of a point X on the payload is r = HE3 + ² g3 , then establish an expression for the velocity vector v of the point X. (d) Suppose after a time interval t 1 − t 0 the point X has returned to its original location in space: r(t1 ) = r(t0 ). 18 Solutions to this problem can be found in Novelia and O’Reilly [206] and O’Reilly and Payen [216].

Some of their work is presented in Section 7.11.

290

Kinematics of Rigid Bodies

E3 g

E1

e2

E2

O

ψ

e1

e2

= e1 ¯

X

e1 φ

e1

= e1

XP

The present configuration of a circular disk moving with one point in contact with a fixed horizontal plane. A set of 3–1–3 Euler angles is being used to parameterize the rotation of the disk shown in the figure.

Figure 7.16

Show that the payload will have rotated through an angle φ(t1 ) − φ(t 0) about the axis g3(t0) during this interval of time. In other words, the rotation tensor Q(t1 )QT (t0 ) corresponds to a rotation of φ (t1 ) − φ(t0 ) about g3 (t0). Exercise 7.3: Consider the circular disk shown in Figure 7.16. The motion of the disk is given by the position vector y of an arbitrary material point Y of the disk and the rotation tensor Q of the rigid disk. (a) Starting from the results that the motions of any points X and Y of the disk have the representations x = QX + d,

y = QY + d,

show that their relative velocity vector and relative acceleration vector satisfy x˙ − y˙ = ω × (x − y) , x¨ − y¨ = ω ˙ × ( x − y ) + ω × ( ω × (x − y)) . (b) To parameterize the rotation tensor of the disk, a set of 3–1–3 Euler angles is used. With the assistance of Figure 7.16, prescribe a reference configuration for the disk. For which orientations of the disk in its present configuration do the singularities of the 3–1–3 Euler angles occur? (c) A sensor is mounted onto the disk and is aligned with the e3 axis so that it measures ω · e3 = ω3 (t). Show that, in general, · t 1 t0

ω3 (t)dt ² = φ (t1 ) − φ (t0 ).

Give a physical interpretation of the case in which the integral of the angle φ.19 19 An application of (7.36) to the navigation of motorcycles can be found in [52].

(7.36) ω3(t)

does yield

7.13

291

Exercises

Exercise 7.4: Consider a rigid body of mass m whose inertia tensors are defined as ·

J = QJ0 QT ,

J0

=

(Π ·

R0

Π) I − Π ⊗ Π

(a) Explain why J0 and J are symmetric. ∑ (b) Show that Ei · (J0 Ek ) = ei · (Jek ), where Q = 3i=1 ei (c) Show that J has the representation J=

3 ³

ρ 0dV.

⊗ Ei .

λi ei ⊗ ei,

(7.37)

i =1

where λ i are the principal values of J0 and Ei J0 . (d) Establish the following identities: J˙ = Ω J − JΩ ,

˙ H

=

=

QT ei are the principal directions of

˙ + ω × (J ω), Jω

˙ · ω. Jω˙· ω = 2H

How do these results simplify if J0 = μ I, where μ is a constant? Note that, in the first of these results, you are showing that the corotational rate of J relative to Q is ◦

zero: J = 0. (e) Suppose one used the representation J=

3 3 ³ ³

Jpn Ep ⊗ En .

p=1 n=1

Starting from (7.37), show that Jpn

3 ³ =

Qpi λi Qni ,

i=1

where Qpi = (QEi ) · Ep. Why are the components Jpn of J not constant? Exercise 7.5: Recall the definitions of the Euler tensors: E

=

QE0QT ,

·

E0

=

R0

Π ⊗ Π ρ 0dV.

(a) Show that J0

=

tr(E 0)I − E0,

J = tr(E)I − E.

(b) Verify that Ei · (E0 Ek ) = ei · (Eek ), where Q = (c) Establish the following results: ◦

E = 0, ◦

∑3

i=1 ei ⊗

Ei .

˙ = ΩE − EΩ, E

where E denotes the corotational derivative of E with respect to Q.

292

Kinematics of Rigid Bodies

(d) What are the Euler tensors for a sphere of mass m and radius R, and a cylinder of mass m, radius R 0, and length ²? You might find it convenient to use the relationships of the form E0 = 21 tr (J0 ) I − J0 that were discussed earlier in this chapter, in Section 7.8. Exercise 7.6: Recall that the angular momentum H and the rotational kinetic energy T rot of a rigid body have the representations H = J ω,

Trot

=

1 ω · J ω. 2

Here, J is the moment of inertia tensor of the rigid body relative to its center of mass. Choosing {Ei } to be the principal directions of J0: J0 = where Q =

∑3

3 ³

λi Ei ⊗

J = QJ0 QT

Ei ,

=

i= 1

3 ³

λi ei ⊗ ei ,

i=1

i=1 ei ⊗ Ei .

(a) Using a set of 3–2–3 Euler angles to parameterize Q,20 show that ω=

(

˙ θ˙ sin(φ ) − ψ

(

)

sin(θ ) cos(φ) e1 )

+ θ˙

cos(φ) + ψ˙ sin(φ) sin( θ ) e2

˙ + ψ

cos( θ ) + φ˙ e3.

(

)

(b) Using a set of 3–2–3 Euler angles, establish expressions for H and Trot as functions of the Euler angles and their time derivatives. Exercise 7.7: Recall the definitions of the angular momenta of a rigid body relative to its center of mass and a point A: ·

H=

R

·

π × vρ dv,

HA

=

R

πA × vρ dv.

(7.38)

Here, π = x − x¯ and πA = x − xA , where xA is the position vector of the point A and x¯ is the position vector of the center of mass. (a) Starting from (7.38)2, and with the assistance of (7.38) 1, show that HA

=

H + (x¯ − xA ) × G,

where G is the linear momentum of the rigid body. (b) Suppose that A is a point of the rigid body: A = XA . Show that πA

=

QΠA ,

where ΠA = X − XA and XA is the position vector of XA in the reference configuration. If vA = 0, show in addition that v = ω × (x − xA ). 20 This set of Euler angles is discussed in Exercise 6.2. In particular, the reader is referred to (6.53) in

Exercise 6.2(g).

7.13

(c) Again supposing that A is a point of the rigid body and that vA HA

=

293

Exercises

=

0, show that

JAω,

where JA is the inertia tensor of the rigid body relative to XA : J

A

·

=

R

(πA · πA )I − πA ⊗ πA ρ dv.

Verify that JA where JA0 =

=

QJ A0 QT ,

·

R0

(( ΠA · ΠA )I −

ΠA

⊗ ΠA ) ρ0 dV.

(d) Using the previous results, prove the parallel axis theorem:21 JA0

=

(

)

¯ − XA ) · (X ¯ − XA )I − (X ¯ − XA) ⊗ (X ¯ − XA ) . J0 + m (X

(7.39)

You should be able to see from this result how a parallel axis theorem relating two Euler tensors could be established. (e) Suppose that the rigid body is a circular cylinder of mass m, length ², and radius R, where ¯ = 0, X XA = −xE1 − zE3 , m² 2 mR2 J0 = (I − E3 ⊗ E3 ) + (I + E3 ⊗ E3 ) . 12 4

What then is JA0 ? An application of this result to a model for a spherical robot can be found in Section 11.12. Exercise 7.8: Consider the rectangular parallelepiped of mass m and dimensions a, b, and c shown in Figure 7.17. Relative to the principal axes {A1, A2 , A3 }, the inertia tensor of this rigid body has the representation

A3 XP

E3

¯

X

c

A2

A1

a b

Figure 7.17

E2

E1

Schematic of a parallelepiped.

21 This representation of the parallel axis theorem is due to Fox [87].

294

Kinematics of Rigid Bodies

J0 =

3 ³

λi Ai ⊗

Ai ,

i =1

where λ1 =

µ m ´ 2 b + c2 , 12

λ2 =

µ m´ 2 a + c2 , 12

λ3 =

µ m ´ 2 a + b2 . 12

The right-handed set of basis vectors {E 1, E 2, E 3} are oriented so that E3 passes diagonally through the parallelepiped. That is, E 3 is parallel to the line connecting X¯ and the point XP : 1 (aA1 − bA2 + cA3) . E3 = √ 2 a + b 2 + c2 In addition, E1

=

cos(α )A1 − sin(α)A3 ,

where α=

´aµ

tan−1

c

.

(a) After calculating an expression for E2 , verify that the transformation from the basis {A1 , A2 , A3 } to the basis { E1 , E2 , E3 } can be written in the form ⎡

⎤

⎡

⎤

E1 A1 ⎣ E2 ⎦ = R ⎣ A2 ⎦ , E3 A3 where

⎡

⎤

R 11 ⎣ R 21 R= R 31

R 12 R 22 R 32

1 ⎣ = 0 0

0 cos( β) − sin(β )

⎡

and

√

cos(β ) =

√

R13 R23 ⎦ R33

⎤⎡

0 cos(α) 0 ⎦ ⎣ sin(β) 0 1 cos(β ) sin(α) 0

a 2 + c2

a2 + b2 + c2

(b) Show that the components J0ik

(7.40)

,

sin(β ) =

= ( J 0Ek ) ·

J0ik

=

3 ³

√

b a2 + b2 + c2

Ei are Rir λr Rkr .

r =1

Using a matrix notation, these equations are equivalent to ⎡

J011 ⎣ J0 12 J013

J012 J022 J023

⎡

⎤

J013 J023 ⎦ J033

=

λ1

R⎣ 0

0

0 λ2

0

⎤

sin(α) ⎦ 0 cos( α)

−

⎤

0 0 ⎦ RT . λ3

.

7.13

295

Exercises

(c) If the body is given an angular velocity ω = ωe3, where ei = QEi , then establish expressions for the angular momentum H of the body relative to its center of mass and the rotational kinetic energy Trot of the body. How do these expressions simplify if a = b = c? Exercise 7.9: In this exercise, you will establish results for a set of Euler angles: the 3–1–2 set. This set of angles is similar to the 3–2–1 set discussed in Section 6.8.1. We will then apply the results of the exercise to solve a navigation problem. We shall decompose the rotation tensor Q into the product of three rotation tensors. The first rotation is about E3 through an angle ψ : L (ψ , E3 ). The second rotation is an ) ( angle θ about e´1 = cos(ψ )E 1 + sin(ψ )E2: L θ , e´1 . The third rotation is through an angle φ about e2 (a) Show that

⎡

=

e´´2

⎤

=

cos(θ )e´2 + sin(θ )e´3 .

⎡

e1 Q11 ⎣ e2 ⎦ = ⎣ Q12 e3 Q13

Q21 Q22 Q23

⎤⎡

⎤

Q31 E1 Q32 ⎦ ⎣ E2 ⎦ , Q33 E3

where Qij = (QEj ) · E i . (b) Show that the Euler basis for the 3–1–2 set of Euler angles has the representation ⎡

⎤

⎡

g1 ⎣ g ⎦=⎣ 2 g3

⎤⎡

⎤

E1 1 ⎦ ⎣ 0 E2 ⎦ . E3 sin(θ )

0 0 cos(ψ ) sin(ψ ) − cos( θ ) sin(ψ ) cos(θ ) cos(ψ )

Using the relations gk · gi = δ ik , show that the dual Euler basis for the 3–1–2 set of Euler angles has the representation ⎡

⎤

g1 ⎣ g2 ⎦ g3

⎡

sin(ψ ) tan(θ ) ⎣ = cos( ψ ) − sec(θ ) sin( ψ )

⎤⎡

⎤

cos(ψ ) tan(θ ) 1 E1 ⎦ ⎣ sin(ψ ) 0 E2 ⎦ . sec( θ ) cos(ψ ) 0 E3

−

(

)

You should notice that the second Euler angle needs to be restricted to θ ∈ − π2 , π2 . (c) By examining the three rotations that compose Q and with the help of the results in (a), show that ⎡

Q11 ⎣ Q21 Q31 where

Q12 Q22 Q32

⎡

cos(φ ) A=⎣ 0 − sin(φ) ⎡

⎤

Q13 Q23 ⎦ Q33

1 0 ⎣ B= 0 cos( θ ) 0 sin(θ )

0 1 0

=

CBA,

⎤

sin(φ) ⎦, 0 cos(φ) ⎤

0 − sin(θ ) ⎦ , cos( θ )

(7.41)

296

Kinematics of Rigid Bodies

⎡

cos(ψ ) ⎣ C= sin(ψ ) 0

⎤

sin(ψ ) 0 cos( ψ ) 0 ⎦ . 0 1

−

(d) Select four different values of the set (φ , θ , ψ ) and determine the axis of rotation q and the angle of rotation of Q. (e) The angular velocity vector associated with the 3–1–2 Euler angles has the representations ω = ψ˙ E3 + θ˙ e´1 + φ˙ e2

= ω1 e 1 + ω2 e 2 + ω3 e 3 .

Show that ⎡

ω1

⎤

⎡

⎣ ω2 ⎦ = ⎣ ω3

and

⎡

˙ ψ

⎤

⎣ φ˙ ⎦ θ˙

⎡ =

⎣

− sin(φ ) cos(θ )

sin(θ ) cos(φ ) cos(θ )

− sec( θ ) sin(φ )

tan(θ ) sin( φ) cos(φ )

0 1 0

0 1 0

⎤⎡

cos( φ) ⎦⎣ 0 sin(φ ) ⎤⎡

sec(θ ) cos(φ ) − tan( θ ) cos(φ ) ⎦ ⎣ sin(φ)

˙ ψ

φ˙

⎤ ⎦

θ˙

ω1 ω2

⎤ ⎦.

(7.42)

ω3

These sets of differential equations are very important in strap-down navigation systems. Indeed, (7.42) are known as the navigation equations: given signals ωi from gyroscopes and initial values of the Euler angles, they can be integrated to determine the orientation (or attitude) of the rigid body that the gyroscopes are mounted to. (f) Suppose that a set of 3–1–2 Euler angles is used to parameterize the rotational ∑ motion of a vehicle: Q = 3i =1 ei ⊗ Ei . Here, e1 is taken to point in the forward direction along the vehicle and E3 is taken to point vertically downward. The first Euler angle ψ is known as the yaw angle, the second angle θ is the roll angle, and the third angle φ is called the pitch angle. (i) A set of three gyroscopes is mounted on the vehicle and provides three signals 22 ωi (t) (see, for example, the signals shown in Figure 7.18). Argue that, with knowledge of the initial orientation of the vehicle, the orientation of the vehicle can be determined by integrating (7.42). (ii) Suppose the vehicle is initially level. As time progresses, the gyroscopes register the following signals: ω1 (t) =

0.02 sin(0.5t),

ω2 (t) =

0.02 sin(0.05t),

ω3 (t) =

2.0 sin(0.5t).

Determine ek (t) for the vehicle. (iii) Suppose the rotation tensor of the vehicle Q = L (ν , q ). How does the axis of rotation q of the vehicle change and is this axis parallel to the instantaneous ω ? axis of rotation i = ±ω ±

7.13

297

Exercises

100 o/s

−100

ω1

+ 40

ω2

− 40

ω3

0

time (s)

100

Time traces of ω3 (t), ω2 (t) − 40, and ω1 (t) + 40 from a set of three gyroscopes mounted on a motorcycle. The motorcycle is moving in a rectangular path and the offsets ³40 are imposed so that the three signals can be distinguished. It is also interesting to observe the amount of noise in the signals for ωi (t).

Figure 7.18

e3 e1

¯

X

κt

E3 g

E3

O

E1

¯

e2

E2

XP

X

κ0

E2

XP

E1 The fixed reference κ0 and present κt configurations of a Poisson top. The surface that the top moves on is taken to be a horizontal plane spanned by E1 and E2 . The image on the left-hand side is a portrait of Siméon Denis Poisson (1781–1840).

Figure 7.19

Exercise 7.10: As shown in Figure 7.19, a Poisson top is an axisymmetric body with a sharp apex that is free to move on a flat surface. The material point of the top in contact with the surface is labeled X P . The position vectors of this point relative to the fixed point O in the present and reference configurations are denoted by xP and XP , respectively. ¯ The material point corresponding to the center of mass of the top is denoted by X, and the position vectors of X¯ relative to the fixed point O in the present and reference ¯ respectively. configurations are denoted by x¯ and X, 22 The principle of operation of a class of gyroscopes is discussed in Section 11.14.

298

Kinematics of Rigid Bodies

(a) A set of 3–2–1 Euler angles is used to describe the orientation of the top. Denoting the first angle by ψ , the second by θ , and the third by φ , give sketches showing how these angles describe the orientation of the top. (b) For which orientations of the top does the set of 3–2–1 Euler angles have singularities? Now, suppose a set of 3–1–3 Euler angles were used to parameterize the rotation tensor of the top. For which orientations of the top does the 3–1–3 set have singularities? (c) Starting from the result that for any point X on a rigid body x = QX + d, show that x = Q (X − XP ) + xP,

x˙ = ω × (x − xP ) + x˙ P .

(d) The moment of inertia tensor of the top in its reference configuration has the representation J0 = λ aE3 ⊗ E3 + λ t (I − E3 ⊗ E3 ) . If ΠP

=

¯ = −² E , XP − X 3 3

ω = ω1 e1 + ω2e2 + ω3 e3 ,

then show that the angular momentum vector of the top relative to the point X P is HP

= λa ω3e3 +

´

λt +

m² 23

µ

˙ P. (ω 1e1 + ω2 e2 ) − ²3 e3 × mx

8

Constraints on and Potential Energies for a Rigid Body

8.1

Introduction

In this chapter, constraints and the forces and moments that enforce them in the dynamics of a single rigid body are discussed. In particular, the constraints associated with pin-jointed rigid bodies, rolling rigid bodies, and sliding rigid bodies, along with prescriptions for the associated constraint forces and moments, are presented. We also discuss Lagrange’s prescription for constraint forces and constraint moments and outline its limitations. The treatment of constraints and the forces and moments that enforce them is adapted from [218–220]. In this chapter, it also proves convenient to discuss potential energies and their associated conservative forces and conservative moments. Our discussion includes as examples springs and central gravitational fields and has obvious parallels to the treatment of constraints and their associated forces and moments.

8.2

Forces and Moments Acting on a Rigid Body

Before discussing constraint forces and constraint moments, we dispense with some preliminaries. The resultant force F acting on a rigid body is the sum of all the forces acting on a rigid body. Similarly, the resultant moment relative to a fixed point O, MO , is the resultant external moment relative to O of all of the moments acting on the rigid body. We denote the resultant moment relative to the center of mass X¯ by M. The moments M and MO may be decomposed into two additive parts: the moment that is due to the individual external forces acting on the rigid body and the applied external moments that are not due to external forces. We refer to the latter as “pure” moments. As an example, consider a system of forces and moments acting on a rigid body. Here, a set of L forces FK (K = 1, . . . , L) acts on the rigid body. The force FK acts at the material point XK which has a position vector xK . In addition, a pure moment Mp , which is not due to the moment of an applied force, also acts on the rigid body (see Figure 8.1). For this system of applied forces and moments, the resultants are

300

Constraints on and Potential Energies for a Rigid Body

FK κt XK

πK

Mp

¯

xK

X

x¯ O

Figure 8.1

A force FK acting at a point XK and a moment Mp acting on a rigid body.

F=

L ±

FK ,

K =1

MO

=

Mp +

M = Mp +

L ± K=1 L ±

xK

×

FK ,

(xK − x¯ ) × FK .

K=1

You should notice how the pure moment M p features in these expressions. The mechanical power P of a force FK acting at a material point X K is defined to be

P = FK · x˙ K . Using the center of mass, we can obtain a different representation of this power. Specifically, we use the identity x˙ K = v¯ + ω × πK , where πK = xK − x. ¯ It then follows that

P

=

FK · x˙ K

=

FK · v¯ + (π K × F K ) · ω.

In words, the power of a force is identical to the combined power of the same force acting at the center of mass and its moment relative to the center of mass. The mechanical power of a pure moment M p is defined to be

P

=

Mp · ω.

You should notice how this expression is consistent with the previous expression for the mechanical power of a force FK . Using the results for the mechanical power of a force FK and a pure moment M p, we find that the resultant mechanical power of the system of L forces and a pure moment discussed previously is

P=

L ± K=1

FK · x˙ K + M p · ω = F · v¯ + M · ω.

8.3

Examples of Constrained Rigid Bodies

301

These results will play a key role in our future discussion of constraint forces and moments and potential energies.

8.3

Examples of Constrained Rigid Bodies

To set the stage for our discussion of constraint forces and moments, we examine a range of problems from rigid-body dynamics. We start with a familiar example and then turn to examples of rolling and sliding rigid bodies and a more complex example of a whirling pendulum. Planar Motion of a Body About a Fixed Point

For our first example, we consider a body which is connected by a revolute joint to the ground at a fixed material point X A (cf. Figure 8.2). The body is free to perform planar rotations about an axis of rotation E3 . However, because XA is fixed and the body cannot rotate about the planar directions E 1 and E2, its motion is subject to five constraints. The most convenient method to express these constraints uses the velocity vector of the point XA and the angular velocity vector of the rigid body: E 1 · vA

=

0,

E 2 · vA

E1 · ω = 0,

=

0,

E3 · vA

E2 · ω = 0.

=

0, (8.1)

To make sure that XA stays fixed, a reaction force RA acts at this point. This reaction force has three independent components, each one restricting the translation of X A in

Fc

E2

E1

XA

Mc e2

e1

A rigid body rotating in a plane about a fixed point X A . The constraint force Fc and constraint moment Mc which ensure that the point XA remains fixed and the axis of rotation is constrained to be E3 by a revolute joint at X A , respectively, are also shown.

Figure 8.2

302

Constraints on and Potential Energies for a Rigid Body

a given direction. The body also cannot rotate about any axis other than E3. Hence, the joint needs to provide a reaction moment that prevents rotation in the E1 and E2 directions. We refer to RA as a constraint force, Fc = RA , and the reaction moment provided by the joint as a constraint moment Mc . In addition to the reaction force’s three independent components, the reaction moment should have two independent components which model the resistance of the revolute joint to rotations in the E1 and E2 directions. In conclusion, we prescribe Fc Mc

3 ± =

µ k Ek

acting at the point XA ,

k=1

= µ 4E1 + µ 5E 2.

(8.2)

Here, the components µ1 , . . . , µ5 remain to be found using the balances of linear and angular momentum (or, equivalently, Lagrange’s equations of motion). For example, in an undergraduate dynamics course it is common to use F = ma to determine the joint reaction force Fc (i.e., µ1 , µ 2, and µ 3). The E3 component of the balance of angular ˙ A , is also used to determine the differential momentum relative to the point X A , MA = H equation governing the motion of the rigid body. Less standard in textbooks is to use the ˙ to determine µ and µ .1 E1 and E 2 components of MA = H A 4 5 It is useful at this stage in our discussion to point out that the prescription (8.2) assumes that the revolute joint (or pin joint) is free of Coulomb friction. If friction were present then a moment −k ±F c ± ω/ ±ω±, where k ≥ 0 is a constant, might be added to the expression for Mc in (8.2)2: Mc

= µ4 E1 + µ 5E2 −

k ±Fc ±

ω ±ω±

.

(8.3)

The reader might also note the similarities between the constraints (8.1) and the expressions (8.2) for Fc and Mc . For example, observe the fact that there are five constraints and five quantities µ 1, . . . , µ 5 and that the vectors associated with the µs can be inferred from the constraints.

Sliding Cylinder

As a second example, consider a cylinder sliding on a smooth horizontal surface as shown in Figure 8.3. The normal to the surface is E3 and the axis of symmetry of the cylinder is e2. Clearly, the center of mass X¯ of the cylinder can only move horizontally and the cylinder cannot rotate into the plane. Following [214], these two constraints can be expressed as 2 E3 · v¯ = 0,

(e2 ×

E3) · ω = 0.

(8.4)

1 An example of such a calculation can be found in [215, Chapter 9]. 2 A derivation of this pair of constraints, starting with the assumption of line contact between the cylinder

and the plane, can be inferred from the discussion of a semicircular cylinder rolling on a horizontal surface in Section 11.13.

8.3

Examples of Constrained Rigid Bodies

303

E3

O

g

Fc E1

E2

¯

X

e2 × E3

Mc

e2

Figure 8.3 A circular cylinder sliding on a smooth horizontal surface. The constraint force F c and constraint moment Mc which ensure that the center of mass X¯ moves in the horizontal plane and the cylinder does not rotate into the plane, respectively, are also shown.

In the second constraint, the vector e2 × E3 has the property that it is always normal to the axis of symmetry of the cylinder and lies in the horizontal plane. If we were to use a 3–2–3 set of Euler angles to describe the motion of the cylinder, then the constraint (8.4)2 could be expressed in the equivalent forms ω · g3

=

0,

φ˙ =

0,

φ − φ0 =

0,

where φ0 is a constant.3 If we consider all possible orientations of the cylinder sliding on the plane, then we conclude that we must allow θ to range from 0 to 2π instead of the usual range from 0 to π . We term the resulting coordinate θe an extended coordinate: ψ ∈

[0, 2π ] ,

θe ∈

[0, 2π ] .

M

2 2 ∼ The configuration manifold for the sliding cylinder is = E × T . Here, the generalized coordinates θe and φ provide a one-to-one covering of the two-torus T 2 that is free of singularities. Along the line of contact of the cylinder and the surface, a vertical (or normal) force field will be present. This force field ensures that the motion of the cylinder satisfies (8.4). The force field appears in the equations of motion of the cylinder as a resultant force (or normal force) and a resultant moment. The resultant force will be in the E3 direction and act at the center of mass and the moment will be perpendicular both to the contact line and the axis of the cylinder. We summarize these observations by noting that the resultant force is a constraint force F c acting at X¯ and the moment is a constraint moment Mc where 3 An explicit expression for g3 can be inferred from Exercise 6.2: g3

= cosec (θ ) (cos (ψ ) E1 + sin (ψ ) E2 ). You should verify that this vector is perpendicular both to the axis of symmetry e 2 of the cylinder and the normal E3 to the plane that the cylinder is moving on.

304

Constraints on and Potential Energies for a Rigid Body

E3 g

E1

e2

E2

O

e1

ψ

= e1

e2

¯

X

e1

φ

e1

XP

Fc = N = µE 3 A circular disk of radius R moving on a smooth horizontal plane. The position vector of the instantaneous point of contact X P of the disk and the plane relative to the center of mass X¯ of the disk is always along e²²2 . A set of 3–1–3 Euler angles is used to parameterize the rotation of the disk. The angle θ is the angle of inclination of the disk from the vertical.

Figure 8.4

Fc Mc

= µ 1E3

¯ acting at the point X,

= µ 2 ( e2 ×

E3) .

The quantities µ1 and µ 2 must be determined from the balance laws.4 Sliding Disk

Consider formulating the equations of motion for the classic problem of a thin circular disk shown in Figure 8.4 that is sliding with a single point XP in contact with a smooth horizontal surface. There is a single constraint on the motion of the disk that can be represented in a variety of manners. If a set of Cartesian coordinates is used to parameterize the position vector of the center of mass, x¯ = xE1 + yE2 + zE3 , and a set of 3–1–3 Euler angles is used to parameterize the rotation tensor Q of the disk, then the constraint on the motion of the disk can be expressed in one of the following equivalent four fashions: z − R sin(θ ) vP · E3

=

0,

=

0,

E3 · v¯ +

z˙ − Rθ˙ cos (θ ) = 0, (

−Re2 × ²²

E3

)

·ω =

0.

To ensure that the constraint is satisfied, a normal force acts at the point XP : Fc

=

N = µE 3 acting at the point XP ,

Mc

=

0.

This system is equipollent to a force Fc acting at the center of mass and a moment ²² The corresponding developments for a rolling disk are discussed in Exercise 8.8. We invite the reader to examine the Poisson top shown in Figure 7.19. The position constraint on this top can be expressed as a velocity constraint: vP · E3 = 0. Furthermore, if the top is in motion on a smooth surface, then Fc = N = µ E3 acting at the point X P and Mc = 0.

−Re2 × Fc .

4 The reader is referred to Exercise 10.3(g) on p. 421 where solutions for F and M can be found. c c

8.3

Examples of Constrained Rigid Bodies

305

E3 g

E2

O

E1

¯

X

XP

Fc = N + Ff A rigid body rolling without slipping on a rough horizontal plane. The constraint force Fc acting at the instantaneous point of contact XP (where vP = 0) can be decomposed into a normal force N and a friction force F f : F c = N + Ff .

Figure 8.5

Rolling and Sliding Rigid Bodies

As a third example, consider a rigid body in motion with a single instantaneous point XP of contact with a surface (see, e.g., Figure 8.5). If the body is sliding, then there is a single constraint: n · vP

=

f (t),

where n is the normal vector to the surface at XP and f (t) is a prescribed function of time which vanishes if the surface is fixed. The vector n depends on the location of the rigid body and time (if the surface is moving). For example, for the sliding disk shown in Figure 8.4, n = E3 and f = 0. At the point of contact, a normal force N is exerted on the sliding body. This normal force is a constraint force and so we prescribe Fc

=

N = µn acting at the point XP ,

Mc

=

0.

Note that we are assuming that the constraint at XP is enforced purely by a force acting at that point. If we assumed dynamic friction was also present at the point, then a friction force would be added to Fc and a frictional moment would be added to Mc : Fc

=

N + Ff

= µn − µd ± µn ±

Mc

3 ± = −

vrel acting at the point XP , ±v rel ±

ki ±µ n± sgn (ω · Ei ) Ei ,

(8.5)

i =1

where ki ≥ 0 are constants that must be determined by experiments and vrel is the velocity vector of XP relative to the surface. For an example of this prescription, see the discussions of the dynamics of Euler’s disk in [146, 165, 166]. For a rigid body rolling on the surface mentioned previously, one has three constraints: Ek · vP − vs · Ek

=

0

(k =

1, 2, 3) .

306

Constraints on and Potential Energies for a Rigid Body

Here, vs is the velocity vector of the point on the surface that is coincident with XP . Rolling contact is maintained by static Coulomb friction Ff , while a normal force N ensures that the rolling body does not penetrate the surface. The resultant of these forces is the constraint force acting on the rigid body: Fc

=

N + Ff

=

3 ±

µk E k

acting at the point X P,

Mc

=

0.

k=1

The three unknowns µ1 , µ 2, and µ3 reflect the fact that the normal force has one unknown component and the friction force has two independent components. It may be helpful to recall that part of the solution procedure when solving for the motion of a rolling rigid body involves solving the unknowns F f = µ 1E 1 + µ2 E²2 and N = µ 3E3 . ² The static friction is limited by the static friction criterion: µs ±N± ≥ ²Ff ². The problems in rigid-body dynamics involving rolling or sliding rigid bodies that have received the most attention involve rigid bodies for which the vectors πP = xP − x¯ have relatively simple forms: specifically, rolling/sliding spheres, the sliding top, and rolling/sliding disks. The most famous example for which π P is very difficult to express in a tractable form is the wobblestone or celt. As first reported by G. T. Walker in 1896 [300], this rigid body has the unique feature of reversing its direction of spin in a counterintuitive manner.5

A Whirling Rigid Body

Consider the rigid body in Figure 8.6 which is attached to a spinning shaft by a revolute joint at XA .6 The rotation of the shaft is parameterized by a set of 3–1–3 Euler angles. The shaft is subject to a prescribed angular speed ±(t) and the rigid body is subject ˙ = Ω

ψ

g

Fc

E2 E1

XA

Mc E3

θ

¯

X

e3 A whirling rigid body. The constraint force F c = RA acting at X A and the constraint moment Mc = µ4g1 + µ5 g3 are also shown.

Figure 8.6

5 For further details on this interesting mechanical system, and more recent research on it, see [14, 25] and

references therein.

6 This example is inspired by the works of Ne˘ımark and Fufaev [201] and Rumyantsev [250], who used

simple systems of this type to explore bifurcations and stability criteria for steady motions of mechanical systems.

8.4

Constraints and Lagrange’s Prescription

307

to five constraints. Although all of these constraints are integrable, the most compact method to express the constraints is as follows: Ek · vA

=

0

(k =

g1 · ω − ± = 0,

1, 2, 3 ) ,

g3 · ω = 0.

Here, we use a set of 3–1–3 Euler angles (ψ , θ , φ) to parameterize the rotation of the rigid body. Note the use of the dual Euler basis to express the constraints in the most compact form possible. Expressions for these basis vectors are recorded in (6.39). To make sure that XA stays fixed, a reaction force RA acts at this point. As in our first example, this reaction force has three independent components, each one restricting the translation of XA in a given direction. The body is only free to rotate in the g2 direction and so we need constraint moments in directions perpendicular to g2 to ensure this freedom is preserved. Because g1 · g2 = 0 and g3 · g2 = 0, we thus prescribe Fc

=

3 ±

µ k Ek

acting at the point XA ,

k=1

Mc

= µ 4g

1

3

+ µ5 g

.

The constraint moment µ 4g1 can be interpreted as the motor torque needed to drive the shaft at its prescribed speed. The power required to achieve this is given by µ4 g1 · ω = 1 µ4 ±. We also note that g is never parallel to g 1.

8.4

Constraints and Lagrange’s Prescription

Traditional prescriptions for Fc and Mc are either couched in terms of physical arguments, as discussed in the examples in the previous section, or involve generalized constraint forces and a virtual work assumption (which is difficult to motivate and even more difficult to teach). An alternative formulation of Lagrange’s prescription can be obtained by combining aspects of both prescriptions as follows. Suppose that a constraint πC = 0 on the motion of a rigid body can be expressed in the form πC = f C · v C + h C · ω + eC,

(8.6)

where vC is the velocity vector of a material point XC on the body, and the functions fC , hC , and eC depend on Q, xC, and t (or equivalently, Q, x¯ , and t). Then, we can define Lagrange’s prescription for Fc and M c as Fc

= µ

Mc

= µ

∂ πC ∂ vC ∂ πC ∂ω

= µ fC

acting at the point XC ,

= µ hC ,

where µ is a function which is determined by the equations of motion.

(8.7)

308

Constraints on and Potential Energies for a Rigid Body

Summary of the constraints and constraint forces Fc and constraint moments Mc for the examples from Section 8.3. The summary excludes prescriptions where dynamic friction is present.

Table 8.1

Pinjointed rigid body

Sliding rigid body

Rolling rigid body

Sliding cylinder

Constraints

Constraint forces and constraint moments

E k · vA = 0 E1 · ω = 0 E2 · ω = 0

Fc = 3k=1 µ k Ek Mc = µ4 E1 + µ 5E2 F c acts at X A

n · vP = 0

Fc = µ n Mc = 0 F c acts at X P

Ek · vP

=

vs · E k

E3 · v¯ = 0 ω=0

(e2 × E 3) ·

Whirling rigid body

E k · vA = 0 g1 · ω = ± g3 · ω = 0

∑

Fc

∑3

k=1 µ k Ek Mc = 0 F c acts at X P =

F c = µ1 E3 Mc = µ 2e 2 × E3 Fc acts at X¯ ∑

Fc = 3k=1 µ k Ek Mc = µ 4 g 1 + µ 5 g 3 F c acts at X A

The prescription (8.7) can be generalized in an obvious manner to systems of constraints and in Chapter 10 we will show how the prescription (8.7) is equivalent to traditional prescriptions for generalized constraint forces. In anticipation of this equivalence result, we used µ in (8.7). As evidenced from the examples discussed in Section 8.3, for each constraint we have a single µ . This ensures that the system of equations governing the motion of the body is sufficient to determine the motion of the body that satisfies the constraints, the constraint force F c , and the constraint moment Mc . We note that an integrable constraint on a rigid body is said to be “ideal” when the constraint force and constraint moment associated with the constraint can be prescribed using Lagrange’s prescription.

A Review of the Examples

To see if the prescription (8.7) makes physical sense, we return to the examples discussed in Section 8.3. The constraints and the constraint forces and constraint moments for the examples are summarized in Table 8.1. It is straightforward to conclude from the results summarized in this table that the prescriptions for Fc and M c are completely compatible with Lagrange’s prescription (8.7). For the examples considered, we emphasize that the Lagrange prescription assumes that dynamic Coulomb friction is absent. If dynamic friction is present, then Coulomb’s prescription must be used. That is, the normal force at the point of contact, XP , must be supplemented by a friction force Ff that opposes the motion of the body relative to the surface that it is in contact with: Fc = N + Ff acting at X P . In addition, a frictional moment can also be prescribed (cf. (8.3) and (8.5)).

8.4

Constraints and Lagrange’s Prescription

309

Power of the Constraint Force and Constraint Moment

The combined mechanical power P of a constraint force Fc and a constraint moment Mc given by Lagrange’s prescription can be computed using (8.7):

P = Fc · vC + M c · ω = µ ( fC · v C +

h C · ω)

= −µeC .

Hence, if eC ³ = 0, we anticipate that P will be nonzero. In this case, the combined effects of Fc and Mc will produce work and change the total energy of the rigid body. An example of this instance occurs in the whirling rigid body discussed in Section 8.3. Choosing a Different Material Point

Consider the sliding disk shown in Figure 8.4 and discussed in Section 8.3. Earlier, we wrote the constraint that the disk is sliding as vP · E3 = 0. Invoking Lagrange’s prescription, we can then prescribe Fc = µ E3 acting at XP and Mc = 0. It is natural to ask what would happen if we wrote the constraint using a different material point. For example, suppose we wrote the constraint using the center of mass: E3 · v¯ +

(

−Re2 × ²²

E3

)

·ω =

0.

Using Lagrange’s prescription with this constraint, we find Fc Mc

= µ E3

(

¯ acting at X,

= µ −Re2 × ²²

)

E3 .

With some insight, we conclude that this force–moment pair is equipollent to a force F = µE3 acting at X P. As a result, one can work with either vP · E3 = 0 or E3 · v¯ + (c ) −Re²² × E · ω = 0 and arrive at equivalent prescriptions for the constraint force and 3 2 the constraint moment. Based on the previous example, we are led to the suspicion that Lagrange’s prescription does not depend on the choice of material point XC that we choose to formulate the constraint. To see that this is true in general, let us choose to express the constraint function (8.6) in terms of another material point, say XA (cf. Figure 8.7). The transformation of the function π C given by (8.6) to the equivalent constraint function πA = fA · vA + hA ·

where πA

=

0 is equivalent to πC vC

=

=

ω + eA ,

(8.8)

0, can be achieved using the identity

vA + ω × (xC − xA ) .

Substituting for vC in (8.6) then yields the correspondences fA

= fC ,

hA

=

hC + (xC − xA ) × fC ,

eA

=

eC.

(8.9)

310

Constraints on and Potential Energies for a Rigid Body

(a)

(b)

Fc = µfC XC

Mc = µhC

xC xA

F c = µfC

XC

xC xA

XA

XA

O

O

Mc = µhC

+ (x C

− xA) × Fc

Equipollence of (a) a constraint force F c acting at X C and a constraint moment and (b) the same force acting at X A and a different constraint moment = µ hC + (xC − xA ) × Fc .

Figure 8.7

Mc Mc

= µ hC

Using (8.9) it is straightforward to see that the constraint force (µ fA acting at X A ) and constraint moment (µh A ) provided by Lagrange’s prescription using (8.8) are equipollent to (8.7): ³

³

Fc

Fc

= µ fC

acting at XC Mc = µ hC

= µfA = µ fC

´

⇐⇒

acting at XA Mc = µ hA = µ hC + (xC − xA ) × Fc

´

.

As shown schematically in Figure 8.7, the difference in the constraint moments can be attributed entirely to the difference in the point of application of the force Fc for the two cases.

8.5

Integrability Criteria

Earlier, in Section 1.10, we examined integrability criteria for constraints of the form f · v + e = 0 on the motion of a single particle. Here, we examine the corresponding criteria for a constraint on the motion of a rigid body. In the dynamics of rigid bodies, situations involving systems of constraints are often inescapable: one only has to think of the case of a rolling rigid body. It is also possible (and common) that, when a sufficient number of constraints are imposed on a rolling rigid body, the resulting system of constraints becomes integrable. Fortunately, there is a theorem, due to Frobenius, that provides a criterion for the integrability of a system of constraints. His criterion will be presented shortly. Before doing so, we discuss the situation of a single constraint on a single rigid body. A Single Constraint

For rigid bodies, what is needed is a generalization of criterion (1.27) for a single particle to constraints of the form πC =

0,

8.5

Integrability Criteria

311

where πC = fC · v ¯ + hC ·

ω + eC .

This constraint is integrable if we can find an integrating factor kC and a function ˙ ² C (xC , Q, t) such that kC ² C = πC . Although the integrability criterion is daunting in the number of algebraic calculations needed, for many problems most of these calculations are trivial (but tedious). The necessary and sufficient conditions for πC = 0 to be integrable involve a set of conditions. To establish these conditions, which are similar to those we discussed earlier for a single particle, we assume that xC is parameterized using a coordinate system ¶ ¶ µ µ q1, q2 , q3 and Q is parameterized using a set of Euler angles ν 1 , ν 2 , ν 3 . Thus, the function πC can be expressed as πC =

3 ± (

fCi q˙ i + hCi ν˙ i

)

+

eC .

i=1

We also define the intermediate functions Wi

= f Ci ,

Wi+3

hCi ,

W7

Ui +3 = ν i ,

U7

=

eC

=

and variables Ui

=

qi ,

=

t,

where i = 1, 2, 3. It remains to define IJKL

=

WJ SLK

+

WK SJL + WL SKJ ,

(8.10)

where SLK are the components of a skew-symmetric matrix: SLK

=

∂ WL ∂ WK − . K ∂U ∂UL

(8.11)

In (8.10) and (8.11), the integer indices J, K, and L range from 1 to 7. With all the preliminaries taken care of, we are now in a position to state a theorem that is due to Frobenius: A necessary and sufficient condition for the constraint πC = 0 to be integrable is that the following 67 (7 − 1) (7 − 2) equations hold for all U1 , . . . , U 7: IJKL

=

0,

for all J, K, L ∈ { 1, . . . , 7}, J

³=

K

³=

L, J ³ = L.

(8.12)

A proof of the theorem can be found in texts on differential equations (see, e.g., [84, Section 161]). We also note that only 10 of the equations IJKL = 0 are independent. Systems of Constraints

When rigid bodies are subject to several constraints, a criterion is available by which the possible integrability of the system of constraints can be evaluated. This criterion is known as Frobenius’ theorem, which we now frame in terms of a single rigid body. It is straightforward to extend this theorem to a system of particles and rigid bodies.

312

Constraints on and Potential Energies for a Rigid Body

Suppose we have a system of R

≤

1 πX¯ =

0,

6 constraints on the motion of a single rigid body: ...,

R πX¯ =

0,

(8.13)

where ¯ + B hX¯ · B πX¯ = B fX¯ · v

ω + B eX¯

µ

(B = 1, . . . , R). ¶

For ease of exposition, we assume that coordinates q , qµ2, q3 have¶ been chosen for the position vector x¯ of the center of mass and parameters ν 1 , ν 2 , ν 3 have been selected for the rotation tensor Q of the rigid body. Thus, each of the B πX¯ s can be rewritten as Bπ =

3 ± (

1

)

i i ˙ + B hi ν ˙ + B e, B fi q

i =1

where we have dropped the subscript X¯ for convenience. Next, we define the following functions and variables: WBi

= B fi ,

WB(i

+3)

= B hi ,

SBLK

=

WB7

= Be

∂ WBK ∂ WBL − ∂UK ∂ UL

(B

=

1, . . . , R),

,

(8.14)

where Ui

=

qi ,

Ui +3 =

In (8.14) and (8.15), i = 1, 2, 3 and L, K used to construct R + 1 matrices: ⎡

W11 ⎢ . W = ⎣ .. WR1

···

..

.

···

=

ν

i

U7

,

SA

t.

(8.15)

1, . . . , 7. The functions WBL and SBLK can be

⎤

W 17 .. ⎥ , . ⎦ WR7

=

⎡

· = −

SA

¸T

0 ⎢ .. = ⎣ . A −S 17

···

..

.

···

⎤

SA17 .. ⎥ , . ⎦ 0

(8.16)

where A = 1, . . . , R. With the help of (8.14), it is easy to see that S1, . . . , SR are skewsymmetric matrices: SBKL = − SBLK . We also define the following seven-dimensional vectors: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ a1 b1 x1 ⎢ . ⎥ ⎢ . ⎥ ⎢ . ⎥ a = ⎣ .. ⎦ , b = ⎣ .. ⎦ , x = ⎣ .. ⎦ . (8.17) a7

b7

x7

We are now in a position to state the F ROBENIUS INTEGRABILITY THEOREM: For a single rigid body, the necessary and sufficient conditions for a system of R < 7 constraints 1 πX¯ = 0, . . . , R πX¯ = 0 to be integrable are for the following equations to hold: ( ) aT S B b = 0

(B

=

1, . . . , R)

for all values of the variables U 1, . . . , U 7 and for all distinct solutions a and b to the equation Wx = 0.

That is, the seven-dimensional vectors a and b lie in the null space of W.

8.6

Potential Energies, Conservative Forces, and Conservative Moments

313

Remarks on Frobenius’ Theorem

The statement of Frobenius’ remarkable theorem is adapted from Forsyth [83], and the reader is referred to this text for a classical proof and to a paper by Hawkins [120] for additional historical perspectives. More modern proofs, using differential forms, are readily found; see, for example, [45, 83, 137, 253]. ( ) As SA are skew-symmetric, aT · SAa = 0 for all possible a. Often this result is a key to using the theorem, for, if W has a one-dimensional null space, then there is only one vector, say b, satisfying Wb = 0 and the theorem is trivially satisfied.

Examples

An example of the use of Frobenius’ theorem is discussed in the exercises at the end of this chapter. There it is used to show that a rolling disk is generally subject to nonintegrable constraints. However, when the disk is further constrained to be vertical and its center of mass moves in a straight line, the constraints on the rolling disk become integrable. As discussed in Exercise 8.9, this is the familiar situation from introductory dynamics classes. A second example of the use of the theorem was mentioned earlier in conjunction with two scleronomic constraints (1.29) on the motion of a single particle: ∑3

f 11 ˙x + f12 y˙ + f13 ˙z = 0,

f 21 x˙ + f22 y˙ + f 23 z˙

∑3

=

0,

where f1 = k=1 f1k Ek and f2 = k=1 f2k Ek are functions of the position vector of the particle, and f2 × f1 ³ = 0. To apply Frobenius’ theorem to this case, some trivial modifications to (8.14)–(8.17) are needed. First, U1

=

U2

x,

U3

y,

=

=

U 4 = t.

z,

For this case, R = 2, and the indices L and K range from 1 to 4. It is left as an exercise to establish that the 2 × 4 matrix W has a two-dimensional null space that is spanned by the vectors a

where gk

=

¹

0

0

= ( f1 × f2 ) · Ek .

⎡

S

A

=

⎢ ⎢ ⎢ ⎢ ⎣

( ) it follows that aT S1 b

0

1

ºT

,

b

=

¹

g1

g2

g3

0

Now, as

∂ fA2 ∂ x1 ∂ fA

0 −

∂ fA 1 ∂ fA 1

∂ x2 ∂ fA 1 3 − ∂ x1 ∂ x3

0

∂ x2

−

∂ fA

0

∂ f A2 ∂ x1

∂ f A2 3 − ∂ x2 ∂ x3

( ) 0 and aT S2b

0

∂ f A1

∂ x3 ∂ f A2 ∂ x3

− −

0 0

∂ fA

3

∂ x1 ∂ fA

3 ∂ x2

0

ºT

,

⎤ ⎥

0 ⎥ ⎥

⎥,

0 ⎦ 0

= = 0. Thus, the hypotheses of Frobenius’ integrability theorem are satisfied, and we conclude that the system of constraints (1.29) is integrable.

8.6

Potential Energies, Conservative Forces, and Conservative Moments

The presence of conservative forces and moments acting on rigid bodies is one of the key features used to solve many problems. Although the definition of a conservative

314

Constraints on and Potential Energies for a Rigid Body

force originally arose in the dynamics of a single particle and is well understood, the same cannot be said for conservative moments (see, e.g., [6, 219]). Indeed, as noted by Ziegler [314], for rigid bodies whose axis of rotation is not fixed, a constant moment is not necessarily conservative. To this end, we start with three well-known examples of forces and their associated moments. These examples are followed by a discussion of Ziegler’s example. After the examples have been presented, a general treatment of conservative forces and moments is given. Following this treatment, you should return to Ziegler’s example and convince yourself that a constant moment is not conservative.

Constant Forces

A constant force P acting on a rigid body is conservative. If this force acts at all material » points of the rigid body, then it is equipollent to a single force R Pρ dv »acting at the center of mass of the rigid body. The potential energy of this force is − R P ρ dv · x. ¯ Notice that there is no moment (relative to the center of mass) associated with this force. The most ubiquitous example of a constant force is a constant gravitational force acting at each material point. This force is equipollent to a force mgg acting at the center of mass. Here, g is the direction of the gravitational force. The potential energy of this force is −mgg · x. ¯

Spring Forces

The next example consists of a linear spring that is suspended between a fixed point B and a point X S on a rigid body B (cf. Figure 8.8). The spring has stiffness K and unstretched length ³0. The spring force acting on the rigid body has a familiar expression: Fs = −K ´

xs − xB acting at Xs , ±x s − xB ±

´ = ±x s − x B± − ³0.

The potential energy associated with the spring is Us =

K 2

(± xs − x B ± − ³0)

With the help of the identity d ±xs − xB± dt

¼

=

xs − xB ±xs − xB ±

2

½

. ¼

· (v s − x ˙ B) =

xs − xB ± xs − x B ±

½ · vs,

it is straightforward to show that U˙ s

= −Fs · v s .

That is, Fs = −

∂ Us ∂ xs

= −K ´

xs − xB . xB ±

±x s −

The force Fs acting at Xs is equipollent to a force F s acting at X¯ and a moment πs × F s .

8.6

Potential Energies, Conservative Forces, and Conservative Moments

315

¯

X

πs

Xs

¯x

xs

B

O

Figure 8.8

point B.

The present configurations of a rigid body B which is connected by a spring to a fixed

Central Gravitational Fields

In celestial and orbital mechanics, a standard problem is to consider the motion of a body subject to a central force field. Such force fields date to Newton and are based on his inverse-square force. You may recall that this conservative force is the force exerted r , where r is the position on a particle of mass m by a particle of mass M: F = − GMm 2 ± r± ±r ± vector of m relative to M and G is the universal gravitational constant. For the force fields of interest, we consider two bodies B1 and B2 with mass densities per unit volume of ρ1 and ρ2, respectively. Every material point of B2 exerts an attractive force on each material point of B1 (see Figure 8.9). If we integrate these forces over all material points in B1 and B2, we will obtain the resultant force exerted by B2 on B1 . Similar integrations apply to the moment and potential energy of these forces. In short, the resultant gravitational force Fn and moment M n on B1 that are due to B2 are ¾

Fn

= −

Mn

= −

¾

R R ¾ 1¾ 2 R1 R2

x1 − x2 ρ dV1ρ2 dV2 2 ±x − x ± 1 ±x 1 − x 2± 1 2 G

(x1 − x ¯ 1) ×

acting at X¯ 1,

G

x1 − x2 ρ1 dV1ρ 2dV2 . 2 ||x1 − x2 || ± x1 − x2 ±

The moment M n is relative to the center of mass X¯ 1 of B1 and is known as the gravity gradient torque. The potential energy associated with this force field is ¾

Un

= −

¾

R1 R

G

±x1 − 2

x2 ±

ρ 1dV1 ρ2dV2 .

316

Constraints on and Potential Energies for a Rigid Body

Rigid body B1 ¯

X1

Rigid body B2 C

¯

X2

¯x1

x¯ x¯ 2

O

Two rigid bodies B 1 and B2 that exert mutual gravitational forces. The point C in this figure denotes the center of mass of this system of rigid bodies.

Figure 8.9

Assuming that B2 is spherical and has mass M, the expressions for the force, moment, and potential energy simplify: ¾

Fn

= −

Mn

= −

R1 ¾ ¾

Un

= −

R1 R1

x1 − x¯ 2 ρ dV 2 ±x − x ± 1 1 ± x1 − x 2 ± 1 2 GM

¯1) × (x1 − x

GM

acting at X¯ 1 ,

x1 − x¯ 2 ρ 1dV1 , x2 ±

||x 1 − x ¯ 2||2 ± x 1 −

GM ρ1 dV1. ± x1 − x 2 ±

It is important to notice that, even in this simplified case, the gravitational force field can exert a moment on the rigid body B1 . As discussed in Exercise 9.15, this moment is the reason why an Earth-based observer always sees the same side of the Moon. To simplify the expressions as much as possible, we now use the fact that B1 is a rigid body: π = x1 − x¯ 1 = QΠ.

In addition, we assume that ±π± is small relative to ±x¯ 1 − x¯ 2 ±. These assumptions allow us to approximate the forces, moments, and potential energy associated with the central force field. After a substantial amount of manipulation, we find the results F n ≈ mg acting at X¯ 1 , ¼ ½ ¼ ½ 3GM 3GM Mn ≈ c × (Jc) = − c × (Ec), R3 R3 ¼ ½ ¼ ½ GMm GM 3GM Un ≈ − − tr(J) + (c · (Jc)) , R 2R 3 2R3

(8.18)

8.6

Potential Energies, Conservative Forces, and Conservative Moments

317

where J is the inertia tensor of B1 relative to its center of mass, E is the Euler tensor of B1 relative to its center of mass, m is the mass of B1 , mg = −

3GM GMm c− (2J + (tr(J) − 5c · Jc) I) c, 2 R 2R4

and R = ±x¯ 1 − x¯ 2 ± ,

c=

x¯ 1 − x¯ 2 . ±x ¯1 − x ¯2±

Notice that c points from the center of mass of the spherically symmetric body B2 to the center of mass of B1 . In addition, because of the presence of the inertia tensor J, (8.18)1 does not correspond to Newton’s gravitational force on a particle of mass m by another particle of mass M. Classic expressions (8.18) are used in the vast majority of works on satellite dynamics (cf. [22, 133, 142]). The expression for the potential energy Un in the form shown in (8.18) is credited to James MacCullagh (1809–1847). 7 In many of the works on the dynamics of a satellite about a fixed point O, it is common c. Such an approximation effectively decouples the to approximate Fn with Fn = − GMm R2 motion of the center of mass X¯ 1 of B1 from its orientation, and one can then conclude that X¯ 1 behaves like a particle in the one-body problem and solve for the orientation of the satellite separately. 8 However, as pointed out by Barkin [17] and Wang et al. [301], this approximation violates angular momentum conservation and energy conservation. Studies on the dynamics of satellites for which Fn = mg can be found in [17, 221, 301, 302]. These works show that steady motions of the satellite are possible where the orbital plane of X¯ 1 does not contain the origin O. This is in contrast to the one-body problem where the orbital plane contains O.

Constant Moments that are Not Conservative

To see that a constant moment is not conservative, we recall Ziegler’s example [314]. He considered the work done by a constant moment M0 E3 during two possible motions of a rigid body. The first motion of the rigid body consists of a rotation about E3 through −180◦ . During this motion, the moment M0 E3 does work equal to −M0π . However, the same final orientation of the body can be achieved by a rotation about E1 through 180◦ , followed by a rotation about E2 through 180 ◦.9 However, the work done by the constant moment during the second motion is zero. Consequently, the work done by the moment M0E3 depends on the “path” taken by the body, and hence M0 E3 cannot be conservative. An alternative proof that a constant moment is generally not conservative can be found in [214]. The proof presented there exploits the dual Euler basis. 7 Notes on his derivation of U and F can be found in Propositions 4 and 5 of [2]. Comments by Allman on n n the derivation of Mn can also be found in [2]. 8 The one-body problem was discussed earlier in Section 2.8. 9 This is an example of the application of the Rodrigues–Hamilton theorem (6.60) discussed in Exercise

6.10 at the end of Chapter 6.

318

Constraints on and Potential Energies for a Rigid Body

In many courses on dynamics, the rotation of the rigid body is constrained to be a fixed-axis rotation and the rotations about E1 and E2 are not permitted. For these cases, a constant moment M0E3 is conservative. General Considerations

It is convenient at this point to give a general treatment of conservative forces and moments in the dynamics of a single rigid body. Our discussion is easily generalized to the case of N rigid bodies. We assume that the most general form of the potential energy is U

=

U (xA, Q) .

Clearly, this function depends on the motion of the material point X A and the rotation of the rigid body. We can calculate the time derivative of this function using (6.50): ˙ U

=

∂U ∂ xA

·

v A + u Q · ω,

where uQ is a vector representing the derivative of U with respect to Q. As discussed in Section 6.11, this vector has numerous representations, and the easiest one to use arises when Q is parameterized by sets of Euler angles. Denoting these angles and their dual µ ¶ µ ¶ Euler basis vectors by γ 1, γ 2, γ 3 and g1 , g2 , g3 , respectively, the vector uQ has the representation uQ

=

3 ± ∂U k ∂γ k

k=1

g .

(8.19)

Consider the conservative force Fcon acting at X A and the conservative moment Mcon associated with this potential. We assume that the work done by these forces and moments is dependent on the initial and final configurations of the rigid bodies but is independent of the motions of the rigid bodies. This implies that ˙ = −U

F con · vA + M con · ω.

˙ and collecting terms, we find that Substituting for U

¼

Fcon +

∂U ∂ xA

½

·

vA

+

(

Mcon + uQ

)

·

ω = 0.

(8.20)

This can be interpreted as an equation for Fcon and M con that must hold for all motions of the rigid body. Assuming that Fcon and Mcon are independent of the linear and angular velocity vectors, we find that in order for (8.20) to hold for all vA and ω it is necessary and sufficient that ∂U acting at XA , F con = − ∂ xA M con = −u Q . (8.21) These are the expressions for the conservative forces and moments associated with a potential energy. It is left as an exercise to verify that these expressions are consistent

8.7

Closing Comments

319

with the results presented earlier for the spring and central gravitational forces. For some of these potential energies, U will not depend on Q and so M con = 0. If a set of Cartesian coordinates is used to parameterize xA and a set of Euler angles is used to parameterize Q: xA

=

x1 E1 + x2E2 + x3 E3 ,

ω=

γ˙

1

g1 + γ˙ 2 g2 + γ˙ 3g3,

then (8.21) have the representations Fcon Mcon

= −

∂U ∂ xA

= −

= −u Q =

3 ± ∂U i=1

∂ xi

Ei acting at XA ,

3 ± ∂U k − g . k=1

∂γ k

If a curvilinear coordinate system is used to parameterize xA , then the representation for Fcon will involve the contravariant basis vectors associated with the curvilinear coordinate system. The representation for Mcon that uses the dual Euler basis is the simplest known representation for a conservative moment.

An Example of a Torsional Spring

We now present a very simple example of a torsional spring acting on a single rigid body. This example illustrates representations (8.21) and makes use of the dual Euler basis. Even in this simple case, the conservative moment associated with the spring has a surprising feature. Consider a single rigid body whose rotation tensor is parameterized by a set of 3–1–3 Euler angles and suppose that a torsional spring acts on the body. The spring is assumed to have a potential energy U = K2t ψ 2, where Kt is a torsional stiffness. Then, with the help of (8.21) and representation (8.19) for uQ in terms of Euler angles, we find that the conservative moment is Mcon

∂U 1 g ∂ψ 1 = −Kt ψ g

= −uQ = −

= −Kt ψ (cot( θ ) ( cos( ψ )E2 −

sin(ψ )E1 ) + E3 ) .

The conservative force Fcon associated with this potential is 0 because the potential is independent of the position vector of any point on the rigid body. It is interesting to notice that the components of Mcon are not entirely in the direction of E3 .

8.7

Closing Comments

The main results in this chapter involved prescriptions for constraint and conservative forces and moments. For the former, we argued that Lagrange’s prescription provided

320

Constraints on and Potential Energies for a Rigid Body

these expressions in cases in which dynamic Coulomb friction was absent. For instance, if a single rigid body is subject to a constraint that can be expressed as fA · vA + h A · ω + eA

=

0,

then Lagrange’s prescription states that Fc

= µ fA

Mc

acting at XA , = µh A .

Examples of Coulomb’s prescription when dynamic friction is present can be found in the discussion on the planar motion of a rigid body about a fixed point and rolling and sliding rigid bodies in Section 8.3 (cf. (8.3) and (8.5) in particular). If we suppose that a single rigid body has a potential energy function ¸

·

U = U x1, x2 , x3 , γ 1 , γ 2 , γ 3 , where xi are the Cartesian coordinates for a point X whose position vector is x and γ i are Euler angles, then our developments showed that the conservative force Fcon and conservative moment Mcon associated with this potential are F con = −

3 ± ∂U i=1

Mcon

∂ xi

= −

E i acting at X,

3 ± ∂U i g, i i=1

∂γ

where gi are the dual Euler basis vectors associated with the Euler angles.

8.8

Exercises

Exercise 8.1: Suppose that the motion of a rigid body is subject to a pair of constraints: ω · g2

=

ω · g3

0,

=

0.

Here, gi are the dual Euler basis for a set of Euler angles of your choice. Give a physical interpretation of these constraints. What are the angular velocity vector, angular momentum vector, and rotational kinetic energy of the resulting constrained rigid body? Exercise 8.2: Consider a force P acting at a point XP of a rigid body. In the present configuration, the point XP has the position vector πP relative to the center of mass X¯ of the rigid body: πP

=

xP

− x. ¯

In addition, vP

= x ˙P = v ¯ +

ω × πP .

8.8

Exercises

321

(a) The mechanical power of P is P · vP . Show that this power has the equivalent representation P · vP

P · v¯ + (πP × P) · ω.

=

Using a free-body diagram, give a physical interpretation of this identity. (b) A force P acting at the point XP is said to be conservative if there exists a potential energy U = U(xP ) such that P=−

∂U ∂ xP

.

Show that this definition implies that ˙ = −U

(c) Show that the potential energy U ¯ and Π : x, ¯ Q, X, P U

=

P · v¯ + (πP =

U(xP )

×

P) · ω.

U(xP ) can also be described as a function of (

¯ ˆ x = U ¯ , Q, ΠP , X

)

.

Here, Q is the rotation tensor of the rigid body and πP equivalence, show that P

= −

ˆ ∂U

∂x ¯

,

πP

×

=

QΠP . Using this

P = −u Q .

(d) Consider the case in which P represents a force that is due to a spring of stiffness K and unstretched length ³0 . One end of the spring is attached to a fixed point O. ( ) ¯ for this force P? ˆ x What are the functions U (xP ) and U ¯ , Q, ΠP , X Exercise 8.3: As shown in Figure 8.10, a tippe top is a body with an axis of symmetry. One of its lateral surfaces can be approximated as a spherical surface of radius R and the other is a cylinder of radius r. The top is designed so that the center of mass X¯ is located below the center of the sphere, and this feature leads to its ability to flip over (see Figure 8.11). 10

e3 XA

E3

g

¯

X

E1

R XP

Figure 8.10

A tippe top moving on a rough horizontal plane.

10 The dynamics of the tippe top has been the subject of several investigations, and the interested reader is

referred to [30, 208, 211, 238, 270, 295] for discussions and references. Among other issues, these papers point out the important role played by friction forces acting at the point of contact.

322

Constraints on and Potential Energies for a Rigid Body

ω0

E3

XA

g

ω1

e3

¯

X

E1

¯

X

XA

XP

e3

The two steady motions of a tippe top: the upright position in which e3 inverted state in which e3 = −E3.

Figure 8.11

=

E3 and the

The instantaneous point of contact of the spherical portion of the top with the horizontal plane is denoted by XP . The position vectors of XP relative to X¯ and the point X A relative to X¯ are xP

−x ¯ = −RE3 +

± e3,

xA − x¯ = he3,

respectively. (a) Using a set of 3–1–3 Euler angles, establish expressions for ωi = ω · ei and ±i = ω · Ei . For which orientations of the tippe top does this set of Euler angles have singularities? (b) Show that the slip velocities of the point X P of the tippe top have the representations vs 1

= x ˙ 1 − ±2 ( R −

± cos(θ )) + ±3 (± sin(θ ) cos(ψ )) ,

vs 2

= x ˙ 2 + ±1 ( R −

± cos(θ )) + ±3 (± sin(θ ) sin(ψ ) ) , ∑

where vP = vs 1 E1 + vs2 E 2 and v¯ = 3k =1 x˙ k Ek . (c) Suppose that the tippe top is sliding. Give a prescription for the constraint force F c acting at X¯ and the moment Mc that enforce the sliding constraint. (d) Suppose that the tippe top is rolling. Give a prescription for the constraint force F c acting at X¯ and the moment Mc that enforce the rolling constraints. (e) With the help of (8.12), show that two of the three constraints on a rolling tippe top are nonintegrable. Exercise 8.4: Consider the mechanical system shown in Figure 8.12. It consists of a rigid body of mass m that is free to rotate about a fixed point XO . The joint at XO does not permit the body to have a spin about e1. A vertical gravitational force mgE1 acts on the body. The inertia tensor of the body relative to its center of mass X¯ is J0

= λ 1E1 ⊗

E1 +

·

2 λ − m³0

¸

(E2 ⊗ E2 +

E3 ⊗ E3 ) .

The position vector of the center of mass X¯ of the body relative to XO is ³0 e1. To parameterize the rotation tensor of the body, we use a set of 1–3–1 Euler angles: g1 = E1, g2 = e²3 , and g3 = e1 . For these angles ω = ψ˙ E1 + θ˙ e²3 + φ˙ e1 =

(

)

˙ cos( θ ) φ˙ + ψ

(

+ θ˙

e1 +

(

˙ θ˙ sin(φ ) − ψ

)

cos(φ) + ψ˙ sin(θ ) sin(φ ) e3 .

)

sin(θ ) cos(φ ) e2

8.8

Exercises

323

˙

ψ g

E3

XO

E2

θ

¯

X

E1

e1 Figure 8.12

A pendulum problem. ¯

X

E3 e3

g

E2

XO

E1 A rigid body that is free to rotate about the fixed point XO . A set of 3–1–3 Euler angles is used to describe the rotation tensor Q of this body.

Figure 8.13

(a) Which orientations of the rigid body coincide with the singularities of the Euler angles? (b) Derive expressions for the unconstrained potential energy U and unconstrained kinetic energy T of the rigid body. (c) Show that the motion of the rigid body is subject to four constraints: ˙ = φ

ω · g3

=

0,

v¯ − ω × (³0e1 ) = 0.

(d) Derive expressions for the constrained potential energy U and kinetic energy T of the rigid body. Exercise 8.5: As shown in Figure 8.13, an axisymmetric rigid body is free to rotate about the fixed point XO . The body, which has mass m, has an inertia tensor J = λa e3 ⊗ e3 + λt (I − e3 ⊗ e3 ) . The position vector of the center of mass of this body relative to XO is x¯ = ³1 e3.

324

Constraints on and Potential Energies for a Rigid Body

(a) Using a set of 3–1–3 Euler angles, show that the angular velocity vector ω also has the representation ω = θ˙e²²1

˙ +ψ

sin(θ )e²²2

˙ +ψ ˙ + (φ

(b) Show that the angular momentum H is H = λ t θ˙ e²²1 + λt ψ˙ sin(θ )e²²2 + λ a

cos(θ ))e²²3 .

(

˙ +ψ ˙ φ

)

cos(θ ) e3.

(c) Show that the angular momentum of the center of mass relative to X O is m³ 21 (ω1 e1 + ω2e2 ) , where ωi = ω · ei . What is the angular momentum of the rigid body relative to O? (d) Show that the kinetic energy T˜ is T˜

2 λt + m³1

·

2 ˙ 2 sin2(θ ) θ˙ + ψ

¸

λa

(

)2

. 2 2 (e) A conservative moment that is due to a torsional spring is applied to the rigid body. As a result, the total potential energy of the rigid body is =

+

˙ +ψ ˙ cos( θ ) φ

K 2 ψ . 2 Here, K is the torsional spring constant. What are the conservative force Fcon acting at X¯ and the moment Mcon acting on the rigid body? U˜

=

mgx¯ · E 3 +

Exercise 8.6: Recall Ziegler’s example of a constant moment M0 E3 that was not con¶ µ servative. After choosing a set of Euler angles γ 1, γ 2, γ 3 , show that you cannot find ∑3 ∂ U i 11 a potential energy function U such that M0E3 = − i=1 ∂ γ i g . Exercise 8.7: Suppose that a linear spring of stiffness K and unstretched length ³0 is attached to a point XA on a rigid body and the other end of the spring is attached to a fixed point O. The position vector of XA relative to the center of mass of the rigid body is πA

=

Re3 .

(a) With the assistance of a set of 3–1–3 Euler angles, show that xA

= ( x1 +

R sin(θ ) sin(ψ )) E1 + (x2 − R sin(θ ) cos(ψ )) E2

+ (x3 +

∑3

R cos( θ )) E3 .

In this equation, x¯ = k=1 xk Ek . (b) With the assistance of a set of 3–1–3 Euler angles, show that the potential energy of the spring is )2 K (√ U = U (x¯ , Q) = u − ³0 , 2 where u = (x1 + R sin(θ ) sin( ψ ))2 + (x2 − R sin(θ ) cos(ψ ))2 11 One solution to this problem can be found in [214].

8.8

+ ( x3 +

325

Exercises

R cos(θ ))2 .

(c) Describe two equivalent methods to find the conservative force Fcon acting at X¯ and the conservative moment M con associated with the spring. Exercise 8.8: As shown in Figure 8.14, a thin rigid circular disk of mass m and radius R rolls (without slipping) on a rough horizontal plane. The rotation tensor of the disk will be described by use of a 3–1–3 set of Euler angles (see Section 6.8.2). This set has the advantage of having the singularities of the Euler angles coincide with the disk lying flat on the horizontal plane (i.e., θ = 0, π ). (a) For the disk, show that the position vector of the instantaneous point of contact X P relative to the center of mass X¯ has the representations πP

= −Re2 = −R sin(φ)e1 − ²²

R cos( φ)e2

= −R ( − cos(θ ) sin(ψ )E1 +

cos(θ ) cos(ψ )E2 + sin(θ )E3) .

(b) By taking the Cartesian components of the rolling condition, vP show that the motion of the disk is subject to three constraints: 0,

1 πX¯ =

2 πX¯ =

0,

3π X ¯ =

= v ¯ +ω×

0,

πP

=

0,

(8.22)

where ˙ cos(ψ ) + Rψ ˙ ˙ 1 + Rφ 1 πX¯ = x

cos(θ ) cos(ψ ) − Rθ˙ sin(θ ) sin( ψ ),

˙ sin(ψ ) + R ψ ˙ cos( θ ) sin(ψ ) + Rθ˙ sin(θ ) cos(ψ ), ˙ 2 + Rφ 2 πX¯ = x ˙ 3 − R θ˙ cos( θ ), 3 πX¯ = x

∑

and x¯ = 3i=1 xi Ei . You should notice that the last of these three constraints integrates to x3 = R sin(θ ) + constant. However, as shown in Exercise 8.9, the system of constraints (8.22) is not integrable.

E3 g

E1

e2

E2

O

ψ

e1

=

e1

¯

X

e2

φ

e1

XP

F c = N + Ff

=

3 i=1

e

µi

Ei

A circular disk of radius R rolling on a rough horizontal plane. The position vector of the instantaneous point of contact XP of the disk and the plane relative to the center of mass X¯ of the disk is always along e²²2 . A set of 3–1–3 Euler angles is used to parameterize the rotation of the disk. The angle θ is the angle of inclination of the disk from the vertical.

Figure 8.14

326

Constraints on and Potential Energies for a Rigid Body

(c) Show that Lagrange’s prescription for the constraint force Fc and the constraint moment Mc acting on the disk is equivalent to assuming that static Coulomb friction forces and a normal force act at X P. Exercise 8.9: Consider the rolling disk discussed in Exercise 8.8 and consider the constraints (8.22) on its motion. Here, we wish to show that this family of constraints is nonintegrable by using Frobenius’ theorem. To use the theorem, we first define the following seven variables: U1 U4

=

U 2 = x2 ,

x1, U5

= φ,

U3

U6

= ψ,

=

x3, U7

= θ,

=

t.

(a) With the help of (8.14), compute the 3 × 7 matrix W. Determine a1 , . . . , a4 that span the null space (kernel) of this matrix. You will find that one of these vectors is a1

=

¹

0

0

0

0

0 0

1

ºT

.

(b) With the help of (8.14), calculate the three 7 × 7 skew-symmetric matrices S1, S2 , and S3 corresponding to 1πX¯ , 2 πX¯ , and 3π X¯ , respectively. You will find that these matrices have at most two nonzero elements. (c) Using the results of (a) and (b), show that Frobenius’ theorem implies that the rolling disk is subject to nonintegrable constraints. You should also indicate how your results can be used to conclude that the sliding disk is subject to an integrable constraint. (d) Suppose the rolling disk is subject to two additional integrable constraints: x2

=

0,

ψ =

0.

That is, the disk is constrained to roll vertically in a straight line. Compute the 5 × 7 matrix W and show that its null space is spanned by a1 and a2 , where a1

=

a2

=

¹ ¹

0

0

0 0

−R cos(θ )

0 0

0 0

1 1

ºT

0

, 0

0

ºT

.

In addition, show that the skew-symmetric matrices S4 and S5 corresponding to the additional constraints are both zero. Finally, invoking Frobenius’ theorem, show that the family of constraints 1 πX¯ , 2πX¯ , 3πX¯ , x˙2 = 0, and ψ˙ = 0 is integrable. Exercise 8.10: A schematic of a Griffin grinding machine is shown in Figure 8.15.12 The grain to be milled is placed in the bin, and a roller is designed to roll without slipping on the inner wall of the grain bin. The normal force generated by the roller is substantial and crushes the grains. The motion of the roller is achieved by two drive shafts. The first shaft (drive shaft I) has an angular velocity vector ωI = ψ˙ E3 . It is coupled by a 12 This figure is adapted from Arnold and Mauder [10] and Webster [304].

8.8

Exercises

327

E3 Drive shaft I O

ψ

E2

E1

U

Universal joint

Drive shaft II γ

φ XP

Grain bin

e3 Roller Figure 8.15

Schematic of a Griffin grinding machine.

universal joint at U to the second shaft. The roller is attached to the second shaft (drive shaft II) by a joint that allows it to have a rotation relative to the second shaft in the direction of e3 . The basis {e1 , e2 , e3} corotates with the roller. (a) Using a set of 3–1–3 Euler angles {ψ , π vector of the roller has the representation ω = ψ˙ ( sin(φ ) sin(π

− γ )e1 +

−γ˙ (cos( φ)e1 −

− γ , φ },

cos(φ ) sin(π

show that the angular velocity − γ )e2 +

cos( π

− γ )e3 )

sin(φ )e2) + φ˙ e3.

For which orientations of the roller does this set of Euler angles have singularities? What is the angular velocity vector ωII of drive shaft II, and what is the angular velocity vector of the roller relative to drive shaft II? (b) The center of mass X¯ of the roller has a position vector relative to the fixed point U of x¯ = He3. ¯ Establish an expression for the velocity vector v¯ of the point X. (c) The instantaneous point X P of contact of the roller with the grain bin has the ¯ following position vector relative to X:

Using the identity vP nents vP · ei .

πP

= −r (cos( φ)e2 +

=

ω

×

πP

+ v, ¯

sin(φ )e1) .

establish expressions for the compo-

328

Constraints on and Potential Energies for a Rigid Body

(d) If vP = 0, then show that related:

γ

˙ = φ

is a constant and the rotational speeds ¼

cos (γ0 ) −

H sin (γ 0) r

½ ˙. ψ

In this equation, γ0 is the constant value of the angle γ .

˙ ψ

and

˙ φ

are

9

Kinetics of a Rigid Body

9.1

Introduction

˙ for a rigid body are discussed. In this chapter, the balance laws F = mv˙¯ and M = H Several component forms of these laws are considered. For instance, the components of ˙ ·e , the balance of angular momentum with respect to the corotational basis ei , M · ei = H i lead to a set of equations that are known as Euler’s equations. In Chapter 10, we shall ˙ · g lead to Lagrange’s equations. show that M · gi = H i Once the balance laws have been discussed, a brief outline of the work–energy theorem is given and it is shown how to establish energy conservation for a rigid body. Our attention then turns to applications. In particular, we discuss the dynamics of a body rotating about a fixed point, moment-free motion of a rigid body, rolling and sliding spheres (including the Chaplygin sphere), and the dynamics of baseballs and footballs. Several other applications are discussed in the exercises. Despite all these examples, our consideration of applications is far from exhaustive. At the conclusion of this chapter, details are provided on some other interesting applications that have been modeled by use of a single rigid body.

9.2

Balance Laws for a Rigid Body

Euler’s laws for a rigid body can be viewed as extensions to Newton’s second law for a single particle. There are two laws, or postulates – the balance of linear momentum and the balance of angular momentum: ˙ = F, G

˙ H O

=

MO .

(9.1)

Here, HO is the angular momentum of the rigid body relative to a fixed point O and MO is the resultant external moment relative to O. As noted in [287, 288], these laws are discussed in several of Euler’s works on rigid-body dynamics and find their definitive form in Euler’s paper [72] that was published in 1776. 1 1 The interested reader is referred to [177, 308], where additional commentaries and perspectives on these

balance laws can be found. For additional background on the problem of the precession of the equinoxes featured in [308], the text of Hestenes [127] is recommended.

330

Kinetics of a Rigid Body

In many cases, it is convenient to give an alternative description of the balance of angular momentum. To do this, we start with the identity HO

=

H + x¯ × G.

Differentiating, and using the balance of linear momentum, we obtain ˙ H O

˙ +v = H ¯ ×

˙ G + x¯ × G

˙ +x = H ¯ ×

F.

˙ Hence, invoking the balance of angular momentum, H O

MO

˙ ˙ = H ¯ × O = H +x

=

M O , we find that

F.

However, the resultant moment relative to a fixed point O, M O , and the resultant moment ¯ M, are related by2 relative to the center of mass X, MO

=

M + x¯ × F.

It follows that ˙ M = H, ¯ which is known as the balance of angular momentum relative to the center of mass X. This form of the balance law is used in many problems for which the rigid body has no fixed point O. In summary, the balance laws for a rigid body are known as Euler’s laws. Two equivalent sets of these laws are used. For the first set, the balance of angular momentum relative to a fixed point O is used: ˙ = F, G

˙ H O

=

MO .

(9.2)

In the second set, the balance of angular momentum is taken relative to the center of ¯ mass X: ˙ = F, G

˙ = M. H

(9.3)

It should be noted that G = mv¯ ,

H = Jω,

HO

=

Jω + x¯ × mv. ¯

Here, O is the origin of the coordinate system used to define x¯ . To determine the motion of the rigid body, it suffices to know x¯ (t) and Q(t). To obtain these results, (9.2) or (9.3) must be supplemented by information on the coordinate system used to parameterize x¯ and the parameterization of Q. 2 This may be seen from our previous discussion of a system of forces and moments acting on a rigid body

in Section 8.2.

9.3

9.3

Work and Energy Conservation

331

Work and Energy Conservation

The work–energy theorem for a rigid body equates the rate of change of kinetic energy to the mechanical power of the external forces and moments acting on the rigid body. There are two equivalent forms of this theorem: T˙

=

F · v¯ + M · ω N ±

=

FK · vK + Mp · ω.

(9.4)

K =1

The second form is the most useful for proving that energy is conserved in a specific problem. After proving the theorem, we will close this section with a discussion of energy conservation. Proving the Work–Energy Theorem

The difficulty in establishing the work–energy theorem lies in dealing with the angular momentum H = Jω. To overcome this hurdle, it is easiest to first show that (

ω · J˙ ω

)

=

0.

(9.5)

This result can be established using an earlier result, J˙ ΩT ω = −ω × ω = 0: (

J˙ ω

)

· ω = ((ΩJ −

Now, as J˙ ω

)

·

ω = 0,

ΩJ − JΩ , and the identity

JΩ) ω) · ω

= (Ω Jω) ·

(

=

ω − (J (ω × ω)) · ω

=

) ( Jω · ΩT ω + (J (0)) · ω

=

0.

²

˙ · ω = J˙ω H

³

· ω = (J ω ˙)·

ω = (Jω) · ω ˙.

In the last step, we invoked the symmetry of J: (Ja) · b (9.6), we can conclude that

=

(

(9.6) )

a · JT b

˙ ·ω = H·ω H ˙,

=

a · (Jb). From (9.7)

a result that we shall presently invoke. To prove the work–energy theorem (9.4), recall the Koenig decomposition for the kinetic energy T of a rigid body: 1 1 mv¯ · v¯ + (Jω) · ω. 2 2 Differentiating T and using (9.5) and (9.7), we find that T=

T˙

) 1 (˙ H · ω + (Jω) · ω ˙ 2 ˙ ·v ˙ · ω. = G ¯ +H ˙ ·v = G ¯ +

332

Kinetics of a Rigid Body

FK κt XK

πK

Mp

¯

xK

X

x¯ O

Figure 9.1

A force FK acting at a point XK and a moment Mp acting on a rigid body.

Invoking the balances of linear and angular momentum, the work–energy theorem (9.4)1 is established: T˙ = F · v¯ + M · ω. To establish the alternative form, (9.4)2 , of the work–energy theorem, we consider a system of N forces FK acting at N material points X K of the rigid body, and a pure moment Mp acting on the rigid body (cf. Figure 9.1). For such a system of forces and moments, we recall that F

N ± =

FK ,

M=

= v ¯ +

µ

(xK − x¯ ) × FK

+

Mp .

K=1

K=1

Noting that vK

´ N ±

ω × (xK

− x), ¯

we find with some rearranging that

F · v¯ + M · ω = Mp · ω +

N ±

FK · vK .

K =1

This result establishes the alternative form, (9.4) 2, of the work–energy theorem. Energy Conservation

In most problems in rigid-body dynamics in which there is no dynamic friction, the total energy E of the body is conserved. To show this, let us assume that the resultant conservative force Fcon and conservative moment (relative to the center of mass) Mcon acting on the body are associated with a potential energy U = U(x¯ , Q): F con = −

∂U ∂ x¯

¯ acting at the point X,

M con = −u Q.

Observe that we used the prescription (8.21) for the conservative force and conservative moment. In addition, we assume that the body is subject to R constraints of the form 1 πX¯ =

0,

...,

R πX¯ =

0,

9.4

333

Additional Forms of the Balance of Angular Momentum

where ¯ + L hX¯ · ω + L eX¯ L πX¯ = L fX¯ · v

(L = 1, . . . , R).

The constraint force and moment are prescribed using Lagrange’s prescription (8.7): Fc

R ± =

L=1

(

µL L fX¯

)

¯ acting at the point X,

Mc

R ± =

L=1

(

µL L h X ¯

)

.

Finally, we assume that the only forces and moments acting on the rigid body are either conservative forces and moments or constraint forces and moments. We now examine the work–energy theorem (9.4) 1 for the rigid body of interest: T˙

=

F · v¯ + M · ω

=

F c · v¯ + Mc · ω + Fcon · v¯ + Mcon · ω.

However, F c · v¯ + Mc · ω =

R ± L=1

F con · v¯ + M con · ω = −

(

µ L L fX¯ · v ¯ + L hX¯ ·

∂U ∂ x¯

·v ¯ −

ω

)

= −

R ± L=1

(

µL L eX¯

)

,

˙ uQ · ω = −U.

Consequently, ˙ − T˙ = −U

R ± L =1

(

µ L L eX ¯

∑

)

. (

)

This implies that E˙ = 0, where E = T + U, if RL=1 µ L L eX¯ = 0. In summary, one situation in which the total energy E is conserved occurs when ¯ = 0, and the constraint forces and constraint moments are prescribed 1 eX¯ = 0, . . . , R eX by Lagrange’s prescription. This situation arises in most of the problems in rigid-body dynamics that are solvable analytically, and we shall see several examples shortly. In the developments above, we have chosen to describe the constraints and the poten¯ It is straightforward to show that we tial energy function using the center of mass X. could have used different material points to describe the constraints and the potential energy function and arrived at identical conclusions.

9.4

Additional Forms of the Balance of Angular Momentum

˙ = M has several component forms and is one of The balance of angular momentum H the most interesting equations in mechanics. In this section, several of these forms are discussed. First, we show that this equation is equivalent to

J

3 ± i=1

ω ˙ i ei + ω × Jω =

M.

334

Kinetics of a Rigid Body

Next, we show that if ei are the principal vectors of J, then we can find Euler’s celebrated equations (9.9). As an intermediate result, we also indicate the corresponding component form of these equations when ei are not principal vectors of J.

A Direct Form

˙ = M can be written as Here, we wish to show that H

Jα + ω × (Jω)

=

M.

˙ To establish this result, we need to examine ω ˙ and J. First, we recall that

H = Jω. Taking the derivative of this expression, we find ˙ = J ˙ ω + Jω H ˙.

To proceed, we need some identities. Specifically, α=ω ˙ =

3 d ± ωi e i dt i=1

3 ± =

=

i =1 3 ±

ω ˙ i ei + ω ×

´ 3 ±

µ ωi e i

i=1 ω ˙ i ei

i =1

and J˙ = ΩJ + JΩT ,

J˙ ω

=

ΩJω − JΩω = ω × Jω.

Using these identities, we find that ˙ = J ˙ ω + Jω H ˙ = ω × Jω + J ω ˙ =

J

3 ±

ω ˙ i ei + ω × J ω.

i=1 ˙ = M is equivalent to In summary, H

J

3 ±

ω ˙ i ei + ω × J ω =

M.

i=1 ˙ = M is very useful when J is a constant tensor – for instance when This form of H dealing with rigid spheres and rigid cubes. It is also used to obtain conservation results for H. ∑ In passing, we note that the result α = 3i =1 ω˙ i ei implies that for ω to be constant it suffices that ωi are constant. This is in spite of the fact that ei may not be stationary.

9.4

Additional Forms of the Balance of Angular Momentum

335

A Component Form

If we choose an arbitrary basis {E1 , E2 , E3 } for E3, then the inertia tensors J0 and J have the representations J0

3 ± 3 ± =

Jik Ei ⊗ Ek ,

J=

i =1 k =1

where ei

=

3 ± 3 ±

Jik ei ⊗ ek ,

i=1 k=1

QEi . Consequently, H = Jω =

3 ± 3 ±

Jik ωk ei .

i=1 k=1

Differentiating H, we find that ˙ = H

˙

Jω =

3 ± 3 ±

Jik ω˙ k ei

+

ω×

i=1 k =1

´ 3 ± 3 ±

µ

Jik ωk ei .

(9.8)

i=1 k=1

˙ = M. HowIf we equate this expression to M, we can find the component forms of H ever, except when ω has a simple form, it is not convenient to consider this component form of the equations. Instances where (9.8) are used include misbalanced rotors, where ω = θ˙ E3 and J13 ± = 0 and/or J23 ± = 0. 3 For cases not involving a fixed axis of rotation, it is prudent to choose {E1 , E2 , E3 } to be the principal directions of J0 .

The Principal Axis Case

If we choose Ei to be the principal directions of J0, then this tensor has the familiar representation J0

=

3 ±

λ i Ei ⊗

Ei ,

i=1

where λi are the principal moments of inertia. The vectors E i are also known as the principal axes of the body in its reference configuration. Defining ei = QEi , we find that the inertia tensor J = QJ0 QT has the representation ∑ J = 3i=1 λ i ei ⊗ ei . Consequently, H = Jω =

3 ±

λi ωi ei .

i=1 ˙ we find Evaluating H, ˙ = Jω H ˙ + ω × Jω =

=

J

3 ±

i=1 3 ±

ω ˙ i ei +

ω×

´ 3 ±

µ λi ωi ei

i=1

λiω ˙ i ei + (λ2 − λ 1)ω1 ω2e3 + (λ 1 − λ3 )ω1ω3 e2 + (λ3 − λ 2)ω3 ω2e1 .

i=1 3 An example of such a system can be found in [215, Section 7, Chapter 9].

336

Kinetics of a Rigid Body

˙ = M has the component form In conclusion, H

λ 1ω ˙ 1 + (λ 3 − λ2 )ω3ω2 =

M · e1 ,

λ 2ω ˙ 2 + (λ 1 − λ3 )ω3ω1 =

M · e2 ,

λ 3ω ˙ 3 + (λ 2 − λ1 )ω1ω2 =

M · e3 .

(9.9)

These equations, known as Euler’s equations, represent three first-order ordinary differential equations for ωi . To determine the rotation tensor Q, it is necessary to supplement (9.9) by the three first-order ordinary differential equations relating ω to Q: (

)

T ˙ . ω = ax QQ

For example, if a set of 3–2–1 Euler angles were used to parameterize Q, then these differential equations would be ⎡

˙ ψ

⎣ θ˙ ˙ φ

⎤

⎤⎡

⎡

0 sin(φ ) sec(θ ) cos( φ) sec(θ ) ⎦⎣ ⎦=⎣ 0 cos( φ) − sin(φ) 1 sin(φ) tan(θ ) cos(φ ) tan(θ )

ω1 ω2

⎤ ⎦.

(9.10)

ω3

You may wish to recall that the 3–2–1 Euler angles were discussed in Section 6.8.1, and differential equations (9.10) can be inferred from the developments there. The corresponding navigation equations for a set of 3–1–2 Euler angles are discussed in Exercise 7.9.

9.5

Moment-Free Motion of a Rigid Body

Moment-free motion of a rigid body occurs when M resolved by solutions of the balance laws ˙ = F, G

=

0. Determining the motion is

˙ = 0 H

for x¯ and Q. It is common to focus exclusively on the balance of angular momentum and determine Q. In addition, although an analytical solution for Q was first found by Carl G. J. Jacobi (1804–1851) in 1849,4 it is usual to focus on ω(t). Another ingenious solution to this problem was presented by Louis Poinsot (1777–1859) in 1834 [229, 230].5 The equations governing the components ωi = ω · ei of the angular velocity vector ˙ · e = 0 (cf. (9.9)): are found from the three equations H i λ1 ω ˙ 1 + (λ3 − λ2)ω3 ω2 =

0,

λ2 ω ˙ 2 + (λ1 − λ3)ω3 ω1 =

0,

λ3 ω ˙ 3 + (λ2 − λ1)ω1 ω2 =

0.

(9.11)

4 Jacobi’s solution is discussed at length in Whittaker [306, Section 69] and Landau and

Lifshitz [164, Section 37].

5 Discussions of Poinsot’s solution can be found in several texts, for instance, Marsden and Ratiu [180] and

Routh [247].

9.5

337

Moment-Free Motion of a Rigid Body

It is easy to see that the solutions to these equations conserve the rotational kinetic energy Trot = 12 H · ω and the angular momentum vector H. Although there are several cases to consider, it suffices to consider three: symmetric body: λ 1 = λ 2 = λ3 ; axisymmetric body: λ1 = λ2 ± = λ3 ; asymmetric body: λ1 < λ 2 < λ 3. For the axisymmetric body, when λ1 < λ3 the body is known as oblate. When λ 1 = the body is known as prolate. For the asymmetric body previously discussed, e1 is known as the minor axis of inertia, e2 is known as the intermediate axis of inertia, and e3 is known as the major axis of inertia. We now turn to discussing the three cases and the solutions for ωi(t).

λ2 > λ3 ,

The Symmetric Body

For the symmetric rigid body, (9.11) simplify to ω˙ i = 0. In other words, the components ∑3 ˙ = ˙ k ek , this implies that ω is constant. If we choose of ω are constant. As ω k=1 ω Q(t0 ) = I, then the axis of rotation q of the body is constant,6 and so we find Q(t)

=

cos(ν )(I − q ⊗ q) + sin(ν )skwt (q ) + q ⊗ q,

where the axis and angle of rotation are q=

ω (t 0) , ²ω (t 0) ²

ν = ² ω (t0 )² (t − t0) ,

and ω(t) = ω (t 0). That is, the constant angular velocity motion is a Type I motion (cf. Section 7.11). For the symmetric body, any axis is a principal axis. Consequently, it is possible to spin such a body at constant speed about any axis. We shall see shortly that there are related results for axisymmetric and asymmetric rigid bodies. The Axisymmetric Body

For an axisymmetric body, it is convenient to define λ t = λ 1 = λ 2 and λ a equations (9.11) governing the components of ω simplify for this case to

= λ3.

The

ω ˙ 1 = kt ±ω2 , ω ˙ 2 = −kt ±ω1 , ω3 = ± ,

(9.12)

where ± = ω3 (t0 ) is a constant and kt

=

λt − λa λt

.

6 As discussed in Section 7.11, other choices of Q ( t ) are possible; however, these may not guarantee that 0

the axis of rotation q of Q is constant.

338

Kinetics of a Rigid Body

Differential equations (9.12) have a simple analytical solution: ¶

ω1(t) ω2(t)

·

¶

=

cos (kt ± (t − t0)) − sin (kt ± (t − t0))

sin (kt ± (t − t0 )) cos (kt ± (t − t0))

·¶

ω1 (t0) ω2 (t0)

·

.

(9.13)

In summary, ωi (t) have been calculated. There are some special cases to consider. First, notice that it is possible to rotate the body at constant speed either about the e3 direction or about any axis in the e1–e2 plane.7 All of these axes are principal axes of the body. Hence, it is possible to spin the body at constant speed about a principal axis. An interesting aspect about the rotation of an axisymmetric body is that the component of ω in the direction of the axis of symmetry e3 is always constant. This occurs even though e3 (t) may be quite complicated and is a consequence of the angular momentum H · e3 being conserved. The conservation of ω3 (t) is one of the key results in rigid-body dynamics and is exploited extensively in designing flywheels. The Asymmetric Body

When the body is asymmetric, its principal moments of inertia are distinct. If we reexamine (9.11) for this case, then we find that if all but one ωi are zero, then the nonzero ωi will remain constant. For instance, if ω2 = 0 and ω3 = 0, then it is possible for ω1 to have any value and for the equations of motion to preserve this value. These results imply that it is possible to rotate the body about a principal axis at constant speed under no applied moment M. Clearly, as with the other two types of rigid bodies, it is possible to spin the body at constant speed about a principal axis. Eigenvectors and Constant Angular Velocity Motions

Arguably, the fastest method to establish the classic result that it is possible to spin a rigid body at constant speed about a principal axis without the application of an external moment is to examine the balance of angular momentum when M = 0 from the perspective of eigenvectors of the inertia tensor.8 When M = 0, Jω ˙ + ω × J ω = 0. As we are interested in the case where ω is constant, we find that ω must satisfy the following identity: ω × Jω = 0.

That is, ω ² Jω. Consequently, ω is either 0 or parallel to an eigenvector of J. As we are interested in the nontrivial case, we arrive at the conclusion that ω is parallel to a principal axis of J. Our arguments here are valid for any rigid body (regardless of whether some or all of the principal moments of inertia are distinct). 7 That is, a vector a e + a e where a2 + a2 = 1 is a principal axis. 1 1 2 2 1 2 8 As discussed by Cayley [43, Section 146] and Truesdell [286], the fact that every rigid body has an axis

about which it can rotate at constant speed without the application of external moments was discovered by the Hungarian scientist Johann A. Segner (1705–1777) in 1755.

9.5

Moment-Free Motion of a Rigid Body

339

In the absence of an external moment, H ² ω when ω is constant. As H is a constant vector, the instantaneous axis of rotation i points in a constant direction in space. This aspect of moment-free motions of a rigid body is one of the signature ideas behind the operation of gyroscopes. The Momentum Sphere

To visualize the solutions of (9.11), a graphical technique is often used. This technique dates to the mid-nineteenth century. It is based on two facts: the solutions ωi (t) to (9.11) preserve the magnitude of H and the rotational kinetic energy T rot. As a result, the solutions hi (t) = H · ei

= λ i ωi(t)

(i = 1, 2, 3)

lie on the intersection of the constant surfaces h = h0 and Trot h2

=

Trot

=

h21 + h22 + h23 , h21 2λ1

+

h22 2λ 2

+

=

T E . Here,

h23 , 2λ 3

and the values of h0 and TE are determined by the initial conditions ωi (t0 ). If we pick a value h0 of h, the surface h = h0 in the three-dimensional space h1 – h2–h3 is a sphere – the momentum sphere. Selecting a value of TE , we find that the surface T rot = T E in the three-dimensional space h1–h2–h3 is an ellipsoid – the energy ellipsoid. The intersection of the ellipsoid with the sphere is either a discrete set of points or a set of curves.9 These intersections are the loci of hi (t). For the axisymmetric body, the intersections are shown in Figure 9.2(a). Corresponding representative intersections for an asymmetric body are shown in Figure 9.2(b).10 These figures are among the most famous in dynamics. The graphical method behind their construction is credited to Poinsot. The first version of these figures can be seen in [230, Figure 25]. A more complete version of Poinsot’s figure is displayed in a paper [181] authored by James C. Maxwell (1831–1879) on an ingenious top, published in 1857. 11 For the case presented in Figure 9.2(a), the energy ellipoid has an axis of revolution (in this case, the third axis). For a symmetric body, the energy ellipsoid degenerates further into a sphere. This sphere coincides with the momentum sphere and so the graphical technique used to determine hi (t) (and ωi (t)) appears to break down. However, for the symmetric case, we found previously that ωi (t) = ωi (t0 ). Consequently, each point on the momentum sphere corresponds to a steady rotational motion of the rigid body. Stability and Instability of the Steady Rotations

We can use the portrait of the trajectories of λ i ωi (t) on the momentum sphere to deduce some conclusions on the nature of the steady rotational motions of a rigid body. Our 9 For a more detailed discussion of these intersections, Synge and Griffith’s text [276] is highly

recommended.

10 These figures were kindly supplied by Patrick Kessler in the spring of 2007. 11 A discussion of Maxwell’s top can be found in [304, Section 87].

340

Kinetics of a Rigid Body

(a)

λ 3ω 3

e2

λ 1ω 1 λ2 ω2

(b)

λ 3ω 3

λ 1ω 1

e2 λ2 ω2

Trajectories of the components λi ωi on the momentum sphere. The curves and points on the sphere are the intersection of the momentum sphere with the energy ellipsoid. For these figures, two distinct rigid bodies are shown: (a) λ1 = λ 2 = 4 and λ3 = 5 and (b) λ 1 = 2, λ 2 = 4, and λ 3 = 5. The figures on the right-hand side show a moment-free motion of the e2 vector that is corotating with a rectangular box motion that corresponds to one of the trajectories on the sphere. For the trajectories and simulations shown in this figure, (9.10) and (9.11) were numerically integrated.

Figure 9.2

discussion is qualitative, and more rigorous presentations of this topic can easily be found elsewhere. For example, analyses of the stability of the steady motions can be found in [112, 133, 180]. For the axisymmetric case, the trajectories shown in Figure 9.2(a) can be used to infer that a steady rotation about the e3 axis is stable. By stability, we mean that, if we perturb the body’s rotation slightly from this steady state, then hi (t) (or equivalently ωi (t)) will remain close to the state (h1(t), h2 (t), h3 (t)) = (0, 0, h3s ), where h3s is the value of h3 corresponding to the steady rotation.12 In contrast, the trajectories in Figure 9.2(a) show that steady rotations about any axis in the e1–e2 plane do not satisfy this condition. Consequently, such steady rotations are unstable. For the asymmetric case, the trajectories shown in Figure 9.2(b) confirm the previous statements that six steady rotations are possible. The trajectories in this figure also illustrate that the four steady rotations about e1 and e3 are stable, whereas the pair of steady 12 Because h i

= λ i ωi ,

it is trivial to ascribe results pertaining to hi to those for ωi.

9.5

341

Moment-Free Motion of a Rigid Body

rotations about the e2 axis is unstable. That is, a rotation about the intermediate axis of inertia is unstable, whereas those about the major (e3) and minor (e1 ) axes are stable. Attitudes of the Rotational Motions

The information we have thus far gleaned from the momentum sphere does not tell the full story about the motion of the rigid body. What is missing is information on the behavior of Q. To find this information, one can use Jacobi’s analytical solutions discussed earlier. Alternatively, one can choose a parameterization for Q and numerically integrate the equations relating ωi to these parameterizations. For example, if a set of 3–2–1 Euler angles is used then, in addition to integrating (9.11), (9.10) would also be integrated to determine φ (t), θ (t), and ψ (t). With the help of these results, ei (t) can be constructed and the motion of the body visualized. Results from two distinct examples of the numerical integrations of (9.10) and (9.11) are shown in Figures 9.3, 9.4, and 9.5. One of these simulations corresponds to a perturbation of the steady rotation of a rigid body rotating about the principal axis corresponding to its maximal moment of inertia. The resulting behaviors of ωi (t) are displayed in the trajectory labeled (i) in Figure 9.3(b), whereas the behaviors of the corotational basis vectors can be seen in Figure 9.4. It should be clear from the former figure that the perturbation to the steady motion does not alter ei (t) appreciably from their steady rotation behaviors. The easiest method of visualizing these results is to toss a book into the air with an initial rotation primarily about e3 and observe that, although the book will wobble, its instantaneous axis of rotation does not wander far from its initial state. In Figure 9.5, the behaviors of ek (t) corresponding to a trajectory of ωi (t) that passes close to the equilibrium (ω1, ω2, ω3 ) = (0, ω0 , 0) that is labeled with a “star” in Figure 9.3(b) are shown. It should be clear from the behavior of e2 (t) shown in Figure 9.5(b) that e2 (t), which is initially close to E2 at time t = 0, makes large excursions from its initial value. This is in contrast to the situation shown in Figure 9.4 and is indicative of

−5

(b)

(a)

E3

5

6

(i) (ii)

¯

X

E1

ω2

ω3

E2 ω0

−5 5

0

ω1

The moment-free motion of a rigid body: (a) a rigid body showing the principal axes and (b) the components ωi (t) = ω · ei corresponding to two different sets of initial conditions: (i) ω(0) = 0.5E2 + 5.0E3 and (ii) ω(0) = 5.0E 2 + 0.1E3 . For these simulations, (9.10) and (9.11) were numerically integrated with the initial condition Q(0) = I and the parameter values λ1 = 2, λ 2 = 4, and λ 3 = 5.

Figure 9.3

342

Kinetics of a Rigid Body

1

−1

e1

e2 · E2

−1

e1 · E3

−1

1

(b)

e1 · E2

(a)

O

e2 · E 3

−1

e1 · E1 1

e2

O

e2 · E 1 1

1

e3 · E2

(c)

e3

−1 e3 · E3

−1

O

e3 · E1 1

Figure 9.4 Simulation results indicating the stability of the steady rotation of the rigid body about the principal axis corresponding to the maximal axis of inertia: (a) the Ei components of e1(t), (b) the Ei components of e2(t), and (c) the Ei components of e3 (t). These results correspond to the trajectory labeled (i) in Figure 9.3(b).

the instability of the steady moment-free motion of a rigid body about its intermediate axis of inertia. The easiest way to see this instability is to take a book and give it an initial angular velocity about e2 . One will see a wobbling motion where ω · e2 will periodically take positive and negative values. Interestingly, it is possible to execute this motion and have the book perform a rotation about e1 or e3 by 180◦. This twisting motion was only recently noted and analyzed by Ashbaugh et al. [13].13 A recent video of a T-handle spinning aboard the International Space Station is also a wonderful demonstration of this instability. 14 13 In [13], two distinct sets of Euler angles are used to avoid the singularities inherent in Euler angle

parameterizations of rotation tensors. For the results presented in Figures 9.4 and 9.5, it was not necessary to introduce a second set. 14 The video can be found at https://www.youtube.com/watch?v=1n-HMSCDYtM. A simulation of the spinning T-handle executing this motion can be seen on the online resource https://rotations.berkeley.edu/a-tumbling-t-handle-in-space/.

9.6

343

The Baseball and the Football

1 1

e1 · E 2

(a)

(b)

e1

−1 e1 · E 3

−1

e2 · E 2

−1 O

e2

e2 · E 3

O

−1

e1 · E 1

e2 · E1

1

1 1

e3 · E 2

(c)

e3

−1 O

e3 · E3

−1 e3 · E1 1 Figure 9.5 Simulation results indicating the instability of the steady rotation of the rigid body about the principal axis corresponding to the intermediate axis of inertia: (a) the Ei components of e1(t), (b) the Ei components of e2 (t), and (c) the Ei components of e3(t). These results correspond to the trajectory labeled (ii) in Figure 9.3(b).

9.6

The Baseball and the Football

Consider a sphere of mass m and radius R that is thrown into space with an initial velocity v(t ¯ 0 ), angular velocity ω(t0 ), and orientation Q(t0). We wish to determine the motion x¯ and Q of the sphere. As discussed by Tait [280], it was known to Isaac Newton that the rotation of the sphere as it moves through the ambient air causes a curvature of the path traced by the center of the sphere. This feature results in interesting dynamics in a variety of sports ranging from golf to baseball and soccer. Our interest here is to examine the curving of the path traced by the center of mass of the sphere. We do this by following several classic works on this problem, most notably Tait [280].

344

Kinetics of a Rigid Body

The Magnus Force

A key force experienced by the sphere is known as the lift or Magnus force (see Figure 9.6):15 FM

=

mB ω × v, ¯

where B is a positive constant. The sign of B is determined by use of Bernoulli’s equation. 16 Clearly, this force models the coupling between rotation and linear velocity. Recent research on free kicks in soccer has shown that there can be a transition in the flow field from turbulent to laminar that causes dramatic changes in the trajectory (see [35]). According to Ireson [136], for some free kicks in soccer, || v¯ || = 25 m/s and mB ≈ 0.15716 kg. Apart from gravity and the Magnus force, the other important force in this problem is the drag force: v¯ 1 . FD = − ρ f ACd (v¯ · v¯ ) 2 ²v ¯² In this expression, Cd is the drag coefficient, ρ f is the density of the fluid that the sphere is moving in, and A is the frontal area of the sphere in contact with the fluid: A = π R2 .

Equations of Motion

For the system at hand, M is constant:

=

0 and H

=

ω(t)

2mR 2 5 ω, =

so we find the important result that ω

ω (t0 ) .

ω

ω v¯

¯

X

FM

v¯

¯

X

FM

Figure 9.6 A rigid sphere whose center of mass is moving to the right with a velocity vector v. ¯ When the ball is rotating clockwise, the velocity of the air moving over the top of the ball is slower than the velocity of the air in contact with the bottom of the ball. From Bernoulli’s equation, the pressure on the top of the ball is greater than the pressure on the bottom of the ball and a net downward force F M results. The opposite occurs when the ball is rotating counterclockwise.

15 Credited to the German scientist Heinrich Gustav Magnus (1802–1870) in 1851 (see [175, 176]). 16 Bernoulli’s equation applies to inviscid fluid flow and states that the sum of the pressure p and 1 ρ U2 is a 2 f constant. Here, U is the fluid flow velocity and ρf is the fluid density.

9.6

345

The Baseball and the Football

In other words, the angular velocity of the sphere does not change. As with the symmetric body discussed earlier, we can easily solve for the rotation tensor of the sphere if we assume that Q (t 0) = I: Q(t)

=

cos(ν ) (I − q ⊗ q ) + sin(ν )skwt (q ) + q ⊗ q,

where the axis and angle of rotation are q=

ω (t 0) , ²ω (t 0) ²

ν = ² ω (t0 )² (t − t0) .

We shall see that solving for the motion of the center of mass in this problem is not trivial. ∑ The rotation tensor Q = 3k =1 ek ⊗ Ek for the rigid body is parameterized by a set of 3–1–3 Euler angles (see Section 6.8.2). The Euler basis vectors for this parameterization have the representations ⎡

⎤

⎡

⎤

⎡

⎤⎡

⎤

g1 E3 sin(φ) sin( θ ) cos( φ) sin( θ ) cos( θ ) e1 ⎣ g ⎦ = ⎣ e³ ⎦ = ⎣ ⎦ ⎣ cos(φ ) − sin(φ ) 0 e2 ⎦ 2 1 g3 e3 0 0 1 e3 ⎡

0 ⎣ cos(ψ ) = sin(θ ) sin( ψ )

0 sin(ψ ) − sin(θ ) cos(ψ )

Further, the Euler angles are subject to the following restrictions: (0, π ), and ψ ∈ [0, 2π ). Using these Euler angles, we find that ω = ±1E1 + ±2E2 + ±3 E3 =

(

)

˙ sin(θ ) sin( ψ ) θ˙ cos(ψ ) + φ

(

˙ +φ ˙ + ψ

)

E1 +

(

θ˙

⎤⎡

⎤

E1 1 ⎦ ⎣ E2 ⎦ . 0 E3 cos(θ ) (9.14) φ ∈

[0, 2π ),

θ ∈

)

sin(ψ ) − φ˙ sin(θ ) cos(ψ ) E2

cos( θ ) E3.

If we use a set of Cartesian coordinates for the position vector of the center of mass, x¯ · Ei = xi , then we would find that ω × v¯ = (x˙ 3±2 − x˙ 2±3 ) E1 + (x˙ 1±3 − x˙ 3 ±1 ) E2 + (x˙ 2 ±1 − x˙ 1 ±2) E 3.

(9.15) From the previous solution to the balance of angular momentum, we know that ±i are constant. The balance of linear momentum for the sphere provides the equations of motion for ˙ in Cartesian coordinates, we find the center of mass. Evaluating F = G

346

Kinetics of a Rigid Body

mx¨ 1 = (F D + F M) · E1 , mx¨ 2 = (F D + F M) · E2 , mx¨ 3 = −mg + (F D + F M ) · E3 . Ignoring the drag force and using (9.15), we find three differential equations for xi (t): mx¨ 1 = mB (x˙ 3±2 − x˙ 2 ±3) , mx¨ 2 = mB (x˙ 1±3 − x˙ 3 ±1) , mx¨ 3 = mB (x˙ 2±1 − x˙ 1 ±2) − mg.

(9.16)

For the general case, these equations can be integrated numerically to determine x¯ (t).

The Path Traced by the Center of Mass of the Ball

Turning our attention to a simple case, suppose ω (t 0) = ±10 E1. From the previous analysis, we know that ω is constant for this rigid body. Consequently, (9.16) simplifies to mx¨ 1

=

0,

mx¨ 2

= −mBx ˙ 3±10 ,

mx¨ 3

=

mBx˙ 2±10 − mg.

The solution to these differential equations, assuming B ±10 ± = 0, is ⎡

where

⎡

⎤

x1(t) − x1 (t0 ) ⎣ x2(t) − x2 (t0 ) ⎦ x3(t) − x3 (t0 )

=

0

⎤

0

⎢ ⎥ ⎢ ⎥ A ⎣ x˙ 2 (t0 ) − B±g10 ⎦ + ⎣ gB( t±−t0 ) ⎦ , 10 ˙x3 ( t0 )

⎡

t − t0 ⎢ A=⎣ 0

⎡

⎤

˙x1 ( t0 )

0

sin( B±10 (t−t 0 )) B±10 1−cos (B±10 (t −t0 )) B±10

0

⎤

0

−

(9.17)

1−cos (B±10 (t −t0 )) ⎥ ⎦. B±10 sin (B±10 (t −t0 )) B±10

From (9.17), the trajectory of the sphere can be determined. Two important features are present. First, the spin ±10 influences the forward speed x˙ 2 of the sphere. Second, the spin also impacts the vertical position and speed. It is also of interest to compare (9.17) with the corresponding solution when the Magnus force is absent. In this case, the path traced by the center of mass is the well-known parabolic trajectory: ⎡

⎤

⎡

⎤

⎡

(t − t0 ) x ˙ 1 ( t0) x1(t) − x1 (t0 ) ⎣ x2(t) − x2 (t0 ) ⎦ = ⎣ (t − t0 ) x˙ 2 (t0) ⎦ − ⎣ ˙ 3 ( t0) x3(t) − x3 (t0 ) (t − t0 ) x

0 0

g t − t0 )2 2(

⎤

⎦.

Representative examples of the trajectories of a point launched from the origin are shown in Figure 9.7. For small values of B±10, we can see from this figure how the trajectory differs from that in which the Magnus force is absent: a spin ( ±10) in one direction will result in the ball “rising,” whereas a “dipping” effect can be observed by

9.7

Motion of a Rigid Body with a Fixed Point

347

15 (vi)

(v)

x3

x2

10

30 (iv) (i)

(ii)

(iii)

−20 Figure 9.7 The trajectories of a sphere that is launched from the origin with an initial velocity v¯ (t0 ) = 10 (E2 + E3) and an initial angular velocity ω (t0) = ±10 E1. The trajectories shown correspond to different values of B±10 : (i) B±10 = −0.5, (ii) B±10 = −0.2, (iii) B ±10 = 0.0, (iv) B±10 = 0.2, (v) B±10 = 0.5, and (vi) B±10 = 1.0. All of the trajectories are displayed for a period of 4 s and g = 9.81 m/s2.

reversing the initial spin. However, we also observe that, for larger values of | B±10| , the behavior of the trajectories becomes unphysical, either through the appearance of a cusp or the reversal in the sign of x˙ 3.17 Thus, the prescription of the Magnus force may have a limited range of physical applicability. Our formulation of the equations of motion for this problem is simplified by the fact that the moment of inertia tensor for the sphere has a simple form. Indeed, it is interesting to compare the flight of a ball predicted by this model with that of a Frisbee. The interested reader is referred to [132, 134] where the equations of motion of a Frisbee are formulated and the lift and drag forces on the Frisbee are computed from experiments.

9.7

Motion of a Rigid Body with a Fixed Point

The problem of a body that is free to rotate about a fixed point XO , which is also a material point of the body, occupies a celebrated place in the history of mechanics. An example of such a body is shown in Figure 9.8, and related systems can be found in the pendula in clocks and several types of spinning tops. In these systems, there are three constraints on the motion of the rigid body and F = ma¯ serves to determine the constraint (reaction) forces at XO that enforce these constraints. The remaining balance ˙ , has an interesting form and is used to determine the rotation tensor Q law, MO = H O of the rigid body. 17 As can be seen from Figure 4 of [280], motions of this type are also present in Tait’s analysis of this

problem.

348

Kinetics of a Rigid Body

Ball-and-socket joint

e1 A

XO

¯ X

g

e3 Linear spring Xs

Figure 9.8 An example of a rigid body that is free to rotate about a fixed point X O , where the fixed point is a material point of the body. In this example, x¯ − xO = he3, where h is a constant, and the body is also subject to conservative forces from the linear spring and gravity.

Kinematics

For the body of interest, the material point XO is fixed at a point that we take to be the origin: xO = 0. We assume that the position vector of the center of mass relative to XO has the representation x¯ − xO

= ² 1e1 + ²2 e2 + ²3e3 ,

where ²i are constants. Differentiating this equation with respect to time, we see that v¯ = ω × (²1 e1 + ²2 e2 + ²3e3 ) .

(9.18)

This relation can be used to establish a convenient representation for the angular momentum HO in terms of an inertia tensor JO for the body relative to XO and the kinetic energy T of the rigid body. To establish the representation for HO , we start with the relation from this quantity in terms of H and G: HO

=

H + x¯ × G

=

Jω + m (²1 e1 + ² 2e2 + ²3 e3 ) × (ω × (²1 e1 + ²2 e2 + ²3e3 ))

=

JO ω.

(9.19)

For the final step in this result, we use the identity a × (b × c) The inertia tensor JO in (9.19) is JO

=

J+m −

²

2 2 2 ²1 + ²2 + ²3

³

=

(a · c)b − (a · b)c.18

I

m (²1e1 + ² 2e2 + ²3 e3 ) ⊗ (²1 e1 + ²2 e2 + ²3e3) .

As expected, this expression for JO is in agreement with the one that we would obtain by using the parallel axis theorem (see (7.39)). Starting with the Koenig decomposition (7.25), a series of standard manipulations provides a convenient expression for the kinetic energy of the rigid body: 18 As can be seen from (7.23), this identity was used earlier to establish the representation H = Jω.

9.7

T

Motion of a Rigid Body with a Fixed Point

=

1² O ³ J ω · ω. 2

349

(9.20)

The details are left as an exercise. We can show that JO is a positive definite symmetric tensor, and consequently it will have a set of principal axes. We now choose ei to be these axes and, as a result, we can write JO

O

= λ 1 e1 ⊗

O e1 + λO 2 e2 ⊗ e2 + λ3 e3 ⊗ e3 .

It is straightforward to show how λ O i are related to the components of J and m, ²1 , O ²2 , and ²3 . We also take this opportunity to note that the principal axes of J are not necessarily parallel to the principal axes of J.

Constraint Forces and Constraint Moments

The motion is subject to three (integrable) constraints: ³1 =

0,

³2 =

0,

³3 =

0.

These constraints arise because the point XO is fixed: ³i = (x ¯ − ²1 e1 − ²2 e2 − ²3e3 ) · Ei .

(9.21)

Differentiating the constraints, we find (9.18). Assuming that the joint at XO is frictionless, we can easily use (9.18) with Lagrange’s prescription to show that Fc

= µ1 E1 + µ2 E2 + µ 3E3

Mc

¯ acting at the point X,

= (−²1 e1 − ²2e2 − ²3 e3) ×

Fc .

Here, F c and Mc are equipollent to the force Fc acting at the joint XO . You may wish to recall that we considered the constraint forces and moments for the situation in which the joint at X O was a pin joint in Section 8.3. For the case where the body was attached by a pin joint at X O , M c would have two additional components.

Equations of Motion

For this problem, it is convenient to follow the procedure of first using F = ma¯ to solve ˙ for the three unknown components of Fc and then using M O = H O to solve for the motion of the body. ˙ The balance law MO = H O can be written in components relative to the basis vectors ei . Recalling that we are choosing this basis to be parallel to the principal axes of JO , we find the component form O O O λ1 ω ˙ 1 + (λ3 − λ2 )ω3 ω2 = O O O λ2 ω ˙ 2 + (λ1 − λ3 )ω3 ω1 = O O O λ3 ω ˙ 3 + (λ2 − λ1 )ω1 ω2 =

MO · e1, MO · e2, MO · e3.

(9.22)

350

Kinetics of a Rigid Body

These equations have obvious parallels to Euler’s equations (9.9). Indeed, if MO = 0, then we can use the solutions for moment-free motion that we discussed previously, with a small number of modifications. Equations (9.22) need to be supplemented by equations relating ω to Q in order to solve for the rotation tensor of the body.19 Once the rotation has been found, we can then use F = ma¯ , where a¯ = α × x¯ + ω × (ω × x¯ ), to determine the reaction force F c at XO .

Euler–Poisson Equations

An alternative formulation of the equations of motion for the case in which a gravitational force −mgE3 is acting on the body is known as the Euler–Poisson equations.20 For these equations, instead of using a set of 3–1–2 Euler angles to parameterize Q and then supplementing (9.22) with (7.42), we work with three of the nine components of Q. ˙ = QΩ in terms of components. Relative to By way of preliminaries, let us write Q 0 the basis Ei ⊗ Ek , we see that ⎡

˙ Q 11

˙ Q 12

˙ Q 13

˙ Q 31

˙ Q 22 ˙ Q 32

˙ Q 23 ˙ Q 33

⎢ ˙ ⎣ Q21

⎤

⎡

Q11 ⎥ ⎣ = Q21 ⎦ Q31

Q12 Q22 Q32

⎤⎡

Q13 Q23 ⎦ ⎣ Q33

0

−ω3

ω3

0

− ω2

ω1

ω2 − ω1

⎤ ⎦.

(9.23)

0

Relations of this form for the time derivatives of the components of the rotation tensor were first found by Poisson in the early nineteenth century21 and are known as the Poisson kinematical relations. It is important to note that E3

=

Q31e1 + Q32e2 + Q33 e3 .

(9.24)

Because of the moment that is due to gravity, we shall subsequently need the components E3 · ei . Note that, by differentiating (9.24) and using the identity e˙ i = ω × ei , we would discover that ˙ ˙ ˙ Q 31 e1 + Q32 e2 + Q33 e3

3 ± = −

Q3k ω × ek .

k=1

These relations constitute three differential equations for Q3i , which are equivalent to those from (9.23). The coefficients Q3i = e3 · Ei are often known as the direction cosines of e3 . The equations of motion for the rigid body consist of three equations governing Q3i and the balance of angular momentum relative to X O . Thus, we combine the balance of 19 An explicit form of the differential equations can be found in Exercise 9.14 at the end of this chapter: see

(9.46).

20 We shall discuss a third alternative, Lagrange’s equations of motion, in Section 10.2 (Chapter 10). 21 See Section 411 of Poisson’s treatise [231].

9.7

Motion of a Rigid Body with a Fixed Point

351

angular momentum that is due to Euler and kinematical relations that are due to Poisson. In component forms, the Euler–Poisson equations are ˙ Q 31 = Q32 ˙ 32 = Q33 Q

hO3 O λ3

hO1 O λ1

−

Q33

−

Q31

hO1

hO2 O λ2

hO3 O λ3

, ,

hO2 + Q31 , O O λ1 λ2 ( O ) O λ2 − λ3 ˙ hO2hO3 + (mgE3 × x¯ ) · e1 , hO1 = O O λ2 λ3 ) ( O O λ3 − λ1 h˙ O2 = hO1hO3 + (mgE3 × x¯ ) · e2 , O O λ1 λ3 ( ) O O λ1 − λ2 ˙ hO1hO2 + (mgE3 × x¯ ) · e3 . hO3 = O λO 1 λ2 ˙ Q 33 = −Q32

(9.25)

Here, we have used the representations HO

3 ± =

hOiei ,

i=1

hOk

O

= λk ωk

(k = 1, 2, 3).

Additional developments and representations of the Euler–Poisson equations (9.25) can be found, for example, in [22, 273].

Conservations

For the problem of interest, we can use representation (9.20) for T to show the following forms of the work–energy theorem: T˙ =

MO · ω =

N ±

FK · vK + Mp · ω.

K=1

If the applied forces and moments acting on the system are conservative then, because Fc acts at a point with zero velocity, it is easy to show that the total energy of the rigid body is conserved. This situation arises when a gravitational force acts on the body. If M O = 0, then HO is conserved. However, in the most common form of this problem a gravitational force −mgE3 acts on the rigid body. In this case, MO ± = 0, but M O has no component in the E3 direction. It is easy to see for this case that HO · E3 is conserved. O If, in addition, the body has an axis of symmetry and λ O 1 = λ 2 , then you should be able to show with the help of (9.22) that HO · e3 is also conserved. The problem in which the body has an axis of symmetry and M O = x¯ × (−mgE3 ) is often known as the (symmetric) Lagrange top.22 The motion of this top conserves E, HO · e3, and HO · E 3 and is one of the most famous mechanical systems. Indeed, 22 In the context of Lagrange’s equations of motion, we will discuss this problem in Sections 10.2 and 10.9.

352

Kinetics of a Rigid Body

as discovered by Lagrange,23 analytical solutions for its equations of motion can be found. This discovery is remarkable, for if we relax the assumption that the body is O symmetric (i.e., λO 1 = λ2 ), then analytical solutions are possible in only a handful of special cases (see [169, 306]). Indeed, it is an interesting exercise to numerically integrate the equations of motion for the case in which MO = x¯ × (−mgE3) and the moments of inertia λO i are distinct.

9.8

Motions of Rolling Spheres and Sliding Spheres

The problem of the homogeneous sphere moving on a flat surface has several applications, bowling and pool being the most famous. The most celebrated classic treatments of this problem are due to Gaspard G. de Coriolis (1792–1843) [54] and Edward J. Routh (1831–1907) [247], and generalizations of their works occupy prominent places in the literature on nonholonomically constrained rigid bodies to this day.24 Indeed, studies on the dynamics of rolling spheres on surfaces of revolution are often known as Routh’s problem. 25 In our treatment, we assume that the surface is rough with a coefficient of static Coulomb friction of µ s and kinetic friction of µd . Of particular interest to us will be the transition between rolling and sliding. Our discussion is heavily influenced by Routh [247] and Synge and Griffith [276]. Consider the sphere moving on the surface shown in Figure 9.9. The radius of the sphere is R and the velocity of the instantaneous point of contact XP of the sphere with the incline is vP

= v ¯ +

ω × (−RE3) .

E3 Sphere of mass m and radius R g

O

E1 ¯

X

Inclined plane XP β

A rigid sphere moving on an inclined plane. The angle of inclination of the plane is β , and a gravitational force −mg cos(β )E3 + mg sin( β)E1 acts on the rigid body.

Figure 9.9

23 See Section IX.34 of Part II of Lagrange’s Mécanique Analytique [159]. This analytical solution is also

discussed by Whittaker [306].

24 See, for example, Borisov and Mamaev [29], Frohlich [89], and Huston et al. [135]. The latter papers

discuss the dynamics of bowling balls.

25 The interested reader is referred to the informative account of Routh’s influence as a tutor and teacher

in [303].

9.8

Motions of Rolling Spheres and Sliding Spheres

353

Note the simple expression for πP here. Because the point XP is the instantaneous point of contact, vP · E3

=

0.

Consequently, this velocity field has the representations vP

¸

=

vs1 E 1 + vs 2 E2

=

uc,

where u = v2s1 + v2s2 is the slip velocity and c = vuP is the slip direction. When the sphere is rolling, there are two additional constraints on vP and, as a result, vs1 = 0 and vs2 = 0. For the rolling sphere, the slip direction is not defined. The resultant force and moment on the sphere are F

= −mg cos(β )E3 +

¯ mg sin(β )E1 + NE3 + Ff acting at X,

M = −RE3 × Ff . When the sphere is rolling, Ff

= µ1 E1 + µ 2E2

acting at XP ,

where µ 1 and µ 2 are unknowns. For the sliding sphere, in contrast, we have the classic prescription Ff

= −µd | N |c

acting at XP .

For convenience, we use the same notation for the friction forces for the rolling sphere and the sliding sphere, but this should not cause confusion.

Rolling Sphere

We determine the motion of the rolling sphere by using the balance laws and the con∑ straints vP = 0. Using Cartesian coordinates for x¯ and setting ω = 3i=1 ±i Ei , we find that these equations are x˙ 1

=

R±2 ,

x˙ 2

= −R±1,

x˙ 3

=

0,

mx¨ 1

=

mg sin(β) + µ 1,

mx¨ 2

= µ2 ,

0 = N − mg cos(β ), 2 2˙ mR ±1 = Rµ 2, 5 2 2˙ mR ±2 = −Rµ1 , 5 2 2˙ mR ±3 = 0. 5

354

Kinetics of a Rigid Body

To solve these equations, it is convenient to first determine the differential equations governing ±i . From the nine equations just listed, we can eliminate several variables to find º

¹

1+ ¹

1+

2 ˙1 mR 2± 5 º 2 ˙2 mR 2± 5 2 2 ˙3 mR ± 5

=

0,

=

mgR sin(β ),

=

0.

These equations are easily solved: ω (t) = ω (t 0) +

¹

g sin(β ) (t − t 0) R

It is left as an exercise to determine x(t). ¯ When β in a straight line at constant speed.

=

1+

2 5

º−1

E 2.

(9.26)

0, you will find that the sphere rolls

Sliding Sphere

For the sliding sphere, it is convenient to examine the differential equations for vP . Differentiating this velocity, we find that v˙ P

˙ s 2 E2 = v ˙ s1 E1 + v ˙¯ + = v

α × (−RE3 ) .

Using the balances of linear and angular momentum, we substitute for v˙¯ and α to find ¹

m˙vs 1

=

º

1+

5 Ff · E1 + mg sin(β ), 2

¹

m˙vs 2

=

º

1+

5 Ff · E2 . 2

(9.27)

After substituting for the friction force, these equations provide two differential equations for the slip velocities. 26 It is convenient to express these equations as differential equations for u and the angle χ , where c = cos(χ )E1 + sin(χ )E2,

cos( χ ) =

vs1 , u

sin(χ )

=

vs 2 . u

When χ is constant, the slip direction c is constant. After some manipulations, we find that (9.27) are equivalent to27 ¹

u˙ = −µd g 1 +

5 2

º

+

g sin(β ) cos(χ ),

uχ˙

= −g sin(β ) sin(χ ).

(9.28)

These differential equations have analytical solutions for χ (t) and u(t). It is left as an exercise to write the five differential equations governing x1 , x2 , and ±i . 26 It is left as an interesting exercise to nondimensionalize and numerically integrate (9.27) to determine the

behavior of the slip velocity components as the ratio of µd to sin(β ) is varied. vs2 2 = v2 s1 + v s2 and sin(χ ) = u and then used (9.27).

27 To obtain these equations, we differentiated u2

9.9

Chaplygin’s Sphere

If we consider the simple case in which the incline is horizontal, then differential equations for the slip velocity simplify considerably to ¹

mv˙ s1

= −µ d mg

1+

5 2

º

¹

vs 1 , u

mv˙ s 2

= −µd mg

1+

5 2

º

β =

355

0. The

vs 2 . u

From these equations, we will find that vs1 and vs2 always tend to zero. To see this, it is best to look at (9.28) and set β = 0: º

¹

u˙ = −µ dg 1 + These equations have the solution

¹

u (t ) = u (t0) − µd g 1 +

5 2

5 , 2

χ ˙ =

0.

(9.29)

º (t − t0) ,

χ (t ) = χ (t 0) .

As a result, u will reach zero in a finite time T and the slip direction remains constant: ¹

T=

5 u (t0 ) 1+ µd g 2

º−1

.

It can be shown that the path traced by the center of the sliding sphere is either a fixed point, a straight line, or a parabolic arc. Once u = 0, the sphere starts rolling. Now, as β = 0, this implies that the sphere will roll at constant speed in a straight line (see Figure 9.10). It is interesting to note that once the sphere starts rolling, it will stay rolling. The transition between the parabolic path during sliding and the straight-line path during rolling is key to hook shots in bowling and massee shots in pool. 28 An example of this transition is shown in Figure 9.10(a). The factor of 52 in the equations of motion for rolling and sliding spheres is also interesting. It is related to the fact that the height of the “center of oscillation” Q of a sphere relative to the center of mass is 2R . As discussed by Coriolis [54], Q is the point 5 one aims for when hitting a cue ball so that it rolls without slipping immediately after the impact of the tip of the cue with the ball.

9.9

Chaplygin’s Sphere

The previous analyses of the rolling sphere assumed that it was homogeneous. Thus, the moment of inertia tensor J was a multiple of the identity. We now consider a rigid body which has an asymmetric tensor J but a spherical outer surface. The dynamics of such a body rolling on a horizontal surface were first considered by Chaplygin [47] in 1903 and the problem is known as Chaplygin’s sphere (or ball) in his honor. With the design and development of spherical robots, interest in the dynamics of Chaplygin’s sphere has increased recently. 29 In contrast to the sphere considered previously, 28 As discussed by Frohlich [89], bowling balls have offset centers of mass and moment of inertia tensors

that are not multiples of I. As a result, some of the intricacies of bowling are not explained by our simple model of rolling spheres and sliding spheres. 29 We refer the interested reader to [24, 27, 28, 255, 259] for references and additional results.

356

Kinetics of a Rigid Body

(a)

s

E2

x¯ E1

O

r

()

u t

( )

(b)

u t0

s

E2

s

x¯ 0

t r

−

t0

O

E1 r

(c)

E2

s

x¯ O

E1

r

Figure 9.10 Plot of slip speed u(t) as a function of time for a sphere that is initially sliding (and eventually rolls) on a rough horizontal plane. For the rolling phases, labeled r, the path traced by the center of mass is a straight line. By way of contrast, for the sliding phases, labeled s, the center of mass can describe (a) a parabolic arc, (b) a straight line, or (c) remain stationary.

E3

¯

X g

E1

O

E2 XP

Rough horizontal plane

Schematic of a rigid body rolling on a rough horizontal plane. The contact surface of the rigid body with the plane is spherical with radius R.

Figure 9.11

the motion of the center of mass can be very intricate (see [147]). In this section, we establish the differential equations governing the motion of the rolling Chaplygin sphere. The corresponding developments for a sliding Chaplygin sphere are discussed by Moshchuk [198]. As shown in Figure 9.11, one method of realizing Chaplygin’s sphere involves a system of six masses rigidly attached to a spherical shell of radius R and mass ms . Referring to Figure 9.12, the masses can be positioned so that the center of mass X¯ of the composite rigid body is at the geometric center of the spherical shell of radius R. The composite rigid body, which has mass m = ms + 2m1 + 2m2 + 2m3 , is assumed to roll on a rough

9.9

Chaplygin’s Sphere

357

m3

r1

r2

m2

r3

m1

The system of masses located on thin rods of negligible mass. The system is rigidly attached to the spherical shell of radius R. Asymmetric placement of the masses can be used to offset the center of mass of the system from the geometric center of the spherical shell.

Figure 9.12

horizontal surface. The inertia tensor J0 of the rigid body relative its center of mass X¯ is J0

= λ1 E1 ⊗ E1 + λ2E 2 ⊗

E2 + λ 3E3 ⊗ E3 ,

where λ1 =

2ms R2 3

+

2m2r22 + 2m3 r23, λ3 =

2ms R2 3

λ2 = +

2ms R2 3

+

2m1 r21 + 2m3r23 ,

2m1r12 + 2m2 r22 .

A vertical gravitational force −mgE3 is assumed to act on the rigid body. A set of Cartesian coordinates is used to parameterize the motion of the center of mass X¯ of the sphere and it is convenient to represent the angular velocity vector ω using a corotational basis: x¯ = xE1 + yE2 + zE3 , (

ω = ω1 e 1 + ω2 e 2 + ω3 e 3 .

)

˙ T , where Q is the rotation tensor of the rigid body: Here, ω = ax QQ

Q=

3 ± i=1

ei

⊗ Ei =

3 ± 3 ±

Qik Ei ⊗ Ek

k =1 i =1

=

3 ± 3 ±

Qik ei ⊗ ek .

(9.30)

k =1 i =1

The position vector πP of the instantaneous point of contact X P of the rigid body with the plane relative to the center of mass X¯ is −RE 3. However, in contrast to the earlier case, we need to keep track of E3 in relation to the corotational basis. Using the representations (9.30), it is straightforward to show that e3 = Q13E1 + Q23 E2 + Q33E3 ,

E3

=

Q31e1 + Q32e2 + Q33 e3 .

358

Kinetics of a Rigid Body

It is also useful at this point to note that for a constant vector A, where A = A 1e1 + A2 e2 + A3 e3 , ˙ = 0 and performing some rearranging, one finds that after computing A

⎡

⎤

⎡

A˙ 1 ⎣ A˙ 2 ⎦ = ⎣ A˙ 3

0

ω3

− ω2

− ω3

0

ω2

−ω1

ω1

0

⎤

⎤⎡

A1 ⎦ ⎣ A2 ⎦ . A3

(9.31)

The identity (9.31) can be applied to the components of a fixed basis vector E k expressed in terms of the corotational basis.30 For instance, for E3 , (9.30)2 and (9.31) imply that ⎡

⎤

⎡

˙ Q 31 ⎣ Q ˙ 32 ⎦ = ⎣ ˙ 33 Q

0

ω3

− ω2

− ω3

0

ω2

−ω1

ω1

0

⎤⎡

⎤

Q31 ⎦ ⎣ Q32 ⎦ . Q33

These equations can be considered as a set of differential equations for E3 · ei . Suppose that the rigid body is rolling on the surface. Starting from the identity v2 − v1 = ω × (x2 − x1 ) for any pair of points X 1 and X2 on a rigid body and using the constraints vP = 0, we quickly find that v¯ = ω × RE3,

a¯ = α × RE3 .

Referring to (7.16), the angular momentum of the rigid body relative to X P is HP For Chaplygin’s sphere, G representations

=

HP

mv¯

= = =

=

=

Jω − πP × G.

ω × RE3 and πP

= −RE 3,

and hence HP has the

Jω + mRE3 × (ω × (RE3 ))

²

³

J + mR 2 (I − E3 ⊗ E3) ω

JPω,

(9.32)

where JP

=

J + mR2 (I − E3 ⊗ E3 )

is the moment of inertia tensor of the body relative to XP .31 For completeness, we note that the tensor JP is invertible:32 ´

−1

JP

=

P

−1

+

2

mR

P−1 E3 ⊗ P −1E3 (

1 − mR2 E3 · P −1E3

µ )

.

30 Such an exercise is similar to the computations that were performed to arrive at (9.25) 1,2,3 previously. A

vector form of (9.31) is used extensively in the Russian literature on the dynamics of rigid bodies.

31 The parallel axis theorem (7.39) can be used to validate this interpretation of J . P 32 The representation for the inverse was computed by exploiting a representation for the inverse of a matrix

that is credited to Bartlett [19].

9.9

359

Chaplygin’s Sphere

1 In this representation for J− P , the tensor P and its inverse have the representations

P = J + mR2I −1

P

=

3 ±

¹

=

i=1

3 ² ±

2 λi + mR

i =1

λi

³

ei

⊗ ei ,

º

1 + mR2

ei

⊗ ei .

For future purposes, we note that P˙ has a representation that is identical to the representation J˙ = ΩJ − JΩ: P˙ = J˙ +

d ² 2 ³ mR I dt

˙ = J =

=

Ω J − JΩ + mR 2 (ΩI − IΩ) »

¼½

=0

¾

Ω P − P Ω.

1 −1 The representation for J− P can be used to establish a representation for the matrix J P that will appear later in the equations of motion (9.35). With the help of the three constraints vP · Ei = 0, Lagrange’s prescription for the constraint forces and constraint moments acting on the system implies that

Fc

= µ 1E1 + µ2 E2 + µ3 E3

acting at XP ,

Mc

=

0.

The components of the constraint force can be identified with a static friction force and a normal force: Ff = µ 1E1 + µ 2E 2 and N = µ3 E3 . Using a balance of linear momentum and imposing the constraints a¯ = α × RE 3, we find that Ff

=

m (α × RE3) ,

N = mgE3 .

(9.33)

˙ where M = π P × Ff , the representation Using a balance of angular momentum M = H (9.32) for HP , and (9.33), we find, after some rearranging, the conservation ˙ H P

where the angular momentum HP

=

=

0,

(9.34)

JPω (cf. (9.32)).

Conservations for Chaplygin’s Sphere

If we impose the constraints v¯ = ω × RE3 , then with the help of (7.24)2 and (9.32), we find that the kinetic energy of the rolling rigid body has the representations T

= = =

m 1 v¯ · v¯ + Jω · ω 2 2 ³ 1² J + mR2 (E1 ⊗ E1 + E2 ⊗ E2) ω · ω 2 1 H P · ω. 2

360

Kinetics of a Rigid Body

Using a work–energy theorem T˙

Fc · vP − mgE3 · v, ¯

=

it is straightforward to show that T˙ = 0, and thus, in addition to the angular momentum HP , the total energy of the Chaplygin sphere is also conserved during a motion of this rigid body. The Governing Equations

The governing equations for the Chaplygin sphere are assembled from the corotational components of (9.34) and the evolution equations (9.32) for the corotational components of E3: ⎡

⎤

⎡

˙ 31 Q ⎣ Q ˙ ⎦=⎣ 32 ˙ Q 33

⎡

ω ˙1

JP ⎣ ω˙ 2 ω ˙3

⎤

0

ω3

− ω3

0

ω2

− ω1

⎡

− ω2

ω1

0

⎤⎡

⎤

Q31 ⎦ ⎣ Q32 ⎦ , Q33

⎤

⎤

⎡

0 (λ3 − λ2)ω3 ω2 ⎦ + ⎣ (λ1 − λ3)ω3 ω1 ⎦ = ⎣ 0 ⎦ , 0 (λ2 − λ1)ω1 ω2

(9.35)

where the matrix JP , which is composed of the ei ⊗ ek components of the tensor JP , has the representation ⎡

JP

=

⎣

(

1 − Q231 −mR Q31 Q32 −mR 2Q31 Q33

2 λ1 + mR 2

)

2

−mR

Q(31Q32 ) 2 2 λ2 + mR 1 − Q32 −mR2 Q32 Q33

2

−mR

⎤

Q31Q33 2 ⎦. −mR Q32 Q33 ( ) λ3 + mR2 1 − Q2 33

Given initial conditions for the orientation of E3 relative to ei (i.e., Q3i (t 0)) and the angular velocity vector components ωk (t0), (9.35) can be solved to determine the components ωi = ω · ei of the angular velocity ω(t). However, to determine x¯ (t), differential equations for Q1k (t) = E1 · ek and Q2k (t) = E2 · ek established using (9.31) (and similar to (9.35)1 ) must be integrated in order that the following differential equations for the x and y coordinates of the center of mass X¯ (and the position of the instantaneous point of contact XP ) can be solved: x˙ = RQ21 ω1 + RQ22 ω2 + RQ23ω3 , y˙ = −RQ11 ω1 − RQ12ω2 − RQ13ω3 .

(9.36)

Thus, to fully solve for the motion of the Chaplygin sphere, initial conditions for Q (t 0), ω (t0 ), and x¯ (t 0) must be provided, nine first-order differential equations (9.23) must be solved for Qik (t), and five first-order differential equations (i.e., (9.35)2 and (9.36)) must be solved for x¯ (t) and ωk (t). The Case where H P is Vertical

As noted by Chaplygin [47], a special case occurs when HP = CE3 , where C is a constant. He observed that the equations of motion simplify in a dramatic fashion in this

9.9

361

Chaplygin’s Sphere

instance.33 To see the simplification, we first note that in addition to the conservation of HP, the total energy T = 0.5HP · ω of the sphere is conserved. Thus, the vertical component of ω is conserved: ω · E3

=

2e , C

where e is the value of the conserved energy. If we next consider Pω, we find that Pω = HP

²

+

¹

=

³

mR2 E ⊗ E3 ω 2

C + mR

¹

2e C

ºº

E3.

In conclusion, P ω is conserved in this case. Indeed, after differentiating P ω and using ˙ we find a familiar set of differential equations for ω: the earlier result for P, Pω ˙ + ω × ( Pω) = 0.

(9.37)

That is, these equations are identical to those for moment-free motion of a rigid body (cf. (9.11)), where λi is replaced with λ i + mR2 . In summary, the equations of motion governing ωi for the Chaplygin sphere when HP = CE3 decouple from those for Qik and can be integrated separately. To determine the motion of the Chaplygin sphere in this case, we solve (9.37) for ωi (t) and then solve (9.23) for Qik and (9.36) for x(t) and y(t). The initial conditions must be chosen so that HP has no horizontal component. That is, 3 ²² ±

λ k + mR

2

³

Q1k (t0 ) ωk (t 0)

³ =

0,

=

0.

k =1 3 ²² ±

λ k + mR

2

³

Q2k (t0 ) ωk (t 0)

³

(9.38)

k =1

A representative sample of results for a Chaplygin sphere is shown in Figures 9.13, 9.14, and 9.15. For the purposes of comparison, the corresponding results for a spherical shell of mass ms and radius R rolling on a horizontal plane are also displayed. For the spherical shell rolling on the horizontal plane with HP = CE3 , it is easy to show that ±1 = ω · E1 and ±2 = ω · E 2 are identically zero and ±3 = ω · E3 is constant. Thus, the shell spins about the vertical with a constant angular velocity vector and its center of mass remains stationary. 34 As a result, the rotational motion of the shell is an example of a constant angular velocity motion as discussed in Section 7.11. As anticipated, the corotational basis vectors trace circular arcs about the instantaneous axis of rotation i (see Figure 9.14(c)). The situation for the Chaplygin sphere is far more complex. The path of the center of mass shows an oscillatory behavior (see Figure 9.13), 33 Our discussion of the special case H P

= CE3 is influenced by the papers of Fedorov [79] and Kilin [147] on the Chaplygin sphere. Analytical expressions for Qik (t), x(t), and y(t) for this case can be found in [147, Section 3.3]. 34 Recall that if ω ˙ = 0, then ωi = ω · ei and ± i = ω · Ei are constant.

362

Kinetics of a Rigid Body

Chaplygin sphere 0.02

Spherical shell

y/R

x/R

−0 05

0 .01

.

−0 04 .

Path (x(t), y(t)) traced by the center of mass of a Chaplygin sphere (λ 1 < λ2 < λ3 ). By way of comparison, the center of mass of a symmetric spherical shell (λ i = 32 mR 2) would remain stationary. The x and y coordinates of the centers of mass of both rigid bodies are 0 at t = 0 and, for the motions of interest, HP = CE3. Figure 9.13

(b)

(a)

(c)

e3

E3 E2

E1

E2

e1

E3

O

E3

O

e3

e2 E2

e2

O

i

e1

E1

E1

Loci of the corotational basis vectors. (a) The vectors e1 (t) and e3(t) for t ∈ [0, 5] and (b) the vector e2 (t) for t ∈ [0, 50] for the Chaplygin sphere. (c) The corresponding results for a spherical shell rolling on a rough horizontal surface with HP = CE3 .

Figure 9.14

the components ωi (t) of ω are not constant (see Figure 9.15), and the loci of ei (t) are no longer simple circular arcs (see Figure 9.14). For the results shown in Figures 9.13, 9.14, and 9.15, the following numerical values were selected: m1 = 3ms ,

m2

=

ω1 (0) =

2ms , 4,

m3 = ms , x (0) = 0,

r1

=

r2

=

r3

=

0.5R,

y (0) = 0.

The initial values Qik (0) were selected by parameterizing Q (0) using a set of 3–2–1 Euler angles where ψ = π/12, θ = π/3, and φ = −π/4. The initial values ω2 (0) and ω3 (0) were then obtained with the help of (9.38). The resulting initial values ensured that HP = CE3. For the results shown for the spherical shell, i = −E3 and ω =

9.10

Closing Comments

363

ω1

4

ω1 ω2

ωi

ω2

ω3

50

t

ω3

−4 Figure 9.15 Corotational components ωi of the angular velocity vector ω for the Chaplygin sphere. The corresponding results for a spherical shell rolling on a rough horizontal surface are denoted by dashed lines.

−4.6188E3.

We close by noting that for all of the simulations shown, the conservation of total energy E = T was verified. Additional Remarks

Additional analyses of the equations of motion for Chaplygin’s sphere and the development of closed-form expressions for Qik (t), x(t), and y(t) for all values of HP can be found in [27, 28, 79, 147, 151]. It was discovered by Chaplygin that when the initial conditions are such that HP is not parallel to E3 , the path of the center of mass of the Chaplygin sphere can be considered as a superposition of a straight-line component and an oscillatory component.35 Examples of such paths are shown in Figure 9.16 for a Chaplygin sphere where λ1 > λ2 > λ3 . As can be seen from this figure, if the initial value of the instantaneous axis of rotation i = ω/ ²ω² is chosen to be close to the intermediate axis ´e2 , then the oscillatory component of the path exhibits an interesting bursting behavior. The behavior of e2 (t) for this case (see Figure 9.16(a)) is reminiscent of the behavior of e2(t) for the moment-free motion of a rigid body shown in Figure 9.5(b). It is also remarkable that the path of the center of mass is a straight line to the naked eye for a wide range of initial conditions and in spite of the complexity of ei (t).

9.10

Closing Comments

We have touched on some problems in rigid-body dynamics. There are several aspects that we have not had either the opportunity or the space to address. Some of these 35 You may wish to recall for the purposes of comparison that the corresponding path traced by the center of

mass of a rolling spherically symmetric body is a straight line.

364

Kinetics of a Rigid Body

(a)

(b)

e2

(c)

e2

O

e2

O

y

O

y x

y x

x

ω1

ω1

ω1 ω3 ω2

t

t

t

ω3 ω2

ω3

ω2

Examples of the simulated path of the center of mass of the Chaplygin sphere for three sets of initials conditions: (a) ω1 (0) = ω3 (0) = 0.01, ω2(0) = 5, and t ∈ [0, 50]; (b) ω1 (0) = 0.01, ω2(0) = ω3 (0) = 5, and t ∈ [0, 10]; and (c) ωk (0) = 5 and t ∈ [0, 10]. The inset images show the corresponding time traces of ωk (t) and e2 (t). For the results shown in this figure, λ1 = 43/6, λ 2 = 17/ 3, and λ 3 = 19/ 6. The components of Q (0) are defined using a set of 3–2–1 Euler angles where ψ = 0, θ = π/4, and φ = 0.

Figure 9.16

additional aspects are discussed in the various exercises and others can be found in the References. The treatises of Appell [9], Papastavridis [226], Routh [247], and Whittaker [306], and the splendid introductory text by Crabtree [55], are particularly recommended. It is important to realize that, although rolling spheres and thrown baseballs have been analyzed for over a century, these problems are very rich. Indeed, a simple change in their kinematical features can lead to dramatically different results. One of the most celebrated instances of this change arises in Chaplygin’s rolling sphere, where J ± = 2 mR2I and πP = −RE3 . As mentioned previously and evidenced in Figure 9.16(a), 5 the path traced by the point of contact of the Chaplygin sphere with the ground can be very intricate. Two other celebrated examples of rigid bodies that exhibit interesting behavior are Euler’s disk and the wobblestone (or celt) [55]. The former consists of a heavy circular cylinder that rolls and slides on a convex mirror. As the disk becomes increasingly horizontal, a whirring sound is heard whose pitch increases. Eventually, the disk comes to a dramatic, abrupt halt accompanied (some believe) by an impact of the disk with the convex mirror. The first analysis of this system was performed by Moffatt [190] and his controversial paper was followed by a series of works promoting alternative mechanisms for the dramatic motion of Euler’s disk (see [146] and references therein). As mentioned earlier, the wobblestone is a rigid body whose curved lateral surface rolls on a horizontal plane. At first glance, the curved surface appears to be symmetric, but this is not the case and, as a result, the wobblestone exhibits an unusual reversal of spin directions. The wobblestone has been the subject of several papers (see, e.g., [170, 228]) and simulations; the paper by Blackowiak et al. [25] is particularly recommended for a lucid explanation of the spin-reversal mechanism.

9.11

365

Exercises

Recent analyses of rigid-body dynamics have focused on the stability and bifurcation of their families of steady motions (see, e.g., [169, 212]). Increasingly, some of these studies are evolving toward an examination of motions of these systems that (although not steady) are asymptotic to a steady motion or connect two steady motions of the rigid body. For the interested reader, recent work on the tippe top [30, 238, 295], the possible jumping behavior of a spinning egg [32, 188, 260], and the attitude of a thrown tennis racket [13] are mentioned as notable examples of modern analyses of rigid-body dynamics. It is hoped that our exposition in this chapter will enable you to explore and appreciate works of this type.

9.11

Exercises

Exercise 9.1: Suppose a rigid body is rolling on a fixed surface under the influence of a gravitational force −mgE3 . Starting from the work–energy theorem for the rigid body, T˙ = F · v¯ + M · ω, prove that the total energy E of the rigid body is conserved. Prove that the total energy is also conserved if the rigid body is sliding on a smooth surface. Exercise 9.2: A rigid body of mass m is moving in space under the influence of an applied force Fa = Fa e3 and an applied moment M = 0. Outline how you would determine the attitude Q(t) and the motion x¯ (t) of the center of mass of the rigid body. Exercise 9.3: The orientation of a rigid body relative to a fixed reference configuration is defined by a rotation tensor Q. At time t = t 0, Q has the value Q (t0) and Q has the value Q (t1 ) at time t = t1. Give a physical interpretation of the rotation tensor Q (t1 ) QT (t0 ). Feel free to make use of the corotational basis in your answer. Exercise 9.4: Consider a rigid body with a fixed point XO . What are the three constraints ˙ on the motion of this rigid body? Why is it sufficient to solve MO = H O to determine the motion of this rigid body? Exercise 9.5: Solutions for ωi (t) have been determined for a rigid-body dynamics problem. How would you determine Q(t) from this solution? (

)

¿

À

i 1 2 3 Exercise 9.6: A rigid body has a potential energy U = U x, ¯ ν , where ν , ν , ν denote the three Euler angles used to parameterize Q. Suppose that the conservative force F con acting at the center of mass and the conservative moment Mcon are such that the following identity is satisfied during all motions of the rigid body: ˙ = −U

F con · x˙¯ + M con · ω.

In this case, verify that Fcon

= −

3 ± ∂U i=1

∂ xi

Ei acting at

¯ X,

Mcon

=

3 ± ∂U i g, − i=1

∂ν i

366

Kinetics of a Rigid Body

where xi = x¯ · Ei and gi are the basis vectors for the dual Euler basis. Show that the gravitational potential energy Un for a rigid body orbiting a fixed spherically symmetric ( ) rigid body has the functional form U = U x¯ , ν i . Exercise 9.7: As shown in Figure 9.9, a rigid sphere of mass m and radius R rolls (without slipping) on an inclined plane. The inertia tensors for the sphere are J = J0 = 2mR 2 µ I, where µ = 5 . (a) What are the three constraints on the motion of the sphere? Show that these constraints imply that ˙ x¨ 1 − R± 2

=

0,

˙ x¨ 2 + R± 1

=

0,

x¨ 3

=

0,

where xi = x¯ · Ei and ±i = ω · Ei . (b) With the help of the balance of linear momentum for the sphere, show that Fc

=

(

)

˙ 2 − mg sin(β ) E1 − mR ± ˙ 1E2 + mg cos(β )E3. mR ±

(c) Show that the balance of angular momentum for the sphere and the results of (b) imply that 7 2 ˙1 mR ± 5

=

0,

7 2 ˙2 mR ± 5

=

mgR sin(β ),

2 2 ˙ 3 = 0. mR ± 5

(d) Starting from the work–energy theorem for a rigid body, prove that the total energy E of the rolling sphere is constant, where E

=

7mR2 2 7mR2 2 mR 2 2 ± + ± + ±3 − mgx1 sin(β ) + mgR cos(β ). 10 1 10 2 5

(e) Why is the angular momentum H of the sphere in the E3 direction conserved? ∑ (f) If, at time t = 0, the sphere is given an initial angular velocity ω(0) = 3i=1 ±i 0 Ei , then show that the angular velocity ω(t) is (9.26). What is the angular acceleration vector α of the sphere? (g) Suppose the sphere is placed on the inclined plane and released from rest with Q (0) = I. Verify that the sphere will start rolling and that the resulting attitude Q of the sphere corresponds to a fixed-axis rotation. Exercise 9.8: In a model for a rigid body flying through the air, the four primary forces on the body are gravity, a lift force, a drag force, and a thrust force: F M

= −mgE3 + =

πt

×

f e1 .

mBω × v¯ + f e1 −

v¯ 1 ¯ ρ f ACd (v¯ · v ¯) acting at X, 2 ²v ¯²

(a) Show that one of the four applied forces is conservative. (b) Show that the lift force does no work. (c) If the thrust force acts at a point whose position vector relative to x¯ is πt , then, with ˙ the assistance of the work–energy theorem, establish an expression for E.

9.11

Rigid body B

Exercises

367

¯

X

¯x

E3 Fixed body of massM

E2 E1 Figure 9.17

mass M.

O

Schematic of a rigid body of mass m that is in orbit about a fixed symmetric body of

Exercise 9.9: This famous problem is discussed in most books on satellite dynamics (see, e.g., [22, 133]). The 24 solutions discussed subsequently date to Lagrange [154, 156] in the late eighteenth century. Among the remarkable aspects about Lagrange’s extraordinary work on this topic in [156] is the (early) use of his celebrated equations of motion in the context of a rigid body and his clear discussion of a set of (what are now known as) 3–1–3 Euler angles. As shown in Figure 9.17, consider a rigid body B of mass m that is in motion in a central gravitational force field about a massive fixed body of mass M. The center of this force field is assumed to be located at a fixed point O. The force, moment, and potential energy of the field are given by approximations (8.18). (a) Verify that Mn = −x¯ × F n. What is the physical relevance of this result? (b) Why are the angular momentum HO and the total energy E of the satellite conserved? (c) Using the balance of linear momentum, show that it is possible for the body to move in a circular orbit x¯ = r0 er about O with a constant orbital angular velocity θ˙0 , which is known as the modified Kepler frequency, ωKm : Á θ˙0 = ωKm = ωK

1+

3 2r20m

(tr(J) − 3er ·

Jer ),

where the Kepler frequency was defined previously (see (2.13)): 2 ωK =

GM r30

.

(9.39)

368

Kinetics of a Rigid Body

In (9.39), er = cos( θ )E1 + sin(θ )E2, and this vector is an eigenvector of J. That is, er is parallel to one of the principal axes of the body. (d) Using the results of (c), show that a steady motion of the rigid body, that is, one in ˙ = 0, is governed by the equation which ω ω × (Jω) = 3ω2K er × (Jer ).

(9.40)

(e) Suppose that the body is asymmetric. That is, the principal values of J0 are distinct. We seek solutions of (9.40) such that ω · er = 0. Show that there are six possible solutions for ω that satisfy (9.40) and four possible solutions for er . Here, you should assume that J is known and as a result Q is known. Hence, there are 6 × 4 possible solutions of (9.40). (f) Suppose that the body is such that J = µI, where µ is a constant. Show that any constant ω satisfies (9.40) and consequently any orientation of the rigid body is possible in this case. (g) Using the results of (e), explain why it is possible for an Earth-based observer to see the same side of a satellite in a circular orbit above the Earth. Exercise 9.10: Referring to Figure 9.18, consider the problem of a sphere of mass m, 2 I moving on a turntable. The contact between the radius R, and inertia tensor J0 = 2mR 5 sphere and the turntable is rough. In addition, the center O of the turntable is fixed and the turntable rotates about the vertical axis E3 with an angular speed ±. This problem was introduced in a classic textbook by Samuel Earnshaw [64] and is discussed by Gersten et al. [93], Lewis and Murray [168], and Pars [227], among many others. (a) Suppose that the sphere is rolling on the turntable. The position vector of the point of contact of the sphere with the turntable is πP = − RE3 . Show that the motion of the sphere is subject to three constraints: v¯ + ω × (−RE3 ) = ±E3 × x¯ .

(9.41)

Sphere of mass m and radius R

E3

Ω

g

E1

O

E2

¯

X

Turntable

Schematic of a rigid sphere of radius R moving on the surface of a horizontal turntable that is spinning with an angular speed ± = ±(t).

Figure 9.18

9.11

Using the representations ω constraints imply that

=

∑3

i=1 ±i Ei

˙ 2 + ±x x¨ 1 − R± ˙ 2 = 0,

and x¯ =

∑3

i=1 xi Ei ,

˙ 1 − ±x x¨ 2 + R± ˙1

=

0,

369

Exercises

show that the three x¨ 3

=

0.

(b) Assuming that a vertical gravitational force acts on the sphere, draw a free-body diagram of the sphere. (c) Using a balance of linear momentum and with the assistance of the constraints, show that the constraint force acting on the sphere is Fc

=

mgE3 − m± (x˙ 2E 1 − x˙ 1E2 ) + mR

(

˙ 2E 1 − ± ˙ 1E2 ±

)

.

(d) Using a balance of angular momentum and with the assistance of the results of (a)–(c), show that the equations governing the motion of the sphere are x¨ 1

¹ ˙1 = ±

=

˙ 2 − ±x R± ˙ 2,

m±R I + mR 2

x¨ 2

º

x˙ 1,

˙ 1 + ±x = −R ± ˙1 ,

¹

˙2 = ±

m± R I + mR 2

º

x˙ 2,

x˙ 3 = 0, ˙3 = ±

0.

(9.42)

2

¯ Here, I = 2mR 5 . Why are (9.42) sufficient to determine x(t) and ω(t) of the sphere? Given a set of Euler angles of your choice, how would you determine Q(t)? (e) For the special case in which ± = 0, show that the center of mass of the sphere will move in a straight line with a constant speed and that the angular velocity vector ω of the sphere will be constant. (f) Numerically integrate (9.42) for a variety of initial conditions. Is it possible for the sphere to fall off a turntable of radius R 0? In choosing your initial conditions (x(t ¯ 0 ), v(t ¯ 0 ), ω(t0 )), make sure they are compatible with rolling condition (9.41). (g) Show that the equations (9.42) can be used to show that

2 2 ±x ˙ 2, x¨ 2 = ±˙x1 . (9.43) 7 7 Provided ± = ±0 is a nonzero constant, verify that the general solution to (9.43) is x¨ 1

x1(t) = x1 (0) + x2(t) = x2 (0) +

7

= −

¹

¹

¹

º

º

¹

ºº

2±0 t 2±0 t −1 +x ˙ 1 ( 0) sin , 2 ±0 7 7 ¹ ¹ ¹ º º ¹ ºº 2 ±0 t 7 2±0 t −˙ x1 (0) cos − 1 +x ˙ 2 ( 0) sin . 2 ±0 7 7 (9.44) x˙ 2 (0) cos

Given the initial position vector x(0) ¯ = x1 (0)E1 + x2(0)E2 and the initial velocity vector v¯ (0) = x˙ 1 (0) E1 + x˙ 2 (0) E2 , what is the initial angular velocity vector of the sphere for the motion (9.44)? Referring to Figure 9.19, show that the path traced by the center of mass of the sphere is a circle of radius R 0 centered at a point C whose position vector is C0: 36 ¹

R0

=

7 2±0

º

²v ¯ (0)²

,

36 For assistance with, and perspectives on, these results, see Gersten et al. [93].

370

Kinetics of a Rigid Body

R0

v(t) ¯

C

x¯ (t)

v¯ (0)

E2

x¯(0) E1

O

Figure 9.19 The path x(t) ¯ of the center of mass X¯ of the sphere rolling without slipping on a rotating turntable. The turntable is rotating in a counterclockwise manner: ±0 > 0.

¹

¹

C0 = x1 (0) −

º

7

º

¹

¹

x˙ 2 (0) E1 + x2 (0) +

2 ±0

7

º

2±0

º

x˙ 1 (0) E2 .

You should observe that the center of mass traverses the circular path in a clockwise (counterclockwise) fashion if ±0 < 0 ( > 0). It is also interesting to note that the radius R 0 of the path x(t) ¯ traced by the center of mass is inversely proportional to the speed of rotation ±0 of the turntable. Exercise 9.11: Recall the definition of the kinetic energy T of a rigid body: T

=

1 2

Â

R

v · vρ dv.

(a) Starting from the definition of T, prove the Koenig decomposition: T

=

1 1 mv¯ · v¯ + H · ω. 2 2

(b) Establish the following intermediate results: J˙ = ΩJ − JΩ,

²

³

J˙ω

·

ω = H·ω ˙,

˙ · ω, Jω˙· ω = 2H

where the angular momentum H = Jω. (c) Using the intermediate results and the balance laws, prove the work–energy theorem: T˙

=

F · v¯ + M · ω.

(d) If F

=

K ± i=1

Fi ,

M

=

K ±

(xi − x¯ ) × Fi

i=1

+

Mp ,

9.11

Exercises

371

then show that T˙

=

K ±

F i · vi

+

Mp · ω.

i=1

Give three examples of the use of this result for a single rigid body. Exercise 9.12: Establish the following theorem of Euler, which can be found in [70, Section 44]: “For any given (rigid) body, we can always find an axis, which passes through its center of gravity, about which the body can rotate freely and uniformly.” How is this result related to Euler’s later result that three distinct axes exist about which a rigid body can rotate freely and at constant angular velocity? This result can be found in [69], where these axes were first termed “principal axes of inertia.” Exercise 9.13: Consider the tippe top shown in Figure 8.10 and discussed in Exercise 8.3. (a) Continuing the earlier exercise, we follow Or [211] and suppose that the point X P slides on the horizontal surface and experiences both Coulomb and viscous friction forces: Ff

= − (µk + µ v ² v P² ) ² N² s

acting at XP ,

where µ k and µv are friction coefficients, N is the normal force, and s is the slip direction. Using a balance of linear momentum, show that the normal force acting on the tippe top is N

²

=

³

m g + l θ¨ sin(θ ) + l θ˙ 2 cos( θ ) E3 acting at X P .

(b) Starting from the work–energy theorem, prove that the total energy of a sliding tippe top decreases with time, whereas that for a rolling tippe top is constant. (c) If the inertia tensor of the tippe top is J = λ t (I − e3 ⊗ e3 ) + λ ae3 ⊗ e3 , what are the differential equations governing the motion of the tippe top? (d) For both rolling and sliding tippe tops, show that the angular momentum H · πP is conserved. This integral of motion, which is known as the Jellett integral, was discovered in the 1870s by J. H. Jellett (1817–1888).37 DISCUSSION :

When the tippe top is rolling on a horizontal surface, it can be shown that, in addition to conservation of total energy and the Jellett integral, the following kinematical quantity is conserved: I1

2

= ω3

²

λa λt + mλa

²

2 2 (πP · e1 ) + (πP · e2)

³

³

+

mλt (π P · e3 )2 .

(9.45)

37 A discussion of the history of this integral (and several others) can be found in Gray and Nickel [101].

372

Kinetics of a Rigid Body

Here, πP is the position vector of the instantaneous point of contact XP (relative to the ¯ of the tippe top with the horizontal surface. The integral of motion I1 center of mass X) was first established by Routh (see [247, Section 243]) and can be described as the Routh integral. It is also known as the Chaplygin integral [143, 153]. To date, the integral I1 has no known physical interpretation. Exercise 9.14: Consider a body that is free to move about one of its material points XO , which is fixed (cf. Figure 9.8). The inertia tensor of the body relative to its center of ∑ mass is J = 3i=1 λi ei ⊗ ei , where {e1 , e2, e3 } is a corotational basis. The position vector of the center of mass X¯ relative to XO is x¯ − xO

=

he3,

where h is a constant. A linear spring of stiffness K and unstretched length ² 0 is attached to the body at the point XS and the other end is attached to a fixed point A: xs − xO

=

s1 e 1 + s3 e 3 ,

xA − xO

= ² AE 1.

In addition, a gravitational force −mgE3 acts on the body. (a) Show that the velocity and acceleration vectors of the center of mass have the representations v¯ = hω2 e1 − hω1 e2 , a¯ = h (ω˙ 2 + ω1 ω3) e1 − h (ω˙ 1 − ω2ω3 ) e2 − h

²

2 2 ω1 + ω2

³

e3 .

∑

Here, ω = 3k=1 ωk ek . (b) Show that the angular momentum of the rigid body relative to X O is HO

²

=

2 λ 1 + mh

³

ω1 e 1 +

²

λ2 + mh

2

³

ω2 e2 + λ3ω3 e3.

Show that the kinetic energy of the rigid body has the representation T

=

1 H O · ω. 2

(c) What are the three constraints on the motion of the rigid body? Give prescriptions for the constraint force Fc acting at XO and the constraint moment Mc that enforce these constraints. (d) Draw a free-body diagram of the rigid body. (e) Using a balance of linear momentum, show that the reaction force at XO is Fc

=

mh (ω˙ 2 + ω1ω3 ) e1 − mh (ω˙ 1 − ω2 ω3 ) e2 −mh

²

2 2 ω1 + ω2

³

e3 + mgE3 − Fs ,

where F s is the spring force. ∑ (f) Assuming that a set of 3–1–3 Euler angles is used to parameterize Q = 3i=1 ei ⊗ Ei , show that the potential energy U = U (x, ¯ Q) of the rigid body in this problem can ˆ (ψ , θ , φ). be expressed as a function of these angles: U = U

9.11

373

Exercises

(g) Verify that the conservative force Fcon acting at X¯ and the conservative moment M con acting on the rigid body satisfy the identity F con · v¯ + M con · ω = M Ocon · ω, where MOcon is the resultant conservative moment acting on the body relative to X O . Using this result, explain why the following prescription for MO con is valid: MO con

= −

ˆ ∂U 1

∂ψ

g

−

ˆ ∂U 2

∂θ

g

−

ˆ ∂U 3

∂φ

g .

(h) Show that the rotational motion of the rigid body is governed by the following equations: ²

λ 1 + mh

2

λ 2 + mh

2

²

⎡

³

³

²

ω ˙1 = ω ˙2 =

λ2 +

²

mh2 − λ 3

λ3 − λ1 − mh

2

³

³

ω2 ω3 +

MO · e1,

ω1 ω3 +

MO · e2,

MO · e3, ⎤⎡ sin(φ )cosec(θ ) cos(φ )cosec(θ ) 0 ⎦=⎣ cos(φ ) − sin(φ ) 0 ⎦⎣ − sin(φ) cot(θ ) − cos( φ) cot( θ ) 1

λ3 ω ˙ 3 = (λ1 − λ2 ) ω1 ω2 + ˙ ψ

⎣ θ˙ ˙ φ

⎤

⎡

ω1 ω2

⎤ ⎦ .(9.46)

ω3

Exercise 9.15: Recall from Section 8.6 that the gravitational force on a particle of mass m orbiting a fixed body of mass M is given by the Newtonian gravitational force Fn =

GMm 3

² r²

r,

where r is the position vector of the particle of mass m relative to the fixed mass M. For a rigid body of mass m, this force is supplemented by a torque on the center of mass of the body of mass m. The resulting torque Mn is known as the gravity gradient torque and it can produce a coupling between the orbit of the body and the attitude (orientation) of the body. Referring to Figure 9.17, we consider the classic problem where the fixed body of mass M is spherical. The standard approximations for the gravitational force and gravity gradient torque acting on the body of mass m in this case are (cf. (8.18)) Fn

=

GMm c, R2

¹

Mn

≈

3GM R3

º

¹

c × (Jc) = −

3GM R3

º

c × (Ec),

where the unit vector c points from the fixed body to the center of mass X¯ of the orbiting body: x¯ , R = ²x¯ ² . c= ²x ¯² (a) If the body of mass m can be approximated as a sphere, then show that the gravity gradient torque is zero. (b) Apart from the case discussed in (a), show that Mn is zero if the vector x¯ is parallel to an eigenvector of J.

374

Kinetics of a Rigid Body

(c) Consider a model for the Moon where this celestial body is considered to be a homogeneous ellipsoid with distinct moments of inertia: J = λ1 e1 ⊗ e1 + λ2 e2 ⊗ e2 + λ3 e3 ⊗ e3. Now suppose the Moon orbits the Earth in a circular orbit of radius R 0 at a constant orbital speed. The Earth is assumed to be spherically symmetric with a fixed center of mass for the purposes of this exercise. Show that the angular velocity (Kepler frequency) ωK of the circular motion of the center of mass of the Moon satisfies the relation GM 2 . ωK = R30 If, in addition, the Moon has a constant angular velocity, show that ω and r are parallel to the principal axes of J and the gravity gradient torque vanishes. (d) How can the results from (c) be applied to explain why an Earth-based observer always sees the same face of the Moon? The first explanation of this phenomenon is attributed to Lagrange [156] in 1782. In your answer, it will be of great help to show that the orbital speed R0 ωK of the center of mass of the Moon is related to the angular velocity of the Moon. DISCUSSION :

The time taken for the Earth to perform a complete revolution about its polar axis is known as a sidereal day and is 23 hours, 56 minutes, and 4 seconds in length. By setting ωK equal to 2π/S, where S is the length of a sidereal day in seconds, substituting for the mass of the Earth and the universal gravitational constant, one can compute the radius R0 for a geostationary orbit: ´

R0 ≈ ≈

2 ωK

µ 1/ 3

GM 4.21633 × 107 m.

The radius of the Earth at the equator is 6, 378 km, so a geostationary satellite can be situated at approximately 35, 785 km above the Earth’s surface at the equator. Such satellites are used for communications. Our numerical results here are approximate, as we are using estimates for G and M. Exercise 9.16: As shown in Figure 9.20, a ball-shaped robot (ballbot or spherical robot) has been designed to help investigate the state of an oil pipeline. The robot operates by moving inside the cylindrical pipeline and sensors, which are mounted inside the outer spherical casing of the robot, monitor the state of the condition of the pipeline. In this problem, we model the robot (as simply as possibly) as a rigid sphere and establish the equations of motion for the sphere. These equations are then used to illustrate a classic result discussed in Littlewood [171]: “Suppose a sphere is started rolling on the inside of a rough vertical cylinder (gravity acting, but no dissipative forces); what happens? The only sensible guess is a spiral descent of increasing steepness; actually the sphere moves up and down between two fixed horizontal planes. Golfers are not so lucky as they think.”

9.11

375

Exercises

Sphere of massm and radius R

E3

¯

X

g

E2

O

Cylinder of radiusH

E1

Figure 9.20

radius H.

Schematic of a rigid sphere of radius R moving on the inner surface of a cylinder of

An example of the motion that John E. Littlewood (1885–1977) is referring to is shown in Figure 9.21(a). A set of cylindrical polar coordinates {r, β, z} is used to parameterize the position of the center of mass of the sphere: x¯ = rer + zE3 ,

er

=

cos (β) E1 + sin (β) E2 .

It is convenient to decompose the rotation tensor of the sphere into the product of two rotations: Q=

3 ±

ei ⊗ Ei

=

Q2 Q1 ,

Q1 = L (β , E3 ) ,

(9.47)

i=1

where Q2 is parameterized by a set of 3–1–3 Euler angles.38 The basis {e1 , e2, e3} corotates with the sphere. Decompositions of the form (9.47) are used in studies on satellite dynamics, as it makes it easier to visualize the steady motions where the center of mass of the satellite describes a circular orbit.39 (a) With the help of the representation Q has the representation Q2 where eβ

=

=

=

∑3

i=1 ei ⊗

e1 ⊗ er + e2 ⊗ eβ

cos (β) E2 − sin (β) E1 .

38 The 3–1–3 set of Euler angles is discussed in Section 6.8.2. 39 An example of such an analysis is discussed in Exercise 10.6.

+

E i , argue that the rotation Q2

e3 ⊗ E3 ,

376

Kinetics of a Rigid Body

(b)

(a)

80

0

τ

=

g Rt

τ

=

g Rt

z/R

z

y

−120

x

(c) 30

g R

Ωr

Ωβ Ωz 80

−10

g R

(a) Representative motion of the center of mass of the sphere as it rolls without slipping on the inner surface of the cylinder. (b) Time trace of z(t) for the motion shown in (a). (c) Time traces of the components ±r , ±β , and ±z of the angular velocity vector ω of the sphere for the motion shown in (a). The results shown in this figure are based on simulations of (9.49).

Figure 9.21

(b) Show that the six components of the angular velocity vector ω with the following representations: ω=

3 ±

ωi ei = ±r er + ±β eβ + ±z E3 ,

i=1

are (

)

(

)

(

)

˙ + β˙ ω1 = ψ ˙ + β˙ ω2 = ψ ˙ + β˙ ω3 = ψ

sin(φ) sin( θ ) + θ˙ cos (φ) , cos(φ ) sin(θ ) − θ˙ sin (φ) , cos (θ ) + φ˙

and ±r = θ˙

cos (φ) + φ˙ sin(θ ) sin( ψ ),

±β = θ˙

sin (φ) − φ˙ sin(θ ) cos(ψ ),

˙ +φ ˙ cos (θ ) . ±z = β˙ + ψ

(c) Suppose that the sphere is rolling without slipping on the inner surface of the cylindrical pipe. If πP = Rer , show that the constraints vP = 0 are equivalent to

9.11

r˙

0,

=

r β˙ + R±z

0,

=

˙z −

R±β

377

Exercises

=

0.

(9.48)

Here, vP = v¯ + ω × πP and r = H − R. It can be shown that the constraints (9.48)2,3 are nonintegrable. You should be able to infer from (9.48) how you would establish the corresponding constraints for a sphere rolling on a horizontal surface or a sphere rolling on a turntable. (d) Give prescriptions for the constraint force Fc and constraint moment M c acting on the sphere. 2 , establish (e) Assuming that the inertia tensor of the sphere is J = λI where λ = 2mR 5 ˙ expressions for the er , eβ , and E3 components of H and H. (f) Establish the following expression for the total energy of the sphere: ³ ² ³ m² 2 λ 2 2 2 z˙ + R 2±2z + ±z + ±β + ±r . 2 2 Prove that the total energy of the sphere is conserved. (g) Using a balance of linear momentum and a balance of angular momentum, establish the following set of differential equations governing the motion of the sphere:

E

⎡ ( ⎣

˙ + β˙ ψ

θ˙ ˙ φ

) ⎤

=

⎡

sin(ψ )cot(θ ) ⎦=⎣ cos(ψ ) sin(ψ )cosec(θ )

cos(ψ )cot(θ ) sin(ψ ) − cos(ψ )cosec(θ )

−

(H −

R) β˙ ˙z =

²

λ + mR

2

²

±r ±β

⎤

⎦,

±z

= −R±z ,

R±β ,

˙ r − λ±β β˙ = λ±

³

⎤⎡

1 0 ⎦⎣ 0

0,

˙ + λ± β˙ = −mgR, ± β r

2 λ + mR

³

˙z = ±

0.

(9.49)

As part of your solution, show that the constraint force acting on the sphere is Fc

= −m ( H −

˙ z eβ R ) β˙ 2er − mR±

+

(

)

˙ β + g E3 acting at XP , m R±

where x¯ = (H − R) er + zE3 and XP is the instantaneous point of contact of the sphere with the cylinder. DISCUSSION :

The equations of motion (9.49) provide nine first-order differential equations to determine the motion of the center of mass and the orientation of the rigid body. If only x¯ (t) and ω(t) are required, then (9.49)1 can be ignored. The equations of motion show that β˙ and ±z are constant: β˙ (t) = β˙ (0) and ±z (t) = ±z (0). In addition, it is straightforward to show from (9.49)2,4,5 that ±β is governed by the differential equation ²

λ+

mR2

³

¨ + λβ˙ ± β

2

(0)±β

=

0.

Thus, ±β (t) oscillates in time with a frequency that depends on β˙ 2(0). In addition, (9.49)3,4 then imply that z(t) and ±r (t) oscillate. Representative examples to illustrate

378

Kinetics of a Rigid Body

our remarks can be seen in Figure 9.21. As Littlewood [171] noted, what is often surprising about this problem is that the rolling sphere oscillates up and down the inner surface of the cylinder, while a particle placed on the rough cylinder would simply fall down the cylinder. This aspect of the dynamics of a rolling rigid body compared to a particle is similar to what we observed with the sphere on the turntable in Exercise 9.10.

10

Lagrange’s Equations of Motion for a Single Rigid Body

10.1

Introduction

In a famous paper [156] on the dynamics of the Moon orbiting the Earth, which was published in 1782, Lagrange used his equations of motion in the form d dt

±

∂T

²

∂q ˙K

−

∂T

= −

∂ qK

∂V

(K =

∂ qK

1, . . . , 6) ,

(10.1)

where V is the potential energy of the Moon and T is its kinetic energy. His interest lay in explaining oscillations (librations) in the attitude of the Moon as seen by an Earth-based observer. What is interesting is that he does not use Euler’s balance laws, although these ˙ were available to him. One might ask what would have happened had he used M = H ˙ and F = mv¯ instead of (10.1)? In this chapter, we will show (among other matters) that his equations are equivalent to Euler’s balance laws and so he would have arrived at the same conclusions. ˙ for a rigid We start this chapter by showing that the balance laws F = mv˙¯ and M = H body are equivalent to Lagrange’s equations of motion: ±

²

d ∂T ∂T − i dt ∂ q˙ ∂ qi ± ² ∂T d ∂T − i dt ∂ ν˙ ∂ νi

=

F · ai ,

=

M · gi .

(10.2)

We illustrate this form of Lagrange’s equations by using a classical problem of a satellite orbiting a fixed body. Indeed, the problem we consider is equivalent to that considered by Lagrange [156]. A second form of Lagrange’s equations of motion is also developed. This form allows broader classes of coordinate choices and, being the most general, is most useful in applications. The form of the celebrated equations is d dt

±

∂T ∂u ˙A

²

−

∂T ∂ uA

where ±A =

F·

= ±A

(A =

1, . . . , 6) ,

¯ ∂v ∂ω +M· . A ∂u ˙ ∂u ˙A

(10.3)

(10.4)

Following the development of (10.3), we show how constraints may be incorporated and illustrate this matter by using examples of rolling disks, sliding disks, and spinning tops.

380

Lagrange’s Equations of Motion for a Single Rigid Body

e3

e1

φ

¯

E3

g

e2

X θ

E1

XO

E2

The Lagrange top: an axisymmetric rigid body that is free to rotate about a fixed point XO . The image on the left-hand side is a portrait of Joseph-Louis Lagrange (1736–1813).

Figure 10.1

When the system of constraints on the system is integrable, the constraint forces and moments are prescribed by use of Lagrange’s prescription (i.e., the constraints are “ideal”), and the coordinates are chosen appropriately; pleasantly, we find that we can decouple (10.3) into a set of equations involving the unconstrained motion and a set of equations for the constraint forces and moments. The former equations are known as reactionless. We then discuss an approach to Lagrange’s equations that we have referred to as Approach II. Much of the material concerning Lagrange’s equations in this chapter is based on Casey [36, 38], supplemented with material on constraint forces and moments from O’Reilly and Srinivasa [219, 220]. A rapid overview of this chapter can be found in [123].

10.2

The Lagrange Top

To motivate the developments in this chapter, we consider the classic example of a symmetric top that is free to rotate about a fixed point X O (cf. Figure 10.1). This problem was discussed by Lagrange as a marvelous illustration of his equations of motion in 1789.1 Discussions of the behavior of the top can be found in several textbooks on dynamics (see, e.g., Arnol’d [11, Section 30], Greenwood [105, Section 8-5], or Whittaker [306, Section 71]) and we leverage these works in our discussion below. The mass of the top is denoted by m and its inertia tensor is J = λt (I − e3 ⊗ e3 ) + λ ae3 ⊗ e3 , where e3 is the axis of symmetry of the top. The position vector of the center of mass relative to XO has the representation x¯ − xO

= ²3 e3.

1 See Section IX.34 of Part II of Lagrange’s Mécanique Analytique [159].

10.2

E2

E3

e2

e3

e1

e2

e2

e2

ψ

e1

θ ψ

381

The Lagrange Top

φ θ

E1

φ

e2

e1

The transformations of various basis vectors induced by the individual angles in a set of 3–1–3 Euler angles. For Lagrange’s top, ψ describes the precession, θ describes the nutation, and φ describes the spinning of the top.

Figure 10.2

M

3 ∼ As XO is fixed, v¯ = ²3ω × e3. The configuration manifold = RP . A set of 3–1–3 Euler angles is used to parameterize Q (cf. Section 6.8.2 and Figure 10.2). The first Euler angle ψ measures the precession of the top, the second Euler angle θ measures the nutation (or inclination to the vertical) of the top, and the final Euler angle φ measures the spin of the top about its axis of symmetry. The angular velocity vector ω of the top has the representation

ω

˙ g + θ˙ g + φ ˙g , = ψ 1 2 3

where g1

=

E3

=

cos (θ ) e3 + sin (θ ) (cos (φ) e2 + sin (φ) e1) ,

g2

=

cos (φ) e1 − sin (φ) e2,

g3

=

cos (θ ) E3 + sin (θ ) (sin (ψ ) E1 − cos (ψ ) E2 )

=

e3.

Because (g 1 × g 2 ) · g 3 = − sin (θ ) ,

the second angle needs to be restricted, θ ∈ (0, π ), in order to avoid singularities. The constrained kinetic and potential energies of the top have the representations T˜

= =

˜ = U

HO · ω λa

(

˙ +ψ ˙ φ

)2

cos (θ )

2 mg²3 cos (θ ) .

+

λt +

m²23 ³

2

2 ˙2 θ˙ + ψ

´

sin2 (θ ) , (10.5)

2 Observe that λO t = λt + m²3 is a mass moment of inertia relative to X O . We assume that Lagrange’s equations of the following form apply to the top with our choice of coordinates:

d dt

µ

˜ ∂L

∂ q˙ i

¶

−

˜ ∂L

∂ qi

=

0

³

q1

= ψ, q

2

= θ, q

3

´ = φ

.

(10.6)

That is, we are assuming that Approach II is valid for this rigid-body dynamics problem and thus, none of the constraint forces acting at X O contribute to the equations of motion

382

Lagrange’s Equations of Motion for a Single Rigid Body

for the generalized coordinates ψ , θ , and φ. Expanding the derivatives of L˜ we find the following equations of motion for the top: ⎡ ⎣

2 λa cos 2(θ ) + λ O t sin ( θ )

0 λa cos( θ )

·

0

λa cos(θ )

O λt

0

0

¸¹

⎤⎡

⎤

¨ ψ

⎡

α1

⎤

⎡

⎦ ⎣ θ¨ ⎦ + ⎣ α2 ⎦ = ⎣

λa

¨ φ

º

˜ − U, ˜ = T

β1 β2

⎡ ⎣

α1 α2

⎤ ⎦

⎡ =

⎣

(10.7) )

(

O ˙ cos (θ ) sin (θ ) − λa φ ˙ θ˙ sin (θ ) λt − λa θ˙ ψ ( ) O 2 ˙ ˙ ˙ λaφ ψ sin (θ ) − λ t − λ a ψ cos (θ ) sin (θ )

2

α3

⎦,

β3

α3

M

where

⎤

⎡

β1

⎣ β2 β3

⎤ ⎦,

˙ θ˙ sin (θ ) −λ a ψ

⎤

⎤

⎡

0 ⎦ = ⎣ mg²3 sin(θ ) ⎦ . 0

The mass matrix M on the left-hand side of (10.7) can also be inferred from the expres2 O sion for the kinetic energy function (10.5)1 . The determinant of M is λO t λ aλt sin (θ ). Thus, in agreement with our earlier remarks about the second Euler angle, the mass matrix will be noninvertible when θ = 0, π . Apart from the relative ease with which (10.7) are obtained for the top, Lagrange’s equations of motion for this system show in a transparent manner the conservations of HO · E 3 = HO · g1 and HO · e3 = HO · g3. These conservations can be traced to the fact that the external moment acting on the system is MO = −² 3e3 × mgE3 . Because ˜ ˜ ∂L ∂L = 0 and = 0, Lagrange’s equations lead to the conclusions that ∂ψ ∂φ d dt

µ

˜ ∂L ˙ ∂ψ

¶

=

HO · E3

=

0,

d dt

µ

˜ ∂L ˙ ∂φ

¶

=

HO · e 3

=

0.

This pair of conservations enables us to reduce (10.7) to a well-known single twoparameter family of ordinary differential equations for θ :2 2 Od θ λt = dt2

(

pψ

−

)(

pφ cos( θ ) pψ cos(θ ) − pφ O λt

sin3(θ )

In this second-order differential equation, pψ conserved momenta: ⎡ ⎣

pψ

=

pφ

=

˜ ∂L ˙ ∂ψ

˜ ∂L

˙ ∂φ

=

HO · E3

=

HO · e 3

⎤

⎦=

»

λO t

=

)

+

mg²3 sin(θ ).

HO · E3 and pφ

sin2(θ ) + λ a cos2( θ ) λa cos( θ )

=

(10.8)

HO · e3 are the

λa cos( θ ) λa

¼»

˙ ψ ˙ φ

¼

.

(10.9) The second-order differential equation (10.8) can be integrated numerically to determine θ (t) in a manner that guarantees that the angular momenta HO · E3 and HO · e3 are ˜ of the conserved. As with (10.7), the solutions to (10.8) conserve the total energy T˜ + U 2 See, for example, [306, Section 71].

10.3

Configuration Manifold of an Unconstrained Rigid Body

383

top. Substituting for φ˙ and ψ˙ using (10.9), the energy can be expressed as a function of θ , θ˙ , pφ , and pψ : (

E˜ =

)2

pψ − pφ cos( θ ) 1 O 2 λ θ˙ + 2 2 t 2λ O t sin ( θ )

+

mg²3 cos (θ ) .

We leave it as an exercise to show that E˙˜ = 0 for the solution θ (t) of (10.8). As we shall see later, the differential equations of motion (10.7) are equivalent to the ˙ O = M O on the Euler basis vectors. projections of the balance of angular momentum H However, in contrast to the methods of Chapter 9, finding the differential equations of motion using Lagrange’s equations is far more direct. We shall return to the Lagrange top later in this chapter, after we have demonstrated why Lagrange’s equations of motion ˙ and why for a rigid body are equivalent to the balance laws F = ma¯ and M = H Lagrange’s prescription for the constraint force justifies the use of Approach II for many problems in rigid-body dynamics.

10.3

Configuration Manifold of an Unconstrained Rigid Body

The motion of a rigid body can be decomposed into a translation x¯ of the center of mass X¯ followed by a rotation Q. Here, the set of all vectors x¯ is a three-dimensional Euclidean space E3 , whereas the set of all rotation tensors Q forms a group known as SO(3).3 As discussed in Section 6.10, the group SO(3) is homeomorphic to RP3. Thus, the configuration manifold of a rigid body is the space of all vectors x¯ and all rotation tensors Q:

M

M

∼ =

E3 × RP 3.

M

S

It should be clear that the dimension of is 6 and that the configuration manifold can be considered as a submanifold of the configuration space : E3 × E9 . Here, the three-dimensional space E3 is the space containing the position vector x¯ and the ninedimensional space E9 is the space containing the nine components of the second-order tensor Q. To parameterize , we can use any curvilinear coordinate system. In one of the forms to follow, we use q1 , q2, and q3 to parameterize the position vector x¯ of the center of mass and any set of Euler angles ν 1 , ν 2 , and ν 3 to parameterize the rotation tensor Q. For the velocity vector of the center of mass, we then have

M

v¯ =

3 ½

q˙ i ai,

i=1

where the covariant basis vectors ai

=

∂x ¯ ∂ qi

.

3 The group O(3) is the group of all orthogonal tensors. The S in SO(3) denotes “special” because rotation

tensors have a determinant of + 1.

384

Lagrange’s Equations of Motion for a Single Rigid Body

You should also recall that {ai } is a basis for E3 and that the Euler angles define the Euler basis {gi }. The latter is also a basis for E3. In particular, we have ω

3 ½ =

ν ˙

i

gi .

i =1

The Euler basis vectors are not linearly independent for certain values of the second Euler angle. It is interesting to note that gi

∂ω

=

∂ν ˙i

,

ai

=

∂ v¯ ∂q ˙i

.

These two identities are easily established. We can calculate the kinematical line element ds for the configuration manifold from the kinetic energy T by using the given choice of Euler angles and curvilinear coordinates: µ¾

ds = µ¾ =

2T m

¶

dt

v¯ · v¯ +

¶

1 ω · (J · ω) dt. m

Shortly, an example will be presented of how to calculate the desired representation for T. As in our discussion of single particles and systems of particles, 4 given a set of coordinates to describe Q and x¯ of a rigid body, we can use the kinetic energy to compute a mass matrix M. A representative particle can also be constructed for the rigid body and a set of covariant basis vectors {a1, . . . , a6} for E6 can be computed. One can then establish the (by now familiar) result that if M is positive definite, then {a1 , . . . , a6 } form a basis for E6 and the coordinate system is free of singularities. When M is not positive definite at a given point, then the coordinate system has a singularity. For rigid bodies, such singularities are commonplace and easily checked by examining the determinant of M. As a consequence, many simulations use two or more sets of curvilinear coordinates and two sets of Euler angles for analyzing a given problem in rigid-body dynamics.

A Representation for the Kinetic Energy

We now consider a specific example. First, we choose a set of spherical polar coordinates R, ³, and ± to parameterize x¯ and a set of 3–1–3 Euler angles to parameterize Q (see Figure 10.2). For these Euler angles, we recall from Section 6.8.2 that the Euler basis { gi } has the representations ⎡

⎤

⎡

⎤

g1 E3 ⎣ g ⎦ = ⎣ e± ⎦ 2 1 g3 e3 4 See Sections 1.8 and 4.8.

⎡

sin(φ ) sin(θ ) cos(φ ) sin(θ ) = ⎣ cos(φ ) − sin(φ) 0 0

⎤⎡

⎤

cos(θ ) e1 ⎦ ⎣ e2 ⎦ . 0 1 e3

10.3

385

Configuration Manifold of an Unconstrained Rigid Body

Using these results, we can readily compute the angular velocity vector: ω = ψ˙ E3 + θ˙ e±1 + φ˙ e3 =

(

˙ ψ

+

)

sin(φ ) sin(θ ) + θ˙ cos(φ ) e1 +

(

˙ ψ

cos(θ )

˙ +φ

)

(

)

˙ cos( φ) sin( θ ) − θ˙ sin(φ ) ψ

e3.

e2 (10.10)

For convenience, we choose Ei to be the principal axes of the body.5 Hence, J0

= λ1 E1 ⊗ E1 + λ2E 2 ⊗

J = QJ 0Q

T

= λ1 e1 ⊗

E2 + λ 3E3 ⊗ E3 ,

e1 + λ2 e2 ⊗ e2 + λ3 e3 ⊗ e3.

When these results are combined, the kinetic energy T of the rigid body has the representations T

= =

m 1 v¯ · v¯ + ω · Jω 2 2 ´ )2 m³ 2 λ1 ( ˙ 2 + R2 sin2( ±)³ ˙2 + ˙ sin(φ ) sin(θ ) + θ˙ cos(φ ) R˙ + R2± ψ 2 2 λ2

(

)2

˙ ψ

cos(φ ) sin(θ ) − θ˙ sin(φ)

λ3

(

+

˙ ψ

cos(θ ) + φ˙

)2

. 2 2 It is left as an exercise to substitute this expression for T into the expression for the kinematical line element ds in order to find a measure of distance that the rigid body travels along the configuration manifold . To examine the possibility of coordinate singularities, we now consider the mass matrix M associated with T . This 6 × 6 matrix is block diagonal for this example: +

M

»

M

where

⎡

m 0 M1 = ⎣ 0 mR2 0 0 ⎡

=

ma44

=

λ1

0 M2

¼

,

⎤

0 ⎦, 0 2 2 mR sin (±)

ma44 M2 = ⎣ λ 1 −2 λ2 sin (2φ) sin (θ ) λ 3 cos (θ ) ³

M1 0

λ1 −λ2

sin (2φ) sin (θ ) 2 2 λ 1 cos (φ) + λ2 sin (φ) 0 2

´

λ3

⎤

cos (θ ) ⎦, 0 λ3

sin2 (φ) + λ2 cos 2 (φ) sin2 (θ ) + λ 3 cos2 (θ ) .

You should notice how the mass matrix simplifies when the body is axisymmetric ( λ1 λ2 ) and spherical (or cubic) (λ1 = λ2 = λ3 ). The determinant of the mass matrix is

=

det (M) = m2 λ1 λ2 λ3 R4 sin2 (±) sin2 (θ ) . From the expression for the determinant, we conclude that singularities in the parameterization of the motion arise when any of the following instances occur: θ = 0, θ = π , 5 If we do not make this choice, then we would find a more complicated representation for the rotational kinetic energy: ω · (Jω) = J 011 ω12 + J022 ω22 + J033 ω23 + 2J 012 ω1 ω2 + 2J 023 ω2ω3 + 2J 013 ω1ω3 , where ωi = ω · e i and J 0 = e i · (Je k ) = Ei · (J0 Ek ). ik

386

Lagrange’s Equations of Motion for a Single Rigid Body

R = 0, ± = 0, or ± = π . That is, the singularities are inherited from the spherical polar coordinate system for x¯ and the 3–1–3 Euler angles for Q.

10.4

Lagrange’s Equations of Motion: A First Form

In this section, we wish to establish Lagrange’s equations for an unconstrained rigid body. Our proof is based on Casey [38], but our developments are not as general as his. The resulting equations (10.11) were first shown by him to be equivalent to the balances of linear momentum and angular momentum for a rigid body.6 There is also a strong suggestion of the equivalence of Lagrange’s equations of motion and the balance of angular momentum for a rigid body in Greenwood [105, Sections 8-2 and 8-6]. 7 We refer to (10.11) below as the first form of Lagrange’s equations of motion. To start, we choose a set of curvilinear coordinates qi to parameterize x: ¯ x ¯ = 1 2 3 i x¯ (q , q , q ). Next, we choose a set of Euler angles ν to parameterize the rotation tensor Q of the rigid body Q = Q( ν 1, ν 2, ν 3). Then, as will subsequently be shown, Lagrange’s equations for the rigid body are ±

²

∂T d ∂T − i dt ∂ q˙ ∂ qi ± ² d ∂T ∂T − i dt ∂ ν˙ ∂ νi

=

F · ai ,

=

M · gi .

(10.11)

Here, gi are the Euler basis vectors and ai are the basis vectors for E3 that are associated with the curvilinear coordinate qi. For the preceding equations, you may wish to recall the results v¯ =

3 ½ i=1

q˙ i ai ,

∂x ¯ ∂ qi

=

∂v ¯ ∂q ˙i

=

ai ,

ω=

3 ½

ν˙

i=1

i

gi ,

∂ω ∂ν ˙i

=

gi .

It should be evident from (10.11) that Lagrange’s equations for a rigid body have similarities to those we encountered earlier with particles. The main difference is the balance of angular momentum. If some of the forces and moments acting on the rigid body are conservative, then for these conservative forces Fcon acting at X¯ and moments M con, we have 8 F con = −

∂U ∂ x¯

M con = −u Q

= −

= −

3 ½ ∂U i

i=1 3 ½ i=1

∂ qi

¯ a acting at X,

∂U i g, ∂ νi

6 See, in particular, Theorems 4.2 and 4.4 of Casey [38]. 7 Greenwood’s exposition is missing the intermediate result

∂T

∂ νi

(10.12)

=

8 These representations were established in Chapter 8 (cf. (8.21)).

H · g˙ i .

10.4

Lagrange’s Equations of Motion: A First Form

where the potential energy function U has the representations U

³

=

387

´

U (x¯ , Q) = U q1 , q2, q3 , ν 1 , ν 2 , ν 3 .

Observe that Fcon · ai

= −

∂U , ∂ qi

Mcon · gi

= −

∂U . ∂ νi

Consequently, it is not necessary to evaluate Fcon and Mcon in Lagrange’s equations; rather, it suffices to evaluate the partial derivatives of U. It is remarkable that the situation with potential energies in rigid bodies is similar to that encountered in systems of particles. It is also possible to write an alternative form of Lagrange’s equations of motion by using the Lagrangian L = T − U. Specifically: ±

²

d ∂L ∂L − i dt ∂ q˙ ∂ qi ± ² ∂L d ∂L − dt ∂ ν˙ i ∂ νi

= (F −

Fcon ) · ai ,

= (M −

Mcon ) · gi .

(10.13)

It is left as an (easy) exercise to show that (10.13) can be established from (10.11). The two main applications of (10.11) that will be considered in this chapter are a satellite problem in Section 10.5 and a cylinder sliding on a smooth horizontal surface in Exercise 10.3 (cf. Figures 10.3 and 10.10). 10.4.1

Proof of Lagrange’s Equations

To prove Lagrange’s equations, we need to exploit the Koenig decomposition and use the angular velocity vector ω0 = QT ω. The proof proceeds quickly after some preliminary results have been addressed. There are four steps in the proof. The first step involves parameterizing x¯ by use of a set of curvilinear coordinates and parameterizing Q by use of a set of Euler angles ν i . These parameterizations imply that the kinetic energy T is a function of these quantities and their time derivatives: ³ ´ 1 1 T = T qi , q˙ i , ν k , ν˙ k = mv¯ · v¯ + J0 ω0 · ω0 . 2 2 Here, the angular velocity vector ω0 = QT ω and the inertia tensor J0 We next consider the partial derivatives of T : ∂T ∂ = i ∂q ˙ ∂q ˙i ∂T = ∂ qi ∂T ∂ν ˙i

=

∂T = ∂ν i

±

=

QT JQ.

²

∂v ¯ 1 mv¯ · v¯ = mv¯ · i = mv¯ · ai , 2 ∂q ˙ ± ² ∂ 1 ∂v ¯ mv¯ · v¯ = mv¯ · i = mv¯ · a˙ i , i ∂q 2 ∂q ± ² ( T ) ∂ 1 ∂ ω0 J ω · ω0 = J 0 ω0 · = J0 ω0 · Q g i , 0 0 ∂ν ˙i 2 ∂ν ˙i ± ² ( T ) ∂ 1 ∂ ω0 J 0 ω0 · ω0 = J 0 ω0 · = J0 ω0 · Q g ˙i . i i ∂ν 2 ∂ν

(10.14)

388

Lagrange’s Equations of Motion for a Single Rigid Body

Here we have used the identities ∂v ¯ ∂ v¯ ai = i , a˙ i = i , ∂q ˙ ∂q

gi

=

Q

∂ ω0 , ∂ν ˙i

g˙ i

=

Q

∂ ω0 . ∂ νi

(10.15)

Shortly, a derivation of these results will be given. For the third step, we examine (10.14)1,2 and note that G = mv¯ . Hence, d dt

±

∂T ∂q ˙i

²

−

∂T = ∂ qi

d (G · ai ) − G · a ˙i dt

˙ · ai = G =

F · ai.

Consequently, the first three Lagrange equations of motion have been established. The fourth step in the derivation is to note that, for any vector b, H · b = Jω · b = QJ0 QT Qω0 · b = J0ω0 · QT b. Observe that we used the identities J = QJ0 QT and ω = QT ω0 to establish the final form of the identity. Using the identity H · b = J0 ω0 · QT b in conjunction with (10.14) 3,4 , we find that ± ² ) ∂T d ( d ∂T − = H · gi − H · g˙ i i i dt ∂ ν˙ ∂ν dt ˙ ·g = H i =

M · gi .

Thus, we have established the last three of Lagrange’s equations of motion. 10.4.2

The Four Identities

The preceding proof of Lagrange’s equations was achieved by use of the four identities (10.15). The first two of these results are similar to those we established for the single particle. You should notice the presence of Q in (10.15)3,4 . Unfortunately, ∂ω ²= g ˙ i. ∂ νi We first establish the easier results: ∂ ∂v ¯ = ∂q ˙i ∂q ˙i

µ 3 ½

¶

k

q˙ ak

=

k=1

and ∂v ¯ ∂ = i ∂q ∂ qi

3 k ½ ∂q ˙ k =1

µ 3 ½

ak ∂q ˙i

¶ k

q˙ ak

k=1

3 ½ =

=

q˙ k

3 ½

δ ik a k =

k=1

∂ ak ∂ qi

k =1 ± ² 3 2¯ ½ ∂ x¯ k ∂ k ∂ x = q˙ = q˙ ∂ qi ∂ q k ∂ qk ∂ qi k=1 k=1 3 ½

3 ½ =

k=1

= a ˙i .

q˙ k

∂ ai ∂ qk

ai

10.4

389

Lagrange’s Equations of Motion: A First Form

Notice that we used the facts that ai are independent of q˙ k and equivalent to the partial derivatives of x¯ with respect to qi . We next consider the easier of the two remaining identities: ∂ ω0

∂

=

∂ ν˙ i

=

∂ ν˙ i T

T ( T ) ∂Q ω + QT Q ω =

±

∂ ν˙ i

∂ω

² =

∂ ν˙ i

( )

Oω + QT gi

Q gi .

Some rearranging gives gi

=

Q

∂ ω0 . ∂ν ˙i

Notice that we used the fact that Q does not depend on ν˙ i to establish this result. For the final result, we again need to be cognizant of the fact that Q depends on the Euler angles ν i but not on ν˙ i . In addition, because the third-order tensor ε is a constant, ∂ε ∂ν ˙i

we have the identity ∂ ∂x

±

∂

(ax (B )) =

∂ε

0,

∂ νi

²

−

∂x

=

1 ε [B] 2

± =

−

=

1 ε 2

0, »

(10.16)

∂B

¼²

± =

∂x

ax

∂B

²

,

∂x

where x represents ν i or ν˙ k and B is any second-order tensor. To continue, we note that ∂ω ∂ gi = = ∂ ν˙ i ∂ν ˙i

± =

ax

∂Q T Q ∂ν i

(

(

˙ ax QQ

T

))

=

µ

∂ ∂ν ˙i

ax

²

µ 3 ½ k=1

k ∂Q T Q ν˙ ∂ν k

¶¶

.

Next we deduce, by differentiating QQT T ∂Q ∂Q T Q = −Q ∂ νi ∂ νi

=

I twice with respect to the Euler angles, that T ∂Q ∂Q = ∂ νi ∂ νk

,

Q

∂Q

T

∂Q T Q . i ∂ ν ∂ νk

The previous results are now used to show the desired identity: ±

g˙ i

=

=

d ax dt 3 ½

µ

ν ˙

k

k =1 3 ½ =

=

k =1 3 ½ k =1

±

ν ˙

k

k

µ

∂

ax µ

ν ˙

∂Q T Q ∂ νi

²²

2

Q

∂ νi∂ ν k

µ

µ

ax Q Q ±

±

ax Q

∂ ∂ νi

T

Q

T

T ∂Q ∂Q + ∂ νi ∂ νk

¶¶

T ∂Q ∂Q ∂ ∂2Q + = i k i k ∂ν ∂ν ∂ν ∂ν ∂ νi

±

QT

∂Q ∂ νk

²

QT

²²

±

∂Q QT k ∂ν

²¶

¶¶

Q

T

390

Lagrange’s Equations of Motion for a Single Rigid Body

µ =

ax Q ±

=

ax Q ±

∂ νi

Q ax

=

Q

Q

∂ Ω0 ∂ νi

3 ½

T

k =1

∂ Ω0 T Q ∂ νi

±

=

µ

∂

k ∂Q ν ˙ = ∂ νk

¶ ˙ = QT Q

Ω0 Q

¶ T

²

²²

∂ ω0 . ∂ νi

In the final stages of the proof we used the fact that ω0 is the axial vector of Ω0 ˙ = QT ΩQ. We also used the identity QT Q (

ax QBQT

)

=

=

det (Q) Q (ax (B )) .

This identity is one method of relating ω0 to ω and it was alluded to earlier when angular velocity vectors were discussed (see (6.8)).

10.5

A Satellite Problem

As an example of a problem from rigid-body dynamics for which there are no constraints, we consider a satellite of mass m that is in orbit about a spherically symmetric body of mass M (cf. Figure 10.3). We assume that the spherically symmetric body is fixed and use its center of mass as the origin of the position vector of the center of mass of the satellite. Preliminaries

To parameterize the motion of the center of mass of the rigid body, we use a spherical polar coordinate system: Rigid body B

e1

Fixed body of massM

E3

x¯

e3

¯

X

e2

β O

E1 Figure 10.3

π

θ

E2 er

A rigid body orbiting a spherically symmetric body of mass M. The angle of latitude

β = 2 − φ.

10.5

E3

E1 = e1

e3

e1

e2 ν

1

ν ν

1

ν

E2

e2

e2

e3

2

391

A Satellite Problem

=

e2 e1

ν

2

3 ν

e3

3

e1

The transformations of various basis vectors induced individual angles in a set ¿ by the À of 1–2–3 Euler angles. For this set of angles, the Euler basis E1 , e±2 , e3 fails to be a basis when ν 2 = ³π/2. This set of Euler angles was also used to describe the knee joint in Chapter 6 (cf. Figure 6.13).

Figure 10.4

x¯ = ReR =

R cos(φ )E3 + R sin(φ )er

=

R cos(φ )E3 + R sin(φ ) (cos(θ )E 1 + sin(θ )E2 )

and ˙ R + R sin(φ )θ˙ eθ v¯ = Re

+

R φ˙ eφ .

You should be able to see from this equation what the covariant basis vectors ai are. Referring to Figure 10.4, we parameterize the rotation tensor Q using a set of 1–2–3 Euler angles: ω = ν˙ 1 E1 + ν˙ 2 e±2 + ν˙ 3e3 .

You should notice that ei representation

=

QEi . The angular velocity vector of the body also has the ω = ω1 e1 + ω2e2 + ω3 e3 ,

where ω1 = ν˙ ω2 = ν˙ ω3 = ν˙

2 2 1

sin cos sin

³ ν

3

³ ν

³ ν

´ + ν ˙

3

2

1

´ −ν ˙

´ + ν ˙

1

3

³

cos

ν

cos

2

³ ν

´

2

cos

´

sin

³ ν

³ ν

3 3

´ ´

, ,

.

You should be able to infer the representations for gi from these results. We shall choose E i to be the principal axes of the body in its reference configuration. Using this specification, it follows that the kinetic energy of the rigid body has the representation T

=

³ ´ ´2 ´ λ ³ m³ 2 1 2 2 R˙ + R2 φ˙ 2 + R2 sin2(φ )θ˙ 2 + ν ˙ sin (ν2 ) + ν ˙1 cos ν cos(ν 3) 2 2 +

λ2

2

³

ν˙

2

cos

³

ν

3

´

−ν ˙

1

cos

³

ν

2

´

sin

³

ν

3

´´ 2

+

λ3

2

³

ν˙

1

sin

³

ν

2

´

+ ν ˙

3

´2

.

If the body has an axis of symmetry so that λ 1 = λ2 , then the expression for the rotational kinetic energy will simplify considerably. The determinant of the mass matrix

392

Lagrange’s Equations of Motion for a Single Rigid Body

(

)

associated with T is m3λ 1λ2 λ3 R4 sin2( φ) cos2 ν 2 . Thus, the singularities in the coordinate system for are identical to those associated with the spherical polar coordinate system and the Euler angles. The sole force and moment acting on the rigid body is due to the central gravitational force exerted on it by the body of mass M. These forces and moments are conservative and are associated with the potential energy

M

U

= −

GMm R

GM 3GM tr(J) + (JeR ) · eR . 3 2R 2R3

−

(10.17)

To express this potential energy in terms of the Euler angles ν i , we need to use the results eR

=

cos(φ )E3 + sin(φ ) cos(θ )E1 + sin(φ ) sin(θ )E2 ,

E1

=

cos

E2

= − sin

E3

=

³

³

ν

sin ³

+

2

´

³ ν

³ ν

sin

2

cos ´

´

1

³ ν

³

cos

sin

1

ν

´

3

´

³ ν

³ ν

cos

3

e1 − cos

3

´

´

ν

3

ν

e1 + sin

−

³

³

´

cos +

³

ν

cos

2

´

³

1

ν

´

³ ν

2

sin

´

sin 1

´

³

ν

sin

³

ν

sin

3

´

³

2

ν

´

³ ν

3

e2 + sin

´

cos 2

´

³ ν

e2 + cos ³

ν

sin

3

´´

³ ν

3

2

´

³ ν

2

e3, ´

e3 ,

e1

´´

e2 + cos

³ ν

1

´

cos

³ ν

2

´

e3 .

Using these results, we can express eR in terms of the corotational basis, and then a useful expression for (JeR ) · eR can be established. We can then derive expressions for the force and moment associated with this potential energy by using representations (10.12). Alternatively, we can appeal to representations (8.18) described in Chapter 8.

The Balance Laws

For the rigid body of interest, we have the balance laws mv˙¯ = F

GMm 3GM eR − (2J + ((λ1 + λ2 + λ3 ) − 5eR · JeR ) I) eR , 2 R 2R 4 ˙ = M H ² ± 3GM eR × (Je R). = R3 = −

Here, we used our earlier representations (8.18) for the conservative force and conservative moment associated with the gravitational potential. You should also notice that if eR is an eigenvector of J, then JeR is parallel to eR . In this case, the so-called gravity– gradient torque vanishes: M = 0. In addition, if R = R 0er and θ˙ = ω0 , where R 0 and ω0 are constant, then the center of mass of the rigid body describes a circular orbit at a constant orbital speed ω0.9 9 See Exercise 9.15.

10.6

Lagrange’s Equations of Motion: A Second Form

393

Lagrange’s Equations of Motion

Lagrange’s equations of motion for the satellite are the ai components of the balance of linear momentum and the gi components of the balance of angular momentum: d dt

±

²

∂T

−

˙ ∂R

∂T ∂R

=

F · eR

= −

d dt

±

∂T

² −

˙ ∂φ

∂T ∂φ

=

±

∂T

² −

∂ θ˙

∂T ∂θ

=

±

∂T ∂ν ˙i

² −

∂T = ∂ νi

) 3GM ( eR · Jeφ , R3

F · R sin(φ )eθ

= −

d dt

3GM ((λ1 + λ 2 + λ3 ) − 3eR · Je R ) , 2R 4

−

F · Reφ

= −

d dt

GMm R2

3GM sin(φ ) (eR · Je θ ) , R3

M · gi ±±

=

3GM R3

²

²

eR × (JeR )

·

gi .

In the last of these equations, i = 1, 2, 3. It is left as an exercise to evaluate the partial derivatives of T with respect to the coordinates and their velocities. Alternatively, one could also use (10.13) to write Lagrange’s equations by using the Lagrangian L = T −U. Conservations

Because the only forces and moments acting on the rigid body are conservative and there are no constraints on the motion of the rigid body, it is easy to see that the total energy E = T + U is conserved. In addition, the angular momentum HO is conserved. To see this, notice that ˙O H

=

MO

=

ReR × F + M.

Substituting for F and M, one finds that MO = 0. Consequently, HO is conserved. It is interesting to note that we cannot conclude that H is conserved for an arbitrary motion of the rigid body, although it is conserved if the gravity–gradient torque M is zero for a specific motion.

10.6

Lagrange’s Equations of Motion: A Second Form

¿

Previously, we assumed that a set of coordinates q1, . . . , q6 parameterize x¯ and Q: ³

´

x¯ = x¯ q1, . . . , q3 ,

Q

³

=

Q q4

= ν

1

, . . . , q6

À

had been chosen to

= ν

3

´

.

(10.18)

394

Lagrange’s Equations of Motion for a Single Rigid Body

We note that the covariant basis vectors associated with this choice of coordinates are ai

=

∂v ¯ ∂q ˙i

,

gi

=

∂ω

(i

∂q ˙ (i+3)

=

1, 2, 3).

With choice (10.18), we showed that Lagrange’s equations of motion have the following form: ±

²

d ∂T ∂T − dt ∂ q˙ i ∂ qi ² ± ∂T d ∂T − (i +3) dt ∂ q˙ ∂ q(i +3)

∂ v¯

=

F·

=

M·

∂ q˙ i

,

∂ω . ∂q ˙ (i+3)

(10.19)

It is not too difficult to see that these equations can be written in a more compact form: d dt

±

∂T ∂ q˙ A

²

−

∂T = ∂ qA

F·

∂ v¯ ∂ω +M· A ∂q ˙ ∂q ˙A

(A =

1, . . . , 6) .

(10.20)

Form (10.20) of Lagrange’s equations is useful in several cases, such as when 1. there are no constraints on the motion of the rigid body, 2. the constraints (and the associated constraint forces and moments) on the rigid body do not couple the rotational and translational degrees of freedom. However, we shall need a more general form of Lagrange’s equations for other applications. We now consider a new choice of coordinates: ´

³

´

³

Q = Q u 1, . . . , u 6 .

x¯ = x¯ u1, . . . , u6 ,

(10.21)

In addition, we suppose that a system of N forces FK , each acting at a material point XK , where K = 1, . . . , N, and a pure moment M p act on the rigid body. With the choice (10.21), we shall see that Lagrange’s equations of motion are d dt

±

∂T ∂ u˙ A

²

−

∂T ∂ uA

= ±A ,

(10.22)

where the generalized forces have the representations ±A =

F·

∂ v¯ ∂ω +M· A ∂ u˙ ∂u ˙A

N ½

=

FK

·

K=1

∂ω ∂ vK + Mp · ∂u ˙A ∂ u˙ A

(A =

1, . . . , 6) .

(10.23)

We shall also discuss shortly examples that use (10.22) to establish the equations of motion for a rigid body. To establish (10.22), we invoke the following identities: 10 d dt

±

∂ v¯ ∂ u˙ A

²

=

∂v ¯ , ∂ uA

d dt

±

∂ω ∂u ˙A

²

±

=

d ∂ ω0 Q A dt ∂u ˙

²

=

Q

∂ ω0 , ∂ uA

10 The proof of these results follows from (10.21) in a manner that is similar to the method by which the

four identities were established in Section 10.4.2.

10.6

Lagrange’s Equations of Motion: A Second Form

395

where ω = Q ω0 ,

and we exploit the fact that J0 is independent of uA and u˙ B . Using the aforementioned identities and the decomposition of the kinetic energy, 11 we can use a straightforward set of manipulations to establish (10.22): d dt

±

∂T

²

−

∂ u˙ A

∂T ∂ uA

=

d dt

±

±

−

∂T ∂u ˙A

=

∂T = ∂ uA

±

mv¯ ·

∂v ¯ ∂u ˙A

±

+

Jω · Q

∂v ¯ mv¯ · A + J0ω0 · ∂u

±

²

∂ ω0 ∂ u˙ A

∂ ω0 ∂ uA

²²

²²

¯ ∂ v¯ ∂v d G· A − G· A dt ∂u ˙ ∂u ± ± ²² ± ² ∂ ω0 d ∂ ω0 H· Q A + −H· Q dt ∂u ˙ ∂ uA ± ² ∂ ω0 ∂v ¯ ˙ · Q ˙ · +H = G A ∂u ˙ ∂u ˙A ∂ v¯ ∂ω = F· +M· A ∂ u˙ ∂u ˙A = ±A . =

(10.24)

This form of Lagrange’s equations is the starting point for most applications and a discussion of Approach II. As with single particles and systems of particles, Lagrange’s equations can be considered as linear combinations of the components of the balance laws. The fact that these combinations can eliminate constraint forces and constraint moments in certain instances is remarkable. Generalized Forces

±A

The generalized force on the right-hand side of Lagrange’s equations of motion has several representations, including the pair shown in (10.23). We now address why these representations are equivalent. Referring to Figure 10.5, suppose that one of the forces acting on a rigid body, say F K , acts at a point XK . The force FK acting at XK is equipollent to a force FK acting at X¯ and a moment MK = (xK − x¯ ) × FK . We also note that vK

= v ¯ +

ω × (xK − x¯ ) ,

∂v ¯ ∂ω ∂ vK = + × ( xK − x ¯) . ∂u ˙A ∂u ˙A ∂u ˙A

In order to establish the second of these equations, we took advantage of the facts that neither xK nor x¯ depend on u˙ A. Now consider FK

·

∂ vK = ∂u ˙A

FK

·

∂ v¯ + FK · ∂u ˙A

±

∂ v¯ ∂ω + (( xK − x ¯ ) × FK ) · A ∂u ˙ ∂ u˙ A ∂ v¯ ∂ω = FK · + MK · . ∂u ˙A ∂u ˙A =

11 Recall that Jω · ω

=

FK

·

QJ 0 ω0 · ω = QJ 0 ω0 · Q ω0

=

²

∂ω × ( xK − x ¯) ∂u ˙A

J 0ω0 · ω0 .

396

Lagrange’s Equations of Motion for a Single Rigid Body

(a)

(b)

FK XK

FK

XK

xK

xK x¯

x¯

¯

X O

O

¯

X

MK = (xK − x¯ ) × FK

Equipollence of (a) a force FK acting at a material point X K and (b) the same force acting at X¯ and a moment MK = (xK − x¯ ) × FK .

Figure 10.5

The identity FK

·

∂ vK = ∂u ˙A

FK ·

∂v ¯ ∂ω + MK · A ∂u ˙ ∂ u˙ A

(10.25)

is the basis for showing that the representations (10.23)1 and (10.23)2 for ±A are equivalent and we leave the remaining details as a straightforward exercise. As we shall see later in our discussion of the dynamics of tops and disks, the identity (10.25) is also useful when incorporating constraint forces acting at a given point of a rigid body into the right-hand side of Lagrange’s equations of motion. A further instance that we shall discuss shortly is the case of a spring force acting at a point Xs (cf. Figure 8.8). We now turn to issues associated with the second form of Lagrange’s equations in the presence of constraints. Changing Coordinates: An Identity

¿

À

¿

À

Recall that we are using two coordinate systems: u1 , . . . , u6 and q1, . . . , q6 . We have tacitly assumed that these coordinate systems are related by invertible functions: qA

³

=

´

qA u1, . . . , u6 .

Then, as q˙ A

=

6 A ½ ∂q u˙ B, B=1

∂ uB

we conclude that ∂q ˙A ∂ qA = B ∂u ∂u ˙B

(A =

1, . . . , 6; B

=

1, . . . , 6) .

(10.26)

These identities are often known as “cancellation of the dots.” Another example of this identity occurs when we examine the expression for the velocity vector: v¯ =

dx¯ dt

6 ½ =

A =1

u˙ A

dx¯ . duA

10.6

Lagrange’s Equations of Motion: A Second Form

397

Differentiating this expression for v¯ with respect to u˙ B , we find another instance of the cancellation of the dots: dx¯ du B

=

dv¯ du˙ B

(B =

1, . . . , 6) .

(10.27)

A related identity can be established when the position vector( of any point, say XP , is ) expressed as a function of the coordinates u1, . . ., u6 : xP = xP u1 , . . . , u6 . It is a good exercise to show that (10.26) and (10.27) also hold when the coordinate transformations are time dependent: qA

³

=

´

qA u1 , . . . , u6 , t .

The resulting identities are used in the literature with Approach II of Lagrange’s equations of motion.

Constraints and Constraint Forces and Constraint Moments

Suppose that an integrable constraint is imposed on the rigid body: ³

´

q1 , . . . , q6 , t

´

=

0. À

¿

We assume that we are at liberty to choose the coordinates u1, . . . , u6 to write this constraint in a simpler form, such as u6 − f (t) = 0. The question we wish to ask now is what are the consequences for the right-hand side of Lagrange’s equations (10.22)? As an example, consider the Lagrange top shown in Figure 10.1 but now assume that the point XO is given a vertical motion: xO = f (t)E3. We can choose u1 = ψ , u2 = θ , u3 = φ, u4 = xO · E1 , u5 = xO · E2 , and u6 = xO · E3. The constraint xO · E 3 = f (t) is equivalent to the constraint u6 − f (t) = 0. The other constraints on the top are u4 = 0 and u5 = 0. ˙ = 0. With some rearranging, The single integrable constraint ´ = 0 implies that ´ ˙ we find that ´ = 0 can be expressed as fX¯ · v¯ + hX¯ · ω + eX¯

=

0,

where fX¯

=

3 ½ ∂´ i i=1

∂q

a, i

hX¯

3 ½ =

i =1

∂´ gi , ∂ q(i+3)

eX¯

=

∂´ ∂t

.

For present purposes, it is most convenient to have representations for fX¯ and hX¯ in terms ¿ À of the coordinates for x¯ and Q, respectively, rather than in terms of u1, . . . , u6 . To answer the question we just posed, suppose that Lagrange’s prescription (8.7) has been used to prescribe the unknown constraint force and constraint moment:

398

Lagrange’s Equations of Motion for a Single Rigid Body

Fc Mc

= µ fX¯ = µ

3 ½ ∂´ i ¯ a acting at X, i

= µ hX¯ = µ

i=1 3 ½

∂q

i=1

∂´ ∂ qi +3

gi .

It is interesting to expand the contributions to ±K of the constraint forces and moments for this case. First, we perform the calculations for the force terms: Fc

∂v ¯ = ∂u ˙A

·

3 ½

µ

i=1

= µ

∂ ´ i ∂ v¯ a · A ∂ qi ∂u ˙

3 3 ½ ½ ∂´ i a i=1 k=1

= µ

∂ qi

3 ½ ∂ ´ ∂ q˙ i i=1

∂ q i ∂ u˙ A

·

3 3 ½ k k ½ ∂´ ∂q ∂q ˙ ˙ k a = µ δi k ∂u ˙A ∂ qi ∂ u ˙A i=1 k =1

.

Second, we consider the contribution from M c: Mc ·

∂ω = ∂u ˙A

3 ½

µ

i =1

= µ

∂´ ∂ω gi · A i +3 ∂q ∂u ˙

3 ½ 3 ½ ∂´ i =1 k =1

= µ

gi · ∂ qi +3

3 ½ ˙ i+3 ∂´ ∂q i =1

∂ qi +3 ∂ u˙ A

3 ½ 3 k +3 k+3 ½ ∂q ˙ ∂ ´ ∂ q˙ k g = µ δi k A i +3 ∂u ˙ ∂q ∂u ˙A i=1 k=1

.

Finally, we combine both sets of calculations: Fc ·

3 3 ½ ½ ∂v ¯ ∂ω ∂ ´ ∂ q˙ i ∂´ ∂ q ˙ i+3 + Mc · = µ +µ ∂u ˙A ∂u ˙A ∂ q i ∂ u˙ A ∂ q i+3 ∂ u ˙A i=1 i=1 3 ± i i +3 ² ½ ∂´ ∂ q ∂´ ∂q + = µ ∂ qi ∂ uA ∂ qi+3 ∂ u A i=1 = µ

ˆ ∂´ . ∂ uA

(10.28)

Notice that we used (10.26) in the penultimate step and also expressed ´ as a function of u1 , . . . , u6 and t: ³

´ =´

q1 , . . . , q6 , t

´

ˆ = ´

³

´

u 1, . . . , u 6, t .

To avoid confusion where any may possibly arise, we ornament ´ with a caret (hat) when it is expressed as a function of u1 , . . . , u6 and t. We are now in a position to draw some important conclusions. For a constraint ´ =

0,

where

³

´ =´

q 1, . . . , q 6, t

´

ˆ = ´

³

´

u 1, . . . , u 6, t ,

10.6

Lagrange’s Equations of Motion: A Second Form

399

Lagrange’s prescription yields Fc ·

¯ ∂v + ∂u ˙A

Mc ·

ˆ ∂ω ∂´ = µ . ∂u ˙A ∂ uA

(10.29)

Thus, in answer to our question, the presence of the constraint ´ = 0 and the constraint forces and moments needed to enforce the constraint using Lagrange’s prescription ˆ introduces a term µ ∂∂u´A on the right-hand side of Lagrange’s equations. This result is (reassuringly) similar to prior results for the cases of a single particle and a system of particles.

Consequences of Lagrange’s Prescription and Integrable Constraints

Let us now explore some of the consequences of (10.29). Suppose we can express the integrable constraint as u6 − f (t) = 0. In addition, the constraint force Fc and constraint moment Mc are prescribed using Lagrange’s prescription (8.7). Then, Fc

·

∂v ¯ ∂u ˙A

+

Mc

·

∂ω ∂ u˙ A

6

= µδA .

As a result, the constraint force and moment will only contribute to the Lagrange’s equation associated with u6 . The familiar decoupling we found with a single particle and a system of particles thus holds for the rigid body!

Potential Energy and Conservative Forces and Moments

Suppose that a potential energy is associated with the rigid body: ³

U = U q 1, . . . , q 6

´

ˆ = U

³

´

u 1, . . . , u 6 .

Recalling (8.20), the conservative forces and moments associated with this potential energy are Fcon

=

3 ½ ∂U i ¯ − a acting at X, i=1

Mcon

∂ qi

= −uQ = −

3 ½ ∂U i =1

∂ qi +3

gi .

Now suppose that we have chosen a new set of coordinates. Then, by setting ´ and µ = 1 in (10.28), we can quickly find that F con ·

ˆ ∂ω ∂U ∂v ¯ + Mcon · = − . A A ∂u ˙ ∂u ˙ ∂ uA

= −U

(10.30)

In summary, the conservative forces and moments appear in a familiar form on the right-hand side of Lagrange’s equations of motion.

400

Lagrange’s Equations of Motion for a Single Rigid Body

As a further demonstration of (10.30), consider a spring force acting at a point Xs (cf. Figure 8.8). As discussed previously in Section 8.6, the spring force and its potential energy function have familiar expressions: Fs

= −K µ

xs − xA acting at X s , ´ xs − x A ´

Us =

K 2 µ , 2

where µ = ´xs − xA ´ − ²0 is the extension of the spring. The contribution of this force to the right-hand side of Lagrange’s equations of motion can be represented in a pair of equivalent fashions using (10.25): Fs ·

∂ v¯ ∂ω + (( xs − x ¯ ) × F s) · = A ∂u ˙ ∂u ˙A

Fs ·

∂ vs . ∂u ˙A

With the help of (10.27), we manipulate the expression for the contribution of Fs : Fs ·

∂ vK ∂ Us ∂ v s · = − A ∂u ˙ ∂ xs ∂u ˙A = −

∂ Us

·

∂ xs ∂ Us . = − ∂ uA

∂ xs ∂ uA

The remarkably simple final expression for the contribution of the spring force is in agreement with (10.30), as expected. Mechanical Power and Energy Conservation

If we have a single integrable constraint on the motion of the rigid body and a potential energy U, then the combined mechanical power P of these forces and moments is

P = F c±· v¯ + M²c · ω − U˙ = μ

−

∂´ ∂t

˙ − U,

(10.31)

where we have assumed that the constraint force and constraint moment are prescribed using Lagrange’s prescription (8.7). Consequently, if the integrable constraint is time independent, ∂∂´t = 0, and the only applied forces and moments are conservative then, with the help of (10.31), we can surmise that T˙

=

˙ P = −U.

As a result, the total energy E = T + U is conserved when the integrable constraint is time independent and all the applied forces and moments are conservative. To arrive at (10.31), a straightforward set of manipulations is needed:

P = (F con + Fc ) · v¯ + (Mcon + Mc ) · ω =

µ ¶ 3 3 ½ ½ ∂´ i ∂´ μ a · v¯ + gi · ω ∂ qi ∂ qi +3 i=1

i =1

¶ µ 3 3 ½ ½ ∂U ∂U i i a · v¯ + g ·ω − i =1

∂ qi

i=1

∂ q i+3

10.6

=

µ 3 ½ ∂´ i q˙ μ

= μ

∂ qi

i=1 µ 6 ½ A=1

˙ − ´

The final result, P 10.6.1

¶

∂´ A

∂ qA

± = μ

+

q˙

∂´

Lagrange’s Equations of Motion: A Second Form

¶

3 ½ ∂´ ∂ q i+3

i=1 −

q˙

i+3

µ ¶ 3 3 ½ ½ ∂U i ∂ U i+3 − q˙ + q˙ i i+3 i=1

µ ¶ 6 ½ ∂U B q˙ B=1

²

401

∂q

i=1

∂q

∂ qB

˙ − U.

∂t

(

˙ − ∂´ = μ ´

)

∂t

˙ − U,

is what was used in writing (10.31).

Summary

Suppose we have a rigid body for which all of the applied forces and moments are conservative: ∂U ¯ acting at X, M = Mc − u Q . F = Fc − ∂x ¯ In addition, suppose that there are two constraints on the motion of the rigid body: ´ =

0,

2πX¯ =

0,

where ´ =

u6 − f (t),

6 ˙ − f˙, 1π X ¯ = u

¯ + h X¯ · 2 πX¯ = fX¯ · v

ω + eX¯ .

The first of these constraints is integrable, whereas the second is nonintegrable. We assume that the constraint forces and moments associated with the constraints are prescribed using Lagrange’s prescription (8.7): Fc

= μ1

3 ½ ∂´ i i=1

a ∂ qi

+ μ2 fX¯

¯ acting at X,

Mc

= μ1

3 ½ ∂´ i=1

∂ q i+3

gi + μ2 hX¯ . (10.32)

Turning to Lagrange’s equations of motion (10.22), we find that the equations would read ± ² d ∂T ∂T ∂U − = − + QA. (10.33) A A dt ∂ u˙ ∂u ∂ uA As the only nonconservative forces acting on the body are due to the constraint forces and constraint moments, QA are simply QA

=

Fc ·

∂v ¯ ∂u ˙A

+

Mc

·

∂ω ∂u ˙A

(A =

1, . . . , 6) .

Using a Lagrangian L = T − U and the prescriptions (10.32) for Fc and Mc , we find that (10.33) can be expressed in the following form: d dt

±

∂L ∂ u˙ B

²

d dt

−

±

∂L = μ2 ∂ uB

∂L ∂u ˙6

²

±

fX¯

·

∂v ¯ ∂ω + hX ¯ · B ∂ u˙ ∂u ˙B

∂L − = μ 1 + μ2 ∂ u6

±

²

∂ v¯ fX¯ · 6 + ∂u ˙

(B = ∂ω hX¯ · 6 ∂u ˙

1, . . . , 5) , ²

.

402

Lagrange’s Equations of Motion for a Single Rigid Body

Notice that the right-hand sides of these forms of Lagrange’s equations are similar to those for a system of particles with the added complication of the moment terms.

10.7

Lagrange’s Equations of Motion: Approach II

We can now examine Approach II applied to Lagrange’s equations for a rigid body whose motion is constrained. Specifically, we parallel the discussion of Section 10.6.1 and assume that the body is subject to one integrable constraint and one nonintegrable constraint. Our discussion is easily generalized to multiple integrable and nonintegrable constraints. The results we discuss were first established by Casey [38]. The resulting form of Lagrange’s equations is the one most commonly used in engineering and physics. Indeed, we used these equations earlier in Section 10.2 to establish the equations of motion for a Lagrange top. ¿ À First, we assume that the six coordinates u1 , . . . , u6 are chosen such that the integrable constraint ´ = 0 can be written simply as ´ =

u6 − f (t).

Imposing this constraint, we can calculate the constrained kinetic energy T˜ and con˜ of the system. The former will be a function of u1 , . . . , u5 , strained potential energy U 1 5 ˜ will be a function of u1 , . . . , u5 and t. The nonintegrable u˙ , . . . , u˙ , and t, whereas U constraint 2π X¯ = 0 can be written in terms of the new coordinates and their velocities: 2 πX¯ =

5 ½

pB u˙ B + e.

B=1

¿

À

Here, p1 , . . . , p5 and e are functions of t and the coordinates u1, . . . , u5 . We will use Lagrange’s prescription to specify the associated constraint forces and constraint moments (see (10.32)). By following the same arguments used to establish Lagrange’s equations for a single particle by use of Approach II, we find that the equations governing the motion of the rigid body are 5 ½

d dt

µ

B =1 ∂ T˜ ∂ u˙ B

pB u˙ B + e = 0, ¶ −

˜ ∂T = ∂ uB

F·

= −

∂v ¯ + ∂u ˙B

M·

∂ω ∂u ˙B

˜ ∂v ¯ ∂U + F ancon · + ∂ uB ∂u ˙B

Mancon ·

∂ω ∂u ˙B

+ μ2 pB .

(10.34)

Here, B = 1, . . . , 5 and the constraint forces and moments associated with the integrable constraint are absent. In these equations, the forces and moments acting on the body have been decomposed:

10.8

F

=

Fancon

M

=

Mancon

+

403

Rolling Disks and Sliding Disks

Fc + Fcon,

+

Mc

+

Mcon .

For instance, Fancon are the applied nonconservative forces. The Magnus force FM = mBω × v, ¯ that is used in studies of the flight path of a baseball, is a good example of such a force. Suppose that the only constraint on the rigid body is integrable (i.e., the constraint π 2 X¯ = 0 is ignored in (10.34)), a single coordinate is used to express the integrable constraint, and the constraint forces and moments are prescribed using Lagrange’s prescription. In this instance, (10.33) shows that Lagrange’s equations of motion decouple into two familiar sets. The firstÀ set gives differential equations of motion for the gener¿ alized coordinates u1, . . . , u5 and the second set provides an equation for μ 1. If one is uninterested in determining the constraint force and constraint moment acting on the rigid body, then Approach II can be used and (10.34) provides the equations of motion for the generalized coordinates. The case of a disk sliding on a smooth surface is one of the classic examples of a system with a single integrable constraint that can be used to demonstrate these comments.

10.8

Rolling Disks and Sliding Disks

As examples of constrained rigid bodies, we consider a rigid circular disk of mass m and radius R that either rolls without slipping on a rough horizontal plane or slides on a smooth horizontal plane (cf. Figure 10.6). It shall be assumed that the inertia tensor J0 has the representation J0 = λ ( E1 ⊗ E1 + E2 ⊗ E2) + λ3E 3 ⊗ E3 .

(10.35)

In other words, we are assuming that E3 is the axis of symmetry of the disk in its 2 reference configuration. The principal moments of inertia are λ 3 = 2λ and λ = mR 4 , but we do not use these substitutions in order to enable an easier tracking of the algebraic manipulations.

E3 g

E2

O

E1

ψ

e1

e2

= e1

¯

X

XP

e2

φ

e1 e1

Fc = N = µ3 E 3 Figure 10.6

A circular disk moving with one point in contact with a smooth horizontal plane.

404

Lagrange’s Equations of Motion for a Single Rigid Body

Preliminaries

The location of the center of mass of the disk can be parameterized by use of a set of Cartesian coordinates: x¯ =

3 ½

xi Ei .

i=1

The rotation tensor of the disk is described by a 3–1–3 set of Euler angles: ω = ψ˙ E 3 + θ˙ e±1 + φ˙ e3 .

For this set of Euler angles we recall, from (6.38) in Section 6.8.2, that the Euler basis vectors gi have the representations ⎤

⎡

⎡

⎤⎡

⎤

sin(φ) sin( θ ) cos( φ) sin( θ ) cos(θ ) e1 g1 = E 3 ⎦ ⎣ e2 ⎦ . ⎣ g = e± ⎦ = ⎣ cos( φ ) − sin( φ ) 0 2 1 g3 = e3 0 0 1 e3 In addition, the Euler angles are subject to the restrictions ψ ∈ [0, 2π ), θ ∈ (0, π ), and φ ∈ [0, 2π ). The angle θ represents the inclination angle of the disk. When this angle is π2 , the disk is vertical. In contrast, when θ = 0 or π , the disk is horizontal. In either of these situations, an entire surface of the disk is in contact with the ground and the equations of motion that will be presented will not be valid.

Constraints

As discussed previously in Exercise 8.8, the motion of the rolling disk is subject to three constraints because the velocity vector of the point of contact XP is zero: vP

0,

=

where vP

= v ¯ +

ω × π P,

πP

= −Re2 = −R sin(φ )e1 − ±±

R cos(φ )e2.

Earlier, we showed how these equations lead to the following three constraints (see (8.22)): 1 πX¯ =

0,

2 πX ¯ =

0,

3πX¯ =

0.

We also found that the third of these constraints was integrable: ´ = x3 − R sin(θ ) =

0.

(10.36)

The remaining constraints were nonintegrable.12 It should be clear that we express all three constraints on the rolling disk in the form fP · vP = 0, and this facilitates the use of Lagrange’s prescription for Fc and Mc . 12 This was established by use of Frobenius’ theorem in Exercise 8.9 at the end of Chapter 8.

10.8

Rolling Disks and Sliding Disks

405

Coordinates and Energies

Motivated by the presence of the integrable constraint x3 set of coordinates: u1 = x1 , 4

u

= θ,

¿

u2

u

5

u3

x2,

=

6

= φ,

u

=

R sin(θ ), we now define a

=

= ψ,

x3 − R sin(θ ).

(10.37)

À

Given values of u1, . . . , u6 , we can uniquely invert relations (10.37) to determine xi and the Euler angles: x1

u 1,

=

x2 = u2 ,

ψ =

u3 ,

¿

³

x3

=

u4 ,

θ =

´

u6 + R sin u4 , φ =

u 5.

(10.38)

À

It should be clear that u1, . . . , u5 are the generalized coordinates for both the rolling disk and the sliding disk. For future reference, we compute that ³

³

v¯ = u˙ 1 E1 + u˙ 2E2 + u˙ 6 + Ru˙ 4 cos u4 ω

3

= u ˙

´´

E3 ,

E3 + u˙ 4e±1 + u˙ 5e3 .

(10.39)

We also note that ω does not depend on u6 or its time derivative. Lengthy expressions for the components of vP in terms of the coordinates and derivatives can be inferred from (8.22) and are not reproduced here. We shall ¿need expressions for the derivatives of v¯ , vP, and ω with respect to the À 1 5 coordinates u , . . . , u . With the help of (10.39), these expressions for v¯ and ω can be obtained in a straightforward manner:

∂v ¯

∂ v¯ = ∂u ˙1 =

∂u ˙4

∂v ¯ = ∂u ˙2

E 1,

∂ v¯

R cos (θ ) E3,

∂v ¯ = ∂u ˙3

E2 ,

∂u ˙5

=

0,

∂v ¯

0,

∂u ˙6

=

E3

(10.40)

and ∂ω = ∂u ˙1

∂ω = ∂ u˙ 2

0,

∂ω ± = e1, ∂u ˙4

∂ω

=

∂u ˙5

∂ω = ∂u ˙3

0,

∂ω

e3 ,

∂ u˙ 6

E 3,

=

0.

(10.41)

The corresponding expressions for vP are found with the help of (10.40) and (10.41) ( ) and the identity vP = v¯ + ω × −Re±±2 : ∂ vP ∂u ˙1

∂ vP ∂u ˙4

=

=

E1 ,

∂ vP ∂u ˙2

=

R cos (θ ) E3 + e±1 × ∂ vP ∂u ˙5

=

e3 ×

(

∂ vP

E2 ,

−Re2

±±

)

(

∂ u˙ 3

−Re2

=

±±

)

Re±±1 ,

=

=

E3 ×

(

−Re2

±±

)

,

R cos (θ ) E3 − Re3 , ∂ vP ∂u ˙6

=

E3 .

(10.42)

406

Lagrange’s Equations of Motion for a Single Rigid Body

For a disk sliding on a smooth surface, F c = μ3 E3 and (10.42) will be used to quickly compute the contribution of this force to the right-hand side of Lagrange’s equations of motion (cf. (10.44)). Given the inertia tensor (10.35), representations (10.39), and the expressions for ω · ek in (10.10), an easy calculation shows that the (unconstrained) kinetic energy of the disk has the representation ±

T

³ ´2 m 2 x˙1 + x˙ 22 + u˙ 6 + Rθ˙ cos (θ ) 2

=

+

λ

2

³

˙ ψ

2

sin2 (θ ) + θ˙ 2

´

+

λ3

2

(

˙ +ψ ˙ φ

In addition, the potential energy of the disk is U

³

=

²

)2

cos(θ )

.

´

mg u6 + R sin(θ ) .

You should notice how the symmetry of the rigid body of interest simplifies the rotational kinetic energy.

Constraint Forces and Constraint Moments

The constraint forces and moments acting on the rolling disk can be prescribed by use of Lagrange’s prescription. With the assistance of the expressions for 1 πX¯ = 0, 2 πX¯ = 0, and 3π X¯ = 0, we find that Fc

= μ 3E 3 + μ1 E1 + μ2 E2

acting at XP ,

Mc

=

0.

(10.43)

This set of constraint forces and moments is equipollent to13 Fc Mc

= μ 3E 3 + μ1 E1 + μ2 E2

³

= μ3

2

−R cos(θ )g

³

+ μ1

³

+ μ2

´

¯ acting at X,

R cos(θ ) cos(ψ )g1 − R sin(θ ) sin(ψ )g2 + R cos(ψ )g3

´

´

R cos(θ ) sin( ψ )g1 + R sin(θ ) cos(ψ )g2 + R sin(ψ )g3 .

That is, Fc and Mc = π P × F c are equipollent to a force F c acting at the instantaneous point of contact XP . The E3 component of Fc is the normal force, whereas the remaining components constitute the static friction force. Thus, Lagrange’s prescription yields a physically reasonable set of constraint forces and constraint moments. For future reference, we compute the contribution of these constraint forces and constraint moments to the right-hand side of Lagrange’s equations (10.33). It is convenient to use the identity (10.25) and the representations (10.42): Q1

=

Fc ·

∂ vP

= μ1 ,

∂u ˙1 ∂ vP Q3 = F c · 3 = ∂u ˙

Q2

(

E3 ×

(

=

Fc

−Re2

±±

·

))

·

∂ vP ∂u ˙2

= μ2 ,

Fc ,

13 The expressions for the dual Euler basis we use here can be inferred from (6.39) in Section 6.8.2.

10.8

Q4 = Fc · Q5

=

∂ vP = ∂ u˙ 5

Fc ·

(

Re±±1

Rolling Disks and Sliding Disks

407

∂ vP = ( R cos (θ ) E3 − Re3 ) · F c , ∂ u˙ 4

)

·

Fc ,

Q6

=

Fc ·

∂ vP = ∂ u˙ 6

E 3 · Fc

= μ3 .

(10.44)

It is a useful exercise to check the above expressions for QA for the case of a disk sliding on a smooth surface. In this instance, F c = μ 3E3 . Consequently, Q1 = . . . = Q5 = 0 and Q6 = μ3 .14 Imposing the Integrable Constraint

If we impose the integrable constraint, then u6 potential and kinetic energies are

=

0 and the resulting constrained

˜ = mgR sin(θ ), U ´ ´ ³ ´ m³ 2 1³ λ 2 2 2 ˙ 2 sin2 (θ ) + ψ T˜ = x˙ 1 + x˙ 22 + λ + mR cos (θ ) θ˙ 2 2 2 +

(

λ3

2

)2

˙+ ψ ˙ cos( θ ) φ

.

It is important to notice that T˜ = T˜ (x1, x2, x˙ 1 , x˙ 2, θ , φ˙ , θ˙, ψ˙ ). In other words, u6 has been eliminated. ∼ The configuration manifolds for rolling disks and sliding disks are identical: = 2 3 E × RP . Both systems have five degrees of freedom and we are using Cartesian coordinates and Euler angles as generalized coordinates for . We can compute the expression for the mass matrix from T˜ and show that the determinant of the mass matrix ( ) is m2 λ3 λ λ + mR2 cos2 (θ ) sin2 (θ ).15 That is, the coordinate system for has a singularity when θ = 0 or θ = π . These singularities correspond to the disk lying flat on the horizontal surface. In this case, the assumption of single-point contact of the disk with the ground plane is violated and so our model also becomes invalid.

M

M

M

The Rolling Disk’s Equations of Motion

Lagrange’s equations of motion for the rolling disk can now easily be obtained. We first note that the resultant force F acting on the disk is composed of a constraint force and a conservative force: F = F c − mgE3 . The resultant moment M consists entirely of the constraint moment: M = M c . Using (10.33), and with the help of (10.40), (10.41), and (10.44), we find that d dt

±

∂T

²

∂x ˙1

d dt d dt

±

±

±

−

∂T

²

˙ ∂φ

²

−

²

∂T ∂ψ

∂T ∂φ

±

= ±1,

∂ x1 −

˙ ∂ψ

∂T

∂T

= ±3 ,

= ±5 ,

²

±

²

d ∂T ∂T − = ±2 , dt ∂ x˙ 2 ∂ x2 ± ² d ∂T ∂T = ± 4, − dt ∂ θ˙ ∂θ ± ² ± ² d ∂T ∂T − = ±6. 6 dt ∂ u˙ ∂ u6

14 For this calculation it is helpful to note that e · E = cos (θ ) and e±± · E 3 3 3 1 15 An explicit expression for M can be seen from (10.50).

=

0.

(10.45)

408

Lagrange’s Equations of Motion for a Single Rigid Body

In these equations, the generalized forces can be computed using several equivalent representations: ±1 =

Q1 −

±2 =

Q2 −

±3 =

Q3 −

±4 =

Q4 −

±5 =

Q5 −

±6 =

Q6 −

∂U ∂ x1 ∂U

∂ x2

∂U

∂ψ ∂U ∂θ ∂U ∂ψ ∂U ∂ x1

=

F · E1 + M · 0 = μ1 ,

=

F · E2 + M · 0 = μ2 ,

=

F · 0 + M · E3

=

F · R cos(θ )E3 + M · e±1

=

= −mgR cos( θ ) +

R cos(θ ) (μ1 cos(ψ ) + μ 2 sin(ψ )) ,

R sin(θ ) (μ2 cos( ψ ) − μ 1 sin(ψ )) ,

=

F · 0 + M · e3 = μ 1R cos( ψ ) + μ 2R sin(ψ ),

=

F · E3 + M · 0 = μ3 − mg.

Evaluating the partial derivatives of T in (10.45) and imposing the integrable constraint u6 = 0 we find, with some minor rearranging, that m¨x1

= μ 1,

mx¨ 2 = μ 2, ´ ( ) d ³ 2 ˙ + λ3 φ ˙ +ψ ˙ cos(θ ) cos(θ ) = ±3 , λ sin (θ )ψ dt ³ ´ 2 2 2 2 λ + mR cos (θ ) θ¨ − mR cos( θ ) sin(θ ) θ˙ +λ3

(

)

˙ +ψ ˙ cos( θ ) ψ ˙ sin(θ ) − λ ψ ˙ φ

d ( dt

2

(

sin(θ ) cos(θ ) = ±4 ,

˙ +ψ ˙ λ3 φ

m

))

cos(θ )

= ±5 ,

d2 (R sin(θ )) = μ 3 − mg. dt2

(10.46)

These equations are supplemented by the constraints16 x3 − R sin(θ )

=

0,

1π X ¯ =

0,

2 πX¯ =

0,

(10.47)

to form a closed system of equations for the nine unknowns xi , φ , θ , ψ , and μi . It is important to note that systems of equations (10.46) and (10.47) are not readily integrated numerically. In particular, they cannot immediately be written in the form y˙ = f(y) that is required for most numerical integrators, such as those based on a Runge– Kutta scheme. Alternative formulations of the equations of motion for the rolling disk can be found in the literature (see, e.g., [212] or [276, Section 14.4]). The procedure is identical to that used in Section 9.8 and Exercise 9.10 with a rolling sphere. In such formulations, ˙. F = ma¯ along with the constraint vP = 0 are used to solve F c as a function of ω and ω ˙ to derive differential equations for the The resulting solution is then used with M = H 16 As discussed in Exercise 8.8, the pair of constraints (10.47) 2,3 is nonintegrable.

10.8

Rolling Disks and Sliding Disks

409

attitude of the disk. Among other issues, the resulting equations can be used to establish stability results for rolling motions of the disk.

Conservations for the Rolling Disk

The easiest method to see that the rolling disk’s total energy E is conserved is to use the alternative form of the work–energy theorem. Two forces act on the rolling disk: the gravitational force −mgE3 and the constraint force Fc that acts at XP . Consequently, T˙ However, vP

=

=

0 and mgE 3 · v¯ =

Fc · vP − mgE3 · v. ¯

d dt (mgx3 );

consequently,

d (T + mgx 3) = 0. dt Because E = T + mgx 3 = T˜ + mgR sin (θ ), this implies that the total energy of the disk is conserved. The proof of energy conservation presented here applies to any rolling rigid body under a gravitational force. Surprisingly, there are two other conserved quantities associated with the rolling disk. These were discovered by Appell [8], Chaplygin [48], and Korteweg [150] in the late nineteenth century (see [29, 56, 143, 212]). Unfortunately, as with the Routh integral (9.45) for the tippe top, their physical interpretation is still an open question.

The Sliding Disk’s Equations of Motion

The equations governing the motion of the sliding disk on a smooth horizontal plane can be obtained from (10.46) and (10.47). Specifically, one sets μ1 = μ2 = 0 and ignores the constraints 1 πX¯ = 0 and 2πX¯ = 0. It is instructive, however, to use Approach II. For the sliding disk there are no nonintegrable constraints, and the constraint force and constraint moment are prescribed by use of Lagrange’s prescription, so that (10.34) simplifies to d dt

µ

˜ ∂T ∂u ˙A

¶

−

˜ ∂ T˜ ∂U ∂v ¯ ∂ω = − + Fc · + Mc · ∂ uA ∂ uA ∂u ˙A ∂u ˙A = −

˜ ∂U

∂ uA

+

FP ·

∂ vP ∂u ˙A

.

Here, A = 1, . . . , 5, Fc = μ3 E3, Mc = −Re±2 × Fc , and we have used the identity (10.25). It is straightforward to verify from (10.44) that FP ·

∂ vP ∂ vP A = μ3 E3 · = μ3 δ6 . A ∂u ˙ ∂ u˙ A

Thus, Lagrange’s equations of motion for the sliding disk can be expressed simply as d dt

µ

˜ ∂L ∂ u˙ A

¶

−

˜ ∂L = ∂ uA

0.

(10.48)

410

Lagrange’s Equations of Motion for a Single Rigid Body

The constrained Lagrangian is easily computed using the earlier expressions for T˜ and ˜ and is recorded here for the reader’s convenience: U L˜ =

´ λ ( )2 m³ 2 3 ˙ 2 2 2 2 ˙ cos(θ ) φ +ψ ˙x1 + x ˙ 2 + R θ˙ cos (θ ) + 2 2 ³

λ

+

2

˙ ψ

2

sin2(θ ) + θ˙2

´

−

mgR sin(θ ).

(10.49)

˜ = mgR sin (θ ) vanishes when the disk is flat on the Observe that the potential energy U horizontal surface. Using (10.49) in conjunction with (10.48), we find that the resulting governing equations can be expressed in the compact form

⎡

⎤

x¨ 1 ⎢ ⎥ ⎢ x¨ 2 ⎥

⎡

⎡

⎤

0 ⎢ ⎥ ⎢ 0 ⎥

⎢ ⎥ ⎥ ⎢ M ⎢ ψ¨ ⎥ + ⎢ α1 ⎥ = ⎢ ⎥ ⎥ ⎢ ⎣ θ¨ ⎦ ⎣ α2 ⎦ ¨ φ

where

⎡

M

=

m 0 0 m 0 0 0 0 0 0

⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎡

α1

⎣ α2 α3

⎤

⎢ ⎢ ⎢ ⎢ ⎢ ⎣

α3

0 0 0 −mgR cos (θ ) 0

⎤

⎥ ⎥ ⎥ ⎥, ⎥ ⎦

0 0

0 0 2 2 λ 3 cos (θ ) + λ sin (θ ) 0 2 0 λ + mR2 sin (θ ) λ3 cos(θ ) 0

(10.50)

0 0 λ 3 cos(θ ) 0 λ3

⎡

⎤ ⎥ ⎥ ⎥ ⎥, ⎥ ⎦

⎤

˙ cos (θ ) sin (θ ) − λ3 φ ˙ θ˙ sin (θ ) 2 (λ − λ 3) θ˙ ψ ( ) ) ⎦ = ⎣ λ 3 ψ˙ 2 + φ˙ ψ˙ − λ ψ˙ 2 − mR2 θ˙ 2 cos (θ ) sin (θ ) ⎦ . ˙ θ˙ sin (θ ) −λ 3 ψ

(

For the sliding disk, we note for completeness that (10.46) 6 can be used to show that Fc = mgE3 + mRd2 sin (θ ) /dt2 . We can also use (10.50) as a starting point to arrive at the equations of motion for the rolling disk. First, we need to supplement the constraints. Second, we need to append the constraint forces and moments associated with the integrable constraints to the right-hand side of (10.50). As with the Lagrange top, Lagrange’s equations of motion can readily be used to show the following four momentum conservations: px1

=

px2

=

∂T ∂ ˙x1

∂T

∂x ˙2

=

G · E1 ,

=

G · E2,

pψ pφ

=

=

∂T ˙ ∂ψ

∂T ˙ ∂φ

=

G·

=

G·

¯ ∂v ˙ ∂ψ

∂v ¯ ˙ ∂φ

+

H·

+

H·

∂ω ˙ ∂ψ

∂ω ˙ ∂φ

=

H · E3 ,

=

H · e3 .

That is, the center of mass of the disk moves at constant speed in a straight line, a result that was to be expected. The remaining conservations can be used to reduce (10.50) to a single two-parameter family of second-order ordinary differential equations for θ (t).17 17 We refer the interested reader to [212] for a discussion of the solutions to this second-order ordinary

differential equation.

10.9

411

Lagrange and Poisson Tops

The parameters in question are the conserved angular momenta H · E3 and H · e3. As with the solutions to (10.50), the solutions θ (t) of the resulting second-order differential equation will conserve the total energy of the disk.

10.9

Lagrange and Poisson Tops

There are two classic problems in rigid-body dynamics involving spinning tops. In the first, known as the Lagrange top, the axisymmetric rigid body is free to rotate about a fixed point XO under the action of a gravitational force (see Figure 10.1). As mentioned previously, this problem was discussed by Lagrange as an illustration of his equations of motion in 1789.18 A few years later, Poisson [231] considered the same problem except now the apex of the top was free to move on a horizontal surface (see Figure 10.7). Poisson’s interest in this problem arose in part because a spinning top can in principle be used to determine the vertical on a ship at sea, and hence was considered to be potentially useful in navigation. In this section, we do not pursue the complete exposition of the equations of motion for the two tops; rather, we focus on choices of coordinates. Our work here will validate why Approach II was used to establish the equations of motion for the Lagrange top in Section 10.2.

e3 e1

¯

X

κt

E3

g

E3

XO

E1

¯

e2

E2

XP

X

κ0

E2

XP

E1 Figure 10.7 The Poisson top: an axisymmetric rigid body that moves with one point XP in contact with a horizontal surface. The image on the left-hand side is a portrait of Siméon Denis Poisson (1781–1840).

18 See Section IX.34 of Part II of Lagrange’s Mécanique Analytique [159].

412

Lagrange’s Equations of Motion for a Single Rigid Body

The Lagrange Top: Coordinates, Constraints, and Energies

For the Lagrange top, we assume that the position vector of its center of mass relative to the point XO is x¯ = xO + ² 1e1 + ²2 e2 + ²3e3 . Here, ²k are constants. Unlike the top considered previously, here we allow the top to be asymmetric. Further, we assume that the rotation tensor of the top is parameterized by a set of 3–1–3 Euler angles.19 That is, the angular velocity vector of the top is ω = ˙ E3 + θ˙ e± + φ ˙ e3. You should notice that the singularities of these Euler angles occur ψ 1 when E3 = ³e3, that is, when θ = 0 or π and the top is vertical. Motivated by the presence of three integrable constraints xO = 0, we now define a set of coordinates: u

1

2

= ψ,

u

= θ,

u

3

= φ,

u

4

=

µ 3 ½

x1 −

¶ ²k ek

·

E1,

k=1

u 5 = x2 −

µ 3 ½

¶ ²k ek

u6

E2,

·

=

x3 −

µ 3 ½

k=1

¶ ²k ek

·

E3.

(10.51)

k=1

In these equations, ek are functions of the Euler angles, but the lengthy expressions are ¿ À not recorded here. Given values of u1 , . . . , u6 , we can uniquely invert relations (10.51) to determine xi and the Euler angles: µ

x1

=

u4 +

3 ½

¶ ²k ek

·

E1,

x2

=

u5 +

µ 3 ½

k=1

x3

=

6

u

+

µ 3 ½

¶ ²k ek

·

E2,

k=1

¶ ²k ek

·

E3 ,

ψ =

u1 ,

θ =

u 2,

φ =

u 3.

(10.52)

k=1

Using (10.52), we can easily compute that vO

= u ˙

4 4

E1 + u˙ 5 E2 + u˙ 6E3 , 5

6

v¯ = u˙ E1 + u˙ E2 + u˙ E3 + ω ×

3 ½

²k ek ,

k=1

ω = u˙ 1E3 + u˙ 2 e±1 + u˙ 3 e3.

(10.53)

You should notice that v¯ will depend on the Euler angles and their time derivatives. This is entirely due to our choice of coordinates. In place of x¯ and the three Euler angles, we are using the Euler angles and the three components of xO as our coordinates. 19 This set of Euler angles has been used extensively in the present chapter. The reader is referred to Section

6.8.2 for additional background.

10.9

413

Lagrange and Poisson Tops

For future reference, we shall need several vectors. To calculate these vectors it is ∂ω ± ∂ω ω = E3 , ˙ = e1, and = e3. First, we calculate convenient to recall that ∂∂ ψ ˙ ∂θ ∂ φ˙ ∂ vO = ∂u ˙1

∂ vO = ∂u ˙4

∂ vO = ∂u ˙2

0,

∂ vO = ∂ u˙ 5

E1,

∂ vO = ∂u ˙3

0,

0,

∂ vO = ∂ u˙ 6

E2 ,

E3

(10.54)

and ∂ω = ∂u ˙1

∂ω ± = e1, ∂u ˙2

E3 ,

∂ω = ∂u ˙4

∂ω = ∂u ˙5

0,

∂ω = e3 , ∂u ˙3

∂ω = ∂u ˙6

0,

0.

For completeness, we also compute the derivatives for v¯ : ¯ ∂v = ∂u ˙1

E3 ×

3 ½

¯ ∂v ± = e1 × ∂ u˙ 2

² k ek ,

k=1

∂v ¯ = ∂u ˙4

∂v ¯ = ∂ u˙ 5

E1 ,

3 ½

3 ½ ¯ ∂v = e3 × ²k ek , ∂u ˙3 k=1

²k ek ,

k =1

∂ v¯ = ∂u ˙6

E2,

E3.

The vectors (10.54) can be used to compute the right-hand sides of Lagrange’s equations of motion. We assume that the top has an axis of symmetry, and hence its inertia tensor J0 has the representation J0

= λ aE3 ⊗

E3 + λt (I − E 3 ⊗ E3 ) .

With the help of (10.53), we find that the kinetic energy of the top has the representation µ

T

=

± ² 3 ³ ´ ½ m ³ 4 ´2 ³ 5 ´2 ³ 6 ´2 4 5 6 u˙ + u ˙ + u ˙ +m u ˙ E1 + u ˙ E2 + u ˙ E3 · ω × ²k ek 2 µ

+

2 λt + m²2 2 + m²3

¶

2

− m²1 ²2 ω1ω2 −

µ

2 ω1 +

2 λt + m²2 1 + m²3

2

¶

µ

2 ω2 +

k =1

2 λ a + m²2 1 + m²2

2

¶

¶

2 ω3

m²2 ²3 ω2ω3 − m²1 ²3 ω1ω3 .

For convenience, we have not substituted for ωi = ω · ei in terms of the Euler angles and their derivatives. The necessary expressions for ωi are given in (10.10) in Section 10.2. The potential energy of the top is µ

U

=

mg u

µ 6

3 ½

+

¶ ²k ek

¶ · E3

.

k =1

This energy is due entirely to the gravitational force. There are three constraints on the motion of Lagrange’s top: u4 = 0, u5 = 0, and u6 = 0. That is, xO = 0. These constraints are integrable and imply that the generalized coordinates for the Lagrange top are the three Euler angles. Thus, the configuration

414

Lagrange’s Equations of Motion for a Single Rigid Body

M

3 ∼ manifold = RP . A straightforward calculation with the three constraints vO · Ek 0, where X O is the fixed point of the top, shows that

Fc

3 ½ =

μk Ek

acting at X O ,

Mc

=

=

0.

k =1

That is, the constraint force is a reaction force at XO . Thus, Lagrange’s prescription gives a physically reasonable prescription for F c , as expected. In addition, with the help of (10.54), we find the following crucial results: Fc · Fc

·

∂ vO = μ 1, ∂u ˙4

∂ vO = ∂u ˙i

Fc ·

0 (i = 1, 2, 3 ) ,

∂ vO = μ2 , ∂ u˙ 4

Fc ·

∂ vO = μ3 . ∂ u˙ 4

These results justify the use of (10.6) to compute Lagrange’s equations of motion (10.6) for the Lagrange top in Section 10.2. The constrained kinetic energy T˜ and the con˜ are readily determined from T and U by setting u4 , u5 , and strained potential energy U 6 u , along with their derivatives, to 0. It is not too difficult to see that 2T˜ = HO · ω and the angular momentum HO can easily be found by use of the parallel axis theorem. We have kept ²k distinct and nonzero to help elucidate some of our kinematical developments. If we now impose the standard assumptions that ²1 = 0 = ² 2, then expressions for T˜ and U˜ = mg²3 cos(θ ) can be found in Section 10.2.

The Poisson Top: Coordinates, Constraints, and Energies

Referring to Figure 10.7, for the Poisson top, the position vector of the center of mass relative to the point X P of contact of the top with the horizontal surface has the representation x¯ − xP

= ² 3e3.

For this problem, it is convenient to choose a different coordinate system from the one used for the Lagrange top. We again use a set of 3–1–3 Euler angles to describe the rotation of the top, but we use two of the Cartesian coordinates of the center of mass and the E 3 coordinate of the point of contact as the other three coordinates:20 u1 u4

=

x1 ,

u2

= ψ,

u5

=

x2,

= θ,

u3

= φ,

u 6 = x3 − ² 3e3 · E 3.

(10.55)

We invite the reader to compare (10.55) with (10.51) and to compute ∂∂u˙v¯A and ∂∂u˙ωA . The coordinate system (10.55) has evident similarities with the one we used earlier for the circular disk (see (10.37)). For future reference, we note that 20 We could also choose u4 = x · E and u5 = x · E instead of u4 = x¯ · E and u5 = x ¯ · E2 . However, P 1 P 2 1 our choice of u4 and u5 will make it easier to establish conservations of linear momentum G · E 1 and

G · E2 from Lagrange’s equations of motion.

10.9

∂ vP ± = −e 1 × ² 3 e 3 , ∂u ˙2

∂ vP = −E3 × ² 3e3, ∂u ˙1 ∂ vP = ∂u ˙4

Lagrange and Poisson Tops

∂ vP = ∂ u˙ 5

E 1,

E2 ,

∂ vP = ∂ u˙ 6

∂ vP = ∂u ˙3

415

0,

E3 .

(10.56)

In contrast to the Lagrange top, the Poisson top is subject to a single integrable constraint u6 = 0. A straightforward calculation with the constraint vP · E3 = 0, where XP is the point of contact of the top with the smooth horizontal surface, shows that the constraint force Fc

= μ1 E3

acting at X P ,

Mc

=

0.

That is, the constraint force is a normal force acting at X P and we have used Lagrange’s prescription. With the help of (10.56), we find that the normal force will only appear in one of Lagrange’s equations of motion: Fc ·

∂ vP = ∂u ˙A

0 (A

=

1, . . . , 5) ,

Fc

·

∂ vP = μ1 . ∂u ˙6

(10.57)

That is, the equations of motion for the Poisson top can be found using the following form of Lagrange’s equations of motion: d dt

µ

˜ ∂L ∂u ˙A

¶ −

˜ ∂L = ∂ u˙ A

0 (A = 1, . . . , 5) ,

˜ The development of expressions for T and U follows in a similar where L˜ = T˜ − U. manner to those discussed previously for the Lagrange top, and so their constrained expressions are merely summarized:

⎞2

⎛

T˜

=

=

λt

2

λt 2 λa 2 2 ω1 + ω2 + ω3 +

2

2

m⎜ ⎟ ⎝ (ω × ²3 e3 ) · E3 ⎠ ¸¹ º 2 · =v ¯ ·E 3 =−² 3 θ˙ sin (θ )

2 2 λt + m²3 sin (θ ) 2 λa ( λt 2 ˙ sin2 (θ ) + ˙ ψ θ˙ + ψ

2

´ m³ 2 x˙1 + x˙ 22 , + 2 ˜ U = mg² 3 cos (θ ) .

2

2

+

´ m³ 2 2 ˙x1 + x ˙2 2

cos (θ ) + φ˙

)2

You should observe from these expressions that the generalized coordinates for the Poisson top are ψ , θ , φ , x1, and x2. Thus, this system has a five-dimensional configuration 3 2 ∼ manifold = E × RP . If we consider the fact that the top might collide with the horizontal surface, then θ ∈ (−θ0 , θ0). That is, the angle of inclination of the top must be restricted to lie in a certain range. As a result, the configuration manifold will have a boundary. The possibility of the top colliding with the horizontal surface is ignored in our model for the top and remains to be incorporated.21

M

21 A related impact phenomenon can also occur with sliding disks and rolling disks (see [146, 212] for a

discussion).

416

Lagrange’s Equations of Motion for a Single Rigid Body

Comments on the Equations of Motion

We can find the equations of motion for the Lagrange and Poisson tops using Approach II and without calculating ∂∂u˙v¯A and ∂∂u˙ωA . There are three reasons for this. First, the constraint forces and moments are compatible with Lagrange’s prescription. Second, the coordinates have been chosen so that the integrable constraints are each described in terms of one coordinate. Third, the applied forces acting on these tops are conservative. Thus, as mentioned previously, the form of Lagrange’s equations of motion using the Lagrangian L˜ = T˜ − U˜ can be employed. The equations of motion for the Lagrange top were presented earlier (see (10.7)). In the interests of brevity, we refrain from writing out the five Lagrange’s equations of motion for the Poisson top. The first, third, fourth, and fifth equation of motion for the Poisson top demonstrate that four momenta are conserved: ⎡

˜ ∂L

˙ ∂ψ

⎢ ⎢ ⎢ ⎢ ⎢ ⎣

˜ ∂L

˙ ∂φ

˜ ∂L

∂ x˙ 1

= = =

˜ ∂L = ∂ x˙ 2

pψ

⎤

⎥ pφ ⎥ ⎥ ⎥ px1 ⎥ ⎦

⎡ ⎢ ⎢ ⎣

=

λt

px2

sin2 (θ ) + λa cos 2(θ ) λa cos( θ ) 0 0

λa cos( θ ) λa

0 0

⎤⎡

⎤

˙ 0 0 ψ ⎥ ⎢ 0 0 ⎥ ⎢ φ˙ ⎥ ⎥. m 0 ⎦ ⎣ x˙ 1 ⎦ 0 m x˙ 2

The momenta correspond to H· E3 , H · e3 , G · E1 , and G · E2 , respectively. We can use the conservations of angular momentum to reduce the Lagrange’s equation corresponding to u2 = θ to a single two-parameter family of ordinary differential equations for θ (t): ³

2 λ t + m²3

´ d 2θ

sin2 (θ )

dt2

=

(pψ

−

pφ cos(θ ))(pψ cos(θ ) − pφ ) λt

sin3(θ )

2 − m²3 cos( θ ) sin(θ )

±

dθ dt

²2

.

+

mg²3 sin(θ ) (10.58)

We invited the reader to compare these equations to (10.8) for a Lagrange top. For both the Lagrange and Poisson tops, it is easy to see that the total energy E is conserved. While the Poisson top has four momenta conservations (G · E1, G · E2 , H · e3, and H · E3 ), the Lagrange top has two angular momenta conservations: HO · E3 and HO · e3.

10.10

Closing Comments

Lagrange’s equations of motion have been illustrated for a variety of systems featuring a single rigid body. When there are no constraints on the rigid body or when the constraints are “ideal,” this formulation of the equations of motion provides a set of differential equations that can be integrated to determine the motion. The most general form of Lagrange’s equations of motion is d dt

±

∂T ∂u ˙A

²

−

∂T = ±A ∂ uA

(A =

1, . . . , 6) ,

10.11

417

Exercises

where ±A =

F·

∂ω ∂ v¯ +M· A ∂u ˙ ∂u ˙A

N ½

FK

=

K =1

·

∂ vK + ∂u ˙A

Mp ·

∂ω . ∂ u˙ A

The identity (10.25) is the basis for showing that the representations for ±A are equivalent. For the constraint forces and constraint moments associated with an integrable constraint ´ = 0, we showed (cf. (10.29)) that Fc ·

∂v ¯ ∂u ˙A

+

Mc ·

∂ω ∂u ˙A

= μ

ˆ ∂´

∂ uA

.

In addition, we also showed that the contributions to ±A by the conservative force Fcon acting at a material point X and a conservative moment Mcon associated with a potential energy U simplify dramatically to a familiar form (cf. (10.30)): F con ·

ˆ ∂ω ∂U ∂v + Mcon · = − . A A ∂u ˙ ∂u ˙ ∂ uA

Thus, Lagrange’s equations of motion for a rigid body share some of the remarkable aspects of the same equations for a single particle and a system of particles. These aspects include the use of a Lagrangian, the possibility of using Approach II, and the notion that a representative particle can be constructed which is free to move on and captures the dynamics of the rigid body. However, for nonintegrably constrained rigid bodies or for rigid bodies when dynamic friction is present, Lagrange’s equations of motion are not very attractive, and it is often best to simply examine the balances of linear momentum and angular momentum. As demonstrated in the examples discussed in Chapter 9, with some insight and patience, ˙ can lead to a tractable set of equations from which the motion F = mv˙¯ and M = H of the rigid body can be inferred. Alternative forms of Lagrange’s equations are available for nonholonomic systems. These equations, of which the most well known are the Gibbs–Appell equations and Boltzmann–Hamel equations, enable further elimination of the constraint forces and moments associated with the nonintegrable constraints. While such equations of motion are beyond the scope of this book, there is a variety of texts and surveys that cover them, for example, Baruh [20], Greenwood [106], Hamel [114], Karapetyan and Rumyantsev [144], Ne˘ımark and Fufaev [202], Papastavridis [224–226], and Udwadia and Kalaba [294], to name but a few. Most other forms of the equations of motion are derivable from Lagrange’s equations of motion (10.3) and so, by use of the developments in this book (and the help of (10.4)), they too can be related to the balances of linear and angular momenta.

M

10.11

Exercises

Exercise 10.1: Consider the mechanical system shown in Figure 10.8. It consists of a rigid body of mass m that is free to rotate about a fixed point XO . A vertical gravitational

418

Lagrange’s Equations of Motion for a Single Rigid Body

˙

ψ g

Fc

E3

XO

Mc

E2

θ

¯

X

E1

e1 Figure 10.8

A whirling axisymmetric rigid body.

force mgE1 acts on the body. The inertia tensor of the body relative to its center of mass X¯ is J0

= λ1 E1 ⊗ E1 + (λ −

m²20)(E2 ⊗ E2 + E3 ⊗ E3 ).

The position vector of the center of mass X¯ of the body relative to XO is ²0 e1. (a) To parameterize the rotation tensor of the body, we use a set of 1–3–1 Euler angles: g1 = E1 , g2 = e±3 , and g3 = e1. Show that ω = ψ˙ E1 + θ˙ e±3 + φ˙ e1 =

(

˙ +ψ ˙ φ

(

)

cos(θ ) e1 +

˙ + θ˙ cos( φ) + ψ

(

˙ θ˙ sin(φ) − ψ

)

)

sin(θ ) cos(φ) e2

sin(θ ) sin(φ) e3 . À

¿

(b) Because the material point XO is fixed, we define the coordinates u4 , . . . , u6 to be the coordinates of X O : x¯ = u4 E1 + u5 E2 + u6 E3 + ²0 e1, and we choose u1 show that ∂v ¯ ∂u ˙1

= ψ,

=

u2

= θ,

E1 × ²0e1 ,

and u3 ∂v ¯ ∂ u˙ 2

= φ.

=

With this choice of the coordinates, ∂v ¯

e±3 × ²0e1 ,

∂ v¯ = ∂u ˙4

E1,

∂ v¯ = ∂u ˙5

∂ω = ∂u ˙1

E1 ,

∂ω ± = e3, ∂u ˙2

E2 ,

∂ u˙ 3

∂v ¯ = ∂u ˙6

=

E3

0, (10.59)

and

∂ω = ∂u ˙4

0,

∂ω = ∂u ˙5

0,

∂ω = e1 , ∂u ˙3 ∂ω = ∂ u˙ 6

0.

These results will not be needed in the remainder of this problem.

(10.60)

10.11

Exercises

419

(c) Derive expressions for the unconstrained potential energy U and unconstrained kinetic energy T of the rigid body. These expressions should be functions of the ¿ À coordinates u1 , . . . , u6 and, where appropriate, their time derivatives. (d) In what follows, the motion of the rigid body is subject to four constraints: ˙ = φ

ω · g3 = 0,

v¯ − ω × (²0 e1) = 0.

Assuming the joint at XO is frictionless, give prescriptions for the constraint force F c and constraint moment Mc acting on the rigid body. (e) What are the generalized coordinates for this system, and why is the configuration manifold a sphere? (f) Prove that the total energy of the rigid body is conserved. (g) Show that the (constrained) kinetic and potential energies of the body are T˜

=

´ 1³ ˙ 2 cos2 (θ ) + λ(θ˙2 + ψ ˙ 2 sin2 (θ )) , λ1 ψ 2

˜ U

= −mg²0

cos( θ ).

(10.61)

In addition, using Approach II, write Lagrange’s equations of motion for the rigid body: d dt

µ

˜ ∂L

∂ θ˙

¶

−

˜ ∂L

=

∂θ

d dt

0,

µ

˜ ∂L

¶

˙ ∂ψ

−

˜ ∂L

∂ψ

=

0,

(10.62)

where L˜ is the Lagrangian. (h) Argue that the solutions ψ (t) and θ (t) to (10.62) determine the motion of the rigid body. You should also discuss why it was not necessary to calculate Lagrange’s equations of motion for all six coordinates. Exercise 10.2: Consider the simple model for an automobile shown in Figure 10.9. It consists of a single rigid body of mass m. The moment of inertia tensor of the rigid ∑ body is J = 3i=1 λi ei ⊗ ei . Here, the inertia tensor J and mass m includes the masses and inertias of the wheels, suspension, engine, and occupants. Interest is restricted to the case where the front wheels are sliding, while the rear wheels are rolling. In other words, the front wheels’ brakes are locked. To model the rolling of the rear wheels in this simple model, it is assumed that v Q · e2

=

0,

(10.63)

e1

e2 E2

ψ

¯

e1

X O

E1

XQ

Rigid body of massm Figure 10.9

A rigid-body model for an automobile moving on a horizontal plane.

E1

420

Lagrange’s Equations of Motion for a Single Rigid Body

where πQ = −²1e1 − ²2e2 is the position vector of X Q relative to the center of mass X¯ of the rigid body. Constraint (10.63) is often known as Chaplygin’s constraint (see [202, 222, 249]). (a) Assume that the rigid body is performing a fixed-axis rotation through an angle ψ about E3 , that the motion of its center of mass is planar, and that (10.63) holds. Using parameterizations of x¯ and Q of your choice, establish expressions for the four constraints on the motion of the rigid body. (b) Verify that one of the constraints in (a) is nonintegrable using Jacobi’s criterion (1.25). (c) Using Lagrange’s prescription, what are the constraint force F c and the constraint moment M c acting on the rigid body? (d) Show that the motion of the rigid body is governed by the equations mx¨ 1

= −μ4

sin(ψ ),

mx¨ 2 =

μ4 cos(ψ ),

¨ = −μ ² , λ 3ψ 4 1

x˙ 1 sin(ψ ) − x˙ 2 cos(ψ ) = −²1ψ˙ ,

(10.64)

where xi = x¯ · Ei . (e) Show that (10.64) allow the center of mass of the rigid body to move in a straight line without the body rotating, and prove that the total energy of the body is conserved. Exercise 10.3: As shown in Figure 10.10, a circular rod of mass m, length ² , and radius R slides on a smooth horizontal plane. We assume that the inertia tensor J0 has the representation J0 = λ ( E1 ⊗ E1 + E3 ⊗ E3) + λ2 E2 ⊗ E2 , λ=

mR2 4

+

m²2 , 12

λ2 =

mR2 . 2

The sole external applied force acting on the rod is gravitational: −mgE3.

E3

O

g

E1

E2 Fc = µ1 E 3 ¯

X

g3 Mc = µ2g 3 Figure 10.10

e2

A circular cylindrical rod moving on a horizontal plane.

10.11

421

Exercises

(a) Assuming that the rotation tensor of the rod is described by a 3–2–3 set of Euler angles, which orientations of the rod coincide with the singularities of this set of Euler angles? (b) For this problem, we use Cartesian coordinates to describe the position of the center of mass. That is, u1 = x1, u2 = x2 , u3 = x3, u4 = ψ , u5 = θ , and u6 = φ . For this set of coordinates, calculate ∂∂u˙v¯A and ∂∂u˙ωA . To write the equations of motion of this rigid body, under which circumstances can you avoid using these 12 vectors? (c) Show that the unconstrained kinetic energy of the rod is T

=

´ )2 m³ 2 λ( 2 2 ˙ sin (θ ) cos( φ) θ˙ sin(φ ) − ψ ˙x1 + x ˙2 + x ˙3 + 2 2 +

λ2

(

˙ θ˙ cos( φ) + ψ

2

sin (θ ) sin(φ)

)2

+

λ

2

(

˙ cos (θ ) φ˙ + ψ

)2

,

where x¯ = x1E 1 + x2E2 + x3 E3. (d) Show that the two constraints on the motion of the rigid body can be written in the form v¯ · E3

=

0,

ω · g3 = 0.

As discussed in Section 8.3, for the constrained rigid body, the range of the second Euler angle θ is extended to the entire unit circle: θe ∈ [0, 2π ) (cf. p. 303). Thus, u5 = θ is replaced with u5 = θe . (e) Using Lagrange’s prescription, what are the constraint force Fc and the constraint moment Mc that enforce the two constraints? With the assistance of a free-body diagram of the sliding rod, give physical interpretations of Fc and Mc . (f) Show that the six Lagrange’s equations of motion for the rod yield the following differential equations for the motion of the rod: m¨x1

=

0,

m¨x2

=

0,

λ2 θ¨e =

0,

¨ = λψ

0.

(10.65)

Lagrange’s equations of motion (10.11) are the easiest to use for this problem. (g) Show that the six Lagrange’s equations of motion for the rod also yield solutions for the constraint force and constraint moment: Fc Mc

=

mgE3 ,

˙ θ˙e sin (θe ) g 3 = −λ 2ψ ˙ θ˙e (cos (ψ ) E1 + = −λ 2 ψ

sin (ψ ) E2)

= λ 2ω ˙.

You might notice that Mc is perpendicular to the axis of symmetry of the cylinder and lies in the horizontal plane. In addition, this moment will vanish unless both θ˙e and ψ˙ are simultaneously nonzero. (h) Give physical interpretations of the solutions to (10.65) and show that they allow the cylinder to have motions where ω is not constant. Exercise 10.4: As shown in Figure 9.9, a sphere of mass m and radius R is free to move on a rough inclined plane. Here, we re-examine this problem, which was discussed

422

Lagrange’s Equations of Motion for a Single Rigid Body

earlier in Section 9.8, by using Lagrange’s equations of motion. The instantaneous point of contact of the sphere with the plane is denoted by XP . You will need to recall that vP

= v ¯ +

ω × πP ,

(10.66)

where v¯ is the velocity vector of the center of mass X¯ of the sphere, and πP is the position ¯ vector of XP relative to X. (a) Using a set of 3–1–3 Euler angles, show that the slip velocities of the sphere are vs 1

= x ˙1 −

R θ˙ sin(ψ ) + Rφ˙ sin(θ ) cos(ψ ),

vs 2

= x ˙2 +

R θ˙ cos( ψ ) + R φ˙ sin(θ ) sin(ψ ), ∑

where vP = vs 1 E1 + vs 2 E2 and v¯ = 3k=1 x˙ k Ek . (b) What is the constraint on a sliding sphere? Give prescriptions for the constraint force Fc and the constraint moment Mc that enforce this constraint. (c) Show that at least one of the three constraints on a rolling sphere is nonintegrable. Give prescriptions for the constraint force F c and the constraint moment Mc that enforce these constraints. (d) If the unconstrained kinetic energy of the sphere is T=

´ m³ ´ mR 2 ³ 2 ˙ + θ˙2 + φ˙ 2 + 2 ψ ˙φ ˙ cos(θ ) + ψ x˙ 21 + x˙ 22 + x˙ 23 , 5 2

what are Lagrange’s equations of motion for the sliding sphere? (e) What alterations need to be made to the Lagrange’s equations of motion for (d) so that they now apply to the rolling sphere? (f) Starting from (10.66) and using a balance of linear momentum and a balance of angular momentum, show that ±

mv˙ s 1

=

1+ ±

mv˙ s 2

=

1+

²

5 Ff · E1 + mg sin(β ), 2 ² 5 Ff · E2, 2

where Ff is the friction force acting on the sphere. If the sphere is rolling, then what is Ff ? Exercise 10.5: Consider a rigid body of mass m that is free to rotate about a fixed point XO . The position vector of the center of mass X¯ of the rigid body relative to XO is x¯ − xO = ²0 e3 , where ²0 is a constant. The inertia tensor of the body relative to its ∑ center of mass is J = 3k=1 λ k ek ⊗ ek . A vertical gravitational force −mgE3 acts on the rigid body. (a) Using a set of 3–1–3 Euler angles, establish an expression for the rotational kinetic energy T rot of the rigid body. (b) What are the three constraints on the motion of the rigid body? (c) Using Lagrange’s prescription, give prescriptions for the constraint force Fc and the constraint moment Mc acting on the rigid body.

10.11

Exercises

423

(d) In terms of the Euler angles and a set of Cartesian coordinates xk = x¯ · Ek , prescribe ¿ À a set of six coordinates u1, . . . , u6 such that the three integrable constraints on the motion of the rigid body can be expressed as ´i =

0

(i =

1, 2, 3) ,

where u4 ,

´1 =

´2 =

u 5,

´3 =

u6 .

Notice that the generalized coordinates for this rigid body are the Euler angles. (e) Calculate the following 12 vectors: ¯ ∂v

∂u ˙A

,

∂ω ∂u ˙A

(A =

(

1, . . . , 6) .

)

What is the potential energy U = U u1 , . . . , u6 of the rigid body? (f) What are Lagrange’s equations of motion for the generalized coordinates? In your solution, show that the contributions to the right-hand sides of these equations that are due to Fc and Mc sum to zero. (g) With the assistance of the balance laws, show that Fc

=

mgE3 + m² 0e¨ 3 acting at XO .

(h) Prove that the total energy E of the rigid body is conserved. Exercise 10.6: In this exercise, we reconsider the satellite problem discussed earlier in Section 10.5. Here, we employ a coordinate system that is popular in the satellite dynamics community (see, e.g., [221, 251, 271]). Referring to Figure 10.3, to parameterize x¯ , we use a spherical polar coordinate system {R, θ , β}, where β is the latitude and θ is the longitude. For this coordinate system, we define the following unit vectors: er = cos(θ )E 1 + sin(θ )E2 , eR = cos( β)er + sin(β )E3 , eβ = cos(β )E3 − sin(β )er , eθ = cos( θ )E2 − sin(θ )E 1. The rotation tensor Q of the satellite is a transformation from the fixed basis {Ei } to the corotational basis {ei }. To parameterize Q, it is standard in satellite dynamics studies to exploit the rotation θ about E3 inherent in the definition of the spherical polar coordinates. To do this, we use the following decomposition: Q = Q2 Q1, where the rotation tensor Q1 is Q1

=

cos(θ )(I − E3 ⊗ E3) + sin(θ )skwt (E3 ) + E3 ⊗ E3 .

The rotation tensor Q2 will be parameterized by using a set of 1–2–3 Euler angles. The first of the Euler angles corresponds to a counterclockwise rotation through an angle 1 ν about the vector er . Similarly, the second rotation corresponds to a counterclockwise ( ) ( ) rotation through an angle ν 2 about cos ν 1 eθ + sin ν 1 E3 , and the last rotation is about e3 through a counterclockwise angle ν 3 .

424

Lagrange’s Equations of Motion for a Single Rigid Body

(a) Starting from the representation x¯ = ReR , show that the velocity vector v¯ of the center of mass of the satellite has the representation ˙ R + R cos(β )θ˙ eθ v¯ = Re

R β˙ eβ .

+

∑

(b) Starting from the representation Q = 3i=1 ei ⊗ Ei , show that e1 = Q2 er , e2 = Q2 eθ , and e3 = Q2 E3 . (c) With the help of three relative angular velocity vectors, show that the angular velocity vector ω of the satellite has the representation ³

ω = θ˙ E3 + ν˙ 1 er + ν˙ 2 cos

³ ν

1

´

eθ

+

sin

³ ν

1

´

E3

´ + ν ˙

3

e3 .

(10.67)

(d) Show that the angular velocity vector of the satellite has the representation ω = ω1 e 1 + ω2 e 2 + ω3 e 3 ,

where ω1 = θ˙

³

+ν ˙

ω2 = θ˙

³

1

cos( ν 2) cos(ν 3 ),

1

+ν ˙

´

sin(ν 1) cos(ν 3 ) + cos(ν 1 ) sin(ν 2 ) sin(ν 3 )

−ν ˙

ω3 = θ˙

´

sin(ν 1) sin( ν 3) − cos(ν 1) sin( ν 2) cos(ν 3 )

+ν ˙

2

sin(ν 3)

2

cos(ν 3 )

cos( ν 2) sin( ν 3),

cos(ν 1 ) cos(ν 2) + ν˙ 1 sin(ν 2 ) + ν˙ 3 .

(e) Suppose the coordinates chosen for the satellite are u1 = R, u2 = θ , u3 = β , u4 = ν 1, u5 = ν 2 , and u6 = ν 3. With this choice, give expressions for the 12 vectors ∂v ¯ ∂ω , (A = 1, . . . , 6) . A ∂u ˙ ∂u ˙A (f) The potential energy associated with the gravitational force and moment can be ¿ À found in (10.17). With the choice of u1 , . . . , u6 , it can be shown that ∂∂Uθ = 0. Using this result, show that ∂∂ θT˙ is conserved. Prove that this conservation corresponds to conservation of HO · E3 . (g) Using Lagrange’s equations, establish the equations governing the motion of the satellite. Exercise 10.7: Consider a body that is free to move about one of its material points, XO , which is fixed (cf. Figure 9.8). This system was discussed previously in Exercises 9.14 and 10.5. In the approach discussed in the present exercise, it is tacitly assumed that the constraints on the rigid body are integrable, that each of these constraints can be expressed in terms of a single coordinate, and that the constraint force Fc and constraint moment Mc that enforce these constraints are prescribed using Lagrange’s prescription. That is, Approach II is valid. However, we never explicitly state (and do not need to specify) the coordinates that are used to express the constraint.

10.11

425

Exercises

(a) Calculate the constrained Lagrangian L˜ and the equations of motion (see (10.13)) d dt

µ

˜ ∂L ∂ γ˙ i

¶

−

˜ ∂L = ∂γ i

0,

(10.68)

where γ 1 = ψ , γ 2 = θ , and γ 3 = φ are the 3–1–3 set of Euler angles used to parameterize Q. ˙ O · g = MO · g , where (b) Show that the equations of motion (10.68) are equivalent to H i i ¿ À g1 , g2, g3 are the Euler basis vectors. (c) Compare the equations of motion (10.68) with (9.46), which were discussed in Exercise 9.14. Exercise 10.8: Recall the definition of the force ±A given by (10.4): ±A =

F·

∂ω ∂v ¯ +M· . A ∂u ˙ ∂u ˙A

Now suppose we have a force P that acts at a point XP on a rigid body. If the position vector of XP is xP, then show that22 P·

∂ vP

∂u ˙A

=

P·

∂v ¯

+ ((x P − x ¯) × ∂u ˙A

P) ·

∂ω ∂u ˙A

.

How can this identity be used to simplify the calculation of ±A ? Exercise 10.9: Consider a prototypical vibrations demonstration shown in Figure 10.11. In this demonstration, a rigid circular disk of mass m and radius R is attached at a point XS on its outer rim by a linear spring to a fixed point O. The spring is assumed to have

E3 O

g

E2

XS

¯ X

−e 3

e2

A disk of mass m and radius R hanging from a fixed point O by a linear spring of stiffness K and unstretched length ²0.

Figure 10.11

22 This identity is used extensively in many treatments of analytical mechanics (see, e.g., [141, 142]).

426

Lagrange’s Equations of Motion for a Single Rigid Body

stiffness K and unstretched length ²0 , and the position of XS relative to the center of mass X¯ of the disk has the representation πS

=

Re3.

The inertia tensor of the disk relative to its center of mass X¯ has the representation J=

mR2 (2e1 ⊗ e1 + e2 ⊗ e2 + e3 ⊗ e3 ) . 4

A vertical gravitational force −mgE3 is assumed to act on the disk. To parameterize the motion of the disk, we use a set of 3–2–1 Euler angles and three Cartesian coordinates: u1

u2 = y = x¯ · E2 ,

x = x¯ · E 1,

=

4

u The configuration manifold

5

= ψ,

M

∼ =

u

= θ,

u3 u

6

=

z = x¯ · E3 ,

= φ.

E3 × RP 3 for this six degree-of-freedom system.

(a) Show that the kinetic energy of the disk has the following representation: T

=

´ m³ 2 2 2 ˙x + y ˙ + ˙z 2 ³ ´´ mR 2 ³ 2 2φ˙ − 4ψ˙ φ˙ sin (θ ) + θ˙ 2 + ψ˙ 2 1 + sin2 (θ ) . + 8

In addition, show that the potential energy of the disk has the representation U

=

mgz +

K 2 µ , 2

where the extension µ of the spring and the components of e3 are Á µ =

(x +

Re31)2 + (y + Re32)2 + (z + Re33 )2 − ²0,

e31 = e3 · E1

=

cos (φ) sin (θ ) cos (ψ ) + sin (φ) sin (ψ ) ,

e32 = e3 · E2

=

cos (φ) sin (θ ) sin (ψ ) − sin (φ) cos (ψ ) ,

e33

=

e3 · E3

=

cos (φ) cos (θ ) .

(b) Show that the equations of motion for the disk can be expressed in the following form: ⎡

⎤

⎡

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ M⎢ ¨ ⎥ ⎢ ψ ⎥ ⎢ ⎥ ⎣ θ¨ ⎦

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

x¨ y¨ z¨

¨ φ

=

−K ∂ x

∂µ

⎤

⎥ ⎥ ∂µ ⎥ −mg − K ⎥ ∂z ⎥ . ⎥ α4 ⎥ ⎦ α5 −K ∂ y

∂µ

α6

(10.69)

10.11

In these equations of motion, the mass matrix M is ⎡

M

=

m 0 0 m 0 0

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 0 ⎢ ⎢ ⎣ 0

0 0 m

0 0

0 0

0

0

0

mR2 4

0 0 0

(

)

1 + sin2(θ ) 0

−

mR2 2

0 0 0 0

mR2 4

sin(θ )

427

Exercises

⎤

0 0 0

⎥ ⎥ ⎥ ⎥ ⎥, mR 2 − 2 sin(θ ) ⎥ ⎥ ⎥ 0 ⎦ mR2 2

0

and α4 = α5 =

) ∂µ mR2 ( ˙ θ˙ cos (θ ) − ψ ˙ θ˙ cos (θ ) sin (θ ) − K φ , 2 ∂ψ

´ mR 2 ³ ∂µ ˙ θ˙ cos (θ ) + ψ ˙ 2 cos (θ ) sin (θ ) − K −2ψ , 4 ∂θ ) ∂µ mR2 ( ˙ θ˙ cos (θ ) − K ψ . α5 = 2 ∂φ

We have refrained from expanding ∂ µ/∂ uK in the interests of brevity. (c) Prove that the total energy E = T + U of the disk is conserved by the solutions to the differential equations (10.69). (d) Show that the determinant of the mass matrix is det (M) = 2m

3

±

mR2 4

²3 ³

´

1 − sin2 (θ ) . ¿

À

Verify that the mass matrix becomes singular when the Euler basis E3, e±2 , e1 fails to be a basis for E3. (e) How would the equations of motion change if uk = xS · Ek for k = 1, 2, 3 were chosen instead of uk = x¯ · Ek ? Exercise 10.10: An axisymmetric rigid body of mass m is suspended by a pin joint at a point XB . As shown in Figure 10.12, the pin joint is attached to a rod of length ²1 that

e1 g

¯

X

˙ = Ω

ψ

θ

XB

E2

O

E3 Figure 10.12

A whirling rigid body.

e1

428

Lagrange’s Equations of Motion for a Single Rigid Body

is being rotated about a vertical axis with an angular speed ¶ of the rigid body of mass m relative to its center of mass X¯ is J = λ 1e1 ⊗ e1 + (λ 2 =

λt −

= ¶(t).

The inertia tensor

md2) (e2 ⊗ e2 + e3 ⊗ e3) .

The position vectors of the center of mass X¯ and the point XB relative to a fixed point O are xB − xO

= ²1 e1 = ² 1 (cos (ψ ) E1 + ±

sin (ψ ) E2) ,

x¯ − xB

=

de1 .

A vertical gravitational force mgE3 acts on the body of mass m. We invite the reader to compare and contrast this problem with Exercise 10.1. To parameterize the rotation tensor of the rigid body of mass m, we use a set of 3–2–1 Euler angles. The three other coordinates of interest are the Cartesian coordinates of X B: u1

u2 = ψ ,

= θ,

xB

u3

= φ,

u4E1 + u5 E2 + u6 E3 .

=

(a) Compute the 12 vectors ∂∂ ˙uvAB and ∂∂u˙ωA , where A = 1, . . . , 6. (b) The motion of the rigid body of mass m is subject to five constraints: vB − ¶E 3 ×

(

)

²1 e1 = ±

0,

˙ = φ

0,

˙ −¶ = ψ

0.

Give prescriptions for the constraint force F c and the constraint moment Mc acting on the rigid body of mass m. (c) Modulo an additive function of time, show that the (constrained) Lagrangian of the body is L˜ =

λ1

2 +

¶

2

sin2 (θ ) +

λt

³ ¶

2

cos2 (θ ) + θ˙ 2

2 md²1 ¶2 cos (θ ) − mgd sin (θ ) .

´ +

m 2 2 ² ¶ 2 1

You should observe that it is possible to compute L˜ in a straightforward manner using the representations (10.70) for xB − xO and ω = θ˙ e2 + ¶E3 without ever having to specify u4, u5 , and u6. Additional observations include the fact that the system has 1 ∼ a single generalized coordinate θ that parameterizes = S in a singularity-free manner. (d) Compute the contributions of the constraint force and constraint moment to the righthand side of Lagrange’s equations of motion:

M

Fc ·

∂ vB ∂ω + Mc · A ∂u ˙ ∂u ˙A

(A =

1, . . . , 6) .

If your expressions in (a) and the prescriptions for Fc and Mc are correct, then the first of these summations should vanish. (e) Determine the second-order differential equation governing the generalized coordinate θ for the rigid body of mass m. (f) For this system, prove that the total energy of the rigid body is conserved if ¶ = 0. You should also notice that when ¶ = 0, the rigid body of mass m behaves as a simple pendulum.

Part IV

Systems of Particles and Rigid Bodies

11

The Dynamics of Systems of Particles and Rigid Bodies

11.1

Introduction

In this chapter, methods to determine the equations of motion of systems of rigid bodies are presented. The methods include the Newton–Euler balance laws and Lagrange’s equations of motion. Apart from problems in celestial and molecular dynamics, most systems involve rigid bodies that are either interconnected by joints or constrained to move on surfaces. Thus, a substantial part of this chapter is devoted to the topics of constraints and the forces and moments that enforce them in the dynamics of systems of rigid bodies. In particular, the constraints associated with interconnected rigid bodies, rolling rigid bodies, and sliding rigid bodies, along with prescriptions for the associated constraint forces and constraint moments, are presented. A complementary discussion of conservative forces and conservative moments is also given. As with a single particle, systems of particles, and a single rigid body, Lagrange’s prescription for constraint forces and constraint moments is emphasized and illuminated with a series of examples. We again recover the remarkable result for a system of rigid bodies that, if the constraints are integrable, each constraint is expressed in terms of a single constraint, and Lagrange’s prescription for the constraint forces and constraint moments is used, then Lagrange’s equations of motion decouple into two sets. The first set can be used to determine the equations of motion for the generalized coordinates and the second set provides expressions that can be used to determine the constraint forces and constraint moments. Examples of the resulting decoupling will be presented throughout this chapter. The presentations here are limited in scope and we do not have the opportunity to discuss many interesting systems featuring several rigid bodies such as the dualspin spacecraft, bicycles, gyrocompasses, and the Dynabee in detail. As discussed in [113], a dual-spin satellite has the ability to reorient itself in an environment where the resultant moment on the satellite is negligible. This ability has been used in communications satellites and was employed in the Galileo spacecraft. This spacecraft was launched in 1989 and some 6 years later began to orbit the planet Jupiter. These orbits were designed so that the spacecraft could capture data on some of the largest moons of Jupiter; Galileo’s mission was a remarkable success.1 The Dynabee (also

1 A history of the Galileo spacecraft’s mission was written by Meltzer [184].

432

The Dynamics of Systems of Particles and Rigid Bodies

known as a Rollerball or Powerball) was invented in the early 1970s by Archie Mishler [187]. The central feature of the device is a rotor that can be spun up to speeds in excess of 5000 rpm by a nontrivial rotation of an outer casing (housing). This novel gyroscopic device is discussed in further detail in Exercise 11.5 at the end of this chapter. After discussing some preliminaries, we immediately examine the application of Lagrange’s equations of motion to two classic systems: a planar double pendulum and a particle on a rotating circular hoop. These examples serve to illustrate how Lagrange’s equations of motion can be used with systems of particles and rigid bodies. We then examine constraints and prescriptions for constraint forces, constraint moments, conservative forces, and conservative moments. This treatment, supplemented by a discussion of Lagrange’s equations of motion, serves to justify our use of Lagrange’s equations with the aforementioned classic systems. We then examine the dynamics of a spherical robot, a cylinder rolling on a cart, and a gyroscope to further illuminate how balance laws or Lagrange’s equations can be used to establish equations of motion for mechanical systems.

11.2

Preliminaries

We are interested in examining the dynamics of a system of N rigid bodies, B1 , . . . , BN (cf. Figure 11.1). For the body BK , we denote its mass by mK , its center of mass by X¯ K , the position of its center of mass by x¯ K , its inertia tensor relative to X¯ K by JK , and its angular velocity by ωK . We also define a corotational basis and rotation tensor QK for each rigid body: {K e1 , K e2, K e3} ,

QK

= K e1 ⊗

E1 + K e2 ⊗ E 2 + K e3 ⊗ E3 ,

where K = 1, . . . , N. Using the aforementioned kinematical quantities, we can define the kinetic energy TK , angular momentum HK , and linear momentum GK in the usual manner: GK

=

mK x˙¯ K ,

HK

=

J K ωK ,

TK

=

1 1 GK · x˙¯ K + HK · ωK . 2 2

Setting K = 1, . . . , N, we can obtain expressions for the kinematic quantities associated with each of the N rigid bodies. For future reference, we can define the center of mass X¯ for a system of rigid bodies. The position vector of this point is defined by x¯ = ∑N

±

1

K =1 mK

N ²

³

mK x¯ K .

K=1

The linear momentum G of the system of rigid bodies is the sum of the individual linear momenta, and G can be related to the momentum of the center of mass: ±

G=

N ² K=1

³

mK x˙¯ =

N ² K =1

mK x˙¯ K .

11.3

A Planar Double Pendulum

433

X1

¯

X1

X2

x2

x1 x¯ 1

B1

¯

X2

x¯ 2

B2

O

The present configurations of a pair of rigid bodies B 1 and B2 showing the position vectors of their centers of mass and representative material points.

Figure 11.1

In many problems featuring systems of rigid bodies, it is necessary to calculate the angular momentum of the system relative to a point, say A: HA . This point is often the center of mass of the system or a fixed point. To compute HA , we use a standard identity (7.16) applied to each individual rigid body: HA

=

N ² (

)

HK + (x¯ K − xA ) × mK x˙¯ K .

K=1

Finally, the kinetic energy of the system of N rigid bodies is simply the sum of the kinetic energies of the individual bodies: T=

N ² K=1

´

1 GK · x˙¯ K 2

µ

+

1 H K · ωK . 2

The resulting expression for T can be tremendously complex, and often symbolic manipulation packages, such as M ATHEMATICA or M APLE, are used. For systems of particles and rigid bodies, we can simply duplicate the previous discussion and ignore the rotation and angular momentum of the particle about its center of mass. In other words, we consider a particle to be equivalent to the motion of the center of mass of a rigid body and ignore the rotation of the rigid body. In the interests of brevity, we leave these developments to the reader. Constraints on, and potential energies for, systems of rigid bodies will be discussed in further detail below. In addition, explicit examples where the application of the definitions of various kinematical quantities discussed above are illuminated will be presented throughout this chapter.

11.3

A Planar Double Pendulum

To motivate some of the developments in this chapter, we consider the classic problem of a planar double pendulum shown in Figure 11.2.2 In this system of rigid bodies, a 2 A version of this system where the masses of the rigid rods were ignored and mass particles were placed at

the ends of the rods was discussed in Exercise 5.6 (Chapter 5).

434

The Dynamics of Systems of Particles and Rigid Bodies

E2

O

θ1 g

m1

1

E1 m2

2

θ2

A planar double pendulum consisting of a pair of rods that are free to move on a vertical plane and are connected by a pin joint.

Figure 11.2

thin rod of length ±1 and mass m1 is free to rotate about a fixed point O with the help of a pin joint. At the end of the rod, a pin joint is used to attach a second thin rod of length ±2 and mass m2. The motion of both rods is planar and each of the motions can be defined using a single angle. We assume that Approach II can be used to establish the equations of motion for this system: d dt

±

³

˜ ∂L

−

∂ θ˙1

˜ ∂L

∂ θ1

=

d dt

0,

±

˜ ∂L

³

∂ θ˙2

−

˜ ∂L

∂ θ2

=

0,

˜ Later, in Section 11.11, a justification for this approach will be where L˜ = T˜ − U. discussed in detail. The system has two degrees of freedom, θ1 and θ2 are the generalized 2 ∼ coordinates, and = T . The constrained potential and kinetic energies for this system can be computed starting with the representations

M

α

e1

=

cos (θα ) E1 + sin (θα ) E2 , x¯ 1 x2

=

=

2x¯ 1 +

H1

=

±1

2 ±2

2

α

e2

=

cos (θα ) E2 − sin (θα ) E1

1 e1 ,

v¯ 1

=

2 e1 ,

v¯ 2

=

m1±21 θ˙1 E3 , 12

H2

±1

2

1, 2) ,

θ˙11 e2 ,

2v¯ 1 +

=

(α =

±2

2

θ˙22 e2 ,

m2± 22 θ˙2E3 . 12

Thus, T˜

=

1 2

˜ = − U

±± ´

m1 ±21 3

m1 g±1 2

³ 2 + m2 ±1

2 θ˙1 +

µ +

m2 ±22 2 θ˙ 3 2

m2 g±1 cos (θ1 ) −

³ +

m2 ±1 ±2θ˙1 θ˙2 cos (θ 2 − θ1 ) ,

m2g±2 cos (θ2 ) . 2

11.3

435

A Planar Double Pendulum

The equations of motion are obtained using the aforementioned form of Lagrange’s equations of motion: ¶

M

θ¨1 θ¨2

·

¶ +

m2 ±1±2 θ˙22 sin (θ1 − θ2) m2 ±1±2 θ˙12 sin (θ2 − θ1)

¹

¸

·

−

=

º

m1 g± 1 2 + m2g ±1 sin (θ1) m g± − 22 2 sin (θ 2)

»

, (11.1)

where the mass matrix M has the representation ¸

M

=

m1 ±21 + m2 ± 2 1 3

m2 ± 1 ± 2 2

m2 ± 1 ±2 2

cos (θ2 − θ1 ) m2 ±22 3

cos (θ2 − θ1 )

»

.

It can be shown that the solutions θα (t) to the equations of motion (11.1) conserve the ˜ of the system. The mass matrix M is positive definite and so the total energy T˜ + U 2 3 ∼ coordinate system used for = T has no singularities. We now consider a slight change to the system and assume that a pair of forces P A and PB act at points XA and XB on the respective rigid bodies. Referring to Figure 11.3,

M

xA

= ± A1 e1,

xB

=

2x¯ 1 + ±B2 e1 .

The equations of motion for this case can be obtained from the following pair of Lagrange’s equations of motion: d dt d dt

± ±

˜ ∂L

∂ θ˙1 ˜ ∂L

∂ θ˙2

³

−

³ −

˜ ∂L

∂ θ1 ˜ ∂L

∂ θ2

=

PA ·

=

PA ·

∂ θ˙1 ∂ vA ∂ θ˙2

+

PB ·

+

PB ·

∂ vB ∂ θ˙1 ∂ vB ∂ θ˙2

, .

E2

O

PA

θ1 g

∂ vA

E1

PB

XA

θ2 XB

Figure 11.3

A planar double pendulum showing a pair of applied forces PA and P B .

3 Here, we are invoking our developments pertaining to coordinate singularities and the mass matrix from

Sections 1.8, 4.8, and 10.3 to a system of particles and rigid bodies. In the interests of brevity, we refrain from constructing a representative particle for the system of particles and rigid bodies. Such a construction would yield explicit expressions for the covariant basis vectors from which the mass matrix M for the system can be constructed.

436

The Dynamics of Systems of Particles and Rigid Bodies

The partial derivatives on the right-hand side of these equations are ∂ vA ∂ vB ∂ θ˙1

∂ θ˙1

∂ vA

= ±A1 e2,

∂ vB

= ± 11 e2 ,

∂ θ˙2

∂ θ˙2

=

0,

= ±B2 e2.

That is, the left-hand side of the equations of motion (11.1) remains the same, but the right-hand side now includes contributions from PA and PB : ¹

¸

−

º

m 1 g±1 2 + m2 g±1 sin (θ1 ) + (±A PA + ±1P B) · 1 e2 m g± − 22 2 sin (θ 2) + (±BP B) · 2 e2

»

.

We will discuss shortly treatments of constraints and constraint forces and constraint moments that will go a long way to our eventual goal of justifying the two forms of Lagrange’s equations we have used to compute the equations of motion.

11.4

A Particle on a Rotating Circular Hoop

For our next example, we consider a particle of mass m that is free to move on a circular hoop of radius R0. A version of this classic problem was discussed earlier in Exercise 3.7 (Chapter 3). In our present discussion, which is based on [123], we assume that the hoop is free to rotate about a vertical axis and the inertia of the circular hoop is considered in the model. In addition to exploring Lagrange’s equations of motion for this system, we also show how including the inertia of the hoop changes the configuration manifold of the system from the unit two-sphere S2 to a torus T 2 and, thus, coordinate singularities are eliminated. Referring to Figure 11.4, we denote the position vector of the particle by r, the position vector of the center of mass of the circular hoop by x¯ , and the rotation tensor of the hoop by Q. The motion of the hoop is constrained so that it performs a fixed-axis rotation about E3 . Let {x1, x2 , x3} be rectangular Cartesian coordinates for x, ¯ { R, θ , φe } ¼ ½ 1 2 3 be spherical polar coordinates for r − x¯ , and ν , ν , ν be a set of 3–2–1 Euler angles for Q. Here, φe is the extension of the ordinary polar angle φ : φe ∈

[0, 2π ],

θ ∈

[0, 2π ].

The reason for extending the range of the coordinate φ from [0, π ] to [0, 2π ] can be explained by the following thought experiment. Imagine placing a particle at a location on the hoop and then rotating the hoop about E3. The coordinate θ determines the configuration of the hoop and φe is needed to determine the unique location of the particle relative to the hoop. In what follows, the reference configuration of the hoop is such that a normal to its face is in the E2 direction. A corotational basis {e1 , e2 , e3} is attached to the hoop. For the unconstrained system, we have the following kinematical quantities: x¯ =

3 ² i=1

xi Ei ,

Q=L

¹ ν

3

º

, e1 L

¹ ν

2

º

, e±2 L

¹ ν

1

º

, E3 ,

r − x¯ = ReR .

11.4

437

A Particle on a Rotating Circular Hoop

E3 ν

2R0

1

O

r

eφ

φe

T

2

eR A particle sliding on a circular hoop. The configuration manifold T 2 when the inertia of the hoop is considered is also shown. Each configuration of the system corresponds to a unique point on the surface of the torus.

Figure 11.4

¼

½

We emphasize that the set eR , eθ , eφ is defined relative to the corotational basis {ei }. The unconstrained velocity vectors for the system are v¯ =

3 ²

˙xi Ei ,

3 ²

ω=

i= 1

ν ˙

i

gi ,

i=1

˙ R + θ˙R sin (φe ) eθ r˙ − v¯ = Re

˙ Re + + φ e φ

ω × (r − x¯ ) , ¼

½

¼

½

where v = r˙. The generalized coordinates for the system are u1, u2 = ν 1, φe ½ ¼ ½ ¼ while the constrained coordinates are u3 , . . . , u9 = x1, x2 , x3, ν 2, ν 3, R, θ . That is, the seven integrable constraints are x1

=

0,

x2 ν

3

=

=

0,

0,

M

x3

R − R0

= =

0,

0,

ν θ =

2

=

0,

0.

2 After the constraints are imposed, = T , the two-torus (cf. Figure 11.4). Lagrange’s prescription for the constraint force and the constraint moment are presented in Exercise 11.7. We henceforth assume that Approach II can be used and a pair of Lagrange’s equations of motion can be used to determine the equations of motion of the system. We thus impose the constraints on the velocities to find the following expressions:

v = ν˙ 1 R0 sin (φe ) eθ

˙e R0 eφ , + φ

v¯ = 0,

ω = ν˙ 1 E3.

We assume that the hoop has a moment of inertia λ3 = m1 k2, where k is the radius of gyration about the center of mass X¯ of the hoop and m1 is the mass of the hoop. We have the following expressions for the constrained kinetic energy and potential energy of the system: T

=

´¹ º µ ¹ º2 2 1 1 m ν˙ 1 R20 sin2 (φe ) + φ˙e2 R20 + m1k2 ν˙ 1 , 2 2

U

=

mgR0 cos (φ e ) .

438

The Dynamics of Systems of Particles and Rigid Bodies

We assume that Lagrange’s equations of motion for the generalized coordinates can be expressed as d dt

±

³

˜ ∂L ∂ ν˙ 1

˜ ∂L − = ∂ ν1

d dt

0,

±

˜ ∂L ˙e ∂φ

³

−

˜ ∂L

∂ φe

=

0,

(11.2)

˜ The equations of motion are found using (11.2): where L˜ = T˜ − U.

¶

M

ν ¨

1

¨ φ e

¸

· +

(

2ν˙ 1φ˙ e R 20 sin (φ e) cos (φ e )

m

) »

) ( 1) 2 ( 2 −R0 sin (φ e ) cos (φe ) ν ˙

The mass matrix M in these equations of motion is ¶

M

¶

mR20 sin2 (φ e ) + m1 k2 0

=

0 mR 20

=

0 0

·

.

(11.3)

·

.

Clearly, when m1 > 0, M is positive definite. Thus, ν 1 and φe form a singularity-free 2 coordinate system for . If the inertia of the hoop is ignored, then → S and ¼ ½ ν 1 , φe become a set of spherical polar coordinates with a two-to-one covering of the unit two-sphere S2 . For this system, M is singular when φe = 0, π . The solutions to ˜ of the system. In addition, the solutions also (11.3) conserve the total energy T˜ + U conserve the angular momentum

M

M

HO · E3

∂ T˜

=

∂ θ˙

=

˜ ∂L

∂ θ˙

.

This conservation can be used to reduce (11.3) to a one-parameter family of secondorder ordinary differential equations for φe (t).4 We next impose an additional constraint on the system and assume that the hoop is being rotated at a constant speed ²0 : 1 ν˙ − ²0 =

0.

If we assume that Approach II can again be used to obtain the equations of motion, then the equation of motion can be found using (11.2)2 by imposing the additional constraint ˜ Omitting details, the resulting equation of motion is on L. ´

¨e = φ

2 ²0 cos (φe ) +

g R0

µ

sin (φe ) .

(11.4)

The phase portrait of the equation of motion is shown in Figure 11.5. To enforce the constraint ν˙ 1 = ²0 , a motor will be needed. The equation of motion (11.2)1 can be used to determine the torque Me applied by this motor: Me where

1

= µg

¹

µ =

= µ E3,

º

2m²0 φ˙e R20 sin (φe ) cos (φ e ) .

4 A closely related reduction was discussed earlier for a single particle in a Newtonian gravitational field in

Section 2.8 and for both the Lagrange top and Poisson top in Chapter 10.

11.5

Constraints

439

φe

10 ˙

φe

0

− 10

1

0

−1 − 1

0

1

Phase portrait of the solutions φe (t) to the differential equation (11.4) plotted on a cylinder. For the portrait shown in this figure, ²20 < g/ R and the system has two equilibria: φe = 0 and φe = π .

Figure 11.5

The power needed to drive the motor can be estimated from the product µ²0 . The electrical power needed will be larger than this value due to losses in the motor.

11.5

Constraints

Constraints on the motions of rigid bodies usually arise in two manners. First, the rigid body is connected to another rigid body in such a way that its relative motion is constrained. The connections in question are usually in the form of joints (e.g., pin joints, ball-and-socket joints, and universal joints). The second manner in which constraints arise occurs when one rigid body is rolling or sliding on the surface of another rigid body. As in our earlier discussions of this topic, the constraints we discuss can be classified as integrable or nonintegrable, and this classification is important in dynamics because it may lead to considerable simplification in the formulation of the equations governing the motion. In our discussion, we consider two rigid bodies, B1 and B2 (see Figure 11.1). To distinguish the rigid bodies, we use a subscript 1 or 2. Thus, for example, the rotation tensor associated with the first body is Q1 and for the second body is Q2. If more than two rigid bodies are involved in the constraint, then it is not too difficult to generalize the treatment we are about to present. It should be noted that the discussion presented here of the various types of joints and contacts is not exhaustive.

Connected Rigid Bodies

As shown in Figure 11.6, consider two rigid bodies B1 and B2 that are connected at the point XP1 of B1 and XP2 of B2 by a joint. The position vector of XP α on Bα relative to X¯ α is denoted by πPα . Because the material points XP1 and XP 2 occupy the same location,

440

The Dynamics of Systems of Particles and Rigid Bodies

¯

X2

B2 πP 2

XP2 X P1

x¯ 2

πP 1

x¯ 1

¯

X1

O

B1

The present configurations of two rigid bodies B1 and B2 that are connected by a joint at the points XP1 and X P2 of the respective bodies.

Figure 11.6

we have the following three constraints that we can represent in a variety of equivalent manners:5 xP1

=

xP2 or equivalently x¯ 1 + πP1

= x ¯2 +

π P2 .

(11.5)

= v ¯2 +

ω2 × πP2 .

(11.6)

These constraints also imply that vP1

=

vP2 or equivalently v¯ 1 + ω1 × πP1

Notice that by integrating (11.6) with respect to time and setting the constants of integration to 0, we will arrive at (11.5). Consequently, constraints (11.6) are integrable. The joint at the material points XP1 and XP2 may have the ability to restrict the rotation of B2 relative to B1 . There are two types of joints to consider: the pin (or revolute) joint and the ball-and-socket joint. The pin joint is similar to the joints used in gyroscopes to connect pairs of gimbals. In contrast, ball-and-socket joints do not place any restriction on the relative angular velocity vector ω2 − ω1 and can be modeled using the constraint (11.5) or (11.6).6 Consider the case of a pin joint. Let s3 be a unit vector that is parallel to the axis of relative rotation permitted by the joint, and let {1 e1, 1 e2, 1e3} be an orthonormal basis 5 By taking the components of this vector equation with respect to a basis, one arrives at three independent

scalar equations. Hence, the vector equation represents three constraints.

6 An example of a ball-and-socket joint can be found in the example of a spherical robot that is discussed in

Section 11.12.

11.5

441

Constraints

that corotates with B1. The pin joint ensures that the relative rotation tensor Q2QT1 is a rotation through an angle ν , say about 1e3 :7 Q2QT1

=

L (ν , 1 e3) ,

ω2 − ω1

= ν ˙ 1 e3 .

This restriction on the relative rotation is most easily expressed in terms of the relative angular velocity vector. Specifically, a pin joint imposes the following pair of constraints: (ω1 − ω2) · 1 e1 =

0,

(ω1 − ω2 ) · 1 e2 =

0.

It should be clear that these two (integrable) constraints are supplemented by (11.6). To see the integrability of these constraints, we can parameterize the relative rotation tensor Q2QT1 using a set of Euler angles. However, as it suffices to use a single angle to parameterize Q2 QT1 and the axis of rotation 1 e3 is determined by Q1 , two of the three Euler angles will be constrained to be zero. Clearly, each of the individual constraints mentioned up to this point can be written in the form π =

0,

(11.7)

where π = f1 · v 1 + f2 · v2 + h 1 · ω1 + h 2 · ω2 + e.

Here, f1 , f2 , h 1, h 2 , and e are functions of x1, x2 , Q1, Q2 , and t. The vector xα is the position vector of a material point Xα of the body Bα . This material point may be the center of mass X¯ α , but is not necessarily so. Furthermore, the velocity constraint (11.7) can be integrated with respect to time to yield a function ³: ³ = ³ (x 1, x2 , Q1 , Q2, t) .

Here, ˙ = π. ³

In other words, for this case, the constraint π = 0 is said to be integrable or holonomic. In general, the constraints associated with connections are usually integrable. This is in marked contrast to the next set of situations we consider.

Rolling and Sliding Rigid Bodies

Consider the situation shown in Figure 11.7. Here, two rigid bodies are instantaneously in contact at the points X P1 and X P2 of the respective bodies. It is assumed that there is a well-defined unit normal n to the surfaces of both bodies at the points of contact. In addition, we use n to define an orthonormal basis {s1 , s2, n}, where s1 and s2 are tangent to the surfaces of both bodies at the points of mutual contact. 7 The angular velocity vector ω − ω can be computed quickly using the notion of a relative angular 2 1

velocity vector discussed in Section 6.7.

442

The Dynamics of Systems of Particles and Rigid Bodies

n ¯

X2

B2

π P2

s2 XP 2

X P1

s1

x¯ 2

π P1 ¯x1

¯

X1

O

B1

The present configurations of two rigid bodies B1 and B2 that are in contact at the points XP1 and X P2 of the respective bodies.

Figure 11.7

As the points of contact are material points of each body, we have xP1

= x ¯1 +

πP1 ,

vP1

= v ¯1 +

ω1 × πP1

(11.8)

xP2

= x ¯2 +

πP2 ,

vP2

= v ¯2 +

ω2 × πP 2 .

(11.9)

and It is important to note that it is not generally possible to differentiate (11.8)1 to arrive at (11.8)2, nor (11.9) 1 to arrive at (11.9)2 . The reason for this is that πPα does not identify the same material point of Bα at each instant of time. In other words, XPα is the instantaneous point of contact and may correspond to a different material point of Bα at each instant of time.

The Sliding Condition

For two bodies in contact, it is assumed that the contact persists and that the two bodies do not interpenetrate. These assumptions lead to the sliding condition: vP1 · n = vP2 · n.

(11.10)

We can express this condition in another form: (v¯ 1 − v¯ 2 ) · n = (ω2 × πP2

−

ω1 × πP1 ) · n.

The sliding condition implies that a certain relative velocity has components only in the tangential directions: vs = vP1 − vP2

=

vs 1 s1 + vs2 s2 .

(11.11)

The velocity vs is often known as the sliding velocity and is important for specifying the friction forces at the points XP1 and XP2 .

11.6

A Canonical Function

443

The Rolling Condition

Rolling occurs when the velocity vectors of the point of contact for each body are identical. In this case, we have the rolling condition: vP1

=

vP2 .

Again, we can express this equation in another form: v¯ 1 + ω1 × π P1

= v ¯2 +

ω2 × πP2 .

(11.12)

These three equations are equivalent to the sliding condition (11.10) and the condition that the sliding velocity vs is zero. The rolling condition (11.12) is equivalent to three scalar equations (cf. (11.7)): π1 =

where, for i

=

0,

π2 =

0,

π3 =

0,

1, 2, 3, π i = fi1 · v ¯ 1 + fi2 · v ¯ 2 + h i1 · =

ω1 + hi 2 · ω2

Ei · vP1 − E i · vP2 .

However, for two of these equations, say π2 = 0 and π3 = 0, it is not possible to find ˙ 2 = π2 and ³ ˙ 3 = π3. In other words, two of the functions ³2 and ³3 such that ³ constraints (11.12) are nonintegrable or nonholonomic. The one constraint of (11.12) that is integrable corresponds to the n component of (11.12). We shall discuss shortly some examples that illustrate this point.

11.6

A Canonical Function

For future purposes, it is convenient to consider two rigid bodies B1 and B2 and construct the general functional form of a possible integrable constraint or potential energy function that couples the motions of both bodies. This function is presented in (11.13) and, for future purposes, we also calculate its derivative. Our treatment here follows [219, 220] with some modifications. As shown in Figure 11.1, the position vector of the center of mass X¯ α of the body Bα is denoted by x¯ α, where α = 1 or 2. Similarly, the rotation tensor of Bα is denoted by Qα . For each rigid body, we can define corotational bases: Q1

3 ² =

1 ei ⊗ E i,

Q2

3 ² =

i=1

2 ei ⊗ Ei .

i=1

It should be noted that, for convenience, we are using the same fixed basis {E1 , E2 , E3 } to define the corotational bases. The angular velocity vectors of the bodies are (

)

˙ QT , ω1 = ax Q 1 1

ω2

=

(

)

˙ QT . ax Q 2 2

444

The Dynamics of Systems of Particles and Rigid Bodies

We also note that the rotation tensor of B2 relative to B1 has a convenient representation in terms of the corotational bases: Q2QT1

3 ² =

2ei ⊗ 1ei .

i=1

This tensor represents the rotation of B2 relative to an observer who is stationary on B1 . A Scalar-Valued Function of the Motions

We now consider a scalar function ´ : ´ = ´ (x 1 , x2 , Q1 , Q2, t) .

Here, x1 is the position vector of a material point X α of the body Bα . The material point could coincide with the center of mass of the rigid body, but this is not necessary. Clearly, the function ´ depends on the motions of both rigid bodies and time. One observes functions of this type when representing integrable constraints on the motions of rigid bodies and potential energies of rigid bodies. ˙ . To calculate this time As in the case of a particle, it is of interest to calculate ´ derivative, we invoke the chain rule: ˙ = ´

∂´ ∂ x1

· v1 +

∂´ ∂ x2

´

·

v2 + tr

∂´ ∂ Q1

T

µ

˙ Q 1

´

+

tr

∂´ ∂ Q2

T

µ

˙ Q 2

+

∂´ ∂t

.

Recalling our earlier discussion in Section 6.11 of derivatives of scalar functions of rotation tensors, it is convenient to define the skew-symmetric tensors Ψ Qα

=

1 2

±

∂´ ∂ Qα

and their associated axial vectors ψQ1

=

T

Qα

´ −

(

Qα

∂´

µT ³

)

ax ΨQ1 ,

ψ Q2

1, 2 )

(α =

∂ Qα

=

(

)

ax ΨQ2 .

Representations for these tensors and vectors, based on the parameterization of Qα , were discussed previously in Section 6.11. In particular, if the Euler angles 1 ν i are used to parameterize Q1 say, then ψ Q1

=

3 ² ∂´ i=1

∂ 1ν

gi , i1

ψQ1 · ω1

=

3 ² ∂´ i=1

∂1 νi

1ν˙

i

,

(11.13)

where 1 gi are the dual Euler basis vectors associated with the Euler angles used to parameterize Q1 . ˙: We can use the aforementioned skew-symmetric tensors to rewrite ´ ˙ = ´

=

∂´ ∂ x1 ∂´ ∂ x1

· v1 + · v1 +

∂´ ∂ x2 ∂´ ∂ x2

·

·

(

v2 + tr ΨQ1 ΩT1

)

+

(

tr ΨQ2 ΩT2

v2 + ψQ1 · ω1 + ψQ2 · ω2 +

)

+

∂´ ∂t

,

∂´ ∂t

11.7

Integrability Criteria

445

where Ωα

˙ QT , = Q α α

ωα

=

ax (Ω α)

(α =

1, 2 )

are the angular velocity tensors and vectors of the rigid bodies. It is more efficient next ˙ above that involves vectors: to use the form of ´ ˙ = ´

∂´ ∂ x1

·

v1 +

∂´ ∂ x2

· v2 +

ψ Q1 · ω1 + ψQ2 · ω2 +

∂´ ∂t

.

(11.14)

With some minor rearrangements, we can eliminate v2 and ω2 in favor of the relative ˙ . The representation (11.14), velocity vectors v2 − v1 and ω2 − ω1 in the expression for ´ which first appeared in [219], will play a key role in examining potential forces and moments. The representation (11.14) is rarely apparent in treatments of rigid-body dynamics. This is partially because specific parameterizations of xα and Qα are used. To elaborate on this point, let x1 =

3 ²

xi E i,

x2

3 ² =

i =1

yi Ei ,

i=1

and suppose that Q1 is parameterized by the Euler angles 1 ν i and the relative rotation tensor Q2QT1 is parameterized by the Euler angles γ i . That is, ω1

3 ² =

i 1ν˙ 1 g i ,

ω2 − ω1

3 ² =

γ˙

i

ni ,

i=1

i=1

where 1 gi and n i are the Euler basis vectors associated with the sets of Euler angles 1ν i and γ i , respectively. Then, ˜ ´ = ´ (x 1, x2 , Q1 , Q2, t) = ´

¹

º

xi , yj, 1ν i , γ k , t .

Furthermore, ˙

˙ = ´ ˜ ´ 3 ² =

i=1

±

˜ ∂´

˜ ∂´

˜ ∂´

˜ ∂´ i x˙i + y˙ i + ˙ + γ˙ 1ν i ∂ xi ∂ yi ∂1ν ∂γ i

i

³ +

˜ ∂´

∂t

.

It is a good exercise to compare this expression with (11.14) and identify the corresponding terms in both expressions.

11.7

Integrability Criteria

Earlier, in Sections 1.10 and 8.5, we examined integrability criteria for constraints on the motion of a single particle and a single rigid body, respectively. Here, we examine the corresponding criteria for a constraint on the motions of two rigid bodies. To begin, we discuss the situation of a single constraint on a pair of rigid bodies. The corresponding developments for Frobenius’ criterion for the integrability of a system of constraints are easily inferred from our earlier discussion in Section 8.5 and are omitted.

446

The Dynamics of Systems of Particles and Rigid Bodies

A Single Constraint

For rigid bodies, what is needed is a generalization of criteria (1.27) and (8.12) to constraints of the form π =

0,

where π = f1 · v 1 + f2 · v2 + h 1 · ω1 + h 2 · ω2 +

e.

As mentioned earlier, this constraint is integrable if we can find an integrating factor ˙ = π . Although the integrability k and a function ³ (x1 , x2, Q1 , Q2 , t) such that k³ criterion is daunting in the number of algebraic calculations needed, for many problems most of these calculations are trivial (but tedious). The necessary and sufficient conditions for π = 0 to be integrable involve a set of up to 66 independent conditions. To establish these conditions, which are similar to those we discussed earlier for a single particle, we assume that x1 is parameterized ¼ 1 2 3½ system using a coordinate system q , q , q , x2 is parameterized using a coordinate ½ ½ ¼ ¼ 1 2 3 4 5 6 q , q , q , Q1 is parameterized using a set of Euler angles denoted by ¼ν , ν , ν ½, and Q2 is parameterized using a second set of Euler angles denoted by ν 4 , ν 5 , ν 6 . Thus, the function π can be expressed as π =

3 ¹ ² i=1

f1i q˙ i

+ f2i q ˙

i+3

+

h1i ν˙ i + h2i ν˙ i+3

º +

e.

We also define the intermediate functions Wi

= f1i ,

Wi+3

= f2i ,

Wi+6

=

h1i ,

Wi+9

=

h2i ,

W13 = e

and variables Ui

=

qi ,

Ui +3 = qi+3 ,

where i

=

1, 2, 3. It remains to define IJKL

Ui +6 = ν i ,

=

WJ SLK

+

U i+9

= ν

i+3

U13

,

WK SJL + WL SKJ ,

=

t,

(11.15)

where SLK are the components of a skew-symmetric matrix: SLK

=

∂ WL

∂UK

−

∂ WK ∂UL

.

(11.16)

In (11.15) and (11.16), the integer indices J, K, and L range from 1 to 13. With all the preliminaries taken care of, we are now in a position to state a theorem that is due to Frobenius: A necessary and sufficient condition for the constraint π = 0 on the motion of a pair of rigid bodies to be integrable is that the following 13 6 (13 − 1)(13 − 2) equations hold for all U 1, . . . , U 13: IJKL

=

0,

for all J, K, L ∈ { 1, . . . , 13}, J

²=

K

²=

L, J

²=

L.

(11.17)

11.8

Constraint Forces and Constraint Moments

447

As discussed previously, a proof of the theorem can be found in texts on differential equations such as [84, Section 161]. We also note that only 66 of the 286 equations IJKL = 0 are independent. Criterion (11.17) can be specialized to the case of a constraint fC · vC + h C · ω +eC = 0 on a single rigid body and, consequently, reduces to the criterion presented in Section 8.5.

11.8

Constraint Forces and Constraint Moments

Given a system of constraints on the motion of one or more rigid bodies, a system of forces and moments is required to ensure that the constraints are enforced for all possible motions of the bodies that are compatible with the constraints. At issue here is the prescription of these forces and moments. Our developments follow [219, 220].

Lagrange’s Prescription

Earlier, we showed that constraints in the motions of systems of a pair of rigid bodies can be expressed in the following canonical form (cf. (11.7)): π =

0,

where π = f1 ·

v1 + f2 · v2 + h 1 · ω1 + h 2 · ω2 + e

and vα is the velocity vector of a material point Xα of the body Bα . As anticipated, this canonical form reduces to that for a single rigid body, a pair of particles, and a single particle. Lagrange’s prescription for the constraint forces and constraint moments that enforce the constraint π = 0 are F c1

= µ f1

acting at X 1,

Mc 1

= µh 1,

F c1

= µ f2

acting at X 2,

Mc 1

= µh 2.

(11.18)

Here, µ is determined using the equations of motion. The combined mechanical power P of these forces and moments is readily computed:

P = F c1 · v1 + Mc1 · ω1 + Fc 2 · v2 + M c2 · ω2 = µ ( f1 · v 1 +

h 1 · ω1 + f2 · v2 + h2 · ω2)

= −µ e.

Consequently, if e = 0, then the constraint forces and constraint moments do no work. We shall see several instances where the constraint forces and constraint moments are workless in the examples that follow. A second aspect of Lagrange’s prescription that is immensely useful arises when the constraint π = 0 is integrable. If we consider a pair of rigid bodies, then a total of

448

The Dynamics of Systems of Particles and Rigid Bodies

¼

½

12 coordinates u1 , . . . , u12 are needed to describe the motion. We suppose that the integrable constraint can be expressed simply as u12 − f (t) = 0. That is, ´ =

u12 − f (t),

12 π = u ˙ − f˙ ,

e = −f˙ .

Now consider the following summations where the constraint forces and moments are prescribed using Lagrange’s prescription (11.18): µA =

Fc1 ·

∂ ω1 ∂ v2 ∂ ω2 ∂ v1 + Mc1 · + F c2 · + Mc 2 · A A A ∂u ˙ ∂u ˙ ∂u ˙ ∂ u˙ A

(A =

1, . . . , 12 ) . (11.19)

Paralleling the developments for a single rigid body (see (10.28) in Chapter 10), we find that µA = μ

∂´ . ∂ uA

(11.20)

Thus, the constraint forces and constraint moments for the integrable constraint ´ = q12 − f (t) will not contribute to 11 of the 12 summations. This result is one of the key components needed for the success of Approach II for systems of rigid bodies. In the case of multiple constraints in a system of two rigid bodies, (11.18) is applied to each constraint separately, each with a distinct μ . For instance, if there are three constraints, then the expressions for the forces and moments will contain μ 1, μ2 , and μ 3. The resultant constraint force Fc α acting at a particular Xα and resultant constraint moment Mcα on each rigid body can then be computed from the prescriptions for the individual constraints. Shortly, we will illuminate this discussion using an example of a pair of pin-jointed rigid bodies. Other examples of systems of rigid bodies with multiple constraints can be found in our discussion of a rigid body rolling on a cart in Section 11.13, a spherical robot in Section 11.12, a gyroscope in Section 11.14, and a Dynabee in Exercise 11.5.

Two Pin-Jointed Bodies and Newton’s Third Law

For example, consider the case of two rigid bodies connected by a pin joint shown in Figure 11.6. We remark that this example can also serve as a model for an anatomical joint in biomechanics. In orthopaedic biomechanics, one often seeks to examine the transmission of forces and moments from one anatomical segment to another via a joint.8 For the pair of pin-jointed bodies, there are five integrable constraints on the relative motion of the bodies: ´1 =

0, . . . , ´5

=

0,

where ˙ = ´ 1

vP1 · E1 − vP2 · E1 ,

˙2 = ´

vP1 · E2 − vP2 · E2 ,

8 The relations (11.21) are very useful in this regard. For additional details on the application to joint

biomechanics see, for example, [243].

11.8

449

Constraint Forces and Constraint Moments

˙ = ´ 3

vP1 · E3 − vP2 · E3 ,

˙4 = ´

ω1 · 1 e 1 − ω2 · 1 e 1 ,

˙5 = ´

ω1 · 1 e 2 − ω2 · 1 e 2 .

We emphasize that the axis of the pin joint is aligned with 1e3 = 2e3 , and its position vectors relative to the centers of mass of the bodies are πP1 and πP2 . Using Lagrange’s prescription (11.18) with the five constraints ´1 = 0, . . . , ´5 = 0, we find that F c1

=

3 ²

μ i Ei

acting at XP1 ,

M c1

= μ 41 e1 + μ51 e2,

μi Ei

acting at X P2 ,

M c2

= −μ 41 e1 − μ51 e2 .

i=1

Fc 2

3 ² = −

i=1

These expressions can be written in an equivalent fashion: Fc1 acting at X¯ 1 , −Fc1

acting at X¯ 2,

M c1 Mc 2

= μ 41 e1 + μ51 e2 +

= −μ41 e1 − μ 51 e2 +

πP1

πP2

×

×

F c1 ,

Fc2 .

(11.21)

As illustrated in Figure 11.8, Fc1 and Fc2 are the equal and opposite reaction forces at the pin joint, and Mr = μ41 e1 + μ 51 e2 and −μ41 e1 − μ 51 e2 are the equal and opposite reaction moments. Relative to the center of mass of each body, the constraint (or reaction) moments Mc 1 and M c2 are not equal and opposite unless π P2 = πP1 .9

Dynamic Friction

For the cases of a frictionless joint between two bodies, for two bodies in rolling contact, and for two bodies in frictionless sliding contact, Lagrange’s prescription provides a

Fc

1

Mr

XP1

¯

πP 1

X2

πP 2

¯

X1

XP2

−Mr Fc

2

Free-body diagrams of the two rigid bodies B1 and B2 illustrating the constraint forces Fc1 and F c2 and the reaction moment Mr . The reaction moment has the prescription Mr = μ41 e1 + μ51 e2 , and the constraint forces are equal and opposite (cf. (11.21)).

Figure 11.8

9 This was first noted in [219] and can be considered as a generalization of Newton’s third law to rigid

bodies.

450

The Dynamics of Systems of Particles and Rigid Bodies

very convenient method of specifying the constraint forces and constraint moments. However, the prescription does not yield physically reasonable forces and moments for joints or contact where dynamic friction is present.

11.9

Potential Energies, Conservative Forces, and Conservative Moments

The prototypical conservative element connecting a pair of rigid bodies is a spring. As a result, we start our discussion of conservative forces and moments acting on a system of rigid bodies by considering a linear spring connecting a pair of rigid bodies. This example helps to motivate a more general treatment that provides prescriptions (11.23) for conservative forces and conservative moments given a potential energy function. Spring Forces

In many mechanical systems, a spring is used to couple the motions of two bodies. Consider the system of two rigid bodies shown in Figure 11.9. Here, a linear spring of stiffness K and unstretched length ±0 is connected to the material point XS1 of the body B1 and the material point X S2 of the body B2. The spring exerts a force Fs1 at the point XS1 and an equal and opposite force F s2 at the point XS2 : ¾ ) x S − x S2 ¾, xS2 ¾ − ±0 ¾ 1 ¾xS − xS ¾ 1 2 (¾ ¾ ) x S2 − x S1 ¾. Fs2 = −K ¾ xS1 − xS2 ¾ − ±0 ¾ ¾x − x ¾ S1 S2

Fs1

= −K

(¾ ¾ xS

1 −

It is easy to see that each of these forces is equipollent to a moment relative to the center of mass and a force acting at the center of mass. The potential energy associated with the spring is ¾ ) K (¾ ¾xS − xS ¾ − ±0 2 . Us = 1 2 2 You should notice that this potential energy depends on the position vectors of both material points XS1 and XS2 . For future reference, U˙ s

=

(¾ K ¾ xS

1 −

( ¾ ) x S1 xS2 ¾ − ±0

−

x S2

¾ ¾xS

) ( ·

1 −

vS1

−

¾

xS2 ¾

v S2

)

.

Hence, Fs 1 · vS1

+

Fs 2 · vS2

˙ s. = −U

It is also convenient to note the identities x S1

=

(

¯ Q 1 XS 1 − X 1

)

+x ¯ 1,

xS2

=

(

Q2 XS2

¯ −X 2

)

+x ¯2,

where XSα is the position vector of XSα in the reference configuration of Bα . Using these identities, one can show that the forces, moments, and potential energy of the spring can

11.9

Potential Energies, Conservative Forces, and Conservative Moments

451

¯

B2

X2

πS 2

XS2

x¯ 2 Linear spring

XS1

x¯ 1

O

πS 1 ¯

X1

B1 Figure 11.9

spring.

The present configurations of two rigid bodies B1 and B2 that are connected by a

be expressed as functions of the rotation tensors of both rigid bodies and the position vectors of their centers of mass. General Considerations

It is convenient at this point to give a general treatment of conservative forces and moments in the dynamics of rigid bodies. Our discussion is in the context of two rigid bodies, but it is easily simplified to the case of one rigid body and a single particle and easily generalized to the case of a system of particles and rigid bodies. We assume that the most general form of the potential energy is U

=

U (x1 , x2, Q1 , Q2 ) .

Clearly, this function depends on the motions of both rigid bodies and time. We can calculate the time derivative of this function by using (11.14): ˙ U

=

∂U ∂ x1

·

v1 +

∂U ∂ x2

·

v2 + u Q1 · ω1 + uQ2 · ω2 ,

where uQα is a vector representing the derivative of U with respect to Qα . As discussed in Section 6.11, this vector has numerous representations, and the easiest to use arises

452

The Dynamics of Systems of Particles and Rigid Bodies

when Q1 and Q2 are parameterized by sets of Euler angles. Denoting these angles and ½ ½ ¼ ½ ¼ ¼ their dual Euler basis vectors by 1ν 1 , 1 ν 2 , 1ν 3 and 1g1, 1g2 , 1g3 , and 2 ν 1, 2ν 2 , 2 ν 3 ½ ¼ 1 and 2 g , 2 g2 , 2 g3 , respectively, the vectors u Q1 and u Q2 have the representations uQ1

=

3 ² ∂U k =1

∂ 1ν

gk , k1

u Q2

=

3 ² ∂U ∂2 νk

k=1

2g

k

.

Consider the conservative forces Fconα acting at Xα and moments Mconα associated with this potential. We assume that the work done by these forces and moments is dependent on the initial and final configurations of the rigid bodies but is independent of the motions of the rigid bodies. This implies that ˙ = −U

F con1 · v1 + F con2 · v2 + M con1 · ω1 + M con2 · ω2 .

˙ and collecting terms, we find that Substituting for U

´

F con1

+

∂U ∂ x1

µ

´

· v1 +

F con2 +

+

(

∂U

µ

∂ x2

M con2 + u Q2

· v2 +

)

·

ω2

(

Mcon1

=

+

uQ1

)

·

ω1

0.

(11.22)

This can be interpreted as an equation for Fconα acting at Xα and M conα that must hold for all motions of the rigid bodies. Assuming that F conα and M conα are independent of the linear and angular velocity vectors, we find that in order for (11.22) to hold for all vα and ωα , it is necessary and sufficient that Fconα

= −

∂U ∂ xα

acting at X α ,

M conα

= −uQα .

(11.23)

These are the expressions for the conservative forces and moments associated with a potential energy. It is left as an exercise for you to verify that these expressions are consistent with the results presented earlier for the spring forces. They can also be applied to establish representations of the form (8.18) for the central gravitational forces and moments between a pair of rigid bodies. With ¼a mind to Lagrange’s equations of motion, we suppose that a set of 12 coor½ dinates u1, . . . , u12 has been chosen to describe the motion of a pair of rigid bodies. The potential energy function of the system can then be expressed as a function of these coordinates: U

ˆ = U

¹

º

u1, . . . , u12 .

By paralleling the developments in Section 10.6 for a single rigid body, 10 we can show that Fcon1 ·

ˆ ∂ ω1 ∂ v2 ∂ ω2 ∂U ∂ v1 + Mcon 1 · + F con2 · + Mcon 2 · = − ∂u ˙A ∂u ˙A ∂u ˙A ∂u ˙A ∂ uA

(A =

1, . . . , 12 ) . (11.24)

This identity enables us to incorporate the contribution of the conservative forces and conservative moments on the right-hand side of Lagrange’s equations of motion into the Lagrangian on the left-hand side of Lagrange’s equations of motion. 10 See p. 399 in particular.

11.10

11.10

453

Lagrange’s Equations of Motion

Lagrange’s Equations of Motion

The governing equations for the system of N rigid bodies are found by using the balances of linear and angular momenta for each of the rigid bodies: m1 x¨¯ 1 = F 1, ˙ H 1

=

¨ . . . , mN x ¯N =

˙ M1 , . . . , H N

=

FN ,

MN .

(11.25)

Here, FK is the resultant force acting on the Kth rigid body and M K is the resultant moment relative to the center of mass X¯ K of the rigid body. When supplemented by the constraints, the balance laws (11.25) yield differential equations for the motion of the bodies and expressions for the constraint forces and moments. As shown in [40], the balance laws (11.25) are equivalent to Lagrange’s¼ equations of½ motion for the system of rigid bodies.11 For these equations, we denote by u1 , . . . , u6N the coordinates chosen to parameterize the position vectors x¯ K and the rotation tensors QK . Then, Lagrange’s equations of motion are d dt

´

∂T ∂u ˙A

µ

−

where µA =

∂T = µA ∂ uA

N ²

´

FK

K= 1

·

(A =

∂v ¯K + ∂u ˙A

MK

·

1, . . . , 6N ) ,

∂ ωK ∂u ˙A

(11.26)

µ

.

(11.27)

It is emphasized that using either (11.25) or (11.26) will lead to equivalent equations of motion for the system of rigid bodies. The position vector xK of a material point X K on BK can be expressed as a function of the 6N coordinates and the following identities hold: xK

¹

=

º

xK u1 , . . . , u6N ,

vK

=

6N ² ∂ xK J u˙ , J =1

∂ uJ

∂ xK ∂ vK = . J ∂u ∂u ˙J

As in the case of a single rigid body, if a force P acts at a point XP of body BK , then the contribution of this force to the right-hand side of (11.26) can be represented simply using the following identities:12 P·

∂ xP

∂ uA

=

P·

∂ vP = ∂ u˙ A

P·

∂v ¯K ∂ ωK + (( xP − x ¯ K ) × P) · . ∂u ˙A ∂u ˙A

In addition, contributions from conservative forces and moments acting on the system of rigid bodies can be replaced by the partial derivative of a potential energy function (cf. (11.24)): N ²

´

K =1

∂ vK F conK · A + ∂u ˙

∂ ωK MconK · ∂u ˙A

µ = −

ˆ ∂U ∂ uA

(A =

1, . . . , 6N ) ,

11 The proof is similar to the one presented in Sections 10.4 and 10.6 for a single rigid body and is not

presented here.

12 A proof of these results was presented earlier: see (10.25) in Chapter 10.

454

The Dynamics of Systems of Particles and Rigid Bodies

where U

ˆ = U =

¹

u1, . . . , u6N

º

¹

U x 1, . . . , x N , 1 ν 1 , 1ν 2, 1 ν 3 , . . . , N ν 1, N ν 2, N ν 3

º

is the potential energy function and K ν 1 , K ν 2, K ν 3 are the Euler angles used to parameterize the rotation tensor QK of the Kth rigid body. As in our discussion of Lagrange’s equations of other systems, it can be shown that if the constraint forces and moments are prescribed by Lagrange’s prescription, the system ½ ¼ of constraints on the system is integrable, and the coordinates u1, . . . , u6N are chosen appropriately, then Lagrange’s equations of motion decouple into two sets of equations: one where the constraint forces and moments are easily determined, and the other from which the motion of the system can be computed. The proof of this can be found in [40], and the interested reader is referred to this paper for details. Here, we are primarily content to use the result, but before doing so it is important to comment on the expressions for the generalized forces that follow from (11.27). Suppose that the system of rigid bodies is subject to C integrable constraints and that the coordinates are chosen so that these constraints can be represented simply as u6N −C+1

=

f 1 (t), . . . , u6N

=

f C(t).

We also suppose that the constraint forces Fc K acting at X¯ K and constraint moments Mc K associated with these constraints are prescribed by Lagrange’s prescription. It can be shown that N ²

´

FcK

·

K=1

∂ v¯ K + ∂u ˙A

McK

·

∂ ωK ∂u ˙A

µ =

0

(A =

1, . . . , 6N − C ) .

(11.28)

That is, the constraint forces and constraint moments do not contribute to the generalized forces µ1, . . . , µ6N −C. As a result, our expressions (11.27) for the forces µA are identical to those found in other treatments of Lagrange’s equations of motion for systems of rigid bodies.13 For our future discussion of gimbal lock, we note that a system of forces FwK acting at material points XK and moments M wK is said to be orthogonal to if N ´ ²

F wK

K =1

·

∂ vK + ∂u ˙A

MwK

·

∂ ωK ∂u ˙A

M

µ =

0

(A =

1, . . . , 6N

−

C) .

(11.29)

Clearly, the system of constraint forces and constraint moments for an integrable constraint that is prescribed using Lagrange’s prescription satisfies (11.29) and thus can be said to be orthogonal to .14 When an applied force or applied moment fails to contribute to the components of the generalized forces associated with the generalized

M

13 See, for example, Baruh [20], Greenwood [106], or Kane [140]. 14 The orthogonality condition can also be demonstrated by paralleling the construction of a representative

M

particle of mass m for a system of particles and rigid bodies in [40]. After examining the condition that the generalized force vector Φ acting on the particle of mass m is orthogonal to , one arrives at (11.29). Explicit examples of such a construction can be found in [123].

11.11

Two Pin-Jointed Rigid Bodies

455

coordinates (i.e., (11.29) holds), then the applied force or applied moment will have no influence on the dynamics of the generalized coordinates. Of particular interest in Section 11.15 will be an instance where a system of applied forces and moments is orthogonal to M and, as a result, gimbal lock is present.

11.11

Two Pin-Jointed Rigid Bodies

As our first example, we return to the case of two bodies, B1 and B2, that are pin jointed. This system was discussed earlier in Sections 11.3 and 11.8 and is illustrated in Figures 11.6 and 11.8. Our goal is to outline how the equations of motion for this system can be established using Lagrange’s equations of motion. We start with the kinematics. As discussed earlier, the pin joint introduces five constraints on the motions of B1 and B2 . Mindful of these constraints, we use Cartesian ¼ ½ coordinates to parameterize x¯ 1 , a set of Euler angles γ 1 , γ 2 , γ 3 to parameterize Q1 , and the angle θ to parameterize the relative rotation tensor Q2QT1 . The angle θ represents the rotation of B2 relative to B1 and the axis of rotation associated with this rotation is e3: Q2 QT1 = L (θ , e3). It is prudent to define the following seven coordinates: uk

= x ¯1 ·

Ek ,

uk +3 =

γ

k

u7

,

= θ,

where k = 1, 2, 3. We also assume that πP1

= ±1 e1,

πP2

= (±2) 2 e1,

where ± 1 and ±2 are constants. We note that the corotational basis vectors for B1 are { e1 = 1 e1, e2 = 1 e2 , e3 = 1e3 }, and the basis vectors { 2e1 , 2 e2 , 2 e3 = e3 } corotate with B2 . From the coordinates chosen, we have the following representations: v¯ 1

=

3 ²

x˙ i Ei ,

v¯ 2

= v ¯1 +

(

ω1 × ±1 e1 − ω1 + θ˙ e3

)

× (±2) 2e1,

i=1

ω1

3 ² =

γ˙

i =1

i

gi ,

ω2

3 ² =

i=1

γ˙

i

gi + θ˙ e3 .

These results can then be used to compute the constrained kinetic energy T˜ of the system. This energy is the sum of the kinetic energies of B1 and B2, and in the interest of brevity, we do not record its full expression here. Expressions for the constraint forces and moments acting on the system that enforce the constraints associated with the pin joint were presented in (11.21). Because these quantities are prescribed by Lagrange’s prescription, and because of our choice of coordinates, we can use Approach II and immediately write the equations of motion for this system: d dt

±

˜ ∂T ∂u ˙A

³

−

∂ T˜ = µA ∂ uA

(A =

1, . . . , 7) ,

(11.30)

456

The Dynamics of Systems of Particles and Rigid Bodies

where µA =

2 ² K =1

´

∂ ωK ∂v ¯K FK · A + MK · A ∂ u˙ ∂u ˙

µ

.

It is emphasized that Fc1 , Fc2 , Mc1 , and Mc2 do not contribute to µ1 , . . . , µ7. Indeed, it is a good exercise to verify that Fc 1 ·

∂ v¯ 1 + ∂u ˙A

Fc2 ·

¯2 ∂v ∂ ω1 ∂ ω2 + Mc1 · + M c2 · = A A ∂u ˙ ∂u ˙ ∂u ˙A

0

(A =

1, . . . , 7) .

You may have noticed that these identities are special cases of (11.28). Once the applied forces and moments are prescribed and T˜ calculated, the seven second-order differential equations of (11.30) suffice to determine the motion of the system of two rigid bodies. The resulting system of differential equations can be formidable, and often additional constraints and geometric simplifications are employed. An alternative formulation of the equations of motion would be to consider the balance of linear and angular momenta for each of the rigid bodies: m1 x¨¯ 1

=

F1 ,

m2x¨¯ 2

=

F2 ,

˙ H 1

=

M1 ,

˙ = M . H 2 2

(11.31)

This would yield a set of coupled equations for the constraint reactions μ1 , . . . , μ5 (cf. (11.21)) and u¨ 1, . . . , u¨ 7. The advantages of Lagrange’s equations of motion are their automatic selection of linear combinations of equations (11.31) should be obvious. Our discussion of the constraint forces and constraint moments at a pin joint can be applied to the planar double pendulum to justify the use of Approach II for this system in Section 11.3. Essentially, the contributions from the constraint forces and constraint moments at both pin joints vanish from the right-hand side of Lagrange’s equations for the generalized coordinates because the constraints are integrable, the constraint forces and constraint moments satisfy Lagrange’s prescription, and the coordinates θ1 and θ2 that we used to describe the motion of the system are not subject to any constraints.

11.12

A Simple Model for a Spherical Robot

As shown in Figure 11.10, a pendulum B2 consisting of a particle of mass mp and a rigid rod of length ±r , radius rr , and mass mr is mounted by a ball-and-socket joint to the geometric center (and center of mass) X¯ 1 of a spherical shell of radius R, mass m1 , and 2 inertia tensor J1 = 2m31 R I. The shell and the massless rod of length 2R constitute a rigid body that we refer to as B1 . The pendulum is free to rotate about X¯ 1 and the spherical shell moves on a rough horizontal plane. When the shell rolls on the plane, the motion of the pendulum enables locomotion of the system. As discussed in the review [313], this simple mechanism is a prototype drive for earlier designs of spherical robots. Modern designs of spherical robots, including the BB-8 droid in the Star Wars film franchise [23, 77], that have piqued the public’s interest in these machines use a different locomotion mechanism. The dynamics and control of spherical robots are an active area of research and we refer the interested reader to [1, 15, 24, 236, 259] and references therein.

11.12

Ball-and-socketjoint

E2 O

457

A Simple Model for a Spherical Robot

Rod of length 2R

Rod of length r

Spherical shell

¯

E1

X1

g

Rough plane

mp XP 1

A spherical shell of radius R and mass m1 rolling on a horizontal plane. The shell can be driven into motion using a suspended pendulum of mass m2 = mr + mp . The ball-and-socket joint coincides with the material points X¯ 1 and X C.

Figure 11.10

e

1 2

E2

e

1 1 g

¯

X1 O

E1 Rough plane

XC

¯

X

e

mp

2 2

e

2 1 X P1

A spherical shell of radius R rolling on a horizontal plane. The point X¯ on the circular rod of length ± r is the center of mass of the system. The ball-and-socket joint ensures that the center of mass X¯ 1 and the point XC ∈ B 2 have the same position vector.

Figure 11.11

In our discussion of this system, we seek to illuminate the previous results on constraints and constraint forces and constraint moments. Indeed, at the conclusion of our presentation on this system, we hope it will be transparent how these forces and moments can be computed using Lagrange’s equations of motion.

Kinematics

As shown in Figure 11.11, a corotational basis {1e1 , 1 e2 , 1 e3} is attached to the cylindrical shell and a corotational basis {2 e1, 2e2 , 2e3 } is attached to the pendulum. The unconstrained system has 12 degrees of freedom. The ball-and-socket joint at X¯ 1 introduces three integrable constraints on the motion of the system, while the rolling contact of the spherical shell introduces one integrable constraint and a pair of nonintegrable constraints. We assume that the rigid mounting rod of length 2R attaching the pin joint to the cylindrical shell is of negligible mass. Consequently, the center of mass X¯ 1 of the spherical shell and mounting rod is at the geometric center of the shell. We choose a set of Cartesian coordinates to describe x¯ 1 and a set of 3–2–1 Euler angles to parameterize the rotation tensor of B1:

458

The Dynamics of Systems of Particles and Rigid Bodies

x¯ 1

=

x1E1 + y1 E2 + z1 E3,

Q1

=

L

¹ ν

3

º

, 1e1 L

¹ ν

2

º

, 1 e±2 L

¹ ν

1

º

, E3 .

The position vector of the center of mass of the composite rigid body B2 consisting of the particle of mass mp and the rod of mass mr has the following representation: x¯ 2

=

xC + ±2e1 ,

´

1

±=

mr + mp

mp ±p +

±r

2

µ

mr ,

where xC is the material point on the rod that is closest to the ball-and-socket joint and m2 = mr + mp . The inertia tensor of the pendulum has the representation J02

=

mr r2r mpmr ± 2r E1 ⊗ E1 (I − E1 ⊗ E1 ) + m2 4 2 µ ´ mr ±2r mr r2r + (I − E1 ⊗ E1 ) . + 4 12

This inertia tensor was computed using the parallel axis theorem (7.39) and the representation for the moment of inertia tensor of a cylinder (7.22). The rotation tensor Q2 of B2 can be parameterized in a variety of manners. Here, we choose to parameterize the relative rotation Q2QT1 using a set of 3–2–3 Euler angles: Q2

=

L

¹

β

3

º

, 2 e3 L

¹

β

2

º

, 2e±2 L

¹

β

1

º

, 1 e 3 Q1 .

We leave it as an exercise to show that ω1

= ν ˙

1

E3 + ν˙ 2 1e±2 + ν˙ 3 1e1 ,

ω2

=

ω1 + β˙ 11 e3 + β˙ 22 e±2 + β˙ 3 2e3 .

The easiest way to prove these representations is to use the notion of a relative angular velocity vector discussed in Section 6.7.

Coordinates and Constraints

The six constraints on the motion of the system are as follows. First, the position vector of the material point X C on B2 and the center of mass X¯ 1 are coincident. Second, the velocity of the instantaneous point of contact XP1 of the shell with the ground plane is zero. These constraints can be expressed in the compact forms v¯ 1 − vC

=

0,

vP1

= v ¯1 +

ω1 × (−RE2 ) = 0.

(11.32)

We note that the first set of constraints (which pertains to the ball-and-socket joint) is integrable and is equivalent to the constraints v¯ 2

= v ¯1 +

ω2 × (−±2e1 ) .

We now choose the coordinates of the system so that the integrable constraints are easily accommodated: u1

=

x1

= x ¯1 ·

u2

E1 , u3

= ν

1

,

=

z1

u4

= x ¯1 · = ν

2

,

u12 = x¯ 1 · E2 − R,

E3, u5

= ν

3

,

11.12

u6 = u9

β

1

u10

= (x C − x ¯ 1 ) · E1 ,

u7

,

459

A Simple Model for a Spherical Robot

= β

2

u8 = β 3 ,

,

= ( xC − x ¯1) ·

u11 = (xC − x¯ 1) · E3.

E2,

If the ball-and-socket joint were disassembled, then the motion of B1 would be described by u1, u2 , u12, u3 , u4, and u5 , while the motion of B2 would be described by all of the coordinates apart from u1 , u2, and u12 . The four integrable constraints on the motion of the system can be expressed simply as u9

=

u10

0,

=

u11

0,

=

¼

u12 = 0.

0,

½

The generalized coordinates for the system are u1 , . . . , u8 and the configuration 3 3 2 manifold of the system is M ∼ = RP × RP × E . The kinetic energy for the system can be computed using the following representation: (

T

=

)

m1 m1R2 mr + mp 1 v¯ 1 · v¯ 1 + ω1 · ω1 + v¯ 2 · v¯ 2 + ω2 · J2 ω2. 2 3 2 2

The moment of inertia tensor J2 of the pendulum can be defined with the help of the expression we established earlier for J02 : ´

J2

=

mr rr2 mr r2r 2 e1 ⊗ 2e1 + 2 4

+

mr ±2r 12

+

mpmr ±2r mr + mp 4

µ

(I − 2 e1 ⊗ 2 e1 ) .

Substituting for the linear velocities and the angular velocities, an expression for T can be found. However, in the interests of brevity, we refrain from recording the final expression for T . The potential energy of the system is due to gravity: U

=

m1gE2 · x¯ 1 + m2 gE2 · (x¯ 1 + ±1e1 ) .

(11.33)

As defined earlier, the constant ± is the location of the center of mass of the pendulum from the point XC . Constraint Forces and Constraint Moments

We now use Lagrange’s prescription (11.18) to prescribe the constraint forces and constraint moments acting on the system. It is easiest to use the velocity forms of the constraints (11.32) for this purpose: F c1

3 ² = −

μi Ei + μ4 E1 + μ 5E2 + μ6 E3

acting at X¯ 1,

i =1

Mc1 Fc2

= −RE2 × (μ4 E1 + μ5 E2 + μ 6E3 ) ,

=

3 ²

μi Ei

acting at XC ,

M c2

=

0.

i=1

Referring to Figure 11.12, the constraint force F c2 corresponds to a reaction force at the ball-and-socket joint. The joint is assumed to be frictionless and thus the constraint moment supplied by the joint is zero. The constraint forces μ 4E1 + μ6 E3 and μ5 E2

460

The Dynamics of Systems of Particles and Rigid Bodies

Fc

XC

¯

X1

−

¯

−Fc E2

X2

mp

2

m1 g

2

−

m2 g

E2

Ff + N = µ4 E 1 + µ5 E2 + µ6 E3 Figure 11.12

Free-body diagrams of the composite rigid body B1 and the attached pendulum B 2.

correspond to a static Coulomb friction force Ff and a normal force N, respectively, at the instantaneous point of contact XP1 of the shell with the ground. In conclusion, Lagrange’s prescription provides a physically reasonable set of constraint forces and constraint moments. The combined power of the constraint forces and constraint moments can be shown to vanish as follows:

P = Fc1 · v¯ 1 + M c1 · ω1 + Fc2 · vC =

Fc2

=

0.

· (vC − v ¯ 1) + (μ4E1 + μ5 E2 + μ6 E3 ) · v P1

Consequently, the total energy of this system is conserved. Lagrange’s Equations of Motion

We close our discussion of this system by examining Lagrange’s equations of motion for this system (cf. (11.26)): d dt

´

∂L ∂u ˙A

µ

−

∂L = µcA ∂ uA

(A =

1, . . . , 12) .

The generalized forces on the right-hand side of this equation only contain contributions from the constraint forces and constraint moments. The equations are supplemented by the six constraints vP1 = 0 and vC − v¯ 1 = 0. In the interests of brevity, we will not expand the left-hand side of these equations. We start by examining the contributions the constraint forces and constraint moments would make to the right-hand sides of Lagrange’s equations of motion (cf. (11.19)) µcA =

Fc1

·

∂ ω1 ∂ vC ∂v ¯1 + M c1 · + Fc2 · A A ∂ u˙ ∂u ˙ ∂u ˙A

∂ vP1 = (μ4 E1 + μ5 E2 + μ 6E3 ) · + Fc2 · ∂u ˙A

(A =

´

∂ ∂u ˙A

1, . . . , 12) , µ

(vC − v ¯ 1)

.

Substituting vP1 = v¯ 1 + ω1 × (−RE2) and computing the partial derivatives of vP1 and vC − v1, we find that µc1 =

E1 · F f

= μ4 ,

µc 2 =

E3 · Ff

= μ5 ,

11.13

µc3 = − (E3 × RE2) ·

(

)

µc4 = − 1 e2 × RE 2 · F f , ±±

µc6 = µc9 =

E 1 · Fc 1

= μ1 ,

0,

461

A Semicircular Cylinder Rolling on a Cart

µc7 =

µc10 =

Ff

=

R μ4 ,

µc5 = − (1e1 × RE2) ·

0,

E2 · Fc1

µc12 = μ5 =

µc 8 = = μ 2,

Ff ,

0,

µc11 =

E 3 · Fc 1

= μ3 ,

N · E2 .

In these expressions, we used the notations Ff = μ4 E1 + μ6 E3 for the friction force and N = μ 5E2 for the normal force acting at X P1 . From the expressions for µc A , we conclude that the right-hand side of the eight Lagrange’s equations of motion for the generalized coordinates will be coupled by constraint forces and constraint moments associated with the two nonintegrable constraints vP1 · E1 = 0 and vP1 · E3 = 0. However, the μ s associated with the four integrable constraints on the system can only be determined using Approach I and will appear in the Lagrange’s equations of motion associated with the coordinates u9 , u10 , u11 , and u12 . These conclusions concerning the constraint forces and constraint moments are anticipated and in agreement with our earlier results for particles and single rigid bodies.

11.13

A Semicircular Cylinder Rolling on a Cart

To apply the previous representations for constraint forces and constraint moments to another concrete example, we now consider a semicircular cylinder that is free to roll on the upper surface of a cart. As shown in Figure 11.13, the cart is being driven into motion on a smooth horizontal track (or rail) by an applied force P(t)E1. By moving the cart back and forth, the friction and normal forces at the line of contact of the semicircular cylinder cause the semicircular cylinder to move. If there is sufficient static friction, then the semicircular cylinder will roll and rock back and forth on the cart. To complement the previous discussion, we seek to establish representations for the constraints and constraint forces and constraint moments for this system. The equations of motion for this system will also be investigated. An elementary treatment of the dynamics of this system was presented in [215, Chapter 10]. A subtle feature of this problem is that the contact between the rigid bodies is not a single point; rather, it is a line. This leads to a one-parameter family of constraints. With some careful consideration and analysis, we are able to reduce the family of constraints to an equivalent set of velocity constraints of the type considered previously. We will also show that the set of 10 constraints on the system constitutes a set of integrable constraints, that the motion of the system is planar, and that the equations of motion can be determined using a pair of Lagrange’s equations of motion. The equations of motion will be integrated numerically and the qualitative behavior of the solutions to the equations of motion explored.

462

The Dynamics of Systems of Particles and Rigid Bodies

e3

E3 g

e1

XC

¯

X P1

X1

E1

R

P = P (t)E 1

¯2 X

cart

XP2

Smooth rail Figure 11.13 A rigid semicircular cylinder of mass m and radius R rolling on a moving cart that has length ± cart . The points X P1 of the semicircular cylinder and XP2 of the cart have identical position vectors.

Preliminary Kinematics

We first dispense with some kinematical preliminaries. The homogeneous semicircular cylinder has mass m1, radius R, length ±, and moment of inertia tensor ´

J0

=

1 1 m1 R2 + m1 ±2 4 12

µ

´

(I − E2 ⊗

2

E2 ) + m1 R

1 2

−

16 9π 2

µ

E2 ⊗ E2 .

Referring to Figure 11.13, XC and X¯ 1 are the geometric center and center of mass of the semicircular cylinder, respectively. The position vector of the center of mass X¯ 2 of the cart is denoted by x¯ 2 . We make the following kinematical assumptions about this system: 1. The cart is assumed to be supported by four wheels, one at each corner. The masses of the wheels are negligible compared to the mass m2 of the cart. The wheels are free to move on a set of smooth rails. These rails prevent the cart from rotating and restrain the translational motion of the cart to be only in the E1 direction. 2. The contact between the semicircular cylinder and the cart is along a line. We denote points belonging to both bodies that are in contact with each other by XL1 and X L2 , respectively. The midpoints of the contact line for the semicircular cylinder and the cart are labeled XP1 and XP2 , respectively. We note that the rails which restrain the motion of the cart are not shown explicitly in Figure 11.13. Additionally, because the cart is constrained to purely translational motions, a corotational basis for the cart is not defined explicitly. The rotation of the semicircular cylinder relative to the cart is parameterized using a set of 3–2–1 Euler angles.15 As the rotation tensor of the cart is assumed to be the identity tensor, it suffices to parameterize Q1 using a set of 3–2–1 Euler angles: Q1

=

L (φ , e1) L

(

θ , e2

±±

)

L (ψ , E3 ) ,

Q2

15 Additional details on this set of Euler angles can be found in Section 6.8.1.

=

I.

11.13

463

A Semicircular Cylinder Rolling on a Cart

As shown in Figure 11.13, because the cart is assumed to only have translational motion, it suffices to use a single corotational basis attached to the semicircular cylinder: {e1, e2, e3 }. The position vectors of X C and the center of mass X¯ 1 have the representations xC

=

x1E 1 + y1 E2 + z1 E3 ,

x¯ 1

=

xC

−

he3 ,

h=

4R . 3π

The position vector of the center of mass X¯ 2 of the cart has the representation x¯ 2

=

x2E1 + y2 E2 + z2 E3.

As we have imposed constraints on the rotation of the cart, we do not need to prescribe a set of Euler angles to parameterize Q2 . For this problem, it is more convenient to describe the system using the coordinates of the geometric center XC than the center of mass X¯ 1. A One-Parameter Family of Constraints

The cart is subject to five integrable constraints. It is convenient to express these constraints in terms of velocities: ω2 = 0,

v¯ 2 · E2

=

0,

v¯ 2 · E3 = 0.

(11.34)

Referring to Figure 11.14, we parameterize the lines L 1 and L 2 of contact with a ¿ À coordinate yP ∈ −±/2, ±/2 . Thus, a point P ∈ L 1 has a position vector ¶

yP

=

xC − RE3 + yP e2, where yP

∈

−

± ±

,

2 2

·

.

E3

L1

e2

E2

yP

= −2

X P1

yP

=

2

E1 XC

ρ

¯

X1

XP1

L2

L1

Aspects of the geometry of the lines of contact L2 of the cart with the line of contact L1 of the semicircular cylinder at a given instant. The position vector of a point X L1 ∈ L1 has

Figure 11.14

the representation xC points on L1.

+

ρ, where ρ = −RE3 + yP e2 . The inset image shows the labeling of

464

The Dynamics of Systems of Particles and Rigid Bodies

We can compute an expression for the velocity of P using

¶

vP1

=

vC

+

ω1 × (−RE3 + yP e2) , where yP

∈

(

−

˙ E3 + θ˙ e±± + φ ˙ e1 = x ˙ 1E1 + y ˙ 1 E2 + z ˙1 E3 + ψ 2

)

± ±

·

, 2 2

× (−RE3 +

yP e2 ) .

Any point P ∈ L 2 has a velocity vector vP2 = x˙2 E2. Thus, the conditions that the semicircular cylinder rolls on the cart can be expressed as a one-parameter family of constraints: ¶

x˙ 2E2

=

vC + ω1 × (−RE3 + yP e2) , where yP

∈

−

± ±

·

, . 2 2

We now consider two members of this family corresponding to the choices yP yP = ±/2: x˙ 2 E2

=

vC + ω1 × (−RE3) , ´

x˙ 2E 2 = vC + ω1 ×

−RE3 +

±

2

(11.35) =

0 and

µ

e2 .

By taking linear combinations of these constraints, we can easily argue that the oneparameter family of constraints (11.35) is equivalent to the following five constraints: ω1 × e 2

x˙ 2 E2

=

=

0,

vC + ω1 × (−RE3) .

Notice that the first of these equations implies that ω1 = θ˙e2 . That is, ψ˙ = 0 and φ˙ = 0. This is a result that is easily observed by rolling a cylinder on a table. In other words, the planar motion of the system that is observed in practice can be reproduced in this model by choosing the Euler angles ψ = 0 and φ = 0. In conclusion, we can express the family of constraints (11.35) in a compact manner using the midpoints of L α : ω1 · e1

=

0,

ω1 · e 3

=

0,

vP1

=

vP2 .

In component form, the constraints can be expressed as ˙ = ψ

x˙1 − x˙ 2 − Rθ˙

=

0,

0,

˙ = φ

0,

y˙ 1 = 0,

˙z1 =

0.

(11.36)

Thus, the one-parameter family of constraints (11.35) is equivalent to a system of five integrable constraints.16 As a result of the 10 integrable constraints on the cart and semicircular cylinder, the 12 degrees of freedom of the unconstrained system are reduced to two degrees of freedom. That is, the system will have two generalized coordinates. It is convenient to choose either x2 and θ or x1 and θ as the generalized coordinates.17 16 Integrability of the constraints (11.36) can be concluded by inspection and fortunately it is not necessary

to examine Frobenius’ integrability theorem, mentioned earlier.

17 We encourage the reader to compare the constraints (11.36) to those for the sliding cylinder shown in

Figure 8.3; they can be found in Section 8.3.

11.13

465

A Semicircular Cylinder Rolling on a Cart

For future purposes, we now use the results from our exploration of the rolling conditions for this problem to establish an equivalent set of constraints for the system from (11.34) and (11.36): vP1 · E1 − vP2 · E1

=

0,

vP1 · E2 − vP2 · E2

( ω1 − ω2 ) · e 1 =

ω2 · E1

=

0,

0, =

0,

vP1 · E3 − vP2 · E3

( ω1 − ω2 ) · e 3 =

ω2 · E2

v¯ 2 · E2

=

0,

=

0,

ω2 · E3

=

0,

0,

=

0,

v¯ 2 · E 3.

(11.37)

In our forthcoming results and analyses, we shall assume that ±cart

, 2 otherwise the semicircular cylinder may have flipped over or fallen off the edge of the cart. −90

◦

0, then the solution α (t) will eventually settle to a constant value α0 . Equation (11.43) can be used to relate this value to the constant angular speed of the platform: ´

ω30 =

KA λ2 β˙

µ

α 0.

(11.44)

This equation is the explanation for the governing principle behind the operation of the gyroscope. Many devices that are similar to this gyroscope can be found: in particular, the gyrocompass, the gyrohorizon, and the dual-axis rate gyroscope. Space prevents us from discussing them here, but excellent treatments can be found in the texts by Arnold and Maunder [10], Crabtree [55], and Ginsberg [96]. Although these mechanical systems can seem dated when compared with modern microelectromechanical gyroscopes, the mechanical design featured in many of their practical implementations is ingenious.

11.15

Orthogonality of Generalized Forces and Gimbal Lock

Coordinate singularities and gimbal lock are two phenomena that appear in models for the dynamics of mechanical systems. The former phenomenon pertains to the coordinates used to parameterize the configuration manifold of the system, while the latter phenomenon has a distinctive physical manifestation that can be attributed to

474

The Dynamics of Systems of Particles and Rigid Bodies

(a)

(b)

3

α

δ

B1

O

γ

1

β

4

B1 O

2 γ

1

B2 E3

B3

B4

3

α

β

2

B3

B2

E1 Example of a platform B 1 mounted in (a) pair (B2 and B 3) and (b) three (B2 , B3 , and B 3) gimbals for a three-axis and a four-axis suspension, respectively. The spacecraft S that supports the bearing for the outermost gimbal is not shown.

Figure 11.18

orthogonality of generalized forces to the configuration manifold.21 As we mentioned previously in Section 6.8, the phenomena of coordinate singularities and gimbal lock are often mistakenly considered as synonymous in the literature and textbooks. Although discussions of gimbal lock can be found in earlier texts such as [240], the phenomenon of gimbal lock rose in prominence during the time of NASA’s Apollo program. Inside the inertial measurement unit (IMU) of the Apollo spacecraft was a stable (inertial) platform about which the spacecraft rotated (see [129, Figure 1] and [196, Figure 15]). The hinged platform was suspended by a pair of gimbals to give it three degrees of freedom (cf. Figure 11.18(a), where the platform is labeled B1 ). The role of the IMU was to give an accurate measure of the spacecraft’s orientation. To achieve this goal, gyroscopes that measure the angular velocity components of B1 were placed on the platform, and motors and sensors to measure angular displacement were placed at the joints between the gimbals.22 The servomotors then provided control torques in response to a feedback mechanism involving the gyroscope signals. The goal of the gimbals and control torques is to isolate the orientation of the platform as much as possible from changes to the orientation of the spacecraft. In this manner, the change in orientation of the spacecraft relative to the fixed orientation of the platform can be used to measure the rotation of the spacecraft. At a particular orientation of the spacecraft, such as the one shown in Figure 11.19, the gimbals became coplanar and control torques became ineffective at stabilizing the platform. Moreover, if the spacecraft is then rotated about an axis normal to the common plane of the gimbals, the platform becomes “locked” and engages in the same rotation

M

21 Our discussion of gimbal lock is based on the recent paper [123]. The notion of a generalized force that is

orthogonal to

was discussed in Section 11.10 (cf. (11.29)).

22 The gyroscopes are similar to those discussed in Section 11.14 and are used to detect the rotation of the

platform relative to a fixed rigid body or inertial reference frame [189].

11.15

Orthogonality of Generalized Forces and Gimbal Lock

475

E 3 = 3 e3 = 1 e3 α

e

2 1

B3

B2

= 3 e1

e

1 3

β

e

1 1

B1

e

1 2 γ

Configuration of the system of three rigid bodies when β2 = 0 and gimbal lock occurs. When β2 = π , E3 = −1 e3 . This figure is based on Figure 15 from a NASA report [196] published in 1962.

Figure 11.19

as the spacecraft, thereby rendering the orientation of the platform useless as an inertial reference. To correct for the locking of the gimbals, the spacecraft would need to pitch away from the problematic orientation and the platform would need to be reoriented relative to the stars. As discussed in [10, Section 11.3], to avoid gimbal lock, three gimbals could be used with a control scheme (cf. Figure 11.18(b)). For the Apollo program, the engineers, who were fully cognizant of gimbal lock, opted to use fewer gimbals to save on weight costs associated with the servomotors and sensors needed to control an added gimbal (cf. [129] and [186, Chapter 5]). In this section of the book, we choose to explore the phenomenon of gimbal lock from a geometric perspective. As discussed below, we explain how this phenomenon is intimately related to the notion of a generalized force being orthogonal to a configuration manifold. We have touched only briefly on the fascinating topic of the IMU for the Apollo spacecraft and refer the reader to Mindell’s book [186] for additional historical perspectives. The interested reader is also referred to [10] for a comprehensive discussion of gyroscopic devices mechanically suspended in gimbals. Additional helpful references on inertial navigation include [80], [209], and [284, Chapter 6]. Several excellent technical overviews and illustrations of the inertial platforms used in the Apollo spacecraft missions can be found in [129, 196, 283]. We also take this opportunity to refer the reader to [20, Section 7.8] for an illuminating discussion of gimbal lock in a Cardan joint. A Pair of Gimbals and Gimbal Lock

As an example of a system synonymous with gimbal lock, consider the system of three rigid bodies shown in Figures 11.18(a) and 11.19. Here, a platform B1 is suspended using a pair of gimbals B2 and B3 by a set of three joints. For the purposes of the present

476

The Dynamics of Systems of Particles and Rigid Bodies

e

α

1 3

B1

γ

B2

e

2 1

B3

B4

δ

E1

β

e

= 4 e1

3 3

The axes of relative rotation for the rigid-body components of the three-axis and four-axis suspensions shown in Figure 11.18.

Figure 11.20

discussion, it suffices to assume that the three bodies are symmetric. We also assume that the centers of mass of the rigid bodies are coincident for the configuration shown in the aforementioned figure. Thus, the angular momenta of the respective rigid bodies are H1

= λ

1

ω1 ,

H2

2

= λ

ω2 ,

H3

3

= λ

ω3 ,

where λ i is the moment of inertia of the body Bi . Referring to Figure 11.18(a), the orientation of the rigid bodies can be described using a set of three angles: Q3 = L (α , E 3 = 3e3) , Q2 = L (β , 2e1 Q1

=

L (γ , 1 e3

= 3e1 ) L (α , E3 = 3 e3 ) ,

ω3

= α ˙ E 3,

ω2

=

= 2 e3) L (β , 2e1 = 3e1 ) L (α , E3 = 3e3 ) ,

ω3 + β˙2 e1, ω1

=

ω2 + γ˙ 1 e3.

The corotational basis in these expressions is illustrated in Figure 11.20. Note that for the purposes of discussion, the rotation of the spacecraft that the gimbaled platform is mounted to is assumed to be in a state of purely translational motion. The constrained kinetic energy of the system of three rigid bodies is ´

T˜

=

µ

¹ º2 1 2 1 3 2 2 2 λ (α) ˙ + λ (α) ˙ + β˙ 2 2 º 1 1 ¹ 2 ( ˙ )2 2 (α) ˙ + β + λ + ( γ˙ ) + 2 α ˙ γ˙ cos (β) . 2

The configuration manifold M of the system is a three-torus T 3. As can be inferred ˜ the angles α , β , and γ , each of which range from the mass matrix M associated with T, from 0 to 2π , form a singularity-free coordinate system for T 3 . To see how gimbal lock manifests in this system, we need to examine the equations of motion. The system of three rigid bodies is subject to 15 integrable constraints. We can prescribe the constraint forces and constraint moments using Lagrange’s prescription (11.18). The equations of motion can be found using Approach II using the constrained kinetic energy T˜ :23 23 At this stage in the book, we assume that the reader is sufficiently versed in applications of Lagrange’s

equations of Ãmotion and that they are able to determine the 15 integrable constraints and coordinates  u4 , . . . , u18 for this system. The reader should also observe that specifying these coordinates is not necessary if one only wishes to determine the equations of motion for the generalized coordinates.

11.15

⎡ 1 λ ⎣ Ä

2

3

+ λ +λ

λ

1

0 cos (β)

477

Orthogonality of Generalized Forces and Gimbal Lock

0

1 2 λ +λ

ÅÆ

⎤⎡ ⎤ 1 λ cos (β) α ¨

−β˙γ˙

⎤

⎡

µ1

⎤

⎦ ⎣β¨ ⎦ + λ1 sin (β) ⎣ α˙ γ˙ ⎦ = ⎣µ2 ⎦ .

0

1 λ

0

⎡

Ç

−α ˙ β˙

γ¨

µ3

M

(11.45) The generalized forces in these equations are composed entirely of components of the applied moments M a1 , Ma2 , and Ma3 acting on the respective rigid bodies: µ1 =

3 ²

M ai

·

M ai

·

M ai

·

i=1 µ2 =

µ3 =

3 ² i=1 3 ² i=1

∂ ωi ∂α ˙ ∂ ωi ∂ β˙ ∂ ωi ∂ γ˙

=

=

=

(

Ma1 + Ma2 + M a3

(

Ma1 + Ma2

)

)

·

E3 ,

· 2e1 ,

Ma1 · 3e3 .

(11.46)

The moments Mai will include contributions from the servomotors at the joints in the system. The set of vectors {E3 = 3 e3 , 2 e1 = 3e1, 2e3 = 1e3 }, each element of which corresponds to a suspension axis, forms a basis for E3 provided β ² = 0, π : (E3 × 2 e1) · 1e3 = − sin (β) .

An example of an instance where β = 0 is shown in Figure 11.19. When β = 0, π , 2 3 e3 = ³1 e3 and the basis { E3, 2e1 , 1 e3 } only span the plane E which has a normal vector 2 e2 = 3e2 . Thus, a moment Ma1 applied to B1 in the 2e2 = 3 e2 direction contributes nothing to µ1, µ2 , or µ3 . Instead, it contributes to µ4, . . . , µ18 and the constraint moments. That is, gimbal lock occurs everywhere on the configuration manifold where β = 0 and β = π . Another way of stating this result is to say that the generalized force components µ1, . . . , µ18 corresponding to a moment Ma1 e2 applied 3 to the platform B1 when β = 0, π are orthogonal to the configuration manifold M ∼ = T (cf. p. 454). Stated in a slightly different manner: suppose gimbal lock occurs and a moment Ma1 = Ma 1e2 is applied to the platform by a disturbance that is transmitted from the spacecraft through the gimbals to the platform. Because this moment is orthogonal to M, it contributes to the constraint moments and gets transmitted through the gimbals to the rigid body (i.e., spacecraft) that is attached to B3. Thus, the applied moment Ma1 = Ma 1e2 serves to “lock” the orientation of the platform to the spacecraft. However, for the IMU to function, the orientation of the platform B1 needs to be independent of the orientation of the spacecraft and so this locking or coupling is highly undesirable.

Three Gimbals

Consider now a system of three nested gimbals (cf. Figures 11.18(b) and 11.20), where an additional (or “redundant”) gimbal B 4 has been added to the system of rigid bodies

478

The Dynamics of Systems of Particles and Rigid Bodies

e

3 3 α

= α0

B3 γ

B2

e

2 1

B1 E1 = 4e1

=

e

3 1

β

δ

B4 Figure 11.21 Configuration of the system of four rigid bodies when the rotation of B 3 is constrained so α = α0 where α0 is a constant. This figure is based on Figure 5 from a NASA report [196] published in 1962.

analyzed previously. The set of angles {δ , α, β, γ } parameterizes the constrained rotations of B4 , B3 , B2 , and B1 , respectively. For this system of four rigid bodies, we have the following rotation tensors and angular velocity vectors: Q4 Q3 Q2

=

Q1

=

=

ω4

L (α , 3 e3 ) Q4 ,

L (β , 2 e1

=

L (δ , E1) ,

= α ˙ 3 e3 +

ω3

= 3 e 1 ) Q3 ,

L (γ , 1e3

= δ˙E1 ,

ω2

= 2 e 3 ) Q4 ,

ω1

=

=

ω4 ,

ω3 + β˙ 2 e1 ,

ω2 + γ˙ 1e3 .

The configuration manifold is the four-torus T 4 and the set of angles {δ , α, β , γ } provides a singularity-free coordinate system for T 4 . Lagrange’s prescription can be used to prescribe the constraint forces and constraint moments enforcing the 20 integrable constraints on the system of four rigid bodies. As shown in Figure 11.21, we further impose the additional constraint α3 = α03. The configuration manifold becomes T 3 and the generalized coordinates, q1 = δ , q2 = β , 3 and q3 = γ , provide a singularity-free coordinate system for M ∼ = T . We do not compute the equations of motion for the gimbaled system of four rigid bodies with three degrees of freedom. However, we note that the mass matrix of the system is invertible if the inertias λ1 , . . . , λ4 are all strictly positive. We now parallel (11.46) and compute µ1 , µ2, and µ3 for the unconstrained (generalized) coordinates q1 = δ , q2 = β , and q3 = γ : µ1 =

(

M a1

+

µ2 =

(

Ma2

+

Ma3

M a2 + M a1

µ3 =

Ma1

+

)

Ma4

· 2e1 ,

· 1 e3 .

)

· E1 ,

11.15

479

Orthogonality of Generalized Forces and Gimbal Lock

The basis vectors on the right-hand side of these expressions for µi dictate whether gimbal lock can occur in this system. That is, if the three axes of rotation (or suspension axes) {E1, 2e1, 1e3 } form a basis for E3, then gimbal lock does not occur. To establish a criterion for this case, we compute the triple product (E1 × 2 e1 ) · 1e3 =

sin (α 0) cos (β) .

Thus, if we choose α0 = π/2, then the gimbal lock that previously occurred in the twogimbal system when β = 0, π will be eliminated. However, gimbal lock in this system can now occur when β = ³π/2. We leave it as an exercise for the reader to show that if, instead of fixing α , we fixed β = β 0, then the basis vectors associated with the generalized forces µi would be E1 , 3 e3 , and 1 e3. In this instance, (E1 × 3 e3) · 1e3 =

cos (α) sin (β 0) .

By choosing β0 ² = 0, π , the gimbal lock observed previously in the two-gimbal system can be avoided, but gimbal lock will now occur when α = ³π/2. We can now conclude that simply adding a third (or redundant) gimbal will not eliminate the possibility of gimbal lock; an additional strategic element is required. One potential method to eliminate gimbal lock involves the development of a scheme to control δ and α and transition to, and from, a gimbaled system where gimbal lock occurs when β → ³π/2 (where α = α0 and δ is free to vary) to one where gimbal lock occurs when β → 0, π (where δ = δ 0 and α is free to vary). An example of a pair of such systems is shown in Figure 11.22. In other words, the control scheme enables a transition between a pair of two-gimbal systems. The key to the success of such a scheme is to ensure that the orientation of the platform B1 is unaffected by the control torques needed to affect such transitions. We close this discussion by emphasizing that gimbal lock can occur even though the generalized coordinates used to parameterize 3 M∼ = T constitute a singularity-free coordinate system. (a)

E3 = 3e3

δ

= 0

E1

3 3 α

B2

B3 B2 B1

e

γ

e

B1

2 1

e

E1 γ

B4

= π2

B3

2 1 β

β

1 3

Figure 11.22

e

(b) α

δ

B4

A three-gimbaled system with three degrees of freedom: (a) the redundant gimbal

B4 is fixed (δ = 0) and gimbal lock occurs when β = 0, π and (b) the outer gimbal B3 is fixed

(α

= π/ 2)

and gimbal lock occurs when β

= ³π/2.

480

The Dynamics of Systems of Particles and Rigid Bodies

11.16

Closing Comments

We have discussed several examples that illustrate how the equations of motion for a system of rigid bodies and particles can be determined. Fortunately, for many physical examples, Lagrange’s prescription is applicable, the constraints on the system are integrable, and each constraint can be expressed in terms of a single coordinate. In this case, Lagrange’s equations can be used to determine the equations of motion by simply restricting attention to the generalized coordinates. The equations of motion for the constrained coordinates can be used to determine the constraint forces and constraint moments on the system if these are needed. This decoupling is remarkable and one of the most beautiful results in mechanics.

11.17

Exercises

Exercise 11.1: Consider a pair of rigid bodies. The rotation tensor of the first rigid body is Q1

3 ² =

1 ei ⊗ E i.

i=1

The rotation tensor of the second rigid body is Q2

=

3 ²

2 ei ⊗ E i.

i=1

Here, 1 ei corotate with the first rigid body and 2ei corotate with the second rigid body. (a) Argue that the rotation tensor of the second body relative to the first body has the representation R = Q2 QT1

3 ² =

2 ei ⊗ 1ei .

i=1

What is the rotation tensor of the first body relative to the second body? (b) Show that the angular velocity vector of the second body relative to the first body can be expressed as ω ˆ = ax

± 3 ²

³

◦

2 ei ⊗ 2 ei

,

i=1

where we have used the corotational derivative with respect to Q1 : ◦

a = a˙ − ω1 × a. (c) Suppose a set of 3–2–1 Euler angles is used to parameterize Q1 and a set of 1–3–1 Euler angles is used to parameterize R. Show that the angular velocity vectors of both bodies and their relative angular velocity vector have the representations

11.17

1

E3 + γ˙ 2 1e±2 + γ˙ 3 1e1 ,

ω1

= γ˙

ω2

˙ ˙ 2 e1 + γ ˙ 2e3 + ν = ν ˙ 1e1 + ν

1

2

3

±

1

3 2 ± 1 ˙ 2 e1 . ˙ 2e3 + ν ω ˆ = ν ˙ 1e1 + ν

Exercises

481

E3 + γ˙ 21 e±2 + γ˙ 31 e1,

(d) Suppose the rotation of the second body relative to the first body is constrained such that 3 ω ˆ = ν ˙ 2 e1 .

Give a physical interpretation of the type of joint needed to enforce this constraint. In addition, give prescriptions for the constraint moments acting on both bodies. Exercise 11.2: As shown in Figure 11.23, a rigid plate of mass m is attached at a point XA by a pin joint to a slender rod of negligible mass. The rod has length ±, one of its ends, O, is fixed, and the rod rotates about the E3 axis with angular speed ψ˙ = ². (a) With the help of a set of 3–1–3 Euler angles, show that the five constraints on the motion of the plate can be written as ˙¯ + ˙ e± = x − ±ψ 1

ω · g1

ω × πA ,

ω · g3

= ²,

=

0,

where x¯ is the position vector of the center of mass of the plate, ω is the angular ¯ velocity of the plate, and π A is the position of XA relative to the center of mass X. (b) Which orientations of the plate coincide with the singularities of the Euler angles? (c) Give prescriptions for the forces Fc and moments M c that ensure that the five constraints are enforced. Illustrate your prescriptions with a free-body diagram.

E3

E1 g

E2

Slender rod of length O

e2 XA

ψ θ

e1 ¯

X

e1 Rectangular plate

e2

A rigid plate that is pin-jointed at X A to a slender rod of negligible mass. The rod is supported by bearings at O that enable it to freely rotate about E3 . The pin joint at X A allows the plate to rotate (relative to the rod) about e±1 through an angle that is denoted by θ . Note that a gravitational force −mgE 3 acts on the system.

Figure 11.23

482

The Dynamics of Systems of Particles and Rigid Bodies

(d) Show that the mechanical power of the constraint forces and moments is nonzero if ² ² = 0. (e) If the vector πA and the moment of inertia tensor J of the plate relative to its center of mass are πA

= −ae2 ,

J=

3 ²

λ i ei ⊗ ei ,

i=1

establish an expression for the angular momentum HA and show that HA unless ² = 0.24 (f) Outline how you would establish the equations of motion of the plate.

²=

JA ω

Exercise 11.3: Following (2.1), the impulse–momentum form of the balances of linear and angular momentum for a rigid body are G (t1 ) − G (t0 ) = H (t1 ) − H (t0 ) =

È t1

t0 È t 1 t0

F(τ )dτ , M(τ )dτ .

(11.47)

Here, we wish to apply these equations to an impact scenario. Thus, t1 − t0 → 0, but G(t) and H(t) can be discontinuous in this limit. We assume that the impact occurs at time t, and so we consider limits where t1 = t + σ ¶ t and t 0 = t − σ ¸ t. As in treatments of discontinuities in continuum mechanics, we introduce the following notation to denote the change (or jump) in a function f (u(t), u(t), ˙ t ) at time t: ¿¿ ÀÀ

f

where σ

>

=

lim f (u(t + σ ), u˙ (t + σ ), t + σ ) − f (u(t − σ ), u˙ (t − σ ), t − σ ) ,

σ →0

0. ½

¼

(a) Suppose that a set of coordinates u1 , . . . , u6 is used to parameterize the motion of the rigid body. What then are the conditions under which ¶¶

∂v ¯ ∂u ˙K

··

¶¶

=

0,

∂ω ∂ u˙ K

··

=

0,

(11.48)

where K = 1, . . . , 6? Give a physical interpretation of the conditions you have found. (b) Assuming that the conditions (11.48) hold, show that by taking appropriate limits, (11.47) can be written as ¶¶ ·· ∂T ˆ , = µ (11.49) K ∂ q˙ K where K ˆ µ

=

K =

1, . . . , 6 and the generalized impulsive forces are lim

σ →0

´È t+σ t−σ

µ

F(τ )dτ

24 You may find the identity a × ( b × c )

∂ v¯ · (t) + lim ∂u ˙K σ →0

= (a · c) ·

´È t +σ t −σ

M(τ )dτ

b − (a · b) · c to be helpful.

µ ·

∂ω (t ) . ∂u ˙K

11.17

Exercises

483

For assistance in establishing (11.49), calculations (10.24) might be helpful. An alternative derivation of (11.49) for a system of particles can be found in [276, Section 15.2]. (c) Show how the six equations of (11.49) can be used to determine the dynamics of a rod impacting a smooth horizontal surface. Exercise 11.4: Consider, once more, the robotic arm shown in Figure 7.15. As discussed, the robotic arm has mass m and moment of inertia tensor relative to its center of mass given by J0

= λ1 E1 ⊗ E1 + λ2E 2 ⊗

E2 + λ 3E3 ⊗ E3 .

A system of motors, which is not shown in Figure 7.15, is used to actuate the rotation of the robotic arm. The rotation consists of 1. a rotation about the E3 axis through an angle ψ , 2. a rotation about g2 = e±1 = cos(ψ )E1 + sin(ψ )E2 through an angle θ . Another system of actuators prescribes the motion of the point X P of the robotic arm. The position vector of the center of mass of the arm relative to X P is x¯ − xP

=

±

2

e3.

(a) Interpreting the angles ψ and θ as members of a 3–1–3 set of Euler angles where φ = 0, show that the dual Euler basis vectors, g 1, g2 , g 3 are not orthonormal. For which positions of the robotic arm are these vectors not defined? (b) Noting that the motion of the point X P is prescribed, xP = u(t), and that the rotation of the arm is prescribed, what are the six constraints on the motion of the arm? (c) Show that the angular momentum H (relative to its center of mass) of the arm is H

˙ = λ1θ˙ e1 + λ2 ψ

sin(θ )e2 + λ 3ψ˙ cos(θ )e3 .

(d) Draw a free-body diagram of the robot arm. In your free-body diagram, include the gravitational force −mgE 3. (e) Show that the force needed to actuate the motion of the center of mass is ) m± ( ˙ ˙ 2ψ θ cos(θ ) + ψ¨ sin(θ ) e1 2 º º m± ¹ 2 m± ¹ 2 ˙ sin(θ ) cos(θ ) − θ¨ e2 − ˙ 2 sin2(θ ) e3 . + ψ θ˙ + ψ 2 2

F act = mgE3 + mu¨ +

(f) Show that the moment M act needed to actuate a slewing maneuver of the robotic arm is Mact

=

m±2 m± ˙ 0θ˙0 cos( θ )e2 e3 × (gE3 + u¨ ) + ψ 2 2 m± 2 ˙ 2 ψ sin(θ ) cos(θ )e1 + ω × H. − 4 0

During the slewing maneuver, θ˙ =

θ˙0

and ψ˙

˙0 . = ψ

Consequently, ω is not constant.

484

The Dynamics of Systems of Particles and Rigid Bodies

et er

3

=

a3

˙

α

T

R

3

˙

γ

et

er

1

= a1

1

et

2

T

R

Schematic of a Dynabee showing the circular track and the rotor . The angles α and γ shown in this figure are used to parameterize the rotation of the rotor relative to the track. This figure is adapted from [109].

Figure 11.24

(g) Show that the time rate of change of the total energy E of the robotic arm is equal to the work done by Fact and Mact : E˙

=

Fact · u˙ + M act · ψ˙ E3 + M act · θ˙ e1 .

Exercise 11.5: As mentioned earlier, a patented wrist exerciser known as a Dynabee, Powerball, or Rollerball features a heavy rotor that can roll without slipping on a circular track [109, 128]. The track is part of a housing that can be held and rotated. As analyzed in [109], effectively rotating the track allows the rotor to spin-up. The resulting spin-up requires that the holder provide an ever-increasing moment and, as a result, the holder exercises several muscles in the arm and wrist. In the problem discussed here, we consider the formulation of the constraints on the rotor and the housing. ½ ¼ As illustrated in Figure 11.24, we define a set of vectors et 1 , et2 , et 3 to corotate with ½ ¼ the track and a set of vectors er 1 , er2 , er 3 to corotate with the rotor. We also define an intermediate set of vectors a1

=

er1 ,

a2

=

et3 × er 1 ,

a3

=

et3 .

We emphasize that these vectors are not corotational. ∑

3 (a) The rotation tensor Q = k=1 etk ⊗ E i of the housing (and track) is assumed to be parameterized by a set of 3–1–3 Euler angles {ψ , θ , φ}. Referring to Figure 11.24, what are the consequences of constraining ψ = −φ and θ to a constant value θ0 ? (b) The rotation tensor R of the rotor relative to the track (housing) is assumed to be parameterized by a set of 3–1–2 Euler angles {α, γ , μ ≈ 0}. We emphasize that

R = er 1 ⊗ et1 + er 2 ⊗ et2 + er 3 ⊗ et3 .

11.17

et

3

= a3

Q

eπ

485

3

R

T β

er eπ

1

¯

X

T

Exercises

= a1

1

P

Schematic showing the two points of contact P and Q of the rotor and the track. For our analysis, the centers of mass of the rotor and housing are assumed to coincide at the ¯ In this figure, the radius R a of the axle of the rotor and the angle μ have been point X. exaggerated.

R

Figure 11.25

T

Using the aforementioned sets of Euler angles, establish representations for the angular velocity vectors of the rotor ωr and the track ωt . (c) Referring to Figure 11.25, we assume that the rotor has two points in contact with ¼ ½ the track. For convenience, we define a fourth set of unit vectors eπ1 , eπ2 , eπ3 such that eπ1 is parallel to the line segment joining P and Q, and eπ2 = a2 . If β is the angle between er 1 and eπ1 , then show that β =

tan

−1

´

Ra Rt

µ

,

where R a is the radius of the rotor’s axle and Rt is the radius of the track. (d) Assuming rolling at the points P and Q, show that these constraints are equivalent to five (scalar) velocity constraints: v¯ r = v¯ t ,

( ω r − ω t ) × e π1 =

0,

where v¯ r and v¯ t are the velocity vectors of the centers of mass of the rotor and housing, respectively. In addition, show that these constraints imply that ´

γ˙ = −

Rt Ra

µ

α ˙.

(11.50)

As in [109], we assume that μ ≈ 0. (e) Assuming that the motion of the housing (and its rigidly attached track) is completely prescribed, show that the system of rigid bodies consisting of the housing and the rotor is subject to 11 independent constraints. Using Lagrange’s prescription, give prescriptions for the constraint forces and moments acting on the two rigid bodies and . (f) Argue that the eπ1 component of the balance of angular momentum for the rotor relative to the center of mass X¯ gives a differential equation for the motion of the rotor. (h) When the rotor is steadily rotating, ψ˙ = α˙ . Explain why (11.50) then implies that the rotor will be spinning faster than the housing.

R

T

486

The Dynamics of Systems of Particles and Rigid Bodies

Exercise 11.6: Consider a rigid body and assume that its rotation tensor Q is parameterized by a set of 3–1–3 Euler angles {ψ , θ , φ }. Suppose that after an interval of time t1 − t 0, e3 (t1 ) = e3 (t 0) .

(11.51)

(a) Show that the rotation of the rigid body during the time interval t1 representation

− t0

has the

Q (t1 ) QT (t0 ) = L (φ (t1 ) − φ (t0 ) , e3 (t0 )) . (b) Show that the measurement of a single rate gyro ω · e3 from a rate gyroscope is insufficient to determine φ (t1) − φ (t 0).25 (c) Give an example of a motion of a rigid body where (11.51) holds.26 (d) For your example in (c), imagine the base point of the vector e3 is fixed, and plot the coordinates of the tip of this vector. The tip lies on the unit sphere. Using a result that can be found in Kelvin and Tait [145, Section 123] and that was rediscovered independently by Goodman and Robinson [98] and Levi [167], show that the area enclosed by the closed curve traced by the tip of e3 (t) on the unit sphere is related to the angle φ (t1 ) − φ (t0 ). (e) Show that the results of (d) are related to Codman’s paradox discussed in Exercise 6.8. Exercise 11.7: Consider a particle of mass m in the rotating circular hoop of mass m1 shown in Figure 11.4 and discussed in Section 11.4. The coordinates for this system were chosen as follows: u1 u u

6

= ν

3

2

=

= ν

x1,

,

u

1

u 7

= ν

u2

= ψe,

x2 ,

u5

, 4

3

=

,

u

8

=

=

x3, u9

R,

= θ.

(a) Show that the constraints on the motion of the particle and the rigid body can be expressed in the following form: v¯ · E1

=

0,

v¯ · E2

=

0,

˙r · eR − v ¯ · eR =

0,

v¯ · E3 r˙ · eθ

=

0,

−v ¯ ·

ω · eθ

eθ

+

=

0,

ω · R 0eφ

ω · er = 0, =

0.

Some of the constraints have been applied to simplify other constraint expressions, for instance, ω · g2 = ω · eθ . (b) With the help of Lagrange’s prescription (11.18), show that the constraint force F c acting on the particle, the constraint force F c1 acting at the center of mass X¯ of the hoop, and the constraint moment Mc1 acting on the hoop have the following representations: Fc

= μ 6eR + μ7 eθ

acting on the particle of mass m,

25 You may wish to recall (7.36) from Exercise 7.3. 26 For assistance with this matter, the papers of Montgomery [191] and O’Reilly [213] might be helpful.

11.17

Fc1

= μ1 E1 + μ2 E2 + μ 3E3 − μ6 eR − μ7 eθ

Mc1

Exercises

487

¯ acting at X,

= μ4 eθ + μ5 er + μ7 R0 eφ .

(c) Show that the generalized forces acting on the system are µ1 =

Fa · R0 sin (φ e ) eθ

µ(2+i) = µ6 =

Ma1 · eθ

µ8 =

M a1 · E3,

Fa · E i + F a1 · E i + μ i

+ μ 4,

Fa · eR

+

+ μ6 ,

µ7 =

µ2 = (i =

Fa · R0 eφ ,

1, 2, 3) ,

Fa · (−R 0 cos (φe ) eθ ) + M a1 · er + μ5,

µ9 =

Fa · R 0 sin (φ e ) eθ

+ μ 7R0

sin (φe ) ,

where Fa1 is the applied force acting at the center of mass of the rigid body, M a1 is the applied moment acting on the rigid body, and F a is the applied force acting on the particle. (d) How do the expressions for the constraint forces, constraint moment, and generalized forces in (b) and (c) change when the additional constraint 1 ν˙ − ²0 =

0

is imposed? (e) Using the results of (c) and (d), verify that both sets of equations of motion (11.3) and (11.4) can be computed using Approach II and (11.2).

Appendix:

A.1

Background on Tensors

Introduction

A tensor is a linear mapping that transforms a vector into another vector. It is common practice to use matrices to represent these transformations. In this Appendix, another quantity, known as a tensor, is introduced to achieve the same purpose. We generally denote tensors by uppercase boldfaced symbols: for example, A, and symbolize the transformation of a vector a by A to a vector b as b = Aa. The advantages of using tensors are that they are often far more compact than matrices, they are easier to differentiate, and their components transform transparently under changes of bases. The latter feature is very important in rigid-body kinematics. To this end, Section 6.2 of Chapter 6 contains an example in which tensor and matrix notations are contrasted and, in the interests of brevity, the reader is referred to this section for further motivation on the advantages of tensors. We also note that the material presented in this Appendix is standard background for courses in continuum mechanics, and in compiling it the primary sources were Casey [36, 38], Chadwick [46], and Gurtin [111].

A.2

Preliminaries: Bases, Alternators, and Kronecker Deltas

Here, Euclidean three-space is denoted by E3 . For this space, we define a right-handed fixed orthonormal basis {E1, E2, E3 }. We also use another right-handed orthonormal basis {p 1, p2 , p 3}. This basis is not necessarily fixed.1 Lowercase italic Latin indices, such as i, j, and k, will range from 1 to 3. You may also wish to recall that a set of vectors {b1 , b 2, b3 } is orthonormal if b i · bk = 0 when i ± = k and bi · bk = 1 when i = k. Further, a set of vectors {b1 , b2 , b 3} is right-handed if the following scalar triple product is positive: [b1 , b2 , b 3] = b3 · (b1 × b 2). 1 This basis serves to represent corotational bases, other time-varying bases such as { e , e , E }, and fixed r θ 3

bases such as { E1 , E2 , E3 }.

A.3

489

The Tensor Product of Two Vectors

Because we will be using copious amounts of dot products, it is convenient to define the Kronecker delta δik : ±

δij =

1 i = j, 0 i ± = j.

Clearly, pi · p k

= δ ik .

We also define the alternating (or Levi-Civita) symbol ±ijk: ± 123 = ± 312 = ± 231 =

1,

± 213 = ± 132 = ± 321 = −1, ±ijk =

0 otherwise.

In words, ±ijk = 1 if ijk is an even permutation of 1, 2, 3; ±ijk = −1 if ijk is an odd permutation of 1, 2, 3; and ±ijk = 0 if either i = j, j = k, or k = i. We also note that [p i , pj , pk ] =

±ijk .

It is easy to verify this result by using the definition of the scalar triple product.

A.3

The Tensor Product of Two Vectors

The tensor (or cross-bun) product of any two vectors a and b in E3 is defined by (a ⊗

b) c = (b · c) a,

(A.1)

where c is any vector in E3 . That is, a ⊗ b projects c onto b and multiplies the resulting scalar by a. Clearly, a ⊗ b transforms c into a vector that is parallel to a. Equivalently, we can define a related tensor product: c (a ⊗ b) = (a · c) b. Both tensor products are used in this book. You should notice that a ⊗ b provides a linear transformation of any vector c that it acts on. For example, if a = E1 , b = E2 , and c = E2 , then (E1 ⊗

E2) E2

=

E1,

E2 (E1 ⊗ E2) = 0.

It is a useful exercise to consider other choices of c here. The tensor product of a and b has some useful properties. First, if two scalars, and a, b, and c are any three vectors, then (α a + β b) ⊗

α

and

β

are any

c = α (a ⊗ c) + β (b ⊗ c) ,

c ⊗ (α a + βb ) = α (c ⊗ a) + β (c ⊗ b ) . These properties follow from the definition of the tensor product. To prove them, one merely shows that the left- and right-hand sides of the identities provide the same transformation of any vector d.

490

Background on Tensors

A.4

Second-Order Tensors

A second-order tensor A is a linear transformation of E3 into itself. That is, for any two vectors a and b and any two scalars α and β , A (α a + β b ) = αAa + β Ab, where Aa and Ab are both vectors in E3. To check if two second-order tensors A and B are equal, it suffices to show that Aa and Ba are identical for all vectors a. The tensor a ⊗ b is a simple example of a second-order tensor. It is standard to define the following composition rules for second-order tensors: (A + B)a = Aa + Ba,

(αA)a = α (Aa),

(AB)a = A(Ba),

where A and B are any second-order tensors, a is any vector, and α is any scalar. We also define the identity tensor I and the zero tensor 0: Ia = a,

0a

=

0,

where a is any vector.

A.5

A Representation Theorem for Second-Order Tensors

It is convenient at this stage to establish a representation for any second-order tensor A. The main result we wish to establish is that A=

3 3 ² ² j=1 k=1

A kj pk

⊗ pj ,

(A.2)

where A ik

=

(Ap k ) · p i ,

aj

3 ² =

A kj pk ,

A=

k=1

3 ²

aj ⊗ pj .

j=1

Here, Aik are known as the components of A relative to the basis {p1 , p 2, p3 }. Initially, ∑ it is convenient to interpret tensors by use of the representation A = 3j=1 aj ⊗ p j. In this light, A transforms p k into ak . Hence, if we know what A does to three orthonormal vectors, we can write its representation immediately. To establish representation (A.2), we note that Ap i is a vector. Consequently, it can be written as a linear combination of the basis vectors p 1, p2 , and p 3: Ap1

=

A 11p1 + A21p 2 + A31p 3,

Ap2

=

A 12p1 + A22p 2 + A32p 3,

Ap3

=

A 13p1 + A23p 2 + A33p 3.

A.5

491

A Representation Theorem for Second-Order Tensors

The order of the indices i and k for the scalars Aik is important. Given any vector b ∑3 i=1 b i pi , we find that Ab =

3 ² j =1

=

3 ²

A(bj pj ) = ³

bj

j =1

=

3 ² k =1

j =1

bj (Ap j )

´

Akj pk

⎛ 3 3 ² ² ⎝ Akj p j =1 k =1

=

3 ²

⎛ 3 3 ² ² ⎝ Akj p j =1 k =1

=

3 ² 3 ²

⎞ k

=

j =1 k =1

⊗p⎠

j

⎞

3 ² i =1

(

bj Akj p k

)

bi p i

pj ⎠ b.

k⊗

In the next-to-last step, we used the definition of the tensor product of two vectors. In summary, we have shown that Ab =

3 3 ² ² (

A kj pk

j =1 k =1

⊗ pj

)

b.

As this result holds for all vectors b, and A is assumed to be a linear transformation, we conclude that (A.2) is a valid representation for A. A Representation for a Linear Transformation

We can use representation theorem (A.2) to establish expressions for the transformation induced by a second-order tensor A. The result will be similar to a familiar matrix multiplication. To proceed, recall that A=

3 ² 3 ² j =1 k =1

Akj p k ⊗ pj .

The transformation induced by A on a vector b is readily computed: Ab =

⎛ ⎞ 3 ² 3 ² ⎝ Akj p k ⊗ pj ⎠ b

3 ² 3 ² =

j =1 k =1

Akj bj pk .

j =1 k=1

If we define c = Ab, then it is easy to see that the components ci ci

=

3 ²

=

A ij bj.

c · pi of c are (A.3)

j =1

Expressed in matrix notation, (A.3) should be familiar: ⎡

⎤

⎡

A11 c1 ⎣ c2 ⎦ = ⎣ A21 A31 c3

A12 A22 A32

⎤⎡

⎤

b1 A 13 ⎦ ⎣ b2 ⎦ . A 23 b3 A 33

(A.4)

492

Background on Tensors

It should be clear from (A.4) that the identity tensor has the representation I = µ ¶ ∑3 i=1 p i ⊗ pi . For familiar choices of the basis p1 , p 2, p 3 , we can use this result to show that I = E1 ⊗ E 1 + E2 ⊗ E 2 + E3 ⊗ E 3 =

er ⊗ er + eθ

=

eR ⊗ eR + eθ

⊗ eθ +

E3 ⊗ E3

⊗ eθ +

Several other examples of the representation I =

eφ

⊗ eφ .

∑3

i=1 p i ⊗

pi appear in this book.

A Representation for a Product of Two Second-Order Tensors

We now turn to the important result of the product of two second-order tensors A and B. The product AB is defined here to be a second-order tensor C. First, let A

=

3 3 ² ² i =1 k =1

Aik p i ⊗ pk ,

B=

3 3 ² ² i=1 k =1

Bik p i ⊗ pk ,

C=

3 3 ² ² i=1 k=1

Cik pi ⊗ pk .

We now solve the equations Ca = (AB)a, where a is any vector for the nine components of C. Using the arbitrariness of a, we conclude that C ik

=

3 ²

A ij Bjk .

j=1

This result is identical to that used in matrix multiplication. Indeed, if we define three matrices whose components are Cik , Aik , and Bik , then we find the representation ⎡

C11

⎢ ⎣ C21

C31

⎡

⎤

C12

C13

C22

C23 ⎦

C32

C33

⎥

=

A 11

⎢ ⎣ A 21

A 31

⎤⎡

⎤

A 12

A13

B11

B12

B 13

A 22

A23 ⎦ ⎣ B21

B22

B 23 ⎦ .

A 32

A33

B31

B32

B 33

⎥⎢

⎥

In this expression, the components of the three tensors are all expressed in the same basis. It is straightforward to establish a representation for the product BA. Finally, we consider the product of two second-order tensors a ⊗ b and c ⊗ d. Using the representation theorem for both tensors, and then taking their product, we find that (a ⊗ b)(c ⊗ d) = a ⊗ d(b · c). This result is the easiest way to remember how to multiply two second-order tensors.

A.6

A.6

Functions of Second-Order Tensors

493

Functions of Second-Order Tensors

Symmetric and Skew-Symmetric Parts of a Second-Order Tensor

The transpose AT of a second-order tensor A is defined, for all vectors a and b, as b · (Aa) = (AT b) · a.

(A.5)

If we consider the second-order tensor c ⊗ d, we can use the definition of the transpose to show that (c ⊗ d)T

=

(a ⊗ b + c ⊗ d)T

d ⊗ c,

=

b ⊗ a + d ⊗ c.

These results will prove to be very useful. Given any two second-order tensors A and B, it can be shown that (AB)T = BT AT . If A = AT , then A is said to be symmetric. In contrast, A is skew-symmetric if A = −AT . Any second-order tensor B can be decomposed into the sum of a symmetric secondorder tensor and a skew-symmetric second-order tensor: B=

) ) 1( 1( B + BT + B − BT . 2 2

Skew-symmetric tensors, which feature prominently in this book, include the angular velocity tensors. These tensors are intimately related to angular velocity vectors. The primary examples of symmetric tensors we encounter are the Euler tensors E0 and E and the inertia tensors J0 and J. We next examine representations for the symmetric and skew-symmetric parts of a second-order tensor A. Using the definition (Aa) · b = (AT b) · a, and the arbitrariness of a and b, it can be shown that AT

=

3 ² 3 ² i=1 k=1

Aki p i ⊗ pk

=

3 ² 3 ² i =1 k =1

Aik p k ⊗ pi ,

where A=

3 3 ² ² i=1 k=1

A ik pi

As a second-order tensor A is symmetric if A A

=

=

AT if Aik

⊗ pk .

AT , we find that =

Aki .

This implies that a symmetric second-order tensor has six independent components. Similarly, A is skew-symmetric if AT = −A: A

= −A

T

if Aik

= −Aki .

Notice that this result implies that a skew-symmetric second-order tensor has three independent components.

494

Background on Tensors

Invariants, Determinants, and Traces

There are three scalar quantities associated with a second-order tensor that are independent of the right-handed orthonormal basis used for E3 . Because these quantities are independent of the basis, they are known as the (principal) invariants of a secondorder tensor. Given a second-order tensor A, the invariants, IA , IIA , and III A , of A are defined as2 [Aa, b, c] + [a, Ab, c] + [a, b, Ac] = IA [a, b, c], [a, Ab, Ac] + [Aa, b, Ac] + [Aa, Ab, c] = IIA [a, b, c], [Aa, Ab, Ac] = IIIA [a, b, c], where a, b, and c are any three vectors. The first invariant is known as the trace of a tensor, and the third invariant is known as the determinant of a tensor: tr(A)

= IA ,

det(A)

=

IIIA .

(A.6)

We shall now see why this terminology is used. First, recall that [Aa, b, c] + [a, Ab, c] + [a, b, Ac] = tr(A)[a, b, c], [Aa, Ab, Ac] = det(A)[a, b, c]. If we choose a = p1 , b = p 2, and c = p 3, where {p1 , p2 , p 3} is a right-handed orthonormal basis for E3 , then we have the intermediate results [p 1, p2 , p 3] = 1, [Ap 1, p2 , p 3] + [p1 , Ap 2, p3 ] + [p 1, p2 , Ap 3] =

3 ²

(Ap i) · pi ,

i=1

[Ap 1, Ap2 , Ap 3] = det(A). Using these results, we find that the trace of A is tr(A)

3 ² =

i=1

(Ap i ) · pi

=

A11 + A22 + A33 .

A similar result holds for the trace of a matrix. Further, we find that det(A) = [Ap1 , Ap 2, Ap3 ] 3 ² 3 ² 3 ²

=

A i1A j2A k3[pi , p j , p k ]

i=1 j=1 k=1 =

3 3 ² 3 ² ² i=1 j=1 k=1

2 Our discussion here closely follows Chadwick [46].

± ijk A i1A j2 Ak3 .

(A.7)

A.6

495

Functions of Second-Order Tensors

Recall that the determinant of a 3×3 matrix whose components are B ik is ⎛⎡

B 11 det ⎝ ⎣ B 21 B 31

B12 B22 B32

⎤⎞

B13 B23 ⎦⎠ B33

3 3 ² 3 ² ² =

± ijk Bi1B j2B k3.

(A.8)

i=1 j=1 k=1

Comparing (A.7) with (A.8), we see that we can calculate the determinant of a tensor A by representing the tensor using a right-handed orthonormal basis and then using a standard result from matrices: ⎛⎡

A11 ⎝ ⎣ A21 det(A) = det A31

⎤⎞

A12 A22 A32

A13 A23 ⎦⎠ . A33

We also note in passing the useful result that tr(a ⊗ b)

=

a · b.

Inverses and Adjugates

The inverse A−1 of a second-order tensor A is the second-order tensor that satisfies A−1 A = AA−1

=

I.

For the inverse to exist, det(A) ± = 0. Taking the transpose of this equation, we find that the inverse of the transpose of A is the transpose of the inverse. The adjugate A∗ of a second-order tensor A is the second-order tensor that satisfies A∗ (a × b) = Aa × Ab, where a and b are arbitrary vectors. If A is invertible, then this definition yields A∗

=

det(A)(A−1 )T .

Notice how this result simplifies if A has a unit determinant.

Eigenvalues and Eigenvectors

The characteristic values (or eigenvalues or principal values) of a second-order tensor A are defined as the roots λ of the characteristic equation of A: det (λI − A) = 0. The three roots of this equation are denoted λ1, λ 2, and λ3 . On expanding the equation, we find that 3 2 λ − IA λ + IIA λ − III A =

0.

That is, det(A) = III A , 1 = (tr(A)2 − tr(A2)) = IIA , 2 = tr(A) = IA .

λ 1λ 2λ3 = λ 1λ 2 + λ2 λ3 + λ 1λ3 λ1 + λ2 + λ3

496

Background on Tensors

If a tensor is symmetric, then the three eigenvalues will be real. The eigenvector (or characteristic direction or principal direction) of a tensor A is the vector u that satisfies Au = λ u,

(A.9)

where λ is a root of the characteristic equation. Clearly, if u is an eigenvector, then −u is also an eigenvector. Modulo this sign, a second-order tensor has three eigenvectors. To determine these eigenvectors, we express (A.9) with the help of (A.4) and then use standard techniques from linear algebra. An explicit example of this type of calculation can be found at the end of Section 7.8 in Chapter 7. Let us now consider a simple example: J0

=

mb2 mc2 ma2 E1 ⊗ E1 + E2 ⊗ E 2 + E3 ⊗ E 3. 12 12 12

By inspection, this tensor has the eigenvalues eigenvectors are E 1, E2, and E3.

A.7

ma2 mb2 , 12 , 12

and

mc 2 12

and the corresponding

Third-Order Tensors

We shall find it necessary to use one example of a third-order tensor. A third-order tensor transforms vectors into second-order tensors and may transform second-order tensors into vectors. We can parallel all the developments for a second-order tensor that we previously performed. However, here it suffices to note that, with respect to a basis { p1 , p 2, p 3} , any third-order tensor A has the representation

A=

3 ² 3 ² 3 ² i=1 j=1 k =1

Aijk pi

⊗ pj ⊗ pk .

We also define the tensor products: (a ⊗ b ⊗ c) [d ⊗ e]

=

a(b · d)(c · e),

(a ⊗ b ⊗ c)d = a ⊗ b(c · d).

(A.10)

Other tensor products can also be defined. The presence of the brackets [·] in (A.10) should be noted. The main example of a third-order tensor we use is known as the alternator ε: ε=

3 3 ² 3 ² ² i=1 j=1 k=1

=

± ijk pi ⊗

pj

⊗ pk

p 1 ⊗ p 2 ⊗ p 3 + p3 ⊗ p1 ⊗ p2 + p 2 ⊗ p 3 ⊗ p 1 −

p2 ⊗ p1 ⊗ p3 − p 1 ⊗ p 3 ⊗ p 2 − p3 ⊗ p2 ⊗ p1 .

This tensor has some useful features. First, if A is a symmetric tensor, then it is a produc∑ tive exercise to show that ε [A] = 0. Second, suppose that c = 3k =1 ck pk is a vector; then

A.7

εc =

3 ² 3 ² 3 ²

± ijk p i ⊗

497

Third-Order Tensors

pj ck

i=1 j=1 k=1 =

c3(p 1 ⊗ p 2 − p2 ⊗ p1 ) + c2 (p3 ⊗ p1 − p 1 ⊗ p 3) + c1(p2 ⊗ p 3 − p 3 ⊗ p 2), (A.11)

which is a skew-symmetric tensor. The fact that ε acts on a vector to produce a skew-symmetric tensor enables us to define a skew-symmetric tensor C for every vector c and vice versa: 1 c = ax (C) = − ε [C] . 2

C = skwt (c) = −εc,

The vector c is known as the axial vector of C. It is a good exercise to verify that, if C has the representation C = c21(p 2 ⊗ p 1 − p1 ⊗ p2 ) + c32 (p3 ⊗ p2 − p 2 ⊗ p 3) + c13(p 1 ⊗ p 3 − p3 ⊗ p1 ), then, with the help of (A.10)1, c = ax (C) = c21 p3 + c13p 2 + c32p1 . We also note the important result that Ca = (−εc)a = c × a. This result allows us to replace cross products with tensor products, and vice versa.

Examples

It is prudent to provide some examples here. First, suppose we are given a skewsymmetric tensor Ω = ² (E 2 ⊗ E1 − E 1 ⊗ E2 ) .

Then ε [Ω]

= −2²E3 ,

and we conclude that the axial vector of Ω is ²E 3: ²E 3 =

ax (² (E2 ⊗ E1 − E1 ⊗ E2)) .

It is useful to verify that (² (E2 ⊗

E1 − E1 ⊗ E2)) a = ²E3 × a

for all vectors a. Similarly, given a vector γ E1 , we can compute that εγ E1

= γ E2 ⊗

E3 − γ E3 ⊗ E2 .

It follows from this result that γ E1 × b = (γ E3 ⊗ E 2 − γ E2 ⊗ E 3) b

498

Background on Tensors

for all vectors b. That is, γ E3 ⊗

A.8

E2 − γ E2 ⊗ E3

=

skwt (γ E1 ) .

Special Types of Second-Order Tensors

There are three types of second-order tensors that play an important role in rigid-body dynamics: proper orthogonal tensors, symmetric positive definite tensors, and skewsymmetric tensors.

Orthogonal Tensors

A second-order tensor L is said to be orthogonal if LLT = LT L = I. That is, the transpose of an orthogonal tensor is its inverse. It also follows that det(L) = ² 1. An orthogonal tensor has the unique property that La · La = a · a, and so it preserves the length of the vector that it transforms. Some examples of orthogonal tensors include I,

−I,

E 1 ⊗ E1 + E 2 ⊗ E2 − E 3 ⊗ E3 .

The last of these examples constitutes a reflection in the plane x3

=

0.

Proper Orthogonal Tensors

A second-order tensor Q is said to be proper orthogonal if QQT = QT Q = I and det(Q) = 1. Euler’s theorem states that this type of tensor is equivalent to a rotation tensor (see Section 7.2). Proper orthogonal second-order tensors are a subclass of the second-order orthogonal tensors. Indeed, it can be shown that any second-order orthogonal tensor is either a rotation tensor or can be obtained by multiplying a rotation tensor by −I. Using a result that is known as Euler’s formula, any rotation tensor can be written as Q = cos( θ )(I − p ⊗ p) − sin(θ )εp + p ⊗ p, where θ is a real number and p is a unit vector. The variable θ is known as the (counterclockwise) angle of rotation, and p is known as the axis of rotation. We examine several examples of rotation tensors in Chapter 6.

Symmetric, Positive Definite, Second-Order Tensors

A tensor A is said to be positive definite if (Aa) · a > 0 for all a ± = 0 and Aa · a = 0 if, and only if, a = 0. A consequence of the definition is that a skew-symmetric secondorder tensor can never be positive definite. Examples of positive definite tensors in mechanics include the Euler tensors E0 and E and the inertia tensors J0 and J.

A.9

499

Derivatives of Tensors

If A is positive definite, then it may be shown that all three of its eigenvalues are positive, and, furthermore, the tensor has the representation A = λ1 u1 ⊗ u1 + λ2u 2 ⊗ u 2 + λ3 u3 ⊗ u3 , where Au i = λi u i . That is, ui is an eigenvector of A with the eigenvalue representation is often known as the spectral decomposition.

A.9

λi .

This

Derivatives of Tensors

We shall often encounter derivatives of tensors. Suppose a tensor A has the representation A=

3 3 ² ² i=1 k=1

A ik pi

⊗ pk

and that the components of A and the vectors pi are functions of time. The timederivative of A is defined as ˙ = A

3 3 ² ² i=1 k=1

A˙ ik pi ⊗ pk

3 3 ² ² +

i=1 k=1

A ik p˙ i

⊗ pk +

3 3 ² ² i= 1 k =1

Aik p i ⊗ p˙ k .

Notice that we differentiate both the components and the basis vectors. For example, consider the tensor A = er ⊗ E1 + eθ

⊗ E2 +

E3 ⊗ E3 .

Then, after expressing er and eθ in terms of E1 , E2 , and E3 and differentiating, we find that ˙ = θ˙ e A θ

⊗ E1 − θ˙ er ⊗

E2 .

˙ T is a skewIt is left as an exercise to show that A is a rotation tensor, and that AA symmetric tensor. We can also define a chain rule and product rules. Suppose A = A(q(t)), B = B(t), and c = c(t). Then, ˙ A

=

∂A ∂q

q˙ ,

d ˙ ˙ (AB) = AB + AB, dt d ˙ + Ac (Ac) = Ac ˙. dt If we have a function ψ = ψ (A), then the derivative of this function with respect to A is defined to be the second-order tensor ∂ψ ∂A

=

3 3 ² ² ∂ψ i =1 k=1

∂ Aik

p i ⊗ pk .

500

Background on Tensors

In addition, if the vectors p i are constant, then ˙ = ψ

3 ² 3 ² ∂ψ i=1 k =1

∂ Aik

·

A˙ ik

=

tr

∂ψ ∂A

˙ A

T

¸

.

This result is used when we consider constraints on the motions of rigid bodies.

A.10

Exercises

Exercise A.1: Consider a = E1 and b = E2 . For any vector c (a ⊗ b)c = c2 E1 and c(a ⊗ b) = c1E 2.

=

Exercise A.2: Consider a = E1 and b = E2 . For any vector c (a ⊗ b − b ⊗ a)c = c2E1 − c1 E2 .

=

∑3

i=1 ci Ei ,

∑3

i=1 ci Ei ,

Exercise A.3: Using the definition of the transpose (A.5), verify that (a ⊗ b)T

show that show that =

b ⊗ a.

Exercise A.4: If a = 10E1 and b = 5E2 , verify that (a ⊗ b)u = 50E1 (E2 · u) and (b ⊗ a) u = 50E2 (E1 · u). Exercise A.5: If a = er and b = eθ , then show that (a ⊗ b)u = er (eθ · u) and (b ⊗ a)u = eθ (er · u). Exercise A.6: Using the definition of the alternator, show that −

εE 3 = E 2 ⊗ E1 − E 1 ⊗ E2 .

Where is this result used in rigid-body kinematics? Verify that E3 (E2 ⊗ E 1 − E1 ⊗ E 2) a for any vector a. Exercise A.7: Show that the matrix multiplication ⎤

⎡

⎡

er cos(θ ) ⎣ eθ ⎦ = ⎣ − sin(θ ) ez 0

sin(θ ) cos(θ ) 0

⎤⎡

×

a

⎤

0 E1 0 ⎦ ⎣ E2 ⎦ E3 1

can be written as er

=

PE 1,

eθ

=

PE 2,

ez = PE 3,

where P = cos( θ ) (E1 ⊗ E1 + E2 ⊗ E2) + sin(θ ) (E2 ⊗ E1 − E1 ⊗ E2) + E3 ⊗ E3 =

cos( θ )(I − E3 ⊗ E3 ) − sin(θ )εE 3 + E3 ⊗ E 3.

To show the second representation of P, it is helpful to first show that P

=

er ⊗ E1 + eθ

⊗

E2 + ez ⊗ E3 .

Finally, verify that P is a proper orthogonal tensor.

=

A.10

501

Exercises

Exercise A.8: Show that the matrix multiplication ⎡

⎤

⎡

⎤⎡

eR cos( θ ) sin(φ ) sin(θ ) sin(φ ) ⎣ eφ ⎦ = ⎣ cos(θ ) cos(φ ) sin(θ ) cos(φ ) eθ − sin(θ ) cos( θ )

⎤

cos(φ ) E1 − sin(φ ) ⎦ ⎣ E2 ⎦ 0 E3

can be written as eR

=

RE3 ,

eφ

=

RE1 ,

eθ

=

RE2 ,

where R = cos( θ ) sin( φ)E1 ⊗ E3 + sin(θ ) sin( φ)E2 ⊗ E3 + cos( φ)E3 ⊗ E3 +

cos(θ ) cos(φ )E1 ⊗ E1 + sin(θ ) cos(φ )E2 ⊗ E1 − sin(φ )E3 ⊗ E1

−

sin(θ )E1 ⊗ E2 + cos(θ )E2 ⊗ E2 .

To see this, it is helpful to first show that R = eR ⊗ E 3 + eφ ⊗ E 1 + eθ

⊗

E2.

Exercise A.9: Verify that the tensor R in the previous exercise is proper orthogonal. Exercise A.10: Give an example that illustrates that tensor multiplication is not commutative (i.e., AB ± = BA). Exercise A.11: Verify that, if B is a second-order tensor, then ax (B) =

) 1 ( ax B − BT . 2

Exercise A.12: Verify that 2 ax (c ⊗ d ) = ax (c ⊗ d − d ⊗ c) = d × c, where c and d are any two vectors. Exercise A.13: Suppose that A and B are skew-symmetric second-order tensors. Verify that ) 1 ( a · b = tr ABT , (A.12) 2 where a and b are the axial vectors of A and B, respectively. Exercise A.14: Suppose that A is a skew-symmetric second-order tensor, a is its axial vector, and C is a symmetric second-order tensor. Verify that ax (AC + CA) = (tr (C) I − C) a. This identity can be used to establish an expression for the angular momentum H of a rigid body in terms of its Euler tensor E and angular velocity vector ω: H = (tr(E)I − E) ω.

502

Background on Tensors

Exercise A.15: Let A be a tensor and b and c be any two vectors. Show that (

)

tr(A)I − AT (b × c) = Ab × c + b × Ac.

Hint: Perhaps the quickest way to prove this result is to first argue that it suffices to pick b = E3 and c = c1 E1 + c2 E2 . Substituting into the identity then yields an easy way to verify the result.

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Index

action line element, 101 Agnesi, M. G., 130 aircraft stability, 235 alternating symbol, 488 alternator, 488, 496 angular acceleration vector, 261 angular momentum particle, 5 rigid body relative to a fixed point O, 269 relative to a point A, 269 relative to center of mass, 269, 276 system of particles, 137 angular velocity tensor, 208 vector, 208, 241, 243, 252, 261 Appell, P., 409 areal velocity, 5, 35, 57, 77 atlas, 97 Atwood’s machine, 196 axial vector, 497 balance of angular momentum rigid body, 329 balance of linear momentum particle, 42, 104 rigid body, 329 barycentric coordinate system, 169 Bernoulli’s equation, 344 Bernoulli, J., 164 Boltzmann, L., 28 Boltzmann–Hamel equations, 417 Bryan angles, 235 Bryan, G. H., 227 cancellation of the dots, 396 Cardan angles, 235 Cardano, G., 227 Cayley, A., 252 center of mass rigid body, 267 system of particles, 139 center of oscillation, 355 changing coordinates, 396 Chaos, 187

Chaplygin integral, 372 Chaplygin sphere, 355, 364 Chaplygin, S. A., 355, 364, 409, 420 chart, 97 Chasles, M., 260 Christoffel symbols, 121, 158 components contravariant, 13 covariant, 13 configuration manifold, 99, 178, 187, 383, 407 configuration space, 148 configurations of a rigid body, 257 connection coefficients, 121, 158, 226 conservation angular momentum, 54, 60, 187, 351, 393, 414, 424, 438 energy, 54, 55, 187, 332, 351, 393, 400, 409, 414, 424, 438 energy-like quantity, 56, 126, 128, 133 Jacobi integral, 56, 126, 133 linear momentum, 54 constraints, 87 ball-and-socket joints, 440, 456 Chaplygin, 420 holonomic, 28, 108 ideal, 118, 158, 308, 379 impulsive, 68 independence, 34 integrability, 32, 310, 445 integrable, 22, 48, 87, 140, 348, 407, 412, 439 rheonomic, 28, 114 scleronomic, 28, 108 multiple, 34, 143, 311, 326, 461, 463 nonholonomic, 28 nonintegrable, 28, 48, 89, 140, 142, 143, 310, 445, 456, 463 one-parameter family of, 463 piecewise integrable, 29 pin joints, 301, 440, 455 positional, 28 revolute joints, 306 rigid bodies, 301, 439 rolling rigid bodies, 304, 441, 456, 457, 461, 484 sliding rigid bodies, 304, 441

Index

static Coulomb friction, 52, 67, 73, 107, 111, 113, 123, 306, 461 system of particles, 151 systems of, 34, 143, 311, 326 velocity, 28 contravariant basis vectors, 11, 149 coordinate curve, 10 coordinate singularities, 21, 85, 88, 92, 95, 101, 121, 153, 157, 181, 197, 198, 382, 384, 385, 407, 426, 435, 438, 468, 473 coordinate surface, 10 coordinates Cartesian, 6, 15, 139, 425, 427 curvilinear, 9, 149 cylindrical polar, 6, 15, 35 generalized, 99, 102, 178–180, 383, 407 helix, 37, 109, 124 parabolic, 36, 119, 127 spherical polar, 7, 15, 36, 123, 139, 390, 423, 436 Coriolis, G. G., 352 corotational basis, 262 derivative, 209, 220, 288 tensor, 288 vector, 264, 268, 288 Coulomb’s prescription, 49, 51, 52, 306, 308 covariant basis vectors, 10, 149 curvilinear coordinates, 97 D’Alembert’s principle, 164 D’Alembert, J. Le Rond, 83, 164 degrees of freedom, 99, 104, 407, 464 dual Euler basis, 226, 230, 233, 250, 317, 320, 406 dual-spin spacecraft, 431 dumbbell satellite, 176 Dynabee, 431, 484 Euler angles 1–2–3 set, 236, 391, 423 1–3–1 set, 255, 322, 480 3–1–2 set, 295, 484 3–1–3 set, 232, 304, 306, 321, 323–326, 375, 380, 403, 411, 421, 422, 481, 483, 484, 486 3–2–1 set, 227, 255, 256, 297, 425, 427, 436, 457, 462, 480 3–2–3 set, 249, 292, 302, 420 angular velocity vector, 223 asymmetric set, 235 body-fixed, 256 extending range, 303, 420 range, 225 singularities, 226, 230, 233, 234, 290, 322 space-fixed, 256 switching sets of, 341 symmetric set, 235 twelve sets, 234 Euler basis, 224

523

Euler tensors, 270, 498 Euler’s disk, 364 Euler’s equations, 336 Euler, L., 43, 211, 255, 258, 371 Euler–Lagrange equations, 165 Euler–Poisson equations, 350 Euler–Rodrigues symmetric parameters, 242 Fick, A. E., 227, 234 forces central, 77 conservative, 45, 144, 151, 313, 321, 399, 450, 453 constraint, 48, 142, 151, 301, 307, 349, 397, 406, 447, 454, 459, 465 drag, 342 friction, 107, 111, 123, 131 generalized, 105, 145, 161, 394, 425, 454 generalized impulsive, 482 gravitational, 46, 169, 176, 315, 392 Magnus, 342, 403 normal, 107, 111, 123, 131 orthogonality condition, 454, 477 spring, 46 Forsyth, A. R., 31 Foucault, L., 3, 471 Frenet triad, 71, 110, 130 Frisbee, 347 Frobenius integrability theorem of, 34, 311, 326, 447 Frobenius, F. G., 32, 311, 447 Galileo spacecraft, 431 Gauss’ principle of least constraint, 164 Gauss, C. F., 164, 241 geodesics, 101 geostationary orbit, 374 Gibbs, J. W., 211 Gibbs–Appell equations, 417 gimbal lock, 473 Goursat, E. J. B., 31 gradient, 12, 24 gravity gradient torque, 315, 373 Griffin grinding machine, 326 gyroscope, 470 Hamilton, Sir W. R., 83, 164, 188, 241, 252 harmonic oscillator, 170 helicoid, 124 Helmholtz, H. L. F. von, 234 Hertz, H., 28, 99 Hoberman sphere, 241 impact of a rigid body, 482 impulse–momentum, 42, 482 inertia tensors, 271, 498 change of bases, 274, 293

524

Index

parallel axis theorem, 293, 348, 358, 457 principal axes of inertia, 272, 274, 371 Jacobi integral, 198 Jacobi’s criterion, 32, 420 Jacobi’s principle of least action, 101 Jacobi, C. G. J., 32, 101, 165, 336 Jellett integral, 371 Jellett, J. H., 371 joint coordinate systems, 235, 448 jumping, 364 Kelvin, Lord, 486 Kepler frequency, 61, 367, 374 Kepler’s Laws, 57 Kepler, J., 57 kinematical line element, 101, 157, 178, 180, 187, 383 kinetic energy Koenig decomposition, 277, 387 particle, 5 rigid body, 277 system of particles, 137 Koenig, J. S., 277 Korteweg, D. J., 409 Kronecker delta, 488 Lagrange multipliers, 164 Lagrange top, 351, 380, 411 Lagrange’s equations of motion rigid body, 387, 393, 401 single particle, 83 contravariant form, 121 covariant form, 84, 120 system of particles, 147, 155 contravariant form, 158, 161 covariant form, 161 Lagrange’s prescription, 49, 51, 52, 140, 142, 151, 164, 307, 397, 447, 454, 459, 465, 476, 486 Lagrange’s principle, 49 Lagrange, J. L., 48, 57, 83, 137, 188, 232, 367, 373, 379, 380, 411 Lagrange–D’Alembert principle, 49 Lagrangian, 85, 154 Lagrangian reduction, 59 Levi-Civita, T., 99, 120 line of nodes, 232 linear momentum particle, 5 rigid body, 268 system of particles, 139 Littlewood, J. E., 375 MacCullagh, J., 317 Magnus, H. G., 344 manifold, 97 Riemannian, 99

mass matrix, 21, 85, 86, 88, 92, 95, 101, 153, 157, 181, 197, 198, 382, 384, 385, 407, 410, 426, 435, 438, 468 Maxwell, J. C., 339 metric tensor, 11 moment potential, 248, 313, 319, 324, 450 moment-free motion of a rigid body, 336, 360 moments conservative, 313, 399, 450, 453 constant, 317 constraint, 301, 307, 349, 397, 406, 447, 454, 459, 465 frictional, 305 gravitational, 315, 392 torsional spring, 324 momentum sphere, 339 motion capture, 262 motion of a rigid body with a fixed point, 347, 372, 424 motor torque, 439 n-body problem, 188, 198 navigation, 242 navigation equations, 296 Newton’s third law, 144, 145, 448 Newton, Sir I., 43, 57, 342 Nondimensionalization, 67 nondimensionalization, 61 Papastavridis, J. G., 29 Pars, L. A., 31 pendulum and cart, 179 phase portrait, 61 planar double pendulum, 193, 433 Poincaré, H., 188 Poinsot, L., 264, 336, 339 Poisson kinematical relations, 350 Poisson top, 297, 304, 411 Poisson, S. D., 264, 411 principal axes, 272, 335, 371 principal moments of inertia, 272 principle of least action, 101 principle of virtual work, 164 Procrustes problem, 243 quaternions, 241 real projective line RP1 , 243 real projective plane RP2 , 243, 286 real projective space RP3 , 243, 383 reference frame, 3 corotational, 268 inertial, 3 relative angular velocity tensor, 219 vector, 219, 225, 440, 458, 471 representative particle, 99, 147, 159

Index

Ricci, M. M. G., 99, 120 Riemann, G. F. B., 99 rigid-body motion, 257 robotic arm, 483 Rodrigues, O., 240, 241 roller coaster, 70 Rollerball (Powerball), 431, 484 rolling disk, 290, 325, 326, 403 rolling sphere, 352, 366, 421 cylinder, 375 turntable, 368 rotation angle of, 214 axis of, 214, 264 Chasles’ theorem, 260 Codman’s paradox, 255, 486 composition of, 212, 261 Rodrigues formula, 243, 252 compound, 221, 238 constant angular velocity motions, 288 direction cosines, 208 Euler parameters, 241, 279 Euler’s formula, 212 Euler’s representation, 211 angular velocity vector, 216 Euler’s theorem, 217, 258 Euler–Rodrigues symmetric parameters, 241, 252, 279 fixed-axis case, 204 Gibbs vector, 240 instantaneous axis of, 264 matrix representation, 214, 229, 238, 241 quaternions, 241, 252, 279 relative, 443 Rodrigues vector, 240 Rodrigues–Hamilton theorem, 255, 317 screw axis of, 264 rotation vector φ r, 245 Routh integral, 372 Routh, E. J., 83, 352 Routh–Voss equations of motion, 83, 90 Routhian reduction, 59 Ruina, A., 29

spherical robot, 375, 456 spinning eggs, 364 stability of motion, 339 static friction criterion, 111, 116 steady orbital motions, 368, 392 Steiner’s surface, 246, 286 Steiner, J., 286 stick–slip, 73, 174 strap-down inertial navigation system, 296 surface of revolution, 90, 119, 124, 127, 129

satellite dynamics, 193, 367, 390, 423 screw motion, 260 Segner, J. A., 338 simple pendulum, 82, 470 Slerp algorithm, 242, 279 sliding cylinder, 302, 420 sliding disk, 290, 304, 403 sliding sphere, 352, 421 slip velocity, 355 SO(3), 243, 383 spectral decomposition theorem, 498 spherical pendulum, 23, 79, 163

vehicle dynamics, 142, 420 Voss, A., 83

Tait angles, 235 Tait, P. G., 227, 342, 486 tensor product second-order and vector, 492 third-order and second-order, 496 third-order and vector, 496 two second-order tensors, 490, 492 two vectors, 489 tensors adjugates, 495 angular velocity, 208 derivatives, 499 determinant, 493 eigenvalues, 495 eigenvectors, 495 invariants, 493 inverses, 495 orthogonal, 498 proper orthogonal, 206, 498 representation theorem for, 490 rotation, 211 second-order, 489 skew-symmetric, 493 symmetric, 493 symmetric positive definite, 498 third-order, 496 trace, 493 three-body problem, 187 tippe top, 321, 371 two-body problem, 57, 169 universal joint, 326

Wahba problem, 242 Walker, G. T., 306 wobblestone, 306, 364 work–energy theorem rigid body, 331 single particle, 55 system of particles, 138, 186 Ziegler, H., 313, 317

525