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P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 This page intentionally left blank ii 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 INTERMEDIATE DYNAMICS FOR ENGINEERS This book has sufficient material for two full-length semester courses in intermediate engineering dynamics. For the first course a Newton– Euler approach is used, followed by a Lagrangian approach in the second. Using some ideas from differential geometry, the equivalence of these two approaches is illuminated throughout the text. In addition, this book contains comprehensive treatments of the kinematics and dynamics of particles and rigid bodies. The subject matter is illuminated by numerous highly structured examples and exercises featuring a wide range of applications and numerical simulations. Oliver M. O’Reilly is a professor of mechanical engineering at the University of California, Berkeley. His research interests lie in continuum mechanics and nonlinear dynamics, specifically in the dynamics of rigid bodies and particles, Cosserat and directed continuua, dynamics of rods, history of mechanics, and vehicle dynamics. O’Reilly is the author of more than 50 archival publications and Engineering Dynamics: A Primer. He is also the recipient of the University of California at Berkeley’s Distinguished Teaching Award and three departmental teaching awards. i 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 ii 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Intermediate Dynamics for Engineers A UNIFIED TREATMENT OF NEWTON–EULER AND LAGRANGIAN MECHANICS Oliver M. O’Reilly University of California, Berkeley iii 21:29 CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521874830 © Oliver M. O’Reilly 2008 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2008 ISBN-13 978-0-511-42336-9 eBook (EBL) ISBN-13 hardback 978-0-521-87483-0 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. P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 This book is dedicated to my adventurous daughter, Anna v 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 vi 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Contents Preface page xi PART ONE DYNAM ICS OF A SINGLE PARTICLE 1 1 Kinematics of a Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 Introduction Reference Frames Kinematics of a Particle Frequently Used Coordinate Systems Curvilinear Coordinates Representations of Particle Kinematics Constraints Classification of Constraints Closing Comments Exercises 3 3 5 6 9 14 15 20 27 27 2 Kinetics of a Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 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 33 33 35 36 37 39 45 47 55 59 60 64 66 vii 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 viii June 9, 2008 Contents 3 Lagrange’s Equations of Motion for a Single Particle . . . . . . . . . . . . 70 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 Introduction Lagrange’s Equations of Motion Equations of Motion for an Unconstrained Particle Lagrange’s Equations in the Presence of Constraints A Particle Moving on a Sphere Some Elements of Geometry and Particle Kinematics The Geometry of Lagrange’s Equations of Motion A Particle Moving on a Helix Summary Exercises PART TWO DYNAM ICS OF A SYSTEM OF PARTICLES 70 71 73 74 78 80 83 87 91 92 101 4 The Equations of Motion for a System of Particles . . . . . . . . . . . . . 103 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 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 The Lagrangian A Constrained System of Particles A Canonical Form of Lagrange’s Equations Alternative Principles of Mechanics Closing Remarks Exercises 103 104 105 107 110 111 113 118 119 122 128 131 131 5 Dynamics of Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . . 134 5.1 5.2 5.3 5.4 5.5 5.6 Introduction Harmonic Oscillators A Dumbbell Satellite A Pendulum and a Cart Two Particles Tethered by an Inextensible String Closing Comments Exercises PART THREE DYNAM ICS OF A SINGLE RIGID BODY 134 134 140 143 147 151 153 161 6 Rotation Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 6.1 6.2 6.3 6.4 Introduction The Simplest Rotation Proper-Orthogonal Tensors Derivatives of a Proper-Orthogonal Tensor 163 164 166 168 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Contents ix 6.5 6.6 Euler’s Representation of a Rotation Tensor Euler’s Theorem: Rotation Tensors and Proper-Orthogonal Tensors 6.7 Relative Angular Velocity Vectors 6.8 Euler Angles 6.9 Further Representations of a Rotation Tensor 6.10 Derivatives of Scalar Functions of Rotation Tensors Exercises 171 176 178 181 191 195 198 7 Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 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 Concluding Remarks Exercises 206 206 211 212 213 215 218 219 223 224 226 226 8 Constraints on and Potentials for Rigid Bodies . . . . . . . . . . . . . . . 237 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 Introduction Constraints A Canonical Function Integrability Criteria Forces and Moments Acting on a Rigid Body Constraint Forces and Constraint Moments Potential Energies and Conservative Forces and Moments Concluding Comments Exercises 237 237 241 243 247 248 256 262 263 9 Kinetics of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 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 Closing Comments Exercises 272 272 274 276 279 285 289 294 297 299 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 x June 9, 2008 Contents 10 Lagrange’s Equations of Motion for a Single Rigid Body . . . . . . . . . 307 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 Introduction Configuration Manifold of an Unconstrained Rigid Body Lagrange’s Equations of Motion: A First Form A Satellite Problem Lagrange’s Equations of Motion: A Second Form Lagrange’s Equations of Motion: Approach II Rolling Disks and Sliding Disks Lagrange and Poisson Tops Closing Comments Exercises PART FOUR SYSTEM S OF RIGID BODIES 307 308 311 315 318 324 325 331 336 336 345 11 Introduction to Multibody Systems . . . . . . . . . . . . . . . . . . . . . . 347 11.1 11.2 11.3 11.4 11.5 Introduction Balance Laws and Lagrange’s Equations of Motion Two Pin-Jointed Rigid Bodies A Single-Axis Rate Gyroscope Closing Comments Exercises 347 347 349 351 355 355 APPENDIX: BACKGROUND ON TENSORS . . . . . . . . . . . . . . . . . . . . . . 362 A.1 A.2 A.3 A.4 A.5 A.6 A.7 A.8 A.9 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 362 362 363 364 364 367 370 372 373 374 Bibliography 377 Index 389 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Preface The writing of this book started more than 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 [205, 207, 208] of Synge and his co-workers 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 [27, 28]. Shades of this result could be seen in an earlier work by Greenwood [79], 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 [169] and Rosenberg [182]. 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 possible throughout this book and have attempted to cite up-to-date references to them and related systems as far as feasible. However, I have not approached the exhaustive treatments by Papastavridis xi 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 xii June 9, 2008 Preface [169] nor its classical counterpart by Routh [184, 185]. 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. Using This Text This book has been written so that it provides sufficient material for two full-length semester courses in 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: 1. Kinematics of a Particle (Section 1.5 can be omitted) 2. Kinetics of a Particle Appendix on Tensors 6. Rotation Tensors 7. Kinematics of Rigid Bodies 8. Constraints on and Potentials for Rigid Bodies 9. Kinetics of a Rigid Body 11. Multibody Systems The second course, in which a Lagrangian approach is used, could be based on the following chapters: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Kinematics of a Particle Kinetics of a Particle Lagrange’s Equations of Motion for a Single Particle Lagrange’s Equations of Motion for a System of Particles Dynamics of Systems of Particles Appendix on Tensors Rotation Tensors (with particular emphasis on Section 6.8) Kinematics of Rigid Bodies Constraints on and Potentials for Rigid Bodies Kinetics of a Rigid Body Lagrange’s Equations of Motion for a Single Rigid Body Multibody Systems 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 Matlab or Mathematica. Completing these exercises is invaluable both in terms of 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 Preface 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 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 4.6 in Subsection 4.11.3 is from the Royal Irish Academy in Dublin, Ireland. I am grateful to Pauric Dempsey, the Head of Communications and 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, Engineering Mechanics III, and ME175, Intermediate 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 undergraduate 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, Todd Lauderdale, Henry Lopez, David Moody, Tom Nordenholz, Jeun Jye Ong, 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 9.1 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 [118] 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 June 9, 2008 xiii 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly xiv 978 0 521 87483 0 June 9, 2008 Preface 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 more than 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. 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 of Victoria Danahy. Despite the assistance of several other 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 PART ONE DYNAMICS OF A SINGLE PARTICLE 1 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 2 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 1 978 0 521 87483 0 June 2, 2008 Kinematics of a Particle 1.1 Introduction One of the main goals of this book is to enable the reader to take a physical system, model it by using particles or 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 in place, the future conclusions on which they are based either do not hold up or lack conviction. Much of the material presented in this chapter will be repeatedly used 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 its 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 motion of particles. In earlier dynamics courses, these topics are intimately related to judicious choices of coordinate systems to solve particle problems. For such problems, a constraint was 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 they 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 on 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. 3 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 4 June 2, 2008 Kinematics of a Particle Path of the particle m A v Figure 1.1. The path of a particle moving in E3 . The position vector, velocity vector, and areal velocity vector of this particle at time t and the position vector of the particle at time t + t are shown. r(t) r(t + t) O 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 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), is 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.∗ 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 this point. 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 [80]. 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 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 E3 . ∗ Discussions of his experiment and their interpretation can be found in [62, 138, 207]. Among his other contributions [215], Foucault is also credited with introducing the term “gyroscope.” 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.3 Kinematics of a Particle 5 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 to be the respective first and second time derivatives of the position vector: v= dr , dt a= dv d2 r = 2. dt dt 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: r (t + t) − r(t) . t→0 t v(t) = lim We also use an overdot to denote the time derivative: v = ṙ and a = r̈. Supplementary to the aforementioned kinematical quantities, we also 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., Moulton [150]). To establish an expression for this vector, we consider the position vector of the particle at time 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)×r(t+t) as t → 0 and using the fact that r(t) × r(t) = 0, we arrive at 2t an expression for the areal velocity vector A (t): r(t) × r (t + t) t→0 2t r (t + t) 1 = r(t) × lim t→0 2 t r (t + t) − r (t) 1 . = r(t) × lim t→0 2 t A (t) = lim That is, A= 1 r × v. 2 (1.1) 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 6 Kinematics of a Particle 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: 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 Frequently Used 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 bi Ei . i=1 For the position vector r, we also have r= 3 xi Ei , i=1 where {x1 , x2 , x3 } are the Cartesian coordinates of the particle. Because Ei are fixed in both magnitude and direction, their time derivatives are zero: Ėi = 0. Cylindrical Polar Coordinates A cylindrical polar coordinate system {r, θ, z} can be defined by a Cartesian coordinate system as follows: x2 , z = x3 , r = x21 + x22 , θ = tan−1 x1 where θ ∈ [0, 2π). Provided r = 0, then we can invert these relations to find that x1 = r cos(θ), x2 = r sin(θ), x3 = z. In other words, given (x1 , x2 , x3 ), a unique (r, θ, z) exists provided (x1 , x2 ) = (0, 0). Otherwise, when r = 0, the coordinate θ is ambiguous. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.4 Frequently Used Coordinate Systems 7 E3 r z O Figure 1.2. Cylindrical polar coordinates r, θ, and z. r E2 eθ θ E1 er Given a position vector r, we can write r = x1 E1 + 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)E2 − sin(θ)E1 , ez = E3 . We also notice that ėr = θ̇eθ , whereas ėθ = −θ̇er . We should also verify that {er , eθ , Ez} is a right-handed orthonormal basis for E3 .∗ Spherical Polar Coordinates A spherical polar coordinate system {R, φ, θ} can be defined by a Cartesian coordinate system as follows: ⎛ ⎞ 2 2 + x x x2 2 1 ⎠, R = x21 + x22 + x23 , , φ = tan−1 ⎝ θ = tan−1 x1 x3 where θ ∈ [0, 2π) and φ ∈ (0, π). Provided φ = 0 or π, we can invert these relations to find x1 = R cos(θ) sin(φ), x2 = R sin(θ) sin(φ), x3 = R cos(φ). Given a position vector r, we can now write r = x1 E1 + x2 E2 + x3 E3 = R sin(φ)(cos(θ)E1 + sin(θ)E2 ) + R cos(φ)E3 = ReR , where, as shown in Figure 1.3, eR = sin(φ) cos(θ)E1 + sin(φ) sin(θ)E2 + cos(φ)E3 . ∗ A basis {p1 , p2 , p3 } is right-handed if p3 · (p1 × 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8 Kinematics of a Particle eR E3 r Figure 1.3. The spherical polar coordinates φ and θ. eφ φ O E2 r eθ θ E1 For future purposes, it is convenient to define the right-handed orthonormal set of vectors {eR , eφ , eθ }: ⎡ eR ⎤ ⎡ cos(θ) sin(φ) sin(θ) sin(φ) cos(φ) − sin(θ) cos(θ) 0 ⎤⎡ E1 ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ eφ ⎦ = ⎣cos(θ) cos(φ) sin(θ) cos(φ) − sin(φ)⎦ ⎣E2 ⎦ . eθ E3 To establish the relations between these vectors and those defined earlier, we first calculate the intermediate relations ⎡ ⎤ ⎡ ⎤⎡ ⎤ er cos(θ) sin(θ) 0 E1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ eθ ⎦ = ⎣− sin(θ) cos(θ) 0⎦ ⎣E2 ⎦ , E3 0 ⎡ ⎤ ⎡ sin(φ) eR ⎢ ⎥ ⎢ ⎣ eφ ⎦ = ⎣cos(φ) eθ 0 0 0 0 1 E3 ⎤⎡ ⎤ cos(φ) er ⎥⎢ ⎥ − sin(φ)⎦ ⎣ eθ ⎦ . 1 (1.2) E3 0 These results enable us to transform among the three distinct sets of basis vectors. 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 ⎡ ⎤ ⎡ ⎤⎡ ⎤ ėR 0 φ̇ θ̇ sin(φ) eR ⎢ ⎥ ⎢ ⎥⎢ ⎥ (1.3) 0 θ̇ cos(φ)⎦ ⎣ eφ ⎦ . ⎣ ėφ ⎦ = ⎣ −φ̇ ėθ −θ̇ sin(φ) −θ̇ cos(φ) 0 eθ 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 should allow us to verify (1.3) rather easily. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.5 Curvilinear Coordinates 9 1.5 Curvilinear Coordinates The preceeding examples of coordinate systems can be considered as specific examples of a curvilinear coordinate system. The development of the vector calculus associated with such a 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 classical 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 that of Simmonds [198]; [139, 201] are also recommended. Consider a curvilinear coordinate system {q1 , q2 , q3 } that is defined by the functions q1 = q̂1 (x1 , x2 , x3 ) , q2 = q̂2 (x1 , x2 , x3 ) , q3 = q̂3 (x1 , x2 , x3 ) . (1.4) We assume that the functions q̂i are locally invertible: x1 = x̂1 q1 , q2 , q3 , x2 = x̂2 q1 , q2 , q3 , x3 = x̂3 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 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 = q̂1 (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 1.4). 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 j q 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 10 Kinematics of a Particle q3 coordinate curve q2 coordinate curve a1 a3 S a2 q1 coordinate surface O Figure 1.4. 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 . 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 x̂i q1 , q2 , q3 Ei . i=1 It is also convenient to define the covariant basis vectors a1 , a2 , and a3 : ai = = ∂r ∂qi 3 ∂ x̂k k=1 ∂qi Ek. 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. You should notice that we can express the relationship between the covariant basis vectors and the Cartesian basis vectors in a matrix form: ⎡ ⎤ ⎡ ∂ x̂1 ∂ x̂2 ∂ x̂3 ⎤ ⎡ ⎤ a1 E1 ∂q1 ∂q1 ∂q1 ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ∂ x̂ ∂ x̂ ∂ x̂ 3⎥ 1 2 ⎣a2 ⎦ = ⎢ ⎣ ∂q2 ∂q2 ∂q2 ⎦ ⎣E2 ⎦ . a3 ∂ x̂1 ∂q3 ∂ x̂2 ∂q3 ∂ x̂3 ∂q3 E3 It is a good exercise to write out the matrix in the preceding equation for various examples of curvilinear coordinate systems, for instance, cylindrical polar coordinates. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.5 Curvilinear Coordinates 11 Contravariant Basis Vectors Curvilinear coordinate systems also have a second set of associated basis vectors: {a1 , a2 , a3 }. These vectors are known as the contravariant basis vectors. One method of defining them is as follows: a1 = 3 ∂ q̂1 i=1 ∂xi a2 = Ei , 3 ∂ q̂2 i=1 ∂xi a3 = Ei , 3 ∂ q̂3 i=1 ∂xi Ei . That is, ak = ∇qk. 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 they form an orthonormal basis for E3 . Using the chain rule of calculus, we can show that ai · a j = δij , where δij is the Kronecker delta. As discussed in the Appendix, δij = 1 if i = j and is 0 otherwise. It is left as an exercise for the reader to show this result.∗ 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 bi ai = i=1 3 bkak. k=1 The components bi are known as the contravariant components, and the components bk are known as the covariant components: b · ai = 3 k=1 b · ai = 3 bkak · ai = 3 bkδki = bi , k=1 bkak · ai = k=1 3 bkδik = bi . k=1 It is very important to note that bk = b · ak in general because ai · ak is not necessarily equal to δik. The trivial case in which xi = qi deserves particular mention. For this case, 3 r = k=1 xi Ei . Consequently, ai = Ei . In addition, ai = Ei , and the covariant and contravariant basis vectors are equal. ∗ The starting point for this exercise is to note that ∂xk ∂x j j = δk . 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 12 Kinematics of a Particle y q =3 q2 = −4 q2 = 2 1 2 q =1 1 q = −1 1 −4 q1 = −3 6 x q2 = 4 −4 Figure 1.5. Projections of the q1 and q2 coordinate surfaces for coordinate system (1.6) on the x–y plane. An Example Although we have met three examples of curvilinear coordinate systems previously, it is useful to introduce an example that features nonorthogonal basis vectors. Consider the following coordinate system for Euclidean three-space: q1 = y, q2 = x − y2 , q3 = z. (1.6) Here, x = x1 , y = x2 , and z = x3 are Cartesian coordinates. Representative projections of the coordinate surfaces for q1 and q2 are shown in Figure 1.5. For this coordinate system, it is straightforward to invert (1.6) to see that x = 2 q2 + q1 and y = q1 . Thus, 2 r = q2 + q1 E1 + q1 E2 + q3 E3 . By taking the derivatives of this representation for r with respect to q1 , q2 , and q3 , we see that a1 = 2q1 E1 + E2 , a2 = E1 , a3 = E3 . This set of vectors comprises the covariant basis vectors. By taking the gradient of qi , we find the contravariant basis vectors: a1 = E2 , a2 = E1 − 2q1 E2 , a3 = E3 . It is interesting to note that a1 · a2 = −2q1 = 0. Further, a1 and a2 do not necessarily have unit magnitudes. By way of illustration, a q1 coordinate curve is shown in Figure 1.6. The vector a1 is tangent to this curve, and a2 and a3 are normal to this curve. To emphasize that a1 is not necessarily parallel to a1 , a q1 coordinate surface is also shown in the figure. It is left as an exercise for the reader to illustrate the tangent vectors a2 and a3 to the q1 coordinate surface shown in the figure. Some Comments on Derivatives Several partial derivatives of functions (q1 , q2 , q3 , q̇1 , q̇2 , q̇3 , t) play a prominent role in this book. When taking the partial derivative of this function with respect 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.5 Curvilinear Coordinates 13 a1 a1 q1 coordinate surface a2 a1 q1 coordinate curve a3 Figure 1.6. A q1 coordinate curve and a q1 coordinate surface showing representative examples of normal (a2 and a3 ) vectors and tangent (a1 ) vectors to the curve. Note that a1 is normal to the q1 coordinate surface and a1 is not parallel to a1 . to q2 say, we assume that t, q1 , q3 , and q̇k are constant. A related remark holds for the partial derivatives with respect to the velocities q̇j and time t. That is, ∂qk = δkj, ∂qj ∂ q̇k = 0, ∂qj ∂t = 0, ∂qj (1.7) ∂qk = 0, ∂ q̇j ∂ q̇k = δkj, ∂ q̇j ∂t = 0. ∂ q̇j (1.8) and In all these equations, j and k range from 1 to 3. You may have noticed that (1.7)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 q̇i are functions of time, whereas the latter assumes that they are constant: ˙ = d ∂ dqi ∂ d2 qk ∂ = + . + dt ∂qi dt ∂ q̇k dt2 ∂t 3 3 i=1 k=1 For example, consider the function = q1 + (q̇3 )2 + 10t. Then, ∂ = 1, ∂q1 ∂ = 2q̇3 , ∂ q̇3 ∂ = 10, ∂t ˙ = It should be clear from this example that ∂ . ∂t ˙ = q̇1 + 2q̇3 q̈3 + 10. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 14 Kinematics of a Particle 1.6 Representations of Particle Kinematics We now turn to establishing expressions for the position, velocity, and acceleration vectors of a particle in terms of the coordinate systems just mentioned. First, for the position vector we have∗ r = x1 E1 + x2 E2 + x3 E3 = rer + zE3 = ReR = 3 x̂i q1 , q2 , q3 Ei . i=1 Differentiating these expressions, we find v = ẋ1 E1 + ẋ2 E2 + ẋ3 E3 = ṙer + rθ̇eθ + żE3 = ṘeR + Rφ̇eφ + R sin(φ)θ̇eθ = 3 q̇i ai . (1.9) i=1 Notice the simplicity of the expression for v when expressed in terms of the covariant basis vectors. For any given curvilinear coordinate system, if we write the position vector as a function of the coordinates q1 , q2 , and q3 , and then differentiate and compare the result with v = 3i=1 q̇i ai , we can read off the covariant basis vectors. For instance, (1.9)2 implies that a1 = er , a2 = reθ , and a3 = E3 for the cylindrical polar coordinate system. A further differentiation yields a = ẍ1 E1 + ẍ2 E2 + ẍ3 E3 = (r̈ − rθ̇2 )er + (rθ̈ + 2ṙθ̇)eθ + z̈E3 = (R̈ − Rφ̇2 − R sin2 (φ)θ̇2 )eR + (Rφ̈ + 2Ṙφ̇ − R sin(φ) cos(φ)θ̇2 )eφ + (R sin(φ)θ̈ + 2Ṙθ̇ sin(φ) + 2Rθ̇φ̇ cos(φ))eθ = 3 i=1 q̈i ai + 3 3 i=1 j=1 q̇i q̇j ∂ai . ∂qj We obtain the final representation for r̈ after noting that ai depend on the curvilinear k i coordinates, which in turn are functions of time: ȧk = 3i=1 ∂a q̇ . ∂qi It is left as an exercise for the reader to establish expressions, using various coordinate systems, for the linear momentum G and the angular momentum HO. ∗ Notice that it is a mistake to assume that r = 3 i=1 qi ai . 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.7 Constraints 15 The kinetic energy T of the particle has a rather elegant representation using the curvilinear coordinates: m T = v·v 2 3 3 m i k = q̇ ai · q̇ ak 2 i=1 = k=1 3 3 m i=1 k=1 2 aikq̇i q̇k, (1.10) where aik = aki = ak · ai . It is also a good exercise to compute aik for a spherical polar coordinate system, and then, with the help of the representation T = 3i=1 3k=1 m2 aikq̇i q̇k, show that m 2 T= Ṙ + R2 φ̇2 + R2 sin2 (φ)θ̇2 . 2 The exercises at the end of this chapter feature this result for other coordinate systems. 1.7 Constraints A constraint is a kinematical restriction on the motion of the particle. They are introduced in problems involving a particle in three manners: either as simplifying assumptions, prescribed motions, or because of rigid connections. The constraints on 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. Classical Examples Consider the four mechanical systems shown in Figure 1.7. The first system is known as the spherical pendulum. Here, a particle of mass m is attached by a rigid rod of length L0 to a fixed point O. The constraint on the motion of the particle in this system can be written as r · eR = L0 . By differentiating this equation, 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 L0 to a fixed point O, but it is also assumed to move on a vertical plane. The constraints on the motion of the particle are r · er = L0 , r · E3 = 0. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 16 Kinematics of a Particle (b) (a) E3 E2 m L0 O O E2 E1 L0 E1 m (c) E3 (d) m E3 moving plane O m α E2 O er E1 spinning conical surface Figure 1.7. Four mechanical systems featuring constraints on the motion of a particle: (a) the spherical pendulum, (b) the planar pendulum, (c) a particle moving on a plane, and (d) a particle moving on a spinning cone. 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 ḟ (t)E3 . The constraint on the motion of the particle is r · E3 = f (t). The final system of interest consists of a particle moving on a spinning cone. The constraint on the motion of the particle can be most easily described with the help of a spherical polar coordinate system: φ + α(t) − π = 0. 2 For all four systems, we have selected a coordinate system in which the constraint(s) on the motion of the particle is 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 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.7 Constraints 17 ∇ m r Surface =0 O Figure 1.8. A particle moving on a surface = 0. The particle in this case is subject to a single 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 1.8). The unit normal vector n to this surface is parallel to ∇ = grad( ) (see Figure 1.8). Depending on the coordinate system used, this vector ∇ has numerous representations: ∂ ∂ = Ei ∂r ∂xi 3 ∇ = i=1 = 3 ∂ i a ∂qi i=1 = ∂ 1∂ ∂ er + eθ + E3 ∂r r ∂θ ∂z = ∂ ∂ 1∂ 1 eR + eφ + eθ . ∂R R ∂φ R sin(φ) ∂θ (1.11) You should notice how simple the expression for the gradient is in curvilinear coordinates.∗ A simple differentiation of the function helps to provide the restriction it imposes on the velocity vector: ˙ = ∗ ∂ ∂ ·v+ . ∂r ∂t To establish this result, we note that ˙ q1 , q2 , q3 = ∇ · v and ˙ q1 , q2 , q3 = 3i=1 tuting for v and comparing both expressions, we arrive at (1.11)2 . ∂ ∂qi q̇i . Substi- 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 18 Kinematics of a Particle ∇ ∇ 2 1 m t r Surface O Surface 2 1 =0 =0 Figure 1.9. A particle subject to two constraints. The dotted curve in this figure corresponds to the curve of intersection of the surfaces tangent vector to this curve. 1 = 0 and 2 = 0, and the vector t is the unit However, if r satisfies the constraint, then (r, t) = 0 and ˙ = 0. Consequently, the constraint = 0 implies that the velocity vector satisfies the restriction ∂ ∂ ·v+ = 0. ∂r ∂t 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 move on the curve to be equivalent to two (simultaneous) constraints: 1 (r, t) = 0, 2 (r, t) = 0. This situation is shown in Figure 1.9. The normal vectors to the two surfaces at a point of their intersection are assumed not to be parallel: ∇ 1 × ∇ 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 be readily 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 ∂ 1 ·v+ = 0, ∂r ∂t ∂ 2 ∂ 2 ·v+ = 0. ∂r ∂t (1.12) If the curve is fixed, then 1 and 2 are not explicit functions of time. In this case, (1.12) can be used to show the expected result that v is tangent to the curve. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.7 Constraints 19 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) = 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 sophisticated the surfaces that the particle is constrained to move on, then 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 (r, t) = 0 can, in principle, be described by the equation q3 = 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 move on the sphere is simply described by R = R0 (t). Here, we are choosing the spherical polar coordinate system to be our coordinate system. The Classical Examples Revisited Returning to the four mechanical systems shown in Figure 1.7, you should convince yourself that the constraint(s) on the motions of the particle in these systems are 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 20 Kinematics of a Particle = 0. Specifically, for the spherical pendulum, individually of the form = r · eR − L0 . 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 − L0 , 2 = r · E3 = 0. In this case, the particle can be visualized as moving on the intersection of a cylinder of radius L0 and a horizontal plane. This intersection defines a circle. If the rod’s length L0 changes with time, then the circle’s radius also changes. For the particle moving on the horizontal surface, = r · E3 − f (t). Notice that in this example = (r, t). For the final system, the particle moving on a cone, the constraint on the motion of the particle can be represented by = 0, where π (r) = φ + α − . 2 If the cone were moving in a manner such that α = α(t), then the function (r, t). For example, suppose α = α0 + A sin(ωt); then (r, t) = φ − = π + α0 + A sin(ωt). 2 You should verify that ˙ = 0, but ∂∂t = Aω cos(ωt). The spinning of the cone has purposefully not been mentioned. This motion will feature 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. 1.8 Classification of Constraints All of the constraints discussed so far can be individually written in the form (r, t) = 0. Thus they are often known as positional constraints. We now define a further type of constraint: π = 0, (1.13) 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 restricts only its velocity vector. Consequently, the constraint π = 0 is often known as a velocity constraint. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.8 Classification of Constraints 21 As we demonstrated earlier, every constraint of the form ferentiated to yield a restriction on the velocity vector: (r, t) = 0 can be dif- ∂ ∂ ·v+ = 0. ∂r ∂t This restriction is of the form (1.13). Thus every constraint (r, t) = 0 provides a 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∗ 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). The terminology here dates to Heinrich Hertz [92] (1857–1894). As noted by Lanczos [124], 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 particularily important when rigid bodies are concerned. However, for pedagogical purposes, it is desirable to introduce them when discussing single particles. We shall shortly 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 1.10). 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. (1.14) These conditions imply that v has no component normal to the hyperbola xy = c. Constraints (1.14) are both clearly of the form f · v + e = 0, where e = 0 and ∗ Further background on integrating factors can be found in most texts on differential equations or differential forms, see, e.g., [61, 64, 114]. It is well known that integrating factors are not unique. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 22 Kinematics of a Particle y = x2 B x = x1 Figure 1.10. 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. A y = x2 B x = x1 Figure 1.11. The motion of a particle subject to the constraints ẏx = 0 and z = 0. The arrows indicate the possible directions of motion of the particle. A f = xE2 + yE1 , and e = 0 and f = E3 , respectively. By construction, constraints (1.14) are both integrable. As a second example, let us examine the following constraints: (xE2 ) · v = 0, z = 0. (1.15) The motions of the particle that satisfy these constraints are shown in Figure 1.11. Notice that it is possible to move between any two points A and B on the x–y plane without violating the constraint ẏx = 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 ẏx = 0 by 1x , for example, we see that, away from the y axis, this constraint is integrable.∗ Considering the possible motions shown in Figure 1.11, it is not surprising to note that we cannot find a smooth function to conclude that the constraint ẏx = 0 is integrable throughout the entirety of E3 . ∗ Here, 1x is an example of the integrating factor k (r, t) mentioned earlier in the definition of a nonintegrable constraint. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.8 Classification of Constraints 23 Instead, we classify the constraint ẏx = 0 as a piecewise-integrable constraint.∗ 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.† The constraint is (−zE1 + E2 ) · v = 0. (1.16) That is, ẏ − zẋ = 0. To demonstrate the type of restrictions ẏ − zẋ = 0 imposes, we choose two points A and B and use the x coordinate to parameterize a path between them. Choosing y = f (x), z= df , dx where f (x) is any sufficiently smooth function, we observe that the constraint −zẋ + ẏ = 0 is satisfied. In order that the particle be able to move between any two points A to B, f (x) is subject to the restrictions yA = f (xA), zA = df (xA), dx yB = f (xB), zB = df (xB), dx where rA = xAE1 + yAE2 + zAE3 and rB = xBE1 + yBE2 + zBE3 . Graphically constructing a function f (x) that meets these restrictions is not difficult, and some examples are presented in Figure 1.12. A specific example of f (x) is discussed in Pars [170], and a slightly modified version of it is presented here: x − xA 2 xB − xA x − xA 3 − (2 (yB − yA) − (xB − xA) (zB + zA)) xB − xA 2 2 2 π (x − xA) + c (x − xA) (xB − x) + d sin xB − xA xB − x + yA, + zA (x − xA) xB − xA f (x) = (3 (yB − yA) − (xB − xA) (zB + zA)) (1.17) 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 ẏ − zẋ = 0. Shortly, we shall verify that this constraint is indeed nonintegrable. ∗ † With the exception of that of Papastavridis [169], this classification is not typically mentioned in the textbooks on classical and analytical mechanics. Further discussion of piecewise-integrable constraints can be found in the interesting paper by Ruina [186]. As discussed in his paper, constraints of this type also arise in many locomotive systems such as passive walking machines that feature impact. A proof of this statement can be found in Section 163 of Forsyth [64]. Our discussion of constraint (1.16) is based on the treatments presented in Goursat [75] and Pars [170]. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 24 Kinematics of a Particle f v Figure 1.12. Three possible motions between two given points A and B of a particle subject to the constraint ẏ − zẋ = 0. The arrows indicate the directions of motion of the particle and the vector f = −zE1 + E2 . The motions presented in this figure were constructed with the assistance of (1.17). E3 E2 A O E1 B 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) . (1.18) 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 {q1 , q2 , q3 } such that f · v + e = 0 is equivalent to the constraint q̇3 + e = 0. The latter constraint in turn is equivalent to the constraint q3 = g, where ġ = e. Using this coordinate system, the dynamics of the particle is easier to analyze. With this in mind, several classical integrability criteria are now presented for single and multiple constraints.∗ The first criterion we examine pertains to constraints of the form f · v = 0, where f is not an explicit function of time. Using a coordinate system {q1 , q2 , q3 }, we can write the constraint π = 0 in the form A SINGLE SCLERONOMIC CONSTRAINT. f 1 q̇1 + f 2 q̇2 + f 3 q̇3 = 0. A necessary and sufficient condition for f · v = 0 to be integrable is that† Ic = 0 ∗ † (1.19) For additional discussion and illustrative examples from mechanics, the texts of Papastavridis [169] and Rosenberg [182] are recommended. Here, the historical remarks on these criteria are based on the paper by Hawkins [91]. Classical proofs of this result can be found in Section 151 of Forsyth [64], Section 442 of Goursat [74], and in Papastavridis [169]. A proof featuring differential forms can be found in Flanders [61], who refers to this result as Frobenius’ integration theorem. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 1.8 Classification of Constraints 25 for all possible choices of qi . Here, ∂f 3 ∂f 1 ∂f 2 ∂f 2 ∂f 3 ∂f 1 − 3 + f2 − 1 + f3 − 2 . Ic = f 1 ∂q2 ∂q ∂q3 ∂q ∂q1 ∂q It is convenient to recall at this point the expression for the curl of a vector field P in Cartesian coordinates: 3 ∂ curl(P) = Ei × P ∂xi i=1 ∂P3 ∂P2 ∂P1 ∂P3 ∂P2 ∂P1 E1 + E2 + E3 , (1.20) = − − − ∂x2 ∂x3 ∂x3 ∂x1 ∂x1 ∂x2 where Pi = P · Ei . With the help of this expression, it is easy to see that criterion (1.19) can also be expressed in the compact form f · (curl (f)) = 0. We refer to (1.19) as Jacobi’s criterion after its discoverer Carl G. J. Jacobi (1804– 1851). Satisfaction of (1.19) 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 xẏ + yẋ = 0 and xẏ = 0 trivially satisfy integrability criterion (1.19), only the former has a continuously defined (r). The function (r) for the latter constraint can be defined in only a piecewise manner (see Figure 1.11). If we use (1.19) to examine the constraint ẏ − zẋ = 0, then we find that∗ Ic = −z (0 − 0) + 1 (−1 − 0) + 0 (0 − 0) . As Ic = −1 = 0, the constraint ẏ − zẋ = 0 is nonintegrable. 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 A SINGLE RHEONOMIC CONSTRAINT. f 1 q̇1 + f 2 q̇2 + f 3 q̇3 + f 4 = 0, and define the variables U 1 = q1 , U2 = q2, U3 = q3, U 4 = t. Clearly, f 4 = e. We next form the functions ∂f L ∂f J ∂f K ∂f K ∂f L ∂f J + f + f . − − − IJKL = f J K L ∂UK ∂UL ∂UL ∂UJ ∂UJ ∂UK ∗ That is, we choose q1 = x, q2 = y, and q3 = z. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 26 Kinematics of a Particle 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 , t: IJKL = 0, for all J, K, L ∈ {1, 2, 3, 4}, L = J = K, K = L. (1.21) For a proof of this theorem, the reader is referred to Section 161 of Forsyth [64] or to Flanders [61]. When particles are subject to several constraints, their independence needs to be examined. For the case of two constraints, we first express them both in the form SYSTEMS OF CONSTRAINTS. f1 · v + e1 = 0, f2 · v + e2 = 0. If f1 × f2 = 0, then the constraints are said to be independent. For integrable constraints, this is equivalent to the condition ∇ 1 × ∇ 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 each of them in the form f1 · v + e1 = 0, f2 · v + e2 = 0, f3 · v + e3 = 0. (1.22) Then, the condition for their independence is that f1 · (f2 × f3 ) = 0. If the constraints are integrable, then this condition is equivalent to ∇ 1 · (∇ 2 × ∇ 3 ) = 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 wellknown instance occurs when two scleronomic constraints are imposed on a particle [169]: f1 · v = 0, f2 · v = 0, (1.23) where the functions f1 and f2 are functions of r, and f1 × f2 = 0. In this case, the system of constraints is integrable. The proof of this result, which is presented in Section 8.4, uses a criterion that is due to Ferdinand G. Frobenius (1849–1917), which we postpone discussion of until Chapter 8. This criterion can also be used to show that if the three constraints (1.22) are imposed on a particle, then the system of constraints is integrable. Consequently, the motion of the particle is prescribed. Other 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 1.1–1.3 27 instances of multiple constraints on the motion of a particle are discussed in the exercises at the end of this chapter. 1.9 Closing Comments In this chapter, we have assembled many of the needed 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. EXERCISES 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. 1.2. Consider a particle whose motion is described in cylindrical polar coordinates as r(t) = 10er , θ(t) = ωt, where ω = 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? 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 −1 x2 2 2 , z = x3 . r = x1 + x2 , θ = tan x1 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 . In addition, show that the contravariant basis vectors are a1 = er , a2 = 1 eθ , r a3 = E3 . 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 m 2 ṙ + r2 θ̇2 + ż2 . T= 2 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 28 Exercises 1.4–1.5 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= x21 + x22 + x23 , −1 x2 , θ = tan x1 ⎛ ⎞ x21 + x22 ⎠. φ = tan−1 ⎝ 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 a1 = eR , a2 = 1 eφ , R a3 = 1 eθ . R sin(φ) 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 u = ± x3 + v = ± −x3 + θ = tan −1 x2 x1 x23 + x21 + x22 , x23 + x21 + 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 plane, 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 1.5–1.6 29 (c) Show that the covariant basis vectors for the parabolic coordinate system are ∂r a1 = = ver + uE3 , ∂u ∂r a2 = = uer − vE3 , ∂v ∂r = uveθ . a3 = ∂θ 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 are a1 = grad(u) = 1 a1 , u2 + v2 a2 = grad(v) = 1 a2 , u2 + v2 a3 = grad(θ) = 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. 1.6. A classical problem is to determine the motion of a particle on a circular helix (see Figure 1.13). In terms of the cylindrical polar coordinates r, θ, z, the equation of the helix is r = R0 , z = αR0 θ, where R0 and α are constants. Here, we use another curvilinear coordinate system to define the motion of particle: q1 = θ, q2 = r, q3 = ν = z − αrθ. 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 + αE3 ), a2 = er + αθE3 , a3 = E3 . Verify that the covariant basis vectors are not orthonormal. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 30 Exercises 1.6–1.8 Bead of mass m g E3 E2 E1 Figure 1.13. A particle moving on a circular helix. (b) Show that the kinetic energy of the particle has the representation T= m (1 + α2 θ2 )ṙ2 + (1 + α2 )r2 θ̇2 + ν̇2 2 m 2ν̇ṙαθ + 2ν̇θ̇αr + 2ṙθ̇α2 rθ . + 2 Calculate the contravariant basis vectors associated with the curvilinear coordinate system. 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 x̂k(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)E1 + sin(α)E2 , a2 = E2 , a3 = E3 , (1.24) and a1 = sec(α)E1 , a2 = − tan(α)E1 + E2 , a3 = E3 , (1.25) 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? 1.8. Given a vector b = 10E1 + 5E2 + 6E3 , calculate its covariant bi and contravariant bi components when the covariant and contravariant basis vectors are defined by (1.24) and (1.25), respectively. In addition, 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 1.8–1.9 31 verify that 3 b= bi ai = i=1 Furthermore, show that b = 1 2 3 a , a , a not a basis? 3 i=1 3 bi ai . i=1 bi ai and b = 3 bi ai . For which values of α is i=1 1.9. This exercise illustrates 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, aik = aik(qr ) = ai · ak. You should notice that aik = aki and aik = aki . The indices i, k, r, and s in this problem range from 1 to 3. (a) For any vector b, show that the covariant and contravariant components are related: bi = 3 bi = aikbk, k=1 3 aikbk. 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 ⎡ ⎤ ⎡ ⎤ ⎡ 1⎤ b1 a11 a12 a13 b ⎢ ⎥ ⎢ ⎥ ⎢ 2⎥ ⎣b2 ⎦ = ⎣a21 a22 a23 ⎦ ⎣b ⎦ , b3 a31 ⎡ ⎤ ⎡ 11 a b1 ⎢ 2 ⎥ ⎢ 21 = ⎣b ⎦ ⎣ a b3 a31 a32 a33 a12 a13 a22 a32 b3 ⎤⎡ ⎤ b1 ⎥⎢ ⎥ 23 ⎦ ⎣ ⎦ . a b2 a33 b3 (b) By choosing b = ar and as , and using the symmetries of akm and ars , show that 3 k=1 aikakj = 3 k=1 aki akj = 3 j aki a jk = δi . k=1 It might be helpful to realize that, by using matrices, one of the preceding results has the following representation: ⎡ ⎤ ⎡ 11 ⎤ ⎡ ⎤ a11 a12 a13 a12 a13 a 1 0 0 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣a21 a22 a23 ⎦ ⎣a21 a22 a23 ⎦ = ⎣0 1 0⎦ . 0 0 1 a31 a32 a33 a31 a32 a33 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 32 Exercises 1.10–1.11 1.10. With the help of integrability criteria (1.19) and (1.21), show that only one of the following constraints is integrable: xẋ + yẏ = −e(t), zẏ + ẋ = 0, cos(z)ẏ − sin(z)ẋ = 0. In addition, show that the integrable constraint corresponds to a particle moving on a cylinder whose radius varies with time. 1.11. With the help of integrability criterion (1.21), show that one of the following constraints is nonintegrable: zẋ + ẏ = −e(t), ż = 0. 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 2 978 0 521 87483 0 June 9, 2008 Kinetics of a Particle 2.1 Introduction In this chapter, the balance law F = ma for a single particle plays a central role. This law is then 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 featured 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 [185] and Whittaker [228] and the more recent texts by Baruh [14], Moon [146], and Sheck [190]. 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 33 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 34 Kinetics of a Particle Euler’s first law. The integral (or impulse momentum) form of this law is t G(t) − G (t0 ) = F(τ)dτ, (2.1) t0 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, it is valid in impact problems, among others. An example of this 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 = Ġ. Assuming that the mass of the particle is constant, we can write F = mr̈. (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.∗ 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, and we shall postpone discussion of several of them until Section 4.11. It is convenient to write the balance law as a set of first-order ordinary differential equations: v = ṙ, F = mv̇. 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 write F = ma using a Cartesian coordinate system, then we find the following three equations: mẍ1 = F · E1 , mẍ2 = F · E2 , mẍ3 = F · E3 . ∗ Discussions of the historical threads from Newton’s Principia [152] that lead to Euler’s explicit formulation of (2.2) in [51] can be found in several works by Clifford A. Truesdell (1919–2000); see, for example, [216, 217]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.3 Work and Power 35 On the other hand, if a cylindrical polar coordinate system is used, we have m r̈ − rθ̇2 = F · er , m rθ̈ + 2ṙθ̇ = F · eθ , mz̈ = F · E3 . (2.3) Finally, if we use a spherical polar coordinate system, we find that m R̈ − Rφ̇2 − R sin2 (φ)θ̇2 = F · eR , m(Rφ̈ + 2Ṙφ̇ − R sin(φ) cos(φ)θ̇2 ) = F · eφ , m(R sin(φ)θ̈ + 2Ṙθ̇ sin(φ) + 2Rθ̇φ̇ cos(φ)) = F · eθ . (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 to be 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, Path of the particle P m v A B Figure 2.1. A force P acting on a particle as it moves from A to B. r rA rB O 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 36 Kinetics of a Particle with respect to time, of the mechanical power: tB WAB = P · vdt. tA 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) = Leαt (cos(ωt)Ex + sin(ωt)Ey ), where L, α, and ω = θ̇ are constant. A straightforward calculation shows that the power of this force is P · v = ωPLeαt , and the work performed by the force is tB ωPL αtB WAB = ωPLeαt dt = (e − eαtA ) , α tA 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 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 (1.11). It is important to notice that, if P is conservative, then its mechanical power is −U̇. To see this, we simply examine U̇ and use the definition of a conservative force: −U̇ = − ∂U · v = −(−P) · v = P · v. ∂r 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 = −U̇ 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.5 Examples of Conservative Forces 37 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 = Pr er + Pθ eθ + PzE3 . 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. 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.∗ Consequently, if curl(P) = 0, then the Cartesian components of P must satisfy the following conditions: ∂P3 ∂P2 = , ∂x2 ∂x3 ∂P1 ∂P3 = , ∂x3 ∂x1 ∂P2 ∂P1 = , ∂x1 ∂x2 where Pi = P · Ei . 2.5 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 −mgE3 , 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 a length L0 . Clearly, the stretched length of the spring is ||r − rA||, and the extension/compression of the spring is = ||r − rA|| − L0 . The potential energy Us associated with the spring is Us = f ( ) ∗ An expression for the curl of a vector field was presented earlier in (1.20). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 38 Kinetics of a Particle m r spring Figure 2.2. A spring whose ends are attached to a particle and a fixed point A. O rA A where f is a function of the change in length of the spring. Evaluating the gradient of Us , we find the spring force Fs : Fs = − ∂Us ∂f r − rA . =− ∂r ∂ ||r − rA|| To establish this result, we use the identity ∂ ∂ r − rA = . (||r − rA|| − L0 ) = ||r − rA|| ∂r ∂r This identity is left as an exercise to establish.∗ The most common spring in engineering dynamics is a linear spring. For this spring, Us = K (||r − rA|| − L0 )2 , 2 Fs = −K (||r − rA|| − L0 ) r − rA . ||r − rA|| 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 polynomial function in . For instance, f( ) = A 2 + B 4, where A > 0 and B are constants. Such a spring is known as hardening if B > 0 and softening if B < 0. 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 is modeled as Fn = mg, where g=− ∗ GM ||r||3 r. To help you with this, it is convenient to first establish that alent to showing that the gradient of ||r|| is eR . √ ∂ x·x ∂x = √x x·x = x ||x|| . This result is equiv- 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.6 Constraint Forces 39 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 as mass particles. Later on, in Sections 4.5 and 8.7, 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 Un = − GMm . ||r|| It should be transparent from the expressions for Un and Fn 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 allows 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 (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 ṙ = v, v̇ = 1 F, m subject to the restrictions on r and v, (r, t) = 0, ∂ ∂ ·v+ = 0. ∂r ∂t 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 (r, t) can be used to determine Fc 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 in O’Reilly and Srinivasa [162], this prescription is necessary and sufficient to ensure that the motion of the particle satisfies the constraint. However, freedom is available to include other arbitrary nonnormal components. This is the reason why prescriptions of constraint forces featuring frictional components are valid. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 40 Kinetics of a Particle ∇ n s2 m r s1 Surface =0 O Figure 2.3. A particle moving on a surface = 0. The vectors s1 and s2 are unit tangent vectors to this surface at the point of contact of the particle and the surface. A Single Constraint Consider the case of a particle subject to a single constraint: (r, t) = 0. Referring to Figure 2.3, we recall that the unit normal vector n to this surface is n= ∇ . ||∇ || 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 righthanded 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) as 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∇ , where we must determine λ = λ(t) by using F = ma. In other words, λ is an undetermined Lagrange multiplier. As discussed in Casey [27], the constraint force Fc is normal to the surface = 0 and is identical to a virtual work prescription that is sometimes known as the Lagrange principle or 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.6 Constraint Forces 41 Ff N=∇ Figure 2.4. The constraint force Fc 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. vrel n Fc = N + F f s2 s1 In the event that the surface is rough, an alternative prescription, which is due to Charles Augustin Coulomb (1736–1806), can be used∗ : Fc = λ∇ where the normal force N = λ∇ + Ff , and the friction force is F f = −µd ||λ∇ || v − vs . ||v − vs || 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 + vrel2 s2 and oppose the motion of the particle relative to the surface (cf. Figure 2.4). The velocity ||v − vs || is sometimes known as the slip speed. The mechanical power of the constraint force Fc is Fc · v = λ∇ · v + Ff · v = −λ ∂ + F f · v, ∂t where we used the identity ˙ =∇ ·v+ ∂ = 0. ∂t For the Lagrange prescription, F f = 0, we can now see that, if the surface that the particle is moving on is fixed, i.e., = (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 L0 whose center is fixed at the origin O. For this surface, the constraint is (r) = r · ∗ This prescription is often known as Amontons–Coulomb friction, after Guillaume Amontons (1663– 1705). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 42 Kinetics of a Particle eR − L0 = 0. Consequently, ∇ conclusion, = eR . In addition, v − vs = L0 φ̇eφ + L0 sin(φ)θ̇eθ . In φ̇eφ + sin(φ)θ̇eθ Fc = λeR − µd |λ| . φ̇2 + sin2 (φ)θ̇2 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, 1 (r, t) = 0 and 2 (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 ∇ 1 and ∇ 2 are both normal to the surfaces that the curve lies on (cf. Figure 1.9). Consequently, t= ∇ ||∇ ×∇ 1×∇ 1 2 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 v − vc = vt. We now turn to prescriptions for the constraint force. The first prescription is the Lagrange prescription: Fc = λ1 ∇ 1 + λ2 ∇ 2, where λ1 = λ1 (t) and λ2 = λ2 (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 ∇ 1 + λ2 ∇ + Ff , 2 where the friction force is F f = −µd ||λ1 ∇ 1 + λ2 ∇ 2 || v − vc . ||v − vc || The friction force opposes the motion of the particle relative to the curve and the normal force N with λ1 ∇ 1 + λ2 ∇ 2 . The mechanical power of the constraint force Fc for this case is Fc · v = λ1 ∇ = −λ1 1 · v + λ2 ∇ 2 · v + Ff · v ∂ 1 ∂ 2 − λ2 + F f · v, ∂t ∂t 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.6 Constraint Forces 43 where we again use the identities ˙1 =∇ 1 ·v+ ∂ 1 = 0, ∂t ˙ 2 = ∇ψ 2 ·v+ ∂ 2 = 0. ∂t For the Lagrange prescription, F f = 0, and we can now see that, if the curve that the particle is moving on is fixed, i.e., 1 = 1 (r) and 2 = 2 (r), then Fc does no work. 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 L0 to a fixed point O. The particle is also constrained to move on a vertical plane. In short, 1 = r · er − L0 = 0 and 2 = r · E3 = 0. With a little work, we find that ∇ 1 = er and ∇ 2 = E3 . For this mechanical system, Lagrange’s prescription is appropriate: Fc = λ1 er + λ2 E3 . For this system, λ1 er can be interpreted as the tension force in the rod and λ2 E3 can be interpreted as the normal force exerted by the plane on the particle. If we let L0 = L0 (t), the prescription for the constraint force will not change. A system that is related to the planar pendulum can be imagined as a particle moving on a rough circle whose radius L0 = L0 (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 θ̇ Fc = λ1 er + λ2 E3 − µd λ21 + λ22 eθ , θ̇ where we used the fact that v − vc = L0 θ̇eθ . 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: i (r, t) = 0, i = 1, 2, 3. As mentioned earlier, when a particle is subject to three constraints, the three equations i (r, t) = 0 can in principle be solved to determine the motion r(t) of the particle. We denote the resulting solution by f(t), i.e., r(t) = f(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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 44 Kinetics of a Particle To examine the equivalence, let us first use Lagrange’s prescription: Fc = λ1 ∇ 1 + λ2 ∇ 2 + λ3 ∇ 3. Here, λ1 , λ2 , and λ3 are functions of time. Because the three constraints are tacitly assumed to be independent, {∇ 1 , ∇ 2 , ∇ 3 } 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 + F f , where the magnitude of F f is restricted by the static friction criterion: F f ≤ µs ||N|| , 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 + mf̈. This solution Fc 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: (r, t) = 0, f · v + e = 0. Using Lagrange’s prescription, we find that the constraint force acting on the particle is Fc = λ1 ∇ + λ2 f. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.7 Conservations 45 Suppose that the applied force acting on the particle is Fa ; then the equations governing the motion of the particle are (r, t) = 0, ṙ = v, f · v + e = 0, v̇ = 1 (Fa + λ1 ∇ m + λ2 f). This set of equations constitutes eight equations for the eight unknowns: λ1 , λ2 , r, and v. 2.7 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 of 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 F(τ)dτ, t0 t we see that G(t) is conserved during an interval of time (t0 , t) if t0 F(τ)dτ = 0. The 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 given vector b(t) being conserved. That is, dtd (G · b) = 0. For this to happen, G ˙· b = Ġ · b + G · ḃ = F · b + G · ḃ = 0. In words, if F · b + G · ḃ = 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 E1 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 46 Kinetics of a Particle 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: ḢO = v × G + r × Ġ = v × mv + r × F = r × F. 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: ḢO = 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 · ċ = 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 by use of F = ma and relates the time rate of change of kinetic energy to the mechanical power of F: Ṫ = 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 = 12 mv · v. Differentiating T, we find d Ṫ = dt 1 1 mv · v = (mv̇ · v + mv · v̇) = mv̇ · v. 2 2 However, we know that mv̇ = F, and so substituting for mv̇, we find that Ṫ = 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 = − ∂U and a nonconservative force Pncon : F = P + Pncon . Here, U is the sum ∂r 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.8 Dynamics of a Particle in a Gravitational Field of the potential energies of the conservative forces acting on the particle. From the work–energy theorem, we find Ṫ = F · v = P · v + Pncon · v =− ∂U · v + Pncon · v ∂r = − U̇ + Pncon · v. Defining the total energy E of the particle by E = T + U, we see that Ė = Pncon · 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 Pncon · v = 0. To see this, let us consider the example of the spherical pendulum whose length L0 = L0 (t). For this particle, P = −mgE3 , Pncon = λeR . Consequently, E = T + mgE3 · r and Pncon · v = λeR · v = L̇0 λ. As a result, when the length of the pendulum is constant, L̇0 = 0, then E is conserved. On the other hand, if L̇0 = 0, then the constraint force λeR does work by giving the particle a velocity in the eR direction. 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.∗ Over 100 years later, it was also discussed in wonderful detail in Lagrange’s famous text Mécanique Analytique.† 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. ∗ † See Section III of Book 1 of [152]. See Section VII of the Second Part of [121]. 47 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 48 Kinetics of a Particle y a m b r θ O A P x a(1 − e) ĥ < 0 ĥ > 0 Figure 2.5. Schematic of a particle of mass m moving about a fixed point O in an elliptical 2 orbit. One of the foci of the ellipse is at O, and the eccentricity e = 1 − ba2 of the ellipse is less than 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. 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 a3 T2 of the orbit, then, for any of two planets, a13 = T12 . 2 2 Concise discussions of Kepler’s laws can be found in [150, 175, 188, 220]. 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 from Chapter 4, this restriction is easily removed and the results presented can be readily applied to deduce the motions of m and M. Many of the results presented feature terminology associated with ellipses, and, for convenience, a summarization of many of the terms associated with an ellipse, such as its eccentricity e and axes a and b, is given in Figure 2.5. The area swept out by the particle can be 1 determined by integrating the areal vector (1.1): A = 2m HO. The third law is remarkable when we note from [188] 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 = 1.00, 0.2412 5.2033 = 1.00, 11.8622 1.5243 = 1.00. 1.8812 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.8 Dynamics of a Particle in a Gravitational Field 49 Following this, we reduce the equations of motion to a single second-order differential equation that we nondimensionalize and integrate numerically.∗ 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 = ma, where F is solely due to Newton’s gravitational force: Fn = − GMm ||r||3 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 Fn = ma. The result will be three differential equations: GMmr m r̈ − rθ̇2 = − √ 3 , r2 + z2 m rθ̈ + 2ṙθ̇ = 0, GMmz mz̈ = − √ 3 . r2 + z2 (2.6) For a given set of six initial conditions,† 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 (2.6) conserve two important kinematical quantities. First, they conserve the total energy E = T + U, where U = Un . Second, ∗ † 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 Gantmacher [67], Chapter 2 of Karapetyan and Rumyantsev [108], and Marsden and Ratiu [138]. The six initial conditions needed are r (t0 ), θ (t0 ), z (t0 ), ṙ (t0 ), θ̇ (t0 ), and ż (t0 ). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 50 Kinetics of a Particle 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 (t0 ) 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 ourselves the freedom to choose E3 , then we can pick this vector such that HO = ĥE3 , where ĥ = mr2 θ̇. Consequently, r and v will have components in only the E1 and E2 directions: z(t) = 0 and ż(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.”∗ We have tacitly ignored the case in which r (t0 ) 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 ż(t) = 0 for the particle. That is, the direction of HO is E3 . As a result, the equations of motion reduce to GMm m r̈ − rθ̇2 = − 2 , r m rθ̈ + 2ṙθ̇ = 0. (2.7) Now the second of these equations can be expressed as d 2 mr θ̇ = 0. dt (2.8) This equation is equivalent to the conservation of HO · E3 . Using this conservation, we can eliminate θ̇ from (2.7) and arrive at a single governing differential equation: mr̈ − ĥ2 GMm =− 2 . 3 mr r Here, ĥ is determined from the initial position and velocity of the particle: ĥ = HO · E3 = (mr (t0 ) × v (t0 )) · E3 . ∗ See, for example, Section 86 of Moulton [150]. (2.9) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.8 Dynamics of a Particle in a Gravitational Field 51 For a particle with no angular momentum, we see from (2.8) that θ̇ = 0 and hence r̈ = − GM . It is not difficult to see that this equation implies that r(t) → 0 as t inr2 creases. This is the collision we discussed earlier. Given r(t0 ) and v(t0 ), we can determine ĥ and then integrate (2.9) to determine ∗ r(t). We can then compute the coordinate θ(t) by integrating θ̇ = ĥ . mr2 (2.10) We can then find expressions for x(t) = r(t) cos (θ(t)) and y(t) = r(t) sin (θ(t)) and construct the orbit of the particle. The easiest solution of (2.9) to compute is the one for which r is constant: r(t) = r0 and ṙ(t) = 0. In this case, (2.9) is satisfied provided r0 = ĥ2 . GMm2 We also show, using (2.10), that θ̇ is constant: ĥ θ̇(t) = ωK = = mr02 (2.11) GM . r03 (2.12) The frequency ωK is known as the Kepler frequency. Physically, the particle is moving in a circular orbit of radius r0 at constant speed r0 ωK about the fixed body of mass M. In numerically integrating (2.9), 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 dr dr the form ṙ = dr = dτ = ωK dτ , we can simplify (2.9) and (2.10) to dt dt dτ 1 1 d2 w = 3 − 2, 2 dτ w w dθ 1 = 2. dτ w (2.13) Notice that we have reduced the problem of determining the motion of the particle to the integration of two differential equations. Differential equation (2.13)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 (2.13)1 is to construct what is known as a phase portrait. In this portrait, dw is plotted as a function of w(τ).† Equilibria dτ of the second-order differential equation correspond to points in the phase portrait 2 where dw = 0 and w(τ) is a constant. To find such points, we set dw = 0 and ddτw2 = 0 dτ dτ 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 feature differential equations with multiple equilibria. ∗ † Differential equation (2.9) 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 [228] for details on how such an integration can be performed. Phase portraits are a standard method for the graphical representation of solutions to ordinary differential equations (see, for example, the texts of [10, 18, 81, 229]). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 52 Kinetics of a Particle 2 h h 1 p c dw dτ e e e −1 −2 1 4 w Figure 2.6. The phase portrait of (2.13)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 orbits. The arrows in this figure correspond to the directions of increasing τ. Returning to the problem at hand, the phase portrait of (2.13)1 is shown in Figure 2.6. There, we see an equilibrium point at w, dw = (1, 0) that corresponds dτ to a circular orbit of the particle. The closed orbits enclosing this point correspond to elliptical orbits of the form shown in Figure 2.7.∗ 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.† 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 (2.7)2 in the form (2.10) to show that ṙ = − ∗ † ĥ d m dθ 1 , r r̈ = − ĥ2 d2 m2 r2 dθ2 1 . r dθ A second integration involving dτ = w12 is needed to construct these orbits, and this is left as an exercise. In the parlance of dynamical systems theory, the homoclinic orbit that passes through the point w, dw dτ = (0.5, 0) connects the fixed point at w, orbit of the particle. dw dτ = (∞, 0) and corresponds to the parabolic 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.8 Dynamics of a Particle in a Gravitational Field y = x2 (a) (b) 53 y = x2 r θ x = x1 O x = x1 O y = x2 ĥ > 0 (c) r θ θP x = x1 O ĥ > 0 y = x2 (d) (e) y = x2 r r ĥ < 0 θP θP O x = x1 O x = x1 ĥ > 0 Figure 2.7. 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) a hyperbolic orbit (e > 1). For each of the orbits (c)–(e), distinct values of θ p are considered. Using the second of these results to rewrite (2.7)1 , we find the differential equation d2 dθ2 1 1 1 + = . r r r0 where r0 was defined earlier [see (2.11)]. (2.14) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 54 Kinetics of a Particle Equation (2.14) is a linear ordinary differential equation for that has the exact solution r = r (θ) = r0 (1 + e cos (θ − θ p))−1 , 1 r as a function of θ (2.15) where e and θ p are constants (which are determined from the initial conditions for 2 the position and velocity of the particle). We can then integrate mrĥ(θ) dθ = dt to determine an analytical expression for θ(t). This is left as an exercise. Solution (2.15) 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 ĥ2 Figure 2.5 for the elliptical orbit, it is easy to see that b = GMm 2. To solve for e and θ p, let us assume that r (t0 ) and v (t0 ) are given. Then we can determine HO and specify E3 and ĥ. To compute θ p, we first calculate ṙ by using (2.15) and the chain rule: GMm e sin (θ(t) − θ p) . ṙ(t) = − (2.16) ĥ We also compute the value E0 of total energy of the particle E0 = m v (t0 ) · v (t0 ) − 2 GMm . r (t0 ) (2.17) Now as the total energy is conserved, the value of this kinematical quantity when θ = θ p is also equal to E0 . With some manipulations of (2.15) and (2.16), it can be shown that E0 = G2 M2 m3 2 e −1 . 2ĥ2 (2.18) We now have a method of determining e and θ p from a given set of initial conditions. The procedure is to compute E0 by use of (2.17) and then use (2.18) to compute e ≥ 0. Once e is known, then (2.16) can be used to compute θ p. 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 2.7). For completeness, we could also have nondimensionalized (2.14): d2 u + u = 1, dθ2 (2.19) where u = w1 = rr0 . The phase portrait of this equation is shown in Figure 2.8. In contrast to the earlier phase portraint, 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 2.8, it is necessary to reconstruct a position vector of the particle corresponding to the particular orbit. Comments For this problem, the rare event occurs that a complete classification of the motions of the particle are possible. For many of the problems that are discussed later on in 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.9 Dynamics of a Particle on a Spinning Cone 2 55 h h h h 1 p c du dθ e e -1 -2 1 u 4 Figure 2.8. The phase portrait of (2.19). 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 orbits. In this figure, the arrow indicates the direction of increasing θ. For the parabolic trajectory, this angle ranges from −π → π. 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, for example, Baruh [14]. 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 exe3 cuting an elliptical orbit. We leave it as an exercise to show that T = √2π a 2 , and, GM consequently, the third law is satisfied. 2.9 Dynamics of a Particle on a Spinning Cone As shown in Figure 1.7(d), 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 L0 . 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 features of its dynamics. Coordinates, Constraints, and Velocities As discussed in Section 1.7, when the particle is moving on the surface of the cone, it is subject to a single constraint = 0. This constraint can be conveniently expressed 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 56 Kinetics of a Particle by use of a spherical polar coordinate system: =φ+α− π . 2 For future reference, we note that the gradient of is ∇ = R1 eφ . Because the cone is rotating with a speed 0 , the velocity of the particle relative to the cone is vrel = ṘeR + R cos(α) θ̇ − 0 eθ . We defer discussion of the case in which the particle is stuck to the cone. Forces The particle is under the influence of a gravitational force −mgE3 and a spring force: Fs = −K (R − L0 ) eR , where K is the stiffness of the spring and L0 is its unstretched length. Assuming that the particle is moving relative to the surface of the cone, we find that the constraint force Fc acting on the particle has the representation Fc = N + F f , where the normal force N is parallel to ∇ : N= λ eφ , R F f = −µd ||N|| vrel . ||vrel || Thus the total force on the particle is F = Fc + Fs − mgE3 . 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 = 0 to find m R̈ − R cos2 (α)θ̇2 = −K (R − L0 ) + F f · eR − mg sin(α), m R cos(α)θ̈ + 2Ṙθ̇ cos(α) = F f · eθ , −mR sin(α) cos(α)θ̇2 = λ + mg cos(α). R (2.20) 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 L0 as a measure of length and L0 g as a measure of time: τ=t g , L0 w= R . L0 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.9 Dynamics of a Particle on a Spinning Cone With the help of identities of the type Ṙ = dR dt 57 = dτ dR dt dτ = g dR , L0 dτ we can rewrite (2.20) in the form 2 d2 w dθ 2 = w cos (α) − ω2 (w − 1) − sin(α) dτ2 dτ − µkn w2 dθ dτ dw dτ − ω0 2 + dw 2 , dτ dθ w dτ − ω0 d dθ 2 2 w cos (α) = −µkn (w cos(α)) 2 dw 2 . dτ dτ dθ 2 w dτ − ω0 + dτ (2.21) In these equations, the constants and dimensionless normal force are L0 KL0 ||N|| h2 2 ω = , ω0 = , n= = cos(α) + tan(α), mg g mg w and we can also show that dimensionless versions of total energy E and angular momentum HO · E3 are 2 E dw 2 1 dθ ω2 2 2 = + w cos (α) + (w − 1)2 + w sin(α), mgL0 2 dτ dτ 2 h= dθ HO · E3 = w2 cos2 (α) . dτ m gL30 Notice that, by nondimensionalizing the equations of motion, we have reduced the number of parameters by two. The Static Friction Case When the particle is stuck to the cone, its velocity vector is v = R0 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 λ1 eθ + λ2 eR . eφ + R0 R0 cos(α) To determine λ, λ1 , and λ2 , we examine F = ma. With some manipulations, we conclude that F f = K (R0 − L0 ) + mg sin(α) − mR0 cos2 (α)20 eR , ˙ 0 eθ , + mR0 cos(α) N = − mg cos(α) + mR0 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 F f ≤ µs ||N||. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 58 Kinetics of a Particle x2 x2 x1 x1 20 1 dR L0 dτ 1 8 R L0 x2 −20 x1 x2 x1 Figure 2.9. The phase portrait of (2.22) and the corresponding planar projections of the trajectories of the particle. For this figure, we assumed that α = 20◦ , h = 5, and ω2 = 10. Consequently the equilbrium point corresponding to a circular trajectory of the particle has the coordinates (w0 , 0) = (1.58896, 0). 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 F f .∗ The Smooth Cone When the particle is moving on a smooth cone, we can simplify (2.21) considerably. Indeed, as in the earlier particle problem, we can exploit the conservation of angular momentum to write a single equation for w: d2 w h2 − ω2 (w − 1) − sin(α). = 3 2 dτ w cos2 (α) (2.22) Integrating this equation, we can find w(τ). Another integration using dθ dθ w2 cos2 (α) dτ = (w0 , 0), where = h provides θ(τ). The equilibrium point at w, dτ w0 is the solution of w30 ∗ h2 − ω2 (w0 − 1) − sin(α) = 0, cos2 (α) 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 ||vvrel || even though vrel = 0. rel 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.10 A Shocking Constraint 59 corresponds to a circular orbit of the particle that has a radius r = L0 w0 cos(α). Some of the other trajectories of w in the w − dw plane are shown in Figure 2.9. In this dτ figure, we have also constructed possible trajectories of the particle corresponding to w(τ). Unlike the problem of the particle subject to Fn that we discussed in Section 2.8, here the classifications of the trajectories for the particle on the cone defy a simple classification. 2.10 A Shocking Constraint We now return to the constraint ẏx = 0, discussed earlier [see (1.15)]. 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 = P1 E1 + P2 E2 . First, we assume that the constraint force that enforces ẏx = 0 has the standard prescription Fc = λxEy , (2.23) 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 xẏ = 0, mẍ = P1 , mÿ = P2 + λx, mz̈ = 0. (2.24) 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 (2.24)1,2,3 have exact solutions that are easy to establish provided the motion is rectilinear: t τ Fy P1 dudτ, λ= (2.25) y(t) = y0 , x(t) = x0 + ẋ0 t + x(t) 0 0 m and x = 0, y(t) = y0 + ẏ0 t + t 0 τ 0 P2 dudτ. m (2.26) Here x0 = x(0), ẋ0 = ẋ(0), y0 = y(0), and ẏ0 = ẏ(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 1.11, we recall that this constraint has the unusual feature that it can be decomposed into two piecewise integrable constraints: ẏ = 0 when x = 0, and x = 0. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 60 Kinetics of a Particle (a) (b) y = x2 y = x2 Ic2 B x = x1 B A x = x1 A Ic1 Figure 2.10. Two possible motions of a particle subject to a constraint ẏx = 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 in which v is not continuous. At the points where the particle makes a transition from one of these integrable constraints to the other, its velocity vector v will be discontinuous and therefore its acceleration vector cannot be defined. At such a transition, the prescription for constraint force (2.23) 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 T+σ Ic = lim mv (T + σ) − mv (T − σ) − Fa dτ . σ→0 T−σ For example, to achieve a motion that goes from point A to point B in Figure 2.10(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 2.10(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 1.11 as a method of realizing the constraint ẏx = 0. 2.11 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.11 A Simple Model for a Roller Coaster E2 61 A y E1 x1 B C Figure 2.11. Schematic of a particle on a cosinusoidal path. 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.∗ 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, 1 = 0 and 2 = 0, where 1 = y − f (x1 ) , 2 = z. These constraints will be enforced by Fc = N + F f . It is a standard exercise to calculate the unit tangent et , the unit normal en , and the binormal eb vectors to this curve [159]: et = 1 E1 + f E2 , 1+ f 2 sgn f E1 + f E2 , en = 1+ f 2 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 ẋ1 > 0. A normal force N = Nen + λ2 E3 , a friction force F f et , and a vertical gravitational force −mgE2 act on the cart (see Figure 2.11). Now, for a particle moving on a curve with a velocity v = vet , the acceleration vector of the particle is a = v̇et + κv2 en , ∗ The work presented on this model was performed in collaboration with Henry Lopez [130]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 62 Kinetics of a Particle where κ is the curvature of the space curve: f κ = 3 . 2 1+ f 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: ẋ1 2 m 1+ f ẍ1 + f f ẋ21 = −mgf − µd 1 + f 2 ||N|| , (2.27) |ẋ1 | where 1 N= 1+ f 2 sgn f mg + f mẋ21 en , mg = en , 2 1+ f if f = 0, when f = 0. (2.28) 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, mgf ≤ µs ||N|| , (2.29) 1 + f 2 the cart remains stationary. With the help of (2.28), (2.29) can be expressed in the simple form (2.30) f ≤ µs . 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 µs . We shall shortly use (2.30) to establish a continuum of points at which the roller coaster can remain in a state of rest. We now choose a cosinusoidal track, πx1 f (x1 ) = A cos . (2.31) L0 In addition, we employ the following nondimensionalizations: g x1 x= , τ= t. L0 L0 Of course, other tracks are possible, and the reader is referred to Shaw and Haddow [193], 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.11 A Simple Model for a Roller Coaster 63 dx dτ 1.5 −2 x 2 −1.5 Figure 2.12. The phase portrait of (2.27)–(2.29) when friction is absent. The points on the x = Lx1 axis labeled by · correspond to equilibria of the cart. For this figure, 0 µs = µd = 0.0. A L0 = 0.25, and 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 2.12. We note the presence of an equilibrium at x = 0, dx = 0 . This point corresponds to the cart’s being stationary dτ at the top of the roller coaster. Referring to Figure 2.13(a), we refer to equilibria of this type as saddles. The two equilibria at x = ±1, dx = 0 represent a stationary dτ cart at the bottom of one of the valleys of the roller coaster. Examining the phase portrait in Figure 2.13(c), we easily see why equilibria of this type are known as centers. The equilibria at x = (−2, 0, 2), dx = 0 correspond to a stationary cart at dτ one of the crests of the roller coaster. When stick–slip friction is present, the phase portrait changes dramatically (see Figure 2.14). First, all of the saddles have split, and between their two split halves we have what we call a sticking region [see Figure 2.13(b)]. Depending on the value of x1 , this region contains either ẋ1 > 0 or ẋ1 < 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 2.13(d)]. 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 (2.30), and a graphical method is shown in Figure 2.15. As µs 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 dx axis will become an equilibrium, and thus a state of rest dτ 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 64 Kinetics of a Particle dx dτ (a) (b) 0.2 0.2 −0.2 0.2 x 0.2 x −0.2 dx dτ (d) 0.2 dx dτ 0.2 x 1.2 0.8 s −0.2 −0.2 (c) dx dτ 0.8 −0.2 s x 1.2 −0.2 Figure 2.13. Expanded views of the phase portraits in the neighborhood of equilibria of (2.27) with f specified by (2.31). For (a) and (c), µd = 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. for the roller coaster.∗ This phenomenon is also easy to explain physically. We note for completeness that, for the present choice of f (x), if µs ≥ πA , L0 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 2.15. 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. ∗ For the phase portrait shown in Figure 2.13(b), the sticking region s is the interval [−0.124755, 0.124755], and for situation shown in Figure 2.13(d), the sticking region s is the interval [1. − 0.124755, 1.124755]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 2.12 Closing Comments 65 dx dτ 1.5 −2 x 2 −1.5 Figure 2.14. The phase portrait of (2.27)–(2.29). Although it is not evident from the figure, the discrete equilibria of the frictionless case shown in Figure 2.12 are now replaced with families of equilibria that correspond to possible resting (sticking) states for the cart. For this figure, A = 0.25 and µd = 0.1. L 0 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. f µs = 0.3 0.2 Figure 2.15. A graphical method to compute the possi- ble sticking regions s of the cart on a smooth roller coaster when µs = 0.3. The method is based on examining (2.30) for the choice (2.31). That is, f = Aπ πx L0 sin L01 . Examples of the sticking regions can be seen in Figure 2.13. s −2 2 s −2 s s s 2 s s s x= x1 L0 x= x1 L0 s 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 66 Exercises 2.1–2.5 EXERCISES 2.1. Which of the following force fields are conservative/nonconservative? P = x1 E1 + x3 E2 , P = x2 E1 + x1 E2 , P = x1 x2 E1 , P = −L0 sin(θ)E1 + L0 cos(θ)E2 , where L0 is a constant. For the conservative force fields, what are the associated potential energies? 2.2. Consider a particle of mass m that is moving in E3 . 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 v of the particle. Give three distinct physical examples of applications of this result. 2.3. In contrast to Exercise 2.2, here 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, give three distinct physical examples of the application of this result. 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 1 = 0 and 2 = 0, where 1 = 1 (r) = x3 , 2 = 2 (r, t) = x1 − f (t). Give a prescription for the constraint force acting on the particle. 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.∗ (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 −mgE3 . One end of the spring is attached to the fixed origin O, and the other is attached ∗ A force P is said to be central if P is parallel to r. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 2.5–2.6 67 to the particle. The spring has a stiffness K and unstretched length L0 . 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 r K τ= t, x= , m L0 and conservation of angular momentum, show that the motion of the particle can be found by integrating the following differential equations: d2 x β2 − 3 = − (x − 1) , dτ2 x dθ β = 2, dτ x (2.32) where β= L20 h √ 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, 20, construct the phase portraits of (2.32)1 . For a selection of the orbits on each of these phase portraits, construct images of the motion of the particle.∗ (e) Verify that the areal velocity vector is conserved for the motions of the particle you found in (d). 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 ∂UP m r̈ − rθ̇2 = − , ∂r mr2 θ̇ = h, (2.33) where h is a constant. ∗ Your results will be qualitatively similar to those presented in Section 2.9 for the particle moving on a smooth cone. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 68 Exercises 2.6–2.7 (e) Show that the solutions to (2.33) conserve the total energy E of the particle. 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 aEx + bEy by a linear spring of unstretched length L0 and stiffness K. A vertical gravitational force −mgE3 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 x · eφ mR0 (φ̈ − sin(φ) cos(φ)θ̇2 ) = mg sin(φ) − K (||x|| − L0 ) ||x|| − µd ||N|| φ̇ , φ̇2 + sin2 (φ)θ̇2 d 1 x · eθ mR20 sin2 (φ)θ̇ = −K (||x|| − L0 ) ||x|| R0 sin(φ) dt − µd ||N|| sin(φ)θ̇ , φ̇2 + sin2 (φ)θ̇2 where x = R0 eR − aEx − bEy . (e) Show that the normal force exerted by the surface on the particle is x · eR eR N = mgE3 · eR + K (||x|| − L0 ) ||x|| − mR0 (φ̇2 + sin2 (φ)θ̇2 )eR . (f) For the case in which the particle is not moving relative to the surface, show that x Fc = mgE3 + K (||x|| − L0 ) . ||x|| 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 · E3 is conserved. Using this conservation, show that the dimensionless equations governing the motion of the particle 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 2.7–2.8 69 simplify to h d2 φ cos(φ) dθ = , + sin(φ). = h2 3 2 dτ dτ2 sin (φ) sin (φ) g R0 1 In these equations, h = mR H · E and τ = t. 0 3 2 g R0 0 (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 surface of the sphere is mg (3 cos(φ) − 2 cos (φ0 )) − mR0 φ̇20 + sin2 (φ0 ) θ̇20 > 0, where φ0 is the value of the initial φ coordinate of the particle and θ̇0 and φ̇0 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. 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. (2.34) (a) Show that one of the constraints is integrable whereas the other is nonintegrable. In addition, for the integrable constraint, specify the function (r, t) = 0. (b) Suppose that, in addition to the constraint force Fc = µ1 (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. (c) With the help of the work–energy theorem Ṫ = 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 . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 3 978 0 521 87483 0 June 2, 2008 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 the late 17th century. As we have seen, this vector-valued equation yields three differential equations from which the motion of a particle can be determined. 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 remains one of the central issues for any student of dynamics. At the end of the 18th century, a formulation of the equations of motion for a single particle appeared in a famous text by Lagrange [121].∗ 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 ∂ q̇i − ∂L = Fncon · ai . ∂qi 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 sufficiently discussed in most textbooks, it can be found in many of the classical texts on dynamics, such as those of Synge and Griffith [207] and Whittaker [228]. A recent paper by Casey [27] explores this equivalence in a transparent manner, and we follow many aspects of his exposition in this chapter. ∗ 70 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 [122]. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.2 Lagrange’s Equations of Motion 71 Casey’s work will also feature later on when we discuss systems of particles and rigid bodies. 3.2 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 [121]. 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 Section 24 in Chapter IV of [183]) and Aurel Voss (1845–1931) [221].∗ Here, an approach is used that is rooted in differential geometry and is contained in some texts on this subject (see, for example, Synge and Schild [208]). It probably migrated from there to the classic text by Synge and Griffith [207] and has been recently revived by Casey [27]. Two Identities We assume that a curvilinear coordinate system has been chosen for E3 . The velocity vector v consequently has the representation v= 3 q̇i ai . i=1 In addition, the kinetic energy has the representations 3 3 m m v·v= ai · akq̇i q̇k. 2 2 i=1 k=1 1 2 3 1 2 3 It is crucial to notice that T = T q , q , q , q̇ , q̇ , q̇ . We now consider in succession the partial derivatives of T with respect to the coordinates and their velocities. We wish to establish the following results: T= ∂T = mv · ȧi , ∂qi ∂T = mv · ai . ∂ q̇i 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 ∂ m v·v = i i ∂q ∂q 2 ∂v . = mv · ∂qi ∗ For further details on the historical development of Lagrange’s equations, see Papastavridis [167, 169]. Equations of motion (3.8)2,3 are examples of what could be referred to as the Routh–Voss equations of motion. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 72 978 0 521 87483 0 June 2, 2008 Lagrange’s Equations of Motion for a Single Particle ∂T To proceed with the goal of concluding that ∂q i = mv · ȧi , we first note that v = 3 ∂ q̇k k k=1 q̇ ak and that ∂qi = 0. As a consequence, 3 ∂ak ∂T k = mv · q̇ . ∂qi ∂qi k=1 ∂r The remaining steps use the fact that ak = ∂q k: 3 ∂ak ∂T k = mv · q̇ ∂qi ∂qi k=1 = mv · 3 k=1 = mv · 3 k=1 ∂2r ∂2r q̇k i k = q̇k k i ∂q ∂q ∂q ∂q 3 k=1 ∂ q̇k k ∂q ∂r = ai ∂qi = mv · ȧi . ∂f k We achieve the last step by noting that ḟ = 3k=1 ∂q for any function f = k q̇ 1 2 3 f q ,q ,q . 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 q̇k. Getting on with the proof, we have 3 ∂ q̇k ∂T = mv · ak ∂ q̇i ∂ q̇i = mv · k=1 3 δkiak k=1 = mv · ai . 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 ∂T d ∂T d − i = (mv · ai ) − mv · ȧi dt ∂ q̇i ∂q dt = ma · ai + mv · ȧi − mv · ȧi = ma · ai = F · ai . 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.3 Equations of Motion for an Unconstrained Particle In conclusion, we have a covariant form of Lagrange’s equations of motion: ∂T d ∂T − i = F · ai . i dt ∂ q̇ ∂q 73 (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 + Fncon , where the potential energy U = U(q1 , q2 , q3 ). As ∇U = 3 ∂U k a , ∂qk k=1 ∂U = 0, ∂ q̇k we find that Lagrange’s equations can be rewritten in the form ∂T d ∂T ∂U ∂U − = Fncon · ai . − − dt ∂ q̇i ∂ q̇i ∂qi ∂qi Introducing the Lagrangian L = T − U, we find an alternative form of Lagrange’s equations: d ∂L ∂L − i = Fncon · ai . (3.2) i dt ∂ q̇ ∂q 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 Lagrange’s equations, we consider the case in which the curvilinear coordinates chosen are the spherical polar coordinates: q1 = R, q2 = φ, and q3 = θ. For these coordinates, we have a1 = eR , a2 = Reφ , a3 = R sin(φ)eθ , and T= m 2 (Ṙ + R2 sin2 (φ)θ̇2 + R2 φ̇2 ). 2 Notice that T does not depend on θ. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 74 978 0 521 87483 0 June 2, 2008 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 = mR sin2 (φ)θ̇2 + mRφ̇2 , ∂R ∂T = 0, ∂θ ∂T = mR2 φ̇, ∂ φ̇ ∂T = mR2 sin(φ) cos(φ)θ̇2 , ∂φ ∂T = mṘ, ∂ Ṙ ∂T = mR2 sin2 (φ)θ̇. ∂ θ̇ Using these results, we find the covariant form of Lagrange’s equations: ∂T d ∂T 2 2 2 = mR sin (φ)θ̇ + mRφ̇ = F · eR , = mṘ − dt ∂ Ṙ ∂R ∂T d ∂T 2 2 2 = mR φ̇ − = mR sin(φ) cos(φ)θ̇ = F · Reφ , dt ∂ φ̇ ∂φ d ∂T ∂T = 0 = F · R sin(φ)eθ . = mR2 sin2 (φ)θ̇ − dt ∂ θ̇ ∂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. Let us now suppose that the only force acting on the particle is gravity: F = −mgE3 , U = mgE3 · r = mgR cos(φ). For this case, the Lagrangian L is L= T−U m = (Ṙ2 + R2 sin2 (φ)θ̇2 + R2 φ̇2 ) − mgR cos(φ). 2 We can calculate Lagrange’s equations of motion using L, ∂L d ∂L − k = 0, dt ∂ q̇k ∂q 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.4 Lagrange’s Equations in the Presence of Constraints 75 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 recommended the first approach. As in the previous section, our exposition follows that of Casey [27]. Preliminaries We assume that the particle is subject to an integrable constraint, (r, t) = 0, and a nonintegrable constraint, f · v + e = 0. Further, we assume that the curvilinear coordinates are chosen such that the integrable constraint has the form (r, t) = q3 − d(t) = 0, and that the constraint forces are prescribed by use of Lagrange’s prescription: 3 3 i Fc = λ1 a + λ2 fia . i=1 Notice that f i = f · ai . 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 . The resultant force acting on the particle is 3 F = λ1 a3 + λ2 f i ai − ∇U + Fancon . i=1 The total nonconservative force acting on the particle is Fncon = Fc + Fancon . The kinetic energy of the particle is m m 1 2 3 i k v·v= aik q , q , q q̇ q̇ , 2 2 3 T= 3 i=1 k=1 where aik = ai · ak. Imposing the integrable constraint on T, we find the constrained kinetic energy: T̃ = T̃2 + T̃1 + T̃0 , (3.4) where m ãikq̇i q̇k, 2 2 T̃2 = 2 i=1 k=1 T̃1 = m 2 i=1 ˙ ãi3 q̇i d, T̃0 = m ã33 d˙ 2 . 2 (3.5) 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 76 June 2, 2008 Lagrange’s Equations of Motion for a Single Particle In these expressions, ãik = ãik(q1 , q2 , t) = aik(q1 , q2 , q3 = d(t)) is the constrained metric tensor. Notice that we use a tilde (˜) to denote imposition of the integrable constraint(s) and that the subscripts on T̃2 , T̃1 and T̃0 refer to the powers of q̇i . A direct calculation shows that∗ ∂T ∂ T̃ ∂T ∂ T̃ = , = 2, 3 1 1 2 3 3 3 ˙ ˙ ∂ q̇ q =d,q̇ =d ∂ q̇ ∂ q̇ q =d,q̇ =d ∂ q̇ ∂T ∂ T̃ ∂T ∂ T̃ (3.6) = 1, = 2, ∂q1 q3 =d,q̇3 =d˙ ∂q ∂q2 q3 =d,q̇3 =d˙ ∂q ∂T ∂ T̃ ∂T ∂ T̃ = = 0, = = 0. 3 3 3 ∂ q̇ q =d,q̇3 =d˙ ∂ q̇ ∂q3 q3 =d,q̇3 =d˙ ∂q3 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 T̃ to obtain the first two Lagrange’s equations of motion, but not the third. Results that are identical in form to (3.6) pertain to the partial derivatives of L and L̃ and U and Ũ. Notice that we did not impose the nonintegrable constraint on the kinetic energy T and the Lagrangian L. It is possible to do this, but the result is not useful to us here. Approach I In the first approach, we evaluate the partial derivatives in Lagrange’s equations of motion (3.2) in the absence of any constraints: d ∂L ∂L − k = Fancon · ak + (Fc = 0) · ak. dt ∂ q̇k ∂q Explicitly, these equations are 3 3 3 m ∂air i r ∂U d ai1 q̇i − q̇ q̇ + 1 = Fancon · a1 , m dt 2 ∂q1 ∂q i=1 i=1 r=1 m ∂air d ∂U ai2 q̇i − q̇i q̇r + 2 = Fancon · a2 , m dt 2 ∂q2 ∂q d m dt 3 i=1 3 3 i=1 r=1 i ai3 q̇ i=1 3 − 3 3 m ∂air i=1 r=1 2 ∂q3 q̇i q̇r + ∂U = Fancon · a3 . ∂q3 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 nonintegrable constraint and the constraint forces. The resulting equations govern the ∗ Suppose g = 10t2 . Then ∂g ∂t t=5 = 2(10)(5) = 100. In words, we evaluate the derivative of g with respect to t and then substitute t = 5 in the resulting function. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.4 Lagrange’s Equations in the Presence of Constraints 77 motion of the particle and the constraint forces: q3 = d, ˙ q̇3 = d, f 1 q̇1 + f 2 q̇2 + f 3 d˙ + e = 0, 3 3 3 d m ∂air i r ∂U i mai1 q̇ − q̇ q̇ + 1 = λ2 f 1 + Fancon · a1 , dt 2 ∂q1 ∂q i=1 i=1 r=1 m ∂air d ∂U ai2 q̇i − q̇i q̇r + 2 = λ2 f 2 + Fancon · a2 , m 2 dt 2 ∂q ∂q 3 i=1 3 3 i=1 r=1 m ∂air d ∂U ai3 q̇i − q̇i q̇r + 3 = λ1 + λ2 f 3 + Fancon · a3 . m 3 dt 2 ∂q ∂q 3 3 i=1 3 (3.7) i=1 r=1 We have refrained from ornamenting U, f i , ai , and aik with a tilde in the last four of these equations. It is crucial to notice that if the nonintegrable constraint were absent, then (3.7) would reduce to two sets of equations. The first of these sets, (3.7)4,5 , would yield differential equations for the unconstrained motion, q1 (t), and q2 (t), of the particle, whereas the second set, (3.7)6 , would provide the constraint force Fc = λ1 a3 acting on the particle. Approach II In Approach II, we work directly with L̃ = T̃ − Ũ. Here, as L̃ = L̃ q1 , q2 , q̇1 , q̇2 , t , the partial derivatives of L̃ with respect to q3 and q̇3 are zero. Consequently, using (3.6) and (3.2), we find that there are only two Lagrange’s equations: d dt d dt ∂ L̃ ∂ q̇1 ∂ L̃ ∂ q̇2 − ∂ L̃ = Fc · ã1 + Fancon · ã1 , ∂q1 − ∂ L̃ = Fc · ã2 + Fancon · ã2 . ∂q2 Introducing the expression for the constraint force Fc and the nonintegrable constraint, we find the equations governing λ2 , q1 (t), and q2 (t) are f 1 q̇1 + f 2 q̇2 + f 3 d˙ + e = 0, d ∂ L̃ ∂ L̃ − 1 = λ2 f 1 + Fancon · ã1 , 1 dt ∂ q̇ ∂q ∂ L̃ d ∂ L̃ − 2 = λ2 f 2 + Fancon · ã2 . 2 dt ∂ q̇ ∂q (3.8) 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 78 978 0 521 87483 0 June 2, 2008 Lagrange’s Equations of Motion for a Single Particle Notice that λ1 does not feature in these equations. In addition, if no nonintegrable constraint were present, then the differential equations provided by Approach II are all that are needed to determine q1 (t) and q2 (t). Equations (3.8)2,3 are examples of the Routh–Voss equations of motion. 3.5 A Particle Moving on a Sphere To clarify the two approaches just discussed, 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 a nonintegrable constraint on the motion of the particle. For the problem at hand, it is convenient to use a spherical polar coordinate system: q1 = θ, q2 = φ, q3 = R. Using this coordinate system, we can write the integrable constraint R = d(t) in the form (r, t) = R − d(t) = 0. As the sphere is smooth, we can use Lagrange’s prescription, Fc = λeR . The kinetic and potential energies of a particle in the chosen coordinate system are m 2 Ṙ + R2 sin2 (φ)θ̇2 + R2 φ̇2 , U = mgR cos(φ). T= 2 The constrained kinetic and potential energies are m ˙2 T̃ = d + d2 sin2 (φ)θ̇2 + d2 φ̇2 , 2 Ũ = mgd cos(φ). (3.9) Finally, the covariant basis vectors are a1 = R sin(φ)eθ , a2 = Reφ , a3 = eR . Their constrained counterparts ãi are easily inferred from these expressions. First, we use Approach II to obtain the equations governing θ(t) and φ(t). There are two equations: d ∂ L̃ ∂ L̃ − = F · ã1 dt ∂ θ̇ ∂θ = Fc · ã1 + Ddeθ · ã1 = Dd2 sin(φ), 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.5 A Particle Moving on a Sphere d dt 79 ∂ L̃ ∂ L̃ = F · ã2 − ∂φ ∂ φ̇ = Fc · ã2 + Ddeθ · ã2 = 0. Evaluating the partial derivatives of the constrained Lagrangian, we find that these equations become d (md2 sin2 (φ)θ̇) = Dd2 sin(φ), dt d 2 md φ̇ − md2 sin(φ) cos(φ)θ̇2 − mgd sin(φ) = 0. (3.10) dt Notice that the constraint force λeR 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 ∂L ∂L 2 2 =0 = mR sin (φ)θ̇ − dt ∂ θ̇ ∂θ d dt d dt = Fancon · R sin(φ)eθ , ∂L ∂L = mR2 sin(φ) cos(φ)θ̇2 + mgR sin(φ) = mR2 φ̇ − ∂φ ∂ φ̇ = Fancon · Reφ , ∂L ∂L = mṘ − = mR sin2 (φ)θ̇2 + mRφ̇2 − mg cos(φ) ∂R ∂ Ṙ = Fancon · eR . Next, we impose the integrable constraint and introduce the constraint force Fc to find the equations of motion: d (md2 sin2 (φ)θ̇) = Dd2 sin(φ), dt d (md2 φ̇) − md2 sin(φ) cos(φ)θ̇2 − mgd sin(φ) = 0, dt d ˙ − (md sin2 (φ)θ̇2 + mdφ̇2 − mg cos(φ)) = λ. (md) dt (3.11) Notice that the first two of these equations are identical to (3.10), whereas the third equation is an equation for the constraint force Fc . We could now introduce an additional constraint: f 1 θ̇ + f 2 φ̇ + f 3 Ṙ + e = 0. Using Lagrange’s prescription, we find that the total constraint force on the particle is f1 f2 Fc = λeR + λ2 eθ + eφ + f 3 eR . R sin(φ) R 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 80 978 0 521 87483 0 June 2, 2008 Lagrange’s Equations of Motion for a Single Particle To obtain the equations of motion for the case in which the nonintegrable constraint is active, we need to introduce only the constraint force associated with the nonintegrable constraint on the right-hand side of (3.11) and to append the nonintegrable constraint to the resulting equations: f 1 θ̇ + f 2 φ̇ + f 3 d˙ + e = 0, d (md2 sin2 (φ)θ̇) = Dd2 sin(φ) + λ2 f 1 , dt d (md2 φ̇) − md2 sin(φ) cos(φ)θ̇2 − mgd sin(φ) = λ2 f 2 , dt d ˙ − (md sin2 (φ)θ̇2 + mdφ̇2 − mg cos(φ)) = λ + λ2 f 3 . (md) dt It is left as an exercise to show what additional simplifications to these equations arise if the nonintegrable constraint were integrable with f 1 = 0, f 2 = 1, f 3 = 0, and e = 0. In this case, one will see that the particle moves on a circle of radius d sin(φ0 ). 3.6 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 needs to be presented. 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: [47, 149, 155, 201]. Reading Chapter 1 of Lanczos [124] for a related discussion on kinetic energy and geometry and the recent paper by Lützen [132] for a historical overview of the interaction between geometry and dynamics in the 19th 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 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.6 Some Elements of Geometry and Particle Kinematics (b) (a) 81 TP S P TP C C P S Figure 3.1. Two examples of manifolds and the tangent spaces to points on them: (a) a curve C and (b) a sphere S. 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. 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 TP M is also n-dimensional. For example, the tangent space TP C is a line for the curve shown in Figure 3.1(a), and the tangent space TP S is a plane for the sphere S shown in Figure 3.1(b). Continuing with the 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 (θ0 ) E2 ) + cos (φ0 ) E3 , eθ = eθ (θ0 ) = − sin (θ0 ) E1 + cos (θ0 ) E2 , form a basis for the tangent space TP S 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 example, for a particle that is free to move in M = E3 , the dimension of TP M is 3. Returning to the example of a sphere, we choose two points P1 and P2 on the sphere of radius R and consider a path V between them (see Figure 3.2). We wish to measure the distance one would travel along V. To do this, we first parameterize the curve with a parameter u, where u = uα at Pα . The curve can then be uniquely described by the functions θ(u) and φ(u).∗ To determine the distance s traveled along the curve, one method would be to evaluate the following integral: u2 dφ dφ dθ dθ + R2 du. (3.12) s = R2 sin2 (φ) du du du du u1 Referring to (3.4), (3.5), and (3.9), 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 ∗ For example, if the curve were a segment of the equator, then θ (u) = θ (u1 ) + (u2 −u1 ) 2πR and φ(u) = π 2. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 82 June 2, 2008 Lagrange’s Equations of Motion for a Single Particle P1 Figure 3.2. A curve V connecting two points on a V sphere S. The velocity vector vrel of a particle moving on this curve would lie in the tangent plane TP S at each point P of the curve. P2 vrel S sphere: u2 s = u1 t2 = 2T̃2 m dt du du 2T̃2 dt = m t1 t2 ||vrel || dt. t1 For the second integral, we changed variables and parameterized the path by using t rather than u. Turning to the example of a circle, the reader is invited to show that a measure of distance corresponding to (3.12) can be established by using the single polar coordinate θ. For M = E3 , the measure of distance can be defined in a standard manner by use of Cartesian coordinates: u2 3 dxk dxk s = du. du du u1 k=1 t It is easy to see that this expression can be rewritten as t12 2T dt. m From the previous discussion, for an n-dimensional manifold M, a measure of distance similar to that provided by (3.12) can be established∗ : s = u2 n n u1 ãik i=1 k=1 ∂qi ∂qk du. ∂u ∂u Parameterizing the path by using time t instead of a variable u, we find t2 n n s = ãikq̇i q̇kdt. t1 ∗ (3.13) i=1 k=1 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.7 The Geometry of Lagrange’s Equations of Motion 83 Following the advocacy of Hertz [92], we can conveniently imagine a (representative) particle of mass m moving on a manifold M. The manifold is known as the configuration manifold. The velocity of the particle relative to M is simply vrel = nk=1 q̇kãk. Thus (3.13) can be expressed as t2 ||vrel || dt. (3.14) s = t1 We see from this expression that, in defining s, we have also defined a measure of the magnitude of a vector vrel ∈ TP M. 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 [179]. Distance measure (3.13) is clearly intimately related to the kinetic energy of a particle, and, following Synge [205], ds = n n ãikq̇i q̇kdt (3.15) i=1 k=1 is known as the kinematical line-element. This measure of distance has a long history with important contributions by several esteemed figures such as Jacobi [103] and Ricci and Levi-Civita [178]. Other choices of ds are available (see, for instance, [123, 124, 149, 205]), and some of them feature prominently in relativistic mechanics. The freedom of selection is similar to the notion that different measures of distance such as meters and feet are possible and is intimately related to Einstein’s theory of relativity. We remarked earlier on a (representative) particle of mass m moving on M. Clearly, for a single particle, such a construction can be easily achieved. Indeed, for a single particle the configuration manifold 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 only rather recently by Casey [27].∗ Subsequently, he extended this construction to a single rigid body [28] and a system of rigid bodies [30]. 3.7 The Geometry of Lagrange’s Equations of Motion Some readers will have gained the perspective that the 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 M. 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 the θ and φ by taking the d sin(φ)eθ and deφ components ∗ We shall examine his construction in Section 4.7. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 84 June 2, 2008 Lagrange’s Equations of Motion for a Single Particle of F = ma. These two vectors, d sin(φ)eθ and deφ , form a basis for the tangent space TP S to a point P of the sphere S. Furthermore, because the constraint force associated with the integrable constraint λeR 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 Casey [27], Lanczos [124], and Synge [205]. 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: (r, t) = q3 − d(t) = 0, where d3 (t) is a known function. When considered in E3 , the constraint =0 represents a moving two-dimensional surface: In this case, a q3 coordinate surface. As mentioned earlier, this surface is known as the configuration manifold M (see Figure 3.3). The velocity of the particle relative to this surface has the representation vrel = q̇1 ã1 + q̇2 ã2 . The coordinates q1 and q2 in this case are known as the generalized coordinates, and 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 the particle constrained to move on the surface has only two. Recall that at each point P of M, {a1 , a2 } evaluated at P is a basis for the tangent plane TP M to M at P. We can also define a relative kinetic energy Trel = T̃2 : m m vrel · vrel = ãi · ãkq̇i q̇k. 2 2 2 Trel = 2 i=1 k=1 We now consider a particle moving on M. To calculate the distance traveled by the particle on the surface in a given interval t1 − t0 of time t, we integrate the magnitude of its velocity vrel with respect to time: s(t1 ) − s(t0 ) = t1 t0 = t0 t1 √ vrel · vrel dt 2Trel dt. m 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.7 The Geometry of Lagrange’s Equations of Motion q2 coordinate curve a3 q1 coordinate curve a2 M a1 q3 = d(t) coordinate surface O Figure 3.3. The configuration manifold M of a particle moving on a surface. Here, the coor- dinates q1 and q2 are known as the generalized coordinates and M is the q3 = d(t) coordinate surface. We can differentiate this result with respect to t to find the kinematical line-element ds: 2 2 2Trel ds = (3.16) ãi · ãkdqi dqk. dt = m i=1 k=1 As emphasized previously in Section 3.6, 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 (3.16) is none other than (3.12). 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 = λã3 . Then Fc · ã1 = Fc · ã2 = 0, and the constraint force does not appear in the first two Lagrange’s equations. The forces Q1 = F · ã1 , Q2 = F · ã2 are known as the generalized forces, and the expressions we use for them are equivalent to those used in other texts on dynamics (for example, [14, 80]). In addition, Approach II can be used to obtain the differential equations governing q1 (t) and q2 (t): ∂ L̃ d ∂ L̃ − 1 = Fancon · ã1 , 1 dt ∂ q̇ ∂q ∂ L̃ d ∂ L̃ − 2 = Fancon · ã2 . (3.17) 2 dt ∂ q̇ ∂q Imposing a nonintegrable constraint on the motion of the particle will not change M. Furthermore, this constraint will, in general, introduce constraint forces into 85 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 86 June 2, 2008 Lagrange’s Equations of Motion for a Single Particle Equations (3.17). These forces will destroy the decoupling that Lagrange’s equations achieve for integrable constraints. A Particle Subject to Two Integrable Constraints We now turn to the case in which a particle is subject to two integrable constraints: 1 (r, t) = q3 − d3 (t) = 0, 2 (r, t) = q2 − d2 (t) = 0. 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 3 1 1 = 0 and 2 = 0 in E defines a curve – in this case a q coordinate curve (see Figure 3.4). The configuration manifold M in this case corresponds to the q1 coordinate curve. This coordinate is the generalized coordinate for the system. We can easily represent the velocity vector of the particle relative to M, which we again denote by vrel : vrel = q̇1 ã1 . It should be clear that ã1 is tangent to M. Indeed, a1 evaluated at P is a basis vector for the one-dimensional space TP M. In addition, we can associate with vrel a relative kinetic energy: Trel = m m vrel · vrel = ã1 · ã1 q̇1 q̇1 . 2 2 Paralleling previous developments [see (3.15)], the kinematical line-element for M is 2Trel ds = dt = ã1 · ã1 dq1 dq1 . m m ã1 r O q3 = d3 (t) coordinate surface q2 = d2 (t) coordinate surface Figure 3.4. A particle moving on a curve. In this figure, ã1 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 M. The vectors ã2 and ã3 , which are not shown, are normal to M. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.8 A Particle Moving on a Helix 87 increasing s & θ eb en et Figure 3.5. A 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 = et × en is the binormal vector. eb en et E3 E1 O E2 With regard to Lagrange’s equations of motion, if the constraint forces are prescribed by use of Lagrange’s prescription, Fc = λ1 ã3 + λ1 ã2 , then we can easily find the differential equation governing q1 (t) by using Approach II: ∂ L̃ d ∂ L̃ − 1 = Fancon · ã1 . dt ∂ q̇1 ∂q Here, L̃ = L̃ q1 , q̇1 , t and F · ã1 is the generalized force for this problem. 3.8 A Particle Moving 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 3.5). 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.∗ Curvilinear Coordinates, Basis Vectors, and Other Kinematics A helix is defined by the intersection of two surfaces: a cylinder r = R and a helicoid z = cθ, where c and R are constants. To conveniently define these surfaces, we define a curvilinear coordinate system: x2 q1 = θ = tan−1 , q2 = r = x21 + x22 , x1 x2 . q3 = η = z − αrθ = x3 − α x21 + x22 tan−1 x1 ∗ This is a coordinate system in which ai are not necessarily parallel to ai . 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 88 June 2, 2008 Lagrange’s Equations of Motion for a Single Particle It is appropriate to notice that r = x1 E1 + x2 E2 + 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 = reθ + αrE3 , a2 = er + αθE3 , a3 = E3 . In addition, using the representation of the gradient in cylindrical polar coordinates, we find that the contravariant basis vectors are a1 = 1 eθ , r a2 = er , a3 = E3 − αθer − αeθ . j You should notice that ai · a j = 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 R is (from [159]) et = √ 1 1 + α2 (eθ + αE3 ) , 1 eb = √ (E3 − αeθ ) . 1 + α2 en = −er , Furthermore, the torsion τ, curvature κ, and arc-length parameter s of the helix are τ= α , R(1 + α2 ) κ= 1 , R(1 + α2 ) s = R 1 + α2 (θ − θ0 ) − s0 . These results also apply to a helix for which α and R 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 = 3 i i=1 q̇ ai . From this result, we can immediately write v = θ̇(reθ + αrE3 ) + ṙ(er + αθE3 ) + η̇E3 . Furthermore, the kinetic energy of the particle is m 2 ṙ + r2 θ̇2 + (η̇ + αṙθ + αrθ̇)2 . T= 2 When the particle is in motion on the helix, it is subject to two constraints and 2 = 0: 1 (r, t) = q2 − R, 2 (r, t) = q3 . 1 =0 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.8 A Particle Moving on a Helix 89 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 = Fa + λ1 ã2 + λ2 ã3 + F f , where θ̇ã1 F f = −µd λ1 ã2 + λ2 ã3 . θ̇ã1 On the other hand, if the particle is not moving relative to the helix, i.e., θ is constant, then F = Fa + λ1 ã2 + λ2 ã3 + λ3 ã1 . The friction force in this case is subject to the static friction criterion: F f ≤ µs ||N|| , where ã1 F f = (λ1 ã + λ2 ã + λ3 ã ) · ||ã1 || 2 3 1 ã1 , ||ã1 || N = λ1 ã2 + λ2 ã3 + λ3 ã1 − F f . Balance of Linear Momentum and Lagrange’s Equations For an unconstrained particle moving in E3 , we have the three Lagrange’s equations: ∂T d ∂T − i = F · ai . i dt ∂ q̇ ∂q For the present coordinate system θ, r, η, these equations read d ∂T = mr2 θ̇ + mαr(η̇ + αṙθ + αrθ̇) dt ∂ θ̇ ∂T = mαṙ(η̇ + αṙθ + αrθ̇) = F · (a1 = reθ + αrE3 ), − ∂θ d ∂T = mṙ + mαθ(η̇ + αṙθ + αrθ̇) dt ∂ ṙ ∂T − = mrθ̇2 + mαθ̇(η̇ + αṙθ + αrθ̇) = F · (a2 = er + αθE3 ), ∂r d ∂T ∂T = m(η̇ + αṙθ + αrθ̇) − = 0 = F · (a3 = E3 ) . dt ∂ η̇ ∂η 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 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 90 June 2, 2008 Lagrange’s Equations of Motion for a Single Particle some algebra, for the case in which the particle is moving relative to the helix, we find three equations: d θ̇ m(1 + α2 )R2 θ̇ = Fa · ã1 − µd λ1 ã2 + λ2 ã3 ||ã1 || , dt |θ̇| θ̇ã1 · ã2 d 2 mα Rθθ̇ − m(1 + α2 )Rθ̇2 = Fa · ã2 + λ1 − µd λ1 ã2 + λ2 ã3 , θ̇ã1 dt θ̇ã1 · ã3 d mαRθ̇ = Fa · ã3 + λ2 − µd λ1 ã2 + λ2 a3 , θ̇ã1 dt where ã1 = Reθ + αRE3 , ã2 = er + αθE3 , ã3 = E3 . These three equations provide a differential equation for the unconstrained motion of the particle and two equations for the unknowns λ1 and λ2 . 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: λ3 = −Fa · ã1 , λ2 = −Fa · ã2 , λ1 = −Fa · ã3 . It remains to use the static friction criterion, but this is left as an easy exercise. The Particle on a Smooth Helix In this case, F = Fa + λ1 ã2 + λ2 ã3 , and because Fc · ã1 = 0, Lagrange’s equations of motion decouple: d m(1 + α2 )R2 θ̇ = Fa · ã1 , dt d 2 mα Rθθ̇ − m(1 + α2 )Rθ̇2 = Fa · ã2 + λ1 , dt d mαRθ̇ = Fa · ã3 + λ2 . dt Consequently, the desired differential equation is m(1 + α2 )R2 θ̈ = Fa · (Reθ + αRE3 ), and the constraint force is Fc = λ1 ã2 + λ2 ã3 = mα2 Rθ̈θ − mRθ̇2 − Fa · ã2 ã2 + mαRθ̈ − Fa · ã3 ã3 . Once θ as a function of time has been calculated from the ordinary differential equation, then Fc as a function of time can be determined. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 3.9 Summary 91 To illustrate the previous equations, consider the case in which the applied force is gravitational, Fa = −mgE3 . Then, from the preceding equations, m(1 + α2 )R2 θ̈ = −mgαR. (3.18) Subject to the initial conditions θ(t0 ) = θ0 and θ̇(t0 ) = ω0 , this equation has the solution gα θ(t) = θ0 + ω0 (t − t0 ) − (t − t0 )2 . 2R(1 + α2 ) Using this result, we find that the constraint force is mgαθ(t) mg 2 Fc = − mRθ̇ (t) ã2 + (E3 − α(θer + eθ )) , 2 1+α 1 + α2 where θ(t) is as previously given. 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: T̃ = m m 2 2 R θ̇ + (αRθ̇)2 = (1 + α2 )R2 θ̇2 . 2 2 Furthermore, ∂ T̃ d ∂ T̃ − = F · ã1 = F · (Reθ + αRE3 ) = Fa · (Reθ + αRE3 ) . dt ∂ θ̇ ∂θ A quick calculation shows that the resulting differential equation is identical to that obtained previously [(3.18)]. Clearly, Lagrange’s equations calculated with Approach II (i.e., with T̃) 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 ∂∂T̃ṙ = ∂∂T̃η̇ = ∂∂rT̃ = ∂∂ηT̃ = 0. Consequently, we cannot recover the other two Lagrange’s equations once we have imposed the constraints. 3.9 Summary 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 ∂T ∂T − i = F · ai . dt ∂ q̇i ∂q In one of the following exercises, 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 (3.22) and a contravariant form (3.23). If we 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 92 Exercise 3.1 decompose the forces acting on the particle into conservative and nonconservative forces, then we can transform (3.1) to (3.2): ∂L d ∂L − i = Fncon · ai . dt ∂ q̇i ∂q 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 Fc = 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 ∂ L̃ ∂ L̃ α = 1, 2. (3.19) − α = Fncon · ãα , α dt ∂ q̇ ∂q These equations feature the constrained Lagrangian L̃ that we obtain from L by imposing the integrable constraint q3 = f (t), and, most important, do not feature λ. That is, equations of motion (3.19) 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 both 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. EXERCISES 3.1. Recall from Exercise 1.5 in Chapter 1 that, for a parabolic coordinate system {u, v, θ}, a1 = ∂r = ver + uE3 , ∂u a3 = a2 = ∂r = uer − vE3 , ∂v ∂r = uveθ , ∂θ and a1 = 1 a1 , u2 + v2 a2 = 1 a2 , u2 + v2 a3 = 1 eθ . uv 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 3.1–3.2 93 (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 m(u2 + v2 )u̇ − m(u̇2 + v̇2 )u − mv2 uθ̇2 = F · a1 , dt d m(u2 + v2 )v̇ − m(u̇2 + v̇2 )v − mu2 vθ̇2 = F · a2 , dt d 2 2 mu v θ̇ = F · a3 . dt (b) Next, we are interested in a particle that is moving on the parabolic surface of revolution: c2 = −z + z2 + r2 , where c is a constant. A vertical gravitational force −mgE3 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 N = − mu̇2 c + mu2 cθ̇2 + mgc a2 . Show that the two second-order differential equations governing the generalized coordinates can be written as a single second-order differential equation: m(u2 + c2 ) ü + mu̇2 u − h2 = −mgu, mu3 c2 (3.20) 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 meters1/2 , what is a dimensionless form of equations of motion (3.20)? (c) Show that the solutions of (3.20) conserve the energy E= h2 mg 2 m 2 (u + c2 )u̇2 + (u − c2 ). + 2 2 2 2mu c 2 How does one arrive at this expression for E? 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 (3.23)].∗ This problem is adapted from the texts of McConnell [139] and Synge and Schild [208]. We take this opportunity to note that (3.23) can be found in an early paper by Ricci and Levi-Civita [178]. 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 ∗ 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 94 Exercise 3.2 force F = 3 i=1 Fi ai = 3 i=1 F i ai ∗ : d ∂T ∂T − k = F · ak, k dt ∂ q̇ ∂q where m aikq̇i q̇k, 2 3 T = T(qr , q̇s ) = 3 (3.21) 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 q̈ + m i i=1 3 3 [si, k]q̇i q̇s = Ġ · ak = Fk, (3.22) i=1 s=1 where a Christoffel symbol of the first kind is defined by [si, k] = ∂as · ak. ∂qi It is important to note that [si, k] = [is, k]. The second form of Lagrange’s equations is known as the contravariant form: mq̈k + m 3 3 ksi q̇i q̇s = Ġ · ak = F k, (3.23) i=1 s=1 where a Christoffel symbol of the second kind is defined by kij = ∂ai k ·a . ∂qj Notice that kij = kji . This form of Lagrange’s equations is used in numerical simulations of mechanical systems.† (a) Show that the Christoffel symbols have the representations 1 ∂aki ∂ask ∂asi [si, k] = + − k , 2 ∂qs ∂qi ∂q and kij = 3 akr [ij, r]. r=1 ∗ † The indices i, j, k, r, and s range from 1 to 3. Most numerical integration packages assume that the differential equations to be integrated are of the form ẋ = f(x, t). By defining the set of variables (states) x1 = q1 , . . . , x3 = q3 , x4 = q̇1 , . . . , x6 = q̇3 , the contravariant form of Lagrange’s equations can be easily placed in the form ẋ = f(x, t). 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 3.2–3.3 95 (b) Starting from Lagrange’s equations, ∂T d ∂T − k = F · ak, k dt ∂ q̇ ∂q derive the following representation for the covariant component form∗ : m 3 aki q̈i + m i=1 3 3 [si, k]q̇i q̇s = Ġ · ak = Fk. i=1 s=1 (c) Starting from Lagrange’s equations in the form m 3 aki q̈i + m i=1 3 3 [si, k]q̇i q̇s = Ġ · ak = Fk, s=1 i=1 derive the following representation for the contravariant component form† : mq̈k + m 3 3 ksi q̇i q̇s = Ġ · ak = F k. s=1 i=1 (d) For which coordinate system do the Christoffel symbols vanish? 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 , and the contravariant basis vectors are a1 = eR , a2 = 1 eφ , R a3 = 1 eθ . R sin(φ) Furthermore, the linear momentum and kinetic energy of a particle of mass m are m 2 G = mṘa1 + mφ̇a2 + mθ̇a3 , Ṙ + R2 φ̇2 + R2 sin2 (φ)θ̇2 . 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. ∗ † Hint : Expand the partial derivatives of T using the representation (3.21). 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. 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 96 Exercises 3.4–3.5 3.4. Consider a particle that is in motion on a rough surface. A curvilinear coordinate system q1 , q2 , q3 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 = q̇1 ã1 + q̇2 ã2 . (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 + F f = 3 λi ai , i=1 where the tildes are dropped for convenience. (i) Show that ⎡ ⎤ ⎡ a11 λ1 ⎢ ⎥ ⎢ ⎣λ2 ⎦ = ⎣a12 a12 a22 ⎥⎢ ⎥ 0⎦ ⎣F f2 ⎦ , a13 a23 1 λ3 0 ⎤⎡ F f1 ⎤ N where N, and F f1 and F f2 uniquely define the normal force N and friction force F f , respectively, and aik = ai · ak with i, k = 1, 2, 3. (ii) For which coordinate systems do F f1 = λ1 , F f2 = λ2 , and N = λ3 ? Give an example to illustrate your answer. (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. 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 = αθ, where α is a constant. A gravitational force −mgE3 acts on the particle. (a) Consider the following curvilinear coordinate system for E3 : q1 = θ, q2 = r, q3 = ν = z − αθ. Show that a1 = reθ + αE3 , a2 = er , a3 = E3 , and that a1 = 1 eθ , r a2 = er , a3 = E3 − α eθ . r 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 3.5–3.6 97 (b) Consider a particle moving on the smooth helicoid: (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) Show that the equations governing the unconstrained motion of the particle are d 2 m r + α2 θ̇ = −mgα, dt d (mṙ) − mrθ̇ 2 = 0. dt (3.24) (iii) Prove that the angular momentum HO · E3 is not conserved. (c) Suppose the nonintegrable constraint rθ̇ + 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 nonintegrable constraint. Indicate how you would solve these equations to determine the motion of the particle and the constraint forces acting on it. 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 3.6). The particle is under the influence of a gravitational force −mgE3 . E3 O E2 g E1 r m Figure 3.6. Schematic of a particle of mass m moving on the inside of a hemisphere of radius R0 . (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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 98 Exercises 3.6–3.7 (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 E0 : 2E0 + 2mg sin (φ) + mg cos (φ) eR . N= − R0 (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. 3.7. As shown in Figure 3.7, consider a bead of mass m that is free to move on a smooth semicircular wire of radius R0 . The wire has a constant angular velocity 0 E3 and whirls about the configuration shown in the figure. The particle is also (t) E3 g O E2 r E1 m (t) Figure 3.7. Schematic of a particle of mass m moving on a semicircular path that is being whirled about the vertical at a speed (t) = 0 . 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 3.7–3.8 99 under the influence of a gravitational force −mgE3 . This is a classical problem that is discussed in several textbooks (see, for example, [78]). (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 particle. (b) Using Lagrange’s equations, establish the equation of motion for the particle: g sin (φ) . (3.25) φ̈ = 20 cos (φ) + R0 After nondimensionalizing (3.25), numerically integrate the resulting differential equation and construct its phase portrait for values of 0.5, 1.0, 1.5 of the parameter R g2 . 0 0 (c) Recall that an equilibrium point x = x0 of the differential equation ẍ = f (x) is such that ẋ = 0 and f (x0 ) = 0. Show that (3.25) has three equilibria: g −1 φ0 = 0, φ0 = π, φ0 = cos − . R0 20 Give physical interpretations for these equilibria and show that the third one is possible if, and only if, 20 is sufficiently large. How do these results correlate to your phase portraits? (d) Starting from the work–energy theorem Ṫ = F · v, prove that the total energy of the particle is not conserved: Ė = Nθ R0 0 sin (φ) , where Nθ is the eθ component of the normal force acting on the particle. 3.8. Consider a particle of mass m moving in E3 . If coordinate system (1.6) 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. (b) As illustrated in Figure 3.8, 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 equations of motion for the particle. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 100 Exercise 3.9 y m q2 = −4 2 −4 6 x Figure 3.8. Schematic of a particle moving on a parabola in the x − y plane. −4 3.9. Consider a particle of mass m moving on a smooth ellipsoid: y2 z2 x2 + + = 1. a2 b2 c2 (a) With the help of a suitable curvilinear coordinate system, establish expressions for the constrained kinetic energy of the particle.∗ (b) Assuming that a gravitational force −mgE3 acts on the particle, establish the two second-order differential equations governing the motion of the particle. (c) Numerically integrate the equations you found in (b) and discuss features of the motion of the particle. ∗ The coordinate systems you need are not discussed in this text but are readily found in the literature. They are often known as “confocal ellipsoidal coordinates.” 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 PART TWO DYNAMICS OF A SYSTEM OF PARTICLES 101 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 102 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 4 978 0 521 87483 0 June 9, 2008 The 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 in Chapter 15 of Synge and Griffith [207].∗ 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 use the construction of a representative single particle by Casey [27]. All 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 4.11. The chapter closes with a discussion of a canonical form of Lagrange’s equations of motion in which time-independent integrable constraints are present. For many specific problems, we 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.† Indeed, once we establish Lagrange’s equations we can also ignore the explicit construction of the single particle. However, for many cases – which are not possible to treat using the approach adopted in most dynamics textbooks – we find that the use of Synge’s and Griffith’s representation of Lagrange’s equations of motion allows us to tremendously increase the range of application of Lagrange’s equations. For instance, as will be shown in some of the examples in Chapter 5, dynamic Coulomb friction is accommodated. ∗ † Related derivations for systems of particles can be found in several texts. For example, Section 6-6 of Greenwood [79] and Section 21 of Whittaker [228]. 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. 103 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 104 The Equations of Motion for a System of Particles Fi E3 mi Figure 4.1. A single particle of mass mi in E3 . The position vector of this particle is ri , and the resultant external force acting on the particle is Fi . ri O E2 E1 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. We consider a system of N particles, each of which is in motion in threedimensional Euclidean space E3 . For the particle of mass mi (see Figure 4.1), the position vector is ri = 3 j xi E j . j=1 We also recall that the kinetic energy of the particle is 1 mi vi · vi . 2 It is also convenient to recall that the linear momentum of the particle of mass mi is Gi = mi ṙi = mi vi . The resultant force acting on the particle of mass mi has the representation Ti = Fi = 3 j Fi E j . j=1 The balance of linear momentum for the particle of mass mi is Fi = mi v̇i . As a consequence of the balance of linear momentum for the particle, we have the angular momentum theorem: ḢOi = ri × Fi , where HOi = 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 Ṫi = Fi · vi for the particle of mass mi . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.3 Coordinates 105 The Kinetic Energy For the system of particles, the combined (total) kinetic energy T is the sum of the kinetic energies: T = T1 + · · · + TN . With the help of the work–energy theorem for each particle, we can determine the corresponding result for the system of particles: Ṫ = 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 r̄, where r̄ = 1 (m1 r1 + · · · + mN rN ) . m1 + · · · + mN It is easy to show from this result that the linear momentum of a system of particles has the representation G = (m1 + · · · + mN ) r̄˙. Next we examine a particle of mass m moving in E3N , and it is important not to confuse this particle with C. 4.3 Coordinates In many problems, Cartesian coordinates for 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 = xE1 + yE2 + zE3 , r2 = xE1 + yE2 + zE3 + R2 eR2 . (4.2) In this equation, R2 = L0 is the length of the rod connecting the particles, and eR2 = sin(φ2 ) (cos(θ2 )E1 + sin(θ2 )E2 ) + cos(φ2 )E3 . m1 E3 Figure 4.2. A particle of mass m1 attached by a rigid rod of length L0 to a particle of mass m2 . g O E1 m2 E2 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 106 June 9, 2008 The Equations of Motion for a System of Particles with related definitions for eφ2 and eθ2 . Notice that eR2 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 q1 , . . . , q3N . As in the case of a single particle, we assume that we can determine the unique Cartesian coordinates for a system of particles once q1 , . . . , q3N are known, and vice versa: qK = q̂K x11 , x21 , x31 , . . . , x1N , x2N , x3N (K = 1, . . . , 3N), j j xi = x̂i (q1 , . . . , q3N ) (i = 1, . . . , N and j = 1, 2, 3). To calculate Lagrange’s equations of motion, we need to calculate the following 3N2 vectors: ∂ri (K = 1, . . . , 3N and i = 1, . . . , N). ∂qK These vectors play the role of the basis vectors ak that we used with the single particle earlier. 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 = R2 , q5 = θ2 , q6 = φ2 . For this coordinate system, and with the help of (4.2), ∂r1 = Ej , ∂qj ∂r1 = 0, ∂q(j+3) and ∂r2 = Ej , ∂qj ∂r2 = eR2 , ∂q4 ∂r2 = R2 sin(φ2 )eθ2 , ∂q5 ∂r2 = R2 eφ2 , ∂q6 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 Fig. 4.3. m1 E3 O E2 linear spring m2 E1 Figure 4.3. A particle of mass m1 attached by a linear spring of stiffness K and unstretched length L0 to a particle of mass m2 . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.4 Constraints and Constraint Forces 107 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 L0 (see Figure 4.2). Then the constraint imposed by the rod on their motion is ||r1 − r2 || − L0 = 0. To enforce this constraint, the particles will be subject to equal and opposite constraint forces: Fc1 = −µt, Fc2 = µt, (4.3) where µ is the tension force in the rod and the unit vector t points from m1 to m2 : t= r2 − r1 . ||r2 − r1 || If we describe r2 − r1 = L0 eR2 , which we did earlier, then t = eR2 . Writing the constraint as = 0, where = r2 − r1 − L0 , then, as ∂ r2 − r1 , =− ∂r1 r1 − r2 ∂ r2 − r1 , = ∂r2 r1 − r2 we observe that (4.3) can be expressed as Fc1 = µ ∂ , ∂r1 Fc2 = µ ∂ . ∂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.∗ 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 = xE1 + yE2 . Further, we need to supplement x and y only by an angle φ to determine the position of any point on the cart. Thus, ∗ For further details on this matter, and references to the vast literature on this prototypical nonintegrably constrained system, see O’Reilly and Tongue [165] and Ruina [186]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 108 The Equations of Motion for a System of Particles m3 e2 E2 e1 m1 e1 φ C E1 E1 O Q Vehicle m2 Figure 4.4. A 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 m3 models a slipping wheel. 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 = µe2 . (4.5) This force ensures that m1 moves only in the e1 direction. Lagrange’s Prescription We now have sufficient motivation to motivate 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: = 0, where = (r1 , r2 , . . . , rN , t). Constraint (4.6) can be differentiated to yield a constraint of the form N i=1 fi · vi + e = 0, (4.6) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.4 Constraints and Constraint Forces 109 where fi = ∂ , ∂ri e= ∂ . ∂t ∂ , for example, we assume that r2 , . . . , rN are fixed. Lagrange’s When evaluating ∂r 1 prescription for the constraint force Fci acting on the particle of mass mi is Fci = µfi = µ ∂ . ∂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.∗ 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 Fci = µfi (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. 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 Fci · 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 arises when ψ is not an explict function of time: = (r1 , . . . , rN ). In a previous example, shown in Figure 4.2, we note that depended on r1 and r2 . If, in addition, L0 were a function of time, then = (r1 , r2 , t). ∗ See, for example, Subsection 8.6.2 in Chapter 8. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 110 The Equations of Motion for a System of Particles Newton’s Third Law For the constraint ||r1 − r2 || − L0 = 0 mentioned earlier, Lagrange’s prescription yields Fc1 = µ r 1 − r2 , ||r1 − r2 || Fc2 = µ r 2 − r1 . ||r1 − r2 || Notice that these constraint forces point along the rod – which is what we would expect from physical grounds. Furthermore, Fc2 = −Fc1 – which is none other than Newton’s third law.∗ 4.5 Conservative Forces and Potential Energies Conservative forces in the dynamics of systems of particles commonly occur. 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 , Un = − Gm1 m2 , ||r2 − r1 || and the potential energy of a spring force between two particles of mass m1 and m2 : Us = K (||r2 − r1 || − L0 )2 . 2 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 rearraging, we find that N ∂U Fconi + · vi = 0. ∂ri (4.9) i=1 ∗ For more details on this interesting result, see Noll [153], O’Reilly and Srinivasa [163], and Subsection 8.6.4 in this book. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.6 Lagrange’s Equations of Motion 111 Assuming that, for each i, Fconi + ∂U is independent of v1 , . . . , vN and that (4.9) is ∂ri true for all possible v1 , . . . , vN , then we conclude that Fconi = − ∂U ∂ri (i = 1, . . . , N). (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 Fcon1 = − r1 − r2 , ||r2 − r1 || ||r2 − r1 || Gm1 m2 Fcon2 = − 2 r2 − r1 , ||r2 − r1 || ||r2 − r1 || Gm1 m2 2 and Fcon1 = − K r 1 − r2 , (||r2 − r1 || − L0 ) ||r2 − r1 || 2 Fcon2 = − K r 2 − r1 , (||r2 − r1 || − L0 ) ||r2 − r1 || 2 respectively. You might again notice that the pairs of forces obey Newton’s third law identically. This interesting result was first pointed out by Lanczos [124] and Noll [153]. 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 r̈i = Fi (i = 1, . . . , N). We now show that these equations are equivalent to Lagrange’s equations of motion: d ∂T ∂T − K = K (4.11) (K = 1, . . . , 3N) , dt ∂ q̇K ∂q where K = N i=1 Fi · ∂ri . ∂qK As there are no constraints on the system, all of the coordinates q1 , . . . , q3N are generalized coordinates and K are the generalized forces. Our results here are adapted from the work of Synge and Griffith. Their derivation is presented in Section 15.1 of [207]. Shortly, an alternative derivation of (4.11) that is due to Casey [27] will be presented.∗ ∗ See Equation (4.20) in Section 4.7. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 112 The Equations of Motion for a System of Particles 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 (i = 1, . . . , N). With these coordinates, we then 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: ∂vi ∂ri = K. K ∂ q̇ ∂q (4.12) This identity is often (fondly) referred to as “canceling the dots.” To verify the identity, we use the chain rule to calculate vi : vi = ṙi = 3N q̇J J =1 ∂ri . ∂qJ Taking the partial derivative of both sides of this expression with respect to q̇K then yields (4.12). We are now in a position to show that ∂T = mi vi · K ∂ q̇ N i=1 ∂ri ∂qK ∂T d = mi vi · K ∂q dt N , i=1 ∂ri ∂qK . (4.13) First, ∂T ∂ = K K ∂ q̇ ∂ q̇ = N N mi N vi · vi mi vi · ∂vi ∂ q̇K mi vi · ∂ri . ∂qK i=1 = 2 i=1 i=1 Notice that we used (4.12) in the final stages of this calculation. The other result follows similarily and so we skip some of the intermediate stages: ∂T ∂vi = mi vi · K K ∂q ∂q N i=1 = N i=1 mi vi · ∂ ∂qK 3N J =1 ∂ri q̇ ∂qJ J 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.7 Construction and Use of a Single Representative Particle = N mi vi · 3N J =1 i=1 = N i=1 d mi vi · dt ∂ q̇J J ∂q ∂ri ∂qK ∂ri ∂qK 113 . In the last stages of this calculation, we used the identities ∂ 2 ri ∂ 2 ri = J K, K J ∂q ∂q ∂q ∂q f˙ = 3N q̇J J =1 ∂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): N N d ∂T ∂T d ∂ri d ∂ri − K = mi vi · K − mi vi · dt ∂ q̇K ∂q dt ∂q dt ∂qK i=1 i=1 N ∂ri mi v̇i · K = ∂q i=1 = N Fi · i=1 ∂ri ∂qK = K , (4.14) where the force K is K = 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 [27]. In this work, he constructs a single representative particle moving in a 3N-dimensional space subject to a force. With the help of this construction, the derivation of Lagrange’s equations of motion follows easily, and tremendous insight can be gained into the geometry of Lagrange’s equations of motion. In this section, we follow Casey [27], 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 Ṫ = · ṙ, = mr̈. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 114 June 9, 2008 The Equations of Motion for a System of Particles In Casey [27], m is chosen (without loss in generality) to be the sum of the masses: m = m1 + · · · + mN . Here, we assume only that m > 0. As we noted earlier in Section 3.6, the interested reader is also referred to Chapter 1 of Lanczos [124] 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 K=1 bK ēK = N 3 b3i+j−3 ē3i+j−3 . i=1 j=1 Here, {ē1 , . . . , ē3N } is a fixed orthonormal basis for E3N . We also define two other sets of basis vectors: mi ē3i+j−3 , e3i+j−3 = m m ē3i+j−3 . e3i+j−3 = mi Here i = 1, . . . , N and j = 1, 2, 3. Notice that eK · eJ = δJK . The particle has a position vector r and a force vector . Both of these vectors have representations similar to that for b. 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: 1 1 mv · v = mi vi · vi . 2 2 N T= i=1 Substituting, v = ṙ = N 3 ṙ3i+j−3 ē3i+j−3 , vi = ṙi = i=1 j=1 3 j ẋi E j , j=1 and equating kinetic energies, we find one solution for r: N N 3 3 mi r= ē3i+j−3 = (ri · E j ) (ri · E j ) e3i+j−3 . m i=1 j=1 (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). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.7 Construction and Use of a Single Representative Particle 115 Prescription for the Force Vector The force vector is prescribed by the requirement that mv̇ = . Substituting (4.15) and using the balance of linear momentum for each particle, one is led to the prescription = N 3 (Fi · E j ) i=1 j=1 m ē3i+j−3 = (Fi · E j ) e3i+j−3 . mi N 3 (4.16) i=1 j=1 mi m Again its interesting to note how the mass ratios vectors e3i+j−3 in (4.15). Next, it is often convenient to note that are subsumed into the basis · e3i+j−3 = Fi · E j , which follows from (4.16). It is also an easy exercise to show that the work–energy theorems for the individual particles, Ṫi = Fi · vi (no sum on i), give a work–energy theorem for the particle of mass m: Ṫ = · v. This theorem can be used to establish energy conservation results for the system of particles. Curvilinear Coordinates We then use the curvilinear coordinates q1 , . . . , q3N to define basis vectors for E3N : ⎛ ⎞ N 3 ∂r ∂ ⎝ j ri · E j = x̂i e3i+j−3 ⎠ aK = K = K ∂q ∂q i=1 j=1 and a = grad(q ) = J J N 3 ∂ q̂J j ∂xi i=1 j=1 e3i+j−3 . It can be shown that aK · aJ = δJK . 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 · aK = N i=1 Fi · ∂ri . ∂qK (4.17) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 116 June 9, 2008 The Equations of Motion for a System of Particles p To establish this result a series of identities and the definition x̂s = rs · E p are invoked: ∂r · aK = · K ∂q ⎛ ⎞ ⎛ ⎞ N N 3 3 ∂ =⎝ (Fi · E j ) e3i+j−3 ⎠ · K ⎝ (rs · E p) e3s+p−3 ⎠ ∂q i=1 j=1 = s=1 p=1 N 3 N 3 i=1 j=1 s=1 p=1 = N 3 N 3 i=1 j=1 s=1 p=1 = N 3 i=1 j=1 = N i=1 = N i=1 p ∂ x̂s (Fi · E j ) ∂qK p ∂ x̂s (Fi · E j ) ∂qK j ∂ x̂i (Fi · E j ) ∂qK e3i+j−3 · e3s+p−3 δis δ jp ⎛ ⎞ 3 ∂ ⎝ j ⎠ Fi · K x̂i E j ∂q j=1 Fi · ∂ri . ∂qK We conclude that the desired identity has been established. 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 = (r, t) = ¯ (r1 , . . . , rN , t) . With some minor manipulations, we find several representations for the derivative of with respect to a curvilinear coordinate: ∂ ¯ ∂ri ∂ ∂ ∂r · K = = · . K ∂q ∂r ∂q ∂ri ∂qK N (4.18) i=1 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 = 0, where [as in (4.6)] (r1 , r2 , . . . , rN , t) = 0, as a constraint on the motion of the single particle of mass m: = ¯ (r, t) = 0. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.7 Construction and Use of a Single Representative Particle This implies that the motion of the single particle is subject to the constraint and, by use of Lagrange’s prescription, a constraint force c , respectively, c = µf, f · v + e = 0, where ∂ ∂ , e= . ∂r ∂t Furthermore, the integrable constraint implies that the particle moves on a configuration manifold M that is a (3N − 1)-dimensional subset of the configuration space E3N . It is important to note that the prescriptions of c and Fci are consistent with each other. Indeed, the equivalence of the constraint forces Fci acting on the system of particles and the constraint force c acting on the single particle of mass m can be inferred from (4.17) and (4.18). f= Conservative Forces and Potential Energies For the 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 ) = Ū(r). Calculating Ū˙ and equating it to −con · v, we find that∗ ∂ Ū . (4.19) ∂r Again, it is important to note that the prescriptions of con and Fconi [see (4.10)] are consistent with each other. As with constraint forces, the equivalence can be inferred from (4.17) and (4.18). con = − 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 ⎧ ⎫ ⎪ ⎨ m1 v̇1 = F1 ⎪ ⎬ is equivalent to mv̇ = , ... ⎪ ⎩m v̇ = F ⎪ ⎭ N N N 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 q̇K aK . i=1 ∗ Essentially, we are solving the equation con + ∂∂rŪ · v = 0 for all possible motions of the system. For this to hold, the terms in the parentheses, assuming they are independent of v, must vanish, and we arrive at (4.19). 117 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 118 June 9, 2008 The Equations of Motion for a System of Particles Consequently, we have the two intermediate results ∂T = mv · aK , ∂ q̇K ∂T = mv · ȧK . ∂qK With the assistance of these results we find that d ∂T d ∂T − K = (mv · aK ) − mv · ȧK K dt ∂ q̇ ∂q dt d (mv) · aK dt = · aK . = We have just established Lagrange’s equations for a system of particles: ∂T d ∂T − K = · aK . K dt ∂ q̇ ∂q (4.20) These equations give the motion r = r(t) of the particle of mass m and hence the motion of the system of particles. We notice that, because of (4.17), Equations (4.20) are identical to those we established earlier. In particular, K = N Fi · i=1 ∂ri = · aK . ∂qK For many problems it is more convenient to use F1 , . . . , FN instead of . However, (4.20) provides wonderful insight into the geometry of Lagrange’s equations for a system of particles. 4.8 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 identity∗ ∂U ∂ri ∂U = · , K ∂q ∂ri ∂qK N i=1 it is easy to see that Lagrange’s equations for a system of particles is ∂L d ∂L − K = QK . K dt ∂ q̇ ∂q ∗ Those readers who read Section 4.7 will have seen this identity earlier as (4.18). (4.21) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.9 A Constrained System of Particles 119 Here, QK are the nonconservative forces acting on the system of particles. The nonconservative forces have the equivalent representations Q K = K + = N ∂U ∂qK Fnconi · i=1 ∂ri ∂qK = ncon · aK . Here, Fnconi = Fi + ∂U , ∂ri ncon = + ∂ Ū . ∂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 (4.21) 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 (4.21) with the kinetic and potential energies being replaced with their constrained counterparts.∗ 4.9 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 − r(t) = 0, N fi · vi + e = 0. i=1 These constraints are equivalent to the following constraints on the motion of the single (representative) particle: q3N − r(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 M. 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 4.5. ∗ That is, in the notation subsequently presented, T is replaced with T̃ and U is replaced with Ũ. As a result, L is replaced with L̃ = T̃ − Ũ. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 120 The Equations of Motion for a System of Particles c m v r Configuration manifold M O Figure 4.5. The representative particle moving on a configuration manifold. For the example shown in this figure, the configuration manifold M is fixed and c is normal to M. 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 = − ∂q3N ∂U + µ1 + µ2 fi + Pi , ∂ri ∂ri where Pi is the resultant of the nonconservative and nonconstraint forces acting on the particle of mass mi . It is important to note that K = N Fi · i=1 =− ∂ri ∂qK ∂U + µ1 δ3N K + µ2 f K + K , ∂qK (4.22) where K = N i=1 Pi · ∂ri , ∂qK fK = From these results, it is easy to construct = N i=1 3N K=1 fi · ∂ri . ∂qK K aK . Approach I For the first approach, we expand the partial derivatives of T and U and then impose the integrable constraint q3N = r(t) on the resulting equations. It can be seen that the resulting Lagrange’s equations of motion decouple into two sets: ( ) d ∂T ∂T ∂U − S = − S + µ2 f S + S , dt ∂ q̇S ∂q ∂q q3N =r,q̇3N =ṙ 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.9 A Constrained System of Particles 121 and ( d dt ∂T ∂ q̇3N ∂T ∂U − 3N = − 3N + µ1 + µ2 f 3N + 3N ∂q ∂q ) , q3N =r,q̇3N =ṙ where S = 1, . . . , 3N − 1, and we emphasize that the constraint q3N = r(t) is imposed after the partial derivatives of T and U have been calculated. 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, T̃ = T̃(q1 , . . . , q3N−1 , q̇1 , . . . , q̇3N−1 , t) = T(q1 , . . . , q3N−1 , q3N = r(t), q̇1 , . . . , q̇3N−1 , q̇3N = ṙ(t)). As in the case of a single particle, T̃ can be used to construct a line-element for the configuration manifold M. We note that, because ∂ T̃ = 0, ∂q3N ∂ T̃ = 0, ∂ q̇3N we are unable to obtain an expression for µ1 . 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 ∂ T̃ ∂ q̇S − ∂ T̃ ∂ Ũ = − S + µ2 f˜ S + ˜ S . S ∂q ∂q Further, no information on the constraint force enforcing the integrable constraint is obtained when this approach is used. Geometric Considerations Once the integrable constraint has been imposed, the single particle of mass m moves on the (3N − 1)-dimensional submanifold of the configuration space E3N . As before, this submanifold is known as the configuration manifold M. A measure of the distance traveled by the single particle on this manifold can be found by use of T̃. Indeed, this energy can be decomposed as T̃ = T̃0 + T̃1 + T̃2 , 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 122 June 9, 2008 The Equations of Motion for a System of Particles where T̃0 = m ã3N3N ṙ2 , 2 T̃1 = m 3N−1 ãK3N q̇K ṙ, K=1 Trel = T̃2 = 3N−1 3N−1 m ãKJ q̇K q̇J . 2 (4.23) K=1 J =1 In the event that the single particle of mass m is determined, we also have the decomposition: aJK = aK · aJ (J, K = 1, . . . , 3N) . The mass m in (4.23) is usually chosen to be m1 + · · · + mN , although other selections such as m = 1 are equally admissible. The kinematical line-element ds for M is 2Trel ds = dt. m This quantity may also be written as 3N−1 3N−1 ds = ã K J KJ dq dq . K=1 J =1 As was the case previously, the imposition of a nonintegrable constraint will not change M or ds. 4.10 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. 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 (4.30) and (4.32)]. Preliminaries We are interested in a system of N particles subject to a set of C (scleronomic) integrable constraints of the form∗ 1 ∗ (r1 , . . . , rN ) = 0, . . . , C (r1 , . . . , rN ) = 0. (4.24) The generalization of our developments to instances in which the constraints are rheonomic can be found elsewhere, for example, Ginsberg [71]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.10 A Canonical Form of Lagrange’s Equations 123 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 , . . . , q3N . 1 = 0, . . . , C = 0 are expressed in terms of the coordinates q The constrained kinetic energy for the system has the representation m ãKJ q̇J q̇K , 2 M T̃ = M K=1 J =1 where we have chosen m = m1 + · · · + mN . We note that ãKJ = ãJK and assume that this matrix is invertible. The inverse is ãJK . For example, if M = 3, then ⎡ ã11 ã12 ⎢ ⎣ã12 ã13 ã22 ã23 ã13 ⎤⎡ ã11 ⎥⎢ ã23 ⎦ ⎣ã12 ã33 ã13 ã12 ã22 ã23 ã13 ⎤ ⎡ 1 ⎥ ⎢ ã23 ⎦ = ⎣0 0 ã33 0 1 0 ⎤ 0 ⎥ 0⎦ . 1 We also define the Christoffel symbols of the first kind as 1 ∂aKJ ∂aKS ∂aSJ + − (J, K, S = 1, . . . , 3N) . [SJ, K] = 2 ∂qS ∂qJ ∂qK (4.25) 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) . (4.26) 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 Christoffel symbols: aJK = aJ · aK , [SI, K] = ∂aS · aK , ∂qI aJK = aJ · aK , JKS = ∂aK J ·a . ∂qS (4.27) Here, all the indices range from 1 to 3N. Using (4.27), the symmetries aJK = aKJ , aJK = aKJ , [SI, K] = [IS, K], and JKS = JSK should be transparent. We also observe that the Christoffel symbols are none other than the covariant and contravariant K components of ∂a . It is a good exercise to show how (4.25) and (4.26) can be used ∂qS to establish (4.27)3,4 when (4.27)1,2 apply. 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 124 The Equations of Motion for a System of Particles constrained kinetic energy, and we now drop the tildes ornamenting the various kinematical quantities. First, we note that ∂T ∂ = R ∂ q̇R ∂ q̇ m aKJ q̇J q̇K 2 M M K=1 J =1 ⎛ m aKJ 2 M = M K=1 J =1 =m M ⎞K δJR δR ⎜ ∂ q̇J ⎟ K 7 7 ∂ q̇ ⎟ ⎜ ⎜ R q̇K + q̇J R ⎟ ⎝∂ q̇ ∂ q̇ ⎠ aRK q̇K . K=1 To arrive at this result, we used the symmetries aRK = aKR and the fact that aRK are independent of q̇S . Similarly, ∂T ∂ = R ∂qR ∂q m aKJ q̇J q̇K 2 M K=1 J =1 m ∂aKJ J K q̇ q̇ . 2 ∂qR M = M M K=1 J =1 K We next differentiate m M K=1 aRK q̇ with respect to time. With some rearranging ∗ and relabeling of the indices, we find d dt ∂T ∂ q̇R =m M aRK q̈K + m K=1 =m M K=1 S=1 ∂qS q̇S q̇K m ∂aRK S K q̇ q̇ 2 ∂qS M aRK q̈K + K=1 M K=1 S=1 m ∂aIR I J q̇ q̇ 2 ∂qJ M + M M ∂aRK M J =1 I=1 =m M K=1 ∗ aRK q̈K + M M m ∂aRK ∂aSR q̇S q̇K . + 2 ∂qS ∂qK K=1 S=1 ∂aRS Although ∂a∂qRK S = ∂qK , because we are summing over the indices S and K, we always have that M M ∂aRK S K M M ∂aRS S K K=1 S=1 ∂qS q̇ q̇ = K=1 S=1 ∂qK q̇ q̇ . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.10 A Canonical Form of Lagrange’s Equations 125 Summarizing, the derivatives of T have the representations ∂T m ∂aKJ J K = q̇ q̇ , R ∂q 2 ∂qR M M K=1 J =1 d dt ∂T ∂ q̇R =m M aRK q̈K + K=1 M M m ∂aRK ∂aSR q̇S q̇K . + 2 ∂qS ∂qK (4.28) K=1 S=1 A Covariant Form of Lagrange’s Equations of Motion A covariant form of Lagrange’s equations of motion was previously established: d dt ∂T ∂ q̇R − ∂T = R ∂qR (R = 1, . . . , M) . (4.29) Here 1 , . . . , M are the generalized forces. We now expand the derivatives of T to establish another form of this equation. With the help of (4.28) 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 q̈K + m M M [SK, R] q̇K q̇S = R (R = 1, . . . , M) . (4.30) 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 geometry,∗ and, in the case of a single particle, was an exercise at the end of Chapter 3. From (4.30), 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 M M ds = a R K RK q̇ q̇ dt. R=1 K=1 Knowledge of ds (which also implies knowledge of aJK ) enables us to write the lefthand side of Lagrange’s equations. 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 in ∗ See, for example, McConnell [139] or Synge and Schild [208]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 126 The Equations of Motion for a System of Particles only the first M equations of motion, we can find only the M contravariant equations from the M covariant equations provided∗ : ⎡ a11 ⎢ ⎢ .. ⎣ . a1M ··· .. . ··· ⎤ ⎡ 11 a1M a ⎥⎢ .. ⎥ ⎢ .. . ⎦⎣ . a1M aMM ··· .. . ··· ⎤ ⎡ a1M 1 ··· ⎥ ⎢ .. ⎥ = ⎢ .. . . . . ⎦ ⎣. aMM 0 ··· ⎤ 0 ⎥ .. ⎥ . .⎦ 1 (4.31) That is, we require several aKJ ’s to be zero: ⎡ a1(M+1) ⎢ ⎢ a2(M+1) ⎢ .. ⎢ ⎣ . a3N(M+1) a1(M+2) ··· a2(M+2) .. . a3N(M+2) ··· .. . ··· a1(3N) ⎤ ⎡ 0 ⎥ ⎢ 0 a2(3N) ⎥ ⎢ ⎥=⎢ ⎢ .. ⎥ ⎢ . . ⎦ ⎣ .. a3N(3N) 0 0 ··· 0 ··· .. . . . . 0 ··· ⎤ 0 ⎥ 0⎥ ⎥ . .. ⎥ ⎥ .⎦ 0 When (4.31) holds, the right-hand side of Lagrange’s equations of motion transforms from the covariant components of to the contravariant components: J = M aJR R . R=1 In summary, the contravariant form of Lagrange’s equations of motion is mq̈J + m M M JKS q̇K q̇S = J (J = 1, . . . , M) . (4.32) K=1 S=1 This form of Lagrange’s equations is very useful in numerical simulations because we have explicit expressions for q̈1 , . . . , q̈M . An Example A simple example with which to explore Equations (4.30) and (4.32) is to consider a single particle moving in E3 under the action of a force F. We describe the motion of this particle by using a spherical polar coordinate system: q1 = R, q2 = φ, and q3 = θ. That is, M = 3. We can also consider our discussion here to be a solution to Exercise 3.3. It should by now be trivial to establish that T= ∗ m 2 Ṙ + R2 sin2 (φ)θ̇2 + R2 φ̇2 . 2 In terms of the representative particle, these restrictions are equivalent to aJ · aK = 0 for all J = 1, . . . , M and all K = M + 1, . . . , 3N. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.10 A Canonical Form of Lagrange’s Equations 127 From the expression for T, we can immediately deduce that ⎡ a11 a13 ⎡ ⎤ 1 0 a22 ⎥ ⎢ a23 ⎦ = ⎣0 R2 a13 a23 a33 a11 a12 a13 ⎢ ⎣a12 ⎡ ⎢ 12 ⎣a a13 d dt a12 0 0 1 0 ⎡ ⎤ a22 ⎥ ⎢ a23 ⎦ = ⎣0 R−2 a23 a33 0 0 ⎤ 0 ⎥ ⎦, 0 R2 sin2 (φ) ⎤ 0 ⎥ ⎦. 0 R−2 sin−2 (φ) We readily deduce the covariant form of Lagrange’s equations of motion from ∂T ∂T − ∂q i = F i by expanding the time derivative: ∂ q̇i ⎡ 1 0 ⎢ m ⎣0 R2 0 0 ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ R̈ F1 C1 ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎦ ⎣ φ̈ ⎦ + ⎣C2 ⎦ = ⎣F2 ⎦ , 0 0 2 R2 sin (φ) θ̈ C3 (4.33) F3 where ⎡ C1 ⎡ ⎤ ⎢ ⎥ ⎢ ⎣C2 ⎦ = ⎢ ⎣ −mR sin2 (φ)θ̇2 + φ̇2 ⎤ ⎥ ⎥, ⎦ −mR sin(φ) cos(φ)θ̇2 + 2mRṘφ̇ 2mR sin2 (φ)Ṙθ̇ + 2mR2 sin(φ) cos(φ)φ̇θ̇ ⎡ ⎤ ⎡ ⎤ F1 F · eR ⎢ ⎥ ⎢ ⎥ ⎣F2 ⎦ = ⎣ F · Reφ ⎦ , C3 F · R sin(φ)eθ ⎤ ⎡ ⎡ 1⎤ F · eR F a11 ⎥ ⎢ ⎢ 2⎥ ⎢ ⎢ 12 1 ⎣F ⎦ = ⎣ F · R eφ ⎥ ⎦ = ⎣a 1 F · R sin(φ) eθ F3 a13 F3 ⎡ a12 a13 ⎤⎡ F1 ⎤ a22 ⎥⎢ ⎥ a23 ⎦ ⎣F2 ⎦ . a23 a33 F3 From (4.33), 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 (4.33) by multiplying both sides of , (4.33) by the inverse of aij : ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ −mR(sin2 (φ)θ̇2 + φ̇2 ) F1 R̈ ⎥ ⎢ ⎢ ⎥ ⎢ 2⎥ m sin(φ) cos(φ) 2 m ⎣ φ̈ ⎦ + ⎢ θ̇ + 2m Ṙφ̇⎥ ⎦ = ⎣F ⎦ . ⎣− R R θ̈ 2m Ṙθ̇ R + 2m cot(φ)φ̇θ̇ F 3 (4.34) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 128 978 0 521 87483 0 June 9, 2008 The Equations of Motion for a System of Particles Again, we can read off the Christoffel symbols of the second kind from the righthand side of these equations. For example, 122 = −R, 233 = − sin(φ) cos(φ), and 323 = cot(φ). 4.11 Alternative Principles of Mechanics Newton’s tremendous contributions to mechanics in the 17th 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 [44], Gauss’ principle of least constraint in 1829 [68], and Hamilton’s principle in 1835 [89].∗ 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. 4.11.1 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 often the basis for treatments of Lagrange’s equations of motion in many texts.† We assume that the particles are subject to a single constraint: f1 · v1 + · · · + fN · vN + e = 0. (4.35) 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 − m1 r̈1 ) · d1 + · · · + (FaN − mN r̈N ) · dN = 0 (4.36) for all possible choices of the vectors d1 , . . . , dN that satisfy the condition f1 · d1 + · · · + fN · dN = 0. (4.37) Notice that the e present in (4.35) is notably absent from (4.37). The vectors dK are known as virtual displacements and are usually denoted by δrK . The virtual work performed by the applied force FaK is defined as FaK · dK ; thus (4.36) states that the combined virtual work of the applied forces FaK and the inertial forces −mK r̈K is zero. Our aim is to obtain the equations of motion of the system of particles from (4.36). To this end, we can introduce Lagrange multipliers to accommodate the ∗ † Further background on these (and many other principles) can be found in Dugas [48], Szabó [209], and Truesdell [216, 217]. See, for example, Section 4.9 of Baruh [14], Section 2.1 of Greenwood [78], or Section 46 of Synge [206]. 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 4.11 Alternative Principles of Mechanics 129 constraint on the vectors dK ∗ : N (FaK − mK r̈N ) · dK + µ K=1 N fK · dK = 0. (4.38) K=1 As a consequence of the Lagrange multiplier µ, the vectors dK can be varied independently. For (4.38) to hold for all such displacements, it is necessary and sufficient that mK r̈K = FaK + µfK (K = 1, . . . , N) . (4.39) However, (4.39) are none other than the balances of linear momenta for a system of particles subject to constraint (4.35), for which the constraint forces are prescribed by use of Lagrange’s prescription: FcK = µfK (K = 1, . . . , N) . (4.40) 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 (4.35) is integrable, we can interpret (4.37) as a normality condition for the vectors dK . Further, it is not too difficult to see that constraint force prescription (4.40) implies that this force is normal to the configuration manifold M in this case. 4.11.2 Gauss’ Principle of Least Constraint Gauss’ principle of least constraint was published in 1829 [68]. 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 (4.35) and suppose that the constraint forces are prescribed by use of Lagrange’s prescription [i.e., (4.40) 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, for example, Hertz [92]). A lucid discussion of the principle can be found in Udwadia and Kalaba [218], and its extension to cases for which Lagrange’s prescription does not hold can be found in O’Reilly and Srinivasa [162]. ∗ When constraint (4.35) is integrable, this is precisely the approach taken by Lagrange in [121, 122] (cf. Section 4 in Chapter 11 of Dugas [48]). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 130 The Equations of Motion for a System of Particles q (t1 ) q q (t0 ) t0 t t1 Figure 4.6. 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 tthe differential equations F = ma. Hamilton’s principle states that this path extremizes I = t01 Ldt [see (4.41)]. The portrait of Hamilton is from the Royal Irish Academy in Dublin, Ireland. 4.11.3 Hamilton’s Principle The remaining principle of interest was discovered by Sir William R. Hamilton (1805–1865) and first published in [89] 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∗ : t1 Ldt, (4.41) I= i i t0 where the Lagrangian L = T q , q̇ − U qi and t0 and t1 are fixed instances of time. As illustrated in Figure 4.6, 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 (4.41) appeared in print in 1835 that the necessary conditions for q1 (t), q2 (t), q3 (t) to extremize I was that q1 (t), q2 (t), q3 (t) satisfy the following differential equations: d ∂L ∂L − k =0 (4.42) (k = 1, 2, 3) . k dt ∂ q̇ ∂q In fact, in the context of extremizing I, (4.42) are known as the Euler–Lagrange equations.† Thus Hamilton’s principle implies that the motion of the system satisfies ∗ † To extremize is to minimize or maximize. Problems featuring the extremization of I can be solved by use of the calculus of variations. In the middle of the 18th century, Euler and Lagrange played seminal roles in the development of this calculus [72]. It was on the subject of this calculus that a 19-year-old Lagrange first wrote to Euler in 1755. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercise 4.1 131 (4.42). At this stage, it should be transparent that (4.42) are equivalent to F = ma 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 included the case in which the system of particles was subject to integrable constraints. Jacobi [102] 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. 4.12 Closing Remarks 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 ∂ri feature enables us to confidently calculate the forces K = N i=1 Fi · ∂qK that appear on the right-hand side of Lagrange’s equations of motion: d dt ∂T ∂ q̇K − ∂T = K ∂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 M 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 turn to examples in which Lagrange’s equations of motion for several systems of particle are established and discussed. EXERCISES 4.1. What are the kinematical line-elements ds, generalized coordinates, and configuration manifolds M of the following systems: (a) a particle attached to a fixed point by a spring; (b) a particle attached to a fixed point by a rod of length L(t); (c) a harmonic oscillator consisting of one particle; (d) a planar double pendulum; (e) a spherical double pendulum. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 132 Exercises 4.2–4.5 4.2. Under which circumstances do constraint forces perform no work? 4.3. Many authors assume that the kinetic energy of a system of particles is a 3N m I J positive-definite function of the velocities q̇K . Letting T = 3N I=1 J =1 2 aIJ q̇ q̇ is equivalent to saying that the matrix whose components are m2 aIJ 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 positive-definiteness breaks down at the singularities of this coordinate system. 4.4. In Casey’s construction of the single particle, what are the distinctions among the bases {eK }, {eK }, and {ēK }? For a given system of two particles, how does one construct these bases for E6 ? 4.5. Consider the following function that depends on the motion of two particles: V = V (||r2 − r1 || , t) . This function is representative of a potential energy function and an integrable constraint. (a) For any vector x, establish the following result: dx x = · ẋ. dt x The easiest way to show this result is to represent x by Cartesian coordinates. (b) Show that V̇ = ∂V ∂V ∂V , · v1 + · v2 + ∂r1 ∂r2 ∂t where ∂V ∂V ∂V x =− =− ∂r1 ∂r2 ∂x x and x = x, x = r2 − r 1 . (c) Consider a system of two particles subject to a constraint = (r2 − r1 , t) . = 0, where (4.43) Using Lagrange’s prescription, show that Fc1 = −Fc2 . Notice that this is Newton’s third law of motion. Give two examples of a physical constraint for which has the form (4.43) and for which the constraint forces that enforce this constraint are equal and opposite. ∗ Recall that a matrix C is positive-definite if, for all nonzero x, xT Cx > 0 and xT Cx = 0 only when x = 0. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 4.5–4.6 133 (d) Consider a system of two particles subject to a conservative force that has a potential energy function: U = U (r2 − r1 ) . (4.44) Using the identity, U̇ = −Fcon1 · v1 − Fcon2 · v2 , show that Fcon1 = −Fcon2 . Notice that this (again) is Newton’s third law of motion. Give two instances in which U has the form (4.44) and demonstrate that the conservative forces are equal and opposite. 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 . 2 −r1 || (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 r2 − r̄ = M (r2 − r1 ) , m+M r1 − r̄ = − m (r2 − r1 ) . m+M Here, r̄ 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∗ : 2 r1 − r̄ m , M r̈1 − r̄¨ = −GMm M + m ||r1 − r̄||3 2 M r2 − r̄ m r̈2 − r̄¨ = −GMm . (4.45) M + m ||r2 − r̄||3 (e) Argue that the results of Section 2.8 can be applied to (4.45) to determine the orbital motions of the particles about their center of mass C. ∗ In celestial mechanics, expressing the equations of motion in this manner is equivalent to using what is known as a barycentric coordinate system (see, for example, [220]), that is, a coordinate system whose origin is at the center of mass. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 5 978 0 521 87483 0 June 9, 2008 Dynamics of Systems of Particles 5.1 Introduction In this chapter, several examples of systems of particles are discussed. We pay particular attention to how the equations of motion for these systems are established by use of Lagrange’s equations. The examples we discuss are classical and 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 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 L1 to a fixed support. It is also connected by a spring of stiffness K2 and unstretched length L2 to a particle of mass m2 . Both particles are constrained to move in the E1 direction: r1 = (L1 + x1 ) E1 + y1 E2 + z1 E3 , r2 = (L1 + L2 + x2 ) E1 + y2 E2 + z2 E3 . Notice that x1 and x2 measure the displacement of the particles from the unstretched spring states. 134 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 5.2 Harmonic Oscillators 135 E2 g m2 m1 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 Here we denote the six coordinates by q1 = x1 , q2 = x2 , q3 = y1 , q4 = y2 , q5 = z1 , q6 = z2 . For the systems of interest, the particles are constrained to move on a line. Thus the system is subject to four constraints: 1 = 0, 2 = 0, 3 = 0, = y2 , 3 = z1 , 4 = 0, where 1 = y1 , 2 4 = z2 . It should be noted that ∂ 1 = E2 , ∂r1 ∂ 2 = 0, ∂r1 ∂ 3 = E3 , ∂r1 ∂ 1 = 0, ∂r2 ∂ 2 = E2 , ∂r2 ∂ 3 = 0, ∂r2 ∂ 4 = 0, ∂r1 and ∂ 4 = E3 . ∂r2 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 m2 2 m1 2 ẋ1 + ẏ21 + ż21 + ẋ2 + ẏ22 + ż22 , 2 2 K1 2 K2 U= x + (x2 − x1 )2 + m1 gy1 + m2 gy2 . 2 1 2 T= (5.1) Imposing the four integrable constraints on these energies, we can determine their constrained counterparts: T̃ = m1 2 m2 2 ẋ + ẋ , 2 1 2 2 Ũ = 1 1 K1 x21 + K2 (x2 − x1 )2 . 2 2 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 136 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 , Fc2 = µ2 E2 + µ4 E3 . Here, µ1 , µ2 , µ3 , and µ4 are normal force components. If the surface is rough, then these prescriptions must be altered to include the friction forces: Fc1 = µ1 E2 + µ3 E3 − µd ||µ1 E2 + µ3 E3 || ẋ1 E1 , |ẋ1 | Fc2 = µ2 E2 + µ4 E3 − µd ||µ2 E2 + µ4 E3 || ẋ2 E1 . |ẋ2 | 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 + L1 ) e1 + y1 e2 + z1 e3 + (x2 + L1 + L2 ) 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 = e4 , a3 = e2 , a4 = e5 , a5 = e3 , a6 = e6 . and If the kinetic energy T = m2 ṙ · ṙ were calculated, it would be identical to (5.1). The constraints on the motion of the single particle have the following representations: 1 = r · e2 , 2 = r · e5 , 3 = r · e3 , 4 = r · e6 . Using these representations, we easily see that the constraint force c that we would prescribe by using Lagrange’s prescription is c = µ1 e2 + µ2 e5 + µ3 e3 + µ4 e6 . When friction is present, this prescription is inadequate. (5.2) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 5.2 Harmonic Oscillators 137 The force vector can be determined by use of prescription (4.16): ẋ1 e1 = −K1 (x1 ) + K2 (x2 − x1 ) − µd ||µ1 E2 + µ3 E3 || |ẋ1 | + (µ1 − m1 g) e2 + µ3 e3 + (µ2 − m2 g) e5 + µ4 e6 ẋ2 + −K2 (x2 − x1 ) − µd ||µ2 E2 + µ4 E3 || e4 . |ẋ2 | Notice that the constraint force contributions to this vector are consistent with our prescription for c in (5.2) when µd = 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 There are four integrable constraints on the system, so the generalized coordinates are x1 and x2 . The configuration manifold M is simply the plane E2 and the kinematical line-element is dx1 2 dx2 2 m1 m2 ds = + dt. m1 + m2 dt m1 + m2 dt Notice that this line-element is easily related tothe standard measure of distance dx 2 dy 2 traveled along a curve (x(τ), y(τ)) on a plane: + dτ dτ. We shall shortly dτ introduce a nonintegrable constraint into the system. This constraint will not affect M or ds. 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 by use of Lagrange’s prescription and ∂ Ũ the other forces acting on the system are conservative. Thus 1 = − ∂q 1 and 2 = ∂ Ũ − ∂q 2 . In conclusion, the equations of motion can be obtained by use of the following form of Lagrange’s equations of motion: d dt ∂ T̃ ∂ q̇α − ∂ T̃ ∂ Ũ = − α, ∂qα ∂q (5.3) where α = 1, 2. With a small amount of work, we find the equations of motion: m1 ẍ1 = −K1 x1 − K2 (x1 − x2 ) , m2 ẍ2 = −K2 (x2 − x1 ) . These equations are classical and are easily seen to be equivalent to F1 · E1 = m1 r̈1 and F2 · E1 = m2 r̈2 , respectively. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 138 Dynamics of Systems of Particles E2 g m2 m1 E1 O P Rough surface Figure 5.2. A system of two particles moving on a rough horizontal surface. The Influences of Dynamic Friction and an Applied Force We now consider the same system for particles, but assume that dynamic Coulomb friction is present and a forcing P = P cos(ωt)E1 is applied to the particle of mass m2 (see Figure 5.2). We have already determined all of the needed 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: d ∂T ∂T − K = K . (5.4) K dt ∂ q̇ ∂q ∂r1 ∂r2 Here, K = F1 · ∂q K + F2 · ∂qK , or equivalently if we use the single representative particle, K = · aK . Clearly we need to append P = P cos(ωt)E1 to F2 and P cos(ωt)e4 to . From (5.4) and the expressions we have established for T and the force, we find the following six equations: m1 ẍ1 = −K1 x1 − K2 (x1 − x2 ) − µd m1 g m2 ẍ2 = −K2 (x2 − x1 ) − µd m2 g ẋ1 , |ẋ1 | ẋ2 + P cos(ωt), |ẋ2 | 0 = N1y − m1 g, 0 = N2y − m2 g, 0 = N1z, 0 = N2z. (5.5) The last four equations yield the constraint forces: N1y = µ1 , N2y = µ2 , N1z = µ3 , and N2z = µ4 . 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 5.2 Harmonic Oscillators 139 oscillations are often termed stick–slip: behavior we encountered earlier in our model for the roller coaster. 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.∗ 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 Fc1 = µ1 E2 + µ3 E3 + µ5 E1 , where µ5 E1 is the static friction force. This force is subject to the static friction criterion: |µ5 | ≤ µs µ21 + µ23 , where µs is the coefficient of static friction. Instead of (5.5), the equations of motion for the oscillator in this case are µ5 = K1 x10 + K2 (x10 − x2 ) , m2 ẍ2 = −K2 (x2 − x10 ) − µdm2 g ẋ2 + P cos(ωt), |ẋ2 | (5.6) where K1 x10 + K2 (x10 − x2 ) ≤ µs m1 g. Thus, if there is sufficient friction to match the spring forces, then x10 will remain stationary. Otherwise, it will tend to slip in the direction of the resultant spring force and (5.5) is then used to determine the motion. Imposing a Nonintegrable Constraint We now consider the introduction of a nonintegrable constraint in the system considered in Section 5.2. The constraint is ẋ1 x2 − ẋ2 = 0. We can also express this constraint as (x2 E1 ) · v1 + (−E1 ) · v2 = 0, or, for the representative particle of mass m, as 1 x2 e − e4 · v = 0. It is left as an exercise to show that the constraint is nonintegrable. Using Lagrange’s prescription, we find that the additional constraint forces on the particles are µ5 x2 E1 and −µ5 E1 . For the representative particle of mass m, c ∗ For further details on numerical schemes for stick–slip oscillators, see [23, 50, 112]. Interesting examples of such oscillators and their application to various fields, including brake squeal, can be found in [21, 111]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 140 Dynamics of Systems of Particles m2 E3 r2 C r̄ O E2 r1 m1 E1 Figure 5.3. A particle of mass m1 attached by linear spring of stiffness K and unstretched length L0 to a particle of mass m2 . Each particle is attracted to the fixed point O by a Newtonian gravitational force field. needs to be augmented by µ5 (x2 e1 − e4 ). Starting from Lagrange’s equations of motion in the form (5.4), we quickly arrive at the equations of motion: m1 ẍ1 = −K1 x1 − K2 (x1 − x2 ) − µd m1 g m2 ẍ2 = −K2 (x2 − x1 ) − µd m2 g ẋ1 + µ5 x2 , |ẋ1 | ẋ2 − µ5 . |ẋ2 | The last four equations are identical to those recorded in (5.5). The two equations of motion are supplemented by the constraint ẋ1 x2 − ẋ2 = 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 [16], 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 extremeties of a spring of stiffness K and unstretched length L0 (see Figure 5.3). The gravitational force exerted on the satellite by the planet of mass M 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 5.3 A Dumbbell Satellite 141 coordinates for the position vector r̄ of the center of mass C of the system and a set of spherical polar coordinates for r2 − r̄ and r1 − r̄. We elect the third choice here. Denoting r2 − r1 by ReR , where R = ||r2 − r1 ||, eR = sin(φ) (cos(θ)E1 + sin(θ)E2 ) + cos(φ)E3 . Some elementary algebra shows that m2 ReR , r1 = r̄ − m1 + m2 r2 = r̄ + m1 m1 + m2 ReR . With our choice of coordinates, r̄ = xE1 + yE2 + zE3 . For future reference, we label our coordinates q1 = x, q2 = y, q3 = z, q4 = R, q5 = φ, q6 = θ. (5.7) We also record the following 12 partial derivatives: ∂r1 ∂r1 ∂r1 = E1 , = E2 , = E3 , ∂x ∂y ∂z ∂r1 m2 m2 ∂r1 eR , Reφ , =− =− ∂R m1 + m2 ∂φ m1 + m2 ∂r1 m2 R sin(φ)eθ , =− ∂θ m1 + m2 and ∂r2 ∂r2 ∂r2 = E1 , = E2 , = E3 , ∂x ∂y ∂z m1 m1 ∂r2 ∂r2 eR , Reφ , = = ∂R m1 + m2 ∂φ m1 + m2 ∂r2 m1 = R sin(φ)eθ . ∂θ m1 + m2 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 a certain relationship in Section 5.3. 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 2 2 m1 m2 m1 + m2 2 ẋ + ẏ2 + ż2 + Ṙ2 + R2 φ̇2 + R2 sin2 (φ) θ̇2 . = 2 2(m1 + m2 ) T= To arrive at this expression, a substantial amount of algebraic cancellations occurred. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 142 June 9, 2008 Dynamics of Systems of Particles The potential energy of the system is due to Newtonian gravitation and the spring kinetic energy: GMm2 K GMm1 − + (R − L0 )2 . U = − 2 2 1 ReR r̄ + m1m+m ReR r̄ − m1m+m 2 2 We could also have included the Newtonian gravitational force that m1 exerts on m2 and vice versa but this will not significantly add 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 ∂L ∂L − S =0 (S = 1, . . . , 6) . S dt ∂ q̇ ∂q Equipped with the expressions for T and U recorded in the previous subsection, we can derive the equations straightforwardly, and 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 = r1 × m1 v1 + r2 × m2 v2 m1 m2 2 = r̄ × mv̄ + R φ̇eθ − θ̇ sin(φ)eφ . m1 + m2 The conservation of HO implies that the linear speeds, ẋ, ẏ, and ż, of the center of mass are coupled to the angular speeds θ̇ and φ̇ of the satellite. We shall observe this coupling later in rigid body models for satellites. The Generalized Coordinates and Configuration Manifold There are no constraints on the system, so the generalized coordinates are x, y, z, R, 6 θ, and φ. The configuration manifold is simply E , and the kinematical line-element is ds = 2T dt. m1 +m2 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 = m1 r̈1 and F2 = m2 r̈2 . 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 = m1 r̈1 and F2 = m2 r̈2 . For instance, Lagrange’s equation of motion for R, d ∂L ∂L = 0, − dt ∂ Ṙ ∂R 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 5.4 A Pendulum and a Cart 143 is equivalent to (F1 = m1 r̈1 ) · ∂r1 ∂r2 + (F2 = m2 r̈2 ) · = 0. ∂R ∂R If we change our selection for the coordinates used for the system, this changes the precise linear combinations of the components of F1 = m1 r̈1 and F2 = m2 r̈2 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 L0 . The motion of both particles is assumed to be planar. 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 : r1 = xE1 + yE2 + z1 E3 , r2 = xE1 + yE2 + rer + (z1 + z2 ) E3 . Smooth horizontal rail E2 m1 O g E1 Linear spring θ r m2 Figure 5.4. A system of two particles connected by a linear spring of stiffness K and unstretched length L0 . The particle of mass m1 is free to move on a smooth horizontal rail and the second particle moves on the x − y plane. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 144 Dynamics of Systems of Particles The angle θ is measured from the E1 direction and is counterclockwise positive. That is, when m2 is directly below m1 , θ = 3π . 2 We label the coordinates as follows: q1 = x, q2 = r, q3 = θ, q4 = y, q5 = z1 , q6 = z2 = (r2 − r1 ) · E3 . (5.8) The constraints on this system of particles are all integrable: 1 = 0, 2 = 0, 3 = 0, where 1 = y = r1 · E2 , 2 = z1 = r1 · E3 , 3 = z2 = (r2 − r1 ) · E3 . At this stage, it is prudent to compute that ∂ 1 = E2 , ∂r1 ∂ 2 = 0, ∂r2 ∂ 1 = 0, ∂r2 ∂ 3 = −E3 , ∂r1 ∂ 2 = E3 , ∂r1 ∂ 3 = E3 . ∂r2 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 is due to gravity and the spring potential energy: U = m1 gy + m2 gy + m2 gr sin(θ) + K (||r2 − r1 || − L0 )2 . 2 Imposing the integrable constraints, we find that Ũ = m2 gr sin(θ) + K (r − L0 )2 . 2 The kinetic energy of the system is also easy to calculate: T= m2 2 m1 + m2 2 ẋ + ẏ2 + ż21 + ṙ + r2 θ̇2 + ż22 2 2 + m2 ẋṙ cos(θ) + ẏṙ sin(θ) − ẋrθ̇ sin(θ) + ẏrθ̇ cos(θ) + m2 ż1 ż2 . (5.9) This expression simplifies when we impose the following constraints: T̃ = m1 + m2 2 m2 2 ẋ + ṙ + r2 θ̇2 + m2 ẋṙ cos(θ) − ẋrθ̇ sin(θ) . 2 2 Generalized Coordinates and Constraints The system has three generalized coordinates: x, r, θ. The configuration manifold M for this system is 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 5.4 A Pendulum and a Cart 145 coordinates parameterize E2 in its entirety. Further, as x ranges from −∞ to ∞, we see that this coordinate parameterizes E. We conclude that M is E3 . We can find the kinematical line-element for the configuration manifold by using T̃: 2T̃ ds = dt, m1 + m2 where we have chosen m = m1 + m2 . Constraint Forces and Resultant Forces The constraint forces on the system can be computed by use of Lagrange’s prescription: Fc1 = 3 µi ∂ i = µ1 E2 + µ2 E3 − µ3 E3 , ∂r1 µi ∂ i = µ3 E3 . ∂r2 i=1 Fc2 = 3 i=1 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 = K (r − L0 ) er + (µ1 − m1 g) E2 + (µ2 − µ3 ) E3 , F2 = −K (r − L0 ) er − m2 gE2 + µ3 E3 , respectively. We shall use these expressions later on to determine the force vector . 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 involving Lagrangian (4.21): d dt ∂L ∂ q̇S ∂L ∂ri − S = QS = Fnconi · S ∂q ∂q 2 (S = 1, . . . , 6) . (5.10) i=1 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. To find the equations of motion, it suffices to examine the first three of Lagrange’s equations of motion: d ∂ L̃ ∂ L̃ − J =0 (J = 1, . . . , 3) . dt ∂ q̇J ∂q Here, L̃ = T̃ − Ũ. Evaluating the partial derivatives of L̃ = T̃ − Ũ and then calculating dtd , we find with some rearrangement that the equations of motion can be 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 146 Dynamics of Systems of Particles expressed in the form ⎡ m1 + m2 ⎢ ⎣ m2 cos(θ) −m2 r sin(θ) m2 cos(θ) −m2 r sin(θ) m2 0 0 m2 r 2 ⎤⎡ ⎤ ⎡ ⎤ ẍ f1 ⎥⎢ ⎥ ⎢ ⎥ ⎦ ⎣ r̈ ⎦ = − ⎣ f 2 ⎦ , θ̈ (5.11) f3 where f 1 , f 2 , and f 3 are quadratic in the velocities: f 1 = −2m2 ṙθ̇ sin(θ) − m2 rθ̇2 cos(θ), f 2 = −m2 rθ̇2 + m2 g sin(θ) + K (r − L0 ) , f 3 = 2m2 rṙθ̇ + m2 gr cos(θ). (5.12) The form (5.11) of the equations of motion is easy to implement numerically. It is also a canonical form for many mechanical systems featuring time-independent (scleronomic) integrable constraints. You should also notice that the matrix on the left-hand side of (5.11) can be obtained by inspection from T̃. 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: ∂L d ∂L − J = QJ (J = 4, . . . , 6) . dt ∂ q̇J ∂q We leave the intermediate steps as an exercise: Fc1 = (m1 + m2 ) gE2 + d m2 ṙ sin(θ) + m2 rθ̇ cos(θ) E2 , dt Fc2 = 0. It is easy to observe from these expressions the expected result that Fc1 · v1 + Fc2 · 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 Ė = 0, where E = T̃ + Ũ. The second conservation can be deduced from (5.11)1 . That is, the linear momentum of the system in the horizontal direction, G · E1 , is conserved. 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 5.5 Two Particles Tethered by an Inextensible String 147 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 ṙ · ṙ will be identical to (5.9). The force vector can be determined by use of prescription (4.16): = K (r − L0 ) cos(θ)e1 + (µ1 − m1 g + K (r − L0 ) sin(θ)) e2 + (µ2 − µ3 ) e3 − K (r − L0 ) cos(θ)e4 + (−m2 g − K (r − L0 ) sin(θ)) e5 + µ3 e6 . We used the expressions for F1 and F2 recorded earlier to compute this vector. If we ∂r1 compute · aS and compare the results with those we obtained by using F1 · ∂q S + F2 · ∂r2 , ∂qS then the two sets of expressions for S should be identical. Remarks The system discussed here is also a good candidate to explore the covariant (4.30) and contravariant (4.32) forms of Lagrange’s equations of motion that we discussed , earlier. Indeed, by using (5.11), we can readily compute the aij matrix 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 L0 , 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 tractible, 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 polar coordinate system {R2 , φ2 , θ2 } is used to describe the motion of the particle of mass m2 : r1 = r1 er1 + z1 E3 , r2 = R2 eR2 . We define six coordinates as follows: q1 = r1 , q2 = θ1 , q3 = φ2 , q4 = θ2 , q5 = x, q6 = z1 . (5.13) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 148 Dynamics of Systems of Particles Rough horizontal plane E3 g m1 O E2 E1 m2 Inextensible string Figure 5.5. A system of two particles connected by an inextensible string of length L0 . Notice that we have introduced a new coordinate x: R2 = L0 − r1 + x. Using the cylindrical polar coordinate system and the spherical polar coordinate systems, we have r1 = r1 er1 + z1 E3 , r2 = (L0 + x − r1 ) eR2 . Hence, ∂r1 = er1 , ∂r1 ∂r1 = r1 eθ1 , ∂θ1 ∂r2 = −eR2 , ∂r1 ∂r1 = 0, ∂φ2 ∂r2 = 0, ∂θ1 ∂r2 = (L0 + x − r1 ) sin(φ2 )eθ2 , ∂θ2 ∂r1 = 0, ∂θ2 ∂r1 = 0, ∂x ∂r1 = E3 , ∂z1 ∂r2 = (L0 + x − r1 ) eφ2 , ∂φ2 ∂r2 = eR2 , ∂x ∂r2 = 0. ∂z1 The Potential and Kinetic Energies The potential energy of the system is due to gravity: U = m1 gE3 · r1 + m2 gE3 · r2 = m1 gz1 + m2 g (L0 + x − r1 ) cos (φ2 ) . 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 = ṙ1 er1 + r1 θ̇1 eθ1 + ż1 E3 . To calculate v2 , we recall the expression for this vector in spherical polar coordinates and then substitute for R2 and Ṙ2 : v2 = (ẋ − ṙ1 ) eR2 + (L0 + x − r1 ) sin (φ2 ) θ̇2 eθ2 + (L0 + x − r1 ) φ̇2 eφ2 . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 5.5 Two Particles Tethered by an Inextensible String 149 The kinetic energy T of the system of particles is m2 m1 v1 · v1 + v2 · v2 2 2 m1 2 ṙ + r12 θ̇21 + ż21 = 2 1 m2 + (ẋ − ṙ1 )2 + (L0 + x − r1 )2 sin2 (φ2 ) θ̇22 + (L0 + x − r1 )2 φ̇22 . 2 T= 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 L0 , 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 : Fc1 = µ1 er1 + µ2 E3 − µd ||µ2 E3 || vrel , ||vrel || Fc2 = µ1 eR2 , where vrel = ṙ1 er1 + r1 θ̇1 eθ1 . The constraint forces associated with the inextensible string can also be prescribed by use of Lagrange’s prescription. Let 1 = ||r2 || + ||r1 || − L0 . Then the inextensibility constraint is 1 Fc1 = µ1 = 0. Using Lagrange’s prescription, we find r1 = µ1 er1 , ||r1 || Fc2 = µ1 eR2 . Notice that we imposed the constraint z1 = 0 to simplify the expression for Fc1 . The constraint forces associated with the constraint 2 = 0, where 2 = z1 , are not prescribed by Lagrange’s prescription because it features a dynamic friction force. 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 150 June 9, 2008 Dynamics of Systems of Particles differential equations are presented: (m1 + m2 ) r̈1 − m1 r1 θ̇21 vrel · er1 + m2 (L0 − r1 ) sin2 (φ2 ) θ̇22 + φ̇22 = −µd ||µ2 E3 || + m2 g cos(φ2 ) , ||vrel || d m1 (L0 − r1 )2 φ̇2 dt d vrel · eθ1 m1 r12 θ̇1 = −µd r1 ||µ2 E3 || , dt ||vrel || − m2 (L0 − r1 )2 θ̇22 sin(φ2 ) cos(φ2 ) = m2 g (L0 − r1 ) sin (φ2 ) , d m2 (L0 − r1 )2 sin2 (φ2 ) θ̇2 = 0. dt (5.14) In addition, the two equations for µ1 and µ2 are µ1 = m2 g cos(φ2 ) − m2 r̈1 − m2 (L0 − r1 ) sin2 (φ2 ) θ̇22 + φ̇22 , µ2 = m1 g. To find the preceding equations, we started with Lagrange’s equations: d dt ∂T ∂ q̇K − ∂T ∂r1 ∂r2 = F1 · K + F2 · K ∂ q̇ ∂q ∂q (K = 1, . . . , 6) . We then substituted for T, F1 , and F2 . 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 · er1 + m2 g cos (φ2 ) , ||vrel || vrel · eθ1 , F1 · r1 eθ1 + F2 · 0 = −µd r1 ||µ2 E3 || ||vrel || F1 · er1 − F2 · eR2 = −µd ||µ2 E3 || F1 · 0 + F2 · (L0 − r1 ) eφ2 = m2 g (L0 − r1 ) sin (φ2 ) , F1 · 0 + F2 · (L0 − r1 ) sin (φ2 ) eθ2 = 0, F1 · 0 + F2 · eR2 = 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 Ṫ = F1 · v1 + F2 · v2 and substitute for the applied and constraint forces: Ṫ = (−m1 gE3 + Fc1 ) · v1 + (−m2 gE3 + Fc2 ) · v2 . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 5.6 Closing Comments 151 Making use of the potential energy function U, we can express this equation as Ė = µ1 (er1 · v1 + eR2 · v2 ) + µ2 E3 · v1 − µd ||µ2 E3 || vrel · v1 , ||vrel || where the total energy E = T + U. Thus, as v1 = vrel and is normal to E3 , and µ2 = m1 g, we surmise that Ė = µ1 (er1 · v1 + eR2 · v2 ) − µd m2 g||v1 ||. However, er1 · v1 + eR2 · v2 = ṙ1 + Ṙ2 , and this sum is zero because r1 + R2 = L0 . In conclusion, Ė = −µd m2 g||v1 ||. Notice that Ė ≤ 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 HO1 · E3 = m1 r12 θ̇1 and HO2 · E3 = m2 (L0 − r1 )2 sin2 (φ2 ) θ̇2 are conserved. That is, the angular momentum of each particle relative to O in the E3 direction is conserved. Configuration Manifold and Its Geometry The configuration manifold for this system is a four-dimensional subspace of E6 parameterized by r ∈ (0, L0 ), θ1 ∈ [0, 2π), θ2 ∈ [0, 2π), and φ2 ∈ (0, π). The kinematical line-element ds for this manifold is given by 2T̃2 ds = dt, m1 + m2 where we find T̃2 from T by imposing the constraints and collecting all those terms that are quadratic in the generalized velocities: T̃2 = m1 2 ṙ1 + r12 θ̇21 2 m2 2 + ṙ1 + (L0 − r1 )2 sin2 (φ2 ) θ̇22 + (L0 − r1 )2 φ̇22 . 2 To visualize the configuration manifold, one would give a two-dimensional picture of a plane with the coordinates r1 cos (θ1 )–r1 sin (θ1 ). This would be supplemented by a three-dimensional image featuring a sphere of radius 1 parameterized by φ2 and θ2 . 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 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 152 Dynamics of Systems of Particles m1 (b) (a) m1 m3 m2 (c) m3 m2 m3 m2 m1 Figure 5.6. Representative orbits from the three-body problem: (a), (b) examples of Lagrange’s equilateral triangle solutions, and (c) the figure-eight solution. For the solutions shown in this figure m1 = m2 = m3 . a Newtonian gravitational force field. That is, the potential energy for the system is [cf. (4.9)] Un = − Gm1 m2 Gm3 m1 Gm2 m3 − − , ||r2 − r1 || ||r1 − r3 || ||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 [117] in 1772∗ to the figure-eight solution that was only recently found numerically by Moore [147] and Moore and Nauenberg [148] and proven to exist by Chenciner and Montgomery [37, 145] (see Figure 5.6).† 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 [4, 13, 45]). 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 ∗ † A discussion of this famous solution can be found in many texts on celestial mechanics, for example, [93, 150, 220]. The interested reader is referred to the on-line article by Casselman [31], where simulations of several three-body problems can be found. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 5.1–5.3 153 equations of motion in [88] and his variational principle in [89]. Unfortunately, we do not have the opportunity to explore the three-body and n-body problems in any detail here; the interested reader is referred to the previously cited texts. The problems just 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 featuring 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 featuring systems of particles of this type are emphasized. EXERCISES 5.1. Consider the systems of particles discussed in Section 5.2. Suppose a timedependent force P(t)E1 acted on the particle m2 .∗ What are the equations of motion for each of these systems? 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.† Then, what are the equations of motion? 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 F2 , respectively. m2 m1 E3 Figure 5.7. A system of two particles. g O E1 (a) 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 convenient to describe its motion with the assistance of the relative position vector r21 = r2 − r1 . We describe this ∗ † This force is not conservative. The forces from these dashpots are not conservative. E2 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 154 Exercise 5.3 vector by using a spherical polar coordinate system {R2 , φ2 , θ2 }. Show that the position vector of the single 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 ) + R2 sin(φ2 ) sin(θ2 ))e5 + (z1 + R2 cos(φ2 ))e6 . (b) Using r and the curvilinear coordinate system it induces on E6 , q1 = r1 , q2 = θ1 , q3 = z1 , q4 = R2 , show that the six covariant basis vectors aJ = ∂r ∂qJ q5 = φ2 , q6 = θ2 , are a1 = ∂r = cos(θ1 )e1 + sin(θ1 )e2 + cos(θ1 )e4 + sin(θ1 )e5 , ∂r1 a2 = ∂r = −r1 sin(θ1 )e1 + r1 cos(θ1 )e2 − r1 sin(θ1 )e4 + r1 cos(θ1 )e5 , ∂θ1 a3 = ∂r = e3 + e6 , ∂z1 a4 = ∂r = sin(φ2 ) cos(θ2 )e4 + sin(φ2 ) sin(θ2 )e5 + cos(φ2 )e6 , ∂R2 a5 = ∂r = R2 cos(φ2 ) cos(θ2 )e4 + R2 cos(φ2 ) sin(θ2 )e5 − R2 sin(φ2 )e6 , ∂φ2 a6 = ∂r = −R2 sin(φ2 ) sin(θ2 )e4 + R2 sin(φ2 ) cos(θ2 )e5 . ∂θ2 (c) Show that the six contravariant basis vectors have the following representations: a1 = cos(θ1 )e1 + sin(θ1 )e2 , a2 = − sin(θ1 ) 1 cos(θ1 ) 2 e + e, r1 r1 a3 = e3 , a4 = sin(φ2 )(cos(θ2 )(e4 − e1 ) + sin(θ2 )(e5 − e2 )) + cos(φ2 )(e6 − e3 ), a5 = sin(φ2 ) 6 cos(φ2 ) cos(θ2 )(e4 − e1 ) + sin(θ2 )(e5 − e2 ) − (e − e3 ), R2 R2 a6 = − sin(θ2 ) cos(θ2 ) (e4 − e1 ) + (e5 − e2 ). R2 sin(φ2 ) R2 sin(φ2 ) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 5.3–5.4 155 (d) Show that the kinetic energy T of the particle of mass m = m1 + m2 is T= m1 + m2 2 ṙ1 + r12 θ̇21 + ż21 2 m2 2 Ṙ2 + R22 φ̇22 + R22 sin2 (φ2 )θ̇22 + 2 + m2 cos(φ2 ) Ṙ2 ż1 + φ̇2 ṙ1 R2 cos(θ21 ) + φ̇2 θ̇1 r1 R2 sin(θ21 ) − m2 sin(φ2 ) R2 φ̇2 ż1 − r1 R2 θ̇1 θ̇2 cos(θ21 ) − ṙ1 Ṙ2 cos(θ21 ) − m2 sin(φ2 ) −r1 Ṙ2 θ̇1 sin(θ21 ) + ṙ1 θ̇2 R2 sin(θ21 ) , where we have used the abbreviation θ21 = θ2 − θ1 . This expression for the kinetic energy follows from the definition T= m m2 m1 v·v= v1 · v1 + v2 · v2 . 2 2 2 (e) If the forces acting on the particles are F1 = −m1 gE3 and F2 = −m2 gE3 , then what are the force and potential energy U associated with this force? (f) What are the six Lagrange’s equations governing the motion of the particle of mass m?∗ 5.4. As shown in Figure 5.8, two particles of mass m1 and m2 are connected by a rigid massless rod of length L2 . 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 L1 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 is in the E1 − E2 plane. E3 g O E2 Figure 5.8. A planar double pendulum. E1 m1 m2 (a) What are the three constraints on the motion of the particle of mass m? (b) Using Lagrange’s prescription, what is the constraint force c acting on the particle of mass m? You should also, if possible, verify that the components of this force are physically realistic. ∗ You should refrain if possible from expanding the time derivative here – it will entail a considerable amount of algebra. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 156 Exercise 5.4–5.5 (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 components of c . (d) Let us now establish some of the equations of (c) by using an equivalent approach. Impose the constraints on T to determine the constrained kinetic energy T̃. In addition, determine the constrained potential energy Ũ. Verify that the following equations correspond to those you obtained from (c)∗ : d ∂ T̃ ∂ T̃ ∂ Ũ − = · a2 = − , dt ∂ θ̇1 ∂θ1 ∂θ1 ∂ T̃ d ∂ T̃ ∂ Ũ − = · a5 = − , dt ∂ φ̇2 ∂φ2 ∂φ2 ∂ T̃ d ∂ T̃ ∂ Ũ − = · a6 = − . dt ∂ θ̇2 ∂θ2 ∂θ2 (e) Suppose a nonintegrable constraint is imposed on the pendulum system discussed in (d): f1 · v1 + f2 · v2 + e = 0. 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 an nonintegrable constraint of your choice. 5.5. As shown in Figure 5.9, a model for an artificial satellite consists of two particles of mass m1 and m2 connected by a rigid massless rod of length L0 . 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. ∗ It is crucial to note that ∂ T̃ ∂ ṙ1 = ∂ T̃ ∂ Ṙ2 = ∂ T̃ ∂ ż1 = ∂ T̃ ∂r1 = ∂ T̃ ∂R2 = ∂ T̃ ∂z1 = 0. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 5.5–5.6 157 m1 E3 m2 Figure 5.9. Schematic of a model for a satellite orbiting a fixed body of mass m3 . O E2 m3 E1 (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 L0 during such motions? 5.6. As shown in Figure 5.10, a particle of mass m1 is connected by a linear spring of stiffness K1 and unstretched length L0 to a fixed point O. A second particle of mass m2 is attached by a rod of length L2 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, we assume that the motions of m1 and m2 are constrained to move on the E1 − E2 plane. E2 g O E1 Figure 5.10. A system of two parti- cles connected by a rigid massless rod of length L2 . linear spring rigid massless rod m1 m2 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 158 Exercise 5.6 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 = r1 er1 + z1 E3 , r2 = r1 + r2 er2 + z2 E3 . We define six coordinates as follows: q1 = θ1 , q2 = θ2 , q3 = r1 , q4 = r2 , q5 = z1 , (a) With the help of (5.16), what are the 12 vectors K = 1, . . . , 6. q6 = z2 . ∂r1 ∂qK and ∂r2 ? ∂qK (5.16) Here, (b) What are the three constraints on the motion of the system of particles? Argue that the constraint forces Fc1 and Fc2 acting on the individual particles have the prescriptions Fc1 = µ1 er2 + µ2 E3 , Fc2 = −µ1 er2 + µ3 E3 . (5.17) Compute the following six components: cK = Fc1 · ∂r1 ∂r2 + Fc2 · K . K ∂q ∂q 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 T̃ and potential energy Ũ of the constrained system of particles? (d) What are Lagrange’s equations of motion for the generalized coordinates of this system of particles? (e) Suppose a nonintegrable constraint∗ r1 θ̇1 + L2 θ̇2 = 0 (5.18) is imposed on the system of particles. After expressing this constraint in the form f1 · v1 + f2 · v2 = 0, argue that Fc1 = µ1 er2 + µ2 E3 + µ4 (eθ1 − eθ2 ) , Fc2 = −µ1 er2 + µ3 E3 + µ4 eθ2 . 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 Ṫ = 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 L1 and nonintegrable constraint (5.18) is removed. In this case, which is the classic ∗ Referring to the discussion of constraint (1.16) in Chapter 1, this constraint is arguably the simplest nonintegrable constraint that we can impose on this system. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 5.6–5.7 159 planar double pendulum, show that the equations governing the motion of the system are d2 θ1 dθ2 2 d2 θ2 sin (θ2 − θ1 ) (1 + α) 2 + αβ cos (θ2 − θ1 ) 2 − αβ dτ dτ dτ = − (1 + α) cos (θ1 ) , d2 θ2 1 1 dθ1 2 d2 θ1 + − θ + sin (θ2 − θ1 ) cos ) (θ 2 1 dτ2 β dτ2 β dτ 1 = − cos (θ2 ) . β (5.19) In writing (5.19), we have used the following dimensionless parameters and time variable: m2 g L2 α= , β= , τ= t. m1 L1 L1 The configuration manifold M for this system is a torus. What is the kinematical line-element for M? (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. 5.7. This problem is adapted from Section 156 of Whittaker [228] and the introduction to [30]. Consider a system of N particles, and, following Lecture 4 from Jacobi [102], define the following function: 1 mk ||rk||2 . 2 N J = k=1 The quantity 2J is often known as the moment of inertia of the system of particles. (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 2 1 mkm j rk − r j , 4 M (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 2 1 mkm j vk − v j . 4 M k=1 j=1 (5.21) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 160 Exercise 5.7 (c) Now suppose that the system of particles is in motion subject to a conservative Newtonian force field: Un = − N N 1 Gmkm j . rk − r j 2 j=1 k=1,k= j (5.22) The presence of the 12 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 balances of linear momenta for each particle, establish Jacobi’s equation: J̈ = 2T + Un . (5.23) This equation is also known as the Lagrange–Jacobi equation (see, for example, [220]). (d) For the orbits of the three-body problem shown in Figure 5.6(b), show that T = − 12 Un . (e) Show that J is a measure of the distance squared from the origin of the configuration space E3N for the representative particle discussed in Section 4.7. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 PART THREE DYNAMICS OF A SINGLE RIGID BODY 161 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 162 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 6 978 0 521 87483 0 June 9, 2008 Rotation Tensors 6.1 Introduction One of the key features of the rigid body dynamics problems that we will shortly examine 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 this subject is intimidating. To quote the mechanician Louis Poinsot (1777–1859), from [172], “. . . 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.” 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, many 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 of 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, Hamilton, and Rodrigues in the early part of the 19th 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 invoked in dynamics texts, and the interested reader is invited to compare our treatment with those in the textbooks cited in the bibliography. 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 163 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 164 Rotation Tensors 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. 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. Here 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 [196]. This material is supplemented with the elegant relative angular velocity vector that was proposed by Casey and Lam [29], the dual Euler basis [160], and a proof of Euler’s theorem by Guo [83]. 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 comprehended. 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 p3 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 rotation on this set of orthonormal right-handed basis vectors p1 , p2 , p3 . 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 vectors p1 , p2 , p3 to {t1 , t2 , t3 } as ⎡ ⎤ ⎡ ⎤ t1 p1 ⎢ ⎥ ⎢ ⎥ ⎣t2 ⎦ = R ⎣p2 ⎦ , t3 (6.1) p3 p2 t1 t2 θ θ p1 θ p3 = t3 Figure 6.1. The transformations of various basis vectors induced by a rotation about p3 through an angle θ. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.2 The Simplest Rotation 165 where ⎡ cos(θ) sin(θ) 0 0 0 1 ⎤ ⎢ ⎥ R = ⎣− sin(θ) cos(θ) 0⎦ . It is easy to see that the matrix R in this equation has a determinant of +1 and that its inverse is its transpose: R−1 = RT . That is, the matrix R is proper-orthogonal. By differentiating (6.1) with respect to time, we find that ⎡ ⎤ ⎡ ⎤⎡ ⎤ ṫ1 − sin(θ) cos(θ) 0 p1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ṫ2 ⎦ = θ̇ ⎣− cos(θ) − sin(θ) 0⎦ ⎣p2 ⎦ . ṫ3 0 0 0 p3 Using the result R−1 = RT , we can easily replace pi in this equation with ti : ⎡ ⎤ ⎡ ⎤⎡ ⎤⎡ ⎤ ṫ1 − sin(θ) cos(θ) 0 cos(θ) − sin(θ) 0 t1 ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ṫ2 ⎦ = θ̇ ⎣− cos(θ) − sin(θ) 0⎦ ⎣ sin(θ) cos(θ) 0⎦ ⎣t2 ⎦ ṫ3 0 ⎡ 0 ⎢ = θ̇ ⎣−1 . 0 0 0 0 0 1 t3 ⎤⎡ ⎤ t1 ⎥⎢ ⎥ 0 ⎦ ⎣t 2 ⎦ . 1 0 0 (6.2) t3 0 0 /0 1 ṘRT Notice that this is equivalent to the familiar results ṫ1 = θ̇ t2 and ṫ2 = −θ̇ t1 . It should also be clear from (6.2) that ṘRT is a skew-symmetric matrix. A vector θ̇ p3 = θ̇t3 can be introduced that has the useful property that ṫk = θ̇ p3 × tk (k = 1, 2, 3) . (6.3) You should notice how the vector θ̇ p3 can be inferred from the components of ṘRT . 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 form∗ tk = Rpk (k = 1, 2, 3) , (6.4) where R is the tensor R = cos(θ) (p1 ⊗ p1 + p2 ⊗ p2 ) − sin(θ) (p1 ⊗ p2 − p2 ⊗ p1 ) + p3 ⊗ p3 . It is left as an exercise to verify that this representation of the rotation is equivalent to (6.1).† Indeed, because I − p3 ⊗ p3 = p1 ⊗ p1 + p2 ⊗ p2 , ∗ † p3 = p1 ⊗ p2 − p2 ⊗ p1 , Background on tensor notation can be found the Appendix. The definition of the tensor product needed to establish this equivalence can be found in (A.1) in the Appendix. The alternating tensor is defined in Section A.7 [see, in particular, (A.11).] 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 166 Rotation Tensors we can express the tensor R entirely in terms of the axis of rotation p3 and the angle of rotation θ: R = cos(θ) (I − p3 ⊗ p3 ) − sin(θ)p3 + p3 ⊗ p3 . (6.5) As we shall see later, this representation 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. Differentiating (6.4), we find that ṫk = Ṙpk + Ṙṗ k 0 = Ṙ RT tk . /0 1 =pk = ṘRT tk. Some straightforward computations are used to show that Ṙ = −θ̇ sin(θ) (p1 ⊗ p1 + p2 ⊗ p2 ) − θ̇ cos(θ) (p1 ⊗ p2 − p2 ⊗ p1 ) , ṘRT = θ̇ (−p1 ⊗ p2 + p2 ⊗ p1 ) . With the help of these results, we conclude that ṫk = ṘRT tk = θ̇ p3 × tk (k = 1, 2, 3) . As expected, this result is in agreement with (6.3). You might have already noticed that θ̇ p3 = θ̇ t3 is the axial vector of ṘRT . We have demonstrated how several results for a familiar rotation can be expressed by using a tensor notation. This notation will prove to be very useful next when we wish to 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 θ̇p3 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 leverage several of these representations in the remainder of this chapter. 6.3 Proper-Orthogonal Tensors To proceed with our treatment of rotations, we appear to regress somewhat and discuss proper-orthogonal tensors. This discussion will lay the foundations for the possibility of three-parameter representations of rotations and subsequently the 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.3 Proper-Orthogonal Tensors 167 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 by three independent parameters. The tensors of interest here are second-order tensors, any one of which has the representation R= 3 3 Rikpi ⊗ pk. i=1 k=1 Let us now consider the transformation induced by R on the basis vectors p1 , p2 , and p3 . We define t1 = Rp1 = 3 Ri1 pi , i=1 t2 = Rp2 = 3 Ri2 pi , i=1 t3 = Rp3 = 3 Ri3 pi . i=1 You should notice that, by using the vectors ti , R has the representation R = t1 ⊗ p1 + t2 ⊗ p2 + t3 ⊗ p3 . We now wish to show that {t1 , t2 , t3 } is a right-handed orthonormal basis. First, let us verify the orthonormality: ti · tk = Rpi · Rpk = RT Rpi · pk = pi · pk = δik. Hence the vectors ti are orthonormal. To establish right-handedness, we use the definition of the determinant [see (A.6)]: [t1 , t2 , t3 ] = [Rp1 , Rp2 , Rp3 ] = det(R)[p1 , p2 , p3 ] = (1)(1) = 1. Hence {t1 , t2 , t3 } is a right-handed orthonormal basis.∗ ∗ It is left as an exercise to verify that this result holds for the example of a simple rotation presented in (6.1). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 168 Rotation Tensors A proper-orthogonal tensor also has a rather unusual representation. To arrive at it, we embark on a series of manipulations: 3 3 R = RRRT = R Rikpi ⊗ pk RT i=1 k=1 = 3 3 RikR (pi ⊗ pk) RT i=1 k=1 = 3 3 Rik (Rpi ⊗ Rpk) i=1 k=1 = 3 3 Rikti ⊗ tk. i=1 k=1 In summary, we have the following representations for R: R= 3 3 i=1 k=1 Rikpi ⊗ pk = 3 3 Rikti ⊗ tk = i=1 k=1 3 ti ⊗ pi . i=1 Notice that the components of R for the first two representations are identical, and the handedness of {p1 , p2 , p3 } is transferred without change by R to {t1 , t2 , t3 }. We observed that the components Rik of R are equal to tk · pi . As this product is equal to the cosine of the angle between tk and pi , each Rik is often referred to as a direction cosine. Consequently the matrix [Rik] is often known as the direction cosine matrix. Clearly, the nine angles whose cosines are tk · pi 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 shortly see, it is possible to arrive at three independent angles to parameterize R, but these angles are not all easily related to the angles between pi and tk. 6.4 Derivatives of a Proper-Orthogonal Tensor Here we consider a proper-orthogonal tensor R that is a function of time: R = R(t). Consider the derivative of RRT : d RRT = ṘRT + RṘT . dt However, İ = O, so the right-hand side of the preceding equation is zero. Hence, T ṘRT = −RṘT = − ṘRT . In other words, ṘRT is a skew-symmetric second-order tensor. We define R = ṘRT , in part because this tensor appears in numerous places later on. The tensor R is known as the angular velocity tensor (of R). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.4 Derivatives of a Proper-Orthogonal Tensor 169 The skew-symmetry of ṘRT allows us to define the angular velocity vector ωR : 1 , ωR = − ṘRT . 2 As mentioned in the Appendix, ωR × a = (ṘRT )a for all vectors a.∗ The most common 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 R = (E2 ⊗ E1 − E1 ⊗ E2 ) . Consequently, we can compute that [R ] = −2E3 . We conclude that the axial vector of R is the angular velocity vector E3 . It also useful to check that ( (E2 ⊗ E1 − E1 ⊗ E2 )) a = E3 × a for all vectors a. In a similar manner, we can also show that RT Ṙ is a skew-symmetric tensor and define an angular velocity tensor 0R and another angular vector ω0R : 0R = RT Ṙ, 1 ω0R = − [RT Ṙ]. 2 (6.7) At a later stage, the reader should be able to show that Rω0R = ω and R0R RT = R . A key to establishing one of these results is the identity, which holds for all orthogonal Q, , QBQT = det (Q) Q ( [B]) . Notice how this identity simplifies when Q is a proper-orthogonal tensor. Corotational Derivatives Recall that, for a proper-orthogonal tensor R, we have the representation R= 3 ti ⊗ pi . i=1 Now suppose that ṗi = 0. Then, 3 3 3 0 T T ˙ 7i R = R = ṘR = ti ⊗ pi + ṫi ⊗ pi RT ti ⊗ ṗ i=1 = 3 i=1 i=1 ṫi ⊗ ti . i=1 ∗ In Section A.7 of the Appendix, several examples featuring the calculation of the axial vector ωR of a given skew-symmetric tensor R are presented. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 170 Rotation Tensors If we now consider R tk, we find a familiar result: ṫi = R ti = ωR × ti . It is left as an exercise to show the less familiar result ṫi = 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 pi , then their time derivatives can be expressed in terms of the angular velocity vector of the rotation tensor and the basis vectors ti . Given any second-order tensor A and any vector a, we have the following representations: a= 3 A= ai ti , i=1 3 3 Aikti ⊗ tk. i=1 k=1 If we assume that a is a function of time, then ȧ = 3 ȧi ti + ai ṫi i=1 = 3 ȧi ti + ai (ωR × ti ) i=1 = 3 ȧi ti + ωR × a. i=1 Similarly, if we assume that A is a function of time, then Ȧ = 3 3 Ȧikti ⊗ tk + i=1 k=1 = 3 3 3 3 Aikṫi ⊗ tk + 3 3 i=1 k=1 Ȧikti ⊗ tk + i=1 k=1 = 3 3 3 3 Aikti ⊗ ṫk i=1 k=1 Aik(R ti ) ⊗ tk + i=1 k=1 3 3 Aikti ⊗ (R tk) i=1 k=1 Ȧikti ⊗ tk + R A − AR . i=1 k=1 o o 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 ti are constant: o A= 3 3 o Ȧikti ⊗ tk, a= i=1 k=1 3 ȧi ti . i=1 o Using the recently established expression for a, we observe that o ȧ = a + ωR × a. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.5 Euler’s Representation of a Rotation Tensor 171 Similarly, for any second-order tensor A, we have o Ȧ = A + R A − AR . 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 explicitly stated which rotation tensor the corotational derivative pertains to in situations in which 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.∗ Using notation introduced over a century later by Gibbs,† 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)(εr) + r ⊗ r, (6.8) where r is a unit vector and φ is a counterclockwise angle of rotation. We refer to (6.8) as Euler’s representation of a rotation tensor and use the function L to prescribe the rotation tensor associated with an axis and 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 shortly examine the role of φ. It is interesting to examine some of the features of representation (6.8). To this end, we define an orthonormal basis {p1 , p2 , p3 } with p3 = r. Using this basis, we have I = p1 ⊗ p1 + p2 ⊗ p2 + p3 ⊗ p3 , r ⊗ r = p3 ⊗ p3 , I − r ⊗ r = p1 ⊗ p1 + p2 ⊗ p2 , −(εr) = p2 ⊗ p1 − p1 ⊗ p2 . Consequently we can write R = L(φ, r = p3 ) = cos(φ)(p1 ⊗ p1 + p2 ⊗ p2 ) + sin(φ)(p2 ⊗ p1 − p1 ⊗ p2 ) + p3 ⊗ p3 . ∗ † This representation can be seen in Section 49 in one of Euler’s great papers on rigid body dynamics from 1775 [56]. 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 = pE1 + qE2 + rE3 . See Section 129 of [231]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 172 Rotation Tensors Using the identity (a ⊗ b)T = b ⊗ a, we find RT = L (φ, r = p3 ) = cos(φ)(p1 ⊗ p1 + p2 ⊗ p2 ) + sin(φ)(p1 ⊗ p2 − p2 ⊗ p1 ) + p3 ⊗ p3 . 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) − sin(φ) 0 ⎢ det(R) = det ⎣ 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 (φ1 , r1 ) represents the composite rotation. It is left as an exercise to show that Rc is a properorthogonal 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 noncommutative, the order in which we consider the composition is important. In general, L (φ2 , r2 ) L (φ1 , r1 ) = L (φ1 , r1 ) L (φ2 , r2 ) . The exceptions arise when r1 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⊥ + a , where a⊥ = a − (a · r)r = (I − r ⊗ r)a, a = (a · r)r = (r ⊗ r)a. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.5 Euler’s Representation of a Rotation Tensor 173 Ra⊥ φ a Figure 6.2. The transformation of a vector a by the rotation tensor Ra a⊥ R = L (φ, r). a r With the assistance of this decomposition, we now compute that Ra = L(φ, r)a = (cos(φ)(I − r ⊗ r) − sin(φ)(εr) + r ⊗ r) a = cos(φ)(I − r ⊗ r)a − sin(φ)(εr)a + (r ⊗ r)a = cos(φ)(I − r ⊗ r)a + sin(φ)r × a + (r · a)r, = cos(φ)a⊥ + sin(φ)r × a⊥ + a . (6.9) 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 (6.9) is known as Euler’s formula. Remarks on Euler’s Representation Euler’s representation (6.8) is unusual in several respects. First, you should notice that L(φ, r) = L(−φ, −r). This implies that there are two different representations for the same rotation tensor. Second, as R is a rotation tensor, R−1 = RT , and this leads to two easy representations for these tensors: 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. 6.5.1 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 θ associated with this tensor. First, one is normally 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 174 Rotation Tensors presented with the matrix components of R with respect to a basis, say {p1 , p2 , p3 }: R= 3 3 Rikpi ⊗ pk. i=1 k=1 If we compare this representation with (6.8), we find that Rik = ((cos(θ)(I − r ⊗ r) − sin(θ)(εr) + r ⊗ r) pk) · pi = cos(θ) (δik − ri rk) + ri rk − 3 sin(θ) jik r j , j=1 where r= 3 ri pi . i=1 Expanding the tation ⎡ R11 R12 ⎢ ⎢R21 R22 ⎣ R31 R32 expressions for the components of R, we find the matrix represenR13 ⎤ ⎡ 1 − r12 ⎥ ⎢ R23 ⎥ = cos (θ) ⎢ −r1 r2 ⎦ ⎣ R33 −r1 r3 ⎡ −r1 r2 ⎢ + sin (θ) ⎣ r3 −r2 r3 0 −r2 ⎤ ⎡ r12 ⎥ ⎢ −r2 r3 ⎥ + ⎢r1 r2 ⎦ ⎣ 1 − r32 r1 r3 1 − r22 −r3 0 −r1 r3 r2 r22 r2 r3 r1 r3 ⎤ ⎥ r2 r3 ⎥ ⎦ r32 ⎤ ⎥ −r1 ⎦ . r1 r1 r2 (6.10) 0 Notice that this matrix can be expressed as the sum of a symmetric and skewsymmetric matrix. The skew-symmetric part is the only part that changes when we transpose the tensor. To determine the angle of rotation, we calculate the trace of R: 1 1 (tr(R) − 1) = (R11 + R22 + R33 − 1) . 2 2 Looking at the skew-symmetric part of R, we find that ⎡ ⎤ ⎡ ⎤ R23 − R32 r1 1 ⎢ ⎥ ⎢ ⎥ ⎣R31 − R13 ⎦ . ⎣r2 ⎦ = − 2 sin(θ) r3 R12 − R21 cos(θ) = (6.11) (6.12) To verify the calculated value of r, we could examine the eigenvectors of R: The eigenvector corresponding to the unit eigenvalue should be parallel to the axis of rotation r. 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 i=1 ri ti . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.5 Euler’s Representation of a Rotation Tensor 175 The proof of this result is based on the observation that R has the same components with respect to the bases pi ⊗ pk and ti ⊗ tk. That is, R = 3i=1 3k=1 Rikpi ⊗ pk = 3 3 i=1 k=1 Rik ti ⊗ tk . AN EXAMPLE. As an example, suppose that the components of a rotation tensor R are ⎡ R11 ⎢ ⎢R21 ⎣ R31 R12 R13 ⎤ ⎡ 0.835959 ⎥ R23 ⎥ = ⎢ ⎦ ⎣0.271321 R33 0.47703 R22 R32 −0.283542 0.957764 −0.0478627 −0.469869 ⎤ ⎥ −0.0952472⎦ . 0.877583 We can compute the angle of rotation θ of this tensor by using (6.11): cos(θ) = 1 (0.835959 + 0.877583 + 0.957764 − 1) . 2 That is, θ = 33.3161◦ . We can calculate the axis of rotation r by using (6.12): r = 0.043135p1 − 0.861981p2 + 0.505103p3 = 0.043135t1 − 0.861981t2 + 0.505103t3 . In writing this result, we are emphasizing that the components of r in the basis p1 , p2 , p3 and {t1 , t2 , t3 } are identical. The Associated Angular Velocity Vector Given Euler’s representation (6.8) we 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 . As a preliminary result, we note that, because r is a unit vector, r · ṙ = 0. In addition, ˙ = 0 and İ = 0. Now, starting from (6.8), R = L(φ, r) = cos(φ)(I − r ⊗ r) − sin(φ)(εr) + r ⊗ r, we differentiate to find Ṙ = −φ̇ sin(φ)(I − r ⊗ r) − φ̇ cos(φ)(εr) + (1 − cos(φ))(ṙ ⊗ r + r ⊗ ṙ) − sin(φ)(εṙ). 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, r = (t1 ⊗ t2 − t2 ⊗ t1 ), ṙ = at1 + bt2 , r × ṙ = at2 − bt1 , ṙ = a(t2 ⊗ t3 − t3 ⊗ t2 ) + b(t3 ⊗ t1 − t1 ⊗ t3 ). (6.13) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 176 Rotation Tensors In these expressions, a and b are the scalar components of ṙ. Using (6.13) along with some manipulations, we find that ṘRT = −φ̇(t1 ⊗ t2 − t2 ⊗ t1 ) + (a(1 − cos(φ)) + b sin(φ))(t1 ⊗ t3 − t3 ⊗ t1 ) + (−b(1 − cos(φ)) + a sin(φ))(t3 ⊗ t2 − t2 ⊗ t3 ) = −φ̇t3 − (a(1 − cos(φ)) + b sin(φ))t2 − (−b(1 − cos(φ)) +a sin(φ))t1 . With the assistance of (6.13), we can now write the desired final result: R = ṘRT = −φ̇r − (1 − cos(φ))(r × ṙ) − sin(φ)ṙ. The associated angular velocity vector is ωR = φ̇r + sin(φ)ṙ + (1 − cos(φ))r × ṙ. (6.14) 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 shown in [161], it is usually possible to choose a fixed ba sis p1 , p2 , p3 where R = 3k=1 tk ⊗ pk so that a constant angular velocity implies a constant φ̇ 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 properorthogonal tensor. However, is the converse true? In other words, is every properorthogonal 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 Zhong-heng Guo [83]. 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] = IA [a, b, c], [Aa, Ab, Ac] = 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 . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.6 Euler’s Theorem: Rotation Tensors and Proper-Orthogonal Tensors 177 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, (6.15) and thus Q has an eigenvalue of 1. As Q has a unit eigenvalue, it has an eigenvector u such that Qu = u. When this equation is multiplied by QT , it follows that u = Qu = QT u. Thus u is a unit eigenvector of both Q and QT . Now consider a vector v ⊥ u. Some manipulations show that Qv · Qu = Qv · u, Qv · Qu = QT 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] = 1. Let us now calculate IQ : IQ = [Qu, v, w] + [u, Qv, w] + [u, v, Qw] = [u, v, w] + [u, Qv, w] + [u, v, Qw] = 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. 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) . (6.16) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 178 Rotation Tensors As mentioned previously, any vector a can be decomposed into two components: a and a⊥ , where a⊥ · u = 0 and a = a + a⊥ . Now, Qa = a , and, from (6.16), 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 (6.9), we conclude that Q has the form (6.8) 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. Thus we use these terms interchangeably. The result that every proper-orthogonal tensor is a rotation tensor is credited to Euler and dates to 1775 [55]: Every proper-orthogonal tensor is a rotation tensor. We shall shortly revisit this result and see why it is also known as Euler’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. We now wish to calculate its angular velocity tensor and vector. To do this, we use the work of Casey and Lam [29], who defined a very useful and intuitive relative angular velocity vector. To discuss the relative angular velocity vector, it is convenient to define three sets of right-handed orthonormal vectors: {1 t1 , 1 t2 , 1 t3 }, {2 t1 , 2 t2 , 2 t3 }, and {p1 , p2 , p3 }. In this section, we assume that ṗi = 0. For a given R1 and R2 , two of these sets can be defined by use of the representations R1 = 3 1 ti ⊗ pi , R2 = i=1 3 2 ti i=1 Notice that R= 3 2 ti ⊗ pi . i=1 In words, R transforms the vector pi into the vector 2 ti . ⊗ 1 ti . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.7 Relative Angular Velocity Vectors 179 Following Casey and Lam [29], let us consider the following relative angular velocity tensor: ˆ R2 = R − R1 . Using the definition of the angular velocity tensors and the fact that Ṙ = Ṙ2 R1 + R2 Ṙ1 , we find that ˆ R2 = R − R1 = ṘRT − Ṙ1 RT1 = Ṙ2 R1 RT1 RT2 + R2 Ṙ1 RT1 RT2 − Ṙ1 RT1 = Ṙ2 RT2 + R2 R1 RT2 − R1 = Ṙ2 RT2 + R2 R1 RT2 + TR1 = Ṙ2 + R2 R1 + TR1 R2 RT2 . However, o T R2 = Ṙ2 + R2 R1 + R1 R2 , o where the corotational derivative R2 is defined to be o R2 = 3 3 Ṙ2ik1 ti ⊗ 1 tk, i=1 k=1 o where R2ik = ((R2 )1 tk) · 1 ti . In words, R2 is the derivative of the tensor R2 assuming that 1 ti are constant. ˆ R2 is In conclusion, the relative angular velocity tensor o ˆ R2 = R − R1 =R2 RT2 . 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 (6.14) to find that the relative angular velocity vector has the representation o o ω̂R2 = ωR − ωR1 = φ̇2 r2 + sin (φ2 ) r2 + (1 − cos (φ2 )) r2 × r2 . (6.17) In this equation, 3 1 2o ω̂R2 = − R2 RT2 , 2 3 and the corotational derivative of r2 = i=1 ri (1 ti ) is (6.18) o r2 = ṙ1 (1 t1 ) + ṙ2 (1 t2 ) + ṙ3 (1 t3 ) . Formula (6.17) 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 180 Rotation Tensors rotation tensors, and we shall invoke (6.17) twice to get a representation for the angular velocity vector corresponding to R. The manner in which we do this is similar to the example now presented. 6.7.1 An Example To illustrate the convenience of relative angular velocity result (6.17), let us consider an example. Suppose we have two rotations R1 and R, where R1 = cos(ψ)(I − E3 ⊗ E3 ) − sin(ψ)E3 + E3 ⊗ E3 , 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)t2 + t2 ⊗ t2 . 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 . In addition, R consists of the rotation R1 followed by a rotation through an angle θ about t2 .∗ To calculate ωR1 we can appeal to (6.14) to find that ωR1 = ψ̇E3 + sin(ψ)Ė3 + (1 − cos(ψ))E3 × Ė3 = ψ̇E3 . To calculate ωR we cannot appeal directly to (6.14) because we do not know the axis and angle of rotation of R.† 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)(t3 ⊗ t3 + t1 ⊗ t1 ) − sin(θ) (t3 ⊗ t1 − t1 ⊗ t3 ) + t2 ⊗ t2 . Calculating the corotational rate of this tensor, we take its derivative, keeping ti fixed: o R2 = −θ̇ sin(θ)(t3 ⊗ t3 + t1 ⊗ t1 ) − θ̇ cos(θ) (t3 ⊗ t1 − t1 ⊗ t3 ) . Hence, ˆ R2 = θ̇ (t1 ⊗ t3 − t3 ⊗ t1 ) , ∗ † In the notation of the previous section, pi is replaced with Ei and 1 ti is replaced with ti . The interested reader might wish to use the famed Rodrigues formula [(6.49)] that is discussed in the exercises to compute these quantities and then one could use (6.14). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.8 Euler Angles 181 and, with the help of (6.18), we conclude that ω̂R2 = θ̇ t2 . (6.19) As an alternative method of calculating (6.19), we can use (6.17) directly. Thus we replace r2 and φ2 in (6.17) with t2 and θ, respectively. We then appeal o to the fact that t2 = 0. In summary, o o ω̂R2 = θ̇ t2 + sin(θ) t2 + (1 − cos(θ))t2 × t2 = θ̇ t2 . This alternative method is clearly equivalent to, but more attractive than, the o method that involved calculating R2 . Combining the expressions for ωR1 and ω̂R2 , we arrive at an expression for the angular velocity vector for R: ωR = θ̇ t2 + ψ̇E3 . The intuitive nature of this result is often surprising. It is very easy to use (6.14) to see that ωR2 = ω̂R2 . 6.8 Euler Angles The most popular representation of a rotation tensor is based on the use of three Euler angles. As discussed in [38, 230], this representation dates to works by Euler [54, 57] that he first presented in 1751.∗ 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 = R̀(γ 1 , γ 2 , γ 3 ) = L(γ 3 , g3 )L(γ 2 , g2 )L(γ 1 , g1 ). (6.20) 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)(εb) + b ⊗ b, ∗ 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 [59, 60], Cheng and Gupta [38], and Wilson [230]. Although [57] dates to the 18th century, it was first published posthumously in 1862. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 182 Rotation Tensors p3 = g1 t3 = g3 t2 φ g2 θ φ ψ p2 ψ p1 θ p1 φ p1 t1 Figure 6.3. 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 ti . The image on the left-hand side is a portrait of Leonhard Euler. 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 shortly see, there are 12 possible choices of the Euler angles. For example, Figure 6.3 illustrates these 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. (6.17)]. For the first rotation, the angular velocity vector is γ̇ 1 g1 [cf. (6.14)]. The angular velocity of the second rotation relative to the first rotation is γ̇ 2 g2 , and the angular velocity of the third rotation relative to the second rotation is γ̇ 3 g3 .∗ Combining the two relative angular velocity vectors with γ̇ 1 g1 , we conclude that ωR = γ̇ 3 g3 + γ̇ 2 g2 + γ̇ 1 g1 . (6.21) 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. ∗ The calculation of these angular velocity vectors is similar to the example discussed in Subsection 6.7.1. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.8 Euler Angles p2 183 t1 t1 t1 t2 ψ ψ t3 t3 t2 t3 θ φ θ p1 φ t3 t2 Figure 6.4. The transformations of various basis vectors induced by the individual angles in a set of 3–2–1 Euler angles. Alternative approaches to establishing (6.21) 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 [131]. Another approach, which can be found in [109, 185] and dates to Euler, features spherical geometry. Finally, a third (lengthy) approach involves directly differentiating (6.20) and then computing R and its axial vector. For the Euler angles to effectively parameterize all rotations, we need to assume that we can find γ k and γ̇ k such that, for any given ωR , (6.21) holds. For this to happen, it is necessary and sufficient that the vectors g1 , g2 , and g3 span E3 . When these vectors are not linearly independent, we say that the Euler angles have a singularity. This singularity is unavoidable for Euler angles and is often known as a “gimbal lock” (cf. [196]). We shall find that this singularity occurs for certain values of γ 2 , and we shall restrict this angle to avoid these singularities. For future purposes, it is also convenient to define the dual Euler basis {g j }: j j j g · gi = δi , where δi is the Kronecker delta.∗ Notice that ωR · gi = γ̇ i . (6.22) To determine this basis, one expresses 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, and g3 · g2 = 0 for the three unknowns a, b, and c. When the Euler angles have singularities, one will find that the two dual Euler basis vectors g1 and g3 cannot be defined. Another point of interest is the observation that g2 = g2 for all possible sets of Euler angles. 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 yaw–pitch–roll behavior, 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. One of the exercises at the end of this chapter ∗ The dual Euler basis vectors are analogous to the contravariant basis vectors and have similar uses in dynamics. We shall use the dual Euler basis later in various contexts, and a rapid summary of these uses can be found in O’Reilly [160]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 184 Rotation Tensors 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 Figure 6.4). These are arguably among the most popular sets of Euler angles.∗ 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)] or the Euler–Cardan angles [after Euler and Girolamo Cardano (1501–1576)]. First, suppose that the rotation tensor has the representation R= 3 ti ⊗ pi , i=1 where {pi } is a fixed Cartesian basis. The first rotation is about p3 through an angle ψ. This rotation transforms pi to ti . The second rotation is about the t2 axis through an angle θ. This rotation transforms ti to ti . The third and last rotation is through an angle φ about the axis t1 = t1 . Thus R = R̀(γ 1 = ψ, γ 2 = θ, γ 3 = φ) = L(φ, t1 )L (θ, t2 ) L(ψ, p3 ). Here, ti = L(φ, t1 )ti , ti = L(θ, t2 = t2 )ti , ti = L(ψ, t3 = p3 )pi . It is not difficult to express the various basis vectors as linear combinations of each other: ⎡ ⎤ ⎡ ⎤⎡ ⎤ t1 cos(ψ) sin(ψ) 0 p1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣t2 ⎦ = ⎣− sin(ψ) cos(ψ) 0⎦ ⎣p2 ⎦ , t3 0 0 p3 1 ⎡ ⎤ ⎡ ⎤⎡ ⎤ t1 cos(θ) 0 − sin(θ) t1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ 1 0 ⎦ ⎣t 2 ⎦ , ⎣t2 ⎦ = ⎣ 0 t3 sin(θ) ⎡ ⎤ ⎡ t1 1 ⎢ ⎥ ⎢ ⎣t2 ⎦ = ⎣0 t3 0 0 0 cos(φ) cos(θ) t3 ⎤⎡ ⎤ t1 ⎥⎢ ⎥ sin(φ) ⎦ ⎣t2 ⎦ . 0 − sin(φ) cos(φ) (6.23) t3 The inverses of these relationships are easy to obtain once you realize that each of the three matrices in (6.23) is orthogonal. As the inverse of an orthogonal matrix is ∗ For example, they are used in Greenwood’s text [79], Rao’s text [176], and numerous texts on vehicle and aircraft dynamics. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.8 Euler Angles 185 its transpose, we quickly arrive at the sought-after results: ⎡ ⎤ ⎡ cos(ψ) − sin(ψ) p1 ⎢ ⎥ ⎢ ⎣p2 ⎦ = ⎣ sin(ψ) cos(ψ) p3 0 0 ⎡ ⎤ ⎡ t1 cos(θ) ⎢ ⎥ ⎢ ⎣t 2 ⎦ = ⎣ 0 t3 0 t3 1 0 sin(θ) 1 0 ⎤⎡ ⎤ t1 ⎥⎢ ⎥ ⎦ ⎣t 2 ⎦ , − sin(θ) 0 cos(θ) ⎡ ⎤ ⎡ t1 1 ⎢ ⎥ ⎢ ⎣t2 ⎦ = ⎣0 t3 ⎤⎡ ⎤ 0 t1 ⎥⎢ ⎥ 0⎦ ⎣t2 ⎦ , t3 ⎤⎡ ⎤ t1 ⎥⎢ ⎥ cos(φ) − sin(φ)⎦ ⎣t2 ⎦ . 0 0 sin(φ) (6.24) t3 cos(φ) Relationships (6.23) and (6.24) can be combined to express ti in terms of pk and vice versa. Later, they will also be used to obtain representations for the Euler basis in terms of pk and ti . By using (6.24) we can find a representation for the components Rij = (Rp j ) · pi . The components are easily expressed by a matrix representation: ⎡ R11 ⎢ ⎢R21 ⎣ R31 R12 R22 R32 ⎤ ⎡ ⎤⎡ ⎤ cos(ψ) − sin(ψ) 0 cos(θ) 0 sin(θ) ⎥ ⎢ ⎢ ⎥ R23 ⎥ = ⎣ sin(ψ) cos(ψ) 0⎥ 1 0 ⎦ ⎦⎣ 0 ⎦ R33 0 0 1 − sin(θ) 0 cos(θ) R13 ⎡ 1 ⎢ × ⎣0 0 0 0 ⎤ ⎥ cos(φ) − sin(φ)⎦ . sin(φ) (6.25) cos(φ) Representation (6.25) is the transpose of what one might naively expect. However, recalling that Rik = pi · tk will hopefully resolve this initial surprise. Indeed, it is useful to note that ⎡ ⎤ ⎡ R11 p1 ⎢ ⎥ ⎢ R ⎣p2 ⎦ = ⎢ ⎣ 21 R31 p3 ⎡ ⎤ ⎡ R11 t1 ⎢ ⎥ ⎢ ⎢ ⎣t2 ⎦ = ⎣R12 R13 t3 R32 ⎤⎡ ⎤ t1 ⎥ ⎥ R23 ⎥ ⎢ ⎦ ⎣t 2 ⎦ , R33 t3 R21 R31 R12 R22 R22 R23 R13 ⎤⎡ ⎤ p1 ⎥⎢ ⎥ ⎥ R32 ⎣p2 ⎦ . ⎦ R33 p3 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 186 Rotation Tensors As mentioned earlier, by examining the individual rotations [cf. (6.23)], we can show that the Euler basis vectors have the representations THE EULER BASIS. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ g1 p3 − sin(θ) sin(φ) cos(θ) cos(φ) cos(θ) t1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ cos(φ) − sin(φ) ⎦ ⎣t2 ⎦ . ⎣g2 ⎦ = ⎣ t2 ⎦ = ⎣ 0 g3 t1 1 0 (6.26) t3 0 Alternatively, we can also express the Euler basis in terms of the basis vectors p1 , p2 , p3 : ⎡ ⎤ ⎡ ⎤ ⎡ g1 p3 ⎢ ⎥ ⎢ ⎥ ⎢ ⎣g2 ⎦ = ⎣ t2 ⎦ = ⎣ g3 0 0 1 − sin(ψ) cos(ψ) 0 ⎤⎡ ⎤ p1 ⎥⎢ ⎥ ⎦ ⎣p2 ⎦ . cos(θ) cos(ψ) cos(θ) sin(ψ) − sin(θ) t1 (6.27) p3 This representation is useful for establishing the components ωR · pk. We can now determine the dual Euler basis vectors gk. We recall the remarks following (6.22) and express each of the dual Euler basis vectors in terms of their components relative to the basis {t1 , t2 , t3 }. That is, THE DUAL EULER BASIS. gk = gk1 t1 + gk2 t2 + gk3 t3 . Combining these results for the dual Euler basis vectors, we observe that ⎡ 1 ⎤ ⎡ 11 g g ⎢ 2 ⎥ ⎢ 21 ⎣g ⎦ = ⎣ g g3 g22 ⎤⎡ ⎤ t1 ⎢ ⎥ ⎥ g23 ⎦ ⎣t2 ⎦ . g32 g33 g12 g31 g13 (6.28) t3 With some manipulation, the relations gi · gk = δki can be expressed as nine equations for the nine unknowns gik: ⎡ 11 ⎤⎡ ⎤ ⎡ ⎤ g21 g31 g − sin(θ) sin(φ) cos(θ) cos(φ) cos(θ) 1 0 0 ⎢ 12 ⎥⎢ ⎥ ⎢ ⎥ g22 g32 ⎦ ⎣ 0 cos(φ) − sin(φ) ⎦ = ⎣0 1 0⎦ . ⎣g g13 g23 g33 1 0 0 0 0 1 , Isolating the matrix gik on the left-hand side of this equation, we find that ⎡ g11 g12 ⎢ 21 ⎣g g31 g13 ⎤ ⎡ 0 g22 ⎥ ⎢ g23 ⎦ = ⎣0 g32 g33 sin(φ) sec(θ) cos(φ) sec(θ) cos(φ) − sin(φ) 1 ⎥ ⎦. 1 sin(φ) tan(θ) cos(φ) tan(θ) That is, the dual Euler basis vectors have the representations ⎤⎡ ⎤ ⎡ 1⎤ ⎡ 0 sin(φ) sec(θ) cos(φ) sec(θ) t1 g ⎥⎢ ⎥ ⎢ 2⎥ ⎢ cos(φ) − sin(φ) ⎦ ⎣t2 ⎦ . ⎣g ⎦ = ⎣0 g3 ⎤ sin(φ) tan(θ) cos(φ) tan(θ) t3 (6.29) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.8 Euler Angles (a) 187 (b) g1 = p3 g1 = p3 1 g t3 θ t3 θ g2 = t2 g2 = t2 p2 ψ θ ψ t1 p1 p2 g3 p1 t1 φ g3 = t1 Figure 6.5. (a) The dual Euler basis vectors g for the 3–2–1 set of Euler angles: g1 t3 , g2 = t2 , k and g3 t1 . (b) The Euler angles ψ and θ serve as coordinates for the Euler basis vector g3 = t1 in a manner that is similar to the role that spherical polar coordinates play in parameterizing eR and eφ . If we had used (6.27) in place of (6.26) to calculate the dual Euler basis vectors, then we would have found the following representations: ⎡ 1⎤ ⎡ ⎤⎡ ⎤ g cos(ψ) tan(θ) sin(ψ) tan(θ) 1 p1 ⎢ 2⎥ ⎢ ⎥⎢ ⎥ cos(ψ) 0⎦ ⎣p2 ⎦ . ⎣g ⎦ = ⎣ − sin(ψ) 3 g cos(ψ) sec(θ) sin(ψ) sec(θ) 0 (6.30) p3 Expressions for the dual Euler basis vectors in terms of tk are easily inferred from (6.30) and are shown in Figure 6.5(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 ⎢ ⎥ ⎢ ⎣g2 ⎦ = ⎣ g3 1 0 ⎤ ⎡ 1⎤ 0 − sin(θ) g ⎥ ⎢ 2⎥ 1 0 ⎦ ⎣g ⎦ . − sin(θ) 0 1 (6.31) 3 g The simplicity of this relationship (related versions of which hold for the other 11 sets of Euler angles) is surprising. If we examine (6.26), we see that the Euler basis fails to be a basis for E3 when θ = ± π2 . One of the easiest ways to see this fact is to consider ψ and θ + π2 to be spherical polar coordinates for t1 [see Figure 6.5(b)]. When θ = ± π2 , g1 = p3 = ±g3 , and the Euler basis does not span E3 . To avoid the aforementioned singularity, it is necessary to place restrictions on the second Euler angle: θ ∈ − π2 , π2 . The other two angles are free to range from 0 to 2π. SINGULARITIES. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 188 Rotation Tensors The angular velocity vector ωR associated with the 3–2–1 Euler angles has several representations: ANGULAR VELOCITY VECTORS. 3 - 1 , T ωR = − ṘR = γ̇ i gi 2 i=1 = φ̇t1 + θ̇ t2 + ψ̇p3 = (−ψ̇ sin(θ) + φ̇)t1 + (ψ̇ sin(φ) cos(θ) + θ̇ cos(φ))t2 + (ψ̇ cos(φ) cos(θ) − θ̇ sin(φ))t3 . To arrive at the primitive representation ω = φ̇t1 + θ̇ t2 + ψ̇p3 , we needed to compute two relative angular velocity vectors. We calculated the first of these, θ̇ t2 , assuming that ti were fixed. The explicit details of this calculation are easily inferred from our earlier example in Subsection 6.7.1. We calculated the second relative angular velocity vector, φ̇t1 , by using the relative rotation tensor L φ, t1 , assuming that ti were fixed. It is also interesting to note that the angular velocity vector ω0R has the representations 1 , ω0R = − RT Ṙ = RT ωR 2 = φ̇RT t1 + θ̇ RT t2 + ψ̇RT p3 = (−ψ̇ sin(θ) + φ̇)p1 + (ψ̇ sin(φ) cos(θ) + θ̇ cos(φ))p2 +(ψ̇ cos(φ) cos(θ) − θ̇ sin(φ))p3 . In establishing this result, we used the fact that RT ti = pi . 6.8.2 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 6.6). These are the set of Euler angles that Lagrange used,∗ and they are also used in Arnol’d [9], Landau and Lifshitz [125], and Thomson [214], 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, are 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 L θ, t1 L(ψ, p3 ), where {p3 , t1 , t3 } is the Euler basis and ti = L (ψ, p3 ) pi , ∗ ti = L θ, t1 ti , ti = L φ, t3 ti . See [118] and Section IX of the Second Part of [121]: His φ, ω, ψ correspond to our φ, θ, ψ, respectively. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.8 Euler Angles p2 189 t3 t1 t1 t2 t3 t2 t2 t2 θ ψ φ θ ψ φ p1 t1 t2 Figure 6.6. The transformations of various basis vectors induced by the individual angles in a set of 3–1–3 Euler angles. 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 t1 is often known as the line of nodes [214]. The angular velocity vector has the representations - i 1 , ωR = − ṘRT = γ̇ gi = φ̇t3 + θ̇ t1 + ψ̇p3 . 2 3 (6.32) i=1 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 set of Euler angles. EULER BASIS. It is not difficult to show that the Euler basis {gi } has the representa- tions ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ g1 p3 sin(φ) sin(θ) cos(φ) sin(θ) cos(θ) t1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ − sin(φ) 0 ⎦ ⎣t 2 ⎦ ⎣g2 ⎦ = ⎣ t1 ⎦ = ⎣ cos(φ) g3 t3 0 ⎡ ⎢ =⎣ 0 t3 1 0 0 1 cos(ψ) sin(ψ) 0 sin(θ) sin(ψ) ⎤⎡ ⎤ p1 ⎥⎢ ⎥ ⎦ ⎣p2 ⎦ . − sin(θ) cos(ψ) cos(θ) (6.33) p3 With these results and (6.32), two other representations for ωR can be obtained, but this is left as an exercise. 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. Fortunately, it is easy to see from Figure 6.7(a) that this is the case when t3 = ±p3 . As a result, restrictions are placed on the second angle: SINGULARITIES. 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 190 Rotation Tensors (b) (a) g1 = p3 g3 = t3 g1 = p3 g3 = t3 θ θ φ g1 t2 φ p2 ψ p2 ψ g3 g2 = t1 p1 g2 = t1 p1 Figure 6.7. (a) The Euler basis and (b) dual Euler basis vectors for the 3–1–3 set of Euler angles. For this set of Euler angles, g1 t2 , g2 = t1 , and g3 t2 . DUAL EULER BASIS. By following the procedure that led to (6.29), we find that the dual Euler basis gi has the representation∗ ⎡ 1⎤ ⎡ g sin(φ)cosec(θ) ⎢ 2⎥ ⎢ cos(φ) ⎣g ⎦ = ⎣ 3 g cos(φ)cosec(θ) − sin(φ) − sin(φ) cot(θ) − cos(φ) cot(θ) ⎤⎡ ⎤ 0 t1 ⎥⎢ ⎥ 0⎦ ⎣t2 ⎦ . 1 (6.34) t3 Similarly, ⎡ 1⎤ ⎡ g − sin(ψ)cot(θ) ⎢ 2⎥ ⎢ cos(ψ) ⎣g ⎦ = ⎣ g3 cos(ψ)cot(θ) sin(ψ) ⎤⎡ ⎤ 1 p1 ⎥⎢ ⎥ 0⎦ ⎣p2 ⎦ . sin(ψ)cosec(θ) − cos(ψ)cosec(θ) 0 p3 The dual Euler basis vectors are shown in Figure 6.7. 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 (6.34), we can show that ωR · g1 = ψ̇, ωR · g2 = θ̇, ωR · g3 = φ̇. We can also establish the following results: ⎡ ⎤ ⎡ g1 ⎢ ⎥ ⎢ ⎣g2 ⎦ = ⎣ g3 1 0 cos(θ) 0 1 0 ⎤ ⎡ 1⎤ g ⎥ ⎢ 2⎥ ⎦ ⎣g ⎦ . cos(θ) 0 1 g3 (6.35) 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. ∗ That is, one calculates the inverse of the transpose of the 3 × 3 matrix in (6.33) that describes gi in terms of tk . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.9 Further Representations of a Rotation Tensor 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 = 3 i=1 ti ⊗ pi : 1–2–3 Set: ωR = ψ̇p1 + θ̇ t2 + φ̇t3 , 3–2–3 Set: ωR = ψ̇p3 + θ̇ t2 + φ̇t3 , 1–2–1 Set: ωR = ψ̇p1 + θ̇ t2 + φ̇t1 , 1–3–1 Set: ωR = ψ̇p1 + θ̇ t3 + φ̇t1 , 1–3–2 Set: ωR = ψ̇p1 + θ̇ t3 + φ̇t2 , 2–3–1 Set: ωR = ψ̇p2 + θ̇ t3 + φ̇t1 , 2–3–2 Set: ωR = ψ̇p2 + θ̇ t3 + φ̇t2 , 2–1–2 Set: ωR = ψ̇p2 + θ̇ t1 + φ̇t2 , 2–1–3 Set: ωR = ψ̇p2 + θ̇ t1 + φ̇t3 , 3–1–3 Set: ωR = ψ̇p3 + θ̇ t1 + φ̇t3 , 2–3–1 Set: ωR = ψ̇p2 + θ̇ t3 + φ̇t1 , 3–1–2 Set: ωR = ψ̇p3 + θ̇ t1 + φ̇t2 . The sets of Euler angles, 121, 131, 232, 212, 313, and 323, are known as the symmetric sets, whereas the other six sets are known as asymmetric sets. The latter 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 [210]. In his seminal text [24] 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 6.8). It is interesting to recall that the Wright brothers first successful flight was in 1903. For all sets of Euler angles, a singularity is present for certain values of the second angle θ. At these values g1 = ±g3 , and the Euler basis fails to be a basis for E3 . To avoid these 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. 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 191 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 192 Rotation Tensors z φ G φ θ φ θ x Figure 6.8. Reproduction of Figure 3 in Bryan’s seminal text [24] on the stability of aircraft. The angles θ and φ in this figure are the pitch and roll 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. Shuster’s review article [196], and we now discuss two of them. These are representations that are due to Olinde Rodrigues [181] in 1840 and the Euler parameter representation. The Rodrigues Vector The Rodrigues representation is based on the vector: φ λ = tan r. 2 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 ) , cos(φ) = 1−λ·λ , 1+λ·λ sin(φ) = 1 + tan2 ( φ2 ) 1 − tan2 ( φ2 ) 1 + tan2 ( φ2 ) , you should be able to verify that cos(φ) = 2λ · r , 1+λ·λ Substituting for r and φ in (6.8), we find the Rodrigues representation: R = R̃(λ) = 1 ((1 − λ · λ)I + 2λ ⊗ λ − 2(ελ)) . 1+λ·λ 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.9 Further Representations of a Rotation Tensor 193 The angular velocity vector associated with this representation is ωR = 2 λ̇ − λ̇ × λ . 1+λ·λ This vector can be calculated by directly substituting into the earlier result associated with Euler representation (6.14). The Euler–Rodrigues Symmetric Parameters One of the most popular four-parameter representations uses the four Euler– Rodrigues symmetric parameters e0 and e.∗ These parameters are often known as the Euler parameters and have very interesting historical connections.† They can be defined as e0 = cos φ , 2 e = sin φ r. 2 As a consequence of their definition, the parameters are subject to what is known as the Euler parameter constraint‡ : 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 (6.8), we find the Euler– Rodrigues symmetric parameter representation: R = R̄(e0 , e) = e20 − e · e I + 2e ⊗ e − 2e0 (εe). ∗ † ‡ The four parameters, e0 and the three components of e, are often considered to be the four components of a quaternion q = e0 + e1 i + e2 j + e3 k, where ei are the components of e relative to a right-handed orthonormal basis, and i, j, and k are bases vectors for the quaternion. Consequently, Euler–Rodrigues symmetric parameters are sometimes referred to as (unit) quaternions. Excellent discussions can be found in Altmann [2, 3] and Gray [77]. The relaxation of this constraint is discussed in O’Reilly and Varadi [166] 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 [69]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 194 Rotation Tensors Suppose R = 3i=1 ti ⊗ pi ; then it is easy to show that e · tk = e · pk. If one writes out expressions for the components of Rik = tk · pi ,∗ ⎡ R11 ⎢ ⎢R21 ⎣ R31 R12 R22 R32 R13 ⎤ ⎡ 1 ⎥ ⎢ R23 ⎥ = e20 − e21 − e22 − e23 ⎢0 ⎦ ⎣ R33 0 ⎡ 0 ⎢ + ⎣ 2e0 e3 −2e0 e2 −2e0 e3 0 2e0 e1 ⎤ ⎡ 2 0 2e1 ⎥ ⎢ 0⎥ + ⎢2e1 e2 ⎦ ⎣ 2e1 e3 1 0 1 0 2e0 e2 2e1 e2 2e22 2e2 e3 2e1 e3 ⎤ ⎥ 2e2 e3 ⎥ ⎦ 2e23 ⎤ ⎥ −2e0 e1 ⎦ , (6.36) 0 then it is easy to see that the rotation tensor is a quadratic function of the four parameters e0 , e1 , e2 , 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 (6.36). The angular velocity vector associated with this representation is ωR = 2 (e0 ė − ė0 e + e × ė) . Again, we can calculate this vector by directly substituting into the earlier result associated with Euler representation (6.14) for ωR . Notice that the angular velocity vector is a relatively simple function of the Euler–Rodrigues symmetric parameters and their derivatives. In one of the exercises at the end of this chapter, 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. Modern applications of these parameters arise in the estimation of the rotation tensor (attitude) of spacecraft and in computer vision and robotics. In these areas, 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 [95, 192] and in the satellite dynamics community as the Wahba problem after Grace Wahba [222]. That is, given the N ≥ 2 measurements a1 , . . . , aN and b1 , . . . , bN , 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 ||bK − RaK − d||2 , 2 (6.37) K=1 where αK is a scalar weight for the Kth measurement. When d = 0, Davenport showed that, by parameterizing R in (6.37) by e0 and e, it is possible to find a ∗ The corresponding matrix representation for Euler’s representation was established earlier; see (6.10). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.10 Derivatives of Scalar Functions of Rotation Tensors 195 very elegant solution to the Wahba problem.∗ The corresponding elegant solution to the Procrustes problem ( i.e., when d = 0) was established later by Horn [95].† 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 γ̇ i gi i=1 = φ̇r + sin(φ)ṙ + (1 − cos(φ))r × ṙ = 2 (e0 ė − ė0 e + e × ė) = 2 λ̇ − λ̇ × λ . 1+λ·λ (6.38) With some minor manipulations of these equations, one can also obtain expressions for ṙ, ė, and λ̇. 6.10 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 be subsequently revealed that a simple expression for the derivative of U can be found in terms of a vector uR and the dual Euler basis. This result is presented in Equation (6.41). To start, we use the simplest representation for R: R= 3 3 Rikpi ⊗ pk, i=1 k=1 where ṗk = 0. Then, as U = U(R) = U (Rik, p j ), we find by using the chain rule that U̇ = 3 3 ∂U Ṙik. ∂Rik i=1 k=1 We can express this result by using the trace operator: U̇ = tr ∗ † ∂U T Ṙ , ∂R Discussions of Davenport’s solution and those of others, along with extensions to Wahba’s original formulation, can be found in [136, 137, 195, 197]. Additional references to other solutions to this problem can be found in [46, 49, 200]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 196 Rotation Tensors where ∂U ∂U = pi ⊗ pk. ∂R ∂Rik 3 3 i=1 k=1 Noting that R = ṘRT , and, after introducing the angular velocity tensor, we can then write ∂U T T U̇ = tr R R . ∂R However, as R is skew-symmetric, tr ATR = 0 for all A = AT . Consequently, ∂U T only the skew-symmetric part of ∂R R contributes to U̇: ∂U T T R R U̇ = tr ∂R = tr UR TR , where we have introduced the skew-symmetric operator: 1 ∂U T ∂U T UR = R −R . 2 ∂R ∂R As UR is a skew-symmetric tensor, we can calculate a vector that corresponds to twice the axial vector of UR ∗ : uR = −ε [UR ] . (6.39) The existence of this vector allows us to establish the following representations for U̇: ∂U T Ṙ U̇ = tr ∂R = tr UR T = uR · ωR . (6.40) Because we have established numerous representations for ωR , we next invoke the final form of the representation for U̇ to determine representations for uR . As an example, suppose U is parameterized by use of Euler angles: U = Û γ k . Then, invoking (6.40)2 , U̇ = 3 3 ∂ Û k γ̇ = u · γ̇ j g j . R ∂γ k k=1 ∗ j=1 The reason for the absence of the factor 12 in (6.39) can be inferred from identity (A.12), which is discussed in the Exercises at the end of the Appendix. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 6.10 Derivatives of Scalar Functions of Rotation Tensors 197 With the help of the dual Euler basis, we conclude that uR = 3 ∂ Û i g. ∂γ i (6.41) i=1 This is the simplest, and most useful, representation that we know of for uR . It first appeared in [163]. We now assume that R is parameterized by use of one of the other three methods mentioned previously. By evaluating U̇, using identity (6.40)2 and representations (6.38), and following the procedure that led to (6.41), we can find three other representations for uR . With details omitted, a summary of the representations is now presented: uR = 3 ∂ Û i g ∂γ i i=1 φ ∂ Ũ ∂ Ũ 1 cot (I − r ⊗ r) − εr = r+ ∂φ 2 2 ∂r 1 ∂ Ū ∂ Ū (e0 I − εe) − = e 2 ∂e ∂e0 = 1 ∂ Ŭ , (I + λ ⊗ λ − ελ) 2 ∂λ (6.42) where U(R) = Ū(e0 , e) = Û(γ 1 , γ 2 , γ 3 ) = Ũ(φ, r) = Ŭ(λ). Representation (6.42)4 was first established by Simmonds [199]. We also note that a representation that is closely related to (6.42)2 is discussed in Antman [5]. Several of the partial derivatives in (6.42) need to be carefully evaluated. For example, because r is a unit vector, the derivative ∂∂rŨ must be evaluated on the sur∂ Ū ∂U face r · r = 1. Related remarks pertain to ∂∂eŪ , ∂e , and ∂R . In other words, these are 0 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 ∂ R̂ T Euler angles. Indeed, (6.42), the chain rule, and the identity εgi = − ∂γ can be iR used to show that ∂ Ū = uR · 2e, ∂e0 ∂ Ū = 2(e0 I + εe)uR , ∂e ∂U = −(εuR )R, ∂R ∂ Ũ = uR · r, ∂φ φ ∂ Ũ = sin(φ)(I − r ⊗ r) + 2 sin2 εr uR . ∂r 2 As discussed in O’Reilly [160] and Simmonds [199], results (6.42) can be used to establish moment potentials associated with conservative moments. We shall use (6.42)1 extensively. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 198 Exercises 6.1–6.2 EXERCISES 6.1. Consider the tensor R: R = cos(θ)E1 ⊗ E1 + sin(θ)E2 ⊗ 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 + E3 ⊗ E3 . (b) Show that R = θ̇ (E2 ⊗ E1 − E1 ⊗ E2 ) = θ̇ (eθ ⊗ er − er ⊗ eθ ) , ωR = θ̇E3 . 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 L θ, t2 L (ψ, E3 ) . This set of Euler angles is used in several texts, for example, Section 4.2 of Ginsberg [71], Kelvin and Tait [109], Routh [184, 185], and Whittaker [228],∗ and is illustrated in Figure 6.3. (a) Draw figures illustrating the relationships between (i) Ei and ti , (ii) ti and ti , and (iii) ti and ti . (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 − sin(θ) cos(φ) sin(φ) sin(θ) cos(θ) t1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ sin(φ) cos(φ) 0 ⎦ ⎣t 2 ⎦ ⎣g2 ⎦ = ⎣ g3 0 ⎡ 0 ⎢ = ⎣ − sin(ψ) 0 1 0 1 cos(ψ) 0 t3 ⎤⎡ ⎤ E1 ⎥⎢ ⎥ ⎦ ⎣E2 ⎦ . cos(ψ) sin(θ) sin(ψ) sin(θ) cos(θ) ∗ E3 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 Section 16 of [228]. Routh’s notation is similar to ours, but his coordinate axes are left-handed. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercise 6.2 199 In addition, show that the Euler angles are such that ⎡ ⎤ ⎡ ⎤⎡ ⎤ cos(φ) sin(φ) 0 cos(θ) 0 − sin(θ) t1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ 1 0 ⎦ ⎣t2 ⎦ = ⎣− sin(φ) cos(φ) 0⎦ ⎣ 0 0 t3 ⎡ 0 1 sin(θ) 0 cos(θ) ⎤⎡ ⎤ 0 E1 ⎢ ⎥⎢ ⎥ × ⎣− sin(ψ) cos(ψ) 0⎦ ⎣E2 ⎦ . cos(ψ) sin(ψ) 0 0 1 (6.43) E3 (d) Using (6.43), derive expressions for the components Rik of R. These components have a variety of representations Rik = (Rtk) · ti = (REk) · Ei = tk · Ei . (e) Recall that every rotation tensor R has an axis of rotation r and angle of rotation. With the help of the results presented in Subsection 6.5.1, select four different values of the set (φ, θ, ψ) and determine the corresponding axis of rotation and the 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 ⎡ 1⎤ ⎡ ⎤⎡ ⎤ g − cos(φ)cosec(θ) sin(φ)cosec(θ) 0 t1 ⎢ 2⎥ ⎢ ⎥⎢ ⎥ sin(φ) cos(φ) 0⎦ ⎣t2 ⎦ . ⎣g ⎦ = ⎣ g3 − sin(φ) cot(θ) 1 cos(φ) cot(θ) t3 With the help of these results, verify the following expressions for the dual Euler basis in terms of the bases {E1 , E2 , E3 } and g1 , g2 , g3 : ⎡ 1⎤ ⎡ − cos(ψ)cot(θ) g ⎢ 2⎥ ⎢ ⎣g ⎦ = ⎣ − sin(ψ) g3 − sin(ψ)cot(θ) cos(ψ) 1 ⎢ =⎣ cosec2 (θ) 0 E1 ⎤ ⎥⎢ ⎥ 0⎦ ⎣E2 ⎦ cos(ψ)cosec(θ) sin(ψ)cosec(θ) 0 ⎡ ⎤⎡ E3 ⎤⎡ ⎤ −cot(θ)cosec(θ) g1 ⎥⎢ ⎥ 1 0 ⎦ ⎣g2 ⎦ . 0 −cot(θ)cosec(θ) 0 cosec2 (θ) g3 (g) For this set of angles, show that the angular velocity vector has the representation ωR = φ̇t3 + θ̇ t2 + ψ̇E3 . (6.44) You will need to use two distinct corotational derivatives to find this representation. The first of these fixes ti whereas the second fixes ti . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 200 Exercises 6.2–6.3 (h) For this set of angles, show that the components of ωR relative to the basis {E1 , E2 , E3 } have the representations 1 = φ̇ sin (θ) cos (ψ) − θ̇ sin (ψ) , 2 = φ̇ sin (θ) sin (ψ) + θ̇ cos (ψ) , 3 = φ̇ cos (θ) + ψ̇, where i = ωR · Ei . These representations can be found in Section 257 of Routh [185]. (i) Suppose that ωk(t) = ωR · tk are known functions. Show that ⎡ ⎤ ⎡ ⎤⎡ ⎤ ψ̇ − cos(φ) cosec(θ) sin(φ) cosec(θ) 0 ω1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ sin(φ) cos(φ) 0⎦ ⎣ω2 ⎦ . ⎣ θ̇ ⎦ = ⎣ cos(φ) cot(θ) φ̇ − sin(φ) cot(θ) 1 (6.45) ω3 (j) This problem involves the numerical integration of (6.45). 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)? 6.3. Recall that a rotation tensor L representing a counterclockwise rotation about an axis p though an angle ν has the representation L = L(ν, p) = cos(ν)(I − p ⊗ p) − sin(ν)p + p ⊗ p, and its associated angular velocity vector has the representation ωL = ν̇p + sin(ν)ṗ + (1 − cos(ν))p × ṗ. Consider two rotation tensors: Q1 = L(θ, E3 ), Q2 = L(φ, e1 )L(θ, E3 ), where e1 = cos(θ)E1 + sin(θ)E2 , e2 = cos(θ)E2 − sin(θ)E1 , e3 = E3 . (a) Show that Q1 has the representation Q1 = e1 ⊗ E1 + e2 ⊗ E2 + e3 ⊗ E3 . (b) Give an example of a system of two rigid bodies for which the rotation tensor of one body is Q1 and the rotation tensor of the second body is Q2 . (c) Given that the relative rotation tensor R2 = Q2 QT1 , show that ωR2 = φ̇e1 + sin(φ)θ̇ e2 + (1 − cos(φ))θ̇ e3 . (d) Explain why ωR2 = ω̂R2 = ωQ2 − ωQ1 . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercise 6.4 201 6.4. The parameterization of the rotation tensor by use of Euler parameters (unit quaternions or symmetric Euler–Rodriques 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 = (f 02 − f · f)I + 2f ⊗ f − 2f 0 (εf). Here, {e0 , e} and {f 0 , f} represent two sets of Euler parameters: φ θ φ θ , e = sin a, f 0 = cos , f = sin b, e0 = cos 2 2 2 2 (6.46) (6.47) where a is the axis of rotation of A and 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)a + a ⊗ a. Verify that A has the representation (6.46)1 . (b) Letting e = 3i=1 ei Ei , what are Aik = (AEk) · Ei ? (c) Show that C = BA = (g20 − g · g)I + 2g ⊗ g − 2g0 (εg), (6.48) where g0 = e0 f 0 − e · f, g = e0 f + f 0 e + f × e. (6.49) This result was first established by Rodrigues [181] in 1840. The earliest English commentary on it is by Cayley [32] in 1845. (d) In terms of e0 , f 0 , f, and e, what are the Euler parameters of the rotation tensors AB and BT AT ? (e) Using the results of (d), show that the compositions of rotations is not, in general, commutative: i.e., AB = BA. (f) Recall that the angular velocity vector associated with A has the representation ωA = 2 (e0 ė − ė0 e + e × ė) , , where ωA = − 12 ȦAT . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 202 Exercises 6.4–6.5 (i) What are ωB and ωC ? (ii) Give an explanation for the following result: ωC = ωB + ωA . (g) If e0 = f 0 = √12 , a = E3 , and b = E2 , then what does the rotation tensor C represent? Illustrate your solution by showing how C transforms the basis {E1 , E2 , E3 }. 6.5. The latitude (λ) and longitude (θ) of a point on the Earth’s surface are illustrated schematically in Figure 6.9. In navigation systems, one uses these angles to define the downward direction ez, the northerly direction ex , and the easterly direction ey : ⎡ ⎤ ⎡ − cos(θ) sin(λ) ex ⎢ ⎥ ⎢ − sin(θ) ⎣ey ⎦ = ⎣ − sin(θ) sin(λ) cos(λ) cos(θ) 0 ⎤⎡ E1 ⎥⎢ ⎥ ⎦ ⎣E2 ⎦ . − cos(θ) cos(λ) − sin(θ) cos(λ) − sin(λ) ez ⎤ (6.50) E3 Here, the triad {E1 , E2 , E3 } is a set of fixed right-handed Cartesian basis vectors. North ex ex ey E3 −ez E2 λ East ey ez θ E1 Down Figure 6.9. The angles of longitude θ and latitude λ. (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 E3 through an angle θ and R2 corresponds to a rotation about ey through an angle − π2 − λ. (b) Given a vector x, x = xx ex + xy ey + xzez = X1 E1 + X2 E2 + X3 E3 . Show that xx = 3 i=1 Ri1 Xi , xy = 3 i=1 Ri2 Xi , xz = 3 i=1 Ri3 Xi , 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercise 6.12 203 where R= 3 3 RikEi ⊗ Ek. i=1 k=1 How are Rik’s related to the matrix in Equation (6.50)? 6.6. Consider two rotation tensors A and B, where A= 3 ti ⊗ Ei , B= i=1 3 ei ⊗ ti , i=1 and {E1 , E2 , E3 } is a fixed, right-handed orthonormal basis for E3 . (a) Show that o Ḃ =B +A B − BA , o where B is the corotational derivative of B assuming that ti are fixed, and A = ȦAT . (b) Consider the rotation tensor C = BA. Using the results of (a), show that o ˆ B =B BT , ˆ B and the angular velocity tenwhere the relative angular velocity tensor sor C are, respectively, ˆ B = C − A , C = ĊCT . ˆ B skew-symmetric, and what does this imply for the product ˆ B b, Why is where b is any vector? (c) Consider the following examples of two tensors: A = cos(ψ)(I − E3 ⊗ E3 ) − sin(ψ)E3 + E3 ⊗ E3 , B = cos(θ)(I − t1 ⊗ t1 ) − sin(θ)t1 + t1 ⊗ t1 , where t1 = cos(ψ)E1 + sin(ψ)E2 . o (i) What are Ȧ and B? (ii) Using Equation (6.14), what are ωA and ω̂B ? (iii) With the help of (6.14), verify that ωB = θ̇ t1 − ψ̇ (cos(θ)E3 − sin(θ)t2 ) + ψ̇E3 . Here, ωB is the angular velocity vector associated with B. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 204 Exercises 6.7–6.11 6.7. Show that the rotation tensor L( π2 , E3 )L( π2 , E1 ) is equivalent to a rotation tensor L( 2π , p), where 3 1 p = √ (E1 + E2 + E3 ) . 3 This result is very useful in analyzing material symmetry groups of crystals. 6.8. Show that∗ L π π π , E2 L , E3 L − , E1 = L , E3 . 2 2 2 2 π This result has a 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 for one explanation of a phenomenon in biomechanics that is known as Codman’s paradox [40]. † 6.9. Examine Sections 1 and 24–30 of Euler [54], in which he introduces the Euler angles (p, q, r). 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 expression for the components R = e1 · ω, P = e2 · ω, and Q = e3 · ω on page 205 of [54] to help with this. Related results, but with a different notation, can be found in [57]. In both of these papers you will find, among other matters, Euler’s discussion of the components of Euler and inertia tensors. 6.10. If p1 , p2 and p3 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. (6.51) This result is discussed in Section 3 of Whittaker [228], and he credits it to Rodrigues [181] and Hamilton [90]. 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. 6.11. Suppose a set of 3–2–1 Euler angles is used to parameterize a rotation tensor R. If we denote the values of these angles by α1 , α2 , and α3 , respectively, then what are the corresponding values of the 3–2–1 Euler angles for the inverse RT of this rotation? ∗ † For the rotation tensors discussed in this problem, you will need to compute Ek . If you need assistance with this, then please see (A.11) in the Appendix. As quoted in Politti et al. [174], 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercise 6.12 205 6.12. In Section 1.7 of Kane et al. [106] two distinct sets of Euler angles are defined. The first corresponds to “body-angles” and the second corresponds to “space-angles.” Thus they define 24 sets of Euler angles. Verify that the 3–2–1 Euler angles discussed in Subsection 6.8.1 are equivalent to the “body-three: 3–2–1” angles in [106]. 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, then, in your solution to this problem, you will find the representation ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ R11 R12 R13 cos (β2 ) 0 sin (β2 ) 1 0 0 ⎢ ⎥ ⎢ ⎢ ⎥ ⎢R21 R22 R23 ⎥ = ⎣0 cos (β ) − sin (β )⎥ 0 1 0 ⎦ 3 3 ⎦× ⎣ ⎣ ⎦ 0 sin (β3 ) cos (β3 ) R31 R32 R33 − sin (β2 ) 0 cos (β2 ) ⎡ cos (β1 ) − sin (β1 ) ⎢ × ⎣ sin (β1 ) 0 cos (β1 ) 0 ⎤ 0 ⎥ 0⎦ . 1 This result should be compared with corresponding representation (6.25) for the “body-three: 3–2–1” angles. For assistance with this problem, the discussion of “body-fixed” and “space-fixed” rotations in Section 3.2 of Ginsberg [71] and a related discussion in Section 7.14 of Pars [170] might be helpful. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 7 978 0 521 87483 0 June 12, 2008 Kinematics of Rigid Bodies 7.1 Introduction This chapter contains results on the three-dimensional kinematics of rigid bodies. We discuss several useful classical 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 configuration manifold for a rigid body. 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 [15], Casey [26, 28], Fox [65], Greenwood [80], and Gurtin [84]. In this book, we follow their work as it leads to transparent developments particularly with regards 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, oneto-one, onto function that has a continuous inverse. It maps material points X of B to points in three-dimensional Euclidean space: x = κt (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 206 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.2 The Motion of a Rigid Body 207 κ0 X Y X̄ π x X X̄ X̄ Y x̄ X Y κt y O Figure 7.1. The reference κ0 and present κt configurations of a body B. reference 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 on the instant of time and the material point of interest. Euler’s Theorem For rigid bodies, the nature of 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 [55, 56] showed that the motion of a body that satisfies (7.1) is such that x1 − x2 = 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 = I, we can pick two points X1 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: the transformation that takes vectors between two points in the body and transforms them into their present state. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 208 Kinematics of Rigid Bodies 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) We can then infer Euler’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 κ0 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 feature the reference configuration κ0 . To do this, we again consider the motion of the body with a fixed point during the interval t ∈ [t0 , t]. We find from (7.3) that x (t) = Q (t) QT (t0 ) x (t0 ) . Thus the motion of the body at the end of the time interval is characterized by the rotation tensor Q (t) QT (t0 ).∗ 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 XP 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.† It is one of the general representations of the rigid body motion that we will often use in our subsequent discussions. ∗ † In general, the tensor Q (t) QT (t0 ) has an axis of rotation and an angle of rotation that differ from those associated with Q (t). It should not come as a surprise that this result was also known to Euler. See his remarks in the introductory sections to [52, 54]. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.2 The Motion of a Rigid Body 209 Q (t) QT (t0 ) κt0 X X x (t0 ) Q(t0 ) X x (t) κt O X Q(t) κ0 Figure 7.2. 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. To discuss an alternative to (7.4) that does not feature 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) . 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 ).∗ 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 [228] 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 θ 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 (t0 ), and φ(t) is this tensor’s angle of rotation. The translational component σs(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 equations: (I − Q(t)) ρ(t) + σ(t)s(t) = z(t). ∗ (7.6) You may notice that we can choose the reference configuration κ0 to be identical to κt0 . Then, Q (t0 ) = I, and representations (7.4) and (7.5) will be identical. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 210 Kinematics of Rigid Bodies φ(t) s(t) X κt x (t) ρ2 Q (t) QT (t0 ) O ρ1 x (t0 ) X κt0 Figure 7.3. 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 σs and rotated about s(t) through an angle φ. Two possible choices of ρ(t) are also shown. For the first, ρ(t) = ρ1 = ρ1 E1 + ρ2 E2 is chosen to be the intercept of the screw axis with the E1 − E2 plane, and, for the second, ρ(t) = ρ2 , where ρ2 is the vector from the origin that intersects the screw axis at a right angle: ρ2 · s = 0. However, (I − Q(t)) is noninvertible,∗ 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): ρ(t) = 1 φ(t) z⊥ (t) + cot s(t) × z(t) , 2 2 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 = I and ρ is indeterminate. ∗ There are many ways to see this result. The first is to refer the reader to the derivation of (6.15). Alternatively, one can use Euler representation (6.8) for a rotation tensor and observe that s is the zero eigenvalue of (I − Q(t)). 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.3 The Angular Velocity and Angular Acceleration Vectors 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 and an angular velocity vector ω∗ : 1 , ω = − Q̇QT . 2 = Q̇QT , The vector ω is the angular velocity vector of the rigid body, and 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 = ω̇. You should notice that 1 , 1 ˙ α = − Q̈QT + Q̇Q̇T = − [], 2 2 where we used the fact that ˙ = O. 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 = ẋ1 − ẋ2 = Q̇ (X1 − X2 ) = Q̇QT Q (X1 − X2 ) = Q (X1 − X2 ) = ω × (x1 − x2 ) . A further differentiation and some manipulations give the relative acceleration vectors: a1 − a2 = v̇1 − v̇2 = ω̇ × (x1 − x2 ) + ω × (ẋ1 − ẋ2 ) = 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 . ∗ As Q̇T (t0 ) = 0, ω is also the angular velocity vector associated with the rotation tensor Q(t)QT (t0 ). Thus representations (7.4) and (7.5) have the same angular velocity vectors. 211 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 212 Kinematics of Rigid Bodies E3 e2 X2 X3 E1 E2 e3 X1 X2 X3 X1 κ0 e1 O κt Figure 7.4. The corotational basis {e1 , e2 , e3 } and the fixed Cartesian basis {E1 , E2 , E3 }. 7.4 A Corotational Basis It is convenient, when discussing the dynamics of rigid bodies, to introduce another basis {e1 , e2 , e3 }, which is known as a corotational basis.∗ Here, we define such a basis and point out some features of its use. Our discussion of the corotational basis follows Casey [26] 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, this is the premise for many navigation 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 E2 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 transparent that Q = e1 ⊗ E1 + e2 ⊗ E2 + e3 ⊗ E3 . ∗ This basis is often referred to as a body-fixed frame or an embedded frame. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.5 Three Distinct Axes of Rotation 213 This result follows from (7.2) and our previous discussions on representations of rotation tensors. Because the corotational basis moves with the body, we can use our previous results for relative velocities and accelerations to see that ėi = ω × ei , ëi = α × ei + ω × (ω × ei ) , where i = 1, 2, 3. As the corotational basis is a basis for E3 , for any vector r we have the representation r= 3 ri ei . 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 ṙ of r has the representations v = ṙ = 3 ṙi ei + i=1 3 ri ėi i=1 o = r + ω × r, o where r is the corotational derivative (with respect to Q) of the vector r. A related expression can be obtained for r̈: a = v̇ = 3 i=1 = 3 r̈i ei + 2 3 ṙi ėi + i=1 3 ri ëi i=1 o r̈i ei + 2ω × r + ω × (ω × r) + α × r. i=1 o 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 o observe that, if r is a corotational vector, then r = 0, and the Coriolis acceleration vanishes. 7.5 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 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 214 Kinematics of Rigid Bodies ω q r 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. r axis of rotation i, can also be defined. This axis is a unit vector parallel to the angular velocity vector ω∗ : 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 ṙ = ω × r for any corotational vector. 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 classical works on rigid body dynamics, for example, Poinsot [172] and Sections 405– 406 of Poisson [173]. It is not universally adapted. For instance, in the literature on kinematics of anatomical joints the terminology instantaneous axis of rotation often refers to s and not i.† In general, the axes q, s, and i are not identical. However, (6.14) can be used to relate the axes and two angles of rotation: ω = ||ω|| i = φ̇s + sin(φ)ṡ + (1 − cos(φ))s × ṡ = θ̇q + sin(θ)q̇ + (1 − cos(θ))q × q̇. (7.8) In addition, Rodrigues formula (6.49) 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 ∗ † You may wish to recall that the angular velocity vector associated with Q (t) and Q (t) QT (t0 ) are identical. The interested reader is referred to Woltring [232] and Woltring et al. [233] for further discussion on this matter. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.6 The Center of Mass and Linear Momentum 215 i = s = E3 2π ν(t) − ν0 e1 E2 E1 e2 π 0 e3 θ(t) q 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 (7.9). A plot of ν(t) is also shown in this figure . where T = 2π ν̇ 0 about E3 through an angle ν, where ν̇ = ν̇0 is constant: Q(t) = L (ν̇0 t, E3 ). In this case, it is easy to compute that s(t) = i(t) = q(t) = E3 and ω = ν̇0 E3 . Now consider an example in which Q describes a rotation about a time-varying axis of rotation at constant speed∗ : Q(t) = 2q(t) ⊗ q(t) − I, where the axis of rotation of Q(t) is ν ν q(t) = cos E1 + sin E2 , 2 2 (7.9) ν(t) = ν̇0 (t − t0 ) + ν0 , and ν̇0 and ν0 are constants. You should notice that, with the possible help of (6.14), ω = ν̇0 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 = q. Indeed, (7.9) 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. 7.6 The Center of Mass and Linear Momentum It is convenient (and traditional) to define a special material point of the rigid body, which we refer to as the center of mass X̄. We then use the velocity vector v̄ of this point in the present configuration to define a very useful expression for the linear momentum G of the rigid body. ∗ This example is adapted from [161]. Other examples of rotations with constant angular velocity vectors but distinct axes s and q can be found in [161]. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 216 Kinematics of Rigid Bodies The Center of Mass The position vectors of the center of mass of the body in its reference and present configurations are defined by R X̄ = 0 Xρ0 dV ρ0 dV R0 , xρdv x̄ = R , R ρdv (7.10) where ρ0 = ρ0 (X) and ρ = ρ(x, t) are the mass densities per unit volume of the body in κ0 and κt , respectively. The regions R0 and R denote the regions of E3 occupied by the body in κ0 and κt , respectively. If a body is homogeneous, then ρ0 is a constant that is independent of X. The principle of mass conservation states that the mass of the body is conserved. That is, dm = ρ0 dV = ρdv, or, equivalently, m= R0 ρ0 dV = ρdv. R It follows immediately from (7.10) that mX̄ = R0 Xρ0 dV, We can also find from these results that (X − X̄)ρ0 dV, 0= R0 mx̄ = R xρdv. 0= R (x − x̄)ρdv. These identities play key roles in establishing expressions for momenta and energies of a rigid body. A special feature of rigid bodies is that the center of mass behaves as if it were a material point, which we denote by X̄. To see this we follow [26] and consider mx̄ = xρdv R = R (QX + d) ρdv = R0 (QX + d) ρ0 dV =Q R0 Xρ0 dV + = mQX̄ + md. R0 ρ0 dV d 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.6 The Center of Mass and Linear Momentum 217 Consequently, x̄ = QX̄ + d. (7.11) 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, such as a rigid circular ring, it does not correspond to a material point. It is convenient in many treatments of rigid bodies to define a reference frame consisting of the center of mass X̄ 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 − x̄ in this frame. Linear Momentum By definition, the linear momentum G of a rigid body is G= vρdv. R 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 x̄: G = mv̄, where v̄ = x̄˙ is the velocity vector of the center of mass. You may recall that a related result holds for a (finite) system of particles. Relative Position Vectors For a material point X of a rigid body, it is convenient to define the relative position vectors π and : π = x − x̄, = X − X̄. Representative examples of these vectors are displayed in Figure 7.1. With the assistance of (7.2) and (7.11), we see that π=Q . (7.12) Using the corotational basis, we also easily see that π · ei = · Ei . This implies that the relative position vectors have the representations = 3 i=1 i Ei , π= 3 i=1 i ei . (7.13) 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 218 Kinematics of Rigid Bodies Furthermore, the corotational derivative (with respect to Q) of π is zero: π̇ = ω × π. That is, π is a corotational vector. 7.7 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 X̄ 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). By definition, the angular momenta of a rigid body relative to its center of mass X̄, H, a fixed point O, HO, and a point A, HA, are H = (x − x̄) × vρdv, HO = HA = R R R x × vρdv, (x − xA) × vρdv. The position vectors in these expressions are relative to the fixed point O, and xA is the position vector of the point A. You should notice that the velocity vector in these expressions is the absolute velocity vector. The aforementioned angular momenta are related by simple and important formulae. To find one of these formulae, we perform some manipulations on HO: HO = x × vρdv = = R R R (π + x̄) × vρdv π × vρdv + = H + x̄ × R R x̄ × vρdv vρdv. That is, HO = H + x̄ × G. (7.14) 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 (7.14), it can also be shown that HA = H + (x̄ − xA) × G, HO = HA + xA × G. These results have obvious similarities to (7.14). (7.15) 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.8 Euler Tensors and Inertia Tensors 219 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 considerably aided 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): ⊗ ρ0 dV, E= π ⊗ πρdv. E0 = R0 R You should notice that E and E0 are symmetric. Using mass conservation, (7.12), and the identity (Aa) ⊗ (Bb) = A(a ⊗ b)BT , we easily see that E = QE0 QT . Furthermore, (Eei ) · ek = (E0 Ei ) · Ek. This implies that E has the representation E= 3 3 Eikei ⊗ 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 constants 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) = a · b, we can restate these definitions as J0 = J= (7.16) (( · )I − ⊗ ) ρ0 dV, ((π · π)I − π ⊗ π) ρdv. R0 R It is easy to see that JT0 = J0 , J = QJ0 QT , The first of these results implies that J= 3 3 i=1 k=1 J ikei ⊗ ek, JT = J. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 220 Kinematics of Rigid Bodies where J ki = (J0 Ei ) · Ek are the constant components of J0 .∗ The symmetry of the inertia tensors also implies that J ki = J ik. Additional Relationships Another set of interesting results follows by inverting the relationships between the Euler and inertia tensors. To do this, 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 (7.17) These results are useful for obtaining the Euler tensors from tabulations of the inertia tensor that are found in numerous undergraduate textbooks on dynamics. It is a good exercise to substitute (7.13) into (7.16) to see that the components of J0 relative to the basis {E1 , E2 , E3 } are volume integrals involving quadratic powers of the components of . For example, J 011 = (J0 E1 ) · E1 = (y2 + z2 )ρ0 dV, R0 J 012 = (J0 E2 ) · E1 = − J 013 = (J0 E3 ) · E1 = − R0 R0 xyρ0 dV, xzρ0 dV, (7.18) where x = π · e1 = · E1 , y = π · e2 = · E2 , z = π · e3 = · E3 . A similar exercise with the components of E0 shows that E011 = (E0 E1 ) · E1 = x2 ρ0 dV, E012 = (E0 E2 ) · E1 = E013 = (E0 E3 ) · E1 = R0 R0 R0 xyρ0 dV, xzρ0 dV. You should notice 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 + λ2 E2 ⊗ E2 + λ3 E3 ⊗ E3 . ∗ Not surprisingly, this is similar to the situation we encountered earlier with the Euler tensors E and E0 . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.8 Euler Tensors and Inertia Tensors 221 Here, λi are known as the principal moments of inertia. As J = QJ0 QT and ei = QEi , we also have 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 J 0ik of the inertia tensor, the principal moments need to satisfy certain inequalities: λ1 + λ2 > λ3 , λ2 + λ3 > λ1 , λ3 + λ1 > λ2 . These inequalities are easy to establish by use of definitions (7.18) and are useful in selecting representative examples of inertias. For instance, (7.19) imply that a rigid body with λ1 = 1, λ2 = 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 + e3 E3 ⊗ E3 . By use of identity (7.17)1 , the constants ei can be related to the principal moments of inertia: e1 = 1 (λ2 + λ3 − λ1 ) , 2 e3 = e2 = 1 (λ1 + λ3 − λ2 ) , 2 1 (λ1 + λ2 − λ3 ) . 2 (7.19) As E = QE0 QT and ei = QEi , we also have E = e1 e1 ⊗ e1 + e2 e2 ⊗ e2 + e3 e3 ⊗ e3 . Notice that the principal axis e j 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. For instance, 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 2mkL2kEk ⊗ Ek. k=1 Consequently, the inertia tensor is J0 = 2 m2 L22 + m3 L23 E1 ⊗ E1 + 2 m1 L21 + m3 L23 E2 ⊗ E2 + 2 m1 L21 + m2 L22 E3 ⊗ E3 . It is left as an exercise to write E and J. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 222 Kinematics of Rigid Bodies m3 m1 L1 E3 L2 m2 X̄ m2 L3 O E2 m1 m3 E1 Figure 7.7. The reference configuration of a dumbell satellite consisting of six masses joined to a center by three massless, rigid rods of lengths 2L1 , 2L2 , 2L3 . The simplest inertia tensor arises when the body is a homogeneous sphere of radius R or a homogeneous cube of length a: J = J0 = 2mR2 I, 5 J = J0 = ma2 I, 6 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 L and radius R has a moment of inertia tensor mR2 mR2 mL2 J0 = E3 ⊗ E3 + + (I − E3 ⊗ E3 ) , 2 4 12 where E3 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 y2 z2 + + = 1. a2 b2 c2 Ellipsoid E3 Figure 7.8. An ellipsoid of mass m. X̄ E1 E2 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.9 Angular Momentum and an Inertia Tensor 223 The inertia tensor of the ellipsoid is J0 = m 2 m 2 m 2 b + c2 E1 ⊗ E1 + a + c2 E2 ⊗ E2 + a + b2 E3 ⊗ E3 . 5 5 5 It is left as an exercise to write the Euler tensor E0 for the ellipsoid.∗ It is crucial to note that for all of the preceding examples we have used 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 a tractable expression for H. We now reconsider the angular momentum H, H = (x − x̄) × vρdv = = = R R R R π × vρdv π × (v̄ + ω × π)ρdv π × v̄ρdv + R π × (ω × π)ρdv. However, because X̄ is the center of mass and the velocity vector v̄ of X̄ is independent of the region of integration, we can take v̄ outside the integral: H= R π × v̄ρdv + = R πρdv × v̄ + = R R ∗ π × (ω × π)ρdv π × (ω × π)ρdv R π × (ω × π)ρdv π × (ω × π)ρdv. Notice that we also used the identity calculation. Summarizing, we have H= R = 0 × v̄ + R πρdv = 0 in the next-to-last step of this R π × (ω × π)ρdv = For assistance, see (7.19). R ((π · π)ω − (π · ω)π) ρdv. (7.20) 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 224 Kinematics of Rigid Bodies In writing this equation, we used the identity a × (b × c) = (a · c)b − (a · b)c. By use 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. 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 mv̄ · v̄ + (Jω) · ω. 2 2 (7.21) Here, a derivation of this result is given. The kinetic energy T of a rigid body is defined to be 1 v · vρdv. T= 2 R We can simplify this expression for the energy by expressing the velocity vector v as v = v̄ + ω × π. Substituting this expression into T, we have 1 v · vρdv T= 2 R 1 1 = v̄ · v̄ρdv + (ω × π) · v̄ρdv + (ω × π) · (ω × π)ρdv. 2 R 2 R R However, we have the following identities v̄ · v̄ρdv = v̄ · v̄ ρdv = mv̄ · v̄, R R R R (ω × π) · v̄ρdv = ω × πρdv · v̄ = (ω × 0) · v̄ = 0, R (ω × π) · (ω × π)ρdv = R = ω· (ω · ω)(π · π) − (ω · π)2 ρdv R = ω · (Jω) . (π · π)I − π ⊗ πρdv ω 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 7.10 Kinetic Energy 225 Substituting these results into the previous expression for T, we find the desired result: T= 1 1 mv̄ · v̄ + (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 Trot = 12 ω · (Jω) of the rigid body. Comments on the Rotational Kinetic Energy The previous development of the Koenig decomposition showed that 2Trot = 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,∗ and we now discuss this representation. Consider the angular velocity vector ω0 : ω0 = QT ω. If ω = 3 i=1 ωi ei , then it follows that ω0 = 3 ωi Ei . i=1 In addition, it can also be shown that ω0 is the axial vector of Q̇T Q: 1 1 ω0 = − [QT Q̇] = − [QT Q]. 2 2 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 2Trot = ω · (Jω) = Qω0 · (JQω0 ) = ω0 · QT JQω0 = ω0 · (J0 ω0 ) . This is the final desired result: Trot = 1 1 ω · Jω = ω0 · J0 ω0 . 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 need to consider only the derivatives of ω0 . ∗ See, for example, [28] and Chapter 15 of [138]. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 226 Exercise 7.1 7.11 Concluding Remarks 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 i Ei , 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. EXERCISES 7.1. Recall that the rotation tensor Q of a rigid body has the representations Q= 3 ei ⊗ Ei = i=1 3 3 Qikei ⊗ ek = i=1 k=1 3 3 QikEi ⊗ Ek. i=1 k=1 We recall that the corotational rates of a vector a and a tensor A relative to the rotation tensor Q are defined as o o A = Ȧ − A + A. a = ȧ − ω × a, In these expressions, = Q̇QT and ω = − 12 [Q̇QT ]. (a) If ai = a · ei and Aik = (Aek) · ei , then show that o a= 3 o A= ȧi ei , i=1 3 3 Ȧikei ⊗ 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: o ω̇ = ω, o Q̇ = QT QQ. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 Exercises 7.1–7.2 227 (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?∗ 7.2. As shown in Figure 7.9, a robotic arm can be used to move a payload. The robot consists of 1. a drive shaft that rotates about the E3 axis through an angle ψ, 2. an axle A that rotates about g2 through an angle θ relative to the drive shaft, 3. 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. g3 Robotic arm Axle φ Payload Figure 7.9. Schematic of a robot consisting θ P g2 of a robotic arm, drive shaft, and axle. The motors used to actuate the robot are not shown. ψ E3 O Drive shaft E2 E1 (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 + Lg3 , then establish an expression for the velocity vector v of the point X. (d) Suppose after a time interval t1 − t0 the point X of has returned to its original location in space: r(t1 ) = r(t0 ). Show that the payload will have rotated through an angle φ(t1 ) − φ(t0 ) about the axis g3 (t0 ) during this interval of time. In other words, the ∗ A solution to this problem can be found in O’Reilly and Payen [161]. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 228 Exercises 7.2–7.3 rotation tensor Q(t1 )QT (t0 ) corresponds to a rotation of φ(t1 ) − φ(t0 ) about g3 (t0 ). 7.3. Consider the circular disk shown in Figure 7.10. 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. E3 g e2 O E1 e2 E2 ψ e1 X̄ e1 φ e1 P Figure 7.10. The present configuration of a circular disk moving with one point in contact with a fixed horizontal plane. (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 ẋ − ẏ = ω × (x − y) , ẍ − ÿ = ω̇ × (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 7.10, 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, t0 t1 ω3 (t)dt = φ(t1 ) − φ(t0 ). (7.22) 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 Exercises 7.3–7.5 229 Give a physical interpretation of the case in which the integral of ω3 (t) does yield the angle φ.∗ 7.4. Consider a rigid body of mass m whose inertia tensors are defined as J0 = ( · )I − ⊗ ρodV. J = QJ0 QT , R0 (a) Explain why J0 and J are symmetric. (b) Show that Ei · (J0 Ek) = ei · (Jek), where Q = 3 i=1 ei ⊗ Ei . (c) Show that J has the representation J= 3 λi ei ⊗ ei , (7.23) i=1 where λi are the principal values of J0 and Ei = QT ei are the principal directions of J0 . (d) Establish the following identities: J̇ = J − J, Ḣ = Jω̇ + ω × (Jω), Jω ˙· ω = 2Ḣ · ω. How do these results simplify if J0 = µI, where µ is a constant? Note that, in the first of these results, you are showing that the corotational rate of J o relative to Q is zero: J = 0. (e) Suppose one used the representation J= 3 3 J pn E p ⊗ En . p=1 n=1 Starting from (7.23), show that J pn = 3 Q pi λi Qni , i=1 where Q pi = (QEi ) · E p. Why are the components J pn of J not constant? 7.5. Recall that a tensor that is intimately related to the familiar inertia tensor is the Euler tensor: E = QE0 QT , E0 = ⊗ ρodV. R0 (a) Show that J0 = tr(E0 )I − E0 , J = tr(E)I − E. (b) Verify that Ei · (E0 Ek) = ei · (Eek), where Q = 3i=1 ei ⊗ Ei . (c) Establish the following results: o E = 0, Ė = E − E, o where E denotes the corotational derivative of E with respect to Q. ∗ An application of (7.22) to the navigation of motorcycles can be found in Coaplen et al. [39]. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 230 Exercises 7.5–7.7 (d) What are the Euler tensors for a sphere of mass m and radius R, and a cylinder of mass m, radius R, and length L? You might find it convenient to use the relationships of the form E0 = 12 tr (J0 ) I − J0 that were discussed earlier in this chapter in Section 7.8. 7.6. Recall that the angular momentum H and rotational kinetic energy Trot of a rigid body have the representations H = Jω, Trot = 1 ω · Jω. 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 , then J0 = 3 λi Ei ⊗ Ei , J = QJ0 QT = 3 i=1 where Q = 3 i=1 λi ei ⊗ ei , i=1 ei ⊗ Ei . (a) Using a set of 3–2–3 Euler angles to parameterize Q,∗ show that ω = (θ̇ sin(φ) − ψ̇sin(θ) cos(φ))e1 + (θ̇ cos(φ) + ψ̇sin(φ) sin(θ))e2 + (ψ̇ cos(θ) + φ̇)e3 . (b) Using a set of 3–2–3 Euler angles, establish an expression for H and Trot as functions of the Euler angles and their time derivatives. 7.7. Recall the definitions of the angular momenta of a rigid body relative to its center of mass and a point A: H= π × vρdv, HA = πA × vρdv. (7.24) R R Here, π = x − x̄ and πA = x − xA, where xA is the position vector of the point A and x̄ is the position vector of the center of mass. (a) Starting from (7.24)2 , and with the assistance of (7.24)1 , show that HA = H + (x̄ − xA) × G, where G is the linear momentum of the rigid body. (b) Suppose that A is a point of the rigid body. Show that πA = Q A, where A = X − XA and XA is the position vector of A in the reference configuration. If vA = 0, show in addition that v = ω × (x − xA). ∗ This set of Euler angles is discussed in Exercise 6.2. In particular, the student is referred to (6.44) in Exercise 6.2(g). 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 Exercises 7.7–7.8 231 (c) Again supposing that A is a point of the rigid body and that vA = 0, show that HA = JAω, where JA is the inertia tensor of the rigid body relative to A: JA = (πA · πA)I − πA ⊗ πAρdv. R Verify that T JA = QJA 0Q , where JA 0 = (( R0 A · A)I − A ⊗ A) ρ0 dV. (d) Using the previous results, prove the parallel axis theorem∗ : JA 0 = J0 + m((X̄ − XA) · (X̄ − XA)I − (X̄ − XA) ⊗ (X̄ − XA)). (7.25) 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 L, and radius R, where X̄ = 0, J0 = XA = −xE1 − zE3 , mL2 mR2 (I − E3 ⊗ E3 ) + (I + E3 ⊗ E3 ) . 12 4 Then, what is JA 0? 7.8. Consider the rectangular parallelepiped of mass m and dimensions a, b, and c shown in Figure 7.11. Relative to the principal axes {A1 , A2 , A3 }, the inertia tensor of this rigid body has the representation JO = 3 λi Ai ⊗ Ai , i=1 where m 2 m 2 m 2 (b + c2 ), (a + c2 ), (a + b2 ). λ2 = λ3 = 12 12 12 The right-handed set of basis vectors {E1 , E2 , E3 } is oriented so that E3 passes diagonally through the parallelepiped. That is, E3 is parallel to the line connecting X̄ and the point P: λ1 = E3 = √ ∗ 1 a2 + b2 + c2 (aA1 − bA2 + cA3 ) . This representation of the parallel axis theorem is due to Fox [65]. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 232 Exercise 7.8 In addition, E1 = cos(α)A1 − sin(α)A3 , where α = tan −1 a . c A3 E3 P E2 X̄ c A2 A1 a b E1 Figure 7.11. Schematic of a parallelepiped. (a) After calculating 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 (7.26) E2 = R A2 , E3 where R11 R = R21 R31 1 = 0 0 and R12 R13 A3 R22 R23 R32 R33 0 0 cos(α) 0 sin(β) cos(β) − sin(β) cos(β) √ a2 + c2 , cos(β) = √ a2 + b2 + c2 0 1 sin(α) 0 3 r=1 cos(α) b sin(β) = √ . 2 a + b2 + c2 (b) Show that the components J 0ik = (J0 Ek) · Ei are J 0ik = − sin(α) 0 , Rir λr Rkr . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 Exercises 7.8–7.9 233 Using a matrix notation, these equations are equivalent to J 011 J 012 J 013 J 012 J 013 λ1 0 0 J 022 J 023 = R 0 λ2 0 RT . J 023 J 033 0 λ3 0 (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? 7.9. In this question, 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 Subsection 6.8.1. We then apply these results 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 e1 = cos(ψ)E1 + sin(ψ)E2 : L(θ, e1 ). The third rotation is through an angle φ about e2 = e2 = cos(θ)e2 + sin(θ)e3 . (a) Show that Q11 e1 Q e2 = 12 Q13 e3 Q21 Q22 Q23 Q31 E1 Q32 E2 , Q33 E3 where Qij = (QE j ) · Ei . (b) Show that the Euler basis for the 3–1–2 set of Euler angles has the representation 0 0 1 g1 E1 cos(ψ) sin(ψ) 0 g2 = E2 . − cos(θ) sin(ψ) cos(θ) cos(ψ) sin(θ) g3 E3 Using the relations gk · gi = δki, show that the dual Euler basis for the 3–1–2 set of Euler angles has the representation 1 sin(ψ) tan(θ) − cos(ψ) tan(θ) 1 g E1 2 cos(ψ) sin(ψ) 0 g = E2 . 3 − sec(θ) sin(ψ) sec(θ) cos(ψ) 0 g E3 You should notice that the second Euler angle needs to be restricted to θ ∈ − π2 , π2 . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 234 Exercise 7.9 (c) By examining the three rotations that compose Q and with the help of the results in (a), show that Q11 Q12 Q13 Q21 Q22 Q23 = CBA, (7.27) Q31 Q32 Q33 where A= cos(φ) 0 sin(φ) 0 1 0 − sin(φ) 0 , cos(φ) 1 0 0 0 sin(θ) cos(θ) B = 0 cos(θ) − sin(θ) , cos(ψ) C = sin(ψ) − sin(ψ) 0 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 = ψ̇E3 + θ̇e1 + φ̇e2 = ω1 e1 + ω2 e2 + ω3 e3 . Show that − sin(φ) cos(θ) 0 ω1 sin(θ) 1 ω2 = ω3 and cos(φ) cos(θ) 0 cos(φ) ψ̇ 0 φ̇ , sin(φ) θ̇ ψ̇ − sec(θ) sin(φ) 0 sec(θ) cos(φ) ω1 φ̇ = tan(θ) sin(φ) 1 − tan(θ) cos(φ) ω2 . θ̇ cos(φ) 0 sin(φ) (7.28) ω3 These sets of differential equations are very important in strap-down navigation systems. (f) Suppose that a set of 3–1–2 Euler angles is used to parameterize the rota tional 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. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 Exercises 7.9–7.10 235 (i) A set of three gyroscopes is mounted on the vehicle and provides three signals ωi (t) (see, for example, the signals shown in Figure 7.12).∗ Argue that, with knowledge of the initial orientation of the vehicle, integrating (7.28), the orientation of the vehicle can be determined. (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) How does the axis of rotation q of the vehicle change and is this axis parallel to the instantaneous axis of rotation i = ||ω ω|| ? 100 ω1 + 40 o /s ω3 ω2 − 40 −100 0 time (s) 100 Figure 7.12. 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 ±40 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). 7.10. As shown in Figure 7.13, 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 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 X̄, and the position vectors of X̄ relative to the fixed point O in the present and reference configurations are denoted by x̄ and X̄, respectively. (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. ∗ The principle of operation of a class of gyroscopes is discussed in Section 11.4. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 236 Exercise 7.10 e3 e1 X̄ E3 e2 κt P g E2 X̄ E1 E3 κ0 P O E2 E1 Figure 7.13. The fixed reference κ0 and present κt configurations of a Poisson top. The surface that the top moves on is taken to be the E1 − E2 plane. (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 , ẋ = ω × (x − xP ) + ẋP . (d) The moment of inertia tensor of the top in its reference configuration has the representation JO = λa E3 ⊗ E3 + λt (I − E3 ⊗ E3 ) . If P = XP − X̄ = −L3 E3 , ω = ω1 e1 + ω2 e2 + ω3 e3 , then show that the angular momentum vector of the top relative to the point P is HP = λa ω3 e3 + λt + mL23 (ω1 e1 + ω2 e2 ) + L3 e3 × mẋP . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 8 978 0 521 87483 0 June 2, 2008 Constraints on and Potentials for Rigid Bodies 8.1 Introduction In this chapter, constraints and the forces and moments that enforce them in the dynamics of rigid bodies are discussed. 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 moments are presented. We also discuss Lagrange’s prescription for constraint forces and constraint moments and outline its limitations. It also proves convenient to discuss potential energies and their associated conservative forces and moments. Our discussion includes as examples springs and central gravitational fields. It has obvious parallels to the treatment of constraints and their associated forces and moments. 8.2 Constraints Constraints in 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. The second manner in which constraints arise occurs when one rigid body is rolling or sliding on the other. As in particles, 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 8.1). We can consider the case of a rigid body interacting with the ground by prescribing the velocity vector of the center of mass and the angular velocity vector of either one of the bodies as functions of time. If more than two rigid bodies are involved in the constraint, then it is not too difficult to generalize the discussion subsequently presented. It should also be noted that the discussion presented here of the various types of joints and contacts is not exhaustive. 237 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 238 Constraints on and Potentials for Rigid Bodies X̄1 x̄2 x̄1 X̄2 O Figure 8.1. The present configurations and centers of mass of two rigid bodies B1 and B2 . Connected Rigid Bodies As shown in Figure 8.2, consider two rigid bodies B1 and B2 that are connected at the point P by a joint. The position vector of P on Bα relative to X̄α is denoted by πPα . Because the point P occupies the same location for both bodies, we have the following three constraints∗ : x̄1 + πP1 = x̄2 + πP2 . (8.1) These constraints also imply that v̄1 + ω1 × πP1 = v̄2 + ω2 × πP2 . (8.2) Notice that by integrating (8.2) with respect to time and setting the constants of integration to 0, we will arrive at (8.1). Consequently, constraints (8.2) are integrable. The joint at P 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 balland-socket joint. The pin joint arises in gyroscopes where it serves to connect the inner and outer gimbals. On the other hand, ball-and-socket joints do not place any restriction on the relative angular velocity vector ω2 − ω1 . 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 {s1 , s2 , s3 } be an orthonormal basis. The pin joint ensures that the relative rotation Q2 QT1 is a rotation about s3 . These restrictions are most easily expressed in terms of the relative angular velocity vector. Specifically, a pin joint imposes the constraints (ω1 − ω2 ) · s1 = 0, (ω1 − ω2 ) · s2 = 0. It should be clear that these two (integrable) constraints are supplemented by (8.2). In addition, it is sufficient to use a single angle to parameterize Q2 QT1 . ∗ 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.2 Constraints 239 X̄2 πP2 P Figure 8.2. The present configurations of two rigid bodies B1 and B2 that are connected by a joint at P. x̄2 πP1 x̄1 X̄1 O Clearly each of the individual constraints mentioned to this point can be written in the form π = 0, (8.3) where π = f1 · v̄1 + f2 · v̄2 + h1 · ω1 + h2 · ω2 + e. Here, f1 , f2 , h1 , h2 , and e are functions of x̄1 , x̄2 , Q1 , Q2 , and t. Furthermore, (8.3) can be integrated with respect to time to yield a function : = (x̄1 , x̄2 , Q1 , Q2 , t) . Here, ˙ = π. 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 8.3. Here, two rigid bodies are instantaneously in contact at the point P. It is assumed that there is a well-defined unit normal n to the surfaces of both bodies at the point P. 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 point of contact P. As the point of contact P coincides with a material point of each body, we have x p1 = x̄1 + π p1 , v p1 = v̄1 + ω1 × π p1 , (8.4) 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 240 Constraints on and Potentials for Rigid Bodies n s2 X̄2 πP2 P s1 x̄2 πP1 X̄1 x̄1 O Figure 8.3. The present configurations of two rigid bodies B1 and B2 that are in contact at the point P. and x p2 = x̄2 + π p2 , v p2 = v̄2 + ω2 × π p2 . (8.5) It is important to note that it is not generally possible to differentiate (8.4)1 to arrive at (8.4)2 , nor (8.5)1 to arrive at (8.5)2 . The reason for this is that π pα does not identify the same material of Bα at each instant of time. In other words, Pα corresponds to a different material point of Bα at each instant of time. 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 π pα have relatively simple forms, specifically, rolling/sliding spheres, the sliding top, and the rolling/sliding disks. The most famous example for which π pα is very difficult to express in a tractable form is the wobblestone or celt. As first reported by G. T. Walker in 1896 [223], this rigid body has the unique feature of reversing its direction of spin in a counterintuitive manner.∗ 8.2.1 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: v p1 · n = v p2 · n. (8.6) We can express this condition in another form: (v̄1 − v̄2 ) · n = (ω2 × π p2 − ω1 × π p1 ) · n. ∗ For further details on this interesting mechanical system, and more recent research on it, see [17] and references therein. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.3 A Canonical Function 241 The sliding condition implies that a certain relative velocity has components in only the tangential directions: vs = v p1 − v p2 = vs1 s1 + vs2 s2 . (8.7) The velocity vs is often known as the sliding velocity and is important for specifying the friction forces at the point P. 8.2.2 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: v p1 = v p2 . Again, we can express this equation in another form: v̄1 + ω1 × π p1 = v̄2 + ω2 × π p2 . (8.8) These three equations are equivalent to sliding condition (8.6) and the condition that the sliding velocity vs is zero. Rolling condition (8.8) is equivalent to three scalar equations [cf. (8.3)]: π1 = 0, π2 = 0, π3 = 0, where, for i = 1, 2, 3, πi = fi1 · v̄1 + fi2 · v̄2 + hi1 · ω1 + hi2 · ω2 + ei . However, for two of these equations, say π2 = 0 and π3 = 0, it is not possible to ˙ 2 = π2 and ˙ 3 = π3 . In other words, two of find functions 2 and 3 such that constraints (8.8) are nonintegrable or nonholonomic. The one constraint of (8.8) that is integrable corresponds to the n component of (8.8). We shall shortly discuss some examples that illustrate this point. 8.3 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 features the motions of both bodies. This function is presented in (8.9) and, for future purposes, we also calculate its derivative. Our treatment here follows [163]. As shown in Figure 8.1, the position vector of the center of mass X̄α of the body Bα is denoted by x̄α , 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 i=1 1 ei ⊗ Ei , Q2 = 3 i=1 2 ei ⊗ Ei . 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 242 Constraints on and Potentials for Rigid Bodies It should be noted that we are using the same fixed basis {E1 , E2 , E3 } to define the corotational bases. The angular velocity vectors of the bodies are 1 , ω1 = − Q̇1 QT1 , 2 1 , ω2 = − Q̇2 QT2 . 2 It should also be noted that the rotation tensor of B2 relative to B1 is Q2 QT1 = 3 2 ei ⊗ 1 ei . 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 : = (x̄1 , x̄2 , Q1 , Q2 , t) . Clearly this 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. As in the case of a particle, it is of interest to calculate ˙ . To calculate this time derivative, we invoke the chain rule: ∂ ∂ ∂ ∂ ∂ T T ˙ = . · v̄1 + · v̄2 + tr Q̇1 + tr Q̇2 + ∂ x̄1 ∂ x̄2 ∂Q1 ∂Q2 ∂t Recalling our earlier discussion in Section 6.10 of derivatives of scalar functions of rotation tensors, it is convenient to define the skew-symmetric tensor T ∂ 1 ∂ T Q − Qα Qα = 2 ∂Qα α ∂Qα and an associated vector ψQα = − [ Qα ] . Representations for these tensors and vectors, based on the parameterization of Qα , were discussed earlier. In particular, if the Euler angles νi are used to parameterize Q1 , say, then ψQ1 = 3 ∂ i=1 gi , ∂νi ψQ1 · ω1 = 3 ∂ i=1 ∂νi ν̇i , (8.9) where gi are the dual Euler basis vectors associated with the Euler angles. We can use the aforementioned skew-symmetric tensors to rewrite ˙ : ˙ = = ∂ ∂ · v̄1 + · v̄2 + tr ∂ x̄1 ∂ x̄2 T Q1 1 + tr T Q2 2 + ∂ ∂ ∂ , · v̄1 + · v̄2 + ψQ1 · ω1 + ψQ2 · ω2 + ∂ x̄1 ∂ x̄2 ∂t ∂ ∂t 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.4 Integrability Criteria 243 where 1 ωα = − [α ], 2 α = Q̇α QTα , (α = 1, 2) , are the angular velocity tensors and vectors of the rigid bodies. It is more efficient next to use the form of ˙ that involves vectors: ∂ ∂ ∂ ˙ = . · v̄1 + · v̄2 + ψQ1 · ω1 + ψQ2 · ω2 + ∂ x̄1 ∂ x̄2 ∂t (8.10) With some minor rearrangements, we can eliminate v̄2 and ω2 in favor of the relative velocity vectors v̄2 − v̄1 and ω2 − ω1 in the expression for ˙ . Equation (8.10), which first appeared in [163], will play a key role in examining potential forces and moments. Result (8.10) is rarely apparent in treatments of rigid body dynamics. This is partially because specific parameterizations of x̄α and Qα are used. To elaborate on this point, let x̄1 = 3 xi Ei , i=1 x̄2 = 3 yi Ei , i=1 and suppose that Q1 is parameterized by the Euler angles νi and the relative rotation tensor Q2 QT1 is parameterized by the Euler angles γ i . That is, ω1 = 3 i=1 ν̇i gi , ω2 − ω1 = 3 γ̇ i ni , i=1 where gi and ni are the Euler basis vectors associated with νi and γ i , respectively. Then, = (x̄1 , x̄2 , Q1 , Q2 , t) = ˜ xi , y j , νi , γ k, t . Furthermore, ˙ = ˙˜ = 3 ∂˜ i=1 ∂˜ ∂˜ ∂˜ i ∂˜ i . ẋi + ẏi + i ν̇ + i γ̇ + ∂xi ∂yi ∂ν ∂γ ∂t It is good exercise to compare this expression with (8.10) and identify the corresponding terms in both expressions. 8.4 Integrability Criteria Earlier, in Section 1.8, 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 motions of two rigid bodies. 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 that, when a sufficient number of constraints are imposed on a rolling rigid body, the resulting system 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 244 Constraints on and Potentials for Rigid Bodies of constraints becomes integrable. Fortunately, there is a theorem, which is 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 pair of rigid bodies. A Single Constraint For rigid bodies, what is needed is a generalization of criterion (1.21) for a single particle to constraints of the form π = 0, where π = f1 · v̄1 + f2 · v̄2 + h1 · ω1 + h2 · ω2 + e. As mentioned earlier, this constraint is integrable if we can find an integrating fac˙ = π. Although the integrability tor k and a function (x̄1 , x̄2 , Q1 , Q2 , t) such that k 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 x̄1 is parameterized by use of a coordinate system {q1 , q2 , q3 }, x̄2 is parameterized by use of a coordinate system {q4 , q5 , q6 }, Q1 is parameterized by use of {ν1 , ν2 , ν3 }, and Q2 is parameterized by use of {ν4 , ν5 , ν6 }. Thus the function π can be expressed as π= 3 f 1i q̇i + f 2i q̇i+3 + h1i ν̇i + h2i ν̇i+3 + e. i=1 We also define the intermediate functions Wi = f 1i , Wi+3 = f 2i , Wi+6 = h1i , Wi+9 = h2i , W13 = e, U i+3 = qi+3 , U i+6 = νi , U i+9 = νi+3 , U 13 = t, and variables U i = qi , where i = 1, 2, 3. It remains to define IJKL = WJ SLK + WK SJL + WLSKJ , (8.11) where SLK are the components of a skew-symmetric matrix: SLK = ∂WL ∂WK − . ∂UK ∂UL (8.12) In (8.11) and (8.12), 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: 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.4 Integrability Criteria 245 A necessary and sufficient condition for the constraint π = 0 to be integrable is that the following 136 (13 − 1)(13 − 2) equations hold for all U1 , . . . , U13 : IJKL = 0, for all J, K, L ∈ {1, . . . , 13}, J = K = L, J = L. (8.13) A proof of the theorem can be found in texts on differential equations (for example, Section 161 of Forsyth [64]). We also note that only 66 of the 286 equations IJKL = 0 are independent. Criterion (8.13) can be specialized to the case of a constraint f · v̄ + h · ω + e = 0 on a single rigid body. There, the indices J, K, L will range from 1 to 7, and the number of independent conditions IJKL = 0 to verify will be 10. 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. To simplify our exposition, we frame this criterion in terms of a single rigid body. Using the developments of the previous section, we can easily extend this to the case of two rigid bodies. Suppose we have a system of R ≤ 6 constraints on the motion of a single rigid body: π1 = 0, . . . , πR = 0, (8.14) where πA = fA · v̄ + hA · ω + eA (A = 1, . . . , R). 1 2 3 Again, we assume that coordinates q , q , q have been chosen for the position vector of the center of mass and parameters ν1 , ν2 , ν3 have been selected for the rotation tensor of the rigid body. Thus each of the πA’s can be rewritten as πA = 3 f Ai q̇i + hAi ν̇i + eA. i=1 Next, we define the following functions and variables: WAi = f Ai , WA(i+3) = hAi , WA7 = eA, ∂WAL ∂WAK − , ∂UK ∂UL where we have used the following notation for the variables A = SLK U i = qi, U i+3 = ν i , U 7 = t. (8.15) (8.16) A can In (8.15) and (8.16), i = 1, 2, 3 and L, K = 1, . . . , 7. The functions WAL and SLK be used to construct two matrices, for example, ⎤ ⎡ W11 · · · W17 ⎢ . .. ⎥ .. ⎥ (8.17) W=⎢ . . ⎦. ⎣ .. WR1 ··· WR7 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 246 Constraints on and Potentials for Rigid Bodies With the help of (8.15), it is easy to see that S1 , . . . , SR are skew-symmetric matrices: A A SKL = −SLK . We also define the following seven-dimensional vectors: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ b1 x1 a1 ⎢.⎥ ⎢.⎥ ⎢.⎥ ⎥ ⎥ ⎥ b=⎢ x=⎢ (8.18) a=⎢ ⎣ .. ⎦ , ⎣ .. ⎦ . ⎣ .. ⎦ , a7 b7 x7 We are now in a position to state the Frobenius integrability theorem: For a single rigid body, the necessary and sufficient conditions for a system of R < 7 constraints π1 = 0, . . . , πR = 0 to be integrable are for the following equations to hold aT · SAb = 0 (A = 1, . . . , R), for all values of the variables U1 , . . . , U7 , 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. The statement of Frobenius’ remarkable theorem is adapted from Forsyth [63], and the reader is referred to this text for a classical proof and to a paper by Hawkins [91] for additional historical perspectives. More modern proofs, featuring differential forms, are readily found; see, for example, [34, 63, 101, 189]. 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 then the theorem is trivially satisfied. REMARKS ON FROBENIUS’ THEOREM. An example of the use of the 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, then 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 (1.23) on the motion of a single particle: EXAMPLES. f 11 ẋ + f 12 ẏ + f 13 ż = 0, f 21 ẋ + f 22 ẏ + f 23 ż = 0, 3 where f1 = k=1 f 1kEk and f2 = k=1 f 2kEk are functions of the position vector of the particle, and f2 × f1 = 0. To apply Frobenius’ theorem to this case, some trivial modifications to (8.15)–(8.18) are needed. First, 3 U1 = x, U2 = y, U3 = z, U 4 = t. 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 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.5 Forces and Moments Acting on a Rigid Body 247 by the vectors a = [0, 0, 0, 1]T , b = [g1 , g2 , g3 , 0]T , where gk = (f1 × f2 ) · Ek. Now, as ⎡ 0 ⎢ ⎢ ⎢ ∂fA2 ∂fA1 ⎢ ⎢ ∂x − ∂x A 2 S =⎢ 1 ⎢ ∂f ∂fA1 ⎢ A3 − ⎢ ⎣ ∂x1 ∂x3 ∂f A1 ∂fA2 − ∂x2 ∂x1 0 ∂fA1 ∂fA3 − ∂x3 ∂x1 ∂fA2 ∂fA3 − ∂x3 ∂x2 ∂fA3 ∂fA2 − ∂x2 ∂x3 0 0 0 0 ⎤ 0 ⎥ ⎥ ⎥ 0⎥ ⎥ ⎥, ⎥ ⎥ 0⎥ ⎦ 0 it follows that aT · S1 b = 0 and aT · S2 b = 0. Thus the hypotheses of Frobenius’ integrability theorem are satisfied, and we conclude that system of constraints (1.23) is integrable. 8.5 Forces and Moments Acting on a Rigid Body Before discussing constraint forces and 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 also denote the resultant moment relative to the center of mass X̄ by M. These moments 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 M p, which is not due to the moment of an applied force, also acts on the rigid body (see Figure 8.4). For this system of applied forces and moments, the resultants are F= L FK , K=1 MO = M p + L xK × FK , K=1 M = Mp + L (xK − x̄) × FK . K=1 You should notice how the pure moment M p features in these expressions. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 248 Constraints on and Potentials for Rigid Bodies FK κt XK πK xK Mp Figure 8.4. A force FK and a moment M p acting on a rigid body. X̄ x̄ O The mechanical power P of a force FK acting at a material point XK is defined to be P = FK · ẋK . Using the center of mass, we can obtain a different representation of this power. Specifically, we use the identity, ẋK = v̄ + ω × πK , where πK = xK − x̄. It then follows that P = FK · ẋK = FK · v̄ + (πK × FK ) · 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 = M p · 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 FK · ẋK + M p · ω = F · v̄ + M · ω. K=1 These results will play a key role in our future discussion of constraint forces and moments and potential energies. 8.6 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 for ensuring 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. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.6 Constraint Forces and Constraint Moments 249 One of the major points of this chapter is the form of Lagrange’s prescription associated with constraints of the form [cf. (8.3)] π = 0, where π = f1 · v̄1 + f2 · v̄2 + h1 · ω1 + h2 · ω2 + e. We showed earlier how most of the commonly used constraints in rigid body dynamics could be written in this form. Pedagogically, it is easiest to present some examples and then give Lagrange’s prescription for the constraint forces and constraint moments. However, this prescription is not universally applicable. For instance, as in particle dynamics, it does not give physically realistic constraint forces and constraint moments when dynamic friction is present. 8.6.1 A Rigid Body Rotating About a Fixed Point Consider the rigid body shown in Figure 8.5. The body is pin-jointed to the fixed point O. As a consequence of the pin joint, the body performs a fixed-axis rotation about E3 = e3 . Further, if the rotation of the body is known, then the motion of the center of mass is also known. As a result of the pin joint, the angular velocity vector ω of the body is subject to two constraints. These constraints can be expressed in a variety of manners. For instance, ω · E1 = 0, ω · E2 = 0, ω · e1 = 0, ω · e2 = 0. (8.19) or, equivalently, Now because O is fixed, there are three additional constraints on the motion of the body: v̄ − ω × x̄ = 0. (8.20) E2 Figure 8.5. A rigid body B that is performing a fixed-axis rotation about the point O. E1 O x̄ X̄ e2 e1 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 250 June 2, 2008 Constraints on and Potentials for Rigid Bodies In other words, the pin joint couples the rotational motion of the body with the motion of its center of mass. To enforce constraints (8.19), we assume that two constraint moments are exerted by the pin joint on the body. These moments have components perpendicular to E3 : Mc1 = µ4 E1 , Mc2 = µ5 E2 , where µ4 and µ5 are functions of time that are determined from the balance laws. These constraint moments are examples of pure moments. Our prescription of the constraint moments assumes that the pin joint is frictionless. If friction were present, then they would have components in the E3 direction. Because the pin joint imposes the constraints that O is fixed, it also supplies a reaction force R acting at O: R= 3 µi Ei , i=1 where µ1 , µ2 , and µ3 are functions of times that are determined from the balance laws. The reaction force and constraint moments are equipollent to a constraint force Fc acting at the center of mass of the rigid body and a constraint moment Mc relative to the center of mass of the rigid body: Fc = 3 µi Ei , Mc = µ4 E1 + µ5 E2 − x̄ × i=1 3 µi Ei . i=1 Notice that there are five constraints on the rigid body that were imposed by the pin joint and five unknown functions µ1 , . . . , µ5 in the expressions for the constraint forces and moments. 8.6.2 A Sphere Rolling or Sliding on an Inclined Plane The problem of a sphere rolling or sliding on a plane has a long history, in part because it is the basis for pool (billiards) and bowling. The main contributors to this problem in the 19th century were Gaspard G. de Coriolis (1792–1843) (see [41]) and Edward J. Routh (1831–1907) (see [184]). Studies on the dynamics of rolling spheres on surfaces of revolution are often known as Routh’s problem.∗ Consider the rigid body shown in Figure 8.6. The body in this case is moving on a fixed inclined plane. In the figure, the body is assumed to be a sphere. For the sphere, it is easy to see that πP = −RE3 , where R is the radius of the sphere. In addition, you should notice that {s1 , s2 , n} = {E1 , E2 , E3 }. ∗ The interested reader is referred to the informative account of Routh’s influence as a tutor and teacher in [226]. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.6 Constraint Forces and Constraint Moments 251 E3 Sphere of mass m and radius R g O E1 X̄ P Inclined plane β Figure 8.6. A rigid sphere moving on an inclined plane. Let us first consider the case in which the sphere is assumed to be sliding. In this case, sliding condition (8.6), vP · n = 0, is simply v̄ · E3 = 0. In addition, the slip velocity vs is vs = vs1 E1 + vs2 E2 = v̄ + ω × (−RE3 ) . To enforce the sliding condition, a force Fc acts at the point P: Fc = µ3 E3 − µd||µ3 E3 || vs , ||vs || where µd is the coefficient of dynamic friction. As in the previous example, µ3 is a function of time that is determined by the balance laws. For the rolling sphere, there are three constraints: vP · E1 = 0, vP · E2 = 0, vP · E3 = 0. These constraints can be expressed in an alternative form: v̄ · Ei − R(ω × E3 ) · Ei = 0, where i = 1, 2, or 3. To enforce the three constraints, we assume that a normal force and a static friction force act at P. The sum of these two forces is a reaction force R: R= 3 µi Ei , i=1 where µi are functions of time that are determined from the balance laws. You should notice that the reaction force R is equipollent to a force Fc = R acting at the center of mass and a moment Mc = −RE3 × R relative to the center of mass. A related comment pertains to the sliding rigid body. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 252 Constraints on and Potentials for Rigid Bodies E3 g e2 O E1 e2 E2 ψ e1 X̄ e1 φ e1 P Figure 8.7. A thin circular disk moving with one point in contact with a rough horizontal surface. 8.6.3 The Rolling Disk As shown in Figure 8.7, 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 Subsection 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). For the disk, the position vector of the instantaneous point of contact P relative to the center of mass X̄ has the representations πP = −Re2 = −R sin(φ)e1 − R cos(φ)e2 = −R (− cos(θ) sin(ψ)E1 + cos(θ) cos(ψ)E2 + sin(θ)E3 ) . Taking the Cartesian components of the rolling condition, vP = v̄ + ω × πP , we find that the motion of the disk is subject to three constraints: π1 = 0, π2 = 0, π3 = 0, (8.21) where π1 = ẋ1 + Rφ̇ cos(ψ) + Rψ̇ cos(θ) cos(ψ) − Rθ̇ sin(θ) sin(ψ), π2 = ẋ2 + Rφ̇ sin(ψ) + Rψ̇ cos(θ) sin(ψ) + Rθ̇ sin(θ) cos(ψ), π3 = ẋ3 − Rθ̇ cos(θ), 3 and x̄ = i=1 xi Ei . You should notice that the last of these three constraints integrates to x3 = R sin(θ) + constant. However, as is shown in one of the exercises at the end of this chapter, system of constraints (8.21) is not integrable. The rolling of the disk is possible because it is assumed that a static friction force Ff acts at P: F f = µ1 E1 + µ2 E2 . 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.6 Constraint Forces and Constraint Moments 253 Furthermore, this force and the normal force N = µ3 E3 are equipollent to Fc = F f + N = 3 µi Ei , Mc = πP × Fc . i=1 Here, µ1 , µ2 , and µ3 are determined from the balance laws. 8.6.4 Lagrange’s Prescription For simplicity, we first discuss Lagrange’s prescription for the constraint forces and constraint moments acting on a single rigid body.∗ Subsequently, the case of multiple constraints in a system of two rigid bodies is discussed. We close this section with an illustration of Lagrange’s prescription by using two pin-jointed rigid bodies. There we also find a generalization of Newton’s third law. Further illustration of the prescription applied to systems of rigid bodies can be found in Sections 11.3 and 11.4 as well as in Exercise 11.4, which features two rolling rigid bodies and the Dynabee. From the examples discussed in the previous sections, we assume that any constraint on a rigid body can be written as [cf. (8.3)] A SINGLE RIGID BODY. π = 0, where π = f · v̄ + h · ω + e. Here, f, h, and e are functions of t, x̄, and Q. Lagrange’s prescription states that the constraint force Fc and constraint moment Mc associated with this constraint is Fc = µf, Mc = µh, where µ is indeterminate. This form of the prescription is an obvious generalization of the corresponding prescription for a single particle and system of particles. For a system of R constraints on the motion of a rigid body, we have R constraints of the form πK = 0, where K = 1, . . . , R, and πK = fK · v̄ + hK · ω + eK . For this system of constraints, Lagrange’s prescription states that Fc = µ1 f1 + · · · + µR fR , Mc = µ1 h1 + · · · + µR hR , where µ1 , . . . , µR are indeterminate. Notice that the prescription introduces R unknowns for the R constraints. ∗ The developments presented in this section are based on the work of O’Reilly and Srinivasa [163]. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 254 Constraints on and Potentials for Rigid Bodies By examining the examples presented in this chapter, you should be able to verify that, apart from situations in which dynamic Coulomb friction is present, Lagrange’s prescription gives physically realistic results.∗ You should also show that, unless e is zero for a single constraint, then the constraint force Fc and the constraint moment Mc will collectively do work. TWO RIGID BODIES. Suppose we have a system of R constraints acting on a system of two rigid bodies: πK = 0, where K = 1, . . . , R. We assume that the constraints are of the form [cf. (8.3)]: πK = fK1 · v̄1 + fK2 · v̄2 + hK1 · ω1 + hK2 · ω2 + eK . Here, fK1 , fK2 , hK1 , hK2 , and eK are functions of x̄1 , x̄2 , Q1 , Q2 , and t. Lagrange’s prescription states that the constraint forces and constraint moments that enforce these R constraints are Fc1 = µ1 f11 + · · · + µR fR1 , Fc2 = µ1 f12 + · · · + µR fR2 , Mc1 = µ1 h11 + · · · + µR hR1 , Mc2 = µ1 h12 + · · · + µR hR2 . (8.22) Here, Fcα is the resultant constraint force acting on Bα , Mcα is the resultant constraint moment relative to the center of mass of Bα acting on Bα , and µ1 , . . . , µR are functions of time that are determined from the balance laws for the system of two rigid bodies. You should notice that, for each of individual R constraints, Lagrange’s prescription introduces a single unknown function of time: µK . Consequently, if there are 10 constraints on the motions of the two rigid bodies, the prescription will introduce the unknowns µ1 , . . . , µ10 . For example, consider the case of two rigid bodies connected by a pin joint shown in Figure 8.8. For this situation, there are five integrable constraints on the relative motion of the bodies: TWO PIN-JOINTED BODIES AND NEWTON’S THIRD LAW. 1 ∗ = 0, . . . , 5 = 0, See, for example, the discussion of the constraint forces and moments on a rolling disk in Subsection 8.6.3. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.6 Constraint Forces and Constraint Moments 255 X̄2 πP2 Figure 8.8. The present configurations of two rigid bodies B1 and B2 that are connected by a pin joint. The axis of the pin joint is aligned with 2 e3 = 1 e3 , and its position vector relative to the centers of mass of the bodies are πP1 and πP2 , respectively. Pin joint x̄2 πP1 x̄1 X̄1 O where ˙ 1 = v̄1 · E1 + (ω1 × πP1 ) · E1 − v̄2 · E1 − (ω2 × πP2 ) · E1 , ˙ 2 = v̄1 · E2 + (ω1 × πP1 ) · E2 − v̄2 · E2 − (ω2 × πP2 ) · E2 , ˙ 3 = v̄1 · E3 + (ω1 × πP1 ) · E3 − v̄2 · E3 − (ω2 × πP2 ) · E3 , ˙ 4 = ω1 · 1 e1 − ω2 · 1 e1 , ˙ 5 = ω1 · 1 e2 − ω2 · 1 e2 . We emphasize that the axis of the pin joint is aligned with 1 e3 = 2 e3 , and its position vectors relative to the centers of mass of the bodies are πP1 and πP2 . Using Lagrange’s prescription, (8.22), with the five constraints 1 = 0, . . . , 5 = 0, we find that Fc1 = 3 µi Ei , i=1 Fc2 = − 3 µi Ei , i=1 Mc1 = µ41 e1 + µ51 e2 + πP1 × 3 µi Ei , i=1 Mc2 = −µ41 e1 − µ51 e2 − πP2 × 3 i=1 µi Ei . (8.23) 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 256 June 2, 2008 Constraints on and Potentials for Rigid Bodies Fc1 Mr X̄2 πP1 πP2 X̄1 −Mr Fc2 Figure 8.9. Free-body diagrams of the two rigid bodies B1 and B2 illustrating the constraint forces Fc1 and Fc2 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. (8.24)]. These expressions can be written in a more revealing fashion: Fc1 = −Fc2 , Mc1 = µ41 e1 + µ51 e2 + πP1 × Fc1 , Mc2 = −µ41 e1 − µ51 e2 + πP2 × Fc2 . (8.24) As illustrated in Figure 8.9, Fc1 and Fc2 are the equal and opposite reaction forces at the pin joint, and Mr = µ41 e1 + µ51 e2 and −µ41 e1 − µ51 e2 are the equal and opposite reaction moments. Relative to the center of mass of each body, the reaction moments Mc1 and Mc2 are not equal and opposite unless πP2 = πP1 .∗ 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 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. A NOTE OF CAUTION. 8.7 Potential Energies and Conservative Forces and 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 force originally arose in the dynamics of a single particle and is well understood, the same cannot be said for conservative moments (see, for example, Antman [5] and O’Reilly and Srinivasa [163]). Indeed, as noted by Ziegler [234], for rigid bodies ∗ This was first noted in [163] and can be considered as a generalization of Newton’s third law to rigid bodies. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.7 Potential Energies and Conservative Forces and Moments 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 Pρdv acting at the center of mass of the rigid body. The potential energy of this force is − R Pρdv · x̄. 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 · x̄ 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 8.10. Here, a linear spring of stiffness K and unstretched length L0 is connected to the material point Xs1 of the body B1 and the material point Xs2 of the body B2 . The spring exerts a force Fs1 at the point Xs1 and an equal and opposite force Fs2 at the point Xs2 : Fs1 = −K(||xs1 − xs2 || − L0 ) xs1 − xs2 , ||xs1 − xs2 || Fs2 = −K(||xs1 − xs2 || − L0 ) xs2 − xs1 . ||xs1 − xs2 || 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 K (||xs1 − xs2 || − L0 )2 . 2 You should notice that this potential energy depends on the position vectors of both material points Xs1 and Xs2 . It is also convenient to note the identities xs1 = Q1 Xs1 − X̄1 + x̄1 , xs2 = Q2 Xs2 − X̄2 + x̄2 , Us = where Xsα is the position vector of Xsα in the reference configuration of Bα . Using these identities, we can see that the forces, moments, and potential energy of the spring can be expressed as functions of the rotation tensors of both rigid bodies and the position vectors of their centers of mass. 257 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 258 Constraints on and Potentials for Rigid Bodies X̄2 πs2 Xs2 x̄2 Linear spring O Figure 8.10. The present configurations of two rigid bodies B1 and B2 that are connected by a spring. Xs1 x̄1 πs1 X̄1 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 r the force exerted on a particle of mass m by a particle of mass M: F = − GMm , ||r||2 ||r|| where r is the position 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 8.11). 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 Mn on B1 that are due to B2 are G x1 − x2 ρ1 dV1 ρ2 dV2 , Fn = − 2 R1 R2 ||x1 − x2 || ||x1 − x2 || x1 − x2 G ρ1 dV1 ρ2 dV2 . Mn = − (x1 − x̄1 ) × 2 ||x1 − x2 || ||x1 − x2 || R1 R2 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.7 Potential Energies and Conservative Forces and Moments Rigid body B1 X̄1 Rigid body B2 C x̄1 X̄2 x̄ x̄2 O Figure 8.11. Two rigid bodies B1 and B2 that exert mutual gravitational forces. The point C in this figure denotes the center of mass of this system of rigid bodies. The moment Mn is relative to the center of mass X̄1 of B1 . The potential energy associated with this force is G ρ1 dV1 ρ2 dV2 . Un = − R1 R2 ||x1 − x2 || Assuming that B2 is spherical and has a mass M, the expressions for the force, moment, and potential energy simplify: GM x1 − x̄2 Fn = − ρ dV1 , 2 ||x − x̄ || 1 ||x − x̄ || 1 2 1 2 R1 x1 − x̄2 GM ρ dV1 , Mn = − (x1 − x̄1 ) × 2 ||x − x̄ || 1 ||x − x̄ || 1 2 1 2 R1 GM ρ1 dV1 . Un = − ||x 1 − x̄2 || R1 It is important to notice that, even in this simplified case, the gravitational force can exert a moment on the rigid body B1 . To simplify the expressions as much as possible, we now use the fact that B1 is a rigid body: π = x1 − x̄1 = Q . In addition, we assume that ||π|| is small relative to ||x̄1 − x̄2 ||. These assumptions allow us to approximate the forces, moments, and potential energy associated with the 259 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 260 June 2, 2008 Constraints on and Potentials for Rigid Bodies central force field. After a substantial amount of manipulation, we find the results Fn ≈ mg, 3GM 3GM c × (Jc) = − c × (Ec), Mn ≈ R3 R3 3GM GM GMm tr(J) + − Un ≈ − (c · (Jc)) , R 2R3 2R3 (8.25) 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, R2 2R4 and R = ||x̄1 − x̄2 || , c= x̄1 − x̄2 . ||x̄1 − x̄2 || 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, (8.25)1 does not correspond to Newton’s gravitational force on a particle of mass m by another particle of mass M. Classical expressions (8.25) are used in the vast majority of works on satellite dynamics (cf. Beletskii [16], Hughes [97], and Kane et al. [106]). The expression for the potential energy Un in the form shown in (8.25) is credited to James Mac Cullagh (1809–1847).∗ In many of the works on the dynamics of a satellite about a fixed point O, it is common to approximate Fn with Fn = − GMm c. Such an approximation effectively R2 decouples the motion of the center of mass X̄1 of B1 from its orientation, and one can then conclude that X̄1 behaves like a particle in the one-body problem and solve for the orientation of the satellite separately.† However, as pointed out by Barkin [12] and Wang et al. [224], this approximation violates angular momentum conservation and energy conservation. Studies on the dynamics of satellites for which Fn = mg can be found in [12, 164, 224, 225]. These works show that steady motions of the satellite are possible where the orbital plane of X̄1 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 [234]. He considered a constant moment ME3 . During a motion of a rigid body consisting of a rotation about E3 through −π rad, this moment does work equal to Mπ. However, the same final orientation of the body can be achieved by a rotation about E1 through π rad, followed by a rotation about E2 through π rad.‡ Now, however, the ∗ † ‡ Notes on his derivation of Un and Fn can be found in Propositions 4 and 5 of [1]. Comments by Allman on the derivation of Mn can also be found in [1]. The one-body problem was discussed earlier in Section 2.8. This is an example of the application of Rodrigues–Hamilton theorem (6.5) discussed in the exercises at the end of Chapter 6. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 8.7 Potential Energies and Conservative Forces and Moments 261 work done by the constant moment is zero! Consequently, the work done by the moment ME3 depends on the “path” taken by the body – and hence ME3 cannot be conservative. 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 ME3 is conservative. An alternative proof that a constant moment is generally not conservative can be found in [160]. The proof presented there exploits the dual Euler basis. 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 easily generalized to the case of N rigid bodies. We assume that the most general form of the potential energy is U = U (x̄1 , x̄2 , 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 (8.10): U̇ = ∂U ∂U · v̄1 + · v̄2 + uQ1 · ω1 + uQ2 · ω2 , ∂ x̄1 ∂ x̄2 where uQα is a vector representing the derivative of U with respect to Qα . As discussed in Section 6.10, this vector has numerous representations, and the easiest to use arises when Q1 and Q2 are parameterized by sets of Euler angles. Denoting these angles and their dual Euler basis vectors by γ 1 , γ 2 , γ 3 and g1 , g2 , g3 , and 1 2 3 ν , ν , ν and h1 , h2 , h3 , respectively, the vectors uQ1 and uQ2 have the representations uQ1 = 3 ∂U k g , ∂γ k k=1 uQ2 = 3 ∂U k h . ∂νk k=1 Consider the conservative forces Fconα and moments Mconα (relative to the respective centers of mass) 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 −U̇ = Fcon1 · v̄1 + Fcon2 · v̄2 + Mcon1 · ω1 + Mcon2 · ω2 . Substituting for U̇ and collecting terms, we find that ∂U ∂U Fcon1 + · v̄1 + Fcon2 + · v̄2 + (Mcon1 + uQ1 ) · ω1 ∂ x̄1 ∂ x̄2 + (Mcon2 + uQ2 ) · ω2 = 0. (8.26) 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 262 Constraints on and Potentials for Rigid Bodies This can be interpreted as an equation for Fconα and Mconα that must hold for all motions of the rigid bodies. Assuming that Fconα and Mconα are independent of the linear and angular velocity vectors, we find that in order for (8.26) to hold for all v̄α and ωα it is necessary and sufficient that Fconα = − ∂U , ∂ x̄α Mconα = −uQα . (8.27) 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 and central gravitational forces. An Example Featuring a Torsional Spring We now present a very simple example featuring a torsional spring acting on a single rigid body. This example illustrates representations (8.27) 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 (8.27) and representation (8.9) for uQ in terms of Euler angles we find that the conservative moment is Mcon = −uQ = − ∂U 1 g ∂ψ = −Kt ψg1 = −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 x̄. It is interesting to notice that the components of Mcon are not entirely in the direction of E3 . 8.8 Concluding 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 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 f · v̄ + h · ω + e = 0, then Lagrange’s prescription states that Fc = µf, Mc = µh. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 8.1–8.2 263 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 x̄ and γ i are Euler angles, then our developments showed that the conservative force Fcon and conservative moment Mcon associated with this potential are Fcon = − 3 ∂U i=1 ∂xi Ei , Mcon = − 3 ∂U i g, ∂γ i i=1 where gi are the dual Euler basis vectors associated with the Euler angles. EXERCISES 8.1. Suppose the motion of a rigid body is subject to two constraints: ω · g3 = 0, ω · g2 = 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? 8.2. Consider a force P acting at a point P of a rigid body. In the present configuration, the point P has the position vector πP relative to the center of mass X̄ of the rigid body: πP = xP − x̄. In addition, vP = ẋP = v̄ + ω × πP . (a) The mechanical power of P is P · vP . Show that this power has the equivalent representation P · vP = P · v̄ + (πP × P) · ω. Using a free-body diagram, give a physical interpretation of this identity. (b) A force P acting at the point P 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 −U̇ = P · v̄ + (πP × P) · ω. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 264 Exercises 8.2–8.3 (c) Show that the potential energy U = U(xP ) can also be described as a function of x̄, Q, X̄ and P : U = U(xP ) = Ũ x̄, Q, P , X̄ . Here, Q is the rotation tensor of the rigid body, and πP = Q equivalence, show that P=− ∂ Ũ , ∂ x̄ P. Using this πP × P = −uQ . (d) Consider the case in which P represents a force that is due to a spring of stiffness K and unstretched length L. One end of the spring is attached to a fixed point O. What are the functions U(xP ) and Ũ(x̄, Q, P , X̄) for this force P? 8.3. Consider two rigid bodies. The rotation tensor of the first rigid body is Q1 = 3 1 ei ⊗ Ei . i=1 The rotation tensor of the second rigid body is Q2 = 3 2 ei ⊗ Ei . i=1 Here, 1 ei corotate with the first rigid body and 2 ei corotate with the second rigid body. (a) Argue that the rotation tensor of the second body relative to the first body is R = Q2 QT1 = 3 2 ei ⊗ 1 ei . 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 is 4 3 5 1 o ω̂ = − 2 ei ⊗2 ei , 2 i=1 where we have used the corotational derivative with respect to Q1 : o a= ȧ − 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 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 8.3–8.4 265 velocity vectors of both bodies and their relative angular velocity vector have the representations ω1 = γ̇ 1 E3 + γ̇ 2 1 e2 + γ̇ 3 1 e1 , ω2 = ν̇1 1 e1 + ν̇2 2 e3 + ν̇3 2 e1 + γ̇ 1 E3 + γ̇ 2 1 e2 + γ̇ 3 1 e1 , ω̂ = ν̇1 1 e1 + ν̇2 2 e3 + ν̇3 2 e1 . (d) Suppose the rotation of the second body relative to the first body is constrained such that ω̂ = ν̇2 2 e3 . 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. 8.4. As shown in Figure 8.12, 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 X̄ is located below the center of the sphere, and this feature leads to its ability to flip over (see Figure 8.13).∗ A Figure 8.12. A tippe top moving on e3 E3 g a rough horizontal plane. X̄ R E1 P The instantaneous point of contact of the spherical portion of the top with the horizontal plane is denoted by P. The position vectors of P relative to X̄ and the point A relative to X̄ are xP − x̄ = −RE3 + le3 , xA − x̄ = 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? ∗ The dynamics of the tippe top has been the subject of several investigations, and the interested reader is referred to [20, 154, 156, 177, 202, 219] for discussions and references. Among other issues, these papers point out the important role played by friction forces acting at the point of contact. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 266 Exercises 8.4–8.5 ω0 ω1 e3 E3 A g X̄ E1 X̄ A P e3 Figure 8.13. The two steady motions of a tippe top: the upright position in which e3 = E3 and the inverted state in which e3 = −E3 . (b) Show that the slip velocities of the point P of the tippe top have the representations vs1 = ẋ1 − 2 (R − l cos(θ)) + 3 (l sin(θ) cos(ψ)) , vs2 = ẋ2 + 1 (R − l cos(θ)) + 3 (l sin(θ) sin(ψ)) , where vP = vs1 E1 + vs2 E2 and v̄ = 3k=1 ẋkEk. (c) Suppose that the tippe top is sliding. Give a prescription for the constraint force Fc and moment Mc that enforce the sliding constraint. (d) Suppose that the tippe top is rolling. Give a prescription for the constraint force Fc and moment Mc that enforce the rolling constraints. (e) With the help of (8.13), show that two of the three constraints on a rolling tippe top are nonintegrable. 8.5. Consider the mechanical system shown in Figure 8.14. It consists of a rigid body of mass m that is free to rotate about a fixed point O. The joint at O does not permit the body to have a spin. A vertical gravitational force mgE1 acts on the body. The inertia tensor of the body relative to its center of mass C is J0 = λ1 E1 ⊗ E1 + λ − mL20 (E2 ⊗ E2 + E3 ⊗ E3 ) . φ O g E2 θ E3 X̄ E1 e1 Figure 8.14. A pendulum problem. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 8.5–8.6 267 The position vector of the center of mass X̄ of the body relative to O is L0 e1 . To parameterize the rotation tensor of the body, we use a set of 1–3–1 Euler angles: g1 = E1 , g2 = e3 and g3 = e1 . For these angles ω = φ̇E1 + θ̇e3 + ψ̇e1 = ψ̇ + φ̇ cos(θ) e1 + θ̇ sin(ψ) − φ̇ sin(θ) cos(ψ) e2 + θ̇ cos(ψ) + φ̇ sin(θ) sin(ψ) e3 . (a) Which orientations of the rigid body coincide with the singularities of the Euler angles? (b) Derive expressions for the unconstrained potential U and kinetic T energies of the rigid body. (c) Show that the motion of the rigid body is subject to four constraints: ψ̇ = ω · g3 = 0, v̄ − ω × (L0 e1 ) = 0. (d) Derive expressions for the constrained potential U and kinetic T energies of the rigid body. 8.6. As shown in Figure 8.15, an axisymmetric rigid body is free to rotate about the fixed point O. The body, which has a 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 O is x̄ = L1 e3 . X̄ E3 Figure 8.15. A rigid body that is free to rotate about the fixed point O. A set of 3–1–3 Euler angles is used to describe the rotation tensor Q of this body. e3 g O E1 (a) Using a set of 3–1–3 Euler angles, show that the angular velocity vector ω also has the representation ω = θ̇e1 + ψ̇ sin(θ)e2 + (φ̇ + ψ̇ cos(θ))e3 . E2 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 268 Exercises 8.6–8.8 (b) Show that the angular momentum H is H = λt θ̇e1 + λt ψ̇ sin(θ)e2 + λa φ̇ + ψ̇ cos(θ) e3 . (c) Show that the angular momentum of the center of mass relative to O is mL21 (ω1 e1 + ω2 e2 ) , 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= λ 2 λt + mL21 2 a θ̇ + ψ̇ 2 sin2 (θ) + φ̇ + ψ̇ cos(θ) . 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 U = mgx̄ · E3 + K 2 ψ . 2 Here, K is the torsional spring constant. What are the conservative force Fcon and moment Mcon acting on the rigid body? 8.7. Recall Ziegler’s example of a constant moment ME3 that was not conserva tive. After choosing a set of Euler angles, γ 1 , γ 2 , γ 3 , show that you cannot find a 3 ∂U i ∗ potential energy function U such that ME3 = − i=1 ∂γ i g . 8.8. Suppose that a linear spring of stiffness K and unstretched length L0 is attached to a point A on a rigid body and the other end of the spring is attached to a fixed point O. The position vector of A 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 + R cos(θ)) E3 . In this equation, x̄ = 3k=1 xkEk. (b) With the assistance of a set of 3–1–3 Euler angles, show that the potential energy of the spring is U = U (x̄, Q) = 2 K √ u − L0 , 2 where u = (x1 + R sin(θ) sin(ψ))2 + (x2 − R sin(θ) cos(ψ))2 + (x3 + R cos(θ))2 . ∗ One solution to this problem can be found in [160]. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 8.8–8.10 269 (c) Describe two equivalent methods to find the conservative force Fcon and conservative moment Mcon associated with the spring. 8.9. Consider the rolling disk discussed in Subsection 8.6.3 and consider constraints (8.21) 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: U 1 = x1 , U 4 = φ, U 2 = x2 , U 5 = ψ, U 3 = x3 , U 6 = θ, U 7 = t. (a) With the help of (8.15), 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 2 3T a1 = 0 0 0 0 0 0 1 . (b) With the help of (8.15), calculate the three 7 × 7 skew-symmetric matrices S1 , S2 , and S3 corresponding to π1 , π2 , and π3 , 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 2 3T a1 = 0 0 0 0 0 0 1 , 2 a2 = −R 0 0 1 0 0 3T 0 . 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 π1 = 0, π2 = 0, π3 = 0, ẋ2 = 0, and ψ̇ = 0 is integrable. 8.10. A schematic of a Griffin grinding machine is shown in Figure 8.16.∗ 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 ∗ This figure is adapted from Arnold and Mauder [8] and Webster [227]. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 270 Exercise 8.10 E3 O Drive shaft I ψ E2 E1 U Universal joint Drive shaft II γ φ P Grain bin Roller e3 Figure 8.16. Schematic of a Griffin grinding machine. first shaft (drive shaft I) has an angular velocity vector ωI = ψ̇E3 . It is coupled by a 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 − γ, φ}, show that the angular velocity vector of the roller has the representation ω = ψ̇ (sin(φ) sin(π − γ)e1 + cos(φ) sin(π − γ)e2 + cos(π − γ)e3 ) −γ̇ (cos(φ)e1 − sin(φ)e2 ) + φ̇e3 . 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 of the roller relative to drive shaft II? (b) The center of mass X̄ of the roller has a position vector relative to the fixed point U of x̄ = He3 . Establish an expression for the velocity vector v̄ of the point X̄. (c) The instantaneous point P of contact of the roller with the grain bin has the following position vector relative to X̄: πP = −r (cos(φ)e2 + sin(φ)e1 ) . Using the identity vP = ω × πP + v̄, establish expressions for the components vP · ei . 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 Exercise 8.10 (d) If vP = 0, then show that γ is a constant and the rotational speeds ψ̇ and φ̇ are related: H φ̇ = cos (γ0 ) − sin (γ0 ) ψ̇. r In this equation, γ0 is the constant value of the angle γ. June 2, 2008 271 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 9 978 0 521 87483 0 June 9, 2008 Kinetics of a Rigid Body 9.1 Introduction In this chapter, the balance laws F = mv̄˙ and M = Ḣ for a rigid body are discussed. Several component forms of these laws are considered. For instance, the components of the balance of angular momentum with respect to the corotational basis ei , M · ei = Ḣ · ei lead to a set of equations [(9.9)] that are known as Euler’s equations. In a subsequent chapter, we shall show that M · gi = Ḣ · gi lead to Lagrange’s equations. Once the balance laws have been discussed, a brief outline is given of the work– energy theorem, 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, 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, some 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: Ġ = F, ḢO = 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 Truesdell [216, 217], these laws are discussed in several of Euler’s works on rigid body dynamics and find their definitive form in Euler’s paper [56] that was published in 1776.∗ ∗ 272 The interested reader is referred to [135, 230], 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 [230], the text of Hestenes [93] is recommended. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.2 Balance Laws for a Rigid Body 273 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 + x̄ × G. Differentiating, and using the balance of linear momentum, we obtain ḢO = Ḣ + v̄ × G + x̄ × Ġ = Ḣ + x̄ × F. Hence, invoking the balance of angular momentum, ḢO = MO, we find that MO = ḢO = Ḣ + x̄ × F. However, the resultant moment relative to a fixed point O, MO, and the resultant moment relative to the center of mass X̄, M, are related by∗ MO = M + x̄ × F. It follows that M = Ḣ, which is known as the balance of angular momentum relative to the center of mass X̄. 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 features Ġ = F, ḢO = MO. (9.2) In the second set, the balance of angular momentum is taken relative to the center of mass X̄: Ġ = F, Ḣ = M. (9.3) It should be noted that G = mv̄, H = Jω, HO = Jω + x̄ × mv̄. Here, O is the origin of the coordinate system used to define x̄. To determine the motion of the rigid body, it suffices to know x̄(t) and Q(t). To obtain these results, Equations (9.2) or (9.3) must be supplemented by information on the coordinate system used to parameterize x̄ and the parameterization of Q. ∗ This may be seen from our previous discussion of a system of forces and moments acting on a rigid body in Section 8.5. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 274 Kinetics of a Rigid Body 9.3 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: Ṫ = F · v̄ + M · ω = N FK · vK + M p · 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, it is easiest to first show that ω · J̇ω = 0. (9.5) This result is established by use of an earlier result, J̇ = J − J, and the identity T ω = −ω × ω = 0: J̇ω · ω = ((J − J) ω) · ω = (Jω) · ω − (J (ω × ω)) · ω = Jω · T ω + (J (0)) · ω = 0. Now, as J̇ω · ω = 0, ˙ · ω = (Jω̇) · ω = (Jω) · ω̇. Ḣ · ω = Jω (9.6) In the last step, we invoked the symmetry of J. From (9.6), we can conclude that Ḣ · ω = H · ω̇, (9.7) a result that we shall presently invoke. To prove work–energy theorem (9.4), recall the Koenig decomposition for the kinetic energy T of a rigid body: T= 1 1 mv̄ · v̄ + (Jω) · ω. 2 2 Differentiating T and using (9.5) and (9.7), we find that Ṫ = Ġ · v̄ + 1 Ḣ · ω + (Jω) · ω̇ 2 = Ġ · v̄ + Ḣ · ω. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.3 Work and Energy Conservation 275 Invoking the balances of linear and angular momentum, work–energy theorem (9.4)1 is established: Ṫ = F · v̄ + 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 XK of the rigid body and a pure moment M p acting on the rigid body. For such a system of forces and moments, we recall that N N F= FK , M= (xK − x̄) × FK + M p. K=1 K=1 Noting that vK = v̄ + ω × (xK − x̄), we find with some rearranging that F · v̄ + M · ω = M p · 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(x̄, Q): ∂U , Mcon = −uQ . ∂ x̄ In addition, we assume that the body is subject to R constraints of the form Fcon = − Lf · v̄ + Lh · ω + Le = 0 (L = 1, . . . , R). The constraint force and moment are prescribed by use of Lagrange’s prescription: Fc = R µL (Lf) , L=1 Mc = R µL (Lh) . L=1 Finally, we assume that the only forces and moments acting on the rigid body are conservative or constraint. We now examine work–energy theorem (9.4)1 for the rigid body of interest: Ṫ = F · v̄ + M · ω = Fc · v̄ + Mc · ω + Fcon · v̄ + Mcon · ω. However, Fc · v̄ + Mc · ω = R µL (Lf · v̄ + Lh · ω) = − L=1 Fcon · v̄ + Mcon · ω = − ∂U · v̄ − uQ · ω = −U̇. ∂ x̄ R L=1 µL (Le) , 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 276 Kinetics of a Rigid Body Consequently, Ṫ = −U̇ − R µL (Le) . L=1 This implies that Ė = 0, where E = T + U, if R L=1 µL (L e) = 0. In summary, one situation in which the total energy E is conserved arises when 1 e = 0, . . . , R e = 0, and the constraint forces and constraint moments are prescribed by Lagrange’s prescription. This situation arises in most of the problems in rigid body dynamics that are solvable analytically, and we shall shortly see several examples. 9.4 Additional Forms of the Balance of Angular Momentum The balance of angular momentum Ḣ = M has several component forms and is one of 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 ω̇i ei + ω × Jω = M. i=1 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 Here, we wish to show that Ḣ = M can be written as Jα + ω × (Jω) = M. To establish this result, we need to examine ω̇ and J̇. First, we recall that H = Jω. Taking the derivative of this expression, we find Ḣ = J̇ω + Jω̇. To proceed, we need some identities. Specifically, α = ω̇ = 3 d ωi ei dt i=1 = 3 ω̇i ei + ω × i=1 = 3 i=1 3 i=1 ω̇i ei ωi ei 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.4 Additional Forms of the Balance of Angular Momentum 277 and J̇ = J + JT , J̇ω = Jω − Jω = ω × Jω. Using the identities, we find that Ḣ = J̇ω + Jω̇ = ω × Jω + Jω̇ =J 3 ω̇i ei + ω × Jω. i=1 In summary, Ḣ = M is equivalent to J 3 ω̇i ei + ω × Jω = M. i=1 This form of Ḣ = M is very useful when J is a constant tensor – for instance when 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 ω̇i 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. 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 J ikEi ⊗ Ek, J= i=1 k=1 3 3 J ikei ⊗ ek, i=1 k=1 where ei = QEi . Consequently, H = Jω = 3 3 J ikωkei . i=1 k=1 Differentiating H, we find that ˙ = Ḣ = Jω 3 3 i=1 k=1 J ikω̇kei + ω × 3 3 J ikωkei . (9.8) i=1 k=1 If we equate this expression to M, we can find the component forms of Ḣ = M. However, except when ω has a simple form, it is not convenient to consider this component form of the equations. Examples of where (9.8) are used include misbalanced rotors, where ω = θ̇E3 and J 13 = 0 and/or J 23 = 0.∗ For cases not involving a fixed axis of rotation, it is prudent to choose {E1 , E2 , E3 } to be the principal directions of J0 . ∗ An example of such a system can be found in Section 7 of Chapter 9 in [159]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 278 Kinetics of a Rigid Body The Principal Axis Case If we choose Ei to be the principal directions of J0 , then this tensor has the familiar representation J0 = 3 λi Ei ⊗ Ei , i=1 where λi are the principal moments of inertia. The vectors Ei 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 represen tation J = 3i=1 λi ei ⊗ ei . Consequently, H = Jω = 3 λi ωi ei . i=1 Evaluating Ḣ, we find Ḣ = Jω̇ + ω × Jω =J 3 ω̇i ei + ω × 3 i=1 = 3 λi ωi ei i=1 λi ω̇i ei + (λ2 − λ1 )ω1 ω2 e3 + (λ1 − λ3 )ω1 ω3 e2 + (λ3 − λ2 )ω3 ω2 e1 . i=1 In conclusion, Ḣ = M has the component form λ1 ω̇1 + (λ3 − λ2 )ω3 ω2 = M · e1 , λ2 ω̇2 + (λ1 − λ3 )ω3 ω1 = M · e2 , λ3 ω̇3 + (λ2 − λ1 )ω1 ω2 = 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, 1 , ω = − Q̇QT . 2 For example, if a set of 3–2–1 Euler angles were used to parameterize Q, then these differential equations would be ⎡ ⎤ ⎡ ⎤⎡ ⎤ ψ̇ 0 sin(φ) sec(θ) cos(φ) sec(θ) ω1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ (9.10) cos(φ) − sin(φ) ⎦ ⎣ω2 ⎦ . ⎣ θ̇ ⎦ = ⎣0 φ̇ 1 sin(φ) tan(θ) cos(φ) tan(θ) ω3 You may wish to recall that the 3–2–1 Euler angles were discussed in Subsection 6.8.1, and differential equations (9.10) can be inferred from the developments there. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.5 Moment-Free Motion of a Rigid Body 279 9.5 Moment-Free Motion of a Rigid Body Moment-free motion of a rigid body occurs when M = 0. Determining the motion is resolved by solutions of the balance laws, Ġ = F, Ḣ = 0, for x̄ 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,∗ it is usual to focus on ω(t). Another ingenious solution to this problem was presented by Poinsot in 1834 [172].† The equations governing the components ωi = ω · ei of the angular velocity vector are found from the three equations Ḣ · ei = 0 [cf. (9.9)]: λ1 ω̇1 + (λ3 − λ2 )ω3 ω2 = 0, λ2 ω̇2 + (λ1 − λ3 )ω3 ω1 = 0, λ3 ω̇3 + (λ2 − λ1 )ω1 ω2 = 0. (9.11) 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: λ1 = λ2 = λ3 ; axisymmetric body: λ1 = λ2 = λ3 ; asymmetric body: λ1 < λ2 < λ3 . For the axisymmetric body, when λ1 < λ3 the body is known as oblate. When λ1 = λ2 > λ3 , 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). The Symmetric Body For the symmetric rigid body, (9.11) simplify to ω̇i = 0. In other words, the com ponents of ω are constant. As ω̇ = 3k=1 ω̇kek, this implies that ω is constant. If we choose Q(t0 ) = I, then the axis of rotation q of the body is constant,‡ and so we find Q(t) = cos(ν)(I − q ⊗ q) − sin(ν)q + q ⊗ q, ∗ † ‡ Jacobi’s solution is discussed at length in Section 69 of Whittaker [228] and Section 37 of Landau and Lifshitz [125]. Discussions of Poinsot’s solution can be found in several texts, for instance, Marsden and Ratiu [138] and Routh [184]. Other choices of Q(t0 ) are possible; however, these may not guarantee that the axis of rotation q of Q is constant. For further details on this matter see [161]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 280 Kinetics of a Rigid Body where the axis and angle of rotation are ω (t0 ) , q = ω (t0 ) ν = ω (t0 ) (t − t0 ) , and ω(t) = ω (t0 ). 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 shortly see that there are related results for axisymmetric and asymmetric rigid bodies. The Axisymmetric Body For an axisymmetric body, it is convenient to define λt = λ1 = λ2 and λa = λ3 . The equations governing the components of angular velocity (9.11) simplify for this case to ω̇1 = kt ω2 , ω̇2 = −kt ω1 , ω3 = , (9.12) where = ω3 (t0 ) is a constant and kt = λt − λa . λt Differential equations (9.12) have a simple analytical solution: 4 5 4 5 54 ω1 (t) cos(kt (t − t0 )) sin(kt (t − t0 )) ω1 (t0 ) = . ω2 (t) − sin(kt (t − t0 )) cos(kt (t − 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. 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 feature about the 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 extensively exploited 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.5 Moment-Free Motion of a Rigid Body 281 two types of rigid bodies, it is possible to spin the body at constant speed about a principal axis. The Momentum Sphere To visualize the solutions of (9.11), a graphical technique is often used. This technique dates to the mid-19th century. It is based on two facts: The solutions ωi (t) to (9.11) preserve the magnitude of H and the rotational kinetic energy Trot . As a result, the solutions hi (t) = H · ei = λi ωi (t) (i = 1, 2, 3), lie on the intersection of the constant surfaces h = h0 and Trot = TE . Here, h2 = h21 + h22 + h23 , Trot = h21 h2 h2 + 2 + 3 , 2λ1 2λ2 2λ3 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 Trot = TE 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.∗ These intersections are the loci of hi (t). For the axisymmetric body, the intersections are shown in Figure 9.1(a). Corresponding representative intersections for an asymmetric body are shown in Figures 9.1(b).† These figures are among the most famous in dynamics. For the case presented in Figure 9.1(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)] breaks 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 λi ωi (t) on the momentum sphere to deduce some conclusions on the nature of the steady rotational motions of a rigid body. Our discussion is very 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 Hahn [85], Hughes [97], and Marsden and Ratiu [138]. ∗ † For a more detailed discussion of these intersections, the text of Synge and Griffith [207] is highly recommended. These figures were kindly supplied by Patrick Kessler in the Spring of 2007. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 282 Kinetics of a Rigid Body (a) λ3 ω3 e2 λ1 ω1 λ2 ω2 (b) λ3 ω3 λ1 ω1 e2 λ2 ω2 Figure 9.1. 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 = λ2 = 4 and λ3 = 5, and (b) λ1 = 2, λ2 = 4, and λ3 = 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. For the axisymmetric case, the trajectories shown in Figure 9.1(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.∗ On the other hand, the trajectories in Figure 9.1(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 9.1(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 ∗ Because hi = λi ωi , it is trivial to ascribe results pertaining to hi to those for ωi . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.5 Moment-Free Motion of a Rigid Body 283 the pair of steady 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 9.2, 9.3, and 9.4. One of these simulations corresponds to a perturbation of the steady rotation of a rigid body rotating about the principal axes corresponding to its maximal moment of inertia. The resulting behaviors of ωi (t) are displayed in the trajectory labeled (i) in Figure 9.2(b), whereas the behaviors of the corotational basis vectors can be seen in Figure 9.3. It should be clear from the former figure that the perturbation to the steady motion does not appreciably alter ei (t) 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 9.4, 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” −5 (b) (a) E3 ω2 5 ω3 (ii) X̄ E1 6 (i) E2 ω0 0 −5 5 ω1 Figure 9.2. The moment-free motion of a rigid body: (a) a rigid body showing the principal axes, (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) = 5.0E2 + 0.1E3 . For these simulations, (9.10) and (9.11) were numerically integrated with the initial conditions Q(0) = I and the parameter values λ1 = 2, λ2 = 4, and λ3 = 5. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 284 Kinetics of a Rigid Body 1 (a) −1 1 (b) e1 · E2 e2 · E2 e1 −1 e1 · E3 O e2 · E3 −1 e2 O −1 e1 · E1 e2 · E1 1 1 1 e3 · E2 (c) e3 −1 e3 · E3 O −1 e3 · E1 1 Figure 9.3. 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 9.2(b). are shown. It should be clear from the behavior of e2 (t) shown in Figure 9.4(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 9.3 and is indicative of 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. [11].∗ ∗ In [11], 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 9.3 and 9.4, it was not necessary to introduce a second set. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.6 The Baseball and the Football 285 1 (a) 1 e1 · E2 −1 (b) e2 · E2 e1 −1 e2 e1 · E3 O e2 · E3 −1 O −1 e1 · E1 e2 · E1 1 1 1 (c) e3 · E2 e3 −1 O e3 · E3 −1 e3 · E1 1 Figure 9.4. 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 9.2(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 v̄(t0 ), angular velocity ω(t0 ), and orientation Q(t0 ). We wish to determine the motion x̄ and Q of the sphere. As discussed by Tait [211], 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 of 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 of the sphere. We do this by following several classical works on this problem: most notably Tait [211]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 286 Kinetics of a Rigid Body ω ω X̄ v̄ X̄ v̄ FM FM Figure 9.5. A rigid sphere whose center of mass is moving to the right with a velocity vector v̄. 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 FM results. The opposite occurs when the ball is rotating counterclockwise. The Magnus Force A key force experienced by the sphere is known as the lift or Magnus force (see Figure 9.5),∗ FM = mBω × v̄, where B is a positive constant. The sign of B is determined by use of Bernoulli’s equation.† 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 Carré et al. [25]). According to Ireson [100], for some free kicks in soccer, ||v̄|| = 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: 1 v̄ FD = − ρ f ACd (v̄ · v̄) . ||v̄|| 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 = πR2 . Equations of Motion For the system at hand, M = 0 and H = ω is constant: 2mR2 ω, 5 so we find the important result that ω(t) = ω (t0 ) . ∗ † Credited to the German scientist Heinrich Gustav Magnus (1802–1870) in 1851 (see [133, 134]). Bernoulli’s equation applies to inviscid fluid flow and states that the sum of the pressure p and 1 2 2 ρ f U is a constant. Here, U is the fluid flow velocity and ρ f is the fluid density. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.6 The Baseball and the Football 287 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 (t0 ) = I: Q(t) = cos(ν)(I − q ⊗ q) − sin(ν)q + q ⊗ q, where the axis and angle of rotation are ω (t0 ) , q = ω (t0 ) ν = ω (t0 ) (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 Subsection 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 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ − sin(φ) 0 ⎦ ⎣e2 ⎦ ⎣g2 ⎦ = ⎣ e1 ⎦ = ⎣ cos(φ) g3 e3 ⎡ ⎢ =⎣ 0 0 0 0 1 cos(ψ) sin(ψ) 0 sin(θ) sin(ψ) e3 1 ⎤⎡ E1 ⎤ ⎥⎢ ⎥ ⎦ ⎣E2 ⎦ . − sin(θ) cos(ψ) cos(θ) (9.14) E3 Further, the Euler angles are subject to the following restrictions: φ ∈ [0, 2π), θ ∈ (0, π), and ψ ∈ [0, 2π). Using these Euler angles, we obtain ω = 1 E1 + 2 E2 + 3 E3 = θ̇ cos(ψ) + φ̇ sin(θ) sin(ψ) E1 + θ̇ sin(ψ) − φ̇ sin(θ) cos(ψ) E2 + ψ̇ + φ̇ cos(θ) E3 . If we use a set of Cartesian coordinates for the position vector of the center of mass, x̄ · Ei = xi , then we would find that ω × v̄ = (ẋ3 2 − ẋ2 3 ) E1 + (ẋ1 3 − ẋ3 1 ) E2 + (ẋ2 1 − ẋ1 2 ) E3 . (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 equation for the motion of the center of mass. Evaluating F = Ġ in Cartesian coordinates, we find mẍ1 = (FD + FM ) · E1 , mẍ2 = (FD + FM ) · E2 , mẍ3 = −mg + (FD + FM ) · E3 . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 288 Kinetics of a Rigid Body Ignoring the drag force and using (9.15), we find three differential equations for xi (t): mẍ1 = mB (ẋ3 2 − ẋ2 3 ) , mẍ2 = mB (ẋ1 3 − ẋ3 1 ) , mẍ3 = mB (ẋ2 1 − ẋ1 2 ) − mg. (9.16) For the general case, these equations can be integrated numerically to determine x̄(t). The Path of the Ball Turning our attention to a simple case, suppose ω (t0 ) = 10 E1 . From the previous analysis, we know that ω is constant for this rigid body. Consequently, (9.16) simplifies to mẍ1 = 0, mẍ2 = −mBẋ3 10 , mẍ3 = mBẋ2 10 − mg. The solution to these differential equations, assuming B10 = 0, is ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 0 x1 (t) − x1 (t0 ) ẋ1 (t0 ) ⎢ g(t−t ) ⎥ ⎢ ⎥ ⎢ g ⎥ 0 ⎥ ⎣x2 (t) − x2 (t0 )⎦ = A ⎣ẋ2 (t0 ) − B10 ⎦ + ⎢ ⎣ B10 ⎦ , x3 (t) − x3 (t0 ) ẋ3 (t0 ) 0 where ⎡ t − t0 ⎢ ⎢ A=⎢ 0 ⎣ 0 0 sin(B10 (t−t0 )) B10 1−cos(B10 (t−t0 )) B10 0 (9.17) ⎤ ⎥ 10 (t−t0 )) ⎥ − 1−cos(B ⎥. B10 ⎦ sin(B10 (t−t0 )) B10 From (9.17), the trajectory of the sphere can be determined. Two important features are present. First, the spin 10 influences the forward speed ẋ2 of the sphere. Second, it also affects 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 of the center of mass is the well-known parabolic trajectory: ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 0 x1 (t) − x1 (t0 ) (t − t0 ) ẋ1 (t0 ) ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥. 0 ⎣x2 (t) − x2 (t0 )⎦ = ⎣(t − t0 ) ẋ2 (t0 )⎦ − ⎢ ⎣ ⎦ 2 g x3 (t) − x3 (t0 ) (t − t0 ) ẋ3 (t0 ) (t − t0 ) 2 Representative examples of the trajectories of a point launched from the origin are shown in Figure 9.6. For small values of B10 , we can see from this figure how the trajectory differs from that in which the Magnus force is absent: A spin (10 ) in 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.7 Motion of a Rigid Body with a Fixed Point 289 15 (vi) (v) x3 x2 30 10 (iv) (i) (ii) (iii) −20 Figure 9.6. 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 B10 : (i) B10 = −0.5, (ii) B10 = −0.2, (iii) B10 = 0.0, (iv) B10 = 0.2, (v) B10 = 0.5, and (vi) B10 = 1.0. All of the trajectories are displayed for a period of 4 s and g = 9.81 m/s/s. one direction will result in the ball “rising,” whereas a “dipping” effect can be observed by reversing the initial spin. However, we also observe that, for larger values of |B10 |, the behavior of the trajectories becomes unphysical: either through the appearance of a cusp or the reversal in the sign of ẋ3 .∗ 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 are 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 [95, 97] where the equations of motion of a Frisbee are formulated and the lift and drag forces on the Frisbee 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 O, 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 9.7, 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 = mā serves to determine the constraint (reaction) forces at O that enforce these constraints. The remaining balance law, MO = ḢO, has an interesting form and is used to determine the rotation tensor Q of the rigid body. ∗ As can be seen in Figure 4 of [211], motions of this type are also present in Tait’s analysis of this problem. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 290 Kinetics of a Rigid Body Ball-and-socket joint A e1 O e3 g X̄ Xs Linear spring Figure 9.7. An example of a rigid body that is free to rotate about a fixed point O where the fixed point is a material point of the body. In this example, x̄ − 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 O 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 O has the representation x̄ − xO = L1 e1 + L2 e2 + L3 e3 , where Li are constants. Differentiating this equation with respect to time, we see that v̄ = ω × (L1 e1 + L2 e2 + L3 e3 ) . (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 O and the kinetic energy T of the rigid body. To establish the representation for HO, we start with the relation for this quantity in terms of H and G: HO = H + x̄ × G = Jω + m (L1 e1 + L2 e2 + L3 e3 ) × (ω × (L1 e1 + L2 e2 + L3 e3 )) = JOω. (9.19) For the final step in this result, we use the identity a × (b × c) = (a · c)b − (a · b)c.∗ The inertia tensor JO in (9.19) is JO = J + m L21 + L22 + L23 I − m (L1 e1 + L2 e2 + L3 e3 ) ⊗ (L1 e1 + L2 e2 + L3 e3 ) . ∗ As can be seen from (7.20), this identity was used earlier to establish the representation H = Jω. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.7 Motion of a Rigid Body with a Fixed Point 291 As expected, this expression for JO is in agreement with the one that we would obtain by using the parallel axis theorem [see (7.25)]. Starting with Koenig decomposition (7.21), a series of standard manipulations provides a convenient expression for the kinetic energy of the rigid body: T= 1 O J ω · ω. 2 (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 O O JO = λO 1 e1 ⊗ e1 + λ2 e2 ⊗ e2 + λ3 e3 ⊗ e3 . It is straightforward to show how λO i are related to the components of J and m, L1 , L2 , and L3 . Constraint Forces and Constraint Moments The motion is subject to three (integrable) constraints: 1 = 0, 2 = 0, 3 = 0. These constraints arise because the point O is fixed: i = (x̄ − L1 e1 − L2 e2 − L3 e3 ) · Ei . (9.21) Differentiating the constraints, we find Equation (9.18). Assuming that the joint at O is frictionless, we can easily use (9.18) with Lagrange’s prescription to show that Fc = µ1 E1 + µ2 E2 + µ3 E3 , Mc = (−L1 e1 − L2 e2 − L3 e3 ) × Fc . Here, Fc and Mc are equipollent to the force Fc acting at the joint O. You may wish to recall that we considered the constraint forces and moments for the situation in which the joint at O was a pin joint in Subsection 8.6.1. For this case, Mc would have two additional components. Equations of Motion For this problem, it is convenient to follow a procedure of using F = mā to solve for the three unknown components of Fc and to use MO = ḢO to solve for the motion of the body. The balance law MO = ḢO can be written in components relative to the basis ei . Recalling that we are choosing this basis to be parallel to the principal axes of JO, we find the component form O O λO 1 ω̇1 + λ3 − λ2 ω3 ω2 = MO · e1 , O O λO 2 ω̇2 + λ1 − λ3 ω3 ω1 = MO · e2 , O O (9.22) λO 3 ω̇3 + λ2 − λ1 ω1 ω2 = MO · e3 . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 292 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.∗ Once the rotation has been found, we can then use F = mā, where ā = α × x̄ + ω × (ω × x̄), to determine R. Euler–Poisson Equations An alternative formulation of the equations of motion for this case in which a gravitational force −mgE3 is acting on the body is known as the Euler–Poisson equations.† For these equations, instead of using a set of 3–1–2 Euler angles to parameterize Q and then supplementing (9.22) with (7.28), we work with three of the nine components of Q. By way of preliminaries, let us write Q̇ = Q0 in terms of components. Relative to the basis Ei ⊗ Ek, we see that ⎡ Q̇11 ⎢ ⎣Q̇21 Q̇31 Q̇12 Q̇13 ⎤ ⎡ Q11 Q̇22 ⎥ ⎢ Q̇23 ⎦ = ⎣Q21 Q̇32 Q̇33 Q31 Q12 Q13 ⎤⎡ 0 Q22 ⎥⎢ Q23 ⎦ ⎣ ω3 Q32 Q33 −ω2 −ω3 0 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 19th century,‡ and are known as the Poisson kinematical relations. It is important to note that E3 = Q31 e1 + Q32 e2 + 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 ėi = ω × ei , we would discover that Q̇31 e1 + Q̇32 e2 + Q̇33 e3 = − 3 Q3kω × 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 O. Thus we combine the ∗ † ‡ An explicit form of the differential equations can be found in an exercise at the end of this chapter: see (9.36). We shall discuss a third alternative, Lagrange’s equations of motion, in Chapter 10. See Section 411 of his treatise [173]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.7 Motion of a Rigid Body with a Fixed Point 293 balance of angular momentum that is due to Euler and kinematical relations that are due to Poisson. In component forms, the Euler–Poisson equations are Q̇31 = Q32 hO3 hO2 − Q33 O , λO λ2 3 Q̇32 = Q33 hO1 hO3 − Q31 O , λO λ3 1 hO1 hO2 + Q31 O , O λ1 λ2 O λ − λO = 2 O O3 hO2 hO3 + (mgE3 × x̄) · e1 , λ2 λ3 O λ3 − λO 1 = hO1 hO3 + (mgE3 × x̄) · e2 , O λO 1 λ3 O λ1 − λO 2 = hO1 hO2 + (mgE3 × x̄) · e3 . O λO 1 λ2 Q̇33 = −Q32 ḣO1 ḣO2 ḣO3 (9.25) Here, we have used the representations HO = 3 hOi ei , hOk = λO k ωk (k = 1, 2, 3). i=1 Additional developments and representations of Euler–Poisson equations (9.25) can be found, for example, in Beletskii [16] and Sudarshan and Mukunda [204]. Conservations For the problem of interest, we can use representation (9.20) for T to show the following forms of the work–energy theorem: Ṫ = MO · ω = N FK · vK + M p · 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 MO = 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 MO has no component in the E3 direction. It is easy to see for this case that O HO · E3 is conserved. If, in addition, the body has an axis of symmetry and λO 1 = λ2 , then you should be able to show with the help of (9.22) that HO · e3 is conserved. The problem in which the body has an axis of symmetry and MO = x̄ × (−mgE3 ) is often known as the (symmetric) Lagrange top.∗ The motion of this top conserves ∗ In the context of Lagrange’s equations of motion, we will discuss this problem in Section 10.8. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 294 Kinetics of a Rigid Body E3 Sphere of mass m and radius R g O E1 X̄ Inclined plane P β Figure 9.8. 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. E, HO · e3 , and HO · E3 , and is one of the most famous mechanical systems. Indeed, as discovered by Lagrange,∗ analytical solutions for its equations of motion can be found. This discovery is remarkable, for if we relax the assumption that the body O is symmetric (i.e., λO 1 = λ2 ), then analytical solutions are possible in only a handful of special cases (see [128, 228]). Indeed, it is an interesting exercise to numerically integrate the equations of motion for the case in which MO = x̄ × (−mgE3 ) and the λO i ’s are distinct. 9.8 Motions of Rolling Spheres and Sliding Spheres The problem of the sphere moving on a flat surface has several applications, bowling and pool being the most famous. The most famous classical treatments of this problem are due to Coriolis [41] and Routh [184], and generalizations of it occupy the literature on nonholonomically constrained rigid bodies to date.† In our treatment, we assume that the surface is rough with a coefficient of static Coulomb friction of µs and kinetic friction of µd. Of particular interest to us will be the transition between rolling and sliding and our discussion is heavily influenced by Routh [184] and Synge and Griffith [207]. Consider the sphere moving on the surface shown in Figure 9.8. The radius of the sphere is R, and the velocity of the point of contact of the sphere with the incline is vP = v̄ + ω × (−RE3 ) . ∗ † See Section IX.34 of the Second Part of Lagrange’s Mécanique Analytique [121]. This analytical solution is also discussed by Whittaker [228]. See, for example, Borisov and Mamaev [19], Frohlich [66], and Huston et al. [99]. The latter papers discuss the dynamics of bowling balls. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.8 Motions of Rolling Spheres and Sliding Spheres 295 Note the simple expression for πP here. Because the point P is the instantaneous point of contact, vP · E3 = 0. Consequently, this velocity field has the representations vP = vs1 E1 + vs2 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 + F f , M = −RE3 × F f . When the sphere is rolling, F f = µ1 E1 + µ2 E2 , where µ1 and µ2 are unknowns. For the sliding sphere, on the other hand, we have the classical prescription F f = −µd |N|c. For convenience, we use the same notation as that of the friction forces for the rolling and sliding spheres, but this should not cause confusion. Rolling Sphere We determine the motion of the rolling sphere by using the balance laws and the constraints vP = 0. Using Cartesian coordinates for x̄ and setting ω = 3i=1 i Ei , we find that these equations are ẋ1 = R2 , ẋ2 = −R1 , ẋ3 = 0, mẍ1 = mg sin(β) + µ1 , mẍ2 = µ2 , 0 = N − mg cos(β), 2 mR2 ˙ 1 = Rµ2 , 5 2 mR2 ˙ 2 = −Rµ1 , 5 2 mR2 ˙ 3 = 0. 5 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 296 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 2 1+ mR2 ˙ 1 = 0, 5 2 1+ mR2 ˙ 2 = mgR sin(β), 5 2 mR2 ˙ 3 = 0. 5 These equations are easily solved: g sin(β) (t − t0 ) 2 −1 ω (t) = ω (t0 ) + E2 . 1+ R 5 (9.26) It is left as an exercise to determine x̄(t). When β = 0, you will find that the sphere rolls in a straight line at constant speed. Sliding Sphere For the sliding sphere, it is convenient to examine the differential equations for vP . Differentiating this velocity, we find that v̇P = v̇s1 E1 + v̇s2 E2 = v̄˙ + α × (−RE3 ) . Using the balances of linear and angular momentum, we substitute for v̄˙ and α to find 5 5 mv̇s1 = 1 + F f · E1 + mg sin(β), F f · E2 . mv̇s2 = 1 + (9.27) 2 2 After substituting for the friction force, these equations provide two differential equations for the slip velocities.∗ 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) = vs2 . u When χ is constant, the slip direction c is constant. After some manipulations, we find that (9.27) are equivalent to† 5 + g sin(β) cos(χ), uχ̇ = −g sin(β) sin(χ). (9.28) u̇ = −µd g 1 + 2 These differential equations have analytical solutions for χ(t) and u(t). It is also left as an exercise to write the five differential equations governing x1 , x2 , and i . ∗ † 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. To obtain these equations we differentiated u2 = v2s1 + v2s2 and sin(χ) = vus2 and then used (9.27). 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 9.9 Closing Comments 297 If we consider the simple case in which the incline is horizontal, then β = 0. The differential equations for the slip velocity simplify considerably to 5 vs1 5 vs2 mv̇s1 = −µd mg 1 + , mv̇s2 = −µd mg 1 + . 2 u 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: 5 , χ̇ = 0. (9.29) u̇ = −µd g 1 + 2 These equations have the solution 5 u (t) = u (t0 ) − µd g 1 + (t − t0 ) , 2 χ (t) = χ (t0 ) . As a result, u will reach zero in a finite time T and the slip direction stays constant: 5 −1 u (t0 ) 1+ . T= µd g 2 It can be shown that the path of the center of the sliding sphere is either fixed, 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 9.9). 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 a key to hook shots in bowling and massee shots in pool.∗ An example of this transition is shown in Figure 9.9(a). The factor of 25 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 in Coriolis [41], Q 5 is the point 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 Closing Comments We have touched on some problems in rigid body dynamics. There are several aspects that we have not had the opportunity to address, and some of them are discussed in the exercises and others in the references at the end of this book. The treatises of Appell [7], Papastavridis [169], Routh [184], and Whittaker [228], and the splendid introductory text by Crabtree [42] 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 a rolling sphere ∗ As discussed in Frohlich [66], bowling balls feature 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 298 Kinetics of a Rigid Body (a) E2 x̄ E1 O u(t) u (t0 ) (b) E2 x̄ 0 t − t0 E1 O (c) E2 x̄ O E1 Figure 9.9. 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. Three representative paths of the center of mass of the sphere are also shown: (a) the path of the sliding center of mass is a parabolic arc; (b) the path of the sliding center of mass is a straight line; and (c) the center of mass is stationary while the sphere is sliding. For the rolling phases, the path of the center of mass is a straight line. The dashed part of these paths denotes the sliding phase of the motion. where J = 25 mR2 I and πP = −RE3 . Such a sphere is often known as a Chaplygin sphere. Partially as a consequence of its asymmetry, the path of the point of contact of the Chaplygin sphere with the ground can be very intricate (see [19, 191] and references therein). Two other celebrated examples of rigid bodies that exhibit interesting behavior are Euler’s disk and the wobblestone (or celt) [42]. 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 [143], and his controversial paper was followed by a series of works promoting alternative mechanisms for the dramatic motion of Euler’s disk (see [110] 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 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 9.1–9.6 299 been the subject of several papers (see, for example, [129, 171]) and simulations; the paper by Blackowiak et al. [17] is particularly recommended for a lucid explanation of the spin reversal mechanism. Recent analyses of rigid body dynamics have focused on the stability and bifurcation of their families of steady motions (see, for example, [128, 157]). Increasingly, some of these studies are evolving toward an examination of motions of these systems that (although not steady) are asymptotic to one or connect two steady motions of the rigid body. For the interested reader, recent work on the tippe top [20, 177, 219], the possible jumping behavior of a spinning egg [22, 142, 194], and the attitude of a thrown tennis racket [11] are mentioned as examples of notable modern analyses of rigid body dynamics. It is hoped that the exposition in this chapter will enable you to explore and appreciate works of this type. EXERCISES 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, Ṫ = F · v̄ + 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. 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 and motion of the center of mass of the rigid body. 9.3. The orientation of a rigid body relative to a fixed reference configuration is defined by a rotation tensor Q. At time t0 this rotation tensor has the value Q(t0 ), and at time t1 this rotation tensor has the value Q(t1 ). Give a physical interpretation of the rotation tensor Q(t1 )QT (t0 ). You should make use of the corotational basis in your answer. 9.4. Consider a rigid body with a fixed point O. What are the three constraints on the motion of this rigid body? Why is it sufficient to solve MO = ḢO to determine the motion of this rigid body? 9.5. Solutions for ωi (t) have been determined for a rigid body dynamics problem. How would you determine Q(t) from this solution? 9.6. A rigid body has a potential energy U = U x̄, γ i , where γ 1 , γ 2 , γ 3 are the Euler angles used to parameterize Q. If the conservative force F and conservative moment M are such that −U̇ = F · x̄˙ + M · ω, (9.30) then verify that F=− 3 ∂U i=1 ∂xi Ei , M=− 3 ∂U i g, ∂γ i i=1 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 300 Exercises 9.6–9.8 where xi = x̄ · 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 x̄, γ i . 9.7. As shown in Figure 9.8, 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 = µI, 2 where µ = 2mR . 5 (a) What are the three constraints on the motion of the sphere? Show that these constraints imply that ẍ1 − R˙ 2 = 0, ẍ2 + R˙ 1 = 0, ẍ3 = 0, where xi = x̄ · Ei , and i = ω · Ei . (b) With the help of the balance of linear momentum for the sphere, show that ˙ 2 − mg sin(β) E1 − mR ˙ 1 E2 + mg cos(β)E3 . Fc = mR (c) Show that the balance of angular momentum for the sphere and the results of (b) imply that 7 ˙ 1 = 0, mR2 5 7 ˙ 2 = mgR sin(β), mR2 5 2 ˙ 3 = 0. mR2 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 mR2 2 1 + 2 + 3 10 10 5 − mgx1 sin(β) + mgR cos(β). (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) = 3 i=1 i0 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. 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 thruster force: 1 v̄ F = −mgE3 + mBω × v̄ + f e1 − ρ f ACd (v̄ · v̄) , 2 ||v̄|| M = πt × f e1 . (a) Show that one of the four applied forces is conservative. (b) Show that the lift force does no work. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 9.8–9.9 301 (c) If the thrust force acts at a point whose position vector relative to x̄ is πt , then, with the assistance of the work–energy theorem, establish an expression for Ė. 9.9. This famous problem is discussed in most books on satellite dynamics (see, for example, Beletskii [16] or Hughes [97]). The 24 solutions subsequently discussed date to Lagrange [116, 118] in the late 18th century. Among the remarkable features about Lagrange’s extraordinary work on this topic in [118] 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 9.10, 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 (8.25). Rigid body B X̄ x̄ Figure 9.10. Schematic of a rigid body of mass m that is in orbit about a fixed symmetric body of mass M. E3 Fixed body of mass M E2 O E1 (a) Verify that Mn = −x̄ × Fn . 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 x̄ = R0 er about O with a constant orbital angular velocity θ̇0 , which is known as the modified Kepler frequency, ωKm : 3 θ̇0 = ωKm = ωK 1 + (9.31) (tr(J) − 3er · Jer ), 2R20 m where the Kepler frequency was defined previously [see (2.12)]: ω2K = GM . r03 In (9.31), 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 302 Exercises 9.9–9.10 (d) Using the results of (c), show that a steady motion of the rigid body, that is, one in which ω̇ = 0, is governed by the equation ω × (Jω) = 3ω2K er × (Jer ). (9.32) (e) Suppose that the body is asymmetric. That is, the principal values of J0 are distinct. We seek solutions of (9.32) such that ω · er = 0. Show that there are six possible solutions for ω that satisfy (9.32) and four possible solutions for er . Here, you should assume that J is known and as a result Q is known. As a result, there are 6 × 4 possible solutions of (9.32). (f) Suppose that the body is such that J = µI, where µ is a constant. Show that any constant ω satisfies (9.32) 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. 9.10. Consider the problem of a sphere of mass m, radius R, and inertia tensor J0 = 2mR2 I moving on a turntable, which is discussed by Gersten et al. [70], Lewis and 5 Murray [127], and Pars [170], among others. The contact between the sphere and the turntable is rough. In addition, the center O of the turntable is fixed and the turntable rotates about the vertical E3 with an angular speed . (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: v̄ + ω × (−RE3 ) = E3 × x̄. Using the representations ω = three constraints imply that ˙ 2 + ẋ2 = 0, ẍ1 − R 3 i=1 i Ei and x̄ = (9.33) 3 i=1 ˙ 1 − ẋ1 = 0, ẍ2 + R xi Ei , show that the ẍ3 = 0. (b) Assuming that a vertical gravitational force acts on the sphere, draw a freebody 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 ˙ 1 E2 . ˙ 2 E1 − Fc = mgE3 − m (ẋ2 E1 − ẋ1 E2 ) + mR 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 9.10–9.11 303 (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 ˙ 2 − ẋ2 , ẍ1 = R ˙ 1 + ẋ1 , ẍ2 = −R ẋ3 = 0, mR ˙ ẋ1 , 1 = µ + mR2 mR ˙ 2 = ẋ2 , µ + mR2 ˙ 3 = 0. (9.34) 2 Here, µ = 2mR . Why are (9.34) sufficient to determine the motion 5 (x̄(t), Q(t)) of the sphere? (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 (9.34) for a variety of initial conditions. Is it possible for the sphere to fall off a turntable of radius R0 ? In choosing your initial conditions (x̄(t0 ), v̄(t0 ), ω(t0 )) make sure that they are compatible with rolling condition (9.33). 9.11. Recall the definition of the kinetic energy T of a rigid body: 1 T= v · vρdv. 2 R (a) Starting from the definition of T, prove the Koenig decomposition: T= 1 1 mv̄ · v̄ + H · ω. 2 2 (b) Establish the following intermediate results: J̇ = J − J, ˙ = H · ω̇, Jω·ω Jω˙· ω = 2Ḣ · ω, where the angular momentum H = Jω. (c) Using the intermediate results and the balance laws, prove the work–energy theorem: Ṫ = F · v̄ + M · ω. (d) If F= K i=1 Fi , M= K i=1 (xi − x̄) × Fi + MP , 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 304 Exercises 9.11–9.13 then show that Ṫ = K Fi · vi + MP · ω. i=1 Give three examples of the use of this result for a single rigid body. 9.12. Establish the following theorem, which can be found in Section 44 of Euler [54]: “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 [53], where these axes were first termed principal axes of inertia. 9.13. Consider the tippe top shown in Figure 8.12 and discussed in Exercise 8.4. (a) Continuing the earlier exercise, we follow Or [156] and suppose that the point P slides on the horizontal surface and experiences both Coulomb and viscous friction forces: F f = − (µk + µv ||vP ||) ||N|| s, where µk 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 = m g + lθ̈ sin(θ) + lθ̇2 cos(θ) E3 . (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 = λt (I − e3 ⊗ e3 ) + λa e3 ⊗ e3 , then 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).∗ (e) When the tippe top is rolling on a horizontal surface, show that, in addition to conservation of total energy, the following kinematical quantity is conserved: I1 = ω23 λa λt + mλa (πP · e1 )2 + (πP · e2 )2 + mλt (πP · e3 )2 . (9.35) ∗ A discussion of the history of this integral (and several others) can be found in Gray and Nickel [76]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises 9.13–9.14 305 Here, πP is the position vector of the instantaneous point of contact P (relative to the center of mass X̄) of the tippe top with the horizontal surface. The integral of motion I1 was first established by Routh (see Section 243 of [184]) and can be described as the Routh integral. It is also known as the Chaplygin integral [107, 115]. 9.14. Consider a body that is free to move about one of its material points O that is fixed (cf. Figure 9.7). 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 X̄ relative to O is x̄ − xO = he3 , where h is a constant. A linear spring of stiffness K and unstretched length L0 is attached to the body at the point XS and the other end is attached to a fixed point A: xs − xO = s1 e1 + s3 e3 , xA − xO = LAE1 . 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 v̄ = hω2 e1 − hω1 e2 , ā = h (ω̇2 + ω1 ω3 ) e1 − h (ω̇1 − ω2 ω3 ) e2 − h ω21 + ω22 e3 . Here, ω = 3k=1 ωkek. (b) Show that the angular momentum of the rigid body relative to O is HO = λ1 + mh2 ω1 e1 + λ2 + mh2 ω2 e2 + λ3 ω3 e3 . Show that the kinetic energy of the rigid body has the representation T= 1 HO · ω. 2 (c) What are the three constraints on the motion of the rigid body? Give prescriptions for the constraint force Fc and 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 O is Fc = mh (ω̇2 + ω1 ω3 ) e1 − mh (ω̇1 − ω2 ω3 ) e2 − mh ω21 + ω22 e3 + mgE3 − Fs , where Fs is the spring force. (f) Assuming that a set of 3–1–3 Euler angles is used to parameterize Q = 3 i=1 ei ⊗ Ei , show that the potential energy U = U (x̄, Q) of the rigid 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 306 Exercise 9.14 body in this problem can be expressed as a function of these angles: U = Û (ψ, θ, φ). (g) Verify that the conservative force Fcon and conservative moment Mcon acting on the rigid body satisfy the identity Fcon · v̄ + Mcon · ω = MOcon · ω, where MOcon is the resultant conservative moment acting on the body relative to O. Using this result, explain why MOcon = − ∂ Û 1 ∂ Û 2 ∂ Û 3 g − g − g . ∂ψ ∂θ ∂φ (h) Show that the rotational motion of the rigid body is governed by the following equations: λ1 + mh2 ω̇1 = λ2 + mh2 − λ3 ω2 ω3 + MO · e1 , λ2 + mh2 ω̇2 = λ3 − λ1 − mh2 ω1 ω3 + MO · e2 , λ3 ω̇3 = ⎡ ⎤ ψ̇ ⎢ ⎥ ⎣ θ̇ ⎦ = φ̇ (λ1 − λ2 ) ω1 ω2 + MO · e3 , ⎡ sin(φ)cosec(θ) cos(φ)cosec(θ) ⎢ cos(φ) − sin(φ) ⎣ ⎤⎡ ⎤ 0 ω1 ⎥⎢ ⎥ 0⎦ ⎣ω2 ⎦ . − sin(φ) cot(θ) − cos(φ) cot(θ) 1 ω3 (9.36) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 10 978 0 521 87483 0 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body 10.1 Introduction In a famous paper [118] 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 ∂ q̇K − ∂T ∂V =− K K ∂q ∂q (K = 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 = Ḣ and F = mv̄˙ 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. We start this chapter by showing that the balance laws F = mv̄˙ and M = Ḣ for a rigid body are equivalent to Lagrange’s equations of motion: d dt d dt ∂T ∂ q̇i ∂T ∂ ν̇i − ∂T = F · ai , ∂qi − ∂T = M · gi . ∂νi (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 [118]. 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 ∂ u̇A − ∂T = A ∂uA (A = 1, . . . , 6) , (10.3) 307 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 308 978 0 521 87483 0 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body where A = F · ∂ v̄ ∂ω + M · A. A ∂ u̇ ∂ u̇ (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. When the system of constraints on the system is integrable, the constraint forces and moments are prescribed by use of Lagrange’s prescription, and the coordinates are chosen appropriately, we pleasantly 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 [26, 28] supplemented with material on constraint forces and moments that is due to O’Reilly and Srinivasa [163]. 10.2 Configuration Manifold of an Unconstrained Rigid Body The motion of a rigid body can be decomposed into a translation x̄ of the center of mass X̄ followed by a rotation Q. Here, the set of all vectors x̄ is a three-dimensional Euclidean space E3 , whereas the set of all rotation tensors Q is a three-dimensional space known as SO(3).∗ Thus the configuration manifold M of a rigid body is the space of all vectors x̄ and all rotation tensors Q. This space is the product of E3 and SO(3): M = E3 ⊕ SO(3). The product ⊕ is a topological product. For example, E3 = E ⊕ E2 = E ⊕ E ⊕ E. It should be clear that the dimension of M is 6. It should also be clear that the configuration manifold can be considered as a submanifold of the configuration space S, which in this case is E3 ⊕ E9 . Here the three-dimensional space is the space containing the position vector x̄, and the nine-dimensional space is the space containing the second-order tensor Q. To parameterize M, 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 x̄ of the center of mass and any set of Euler angles, ν1 , ν2 , and ν3 , to parameterize the rotation tensor Q. For the velocity vector of the center of mass, we then have v̄ = 3 q̇i ai , i=1 ∗ The space O(3) is the space of all orthogonal tensors. Hence, the S in SO(3) denotes “special” because rotation tensors have a determinant of +1. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.2 Configuration Manifold of an Unconstrained Rigid Body where the covariant basis vectors ai = ∂ x̄ . ∂qi You should also recall that {ai } is a basis for E3 and that the Euler angles define the Euler basis {gi }, which is also a basis for E3 . In particular, we have ω= 3 ν̇i 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 , ∂ ν̇i ai = ∂ v̄ . ∂ q̇i These two identities are easily established. We can calculate the kinematical line-element ds for the configuration manifold by using the given choice of Euler angles and curvilinear coordinates from the kinetic energy T: ds = = 2T m dt 1 v̄ · v̄ + ω · (J · ω) dt. m Shortly an example will be presented of how to calculate the desired representation for T. As we are using curvilinear coordinates and Euler angles to parameterize the motion of the rigid body, for certain points on the configuration manifold singularities in the parameters will arise. At these points, it is possible for the body to be in motion, yet the value we will get for T will be zero. This situation violates one of the chief attributes of T, namely that T is zero if and only if v̄ = 0 and ω = 0, i.e., the rigid body is instantaneously at rest. The singularities will also result in errors when the equations of motion are being integrated numerically. For this reason, many computer codes 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, , and to parameterize x̄ and a set of 3–1–3 Euler angles to 309 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 310 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body parameterize Q. For these Euler angles, we recall from Subsection 6.8.2 that the Euler basis {gi } has the representations g1 E3 g2 = e1 g3 e3 = sin(φ) sin(θ) cos(φ) sin(θ) cos(θ) cos(φ) − sin(φ) 0 e2 . 0 0 1 e3 e1 Using these results, we can readily compute the angular velocity vector: ω = ψ̇E3 + θ̇e1 + φ̇e3 = ψ̇sin(φ) sin(θ) + θ̇ cos(φ) e1 + ψ̇cos(φ) sin(θ) − θ̇ sin(φ) e2 + ψ̇cos(θ) + φ̇ e3 . (10.5) For convenience, we choose Ei to be the principal axes of the body.∗ Hence, J0 = λ1 E1 ⊗ E1 + λ2 E2 ⊗ E2 + λ3 E3 ⊗ E3 , J = QJ0 QT = λ1 e1 ⊗ e1 + λ2 e2 ⊗ e2 + λ3 e3 ⊗ e3 . When these results are combined, the kinetic energy T of the rigid body has the representations m 1 v̄ · v̄ + ω · Jω 2 2 λ 2 m 2 1 Ṙ + R2 ˙ 2 + R2 sin2 ()˙ 2 + ψ̇sin(φ) sin(θ) + θ̇ cos(φ) = 2 2 2 λ2 λ3 2 ψ̇cos(φ) sin(θ) − θ̇ sin(φ) + ψ̇cos(θ) + φ̇ . + 2 2 T= 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 M. Singularities in the parameterization of the motion arise when θ = 0, π and = 0, π. To see the effects of these singularities on T, let us consider an instant in which = 0 and θ = 0. At this instant, the preceding expression for T simplifies to T= ∗ m 2 Ṙ + R2 ˙ 2 + 0 2 2 λ2 2 λ3 2 λ1 0 + θ̇ cos(φ) + 0 − θ̇ sin(φ) + ψ̇ + φ̇ . + 2 2 2 If we do not make this choice, then we would find a more complicated representation for the rotational kinetic energy: ω · (Jω) = J 011 ω21 + J 022 ω22 + J 033 ω23 + 2J 012 ω1 ω2 + 2J 023 ω2 ω3 + 2J 013 ω1 ω3 , where ωi = ω · ei and J 0ik = ei · (Jek ) = Ei · (J0 Ek ). 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.3 Lagrange’s Equations of Motion: A First Form 311 If, in addition, ψ̇ = −φ̇, θ̇ = Ṙ = ˙ = 0, and ˙ = 0, then the preceding expression for T = 0, but the actual kinetic energy of the body is nonzero. 10.3 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 [28], but our developments are not as general as his. Resulting equations (10.6) were first shown by him to be equivalent to the balances of linear and angular momentum for a rigid body.∗ 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 Sections 8-2 and 8-6 of Greenwood [79].† We refer to (10.6) as the first form of Lagrange’s equations of motion. To start, we choose a set of curvilinear coordinates qi to parameterize x̄: x̄ = 1 x̄(q , q2 , q3 ). Next, we choose a set of Euler angles νi to parameterize the rotation tensor Q of the rigid body Q = Q(ν1 , ν2 , ν3 ). Then, as will be subsequently shown, Lagrange’s equations for the rigid body are d ∂T ∂T − i = F · ai , dt ∂ q̇i ∂q d ∂T ∂T (10.6) − i = M · gi . i dt ∂ ν̇ ∂ν 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 v̄ = 3 i q̇ ai , i=1 ∂ v̄ = ai , ∂ q̇i ω= 3 ∂ω = gi . ∂ ν̇i ν̇i gi , i=1 It should be evident from (10.6) 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 and moments Mcon , we have Fcon = − 3 ∂U i a, ∂qi Mcon = − i=1 3 ∂U i=1 ∂νi gi , (10.7) where the potential energy function U has the representations U = U (x̄, Q) = U q1 , q2 , q3 , ν1 , ν2 , ν3 . ∂U ∂U Notice that Fcon · ai = − ∂q i and Mcon · gi = − ∂ν i . Consequently, it is not necessary to evaluate Fcon and Mcon in Lagrange’s equations; rather it suffices to evaluate the ∗ † See, in particular, Theorems 4.2 and 4.4 of Casey [28]. Greenwood’s exposition is missing the intermediate result ∂T ∂ γ̇ i = H · ġi . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 312 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body 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, d dt d dt ∂L ∂ q̇i ∂L ∂ ν̇i − ∂L = (F − Fcon ) · ai , ∂qi − ∂L = (M − Mcon ) · gi . ∂νi (10.8) It is left as an (easy) exercise to show that (10.8) can be established from (10.6). 10.3.1 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 x̄ by use of a set of curvilinear coordinates and parameterizing Q by use of a set of Euler angles νi . These parameterizations imply that the kinetic energy T is a function of these quantities and their time derivatives: 1 1 T = T qi , q̇i , νk, ν̇k = mv̄ · v̄ + J0 ω0 · ω0 . 2 2 Here, the angular velocity vector ω0 = QT ω and the inertia tensor J0 = QT JQ. We next consider the partial derivatives of T: 1 ∂ v̄ mv̄ · v̄ = mv̄ · i = mv̄ · ai , 2 ∂ q̇ ∂T 1 ∂ ∂ v̄ = i mv̄ · v̄ = mv̄ · i = mv̄ · ȧi , i ∂q ∂q 2 ∂q ∂T 1 ∂ ∂ω0 J0 ω0 · ω0 = J0 ω0 · i = J0 ω0 · QT gi , = i ∂ ν̇i ∂ ν̇ 2 ∂ ν̇ ∂T 1 ∂ ∂ω0 J0 ω0 · ω0 = J0 ω0 · i = J0 ω0 · QT ġi . = i ∂νi ∂ν 2 ∂ν ∂T ∂ = i i ∂ q̇ ∂ q̇ (10.9) Here we have used the identities ai = ∂ v̄ , ∂ q̇i ȧi = ∂ v̄ , ∂qi gi = Q ∂ω0 , ∂ ν̇i Shortly a derivation of these results will be given. ġi = Q ∂ω0 . ∂νi (10.10) 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.3 Lagrange’s Equations of Motion: A First Form 313 For the third step, we examine (10.9)1,2 and note that G = mv̄. Hence, ∂T d d ∂T − i = (G · ai ) − G · ȧi dt ∂ q̇i ∂q dt = Ġ · ai = 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 = J0 ω0 · QT b. Using this result in conjunction with (10.9)3,4 , we find that ∂T d d ∂T − i = (H · gi ) − H · ġi dt ∂ ν̇i ∂ν dt = Ḣ · gi = M · gi . Thus we have established the last three of Lagrange’s equations of motion. 10.3.2 The Four Identities The proof of Lagrange’s equations is achieved by use of the four identities, (10.10). The first two of these results are similar to those we established for the single particle. You should notice the presence of Q in (10.10)3,4 . Unfortunately, ∂∂νωi = ġi . We first establish the easier results: 3 3 ∂ v̄ ∂ k = q̇ a δikak = ai = k ∂ q̇i ∂ q̇i k=1 k=1 and ∂ v̄ ∂ = i i ∂q ∂q = 3 k=1 = 3 k=1 3 q̇ ak = k k=1 3 k=1 q̇k ∂ak ∂qi ∂ 2 x̄ ∂ q̇ = q̇k k ∂qi ∂qk ∂q 3 k k=1 q̇k ∂ x̄ ∂qi ∂ai ∂qk = ȧi . Notice that we used the facts that ai are both independent of q̇k and the derivatives of x̄ with respect to qi . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 314 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body We next consider the easier of the two remaining identities: ∂ω0 ∂ T ∂QT T ∂ω = 0ω + QT (gi ) = ω = ω + Q Q ∂ ν̇i ∂ ν̇i ∂ ν̇i ∂ ν̇i = QT gi . Some rearranging gives gi = Q ∂ω0 . ∂ ν̇i Notice that we used the fact that Q does not depend on ν̇i to establish this result. For the final result, we again need to be cognizant of the fact that Q depends on the Euler angles νi but not on ν̇i . First, we note that 5 4 3 1 , ∂ ∂ω ∂ 1 k ∂Q T T gi = i = i − Q̇Q = i − ν̇ Q ∂ ν̇ ∂ ν̇ 2 ∂ ν̇ 2 ∂νk k=1 ) ( 1 ∂Q T =− Q . 2 ∂νi Next, we deduce, by differentiating QQT = I twice with respect to the Euler angles, that ∂Q T ∂QT Q = −Q , ∂νi ∂νi ∂Q ∂QT ∂QT ∂Q T = Q Q . ∂νi ∂νk ∂νi ∂νk The previous results are now used to show the desired identity: ) ( 1 ∂Q T d − ġi = Q dt 2 ∂νi ) ( 2 3 1 ∂ Q T ∂Q ∂QT ν̇k − Q + = 2 ∂νi ∂νk ∂νi ∂νk k=1 = 3 k=1 = 3 k=1 ) ( 1 ∂2Q ∂QT ∂Q T Q ν̇k − Q QT i k + 2 ∂ν ∂ν ∂νi ∂νk ) ( 1 ∂ ∂Q ν̇k − Q i QT k QT 2 ∂ν ∂ν 4 5 3 1 ∂ T k ∂Q T =− Q i Q ν̇ Q 2 ∂ν ∂νk k=1 ) ( 1 ∂0 = − Q i QT 2 ∂ν ) ( 1 ∂0 =Q − 2 ∂νi =Q ∂ω0 . ∂νi 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 10.4 A Satellite Problem June 12, 2008 315 In the final stages of the proof we used the facts that ω0 is the axial vector of 0 = QT Q̇ = QT Q. We also used the identity , QBQT = det (Q) Q ( [B]) . This identity is one method relating ω0 to ω, and it was alluded to earlier when angular velocity vectors were discussed [see (6.7)]. 10.4 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 10.1). 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: x̄ = ReR = R cos(φ)E3 + R sin(φ)er = R cos(φ)E3 + R sin(φ) (cos(θ)E1 + sin(θ)E2 ) and v̄ = ṘeR + R sin(φ)θ̇eθ + Rφ̇eφ . You should be able to see from this equation what the covariant basis vectors ai are. We parameterize the rotation tensor Q by using a set of 1–2–3 Euler angles: ω = ν̇1 E1 + ν̇2 e2 + ν̇3 e3 . You should notice that ei = QEi . The angular velocity vector of the body also has the representation ω = ω1 e1 + ω2 e2 + ω3 e3 , where ω1 = ν̇2 sin(ν3 ) + ν̇1 cos(ν2 ) cos(ν3 ), ω2 = ν̇2 cos(ν3 ) − ν̇1 cos(ν2 ) sin(ν3 ), ω3 = ν̇1 sin(ν2 ) + ν̇3 . You should be able to infer the representations for gi from these results. We shall choose Ei 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 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 316 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body Rigid body B e3 e1 Fixed symmetric body of mass M E3 X̄ x̄ e2 β O E2 θ E1 er Figure 10.1. A rigid body orbiting a spherically symmetric body of mass M. The angle of lati- tude β = π 2 − φ. has the representation T= m 2 Ṙ + R2 φ̇2 + R2 sin2 (φ)θ̇2 2 λ1 + (ν̇2 sin(ν3 ) + ν̇1 cos(ν2 ) cos(ν3 ))2 2 λ2 + (ν̇2 cos(ν3 ) − ν̇1 cos(ν2 ) sin(ν3 ))2 2 λ3 + (ν̇1 sin(ν2 ) + ν̇3 )2 . 2 If the body has an axis of symmetry so that λ1 = λ2 , then the expression for the rotational kinetic energy will simplify considerably. 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 U=− GMm GM 3GM − tr(J) + (JeR ) · eR . R 2R3 2R3 (10.11) To express this potential energy in terms of the Euler angles νi , we need to use the results eR = cos(φ)E3 + sin(φ) cos(θ)E1 + sin(φ) sin(θ)E2 , E1 = cos(ν2 ) cos(ν3 )e1 − cos(ν2 ) sin(ν3 )e2 + sin(ν2 )e3 , E2 = − sin(ν2 ) cos(ν3 )e1 + sin(ν2 ) sin(ν3 )e2 + cos(ν2 )e3 , E3 = (sin(ν1 ) sin(ν3 ) − cos(ν1 ) sin(ν2 ) cos(ν3 )) e1 + (sin(ν1 ) cos(ν3 ) + cos(ν1 ) sin(ν2 ) sin(ν3 )) e2 + cos(ν1 ) cos(ν2 )e3 . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 10.4 A Satellite Problem 317 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 the expressions for the force and moment associated with this potential energy by using representations (10.7). Alternatively, we can appeal to representations (8.25) described in Chapter 8. The Balance Laws For the rigid body of interest, we have the balance laws mv̄˙ = F =− GMm 3GM eR − (2J + ((λ1 + λ2 + λ3 ) − 5eR · JeR ) I) eR , 2 R 2R4 Ḣ = M 3GM = eR × (JeR ). R3 Here, we used our earlier representations, (8.25), 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 M = 0. In addition, if R = R0 er and θ̇ = ω0 , where R0 and ω0 are constant, then the center of mass of the rigid body describes a circular orbit at a constant orbital speed ω0 . 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: ∂T d ∂T − = F · eR dt ∂ Ṙ ∂R =− d dt ∂T ∂ φ̇ − ∂T ∂ θ̇ − ∂T ∂ ν̇i − 3GM (eR · Jeφ ) , R3 ∂T = F · R sin(φ)eθ ∂θ =− d dt GMm 3GM − ((λ1 + λ2 + λ3 ) − 3eR · JeR ) , R2 2R4 ∂T = F · Reφ ∂φ =− d dt June 12, 2008 3GM sin(φ) (eR · Jeθ ) , R3 ∂T = M · gi ∂νi 3GM · gi . e = × (Je ) R R R3 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 318 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body In the last of these equations, i = 1, . . . , 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 (10.8) 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 ḢO = 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. 10.5 Lagrange’s Equations of Motion: A Second Form Previously, we assumed that a set of coordinates q1 , . . . , q6 had been chosen to parameterize x̄ and Q: x̄ = x̄ q1 , . . . , q3 , Q = Q q4 = ν1 , . . . , q6 = ν3 . (10.12) We note that the covariant basis vectors associated with this choice of coordinates are ai = ∂ v̄ , ∂ q̇i gi = ∂ω , ∂ q̇(i+3) i = 1, 2, 3. With choice (10.12), we showed that Lagrange’s equations of motion have the following form: d ∂T ∂T ∂ v̄ − i = F· i, i dt ∂ q̇ ∂q ∂ q̇ d ∂T ∂T ∂ω − (i+3) = M · (i+3) . (10.13) (i+3) dt ∂ q̇ ∂q ∂ q̇ It is not too difficult to see that these equations can be written in a more compact form: d ∂T ∂T ∂ v̄ ∂ω − A =F· A +M· A (10.14) (A = 1, . . . , 6) . dt ∂ q̇A ∂q ∂ q̇ ∂ q̇ Form (10.14) of Lagrange’s equations is useful in several cases. Among them, 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. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.5 Lagrange’s Equations of Motion: A Second Form 319 However, we shall need a more general form of Lagrange’s equations for other applications. We now consider a new choice of coordinates: x̄ = x̄ u1 , . . . , u6 , Q = Q u1 , . . . , u6 . (10.15) With the choice (10.15), we shall see that Lagrange’s equations of motion are ∂T d ∂T − A = A , (10.16) dt ∂ u̇A ∂u where the generalized forces are A = F · ∂ v̄ ∂ω +M· A A ∂ u̇ ∂ u̇ (A = 1, . . . , 6) . We shall also shortly discuss examples that use this form. To establish (10.16), we invoke the following identities∗ : ∂ v̄ d ∂ω0 ∂ω0 d ∂ v̄ d ∂ω = = = Q A, Q , dt ∂ u̇A ∂uA dt ∂ u̇A dt ∂ u̇A ∂u where ω = Qω0 . Using these identities and the decomposition of the kinetic energy,† we can use a straightforward set of manipulations to establish (10.16): d ∂T d ∂ω0 ∂T ∂ v̄ − A = mv̄ · A + Jω · Q A dt ∂ u̇A ∂u dt ∂ u̇ ∂ u̇ ∂T ∂ v̄ ∂ω0 − = mv̄ · A + J0 ω0 · ∂uA ∂u ∂uA ∂ v̄ d ∂ω0 G· A +H· Q A = dt ∂ u̇ ∂ u̇ ∂T ∂ v̄ ∂ω0 − =G· A +H· Q A ∂uA ∂u ∂u ∂ v̄ ∂ω0 = Ġ · A + Ḣ · Q A ∂ u̇ ∂ u̇ = F· ∂ v̄ ∂ω + M · A. A ∂ u̇ ∂ u̇ (10.17) This form of Lagrange’s equations is the starting point for most applications and a discussion of Approach II. We now turn to issues associated with the second form of Lagrange’s equations in the presence of constraints. ∗ † The proof of these results follows from (10.15) in a manner that is similar to the method by which the four identities were established in Subsection 10.3.2. Recall that Jω · ω = QJ0 ω0 · ω = QJ0 ω0 · Qω0 = J0 ω0 · ω0 . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 320 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body 10.5.1 Changing Coordinates: An Identity 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 = qA u1 , . . . , u6 . Then, as q̇A = 6 ∂qA B=1 ∂uB u̇B, we conclude that ∂qA ∂ q̇A = B B ∂u ∂ u̇ (A = 1, . . . , 6, B = 1, . . . , 6) . (10.18) These identities are often known as “cancellation of the dots.” It is a good exercise to show that (10.18) also hold when the coordinate transformations are time dependent: qA = qA u1 , . . . , u6 , t . The resulting identities are used in the literature with Approach II of Lagrange’s equations of motion. 10.5.2 Constraints and Constraint Forces and Moments Suppose that an integrable constraint is imposed on the rigid body: 1 q , . . . , q6 , t = 0. 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? The constraint = 0 implies that ˙ = 0. With some rearranging we find that ˙ = 0 can be expressed as f · v̄ + h · ω + e = 0, where f= 3 ∂ i a, ∂qi i=1 h= 3 i=1 ∂ gi , ∂q(i+3) e= ∂ . ∂t For the present, it is most convenient to have representations for f and h in terms of the coordinates for x̄ and Q, respectively, rather than in terms of u1 , . . . , u6 . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.5 Lagrange’s Equations of Motion: A Second Form 321 To answer the question we just posed, we suppose that Lagrange’s prescription has been used to prescribe the unknown constraint force and constraint moment: Fc = µf = µ 3 ∂ i a, ∂qi i=1 Mc = µh = µ 3 ∂ gi . ∂qi+3 i=1 It is interesting to expand the constraint forces and moments for this case: 3 3 ∂ v̄ ∂ω ∂ i ∂ v̄ ∂ ∂ω Fc · A + Mc · A = µ ia · A + µ i+3 gi · A ∂ u̇ ∂ u̇ ∂q ∂ u̇ ∂q ∂ u̇ i=1 =µ i=1 3 3 i=1 k=1 +µ ∂ i ∂ q̇k a · A ak ∂qi ∂ u̇ 3 3 ∂ ∂ q̇k+3 gi · gk i+3 ∂q ∂ u̇A i=1 k=1 =µ 3 3 i=1 k=1 =µ 3 i=1 k=1 i=1 3 3 ∂ ∂qi ∂ ∂qi+3 + µ i A i+3 ∂q ∂u ∂q ∂uA i=1 =µ 3 3 3 ∂ ∂ q̇i ∂ ∂ q̇i+3 + µ i A i+3 ∂q ∂ u̇ ∂q ∂ u̇A i=1 =µ ∂ ∂ ∂ q̇k k ∂ q̇k+3 k δi + µ δ i A i+3 ∂q ∂ u̇ ∂q ∂ u̇A i i=1 ∂ˆ . ∂uA Notice that we used (10.18) in the penultimate step, and also expressed tion of u1 , . . . , u6 and t: 1 = q , . . . , q6 , t = ˆ u1 , . . . , u6 , t . (10.19) as a func- 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 make some important conclusions. For a constraint 1 = 0, where = q , . . . , q6 , t = ˆ u1 , . . . , u6 , t , Lagrange’s prescription yields ∂ v̄ ∂ω ∂ˆ + M · = µ . (10.20) c ∂ u̇A ∂ u̇A ∂uA Thus, in answer to our question, the presence of the constraint = 0 and the con∂ˆ straint forces and moments it requires introduces a term µ ∂u A on the right-hand side Fc · 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 322 978 0 521 87483 0 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body of Lagrange’s equations. This event is (reassuringly) similar to what arose previously in the case of a single particle and a system of particles. 10.5.3 Consequences of Lagrange’s Prescription and Integrable Constraints Let us now explore some of the consequences of (10.20). Suppose we can express the integrable constraint as u6 − f (t) = 0. Then, Fc · ∂ v̄ ∂ω + Mc · A = µδ6A. A ∂ u̇ ∂ u̇ As a result, the constraint force and moment will contribute to only 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! 10.5.4 Potential Energy and Conservative Forces and Moments Suppose that a potential energy is associated with the rigid body: U = U q1 , . . . , q6 = Û u1 , . . . , u6 . The conservative forces and moments associated with this potential energy are Fcon = − 3 ∂U i a, ∂qi i=1 Mcon = −uQ = − 3 ∂U i g. ∂qi+3 i=1 Now suppose that we have chosen a new set of coordinates. Then, by setting −U and µ = 1 in (10.19), we can show that Fcon · = ∂ v̄ ∂ω ∂ Û + Mcon · A = − A . A ∂ u̇ ∂ u̇ ∂u In summary, the conservative forces and moments appear in a familiar form on the right-hand side of Lagrange’s equations of motion. 10.5.5 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 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.5 Lagrange’s Equations of Motion: A Second Form 323 moments is P = Fc · v̄ + Mc · ω − U̇ ∂ − U̇, =µ − ∂t (10.21) where we have assumed that the constraint force and constraint moment are prescribed by using Lagrange’s prescription. Consequently, if the integrable constraint is time independent, ∂∂t = 0, and the only applied forces and moments are conservative, then, with the help of (10.21), we can surmise that Ṫ = P = −U̇. 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 (10.21), a straightforward set of manipulations is needed: P = (Fcon + Fc ) · v̄ + (Mcon + Mc ) · ω 3 3 ∂ ∂ i i a · v̄ + g ·ω =µ ∂qi ∂qi+3 i=1 i=1 i=1 i=1 3 3 ∂U i ∂U i − a · v̄ + g ·ω ∂qi ∂qi+3 3 3 3 3 ∂U ∂ i ∂ ∂U i+3 i+3 i =µ q̇ + q̇ q̇ + q̇ − ∂qi ∂qi+3 ∂qi ∂qi+3 i=1 i=1 6 ∂U B q̇ − ∂qB i=1 i=1 6 ∂ q̇A ∂qA A=1 B=1 ∂ − U̇. =µ ˙ − ∂t The final result, P = µ ˙ − ∂∂t − U̇, is what was used in writing (10.21). =µ 10.5.6 Summary Suppose we have a rigid body for which all of the applied forces and moments are conservative: F = Fc − ∂U , ∂ x̄ M = Mc − uQ . In addition, suppose that there are two constraints on the motion of the rigid body: = u6 − f (t) = 0, f · v̄ + h · ω + e = 0. The first of these constraints is integrable, whereas the second is nonintegrable. We assume that the constraint forces and moments associated with them are prescribed 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 324 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body using Lagrange’s prescription: Fc = µ1 3 ∂ i a + µ2 f, ∂qi Mc = µ1 i=1 3 ∂ gi + µ2 h. ∂qi+3 (10.22) i=1 Turning to Lagrange’s equations of motion (10.16), we find that the equations would read ∂T d ∂T ∂U − A = − A + QA. (10.23) dt ∂ u̇A ∂u ∂u As the only nonconservative forces acting on the body are due to the constraint forces and constraint moments, QA are simply QA = Fc · ∂ v̄ ∂ω + Mc · A ∂ u̇A ∂ u̇ (A = 1, . . . , 6) . Using a Lagrangian L = T − U and using prescriptions (10.22) for Fc and Mc , we find that (10.23) can be expressed in the following form: d ∂L ∂L ∂ v̄ ∂ω − B = µ2 f · B + h · B , dt ∂ u̇B ∂u ∂ u̇ ∂ u̇ d ∂L ∂L ∂ v̄ ∂ω − 6 = µ1 + µ2 f · 6 + h · 6 . dt ∂ u̇6 ∂u ∂ u̇ ∂ u̇ In these equations, B = 1, . . . , 5. 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. 10.6 Lagrange’s Equations of Motion: Approach II We examine Approach II applied to Lagrange’s equations for a rigid body whose motion is constrained. Specifically, we parallel the discussion of Subsection 10.5.6 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 [28]. The resulting form of Lagrange’s equations is the one most commonly used in engineering and physics. First, we assume that the six coordinates u1 . . . , u6 are chosen such that the integrable constraint = 0 can be simply written as = u6 − f (t). Imposing this constraint, we can calculate the constrained kinetic T̃ and potential Ũ energies of the system. The former will be a function of u1 , . . . , u5 , u̇1 , . . . , u̇5 , and t, whereas the latter will be a function of u1 , . . . , u5 and t. The nonintegrable constraint π2 = 0 can be written in terms of the new coordinates and their velocities: π2 = 5 B=1 pBu̇B + e. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.7 Rolling Disks and Sliding Disks 325 Here p1 , . . . , p5 and e are functions of t and u1 , . . . , u5 . We will use Lagrange’s prescription to specify the associated constraint forces and constraint moments [see (10.22)]. 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 d dt B=1 ∂ T̃ ∂ u̇B pBu̇B + e = 0, − ∂ T̃ ∂ v̄ ∂ω = F· B +M· B B ∂u ∂ u̇ ∂ u̇ =− ∂ Ũ ∂ v̄ ∂ω + Fanc · B + Manc · B + µ2 pB. ∂uB ∂ u̇ ∂ u̇ (10.24) 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: F = Fanc + Fc + Fcon , M = Manc + Mc + Mcon . For instance, Fanc are the applied nonconservative forces. The Magnus force FM = mBω × v̄, which is used in studies of the flight path of a baseball, is a good example of such a force. 10.7 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 (cf. Figure 10.2) or slides on a smooth horizontal plane. It shall be assumed that the inertia tensor J0 has the representation JO = λ (E1 ⊗ E1 + E2 ⊗ E2 ) + λ3 E3 ⊗ E3 . (10.25) 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 λ3 = 2λ and λ = mR , 4 but we do not use these substitutions in order to enable an easier tracking of the algebraic manipulations. Preliminaries The location of the center of mass of the disk can be parameterized by use of a set of Cartesian coordinates: x̄ = 3 i=1 xi Ei . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 326 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body E3 g e2 O ψ E1 e2 E2 e1 e1 X̄ φ e1 P Figure 10.2. A circular disk moving with one point in contact with a horizontal plane. The rotation tensor of the disk is described by a 3–1–3 set of Euler angles: ω = ψ̇E3 + θ̇e1 + φ̇e3 . For this set of Euler angles, we recall, from (6.33) in Subsection 6.8.2, that the Euler basis vectors gi have the representations g1 = E3 sin(φ) sin(θ) cos(φ) sin(θ) cos(θ) e1 − sin(φ) 0 e2 . g2 = e1 = cos(φ) g3 = e3 0 0 1 e3 (10.26) In addition, the Euler angles are subject to the restrictions ψ ∈ [0, 2π), θ ∈ (0, π), and φ ∈ [0, 2π). Physically, the angle θ represents the inclination angle of the disk. When this angle is π2 , the disk is vertical. On the other hand, 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 Subsection 8.6.3, the motion of the rolling disk is subject to three constraints because the velocity vector of the point of contact P is zero: vP = v̄ + ω × πP = 0. (10.27) Earlier, we showed how these equations lead to the following three constraints [see (8.21)]: π1 = 0, π2 = 0, π3 = 0. We also found that the third of these constraints was integrable: = x3 − R sin(θ) = 0, (10.28) 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.7 Rolling Disks and Sliding Disks 327 whereas the remaining constraints were nonintegrable.∗ It should be clear that we express all three constraints on the rolling disk in the form f · v̄ + h · ω = 0, and this facilitates the use of Lagrange’s prescription for Fc and Mc . Coordinates and Energies Motivated by the presence of the integrable constraint x3 = R sin(θ), we now define a set of coordinates: u1 = x1 , u4 = θ, u2 = x2 , u5 = φ, u3 = ψ, u6 = x3 − R sin(θ). (10.29) Given values of u1 , . . . , u6 , we can uniquely invert relations (10.29) to determine xi and the Euler angles: x2 = u2 , x3 = u6 + R sin u4 , x1 = u1 , ψ = u3 , θ = u4 , φ = u5 . (10.30) 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 v̄ = u̇1 E1 + u̇2 E1 + u̇6 + Ru̇4 cos u4 E3 , ω = u̇3 E3 + u̇4 e1 + u̇5 e3 . (10.31) We also note that ω does not depend on u6 or its time derivative. We shall need the derivatives of v̄ and ω with respect to the coordinates u1 , . . . , u5 . With the help of (10.31), these can be obtained in a straightforward manner: ∂ v̄ = E1 , ∂ u̇1 ∂ v̄ = R cos (θ) E3 , ∂ u̇4 ∂ v̄ = E2 , ∂ u̇2 ∂ v̄ = 0, ∂ u̇5 ∂ v̄ = 0, ∂ u̇3 ∂ v̄ = E3 , ∂ u̇6 (10.32) and ∂ω = 0, ∂ u̇1 ∂ω = e1 , ∂ u̇4 ∂ω = 0, ∂ u̇2 ∂ω = e3 , ∂ u̇5 ∂ω = E3 , ∂ u̇3 ∂ω = 0. ∂ u̇6 (10.33) These vectors will soon feature in Lagrange’s equations of motion. Given the inertia tensor (10.25), representations (10.31), and the expressions for ω · ek given in (10.5), an easy calculation shows that the (unconstrained) kinetic ∗ This was established by use of Frobenius’ theorem in one of the exercises at the end of Chapter 8. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 328 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body energy of the disk has the representation 2 m 2 ẋ1 + ẋ22 + u̇6 + Rθ̇ cos (θ) T= 2 λ 2 λ 2 2 3 ψ̇ sin (θ) + θ̇2 + φ̇ + ψ̇ cos(θ) . + 2 2 In addition, the potential energy of the disk is U = 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 πi , we find Fc = µ3 E3 + µ1 E1 + µ2 E2 , Mc = µ3 −R cos(θ)g2 + µ1 R cos(θ) cos(ψ)g1 − R sin(θ) sin(ψ)g2 + R cos(ψ)g3 + µ2 R cos(θ) sin(ψ)g1 + R sin(θ) cos(ψ)g2 + R sin(ψ)g3 . The expressions for the dual Euler basis we use here can be inferred from (6.34) in Section 6.8. If we express the dual Euler basis vectors gi in terms of ei and expand the expressions for Mc , we will find that Mc = πP × Fc . This implies, not surprisingly, that Fc and Mc are equipollent to a force R = Fc acting at the instantaneous point of contact P. The E3 component of this force is the normal force, whereas the remaining components constitute the friction force. Thus Lagrange’s prescription yields a physically reasonable set of constraint forces and constraint moments. Imposing the Integrable Constraint If we impose the integrable constraint, then u6 = 0 and the resulting constrained potential and kinetic energies are Ũ = mgR sin(θ), T̃ = λ 2 2 1 m 2 ẋ1 + ẋ22 + ψ̇ sin (θ) + λ + mR2 cos2 (θ) θ̇2 2 2 2 2 λ3 φ̇ + ψ̇ cos(θ) . + 2 It is important to notice that T̃ = T̃(x1 , x2 , ẋ1 , ẋ2 , θ, φ̇, θ̇, ψ̇). In other words, u6 has been eliminated. The Rolling Disk’s Equations of Motion Lagrange’s equations of motion for the rolling disk can now be easily obtained. We first note that the resultant force F acting on the disk is composed of a constraint 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.7 Rolling Disks and Sliding Disks 329 force and a conservative force: F = Fc − mgE3 . The resultant moment M consists entirely of the constraint moment: M = Mc Using (10.6), and with the help of (10.32) and (10.33), we find that d dt ∂T ∂ ẋ1 − ∂T ∂x1 = 1 , ∂T ∂T − = ∂ ẋ2 ∂x2 d ∂T ∂T = − dt ∂ ψ̇ ∂ψ ∂T d ∂T − = dt ∂ θ̇ ∂θ d ∂T ∂T = − dt ∂ φ̇ ∂φ d ∂T ∂T − = dt ∂ u̇6 ∂u6 d dt 2 , 3 , 4 , 5 , (10.34) 6 . In these equations, 1 = F · E1 + M · 0 = µ1 , 2 = F · E2 + M · 0 = µ2 , 3 = F · 0 + M · E3 = R cos(θ) (µ1 cos(ψ) + µ2 sin(ψ)) , 4 = F · R cos(θ)E3 + M · e1 = −mgR cos(θ) + R sin(θ) (µ2 cos(ψ) − µ1 sin(ψ)) , 5 = F · 0 + M · e3 = µ1 R cos(ψ) + µ2 R sin(ψ), 6 = F · E3 + M · 0 = µ3 − mg. Evaluating the partial derivatives of T in (10.34) and imposing the integrable constraint u6 = 0, we find, with some minor rearranging, that mẍ1 = µ1 , mẍ2 = µ2 , d λ sin2 (θ)ψ̇ + λ3 φ̇ + ψ̇ cos(θ) cos(θ) = 3 , dt λθ̈ + λ3 φ̇ + ψ̇ cos(θ) ψ̇ sin(θ) − λψ̇ 2 sin(θ) cos(θ) = 4 , d λ3 φ̇ + ψ̇ cos(θ) = 5 , dt m d2 (R sin(θ)) = µ3 − mg. dt2 (10.35) 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 330 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body These equations are supplemented by the constraints, x3 − R sin(θ) = 0, π1 = 0, π2 = 0, (10.36) 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 (10.35) and (10.36) are not readily integrated numerically. In particular, they cannot immediately be written in the form ẏ = f(y), which is required for most numerical integrators – such as those based on a Runge–Kutta scheme. 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 P. Consequently, Ṫ = Fc · vP − mgE3 · v̄. However, vP = 0, and mgE3 · v̄ = d dt (mgx3 ); consequently, d (T + mgx3 ) = 0. dt Because E = T + mgx3 , 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 [6], Chaplygin [36], and Korteweg [113] in the late 19th century (see [19, 43, 107, 157]). Unfortunately, as with Routh integral (9.35) 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 (10.35) and (10.36). Specifically, one sets µ1 = µ2 = 0 and ignores the constraints π1 = 0 and π2 = 0. It is instructive, however, to use Approach II. For the sliding disk there is no nonintegrable constraint, and the constraint force and constraint moment are prescribed by use of Lagrange’s prescription, so (10.24) simplifies to d dt ∂ T̃ ∂ u̇A − ∂ T̃ ∂ Ũ ∂ v̄ ∂ω = − A + Fc · A + Mc · A . A ∂u ∂u ∂ u̇ ∂ u̇ Here, A = 1, . . . , 5, Fc = µ3 E3 , and Mc = −Re1 × Fc . (10.37) 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.8 Lagrange and Poisson Tops 331 Using (10.29) in conjunction with (10.37), we find that the resulting governing equations are mẍ1 = 0, mẍ2 = 0, d λ sin2 (θ)ψ̇ + λ3 φ̇ + ψ̇ cos(θ) cos(θ) = 0, dt sin(2θ) = −mgR cos(θ), 2 d λ3 φ̇ + ψ̇ cos(θ) = 0. dt λθ̈ + λ3 φ̇ψ̇ sin(θ) + (λ3 − λ) ψ̇ 2 (10.38) Notice that we could also use these equations to arrive at the equations of motion for the rolling disk. First, we need to supplement the constraints and, second, we need to append the constraint forces and moments associated with the integrable constraints to the right-hand side of (10.38). The equations governing the motion of the sliding disk are clearly far simpler than those for the rolling disk. Indeed, these equations can be written in the form ẏ = f(y), where y is a column vector with 10 rows. These equations can then be integrated numerically by use of a standard numerical integrator. Any numerical simulations of these equations should ensure that the linear momentum G, angular momenta H · e3 and H · E3 , and total energy E of the sliding disk are conserved. Configuration Manifold The configuration manifold M for the rolling disk and the sliding disk are identical: M = E2 ⊕ SO(3). Here, x1 and x2 are coordinates for E2 and the Euler angles are coordinates for SO(3). That is, both disks have five degrees-of-freedom and five generalized cordinates. The kinematical line-element ds is 2 T̃ dt, ds = m where T̃ is the constrained kinetic energy defined earlier. You should notice that the nonintegrable constraints on the rolling disk do not affect the configuration manifold. This is identical to a situation we encountered earlier with the single particle subject to nonintegrable constraints. 10.8 Lagrange and Poisson Tops There are two classical problems in rigid body dynamics involving spinning tops. In the first, known as the Lagrange top, the axisymmetric rigid body is free to rotate 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 332 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body e3 e1 e2 E3 O g X̄ E2 E1 Figure 10.3. The Lagrange top: an axisymmetric rigid body that is free to rotate about a fixed point O. The image on the left-hand side is a portrait of Joseph-Louis Lagrange. about a fixed point O under the action of a gravitational force (see Figure 10.3). This problem was discussed by Lagrange as an illustration of his equations of motion in 1789.∗ A few years later, Poisson [173] considered the same problem except now the apex of the top was free to move on a horizontal surface (see Figure 10.4). 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 motions for the two tops; rather, we focus on choices of coordinates for them and briefly discuss how their equations of motion can be found. 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 O is x̄ = x0 + L1 e1 + L2 e2 + L3 e3 . Here, Lk are constants. Further, we assume that the rotation tensor of the top is parameterized by a set of 3–1–3 Euler angles.† That is, the angular velocity vector of the top is ω = ψ̇E3 + θ̇e1 + φ̇e3 . You should notice that the singularities of these Euler angles occur when E3 = ±e3 , that is, when θ = 0 or π and the top is vertical. ∗ † See Section IX.34 of the Second Part of Lagrange’s Mécanique Analytique [121]. This set of Euler angles has been used extensively in the present chapter and is discussed in Subsection 6.8.2. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.8 Lagrange and Poisson Tops 333 e3 g e1 X̄ e2 E3 E2 O P E1 Figure 10.4. The Poisson top: an axisymmetric rigid body that moves with one point P in contact with a horizontal surface. Motivated by the presence of three integrable constraints xO = 0, we now define a set of coordinates: 3 1 2 3 4 u = ψ, u = θ, u = φ, u = x1 − Lkek · E1 , k=1 u = x2 − 5 3 Lkek · E2 , u = x3 − 6 3 k=1 Lkek · E3 . (10.39) k=1 In these equations, ek are functions of the Euler angles, but the lengthy expressions are not recorded here. Given values of u1 , . . . , u6 , we can uniquely invert relations (10.39) to determine xi and the Euler angles: x1 = u4 + 3 x2 = u5 + Lkek · E1 , 3 k=1 x3 = u + 6 3 Lkek · E2 , k=1 Lkek · E3 , ψ = u1 , θ = u2 , φ = u3 . (10.40) k=1 Using (10.40), we can easily compute that v̄ = u̇4 E1 + u̇5 E2 + u̇6 E3 + ω × 3 Lkek. k=1 ω = u̇1 E3 + u̇2 e1 + u̇3 e3 . (10.41) You should notice that v̄ will depend on the Euler angles and their time derivatives. This is entirely due to our choice of coordinates. In place of x̄ and the three Euler angles, we are using the Euler angles and the three components of xO as our coordinates. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 334 June 12, 2008 Lagrange’s Equations of Motion for a Single Rigid Body For future reference, we shall need several vectors. To calculate these vectors it is convenient to recall that ∂∂ω = E3 , ∂∂ωθ̇ = e1 , and ∂∂ω = e3 . First, we calculate ψ̇ φ̇ 3 ∂ v̄ = E3 × Lkek, ∂ ψ̇ k=1 3 ∂ v̄ Lkek, = e1 × ∂ θ̇ k=1 3 ∂ v̄ = e3 × Lkek, ∂ φ̇ k=1 ∂ v̄ = E1 , ∂ u̇4 ∂ v̄ = E2 , ∂ u̇5 ∂ v̄ = E3 , ∂ u̇6 ∂ω = E3 , ∂ u̇1 ∂ω = e1 , ∂ u̇2 ∂ω = e3 , ∂ u̇3 ∂ω = 0, ∂ u̇4 ∂ω = 0, ∂ u̇5 ∂ω = 0. ∂ u̇6 (10.42) and (10.43) These vectors 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 = λa E3 ⊗ E3 + λt (I − E3 ⊗ E3 ) . With the help of (10.41), we find that the kinetic energy of the top has the representation m 4 2 5 2 6 2 T= u̇ + u̇ + u̇ 2 3 4 5 6 + m u̇ E1 + u̇ E2 + u̇ E3 · ω × Lkek + + λt + mL22 + mL23 2 λa + mL21 + mL22 2 ω21 + k=1 λt + mL21 + mL23 2 ω22 ω23 . For convenience, we have not substituted for ωi = ω · ei in terms of the Euler angles and their derivatives. The needed expressions for ωi are given in (10.5) in Section 10.2. The potential energy of the top is 3 3 U = mg u + Lkek · 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 6 u = 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 manifold M for the top is SO(3). A straightforward calculation with the three 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 10.8 Lagrange and Poisson Tops 335 constraints vO · Ek = 0, where O is the fixed point of the top, shows that the con 3 3 straint force Fc = k=1 µkEk and constraint moment Mc = − k=1 Lk ek × Fc . As expected, this system of forces and moments is equipollent to a force Fc acting at O. The constrained kinetic and potential energies are λt + mL21 + mL23 λt + mL22 + mL23 2 T̃ = ω1 + ω22 2 2 λa + mL21 + mL22 + ω23 , 2 3 Ũ = mg Lkek · E3 . k=1 It is not too difficult to see that 2T̃ = HO · ω and the angular momentum HO can be easily found by use of the parallel axis theorem. We have kept Lk distinct and nonzero to help elucidate some of our kinematical developments. If we now impose the standard assumptions that L1 = 0 = L2 , then our results for T̃ and Ũ = mgL cos(θ) are in accord with those found in many dynamics textbooks. The Poisson Top: Coordinates, Constraints, and Energies For the Poisson top, it is convenient to choose a different coordinate system: u1 = ψ, u = x1 , 4 u = x2 , 5 u2 = θ, u3 = φ, u = x3 − 6 3 Lkek · E3 . (10.44) k=1 It is left to the reader to compare (10.44) with (10.39) and to compute ∂∂u̇v̄A and ∂∂u̇ωA . 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 P is the point of contact of the top with the horizontal surface, shows that the 3 constraint force Fc = µ1 E3 and constraint moment Mc = − k=1 Lk ek × µ1 E3 . As expected, this force and this moment are equipollent to a force Fc acting at P. The development of expressions for T and U follow in a similar manner to those discussed previously for the Lagrange top, and so their constrained expressions are merely summarized: λt + mL21 + mL23 λt + mL22 + mL23 2 T̃ = ω1 + ω22 2 2 λa + mL21 + mL22 m 2 ẋ + ẋ22 , + ω23 + 2 2 1 3 Ũ = U = mg Lkek · E3 . k=1 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 336 June 12, 2008 Exercise 10.1 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 manifold M. Indeed, the configuration manifold of this top is identical to that for the sliding disk. Comments on the Equations of Motion We can find the equations of motion for the Lagrange and Poisson tops by using Approach II and without calculating ∂∂u̇v̄A and ∂∂u̇ωA . 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, and finally the applied forces acting on these tops are conservative. Thus the form of Lagrange’s equations of motion featuring the Lagrangian L̃ = T̃ − Ũ can be used. In the interest of brevity, explicit equations of motion are not given here: They will be similar to the earlier results for the disks. For both tops, it is easy to see that the total energy E is conserved. The Poisson top also features linear momenta conservations: G · E1 and G · E2 . If we now impose the conditions, L1 = L2 = 0 for both tops, then we find that the Lagrange top features two angular momenta conservations: HO · E3 and HO · e3 , whereas the Poisson top features the angular momenta conservations H · E3 and H · e3 . 10.9 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. 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 and angular momentum. As demonstrated in the examples discussed in Chapter 9, with some insight and patience, F = mv̄˙ and M = Ḣ can lead to a tractable set of equations from which the motion of the rigid body can be inferred. EXERCISES 10.1. Consider the mechanical system shown in Figure 10.5. It consists of a rigid body of mass m that is free to rotate about a fixed point O. A vertical gravitational force mgE1 acts on the body. The inertia tensor of the body relative to its center of mass C is J0 = λ1 E1 ⊗ E1 + (λ − mL20 )(E2 ⊗ E2 + E3 ⊗ E3 ). The position vector of the center of mass X̄ of the body relative to O is L0 e1 . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 Exercise 10.1 337 ψ O g E2 θ E3 X̄ E1 e1 Figure 10.5. A whirling axisymmetric rigid body. (a) To parameterize the rotation tensor of the body, we use a set of 1–3–1 Euler angles: g1 = E1 , g2 = e3 , and g3 = e1 . Show that ω = ψ̇E1 + θ̇e3 + φ̇e1 = φ̇ + ψ̇ cos(θ) e1 + θ̇ sin(φ) − ψ̇ sin(θ) cos(φ) e2 + θ̇ cos(φ) + ψ̇ sin(θ) sin(φ) e3 . (b) Because the point O is fixed, we define the coordinates u4 , . . . , u6 to be the coordinates of O: x̄ = u4 E1 + u5 E2 + u6 E3 + L0 e1 , and we choose u1 = ψ, u2 = θ, and u3 = φ. With this choice of the coordinates, show that ∂ v̄ = E1 × L0 e1 , ∂ u̇1 ∂ v̄ = e3 × L0 e1 , ∂ u̇2 ∂ v̄ = 0, ∂ u̇3 ∂ v̄ = E1 , ∂ u̇4 ∂ v̄ = E2 , ∂ u̇5 ∂ v̄ = E3 , ∂ u̇6 ∂ω = E1 , ∂ u̇1 ∂ω = e3 , ∂ u̇2 ∂ω = e1 , ∂ u̇3 ∂ω = 0, ∂ u̇4 ∂ω = 0, ∂ u̇5 ∂ω = 0. ∂ u̇6 (10.45) and (10.46) These results won’t be needed in the remainder of this problem. (c) Derive expressions for the unconstrained potential U and kinetic T energies of the rigid body. These expressions should be functions of the coordinates u1 , . . . , u6 and, where appropriate, their time derivatives. (d) In what follows, the motion of the rigid body is subject to four constraints: φ̇ = ω · g3 = 0, v̄ − ω × (L0 e1 ) = 0. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 338 Exercises 10.1–10.2 Assuming the joint at O is frictionless, give prescriptions for the constraint force and moment 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 T̃ = 1 λ1 ψ̇ 2 cos2 (θ) + λ(θ̇2 + ψ̇ 2 sin2 (θ)) , 2 Ũ = mgL0 cos(θ). (10.47) In addition, using Approach II, write Lagrange’s equations of motion for the rigid body: ∂ L̃ d ∂ L̃ ∂ L̃ d ∂ L̃ − − = 0, = 0, (10.48) dt ∂ θ̇ ∂θ dt ∂ ψ̇ ∂ψ where L̃ is the Lagrangian. (h) Argue that the solutions ψ(t) and θ(t) to (10.48) 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. 10.2. Consider the simple model for an automobile shown in Figure 10.6. 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 include the masses and inertias of the wheels, suspension, engine, and occupants. Interest is restricted to the case in which 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 vQ · e2 = 0, (10.49) where πQ = −L1 e1 − L2 e2 is the position vector of Q relative to the center of mass X̄ of the rigid body. Constraint (10.49) is often known as Chaplygin’s constraint (see [151, 165, 186]). e2 e1 E2 e1 φ E1 X̄ O E1 Q Rigid body of mass m Figure 10.6. A rigid body model for an automobile moving on a horizontal plane. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 Exercises 10.2–10.3 339 (a) Assume that the rigid body is performing a fixed-axis rotation through an angle φ about E3 , that the motion of its center of mass is planar, and that (10.49) holds. Using parameterizations of x̄ 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. (c) Using Lagrange’s prescription, what are the constraint forces Fc and moments Mc acting on the rigid body? (d) Show that the motion of the rigid body is governed by the equations mẍ1 = −µ4 sin(φ), mẍ2 = µ4 cos(φ), λ3 φ̈ = −µ4 L1 , ẋ1 sin(φ) − ẋ2 cos(φ) = −L1 φ̇, (10.50) where xi = x̄ · Ei . (e) Show that (10.50) 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. 10.3. As shown in Figure 10.7, a circular rod of mass m, length L, and radius R slides on a smooth horizontal plane. We assume that the inertia tensor J0 has the representation JO = λ (E1 ⊗ E1 + E3 ⊗ E3 ) + λ2 E2 ⊗ E2 . The sole external applied force acting on the rod is gravitational: −mgE3 . E3 g E2 O e3 Circular cylinder X̄ E1 e1 e2 Horizontal plane Figure 10.7. A circular rod moving on a horizontal plane. (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? 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 340 Exercises 10.3–10.4 (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 ∂∂u̇v̄A and ∂∂u̇ωA . 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 λ m 2 ẋ1 + ẋ22 + ẋ23 + θ̇ sin(φ) − ψ̇ sin(θ) cos(φ) 2 2 2 λ2 θ̇ cos(φ) + ψ̇ sin(θ) sin(φ) + 2 2 λ φ̇ + ψ̇ cos(θ) , + 2 where x̄ = x1 E1 + x2 E2 + x3 E3 . (d) Show that the two constraints on the motion of the rigid body can be written in the form v̄ · E3 = 0, ω · g3 = 0. (e) Using Lagrange’s prescription, what are the constraint force Fc and 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: mẍ1 = 0, mẍ2 = 0, λ2 θ̈ = 0, λψ̈ = 0. (10.51) (g) Show that the six Lagrange’s equations of motion for the rod also yield solutions for the constraint force and constraint moment: Fc = mgE3 , Mc = −λ2 ψ̇ θ̇ sin(θ)g3 . By expressing g3 in terms of e1 , you should also be able to establish a simple expression for Mc . (h) Give physical interpretations of the solutions to (10.51) and show that they allow the cylinder to have motions where ω is not constant. 10.4. As shown in Figure 9.8, a sphere of mass m and radius R is free to move on a rough inclined plane. Here, we reexamine this problem, which was discussed 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 P. You will need to recall that vP = v̄ + ω × πP , (10.52) where v̄ is the velocity vector of the center of mass X̄ of the sphere, and πP is the position vector of P relative to X̄. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 Exercises 10.4–10.5 341 (a) Using a set of 3–1–3 Euler angles, show that the slip velocities of the sphere are vs1 = ẋ1 − Rθ̇ sin(ψ) + Rφ̇ sin(θ) cos(ψ), vs2 = ẋ2 + Rθ̇ cos(ψ) + Rφ̇ sin(θ) sin(ψ), where vP = vs1 E1 + vs2 E2 and v̄ = 3k=1 ẋkEk. (b) What is the constraint on a sliding sphere? Give prescriptions for the constraint force Fc and 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 forces Fc and moments Mc that enforce these constraints. (d) If the unconstrained kinetic energy of the sphere is T= m 2 mR2 2 ψ̇ + θ̇2 + φ̇2 + 2ψ̇ φ̇ cos(θ) + ẋ + ẋ22 + ẋ23 , 5 2 1 then 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 (10.52) and using balances of linear and angular momentum, show that 5 mv̇s1 = 1 + F f · E1 + mg sin(β), 2 5 mv̇s2 = 1 + F f · E2 , 2 where F f is the friction force acting on the sphere. If the sphere is rolling, then what is F f ? 10.5. Consider a rigid body of mass m that is free to rotate about a fixed point O. The position vector of the center of mass X̄ of the rigid body relative to O is x̄ − xO = L0 e3 , where L0 is a constant. The inertia tensor of the body relative to its center of mass is J = 3k=1 λkek ⊗ 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 Trot 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 moment Mc acting on the rigid body. (d) In terms of the Euler angles and a set of Cartesian coordinates xk = x̄ · Ek, prescribe 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) , 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 342 Exercises 10.5–10.6 where 1 = u4 , 2 = u5 , 3 = u6 . Notice that the generalized coordinates for this rigid body are the Euler angles. (e) Calculate the following 12 vectors: ∂ v̄ , ∂ u̇A ∂ω ∂ u̇A (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 + mL0 ë3 . What is the corresponding result for Mc ? (h) Prove that the total energy E of the rigid body is conserved. 10.6. In this exercise, we reconsider the satellite problem discussed earlier in Section 10.4. Here, we employ a coordinate system that is popular in the satellite dynamics community (see, for example, [164, 187, 203]). Referring to Figure 10.1, to parameterize x̄, we use a spherical polar coordinate system, R, θ, and β, where β is the latitude and θ is the longitude. For this coordinate system, we define the following unit vectors: er = cos(θ)E1 + sin(θ)E2 , eR = cos(β)er + sin(β)E3 , eβ = cos(β)E3 − sin(β)er , eθ = cos(θ)E2 − sin(θ)E1 . 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)E3 + E3 ⊗ E3 . The rotation tensor Q2 will be parameterized by a set of 1–2–3 Euler angles. The first of the Euler angles corresponds to a counterclockwise rotation through an angle ν1 about the vector er . Similarly, the second rotation corresponds to a counterclockwise rotation through an angle ν2 about cos ν1 eθ + sin ν1 E3 and the last rotation is about e3 through a counterclockwise angle ν3 . 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 Exercises 10.6–10.7 June 12, 2008 343 (a) Starting from the representation x̄ = ReR , show that the velocity vector v̄ of the center of mass of the satellite has the representation v̄ = ṘeR + R cos(β)θ̇eθ + Rβ̇eβ . (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 ω = θ̇E3 + ν̇1 er + ν̇2 cos ν1 eθ + sin ν1 E3 + ν̇3 e3 . (10.53) (d) Show that the angular velocity vector of the satellite has the representation ω = ω1 e1 + ω2 e2 + ω3 e3 , where ω1 = θ̇ sin(ν1 ) sin(ν3 ) − cos(ν1 ) sin(ν2 ) cos(ν3 ) + ν̇2 sin(ν3 ) + ν̇1 cos(ν2 ) cos(ν3 ), ω2 = θ̇ sin(ν1 ) cos(ν3 ) + cos(ν1 ) sin(ν2 ) sin(ν3 ) + ν̇2 cos(ν3 ) − ν̇1 cos(ν2 ) sin(ν3 ), ω3 = θ̇ cos(ν1 ) cos(ν2 ) + ν̇1 sin(ν2 ) + ν̇3 . (e) Suppose the coordinates chosen for the satellite are u1 = R, u2 = θ, u3 = β, u4 = ν1 , u5 = ν2 , and u6 = ν3 . With this choice, give expressions for the 12 vectors ∂ v̄ ∂ω , (A = 1, . . . , 6) . ∂ u̇A ∂ u̇A (f) The potential energy associated with the gravitational force and moment can be found in Equation (10.11). With the choice of u1 , . . . , u6 , it can be shown that ∂U = 0. Using this result, show that ∂T is conserved. Prove that ∂θ ∂ θ̇ this conservation corresponds to conservation of HO · E3 . (g) Using Lagrange’s equations, establish the equations governing the motion of the satellite. 10.7. Consider a body that is free to move about one of its material points O, which is fixed (cf. Figure 9.7). This system was discussed previously in Exercises 9.15 and 10.5. (a) Calculate the constrained Lagrangian L̃ and the equations of motion [see (10.8)]: d ∂ L̃ ∂ L̃ − i = 0, (10.54) i dt ∂ γ̇ ∂γ where γ 1 = ψ, γ 2 = θ, and γ 3 = φ are the 3–1–3 set of Euler angles used to parameterize Q. 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 12, 2008 344 Exercises 10.7–10.8 (b) Show that equations of motion (10.54) are equivalent to ḢO · gi = MO · gi , where g1 , g2 , g3 are the Euler basis vectors. (c) Compare equations of motion (10.54) with (9.36), which were discussed in Exercise 9.14. 10.8. Recall the definition of the force A given by (10.4): A = F · ∂ v̄ ∂ω + M · A. A ∂ u̇ ∂ u̇ 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 that∗ P· ∂vP ∂ v̄ ∂ω = P · A + ((xP − x̄) × P) · A . A ∂ u̇ ∂ u̇ ∂ u̇ How can this identity be used to simplify the calculation of A? ∗ This identity is used extensively in many treatments of analytical mechanics (see, for example, Kane and Levinson [105] and Kane et al. [106]). 5:0 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 PART FOUR SYSTEMS OF RIGID BODIES 345 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 346 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 11 978 0 521 87483 0 June 2, 2008 Introduction to Multibody Systems 11.1 Introduction In this chapter, we examine systems of rigid bodies. Our goal is simply to discover how our previous developments can be used to obtain the equations of motion for these systems. As you might imagine, the equations of motion can be very complex, and judicious component selections from the balance laws are often needed to extract the equations of motion. This is illustrated with the example of a rate gyroscope. 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 [86], 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 its orbits of 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.∗ The Dynabee (or Rollerball) was invented in the early 1970s by Archie Mishler [141] and features spinning a rotor to speeds in excess of 5000 rpm by carefully rotating an outer casing (housing). This novel gyroscopic device is discussed in an exercise at the end of this chapter. 11.2 Balance Laws and Lagrange’s Equations of Motion We are interested in examining the dynamics of a system of N rigid bodies, B1 , . . . , BN . For the body BK , we denote its mass by mK , its center of mass by X̄K , the position of its center of mass by x̄K , its inertia tensor relative to X̄K by JK , and its angular velocity by ωK . Using these quantities, we can define the kinetic energy TK , angular momentum HK , and linear momentum GK in the usual manner: 1 1 GK = mK x̄˙ K , HK = JK ωK , TK = GK · x̄˙ K + HK · ωK . 2 2 ∗ A history of the Galileo spacecraft’s mission was recently written by Meltzer [140]. 347 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 348 Introduction to Multibody Systems Setting K = 1, . . . , N we can obtain expressions for the kinematic quantities associated with each of the N rigid bodies. Constraints on, and potential energies for, the system of rigid bodies can be inferred from our developments in Chapter 8 and we shall adopt them here. For the system of rigid bodies, we can define the center of mass X̄. The position vector of this point is defined by N 1 x̄ = N mK x̄K . K=1 mK 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: N N ˙ G= mK x̄ = mK x̄˙ K . K=1 K=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 [(7.15)] applied to each individual rigid body: HA = N (HK + (x̄K − xA) × mK x̄˙ 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 = T1 + · · · + TN . The resulting expression for T can be tremendously complex, and often symbolic manipulation packages, such as Mathematica or Maple, are used. We find the governing equations for the system of N rigid bodies by using the balances of linear and angular momenta for each of the rigid bodies: m1 x̄¨ 1 = F1 , . . . , mN x̄¨ N = FN , Ḣ1 = M1 , . . . , ḢN = MN . (11.1) Here, FK is the resultant force acting on the Kth rigid body and MK is the resultant moment relative to the center of mass X̄K of the rigid body. Equations (11.1) yield differential equations for the motion of the bodies and expressions for the constraint forces and moments. As shown in [30], balance laws (11.1) are equivalent to Lagrange’s equations of motion for the system of rigid bodies. For these equations, we denote by u1 , . . . , u6N the coordinates chosen to parameterize the position vectors x̄K and rotation tensors QK . Then Lagrange’s equations of motion are ∂T d ∂T − A = A (11.2) (A = 1, . . . , 6N) , A dt ∂ u̇ ∂u 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 11.3 Two Pin-Jointed Rigid Bodies 349 where v̄ = x̄˙ and A = N FK · K=1 ∂ v̄K ∂ωK + MK · A . ∂ u̇A ∂ u̇ (11.3) It is emphasized that using either (11.1) or (11.2) will lead to equivalent equations of motion for the system of rigid bodies. 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 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 from which the constraint forces and moments are easily determined, and the other from which motion of the system can be computed. The proof of this can be found in [30], 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 (11.3). 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 simply represented as u6N−C+1 = f 1 (t), . . . , u6N = f C(t). We also suppose that the constraint forces FcK and constraint moments McK associated with these constraints are prescribed by Lagrange’s prescription. It can be shown that N K=1 ∂ v̄K ∂ωK FcK · A + McK · A = 0 ∂ u̇ ∂ u̇ N (A = 1, . . . , 6N − C) . (11.4) K=1 That is, the constraint forces and constraint moments do not contribute to the generalized forces 1 , . . . , 6N−C. As a result, our expressions (11.3) for the forces A are identical to those found in other treatments of Lagrange’s equations of motion for systems of rigid bodies that are available.∗ 11.3 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 Subsection 8.6.4 and is illustrated in Figure 8.8. Our goal is to outline how the equations of motion for this system can be established by 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 coordinates to parameterize x̄1 , a set of Euler angles γ 1 , γ 2 , γ 3 to parameterize Q1 , and the angle θ to parameterize the relative rotation tensor Q2 QT1 . The angle θ represents the rotation of B2 relative to B1 and the axis of rotation ∗ See, for example, Baruh [14], Greenwood [80], and Kane [104]. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 350 Introduction to Multibody Systems associated with this rotation is e3 : Q2 QT1 = L (θ, e3 ). It is prudent to define the following seven coordinates: uk = x̄1 · Ek, uk+3 = γ k, u7 = θ, where k = 1, 2, 3. We also assume that πP1 = L1 e1 , πP1 = (L2 ) 2 e1 , where L1 and L2 are constants. We also note that the corotational basis vectors for B1 are {e1 = 1 e1 , e2 = 1 e2 , e3 = 1 e3 }, and the basis vectors {2 e1 , 2 e2 , 2 e3 = e3 } corotate with B2 . From the coordinates chosen, we have the following representations: v̄1 = 3 ẋi Ei , i=1 v̄2 = v̄1 + ω1 × L1 e1 + ω1 + θ̇e3 × (L2 ) 2 e1 , ω1 = 3 γ̇ i gi , i=1 ω2 = 3 γ̇ i gi + θ̇e3 . i=1 These results can then be used to compute the constrained kinetic energy T̃ 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 Equation (8.23). 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 ∂ T̃ ∂ T̃ (11.5) − A = A (A = 1, . . . , 7) , A dt ∂ u̇ ∂u where A = 2 K=1 FK · ∂ v̄K ∂ωK + MK · A . ∂ u̇A ∂ u̇ It is emphasized that Fc1 , Fc2 , Mc1 , and Mc2 do not contribute to 1 , . . . , 7 . Indeed, it is a good exercise to verify that Fc1 · ∗ ∂ v̄1 ∂ v̄2 ∂ω1 ∂ω2 + Fc2 · A + Mc1 · A + Mc2 · A = 0 A ∂ u̇ ∂ u̇ ∂ u̇ ∂ u̇ (A = 1, . . . , 7) . Technically, we have shown that Approach II works only for a single rigid body, a single particle, and a system of particles. As mentioned previously, the corresponding result for a system of rigid bodies was established in [30], and we make use of it here. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 11.4 A Single-Axis Rate Gyroscope 351 You may have noticed that these identities are special cases of (11.4). Once the applied forces and moments are prescribed and T̃ calculated, the seven second-order differential equations of (11.5) 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 ẍ1 = F1 , m2 ẍ2 = F2 , Ḣ1 = M1 , Ḣ2 = M2 . (11.6) This would yield a set of coupled equations for the constraint reactions µ1 , . . . , µ5 [cf. (8.23)] and ü1 , . . . , ü7 . The advantages of Lagrange’s equations of motion in its automatic selection of linear combinations of Equations (11.6) should be obvious. 11.4 A Single-Axis Rate Gyroscope As shown in Figure 11.1, a single-axis rate gyroscope is mounted on a platform P. The gyroscope consists of a gimbal G that is free to rotate relative to P through an angle α about the axis e1 and a rotor R that is free to rotate relative to G through an angle β about the axis e2 . In addition to the pair of bearings connecting the gimbal and platform, a spring of stiffness K and a dashpot with a damping coefficient c are suspended between G and P. t3 r3 R G r2 β t1 r1 e1 t2 e2 α K c P Figure 11.1. Schematic of a single-axis rate gyroscope. This devices consists of a rotor R that is suspended by a gimbal G over a platform P. When the platform rotates about t3 the rotor will want to dip, and this tendency is resisted by the spring and dashpot shown in the figure. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 352 Introduction to Multibody Systems The principle of operation of the gyroscope is as follows: Suppose that the platform is given an angular velocity about the vertical t3 direction; then the rotor will seek to rotate about a horizontal direction e1 . This tendency to rotate will be counterbalanced by the spring, but will result in an angle of deflection α = α0 . As we shall subsequently see, ∗ α0 ∝ . Consequently, if the gyroscope is calibrated, then a measurement of α0 can be used to determine . In practice, however, α0 is sensitive to many other effects that can often serve to corrupt the measurement of . A wonderful discussion of these and other matters on rate gyroscopes can be found in Arnold and Maunder [8]. Here, we follow some of their developments and outline how (11.7) can be established. Before doing so, we note that three single-rate gyroscopes mounted at right angles to each other on a rigid body B can be used to measure ωi = ω · ei . Integrating ωi (t) with the help of equations of the form (7.28) can then be used to determine the attitude Q of B. Such a system is known as a “strap-down inertial navigation system.” Kinematics By way of notation, the bases vectors {t1 , t2 , t3 }, {e1 , e2 , e3 }, and {r1 , r2 , r3 } corotate with P, G, and R, respectively. All of these bases are right-handed, and, with the help of a fixed Cartesian basis {E1 , E2 , E3 }, can be used to define the rotation tensor of each of the three rigid bodies. In the analysis that follows, it is assumed that there is negligible friction between the rotor and the gimbal. The position vectors of the center of mass of the gimbal and rotor are assumed to be coincident and denoted by x̄. The rotation tensor QP = 3i=1 ti ⊗ Ei of the platform is parameterized by a set of 3–1–3 Euler angles. Denoting the angular velocity vector of the platform by ω p, the gimbal by ωg , and the rotor by ωr , with the help of (6.32), we can easily establish representations for these three angular velocity vectors: ω p = φ̇t3 + θ̇t1 + ψ̇E3 , ωg = ω p + α̇e1 , ωr = ωg + β̇e2 . We denote the inertia tensors of the rotor and gimbal relative to their centers of mass by Jr = λ1 (I − r2 ⊗ r2 ) + λ2 r2 ⊗ r2 , Jg = ρ1 e1 ⊗ e1 + ρ2 e2 ⊗ e2 + ρ3 e3 ⊗ e3 , ∗ See, in particular, (11.9). 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 11.4 A Single-Axis Rate Gyroscope 353 respectively. Notice that the symmetry of the rotor allows its inertia tensor to be easily written in either the corotational basis for the gimbal or rotor: Jr = λ1 (I − r2 ⊗ r2 ) + λ2 r2 ⊗ r2 = λ1 (I − e2 ⊗ e2 ) + λ2 e2 ⊗ e2 . The angular momentum of the rotor relative to its center of mass and the angular momentum of the rotor and gimbal relative to their coincident centers of mass are Hr = Jr ωr , H = Hr + Hg = Jg ωg + Jr ωr = ((ρ1 + λ1 ) e1 ⊗ e1 + (ρ2 + λ2 ) e2 ⊗ e2 + (ρ3 + λ3 ) e3 ⊗ e3 ) ωg + λ2 β̇e2 . The symmetry of the rotor is partially responsible for the simplicity of the expression for H. Constraint Forces and Constraint Moments There are five constraints on the motion of the gimbal relative to the rotor: v̄r = v̄g , (ωr − ωg ) · r1 = 0, (ωr − ωg ) · r3 = 0. We assume that the bearings on the gimbal supporting the rotor are frictionless, and this allows us to use Lagrange’s prescription (8.22) to prescribe the constraint forces and constraint moments: Fcr = −Fcg = 3 µkek, k=1 Mcr = −Mcg = µ4 e1 + µ5 e3 . It is a good exercise to show that the force Fcr and moment Mcr are equipollent to two reaction forces R1 and R2 that act on opposite ends of the rotor. This implies that Lagrange’s prescription yields physically reasonable results. Governing Equations for the Gyroscope We can use a balance of angular momentum of the rotor relative to its center of mass to establish an interesting conservation. First, we note that Mr = Mcr and this 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 354 Introduction to Multibody Systems moment has no component in the e2 = r2 direction. It is also easy to see that Ḣr · e2 = 0. Consequently, h = Hr · e2 is conserved, where h = Jr ωr · r2 = λ2 β̇ + λ2 ω p · e2 . (11.7) You should notice from this conservation that we can change the relative speed β̇ of the rotor by changing the angular velocity of the platform in the e2 = r2 direction. Generally, it is assumed that β̇ ω p · e2 . Consequently, β̇ is assumed to be constant during the operation of the gyroscope. Examining the resultant force and moment on the gimbal–rotor system, we easily see that the constraint moments at the bearings where the gimbal is attached to the platform have no moment in the e1 direction. Consequently, it is prudent to consider the balance of angular momentum Ḣ = M for the gimbal–rotor system and examine the e1 component of this equation. After approximating the moment in the e1 direction that is due to the spring and dashpot as −(Kα + cα̇)Ae1 , where A is a constant whose dimensions are length squared, we find a differential equation governing the angle α: (λ1 + ρ1 ) α̈ + KAα + cAα̇ = λ2 β̇ω3 − (λ1 + ρ1 ) ω̇1 where ω p = 3 i=1 + (λ2 + ρ2 − (λ3 + ρ3 )) ω2 ω3 , (11.8) ωi ti . Operation of the Gyroscope Differential equation (11.8) informs us how α changes when the platform is moved. It should be transparent that the angle α is sensitive to all of the components of ω p. For an operational gyroscope, the rotor is usually spun to a sufficiently high speed β̇, that λ2 β̇ω3 dominates the right-hand side of (11.8). Consequently, we approximate (11.8) by (λ1 + ρ1 ) α̈ + KAα + cAα̇ = λ2 β̇ω3 . (11.9) Now suppose initially that α = 0 (the rotor is horizontal), β̇ is large, and ω3 = 0. It should be easy to see from (11.9) that the rotor will remain horizontal. Now if ω3 is given a constant nonzero value ω30 , then α(t) changes from its rest value and will vary with time. It is left as an exercise to integrate (11.9) and to observe that, if λ2 β̇ω3 is constant and cA > 0, then the solution α(t) will eventually settle to a constant value α0 . Equation (11.9) can be used to relate this value to the constant angular speed of the platform: KA α0 . ω30 = (11.10) 2λ2 β̇ This equation is the explanation for the governing principle behind the operation of the gyroscope. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercise 11.1 355 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 [8], Crabtree [42], and Ginsberg [71]. 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. 11.5 Closing Comments We have touched here on two examples that illustrate how the equations of motion for a system of rigid bodies can be determined. Fortunately, for many physical examples, Lagrange’s prescription is applicable and Lagrange’s equations can be used to determine the equations of motion: a theme that has been a central thread throughout this book and is extensively exploited in formulations of multibody systems. 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 a further elimination of the constraint forces and moments associated with the nonintegrable constraints. Such equations of motion are beyond the scope of this book, and there is a variety of texts and surveys that cover them, for example, Baruh [14], Greenwood [80], Hamel [87], Karapetyan and Rumyantsev [108], Neı̆mark and Fufaev [151], Papastavridis [167–169], and Udwadia and Kalaba [218], to name but a few. Most other forms of the equations of motion are derivable from Lagrange’s equations of motion (11.2) and so, by use of the developments in this book [and the help of (11.3)], they too can be related to the balances of linear and angular momenta. EXERCISES 11.1. As shown in Figure 11.2, a rigid plate of mass m is attached at a point A by a pin joint to a slender rod. The rod has a length L, one of its ends, O, is fixed, and it rotates about the E3 axis with an angular speed ψ̇ = . (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 −Lψ̇e1 = x̄˙ + ω × πA, ω · g1 = , ω · g3 = 0, (11.11) where x̄ is the position vector of the center of mass of the plate, ω is the angular velocity of the plate, and πA is the position of A relative to the X̄. (b) Which orientations of the plate coincide with the singularities of the Euler angles? (c) Give prescriptions for the forces Fc and moments Mc that ensure that the five constraints are enforced. Illustrate your prescriptions with a free-body diagram. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 356 Exercises 11.1–11.2 E3 Slender rod of length L O E1 g E2 A ψ X̄ e1 e2 Rectangular plate Figure 11.2. A rigid plate that is pin jointed at A to a slender rod. The rod is supported by bearings at O that enable it to freely rotate about E3 . Note that a gravitational force −mgE3 acts on the plate. (d) Show that the mechanical power of the constraint forces and moments is nonzero if = 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 λi ei ⊗ ei , (11.12) i=1 then establish an expression for the angular momentum HA and show that HA = JAω unless = 0.∗ (f) Outline how you would establish the equations of motion of the plate. 11.2. Following (2.1), the impulse momentum form of the balances of linear and angular momentum for a rigid body are t1 G (t1 ) − G (t0 ) = F(τ)dτ, t0 t1 H (t1 ) − H (t0 ) = M(τ)dτ. (11.13) t0 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 t0 = t − σ t. ∗ Hint: You may find the identity a × (b × c) = (a · c) · b − (a · b) · c helpful. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercises 11.2–11.3 357 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), u̇(t), t) at time t: [[f ]] = lim f (u(t + σ), u̇(t + σ), t + σ) − f (u(t − σ), u̇(t − σ), t − σ) , σ→0 where σ > 0. (a) Suppose a set of coordinates u1 , . . . , u6 is used to parameterize the motion of the rigid body. Then, what are the conditions under which (( )) (( )) ∂ v̄ ∂ω = 0, = 0, (11.14) ∂ u̇K ∂ u̇K where K = 1, . . . , 6? Give a physical interpretation of the conditions you have found. (b) Assuming that (11.14) holds, by taking appropriate limits show that (11.13) can be written as (( )) ∂T ˆ K, = (11.15) ∂ q̇K where K = 1, . . . , 6 and the generalized impulsive forces are t+σ t+σ ∂ v̄ ∂ω ˆ K = lim F(τ)dτ · K (t) + lim M(τ)dτ · K (t) . σ→0 σ→0 ∂ u̇ ∂ u̇ t−σ t−σ For assistance in establishing (11.15), calculations (10.17) might be helpful. An alternative derivation of (11.15) for a system of particles can be found in Section 15.2 of Synge and Griffith [207]. (c) Show how the six equations of (11.15) can be used to determine the dynamics of a rod impacting a smooth horizontal surface. 11.3. Consider, once again, the robotic arm shown in Figure 7.9. As discussed in an earlier exercise, the robotic arm has a mass m and moment of inertia tensor relative to its center of mass, J0 = λ1 E1 ⊗ E1 + λ2 E2 ⊗ E2 + λ3 E3 ⊗ E3 . A system of motors, which is not shown in Figure 7.9, 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 = e1 = cos(ψ)E1 + sin(ψ)E2 through an angle θ. Another system of actuators prescribes the motion of the point P of the robotic arm. The position vector of the center of mass of the arm relative to P is x̄ − xP = L e3 . 2 (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, g1 , g2 , and g3 , are not 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 358 Exercises 11.3–11.4 orthonormal. For which positions of the robotic arm are these vectors not defined? (b) Noting that the motion of the point 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 = λ1 θ̇e1 + λ2 ψ̇ sin(θ)e2 + λ3 ψ̇ cos(θ)e3 . (d) Draw a free-body diagram of the robot arm. In your free-body diagram, include the gravitational force −mgE3 . (e) Show that the force needed to actuate the motion of the center of mass is mL 2ψ̇ θ̇ cos(θ) + ψ̈ sin(θ) e1 2 mL 2 mL 2 ψ̇ sin(θ) cos(θ) − θ̈ e2 − (θ̇ + ψ̇ 2 sin2 (θ))e3 . + 2 2 (f) Show the moment Mc needed to actuate a slewing maneuver of the robotic arm is Fact = mgE3 + mü + mL mL2 e3 × (gE3 + ü) + ψ̇ 0 θ̇0 cos(θ)e2 2 2 Mact = − mL2 2 ψ̇ 0 sin(θ) cos(θ)e1 + ω × H. 4 During the slewing maneuver, θ̇ = θ̇0 , ψ̇ = ψ̇ 0 , and consequently ω is constant. (g) Show that the time rate of change in the total energy E of the robotic arm is equal to the work done by Fact and Mact : Ė = Fact · u̇ + Mact · ψ̇E3 + Mact · θ̇e1 . 11.4. As mentioned earlier, a recently invented wrist exerciser, which is known as the Dynabee, Powerball, or Rollerball, features a heavy rotor that can roll without slipping on a circular track [82, 94]. The track is part of a housing that can be held and rotated. As analyzed in [82], 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. As illustrated in Figure 11.3, we define a set of vectors {et1 , et2 , et3 } to corotate with the track and the set of vectors {er1 , er2 , er3 } to corotate with the rotor. We also define an intermediate set of vectors a1 = er1 , a2 = et3 × er1 , a3 = et3 . We emphasize that these vectors are not corotational. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 Exercise 11.4 359 et3 = a3 α̇ er3 T R γ̇ er1 = a1 et1 et2 Figure 11.3. Schematic of a Dynabee showing the circular track T and the rotor R. 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 [82]. (a) The rotation tensor Q = 3k=1 etk ⊗ Ei of the housing is assumed to be parameterized by a set of 3–1–3 Euler angles (ψ, θ, φ). Referring to Figure 11.3, 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 = er1 ⊗ et1 + er2 ⊗ et2 + er3 ⊗ et3 . (11.16) Using the aforementioned sets of Euler angles, establish representations for the angular velocity vectors of the rotor ωr and the track ωr . (c) Referring to Figure 11.4, we assume that the rotor has two points in contact with 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 er1 and eπ1 , then show that Ra , β = tan−1 Rt where Ra 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 imply that v̄r = v̄t , (ωr − ωt ) × eπ1 = 0, 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 2, 2008 360 Exercises 11.4–11.5 et3 = a3 eπ3 R Q T β er1 = a1 X̄ P T eπ1 Figure 11.4. 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 R and housing T are assumed to coincide at the point X̄. In this figure, the radius Ra of the axle of the rotor and the angle µ have been exaggerated. where v̄r and v̄t are the velocity vectors of the centers of mass of the rotor and housing, respectively. In addition, show that these constraints imply that Rt γ̇ = − Ra α̇. (11.17) As in [82], we assume that µ ≈ 0. (e) Assuming that the motion of the housing 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 R and T . (f) Argue that the eπ1 component of the balance of angular momentum for the rotor relative to center of mass X̄ gives a differential equation for the motion of the rotor. (g) When the rotor is steadily rotating, ψ̇ = α̇. Explain why (11.17) then implies that the rotor will be spinning faster than the housing. 11.5. 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 − t0 , e3 (t1 ) = e3 (t0 ) . (11.18) 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 Exercise 11.5 361 (a) Show that the rotation of the rigid body during the time interval t1 − t0 has the representation 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 (t0 ).∗ (c) Give an example of a motion of a rigid body where (11.18) holds.† (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 Section 123 of Kelvin and Tait [109] and that was independently rediscovered by Goodman and Robinson [73] and Levi [126], 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. ∗ † June 2, 2008 You may wish to recall (7.22) from Exercise 7.4. For assistance with this matter, the papers of Montgomery [144] and O’Reilly [158] might be helpful. 13:58 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Appendix: Background on Tensors A.1 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, the introductory 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 them, the primary sources were Casey [26, 28], Chadwick [35], and Gurtin [84]. A.2 Preliminaries: Bases, Alternators, and Kronecker Deltas Here, Euclidean three-space is denoted by E3 . For this space, we define a righthanded fixed orthonormal basis {E1 , E2 , E3 }. We also use another right-handed orthonormal basis {p1 , p2 , p3 }. This basis is not necessarily fixed.∗ 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 , b2 , b3 } is orthonormal if bi · bk = 0 when i = k and bi · bk = 1 when i = k. Further, a set of vectors {b1 , b2 , b3 } ∗ 362 This basis serves to represent corotational bases, other time-varying bases such as {er , eθ , E3 }, and fixed bases such as {E1 , E2 , E3 }. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 A.3 The Tensor Product of Two Vectors 363 is right-handed if the following scalar triple product is positive: [b1 , b2 , b3 ] = b3 · (b1 × b2 ). Because we will be using copious amounts of dot products, it is convenient to define the Kronecker delta δik: 6 1, i = j δij = . 0, i = j Clearly, pi · pk = δik. We also define the alternating (or Levi–Civita) symbol 123 = 312 = 231 = 1, 213 = 132 = 321 = −1, ijk 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 [pi , p j , 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 α and β are any two scalars, and a, b, and c are any three vectors, then (αa + βb) ⊗ c = α (a ⊗ c) + β (b ⊗ c) , c ⊗ (αa + βb) = α (c ⊗ a) + β (c ⊗ b) . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 364 Background on Tensors 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. 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 (αa + βb) = αAa + βAb, 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 O: Ia = a, Oa = 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 Akj pk ⊗ p j , (A.2) j=1 k=1 where Aik = (Apk) · pi , aj = 3 Akj pk, A= k=1 3 aj ⊗ pj . j=1 Here, Aik are known as the components of A relative to the basis {p1 , p2 , p3 }. Initially, it is convenient to interpret tensors by use of the representation A = 3 j=1 a j ⊗ p j . In this light, A transforms pk 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 Api is a vector. Consequently, it can be written as a linear combination of the basis vectors p1 , p2 , and p3 : Ap1 = A11 p1 + A21 p2 + A31 p3 , Ap2 = A12 p1 + A22 p2 + A32 p3 , Ap3 = A13 p1 + A23 p2 + A33 p3 . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 A.5 A Representation Theorem for Second-Order Tensors 365 The order of the indices i and k for the scalars Aik is important. Given any vector b = 3i=1 bi pi , we find that Ab = 3 A(bj p j ) = j=1 = 3 3 bj (Ap j ) j=1 bj 3 j=1 Akj pk = k=1 3 3 bj (Akj pk) j=1 k=1 ⎛ ⎞ 3 3 3 =⎝ Akj pk ⊗ p j ⎠ bi pi j=1 k=1 i=1 ⎞ ⎛ 3 3 =⎝ Akj pk ⊗ p j ⎠ b. j=1 k=1 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 (Akj pk ⊗ p j ) b. j=1 k=1 As this result holds for all vectors b, and A is assumed to be a linear transformation, we conclude (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 3 Akj pk ⊗ p j . j=1 k=1 The transformation induced by A on a vector b is readily computed: ⎛ Ab = ⎝ 3 3 ⎞ Akj pk ⊗ p j ⎠ b = j=1 k=1 3 3 Akj bj pk. j=1 k=1 If we define c = Ab, then it is easy to see that the components ci = c · pi of c are ci = 3 j=1 Aij bj . (A.3) 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 366 Background on Tensors Expressed in matrix notation, (A.3) should be familiar: ⎡ ⎤ ⎡ A11 c1 ⎢ ⎥ ⎢ ⎣c2 ⎦ = ⎣A21 c3 A31 A22 ⎤⎡ ⎤ b1 ⎥⎢ ⎥ A23 ⎦ ⎣b2 ⎦ . A32 A33 A12 A13 (A.4) b3 It should be clear from (A.4) that the identity tensor has the representation I = 3 i=1 pi ⊗ pi . For familiar choices of the basis p1 , p2 , p3 , we can use this result to show that I = E1 ⊗ E1 + E2 ⊗ E2 + E3 ⊗ E3 = er ⊗ er + eθ ⊗ eθ + E3 ⊗ E3 = eR ⊗ eR + eθ ⊗ eθ + eφ ⊗ eφ . Several other examples of this representation I = 3i=1 pi ⊗ 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 Aikpi ⊗ pk, i=1 k=1 B= 3 3 Bikpi ⊗ pk, C= i=1 k=1 3 3 Cikpi ⊗ pk. i=1 k=1 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 Cik = 3 Aij 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 C12 C13 A11 A12 A13 B11 B12 B13 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣C21 C22 C23 ⎦ = ⎣A21 A22 A23 ⎦ ⎣B21 B22 B23 ⎦ . C31 C32 C33 A31 A32 A33 B31 B32 B33 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, 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 A.6 Functions of Second-Order Tensors 367 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 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 = d ⊗ c, (a ⊗ b + c ⊗ d)T = 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 = T T B A . If A = AT , then A is said to be symmetric. On the other hand, A is skewsymmetric if A = −AT . Any second-order tensor B can be decomposed into the sum of a symmetric second-order tensor and a skew-symmetric second-order tensor: 1 1 B + BT + B − BT . 2 2 Skew-symmetric tensors that 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 B= AT = 3 3 Aki pi ⊗ pk = i=1 k=1 3 3 Aikpk ⊗ pi i=1 k=1 where A= 3 3 Aikpi ⊗ pk. i=1 k=1 As a second-order tensor A is symmetric if A = AT , we find that A = AT if Aik = Aki . This implies that a symmetric second-order tensor has six independent components. Similarly, A is skew-symmetric if AT = −A: A = −AT if Aik = −Aki . 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 368 Background on Tensors Notice that this result implies that a skew-symmetric second-order tensor has three independent components. 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 second-order tensor. Given a second-order tensor A, the invariants, IA , IIA , and IIIA , of A are defined as∗ [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 = p2 , and c = p3 , where {p1 , p2 , p3 } is a right-handed orthonormal basis for E3 , then we have the intermediate results [p1 , p2 , p3 ] = 1, [Ap1 , p2 , p3 ] + [p1 , Ap2 , p3 ] + [p1 , p2 , Ap3 ] = 3 (Api ) · pi , i=1 [Ap1 , Ap2 , Ap3 ] = det(A). Using these results, we find that the trace of A is tr(A) = 3 (Api ) · pi = A11 + A22 + A33 . i=1 ∗ Our discussion here closely follows Chadwick [35]. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 A.6 Functions of Second-Order Tensors 369 A similar result holds for the trace of a matrix. Further, we find that det(A) = [Ap1 , Ap2 , Ap3 ] = 3 3 3 Ai1 Aj2 Ak3 [pi , p j , pk] i=1 j=1 k=1 = 3 3 3 ijk Ai1 Aj2 Ak3 . (A.7) i=1 j=1 k=1 Recall that the determinant of a 3×3 matrix whose components are Bik is ⎛⎡ B11 ⎜⎢ det ⎝⎣B21 B31 B12 B22 B32 B13 ⎤⎞ ⎥⎟ B23 ⎦⎠ = 3 3 3 ijk Bi1 Bj2 Bk3 . (A.8) i=1 j=1 k=1 B33 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 ⎜⎢ det(A) = det ⎝⎣A21 A31 A12 A13 ⎤⎞ A22 ⎥⎟ A23 ⎦⎠ . A32 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 370 Background on Tensors 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 , λ2 , and λ3 . On expanding the equation, we find that λ3 − IA λ2 + IIA λ − IIIA = 0. That is, λ1 λ2 λ3 = det(A) = IIIA , λ1 λ 2 + λ2 λ 3 + λ1 λ 3 = 1 (tr(A)2 − tr(A2 )) = IIA , 2 λ1 + λ2 + λ3 = tr(A) = IA . It should be noted that if a tensor is positive-definite, then all of its eigenvalues are strictly positive. The eigenvector (or characteristic direction or principal direction) of a tensor A is the vector u that satisfies Au = λu, (A.9) where λ is a root of the characteristic equation. 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. Let us now consider a simple example: J0 = ma2 mb2 mc2 E1 ⊗ E1 + E2 ⊗ E2 + E3 ⊗ E3 . 12 12 12 2 2 2 By inspection, this tensor has the eigenvalues ma , mb , and mc and the correspond12 12 12 ing eigenvectors are E1 , E2 , and E3 . We also note that, if a, b, and c are nonzero, then J0 is positive-definite. A.7 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 secondorder 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 , p2 , p3 }, any third-order tensor A has the representation A= 3 3 3 i=1 j=1 k=1 Aijkpi ⊗ p j ⊗ pk. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 A.7 Third-Order Tensors 371 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 : = 3 3 3 ijk pi ⊗ p j ⊗ pk i=1 j=1 k=1 = p1 ⊗ p2 ⊗ p3 + p3 ⊗ p1 ⊗ p2 + p2 ⊗ p3 ⊗ p1 − p2 ⊗ p1 ⊗ p3 − p1 ⊗ p3 ⊗ p2 − p3 ⊗ p2 ⊗ p1 . This tensor has some useful features. First, if A is a symmetric tensor, then it is a productive exercise to show that [A] = 0. Second, suppose that c = 3k=1 ckpk is a vector; then c = 3 3 3 ijk pi ⊗ p j ck i=1 j=1 k=1 = c3 (p1 ⊗ p2 − p2 ⊗ p1 )+c2 (p3 ⊗ p1 − p1 ⊗ p3 ) + c1 (p2 ⊗ p3 − p3 ⊗ p2 ), (A.11) which is a skew-symmetric tensor. The fact that 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 = − [C] . 2 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 = −c, C = c21 (p2 ⊗ p1 − p1 ⊗ p2 ) + c32 (p3 ⊗ p2 − p2 ⊗ p3 ) + c13 (p1 ⊗ p3 − p3 ⊗ p1 ), then, with the help of (A.10)1 , 1 c = − [C] = c21 p3 + c13 p2 + c32 p1 . 2 We also note the important result that Ca = (−c)a = c × a. This result allows us to replace cross products with tensor products, and vice versa. It is prudent to provide some examples here. First, suppose we are given a skew-symmetric tensor EXAMPLES. = (E2 ⊗ E1 − E1 ⊗ E2 ) . Then, [] = −2E3 , 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 372 Background on Tensors and we conclude that the axial vector of is E3 . It is useful to verify that ( (E2 ⊗ E1 − E1 ⊗ E2 )) a = E3 × a for all vectors a. Similarly, given a vector γE1 , we can compute that γE1 = γE2 ⊗ E3 − γE3 ⊗ E2 . It follows from this result that γE1 × b = (γE3 ⊗ E2 − γE2 ⊗ E3 ) b for all vectors b. A.8 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 skew-symmetric 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) = ±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, E1 ⊗ E1 + E2 ⊗ E2 − E3 ⊗ 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)p + 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 A.9 Derivatives of Tensors 373 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 and the inertia 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 + λ2 u2 ⊗ u2 + λ3 u3 ⊗ u3 , where Aui = λi ui . That is, ui is an eigenvector of A with the eigenvalue λi . This representation is often known as the spectral decomposition. A.9 Derivatives of Tensors We shall often encounter derivatives of tensors. Suppose a tensor A has the representation A= 3 3 Aikpi ⊗ pk i=1 k=1 and that the components of A and the vectors pi are functions of time. The time derivative of A is defined as Ȧ = 3 3 i=1 k=1 Ȧikpi ⊗ pk + 3 3 Aikṗi ⊗ pk + i=1 k=1 3 3 Aikpi ⊗ ṗk. i=1 k=1 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 Ȧ = θ̇eθ ⊗ E1 − θ̇er ⊗ E2 . It is left as an exercise to show that A is a rotation tensor and that ȦAT is a skewsymmetric tensor. We can also define a chain rule and product rules. Suppose A = A(q(t)), B = B(t), and c = c(t). Then, Ȧ = ∂A q̇, ∂q d (AB) = ȦB + AḂ, dt d (Ac) = Ȧc + Aċ. dt 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 374 Exercises A.1–A.7 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 ∂ψ = pi ⊗ pk. ∂A ∂Aik 3 3 i=1 k=1 In addition, if the vectors pi are constant, then 3 3 ∂ψ ∂ψ T ψ̇ = Ȧik = tr Ȧ . ∂Aik ∂A i=1 k=1 This result is used when we consider constraints on the motions of rigid bodies. EXERCISES A.1. Consider a = E1 and b = E2 . For any vector c = 3i=1 ci Ei , show that (a ⊗ b)c = c2 E1 and c(a ⊗ b) = c1 E2 . A.2. Consider a = E1 and b = E2 . For any vector c = 3i=1 ci Ei , show that (a ⊗ b − b ⊗ a)c = c2 E1 − c1 E2 . A.3. Using the definition of the transpose (A.5), verify that (a ⊗ b)T = b ⊗ a. A.4. If a = 10E1 and b = 5E2 , verify that (a ⊗ b)u = 50E1 (E2 · u) and (b ⊗ a)u = 50E2 (E1 · u). A.5. If a = er and b = eθ , then show that (a ⊗ b)u = er (eθ · u) and (b ⊗ a)u = eθ (er · u). A.6. Using the definition of the alternator, show that −E3 = E2 ⊗ E1 − E1 ⊗ E2 . Where is this result used in rigid body kinematics? Verify that E3 × a = (E2 ⊗ E1 − E1 ⊗ E2 ) a for any vector a. A.7. Show that the following matrix multiplication, ⎡ ⎤ ⎡ ⎤⎡ ⎤ cos(θ) sin(θ) 0 er E1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣eθ ⎦ = ⎣− sin(θ) cos(θ) 0⎦ ⎣E2 ⎦ , ez 0 0 1 E3 can be written as er = PE1 , eθ = PE2 , ez = PE3 , where P = cos(θ) (E1 ⊗ E1 + E2 ⊗ E2 ) + sin(θ) (E2 ⊗ E1 − E1 ⊗ E2 ) + E3 ⊗ E3 = cos(θ)(I − E3 ⊗ E3 ) − sin(θ)E3 + E3 ⊗ E3 . 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Exercises A.8–A.14 375 A.8. Show that the matrix multiplication ⎡ ⎤ ⎡ ⎤⎡ ⎤ eR cos(θ) sin(φ) sin(θ) sin(φ) cos(φ) E1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ eφ ⎦ = ⎣cos(θ) cos(φ) sin(θ) cos(φ) − sin(φ)⎦ ⎣E2 ⎦ eθ − sin(θ) cos(θ) 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 ⊗ E3 + eφ ⊗ E1 + eθ ⊗ E2 . A.9. Verify that the tensor R in the previous exercise is proper-orthogonal. A.10. Give an example that illustrates that tensor multiplication is not commutative, i.e., AB = BA. A.11. Verify that, if B is a second-order tensor, then [B] = 1 , B − BT . 2 A.12. Verify that 1 −ε[c ⊗ d] = − ε[c ⊗ d − d ⊗ c] = d × c, 2 where c and d are any two vectors. A.13. Suppose that A and B are skew-symmetric second-order tensors. Verify that a·b= 1 tr ABT , 2 (A.12) where a and b are the axial vectors of A and B, respectively. 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 1 − ε [AC + CA] = (tr (C) I − C) a. 2 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. 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 376 June 9, 2008 Exercise A.15 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. 21:29 P1: FYX mastercup cuus273/Oliver M. 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O’Reilly 978 0 521 87483 0 June 9, 2008 Index Aircraft stability, 190 Alternating symbol, 363 Alternator, 363, 370 Angular acceleration vector, 211 Angular momentum particle, 5 rigid body relative to a fixed point O, 218 relative to a point A, 218 relative to center of mass, 218, 223 system of particles, 104 Angular velocity tensor, 168 vector, 168, 193, 195, 201, 211 Appell, P., 330 Areal velocity, 5, 27, 47, 66 Atlas, 80 Axial vector, 371 Balance of angular momentum rigid body, 272 Balance of linear momentum particle, 33, 84 rigid body, 272 Barycentric coordinate system, 133 Bernoulli’s equation, 286 Bernoulli, J., 128 Boltzmann, L., 20 Boltzmann–Hamel equations, 355 Bryan angles, 190, 191 Bryan, G. H., 184 Cardan angles, 190, 191 Cardano, G., 184 Cayley, A., 201 Center of mass rigid body, 215 system of particles, 105 Center of oscillation, 297 Changing coordinates, 320 Chaos, 151 Chaplygin integral, 305 Chaplygin sphere, 298 Chaplygin, S. A., 298, 330, 338 Chart, 80 Chasles, M., 209 Christoffel symbols, 94, 123 Components contravariant, 11 covariant, 11 Configuration manifold, 82, 142, 151, 308, 331 Configuration space, 114 Configurations of a rigid body, 206 Conservation angular momentum, 46, 49, 151, 293, 318, 335, 343 energy, 45, 46, 151, 275, 293, 318, 322, 330, 335, 343 linear momentum, 45 Constraints, 74 ball-and-socket joints, 238 Chaplygin, 338 holonomic, 20 ideal, 92 impulsive, 59 independence, 26 integrability, 243 integrability criteria, 23, 32 integrable, 15, 39, 75, 107, 238, 290, 328, 333 rheonomic, 20 scleronomic, 20 multiple, 26, 109, 245, 269 nonholonomic, 21 nonintegrable, 21, 39, 76, 107, 109, 243, 252 piecewise integrable, 23 pin joints, 238 rigid bodies, 237 rolling rigid bodies, 239, 358 sliding rigid bodies, 239 static Coulomb friction, 43, 57, 63, 95 system of particles, 116 systems of, 26, 109, 245, 269 389 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 390 Index Contravariant basis vectors, 11, 115 Coordinate curve, 9 Coordinate surface, 9 Coordinates Cartesian, 6, 105 curvilinear, 9, 115 cylindrical polar, 6, 29 ellipsoidal, 99 generalized, 83, 142–144, 308, 331 helix, 29, 87, 96 parabolic, 28, 92, 99 spherical polar, 7, 28, 95, 105, 315, 342 Coriolis, G. G., 250, 294 Corotational basis, 212 derivative, 169, 178, 226 tensor, 226 vector, 213, 217, 226 Covariant basis vectors, 10, 115 Curvilinear coordinates, 80, 115 D’Alembert, J. Le Rond, 71, 128 D’Alembert’s principle, 103, 128 Degrees-of-freedom, 84, 331 Dual Euler basis, 183, 186, 189, 199, 261, 263, 328 Dual-spin spacecraft, 347 Dumbbell satellite, 140 Dynabee, 347, 358 Euler angles 1–3–1 set, 204, 264, 267 3–1–2 set, 233 3–1–3 set, 188, 251, 265, 267–269, 325, 332, 340, 341, 357, 358 3–2–1 set, 184, 204, 235, 264 3–2–3 set, 198, 230, 339 angular velocity vector, 181 asymmetric set, 191 body-fixed, 204 singularities, 183, 187, 189, 190, 228, 267 space-fixed, 204 switching sets of, 283 symmetric set, 191 twelve sets, 190 Euler basis, 181 Euler tensors, 219, 372 Euler’s disk, 298 Euler’s equations, 278 Euler, L., 34, 163, 171, 204, 207, 304 Euler–Lagrange equations, 130 Euler–Poisson equations, 292 Forces central, 66 conservative, 36, 110, 117, 256, 264, 322 constraint, 39, 108, 116, 248, 291, 320, 328 drag, 285 generalized, 85, 111, 125, 319, 344, 349 generalized impulsive, 357 gravitational, 37, 133, 140, 257, 316 Magnus, 286, 325 spring, 37 Foucault, J. B. L., 4 Frenet triad, 60, 87 Frisbee, 289 Frobenius integrability theorem of, 26, 246, 269 Frobenius, F. G., 24, 26, 245 Galileo spacecraft, 347 Gauss’ principle of least constraint, 103, 129 Gauss, C. F., 128, 193 Gibbs, J. W., 171 Gibbs–Appell equations, 355 Griffin grinding machine, 269 Gyroscope, 351 Hamilton, Sir W. R., 71, 130, 152, 193 Harmonic oscillator, 134 Helicoid, 96 Hertz, H., 20, 82 Hoberman sphere, 193 Impact of a rigid body, 356 Impulse momentum, 33, 356 Inertia tensors, 219, 372 change of bases, 231 parallel axis theorem, 231, 290 principal axes of inertia, 220, 304 Jacobi integral, 159 Jacobi’s criterion, 25 Jacobi, C. G. J., 24, 82, 130, 279 Jellett integral, 304 Jellett, J. H., 304 Jumping, 298 Kelvin, Lord, 361 Kepler frequency, 51, 301 Kepler’s laws, 47 Kepler, J., 47 Kinematical line-element, 121, 142, 144, 151, 308, 331 Kinetic energy, 135 Koenig decomposition, 224, 312 particle, 5 rigid body, 224 system of particles, 104 Koenig, J. S., 224 Korteweg, D. J., 330 Kronecker delta, 362 Lagrange principle of, 40 Lagrange multipliers, 128 Lagrange top, 331 Lagrange’s equations of motion rigid body, 312, 318, 324 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 Index single particle, 71 contravariant form, 94 covariant form, 72, 93 system of particles, 113, 119 contravariant form, 125 covariant form, 125 Lagrange’s prescription, 40, 107, 108, 116, 128, 253, 320 Lagrange, J.-L., 39, 47, 71, 103, 152, 188, 301, 307, 332 Lagrange–D’Alembert principle, 40 Lagrangian, 73, 118 Lagrangian reduction, 48 Levi–Civita, T., 82, 93 Line of nodes, 188 Linear momentum particle, 5 rigid body, 217 system of particles, 105 Mac Cullagh, J., 260 Magnus, H. G., 286 Manifold, 80 Riemannian, 82 Moment potential, 197, 256, 262, 268 Moment-free motion of a rigid body, 279 Moments conservative, 256, 322 constant, 260 constraint, 248, 291, 320, 328 gravitational, 257, 316 torsional spring, 268 Momentum sphere, 281 Motion of a rigid body with a fixed point, 289, 305, 343 N-body problem, 159 Navigation, 194 Newton’s third law, 109, 110, 254 Newton, Sir I., 34, 47, 285 Nondimensionalization, 51, 56 Pendulum and cart, 143 Phase portrait, 51 Planar double pendulum, 157 Poincaré, H., 152 Poinsot, L., 163, 213, 279 Poisson kinematical relations, 292 Poisson top, 235, 331 Poisson, S. D., 213, 331 Principal axes, 220, 277, 304 Principal moments of inertia, 220 Principle of virtual work, 128 Procrustes problem, 194 Quaternions, 193 Reference frame, 3 corotational, 217 391 inertial, 3 Relative angular velocity tensor, 178 vector, 178, 182 Representative particle, 113, 123 Ricci, M. M. G., 82, 93 Riemann, G. F. B., 82 Rigid body motion, 206 Robotic arm, 357 Rollerball (Powerball), 347, 358 Roller coaster, 60 Rolling disk, 228, 251, 269, 325 Rolling rod, 339 Rolling sphere, 250, 294, 300, 340 turntable, 302 Rotation angle of, 173 axis of, 173, 213 Chasles’ theorem, 209 Codman’s paradox, 204, 361 composition of, 172, 211 Rodrigues formula, 201 constant angular velocity motions, 227 direction cosines, 168 Euler parameters, 193 Euler’s formula, 172 Euler’s representation, 171 angular velocity vector, 175 Euler’s theorem, 176, 207 Euler–Rodrigues symmetric parameters, 193, 201 fixed-axis case, 164 Gibbs vector, 192 instantaneous axis of, 213 matrix representation, 173, 185, 193 quaternions, 193 relative, 241, 264 Rodrigues vector, 192 Rodrigues–Hamilton theorem, 204, 260 screw axis of, 213 Routh integral, 305 Routh, E. J., 71, 250, 294 Routh–Voss equations of motion, 71, 78 Routhian reduction, 48 Satellite dynamics, 156, 301, 315, 342 Screw motion, 209 Sliding disk, 228, 325 Sliding sphere, 250, 294, 340 Slip velocity, 297 SO(3), 308 Spectral decomposition theorem, 372 Spinning eggs, 298 Stability of motion, 281 Static friction criterion, 89 Steady orbital motions, 302, 317 Stick–slip, 63, 138 Strap-down inertial navigation system, 351 21:29 P1: FYX mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 392 Index Tait angles, 190, 191 Tait, P. G., 184, 285, 361 Tensor product second-order and vector, 366 third-order and second-order, 370 third-order and vector, 370 two second-order tensors, 364, 366 two vectors, 363 Tensors adjugates, 369 angular velocity, 168 derivatives, 373 determinant, 368 eigenvalues, 370 eigenvectors, 370 invariants, 368 inverses, 369 orthogonal, 372 proper-orthogonal, 166, 176, 372 representation theorem for, 364 rotation, 163, 171 second-order, 364 skew-symmetric, 169, 367 symmetric, 367 symmetric positive-definite, 372 third-order, 370 trace, 368 Three-body problem, 151 Tippe top, 265, 304 Two-body problem, 47, 133 Universal joint, 269 Vehicle dynamics, 107, 338 Voss, A., 71 Wahba problem, 194 Walker, G. T., 240 Wobblestone, 240, 298 Work–energy theorem rigid body, 274 single particle, 46 system of particles, 104 Ziegler, H., 256, 260 21:29