Dynamics of Particles & Rigid Bodies

8/13/2019 Dynamics of Particles & Rigid Bodies

http://slidepdf.com/reader/full/dynamics-of-particles-rigid-bodies 1/225

Lecture Notes on the Dynamics of Particles

and Rigid Bodies

Class Notes for Engineering Mechanics III ME170

and Intermediate Dynamics ME175

Oliver M. O’Reilly

August 18, 2004

Department of Mechanical Engineering

University of California at Berkeley

Berkeley

California 94720-1740

U. S. A.





Preface

These notes are a partial compilation of class notes and exercises for two courses on dynamics,ME170: Engineering Mechanics III and ME175: Intermediate Dynamics, that I havetaught at the Department of Mechanical Engineering at the University of California at Berkeley

over the past decade. Some of the aims of these courses are to give senior and first–year graduatestudents in Mechanical Engineering requisite skills in the area of dynamics of rigid bodies. Thenotes are intended to be a sequel to my book Engineering Dynamics: A Primer which waspublished by Springer–Verlag in 2001.

As the notes are partial, many topics and applications covered in the courses ME170 andME175 are not covered in these notes. I am continually expanding and correcting these notesand warmly welcome your comments and corrections. From ME175 in the Spring Semester 2004,thanks to Joshua Coaplen, Philip Stephanou and Xiao Xiao for their corrections and comments.Thanks are also due to Wayne Huang for providing Figures 8.1 and 8.2.

Oliver M. O’[email protected]

Berkeley

August 2004





Contents

Part I Dynamics of a Single Particle

1 Kinematics of a Single Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Kinematics of a Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Coordinate Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3.1 Cartesian Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3.2 Cylindrical Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3.3 Spherical Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3.4 Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3.5 Some Comments on Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.6 Parameterizations of Particle Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4.1 Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4.2 The Canonical Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4.3 Curvilinear Coordinates and Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4.4 Further Examples of Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Integrable and Non-Integrable Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Kinetics of a Single Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 The Balance Law for a Single Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3 Work and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.4 Conservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.5 Examples of Conservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5.1 Constant Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.5.2 Spring Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.6 Constraint Forces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.6.1 A Single Constraint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.6.2 Two Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27



VIII Contents

2.6.3 Three Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.6.4 Non–Integrable Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.7 Conservations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.7.1 Conservation of Linear Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.7.2 Conservation of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.7.3 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.8 Lagrange’s Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.8.1 Two Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.8.2 A Covariant Form of Lagrange’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.8.3 The Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.8.4 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.9 Lagrange’s Equations in the Presence of Constraints . . . . . . . . . . . . . . . . . . . . . . . . . 362.9.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.9.2 Approach I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.9.3 Approach II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.9.4 A Particle Moving on a Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.10 The Geometry of Lagrange’s Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.10.1 A Particle Subject to a Single Integrable Constraint . . . . . . . . . . . . . . . . . . . 422.10.2 A Particle Subject to Two Integrable Constraints . . . . . . . . . . . . . . . . . . . . . . 432.10.3 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.11 A Particle Moving on a Helix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.11.1 Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.11.2 Various Basis Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.11.3 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.11.4 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.11.5 Balance of Linear Momentum and Lagrange’s Equations . . . . . . . . . . . . . . . 472.11.6 Equations of Motion for the Particle on the Helix. . . . . . . . . . . . . . . . . . . . . . 482.11.7 Equations of Motion for the Particle on a Smooth Helix . . . . . . . . . . . . . . . . 482.11.8 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.11.9 Some Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Part II Dynamics of a System of Particles

3 Kinematics and Dynamics of a System of Particles . . . . . . . . . . . . . . . . . . . . . . . . . 593.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.2 The System of N Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.3 The Single Particle’s Position and Force Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.3.1 Prescription for the Position Vector r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.3.2 Prescription for the Force Vector Φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.4 Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.4.1 Some Correspondances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.5 Lagrange’s Equations: An Unconstrained System of Particles. . . . . . . . . . . . . . . . . . 643.6 Constraints and Constraint Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64



Contents IX

3.7 Conservative Forces and Potential Energies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 663.8 The Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 663.9 Lagrange’s Equations: Constrained System of Particles . . . . . . . . . . . . . . . . . . . . . . . 67

3.9.1 Approach I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.9.2 Approach II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 683.10 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

3.10.1 System I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693.10.2 System II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703.10.3 System III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.10.4 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

3.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

Part III Kinematics and Dynamics of a Single Rigid Body

4 Background on Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 794.1 Preliminaries. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.2 The Tensor Product of Two Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.3 Second–Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

4.3.1 Invariants, Determinants and Traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.3.2 Inverses and Adjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.3.3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.4 A Representation Theorem for Second–Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . 824.4.1 Uses of the Representation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4.5 Third–Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.6 Additional Types of Second–Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

4.6.1 Orthogonal Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 874.6.2 Proper–Orthogonal Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 874.6.3 Symmetric, Positive–Definite Second–Order Tensors . . . . . . . . . . . . . . . . . . . 88

4.7 Derivatives of Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5 Rotation Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.2 Rotation Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.3 Derivatives of a Rotation Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

5.3.1 Corotational Derivatives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.4 Relative Angular Velocity Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955.5 Euler’s Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.5.1 Euler’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.5.2 Remarks on Euler’s Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 985.5.3 The Associated Angular Velocity Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

5.6 Other Representations of a Rotation Tensor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

5.6.1 The Rodrigues Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 995.6.2 The Euler–Rodrigues Symmetric Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . 99



X Contents

5.6.3 Euler Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1005.6.4 Summary of the Angular Velocity Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

5.7 Derivatives of Scalar Functions of Rotation Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . 1065.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

6 Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.2 General Considerations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.3 The Angular Velocity and Angular Acceleration of a Rigid Body . . . . . . . . . . . . . . 1176.4 A Corotational Basis and the Motion of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . 1186.5 Center of Mass and Linear Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

6.5.1 The Center of Mass. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1196.5.2 The Linear Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.5.3 Relative Position Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

6.6 Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1226.7 Euler Tensors and Inertia Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

6.7.1 Euler Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

6.7.2 Inertia Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1246.7.3 Additional Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1246.7.4 Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

6.8 Angular Momentum and an Inertia Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1276.9 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

6.9.1 Comments on the Rotational Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 1296.10 Configuration Manifold of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

6.10.1 A Representation for the Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1316.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

7 Constraints on and Potentials for Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1397.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1397.2 Preliminaries Concerning Two Rigid Bodies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

7.2.1 A Scalar Valued Function of the Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1407.3 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

7.3.1 Inter–Connected Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1427.3.2 Rolling and Sliding Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

7.4 Forces and Moments Acting on a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1457.5 Constraint Forces and Constraint Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

7.5.1 A Rigid Body Rotating about a Fixed Point . . . . . . . . . . . . . . . . . . . . . . . . . . 1477.5.2 A Sphere Rolling or Sliding on an Inclined Plane . . . . . . . . . . . . . . . . . . . . . . 1487.5.3 The Rolling Disk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1497.5.4 The Normality Prescription . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

7.6 Potential Energies and Conservative Forces and Moments . . . . . . . . . . . . . . . . . . . . . 1527.6.1 Constant Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1527.6.2 Spring Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

7.6.3 Central Gravitational Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1547.6.4 Constant Moments Which Are Not Conservative . . . . . . . . . . . . . . . . . . . . . . 156



Contents XI

7.6.5 General Considerations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1567.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

8 Kinetics of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1658.2 Balance Laws for a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1658.3 Work and Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

8.3.1 Proving the Work–Energy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1678.3.2 Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

8.4 Additional Forms of the Balance of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . 1698.4.1 A Direct Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1698.4.2 A Component Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1708.4.3 The Principal Axis Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

8.5 Moment-Free Motion of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1728.5.1 The Symmetric Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1738.5.2 The Axisymmetric Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1738.5.3 The Asymmetric Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

8.5.4 The Momentum Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1748.6 The Baseball and the Football . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1758.7 Motions of Rolling Spheres and Sliding Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1788.8 Closing Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1818.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

9 Lagrange’s Equations of Motion for a Single Rigid Body . . . . . . . . . . . . . . . . . . . 1879.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1879.2 Lagrange’s Equations of Motion: A First Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

9.2.1 Proof of Lagrange’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1889.2.2 The Four Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

9.3 A Satellite Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1919.3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

9.3.2 The Balance Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1939.3.3 Lagrange’s Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1939.3.4 Conservations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

9.4 Lagrange’s Equations of Motion: The Second Form . . . . . . . . . . . . . . . . . . . . . . . . . . 1949.4.1 Changing Coordinates: An Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1969.4.2 Constraints and Constraint Forces and Moments . . . . . . . . . . . . . . . . . . . . . . 1969.4.3 Consequences of Normality and Integrable Constraints . . . . . . . . . . . . . . . . . 1979.4.4 Potential Energy and Conservative Forces and Moments . . . . . . . . . . . . . . . . 1989.4.5 Mechanical Power and Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . 1989.4.6 A Short Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

9.5 Lagrange’s Equations of Motion: A Third Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2009.6 Rolling Disks and Sliding Disks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

9.6.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

9.6.2 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2029.6.3 Imposing the Integrable Constraint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203



XII Contents

9.6.4 The Rolling Disk’s Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2049.6.5 Conservations for the Rolling Disk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2059.6.6 The Sliding Disk’s Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2059.6.7 Configuration Manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

9.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211



Part I

Dynamics of a Single Particle





1

Kinematics of a Single Particle

1.1 Introduction

In this chapter, several kinematical quantities pertaining to the motion of a particle in Euclideanthree-space are introduced. After supplementing these definitions with some results on coordinatesystems, various representations are obtained. The chapter closes with a discussion of constraints.

1.2 Kinematics of a Particle

Consider a single particle of mass m which is moving in Euclidean three–space E . The positionvector of the particle relative to a fixed origin O is denoted by r. This vector is usually consideredto 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, firstand second time derivatives of the position vector:

v = dr

dt , a = dv

dt = d2r

dt2 . (1.1)

It is crucial to note that because r is measured relative to a fixed origin, v and a are the absolutevelocity and acceleration vectors. We shall also use a superposed dot to denote the time derivative.For example, v = r and a = r.

Supplementary to the aforementioned kinematical quantities, we also have the linear momen-tum G of the particle:

G = mv. (1.2)

Further, one has the angular momentum HO of the particle relative to O :

HO = r × mv. (1.3)

Finally, we recall the definition of the kinetic energy T of the single particle:

T = 1

2mv · v. (1.4)



2 1 Kinematics of a Single Particle

The definition of the kinematical quantities that we have introduced are independent of thecoordinate system that is used for E . However, in solving most problems it is crucial to haveexpressions for these quantities in terms of the chosen coordinate system. It is to this issue thatwe now turn.

1.3 Coordinate Systems

Depending on the problem of interest, there are several suitable coordinate systems for E . Themost commonly used systems are Cartesian coordinates x1, x2, x3, cylindrical polar coordi-nates r,θ,z, and spherical polar coordinates R ,φ ,θ. All of these coordinate systems can beconsidered as specific examples of a curvilinear coordinate system q 1, q 2, q 3 for E .

1.3.1 Cartesian Coordinate System

For the Cartesian coordinate system x1, x2, x3 a set of right–handed orthonormal vectors aredefined:

E1, E2, E3

. Given any vector b in

E , this vector has the representation

b =3

i=1

biEi. (1.5)

For the position vector r, we also have (with an abuse of notation)

r =3

i=1

xiEi, (1.6)

where x1, x2, x3 are the Cartesian coordinates of the particle. Because Ei are fixed in bothmagnitude and direction, their time derivatives are zero.

E

2

E3

eθ

er

E

1

r

θr

z

O

Fig. 1.1. Cylindrical polar coordinates r , θ , and z .



1.3 Coordinate Systems 3

1.3.2 Cylindrical Polar Coordinates

A cylindrical polar coordinate system r,θ,z can be defined using a Cartesian coordinate systemas follows:

r =

x21 + x22, θ = tan−1x2

x1

, z = x3, (1.7)

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. (1.8)

In other words, given (x1, x2, x3) then a unique (r,θ,z) exists provided (x1, x2) = (0, 0).Given a position vector r, we can write

r = x1E1 + x2E2 + x3E3

= r(cos(θ)E1 + sin(θ)E2) + zE3

= rer + zE3, (1.9)

where er = cos(θ)E1 + sin(θ)E2.It is convenient to define set of unit vectors er, eθ, Ez:

er = cos(θ)E1 + sin(θ)E2, eθ = cos(θ)E2 − sin(θ)E1, ez = E3. (1.10)

We also notice that er = θeθ, while eθ = −θer. You should also verify that er, eθ, Ez is aright-handed orthonormal basis for E .1

E

2

E3

eθ

eφφ

e R

E

1

r

θr

O

Fig. 1.2. Spherical polar coordinates R, φ, and θ .

1 A basis a1, a2, a3 is right-handed if a3 · (a1 × a2) > 0, and is orthonormal if the magnitude of eachof the vectors ai is 1 and they are mutually perpindicular: a1 · a2 = 0, a2 · a3 = 0 and a1 · a3 = 0.




1.3.3 Spherical Polar Coordinates

A spherical polar coordinate system R,φ,θ can be defined using a Cartesian coordinate systemas follows:

R =

x21 + x2

2 + x23, θ = tan−1

x2

x1

, φ = tan−1

x21 + x2

2

x3

, (1.11)

where θ ∈ [0, 2π) and φ ∈ (0, π). Provided φ = 0, π, we can invert these relations to find

x1 = R cos(θ) sin(φ), x2 = R sin(θ) sin(φ), x3 = R cos(φ). (1.12)

Given a position vector r, we can now write

r = x1E1 + x2E2 + x3E3

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

= ReR, (1.13)

where eR = sin(φ) cos(θ)E1 + sin(φ)sin(θ)E2 + cos(φ)E3.For future purposes, it is convenient to define the right-handed orthonormal set of vectors

eR, eφ, eθ: eR

eφ

eθ

=

cos(θ) sin(φ) sin(θ) sin(φ) cos(φ)

cos(θ) cos(φ) sin(θ) cos(φ) −sin(φ)− sin(θ) cos(θ) 0

E1

E2

E3

. (1.14)

By assuming that θ and φ are functions of time, we find that eR

eφ

eθ

=

0 φ θ sin(φ)

−φ 0 θ cos(φ)

−θ sin(φ) −θ cos(φ) 0

eR

eφ

eθ

. (1.15)

These relations have an interesting form: notice that the matrix is skew–symmetric. We shall seenumerous examples of this later on.

1.3.4 Curvilinear Coordinates

The preceeding examples of coordinate systems can be considered as specific examples of acurvilinear coordinate system. To this end, consider a coordinate system q 1, q 2, q 3 which isdefined by the functions

q 1 = q 1 (x1, x2, x3) ,

q 2 = q 2 (x1, x2, x3) ,

q 3 = q 3 (x1, x2, x3) . (1.16)

We assume that the functions q i are locally invertible:




x1 = x1

q 1, q 2, q 3

,

x2 = x2

q 1, q 2, q 3

,

x3 = x3 q 1, q 2, q 3

. (1.17)

This invertibility implies that given the curvilinear coordinates of any point in E , there is a uniqueset of Cartesian coordinates for this point and vice versa.

The invertibility mentioned earlier usually breaks down at several points in E . For instance,the cylindrical polar coordinate system is not defined when x2

1 + x22 = 0. This set of points

corresponds to the x3 axis.Fixing the value of one of the curvilinear coordinates q 1 say to equal q 10 , we can determine

the values of x1, x2 and x3 such that the equation

q 10 = q 1 (x1, x2, x3) (1.18)

is satisfied. The union of all the points represented by these Cartesian coordinates defines asurface which is known as the q 1 coordinate surface (cf. Figure 1.3). If we move on this surfacewe find that the coordinates q 2 and q 3 will vary. Indeed, the curves on the q 1 coordinate surfacewhich are found by varying q 2 while keeping q 3 fixed are known as q 2 coordinate curves.

More generally, the surface corresponding to a constant value of a coordinate q j is known asa q j coordinate surface. Similarily, the curve obtained by varying the coordinate q j , while fixingthe remaining two curvilinear coordinates, is known as a q j coordinate curve.

O

E

2

E3

E

2a

3a

1

a13 q coordinate curve2 q coordinate curve

1 q coordinate surface

Fig. 1.3. An example of a q 1 coordinate surface. At a point on this surface a1 is normal to the surface,while a2 and a3 are tangent to the surface. The q 1 coordinate surface is foliated by curves of constant q 2

and q 3.

The aforementioned invertibility also implies that the position vector r of any point can beexpressed as a function of the curvilinear coordinates:

r =3

i=1

xi

q 1, q 2, q 3

Ei. (1.19)




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

ai = ∂ r

∂q i

=

3k=1

∂ xk

∂q i Ek. (1.20)

These basis vectors are extremely important. Mathematically, when we take the derivative withrespect to q 2 we fix q 1 and q 3, consequently, a2 points in the direction of increasing q 2. As a result,a2 is tangent to a q 2 coordinate curve. Generalizing this statement, we see that ai is tangent toa q i coordinate curve.

You should notice that we can express the relationship between the covariant basis vectorsand the Cartesian basis vectors in a matrix form:

a1

a2

a3

=

∂ x1∂q1

∂ x2∂q1

∂ x3∂q1

∂ x1∂q2

∂ x2∂q2

∂ x3∂q2

∂ x1∂q3

∂ x2∂q3

∂ x3∂q3

E1

E2

E3

. (1.21)

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

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 asfollows:

a1 =

3i=1

∂ q 1

∂xi

Ei, a2 =

3i=1

∂ q 2

∂xi

Ei, a3 =

3i=1

∂ q 3

∂xi

Ei. (1.22)

Geometrically, ai is normal to a q i coordinate surface. However, as in the case of the covariantbasis vectors, the contravariant basis vectors are not necessary unit vectors, nor do they form anorthonormal basis for E .

Using the chain rule of calculus, it can be shown that ai

·aj = δ ij , where δ ij is the Kronecker

delta: δ ij = 1 if i = j and is 0 otherwise. We leave it as an exercise to show this result.As a1, a2, a3 and a1, a2, a3 form bases for E , any vector b can be described as linear

combinations of either sets of vectors:

b =

3i=1

biai =

3k=1

bkak. (1.23)

The components bi are known as the contravariant components, while the components bk areknown as the covariant components:

b · ai =

3k=1

bkak

· ai =

3k=1

bkδ ki = bi,

b · ai =

3k=1

bkak

· ai =

3k=1

bkδ ik = bi. (1.24)




It is very important to note that bk = b · ak in general because ai · ak is not necessarily equal toδ ik.

The trivial case where xi = q i deserves particular mention. For this case r =

3k=1 xiEi.

Consequently, ai = Ei. In addition, ai = Ei, and the covariant and contravariant basis vectors

are equal.

1.3.5 Some Comments on Derivatives

Throughout these notes the derivatives of functions Φ

q 1, q 2, q 3, q 1, q 2, q 3, t

will feature. In par-

ticular, many readers will be confused between the derivative ddt

and the derivative ∂ ∂t

. The formerderivative assumes that q i and q i are functions of time, while the latter assumes that they areconstant:

Φ = dΦ

dt =

3i=1

∂Φ

∂q idq i

dt +

3k=1

∂Φ

∂ q kd2q i

dt2 +

∂Φ

∂t . (1.25)

For example, consider the function

Φ = q 1 + (q 3)2 + 10t. (1.26)

Then,∂Φ

∂q 1 = 1,

∂Φ

∂ q 3 = 2q 3,

∂ Φ

∂t = 10, Φ = q 1 + 2q 3q 3 + 10. (1.27)

1.3.6 Parameterizations of Particle Kinematics

We now turn to establishing expressions for the position, velocity, and acceleration vectors of aparticle in terms of the coordinate systems just mentioned.

First, for the position vector we have2

r = x1E1 + x2E2 + x3E3

= rer + zE3

= ReR

=

3i=1

xi

q 1, q 2, q 3

Ei. (1.28)

Differentiating these expressions we find

v = x1E1 + x2E2 + x3E3

= rer + rθeθ + zE3

= ReR + R φeφ + R sin(φ)θeθ

=

3

i=1

q i ∂ r

∂q i

=

3

i=1

q iai. (1.29)

2 Notice that it is a mistake to assume that r =

3

i=1q iai.




Notice the simplicity of the expression for v when expressed in terms of the covariant basisvectors.

A further differentiation yields

a = x1E1 + x2E2 + x3E3= (r − rθ2)er + (rθ + 2rθ)eθ + zE3

= ( R − R φ2 − R sin2(φ)θ2)eR + (Rφ + 2 R φ − R sin(φ)cos(φ)θ2)eφ

+ (R sin(φ)θ + 2 Rθ sin(φ) + 2Rθ φ cos(φ))eθ

=

3i=1

q iai +

3i=1

3j=1

q iq j∂ ai

∂q j. (1.30)

The final representation for r is obtained after noting that ai depend on the curvilinear coordi-nates which in turn are functions of time.

We leave it as an exercise to establish expressions, using various coordinate systems, for thelinear momentum G and the angular momentum HO. The kinetic energy T of the particle has arather elegant representation using the curvilinear coordinates:

T = m

2 v · v

= m

2

3i=1

q iai

·

3k=1

q kak

= m

2 aik q iq k, (1.31)

where aik = aki = ak · ai. It is a good exercise to show that this representation can be used toshow that, for spherical polar coordinates,

T = m

2

R2 + R2 φ2 + R2 sin2(φ)θ2

. (1.32)

The exercises at the end of this chapter feature other coordinate systems.

1.4 Constraints

A constraint is a kinematical restriction on the motion of the particle. They are introduced inproblems involving a particle in three manners; either as simplifying assumptions, prescribedmotions, or due to rigid connections. The constraints on the motion of a particle dictate, to alarge extent, the coordinate system used to solve the problem of determining the motion of theparticle.

1.4.1 Some Examples

It is helpful to consider some examples. Consider the three mechanical systems shown in Figure1.4. The first system is known as the spherical pendulum. Here, a particle of mass m is attached



1.4 Constraints 9

O

1

E2

E

m

O

rigid rod of length L

O

E

2

E3

E

1

m

moving plane

E

2

E3

E

1

m

rigid rod of length L

Fig. 1.4. Three mechanical systems: the spherical pendulum, the planar pendulum, and a particle movingon a surface.

by a rigid rod of length L to a fixed point O. The constraint on the motion of the particle in thissystem can be written as

r · eR = L. (1.33)

The next system is the planar pendulum. Again, the particle is attached by a rigid rod of lengthL to a fixed point O , but it is also assumed to move on a vertical plane. The constraints on themotion of the particle are

r · er = L, r · E3 = 0. (1.34)

The third system involves a particle moving on a horizontal surface which is moving with avelocity v(t)E3. The constraint on the motion of the particle is

(r(t) − r(t0)) · E3 =

t

t0

v(τ )dτ. (1.35)




Notice that by using an appropriate coordinate system, we can easily write down the constraint(s)on the particle.

O

E

r

2

E3

E

1

m

surface ψ = 0

∇ψ

Fig. 1.5. A particle moving on a surface ψ = 0.

1.4.2 The Canonical Case

Turning to a more abstract case. Consider a particle moving on a surface ψ = 0 (cf. Figure 1.5).The constraint on the motion of the particle can be written as

ψ(r, t) = 0 (1.36)

This is a single equation. By fixing t and varying r it is easy to see that it defines a two–

dimensional surface. The normal to this surface is ∇ψ = grad(ψ). Dependent on the coordinatesystem used, this vector has numerous representations:

∇ψ = ∂ψ

∂ r =

3i=1

∂ψ

∂xi

Ei

=

3i=1

∂ψ

∂q iai

= ∂ψ

∂r er +

1

r

∂ψ

∂θ eθ +

∂ψ

∂z E3

= ∂ψ

∂ReR +

1

R

∂ψ

∂φeφ +

1

R sin(φ)

∂ψ

∂θ eθ. (1.37)

You should notice how simple the expression for the gradient is in curvilinear coordinates. Theunit normal n to the surface is



1.4 Constraints 11

n = ∇ψ

||∇ψ|| . (1.38)

It is also useful to notice that if ψ is a function of r only, then the surface is fixed. We alsonote that a simple differentiation shows that

ψ = ∂ψ

∂ r · v +

∂ ψ

∂t . (1.39)

However, if r satisfies the constraint, then ψ (r, t) = 0, and ψ = 0. Consequently,

∂ψ

∂ r · v +

∂ψ

∂t = 0. (1.40)

This result will be important in our discussion of the mechanical power of the constraint forces.

O

E

2

2

E

r

3

E

1

1

m

surface ψ = 0

surface ψ = 0

1∇ψ 2

∇ψ

Fig. 1.6. A particle subject to two constraints. The dotted curve in this figure corresponds to the curveof intersection of the surfaces ψ1 = 0 and ψ2 = 0.

We now consider the more complex case of a particle moving on a curve. A curve can beconsidered as the intersection of two surfaces, and so the particle is subject to two constraints:

ψ1(r, t) = 0, ψ2(r, t) = 0. (1.41)

This situation is shown in Figure 1.6. The normal vectors to these surfaces at a point of intersec-tion are assumed not to be parallel.

If the motion of a particle is completely constrained, then it is subject to three constraints

(cf. Figure 1.6). This is equivalent to the particle lying on the intersection of three intersectingsurfaces. In other words, the particle is subject to three constraints:




ψ1(r, t) = 0, ψ2(r, t) = 0, ψ3(r, t) = 0. (1.42)

The three normal vectors to these surfaces at the point of their intersection are assumed to forma basis for E and, from the inverse function theorem of calculus, this means that given the three

surfaces and the time t, we can uniquely find r.

1.4.3 Curvilinear Coordinates and Constraints

The constraints we have considered on the motion of the particle have been described in termsof surfaces that the motion of the particle is restricted to. These surfaces can be described interms of the coordinate system used for E . The description is greatly facilitated by a judiciouschoice of coordinates. For instance, if a particle is constrained to move on a fixed plane, then onecan always choose the origin O and the Cartesian coordinates such that the constraint is easilydescribed by the equation x3 = constant. Similarily, if a particle is constrained to move on asphere, then spherical polar coordinates are an obvious choice.

The more sophisticated the surfaces that the particle is constrained to move on, then themore difficult it becomes to choose an appropriate coordinate system. However, help is at hand,

because any surface ψ(r) = 0 can be described in an appropriate curvilinear coordinate systemby a simple equation q 3 = constant. Furthermore, a moving surface ψ (r, t) = 0 can be describedby the equation q 3 = f (t), where f is a function of time t. For example, suppose a particle ismoving on a sphere whose radius is a known function L(t). Then the constraint that the particlemoves on the sphere is simply described by

R = L(t). (1.43)

Here, we are choosing the spherical polar coordinate system to be our curvilinear coordinatesystem.

1.4.4 Further Examples of Constraints

Returning to the three mechanical systems shown in Figure 1.4, you should convince yourself that the constraint(s) on the motions of the particle in these systems are of the form ψ = 0.Specifically, for the spherical pendulum,

ψ = r · eR − L = 0. (1.44)

We may imagine the particle in a spherical pendulum as moving on a sphere. For the planarpendulum, we have

ψ1 = r · er − L = 0, ψ2 = r · E3 = 0. (1.45)

In this case, the particle can be visualized as moving on the intersection of a cylinder of radius L

and a horizontal plane. This intersection defines a circle. If the rod’s length L changes with time,then the circle’s radius changes also. Finally, for the particle moving on the horizontal surface,

ψ = (r(t) − r(t0)) · E3 −

t

t0

v(τ )dτ = 0. (1.46)



1.5 Integrable and Non-Integrable Constraints 13

E3

er O

g

αRough cone

m

Linear spring

Fig. 1.7. A particle moving on the inner surface of a cone.

Notice that in this example ψ = ψ(r, t). It is a good exercise to calculate ∇ψ and ∂ψ∂t

for each of these examples.

The aforementioned considerations are independent of the forces acting on the particle. Forinstance, consider the particle moving on the surface of a cone shown in Figure 1.7. For thissystem, the constraint on the motion of the particle is

π

2 − φ = α. (1.47)

That isψ(r) = φ + α − π

2 = 0. (1.48)

If the cone were moving in a manner that α = α(t), then the function ψ = ψ(r, t). For example,suppose α = α0 + A sin(ωt), then

ψ(r, t) = φ − π

2 + α0 + A sin(ωt) = 0. (1.49)

You should verify that ψ = 0, but ∂ψ∂t

= Aω cos(ωt).The forces acting on the particle which ensure that the constraint is enforced are known as

constraint forces. We shall address these forces in the next chapter.

1.5 Integrable and Non-Integrable Constraints

All of the constraints discussed so far can be written in the form

ψ(r, t) = 0. (1.50)

In other words, they impose a time dependent restriction on the position of the particle. Differ-entiating the constraint with respect to time, we find that

ψ =

∂ψ

∂ r

· v +

∂ψ

∂t = 0. (1.51)

Notice that the constraint also imposes a restriction on the velocity of the particle.A type of constraint which is more general than ψ (r, t) = 0 is the following




f · v + h = 0, (1.52)

where f = f (r, t) and h = h(r, t). Such a constraint does not restrict the position of the particle- it only restricts its velocity.

If we can find a function ψ(r, t) such that f = ∂ψ

∂ r and h = ∂ψ

∂t , then the constraint f

·v + h = 0

is said to be integrable (or holonomic), otherwise, the constraint f · v + h = 0 is said to be non–integrable (non–holonomic).

Suppose a constraint f ·v + h = 0 is imposed on the motion of the particle, and suppose thatthis constraint is integrable. The integrability implies that we can find a curvilinear coordinatesystem such that f · v + h = 0 is equivalent to the constraint q 3 + h = 0. This constraint in turnis equivalent to the constraint q 3 = g where g = h.

There are several criteria to examine whether or not a constraint f ·v + h = 0 is integrable. Anexcellent survey of these criteria is contained in Rosenberg [46]. To discuss one of these criteria,suppose that we use a curvilinear coordinate system to describe the motion of the particle, thenwe have the representations

f =

3

i=1

f iai, v =

3

i=1

q iai. (1.53)

That is the constraint can be expressed as

f 1 q 1 + f 2 q 2 + f 3 q 3 + h = 0. (1.54)

Next, suppose that the constraint were integrable. Then we could find a function ψ = ψ(q 1, q 2, q 3, t)such that

ψ = f · v + h, (1.55)

that is,∂ψ

∂q i = f i,

∂ψ

∂t = h. (1.56)

For such a φ to exist, we would need the following equalities to hold:

∂f i

∂q j =

∂f j

∂q i ,

∂f i

∂t =

∂h

∂q i , (i = 1, 2, 3) and ( j = 1, 2, 3). (1.57)

These equalities are necessary for a constraint to be integrable. If any of them are violated, thenthe constraint f · v + h = 0 is non–integrable.

The distinction between integrable and non-integrable constraints becomes particularily im-portant when rigid bodies are concerned. However, for pedagogical purposes, it is desirable tointroduce them when discussing single particles.

1.6 Exercises

1.1 Recall that the cylindrical polar coordinates r,θ,z are defined using Cartesian coordinates

x = x1, y = x2, z = x3

by the relations

r =

x21 + x2

2, θ = tan−1

x2

x1

, z = x3. (1.58)



1.6 Exercises 15

Show that the covariant basis vectors associated with the curvilinear coordinate system,q 1 = r, q 2 = θ, and q 3 = z, are

a1 = er, a2 = reθ, a3 = E3. (1.59)

In addition, show that the contravariant basis vectors are

a1 = er, a2 = 1

reθ, a3 = E3. (1.60)

It is a good exercise to convince yourself with an illustration that a2 is tangent to a θ

coordinate curve, while a2 is normal to a θ coordinate surface. Finally, for this coordinatesystem, show that

T = m

2

r2 + r2 θ2 + z2

. (1.61)

1.2 Recall that the spherical polar coordinates R,φ,θ are defined using Cartesian coordinatesx = x1, y = x2, z = x3 by the relations

R =

x21 + x

22 + x

23, θ = tan−

1 x2

x1

, φ = tan−

1 x21 + x2

2

x3

. (1.62)

Show that the covariant basis vectors associated with the curvilinear coordinate system,q 1 = R, q 2 = φ, and q 3 = θ, are

a1 = eR, a2 = Reφ, a3 = R sin(φ)eθ. (1.63)

In addition, show that the contravariant basis vectors are

a1 = eR, a2 = 1

Reφ, a3 =

1

R sin(φ)eθ. (1.64)

1.3 The parabolidal coordinates u ,v ,θ are defined using Cartesian coordinates x = x1, y =x2, z = x3

by the relations

u = ±

x3 +

x23 + (x2

1 + x22),

v = ±

−x3 +

x23 + (x2

1 + x22),

θ = tan−1

x2

x1

. (1.65)

In addition, the inverse relations can be defined:

x1 = uv cos(θ), x2 = uv sin(θ), x3 = 1

2(u2 − v2). (1.66)

(a) In the r − x3 plane, draw several representative examples of the projections of the u andv 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 coordinatecurves foliate this surface.(c) Show that the covariant basis vectors for the parabolidal coordinate system are

a1 = ∂ r

∂u = v er + uE3,

a2 = ∂ r

∂v = uer − vE3,

a3 = ∂ r

∂θ = uv eθ. (1.67)

Draw representative examples of these vectors.(d) Show that the contravariant basis vectors for the parabolidal coordinate system are

a1 = grad(u) = 1

u2 + v2a1,

a2 = grad(v) = 1

u2 + v2a2,

a3 = grad(θ) = 1uv

eθ. (1.68)

Again, draw representative examples of these vectors.(e) Where are the singularities of the parabolidal coordinate system. Verify that at thesesingularities, the contravariant basis vectors are not defined.(f ) For a particle of mass m which is moving in E 3, establish expressions for the kinetic energyT and linear momentum G in terms of u ,v ,θ and their time derivatives.

E2

E3

E1

Circular Helix

r = L and z = Lαθ

Bead of mass m

g

Fig. 1.8. A particle moving on a circular helix.

1.4 A classical problem is to determine the motion of a particle on a circular helix (see Figure1.8). In terms of the cylindrical polar coordinates r, θ,z, the equation of the helix is



1.6 Exercises 17

r = L, z = αLθ, (1.69)

where L and α are constants. Here, we will use another curvilinear coordinate system todefine the motion of particle:

q 1 = θ, q 2 = r, q 3 = ν = z − αrθ. (1.70)

(a) Show that the covariant basis vectors associated with this coordinate system are

a1 = r(eθ + αE3), a2 = er + αθE3, a3 = E3. (1.71)

Verify that the covariant basis vectors are not orthonormal.(b) Show that the kinetic energy of the particle has the representation

T = m

2

(1 + α2θ2)r2 + (1 + α2)r2 θ2 + ν 2 + 2ν rαθ + 2ν θαr + 2rθα2rθ

. (1.72)

Calculate the contravariant basis vectors associated with the curvilinear coordinate system.1.5 Consider the following curvilinear coordinate system:

q 1 = x1 sec(α), q 2 = x2 − x1 tan(α), q 3 = x3. (1.73)

Draw the coordinate curves and surfaces for this coordinate system and then show that thecovariant and contravariant basis vectors are

a1 = cos(α)E1 + sin(α)E2, a2 = E2, a3 = E3, (1.74)

anda1 = sec(α)E1, a2 = − tan(α)E1 + E2, a3 = E3, (1.75)

respectively. Illustrate these vectors on the coordinate curves and surfaces you previouslydrew. You should notice that the coordinate system is not orthogonal. When is ai not abasis?

1.6 Given a vector b = 10E1 + 5E2 + 6E3, calculate its covariant bi and contravariant bi

components when a1 = cos(α)E1 + sin(α)E2, a2 = E2 and a3 = E3. In addition, verify

that b = 3

i=1 biai = 3

i=1 biai. Furthermore, show that b =3i=1 biai and b = 3

i=1 biai.When is ai not a basis?

1.7 This exercise is related to examining how the covariant and contravariant components of avector are related. To start, we define the following scalars using the covariant and contravari-ant basis vectors:

aik = aik(q r) = ai · ak, aik = aik(q r) = ai · ak. (1.76)

You should notice that aik = aki and aik = aki. The indices i, k , m, r , and s in this problemrange from 1 to 3.(a) For any vector b, show that the covariant and contravariant components are related

bi =3

k=1

aikbk, bi =3

k=1

aikbk. (1.77)




In other words, the covariant components are linear combinations of the contravariant com-ponents and vice-versa.(b) By choosing b = ar and as, and using the symmetries of akm and ars show that

3k=1

aikakj =3

k=1

akiakj =3

k=1

akiajk = δ ji . (1.78)

1.8 Show that only the first of the following three constraints are integrable:

xx + yy = −h(t), y + zx = 0, xx + zyy + zyz = 0. (1.79)

Here, x = x1, y = x2, and z = x3. In addition, show that the integrable constraint correspondsto a particle moving on a cylinder whose radius varies with time.



2

Kinetics of a Single Particle

2.1 Introduction

In this chapter, the balance law for a single particle is introduced. We then turn to discussing

the two most important catagories of forces acting on a particle; namely, conservative forcesand constraint forces. This chapter also discussions on energy and momentum conservations. Theconcluding sections of this chapter discuss Lagrange’s equations of motion for a single particle. Theexposition of Lagrange’s equations is based on a paper by Casey [9]. In particular, a differentialgeometric approach to Lagrange’s equations is taken.

2.2 The Balance Law for a Single Particle

As usual, consider a single particle of mass m which is moving in Euclidean three–space E . Theposition 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’ssecond law or Euler’s first law. The integral (or impulse–momentum) form of this law is

G(t) − G(t0) =

tt0

F(τ )dτ, (2.1)

where F is the resultant force acting on the particle. Notice that this form of the balance lawdoes not assume that v is differentiable with respect to time t.

If we assume that G is differentiable with respect to time, then we can differentiate both sidesof the integral form of the balance of linear momentum to obtain the local form:

F = G, (2.2)

where G = mv = mr is the linear momentum of the particle. Assuming that the mass of theparticle is constant, we can write

F = mr. (2.3)

This law represents three (scalar) equations which relate F and the rate of change of linearmomentum of the particle. We shall refer to F = ma as the balance of linear momentum.



20 2 Kinetics of a Single Particle

It is convenient to write the balance law as a set of first-order ordinary differential equations:

v = r, F = mv. (2.4)

In the absence of constraints, these represent six scalar (differential) equations for the six un-

knowns r(t) and v(t). To solve these equations, six initial conditions r(t0) and v(t0) must bespecified. Alternatively, if the problem is formulated as a boundary–value problem, then a com-bination 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 three equations:

mx1 = F · E1,

mx2 = F · E2,

mx3 = F · E3. (2.5)

On the other hand, if a cylindrical polar coordinate system is used, we have

m(r − rθ2) = F · er,

m(rθ + 2rθ) = F

·eθ,

mz = F · E3. (2.6)

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

m( R − R φ2 − R sin2(φ)θ2) = F · eR,

m(Rφ + 2 R φ − R sin(φ) cos(φ)θ2) = F · eφ,

m(R sin(φ)θ + 2 Rθ sin(φ) + 2Rθ φ cos(θ)) = F · eθ. (2.7)

Notice that these equations are projections of F = ma onto a set of basis vectors for E .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 easymanner. We now turn to these equations.

2.3 Work and Power

Consider a force P which is acting on a particle of mass m. The mechanical power P of the forceis defined to be

P = P · v. (2.8)

Clearly, if P is perpindicular 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. Similarily, when t = tB , the particleis at B : r(tB) = rB. During the interval of time that the particle moves from A to B , we supposethat a force P, among others, acts on the particle. The work W AB performed by P during thistime interval is defined to be the integral, with respect to time, of the mechanical power:

W AB =

tBtA

P · vdt. (2.9)



2.3 Work and Power 21

r

r r

v

P

O

path of the

particle

A

A B

B

m

Fig. 2.1. A force P acting on a particle as it moves from A to B .

Notice that this is a line integral, and we are using t to parameterize the path of the particle.

Dependent on the choice of coordinate system, the integral in this expression has severalequivalent representations, for example, tBtA

P · vdt =

rBrA

P 1dq 1 + P 2dq 2 + P 3dq 3

=

rBrA

P rdr + P θrdθ + P zdz

=

rBrA

P RdR + P θR sin(φ)dθ + P φRdφ (2.10)

where

P =3

i=1

P iai = P rer + P θeθ + P zE3 = P ReR + P θeθ + P φeφ, (2.11)

and

v =3

i=1

q iai = rer + rθeθ + zE3 = ReR + R sin(φ)θeθ + R φeφ. (2.12)

You should also notice that identities of the following form were used: θdt = dθ and q idt = dq i.It is useful to consider an example. Suppose a force P = P eθ 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 = θ are constant. Astraightforward calculation shows that the power of this force is

P · v = ωPLeαt, (2.13)

and the work performed by the force is

W AB = tB

tA ωPLe

αt

dt =

ωP L

α

e

αtB

− e

αtA, (2.14)

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 motionof the particle is independent of the path of particle. Using a result from vector calculus, the path

independence implies that P is the gradient of a scalar function U = U (r):

P = −∇U = −3

i=1

∂U

∂q iai. (2.15)

The function U is known as the potential energy associated with the force P, and the minus signis a historical convention.

It is important to notice that if P is conservative, then its mechanical power is − U . To seethis, we simply examine U and use the definition of a conservative force:

− U = −∂U

∂ r · v = −(−P) · v = P · v. (2.16)

This result holds for all motions of the particle.To check if a given force P is conservative, one approach is to find a potential function U such

thatP · v = − U (2.17)

holds for all motions of the particle. This approach reduces to solving a set of coupled partialdifferential equations for U . For example, if curvilinear coordinates are used, one needs to solvethe following three partial differential equations for U (q 1, q 2, q 3):

P 1 = − ∂U

∂q 1, P 2 = − ∂U

∂q 2, P 3 = − ∂U

∂q 3. (2.18)

You might notice that the solution to this equation will yield a potential energy U (q 1, q 2, q 3)modulo an additive constant. This constant is usually set by the condition that U = 0 when thecoordinates q i are zero.

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 P is conservative, then curl(P) = 0. In Cartesian coordinates,

curl(P) =

3i=1

∂

∂xi

Ei

× P

=

∂P 3

∂x2− ∂P 2

∂x3

E1 +

∂P 1

∂x3− ∂P 3

∂x1

E2 +

∂P 2

∂x1− ∂ P 1

∂x2

E3.

(2.19)

Here P i = P · Ei. Consequently, if curl(P) = 0, then the Cartesian components of P must satisfythe following conditions:

∂P 3

∂x2 =

∂P 2

∂x3 ,

∂P 1

∂x3 =

∂P 3

∂x1 ,

∂P 2

∂x1 =

∂P 1

∂x2 . (2.20)

These results have representations for any coordinate system, but we do not pursue them here.



2.5 Examples of Conservative Forces 23

2.5 Examples of Conservative Forces

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

2.5.1 Constant Forces

All constant forces are conservative. To see this let C denote a constant force, and let U c = −C·r.Now, ∇U c = −C, and, consequently, U c is the potential energy associated with C. The mostcommon example of constant forces are the gravitational forces −mgE2 and −mgE3 and theirassociated potentials are mg E2 · r and mgE3 · r, respectively.

r

r

O

D

D

spring

m

Fig. 2.2. A spring attached to a particle and a fixed point D.

2.5.2 Spring Forces

Consider the spring shown in Figure 2.2. One end of the spring is attached to a fixed point D,while the other end is attached to a particle of mass m. When the spring is unstretched, it has alength L. Clearly, the stretched length of the spring is ||r − rD||, and the extension/compressionof the spring is

= ||r − rD|| − L (2.21)

The potential energy U s associated with the spring is

U s = f () (2.22)

where f is a function of the change in length of the spring. Evaluating the gradient of U s, we findthe spring force Fs:

Fs = −∂U s

∂ r = −∂f

∂

r

−rD

||r − rD|| . (2.23)

To establish this result, we used the identity




∂

∂ r =

∂

∂ r (||r − rD|| − L) =

r − rD

||r − rD|| . (2.24)

We leave this identity as an exercise to establish.1

The most common spring in engineering dynamics is a linear spring. For this spring,

U s = K

2 (||r − rD|| − L)

2, Fs = −K (||r − rD|| − L)

r − rD

||r − rD|| . (2.25)

In words, the potential energy of a linear spring is a quadratic function of its change in length.Examples of nonlinear springs include those where f is a polynomial function in . For instance,

f () = A2 + B4, (2.26)

where A > 0 and B are constants. Such a spring is known as hardening if B > 0 and softening if B < 0.

2.6 Constraint Forces

A constraint force Fc is a force which ensures that a constraint is enforced. Examples of theseforces include reaction forces, normal forces, and tension forces in inextensible strings.

Given a constraint ψ(r, t) = 0 on the motion of the particle, there is no unique prescriptionfor the associated constraint force. Choosing the correct prescription depends on the physicalsituation that the constraint represents. However, when we turn to solving for the motion of theparticle using F = ma, we see that we need to solve the six equations

r = v, v = 1

mF, (2.27)

subject to the restrictions on r and v

ψ(r, t) = 0, ∂ψ∂ r

· v + ∂ψ∂t

= 0. (2.28)

To close this system of equations, an additional unknown is introduced in the form of the con-straint force Fc. The prescription of Fc must be such that F = ma and ψ(r(t) can be used todetermine Fc and r(t).

There are no unique prescriptions for constraint forces. Indeed as shown in O’Reilly andSrinivasa [38], the prescription most commonly used in the literature, which dates to Lagrangeand is equivalent to the normality prescription, is necessary and sufficient to ensure that themotion of the particle satisfies the constraint. However, there is a freedom available to includeother arbitrary non-normal components. This is the reason why Coulomb’s prescription is alsovalid.

1 To help you with this, it is convenient to first establish that ∂ √ x·x

∂ x = x

√ x·x = x

||x||. This result is

equivalent to showing that the gradient of ||r|| is eR.



2.6 Constraint Forces 25

2.6.1 A Single Constraint

Consider the case of a particle subject to a single constraint:

ψ(r, t) = 0. (2.29)

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

n = ∇ψ

||∇ψ|| . (2.30)

Knowing n, we can construct a unit tangent vector t1 to the surface. In addition, by defininganother unit tangent vector t2 = n × t1, one has constructed a right-handed orthonormal basist1, t2, n for E . This is not in general a constant set of vectors, rather it changes as one movesfrom point to point along the surface. The final ingredient we need is to denote the velocity vectorof the point of the surface (that the particle is in contact with) as vs.

O

r

n

t

t1

2m

surface

Fig. 2.3. A particle moving on a surface ψ = 0. The vectors t1 and t2 are tangent vectors to this surfaceat r.

With this background in mind, we now consider two prescriptions for the constraint force Fc.The first prescription is known as the normality prescription and is due to James Casey:2

Fc = λ∇ψ, (2.31)

where λ = λ(t) must be determined using F = ma. In other words, λ is a Lagrange multiplier. Inthis prescription, Fc is parallel to the normal vector n. On physical grounds, this prescription is

2 This prescription is identical to the virtual work prescription used in dynamics. The virtual workprescription is sometimes known as D’Alembert’s principle or the Lagrange–D’Alembert principle.




justified if the surface, that the particle is constrained to move on, is smooth. In the event thatthe surface is rough, an alternative prescription, due to Charles Augustin Coulomb, can be used:

Fc = λ∇ψ + Ff , (2.32)

where the normal force N = λ∇ψ and the friction force is

Ff = −µd||λ∇ψ|| v − vs

||v − vs|| . (2.33)

Here, µd is known as the coefficient of dynamic friction. Notice that the tangential componentsof Fc oppose the motion of the particle relative to the surface (cf. Figure 2.4).

n

t

t1

2

Fc

F f

F f

vrel

N =

Fig. 2.4. The constraint force Fc acting on a particle moving on a rough surface. The velocity vectorvrel = v − vs is the velocity vector of the particle relative to the point of its contact with the surface.

The mechanical power of the constraint force Fc is

Fc · v = λ∇ψ · v + Ff · v

= −λ∂ψ

∂t + Ff · v, (2.34)

where we used the identity

ψ = ∇ψ · v + ∂ ψ

∂t = 0. (2.35)

For the normality prescription, Ff = 0 we can now see that if the surface that the particle ismoving on is fixed, i.e., ψ = ψ(r), then Fc does no work. Otherwise, this constraint force isexpected to do work because its normal component ensures that part of the velocity vector of theparticle is vs. For the Coulomb prescription, except when vs = 0, it is not possible to predict if work is done on the particle by this force.

As a first example, consider a particle moving on a rough sphere of radius L which center isfixed at the origin O. For this surface, the constraint is ψ(r) = r · eR − L = 0. Consequently,∇ψ = eR. In addition, v − vs = L φeφ + L sin(φ)θeθ. In conclusion,




Fc = λeR − µd|λ| φeφ + sin(φ)θeθ φ2 + sin2(φ)θ2

. (2.36)

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 prescribed by normality: Fc = λeR. In thespherical pendulum, −λ is the tension in the rod connecting the particle to the fixed point O .

2.6.2 Two Constraints

When a particle is subject to two constraints, ψ1(r, t) = 0 and ψ2(r, t) = 0, then it can beconsidered as constrained to move on a curve. The curve in question is formed by the instantaneousintersection of the surfaces defined by the constraints.

At each point on the curve there is a tangent vector t. This vector can be defined by firstobserving that ∇ψ1 and ∇ψ2 are both normal to the surfaces that the curve lies on (cf. Figure2.5). Consequently,

t = ∇ψ1 × ∇ψ2

||∇ψ1 × ∇ψ2||. (2.37)

For each instant in time, the point of the curve which is in contact with the particle has a velocity.We shall denote this velocity by vc. The velocity vector of the particle relative to the curve isv − vc = vt.

O2

r

t

1

m

surface

surface

1 2

Fig. 2.5. A particle subject to two constraints. The dotted curve in this figure corresponds to the curveof intersection of the surfaces ψ1 = 0 and ψ2 = 0.

We now turn to prescriptions for the constraint force. The first prescription is the normalityprescription:Fc = λ1∇ψ1 + λ2∇ψ2, (2.38)




where λ1 = λ1(t) and λ2 = λ2(t) are both determined using F = ma. As in the case of asingle constraint, the normality prescription is valid when the curve that the particle moves onis smooth. For the rough case, we use Coulomb’s prescription:

Fc = λ1∇ψ1 + λ2∇ψ2 + Ff , (2.39)

where the friction force is

Ff = −µd||λ1∇ψ1 + λ2∇ψ2|| v − vc

||v − vc|| . (2.40)

Again, µd is the coefficient of dynamic friction, the friction force opposes the motion of the particlerelative to the curve, and the normal force N with λ1∇ψ1 + λ2∇ψ2.

The mechanical power of the constraint force Fc for this case is

Fc · v = λ1∇ψ1 · v + λ2∇ψ2 · v + Ff · v

= −λ1∂ψ1

∂t − λ2

∂ψ2

∂t + Ff · v, (2.41)

where we again used the identities

ψ1 = ∇ψ1 · v + ∂ψ1

∂t = 0, ψ2 = ∇ψ2 · v +

∂ψ2

∂t = 0. (2.42)

For the normality prescription, Ff = 0 and we can now see that if the curve that the particleis moving on is fixed, i.e, ψ1 = ψ1(r) and ψ2 = ψ2(r), then Fc does no work. Otherwise, thisconstraint force is expected to do work because its normal components force part of the velocityvector 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 which is attached by a rod of length L to a fixed point O. The particle is also constrainedto move on a vertical plane. In short, ψ1 = r

·er

−L = 0 and ψ2 = r

·E3 = 0. With a little work,

we find that ∇ψ1 = er and ∇ψ2 = E3. For this mechanical system, the normality prescription isappropriate:

Fc = λ1er + λ2E3, (2.43)

where λ1er can be interpreted as the tension force in the rod and λ2E3 can be interepreted asthe normal force exerted by the plane on the particle. If we let L = L(t), the prescription for theconstraint force will not change.

A system which is related to the planar pendulum is to imagine a particle moving on a roughcircle whose radius L = L(t). The particle is subject to the same constraints as it is in the planarpendulum, however, the normality prescription is not valid. Instead, we now have

Fc = λ1er + λ2E3 − µd

λ21 + λ2

2

θ

|θ|eθ, (2.44)

where we used the fact that v − vc = Lθeθ.




2.6.3 Three Constraints

The reader may have notice that our expressions for the constraint force when we employedCoulomb’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 tothe motion of the particle subject to three constraints: ψi(r, t) = 0, i = 1, . . . , 3. As mentionedearlier, when a particle is subject to three constraints, the three equations ψi(r, t) = 0 can inprinciple be solve to determine the motion r(t) of the particle. We denote the resulting solutionby f (t), i.e., r(t) = f (t). In other words, the motion is completely prescribed. In this case, thesole purpose of F = ma is to determine the constraint force Fc.

For the case where the particle is subject to three constraints, the normality prescription anda prescription based on static Coulomb friction are equivalent. This equivalence holds inspite of the distinct physical situations these prescriptions pertain to.

To examine the equivalence, let’s first use the normality prescription:

Fc = λ1∇ψ1 + λ2∇ψ2 + λ3∇ψ3. (2.45)

Here, λ1

, λ2

, and λ3

are functions of time. Because the three constraints are tacitly assumed to beindependent, ∇1ψ1, ∇2ψ2, ∇3ψ3 forms a basis for E . Consequently, the normality prescriptionprovides a vector Fc with three independent components. Coulomb’s static friction prescriptionfor a particle which is not moving relative to the curve or surface on which it lies is

Fc = N + Ff , (2.46)

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

||Ff || ≤ µs||N||, (2.47)

where µs is the coefficient of static friction. Again, the Coulomb prescription provides a vectorFc with three independent components. In other words, both prescriptions state that Fc consistsof 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 non–constraint forces, then we see how Fc is determined from F = ma:

Fc = −Fa + mf . (2.48)

This solution Fc will be the same regardless of whether or not one uses the normality prescriptionor Coulomb’s prescription.

2.6.4 Non–Integrable Constraints

Our discussion of constraint forces has focused entirely on the case of integrable constraints. If anon–integrable constraint,

f

·v + h = 0, (2.49)

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 normality prescription:




Fc = λf . (2.50)

The main reason for adopting this prescription is as follows: In the event that the non–integrableconstraint turns out to be integrable, then the prescription we employ will agree with the nor-

mality prescription we discussed earlier.As a further example, suppose the motion of the particle is subject to two constraints, one of which is integrable:

ψ(r, t) = 0, f · v + h = 0. (2.51)

Using the normality prescription, we find that the constraint force acting on the particle is

Fc = λ1∇ψ + λ2f . (2.52)

Suppose that the applied force acting on the particle is Fa, then the equations governing themotion of the particle are

ψ(r, t) = 0, f · v + h = 0,

r = v, v = 1m

(Fa + λ1∇ψ + λ2f ) . (2.53)

This set of equations constitutes 8 equations for the 8 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 tobe conserved if it is constant during the motion of the particle. The conserved quantities are oftenknown as integrals of motion . The solutions of many problems in particle mechanics are basedon the observation that either a momentum and/or an energy is conserved. At this stage in thedevelopment 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. Weshall consider numerous examples of these conservations later on.

2.7.1 Conservation of Linear Momentum

The linear momentum G of a particle is defined as G = mv. Recalling the integral form of thebalance of linear momentum,

G(t) − G(t0) =

tt0

F(τ )dτ, (2.54)

we see that G(t) is conserved during an interval of time (t0, t) if tt0

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 ddt

(G · b) = 0. For this to happen,

˙G · b = G · b + G · b = F · b + G · b = 0. (2.55)

In words, if F · b + G · b = 0, then G · b is conserved.



2.7 Conservations 31

Examples of conservation of linear momentum arise in many problems. For example, considera 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 asmooth vertical wall. Then, the components of G in the two tangential directions are conserved.

For these two examples, the vector b is constant.

2.7.2 Conservation of Angular Momentum

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

HO = v × G + r × G = v × mv + r × F = r × F. (2.56)

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

HO = r × F. (2.57)

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, secondly, a component, say c(t) is conserved. For the first case, we see from theangular momentum theorem that HO is conserved if F is parallel to r. Problems where this arisesare known as central force problems. Dating to Newton, they occupy an important place in thehistory of dynamics. Using the angular momentum theorem, it is easy to see that the second formof conservation, HO · c is constant, arises when r × F · c + HO · c = 0.

2.7.3 Conservation of Energy

As a prelude to discussing the conservation of energy, we first need to discuss the work–energytheorem. This theorem is a result that is established using F = ma and relates the time rate of

change of kinetic energy to the mechanical power of F:

T = F · v. (2.58)

This theorem is the basis for establishing conservation of energy results for a single particle.The proof of the work–energy theorem is very straightforward. First, recall that T = 1

2mv · v.

Differentiating T we find

T = d

dt

1

2mv · v

=

1

2 (mv · v + mv · v) = mv · v. (2.59)

However, we know that mv = F, and so substituting for mv, we find that T = F · v, as required.To examine situations where 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

∂ r and a non–conservative force Pncon: F = P + Pncon. Here, U is the sum of the potential energies of theconservative forces acting on the particle. From the work–energy theorem, we find




T = F · v

= P · v + Pncon · v

= −∂U

∂ r · v + Pncon · v

= − U + Pncon · v. (2.60)

Defining the total energy E of the particle by

E = T + U, (2.61)

we see, from the work–energy theorem, that

E = Pncon · v. (2.62)

This result states that if, during a motion of the particle, the non-conservative forces do no work,then the total energy of the particle is conserved.

To examine whether or not energy is conserved, it usually suffices to check whether or notPncon

· v = 0. To see this let’s consider the example of the spherical pendulum whose length

L = L(t). For this particle:P = −mgE3, Pncon = λeR. (2.63)

Consequently,E = T + mgE3 · r, (2.64)

andPncon · v = λeR · v = Lλ. (2.65)

As a result, when the length of the pendulum is constant, L = 0, then E is conserved. On theother hand, if L = 0, then the constraint force λeR does work by giving the particle a velocity inthe eR direction.

2.8 Lagrange’s Equations of Motion

There are several approaches to deriving Lagrange’s equations of motion that appear in theliterature. Among them, a variational principle known as Hamilton’s principle is arguable the mostpopular. Here, we use an approach based on differential geometry. This approach is contained innumerous texts on differential geometry (see, for example, Synge and Schild [53]). It migrated fromthere to the classic text by Synge and Griffith [52], and has been recently revived by Casey [9].Our exposition follows Casey [9].

2.8.1 Two Identities

We assume that a curvilinear coordinate system has been chosen for E . The velocity vector vconsequently has the representation

v =3

i=1

q iai. (2.66)



2.8 Lagrange’s Equations of Motion 33

In addition, the kinetic energy has the representations

T = m

2 v · v =

m

2

3

i=1

3

k=1

ai · ak q iq k. (2.67)

It is crucial to notice that T = T

q 1, q 2, q 3, q 1, q 2, q 3

.We now consider in succession the partial derivatives of T with respect to the coordinates and

their velocities. We wish to establish the following results:

∂T

∂q i = mv · ai,

∂T

∂ q i = mv · ai. (2.68)

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

∂T

∂q i =

∂

∂q i

m

2 v · v

= mv · ∂ v

∂q i

. (2.69)

To simplify this expression further, we note that v =3

i=1 q kak and that ∂ qk

∂qi = 0, consequently,

∂T

∂q i = mv ·

3k=1

q k∂ ak

∂q i

. (2.70)

The final steps use the fact that ak = ∂ r∂qi

:

∂T

∂q i = mv ·

3k=1

q k∂ ak

∂q i

= mv · 3

k=1

q k ∂ 2r∂q k∂q i

= mv ·

3k=1

q k ∂

∂q k

∂ r

∂q i

= mv ·

3k=1

q k ∂

∂q k (ai)

= mv · ai. (2.71)

This is the final desired result. The last step was achieved by noting that f = 3

k=1∂f ∂qk

q k for

any function f = f (q 1, q 2, q 3).

The next result, which is far easier to establish, involves the partial derivative of T withrespect to a velocity. The reason this result is easier to establish is because the basis vectors ai

do not depend on q k. Getting on with the result:




∂T

∂ q i = mv ·

3k=1

∂ q k

∂ q i ak

= mv·

3

k=1

δ ki ak= mv · ai. (2.72)

This completes the proofs of both identities.

2.8.2 A Covariant Form of Lagrange’s Equations

It is crucial to note that the 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 itscovariant components, i.e., dotting it with ai.

To start, we consider

d

dt ∂T

∂ q i−

∂T

∂q i = d

dt (mv · ai) − mv · ai

= ma · ai + mv · ai − mv · ai

= ma · ai

= F · ai. (2.73)

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

d

dt

∂T

∂ q i

− ∂T

∂q i = F · ai. (2.74)

The beauty of this equation is best appreciated by using it to establish component forms of F = ma for various curvilinear coordinate systems.

2.8.3 The Lagrangian

Another form of Lagrange’s equations arises when one decomposes the force F into its conservativeand non–conservative parts:

F = −∇U + Fncon, (2.75)

where the potential energy U = U (q 1, q 2, q 3). As

∇U =

3k=1

∂U

∂q kak,

∂U

∂ q k = 0, (2.76)

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

d

dt

∂T

∂ q i − ∂U

∂ q i

−

∂T

∂q i − ∂U

∂q i

= Fncon · ai. (2.77)



2.8 Lagrange’s Equations of Motion 35

Introducing the Lagrangian L = T − U we find an alternative form of Lagrange’s equations:

d

dt

∂L

∂ q i

− ∂L

∂q i = Fncon · ai. (2.78)

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

2.8.4 An Example

To illustrate the ease of Lagrange’s equations we consider the case where the curvilinear co-ordinates chosen are the spherical polar coordinates: q 1 = R, q 2 = φ, and q 3 = θ. For thesecoordinates, we have

a1 = eR, a2 = Reφ, a3 = R sin(φ)eθ, (2.79)

andT =

m

2 R2 + R2 sin2(φ)θ2 + R2 φ2

. (2.80)

Notice that T does not depend on θ .Lagrange’s equations of motion for the spherical polar coordinate system are obtained by first

calculating the six partial derivatives of T :

∂T ∂R

= mR sin2(φ)θ2 + mR φ2, ∂T ∂φ

= mR2 sin(φ) cos(φ)θ2,∂T ∂θ

= 0, ∂T

∂ R = m R,

∂T

∂ φ = mR2 φ, ∂T

∂ θ = mR2 sin2(φ)θ.

(2.81)

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

d

dt

∂T

∂ R= m R

−

∂T

∂R = mR sin2(φ)θ2 + mR φ2

= F · eR

d

dt∂T

∂ φ = mR2 φ−∂T

∂φ = mR2 sin(φ) cos(φ)θ2

= F · Reφ,

d

dt

∂T

∂ θ= mR2 sin2(φ)θ

−

∂T

∂θ = 0

= F · R sin(φ)eθ.

(2.82)

Clearly, these equations were far easier to calculate than an alternative approach which involvesdifferentiating 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). (2.83)

For this case, the Lagrangian L is

L = T − U

= m

2

R2 + R2 sin2(φ)θ2 + R2 φ2

− mgR cos(φ). (2.84)




We can calculate Lagrange’s equations of motion using L,

d

dt

∂L

∂ q k

− ∂L

∂q k = 0, (2.85)

or by substituting for F in (2.82). We leave it as an exercise to show that both approaches areequivalent.

2.9 Lagrange’s Equations in the Presence of Constraints

The previous discussion of Lagrange’s equations did not address situations where constraints onthe motion of the particle was present. It is to this matter that we now turn our attention. Withintegrable constraints, whose constraint forces are prescribed using normality, the beauty andpower of Lagrange’s equations manifests. In this case, it is possible to choose the curvilinearcoordinates q i such that the equations of motion decouple into two sets. The first set describesthe unconstrained motion of the particle, while the second set yields the constraint forces as

functions of the unconstrained motion.There are two approaches to obtaining Lagrange’s equations. We shall refer to them through-out these notes as Approach I and Approach II. For the novice, we highly recommended the firstapproach. As in the previous section, our exposition follows Casey [9].

2.9.1 Preliminaries

We assume that the particle is subject to an integrable constraint,

ψ(r, t) = 0, (2.86)

and a non–integrable constraint,f · v + h = 0. (2.87)

We assume that the curvilinear coordinates are chosen such that the integrable constraint hasthe form

ψ(r, t) = q 3 − d(t) = 0. (2.88)

Furthermore, we assume that the constraint forces are prescribed using the normality prescription:

Fc = λ1a3 + λ2

3i=1

f iai

. (2.89)

Notice that f i = f · ai.We assume that there is an applied force Fa acting on the particle. This applied force can be

decomposed into conservative and non-conservative part:

Fa = −∇U + Fancon. (2.90)

The resultant force acting on the particle is



2.9 Lagrange’s Equations in the Presence of Constraints 37

F = λ1a3 + λ2

3i=1

f iai

− ∇U + Fancon. (2.91)

The total non–conservative force acting on the particle is Fncon = Fc + Fancon.

The kinetic energy of the particle is

T = m

2 v · v =

m

2

3i=1

3k=1

aik(q 1, q 2, q 3)q iq k, (2.92)

where aik = ai · ak. Imposing the integrable constraint on T we find the constrained kineticenergy:

T = m

2

2i=1

2k=1

aik q iq k + m

2i=1

ai3 q i d + m

2 a33

d2, (2.93)

whereaik = aik(q 1, q 2, t) = aik(q 1, q 2, q 3 = d(t)) (2.94)

is the constrained metric tensor. Notice that we use a tilde to denote imposition of the integrableconstraint.

A direct calculation shows that3

∂T ∂ q1

q3=d,q3= d

= ∂ T ∂ q1 , ∂T

∂ q2

q3=d,q3= d

= ∂ T ∂ q2 ,

∂T ∂q1

q3=d,q3= d

= ∂ T ∂q1

, ∂T ∂q2

q3=d,q3= d

= ∂ T ∂q2

,

∂T ∂ q3

q3=d,q3= d

= ∂ T ∂ q3 = 0, ∂T

∂q3

q3=d,q3= d

= ∂ T ∂q3

= 0.

(2.95)

In these relations, the partial derivative of T is evaluated prior to imposing the constraint q 3 =d(t). These relations imply that we can use T to obtain the first two Lagrange’s equations of motion, but not the third. Results which are identical in form to (2.95) pertain to the partial

derivatives of L and ˜L and U and

˜U .Notice that we didn’t impose the non-integrable 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.

2.9.2 Approach I

In the first approach, we evaluate the partial derivatives in Lagrange’s equations of motion (2.78)in the absence of any constraints:

d

dt

∂L

∂ q k

− ∂L

∂q k = Fancon · ak + (Fc = 0) · ak. (2.96)

3 Our notation here is standard but may need some explanation. 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 = 5in the resulting function.




Explicitly, these equations are

d

dt

m

3

i=1

ai1 q i

−

3

i=1

3

r=1

m∂air

∂q 1 q iq r +

∂U

∂q 1 = Fancon · a1,

d

dt

m

3i=1

ai2 q i

−

3i=1

3r=1

m∂air

∂q 2 q iq r +

∂U


d

dt

m

3i=1

ai3 q i

−

3i=1

3r=1

m∂air

∂q 3 q iq r +

∂U

∂q 3 = Fancon · a3.

(2.97)

Notice that we haven’t introduced the constraint forces on the right–hand side of these equations.That is, in the above equations F = −∇U + Fancon.

We now impose the integrable constraint q 3 = d(t), and introduce the non-integrable con-straint, and the constraint forces. The resulting equations governing the motion of the particleand the constraint forces

q 3 = d

q 3 = d,

f 1 q 1 + f 2 q 2 + f 3 d + h = 0,

d

dt

3i=1

mai1 q i

−

3i=1

3r=1

m∂air

∂q 1 q iq r +

∂U

∂q 1 = λ2f 1 + Fancon · a1,

d

dt

m

3i=1

ai2 q i

−

3i=1

3r=1

m∂air

∂q 2 q iq r +

∂U

∂q 2 = λ2f 2 + Fancon · a2,

d

dt

m

3

i=1

ai3 q i

−

3

i=1

3

r=1

m∂air

∂q 3 q iq r +

∂U

∂q 3 = λ1 + λ2f 3 + Fancon · a3.

(2.98)

In the interests of brevity, we have refrained from ornamenting U , f i, ai, and aik with a tilde inthe last four of these equations.

It is crucial to notice that if the non–integrable constraint were absent, then (2.98) wouldreduce to two sets of equations. The first of these sets, (2.98)4,5 would yield differential equationsfor the unconstrained motion, q 1(t), and q 2(t), of the particle, while the second set (2.98)6 wouldprovide the constraint force Fc = λ1a3 acting on the particle.

2.9.3 Approach II

In Approach II, we work directly with L = T

− U . Here, as

L = L

q 1, q 2, q 1, q 2, t

(2.99)



2.9 Lagrange’s Equations in the Presence of Constraints 39

the partial derivatives of L with respect to q 3 and q 3 are zero. Consequently, using (2.95) and(2.78) there are only two Lagrange’s equations:

d

dt ∂ L

∂ q 1−

∂ L

∂q 1 = Fc

·a1 + Fancon

·a1,

d

dt

∂ L

∂ q 2

− ∂ L

∂q 2 = Fc · a2 + Fancon · a2. (2.100)

Introducing the expression for the constraint force Fc and the non–integrable constraint, we findthe equations governing λ2, q 1(t) and q 2(t) are

f 1 q 1 + f 2 q 2 + f 3 d + h = 0,

d

dt

∂ L

∂ q 1

− ∂ L

∂q 1 = λ2f 1 + Fancon · a1,

d

dt ∂ L

∂ q 2−

∂ L

∂q 2 = λ

2f 2 + Fancon ·

˜a2

. (2.101)

Notice that λ1 does not feature in these equations. In addition, if no non–integrable constraintwere present, then the differential equations provided by Approach II are all that is needed todetermine q 1(t) and q 2(t).

2.9.4 A Particle Moving on a Sphere

To clarify the two approaches discussed above, we now consider the example of a particle movingon a smooth sphere whose radius R is a known function of time: R = d(t). The particle is subjectto a conservative force −mgE3 and a non–conservative force DReθ, where D is a constant. Lateron, we shall impose a non–integrable constraint on the motion of the particle.

For the problem at hand, it is convenient to use a spherical polar coordinate system:

q 1 = θ, q 2 = φ, q 3 = R. (2.102)

Using this coordinate system, the integrable constraint R = d(t) can be written in the form

ψ(r, t) = R − d(t) = 0. (2.103)

As the sphere is smooth, we can use the normality prescription:

Fc = λeR. (2.104)

The kinetic and potential energy of a particle in the chosen coordinate system are

T =

m

2 ˙

R

2

+ R

2

sin

2

(φ)˙

θ

2

+ R

2 ˙φ

2, U = mgR cos(φ). (2.105)

The constrained kinetic and potential energies are




T = m

2

d2 + d2 sin2(φ)θ2 + d2 φ2

, U = mgd cos(φ). (2.106)

Finally, the covariant basis vectors are

a1 = R sin(φ)eθ, a2 = Reφ, a3 = eR. (2.107)

Their constrained counterparts ai are easily inferred from these expressions.First, we shall use Approach II to obtain the equations governing θ(t) and φ(t). There are

two equations

d

dt

∂ L

∂ θ

− ∂ L

∂θ = F · a1

= Fc · a1 + Ddeθ · a1

= Dd2 sin(φ),

d

dt

∂ L

∂ φ

− ∂ L

∂φ = F · a1

= Fc · a2 + Ddeθ · a2

= 0. (2.108)

Evaluating the partial derivatives of the constrained Lagrangian, we find that these equationsbecome

d

dt

md2 sin2(φ)θ

= Dd2 sin(φ),

d

dt

md2 φ

− md2 sin(φ) cos(φ)θ2 − mgd sin(φ) = 0. (2.109)

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. (2.82)):

d

dt

∂L

∂ θ= mR2 sin2(φ)θ

−

∂L

∂θ = 0

= Fancon · R sin(φ)eθ,

d

dt

∂L

∂ φ= mR2 φ

−

∂L

∂φ = mR2 sin(φ) cos(φ)θ2 + mgR sin(φ)

= Fancon · Reφ,

d

dt

∂L

∂ R= m R

−

∂L

∂R = mR sin2(φ)θ2 + mR φ2 − mg cos(φ)

= Fancon · eR. (2.110)

Next, we impose the integrable constraint and introduce the constraint force Fc to find theequations of motion:



2.10 The Geometry of Lagrange’s Equations of Motion 41

d

dt

md2 sin2(φ)θ

= Dd2 sin(φ).

d

dt

md2 φ

− md2 sin(φ)cos(φ)θ2 − mgd sin(φ) = 0,

ddt

m d − md sin2(φ)θ2 + md φ2 − mg cos(φ)

= λ. (2.111)

Notice that the first two of these equations are identical to (2.109), while the third equation isan equation for the constraint force Fc.

We could now introduce an additional constraint:

f 1 θ + f 2 φ + f 3 R + h = 0. (2.112)

Using the normality prescription, the total constraint force on the particle is

Fc = λeR + λ2

f 1

R sin(φ)eθ +

f 2

Reφ + f 3eR

. (2.113)

To obtain the equations of motion for the case where the non–integrable constraint is active,we only need to introduce the constraint force associated with the non–integrable constrainton the right–hand side of (2.111), and to append to the resulting equations the non–integrableconstraint:

f 1 θ + f 2 φ + f 3 d + h = 0,

d

dt

md2 sin2(φ)θ

= Dd2 sin(φ) + λ2f 1.

d

dt

md2 φ

− md2 sin(φ) cos(φ)θ2 − mgd sin(φ) = λ2f 2,

d

dt

m d

−

md sin2(φ)θ2 + md φ2 − mg cos(φ)

= λ + λ2f 3. (2.114)

We leave it as an exercise to show what additional simplifications to these equations arise if thenon–integrable constraint were integrable with f 1 = 0, f 2 = 1, f 3 = 0 and h = 0. In this case,one will see that the particle moves on a circle of radius d sin(φ0).

2.10 The Geometry of Lagrange’s Equations of Motion

Some readers will have gained the perspective that the Lagrange’s equations of motion obtainedusing Approach II are projections of F = ma onto the covariant basis vectors for the unconstrainedcoordinates.

For those who haven’t yet found this perspective, let us recall the example of the particlemoving on the sphere of radius R = d(t). There, the two Lagrange’s equations for the θ andφ were obtained by taking the d sin(φ)eθ and deφ components of F = ma. These two vectors,

d sin(φ)eθ and deφ, form a basis for the tangent space to a point of the sphere. Furthermore,because the constraint force associated with the integrable constraint λeR is perpindicular to thesphere, this force didn’t appear in the two Lagrange’s equations.




An important feature of non–integrable constraints is that the constraint force associated withthese constraints are not decoupled from the equations governing the unconstrained motion. Thisdeficiency in Lagrange’s equations of motion can be removed by using alternate forms of F = mawhich are suited to non–integrably 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’s paper [9]. We also highly recommend that you readChapter 1 of Lanczos [29] for a related discussion on kinetic energy and geometry.

2a

1a

a3

O

1 q coordinate curve2 q coordinate curve

3 q = d(t) coordinate surface

Fig. 2.6. The configuration manifold M of a particle moving on a surface.

2.10.1 A Particle Subject to a Single Integrable Constraint

First, let us consider the case where the particle is subject to a single integrable constraint:

ψ(r, t) = q 3 − d(t) = 0, (2.115)

where d3(t) is a known function. When considered in E , the constraint ψ = 0 represents a movingtwo–dimensional surface - in this case a q 3 coordinate surface. This surface is known as theconfiguration manifold M (see Figure 2.6). The velocity of the particle relative to this surfacehas the representation

vrel = q 1a1 + q 2a2. (2.116)

At each point P of M, a1, a2 evaluated at P is a basis for the tangent plane to M at P . Wedenote this plane by T P M. We can also define a relative kinetic energy:

T rel = m

2 vrel · vrel =

2i=1

2k=1

m

2 ai · ak q i

q k

. (2.117)



2.10 The Geometry of Lagrange’s Equations of Motion 43

We now consider a particle moving on M. To calculate the distance travelled by the particle onthe 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√ vrel · vreldt

=

t1t0

2T rel

m dt. (2.118)

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

ds =

2T rel

m dt =

2i=1

2k=1

ai · akdq idq k. (2.119)

Notice that the measure of distance is defined by the kinetic energy T rel .With regard to Lagrange’s equations of motion, suppose that the constraint forces associated

with the integrable constraint are prescribed using normality: Fc = λa3. Then, Fc ·a1 = Fc ·a2 =0, and the constraint force does not appear in the first two Lagrange’s equations. In addition,Approach II can be used to obtain the differential equations governing q 1(t) and q 2(t):

d

dt

∂ L

∂ q 1

− ∂ L


d

dt

∂ L

∂ q 2

− ∂ L

∂q 2 = Fancon · a2. (2.120)

Imposing a non–integrable constraint on the motion of the particle will not change M. Further-more, this constraint will, in general, introduce constraint forces into the equations (2.120). Theseforces will destroy the decoupling that Lagrange’s equations achieves for integrable constraints.

It is a good exercise to convince yourself that the configuration manifold of (i) a particlemoving on a sphere of radius R = d(t) is the sphere, (ii) a particle moving on a cone is the cone,and (iii) a particle moving on a cylinder of radius r = d(t) is the cylinder. We also leave it as anexercise for you to prescribe the kinematical line–element ds for these manifolds.

2.10.2 A Particle Subject to Two Integrable Constraints

We now turn to the case where a particle is subject to two integrable constraints:

ψ1(r, t) = q 3 − d3(t) = 0, ψ2(r, t) = q 2 − d2(t) = 0. (2.121)

Notice that we have chosen the curvilinear coordinates so that the constraints are easily repre-sented. At each instant of time, the intersection of the two surfaces ψ1 = 0 and psi2 = 0 in

E defines a curve - in this case a q 1 coordinate curve (see Figure 2.7). The configuration manifoldM in this case corresponds to the q 1 coordinate curve. This curve.




O

r

a1

m

22q = d (t) coordinate surface

33q = d (t) coordinate surface

Fig. 2.7. The configuration manifold M of a particle moving on a curve. In this figure, a1 is tangent to

the q 1

coordinate curve corresponding to q 2

= d2

(t) and q 3

= d3

(t). The vectors a2

and a3

, which arenot shown, are normal to this q 1 coordinate curve.

We can easily represent the velocity vector of the particle relative to M, which we againdenote by vrel :

vrel = q 1a1. (2.122)

It should be clear that a1 is tangent to M. Indeed, denoting the tangent line to any point P of M by T P M, a1 evaluate at P is a basis vector for the one–dimensional space T P M. In addition,we can associate with vrel a relative kinetic energy:

T rel = m

2 vrel · vrel =

m

2 a1 · a1 q 1q 1. (2.123)

Paralleling the previous developments, the kinematical line–element for

M is

ds =

2T rel

m dt =

a1 · a1dq 1dq 1. (2.124)

With regard to Lagrange’s equations of motion, if the constraint forces satisfy normality Fc =λ1a3 + λ1a2, then we can easily find the differential equation governing q 1(t) using Approach II:

d

dt

∂ L

∂ q 1

− ∂ L

∂q 1 = Fancon · a1. (2.125)

Here, L = L

q 1, q 1, t

.

2.10.3 Concluding Remarks

We have very briefly touched on geometry here, and there are many more issues and ideas toexplore. For additional references, we recommend the papers of Casey [9, 10], and the texts of McConnell [32], Lanczos [29], Simmonds [49], and Synge and Schild [53].



2.11 A Particle Moving on a Helix 45

2.11 A Particle Moving on a Helix

As an illustrative example, we turn our attention to establishing results for a particle which isin motion on a helix (see Figure 2.8). 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 aprototypical problem to illustrate how the Serret-Frenet formulae and the Frenet triad et, en, ebcan be used to determine the motion of a particle on a space curve. Second, we can use thisexample to illustrate a non-orthogonal curvilinear coordinate system.4

E2

E3

E1

e t

en

eb

et

en

eb

Direction of

increasing

s and θ

Fig. 2.8. A helix and its associated Frenet triad et, en, eb. Here, et is the unit tangent vector, en isthe unit principal normal vector and eb = et × en is the binormal vector.

2.11.1 Curvilinear Coordinates

A helix is defined by the intersection of two surfaces: a cirle r = R and a helicoid z = cθ , wherec and R are constants. To conveniently define these surfaces, we define a curvilinear coordinatesystem:

q 1 = θ = tan−1

x2

x1

, q 2 = r =

x21 + x2

2,

q 3 = η = z − αrθ = x3 − α

x21 + x2

2 tan−1

x2

x1

. (2.126)

It is appropriate to notice that

4

This is a coordinate system where ai are not necessarily parallel to ai.




r = x1E1 + x2E2 + x3E3

= r cos(θ)E1 + r sin(θ)E2 + (η + αrθ) E3. (2.127)

Our labelling 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 asthe cylindrical and spherical polar coordinate systems.

2.11.2 Various Basis Vectors

The coordinates θ , r and η can be used to define bases for Euclidean three-space:

a1 = reθ + αrE3, a2 = er + αθE3, a3 = E3. (2.128)

In addition, using the representation of the gradient in cylindrical polar coordinates, one findsthat the contravariant basis vectors are

a1 = 1

reθ, a2 = er, a3 = E3 − αθer − αeθ. (2.129)

You should notice that ai · aj = δ ji - as expected. Further, neither the covariant basis nor the

contravariant basis are orthogonal.You may recall that the Frenet triad for the helix of radius R is

et = 1√ 1 + α2

(er + αE3) , en = −er, eb = 1√ 1 + α2

(E3 − αeθ) . (2.130)

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. (2.131)

These results also apply to a helix where α and R are functions of time. You should verify that

a1 is parallel to et, while a2

and a3

are in the plane formed by en and eb.

2.11.3 Kinematics

For a particle moving in Euclidean three-space, we have

v =3

i=1

q iai. (2.132)

From this result, we can immediately write

v = θ(reθ + αrE3) + r(er + αθE3) + ηE3. (2.133)

Furthermore, the kinetic energy of the particle is

T = m

2

r2 + r2 θ2 + (η + αrθ + αrθ)2

. (2.134)




When the particle is in motion on the helix, it is subject to two constraints:

φ1(r, t) = q 2 − R = 0,

φ2(r, t) = q 3 = 0. (2.135)

In preparation for writing down expressions for the constraint forces acting on a particle movingon the helix, you should calculate the gradient of these two functions.

2.11.4 Forces

We assume that an applied force Fa acts on the particle. In addition, we assume that the frictionis of the Coulomb-type. Consequently, if the particle is moving relative to the helix:

F = Fa + λ1a2 + λ2a3 + Ff , (2.136)

where

Ff = −µd||λ1a2 + λ2a3|| θa1

||θa1

||. (2.137)

On the other hand, if the particle is not moving relative to the helix, i.e., θ is constant, then

F = Fa + λ1a2 + λ2a3 + λ3a1. (2.138)

The friction force in this case is subject to the static friction criterion:

||Ff || ≤ µs||N||, (2.139)

where

Ff =

(λ1a2 + λ2a3 + λ3a1) · a1

||a1||

a1

||a1|| ,

N = λ1a2 + λ2a3 + λ3a1 − Ff . (2.140)

2.11.5 Balance of Linear Momentum and Lagrange’s Equations

For an unconstained particle moving in E 3, we have the three Lagrange’s equations:

d

dt

∂T

∂ q i

− ∂T

∂q i = F · ai. (2.141)

For the present coordinate system θ,r, η, these equations read:

d

dt

∂T

∂ θ= mr2 θ + mαr(η + αrθ + αrθ)

−∂T

∂θ = mαr(η + αrθ + αrθ) = F · (a1 = reθ + αrE3),

d

dt

∂T

∂ r = mr + mαθ(η + αrθ + αrθ)




−

∂T

∂r = mrθ2 + mαθ(η + αrθ + αrθ)

= F · (a2 = er + αθE3),

d

dt ∂T

∂ η = m(η + αrθ + αrθ)

−

∂T

∂η = 0

= F · (a3 = E3).

(2.142)

2.11.6 Equations of Motion for the Particle on the Helix

The equations of motion for the particle on the helix are obtained from the above equations bysubstituting for the resultant force and imposing the constraints. With some algebra, for the casewhere the particle is moving relative to the helix, we find three equations:

d

dt

m(1 + α2)R2 θ

= Fa · a1 − µd||λ1a2 + λ2a3|| θa1 · a1

||θa1||

= Fa · a1 − µd||λ1a2 + λ2a3||||a1|| θ

|θ

|,

d

dt

mα2Rθθ

− m(1 + α2)Rθ2 = Fa · a2 + λ1

− µd||λ1a2 + λ2a3|| θa1 · a2

||θa1||,

d

dt

mαRθ

= Fa · a3 + λ2

− µd||λ1a2 + λ2a3|| θa1 · a3

||θa1||,

(2.143)

where now

a1 = Reθ + αRE3, a2 = er + αθE3, a3 = E3. (2.144)These three equations provide a differential equation for the unconstrained motion of the particleand two equations for the unknowns λ1 and λ2.

For the case, where the motion of the particle is specified - i.e., the particle is not movingrelative to the helix, we find, from F = ma, three equations for the three unknowns:

λ3 = −Fa · a1, λ2 = −Fa · a2, λ1 = −Fa · a3. (2.145)

It remains to use the static friction criterion, but we leave this as an easy exercise.

2.11.7 Equations of Motion for the Particle on a Smooth Helix

In this case,

F = Fa + λ1a2 + λ2a3, (2.146)

and because Fc · a1 = 0, Lagrange’s equations of motion decouple:




d

dt

m(1 + α2)R2 θ

= Fa · a1,

d

dt

mα2Rθθ

− m(1 + α2)Rθ2 = Fa · a2 + λ1,

ddt

mαRθ

= Fa · a3 + λ2.

(2.147)

Consequently, the desired differential equation is

m(1 + α2)R2θ = Fa · (Reθ + αRE3), (2.148)

while the constraint force is

Fc = λ1a2 + λ2a3

=

mα2Rθθ − mRθ2 − Fa · a2

a2 +

mαRθ − Fa · a3

a3.

(2.149)

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.

2.11.8 An Example

To illustrate the previous equations, consider the case where the applied force is gravitationalFa = −mgE3. Then, from the above equations,

m(1 + α2)R2θ = Fa · (Reθ + αRE3) = −mgαR. (2.150)

Subject to the initial conditions θ(t0) = θ0 and θ(t0) = ω0, this equation has the solution

θ(t) = θ0 + ω0(t − t0) − gα2R(1 + α2)

(t − t0)2. (2.151)

Using this result, we find that the constraint force is

Fc =

mgαθ(t)

1 + α2 − mRθ2(t)

a2 +

mg

1 + α2 (E3 − α(θer + eθ)) (2.152)

where θ(t) is given above.

2.11.9 Some Observations

Suppose, one is only interested in determining the differential equation governing the uncon-

strained motion of the particle moving on a smooth helix. In other words, the constraint forcesare of no concern. One can obtain this differential equation, by imposing the constraints on theexpression for T :




T = m

2

R2 θ2 + (αRθ)2

=

m

2

(1 + α2)R2 θ2

. (2.153)

Furthermore,

ddt

∂

˜T

∂ θ

− ∂ ˜T

∂θ = F · a1 = F · (Reθ + αRE3) = Fa · (Reθ + αRE3). (2.154)

A quick calculation shows that the resulting differential equation is identical to that obtainedpreviously (2.150).

Clearly, Lagrange’s equations calculated using Approach II (i.e., using T ) has its advantages,but it cannot accommodate dynamic friction forces. It is, however, the standard approach to

Lagrange’s equations in the literature and textbooks. You should note that ∂ T ∂ r

= ∂ T ∂ η = ∂ T

∂r = ∂ T

∂η

= 0. Consequently, we cannot recover the other 2 Lagrange’s equations once we have imposedthe constraint.

2.12 Exercises

2.1 Which of the following force fields are conservative/nonconservative?

P = x1E1 + x3E2,

P = x2E1 + x1E2,

P = x1x2E1,

P = −L sin(θ)E1 + L cos(θ)E2. (2.155)

For the conservative force fields, what are the associated potential energies?2.2 Consider a particle of mass m which is moving in E . Suppose the only forces acting on the

particle are conservative. Starting from the work–energy theorem, prove that the total energyE of the particle is conserved.

Suppose during a motion, where the initial conditions r0 and v0 are known, the positionr(t1) at some later time t1 is known. Argue, that the conservation of energy can be used todetermine the speed v of the particle. Give three distinct physical examples of applicationsof this result.

2.3 In contrast to Exercise 2.2, consider a particle which is moving on a smooth surface. Theconstraint force acting on the particle is prescribed using normality, and the applied forcesacting on the particle are conservative. Prove that E is again conserved. In addition, showthat the speed of the particle can be determined at a known position r(t1) if the initialposition and velocity vectors are known. Finally, give three distinct physical examples of theapplication of this result.

2.4 A particle is free to move on a smooth horizontal surface x3 = 0. At the same time, a roughplane propels the particle in the E1 direction. That is, the constraints on the motion of theparticle are

ψ1 = ψ1(r) = x3 = 0, ψ2 = ψ2(r, t) = x1 − f (t) = 0. (2.156)

Give a prescription for the constraint force acting on the particle.



2.12 Exercises 51

2.5 A particle of mass m is free to move on the inner surface of a rough sphere of constant radiusR0. The center of the sphere is located at the origin O . The particle is attached to a point A

whose position vector is aEx + bEy by a linear spring of unstretched length L and stiffnessK . A vertical gravitational force −mgE3 also acts on the particle.

(a) Using a spherical polar coordinate system, derive expressions for the acceleration vectora 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) What is the constraint force Fc acting on the particle?(d) If the particle is moving relative to the surface, show that the equations governing themotion of the particle are

mR0(φ − sin(φ) cos(φ)θ2) = mg sin(φ) − K (x − L)x · eφ

x

−µdN φ φ2 + sin2(φ)θ2

,

1

R0 sin(φ)

d

dtmR2

0

sin2(φ)θ = −

K (

x −

L)x · eθ

x−µdN sin(φ)θ

φ2 + sin2(φ)θ2,

(2.157)

where x = R0eR − aEx − bEy.(e) Show that the normal force exerted by the surface on the particle is

N = −mR0( φ2 + sin2(φ)θ2)eR +

mgE3 · eR + K (x − L)

x · eR

x

eR. (2.158)

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

Fc = mgE3 + K (x − L) x

x . (2.159)

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 relativeto the surface.(h) When is it possible for the particle to loose contact with the surface?(i) If the spring is removed, and the surface is assumed to be smooth, prove that the angularmomentum HO · E3 is conserved. Using this conservation, show that the equations governingthe motion of the particle simplify to

θ = h

mR2

0 sin2

(φ)

, mR0φ

−

cos(φ)

sin3

(φ)

h

mR2

02

= mg sin(φ). (2.160)

In these equations, h = H0 · E3.




2.6 The balance of linear momentum for a particle whose mass m varies as a function of time t is

d

dt (mv) = F, (2.161)

where F is the resultant force acting on the particle.(a) Suppose that a curvilinear coordinate system q i has been selected for Euclidean three-space E 3. Starting from the results

ai = ∂ r

∂q i, v =

3i=1

q iai, (2.162)

establish, Lagrange’s equations of motion for the particle of varying mass:

d

dt

∂T

∂ q i

− ∂T

∂q i = G · ai = F · ai. (2.163)

Here, G = mv.

(b) For a particle of varying mass, establish the work–energy theorem:

d

dt (mT ) = F · G. (2.164)

(c) A rocket of mass m = m(t) is in motion about a fixed planet. The external force actingon the rocket is assumed to be composed of a conservative force P and a thrust force K. Thepotential energy for the conservative force is

U = −αm

||r|| , (2.165)

where r is the position vector of the rocket relative to the fixed center of the planet. Usinga spherical polar coordinate system, what are the equations governing the motion for the

rocket? Suppose it is desired to fly the rocket radially outward from the planet such that

R = 10t2 + R0, θ = θ0, φ = φ0. (2.166)

What is the thrust force K needed to execute this motion?2.7 Recall that, for a parabolidal coordinate system u ,v ,θ,

a1 = ∂ r

∂u = ver + uE3, a2 =

∂ r

∂v = uer − vE3, a3 =

∂ r

∂θ = uveθ, (2.167)

and

a1 = 1

u2 + v2a1, a2 =

1

u2 + v2a2, a3 =

1

uveθ. (2.168)

(a) Consider a particle of mass m which is acted upon by a force F and is free to move in

E .

Show that the equations of motion of the particle are



2.12 Exercises 53

d

dt

m(u2 + v2)u

− m(u2 + v2)u − mv2uθ2 = F · a1,

d

dt m(u2 + v2)v

− m(u2 + v2)v − mu2vθ2 = F · a2,

ddt

mu2v2 θ

= F · a3. (2.169)

(b) Next, we are interested in a particle which is moving on the parabola of revolution:

c2 = −z +

z2 + r2, (2.170)

where c is a constant. A vertical gravitational force −mgE3 acts on the particle. Using theresults of (a), derive the equations governing the unconstrained motion of the particle andshow that the normal force acting on the particle is

N = −

mu2c + mu2cθ2 + mgc

a2. (2.171)

Show that the two second-order differential equations governing the motion of the particlecan be written as a single second–order differential equation:

m(u2 + c2)

u + mu2u − h2

mu3c2 = −mgu, (2.172)

where h is a constant. (This constant is none other than HO · E3 which is conserved).(c) Show that the solutions of (2.172) conserve the energy

E = m

2 (u2 + c2)u2 +

h2

2mu2c2 +

mg

2 (u2 − c2). (2.173)

Where does this energy come from?2.8 This problem is adapted from the texts of McConnell [32] and Synge and Schild [53]. It tests

your familiarity with partial differentiation.Recall the covariant component forms of Lagrange’s equations of motion for a particle whichis in motion in Euclidean three space under the influence of a resultant external force F =3

i=1 F iai = 3

i=1 F iai:5

d

dt

∂T

∂ q k

− ∂T

∂q k = F · ak, (2.174)

where

T = T (q r, q s) = m

2

3i=1

3k=1

aik q iq k, (2.175)

andaik = aik(q r) = ai · ak, aik = aik(q r) = ai · ak. (2.176)

You should notice that aik = aki and a

ik

= a

ki

.5 The indices i, j, k, r, and s range from 1 to 3.




Here, we wish to show that Lagrange’s equations can be written in two other equivalentforms. The first one is the covariant form:

m

3

i=1

aki q i + m

3

i=1

3

s=1

[si,k]q iq s = G

·ak = F k , (2.177)

where the Christoffel symbol of the first kind is defined by

[si,k] = ∂ as

∂q i · ak. (2.178)

It is important to note that [s i, k] = [i s, k]. The second form of Lagrange’s equations isknown as the contravariant form:

mq k + m

3i=1

3s=1

Γ ksi q iq s = G · ak = F k, (2.179)

where the Christoffel symbol of the second kind is defined by

Γ kij = ∂ ai

∂q j · ak. (2.180)

Notice that Γ kij = Γ kji . This form of Lagrange’s equations is used in numerical simulations of

equations of motion of numerous mechanical systems.6

(a) First, we dispense with a crucial preliminary result. For any vector b, show that thecovariant and contravariant components are related by

b1b2b3

=

a11 a12 a13

a12 a22 a23a13 a23 a33

b1

b2

b3

, (2.181)

and b1

b2

b3

=

a11 a12 a13

a12 a22 a23

a13 a23 a33

b1

b2b3

. (2.182)

In other words, the covariant components are linear combinations of the contravariant com-ponents and vice versa.(b) Show that

3k=1

aikakj =3

k=1

akiakj =3

k=1

akiajk = δ ji . (2.183)

(c) Show that the Christoffel symbols have the representations

6 Most numerical integration packages assume that the differential equations to be integrated are of the

form x = f (x, t). By defining the set of variables (states) x1 = q 1

, . . . x3 = q 3

, x4 = q 1

, . . . x6 = q 3

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



2.12 Exercises 55

[si,k] = 1

2

∂aki

∂q s +

∂ask

∂q i − ∂asi

∂q k

, (2.184)

and

Γ kij =

3r=1

akr [ij,r]. (2.185)

(d) Starting from Lagrange’s equations,

d

dt

∂T

∂ q k

− ∂T

∂q k = F · ak, (2.186)

derive the following representation for the covariant component form:

m

3i=1

aki q i + m

3i=1

3s=1

[si,k]q iq s = G · ak = F k. (2.187)

Hints: Expand the partial derivatives of T using the representation for this energy given

previously. Then take the appropriate time derivative and reorganize the resulting equationusing the aforementioned symmetries. You may need to relabel certain indices to obtain thedesired results.(e) Starting from Lagrange’s equations in the form:

m

3i=1

aki q i + m

3s=1

3i=1

[si,k]q iq s = G · ak = F k, (2.188)

derive the following representation for the contravariant component form:

mq k + m

3

s=1

3

i=1

Γ ksi q iq s = G · ak = F k. (2.189)

Hints: Multiply the covariant form by ask and sum over k. After some rearranging andrelabelling of the indices you should get the final desired result.Notice that the covariant component form and contravariant component forms of these equa-tions can be viewed as linear combinations of each other.(f ) For which coordinate system do the Christoffel symbols vanish?

2.9 Recall that for spherical polar coordinates, R,φ,θ, the covariant basis vectors are

a1 = eR, a2 = Reφ, a3 = R sin(φ)eθ, (2.190)

while the contravariant basis vectors are

a1 = eR, a2 = 1

Reφ, a3 =

1

R sin(φ)eθ. (2.191)

Furthermore, the linear momentum and kinetic energy of a particle of mass m are




G = m Ra1 + m φa2 + mθa3, T = m

2

R2 + R2 φ2 + R2 sin2(φ)θ2

. (2.192)

(a) For a particle of mass m which is in motion in E 3 under the influence of a resultantforce 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 which is in motion in E 3 under the influence of a resultant forceF, establish the three contravariant components of Lagrange’s equations of motion. In yoursolution avoid explicitly calculating the 27 Christoffel symbols of the second kind.



Part II

Dynamics of a System of Particles





3

Kinematics and Dynamics of a System of Particles

3.1 Introduction

In this chapter we establish Lagrange’s equations for a system of particles. The derivation of theseequations is based on the construction of a single particle. This construction follows Casey [9]and it explicitly shows how the balances of linear momentum for each of the particles are equiv-alent to Lagrange’s equations. An alternative derivation, which also illustrates this equivalenceis presented in Chapter 15 of Synge and Griffith [52].

For many specific problems, one can obtain Lagrange’s equations by merely calculating thekinetic and potential energies of the system. This approach is used in most dynamics textbooksand a construction of a single particle is never mentioned.1 Indeed, once we establish Lagrange’sequations we can also ignore the explicit construction of the single particle. However, for manycases - which are not possible to treat using the approach adopted in most dynamics textbooks- we find that the construction of a single particle allows us to tremendously increase the rangeof application of Lagrange’s equations. For instance, in one of the examples discussed below,dynamic Coulomb friction is accommodated (see Section 3.10).

3.2 The System of N Particles

Here, we are interested in establishing Lagrange’s 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 three-dimensional Euclideanspace E 3. For the particle of mass mi (see Figure 3.1), the position vector is

ri =3

j=1

xji Ej . (3.1)

We also recall that the kinetic energy of the particle is

1 These texts either use Hamilton’s principle (also known as the principal of least action) to deriveLagrange’s equations or the principal of virtual work.



60 3 Kinematics and Dynamics of a System of Particles

O

E

2

i E3

E

ir

iF

1

m

Fig. 3.1. A single particle of mass mi in E 3. The position vector of this particle is ri and the resultantexternal force acting on the particle is Fi.

T i = 1

2mivi · vi, (no sum on i). (3.2)

It is also convenient to recall that the linear momentum of the particle of mass m

i is Gi = m

iri =mivi (again, no sum on i).The resultant force acting on the particle of mass mi has the representation

Fi =3

j=1

F ji Ej . (3.3)

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

Fi = mi vi, (i = 1, . . . , N (no sum on i)). (3.4)

For the system of particles, we can define a center of mass C . This point, which lies in E 3,has the position vector r, where

r = 1

m1 + . . . mN

(m1r1 + . . . + mN rN ) . (3.5)

In the sequel, we will examine a particle of mass m moving in E 3N and it is important not toconfuse this particle with C .

3.3 The Single Particle’s Position and Force Vectors

We now follow Casey [9], and construct a single particle of mass m which is moving in E 3N . Theposition vector of this particle is r. The kinetic energy T of this particle is the same as the kineticenergy of the system of particles that it represents and the force on the particle is such that

T = Φ · r, Φ = mr. (3.6)



3.3 The Single Particle’s Position and Force Vectors 61

In Casey [9], m is chosen (without loss in generality) to be the sum of the masses: m = m1 +. . . + mN . Here, we only assume that m > 0. The interested reader is also referred to Chapter 1of Lanczos [29] for a related discussion of the geometry of a mechanical system.

The space E 3N , which is known as the configuration space, is equipped with a Cartesian

coordinate system. Consequently, for any vector b,

b =3N

K =1

bK eK =N

i=1

3j=1

b3i+j−3e3i+j−3. (3.7)

Here, e1, . . . , e3N is a fixed orthonormal basis for E 3N . We also define two other sets of basesvectors:

e3i+j−3 =

mi

m e3i+j−3,

e3i+j−3 =

m

mi

e3i+j−3. (3.8)

Here i = 1, . . . , N and j = 1, 2, 3. Notice that

eK · eJ = δ J K . (3.9)

The particle has a position vector r and a force vector Φ. Both of these vectors have representa-tions similar to that for b.

3.3.1 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 of the system of particles:

T = 1

2 mv · v = 1

2

N i=1

mivi · vi. (3.10)

Substituting,

v = r =N

i=1

3j=1

r3i+j−3e3i+j−3, vi = ri =3

j=1

xji Ej , (3.11)

and equating kinetic energies, we find one solution for r:

r =N

i=1

3j=1

(ri · Ej)

mi

m e3i+j−3 =

N i=1

3j=1

(ri · Ej) e3i+j−3. (3.12)

It’s a good exercise to show that the other solutions for r can be shown to be equivalent to (3.12)

modulo a translation of the origin and a relabelling of axes of E 3N . Notice how the mass ratiosmi

m are subsumed into the basis vectors e3i+j−3 in (3.12).




3.3.2 Prescription for the Force Vector Φ

The force vector Φ is prescribed by the requirement that

mv = Φ. (3.13)

Substituting (3.12) and using the balance of linear momentum for each particle, one is lead tothe prescription

Φ =N

i=1

3j=1

(Fi · Ej)

m

mi

e3i+j−3 =N

i=1

3j=1

(Fi · Ej) e3i+j−3. (3.14)

Again its interesting to note how the mass ratios mi

m are subsumed into the basis vectors e3i+j−3

in (3.12).For the sequel it is often convenient to recall the prescription

Φ · e3i+j−3 = Fi · Ej , (3.15)

which follows from (3.14). It is also an easy exercise to show that the work-energy theorems forthe individual particles, T i = Fi · vi, (no sum on i), gives a work-energy theorem for the particleof mass m:

T = Φ · v. (3.16)

This theorem can be used to establish energy conservation results for the system of particles.

3.4 Curvilinear Coordinates

In many problems, Cartesian coordinates for ri and r are not very useful. For example, to specifyintegrable constraints on the motions of the N particles, we normally use a curvilinear coordinatesystem:

q K = q K (x11, x2

1, x31,...,x1

N , x2N , x3

N ), (K = 1, . . . , 3N ),

xji = x

ji (q 1,...,q 3N ), (i = 1, . . . , N and j = 1, 2, 3). (3.17)

We then use the curvilinear coordinates q 1, . . . , q 3N to define basis vectors for E 3N :

aK = ∂ r

∂q K =

∂

∂q K

N

i=1

3j=1

ri · Ej = xj

i

e3i+j−3

, (3.18)

and

aJ = grad(q J ) =

N

i=1

3

j=1

∂ q J

∂xji

e3i+j−3. (3.19)

It can be shown that aK · aJ = δ J K . For most problems, it is not necessary to explicitly calculate

the contravariant basis vectors aJ .



3.4 Curvilinear Coordinates 63

3.4.1 Some Correspondances

For future, reference it is interesting to note that

Φ · aK =

N i=1

Fi · ∂ ri∂q K . (3.20)

To establish this result a series of identities is invoked:

Φ · aK = Φ · ∂ r

∂q K

=

N

i=1

3j=1

(Fi · Ej) e3i+j−3

· ∂

∂q K

N s=1

3 p=1

(rs · E p = x ps) e3s+ p−3

=N

i=1

3j=1

N s=1

3 p=1

(Fi · Ej)

∂ x p

s

∂q K

e3i+j−3 · e3s+ p−3

=N

i=1

3j=1

N s=1

3 p=1

(Fi · Ej)

∂ x p

s

∂q K

δ isδ j p

=

N i=1

3j=1

(Fi · Ej)

∂ x

ji

∂q K

=N

i=1

Fi · ∂

∂q K

3

j=1

xji Ej

=

N i=1

Fi · ∂ ri

∂q K . (3.21)

We conclude that the desired identity has been establish.The identity (3.20) is similar to several other results pertaining to the system of particles and

the single particle of mass m. For instance, consider a function

Γ = Γ (r, t) = Γ (r1, . . . , rN , t) . (3.22)

With some minor manipulations, we find several representations for the derivative of Γ withrespect to a curvilinear coordinate:

∂Γ

∂q K =

∂Γ

∂ r · ∂ r

∂q K =

N i=1

∂ Γ

∂ ri

· ∂ ri

∂q K . (3.23)

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.




3.5 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 v1 = F1. . .

mN vN = FN

is equivalent to mv = Φ, (3.24)

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 i=1

q K aK . (3.25)

Consequently, we have the two intermediate results

∂T

∂ q K = mv · aK ,

∂T

∂q K = mv · aK . (3.26)

With the assistance of these results one finds that

d

dt

∂T

∂ q K

− ∂T

∂q K =

d

dt (mv · aK ) − mv · aK

= d

dt (mv) · aK

= Φ · aK . (3.27)

We have just established Lagrange’s equations for a system of particles:

d

dt

∂T

∂ q K

− ∂T

∂q K = Φ · aK . (3.28)

These equations give the motion r = r(t) of the particle of mass m and, hence, the motion of thesystem of particles.

3.6 Constraints and Constraint Forces

Now suppose that the system of particles is subject to an integrable constraint:

φ(r1, r2, . . . , rN , t) = 0. (3.29)

To motivate this functional form, consider two particles m1 and m2 which are connected by arigid rod of length R. Then the constraint imposed by the rod on their motion is ||r1−r2||−R = 0.

This constraint is of the form (3.29).The constraint (3.29) can be differentiated to yield a constraint of the form:



3.6 Constraints and Constraint Forces 65

N i=1

f i · vi + g = 0, (3.30)

where

f i = ∂φ

∂ ri

,

g = ∂φ

∂t . (3.31)

Prescribing the constraint forces using normality, we find that the constraint force acting on theith particle is

Fci = λf i = λ∂φ

∂ ri

. (3.32)

Clearly, this prescription can also be applied to non-integrable constraints.For the constraint ||r1 − r2|| − R = 0 mentioned earlier, the normality prescription yields:

Fc1 = λ r1

−r2

||r1 − r2|| ,

Fc2 = λ r2 − r1||r1 − r2|| . (3.33)

Notice that these constraint forces point along the rod - which is what we would expect fromphysical grounds. Furthermore, Fc2 = −Fc1 - that is none other than Newton’s third law.2

For the single particle, we can express the constraint φ = 0 as a constraint on the motion of the single particle of mass m:

φ = φ(r, t) = 0. (3.34)

This implies that the motion of the single particle is subject to the constraint and, using thenormality prescription, a constraint force Φc, respectively,

f · v + g = 0, Φc = λf (3.35)

where

f = ∂φ

∂ r, g =

∂φ

∂t . (3.36)

Furthermore, the integrable constraint implies that the particle moves on a configuration manifoldM which is 3N − 1 dimensional subset of the configuration space E 3N .

It is important to note that the prescriptions of Φc and Fci are consistent with each other.Indeed, the equivalence of the constraint forces Fci acting on the system of particles and theconstraint force Φc acting on the single particle of mass m can be inferred using (3.20) and(3.23).

2 For more details on this interesting result see Noll [35] and O’Reilly and Srinivasa [39].




3.7 Conservative Forces and Potential Energies

Suppose that the conservative forces Fcon1, . . . , FconN acting on the each respective particle of the system of particles have a potential energy

U (r1, r2,..., rN ) . (3.37)

Notice that we are presuming that this is the most general form of the potential energy of conservative forces in a system of particles. You should check that it encompasses the inversegravitational law between two particles and the potential energy of a spring force between twoparticles.

Equating the time derivative of U to the negative power of the conservative forces, we findthat

Fconi = −∂U

∂ ri

. (3.38)

For the single particle, we can express the potential energy U as a function of the motion of thesingle particle of mass m:

U (r1, r2,..., rN ) = U (r). (3.39)

Calculating ˙U and equating it to −Φcon · v, we find that3

Φcon = −∂U

∂ r . (3.40)

Again, it is important to note that the prescriptions of Φcon and Fconi are consistent with eachother. As with constraint forces, the equivalence can be inferred using (3.20) and (3.23).

3.8 The Lagrangian

In many mechanical systems, the sole forces which 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. (3.41)

Using the Lagrangian, its easy to see that Lagrange’s equations for a system of particles is

d

dt

∂L

∂ q K

− ∂L

∂q K = Φncon · aK . (3.42)

Here, Φncon are the non-conservative forces acting on the system of particles:

Φncon = Φ−Φ p. (3.43)

3 Essentially, one is solving the equation Φcon + ∂U ∂ r · v = 0 for all possible motions of the system. In

order for this to hold, the terms in the paranthesis must cancel and we arrive at (3.40).



3.9 Lagrange’s Equations: Constrained System of Particles 67

Notice that if Φncon = 0 - as it is in many celestial mechanics problems - one can write down theequations of motion for a system of particles without every having to calculate an accelerationvector!

When constraints are present, a form of Lagrange’s equations similar to (3.42) can also be

obtained provided the constraints are integrable and the constraint forces satisfy the normaliltyprescription. In this case, the equations are identical to (3.42) with the kinetic and potentialenergies being replaced by their constrained counterparts.4

3.9 Lagrange’s Equations: Constrained System of Particles

Consider a system of particles which are subject to one integrable and one non-integrable con-straint. We choose the curvilinear coordinates to express these constraints as

φ(r, t) = q 3N − r(t) = 0,

f · v + g = 0. (3.44)

Assuming that the constraint forces associated with these constraints satisfy the normality pre-scription, the force vector Φ has the decomposition

Φ = −3N

K =1

∂U

∂q K aK + λ1a3N + λ2f + Φo. (3.45)

Notice that the contributions due to the conservative forces have been subsumed in the potentialenergy U .

3.9.1 Approach I

Lagrange’s equations of motion decouple into two sets: d

dt

∂T

∂ q S

− ∂T

∂q S = − ∂U

∂q S + λ2f · aS + Φo · aS

q3N =r,q3N =r

, (3.46)

and d

dt

∂T

∂ q 3N

− ∂T

∂q 3N = − ∂U

∂q 3N + λ1 + λ2f · a3N + Φo · a3N

q3N =r,q3N =r

, (3.47)

where S = 1, ..., 3N − 1, and the constraint q 3N = r(t) is imposed after the partial derivativesof T and U have been calculated.

4 That is, in the notation presented below, T is replaced by T and U is replaced by U . As a result, L isreplaced by L = T − U .




3.9.2 Approach II

For comparison, we now use the second approach. First, we impose the integrable constraint onthe kinetic and potential energies, in addition to the basis vectors and constraints. For instance,

T = T (q 1, . . . , q 3N −1, q 1, . . . , q 3N −1, t)

= T (q 1, . . . , q 3N −1, q 3N = r(t), q 1, . . . , q 3N −1, q 3N = r(t)). (3.48)

As in the case of a single particle, T can be used to construct a metric for the configurationmanifold M

Because we have eliminated the coordinate associated with the integrable constraint, we canonly obtain 3N − 1 Lagrange’s equations:

d

dt

∂ T

∂ q S

− ∂ T

∂q S = − ∂ U

∂q S + λ2f · aS + Φo · aS . (3.49)

Further, no information on the constraint force enforcing the integrable constraint is obtained

using this approach.Once the integrable constraint has been imposed,s the single particle of mass m moves on the3N −1 dimensional submanifold of E 3N . As before, this submanifold is known as the configurationmanifold M. A measure of the distance travelled by the single particle on this manifold can befound using T . Indeed, this energy can be decomposed as

T = T 0 + T 1 + T 2, (3.50)

where

T 0 = m

2 a3N · a3N r

2,

T 1 = m

3N −1

K =1

aK · a3N q K r,

T rel = T 2 = m

2

3N −1K =1

3N −1J =1

aK · aJ q K q J . (3.51)

The inertia metric for M is

ds =

2T rel

m

dt. (3.52)

This metric may also be written as

ds =

3N −1K =1

3N −1J =1

aK · aJ dq K dq J . (3.53)

As was the case previously, the imposition of a non-integrable constraint won’t change M or itsmetric.



3.10 Examples 69

3.10 Examples

For illustrate the construction of a single particle and the derivation of Lagrange’s equations fora system of particles, we now consider three relatively simple examples. The examples all feature

a system of two particles.

3.10.1 System I

Consider the system of particles shown in Figure 3.2. A particle of mass m1 is connected by aspring of stiffness K 1 and unstretched length L1 to a fixed support. It is also connected, by aspring of stiffness K 2 and unstretched length L2 to a particle of mass m2. Both particles areconstrained to move in the E1 direction:

r1 = (L1 + x1)Ex, r2 = (L1 + L2 + x2)Ex. (3.54)

Notice that x1 and x2 measure the displacement of the particles from the unstretched springstates. In the sequel, we denote

q 1 = x1,

q 2 = x2,

q 3 = r1 · E2,

q 4 = r1 · E3,

q 5 = r2 · E2,

q 6 = r2 · E3. (3.55)

It is easy to see that aK = eK and aK = eK for K = 1, . . . , 6.

g

1

Smooth surface

E2

E O

m1

m2

Fig. 3.2. The system of two particles moving on a smooth horizontal surface. The particles are connectedby linear springs.

We now assume that the line that the masses move on is smooth. Further, it is easy to seethat the constrained kinetic and potential energies of the system are

T = 1

2m1 x2

1 + 1

2m2 x2

2, U = 1

2K 1x2

1 + 1

2K 2(x2 − x1)2. (3.56)




Because the surface that the individual particles move on is smooth, it is not too difficult tosee that the 4 constraint forces acting on the single particle of mass m satisfy the normalityprescription:

Φc = λ2a3 + λ3a4 + λ4a5 + λ5a6. (3.57)

We now have all the kinematical and kinetic ingredients in place.Lagrange’s equations of motion for the unconstrained motion of the system are:

d

dt

∂ T

∂ q α

− ∂ T

∂q α = Φ · aα = − ∂ U

∂q α, (3.58)

where α = 1, 2. Explicitly, these equations have the form

m1x1 = −K 1x1 − K 2(x1 − x2),

m2x2 = −K 2(x2 − x1).

It should be apparently that for this example, we didn’t explicitly use the construction of thesingle particle.

3.10.2 System II

Consider the system of particles shown in Figure 3.2. A particle of mass m1 is connected by aspring of stiffness K 1 and unstretched length L1 to a fixed support. It is also connected, by aspring of stiffness K 2 and unstretched length L2 to a particle of mass m2. Both particles areconstrained to move in the E1 direction:

r1 = (L1 + x1)Ex, r2 = (L1 + L2 + x2)Ex. (3.59)

Notice that x1 and x2 measure the displacement of the particles from the unstretched springstates.

In the sequel, we again denote

q 1 = x1,

q 2 = x2,

q 3 = r1 · E2,

q 4 = r1 · E3,

q 5 = r2 · E2,

q 6 = r2 · E3. (3.60)

We now assume that the line that the masses move on is smooth. Furthermore, the system issubject to a non-integrable constraint:

x22 x1 + x1 x2 − f (t) = 0 . (3.61)

As before, the 4 constraint forces enforcing the integrability constraints satisfy the normalityprescription. The constraint force acting on the single particle is



3.10 Examples 71

Φc = λ1(x22a1 + x1a2) + λ2a3 + λ3a4 + λ4a5 + λ5a6. (3.62)

The first constraint force in this expression represents the constraint force associated with thenon-integrable constraint.

It is easy to see that the constrained kinetic and potential energies of the system are indenticalto those for the previous example:

T = 1

2m1 x2

1 + 1

2m2 x2

2, U = 1

2K 1x2

1 + 1

2K 2(x2 − x1)2. (3.63)

Lagrange’s equations of motion for the unconstrained motion of the system are

d

dt

∂ T

∂ q α

− ∂ T

∂q α = Φ · aα = − ∂ U

∂q α + λ1(x2

2a1 + x1a2) · aα. (3.64)

where α = 1, 2. Explicitly,

m1x1 =

−K 1x1

−K 2(x1

−x2) + λ1x2

2,

m2x2 = −K 2(x2 − x1) + λ1x1.

3.10.3 System III

Consider the system of particles shown in Figure 3.3. Many of its features were described earlier.In the sequel, we denote the coordinates q 1, . . . , q 6 as in the previous two examples.

g

1

Rough horizontal surface

E2

E O

m1

m2

Fig. 3.3. A system of two particles moving on a rough horizontal surface.

We now assume that the line that the masses move on is rough. The constraint forces actingon each particle are

Fc1 = N 1yE2 + N 1zE3 − µd||N 1yE2 + N 1zE3|| x1E1

||x1E1|| ,

Fc2 = N 2yE2 + N 2zE3 − µd||N 2yE2 + N 2zE3|| x2E1||x2E1|| .




Using the correspondance between Fi and Φ, the constraint force acting on the single particlecan be calculated:

Φc = −µd N 21y + N 21z

x1

|x1

|

e1 − µd N 22y + N 22z

x2

|x2

|

e2 + N 1ye3 + N 1ze4 + N 2ye5 + N 2ze6. (3.65)

It is a good exercise to verify this result and compare it to those for the previous two examples.Further, the kinetic energy and constrained potential energy of the system are, as before,

T = 1

2m1(x2

1 + y21 + z21) + 1

2m2(x2

2 + y22 + z22), U = 1

2K 1x2

1 + 1

2K 2(x2 − x1)2. (3.66)

Lagrange’s equations of motion for the system are

d

dt

∂T

∂ q K

− ∂T

∂q K = Φ · aK . (3.67)

Explicitly, we now have the six equations

m1x1 = −K 1x1 − K 2(x1 − x2) − µdm1g x1

|x1| ,

m2x2 = −K 2(x2 − x1) − µdm2g

x2

|x2| ,

0 = N 1y − m1g,

0 = N 1z,

0 = N 2y − m2g,

0 = N 2z. (3.68)

The last four equations yield the constraint forces.

3.10.4 Remarks

Notice that for Systems I and II, it was not necessary to calculate aK or aK . In otherwords, noexplicit mention of the single particle is necessary. Finally, because the constraint forces in SystemIII were not precribed by normality, it was necessary to calculate all 6 of Lagrange’s equations.

3.11 Exercises

3.1 What are the kinematical line elements ds, generalized coordinates and configuration mani-folds 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.

3.2 Why is it a mistake to say that constraint forces do no work?3.3 Consider Systems I, II and III discussed in Section 3.10. Suppose a time-dependent force

P (t)E1 acted on the particle m2.5 Then, what are the equations of motion of the system?5 This force is not conservative.



3.11 Exercises 73

3.4 Again, consider Systems I, II and III discussed in Section 3.10. Suppose, in addition to thesprings, there are viscous dashpots in these systems.6 Then, what are the equations of motionof the system?

3.5 Many authors assume that the kinetic energy of a system of particles is a positive definite

function of the velocities q K

. Letting T =3N

I =13N

J =1m

2 aIJ q I

q J

, this is equivalent to sayingthat the matrix whose components are m

2 aIJ is positive definite. You probably recall off the

top of your head that a matrix C is positive definite if, for all nonzero x, xT Cx > 0 andxT Cx = 0 only when x = 0.Using the spherical pendulum as an example, show that certain representations of T arenot always positive-definite. Specifically, if one uses spherical polar coordinates, then thepositive-definitness breaks down at the singularities of this coordinate system.

3.6 In Casey’s construction of the single particle what are the distinctions between the baseseK , eK and eK ? For a given system of two particles, how does one construct thesebases for E 6?

3.7 Here, we are interested in establishing a particular representation for the equations governingthe motion of two unconstrained particles. In a subsequent exercise, one can use imposeconstraints to yield the equations of motion of a pendulum system.

g

E

2

E3

E

1

O

2m1m

Fig. 3.4. A system of two particles

Consider the system of particles shown in Figure 3.4. The particles are free to move in E 3under the influences of resultant external forces F1 and F2, respectively.(a) To establish the equations of motion for the single particle, we use a cylindrical polarcoordinate system r1, θ1, z1 for the particle of mass m1. For the second particle, it is con-venient to describe its motion with the assistance of the relative position vector r21 = r2 -r1. We describe this vector using a spherical polar coordinate system R2, φ2, θ2. Show thatthe position vector of the single particle is

r = r1 cos(θ1)e1 + r1 sin(θ1)e2 + z1e3 + (r1 cos(θ1) + R2 sin(φ2) cos(θ2))e4

+(r1 sin(θ1) + R2 sin(φ2) sin(θ2))e5 + (z1 + R2 cos(φ2))e6. (3.69)

(b) Using r and the curvilinear coordinate system it enduces on E 6,

q 1 = r1, q 2 = θ1, q 3 = z1, q 4 = R2, q 5 = φ2, q 6 = θ2, (3.70)6 The forces from these dashpots are not conservative.




show that the six covariant basis vectors aJ = ∂ r∂qJ

are

a1 = ∂ r

∂r1= cos(θ1)e1 + sin(θ1)e2 + cos(θ1)e4 + sin(θ1)e5,

a2 = ∂ r

∂θ1= −r1 sin(θ1)e1 + r1 cos(θ1)e2 − r1 sin(θ1)e4 + r1 cos(θ1)e5,

a3 = ∂ r

∂z1= e3 + e6,

a4 = ∂ r

∂R2= sin(φ2) cos(θ2)e4 + sin(φ2) sin(θ2)e5 + cos(φ2)e6,

a5 = ∂ r

∂φ2= R2 cos(φ2) cos(θ2)e4 + R2 cos(φ2) sin(θ2)e5 − R2 sin(φ2)e6,

a6 = ∂ r

∂θ2= −R2 sin(φ2) sin(θ2)e4 + R2 sin(φ2) cos(θ2)e5. (3.71)

(c) Show that the six contravariant basis vectors have the following representations:

a1 = cos(θ1)e1 + sin(θ1)e2,

a2 = −sin(θ1)

r1e1 +

cos(θ1)

r1e2,

a3 = e3,

a4 = sin(φ2)(cos(θ2)(e4 − e1) + sin(θ2)(e5 − e2)) + cos(φ2)(e6 − e3),

a5 = cos(φ2)

R2

cos(θ2)(e4 − e1) + sin(θ2)(e5 − e2)

− sin(φ2)

R2(e6 − e3),

a6 =

− sin(θ2)

R2 sin(φ2)

(e4

−e1) +

cos(θ2)

R2 sin(φ2)

(e5

−e2). (3.72)

(d) Show that the kinetic energy T of the particle of mass m = m1 + m2 is

T = m1 + m2

2

r21 + r21

θ21 + z21

+

m2

2

R22 + R2

2 φ22 + R2

2 sin2(φ2)θ22

+m2 cos(φ2)

R2 z1 + φ2 r1R2 cos(θ2 − θ1) + φ2

θ1r1R2 sin(θ2 − θ1)

−m2 sin(φ2)

R2 φ2 z1 − r1R2

θ1 θ2 cos(θ2 − θ1) − r1 R2 cos(θ2 − θ1)

−m2 sin(φ2)−r1 R2

θ1 sin(θ2 − θ1) + r1 θ2R2 sin(θ2 − θ1)

. (3.73)

This expression for the kinetic energy follows from the definition

T = m2 v · v = m12 v1 · v1 + m2

2 v2 · v2. (3.74)



3.11 Exercises 75

(e) If the forces acting on the particles are F1 = - m1gE3 and F2 = - m2gE3, then what isthe force Φ and what is the potential energy associated with this force?(f ) What are the six Lagrange’s equations governing the motion of the particle of mass m.7

3.8 As shown in Figure 3.5, 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, theparticle 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 themotion of m1 is in the E1 − E2 plane.

g

E

2

m2

m1

E3

E

1

O

Fig. 3.5. A planar double pendulum.

(a) What are the three constraints on the motion of the particle of mass m?(b) Using the normality prescription, what is the constraint force Φc acting on the particle of mass m. You should also, if possible, verify that the components of this force are physicallyrealistic.

(c) Starting from the final results of Exercise 3.7, establish Lagrange’s equations of motionfor the pendulum system. In your solution, clearly distinguish the equations governing themotion of the particle and the equations giving the components of Φc.(d) Let us now establish some of the equations of (c) using an equivalent approach. Imposethe constraints on T to determine the constrained kinetic energy T . In addition, determinethe constrained potential energy U . Verify that the following equations correspond to thoseyou obtained from (c):8

d

dt

∂ T

∂ θ1

− ∂ T

∂θ1= Φ · a2 = − ∂ U

∂θ1,

7 You should refrain if possible from expanding the time derivative here - it will be a considerableamount of algebra.

8

It is crucial to note that ∂ ˜

T ∂ r1 = ∂ ˜

T ∂ R2 = ∂ ˜

T ∂ z1 = ∂ ˜

T ∂r1 = ∂ ˜

T ∂R2 = ∂ ˜

T ∂z1 = 0.




d

dt

∂ T

∂ φ2

− ∂ T

∂φ2= Φ · a5 = − ∂ U

∂φ2, (3.75)

and

ddt

∂ T ∂ θ2

− ∂ T

∂θ2= Φ · a6 = − ∂ U

∂θ2. (3.76)

(e) Suppose a non–integrable constraint is imposed on the pendulum system discussed in (d):

f 1 · v1 + f 2 · v2 + h = 0. (3.77)

Show that this constraint can be expressed as

f · v + h = 0. (3.78)

In addition, what are the equations governing the motion of the non–integrably constrainedsystem? Illustrate your solution with an non–integrable constraint of your choice.

3.9 For a system of unconstrained particles, establish the following result due to Synge andGriffith [52]:

d

dt

∂T

∂ q K

− ∂T

∂q K =

N i=1

Fi · ∂ ri

∂q K , (K = 1, . . . , 3N ) . (3.79)

You might find it convenient to use the identity (3.20) discussed previously. Synge and Grif-fith’s derivation is presented in Section 15.1 of [52]. How would you incorporate constraintsand constraint forces into this formulation of Lagrange’s equations?



Part III

Kinematics and Dynamics of a Single Rigid Body





4

Background on Tensors

4.1 Preliminaries

In this chapter, preliminary background on tensors is presented. The material is primarily based

on Casey [8], Chadwick [15], and Gurtin [23].In this chapter, E denotes Euclidean three–space. For this space, we define a right–handed

fixed orthonormal basis E1, E2, E3. We shall also use another right–handed orthonormal basisp1, p2, p3. This basis is not necessarily fixed.1 Finally, in this chapter, lower–case 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 is right–handed if the following scalar triple product is positive:

[b1, b2, b3] = b3 · (b1 × b2). (4.1)

Because we will be using specious amounts of dot products, it is convenient to define theKronecker delta δ ik:

δ ij = 1 when i = j and δ ij = 0 when i

= j. (4.2)

Clearly,pi · pk = δ ik, (4.3)

We also define the alternating (or Levi–Civita) symbol ijk :

123 = 312 = 231 = 1,

213 = 132 = 321 = −1,

ijk = 0 otherwise. (4.4)

In words, ijk = 1, if ij k is an even permutation of 1, 2, 3, ijk = −1 if ij k is an odd permutationof 1, 2, 3, and ijk = 0 if either i = j , j = k, or k = i. We also note that

[pi, pj , pk] = ijk . (4.5)

You should verify this result using the definition of the scalar triple product.

1 This basis serves to represent corotational bases, other time-varying bases such as er, eθ, E3, andfixed bases such as E1, E2, E3.



80 4 Background on Tensors

4.2 The Tensor Product of Two Vectors

The tensor (or cross-bun) product of any two vectors a and b in E is defined to be

(a⊗

b) c = (b·

c) a (4.6)

where c is any vector in E . In words, a⊗b projects c onto b and multiplies the resulting scalar bya. You should also notice that 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. (4.7)

Both tensor products will be used in the sequel. You should notice that a ⊗ b provides a lineartransformation of any vector c that it acts upon.

The tensor product of a and b has some nice properties. First, if α and β are any two scalars,and a, b, and c are any three vectors, then

(αa + β b) ⊗ c = α (a ⊗ c) + β (b ⊗ c) ,

c ⊗ (αa + β b) = α (c ⊗ a) + β (c ⊗ b) . (4.8)

These properties follow from the definition of the tensor product. To prove them, one merelyshows that the left– and right–hand sides of the identities provide the same transformation of any vector d.

4.3 Second–Order Tensors

A second–order tensor A is a linear transformation of E into itself. That is, for any two vectorsa and b and any two scalars α and β ,

A (αa + β b) = αAa + β Ab, (4.9)

where Aa and Ab are both vectors in E . 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. It should be clear that a ⊗ bis an 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), (4.10)

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, (4.11)

where a is any vector.

The transpose AT

of a second–order tensor A is defined, for all vectors a and b, as

b · (Aa) = (AT b) · a. (4.12)



4.3 Second–Order Tensors 81

If we consider the second–order tensor c ⊗ d, we can use the definition of the transpose to showthat

(c ⊗ d)T = d ⊗ c. (4.13)

This result is very easy to use.

Given any two second–order tensors A and B, it can be shown that (AB)T = BT AT . If A = AT , then A is said to be symmetric. On the other hand, A is skew–symmetric if A = −AT .Any second–order tensor B can be decomposed into the sum of a symmetric second–order tensorand a skew–symmetric second–order tensor:

B = 1

2

B + BT

+

1

2

B − BT

. (4.14)

This result will prove to be very useful in the sequel.

4.3.1 Invariants, Determinants and Traces

There are three scalar quantities associated with a second–order tensor that are independent of the orthonormal basis used for

E . Because these quantities are independent of the basis, they are

known as the invariants of a second–order tensor. Given a second–order tensor A, the principalinvariants, I A, I I A, and I II A, of A are defined as

[Aa, b, c] + [a, Ab, c] + [a, b, Ac] = I A[a, b, c],

[a, Ab, Ac] + [Aa, b, Ac] + [Aa, Ab, c] = I I A[a, b, c],

[Aa, Ab, Ac] = I II A[a, b, c], (4.15)

where a, b, and c are any three vectors.The first invariant is known as the trace of a tensor, while the third invariant is known as the

determinant of a tensor:tr(A) = I A, det(A) = I II A. (4.16)

We shall shortly see why this terminology is used.

4.3.2 Inverses and Adjugates

The inverse A−1 of a second–order tensor A is the second–order tensor which satisfies

A−1A = AA−1 = I. (4.17)

For the inverse to exist, det(A) = 0. Taking the transpose of this equation, we find that theinverse 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 which satisfies

A∗(a × b) = Aa × Ab, (4.18)

where a and b are arbitrary vectors. If A is invertible, then this definition yields,

A∗ = det(A)(A−1)T . (4.19)

Notice how this result simplifies if A has a unit determinant.




4.3.3 Eigenvalues and Eigenvectors

The characteristic values (or eigenvalues or principal values) of a second–order tensor A aredefined as the roots λ of the characteristic equation of A:

det(λI − A) = 0. (4.20)

The three roots of this equation are denoted λi. If we expand the equation we find that

λ3 − I Aλ2 + II Aλ − III A = 0. (4.21)

That is

λ1λ2λ3 = det(A) = I II A,

λ1λ2 + λ2λ3 + λ1λ3 = 1

2(tr(A)2 − tr(A2)) = I I A,

λ1 + λ2 + λ3 = tr(A) = I A. (4.22)

The eigenvector (or characteristic direction or principal direction) of a tensor A is the vectoru which satisfies

Au = λu, (4.23)

where λ is a root of the characteristic equation. A second–order tensor has three eigenvectors.

4.4 A Representation Theorem for Second–Order Tensors

It is convenient at this stage to establish a representation for any second–order tensor A. Themain result we wish to establish is that

A =3

i=1

3

k=1

Aikpi

⊗pk, (4.24)

whereAik = (Apk) · pi. (4.25)

Here, Aik are known as the components of A relative to the basis p1, p2, p3.To establish the representation, we note that Api is a vector. Consequently, it can be written

as a linear combination of the vectors p1, p2, and p3:

Ap1 = A11p1 + A21p2 + A31p3,

Ap2 = A12p1 + A22p2 + A32p3,

Ap3 = A13p1 + A23p2 + A33p3. (4.26)

The order of the indices ik for the scalars Aik is important. You should also notice that

Aik = (Apk) · pi. (4.27)



4.4 A Representation Theorem for Second–Order Tensors 83

Given any vector b =3

i=1 bipi, we find that

Ab =

3

j=1

A(bjpj)

=3

j=1

bj(Apj)

=

3j=1

bj

3k=1

Akjpk

=3

j=1

3k=1

bj (Akjpk)

=

3j=1

3k=1

(Akj pk ⊗ pj)

3i=1

bipi

=3

j=1

3k=1

(Akj pk ⊗ pj) b. (4.28)

In the next to last step we used the definition of the tensor product of two vectors. In summary,we have that

Ab =

3j=1

3k=1

(Akjpk ⊗ pj) b. (4.29)

As this result holds for all vectors b and A is assumed to be a linear transformation, we concludethat

A =

3

j=1

3

k=1

Akjpk ⊗ pj . (4.30)

This is the representation theorem for second–order tensors.

4.4.1 Uses of the Representation Theorem

We can use the representation theorem to establish expressions for the transformation inducedby a second–order tensor A. To this end, recall that

A =3

j=1

3k=1

Akjpk ⊗ pj . (4.31)

The transformation induced by A on a vector b is

Ab = 3

j=1

3k=1

Akj pk ⊗ pj

b =3

j=1

3k=1

Akjbjpk. (4.32)




If we define c = Ab, then it is easy to see that the components ci = c · pi of c are

ci =

3

j=1

Aijbj . (4.33)

It should be clear from these results that the identity tensor I =3

i=1 pi ⊗ pi.Let us next examine the determinant and trace of A. Recall that

[Aa, b, c] + [a, Ab, c] + [a, b, Ac] = tr(A)[a, b, c],

[Aa, Ab, Ac] = det(A)[a, b, c]. (4.34)

If we choose a = p1, b = p2 and c = p3, where p1, p2, p3 is a right–handed orthonormal basisfor E , then we have the intermediate results

[p1, p2, p3] = 1,

[Ap1, p2, p3] + [p1, Ap2, p3] + [p1, p2, Ap3] =

3

i=1

(Api) · pi,

[Ap1, Ap2, Ap3] = det(A). (4.35)

Using these results, we find that the trace of A is

tr(A) =

3i=1

(Api) · pi = A11 + A22 + A33. (4.36)

In addition, we find that

det(A) = [Ap1, Ap2, Ap3]

=3

i=1

3

j=1

3

k=1

Ai1Aj2Ak3[p1, pj , pk]

=

3i=1

3j=1

3k=1

ijkAi1Aj2Ak3. (4.37)

We recall that the determinant of a 3x3 matrix whose components are Bik is

det

B11 B12 B13

B21 B22 B23

B31 B32 B33

=

3i=1

3j=1

3k=1

ijk Bi1Bj2Bk3. (4.38)

Comparing (4.37) to 4.38) we see that the determinant of a tensor A can be calculated byrepresenting the tensor using a right–handed orthonormal basis and then using a standard resultfrom matrices:

det(A) = detA11 A12 A13

A21 A22 A23

A31 A32 A33

. (4.39)



4.4 A Representation Theorem for Second–Order Tensors 85

As a next result, we examine the symmetric and skew–symmetric parts of a second–ordertensor A. As a preliminary, recall that the transpose AT of this second–order tensor is definedas

(Aa) · b = (AT b) · a. (4.40)

Using this definition and the arbitrariness of a and b, it can be shown that

AT =

3i=1

3k=1

Akipi ⊗ pk =

3i=1

3k=1

Aikpk ⊗ pi (4.41)

where

A =

3i=1

3k=1

Aikpi ⊗ pk. (4.42)

Recall that a second–order tensor A is symmetric if A = AT . From the representations justmentioned, we find that

A = AT if Aik = Aki. (4.43)

This implies that a symmetric second–order tensor has 6 independent components. Similarily, Ais skew–symmetric if AT = −A:

A = −AT if Aik = −Aki. (4.44)

Notice that this result implies that a skew–symmetric second–order tensor has 3 independentcomponents.

We now turn to the important result of the product of two second–order tensors A and B.The product AB is a second–order tensor C. First, let

A =

3i=1

3k=1

Aikpi ⊗ pk, B =

3i=1

3k=1

Bikpi ⊗ pk, C =

3i=1

3k=1

C ikpi ⊗ pk. (4.45)

We now solve the equationsCa = (AB)a, (4.46)

where a is any vector, for the 9 components of C. Using the arbitrariness of a, we find that

C ik =3

j=1

AijBjk . (4.47)

This result is similar in form to one used in matrix multiplication. Indeed, if we define threematrices whose components are C ik, Aik and Bik, then we find that

C 11 C 12 C 13C 21 C 22 C 23C 31 C 32 C 33

=

A11 A12 A13

A21 A22 A23

A31 A32 A33

B11 B12 B13

B21 B22 B23

B31 B32 B33

. (4.48)

You should notice that in this expression the components of the three tensors are all expressedin the same basis.




It is illuminating to consider the product of two second–order tensors a ⊗ b and c ⊗ d. Usingthe representation theorem for both tensors, and then taking their product, we find that

(a ⊗ b)(c ⊗ d) = a ⊗ d(b · c). (4.49)

This result is the easiest way to remember how to multiply two second–order tensors. We alsonote in passing that tr(a ⊗ b) = a · b.

4.5 Third–Order Tensors

We shall find it necessary to use one example of a third–order tensor. A third–order tensor trans-forms vectors into second–order tensors and may transform second–order tensors into vectors.One 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 tensorA has the representation

A =

3

i=1

3

j=1

3

k=1

Aijkpi ⊗ pj ⊗ pk

. (4.50)

We also define the tensor products:

(a ⊗ b ⊗ c)[d ⊗ e] = a(b · d)(c · e), (a ⊗ b ⊗ c)d = a ⊗ b(c · d). (4.51)

Other tensor products can also be defined.The main example of a third–order tensor we will use is known as the alternator :

=

3i=1

3j=1

3k=1

ijk pi ⊗ pj ⊗ pk

= p1 ⊗ p2 ⊗ p3 + p3 ⊗ p1 ⊗ p2 + p2 ⊗ p3 ⊗ p1

− p

2 ⊗p1 ⊗

p3 −

p1 ⊗

p3 ⊗

p2 −

p3 ⊗

p2 ⊗

p1

. (4.52)

This tensor has some useful features. First, if A is a symmetric tensor, then [A] = 0. Secondly,suppose that c is a vector, then

c =3

i=1

3j=1

3k=1

ijk pi ⊗ pjck

= c3(p1 ⊗ p2 − p2 ⊗ p1) + c2(p3 ⊗ p1 − p1 ⊗ p3)

+ c1(p2 ⊗ p3 − p3 ⊗ p2), (4.53)

which is a skew–symmetric tensor.The fact that acts on a vector to produce a skew–symmetric tensor enables one to define a

skew–symmetric tensor C for every vector c and vice versa:

C = −c, c = −1

2[C]. (4.54)



4.6 Additional Types of Second–Order Tensors 87

The vector c is known as the axial–vector of C. It is a good exercise to verify that if C has therepresentation

C = c21(p2 ⊗ p1 − p1 ⊗ p2) + c32(p3 ⊗ p2 − p2 ⊗ p3)

+ c13(p1 ⊗ p3 − p3 ⊗ p1), (4.55)

then,

c = −1

2[C] = c21p3 + c13p2 + c32p1. (4.56)

We also note the important result that

Ca = (−c)a = c × a. (4.57)

This result enable us to replace cross products with tensor products.

4.6 Additional Types of Second–Order Tensors

There are three types of second–order tensors which play an important role in rigid body dynam-ics: proper–orthogonal tensors, symmetric positive–definite tensors, and skew–symmetric tensors.

4.6.1 Orthogonal Tensors

A second–order tensor L is said to be orthogonal if LLT = LT L = I. In other words, the transposeof an orthogonal tensor is its inverse. It also follows that det L = ±1. An orthogonal tensor hasthe unique property that La · La = a · a. In other words, it preserves the length of the vectorthat it transforms.

Some examples of orthogonal tensors include

I, −I, E1 ⊗ E1 + E2 ⊗ E2 − E3 ⊗ E3. (4.58)

The last of these examples constitutes a reflection in the plane x3 = 0.

4.6.2 Proper–Orthogonal Tensors

A second–order tensor Q is said to be proper–orthogonal if QQT = QT Q = I and det(Q) = 1.This type of tensor is also known as a rotation tensor. Proper–orthogonal second–order tensorsare 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 rotationtensor by −I.

Using a result which is known as Euler’s formula, any rotation tensor can be written as

Q = cos(θ)(I − p ⊗ p) − sin(θ)p + p ⊗ p, (4.59)

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 shall shortly examine several examplesof rotation tensors.




4.6.3 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 onlyif a = 0. It should be clear from the definition that a skew–symmetric second–order tensor can

never be positive–definite.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 = λ1u1 ⊗ u1 + λ2u2 ⊗ u2 + λ3u3 ⊗ u3, (4.60)

where Aui = λiui. This representation is often known as the spectral decomposition.

4.7 Derivatives of Tensors

We shall often encounter derivatives of tensors. Suppose a tensor A has the representation

A =

3i=1

3k=1

Aikpi ⊗ pk. (4.61)

We suppose here that the components of A and the vectors pi are functions of time.The time–derivative of A is defined as

A =

3i=1

3k=1

Aikpi ⊗ pk

+

3i=1

3k=1

Aik pi ⊗ pk +

3i=1

3k=1

Aikpi ⊗ pk. (4.62)

Notice that we differentiate both the components and the basis vectors.We can also define a chain rule and product rules. Suppose A = A(q (t)), B = B(t), and

c = c(t). Then

A = ∂ A

∂q q,

d

dt (AB) = AB + A B,

d

dt (Ac) = Ac + Ac. (4.63)

If we have a function ψ = ψ(A), then the derivative of this function with respect to A is definedto be the second–order tensor

∂ψ

∂ A

=

3

i=1

3

k=1

∂ψ

∂Aik

pi

⊗pk. (4.64)

In addition, if the vectors pi are constant, then



4.8 Exercises 89

ψ =3

i=1

3k=1

∂ψ

∂Aik

Aik = tr

∂ψ

∂ AAT

. (4.65)

This result will be used when we consider constraints on the motions of rigid bodies.

4.8 Exercises

4.1 Consider a = E1 and b = E2. For any vector c =3

i=1 ciEi, show that (a ⊗ b)c = c2E1 andc(a ⊗ b) = c1E2.

4.2 Consider a = E1 and b = E2. For any vector c = 3

i=1 ciEi, show that (a ⊗ b − b ⊗ a)c =c2E1 − c1E2 .

4.3 Using the definition of the transpose, verify that (a ⊗ b)T = b ⊗ a.4.4 If a = 10E1 and b = 5E2, then show that (a⊗b)u = 50E1(E2 ·u) and (b⊗a)u = 50E2(E1 ·u).4.5 If a = er and b = eθ, then show that (a ⊗ b)u = er(eθ · u) and (b ⊗ a)u = eθ(er · u).4.6 Show that the following matrix multiplication

er

eθ

ez

=

cos(θ) sin(θ) 0

− sin(θ) cos(θ) 00 0 1

E1

E2

E3

, (4.66)

can be written aser = PE1 , eθ = PE2 , ez = PE3, (4.67)

where

P = cos(θ)E1 ⊗ E1 + sin(θ)E2 ⊗ E1

+ cos(θ)E2 ⊗ E2 − sin(θ)E1 ⊗ E2

+ E3 ⊗ E3

= cos(θ)(I−

E3

⊗E3)

−sin(θ)E3 + E3

⊗E3. (4.68)

To show the final result, it is helpful to first show that

P = er ⊗ E1 + eθ ⊗ E2 + ez ⊗ E3. (4.69)

Finally, verify that P is a proper–orthogonal second order tensor.4.7 Show that the following matrix multiplication

eR

eφ

eθ

=

cos(θ) sin(φ) sin(θ) sin(φ) cos(φ)

cos(θ) cos(φ) sin(θ) cos(φ) − sin(φ)− sin(θ) cos(θ) 0

E1

E2

E3

, (4.70)

can be written as

eR = RE3 , eφ = RE1 , eθ = RE2, (4.71)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.

(4.72)To show the final result, it is helpful to first show that

R = eR ⊗ E3 + eφ ⊗ E1 + eθ ⊗ E2. (4.73)

4.8 Verify that the tensor R in the previous exercise is proper–orthogonal.4.9 Give an example which illustrates that tensor multiplication is not commutative, i.e., AB =

BA.4.10 Verify that if B is a second–order tensor, then

[B] = 1

2[B − BT ]. (4.74)

4.11 Let c and d be any two vectors. Verify that

−ε[c ⊗ d] = −1

2ε[c ⊗ d − d ⊗ c] = d × c. (4.75)

4.12 Let A and B be skew-symmetric second–order tensors. Verify that

a · b = 1

2 tr(ABT ), (4.76)

where a and b are the axial vectors of A and B, respectively.4.13 Let A be a skew-symmetric second–order tensor, a be its axial vector, and C be a symmetric

second–order tensor. Verify that

−1

2ε[AC + CA] = (tr(C)I

−C)a. (4.77)

This identity is used to establish an expression for the angular momentum of a rigid body.4.14 Let A be a tensor, and b and c be any two vectors. Then show that

tr(A)I − AT

(b × c) = Ab × c + b × Ac. (4.78)

Hint: perhaps the quickest way to prove this result is to first argue that it suffices to pickb = E1, c = c1E1 + c2E2. Substituting into the identity then yields an easy to verify result.



5

Rotation Tensors

5.1 Introduction

This chapter present several results on rotation tensors. Most of the results we provide are well-known and can be found in Beatty [3], Casey [8, 10], Casey and Lam [11], and Shuster [48].However, the notations in many of these references differ significantly and combining the resultsin a single notation is necessary.

For this chapter, we emphasize that E denotes Euclidean three–space. For this space, we definea right–handed fixed orthonormal basis E1, E2, E3. We shall also use another right–handedorthonormal basis p1, p2, p3. This basis is not necessarily fixed. Lower–case Latin indices, suchas i, j, and k , will range from 1 to 3.

5.2 Rotation Tensors

Recall that a proper–orthogonal second–order tensor (or a rotation tensor) R is a tensor which

has a unit determinant and whose inverse is its transpose:RT R = RRT = I, det(R) = 1. (5.1)

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 rotation tensorcan be parameterized using three independent parameters.

As a rotation tensor is a second–order tensor, it has the following representation:

R =3

i=1

3k=1

Rikpi ⊗ pk. (5.2)

Let us now consider the transformation induced by R on the basis vectors. We define,

t1 = Rp1 =3

i=1

Ri1pi,



92 5 Rotation Tensors

t2 = Rp2 =3

i=1

Ri2pi,

t3 = Rp3 =

3

i=1

Ri3pi. (5.3)

You should notice that by using the vectors ti, R has the representation

R = t1 ⊗ p1 + t2 ⊗ p2 + t3 ⊗ p3. (5.4)

We now wish to show that t1, t2, t3 is a right–handed orthonormal basis. First, let’s verify theorthonormality:

ti · tk = Rpi · Rpk

= RT Rpi · pk

= pi · pk

= δ ik. (5.5)

Hence, the vectors ti are orthonormal. To establish right-handedness, we use the definition of thedeterminant:

[t1, t2, t3] = [Rp1, Rp2, Rp3]

= det(R)[p1, p2, p3]

= (1)(1) = 1. (5.6)

Hence, t1, t2, t3 is a right–handed orthonormal basis.The rotation tensor also has a rather unusual representation. Consider

R = RRRT

= R

3

i=1

3

k=1

Rikpi ⊗ pk

RT

=3

i=1

3k=1

RikR (pi ⊗ pk) RT

=

3i=1

3k=1

Rik (Rpi ⊗ Rpk)

=

3i=1

3k=1

Rikti ⊗ tk. (5.7)

In summary, we have the following representations for R:

R =3

i=1

3

k=1

Rikpi

⊗pk =

3

i=1

3

k=1

Rikti

⊗tk =

3

i=1

ti

⊗pi. (5.8)

Notice that the components of R for the first two representations are the same.



5.3 Derivatives of a Rotation Tensor 93

5.3 Derivatives of a Rotation Tensor

Here, we consider a rotation tensor R which is a function of time: R = R(t). Consider, thederivative of RRT :

d

dt

RRT

= RRT + R RT . (5.9)

However, I = O, so the right–hand side of the above equation is zero. Hence,

RRT = −R RT = −

RRT T

. (5.10)

In other words, RRT is a skew–symmetric second–order tensor. We shall define

Ω R = RRT , (5.11)

in part because this tensor appears in numerous places later on. The tensor Ω R is known as theangular velocity tensor (of R).

The skew–symmetry of RRT allows us to define the angular velocity vector ωR:

ωR = −1

2[ RRT ]. (5.12)

As mentioned earlier, ωR × a = ( RRT )a for all vectors a.In a similar manner, we can also show that RT R is a skew–symmetric tensor, and define an

angular velocity tensor Ω 0R and another angular vector ω0R :

Ω 0R = RT R, ω0R = −1

2[RT R]. (5.13)

At a later stage, you should show that Rω0R = ω and RΩ 0RRT = Ω R. A key to establishingone of these results is the identity, which holds for all orthogonal Q,

QBQT

= det(Q) Q ( [B]) . (5.14)

Notice how this identity simplifies when Q is a rotation tensor.

5.3.1 Corotational Derivatives

Recall that for a rotation tensor R, we have the representation

R =3

i=1

ti ⊗ pi. (5.15)

Now suppose that pi = 0. Then,




Ω R = RRT =

3i=1

ti ⊗ pi +3

i=1

ti ⊗ pi

RT

= 3

i=1

ti

⊗piRT

=

3i=1

ti ⊗ ti. (5.16)

If we now consider Ω Rtk, we find a familiar result:

ti = Ω Rti = ωR × ti. (5.17)

We leave it as an exercise to show the less familiar result

ti = R(ω0R × pi). (5.18)

It is important to note that if ti are defined using a rotation tensor R and a fixed basis pi, then

their time derivatives can be expressed in terms of the angular velocity vector of the rotationtensor and the basis vectors ti.Given any second–order tensor A and any vector a, we have the representations:

a =3

i=1

aiti, A =3

i=1

3k=1

Aikti ⊗ tk. (5.19)

If we assume that a and A are functions of time, then

a =

3i=1

aiti + aiti

=3

i=1

aiti + ai(ωR

×ti)

=

3i=1

aiti + ωR × a,

A =

3i=1

3k=1

Aikti ⊗ tk

3i=1

3k=1

Aikti ⊗ tk +

3i=1

3k=1

Aikti ⊗ tk

=

3i=1

3k=1

Aikti ⊗ tk +

3i=1

3k=1

Aik(Ω Rti) ⊗ tk +

3i=1

3k=1

Aikti ⊗ (Ω Rtk)

=

3i=1

3k=1

Aikti ⊗ tk + Ω RA − AΩ R. (5.20)

The derivativeso

A and oa are known as the corotational derivatives. They are the respective

derivatives of A and a if the vectors ti were constant:



5.4 Relative Angular Velocity Vectors 95

o

A=3

i=1

3k=1

Aikti ⊗ tk, o

a=3

i=1

aiti. (5.21)

Notice that for any vector a, we have

a =oa +ωR × a. (5.22)

Similarily, for any second–order tensor A, we have

A =o

A +Ω RA − AΩ R. (5.23)

The terms involving the angular velocity vectors and tensors in these expressions are the resultof the orthonormal vectors ti changing with time.

5.4 Relative Angular Velocity Vectors

Consider two rotation tensors: R1 = R1(t) and R2 = R2(t). The product R = R2R1 of thesetwo tensors is also a rotation tensor. We now wish to calculate its angular velocity tensor andvector.

To this end, it is convenient to define three sets of right–handed orthonormal vectors:1t1, 1t2, 1t3, 2t1, 2t2, 2t3, and p1, p2, p3. These three sets are defined by

R1 =3

i=1

1ti ⊗ pi, R2 =3

i=1

2ti ⊗ 1ti. (5.24)

Notice that

R =

3

i=1

2ti ⊗ pi. (5.25)

In words, R transforms the vector pi into the vector 2ti.Following Casey and Lam [11], let us consider the following relative angular velocity tensor:

Ω R2 = Ω R −Ω R1

. (5.26)

Using the defintion of the angular velocity tensors, we find that

Ω R2 = Ω R −Ω R1

= RRT − R1RT 1

= R2R1RT 1 RT

2 + R2 R1RT

1 RT 2 − R1RT

1

= R2RT 2 + R2Ω R1

RT 2 − Ω R1

= R2RT 2 + R2Ω R1

RT 2 +Ω T

R1

= R2

+ R2Ω R1 + Ω T

R1

R2RT

2. (5.27)

However,




o

R2= R2 + R2Ω R1 + Ω T

R1R2, (5.28)

where the corotational derivativeo

R2 is defined to be

oR2=3

i=1

3k=1

R2ik1ti ⊗ 1tk, (5.29)

where R2ik = ((R2)1tk) · 1ti. In words,o

R2 is the derivative of the tensor R2 assuming that R1

is constant.In conclusion, the relative angular velocity tensor Ω R2

is

Ω R2 = Ω R −Ω R1

=o

R2 RT 2 . (5.30)

This tensor has an associated angular velocity vector:

ωR2 = ωR −ωR1

= −1

2[

o

R2 RT 2 ]. (5.31)

This formula will prove to be extremely useful when calculating the angular velocity vectorassociated with various representations of a rotation tensor.

5.5 Euler’s Representation

Euler showed that any rotation tensor can be represented as follows:

R = R(φ, r) = cos(φ)(I − r ⊗ r) − sin(φ)(εr) + r ⊗ r, (5.32)

where r is a unit vector and φ is a counterclockwise angle of rotation. The three independentparameters here are the angle of rotation and the two independent components of the unit vector

r.To see that Euler’s representation is valid, we need to show that it satisfies (5.1). To this end,

we define an orthonormal basis p1, p2, p3, where p3 = r. Using this basis,

I = p1 ⊗ p1 + p2 ⊗ p2 + p3 ⊗ p3,

r ⊗ r = p3 ⊗ p3,

I − r ⊗ r = p1 ⊗ p1 + p2 ⊗ p2,

−(εr) = p2 ⊗ p1 − p1 ⊗ p2. (5.33)

Consequently, we can write

R = R(φ, r = p3) = cos(φ)(p1 ⊗ p1 + p2 ⊗ p2)

+ sin(φ)(p2 ⊗ p1 − p1 ⊗ p2) + p3 ⊗ p3. (5.34)Using the identity (a ⊗ b)T = b ⊗ a, we find



5.5 Euler’s Representation 97

RT = R(φ, r = p3) = cos(φ)(p1 ⊗ p1 + p2 ⊗ p2)

+ sin(φ)(p1 ⊗ p2 − p2 ⊗ p1) + p3 ⊗ p3. (5.35)

We are now in a position to verify that R is a rotation tensor. First, we check that RRT is indeed

the identity tensor. Next, we examine the determinant of R:

det(R) = det

cos(φ) −sin(φ) 0

sin(φ) cos(φ) 00 0 1

= 1. (5.36)

In conclusion, R is a rotation tensor.

r a

Ra

Ra

aa||

| |

| |

φ

Fig. 5.1. The transformation by R of a vector a.

5.5.1 Euler’s Formula

We now examine the action of R on a vector a. As shown in Figure 5.1, the part of a which isparallel to r is unaltered by the transformation, while the part of a which is perpindicular to r isrotated through an angle φ counter–clockwise about r. To see this it is convenient to decomposea:

a = a⊥ + a, (5.37)

wherea⊥ = a − (a · r)r = (I − r ⊗ r)a, a = (a · r)r = (r ⊗ r)a (5.38)

With the assistance of this decomposition, we now calculate that

Ra = R(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. (5.39)




Noting that cos(φ)a⊥ +sin(φ)r×a⊥ is a rotation of a⊥ about r we obtain the desired conclusion.The expression for Ra in (5.39) is known as Euler’s formula.

5.5.2 Remarks on Euler’s Representation

Euler’s representation (5.32) is unusual in several respects. First of all notice that

R(φ, r) = R(−φ, −r). (5.40)

This implies that there are two different representations for the same rotation tensor. Anotherpeculiarity, is that R(φ = 0, r) = I holds for all vectors r. Finally, we note that Euler’s represen-tation is used to define other representations of rotation tensors in these notes.

5.5.3 The Associated Angular Velocity Vector

Given Euler’s representation (5.32) we assume that R = R(t). This implies, in general, thatφ = 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 · r = 0. In addition, = 0

and I = 0. Now, starting from (5.32),

R = R(φ, r) = cos(φ)(I − r ⊗ r) − sin(φ)(εr) + r ⊗ r, (5.41)

we differentiate to find

R = −φ sin(φ)(I − r ⊗ r) − φ cos(φ)(εr) + (1 − cos(φ))(r ⊗ r + r ⊗ r) − sin(φ)(εr). (5.42)

To proceed further, we define a right-handed orthonormal basis t1, t2, t3, such that t3 = r ata given instant in time. Then, at the same instant in time

r = (t1 ⊗ t2 − t2 ⊗ t1),

r = at1 + bt2,

r × r = at2 − bt1,

r = a(t2 ⊗ t3 − t3 ⊗ t2) + b(t3 ⊗ t1 − t1 ⊗ t3). (5.43)In these expressions, a and b are the scalar components of r. Using (5.43) along with somemanipulations, we find that

RRT = −φ(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)

= −φt3 − (a(1 − cos(φ)) + b sin(φ))t2 − (−b(1 − cos(φ)) + a sin(φ))t1. (5.44)

With the assistance of (5.43), we can now write the desired final result:

Ω R = RRT = −φr − (1 − cos(φ))(r × r) − sin(φ)r. (5.45)

The associated angular velocity vector is

ωR = φr + sin(φ)r + (1 − cos(φ))r × r. (5.46)

If r is constant, then the expression for the angular velocity vector simplifies considerably.



5.6 Other Representations of a Rotation Tensor 99

5.6 Other Representations of a Rotation Tensor

There are several other representations of a rotation tensor. Most of them are discussed in Shus-ter’s review article [48]. We now discuss three other representations: the Rodrigues representation,

the Euler parameter representation, and the Euler angle representation.

5.6.1 The Rodrigues Vector

The Rodrigues representation is based on the Rodrigues vector:

λ = tan

φ

2

r. (5.47)

This vector is sometimes called the Gibbs vector. Clearly, the Rodrigues vector λ is not a unitvector. Indeed, when φ = 0, λ = 0, and when φ = π, λ is undefined. Consequently, if φ variesthrough π , then we cannot use the Rodrigues representation discussed below.

With the assistance of the identities

sin(φ) = 2 tan(φ

2 )

1 + tan2(φ2 )

, cos(φ) = 1 − tan2( φ

2 )

1 + tan2( φ2 )

, (5.48)

you should notice that

cos(φ) = 1 − λ · λ1 + λ · λ , sin(φ) =

2λ · r

1 + λ · λ , (5.49)

Subsitituting for r and φ in (5.32), we find the Rodrigues representation:

R = R(λ) = 1

1 + λ · λ ((1 − λ · λ)I + 2λ⊗ λ− 2(ελ)) . (5.50)

The angular velocity vector associated with this representation is

ωR = 21 + λ · λ

λ− λ× λ

. (5.51)

This vector can be calculated by directly substituting into the earlier result associated with theEuler representation (5.46).

5.6.2 The Euler–Rodrigues Symmetric Parameters

One of the most common four–parameter representations uses the four Euler–Rodrigues symmet-ric parameters e0 and e.1 These parameters are often known as the Euler parameters. They canbe defined as

1 The four parameters, e0 and the three components of e, are often considered to be the four componentsof a quaternion q = e0 + e1i + e2 j + e3k, where ei are the components of e relative to a right–handedorthonormal basis, and i, j , and k are bases vectors for the quaternion. Consequently, Euler–Rodriguessymmetric parameters are sometimes referred to as (unit) quaternions.




e0 = cos

φ

2

, e = sin

φ

2

r. (5.52)

As a consequence of their definition, the parameters are subject to what is known as the Eulerparameter constraint:

e20 + e · e = 1. (5.53)

We also note thatλ =

e

e0. (5.54)

It is possible to express r and φ in terms of the parameters e0 and e, but we leave this as anexercise.

Substituting for r and φ in the Euler representation (5.32), the Euler–Rodrigues symmetricparameter representation is found:

R = R(e0, e) = (e20 − e · e)I + 2e ⊗ e − 2e0(εe). (5.55)

Notice that the rotation tensor is a quadratic function of the four parameters. As a result, thisrepresentation has several computational advantages over other representations.

The angular velocity vector associated with this representation is

ωR = 2 (e0e − e0e + e × e) . (5.56)

Again, this vector can be calculated by directly substituting into the earlier result associatedwith the Euler representation (5.46). Notice that the angular velocity vector is a relatively simplefunction 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 Rodrigues in 1840, are remarkably elegant.

5.6.3 Euler Angles

The most popular three–parameter representation is based on the Euler angles γ i

:

R = R(γ 1, γ 2, γ 3) = F(γ 3, g3)F(γ 2, g2)F(γ 1, g1). (5.57)

Here, the set of unit vectors gi is known as the Euler basis. The function F(θ, b) is definedusing the Euler representation:

F(θ, b) = cos(θ)(I − b ⊗ b) − sin(θ)(εb) + b ⊗ b, (5.58)

where b is a unit vector and θ is the counter–clockwise angle of rotation. In general, g3 is afunction of γ 2 and γ 1, while g2 is a function of γ 1. As we shall shortly see, there are 12 possiblechoices of the Euler angles.

If we assume that g1 is constant, then the angular velocity vector associated with the Eulerangle representation can be established using the relative angular velocity vector.2 In this case,

2 An alternative approach is used in most textbooks. In this approach, one considers each of the Eulerangles to be infinitesimal. A good example of this approach can be found in Section 2.9 of Lurie [31].




there are two relative angular velocity vectors (cf. (5.31)). For the first rotation, the angularvelocity vector is γ 1g1 (cf. (5.46). The angular velocity of the second rotation relative to the firstrotation is γ 2g2, and the angular velocity of the third rotation relative to the second rotation isγ 3g3. Combining the two relative angular velocity vectors with γ 1g1, we conclude that

ωR = γ 3g3 + γ 2g2 + γ 1g1. (5.59)

If the rotation tensor R transforms the vectors pi into the set ti, then it is possible to expressthe Euler basis in terms of either set of vectors.

For future purposes, it is also convenient to define the dual Euler basis gj: gj · gi = δ ji

where δ ji is the Kronecker delta. Notice that

ωR · gi = γ i. (5.60)

To determine this basis, one expresses gi in terms of a right–handed basis, say ti. Then, todetermine 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.

t3

t2

φ

φ

t1

θ

θ

t3

1

E2

Eψ

ψ

t1’

t1’

t2’

t3’

"

"

t3"

t2"

Fig. 5.2. The transformation of basis vectors induced by the 3-2-1 Euler angles.




3-2-1 Euler Angles

To elaborate further on the Euler angles, we now consider the 3-2-1 set of Euler angles (see Figure5.2).3 First, suppose that the rotation tensor has the representation

R =3

i=1

ti ⊗ Ei, (5.61)

where Ei is a fixed Cartesian basis. The first rotation, is about E3 through an angle ψ. Thisrotation transforms Ei to t

i. The second rotation is about the t

2 axis through an angle θ. This

rotation transforms t

i to t”i . The third and last rotation is through an angle φ about the axist”1 = t1. In short,

R = R(γ 1 = ψ, γ 2 = θ, γ 3 = φ) = F(φ, t1)F(θ, t

2)F(ψ, E3). (5.62)

You should also notice that

ti = F(φ, t1)t”i , t”i = F(θ, t”2 = t

2)t

i, t

i = F(ψ, t

3 = E3)Ei. (5.63)

It is not too difficult to express the various basis vectors as linear combinations of one another: t

1

t

2

t

3

=

cos(ψ) sin(ψ) 0

− sin(ψ) cos(ψ) 00 0 1

E1

E2

E3

,

t”1

t”2t”3

=

cos(θ) 0 −sin(θ)

0 1 0sin(θ) 0 cos(θ)

t

1

t

2

t

3

,

t1t2t3

=

1 0 00 cos(φ) sin(φ)0 −sin(φ) cos(φ)

t”1t”2t”3

, (5.64)

These relationships can be combined to express ti in terms of Ek, and a representation for thecomponents Rij = (REj) · Ei can be obtained. However, we do not present this representationhere.

By examining the individual rotations (cf. (5.64)), we can show that the Euler basis has therepresentation

g1

g2

g3

=

E3

t

2

t1

=

− sin(θ) sin(φ) cos(θ) cos(φ)cos(θ)

0 cos(φ) − sin(φ)1 0 0

t1

t2t3

. (5.65)

In addition, we can also determine the dual Euler basis vectors gk (see remarks following (5.60)).To start this calculation, we express each of the dual Euler basis vectors in terms of their com-ponents relative to the basis

t1, . . . , t3

. That is, gk = gk1t1 + gk2t2 + gk3t3. Combining these

results for the dual Euler basis vectors, we observe that3 These are the Euler angles used in D. T. Greenwood’s text [22] and Casey’s papers [8, 10].



5.6 Other Representations of a Rotation Tensor 103g1

g2

g3

=

g11 g12 g13

g21 g22 g23

g31 g32 g33

t1

t2t3

. (5.66)

With some manipulation, the relations gi

·gk = δ k

i

can be expressed as nine equations for thenine unknowns g ik:

g11 g21 g31

g12 g22 g32

g13 g23 g33

− sin(θ) sin(φ)cos(θ) cos(φ) cos(θ)

0 cos(φ) − sin(φ)1 0 0

=

1 0 0

0 1 00 0 1

. (5.67)

Inverting the matrix

gikT

, we find that g11 g12 g13

g21 g22 g23

g31 g32 g33

=

0 sin(φ) sec(θ) cos(φ) sec(θ)

0 cos(φ) − sin(φ)1 sin(φ) tan(θ) cos(φ) tan(θ)

. (5.68)

That is, the dual Euler basis gi has the representationg1

g2

g3

=

0 sin(φ)sec(θ) cos(φ) sec(θ)

0 cos(φ) − sin(φ)1 sin(φ) tan(θ) cos(φ) tan(θ)

t1

t2t3

. (5.69)

If we examine (5.65), we see that the Euler basis fails to be a basis for E when θ = ±π2 . One of

the easiest ways to see this fact is to consider ψ and θ to be spherical polar coordinates for t1(see Figure 5.3). When θ = ±π

2 , g1 = E3 = ±g3, and the Euler basis does not span E . To avoidthe aforementioned singularity, it is necessary to place restrictions on the second Euler angle:θ ∈ (−π

2 , π2 ). The other two angles are free to range from 0 to 2 π.

E

2

E3

t1θ

t1

E

1

ψ ’

Fig. 5.3. Visualizing the first two Euler angles ψ and θ of the 3-2-1 set of Euler angles as spherical polarcoordinates for the third Euler basis vector t1.

The angular velocity vector ωR associated with the 3-2-1 Euler angles has the representations

ωR = −1

2[ RRT ] =

3i=1

γ igi




= φt1 + θt

2 + ψE3

= (− ψ sin(θ) + φ)t1 + ( ψ sin(φ)cos(θ) + θ cos(φ))t2

+ ( ψ cos(φ) cos(θ) − θ sin(φ))t3. (5.70)

It is also interesting to note that the angular velocity vector ω0R has the representation

ω0R = −1

2[RT R] = RT ωR

= φRT t1 + θRT t

2 + ψRT E3

= (− ψ sin(θ) + φ)E1 + ( ψ sin(φ) cos(θ) + θ cos(φ))E2

+ ( ψ cos(φ) cos(θ) − θ sin(φ))E3. (5.71)

In establishing this result, we used the fact that RT ti = Ei.

3-1-3 Euler Angles

We can parallel the developments of the previous section for another set of Euler angles, the 3-1-3Euler angles. Paralleling the work for the 3-2-1 Euler angles:

R = R(φ, t3 = t

3 )R(θ, t

1)R(ψ, p3), (5.72)

where p3, t

1, t3 is the Euler basis and

t

i = R(ψ, p3)pi, t

i = R(θ, t

1)ti, ti = R(φ, t

3 )t

i . (5.73)

The angular velocity vector has the representations

ωR = −1

2[ RRT ] = γ igi = φt3 + θt

1 + ψp3. (5.74)

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 ψ represent different angles of rotation for these two set of Eulerangles.

It is not too difficult to show that the Euler basis gi has the representationsg1

g2

g3

=

p3

t

1

t3

=

sin(φ) sin(θ) cos(φ) sin(θ) cos(θ)

cos(φ) − sin(φ) 00 0 1

t1

t2t3

= 0 0 1

cos(ψ) sin(ψ) 0sin(θ) sin(ψ) − sin(θ) cos(ψ) cos(θ)

p1

p2

p3 .

(5.75)




In addition, the Euler angles are subject to the restrictions:

φ ∈ [0, 2π), θ ∈ (0, π), ψ ∈ [0, 2π). (5.76)

These restrictions are designed to ensure that the Euler basis remains a basis for

E 3.

By following the procedure that lead to (5.69) that the dual Euler basis gi has the repre-sentation4

g1

g2

g3

=

sin(φ)cosec(θ) cos(φ)cosec(θ) 0

cos(φ) − sin(φ) 0− sin(φ) cot(θ) −cos(φ) cot(θ) 1

t1

t2t3

. (5.77)

Using these results, it may be shown that

ωR · g1 = ψ, ωR · g2 = θ, ωR · g3 = φ. (5.78)

You should notice the similarities and differences in the results for 3-1-3 Euler angles and the3-2-1 set.

Other Sets of Euler Angles

There are 12 sets of Euler angles. For the Euler basis, one has three choices for g1, and becauseg1 = 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 anglesare being used is to specify the angular velocity vector. Here, we give expressions for each of the12 sets of Euler angles for a rotation tensor R =

3i=1 ti ⊗ Ei:

123 Set: ωR = ψE1 + θt

2 + φt3,


2 + φt3,


2 + φt1,


3 + φt1,

132 Set: ωR = ˙ψE1 +

˙θt

3 + ˙φt2,


3 + φt1,


3 + φt2,


1 + φt2,


1 + φt3,


1 + φt3,


3 + φt1,


1 + φt2. (5.79)

The sets of Euler angles, 121, 131, 232, 212, 313, and 323 are known as the symetric sets, whilethe other six sets are known as asymmetric sets, Cardan angles, Tait angles or Bryant angles.

4 That is one calculates the inverse of the transpose of the three by three matrix in (5.75) which describesgi in terms of tk.




For all sets of Euler angles, a singularity is present for certain values of the second angle θ. Atthese values g1 = ±g3, and the Euler basis fails to be a basis for E . To avoid these singularitiesit is often necessary to use two different sets of Euler angles and to switch from one set to theother as a singularity is approached.

5.6.4 Summary of the Angular Velocity Vectors

Pertaining to the representations discussed above, it was shown that

ωR =

3i=1

γ igi

= φr + sin(φ)r + (1 − cos(φ))r × r

= 2 (e0e − e0e + e × e)

= 2

1 + λ · λλ− λ× λ

. (5.80)

With some minor manipulations of these equations, one can also obtain expressions for r, e andλ.

5.7 Derivatives of Scalar Functions of Rotation Tensors

Consider a function U (R). We wish to calculate the time-derivative of this function. Some simplemanipulations gives three representations for this derivative:

U = tr

∂U (R)

∂ RRT

= trU R( RRT )T

= uR ·ωR, (5.81)

where we have found it convenient to define the operators uR and U R:

U R = 1

2

∂U

∂ RRT − R

∂U

∂ R

T

, uR = −ε[U R]. (5.82)

Notice that after finding ∂U (R)

∂ RRT , we then replaced R with Ω RR to obtain the second repre-

sentation.We now assume that R is parameterized using one of the four methods mentioned in the

previous section. By evaluating the derivative with respect to t of U (e0, e) for example and using(5.80), we find the representations

uR =3

i=1

∂ U

∂γ igi



5.8 Exercises 107

= ∂ U

∂φr +

1

2

cot

φ

2

(I − r ⊗ r) − εr

∂ U

∂ r

= 1

2 (e0I − εe)∂ U

∂ e − ∂ U

∂e0e

= 1

2 (I + λ⊗ λ− ελ)

∂ U

∂ λ. (5.83)

where U (R) = U (e0, e) = U (γ 1, γ 2, γ 3) = U (φ, r) = U (λ).Several of the partial derivatives in (5.83) need to be carefully evaluated. For example, because

r is a unit vector, the derivative ∂ Σ ∂ r

must be evaluated on the surface r · r = 1. Related remarks

pertain to ∂ U ∂ e

, ∂ U ∂e0

and ∂U ∂ R

. In other words, these are tangential or surface derivatives. Onemethod of evaluating them is to parameterize R by the Euler angles, and then transform fromthe parameters of interest to the Euler angles. Indeed, (5.83), the chain rule, and the identity εgi

= - ∂ R∂γ i

RT can be used to show that

∂ U

∂e0 = uR · 2e,∂ U

∂ e = 2(e0I + εe)uR,

∂U

∂ R = −(εuR)R,

∂ U

∂φ = uR · r,

∂ U

∂ r =

sin(φ)(I − r ⊗ r) + 2 sin2

φ

2

εr

uR. (5.84)

These results can be used to establish moment potentials associated with conservative moments.

5.8 Exercises

5.1 Recall that three Euler angles ν 1, ν 2, and ν 3 can be used to parameterize a rotation tensorR. Consider the 3-2-3 set of Euler angles:

R = F(ψ, e3 = e

3 )F(θ, e

2)F(φ, E3). (5.85)

(a) Draw figures illustrating the relationships between (i) Ei and e

i, (ii) e

i and e

i , and (iii)

e

i and ei.(b) Explain why the second angle of rotation ν 2 = θ is restricted to lie between 0 and π .(c) For this set of Euler angles, show that the Euler basis has the representation

g1g2

g3

=E3

t

2

t3




=

− sin(θ)cos(ψ) sin(ψ) sin(θ) cos(θ)

sin(ψ) cos(ψ) 00 0 1

t1

t2t3

. (5.86)

(d) For this set of Euler angles, show that the dual Euler basis has the representationg1

g2

g3

=

− cos(ψ)cosec(θ) sin(ψ)cosec(θ) 0

sin(ψ) cos(ψ) 0cos(ψ) cot(θ) − sin(ψ) cot(θ) 1

t1

t2t3

. (5.87)

(e) For this set of angles, show the angular velocity vector has the representation

ωR = −1

2[ RRT ] = ν igi = ψe3 + θe

2 + φE3. (5.88)

(f ) Suppose that ωi(t) = ω · ei are known functions. Show that

φ

θ˙

ψ

=

− cos(ψ)cosec(θ) sin(ψ)cosec(θ) 0sin(ψ) cos(ψ) 0

cos(ψ)cot(θ) − sin(ψ) cot(θ) 1

ω1

ω2

ω3

. (5.89)

(g) Why can the solutions to (5.89) be used to find the rotation tensor R? Illustrate youranswer with the specific example, ω1 = 0, ω2 = 0, ω3 = constant.

5.2 Recall that a rotation tensor F representing a counterclockwise rotation about an axis pthough an angle ν has the representation

F = F(ν, p) = cos(ν )(I − p ⊗ p) − sin(ν )p + p ⊗ p. (5.90)

The angular velocity vector associated with this rotation is

ωF = −1

2

FFT

= ν p + sin(ν )p + (1 − cos(ν ))p × p. (5.91)

Consider two rotation tensors:

Q1 = F(θ, E3), Q2 = F(φ, e1)F(θ, E3), (5.92)

wheree1 = cos(θ)E1 + sin(θ)E2, e2 = cos(θ)E2 − sin(θ)E1, e3 = E3. (5.93)

(a) Show that Q1 has the representation

Q1 = e1 ⊗ E1 + e2 ⊗ E2 + e3 ⊗ E3. (5.94)

(b) Give an example of a system of two rigid bodies where the rotation tensor of one bodyis Q1, while the rotation tensor of the second body is Q2.(c) Given that the relative rotation tensor R = Q2QT

1 , show that

ωR = φe1 + sin(φ)θe2 + (1 − cos(φ))θe3. (5.95)

(d) Explain whyωR = ωQ2

−ωQ1. (5.96)



5.8 Exercises 109

5.3 The parameterization of the rotation tensor using Euler parameters (unit quaternions orsymmetric Rodriques-Euler parameters) has the beautiful consequence that the formula forthe composition of two rotations is extremely elegant. Indeed, the same results for two tensorsdescribed 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 20 − f · f )I + 2f ⊗ f − 2f 0(εf ). (5.97)

Here, e0, e and f 0, f represent two sets of Euler parameters:

e0 = cos

φ

2

, e = sin

φ

2

a, f 0 = cos

θ

2

, f = sin

θ

2

b, (5.98)

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 aclockwise rotation of φ about a.

(a) Recall the representation for a rotation tensor A in terms of the angle of rotation φ andthe axis of rotation a:

A(φ, a) = cos(φ)(I − a ⊗ a) − sin(φ)a + a ⊗ a.

Verify that A has the representation (5.97)1.

(b) Letting e =3

i=1 eiEi, what are Aik = (AEk) · Ei?(c) Show that

C = BA = (g20 − g · g)I + 2g ⊗ g − 2g0(εg), (5.99)

whereg0 = e0f 0 − e · f , g = e0f + f 0e + f × e. (5.100)

This result was first established by Rodrigues [45] in 1840. The earliest English commentaryon it is by Cayley [13] in 1845.(d) In terms of e0, f 0, f , and e, what are the Euler parameters of the rotation tensors ABand 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 (e0e − e0e + e × e) , (5.101)

where ωA = −12[ AAT ].

(i) What are ωB and ωC?(ii) Give an explanation for the following result:

ωC = ωB + ωA. (5.102)

(g) If e0 = f 0 = 1√ 2

, 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.




5.4 Using the relative angular velocity vectors, show that the angular velocity vector associatedwith the tensor R = R2R1 is the D’arboux vector:

ωSF = τ et + κeb, (5.103)

where

τ = dν

ds +

dθ

ds cos(φ), κ =

dφ

ds

2

+

dθ

ds sin(φ)

2

. (5.104)

Here, the tensor R transforms a fixed Cartesian basis to the Serret–Frenet triad:

en = RE1 , eb = RE2 , et = RE3, (5.105)

where R = R2R1, and

R1 = (cos(θ) cos(φ)E1 + sin(θ)cos(φ)E2 − sin(φ)E3) ⊗ E1

+ (− sin(θ)E1 + cos(θ)E2) ⊗ E2

+ (cos(θ) sin(φ)E1 + sin(θ) sin(φ)E2 + cos(φ)E3) ⊗ E3,

R2 = cos(ν )en ⊗ en + sin(ν )eb ⊗ en

+ cos(ν )eb ⊗ eb − sin(ν )en ⊗ eb + et ⊗ et. (5.106)

To solve this problem, it is helpful to first show that R1 is a composition of two rotations:

R1 = F(φ, eθ)F(θ, E3) (5.107)

where R1 = eφ ⊗ E1 + eθ ⊗ E2 + et ⊗ E3. After calculating ωR1 and ωR, you should then

appeal to the result:

eb = 1

dφds

2+

dθds

sin(φ)

2

dφ

dseθ − sin(φ)

dθ

dseφ

. (5.108)

This result follows from the definitions eb = et × en and detds = κen. If you don’t use (5.108),

but instead take a longer path, which involves expressing ωSF in terms of its components inthe Serret–Frenet triad, from the en component, you will find that

tan(ν ) =dθds

sin(φ)dφds

. (5.109)

On your way to finding ωSF , this equation can then be used to find cos(ν ) and sin(ν ) interms of dθ

ds sin(φ) and dφ

ds.

5.5 The latitude (λ) and longitude (θ) of a point on the Earth’s surface are illustrated schemat-ically in Figure 5.4. In navigation systems, one use these angles to define the downwarddirection ez, the northerly direction ex and the easterly direction ey:

exey

ez

=− cos(θ) sin(λ) − sin(θ)sin(λ) cos(λ)− sin(θ) cos(θ) 0

− cos(θ) cos(λ) −sin(θ)cos(λ) −sin(λ)

E1E2

E3

. (5.110)



5.8 Exercises 111

E

2

E3

ey

ex

ez

λ E

1

ex

ey

θ

e z -

Down

East

North

Fig. 5.4. The angles of longitude θ and latitude λ.

Here, the triad E1, E2, E3 is a set of fixed right–handed Cartesian basis vectors.(a) Given that R = ex ⊗ E1 + ey ⊗ E2 + ez ⊗ E3, verify that this rotation tensor can becomposed 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 = xxex + xyey + xzez = X 1E1 + X 2E2 + X 3E3. (5.111)

Show that

xx =3

i=1

Ri1X i, xy =3

i=1

Ri2X i, xz =3

i=1

Ri3X i, (5.112)

where

R =

3i=1

3k=1

RikEi ⊗ Ek. (5.113)

How are Rik’s related to the matrix in equation (5.110)?5.6 Consider two rotation tensors A and B where

A =3

i=1

ti ⊗ Ei, B =3

i=1

ei ⊗ ti, (5.114)

where E1, E2, E3 is a fixed, right-handed orthonormal basis for E 3.(a) Show that

B =o

B +Ω AB − BΩ A, (5.115)

whereo

B is the corotational derivative of B assuming that A is fixed, and

Ω A = AAT . (5.116)

(b) Consider the rotation tensor C = BA. Using the results of (a), show that




Ω B =o

B BT , (5.117)

where the relative angular velocity tensor Ω B and the angular velocity tensor Ω C are, re-spectively,

ˆΩ B = Ω C −Ω A, Ω C =

˙CC

T

. (5.118)Why is Ω B skew-symmetric, and what does this imply for the product Ω Bg where g is anyvector?(c) Consider the example

A = cos(ψ)(I − E3 ⊗ E3) − sin(ψ)E3 + E3 ⊗ E3,

B = cos(θ)(I − t1 ⊗ t1) − sin(θ)t1 + t1 ⊗ t1, (5.119)

wheret1 = cos(ψ)E1 + sin(ψ)E2. (5.120)

You also may need to recall here that any rotation tensor R has the following representation:

R = cos(ν )(I

−p

⊗p)

−sin(ν )p + p

⊗p,

where p is a unit vector and ν is a counter-clockwise angle of rotation. The angular velocityassociated with this rotation is

ωR = −1

2

RRT

= ν p + (1 − cos(ν )) p × p + sin(ν )p. (5.121)

(i) What are A ando

B?(ii) Using equation (5.121), what are ωA and ωB?

(iii) Again with the assistance of equation (5.121), verify that

ωB = θt1 − ψ (cos(θ)E3 − sin(θ)t2) + ψE3. (5.122)

Here, ωB is the angular velocity vector associated with B.

5.7 Consider the tensor R:

R = cos(θ)E1 ⊗ E1 + sin(θ)E2 ⊗ E1

+ cos(θ)E2 ⊗ E2 − sin(θ)E1 ⊗ E2 + E3 ⊗ E3. (5.123)

(a) Show that R 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. (5.124)

(b) Show that

Ω = θ (E2

⊗E1

−E1

⊗E2) ,

= θ (eθ ⊗ er − er ⊗ eθ) ,

ω = θE3, (5.125)



5.8 Exercises 113

where

Ω = RRT , ω = −1

2[Ω ]. (5.126)

5.8 Show that the rotation tensor F(π2 , E3)F(π

2 , E1) corresponds to F( 2π3 , p) where

p = 1√

3(E1 + E2 + E3) . (5.127)

This result is very useful in analyzing material symmetry groups of crystals.5.9 Show that the rotation tensor F(π, E2)F(π, E1) corresponds to F(π, E3). This result has a

interpretation which is sometimes used to show that successive rotations about two perpindic-ular axes can be reproduced by a single rotation about an axis which is perpindicular to bothaxes. It is also the source for one explanation of a paradox in biomechanics.





6

Kinematics of Rigid Bodies

6.1 Introduction

This chapter contains results on the three–dimensional kinematics of rigid bodies. In particular,we show how to establish certain useful representations for the velocity and acceleration vectorsof any material point of a rigid body. We also discuss the angular velocity vector of a rigid body.Next, we discuss the linear momentum G, angular momenta H, HO, and HA, and kinetic energyT of rigid bodies and the inertias that accompany them. The chapter concludes with a discussionof the configuration manifold for a rigid body.

The principal references for this chapter are Beatty [3], and Casey [8, 10].

6.2 General Considerations

To discuss the kinematics of rigid bodies, it is convenient to follow some developments in contin-uum mechanics and define the reference and present configuration of a rigid body.

First, a body B is a collection of material points (mass particles or particles). We denote amaterial point of B by X . The position of the material point X , relative to a fixed origin, at timet is denoted by x (see Figure 6.1). The present (or current) configuration κt of the body is asmooth, one-to-one, onto function that has a continuous inverse. It maps material points X of Bto 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 notethat the present configuration defines the state of the body at time t.

It is convenient to define a fixed reference configuration κ0 of the body. This configurationis defined by the invertible function X = κ0(X ). Using the invertibility of this function, we canuse the position vector X of a material point X in the reference configuration to uniquely definethe material point of interest. Later, we will use the reference configuration to determine manyof the properties of a body, such as its mass m and inertias.

Using the reference configuration, we can define the motion of the body as a function of X

and t:x = χ(X, t). (6.1)





6.3 The Angular Velocity and Angular Acceleration of a Rigid Body 117

necessarily the case. Perhaps, the most useful interpretation of Q is to interpret it as the trans-formation which takes vectors between two points in the body and transforms them into therepresent state.

6.3 The Angular Velocity and Angular Acceleration of a Rigid Body

The motion of a rigid body has the form (6.3). Because Q is a rotation tensor, we can define anangular velocity tensor Ω and an angular velocity vector ω:

Ω = QQT , ω = −1

2[ QQT ]. (6.5)

The vector ω is the angular velocity vector of the rigid body, while Ω is the angular velocitytensor of the rigid body. The rotation tensor Q can be represented in a variety of manners, forinstance, 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 therigid body:

α = ω. (6.6)

You should notice that

α = −1

2[QQT + Q QT ] = −1

2[ Ω ]. (6.7)

where we used the fact that = O.We can use the result (6.4) to determine the relative velocity and acceleration vectors of any

two points X 1 and X 2 of the rigid body:

v1 − v2 = x1 − x2

= Q (X1 − X2)

= QQT Q (X1

−X2)

= Ω Q (X1 − X2)

= ω × (x1 − x2) . (6.8)

A further differentiation and some manipulations gives the relative acceleration vectors:

a1 − a2 = v1 − v2

= ω × (x1 − x2) + ω × (x1 − x2)

= α × (x1 − x2) + ω × (v1 − v2) . (6.9)

You should notice that the final forms of the relative velocity and acceleration vectors are ex-pressed as functions of t, x1, and x2. They can also be expressed as functions of t, X1, andX2.



118 6 Kinematics of Rigid Bodies

6.4 A Corotational Basis and the Motion of a Rigid Body

It is convenient, when discussing the dynamics of rigid bodies, to introduce another basise1, e2, e3 which is known as a corotational basis.2 This section discusses such a basis and

points out some features of its use. Our discussion of the corotational basis follows Casey [8].

O

X1

X1

E1

E3

E2

e1

e3

e2X4

X4

X3

X3X2

X2

Present configuration

Reference configuration

Fig. 6.2. The corotational basis e1, e2, e3 and the fixed Cartesian basis E1, E2, E3.

Referring to Figure 6.2, we start by picking 4 material points X 1, X 2, X 3, and X 4 of the body.These points are chosen, such that the vectors

E1 = X1 − X4, E2 = X2 − X4, E3 = X3 − X4 (6.10)

form a fixed, right-handed, Cartesian basis. We next consider the present relative locations of the 4 material points. Because Q preserves lengths and orientations, the three vectors x1 − x4,

x2 − x4, and x3 − x4 will also form a right-handed orthonormal basis. As a result, using (6.4),we define the corotational basis to be

e1 = x1 − x4, e2 = x2 − x4, e3 = x3 − x4. (6.11)

You should notice thatQ = e1 ⊗ E1 + e2 ⊗ E2 + e3 ⊗ E3. (6.12)

This result follows from (6.4) and our previous discussions on representations of rotation tensors.Since the corotational basis moves with the body, we can use our previous results for relative

velocities to see thatei = ω × ei, ei = α× ei + ω × (ω × ei) , (6.13)

where i = 1, 2, 3.As the corotational basis is a basis for Euclidean three-space

E 3, for any vector r we have the

representation2 This basis is often referred to as a body-fixed frame or an embedded frame.



6.5 Center of Mass and Linear Momentum 119

r =3

i=1

riei. (6.14)

The time derivative r of r has the representations:

v = r

=

3i=1

riei +

3i=1

riei

= or +ω × r, (6.15)

where or is the corotational derivative of the vector r. A related expression can be obtained for r:

a = v

=3

i=1

riei + 23

i=1

riei +3

i=1

riei

=

3i=1

riei + 2ω× o

r +ω × (ω × r) + α× r. (6.16)

This result should be familiar to you from other courses.

6.5 Center of Mass and Linear Momentum

In some of the previous developments, we defined the motion of one material point relative toanother material point of the same body. It is convenient for later purposes to now define aparticular point: the center of mass C .

We first dispense with some preliminaries. Let R0 and R denote the regions of Euclidean three–space occupied by the body in its reference and present configurations, respectively. Further, let

Y and y be the position vectors of a material point Y of the body in its reference and presentconfigurations, respectively (see Figure 6.3).

6.5.1 The Center of Mass

The position vectors of the center of mass of the body in its reference and present configurationsare defined by

X =

R0

Xρ0dV R0

ρ0dV , x =

R xρdv R ρdv

, (6.17)

where ρ0 = ρ0(X) and ρ = ρ(x, t) are the mass densities per unit volume of the body in thereference and present configurations. If a body is homogeneous, then ρ0 is a constant that isindependent of X.

We assume that the mass of the body is conserved. That is,

dm = ρ0dV = ρdv, (6.18)




O

x

y

Y

X

Present

configuration

Reference

configuration

Y

X Y

X

Fig. 6.3. Position vectors of two material points X and Y of a body.

or, equivalently,

m =

R0

ρ0dV =

R

ρdv. (6.19)

This is the principle of mass conservation. Hence,

mX =

R0

Xρ0dV, mx =

R

xρdv. (6.20)

In addition, one has the useful identities

0 =

R0

(X − X)ρ0dV, 0 =

R(x − x)ρdv. (6.21)

These identities play key roles in establishing expressions for momenta and energies of a rigidbody.

A special feature of rigid bodies is that the center of mass behaves as if it were a materialpoint: which we denote by C . To see this, consider

mx =

R

xρdv

=

R

(QX + q) ρdv

=

R0

(QX + q) ρ0dV

= Q

R0Xρ0dV

+ R0

ρ0dV q

= mQX + mq. (6.22)



6.5 Center of Mass and Linear Momentum 121

Consequently,x = QX + q. (6.23)

Recalling (6.3), this implies that the center of mass of the rigid body behaves as if it were amaterial point of the rigid body.

For many bodies, such as a rigid homogeneous sphere, the center of mass corresponds tothe geometric center of the sphere, while for others, such as a rigid circular ring, it does notcorrespond to a material point.

6.5.2 The Linear Momentum

We next turn to the linear momentum G of a rigid body. By definition, this momentum is

G =

R

vρdv. (6.24)

That is, the linear momentum of a rigid body is the sum of the linear momenta of its constituents.We can establish an alternative expression for G using the center of mass:

G =

R

vρdv

=

R

dx

dt ρdv

=

R0

dx

dt ρ0dV

= d

dt

R0

xρ0dV

= d

dt

R

xρdv

= ddt (mx)

= mv, (6.25)

where v = ˙x is the velocity vector of the center of mass. Hence,

G = mv. (6.26)

A related result holds for a (finite) system of particles.

6.5.3 Relative Position Vectors

For a material point X of a rigid body, it is convenient to define the relative position vectors π

and Π : Π = X − X, π = x − x. (6.27)




Representative examples of these vectors are displayed in Figure 3.4. With the assistance of (6.4)and (6.23), we see that

π = QΠ . (6.28)

Using the corotational basis, it is also easy to see that

π · ei = Π · Ei. (6.29)

This implies that the relative position vectors have the representations

Π =3

i=1

Π iEi, π =3

i=1

Π iei. (6.30)

Furthermore, the corotational derivative of π is zero: π = ω × π.

6.6 Angular Momenta

The angular momenta of a rigid body are its most important distinctive feature when comparedto a particle. In particular, the angular momentum relative to two points, the center of mass C

and a fixed point O, is of considerable importance For convenience, we shall assume that thefixed point O is also the origin (see Figure 6.4).

O

x

x

X

X

Present

configuration

Reference

configuration

C

X C

X

Fig. 6.4. Relative position vectors Π and π of a material point X of a body.

By definition, the angular momenta of a rigid body relative to its center of mass C , H, a fixedpoint O , HO, and a point A, HA, are



6.7 Euler Tensors and Inertia Tensors 123

H =

R

(x − x) × vρdv,

HO =

R

x × vρdv,

HA = R

(x − xA) × vρdv. (6.31)

where the position vectors in these expressions are relative to the fixed point O, and xA is theposition vector of the point A. You should notice that the velocity vector in these expressions isthe absolute velocity vector.

The aforementioned angular momenta are related by simple and important formulae. To findone of these formula, we perform some manipulations on HO:

HO =

R

x × vρdv

=

R(π + x) × vρdv

= Rπ × vρdv +

R

x × vρdv

= H + x × R

vρdv. (6.32)

That is,HO = H + x × G. (6.33)

In words, the angular momentum of a rigid body relative to a fixed point O is the sum of theangular momentum of the rigid body about its center of mass and the angular momentum of itscenter of mass relative to O. Paralleling the establishment of (6.33), it can also be shown that

HA = H + (x − xA) × G, HO = HA + xA × G. (6.34)

These results have obvious similarities to (6.33).

6.7 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 momemtum H. These developments are considerably aided using the Euler tensorsE0 and E and the inertia tensors J0 and J.

6.7.1 Euler Tensors

We next define the Euler tensors (relative to the center of mass of the rigid body):

E0 =

R0

Π ⊗Π ρ0dV, E =

Rπ ⊗ πρdv. (6.35)




You should also notice that E and E0 are symmetric.Using mass conservation, (6.28), and the identity (Aa) ⊗ (Bb) = A(a ⊗ b)BT , it is easy to

see thatE = QE0QT . (6.36)

Furthermore,(Eei) · ek = (E0Ei) · Ek. (6.37)

This implies that E has the representation

E =3

i=1

3k=1

E ikei ⊗ ek, (6.38)

where E ik = (E0Ek) · Ei are the constant components of E0. In other words, although E is afunction 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.

6.7.2 Inertia Tensors

The inertia tensors J0 and J can be defined using the Euler tensors:

J0 = tr(E0)I − E0, J = tr(E)I − E. (6.39)

Using the definitions of the Euler tensors and the identity tr(a ⊗ b) = a · b, these definitions canbe restated as

J0 =

R0

(Π ·Π )I −Π ⊗Π ρ0dV, J =

R

(π · π)I − π ⊗ πρdv. (6.40)

It is easy to see thatJ = QJ0QT , JT

0 = J0, JT = J. (6.41)

The first of these results implies that

J =

3i=1

3k=1

J ikei ⊗ ek, (6.42)

where J ik = (J0Ei) · Ek are the constant components of J0. Not surprisingly, this is similar tothe situation that arose with the Euler tensor.

6.7.3 Additional Relationships

Another set of interesting results follows by inverting the relationships between the Euler andinertia tensors:

E0 = 1

2tr(J0)I

−J0, E =

1

2tr(J)I

−J. (6.43)

These results are useful for obtaining the Euler tensors from tabulations of the inertia tensorwhich are found in numerous textbooks.



6.7 Euler Tensors and Inertia Tensors 125

It is a good exercise to substitute (6.30) into (6.40) to see that the components of J0 relativeto the basis E1, E2, E3 are volume integrals involving quadratic powers of the components of Π . For example,

J 011 = (J0E1) · E1 = R0 y

2

+ z

2

ρ0dV,

J 012 = (J0E2) · E1 = − R0

xyρ0dV,

J 013 = (J0E3) · E1 = − R0

xzρ0dV, (6.44)

wherex = π · e1 = Π · E1, y = π · e2 = Π · E2, z = π · e3 = Π · E3. (6.45)

A similar exercise with the components of E0 shows that

E 011 = (E0E1) · E1 =

R0

x2ρ0dV,

E 012 = (E0E2) · E1 = R0

xyρ0dV,

E 013 = (E0E3) · E1 =

R0

xzρ0dV. (6.46)

You should notice the simple relationship between the off–diagonal components of E0 and J0.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 = λ1E1 ⊗ E1 + λ2E2 ⊗ E2 + λ3E3 ⊗ E3. (6.47)

Here, λi are known as the principal moments of inertia. As J = QJ0QT and ei = QEi, we alsohave

J = λ1e1 ⊗ e1 + λ2e2 ⊗ e2 + λ3e3 ⊗ e3. (6.48)It is common to refer to ei as the principal axes of the rigid body.

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 = e1E1 ⊗ E1 + e2E2 ⊗ E2 + e3E3 ⊗ E3. (6.49)

Using the identity J0 = tr(E0)I − E0, the constants ei can be related to the principal momentsof inertia:

e1 = 1

2 (λ2 + λ3 − λ1) , e2 =

1

2 (λ1 + λ3 − λ2) , e3 =

1

2 (λ1 + λ2 − λ3) . (6.50)

As E = QE0QT and ei = QEi, we also have

E = e1e1⊗

e1 + e2e2⊗

e2 + e3e3⊗

e3. (6.51)

Notice that the principal axis ej corresponding to the maximum value of λi corresponds to theminimum value of ei.




Ellipsoid

2

E3

E

E1

Fig. 6.5. An ellipsoid of mass m.

6.7.4 Some Examples

The simplest inertia tensor arises when the body is a homogeneous sphere of radius R or ahomogeneous cube of length a:

J = J0 = 2mR2

5 I, J = J0 =

ma2

6 I, (6.52)

respectively. For these bodies, any three mutually perpindicular unit vectors are principal axes.The next class of bodies are those with an axis of symmetry. For instance, a homogeneous

circular rod of length L and radius R, has a moment of inertia tensor

J0 = mR2

2 E3 ⊗ E3 +

mR2

4 +

mL2

12

(I − E3 ⊗ E3) , (6.53)

where E3 is the axis of symmetry of the circular rod in its reference configuration.

Most bodies, however, don’t have an axis of symmetry. Consider the homogeneous ellipsoidshown in Figure 6.5. The equation for the lateral surface of the ellipsoid is

x2

a2 +

y2

b2 +

z2

c2 = 1. (6.54)

The inertia tensor of the ellipsoid is

J0 = m

5

b2 + c2

E1 ⊗ E1 +

m

5

a2 + c2

E2 ⊗ E2 +

m

5

a2 + b2

E3 ⊗ E3. (6.55)

We leave it as an exercise to write down the Euler tensor E0 for the ellipsoid.It is crucial to note that for all of the above examples we have used the property that any

body has three principal axes. Writing the inertia and Euler tensors with respect to these axesprovides their simplest possible representations.



6.8 Angular Momentum and an Inertia Tensor 127

6.8 Angular Momentum and an Inertia Tensor

The result we now wish to establish is that H = Jω. This is arguably one of the most importantresults in rigid body dynamics. In particular, with the assistance of J0, for a particular rigid body

it allows us to write down a tractible expression for H.We now reconsider the angular momentum H,

H =

R

(x − x) × vρdv

=

Rπ × vρdv

=

Rπ × (v + ω × π)ρdv

=

Rπ × vρdv +

Rπ × (ω × π)ρdv. (6.56)

However, since C is the center of mass and the velocity vector of C is independent of the region

of integration we can take v outside the integral:

H =

Rπ × vρdv +

Rπ × (ω × π)ρdv

=

Rπρdv × v +


= 0 × v +


=

Rπ × (ω × π)ρdv. (6.57)

Notice that we also used the identity R πρdv = 0 in the next to last step of this calculation.

Summarizing, we have

H =

Rπ × (ω × π)ρdv =

R

((π · π)ω − (π · ω)π) ρdv. (6.58)

In writing this equation, we used the identity a × (b×c) = (a ·c)b− (a ·b)c. Using the definitionof the inertia tensor, J, it should now be apparent that

H = Jω. (6.59)

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 notbe parallel.




6.9 Kinetic Energy

The kinetic energy of a rigid body has a very convenient representation which was first establishedby Koenig:

T = 12 mv · v + 12 (Jω) · ω. (6.60)

Here, we give a derivation of this result.The kinetic energy T of a rigid body is defined to be

T = 1

2

R

v · vρdv. (6.61)

This expression for the energy can be simplified by expressing the velocity vector v as

v = v +ω × π. (6.62)

Substituting this expression into T :

T = 12

R

v · vρdv

= 1

2

R

v · vρdv +

R

(ω × π) · vρdv

+ 1

2

R

(ω × π) · (ω × π)ρdv. (6.63)

However, we have the following identities R

v · vρdv = v · v R

ρdv = mv · v,

R(ω × π) · vρdv = ω × R

πρdv · v = (ω × 0) · v = 0, R

(ω × π) · (ω × π)ρdv =

R

(ω ·ω)(π · π) − (ω · π)2ρdv

= ω ·

R(π · π)I − π ⊗ πρdv

ω

= ω · (Jω) . (6.64)

Substituting these results into the previous expression for T , we find that

T = 1

2mv · v +

1

2 (Jω) · ω. (6.65)

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 massand the rotational kinetic energy T rot = 1

2ω · (Jω) of the rigid body. The kinetic energy of thecenter of mass is often referred to as the translational kinetic energy.






The product ⊕ is a topological product. For example E 3 = E ⊕ E 2 = E ⊕ E ⊕ E . It should beclear that the dimension of M is 6. It should also be clear that the configuration manifold can beconsidered as a sub-manifold of the configuration space S which in this case is E 3 ⊕E 9. Here, thethree-dimensional space is the space containing the position vector x, while the nine-dimensional

space is the space containing the second-order tensor.To parameterize M we can use any curvilinear coordinate system, q 1, q 2, and q 3, to parametrize

the position vector x of the center of mass, and any set of Euler angles, γ 1, γ 2, and γ 3, toparametrize the rotation tensor Q. For the velocity vector of the center of mass, we then have

v =

3i=1

q iai, (6.72)

where the covariant basis vectors

ai = ∂ x

∂q i. (6.73)

You should also recall that ai is a basis for E 3. The Euler angles define the Euler basis gi,which is also a basis for

E 3. In particular, we have

ω = γ igi. (6.74)

The Euler basis vectors aren’t linearly independent for certain values of the second Euler angle.It is interesting to note that

gi = ∂ ω

∂ γ i, ai =

∂ v

∂ q i, (6.75)

two identities which are easily established.In most treatments, it is normal to choose the basis Ei which used to define the Euler angles

to be the principal axes of J0 and E0. Except, where explicitly mentioned, we shall assume thatthis is the case.

The kinematical line–element ds for the configuration manifold can be calculated, using thegiven choice of Euler angles and curvilinear coordinates, from the kinetic energy T :

ds =

2T

m

dt

=

v · v +

1

mω · (J · ω)

dt. (6.76)

We shall shortly present an example of how to calculate the desired representation for T .As we are using curvilinear coordinates and Euler angles to parametrize the motion of the

rigid body, for certain points on the configuration manifold singularities in the parameters willarise. At these points, it is possible for the body to be in motion, yet the value we will get forT will be zero. This situation violates one of the chief attributes of T , namely T is zero if andonly if v = 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 set of curvilinear coordinates and two or more sets of Euler angles for a given rigid body.



6.10 Configuration Manifold of a Rigid Body 131

6.10.1 A Representation for the Kinetic Energy

We now consider a specific example. First, we choose a set of spherical polar coordinates R, Θ

and Φ, to parametrize x, and a set of 3-1-3 Euler angles to parametrize Q. For these Euler angles:

the Euler basis gi has the representationsg1

g2

g3

=

E3

e

1

e3

=

sin(ψ) sin(θ) cos(ψ) sin(θ) cos(θ)

cos(ψ) − sin(ψ) 00 0 1

e1

e2e3

. (6.77)

The associated angular velocity vector has the representations:

ω = φE3 + θe

1 + ψe3

= φ sin(ψ) sin(θ) + θ cos(ψ) e1 + φ cos(ψ) sin(θ)

− θ sin(ψ) e2

+

φ cos(θ) + ψ

e3. (6.78)

We choose Ei to be the principal axes of the body.4 Hence,

J0 = λ1E1 ⊗ E1 + λ2E2 ⊗ E2 + λ3E3 ⊗ E3,

J = QJ0QT = λ1e1 ⊗ e1 + λ2e2 ⊗ e2 + λ3e3 ⊗ e3. (6.79)

Combining these results, the kinetic energy T of the rigid body has the representation:

T = m

2 v · v +

1

2ω · Jω

= m

2 R2 + R2 Φ2 + R2 sin2(Φ) Θ2

+ λ1

2

φ sin(ψ) sin(θ) + θ cos(ψ)

2+

λ2

2

φ cos(ψ)sin(θ) − θ sin(ψ)

2+

λ3

2

φ cos(θ) + ψ

2.

(6.80)

We leave it as an exercise to substitute this expression for T into the expression for the kinematicalline–element ds in order to find a measure of distance that the rigid body travels along theconfiguration manifold M .

Singularities in the parametrization of the motion arise when θ = 0, π, and Φ = 0, π. To seethe effects of these singularities on T , let us consider an instant, where Φ = 0 and θ = 0. At thisinstant, the above expression for T simplifies to

4 If we don’t make this choice, then we would find a more complicated representation for the rotationalkinetic energy: ω · (Jω) = J 011ω2

1 + J 022ω2

2 + J 033ω2

3 + 2J 012ω1ω2 + 2J 023ω2ω3 + 2J 013ω1ω3, whereωi = ω · ei and J 0ik = ei · (Jek) = Ei · (J0Ek).




T = m

2

R2 + R2 Φ2 + 0

+

λ1

2

0 + θ cos(ψ)

2+

λ2

2

0 − θ sin(ψ)

2+

λ3

2

φ + ψ

2. (6.81)

If, in addition, φ = − ψ, θ = R = Φ, and Θ = 0, then the above expression for T = 0, but theactual kinetic energy of the body is non–zero.

6.11 Exercises

6.1 Recall that the angular momentum of a rigid body relative to a point A is related to itsangular momentum relative to the center of mass by

HA = H − (rA − x) × G. (6.82)

Establish an expression for HA. Why is the balance of angular momentum relative to amoving point A not normally used?

6.2 As shown in Figure 6.6, a robotic arm is used to move a payload. The robotic arm consists of (i) a drive shaft which rotates about the E3 axis through an angle ψ ,

(ii) an axle A which rotates about g2 through an angle θ relative to the drive shaft,(iii) an axle B which rotates about g3 through an angle φ relative to axle A.

The payload is rigidly attached to the axle B. In this question, the drive shaft, axles A andB, and the payload are assumed to be rigid.

E1

E2

E3

2g

3g

Drive shaft

Payload

Axle B

Axle A

ψ

φ

θ

Fig. 6.6. A robotic arm.

(a) What are the angular velocity vectors of the drive shaft, the axle A, the axle B , and thepayload?



6.11 Exercises 133

(b) Which set of Euler angles is being used to parameterize the rotation tensor Q of thepayload?(c) If the position vector r of a point X on the payload is

r = LE3 + H g3. (6.83)Then in terms of the angles θ , φ and ψ and their rates of change, establish an expression forthe velocity vector v of the point X .(d) Suppose after a time interval t1 − t0 the point X of (c) has returned to its originallocation in space:

r(t1) = r(t0). (6.84)

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 rotation tensor Q(t1)QT (t0) corresponds toa rotation of φ(t1) − φ(t0) about g3(t0).

6.3 Consider the circular disk shown in Figure 6.7. The motion of the disk is given by the positionvector y of an arbitrary material point Y of the disk and the rotation tensor Q of the rigiddisk.

Circular disk

2e2e

1e

1e

φ

P

X

E1 ψ

E

2

E3

E

1

O

"

"

1e"

Fig. 6.7. The present configuration of a circular disk moving with one point in contact with a fixedhorizontal plane.

(a) Starting from the results that the position vectors x and y of any points X and Y of thedisk are

x = QX + q, y = QY + q, (6.85)

show that their relative velocity vector and acceleration vector satisfy

x − y = ω × (x − y) ,

x − y = ω × (x − y) + ω × (ω × (x − y)) . (6.86)

(b) If ei = QEi (i = 1, 2, 3) are the corotational basis vectors for the disk and ω =3

i=1 ωiei,show that




ω =3

i=1

ωiei. (6.87)

(c) To parameterize the rotation tensor of the disk, a set of 3-1-3 Euler angles are used. With

the assistance of the figure above, prescribe a reference configuration for the disk. For whichorientations of the disk in its present configuration do the singularities of the 3-1-3 Eulerangles occur?(d) A sensor is mounted onto the disk and is aligned with the e3 axis so that it measuresω · e3 = ω3(t). Show that, in general, t1

t0

ω3(t)dt = φ(t1) − φ(t0). (6.88)

Give a physical interpretation of the case where the integral of ω3(t) does yield the angle φ.6.4 Consider a rigid body of mass m which is in motion under a resultant moment M and a force

F. The inertia tensor J of the body is defined by

J = QJ0QT

, J0 = R0

(Π ·Π )I −Π ⊗Π ρodV. (6.89)

(a) Why are J0 and J symmetric?

(b) Show that Ei · (J0Ek) = ei · (Jek) , where Q =3

i=1 ei ⊗ Ei.(c) Show that J has the representation

J =3

i=1

λiei ⊗ ei, (6.90)

where λi are the principal values of J0 and Ei = QT ei are the principal directions of J0.(d) Establish the following results:

J = Ω J

−JΩ , H = Jω +ω

×(Jω). (6.91)

How do these results simplify if J0 = µI where µ is a constant?(e) From the balance of angular momentum (relative to the center of mass) of this body, showthat one obtains three differential equations for the corotational components of the angularvelocity vector ω:

λ1 ω1 = (λ2 − λ3) ω2ω3 + M · e1,

λ2 ω2 = (λ3 − λ1) ω1ω3 + M · e2,

λ3 ω3 = (λ1 − λ2) ω1ω2 + M · e3. (6.92)

Here, ωi = ω · ei.(f ) Suppose one used the representation

J =3

p=1

3n=1

J pnE p ⊗ En. (6.93)



6.11 Exercises 135

Starting from (6.90), show that

J pn =

3i=1

Q piλiQni, (6.94)

where Q pi = (QEi) · E p. Why are the components J pn of J not constant?6.5 Recall the definition of the inertia tensor J:

J = QJ0QT , J0 =

R0

(Π ·Π )I −Π ⊗Π ρodV. (6.95)

(i) Why are J0 and J symmetric?

(ii) Show that Ei · (J0Ek) = ei · (Jek) , where Q =3

i=1 ei ⊗ Ei.(iii) If Ei are chosen to be principal axes of J0, then what are the principal axes of J. Why is

such a choice important?(iv) Establish the following results:

J = Ω J − JΩ , ˙Jω · ω = 2 H · ω ,

where the angular momentum H = Jω.(v) What are J0 and J for a rigid cube, a rigid sphere, a rigid cylinder and a rigid rectangle?

In your solution clearly indicate what your choice of the reference configuration (i.e., thebasis E1, E2, E3) is.

6.6 A tensor which is intimately related to the inertia tensor is the Euler tensor:

E = QE0QT , E0 =

R0

Π ⊗Π ρodV. (6.96)

(i) Show thatJ0 = tr(E0)I − E0, J = tr(E)I − E.

(ii) Show that Ei · (E0Ek) = ei · (Eek), where Q = 3i=1 ei ⊗ Ei.

(iii) Establish the following result: E = Ω E − EΩ . (6.97)

(iv) What are the Euler tensors for a sphere of mass m and radius R, and a cylinder of massm, radius R, and length L.

6.7 Recall that the angular momentum H and rotational kinetic energy T rot of a rigid body havethe representations

H = Jω , T rot = 1

2ω · Jω. (6.98)

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

i=1

λiEi ⊗ Ei , J = QJ0QT =3

i=1

λiei ⊗ ei , (6.99)

where Q =3

i=1 ei ⊗ Ei.




(i) Using a set of 3-2-3 Euler angles to parameterize Q, show that

ω =

θ sin(φ) − ψ sin(θ) cos(φ)

e1 +

θ cos(φ) + ψ sin(φ)sin(θ)

e2

+ ˙

ψ cos(θ) + ˙φ

e3. (6.100)

(ii) Using a set of 3-2-3 Euler angles, establish an expression for H and T rot as functions of the Euler angles and their time-derivatives.

(iii) Suppose the motion of a rigid body is subject to 2 constraints:

ω · g3 = 0, ω · g2 = 0. (6.101)

Give a physical interpretation of these constraints. What are the angular velocity vector,angular momentum vector and rotational kinetic energy of the resulting constrained rigidbody? Why is this case so ubiquituous?

6.8 Recall the definitions of the angular momenta of a rigid body relative to its center of massand a point A:

H =

Rπ × vρdv =

Rπ × (ω × π)ρdv, HA =

RπA × vρdv. (6.102)

Here,π = x − x, πA = x − xA, (6.103)

where xA is the position vector of the point A, and x is the position vector of the center of mass.(a) Starting from (6.102)2, and with the assistance of (6.102)1, show that

HA = H + (x − xA) × G, (6.104)

where G is the linear momentum of the rigid body.(b) Suppose that A is a point of the rigid body. Show that

πA = QΠ A, (6.105)

where Π 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). (6.106)

(c) Again supposing that A is a point of the rigid body and that vA = 0, show that

HA = JAω, (6.107)

where JA is the inertia tensor of the rigid body relative to A:

JA = R(πA

·πA)I

−πA

⊗πAρdv. (6.108)

Verify that



6.11 Exercises 137

JA = QJA0 QT , (6.109)

where

JA0 = R0

(Π A ·Π A)I −Π A ⊗Π Aρ0dV. (6.110)

(d) Using the previous results, show the parallel–axis theorem:

JA0 = J0 + m

(X − XA) · (X − XA)I − (X − XA) ⊗ (X − XA)

. (6.111)

(e) Suppose that the rigid body is a circular cylinder of mass m, length L, and radius R,where

X = 0, XA = −xE1 − zE3,

J0 = mL2

12 (I − E3 ⊗ E3) +

mR2

4 (E1 ⊗ E1 + E2 ⊗ E2 + 2E3 ⊗ E3). (6.112)

Then, what is JA0 ?





7

Constraints on and Potentials for Rigid Bodies

7.1 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 inter–connected rigid bodies,rolling rigid bodies and sliding rigid bodies are present along with prescriptions for the associatedconstraint forces and moments are also presented. We also discuss the normality prescription forconstraint forces and constraint moments and outline its limitations.

It also proves convenient to discuss potential energies and their associated conservative forcesand moments. Our discussion includes as examples springs and central gravitational fields. It hasobvious parallels to the treatment of constraints and their associated forces and moments.

Much of the material in these notes is based on Antman [1], Casey [10], Hughes [25], O’Reillyand Srinivasa [39], and Simmonds [50].

7.2 Preliminaries Concerning Two Rigid Bodies

For future purposes it is convenient to consider two rigid bodies B1 and B2. As shown in Figure7.1, the position vector of the center of mass X α of the body Bα is denoted by xα where α = 1or 2. Similarily, the rotation tensor of Bα is denoted by Qα. For each rigid body, we can definecorotational bases:

Q1 =3

i=1

1ei ⊗ Ei, Q2 =3

i=1

2ei ⊗ Ei. (7.1)

It should be noted that we are using the same fixed basis E1, E2, E3 to define the corotationalbases. The angular velocity vectors of each body are

ω1 = −1

2

Q1QT 1

, ω2 = −1

2

Q2QT 2

. (7.2)

It should also be noted that the rotation tensor of B2 relative to B1 is

Q2QT 1 =

3i=1

2ei ⊗ 1ei. (7.3)



140 7 Constraints on and Potentials for Rigid Bodies

This tensor represents the rotation of B2 relative to an observer who is stationary on B1.

x

O

X2x

1

2

X1

Fig. 7.1. The present configurations and centers of mass of two rigid bodies B 1 and B 2.

7.2.1 A Scalar Valued Function of the Motions

We now consider a scalar function Φ:

Φ = Φ (x1, x2, Q1, Q2, t) . (7.4)

Clearly, this function depends on the motions of both rigid bodies and time. We shall see functionsof this type when we represent integrable constraints on the motions of rigid bodies and potentialenergies of rigid bodies.

As in the case of a particle, it is of interest to calculate Φ. To calculate this time derivative,we invoke the chain-rule:

Φ = ∂Φ

∂ x1· v1 +

∂Φ

∂ x2· v2 + tr

∂Φ

∂ Q1QT 1

+ tr ∂Φ

∂ Q2QT 2

+ ∂Φ

∂t . (7.5)

Recalling our earlier discussion of derivatives of scalar functions of rotation tensors, it is conve-nient to define the skew–symmetric tensor

ΦQα =

1

2

∂Φ

∂ Qα

QT α − Qα

∂Φ

∂ Qα

T

(7.6)

and an associated vectorφQα

= −[ΦQα]. (7.7)

Representations for these tensors and vectors, based on the parameterization of Qα were discussedearlier. In particular, if the Euler angles ν i are used to parameterize Q1, say, then

φQ1 =

3i=1

∂Φ

∂ν igi, φQ1

·ω1 =3

i=1

∂Φ

∂ν i ν i, (7.8)



7.2 Preliminaries Concerning Two Rigid Bodies 141

where gi is the dual Euler basis associated with the Euler angles.We can use the aforementioned skew–symmetric tensors to rewrite Φ:

Φ = ∂Φ

∂ x1 · v1 +

∂Φ

∂ x2 · v2 + tr ΦQ1

Ω T 1 + trΦQ2

Ω T 2 +

∂Φ

∂t=

∂Φ

∂ x1· v1 +

∂Φ

∂ x2· v2 + φQ1

·ω1 + φQ2 ·ω2 +

∂ Φ

∂t , (7.9)

where

Ω α = QαQT α , ωα = −1

2[Ω α], (7.10)

are the angular velocity tensors and vectors of the rigid bodies.It is more efficient in the sequel to use the form of Φ which involves vectors:

Φ = ∂Φ

∂ x1· v1 +

∂Φ

∂ x2· v2 + φQ1

·ω1 + φQ2 ·ω2 +

∂Φ

∂t . (7.11)

With some minor rearrangements, we can eliminate v2 and ω2 in favor of the relative velocity

vectors v2 − v1 and ω2 −ω1 in the expression for Φ. The equation (7.11) will play a key role inexamining potential forces and moments.

The result (7.11) is rarely apparent in treatments of rigid body dynamics. This is partiallybecause specific parameterizations of xα and Qα are used. To elaborate on this point, let

x1 =

3i=1

xiEi, x2 =

3i=1

yiEi, (7.12)

and suppose that Q1 is parameterized by the Euler angles ν i, while the relative rotation tensorQ2QT

1 is parameterized by the Euler angles γ i. That is,

ω1 =3

i=1

ν igi, ω2

−ω1 =

3

i=1

γ ini, (7.13)

where gi and ni are the Euler basis associated with ν i and γ i, respectively. Then,

Φ = Φ(x1, x2, Q1, Q2, t) = Φ(xi, yi, ν i, γ i, t). (7.14)

Furthermore,

Φ = ˙Φ

=3

i=1

∂ Φ

∂xi

xi + ∂ Φ

∂yi

yi + ∂ Φ

∂ν i ν i +

∂ Φ

∂γ i γ i +

∂ Φ

∂t . (7.15)

It is good exercise to compare this expression with (7.11) and identify the corresponding terms

in both expressions.




7.3 Constraints

Constraints in the motions of rigid bodies usually arise in two manners. First, the rigid bodyis 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 constraintsarise occurs when one rigid body is rolling or sliding on the other. The constraints we discusscan be classified as integrable or non–integrable. This classification is important in dynamicsbecause it may lead to considerable simplification in the formulation of the equations governingthe motion.

In our discussion, we consider two rigid bodies, B1 and B2. The case of a rigid body interactingwith the ground can be considered by prescribing the velocity vector and angular velocity vectorof one of the bodies as a function of time. If more than two rigid bodies are involved in theconstraint, then it is not too difficult to generalize the discussion presented below. We also notethat the discussion presented here of the various types of joints and contacts is not exhaustive.

7.3.1 Inter–Connected Rigid Bodies

As shown in Figure , consider two rigid bodies B1 and B2 which are connected at the point P by a joint. The position vector of P on Bα relative to X α is denoted by pα. Because the point P

occupies the same location for both bodies, we have the three constraints:1

x1 + p1 = x2 + p2. (7.16)

These constraints also imply that

v1 + ω1 × p1 = v2 + ω2 × p2. (7.17)

Notice that by integrating (7.17) with respect to time and setting the constant of integration to0, we will arrive at (7.16). Consequently, the constraints (7.17) 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 ball and socket joint. The pin jointarises in gyroscopes where it serves to connect the inner and outer gimbals. On the other hand,ball and socket joints don’t place any restriction on the relative angular velocity vector ω2 −ω1.

Consider the case of a pin joint. Let s3 be a unit vector which is parallel to the rotation axis of the joint, and s1, s2, s3 be a orthonormal basis. The pin joint ensures that the relative rotationQ2QT

1 is a rotation about s3. These restrictions are most easily expressed in terms of the relativeangular velocity vector. Specifically, a pin joint imposes the constraints:

(ω1 −ω2) · s1 = 0, (ω1 − ω2) · s2 = 0. (7.18)

It should be clear that these two (integrable) constraints are supplemented by (7.17). In addition,it is sufficient to use a single angle to parameterize Q2QT

1 .Clearly each of the individual constraints mentioned to this point can be written in the form

1 By taking the components of this vector equation with respect to a basis, one arrives at three inde-pendent scalar equations. Hence, the vector equation represents three constraints.



7.3 Constraints 143

x

O

X

P

p1

p2

1

1

x2

X2

Fig. 7.2. The present configurations of two rigid bodies B 1 and B 2 which are connected by a joint at P .

7.3.1

π = 0, (7.19)

whereπ = f 1 · v1 + f 2 · v2 + h1 · ω1 + h2 · ω2 + e. (7.20)

Here, f 1, f 2, h1, h2, and e are functions of x1, x2, Q1, Q2, and t. Furthermore, (7.19) can be inte-grated with respect to time, to yield a function Π :

Π = Π (x1, x2, Q1, Q2, t). (7.21)

Here,Π = π. (7.22)

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 tothe next set of situations we consider.

7.3.2 Rolling and Sliding Rigid Bodies

Consider the situation shown in Figure 7.3. Here, two rigid bodies are instantaneously in contactat the point P . It is assumed that there is a well–defined unit normal n to the surfaces of bothbodies at the point P . In addition, we use n to define an orthonormal basis t1, t2, n where t1and t2 are tangent to the surfaces of both bodies at the point of contact P .

As the point of contact P concides with a material point of each body, we have

x p1 = x1 + π p1, v p1 = v1 + ω1 × π p1, (7.23)

and





7.4 Forces and Moments Acting on a Rigid Body 145

The Rolling Condition

Rolling occurs when the velocity vectors of the point of contact for each body are identical. Inthis case, we have the rolling condition:

v p1 = v p2. (7.28)

Again, we can express this equation in another form:

v1 + ω1 × π p1 = v2 + ω2 × π p2. (7.29)

These three equations are equivalent to the sliding condition (7.25) and the condition that thesliding velocity vs is zero.

The rolling condition (7.29) is equivalent to three scalar equations (cf. (7.19)):

πi = 0, (7.30)

whereπi = f i1 · v1 + f i2 · v2 + hi1 ·ω1 + hi2 · ω2 + ei. (7.31)

However, for two of these equations, say π

2 = 0 and π

3 = 0, it is not possible to find functionsΠ 2 and Π 3 such that Π 2 = π2 and Π 3 = π3. In other words, two of the constraints (7.29) arenon–integrable or non–holonomic. The one constraint of (7.29) that is integrable corresponds tothe n component of (7.29). We shall shortly discuss some examples which illustrate this point.

7.4 Forces and Moments Acting on a Rigid Body

Before discussing constraint forces, we dispense with some preliminaries. The resultant force Facting on a rigid body is the sum of all the forces acting on a rigid body. Similarly, the resultantmoment 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 centerof mass X by M. These moments may be decomposed into two additive parts, the moment dueto the individual external forces acting on the rigid body and the applied external moments thatare not due to external forces. We shall 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) act on the rigid body. The force FK acts at the material point X K ,which has a position vector xK . In addition, a pure moment M p, which is not due to the momentof an applied force, also acts on the rigid body (see Figure 7.4). For this system of applied forcesand moments, the resultants are

F =L

K =1

FK ,

MO = M p +

LK =1

xK × FK ,

M = M p +L

K =1

(xK − x) × FK . (7.32)




x

O

X

X

present

configuration

of

F K

K

x K

K M p

Fig. 7.4. A force FK and a moment Mp acting on a rigid body.

You should notice how the pure moment M p features in these expressions.The mechanical power P of a force FK acting at a material point X K is defined to be

P = FK · xK . (7.33)

Using the center of mass, we can obtain a different representation of this power. Specifically, weuse the identity, xK = v + ω × πK , where πK = xK − x. It then follows that

P = FK · xK = FK · v + (πK × FK ) ·ω. (7.34)

In words, the power of a force is identical to the combined power of the same force acting at thecenter of mass, and its moment, relative to the center of mass. The mechanical power of a puremoment M p is defined to be

P = M p ·ω. (7.35)

You should notice how this expression is consistent with the previous expression for the mechanicalpower 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 previouslyis

P =

LK =1

FK · xK + M p · ω = F · v + M · ω. (7.36)

These results will play a key role in our future discussion of constraint forces and moments andpotential energies.

7.5 Constraint Forces and Constraint Moments

Given a system of constraints on the motion of one or more rigid bodies, a system of forces andmoments is required to ensure that the constraints are enforced for all possible motions of the



7.5 Constraint Forces and Constraint Moments 147

bodies which are compatible with the constraints. At issue here is the prescription of these forcesand moments.

One of the major points of this section is the normality prescription associated with constraintsof the form (cf. (7.19))

π = 0, (7.37)where

π = f 1 · v1 + f 2 · v2 + h1 · ω1 + h2 · ω2 + e. (7.38)

We showed earlier, how most of the commonly used constraints in rigid body dynamics could bewritten in this form.

Pedagogically, it is easiest to present some examples and then give the normality prescription.However, this prescription is not universally applicable. For instance, as in particle dynamics, itdoes not give physically realistic constraint forces and moments when dynamic friction is present.

7.5.1 A Rigid Body Rotating about a Fixed Point

Consider the rigid body shown in Figure 7.5. 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.

X

O

x

E2

E1

Pin joint at the

fixed point O

e2 e1

Fig. 7.5. A rigid body B which is performing a fixed–axis rotation about the point O.

As a result of the pin joint, the angular velocity vector ω of the body is subject to twoconstraints. These constraints can be expressed in a variety of manners. For instance,

ω · E1 = 0, ω · E2 = 0, (7.39)

orω · e1 = 0, ω · e2 = 0. (7.40)

Now because O is fixed, there are three additional constraints on the motion of the body:

v − ω × x = 0. (7.41)




In other words, the pin joint couples the rotational motion of the body with the motion of itscenter of mass.

To enforce the constraints (7.39), we assume that two constraint moments are exerted by thepin joint on the body. These moments have components perpindicular to E3:

Mc1 = µ4E1, Mc2 = µ5E2, (7.42)

where µ4 and µ5 are functions of time which are determined from the balance laws. Theseconstraint moments are examples of pure moments. Our prescription of the constraint momentsassumes that the pin joint is frictionless. If friction were present, then they would have componentsin the E3 direction. Because the pin joint also imposes the constraints that O is fixed, it alsosupplies a reaction force R acting at O :

R =

3i=1

µiEi, (7.43)

where µ1, µ2, and µ3 are functions of times which are determined from the balance laws.

The reaction force and constraint moments are equipollent to a constraint force Fc acting atthe center of mass of the rigid body and a constraint moment Mc relative to the center of massof the rigid body:

F =

3i=1

µiEi, Mc = µ4E1 + µ5E2 − x ×

3i=1

µiEi

. (7.44)

Notice that there are 5 constraints on the rigid body which were imposed by the pin joint and 5unknown functions µ1, . . . , µ5 in the expressions for the constraint forces and moments.

7.5.2 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 thebasis for pool (billiards) and bowling. The main contributors to this problem in the 19th centurywere the French mechanician Gaspard G. de Coriolis (1792–1843) (see [17] and Edward J. Routh(1831–1907) (see [47]). Studies on the dynamics of rolling spheres on surfaces of revoluation areoften known as Routh’s problem.

Consider the rigid body shown in Figure 7.6. The body in this case is moving on a fixedinclined plane. In the figure, the body is assumed to be a sphere. For the sphere, it is easy tosee that πP = −RE3, where R is the radius of the sphere. In addition, you should notice thatt1, t2, n = E1, E2, E3.

Let us first consider the case where the sphere is assumed to be sliding. In this case, the slidingcondition (7.25) vP · n = 0 is simply:

v · E3 = 0. (7.45)

In addition, the slip velocity vs is

vs = vs1E1 + vs2E2 = v − ω × (−RE3) . (7.46)




x

x P

E3

E1

Sphere of radius R

Inclined plane P

X

O

Fig. 7.6. A rigid sphere moving on an inclined plane.

To enforce the sliding condition, a force Fc acts at the point P :

Fc = µ3E3 − µd||µ3E3|| vs

||vs|| , (7.47)

where µd is the coefficient of dynamic friction. As in the previous example, µ3 is a function of time which is determined by the balance laws.

For the rolling sphere, there are three constraints:

vP · E1 = 0, vP · E2 = 0, vP · E3 = 0. (7.48)

These constraints can be expressed in an alternative form:

v·

Ei −

R(ω×

E3

)·

Ei, (7.49)

where i = 1, 2, or 3. To enforce the three constraints, we assume that a normal force and a staticfriction force acts at P . The sum of these two forces is a reaction force R:

R =3

i=1

µiEi, (7.50)

where µi are functions of time which are determined from the balance laws. You should noticethat the reaction force R is equipollent to a force Fc = R acting at the center of mass and amoment Mc = −RE3 × R relative to the center of mass. A related comment pertains to thesliding rigid body.

7.5.3 The Rolling Disk

As shown in Figure 7.7, 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 using a 3-1-3 set of




Euler angles. This set has the advantage of having the singularities of the Euler angles coincidewith the disk lying flat on the horizontal plane (i.e., θ = 0, π).

g

E

2

Circular disk of mass m

and radius R

E3

E

2e2e

1e

1e

1 φ ψ

O

P

C

"

"

Fig. 7.7. A thin circular disk rolling on a rough horizontal surface.

For the disk, the position vector of the instantaneous point of contact P relative to the centerof mass C has the representations

πP = −Re

2 = −R sin(ψ)e1 − R cos(ψ)e2

= −R(− cos(θ)sin(φ)E1 + cos(θ) cos(φ)E2 + sin(θ)E3). (7.51)

Using the rolling condition, vP = v + ω × πP , we find that the motion of the disk is subject tothree constraints:

x1 + R ψ cos(φ) + R φ cos(θ) cos(φ) − Rθ sin(θ) sin(φ) = 0,

x2 + R ψ sin(φ) + R φ cos(θ) sin(φ) + Rθ sin(θ) cos(φ) = 0,

x3 − Rθ cos(θ) = 0. (7.52)

where x = 3

i=1 xiEi. You should notice that the last of these three constraints integrates tox3 = R sin(θ) + constant. However, the first two constraints are not integrable.

The rolling of the disk is possible because it is assumed that a static friction force Ff acts atP :

Ff = µ1E1 + µ2E2. (7.53)

Furthermore, this force and the normal force N = µ3E3 is equipollent to

Fc = Ff + N =

3i=1

µiEi, Mc = πP × Fc. (7.54)

Here, µ1, µ2, and µ3 are determined from the balance laws.




7.5.4 The Normality Prescription

The normality prescription is due to Casey [10], and is equivalent to the virtual work prescrip-tion commonly used in dynamics. The developments we present in this section can be found in

O’Reilly and Srinivasa [39]. For simplicity, we first discuss this prescription for a single rigid body.Subsequently, the case of multiple constraints in a system of two rigid bodies is discussed.

A Single Rigid Body

Based on the examples discussed in the previous sections, we assume that any constraint on arigid body can be written as (cf. (7.19))

π = 0, (7.55)

whereπ = f · v + h ·ω + e = 0. (7.56)

Here, f , h, and e are functions of t, x, and Q. The normality prescription states that the constraintforce Fc and constraint moment Mc associated with this constraint is

Fc = µf , Mc = µh. (7.57)

where µ is indeterminate. This form of the normality 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 theform

πK = 0, (7.58)

where K = 1, . . . , R, andπK = f K · v + hK ·ω + eK = 0. (7.59)

For this system of constraints, the normality prescription states that

Fc = µ1f 1 + . . . + µRf R, Mc = µ1h1 + . . . + µRhR. (7.60)

where µ1, . . . , µR are indeterminate. Notice that the prescription introduces R unknowns for theR constraints.

By examining the examples presented in this chapter, you should be able to verify that, apartfrom situations where dynamic Coulomb friction is present, the normality prescription givesphysically realistic results. You should also show that unless e is zero for a single constraint, thenthe 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, (7.61)

where K = 1, . . . , R. We assume that the constraints are of the form (cf. (7.19)):




πK = f K 1 · v1 + f K 2 · v2 + hK 1 · ω1 + hK 2 ·ω2 + eK . (7.62)

Here, f K 1, f K 2, hK 1, hK 2, and eK are functions of x1, x2, Q1, Q2, and t. The normality prescriptionstates that the constraint forces and constraint moments which enforce these R constraints are

Fc1 = µ1f 11 + . . . + µK f K 1,

Fc2 = µ1f 12 + . . . + µK f K 2,

Mc1 = µ1h11 + . . . + µK hK 1,

Mc2 = µ1h12 + . . . + µK hK 2. (7.63)

Here, Fcα is the resultant constraint force acting on Bα, Mcα is the resultant constraint momentrelative to the center of mass of Bα acting on Bα, and µ1, . . . , µR are functions of time which aredetermined from the balance laws for the system of two rigid bodies.

You should notice that for each of individual R constraints, the normality prescription in-troduces a single unknown function of time: µK . Consequently, if there are 10 constraints onthe motions of the two rigid bodies, the normality prescription will introduce the unknownsµ1, . . . , µ10.

For the cases of a frictionless joint between two bodies, for two bodies in rolling contact,and for two bodies in frictionless sliding contact or in rolling contact, the normality prescriptionprovides a very convenient method of specifying the constraint forces and constraint moments.However, the normality prescription does not yield physically reasonable prescriptions for jointsor contact where dynamic friction is involved.2

7.6 Potential Energies and Conservative Forces and Moments

The presence of conservative forces and moments acting on rigid bodies is one of the key featuresused to solve many problems. Although the definition of a conservative force originally arose inthe dynamics of a single particle and is well understood, the same cannot be said for conservative

moments (see, Antman [1] and O’Reilly and Srinivasa [39]). Indeed, as noted by Ziegler [56], forrigid bodies whose axis of rotation is not fixed, a constant moment is not necessarily conservative.To this end we start with three well-known examples of forces and their associated moments.These examples are followed by a discussion of Ziegler’s example. After the examples have beenpresented, we will give a general treatment of conservative forces and moments. Following thistreatment, you should return to Ziegler’s example and convince yourself that a constant momentisn’t conservative.

7.6.1 Constant Forces

A constant force P acting on a rigid body is conservative. If this force acts at all material pointsof 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

·x. Notice that there is no moment

(relative to the center of mass) associated with this force.2 It is a good exercise to verify our comments for the examples discussed earlier.



7.6 Potential Energies and Conservative Forces and Moments 153

The most ubiquituous example of a constant force is a constant gravitational force acting ateach material point. This force is equipollent to a force mgg acting at the center of mass. Hereg is the direction of the gravitational force. The potential energy of this force is −mgg · x

7.6.2 Spring Forces

In many mechanical systems, a spring is used to couple the motions of two bodies. Consider thesystem of two rigid bodies shown in Figure 7.8. Here, a linear spring of stiffness K and unstretchedlength L is connected to the material point X s1 of the body B1 and the material point X s2 of the body B2.

x

O

s2

1

x2

X

s1

1

Xs1

Xs2

X2

Linear spring

Fig. 7.8. The present configurations of two rigid bodies B 1 and B 2 which are connected by a spring.

The spring exerts a force Fs1 at the point X s1 and an equal and opposite force Fs2 at thepoint X s2:

Fs1 =

−K (

||xs1

−xs2

|| −L)

xs1 − xs2

||xs1 − xs2||,

Fs2 = −K (||xs1 − xs2|| − L) xs2 − xs1

||xs2 − xs1|| . (7.64)




It is easy to see that each of these forces is equipollent to a moment relative to the center of massand a force acting at the center of mass. The potential energy associated with the spring is

U s = K

2

(

||xs1

−xs2

|| −L)

2. (7.65)

You should notice that this potential energy depends on the position vectors of both materialpoints X s1 and X s2.

It is also convenient to note the identities

xs1 = Q1

Xs1 − X1

+ x1, xs2 = Q2

Xs2 − X2

+ x2, (7.66)

where Xsα is the position vector of X sα in the reference configuration of Bα. Using these identities,it can be seen that the forces, moments and potential energy of the spring can be expressed asfunctions of the rotation tensors of both rigid bodies and the position vectors of their centers of mass.

7.6.3 Central Gravitational Fields

In celestial and orbital mechanics, a standard problem is to consider the motion of a body subjectto a central force field. Such force fields date to Newton and are based on his inverse–squareforce. You may recall that this conservative force is the force exerted on a particle of mass m bya particle of mass M : F = −GMm

||r||2r||r|| , where r is the position vector of m relative to M , and G

is the universal gravitational constant.

C

center of massof the system

rigid body

rigid body

x

O

x1

1

x2

2

X2

X1

Fig. 7.9. Two rigid bodies B 1 and B 2 which exert mutual gravitational forces.



7.6 Potential Energies and Conservative Forces and Moments 155

For the force fields of interest, we consider two bodies B1 and B2. Every material point of B2 exerts an attractive force on each material point of B1. If we integrate these forces over allmaterial points in B1 and B2, we will obtain the resulant force exerted by B2 on B1. Similarintegrations apply to the moment and potential energy of these forces. In short, the resultant

gravitational force Fn and moment Mn on B1 due to B2 are:

Fn = − R1

R2

G

||x1 − x2||2x1 − x2

||x1 − x2||ρ1dV 1ρ2dV 2,

Mn = − R1

R2

(x1 − x1) × G

||x1 − x2||2x1 − x2

||x1 − x2||ρ1dV 1ρ2dV 2.

(7.67)

The moment Mn is relative to the center of mass X 1 of B1. The potential energy associated withthis force is

U n = − R1

R2

G

||x1 − x2||ρ1dV 1ρ2dV 2. (7.68)

Assuming that

B2 is spherical and has a mass M , the expressions for the force, moment, and

potential energy simplify: simplifies to

Fn = − R1

GM

||x1 − x2||2x1 − x2

||x1 − x2||ρ1dV 1,

Mn = − R1

(x1 − x1) × GM

||x1 − x2||2x1 − x2

||x1 − x2||ρ1dV 1,

U n = − R1

GM

||x1 − x2||ρ1dV 1. (7.69)

It is important to notice, that even in this simplified case, the gravitational force can exert amoment 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 − x1 = QΠ . (7.70)

In addition, we assume that ||π|| is small relative to ||x1 − x2||. These assumptions allow us toapproximate the forces, moments and potential energy associated with the central force field.After a substantial amount of manipulation, we find that

Fn ≈ mg,

Mn ≈

3GM

R3

c × (Jc) = −

3GM m

R3

c × (Ec),

U n ≈ −GM m

R −

GM

2R3

tr(J) +

3GM

2R3

(c · (Jc)) . (7.71)

where J is the inertia tensor of B

1 relative to its center of mass, m is the mass of B

1,

mg = −GM m

R2 c − 3GM

2R4 (2J + (tr(J) − 5c · Jc) I) c, (7.72)




and

R = ||x1 − x2||, c = x1 − x2

||x1 − x2|| . (7.73)

Notice that c points from the center of mass of the spherically symmetric body B2 to the center

of mass of B1. In addition, (7.71)1 does not correspond to the Newton’s gravitational force on aparticle of mass m by another particle of mass M .

The expressions (7.71) are used in the vast majority of works on satellite dynamics (cf. Belet-skii [4] and Hughes [25]). They can also be traced to D’Alembert, Euler and Lagrange. Indeed,Lagrange used these expressions to examine the rotational dynamics of the moon under the in-fluence of the Earth’s gravitational field. Many researches dating from the early 1960’s have usedthese expressions in analyzing the dynamics of earth orbiting satellites.

In many of these researches, it is common to approximate Fn with Fn = −GMmR2 c. Although

this approximation effectively decouples the motion of the center of mass of B1 from its orienta-tion, it violates angular momentum conservation and energy conservation. Consequently, we keepthe more elaborate expression Fn = mg. This deficiency was recently pointed out by Wang et

al. [54].

7.6.4 Constant Moments Which Are Not Conservative

To see that a constant moment isn’t conservative, we recall Ziegler’s example [56]. He considereda constant moment M E3. During a motion of a rigid body consisting of a rotation about E3

through π radians, this moment does work equal to M π. However, the same final orientation of the body can be achieved by a rotation about E1 through π radians, followed by a rotation aboutE2 through π radians. Now however the work done by the constant moment is zero! Consequently,the work done by the moment M E3 depends on the “path” taken by the body - and it cannotbe conservative.

In other dynamics courses, the rotation of the rigid body is constrained to be a fixed-axisrotation and the rotations about E1 and E2 aren’t permitted. For these cases, a constant momentM E3 is conservative.

7.6.5 General Considerations

It is convenient at this point to give a general treatment of conservative forces and moments inthe dynamics of rigid bodies. Our discussion is in the context of two rigid bodies, but it is easilysimplified 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 (x1, x2, Q1, Q2) . (7.74)

Clearly, this function depends on the motions of both rigid bodies and time. We can calculatethe time derivative of this function using (7.11):

U = ∂U

∂ x1 · v

1 +

∂U

∂ x2 · v2

+ uQ1 ·ω1

+ uQ2 ·ω2

, (7.75)

where



7.7 Exercises 157

uQα = −

1

2

∂U

∂ Qα

QT α − Qα

∂U

∂ Qα

T

. (7.76)

Consider the conservative forces Pα and moments M pα (relative to the respective centers of mass)

associated with this potential. We assume that the work done by these forces and moments isdependent on the initial and final configurations of the rigid bodies but is independent of themotions of the rigid bodies. This implies that

− U = P1 · v1 + P2 · v2 + M p1 · ω1 + M p2 · ω2. (7.77)

Substituting for U and collecting terms, we find thatP1 +

∂U

∂ x1

· v1 +

P2 +

∂U

∂ x2

· v2 + (M p1 + uQ1

) ·ω1

+ (M p2 + uQ2) ·ω2 = 0. (7.78)

This can be interpreted as an equation for Pα and M pα which must hold for all motions of therigid bodies. In other words, it must hold for all v

α and ω

α. In order for this to arise, it is

necessary and sufficient that

Pα = − ∂ U

∂ xα

, M pα = −uQα. (7.79)

These are the expressions for the conservative forces and moments associated with a potentialenergy. We leave it as an exercise for you to verify that these expressions are consistent with theresults presented for the spring and central gravitational forces.

7.7 Exercises

7.1 Suppose the motion of a rigid body is subject to 2 constraints:

ω · g3 = 0, ω · g2 = 0. (7.80)

Here, gi are the dual Euler basis for a given set of Euler angles. Give a physical interpretationof these constraints. What are the angular velocity vector, angular momentum vector androtational kinetic energy of the resulting constrained rigid body?

7.2 Consider a force P acting at a point P of a rigid body. In the present configuration, the pointP has the position vector πP relative to the center of mass X of the rigid body:

πP = xP − x. (7.81)

In addition,vP = xP = v +ω × πP . (7.82)

Here, ω is the angular velocity vector of the rigid body, and v is the velocity vector of thecenter of mass.




(a) The mechanical power of P is P · vP . Show that this power has the equivalent represen-tation

P · vP = P · v + (πP × P) ·ω. (7.83)

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 potentialenergy U = U (xP ) such that

P = − ∂U

∂ xP

. (7.84)

Show that this definition implies that

− U = P · v + (πP × P) · ω. (7.85)

(c) Show that the potential energy U = U (xP ) can also be described as a function of x, Q,X, and Π P :

U = U (xP ) = U (x, Q,Π P ). (7.86)

Here, Q is the rotation tensor of the rigid body, and πP = QΠ P . Using this equivalence,

show thatP = −∂ U

∂ x, πP × P = −uQ, (7.87)

where

uQ = −1

2

∂ U

∂ QQT − Q

∂ U

∂ Q

T . (7.88)

(d) Consider the case where P represents a force due to a spring of stiffness K and un-stretched length L. One end of the spring is attached to a fixed point O. What are thefunctions U (xP ) and U (x, Q,Π P ) for this force P?

7.3 Consider two rigid bodies. The rotation tensor of the first rigid body is

Q1 =

3i=1

1ei ⊗ Ei. (7.89)

The rotation tensor of the second rigid body is

Q2 =

3i=1

2ei ⊗ Ei. (7.90)

Here, 1ei corotate with the first rigid body, while 2ei corotate with the second rigid body.(a) Argue that the rotation tensor of the second body relative to the first body is

R = Q2QT 1 =

3

i=1

2ei ⊗ 1ei. (7.91)

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



7.7 Exercises 159

ω = −1

2

3i=1

2oei ⊗2ei

, (7.92)

where we have used the corotational derivative

oa= a −ω1 × a, (7.93)

and

ω1 = −1

2

3i=1

1ei ⊗ 1ei

= −1

2

Q1QT 1

. (7.94)

(c) Suppose a set of 3-2-1 Euler angles is used to parameterize Q1 and a set of 1-3-1 Eulerangles is used to parameterize R. Show that the angular velocity vectors of both bodies andtheir relative angular velocity vector have the representations

ω1 = γ 1E3 + γ 21e

2 + γ 31e1,

ω2 = ν 11e1 + ν 22e

3 + ν 32e1 + γ 1E3 + γ 21e

2 + γ 31e1,

ω = ν 1

1e1 + ν 2

2e

3 + ν 3

2e1. (7.95)

(d) Suppose the rotation of the second body relative to the first body is constrained suchthat

ω = ν 22e

3. (7.96)

Give a physical interpretation of the type of joint needed to enforce this constraint. In addi-tion, give prescriptions for the constraint moments acting on both bodies.

7.4 Consider the mechanical system shown Figure 7.10. It consists of a rigid body of mass m

which is free to rotate about a fixed point O. The joint at O does not permit the body tohave a spin. A vertical gravitational force mgE1 acts on the body. The inertia tensor of thebody relative to its center of mass C is

J0 = λ1E1 ⊗ E1 + (λ − mL2)(E2 ⊗ E2 + E3 ⊗ E3) .

The position vector of the center of mass C of the body relative to O is Le1.

E

2

E3

E

1

r

O

C

g

Fig. 7.10. A pendulum problem.

To parameterize the rotation tensor of the body, we will use a set of 1-3-1 Euler angles: g1 =E1, g2 = e3 and g3 = e1. For these angles




ω = φE1 + θe3 + ψe1

=

ψ + φ cos(θ)

e1 +

θ sin(ψ) − φ sin(θ) cos(ψ)

e2

+θ cos(ψ) + φ sin(θ) sin(ψ) e3. (7.97)

(a) Which orientations of the rigid body coincide with the singularities of the Euler angles?(Recall that the singularities in the Euler angles occur in the second Euler angle).(b) Derive expressions for the unconstrained potential U and kinetic T energies of the rigidbody.(c) Show that the motion of the rigid body is subject to four constraints:

ψ = ω · g3 = 0, v −ω × (Le1) = 0. (7.98)

In your solution, give a physical interpretation of these constraints.(d) Derive expressions for the constrained potential U and kinetic T energies of the rigidbody.

7.5 Consider the robotic arm shown in Figure 7.11. The robotic arm has a mass m and moment

of inertia tensor, relative to its center of mass,

JO = λ1E1 ⊗ E1 + λ2E2 ⊗ E2 + λ3E3 ⊗ E3. (7.99)

A system of motors, which are not shown in Figure 7.11, are used to actuate the rotation of the robotic arm. The rotation consists of (i) a rotation about the E3 axis through an angle ψ ,

(ii) a rotation about g2 = e

1 = cos(ψ)E1 + sin(ψ)E2 through an angle θ .Another system of actuators prescribes the motion of the point A of the arm. The positionvector of the center of mass of the arm relative to A is

x − xA = Le3. (7.100)

(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 orthonormal. For whichpositions of the robot arm are these vectors not defined?(b) Noting that the motion of the point A is prescribed: xA = s(t), and that the rotationof 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 θe1 + λ2

ψ sin(θ)e2 + λ3 ψ cos(θ)e3. (7.101)

(d) Draw a free–body diagram of the robot arm. In your free–body diagram, include agravitational force −mgE3.(e) Show that the force needed to actuate the motion of the center of mass is

Fact = mgE3 + ms + mL(2 ψθ cos(θ) + ψ sin(θ))e1

+ mL( ψ2 sin(θ)cos(θ) − θ)e2 − mL(θ2 + ψ2 sin2(θ))e3.

(7.102)



7.7 Exercises 161

E1

E2

E3

2g

2g

3e

Robotic arm

ψ

ψ

θg

A

Fig. 7.11. A robotic arm.

In your solution, relate the force Fact to your solution for (d)(f ) Show the moment Mc needed to actuate a slewing maneuver of the robotic arm is

Mact = mLe3 × (gE3 + s) + 2mL2 ψ0 θ0 cos(θ)e2

− mL2 ψ20 sin(θ)cos(θ)e1 + ω × H. (7.103)

During the slewing maneuver, θ = θ0, ψ = ψ0, and consequently ω is constant. In yoursolution, relate the moment Mact to your solution for (d)(g) Show that the change in the total energy E of the robotic arm is equal to the work doneby Fact and Mact:

E = Fact

· s + Mact

· ψE3 + Mact

· θe1. (7.104)

7.6 As shown in Figure 7.12, an axisymmetric rigid body is free to rotate about the fixed pointO. The body, which has a mass m, has an inertia tensor

J = λae3 ⊗ e3 + λt (I − e3 ⊗ e3) . (7.105)

The position vector of the center of mass of this body relative to O is

x = Le3. (7.106)

(a) Using a set of 3-1-3 Euler angles, show that the angular velocity vector ω of the rigidbody has the representation

ω =

ψ sin(θ) sin(φ) + θ cos(φ)

e1

+

ψ sin(θ) cos(φ) − θ sin(φ)

e2

+ ( φ + ψ cos(θ))e3. (7.107)




2

E3

e3

E

1E

O

C

g

Fig. 7.12. A rigid body which is free to rotate about the fixed point O. A set of 3-1-3 Euler angles isused to describe the rotation tensor Q of this body.

In your solution, feel free to use the results given in the Appendix.(b) Using the same set of 3-1-3 Euler angles as in (a), show that the angular velocity vectorω also has the representation

ω = θe

1 + ψ sin(θ)e

2 + ( φ + ψ cos(θ))e

3 . (7.108)

(c) Show that the angular momentum H of the arm is

H = λt θe

1 + λt ψ sin(θ)e

2 + λa

φ + ψ cos(θ)

e3. (7.109)

(d) Show that the angular momentum of the center of mass relative to O is

mL2 (ω1e1 + ω2e2) (7.110)

where ωi = ω · ei. What is the angular momentum of the rigid body relative to O?(e) Show that the kinetic energy T of the rigid body is

T = λt + mL2

2

θ2 + ψ2 sin2(θ)

+

λa

2

φ + ψ cos(θ)

2. (7.111)

(f ) A conservative moment due to a torsional spring is applied to the rigid body. As a result,the total potential energy of the rigid body is

U = mgx · E3 + K

2 ψ2. (7.112)

Here, K is the torsional spring constant. What are the conservative force Fcon and moment

Mcon acting on the rigid body?(g) Suppose that a moment GE3 acts on the rigid body - where G is constant. By examiningthe mechanical power GE3 ·ω of this moment, show that it is not conservative.



7.7 Exercises 163

7.7 Recall Ziegler’s example of a constant moment M E3 which wasn’t conservative. After choos-ing a set of Euler angles, ν 1, ν 2, ν 3 show that one cannot find U such that M E3 =−3

i=1∂U ∂ν i

gi.





8

Kinetics of a Rigid Body

8.1 Introduction

In this chapter, the balance laws F = m ˙v and M = H for a rigid body are discussed. Severalcomponent 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 = H · ei, lead to a set of equations which are known as Euler’s equations. In a subsequent chapter, we shall show thatM · gi = H · gi lead to Lagrange’s equations.

Once the balance laws have been discussed, we give a brief outline of the work-energy theoremand show how to establish energy conservation for a rigid body.

8.2 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 particle.There are two laws, or postulates, the balance of linear momentum and the balance of angular

momentum: G = F, HO = MO. (8.1)

Here, HO is the angular momentum of the rigid body relative to a fixed point O, and MO is theresultant external moment relative to O .

In many cases it is convenient to give an alternative description of the balance of angularmomentum. To do this we start with the identity

HO = H + x × G. (8.2)

Differentiating, and using the balance of linear momentum, we find that

HO = H + v × G + x × G

= H + x

×F. (8.3)

Hence, invoking the balance of angular momentum, we find that



166 8 Kinetics of a Rigid Body

MO = HO = H + x × F. (8.4)

However, the resultant moment relative to a fixed point O, MO, and the resultant moment relativeto the center of mass X , M, are related by1

MO = M + x × F. (8.5)

It follows thatM = H, (8.6)

which is known as the balance of angular momentum relative to the center of mass X . This formof the balance law is used in many problems where 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 setsof these laws can be established. For the first set, the balance of angular momentum relative toa fixed point O features

G = F, HO = MO, (8.7)

In the second set, the balance of angular momentum is taken relative to the center of mass X :

G = F, H = M, (8.8)

It should be noted that

G = mv, H = Jω, HO = x × mv + Jω. (8.9)

Here O is the origin of the coordinate system used to define x.To determine the motion of the rigid body, it suffices to know x(t) and Q(t). To obtain these

results, the equations (8.7) or (8.8) must be supplemented by information on the coordinatesystem used to parameterize x and the parameterization of Q.

8.3 Work and Energy Conservation

The work–energy theorem for a rigid body equates the rate of change of kinetic energy to themechanical power of the external forces and moments acting on the rigid body. There are twoequivalent forms of this theorem:

T = F · v + M · ω

=N

K =1

FK · vK + M p ·ω. (8.10)

The second form is the most useful for proving that energy is conserved in a specific problem.We close this section with a discussion of energy conservation.

1 This may be seen from our previous discussion of a system of forces and moments acting on a rigidbody.



8.3 Work and Energy Conservation 167

8.3.1 Proving the Work–Energy Theorem

The difficulty in establishing the work–energy lies in dealing with the angular momentum H = Jω.To overcome, this difficulty, it is easiest to first show that

ω · (Jω) = 0. (8.11)

This result is established using an earlier result, J = Ω J−JΩ , and the identity Ωω = ω×ω = 0:

(Jω) · ω = ((Ω J − JΩ )ω) · ω= (Ω Jω) · ω − (J(ω × ω)) ·ω= Jω · (Ω T ω) + (J(0)) · ω= 0 + 0. (8.12)

In establishing the identity, we also used the skew–symmetry of Ω . Because Jω · ω = 0,

( ˙Jω)

·ω = (Jω)

·ω, (8.13)

which we shall presently invoke.To establish the work–energy theorem, recall the Koenig decomposition for the kinetic energy

T of a rigid body:

T = 1

2mv · v +

1

2 (Jω) · ω. (8.14)

Differentiating T , we find

T = m ˙v · v + 1

2

( ˙Jω) ·ω + (Jω) · ω

= m ˙v · v +

1

2

( ˙Jω) ·ω + (Jω) · ω + (Jω) · ω

= m ˙v

· v +

1

2( ˙Jω)

·ω +ω

·( ˙Jω)

= m ˙v · v + ( ˙Jω) · ω. (8.15)

Invoking the balances of linear and angular momentum, the work–energy theorem is obtained:

T = F · v + M · ω. (8.16)

To establish the alternative form of the work–energy theorem, we consider a system of N

forces FK acting at N material points X K of the rigid body, and a pure moment M p acting onthe rigid body. For this system of forces and moments:

F =

N

K =1

FK , M =

N

K =1

(xK − x) × FK

+ M p. (8.17)

Noting that vK = v + ω × (xK − x), we find with some rearranging that




F · v + M · ω =N

K =1

FK · v +

N K =1

((xK − x) × FK ) · ω

+ M p · ω

= N

K =1

FK

·(v + ω

×(xK

−x))+ M p

·ω

=

N K =1

FK · vK + M p · ω. (8.18)

This result establishes the alternative form of the work–energy theorem.

8.3.2 Energy Conservation

In most problems in rigid body dynamics where there is no dynamic friction, the total energy E

of the body is conserved. To show this, let’s assume that the resultant conservative force Fcon

and conservative moment (relative to the center of mass) Mcon acting on the body are associatedwith a potential energy U = U (x, Q):

Fcon = −∂U

∂ x, Mcon = −uQ. (8.19)

In addition, we assume that the body is subject to R constraints of the form

Lf · v + Lh · ω + Le, (L = 1, . . . , R). (8.20)

The constraint force and moment due to these constraints are prescribed using normality:

Fc =R

L=1

µLLf , Mc =R

L=1

µLLh. (8.21)

Finally, we assume that the only forces and moments acting on the rigid body are conservativeor constraint.

We now examine the work-energy theorem for the rigid body of interest:

T = F · v + M · ω= Fc · v + Mc · ω + Fcon · v + Mcon · ω. (8.22)

However,

Fc · v + Mc ·ω =

RL=1

µL (Lf · v + Lh · ω)

= −R

L=1

µLLe,

Fcon · v + Mcon ·ω = −∂U

∂ x · v − uQ · ω= − U . (8.23)



8.4 Additional Forms of the Balance of Angular Momentum 169

Consequently,

T = − U −R

L=1

µLLe, (8.24)

This implies that E = 0, where E = T + U , if

RL=1 µLLe = 0.In summary, one situation where the total energy E is conserved arises when Le = 0, and the

constraint forces and constraint moments are prescribed by normality. This situation arises inmost of the solved problems in rigid body dynamics, and we shall shortly see several examples.

8.4 Additional Forms of the Balance of Angular Momentum

The balance of angular momentum H = M has several component forms and is one of the mostinteresting equations in mechanics. In this section, several of these forms are discussed. First, weshow that this equation is equivalent to

J

3i=1

ωiei + ω × Jω = M. (8.25)

Next, we show that, if ei are the principal vectors of J,

λ1 ω1 + (λ3 − λ2)ω3ω2 = M · e1,

λ2 ω2 + (λ1 − λ3)ω3ω1 = M · e2,

λ3 ω3 + (λ2 − λ1)ω1ω2 = M · e3. (8.26)

As an intermediate result, we also indicate the corresponding component form when ei are notprincipal vectors of J.

8.4.1 A Direct Form

Here, we wish to show that H = M can be written as

Jα+ ω × (Jω) = M. (8.27)

To establish this result, we need to examine ω and J.First, we recall that

H = Jω. (8.28)

Taking the derivative of this expression, we find

H = Jω + Jω. (8.29)

To proceed, we need some identities. Specifically,




α = ω = d

dt

3i=1

ωiei

=

3

i=1

ωiei +

3

i=1

ωiei

=

3i=1

ωiei +

3i=1

ωi (ω × ei)

=3

i=1

ωiei + ω ×3

i=1

ωiei

=

3i=1

ωiei + ω ×ω

=3

i=1

ωiei, (8.30)

andJ = Ω J + JΩ T , Jω = Ω Jω − JΩω = ω × Jω. (8.31)

Using the identities, we find that

H = Jω + Jω

= Jω + ω × Jω

= J3

i=1

ωiei + ω × Jω. (8.32)

In summary, H = M is equivalent to

J3

i=1

ωiei + ω × Jω = M. (8.33)

This form of H = M is very useful when J is a constant tensor - for instance when dealing withrigid spheres and rigid cubes. It is also used to obtain conservation results for H.

8.4.2 A Component Form

If we choose an arbitrary basis E1, E2, E3 for E 3, then the inertia tensors JO and J have therepresentations

J0 =3

i=1

3

k=1

J ikEi ⊗ Ek, J =3

i=1

3

k=1

J ikei ⊗ ek, (8.34)

where ei = QEi. Consequently,



8.4 Additional Forms of the Balance of Angular Momentum 171

H = Jω =3

i=1

3k=1

J ikωkei. (8.35)

Differentiating H, we find that

H = ˙Jω =3

i=1

3k=1

J ik ωkei + ω ×

3i=1

3k=1

J ikωkei

. (8.36)

If we equate this expression to M, we can find the component forms of H = M. However, exceptwhen ω has a simple form, it is normal not to consider this component form of the equations.Examples of where (8.36) are used include misbalanced rotors where ω = θE3, and J 13 = 0 and/orJ 23 = 0. For cases not involving a fixed axis of rotation, it is prudent to choose E1, E2, E3 tobe the principal directions of J0.

8.4.3 The Principal Axis Case

If we choose Ei to be the principal directions of J0, then this tensor has the representation

J0 =3

i=1

λiEi ⊗ Ei, (8.37)

where λi are the principal moments of inertia. The vectors Ei are also known as the principalaxes of the body in its reference configuration.

Defining ei = QEi, we find that the inertia tensor J = QJ0QT has the representation

J =

3i=1

λiei ⊗ ei. (8.38)

Consequently,

H = Jω =3

i=1

λiωiei. (8.39)

Evaluating H, we find

H = Jω +ω × Jω

= J3

i=1

ωiei + ω ×

3i=1

λiωiei

=3

i=1

λi ωiei + (λ2 − λ1)ω1ω2e3 + (λ1 − λ3)ω1ω3e2 + (λ3 − λ2)ω3ω2e1. (8.40)

Consequently, H = M can be written in the form




λ1 ω1 + (λ3 − λ2)ω3ω2 = M · e1,

λ2 ω2 + (λ1 − λ3)ω3ω1 = M · e2,

λ3 ω3 + (λ2 − λ1)ω1ω2 = M · e3. (8.41)

These equations represent 3 first-order ordinary differential equations for ωi. They are supple-mented by the 3 first-order ordinary differential equations relating ω to Q,

ω = −1

2[ QQT ], (8.42)

in order to determine the orientation of the rigid body.

8.5 Moment-Free Motion of a Rigid Body

Moment-free motion of a rigid body occurs when M = 0. Determining the motion is resolved bysolving the balance laws,

G = F, H = 0, (8.43)for x and Q. It is common to focus exclusively on the balance of angular momentum and determineQ. In addition, although an analytical solution for Q was first found by the German mathematicanCarl G. J. Jacobi (1804–1851) in 18492, it is usual to focus on ω(t). Another ingenious solutionto this problem was presented by the French mathematican Louis Poinsot (1777–1859) in 1834[44].

The equations governing the components ωi = ω · ei of the angular velocity vector are foundfrom the three equations H · ei = 0 (cf. (8.41)):

λ1 ω1 + (λ3 − λ2)ω3ω2 = 0,

λ2 ω2 + (λ1 − λ3)ω3ω1 = 0,

λ3 ω3 + (λ2 − λ1)ω1ω2 = 0. (8.44)

It is easy to see that the solutions to these equations conserve the rotational kinetic energyT rot = 1

2H · 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, thebody is known as prolate. For the asymmetric body discussed above, e1 is known as the minoraxis of inertia, e2 is known as the intermediate axis of inertia, and e3 is known as the major axisof inertia. We now turn to discussing the three cases and the solutions for ωi(t).

2 Jacobi’s solution is discussed at length in Section 69 of Whittaker’s treatise [55].



8.5 Moment-Free Motion of a Rigid Body 173

8.5.1 The Symmetric Body

For the symmetric rigid body, (8.44) simplify to ωi = 0. In otherwords, the components of ω are

constant. As ω = 3k=1 ωkek, this implies that ω is constant. This in turn implies that the axis

of rotation p of the body is constant and so we find

Q(t) = cos(ν )(I − p ⊗ p) − sin(ν )p + p ⊗ p, (8.45)

where the axis and angle of rotation are

p = ω(t0)

||ω(t0)|| , ν = ||ω(t0)||(t − t0), (8.46)

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 of its principal axes. We shall shortly see that this resultalso applies for axisymmetric and asymmetric rigid bodies.

8.5.2 The Axisymmetric Body

For an axisymmetric body, it’s convenient to define λt = λ1 = λ2 and λa = λ3. The equationsgoverning the components of angular velocity (8.44) simplify for this case to

ω1 = ktΩω2,

ω2 = −ktΩω1,

ω3 = Ω . (8.47)

where Ω = ω3(t0) is a constant and

kt = λt − λa

λt

. (8.48)

The differential equations (8.47) have a simple solution:ω1(t)ω2(t)

=

cos(ktΩ (t − t0)) sin (ktΩ (t − t0))− sin(ktΩ (t − t0)) cos (ktΩ (t − t0))

ω1(t0)ω2(t0)

. (8.49)

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 speedabout a principal axis.

An interesting feature about the axisymmetric body is that the component of ω in the di-rection of the axis of symmetry e3 is always constant. This occurs even though e3(t) may bequite 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 exploitedin designing flywheels.




8.5.3 The Asymmetric Body

When the body is asymmetric, its principal moments of inertia are distinct. If we reexamine(8.44) for this case, then we find that if all but one ωi is zero, then the non-zero ωi will remain

constant. For instance, if ω2 = 0 and ω3 = 0, then it is possible for ω1 to have any value and forthe equations of motion to preserve this value. These results imply that it is possible to rotatethe body about a principal axis at constant speed under no applied moment M. Clearly, as withthe other two types of rigid bodies, it is possible to spin the body at constant speed about aprincipal axis.

8.5.4 The Momentum Sphere

To visualize the solutions of (8.44), a graphical technique is often used. This technique dates tothe mid-nineteeth century. It is based on two facts: (8.44) preserves the magnitude of H and therotational kinetic energy T rot. As a result, the solutions

h1(t) = H

·e1 = λ1ω1(t), h2(t) = H

·e2 = λ2ω2(t), h3(t) = H

·e3 = λ3ω3(t). (8.50)

lie on the intersection of the constant surfaces h = h0 and T rot = T E . Here,

h2 = h21 + h2

2 + h23,

T rot = h2

1

2λ1+

h22

2λ2+

h23

2λ3, (8.51)

and the values of h0 and T E are determined by the initial conditions ωi(t0).

Fig. 8.1. The momentum sphere for an axisymmetric rigid body. The curves and points on the sphere arethe intersection of the momentum sphere with the energy ellipsoid. The points on the equator correspondto the body rotating steadily about a principal axis, while the circles correspond to the solutions describedby (8.49).

If we pick a value of h the surface h = h0 in the three-dimensional space h1 − h2 − h3 is asphere - the momentum sphere . Selecting a value of T E , we find that the surface T rot = T E in the



8.6 The Baseball and the Football 175

Fig. 8.2. The momentum sphere for an asymmetric rigid body. The curves and points on the sphere arethe intersection of the momentum sphere with the energy ellipsoid. The points correspond to the steadyrotating solutions, while the curves are non-trivial solutions to (8.44). Notice that 4 of the curves are notclosed. For this figure, λ1 < λ2 < λ3.

three-dimensional space h1 − h2 − h3 is an ellipsoid - the energy ellipsoid . The intersection of theellipsoid with the sphere is either a discrete set of points, or a set of curves. 3 These intersectionsare the locii of hi(t). For the axisymmetric body, the intersections are shown in Figure 8.1. Thecorrsponding intersections for the asymmetric body are shown in Figure 8.2.4 The latter figureis one of the most famous in dynamics.

For the case presented in Figure 8.1, the energy ellipoid has an axis of revolution (in this casethe third axis). For a symmetric body, the energy ellipsoid degenerates further into a sphere. Thissphere coincides with the momentum sphere and so the graphical technique used to determinehi(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 rotationalmotion of the rigid body.

8.6 The Baseball and the Football

Consider a sphere of mass m and radius R which is thrown into space with an initial velocityv(t0), angular velocity ω(t0) and orientation Q(t0). We wish to determine the motion x and Qof the sphere.

This problem is that encountered by a soccer player or a baseball pitcher. A key force in thisproblem is known as the lift or Magnus force (see Figure 8.3),5

FM = mBω × v, (8.52)

3 For a more detailed discussion of these intersections, Synge and Griffith’s text [52] is highly recom-mended.

4 These figures were kindly supplied by Wayne Huang in the Summer of 2004.5 Due to the German scientist Heinrich Gustav Magnus (1802–1870). He is credited with developing this

force in 1853 - but I haven’t been able to track down the paper. Good sources for further applicationswhere this force is important are Bloomfield [5] and Ireson [26].




FM

X

FM

X v

Fig. 8.3. A rigid sphere whose center of mass is moving to the right with a velocity vector v. When theball is rotating clockwise the velocity of the air moving over the top of the ball is slower than the velocityof 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 clockwise.

where B is a positive constant. The sign of B was determined using Bernoulli’s equation.6 Re-cent research on free-kicks in soccer has shown that there can be a transition in the flow fieldfrom turbulent to laminar that causes dramatic changes in the trajectory (see Carre et al. [7]).According to Ireson [26], for some free-kicks in soccer ||v|| = 25 meters/second and mB ≈ 0.15716kg.

Apart from gravity and the Magnus force, the other possibly important force in this problemis the drag force:

Fd = −1

2ρf AC d (v · v)

v

||v|| . (8.53)

In this expression, C d is the drag coefficient, ρf is the density of the fluid that the sphere ismoving in, and A is the frontal area of the sphere in contact with the fluid: A = πR2.

For the system at hand, M = 0 and H =

2mR2

5 ω

, so we find the important result that ω

isconstant:ω(t) = ω (t0) . (8.54)

In other words, the angular velocity of the sphere doesn’t change. As with the symmetric bodydiscussed earlier, we can easily solve for the rotation tensor of the sphere:

Q(t) = cos(ν )(I − p ⊗ p) − sin(ν )p + p ⊗ p, (8.55)

where the axis and angle of rotation are

p = ω(t0)

||ω(t0)|| , ν = ||ω(t0)||(t − t0). (8.56)

We shall see that solving for the motion of the center of mass in this problem is not so trivial.

6 Bernoulli’s equation applies to inviscid fluid flow and states that the sum of the pressure p and 1

2ρf U 2

is a constant. Here, U is the fluid flow velocity and ρf is the fluid density.



8.6 The Baseball and the Football 177

The rotation tensor Q =3

k=1 ek ⊗ Ek for the rigid body is parameterized by a set of 3-1-3Euler angles. The Euler basis vectors for this parameterization have the representations

g1

g2g3

=E3

e

1e3

= sin(ψ) sin(θ) cos(ψ) sin(θ) cos(θ)

cos(ψ) − sin(ψ) 00 0 1e1

e2e3

=

0 0 1

cos(φ) sin(φ) 0sin(θ) sin(φ) −sin(θ)cos(φ) cos(θ)

E1

E2

E3

. (8.57)

Further, the Euler angles are subject to the restrictions: φ ∈ [0, 2π), θ ∈ (0, π), and ψ ∈ [0, 2π).Using these Euler angles,

ω = Ω iEi

=

θ cos(φ) + ψ sin(θ) sin(φ)

E1 +

θ sin(φ) − ψ sin(θ) cos(φ)

E2 +

φ + ψ cos(θ)

E3.

(8.58)

If we use a set of Cartesian coordinates for the position vector of the center of mass, x · Ei = xi,then we would find that

ω × v = (x2Ω 3 − x3Ω 2) E1 + (x1Ω 3 − x3Ω 1) E2 + (x2Ω 1 − x1Ω 2) E3. (8.59)

From the previous solution to the balance of angular momentum, we know that Ω i are constant.The balance of linear momentum for the sphere provides the equation for the motion of the

center of mass. Evaluating F = G in Cartesian coordinates, we find

mx1 = (FD + FM ) · E1,

mx2 = (FD + FM ) · E2,

mx3 = −mg + (FD + FM ) · E3. (8.60)

Ignoring the drag force and using (8.59), we find three differential equations for xi(t):

mx1 = mB (x2Ω 3 − x3Ω 2) ,

mx2 = mB (x1Ω 3 − x3Ω 1) ,

mx3 = mB (x2Ω 1 − x1Ω 2) − mg. (8.61)

For the general case, these equations can be integrated numerically to determine x(t).Turning our attention to a simple case, suppose ω (t0) = Ω 10E1. From the previous analysis,

we find that Ω 1 = Ω 10 and Ω 2 = Ω 3 = 0. Consequently, (8.61) simplifies to

mx1 = 0,

mx2 = −mBx3Ω 10,

mx3 = mBx2Ω 10 − mg. (8.62)

The solution to these differential equations, assuming B Ω 10 = 0, is




x1 = x1(t0) + x1 (t0) (t − t0) ,

x2(t) = 1

BΩ 10

x2 (t0) − g

BΩ 10

sin(BΩ 10 (t − t0)) − x3 (t0) (1 − cos(BΩ 10 (t − t0)))

+x2 (t0) +

g (t

−t0)

BΩ 10 ,

x3(t) = 1

BΩ 10

x2 (t0) − g

BΩ 10

(1 − cos(BΩ 10 (t − t0))) + x3 (t0)sin(BΩ 10 (t − t0))

+x3 (t0) . (8.63)

From these expressions, the trajectory of the sphere can be determined. Two important featuresare present. First, the spin Ω 10 influences the forward x2 speed of the sphere. Second, it alsoeffects the vertical position and speed. We leave it as an exercise to determine these effects forvarious examples of Ω 10.

8.7 Motions of Rolling Spheres and Sliding Spheres

The problem of the sphere moving on a flat surface has several applications, bowling and poolbeing the most famous. The most famous classical treatments of this problem are due to Coriolis[17] and Routh [47], and generalizations of it occupy the literature on non-holonomically con-strained rigid bodies to this date. In our treatment, we assume that the surface is rough witha coefficient of static Coulomb friction of µs and kinetic friction of µd. Of particular interest tous will be the transition between rolling and sliding and our discussion is heavily influenced bySynge and Griffith’s discussion in [52].

E3

E1

Sphere of radius R

Inclined plane P

X

O

g

Fig. 8.4. A rigid sphere moving on an inclined plane. The angle of inclination of the plane is β , and agravitational force −mg cos(β )E3 + mg sin(β )E1 acts on the rigid body.



8.7 Motions of Rolling Spheres and Sliding Spheres 179

Consider the sphere moving on the surface shown in Figure 8.4. The radius of the sphere isR and the velocity of the point of contact of the sphere with the incline is

vP = v +ω × (−RE3) . (8.64)

Note the simple expression for πP here. Because the point P is the instantaneous point of contact,

vP · E3 = 0. (8.65)

Consequently, this velocity field has the representations

vP = vs1E1 + vs2E2 = us (8.66)

where u =

v2s1 + v2s2 is the slip velocity and s = vP u

is the slip direction. When the sphere isrolling, there are two additional constraints on vP and, as a result, vs1 = 0 and vs2 = 0. For therolling 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 + N E3 + Ff , M = −RE3 × Ff . (8.67)When the sphere is rolling,

Ff = µ1E1 + µ2E2, (8.68)

where µ1 and µ2 are unknowns. For the sliding sphere on the other hand, we have

Ff = −µd|N |s. (8.69)

For convenience, we have used the same notation for the friction forces for the rolling and slidingspheres, but this shouldn’t cause confusion.

The motion of the rolling sphere is determined using the balance laws and the constraintsvP = 0. Using Cartesian coordinates for x and setting ω =

3i=1 Ω iEi, these equations are

x1

= RΩ 2

,

x2 = −RΩ 1,

x3 = N − mg cos(β ),

mx1 = mg sin(β ) + µ1,

mx2 = µ2,

0 = N − mg cos(β ),

2

5mR2 Ω 1 = Rµ2,

2

5mR2 Ω 2 = −Rµ1,

2

5mR2 Ω 3 = 0. (8.70)

To solve these equations, it’s convenient to first determine the differential equations governingΩ i. From the nine equations listed above, we can eliminate several variables to find



180 8 Kinetics of a Rigid Body1 +

2

5

mR2 Ω 1 = 0,

1 +

2

5mR2 Ω 2 = −mgR sin(β ),

25

mR2 Ω 3 = 0. (8.71)

These equations are easily solved:

Ω (t) = Ω (t0) − g sin(β ) (t − t0)

R

1 +

2

5

−1

E2. (8.72)

We leave it as an exercise to determine x(t). When β = 0, you will find that the sphere rolls ina straight line at constant speed.

For the sliding sphere, it is convenient to examine the differential equations for vP . Differen-tiating this velocity we find that

vP = vs1E1 + vs2E2

= ˙v + α× (−RE3) . (8.73)

Using the balances of linear and angular momentum, we substitute for ˙v and α to find

mvs1 =

1 +

5

2

Ff · E1 − mg sin(β ), mvs2 =

1 +

5

2

Ff · E2. (8.74)

After substituting for the friction force, these equations provide two differential equations for theslip velocities. We leave it as an exercise to write down the five differential equations governingx1, x2 and Ω i. Solving for x(t), you will find that when β = 0, the path of the center of mass iseither an arc of a parabola or a straight line.

If we consider the simple case when the incline is horizontal, then β = 0. The differentialequations for the slip velocity simplify considerably to

mvs1 = −µdmg

1 + 5

2 v

s1u , mvs2 = −µdmg

1 +

5

2 v

s2u . (8.75)

From these equations, we will find that vs1 and vs2 always tend to zero. To see this, it’s best tocalculate u:

u = −µdg

1 +

5

2

. (8.76)

This equation has the solution

u (t) = u (t0) − µdg

1 +

5

2

(t − t0) . (8.77)

As a result, u will reach zero in finite time. Once it does, the sphere starts rolling. Now as β = 0,this implies that the sphere will roll at constant speed in a straight line. It is interesting to note

that once the sphere starts rolling it will remain rolling. The transition between the parabolicpath during sliding and the straight line path during rolling is a key to hooks in bowling andmassee shots in pool.



8.9 Exercises 181

8.8 Closing Comments

We have touched on some problems in rigid body dynamics. There are several aspects of theseproblems that we haven’t had the opportunity to address and some of them are discussed in

the exercises and others in the references at the end of these notes. It is important to realizethat although rolling spheres and thrown baseballs have been analyzed for over a century, theseproblems are very rich - a simple change in an assumption can lead to dramatically differentresults. A good example of this is the Chaplygin sphere. This is a rolling sphere where J = 2

5mR2I.

Partially as a consequence of its asymmetry the path of the point of contact of the Chaplyginsphere with the ground can be very intricate.

8.9 Exercises

8.1 Suppose a rigid body is rolling on a fixed surface. A gravitational force −mgE3 acts on therigid body. Starting from the work-energy theorem for the rigid body,

T = F · v + M · ω, (8.78)

prove that the total energy E of the rigid body is conserved. Prove that the total energy isalso conserved if the rigid body is sliding on a smooth surface.

8.2 A rigid body of mass m is moving in space under the influence of an applied force Fa = F ae3and an applied moment M = 0. Outline how would you determine the attitude Q and motionof the center of mass of the rigid body?

8.3 The orientation of a rigid body relative to a fixed reference configuration is defined by arotation tensor Q. At time t0 this rotation tensor is Q(t0) and at time t1 this rotation tensoris Q(t1). Give a physical interpretation of the rotation tensor Q(t1)QT (t0). You should makeuse of the corotational basis in your answer.

8.4 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 = HO to determine the motion of this rigid

body?8.5 A solution for ωi(t) has been determined for a rigid body dynamics problem. How would you

determine Q from this solution?8.6 A rigid body has a potential energy U = U

x, γ i

, where γ 1, γ 2, γ 3 are the Euler angles used

to parametrize Q. If the conservative force F and conservative moment M are such that

− U = F · ˙x + M · ω, (8.79)

then verify that

F = −3

i=1

∂U

∂xi

Ei, M = −3

i=1

∂U

∂γ igi, (8.80)

where xi = x

· Ei and gi are the basis vectors for the dual Euler basis. Show that the

gravitational potential energy U n for a rigid body orbiting a fixed spherically symmetric rigidbody is a function of the form U = U

x, γ i

.




8.7 As shown in Figure 8.4, 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, where µ = 2mR2

5 .(a) What are the three constraints on the motion of the sphere? Show that these constraintsimply that

x1 − R Ω 2 = 0, x2 + R Ω 1 = 0, x3 = 0. (8.81)where xi = x · Ei, and Ω i = ω · Ei.(b) Draw a free-body diagram of the sphere. Clearly indicate the constraint forces Fc actingon the sphere.(c) Show that balance of linear momentum for the sphere implies that

Fc =

mR Ω 2 − mg sin(β )

E1 − mR Ω 1E2 + mg cos(β )E3. (8.82)

(d) Show that the balance of angular momentum for the sphere and the results of (c) canbe used to show that

7

5mR2 Ω 1 = 0,

7

5mR2 Ω 2 = mgR sin(β ),

2

5mR2 Ω 3 = 0. (8.83)

(e) Starting from the work-energy theorem for a rigid body, prove that the total energy E

of the rolling sphere is constant. In addition, show that

E = 7mR2

10 Ω 21 +

7mR2

10 Ω 22 +

mR2

5 Ω 23 − mgx1 sin(β ) + mgR cos(β ). (8.84)

(f ) Why is the angular momentum H of the sphere in the E3 direction conserved?

(g) If, at time t = 0, the sphere is given an initial angular velocity ω(0) =3

i=1 Ω i0Ei, thenshow that the angular velocity ω(t) is

ω(t) = ω(0) + 5g sin(β )t

7R E2. (8.85)

What is the angular acceleration vector α of the sphere?(h) Suppose the sphere is placed on the inclined plane and released from rest. Verify thatthe sphere will start rolling and that the resulting attitude Q of the sphere corresponds to afixed-axis rotation.

8.8 Consider a rigid body of mass m which is in motion under a resultant moment M and a forceF. We also suppose that the inertia tensor J0 has the representation:

JO = λ1E1 ⊗ E1 + λ2E2 ⊗ E2 + λ3E3 ⊗ E3. (8.86)

(a) From the balance of angular momentum (relative to the center of mass) of this body, showthat one obtains three differential equations for the corotational components of the angularvelocity vector:

λ1 ω1 = (λ2

−λ3) ω2ω3 + M

·e1,

λ2 ω2 = (λ3 − λ1) ω1ω3 + M · e2,

λ3 ω3 = (λ1 − λ2) ω1ω2 + M · e3. (8.87)



8.9 Exercises 183

Here, ωi = ω · ei where ei is a corotational basis. These equations, when M = 0 are knownas Euler’s equations.(b) Argue that, in the absence of constraints, one obtains a closed system of equations fromthe balance laws and the definition of the angular velocity vector which provide the motion

of the center of mass of the rigid body and the rotation tensor of the rigid body.(c) Consider a sphere of mass m and radius R. Suppose the sphere is lofted into the airwith an initial angular velocity vector ω(t0) and rotation tensor Q(t0) = I. In addition, theposition and velocity vectors of its center of mass at this instant are x0 and v0. During thesubsequent motion, a gravitational force −mgE3 acts on the body.Prove that7

ω(t) = ω(t0), (8.88)

and, consequently,Q(t) = cos(ν )(I − p ⊗ p) − sin(ν )p + p ⊗ p, (8.89)

where

p = ω(t0)

||ω(t0)|| , ν = ||ω(t0)||(t − t0). (8.90)

Finally, show thatx(t) = x0 + v0(t − t0) − g

2(t − t0)2E3. (8.91)

Give a physical interpretation of these results. Furthermore, if the body was not symmetricwould the angular velocity vector necessarily remain constant? In this problem, we could alsohave chosen a set of Euler angles to parameterize Q. Why would this approach have beenmore complicated for the problem of the lofted sphere?

8.9 This famous problem is discussed in most books on satellite dynamics (see, for example,Hughes [25]). The 24 solutions discussed below date to Lagrange in the late 1700’s.Consider a rigid body B which is in motion in a central gravitational force field. The center of this force field is assumed to be located at a fixed point O. The force, moment, and potentialenergy of the field are given by the approximations

Fn ≈ −

GM m||x||2

x||x||

−

3GM

2||x||4

2J +

tr(J) − 5

x

||x|| · J x

||x||

I

x

||x|| ,

Mn ≈

3GM

R3

x

||x|| ×

J x

||x||

,

U n ≈ −GM m

R −

GM

2||x||3

tr(J) +

3GM

2||x||3

x

||x|| ·

J x

||x||

, (8.92)

where J is the inertia tensor of B relative to its center of mass, and m is the mass of B.(a) Verify that Mn = −x × 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?

7 To help with the solution of this problem, use the formula for the angular velocity vector associatedwith a rotation tensor corresponding to a counterclockwise rotation of ν about an axis p.




(c) Henceforth, we approximate Fn as

Fm ≈ −

GM m

||x||2

x

||x|| . (8.93)

Using the balance of linear momentum and conservation of x × mv, show that it is possiblefor the body to move in a circular orbit x = Rer about O with an orbital angular velocity,which is known as the Kepler frequency, ωK :

ω2K =

θ2 = GM

R3 . (8.94)

Here, er = cos(θ)E1 + sin(θ)E2.(d) Using the results of (a), show that a steady motion of the rigid body, that is, one whereω = 0, is governed by the equation

ω × (Jω) = 3ω2K er × (Jer). (8.95)

(e) Suppose that the body is asymmetric. That is, the principal values of J0 are distinct. Weseek solutions of (8.95) such that ω · er = 0. Show that there are 6 possible solutions for ωwhich satisfy (8.95) and 4 possible solutions for er. Here, you should assume that J is knownand as a result Q is known. As a result there are 6 × 4 possible solutions of (8.95).(f ) Suppose that the body is such that J = µI where µ is a constant. Show that any constantω satisfies (8.95) 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 thesame side of a satellite in a circular orbit above the Earth.

8.10 A sphere of mass m and radius R moves on a turntable. The contact between the sphere andthe turntable is rough. In addition, the center O of the turntable is fixed and the turntablerotates about the vertical E3 with an angular speed Ω . This problem is adopted from Lewisand Murray [30].(a) If the sphere is rolling on the turntable, show that the motion of the sphere is subject to

three constraints: v + ω × (−RE3) = Ω E3 × x. (8.96)

Using the representations ω = 3

i=1 ωiEi, and x = 3

i=1 xiEi, show that the three con-straints imply that

x1 − Rω2 + Ω x2 = 0, x2 + Rω1 − Ω x1 = 0, x3 = 0. (8.97)

(b) Assuming that a vertical gravitational force acts on the sphere, draw a free–body diagramof the sphere.(c) Using a balance of linear momentum and with the assistance of the constraints, show thatthe constraint force acting on the sphere is

Fc = mgE3 − mΩ (x2E1 − x1E2) + mR(ω2E1 − ω1E2). (8.98)

(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



8.9 Exercises 185

x1 = Rω2 − Ω x2,

x2 = −Rω1 + Ω x1,

x3 = 0,

ω1 = mΩR

µ + mR2

x1,

ω2 =

mΩR

µ + mR2

x2,

ω3 = 0. (8.99)

Why are these equations sufficient to determine the motion (x(t), Q(t)) of the sphere.(e) For the special case where Ω = 0, show that the center of mass of the sphere will movein a straight line with a constant speed, and that the angular velocity vector ω of the spherewill be constant.(f ) Numerically integrate (8.99) for a variety of initial conditions. Is it possible for the sphereto fall off a turntable of radius R0? In choosing your initial conditions (x(t0), v(t0),ω(t0))make sure that they are compatible with the rolling condition.

8.11 Recall the definition of the kinetic energy T of a rigid body:

T = 1

2

R

v · v ρdv. (8.100)

(a) Starting from the definition of T , prove the Koenig decomposition:

T = 1

2mv · v +

1

2H ·ω. (8.101)

(b) Establish the following intermediate results:

J = Ω J − JΩ , ˙Jω·ω = H · ω, ˙Jω · ω = 2 H ·ω, (8.102)

where the angular momentum H = Jω.(c) Using the intermediate results and the balance laws, prove the work-energy theorem:

T = F · v + M · ω. (8.103)

(d) If

F =K

i=1

Fi , M =K

i=1

(xi − x) × Fi + MP , (8.104)

then show that

T =

K i=1

Fi · vi + MP · ω. (8.105)

Give three examples of the use of this result for a single rigid body.





9

Lagrange’s Equations of Motion for a Single Rigid Body

9.1 Introduction

In this chapter, the balance laws F = m ˙v and M = H for a rigid body are shown to be equivalent

to Lagrange’s equations of motion. Three separate forms of these equations are developed. Forinstance the second form we establish is

d

dt

∂T

∂ uA

− ∂T

∂uA = F · ∂ v

∂ uA + M · ∂ ω

∂ uA, (A = 1, . . . , 6) . (9.1)

This form of Lagrange’s equations of motion is valid in the absence of constraints. We then turnto showing show how constraints can be imposed and conservative forces and moments can beintroduced. If the constraint forces and moments are prescribed using normality, we pleasantlyfind that we can decouple (9.1) into a set of equations involving the unconstrained motion anda set of equations for the constraint forces and moments. The former equations are known asreaction-less. We then discuss an approach to Lagrange’s equations which we have referred to asApproach II. The two approaches are illustrated using the examples of rolling disks and sliding

disks.Much of the material concerning Lagrange’s equations in these notes are based on Casey [8, 10]supplemented with material on constraint forces and moments due to O’Reilly and Srinivasa [39].

9.2 Lagrange’s Equations of Motion: A First Form

In this section, we wish to establish Lagrange’s equations for an unconstrained rigid body. Ourproof is based on Casey [10], but our developments are not as general as his. The resultingequations are known as the first form of Lagrange’s equations of motion.

To start, we choose a set of curvilinear coordinates q i to parametrize x: x = x(q 1, q 2, q 3).Next, we choose a set of Euler angles ν i to parameterize the rotation tensor Q of the rigid bodyQ = Q(ν 1, ν 2, ν 3). Then, as will be shown below, Lagrange’s equations for the rigid body are

d

dt

∂T

∂ q i

− ∂T

∂q i = G · ai = F · ai,



188 9 Lagrange’s Equations of Motion for a Single Rigid Body

d

dt

∂T

∂ ν i

− ∂T

∂ν i = H · gi = M · gi. (9.2)

Here, gi are the Euler basis vectors and ai are the basis vector of E 3 associated with the curvilinear

coordinate q i

. For the above equations, you may wish to recall the results

v =3

i=1

q iai, ∂ v

∂ q i = ai, ω =

3i=1

ν igi, ∂ ω

∂ ν i = gi. (9.3)

You should notice from that the form of Lagrange’s equations are similar to those you encounteredwith a single particle. 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 theseconservative forces Fcon and moments Mcon, we have

Fcon = −3

i=1

∂U

∂q iai, Mcon = −

3i=1

∂U

∂ν igi, (9.4)

where the potential energy function U has the representations:

U = U (x, Q) = U

q i, ν i

. (9.5)

Notice that Fcon · ai = − ∂U ∂qi

, and Mcon · gi = − ∂U ∂ν i

. Consequently, it is not necessary to evaluateFcon and Mcon in Lagrange’s equations, rather it suffices to evaluate the partial derivatives of U . It is remarkable that the situation with potential energies in rigid bodies is similar to thatencountered in systems of particles.

It is also possible to write an alternative form of Lagrange’s equations of motion using theLagrangian L = T − U . Specifically,

d

dt

∂L

∂ q i

− ∂L

∂q i = (F − Fcon) · ai,

ddt

∂L∂ ν i

− ∂L

∂ν i = (M − Mcon) · gi. (9.6)

We leave it as an (easy) exercise to show that (9.6) can be established from (9.2).

9.2.1 Proof of Lagrange’s Equations

To prove Lagrange’s equations, we need to exploit the Koenig decomposition, and use the angularvelocity vector ω0 = QT ω. The proof proceeds quickly after some preliminary results have beenaddressed.

There are four steps in the proof. The first step involves parametrizing x using a set of curvi-linear coordinates and parametrizing Q using a set of Euler angles ν i. These parameterizationsimply that the kinetic energy T is a function of these quantities and their time derivatives:

T = 1

2mv · v +

1

2J0ω0 · ω0. = T

q i, q i, ν i, ν i

. (9.7)



9.2 Lagrange’s Equations of Motion: A First Form 189

Here, ω0 = QT ω.We next consider the partial derivatives of T :

∂T

∂ q

i =

∂

∂ q

i 1

2

mv

· v = mv

·

∂ v

∂ q

i = mv

·ai,

∂T

∂q i =

∂

∂q i

1

2mv · v

= mv · ∂ v

∂q i = mv · ai,

∂T

∂ ν i =

∂

∂ ν i

1

2J0ω0 ·ω0

= J0ω0 · ∂ ω0

∂ ν i = J0ω0 · QT gi

,

∂T

∂ν i =

∂

∂ν i

1

2J0ω0 ·ω0

= J0ω0 · ∂ ω0

∂ν i = J0ω0 · QT gi

.

(9.8)

Here, we have used the identities,

ai = ∂ v

∂ q i, ai =

∂ v

∂q i, gi = Q

∂ ω0

∂ ν i , gi = Q

∂ ω0

∂ν i . (9.9)

We shall shortly give a derivation of these results.For the third step, we examine (9.8)1,2 and note that G = mv. Hence,

d

dt

∂T

∂ q i

− ∂T

∂q i =

d

dt (G · ai) − G · ai

= G · ai

= F · ai. (9.10)

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 = QJ0QT Qω0

·b = J0ω0

·QT b. (9.11)

Using this result in conjunction with (9.8)3,4, we find that

d

dt

∂T

∂ ν i

− ∂T

∂ν i =

d

dt (H · gi) − H · gi

= H · gi

= M · gi. (9.12)

Thus, we have established the last three of Lagrange’s equations of motion.

9.2.2 The Four Identities

The proof of Lagrange’s equations is facilitated by the following four results:

ai = ∂ v

∂ q i, ai =

∂ v

∂q i, (9.13)




and

gi = Q∂ ω0

∂ ν i , gi = Q

∂ ω0

∂ν i , (9.14)

The first two results are similar to those we established for the single particle. You should notice

the presence of Q in the results for the Euler basis. Unfortunately, ∂ ω∂ν i = gi.We first establish the easier results:

∂ v

∂ q i =

∂

∂ q i

3k=1

q kak

= δ ikak = ai,

∂ v

∂q i =

∂

∂q i

3k=1

q kak

=

3k=1

q k∂ ak

∂q i

=3

k=1

q k ∂ 2x

∂q i∂q k =

3k=1

q k ∂

∂q k

∂ x

∂q i

=

3

k=1

q k∂ ai

∂q k

= ai. (9.15)

Notice that we used the facts that ai are both independent of q k, and the derivatives of x withrespect to q i.

We next consider the easier of the two remaining identities:

∂ ω0

∂ ν i =

∂

∂ ν i

QT ω

=

∂ QT

∂ ν i ω + QT

∂ ω

∂ ν i

= 0ω + QT (gi)

= QT gi. (9.16)

Some rearranging gives

gi = Q∂ ω0

∂ ν i , (9.17)

Notice that we used the fact that Q does not depend on ν i to establish this result.For the final result, we need to be cognisant of the fact that Q depends on the Euler angles

ν i but not on ν i. First, we note that

gi = ∂ ω

∂ ν i =

∂

∂ ν i

−1

2

QQT

= ∂

∂ ν i

−1

2

3k=1

ν k ∂ Q

∂ν kQT

= −1

2

∂ Q

∂ν iQT

. (9.18)

Next, we note that because QQT = I,

∂ Q

∂ν i QT

= −Q

∂ QT

∂ν i ,

∂ Q

∂ν i

∂ QT

∂ν k = Q

∂ QT

∂ν i

∂ Q

∂ν k QT

. (9.19)

The previous results are now used to show the desired identity:



9.3 A Satellite Problem 191

gi = d

dt

−1

2

∂ Q

∂ν iQT

=

3

k=1

ν k −1

2

∂ 2Q

∂ν i∂ν kQT +

∂ Q

∂ν i∂ QT

∂ν k =

3k=1

ν k

−1

2

Q

QT ∂ 2Q

∂ν i∂ν k +

∂ QT

∂ν i∂ Q

∂ν k

QT

=3

k=1

ν k

−1

2

Q

∂

∂ν i

QT ∂ Q

∂ν k

QT

= −1

2

Q

∂

∂ν i

QT

3k=1

ν k ∂ Q

∂ν k

QT

= −1

2

Q

∂ Ω 0

∂ν i QT

= Q

−1

2

∂ Ω 0

∂ν i

= Q

∂ ω0

∂ν i . (9.20)

In the final stages of the proof we used the facts that ω0 is the axial vector of Ω 0 = QT Q =QT Ω Q. We also used the identity

QBQT

= det(Q) Q ( [B]) . (9.21)

This identity is one method relating ω0 to ω and it was alluded to earlier when discussing angularvelocity vectors (see (5.13)).

9.3 A Satellite Problem

As an example of a problem from rigid body dynamics where there are no constraints, we considera satellite of mass m which is in orbit about a spherically symmetric body of mass M (cf. Figure9.1). We assume that the spherically symmetric body is fixed and use its center of mass as theorigin of the position vector of the center of mass of the satellite.

9.3.1 Preliminaries

To parametrize the motion of the center of mass of the rigid body we use a spherical polarcoordinate system. For this coordinate system

x = ReR

= R cos(φ)E3 + R sin(φ)er

= R cos(φ)E3 + R sin(φ) (cos(θ)E1 + sin(θ)E2) , (9.22)

and




Rigid body

Spherically symmetric body of mass M

O

x

X

1E

2E

3E

Fig. 9.1. A rigid body orbiting a spherically symmetric body of mass M .

v = ReR + R sin(φ)θeθ + R φeφ. (9.23)

You should be able to see from this equation what the covariant basis vectors ai are.We parameterize the rotation tensor Q using a set of 1-2-3 Euler angles:

ω = ν 1E1 + ν 2e

2 + ν 3e3. (9.24)

You should notice that ei = QEi. The angular velocity vector of the body also has the represen-tation

ω =

3i=1

ωiei (9.25)

where,

ω1 = ν 2 sin(ν 3) + ν 1 cos(ν 2) cos(ν 3)ω2 = ν 2 cos(ν 3) − ν 1 cos(ν 2) sin(ν 3)

ω3 = ν 1 sin(ν 2) + ν 3. (9.26)

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 has the representation

T = m

2

R2 + R2 φ2 + R2 sin2(φ)θ2

+

λ1

2 (ν 2 sin(ν 3) + ν 1 cos(ν 2) cos(ν 3))

2

+ λ2

2 (ν 2 cos(ν 3)

− ν 1 cos(ν 2) sin(ν 3))

2

+ λ3

2 (ν 1 sin(ν 2) + ν 3)

2(9.27)



9.3 A Satellite Problem 193

If the body has an axis of symmetry where λ1 = λ2, then the expression for the rotational kineticenergy will simplify.

The sole force and moment acting on the rigid body is due to the central gravitational forceexerted upon it by the body of mass M . These forces and moments are conservative and are

associated with the potential energy

U = −GM m

R − GM

2R3 tr(J) +

3GM

2R3 (JeR) · eR. (9.28)

To express this potential energy in terms of the Euler angles ν i, we need to use the results,

eR = cos(φ)E3 + sin(φ) cos(θ)E1 + sin(φ) sin(θ)E2

E1 = cos(ν 2) cos(ν 3)e1 − cos(ν 2) sin(ν 3)e2 + sin(ν 2)e3,

E2 = − sin(ν 2)cos(ν 3)e1 + sin(ν 2) sin(ν 3)e2 + cos(ν 2)e3,

E3 = (sin(ν 1)sin(ν 3) − cos(ν 1)sin(ν 2) cos(ν 3)) e1

+ (sin(ν 1)cos(ν 3) + cos(ν 1) sin(ν 2)sin(ν 3)) e2

+ cos(ν 1) cos(ν 2)e3. (9.29)

The expressions for the force and moment associated with this potential energy can be inferredfrom the developments in the previous set of notes.

9.3.2 The Balance Laws

For the rigid body of interest, we have from the balance laws,

m ˙v = F

= −GM m

R2 eR − 3GM

2R4 (2J + ((λ1 + λ2 + λ3) − 5eR · JeR) I) eR,

H = M

=

3GM R3

eR × (JeR). (9.30)

You should notice that if eR is an eigenvector of J then the so-called gravity-gradient torque

M = 0. In addition, if R = R0er and θ = 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.

9.3.3 Lagrange’s Equations of Motion

Lagrange’s equations of motion for the satellite, are the ai components of the balance of linearmomentum and the gi components of the balance of angular momentum:

d

dt∂T

∂ R−

∂T

∂R

= F

·eR

= −GM m

R2 − 3GM

2R4 ((λ1 + λ2 + λ3) − 3eR · JeR) ,




d

dt

∂T

∂ φ

− ∂ T

∂φ = F · Reφ = −3GM

R3 (eR · Jeφ) ,

d

dt ∂T

∂ θ − ∂ T

∂θ = F · R sin(φ)eθ = −3GM sin(φ)

R3 (eR · Jeθ) ,

d

dt

∂T

∂ ν i

− ∂T

∂ν i= M · gi =

3GM

R3

eR × (JeR) · gi. (9.31)

In the last of these equations, i = 1, . . . , 3. We leave it as an exercise to evaluate the partialderivatives of T with respect to the coordinates and their velocities.

One could also use (9.6) to write down Lagrange’s equations using the Lagrangean L = T −U .Specifically,

d

dt

∂L

∂ R

− ∂L

∂R = 0,

d

dt

∂L

∂ φ

− ∂L

∂φ = 0,

ddt

∂L∂ θ

− ∂L∂θ

= 0,

d

dt

∂L

∂ ν i

− ∂L

∂ν i= 0. (9.32)

Again, we leave evaluating the partial derivatives of L as an exercise.

9.3.4 Conservations

Because the only forces and moments acting on the rigid body are conservative, and there are noconstraints on the motion of the rigid body, it is easy to see that the total energy E = T + U isconserved.

In addition, the angular momentum HO

is conserved. To see this notice that

HO = MO = ReR × F + M. (9.33)

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 specificmotion.

9.4 Lagrange’s Equations of Motion: The Second Form

Previously, we assumed that a set of coordinates q 1, . . . , q 6 had been chosen to parameterize x

and Q: x = x

q 1, . . . , q 3

, Q = Q

q 4 = ν 1, . . . , q 6 = ν 3

. (9.34)



9.4 Lagrange’s Equations of Motion: The Second Form 195

Here, q 4, . . . , q 6 are a set of Euler angles. We note that the covariant basis vectors associated withthis choice of coordinates is

ai = ∂ v

∂ q i, gi =

∂ ω

∂ q (i+3), i = 1, 2, 3. (9.35)

With the choice (9.34), we showed that Lagrange’s equations of motion have the form:

d

dt

∂T

∂ q i

− ∂T

∂q i = F · ai = F · ∂ v

∂ q i,

d

dt

∂T

∂ q (i+3)

− ∂T

∂q (i+3) = M · gi = M · ∂ ω

∂ q (i+3). (9.36)

It’s not too difficult to see that these equations can be written in a more compact form:

d

dt

∂ T

∂ q A

− ∂T

∂q A = F · ∂ v

∂ q A + M · ∂ ω

∂ q A, (A = 1, . . . , 6) . (9.37)

The form (9.36) of Lagrange’s equations is useful in several cases. Among them:

(i) There are no constraints on the motion of the rigid body.(ii) The constraints (and the associated constraint forces and moments) on the rigid body don’t

couple the rotational and translational degrees of freedom.

However, we need a more general form of Lagrange’s equations for other applications.We now consider a new choice of coordinates:

x = x

u1, . . . , u6

, Q = Q

u1, . . . , u6

. (9.38)

With the choice (9.38), we shall see that Lagrange’s equations of motion have the form:

d

dt ∂T

∂ uA−

∂T

∂uA

= RA, (9.39)

where

RA = F · ∂ v

∂ uA + M · ∂ ω

∂ uA, (A = 1, . . . , 6) . (9.40)

We shall shortly discuss examples which use this form.To establish (9.39), we invoke the following identities:1

d

dt

∂ v

∂ uA

=

∂ v

∂uA,

d

dt

∂ ω

∂ uA

=

d

dt

Q

∂ ω0

∂ uA

= Q

∂ ω0

∂uA, (9.41)

whereω = Qω0. (9.42)

1

The proof of these results follows from (9.38) in a manner which is similar to the method by whichthe four identities were established in Section 9.2.2.




Using these identities and the decomposition of the kinetic energy2, a straight forward set of manipulations can be used to establish (9.39):

d

dt ∂T

∂ uA−

∂T

∂uA

= d

dtmv

·

∂ v

∂ uA

+ Jω

· Q

∂ ω0

∂ uA−

∂T

∂uA

= mv

·

∂ v

∂uA

+ J0ω0

· ∂ ω0

∂uA

= d

dt

G · ∂ v

∂ uA + H ·

Q

∂ ω0

∂ uA

−

∂T

∂uA = G · ∂ v

∂uA + H ·

Q

∂ ω0

∂uA

= G · ∂ v

∂ uA + H ·

Q

∂ ω0

∂ uA

= F · ∂ v

∂ uA + M · ∂ ω

∂ uA. (9.43)

This form of Lagrange’s equations is the starting point for most applications and a discussion of Approach II.

We shall now turn to issues associated with the second form of Lagrange’s equations in thepresence of constraints.

9.4.1 Changing Coordinates: An Identity

Recall that we are using two coordinate systems: u1, . . . , u6 and q 1, . . . , q 6. We have tacitly as-sumed that these coordinate systems are related by invertible functions:

q A = q A

u1, . . . , u6

. (9.44)

Then, as

q A =6

B=1

∂q A

∂uB uB, (9.45)

we have that∂q A

∂uB =

∂ q A

∂ uB , (A = 1, . . . , 6, B = 1, . . . , 6) . (9.46)These identities are often known as “cancellation of the dots”.

It’s a good exercise to show that (9.46) also hold when the coordinate transformations aretime-depenedent:

q A = q A

u1, . . . , u6, t

. (9.47)

The resulting identities are used in the literature with Approach II of Lagrange’s equations of motion.

9.4.2 Constraints and Constraint Forces and Moments

Suppose that an integrable constraint is imposed on the rigid body

Φ

q 1, . . . , q 6, t

= 0. (9.48)2 Recall that Jω · ω = QJ0ω0 · ω = QJ0ω0 · Qω0 = J0ω0 · ω0.




We suppose that we can choose the coordinates u1, . . . , u6 to write this constraint in a simplerform such as

u6 − f (t) = 0. (9.49)

The question we wish to ask now is what are the consequences for the right-hand-side Lagrange’s

equations?To answer the question, we suppose that the normality prescription has been used to prescribe

the unknown constraint force and constraint moment:

Fc = µ

3i=1

∂Φ

∂q iai,

Mc = µ

3i=1

∂Φ

∂q i+3gi. (9.50)

It is interesting to expand the constraint forces and moments for this case,

Fc · ∂ v

∂ uA + Mc · ∂ ω

∂ uA =

3i=1

µ

∂Φ

∂q i ai

· ∂ v

∂ uA +

3i=1

µ

∂Φ

∂q i+3 gi

· ∂ ω

∂ uA

= µ

3i=1

3k=1

∂Φ

∂q iai · ∂ q k

∂ uAak +

3i=1

3k=1

∂Φ

∂q i+3gi · ∂ q k+3

∂ uA gk

= µ

3i=1

3k=1

∂Φ

∂q i∂ q k

∂ uAδ ki + µ

3i=1

3k=1

∂Φ

∂q i+3

∂ q k+3

∂ uA δ ki

= µ

3i=1

∂Φ

∂q i∂ q i

∂ uA + µ

3i=1

∂Φ

∂q i+3

∂ q i+3

∂ uA

= µ

3

i=1

∂Φ

∂q i∂q i

∂uA + µ

3

i=1

∂Φ

∂q i+3

∂q i+3

∂uA

= µ ∂Φ

∂uA (9.51)

Notice that we used (9.46) in the next-to-last step. In summary, for a constraint

Φ = 0, where Φ = Φ

q 1, . . . , q 6, t

, (9.52)

the normality prescription yields

Fc · ∂ v

∂ uA + Mc · ∂ ω

∂ uA = µ

∂Φ

∂uA. (9.53)

9.4.3 Consequences of Normality and Integrable Constraints

Let’s now explore some of the consequences of (9.53). Suppose we can express the integrableconstraint as




u6 − f (t) = 0, (9.54)

Then

Fc · ∂ v

∂ uA + Mc · ∂ ω

∂ uA = µδ 6A. (9.55)

As a result the constraint force and moment will only contribute to the Lagrange’s equationassociated with u6. The familiar decoupling we found with a single particle and a system of particles thus holds for the rigid body!

9.4.4 Potential Energy and Conservative Forces and Moments

Suppose that a potential energy is associated with the rigid body

U = U

q 1, . . . , q 6

= 0. (9.56)

Recall that the conservative forces and moments associated with this potential energy are

Fcon = −3

i=1

∂U

∂q iai,

Mcon = −3

i=1

∂U

∂q i+3gi. (9.57)

We suppose that we have chosen a new set of coordinates. Then by setting Φ = −U and µ = 1in (9.51) it can be shown that

Fcon · ∂ v

∂ uA + Mcon · ∂ ω

∂ uA = − ∂ U

∂uA. (9.58)

In summary, the conservative forces and moments appear in a familiar form on the right-hand-sideof Lagrange’s equations of motion.

9.4.5 Mechanical Power and Energy Conservation

Suppose we have a single integrable constraint on the motion of the rigid body and a potentialenergy U . Then the mechanical power of these forces and moments is

P = (Fcon + Fc) · v + (Mcon + Mc) · ω

= µ

3i=1

∂Φ

∂q iai · v +

3i=1

∂Φ

∂q i+3gi · ω

−

3i=1

∂U

∂q iai · v +

3i=1

∂U

∂q i+3gi ·ω

= µ

3i=1

∂Φ

∂q i q i +

3i=1

∂Φ

∂q i+3 q i+3

−

3i=1

∂U

∂q i q i +

3i=1

∂U

∂q i+3 q i+3

= µ

6

A=1

∂Φ

∂q A q A

−

6

B=1

∂U

∂q B q B

= µ

Φ − ∂Φ

∂t

− U . (9.59)




Consequently, if the integrable constraint is time-independent ∂Φ∂t

= 0, and the only applied forcesand moments are conservative, then

T = P = µ (0 − 0) − U . (9.60)

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.

9.4.6 A Short Summary

Suppose we have a rigid body where all of the applied forces and moments are conservative:

F = Fncon − ∂U

∂ x, M = Mncon − uQ. (9.61)

In addition, suppose that there are two constraints on the motion of the rigid body:

Ψ = u6

−f (t) = 0, f

· v + h

·ω + e = 0. (9.62)

The first of these constraints is integrable, while the second is non-integrable. We assume thatthe constraint forces and moments associated with them are prescribed using normality:

Fc = µ1∂ v

∂ u6 + µ2f , Mc = µ1

∂ ω

∂ u6 + µ2h. (9.63)

The presence of µ1 and µ2 should be noted.Turning to Lagrange’s equations of motion (9.39), the equations would read

d

dt

∂T

∂ uA

− ∂T

∂uA = − ∂ U

∂uA + QA (9.64)

where

QA = Fncon · ∂ v∂ uA + Mncon · ∂ ω

∂ uA , (A = 1, . . . , 6) . (9.65)

Using a Lagrangian L = T −U and using the prescriptions for the constraint forces and moments,we find that (9.64) can be expressed in the following form:

d

dt

∂L

∂ uB

− ∂L

∂uB = µ1

f · ∂ v

∂ uB + h · ∂ ω

∂ uB

,

d

dt

∂L

∂ u6

− ∂L

∂u6 = µ2. (9.66)

In these equations B = 1, . . . , 5. Notice that the right-hand side of these forms of Lagrange’sequations are similar to those for a system of particles with the added complication of the momentterms.




9.5 Lagrange’s Equations of Motion: A Third Form

We now turn to Approach II for Lagrange’s equations for a rigid body whose motion is con-strained. Specifically, we parallel the discussion of Section 9.4.6 and assume that the body is

subject to one integrable constraint and one non-integrable constraint. Our discussion is easilygeneralized to multiple integrable and non-integrable constraints. The results we discuss were firstestablished by Casey [10]. The resulting form of Lagrange’s equations is the one most commonlyused in engineering and physics.

First, we assume that the six coordinates u1 . . . , u6 are chosen such that the integrable con-straint Φ = 0 can be simply expressed:

Φ = Φ = u6 − f (t). (9.67)

Imposing this constraint we can calculate the constrained kinetic T and potential U energies of the system. The former will be a function of u1, . . . , u5, u1, . . . , u5, and t, while the latter will bea function of u1, . . . , u5 and t. The non-integrable constraint π2 = 0 can be written in terms of the new coordinates and their velocities:

π2 =

5B=1

pB uB

+ e. (9.68)

Here p1, . . . , p5 and e are functions of t and u1, . . . , u5. We will use the normality prescription tospecify the associated constraint forces and moments (see (9.63)).

By following the same arguments used to establish Lagrange’s equations for a single particleusing Approach II, we find that the equations governing the motion of the rigid body are

5B=1

pA uA + e = 0,

d

dt

∂ T

∂ uB

− ∂ T

∂uB = F ·

3k=1

∂q k

∂uBak

+ M ·

3k=1

∂ν k

∂uBgk

= − ∂ U ∂uB

+ Fanc · 3k=1

∂q k

∂uBak

+ Manc · 3

k=1

∂ν k

∂uBgk

+ µ2 pB.

(9.69)

Here, B = 1, . . . , 5, and the constraint forces and moments associated with the integrable con-straint are absent. In these equations, the forces and moments acting on the body have beendecomposed:

F = Fanc + Fc + Fcon,

M = Manc + Mc + Mcon. (9.70)

For instance, Fanc are the applied non-conservative forces. The Magnus force FM = mBω × v,which used in studies of the flight path of a baseball is a good example of such a force.3

3

The German scientist Heinrich Gustav Magnus (1802–1870) is credited with developing this force in1853 - but I haven’t been able to track down his original paper. A good source for further applicationsof the Magnus effect is Bloomfield [5].



9.6 Rolling Disks and Sliding Disks 201

9.6 Rolling Disks and Sliding Disks

As examples of constrained rigid bodies, we consider a rigid circular disk of mass m and radiusR which either rolls without slipping on a rough horizontal plane (cf. Figure 9.2), or slides on a

smooth horizontal plane. We assume that the contact between the disk and the surface is suchthat rolling is always possible.We assume that the inertia tensor J0 has the representation

JO = λ (E1 ⊗ E1 + E2 ⊗ E2) + λ3E3 ⊗ E3. (9.71)

In other words, we are assuming that E3 is the axis of symmetry of the disk in its referenceconfiguration.

g

E

2

Circular disk of mass m

and radius R

E3

E

2e2e

1e

1e

1 φ ψ

O

P

C

"

"

Fig. 9.2. A circular disk moving with one point in contact with a horizontal plane.

9.6.1 Preliminaries

The center of mass of the disk is parametrized using a set of Cartesian coordinates:

x =

3i=1

xiEi. (9.72)

Consequently, ai = Ei.We assume that the inertia tensor J0 has the representation

JO = λ (E1 ⊗ E1 + E2 ⊗ E2) + λ3E3 ⊗ E3. (9.73)

In other words, we are assuming that E3 is the axis of symmetry of the disk in its referenceconfiguration.

The rotation tensor of the disk will be described using a 3-1-3 set of Euler angles:

ω = φE3 + θe

1 + ψe3. (9.74)




For this set of Euler angles, the Euler basis vectors gi have the representations

g1 = E3

g2 = e

1

g3 = e3

=

sin(ψ) sin(θ) cos(ψ) sin(θ) cos(θ)cos(ψ) − sin(ψ) 0

0 0 1

e1e2

e3

=

0 0 1

cos(φ) sin(φ) 0sin(θ) sin(φ) −sin(θ)cos(φ) cos(θ)

E1

E2

E3

.

(9.75)

In addition, the Euler angles are subject to the restrictions: φ ∈ [0, 2π), θ ∈ (0, π), and ψ ∈ [0, 2π).The (unconstrained) kinetic energy of the disk has the representation

T = m

2

x21 + x2

2 + x23

+

λ

2

φ2 sin2(θ) + θ2

+

λ3

2

ψ + φ cos(θ)

2. (9.76)

In addition, the potential energy of the disk is U = mgx3. You should notice how the symmetry

of the rigid body of interest simplifies the rotational kinetic energy.

9.6.2 Constraints

The motion of the rolling disk is subject to three constraints because the velocity vector of thepoint of contact P is zero:

vP = v + ω × πP = 0. (9.77)

For the disk, the position vector of the instantaneous point of contact P relative to the center of mass X is

πP = −Re

2 = −R sin(ψ)e1 − R cos(ψ)e2

= −R(− cos(θ)sin(φ)E1 + cos(θ) cos(φ)E2 + sin(θ)E3).

(9.78)

Substituting the expression for πP into (9.77) and taking the Ei components of the resultingvector, we find that (9.77) is equivalent to three constraints

x1 + R ψ cos(φ) + R φ cos(θ) cos(φ) − Rθ sin(θ) sin(φ) = 0,

x2 + R ψ sin(φ) + R φ cos(θ) sin(φ) + Rθ sin(θ) cos(φ) = 0,

x3 − Rθ cos(θ) = 0. (9.79)

The last constraint is integrable. It implies that x3 = R sin(θ)+constant. Unfortunately, the firsttwo equations are not integrable.

Combining our previous results, we find that the rolling disk is subject to an integrableconstraint:

Φ = x3 − R sin(θ) + constant = 0, (9.80)

and two non-integrable constraints




π2 = 0, π3 = 0, (9.81)

where

π2 = (E1)

· v + R cos(θ) cos(φ)g1

−R sin(θ) sin(φ)g2 + R cos(φ)g3 ·

ω,

π3 = (E2) · v +

R cos(θ) sin(φ)g1 + R sin(θ) cos(φ)g2 + R sin(φ)g3 · ω.

(9.82)

Notice that we can define a constraint π1 = 0 which is equivalent to the integrable constraint:

π1 = Φ = (E3) · v +−R cos(θ)g2

· ω. (9.83)

It should be clear that we have written all of the constraints in the form f · v + h · ω = 0 tofacilitate our use of the normality prescription.

The constraint forces and moments acting on the rolling disk can be prescribed using normal-ity. With the assistance of the expressions for πi we find

Fc = µ1E3 + µ2E1 + µ3E2,

Mc = µ1−R cos(θ)g2

+ µ2

R cos(θ) cos(φ)g1 − R sin(θ) sin(φ)g2 + R cos(φ)g3

+ µ3

R cos(θ) sin(φ)g1 + R sin(θ) cos(φ)g2 + R sin(φ)g3

.

(9.84)

If we express the dual Euler basis vectors gi in terms of e

i and expand the expressions for Mc,we will find that Mc = πP ×Fc. This implies, that Fc and Mc are equipollent to a force R = Fc

acting at P . The E3 component of this force is the normal force, while the remaining componentsconstitute the friction force.

9.6.3 Imposing the Integrable Constraint

If we impose the integrable constraint, the resulting constrained potential and kinetic energiesare

U = mgR sin(θ),

T = m

2

x21 + x2

2

+

λ

2

φ2 sin2(θ)

+

λ + mR2 sin2(θ)

2

θ2 +

λ3

2

ψ + φ cos(θ)

2. (9.85)

It is crucial to notice that T = T (x1, x2, x1, x2,φ ,θ,ψ , φ, θ, ψ). In other words, x3 has beeneliminated.

In terms of Approach II, we now define a coordinate system uA:

u1

= x1, u2

= x2, u3

= φ, u4

= θ, u5

= ψ, u6

= x3 − R sin(θ). (9.86)

It should be clear that the integrable constraint on the motion of the rolling disk is simply u6 = 0.




9.6.4 The Rolling Disk’s Equations of Motion

Lagrange’s equations of motion for the rolling disk can now be easily obtained. We first note thatthe resultant force F acting on the disk is composed of a constraint force and a conservative force

−mg

E3. The resultant moment M consists entirely of the constraint moment.Using (9.2), we find that

d

dt

∂T

∂ x1

−

∂T

∂x1

= F · E1 = µ2,

d

dt

∂T

∂ x2

−

∂T

∂x2

= F · E2 = µ3,

d

dt

∂T

∂ x3

−

∂T

∂x3

= F · E3 = µ1 − mg,

d

dt

∂T

∂ φ

− ∂T

∂φ = M · g1

= R cos(θ) (µ2 cos(φ) + µ3 sin(φ)) ,

ddt

∂T ∂ θ

− ∂T

∂θ = M · g2

= −µ1R cos(θ) + R sin(θ) (µ3 cos(φ) − µ2 sin(φ)) ,

d

dt

∂T

∂ ψ

− ∂T

∂ψ = M · g3

= µ2R cos(φ) + µ3R sin(φ). (9.87)

Evaluating the partial derivatives of T in these equations, we find

mx1 = µ2,

mx2 = µ3,

mx3 = µ1

−mg,

ddt

λ sin2(θ) φ + λ3

ψ + φ cos(θ)

cos(θ)

= M · g1,

d

dt

λθ

+ λ3

ψ + φ cos(θ)

φ sin(θ) = M · g2,

d

dt

λ3

ψ + φ cos(θ)

= µ2R cos(φ) + µ3R sin(φ).

(9.88)

These equations are supplemented by the three constraints,

π1 = 0, π2 = 0, π3 = 0. (9.89)

to form a closed system of equations for the unknowns xi, φ, θ , ψ and µi.

It is important to note that the system of equations (9.88) and (9.89) are not readily integratednumerically. In particular, they cannot be written in the form y = f (y) which is required for mostnumerical integrators - such as those based on a Runge-Kutta scheme.




9.6.5 Conservations for the Rolling Disk

The easiest method to see that the rolling disk’s total energy E is conserved is to use the alterna-tive form of the work energy theorem. Two forces act on the rolling disk: the gravitational force

−mgE3 and the constraint force R = Fc which acts at P . Consequently,

T = Fc · vP − mgE3 · v. (9.90)

However, vP = 0, and mgE3 · v = ddt

(mgx3), consequently,

d

dt (T + mgx3) = 0. (9.91)

Because E = T + mgx3, this implies that the total energy of the disk is conserved.The above proof of energy conservation 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, Chaplygin, and Korteweg in the late nineteenth century (see Borisov

and Mamaev [6] and O’Reilly [36]). Unfortunately, their physical interpretation is still an openquestion.

9.6.6 The Sliding Disk’s Equations of Motion

The equations governing the motion of the sliding disk on a smooth horizontal plane can beobtained from (9.88) and (9.89). Specifically, one sets µ2 = µ3 = 0 and ignores the constraintsπ2 = 0 and π3 = 0.

It is instructive however to use Approach II. For the sliding disk there is no non-integrableconstraint, so (9.69) simplifies to

d

dt

∂ T

∂ uA−

∂ T

∂uA

= F

· 3

k=1

∂q k

∂uA

ak+ M

· 3

k=1

∂ν k

∂uA

gk ,

= Fnc ·

3k=1

∂q k

∂uAak

+ Mnc ·

3k=1

∂ν k

∂uAgk

.

(9.92)

Here, A = 1, . . . , 5.Using (9.86) in conjunction with (9.92), we find that the resulting governing equations are

mx1 = 0,

mx2 = 0,

d

dt λ sin2(θ) φ + λ3 ψ + φ cos(θ) cos(θ) = 0,

d

dt

(λ + mR2 sin2(θ))θ

+ λ3

ψ φ sin(θ)




+

λ3 φ2 − mR2 θ2

sin(2θ)

2= −mgR cos(θ),

d

dtλ3 ψ + φ cos(θ) = 0. (9.93)

Notice that we could also use these equations to arrive at the equations of motion for the rollingdisk. First, we need to supplement the constraints and, secondly, one needs to append the con-straint forces and moments associated with the integrable constraints to the right–hand side of (9.93).

The equations governing the motion of the sliding disk are clearly far simpler than those forthe rolling disk. Indeed, these equations can then be written in the form y = f (y) where y is acolumn vector with 10 rows. These equations can then be integrated numerically using a standardnumerical integrator.

9.6.7 Configuration Manifold

The configuration manifold M for the rolling disk and the sliding disk are identical:

M = E 2 ⊕ SO(3). (9.94)

Here, x1 and x2 are coordinates for E 2, while the Euler angles are coordinates for SO(3). Thekinematical line-element ds is

ds =

2T

m

dt, (9.95)

where T is the constrained kinetic energy defined earlier.You should notice that the non-integrable constraints on the rolling disk do not effect the

configuration manifold. This is a situation we encountered earlier with the single particle.

9.7 Exercises

9.1 Consider the mechanical system shown in Figure 9.3. It consists of a rigid body of mass m

which is free to rotate about a fixed point O. A vertical gravitational force mgE1 acts on thebody. The inertia tensor of the body relative to its center of mass C is

J0 = λ1E1 ⊗ E1 + (λ − mL2)(E2 ⊗ E2 + E3 ⊗ E3). (9.96)

The position vector of the center of mass C of the body relative to O is Le1.(a) To parameterize the rotation tensor of the body, we will use a set of 1-3-1 Euler angles:p1 = E1, p2 = e3 and p3 = e1. Show that

ω = φE1 + θe3 + ψe1

=

ψ + φ cos(θ)

e1



9.7 Exercises 207

E

2

E3

E

1

r

O

C

g

Fig. 9.3. A pendulum problem.

+


e2

+

θ cos(ψ) + φ sin(θ) sin(ψ)

e3.

(9.97)

(b) Derive expressions for the unconstrained potential U and kinetic T energies of the rigidbody.(c) In what follows, the motion of the rigid body is subject to four constraints:

ψ = ω · p3 = 0, v − ω × (Le1) = 0. (9.98)

Give a physical interpretation of these constraints, and, using the normality prescription, giveexpressions for the constraint forces and moments acting on the rigid body.(d) Prove that the total energy of the rigid body is conserved.(e) Show that the (constrained) kinetic and potential energies of the body are

T = 1

2

λ1

φ2 cos2(θ) + λ(θ2 + φ2 sin2(θ))

, U = −mgL cos(θ). (9.99)

In addition, write out Lagrange’s equations of motion for the rigid body:

ddt

∂ T ∂ θ

− ∂ T

∂θ + ∂ U

∂θ = 0, d

dt

∂ T ∂ φ

− ∂ T

∂φ + ∂ U

∂φ = 0. (9.100)

Argue that the solution to these equations determines the motion of the rigid body.9.2 Consider the simple model for an automobile shown in Figure 9.4. It consists of a single rigid

body of mass m. The moment of inertia tensor of the rigid body is J =3

i=1 λiei ⊗ ei. Here,the inertia tensor J and mass m includes the masses and inertias of the wheels, suspension,engine, and occupants.Interest is restricted to the case where the front wheels are sliding, while the rear wheels arerolling. In other words, the front wheels’ brakes are locked. To model the rolling of the rearwheels in this simple model, it is assumed that

vQ · e2 = 0. (9.101)

where πQ = −L1e1 − L2e2 is the position vector of Q relative to the center of mass X of therigid body. The constraint (9.101) is often known as Chaplygin’s constraint (see [34, 41]).




Rigid body of mass m

1

E2

E

1E

2e

1e

1e

O X

Q

Fig. 9.4. A rigid body model for an automobile moving on a horizontal plane.

(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 (9.101) holds. Usingparameterizations of x and Q of your choice, establish expressions for the 4 constraints onthe motion of the rigid body.

(b) Verify that one of the constraints in (a) is non–integrable.(c) Using the normality 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

mx1 = −µ4 sin(φ), mx2 = µ4 cos(φ), λ3φ = −µ4L1,

x1 sin(φ) − x2 cos(φ) = −L1 φ, (9.102)

where xi = x · Ei.(e) Show that (9.102) allow the center of mass of the rigid body to move in a straight linewithout the body rotating, and prove that the total energy of the body is conserved.

9.3 As shown in Figure 9.5, a circular rod of mass m, length L, and radius R slides on a smoothhorizontal plane. We assume that the inertia tensor J0 has the representation

JO = λ (E1 ⊗ E1 + E3 ⊗ E3) + λ2E2 ⊗ E2. (9.103)

The sole external applied force acting on the rod is gravitational: −mgE3.(a) Assuming that the rotation tensor of the rod is described using a 3-2-3 set of Eulerangles, which orientations of the rod coincide the singularities of this set of Euler angles?(b) Show that the unconstrained kinetic energy of the rod is

T = m

2

x21 + x2

2 + x23

+

λ

2


2+

λ2

2

θ cos(ψ) + φ sin(θ)sin(ψ)

2+

λ

2

ψ + φ cos(θ)

2,

(9.104)

where x = xiEi.(c) Show that the two constraints on the motion of the rigid body can be written in theform



9.7 Exercises 209

E

2

E3

E

2e

3e

1e

1

g

Circular cylinder

Horizontal plane

O

X

Fig. 9.5. A circular rod moving on a horizontal plane.

v

·E3 = 0, ω

·g3 = 0. (9.105)

(d) Using the normality prescription, what are the constraint force Fc and constraint momentMc which enforce the two constraints? With the assistance of a free-body diagram of thesliding rod, give physical interpretations of Fc and Mc.(e) Show that the 6 Lagrange’s equations of motion for the rod yield the following differentialequations for the motion of the rod:

mx1 = 0, mx2 = 0, λ2θ = 0, λφ = 0. (9.106)

(f ) Show that the 6 Lagrange’s equations of motion for the rod also yield solutions for theconstraint force and constraint moment:

Fc = mgE3, Mc = −λ2 φθ sin(θ)g3. (9.107)

(g) Give a physical interpretation of the solutions to (9.106), and show that they allow thecylinder to have motions where ω = constant.





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