A Concise Introduction to Mechanics of Rigid Bodies
L. Huang
A Concise Introduction to Mechanics of Rigid Bodies Multidisciplinary Engineering
123
L. Huang School of Engineering Auckland University of Technology Private Bag 92006 1142 Auckland New Zealand
[email protected]
ISBN 978-1-4614-0471-2 e-ISBN 978-1-4614-0472-9 DOI 10.1007/978-1-4614-0472-9 Springer New York Dordrecht Heidelberg London Library of Congress Control Number: 2011940828 © Springer Science+Business Media, LLC 2012 All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)
For Xin, Zhongjia, and Zhongjin
Preface
Mechanics is a mature engineering subject. Why do we need another mechanics book? This book presents rigid-body mechanics in a compact form for a multidisciplinary engineering program. Typically, such programs include mechatronics, which covers mechanical engineering, electrical and electronic engineering, and computer engineering, and biomechanics, which is related to human physiology, sports, and engineering mechanics. In terms of course structure, they have the following features: • A wide range of subjects in various areas is taught. • Limited teaching time is allocated to each subject. • There is a sharp transition from fundamental to specialized subjects. Mechanics is one of the most important subjects in traditional mechanical engineering programs. It usually takes two semesters to cover all the relevant topics, arranged in the following order: mechanics of particle motion (one dimension), planar motion (two dimensions), and rigid body motion in space. In comparison, the time allotted to the teaching of mechanics is much shorter in a multidisciplinary engineering program. It is hard to find a concise presentation of the material that covers the essential principles of mechanics and that can be taught in a limited time. Many existing textbooks focus almost exclusively on the mechanics of particle and planar motions, which are easily visualized and can be readily analyzed with geometrical and graphical methods. On the other hand, multibody dynamics in three-dimensional space is the cornerstone of many specialized courses in multidisciplinary engineering programs (e.g., robotics). There is a huge gap between what is covered in traditional textbooks on mechanics and the requirements for specialized courses (e.g. [7] and [8]) on the motions of rigid bodies and complex systems in three-dimensional space. Thus a book is needed that narrows the gap by focusing on the mechanics of rigid bodies, with particle motion and planar motion as special cases.
vii
viii
Preface
The main motivation for writing this book was to produce a work that is compact but comprehensive in its coverage of the essential principles of rigid-body dynamics. To achieve this, the following approaches were adopted in the book: • The three-dimensional dynamics of a rigid body is dealt with from the beginning of the book to the end. Mechanics in relation to particle motion and planar motion are treated as special cases. This is in contrast to the traditional, sequential coverage of topics: particle motion ! motion in a line ! planar motion ! threedimensional motion. • Matrix and vector manipulations are used extensively to make the presentation concise and clear. This is in contrast to the traditional approach, which is dominated by scalar and vector manipulations. This is fine for low-dimensional motion but too clumsy for describing motions and mass properties with respect to different reference frames. Matrix and vector notations can greatly simplify mathematical expressions and reveal the essential principles governing motions in high dimensions; this is also in line with the general approach taken in advanced specialized courses like robotics and mechanisms. In this regard, the reader is expected to be familiar with concepts and manipulations of vectors and matrices, which are covered in undergraduate courses like linear algebra or engineering mathematics. Chapter 1 of this book is designed for readers who might need to review the basics in the relevant areas. • The concepts of observation frame and description frame are introduced and are reflected by a set of new vector notations to define kinematic variables more clearly. The main parts of the book concern the description and analysis of position, velocity and acceleration, inertial properties, and the establishment of equations of motion through three main methods: Newton–Euler formulation, D’Alembert’s principle, and Lagrange equations. Some examples are taken from existing books on engineering mechanics and biomechanics. They are addressed with the methodologies and approaches presented in the book, and each step to reach the solution is painstakingly and rigorously explained. There are no question sets for practice after each chapter since rich sources of questions can be found in many existing books on mechanics (e.g. [3] and [5]). Though some important topics like the impulse-momentum method, collisions, vibrations, and forward dynamics (derivation of a rigid body’s motion from forces or torques) are not covered due to space constraints, what is presented in the book should provide a sufficient foundation for dealing with those topics. Statics is not covered here either since it can be treated as a special case of dynamics. As a note of clarification, dynamics in this book means the study of the relation between changes in motion and the causes of those changes (forces and torques). In some books, this is called kinetics, and dynamics includes both kinematics and kinetics.
Preface
ix
The book comprises the following chapters. Chapter 1 covers preliminary mathematical knowledge needed to study the book. It focuses on vector and matrix operations which has been discussed in [1] and [2]. Chapter 2 is on kinematics, Chap. 3 is on dynamics, and Chap. 4 presents case studies. The book is suitable for undergraduates and postgraduate students in multidisciplinary engineering programs like mechatronics and biomechanics. It is also suitable for students in mechanical engineering and engineers and researchers interested in rigid-body dynamics.
Contents
1
Preliminary Remarks on Vectors and Matrices . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1 Vector .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1.1 Definition.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1.2 Operations .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2 Matrix.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.1 Definition.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.2 Operations .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
1 1 1 3 8 8 10
2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1 Coordinate Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2 Position and Orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2.1 Vector Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2.2 Orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2.3 Position .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3 Velocity .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3.1 Linear Velocity .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3.2 Angular Velocity .. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3.3 Relation Between Linear and Angular Velocities . . . . . . . . . . . . . 2.4 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
19 19 22 23 25 28 34 34 36 40 42
3 Dynamics .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1 Inertial Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.2 Theorems and Rules . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2 Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2.1 Linear Momentum . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2.2 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3 Force, Moment of Force, Torque, Work, and Power . . . . . . . . . . . . . . . . . . 3.4 Impulse, Work, and Power .. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.5 Kinetic Energy, Potential Energy, and Mechanical Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
61 61 61 68 73 74 74 80 82 84
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Contents
3.6 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 90 3.6.1 Newton–Euler Formulation . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 91 3.6.2 D’Alembert’s Principle .. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 110 3.6.3 Lagrange Equations.. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 117 4 Case Studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1 Two-Link Planar Robotic Arm . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2 Human Body Doing Twisting Somersaults . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.3 Bicycle Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
131 131 138 143
References .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 151 Index . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 153
Chapter 1
Preliminary Remarks on Vectors and Matrices
1.1 Vector 1.1.1 Definition A vector is a quantity determined by both magnitude and direction. Figure 1.2 shows a force f applied at a block moving at velocity v. Force f and velocity v can be completely described only by their magnitudes and directions. Geometrically, ! a vectoris a directed line segment as shown in Fig. 1.1. In the figure, p D AB is a vector from point A to point B with a magnitude equal to the length of the line, ! kABk. In this book, a point is indicated by uppercase (e.g., A and B), whereas a vector or scalar is written in lowercase (e.g., p). Whether a lowercase character describes a vector or a scalar should be clear from the context of the discussion. For the purpose of analysis, a vector is defined by the coordinates of any two points on the corresponding directed line in a predefined rectangular (Cartesian) coordinate frame. Figure 1.2 shows a coordinate frame O xO yO O z, where O is the origin, and x, O y, O and zO are three unit vectors along the directions of the coordinate axes. A unit vector has a unit magnitude and is identified by a hat symbol (O) above its name. If the coordinates of points A and B are .ax ; ay ; az / and .bx ; by ; bz / respectively, ! ! ! then vectors OA, OB, and AB can be written as follows: ! a D OA D Œax ay az T 2 R3 ; ! b D OB D Œbx by bz T 2 R3 ; ! p D AB D Œbx ax by ay bz az T 2 R3 ;
L. Huang, A Concise Introduction to Mechanics of Rigid Bodies: Multidisciplinary Engineering, DOI 10.1007/978-1-4614-0472-9 1, © Springer Science+Business Media, LLC 2012
1
2
1 Preliminary Remarks on Vectors and Matrices
Fig. 1.1 Force as a vector
f yˆ v xˆ
O
Fig. 1.2 Definition of a vector in a coordinate frame
zˆ bz B(bx,by,bz)
OA
p
= ra
b qz
A(bx,by,bz)
qx a
bx
qy
by
yˆ
O
xˆ
where
2 3 x T 4 Œx y z D y 5: z
The symbol T stands for “transposing” a row (column) vector into a column (row) vector and R3 means that the vector contains three real numbers. The number of elements in a vector is called its size or dimension. Here the size of the vector is three. The analytic representation of a vector can be derived from its magnitude and ! direction. As shown in Fig. 1.2, if vector a D OA is given by its magnitude ra and the angles x , y , and z it makes with axes x, O y, O and zO, respectively, (so that cos2 x C cos2 y C cos2 z D 1), then its analytical form is a D Œra cos x ra cos y ra cos z T : Take the force in Fig. 1.1 as an example. If the force has a magnitude of 10 N and an angle of 6 with axis x, O then h iT h p iT f D 10 cos 10 sin D 5 35 : 6 6
1.1 Vector
3
Special vectors iO , jO, and kO are the unit vectors along the axes of a base coordinate frame and are called base vectors. They are defined as follows: iO D Œ1 0 0T ; jO D Œ0 1 0T ; kO D Œ0 0 1T : Any vector in a three-dimensional space can be formed by a linear combination of O For example, iO , jO, and k. O Œa b cT D aiO C b jO C c k; where a, b, and c are real numbers.
1.1.2 Operations To discuss operations on vectors, the following vectors are first defined: O u D Œux uy uz T D ux iO C uy jO C uz k; O v D Œvx vy vz T D vx iO C vy jO C vz k; O w D Œwx wy wz T D wx iO C wy jO C wz k: • Norm (length) kuk D
q u2x C u2y C u2z I
• Addition and subtraction u ˙ v D Œux ˙ vx uy ˙ vy uz ˙ vz T I • Multiplication with a scalar, for k 2 R ku D Œkux kuy kuz T I • Dot product (inner product) u v D< u; v >D uT v D ux vx C uy vy C uz vz : For base vectors: iO jO D iO kO D jO kO D 0; iO iO D jO jO D kO kO D 1
(1.1)
The three expressions for a dot product, u v, < u; v >, and uT v, will be used interchangeable.
4
1 Preliminary Remarks on Vectors and Matrices
It can be proved that uT v D kukkvk cos ; where is the angle between u and v. If u and v are orthogonal to each other, then D 2 and uT v D 0. This formula can also be used to find the angle between two vectors and the norm (length) of a vector: T u v 1 ; D cos kukkvk p kuk D uT u: The dot product has the following properties: k.uT v/ D .ku/T v D k.uT v/; uT .v C w/ D uT v C uT w; where k is a scalar. • Cross product O u v D .uy vz uz vy /iO C .uz vx ux vz /jO C .ux vy uy vx /k:
(1.2)
O For base vectors iO , jO, and k: O iO jO D k;
kO iO D jO;
jO k D iO ;
O jO iO D k;
kO j D iO ;
iO kO D jO;
iO iO D jO j D kO kO D 0:
(1.3)
The cross product of u and v is a vector perpendicular to the plane formed by u and v, and its length is given by ku vk D kukkvk sin ; where is the angle between u and v. The magnitude ku vk is the area of the parallelogram formed by vectors u and v. If u and v are the axes of a Cartesian coordinate frame, then u v is another axis formed by u and v according to the right-hand rule. From the definition of cross product,: – If u D v or u is aligned with or parallel to v, then u v D 0; – u v D .v u/. Cross product u v can also be treated as the following determinant of a matrix consisting of the elements u and v and the base vectors ˇ ˇ iO ˇ u v D ˇˇ ux ˇv x
jO uy vy
which can be proved to be equivalent to (1.2).
ˇ kO ˇˇ uz ˇˇ ; vz ˇ
1.1 Vector
5
With knowledge of the operations on the base vectors such as those listed in (1.1) and (1.3), we can readily obtain the results of various operations on any vectors that can be expressed as linear combinations of the base vectors. For example, O ˙ .vx iO C vy jO C vz k/ O u ˙ v D .ux iO C uy jO C uz k/ O D .ux ˙ vx /iO C .uy ˙ vy /jO C .uz ˙ vz /k; O uT v D .ux iOT C uy jOT C uz kO T /.vx iO C vy jO C vz k/ O D ux vx .iO T iO / C ux vy .iO T jO/ C ux vz .iO T k/ O Cuy vx .jOT iO / C uy vy .jOT jO/ C uy vz .jOT k/ O Cuz vx .kO T iO/ C uz vy .kO T jO/ C uz vz .kO T k/ D u x v x C u y v y C uz v z ; O u v D ux vx .iO iO / C ux vy .iO jO/ C ux vz .iO k/ O Cuy vx .jO iO / C uy vy .jO jO/ C uy vz .jO k/ O Cuz vx .kO iO / C uz vy .kO jO/ C uz vz .kO k/ D ux vy kO ux vz jO uy vx kO C uy vz iO C uz vx jO uz vy iO O D .uy vz uz vy /iO C .uz vx ux vz /jO C .ux vy uy vx /k: The operations of vectors have very clear geometric meanings, which are summarized in Figs. 1.3–1.5. • Triple products For three vectors, u; v; w, u .v w/ D .uT w/v .uT v/w;
(1.4)
.u v/ w D .u w/v .v w/u;
(1.5)
T
T
.u v/T w D uT .v w/:
(1.6)
• Differentiation and Integration If a vector is a function of another variable (e.g., time t), its differentiation and integration with respect to that variable can be done elementwise. For example, if u.t/ D Œu1 .t/ u2 .t/ : : : un .t/T , then du.t/ 0 D u D ŒPu1 .t/ uP 2 .t/ : : : uP n .t/T ; dt T Z Z Z Z u2 .t/dt : : : un .t/dt : u.t/ D u1 .t/dt uP .t/ D
6
1 Preliminary Remarks on Vectors and Matrices
Fig. 1.3 Geometric interpretation of vector addition and subtraction
z
kˆ
u+v
u−v
v
v u
u jˆ
O
y
iˆ x Fig. 1.4 Geometric interpretation of dot (inner) product of vectors
z
Projection of v onto u: kˆ u ab =|| ab|| || u || =|| v||cos θ =
|| uTv|| || u||2
cos θ =
v
||uTv|| ||u||||v||
θ b
u || u||
u a
ˆj
u
O
y
iˆ x
Common rules of differentiation can be extended to vector functions: 0
0
0
0
0
0
0
0
0
.u ˙ v/ D u ˙ v ; .u v/ D u v C u v ; .u v/ D u v C u v : If a vector is a multivariable function, e.g., u.q1 ; q2 ; : : : ; qm /DŒu1 .q1 ; q2 ; : : : ; qm / u2 .q1 ; q2 ; : : : ; qm / : : : un .q1 ; q2 ; : : : ; qm /T , then @u D @qj
"
@un @u1 @u2 ::: @qj @qj @qj
#T :
1.1 Vector
7
Fig. 1.5 Geometric interpretation of cross product of vectors
z
kˆ area =||u ×v||
Right−hand rule v u×v
θ u ˆj
O
y
iˆ x
• Jacobian A Jacobian is a concept to reflect the relation between the changes in two vectors. Assume there are two vectors, x D Œx1 ; x2 ; : : : ; xm T 2 Rm and y D Œy1 ; y2 ; : : : ; xn T 2 Rn , and y is a function of x: y D y.x/; which is equivalent to y1 D y1 .x1 ; x2 ; : : : ; xm /; y2 D y2 .x1 ; x2 ; : : : ; xm /; :: :
:: :
yn D yn .x1 ; x2 ; : : : ; xm /: Taking the differential of yi (i D 1; 2; : : : ; n), we have ıy1 D
@y1 @y1 @y1 @y1 ıx1 C ıx2 C C ıxj C C ; @x1 @x2 @xj @xm
(1.7)
ıy2 D
@y2 @y2 @y2 @y2 ıx1 C ıx2 C C ıxj C C ; @x1 @x2 @xj @xm
(1.8)
:: :
(1.9)
:: : ıyn D
@yn @yn @yn @yn ıx1 C ıx2 C C ıxj C C : @x1 @x2 @xj @xm
(1.10)
8
1 Preliminary Remarks on Vectors and Matrices
Define vectors
ıy D Œıy1 ıy2 : : : ıyn T
ıx D Œıx1 ıx2 : : : ıxn T as the differentials for vectors y and x, respectively. Equations (1.7)–(1.10) show that ıy and ıx are related by the following array of n rows and m columns formed by the coefficients of ıxj in the expression for yi (i D 1; 2; : : : ; n, j D 1; 2; : : : ; m): 2 3 @y1 @y1 @y1 6 @x1 @x2 : : : @xm 7 6 @y @y 7 @y2 7 6 2 2 6 @x1 @x2 : : : @xm 7 (1.11) J D 6 : : : : 7; 6 : : : : 7 6 : : : : 7 4 5 @yn @yn @yn : : : @x1 @x2 @xm which is called the Jacobian linking the change in x to the change in y. Jacobian J is a collection of m vectors that are the partial derivatives of y with respect to the elements of x: h @y @y @y i : (1.12) J D ::: @x1 @x2 @xm Following the notation of partial differentiation for a single variable function, a Jacobian can also be expressed as @y : (1.13) @x The Jacobian defined above is actually an nm matrix, which will be discussed in the next section. Following the rule of matrix multiplication to be defined in the next section, the relation between the differentials of vectors y and x (1.10) is rewritten in a more compact form: ıy D J ıx: (1.14)
J D
1.2 Matrix 1.2.1 Definition An m m matrix A is an array of mn elements (entries) arranged in m rows and n columns: 2 3 a11 a12 : : : a1n 6 7 6 a21 a22 : : : a2n 7 AD6 : (1.15) 7 2 Rmn ; 4 :: 5 am1 am2 : : : amn
1.2 Matrix
9
where aij (i; j D 1; 2; : : : ; m) denotes the element at the location .i; j / (junction of row i and column j ) and Rmn means that ai;j are real numbers. Sometimes matrix A can be simply represented by Œaij . mn is the dimension of the matrix. When m D n, the matrix is called a square matrix. The Jacobian defined in (1.11) is one example of a matrix. Matrix A can also be viewed as a collection of m column vectors, ai (i D 1; : : : ; n): A D Œa1 a2 : : : an 2 3 a1i 6 a2i 7 6 7 ai D 6 : 7 4 :: 5 ami or as a collection of n row vectors 2
a1 6 a2 6 AD6 : 4 ::
3 7 7 7 5
am aj D Œaj1 aj 2 : : : aj n ;
j D 1; : : : ; m:
By using matrix A, some commonly used special types of matrix are described below. • Diagonal matrix: This is a square matrix (m D n) where all the off-diagonal elements are 0, i.e., ai;j D 0 if i ¤ j . It is represented as diag.a11 ; a22 ; : : : ; ann /. • Identify or unit matrix: This is a diagonal matrix where all the diagonal elements are one, i.e., ai i D 1. It is represented by I m , where m is the number of rows (columns) of the matrix. • Zero matrix: This is a matrix where all the elements are zero, i.e., aij D 0. It is represented by Om;n . • Symmetric matrix: This is a square matrix where the elements whose positions are symmetrical to the diagonal of the matrix are equal to each other, i.e., aij D aj i . • Skew-symmetric matrix: This is similar to a symmetric matrix in pattern except that aij D aj i and the diagonal elements are zero, i.e., ai i D 0. • Orthogonal matrix: This refers to a square matrix in which column (row) vectors are the unit vectors and orthogonal to each other, i.e., aiT aj D 0 for i ¤ j and aiT ai D 1.
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1 Preliminary Remarks on Vectors and Matrices
The following are some examples of these matrices: 2
3 2 0 0 1 0 2 0 5 ; I3 D 4 0 1 0 3 0 0 2 3 1 5 4 Symmetric matrix W 4 5 2 7 5 I 4 7 3 2 3 0 5 4 Skew-symmetric matrix W 4 5 0 7 5 I 4 7 0 2 3 cos sin 0 Orthogonal matrix W 4 sin cos 0 5 : 0 0 1 1 diag.1 2 3/ D 4 0 0
3 0 0 5; 1
O2;3 D
000 I 000
1.2.2 Operations • Transposition Given matrix A defined in (1.15), its transposition is defined as 2 6 6 AT D 6 4
a11 a12 :: :
a21 a21
::: :::
3 am1 am2 7 7 7: 5
a1n
a2n
:::
amn
If AT D A, then A is a symmetric matrix. • Addition and Subtraction Addition or subtraction is the operation of two matrices that have the same numbers of rows and columns. Given A 2 Rmn and B 2 Rmn , the result is C D A ˙ B 2 R mn whose element is the addition or subtraction of the corresponding elements of A and B: cij D aij ˙ bij ;
i D 1; : : : ; m; j D 1; : : : n:
• Multiplication To multiply matrix A by matrix B, the number of columns of A must equal the number of rows of B. If A 2 Rmn , then B 2 Rnl . The entry at the i th row and the j th column of their product C D AB is defined as cij D
n X kD1
ai k bkj ;
1.2 Matrix
11
which is derived by multiplying elementwise the i th row of A by the j th column of B and then summing up the products. Given vector ai D Œai1 ai 2 : : : ai n T formed by the entries of the i th row of A, and vector bj D Œb1j b2j : : : bnj T formed by the entries of the j th column of B, cij is the inner product of ai and bj , cij D aiT bj . In a special case where A has only a row vector and B has only a column vector of the same dimension, the product AB is the inner (dot) product of two vectors. For example, given AD
132 ; 012
2
3 1 4 B D 4 0 1 5; 3 2
then AB D
7 11 11C30C23 14C31C22 : D 6 5 01C10C23 04C11C22
Note that generally AB ¤ BA, and .AB/T D B T AT . If matrix B contains only one column vector (e.g., b), then the product AB becomes the product of a matrix and a vector (Ab). The result is another vector whose size (dimension) is identical to the number of rows of A. For A 2 R23 and b D Œb1 b2 b3 T : 2 3 b a11 a12 a13 4 1 5 a11 b1 C a12 b2 C a13 b3 Ab D : b2 D a21 a22 a23 a21 b1 C a22 b2 C a23 b3 b3
In equation (1.12), the differential of one vector is the product of the Jacobian with the differential of another vector. There are relations between matrix multiplication and cross (inner) product of vectors that are important for the study of the mechanics of rigid bodies. – The cross product of two vectors can be expressed as the product of a matrix and a vector. For vectors u D Œux uy uz T and v D Œvx vy vz T : u v D S.u/v; where
3 0 uz uy S.u/ D 4 uz 0 ux 5 : uy ux 0 2
Note S.u/ is a skew-symmetric matrix consisting of the elements of u. In the next chapter, S.u/ will be used to define the angular velocity of a rigid
12
1 Preliminary Remarks on Vectors and Matrices
body and the above cross product will indicate the linear velocity of a point whose position is specified by vector v on a rigid body rotating with an angular velocity u. This expression can also be used to define some other key concepts of mechanics like moment of force and moment of inertia. – If A 2 R33 is an orthogonal matrix with determinant 1, then [4] .Au/ .Av/ D A.u v/;
(1.16)
where u 2 R3 and v 2 R3 . It can also be verified that .Au/T .Av/ D uT v:
(1.17)
The multiplication of a matrix by a scalar is equivalent to each entry of the matrix being multiplied by the scalar. • Partitioning Partitioning a matrix means dividing it into submatrices with suitable dimensions. This method is useful for the multiplication of matrices with high dimensions. Consider matrices A 2 Rmn and B 2 Rnp . Matrix A is partitioned as AD
A11 A12 ; A21 A22
where A11 (A21 ) and A12 (A22 ) are submatrices with the same number of rows, and A11 (A12 ) and A21 (A22 ) have the same number of columns. For example, A11 2 Rl;k , A12 2 Rl;nk , A21 2 Rml;k , and A22 2 Rml;nk . Accordingly, matrix B is partitioned as
B11 B12 BD B21 B22
where B11 2 Rk;q , B12 2 Rk;pq , B21 2 Rnk;q and B22 2 Rnk;pq . Then A11 A12 B11 B12 AB D A21 A22 B21 B22 A11 B11 C A12 B21 A11 B12 C A12 B22 : D A21 B11 C A22 B21 A21 B12 C A22 B22 If B 2 Rnp is partitioned into its column vectors, B D b1 b2 bp ;
1.2 Matrix
13
then
AB D Ab1 Ab2 : : : Abp ;
which means that each column vector of AB is the multiplication of A by the corresponding column vector of B. This partition method is used to define the Jacobian in (1.12). Example 1.1 2
:: : :: :
6 1 2 6 6 AD6 3 0 6::: ::: ::: 4 : 0 3 :: and
3
2
3 1 7 7 0 2 7 7; ::: :::7 5
6 6 6 B D6 6 4
1 1
1 0 ::: 0 1
2 1 ::: 1 1
3 2 01 31 12 12 C 2 6 30 11 7 02 01 7 4 6 7 D AB D 6 5 01 5 12 4 1 C 1 1 03 11 01 2
3 7 7 7 7 7 5
3 8 8 5: 3
• Determinant of a Matrix In this book, the most frequently encountered determinants i are for 2 2 and 3 3 h a11 a12 matrices. The determinant of a 2 2 matrix A D a a is defined as 21
22
ˇ ˇ ˇ a11 a12 ˇ ˇ ˇ D a11 a22 a12 a21 : detA D jAj D ˇ a21 a22 ˇ For a 3 3 matrix,
its determinant is ˇ ˇ a11 ˇ detA D jAj D ˇˇ a21 ˇ a 31
2
a11 A D 4 a21 a31
a12 a22 a32
a13 a23 a33
a12 a22 a32
3 a13 a23 5 ; a33
ˇ ˇ ˇ ˇ ˇ ˇ
D a11 a22 a33Ca13 a21 a32Ca12 a23 a31a13 a22 a31a12 a21 a33a11 a23 a32 : • Minor and cofactor The minor, Mij , corresponding to the element aij of matrix A, is a new matrix with a reduced dimension produced by deleting the i th row and j th column of
14
1 Preliminary Remarks on Vectors and Matrices
matrix A. For matrix A defined in (1.15), the minors for the elements in the first row are a22 a23 a21 a23 a21 a22 ; M12 D ; M13 D : (1.18) M11 D a32 a33 a31 a33 a31 a32 The cofactor of the minor Mij is defined as Aij D .1/i Cj jMij j: For M12 in (1.18), A12
ˇ ˇ ˇ a21 a23 ˇ ˇ ˇ D .a21 a33 a31 a23 /: D .1/ ˇ a31 a33 ˇ 3
The determinant of matrix A 2 Rnn can be defined with minors and cofactors corresponding to any of its rows or columns. For example, jAj D
n n X X .1/i Cj aij jMij j D aij Aij i D1
i D1
is determined by the minors and cofactors of the j th column. The determinant of matrix A in (1.15) can be calculated as ˇ ˇ ˇ ˇ ˇ ˇ ˇa a ˇ ˇa a ˇ ˇa a ˇ jAj D a11 A11 Ca12 A12 Ca13 A13 D a11 ˇˇ 22 23 ˇˇ a12 ˇˇ 21 23 ˇˇ Ca13 ˇˇ 21 22 ˇˇ a32 a33 a31 a33 a31 a32 • Adjoint The adjoint of a matrix (e.g., A 2 Rnn ) is defined as 2 3 A11 A21 : : : An1 6 A12 A22 : : : An2 7 6 7 adjA D 6 : :: 7; 4 :: : 5 A1n A2n : : : Ann which is a matrix consisting of the cofactors of matrix A in a transposed order. • Inverse The inverse of a square matrix A 2 Rmm is denoted as A1 . Its product with A is an identity matrix I n , AA1 D A1 A D I n ; and thus A1 D
adjA : jAj
It can be proved that .AB/1 D B 1 A1 , and if AT D A1 or AT A D AAT D I n , then A is an orthogonal matrix.
1.2 Matrix
15
For A D
h
a11 a12 a21 a22
i , its inverse is
A1 D
a22 a12 a21 a11 jAj
a22 a12 a21 a11 : D a11 a22 a12 a21
• Rank The rank of a matrix is the maximum number of column (row) vectors that are linearly independent. A group of vectors is linearly independent if none of the vectors can be expressed as the linear combination of other vectors in the same group. • Eigenvalue and Eigenvector Given A D Œaij , if Ax D x (x ¤ 0, where is a scalar and x is a vector), then is called an eigenvalue and x ¤ 0 is called an eigenvector corresponding to . The equation Ax D x can also be expressed as .A I /x D 0:
(1.19)
The above linear simultaneous equation in x has a nontrivial solution if and only if jA I j D 0; which is called the characteristic equation of A and can be used to solve the eigenvalue . Putting the value of into (1.19), eigenvector x can then be solved. Example 1.2 Given
AD
5 2 ; 2 2
(1.20)
then ˇ ˇ 5 jA I j D ˇˇ 2
ˇ ˇ 2 ˇ D 2 C 7 C 6 D 0; 2 ˇ
which has two solutions: 1 D 6 and 2 D 1. For 1 D 6, the equation .A I /x D 0 can be expressed as
4x1 C 2x2 D 0; 2x1 x2 D 0:
Its solution is x D k1 Œ1 2T , where k can be any real number. Following the same procedure, the eigenvector corresponding to 2 can be found to be x D k2 Œ2 1T , where k2 is any real number.
16
1 Preliminary Remarks on Vectors and Matrices
• Quadratic Form A quadratic form is a scalar function of a square matrix A 2 Rnn and a vector of a compatible dimension x 2 Rn ; it is defined as an inner product of x and Ax, x T Ax. If a11 a12 AD and x D Œx1 x2 x3 T ; a21 a22 then x T Ax D a11 x12 C a22 x22 C .a21 C a12 /x1 x2 : As x T Ax is a scalar, so x T Ax D .x T Ax/T D x T AT x
or x T .A AT /x D 0:
If A is a skew-symmetric matrix, AT D A, then x T Ax D 0. If for any x ¤ 0, x T Ax > 0, A is positive definite. It is negative definite if T x Ax < 0. It can be proved that the eigenvalues of a positive definite matrix are positive, and those of a negative definite matrix are negative. For A in (1.20),
5 2 x Ax D Œx1 x2 2 2
T
x1 x2
D 3x12 0:
If x ¤ 0, then x T Ax < 0 and A is negative definite. A has negative eigenvalues 6 and 2. • Similar Transformation Matrices A 2 Rnn and B 2 Rnn are similar to each other if A D CBC 1 ;
(1.21)
where C 2 Rnn is an invertible matrix. If C is an orthogonal matrix, then C 1 D C T and A D CBC T: (1.22) Similar matrices have the same rank, determinant, trace (sum of diagonal elements), and eigenvalues. • Differentiation and Integration If a matrix is a function of another variable (e.g., time t), then its differentiation and integration with respect to that variable can be done elementwise. For matrix A, which is a function of time t, 2 6 6 A.t/ D 6 4
a11 .t/ a21 .t/ :: :
a12 .t/ a22 .t/ :: :
::: ::: :: :
a1n .t/ a2n .t/ :: :
am1 .t/
am2 .t/
:::
amn .t/
3 7 7 7; 5
1.2 Matrix
17
2 6 dA.t/ P D6 D A.t/ 6 4 dt
aP 11 .t/ aP 21 .t/ :: :
aP 12 .t/ aP 22 .t/ :: :
::: ::: :: :
aP 1n .t/ aP 2n .t/ :: :
3 7 7 7; 5
aP m1 .t/ aP m2 .t/ : : : aP mn .t/ 2 R 3 R R a11 .t/dt a12 .t/dt : : : a1n .t/dt 6 R 7 R R 6 7 Z 6 a21 .t/dt a22 .t/dt : : : a2n .t/dt 7 7; A.t/dt D 6 6 :: :: :: :: 7 6 7 : : : : 4 5 R R R am1 .t/dt am2 .t/dt : : : amn .t/dt The rules found in calculus such as the product rule and the composite rule also apply here. For example, given A.t/ 2 Rmn , B.t/ 2 Rnp , and x.t/ 2 Rn (a vector), then d.AB/ P C AB; P D AB dt d.Ax/ P C Ax; D Ax P dt d.x T Ax/ P C x T AxP D xP T .A C AT /x C x T Ax: P (1.23) D xP T Ax C x T Ax dt If A is a symmetrical matrix (AT D A/, then d.x T Ax/ P D 2xP T Ax C x T Ax: dt
Chapter 2
Kinematics
2.1 Coordinate Frames The first step in the study of the kinematics and dynamics of a rigid body is to have a clear description of its position and orientation and their changes with time. This is achieved by using coordinate frames, vectors, matrices, and other mathematical tools. Several standard coordinate frames exist including • Cartesian coordinate frames • Spherical coordinate frames and • Cylindrical coordinate frames. A Cartesian or rectangular coordinate frame consists of an origin and three mutually perpendicular axes arranged according to the right-hand rule. In this book, without an explicit explanation, a coordinate frame or simply a frame means a Cartesian coordinate frame. As depicted in Fig. 2.1, coordinate frame fAg W OA xOA yOA zOA is attached to a rigid body (A). It has an origin OA and three axes along unit vectors xOA , yOA , and zOA . Frame fBg W OB xOB yOB zOB is attached to another body (B). It has an origin OB and three axes along unit vectors xOB , yOB , and zOB . Frames like fAg and fBg that are attached to rigid bodies are usually called body frames or simply frames. A special coordinate frame is the so-called universe frame or inertial frame of reference, fU g W OX Y Z, in which Newton’s law of motion is valid. In the context of engineering applications, the universe frame is always assumed to be fixed on the Earth and its axes are O denoted by the special base vectors iO , jO, and k. Though the Cartesian coordinate frame is the most popular one for studying rigid-body motions, cylindrical and spherical coordinate frames are more convenient for describing circular motions. Figure 2.2 shows a cylindrical coordinate frame fAg defined in universe frame fU g. It consists of an origin OA and three mutually perpendicular axes (called
L. Huang, A Concise Introduction to Mechanics of Rigid Bodies: Multidisciplinary Engineering, DOI 10.1007/978-1-4614-0472-9 2, © Springer Science+Business Media, LLC 2012
19
20
2 Kinematics
Fig. 2.1 Coordinate frames
{B}
xˆB
zˆA Ap
{A}
Z kˆ {U} iˆ
pO
OB
zˆB
OB pO
B
yˆ B
yˆA
OA
A
xˆA O
ˆj
Y
X Fig. 2.2 Cylindrical coordinate frame
Z
eˆz
kˆ
{A} eˆφ
z {U } OA O
φ iˆ
pO
ˆj
A
r
Y eˆr
X
cylindrical axes), eOz , eOr , and eO : eOr D cos iO C sin jO; O eOz D k; eO D eOz eOr D sin iO C cos jO; right-hand rule; pOA D r eOr C zeOz ;
(2.1)
where is the angle between the plane formed by axis Z and pOA and the plane XZ, r is the length of the projection of pOA on the X Y plane or the radius of the cylinder, and z is the z coordinate of point OA in fU g. r, , and z are called the cylindrical coordinates of point OA .
2.1 Coordinate Frames
21
Fig. 2.3 Spherical coordinate frame
Z
kˆ
eˆr
pO
{U}
θ
eˆφ A
{A} ˆj
OA r
O
φ
eˆθ
Y
iˆ X
A unique feature of a cylindrical coordinate frame is that its axes change with the position of its origin, whereas in a Cartesian coordinate frame, the origin has no effect on the directions of the axes of the frame. There is a one-to-one mapping between the Cartesian coordinates and the cylindrical coordinates of a point. For point OA , the mapping from its cylindrical coordinates to its Cartesian coordinates can be derived from expanding (2.1): O pOA D r cos iO C r sin jO C zk; xOA D r cos ; yOA D r sin : zOA D z Given the Cartesian coordinates .xOA ; yOA ; zOA / of point OA , the corresponding cylindrical coordinates are q r D xO2A C yO2A ; D tan1
yOA ; xOA
xOA ¤ 0;
z D zOA : If xOA D 0 and yOA ¤ 0, then D ˙ 2 . If xOA D yOA D 0, then OA is located on the Z axis and is not defined. Figure 2.3 shows a spherical frame fAg W OA eO eO er O in the universe coordinate frame fU g. The frame is defined by pOA D r eOr ; O eOr D sin cos iO C sin sin jO C cos k; O eO D cos cos iO C cos sin jO sin k; eO D eOr eO D sin iO C cos jO; right-hand rule:
22
2 Kinematics
The spherical coordinates include r, , and , where r is the radius of the sphere, is the colatitude, and is the azimuth angle. Like a cylindrical coordinate frame, the axes of a spherical coordinate frame depend on its origin. The spherical coordinates and the Cartesian coordinates of origin OA are related through rD
q
xO2A C yO2A C zO2A ;
D tan1 q D tan1
zOA xO2A C yO2A
yOA ; xOA
;
xOA ¤ 0;
yOA ¤ 0;
xOA ¤ 0:
2.2 Position and Orientation The positions and orientations of rigid bodies A and B are fully described byfBg and fAg, the frames attached to them. When OB , xOB , yOB , and zOB are observed and described in fAg, they are denoted as vectors ApOB , AxOB , AyOB , and AzOB . Many key quantities describing the motion of a rigid body like position, velocity, and acceleration are presented in the form of a vector. The vector should contain the following information: • The coordinate frame, called the observation frame, in which the vector is observed. In this frame, the geometrical entity of the vector, a directional line, is formulated and observed by one who keeps a fixed position and orientation with respect to the frame. • The coordinate frame, called the description frame, in which the vector is described. In this frame, the directional line observed in the observation frame is presented by its projections (x, y, and z components) along the axes of the frame. In other words, a numerical or algebraic expression of the vector, a row or a column of numbers or variables, is formulated. Take the position vector of the origin of fBg, OB in Fig. 2.1, as an example. If the observation frame is fAg, then the vector is observed as the directional line OA OB . If the observation frame changes to fU g, then the vector is observed as the directional line OOB . What happens if the observation frame is fBg? The directional line is now degenerated to a point, and a zero (null) vector is formed. For each directional line observed, its x, y, and z components along the axes of description frame fAg are different from those along the axes of another description frame fU g, generating different vector expressions. It is clumsy to explain in words the observation frame and the description frame for the definition of a vector. We must find a way to embed them in a neat and compact notation for the vector.
2.2 Position and Orientation
23
2.2.1 Vector Notations Let pB be the position vector of point B. Define pB / A
C
as the vector pB observed in frame fAg and described in frame fC g. This notation consists of three parts: • pB : the name of the vector that is at the center of the notation. It refers to the (kinematic or dynamic) property that the vector stands for and the physical entity with which that property is associated. Here, “p” stands for position, and “B” refers to point B. Depending on the way the vector is defined, the number of symbols in the name varies. • A: the name of the observation frame that forms the right subscript of the vector name. It is separated from the vector name with a slash. • C : the name of the observation frame that is the left superscript of the vector name. The notation can be simplified in the following special cases: • When a universe frame is involved, its name is not shown in the notation. For example: – pB/ A means the position of B observed in frame fAg but described in universe frame fU g. – pB means the position of B observed and described in universe frame fU g. – A pB/ means the position of B observed in universe frame fU g but described in frame fAg. • When the observation frame and the description frame are the same, the right subscript (the name of the observation frame) is not shown. For example: – ApB means the position of B observed and described in frame fAg. Remark. The above notations are defined using position vector as an example, but they can be used to represent any other types of vectors. If p is replaced by v, !, or a, it represents a linear velocity, an angular velocity, or a linear acceleration, respectively. In some cases when no confusion would be caused without explicitly listing the observation frame and the description frame, it is acceptable to omit the names of the frames to simplify the notation. The following example is presented to show the application of the notations defined above.
24
2 Kinematics
Fig. 2.4 Observation and description of a vector in different frames
yˆA
xˆB
Z kˆ {A}
{U} O O A zˆA
iˆ
{B}
C x ˆj
xˆA
OB
zˆB
Y a
X
b
yˆB
Example 2.1. Universe frame fU g and two body frames fAg and fBg are set up as O and shown in Fig. 2.4. fAg and fU g share the same origin, and xOA D jO, yOA D k, O zOA D iO . The origin of frame fBg is on axis jO with a distance a from OA , and xOB D k, yOB D iO , and zOB D jO. Point C is on axis zOB . The expressions for the positions of OB and C are determined by the choice of observation frame and description frame. For the position of point OB : pOB D Œa 0 0T ; observed in fAg and described in fAgI
A
pOB/ A D Œ0 a 0T ; observed in fAg and described in fU gI pOB D Œ0 a 0T ; observed in fU g and described in fU gI B
pOB / D Œ0 0 aT ; observed in fU g and described in fBgI
B
pOB D Œ0 0 0T ; observed in fBg and described in fBg. This is the origin of fBg!
For the position of point C : pC D Œb 0 0T ; observed in fBg and described in fBgI
B
pC/ B D Œ0 0 bT ; observed in fBg and described in fU gI pC D Œ0 a bT ; observed in fU g and described in fU g: Let x be the vector corresponding to the directional line OC ; then x D Œa b 0T ; observed and described in fAgI
A
x D Œ0 a bT ; observed and described in fU gI x/A D x/OA D Œ0 a bT ; observed in fAg and described in fU gI B
x/A D B x/OA D Œb 0 aT ; observed in fAg and described in fBgI x/ D Œb 0 aT ; observed in fU g and described in fBg:
B
2.2 Position and Orientation
25
2.2.2 Orientation Referring to Fig. 2.1, the orientation of bodies A and B with respect to universe frame fU g are governed by the directions of the unit vectors along their orthogonal axes. Using them as column vectors, rotation matrices are formed: RA D ŒxOA yOA zOA 2 R33 ; RB D ŒxOB yOB zOB 2 R33 : As the column vectors are orthogonal to each other, RA and RB are orthogonal matrices, RAT RA D I 3 , and RBT RB D I 3 . Similarly, the orientation of fBg with respect to fAg can be described compactly by putting base vectors AxOB , AyOB , and AzOB in a rotation matrix: R D ŒAxOB AxOB AzOB 2 R 33 :
A B
(2.2)
Note AxOB , AyOB , and AzOB are the base vectors of fBg observed and described in fAg, and they are presented in the notations introduced in Sect. 2.2.1. Treating AxOB as a combination of the projections of xOB onto the three axes of frame fAg, we have xOB D .xOBT xOA /xOA C .xOBT yOA /yOA C .xOBT zOA /OzA :
A
This also applies to AyOB and AzOB such that yOB D .yOBT xOA /xOA C .yOBT yOA /yOA C .yOBT zOA /OzA ;
A
zOB D .OzBT xOA /xOA C .OzBT yOA /yOA C .OzBT zOA /OzA :
A
Substituting them into (2.2), we have 2
xOBT xOA yOBT xOA zOBT xOA
3
6 7 T T T 7 RD6 4 xOB yOA yOB yOA zOB yOA 5 : xOBT zOA yOBT zOA zOBT zOA
A B
What follows is an important relation between RA , RB , and BAR: RB D RA BAR:
(2.3)
26
2 Kinematics
zˆA
Fig. 2.5 Relation between a rotation matrix with corresponding rotation axis and angular displacement
xˆS OA
{U} → {A}
OS kˆ rz
{D} → {S}
Z
xˆA
ry
yˆ D
ˆj
rx
X
zˆS
θ
xˆD
OD rˆ
O
yˆS
yˆA
Y
zˆD
iˆ
This can be proved by expanding the right-hand side of (2.3): 2
xOBT xOA yOBT xOA zOBT xOA
3
6 7 T T T 7 RA BAR D ŒxOA yOA zOA 6 4 xOB yOA yOB yOA zOB yOA 5 xOBT zOA yOBT zOA zOBT zOA D xOA xOAT C yOA yOAT C zOA zOAT ŒxOB yOB zOB D ŒxOA yOA zOA ŒxOA yOA zOA T RB D .RA RAT /RB D I 3 RB D RB : The rotation matrix provides a compact and systematic way to describe a body’s rotation, but it contains nine parameters that are not independent of each other as they need to meet the requirement of an orthogonal matrix. In what follows, an alternative expression for the orientation of a body is presented by specifying the angular displacement and the axis of rotation. As shown in Fig. 2.5, which is derived from Fig. 2.1, assume the orientation of fAg is generated by fU g rotating about axis rO D Œrx ry rz T at angle . Define a new frame fDg such that its origin is at O (origin of fU g) and its three axes are xOD D rO ; yOD D
rO kO ; O kOr kk
zOD D xOD yOD D
O rO .Or k/ : O kOr kk
O otherwise yOD and zOD would not be defined. Note that rO cannot align with k;
2.2 Position and Orientation
27
The rotation matrix of frame fDg with respect to fU g is R D ŒxOD yOD zOD ;
U D
(2.4)
and its inverse is the rotation matrix describing the orientation of U with respect to fDg, D R D DUR1 D DURT : U After fU g rotates around vector rO at an angle , fU g changes to fAg, and fDg to fS g. As fDg is fixed with respect to fU g, the relative orientation between fAg and fS g is the same as that between fU g and fDg: R D DUR;
A S S A
R D UDR:
(2.5)
But RA D RS AS R: From (2.5), RA D SURUDR D .DUR SDR/ UDR:
(2.6)
As fS g is produced by rotating fDg around its X axis by , 2
3 1 0 0 D R D Rx ./ D 4 0 cos sin 5 : S 0 sin cos
(2.7)
Substituting (2.4) and (2.7) into (2.6), we obtain 2
rx
6 RA D 6 4 ry rz
p ry
1rz2 prx 2 1rz
0
prx rz
1rz2 ry rz
32 r x 1 0 0 6 p ry 76 6 7 4 0 cos sin 7 5 4 1rz2 5 prx rz 0 sin cos 2 2 32
p 2 p 1rz 1 rz
1rz
ry prx
1rz2
pry rz
1rz2
3 rz 7 0 7 5 p 2 1 rz
3 r 2 .1 cos / C cos rx ry .1 cos / rz sin rx rz .1 cos / C ry sin 7 6 x D 4 rx ry .1 cos / C rz sin ry2 .1 cos / C cos ry rz .1 cos / rx sin 5 : rz2 .1 cos / C cos rx rz .1 cos / ry sin ry rz .1 cos / C rx sin 2
Note that rx2 C ry2 C rz2 D 1 is used to simplify the above expressions. To show the axis and the angle associated with the rotation, the matrix is denoted as Rr ./, which is in the same form as that for basic rotation matrices (Rx ./, Ry ./, and Rz ./). To make the mathematical expressions more compact, the following abbreviations of some common trigonometric functions are used:
c D cos
s D sin
28
2 Kinematics
and then 2
3 rx ry .1 c / rz s rx rz .1 c / C ry s rx2 .1 c / C c RA DRr ./4 rx ry .1 c / C rz s ry2 .1 c / C c ry rz .1 c / rx s 5 : rx rz .1 c / ry s ry rz .1 c / C rx s rz2 .1 c / C c (2.8) Rr ./ meets all the requirements of a rotation matrix, and it becomes a basic rotation O of fU g. matrix if rO is replaced by one of the principal axes (iO , jO, or k) It can be shown that rO is the eigenvector of Rr ./, just as xO is the eigenvector of Rx ./. This can be verified from the fact that the rotation axis does not change during rotation. The relation in (2.8) can also be used to find rO and associated with the rotation matrix 2 3 r11 r12 r13 RA D 4 r21 r22 r23 5 : (2.9) r31 r32 r33 Equalizing (2.9) and (2.8) elementwise, we have D cos1 rO D
r C r C r 1 11 22 33 ; 2
Œr32r23 r13r31 r21r12 T ; sin ¤ 0: 2 sin
(2.10) (2.11)
2.2.3 Position Referring to Fig. 2.1, the position of body fAg with respect to universe frame fU g is given by vector pOA . Although any point in the body can be chosen as the origin (OA ) of the body frame, it will be made clear later that the choice of OA affects the description of dynamics of the body, and a good candidate for OA is the center of mass (COM). The relation between the positions of two bodies are specified by that of the positions of the origins of their body frames. The position relation of frames fAg and fBg in Fig. 2.1 is given by pOB D pOA C pOB/A ;
(2.12)
which follows the rule of vector addition. Note all the vectors are described in universe frame fU g. Express ApOB with its column vectors: pOB D ŒxA yA zA T
A
2.2 Position and Orientation
29
ˆzA
Fig. 2.6 Position of a point in body frame
C {A} Ap
C
pC kˆ
iˆ
OA
Z pO
{U}
xˆA
A
ˆj O
yˆA
Y
X
then pOB/A D xA xOA C yA yOA C zA zOA D ŒxOA yOA zOA ŒxA yA zA T D RA ApOB :
(2.13)
Substituting (2.13) into (2.12), we have pOB D pOA C RA ApOB :
(2.14)
The position of any point in the body frame can be obtained in the same way. Referring to Fig. 2.6, the position of point C is given by pC DpOA C pC/ A ; where pC/A D RA pC . A
Remark. Equation (2.13) can be generalized to describe the relation between two vectors observed in the same frame (e.g., fAg), but described in different frames (e.g., fAg and fU g). One vector is transformed into another by the rotation matrix (e.g., RA ) between the description frames. If the universe frame in (2.13) is replaced with another frame fDg, then pOB/ A D ADR ApOB :
D
(2.15)
In what follows, several examples are provided to show the applications of the concepts and methods presented above. They include the establishment of basic rotation matrices, the steps to set up coordinate frames and establish their relations in a mechanism consisting of multiple rigid bodies, and position and orientation analysis for rigid bodies.
30
2 Kinematics
Fig. 2.7 Example 2.2
zˆA
OA
cosθ
xA
zˆB − sin θ OB
θ
sinθ
θ
yˆ B cosθ
yˆA
xˆ B
Example 2.2. In this example, the rotation matrix (basic rotation matrix) for a frame rotating about one of its principal axes is to be obtained. As shown in Fig. 2.7, frame fAg W OA xOA yOA zOA rotates around axis zOA at angle . As a result, a new frame, fBg W OB xOB yOB zOB , is generated. Before proceeding further, note the following rules about the measurement of a rotation angle: • The angle is positive for an anticlockwise rotation. • The unit of the angle is radian. • The range of the angles is normalized in the interval Œ . It is clear that OB D OA ; xOB D cos xOA C sin yOA ; yOB D sin xOA C cos yOA ; zOB D zOA : As such
2
cos A 4 R D R ./ D sin z B 0
3 sin 0 cos 0 5 : 0 1
(2.16)
Note Rz ./ is used to denote any rotation matrix produced by rotating the frame around its z axis. Following the same procedure, the rotation matrices corresponding to rotations around the X and Y axes are derived: 2
3 1 0 0 Rx ./ D 4 0 cos sin 5 ; 0 sin cos
2
3 cos 0 sin Ry ./ D 4 0 1 0 5: sin 0 cos
(2.17)
2.2 Position and Orientation
31
H
zˆB
{B} E kˆ
pE
β
2d
OB
b
G 2c
Z zˆA
a
{A}
ω1
xˆB
iˆ
O xˆA
ω2
F
yˆ A
OA X
yˆB
{U}
θ ˆj
Y
Fig. 2.8 Example 2.3
Example 2.3. As schematically shown in Fig. 2.8, a mechanism consists of a rod (OA OB ) and a rectangle box (EF GH ). The rod rotates with speed !1 about the fixed axis Z perpendicular to the ground and the box rotates with speed !2 about the joint axis with the rod, xB , which goes through the center of the box. The dimensions of the rod and the box are shown in the figure. Our task is to determine the positions and orientation of all the rigid bodies and their relations as the functions of the angular displacements of the rod and the box about their rotation axes, respectively. Note in this example and the rest of the book, ISO units are used for measurements without explicit explanation. First, the following coordinate frames are set up: • Universe coordinate frame fU g W OX Y Z. Its origin O is at the bottom of the rod (OA ). Three fixed principal axes X , Y , and Z are along the base vectors iO , jO, and k. • Body frame of the rod fAg W OA xOA yOA zOA . Its origin OA coincides with O, and zOA coincides with the Z axis. The yOA axis is the projection of the rod on the XOY plane. Its angle with the Y axis describes the angular displacement of the rod around the Z axis. The xOA axis is then decided by the right-hand rule. • Body frame of the box fBg W OB xOB yOB zOB . Its origin OB is at the joint at the end of the rod, OB . The xOB axis is along the rotation axis, which is parallel to xOA . The yOB and zOB axes are perpendicular to the sides GH and EH of the rectangular box, respectively.
32
2 Kinematics
zˆA
zˆB β
cos β
β
-sin β
Z
X
iˆ
zˆA
O
cosθ
θ
xˆ B
kˆ
sinθ
-sin θ
cos θ
θ ˆj
xˆ A
yˆ B
OB
sin β
cos β
yˆ A
xˆ A
yˆ A
Y
Fig. 2.9 Example: relations between the axes of the frames
The relations between the axes of fAg, fBg, and fU g are sketched in Fig. 2.9. Observing and describing fAg in fU g, we have pOA D Œ0 0 0T ;
(2.18)
xOA D cos iO C sin jO C 0kO D Œcos sin 0T ; yOA D sin iO C cos jO C 0kO D Œ sin cos 0T ; zOA D kO D Œ0 0 1T ; and 2 3 cos sin 0 RA D ŒxOA yOA zOA D 4 sin cos 0 5 : 0 0 1
(2.19)
Observing and describing fBg in fAg, we have pOB D Œ0 b aT ;
A
(2.20)
xOB D Œ1 0 0T ; xOB ==xOA ;
A
yOB D Œ0 cos ˇ sin ˇT
A
zOB D Œ0 sin ˇ cos ˇT ; and 2 3 1 0 0 A R D ŒAxOB AyOB AzOB D 4 0 cos ˇ sin ˇ 5 : B 0 sin ˇ cos ˇ A
(2.21)
2.2 Position and Orientation
33
Note rotation matrices RA and BAR can also be obtained using basic rotation matrices defined by (2.16) and (2.17): RA D Rz ./; BAR D Rx .ˇ/ Now examine the description of frame fBg in universe frame fU g: pOB D RA ApOB D Œb sin b cos aT ;
(2.22)
xOB D xOA D RA xOB ; A
yOB D 0 xOA C cos ˇ yOA C sin ˇ zOA D RA Œ0 cos ˇ sin ˇT D RA AyOB ; zOB D 0 xOA sin ˇ yOA C cos ˇ zOA D RA Œ0 sin ˇ cos ˇT D RA AzOB ; and 2 3 cos sin cos ˇ sin sin ˇ RB D ŒxOB yOB zOB D RA BAR D 4 sin cos cos ˇ cos sin ˇ 5 : (2.23) 0 sin ˇ cos ˇ Thus (2.18), (2.19), (2.22), and (2.23) give the positions and orientations of the rod and the box as the functions of the rotation angles and ˇ, respectively. Example 2.4. This example is the continuation of Example 2.2.3. Our task is to (1) find the position of point E and an end of edge EH of the box and (2) derive the rotational axis of frame fBg and the angular displacement of the frame around this axis with respect to the universe frame fU g W OX Y Z shown in Fig. 2.8. From Fig. 2.8 the expressions for the position of E in frames fBg and fU g respectively are pE D Œ0 c d T ;
(2.24)
pE D pOB C RB pE :
(2.25)
B
B
Terms pOB and RB are derived in (2.22) and (2.23), respectively, and BpE is given in (2.24). Substituting them into (2.25), we have 2 3 .b C c cos ˇ d sin ˇ/ sin pE D 4 .b c cos ˇ d sin ˇ/ cos 5 : a c sin ˇ C d cos ˇ The equivalent rotational axis (OrB ) and the angular displacement (B ) for frame fBg with respect to frame fU g can be found from applying (2.10) and (2.11) on the rotation matrix RB in (2.23): B D cos1 rO D
cos .1 C cos ˇ/ C cos ˇ 1 2
Œsin ˇ.1Csin / sin ˇ sin sin .1Ccos ˇ/T ; sin B ¤ 0: 2 sin B
34
2 Kinematics
2.3 Velocity In the previous sections, the methods for determining the position and orientation of a rigid body were discussed. If position and orientation change with time, two types of velocities are defined to quantify their rate of change: • Linear velocity: This is the velocity of a fixed point in a body. It is normally referred to as the velocity of the origin of the body frame. • Angular velocity: This refers to the angular velocity of the body around an instantaneous rotation axis. It is related to the rate of change of the a frame’s rotation matrix with respect to time. The magnitude of the velocity is called the speed. For the linear velocity v and the angular velocity !, the corresponding linear and angular speeds are kvk and k!k, respectively.
2.3.1 Linear Velocity Referring to Fig. 2.10, vA is the linear velocity of a body represented by body frame fAg. Using the notations introduced in Sect. 2.2.1, it is observed and described in universe frame fU g. By definition, dpOA : vA D pPOA D dt By default, velocity means instantaneous velocity, which is the velocity at the instant of concern. Sometimes we might only be concerned with the average
aC
vC
zˆA
C
kˆ {U} Fig. 2.10 Linear and angular velocity of a rigid body
iˆ X
yˆA
OA
pC pO
A
ˆj O
aA
pC
A
vA Z
{A}
Y
xˆA
ωA
αA
2.3 Velocity
35
velocity, the ratio between the displacement and the time duration. For example, the average velocity of frame fAg from time instant ti to tf is defined as vA D
pOA .tf / pOA .ti / : tf ti
The velocity of a point (C ) in a body is given by vC D pPC D
dpC dt
As pC D pOA C pC/A ; pC/A D RA ApC ; then vC D pPOA C RA AvC C RPA ApC D vA C vC/A C RPA ApC :
(2.26)
vC consists of the following three parts: • vA D pPOA : due to the linear motion (translation) of frame fAg; • vC/A D RA AvC : due to the motion of point C with respect to frame fAg; • RPA ApC : due to the rotation of frame fAg. In some books, this is called the apparent velocity. RPA is the rate of change with time of the orientation of frame fAg. It will be shown later that it is a skew-symmetric matrix from which the angular velocity of a body can be obtained. There are two special cases regarding the velocity of point C . • When RPA D 0 and vA ¤ 0: vC D vA C RA AvC D vA C vC/A : In this case, frame fAg only translates with velocity vA in fU g. The velocity of C is the sum of the linear velocity of frame fAg and velocity of C with respect to fAg. • When RPA ¤ 0 and vA D 0: vC D RA AvC C RPA ApC D vC/A C RPA ApC : In this case, frame fAg only rotates in fU g. The velocity of C is the sum of the linear velocity caused by the rotation of frame fAg and the velocity of C with respect to fAg.
36
2 Kinematics
Fig. 2.11 A rigid body rotating around a fixed axis
kˆ
Z b a
O iˆ
ω
θ
ˆj Y
X
2.3.2 Angular Velocity Angular velocity is a concept unique to rigid bodies. It can be easily understood if the rigid body rotates about a fixed axis. As shown in Fig. 2.11, a rigid body rotates about the X axis of the universe frame. Its angular displacement () is represented by the angle between any fixed line (e.g., ab) on the body projected on the plane perpendicular to the X axis (Y Z plane) and a principal axis (y) O on the same plane. The angular velocity ! is then given by !D
d O i D P iO: dt
However, it is not so straightforward to describe the angular velocity if the rotation axis also changes with time. As shown in (2.26), angular velocity is related to the rate of change of the rotation matrix with respect to time. This provides a mechanism to find it. According to the first principle of differentiation, the derivative of RA at time instant t is RA .t C t/ RA .t/ RPA .t/ D lim : (2.27) t !0 t Assuming that the instantaneous rotation axis is rO and the angular displacement is during time duration t, then RA .t C t/ D Rr ./RA .t/;
(2.28)
where 2
3 2 3 rx2 rx ry rx rz cos rz sin ry sin 6 7 Rr . / D .1 cos /4 rx ry ry2 ry rz 5C4 rz sin cos rx sin 5 2 ry sin rx sin cos rx rz ry rz rz
as defined in (2.8).
2.3 Velocity
37
When t ! 0, ! 0, cos 1;
sin ;
and Rr ./ is rewritten as 3 1 rz ry Rr ./D 4 rz 1 rx 5 : ry rx 1 2
(2.29)
Substituting (2.28) and (2.29) into (2.27), we have 3 ry rz t t 5 RA .t/ RPA .t/ D lim 4 rz 0 rx t t t !0 ry t rx t 0 2 3 0 rz P ry P D 4 rz P 0 rx P 5 RA .t/: ry P rx P 0 2
0
(2.30)
Define !A D P rO D Œ!x !y !z T ;
(2.31)
where P !x D rx ; and
!y D ry P ;
!z D rz P
3 0 !z !y ˝A D 4 !z 0 !x 5 : !y !x 0 2
(2.32)
Equation (2.30) is rewritten as RPA .t/ D ˝A RA .t/
(2.33)
˝A D RPA .t/RA1 .t/ D RPA .t/RAT .t/:
(2.34)
and Note that RA1 .t/ D RAT .t/ is used in the above derivation. Also note that ˝A is a skew-symmetric matrix: ˝AT C ˝A D 0
38
2 Kinematics
Matrix ˝A is fully determined by the components of !A . Usually this relationship is denoted as
˝A D Œ!A
or ˝A D S.!A /:
The second notation stands for a skew-symmetric matrix determined by !A as defined in (2.31). Consider a body rotating around the X axis as shown in Fig. 2.11. The rotation matrix is given by (2.17), which is reproduced below: 2
3 1 0 0 RA D Rx ./ D 4 0 cos sin 5 : 0 sin cos Then 2
32 3 0 0 0 1 0 0 ˝A D RPA RAT D 4 0 P sin P cos 5 4 0 cos sin 5 0 P cos P sin 0 sin cos 2 3 0 0 0 D 4 0 0 P 5 : 0 P 0 So P 0 0T !A D Œ1
or !A D P iO ;
which is exactly the angular velocity around the X axis. Equation (2.34) can also be derived from the fact that RA .t/RAT .t/ D I 3 : Differentiating it with time t, we have RPA .t/RAT .t/ C RA .t/RPAT .t/ D 0 or .RPA .t/RAT .t//T C RPA .t/RAT .t/ D 0: Define
˝A D RPA .t/RAT .t/I then ˝AT C ˝A D 0: Thus ˝A is a skew-symmetric matrix.
2.3 Velocity
39
Next, we discuss the angular velocity of one frame relative to another frame. Taking frames fAg and fBg as an example, we have ˝B D ŒA!B D BARP BART
A
RP D A˝B BAR;
A B
(2.35)
where A!B is the angular velocity of fBg with respect to fAg, and A˝B is the corresponding skew-symmetric matrix. There is an interesting relation among !B , !A , and A!B : !B D !A C !B/A D !A C RA A!B :
(2.36) (2.37)
This means that if the relative angular velocity of body B with respect to body A and the angular velocity of the latter have the same description frame (fU g), a simple addition of these velocities is the angular velocity of body B with respect to the description frame (fU g). The following is the proof of this relation. Note RB D RA BAR: Differentiating the above and multiplying it by RBT , we have P BT RPB RBT D ˝B D .RPA BAR C RA BAR/R D .˝A RA BAR C RA A˝B BAR/RBT D .˝A RB C RA A˝B BAR/RBT D ˝A C RA A˝B .RB ABR/T D ˝A C RA A˝B RAT : From (1.16), for any vector x 2 R3 , RA A˝B RAT x D RA .A!B .RAT x// D .RA A!B / .RA RAT x/ D .RA A!B /x D S.RA A!B /x; which means that RA A˝B RAT D RA A!B . Substituting it into (2.38), we have Œ!B D Œ!A C ŒRA A!B ; and then (2.36) and (2.37) are derived.
(2.38)
40
2 Kinematics
Fig. 2.12 Relation between angular velocities
{B}
zˆA
zˆB B
OB
A
ω1 Z
yˆB
xˆB
{A}
ω2
yˆA
OA
{U}
xˆA Y
O X
This relation is also valid if the description frame is any frame other than the universe frame fU g. For example, if the description frame is fDg, then !B D D!A CD !B/A :
D
(2.39)
What follows is an example to show the application of this relation. Referring to Fig. 2.12, a plate is rotating with an angular speed of !2 about bar AB of the frame. At the mean time, the frame rotates with an angular speed of !1 about axis zOA . The bar and the plate are represented by body frames fAg and fBg, respectively. Assume that AB is parallel to yOA ; then !A D !1 zOA !B/A D !2 xOB D !2 yOA and !B D !A C !B/A D !1 zOA C !2 yOA :
2.3.3 Relation Between Linear and Angular Velocities Consider the velocity of point C in Fig. 2.10. From its expression given in (2.26) and the definition of angular velocity, we have vC D vA C RA AvC C ˝A RA ApC D vA C vC/A C !A pC/A :
(2.40)
2.3 Velocity Fig. 2.13 Relation between linear and angular velocities in a rotating body
41
C zˆA = ωA × zˆA
vC = ωA × p C/A = ωA × ( RAApc) pC
A
zˆA
ωA
yˆA OA Z
yˆ A = ωA × yˆ A
xˆA xˆA = ωA × xˆ A O
Y
X
If C is fixed on a body that only rotates, ApC is a constant, AvC D 0, vA D 0, and vC D pPC/A D !A pC/A :
(2.41)
This shows that the linear velocity of a point fixed on a rotating body is a cross product of the body’s angular velocity and the position vector of that point. This conclusion can also be found in (2.30) and (2.33), where xPOA D ˝A xOA D !A xOA ;
(2.42)
yPOA D ˝A yOA D !A yOA ;
(2.43)
zPOA D ˝A zOA D !A zOA :
(2.44)
Each of them corresponds to the linear velocity of the tip of the corresponding principal axis rotating with angular velocity !A . The above relations are also depicted schematically in Fig. 2.13. An interesting relation between a vector and its derivative can be found from (2.41)–(2.44). Vectors pC/A , xOA , yOA , and zOA are all fixed with respect to frame fAg. They are all described in universe frame fU g. Using b to represent those vectors, their derivatives have the same expression: bP D ˝A b D !A b: (2.45) This relation is also schematically depicted in Fig. 2.14. In a more general case when the motion of vector b with respect to frame fAg is considered, the derivative of b corresponds to the last two terms of (2.40) if treating b as pC/A : (2.46) bP D bP/A C !A b:
42
2 Kinematics
Fig. 2.14 Derivative of a vector fixed in a frame
ωA
zˆA
EF = ωA × EF
yˆA F
OA E
Z xˆA O
Y
X
According to the vector notations defined in Sect. 2.2.1, bP/A refers to the derivative of b observed in fAg but described in fU g. In some books, it is denoted as ıb to ıt db P differentiate it from b D dt . For b D pC/A in (2.40), bP/A D pPC/A D vC/A D RA ApPC : The relations presented in (2.45) and (2.46) play a very important role in studying the kinematics and dynamics of rigid bodies.
2.4 Acceleration Velocity is the rate of change of position or orientation with time. Likewise, acceleration is the measurement of the rate of change of velocity with time. There are two types of acceleration, linear acceleration corresponding to linear velocity and angular acceleration corresponding to angular velocity. Take frame fAg in Fig. 2.10 as an example. Its linear and angular accelerations, represented by aA and ˛A , respectively, are the time derivatives of its linear and angular velocities, respectively: aA D vPA ; ˛A D !PA : Consider !B/A D RA A!B :
2.4 Acceleration
43
Differentiating it with respect to time t, we have !PB/A D RPA A!B C RA A!PB D !A .RA A!B / C RA A˛B D !A !B/A C ˛B/A ;
(2.47)
where
˛B D A!PB ;
A
˛B/A D RA A˛B : Differentiating (2.37) with respect to time t and considering (2.47), we have the angular acceleration of frame fBg: ˛B D !PB D ˛A C !A !B/A C ˛B/A : Next let us examine the acceleration of point C with respect to fU g as shown in Fig. 2.10. From the definition, its acceleration is obtained by differentiating vC with respect to time t: aC D vPC : vC is given in (2.40), which is reproduced below for completeness: vC D vA C RA AvC C !A .RA ApC /: Taking the time derivative of vC , we have aC D aA C RA AvPC C RPA AvC C !PA .RA ApC / C !A .RA ApPC C RPA ApC /:
(2.48)
Defining
aC D AvPC
A
as the acceleration of point C relative to fAg, and noting that RPA AvC D ˝A RA AvC D !A .RA AvC /; RPA ApC D ˝A RA ApC D !A .RA ApC /; pPC D AvC ;
A
(2.48) can be rewritten as aC D aA C RA AaC C 2!A .RA AvC / C ˛A .RA ApC / C !A .!A .RA ApC // D aA C aC/A C 2!A vC/A C ˛A pC/A C !A .!A pC/A /:
(2.49)
44
2 Kinematics
This is a comprehensive expression of the acceleration of a point that consists of the following parts: • aA : acceleration of frame fAg; • aC/A D RA AaC : acceleration of C with respect to fAg but described in universe frame fU g; • 2!A vC/A : Coriolis acceleration reflecting the effect of interaction between the linear motion of point C and the rotation of frame fAg; • ˛A pC/A : acceleration resulting from the angular acceleration of frame fAg; it corresponds to the tangent acceleration of a particle in circular motion; • !A .!A pC/A /: acceleration related to the angular velocity of frame fAg; it corresponds to the centripetal acceleration of a particle in circular motion. If C is the origin of a frame, then (2.49) can also be used to calculate the linear acceleration of the frame. Up to now, we have covered all the essential concepts of position, velocity, and acceleration of a body frame and a point. As a summary, our steps taken for introducing these concepts are listed below. Step 1: Set up coordinate frames for observing and describing the position and orientation of the body and the position of the point (Sects. 2.1 and 2.2.1). Step 2: Obtain the position of a point, or position and orientation of a body, as functions of time and system parameters (Sect. 2.2). Step 3: Determine the linear velocity from the differentiation of position with respect to time (Sect. 2.3). Step 4: Determine the angular velocity from the product of the differentiation of the rotational matrix with respect to time and its transpose (Sect. 2.3). Step 5: Determine the linear acceleration and angular acceleration differentiating the linear and angular velocities, respectively (Sect. 2.4). The above steps are based on the first principle of differential relations among position, velocity, and acceleration and can be used to analyze the kinematics of rigid bodies. Alternatively, the formulas developed in the above sections can be readily used. They are grouped below for easy reference. • For point C in frame fAg: pC DpOA C pC/A ;
(2.50)
vC D vA C vC/A C !A pC/A ;
(2.51)
aC D aA C aC/A C 2!A vC/A C ˛A pC/A C !A .!A pC/A /:
(2.52)
• For frame fBg relative to frame fAg: – Angular motion: RPA D ˝A RA ;
(2.53)
˝A D RPA RA ;
(2.54)
T
2.4 Acceleration
45
3 0 !z !y ˝A D 4 !z 0 !x 5 ; !y !x 0
(2.55)
!A D Œ!x !y !z T ;
(2.56)
RB D R R;
(2.57)
!B/A D RA A!B ;
(2.58)
2
A A B
!B D !A C !B/A ;
(2.59)
˛B D A!PB ;
(2.60)
A
˛B/A D RA ˛B ;
(2.61)
!PB/A D !A !B/A C ˛B/A ;
(2.62)
A
˛B D !PB D ˛A C !A !B/A C ˛B/A :
(2.63)
– Linear motion: This is also governed by (2.50)–(2.52), replacing point C with OB . • For vector b in frame fAg: – When b is fixed with respect to fAg: bP D !A b:
(2.64)
– When b changes with respect to fAg: bP D bP/A C !A b; P bP/A D RA Ab:
(2.65) (2.66)
Remark. • The vector notations introduced in Sect. 2.2.1 are used for kinematic analysis in this chapter. For example, pC/A stands for the position of C observed in fAg but described in fU g. It is related to the position of C observed and described in fAg through the following transformation: pC/A D RA ApC ; where RA is the rotation matrix of fAg with respect to fU g. This rule is also applicable for other vectors like vC/A , aC/A , !B/A , and ˛B/A . • The terms in these formulas are related to physical properties like rotational axes and basic kinematic variables (relative position, velocities, and accelerations, ApC , A vC , AaC , A!B , A˛B , etc.), which in many applications can be derived from analyzing simple linear or circular motions.
46
2 Kinematics
Special Cases • Circular motion Referring to Fig. 2.2, a cylindrical coordinate frame is set up to describe circular motions. The position of OA is pOA D r eOr C zeOz ; where eOr D cos iO C sin jO; O eOz D k; eO D eOz eOr D sin iO C cos jO
(2.67)
are the axes of the cylindrical frame attached to point OA . The following relations between ePOr and ePO can be derived directly from the above equations: ePOr D P sin iO C P cos jO D P eO ;
(2.68)
ePO D P cos iO P sin jO D P eOr :
(2.69)
So the derivative of eOr is a vector aligning with eO , and vice versa. Differentiating pOA with respect to time t and considering relations in (2.68) and (2.69), we have the velocity and acceleration of OA : vA D rP eOr C r P eO C zPeOz ; P eO C zReOz : aA D .Rr r P 2 /eOr C .r R C 2rP /
(2.70)
In a circular motion, r is constant and rP D 0, and the above equation is simplified to vA D r P eO C zPeOz ;
(2.71)
aA D r P 2 eOr C r R eO C zReOz :
(2.72)
Velocity vA contains two components: tangent velocity (r P eO ) and velocity along eOz (PzeOz ). Acceleration aA contains three components: tangent acceleration (r R eO ), normal or centripetal acceleration (r P 2 eOr ), and acceleration along eOz (RzeOz ). The unit tangent and normal vectors correspond to eOr and eO , respectively. Obviously (2.70) and (2.72) are similar to (2.52) in terms of format.
2.4 Acceleration
47
zˆA = kˆ
Fig. 2.15 A rigid body in planar motion
{A} Ap
c
OA
Z ˆ k
pO
{U } iˆ
C yˆ A
A
θ
ˆj
xˆA
Y
O X
Since eOz is the fixed rotational axis of frame A, !A D P eOz ; ˛A D R eOz : • Planar motion In many engineering applications, a rigid body’s motion is constrained in a twodimensional plane. The body’s rotation axis is fixed and is perpendicular to the plane. Its angular velocity and angular acceleration are along the rotation axis, and position, linear velocity, and linear acceleration lie on the same plane. Referring to Fig. 2.15, axis zOA of the body frame is chosen to be in the same direction of axis Z of the universe frame, and plane xOA yOA is parallel to plane X Y . O axis to The angle, , is the angular displacement of axis X (Y ) around the zOA (k) make itself align with xOA (yOA ). The rotation matrix of the body is 2
cos RA D Rz ./ D 4 sin 0
3 sin 0 cos 0 5 : 0 1
The angular velocity and the angular acceleration of the body are P T D P k; O !A D Œ0 0 R T D R k: O ˛A D !PA D Œ0 0 • Motion around a fixed point In this case, the origin of the body frame is chosen to be the fixed point, vA D 0 and aA D 0. In what follows, several examples will be provided to apply the kinematic analysis methods described in this section.
48
2 Kinematics
Example 2.5. Referring to Fig. 2.8, the rod rotates about axis zOA (fixed) with angular speed !1 , which increases at rate !P 1 , and the box rotates around axis xOB at constant speed !2 . Our task is to find • The linear and angular velocities and linear and angular accelerations of the box and • The velocity and acceleration of point E with respect to universe frame fU g. Here the rigid body of concern is fBg and the point of concern is E. The first step is to derive the descriptions of the orientation of fBg and the position of E with respect to frame fU g. This was completed in Examples 2.2.3 and 2.2.4. The relevant terms were derived there and are reproduced below for completeness: • Rotation matrix of fBg with respect to fU g 2
3 cos sin cos ˇ sin sin ˇ RB D 4 sin cos cos ˇ cos sin ˇ 5 I 0 sin ˇ cos ˇ
(2.73)
• Position of fBg with respect to fU g: pOB D Œb sin b cos aT I
(2.74)
• Position of frame fBg with respect to fAg: pOB D Œ0 b aT I
A
(2.75)
• Position of fAg with respect to fU g: pOA D Œ0 0 0T I
(2.76)
• Rotation matrix of frame fAg with respect to fU g: 2
3 cos sin 0 RA D 4 sin cos 0 5 I 0 0 1
(2.77)
• Rotation matrix of frame fBg with respect to fAg: 2
3 1 0 0 A R D 4 0 cos ˇ sin ˇ 5 I B 0 sin ˇ cos ˇ
(2.78)
• Position of E with respect to fBg: pE D Œ0 c d T I
B
(2.79)
2.4 Acceleration
49
• Position of E with respect to fU g: 2
3 .b C c cos ˇ d sin ˇ/ sin pE D 4 .b c cos ˇ d sin ˇ/ cos 5 : a c sin ˇ C d cos ˇ
(2.80)
From the given conditions, !1 D P ;
(2.81)
P !2 D ˇ:
(2.82)
The linear velocity of frame fBg (box) is derived by differentiating pOB (2.74) with respect to time: P vB D pPOB D b Œcos sin 0T D b!1 Œcos sin 0T ; and the linear acceleration of the box is aB D vPB D b !P 1 Œcos sin 0T b!1 ŒP sin P cos 0T D b !P 1 Œcos sin 0T b!1 Œ!1 sin !1 cos 0T D bŒ!P 1 cos C !12 sin !P 1 sin !12 cos 0T : First examine the angular velocities and acceleration of fBg. From the definition of angular velocity, ˝B D Œ!B D RPB RBT ; (2.83) where RB is given in (2.73). Differentiating RB with respect to time and noting P D !1 and ˇP D !2 , we have 2
3 !1 sin !1 cos cosˇ C !2 sin sinˇ !1 cos sinˇ C !2 sin cosˇ RPB D4 !1 cos !1 sin cosˇ !2 cos sinˇ !1 sin sinˇ !2 cos cosˇ 5 : 0 !2 cosˇ !2 sinˇ Note (2.81) and (2.82) are used in the above derivation. Substituting the above into (2.83), we have 3 !2 sin 0 !1 ˝B D Œ!B D 4 !1 0 !2 cos 5 : !2 sin !2 cos 0 2
From the definition (2.31) of angular velocity, !B D Œ!2 cos !2 sin !1 T :
(2.84)
50
2 Kinematics
Differentiating !B and noting that !2 is a constant, the angular acceleration of fBg is obtained: P 2 cos !P 1 T ˛B D !PB D Œ!2 P sin ! D Œ!1 !2 sin !1 !2 cos !P 1 T : Next, examine the velocity and acceleration of point E. Its velocity is derived directly by differentiating its position, pE (2.80), with respect to time: 3 .b C c cos ˇ d sin ˇ/ cos !1 C .c sin ˇ C d cos ˇ/ sin !2 vE D pPE D 4 .b c cos ˇ d sin ˇ/ sin !1 C .c sin ˇ d cos ˇ/ cos !2 5 : .d sin ˇ C c cos ˇ/!2 2
The acceleration of point E is obtained by differentiating vE with respect to time: aE D ŒaEx aEy aEz T ; where aEx D 2!1 !2 .d cos ˇ C c sin ˇ/ C !22 sin .c cos ˇ d sin ˇ/ C.!12 sin !P 1 cos /.b C c cos ˇ d sin ˇ/; aEy D 2!1 !2 sin .d cos ˇ c sin ˇ/ C !22 cos .c cos ˇ C d sin ˇ/ C.!1 cos C !P 1 sin /.c cos ˇ b C d sin ˇ/; aEz D !22 .d cos ˇ c sin ˇ/: Note that P D !1 , ˇP D !2 , and !P2 D 0 are used in the above derivation. In the above, all the quantities in the task list are derived mainly based on the first principles of differentiation of the expressions for positions or rotation matrices. In what follows, the velocities and accelerations of frame fBg will be obtained from studying its relations relative to other bodies in the mechanism, and the formula listed from (2.50) to (2.52) will be used for this purpose. From the conditions given, the following terms are obtained: • Angular velocity of the rod (fAg) with respect to fU g: !A D Œ0 0 !1 T I
(2.85)
• Angular acceleration of the rod (fAg) with respect to fU g: ˛A D Œ0 0 !P 1 T I
(2.86)
• linear velocity of the rod (fAg) with respect to fU g: vA D 0I
(2.87)
2.4 Acceleration
51
• linear acceleration of the rod (fAg) with respect to fU g: aA D 0I
(2.88)
• Angular velocity of the box (fBg) with respect to the rod (fAg): !B D Œ!2 0 0T I
A
(2.89)
• Angular acceleration of the box (fBg) with respect to the rod (fAg): ˛B D 0:
A
(2.90)
As vA D 0 and AvB D 0, then vB D !A .RA ApOB /:
(2.91)
Substituting (2.85), (2.77), and (2.75) into the above equations, and after some matrix and vector manipulations, we have vB D b!1 Œcos sin 0T : From (2.52), the linear acceleration of fBg is given by aB D aA C RA AaB C 2!A .RA AvB / C ˛A .RA ApOB / C!A .!A .RA ApOB //:
(2.92)
As aA D 0, AvB D 0, and AaB D 0, aB D bŒ!P 1 cos C !12 sin !P 1 sin !12 cos 0T : Note that OB is in a circular motion with zO being the rotational axis and b the radius. vB and aB can also be solved using (2.71) and (2.72): vB D b!1 eO ; aB D b !P 1 eO
(2.93) b!12 eOr :
(2.94)
Here eO is the unit tangent vector corresponding to angular displacement . It is equivalent to eO in (2.67): eO D xOA D Œcos sin 0T ; eOr D yOA D Œ sin cos 0T : Substituting them into (2.93) and (2.94), vB and aB are derived. Describe in the universe frame the angular velocities of frame fAg, and the relative angular velocity of fBg with respect to fAg: O !A D !1 k; !B/A D RA A!B D !2 xOB I
52
2 Kinematics
Fig. 2.16 Example 2.6
Z kˆ
ω p ,α p
zˆN q
{N} xˆP iˆ
{S} yˆB ω s ,α s
zˆP yˆ N
O xˆ N
zˆB
xˆB
yˆP
ˆj
Y
{P}
X ω n ,α n
then !B can be obtained through application of (2.59): !B D !A C !B/A D !1 kO C !2 xOB D Œ!2 cos !2 sin !1 T : The angular acceleration of fBg is given by ˛B D ˛A C !A .RA A!B / C RA A˛B :
(2.95)
Noting that A˛B D 0, and substituting them together with other known terms into (2.95), we have aB D bŒ!P 1 cos C !12 sin !P 1 sin !12 cos 0T ; ˛B D Œ!1 !2 sin !1 !2 cos !P 1 T : Example 2.6. This example is about the motion of a gyrotop (Fig. 2.16), which consists of three rotations called spin, nutation, and precession. At the instant of interest, the spin axis is on the YOZ plane and has an angle with respect to the Z axis of the universe frame fU g W OX Y Z. The direction of precession and nutation are along the Z and X axes. The spin axis is chosen as a principal axis (yOB ) of the body frame of the gyrotop. The angular speeds for the three rotations are !s , !n , and !p , respectively, and the magnitudes of their angular accelerations are ˛s , ˛n , and !p , respectively. Our task is to determine the angular velocity and angular acceleration of the gyrotop with respect to the universe frame. Assume that the rotation of the gyrotop proceeds in the order precession (P), nutation (N), and spin (S). At the start, the body frame of the gyrotop is the universe frame. The frame generated from each rotation can be described with respect to that generated from the previous rotation, and they are given as RP D RotZ .P / (for frame fP g generated from precession), NPR D RotX .N / (for frame fN g generated from nutation), and SNR D RotY .S / (for frame fS g generated from spin), where
2.4 Acceleration
53
O iO , and yOB , P , N , and S are the angular displacements about the rotational axes k, respectively. Frame fS g is actually the body frame of the gyrotop where the axis yOB aligns with the spin axis. The corresponding angular velocities described in the universe frames are denoted by !P , RP P!N , and RN N!S , respectively. From the given conditions: yOB D Œ0 sin cosI
(2.96)
!P D Œ0 0 !p ; angular velocity for precessionI
(2.97)
!PP D Œ0 0 ˛p ; angular acceleration for precessionI
(2.98)
T
T
!N/P D RP !N D Œ!n 0 0 ; angular velocity for nutationI T
P
(2.99)
!PN/P D RP !PN D Œ˛n 0 0 ; angular acceleration for nutationI
(2.100)
!S/N D RN !S D !s yOB ; angular velocity for spinI
(2.101)
!PS/N D RN !PS D ˛s yOB ; angular acceleration for spin:
(2.102)
T
P
N
N
From (2.36) and (2.39) while considering (2.13) and (2.15): !S D !P C !S/P D !P C RP P!S ;
(2.103)
!S D P!N C P!S/N D P!N C NPR N!S :
(2.104)
P
Substituting (2.104) into (2.103), we have !S D !P C !N/P C !S/N :
(2.105)
Using the terms in (2.97), (2.99), and (2.101), we have !S D Œ!n !s sin !p C !s cosT ; which is the angular velocity of the gyrotop with respect to the universe frame. Differentiating (2.103) with respect to time t gives the angular acceleration of the gyrotop: ˛S D !PP C !PS/P C !P !S/P : Substituting (2.104) into the above equation and noting that !S/P D RP P!S , ˛S D !PP C !PS/P C !P .!N/P C !S/N /:
(2.106)
All the terms except for !PS/P on the right-hand side of the above equation are known. Differentiating (2.104) with respect to time t and premultiplying it by RP , we have !PS/P D RP P!PS D RP .P!PN C NPR N!PS C P!N .NPR N!S // D !PN/P C !PS/N C RP .P!N .NPR N!S //:
(2.107)
54
2 Kinematics
According to (1.16), RP .P!N .NPR N!S // D .RP P!N / .RP NPR N!S / D .RP P!N / .RN N!S / D !N/P !S/N ; and (2.107) becomes !PS/P D !PN/P C !PS/N C !N/P !S/N : Substituting it into (2.106), we have ˛S D !PP C !PN/P C !PS/N C !N/P !S/N C !P .!N/P C !S/N /:
(2.108)
Inserting the values listed from (2.97) to (2.102) into the above equation, we have the angular acceleration of the gyrotop with respect to the universe frame: ˛S D Œ˛Sx ˛Sy ˛Sz T ; where ˛Sx D ˛n C ˛s !p !s sin ; ˛Sy D !p .!n C !s / C ˛s sin !n !s cos ; ˛Sz D ˛p C ˛s cos C !n !s sin : Remark. • Our approach for the above velocity and acceleration analysis is based mainly on the manipulations of vectors and rotation matrices related to rigid-body motions. They are denoted with the notations defined in Sect. 2.2.1. The advantage of this method is that once the coordinate frames are set up, kinematic analysis can be done systematically either through the existing procedure and formulas or the first principle of relating the derivative of a rotation matrix to the angular velocity and acceleration of the body. • Equations (2.105) and (2.108) in this example are the kinematic equations governing the motion of a generic gyrotop. Example 2.7. A slider crank mechanism is schematically shown in Fig. 2.17. Given that the crank (Link 1) is at angle ' with angular speed !, determine what its angular acceleration should be in order for the piston (slider) to accelerate to the left at ˛x . The dimensions of the mechanism are shown in the figure. Set up the universe coordinate frame fU g W OX Y Z, where axes X and Y form the plane for the motion of the mechanism and axis Z is perpendicular to the plane according to the right-hand rule. The X Y plane is on the paper, and thus axis Z (denoted as a black dot) points out of the paper. From the given conditions, the angular velocity of Link 1 is O !1 D ! k: (2.109)
2.4 Acceleration
55
Fig. 2.17 Example 2.7
A
Link 1
Y ˆj Z (kˆ)
O
a
ω ϕ X
Link 2
b ωb
θ B
vB,aB
iˆ
The position of joint A is pA D a cos ' iO C a sin ' jO:
(2.110)
Link 1 is in a circular motion, so the velocity of A is O vA D !1 pA D ! kO .a cos ' iO C a sin ' jO/ D a!.cos ' jO sin ' i/: The velocity of joint B is vB D vA C !2 pB/A ;
(2.111)
where !2 D !b kO is the angular velocity of Link 2 and pB/A is the position vector from A to B. From the figure, pB/A D b cos iO C b sin jO;
(2.112)
pB D pA C pB/A D .a cos ' C b cos /iO C .a sin ' C b sin /jO: Substituting (2.112) into (2.111), we have vB D .a!1 sin ' C b!b sin /iO C .a! cos ' C b!b cos '/jO: Next we will solve !b and from the available conditions. Considering that the slider can only travel along the X axis, the Y elements of both vB and pB are zero: a sin ' C b sin D 0; a! cos ' C b!b cos ' D 0: Solving for and !b from the above equations, we have a sin ' ; D arcsin b a cos ' !b D !: b cos
(2.113) (2.114)
Now we are ready to perform acceleration analysis. Assume the acceleration of Link 1 is O (2.115) ˛1 D ˛ k:
56
2 Kinematics
As Link 1 is in a circular motion around axis Z, the acceleration of A is ˛A D ˛1 pA C !1 .!1 pA /: Substituting expressions for pA (2.110), !1 (2.109), and ˛1 (2.115) into the above equation, we have ˛A D a.˛ sin ' C ! 2 cos '/iO C a.˛ cos ' ! 2 sin '/jO:
(2.116)
The acceleration of the slider, aB , is related to the acceleration of A (aA ) through aB D aA C aB/A ;
(2.117)
where aB/A D ˛2 pB/A C !2 .!2 pB/A / is the acceleration of B with respect to the body frame at A. Considering that ˛2 D ˛b kO is the angular acceleration of Link 2, and substituting expressions of !2 and pB/A into the above equation, we have aB/A D b.˛b sin C !b2 cos /iO C b.˛b cos !b2 sin /jO:
(2.118)
Then from (2.116) to (2.118) we have aB D .a˛ sin ' C a! 2 cos ' C b˛b sin C b!b2 cos /iO C .a˛ cos ' a! 2 sin ' C b˛b cos b!b2 sin /jO; where, except for ˛ and ˛b , all the terms on the right-hand side of the equation are known. From aB D ˛x iO , it follows that a˛ sin ' C a! 2 cos ' C b˛b sin C b!b2 cos D ˛x ; a˛ cos ' a! 2 sin ' C b˛b cos b!b2 sin D 0; and ˛D
p2 sin p1 cos ; a sin. C '/
p1 D ˛x a! 2 cos ' b!b2 cos ; p2 D a! 2 sin ' C b!b2 sin ; where and !b are given in (2.113) and (2.114), respectively. This is the angular acceleration needed for the crank (Link 1) to make the slider accelerate at ax to the left.
2.4 Acceleration
57
xˆB
Z
ω1
ω2
zˆB
A
θ α
O
X
B
zˆB
yˆ B B A Y β
O
X Initial Position
Fig. 2.18 Example 2.8
Example 2.8. As shown in Fig. 2.18, axis OA is driven by a motor (not shown) to rotate at a constant angular speed !1 about vertical axis Z. The wheel of radius r rotates around axis OA and travels along the cone surface parallel to axis OA. Point B is the contact point between the wheel and the surface. kOAk D l, the angle between axes OA and Z, is , and OB formsan angle ˛ with OA. The plane formed by OA and OB has an angle displacement ˇ from its initial position. Find the wheel’s angular velocity and its angular speed about axis OA. To describe the motion of the system, set up universe frame OX Y Z at joint O and body frame AxOB yOB zOB at point A shown in the figure. From the assumption, the angular velocity of OA around the Z axis is O ! D !1 k:
58
2 Kinematics
The velocity of point A is thus vA D !1 kO OA: Assume the angular velocity of the wheel is !w ; then the velocity of point A can also be expressed as vA D vB C !w BA; where vB is the velocity of contact point B between the wheel and the surface. Since normally there is no slippage between the wheel and the surface, vB D 0, and thus vA D !w BA and !1 kO OA D !w BA:
(2.119)
Let the angular speed of the wheel about axis OA be !2 . From (2.59) we have !w D !1 kO C !2 zOB D !1 kO C !2
OA : l
(2.120)
OA D OA is used in the above equation. Note that zOB D kOAk l Substituting the above into (2.119), we have
OA BA !1 kO .OA BA/ D !1 kO OB D !2 : l Taking the norm of each side of the above equation, and considering that OA is perpendicular to BA, kOAk D l, and kBAk D r, and the angle between axis Z and OB is ˛ C ˇ, we have !1 kOBk sin.˛ C / D !2 r: But kOBk sin ˛ D r, so from the above equation we have !2 D
sin.˛ C / !1 ; sin ˛
˛ ¤ 0; :
Note that ˛ cannot be zero or when the wheel has a nonzero radius. Then from (2.120) to (2.121) we have the angular velocity of the wheel: sin.˛ C / zOB : !w D !1 kO C sin ˛
(2.121)
2.4 Acceleration
59
Consider that O zOB D sin cos ˇ iO C sin sin ˇ jO C cos kI then O !w D !wx iO C !wy jO C !wz k; where sin.˛ C / sin cos ˇ !1 ; sin ˛ sin.˛ C / sin sin ˇ D !1 ; sin ˛ D .1 C cos /!1 :
!wx D !wy !w z
Chapter 3
Dynamics
3.1 Inertial Properties 3.1.1 Definitions The mass of a particle is the measurement of its resistance against changes in its velocity, a property called inertia. Similarly, a rigid body’s inertia is a measurement of its resistance against changes in its linear and angular velocities.
3.1.1.1 Inertial Properties for Linear Motion The linear motion of a body is represented by that of a point in the body that is normally the COM. As such, the inertia corresponding to linear motion is related to two factors: the mass and the position of the COM. Consider the rigid body in Fig. 3.1, which is a continuum (volume V ) of an infinite number of particles, and its density, .x; y; z/, is a function of the position of the point. The mass (m) of the body is Z Z mD .x; y; z/dV D dm; V
m
and the coordinates of its COM are R R xdm V x.x; y; z/dV D m ; xc D m m R R ydm V y.x; y; z/dV D m ; yc D m m R R z.x; y; z/dV zdm D m : zc D V m m
L. Huang, A Concise Introduction to Mechanics of Rigid Bodies: Multidisciplinary Engineering, DOI 10.1007/978-1-4614-0472-9 3, © Springer Science+Business Media, LLC 2012
61
62
3 Dynamics
Z kˆ
Fig. 3.1 Mass and center of mass (COM) of a rigid body
V
z
dV C(xc, yc,zc )
p pC
{U}
p sin θ
θ
y
O
X
If .x; y; z/ D , a constant, then mD
rz = x + y 2
x
uˆ
P(x, y,z)
2
− gkˆ
Y ˆj
iˆ
Z dV; V
R R xdV V xdV R D V ; xc D V dV V R R ydV ydV V yc D R D V ; V V dV R R zdV zdV V zc D R D V ; V V dV where V is the volume of the body. Define the following vectors to represent the position of point p and the COM: p D Œpx py pz T ; pC D Œxc yc zc T : The COM can be expressed in compact form as R R pdm V pdV pC D D m : V m and it leads to
Z
Z V
.pC p/dV D
m
.pC p/dm D 0:
(3.1)
3.1 Inertial Properties
63
3.1.1.2 Inertial Properties in Relation to Angular Motion The mass moment of inertia and mass product of inertia are the inertial properties in relation to the angular motion of a rigid body. For brevity, they are usually called the moment of inertia and product of inertia, respectively. The moment of inertia is defined with respect to an axis and is related to the mass at a point and its distance to the axis. The moments of inertia with respect to the principal axes of the coordinate frame fU g W OX Y Z are defined as Z Z Ixx D .x; y; z/.y 2 C z2 /dV D .y 2 C z2 /dm; Z Iyy D Z Izz D
V
V
V
.x; y; z/.x 2 C z2 /dV D .x; y; z/.x 2 C y 2 /dV D
Z
m
m
.x 2 C z2 /dm;
Z
m
.x 2 C y 2 /dm:
The product of inertia is defined with respect to a plane and is related to the mass at a point and the product of its coordinates on the principal axes defining the plane. The products of inertia with respect to the planes formed by the principal axes of the coordinate frame fU g W OX Y Z are Z Z Ixy D Iyx D .x; y; z/xydV D xydm; V
Z Ixz D Izx D
V
Z .x; y; z/xzdV D
Z Iyz D Izy D
V
m
m
xzdm;
Z .x; y; z/yzdV D
m
yzdm:
Though generally a triple integral is needed to obtain the inertial properties of a body, this can be avoided if the body is in a special shape. For example, if the body is symmetrical to the Y Z plane, then xc D 0, Ixy D Ixz D 0; if the body is symmetrical to both Y Z and XZ planes, then xc D yc D 0 and Ixy D Ixz D Iyz D 0. In the next section, some rules will also be presented to simplify the calculation of inertial properties. When the axes of the coordinate frame are selected such that Ixy D Ixz D Iyz D 0, they are called the principal axes of inertia of the body. The corresponding moments of inertia Ixx , Iyy , and Izz are called the principal moments of inertia. The moments of inertia and products of inertia can be combined compactly in the so-called inertia matrix: 3 Ixx Ixy Ixz I D 4 Ixy Iyy Iyz 5 : Ixz Iyz Izz 2
It can be seen that the inertia matrix is symmetric.
(3.2)
64
3 Dynamics
When the products of inertia are zero, the inertia matrix is a diagonal matrix: 3 Ixx 0 0 I D 4 0 Iyy 0 5 : 0 0 Izz 2
In this case, the principal axes of the coordinate frame are the principal axes of inertia, and the diagonal elements of the inertia matrix are the principal moments of inertia of the body. Among the three principal moments of inertia are the maximum and the minimum moments of inertia of a body. When a body is degenerated to a thin plate on a two-dimensional plane (X Y plane), there is only the moment of inertia with respect to the axis perpendicular to the plane (Z) and the product of inertia with respect to the plane (X Y plane). The COM is located on the X Y plane. Using the moments of inertia with respect to the axes of the frame being defined, the moment of inertia about an arbitrary axis can be derived. Referring to Fig. 3.1, uO is an arbitrary axis going through a body. The distance from an arbitrary point P to uO is d.x; y; z/ D kpk sin D kOu pk: Therefore, the moment of inertia of the body about uO is Z Iu D
V
Z
.x; y; z/d 2 .x; y; z/dV D
V
Z D
V
.x; y; z/kOu pk2 dV
.x; y; z/.Ou p/T .Ou p/dV:
Substituting rO D Œux uy uz T and p D Œx y zT into the above equation, we have Z Iu D
V
D u2x
.x; y; z/Œ.ux z uz y/2 C .uz x ux z/2 C .ux y uy x/2 dV Z V
Cu2z
.x; y; z/.y 2 C z2 /dV C u2y
Z V
2ux uz
Z V
.x; y; z/.x 2 C z2 /dV
.x; y; z/.x 2 C y 2 /dV 2ux uy Z
Z V
.x; y; z/xzdV 2uy uz
V
Z V
.x; y; z/xydV
.x; y; z/yzdV
and Ir D Ixx u2x C Iyy u2y C Izz u2z 2Ixy ux uy 2Ixz ux uz 2Iyz uy uz ;
(3.3)
3.1 Inertial Properties
65
which is determined by the moments of inertia about the principal axes of the coordinate frame and the products of inertia about the coordinate planes formed by the principal axes. Define a set of new variables: ux XDp ; Ir uy Y Dp ; Ir uz ZDp I Ir
(3.4) (3.5) (3.6)
then from (3.3), Ixx X 2 C Iyy Y 2 C Izz Z 2 2Ixy X Y 2Ixz XZ 2Iyz Y Z D 1:
(3.7)
This is the equation of an ellipse so called inertia ellipse formed with the “points” K.X; Y; Z/ determined by the axis (u) and the body’s moment of inertia about it (Iu ). Though there is no geometric entity corresponding to point K, the vector from the origin (O) of the coordinate frame to K is in the direction of the axis rO D Œux uy uz T . Given any point, K, on the inertia ellipse, the moment of inertia about the axis r D OK is given by IOK D Ir D
1 kOKk2
:
The longest and shortest axes of the ellipse correspond to the minimum and maximum moments of inertia, respectively. If the principal axes of the coordinate frames are the principal axes of inertia (Ixy D Ixz D Iyz D 0) and Ixx D Iyy D Izz , then X 2 C Y 2 C Z2 D
1 1 1 D D : Ixx Ixx Ixx
In this case, the moment of inertia with respect to any of the axes going through the origin of the coordinate frame is the same, and the inertia ellipse now becomes an inertia sphere. The radius of the sphere is the inverse of the square root of the moment of inertia. For example, the moment of inertia of a uniform sphere or a cube with respect to any axis through its center is a constant and can be represented by an inertia sphere. The following example shows how to derive the inertial property of a rigid body. Example 3.1. In this example, we will derive the mass, the COM, the moments of inertia, and the products of inertia of the rigid body in Fig. 3.2. The body is a paraboloid formed by the revolution of a parabola around the Y axis of the coordinate frame. The mass density of the body is assumed to be a constant, .
66
3 Dynamics
Z
Z
z 21 = cy ( On the YZ plane ) ca z1
x2 + z2 = cy
cy
cy − z2
− cy − z2
O
a
r y
O
Y
r
X
p(x,y,z)
p(x, y,z)
− cy
X Fig. 3.2 Example 3.1
As the body is symmetrical to both the X Y and Y Z planes, the COM is located at the Y axis (zc D xc D 0) and the products of inertia are zero (Ixz D Iyz D Ixy D 0). The axes of the coordinate frame are the principal axes of inertia. The remaining terms that need to be calculated are the y coordinate of the COM (yc ) and the moments of inertia Ixx , Iyy , and Izz . Consider a point p.x; y; z/ in the body. It is inside a circular disc with the center at .0; y; 0/ and radius p r D z1 D cy: Then x 2 C z2 r 2 D cy; where 0 y a: Therefore, the volume of the body is ZZZ V D
Z dV D 0
a
Z
p cy p cy
Z pcyz2 p
cyz2
dxdzdy D
2 ca ; 2
and the mass of the body is m D V D
ca2 : 2
The coordinate of the COM at the Y axis is p R a R pcy R cyz2 RRR p p 0 cy cyz2 ydxdzdy ydV 2 D a: yc D D 2 V ca 3 2 Let zD
p cy sin ;
h i : 2 ; 2 2
(3.8)
3.1 Inertial Properties
67
Then, from (3.8), ZZZ
Z dV D
a
Z
0
ZZZ
Z ydV D
a 0
Z
2
Z
p cy cos
2
p cy cos p Z cy cos
2
p cy cos
2
p
cy cos dxddy;
p y cy cos dxddy:
Following the definition of moment of inertia, ZZZ Iyy D
2
Z
2
.x C z /dV D
a
p cy p cy
0
1 c 2 a3 ; 6 ZZZ Z 2 2 D .y C z /dV D
Z
Z pcyz2
p
cyz2
.x 2 C z2 /dxdzdy
D Ixx
a
p cy p cy
0
3a C c ca3 ; 12 ZZZ Z Izz D .x 2 C y 2 /dV D
Z
Z pcyz2 p
cyz2
.y 2 C z2 /dxdzdy
D
0
D
a
Z
p cy p cy
Z pcyz2 p
cyz2
.x 2 C y 2 /dxdzdy
3a C c ca3 : 12
In summary, the mass of the body is mD the COM of the body is
ca2 ; 2
h 2 iT pC D 0 a 0 ; 3
and the inertia matrix is 2 3aCc I D4
12
ca3 0 0
0 1 2 3 c a 6 0
3 0 5: 0 3aCc 3 ca 12
68
3 Dynamics
zˆ
Fig. 3.3 Parallel axes and parallel planes between two coordinate frames
zˆ
zc yˆ
C xˆ O
xc
yc
yˆ
xˆ
3.1.2 Theorems and Rules • Parallel-axis and parallel-plane theorems 0 0 0 Refer to Fig. 3.3. Axes xO , yO , and zO and the COM (C ) form the body coordinate 0 0 0 frame. Another set of axes, xO , yO , and zO , are parallel to x, O y, O and zO, respectively, and pass through point O. Then there are the following relations between the moments of inertia with respect to these two sets of axes: Ixx D Ix 0 x 0 C m.yc2 C z2c /;
(3.9)
z2c /;
(3.10)
Izz D Iz0 z0 C m.xc2 C yc2 /;
(3.11)
Iyy D Iy 0 y 0 C
m.xc2
C
where m is the mass of the body and xc , yc , and zc are the coordinates of the COM in the frame O xO yO O z. This is called the parallel-axis theorem. The parallel-plane theorem concerns the product of inertia of a body with respect to a plane that is parallel to the plane passing through the COM of the body. According to this theorem, the following relations exist between the products of inertia with respect to different planes: Ixy D Ix 0 y 0 C mxc yc ; Ixz D Ix 0 z0 C mxc zc ; Iyz D Iy 0 z0 C myc zc : Let us prove the above theorems. Consider a point in a body, p, whose 0 0 0 0 0 0 coordinates in the frame C xO yO zO are x , y , and z . Its coordinates in the frame
3.1 Inertial Properties
69
0
0
0
O xO yO O z are x C xc , y C yc , and z C zc . From the definitions of moment of inertia and product of inertia, Z Z h i 0 0 2 2 .y C yc /2 C .z C zc /2 dm (3.12) Ixx D .y C z /dm D m
m
Z Z 0 0 D Ix 0 x 0 C m.yc2 C z2c / C 2 yc y dm C zc z dm ; Z Iyy D
2
m
2
.x C z /dm D
Z h m
m
i 0 0 .x C xc /2 C .z C zc /2 dm
D Iy 0 y 0 C m.xc2 C z2c / C 2 xc Z Izz D
m
.x 2 C y 2 /dm D
D Iz z C 0 0
m.xc2
Z Ixy D
m
C
yc2 /
Z xydm D
Ixz D
m
(3.16)
Z Z 0 0 C 2 xc x dm C yc y dm ;
(3.17)
m
Z
Z
m
m
(3.18)
Z
0
x dm C xc
0
m
y dm;
0
Z m
Z
0
x dm C xc
0
m
m
.x C xc /.z C zc /dm
Z yzdm D
m
0
D Ix 0 z0 C mxc zc C zc Iyz D
m
m
0
m
0
z dm;
m
Z
0
y dm C yc 0
(3.21)
m
0
Z
(3.19) (3.20)
.y C yc /.z C zc /dm
D Iy 0 z0 C myc zc C zc
(3.14)
i 0 0 .x C xc /2 C .y C yc /2 dm
Z h
0
x dm C zc
.x C xc /.y C yc /dm
Z xzdm D
Z
(3.15)
D Ix 0 y 0 C mxc yc C yc Z
Z
0 z dm ;
0
m
(3.13)
m
(3.22) 0
z dm:
(3.23)
m
0 0
The COM is at the origin of the frame C xO yO zO . From the definition of COM, this means that Z Z Z 0 0 0 x dm D 0; y dm D 0; and z dm D 0: m
m
m
Substituting them into (3.12)–(3.23), the parallel-axis and parallel-plane theorems are then proved. • Composite-bodies rule. The moment (product) of inertia of a body about an axis (plane) is the sum of the moments of inertia of all its composite bodies about the same axis (plane). This rule is obvious from the definitions of moment of inertia and product of inertia.
70
3 Dynamics
zˆ
Fig. 3.4 Composite bodies
B3
B2 C2
C3
B1 C1
yˆ
O
xˆ For example, as shown in Fig. 3.4, a rigid body consists of three bodies denoted by B1, B2, and B3. Their moments of inertia and the products of inertia with respect to the coordinate frames OX Y Z are Iixx , Iiyy , Iizz , Iixy , Iixz , and Iiyz (i D 1; 2; 3), respectively. The moments of inertia and products of inertia of the body with respect to the same frame are Ixx D
Ixy D
3 X
Iixx ;
Iyy D
3 X
Iiyy ;
Izz D
3 X
i D1
i D1
i D1
3 X
3 X
3 X
Iixy ;
Ixz D
i D1
Iixz ;
i D1
Iyz D
Iizz I
Iiyz :
i D1
The mass of the body is the sum of the masses of the composite bodies, and each coordinate of its COM is the weighted average of the corresponding coordinates of the COMs of the composite bodies. Assume the masses of the composite bodies in Fig. 3.4 are mi and the COMs are Ci .xci ; yci ; zci / (i D 1; 2; 3), respectively. Then the mass (m) of the body is mD
3 X
mi ;
i D1
and the coordinates of the COM (C.xc ; yc ; zc /) are P3 P3 P3 mi xci mi yci mi zci xc D i D1 ; yc D i D1 ; zc D i D1 : m m m The application of the above theorems and rules will be shown in the following examples. Example 3.2. As shown in Fig. 3.5, a uniform rod AB has a length l and mass m. There are two parallel axes perpendicular to the rod, zO1 at point A and zO2 at point D. The distance between A and D is d D 34 l. Calculate the moments of inertia of the rod with respect to zO1 and zO 2 . As the mass of the rod is uniformly distributed, its COM, C , is in the middle of the line AB, AC D CB D 12 l.
3.1 Inertial Properties
71
Fig. 3.5 Example 3.2
zˆ2
zˆ
zˆ1
s A
xˆ C
D
B
At C , set up axis zO parallel to zO1 (Oz2 ) and axis xO along the rod. Assume the area of the base of the rod is s; then the mass density of the rod is D
m : sl
(3.24)
The moment of inertia of the rod with respect to zO is given by Z Izz D
2
m
x dm D
Z
Z
2
V
x dV D
1 2l
12 l
2
x sdx D s
Z
1 2l
12 l
x 2 dx:
Substituting in (3.24) into the above equation, we have Izz D
1 ml 2 : 12
Using the parallel-axis theorem, the moments of inertia with respect to zO1 and zO 2 are, respectively, 1 2 1 Iz1 z1 D Izz C m l D ml 2 ; 2 3 3 2 1 2 1 7 ml 2 : Iz2 z2 D Izz C m d l D Izz C m l l D 2 4 2 48 Example 3.3. In Fig. 3.6, a rigid body consists of a cylinder and a paraboloid that have the same mass density . The inertial properties of the paraboloid were discussed in Example 3.1. Calculate the inertial properties of the body: mass, COM, moments of inertia, and products of inertia with respect to the axes and planes of the coordinate frame OX Y Z. From the symmetry of a body, the products of inertia are zero (Ixy D Ixz D Iyz D 0) and the COM is on the Y axis (xc D zc D 0). For a cylinder, its COM (C2 ), volume (V2 ), and mass (m2 ) are derived as follows: aCb ; 2 p V2 D R2 .b a/ D . ca/2 .b a/ D ca.b a/;
xc2 D zc2 D 0;
yc2 D
m2 D V2 D ca.b a/:
72
3 Dynamics
Z
zˆ′
z21 = cy z1 R = ca O y
a
C1
yˆ ′ C2
b
Y
xˆ ′ X Fig. 3.6 Example 3.3
The moments of inertia of a cylinder with respect to the axes of its body frame at 0 0 0 the COM, C2 x y z , can be derived from existing formulas [5]: p 1 1 1 m2 R2 D ca.b a/. ca/2 D c 2 a2 .b a/; 2 2 2 1 1 m2 Œ3R2 C .b a/2 D ca.b a/Œ3ca C .b a/2 : I2x0x0 D I2z0z0 D 12 12 Using the parallel-axis and parallel-plane theorems, the cylinder’smoments of inertia with respect to coordinate frame OX Y Z are I2
0 yy0
D
I2yy D I2 0 0 D yy
1 c 2 a2 .b a/; 2
I2xx D I2zz D I2 0 0 C m2 yc22 zz
1 .a C b/2 ca.b a/Œ3ca C .b a/2 C ca.b a/ 12 4 1 ca.b a/Œ3ca C .b a/2 C 3.a C b/2 : D 12 According to the composite-bodies rule, the inertia properties of a rigid body are obtained by D
m D m1 C m 2 ; m1 yc1 C m2 yc2 m1 m2 D yc C yc ; yc D m1 C m 2 m1 C m 2 1 m1 C m 2 2
3.2 Momentum
73
Fig. 3.7 Angular momentum
Z kˆ
zˆB {B} vP
dm
p
C
rA pA
pC
{U}
ω
v
P rC
pC/A
yˆB
A vA
xˆB O
ˆj Y
iˆ X Ixx D I1xx C I2xx ; Iyy D I1yy C I2yy ; Izz D I1zz C I2zz ; where m1 is the mass, yc1 is the y coordinate of the COM, and I1xx , I1yy , and I1zz are the moments of inertia of the paraboloid in Example 3.1.
3.2 Momentum Having defined inertia properties,we are ready to define momentum, an important concept of rigid-body dynamics. Linear momentum and angular momentum correspond to linear and angular motions, respectively. Consider a rigid body as shown in Fig. 3.7. To define momentums,two coordinate frames are set up: fU g W OX Y Z is the universe frame and fBg W C xOB yOB zOB is the body frame whose origin is at C , the COM of the body. Point A is the point with respect to which the angular momentum will be defined, point P is an arbitrary point with a differential mass dm, pA , and pC are the position vectors of A and C with respect to the universe frame, respectively. rA and rC are the position vectors from A to P and from C to P , respectively, pC/A is the position vector from A to C , vA and vP are the velocities of points A and P , respectively, and v and ! are the linear and angular velocities of the body. According to the definition of the linear velocity of a body, v is the velocity of point C , the COM of the body. The mass of the body is assumed to be m. Note that the above-defined vectors are observed and described in the universe frame. The notations introduced in Sect. 2.2.1 are used to denote the vectors.
74
3 Dynamics
3.2.1 Linear Momentum The linear momentum of a body is defined as
L D mv;
(3.25)
which is equivalent to the linear momentum of a particle of mass m at the COM. Linear momentum L can be treated as the integration of the linear momentums of all the particles contained in a body: Z LD pdm; P m
where p is the position vector of an arbitrary point and dm is the differential mass at that point. The above equation can be rearranged as R R d.pC m dm/ d. m pdm/ d.pC m/ D D D mpPC D mv; LD dt dt dt since pC is the position vector of the COM and pPC D v.
3.2.2 Angular Momentum Refer to Fig. 3.7. The angular momentum of a body is defined with respect to a point (A) and is the integration of the moments of the linear momentum of the differential masses in the body with respect to that point. The linear momentum (dL) of the differential mass at any point (P ) is dL D vP dm; vP D vA C ! rA : Thus dL D dm.vA C ! rA /: The moment of dL with respect to A is the cross product of position vector rA and dL: dHA D rA dL D rA vP dm D rA .vA C ! rA /dm: The angular momentum of a body with respect to A is the integration of dHA over the mass of the whole body: Z HA D
m
rA vP dm D
Z m
Z rA dm vA C rA .! rA /dm: m
(3.26)
3.2 Momentum
75
Replacing point A with the COM C in (3.26), we have Z Z rC dm v C rC .! rC /dm: HC D m
m
According to the definition of COM: Z
Z m
rC dm D
Z
m
.pC p/dm D mpC
then
m
pdm D 0I
Z HC D
m
rC .! rC /dm:
(3.27)
Describing rC and ! in frame fBg, we have rC D ŒxB yB zB T ;
B
! D Œ!xB !yB !zB T :
B
Then rC D xB xOB C yB yOB C zB zOB ; ! D !xB xOB C !yB yOB C !zB zOB : Substituting them into (3.27), we have HC D HxB xOB C HyB yOB CHzB zOB ; where
Z HxB D !xB
2
m
Z
2
.yB C zB /dm !yB
Z
HyB D !xB Z HzB D !xB
m
xB yB dm C !yB Z xB zB dm !yB
m
m
Z m
Z m
(3.28)
xB yB dm !zB
m
.xB2 C zB2 /dm !zB Z
yB zB dm C !zB
m
xB zB dm;
Z m
yB zB dm;
.xB2 C yB2 /dm:
From the definitions of moment and product of inertia, Z IxB xB D
m
Z IzB zB D
m
.yB2 C zB2 /dm; 2
2
.xB C yB /dm;
IxB zB D IzB xB D
Z IyB yB D
m
.xB2 C zB2 /dm; Z
IxB yB D IyB xB D
Z m
m
xB yB dm;
Z
xB zB dm;
IyB zB D IzB yB D
m
yB zB dm:
76
3 Dynamics
Then HxB D IxB xB !xB IxB yB !yB IxB zB B!z ; HyB D IyB xB !xB C IyB yB !yB IyB zB B!z ; HzB D IzB xB !xB IzB yB !yB C IzB zB B!z : Letting HC D HxB HyB HzB ;
B
(3.29)
! D Œ!xB !yB !zB T ; 2 3 IxB xB IxB yB IxB zB B I D 4 IyB xB IyB yB IyB zB 5 IzB xB IxB yB IzB zB
B
be the angular momentum, the angular velocity, and the inertia matrix defined with respect to frame fBg, (3.29) can be written in the following compact form: HC D BI B !:
B
(3.30)
Note the notations for the vectors in Sect. 2.2.1 are also used for the definition of inertia matrix. BI means the inertia matrix defined and described with respect to frame fBg, whereas I is the inertia matrix defined and described with respect to universe frame fU g. If the axes of fBg are the principal axes of inertia of a body, then 2 3 IxB xB 0 0 B I D 4 0 BIyB yB 0 5 0 0 BIzB zB and HC D ŒIxB xB !xB
B
IyB yB !yB
IzB zB !zB T :
The angular momentum with respect to the COM can also be described with respect to universe frame fU g. Note the rotation matrix of fBg with respect to fU g is given by RB D ŒxOB yOB zOB I then (3.28) is rearranged as HC D RB B HC : Substituting (3.30) into the above equation and considering that RB T RB D RB RBT D I 3 , we have HC D .RB BI RBT /.RB B !/: Define I/B D RB BI RBT ;
(3.31)
3.2 Momentum
77
and note that ! D RB B !; then HC D I/B !:
(3.32)
Like BI , I/B is a symmetric matrix. It is the result of transforming BI through the similar transformation RB , or, in other words, I/B and BI are similar. BI is a constant after the body frame is defined, whereas I/B changes with RB with the rotation of the body. Note that I/B is different from the I defined in (3.2), which is the inertia matrix of a body with respect to universe frame fU g. HC and BHC are the angular momentum about the COM. They can be used to find angular momentums with respect to any other points. Let us study a body’s angular momentum about point A. As shown in Fig. 3.7, rA D rC C pC/A ; vA D v C ! .pC/A / D v ! pC/A : Substituting them into (3.26), we have Z .rC C pC/A /dm .v ! pC/A / HA D m
Z C
m
Noting that
.rC C pC/A / .! .rC C pC/A //dm:
(3.33)
Z
Z m
dm D m;
m
rC dm D 0;
(3.33) is simplified to Z HA D
m
rC .! rC /dm C pC/A .mv/:
From the definition of linear momentum L (3.25) and HC (3.28), HA D HC C pC/A .mv/ D HC C pC/A L:
(3.34)
This concise expression shows that the angular momentum of a body with respect to any point is the sum of the moment of the linear momentum with respect to that point and the angular momentum with respect to the COM. The most computationally intensive part is HC , which can be obtained through (3.30)–(3.32). In summary, what follows are the key concepts and relations of the linear and angular momentums of a rigid body. • Linear momentum
L D mvI
78
3 Dynamics
• Angular momentum – With respect to the COM and described in body frame fBg: HC D BI B !I
B
– With respect to the COM and described in universe frame fU g: HC D I/B !; I/B D RB BI RBT ; ! D RB B!I – With respect to a point A and described in the universe frame: HA D HC C pC/A .mv/ D I/B ! C pC/A .mv/: Remark. If a body can be divided into several small bodies, its momentum (linear or angular) is the sum of those of all the composite bodies. Several examples are provided below to demonstrate the applications of these concepts and relations. Example 3.4 Swinging of arm. This example is based on Example 3.8 in [6]. While standing straight, a man begins swinging his arm at constant frequency (Fig. 3.8). In Fig. 3.8, h is the vertical distance between the COM (C ) and the line connecting the shoulders, d is half the length between the shoulder, and l and m are the length and mass, respectively, of each arm. Compute the angular momentum of the man about his COM in the standard standing position. Define the universe frame fU g W CX Y Z with the origin at the COM of the man. Axis X points outside the paper and is denoted by a black dot. The COMs of the two arms are C1 and C2 , each of which is assumed to be at the middle of the arm. The body frame, C1 xO 1 yO1 zO1 , is shown in the figure. Note that the yO1 axis is parallel to the Y (jO) axis, but xO 1 and zO1 are not necessarily parallel to the X and Z axes, respectively. During the swinging movement, the left and right arms are symmetrical to the X axis. Assume their angular displacements about the X axis are and , and the corresponding angular velocities are P and P , respectively. The angular momentum of the left arm with respect to the COM of the man is HlC D Iy1 y1 P jO C pC1 .mv1 /;
(3.35)
where Iy1 y1 is the moment of inertia of the arm with respect to the yO1 axis, and pC1 and v1 are the position and velocity of C1 , respectively. As the left arm rotates around joint A1 , v1 D P jO r1 ;
(3.36)
3.2 Momentum
79
Z
d
A2
C2
A1
A1 r1
zˆ1
r2 h C
kˆ
Z
kˆ
C1 xˆ 1
l
yˆ 1 Y
iˆ X
ˆj
iˆ X
C2
pC 2 C
−θ
ˆj Y
C1 pC
1
θ
Fig. 3.8 Example 3.4 [6]
where r1 is the vector from A to C1 , which is a function of :
Substituting
l O r1 D .sin iO C cos k/: 2
(3.37)
pC1 D hkO C d jO C r1
(3.38)
together with (3.36)–(3.38) into (3.35), we have P C .jO r1 // HlC D Iy1 y1 P jO C m.p 1 2 h i mdl P O C P Iy y C m l hl cos jO: D .sin iO C cos k/ 1 1 2 4 2 For the right arm, l O .sin iO cos k/; 2 pC2 D hkO d jO C r2 ; r2 D
v2 D P jO r2 :
(3.39) (3.40)
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3 Dynamics
The angular momentum of the right arm with respect to the COM of the man is HrC D Iy2 y2 P jO C pC2 .mv2 / D
2 h i mdl P O P Iy y C m l hl cos jO: .sin iO cos k/ 1 1 2 4 2
Thus the total angular momentum of the man with respect to his COM is O HC D HlC C HrC D mdl P cos k: Note the direction of the angular momentum is along the upward vertical direction (Z axis). This matches our own experience when swinging our arms.
3.3 Force, Moment of Force, Torque, Work, and Power Rigid-body dynamics is about the relation between the change of a rigid body’s motion and the sources of that change. This relation is called an equation of motion, which will be discussed in later parts of this chapter. In previous sections, various terms for the description of rigid-body motions were covered. This section will address the sources of changes in motion – force, moment of force, and important derived quantities like impulse, work, and power. As shown in Fig. 3.9, forces f1 ; f2 ; : : : ; fi ; : : : ; fn act at various locations on a rigid body. Their positions relative to the COM, C , are denoted by r1 ; r2 ; : : : ; ri ;
Z kˆ
f3 f1
fn r3
rn
r1
{B}
r2
C ri
f2
{U} fi
O
Fig. 3.9 Forces acting on a rigid body
X
iˆ
Y ˆj
3.3 Force, Moment of Force, Torque, Work, and Power
81
Z kˆ
Fig. 3.10 Moment, work, and power of force
Line of action
r1
C
ri
Treajectory
ri′
r2 vi
θ
{U}
fi O
X
Y ˆj
iˆ : : : ; rn , respectively. The resultant force is simply the sum of all the forces: F D
n X
fi :
(3.41)
i D1
The moment of force is defined with respect to a point in space. In Fig. 3.9, this point is chosen as the COM of the body. The moment of force fi about C is the cross product of ri and fi : MCi D ri fi ; and the resultant moment of force about C is the sum of all the moments of force about C : n n X X MCi D r i fi : MC D i D1
i D1
When calculating MCi , ri can be a vector from C to any point along the line of action of fi . We will prove that the moment of force is independent of the choice 0 of point on this line. Consider another point whose position vector is ri (Fig. 3.10). Note that 0 0 ri fi ri fi D .ri ri / fi : 0
As ri ri is along the direction of fi , 0
.ri ri / fi D 0 and 0
ri fi D ri fi D MCi :
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3 Dynamics
Z kˆ
Fig. 3.11 Moment of a couple
mC = rBA × fi
{B} rA
D
rBA
A {U} X
rB − f i
fi O
ω
B
d
Y ˆj
iˆ The direction of MCi is along the line perpendicular to the plane formed by ri and fi according to the right-hand rule, and the magnitude of MCi is kMCi k D kfi kkri k sin : A special type of moment of force is the moment of a couple, also called the couple moment or simply torque. It is the moment of force caused by a couple that consists of a pair of forces with the same magnitude but in opposite directions. As the sum of the two forces is zero, the couple has no contribution to the resultant force on the body. The moment of a couple is the sum of the moments of forces in the couple. It can be proved that it is a free vector independent of the point with respect to which the moment of force is defined. In Fig. 3.11, forces fi and fi form a couple. The moment of couple due to fi and fi with respect to point C is D rA fi C rB .fi / D .rA rB / fi D rBA fi ; where rBA D rA rB and A and B are any two different points along the line of action of fi and fi , respectively. Obviously, is independent of C . If the perpendicular distance between fi and fi is d , then the magnitude of the torque is kk D kfi kd:
3.4 Impulse, Work, and Power • Linear impulse and angular impulse Linear impulse is the integration of the resultant force on a body over a period of time. The linear impulse of force f from t1 to t2 is
Impl D
Z
t2
t1
f dt:
3.4 Impulse, Work, and Power
83
Angular impulse is the integration of the moment of force with respect to a point (e.g., C , the COM of a body) over a period of time. The angular impulse of the moment of force MC from t1 and t2 is Z t2 MC dt: ImpA D t1
• Work and power Referring to Fig. 3.9, the work done by a force (fi ) along the trajectory from point r1 to point r2 is Z r2 Z r2 Z s fi dri D fiT dri D kfi k cos ds; (3.42) Wi D r1
0
r1
where s is the length of the trajectory measured from the starting point. If the work is determined by the starting and ending position of the point of action of force only, or, in other words, the work is independent of the path traced by the point of action of force, the force is called a conservative force. Gravitational force is an example of a conservative force. The power is the time derivative of Wi : Pi D
dWi dri D fi T D fi T vi ; dt dt
where vi is the velocity of the point of action of the force. The work and power of a force (fi ) has the following relation: Z T Z T Wi D Pi dt D fiT vi dt: 0
0
The work of a couple, denoted by WC , is the sum of the work of a pair of forces in the couple. Take the couple in Fig. 3.11 as an example. Assume that the points of action for the pair of forces (fi and fi ) have differential displacements drA and drB , respectively, at their positions of interest (rA and rB ). The resulting differential work is dWc D fiT drA fiT drB D fiT .drA drB / D fiT drBA :
(3.43)
Note that the positions of action points A and B are specified by the vectors from an arbitrary point (D) to them, and rBA is the vector from point B to point A. From (3.43), Z rBA 2 Wc D fiT drBA : rBA1
The power of a couple is the time derivative of Wc : Pc D
dWc drBA D fiT D fiT rPBA : dt dt
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3 Dynamics
As rBA is a constant vector with respect to the body frame, rPBA D ! rBA and Pc D fi T .! rBA / D .rBA fi /T ! D mTc !: This means that the power of a couple is the inner product of the moment of the couple and the angular velocity of the body. Given the power of a couple, the work of a couple from t1 to t2 is given by Z Wc D
t2
t1
Pc dt:
3.5 Kinetic Energy, Potential Energy, and Mechanical Energy Conservation Like linear and angular momentum, kinetic energy is a quantity derived from the motion of a rigid body. Consider the kinetic energy of differential mass d m at point P in the body shown in Fig. 3.7: 1 1 dm kvP k2 D dm vPT vP : 2 2
dK D
The kinetic energy of the body is Z
Z dK D
KD
m
1 2 v vP dm: 2 P
(3.44)
Considering that vP D v C ! rC ; and substituting it into (3.44), we have Z
Z KD
dK D m
1 .v C ! rC /T .v C ! rC /dm: 2
From the definition of COM, Z m
then KD
1 1 T mv v C 2 2
rC dm D 0I Z m
.! rC /T .! rC /dm:
(3.45)
3.5 Kinetic Energy, Potential Energy, and Mechanical Energy Conservation
85
From the rule of triple products of vectors (1.6), .! rC /T .! rC / D ! T ŒrC .! rC /: Substituting the above into (3.45), we have KD D
1 T 1 mv v C ! T 2 2
Z rC .! rC /dm D
m
1 T 1 mv v C ! T HC 2 2
1 1 mkvk2 C ! T I/B !: 2 2
(3.46) (3.47)
It can be seen that the kinetic energy of a body consists of a part due to linear motion ( 21 mkvk2 ) and another part due to angular motion ( 12 ! T I/B !). In (3.47), the angular velocity used for deriving the kinetic energy is described in the universe frame. Alternatively, the angular velocity described in the body frame can also be used for the same purpose. Consider HC D RB BHC ; ! D RB B !; RBT RB D RB RBT D I 3 : From (3.46) and (3.47), KD
1 1 1 1 T mv v C B! T BHC D mvT v C B! T BI B!: 2 2 2 2
If the axes of the body frame are the principal axes of inertia, then KD
1 1 mkvk2 C .IxB xB !x2B C IyB yB !y2B C IzB zB !z2B /: 2 2
The potential energy of a body is a function of its position relative to a reference frame. Gravitational potential energy and elastic potential energy are two types of potential energy in rigid-body dynamics. The gravitational potential energy of the body shown in Fig. 3.1 with reference to the X Y plane is defined as Z Ug D
m
zgdm D mgzC ;
where g is the magnitude of the gravitational acceleration, which is in the direction O z is the Z-coordinate of the position of the differential mass dm, m is the of k, mass of the body, and zC is the Z-coordinate of the COM.
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3 Dynamics
Elastic potential energy is related to the spring element when it is elongated or compressed by a distance from its undeformed position. Assume the distance is s; then the elastic potential energy is Ue D
1 2 ks ; 2
where k is the spring constant. Mechanical energy is the sum of kinetic energy and potential energy: E D K C U; where U represents the gravitational potential energy (Ug ) or elastic potential energy (Ue ). The mechanical energy of a system is constant if only the conservative forces like gravitational force and spring elastic force do the work: E D constant: This is the so-called mechanical energy conservation theorem, a very powerful tool for solving dynamics problems. Potential energy is equal to the negative value of the work done by the associated conservative force (e.g., gravitational force or spring force). From the definition of work (3.42), the following relation between the conservative force and the potential energy is obtained: fi D
@Wi T @U T D : @ri @ri
(3.48)
This means that the conservative force can be obtained by differentiating the potential energy with respect to the position of the point of action of the force. Take gravitational energy and gravitational force as an example. A mass m at the position ri D Œxi yi zi has a gravitational energy of Ug D mgzi . Differentiating Ug with respect to ri , we have
@Ug T O D Œ0 0 mg D mg k; @ri
which is exactly the gravitational force. Example 3.5. In this example, we will find the kinetic energy and gravitational potential energy of the man doing swings as described in Example 3.4. For analysis purposes, the man’s body is divided into three parts: left arm, right arm, and other parts combined in the main body. The main body is assumed to be stationary when the arms swing. The frame CX Y Z is used as the reference frame for gravitational energy.
3.5 Kinetic Energy, Potential Energy, and Mechanical Energy Conservation
87
The man’s gravitational energy is Ug D mgpC1z C mgpC2z ; where pC1z and pC2z are the Z elements of vectors pC1 and pC2 , respectively, pC1z D kO T pC1 ; pC2z D kO T pC2 : Substituting the expressions of pC1 and pC2 in (3.38) and (3.39) into the above equations, we have l pC1z D pC2z D h cos ; 2 and thus l Ug D 2mg h cos : 2 As the main body is stationary, the kinetic energy is contributed from the movements of the two arms: KD
1 1 m.vT1 v1 C vT2 v2 / C .Iy1 y1 P 2 C Iy2 y2 P 2 /: 2 2
The expressions v1 and v2 are given in (3.36) and (3.40), respectively. Substituting them into the above equation, we have h1 i 1 K D ml 2 C .Iy1 y1 C Iy2 y2 / P 2 : 4 2 Example 3.6. As shown in Fig. 3.12, a uniform bar with mass m and length 2l is supported by the frictionless horizontal and vertical planes. At its initial position, the bar is stationary and has an angle of 0 with the vertical plane. Calculate the bar’s linear and angular velocities after it is released from the initial position and it has an angle with the vertical plane. Assume the bar retains contact with the plane during the motion. Set up the coordinate frame OX Y Z, where the origin O is at the junction between the horizontal and vertical planes and the X and Y axes are along the directions of the two planes, respectively. According to the right-hand rule, the Z axis (black dot) points outside of the X Y plane. Inertial Property 0
In Example 3.2, the bar’s moment of inertia with the z axis is obtained: Iz 0 z 0 D
1 1 m.2l/2 D ml 2 : 12 3
88
3 Dynamics
Fig. 3.12 Example 3.6
θ0 A
nA
θ
C Y mg
O
X
nB
B
Z
Forces As planes are frictionless, the reaction forces from the planes at the contact points, NA and NB , are parallel to the X and Y axes, respectively: NA D nA iO ; NB D nB jO; where nA and nB are the magnitudes of the forces. The gravitational force at COM C is mg jO. Positions, Velocities, and Accelerations The linear motion of the bar is within the X Y plane and its rotation is about the Z axis. The bar’s linear velocity (v), linear acceleration (a), angular velocity (!), and angular acceleration (˛) are v D vx iO C vy jO; a D ax iO C ay jO; O ! D !z kO D P k; O ˛ D ˛z kO D R k: Note the linear velocity and acceleration refer to those of the COM, C.xC ; yC ; 0/, which are determined by xC D l sin ; yC D l cos :
3.5 Kinetic Energy, Potential Energy, and Mechanical Energy Conservation
89
Then vx D xPC D l cos P D l cos !z ;
(3.49)
vy D yPC D l sin P D l sin !z :
(3.50)
Energy Conservation As NA and NB are perpendicular to the directions of motion of the points of action, A and B, respectively, they do not do the work. The only force doing the work is the gravitational force, mg jO. According to the mechanical energy conservation theorem, the mechanical energy of the bar is conserved. At the initial position, the bar’s mechanical energy is E0 D mgyC0 D mgl cos 0 : When the bar moves to a position where it has an angle with the vertical plane, its kinetic energy is 1 1 mkvk2 C Iz0 z0 !z2 2 2 1 1 1 2 2 D m.v2x C v2y / C ml !z 2 2 3 1 1 D ml 2 !z2 C ml 2 !z2 2 6 2 D ml 2 !z2 ; 3
KD
and its potential energy is Ug D mgyC D mgl cos : The bar’s mechanical energy at this position is thus ED
2 2 2 ml !z C mgl cos : 3
As the mechanical energy is conserved, E D E0 , 2 2 2 ml !z C mgl cos D mgl cos 0 3 and
r !z D
3g .cos cos 0 /: 2l
90
3 Dynamics
Substituting !z into (3.49) and (3.50), we have r 3gl .cos cos 0 /; vx D cos 2 r 3gl vy D sin .cos cos 0 /: 2
(3.51)
So the linear and angular velocities of the bar are r 3gl .cos cos 0 /.cos iO sin jO/; v D vx iO C vy jO D 2 r 3g O .cos cos 0 /k: !D 2l
3.6 Equations of Motion An equation of motion is a mathematical expression of the relation between forces and torques and the changes in motion of a rigid body. Its foundation is Newton’s second law governing the motion of a particle. Assume a force fi acts on the i th particle with mass mi in a rigid body. According to Newton’s second law, the particle’s acceleration ai is determined by fi D mi ai :
(3.52)
This means that the particle accelerates in the direction of the acting force, and the magnitude of the acceleration is proportional to the magnitude of the force. Mass mi is a measurement of the effect of the force on the particle’s acceleration. Given a force fi , the larger the mass, the smaller the acceleration will be. In other words, the mass reflects the particle’s resistance to the change in velocity, a property called inertia. Equation 3.52 can be viewed as an equation of motion of a particle. It is only valid when fi and ai are observed in a so-called inertial frame of reference. In engineering applications, the Earth is an acceptable approximation to the inertial frame of reference, which is also referred to as the universe frame (e.g., fU g as shown in Fig. 2.1) in this book. Several methods for establishing a rigid body’s equations of motion are developed from Newton’s second law. They mainly include: • The Newton–Euler formulation. The equations of motion are established from studying the rate of change the (linear and angular) momentums of a rigid body with respect to time. The relations among inertial properties, forces and moments of forces, velocity, and accelerations for linear and angular motions, are expressed with the equations in a compact form.
3.6 Equations of Motion
91
• D’Alembert’s principle. The concept of inertial force is introduced to formulate the equations of motion for a rigid body in a form similar to an equation of equilibrium in static analysis. • Lagrange’s equations. The equations of motion are established from studying the relations among the motions and associated energies of rigid bodies. Concepts like configuration space, generalized coordinates, generalized forces, and Lagrange function are introduced for the development of the equations of motion.
3.6.1 Newton–Euler Formulation First, let us examine some important definitions and relations regarding the linear momentum and angular momentum of the body shown in Fig. 3.7. The linear momentum of the body is given in (3.25), which is reproduced below for completeness: L D mv; (3.53) where m and v are the mass and the linear velocity of the body, respectively. Note the linear velocity v is the velocity of the COM. Differentiating L with respect to time t, we have LP D mPv D ma; where a D vP is the linear acceleration of the body, which is the acceleration of the COM. From the definition of the COM in (3.1), Z mv D mpPC D pdm; P Z ma D mpRC D
m
m
pdm; R
(3.54)
where m is the mass, pC is the position, v is the linear velocity, and a is the linear acceleration of the body. In (3.54), dm can be viewed as the differential mass of the particle at position p that has acceleration p. R According to Newton’ second law, pdm R equals the total force acting on the particle, which includes the internal forces from other particles. Through the integration of the right-hand side of (3.54), the internal forces that appear in pairs with the same magnitude and opposite directions cancel each other out, leaving only the external resultant force, F , acting on the body: F D LP D ma D mpRC : A detailed definition of F is given by (3.41).
(3.55)
92
3 Dynamics
Equation (3.55) is the equation of the linear motion of a rigid body under the resultant force F . When F D 0, LP D 0; (3.56) which means that the linear momentum of the body is conserved if the resultant force is zero. This is the principle of conservation of linear momentum. Next, we will develop the equation of angular motion of a rigid body under a moment of force. For the rigid body depicted in Fig. 3.7, its angular momentum and its relation to the inertia matrix and the momentum of force are established in (3.31), (3.32), and (3.34). They are reproduced below for completeness: HA D pC/A L C HC ;
(3.57)
HC D I/B !;
(3.58)
I/B D RB I RBT ;
(3.59)
B
where HA and HC are the body’s angular momentums with respect to point A and the COM, respectively, RB is the rotation matrix of the body frame fBg with respect to universe frame fU g, BI is the inertia matrix of the body with respect to its body frame fBg, I/B is similar to fBg through a similar transformation RB , pC/A is the position vector from point A to point C , and ! is the angular velocity of the body. Differentiating (3.57) with respect to time t, we have HPA D pPC/A L C pC/A LP C HPC :
(3.60)
From (3.58) and (3.59), HPC D I/B ˛ C IP/B ! D I/B ˛ C .RPB BI RBT C RB BI RPBT /!;
(3.61)
where ˛ D !P is the angular acceleration. From the relation between the rotation matrix and its time derivative in (2.35), RPB D ˝B RB ; where
2
3 0 !z !y ˝B D S.!/ D 4 !z 0 !x 5 ; !y !x 0
as defined in (2.32). Substituting (3.62) into (3.61), we have HPC D I/B ˛ C ˝B .RB BI RBT /! C .RB BI RBT /˝BT ! D I/B ˛ C ˝B I/B ! I/B .˝B !/
(3.62)
3.6 Equations of Motion
93
D I/B ˛ C ! .I/B !/ I/B .! !/ D I/B ˛ C ! .I/B !/ D I/B ˛ C ! HC :
(3.63)
Substituting (3.63) into (3.60), and considering (3.53) and (3.55), we have HPA D pPC/A .mv/ C pC/A .ma/ C I/B ˛ C ! HC :
(3.64)
Considering that pPC/A .mv/ D .v vA / .mv/ D mv v vA .mv/ D vA .mv/; and substituting it into (3.64), we have HPA D vA .mv/ C pC/A .ma/ C I/B ˛ C ! HC :
(3.65)
HA is defined in (3.26), which is reproduced below: Z rA vP dm: HA D m
Differentiating it with respect to time t and noting that rPA D pP pPA D vP vA ; we have HPA D
Z Z
Z m
D m
rPA vP dm C Z vP vP dm
D vA
Z
rA vPP dm
m
m
Z vA vP dm C
Z
m
vP dm C
m
rA vPP dm:
m
rA vPP dm (3.66)
According to Newton’s second law, vPP dm is the total force acting on particle dm and rA vPP dm is the moment of that force with respect to point A. Then Z MA D
m
rA vPP dm
(3.67)
is the resultant moment of forces with respect to point A. Note the moments of internal forces cancel each other out, and MA is contributed by external forces only. Consider the definition of COM: Z vP dm: (3.68) mv D m
94
3 Dynamics
Substituting (3.67) and (3.68) into (3.66) and considering (3.65), we have MA D HPA C vA .mv/ D pC/A .ma/ C I/B ˛ C ! HC ;
(3.69)
which is the equation of motion governing the angular motion of a rigid body. The equations of motion are simplified in the following special cases. • Point A is the COM (point C ). Now vA D v and pC/A D 0; then, from (3.69), MC D HPC D I/B ˛ C ! HC :
(3.70)
• Point A is the origin of universe frame fU g. In this case, MO D HPO D pC .ma/ C I/B ˛ C ! HC :
(3.71)
• The resultant moment of force is zero. From (3.70) and (3.69), we have HPC D 0;
(3.72)
HPA D 0;
(3.73)
which means that the angular momentum does not change with time. In other words, the angular momentum is conserved if the resultant moment of force is zero. This is the principle of conservation of angular momentum. Equations (3.69)–(3.73), together with (3.55) and (3.56), form a complete set of equations of motion governing the motion of a rigid body. Remark. • Equations (3.56), (3.72) and (3.73) concern the conservation of linear momentum when the resultant force is zero, and the conservation of angular momentum when the resultant moment of force is zero. They provide a powerful tool for dynamic analysis of a system subjected to zero resultant force or zero resultant moment of force. • The instant of interest for dynamic analysis can be chosen such that the body frame and the universe frame are parallel each other. In this case, rotation matrix RB becomes an identity matrix, which will greatly simplify mathematical manipulations and calculations. • When the body is in a planar motion, the equation of motion can be much simplified. Take the planar motion shown in Fig. 2.15 as an example. In (3.55), the terms F and a contain X and Y elements only. In (3.69) and (3.70), the terms !, ˛, HC , MA , and MC contain their Z elements only: !z , HCz , MAz , and
3.6 Equations of Motion
95
yˆB
Fig. 3.13 Equation of motion for planar motion
{B}
eˆφ
Y
xˆB
zˆB
ˆj
C pC
{U } eˆr
φ
Z O kˆ
iˆ
X
ω ,α
MCz , respectively. Since ! HC D 0, (3.55) is in a much simpler form: O MAz kO D pC/A .ma/ C IzB zB ˛z k: • The simplest case is where the body rotates with angular velocity ! and ˛ around a fixed axis as shown in Fig. 3.13, which is redrawn from Fig. 2.15. The X Y plane is on the paper, and the direction of the fixed axis points outside of the paper and is denoted by a black dot. Choose the fixed axis to be the Z axis, and the crossing point between the fixed axis and the X Y plane as the origin (O) of the universe frame. From the kinematic equations for circular motion (2.71) and (2.72), the linear velocity and acceleration of the body are v D r!z eO ; a D r!z2 eOr C r˛z eO ;
(3.74)
where r D kpC k D constantI pC eOr D ; unit radial vectorI r eO D kO eOr ; unit tangent vector: Substituting (3.74) into (3.71) and considering that pC/A D pC D r eOr , we have MOz D .mr 2 C IzB zB /˛z : Since zOB ==Z, from the parallel-axis theorem, Izz D mr 2 C IzB zB I
(3.75)
96
3 Dynamics
then (3.75) can be simplified to MOz D Izz ˛z :
(3.76)
Example 3.7. Referring to the system described in Example 3.6, the assumption that the contacts between the bar and the supporting planes are always maintained is not tenable.Find the angular acceleration of the bar when it is in contact with the vertical plane, and find also its position where it begins to lose contact with the vertical plane. The bar’s acceleration (ax ) when it is in contact with the vertical plane is obtained by differentiating its velocity [vx in (3.51)] with respect to time t: .3 cos 2 cos 0 / P ax D vP x D p sin ; 2 .cos cos 0 /
(3.77)
. where D 3gl 2 From the Newton–Euler formulation, nA D max : When the bar loses contact with the vertical plane, the magnitude of the normal force, nA , becomes zero and, accordingly, ax D 0. Letting ax D 0 in (3.77), and excluding trivial cases like P D 0 and D 0 , we have 3 cos 2 cos 0 D 0 or sin D 0: From the above equation, we have D k (k D 0; 1; : : :), or D cos1 . 23 cos 0 /. These are the angles between the bar and the vertical plane when the bar begins to lose contact with the vertical plane. Example 3.8. This example is based on the example in [6]. A gymnast holding onto a bar rotates from a horizontal position under the action of gravitational force while keeping his body in a straight line, as shown in Fig. 3.14. Assume the gymnast has mass m and length l when the body is fully extended and the mass is uniformly distributed. The gravitational acceleration is g. Determine the equations of motion of the gymnast. First set up the universe frame fU g W OX Y Z and the body frame fBg W C xOB yOB zOB to describe the motion of the gymnast. The origin of fU g, O, is the contact point the gymnast makes with the horizontal bar, and the origin of fBg, C , is the gymnast’s COM. His X and Y axes are at the vertical plane (paper) and his Z axis is perpendicular to the vertical plane and points outside the paper.
3.6 Equations of Motion
97
yˆ
Fig. 3.14 Example 3.8 [6]
ˆj {U}
fy kˆ
zˆ
fx
O
iˆ
xˆ
θ
yˆ B
pC C
l
{B}
zˆB
xˆ B
mg
Kinematic Analysis The position of the gymnast is described by the position of C (COM), pC D Œx y 0T , and his orientation is described by the angle, , between his body with the axis zO. From the given assumptions, pC D
1 l.sin iO cos jO/: 2
The gymnast’s angular and linear velocities and accelerations are O ! D P k; O ˛ D ˛ D R k; 1 P l .cos iO C sin jO/; 2 1 a D vP D pRC D lŒ.R cos P 2 sin /iO C .R sin C P 2 cos /jO: 2 v D pPC D
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3 Dynamics
Force and Moment of Force Analysis The forces acting on the gymnast include the gravitational force, mg jO, and the reaction force from the bar, f D fx iO Cfy jO. The total force acting on the gymnast is F D fx iO C .fy mg/jO: The total moment of force with respect to O is MO D pC .mg jO/ D
l 1 O .sin iO cos jO/ .mg jO/ D mgl sin k: 2 2
As f goes through O, it makes no contribution to the moment of force. The gymnast’s body can be approximated by a uniform slender rod as discussed in Example 3.2. Its moment of inertia with respect to the Z axis is Izz D
1 2 ml : 3
Equations of Motion Based on (3.55) and (3.76), we have the following equations of motion of the gymnast: 1 F D ma ) fx iO C .fy mg/jO D mlŒ.R cos P 2 sin /iO 2 C.R sin C P 2 cos /jO; 1 1 R MOz D Izz ˛z ) mgl sin D ml 2 : 2 3 Rearranging the above equations, we have 3g R D sin ; 2l 1 fx D ml.R cos P 2 sin /; 2 1 fy D mg C ml.R sin C P 2 cos /: 2
(3.78) (3.79) (3.80)
Equations (3.79) and (3.80) can also be used to calculate the reaction forces from the horizontal bar. Example 3.9. The rotor of an airplane engine shown in Fig. 3.15 has mass m D 90 kg. Its moments of inertia about the rotational axis and the axis perpendicular to the rotational axis are 4.76 and 10:8 kgm2, respectively. It has a constant angular
3.6 Equations of Motion
99
zˆA
yˆA
zˆB
NA
A yˆB
l
ω1
C
n
xˆ A
xˆ B
NB
kˆ Z
ˆj
O r
B
iˆ
Y X
Fig. 3.15 Example 3.9
speed of n D 12;000 rpm. The shaft of the rotor is along the rotation axis and is supported by the bearings A and B. The distance between the bearings is l D 1:2 m. Assume the airplane travels at a constant speed of v D 720 km=h in a circle with radius of r D 1;200 m on the horizontal plane. Calculate the reaction forces at both bearings. The body frame fBg W C xOB yOB zB of the rotor is set up at its COM. The yOB axis is along the rotation axis of the rotor, and zOB and xOB are on the plane perpendicular to the rotation axis. The body frame fAg W AxOA yOA zOA of the airplane is set at A. The universe coordinate frame fU g W OX Y Z is also set at a fixed point O on the horizontal plane, where the airplane is traveling in a circle. The Z axis runs along the angular velocity of the airplane. Assume that at the instant of interest, the axes of fBg, fAg, and fU g are parallel to each other, as shown in the figure. Thus, the rotation matrices are RB D RA D BAR D I 3 : From the given conditions, the angular speed of the rotor is !1 D
2 n D 1;256 .rad=s/: 60
The angular velocity of the rotor with respect to the airplane is !B D Œ0 !1 0T D Œ0 1;256 0T .rad=s/:
A
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3 Dynamics
The angular speed of the airplane is !2 D
v D 0:167 .rad=s/; r
and the angular velocity of the airplane is !A D Œ0 0 !2 T D Œ0 0 0:167T .rad=s/: From (2.59), the angular velocity of the rotor is ! D Œ!x !y !z T D !A C RA A!B D Œ0 1; 256 0:167T :
(3.81)
From (2.63), the angular acceleration of the rotor is ˛ D ! D Œ˛x ˛y ˛z T D ˛A C !A .RA A!B / D 0 C !1 kO !2 jO D !1 !2 iO D 209:8iO D Œ209:8 0 0T :
(3.82)
Alternatively, the angular velocity of the rotor can be written directly as ! D !2 kO C !1 yOB :
(3.83)
Differentiating it with respect to time t, we have the rotor’s angular acceleration as
˛ D ˛2 kO C ˛1 yOB C !1 yPOB : Note that ˛1 D !P2 D 0 and yPOB D !A yOB I
then
˛ D !1 !A yOB D !1 !2 kO yOB :
(3.84)
Substituting yOB D jO and !1 and !2 into (3.83) and (3.84), the angular velocity and acceleration of the rotor are obtained. They are identical to the results from (3.81) and (3.82), respectively. As the COM of the rotor is in a uniform circular motion, the linear acceleration of the rotor is a D Œax ay az T D
v2 O i D 33:3iO D Œ33:3 0 0T : r
It is actually the centripetal acceleration associated with the circular motion of the rotor.
3.6 Equations of Motion
101
The axes of the body frame are the principal axes of inertia of the rotor, and thus the inertia matrix of the rotor is 3 2 Ixx 0 0 B I D 4 0 Iyy 0 5 : 0 0 Izz From the conditions given, Iyy D 4:76 kgm2 and Ixx D Izz D 10:8 kgm2. Due to the definitions of the coordinate frames, I/B D BI . Assume the forces at bearings A and B are NA and NB , respectively. From the Newton–Euler formulation, F D NA C NB mg kO D max iO ; MC D
l O l j NA jO NB D HPC D I/B ˛ C ! .I/B ; !/ 2 2
(3.85) (3.86)
where g D 9:8 m=s2 is the gravitational acceleration and HC is the angular momentum of the rotor with respect to its COM. As NA and NB are both perpendicular to yOB , O NA D NAx iO C NAz k; O NB D NBx iO C NBz k: Substituting them into (3.85) and (3.86), we have NAz C NBz D mg; NAx C NBx D max ; l .NA NBx / D 0; 2 x l .NA NBz / D Ixx ˛x C .Izz Iyy /!y !z : 2 z Substituting the values of the terms in the above equations and solving for four unknowns, NAx , NAz , NBx , and NBz , we have NAx D 1;498:5 .N /; NBx D 1;498:5 .N /;
NAz D 391:5 .N /; NBz D 1;273:5 .N /;
which are the elements of the reaction forces of bearings A and B along axes iO O respectively. Their projections along axis jO are zero. To express the reaction and k, forces in vector form, we have NA D Œ1;498:5 0 391:5T .N /; NB D Œ1;498:5 0 1;273:5T .N /:
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3 Dynamics
Note that NA and NB are combined to balance the weight of the rotor and to produce the moment of force required for the rate of change of the angular momentum, HPC , of the rotor. This moment of force is normally called the gyroscopic moment. Another interesting fact is the huge difference between the angular speed of the rotor (!1 D 1;256 .rad=s/ and that of the airplane (!2 D 0:167 .rad=s/. It affects their contribution to the rotor’s angular velocity, O ! D !1 jO C !2 k; and its angular momentum, O HC D I/B ! D Iyy !1 jO C Izz !2 k: As !2 !1 , the part of HC contributed by !2 can be ignored, and HC can be approximated by HC D Iyy !1 jOI then O Iyy !1 jO ! .HC / D ! .I/B !/ D .!1 jO C !2 k/ D !2 kO .Iy y!1 jO/ D Iyy !1 !2 iO: Substituting this into (3.86), the reaction forces at bearings A and B are derived. It can be seen that they are very close to the previous results. Example 3.10. As shown in Fig. 3.16, a man with mass mm walks on a boat with mass mb . Find the horizontal displacement of the boat on the river after the man travels a distance l in the boat. Assume that both the man and the boat are stationary initially and ignore the resistive forces of the water. Set up the universe frame with origin O at the end of the boat when it is at its starting position and the X axis along the horizontal direction. The system, consisting of the man and the boat, is not subject to external forces along the X axis. From the principle of conservation of linear momentum, .mm C mb /xPC0 D .mm C mb /xPC ; where xPC0 and xPC are the velocities of the COM of the whole system at the moment when the man starts to work and any time instant afterward. Based on the given assumption, xPC0 D 0, and thus xPC D 0. Let xm0 , xb0 , xC0 be, respectively, the position of the COMs of the man, the boat, and the whole system when the man starts to walk. They become xm1 , xb1 , and xC1 at the end of the walk. From the principle of composite bodies, mmxm0 C mb xb0 ; mm C m b mmxm1 C mb xb1 : xC1 D mm C m b xC0 D
3.6 Equations of Motion
103
xb0
Fig. 3.16 Example 3.10
xm 0 Δx
Cm A Cb
O
xm 1
xb 1
X Cm
l
A
Start
Cb
B
End
Δx
As xPC D 0, the COM of the whole system remains the same: xC0 D xC1 ) mm xm0 C mb xb0 D mm xm1 C mb xb1 : Rearranging the above equation, we have mm .xm1 xm0 / D mb .xb1 xb0 / D mb x;
(3.87)
where x D xb1 xb0 is the horizontal displacement of the boat. x is also the displacement of point A, from which the man starts to walk; thus xm1 xm0 D l C x: Substituting this into (3.87) and solving for x, we have x D
mm l: mm C m b
This means that the boat moves in a direction opposite to the man’s walking direction. The boat’s displacement will be very small if the mass of the boat is very large compared to the mass of the man. Example 3.11. In Example 2.8, the motion of a wheel traveling along a cone surface is studied. In this example, we will find the normal force acting on the wheel during its motion. Assume the mass of the wheel is m and its moment of inertia about the OA axis is Iw . Ignore the masses of the links and the frictional force between the wheel and the cone surface. For completeness, the main part of Fig. 2.18 is
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3 Dynamics
xˆ B
X iˆ fo
ω1
O
ω2
A
a
θ
α
zˆB
kˆ
Z
fg B
N
Fig. 3.17 Example 3.11
redrawn with some modifications in Fig. 3.17. To simplify calculation, the axes of the universe frame at the instant of interest are set to be parallel to those of the body frame, xOB ==X , yOB ==Y , and zOB ==Z. As shown in the figure, the wheel and the OA axis experience the following external forces: fg (the gravitational force due to the mass of the wheel), N (the normal force at contact point B), and fO (the force at joint O). The following relations exist: fg D mg. sin xOB cos zOB /;
(3.88)
N D nxOB ;
(3.89)
OA D l zOB ;
(3.90)
BA D r xOB ; OB D OA BA D l zOB r xOB ; where n is the magnitude of the force N and g is the gravitational acceleration. Note the universe frame here is different from that used in Example 2.8, and this results in a different expression of the same vector. The total moment of force with respect to point O is MO D OA fg C OB N D l zOB mg. sin xOB cos zOB / C .l zOB r xOB / .nxOB / D .n mg sin /l yOB D .n mg sin /l jO: Note that yOB D jO at the instant of interest is used in the above derivation. The angular velocity of the wheel is now expressed as O C !2 zOB ; !w D !1 .sin iO C cos k/ where !2 is given in Example 2.8, which is reproduced below: !2 D
sin.˛ C / !1 ; sin ˛
˛ ¤ 0; :
(3.91)
3.6 Equations of Motion
105
Thus O C !w D !1 .sin iO C cos k/
sin.˛ C / !1 zOB : sin ˛
(3.92)
Except for zOB , the terms on the right-hand side of the above equation are constant. Thus sin.˛ C / !1 zPB : !P w D sin ˛ From (2.64), O zB zPB D !1 .sin iO C cos k/ O kO D !1 .sin iO C cos k/ D !1 sin jO; and thus !P w D
h sin.˛ C / sin 2 sin.˛ C / sin 2 iT !1 jO D 0 !1 0 : sin ˛ sin ˛
(3.93)
Note that zB D kO at the instant of interest is used in the above derivation, and at this instant, O C sin.˛ C / !1 kO !w D !1 .sin iO C cos k/ sin ˛ sin.˛ C / O k: D !1 Œsin iO C .cos C sin ˛ O the As OA rotates at a constant velocity !1 around the fixed axis sin iO C cos k, acceleration of point A is O D l sin ! 2 Œcos 0 sin T : a D l!12 sin .cos iO sin k/ 1
(3.94)
From the Newton–Euler equation for rotational motion, (3.71), MO D OA .ma/ C I/B !P w C !w .I/B !w /;
(3.95)
where I/B D RB BI RBT , RB is the rotation matrix of the body frame with respect to the universe frame, and BI is the inertia matrix of the body with respect to the body frame. As the universe frame is parallel to the body frame of the wheel at the instant of interest, so RB D I 3 , and then 3 Ixx 0 0 I/B D BI D 4 0 Iyy 0 5 ; 0 0 Iw 2
(3.96)
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3 Dynamics
f3
Zˆ
Zˆ
A
zˆB
A D
D C
C
ω0
{U}
θ
f1
θ
mg xˆB
3 l l 4
O
Yˆ φ
O (X )
yˆ B {B}
Yˆ
f2
Xˆ
yˆB A′
Fig. 3.18 Example 3.12 [6]
where Ixx , Iyy , and Izz are the moments of inertia of the wheel about the axes of the body frame, which are the principal axes of the wheel. Considering the symmetry of the structure of the wheel, it can be assumed that Ixx D Iyy . Substituting (3.92)–(3.94) into (3.95) and considering the definitions of OA (3.90) and I/B (3.96), we have 1 2 2 2 MO D ml !1 sin C Ixx !1 sin.2/ Izz !1 sin .!1 cos C !2 / jO: (3.97) 2 Note that Ixx D Iyy is used in the above derivation. From (3.91) and (3.97), we have 1 .n mg sin /l D ml 2 !12 sin C Ixx !12 sin.2/ Izz !1 sin .!1 cos C !2 /; 2 and thus 1 n D mg sin C ml!12 sin C l 1 Ixx !12 sin.2/ Izz l 1 !1 sin .!1 cos C !2 /; 2 which is the normal force acting on the wheel at contact point B. Example 3.12. This example is taken from Example 9.6 in [6]. Consider a dancer rotating around a pole while holding onto it (Fig. 3.18). The dancer’s feet are placed next to the pole as he rotates around the pole at a constant rate !0 . The angle of inclination with the vertical axis remains constant. Assume the height of the dancer is l. His hands, used to grab the pole, are parallel to the ground and have a
3.6 Equations of Motion
107
distance of 34 l from the feet along the direction of his body. Determine the reaction forces exerted on the dancer at his feet and hands. O with origin O being the Set up the universal coordinate frame fU g W O XO YO Z, contact point of the dancer’s feet with the ground. The ZO axis runs along the pole and the XO and yO axes are on the ground. The dancer’s body frame fBg W C xOB yOB zOB is placed at his COM. The zOB axis runs in the direction from the feet to the head (A) of the dancer, and the xO axis is perpendicular to zOB and lies on the plane formed by 0 OA and OA0 . A is the projection of A on the X Y plane. The yOB axis is determined from the right-hand rule. Assume the dancer has a height l, mass m, and center of gravity C . For simplicity and without loss of generality, the dancer is treated as a slender cylinder with length l. • Kinematic analysis – Positions From the definition of the dancer’s body frame, O zOB D sin rO C cos k;
(3.98)
O xOB D sin rO sin k; where rO is the unit vector along the projection of OA on the X Y plane: rO D
OA0 kOA0 k
D cos iO C sin jO:
(3.99)
The yOB axis can be derived by the right-hand rule: yOB D zOB xOB D sin iO C cos jO: The rotation matrix of the body with respect to the universe frame is 2
3 cos cos sin sin cos RB D ŒxOB yOB zOB D 4 cos sin cos sin sin 5 : sin 0 cos
(3.100)
The position of the COM is pC D
1 1 O l zOB D l.sin rO C cos k/: 2 2
(3.101)
– Velocities The angular velocity of the dancer is O ! D !0 k:
(3.102)
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3 Dynamics
The linear velocityof the dancer is 1 O v D ! pC D !0 kO l.sin cos iO C sin sin jO C cos k/ 2 1 1 (3.103) D l!0 sin . sin iO C cos jO/ D l!0 sin yOB : 2 2 – Accelerations As ! is a constant, the angular acceleration of the dancer is zero: ˛ D !P D 0:
(3.104)
The linear acceleration can be obtained by differentiating v in (3.103) with respect to t while considering that P D !0 : 1 1 a D vP D l!02 sin .cos iO C sin jO/ D l!02 sin rO : 2 2
(3.105)
• Inertial properties As the dancer’s body is treated as a uniform slander rod with mass m, its inertia matrix with respect to the body frame fBg is (Example 3.3.2) 2 3 2 3 Ixx 0 0 1 0 0 1 B I D 4 0 Iyy 0 5 D ml 2 4 0 1 0 5 : (3.106) 12 0 0 0 0 0 0 From (3.59), I/B D RB BI RBT : Substituting RB (3.100) and BI (3.106) into the above equation, we have 2 3 .sin.2/ sin2/=2 .sin.2/ cos /=2 1 cos2 sin2 1 I/B D ml 2 4 .sin.2/ sin2/=2 1 sin2 sin2 .sin.2/ sin /=2 5: 12 .sin.2/ cos /=2 .sin.2/ sin /=2 sin2 (3.107) • Force and moment of force analysis O In The dancer experiences gravitational force acting at his COM, mgk. addition, there are reaction forces acting at his feetand hands. O and the friction The reaction force at his feet consists of the normal force, f1 k, 0 force that is assumed to run in the direction of OA , f2 r. O f1 and f2 are the magnitudes of the forces. The reaction force at the hands acts along the direction opposite to the direction of the friction force at the feet. It can be written as f3 rO , where f3 is its magnitude. The total force acting on the dancer is F D mg kO C f1 kO C f2 rO f3 rO D .f1 mg/kO C .f2 f3 /Or :
(3.108)
3.6 Equations of Motion
109
The moment of force with respect to O is O O C OC .mg k/: MO D OD .f3 r/ Note that OD D
3 l zOB ; 4
OC D
(3.109)
1 l zOB : 2
From (3.109), 3 1 O MO D f3 l zOB rO mgl zOB k: 4 2 Substituting zOB (3.98) and rO (3.99) into the above equation, we have 1 3 MO D mg sin f3 cos l yOB : 2 4
(3.110)
• Application of the Newton–Euler formulation According to the Newton–Euler formulation, F D ma; MO D pC .ma/ C I/B ˛ C ! .I/B !/; where pC , !, a, ˛, I/B , F , and MO are, respectively, given in (3.101), (3.102), (3.105), (3.104), (3.107), (3.108), and (3.110). Substituting them into the above equation, we have f1 mg D 0; 1 f2 f3 D ml!02 sin ; 2 1 1 3 f3 cos mg sin D ml!02 sin 2: 4 2 6 Solving f1 , f2 , and f3 from the above equations, we have f1 D mg; 2 1 mg tan ml 2 !02 sin ; 3 18 2 4 f3 D mg tan C ml 2 !02 sin : 3 9 f2 D
Remark. • To simplify calculations, universe frame fU g can be chosen to be parallel to body frame fBg at the instant of interest, as we did in Example 3.9. In this case, the rotation matrix RB D I 3 . This method can be used if we are only concerned with the magnitudes of the forces that can be proved to be independent of the direction of the universe frame.
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3 Dynamics
• As the dancer rotates around a fixed axis, his motion can be described by using cylindrical coordinate frames, described in Sect. 2.1, and steps to derive the equations of motion will be simplified.
3.6.2 D’Alembert’s Principle D’Alembert’s principle is named after its inventor, French scientist Jean le Rond d’Alembert. In general form, the principle states the following relation among the forces, accelerations, and “virtual” displacements of a group of particles: n X .fi mi ai /T ıri D 0;
(3.111)
i D1
where fi , mi , ai , and ıri are the applied force, mass, acceleration, and virtual displacement of the i th particle, respectively. The “virtual” displacement is an imagined displacement of the particle that is consistent with the constraint of the system of particles. The term mi ai is called the inertial force to differentiate it from the applied force fi . Fi D fi C.mi ai / can be seen as the total force experienced by a particle. As a result, (3.111) can be rewritten as n X i D1
FiT ıri D
n X
ıWi D 0;
i D1
which means that the sum of work (ıWi , called virtual work) of Fi along the virtual displacement ıi is zero. This is analogous to the virtual work principle for a static system. When the constraints on a particle are rigid, (3.111) is equivalent to fi mi ai D 0: This means that the particle is in equilibrium under the applied force fi and the inertial force mi ai , and the above equation is the equation of equilibrium for the particle. After introducing the concept of inertial force, the principles of static equilibrium can be used to analyze the dynamics of particles. This method can be extended to analyze the dynamics of a rigid body. Rewrite the equation of linear motion of a rigid body (3.55): F C .ma/ D 0;
(3.112)
where F is the applied force, a D pRC is the linear acceleration of the body, and ma is the inertial force acting at the COM.
3.6 Equations of Motion
111
f3
Fig. 3.19 Application of D’Alembert principle in Example 3.12
Zˆ
zˆB
A D
− ma C f1
θ
yˆB {B} mg xˆB
O rˆ φ
Yˆ f2
Xˆ
yˆ B A′
The equation of motion for general angular motions (3.69) can be rearranged such that the terms containing the inertial force are on the left-hand side of the equation: MA C pC/A .ma/ D I/B ˛ C ! HC ; (3.113) where pC/A .ma/ is the moment of inertial force .ma/ with respect to point A. If a rigid body rotates around one of its principal axes of inertia, then ! HC D 0. Defining I/B ˛ as the inertial torque due to angular acceleration, (3.113) can be rewritten as MA C pC/A .ma/ C .I/B ˛/ D 0;
(3.114)
which is the equilibrium equation concerning with moments of force. Next the D’Alembert principle is applied to solve the problems in Example 3.12. O f2 rO , The inertial force ma is depicted together with other applied forces, f1 k, O f3 r, O and mg k, in the free body diagram shown in Fig. 3.19. The linear acceleration of the dancer (a) is derived in (3.105). The equation of equilibrium for all the forces is f1 kO C f2 rO f3 rO mg kO ma D 0:
(3.115)
The total moment of force (contributed by both applied forces and inertial force) with respect to O is O C OC .ma/: MO D OD .f3 rO / C OC .mg k/ Applying the Newton–Euler formulation, we have O C OC .ma/ D I/B ˛ C ! .I/B !/: MO D OD .f3 rO / C OC .mg k/ The force magnitudes f1 , f2 , and f3 can then be solved from (3.115) and (3.6.2).
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3 Dynamics
ω0
NA
NB
r yˆ B
C
A
O
τ
B
zˆB xˆB {B}
Y
θ O e mg C
{U}
X
l2
l1
Z Fig. 3.20 Example 3.13
Example 3.13. As shown in Fig. 3.20, a uniform and symmetric disc with mass m is fixed at a horizontal rod supported by two bearings, A and B. The COM of the disc is C , which has a distance e from the rod. Driven by the toque in the direction of rod, the rod and the disc rotate at a constant angular speed of !0 . Find the torque and the forces acting on the bearings when the discs are at the positions as shown in the figure, and identify the position of the disc at which the magnitudes of the reaction forces at the bearings reach the maximum. Assume the inertia matrix of the disc with respect to its body coordinate frame is available and ignore the mass of the rod. First set up the universe coordinate frame fU g W OX Y Z, where the origin of the frame is at the rotation center O, the Y axis is along the direction of the rod, the X axis aligns with the initial position of a fixed radius line on the disc (for angular position measurement) and points to the front view plane, and the Z axis points down in the direction of the gravitational acceleration g. The body frame of the disc, fBg; C xOByOB zOB , is set up at its COM C . In the initial position ( D 0), axes xOB , yOB , and zOB are parallel to axes X , Y , and Z, respectively. As the axes of the body frame fBg are the principal axes of inertia of the disc, the inertia matrix of the disc is I D diag.IxB xB ; IyB yB ; IzB zB /;
B
(3.116)
where diag means a diagonal matrix formed by the elements in the brackets. The rotation matrix of the disc when it has angular displacement around the Y axis is 2 3 cos 0 sin RB D ŒxOB yOB zOB D 4 0 1 0 5: sin 0 cos Then 3 1 0 .I IxB xB / sin.2/ IxB xB cos2 C IzB zB sin2 2 zB zB 5: I/B D RB BI RBT D 4 0 IyB yB 0 2 1 2 .I I / sin.2/ 0 I sin C I cos xB xB xB xB zB zB 2 zB zB 2
3.6 Equations of Motion
113
As the uniform disc is symmetric with respect to the axes of the body frame, IxB xB D IzB zB , and I/B D diag.IxB xB ; IyB yB ; IxB xB / D BI : The following torques and forces act on the rod and the disc: the applied torque O and the reaction forces from bearings A D o jO, the gravitational forces mg k, and B, NA and NB . The angular velocity of the disc is ! D !0 jO D Œ0 !0 0T : The disc’s angular acceleration is ˛ D 0 and its linear acceleration is a D !02 OC ;
(3.117)
O OC D e.cos iO sin k/:
(3.118)
where As a result of the acceleration of the COM C , the inertial force ma is added to the system. According to D’Alembert’s principle, under the inertial force and the applied forces together, the rod and the disc are in a “static equilibrium” state governed by the following equilibrium equation: NA C NB C mg kO ma D 0:
(3.119)
Choosing O as the point with respect to which the moments of force are defined, the following equation for angular motion is obtained from the Newton–Euler formulation: l1 jO NA C l2 jO NB C OC .mgkO ma/ C 0 jO D I/B ˛ C ! .I/B !/: As ˛ D 0 and ! .I/B !/ D 0, then l1 jO NA C l2 jO NB C OC .mg kO ma/ C 0 jO D 0:
(3.120)
Since no reaction forces occur along the rotational axis at the bearings, then O NA D NAx iO C NAz k; O NB D NBx iO C NBz k: Substituting them into (3.117)–(3.120), we have NAx C NBx C me! 2 cos D 0; NAz C NBz C mg me! 2 sin D 0; NAx l1 NBx l2 D 0; NAz l1 NBz l2 D 0; mge cos 0 D 0:
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3 Dynamics
From the above equations, 0 D mge cos ; l2 cos me!02 ; l l1 cos NBx D me!02 ; l l2 l2 sin NAz D mg C me!02 ; l l l1 l1 sin NBz D mg C me!02 ; l l
NAx D
where l D l1 C l2 is the length of the rod between the two bearings. The magnitudes of NA and NB are q q l2 kNA k D NA2x C NA2z D m g 2 C e 2 !04 2ge!02 sin ; l q q l 1 kNB k D NB2x C NB2z D m g 2 C e 2 !04 2ge!02 sin : l Their maximums are l2 m.g C e!02 /; l l1 D m.g C e!02 /; l
kNA kmax D kNB kmax
when D 2 or the COM C reaches its lowest point in the Z direction. Note that the term me! 2 is the centripetal force caused by the displacement of the mass center C from the rotational center O. If the angular speed is very high, the effect of this force on the bearings is quite substantial. It is thus important to put the disc’s COM at the center of rotation. The mass distribution of the disc can also be varied by reducing mass from or adding mass to the disc in a specific way to neutralize the effects of the centripetal force and its moment caused by the misalignment of the COM of the disc. This method is called dynamic balance. Example 3.14. As shown in Fig. 3.21, the system to be studied is the same as that in Example 3.13, except that the COM of the disc now is at the rotation center, but the disc has an angle with the rod. Find the applied torque and the reaction forces in the bearings when the rod and the disc are in the positions shown in the figure. To make it easier to view the system, a perspective view is used to schematically show the system (Fig. 3.21). The universe frame is the same as that in the last example. The origin of the body frame is at the center of the disc O, the yOB axis is perpendicular to the disc and has an angle with the Y axis. The xOB axis aligns with the X axis when D 0, and zOB is determined by the right-hand rule.
3.6 Equations of Motion
115
xˆB
Fig. 3.21 Example 3.14
θ
NA r O
A
X yˆB
NB
φ mg
l1
zˆB l2
B
ω0 Y
τ
Z At the initial state ( D 0), the axes of the body frame are xOB0 D iO D Œ1 0 0T ; yOB0 D cos jO sin kO D Œ0 cos sin T ; zOB0 D xOB yOB D cos kO C sin jO D Œ0 sin cos T : After the rod rotates around the Y axis rotation matrix is 2 cos 4 Ry ./ D 0 sin
by angle , the corresponding basic 3 0 sin 1 0 5: 0 cos
The axes of the body frames are xOB D Ry ./xOB0 D Œcos 0 sin T ; yOB D Ry ./yOB0 D Œ sin sin cos cos sin T ; zOB D Ry ./OzB0 D Œsin cos sin cos cos T ; and the rotation matrix of the body frame is 2
3 cos sin sin sin cos RB D ŒxOB yOB zOB D 4 0 cos sin 5 : sin cos sin cos cos The inertia matrix of the disc with respect to its body frame (BI ) is given by (3.116), and I/B D RB BI 2 a11 4 D a21 a31
RBT
3 a12 a13 a22 a23 5 ; a32 a33
(3.121)
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3 Dynamics
where a11 D IxB xB .cos2 C sin2 cos2/ C IyB yB sin2 sin2; 1 .Ix x IyB yB / sin sin 2; 2 BB 1 D .IyB yB IxB xB / sin 2 sin2; 2
a12 D a21 D a13 D a31
a22 D IxB xB sin2 C IyB yB cos2; a23 D a32 D
1 .Ix x IyB yB / cos sin 2; 2 BB
a33 D IxB xB .sin2 C cos2 cos2/ C IyB yB cos2 sin2: Note that IxB xB D IzB zB is used in the above derivation. The forces and torques applied in the system include the applied torque, D O and the reaction forces from bearings A and B, o jO, the gravitational force, mg k, O O O NA D NAx i C NAz k and NB D NBx iO C NBz k. The linear acceleration of the disc is zero, as the COM is at the rotation axis (a D 0). The angular acceleration of the disc is also zero (˛ D 0) since its angular velocity is constant. Choosing O as the point with respect to which the moments of force are calculated and following the D’Alembert principle, we have the following equilibrium equations for forces and moments of forces: NA C NB C mg kO ma D 0;
(3.122)
l1 jO NA C l2 jO NB C 0 jO D I/B ˛ C ! .I/B !/:
(3.123)
Substituting a D 0, ˛ D 0, ! D !0 jO, and I/B in (3.121) into (3.122) and (3.123), we have NAx C NBx D 0; NAz C NBz C mg D 0; 1 .Iy y IxB xB /!02 sin sin 2; 2 BB 1 NAz l1 NBz l2 D .IyB yB IxB xB /!02 cos sin 2; 2 0 D 0:
NAx l1 NBx l2 D
Solving the above equations for NA and NB , we have NAx D
IyB yB IxB xB 2 !0 sin 2 sin ; 2l
3.6 Equations of Motion
117
Y
Fig. 3.22 Two-link mechanism
B
θ
A
(x2, y2)
(x1, y1)
β O
α
X
IyB yB IxB xB 2 !0 sin 2 sin ; 2l Iy y IxB xB 2 l2 NAz D mg C B B !0 sin 2 cos ; l 2l Iy y IxB xB 2 l1 !0 sin 2 cos ; NBz D mg B B l 2l 0 D 0;
NBx D
where l D l1 C l2 . Note that the X components of the reaction forces at bearings A and B are due to the angle between the disc and the rod. The Y components of the forces are due to both the gravitational force mg and the angle . The forces due to angle disappear when it is zero and reach a maximum when it is 4 . It is thus important to make the disc perpendicular to the rod to reduce the force exerted at the bearings.
3.6.3 Lagrange Equations Equations of motion developed with the Newton–Euler formulation are described with a rigid body’s Cartesian coordinates. Cartesian coordinates may not always be the best choice for the description of motions. For example, circular motions can be better described with polar or cylindrical coordinates. Coordinates are a set of independent variables to determine the configuration (position and orientation) of a single or a group of rigid bodies. The minimum number of those coordinates is called the degrees of freedom of a rigid body or a system formed by multiple rigid bodies. Refer to the planar two-link mechanism shown schematically in Fig. 3.22. It consists of two uniform links, OA and AB, that are linked by a rotational joint. One end of the link OA, O, is attached to the ground through a rotational joint. The lengths of the links are kOAk D l1 and kABk D l2 . A universe coordinate frame, OX Y , is set up as shown in the figure. The configuration of the mechanism is determined by the coordinates of the centers of masses of the
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3 Dynamics
two links, .x1 ; y1 / and .x2 ; y2 /, and their angles with respect to X axis, ˛ and ˇ, respectively. In total there are six variables, x1 , y1 , x2 , y2 , ˛, and ˇ, but they meet the following constraints: tan ˛ D
y1 ; x1
y2 2y1 ; x2 2x1 q l1 D 2 x12 C y12 ; p l2 D 2 .x2 2x1 /2 C .y2 2y1 /2 :
tan ˇ D
(3.124)
As a result, only two independent variables are needed to determine the configuration of the mechanism. The remaining coordinates can be obtained by solving the above equations. This corresponds to the commonsense notion that a planar mechanism has two degrees of freedom. An immediate question arises as to which two variables should be selected as independent variables. This question is concerned with the concept of generalized coordinates, a set of independent variables (coordinates) to determine the configuration of one or a group of interconnected rigid bodies. The generalized coordinates are usually selected by intuition and are not necessarily Cartesian, polar, or cylindrical coordinates. In the above example, x1 and y1 are ruled out as they are related through (3.124). A good choice of generalized coordinates are ˛ and ˇ, or ˛ and , where is the angle between the two links. After the generalized coordinates are selected, they can be used to formulate the equations of motion. One way to achieve this is to obtain the equations of motion through the Newton–Euler formulation, and then to express it in generalized coordinates through coordinate transformation. A more convenient way is to use generalized coordinates to set up the equations of motion directly. The Lagrange equations, named after their inventor, are designed for this purpose. According to D’Alembert’s principle, the following relation exists in a system consisting of n particles: n X .fi mi ai /T ıri D 0;
(3.125)
i D1
which is reproduced from (3.111). Here fi is the applied force at the i th particle, mi , ai , and ri are the mass, acceleration, and “virtual” displacement of the particle, respectively. Denote generalized coordinates as q D Œq1 q2 : : : ; qm T . Generally, ri D ŒXi Yi Zi T is a function of q and time t: xi D xi .q1 ; q2 ; : : : ; qn ; t/; yi D yi .q1 ; q2 ; : : : ; qn ; t/; zi D zi .q1 ; q2 ; : : : ; qn ; t/:
3.6 Equations of Motion
119
They are written in the following compact form: ri D ri .q; t/:
(3.126)
The differential of ri is ıri D Ji .q/ıq C where
2 @xi
@xi @q1 @q2
@ri ıt; @t
:::
@xi @qm
@yi @q2
:::
@yi @qm
@zi @zi @q1 @q2
:::
@zi @qm
6 6 @ri 6 @yi Ji .q/ D D 6 @q 6 1 @q 4
3 7 7 7 7 2 R3m 7 5
is the Jacobian between q and ri . For the definition of Jacobian, refer to (1.11)– (1.13) in Chap.1. According to the definition of virtual displacement, ıq is allowed when t is constant (ıt D 0). Thus, ıri D Ji .q/ıq: (3.127) Substituting (3.127) into (3.125), we have n X .fi mi ai /T Ji .q/ıq D 0: i D1
For simplicity without causing ambiguities, sometimes the arguments (q) in Ji .q/ and other functions are omitted. Define a new variable Qj D
n X
fiT
i D1
Note that
@ri @qj
n X
@ri ; @qj
j D 1; 2; : : : ; m:
is the j th column vector of Ji and
fi Ji ıq D T
i D1
n X i D1
D
m X j D1
fi
T
@ri @ri @ri ::: Œıq1 ıq2 : : : ıqm T @q1 @q2 @qm
Qj ıqj :
(3.128)
120
3 Dynamics
This is the total virtual work done by all the applied force on the particles. The term Qj plays the role of a “force” corresponding to the virtual displacement ıqj and is called a generalized force. The generalized force can be a force or a moment of force, or a mathematical term without a concrete physical counterpart. From the definition of the generalized force Qj and ai D rRi , (3.6.3) can be rewritten as m n X X Qj ıqj C .mi rRi /T Ji ıq D 0: (3.129) j D1
i D1
The second term of the left-hand side of the above equation is the total virtual work done by the inertial forces. Define another variable Qj D
n n X X @ri @ri .mi rR /T D mi rR T ; @q @q j j i D1 i D1
j D 1; 2; : : : ; m;
(3.130)
which is in the same form as that of Qj , except that fi is replaced by the inertial force mi ai . Qj is called the generalized inertial force. The total virtual work done by the inertial forces is expressed in the following compact form: n m X X .mi rRi /T Ji ıq D Qj ıqj ; i D1
and (3.129) becomes
j D 1; 2; : : : ; m;
j D1 m X
.Qj C Qj /ıqj D 0:
(3.131)
j D1
For most engineering systems, ıqj are independent of each other and the above equation is equivalent to (3.132) Qj C Qj D 0: Qj is related to the second-order derivative of ri with respect to time t. Since ri is the function of the generalized coordinates (q), Qj can be expressed as a function of the generalized coordinates and their derivatives. The velocity of the i th particle is obtained by differentiating ri (3.126) with respect to time t: @ri @ri @ri rPi D qP C D Ji qP C ; (3.133) @q @t @t where Ji D
@Pri @ri D : @qP @q
(3.134)
3.6 Equations of Motion
121
Expressing Ji with its column vectors, we have
@Pri @Pri @Pri ::: @qP 1 @qP 2 @qP m @ri @ri @ri ::: : D @q1 @q2 @qm
Ji D
From (3.133),
@Pri @ @ri @ P C : D .Ji q/ @q @q @q @t
(3.135)
(3.136)
P we have Expanding Ji q,
@ri @ri @ri ::: Ji qP D ŒqP1 qP2 : : : qP m T @q1 @q2 @qm D
m X
qPj
j D1
@ri : @qj
Substituting it into (3.136), we have X @ @ri @Pri @ @ri qPj D C @q @q @qj @q @t j D1 m
D
m X j D1
D Consider
qPj
@ @ri @ @ri C @qj @q @t @q
d @ri : dt @q
d T @ri @ri d @ri C rPiT rPi D rRiT : dt @qj @qj dt @qj
From (3.134) and (3.137), T @ d T @Pri d rPi rPi d T @ri rPi rPi ; D D dt @qj dt @qP j dt @qP j 2 @ rPiT rPi d @ri @Pri D rPiT : D rPiT dt @qj @qj @qj 2 Substituting these into (3.138), we have T T @ @ d @ri rPi rPi rPi rPi rRiT : D @qj dt @qP j 2 @qj 2
(3.137)
(3.138)
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3 Dynamics
With the above equation, the generalized inertial force in (3.130) is written as Qj D
n X
mi
i D1
d D dt
d dt
@ @qPj
@ @qP j
rPiT rPi 2
n X mi rP T rPi
C !!
i
i D1
2
@ @qj
@ C @qj
rPiT rPi 2
n X mi rP T rPi i
i D1
2
! :
From the definition of kinetic energy in Sect. 3.5, the kinetic energy of all the particles is n X mi rPiT rPi ; KD 2 i D1 and Qj can be written as Qj D
d @K @K : C dt @qPj @qj
Substituting it into (3.132), we have the Lagrange equations: Qj D
d @K @K ; dt @qP j @qj
j D 1; 2; : : : ; m:
(3.139)
With the definition of the generalized force vector Q D ŒQ1 Q2 ; : : : ; Qm T ; the Lagrange equations in a vector form is obtained, QT D
d @K @K ; dt @qP @q
j D 1; 2; : : : ; m:
(3.140)
@K Note that @K @qP and @q are row vectors, and thus the transpose of Q is needed in the equation. This is different from the notations used in some books. Though the Lagrange equations are derived for a system that is a collection of discrete particles, they are applicable for a system consisting of continuous masses like rigid bodies. Though a rigid body has an infinite number of particles, their positions can be specified by the limited number of generalized coordinates. For example, for a rigid rod rotating around an axis, its rotation angle is the sole generalized coordinate to determine the positions of all the particles in the rod. When all the applied forces are conservative forces, they can be expressed as
QT D
@U ; @q
where U is the potential energy of the system.
(3.141)
3.6 Equations of Motion
123
As U is only determined by the generalized coordinates, d @U D 0: dt @qP
@U D 0; @qP Define
L D K U; which is called the Lagrange function; then the Lagrange equations represented by (3.139) can be simplified to d @L @L D 0: dt @qP @q
(3.142)
If the generalized forces consist of conservative force Q and nonconservative force F , the Lagrange equations (3.139) can be written as d @L @L FT D : (3.143) dt @qP @q With the introduction of generalized coordinates and general forces in the Lagrange equations, force and motion analysis of the system is simplified. After the general coordinates are selected and the kinetic and potential energies of the system are specified, the dynamic equations of a system in terms of generalized coordinates can be established systematically. Note that the generalized coordinates are independent of each other and the forces associated with the constraints (called internal forces) disappear in Lagrange equations. To solve those forces, force analysis with a free body diagram should be used and the Newton–Euler formulation can be applied. As an example, consider a particle with mass m moving along the X axis of global coordinates. Its generalized coordinate is naturally the X coordinate x. Its kinetic energy is K D 12 mxP 2 and potential energy U D 0. The Lagrange function is L D K U D 12 mxR 2 . Using the Lagrange equations, the force acting on a particle along the X axis is 1 d @xP 2 @xP 2 D mx: R f D m 2 dt @xP @x The above relation between force f and acceleration xR is exactly Newton’s second law! In what follows, several examples are presented to show the application of the Lagrange equations. Example 3.15. As shown in Fig. 3.23, an inverted pendulum is mounted on and balanced by a moving block. It rotates freely on the vertical plane around the hinge at the contact point O. The length and mass of the pendulum are 2l and m, respectively. The moving platform has mass M . Derive the dynamic equations for the pendulum
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3 Dynamics
Y
θ
c x O
mg
rc
2l
X
h P
u
Mg Fig. 3.23 Example 3.15
and the platform given that the force u acts horizontally at the block. The angular displacement of the pendulum is measured from the vertical line going through point P . Ignore friction forces. First, set up the two-axis world coordinate OXZ for the description of the planar motion of the pendulum. The origin O is a fixed point at the same level as that of P , and OX and OY are along the horizontal and vertical directions, respectively. The system has two degrees of freedom corresponding to the X coordinate of point P and angular displacement . The generalized coordinates are set as q D Œx T : The generalized force vector corresponding to q is F D Œu 0T : The position of the COM of the pendulum (c) is rc D Œx C l sin l cos : The linear velocity and the speed of the pendulum are, respectively, rPc D ŒxP C l P cos l P sin T ; q kPrc k D xP 2 C l 2 P 2 C 2l xP P cos : Based on the results of Example 3.2, the moment of inertia of the pendulum about the axis going through its COM and being parallel to the rotating axis through P is I D
1 1 m.2l/2 D ml 2 : 12 3
3.6 Equations of Motion
125
The kinetic energy of the inverted pendulum system is 1 1 1 M xP 2 C mkPrc k2 C I P 2 2 2 2 1 2 D .M C m/xP 2 C ml 2 P 2 C ml xP P cos : 2 3
KD
The potential energy of the system is contributed from the masses of the pendulum (m) and the moving block (M ): U D mgl cos mgh; where g is the gravitational acceleration and h is the distance of the COM of the moving block from the OX axis. The Lagrange function is LDKU D
2 1 .M C m/xP 2 C ml 2 P 2 C ml xP P cos mgl cos C Mgh: 2 3
From the Lagrange equations, d @L @L D .M C m/xR C ml R cos ml P 2 sin ; uD dt @xP @x 4 d @L @L D ml 2 R C ml xR cos mgl sin : 0D dt @P @ 3
(3.144) (3.145)
These two equations completely describe the dynamic behavior of the system. As l ¤ 0 and m ¤ 0, equation of motion (3.145) can be simplified to 4 R l C xR cos g sin D 0: 3
(3.146)
Example 3.16. In this example, the inverted pendulum in Example 3.15 is driven by a spring, as shown in Fig. 3.24. The spring constant is k. Assume the spring oscillates according to
D a sin !t; where a and ! are constants representing the magnitude and the frequency of the oscillation, respectively. All other conditions are the same as in Example 3.15. Establish the dynamic equations of motion for the pendulum when it is near the vertical position. Now there are three variables to determine the configuration of the system, , x, and . As is given as a function of time, the system still has two degrees of freedom and the generalized coordinates are the same as in Example 3.15: q D Œx T : Assume that the length of the spring is lo when it is in the equilibrium state. The system’s potential energy is the sum of the elastic potential energy of the spring and
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3 Dynamics
θ
Y
c
O
k
ξ
mg
rc
x
2l
X
h P
Mg
Fig. 3.24 Example 3.15
gravitational potential energy of masses M and m: 1 k.kx k l0 /2 C mgl cos Mgh; 2 1 D k.kx a sin !tk l0 /2 C mgl cos Mgh: 2 Then the Lagrange function of the system is U D
1 2 .M C m/xP 2 C ml 2 P 2 C ml xP P cos 2 3 1 k.kx a sin !tk l0 /2 mgl cos C Mgh: 2 As all the forces in the system are conservative, applying the Lagrange equations with the assumption that x a sin !t (thus the absolute sign of kx a sin !tk can be omitted), we have L DK U D
d @L @L D .M C m/xR C ml R cos ml P 2 sin k.x a sin !t l0 / D 0: dt @xP @x It can be verified that when x a sin !t, the above equation is still valid. As the terms related to and P in the Lagrange equation do not change, the dynamic equation for angular movement is identical to (3.146), which is produced below for completeness: 4 R l C xR cos g sin D 0: 3
(3.147)
When the pendulum is near the vertical position, ! 0, cos ! 1, and sin ! . The above equations are simplified to
3.6 Equations of Motion
127
Y
Fig. 3.25 Example 3.17
{U}
ω2
ˆj
A r R
τ
ω 1C
(Z, kˆ ) O
B
θ
iˆ
X
.M C m/xR C ml R ml P 2 kx D k.a sin !t C l0 /; 4 R l C xR g D 0: 3 Example 3.17. A planar satellite gear system is shown in Fig. 3.25. The uniform rod OA rotates around point O. A gear with radius r rotates around point A and moves along the surface of a large gear. The large gear’s center is at O and its radius is R. The masses of the rod OA and the small gear are m1 and m2 , respectively. If a torque with magnitude is applied at rod OA, calculate the rod’s acceleration as a function of . Assume that there is no slippage between the gears and ignore frictional forces. Set up a universe coordinate frame fU g W OX Y Z where the OX and OY axes point along the horizontal and vertical directions, respectively, and OZ (not shown) is perpendicular to the X Y plane. The system has one degree of freedom specified by the angular displacement of rod OA about the OX axis. The motions of the rod and the small gear are constrained on the X Y plane. The COM of the rod and the contact point between the gears are denoted by C and B, respectively. The position vectors of points C , B, and A are, respectively, 1 .R C r/.iO cos C jO sin /; 2 OB D R.iO cos C jO sin /;
OC D
OA D .R C r/.iO cos C jO sin /: As the angular velocities of the rod and the small gear are in the direction of O respectively. !1 is the time derivative of the O k, they are denoted by !1 kO and !2 k, angular displacement : P !1 D :
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3 Dynamics
The velocities of points C and A are, respectively, 1 P iO sin C jO cos /; vC D !1 kO OC D .R C r/. 2 P iO sin C jO cos /: vA D !1 kO OA D .R C r/.
(3.148) (3.149)
The velocity of contact point B at the small gear is vB D vA C !2 kO AB D vA !2 kO r.iO cos C jO sin / D .R C r/P r!2 /.iO sin C jO cos /:
(3.150)
As there is no slippage between the gears and the big gear is fixed, vB D 0. From (3.150), RCr P : .R C r/P r!2 D 0 ! !2 D r It is now obvious that can be chosen as the generalized coordinate and is the corresponding generalized force. Next study the kinetic and potential energies of the system. As rod OA and the O small gear rotate around one of their principal axes of inertia in the direction of k, their kinetic energies are, respectively, 1 m1 kvC k2 C 2 1 K2 D m2 kvA k2 C 2
K1 D
1 I1 !12 ; 2 1 I2 !22 ; 2
(3.151) (3.152)
where I1 and I2 are the moments of inertia of the rod and the small gear with respect to their rotational axes. The rod and the gear can be viewed as a uniform slander bar and a thin disc, respectively, and thus 1 m1 .R C r/2 ; 12 1 I2 D m2 r 2 : 2 I1 D
(3.153) (3.154)
Substituting (3.148), (3.149), (3.153), and (3.154) into (3.151) and (3.152), we have 1 m1 .R C r/2 P 2 ; 6 3 K2 D m2 .R C r/2 P 2 : 4 K1 D
Then the total kinetic energy is K D K1 C K 2 D
1 6
m1 C
3 .R C r/2 P 2 : 4
(3.155) (3.156)
3.6 Equations of Motion
129
The potential energy is due to the masses of the rod and the small gear. Choosing plane XZ as a reference plane, the potential energy is U D
1 1 m1 g.R C r/ sin C m2 g.R C r/ sin D m1 C m2 .R C r/g sin ; 2 2
where g is the gravitational acceleration. The Lagrange function is L DK U D
1 3 1 .R C r/2 P 2 m1 C m1 C m2 .R C r/g sin : 6 4 2
According to the Lagrange equations, D which yields D
d @L @L ; dt @P @
1
3 R m1 C m2 .R C r/2 : 3 2
Then the angular acceleration of the rod is R D
6 : .2m1 C 9m2 /.R C r/2
Chapter 4
Case Studies
4.1 Two-Link Planar Robotic Arm A two-link planar robotic arm is schematically shown in Fig. 4.1. The arm consists of two uniform links rotating around joints O1 and O2 on the vertical plane. The lengths of the links are l1 and l2 , their masses are m1 and m2 , and their centers of mass (COM) are C1 and C2 , respectively. The links are driven by the motors at the joints with output torques 1 and 2 , respectively. Our task is to find the equations of motion of the robotic arm. First, set up the universe coordinate frame O1 X Y whose origin is at O1 and axes O1 X and O1 Y are along the horizontal and vertical directions, respectively. Their unit vectors are iO and jO, respectively. Though it is not shown in the figure, another axis, O1 Z (unit vector kO D iO jO), perpendicular to the plane is used to describe the angular velocities and accelerations of the links.
Positions • Link O1 O2 The angular position of Link O1 O2 is 1 measured from the OX1 axis. As the link is uniform in mass distribution, its COM (C1 ) is at the middle of O1 O2 . Thus the link’s linear position is rC1 D O1 C1 D
1 l1 .cos 1 iO C sin 1 jO/: 2
L. Huang, A Concise Introduction to Mechanics of Rigid Bodies: Multidisciplinary Engineering, DOI 10.1007/978-1-4614-0472-9 4, © Springer Science+Business Media, LLC 2012
(4.1)
131
132
4 Case Studies
Fig. 4.1 Two-link robotic arm
Y
a2
E (xE, yE) c2
l2
ˆj a1 l1
O1
c1
m2, I2
θ2
O2 m1, I1 θ1 iˆ
X
• Link O2 E The angular position of Link O2 E is 2 measured from the O1 X axis: 2 D 1 C 2 ;
(4.2)
where 2 is the angle between Links O1 O2 and O2 E. The linear position of the link is rC2 D OC2 D O1 O2 C O2 C2 ; where O1 O2 D l1 .cos 1 iO C sin 1 jO/; O2 C2 D
(4.3)
1 1 l2 .cos 2 iO C sin 2 jO/ D l2 .cos.1 C 2 /iO C sin.1 C 2 /jO/: (4.4) 2 2
Then 1 rC2 D l1 .cos 1 iO C sin 1 jO/ C l2 .cos.1 C 2 /iO C sin.1 C 2 /jO/ 2 1 1 D l1 cos 1 C l2 cos.1 C 2 / iO C l1 sin 1 C l2 sin.1 C 2 / jO: (4.5) 2 2 Velocities and Accelerations The links’ velocities and accelerations can be derived directly by differentiating their position functions with respect to time. Their linear velocities are denoted by v1 and v2 , linear accelerations by a1 and a2 , angular velocities !1 and !2 , and angular accelerations by ˛1 and ˛2 .
4.1 Two-Link Planar Robotic Arm
133
• Link O1 O2 Differentiating positions 1 and rC1 (4.1) with respect to time, the angular and linear velocities of Link O1 O2 are obtained, respectively, as O !1 D P1 k;
(4.6)
1 v1 D rPC1 D l1 P1 .sin 1 iO cos 1 jO/: 2
(4.7)
The linear and angular accelerations are obtained by differentiating the above velocity functions with respect to time: O ˛1 D !P 1 D R1 k;
(4.8)
1 a1 D vP 1 D l1 Œ.P12 cos 1 C R1 sin 1 /iO C .P12 sin 1 R1 cos 1 /jO: (4.9) 2 • Link O2 E Following the same procedure for obtaining the velocities of Link O1 O2 , the velocities and accelerations of Link O2 E are obtained: O !2 D P 2 kO D .P1 C P2 /k; 1 P P P v2 D rPC2 D l1 1 sin 1 C l2 .1 C 2 / sin.1 C 2 / iO 2 1 P P P C l1 1 cos 1 C l2 .1 C 2 / cos.1 C 2 / jO; 2
(4.10)
(4.11)
O ˛2 D !P 2 D .R1 C R2 /k; a2 D vP 2 D a2x iO C a2y jO; where a2x D l1 .P12 cos 1 C R1 sin 1 /
a2y
1 l2 Œ.R1 C R2 / sin.1 C 2 / C .P1 C P2 /2 cos.1 C 2 /; 2 D l1 .P 2 sin 1 R1 cos 1 / 1
1 C l2 Œ.R1 C R2 / cos.1 C 2 / .P1 C P2 /2 sin.1 C 2 /: 2 Alternatively, the formulas given by (2.50)–(2.66) can also be used for the kinematic analysis of a robotic arm. For the sake of revision, the kinematic analysis based on this method is presented below.
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4 Case Studies
Velocities • Link O1 O2 Link O1 O2 rotates around the fixed point O1 , so its angular and linear velocities are given by O !1 D P1 k;
(4.12)
1 v1 D !1 rC1 D l1 P1 .sin 1 iO cos 1 jO/: 2
(4.13)
• Link O2 E Link O2 E rotates around O2 , and its angular velocity with respect to Link O1 O2 is O !2/ 1 D P2 k: Thus, the angular velocity of Link O2 E (with respect to the universe frame) is O !2 D !1 C !2/ 1 D .P1 C P2 /k:
(4.14)
The linear velocity of the link is v2 D vO2 C !2 O2 C2 D !1 O1 O2 C !2 O2 C2 :
(4.15)
Substituting O2 C2 , O1 O2 , !1 , and !2 in (4.3), (4.4), (4.12), and (4.14) into the above equation, we have 1 O cos.1 C 2 /jO v2 D l1 P1 . sin 1 iO C cos 1 jO/C l2 .P1 C P2 /Œ sin.1 C 2 /iC 2 1 D l1 P1 sin 1 l2 .P1CP2 / sin.1C2 / iO 2 1 C l1 P1 cos 1C l2 .P1CP2 / cos.1C2 / jO: 2 Accelerations • Link O1 O2 The angular acceleration of Link O1 O2 is O ˛1 D R1 k:
(4.16)
a1 D ˛1 rC1 C !1 .!1 rC1 /:
(4.17)
Its linear acceleration is
4.1 Two-Link Planar Robotic Arm
135
Substituting rC1 , !1 , and ˛1 in (4.1), (4.12), and (4.16) into the above equation, we have 1 1 a1 D l1 R1 . sin 1 iO C cos 1 jO/ l1 P12 .cos 1 iO C sin 1 jO/ 2 2 1 D l1 Œ.P12 cos 1 C R1 sin 1 /iO C .P12 sin 1 R1 cos 1 /jO/: 2 Note that C1 does not move with respect to Link O1 O2 , and thus there is no Coriolis acceleration contribution to a1 . • Link O2 OE The angular acceleration of Link O2 OE with respect to Link O1 O2 is O ˛2/ 1 D R2 k: Its angular acceleration is O ˛2 D ˛1 C ˛2/ 1 D .R1 C R2 /kO D R 2 k: To calculate the linear acceleration of the link, first find the acceleration of point O2 : aO2 D ˛1 O1 O2 C !1 .!1 O1 O2 / D l1 Œ.P12 cos 1 C R1 sin 1 /iO C .P12 sin 1 R1 cos 1 /jO/: Then the linear acceleration of the link is a2 D aO2 C ˛2 O2 C2 C !2 .!2 O2 C2 /: The expressions of the terms on the right-hand side of the equation are all available. Substituting them into the above equation, we have 1 a2 D l1 .P12 cos 1CR1 sin 1 / l2 ..R1CR2 / sin.1C2 /C.P1CP2 /2 2 cos.1C2 // iO 1 l1 .P12 sin 1R1 cos 1 / l2 ..R1CR2 / cos.1C2 /C.P1CP2 /2 2 sin.1C2 // jO:
Equations of Motion For the two-link robotic arm, it is clear that the generalized coordinates are 1 and 2 and the corresponding generalized forces are 1 and 2 . Lagrange equations can be used to set up its equations of motion.
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4 Case Studies
The kinetic energies of the links are 1 m1 kv1 k2 C 2 1 K2 D m2 kv2 k2 C 2 K1 D
1 I1 k!1 k2 ; 2 1 I2 k!2 k2 ; 2
(4.18) (4.19)
where I1 and I2 are the moments of inertia of the corresponding links with respect O As the link to the axes that go through the COMs and are along the direction of k. is uniform in terms of mass distribution, from Example 3.2: I1 D
1 m1 l12 ; 12
I2 D
1 m2 l22 : 12
Substituting these into (4.18) and (4.19) and considering !1 , v1 , !2 , and v2 given in (4.12), (4.13), (4.10), and (4.11), we have 1 m1 l12 P12 ; 6 1 1 K2 D m2 l12 P12 C m2 l2 Œl2 .P1 C P2 /2 C 3l1 P1 .P1 C P2 / cos 2 : 2 6
K1 D
Taking the ground as a reference, the potential energies of the links are 1 m1 gl1 sin 1 ; 2 1 U2 D m2 g l1 sin 1 C l2 sin.1 C 2 / ; 2 U1 D
where g is the magnitude of the gravitational acceleration. So the Lagrange function is L D K1 CK2 .U1 CU2 / D f1 .2 /P12 Cf2 .2 /P22 Cf3 .2 /P1 P2 Cf4 .1 ; 2 /; (4.20) where f1 .2 / D
1 1 1 m1 C m2 l12 C m2 l2 .l2 C 3l1 cos 2 /; 6 2 6
1 m2 l22 ; 6 1 1 f3 .2 / D m2 l22 C m2 l1 l2 cos 2 ; 3 2 1 1 m1 C m2 gl1 sin 1 m2 gl2 sin.1 C 2 /: f4 .1 ; 2 / D 2 2 f2 .2 / D
4.1 Two-Link Planar Robotic Arm
137
Applying the Lagrange equations 1 D
d dt
2 D
d dt
@L @P1 @L @P2
@L ; @1
@L ; @2
we have 1 D
2 D
1 1 .m1 l12 C 3m2 l12 C m2 l22 C 3m2 l1 l2 cos 2 /R1 C m2 l2 .2l2 C 3l1 cos 2 /R2 3 6 1 1 m2 l1 l2 sin 2 .2P1 C P2 /P2 C m1 gl1 cos 1 2 2 1 Cm2 g l1 cos 1 C l2 cos.1 C 2 / ; 2 1 1 m2 l2 .2l2 C 3l1 cos 2 /R1 C m2 l22 R2 6 3 1 1 C m2 l1 l2 sin 2 P12 C m2 gl2 cos.1 C 2 /; 2 2
which are the equations of motion of the robotic arm. We define the following matrices and vectors: q D Œ1 2 T ; qP D ŒP1 P2 T ; D Œ1 2 T ; # " 1 .m1 l12 C 3m2 l12 C m2 l22 C 3m2 l1 l2 cos 2 / 16 m2 l2 .2l2 C 3l1 cos 2 / 3 ; M.q/ D 1 1 2 6 m2 l2 .2l2 C 3l1 cos 2 / 3 m2 l2 " # P2 .P1 C P2 / 1 C.q; q/ P D m2 l1 l2 sin 2 ; 2 P1 0 " # 1 m gl cos 1 C m2 g.l1 cos 1 C 12 l2 cos.1 C 2 // 2 1 1 G.q/ D : 1 2 m2 gl2 cos.1 C 2 /
Then the equation of motion of the robotic arm is written in the following compact form: D M.q/qR C C.q; q/ P qP C G.q/: (4.21) Note that M.q/ is the function of the inertial parameters and the position of the manipulator and is called the inertia matrix. C.q; q/ P qP represents the contributions of centrifugal forces and Coriolis forces, and G.q/ is related to the gravitational forces.
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The above equations of motion have the following properties: • M.q/ is symmetric and positive definite. • MP .q/ 2C.q; qP is skew symmetric, that is, given a vector x 2 R2 , x T .MP .q/ 2C.q; q/x P D 0. This can be verified by observing that # " 0 2P1 C P2 / 1 : MP .q/ 2C.q; q/ P D m2 l1 l2 sin 2 2 .2P1 C P2 / 0 These properties are important to the controller design of robotic arms. Note that the choice of matrix C.q; q/ P is not unique. For example, it can be written as " # P2 12 P2 C.q; q/ P D m2 l1 l2 sin 2 ; 1 P 1 0 2
but in this case, " MP .q/ 2C.q; q/ P D m2 l1 l2 sin 2
P2 1 .2P1 C P2 / 2
1 P 2 2
0
# ;
which is not skew symmetric.
4.2 Human Body Doing Twisting Somersaults This case study is adopted from Example 9.8 in [6]. As shown in Fig. 4.2, a diver dives off a platform from an upright standing position at an angular speed !0 in the direction of vector yOB , which is parallel to his shoulder. Immediately after the jump, he moves his left arm to the horizontal position. Determine his angular velocity at the moment when his left arm is raised. The diver weighs m D 80 kg, and each of his arms weighs ma D 4 kg. His height (h), width (w), and depth (d ) are 1.8, 0.35, and 0:2 m, respectively, and the length of his arm is la D 0:9 m. Considering the physical structure of the diver, several assumptions can be made: • The diver’s body excluding his arms is treated as a uniform rectangular cube with sides h, w, and b and a mass mb D m 2ma D 72 kg. • The diver’s arms are represented by slender rods with mass ma and length la . • The contribution of the masses of the arms (which are very insignificant relative to the body of the diver) to the position of the diver’s COM is ignored. • The moments of inertia of the arms with respect to their own body frames are ignored. From these assumptions, the diver before and after jumping and lifting his left arm is schematically shown in Fig. 4.3. The body frame fBg W C xOB yOB zOB is set up at the COM (C ) of the diver. Axis yOB is parallel to the diver’s shoulder, and xOB and zOB are along
4.2 Human Body Doing Twisting Somersaults
139
zˆB
C ω1
yˆ B mg
Z
xˆ B
ω2
kˆ
jˆ
Y
O iˆ
X
Fig. 4.2 Twisting somersaults in diving [6] C1
rˆ
a
b zˆB
h
C1
yˆB
C
d
zˆB C2
C2
h 2
α
yˆB
C mg
xˆB
ω1
Z
w
xˆB
ω1
ˆ Before the arm is lifted up k
ˆj
Y
After the arm is lifted up
O iˆ
X
Fig. 4.3 Rigid-body model of diver
the upward and front directions relative to the diver’s body. The universe coordinate frame fU g W OX Y Z is located at the diving platform. The origin O is a point on the platform, the X , Y , and Z axes are parallel to xOB , yOB , and zOB , respectively, when the diver stands on the platform before jumping. Refer to Fig. 4.3. Let us examine the inertial properties of the diver before and after his left arm is raised.
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4 Case Studies
• Before the left arm is raised: According to the formulas of the moments of inertia of a cube [5], the inertia matrix of the diver’s body excluding his arms with respect to his body frame is 2 6 Ib D 4
B
Ib x B x B
0
0
0
Ib y B y B
0
0
0
Ib z B zB
3 7 5;
where 1 mb .w2 C h2 / D 20:18 kgm2 ; 12 1 D mb h2 D 19:44 kgm2 ; 12 1 D mb w2 D 0:74 kgm2 : 12
Ib x B xB D Ib y B y B Ib B z B
The linear positions of the diver’s two arms with respect to the body frame are 1 1 T pC1 D 0 w h D Œ0 0:175 0:45T .m/ D ŒxC1 yC1 zC1 T ; 2 4 1 1 T B pC2 D 0 w h D Œ0 0:175 0:45T .m/ D ŒxC2 yC2 zC2 T : 2 4 B
As the moments of inertia of the arms with respect to their own body frames are ignored, the moments and products of inertia of the left and right arms with respect to the diver’s body frame are due to their masses at their COMs (C1 and C2 ). They are as follows:
Left Arm IlxB xB D ma .yC21 C zC21 / D 0:933 kgm2 ; IlyB yB D ma .xC21 C zC21 / D 0:81 kgm2 ; IlzB zB D ma .xC21 C yC21 / D 0:123 kgm2 ; IlxB yB D IlxB zB D 0:00 kgm2 ; IlyB zB D ma .yC1 zC1 / D 0:315 kgm2 I
4.2 Human Body Doing Twisting Somersaults
141
Right Arm IrxB xB D ma .yC22 C zC22 / D 0:933 kgm2 ; IryB yB D ma .xC22 C zC22 / D 0:81 kgm2 ; IrzB zB D ma .xC22 C yC22 / D 0:123 kgm2 ; IrxB yB D IlxB zB D 0:00 kgm2 ; IryB zB D ma .yC2 zC2 / D 0:315 kgm2 : Note that the dimensions of the arms are ignored as they are very small relative to those of the diver’s body. According to the composite-bodies rule, the inertia matrix of the body and the arms with respect to the body frame is 2 3 0 I xB xB 0 B I D 4 0 I yB yB 0 5 ; (4.22) 0 0 IzB zB where I xB xB D IbxB xB C IlxB xB C IrxB xB D 22:04 kgm2 ; I yB yB D IbyB yB C IlyB yB C IryB yB D 21:06 kgm2 ; IzB zB D IbzB zB C IlzB zB C IrzB zB D 0:98 kgm2 : • After the left arm is raised: The position of the COM of the left arm changes with the motion of the arm. The position of C1 with respect to the body frame is h 0 0 0 iT 1 1 1 T 0 B pC1 D 0 w C h h D Œ0 0:625 0:9T .m/ D xC1 yC1 zC1 : 2 4 2 As the mass of the arm is very small in comparison to the mass of the diver, its effects on the position of the COM (C ) of the diver’s body and the positions, orientations, and inertial properties of other parts of the diver are ignored. Thus only the moments and the products of inertia of the left arm need to be recalculated: 0
0
0
0
0
0
0
0
IlxB xB D ma .yC12 C zC21 / D 4:81 kgm2 ; IlyB yB D ma .xC12 C zC21 / D 3:24 kgm2 ; 0
IlzB zB D ma .xC12 C yC12 / D 1:56 kgm2 ; 0
0
IlxB yB D IlxB zB D 0:00 kgm2 ; 0
0
0
IlyB zB D ma .yC1 zC1 / D 2:25 kgm2 :
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The new inertia matrix of the body and the arms is 2 0 3 I xB x B 0 0; 0 0 0 6 7 B I D 4 0 I yB yB IyB zB ; 5; 0
0
(4.23)
0
IyB zB IzB zB
where 0
0
0
0
I xB xB D IbxB xB C IlxB xB C IrxB xB D 25:91 kgm2 ; I yB yB D IbyB yB C IlyB yB C IryB yB D 23:49 kgm2 ; 0
0
0
0
IzB zB D IbzB zB C IlzB zB C IrzB zB D 2:42 kgm2 ; IyB zB D IbyB zB C IlyB zB C IryB zB D 1:935 kgm2 : At the moment when the diver jumps, the angle of the diver’s body to the vertical plane, ˛, is almost zero. The body frame fBg can be treated as parallel to the global frame fU g and rotation matrix RB D I 3 . As a result, inertia matrices 0 B I and BI remain the same through a similar transformation RB : 0
0
0
I/B D RBT BI RB D BI ; I/B D RBT BI RB D BI :
(4.24)
When the diver is in the air, the only force he experiences is the gravitational force going through the COM (C ). The total moment of force with respect to C is zero, and the angular momentum of the diver with respect to C is conserved before and after the diver’s left arm is raised. This fact can be used to find the angular velocity of the diver. Before the left arm is raised, the diver’s angular velocity is !1 D !0 yOB D !0 jO D Œ0 !0 0T ;
(4.25)
and the diver’s angular momentum with respect to C is HC D I/B !1 : Substituting (4.24) and (4.25) into the above equation, HC D Œ0 I yB yB !0 0T :
(4.26)
After the left arm is raised in the horizontal direction, the diver’s angular velocity can be written as !2 D Œ!x !y !z T : The diver’s angular moment with respect to C is 0
0
0
HC D I/B !2 D I/B Œ!x !y !z T :
4.3 Bicycle Balance
143
Substituting (4.24) into the above equation, 0
0
0
0
0
0
HC D ŒI xB xB !x I yB yB !y IyB zB !z IyB zB !y C IzB zB !z T :
(4.27)
From the conservation of angular momentum, 0
HC D HC : From (4.26) and (4.27), 0
I xB xB !x D 0; 0
0
I yB yB !y IyB zB !z D I yB yB !0 ; 0
0
IyB zB !y C IzB zB !z D 0: 0
0
0
0
Obtaining the values of I yB yB , I xB xB , I yB yB , IyB zB , and IzB zB !z from (4.22) and (4.23), !x , !y , and !z are solved from the above equations: !x D 0;
!y D 0:975!0 ;
!z D 0:78!0
or !2 D 1:23!0 rO ; where rO D Œ0 0:78 0:63T : This means that when the left arm is raised at angular speed !0 , the diver rotates with speed 1:23!0 around the rO axis on the Y Z plane – a typical twisting rotation observed in diving. The rigid-body dynamics analysis presented here provides a rigorous explanation of this phenomenon. A more comprehensive and accurate study could be done by treating the diver as a multibody system consisting of his body, arms, legs, and joints, but the assumptions used in this example are realistic and greatly simplify the analysis. The result obtained is a good approximation to the solution obtained from multibody dynamic analysis based on accurate models.
4.3 Bicycle Balance This case study addresses the kinematics and equations of motion for a bicycle near its balance state [9]. Figure 4.4 schematically shows the side view of a bicycle standing on horizontal ground. According to [9], the bicycle is divided into the following four main parts: rear wheel, frame (including the rider), front fork, and front wheel. The rear wheel and the frame form a plane called the rear plane,
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4 Case Studies Steering axis
C ω1
Frame
mg
Front fork
Rear wheel
h
C1
r1
Front wheel
v1
r2 f1
P1
a b
ω2
C2
P2
γ
f2
P3
d
Fig. 4.4 Bicycle – side view
and the front fork and the front wheel form another plane called the front plane. In the figure, the wheels, frame, and front fork are all on the same vertical plane. The COMs of the frame plane (C ), the centers of the wheels (C1 and C2 ), and the distances among them are shown. Point C2 is treated as the COM of the front plane. Points P1 and P2 are the contact points of the rear and front wheels on the ground, respectively. Point P3 is the intersection between the steering axis of the front fork and the ground. The angle ( ) between the steering axis and the ground is called the head angle, the distance (d ) between P2 and P3 is called the trail, and the distance (b) between P2 and P3 is called the wheel base. The head angle, the train, and the wheel base are the key parameters of a bicycle. The angular speed of the rear and front wheels are constant and denoted by !1 and !2 , respectively. Figure 4.5 shows the top view of a bicycle in a more general configuration. The universe frame fU g W OX Y Z is set up at a fixed point on the ground. The Z axis is perpendicular to the ground and is shown as a black dot. The base vectors for the O The body frame for the rear axes of the universe frame are denoted by iO , jO, and k. plane is fB1 g W C xO 1 yO1 zO1 , which has its origin at its COM (C ). The xO 1 axis is on the rear plane and is parallel to the ground. The yO1 axis is along the direction of the rotation axis of the rear wheel and is perpendicular to the rear plane and zO1 D xO 1 yO1 . The body frame for the front plane, fB2 g W C2 xO 2 yO2 zO2 , has its origin at C2 . Its axes (xO 2 , yO2 , and zO2 ) are defined in the same way as for the definition of the xO 1 , yO1 , and zO1 axes. Assume that the mass of the rear and front planes are m1 and m2 , respectively. The axes of their body frames are their principal axes of inertia, and the inertia matrices are B1I D diag (Ix1 x1 ; Iy1 y1 ; Iz1 z1 ) and B2I D diag (Ix2 x2 ; Iy2 y2 ; Iz2 z2 /, respectively. Considering the symmetry in the structure of the bicycle, it can be assumed that Iy1 y1 D Iz1 z1 and Iy2 y2 D Iz2 z2 .
4.3 Bicycle Balance
145
yˆ2 yˆ1
Y ˆj P1
δ1
Z
ˆ (k)
O
iˆ
φ1
zˆ1 xˆ1 C
P3 P2(C2)
xˆ2 θ
φ2
zˆ2
X
Fig. 4.5 Bicycle – top view
A bicycle is a complex multibody system if each part of it is modeled accurately. Considering the unique features of a bicycle’s structure and motion when it is near the state of balance, the following assumptions are made: • • • • • •
The bicycle moves on smooth horizontal ground. There is no slippage between the wheels and the ground. The steering angle is very small: ! 0, sin , and cos 1. The head angle D 90ı and the trail c D 0. The rider is fixed with respect to the frame. The masses of the wheels are ignored relative to the masses of other parts of the bicycle. • The front plane is always in an upright position. • The diameters of the two wheels are the same, r1 D r2 .
Kinematic Analysis Positions and Orientations O where x1 , y1 , and z1 are the The rear plane’s linear position is r1 D x1 iO Cy1 jO Cz1 k, coordinates of the COM (C ). Its orientation varies with 1 (yaw), the angle between the xO 1 and X axes, and ı1 (roll), the angular displacement it has around the xO 1 axis. O where x2 , y2 , and z2 are the The front plane’s linear position is r2 D x2 iO Cy2 jO Cz2 k, coordinates of point C2 . Its orientation only varies with 2 (yaw), the angle between axes xO 2 and X , as the bicycle is always kept in an upright position. It is assumed that at the bicycle’s initial position, as shown in Fig. 4.4, 1 D 2 D ı1 D 0.
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4 Case Studies
The rotation matrices of the rear plane and front planes, respectively, are 2 3 32 cos 1 sin 1 0 1 0 0 R1 D Rz .1 /Rx .ı1 / D 4 sin 1 cos 1 0 5 4 0 cos ı1 sin ı1 5 0
0
1
0 sin ı1 cos ı1
3 cos 1 sin 1 cos ı1 sin 1 sin ı1 D 4 sin 1 cos 1 cos ı1 cos 1 sin ı1 5; 0 sin ı1 cos ı1 2 3 cos 2 sin 2 cos ı2 sin 2 sin ı2 R2 D Rz .2 / D 4 sin 2 cos 2 cos ı2 cos 2 sin ı2 5; 0 sin ı2 cos ı2 2
(4.28)
(4.29)
where Rz ./ and Rx ./ are the basic rotation matrices introduced in Example 2.2. Noting that 2 D 1 C and ! 0; R2 is rewritten as 3 2 cos 1 sin 1 sin 1 cos 1 o R2 D 4 sin 1 C cos 1 cos 1 sin 1 0 5: 0 0 1 From R1 and R2 , the expressions of the principal axes of the body frames fB1 g and fB2 g in the universe frame are obtained: xO 1 D Œcos 1 sin 1 0T ;
(4.30)
yO1 D Œ sin 1 cos ı1 cos 1 cos ı1 sin ı1 ;
(4.31)
zO1 D Œsin 1 sin ı1 cos 1 sin ı1 cos ı1 T ;
(4.32)
xO 2 D Œcos 2 sin 2 0 ;
(4.33)
T
T
yO2 D Œ sin 2 cos ı2 cos 2 cos ı2 sin ı2 T ;
(4.34)
zO2 D Œsin 2 sin ı2 cos 2 sin ı2 cos ı2 :
(4.35)
T
Velocities The angular velocities of the rear plane, !R , are obtained from rotation matrix R1 according to (2.36): O !R D P1 kO C ıP1 xO 1 D ıP1 cos 1 iO C ıP1 sin 1 jO C P 1 k:
(4.36)
Note the expression of xO 1 in (4.30) is used in the above derivation. Similarly, the angular velocity of the front plane, !F , is O !F D P 2 kO C ıP2 xO 2 D ıP2 cos 2 iO C ıP2 sin 2 jO C P 2 k: The angular velocities of the two wheels are !1 yO1 and !2 yO2 , respectively.
(4.37)
4.3 Bicycle Balance
147
As there is no slippage between the wheels and the ground, the velocities of points P1 and P2 with respect to the universe frame are zero: vP1 D vP2 D 0: The velocities of C1 and C2 are vC1 D vP1 C !1 yO1 P1 C1 D !1 yO1 r1 zO1 D r1 !1 xO 1 ; vC2 D vP2 C !2 yO2 P2 C2 D !2 yO2 r2 zO2 D r2 !2 xO 2 : As C1 and C2 are the points on the rear plane, their velocities are related by vC2 D vC1 C !R C1 C2 D r1 !1 xO 1 C !R .b xO 1 C .r2 r1 /Oz1 /: Substituting (4.36) into the above equation and considering the assumption that r1 D r2 , we have vC2 D r2 !2 xO 2 D r1 !1 xO 1 C b P 1 yO1 : Because .r2 !2 xO 2 / xO 2 D 0, .r1 !1 xO 1 C b P1 yO1 / xO 2 D 0 and
r1 !1 kxO 1 xO 2 k D r1 !1 sin D b P 1 kyO1 xO 2 k:
As it is assumed that ! 0, sin , then yO1 is nearly perpendicular to xO 2 , kyO1 xO 2 k 1. As a result, r1 !1 D b P 1 ; r1 !1 : P1 D b
(4.38)
Then the angular velocity of the rear plane can be expressed as a function of !1 , , and ıP1 : !R D
r1 !1 O P r1 !1 O k C ı1 xO 1 D ıP1 cos 1 iO C ıP1 sin 1 jO C k: b b
(4.39)
The relation between !1 and !2 can also be obtained from (4.3). Taking the inner products of both sides of the equation with xO 1 , we have r2 !2 xO 1T xO 2 D r1 !1 C b P 1 xO 1T yO1 D r1 !1 : As xO 1T xO 2 D cos , then !1 D
r2 !2 cos : r1
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4 Case Studies
The linear velocities of the rear plane and the front plane, respectively, are vR D vC1 D v1 xO 1 D v1 .cos 1 iO C sin 1 jO/; vF D vC2 D v2 xO 2 D v2 .cos 2 iO C sin 2 jO/; where v1 D r1 !1 and v2 D r2 !2 are the speeds of the rear and front wheels, respectively.
Accelerations Assume that !1 is a constant; the angular acceleration of the rear plane is ˛R D !P R D
v1 P O R k C ı1 xO 1 C ıP1 xPO 1 ; b
(4.40)
where
xPO 1 D !R xO 1 : Substituting (4.39) and (4.30) into the above equation, we have v1 xPO 1 D . sin 1 iO C cos 1 jO/: b
(4.41)
Substituting this into (4.40) and considering (4.30), we have O ˛R D ˛x iO C ˛y jO C ˛z k;
(4.42)
where v1 ˛x D ıR1 cos 1 ıP1 sin 1 ; b v 1 ˛y D ıR1 sin 1 C ıP1 cos 1 ; b v1 P ˛z D : b The linear acceleration of the rear plane is aR D vP1 xO 1 C v1 xPO 1 :
(4.43)
Considering that v1 (the linear speed of the rear wheel) is constant and substituting (4.41) into the above equation, we have aR D
v21 . sin 1 iO C cos 1 jO/: b
(4.44)
This is the centripetal acceleration due to the instantaneous circular motion of C1 .
4.3 Bicycle Balance
149
Equations of Motion Assume that the rider keeps the bicycle in balance by adjusting steer angle only. Ignoring the reaction force from the front plane, the rear plane is subject to the following forces. O at the COM C , where g D 9:8 m=s2 is the magnitude • Gravitational force: mg k, of gravitational acceleration. • Static friction forces f1 xO 1 at the contact point P1 between the wheel and the ground. The moment of force with respect to point P1 is thus O MP1 D P1 C1 .mg k/: Noting that P1 C1 D axO 1 C hOz1 and the expressions xO 1 and zO1 given by (4.30) and (4.32), respectively, we have MP1 D mgŒ.a sin 1 h cos 1 sin ı1 /iO C .a cos 1 C h sin 1 sin ı1 /jO: (4.45) The angular momentum of the rear plane with respect to C1 is HC1 D R1 B1I R1T !R :
(4.46)
Applying the Newton–Euler formulation, we have the following equation of motion: MP1 D R1 B1I R1T ˛R C P1 C1 .maR / C !R HC1 :
(4.47)
The equation will become very complex after expansion. It can be simplified by considering that in a normal situation, the angle ı1 is usually small. It is reasonable to assume that ı1 sin ı1 and cos ı1 1. Furthermore, as we are only concerned with the balance of the bicycle, the X axis of the universe frame can be chosen such that it is aligned with xO 1 at the instant of interest and as such 1 D 0 and iO D xO 1 . With these assumptions, the terms R1 (4.28), !R (4.39), ˛R (4.42), aR (4.44), and MP1 (4.45)used in the equation can be simplified such that 2
3 1 0 0 R1 D 4 0 1 ı1 5; 0 ı1 1 v1 O k; b v1 v1 O ˛R D ıR1 iO C ıP1 jO C P k; b b
!R D ıP1 iO C
aR D
v21 jO; b
MP1 D mg.hı1 iO C ajO/;
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4 Case Studies
and xO 1 D D Œcos 1 sin 1 0T D Œ1 0 0T ; zO1 D Œsin 1 sin ı1 cos 1 sin ı1 cos ı1 T Œ0 ı1 1T ; P1 C1 D axO 1 C hOz1 D Œa hı1 hT : Substituting the above terms into (4.47), and noting the definitions of B1I and HC1 , and Iy1 y1 D Iz1 z1 , we have mv21 h D 0; Ix1 x1 ıR1 mghı1 b v1 Ix1 x1 ıP1 mga D 0; b 2 .1 C ı /Iy y P C mv1 a D 0: 1
1 1
(4.48) (4.49) (4.50)
The above equations of motion for the bicycle near its balance state are developed given the assumptions made. The equations would be much more complex without those assumptions. From those equations, the steering angle and the rear wheel’s speed v can be determined to make ı1 close to the value needed for the balance of the bicycle. The effects of the design parameters (a, b, h, m1 , Ix1 x1 , etc.) on the dynamic behavior of the bicycle can also be studied. These interesting and challenging issues are beyond the scope of this book.
References
1. Kreyszig, E.: Advanced Engineering Mathematics, 7th Edition, John Wiley & Sons, Inc. (1993) 2. Johnson Lee, W., Dean Riess, R. and Jimmy Arnold, T.: Introduction to Linear Algebra, Addison Wesley Boston (2002) 3. Malvern, L. E.:Engineering Mechanics, Prentice Hall Eaglewood Cliffs, New Jersey (1976) 4. Massey W. S.: “Cross products of vectors in higher dimensional Euclidean spaces”, The American Mathematical Monthly, Vol.90, No.10, pp. 697–701 (Dec. 1983) 5. Hibbeler, R. C.: Engineering Mechanics-Dynamics, 12th Ed., Prentice Hall, Pearson Education South Asia Pte Ltd, Singapore (2010) 6. Tozeren, A.: Human Body Dynamics – Classical Mechanics and Human Movement, SpringerVerlag New York, Inc. (2000) 7. John Craig, J.:Introduction to Robotics: Mechanics and Control, Pearson Education, Inc, Upper Saddle River, New Jersey (2005) 8. Sciavicco, L. and Siciliano, B.: Modeling and Control of Robot Manipulator, International Editions, McGraw-Hill Book Co.-Singapore (1996) 9. K. J. Astrom, R. E. Klein and A. Lennartsson, “ Bicycle dynamics and control”, IEEE Control Systems Magazine, pp. 26–47, Vol. 26, Issue 4 (2005)
L. Huang, A Concise Introduction to Mechanics of Rigid Bodies: Multidisciplinary Engineering, DOI 10.1007/978-1-4614-0472-9, © Springer Science+Business Media, LLC 2012
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Index
A Acceleration, 42 Acceleration Angular acceleration, 42, 92 Centripetal acceleration, 44, 46, 100 Coriolis acceleration, 44 Linear acceleration, 42 Tangent acceleration, 44, 46 Azimuth angle, 22
Couple, 82 Couple moment, 82
B Base vector, 19 Basic rotation matrix, 28, 30
E Energy Mechanical energy, 86 Mechanical energy conservation, 86, 89 Equation of equilibrium, 110 Equation of motion, 90 External force, 91, 93
C Cartesian coordinate frame, 4 Center of mass, 28, 61 Circular motion, 46 Colatitude, 22 Composite-bodies rule, 69 Configuration space, 91 Conservation of angular momentum, 94 Conservation of linear momentum, 92 Conservative force, 83, 86, 122, 123 Coordinate frame, 19 Coordinate frame Body frames, 19 Cartesian coordinate frame, 19 Cylindrical coordinate frame, 20 Description frame, 22 Inertial frame of reference, 19, 90 Observation frame, 22, 23 Right-hand rule, 19 Spherical coordinate frame, 21 Universe frame, 19, 23, 90
D D’Alembert’s principle, 91, 110, 117 Degrees of freedom, 117 Directed line, 1 Dynamic balance, 114
F Force, 81
G Generalized coordinates, 91, 118 Generalized force, 91, 120 Generalized inertial force, 120 Gyroscopic moment, 102
I Impulse, 82 Impulse Angular impulse, 83 Linear impulse, 82 Inertia, 90
L. Huang, A Concise Introduction to Mechanics of Rigid Bodies: Multidisciplinary Engineering, DOI 10.1007/978-1-4614-0472-9, © Springer Science+Business Media, LLC 2012
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154 Inertia ellipse, 65 Inertia matrix, 63 Inertia sphere, 65 Inertial force, 91, 110, 120 Internal force, 91, 93, 110, 123 J Jacobian, 119 K Kinematics, 19 Kinetic energy, 84, 85, 122 L Lagrange function, 91, 123 Lagrange’s equations, 91, 123 M Mass, 61 Mass moment of inertia, 63 Mass product of inertia, 63 Matrix, 8 Matrix Addition and subtraction, 10 Adjoint, 14 Characteristic equation, 15 Cofactor, 13 Determinant, 4 Diagonal matrix, 9, 64 Differentiation, 16 Eigenvalue, 15 Eigenvector, 15, 28 Identify matrix, 9 Integration, 16 Inverse, 14 Jacobian, 7, 8, 119 Minor, 13 Multiplication, 10 Negative definite, 16 Orthogonal matrix, 9 Partitioning, 12, 13 Positive definite, 16 Quadratic form, 16 Rank, 15 Similar transformation, 16 Skew-symmetric matrix, 9, 38 Square matrix, 9 Symmetric matrix, 9, 63 Transposition, 10 Unit matrix, 9 Zero matrix, 9
Index Moment of a couple, 82 Moment of force, 81 Moment of inertia, 63, 65 Momentum, 73 Momentum Angular momentum, 73, 74, 77, 78 Linear momentum, 73, 74, 78, 92
N Newton’s second law, 90, 91 Newton–Euler formulation, 90–92, 94 Nonconservative force, 123
O Orientation, 26
P Parallel-axis theorem, 68 Parallel-axis theorem, 69, 95 Parallel-plane theorem, 68, 69 Planar motion, 47, 95 Position Description, 28 Notation, 23 Relation, 28 Potential energy, 85 Potential energy Elastic potential energy, 86 Gravitational potential energy, 85 Power, 83 Power of a couple, 83, 84 Principal axes of inertia, 63, 76, 85 Product of inertia, 63
R Resultant force, 81 Resultant moment of force, 81 Rotation Axis, 26 Rotation matrix, 25
S Speed, 34 Speed Angular speed, 34 Linear speed, 34
T Torque, 82
Index V Vector, 1 Vector Addition and subtraction, 3 Base vectors, 3 Column vector, 9 Cross product, 4, 11, 74, 81 Differential, 8 Differentiation, 5 Dot product, 3 Integration, 5 Linearly independent, 15 Norm, 4 Notation, 76 Orthogonal vectors, 4 Partial derivatives, 8 Row vector, 9
155 Triple products, 5 Unit vector, 1 Velocity, 34 Velocity Angular velocity, 34, 36, 38–40 Apparent velocity, 35 Average velocity, 35 Instantaneous velocity, 34 Linear velocity, 34, 40 Tangent velocity, 46 Virtual displacement, 110 Virtual work principle, 110
W Work, 83 Work of a couple, 83
