CHAPTER 7 Basic Concepts and Kinematics of Rigid Body Motion

7.1 Degrees-of-freedom: F2 F1 (of a rigid body) m f m2 1 12 f Consider three 21 Z f r 23 unconstrained particles 1 C m f32 rC 3 O m r3 3

X F3 Y 1 The positions are defined by

ri  xi i  yi j zi k i = 1, 2, 3.  degrees of freedom = n = 9 = 3N Now, constrain the particles  (three particles placed l m m1 1 2 at the corners of a Z triangle whose sides r1 l2 are formed by rigid l3 massless rods) O m r3 3 X

Y 2 Now: there are three constraints

r1 r 2  l 1;; r 3  r 2  l 2 r 1  r 3  l 3 2 2 2 2 or(x2 x 1 )(  y 2  y 1 )(  z 2  z 1 )  l 1  0, etc .  n = 3(N) - 3 = 6  degrees of freedom. Now: as another particle is introduced, its position is specified by 3 additional coordinates, but also have 3 additional constraints.  four rigidly connected particles also have only 6 degrees of freedom.

3 In general: a rigid body (any collection of particles whose relative positions are fixed) has 6 degrees of freedom.

• translational motion of a point on the body - specified by 3 translational degrees of freedom. • rotational motion about the specified point - 3 rotational degrees of freedom.

4 Laws of motion for a system of particles (extended to a rigid body) a) F mrc  translational motion of the C.M. (3 degrees of freedom)

d b) MH about C.M. or an inertially dt fixed point  rotational motion about the C.M. (3 rotational degrees of freedom) 5 7.2 Moments of Inertia Z Recall: the notation P and definitions rP  C i C m fi2 r mi The equation for O C ri moment about an X Fi arbitrary point P is: Y d M  H   mr p p c p dt 6 Now, the angular momentum about the point P is N H   m  p  i i i1 i where   - velocity of as viewed by a i mi non-rotating observer translating with P. (relative velocity in an inertial frame)

For a rigid body - suppose that P is fixed in the body   = constant, and       i i i where  - angular velocity of the body.

7 Thus, the angular momentum about P is  N Z H  m       p  i i  i  t1 dm By analogy: for a rigid body r  rotating with r angular velocity  O P P X V Y H  (r) (  )dV ; dm  (r)dV   V 8 We now consider the various cases: Reference point P is at origin: Z  x i  yj  zk  dm  xi   y j   z k Then 22  (    )  [(yz  ) x O=P xyy  xz  z][ i   xy  x 22 X V ()]x  zyz  yz j Y 22 [()] xzx  yz  y  x  y  z k 9 Let us define: I  (y2  z2 )dV xx  V I  (x2  z2 )dV ; I  (x2  z2 )dV yy  zz  V V

These are the moments of inertia

Ixx  about x axis   Iyy  about y axis through the reference point O  P  Izz  about z axis 

10 Similarly, we define products of Inertia: I  I    xy dV xy yx  V I  I    xz dV xz zx  V I  I    yzdV yz zy  V

 Hp  Hx i  Hy j Hz k (angular momentum vector for the body, or angular momentum about P) 11 Here, HIIIx xx x  xy  y  xz  z

HIIIy yx x  yy  y  yz  z

HIIIz zx x  zy  y  zz  z In compact notation: 33 HP   Iij j ei i 1 j 1 where e1  i , e2  j, e3  k and symbols 1, 2, 3  x, y, z.

12 The components of the angular momentum

are 3 Hi I ij j , i 1,2,3 j1

Note that I ii is the second moment of the mass distribution with respect to a Cartesian axis. Radius of gyration Iii (effective location of kii , 1,2,3 m a mass point) or 2 Iii mk i 13 Three moments of inertia and three products of inertia specify the inertia properties of a rigid body with regards to rotational motion.

• Note that P is fixed in the rigid body and  is the angular velocity of the rigid body. • No assumption is made concerning the rotational motion of the xyz system. The angular velocity in terms of the xyz system,

 x i  y j z k is valid at the instant considered. 14 • as well as are, in general, x , y ,z Iij functions of time depending on the orientation of xyz relative to the rigid body.

• To avoid the difficulties associated with treating as functions of time, often one chooses xyz system that is fixed to the rigid body and rotates with it.

• Called a body-fixed coordinate system 15 7.3 MATRIX NOTATION:

Consider H Hx i  H y j  H z k If i , j , k are known, the three scalar components H x , H y , H z can be used to represent H H x    • We write  H    H y  as a column vector.   H z  • A force F can be represented as , a row vector. F  Fx , Fy , Fz  16 Fx    or  F    F y  as a column vector.   Fz  A square matrix is a n  n array of elements: e.g. the elements of inertia I ij can be written as a square matrix

I xx I xy I xz    I   I yx I yy I yz    I zx I zy I zz  This is called an inertia matrix for the body. 17 The angular velocity vector can be represented as x   ,   y a column vector.  z T  •   x  y  z ,the transpose of a column

vector gives a row vector, etc.

   

 H • consider , the product of a row vector with a column vector

HHHHH  x x   y y   z z   

18 In scalar components, it is easy to see that

H  Iij j ei can be expressed as H I Clearly, multiplication of with I transforms    into the vector  H  , usually with a different magnitude as well as direction. In general, The angular velocity and the angular momentum vectors for a rigid body are in different directions 19 Assignment: Complete the review of matrix operations in 7.3. 7.4 Kinetic Energy For a system of n particles, the kinetic energy n is 112 T mvc  m i i  i 22i1 where

• vC - speed of the center of mass  th • i - velocity of the i particle as viewed from the C.M. 20 For a set of particles rigidly connected and the assemblage rotating with angular velocity  ,

ii  

i  i()()()    i    i   i    i 11nn Trot  m i i  i    i  m i  i 22ii11 (using permutation in scalar triple product)

Now, for a continuous mass distribution

21 11 T     dV       dV rot  22VV

HC  1 dm or T  H rot2 C If P is a fixed point: 1  TH  rot2 P O=P In vector matrix notation V 11T THHrot P    P 22 22 1 T Since HITI  ,      P rot 2 If one uses xyz coordinate system located at the center of mass of the rigid body, 1 T TIIII    [ 2   2   2 rot2 xx x yy y zz z

2IIIxyxy   2 xzxz    2 yzyz   ]/ 2 In summation, notation 1 33 TIrot  ij i j 2 ij11 23 Ex: If  has the direction of one of the coordinate axes at an instant, 1 TI  2 rot 2 Here, I - moment of inertia about the axis of rotation,  - instantaneous angular velocity of the rigid body.

24 7-6 Translation of Coordinate Axes

Z’ Z O’x’y’z’- coordinate system located at the C=O’ centroid of the body; Y’ O Oxyz-any other coord Y system; X X’ O’=C-the centroid O: center of mass: the coordinates of the center of mass in Oxyz system are (xc , yc , zc )

25 Let I xx , I yy , I zz - moments of inertia about xyz axes system

I xx , I yy , I zz = moments of inertia about centroidal axes m - mass of the body Easy to show (read 7.6) that for moments of inertia 2 2 I xx  I xx  m(yc  zc ) 2 2  I yy  I yy  m(xc  zc )parallel axis theorem 2 2  I zz  I zz  m(xc  yc )  26 For products of inertia

I xy  I xy  mxc yc

I xz  I xz  mxc zc

I yz  I yz  myc zc

• Note that the moments of inertia about the centroidal axes are the smallest. • The products of inertia may increase or decrease compared to those about centroidal axes depending on the particular case. 27 7.7 Rotation of Coordinate Axes Consider two different coordinate systems. We assume that the origins for the two systems coincide. z’ z P r consider a vector:

O y’ r  xi  y j  zk x  xi y j zk x’ y These are two ways of expressing the same vector r . 28 The two systems are characterized by unit vectors i, j,k and i, j,k

Let i  lxx ilyx jlzx k • are the cosines of the angles lxx , lyx , lzx made by the x axis with the x, y and z directions, respectively. Note that,

i lx x i  l y  x j   l z  x k  1 l2  l 2  l 2 1 xx y'' y z z 29 Consider the example: z’(k’)  xx'  direction x(i)  z’x  angles of yx'  x’x  x-axis y’x   O zx'  x’(i’) y’(j’)

Also, lx'' x cosx'' x , l y x cos y x ,

lzx'  coszx' 30 Similarily, we can write

j lx y i  l y  y j   l z  y k 

k lx z i  l y  z j   l z  z k  Thus, the vector r  x i  y j  z k can be written as r x() lx x i  l y  x j   l z  x k 

y() lx y i  l y  y j   l z  y k 

z() lx z i  l y  z j   l z  z k  x i   y  j   z  k 

31 In vector-matrix notation

x'  lx''' x l x y l x z  x      y'   ly''' x l y y l y z  y      z'  lz''' x l z y l z z  z  In more compact notation r lr where  r  and  r   are representations of r and r in the two coordinate systems. It provides relation between components of the same vector in two coordinate systems. 32 Ex: As another example, consider the Kinetic energy of a rigid body

z’ z O =C, the center of mass  of the body. There are two coordinate axes, xyz, O y’ x’y’z’ x   i  j  k x’ y x y z x i y jz k Here    and    are two representations of the same angular velocity vector. 33 Then, the kinetic energy expressions in the two coordinate systems are 11TT TII          rot 22 Also,   l  or TT   lT

11TTT T   I      l  I l  rot 22 and I   lT  I l In component form, this transformation of 33 inertia matrices is 1 Iij  l m'''' i l n j I ' m n 2 mn' 1 ' 1 34 Some properties of the matrix [l]. It is an orthogonal matrix, i.e., 1  llT   (the identity matrix) or lT   l1

Since det  ll   det(   T ) , (true for any matrix) (detl )2  1 the matrix operation with [l] only rotates a given vector.

35 T The relation  1    ll    can be explicitly written as 3 1 if i j llm'' i m j   m'1 0 if i j These are nine relations in the nine components of the matrix [l]. Six of the these equations are linearly independent.  There are 9 - 6 = 3 rotational degrees of freedom for a set of orthogonal coordinate axes (or for a rigid body). 36 Ex: Rotation of axes Consider a vector r z’ z that is represented 30 in xyz and x’y’z’ y’ axes. The two 30 O systems are rotated y 30° r x by 30° with respect z to each other. x,x’ y Then, r x i   y  j   z  k  r'   l r 37 z’ z 30 r'   l r y’  O Now x’z 30 y x  l x  l y  l z 30° x x x  y x  z x’y x,x’ lxx cos0 1, lx ycos x y  cos90  0 lx zcos x z  cos90  0 y  ly x x  l y  y y  l y  z z 38 ly xcos y x  cos90  0, ly ycos y y  cos30  3 / 2 ly zcos y z  cos60  1/ 2 z  lz x x  l z  y y  l z  z z llz x0, z y  cos120   1/ 2 lzz cos30 3 / 2 xx 1 0 0       yy 0 3 / 2 1/ 2       zz 0 1/ 2 3 / 2   39 7.8 Principal Axes consider the inertia properties of a rigid body I (y22 z )dV z’ z xx  dm V I (x22 z )dV  yy  O y’ V x I (x22y )dV x’ y zz  V Then, it is easy to see that I +I +I 2 (r 2 )dV xx yy zz  V 2 wherer2 =x 2 y 2  z 2  x 2  y  2  z  2   40 Ixx +I yy +I zz I xx  +I yy   +I zz   i.e., the sum of moments of inertia is invariant to coordinate system rotation. More generally, tr[I]sum of diagonal terms is unchanged due to coordinate rotation (an orthogonal transform). Consider products of inertia: As an example: I    xy dV xy  V A 180  rotation about the x-axis  41 z

 Ix y =-I xy 180 O Ix z =-I xz y’ y 180° IIy z yz x,x’ z’

• In general, the products of inertia have no preferred sign; the sign depends on the orientation of the body with respect to the coordinate system. 42 For a body with random orientation, positive and negative values of products of inertia are equally likely to occur.

• The moments and products of inertia are a smooth function of the orientation of the coordinate system orientation since I   l I lT relates the inertia properties in two systems  It is possible to find a coordinate system in which the products of inertia vanish simultaneously 43  Such a coordinate system is called the principal axes of the rigid body. • Consider the relation (alternate way to think)

Q: Is it possible to find a coordinate system in which the angular momentum vector is instantaneously parallel to the angular velocity vector? i.e., can we write  H   I    for some system?

44 If one can do that, then I I I 1 or ()IIIIxx xy xz x  IIIIxy( yy ) yz y 0  IIIIzx zy() zz  z (This is really the eigenvalue problem for the inertia matrix [I] ). If     0 , i.e. ,   0 , then det ([I] - I [1]) = 0

45 Or ()IIIIxx xy xz

IIIIxy( yy ) yz 0

IIIIzx zy() zz 

(this is a characteristic equation, a cubic in I

with coefficients I ij : 3 2 aoI  a1I  a2I  a3  0. Let I 1 , I 2 , I 3 be the roots of the cubic and 1,,  2  3 be the eigenvectors associated with the eigenvalues: 46 They satisfy ii, i = 1, 2, 3. []II  i   Clearly, the eigenvectors are known only up to an arbitrary constant, i.e., only the ratio of the components are fixed; in particular, if we assume  x to be arbitrary and non-zero, the ratios satisfy

()IIIyy i yz y/ x  I xy      III() zy zz i z/ x  I zx  47 If one chooses the constraint 2 2 2 x  y z 1, then the values of  x ,  y and  z are direction cosines of   they determine the direction of the corresponding principal axis at the reference point P. It is easy to show that if two principal moments of inertia are distinct, the corresponding eigenvectors are orthogonal, i.e., T 120 ifII .     12 48 THE ELLPSOID OF INERTIA Read from the text. Ex. 7.4 (text) Consider a given body with a coordinate system located at its centroid. z dm In this coordinate  system, the inertia C properties of the body x y are given by a matrix [I]. 49 150 0 100 2 I 0 250 0 kg m 100 0 300 We want to find the principal moments of inertia and the associated directions; also, the coordinate transformation [l], specified by the rotation matrix, which diagonalizes the inertia matrix. Essentially, we are considering the eigenvalue problem I I for I. 50 The characteristic equation for the matrix [I] is 150I 0 100  0 250I 0 0 100 0 300 I  (250  I) (I2  450I 3.5104 )  0 The roots, the principal moments of inertia, are 2 2 2 I1 100kg  m ,I2  250kg  m ,I3  350kg  m We now need to find the eigenvectors, which give directions of the principal axes.

51 The eigenvalue problem is   I  I[1]  0

or (150  I i )  x  100 z  0 (1) (2) (250 Iiy ) 0 (3) 100x  (300 I i ) z  0 2 a) Consider the 1st eig.val.: I 1  100 kg  m :

(2)   y  0 (1)  50x 100z or z /x 1/ 2. Thus 1 T   x 1 , 0 , 1/ 2 . 52 2 a) Consider the 2nd eig.val.: I 2  250 kg m :

(2)  y is arbitrary

(1)  100xz  100  0 (3)  100xz  50  0 The only possible solution to these equations is 2 T x  z 0     y  0 , 1 , 0 . 2 c) Consider the 3rd eig.val.: I 3  350 kg m :

(2)  Then (1), (3)  zx/ 2.  y  0 Thus, 3 T  x 1, 0 , 2 . 53 Note: The three eigenvectors are, as expected, orthogonal to each other. iT  j 0,ij .

3 z {} Note that z’ 13,  

3 are in xz plane {}1 C x’ {}2 1 y, y’ x Clearly, x’y’z’ is the principal coordinate system 54 The direction cosines for the x  y  z  system are:

lx xcos x x  cos1  2 / 5

lx ycos x y  cos90  0

lx zcos x z  sin1  1/ 5

Similarly, ly x cos y x  cos90  0,

lly ycos y  y  cos0  1, y  z  cos y  z  cos90  0 and lz x 1/ 5, l z  y  0, l z  z  2 / 5

These allow us to construct the rotation matrix 55 Thus, 2 / 5 0 1/ 5  l  0 1 0  1/ 5 0 2 / 5

• It is easy to check that

100 0 0  T 2 []I l [][] I l  0 250 0 kg  m 0 0 350

56 7.9 Displacements of a Rigid Body Euler’s Theorem: The most general displacement of a rigid body with one point fixed is equivalent to a single rotation about some axis through that fixed point Chasles’ Theorem: The most general displacement of a rigid body is equivalent to a screw displacement, i.e., translational motion of a reference point followed by rotation about an axis through the ref. point Assignment: Read the Section 7.9 57 7.10 Axis and Angle of Rotation • We know that the components of a vector in two different coordinate systems are obtained by the application of a rotation matrix [l] whose elements are the direction z cosines. z’ P r

O y’ x y x’ 58 Then, the vector r can be written as r xi  yj  zk  xi   yj    zk  

x'  lx''' x l x y l x z  x      y'  ly''' x l y y l y z  y      z'  lz''' x l z y l z z  z  A natural question: Is there a vector whose coordinates remain unchanged as the coordinate system xyz is rotated to the coordinate system xyz? 59 If such a vector r exists, r xi  yj  zk  xi  yj   zk  where i ,,,, j k and i  j  k  are the basis vectors that define the two coordinate systems, then, given the rotation matrix [l], the relation  r '   l   r  gives

x  lx''' x l x y l x z  x      y   ly''' x l y y l y z  y      z  lz''' x l z y l z z  z 

60 or1 r l r 0 must be satisfied.

This is an eigenvalue question for matrix [l]. • Thus, the existence of such a vector  r  (or r ) is associated with matrix [l] having an eigenvalue of 1. The corresponding eigenvector will then define the direction which remains unchanged due to rotation, and hence, represents the axis of rotation.

61 7.10 Axis and Angle of Rotation

7.11 Reduction of Forces - equivalent forces and couples Reading Assignments

62 7.12 Infinitesimal Rotations • Consider a sequence of rotations: the new and old representations of the vector r are related by r'  l r   r z z’ P (x,y,z) r or (x’,y’,z’) O y’ x y x’

63 Ex: Consider a counterclockwise rotation about the x axis. z’ z P 1 r Then, y’ 1 rr'  1   y 1 O where, x,x’ the rotation matrix is 1 0 0 0 cos sin  1  1 1 0 sin11 cos 64 Suppose we perform two rotations in a sequence:

r  r  1  r

r  r   2 r    2  1  r combining r  r   r where  21   (combined rotation matrix)

65 Since matrix multiplication does commute,

  2  1   1  2 

Thus, the order in which rotations are accomplished is crucial to know for finite rotations.

We now show that infinitesimal rotations commute!!

66 Ex: Consider a sequence of two rotations given below: 1. xyzx’y’z’. 2. x’y’z’ x”y”z”

Let us call it 1  Let us call it  2  z’ z z’ z”

y z

y y’ y’ y” z y x  y z x” x’ x’ 67 cosz sin  z 0   1  z 0     sin  cos  0      1 0   1   z  z z   z  0 0 1   0 0 1 

[1] [1 ]

cosy 0 sin  z 1 0  y    0 1 0  0 1 0  2  y   sinz 0 cos  z  y 0 1

[1] [2 ]

68 Thus, the two sequences give:

2  1 ([1]  [ 2 ])([1]  [ 1 ])

[1]  [1 ]  [  2 ]  [  2 ][  1 ]

[1]  [12 ]  [ ]

1  2 ([1]  [ 1 ])([1]  [ 2 ])

[1]  [1 ]  [  2 ]  [  1 ][  2 ]

[1]  [12 ]  [ ] Order not important and rotations can be added vectorially. Infinitesimal rotations  angular velocities add as vectors 69 Ex: Consider now a sequence of three infinitesimal rotations:

1. x - about x axis

r  r x  r

2. y - about y’ axis       r  r y  r    y   x  r 3. z - about z” axis     r  r z r    z  y  x  r rr [1] ˆ   ˆ   ˆ    z  y x    70 Here,

0 0 0 0 0  y  ˆˆ0 0  ,  0 0 0  x  x y   0xy 0 0 0

00 z ˆ 00  zz  0 0 0

71 Let ˆ ˆ ˆ ˆ [] xyz     

0 zy Then, [ˆ ]  0  zx zx 0 and the complete relation is r [1] ˆ   ˆ   ˆ r  [1]   ˆ r    z  y x       

72 Now, apply an infinitesimal rotation to a vector r of constant length. e.g.: consider a y rotation about z axis. r r  Then, z r’ rr [1]     O x or r  r   r

73 Note: in the above, we considered rotation of coordinate axes with vector fixed in space  any changes in components of r are entirely due to coordinate axes rotations. Now: Consider coordinate system fixed and let the vector rotate in opposite direction. Suppose that this rotation takes place in time t. Then, rr   r  lim t 0 t

74 [] r or r lim [][] r or r   r t 0 t [] Here []   the angular lim t t 0 velocity matrix The negative sign is introduced so that [] refers to rotation of r and not to that of the coordinate system. Recall: for a vector of constant length, rr

75 Thus: rr and rr  []   are statements of the same fact in different forms. Note: []  is a skew symmetric matrix

0 zy  consider the matrix: [ ] zx 0   yx0

76 Then: ()zyy z x     r []() r  zx x   z    y      ()yxx y z Also: i j k

rr   x  y  z x y z

()() zy   y z i   z x   x z j

() yxx  y k 77