Chapter 21. 3-D Kinetics of a Rigid Body
 Basic Questions
1. What is the origin of 3D motion of a rigid body?
 
 
- m, p , F (Translation) and I, H , M (Rotation)
2. How to establish the equations of motion of a body in 3-D motion?
3. How to establish the energy and momentum principles?
 Equations of motion for a rigid body in 3-D motion
 d mv  dp

1. Translation:  F 
dt
dt

Mass m: Resistance to F

 d I  dH

2. Rotation:  M 
dt
dt

Moment of inertia I: Resistance to M (2-D motion)
or
Inertia tensor I in 3-D motion
 Inertia Tensor (Moments of inertia + Products of inertia)
: Depending on the shape of body, axes and planes of rotation
1. Moments of Inertia (Mass distribution about a reference axis)
Moment of inertia for a differential element dm
 dm  r 2 (r: Shortest distance from dm to the reference axis)
e.g. About x axis,
dI xx  rx2dm  ( y 2  z 2 )dm
z
where rx 2  y 2  z 2
: Distance from dm to x axis
dm
z
y
x
 Moment of inertia of a whole body about x, y, z axes
I xx   rx2 dm   ( y 2  z 2 )dm > 0 (Positive)
m
m
I yy   ry2 dm   ( x 2  z 2 )dm > 0 (Positive)
m
m
I zz   rz2 dm   ( x 2  y 2 )dm > 0 (Positive)
m
m
x
y
2. Products of Inertia (Mass distribution about reference planes)
Product of inertia for a differential element dm
 dm r1r2
( r1, r2 : Shortest distances from 2 orthogonal planes to dm)
e.g. About y-z and x-z planes,
dI xy  xydm  dI yx
where x (y): Distance from dm to y-z (x-z) plane z
dm
z
y
x
x
y
 Products of inertia Ixy, Iyz, Izx of total body about x-y, y-z, z-x planes
I xy  I yx   xy dm > 0 or < 0 or = 0
m
I yz  I zy   yz dm > 0 or < 0 or = 0
m
I xz  I zx   xz dm > 0 or < 0 or = 0
m
 The sign of a Product of inertia depends on the symmetry of mass
distribution about its orthogonal planes.
3. Inertia tensor - Inertial property of a rotating body in 3D motion
 Definition:
 I xx

I    I yx
 I
 zx
 I xy
I yy
 I zy
 I xz 

 I yz 
I zz 
If the body shape is symmetry about x-y, y-z, and x-z planes,
 All products of inertia = 0,
i.e.
 I xx

I  0
 0

0
I yy
0
0 

0  : Diagonalized tensor
I zz 
 Ix = Ixx, Iy = Iyy, Iz = Izz : Principal moments of inertia
a
4. Moment of inertia about an arbitrary axis of rotation
z
Z
If, tensor I about xyz reference frames: Known
dm
Q: What is the moment of inertia about an
r 
arbitrary Oa axis along an unit vector û a ?
ua
b = r sin 
O
x
Moment of inertia for a differential element dm,
= dm  b 2 [b: the shortest (perpendicular) distance from dm to Oa]

( b  r sin   uˆa  r )
By definition of moment of inertia,
2
I Oa  m b 2 dm   uˆa  r dm 
m


 uˆa  r   uˆa  r dm
m

where uˆa  u xiˆ  u y ˆj  u z kˆ and r  xiˆ  yˆj  zkˆ

 uˆa  r  (u y z  u z y )iˆ  (u z x  u x z ) ˆj  (u x y  u y x)kˆ
Then, I Oa   [(u y z  u z y ) 2  (u z x  u x z ) 2  (u x y  u y x) 2 ]dm
m
 u x 2  ( y 2  z 2 )dm  u y 2  ( z 2  x 2 )dm  u z 2  ( x 2  y 2 )dm
m
m
m
 2u x u y  xydm  2u y u z  yzdm  2u z u x  zxdm
m
m
m
 I Oa  I xxu x 2  I yy u y 2  I zz u z 2  2 I xyu xu y  2 I yz u y u z  2 I zx u z u x
where u x , u y , u z : Direction cosines of x y z axes
y
5. Parallel-axis and Parallel-plane theorems
If Moments and Products of inertia about G (Mass center)
& Position of G (xG, yG, zG) about the axes of interest: Known
Moments of Inertia
Ixx = (Ix’x’)G + m(yG2 + zG2)
Iyy = (Iy’y’)G + m(xG2 + zG2)
: Parallel-axis theorem
Izz = (Iz’z’)G + m(xG2 + yG2)
Products of Inertia
Ixy = (Ix’y’)G + mxGyG
Iyz = (Iy’z’)G + myGzG
: Parallel-plane theorem
Izx = (Iz’x’)G + mzGxG
z’
z
G
y’
x’
zG
y
xG
x
yG

 Angular momentum, H


z

Consider a rotating rigid body with 

v
x’

rA
Q: What is the angular momentum
about a point A?
G
dm 
A
A
y
x
Let XYZ be the inertial frame of reference
 Differential angular momentum of a differential element dm about A


 
= dH A   A  v dm
where  A : Position vector from A to dm

v : Velocity of dm in the XYZ
 
 
Since v  v A     A from 3-D kinematics,

 

 

 

dH A   A  (v A     A )dm   A  v Adm   A  (   A )dm
 Total angular momentum of a whole body

HA 





  A  v A dm    A  (   A )dm
m
m


 

   Adm  v A    A  (   A )dm
m
m
: General expression of Angular momentum
 Angular momentum of a body for special cases

(Point O) i.e. v A  0
(1) Point A is Fixed


 
H O   O  (  O )dm
m
(2) Point A is a Mass Center

(Point G) i.e.  G dm = 0
m


 
H G   G  (  G )dm
m
(3) Arbitrary Point A



 

H A    Adm  v A    A  (   A )dm
m
m



 G / A  mvG  H G
(See Prob. 21-22)

where  G / A : Position vector from A to G

vG : Velocity of G in the XYZ

 Rectangular components of H

  
H     (   )dm
For a fixed point A,
m
 

Q: Express H ,  and  in terms of x, y, z components.
H xiˆ  H y ˆj  H z kˆ   ( xiˆ  yj  zk )  [( xiˆ   y ˆj   z kˆ)  ( xiˆ  yˆj  zkˆ)]dm
m
= [ x  ( y 2  z 2 )dm   y  xydm   z  xzdm] iˆ
m
m
m
+ [ x  xydm   y  ( x 2  z 2 )dm   z  yzdm] ˆj
m
m
m
+ [ x  zxdm   y  yzdm   z  ( x 2  y 2 )dm]kˆ
m
m
m
 ( I xxx  I xy y  I xzz )iˆ  ( I yxx  I yy y  I yzz ) ˆj
 ( I zx x  I zy y  I zz z )kˆ

 Rectangular components of H
x component :
Hx = Ixxx – Ixyy - Ixzz
y component :
Hy = -Iyxx + Iyyy - Iyzz
z component : Hz = -Izxx – Izyy + Izzz
If x y z axes: Principal axes of inertia, i.e. all products of inertia = 0
 Hx = Ixx
Hy = Iyy
Hz = Izz
(Same as 2-D motion)