 Equations of motion
1. Equations of Translational Motion
: Typically defined in terms of the acceleration of G

 dL

(1 Vector equation or 3 Scalar equations)
 maG
F 
dt
2. Equations of Rotational motion
z
XYZ: Fixed reference frame
Z
xyz : Moving frame (Origin at G)

Question: What is the relation between
G ri/G
mass mi
x
rG
y
ri
External force and Moment?
X
C
Let’s start with angular momentum of the i-th particle mi about G
(accelerating: noninertial) with respect to the xyz frame,
Linear Momentum

H i G  ri / G  mi vi / G


( ri / G & vi / G : vectors w.r.t. G)
Position of i-th particle from G
Time rate of change of angular momentum (Time derivative)

d 




( H i )G  ( H i )G  ri / G  mi vi / G  ri / G  mi vi / G
dt




=
0,
because
r

v
i/G
i/G
 ri / G  mi ai / G
Y
By summing up for all the particles,



H G   ri / G  mi ai / G 
i
: Time rate of change of angular momentum relative to G

 What is the ai / G ?



ai / G (Measured about noninertial G)  ai (absolute)  aG (motion of G)


where ai & aG : Vectors w.r.t. inertial XYZ frame)


 
Then, ( H i )G   ri / G  mi (ai  aG ) 
=0, since

 mi ri / G  0
i
i




  (ri / G  mi ai )  ( mi ri / G )  aG
i
i



Denoting  M G   (ri / G  Fi )
i

i

 M G  H G
i


where Fi  mai (External force)
or


 M O  H O
(if O is a fixed point)
i
: Equations of rotational motion about a translating xyz

(Net moments acting on a body = Time derivative of H )
Question. What is the relation between Moment and Change in Angular
momentum?

 Time derivative of H from fixed XYZ
= Time derivative measured from the moving xyz frame
+ Motion of xyz


Consider a rotating rigid body having  and xyz frame having  (  )

From Chap. 20 (Time derivative of a vector in a rotating frame)



 
M

H

(
H
)


 HO
 O
O
O xyz



 
M

H

(
H
)


 HG
 G
G
G xyz


where ( H ) xyz : Time derivative of H w.r.t. xyz frame

 Motion analysis of a rotating rigid body (with  )
using the rotational equations of motion

: Depending on the choice of rotating xyz (with Ω )

Choice 1. x y z frame having only a translational motion (   0 )

e.g. xyz with an origin at G and   0 , then

M

(
H
 G
G ) xyz
: Looked simple? but its analysis is NOT,

Direction of  of a body about xyz: Not constant of time
 Moment and Products of inertia: Function of time
 
Choice 2. xyz axes having a motion of   
: Fixed in and moved with the body

Since the direction of  of a body about xyz: Constant of time
 Moments and products of inertia about xyz: Constant
From the rotational equations of motion,


 
M

(
H
)


 H (Neglecting the subscripts O and G)

xyz


where   x iˆ  y ˆj  z kˆ & H  H xiˆ  H yˆj  H zkˆ
1) x component
 M x  ( H x ) xyz  ( y H z   z H y )
 ( I xx x  I xy y  I xz z )  [ y ( I zx x  I zy y  I zz z )
  z ( I yx x  I yy y  I yz z )]
 I xx x  ( I yy  I zz ) y y  I xy ( y   z x )
 I yz ( y 2   z 2 )  I zx ( z   x y )
2) y component
 M y  I yy y  ( I zz  I xx ) z x  I yz ( z  x y )
 I zx ( z 2   x 2 )  I xy ( x   y z )
3) z component
 M z  I zz z  ( I xx  I yy )x y  I zx ( x   y z )
 Ix xy ( x 2   y 2 )  I yz ( y   z x )
 In special cases
If x, y, z axes: Principal axes of inertia
(i.e. All products of inertia = 0 & I xx  I x , I yy  I y , I zz  I z )
 M x  I x x  ( I y  I z ) y y
 M y  I y y  ( I z  I x )zx
 M z  I z z  ( I x  I y ) x y
 Simple ways to find  x ,  y ,  z
: Euler equations of motion
(observed from moving xyz frame!)
 
Since    ,

 


 from XYZ frame  ( ) xyz      ( ) xyz
: Independent to the frame of axes


 Method 1. Find the time derivative of  w.r.t. fixed XYZ axes (  )

and Determine the components of  along xyz (  x ,  y ,  z )

Method 2. Find the components  along xyz axes (  x ,  y ,  z )
and take the time derivatives of the components (  x ,  y ,  z )
 
Choice 3. xyz axes having a motion of   
: Particularly suitable if the body is symmetrical about its
spinning axes. (i.e. xyz: principal axes)
e.g. Gyroscopes, spinning tops

Since all products of inertia = 0, ( H  I x xiˆ  I y y ˆj  I z z kˆ )
 M x  ( H x ) xyz   y H z   z H y  I x x  I z  y z  I y  z y
 M y  ( H y ) xyz   z H x   x H z  I y y  I x  z  I z  x z x
 M z  ( H z ) xyz   x H y   y H x  I z z  I y  x y  I x  y x

where  x( y, z ) : x (y, z) component of  from XYZ frame
&  x ,  y : Determined relative to the rotating x y z axes
 Effect of Forces and Moments to Linear and Angular momenta
- Principle of impulse and momentum
Application of forces 
 dp G d (mvG )
 F  dt  dt
tf



F
  dt  m(vG ) f  m(vG )i
: 3 scalar equations
ti
: Sum of all the impulses by the external force
=
Change in the linear momentum


dH O
Application of moments   M O 
dt
tf



M
dt

(
H
)

(
H
 O
O f
O )i
: 3 scalar equations
ti
: Sum of all the angular impulses by the external moments
=
Change in the angular momentum
Force (Moment) felt by object

Change of Linear (Angular) momentum
Time duration of Force applicatio n
e.g. For a same amount of momentum change,
Long time duration

Weak force

Soft impact
Short time duration

Strong force

Hard impact


z
 Kinetic Energy of a body in 3D motion

v
Kinetic energy of a rigid body
= Sum of the kinetic energy of all
dm

rA
differential element dm
G

A
A
x

 Kinetic energy of dm with v about from XYZ
  
 
1
1
  1

dT  dmv 2  dm(v  v )  dm[(v A     )  (v A     )]
2
2
2
 
 
1
1
 
  
 dm(v A  v A )  dmv A     A  dm(   A )  (   A )
2
2
 Kinetic energy of a whole body


 
1  
1  

T  m(v A  v A )  v A  (    Adm)   (   A )  (   A )d m
2
2m
m


 
1  
1  

 m(v A  v A )  v A  (    Adm)      A  (   A )d m
2
2 m
m
: General expression of the kinetic energy of a rigid body
y
For special cases,
(1) Point A is Fixed
(Point O)

i.e. v A  0
 
1  
1  
T     O  (  O )d m    H O
2 m
2
Furthermore, if x, y, z axes are the principal axes,

(i.e. H O  I x xiˆ  I y y ˆj  I z z kˆ )
T 
1
1
1
I x x 2  I y y 2  I z z 2
2
2
2
ii) Point A is the Mass Center G
i.e. A = G, then
1  
1  
1 2 1  
T  m(vG  vG )    H G  mvG    H G
2
2
2
2

 G dm  0
m
Furthermore, if x, y, z axes are the principal axes,
1
1
1
1
 T  mvG2  I x x 2  I y y 2  I z z 2
2
2
2
2
 Effect of Forces and Moments to Kinetic Energy
(Principle of work and energy)
U i  f
f
 
   F  ds  T f  Ti
i
: Work done by all external forces from the initial to final position
= Change in translational and rotational kinetic energy