ME 5550 Intermediate Dynamics
Equations of Motion of Example System II
In previous notes for Example System II, we
found R B the angular velocity of the bar and
 IG e that the inertia matrix (associated with I G )
resolved in bar-fixed directions B : (e1, n2 , e3 ) to be
 B  (S ) e1   n2  (C ) e3
R
and
0 0 0
m 2
0 1 0
 IG  


12
0 0 1 
Reference Frames
The angular momentum of B was also found to be
H G  I G  R B  m12  n2  C e3 
2
The EOM (equations of motion) of B can be found using the Newton/Euler equations
along with the free-body diagrams shown at the right.
F  m a
M  I 
R
G
G
G
 B    R B  H G 
R
Given that   constant , the terms on the right side of
the moment equation may be calculated as follows
e1
2
m
R
B  HG 
S
12
0
n2


e3
m 2 2
C 
  S C n2  S e3 
12
C
0 0 0 C 
m 2

 m 2

IG   B 
0 1 0   
 n2  S e3 

12 
12

0 0 1  
 S 
R
Kamman – ME 5550 – page: 1/2
Inverse Dynamics (assuming  and  are constant)
F1  F3  0
F2  md  2
T1  0
T2  121 m 2 2 S C
T3   16 m 2  S
Forward Dynamics of Bar B (assuming   constant ,    , and    )
F1  F3  0
F2  md  2
T1  0
   2 S C  12T2 m
2
T3   16 m 2  S
Note that the fourth equation is a differential equation that describes the changes in the
angle  .
Equilibrium Positions for the Bar
If T2 (t )  0 , the bar exhibits equilibrium positions. These positions may be calculated by
setting all time-varying parts of the differential equation to zero. That is,
2 S C  0
This equation is satisfied when   0, / 2 . In general, these equilibrium positions may be
stable or unstable.
Kamman – ME 5550 – page: 2/2