ME 6590 Multibody Dynamics
Examples for D'Alembert's Principle
Examples
1. Double Pendulum: Find the EOM of the
double pendulum shown at the right using
D'Alembert's principle and 1 and  2 as the
generalized coordinates. Both bodies have mass
m and length L.
D'Alembert's Principle:
2
vGi 
 Bi

m
a


 Fqk

 i Gi
   I Gi   Bi   Bi  H Gi  




i 1 
i 1
k 
k

2

(k  1, ,2)
where
 B  1k
 B1 1  k
 B1  2  0
 B   2k
 B2 1  0
 B2  2  k
vG1  12 L1e1
vG1 1  12 Le1
vG1  2  0
vG2  L1e1  12 L 2e2
vG2 1  Le1
vG2  2  12 Le2
1
2
aG1  12 L1e1  12 L12er1

 
aG2  L1e1  L12er1 

  v
  v
1
2
L 2e2  12 L 22er2

m a   v

   m  L   L  C  L  S 
   m  L   L  C  L  S 
m1aG1  vG1 1  14 mL21
m2aG2
m2aG2
G2
G2
1 G1
2
1
1
2
1
1
4
2
2
2
2
2
1
2
1
2
21
2
1
2
1
2
21
G1

 2  0
2
2 21
2
2
1 21
I G1   B1   121 mL2 1k
I G2   B2   121 mL2  2k
 
  mg j    v
 

    mg j    v

F1  mg j  vG1 1  mg j  vG2 1   32 mgLS1
F2
G1
2
G2

 2   12 mgLS2
1/4
Substituting into D'Alembert's principle gives the two differential EOM


4
3
mL2 1   12 mL2C21  2   12 mL2 S21  22   23 mgL  S1  0
1
2
mL2C21 1   13 mL2  2   12 mL2 S21 12   12 mgL  S2  0
2. Example System II (from ME 555): Find the
EOM of the bar using D'Alembert's principle,
given that the frame F is light (massless), the
bar B has mass m and length L, the motor
torque M (t ) is applied between the frame and
the bar, and that the motor torque M  (t ) is
applied between the ground and the frame F.
Use  and  as the generalized coordinates
Reference frames: F : (n1, n2 , k ) , B : (e1, n2 , e3 )
D'Alembert's Principle:
vGB 

 B
m
a

 F
 B GB
   I GB   B   B  H GB   






vGB 

 B
 F
 mB aGB 
   I GB   B   B  H GB   










where
F   k
 F   k
 F   0
 B   n2   k
 B   k
 B   n2  e2
vG  d n1
vG   dn1
vG   0
aG  d n1  d 2n2
F   k
 B   n1   n2   k  ( S  C )e1   e2  ( C   S )e3
maG  (vG  )  md 2
maG  (vG  )  0
HG  121 mL ( e2  C e3 )
2

IG   B  121 mL2  e2  ( C   S )e3

2/4
 B  HG  121 mL2  ( 2 S C )e2  ( S )e3 
I
I

  


   mL 
  H         mL  S C
  H        mL  S C
F   M n         M n    
F   M n         M n    
G
  B   B   121 mL2 ( C   S )C
G
B
2
1
12
B
B
G
B
B
G
B
2
1
12
1
12
2
2





 2
B
 2
F

 2
B
 2
F


    M k    

  
   M k    F   M

F
M
Substituting into D'Alembert's principle give the two differential EOM
 md
2
 121 mL2C2    16 mL2 S C   M 

1
12
mL2    121 mL2 S C  2  M 
3. Single Link Pendulum with Constraints: Find the EOM
of the pendulum using a dependent set of generalized
coordinates and D'Alembert's principle with Lagrange
multipliers. The bar has mass m and length L and is driven
by a motor torque M. The dependent set of generalized
coordinates is (q1, q2 , q3 )  ( x, y, ) .
D'Alembert's Principle:
2
vG

maG 
 IG   
 Fqk    j a jk
qk
qk
j 1

where
 k

(k  1, ,2)
vG  xi  yj
   k
vG x  i
vG y  j
  k
vG   0
aG  xi  yj
maG  vG x  mx
maG  vG y  my
maG  vG   0
I
G

 x   y  0

       121 mL2 
3/4




F  mgj  vG   Mk     M
Fx  mgj   vG x   Mk    x   0
Fy  mgj   vG y   Mk    y   mg
Constraints:
x  L2 S  0
or
y  L2 C  0
or
x   L2 C   0
y   L2 S   0
a11  1, a12  0, a13   L2 C
a21  0, a22  1, a23   L2 S
Substituting into D'Alembert's principle give the three differential/algebraic EOM
mx  1

my  mg  2
1
12
mL2   M   L2 C  1   L2 S  2
The two differentiated constraint equations are
x   L2 C    L2 S  2  0
y   L2 S    L2 C  2  0
4/4