ME 3600 Control Systems
Equations of Motion of a Simple Pendulum and Accelerating Simple Pendulum
Simple Pendulum
o A simple pendulum with end mass m is shown
in the figure to the right. The end mass is
assumed to be connected to the support with a
light, slender, rigid rod of length .
o The differential equation of motion of the
pendulum may be found by summing forces in
the tangential (e t ) direction. Referring to the
free body diagram, we write
 Ft  mat
  mg sin( )  m 
or
  g sin( )  0
(1)
o The solution of this differential equation describes the movement of the pendulum for
any initial conditions.
o Equilibrium Positions: These are found by setting   0 in the differential equation of
motion (1). Hence, the pendulum has two equilibrium positions: eq  0,  .
o Linearized Equation of Motion: eq  0
To study small motions of the pendulum
about the equilibrium position eq  0 , we let   eq   and linearize the equation of
motion (1) about that position. This is done by linearizing the function f ( )  sin( )
about   0 .
 df 
f  d
 0
  1    
o Hence, the approximate linear equation of motion for small motions about   0 is
 
 g    0
o The solution of this differential equation describes small movements of the pendulum
about   0 .
Kamman – ME 3600 – page: 1/2
Accelerating Simple Pendulum
o An accelerating simple pendulum with end mass m
is shown in the figure to the right. The end mass is
assumed to be connected to the accelerating base with
a light, slender, rigid rod of length .
o The base accelerates at a constant rate a0 to the left.
The acceleration of mass m may be found using the
relative acceleration equation as follows:
 e   e 
    a cos( )  e     a sin( )  e
am  a A  am / A  a0 i 
2
t
n
2
0
t
0
n
o The differential equation of motion of the pendulum may be found by summing forces
in the tangential (e t ) direction. Referring to the free body diagram, we write
 Ft  mat

  mg sin( )  m   a0 cos( )

 
 g sin( )  a cos( )   0
0
o The solution of this differential equation describes the movement of the pendulum for
any initial conditions.
o Equilibrium Positions: As before, these are found by setting   0 . Hence, the
a 
accelerating pendulum has two equilibrium positions: eq  tan 1 g0 . For example, if
a0  0.5 g , then eq  tan 1 (0.5)  26.57o ,206.57o .
o Linearized Equation of Motion: a0  0.5g and eq  26.57o
To study small motion
of the pendulum about eq  26.57o , we let   eq   and linearize the equation of
motion about that position. This is done
g
f ( )   sin( )  12 cos( )  about eq  26.57o .
 df
f ( )  
 d


g
    cos( )  12 sin( ) 

 eq 


 eq
by
linearizing

  1.118
g
the
function
 
o The approximate linear equation of motion for small motion about   26.57o is

  1.118
g
   0
o The solution of this differential equation describes small movements of the pendulum
about   26.57o .
Kamman – ME 3600 – page: 2/2