Page 200 - Intro to Tensor Calculus

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EXAMPLE 2.2-4. (Compound pendulum) Find the equations of motion for the compound pendulum
illustrated in the figure 2.2-4.
1
2
Solution: Choose for the generalized coordinates the angles x = θ 1 and x = θ 2 illustrated in the figure
2.2-4. To find the potential function V for this system we consider the work done as the masses m 1 and
m 2 are moved. Consider independent motions of the angles θ 1 and θ 2 . Imagine the compound pendulum
initially in the vertical position as illustrated in the figure 2.2-4(a). Now let m 1 be displaced due to a change
in θ 1 and obtain the figure 2.2-4(b). The work done to achieve this position is

W 1 = −(m 1 + m 2 )gh 1 = −(m 1 + m 2 )gL 1(1 − cos θ 1 ).

Starting from the position in figure 2.2-4(b) we now let θ 2 undergo a displacement and achieve the configu-
ration in the figure 2.2-4(c).

Figure 2.2-4 Compound pendulum

The work done due to the displacement θ 2 can be represented

W 2 = −m 2 gh 2 = −m 2 gL 2(1 − cos θ 2 ).

Since the potential energy V satisfies V = −W to within an additive constant, we can write

V = −W = −W 1 − W 2 = −(m 1 + m 2 )gL 1 cos θ 1 − m 2 gL 2 cos θ 2 + constant,

where the constant term in the potential energy has been neglected since it does not contribute anything to
the equations of motion. (i.e. the derivative of a constant is zero.)
The kinetic energy term for this system can be represented
2 2
1 ds 1 1 ds 2
T = m 1 + m 2
2 dt 2 dt
(2.2.27)
1 2 2 1 2 2
T = m 1 (˙x +˙y )+ m 2 (˙x +˙y ),
2
1
1
2
2 2