Page 198 - Intro to Tensor Calculus

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Figure 2.2-2 Simply pulley system

EXAMPLE 2.2-2. (Simple pulley system) Find the equation of motion for the simply pulley system
illustrated in the figure 2.2-2.
Solution: The given system has only one degree of freedom, say y 1 . It is assumed that

y 1 + y 2 = ` =a constant.

The kinetic energy of the system is
1 2
T = (m 1 + m 2 )˙y .
1
2
Let y 1 increase by an amount dy 1 and show the work done by gravity can be expressed as

dW = m 1 gdy 1 + m 2 gdy 2

dW = m 1 gdy 1 − m 2 gdy 1
dW =(m 1 − m 2 )gdy 1 = Q 1 dy 1 .

Here Q 1 =(m 1 − m 2 )g is the external force acting on the system where g is the acceleration of gravity. The
Lagrange equation of motion is

d ∂T ∂T
− = Q 1
dt ∂ ˙y 1 ∂y 1
or
(m 1 + m 2 )¨y 1 =(m 1 − m 2 )g.

Initial conditions must be applied to y 1 and ˙y 1 before this equation can be solved.