Chapter 5, Problem 30RE

Spring pendulum The rotational form of Newton’s second law of motion is:

The time rate of change of angular momentum about a point is equal to the moment of the resultant force (torque).

In the absence of damping or other external forces, an analogue of (14) in Section 5.3 for the pendulum shown in Figure 5.3.3 is then

$\frac{d}{dt}\left(m{l}^2\frac{d\theta }{dt}\right)=-mgl\sin \theta .(1)$

1. (a) When m and l are constant show that (1) reduces to (6) of Section 5.3.
2. (b) Now suppose the rod in Figure 5.3.3 is replaced with a spring of negligible mass. When a mass m is attached to its free end the spring hangs in the vertical equilibrium position shown in Figure 5.R.4 and has length l₀. When

Figure 5.R.4 Spring pendulum in Problem 30

the spring pendulum is set in motion we assume that the motion takes place in a vertical plane and the spring is stiff enough not to bend. For t > 0 the length of the spring is then l(t) = l₀ + x(t), where x is the displacement from the equilibrium position. Find the differential equation for the displacement angle θ(t) defined by (1).