Question

A merry-go-round rotates at the rate of
0.26 rev/s with an 78 kg man standing at
a point 1.9 m from the axis of rotation.
What is the new angular speed when the
man walks to a point 0 m from the center?
Consider the merry-go-round is a solid 99 kg
cylinder of radius of 1.9 m.
Answer in units of rad/s.

Answer 1

The law of conservation of angular momentum gives

$$J_i\omega_i=J_f\omega_f$$

The initial moment of inertia of the system is given by

$$J_i=\frac{1}{2}MR^2+mr^2$$

The final moment of inertia

$$J_f=\frac{1}{2}MR^2$$

Hence

$$\omega_f=\frac{J_i}{J_f}\omega_i=\frac{\frac{1}{2}MR^2+mr^2}{\frac{1}{2}MR^2}\omega_i$$ $$=\frac{\frac{1}{2}\times 99\times (1.9)^2+78\times (1.9)^2}{\frac{1}{2}\times 99\times (1.9)^2}\times 0.26=0.67\:\rm rev/s$$