Answer to Question #293320 in Physics for torque

Question

1. A merry-go-round rotates at the rate of 0.24 rev/s with an 71 kg man standing at a point 2.7 m from the axis of rotation. What is the new angular speed when the man walks to a point 1 m from the center? Consider the merry-go-round is a solid 79 kg cylinder of radius of 2.7 m
2. What is the change in kinetic energy due to this movement?

Answer 1

1. Apply law of conservation of angular momentum:

$\bigg(\dfrac 12MR^2+mr_1^2\bigg)\omega_1=\bigg(\dfrac12 MR^2+mr_2^2\bigg)\omega_2,\\\space\\ \omega_2=\omega_1\dfrac{MR^2+2mr_1^2}{MR^2+2mr_2^2}=2\pi n\dfrac{MR^2+2mr_1^2}{MR^2+2mr_2^2},\\\space\\ \omega_2=3.38\text{ rad/s}.$

2. The change in kinetic energy:

$\Delta E_K=\dfrac12m(r_2^2\omega_2^2-r_1^2\omega_1^2)=-183\text{ J}.$