Answer to Question #164367 in Physics for Tewo

Question

1. A merry-go-round rotates at the rate of 0.47 rev/s with an 98 kg man standing at a point 2 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 57 kg cylinder of radius of 2 m. Answer in units of rad/s. 
2. What is the change in kinetic energy due to this movement? Answer in units of J.

Answer 1

Moment of inertia of the merry-go-round:

$$I=\frac 12 MR^2.$$

Moment of inertia of the man initially:

$$i=\frac12mr^2$$

Finally (when the man is at the center): 0.

Initially, both bodies rotate at the same angular velocity creating the angular momentum of:

$$L_1=(I+i)\omega_i$$

After the man (we treat him as a point without any size) crawled to the center, the angular momentum is the same in magnitude but what different now is the moment of inertia of the system:

$$L_2=I\omega_f.\\ L_1=L_2,\\ (I+i)\omega_i=I\omega_f,\\\space\\ \omega_f=\bigg(1+\frac iI\bigg)\omega_i=\bigg(1+\frac{mr^2}{MR^2}\bigg)\omega_i=\\\space\\ =8.03\text{ rad/s}.$$

The change in kinetic energy:

$$T_1=(I+i)\omega_i^2,\\ T_2=I\omega_f^2,\\ \Delta T=T_2-T_1=I(\omega_f^2-\omega_i^2)-i\omega^2_i=9295\text{ J}.$$