A 25kg merry go round of radius 3.0 m rotates at the rate of 0.50 rev/s. Standin

Question

A 25kg merry go round of radius 3.0 m rotates at the rate of 0.50 rev/s. Standing on the merry go round at a point 1.0 m from the axis of rotation is an 80kg clown.

1. Determine the new angular speed (in rad/s) when the clown walks to a point 2.0 m from the center. Assume the merry go round is a solid disk

2. Calculate the change in the kinetic energy (in Joules) due to this movement. How do you account for this change in kinetic energy

3. Determine the new angular speed (in rad/s) of the merry go round if the clown falls off. Assume that the torque the clown exerts on the merry go round is negligible

Explanation / Answer

1)
moment of inertia of disk = 0.5*M*R^2 = 0.5*25*3^2 = 112.5 Kg.m^2
moment of inertia of clown = m*r^2

wi = 0.5 rev/s = 0.5*2*pi rad/s = 3.14 rad/s

use conservation of angular momentum:
(I1+I2)* wi = (I1+I2')*wf
(112.5 + 80*1^2)* 3.14 = (112.5 + 80*2^2)*wf
(112.5 + 80*1^2)* 3.14 = (112.5 + 80*2^2)*wf
604.45 = 432.5 * wf
wf = 1.4 rad/s
Answer: 1.4 rad/s

2.
Cange in kinetic energy = initial kinetic energy - final kinetic energy
= 0.5* (I1+I2)* wi^2 - 0.5* (I1+I2)* wf^2
= 0.5* (112.5 + 80*1^2)* 3.14^2 - 0.5* (112.5 + 80*2^2)*1.4^2
=949 - 423.9
= 525.1 J

This is the decrease in kinetic energy
This energy has been lost has heat due to walking

3.
use conservation of angular momentum:
(I1+I2)* wi = (I1+I2')*wf
(112.5 + 80*1^2)* 3.14 = (112.5 + 0)*wf
(112.5 + 80*1^2)* 3.14 = 112.5 *wf
604.45 = 112.5 * wf
wf = 5.37 rad/s
Answer: 5.37 rad/s

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