# Angular momentum problem

1. Jul 12, 2014

### BrainMan

1. The problem statement, all variables and given/known data
A merry-go-round rotates at an angular velocity of 0.2 rev/s with an 80 kg man standing at a point 2 m form the axis of rotation. what is the new angular velocity when the man walks to a point 1 m from the center? Assume the merry-go-round is a solid cylinder of mass 25 kg and radius 2 m.

2. Relevant equations
I= MR^2
L = Iω

3. The attempt at a solution

First I found the moment of inertia of the platform
25 x 4 = 100
Then I found the original moment of inertia of the man
80(4) = 320

Then I found the total angular momentum
L = (100 + 320)0.2/2pi
L = 42/pi

Then I found the moment of inertia after the man moved
I = 80(1)
I = 80

Then I compared the old momentum to the new momentum
42/pi = 180ω
.074 x 2pi = .467 rev/sec

The correct answer is .569 rev/sec

2. Jul 12, 2014

### Nathanael

The problem said the merry-go-round was a solid cylinder, this is the incorrect moment of inertia.

Edit:
All your steps are good though; I walked through your own steps with the correct moment of inertia and got the correct answer.

Last edited: Jul 12, 2014
3. Jul 30, 2014

### BrainMan

I got it right. Thanks!