What is the new angular velocity when a man moves on a rotating merry-go-round?

[BrainMan]

Homework Statement

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.

Homework Equations

I= MR^2
L = Iω

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
ω = .074 rad/sec
.074 x 2pi = .467 rev/sec

The correct answer is .569 rev/sec

 

[Nathanael]

Assume the merry-go-round is a solid cylinder of mass 25 kg and radius 2 m....

I= MR^2

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.

 

[BrainMan]

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.

I got it right. Thanks!