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## 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