# Rotational Inertia of merry go round

## Homework Statement

a. A merry go round is rotating in the counter-clockwise direction. Initially, a 50 kg child is sitting on the edge of the merry-go-around, which rotates at 0.5 radians per second. The moment of inertia of the merry-go-round is 2150 kg-m2. The radius is 2.50 meters. The child can be treated as a point mass. What is the total angular momentum?

b. If the child crawls to the center of the merry-go-round, will the speed at which it rotates change? If so, what is the new rotation speed?

## Homework Equations

$$\vec{L}$$ = I$$\vec{w}$$
I = m1r12 + m2r22

## The Attempt at a Solution

The moment of inertia of the merry-go-round is 2150 kg-m2. We need to add to that value the moment of inertia of the child.

Ichild = (2.50 m) * (50 kg) = 125 kg-m2
Itotal = 2150 + 125 = 2275 kg-m2

$$\vec{L}$$ = I$$\vec{w}$$ = (2275)(0.5) = 1137.5