# Angular change in velocity (momentum)

## Homework Statement

A merry-go-round has a radius of 3.0 m and a rotational inertia of 600 kg m2. The merry-go-round is initially at rest. A 20 kg child is running at 5.0 m/s along a line tangent to the rim. Find the angular velocity of the merry-go-round after the child jumps on.

a. 0.38 rad/s
b. 0.45 rad/s
c. 0.71 rad/s
d. 0.56 rad/s
e. 1.2 rad/s

p(angular)=Iω
p=mv
p(angular)=p*r

## The Attempt at a Solution

i tried to use conservation of momentum by saying:

r*mv+0=(I+mr)ω

and i got ω=0.45 but the answer key says the answer is a. 0.38 rad/s

## Answers and Replies

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i tried to use conservation of momentum by saying:

r*mv+0=(I+mr
Right idea, but you used the wrong formula for the rotational inertia of the child. What's the rotational inertia of a point mass (we can treat the child as a point mass) that is some distance from the axis?

(Always check units. What are the units of I? Does mr have those units?)

Ohh so it will be mr^2 right? :)
How is it derived anyways?