Calculating Angular Speed of Playground Merry-Go-Round

[Husker70]

Homework Statement

A playground merry-go round of Radius R=2.00m has a moment of
inertia I = 250 kg.m^2 and is rotating at 10.0 rev/min about a frictionless
vertical axle. Facing the axle a 25.0kg child hops onto the merry go round
and manages to sit down on the edge. What is the new angular speed of
the merry go round?

Homework Equations

I = 250 kg.m^2
10.0 rev/min
25.0kg child
radius = 2.00m

I=r^2m
L = r x p
L = Iw

The Attempt at a Solution

I = r^2m
m = I/r^2
m = 250kg.m^2/(2.00m)^2 = 62.5kg

I added the 62.5kg + 25.0kg = 87.5kg

I = r^2m
= (2.00m)^2(87.5kg) = 350kg.m^2

I don't know what to do with this from here.
Thanks for any help,
Kevin

 

[rock.freak667]

Use the conservation of angular momentum