Calculating Angular Speed of a Merry-Go-Round After Adding a Child

[phyhelpme]

What do I need to do with this problem?

Homework Statement

A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kgm2 and is rotating at 15.0 rev/min about a frictionless vertical axle. Facing the axle, a 35.0 kg child hopes onto the merry-go-round. What is the new angular speed of the merry-go-round?

 

[DaveC426913]

Again, show your work.

 

[phyhelpme]

How does this look?

Ok so moment of intertia =35kg(2^2)=140
15rev/m*pi/30=1.57 rad/s

Linitial=Lfinal
Linitial=I*angular speed
250(1.57)=263.12w
392.5=263.12w
1.49 rad/s=w

Converting back to rev/min:
(1.49*60)/2pi=14.23 rev/min

 

[phyhelpme]

bump.

 

[phyhelpme]

I think this now is the correct answer.
Final moment of inertial (if)= 250kgm^2+(35kg)(2m)^2=390kgm^2

IiWi=IfWf
(250)(15)=390wf
3750=390wf
= 9.62 rev/min