# Homework Help: Merry Go Round

1. Apr 5, 2006

### Jacob87411

A playground merry-go-round of radius R = 1.20 m has a moment of inertia I = 230 kgm2 and is rotating at 10.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round?

Little bit of a problem with this one. How does the moment of inertia change when the 23 kg child steps on...Do you need to use angular momentum conservation?

2. Apr 5, 2006

### Staff: Mentor

Treat the child as a particle. (What's the moment of inertia of particle at a distance from an axis?) Just add the child's moment of inertia to that of the merry-go-round.

Yes.

3. Apr 5, 2006

### Jacob87411

So the moment of inertia for a particle is Mi*ri = (23)(1.2)=27.6

So I of the system is (230+27.6)=257.6. So now we take conservation of angular momentum:

L initial = L Final
L Initial = I*angular speed = 230*1.05 (10 rev/min = 20pi/min = 1.05rad/s?)
L Final = 257.6*angular speed, so we set this equal to each other

230*1.05 = 257.6w
w=.9375?

4. Apr 5, 2006

### Staff: Mentor

No. Does this expression even have the right units for moment of inertia?

5. Apr 5, 2006

### Jacob87411

whoops, r should be squared right...so (23)(1.2^2) = 33.12

So 230(1.05)=263.12w
w=,874?

6. Apr 5, 2006

### Jacob87411

Right sorry I always forget..it says I am off by a power of 10, why is this

7. Apr 5, 2006

### Staff: Mentor

Right. But be sure to include proper units when stating a physical quantity.

Check your arithmetic. Also, why not express the new angular speed in the same units as the original?

8. Apr 5, 2006

### Jacob87411

oh wow whoops so w=.917 so converting back we take (.17 * 60)/2*pi giving 8.75 rev/min. Thanks for the help