# Work done in rotational motion

1. Nov 14, 2007

### thebest100

1. The problem statement, all variables and given/known data

A turntable of radius 25 cm and rotational inertia 0.0154 kg m^2 is spinning freely at 22.0 rpm about its central axis, with a 19.5-g mouse on its outer edge. The mouse walks from the edge to the center. Find (a) the new rotation speed and (b) the work done by the mouse.

2. Relevant equations

W = Kf - Ki
Inertia for a disk is MR^2
Intertia for a ring is (1/2)MR^2

3. The attempt at a solution
solution to part A) 23.7 rpm

i have tried the following but they all seem to be wrong (303 mJ, 3.3 mJ, 127 mJ)

the answer to part B must be in mJ

please give me an idea on how to solve part B.

Last edited: Nov 14, 2007
2. Nov 14, 2007

### Shooting Star

I haven’t checked your calculations for part A, so I’m writing the method anyway.

The initial angular momentum comprises of the ang mom of the disk and ang mom of the mouse. The final ang mom is only of the disk, since the mouse is at the centre. Equate initial and final angular momenta.

Initial KE of disk plus mouse < Final KE of same. The increase is because of the work done by the mouse.

Now plug in the numbers.

3. Nov 14, 2007

### thebest100

by "Final KE of same" you mean KE of the disk and the mouse

would the KE of the mouse at the center of the disk be 0, so only the KE of disk would be involved in the final right

4. Nov 14, 2007

### Staff: Mentor

Right.

Yes. Just treat the mouse as a point mass (ignore its rotation).

5. Nov 14, 2007

### thebest100

Thank You Shooting Star and Doc Al. I found my mistake, I was using 1/2MR^2 for the Inertia of the mouse, instead of MR^2.

Thank you