# Rotational Kinetic Energy

1. Apr 28, 2009

### unteng10

1. The problem statement, all variables and given/known data
A skater spins with an angular speed of 17.4 rad/s with his arms outstretched. He lowers his arms, decreasing his moment of inertia from 43 kg/m^2 to 37 kg/m^2. Calculate his initial and final rotational kinetic energy.

2. Relevant equations
L=I$$\omega$$
KE=1/2I$$\omega$$^2

3. The attempt at a solution
Not sure if im on the right track here, for initial kinetic energy I came up with 3.53 J. I manipulated L=I$$\omega$$ to get $$\omega$$=L/I to find my angular velocity. Then plugged that in the KE=(1/2)(43 kg/m^2)(.405^2) to get 3.53 J. Did I do this correctly?

2. Apr 28, 2009

### rl.bhat

A skater spins with an angular speed of 17.4 rad/s
This is not the angular momentum L, but is is the angular velocity w.

3. Apr 28, 2009

### unteng10

Okay, so the speed would be $$\omega$$?

4. Apr 28, 2009

### rl.bhat

Yes. Angular speed is w.