How Does a Skater's Rotational Kinetic Energy Change When He Lowers His Arms?

[unteng10]

Homework Statement

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.

Homework Equations

L=I$$\omega$$
KE=1/2I$$\omega$$^2

The Attempt at a Solution

Not sure if I am 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?

 

[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.

 

[unteng10]

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

 

[rl.bhat]

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

Yes. Angular speed is w.