Moment of Inertia/ Kinetic Energy of ice skater

[klopez]

An ice skater starts a spin with her arms stretched out to the sides. She balances on the tip of one skate to turn without friction. She then pulls her arms in so that her moment of inertia decreases by a factor of two. In the process of her doing so, what happens to her kinetic energy?
-It undergoes a change by an amount that obviously depends on how fast the skater pulls her arms in.
-It decreases by a factor of two.
-It is zero because her center of mass is stationary.
-It increases by a factor of four.
-It decreases by a factor of four.
-It remains constant.
-It increases by a factor of two.

I know that Ki = (1/2)Iω2 , so isn't her Kf = (1/2)(I/2)(2ω2) = (1/2)Iω2 = Ki ?
Doesn't the angular speed increase by a factor of two as well?
My guess would be "It remains constant" (And I can only choose one)

Can anyone give me advice on this problem? Thank you

Kevin

 

[Doc Al]

The KE = (1/2) I \omega^2. You know how I changes, but how does \omega change?

Hint: What's conserved?

 

[klopez]

Tell me if this makes sense...

Conservation of angular momentum Li = Lf

wi = omega_i
wf = omega_fLi = I*wi
Lf = I/2 * wf

2wi = wf

Now in terms of kinetic energy

Ki = (1/2)I*wi^2

Kf = (1/2)*(I/2)*wf^2 (I plug in my wf from above)
Kf = (I/4)(4wi^2)
Kf = I*wi^2

When I compare both Ki and Kf now, Kf is twice as large as Ki because it does not have a (1/2) in its equation like Ki does.

Is this correct?

 

[Doc Al]

Perfect!

 

[Gear300]

You seem to have got it right...but before you continue on...remember the conservation of energy principle. If Kf is twice as large as Ki, where did that additional energy come from??