How Does Changing Arm Position Affect a Spinning Professor's Kinetic Energy?

[tandoorichicken]

A physics teacher stands on a freely rotation platform. He holds a dumbbell in each hand of his outstretched arms while a student gives him a push until his angular velocity reaches 1.5 rad/sec. When the freely spinning professor pulls his hands in close to his body, his angular velocity increases to 5.0 rad/sec. What is the ratio of his final kinetic energy to his initial energy?

Basically what I have done is formed the professor into a model of 1) a rod with two spinning dumbells and 2) a cylinder. Using that I was able to use the equations for rotational KE:
$$KE_0 = \frac{1}{2} I \omega_{0}^{2} = \frac{1}{2} (mr^2) 1.5^2 = 1.125mr^2$$ and
$$KE_f = \frac{1}{2} I \omega_{f}^{2} = \frac{1}{2} (\frac{1}{2} mr^2) 5^2 = \frac{25}{4} mr^2$$.
Dividing the second equation by the first I got a ratio of 25:4.5
However, it never occurred to me that the weight of the teacher and the weight of the dumbbells might have an effect on the outcome of this problem. So basically, how do I do this problem?

 

[Kurdt]

Angular momentum is given by $$L=I\omega$$

Basically this is a quantity that must be conserved so if his initial $$E_k=\frac{L^2}{2I}$$ must be equal to his final kinetic energy and thus a ratio of 1:1 would be the answer as far as I can see.

 

[HallsofIvy]

"Weight" has nothing to do with this problem- everything is happening horizontally, not vertically. You do, of course, have to use the mass, m. It cancels out in the final calculation as you saw.

However, you seem to be using the same "r" in both problems and that is not correct. r is the distance from the center of rotation (i.e. the teacher's central axis) to the center of mass. That changes when the dumbbells are drawn in toward the body.

Since you are not told the different radii (you might estimate them based on a typical person but you can't get them precisely), the best thing to do is use Kurdt's hint. Can you say "conservation of energy"?