Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
[tex]E_{linear}= \frac{1}{2}mv^2 [/tex]
where
[tex]v=r \omega[/tex]
giving
[tex]E_{linear}= \frac{1}{2}mr^2 \omega ^2[/tex]
The rotational part requires the moment of inertia of a solid cylinder
[tex]I_{cyl} = \frac{1}{2}mr^2 [/tex]
Then the rotational kinetic energy is
[tex]E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2[/tex]
Adding the two types of energy and factoring out common terms gives
[tex] \frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2}) [/tex]
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.
