Question

Master of physics needed

Answer 1

Hey JayDilla, I get 1/3.  Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
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.