Question

Master of physics needed

A solid uniform cylinder is rolling without slipping. What fraction of its kinetic energy is rotational?
1/2
1/3
1/4
2/3
3/4

Answer 1

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:

`E_(linear)= (1)/(2)mv^2`
where

`v=r \omega`
giving

`E_(linear)= (1)/(2)mr^2 \omega ^2`

The rotational part requires the moment of inertia of a solid cylinder

`I_(cyl) = (1)/(2)mr^2`
Then the rotational kinetic energy is

`E_(rot)= (1)/(2)I \omega ^2= (1)/(4)mr^2 \omega ^2`
Adding the two types of energy and factoring out common terms gives

`(1)/(2)mr^2 \omega ^2(1+ (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.