# Linear and Rotational Kinetic Energy

1. Oct 29, 2013

### woaini

1. The problem statement, all variables and given/known data

A solid cylinder is rolling without slipping. What fraction of its kinetic energy is linear?

2. Relevant equations

Ke=$\frac{mv^2}{2}$+$\frac{I(v/r)^2}{2}$

3. The attempt at a solution

Ke=$\frac{mv^2}{2}$+$\frac{(v/r)^2}{2}$*$\frac{(mr)^2}{2}$

Ke=$\frac{3(mv)^2}{4}$

Linear Ke =$\frac{(mv)^2}{4}$

Fraction of Linear Ke = $\frac{linear Ke}{Ke}$ = $\frac{(1/4)}{(3/4)}$ = $\frac{1}{3}$

2. Oct 29, 2013

### tiny-tim

hi woaini!
uhh?

3. Oct 29, 2013

### haruspex

The m should not be getting squared. You need to be more careful with the parentheses.

4. Oct 29, 2013

### woaini

Nvm it should be 1/2. Therefore I should get a final answer of 2/3 linear kinetic energy?

5. Oct 30, 2013

### tiny-tim

yup!

btw, a way of checking this is to say the mass is m, the "rolling mass" (= I/r2) = m/2, so the "effective mass" is 3m/2, and eg the acceleration down a slope must be multiplied by 2/3 (but don't use that in the exam )

(for a sphere, it's m + 2m/5 = 7m/5, and the acceleration must be multiplied by 5/7)