# Relationship between linear and rotational kinetic energy

1. Jul 31, 2014

### BrainMan

1. The problem statement, all variables and given/known data
Consider a solid disk of mass m and radius R rolling along a horizontal surface with the center of the disk moving horizontally with a speed v. The total kinetic energy of the disk is a combination of translational kinetic energy , KEt and rotational kinetic energy, KEr. Find the ration KEt/KEr

2. Relevant equations

3. The attempt at a solution
I basically broke down the equations into their simplest forms to try to find the relationship. So
E = 1/2mv^2 + 1/2IW^2
E = 1/2mv^2 + 1/2 (1/2mR^2)w^2
E = 1/2mv^2 + 1/4mR^2 * W^2
I am really not sure how to how about solving this problem. I know I could make up some imaginary numbers for the variables and just guess and check for the relationship but I would prefer to solve this algebraically. The correct answer is 2.

2. Jul 31, 2014

### Orodruin

Staff Emeritus
You are after the translational and rotational kinetic energies (or rather, their ratio), not the total kinetic energy.

What is the ratio? How can you simplify it?

3. Jul 31, 2014

### CAF123

Using the fact that the disk does not slip or travels in a 'pure roll', you can substitute for $\omega$ in terms of other variables. Then take the ratio.

4. Jul 31, 2014

### BrainMan

OK so I figured it out by breaking down the velocity into its parts. So
E = 1/2mv^2 + 1/4mR^2 * W^2
E = 1/2m(Rw)2 + 1/4mR^2 *w^2
E = 1/2 mR^2 * w^2 + 1/4mR^2 * w^2
Since everything in the translational and rotational kinetic energies is the same except 1/2 and 1/4 I can conclude that the translational kinetic energy is twice as big as the rotational kinetic energy. What was your approach?

5. Jul 31, 2014

### Orodruin

Staff Emeritus
Correct. There really is not much more to it.
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