# What fraction of its total kinetic energy is rotational kinetic energy

1. Nov 23, 2013

### jimmyboykun

1. The problem statement, all variables and given/known data
A spherical object with moment of inertia 0.57mr2 rolls without slipping down an incline. At the bottom of the incline.

2. Relevant equations
For a sphere (or cylinder) rolling without slipping ω = v/r.

Rotational KE = ½Iω²

The translational kinetic energy = ½mv²

3. The attempt at a solution
Rotational KE = ½Iω²
= ½ * 0.57mr² (v/r)²
= 0.285mv²

The total kinetic energy = 0.285mv² + ½mv² = 0.785mv²

Rotational kinetic energy as a fraction of the total energy is:
(0.285mv²) / (0.785mv²) = 0.363

the question asked to answer in percentage so would it be 36.3%

2. Nov 23, 2013

### haruspex

Yes.

3. Nov 25, 2014

### phyllis

HI I'm trying to work out the same problem but I dont get where the total kinetic energy = 0.285mv² + ½mv² = 0.785mv²

4. Nov 25, 2014

### haruspex

What's the linear KE in terms of m and v? What's the rotational KE in terms of I and $\omega$? What is the relationship between v, r and $\omega$?

5. Nov 25, 2014

### phyllis

liner KE in terms of m and v would be 1/2mv^2, rotational KE in terms of 1/2Iω^2, and I know that v= ωr

6. Nov 25, 2014

### haruspex

And you know I here in terms of m and r. Substitute to find rotational KE in terms of m and v.

7. Nov 25, 2014

### phyllis

Rotational KE = ½Iω²
In my problem moment of inertia 0.6mr^2
= 1/2 * 0.6mr^2 (v/r)^2
= 0.3mv^2

8. Nov 25, 2014

### haruspex

OK, so what's the total KE in terms of m and v? What fraction of that is rotational?

9. Nov 25, 2014

### phyllis

This is where I get lost, I understand that total KE is 1/2mv^2+.3mr^2

10. Nov 25, 2014

### haruspex

No, the .3mr^2 is wrong. Go through the steps again carefully. If you still get that, please post all steps.
(The .3 is right, it's the mr^2 bit)

Last edited: Nov 25, 2014
11. Nov 25, 2014

### phyllis

1. The problem statement, all variables and given/known data
A spherical object with moment of inertia 0.6mr2 rolls without slipping down an incline. At the bottom of the incline.

2. Relevant equations
Rotational KE = ½Iω²
The translational kinetic energy = ½mv²

3. The attempt at a solution
Rotational KE = ½Iω²
= ½ * 0.6mr² (v/r)²
= 0.3mv²

The total kinetic energy = 0.3mv² + ½mv²

12. Nov 25, 2014

### haruspex

Right. So what fraction of that is rotational?

13. Nov 25, 2014

### phyllis

3/10

14. Nov 25, 2014

### haruspex

No.
Simplify that to get the total KE.

15. Nov 25, 2014

### phyllis

total KE = 0.8mv²

16. Nov 25, 2014

### haruspex

Right. What fraction is 0.3mv² of that?

17. Nov 25, 2014

### phyllis

0.375

18. Nov 25, 2014

### haruspex

Er.. no. How did you get that?

19. Nov 25, 2014

### phyllis

sorry I meant .375

20. Nov 25, 2014

Right.