# Velocity of Center of Mass for a Downwardly-Rotating Sphere

1. Mar 4, 2015

### CARNiVORE

1. The problem statement, all variables and given/known data
mass = M
radius = r
rot. inertia = i
height = h
Sphere of mass M is released from rest at the top of an inclined plane.
The speed of the center of mass at the bottom of the incline, without friction, is sqrt(2gh).

I need to find the velocity of the center of mass assuming the plane has friction and the sphere rolls without slipping.

2. Relevant equations

I'm perplexed. My only guess would be vcm = (Σmivi)/(Σmi)

3. The attempt at a solution

The correct answer is √(2MgHr2)/(I+Mr2).

2. Mar 4, 2015

### haruspex

A more useful relevant equation would be work conservation. What does that have to say?

3. Mar 4, 2015

### CARNiVORE

Thanks for the response. That's a good idea. Let's see what I can do with that. It starts from rest, so Ki = 0. It ends at h=0, so Uf = 0.
Wi = Ki
Ui = Kf + Wƒ
mgh = .5mv2 + μmgh
2gh = v2 + 2μgh
√2gh - 2μgh = v

Definitely incorrect. Since μ isn't even in the answer, I'm guessing that I either need to solve for μ, or I approached this completely incorrectly. Maybe both. What do you think?

4. Mar 4, 2015

### haruspex

It is rolling now, not sliding, so the friction does no work. On the other hand, you have not counted all the KE of the ball.

5. Mar 4, 2015

### CARNiVORE

Oops, that was a silly mistake. I also forgot to add .5iω2 to the final kinetic energy. I've figured it out now, though:

mgh = .5mv2 + .5iω2
2mgh = mv2 + i(v2/r2)
2mghr2 = mv2r2 + iv2
2mghr2 = v2(mr2 + i)
2mghr2/(mr2 + i) = v2

Then just take the square root of both sides. Thanks for your help!

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