Velocity of Center of Mass for a Downwardly-Rotating Sphere

[CARNiVORE]

Homework Statement

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.

Homework Equations

I'm perplexed. My only guess would be v_cm = (Σm_iv_i)/(Σm_i)

The Attempt at a Solution

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

 

[haruspex]

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

 

[CARNiVORE]

Thanks for the response. That's a good idea. Let's see what I can do with that. It starts from rest, so K_i = 0. It ends at h=0, so U_f = 0.
W_i = K_i
U_i = K_f + W_ƒ
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?

 

[haruspex]

Thanks for the response. That's a good idea. Let's see what I can do with that. It starts from rest, so K_i = 0. It ends at h=0, so U_f = 0.
W_i = K_i
U_i = K_f + W_ƒ
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?

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.

 

[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!