Question about objects rolling on a incline plane

[r1p4c3]

Before I start the problem, I would like to apologize for not using the formating. I am posting this from my high school's network and they have an absurd filtering algorithm that blocks css and other web scripts so I can't use the normal formatting since I can't remember the format tags for vBulletin.

This problem is from my AP Physics rotation test, and of my peers that I talked to, we can't agree what the answer is.

Homework Statement

A ball of uniform density is placed at the top of the left side of a U shaped track. The left half of the track is friction less and the right half has friction. The top of the track is a height h above the bottom of the track. The ball has a mass of m and a radius of r.

What is the maximum height the ball will reach on the left side??

Homework Equations

v = r x w

I = 2/5 m*r^2

GPE = mgh

KE = 1/2 m*v^2 + 1/2 I*w^2

The Attempt at a Solution

Energy is conserved throughout the path of the ball. Since the left half of the track is friction less, so it won't roll but slide.
GPE = KE ==> mgh = 1/2 m*(v0)^2 ==> v0 = sqrt(2gh)

Energy before entering the friction part should be the same after entering it.
1/2 m*(v0)^2 = 1/2 m*v^2 + 1/2 I*w^2 ==> 1/2 m*(v0)^2 = 7/5 m*v^2 ==> v = (v0)*sqrt(5/7) = sqr (10gh/7)

Finally, energy is conserved while it rolls up the right side
1/2 m*v^2 + 1/2 I*w^2 = mgh
7/10 m*v^2 = mgh
7/10 v^2 = gh
h = 7/(5g) v^2 = 7/5g * 10gh/7 = h

The math says the final height will be the same as the initial height, but that answer seems too simple for this question.

Edit: I pressed submit too early. Is the math and assumptions I posted correct? ?

 

[Doc Al]

What would happen if the ball were released from the top of the right side?

 

[Abhinav]

Wont friction do some work in stopping the object and hence some energy would be lost ?? Can conservation of energy be applied ??

 

[HallsofIvy]

No, if the ball is rolling without sliding there will be no loss of energy. Your "v0" is the speed of the ball at the bottom of the track, right? You should say that. Yes, at the bottom of the track, after sliding down the frictionless side, all potential energy will have been converted to kinetic energy. As soon as it starts up the right side, where there is enough friction to cause the ball to roll, potential energy will be converted to both the linear kinetic energy of its forward motion and the rolling energy. However, when the ball stops both its linear kinetic energy and its rolling energy will be 0 so all energy will have been converted back to potential energy. It will now have the same potential energy as initially so will have the same height.

The situation is different if the ball starts on the right, from the same height, of course. As the ball goes down the right side, it will gain both linear kinetic energy and rolling energy. At the bottom, its linear kinetic energy will not be much as as before because only part of the potential energy will have been converted to linear kinetic energy. As the ball goes up the right side, there will be NO friction to stop the rolling so the ball will go up until its linear kinetic energy has been converted back to potential energy so it will still be spinning even after it has stopped going up. Since only part of the initial potential energy has been converted back to potential energy, the ball will not go as high on the right.