Conservation of energy of ball in a half pipe

[rasputin66]

A ball is released on the inside surface of a half-pipe. The left side of the pipe is rough, so that the ball rolls there without slipping. The right side of the pipe is coated with frictionless ice. If the ball is released from a height h on the left side, how high will it go on the right side? Neglect air resistance.

A. h
B. Higher than h
C. Lower than h
D. There is no way to answer without knowing the mass of the ball.

I see the answer is C, but I don't understand why. Doesn't the law of conservation turn PE into enough KE to get back up to h? Why not?

 

[Pengwuino]

The law of conservation of energy does mean any change in potential energy will generally become a kinetic energy, but this problem is dealing with more than one type of kinetic energy. The key is the wording "the ball rolls there without slipping". What this means is that the ball rotates. Some of the potential energy goes into the ball moving in the normal way down the hill like problems you probably did in a previous part of the course. By implying a finite sized ball that has the ability to rotate, you're adding a new way for that potential energy to be spent.