Posted by **Ree** on Tuesday, November 23, 2010 at 2:52pm.

If the ball is released from rest at a height of 0.71 above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track? Assume the ball is a solid sphere of radius 3.4 and mass 0.14 .

How high does the ball rise on the frictionless side?

- Physics-units required -
**MathMate**, Tuesday, November 23, 2010 at 7:21pm
The potential energy of the ball falling from the no-slip side will be converted into two kinetic energies, one of linear motion, given by (1/2)mv², and the other of rotational velocity of the ball, given by (1/2)Iω².

Note that the axis of rotation is about the sphere's outer surface. Use the parallel axis theorem to convert the moment of inertia about the centre to that about the outer surface.

Thus,

mgh=(1/2)mv²+(1/2)Iω².

Note that for no-slip, the rotational velocity (ω) is related to the linear velocity.

On the frictionless side, the ball will continue to rotate, but only the translational kinetic energy (1/2)mv² will be converted to gravitational energy as it rises.

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