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## Homework Statement

A uniform solid sphere of radius r is placed on the inside surface of a hemispherical bowl with much larger radius R. The sphere is released from rest at an angle theta to the vertical and roles without slipping. Determine the angular speed of the sphere when it reaches the bottom of the bowl.

## Homework Equations

Kinetic energy of rolling motion = Iw^2/2 + mw^2r^2/2

Potential energy of sphere = mgh.

I=2mr^2/5 for uniform sphere.

## The Attempt at a Solution

I am not sure what to use for h. At first I thought that I should use the height of the center of mass of the sphere, so h = (R-r)(1-cos(theta)). Doing all the substitutions and solving for omega:

w^2 = [10g(R-r)(1-cos theta)]/7r^2

Then I thought that since the angular velocity should be the same at all points of the disk, I can use the height of the bottom of the sphere, h=R(1-cos theta):

w^2 = [10gR(1-cos theta)]/7r^2

Then I started wondering, why, if the angular speed is the same, do I get two different answers for it?

Then I realized that this makes sense, since the point on the rim of the sphere would be rotating around the axis of the sphere as well as around the center of the bowl. That would make the angular speed greater, as we see here. So, obviously, the first answer is the one to use, since the CM is not rotating, just moving parallel to the surface of the bowl, and so its angular velocity around the center of the bowl would give the right speed.

So, very pleased with my beautiful line of reasoning, I checked the answer, only to discover that I am wrong, since the given answer is

w^2 = [10gR(1-cos theta)]/7r^2

which is the same as my second answer, using the point of the ball resting on the surface of the bowl, not the CM.

However, I can't come up with the reason why this is the better solution

Can someone help me understand this, please?

Thank you,

Dorothy