# Moment of inertia

1. Apr 24, 2005

### Frod

Hello, I have been given a loop-the-loop problem to do. Now the problem I have isn't with conservation of mechanical or rotational energy, it's with the moment of inertia of the solid sphere. Usually it would be (2/5)*M*r^2 however the sphere is rolling on a "V" track (see image). I get the correct answer if I leave the moment of inertia as (2/5)*M*r^2, however, shouldn't it be different as it is no longer rolling on r, instead it is rolling on r' (see image)?

I have worked out that the angular velocity will be [(2)^0.5*v]/r, where r is the radius of the sphere and v the velocity of the sphere. Any help would be greatly appreciated. Thank you.

Frod.

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2. Apr 24, 2005

### jdavel

Frod,

If I understand this problem (and I'm not sure I do) the moment of inertia of inertia of the sphere doesn't change, but the torque on the sphere does, because it's applied at a smaller radius.

3. Apr 24, 2005

### PBRMEASAP

The fact that the ball makes contact with the track at a smaller distance from the axis of rotation does not change anything. The ball is still rotating about an axis through its center of mass, and the moment of inertia of a ball about any axis through its CM is what you said, (2/5)*M*R^2. In this way, the motion of the ball can be separated neatly into two parts: 1) translation of the center of mass, and 2) rotation about an axis through the CM. The total kinetic energy is the sum of kinetic energies associated with the two motions.

edit: what jdavel says is true, the torque applied by the track is smaller. What this means in terms of calculation is that your condition for rolling without slipping will take into account the smaller radius. So I shouldn't have said that it doesn't change ANYTHING. Your answer for angular velocity looks good.

Last edited: Apr 24, 2005