- #1

Vitani11

- 275

- 3

## Homework Statement

A uniform spherical ball of mass M and radius R, initially spinning about a horizontal axis with angular speed ω0, is placed gently on the floor. The initial center-of-mass velocity of the ball is zero. Derive that the moment of inertia ICM of the ball about an axis passing through its center of mass is given by ICM = (2/5)MR2 .

I = ∫ρ

^{2}dm

## Homework Equations

I = ∫∫∫(ρ

^{2})ρ

^{2}dρsinθdθdΦ

## The Attempt at a Solution

I get I = (3/5)MR

^{2}. I noticed that this is almost correct apart from a moment of inertia for a point particle, so I subtracted MR

^{2}. I reasoned that since the ball is touching the ground and rotating about its axis doesn't it have "two moments"? One touching the ground and the other on its axis. And so to not include the part touching the ground would make sense to get just a sphere. Or is this not right? I saw the "correct" way of adding up disks but does this also work? Also another problem asks me to do the same but for a disc rolling down an incline(except this time the moment of inertia was to be found about the point where the ball touches the ground). I did the same thing as for the sphere but obviously with a different integral and then I added MR

^{2}to MR

^{2}/2 to get the correct answer of 3MR

^{2}/2.