# Moment of Inertia of a Hemisphere

1. Oct 18, 2008

### hp-p00nst3r

1. The problem statement, all variables and given/known data
The moment of inertia of a uniform sphere of mass m and radius r about an axis passing through its center of mass is (2/5) mr^2. If half of the sphere is removed, what is the moment of inertia about the same axis?

2. Relevant equations
integral of r^2 dm

3. The attempt at a solution
http://img407.imageshack.us/img407/5250/mech221ps4moiim8.jpg [Broken]
I think this is wrong since I've looked up the actual I for this and it gives me (2/5) mr^2 still. I also tried using the parallel axis theorem on this and it also gives me the (2/5) mr^2. What am I doing wrong?

Last edited by a moderator: May 3, 2017
2. Oct 18, 2008

### alphysicist

Hi hp-p00nst3r,

About the third thing you have written down is this:

$$dI = \frac{1}{2} z^2 dm$$

This should represent the moment of inertia of a representative disk, like the one you have in the diagram. However, the moment of inertia of a disk is (1/2) M R2, where R is the radius of the disk.

From the figure, you can see that the radius of the disk is not z, so your expression for dI is not correct. The radius is actually y, so once you have used the expression for y in dI, I think you should get the right answer.

Last edited by a moderator: May 3, 2017