# Moment of inertia help needed

1. Feb 24, 2004

### ilikephysics

I'm really having problems understanding how to do moment of inertia. Can someone please help me with this problem? Explain it to me please. Thanks so much.

Question:

Find the moment of inertia of a box of sides a, b, and c, mass M, and uniform density for rotations about an axis passing through its center and perpendicular to the two faces of sides a and b.

Find the moment of inertia for rotations about an axis passing along one edge of length c.

2. Feb 25, 2004

### ilikephysics

3. Feb 25, 2004

### paul11273

I think this question sounds a little vague. Can you clairfy it a little better?

Also, show some work that you have done so we can see where you are having trouble. That will also help us (atleast me) visualize what is going on.

4. Feb 25, 2004

### HallsofIvy

Staff Emeritus
The "moment of inertia" of an object around an axis of rotation is the integral of (distance of each point from the axis of rotation)2 times the density. The integral is taken over the volume of the object. The fact that this is not circularly symmetric makes it a little harder. Take the (uniform) density to be the constant &delta; Set up a coordinate system so that center of one "a by b" face is at (0,0,0) and the center of the other face is at (0,0,c). Then the distance from a point (x,y,z) to the nearest point on the axis of rotation, (0, 0, z), is &radic:(x2+ y2and its square is, of course, simply x2+y2. The moment of inertia is:
$$\int_{x=-a/2}^{a/2}\int_{y=-b/2}^{b/2}\int_{z=0}^{c} \delta(x^2+ y^2)dzdydx$$