# Moment of inertia of hemispherical shell

1. Oct 20, 2011

### mishima

1. The problem statement, all variables and given/known data
The distribution of mass on the hemispherical shell:

z=(R2-x2-y2)1/2

is given by

σ(x,y,z)=(σ0/R2)(x2+y2)

where σ0 is a constant. Find the moment of inertia about the z-axis of the hemispherical shell.
2. Relevant equations
I=∫r2dm

3. The attempt at a solution
r2 is just R2. ∫dm in spherical coordinates is:

σ0R2∫∫sin3θ dθ d(phi)

with boundaries 0 to pi/2 for θ, and 0 to 2pi for phi. The completed definite integral representing the total mass is then:

(4pi/3) σ0R2.

I feel pretty confident about that, since it is given in the back of the book as the first step in the problem. What I don't understand is why the moment of inertia is anything more than just this value times another R2 to give:

(4pi/3) σ0R4.

The answer is supposed to be:

(16pi/15) σ0R4.

So, I'm guessing there's something more to the calculation of moment of inertia. I've been looking through the different derivations of moments but can't see how to get the right factor. I could use a hint.

2. Oct 20, 2011

### mishima

Oh, I understand why I can't do it that way. r in the moment of inertia formula is measured from the center of mass and is not equal to R in this case.

3. Oct 20, 2011

### LawrenceC

What are the units of sigma0?

4. Oct 20, 2011

### mishima

None given, but the sigma function is a mass per area so I guess it represents an average mass density, or "baseline" mass density that then varies according to the more general function.

5. Oct 20, 2011

### mishima

Yay, got it.

∫∫r2σdA

r in the formula for moment of inertia can be put into spherical coordinates and integrated over the whole hemisphere (also in spherical).

σ0R4∫∫sin5dθ d(phi)

The first integral makes the 8/15 factor needed and the second of course is 2 pi. I guess this was a fairly simple problem after all.

6. Dec 28, 2011

### sriracha

I follow this, but do you mean to say rotational axis, not center of mass? The center of mass would be a point. If r is distance from the rotational axis then we can let r=p*sin$\phi$, or in this case r=R*sin$\phi$.

I follow this and it makes sense, but when I integrated sin^5$\phi$, I got

-5 Cos[x] 5 Cos[3 x] Cos[5 x]
--------- + ----------- - ----------
8.................48.................80

which, when integrated from [o,∏/2], equals zero.