# Deriving the moment of inertia for a sphere

1. Dec 2, 2007

### amolv06

1. The problem statement, all variables and given/known data

Derive the moment of inertia for a solid sphere with a uniform mass

2. Relevant equations

[Tex]I= \sum mr^{2}[/tex]

3. The attempt at a solution

I decided to change everything to polar coordinates. Since the polar coordinate substitution is

$$\int\int\int_{v} Fr^{2}sin(\phi)drd\phi d\theta$$

I figured that all you should do is plug in the moment of inertia equation into the integral giving you:

$$m\int^{2\pi}_{0}\int^{\pi}_{0}\int^{R}_{0} r^{4}sin(\phi)drd\phid\theta =\frac{4\pi R^{5}}{5}$$

however this does not seem to be the correct answer. Can anyone tell me what I'm doing wrong? Thanks in advance for your time and any help.

Last edited: Dec 2, 2007
2. Dec 2, 2007

### amolv06

For some reason the images are coming up in the wrong places, and one image is actually missing. I don't know how to fix that. Sorry.

3. Dec 2, 2007

### CompuChip

The last image is probably not showing because it has a syntax error: there is something like
d\phid\theta
in it, and LaTeX doesn't know the symbol \phid. Insert a space and it will work.
In the first formula you quoted, what is this F and where does the sine come from? Also, did you put in the mass density
$$\rho = \frac{m}{V} = \frac{m}{\tfrac43 \pi R^3}$$
correctly (I think not, which would account for the R^5 instead of R^2 as you should have found).

I think I see where the $r^2 \sin\theta$ comes from, is that the square of the shortest distance from a point on the sphere to some axis? I calculate that distance to be r sin(theta), so in the formula there is also a square missing probably, and the integration there is over the volume dV, the dr dtheta dphi only come in when you switch to polar coordinates[/edit]

Last edited: Dec 2, 2007
4. Dec 2, 2007

### cristo

Staff Emeritus
I've corrected your tex to make it easy to read. Your problem was that that the first tags were [ Tex] [ /tex] (plus the problems that compuchip mentioned). See the quote.

Last edited: Dec 2, 2007
5. Dec 2, 2007

### Mindscrape

I too am not so sure about what you are doing. If you use an inertia tensor the formula goes as

$$I_{ij} = \int_V \rho(\mathbf{r}) (\delta_{ij} \sum_k x_k^2 -x_i x_j) dv$$

so, for example, the first element would be

$$I_{11} = \int_V \frac{M}{4/3 \pi R^3} (x^2 + y^2 + z^2 - x^2) dv$$

where you can use coordinate and jacobian (for the volume differential) conversions to get it into spherical form (which I think is what you meant because you certainly didn't show a polar jacobian). From symmetry all the off diagonals will be zero, and all the diagonals will be the same.