# Deriving the moment of inertia for a sphere

by amolv06
Tags: deriving, inertia, moment, sphere
 P: 47 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.
 P: 47 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.
 HW Helper Sci Advisor P: 4,282 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]
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P: 8,262

## Deriving the moment of inertia for a sphere

 Quote by 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 $$I= \sum mr^{2}$$ 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: $$m\int^{2\pi}_{0}\int^{\pi}_{0}\int^{R}_{0} r^{4}sin(\phi)drd\phi d\theta =\frac{4\pi R^{5}}{5}$$ 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.
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.
 P: 1,877 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.

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