# Moment of Inertia of a solid sphere

1. Nov 3, 2014

### cwbullivant

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

Taylor, Classical Mechanics Problem 10.11 **
a) Use the result of problem 10.4 (derivation of the general integral for a moment of inertia of a continuous mass distribution in spherical coordinates, using point particles) to find the moment of inertia of a uniform solid sphere for rotation about a diameter.
b) Do likewise for a uniform hollow sphere whose inner and outer radii are a and b. [One slick way to do this is to think of the hollow sphere as a solid sphere of radius b from which you have removed a sphere of the same density but radius a.]

2. Relevant equations

$I = \int r^2 dm$ - 1

$dm = \rho dV$ - 2

$\rho = \frac{M}{\frac{4\pi R^3}{3}} = \frac{3M}{4\pi R^3}$ - 3

$dV = r^2 sin \theta dr d\theta d\phi$ - 4

3. The attempt at a solution

I'm limiting it to part a for now, since that's where I got stuck. The problem statement is "any diameter", and I'm going to center the sphere on the origin and have it rotate about the z-axis, for convenience. The previous problem it references is a derivation of the general integral in spherical coordinates, so that's the system I'll be using here.

First, combining the equations to make a volume integral:

$I = \frac{3M}{4\pi R^3} \iiint r^4 sin \theta dr d\theta d\phi$

Where the integration bounds are:

$r: 0 \rightarrow R$
$\theta: 0 \rightarrow \pi$
$\phi: 0 \rightarrow 2 \pi$

First integrating phi:

$I = \frac{3M}{2 R^3} \iint r^4 sin \theta dr d\theta$

And then theta:

$I = \frac{3M}{R^3} \int r^4 dr$

And finally r:

$I = \frac{3M}{R^3} \frac{R^5}{5} = \frac{3}{5}MR^2$

But I know this is wrong, as the answer I learned in freshman mechanics was

$I = \frac{2}{5} MR^2$

And I can't seem to figure out why the answer is incorrect.

2. Nov 3, 2014

### Orodruin

Staff Emeritus
$r$ in the definition of the moment of inertia is the distance from the axis of rotation, not the spherical coordinate $r$.