Moment of Inertia for a Thick Spherical Shell

[b100c]

Homework Statement

A) [/B]Consider a hollow sphere of uniform density with an outer radius R and inner radius \alpha R, where 0\leq\alpha\leq1. Calculate its moment of inertia.
B) Take the limit as \lim_{\alpha\to1} to determine the moment of inertia of a thin spherical shell.

Homework Equations

Moment of Inertia: I = \int r^2 dm

The Attempt at a Solution

dm = \rho dV. Where rho is density. The volume element for a sphere is dV=r^2sin\theta d\theta d\varphi dr
So I think I would integrate over a sphere but instead from inner radius to the outer radius? I = \rho \int_{\alpha R}^{R} r^4 dr \int_{0}^{\pi} sin\theta d\theta \int_{0}^{2\pi} d\varphi
Which yields \frac{4\pi}{5} \rho (R^5 - \alpha R^5)
If \rho = \frac{m}{V} = \frac{m}{\frac{4\pi (R^3 - \alpha R^3)}{3}}
Then the equation for moment of inertia becomes
I = \frac{3}{5}mR^2 \frac{1-\alpha^5}{1-\alpha^3}

The problems is now when I take the limit as alpha approaches 1, and apply l'Hopital's rule, I get that moment of inertia is mR^2, when there should be a factor of \frac{2}{3}?

 

[BvU]

In your I calculation, the r in your relevant equation is not the distance from the origin, but the distance from the axis of rotation ...

 

[b100c]

Thanks BvU, that was a stupid mistake on my part. I replaced r with rsin(theta) in spherical coordinates and I got the correct answer.

 

[BvU]

well done!