# Moment of inertia of a sphere

1. Jan 12, 2006

### Benny

I am having trouble with the following question.

Q. Derive the expression for the moment of inertia of a spherical ball of mass m and radius r about the axis z through its centre of gravity.

Ans: $I_G = \frac{2}{5}mr^2$.

I did the following.

$$I_G = \int\limits_m^{} {r^2 } dm$$

$$dm = \rho dV$$ where rho is the constant density.

$$V = \frac{4}{3}\pi r^3 \Rightarrow dV = 4\pi r^2 dr$$

$$I_G = 4\pi \rho \int\limits_0^r {r^4 dr} = \frac{{4\pi r^5 \rho }}{5} = \frac{3}{5}\left( {\frac{4}{3}\pi r^3 \rho } \right)r^2 = \frac{3}{5}mr^2 \ne \frac{2}{5}mr^2$$

I can get the answer using a triple integral but the question seems to only require a single integral. At the moment I can't see where I've gone wrong. Can someone help me out? Thanks.

2. Jan 12, 2006

### Tide

To calculate the moment of inertia you need to integrate over the square of the distance from the axis of rotation which is $r^2 \sin^2 \theta$ and not just $r^2$. This also requires you to integrate over the polar angle.

3. Jan 12, 2006

### Benny

Oh ok. I just looked up the definition on mathworld and it appears to say that a perpendicular distance needs to be considered. Hopefully these will become more easy to do as I get more practice. Thanks for the help.