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.