# Moment of Inertia of a Disk

1. Dec 25, 2005

### amcavoy

I am having a bit of trouble deriving the moment of inertia for a disk with uniform density:

$$I=\int r^{2}\,dm=\int \rho r^{2}\,dV$$

For a disk, I just used dA instead of dV. Now, to calculate the density:

$$\rho = \frac{\text{mass}}{\text{volume}}=\frac{m}{\pi r^{2}}$$

So now we have:

$$I=\int \rho r^{2}\,dA=\frac{m}{\pi}\int \,dA=\boxed{mr^{2}}$$

However, I know that the moment of inertia for a disk is $\frac{1}{2}mr^{2}$. Where did I go wrong?

Thank you.

2. Dec 25, 2005

### amcavoy

Nevermind I see where I went wrong (I used r twice, whereas I should have used R and r, where r ranges from 0 to R).

3. Dec 25, 2005

### Staff: Mentor

In the integral, one would have $r^2\,\frac{m}{\pi\,r^2} r dr d\theta$ the r2 terms cancel, and that leaves dA = r dr d$d\theta$.