# Deriving moment of Inertia

1. Nov 12, 2009

### kuahji

I'm attempting to derive the moment of inertia for a cylindrical object.

I know that I=$$\int r^2 dm$$

which equals =$$\int r^2 p dV$$

My question begins here, the derivations I seen pull p out of the integral, which makes sense to do, because in this case it's a constant. p=M/($$\pi$$r^2L). So if I don't pull p out before integrating I get I=Mr^2, if I do pull it out, I get I=M/2r^2. I know the answer should be I=M/2r^2 because I have a solid cylindrical object. So why am I getting a different result when I leave p in, & a different result when I pull p out or am I just making a silly math error?

Below is my work when I leave p inside the integral

I=$$\int r^2*p*(2\pi*r)dr$$
=2M$$\int r dr$$ (replacing p with M/($$\pi$$r^2L) before integrating)
=Mr^2

2. Nov 12, 2009

### ApexOfDE

I believe that density is constant for each material.

p/s: you replace m = DV, then you replace D = M/V... I dont get it :(

3. Nov 12, 2009

### kuahji

I replaced dm with pdV & then p with M/V.