# Deriving moment of Inertia

by kuahji
Tags: deriving, inertia, moment
 P: 395 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