## Deriving moment of Inertia

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
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 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 :(
 I replaced dm with pdV & then p with M/V.

 Similar discussions for: Deriving moment of Inertia Thread Forum Replies Introductory Physics Homework 1 Calculus & Beyond Homework 4 Introductory Physics Homework 2 Classical Physics 1 Introductory Physics Homework 1