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Deriving moment of Inertia

  1. Nov 12, 2009 #1
    I'm attempting to derive the moment of inertia for a cylindrical object.

    I know that I=[tex]\int r^2 dm[/tex]

    which equals =[tex]\int r^2 p dV[/tex]

    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/([tex]\pi[/tex]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=[tex]\int r^2*p*(2\pi*r)dr[/tex]
    =2M[tex]\int r dr[/tex] (replacing p with M/([tex]\pi[/tex]r^2L) before integrating)
  2. jcsd
  3. Nov 12, 2009 #2
    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 :(
  4. Nov 12, 2009 #3
    I replaced dm with pdV & then p with M/V.
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