Deriving moment of Inertia


by kuahji
Tags: deriving, inertia, moment
kuahji
kuahji is offline
#1
Nov12-09, 05:39 PM
P: 394
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)
=Mr^2
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ApexOfDE
ApexOfDE is offline
#2
Nov12-09, 07:33 PM
ApexOfDE's Avatar
P: 122
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 :(
kuahji
kuahji is offline
#3
Nov12-09, 07:50 PM
P: 394
I replaced dm with pdV & then p with M/V.


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