- #1
kuahji
- 394
- 2
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
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