
#1
Nov1209, 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 



#2
Nov1209, 07:33 PM

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 :( 



#3
Nov1209, 07:50 PM

P: 394

I replaced dm with pdV & then p with M/V.



Register to reply 
Related Discussions  
Help Deriving formula for moment of inertia lab  Introductory Physics Homework  1  
Deriving the moment of inertia for a sphere  Calculus & Beyond Homework  4  
Deriving Moment of Inertia  Introductory Physics Homework  2  
Second Moment of Inertia  Area Moment of Inertia  Classical Physics  1  
What are moment of inertia, mass moment of inertia, and radius of gyration?  Introductory Physics Homework  1 