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Using primitives to integrate moments of Inertia

  • Thread starter Nikolas
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  • #1
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using "primitives" to integrate moments of Inertia

My classmates are lost and I just can't think the way my proffessor does when approaching these problems. He gave us about 50 shapes to find moments of inertia for this weekend and I'm not having any trouble doing them... my way. I can use a point particle primitive (mr^2) and setup a double or triple integral for 2 and 3 dimensional objects and get the right answer without any trouble, but he's making us to use other "primitives." For example to get the moment of Inertia for a sphere we have to use a stack of disks, the primitive being (1/2)MR^2 and dIprim = (1/2)(r^2)dm. But I don't see how that can work with dm being ρdV or ρ(r^2)sinΦdρdΦdθ. The closest I can come to getting his way to work is setting dm to ρrdzdr and running z from -r to r but I end up with (3M(R^2))/(20π) when it should be (3M(R^2))/(5). I don't see how I can get rid of that π since ρ = M/((4/3)π(R^3)).

I think the main incompatibility is that I'm using multiple integrals and whatever coordinate system seems to fit best since that's how I visualize things and how it makes sense to me but since calc 3 isn't a prerequesite for the course he uses a singular integral, and I just can't see how it works.

Thanks,

Nik
 

Answers and Replies

  • #2
lightgrav
Homework Helper
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What? He's just done the integral in dΦ and the integral in dr already.
You still have to find the correct radius of each disk as function of z,
before you integrate along dz (from -R to +R ... sound familiar?)
Using Pythagoras, r^2 + z^2 = R^2
 
Last edited:
  • #3
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http://members.cox.net/mr2host2/integral1.gif
in (1/2)(r^2)dm, dm = (rho)(pi)(R^2)dz, (r^2)=(R^2)-(z^2), and rho=(3M)/(4piR^3).... but it gives me I=(1/2)M(R^2) which is wrong. Where am I going wrong in my thinking? I'm closer now than I am before though...
 
Last edited by a moderator:
  • #4
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nevermind I got it finally, the R^2 in dm needed to be r^2 and replaced by R^2-z^2
 

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