Using primitives to integrate moments of Inertia

AI Thread Summary
The discussion focuses on using "primitives" to calculate moments of inertia, specifically for a sphere. The original poster struggles to align their approach using multiple integrals with the professor's requirement for a singular integral method. They initially find success with their method but encounter discrepancies in results when applying the professor's approach. Key to resolving the issue is correctly identifying the radius of each disk as a function of z and substituting the correct expressions for dm. Ultimately, the poster realizes the mistake in their calculations and successfully arrives at the correct moment of inertia.
Nikolas
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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
 
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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
 
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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...
 
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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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