How Do You Calculate the Moment of Inertia for a Sphere Using Shell Integration?

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for a sphere using shell integration. The original poster attempts to derive the moment of inertia by decomposing the sphere into infinitesimal shells and integrating over them.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration process and the substitution of mass elements. There are questions regarding the correct expression for density and the proper setup of the integral.

Discussion Status

Some participants have pointed out potential errors in the original poster's approach, particularly concerning the replacement of the mass element under the integral. The conversation is ongoing, with attempts to clarify the integration steps and assumptions being made.

Contextual Notes

There is a mention of confusion regarding the expression for density and how it affects the integration process. The original poster also notes a formatting issue with symbols in their post.

Zamba
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Homework Statement


calculate the moment of inertia of a sphere of mass M and radius R by integrating over thin shells

Homework Equations


Ishell=(2/3)mR2

The Attempt at a Solution


this is what i have so far
the sphere is decomposed into infinitesimal shells with surface area 4[tex]\pi[/tex]r2
the mass of each shell is dm=[tex]\rho[/tex](4[tex]\pi[/tex]r2)dr
after expanding rho and canceling terms I get
dm=(3m/r)dr

I=[tex]\int[/tex](2/3)(dm)r2dr from 0 to R.
I=2mrdr from 0 to R
this gives me I=mr2

does anyone see what I did wrong?
PS. those "pi" are not supposed to be powers. sorry, I'm not sure how to change them
 
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You did not replace dm with ρ(4πr2) under the integral sign. Instead, you integrated dm into m and then integrated over dr once more.
 
i'm sorry, i didn't write it all out

I=[tex]\int[/tex](2/3)dmr2 dr
from here I replaced dm with (3m/r)dr so i got

I=[tex]\int[/tex](2/3)(3m/r)r2 dr
then i got
I=[tex]\int[/tex]2mr dr from 0 to R
which gave me mr2
 
Zamba said:
... from here I replaced dm with (3m/r)dr...
And why is that correct? What expression did you use for the density?
 

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