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Moment of inertia of a sphere

  • Thread starter Zamba
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  • #1
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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
 

Answers and Replies

  • #2
kuruman
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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.
 
  • #3
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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
 
  • #4
kuruman
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... 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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