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Moment of inertia of a sphere

  1. Apr 3, 2010 #1
    1. The problem statement, all variables and given/known data
    calculate the moment of inertia of a sphere of mass M and radius R by integrating over thin shells

    2. Relevant equations
    Ishell=(2/3)mR2


    3. 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
     
  2. jcsd
  3. Apr 3, 2010 #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.
     
  4. Apr 3, 2010 #3
    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
     
  5. Apr 4, 2010 #4

    kuruman

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    And why is that correct? What expression did you use for the density?
     
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