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Moment of Inertia of Hollow Cylinder Derivation

  1. Aug 19, 2014 #1
    For a uniform, hollow cylinder, why is this derivation wrong?

    M = mass of whole solid cylinder
    m = mass of missing cylindrical piece
    R = radius of whole cylinder
    r = radius of missing cylindrical piece

    moment of inertia = moment of inertia of whole cylinder - moment of inertia of missing cylindrical piece

    I = MR2/2 - mr2/2

    m/M = pi*r2*h/pi*R2*h = r2/R2

    m = M*r2/R2

    I = MR2/2 - M*r4/2R2

    I = MR4/2R2 - M*r4/2R2

    I = M/2R2*(R4 - r4)
     
  2. jcsd
  3. Aug 19, 2014 #2
    It seems correct, it is only that the result contains in its expression the mass M of the whole cylinder and not the mass of the corresponding hollow cylinder. If you substitute [itex]M=\frac{M_{h}R^2}{R^2-r^2}[/itex] you ll get the usual expression for the inertia of a hollow cylinder [itex]I=\frac{1}{2}M_h(R^2+r^2)[/itex].
     
    Last edited: Aug 19, 2014
  4. Aug 19, 2014 #3

    olivermsun

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    It's fine as written. If you want to express ##I## using the mass of the outer shell only, call it ##M^\prime,## then you have to use ##M^\prime = M - m = M (1 - r^2/R^2)## to get the usual form for ##I##.
     
  5. Aug 21, 2014 #4
    Thanks guys. Just a matter of confusion due to the specific application of the formula.
     
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