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I need help finding my mistake moment of inertia

  1. Nov 3, 2008 #1
    How do i prove,using integration, that the moment of inertia of a hollow cylinder that has mass M, an outside radius R2, and an inside radius R1 is given by...
    I=.5(M)(R2^2+R1^2)

    here is the work I have done...
    I am really close but do not see where I made my mistake
    Can someone help me find it?

    If you look at the cylinder like a bunch of hoops stacked together then

    I(hoop)=MR^2

    dV=(2*pi*R)(dR)(L)

    P=sigma
    P=dM/dV
    dM=(2*pi*R)(dR)(L)(P)

    I=int(R^2*dM)
    I=int[a,b](R^2)(2*pi*R)(dR)(L)(P)
    I=(2*pi*L*P)*int[R1,R2](R^3dR)
    I=(2*pi*L*P)*[(R2^4-R1^4)/4]
    I=(pi*L*P)*[(R2^2-R1^2)(R2^2+R1^2)/2]
    V=pi*L(R2^2-R1^2)
    M=pi*L*P(R2^2-R1^2)
    (R2^2-R1^2)=(pi*L*P)/M
    I=(R2^2+R1^2)/2M

    I have most of the equation correct but M is in the denominator when if needs to be just the opposite

    Thank you in advance
     
  2. jcsd
  3. Nov 4, 2008 #2
    Hi there,

    Everything is fine up to the last two steps.

    From M = pi*L*P(R2^2 - R1^2) you should get M/(pi*L*P) = (R2^2 - R1^2) and not (pi*L*P)/M = (R2^2 - R1^2).

    When you change this you should get the right answer.

    Don't worry about it, it's just a small mistake. So, you shouldn't get penalized heavily in an exam for it.

    I make mistakes like these all the time. The way to avoid them is to always double-check all the steps in your calculations.

    Hope this helps,

    Wynand.
     
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