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Moment of inertia of a solid of rotation

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data
    The first quadrant area bounded by the curve
    [tex]x^3+y^3=8[/tex]
    is rotated around y-axis to give a solid of rotation. The question asks for an integral which represents the solid's moment of inertia around the axis.

    My answer is:

    [tex]I_y=M\frac{\int_0^2x^2ydx}{\int_0^2x^2dx}[/tex]

    but seems like I'm wrong. Could someone help me out, please?
     
  2. jcsd
  3. Jul 1, 2011 #2

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  4. Jul 1, 2011 #3

    SammyS

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    The Moment of Inertial of a cylindrical shell of mass, m, and radius, R, is: m(R2).
     
  5. Jul 1, 2011 #4
    We need to integrate over the height, that is from 0 to 2, and for each y we need to account for a disk of radius
    [tex]x=\sqrt[3]{(8-y^3)}[/tex]
    which moment of inertia is:
    [tex]dI=\rho x^2 dy=\rho (8-y^3)^{2/3}dy[/tex]
    For the whole body this is:
    [tex]I=\int_0^2\rho (8-y^3)^{2/3} dy[/tex]
    [tex]I=\frac{m}{V}\int_0^2 (8-y^3)^{2/3} dy[/tex]

    I think this is wrong.
     
  6. Jul 1, 2011 #5

    SammyS

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    The Moment of Inertial of a disk of mass, m, and radius, R, is: (½)m(R2).

    However, I think it would be better to use cylindrical shells as tiny-tim recommended.

    The last integral you give, [itex]\displaystyle \int_0^2 (8-y^3)^{2/3} dy[/itex] , gives the volume of your solid of revolution.
     
    Last edited: Jul 1, 2011
  7. Jul 1, 2011 #6

    tiny-tim

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    Hi Grand! :smile:
    This is wrong for two reasons:

    i] the moment of inertia of a disc is 1/2 mr2, not mr2

    ii] you can't just use ρ instead of m … you need m for the whole disc, which is ρr2dy, so the whole formula for moment of inertia of the disc is 1/2 ρr4dy

    But why are you making things so difficult for yourself? As SammyS :smile: says, cylindrical shells would be much easier.

    Try again. :smile:
     
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