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Moment of inertia of a cone

  1. Jan 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the moment of inertia of a solid cone about its longitudinal axis (z-axis)

    The cone: [itex]x^2+y^2<=z^2, 0<=z<=h[/itex]

    [itex]I_z = \int\int\int_T(x^2+y^2)dxdydz[/itex]


    2. Relevant equations
    Representing the cone in cylindrical coords:

    [itex] x=zcos\theta [/itex]
    [itex] y=zsin\theta [/itex]
    [itex] z=z [/itex]

    3. The attempt at a solution
    The integral in cylindrical coords is:

    [itex]I_z = \int_0^h \int_0^{2\pi} \int_0^z (z^2)rdrd\theta dz[/itex]

    Evaluating the triple integral gives:
    [itex]\pi h^5/5[/itex]

    But the answer in the book is:
    [itex]\pi h^5/10[/itex]

    I don't see what I did wrong
     
  2. jcsd
  3. Jan 11, 2015 #2

    LCKurtz

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    The moment arm for the second moment from the z axis is ##r^2##, not ##z^2##.

     
  4. Jan 12, 2015 #3
    I see:

    [itex] x=rcos\theta, y=rsin\theta[/itex] where [itex]0<=r<=z[/itex]

    Thanks alot
     
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