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Homework Help: Moment of Inertia of a Triangular Prism

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data
    A triangular prism of mass M, whose two ends are equilateral triangles parallel to the xy plane with side 2a, is centered on the origin with its axis on the z axis. Find its moment of inertia for rotation about the z axis. Without doing any integrals write down and explain its two products of inertia for rotation about the z axis.


    2. Relevant equations



    3. The attempt at a solution
    I've arrived at a soultion, but I'd like to see what other people think, if it is right or not.

    I know that the moment of inertia (here, just I), is this:

    I = integral of [ (x^2 + y^2) dm]
    and
    p = M/V, so
    dm = pdV, so
    I = p * integral of [ (x^2 + y^2) dV], so
    I = p * triple integral of [ (x^2 + y^2) dx dy dz]

    This will need to be done as the sum of two integrals, since there is a discontinuity in the change in y.

    For the first integral:
    The limits of the x integral are -a to 0.
    The limits of the y integral are -a/√3 to 2a/√3.
    The limits of the z integral are, let's say, 0 to h.

    I evaluated this to be p*h*(-4/3)(a^4)/(√3)
    Since p = M/V and V = (h*a^2)/(√3), the first integral equals
    M*(-4/3)(a^2)

    For the second integral:

    The limits of the x integral are 0 to a.
    The limits of the y integral are 2a/√3 to -a/√3.
    The limits of the z integral are 0 to h.

    I evaluated this to be, of course, the same thing,
    M*(-4/3)(a^2)

    So,
    I = (-8/3)(a^2)M

    That should be the moment of inertia for rotation about the z axis.

    The products of inertia should both be zero (because of symmetry).

    Does this seem right? More importantly, is it right? Thank you for looking at it.
     
  2. jcsd
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