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Moment of Inertia (Triangular Prism)

  1. Nov 21, 2013 #1
    1. The problem statement, all variables and given/known data
    A triangular prism (like a box of toblerone) of mass M, whose ends are equilateral triangles parallel to the xy plane with side 2a, is centered on the origin with its axis along 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
    I=∫r2dm
    ρ=M/V
    r2=x2+y2

    3. The attempt at a solution
    ρ=M/(.5bhH);H is total length along z-axis
    ρdV=dM=.5ρdxdydz

    I=∫∫∫x2+y2dxdydz

    I evaluated this with integration limits
    for x: 0 to a
    for y: 0 to √3a/2
    for z: 0 to h/2

    after doing it all out and substituting M/V back in for ρ, I got an answer of 7Ma2/96. I muliplied by 2 since that integral was only for half of the object, so my final answer would be 7Ma2/48. I have an intuition that this is incorrect. Can anyone point me in the right direction? Are my limits of integration correct? Any help would be much appreciated. Thank you
     
  2. jcsd
  3. Nov 21, 2013 #2

    vela

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    Where exactly is the axis of the prism?


    Where did the 0.5 come from in the last line?

    I don't understand your limits at all. Why are you integrating z from 0 to half the height of the triangle?

     
  4. Nov 21, 2013 #3
    Not sure I really understand what you're asking. It says it's along the z-axis


    ρ=M/V
    ρV=M; V=A*H; A=.5bh
    ρ(.5bhH)=M
    ρdV=dM
    I substituted in .5bhH for V but looking back on it I don't think I should have.
    would dM=ρdxdydz be correct?


    Sorry that was a typo. "h" should have been "H"
     
  5. Nov 21, 2013 #4

    vela

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    I'll assume one of the edges of the prism lies along the z-axis then. Is that right?



    Yes, that's generically correct because in Cartesian coordinates, the volume element is given by dV = dx dy dz.



    I still don't understand why you're only integrating to H/2. It's not like dividing by 2 and then subsequently multiplying by 2 simplifies the calculation any.
     
  6. Nov 21, 2013 #5
    The way I interpret it is that the axis of rotation is along the z-axis, and that the z-axis runs through the center of mass. Basically, the triangle is centered on the z-axis. The wording is very vague, and I am a little confused by what is meant.


    So I should integrate from -H/2 to H/2 is what your saying?
     
  7. Nov 21, 2013 #6

    vela

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    That sounds reasonable. So the z-axis lies somewhere in the middle of the prism. Doesn't this mean that x and y should take on both positive and negative values? Have you drawn a sketch?



    Yes.

    You'll see that the integration over the z-coordinate simply amounts to multiplying by H. Rather than using dV = dx dy dz, you could write dV = H dx dy and forgo integrating over z altogether. The infinitesimal volume element would be a rectangular solid with sides of length dx, dy, and H. You can do this because every piece of that element is the same distance r from the axis of rotation.
     
  8. Nov 21, 2013 #7
    I have drawn a sketch. I'm going to try and work it out with the z-axis in the center of the triangle, and then with the z-axis as one of the prism edges, and use the parallel-axis theorem. I can take a picture of my work then. I know x and y would have positive and negative values, but if I integrated from them, wouldn't they just cancel out because of the symmetry?
     
  9. Nov 21, 2013 #8

    vela

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    No, it wouldn't cancel out because r2 is positive regardless of the signs of x and y.
     
  10. Nov 21, 2013 #9

    haruspex

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    Why not 0 to H? It won't make any difference. In fact, forget z and just treat it as a flat triangle.
    I suggest using the symmetry of the triangle. Cut it into six identical right angled triangles, each with a corner at the z axis. That will simplify the integration bounds.
     
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