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Moment of inertia of triangular prism

  1. Dec 13, 2006 #1
    1. The problem statement, all variables and given/known data
    Find the moment of inertia of triangular prism (equilateral triangles with side 2a, parallel to xy-plane), mass M. It's centered at the origin with long side parallel to z-axis. Find moment of inertia about the z-axis, and without doing integrals explain two products of inertia (Ixy, Ixz, I suppose)
    2. Relevant equations
    The integral equation for Izz=int[(M/V)*(x^2+y^2)dV]
    where V=volume, or in cylindrical coords. we can write (x^2+y^2)=r^2, where r=distance from z-axis
    3. The attempt at a solution
    I just can't figure out - if I do it in cartesian coords - what limits of x and y should I substitute? It doesn't look like it should be -a to a, since that wouldn't be geometrical center, and plus to that - it's a triangle, not square - so I should connect x to y somehow.... Or should I do it in different coord. system?...

    Thanks a lot in advance!
    I understand that this is easy question, but I just can't figure it out for some reason....
     
  2. jcsd
  3. Dec 13, 2006 #2

    OlderDan

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    Homework Helper

    I don't think you want to use cylindrical coordinates for this. Cartesian coordinates will do nicely. You need to know where the centroid of the tiangle is to put it at the origin. By symmetry, it is at the intersection of the bisectors of the angles. I assume you will want one side parallel to the x axis (or y axis; your choice). The limits of integration are determined from the equations of the boundary lines forming the triangle. Draw the trianngle in the x-y plane and write the equations of the lines.
     
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