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Moment of Inertia in Equilateral Triangle

  1. Apr 21, 2012 #1
    The problem is to find the moment of inertia for a solid equilateral triangle about an axis at the triangle's vertex which is perpendicular to the plane of the triangle. Mass is given as M.

    I was wanting to treat it like 3 point masses at each vertex with 1/3 mass. Then I could use the sum definition of moment of inertia instead of the integral. The distance from axis at center of the equilateral triangle to each vertex is L/√3.

    Ʃ mr2= 3(1/3 M * L2/9) = ML2/9

    Then by the parallel axis theorem I can shift this out to a vertex using that same distance for a final result of

    2ML2/9

    I know it can be done by treating dA like a thin rod and using calculus, I was just curious why the above doesn't get the same answer. Isn't it dynamically the same to treat it like 3 1/3 masses on each vertex?
     
  2. jcsd
  3. Apr 22, 2012 #2

    Filip Larsen

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    Gold Member

    You cannot just "move" all the mass to the vertices without getting a different moment of inertia. It is possible to make a 3-point-mass triangle that has an equivalent moment of inertia around the center of mass perpendicular to the plate, but each 1/3 of the mass would then have to be placed L/(2√6) from the center of mass and not at geometric vertex at distance L/√3. You can think of this as a partitioning the triangle into 3 identically diamond-shaped parts.

    However, the moment of inertia around center of mass can be found fairly easy by standard integrals so you should probably be able to show how to get the moment of inertia following this approach.
     
    Last edited: Apr 22, 2012
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