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Moment of Inertia of a regular hexagonal plate?

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data
    How do you calculate the moment of inertia of a regular hexagonal plate of side a and mass M along an axis passing through its opposite vertices?


    2. Relevant equations
    Moment of inertia for a right triangle with an axis running along its base would be I = m h2 /6 where h and m are the height and the mass of the triangle respectively.


    3. The attempt at a solution
    I tried breaking up the hexagon into 4 congruent right triangles and a rectangle. So there would be 2 triangles at the top, the rectangle in the middle and the two triangles below.
    The moment of inertia of the triangle would be = [itex] \frac{m a^2} {8} [/itex] because the height will be = asin60
    Mass of this triangular bit = M/12
    Density of the hexagon = 2M/(a23√3)
    So the rectangular bit in the middle will have moment of inertia equal to this density multiplied by the double integral of x^2 with x running from a√3 to -a√3 and y running from a/2 to -a/2.
    This gives moment of inertia for the rectangular bit = 4Ma2/3.
    I thought then you finally add all these up to get moment of inertia for the hexagon because integration is a linear operator or whatever. But I get the wrong answer.
     
    Last edited: May 21, 2013
  2. jcsd
  3. May 21, 2013 #2

    haruspex

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    Seems too much. Pls post your working.
     
  4. May 22, 2013 #3
    Silly silly me. The integral for x runs from (a√3)/2 to -(a√3)/2 and this gives the right answer.
     
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