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Moment of inertia of a right triangle

  1. Apr 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A right triangle has height 'h' and width 'b.' The right triangle has a constant area density. Calculate the moment of inertia of the triangle rotated around an axis that runs along side 'h.'


    2. Relevant equations

    I = integral(r^2*dm) where 'r' is distance from the axis

    3. The attempt at a solution

    equation of hypotenuse is (h/b)

    r^2 dm = r^2 * p * dA where p is area density and dA is the area of the rectangles.

    = r^2 * p * (h/b)r * dr = r^3 * p * (h/b) * dr --> integrate =
    ( (ph)/(3b) )r^3 with the limits of integration being from r=0 to r=b so:
    ( (ph)/3 )b^2 - 0 so I = ( (ph)/3 )b^2

    But all my friends said I was wrong so can someone please tell me why?
     
  2. jcsd
  3. Apr 2, 2009 #2

    tiny-tim

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    Hi homeslice64! Welcome to PF! :smile:

    (have an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
    erm :redface: … ∫r3dr isn't r3/3 :wink:
     
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