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Moment of inertia for thin uniform rectangle/hexagon

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data
    the first problem is i have a rectangle split into four triangles the long side labeled a the short side labeled b and the inner angles of the triangles are [tex]\theta[/tex] [tex]_{1}[/tex] and [tex]\theta[/tex] [tex]_{2}[/tex]
    (since the traingles are made from a rectangle their are 2 sets of triangles with different [tex]\theta[/tex])
    also not [tex]\theta[/tex] is the angle towards the center of the mass not an outer ridge angle(all together 4 [tex]\theta[/tex] representing all 4 triangles)

    2. Relevant equations

    im given the eqaution
    I=moment of inertia= [tex](1/12) [/tex]M(a[tex]^{2}[/tex] + b[tex]^{2}[/tex])
    and also earlier in class worked out for isosceles trianlges

    3. The attempt at a solution
    given those 2 eqautions i have to prove that I=[tex](1/12) [/tex]M(a[tex]^{2}[/tex]b[tex]^{2}[/tex]) really gives the moment of inertia for full mass.
    I started working backwards on the problems replacing M with [tex]\sigma h^{2} tan((1/2)\theta[/tex])
    being that [tex]\sigma = M/A [/tex] and A = h[tex]^{2}[/tex][tex]\ tan((1/2)\theta[/tex])
    btw M=total mass of object, A=total area
    (sorry trying my best to not confuse on problem)
    since the height of one triangle is 1/2b or 1/2a depending on which triangle picked
    i also replaced the terms for a[tex]^{2}[/tex] and b[tex]^{2}[/tex]

    but im at a lost at this point and dont know where to go past this
    i figured if i reverse engineered the problem i may be able to figure the 4 triangles relationship to the total moment of inertia so then i could return the sumed up eqaution into the original eqaution for an isosceles triangle so i can understand how to use the triangles for a hexagon and octagon.

    please help if possible with this problem.
  2. jcsd
  3. Oct 4, 2007 #2


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    Staff Emeritus
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    Gold Member

    When you specify a moment of inertia for an object, you need to also specify the axis about which the moment is calculated.

    You basically need to write down h and [itex]\theta[/itex] in terms of the sides a,b, and use the fact that m(triangle) = (M/ab)*(area of triangle). If necessary you may need to use the Parallel Axis theorem to have all moments about the same point. Then it's just plugging and adding up the 4 moments to get the total.
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