1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook