# Moment of inertia for thin uniform rectangle/hexagon

1. Oct 4, 2007

### rundream

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 $$\theta$$ $$_{1}$$ and $$\theta$$ $$_{2}$$
(since the traingles are made from a rectangle their are 2 sets of triangles with different $$\theta$$)
also not $$\theta$$ is the angle towards the center of the mass not an outer ridge angle(all together 4 $$\theta$$ representing all 4 triangles)

2. Relevant equations

im given the eqaution
I=moment of inertia= $$(1/12)$$M(a$$^{2}$$ + b$$^{2}$$)
and also earlier in class worked out for isosceles trianlges
I=($$1/2)$$M[1+($$1/3$$)tan$$^{2}$$((1/2)$$\theta$$)]h$$^{2}$$

3. The attempt at a solution
given those 2 eqautions i have to prove that I=$$(1/12)$$M(a$$^{2}$$b$$^{2}$$) really gives the moment of inertia for full mass.
I started working backwards on the problems replacing M with $$\sigma h^{2} tan((1/2)\theta$$)
being that $$\sigma = M/A$$ and A = h$$^{2}$$$$\ tan((1/2)\theta$$)
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$$^{2}$$ and b$$^{2}$$

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.

You basically need to write down h and $\theta$ 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.