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I was reading about moment of inertia on Wikipedia and thought it was weird that it had common values for shapes like tetrehedron and cuboids but not triangular prisms or triangular plates, so I tried working it out myself. I will post my attempt below, but for some reason I cannot find any source online that confirms or denies my solution. Please let me know if you find anything wrong with it. Thanks.

Q: What is the moment of inertia of an equilateral triangular plate of uniform density $\rho$, mass $M$, side length $L$, rotating about an axis perpendicular to the triangle's plane and passing through its center?

  1. First I modeled an equilateral triangle using three lines with its center of geometry at the origin as follows: $x=\frac{1}{\sqrt{3}}y-\frac{1}{3}L \\ x=\frac{1}{3}L-\frac{1}{\sqrt{3}}y \\ y=-\frac{\sqrt{3}}{6}L$

I used the fact that the circumradius of an equilateral triangle is $\frac{\sqrt3}{3}L$ and that its height is $\frac{\sqrt{3}}{2}L$ .

  1. Next, using the definition of moment of inertia ($I$) and with the help of Wolfram Alpha, I obtained the following result:

$$I=\int r^2 dm=\rho \int r^2 dA\\ =\rho \int_{-\frac{\sqrt{3}}{6}L}^{\frac{\sqrt{3}}{3}L} \int_{\frac{1}{\sqrt{3}}y-\frac{1}{3}L}^{\frac{1}{3}L-\frac{1}{\sqrt{3}}y} x^2+y^2 dxdy\\ =\frac{\rho}{16 \sqrt{3}}L^4=(\frac{4M}{\sqrt{3} L^2})(\frac{L^4}{16\sqrt{3}})\\ =\frac{1}{12}ML^2$$

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closed as off-topic by sammy gerbil, ZeroTheHero, ACuriousMind Jun 24 '18 at 13:52

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I can confirm your result. I can also suggest you a neater way to derive it inspired by David Morin - Introduction to Classical Mechanics, check it out in a library if you have access.

The main idea is to use the symmetry of the equilateral triangle and split it into 4 smaller equilateral triangles like this

enter image description here

Now analyse how the moment of inertia changes when we rescale its mass and sidelength, i.e. if $ i = \alpha ml^2$ and $L = 2l, M = 4m$, then $I = \alpha (4m) (2l)^2 = 16i$, where $m = mass \ of\ a \ small \ triangle$, $l = sidelength \ of\ a\ small\ triangle$ and the capital $M, L$ correspond to the larger triangle. But the moment of inertia of the big triangle can be also split into $4$ moments of inertia. Be aware that we need to use the parallel axis theorem for the $3$ triangles which enclose the central triangle. Hence, $$I = 16i = 4i + 3m(\frac{l\sqrt{3}}{3})^2.$$ Which reduces to $i = \frac{1}{12} m l^2.$

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Another solution is to integrate the triangle from an apex to the base using the double integral of r^2dm, which becomes (x^2+y^2)dxdy. Using the limits of x to be 0 to h, and the limits of y to be -xtan30 and +xtan30, you get the moment of inertia about an apex to be 0.32075h^4M/AL, where h is the height of the triangle and L is the area. Using the parallel axis theorem, you can find the moment of inertia about the center by subtracting Mr^2, where r is 2/3h. The final result is ML^2/12.

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