What is the moment of inertia of a thin equilateral triangular plate of mass $m$ and side $a$ about an axis through its center (perpendicular to the plane)?
I expressed $r$ as $r = \sqrt{x^2 + y^2}$, and $dm = \rho dS$ so I get:
$$I = \int r^2\, dm = \rho \int (x^2 + y^2)\, dS = \rho \int \int (x^2 + y^2)\, dx dy.$$
Now for the intervals, I expressed the height the triangle as $v = \frac{a \sqrt{3}}{2}$ and because I know the center of the triange is $\frac{2}{3}$ from the top point and $\frac{1}{3}$ from the base, my interval will be $y \in (-\frac{a \sqrt{3}}{6}; \frac{a \sqrt{3}}{3})$. For the $x$ interval it is just $x \in (-\frac{a}{2}; \frac{a}{2})$. When I evaluate the integral, I get: $$ I = \frac{1}{3}ma^2.$$
My questions are: First, is this even correct? Second, isn't there a simpler way to find the moment of inertia (using integrals)?
Thanks.
Best Answer
Another way is to do the integration in polar coordinates. We can focus on one sixth of the triangle, with $\phi=0$ corresponding to a midpoint and $\phi=\frac\pi3$ corresponding to a vertex and the side between them described by $r=\frac h{3\cos\phi}$:
\begin{eqnarray} \int_0^\frac\pi3\mathrm d\phi\int_0^\frac h{3\cos\phi}r\mathrm drr^2 &=& \frac{h^4}{4\cdot3^4}\int_0^\frac\pi3\frac{\mathrm d\phi}{\cos^4\phi} \\ &=& \frac{h^4}{4\cdot3^4}\left[\frac{(2+\cos2\phi)\sin\phi}{3\cos^3\phi}\right]_0^\frac\pi3 \\ &=& \frac{\sqrt3h^4}{2\cdot3^4}\;. \end{eqnarray}
With $h=\frac{\sqrt3}2a$ and the area $A=\frac{\sqrt3}4a^2$, and taking into account the factor $6$, this yields the moment of inertia
$$6\cdot\frac{\sqrt3}{2\cdot3^4}\left(\frac{\sqrt3}2a\right)^4\left(\frac{\sqrt3}4a^2\right)^{-1}m=\frac1{12}ma^2\;, $$
in agreement with the accepted answer to this physics.SE question, which, by the way, shows a nice way to use symmetry to get the moment of inertia without evaluating any integrals.
From your description of your own calculation, it sounds as if you didn’t take into account that the limits for the inner integral over $x$ depend on $y$.