21 August 2014

Mass moment of inertia of an equilateral triangle

As in previous posts, I would like to determine the moments of inertia of a solid body, this time an equilateral triangular prism. I will start in this post by a (very thin) equilateral triangle. The challenge is getting the result in the simplest way, making the most of the symmetry elements and taking advantage of the parallel axis theorem.

Around the $$z$$ axis

The $$z$$ axis goes through the center of mass of the triangle of interest (gray central area of side $$L$$ in the illustration above) and is perpendicular to its plane. We denote the corresponding moment by $$I_z(L)$$. The moment of the large triangle, with side $$2L$$, is $$I_z(2L)$$. We can relate these two parameters in two ways:
• For a given shape and surface mass density, the moment of inertia scales as the size to the fourth power, on dimensional grounds. Thus, $$I_z(2L) = 16 I_z(L)$$.
• The large triangle can also be described as the rigid assembly of the small central triangle and the three adjacent ones. The parallel axis theorem yields:
$I_z(2L) = I_z(L) + 3 [I_z(L) + m(L) d^2]$
where $$m(L)= \mu L^2 \sqrt{3}/4$$ is the mass of the small triangle, with $$\mu$$ the surface mass density, and $$d=L/\sqrt{3}$$ is the distance between the centers of mass of the side triangles and the $$z$$ axis.

Combining these two expressions for $$I_z(2L)$$ immediately yields:
$I_z(L) = \mu L^4 \frac{\sqrt{3}}{48} \tag{1}$

Around the $$y$$ axis

The $$y$$ axis is contained in the plane of the triangle and goes through its center of mass and one vertex. Using the same strategy as above, we get:
$\left\{ \begin{array}{ll} I_y(2L) &= 16 I_y(L)\\ I_y(2L) &= 2 I_y(L) + 2 [I_y(L) + m(L) (L/2)^2] \end{array} \right.$
where on the right-hand side of the second equality the first term corresponds to the central and top triangles (both their centers of mass are on axis $$y$$) and the second one to the side triangles, whose centers are shifted by $$L/2$$. Finally:
$I_y(L) = \mu L^4 \frac{\sqrt{3}}{96} = \frac{I_z(L)}{2} \tag{2}$

How about the $$x$$ axis? To answer this question, we start by noting that there are three equivalent directions within the plane of the triangle: $$y$$ and the axes (say, $$y'$$ and $$y''$$) going through the other two vertices: $$I_y(L) = I_{y'}(L) = I_{y''}(L)$$. This third-order symmetry in a two-dimensional space means that the inertia tensor is in fact isotropic in the plane of the triangle, with the same value $$I_{\bot}(L) = I_{y}(L) = I_{x}(L)$$ for any axis in this plane. The inertia tensor is then:
$\mathrm{I} = \left ( \begin{matrix} I_{\bot}(L) & 0 & 0 \\ 0 & I_{\bot}(L) & 0 \\ 0 & 0 & I_z(L) \end{matrix} \right )$
This isotropy of a tensorial property for a system that does not in fact have full rotational symmetry is a very useful result (albeit somewhat counterintuitive). To give only one example from a completely different area of physics: a cubic crystal cannot be birefringent!