## 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!