## 23 August 2014

### Moment of inertia of a clipped triangle

After calculating the moment of inertia for an equilateral triangle, let us consider the same shape, but with clipped corners, as in the drawing below:

We will preserve the notations of the previous post, adding the superscript "tr" for the truncated shape: $$I_{z}^{\text{tr}} (L,a)$$ is the moment about the $$z$$ axis of the equilateral triangle with side $$L$$, clipped by $$a$$ at each corner (with $$a \leq L/2$$). We will also use the same strategy, writing the moments of the complete shape as a combination of its four fragments:
$I_{z}(L) = I_{z}^{\text{tr}} (L,a) + 3[I_{z}(a)+m(a)d^2] ,$
where $$d=(L-a)/\sqrt{3} .$$ Using the results obtained for the full triangle immediately yields:
$I_{z}^{\text{tr}} (L,a) = \frac{\sqrt{3}}{48} \mu \left [ L^4 - 3a^4 - 12 a^2 (L-a)^2\right ] \tag{3}$
Similarly, from:
$I_{y}(L) = I_{y}^{\text{tr}} (L,a)+I_{y}(a) + 2[I_{y}(a)+m(a)(L-a)^2/4]$
we get:
$I_{y}^{\text{tr}} (L,a) = \frac{\sqrt{3}}{96} \mu \left [ L^4 - 3a^4 - 12 a^2 (L-a)^2\right ] = I_{z}^{\text{tr}} (L,a)/2 \tag{4}$
The clipped shape preserves the threefold symmetry of the original one, so the same conclusion as to the in-plane isotropy of the inertia tensor holds. Also, $$I_{z} = 2 I_{y}$$ in both cases; I'm sure there is some elegant way to explain this, but I can't find it.
A quick check of results (3) and (4) is that $$I_{y,z}^{\text{tr}} (2L,L) = I_{y,z}(L) .$$ In this case, one retrieves the situation shown in the illustration to the previous post.