2.13: Momental Ellipse

Consider a plane lamina such that its radius of gyration about some axis through the centre of mass is $k$. Let P be a vector in the direction of that axis, originating at the centre of mass, given by

$${\bf P} = \frac{a^2}{k} {\bf\hat{r}} \label{eq:2.13.1}$$

Here $\bf \hat{r}$ is a unit vector in the direction of interest; $k$ is the radius of gyration, and $a$ is an arbitrary length introduced so that the dimensions of $\bf P$ are those of length, and the length of the vector$\bf P$ is inversely proportional to the radius of gyration. The moment of inertia is $Mk^2 = \frac{Ma^4}{ P^2}$. That is to say

$$\frac{Ma^4}{P^2} = A \cos ^2 \theta - 2 H \sin \theta \cos \theta + B \sin^2 \theta, \tag{2.13.2}\label{eq:2.13.2}$$

where $A, H$ and $B$ are the moments with respect to the $x$- and $y$-axes. Let $(x , y)$ be the coordinates of the tip of the vector $\bf P$, so that $x = P\cos \theta$ and $y = P\sin \theta$. Then

$$Ma^4 = Ax^2 -2Hxy + By^2 .\label{eq:2.13.3}$$

Thus, no matter what the shape of the lamina, however irregular and asymmetric, the tip of the vector $\bf P$ traces out an ellipse, whose axes are inclined at angles $\frac{1}{2} \tan^{-1} (\frac{2H}{B-A} )$ to the $x$ - axis.

This is the momental ellipse, and the axes of the momental ellipse are the principal axes of the lamina.

Example $\PageIndex{1}$

Consider a regular $n$-gon. By symmetry the moment of inertia is the same about any two axes in the plane inclined at $2 \pi / n$ to each other. This is possible only if the momental ellipse is a circle. It follows that the moment of inertia of a uniform polygonal plane lamina is the same about any axis in its plane and passing through its centroid.

Exercise $\PageIndex{1}$

Show that the moment of inertia of a uniform plane $n$ - gon of side $2a$ about any axis in its plane and passing through its centroid is $\frac{1}{12} ma^2 (1+3\cot^2 ( \pi /n))$.

What is this for a square? For an equilateral triangle?