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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?