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Physics LibreTexts

2.13: Momental Ellipse

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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

P=a2kˆr

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Here ˆr 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 P are those of length, and the length of the vectorP is inversely proportional to the radius of gyration. The moment of inertia is Mk2=Ma4P2. That is to say

Ma4P2=Acos2θ2Hsinθcosθ+Bsin2θ,

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 P, so that x=Pcosθ and y=Psinθ. Then

Ma4=Ax22Hxy+By2.

Thus, no matter what the shape of the lamina, however irregular and asymmetric, the tip of the vector P traces out an ellipse, whose axes are inclined at angles 12tan1(2HBA) to the x - axis.

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

Example 2.13.1

Consider a regular n-gon. By symmetry the moment of inertia is the same about any two axes in the plane inclined at 2π/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 2.13.1

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 112ma2(1+3cot2(π/n)).

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


This page titled 2.13: Momental Ellipse is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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