Finding Moment of Inertia of ellipse.

Question

Show that the moment if inertia of an elliptic area of mass M and semi-axis a and b about a semi-diameter of length r is $$\frac{Ma^2b^2}{4r^2}$$. My attempt.

1. I know that MI about ox is ${Mb^2 \over 4}$.
2. MI about oy axis is ${Ma^2 \over 4}$. How to proceed further....

Answer 1

Let $\theta$ be the angle between the axis you wish to compute MI and the $x$-axis.

For any point $(x,y)$ in the plane, its distance to the axis equals to $|x\sin\theta - y\cos\theta|$. The desired MI is given by following integral

$${\rm MI}_\theta \stackrel{def}{=} \rho\int_{\Omega} (x\sin\theta - y\cos\theta)^2 dx dy$$ where $\rho$ is mass density and $\Omega$ is the region for the ellipse.

Expand the integrand and notice under transform $y \mapsto -y$, $\Omega$ remains invariant while the cross term proportional to $xy$ pickup a minus sign. The cross term will not contribute anything to the integral. This leads to

$$MI_\theta = \left(\rho \int_{\Omega} x^2 dxdy\right) \sin^2\theta + \left(\rho\int_{\Omega} y^2 dx dy\right) \cos^2\theta$$ The two coefficients inside the parentheses are nothing but the MI with respect to $y$ and $x$-axes. This means

$$MI_{\theta} = \frac{M}{4}((a\sin\theta)^2 + (b\cos\theta)^2) = \frac{Ma^2b^2}{4}\left(\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}\right) $$ Since the axis has semi-diameter $r$, when you start at origin and move forward for a distance $r$, you will hit the circumference of ellipse. This implies

$$\frac{(r\cos\theta)^2}{a^2} + \frac{(r\sin\theta)^2}{b^2} = 1 \quad\implies\quad \frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2} = \frac{1}{r^2}$$

Substitute this back into above expression of ${\rm MI}_\theta$ and you are done.

Answer 2

Partial answer, based on the fact that the image of any ellipse under affine transformation is another ellipse.

Let the equation of the ellipse be $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ or $$b^2 x^2 + a^2 y^2 = 1$$ and let the desired semi-diameter intersect the ellipse at $(h, k)$ where $h^2 + k^2 = r^2$.

I claim that there is an affine transformation consisting solely of a scaling along the semi-diameter that can transform this ellipse into a circle. Specifically, write $T = PAP^{-1}$, where $$P = \begin{bmatrix}h & -k \\ k & h \end{bmatrix}$$ and $$P^{-1} = \dfrac{1}{r^2} \begin{bmatrix} h & k \\ -k & h \end{bmatrix}$$ transform coordinates between the standard basis and the basis vectors $(h, k)$ and $(k, -h)$, and $$A = \begin{bmatrix} \lambda & 0 \\ 0 & 1 \end{bmatrix}.$$

As $\det T = \lambda$ and $T$ preserves distances from the axis of rotation, the rotational inertia of the resulting shape is $\lambda$ times that of the original shape. The matrix expression for $T$ is $$T = \frac{1}{r^2} \begin{bmatrix} \lambda h^2 + k^2 & hk - \lambda h k \\ hk - \lambda hk & h^2 + \lambda k^2 \end{bmatrix}$$ and the squared radius $R(x, y) := ||T(x, y)||^2$ of the image of some point $(x, y)$ is thus (omitting a bunch of tedious steps in the algebra) \begin{align*} R &= \frac{1}{r^4} \left[ (\lambda h^2 + k^2) x + (hk - \lambda hk) y \right]^2 + \frac{1}{r^4} \left[(hk - \lambda hk) x + (hk - \lambda hk) y \right]^2 \\ &= \frac{1}{r^2} \left( \lambda^2 (hx - ky)^2 + (hy + kx))^2 \right) \end{align*} and finding the value of $\lambda$ for which this expression is constant (and the radius of the corresponding circle) is just a straightforward tedious exercise in algebra or differential calculus from which the final answer would follow readily.

Answer 3

This is a comment on https://math.stackexchange.com/a/3371113/349530 answer by @achille-hui because I don't have enough reputations to comment.

How are the angle of diameter with x-axis and the parameter for the extremity of the diameter same value $\theta$?

In general, a point $P$ on ellipse corresponding to parameter $\theta$, will have co-ordinates $(a \cos\theta,b\sin\theta)$ and thus $OP$ will have an angle $\psi$ with x-axis given by $\tan\psi=\frac{b}{a}\tan\theta$ and hence will be different than $\theta$.