Moment of inertia about center of mass of a curve that is the arc of a circle.

Question

Let $(x(s),y(s))$ be a smooth 2-d plane curve which is an arc of a circle of a certain radius $r$. Assume it is represented by an inelastic string $S$ of finite length, lying in a 2-d plane. Let there be an infinitesimally small displacement in the inelastic string so that its new position is given by another smooth curve $(x'(s),y'(s))$. Let $$\delta = \max(\sup_s(|x(s)-x'(s)|), \sup_s(|y(s)-y'(s)|))$$

Consider the set $K_{\epsilon}$ of all curves $(x'(s),y'(s))$ such that their displacement $\delta \le \epsilon$. It needs to be shown that we can consider small enough $\epsilon$ such that the curve $(x(s),y(s))$ has the highest moment of inertia about center of mass, among all other curves that belong to $K_{\epsilon}$.

I'd like to know how to prove this statement if its true.

PS

The moment of inertia of a curve $(x(s),y(s))$ of finite length is given as $\int_0^s ((x(s)-x_c)^2 + (y(s)-y_c)^2) ds$. where $(x_c,y_c)$ is the center of mass of the curve.

Answer 1

Consider dilating the curve by a constant $c$. The moment of inertia of the dilated curve is proportional to $c^2$. The distance between the original curve and the dilated curve is proportional to $c-1$. What does this imply for the statement?

Edit: Okay, if we're not allowed to dilate the string, we can still change its curvature. Define a family of curves $\gamma_r$ for $r>0$ by $$\gamma_r(s) = \left(r\cos(s/r), r\sin(s/r)\right).$$ Then you can check that $\gamma_r$ is as continuous as you need it to be, and the moment of inertia for any given length is a strictly increasing function of $r$; it has no local extrema.

Answer 2

Taking the variation of the moment of inertia gives $$ \begin{align} \delta\int_0^L\left((x-x_0)^2+(y-y_0)^2\right)\,\mathrm{d}s &=\int_0^L\left(2(x-x_0)\,\delta x+2(y-y_0)\,\delta y\right)\,\mathrm{d}s\\ &=0 \end{align} $$ Since $\left(\frac{\mathrm{d}x}{\mathrm{d}s}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}s}\right)^2=1$, we can also take the variation of this and integrate by parts to get $$ \begin{align} 0 &=\delta L\\[9pt] &=\delta\int_0^L1\,\mathrm{d}s\\ &=\delta\int_0^L\left(\left(\frac{\mathrm{d}x}{\mathrm{d}s}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}s}\right)^2\right)\,\mathrm{d}s\\ &=2\int_0^L\left(\frac{\mathrm{d}x}{\mathrm{d}s}\frac{\mathrm{d}\delta x}{\mathrm{d}s}+\frac{\mathrm{d}y}{\mathrm{d}s}\frac{\mathrm{d}\delta y}{\mathrm{d}s}\right)\,\mathrm{d}s\\ &=-2\int_0^L\left(\frac{\mathrm{d}^2x}{\mathrm{d}s^2}\delta x+\frac{\mathrm{d}^2y}{\mathrm{d}s^2}\delta y\right)\,\mathrm{d}s \end{align} $$ So now you have two equations constraining $\delta x$ and $\delta y$: $$ \begin{align} 0&=\int_0^L\left(\frac{\mathrm{d}^2x}{\mathrm{d}s^2}\delta x+\frac{\mathrm{d}^2y}{\mathrm{d}s^2}\delta y\right)\,\mathrm{d}s\tag{1}\\ 0&=\int_0^L\left((x-x_0)\,\delta x+(y-y_0)\,\delta y\right)\,\mathrm{d}s\tag{2} \end{align} $$ $(1)$ says that the variation does not change the length of the curve. $(2)$ says the moment of inertia is critical. Therefore, we want any variation that satisfies $(1)$ to satisfy $(2)$. That is, any variation that is orthogonal to $\left(\frac{\mathrm{d}^2x}{\mathrm{d}s^2},\frac{\mathrm{d}^2y}{\mathrm{d}s^2}\right)$ needs to be orthogonal to $(x-x_0,y-y_0)$. Thus, there must be a constant $\lambda$ so that $$ (x-x_0,y-y_0)=\lambda\left(\frac{\mathrm{d}^2x}{\mathrm{d}s^2},\frac{\mathrm{d}^2y}{\mathrm{d}s^2}\right)\tag{3} $$ If $\lambda=0$, the curve degenerates to the point $(x_0,y_0)$, so let's assume that $\lambda\ne0$.

Since $\left(\frac{\mathrm{d}x}{\mathrm{d}s}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}s}\right)^2=1$, we have, using $(3)$, $$ \begin{align} 0&=\lambda\left(\frac{\mathrm{d}x}{\mathrm{d}s}\frac{\mathrm{d}^2x}{\mathrm{d}s^2}+\frac{\mathrm{d}y}{\mathrm{d}s}\frac{\mathrm{d}^2y}{\mathrm{d}s^2}\right)\\ &=(x-x_0)\frac{\mathrm{d}x}{\mathrm{d}s}+(y-y_0)\frac{\mathrm{d}y}{\mathrm{d}s}\\[6pt] r^2&=(x-x_0)^2+(y-y_0)^2\tag{4} \end{align} $$ Thus, assuming the length is constant and the moment of inertia is critical, we get that the curve is an arc of a circle centered at $(x_0,y_0)$.