1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Please note that by inelastic string, I means all the displacements are such that the length of the string/curve is always same. So there cannot be any dilations. $\endgroup$
    – Rajesh D
    Aug 18, 2013 at 6:31
  • $\begingroup$ Do you mean $$ \delta = \max(\sup_s(|x(s)-x'(s)|), \sup_s(|y(s)-y'(s)|)) $$ ?? $\endgroup$
    – robjohn
    Aug 18, 2013 at 16:17
  • $\begingroup$ yes. It was a mistake. Thanks for pointing out. $\endgroup$
    – Rajesh D
    Aug 18, 2013 at 16:19
  • $\begingroup$ Is the question whether a circular arc is a local maximum of the moment of inertia? $\endgroup$
    – robjohn
    Aug 18, 2013 at 16:20
  • $\begingroup$ @robjohn : yes, it is precisely. $\endgroup$
    – Rajesh D
    Aug 18, 2013 at 16:39

2 Answers 2

2
$\begingroup$

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 $(x_0,y_0)$.

$\endgroup$
4
  • $\begingroup$ Thank you for the nice answer. The moment of inertia is about the center of mass of the curve rather than any given point. So in this case, If we invoke Parallel axis theorem does the result still hold. I guess it is still valid. $\endgroup$
    – Rajesh D
    Aug 19, 2013 at 2:17
  • $\begingroup$ If the center of the circle is not the center of mass, then the curve can be perturbed to increase and decrease the moment of intertia. $\endgroup$
    – robjohn
    Aug 19, 2013 at 3:09
  • $\begingroup$ Is it necessary that even after perturbation, it should always remain as an arc of some circle or it can take any shape? I mean, if we look at answer by Chris, I want to know that his answer is one off counter example? Is there any more constraints on perturbations that when met will make it a stationary point. $\endgroup$
    – Rajesh D
    Aug 19, 2013 at 4:07
  • $\begingroup$ What this says is that for a given length of curve, the minimum moment occurs when that curve is a circle around the point $(x_0,y_0)$; then the center of mass is $(x_0,y_0)$. $\endgroup$
    – robjohn
    Aug 19, 2013 at 9:19
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ The displacement is such that the length of the string/curve does not change. That's why I had mentioned inelastic string. $\endgroup$
    – Rajesh D
    Aug 18, 2013 at 6:30
  • $\begingroup$ How does $\delta$ (distance) vary with $r$ in this case? $\endgroup$
    – Rajesh D
    Aug 18, 2013 at 9:26
  • $\begingroup$ Agreed that it is monotonously increasing. But is there any mathematical way where in I can avoid this type of displacement (always being the arc of some circle). $\endgroup$
    – Rajesh D
    Aug 18, 2013 at 13:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .