When is the moment of inertia of a smooth plane curve is maximum?

Given a smooth plane curve $(x(s),y(s))$, parameterized in arc length $s$, of fixed finite length $L$, its moment of inertia about its center of mass (axis perpendicular to the plane) is given as $$MI = \int_0^L ((x(s)-x_{cm})^2 + (y(s)-y_{cm})^2) ds$$. What I predict from earlier discussions, and almost convinced is that if we fix length $L$, $MI$ is maximum when the curve is a straight line. I lack the faculty of mathematical machinery (I guess calculus of variations) to prove it, hence is my gentle request to help me out in proving it and thoroughly understanding the situation and all the corollaries and nuances. This not just the result I need, but I want to do more with it and hence would like understand all the things that are making this result and even more general ones (only to plane curves though).

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There is no need for any calculus of variation. Ordinary calculus is enough.

For simplicity of derivation, we will use complex numbers to represent points on the plane.
Let $z(s) = x(s) + i y(s)$ and WOLOG, we will assume $z(0) = 0$. We can express the
position on our curve as an integral:

$$z(t) = \int_0^t z'(s)\;ds$$

Let $\theta(t) = \begin{cases} 1 & t > 0\\ 0 & t \le 0\end{cases}$ be the step function.
The center of mass is given by

\begin{align}z_{cm} &= \frac{1}{L} \int_0^L z(t) dt = \frac{1}{L} \int_0^L \int_0^t z'(s)\;ds\; dt\\ &= \frac{1}{L} \iint_{[0,L]^2} \theta(t-s) z'(s)\;ds\; dt = \int_0^L \left(1-\frac{s}{L}\right) z'(s)\;ds \end{align}

The moment of inertia w.r.t. $z(0)$, the origin, is given by

\begin{align} \mathcal{M}_0 &= \int_0^L |z(s)|^2 ds = \int_0^L \left(\int_0^t z'(s_1)ds_1\right)\left(\int_0^t \bar{z}'(s_2) ds_2\right) dt\\ &=\iiint_{[0,L]^3} \theta(t-s_1)\theta(t-s_2) z'(s_1) \bar{z}'(s_2) ds_1 ds_2 dt\\ &=\iiint_{[0,L]^3} \theta(t-\max(s_1,s_2)) z'(s_1) \bar{z}'(s_2) ds_1 ds_2 dt\\ &= \iint_{[0,L]^2} \left(L – \max(s_1,s_2)\right) z'(s_1) \bar{z}'(s_2) ds_1 ds_2 \end{align}
and hence the moment of inertia w.r.t. the center of mass is

$$\mathcal{M}_{cm} = \mathcal{M}_0 – L |z_{cm}|^2 = L \iint_{[0,L]^2} \Lambda(s_1,s_2) z'(s_1) \bar{z}'(s_2) ds_1 ds_2$$

where
\begin{align}\Lambda(s_1,s_2) &= 1 – \max\left(\frac{s_1}{L},\frac{s_2}{L}\right) – \left(1-\frac{s_1}{L}\right)\left(1-\frac{s_2}{L}\right)\\ &= \left( 1 – \max\left(\frac{s_1}{L},\frac{s_2}{L}\right)\right)\min\left(\frac{s_1}{L},\frac{s_2}{L}\right) \end{align}

Since $|z'(s)| \equiv 1$ and $\Lambda(s_1,s_2) > 0$ for $(s_1,s_2) \in (0,L)^2$, we can bound $\mathcal{M}_{cm}$ as

$$\mathcal{M}_{cm} \le L \iint_{[0,L]^2} \Lambda(s_1,s_2) |z'(s_1)||z'(s_2)| ds_1 ds_2 = L \iint_{[0,L]^2} \Lambda(s_1,s_2) ds_1 ds_2$$
Notice the equality in above inequality is achieved when and only when $z'(s)$ is a constant.
We can conclude $\mathcal{M}_{cm}$ is largest for straight lines.

Assumptions:

1. the center of mass is $0$
$$0=\int_0^L\delta f\,\mathrm{d}s\tag{1}$$
2. $f$ is parametrized by arc length: $f’\cdot f’=1$
\begin{align} 0&=\int_0^Lf’\cdot\delta f’\,\mathrm{d}s\\ &=\int_0^Lf’\cdot\,\mathrm{d}\delta f\\ &=\Big[\,f’\cdot\delta f\,\Big]_0^L-\int_0^Lf”\cdot\delta f\,\mathrm{d}s\\ &=-\int_0^Lf”\cdot\delta f\,\mathrm{d}s\tag{2} \end{align}
Note that $\delta f$ can be adjusted in the direction of $f’$ near $0$ and $L$ without affecting the integral of $f”\cdot\delta f$ since $f’\cdot f”=0$. Thus, we can assume $\Big[\,f’\cdot\delta f\,\Big]_0^L=0$.

3. the moment of inertia is stationary
$$0=\int_0^Lf\cdot\delta f\,\mathrm{d}s\tag{3}$$
Conclusions:

For an extreme $f$, any $\delta f$ that satisfies $(1)$ and $(2)$ also satisfies $(3)$; thus, linearity says that there are constants $a$ and $b$ so that
$$f=a+bf”\tag{4}$$
Since $f$ is parameterized by arclength $f’\cdot f”=0$, integrating the dot product of $(4)$ with $f’$ yields
$$\frac12f\cdot f=a\cdot f+c\tag{5}$$
Equation $(5)$ represents an arc of the circle
$$|f-a|=\left(2c+|a|^2\right)^{1/2}\tag{6}$$
or in the extreme case, a line segment.

Checking Possible Arcs:

The equation of an arc of radius $r$ is
$$f=r(\cos(s/r),\sin(s/r))\tag{7}$$
The center of mass is
$$\frac1L\int_{-L/2}^{L/2}r(\cos(s/r),\sin(s/r))\,\mathrm{d}s=\left(\frac{2r^2}{L}\sin\left(\frac{L}{2r}\right),0\right)\tag{8}$$
The moment of inertia is
\begin{align} &\frac{r^2}{L}\int_{-L/2}^{L/2}\left[\left(\cos(s/r)-\frac{2r}{L}\sin\left(\frac{L}{2r}\right)\right)^2+\sin^2(s/r)\right]\,\mathrm{d}s\\ &=\frac{r^2}{L}\int_{-L/2}^{L/2}\left[1-\frac{4r}{L}\sin\left(\frac{L}{2r}\right)\cos(s/r)+\frac{4r^2}{L^2}\sin^2\left(\frac{L}{2r}\right)\right]\,\mathrm{d}s\\ &=r^2-\frac{4r^4}{L^2}\sin^2\left(\frac{L}{2r}\right)\\ &=r^2\left(1-\frac{\sin^2\left(\frac{L}{2r}\right)}{\left(\frac{L}{2r}\right)^2}\right)\tag{9} \end{align}
$(9)$ increases to $\frac{L^2}{12}$ as $r\to\infty$. Thus, the maximal moment of inertia would be at the extreme case of a line segment.

There is no maximum ( monotonous increase) , but only a minimum.

Using polar coordinates, when object and constraint functions are together, variational problem is

$\int r^2 ds – \lambda^2 \int ds,$ where $ds= \sqrt{(r^2 + r^{‘2})} d \theta$

Lagrangian $(r^2 – \lambda^2)$ is independent of $r^{‘}$ when considered with respect to arc $s$. So it is not a variational problem.

Minimum M of I when $r$ is independent of $\theta$, or when $r$ is a constant. If arc length L is given, minimizing solution is for constant radius loop $L/ (2 \pi).$( Cowboy lasso)

I am not forgetting about he previous post, but the attached image proves that circles are also extrema of the momentum of inertia.

Although a few minutes testing will be sufficient to convince that probably circles are unstable extrema (minimum), while straight lines are stable extrema (maximum).

This is a classical problem of the calculus of variations, and its is surprising it is not in the collections of classical applications.

Intuitively, it is the shape taken by a lasso, a rope in rotation around a fixed point (which will also be the center of mass, as will be able to check on the result). In facts, playing with a lasso shows that we can expect circles as unstable solutions and straight lines as stable solutions.

The computation is hard to understand and involves sophisticated use of the calculus of variation. Lets, after choosing appropriate axes, maximize
$$M = \int_a^b (x^2 + y^2) ds, \text{ subject to } L= \int_a^b ds \text{, in which }ds=\sqrt{1+y’^2}.$$

The first thing to do is to define the derivative of $M$ and $L$ when looked as function of the curve $y=y(x)$. The so called functional derivative of $$F = \int_a^b f(x,y,y’) dx,$$ is shown to be $$\delta F = \frac {\partial f} {\partial y} – \frac d {dx} \frac{\partial f} {\partial y’}.$$

A complete demonstration can be found here. Basically, you replace $y$ by $y+\epsilon \: \eta$ where $\epsilon \rightarrow 0$ is a number and $\eta$ a fixed function. Then you compute the derivative is the usual way: $(f(x,y+\epsilon \: \eta, (y+\epsilon \: \eta)’) – f(x,y, y’)) / \epsilon$. Integrating by part replaces the $(y+\epsilon \: \eta)’$ term by $-\frac d {dx} \frac{\partial f} {\partial y’}$ and a term outside the integral sign, which vanishes because the end points $a$ and $b$ are fixed. Because the formulas are linear, the $\epsilon$ cancels, giving a result $\int {\delta F} \eta dx$, valid for any $\eta$ thus the result.

The functional derivatives are second order differentials that can be computed formally. For $L$ we get:
$$\delta L = \frac \partial {\partial y} \sqrt{1+y’^2} – \frac d {dx} \left[ \frac \partial {\partial y’} \sqrt{1+y’^2}\right]= 0 – \frac d {dx} \left[ \frac{y’}{\sqrt{1+y’^2}} \right],$$
the first term is null because $ds$ depend only on $y’$ and not on $y$, the second is a derivative when $ds$ is looked as a function of on $y’$. We can the pursue with the usual derivative as a function of $x$:
$$\delta L=- \frac{y” \sqrt{1+y’^2} – y’ (\sqrt{1+y’^2})’}{1+y’^2} = \cdots = \frac{y”}{(1+y’^2)^{3/2}}.$$

For $M$ we have:

$$\delta M = \frac \partial {\partial y} \left[ (x^2+y^2) \sqrt{1+y’^2}\right]- \frac d {dx} \frac \partial {\partial {y’}} \left[ (x^2+y^2) \sqrt{1+y’^2}\right],$$
thus,
$$\delta M = 2y \sqrt{1+y’^2} – \frac d {dx} \left[ (x^2+y^2) \frac \partial {\partial {y’}} \sqrt{1+y’^2}\right] \\= 2y \sqrt{1+y’^2} – \frac d {dx} \left[ (x^2+y^2) \frac {y’} {\sqrt{1+y’^2}}\right] \\= \frac{2y (1+y’^2)}{\sqrt{1+y’^2}} – (2x+2yy’) \frac {y’} {\sqrt{1+y’^2}} – (x^2+y^2) \frac {y”} {(1+y’^2)^{3/2}} \\= \frac{2y-2x y’}{\sqrt{1+y’^2}} – \frac {(x^2+y^2)y”}{(1+y’^2)^{3/2}}$$

To solve the question of the function of a given length with the highest moment of inertia, we have to introduce a Lagrange multiplier. It express at the extremum of a function $M$ subject to a condition $L = C^{te}$, the tangents planes of $M$ and $L$ are parallels. Here, the condition means that is exist a constant $\mu$, called the Lagrange-multiplier, such that $\delta M = \mu \delta L$.

This equation, known as the Euler-Lagrange equation, says there exist $\mu$ such that
$$\frac{2y-2x y’}{\sqrt{1+y’^2}} – \frac {(x^2+y^2)y”}{(1+y’^2)^{3/2}} = \mu \frac{y”}{(1+y’^2)^{3/2}},$$
which is the same as
$$(x^2+y^2+\mu)y” = 2(y-x y’)(1+y’^2)$$

Which needs some checks to be continued