# 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).

• It certainly is essential that your curve be connected (which could be surmised from your explicit parametrization statement). Dec 23, 2013 at 5:03
• @Ted Shifrin : yes, agreed. thanks for pointing. Dec 23, 2013 at 5:21
• Can you explain the relation between this question and a similar one ? Dec 23, 2013 at 16:54
• @TonyPiccolo : In Chris Culter 's answer of that question, If the curve is always an arc of a circle, he shows that the moment of inertia is an increasing function of $r$, the radius of curvature. But my question now here is that, the straight line has infinite radius of curvature, hence I wonder if the straight line is the global maximum? Dec 23, 2013 at 17:13
• F.A.Valentine has written Curves of given length and minimum or maximum moments of inertia (1934) but I cannot read it. Can you ? Jan 3, 2014 at 8:33

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.

• Impressive! The kernel $\Lambda$ would look much better if you took $L=1$ (obviously WOLOG). Dec 26, 2013 at 6:43
• @achille : Its actually mind blowing how you arrived at this using algebra. I wonder whether there are any geometric interpretations of the expression for $\mathcal(M_{cm}$, or can we arrive at it using geometrical arguments. It would be very interesting to know about such a thing. Dec 26, 2013 at 11:37
• @RajeshD I don't have any geometric interpretation for $\mathcal{M}_{cm}$. This approach is motivated by visualizing the curve as a chain of line segments with small fixed lengths. Since the only degree of freedoms are the directions of the line segments and we know in certain sense, the moment of inertia $\mathcal{M}_{cm}$ is a non-negative quadratic functions of these degree of freedoms. I try to express the dependence explicitly and see what can be done. Dec 26, 2013 at 11:56
• @Robjohn and achillehui : As per my perception Robjohn's answer seems to be elegant, but I somehow like the expression for moment of inertia $MI_cm$ derived in Achille's answer which explicitly shows the only degree of freedom angles of the tangents and says it is maximum when the angle is a constant function. Jan 2, 2014 at 9:07

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.

• Thanks for the answer. There are some things I don't understand. In the begining what you mean by $\delta f$ and other things, I need a bit explanation on your considerations. How you claim that $f$ need to be an arc? etc,. Also at the end I am not able to understand how we get $L^2/12$ as $r$ goes to $\infty$. Dec 27, 2013 at 3:03
• @RajeshD: $\delta f$ is a small instantaneous perturbation of $f$. More precisely, $$\delta\int F(f(x))\,\mathrm{d}x=\int F'(f(x))\delta f(x)\,\mathrm{d}x$$
– robjohn
Dec 27, 2013 at 9:21
• \begin{align}\lim_{r\to\infty}r^2\left(1-\frac{\sin^2\left(\frac{L}{2r}\right)} {\left(\frac{L}{2r}\right)^2}\right) &=\lim_{r\to\infty}\frac{\frac{L^2}{4}} {\left(\frac{L}{2r}\right)^2} \left(1-\frac{\sin^2\left(\frac{L}{2r}\right)} {\left(\frac{L}{2r}\right)^2}\right)\\ &=\lim_{x\to0}\frac{L^2/4}{x^2}\left(1-\frac{\sin^2(x)} {x^2}\right)\\ &=\lim_{x\to0}\frac{L^2/4}{x^2}\left(1-\frac{\sin(x)} {x}\right)\left(1+\frac{\sin(x)} {x}\right)\\ &=\lim_{x\to0}\frac{L^2}{2}\left(\frac{x-\sin(x)} {x^3}\right)\\ &=\lim_{x\to0}\frac{L^2}{2}\left(\frac{\cos(x)} {6}\right)\\&=\frac{L^2}{12}\end{align}
– robjohn
Dec 27, 2013 at 9:52

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

• How did you assume that the moment of inertia of the curve about its center of mass to be $$M = \int_a^b y^2 dx$$, ofcourse you might have chosen origin as center of mass, but it is still wrong as per the definition given in the question. Dec 31, 2013 at 20:56
• I really don't get what you are trying to say, as per my understanding both the answers given by Robjohn and Achille are correct. They explicitly use arc length parameterization and also assume curve to be of fixed length $L$. Dec 31, 2013 at 20:59
• Of course, I am choosing the origin at the center of mass, computations are difficult enough. Jan 1, 2014 at 11:47
• b) You are write, I am on the wrong question. I was looking for the curve minimizing the moment on inertia about an axis ("chosen" as Ox). I'll edit in the post. Jan 1, 2014 at 11:54
• c) No Robjohn and Achille did not took into account that the curve length is constant. They use ∫ds = L, not d(∫ds)=0. In facts, if you follow their reasoning to search the curve maximizing the include area (or the lowest center of gravity), you also find straight lines, not circles (or catenaries). Jan 1, 2014 at 12:05