I'm trying to solve a physics problem that I heard ~10 years ago in undergrad that was casually posed to me without a solution in mind; it has been bothering me ever since! Please let me know if this is the wrong place to ask this question. I figured it was either physics or math and decided to start here!


A rope of length $L$ connects two sides of a gap of width $w$. Assuming $L > w$, when an object is hung from one end of the gap, it will slide across the gap to the other side under the influence of gravity. Assume the mass of the object is negligible so that it does not deform the shape of the rope, and also ignore any friction/air resistance. Which value of $L$ will result in the object sliding across the gap the fastest?


Intuitively, some value of $L$ must be optimal; if $L=w$, then the object does not slide at all, and if $L>>w$, then the object will spend too much time in almost-free-fall before reaching the bottom of the rope.

My attempt:

The shape of the rope is a catenary $y(x)=a\cosh(\frac{x}{a})$ for some value of $a$. To make things easier, I've assumed $w=2$ with the ends of the gap at $(-1,0)$ and $(1,0)$, so that we can model the shape of the rope by the equation $y(x)=a\cosh(\frac{x}{a})-a\cosh(\frac{-1}{a})$.

y(x) for different values of a

This lets us directly convert the $y$-value into change in potential energy, so the kinetic energy at some position $x$ is $-mgy(x)$. Solving for velocity, we get that $v(x) = \sqrt{-2gy(x)}$. Since I only care about the $x$-component of velocity, I multiplied by $\cos(\arctan(y'(x)))=\frac{1}{\sqrt{1+y'(x)^2}}$ to get $v_x(x) = \sqrt{\frac{-2gy(x)}{1+y'(x)^2}}$. The time to cross the gap should then be $t = \int_{-1}^1 \frac{1}{v_x(x)} \,dx$. The question asks us to minimize this value $t$ with respect to the rope length $L$, which is essentially determined by the shape constant $a$.

I don't know a lot about these integrals, but using Mathematica, I've found that

  • $$y'(x) = \sinh\left(\frac{x}{a}\right)$$
  • $$v_x(x) = 2\sqrt{\frac{a\cdot g\cdot (\cosh(1/a)-\cosh(x/a))}{1+\cosh(2x/a)}}$$
  • $$t = i\sqrt{\frac{a}{g}}\text{csch}\left(\frac{1}{2a}\right)\left[\left(\cosh\left(\frac{1}{a}\right)-1\right)E\left(\frac{i}{2a} \Bigm\vert -\text{csch}\left(\frac{1}{2a}\right)^2\right)-\cosh\left(\frac{1}{a}\right)F\left(\frac{i}{2a} \Bigm\vert -\text{csch}\left(\frac{1}{2a}\right)^2\right)\right]$$

Here, $E$ and $F$ are the elliptic integrals of the second and first kind, respectively (Mathematica calls them EllipticE and EllipticF). Mathematica has a hard time minimizing $t$ with respect to $a$, but I can compute that the derivative $\frac{dt}{da}$ is $$ \frac{dt}{da} = -\frac{1}{2\sqrt{a^3g}} \cdot i\left[ \text{csch}\left(\frac{1}{2a}\right)F\left( \frac{i}{2a} \Bigm\vert -\text{csch}\left(\frac{1}{2a}\right)^2 \right)\left(a\cdot \cosh\left(\frac{1}{a}\right)-\sinh\left(\frac{1}{a}\right)\right)+\text{sech}\left(\frac{1}{2a}\right)E\left(\frac{i}{2a} \Bigm\vert -\text{csch}\left(\frac{1}{2a}\right)^2\right)\left(\cosh\left(\frac{1}{a}\right)-a\cdot \sinh\left(\frac{1}{a}\right)\right)\right]$$

Plots of (t vs. a) and (dt/da vs. a)

$g$ thankfully ends up not affecting the optimal value of $a$, so the above plot just sets $g=1$. Mathematica's FindRoot still seems to struggle with figuring out where exactly $\frac{dt}{da}=0$, but it seems to be somewhere around $0.76$ish. Specifically, I usually get the same first 9 digits of $0.761797489\dots$. This corresponds to an arc length of around 2.62, or roughly 31% longer than the gap length $w = 2$. That's already kind of the answer; I think intuitively this should scale to any gap length, although weirdly enough I couldn't even get most of the above calculus to work if I set the endpoints of the gap anywhere other than $\pm 1$. I think it would be cool if this were related to some already-known mathematical constants, but the only way I know of other than explicitly solving these equations is making WolframAlpha guess at a closed form, and it doesn't have any particularly tempting guesses.

My questions:

  • Does anybody see any errors in my work?
  • Is there a different way to approach this problem that is more likely to result in an explicit answer?
  • Is there a better way to solve $\frac{dt}{da} = 0$ other than Mathematica's FindRoot?
  • $\begingroup$ You are assuming the two sides of the gap are the same height. That condition doesn't appear in the statement of the problem. $\endgroup$ Commented Jun 16 at 0:40

2 Answers 2


Trying your problem, what I would suggest first is to write $$\frac{dt}{da} = -\frac{1}{2\sqrt{a^3g}}\,f(a)$$ As your plot shows, there are values for which $f(a)$ has a complex value and this does make Mathematica very happy when trying to find the zero of $f(a)$.

Better is to work with $\Re(f(a))$ and to avoid places where $a$ is small.

Concerning the starting point, let $a=\frac 1x$ and expand as a series $$f\left(\frac{1}{x}\right)=\left(\frac{2 i}{x^2}-\frac{13 i}{12}+O\left(x^2\right)\right) F\left(\frac{i x}{2}|-\text{csch}^2\left(\frac{x}{2 }\right)\right)+O\left(x^2\right) E\left(\frac{i x}{2}|-\text{csch}^2\left(\frac{x}{2 }\right)\right)$$

This gives as an estimate $$x=2 \sqrt{\frac{6}{13}}\quad \implies \quad a=\sqrt{\frac{13}{24} }=0.73598$$


gives the result

Newton iterates are $$\left( \begin{array}{cc} n & x_{(n)} \\ 0 & 1.35873244097 \\ 1 & 1.31206518221 \\ 2 & 1.31268470015 \\ 3 & 1.31268481976 \\ \end{array} \right)$$


Too long for a comment.

A way to solve your problem. The considered system is holonomic and can be modeled considering the lagrangian

$$ L = \frac 12 m \|\dot p\|^2-m g y + \lambda\left(y-a\cosh\left(\frac xa\right)\right) $$

with $p = (x,y)$. This lagrangian describes the object movement along the rope, under the gravitation field. The obtained movement equations are:

$$ \cases{ \dot x = v_x\\ \dot y = v_y\\ \dot v_x = -\frac{\tanh \left(\frac{x}{a}\right) \left(a g \text{sech}\left(\frac{x}{a}\right)+v_x^2\right)}{a}\\ \dot v_y = \frac{v_x^2 \text{sech}\left(\frac{x}{a}\right)}{a}-g \tanh ^2\left(\frac{x}{a}\right) } $$

Unfortunately, the ode does not seem to have a solution in closed form, but supposing we have a closed form for $v_a=(v_x, v_y)$ then

$$ t_a = \int \frac{ds}{\|v_a\|} $$

choosing then the optimum $a$. Those final steps can be accomplished with the help of MATHEMATICA.


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