First of all, happy Pi Day!!! Time to get to the physics. I found this Princeton problem looking at old quals for fun.

It turned out to be really interesting, so here we are.

Princeton Physics Preliminary Exams J98M.1 - Hanging Rope

Problem found here.

Problem Statement

A piece of thin uniform unstretchable rope has length $2L$ and mass $M$. Its ends are attached to points at the same height separated by distance $2 w$ and the rope hangs between them under the influence of gravity (of course $w < L$). Let us set up coordinates $(x,y)$ in the plane of the rope, so that the end points have equal values of $y$ and $x = \pm w$. You will be asked to determine the vertical coordinate of the rope, $y$, as a function of $x$.

  1. Write down the functional of $y(x)$ that has to be minimized. What is the form of the constraint?
  2. One may think of the functional to be minimized as the action as an for a 1-dimensional particle with coordinate $y$ and time $x$. Find a conserved quantity.
  3. For a given value of the conserved quantity, find $y(x)$. What is the equation relating the conserved quantity to $w$ and $L$?

Solution

Part 1

The potential energy must be minimized. Letting the reference height be $y=0$, for a small piece of the rope with horizontal projection $dx$ and length

\[ds = dx \, \sqrt{1 + \left (\frac{dy(x)}{dx} \right)^2}\]

Defining the linear mass density of the rope to be $\rho = \frac{M}{2L}$, the potential energy of this piece is

\[dU = g \, y(x) \, dm = \rho g \, y(x) \sqrt{1 + \left (\frac{dy(x)}{dx} \right)^2} \, dx\]

The total potential energy is then

\[U = \rho g \int \limits_{-w}^{w} y(x) \sqrt{1 + \left (\frac{dy}{dx} \right)^2} \, dx\]

Finally, we have a constraint on the length. By definition, $y(x)$ must satisfy

\[2L = \int \limits_{-w}^{w} \sqrt{1 + \left (\frac{dy}{dx} \right)^2} \, dx\]

Part 2

In order to minimize the potential energy, we use the Euler-Lagrange equations. The Lagrangian is then

\[\mathcal{L} = y(x) \sqrt{1 + y'(x)^2}\]

Note that the Lagrangian is independent of $x$. Its a well-known fact that when this occurs, this quantity is constant:

\[\mathcal{L} - y'(x) \frac{\partial \mathcal{L}}{\partial y'(x)} = C\]

Part 3

Let the constant be $a$. Substituting the Lagrangian we’ve already found into this equation, we see that it saves us some time compared to the Euler-Lagrange equation. Evaluating and simplifying, we see that

\[\left ( \frac{y'(x)}{a} \right ) - (y(x))^2 = 1\]

We recognize the similiarity with the hyperbolic trigonometric functions $\sinh{(x)}$ and $\cosh{(x)}$ due to them being derivatives of each other and the identity $\sinh^2 {x} - \cosh^2 {x} = 1$. We guess a solution in the form of $y(x) = a \cosh{bx}$. This works as long as $a = \frac{1}{b}$.

We first consider our symmetry constraint. Since a hyperbolic cosine is an even function $y(w) = y(-w)$.

Finally, we must consider our length constraint. Evaluating the length integral

\[2L = 2a \sinh{\left (\frac{w}{a} \right)}\]

This is a transcendental equation that $a$, $L$, and $w$ must satisfy.

Aftermath

This was a fun problem!!!

The curve we just solved for is called a catenary. Also, using hyperbolic trig functions are a common thing in physics, especially in special relativity.