By 苏剑林 | July 06, 2019
A few days ago, several math/physics groups were forwarding a problem that teacher Li Yongle posted on his Weibo:
The curve problem of a rope fixed on a pole and rotating.
Realizing I hadn't done any math or physics problems in a while, I thought about it and searched for some materials, and now I'd like to share them with you.
In fact, the way Li Yongle presents it makes it feel like it's just a pure physics exercise. However, this problem is actually related to the shapes of curves in several real-life examples. For instance, when jumping rope quickly (and the rope is already taut), what is the shape of the rope?
The shape of a skipping rope and the original problem proposed by Teacher Li Yongle are essentially equivalent. The basic characteristics are:
1. There is a thin rope with uniform linear density;
2. The rope is rotating at high speed, and the angular velocity \(\omega\) can be regarded as a constant;
3. The rope is also subject to the action of gravity;
4. Find the curve shape of the rope when it reaches (relative) static equilibrium.
Of course, in real life, there is also air resistance, which makes the equilibrium state not necessarily a planar curve. The most typical example is the mosquito-repelling fans found on the market:
In this article, we are only concerned with the equilibrium formed within a plane. In this section, we derive the corresponding differential equations. As an amateur physics enthusiast, I am not proficient in professional force analysis; instead, I will base my derivation on the principle of least action in the calculus of variations (refer to the early articles on physics and mathematics in this blog, such as "Natural Extremes Series—8. Extremal Analysis", "Revisiting the 'Rotating Spring Extension' Problem (Variational Solution)", "Corrections and Reflections on the 'Equilibrium State Axiom'", "An Overview of Variations and Theoretical Mechanics", etc.).
The principle of least action essentially requires us to find the variation of the difference between kinetic energy and potential energy. For Li Yongle's original problem, the rotating motion is horizontal, while gravity is vertical. Assuming the linear density is 1, the kinetic energy and potential energy are respectively: \begin{equation}T = \int \frac{1}{2}\omega^2 x^2 \sqrt{dx^2+dy^2},\quad U = \int gy \sqrt{dx^2+dy^2}\end{equation} So, it is essentially the variation of \begin{equation}S = \int \left(\frac{1}{2}\omega^2 x^2 - gy\right)\sqrt{dx^2+dy^2} = \int \left(\frac{1}{2}\omega^2 x^2 - gy\right) ds\label{eq:t}\end{equation}
In the case of jumping rope, both rotation and gravity are in the vertical direction (considering only the shape when the rope is at its lowest point), so the target of variation is \begin{equation}S = \int \left(\frac{1}{2}\omega^2 y^2 - gy\right)ds\end{equation} This looks similar to $\eqref{eq:t}$, but there is an essential difference in difficulty. By simply making a transformation \(z = y - g/\omega^2\), it becomes equivalent to equation $\eqref{eq:t}$ in the case where \(g=0\). As we will see later, analytical solutions can be found for this special case, whereas they cannot for the original equation $\eqref{eq:t}$.
The variational method is used when the two boundary points of the rope are fixed. However, in Li Yongle's problem, only one boundary point seems to be fixed. Can the variational method still be used? Yes, we can imagine that the other boundary point is also fixed (even though we don't know its coordinates); this does not change the original shape and trajectory of motion.
Without further ado, let's formally perform the variation. Denote \begin{equation}\varphi(x, y) = \frac{1}{2}\omega^2 x^2 - gy\end{equation} We get \begin{equation}\begin{aligned}&\delta \int \varphi(x, y) ds \\ =& \int \left(\omega^2 x \delta x - g \delta y\right) ds + \varphi(x, y) \delta ds\\ =& \int \left(\omega^2 x \delta x - g \delta y\right) ds + \varphi(x, y) \frac{dx \delta dx + dy \delta dy}{ds}\\ =& \int \left(\omega^2 x \delta x - g \delta y\right) ds + \varphi(x, y) \left(\frac{dx}{ds}d\delta x + \frac{dy}{ds}d\delta y\right)\\ =& \int \left(\omega^2 x \delta x - g \delta y\right) ds - d\left(\varphi(x, y) \frac{dx}{ds}\right)\delta x - d\left(\varphi(x, y) \frac{dy}{ds}\right)\delta y\end{aligned}\end{equation} Where the final equality uses integration by parts and the convention that variations at boundaries are zero, which are standard operations in the calculus of variations. Then, we rearrange to get: \begin{equation}\begin{aligned}&\delta \int \varphi(x, y) ds \\ =& \int \left[\left(\omega^2 x - \frac{d}{ds}\left(\varphi(x, y) \frac{dx}{ds}\right)\right)\delta x + \left(-g - \frac{d}{ds}\left(\varphi(x, y) \frac{dy}{ds}\right)\right)\delta y\right]ds\end{aligned}\label{eq:var_result}\end{equation} Setting the terms for \(\delta x\) and \(\delta y\) to zero, we obtain the system of differential equations: \begin{equation}\left\{\begin{aligned}&\frac{d}{ds}\left(\varphi(x, y) \frac{dx}{ds}\right)=\omega^2 x\\ &\frac{d}{ds}\left(\varphi(x, y) \frac{dy}{ds}\right) = -g \end{aligned}\right.\end{equation} साथ ही the condition \begin{equation}\left(\frac{dx}{ds}\right)^2 + \left(\frac{dy}{ds}\right)^2 = 1\label{eq:ode1}\end{equation} From the second differential equation, we directly obtain \begin{equation}\varphi(x, y) \frac{dy}{ds} = -gs + C_1\label{eq:ode2}\end{equation} By using $\eqref{eq:ode1}$ and $\eqref{eq:ode2}$, along with the boundary conditions, the shape of the curve can theoretically be solved.
Currently, it seems impossible to solve $\eqref{eq:ode1}$ and $\eqref{eq:ode2}$ analytically. Therefore, we can analyze some special cases, the simplest two being \(\omega=0\) or \(g=0\).
For \(\omega=0\), if it is Li Yongle's problem, the answer is trivial: it is just a vertical straight line pointing downward. However, we can add additional boundary conditions to make the result non-trivial—specifically, it becomes a "catenary". We have solved the catenary before, refer to "Natural Extremes Series—7. The Catenary Problem"; here we re-derive it using $\eqref{eq:ode1}$ and $\eqref{eq:ode2}$. When \(\omega=0\), $\eqref{eq:ode2}$ becomes \begin{equation}-gy\frac{dy}{ds} = -gs + C_1\end{equation} Or equivalently \begin{equation}\frac{1}{2}\frac{d\left(y^2\right)}{ds} = s + C_1\end{equation} (Letting the original \(-C_1/g\) be the new \(C_1\)), then solving for: \begin{equation}y^2 = s^2 + 2C_1 s + C_2\end{equation} Without loss of generality, let \(C_1=0\), then (choosing an appropriate coordinate system ensures \(C_2\) is non-negative, so we can denote \(C_2=a^2\)) \begin{equation}s = \sqrt{y^2 - a^2}\quad \Rightarrow\quad ds = \frac{y dy}{\sqrt{y^2 - a^2}}\end{equation} Therefore \begin{equation}\frac{\sqrt{dx^2 + dy^2}}{dy} = \frac{y}{\sqrt{y^2 - a^2}}\quad \Rightarrow\quad \frac{dx}{dy}=\frac{1}{\sqrt{y^2 - a^2}}\end{equation} Finally, we solve to get \begin{equation}x = \text{arccosh}\left(\frac{y}{a}\right) + b\quad \Rightarrow\quad y = a\cosh \,(x - b)\end{equation} This is the catenary equation.
As discussed earlier, when \(g=0\), this corresponds to the shape of the rope when skipping, also known as the Rope skipping curve or Troposkien curve. In this case, $\eqref{eq:ode2}$ becomes \begin{equation}\frac{1}{2}\omega^2 x^2 \frac{dy}{ds} = C_1\quad \Rightarrow\quad \frac{\sqrt{dx^2 + dy^2}}{dy} = \beta x^2\end{equation} Which is \begin{equation}y=\int \frac{1}{\sqrt{\beta^2 x^4 - 1}}dx\end{equation} This leads into the territory of elliptic integrals.
When neither \(\omega\) nor \(g\) equals 0, the search keywords in English are likely Shape of rotating rope / spinning lasso. There is a discussion on stackexchange about it. Friends who can access Google can also look here.
Although an analytical solution is elusive, it does not prevent a numerical solution. For example:
The reference code (Python) is as follows:
import numpy as np
from scipy.integrate import solve_ivp
import matplotlib.pyplot as plt
omega = 2
g = 3
T = 0.5
h = 0.0001
C1 = 0
def F(t, x):
x, y = x
# G represents dy/ds
G = (C1 - g * t) / (1 / 2. * omega**2 * x**2 - g * y)
return [G, - np.sqrt(1 - G**2)]
x0, y0 = 1, 0.
ts = np.arange(0, T, h)
xs = solve_ivp(F, (0, T), [x0, y0], method='RK23', t_eval=ts)
x, y = xs.y
ts = xs.t
plt.clf()
plt.figure(figsize=(6, 6))
plt.plot(x, y, linewidth=2)
# plt.xlim(0.9, 1.15)
# plt.ylim(-0.25, 0)
plt.savefig('test.png')
As mentioned before, the most general formulation of this problem usually features two fixed endpoints. By adjusting \(C_1\), we might get a curve like this (corresponding to \(C_1=1\), the so-called "non-trivial solution"):
For more related discussions, you can refer to Zhihu: https://www.zhihu.com/question/332446554/answer/746737923
Taking advantage of the popularity of Teacher Li Yongle's problem, I took the opportunity to review some of my old physics knowledge and work on this problem. Of course, I did not solve it completely for the general case, having to return to numerical methods. However, the gains from searching for materials during this process were quite substantial.
Once, I had a dream of becoming a mathematician/physicist/astronomer, but I ended up doing odd jobs in the field of machine learning~~