Differential Equation: "I love Physics" - Pendulum meets Capstan equation | Approximation

Question

I came across an amusing YouTube short video (https://www.youtube.com/shorts/Stu0-EiK2Zs) and thought: Let's try to describe this problem physically!

My main system is the pendulum with the weight. Here is $l_w$ the length of the pendulum, $R$ the radius of the pipe, $\theta$ the angle, $g$ the gravitational acceleration constant and $m_w$ the mass of the weight.

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Pendulum:

Coordinates:

$$x = l_w \sin \theta - R \cos \theta, \hspace{20pt} y = l_w \cos \theta + R \sin\theta$$

Velocities:

$$\dot{x} = \dot{l_w} \sin \theta + l_w\dot{\theta}\cos\theta + R \dot{\theta}\sin \theta, \hspace{20pt} \dot{y} = \dot{l_w} \cos \theta - l_w\dot{\theta}\sin\theta + R\dot{\theta} \cos\theta$$

Kinetic Energy + Potential:

$$T = \frac{m_w}{2}(\dot{x}^2+\dot{y}^2) = \frac{m_w}{2}(\dot{l_w}^2 + l_w^2\dot{\theta}^2 + R^2\dot{\theta}^2 + 2R\dot{l_w}\dot{\theta})$$

$$V = m_wg(h-y) = m_wgh - m_wg (l_w \cos \theta + R \sin\theta)$$

Lagrange Function:

$$\mathcal{L} = T - V = \frac{m_w}{2}(\dot{l_w}^2 + l_w^2\dot{\theta}^2+ R^2\dot{\theta}^2 + 2R\dot{l_w}\dot{\theta}) - m_wgh + m_wg (l_w \cos \theta + R \sin\theta)$$

Equation formula:

$$\frac{d}{dt}\left(\frac{d\mathcal{L}}{d\dot{l_w}}\right) - \frac{d\mathcal{L}}{dl_w} = Q_{l_w}, \hspace{30pt} \frac{d}{dt}\left(\frac{d\mathcal{L}}{d\dot{\theta}}\right) - \frac{d\mathcal{L}}{d\theta} = Q_\theta$$

Differential equations:

$$\ddot{l_w} + R\ddot{\theta} - l_w \,\dot{\theta}^2 - g\cos\theta = \frac{Q_{l_w}}{m_w}$$

$$(R^2 + l_w^2) \ddot{\theta} + R\ddot{l_w} + 2\,l_w\dot{l_w} \dot{\theta} + l_wg\sin\theta - Rg\cos\theta = \frac{Q_\theta}{m_w}$$

Approximation: $R \ll l_w$

$$\ddot{l_w} - l_w \,\dot{\theta}^2 - g\cos\theta = \frac{Q_{l_w}}{m_w}$$

$$l_w^2 \ddot{\theta} + 2\,l_w\dot{l_w} \dot{\theta} + l_wg\sin\theta = \frac{Q_\theta}{m_w}$$

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The problem are the generalized forces $Q_{l_w}$ and $Q_{\theta}$.

The length of the pendulum will decrease with time because of an acceleration through the falling human and because of the rope wrapping around the pipe. I therefore came up with the following relationship:

$$l_w(t) = L - \lambda(t) - l_h(t)$$

with total length $L$, length of human side $l_h$ and wrapped length $λ$ around the pipe with radius $R$.

The length $l_h$ changes by an acceleration $a(t)$, which results from the gravitational acceleration and the deceleration due to friction and the bending of the rope around the pipe. We use the Capstan equation for this.

Force by weight:

$$F_w = F_{w0} \cdot e^{μ \cdot \alpha}, \hspace{30pt} \alpha = \phi + θ(t)$$

with $θ(t)$ as the pendulum angle and $μ$ the friction constant. The angle $\phi$ is a constant angle that is used to correct the angle $\alpha$ to $\theta$. The angle $\alpha$ describes the wrapping of the rope on the pipe.

If the pendulum is deflected to $\theta(t=0) = \theta_0 \in (-\pi/2,0)$, the pendulum begins to move in the direction of $+\theta$ due to gravity. We assume here that the pendulum receives so much energy that no backward movement takes place. It holds $\dot{\theta} > 0$.

If the pendulum angle is zero (straight down in the direction of the gravitational force), the pendulum hangs down on one side of the axle, while the person hangs straight down on the other side. At $\theta = 0$, half of the pipe is covered with rope. It follows for $\phi = \pi$, and we get:

$$F_w = F_{w0} \cdot e^{μ \cdot (\pi + \theta)}$$

Force due to human:

$$F_h = m_h \cdot g$$

Resulting acceleration for human:

$$a(t) = \frac{F_h - F_w}{m_h} = \left(1 - \frac{F_{w0}}{m_hg}\right)g \, e^{\mu(\pi+\theta)}$$

$$F_{m0} = m_wg + m_w\dot{\theta}^2l_w$$

The length of the wrapped rope is calculated as follows

$$λ(t) = R \cdot (\pi + \theta(t))$$

The length of $l_h(t)$ is changing with the acceleration. We get:

$$\ddot{l_h} = a(t)$$

If we now caluclate the 2nd derivative of $l_w(t)$, we obtain:

$$\ddot{l}_w^\text{(external)} = -R\ddot{θ} - a(t)$$

I used this expression for the external force $F_{l_w} = m_w \ddot{l}_w^\text{(external)}$, acting against the direction of the rope. I got:

$$\mathbf{F} = F_{l_w} \cdot (-\sin\theta, -\cos\theta)$$

The generalized forces are given by the projections of the external force $\mathbf{F}$ onto the directions of the generalized coordinates $l_w$ and $\theta$.

For a generalized coordinate $q_i$, the generalized force $Q_i$ is given by:

$$Q_i = \mathbf{F} \cdot \frac{\partial \mathbf{r}}{\partial q_i}$$

where $\mathbf{r}$ is the position vector $(x, y)$.

The position vector is:

$$\mathbf{r} = (x, y) = (l \sin(\theta) - R \cos(\theta), l \cos(\theta) + R \sin(\theta))$$

For $l$:

$$\frac{\partial \mathbf{r}}{\partial l} = \left( \sin(\theta), \cos(\theta) \right)$$

Thus, the generalized force in the direction of $l$ is:

$$Q_l = \mathbf{F} \cdot \frac{\partial \mathbf{r}}{\partial l} = -F_{l_w}$$

For $\theta$:

$$\frac{\partial \mathbf{r}}{\partial \theta} = \left( l_w \cos\theta + R \sin\theta, -l_w \sin\theta + R \cos\theta \right)$$

Thus, the generalized force in the direction of $\theta$ is:

$$Q_\theta = \mathbf{F} \cdot \frac{\partial \mathbf{r}}{\partial \theta} = RF_{l_w}$$

This looks way too wrong... I'm not sure about that. With $R\ll l_w$ I could set $Q_{\theta} = 0$.

I tried to get a numerical analysis of the behaviour, but my code doesn't work right:

import numpy as np
from scipy.integrate import solve_ivp
import matplotlib.pyplot as plt
parameters
g = 9.81 # gravitational acceleration
r = 0.15 # radius of pipe
mu = 0.2 # friction
m1 = 20  # mass of weight
m2 = 70  # mass of human
Target value for theta
theta_target = (1/mu) * np.log(m2/m1) - np.pi
Define the system of ODEs
def system(t, y):
    l_w, dl_w, theta, dtheta = y
    d2theta = - (1/l_w) * (2 * dl_w * dtheta + g * np.sin(theta))
    d2l_w = l_w * dtheta*2 - g  (1 - np.cos(theta)) - r * (d2theta) + g * (m1/m2) * np.exp(mu * (np.pi + theta))
    return [dl_w, d2l_w, dtheta, d2theta]
Initial conditions
l_w0 = 10 
dl_w0 = -0.1
theta0 = -np.pi/3 
dtheta0 = 0.1
y0 = [l_w0, dl_w0, theta0, dtheta0]
Event function to detect when theta reaches the target value
def event(t, y):
    return y[2] - theta_target
event.terminal = True
event.direction = 0
Solve the system
t_start = 0
t_end = 20
sol = solve_ivp(system, [t_start, t_end], y0, events=event)
Extract the results
t = sol.t
l_w = sol.y[0]
theta = sol.y[2]
Plot (t, theta)
plt.figure()
plt.plot(t, theta, label='θ(t)')
plt.axhline(theta_target, color='r', linestyle='--', label='θ_target')
plt.xlabel('Time (t)')
plt.ylabel('θ (theta)')
plt.title('θ(t) vs Time')
plt.legend()
plt.grid()
plt.show()
Plot (t, l_w)
plt.figure()
plt.plot(t, l_w, label='l_w(t)')
plt.xlabel('Time (t)')
plt.ylabel('l_w (length of wire)')
plt.title('l_w(t) vs Time')
plt.legend()
plt.grid()
plt.show()
def acceleration(t,theta):
    a = g - g * m1/m2 * np.exp(mu * (np.pi + theta))
    return a
Plot (t, acceleration(t,theta))
plt.figure()
plt.plot(t, acceleration(t,theta), label='a(t)')
plt.xlabel('Time (t)')
plt.legend()
plt.grid()
plt.show()
Print t0 if the event was detected
if sol.t_events[0].size > 0:
    t0 = sol.t_events[0][0]
    print(f"t0 = {t0}")
else:
    print("No event found where θ(t) reaches the target value.")

But ultimately I'm stuck right here. I don't know if I made a mistake when setting up the differential equation. I also don't have a good idea how to come up with a sensible approximation to determine how long the fall of the person lasts in order to ultimately determine how long the rope must be or how heavy the mass must be. I hope that I can arouse someone's interest in helping me with this summary.

Answer 1

I will present a more streamlined approach based on an application of Newton's laws. Lagrange equations are difficult to apply here due to the fact that friction on the pole around which the ball is winding plays a significant role, and hence the system is technically nonconservative. Newton's laws allow for a clean, systematic derivation. (As seen in the edit below however, the external forces can be taken into account as generalized forces, and yield the same equations of motion).

Consider a pole with circular cross-section of radius $R$ centered at the origin. Human of mass $M$ is hanging from a massless rope of total length $\ell$ in the direction of Earth's gravity and the ball of mass $m$ on the other side of the rope is released with the rope taut. The piece of the rope between the ball and pole has length $w$, the one between the human and the pole $z$ and the remaining rope wound around the pole is of length $R\theta$, where $\theta$ is the winding angle of the rope around it. Assuming the rope is inextensible this yields the constraint $$z+R\theta+w=\ell$$

We can assign coordinates to the moving bodies. Obviously for the human $$\mathbf{r}_h=R\hat{x}-z\hat{z}$$ while the position of the weight is described by $$\mathbf{r}_w=R\hat{n}+w\hat{t}$$ where $\hat{n}, \hat{t}$ are the normal and tangent vector to the circle at the winding point: $$\hat{n}=(\cos\theta, \sin\theta)~~,~ \hat{t}=(-\sin\theta, \cos\theta)$$ We will use this comoving frame to describe the motion. To compute the second derivative first we note $\dot{\hat{t}}=-\dot{\theta}\hat{n}$ and $\dot{\hat{n}}=\dot{\theta}\hat{t}$ and we find

$$\ddot{\mathbf{r}}_{w}=(-R\dot{\theta}^2-2\dot{w}\dot{\theta}-w\ddot{\theta})\hat{n}+(R\ddot{\theta}+\ddot{w}-\dot{w}\dot{\theta}^2)\hat{t}$$

Total forces are given by

$$\mathbf{F}_h=(T_h-Mg)\hat{z}~, \mathbf{F}_w=-(T_w+mg\cos\theta)\hat{t}-mg\sin\theta\hat{n}$$ and the final ingredient is the constant friction constitutive relation $T_h=e^{\mu\theta}T_w$ which is precisely the reason why this phenomenon occurs. Writing out Newton's laws, we obtain a total of 3 equations

$$-R\dot{\theta}^2-2\dot{w}\dot{\theta}-w\ddot{\theta}=-g\sin\theta$$

$$R\ddot{\theta}+\ddot{w}-w\dot{\theta}^2=-\frac{T_w}{m}-g\cos\theta$$

$$-M\ddot{z}=-Mg+T_h$$

From here one can eliminate the unknown string tension forces and $z$ using the constraint equation and obtain two coupled differential equations for $w,\theta$.

$$R\dot{\theta}^2+2\dot{w}\dot{\theta}+w\ddot{\theta}=g\sin\theta$$

$$\left(1+\frac{Me^{-\mu\theta}}{m}\right)(R\ddot{\theta}+\ddot{w})-w\dot{\theta}^2=-\frac{M}{m}ge^{-\mu\theta}-g\cos\theta$$

EDIT:

I'm going to use Lagrangian formalism to derive the equations of motion per the request in the comments. The string tensions are difficult to take into account as purely constraint forces, as this would require a detailed treatment of the rope degrees of freedom. Instead, we will include them as generalized forces. The Lagrangian can be written as

$$\mathcal{L}=\frac{1}{2}m\dot{\mathbf{r}}_w^2+\frac{1}{2}M\dot{\mathbf{r}}_h^2-V$$

Expressing that in terms of the generalized coordinates $\mathbf{q}=(w,\theta,z)$ yields the expression

$$\mathcal{L}=\frac{1}{2}m(R\dot{\theta}+\dot{w})^2+w^2\dot{\theta}^2-mg(R\cos\theta-w\sin\theta)+\frac{1}{2}M\dot{z}^2+Mgz$$

We also define the external tension forces $\mathbf{T}_h=T_h\hat{z}$, $\mathbf{T}_w=-T_w\hat{t}$. We can compute the generalized forces using the definition:

$$Q_w=\mathbf{T}_w\cdot\frac{\partial \mathbf{r}_w}{\partial w}=-T_w$$

$$Q_\theta=\mathbf{T}_w\cdot\frac{\partial \mathbf{r}_w}{\partial \theta}=-RT_w$$

$$Q_z=\mathbf{T}_h\cdot\frac{\partial \mathbf{r}_h}{\partial z}=-T_h$$

Finally, computing the derivatives of the Lagrangian leads to the 3 equations

$$R\ddot{\theta}+\ddot{w}-w\dot{\theta}^2=-g\cos\theta-\frac{T_w}{m}$$

$$R(R\ddot{\theta}+\ddot{w})+w^2\ddot{\theta}+2w\dot{w}\dot{\theta}=-g(R\cos\theta-w\sin\theta)-\frac{RT_w}{m}$$

$$\ddot{z}=g-\frac{T_h}{M}$$

These equations are equivalent to the ones derived with Newton's law (subtract $R$ times the first equation from the second to derive Newton's law second equation).

After subtraction, we get the equivalent set of ODEs

$$R\ddot{\theta}+\ddot{w}-w\dot{\theta}^2=-g\cos\theta-\frac{T_w}{m}$$

$$w\ddot{\theta}+2\dot{w}\dot{\theta}+R\dot{\theta}^2=g\sin\theta$$

$$\ddot{z}=g-\frac{T_h}{M}$$

Numerical Analysis:

Using the following Mathematica code, one can play around with the system numerically;

R = 0.2;
g = 9.81;
mu = 0.01;
m = 2;
M = 70;
temp = NDSolve[{f'[x] == F[x],
y'[x] == W[x], 
y[x]*F'[x] + 2 W[x]*F[x] + R*(F[x])^2 - g*Sin[f[x]] == 0, 
R*F'[x] + W'[x] - y[x]*F[x]^2 + g*Cos[f[x]] + 
 M*(R*F'[x] + W'[x] + g)*Exp[-mu*(f[x])]/m == 0, y[0] == 25, 
f[0] == Pi/4, W[0] == 0, F[0] == 0}, {y, f, W, F}, {x, 0, 10}]
Plot[{(f /. temp[[1]])[x], (y /. temp[[1]])[x]} , {x, 0, 2.3}]

As it turns out, the ODE system can indeed reproduce the behavior shown in the video, but this requires fine-tuning of the initial conditions. The system becomes stiff near $w=0$, and indeed, this is the behavior we are seeking; the counterweight at that point is located near the origin and the motion there is no longer described by our equations, so we tune our initial conditions to find stiffness and only examine the motion before $w\approx 0$. Using the conditions shown above, I can get the counterweight to perform a modest 3.2 rotations before the system collapses, and the whole trajectory lasts about 2.3 secs. Note that the mass of the object or $w(0)$ cannot be too large: when the mass is too large, $w$ may explode at infinity and clip the human's head; while if the ball is located too far the system gains too much gravitational energy before friction can dissipate enough of it. As a result, the system starts oscillating. I hope this clears up the situation regarding this model!