Shape of a rotating rope with one free-end [on hold]

Question

One end of a uniform rope (with total mass $M$) is fixed on the edge of a cylinder. The cylinder has a radius $R$ and rotates with angular velocity $\omega$. The axis is vertical in a gravitational field. Air drag is neglected. What is the shape of the rope?

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I tried to use the Euler-Lagrange equation to solve this problem but I don't know how to handle free-end boundary condition. Maybe some other method?

Answer 1

Before developing the theory, I decided to first make an experiment in order to understand, what we are dealing with. A cylinder with a diameter of 11.5 cm is mounted on the motor shaft (I used an old popcorn machine). I attached a 12.5 cm length of clothesline with a screw, so that exactly 11.5 cm leaves the cylinder. When the rope hangs freely, it forms a specific figure, which must be described first of all in order to find the parameters of the model (see Figure 1 left, center). When the cylinder rotated with an angular velocity of $\omega = 8 \pi$, the rope became almost horizontal. In this case, the rope was slightly bent in the horizontal plane due to the aerodynamic drag (See fig.1 right).

I used the theory of elastic rods described in the book L.D. Landau, E.M. Lifshitz, Theory of Elasticity. From this theory, I derived a system of equations: $$-\frac {EIL}{M}\theta ''=F_x\cos (\theta (l))-F_y\sin(\theta (l))$$ $$F_x'=-x\omega^2, F_y'=-g$$ $$x'=\sin(\theta (l)), y'=\cos (\theta (l))$$ This system of equations describes the equilibrium of a round rod under the action of distributed forces and torques. Here $E$ is Young's modulus, $I$ is the moment of inertia, $L$ is the length of the rope, $M$ is the mass of the rope. All derivatives are calculated by the length parameter $l$. The $\theta $ angle is measured from the vertical axis $y$. Boundary conditions are as following: $$\theta (0)=\frac {\pi}{2}, \theta '(L)=0, x(0)=R, y(0)=0, F_x(L)=0, F_y(L)=0$$ Here $R$ is the radius of cylinder. We set $g = 9.81, \omega = 0, \frac {EIL}{M} = 0.00012, L=0.115 m$, then the calculation curve in Fig. 2 above, qualitatively corresponds to the free-hanging rope in Fig. 1 on the left. We set $\omega = 8 \pi $, then the calculated curve in Fig.2 below, qualitatively corresponds to the rotating rope in Fig.1 on the right. Some intermediate cases we will consider as the implementation of experiments. I took a short rope of 6 cm outside the cylinder. This rope (green) has a different texture and thickness. However, in the free state with $\omega = 0$, it takes the form as the first rope, and with $\omega = 40 rad / s$ rises horizontally as the first long rope - see Figure 3 at the top. In this case, at $\frac {EIL}{M} =6*10^{-6}$, the model describes both states of the rope - see Figure 3 below.

In the third experiment, I took a 16 cm long rope that was more rigid than the first two. In the absence of rotation, the rope had an incline of about 34 degrees to horizon line at the free end, see fig.4 at the top left. In the presence of rotation with the speed of $\omega = 50 rad / s$, the rope takes a horizontal position and even slightly above the horizon line - fig.4 in the upper right. In this case, at $\frac {EIL}{M} =0.01$, the model describes both states of the rope - see Figure 4 below.

In the fourth experiment, I took a rope the same as in the third, but 20 cm long. I wanted to check whether standing waves with an amplitude of 1-2 millimeters were formed on the rope. These waves are clearly visible in Figure 2-4 on the calculated curves with $\omega > 0$. I photographed with a flash a rotating rope with $\omega = 125$ opposite the screen, so that the shadow from the rope was visible. In Figure 5, this photo is shown at the top right. Top left is the exact same rope in a free state, and below are calculated curves for two states (rest and rotation).

Finally, in Figure 6 shows photographs of various ropes at a rotation speed of $\omega=3-6 rad / s$. The right photo shows the general view of the ropes used in the experiments. It can be seen that the shape of the ropes is not similar to that obtained in the calculations on the theory of chains. This is due to the fact that all the ropes start from a horizontal surface to which they are attached with a screw. In the lower part of Fig. 6 shows the calculated curves describing the experimental data.

Experiment with soft rope. I took two soft ropes 20 cm and 30 cm long and rotated them at high and low speed. As it turned out, a soft rope of such length at a low speed of rotation bends like a chain. Figures 7 and 8 show the shape of the rope with a length of 20 and 30 cm, respectively, at different speeds of rotation. The calculations are made on the model presented above with various parameters of the stiffness and the angle of contact of the rope with the cylinder.

Soft rope 20 cm long.

Soft rope 30 cm long at different speeds of rotation.

Answer 2

In cylindrical coordinates, let the shape of the rope be parameterized by r = r(s), $\theta=\theta(s)$, and z = z(s), where s is distance measured along the rope. Then a unit vector along the rope is given by: $$\mathbf{i_s}=\frac{dr}{ds}\mathbf{i_r}+r\frac{d\theta}{ds}\mathbf{i_{\theta}}+\frac{dz}{ds}\mathbf{i_z}$$For an inextensible rope, we must have:$$\left(\frac{dr}{ds}\right)^2+\left(r\frac{d\theta}{ds}\right)^2+\left(\frac{dz}{ds}\right)^2=1$$Letting T(s) represent the tension in the rope at location s along the rope, the force balance on the section of rope between s and s + ds is given by:$$\frac{d(T\mathbf{i_s})}{ds}-\rho g\mathbf{i_z}=-\rho \omega^2r\mathbf{i_r}$$where $\rho=M/L$. In component form, this becomes:$$\frac{d}{ds}\left(T\frac{dz}{ds}\right)=\rho g$$$$\frac{d}{ds}\left(T\frac{dr}{ds}\right)-r\left(\frac{d\theta}{ds}\right)^2T=-\rho\omega^2r$$$$\frac{d}{ds}\left(Tr\frac{d\theta}{ds}\right)+T\frac{dr}{ds}\frac{d\theta}{ds}=0$$The z equation can be integrated once immediately to yield: $$T\frac{dz}{ds}=\left[T\frac{dz}{ds}\right]_{s=0}+\rho g s$$ Similarly, the $\theta$ equation can be integrated to yield: $$Tr^2\frac{d\theta}{ds}=\left[Tr^2\frac{d\theta}{ds}\right]_{s=0}$$ We have 4 equations in the four unknowns r, z, $\theta$, and T, but integrating these equations seems daunting.

ADDENDUM

After further consideration, I see no reason why $\theta$ needs to be anything other than zero over the entire length of the rope (i.e., no variations in $\theta$). So that's what I'm going to assume from this point on. With this assumption, the in extensibility condition now becomes: $$\left(\frac{dr}{ds}\right)^2+\left(\frac{dz}{ds}\right)^2=1$$If we let $\phi(s)$ represent the contour angle of the rope with respect to the horizontal at location s along the rope, then we can write: $$\frac{dr}{ds}=\cos{\phi}\tag{A}$$ $$\frac{dz}{ds}=-\sin{\phi}\tag{B}$$These equations satisfy the in extensibility condition exactly. Once the function $\phi(s)$ is established, this determines the shape of the rope over its entire length.

In terms of $\phi$, the unit tangent vector along the rope is given by: $$\mathbf{i_s}=\cos{\phi}\mathbf{i_r}-\sin{\phi}\mathbf{i_z}$$ and the derivative with respect to s of the unit tangent vector (i.e., the unit normal vector times the curvature) is given by:$$\frac{d\mathbf{i_s}}{ds}=-(\sin{\phi}\mathbf{i_r}+\cos{\phi}\mathbf{i_z})\frac{d\phi}{ds}$$ If we substitute these equations into the differential force balance equation, we obtain: $$(\cos{\phi}\mathbf{i_r}-\sin{\phi}\mathbf{i_z})\frac{dT}{ds}-T(\sin{\phi}\mathbf{i_r}+\cos{\phi}\mathbf{i_z})\frac{d\phi}{ds}-\rho g\mathbf{i_z}=-\rho \omega^2r\mathbf{i_r}$$If we dot this equation with the unit tangent vector and then also with respect to the unit normal vector, we obtain: $$\frac{dT}{ds}=-\rho g \sin{\phi}-\rho\omega^2r\cos{\phi}=\rho g\frac{dz}{ds}-\rho g \omega^2 r\frac{dr}{ds}\tag{1}$$and$$T\frac{d\phi}{ds}=-\rho g\cos{\phi}+\rho \omega^2r\sin{\phi}\tag{2}$$

Eqn. 1 can be integrated immediately to yield the tension T: $$T=T(0)+\rho g z-\frac{\rho \omega^2 (r^2-R^2)}{2}\tag{3}$$If we combined Eqns. 2 and 3, we obtain an equation for the derivative of $\phi$ with respect to s: $$\frac{d\phi}{ds}=\frac{-\rho g\cos{\phi}+\rho \omega^2r\sin{\phi}}{T(0)+\rho g z-\frac{\rho \omega^2 (r^2-R^2)}{2}}\tag{4}$$

This equation could be integrated numerically together with equations A and B to get the rope shape if we knew that initial values for T and $\phi$. The initial tension must be such that the value of T at s = L is zero. Also, since the denominator must be equal to zero at s = L, the numerator must also be zero at this location in order for the curvature to be finite. So, at s = L, we must have $$r(L)\tan{\phi(L)}=\frac{g}{\omega^2}$$This is a pretty nasty boundary condition that would have to be satisfied. But, conceptually, we could solve the problem by using the shooting method, and adjusting the initial values of T and $\phi$ until the required conditions are satisfied at s = L.

CONTINUATION

Before continuing and presenting a method for solving the differential equations for the shape of the rope, I'm first going to follow @Hussein's recommendation, and reduce the equations to dimensionless form. This is done simply by scaling all the spatial parameters r, z, s, and L by the radius R of the drum. In terms of the new dimensionless variables, our equations now become:

$$\frac{dr}{ds}=\cos{\phi}\tag{5}$$ $$\frac{dz}{ds}=-\sin{\phi}\tag{6}$$ $$\frac{d\phi}{ds}=\frac{-\cos{\phi}+\beta r\sin{\phi}}{[z-z(L)]-\beta\frac{(r^2-r^2(L))}{2}}\tag{7}$$where $$\beta=\frac{\omega^2R}{g}\tag{8}$$and the dimensionless tension is given by $$\tau=\frac{T}{\rho g R}=[z-z(L)]-\beta\frac{(r^2-r^2(L))}{2}\tag{9}$$ and our zero-tension boundary condition at s = L now becomes $$r(L)\tan{\phi(L)}=\frac{1}{\beta}\tag{10}$$ In our subsequent development, we are going to also need to know the value of the dimensionless curvature $d\phi/ds$ at s = L. Because of the zero-tension boundary condition (Eqn. 10) at s = L, both the numerator and denominator of Eqn. 7 for $d\phi/ds$ approach zero at this location. However, we can still obtain the value for $d\phi/ds$ by applying l'Hospital's rule; this yields:$$\left[\frac{d\phi}{ds}\right]_{s=L}=-\frac{\beta^2r(L)}{2[1+(\beta r(L))^2]^{3/2}}\tag{11}$$

METHOD OF SOLUTION

The differential equation can be integrated, subject to the prescribed boundary conditions, by either stating at s = 0 and integrating forward to increasing radii, or by starting at s = L and integrating backward toward lower radii. For various reasons that I won't get into here, it is more straightforward to start at s = L and to integrate backwards.

To integrate backwards, we make a change of variable according to $$S=L-s$$ Our differental equation and initial conditions in terms of S then become:

$$\frac{dr}{dS}=-\cos{\phi}\tag{5a}$$ $$\frac{dz}{dS}=\sin{\phi}\tag{6a}$$ $$\frac{d\phi}{dS}=\frac{\cos{\phi}-\beta r\sin{\phi}}{[z-z(0)]-\beta\frac{(r^2-r^2(0))}{2}}\tag{7a}$$where the dimensionless tension is now given by $$\tau=\frac{T}{\rho g R}=[z-z(0)]-\beta\frac{(r^2-r^2(0))}{2}\tag{8a}$$Eqn. 7a applies at all values of S except S = 0, where $$\left[\frac{d\phi}{dS}\right]_{S=0}=+\frac{\beta^2r(0)}{2[1+(\beta r(0))^2]^{3/2}}\tag{11a}$$In addition, at S = 0, we have the initial condition of $\phi$ as: $$r(0)\tan{\phi(0)}=\frac{1}{\beta}\tag{10a}$$And, with no loss of generality we can take $$z(0)=0$$

Prior to carrying out the integration of these equations as an initial value problem, we don't know the value of r(0) that will be necessary for r(L) to be unity at S = L. So we can choose various values of r(0) and perform the integration, iterating on r(0) until we obtain a solution where r(L) = 1.0. Or we can just choose different value of r(0) and generate an array of solutions for the values of L that each of them implies at S = L.

The easiest way to integrate these equations numerically as an initial value problem is to employ forward Euler with a small step size for good accuracy.

RESULTS OF SAMPLE CALCULATION

I have carried out a numerical solution of the model differential equations on an Excel Spreadsheet using the approach described above. The objective was to compare with @rob's results. The case considered was with L=10 R and $\beta=0.25$, where $\beta = 0.25$ corresponds to rob's case of $\omega= 0.5 \omega_0$.

This shows the dimensionless vertical coordinate vs the dimensionless radial coordinate for the rope. To the eye, the results are a very close match to rob's results for the same case in his figure. In particular, the dimensionless vertical drop is about 4.75 and the dimensionless radial location of the rope tail is at about 9.75. The dimensionless rope tension at the drum for this case was about 16.5

RESULTS FOR CASE REQUESTED BY Alex Trounev

Alex Trounev has requested that I perform the calculation for the following case: $\omega=2\pi$, R = 0.1 meters, L = 1 meter, and $g = 9.81/ m^2/sec$. For these parameter values, we have that the dimensionless radial acceleration $\beta$ is given by $$\beta=\frac{\omega^2R}{g}=\frac{(2\pi)^2(0.1)}{9.81}=0.4024$$and the dimensionless length of the rope is $L/R=10$. The calculated shape of the rope for this case is shown in the figure below:

The vertical drop of the rope from the drum to the free end is predicted to be about 0.3 meters, and the radial extent of the rope from the drum to the free end is predicted to run from 0.1 meters to 1.053 meters.

The predicted dimensionless tension in the rope at the drum is predicted to be $\tau=25.1$. The actual dimensional tension is related to the dimensionless tension by $$T=\rho g R \tau=\rho g L\frac{R}{L}\tau=W\frac{R}{L}\tau$$where W is the weight of the rope. So, in this case, $$T=(0.1)(25.1)W=2.51W$$That is 2.51 times the weight of the rope. Of course, the vertical component of the tension at the drum must be equal to the weight of the rope. So the remainder of the tension in the rope is the effect of the horizontal component associated with the angular acceleration.

Answer 3

Here's a steady-state solution for a constant-length chain made of a large (but finite) number of links, found via the Euler-Lagrange method, in the form of a recursive set of equations. Since the chain can't be stretched there isn't any potential energy stored in the length-wise degrees of freedom, so the question of how to handle the zero-tension boundary condition at the free end doesn't arise: the Euler-Lagrange method depends only on the kinetic and potential energies of the bodies being modeled.

Let's model the chain as $N+1$ points of mass $m$, each separated from its nearest neighbors by $\ell=L/N$, all lying in the $(r,z)$ plane. The zeroth point is at $(r_0,z_0)=(R,0)$, and the line from the $(n-1)$-th point to the $n$-th makes an angle $\theta_n$ with the vertical. The location of the $n$-th point is therefore

\begin{align} r_n &= R + \sum_{i=1}^n \ell\sin\theta_i & z_n &= -\sum_{i=1}^n \ell\cos\theta_i \end{align}

The kinetic energy for the $n$-th particle has terms for rotational kinetic energy and for possible motion in $r$ or $z$:

$$ T_n = \frac12m \left( \omega^2 r_n^2 + \dot r_n ^2 + \dot z_n ^2 \right) $$

where

\begin{align} \dot r_n &= \sum^n \ell \dot\theta_i \cos\theta_i & \dot z_n &= \sum^n \ell \dot\theta_i \sin\theta_i \end{align}

Similarly, the gravitational potential for the $n$-th point on the chain also depends on the positions of all the points connecting it to the axis:

$$ U_n = mgz_n = -mg\ell\sum^n \cos\theta_i $$

It's a little tedious to write down $\partial L_n/\partial\theta_i$ and $\frac{\mathrm d}{\mathrm dt}\partial L_n/\partial\dot\theta_i$ to construct the Euler-Lagrange equations, depending on how fastidious you are about expanding the expressions like $r_n$ and $\dot z_n$ versus trusting your dexterity with the chain rule for derivatives. But we're interested (for now) only in the steady-state solution, for which all the time derivatives will vanish. Dimensional analysis suggests (and an explicit calculation confirms) that the terms which survive are the ones where the units of $\mathrm s^{-2}$ come from $\omega^2$ and $g$:

\begin{align} 0 &= \frac{\partial L_n}{\partial\theta_i} - \frac{\mathrm d}{\mathrm dt}\frac{\partial L_n}{\partial\dot\theta_i } \\ &= \begin{cases} m\ell \cdot (\omega^2 r_n \cos\theta_i - g\sin\theta_i) + \text{terms with dots} & \text{if } 1 \leq i \leq n \\ 0 & \text{otherwise} \end{cases} \end{align}

Note the funny limits. We want one equation of motion (solved for the steady state, by zeroing out the terms with dots) for each of the $\theta_i$, but each equation of motion depends on the entire Lagrangian $L = \sum_{n=1}^N L_n$:

\begin{align} 0 &= \left( \frac{\partial }{\partial\theta_i} - \frac{\mathrm d}{\mathrm dt}\frac{\partial }{\partial\dot\theta_i } \right) \sum_{n=1}^N L_n = \sum_{n=i}^N \left( \frac{\partial }{\partial\theta_i} - \frac{\mathrm d}{\mathrm dt}\frac{\partial }{\partial\dot\theta_i } \right) L_n \end{align}

The change in the lower limit of the sum just removes all of the terms with $n<i$. This leads to a funny-looking recursion relation:

\begin{align} \frac{(N-i+1)g}{\omega^2} \tan\theta_i &= \sum_{n=i}^N r_n \\ \frac{Ng}{\omega^2}\tan\theta_1 &= r_N + r_{N-1} + \cdots + r_2 + r_1 & r_1 &= R + \ell\sin\theta_1 \\ \vdots && \vdots \\ \frac{g}{\omega^2}\tan\theta_{N} &= r_N & r_{N} &= r_{N-1} + \ell\sin\theta_{N} \end{align}

So the equilibrium angle of the connection to the final link in the chain depends only the radial location of the final link, while the equilbrium angles of the centerward connections depend on the radial locations of all the outer links in the chain that they are supporting. That makes a kind of intuitive sense: if the mass of the outer part of the chain is further from the axis of rotation, the link has to provide more centripetal force to hold it in.

Unfortunately the recursion in this case goes the wrong way: you can't start at the axis and predict what the chain will do. (If we hadn't thrown away all of the terms with dots, we could start with an initial shape and predict how it would evolve, but that's a different project.) I solved the recursion relation for the equilibrium shape by guessing some values of $r_N$ for the outer end of the chain and choosing the one that puts the zeroth point closest to the assumed location of $r_0=R$.

Here are a couple of plots showing numerical solutions for the equilibrium shape of the line as the rotational frequency $\omega$ and the total length $L$ are varied. It is natural to measure $L$ in units of $R$, and $\omega$ in units of $\omega_0 = \sqrt{g/R}$. Note that the shape does change as the total length of the chain increases, contrary to the solution of Thomas Fritch.

A future revision to this answer will either have some literature references or some experimental photos, depending on whether I can get my kids excited about it.

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Experiments with a two-meter chain and a swivel chair in my yard suggest that this shape is qualitatively correct, but I had a little trouble with the photography.

The total length of the chain is about twice the length from my sternum to my arm; I'm holding it a quarter of the way along, so that the two segments have $L/R = 1/2, 3/2$. The long segment really does like to sit higher than the short segment, but it's hard to spin my self with a stable enough frequency to tamp down the vibrations in the two chains. I calculate from the photo timestamps that I was spinning with a period of about two seconds, which (living on Earth and having roughly a two-meter armspan) is pretty close to $\omega=\omega_0/2$. There's some foreshortening in the photo: the chains are pretty much parallel to my arm, but they're not perpendicular to the camera. I was prevented from getting a better photo because I got dizzy from the spinning and my photographer (age 11) thought the weather was too hot to sit outdoors and push the shutter button on the camera.

But the qualitative features of this model --- a concave-up chain curvature, and longer chains more horizontal than shorter ones at the same frequency, and the order of magnitude of the angle chain angle at this scaled frequency --- are all supported by the experiment.

Answer 4

This whole derivation assumes the gravity-less case is a straight line, but I'm having issues with that. Until I have a better understanding, this is my answer:

The problem in trying to use Euler-Lagrange, in my opinion, is that we have a nonrigid body and hence an infinite number of generalized coordinates (you can't characterize any curve by a finite set of numbers), which lead to an infinite number of Euler-Lagrange equations.

The method that made more sense to me is assuming that in a stationary state (a state where the rope retains its shape) the net Force acting on an infinitesimal length of the rope points radially. Then, by dividing the rope in discrete chunks we can calculate the tension between neighboring chunks. Then we take the continuous limit and derive an expression for the tension at every point of the rope. Finally, by demanding that the tension be always parallel to the rope, we can get the expression for the curve the rope makes.

If the rope has N chunks of length $\frac{L}{N}$ and mass $\frac{m}{N}$, and the tension force the $N$th chunk applies to the $N+1$th one is $\vec{T_n}$, we require that at the end of the rope we have $ \vec{T_{N-1}} + \vec{F_N} =0$, where $\vec{F_n} = m_n \omega^2 \vec{r_n} + m_n \vec{g} $ are the other forces acting on the chunk. The distance of the chunk to the rotating axis is given by $\vec{r_n} = (R + \frac{nL}{N})\hat{x}$, so the term that contain this is due to Centrifugal 'Force'. ($\vec{g}$ points down)

In general, we require

$$\vec{T_n} - \vec{T_{n+1}} + \vec{F_n} = 0$$

Which implies in (defining $\delta = \frac{L}{N}$)

$$ \frac{ \vec{T_{n+1}} - \vec{T_n}} {\delta} = \frac{m}{L} ( \omega^2 \vec{r_n} + \vec{g}) $$

Taking the continuous limit, with $\lambda$ as a parametrization of the curve, from $0$ to $L$:

$\frac{d}{d\lambda} \vec{T(\lambda)} = \frac{m}{L} (\omega^2(R+\lambda)\hat{x} - g\hat{y})$

$ \vec{T(\lambda)} = \frac{m}{L} (\omega^2(R\lambda+\frac{\lambda^2}{2})\hat{x} - g\lambda \hat{y}) + \vec{c}$, where $\vec{c}$ is the integration constant.

Remembering that the Tension must be parallel to the rope we get the relationship between the coordinates of the curve:

$\frac{dy}{dx} = \frac{T_y(\lambda)}{T_x(\lambda)}$

We also have $d\lambda^2 = dx^2 + dy^2$, so $\frac{dx}{d\lambda} = (1 + y'^2)^{1/2}$. Solving it for $y'$ and using the last equation we arrive at an expression for $x(\lambda)$, which we invert to get $\lambda(x)$ and finally integrate $\frac{dy}{dx} = \frac{T_y(\lambda(x))}{T_x(\lambda(x)}$ to get $y(x)$.

If I didn't make a mistake, $\frac{dy}{dx} = -\frac{2g}{\omega^2} \frac{1}{R + \sqrt{(x-k)^2 - \frac{4g^2}{\omega^4}}}$,

where $k = -\sqrt{(R+L)^2 + \frac{4g^2}{\omega^4}}$.

You can plug this in Wolfram Alpha to integrate or plot it.

Answer 5

Doing the balance of forces and using $s\in [0, L]$ as parameterization o the rope, you have the following differential equation (see references 1 and 2):

$$\frac{\partial}{\partial s}(T(s) \mathbf{x}_s) + \begin{pmatrix}\omega^2 \rho x\\ -\rho g\end{pmatrix} = \begin{pmatrix}0\\ 0\end{pmatrix}\, ,$$

subject to the following boundary conditions

\begin{align} &\mathbf{x}(0) = \begin{pmatrix}0\\ 0\end{pmatrix}\\ &T(L)\mathbf{x}_s(L) = \begin{pmatrix}0\\ 0\end{pmatrix} \end{align}

That means that it is fixed at one end and that the tension is zero at the other end. This is not enough to solve the equations, we need to find the tension. Two options for this are:

1. Elastic behavior for the rope:

$$T(s) = E (\Vert \mathbf{x}_s\Vert - 1) \, .$$

2. Inextensible rope:

$$ \Vert \mathbf{x}_s\Vert = 1\, .$$

References

1. Yong, Darryl. "Strings, chains, and ropes." SIAM review 48.4 (2006): 771-781.

2. Antman, Stuart S. "The equations for large vibrations of strings." The American Mathematical Monthly 87.5 (1980): 359-370.

Answer 6

Since all the answers above only give partial information about the solution to the problem, this is an attempt at summarizing and putting a tombstone to the problem of the continuum string attached on a rotating cylinder by providing an ODE to be solved for the equilibrium shape of the string.

First, let's start with a Newtonian treatment of the string. A Lagrangian formulation will be given soon thereafter.

The shape of the string at equilibrium is $\mathbf{r}=(r(t), z(t))$, where t is the arclength. We partition the string in two pieces ($t\leq s$ and $t> s$), and we define the force of tension exerted by the upper piece ($t\leq s$) to the lower piece ($t> s$) to be $-\mathbf{T}(s).$ Obviously the tension experienced by the upper piece due to the lower piece is $\mathbf{T}(s).$

Now partition the string in three pieces ($t\leq s, s<t\leq s+ds,t>s+ds$) and consider the forces exerted on the middle piece. Since the middle infinitesimal string piece equilibrates:

$$\mathbf{T}(s+ds)-\mathbf{T}(s)+\mathbf{g}dm+\omega^2\mathbf{r}dm=0$$

Now observe that tension for a rigid curved string has to be exerted along the tangent at the point where it's applied, so for example $\mathbf{T}(s)$ is parallel to the tangent at the point $(r(s), z(s))$. Then we can write

$$\mathbf{T}(s)=T(s)(\cos\theta(s)\mathbf{\hat{r}}+\sin\theta(s)\mathbf{\hat{z}})$$

where $\tan\theta(s)=\frac{dz}{ds}\Big(\frac{dr}{ds}\Big)^{-1}$ is the angle of the tangent with the r axis. Thus we obtain the set of equations:

$$\frac{d}{ds}(T(s)\cos\theta(s))=-\frac{dm}{ds}\omega^2r(s)~~~~~(1)\\ \frac{d}{ds}(T(s)\sin\theta(s))=\frac{dm}{ds}g~~~~~~(2)\\$$

Since the shape of the string is parametrized by it's arclength we have that $dm/ds=\rho$ (this wouldn't be true had we chosen any other variable to parametrize the shape). The second equation maybe integrated immediately with the boundary condition that $T(L)=0$ to yield $$T=\frac{\rho g(s-L)}{\sin\theta} ~~(3)$$ Substituting this relation in the first equation along with the restriction $(\frac{dr}{ds})^2+(\frac{dz}{ds})^2=1$ and the fact that $dz/ds<0$ we obtain an equation for $dr/ds\equiv\dot{r}$:

$$\frac{d}{ds}\Big((s-L)\frac{\dot{r}}{\sqrt{1-\dot{r}^2}}\Big)=\frac{\omega^2}{g}r$$

To solve this equation we need two boundary conditions. The first comes from the point of attachment which requires: $r(0)=R$. I have failed to identify a second boundary condition and I believe that the problem is incomplete, in the sense that more details about the way the string has been attached to the cylinder are necessary (is the string looped vertically/horizontally around a hook for example? This is going to make a difference since to me it sounds like a vertical loop is going to yield $\dot{z}(0)=0$ while a horizontal one $\dot{r}(0)=0$). I also failed in analytically solving the above equation. However, the analysis above is strongly supported by the fact that these equations arise from extremizing an energy functional under a constraint. The functional is given by the potential energy of the string but with an appropriate Lagrange multiplier which reflects the fact that the string has constant length:

$$-E[r(s),z(s), T(s)]=\int_{0}^{L} ds~\Big(\frac{1}{2}\rho\omega^2 r^2(s)-\rho gz(s)\Big)-\int_{0}^{L}ds~T(s)(\sqrt{\dot{r}^2(s)+\dot{z}^2(s)}-1)$$

Variation of the functional and application of the constraint yields the exact same equations as above.

EDIT:

It looks like that, actually, the problem is not ill-defined, but the boundary condition imposed is fairly complicated. Forming the linear combination $(1)\cos\theta+(2)\sin\theta$ we can prove that:

$$T(s)=T(0)-\frac{1}{2}\rho\omega^2(r^2(s)-R^2)+\rho gz(s)=T(L)-\frac{1}{2}\rho\omega^2(r^2(s)-r^2(L))+\rho g(z(s)-z(L))$$

which yields the condition:

$$T(0)=\frac{1}{2}\rho\omega^2 (r^2(L)-R^2)-\rho gz(L)$$

but also

$$T(s)\frac{d\theta}{ds}=\rho g \cos\theta+\rho\omega^2 r\sin\theta$$

from which we obtain

$$r(L)\dot{z}(L)+\frac{g}{\omega^2}\dot{r}(L)=0$$

unless $\frac{d\theta}{ds}|_{s=L}$ is allowed to approach infinity, which is in principle conceivable.

Also, in conjunction with equation $(3)$ yields the following complicated boundary condition that mixes the two endpoints of the inextensible string:

$$(\omega^2 (r^2(L)-R^2)-gz(L))\dot{z}(0)+gL=0$$

It turns out that this boundary condition is in principle sufficient to fully determine the solution of the system of ODE's:

$$\frac{d}{ds}\Big((s-L)\frac{\dot{r}}{\sqrt{1-\dot{r}^2}}\Big)=\frac{\omega^2}{g}r\\\dot{r}^2+\dot{z}^2=1\\r(0)=R~~,~~z(0)=0~~,\\ ~~[\omega^2 (r^2(L)-R^2)-gz(L)]\dot{z}(0)+gL=0~~,~~ r(L)\dot{z}(L)+\frac{g}{\omega^2}\dot{r}(L)=0 $$

However these boundary conditions are one too many and one might need to understand how all of these fit together in one scheme without clashing.

Answer 7

In the absence of an exact analytical solution, it may be interesting to look at approximate solutions in limiting cases. So I decide to add to this discussion. A figure below explains notations.

A form of the rope $(x(l),y(l))$ is parametrized by its length $l$. Projections of tension force $T_x$ and $T_y$ are $$ T_x = \rho \omega^2 \int_l^L\ dl'\ x(l'),\quad T_y = \rho g \int_l^L\ dl' = \rho g (L-l), \quad (1) $$ where $\rho = M/L$. Tension force is tangent to the rope, hence the differential equation: $$ \frac{y'(l)}{x'(l)} = y'(x) = -\frac{T_y}{T_x}.\qquad (2) $$ I was able to obtain a second-order non-linear differential equation for $y'(x)$ from the system of equations (1,2). I see no perspective to obtain a solution of this equation. Instead, let's consider two limiting cases.

1. Small frequency: $\omega^2 R \ll g$. In this case, the rope is almost vertical and an initial approximation is $x(l)\approx R$, $y(l) \approx -l$. Equations (1) become $$ T_x \approx \rho\omega^2(L-l),\qquad T_y = \rho g (L-l). $$ Now, due to $y'(l)\approx -1$ equation (2) gives $$ \frac{1}{x'(l)}\approx\frac{g}{\omega^2 R}\longrightarrow x(l)\approx R+\frac{\omega^2R}{g}l. $$ Finally, we obtain the approximate form of the rope in this limiting case: $$ y(x) \approx -\frac{g}{\omega^2R}(x-R).\qquad (3) $$

2. Large frequency: $\omega^2 R \gg g$. In this case, the rope is almost horizontal and initial approximation is $x(l)\approx R+l$, $y(l) \approx 0$. Equations (1) become $$ T_x \approx \rho\omega^2\left(R(L-l)+(L^2-l^2)/2\right),\qquad T_y = \rho g (L-l). $$ Now, due to $x'(l)\approx 1$ equation (2) gives $$ y'(l)\approx-\frac{g}{\omega^2}\frac{1}{R+(L+l)/2}\longrightarrow y(l)\approx -\frac{2g}{\omega^2}\ln\left(\frac{R+(L+l)/2}{R+L/2}\right). $$ Finally, we obtain the approximate form of the rope in this limiting case: $$ y(x) \approx -\frac{2g}{\omega^2}\ln\left(\frac{R+L+x}{2R+L}\right).\qquad (4) $$

As far as I can see, these approximate solutions satisfy @Chet Miller's boundary condition. In my notations, this condition takes form $x(L)y'(x(L)) = -g/\omega^2$

Update. The approximate solution obtained before for the small frequency case ($\omega^2R \ll g$) is too rude. It doesn't even show if the rope is bending up or down. Next iteration gives $$ x(l) \approx R + \frac{\omega^2R}{g} l +\left(\frac{\omega^2R}{g}\right)^2 \frac{Ll+l^2/2}{2R},\quad y(l)\approx -l+\frac12\left(\frac{\omega^2R}{g}\right)^2 l.\quad (5) $$ And the approximate form of the rope now is $$ y(x) \approx -\frac{g}{\omega^2R}(x-R)\left(1-\frac12\frac{\omega^2R}{g}\frac{L}{R}\right) + \frac1{4R}\frac{\omega^2R}{g}\left(\frac{g}{\omega^2R}(x-R)\right)^2. \quad (6) $$ These formulas probably are applicable when $\omega^2R\ll g$ and the ration $L/R$ is not too large. This solution also satisfies the boundary condition $x(L)y'(x(L)) = -g/\omega^2$ with correspondent accuracy.

Answer 8

The Equation for a peach of rope :

Forces summation: $$\sum_r=F_r+dF_r-F_r-q_r\,ds=dF_r-q_r\,ds=0$$

$\Rightarrow$

$$\frac{d}{ds}F_r(s)=q_r\tag 1$$

$$\sum_z=F_z+dF_z-F_z+q_z\,ds=dF_r-q_z\,ds=0$$

$\Rightarrow$

$$\frac{d}{ds}F_z(s)=-q_z\tag 2$$

Torque sum in Point A:

$$\sum \tau=F_z\,dr-F_r\,dz=0$$

$\Rightarrow$

$$\frac{dr}{dz}=\frac{F_r(z)}{F_z(z)}\tag 3$$

with:

$\frac{dF_r}{dz}=\frac{dF_r}{ds}\frac{ds}{dz}=\frac{dF_r}{ds}\,\frac{ds}{dr}\, \frac{dr}{dz}$

$\frac{ds}{dr}=\sqrt{1+\left(\frac{dz}{dr}\right)^2}$

$q_r=\rho\,g\,A$

$q_z=\rho\,\omega^2\,z\,A$

we obtain for equation (1)

$${\frac{d}{dz}}{\it Fr} \left( z \right) ={\frac {\rho\,gA\sqrt { \left( {\it Fz} \left( z \right) \right) ^{2}+ \left( {\it Fr} \left( z \right) \right) ^{2}}}{{\it Fz} \left( z \right) }} \tag 4$$

for equation (2)

$${\frac {d}{dz}}{\it Fz} \left( z \right) ={\frac {\rho\,{\omega}^{2} \,z A\sqrt { \left( {\it Fz} \left( z \right) \right) ^{2}+ \left( {\it Fr} \left( z \right) \right) ^{2}}}{{\it Fz} \left( z \right) }} \tag 5$$

where:

$q_r$ is weight per length

$q_z$ is the centrifugal force per length

$A$ rope area

$R$ zylinder radius

$\rho$ rope density

$L$ Rope length

the numerical solution of equations (3), (4) and (5) solve the problem. the shape of the rope is the solution $r(z)$

Simulation Data:

Initial condition: $F_z(0)=m\,\omega^2\,R$

$F_r(0)=-m\,g$

$r(0)=0$

Parameters unit $[m]\,,[kg]\,,[s]$

$\omega=\frac{n\pi}{30}$

$R=0.10$

$\rho=7.85 10^3$

$g=9.81$

$L=2$

$d=0.5 10^{-2}$

$A=\frac{\pi\,d^2}{4}$

$m=\rho\,A\,L$

$n=50$ Rpm