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Let's take a wire or a rope. I usually do this with a chain or my scarf.

I fixate one end in my hand and apply rotation (by subtle movements of this endpoint like spinning a lasso). The rope gets into rotation and obtains certain bent shape: enter image description here

Some part is missing because cellphone cameras ain't great for high-speed photos but I hope you can imagine all the scarf.

The question is: how can I calculate/predict this shape?

Although this problem doesn't seem that bizarre, I have never seen any solution. Nor I have found this question asked anywhere on internet... Must be because I just don't know how to formulate it without pictures.

Also by taking longer rope I get more than one bend: enter image description here

I apoligize for the quality again. It is even harder to rotate this while taking picture. The form is not spiral, it is more like a shape in a plane that's rotating.

I'll be thankful for explanations, solutions, links or at least a correct formulation of this problem.

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    $\begingroup$ Does this help? $\endgroup$ Mar 12 '12 at 14:51
  • $\begingroup$ Didn't read through all but seems it's not a solution to find the curve. Instead they choose the curve to be piecewise (2 pieces) linear and just search for some angles. They also have honda as the joint, but in this case, even if we choose the 2-piece-linear model, the lenghts of pieces would remain unknown... $\endgroup$
    – Džuris
    Mar 12 '12 at 16:13
  • $\begingroup$ This shape is described in C. Mack's 'Theory of the spinning balloon' (Q. J. Mech. Appl. Math. 11 no. 2, 196–207 (1958)) and also in J.A. Hanna's 'Rotating Strings' (J. Phys. A: Math. Theor. 46 no. 23, 235201 (2013), arXiv:1204.4941). $\endgroup$ Nov 6 '15 at 20:16
  • $\begingroup$ This is an extremely complicated dynamics problem that can't be solved in the minutes or even hours that people take to post an answer here. Be careful about trusting any solutions anyone offers regardless of how sophisticated the equations look. In other words, don't get baffled by BS. $\endgroup$ Jan 15 '17 at 14:50
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It's an interesting problem. I've tried to study it neglecting the air friction, which is likely to be nonnegligible. So let's parametrize the rope of length $L$ by the vector $(z(l), r(l))$, with $l$ being the curviligne coordinate along the rope and $(z,r)$ the cylindrical coordinates of the rope (I've omitted $\theta$, which can be assumed to be constant in the rotating frame [see my comment below this post for a justification of this omission]). In the rotating frame, the potential energy of the rope is $$E_p=\int_0^L \mu dl\left(gz-\frac12\Omega^2r^2\right)$$ and the definition of the curviligne integral gives the constraint $r'^2+l'^2=1$. So the problem can be translated into minimizing the following quantity : $$\begin{align} \int_0^Ldl \mathcal L(z,r;z',r';l) & & \text{with } \mathcal L(z,r;z',r';l)=\mu gz - \tfrac12\mu\Omega^2r^2 - \lambda(z'^2+r'^2). \end{align}$$ One should not forget (as I did in a first tentative answer) that the Lagrange multiplier $\lambda$ depends on $l$/ It is a standard Euler-Lagrange problem and its solution is given by $$\begin{align} \frac{\partial\mathcal L}{\partial z}-\frac{d}{dl}\frac{\partial\mathcal L}{\partial z}&=0& &\text{and}& \frac{\partial\mathcal L}{\partial r}-\frac{d}{dl}\frac{\partial\mathcal L}{\partial r}&=0& \end{align}$$ The equation in $z$ becomes $$\begin{align} \mu g&=-2\frac{d}{dl}(\lambda z')& \lambda z'&= -\tfrac12 \mu g(l-l_0), \end{align}$$ where $l_0$ is an integration constant. For the optimal solution $z'\le 0$, since replacing $z' by -|z'|$ can only decrease $\mathcal L$, ans we have $z'=-\sqrt{1-r'^2}$. One has therefore $$\lambda=\frac{\mu g(l-l_0)}{2\sqrt{1-r'^2}}$$. The equation in $r$ is then $$-\mu\Omega^2r=-2\frac{d}{dl}(\lambda r') =-\frac{d}{dl}\frac{\mu g(l-l_0)r'}{\sqrt{1-r'^2}}$$ We have then the following equation to integrate $$0=\frac{\Omega^2}{g}r(1-r'^2)^{3/2}-r'(1-r'^2)-(l-l_0)r''.$$ I don't know how to integrate it analytically, but when $l-l_0<0$, it seems that we can have oscillations.

Edited to add :

If one looks at the signs of $r'$ and $r''$ as function of $r$ and $r'$ for a fixed $l$, one can quickly draw some arrows in phase space (i.e.) and see that the flow goes "in circles" when $l<l_0$ and tends asymptotically to the curve $\frac{\Omega^2}{g}r=\frac{r'}{\sqrt{1-r'^2}}$ when $l>l0$. So one can expect a finite number of oscillations of the rope. But the whole question is then how $l_0$ relates to the parameters of the system.

Edit: In the limit of almost vertical rope ($r'\ll1$), the equation becomes $$0=\frac{\Omega^2}{g}r-r'-(l-l_0)r''$$ which is simpler. And indeed, Wolfram Alpha has an analytical solution using modified Bessel functions.

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  • $\begingroup$ The Lagrange condition should be $z'^2+r'^2+(r\phi ')^2 =1$, applied at all points on the curve, so there should be a Lagrange multiplier function $\lambda(l)$. Putting all of this into Mathematica gives horribly coupled, non-linear differential equations which seem to be difficult to solve. Also, I'm not 100% sure what are the correct boundaries conditions to apply (or even how many of them I need...) $\endgroup$
    – genneth
    Mar 12 '12 at 19:32
  • $\begingroup$ Nice try so far, I didn't even know which tools to choose here... If I got it right, the $r=\frac{1}{\Omega}f(z)$ already shows this is going wrong as this $r(z)$ has only one zero while the actual solution show at least two. $\endgroup$
    – Džuris
    Mar 12 '12 at 22:04
  • $\begingroup$ @genneth Thanks for the hint on $\lambda(l)$. I've corrected the answer accordingly and it seems more tractable. $\endgroup$ Mar 13 '12 at 11:20
  • $\begingroup$ You're still missing the tangential part of the problem though. Your set up does not allow a rope which coil around the axis (like the OP's picture does)! $\endgroup$
    – genneth
    Mar 17 '12 at 20:31
  • $\begingroup$ Also, you should note that the Lagrange multiplier field is actually the tension along the rope. I believe the complete solution is actually in terms of a differential-integral equation, as the boundary condition are such that you don't know the tension required at the top of the rope without knowing the configuration --- you only know that the tension must be zero at the tip of the rope and need to integrate back up. $\endgroup$
    – genneth
    Mar 17 '12 at 20:33
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I think the first part of the curve (the part between your hand and the stationary "node" where the curve crosses back over the vertical axis) would be described by the Troposkein Curve, or at least be extremely similar.

This curve would, I imagine, either fully describe, or at least be related to, any other segments between nodes if you had more than one.

The last "tail" bit at the bottom of the rope would be different, obviously, but you could probably get the shape using a similar method.

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I solved equations written by @genneth numerically (for initial condition I used a shooting method). Crucial part is that I did not use equation $$(Tr^2\phi')' = 0 $$ I.e. I supposed that curve is flat. My results is below. $\omega=10$$\omega=20$$\omega=40$$l=100$

NOTE that in 3 first pictured only $\omega$ changes. In the last picture I changed length $l$. This pictures and a lot of others let me to conclude that number of ``modes'' $N_a$ i.e. points where $r'=0$ is $$N_a \propto \sqrt{l}\ln{\omega}$$ for large $\omega$.

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This is not a complete solution but I'm making it a community wiki so that others can supplement and correct it.

Frédéric Grosshans is almost correct in the above, but misses a crucial part of the problem: the rope may coil around the axis. The general method of assuming a stable equilibrium shape and minimising the (effective) potential energy in a rotating frame is a good one, but perhaps one should be wary of assuming that such a stable equilibrium even exists.

Pressing on, we parametrise the rope as $(z(l), r(l), \phi(l))$, and the potential energy is: $$U = \int_0^L dl \; \rho \left(\frac{1}{2} \omega^2 r^2 - g z\right)$$ where $\rho$ is the density per length (apparently irrelevant in the end, but we will keep it to make sure that the units are sane) and we take the convention that $z$ increases downwards.

The constraint is that each small element of the rope must be at a fixed length, so we introduce a Lagrange multiplier field (this is a field theory in 1D!): $$L = \rho \left(\frac{1}{2} \omega^2 r^2 - g z\right) + \frac{1}{2}T\left(r'^2 + z'^2 + r^2 \phi'^2 - 1\right)$$ where $T$ is the Lagrange multiplier field (the name will become obvious, as will the random factor of 1/2).

Pushing through the Euler-Lagrange equations yields: $$\begin{align*} \rho g + (Tz')' &= 0 \\ \rho \omega^2 r + T r \phi'^2 - (Tr')' &= 0 \\ (T r^2 \phi ')' &= 0 \\ r'^2 + z'^2 + r^2 \phi'^2 &= 1 \end{align*}$$

The first equation encodes the meaning of $T$ as the tension in the rope. We may integrate it and obtain the quite obviously sensible $$Tz' = \rho g (L - l)$$ (using the boundary condition that $T = 0$ at $l = L$) which simply encodes that the vertical component of the tension must balance the weight of the rope below.

The remaining equations are 1st order in $z$ and 2nd order in $\phi$ and $r$. We have the coordinate setting equations that $z(0) = 0$ and $\phi(0) = 0$. There is a genuine extra parameter $r(0)$ which dictates the radius of the driving force at the top. However, one still needs another boundary condition for $r$ and $\phi$ (or possibly their derivatives) for the problem to be well posed. In addition, the condition for $T$ leads to a tricky integral equation, since the natural boundary condition is that $T(L) = 0$, and one would then need to integrate this to obtain the boundary at $l = 0$.

Finally, I should mention that this partial solution as it stands has one serious deficiency which may or may not be connected to the boundary condition problem above: the equations are symmetric for $\omega \rightarrow - \omega$, so apparently an available solution would have the rope pointing forwards of the rotating force. But perhaps this is a maximum of the energy rather than the minimum.

In any case, the problem seems highly non-linear, and I would be surprised (and delighted) if an analytic solution is possible. One might hope however that there are easy to understand regimes (apart from the trivial ones at $\omega = 0$ or $\infty$).

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  • $\begingroup$ The symmetry $\pm\omega$ is linked to the fact that the centrifugal force does not depends on the rotation direction. I think the only way to break this symmetry is to add friction. $\endgroup$ Mar 19 '12 at 14:58
  • $\begingroup$ It may be as you say in a comment to your answer above: the rope may just lie in a plane. Experimentally this seems true (just tried it :) ) but I haven't seen it yet from the maths alone... Any ideas? I'm thinking that $Tr^2 \phi '$ must have some physical interpretation which will help. $\endgroup$
    – genneth
    Mar 19 '12 at 16:38
  • $\begingroup$ Let's have a look at equation $(Tr^2\phi')'$ = 0 when integrated: $Tr^2\phi'$ = C = 0. Seems like identity. $\endgroup$
    – LRDPRDX
    Nov 25 '16 at 16:08
  • $\begingroup$ I think that your expression for potential energy is incorrect, more precisely the centrifugal term. It says that centrifugal force has direction against $r$ which is not true. $\endgroup$
    – LRDPRDX
    Apr 10 '17 at 5:47
  • $\begingroup$ There is an analytic solution in a closed form, see my solution below. $\endgroup$
    – bkocsis
    Jul 7 '21 at 19:29
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I have a probable guess that this phenomenon is somehow linked to (or is similar to) standing waves.

If you hold one end of the rope and do a horizontal simple harmonic motion (i.e. oscillate the end horizontally), a plane transverse wave will be created and when it travels down to the other end it will reflect back (as waves reflect at one free end of the medium), thus a standing wave is formed, with the free end being an anti-node. Depending on the length of the rope and the angular velocity of the oscillation, none or one or several nodes will form along the rope.

Consider the motion of the rotating rope as the result of two transverse waves traveling in two planes perpendicular to each other, and the standing wave explanation does make some sense.

I'm not sure about it -- just a guess.

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  • $\begingroup$ Indeed it is related to a standing wave, see solution below. The solutions are parameterized by the number of nodes. $\endgroup$
    – bkocsis
    Jul 7 '21 at 19:30
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We will show that the shape of the rotating rope is the Bessel function.

Let us adopt a rotating coordinate system in which the rope is at rest. We shall use cylindrical coordinates, $r$ and $z$ are the radial and the vertical coordinates, where $(r,z)=(0,0)$ denotes the point of suspension and $z$ increases downwards. The rope runs along the points $(r(l),z(l))$, where $l$ is the length along the rope measured from the free end, i.e. $l=L$ at the point of suspension.

The forces acting on an infinitesimal rope element of mass $dm$ are the gravitational and the centrifugal forces ($dm\, g$ and $dm\, \omega^2 r$) and the constraint forces that arise due to the tension in the rope that make sure that the line density remains $\eta = m/L$ everywhere along the rope, where $m$ and $L$ are respectively the total mass and length. The tension in the rope at $l$ exerts a force $\vec{F}_T(l)$ which is parallel with the tangent vector of the rope. For a rope element between $l-dl$ and $l+dl$, forces arise at the bounding points i.e. $\vec{F}_T(l+dl)$ and at $\vec{F}_T(l-dl)$.

The radial and vertical equations of motion are \begin{align} dm \frac{d^2r}{dt^2} &= dm \,\omega^2 r - F_T(l+dl) \sin\theta(l+dl) + F_T(l-dl) \sin\theta(l-dl) = 0 ,\\ dm \frac{d^2z}{dt^2} &= dm\, g - F_T(l+dl) \cos\theta(l+dl) + F_T(l-dl) \cos\theta(l-dl) = 0 .\\ \end{align} Here $\theta$ denotes the angle subtended by the tangent of the rope with the vertical, i.e. $\tan \theta = -dr/dz$. This generally changes with the position along the rope which we write as $\theta(l)$. The right hand side is zero given that the rope is at rest in the rotating frame. Note that the rope must be planar at rest, since no forces act in the azimuthal direction and so the rope tension cannot have an azimuthal component, the rope must have a zero derivative along the azimuthal direction.

Divide the equations with $2dl$: \begin{align} \frac{d}{dl} (F_T \sin\theta) &= \eta \omega^2 r, \\ \frac{d}{dl} (F_T \cos\theta) &= \eta g. \end{align} The vertical equation may be integrated as $$F_T \cos\theta = \eta g l. $$ This satisfies the boundary conditions since at the lower end $l=0$ so $F_T=0$, and at $l=L$ we have $\cos \theta=1$ implying that $F_T=\eta g L = m g$. Substituting in the radial equation and using that $\sin \theta = \tan \theta \cos \theta = -(dr/dz)\cos \theta$, gives $$\frac{d}{dl} \left(\eta g l \frac{dr}{dz}\right) = -\eta \omega^2 r. $$

In the limit that $\theta\ll 1$, we can approximate $l\approx L-z$. To simplify the equations, redefine the coordinate system with the substitution $z\rightarrow L-z$, so that $z=0$ is the lower free end of the rope and $z=L$ is the upper fixed end. This yields $$\frac{d}{dz} \left(\eta g z \frac{dr}{dz}\right) \approx -\eta \omega^2 r. $$ This approximation allows us to obtain an exact analytic solution in a closed form. In practice we find numerically that the exact solution is very well described by this approximation.

We now manipulate this equation to arrive at the Bessel equation. Introduce a new variable $x = 2 \omega \sqrt{z/g}$ so that $z = x^2 g/(4\omega^2)$. The chain rule implies that the derivative with respect to $z$ may be written as $$\frac{d}{dz} = \frac{dx}{dz}\frac{d}{dx} = \frac{1}{dz/dx}\frac{d}{dx} = \frac{2\omega^2}{g x} \frac{d}{dx}.$$ The constants drop out after substitution and we get $$\frac{1}{x}\frac{d}{dx} \left(x\frac{dr}{dx}\right) = r $$ or $$x^2 \frac{d^2 r}{dx^2} + x\frac{dr}{dx} + x^2 r = 0. $$ This is Bessel's differential equation defining the Bessel functions $J_0(x)$ and $Y_0(x)$, the general solution is a linear combination of these functions. Since $Y_0(x)$ diverges at $x=0$ and $J_0(0)=1$, the physical solution is proportional to the Bessel function of the first kind $J_0(x)$. Thus, the shape of the rotating rope is

$$r(z) = r_{\mathrm{end}}\, J_0(x(z)) = r_{\mathrm{end}} \,J_0\left( 2 \omega \sqrt{\frac{z}{g}} \right).$$

Here $r_{\mathrm{end}}$ is the radial coordinate at the loose end. The boundary condition at the suspension is that at $z = L$ the radial coordinate satisfies $r=0$, so we must require that $J_0\left( 2 \omega \sqrt{L/g} \right) = 0$. The zeros of the Bessel functions are well known, they are tabulated. Let us denote the $n$th zero with $x_n = \{2.4048, 5.5201, 8.6537, 11.7915, \dots\}$. The boundary condition is satisfied when $\omega = \frac{x_n}{2} \sqrt{g/L}$ for any $n$. Thus

$$r(z) = r_{\mathrm{end}}\, J_0(x(z)) = r_{\mathrm{end}}\, J_0\left( x_n \sqrt{\frac{z}{L}} \right).$$

For larger frequencies the rope has multiple nodes, i.e. places where it crosses $r=0$. The rope has $k$ nodes if the angular frequency is $\omega = \frac{x_{k+1}}{2} \sqrt{g/L}$, one node for $\omega = \frac{x_2}{2} \sqrt{g/L} = 2.26\sqrt{g/L}$, two nodes if $4.33\sqrt{g/L}$, and three nodes if $5.896 \sqrt{g/L}$.

In the limit, $\omega \gg \sqrt{g/L}$ we can use the asymptotic formula for the zeros of Bessel functions $x_{k+1} \approx (k- \frac{3}{4}) \pi $, so it has $k$ number of nodes when $$\omega = \frac{\pi}{2}\left(k + \frac{3}{4}\right) \sqrt{\frac{g}{L}}.$$ Solving for $k$ shows that the number of nodes for $\omega \gg \sqrt{g/L}$ is $$k=\frac{2\omega}{\pi}\sqrt{\frac{L}{g}}-\frac{3}{4}.$$ In this case, the asymptotic form of the Bessel function $$ J_0(x) = \sqrt{\frac{2}{\pi x}} \cos\left(x - \frac{\pi}{4}\right) + \mathcal{O}(x^{-3/2}) $$ implies that for large $\omega$ and large $z$ $$r(z) \approx r_{\mathrm{end}}\, \sqrt{\frac{1}{\pi \omega} \sqrt{\frac{g}{z}}} \cos\left( 2 \omega \sqrt{\frac{z}{g}} - \frac{\pi}{4}\right).$$

Note that we assumed that the top end fixed at $z=L$ satisfies $r(L)=0$. In the more general case, the boundary conditions are $r(L)=r_0$ and $r(0)=r_{\mathrm{end}}$. This implies that $J_0(2\omega\sqrt{L/g}) = r_0/r_{\mathrm{end}}$, and we get a similar solution for large $\omega$ if $r_0/r_{\mathrm{end}}\ll 1$.

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    $\begingroup$ That's a strong result! $\endgroup$
    – Džuris
    Jul 7 '21 at 22:16
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It will be some sort of hyperbolic function, and the force and length of rope will determine the frequency to predict the shape. The initial 'circle' at your wrist will define the constant force to explain the sin curve. This is not a very good explanation but if you were to define a hyperbolic function using a z axis it could be sinh z=(e^z-e^-z)/2.

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    $\begingroup$ The second curve doesn't seem extremely hyperbolic to me, would that be a function made of sinh and/or cosh pieces? $\endgroup$
    – Džuris
    Mar 12 '12 at 16:08

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