Shape of rotating rope (lasso problem?)

Question

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:

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:

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.

Answer 1

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.

Answer 2

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 $\infty$).

Answer 3

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.