# Shape of rotating rope (lasso problem?)

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.

-
Does this help? –  Vijay Murthy Mar 12 '12 at 14:51
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... –  Juris Mar 12 '12 at 16:13

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.

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.

-
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...) –  genneth Mar 12 '12 at 19:32
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. –  Juris Mar 12 '12 at 22:04
@genneth Thanks for the hint on $\lambda(l)$. I've corrected the answer accordingly and it seems more tractable. –  Frédéric Grosshans Mar 13 '12 at 11:20
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)! –  genneth Mar 17 '12 at 20:31
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. –  genneth Mar 17 '12 at 20:33
show 1 more comment

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.

-

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 $\infty$).

-
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. –  Frédéric Grosshans Mar 19 '12 at 14:58
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. –  genneth Mar 19 '12 at 16:38