## Tuesday, June 8, 2021

### Enforced Rotation of Tarzan Rope (Solution)

The semester has ended, and alas, nobody posted a solution to the difficult problem I posed months ago (see problem here).  In short, we have a rope that is suspended from the top and is being moved along a circular path in the horizontal plane with constant angular velocity.  Aerodynamic effects shall be neglected.  We are seeking a lateral deflection function.  Here is my solution...

With a problem such as this, we must begin with a physical model.  My hand drawing is seen below (I apologize for the crude sketch, but the summer me exerts less effort):

The solid blue line represents the rope whose profile we aim to determine.  At some location (x, y), we will apply Newton's second law to a single mass element dm.  My free body diagram is on the right side.  There are two external forces acting on the element; one is real and the other, a pseudo-force.  The real force, dFg, is gravitational, while the centrifugal load, dFc, is a pseudo-force as it is effectively an inertial term.  Finally, tension acts internally, pulling this element in both directions tangent to the rope's profile at (xy).  The upward pointing tension is (correctly) assumed slightly higher than the downward one, by some amount dT.  One useful, though limiting facet of the assumed model, is that, at a given vertical location x, each element simply displaces horizontally - in reality, it also shifts up vertically, ever so slightly.  This simplification allows an elegant solution, but whose accuracy is limited as we shall see.

Applying Newton's second law to that element on both axes, we get:

dFc = dTsinθ                                                                                                    (1)

dFg = dTcosθ                                                                                                   (2)

We can express the elemental forces as:

dFc = dm(ω2y)                                                                                                 (3)

dFg = dm(g)                                                                                                     (4)

The angular velocity of the enforced circular motion is denoted by ω.  If we divide equation 1 by equation 2 and then divide equation 3 by equation 4, we get the relationship

tanθ = ω2y/g                                                                                                     (5)

The key realization to move forward is that the derivative dy/dx = tanθ.  This yields the governing equation:

dy/dx = ω2y/g                                                                                                    (6)

The particular solution to equation 6, after having applied the boundary condition y(0) = R0, the radius of the enforced circular path, is given by:

y(x) = R0exp(xω2/g)                                                                                          (7)

This solution is quite interesting.  We first notice that the density and area of cross-section of the rope have no effect on the shape it takes.  This is not surprising because both external forces were proportional to the elemental mass.  The more important takeaway here is that the lateral deflection becomes exponential.  The faster we spin the top of the rope, the more dramatic the curve.  This makes sense, but there is a serious flaw: the rope has a finite length.  As this function is exponential, there is no limit to the lateral deflection it describes.  As the imposed angular velocity increases, the lateral deflection can quickly become greater than the total length of the rope, which is physically impossible.

I suspect that I ran into this problem because, in my original model, I neglected the gain in altitude that a particle driven laterally inevitably experiences.  For fun, I included this effect in a subsequent attempt.  After a page of work, I saw that numerical tools would be required to solve.  Again, it's summer, and I am content to move on and not pursue this problem further, especially when a closed solution appears impossible.

Equation 7 may be a good approximation of the rope's profile for fairly slow rotation rates.  An experiment is difficult to conduct for multiple reasons. While air effects lead to a three dimensional profile, so to would inertial effects when it comes to establishing planar motion.  In principle, it may be possible to enforce the theoretical equilibrium configuration as well as a uniform angular velocity for all string elements, but it is not practical.  Failure to do this would inevitably lead to a helical 3D profile.

You may be thinking I did all that work for nothing.  It is important to realize that simplified approaches teach us a lot about complex problems.  They give confidence to the more strenuous, complex solutions that follow them.

And now, out of my cave.  Summer beckons.