On a space station a person wants to play skipping-rope. Although they cannot jump, they turn the rope at a constant angular velocity.
What would the shape of the skipping rope be, and could we set up an equation by taking a generalised case?
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2$\begingroup$ Possible duplicate of Shape of rotating rope (lasso problem?) $\endgroup$– sammy gerbilOct 14, 2016 at 19:36
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1$\begingroup$ @sammygerbil Nope, this was a question that I asked my professor last week. And I'm thankful you have pointed this out now, I'll take a look at it as it would be more helpful. $\endgroup$– Spoilt MilkOct 14, 2016 at 19:42
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$\begingroup$ Have you made an attempt to solve this already, Naveen? If so, please post your calculation, either in the Question or as an Answer.... Yes, the lasso problem includes gravity and has one end loose. In the case of the skipping rope both ends are fixed and gravity is neglected. $\endgroup$– sammy gerbilOct 14, 2016 at 19:46
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$\begingroup$ @sammygerbil I did attempt to solve by initially considering a Troposkein equation and tried to make approximations to it by circular arc segments. I then found this paper online, where a similar calculation has been done, prod.sandia.gov/techlib/access-control.cgi/1974/740100.pdf $\endgroup$– Spoilt MilkOct 14, 2016 at 20:12
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2 Answers
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For obvious reasons the curve is called the Skipping Rope Curve. It is also known as the Troposkein . A mathematical expression for it is more difficult to derive than in the case of a stationary rope sagging under gravity, for which the result is the Catenary.
For the catenary, the external force on each element of the rope is gravity, which is the same on each element. For the rotating rope, in the absence of gravity, the "external force" is the fictitious centrifugal force, which is proportional to the distance of the element from the rotation axis. The result is the Jacobi Elliptic Function denoted 'sn'.
answered Oct 14, 2016 at 0:05
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$\begingroup$ DOI: dx.doi.org/10.1088/0143-0807/28/2/009 $\endgroup$– Qmechanic ♦Oct 14, 2016 at 20:13
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$\begingroup$ @Qmechanic : Same as the ResearchGate paper cited in my answer, I think. The latter version is potentially available on request; the IOP version has restricted access. $\endgroup$ Oct 14, 2016 at 20:17
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$\begingroup$ Well, RG needs login, which is not convenient. $\endgroup$– Qmechanic ♦Oct 14, 2016 at 20:23
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$\begingroup$ IOP also requires individual or institutional login, or payment of £24 or $33 fee. RG requires only registration, no fee. $\endgroup$ Oct 14, 2016 at 20:27
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$\begingroup$ Yeah, but RG spams. More importantly, it is not a permalink, and hence susceptible to link rot. $\endgroup$– Qmechanic ♦Oct 14, 2016 at 20:29
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The situation is not quite the same as a catenary. In a catenary, the rope is in a uniform gravitational field. In this situation, once steady state is reached, the rope could have a centripetal force outwards, which would mean it's "gravity" increases linearly with distance from the center.
Consider the derivation given in the wikipedia article. You'd need to change $T\sin\phi = \lambda gs$. But it changes to a number that's proportional to the distance of the rope from the astronaut's hand. That's actually a hard number to parameterize, because it's needs the solution we are trying to obtain!
I'll leave it to others to figure out the solution, but qualitatively, you would expect it to be a little more pointy than a normal catenary, since the middle of the rope would be effectively heavier.
