It is easy to be a lord of the rings

in #mathematics6 years ago (edited)

I am currently still on holidays but today I am at my partner's place to take a break from travelling. I wrote about half of this before I went away so I thought I just finish it today.

I saw this awesome gif a bit over a week ago: Lord of the rings (I didn't embed it because of possible copyright issues). In case you are too afraid of hitting links here is short description: He swings a ring on a string in such a way that after a couple of (quasi-)rotations it hooks on to the hook. It is quite impressive what he does. But it might be a little bit less impressive than you might think.

Mathematicians like models

To get a better understanding let's create a mathematical model of the ring-string motion. Observe that the string on which the ring is attached is almost straight. So we might as well model it as a rigid stick. The ring at the bottom is much more heavy as the string. Hence, we can just assume that there is only a mass puling at the bottom of this stick at that the stick itself has zero mass. In short we have the following set-up:


pendu.png

In addition, we neglect air resistance, friction, spinning of the ring. The equations describing the motion are then given by the governing equations of the so-called spherical pendulum since the motion is restricted to a sphere centered at the support of the pendulum [1]. It is a well-studied system in most university math curriculums.

We have the model but what is the question?

We want to answer the following question: how easy is it to navigate the pendulum to a certain location. Recall that all the motion is actually restricted to a sphere which is centered at the support of the pendulum. So we can only consider locations on this sphere. So the question then becomes: how easy is it to navigate to a certain location on the sphere.

The computer, your best friend

With the magic of the computer we can trace the motion that the pendulum makes. To do that we need to specify the starting point of the pendulum and its starting velocity. Below you can see the traced motion of the end point of the pendulum for a specific starting point and velocity after 10 seconds and 100 seconds.


g4260.png

So it gets close to all the points on a certain surface of the sphere. Furthermore, if you change the starting point and velocity a bit then the traced motion will look very similar. So what does this mean? Well, it means that precision is not required to send the pendulum to a specific location on this blue colored part of the sphere. Or in terms of the original problem, to get the ring on the hook you don't have to release it with a very specific initial velocity or at a specific position.

A predictable conclusion

In conclusion, this system behaves in a nice predictable way which allows you to control the outcome easily. There are systems for which a small change in the initial set-up can have completely different outcomes (when you wait sufficiently long). These systems are called chaotic and can be immensely difficult to control. However, sometimes patterns immerge for chaotic systems which create a certain predictability which is a bit different from our ordinary understanding. But I will leave this to another time :)


Technical appendix

So the underlying math here is mostly hamiltonian mechanics with a bit of topology and analysis. It is quite technical so only read this part of you enjoy that type of thing.

Almost all orbits (in a Lebesgue measure theoretical sense) for the spherical pendulum are topologically equivalent to a dense set of a 2-torus. Observe that these orbits are not periodic. To see this requires a special normal form transformation, see for example [1]. To hit a point on a 2-torus with a dense set can be pretty difficult. So instead consider a open neighbourhood of a point on the 2-torus. This set will interset the dense set by the definition of dense set. The set given by all periodic orbits forms a dense set in the phase space. But this dense set forms a set of Lebesgue measure zero. So dense orbits are in the majority. All of this is connected to the topological versus the analytical properties of the rational and irrational numbers in relation to the real line.


References

[1] R. Cushman, L. Larry, Global Aspects of Classical Integrable Systems. Birkhauser, 2015



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I'm not a mathematician, and i never studies physics, but I do enjoy your explanation about the "lord of the rings", for us, it looks cool and we might attempt to do the same thing blindly. But for you,it's all physics. Love the things you write, it's not too difficult to understand too since I've learned some basics in science.

Thank you for the educational post

Thankies for the support and the kind words. I am always happy to teach math/physics :)



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How anyone can hate math...

Since seven hate nine (789), seven hate math :D

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