Revolving pendulum contradicts laplace determinism

Question

A question came once into my mind

There is a pendulum having length of string 1metre and was initially at rest. Now the point of suspension suddenly starts to move in uniform horizontal circular motion of radius 10 cm and angular velocity 1 second inverse. The bob of the pendulum will initially show some erratic motion but after some time it itself will also begin to move in a uniform horizontal circular motion with same angular velocity. Your task is to find the radius of this circular motion of bob.

The question appeared simple at first sight.

This is how I thought the motion will be. It looks like a part of a conical pendulum.It is similar to that seen while riding in merry-go-round.(centrifugal causes the horse to move slightly away from the axis). Note that at any instant the bob and point of suspension has same angular position. The surface of revolution of the string is that of a frustum. Now I can easily apply Newton's laws of motion and solve it. I did got an answer but I don't remember it and it's irrelevant, if you are convinced that this is a possible solution.

I asked this question to my friend and he solved and gave a different answer. I thought he did mistake but I was surprised to see his approach. This is how he thought the motion should occur

Here the bob has a slightly smaller radius. The surface of revolution of string is in the shape of 2 inverted cones joined at vertex. At any instant the angular position of bob is opposite to the point of suspension. Here also we got a second (also valid) answer to the question.

You can try to perform this experiment practically by attaching some heavy weight to the end of a rope and then revolving it by hand. Someone can make a good multiple option correct question from this giving both the answers in the option but most of the people will only be able to think of 1 answer.

Anyways,my query is regarding laplace determinism which states that if all the conditions and parameters of a system are known at an instant then the values of these parameters can be accurately predicted at any other instant. There seems to be a violation of that here because initial conditions were known and we are getting 2 possible answers. How is this possible?

Answer 1

Petr's answer is making some good points, but there is actually a much simpler, philosophy-less answer that largely resolves the issue (with a caveat that I'll get into later). Essentially, what happens is this: if the exciting motion is slower than the (linearized) natural frequency of the pendulum, then it will settle into the merry-go-round configuration. If it's faster, it'll go into the double-cone configuration.

(That's assuming there is some kind of friction in the system, which there always is. If there was no friction, then it would never settle into either of these configurations, but instead keep forever wobbling irregularly!)

A simple qualitative reason why it behaves like this is an energy argument: at slow frequency, the merry-go-round has the pendulum hanging almost straight down, i.e. the mass is low down (low potential energy), and the kinetic energy is anyways low at slow speed. Whereas at high frequency, the merry-go-round has the bob moving on a longer path, thus with faster speed and higher kinetic energy.
Meanwhile, the double-cone always requires the pendulum to cross over the center, which requires some potential energy even at very low frequency. But at high frequency it can make the bob move at an arbitrarily small radius around the center, and thus minimise kinetic energy.

The system will tend towards the configuration with lower total energy, because any air friction etc. losses will consistently remove energy. Although the exciting circular motion can both add and remove energy, it has no inherent preference, and the state where it only replentishes energy lost to friction around the lowest-energy steady state is most stable.

All of this rather hand-wavey argument can be easily made more rigorous in the special case where the pendulum is much longer than the scale of the initial motion, because then the pendulum behaves as a simple harmonic oscillator, and the solution is that of the classic driven oscillator, known by different names in different applications. The electronic analogue is the LC filter, which has a well-known behaviour of flipping the phase as you cross the resonance frequency. This transition is exactly analogue to the pendulum ending up in the two different states, depending on the frequency with which you excite it.

Now, the caveat: there is an important difference with the pendulum, namely that it's (for non-infinitesimal displacements) not linear, and the nonlinearity can actually stabilise the merry-go-round even at high frequency. The extreme case is where it's centrifugeing the bob outwards at almost horizontal angle. This will happen in particular if you start at low frequency and make it ever faster, provided friction is sufficiently small (as it is with a heavy bob in air, but not with e.g. a feather or in water). I believe the nonlinearity can also stabilize the double-cone at frequencies below resonance, but not sure about this. These effects are what the other answers mean when they say it depends on the initial condition. Indeed, in the edge case at almost exactly the resonance frequency I would expect that tiny changes in the initial state can cause the pendulum to end up in the less energetically favourable configuration. But still, the movement is deterministic, and with reasonably accurate start conditions you can reliably achieve long-term predictable movement, unlike with the textbook chaotic systems like the Lorenz attractor.

Answer 2

There may be at least two different reasons for this difference. (I intentionally will not discuss which of them is applicable to this particular setup, but will present two reasons that may be at work in many similar situations.)

The first reason is that you consider only the stabilized part of the movement, while omitting the "[t]he bob of the pendulum will initially show some erratic motion" part. It may be that the resulting stabilized motion depends on the nature of suspension point movement during this phase.

You say that "[y]ou can try to perform this experiment practically by attaching some heavy weight to the end of a rope and then revolving it by hand", I presume that you mean that it is possible to intentionally get both configurations, and probably, with some experience you will learn how exactly you need to start rotation with your hand to get a selected configuration. Then this "how exactly you need to start rotation" will represent exactly the difference that you are looking for. You will have two different ways of moving your hand, and two different outcomes, and you will be able to intentionally choose one of them.

So the first reason is that there may be some observable differences in suspension point movement during initial stage that results in different configuration. (In particular, the acceleration of the suspension point when it moves from rest to circular movement may be important.)

Of course, this leave open the question "what will happen if the suspension points moves in an ideal circular trajectory". It is possible that only one specific configuration can be achieved, but to understand which one you will need to analyze this starting period. Nor your analysis (that looks only at stabilized movement), nor your experiment with hand (where you don't have precise control on suspension point movement) can be used to answer this question.

A simple example to illustrate this: consider an ideal sphere on an ideal horizontal surface. Let's mark some point on a sphere, and then push the sphere to start rolling. The sphere will stop at some point due to friction; let's ask: where the marked point will be on the sphere when it stops?

We can do a stationary analysis, but (assuming the mark itself is of zero weight) we will not get any definite answer; any orientation of the sphere will be a possible stationary solution. But if we analyze the movement and the initial position of the mark, then it will be rather easy to find where the mark will be located at the end.

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And the second reason is deterministic chaos. In some systems, even minor perturbances in the initial conditions and/or external conditions during system motion can result in radically different results. In this case, the choice of specific configuration of resulting movement will depend on minor perturbances in initial state of the pendulum, or on minor external influences (wind, etc.) that we usually neglect in our analysis. So if you know the initial conditions and external influences with a very very very good precision, you know the result, but if you don't know them to a needed very high precision, the result can be seen as random.

Like above, this leaves open the question "what will happen in an ideal situation", but the difference from the first reason described above is that this question becomes unpractical in this case, because you never have an absolutely precise situation.

(Not to mention that the required "very high precision", if for example we talk about spatial position, can easily be much less than the size of an atom, which is beyond the limits of applicability of classical mechanics. That is, the result may radically change if you change the initial position by a fraction of atom size; but you can not require a classical mechanics initial state to be specified to that precision.)

This is very similar to the flipping coin problem. While the movement of the coin can be seen as absolutely deterministic, still the outcome can not be predicted in a reasonable way, because small alterations in the initial state can grow large enough to change the result.

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UPD: I would also add a third possibility, namely that one of the configurations can be unstable. That is, it will be a valid solution of the equations, but any small variation will lead to the system leaving that configuration. Moreover, this means that such a configuration will not be reachable in the first place. You did not do any stability analysis, but if your manual experiment shows that both configurations are possible, then most probably both are stable.

Answer 3

if all the conditions and parameters of a system are known at an instant then the values of these parameters can be accurately predicted at any other instant. There seems to be a violation of that here because initial conditions were known and we are getting 2 possible answers. How is this possible?

The initial conditions for the two cases are different. There is no requirement that the prediction be the same for two different initial conditions.

Answer 4

Here is an answer that's supposed to be short and to the point.

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While the pendulum has two different, stable modes of operation, you need to distinguish that from the determinism question: While both modes are stable once the pendulum enters them, they require different initial movements of the suspension points to be entered. The later is what determinism is all about: Same input => same output. It's irrelevant that you can get a different stable mode of operation with a different initial(!) input, determinism only applies if the entire inputs and setups are equal.

More specifically to the question, with the provided input instructions, only the merry-go-round mode can be reached. Why? Well, the pendulum has a frequency of about 1Hz, and the angular velocity is 1/s which translates to one revolution in Tau seconds ($6.28s$). That means that the bob will basically follow the motion of the suspension point during the first revolution and move outwards as it picks up momentum. The key here is really the magnitude of the involved numbers. If the pendulum were longer, or the rotation frequency higher, we couldn't be as sure which mode is reached. But with the pendulums' frequency being significantly higher than the rotation frequency, the initial motions of the pendulum are qualitatively unambiguous.

Answer 5

Now the point of suspension suddenly starts to move in uniform horizontal circular motion

There's no any suddenly, that's the problem which you have missed. There can be many suspension point transitions from a steady state to a circular-motion one, some of them :

Depending on how suspension point transits from a steady to circular motion state, transition speed dynamics, etc. ,- you can get different pendulum movements outcome. Each and every slightest movements of your hand - matters.