Explanation of the 'chain fountain': some doubts

[voko]

I don't disagree with your math given those assumptions, but ignoring the weight of the chain in the circular part of the motion seems just wrong. You could estimate the relative size of the weight and the tension, from the kinematics of the real world experiment.

From the video, the radius of curvature is about 3 cm, and the speed is about 5 m/s. That makes normal acceleration in the arc about 833 m/s², a lot more than the gravitational acceleration. The uncertainty in the radius of just 1 cm is greater than the contribution of gravity, and so is the uncertainty of 1 m/s in speed. The fluctuation in the radius of curvature due to the clearly visible instabilities in motion will have a greater effect than gravity, so taking the smaller effect into consideration while ignoring the bigger seems strange to me.

 

[JollyOlly]

Since the actual shape of the chain is a continuous curve, not an arc joined to two straight lines, it makes no physical sense to suddenly ignore the weight at the point where the curved part is assumed to start.

I think you have an interesting point here. If you look at Mould's video closely you will see that the rising part of the chain is fairly straight; then the chain makes a sharp right angle bend followed by a more gradual arc into the falling section. Now all of us (myself included) have been assuming that the chain can be divided into three sections - straight up, arc, straight down. Perhaps we should consider a model which is divided into four sections - straight up, sharp 90deg curve, moderate 90deg curve, straight down. Would that make any difference to your analysis Voko?

 

[AlephZero]

From the video, the radius of curvature is about 3 cm, and the speed is about 5 m/s.

OK, so we can use that as a basis for formulating a more controlled situation which could also be done as an experiment.

Consider the chain running over a frictionless pulley of radius r, at constant speed v, and find the conditions for the chain to "lift off" from the pulley. I expect that would also shed light on the difference between the "macaroni" and "balls on a string" experiments.

 

[voko]

Consider the chain running over a frictionless pulley of radius r, at constant speed v, and find the conditions for the chain to "lift off" from the pulley.

Lift off meaning it is no longer acted on by any force from the pulley? How is that different from my previous analysis?

 

[voko]

Perhaps we should consider a model which is divided into four sections - straight up, sharp 90deg curve, moderate 90deg curve, straight down. Would that make any difference to your analysis Voko?

If we keep ignoring gravity, that does not affect anything. Tension is still constant. If we want to take gravity into account, that is no longer true. Tension will depend on location in the chain. The important question then is whether tension at the ends of the arc, whatever the arc is, but provided that the ends are at the same height, is the same. That I am not sure about generically yet (even though I am inclined to think so, intuitively). Perhaps we could agree on some particular shape for the arc and see whether that is the case for that shape.

 

[AlephZero]

Lift off meaning it is no longer acted on by any force from the pulley? How is that different from my previous analysis?

If your analysis is correct, there is no difference. It could also be extended to the "balls on a string" experiment by taking the pulley as a polygon instead of a circle.

But it is a much more controllable system to experiment with. Given the mass per unit length of chain, you can set the chain velocity and tension to any values independently, by choosing the chain lengths on each side of the pulley. It removes the unknown initial conditions caused by "throwing" the end of the chain out of the pot.

The fluctuation in the radius of curvature due to the clearly visible instabilities in motion will have a greater effect than gravity, so taking the smaller effect into consideration while ignoring the bigger seems strange to me.

There are (at least) two ways to interpret the instabilities. The simple way is to assume they are stable perturbations from a steady state solution, and start by looking at the steady state only.

Another possibility is that this is an inherently chaotic system, and the "steady state" is just the visual appearance of an attractor. That might lead to a model which is fundamentally different from a chain over a pulley.

 

[voko]

Some more analysis with gravity taken into account. Using the equations from #26. Dot-multiplying both sides of the second equation with itself, the result is $$ T'^2 + \left[ T - \lambda v^2 \right]^2 \kappa^2 = \lambda^2 g^2 $$ where the prime denotes differentiation w.r.t. $s$. It follows immediately that constant tension under gravity is possible only when curvature is constant. If curvature is zero (straight line), tension is arbitrary. If it is not zero (circular motion), $T = \lambda (v^2 \pm gR)$, where $R$ is the radius of curvature.

To investigate more general motion, let $S = T - \lambda v^2$ and that becomes $$ S'^2 + \kappa^2 S^2 = \lambda^2 g^2 $$

In the case of a chain "lifting up" while moving uniformly over a pulley, the curvature is constant, which yields $$ S'^2 + {1 \over R^2} S^2 = \lambda^2 g^2 $$ where $R$ is the radius of the pulley.

The general solution is given by $$ S = \lambda g R \sin \left( {s \over R} + \alpha \right) $$ Let $s = s_1$ be the right-most point of the pulley; $s = s_2 > s_1$ the left-most. Obviously, $s_2 - s_1 = \pi R$ hence $$ S(s_2) = \lambda g R \sin \left( {s_1 + \pi R \over R} + \alpha \right) = - S(s_1) $$ Let $T_0 = \lambda v^2$ be the tension in the gravity-free environment. Then $$T_1 = T_0 + S(s_1) \\ T_2 = T_0 - S(s_1) \\ |T_1 - T_2| = |2 S(s_1) | \le 2 \lambda g R $$ Under the condition that $gR \ll v^2$ we obtain that the difference in the tension on the opposite sides of the pulley is negligible as expected.

 

[JollyOlly]

I have been giving your arguments serious consideration but I remain convinced that the impressive mathematics you quote is not applicable to the dynamic situation which we are trying to explain and that the tensions at the top of the two straight sections of the fountain do not have to be equal.

Lets imagine a simple thought experiment.

Imagine a very long chain like Steven Mould's situated in deep space. The chain is straight, stationary and has no tension in it. Imagine now that you give the free end of the chain a violent wiggle. I am sure you will agree that some sort of wave will be generated which will travel at a finite speed v down the chain. (I do not know a formula for this speed but it is obvious that it cannot be the classical √T/λ because T is zero)

Now for the clever bit.

Imagine taking hold of the free end of the chain and move it parallel to and in the direction of the rest of the chain at a constant speed of 2v. This will create a dynamic loop which travels along the chain at a speed v. Since waves cannot travel faster than v, no influence can reach the stationary bits of the chain before the loop gets there so the chain enters the loop with tension T₁ = 0.

It is clear, though, that momentum is being created at a rate of 2λv² so the tension in the moving part of the chain T₂ must be equal to this. (It is also worth pointing out that the work done by this force is exactly equal to the KE gained by the chain so there is no loss of energy in this process - as you would expect.)

The connection with the chain fountain will become clear if you imagine yourself traveling along beside the loop at a speed v. The loop in the chain will appear to be stationary; chain will enter the loop with zero tension and exit with tension 2λv².

A complete explanation of the dynamics of the chain fountain must include an analysis of the way waves travel along an unsupported chain and such an analysis will not only generate the correct relation between the various heights involved and the speed of the chain, it will also explain how those fascinating curly loops form in the chain and why they are so stable.

 

[John Biggins]

Brinx - you analysis looks clear and correct to me. I think it would be cleaner with h_3=0, so that the black-box entails the entire portion of the fountain above the pot. Taking that forward, you finish with

"To solve for v or L independently, we would need a separate piece of information besides what I've discussed here it seems."

To do this you need to work to treat the pot as a second black box to work out the tension required just above the pot to draw the chain out at velocity v. The pot contains a long section of chain of length W. The forces on the black box are thus the tension (T) pulling up, gravity (\lambda g W) pulling down, and R the reaction force from the table/hand supporting the pot. The normal assumption is that the reaction force supporting the pot equals the weight of chain in the pot, so R=\lambda g W and the net force is given by T. There is only traffic out of the box, so a the momentum out of the box in unit time is \lambda v^2, so f=dp/dt requires T=\lambdda v^2.

Combining that with your equation

2 * lambda * v^2 = 2T + L * lambda * g

gives L=0, i.e. no fountain.

To fix this you need to revisit R=\lambda g W. If you set R>\lambda g W you will get a fountain. That is the anomalous push from the pot.

In energy terms, if we maintain R=\lambda g W and hence L=0, then taking your other equation T=\lambda g h_1 and combining it with my T=\lambda v^2 gives v as
v^2=g h_1

If you consider a unit length of chain going round the fountain, it releases gravitational energy \lambda g h_1=\lambda v^2, but only gains kinetic energy (1/2)\lambda v^2. Thus half the gravitational energy released is dissipated into modes other than KE. The anomalous reaction force harnesses this energy, turning it back into linear KE and enhancing the speed of the fountain.

 

[John Biggins]

JohnOlly

Your thought experiment is a nice one, but you've made a mistake in your analysis. As you say, you pull the end at a speed 2v, so after a time t, the end has moved a distance 2vt. Also as you say, the loop moves at v, so the amount of chain that is moving after a time t is 2vt-vt=vt. It has momentum (\lambda v t)*(2v)=2\lambda v^2 t, so as you say the force must have magnitude 2\lambda v^2. It has KE (1/2)(\lambda v t)*(2v)^2=2 \lambda v^3 t. However, the total work done by the force is given by force*distance-moved-by-force = (2\lambda v^2)*(2 vt)=4 \lambda v^3 t. Thus half the work done by the force has vanished in the "picking up" process.

Your setting the tension in the straight part of the chain to 0 is, I think, the source of your problem. The argument you quote is for transverse wave speeds, but tension is transmitted by longitudinal waves, and the wave-speed for longitudinal waves is infinite in an inextensible string.

Finally, a quick note on "those fascinating curly loops form in the chain and why they are so stable". This is easy to explain. The tension near the top of the fountain is T=\lambda v^2. Transverse wave-speed is Sqrt[T/\lambda]=v. Therefore waves propagating backwards along the fountain appear frozen in space.

 

[JollyOlly]

JohnOlly

Your thought experiment is a nice one, but you've made a mistake in your analysis. As you say, you pull the end at a speed 2v, so after a time t, the end has moved a distance 2vt. Also as you say, the loop moves at v, so the amount of chain that is moving after a time t is 2vt-vt=vt. It has momentum (\lambda v t)*(2v)=2\lambda v^2 t, so as you say the force must have magnitude 2\lambda v^2. It has KE (1/2)(\lambda v t)*(2v)^2=2 \lambda v^3 t. However, the total work done by the force is given by force*distance-moved-by-force = (2\lambda v^2)*(2 vt)=4 \lambda v^3 t. Thus half the work done by the force has vanished in the "picking up" process.

Yes - how silly of me! I knew when I wrote the post that I shouldn't have mentioned energy as that was not my point at all. All I wished to demonstrate was that you can devise situations in which the tensions at the two ends of a freely flowing loop of chain can differ. (In fact, if you consider the situation from inertial frames other than the one in which the chain is initially stationary, you can get any answer you like for the energy efficiency of the process - but please let us not go down that road!)

You yourself have admitted that the tension in the moving arm is 2\lambda v^2 and that the rate of change of momentum is the same. This can only be true if the tension in the other arm is zero - which is my point.

I don't think your point about transverse and longitudinal waves is fatal to my argument either. I never said that the chain had to be inextensible but since the motion that I am interested in is transverse, I don't really care how fast longitudinal waves travel down the chain.

 

[John Biggins]

JohnOlly

I think we can all agree that chain flowing round a loop should not spontaneously loose energy, yet if you look at your situation, in either frame, work done by the force is being lost, so something is wrong. To fix it, you need to let there be a tension in the stationary part of the chain. Indeed, an equal tension to that in the moving part. This is allowed because tensions can propagate infinitely quickly as longitudinal waves.

"You yourself have admitted that the tension in the moving arm is 2\lambda v^2 "

I didn't mean to. It is if you set the tension in the stationary part to zero, but then, as I showed, you have an energy problem. If you instead ask what must that tension be to conserve energy, you will get the by now familiar result T=\lambda v^2 throughout the chain.

 

[voko]

Moving parts of the chain at 2v, where v is assumed to be the speed of propagation of a material wave in the chain makes the material wave a shock wave. Which may be interesting in itself, but is irrelevant for the problem at hand.

 

[CWatters]

It's also interesting that the pile of chain ends up displaced horizontally relative to it's starting position (despite gravity only acting vertically). Cue someone to claim this violates one of Newton's laws :-)

 

[John Biggins]

voko

you might be interested in some extra work I have lodged at

http://arxiv.org/pdf/1401.5810v1.pdf

which does the analysis balancing gravity, tension and centripetal acceleration more carefully. It turns out, rather beautifully in my view, that the stable trajectory of a chain moving along its own length under gravity is a catenary, much like the one it would hang in without motion, but with the additional possibility of an inverted catenary, which is what is observed in the chain fountain, particularly if the pot is tilted.

 

[voko]

John, thanks for sharing that, very interesting.

The funniest thing is that just yesterday I realized that the second equation in #23 could be analyzed in the Frenet frame. When dot-multiplied with $\vec \tau$, it yields $$ T' = - \lambda \vec g \cdot {\partial \vec r \over \partial s} = \lambda g y' $$ When dot-multiplied with $\vec n$, it yields $$ \left[ T - \lambda v^2 \right] \kappa = - \lambda \vec g \cdot \vec n $$ Which are the same equations you obtained by a more direct mechanical consideration. I should have analyzed that further :)

 

[John Biggins]

voko

I think those are just beautiful equations. Consider what happens in the absence of gravity (g=0) -> they tell you a chain flowing along its own length in the absence of external forces has constant tension T=\lambda v^2, and that it can flow around any shape with a perfect balance of tensile forces and centripetal acceleration. I find that very unintuitive - I'd have expected centrifugal effects to try and smooth out any sharp bends in a flowing chain, but apparently not. Then try re-imposing gravity and eliminating v. You now have the statics equations for a chain hanging under gravity, which is well known to be solved by a catenary shape and a non-constant tension. Then put velocity back in, and you see that the constant velocity just adds a constant offset to T of \lambda v^2, and the old solution still works, so the chain flows round a catenary. Finally, consider replacing gravity by any position dependent force and eliminating velocity. We now have a statics problem we can't solve for a chain hanging in a general force-field, but whatever shape the solution is, if the chain starts flowing along its own length the same offsetting trick will apply, the tension will rise everywhere by \lambda v^2 but the chain will flow around the same shape it adopted whilst stationary.

Curiously this offsetting trick was first seen in the 1854 cambridge mathematics exam!

 

[JollyOlly]

OK I give in!

The tension in a chain flowing along its length in an arbitrary curve is constant and equal to λv².

It wasn't Voko's mathematics that persuaded me but John's refutation of my thought experiment. When you start pulling the chain along its length, each bead in the loop has to be pulled sideways. But this sideways motion will inevitably result in a force acting on the next bead which has a large component parallel to the chain. The next bead (which I am claiming is still stationary because the loop hasn't reached it yet) can only resist this force if it is held in place by a force from the next bead further along - ie there must be a tension in the stationary chain.

I am grateful to you both for putting me right. I am still, however, finding it difficult to believe because the implication is that John's 'anomalous force' must support the whole weight of the rising section of the fountain = λh₂g and none of the proposed mechanisms for this force seem to me to be at all convincing.

John, do you think that a perfectly flexible and frictionless snake carefully coiled into a helical coil inside a smooth vertical cylinder would produce a fountain?

 

[Brinx]

We did some more experiments here at the department, and we got a couple of interesting results.

What we did: we used a ball chain (2.4mm ball diameter) of several meters in length, and piled it into a compact heap onto the middle of a table. We let the end of the chain dangle over the table edge, in such a way that there was a horizontal chain segment of ~30 cm between the piled chain segment and the edge of the table. The end of the chain was pulled gently down until the chain started to drop by itself.

What we observed: While initially being pulled away from the heap sideways, the chain still exhibited the same behaviour as has been seen before: it formed an arc straight up from the heap, came back to the table surface and traveled the remaining distance to the table's edge horizontally before falling over the edge.

In an alternative setup we again placed the chain bunched together on the table, but now in such a way that every single link touched the table surface (all of the chain was in a horizontal plane). When letting the chain fall over the edge of the table in this case, no vertical arc formed but all wavelike behaviour was confined to the horizontal plane.

See this short Youtube video.

 

[JollyOlly]

That's really interesting and confirms Biggins analysis.

The reason why the phenomenon seems to be so counter-intuitive is that the table has to provide an upwards force on the chain (over and above the normal reaction force on the chain) and, bizarrely, the chain seems to be using this upwards force to push itself into the air!

It follows that if the table is somehow pushing the chain upwards, it must be because the chain is somehow pushing the table downwards - and if we can explain the latter force, we have explained the former.

Now when the chain is piled into a heap, the moving chain is dragged over the the remaining pile and (as has been explained in Biggins'paper) the collisions between the moving beads and the stationary ones below result in a series of downward impulses being imparted to the stationary beads on the table. This gives the moving beads an extra impulse upwards and effectively causes the beads to have a bit of upwards momentum before they actually leave the pile. The tension in the chain immediately above the table is therefore a little bit less than the expected λv² which means that there is a little bit of force left over to support the weight of the vertical section.

If the chain is laid out neatly on the table, the beads do not have to roll over each other and therefore receive no upward impulses and the arc fails to materialize. Neat!

I was surprised to learn that it is the nature of the chain itself which makes all the difference. A smooth rope will never work.

Biggins mentions in his paper the remarkable effect that a chain hanging vertically and allowed to fall onto a horizontal surface will fall faster than one hanging freely! I am finding this difficult to believe. It implies that the floor exerts a downward force on the chain and that therefore the chain exerts an upwards force on the floor. Can anyone think of a mechanism whereby this might come about?

 

[Gruxg]

In an alternative setup we again placed the chain bunched together on the table, but now in such a way that every single link touched the table surface (all of the chain was in a horizontal plane). When letting the chain fall over the edge of the table in this case, no vertical arc formed but all wavelike behaviour was confined to the horizontal plane.

There are some youtube videos showing the formation of a vertical arc in those conditions:

How can it be explained with Biggin's theory? Or am I missunderstanding something?

 

[A.T.]

How can it be explained with Biggin's theory? Or am I missunderstanding something?

Don't know about Biggin's theory, but it seems obvious that a fast chain cannot make sharp turns, because the internal tension providing the centripetal force will redistribute along the chain. It also cannot fall off in a smooth inverted parabola like independent beads would, because its horizontal speed at the edge is increasing, and thus higher than that of the previously dropped chain below. The fall path would have a sharp turn down, which is redistributed to a rounder curve around the edge in all directions, including lifting upwards.