Dropping an extended Slinky -- Why does the bottom of the Slinky not fall?

[DaTario]

Yes, and in Newtonian physics, when appropriate, we use the approximation of extended bodies as perfectly rigid objects (infinite propagation speed of mechanical disturbances). But this approximation definitely not appropriate for a slinky.

But who is proposing the slinky as a rigid body?

 

[A.T.]

But who is proposing the slinky as a rigid body?

To have instantiations motion at the bottom, you would need infinite propagation speed of mechanical disturbances, which implies a perfectly rigid body.

 

[DaTario]

My present position (and proposition) was described in post #60. Initially I proposed that the fall of the first loop would produce a local departure from equilibrium that would lead to a rapid chain reaction producing some disequilibrium effect in the lowest loop. However, I later realized that throughout the process, masses were leaving rest and starting to form an accelerated subsystem. This accelerated subsystem of increasing mass can provide the necessary support to maintain the balance of the lower turns.

I would like to know if you agree with this line of reasoning.

 

[A.T.]

This accelerated subsystem of increasing mass can provide the necessary support to maintain the balance of the lower turns.

The balance of forces on the lower turns depends on the force and thus deformation of the turn right above them. It does not depend of whether some turns way above them are accelerating or not.

 

[DaTario]

The balance of forces on the lower turns depends on the force and thus deformation of the turn right above them. It does not depend of whether some turns way above them are accelerating or not.

It seems we have a chain of interdependent subsystems, haven't we?

 

[PeroK]

Let me try another analogy. Imagine there is an earthquake under the ocean that sets off a tsunami. The water wave moves at a finite speed and could take an hour or more to reach the shore. The whole ocean of water does not move uniformly at once. Or, look at the time sound waves take to move through the air. They are not instantaneous either.

Initially I proposed that the fall of the first loop would produce a local departure from equilibrium that would lead to a rapid chain reaction producing some disequilibrium effect in the lowest loop. However, I later realized that throughout the process, masses were leaving rest and starting to form an accelerated subsystem. This accelerated subsystem of increasing mass can provide the necessary support to maintain the balance of the lower turns.

I would like to know if you agree with this line of reasoning.

By "a local departure from equilibrium that would lead to a rapid chain reaction producing some disequilibrium effect in the lowest loop", I guess you mean a wave. It's nothing more and nothing less. Waves in different objects move at different speeds. The slinky is a visual representation of this. It's not like you are being asked to accept something abstract. What is the point of this thread?

 

[DaTario]

Let me try another analogy. Imagine there is an earthquake under the ocean that sets off a tsunami. The water wave moves at a finite speed and could take an hour or more to reach the shore. The whole ocean of water does not move uniformly at once. Or, look at the time sound waves take to move through the air. They are not instantaneous either.



By "a local departure from equilibrium that would lead to a rapid chain reaction producing some disequilibrium effect in the lowest loop", I guess you mean a wave. It's nothing more and nothing less. Waves in different objects move at different speeds. The slinky is a visual representation of this. It's not like you are being asked to accept something abstract. What is the point of this thread?

I apologize to everyone for having used user Notinuse's post to ask anyone interested in this topic if a more Newtonian explanation (with elastic forces decreasing and masses becoming accelerated) could be accepted. Apparently no one that contributed here found the idea interesting. Maybe it's a good time to stop. I don't want to make anyone sad.

 

[Dale]

if a more Newtonian explanation (with elastic forces decreasing and masses becoming accelerated) could be accepted. Apparently no one that contributed here found the idea interesting.

For my part, that is not what I objected to. This is a scenario that is well within the scope of Newtonian mechanics. And discrete element models are well studied with known uses and limitations.

I personally objected to

1) the claim that such a discrete element model is more fundamental than a continuum explanation

2) the claim that a discrete explanation would result in instantaneous movement at the bottom

The first objection is a matter of opinion, but the second objection is not. A correct Newtonian discrete element explanation leads to waves and has the same feature that the bottom doesn’t start moving immediately.

 

[DaTario]

For my part, that is not what I objected to. I personally objected to

1) the claim that such a lumped element model is more fundamental than a continuum explanation

2) the claim that a discrete explanation would result in instantaneous movement at the bottom

The first objection is a matter of opinion, but the second objection is not. A correct Newtonian discretized explanation leads to discretized waves.

Hi,

Regarding your point number 1, if what you call continuum explanation is the well known wave equation, I would say that a 'lumped element model' is likely to be free from aproximations tipically used in the first approach such as small-amplitude, continuous medium, uniformity and isotropy, neglecting of damping and dispersion, quasi-static approximation, linear approximation and homogeneous boundary conditions.

Furthermore, the slinky, falling, that is, accelerated, vertically, with both ends free, does not exactly constitute an example of a familiar and safe application of the wave equation and the notions pertinent to this context (wave physics).

Regarding the point number 2, I abandoned this view when in post #60 I exposed the idea that along with the systematic decrease, over time, of the elastic forces between the upper turns, there was an increase in the number of particles having acceleration in the system, and that these two processes together would explain ( or would probably explain) the maintenance of the equilibrium condition of the lower turns.

 

[Orodruin]

Regarding your point number 1, if what you call continuum explanation is the well known wave equation, I would say that a 'lumped element model' is likely to be free from aproximations tipically used in the first approach such as

Let us consider these one at a time:

small-amplitude,

… is typical for transversal waves, not longitudinal waves. Furthermore, the discrete model suffers similarly from possible non-linearities for large displacements in an actual material if you create a model with masses and springs.

continuous medium,

… is no more of an approximation than a discrete model with a chain of masses and springs is. If you really want to make a molecular level approximation, a spring is far from a single chain of molecules.

uniformity and isotropy,

… is not better handled by a discrete model.

neglecting of damping and dispersion,

… is again not better handled by a discrete model. It is as easy to introduce in a discrete model as it is in a continuous one so that is nothing particular either.

quasi-static approximation,

… is not in either model.

linear approximation

… is in both models. Can be taken out of either but then leads to a harder problem to solve.

and homogeneous boundary conditions.

… in in both models (or neither, depending on the actual boundary conditions).

To me, your distinction simply does not hold up to scrutiny.

 

[DaTario]

To me, your distinction simply does not hold up to scrutiny.

Hi, Orodruin. I think you show signs that you are happy with the explanations you give about this system. I was looking to see if there are people out here who aren't happy with this explanation. The thesis that the two paths of explanation are not distinct seems unlikely to me and I do not think it has been sufficiently well defended.

 

[Orodruin]

Hi, Orodruin. I think you show signs that you are happy with the explanations you give about this system. I was looking to see if there are people out here who aren't happy with this explanation. The thesis that the two paths of explanation are not distinct seems unlikely to me and I do not think it has been sufficiently well defended.

I mean, the effect is well known and it is quite clear that regardless of the description you use you end up with the same result: Making a free body diagram of the lower part of the slinky will have it hanging still until what is just above starts contracting. There really is nothing more to it and it works the same irrespective of the model you use.

If you disagree with this then I believe you simply have not worked enough with the discrete model. In both models you have a propagation of the disturbance “down-spring” and the bottom doesn’t move until that disturbance reaches the end.

 

[DaTario]

I mean, the effect is well known and it is quite clear that regardless of the description you use you end up with the same result: Making a free body diagram of the lower part of the slinky will have it hanging still until what is just above starts contracting. There really is nothing more to it and it works the same irrespective of the model you use.

If you disagree with this then I believe you simply have not worked enough with the discrete model. In both models you have a propagation of the disturbance “down-spring” and the bottom doesn’t move until that disturbance reaches the end.

Believe me when I say I'm not getting off topic. But how do you explain the electric field produced by a uniformly charged plane?

 

[Orodruin]

Believe me when I say I'm not getting off topic. But how do you explain the electric field produced by a uniformly charged plane?

The answer is still that the continuous model prediction is just as good as the discrete model prediction here. Besides, you are missing the point of the above discussion, which is that both models - continuous and discrete - have the same fundamental explanation of why the bottom does not fall until some time has passed. Just as the uniformly charged plane has an electric field produced by non-zero charge - whether that charge be continuously or discretely distributed. Changing from a continuous to a discrete model does not affect the actual physics in the limit where the discrete model goes to the continuum limit.

Also in the case of the electric field, both models are wrong to some extent.

 

[Orodruin]

To add to that, there is actually quite a set of literature on this subject and the predictions of the models - continuous and discrete alike - closely reproduce the behaviour of a real falling slinky.

 

[Orodruin]

Waves in different objects move at different speeds. The slinky is a visual representation of this.

It should be pointed out though, that what we have in the slinky drop experiment is not your typical wave moving at the wave speed of the medium. It is moving faster than the wave speed of the medium along with an abrupt change in the coil density. These are characteristics of a shock wave rather than your normal wave propagating at a speed set by the medium.

 

[DaTario]

Just as the uniformly charged plane has an electric field produced by non-zero charge - whether that charge be continuously or discretely distributed. Changing from a continuous to a discrete model does not affect the actual physics in the limit where the discrete model goes to the continuum limit.

I did not invoke the problem of the electric field produced by a uniformly charged plane, to discuss discrete or continuous approaches. Sorry if I gave you this impression. What I want to exemplify is taking advantage of an opportunity to observe something unique that both contexts seem to offer. Regarding the uniformly charged infinite plane, let's see the following:

We know that considering any point charge, when we move away from it the field in our position decreases in magnitude. When we take a step back, moving away from a plane, we are at the same time moving away from all the area elements of this plane. The total field is the vector sum of the fields produced by all elements of area in the plane. What keeps the total field constant regardless of the distance from the plane is a perfect compensation between the decrease in the modulus of the electric field vectors originating from each charge element and the geometric effect of greater alignment between these vectors. Thus, if in a position further away from the plane the vectors have a smaller module, they are also more aligned, that is, they are closer to parallelism, so that this geometric condition perfectly compensates for the effect of distance.

In the slinky problem I have the impression that something similar happens. Elastic forces start decreasing from the upper part of the slinky but at the same time we see masses being accelerated so that balance is maintained for the remaining part of the spring. It is more or less like we set initially $y$ as a vertical coordinate for a mass and spring system, and write Newton's second law:
$$m \frac{d^2y}{dt^2} = - k \Delta y - mg.$$
Now we write it in the form:
$$m \frac{d^2y}{dt^2} + k \Delta y = - mg.$$
so that we can see in the first term an informal suggestion of the compensatory mechanism to which I refer. (* Informal mainly because the slinky is not such a simple mass and spring system *). If the elastic tension decrease but some of the masses start accelerating, then we can still have the equilbrium.

But I recognize that I don't have a closed method to express clearly and consistently this compensation, so I decide not to proceed with this debate. I thank you all.

 

[Orodruin]

Sorry, but I do not see the analogy here. There are no forces from the upper part of the string on the lower end. All of the forces are contact forces and the force on the lower part is the force from the part immediately above it. In a discrete model, the only elongation that matters for the bottom mass is the very last spring. It is irrelevant what the elongations of the other springs are.

You can of course do a harmonic series expansion in terms of standing waves, but that also won’t give you a good description because, as has been discussed elsewhere, what occurs in the slinky drop experiment is a shock wave rather than a wave propagating at the characteristic wave speed of the slinky.

 

[A.T.]

It seems we have a chain of interdependent subsystems, haven't we?

The instantaneous balance of forces on each chain element depends only on the current deformation of the neighboring elements. It does not depend on what other elements further up are currently doing.

Explaining the local force balance at the bottom by acceleration of the elements at the top make no sense. It's only relevant for the global momentum conservation, as the center of mass must accelerate down at g.

 

[DaTario]

Orodruin and A.T.

I understand your points and they seem ok to me. As I said, my point is weakly supported, I will no longer defend it here. Thank you very much for this debate. I would like to point out some minor divergencies I have relatvely to what you said in these last posts:

A.T. said:

"The instantaneous balance of forces on each chain element depends only on the current deformation of the neighboring elements."

I would note that, besides gravity, at each time there is an element of mass in the slinky which is on the moving interface between those accelerated masses (above) and those static masses (below). For this particular mass element (at this particular time), it seems that the balance or unbalance of forces depends also on the acceleration of neighboring elements.

Orodruin said:

1) "All of the forces are contact forces"

I do not agree with this. As I said, there is gravity (field force) and there is inertia, which reponds by the term $m y''(t)$.

2) "You can of course do a harmonic series expansion in terms of standing waves"

I would find it enoumously anti intuitive or non pratical, to try harmonic series expansion, as the spring is changing size which makes its harmonic family to change continuously over time. I would not say it is impossible, but it seems not to be a good idea.

 

[A.T.]

it seems that the balance or unbalance of forces depends also on the acceleration of neighboring elements.

Why? In the simple mass-spring model the force of the spring depends only on its current length, not on the acceleration of its ends.

 

[DaTario]

Why? In the simple mass-spring model the force of the spring depends only on its current length, not on the acceleration of its ends.

The problem is that with the slinky, the mass is the spring.

 

[A.T.]

The problem is that with the slinky, the mass is the spring.

In a mass-spring model you represent a massive spring as masses connected by massless idealized springs. Isn't that what you wanted to use?

 

[Orodruin]

1) "All of the forces are contact forces"

I do not agree with this. As I said, there is gravity (field force) and there is inertia, which reponds by the term my″(t).

I am talking about the internal forces of the slinky. Inertia is not a force.

2) "You can of course do a harmonic series expansion in terms of standing waves"

I would find it enoumously anti intuitive or non pratical, to try harmonic series expansion, as the spring is changing size which makes its harmonic family to change continuously over time. I would not say it is impossible, but it seems not to be a good idea.

No, not like that. The harmonic series is made not in physical position, but in slinky fraction - wgg ha uch alwaya varies from 0 to 1. Physical position is the dependent variable, not the domain of the function!

Regardless, it is a bad idea for the other reasons I have mentioned, but what you claim is not one of them.

it seems that the balance or unbalance of forces depends also on the acceleration of neighboring elements.

The acceleration of that element is a matter of an essentially inelastic collision between the falling and stationary parts of the slinky. You do not really need to balance the forces of the parts to perform the analysis. It is sufficient to consider the total momentum of the slinky.

 

[pbuk]

It should be pointed out though, that what we have in the slinky drop experiment is not your typical wave moving at the wave speed of the medium. It is moving faster than the wave speed of the medium along with an abrupt change in the coil density. These are characteristics of a shock wave rather than your normal wave propagating at a speed set by the medium.

That is a very good point - I will correct my post #3.

 

[sophiecentaur]

That is not a good explanation: introducing the CoM doesn't help.

The motion of the CM will be downwards at g. Surely that's relevant to a verbal explanation, at least. The bottom end will be in equilibrium and the top part of the slinky will initially be accelerating at 2g.
Once the spring starts to collapse, the upward force on the bottom bit will be less so that would seem to imply a net downwards acceleration after the top is released. Why would the restoring force not gradually reduce as the spring collapses (despite any delay due to the longitudinal wave speed)? (Answer, probably is that the change in force at the top is a step function)

BTW, has this video been verified? It would be very easy to arrange for the bottom of the slinky to be supported from behind until it collapses (or with a light thread inside it). There is a bit of a caffuffle in the few frames where the spring all comes together.

This is like the old chestnut about what would happen to the Moon if the Earth suddenly disappeared - except no energy is required to release the slinky from your hand. But the connecting argument could involve gravitational waves.

 

[Orodruin]

The motion of the CM will be downwards at g. Surely that's relevant to a verbal explanation, at least. The bottom end will be in equilibrium and the top part of the slinky will initially be accelerating at 2g.

The very top of the slinky accelerates instantaneously to a finite speed. The CM motion is indeed useful when you consider the description of the drop in that it tells you what the total momentum (and therefore the momentum in the moving part of the slinky) must be. See https://www.physicsforums.com/insights/the-slinky-drop-experiment-analysed/

Once the spring starts to collapse, the upward force on the bottom bit will be less so that would seem to imply a net downwards acceleration after the top is released. Why would the restoring force not gradually reduce as the spring collapses (despite any delay due to the longitudinal wave speed)? (Answer, probably is that the change in force at the top is a step function)

Because it cannot react instantaneously, there is a delay. The wave speed in the slinky is also slower than that of the shock wave, resulting in that no part of the slinky actually starts contracting until it is hit by the shock - making the change in force a step function everywhere.

BTW, has this video been verified? It would be very easy to arrange for the bottom of the slinky to be supported from behind until it collapses (or with a light thread inside it). There is a bit of a caffuffle in the few frames where the spring all comes together.

I have performed this experiment myself several times. It is very easy to do with a slinky and a mobile phone camera.

This is like the old chestnut about what would happen to the Moon if the Earth suddenly disappeared - except no energy is required to release the slinky from your hand. But the connecting argument could involve gravitational waves.

I have always despised that example, it is badly formulated. The Earth simply cannot suddenly disappear, it would violate the local conservation of stress energy, which is a direct implication of the Einstein field equations. This is different, you can let go of the slinky at any time you wish.

Edit: The slinky drop is also different because the signal - being a shock wave - reaches the end faster than the local wave speed would imply.

 

[sophiecentaur]

the signal - being a shock wave

Hmm. I thought a shock wave was caused by a driving source that's faster than the wave speed. The applied force can't be greater than the tension in the spring; however, it is applied as a step function (i.e. instantaneous negative tension force). Initial downwards acceleration of the top coil will be 2g.

 

[sophiecentaur]

The Earth simply cannot suddenly disappear, it would violate the local conservation of stress energy,

Of course but a high speed impact with a large body in a direction nearly on the line of centres away from the Moon would move Earth away pretty fast with a very small impulse directly to the Moon (short interaction time as the body goes past the Moon). But it's not a good scenario even though it's a very popular question. The only thing wrong with it in principle is the huge masses involved, even in a scale model - like two asteroids with one of them pulled away with a cable. In that case, the 'satellite' would carry on in a straight line ( hyperbola, perhaps), at less than 90 degrees from the line of centres. But that semi practical model would lack drama and annoy everyone so you'd be unlikely to hear that from a class of year 10s.

 

[Orodruin]

Hmm. I thought a shock wave was caused by a driving source that's faster than the wave speed.

This is the point, the top part moves faster than the wave speed.

The applied force can't be greater than the tension in the spring; however, it is applied as a step function (i.e. instantaneous negative tension force). Initial downwards acceleration of the top coil will be 2g.

(My emphasis) It will not. The tension at the top in the initial configuration is equal to $mg$, the mass of the top coil is significantly smaller than $m/2$. If you consider the top element $dm$ of the slinky, its acceleration is tension/mass = $mg/dm$, which diverges as $dm \to 0$, ie, infinite acceleration.

Of course but a high speed impact with a large body in a direction nearly on the line of centres away from the Moon would move Earth away pretty fast with a very small impulse directly to the Moon (short interaction time as the body goes past the Moon).

The speed cannot exceed $c$. ”Pretty fast” is not fast enough. If it is fast enough and massive enough to move the Earth away, its own gravitational impact on the Moon will definitely not be negligible. I think you really need to think this through because this suggestion severely complicates the scenario.

But it's not a good scenario even though it's a very popular question. The only thing wrong with it in principle is the huge masses involved

No, that is certainly not the only thing wrong with it. The point is that the popular question, as it is usually stated, is nonsensical within the theory. You need a scenario that is viable within the theory. Your suggested remedy severely complicates everything.

 

[A.T.]

BTW, has this video been verified?

They are articles about it, but I cannot access them:
https://pubs.aip.org/aapt/pte/article-abstract/39/2/90/272881/Analysis-of-Slinky-levitation
https://pubs.aip.org/aapt/ajp/artic...inky-oscillations-and-the-notion-of-effective

 

[Orodruin]

They are articles about it, but I cannot access them:
https://pubs.aip.org/aapt/pte/article-abstract/39/2/90/272881/Analysis-of-Slinky-levitation
https://pubs.aip.org/aapt/ajp/artic...inky-oscillations-and-the-notion-of-effective

There are several references listed in my Insight. I know the paper by Unruh is on arXiv so that at least is freely available.

 

[DaTario]

I remembered yesterday one argument that can clarify the kind of explanation I am seeking. It is related to the plucked string problem (there is no gravity here). It can be solved by more than one method. One of such methods of solving consists in considering its dynamics (fig below) as a bar which is growing in size and mass while it is pulled by a constant tension (initially downward -- a,b,c,d).

The role of the mass (specifically the accretion process) seems to be potentially of good help in understanding the reason why those parts of the string close to the fixed points, which were initally at rest, remain at rest during the first steps of the dynamics.
I see some important correlation between these two problems. And this correlation seems to have something to do with the accretion process in the accelerated body (subsystem) as a means to maintain an equilibrium condition in the moving vicinity of this subsystem.

 

[Orodruin]

I remembered yesterday one argument that can clarify the kind of explanation I am seeking. It is related to the plucked string problem (there is no gravity here). It can be solved by more than one method. One of such methods of solving consists in considering its dynamics (fig below) as a bar which is growing in size and mass while it is pulled by a constant tension (initially downward -- a,b,c,d).
View attachment 342918
The role of the mass (specifically the accretion process) seems to be potentially of good help in understanding the reason why those parts of the string close to the fixed points, which were initally at rest, remain at rest during the first steps of the dynamics.
I see some important correlation between these two problems. And this correlation seems to have something to do with the accretion process in the accelerated body (subsystem) as a means to maintain an equilibrium condition in the moving vicinity of this subsystem.

The staying in place part is the same type of phenomenon. Nothing changes until the signal arrives. However, that’s where the similarity ends. In this case the evolution of the string is governed by the wave equation. The signal propagates at the wave speed in the steing. In the slinky drop, the signal is a shock wave, propagating faster than the wave speed in the slinky.

 

[DaTario]

In both systems we have:

1) a subsystem whose mass and size increases in time.
2) another disjoint and complementary subsystem containing elastic tensions which are decreasing in size and naturally in tensions.
3) this last subsystem retains an equilibrium configuration which was set up at the beginning of the experiment.

As I believe in the importance of the role of the subsystem I described in the item 1 above, let me ask you a question:

Consider two identical springs A and B with 100 loops, the loops are numbered in both springs so that the loop 1 in each spring is the closest to the ground and the loop 100 is the furthest from the ground. We start two identical experiments. They start falling identically as expected. At the time the loop number 45 will start moving we use a very clever mechanism to cut, with minimum disturbance, the spring A so that it becomes two falling pieces: the subsystem 100-46 and the subsystem 45-1.
Do you believe loop 1 of spring A will move at the same time of loop 1 in spring B?

In my opinion, the loop 1 of spring A will move first.