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

[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.

 

[Orodruin]

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

Physics is not about opinion and yours is wrong. It indicates you simply do not understand the physics in play.

As long as the motion is one-dimensional, the upper system will keep crashing into the lower one. Cutting the spring will not change this. The force in the slinky is one of compression, not of tension.

If you somehow managed to make the two pieces not collide, it will take longer for the lower coil of A to move because the lower part is not receiving any monentum from being crashed into by the upper part.

 

[A.T.]

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?

The level of idealization and the time point of cutting the spring need to be defined more precisely here. If the tension between loop 45 and 46 has already dropped to zero, then cutting the loop does nothing.

 

[Orodruin]

The level of idealization and the time point of cutting the spring need to be defined more precisely here. If the tension between loop 45 and 46 has already dropped to zero, then cutting the loop does nothing.

Not really. In the idealised shock wave description, the tension drop is a step function. Cutting at the time of the step function arriving is a sufficient level of description.

To be more specific, if you can satisfy this:

If you somehow managed to make the two pieces not collide, it will take longer for the lower coil of A to move because the lower part is not receiving any momentum from being crashed into by the upper part.

Then it is still easy to compute the times. The relation between the shock wave travel fraction $\sigma$ and time taken is given in my Insight:
$$
\frac{gt^2}{2L} = \sigma^2(1-2\sigma/3)
$$
Here, $L$ is the slinky extension under gravity $g$. In particular, the full ($\sigma = 1$) drop time is
$$
t_0 = \sqrt{\frac{2L}{3g}}.
$$
If we cut when the shock wave reaches $\sigma = 1/2$ for cleaner math, the time until the cut is given by
$$
\frac{gt_1^2}{2L} = \frac{1}{6}
$$
ie
$$
t_1 = \sqrt{\frac{L}{3g}}.
$$
The time after the cut is given by the original formula relpacing $L\to L/4$ and so
$$
t_2 = \sqrt{\frac{L}{6g}}.
$$
We are thus comparing $\sqrt 2$ to $1 + \sqrt{1/2}$. It is elementary math to conclude the second - corresponding to the cut slinky - is larger.

 

[DaTario]

I think I understand what you mean when you say physics is not a matter of opinion. I had not done the calculus of this problem. All I had was my physics intuition and I have no problem with your having a better intuition than me in this problem. Congratulations. I am happy to have my question analyzed by you and others here. My interest is to gain some insight on what is the role of the mass acceleration in this process of maintaining the equilibrium of subsystems in these two problems. One of the fundamental notions for me in these problems is: if a falling person pulls down, with a certain force F, a mass m, he/she can prolong the falling time a little. In this case, we are not dealing with a person but with an inanimate mechanism. It's also somewhat analogous to what happens in rocket propulsion.

I suppose a slinky which cannot compress, just having traction (or tension). But the collision between loops is a problem I wouldn't know how to solve.

I would like you to recognize that students in a physics degree program are not accustomed to converting sentences involving the concepts of information and signal into familiar Newtonian interactions. The explanation in these terms seems to be essentially vague and cold.

Your insight on this subject seems to be a possibly valid effort in the direction of solving this problem step by step.

It's worth noting that among all the footage I've seen of this slinky crash, none of it seemed organized enough to resemble what we're trying to idealize.

In your insight, what is the definition of $u'(s)$ in the third equation?

 

[Orodruin]

I would like you to recognize that students in a physics degree program are not accustomed to converting sentences involving the concepts of information and signal into familiar Newtonian interactions.

I would say that being able to do such things is exactly what should be expected from students in a physics degree program.

But the collision between loops is a problem I wouldn't know how to solve.

It has to be modelled. My assumption was, after looking at the footage, that the collisions are fully inelastic to a good approximation, simply because the upper part sticks together without recoiling. This works out to a pretty good model.

Your insight on this subject seems to be a possibly valid effort in the direction of solving this problem step by step.

Please see the regerences given in the Insight. Although I started out by doing the modelling myself, others have done exactly the same analysis before. Perhaps most notably Unruh who is a well known theoretical physicist.

It's worth noting that among all the footage I've seen of this slinky crash, none of it seemed organized enough to resemble what we're trying to idealize.

If faced with fuzzy data, you can create your own data by doing the experiment yourself. This particular experiment is easy enough to perform using only a slinky (I bought one recently for about €8) and a smart phone camera. You can download some video editing tools for free to analyse the footage frame by frame.

In your insight, what is the definition of u′(s) in the third equation?

Standard notation for the derivative $du/ds$.

 

[DaTario]

I would say that being able to do such things is exactly what should be expected from students in a physics degree program.

At the end of it, may be, but IMO not at the stage this problem is usually presented. In fact, I think that the vast majority of physics graduates don't do this translation well. I think they may be able to relate concepts from Newtonian physics to the concept of memory in a system, which is related to the information context. But I don't think they are capable of understanding, for example, how and in what passage, in the process of deducing the wave equation, the finiteness of the wave propagation speed emerges. I mean going beyond basic dimensional analysis, which consists, for example, of saying that the square root of the ratio between tension and linear mass density gives us a velocity.

 

[A.T.]

Not really. In the idealised shock wave description, the tension drop is a step function. Cutting at the time of the step function arriving is a sufficient level of description. To be more specific, if you can satisfy this:

If you somehow managed to make the two pieces not collide, ....

Yes, I agree that without collision the cut at zero tension will make a difference. But with collision it would not, because the tension is gone already, and the compression during collision doesn't care about the cut. That is what I meant by level of idealization.

 

[DaTario]

Dear Orodruin, I am a physicst and I am ashamed of the gaps I allowed to exist in my training as a physicist. But could you please explain the equation $T(s) = \alpha u'(s)$ and how it relates to Hooke's law? I have never seen this connection between Hooke's law and velocity. I would also like to ask you if the spring, having mass uniformly distributed and distending non-uniformly due to the action of the gravitational field, would not lead to the conclusion that $\alpha$ in reality should be an $\alpha(s)$.

Thank you in advance for your attention and kindness.

 

[DaTario]

When we consider the physics of toppling dominoes we clearly see a set of dead times which are basically those intervals between the beginning of the fall and the instant of collision with the next piece. This intervals of time give basis for us to understand the finitiness of the velocity in this propagation. Something like this does not seem to be available neither with the slinky nor with the plucked string. Furthermore, it is kind of weird to say that the system of dominoes is falling.

 

[Orodruin]

Dear Orodruin, I am a physicst and I am ashamed of the gaps I allowed to exist in my training as a physicist. But could you please explain the equation $T(s) = \alpha u'(s)$ and how it relates to Hooke's law?

Gladly. It does not relate to Hooke’s law - it is Hooke’s law

The more commonly shown form of Hooke’s law is $F = kx$ where $F$ is the force in an ideal spring, $k$ is a constant, and $x$ is the elongation of the spring.

Now this is not very satisfying when we deal with a non-ideal spring that has mass or is otherwise unevenly extended, because we know that the force/tension will change throughout the spring. Instead, we would like the local version of Hooke’s law. We want to express the tension at a position only in terms of local quantities. We can rewrite the global Hooke’s law by letting $\epsilon = x/\ell$, where $\ell$ is the rest length of the spring. It is now on the form $F = k\ell \epsilon = \alpha \epsilon$. Here, $\alpha = k\ell$ is a constant that does not depend on the length of the full string. You can convince yourself that if you have two identical springs in series, then $k$ halves, corresponding to $k \ell = \alpha$ remaining constant. Consequently, the strain $\epsilon = x/\ell$ = extension/total length is what actually gives you the force in the spring when multiplied with the spring-length independent quantity $\alpha$. If you look at a small part $\delta s$ of the spring around (rest) position $s$ and the displacement from rest is $u(s)$, then the extension of this segment is $u(s+\Delta s) - u(s)$. Dividing by the segment rest length $\Delta s$ to obtain $\epsilon$ and letting $\Delta s \to 0$ results in $\epsilon = du/ds = u’(s)$. This is a derivative with respect to the spring rest position $s$, not with respect to time. The quantity $\epsilon = u’(s)$ is a strain, not a velocity. It represents how stretched the spring is at the part of the spring which is at position $s$ when the spring is unstretched.

We therefore have that the for e in the spring, ie, the tension $T$, is given by the local Hooke’s law $T = \alpha u’(s)$.

I have never seen this connection between Hooke's law and velocity.

See above. There is no velocity in the expression.

I would also like to ask you if the spring, having mass uniformly distributed and distending non-uniformly due to the action of the gravitational field, would not lead to the conclusion that $\alpha$ in reality should be an $\alpha(s)$.

No. $s$ refers to the position at rest. That $\alpha$ is constant is true as long as we remain in the linear regime to good approximation. All of those effects are taken into account in the model.

 

[DaTario]

Thank you. In fact, you had already informed me before (post #110) that $u'$ was $du/ds$.

 

[DaTario]

This is a derivative with respect to the spring rest position $s$, not with respect to time.

but strictly speaking $s$ is not position, but rather a dimensionless quantity that points to the given position, isn't it?

 

[Orodruin]

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 captured this in my office today:


Simple slow motion capture using an iPhone 11 camera.

If you look closely you might see an effect that we have not discussed here, but that is well known and described in the literature. See if you can figure it out!

 

[A.T.]

If you look closely you might see an effect that we have not discussed here, but that is well known and described in the literature. See if you can figure it out!

This?

... some torsional movement at the bottom does start much earlier than the vertical movement,

It seems that it causes some relaxation that reflects at the bottom and goes up again.

 

[Orodruin]

This?

It seems that it causes some relaxation that reflects at the bottom and goes up again.

Yes! The disturbance is quite visible once you see it. In the original capture you also see it move down the string and the bottom of the string actually moves a few cm when it reflects due to the mixing of the longitudinal and twisting modes. Then you see the reflected signal start moving up the spring.

Unlike the longitudinal waves, the twist mode speed visibly outruns the shock wave.