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

[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.

 

[A.T.]

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.

Yes, for dominoes the finitness of the propagation speed is somewhat simpler to understand, than for a linear mass-spring system. There is no "dead times", here but instead a double integration at each finite element: the position of the element n-1 determines the acceleration of element n (along with two constants).

 

[DaTario]

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

So if we filmed the slinky from below from the moment it was released with its lower metal end painted green as in the picture below, we would see this end rotate as it descends, is that it?

fig.: Slinky seen from below.

 

[Orodruin]

So if we filmed the slinky from below from the moment it was released with its lower metal end painted green as in the picture below, we would see this end rotate as it descends, is that it?

View attachment 343044
fig.: Slinky seen from below.

Not very much. It is a twist wave signal. This signal will also take some time to reach the bottom, but it will be faster than the longitudinal shock wave. As the twist signal reaches the bottom, it will reflect off the end of the spring. This will result in some twisting motion and typically also excite some longitudinal signal from the bottom up. If you look closely you will see the reflected twist signal as well as the lower end actually falling a little bit as the twist signal reaches it.

 

[DaTario]

Yes, I saw a subtle longitudinal wave rising from the lower end.

It seems to behave as Wilberforce pendulum with almost zero moment of inertia.

 

[AlexisBlackwell]

A falling spring, like an ordinary object, is subject to gravity. However, due to its structure and elasticity, the bottom of the spring is pulled upward, following the center of mass. This happens because when the top part of the spring falls, it pulls the bottom part with it, creating a balance between the force of gravity and the elasticity of the spring. The lower part appears to be "hovering" in the air, although in fact it is supported by the elastic force of the spring.

 

[Orodruin]

A falling spring, like an ordinary object, is subject to gravity. However, due to its structure and elasticity, the bottom of the spring is pulled upward, following the center of mass. This happens because when the top part of the spring falls, it pulls the bottom part with it, creating a balance between the force of gravity and the elasticity of the spring. The lower part appears to be "hovering" in the air, although in fact it is supported by the elastic force of the spring.

This has already been discussed extensively in this thread. What is your point?