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

[Notinuse]

Please could someone explain this to me?

GRAVITY IS A MYTH (59 sec)

Summary; in slow-mo a man drops a virtually extended free hanging slinky; the lower part appears to hang in the air for a moment.

In the comments: Rob; “..the overall mass of the slinky is falling but the bottom part is being pulled up by the springiness at the same rate the centre of mass is falling.”

What are your thoughts?

 

[kuruman]

The video that you posted is an excerpt from the complete explanation posted here. Its asinine conclusion in the end that "If gravity existed the force would pull the slinky down at the same rate" fails to understand that, while extended, the slinky is a deformable body and becomes a point-like mass in its fully-compressed state at which time all points on the slinky accelerate at the same rate. I bet that the person who wrote that deliberately misleading statement would never consider jumping off the roof of a building if asked.

 

[pbuk]

Why does the bottom of the Slinky rise?

It doesn't. The video clearly shows the bottom of the Slinky staying still.

One way to understand why this happens is to realise that the information that the top of the Slinky has been released travels at the rate of propogation of a compression wave in the Slinky (the "speed of sound")as a shock wave in the Slinky at a finite speed. [edited see #86]

“..the overall mass of the slinky is falling but the bottom part is being pulled up by the springiness at the same rate the centre of mass is falling.”

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

 

[phinds]

If you hold the top of a slinky and release the rest of it, and continue holding the top until JUST AFTER the bottom starts to spring back up again (which is NOT what he is doing in the video), the bottom will continue to rise, briefly, before starting to fall again.

 

[pbuk]

If you hold the top of a slinky and release the rest of it, and continue holding the top until JUST AFTER the bottom starts to spring back up again (which is NOT what he is doing in the video), the bottom will continue to rise, briefly, before starting to fall again.

Yes, anything you do at one end can only affect the other end at the rate of propogation of the appropriate wave.

 

[phinds]

Yes, anything you do at one end can only affect the other end at the rate of propogation of the appropriate wave.

Exactly

 

[vanhees71]

https://arxiv.org/abs/1110.4368v2

 

[Dale]

What are your thoughts?

The claim of the video is pretty stupid. This is well understood by scientists, even if it is not understood by random YouTube people.

The whole slinky is in free fall because there is no external force other than gravity. The top of the slinky is pulled down by the rest of the slinky, so it accelerates more than free fall. The bottom of the slinky is pulled up by the rest of the slinky, so it accelerates less than free fall. On average it accelerates at free fall.

 

[Notinuse]

@kuruman - thank-you for the original source; that is interesting.

@pbuk - I did not suggest a rise; someone has edited the post title.

 

[berkeman]

@pbuk - I did not suggest a rise; someone has edited the post title.

Apologies, I misinterpreted this part of your OP:

In the comments: Rob; “..the overall mass of the slinky is falling but the bottom part is being pulled up by the springiness at the same rate the centre of mass is falling.”

I have edited the thread title to remove my "rise" comment.

 

[kuruman]

It might be helpful to imagine what the person holding the stretched slinky would observe if he jumped out the window still holding the stretched out slinky. If you try this at home, please make prior arrangements for a soft landing.

Spoiler

At the jump-off moment, all parts of the person + slinky system are initially at rest. At any later time, the stretched slinky will just contact to its minimum length.

That's exactly what the slinky would do if one placed it on a horizontal surface (gravity is turned off), stretched it with both hands and then let go one end.

 

[Drakkith]

I was imagining doing this with a rubber band and having the end snap back upwards, then realized the entire idea was that gravity and the upwards force on the end of the slinky are equal. Thus when the slinky is let go the bottom end doesn't move because the forces are still balanced.

 

[kuruman]

I was imagining doing this with a rubber band and having the end snap back upwards, then realized the entire idea was that gravity and the upwards force on the end of the slinky are equal. Thus when the slinky is let go the bottom end doesn't move because the forces are still balanced.

Right. Unlike the rubber band that is stretched by you, it's gravity that provides the stretching force. When you let go, the slinky is in free fall and, as far as it is concerned, there is no gravity to stretch it. Thus, it collapses upon itself and comes to rest in the accelerated frame.

 

[Dale]

The other difference between a slinky and a rubber band is the speed of the wave. The longitudinal wave in a slinky is very slow. The longitudinal wave in a rubber band is much faster.

 

[DaTario]

Hi Notinuse and all,

I feel sympathy for this explanation involving waves, but I think it represents a more mature and less primitive narrative, relative to the explanation that I would consider more basic and complete. I tend to think that an explanation in terms of many coupled systems of the type mass & spring, forming the slinky, would bring a more fundamental certification to this result (or would better show what happens in fact).

In this sense, when I begin to imagine what would happen to the equations of motion of all those masses, instead of concluding that the lower part would remain stationary, it seems to me plausible that another conclusion would emerge, namely, that the movement of the lower part starts as soon as the spring is released, but, due to the coupling of the many systems, the movement of the coil closest to the ground is many orders of magnitude smaller than the movement of the turns furthest from the ground.

Something like this gear system below, recently posted on the internet.



I must say that this system makes me uncomfortable even when I consider the idea of a wave propagating "the information that something or someone has released the upper end."

 

[Drakkith]

I must say that this system makes me uncomfortable even when I consider the idea of a wave propagating "the information that something or someone has released the upper end."

Why's that?

In this sense, when I begin to imagine what would happen to the equations of motion of all those masses, instead of concluding that the lower part would remain stationary, it seems to me plausible that another conclusion would emerge, namely, that the movement of the lower part starts as soon as the spring is released, but, due to the coupling of the many systems, the movement of the coil closest to the ground is many orders of magnitude smaller than the movement of the turns furthest from the ground.

Except that's not a correct explanation.

 

[DaTario]

Why's that?

Hi Drakkith, if the highest mass element $dm$ falls $dy$ in a time $dt$, I think it is reasonable to propose that the equilibrium condition for the second mass element is no longer guaranteed. If we divide $dt$ by two, we conclude that the highest mass element $dm$ has fallen by a fraction of $dy$. From this it seems to follow that the second element of mass $dm$ was no longer in equilibrium (at rest) when the time $dt/2$ elapsed and was therefore in motion at $dt$. This reasoning seems to apply to the following mass elements. I think it is likely that the coupling leads to increasingly smaller responses (displacements) of each mass element over the course of $dt$.

This is not explict in the wave formalism. Perhaps it is not even accessible.

 

[Nugatory]

an explanation in terms of many coupled systems of the type mass & spring, forming the slinky, would bring a more fundamental certification to this result

That’s where the wave equation comes from

 

[DaTario]

That’s where the wave equation comes from

Hi, Nugatory. I know this, but on the way until the wave equation was ready, many approximations were made that, in my opinion, took away from the dynamics it provides the ability to correctly describe the details we are interested in here.

 

[A.T.]

that the movement of the lower part starts as soon as the spring is released

That would violate relativity. But some torsional movement at the bottom does start much earlier than the vertical movement, because the speed of sound in the material is much higher than the vertical wave speed of the slinky.

 

[Dale]

Hi, Nugatory. I know this, but on the way until the wave equation was ready, many approximations were made that, in my opinion, took away from the dynamics it provides the ability to correctly describe the details we are interested in here.

Your coupled system approach will make more coarse approximations. Not a better approximation.

I think that you are incorrect on the behavior of the coupled system. A correctly set up system will just give a discretized wave. If you get something else then the approach is wrong because it fails to replicate the observed behavior.

To me, the question to ask is why does the bottom end hang there when the top end is being held. The answer is tension. At each point along the spring the section of spring below is held by two equal forces: gravity and tension. Gravity doesn’t change, so the spring can only move as tension changes. The speed at which tension changes is the speed of the longitudinal wave.

 

[PeroK]

You can generalise this idea. Consider two masses on a vertical extended spring released to fall under gravity. The internal dynamics of the system will continue, with the masses executing mutual SHM while in free fall. The centre of mass will accelerate downwards at $g$, but the masses will have accelerations that vary from greater than $g$ to less than $g$.

Likewise, the slinky's centre of mass accelerates at $g$. But, due to the internal tension, the top accelerates greater than $g$ and the bottom less than $g$. At least until the slinky has completely collapsed.

 

[DaTario]

That would violate relativity. But some torsional movement at the bottom does start much earlier than the vertical movement, because the speed of sound in the material is much higher than the vertical wave speed of the slinky.

Thank you A.T., for your contribution, but regarding your comment on relativity, this is a Classical Physics' discussion, isn't it?

 

[DaTario]

Your coupled system approach will make more coarse approximations. Not a better approximation.

If you take into account that we are comparing the N coupled systems approach with an approach in which the leap to the continuum occurred with the adoption of certain approximations, then, the victory in this battle seems to be an open question.

The speed at which tension changes is the speed of the longitudinal wave.

I really accept this notion in my everyday life and I confess I use this as a tool. But in reality I have a problem with this approach when the arguments in my post #17 come to my mind. Why is it that if a differential interval of $dt$ has passed, the first system is no longer in equilibrium, should we maintain that the second is still in equilibrium? Is there a kind of a dead time (or waiting time) between two consecutive subsystem's responses? Is there a quantum of waiting in these dynamics?

 

[PeroK]

Why is it that if a differential interval of $dt$ has passed,

$dt$ is not a finite time interval.

 

[DaTario]

$dt$ is not a finite time interval.

But there is no problem in considering the state of a physical system in $dt/2$, is there? Are you proposing some quantum jump? That would be interesting.

 

[DaTario]

But some torsional movement at the bottom does start much earlier than the vertical movement, because the speed of sound in the material is much higher than the vertical wave speed of the slinky.

Maybe it's a solution at least for me and at least for now to accept that more immediate consequences can be noticed on the lowest coil due to spring torsion.

I also have no intention of proving relativity wrong. My point is that the pattern of immediately noticeable consequences in the lowest loop when the highest loop is released may involve extremely small displacements (which would even make the approximations made in wave theory justified).

 

[PeroK]

But there is no problem in considering the state of a physical system in $dt/2$, is there? Are you proposing some quantum jump? That would be interesting.

It's a common mistake to use $dt$ as a differential and then to assume it's a finite time interval and thereby reach a contradiction (such as the impossibility of circular motion).

 

[Dale]

If you take into account that we are comparing the N coupled systems approach with an approach in which the leap to the continuum occurred with the adoption of certain approximations, then, the victory in this battle seems to be an open question.

Discretization involves its own set of approximations. The claim that it is “a more fundamental certification to this result” is not valid, regardless of the open question.

Why is it that if a differential interval of dt has passed, the first system is no longer in equilibrium, should we maintain that the second is still in equilibrium?

Yes. That is exactly what we should maintain. The only way for the second system to not be in equilibrium is for the tension to change. The tension does not change instantaneously. Therefore there must be some time where the one discrete element is not in equilibrium but the next is.

I have a problem with this approach when the arguments in my post #17 come to my mind

Your arguments in post 17 are simply wrong:

I think it is reasonable to propose that the equilibrium condition for the second mass element is no longer guaranteed

If that is the case then your discretization time step is far too coarse and your solution will be grossly inaccurate.

Is there a kind of a dead time (or waiting time) between two consecutive subsystem's responses?

Yes. This is the time that it takes for the wave to cross the subsystem’s length.

 

[DaTario]

It's a common mistake to use $dt$ as a differential and then to assume it's a finite time interval and thereby reach a contradiction (such as the impossibility of circular motion).

But is the present situation such a case? If positive, how so?

 

[DaTario]

Discretization involves its own set of approximations. The claim that it is “a more fundamental certification to this result” is not valid, regardless of the open question.

You seem to be just posing and axiom here.
I may agree with you in that both the wave and the many subsystem approaches lead to imperfect descriptions.

The tension does not change instantaneously. Therefore there must be some time where the one discrete element is not in equilibrium but the next is.

where is this 'dead time' (or 'waiting time') defined? I think the tension is a function of the position of the masses (i.e., the length of the spring). Once the equilibrium configuration is changed the movement starts -- no waiting time.

Your arguments in post 17 are simply wrong:

Okay, maybe so. I'm not here trying to defend a closed theory that I think is already free from flaws. But how do you refute my arguments from post #17?

 

[PeroK]

Okay, maybe so. I'm not here trying to defend a closed theory that I think is already free from flaws. But how do you refute my arguments from post #17?

It's a common mistake to use $dt$ as a differential and then to assume it's a finite time interval and thereby reach a contradiction (such as the impossibility of circular motion).

Or, in this case, if I understand your argument, the impossibility of a wave pulse.

 

[DaTario]

Yes. This is the time that it takes for the wave to cross the subsystem’s length.

you are referring to a time interval that is a global parameter of the dynamics. But what I would like to know is how those micro intervals of time, related to elementary occurrences of waiting, are distributed on the temporal axis.

 

[DaTario]

It's a common mistake to use $dt$ as a differential and then to assume it's a finite time interval and thereby reach a contradiction (such as the impossibility of circular motion).

By dividing a differential time interval by two one is not necessarily commiting a mistake, is he?

Or, in this case, if I understand your argument, the impossibility of a wave pulse.

I believe in the possibility of a wave pulse, but I also believe in a probable underlying dynamic that is hidden in formalism due to approximations. Consider the video I sent in post #15 as inspiration.

 

[Dale]

where is this 'dead time' (or 'waiting time') defined?

I already answered this:

This is the time that it takes for the wave to cross the subsystem’s length

But how do you refute my arguments from post #17?

I already answered this too:

If that is the case then your discretization time step is far too coarse and your solution will be grossly inaccurate.

Your whole reasoning is based on the premise that

if the highest mass element dm falls dy in a time dt, I think it is reasonable to propose that the equilibrium condition for the second mass element is no longer guaranteed

That premise is wrong. In any case where it holds it indicates that your time step is too coarse. This will result in a pathologically wrong solution, one that reflects the incorrect numerical method used to solve it rather than one that is faithful to the system being modeled.