A Spinning a Slinky: Momentum & Angular Momentum Explained

[a1call]

Hi all,

"When the top end of the Slinky is dropped, the information of the tension change must propagate to the bottom end before both sides begin to fall; the top of an extended Slinky will drop while the bottom initially remains in its original position, compressing the spring.[12] This creates a suspension time of ~0.3 s for an original Slinky,[13][14] but has potential to create a much larger suspension time."

https://en.m.wikipedia.org/wiki/Slinky#Levitation

https://en.m.wikipedia.org/wiki/Hooke's_law#Rotation_in_gravity-free_space

Suppose you are spinning a Slinky around some point A in absence of gravity. The links above seem to imply that perhaps the far end of the Slinky will momentarily maintain a constant distance from point A after its release and perhaps even maintain its angular momentum. This seems wrong to me.
The law of conservation of Momentum implies that the Center-of-Mass of the slinky would immediately follow a straight line after being released. Furthermore the COM would maintain (temporarily) a constant position relative to the far end before the slinky contracts. All this should result in the far end not only not maintaining an angular momentum around point A, but also immediately increasing its distance from point A.
The difference here being that unlike dropping a slinky which would entail the COM staying aligned to the gravity vector throughout the free-fall, in the spinning scenario the COM will immediately move away from the radial path to a tangent path.
A video and discussion of the free fall version can be found at

http://blogs.discovermagazine.com/badastronomy/2011/09/26/slinky-drop-physics/

Link credit to:

01101001

Am I completely off?

Thank you in advance for any comments and insights.

 

[A.T.]

The law of conservation of Momentum implies that the Center-of-Mass of the slinky would immediately follow a straight line after being released.

How does conservation of momentum imply that? Instantaneous information would violate relativity.

 

[a1call]

AFAIK the law of conservation of momentum is a classical mechanics law. As such if from t0 (after release) till some t1, the COM does not follow a straight path and then from t1 onward the COM does follow a straight path, then there is a change in momentum which would be a reaction without an action and impossible from a classical vantage point.
But I may be corrected on that.

 

[A.T.]

But I may be corrected on that.

Sorry, I cut the quote too early. This is what I don't get:

Furthermore the COM would maintain (temporarily) a constant position relative to the far end before the slinky contracts.

Why can't the far end still move on a circle, while the COM moves on a straight line?

 

[a1call]

That is a good question. Not 100% sure. But assuming that the COM starts on a straight path immediately after release, then relative (classically speaking) to the COM before the contraction reaches and surpasses the COM (at slinky's wire midpoint irrelevant of the spring length ), the COM will maintain a constant distance to the far end. It would be inconsistent for the far end to maintain an angular (circular) path and equal distance, around and from point A, while keeping equal distance from the COM which is moving tangentially in a straight path relative to point A. Or so I figure.

 

[a1call]

The only alternative I can think of is that the near end goes into such crooked formations that would shift the COM away from the wire midpoint and in such form to maintain the angular momentum of the far end briefly.
So I just don't know.

 

[a1call]

I actually do have a Slinky at hand, but do not have the means to record or room to spin it.
I can visually observe the "levitation" in the drop scenario. I also tried a partially spin scenario and the far end did seem to move away immediately, but the test was far unreliable for a definite assessment.

 

[A.T.]

the COM (at slinky's wire midpoint irrelevant of the spring length )

Is that really true, given a non-uniform compression of the spring?

 

[vanhees71]

Check this:

http://www.princeton.edu/~rvdb/WebGL/Slinky.html

and a great paper on the falling slinky by the same author:

https://www.princeton.edu/~rvdb/WebGL/slinky.pdf

 

[a1call]

Is that really true, given a non-uniform compression of the spring?

I would think yes if the compression is linear and along a straight line (which may not be the case here). In a linear compression, the two halves of the slinky's mass will be on the opposing sides of the wire midpoint. If the compression is not along a straight line however, the COM can be anywhere in or outside the space occupied by the slinky.
To further complicate things, the delay in the release signal reaching the far end is inversely related to the slinky's stretch while it is directly related to its wire length.

 

[a1call]

Check this:

http://www.princeton.edu/~rvdb/WebGL/Slinky.html

and a great paper on the falling slinky by the same author:

https://www.princeton.edu/~rvdb/WebGL/slinky.pdf

Thank you for that.
In the slow motion there is a delay in the wire midpoint reacting to the release, which would falsify my argument that the COM would have to react immediately to the release.
So, is the angular momentum preserved briefly in the spinning scenario?

 

[A.T.]

the two halves of the slinky's mass will be on the opposing sides of the wire midpoint

That doesn't make the wire midpoint the COM.

 

[a1call]

That doesn't make the wire midpoint the COM.

I can't see why not. If you froze the slinky stiff with one half composed along a straight line and the other half uncompressed but straight along the same line, the wire midpoint would be the balancing point. To make sure we are talking about the same thing by wire-midpoint I am referring to the midpoint along an uncoiled wire which would be used to fabricate the slinky by coiling that wire. in other words the invariable midpoint of an uncoiled slinky.

Unless you are talking about the relativistic mass of the potential energy of the compression which would be again out of the classical mechanics domain, and immeasurable for all practical purposes.

 

[a1call]

To be more precise the COM would be on the slinky axis point closest to the wire-midpoint, but I think that is inconsequential to the argument.
It seems my problem arose from the false assumption that the COM would have to react instantaneously to a release and without delay. But as shown in the slow motion posting by V71, that is clearly not the case.
It would be difficult to verify this in practice, because at presence of gravity the spinning will have to be fast enough to overcome gravity and thus the delay would be very short and difficult to observe. A highly stretched slinky would show very short delay.

 

[A.T.]

If you froze the slinky stiff with one half composed along a straight line and the other half uncompressed but straight along the same line, the wire midpoint would be the balancing point.

No. The COM would be somewhere in the uncompressed half.

It seems my problem arose from the false assumption that the COM would have to react instantaneously to a release and without delay.

That assumption is correct due to momentum conservation. Your problem is a misunderstanding of COM (see above).

 

[a1call]

Yes I understand now, with the leverage playing a role. Correct?

 

[A.T.]

Yes I understand now, with the leverage playing a role. Correct?

Yes, in terms of balancing, the COM of the uncompressed half has a longer lever arm than the COM of the compressed half.

When you release the slinky at the center, the COM will start moving linearly without delay. But the far end will initially continue on a circle.

 

[a1call]

Got it.
Thank you.

 

[A.T.]

Got it.
Thank you.

You're welcome.

Also note that torsion travels much faster along the slinky, than compression. In some slow motion videos of the slinky drop you can see, that the bottom starts to spin long before it starts to fall.

 

[A.T.]

Suppose you are spinning a Slinky around some point A in absence of gravity.

Here the experiment at time around 5:50:

 

[a1call]

Nice find.
It appears the slinky is overstretched compared to 1 g, making the delay over-shortened. Also the far end never gets stable. Perhaps the free fall duration is too short for the proper conduction of this experiment. Perhaps someone will try it aboard the International Space Station at some point.

 

[A.T.]

Perhaps someone will try it aboard the International Space Station at some point.

Inside? They have even less space there. A computer simulation could help here.

 

[bahamagreen]

Looks to me that for the Slinky held still in gravity, there is a tension gradient increasing from the bottom of the Slinky to the top (because each coil is holding the weight of the coils below it).
That means the COM is initially lower than the midpoint distance between the bottom and top (the coils below the midpoint are closer together than the ones above the midpoint).

Upon release, the contraction rate is also a gradient increasing from bottom to top (the top has greater tension and approaches the midpoint distance faster than the bottom, so the COM of the Slinky is actually falling from the moment of release of the top (delayed by the speed of sound in the metal path of the coils).

The bottom does not begin to fall until the increased tension above the midpoint reduces to match the tension below the midpoint... even though the COM is already moving down...