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A spring, a spring, a marvelous thing!
June 19, 2012 3:52 PM   Subscribe

Modeling a Falling Slinky

"How does a slinky fall when extended by its own weight and then released? We discover the surprising answer using a slow motion camera that records 300 frames per second."
posted by Blazecock Pileon (22 comments total) 3 users marked this as a favorite

 
Man, I already had a headache, now you've gotta drag senior-year Multivariable Vibrations class into things?

Hey, this is pretty cool.
posted by notsnot at 3:58 PM on June 19, 2012


previously.
posted by King Bee at 4:05 PM on June 19, 2012


I haven't looked much at that old post, but there seems to be a bit more to this post, probably. Carry on.
posted by Blazecock Pileon at 4:08 PM on June 19, 2012


A slinky dress.
posted by chavenet at 4:19 PM on June 19, 2012 [2 favorites]


This thread is worthless without Laplace transforms.
posted by LordSludge at 4:30 PM on June 19, 2012 [2 favorites]


And no damping component?? Egads, what are you kids learning in school these days?
posted by LordSludge at 4:32 PM on June 19, 2012


A slinky-based perpetual motion machine

disclaimer: the artist is a friend's brother
posted by qxntpqbbbqxl at 4:36 PM on June 19, 2012 [1 favorite]


A different model.
posted by Vectorcon Systems at 4:38 PM on June 19, 2012


Yeah, these are the exact same videos that were already posted, and I did a little video analysis and found the claim was incorrect. In this post, this graph showing the horizontal red line indicating the motionless bottom of the slinky, should start a slight downward slope immediately.
posted by charlie don't surf at 4:39 PM on June 19, 2012


Really, it depends on the spring constant and weight. If you dropped the valve spring from a car like this, the whole mass would move downward.
posted by notsnot at 4:51 PM on June 19, 2012 [1 favorite]


notsnot, I think that's not true. For one, consider the odds that a random Slinky has been fine-tuned to exactly fall in the "falls top first" parameter space. Second, note that the simulation didn't need to measure a Slinky's weight or spring constant.

When an ideal spring has been fully extended due to gravity, the displacement of each link of the spring from its rest position is the exact displacement needed to generate enough force to counteract gravity (otherwise it wouldn't be at rest). Thus, when the spring falls, the spring will necessarily contract with an acceleration matching that of gravity.
posted by 0xFCAF at 5:09 PM on June 19, 2012 [1 favorite]


I don't know, charlie don't surf... in these somewhat better videos, the bottom really does look completely motionless.
posted by gilrain at 5:13 PM on June 19, 2012 [1 favorite]


Yeah, the effect would be almost imperceptible in a stiff spring, but it should still occur. The researcher in the video I linked above speculates that it should even occur, to a tiny degree, in, say, a steel rod.
posted by gilrain at 5:14 PM on June 19, 2012


I don't have time to look into the physics tonight, but I'll mention a common physics demonstration about inertia and center of mass that is stuck in the back of my mind and somehow seems related.

Get a big weight, like 10 pounds, with an eye bolt at the top and bottom. Hang it from an eyebolt on a solid beam, using a foot of string capable of bearing like 20 pounds. Now tie another foot of string on the eyebolt on the bottom. So top to bottom you see a beam, eyebolt, string, eyebolt weight eyebolt, string.

If you pull slowly on the bottom string, the top string will break. If you yank down rapidly, forcefully, the bottom string will break.
posted by charlie don't surf at 5:43 PM on June 19, 2012


I guessed, was wrong and am so glad I chose computer science over mechanical engineering.
posted by double block and bleed at 7:02 PM on June 19, 2012


For some reason I read this as "model falling slink-ily" and clicked on it expecting to see thin women in glitzy clothing gracefully falling off catwalks.
posted by geegollygosh at 7:08 PM on June 19, 2012 [1 favorite]


This is my gut understanding of what's happening:

It's not that you dropped the slinky, you dropped the section you were holding..those coils have to fall a long way before the bottom gets the "kinetic information" that there's downward motion..before that point the bottom coils are suspended from the top coils as if they were still being held.

It's kind of a bandwidth issue, really..the spring is so thin and long it takes quite a while to communicate motion through it. Unlike the very high bandwidth of say, a baseball bat.

As the top coils accelerate towards the bottom the bottom gets increasing "pushed" downward until the whole mass is moving together.

I suspect in zero gravity you could perform the experiment sideways or upside down and it would work just the same.
posted by chronkite at 7:53 PM on June 19, 2012


Interesting article (and great videos), but I'm not sure I agree with the Wired guy's interpretation:
In this center of mass frame, both ends of the slinky do move. Why isn’t the motion of the ends symmetrical? I am not quite sure. I suspect it is because I am only plotting the relative positions of the top and bottom mass and not the ones in between. Actually, it is probably caused by the drag force I added in there. Anyway, I think you can still see that the bottom of the slinky is interacting with the rest of the slinky. The idea that “it doesn’t know the slinky has dropped” isn’t really needed here.
Isn't it obvious that the motion of the ends is asymmetrical because the top part of the spring is much more stretched out, and so contracts more quickly towards the centre of mass when the slinky is let go? I don't think it's clear at all that the bottom of the slinky is interacting with the rest of the slinky until the end. The video-physicist's explanation involving information (which is really another way of describing the relationship between cause and effect in a system where a cause can only propagate through a particular medium at a certain speed) may not be the most intuitive explanation, but it's as good as any other.

I suspect in zero gravity you could perform the experiment sideways or upside down and it would work just the same.

In zero gravity the slinky wouldn't stretch out at all, so it also wouldn't contract when let go.
posted by A Thousand Baited Hooks at 4:44 AM on June 20, 2012


In zero gravity the slinky wouldn't stretch out at all, so it also wouldn't contract when let go.

There is still mass and inertia in zero-g. Gravity wouldn't be doing the work for you, but I suspect if you pulled one side it would stretch out then subsequently contract because of work and energy storage.

Also this.
posted by pashdown at 7:17 AM on June 20, 2012


In zero gravity the slinky wouldn't stretch out at all, so it also wouldn't contract when let go.

I will bet one hundred million dollars that you are wrong.
posted by chronkite at 8:19 AM on June 20, 2012


Okay, obviously the slinky would stretch out if something was holding it at both ends. But an essential part of the experiment is that the slinky is being stretched out by the constant force of gravity acting on all parts of the slinky at once (acceleration would also work).

If, in zero gravity, you stretched the slinky out by holding it at both ends then let go of both at once, the slinky would contract inwards from both ends so that its centre of mass stayed still. If you only let go of one end, the motion of the slinky would depend completely on the force being exerted at the other end.
posted by A Thousand Baited Hooks at 3:04 PM on June 20, 2012


It would also stretch out if you held one end and "threw" the other..if you then brought the part you were holding to a complete stop just as the slinky reached its full extension and let go of it, you'd see the same behavior as in the video, sideways or upside down or whatever. Point is gravity has nothing to do with the effect.
posted by chronkite at 4:54 PM on June 20, 2012


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