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Explaining an astonishing slinky
- colanderman 13y agoThis could make for a fun amusement park ride: hang from a tower a few feet off the ground using a bungee cord; detach bungee cord while simultaneously pushing horizonally; bam you're Superman until the bungee cord comes crashing on your head. Actually that last part needs work :)
- wtallis 13y agoThis derivation assumed a uniform distribution of mass in an unstretched slinky, and that all the mass is stretchable. This probably means that in order for this to work with a human attached to the bottom of the bungee cord, the human's mass needs to be negligible compared to the mass of the bungee cord. Thus, the human getting crushed by the bungee cord is not a problem to be dealt with, it is a requirement.
- StavrosK 13y agoHmm, are you sure? The disturbance at the top of the cord has to travel downwards, and the speed at which it travels is the speed of sound in the cord itself. No matter what the mass at the bottom is, it can't "know" it's been let go of until the information propagates in the cord.
- ars 13y agoIt feels it almost instantly (speed of sound in matter), because the tension in the cord is lower. In a slinky the inertia of the slinky itself resists things, but not so in a light bungee cord. You can try a weight at the top and bottom maybe.
- jlgreco 13y agoIf you watch the second Veritasium video, you'll see that adding a weight to the bottom of the slinky does not change any of the behavior; it still does not fall until the 'pressure front' reaches it. A heavier weight at the top will accelerate just as fast due to gravity as a small weight. As far as I can tell the only possible effect of a weight at the top would be to ensure that the elasticity of the bungee cord does not accelerate the top much faster than the speed of gravity; you should be able to just use a less elastic cord.
- plaguuuuuu 13y agoYou could use a parachute to arrest the rate of fall. The force of the parachute acting against air resistance would resist the spring's compressive force and thus the spring would recompress more slowly, lengthening the time for the top to fall to the bottom.
- judk 13y agoI think I that would also tend to lift the rider up to compress the spring, and since the parachute is lifting the system.
- colanderman 13y agoMm, no, the opposite I think – the rider will fall slowly. The spring remains extended due to the weight of the rider – being held aloft by a tower vs. a parachute makes no difference here. However the descending parachute shortens the spring at a rate slow enough that it can be transmitted to the rider well before the parachute reaches the ground, so the rider will fall a little bit. (This is analogous to simply lowering the hand holding the slinky – the bottom eventually follows suit.)
- plaguuuuuu 13y agoI'm pretty sure the rider stays in the same place O_o Oh god. This is going to be the next airplane-on-treadmill isn't it.
- tedsanders 13y agoVery edited: Ah, so the human's mass can be heavy compared to the bungee cord, but if that's the case then the duration of the superman effect would be short.
- wtallis 13y agoIn the limiting case of an ultralight bungee cord, the human has to start falling almost immediately after the cord is cut, because the center of mass cannot fall as required without movement of the human as well. This incidentally gives us lower bounds on the speed of sound and stiffness of such a bungee cord. In order for the human to be held aloft for any significant period of time by the recoiling bungee cord after it is cut, the cord must be relatively heavy - enough to move the center of mass of the cord+human system well away from the human.
- carlob 13y agoYou also need to account for the larger mass at the bottom…
- bdegman 13y agoAs long as you don't hit the ground while fully at rest it'll be fine.
- cjg 13y agoBut it wouldn't feel like flying. It would just feel like you were being held up by the bungee cord. You might as well have the same ride but not actually drop the bungee - sounds safer too.
- plaguuuuuu 13y agoPresumably you would travel in an arc, because the spring's compressive force would be pulling you towards the top of the spring... the spring would then not be perpendicular to the ground any more So, that arc would curve upwards away from the ground. You'd probably get hurt falling down!
- judk 13y agoThe compressive force is already canceled by gravity, unless some other device is adding a extra downward force temporarily.
- elipsey 13y agoAmazingly, the slinky can also time travel.
- losethos 13y agoNigger posts this shit he doesn't understand and what does he want? I asked my little brother if the chairs would float if we removed all the air in the room. LOL I got a new question. If a helium balloon didn't pop, would it float or sink if you removed all the air in the room. Heavy objects do, in fact, hit the Earth sooner. F=Gm1m2/r^2. (applied to object AND to Earth)
- britta 13y agoMy friend writes this blog and I edit it, so if people have questions, feel free to ask. More explorations of falling slinkys: * http://www.wired.com/wiredscience/2011/09/modeling-a-falling-slinky/ http://www.wired.com/wiredscience/2011/09/modeling-a-falling... - a different way to model it. * http://www.youtube.com/watch?v=b9-XgSYLxDk http://www.youtube.com/watch?v=b9-XgSYLxDk - video analysis. * http://wamc.org/post/dr-mike-wheatland-university-sydney-physics-falling-slinky http://wamc.org/post/dr-mike-wheatland-university-sydney-phy... + http://www.physics.usyd.edu.au/~wheat/slinky/ http://www.physics.usyd.edu.au/~wheat/slinky/ - more experimenting and a formal paper.
- callesgg 13y agoDoes the people how think this is astonishing also think it is astonishing than one will not move forward when walking backwards on an Escalator?
- saurik 13y agoThis doesn't seem to be quite the same effect as the entire body of the slinky is not contracting. So, it isn't just that the bottom and top are moving together as fast as the entire slinky is falling, causing the bottom to remain stationary, but that the entire bottom of the slinky is remaining stationary (not just slowed down, but exactly stationary), in its stretched state, until the top of the slinky reaches that point. I don't think you are giving the phenomenon enough credit.
- judk 13y agoAcceleration is zero on an escalator, which is intuitively simpler.
- aqme28 13y agoI think you're really overcomplicating this. 1. Since gravity acts on all points of the slinky equally, you can aggregate this by saying that gravity acts on the slinky's center of mass. 2. The slinky acts like a spring. Since it is being held stationary, the forces on the bottom part of the slinky equal out. There is a force of gravity going down which equals an upward spring force. Therefore when it is dropped, the center of mass falls at g=9.8 m/s^2, while the bottom part initially experiences no net forces. You can also show why the net forces on the bottom of the spring will remain 0 (0 = mg - F_spring) for a spring obeying Hooke's law (F = k*d), where d falls with gravity.
- bentcorner 13y agoI don't think it's complicated either, but it's still (at first) unintuitive. Interesting things tend to happen when you have forces that cancel/counteract each other (e.g., spinning a bucket of water over your head, dropping a magnet in a metal pipe)
- beloch 13y agoIt's more intuitive if you think of what a much tighter spring (i.e. higher spring constant) would do. First, you'd have to hold it open because gravity wouldn't be sufficient to stretch it out. If you released both ends simultaneously, it would fully contract in much less time than it would take to hit the ground. As a result, the bottom would move up fast before moving down as the fully contracted spring falls. The interesting thing worth noting is that you don't have to carefully choose a spring such that the force of contraction is equal to the force of gravity to observe a stationary bottom end, as in the gif. You just have to let any spring hang freely. Prior to release, the upwards force from the spring on it's own bottom exactly balances the downwards force from gravity. For a few moments after release, until the point when the spring contracts enough that it no longer applies the same upwards force to it's own bottom end, the forces on the bottom of the spring will remain balanced and no acceleration will be observed. This is, of course, much easier for the human eye to observe in slow motion and with a relatively large spring with a low spring constant, such as a slinky!
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- nerfhammer 13y agosimpler explanation? when you hold the slinky in the air by the top, the weight of the bottom is equal to the force of the tension of the spring, otherwise the bottom wouldn't remain stationary. once you let go, the force from the spring should decrease as the top falls downward. But the top gathers up more of the bottom as it falls, so the smaller bottom needs less force to hold it in place. Apparently the decrease in mass of the bottom and decrease in tension of the spring exactly cancel each other out.
- afreak 13y agoThe way that a slinky moves also demonstrates the question that some ask about the giant stick and the speed of light. To explain, if you had a button a light year away and had the option to press it via a remote hand, the fastest we could tell the remote hand to press it would be one light year. However, there has been the question of whether a long stick that is one light year in distance in lieu of sending a signal to the remote hand could be faster. The way the slinky moves would demonstrate that the giant stick would not move faster than the speed of light as the motion exerted on one end would actually travel much slower. The slinky is quite useful for demonstrating movement of objects. :)
- jessaustin 13y ago...the fastest we could tell the remote hand to press it would be one light year. Time is measured in years, not light years! b^)
- ahelwer 13y agoIt would travel at the speed of sound in the stick material, since you are transmitting a compression.
- pbhjpbhj 13y agoEven having studied relativity and such at Uni I haven't come across this particular illustration before. It should be called the Fing-long-er Gedanken. Thanks for sharing.
- jlgreco 13y agoI encountered it in university, though the example was given in a freshman level course by some kid who thought he was cleverly circumventing information's speed limit. The professor seemed quite pleased with it.
- danbmil99 13y agoMy granddad was an inventor. One time he had a long slinky (I think he must have joined a few together) that was hanging, stretched out parallel to the floor, on a number of closely spaced fishing lines, attached to a 4x4 overhead. At one end he had a geared down variable-speed motor that pushed the slinky with a sine wave motion. I believe the other end was fixed. In between, he had painted various parts of the slinky blue. The blue parts seemed to be completely still, even as the areas between the blue pulsated back and forth. He said it was a demonstration he built "to show my ninkompoop investors about standing waves".
- mikegriff 13y agoThat's a good way to show standing waves, but did they understand what they were looking at, or did they just think it was some sort of a trick?
- danbmil99 13y agoWell I think the "ninkompoops" invested about $10M in the 60's so they must have thought he had something going on. I guess it was that era's version of a dog/pony show.
- pbhjpbhj 13y agoWhat were they investing in?
- danbmil99 13y agoGramp's machine was called "Big George". It vibrated stuff at insane frequencies and amplitudes. At various times he claimed it 1) Made paper stronger 2) Aged wine 3) Could filter industrial smokestack pollution But after many years working with engineers, I finally realized that gramps just wanted to make the loudest, highest frequency sound possible. His patent (one of many, no he wasn't a troll, he built everything he patented) http://patent.ipexl.com/US/3708033.html http://patent.ipexl.com/US/3708033.html