I never could make my slinky work
I was sending it the wrong signal
Researchers from the University of Sydney have explained why a spring dropped from a height - in this case the toy “slinky” – appear to ignore the force of gravity for a time. The very odd thing is that “if a slinky is hanging vertically under gravity from its top (at rest) and then released, the bottom of the slinky does not …
Well the plastic ones are probably not real slinkys (slinkies?)
I bought a new "original" slinky ~10 years ago that says on the box something to the effect that all official slinkys ever made have been made on the same machine. I assume they're still doing the same.
No what they're talking about is similar to Wile E Coyote running off the edge of a cliff and only falling when he realises he's not on the ground anymore. Then instead of his head staring at the camera as his bottom half goes shooting down, the opposite happens. Thinking about it again, you might be right, the bottom end of the slinky isn't just sitting in the air, it's trying to collapse up towards its other end as the other end is collapsing towards it. I need a coffee.
"Thinking about it again, you might be right, the bottom end of the slinky isn't just sitting in the air, it's trying to collapse up towards its other end as the other end is collapsing towards it. "
I must confess to agreeing with you two - that is precisely what I always thought was happening. Running a thought experiment of a spring lying on a table, held at both ends - release the ends and the spring contracts towards the centre of the spring (due to the tension in the spring). Make the force used to hold the spring equal to the force experienced by the end of the slinky when hanging in equilibrium, and the same thing happens obviously.
Back to the hanging slinky, if we could arrange to be in a free-falling environment, observing the centre of the slinky in front of us and starting from rest when the top of the slinky is released (say in a tiny lift with a port-hole) - surely we should see the same as the spring release on the table above ? (modulo friction, air resistance etc).
Or would we ? The paper seems to say no - but i've not read it yet - but i'd be intrigued to know more, and to have someone pick holes in the analogy above.
It's actually relatively simple if you calculate stuff from the center of mass.
In a slinky, anything below the center of mass will suddely retract as that part of it becomes essentially weightless as the spring starts to fall, giving the illusion of the bottom having "hang time".
The center of mass still drops down according to our old pal Newton.
I always thought about it in terms of centre of mass as well.
I think what the boffins are getting at is a bit more detail as to teh actual molecular mechanism of how a centre of mass works through communicating molecular bindings. While the slinky is attached at the top, it's part of a combined structure that has a completely different centre of mass somewhere else. The slinky only starts acting independently with it's own centre of mass once it's released. And since nothing can travel faster than light, the information that the top of the slinky has been released cannot be instantaneously transmitted to teh rest of the slinky, this 'information' is passed down the slinky as a release of tension between the molecular bindings making up the spring.
Or something like that.
Has it been proved that information is constrained by the speed of c.
What I've read there are theories of that information is instant independent of c. I guess this is quantum physics and I have very little understanding of them.
But I do not believe that this is applicable on this experiment, it should be enough with newton physics. Just break it down to molecular level with their own centre of masses.
Information is indeed constrained by the speed of light, or rather: if information can be transmitted faster than c then it is relatively easy to build a time machine, and that's assumed not to be possible.
However as you rightly say, the speed of light has nothing to do with this experiment at all. The information that tells the bottom of the slinky to start moving is transmitted by a longitudinal wave travelling down the slinky, which is essentially travelling at the speed of sound in the slinky, which is remarkably slow. This is, of course, the same for any object: if you drop a vertical metal rod then the bottom will not start falling until the information has reached it, by the same mechanism. In this case you don't see it because the speed of sound in the rod it rather high. You might *hear* it: as the signal is reflected from the end of the rod it may audibly ring.
"The center of mass still drops down according to our old pal Newton."
The slinky is both falling and contracting at the same time. It has two forces acting on it; gravity and the tension within the spring. The speed it contracts up at the bottom end exactly equals the speed the centre of slinky mass is dropping down. Hence the far end does not move until the centre of mass reaches it.
The force *up* stored within the spring is exactly that obtain from gravity when the spring is first stretched out. So it is no surprise that it precisely balances the gravity force *down* when the spring is dropped. At least for the time it lasts and the spring is fully contracted.
If someone was to stretch the spring out further at the bottom end, and let go at the same time as it was dropped, the force up would be greater for a while, and we would see the bottom end *rise* at first. No-one would think this unusual. So why all this nonsense when the forces are balanced?
All this talk about "messages" and "waves" over complicates what is quite simple. And I'm not a physicist.
That was my thinking, but, watching the video, the gaps between the rings towards the bottom don't seem to be contracting while the top rings are and the top falls. It does seem the point to which the top and bottom would collapse is moving from top to bottom as well, some other lower point is also moving in the same direction; above that point it is a contracting spring, below that it remains non-collapsing spring.
No matter how it actually works; the spring explanation seems far more reasonable to me than 'hangs around defying gravity until it receives a signal not to'. However; superb videos, so an overall thumbs-up.
Why would it contract at the bottom? The downwards force pulling it down is constant and until there is a change in the stretch at that particular point the upwards force will not change either. So only when contraction above has occurred will any change happen (which will be downwards acceleration as the contracting spring provides a reduced upwards force.
You do understand that the center of mass varies? It's just the summation (integral) of point-mass positions (here altitudes above ground) of the body's (slinky) divided by the total mass of slinky. What simply happens is that both 2nd and 3d Newton's Laws apply to point masses result in technically infinitely many equations. Those equations being added together result in one equation where you have \bar x\times M (total mass times center of mass) as a (new) variable, all interior forces of the slinky's particles will cancel out. The only exterior force will matter, that's gravity. So here we go
d^2(M\bar x )\over(d^2t)=-Mg, or
d(M\bar v)=-Mg dt (in the differential form, Sir Isaac was actually using)
PS So the original commenter was right. Things do get more interesting indeed with springs and slinkies is that center of mass is clear to vary w/r to the slinky itself. Mechanics is more complex, or rather cannot be turned to simpler model . Different slinky's particles will have different velocities and accelerations at different times. What is more interesting, that not only the lower end will have a near zero acceleration, the upper one will have more than g at some point (2g for a spring, as I believe).
Sir Isaac only (while D'Alembert, Leibnitz, Lagrange, Laplace, the Bernoulli's and other titans to be mentioned as well) I forgot to say about another great rival of Sir. Isaac's, Robert Hooke. That law named after him is what makes springs and slinkies appear out of the ordinary in this other examples, those interior forces (the tension) to be canceled make the particles move within the system in such a bazaar way.
It's not the same, no.
if it's sat on a table horizontally and stretched, then it will spring towards its centre again when released.
If it's hanging, then the bottom rings have round equilibrium between the "up" pull and gravity. If you turned off gravity, it would spring up from the bottom, much like when sat on the table.
But they're not switching off gravity, they're releasing the top. Therefore, you would expect that on the instant that the top is released, the bottom (which now has the same gravity pull down, but less force pulling it UP therefore balancing the gravity), should start to fall immediately.
I'm not pretending I understand this fully, but It intuitively makes sense to me that that the "top" rings, which were always being pulled by gravity and have now lost their balancing force imbued by being tethered, would "spring" down at a faster rate than the bottom rings which fall under gravity alone ....
"If it's hanging, then the bottom rings have round equilibrium between the "up" pull and gravity. If you turned off gravity, it would spring up from the bottom, much like when sat on the table.
But they're not switching off gravity, they're releasing the top."
Indeed - hence the description of viewing it from a free-falling reference frame (same as the spring itself).
"Therefore, you would expect that on the instant that the top is released, the bottom (which now has the same gravity pull down, but less force pulling it UP therefore balancing the gravity), should start to fall immediately."
The bottom of the spring is being pulled up by the spring tension and down by gravity (along with all the rest of the spring), so it should start to fall but slowly initially (in the static reference frame). In the free-fall reference frame of the centre of the spring it will appear to contract toward the centre - or at least that is what I was pondering ! I've been talking myself around a few - mutually contradictory - scenarios ever since :)
Yes .... I don't doubt that the centre of gravity still falls as expected (since the top can fall faster than gravity, and the bottom of the spring slower than gravity...)
I'm starting to think that (on the instant after release) each "ring" is still holding up the ring below. i.e. that each small component in the chain is effectively still in equilibrium ....... each bundle of springy atoms is still holding up the next bundle of springy atoms .... and that the really slow "propagation" inherent to the slinky means that the "release" of tension takes ages to propagate down ....
After all, if you're sky diving, you could still push and pull on objects, regardless of the fact that you're not anchored to any thing.
In free-fall, the spring collapses towards the centre. This does not contradict the article: if you arranged this in a free-falling environment, you would have one fundamental difference - you would have to hold both sides of the slinky. Therefore, the "release signal" starts from both ends at the same time.
The way to understand what is going on is to think about it from the point of view of the bottom of the slinky. Initially it's at rest: when the top of the slinky is released what tells it that it should start moving? The answer is: nothing does, so it stays still, until the signal moving down through the slinky reaches it.
"when the top of the slinky is released what tells [the bottom] that it should start moving?"
For that matter, what "tells" the top that it should start moving? Gravity right? I don't understand why gravity can't tell the top and bottom to start moving at the same time @_@
Hence, my upvotes of the tension/contraction theories promulgated elsewhere.
What tells the top it should start moving is that nothing is now holding it up.
Mechanics (Physics), Math (the three in one) do :) To simplify, apply 2nd, 3d Newtons' Laws to every particle of the slinky, say spring. Apply Hooke's Law. When you hold the upper end of the spring and the spring doesn't vibrate, the system is at rest, the sum all forces is zero. When the upper end is being released, you find all particles to endure two forces, the gravity and the Hooke's tension of the spring. So, at some point the acceleration of top is about 2g (due to gravity + tension) the bottom part has (gravity-tension) hence near zero.
"why gravity can't tell the top and bottom to start moving at the same time"
It doesn't even have to do this at the same time. Gravity is telling all of it *all* the time to start moving. It's just that until the top is released there is nothing counteracting that.
What the slinky is doing is probably no different from what a solid iron rod would do, it's just that the spring tension stored in the iron rod, and the time the bottom of it would spend "hanging" is infinitesimally small in comparison.
I'm not a physicists. I may not have a clue what I'm talking about. But it makes sense to me.
Hang the slinky from a hook, then start pushing up the bottom with your hand.. Notice that now it is the top that doesn't seem to react to the force being applied to the bottom, and the tension has to be removed from the slinky before the top starts to move up. The paper is saying, I believe, that the release of tension itself acts as a signal.
Exactly, which means the top part should be falling with the acceleration generated by a force equivalent to 2g compared with a "non springy" item being dropped. This state may only be true in the very beginning or until the whole spring has collapsed.
I would guess the initial acceleration is comparable to being pulled with 2g and it will decrease towards 1g
In that state the bottom part will have an acceleration of equivalent to being pulled with 0g.
While collapsing. It would be interesting to see a item being dropped at the same time.
The information is however instant/(or travels with speed of c) over the whole physical object. When released it got released. Remember it's 1 g that pulls bottom end of the spring in it's first place.
What's new in this? All newton physics. Expect that you cannot calculate with centre of mass. Or you can if you wan't to know when it reaches the ground. Centre of the length of the spring should give you the calculation point for when it will reach the ground.
Bingo. The energy it takes to hold lower end up is stored when we stretch it.
This might be run in reverse horizontally. One end doesn't move until the other end pulls it. In that case,however, we'll need to move faster, or attach a mass to the stationary end, as a Slinky(tm) has much less inertia than is needed reach maximum elongation at ordinary speeds.
"The bottom doesn't move because it's being held in place by spring tension, which is only released when the adjacent coils collapse."
In part. Guesses.
Loss in stored Gravitational Energy, GE, from the bits at the top and in between plus stored Spring Energy, SE, from the bits in the middle act to compensate for the bit at the bottom 'thinking' about gaining Kinetic Energy, KE, from the above.
I note the misleading shiftyware demonstration was done on an Apple.
ARGHHHH ARGHHHHH ARGHHHHHH
If the top of the spring is accelerating at the speed of gravity (just keeping the term simple, it is in free fall as the attraction of the spring to itself (its collapse) is equal and opposite) the reaction to that (tension / intention) is equal and opposite acceleration up towards the top of the spring (with an equal and opposite force.)
The bottom of the spring is thus accelerating upwards at the speed of gravity.
At about 1:12 to 1:23, you can see small disturbances near the bottom of the slinky being propagated towards the bottom. These seem to be mechanical vibrations being propagated along the length of the slinky material (i.e. high speed travel through a metal) as the coil above bash into each other. As might be expected, these disturbances travel faster than the compression wave of the slinky structure itself.
...connected to the fact that nothing can travel faster then light - not even information. So if you had a metal rod between the Earth and Alpha Centuri and there was a person on each end and you tried to comminucate by pulling or twisting the rod (thinking the other end would move in the same way at the same time )and you could use this to communicate instantly over 4 light years, the movement wouldn't arrive at the other end until over 4 years later?
Again, icon, apologies etc. The question may not warrent it, but I think the answer may well do.
Not really, actually.
The point you miss is that even if you have a 4-light-year-long device, it's still a physical device, made up of atoms. Those atoms have to impart force on each other and literally MOVE in order to propagate the force to the next atom. It's nothing weird or special, just the sheer length of the thing means it will take a while for the atoms to compress up to the point that they push the next atom, push the next atom, push the next atom, etc. until you have a wavefront moving towards the other end as the action takes place. Think of the rod as, say, a sponge and you'll get the idea, no special physics here, you just have to push enough for the material to see the effect all the way along (and we don't tend to deal in single materials longer than, say, a couple of hundred meters on Earth, ever, so we never "see" this effect but it's there even on the most exotic of materials and when you're talking 3.8 x 10^ 16 meters, those effects would be a little more visible).
You would have to have an entire incompressible material and absolute zero for any weird physical effect to do with the speed of light (and we already have an implausibly long, straight, perfect rod that stays as such when subjected to forces necessary to move a 4-light-year-long piece of material, so we're way out of the bounds of "practical" physics here). But the effect is inherent and visible with just simple Newtonian physical explanations too.
Here there is no "gravity signal", it's just that the bottom of the spring is preventing from falling because it is suspended by the bit above it. The very top of the spring may be released but that takes a fraction of a second to move any appreciable distance and yet all the parts underneath are still suspended to the atom above them (which has gravity acting down, and an attractive force from the atom above it). There is no "speed-of-light" or sub-atomic effect acting here, though the principle is similar. It's literally just going to take a little while for the top of the spring to compress the spring and impart physical forces on the atoms below it that overcome their attractive forces to each other. Again - happens in all materials, just this one is particularly pretty to watch because it takes long enough that we can see it because of the springy nature of it.
(The speed-of-light effect, for instance, of the material "realising" that nothing was holding it up and thus the entire material was subject to gravity would travel a 30cm slinky in 100 picoseconds, not 0.3 seconds - one 3,000,000,000th of the time).
This is a simple, Newtonian effect of having a solid material made of atoms imparting forces on each other and nothing "fancy" at all. In fact, the whole "signal" speak is very dubious from a physical point of view and I think is being misinterpreted to make it sound more interesting. The "signal" is just physics taking effect as the atoms "catch up" with their neighbours.
If you tried to communicate by pulling or twisting a metal rod, the signal would propagate at the speed of sound through the rod. Twisting would be a shear wave, and pulling a longitudinal wave - they would have a different speed of propagation (S-waves and P-waves). It would be very slow compared to light speed, and would take lots longer than 4 years.
For steel, the longitudinal wave speed is ~ 6000m/s, and the shear wave speed ~3200m/s.
4 light years / (6000 m/s) ~ 200 000 years
It would go across the rod at the speed of sound rather than the speed of light. I don't know what the speed of sound in the metal rod would be, as it depends on the type of metal, but probably somewhere in the order of 6000 m/s rather than 299 792 458 m/s for light.
It is highly suspicious that this slinky does not fall at the speed of gravity! Looking at the movie, it is clear that the slinky could only be collapsed this way by preinstalled tension packages distributed all over the coil. Moreover, professor Crank for the University of Lower Uppsala has found nanothermite particles in the slinky debris, a clear indication that there was fowl play.
North Korea has just announced a new anti-gravity drive based on using just the bottom few coils of a slinky. This new technology will have them on mars in a few months it is claimed. The astronauts will return the following day when they will collect some more supplies before returning to Mars, and back the following day for news conference, followed by a return to Mars and......the North Korean government admits there are still a few kinks to iron out!
Exactly what I was thinking. In essence, the rocket stack compresses ever so slightly during the initial push. A fully-kitted Saturn V stood some 363 feet tall. For simplicity, let's assume a propagation speed equal to the speed of sound through aluminum (the rocket stacks were mostly aluminum while the fuel was primarily liquid and therefore would have a slower speed of sound): about 21000 feet per second. Meaning, the rocket would compress for about 17 thousandths of a second before the top of the rocket stated moving as well.
I, too, agree with earlier posts putting a lot of this down to spring tension and gravity.
IANAS so some of my reasoning may be faulty, please point it out.
So an object has potential energy whilst being supported preventing it from falling. When you release the bottom of the slinky, the allowable mass that falls is down to the gravity acting on the mass that pulls against the spring tension. So when it stops, you effectively have the energy of the falling portion of the slinky to transfer that energy into the spring, (storing it in the spring) so the bottom position is still because the force of gravity vs the spring tension (and the spring tension only exists, don't forget, because one end of the spring is fixed by being held). Once you let go of the top, that potential energy is released as kinetic(?) energy, pulling the spring together. At the same time, the entire mass is now being drawn down by gravity, which hasn't changed, obviously, so the spring doesn't move from it's lowest position because gravity dictated it's position in relation to the spring tension initially, and now that spring tension is allowed to act by releasing the top, the bottom is effectively being pulled down, whilst the tension pulls it up which is a released of the previously stored energy.
Phew. Sorry about the badly worded paragraph.
However, with this in mind, I can see the flaw. I would expect the rings to collapse all evenly at a constant rate. So maybe this is why they are paid for it, whereas I work in IT.
Nah, your right, but's still a spring the top part collapse forms more or less a "solid" cylinder This is the part that actually will be moving, the one with no stored energy left in it.
And as I tried to point out in a post here the acceleration should be in the beginning equivalent to the acceleration of an non springy object being attracted with 2 g. In a sense the top part will "fall" faster than e.g. e pen should.
I really would like to see this experiment with a solid object next to the top part being dropped at the same time.
I also think that if you drop a solid object from the height of the mid point of that spring they should hit ground at the same time.
It turns into heat. The atoms in a stretched spring are alightly further apart than they would be in an unstretched spring, and therefore there is slightly more energy in the inter-atomic bonds that are destroyed one by one by the acid.
A better puzzle is this one
You have a battery connected through a timer switch to an electromagnet. This is supported on a disk which is at rest but free to rotate on very good bearings.
The timer clicks, the magnetic field decays, the electromagnet kicks, and the disk starts rotating. Where did the angular momentum come from?
It's from the Feynman lectures (of course).
ANSWER: The energy would be released during the spring's dissolution. Since it wouldn't happen all at once, parts of the spring would find ways to break free even as it dissolves. So once you dropped that spring and clamp in the acid, I'd slam a lockable lid on it and find some cover.
A variant to what Omni describes would be to smash a hard candy with a sledgehammer. In this case, you're compressing the candy very hard and very fast, well past its ability to absorb it, so it breaks apart and finds a way to release the energy (out the gap between the hammer and the surface). Same idea: the energy gets released at some point.
I think it's both: if the speed of sound/signal in the spring were infinite, you would still see the bottom of the suspended spring remain apparently still for a bit, but the coils would collapse absolutely uniformly. We don't see that.
The finite speed of (different kinds of) signal in the spring means the coils near the bottom don't collapse much at all until the wavefront of "going slack" approaches. But twist in the coil can get there much earlier because its signal speed is greater being through the stiff solid not via the spring property.
It's amusing but obvious that the top of the spring initially accelerates way harder than gravity since it's under tension <=> holding up the weight of the whole spring. That acceleration is because the tension is still there; you can view that as the weight still being there even though the whole is in freefall, because the rest of the spring doesn't yet know it's in freefall. Actually it's due to the mass/intertia of the spring which makes it unable to instantly collapse once the tension is removed, which itself is exactly what causes the speed of a signal to be finite.
I think the right way to view it isn't signal theory. It's that there is tension in the spring. When the top is released, it accelerates downwards at a greater acceleration than G, because it is also being pulled down by the tension in the spring.
For every action there is an equal and opposite reaction. At the bottom of the spring, the tension is pulling upwards, oposing gravity. And of course, the Earth itself moves towards the Slinky, by some unimaginably minuscule amount.
If you position the unstretched slinky such that its center of gravity is the same height above the ground as the stretched one, then yes the bottom should move about the same time as the unstretched bottom passed, and both would proceed to hit the ground pretty close to simultaneously.
For the obvious answer, look at the geography ... It's Australian so, as the Aussies are upside down relative to the UK, the bottom of the slinky is actually the top and you see a simple pulse run along the spring. Try it - hold the slinky, tap the bottom of the spring in the UK and the pulse runs. According to my calculations, if they tried the same thing standing on their heads it wouldn't work. QED
If you model the static solution with lumped masses you get tension at a distance X links from the bottom of the spring is proportional to X. e.g. for 3 masses of 1kg connected by two massless springs and held up by a hand, tension in the bottom spring is 10N, the top spring is 20N, and the hand applies 30N (to balance the 30N weight of the device as a whole). So the bottom link is say 1m separate while the top link is only 1/2m separate. Effectively, each point is pulled down by the weight of the spring below, and that's more weight at the top and less at the bottom. It works just like undersea water pressure.
This is why you see in the video the spring starts out more stretched at the top and less at the bottom.
When that hand force disappears the spring will *not* compress uniformly towards the center of mass. In fact, the forces continue to balance for the bottom two masses while the top mass has a 30N downward force on it. Literally, acceleration of top mass is 3m/s and acceleration of all the other masses is zero. As the top link shrinks by X=1.5t^2 the second mass starts to experience acceleration of k.1.5.t^2 due to the created imbalance in the spring forces there. So that moves, and the bottom link shrinks, as Y=k/8.t^4 (integrate X twice). In other words, the acceleration of the second mass is zero, the rate of change of acceleration is zero, but the rate of change of the rate of change of acceleration is non-zero. The acceleration of the third mass has the 6th derivative of X be non-zero. For larger numbers of masses the final expression for the bottom link is something like k/N! * t^2N, which is "motionless" in anyone's book.
Once the top mass reaches the 2nd mass the spring force from above is now zero, so it's like the 2nd mass has been "released" in the same way the 1st one was, plus it gets an impulse from the collision. So now the situation is about the same, but one mass lower. Thus, we expect to see what the video shows - that the top mass is the only one which moves, until the spring force is eliminated, then the next mass starts to move, and so on. It's a highly non-linear situation.
So it can be deduced from Newtonian physics, but it's not a uniformly extended spring so the simple center-of-mass calculations don't work (and they predict something else anyway - that the bottom rises up). The center-of-mass will as usual obey Newton's law of gravitation but the behaviour around the CoM is not what you might expect. Anyone who claims this is not counter-intuitive isn't thinking about it hard enough ;) OTOH, papers describing what a game physics engine could figure out don't seem to be genuine "new physics" :)
"So the bottom link is say 1m separate while the top link is only 1/2m separate"
Swap those two. The bottom is *less* separate; the top is *more* separate.
"(and they predict something else anyway - that the bottom rises up)"
Or it may fall :) It definitely doesn't seem to stay where it is for a long period of time.
This is an excelent physical demonstration of the difference between Wave Speed and Electron Speed in a power transmission line. Not because it is a good analog -- it is not a good analog -- but because it demonstrates two things: the wave speed is finite, and the result is unexpected.
Those two ideas are both difficult for many non-engineers dealing with any kind of transmission line. They expect wave speed to be non observable, and they expect that their ideas based on infinite wave speed will be correct.
So what you're telling me is if we could prevent the bottom of the slinky from ever finding out the top of the has been dropped, it would just sit there until otherwise informed? Sounds like the way we run our helpdesk..
Imagine the cost savings in travel if we could work out a way of implementing this slinky theory into the aircraft manufacturing process..
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