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.
I still think that's blurring in incidental factors. The minimal explanation IMHO would be:
Every solid has a vibration propation speed, or "speed of sound". The more compressible, the lower the speed. Slinkies are very compressible.
When the slinky is released, the bottom has nothing to hold it up anymore. However (the key part), the affect is delayed by how the "signal" of the lost support can only travel at the (very low) speed of sound in the slinky.
If/when the object is a lot stiffer (say, a broomstick), the bottom "learns" of the loss of support a lot faster and you couldn't so easily perceive the lag between dropping and the start if the bottom's fall. Plus, the % length contraction is a lot smaller for a broomstick than a slinky.
If you had supported it only from the bottom, there would likewise be a delay in the signal reaching the top, although the slinky would already be closed and no longer have the compressibility of when it has room to contract further.
Even shorter: it's all about signal propagation and the slinky permits a really slow speed of it.
The following is a response from Dan (who hasn't registered an account yet):
Thank you for your comment. I wrote an explanation of why I'm not overcomplicating the problem - allow me to address your concerns:
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."
Well, not quite - a non-rigid body behaves differently in the case of uniform gravitational field compared to the case of an equal force localized at its center. To see this, take the limit where the spring constant goes to zero. This physically represents a cloud of uncoupled particles, all but one of which would remain motionless if the force were applied only to the central particle. It is true, however, that the center of mass of the falling spring will accelerate at a rate g when the spring is dropped. This is clear since the force of gravity is the only force acting on the spring after release, but it's also interesting to look more closely at the complex motion of all the other parts of the spring.
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."
Yes, this is true, and I use a more general version of this principle (not restricted to the bottom point only).
"Therefore when it is dropped, [...] the bottom part initially experiences no net forces."
This is also true, but more limited than I described. Showing that there are initially no net forces on the bottom of the spring only shows that its acceleration at time zero is zero. A priori it is possible that the bottom's displacement is only zero instantaneously. If you think this is absurd, consider the problem with n masses connected by n-1 ideal, massless springs, and you'll see that the displacement of the bottom particle has non-zero (but possibly small) displacement even instantaneously after the top mass is released. So it is not obvious. The finite propagation speed emerges when we pass to the continuum system.
I hope this helps explain my post - I tried to look deeper into the problem, by extracting the motion of the entire spring rather than just its bottom at the instant it is dropped.
I 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)
It'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!
The 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. :)
I 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.
This 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 :)
This 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.
Hmm, 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.
If 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.
You 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.
Mm, 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.)
Consider the slinky again. If I dropped it with a parachute to slow it.... no, replace "parachute" with "my hand", but assume my hand follows the same trajectory the parachute would... the end of the slinky would not wait for my hand to approach it. The information about a change (the drop) would take the same amount of time to reach the end of the slinky. The information travels just as fast, even though the information is "we're falling slowly" instead of "we're falling".
Very 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.
In 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.
Presumably 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!
But 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.
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.
My 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".
This 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.
More explorations of falling slinkys:
* http://www.wired.com/wiredscience/2011/09/modeling-a-falling... - a different way to model it.
* http://www.youtube.com/watch?v=b9-XgSYLxDk - video analysis.
* http://wamc.org/post/dr-mike-wheatland-university-sydney-phy... + http://www.physics.usyd.edu.au/~wheat/slinky/ - more experimenting and a formal paper.