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!
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.