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 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.