Why does the base of this slinky not fall immediately to gravity? My guess is tension in the springs is a force > mass*gravity but even then it is dumbfounding.
Answer
What an awesome question! By the way, as far as I know, the original video is here for those interested.
One key to understanding this is the following fact from classical mechanics that is a version of Newton's second law for systems of particles:
The net external force acting on a system of particles equals the total mass $M$ of the system times the acceleration of its center of mass $$ \mathbf F_{\mathrm{ext},\mathrm{net}} = M\mathbf a_\mathrm{cm} $$ In the case of the slinky, which we can model as a system of many particles, the net external force on the system is simply the weight of the slinky. This is just given by its mass multiplied by $\mathbf g$, the acceleration due to gravity, so from the statement above, we get $$ M\mathbf g = M\mathbf a_\mathrm{cm} $$ so it follows that $$ \mathbf a_\mathrm{cm} = \mathbf g $$ In other words we have shown that
The center of mass of the slinky must move as if it is a particle falling under the influence of gravity.
However, there is nothing requiring that the individual particles in the system must move as though they are each falling freely under influence of gravity. This is the case because there are interactions between the particles that affect their motion in addition to the force due to gravity. In particular, there is tension in the slinky, as you point out.
You are absolutely correct that the bottom of the slinky does not move because the tension of the rest of the slinky pulling up balances the force due to gravity pulling down until the moment that the slinky is fully compressed and the whole thing falls with the acceleration due to gravity. Regardless, the center of mass is moving as though it is freely falling the whole time.
By the way, there are some nice comments about this experiment from the angle of wave propagation on physics.SE user @Mark Eichenlaub's blog which can be found here.
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