Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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.

share|cite|improve this question
It helps to know that the spring forces between elements of the slinky are local. And you should note that the system was in equilibrium just before release. From there you should be able to deduce the behavior of each part of the slinky in the instant of release. And for the "next" moment and so on... It is worth thinking this through. – dmckee Mar 14 '13 at 14:19
Here's a WebGL animation of the numerical integration of the differential equation that describes the motion... – Robert J. Vanderbei Sep 20 at 13:42

2 Answers 2

up vote 11 down vote accepted

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.

share|cite|improve this answer
Just found this, thanks for the shout out, and nice answer! – Mark Eichenlaub May 27 '13 at 8:37
@MarkEichenlaub Sure thing! – joshphysics May 28 '13 at 2:05
This was the best answer I found :… – daaxix Oct 30 at 15:35

The force on the centroid is NOT = mg. It is m(g - T) where T is the tension in the spring at the position of the centroid. I think that a more rigorous treatment will find that the topmost coil experiences F = ma = mg since T has a purely downward component.

The way to sort this out is to conduct an experiment comparing the fall of a slinky and a freely-falling rigid body and see if it's the centroid that accelerates at g or the topmost coil as I predict.

share|cite|improve this answer
No! The force on the centroid is $m \cdot \vec{g}$. Once you've concentrated the body at the centroid, there is nothing else to exert the tensile force on the centroid – Pranav Hosangadi Dec 25 '14 at 18:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.