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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.

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    $\begingroup$ 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. $\endgroup$ – dmckee --- ex-moderator kitten Mar 14 '13 at 14:19
  • $\begingroup$ Here's a WebGL animation of the numerical integration of the differential equation that describes the motion... princeton.edu/~rvdb/WebGL/Slinky.html $\endgroup$ – user93299 Sep 20 '15 at 13:42
  • $\begingroup$ For best results, how is one supposed to make a slinky move down stairs? I am totally blind, so I am not sure if one is supposed to roll it like a barrel, or if you have to orient it so that it moves down head first, if it had a head. $\endgroup$ – HeavenlyHarmony Jun 27 '18 at 13:24
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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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  • $\begingroup$ Just found this, thanks for the shout out, and nice answer! $\endgroup$ – Mark Eichenlaub May 27 '13 at 8:37
  • $\begingroup$ @MarkEichenlaub Sure thing! $\endgroup$ – joshphysics May 28 '13 at 2:05
  • $\begingroup$ This was the best answer I found : danielwalsh.tumblr.com/post/11566016253/… $\endgroup$ – daaxix Oct 30 '15 at 15:35
  • $\begingroup$ I choice a different way, when slinky becoming short , the tension act on the bottom will become smaller, but the gravity act on the bottom is not change, so the bottom will falling after release. Where is my mistake? $\endgroup$ – lanse7pty Nov 29 '18 at 12:59
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This problem has been considered before in the literature, the first instance of which seems to be [1] in which an exact solution is found for the motion of an ideal spring in this configuration. It also has some discussion of subtle differences between a slinky and the spring considered below which will not qualitatively change our conclusions. Notably absent is any discussion of collisions; spring elements are assumed to pass through each other freely. This assumption is mostly irrelevant for the range of times we are concerned with.

The first thing to note here is that most of our intuition about systems with springs comes from the idealized case of a massless spring holding a mass on its end, so that the tension is constant throughout the spring; in that case the result is simple harmonic motion. The initial condition here is quite different: the mass which is extending the spring is the mass of the spring itself, which is distributed along the full length of the spring. This is visible in the image as there are wider gaps between the coils at the top. Hence we should not expect simple harmonic motion for the spring.

The motion of the spring is instead described by the differential equation

$$m \frac{\partial^2 }{\partial t^2} y(t,\xi)= k \frac{\partial^2}{\partial \xi^2}y(t,\xi) + mg.$$

Here $\xi$ ranges from $0$ to $1$ and $y(t,\xi)$ is the vertical position of the point on the spring at time $t$ which (initially) has mass $\xi m$ above that point and $(1-\xi)m$ below it. In particular $y(t,1)$ is the position of the bottom of the spring at time $t$.

Other than the $mg$ term this is just the wave equation. The $mg$ does not actually complicate things much here; we can remove it by the equivalence principle by going to an accelerated frame. An important property of the wave equation is that waves propagate at a finite speed $v =\sqrt{k/m}$. This is necessary for e.g. causality to be preserved in electrodynamics.

Until the waves propagate from the location of the disturbance $\xi =0$ to the bottom, the bottom does not have any access to the information that the top has been released. Thus it continues moving exactly the same as it was moving while the top of the spring was being held up, which is just to say that it remains (exactly!) stationary. Once the wave hits $\xi=1$ at $t=1/v=\sqrt{m/k}$ the bottom begins moving. The key facts here are that the tension in the spring is local (this is what allows us to write the equation above), that the spring at $t<0$ has arranged itself in such a way as to balance all the forces to remain stationary, and that the disturbance originating at $\xi=0$ propagates only with finite speed.

References:

[1] M. G. Calkin, “Motion of a falling spring,” Am. J. Phys. 61(3), 261–264 (1993).

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Cool Question.

It's all based on Newton's second law. Take a look at this part of the slinky

enter image description here

On the part above, there are three forces acting on it. The tension from the upper part of the slinky, the tension from the lower part of the slinky and $mg$. All of them are balanced.

But if you talk about this part of the slinky:

enter image description here

The forces are Tension, $mg$ and that force by the hand. Note that $mg$ and tension are in the same direction i.e downwards and they are balanced too, obviously.

When you release the slinky, the net force on the topmost element becomes $tension+mg$ and is pointed downward but on the rest of all the parts of the slinky, the forces are STILL balanced.

So what you see is the topmost part rushing down while the others just stay in their position.

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Pretty interesting, it seems to defy logic. But here's what's happening. Normally, a horizontal slinky held stretched in two hands and released, both ends would move toward the center. But held vertically, and where gravity is the invisible other hand stretching the slinky down, when held from the top, the bottom comes to rest at a point of equilibrium where the spring can't be stretched more due to gravity, and due to gravity the spring does not pull the bottom up any higher than this balanced length.

Now when you release the top end, remember normally both ends of a spring will usually move toward each other, unless one end is still being held, in this case by gravity. The top comes down faster than you'd expect from free fall because you have the effect of gravity plus the effect of the spring wanting to compress working at the same time. The bottom wants to go up, but can't because it would have to overcome gravity, and every instant of the fall the spring is becoming less stretched, and therefore the spring force is diminishing.

So even though the bottom gets lighter, the stretched portion of the spring gets shorter, and at the exact same time it is becoming easier to rise to meet the top, the spring force is decreasing, never able to lift the bottom at all. The bottom never falls as long as it is stretched at all, because if it's stretched at all it is hanging rather than falling. It's a matter of balancing forces, spring pulling up, gravity pulling down, a constantly changing ─ getting lighter ─ weight (the still stretched portion) and a constantly changing ─ getting less strong ─ spring force (the degree of extension between coils).

And these two forces remaining ever changing but always in balance which offset each other in such a way as to give the "hanging" portion of the slinky the appearance of being weightless. But more weightless or at a constant altitude much like an airplane would be considered weightless; balanced forces would be a better explanation than weightless. A sheet of paper can hover in the air with the perfect air pressure underneath it that offsets the weight of the paper.

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    $\begingroup$ Welcome on Physics SE and thank you for your answer :) you might want to consider adding paragraphs to make your answer more easy to read though - as it is, I fear that a lot of people will not be attracted to reading it. $\endgroup$ – Sanya Jan 19 '17 at 20:33
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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.

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    $\begingroup$ 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 $\endgroup$ – Pranav Hosangadi Dec 25 '14 at 18:28

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