Tuesday, June 26, 2012

The falling slinky problem

Let's take a break from statistics and do some physics!

My friend Ted Bunn recently wrote about the falling slinky problem in his blog.  He points to this video, which shows a falling slinky in slow motion.  After the top of the slinky is released, the bottom seems to hover until the top reaches it.  The effect is particularly strange because if you look carefully, the top of the slinky does not accelerate as we expect for an object in free fall.  Rather, it falls at a constant rate.

Ted explains:
...the information that the top end has been dropped can’t propagate down the slinky any faster than the speed of sound in the slinky (i.e., the speed at which waves propagate down it), so there’s a delay before the bottom end “knows” it’s been dropped. But it’s surprising (at least to me) to see how long the delay is.
This explains why there is a delay, but to me it doesn't explain why the delay is the same as the time it takes for the top of the slinky to reach the bottom.  There are lots of models out there that explain parts of this behavior, but the ones I found are either complicated or wrong.

Here's my take on it.  First, let's assume that what we see in the video is correct: the slinky collapses from top to bottom, so that each coil doesn't move until the one above it comes down and (nearly) hits it.

Let's call the initial length L and the mass m.  After some time, a fraction of the slinky, x, has collapsed.  At that point, the collapsed part of the slinky has mass xm at height (1-x)L.  The rest of the slinky is spread uniformly [EDIT: this assumption is not right...see Ted's comment below] between height 0 and (1-x)L.  So the center of mass is

x(1-x)L + (1-x)(1-x)L/2

Since the slinky is in free fall, we know the center of mass as a function of time:

L/2 - g/2 t^2

If we set those equal and type them into WolframAlpha, we get

x = sqrt(g/L) t

Which means that the top of the slinky is moving at constant speed.  Remember that x is the fraction of the slinky that collapsed; to get the distance traveled, we multiply by L:

d = xL = sqrt(gL) t

So the speed of the top of the slinky is sqrt(gL).

We can get to the same result a different way by using the formula for wave speed in a vibrating string: sqrt(T/μ), where T is tension and μ is mass per linear measure.  In this case T=mg and μ=m/L.  Plug that in and get wave speed sqrt(gL).

I think this analysis is useful, but to be rigorous, I haven't really explained why the slinky behaves the way it does.  I have only shown that if the slinky collapses from top to bottom (as it appears to), then the top moves at a constant speed (as it appears to).

[UPDATE: Provoked by my amateurish attempts at Physics, Ted Bunn wrote up a version of this model that deals correctly with the change in the density of the spring from top to bottom.  The result is that the speed of the top of the slinky is almost constant -- it slows down a bit at the end. ]


  1. I don't think this analysis is quite right, although it's got the large-scale features of the motion qualitatively correct. The problem is that phrase "spread uniformly." The initial state of the slinky has a far-from-uniform linear mass density: it's much more stretched at the top than at the bottom.

    1. Sadly, you're right. I did a little work to see if this model can be salvaged, but it doesn't look good...

  2. I think the analysis is messy but doable: http://blog.richmond.edu/physicsbunn/2012/06/26/more-on-the-slinky/ . Although the final result is a very messy formula, it turns out to lead to very nearly the same behavior as your original model -- that is, the top end of the slinky moves with nearly uniform velocity.

  3. At the beginning,the center of the spring mass is not at its center. It is somewhere below its center.

    1. That's correct, and a corollary to Ted's objection to my assumption that the mass is spread uniformly.