## Tagged: Canadian Centre for Astrophysics

# Slinky physics (cont.)

Posted by Parker on 29 June 2012 at 11:35 · Email a comment · Report a tpyo

In previous installments, we brought you video of the amazing levitating Slinky, and Peter Barss wondered how the Slinky had been calibrated to work exactly this way. I asked physicists to come forth, and they have—not just physicists, but an *astro*physicist. (Who better to explain levitation?)

Saint Mary’s grad Jonathan Dursi, now a senior research associate with the Canadian Centre for Astrophysics, furnished this detailed by elegant explanation:

Sometimes you hear that there’s three things taught in first year Engineering (or Physics, or whatever); things fall down; F=Ma; and you can’t push on a string.* It’s exactly those three things at play here, and it’s fun to see how they play out.

“How does this happen without calibrating the force exerted by the energy stored in the “spring” of the slinky to match the force of gravity?”

This calibration is done by the slinky itself, or in fact, any spring (or string, or…)

The spring stretches to the point that the tension in the spring — the force pulling the lowest link up — is exactly equal to the force gravity exerts on it to pull the link down. If those weren’t in balance (say, gravity was pulling more), then the lower parts of the spring would stretch out, and the tension would grow, and they would balance. (Hooke’s law, that the tension in the spring is directly proportional to how far it is stretched: F = k x, larger k means harder to stretch out the spring, is often a good approximation for these sorts of problems; but those are details. As long as the tension increased in any manner as you stretched out the spring, this would play out the same way). If it was the spring force which exceeded that of gravity, the spring would pull back, tension would reduce, and again things would end up calibrated.

This is F=ma; when you’re holding the top of the slinky, the fact that the parts aren’t accelerating upwards or downwards (a=0) means that at every point in the spring, the tension pulling upwards is the same as the gravitational force pulling downwards, so that the net force is zero (F=0, so m g = k x). If you went somewhere with a different gravitational acceleration (g), or placed a weight at the end of the spring, the spring would stretch out to achieve the same balance under the new force — the final length would be different, but the balance would again be met.

That much is true of really anything – an iron rod, or a slinky. But they all behave pretty differently when you hold them and drop them. If you hold up an iron rod vertically, it is also true that the bottom of the rod is being pulled upwards by a tension inside the rod and downwards by gravity, each perfectly balanced: but when you let go of it, the bottom does *not* stay in place until the top catches up!

This is where the “you can’t push on a string” bit comes in. The iron rod has a lot of internal rigidity; just try and squish one. When the top starts falling, not being suspended from above by any tension, it can’t catch up with what’s below it; it pushes what’s below it down with it, and the whole thing falls as a single rigid body.

But back to springs — the slinky *is* carefully built to _not_ be rigid. (It has to be able to flop over itself, after all, to climb down stars, alone or in pairs, and make a slinkity sound).

So imagine a three-link slinky, each link having a very small mass, and the whole thing is held up by the top, and there’s no internal rigidity. Each link is feeling a force of gravity down (m g) and a tension from the spring above it (k x) where x is the distance between links. The spring constant of a slinky, k, is really small — you can pull a slinky apart without exerting a lot of force.

Now let go of the top link. It immediately starts accelerating downwards, under the force of gravity. (The acceleration is g = kx/m). This *doesn’t* start _pushing_ on the bottom link — there’s no way to push! — but it slowly starts reducing the upwards tension on the second link, because the distance between them is reducing.

*But*, because the spring force is so small, the distance between links is large, and we’re accelerating from a full stop, this whole process takes a while. So it slowly starts falling, and as it falls, the tension pulling the second link upwards starts lessening — but only proportional to the amount of that distance the top link has fallen so far. Let’s say it would take the top link 1 second to fall the whole distance to the second link. It will take 0.5 second for it to drop even 1/4 of that distance, lessening the upwards force on that second link by only 25%. It will take 0.7 seconds for it to drop 1/2 of that amount, lessening the upwards force by only 50%. It’s only as the top link gets quite close to the second link that the second link looses most of its upwards support, and itself begins accelerating downwards in earnest, repeating this pattern to the lower link.

So, roughly speaking, the n’th link in the chain doesn’t start moving much until the n-1’st link has caught up to it.

Now imagine a more tightly wound slinky, so that it’s significantly harder to stretch it out. That means when you hold it up, it’s much shorter (harder to stretch it) so there’s a much smaller distance between links. There’s still no internal rigidity (say), but this whole process happens much faster, because the distance between links is much faster; the bottom “levitates” still but for a much briefer period of time. Keep tightening your mental slinky, and it happens faster and faster and faster until there’s no obvious moment of levitation at all.

This is what is meant by the “bulltwaddle” about information transfer and signals. The “We’re falling now” signal is sent from one link to the next by removing the upwards tension force. That signal travels at the same speed as the wave of now-collapsed top links moving downwards.

Quite generally, a wave travels at a speed proportional to the square root of the force which generates the wave (here, the spring tension) divided by the density of the medium. This is what sets the speed of sound in water or air (the square root of pressure over density); the speed at which a guitar string vibrates when plucked (square root of string tension over string mass), ripples on a pond, etc. Here, the wave travels at a speed proportional to the square root of the tension (k x) over the density of the spring (m/x). Because the spring adjusts itself initially so that k x = m g, this means that the wave speed is proportional to g * sqrt( m / k ), and it takes a time x over velocity, or sqrt(m/k), for the signal to travel. The looser the spring (small k), the longer it takes this signal to travel downwards. In a tighter slinky (larger k) this signal speed will get faster and faster and there will be a briefer and briefer moment of “levitation.”

(I will choose to overlook those cautious quotes Dr. Dursi placed around “levitation.”)

A reader who styles himself “Krackalakin” offers both a shorter explanation…

Consider that by holding a suspended Slinky by one end, you are holding a spring in tension… The force of gravity is the tension, or what is forcing the spring to extend and is directly proportional to that of the force puling it down – gravty! So there is no collusion or “transfer of secrets” that happen.

…and a link to an MIT classroom video wherein Prof. Walter Lewin explains the underlying principle.

The spring stuff starts about 1:30 in, but if you watch the whole thing, you can imagine what it would be like to be an MIT student.

Much obliged to both readers.

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* I’ve heard this rule applied in a different context, but this is a family blog.

Filed under: Physics · Tagged with: Canadian Centre for Astrophysics, Jonathan Dursi, levitation, Peter Barss, Slinky