

Explaining an astonishing slinky - scott_s
http://danielwalsh.tumblr.com/day/2011/10/17

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aqme28
I think you're really overcomplicating this.

1\. Since gravity acts on all points of the slinky equally, you can aggregate
this by saying that gravity acts on the slinky's center of mass.

2\. The slinky acts like a spring. Since it is being held stationary, the
forces on the bottom part of the slinky equal out. There is a force of gravity
going down which equals an upward spring force.

Therefore when it is dropped, the center of mass falls at g=9.8 m/s^2, while
the bottom part initially experiences no net forces.

You can also show why the net forces on the bottom of the spring will remain 0
(0 = mg - F_spring) for a spring obeying Hooke's law (F = k*d), where d falls
with gravity.

~~~
bentcorner
I don't think it's complicated either, but it's still (at first) unintuitive.

Interesting things tend to happen when you have forces that cancel/counteract
each other (e.g., spinning a bucket of water over your head, dropping a magnet
in a metal pipe)

~~~
beloch
It's more intuitive if you think of what a much tighter spring (i.e. higher
spring constant) would do. First, you'd have to hold it open because gravity
wouldn't be sufficient to stretch it out. If you released both ends
simultaneously, it would fully contract in much less time than it would take
to hit the ground. As a result, the bottom would move up fast before moving
down as the fully contracted spring falls.

The interesting thing worth noting is that you don't have to carefully choose
a spring such that the force of contraction is equal to the force of gravity
to observe a stationary bottom end, as in the gif. You just have to let _any_
spring hang freely. Prior to release, the upwards force from the spring on
it's own bottom exactly balances the downwards force from gravity. For a few
moments after release, until the point when the spring contracts enough that
it no longer applies the same upwards force to it's own bottom end, the forces
on the bottom of the spring will remain balanced and no acceleration will be
observed. This is, of course, much easier for the human eye to observe in slow
motion and with a relatively large spring with a low spring constant, such as
a slinky!

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britta
My friend writes this blog and I edit it, so if people have questions, feel
free to ask.

More explorations of falling slinkys:

* [http://www.wired.com/wiredscience/2011/09/modeling-a-falling...](http://www.wired.com/wiredscience/2011/09/modeling-a-falling-slinky/) \- a different way to model it.

* [http://www.youtube.com/watch?v=b9-XgSYLxDk](http://www.youtube.com/watch?v=b9-XgSYLxDk) \- video analysis.

* [http://wamc.org/post/dr-mike-wheatland-university-sydney-phy...](http://wamc.org/post/dr-mike-wheatland-university-sydney-physics-falling-slinky) \+ [http://www.physics.usyd.edu.au/~wheat/slinky/](http://www.physics.usyd.edu.au/~wheat/slinky/) \- more experimenting and a formal paper.

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afreak
The way that a slinky moves also demonstrates the question that some ask about
the giant stick and the speed of light.

To explain, if you had a button a light year away and had the option to press
it via a remote hand, the fastest we could tell the remote hand to press it
would be one light year.

However, there has been the question of whether a long stick that is one light
year in distance in lieu of sending a signal to the remote hand could be
faster. The way the slinky moves would demonstrate that the giant stick would
not move faster than the speed of light as the motion exerted on one end would
actually travel much slower.

The slinky is quite useful for demonstrating movement of objects. :)

~~~
pbhjpbhj
Even having studied relativity and such at Uni I haven't come across this
particular illustration before.

It should be called the Fing-long-er Gedanken.

Thanks for sharing.

~~~
jlgreco
I encountered it in university, though the example was given in a freshman
level course by some kid who thought he was cleverly circumventing
information's speed limit. The professor seemed quite pleased with it.

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colanderman
This could make for a fun amusement park ride: hang from a tower a few feet
off the ground using a bungee cord; detach bungee cord while simultaneously
pushing horizonally; bam you're Superman until the bungee cord comes crashing
on your head. Actually that last part needs work :)

~~~
wtallis
This derivation assumed a uniform distribution of mass in an unstretched
slinky, and that all the mass is stretchable. This probably means that in
order for this to work with a human attached to the bottom of the bungee cord,
the human's mass needs to be negligible compared to the mass of the bungee
cord. Thus, the human getting crushed by the bungee cord is not a problem to
be dealt with, it is a _requirement_.

~~~
StavrosK
Hmm, are you sure? The disturbance at the top of the cord has to travel
downwards, and the speed at which it travels is the speed of sound in the cord
itself. No matter what the mass at the bottom is, it can't "know" it's been
let go of until the information propagates in the cord.

~~~
ars
It feels it almost instantly (speed of sound in matter), because the tension
in the cord is lower.

In a slinky the inertia of the slinky itself resists things, but not so in a
light bungee cord.

You can try a weight at the top and bottom maybe.

~~~
jlgreco
If you watch the second Veritasium video, you'll see that adding a weight to
the bottom of the slinky does not change any of the behavior; it still does
not fall until the 'pressure front' reaches it.

A heavier weight at the top will accelerate just as fast due to gravity as a
small weight. As far as I can tell the only possible effect of a weight at the
top would be to ensure that the elasticity of the bungee cord does not
accelerate the top _much_ faster than the speed of gravity; you should be able
to just use a less elastic cord.

~~~
plaguuuuuu
You could use a parachute to arrest the rate of fall. The force of the
parachute acting against air resistance would resist the spring's compressive
force and thus the spring would recompress more slowly, lengthening the time
for the top to fall to the bottom.

~~~
judk
I think I that would also tend to lift the rider up to compress the spring,
and since the parachute is lifting the system.

~~~
colanderman
Mm, no, the opposite I think – the rider will fall slowly. The spring remains
extended due to the weight of the rider – being held aloft by a tower vs. a
parachute makes no difference here. However the _descending_ parachute
shortens the spring at a rate slow enough that it can be transmitted to the
rider well before the parachute reaches the ground, so the rider will fall a
little bit.

(This is analogous to simply lowering the hand holding the slinky – the bottom
eventually follows suit.)

~~~
plaguuuuuu
I'm pretty sure the rider stays in the same place O_o

Oh god. This is going to be the next airplane-on-treadmill isn't it.

~~~
jlgreco
I am very confident that it would fall.

Consider the slinky again. If I dropped it with a parachute to slow it.... no,
replace "parachute" with "my hand", but assume my hand follows the same
trajectory the parachute would... the end of the slinky would not wait for my
hand to approach it. The information about a change (the drop) would take the
same amount of time to reach the end of the slinky. The information travels
just as fast, even though the information is _" we're falling slowly"_ instead
of _" we're falling"_.

~~~
judk
I was thinking of a parachute unfurling, which generates an upward force as it
catches air, counteracting gravity temporarily.

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nerfhammer
simpler explanation?

when you hold the slinky in the air by the top, the weight of the bottom is
equal to the force of the tension of the spring, otherwise the bottom wouldn't
remain stationary.

once you let go, the force from the spring should decrease as the top falls
downward. But the top gathers up more of the bottom as it falls, so the
smaller bottom needs less force to hold it in place. Apparently the decrease
in mass of the bottom and decrease in tension of the spring exactly cancel
each other out.

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danbmil99
My granddad was an inventor. One time he had a long slinky (I think he must
have joined a few together) that was hanging, stretched out parallel to the
floor, on a number of closely spaced fishing lines, attached to a 4x4
overhead.

At one end he had a geared down variable-speed motor that pushed the slinky
with a sine wave motion. I believe the other end was fixed. In between, he had
painted various parts of the slinky blue. The blue parts seemed to be
completely still, even as the areas between the blue pulsated back and forth.

He said it was a demonstration he built "to show my ninkompoop investors about
standing waves".

~~~
mikegriff
That's a good way to show standing waves, but did they understand what they
were looking at, or did they just think it was some sort of a trick?

~~~
danbmil99
Well I think the "ninkompoops" invested about $10M in the 60's so they must
have thought he had something going on.

I guess it was that era's version of a dog/pony show.

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elipsey
Amazingly, the slinky can also time travel.

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callesgg
Does the people how think this is astonishing also think it is astonishing
than one will not move forward when walking backwards on an Escalator?

~~~
saurik
This doesn't seem to be quite the same effect as the entire body of the slinky
is not contracting. So, it isn't just that the bottom and top are moving
together as fast as the entire slinky is falling, causing the bottom to remain
stationary, but that the entire bottom of the slinky is remaining stationary
(not just slowed down, but exactly stationary), in its stretched state, until
the top of the slinky reaches that point. I don't think you are giving the
phenomenon enough credit.

