Dan’s Geometrical Curiosities

Vi Hart’s “Doodle Music” video is a beautiful combination of art, math and music, illustrating parallels between symmetries in curves and symmetries in sound:

I particularly enjoyed the braid melody at 2:35, so it made me want to discover a mathematical function governing the vibrations in the air that define a “musical braid”.

Looking at the braid

First, let’s investigate what Vi’s musical braid looks like. If we look at a section of the spectrogram of the melody, here is what we get:

Even with this visualization, it is not obvious where the braid is. But Vi’s voice is luckily not a pure sine tone, meaning that there are many harmonics visible. Vi also chose to quantize the braid by discretely changing the pitch of her voice to make the result sound more musical, sticking to western musical tones. A pure sine tone usually does not sound very pleasing, and graphically looks like a flat line in the spectrogram; it appears similar to a the spectrogram of a whistle. To mathematically reproduce Vi’s musical braid so that it appears in the spectrogram, we need three independent sine tones, whose frequency varies in accordance with the drawing she made with sharpies.

Making the first thread of the braid

To mathematically “draw” Vi’s braid, we need to determine how to construct each thread individually, and then add them together at the end after erasing the parts that pass behind other threads. Let’s say that the shape of each thread is a sine wave. Then we need to create a frequency modulated signal, that is, a sine wave that changes pitch in a sinusoidal manner. A pure tone at a frequency \(f\) can be generated as $$y(t)=A \sin(2 \pi f t+\phi),$$ where \(t\) is in seconds, and \(\phi\) is the phase. In our case, we need the frequency itself to change as a function of time, in particular, we want our frequency to be $$f(t)=\bar{f}+\tilde{f}\sin(2\pi g t),$$

where 

$$\bar{f}=\frac{f_\mbox{max}+f_\mbox{min}}{2},\;\tilde{f}=\frac{f_\mbox{max}-f_\mbox{min}}{2}$$

It may be tempting to plug this result for \(f(t)\)directly into our pure sine formula, but this doesn’t quite work. To see this, imagine our answer is \(y(t)=A \sin(a(t))\), with \(a(t)\) some function to be determined. Recognize that in order for our frequency modulated signal to be locally identical to a pure sine wave with some phase, we require that the argument of the pure sine, \(2 \pi f t+\phi\), be the best approximation to the argument of our frequency modulated signal, that is, the tangent line. So we require that the slope of the tangent line (\(a’(t)\)) be equal to our desired frequency (\(f(t)\)), indicating that our undetermined function \(a(t)\) is essentially just the indefinite integral of our desired frequency function \(f(t)\):

$$y(t)=A \sin\left(2\pi \int f(t) \; dt+\phi\right).$$

Let us determine \(a(t)\):

$$a(t)=2\pi \int \bar{f}+\tilde{f}\sin(2\pi g t) \; dt=2\pi\bar{f}t-\frac{\tilde{f}\cos(2 \pi g t)}{g},$$

so our desired function is

$$A \sin\left(2\pi\bar{f}t-\frac{\tilde{f}\cos(2 \pi g t)}{g}\right).$$

This gives an output spectrogram of:

Perfect!

Weaving the second and third threads of the braid

Now we need two more sine waves, each offset by a different amount. Since there are \(2 \pi\) radians in a full cycle, and we have three waves that we want to be spread out evenly, \(\phi=\{0,\frac{2\pi}{3},\frac{4\pi}{3}\}\).

By adding several of these warbly sine tones together (call them “metasines”), we can create Vi’s musical braid. To make it look like they are passing over and under each other, we have to increase and decrease their respective volumes at the appropriate moments to give us the sense of depth. When this is done, the spectrogram looks like this:

The outcome

Here is the result played as a sound:

Sonified Braid by danielkwalsh

Interesting, but not as beautiful as the original. My friend Britta said it reminded her of a fire alarm.

A friend recently sent me a video that demonstrates a remarkable slinky-like object called a toroflux, designed by Jochen Valett:

It is a long ribbon of spring metal wrapped into a coil, except instead of coiling into a helix, as most springs do, this one coils through itself. It coils again and again, for the entire extent of the ribbon, where it is finally joined to the other end, creating a single loop of metal ribbon. Because the loop is passed through itself many times, the resulting shape is non-trivial in the knot-theoretic sense; it takes the form of a torus knot.

This spring behaves in an extraordinary way when manipulated. Because the spring is under tension, and is seeking the lowest energy state possible, when it is released, it prefers to open into a torus with the largest “minor radius” possible. That is, it becomes a horn torus, where the “hole” in the doughnut vanishes.

The force responsible for this then has a tendency to “grab” onto anything placed through the hole of the torus (like a stick), and if the object has enough friction, the toroflux will have good traction. The result is remarkable; the toroflux begins spinning, while simultaneously falling, creating the illusion of a silver bubble-like structure.

I was so excited by this that I had to purchase one for myself. If I had a ribbon of spring metal, I would have wanted to make my own, but unfortunately I couldn’t find anything that would suffice. Before it arrived, I made predictions about how such an object could work, and I have enjoyed playing with it. I would like to explain here how the toroflux works using an informal analogy, hopefully making it understandable.

Capturing the idea of “rolling” down

Let’s say we want to build a device that rolls down a vertical rope. When we think of rolling, the first thing that comes to mind is wheels. So we could accomplish the task by placing a number of vertical bicycle wheels radially in contact with the rope, each wheel connected to its neighbor via some framework. Here I have pictured two wheels for simplicity, but the principle could be generalized to many wheels. Provided that the device doesn’t slip off, and traction is maintained between the wheels and the rope, the assembly will fall, and the wheels will simultaneously turn as indicated by the blue arrows (relative to the falling frame).

This allows our apparatus to roll down straight. But what if we want the device to spin like the toroflux does? Well, just like when driving a car, you can guide a wheeled device in a certain direction by turning the wheels. In our case, we want the device to begin rotating. What would happen if we took these wheels and rotated the axis of each wheel some fixed angle about the radial direction? The adjacent diagram illustrates this with a top-down view of the wheel system, with the rope in cross-section. As the wheels “drive” down the rope, they have no choice but to make the whole assembly spin because of the way they are angled.

Now we have something that is much more similar to the toroflux. It not only rolls down a rope, but the whole system also rotates about the axis of the rope as it does so. The ratio of the downward velocity and the rotational velocity is determined by the angle of the wheels.

Doing away with the chassis

We now have a number of wheels held together by the support struts and also held tight to the rope with spring mechanisms to maintain traction. The support chassis is necessary for a simple reason: the wheels are not connected to one another. In the topological sense, the set of wheels is comprised of more than one connected component. We can rid ourselves of the chassis if we create a single wheel that locally (relative to the rope) looks identical to the wheels. That is, the rope should be unable to tell the difference between many tires touching it and a single, serpentine tire touching it and curving inward toward the rope in exactly the same way that the individual wheels did.

Here is a way to accomplish this. If you follow the curvature of a wheel, instead of curving around in a circle and reconnecting directly with itself, it must arc over toward the next wheel and join with it! This wheel will also do the same thing with its other neighbor, and this continues all the way around until we have only one spiraling wheel, all one piece. This does not interfere with the rolling of the wheels, because all the wheels spin in the same direction, so the “excess” tread from one wheel goes into the tread that is moving away on the neighboring wheel. Thus the physical result is essentially the same. (This forms a torus “knot” classified by {1,n}, signifying that in a single rotation around the rope, the coils spiral n times.) The above animation shows how the separate wheels are deformed and connected to each other to form a single coil-tire.

This is all fine and good, but we still require a great amount of tension in the spring to keep it pressed firmly enough on the rope to prevent slipping. As the device rotates faster, centrifugal force will pull the spring-tire apart, and it will begin to lose traction with the rope. We will see in the next section how to prevent this from happening.

Using centrifugal force to our advantage

Our device runs into some trouble when it starts spinning really fast. As the device spins, the coils pull apart and lose traction with the rope. We can put the coils under more tension, but there are practical limitations. There is a much more clever way to solve this problem, and it involves using centrifugal force against itself to achieve what we are looking for.

Instead of running the coil in the way I described earlier (as a {1,n} torus), we can loop the torus through itself, so that each coil wraps around the other side of the rope, so when the toroflux begins spinning rapidly, centrifugal force only pulls it tighter around the rope. In this way, we form a {n-1,n} knot, which is a true knot, so that upon orbiting the rope n-1 times, the coil winds n times. This knot resembles but is not actually equivalent to a single torus of circles of the Hopf fibration.

Hopefully this explanation makes clear how the toroflux works, and also helps illuminate the ingenuity and utter elegance of this invention. I also made a video to demonstrate it:

My friend Cody drinks a lot of Red Juice. When bought from Costco, two 96 fluid ounce bottles of Red Juice come paired like this:

That red connector between the bottles creates a handle for easy transportation. Most people dispose of these connectors and consume the Red Juice with the help of friends. Cody manages to drink this quantity of juice in a few days and sets the connectors aside, initially out of laziness but soon building a collection of them.

One day Cody asked me what the coolest kind of polyhedron was that one could build with 121 edges. Well, the rhombicosidodecahedron can be built with 120 same-sized edges. I had no idea what Cody had in mind — I expected he had obtained some kind of polyhedral construction kit — but instead he responded with a photo of his large collection of Red Juice connectors. After seeing the photo, I was genuinely worried for Cody’s well-being and what his dentist might have to say. But my concern for Cody’s health was soon outweighed by my excitement to begin the construction of a new rhombicosidodecahedron made from these edges!

What is a rhombicosidodecahedron?

A rhombicosidodecahedron is made of triangular (20), square (30), and pentagonal (12) faces. This animated gif (from Wikipedia) demonstrates the construction of one. First imagine placing a small explosive charge at the center of an icosahedron made by gluing triangles together — when the charge explodes, the faces blow outward. Then freeze time at the moment the shared edge of two neighboring triangles separates to a distance equal to one side of these triangles. Now connect these edges up again to form squares joining the triangles. At the same time, you will create 12 pentagons coming from where the vertices of the triangles used to be. The resulting solid is a rhombicosidodecahedron, and it is therefore no coincidence that it has the same number of triangles as the icosahedron, the same number of pentagons as the dodecahedron (the icosahedron’s evil arch-enemy), and the same number of squares as the number of edges on either of these polyhedra.

How many edges do we need?

Let’s find the number of connectors required to build such a structure, with the connectors acting as edges in the design.

• 20 triangular faces (from icosahedron)
• 12 pentagonal faces (from dodecahedron)
• E squares, where E is the number of edges on either the dodecahedron or the icosahedron. Sometimes I forget this, but it can be obtained by Euler’s formula:

\( \chi=V-E+F=2,\)

where \(\chi\) is the “Euler characteristic”, which is 2 for all convex polyhedra. This means that the number of edges for either the icosahedron or the dodecahedron is

\(E=12+20-2=30.\)

So both the dodecahedron and the icosahedron have 30 edges, so the rhombicosidodecahedron also has 30 squares. Then the number of edges on the rhombicosidodecahedron can be computed by adding the total number of edges belonging to each face:

\(12\times 5+20\times 3+30\times 4=240,\)

but we have actually counted the edges exactly twice, since each edge is counted from both of the neighboring faces. This tells us that the rhombicosidodecahedron has 120 edges. Another way to compute the number of faces without needing Euler’s formula at all is to notice that every edge is a neighbor to either a pentagon and a square or a triangle and a square. So right away, the number of edges must be equal to

\(12\times 5+20\times 3=120,\)

which gives the same answer. In order to have a collection of 120 connectors, Cody must have consumed 240 Red Juices, since each connector holds two bottles. This comes down to about 180 gallons of juice. Granted, he has been collecting for quite some time now.

An interesting property of the rhombicosidodecahedron is that each vertex connects exactly four edges. I’ll talk about this later.

How to build the rhombicosidodecahedron?

While discussing our plan of attack, Cody brought up a concern: because these connectors would be stacked at the circular part that holds the neck of the bottle, some Red Juice connectors would stack in asymmetrical ways. That is, a particular connector might be stacked third-out from the center on one side, but only the first-out on the other side. Aside from the poor aesthetics of angled connections, this would lead to asymmetries in the final product, as some connectors would be skewed and therefore not extending their true distance. A mathematical question comes to mind: is it possible to connect these in such a way that there isn’t any warping? Above is an illustration of a single vertex, showing which layer each connector occupies. If one of these connectors is on a different layer on a neighboring vertex, there will be warping.

My solution to this problem was to reduce it to another common mathematical problem: edge coloring. Since each vertex has exactly four edges radiating from it, we can ask if there is a way to color the edges of the rhombicosidodecahedron with exactly four colors in such a way that no two colors are adjacent. In this way, each of the four colors represents one of the four layers. It is tempting at this point to exclaim that it can certainly be done according to the four color theorem, but unfortunately we cannot apply that immediately here, because we are coloring edges and not vertices of a graph. We could also try to use Vizing’s Theorem, which states that the chromatic index of a graph is either \(\Delta\) or \(\Delta+1\), where \(\Delta\) is the largest vertex degree of the graph. Unfortunately, this does not help us here, since our maximum vertex degree is already four, so we cannot say right away whether we can color our graph with only four edges.

I decided that since I would end up needing a constructive solution anyway, I might as well find one. I began with a planar projection of the rhombicosidodecahedron (as seen on the right), beginning from the center and working my way outward.

Occasionally, I would come to an edge where I simply could not give it any color; all of my four colors were already taken by neighboring edges. At this point, I would just decide on a color for the new edge, and fix whichever edge was now conflicting. After fixing this edge, there would be exactly one other edge with a problem. This created a cascading effect that was relatively managable because the number of areas did not branch off exponentially. I continued following the chain reaction, edge by edge, until the problem got pushed off to the frontier of the area on which I had been working, at which point there were no more problems remaining. It was then safe to continue. After a bit of tedious work, I was happy to find that the entire graph was properly colored, and no edges remained. I was left with this:

At this point, I couldn’t help but make a 3D model of this, so it would be easier to see how this wrapped onto the polyhedron:

The construction process

Finally Cody visited me and remembered to bring the Red Juice connectors. We purchased some cable ties to hold it all together.

Success!

A friend recently sent me an animated gif depicting a man dropping a stretched slinky in slow motion:

The first time you watch it, it’s hard to resist calling it fake. The slow motion footage clearly shows that the bottom of the slinky doesn’t move until the whole slinky collapses to the bottom. But common sense tells us that this is impossible; of course all parts of the slinky should fall downward under the influence of gravity. (For an extended version of this gif, you can also watch the source video by Veritasium on YouTube, which includes a brief explanation.) My friend was especially skeptical, so I wanted to analyze this problem with cool illustrations and animated gifs…and of course some MATH. If you want to cheat, you can scroll to the end of this post to see the solution and its visualization.

I decided to idealize the problem like this: the slinky is an ideal spring with mass distributed uniformly throughout. It is also a spring that can pass through itself. These assumptions make analyzing the problem easier.

Initial Shape of the Slinky

Before we can figure out how the slinky bounces around as time goes on, we first must figure out what the slinky looks like before it is released. Springy things, we know, always try to un-spring themselves (that is, they try to release stress to seek out a “most relaxed state”, or static equilibrium).

(Many things have this behavior. For example, consider water in a pool. Water molecules are trying to “relax” in the earth’s gravitational field — they all want to fall as low as possible — but they cannot all fall to the bottom of the pool, because they all take up some finite volume. If you make waves in the pool, the molecules shuffle about in an effort to try to minimize the total energy in the pool. It’s not until the pool becomes still that you can see that all of the water on the surface has reached a compromise: it becomes flat — equal opportunity for all molecules.)

In our case, we have gravity pulling down, but we also have the stretchiness of the spring playing a role. How can we write down the total energy in the spring due to both the springiness and also the force of gravity?

First, we need a way to keep track of different points on the slinky. Position from the bottom is no good, because the slinky stretches, and points that were once one inch from the bottom now might be two inches. We need a naming system that always assigns the same name to any particular atom on the slinky, so I name an atom by how much mass is below it. While the slinky can stretch around, the amount of mass below a certain point on the slinky remains fixed, so the name of the bottom of the slinky will be zero, while the name of the top will be \(M\), where \(M\) is the total mass of the spring. We also need to know where all of these points on the slinky are — how high off the ground they are. We will call the height of a given point (parametrized by \(m\), the mass below the point) \(y(m)\).This allows us to write down a formula for the potential energy due to gravity using an integral:

$$U_{gravity}=\int_0^M g y\;dm$$

How about the potential energy in the stretch of the spring? Let’s look at a small region of spring. We can get an idea of how “stretched” the spring is at this point by computing the derivative \(\frac{dy}{dm}\). If the whole spring were stretched at this same stretch factor, then the length of such a spring would be \(\frac{dy}{dm} M\). However, the whole spring is not stretched linearly like this, but we know that the small region we’re looking at looks identical to any small region of the same size on the uniformly stretched spring. This means that the amount of stretch energy in this region is equal to the amount of stretch energy in our entire uniformly stretched spring, multiplied by the percentage of region we’re actually looking at. Mathematically, this means if the energy in the uniform spring is \(\frac{1}{2} k(\frac{dy}{dm} M)^2\), then we must multiply this by a factor \(\frac{dm}{M}\) to extract the potential energy due to the small region alone. This amounts to

$$U_{spring}=\int_0^M \frac{1}{2} k y’^2 M\;dm.$$

All together, this becomes:

$$U_{total}=\int_0^M g y+\frac{M}{2} k y’^2\;dm.$$

We want to find a spring configuration that minimizes this energy. This amounts to solving the Euler-Lagrange equations

$$\frac{d}{dm}\left[\frac{\partial L}{\partial y’}\right]-\frac{\partial L}{\partial y}=0,$$

where \(L=g y+\frac{M}{2} k y’^2\) is our potential energy functional. Take note that the primes here represent derivatives with respect to \(m\), not with respect to \(x\), which is unconventional.

We find the partial derivatives to be \(\frac{\partial L}{\partial y’}=k y’ M\), and \(\frac{\partial L}{\partial y}=g\). The resulting differential equation becomes \(k y” M=g\), which integrates to

$$y(m)=\frac{g m^2}{2 k M}.$$

Great. Now we’ve found the initial configuration of the dangling spring. An interesting thing to notice here is that the length of the dangling slinky is exactly half that of a weightless spring (with the same spring constant) with all of its mass concentrated at the point at the bottom. I promised there would be some neat graphics, so on the right is what the spring looks like before dropping, with mathematical precision.

How the Slinky Wiggles

Now that we know what the spring looks like initially, let’s work out how it moves over time. Looking again at the resulting differential equation, and writing it in the suggestive form

$$k M y^{\prime\prime} -g=0,$$

we can conclude that the left hand side of the equation represents the accelerations of a particular point on our spring, and as they’re being set to zero, it’s no surprise that we were looking for the equilibrium condition. This allows us to extend our result to a dynamic one by replacing the zero with the acceleration

$$k M y^{\prime\prime}-g=\ddot{y},$$

where the double dot over the variable y represents two time derivatives. It’s clear that the first term on the left hand side represents accelerations due to internal tensions in the spring, and the second term represents the acceleration due to gravity. If we move our analysis into a freely-falling reference frame, we see that the g term vanishes, so we are left with 

$$k M y^{\prime\prime}=\ddot{y},$$

which is the wave equation in one dimension, which might be more familiar in the form

$$\frac{\partial^2 u}{\partial t^2}=v^2 \frac{\partial^2 u}{\partial x^2},$$

where \(v\) is the wave velocity. In our case our “velocity” is \(\sqrt{k M}\), which does not have units of velocity in the standard sense. We want to figure out how the wave will propagate over time given the initial condition we found above.

Solving the Wave Equation

Using d’Alembert’s method for solving the wave equation, we can attempt to express the solution as a sum of two oppositely propagating waves:

$$y(m,t)=f(m-vt)+g(m+vt).$$

This wave must satisfy initial conditions, and once the spring is released, it must also satisfy certain boundary conditions on the ends of the spring. Once the spring is released, there cannot any longer be any tension on the edges of the spring. Thus, the stretching at the boundaries must go to zero, and hence \(\frac{dy}{dm}=0\) at these boundaries. These are equivalent to saying that the wave has “loose” ends; when a wave hits these loose ends, it reflects back off of them unchanged. If the wave was instead held fixed at both ends, like a string on a piano, the wave bounces back inverted, in order to ensure that the string remains fixed at the ends. In the same way, when our wave reflects, it reflects unchanged to ensure that the derivative at the boundaries remains zero.

Secondly, there is also an initial velocity condition. At the moment the slinky is released, it is at rest. How can two waves travelling in opposite directions sum to a motionless wave? Simple: if at a given position one wave (say \(f(m)\)) is moving up (due to the fact that it’s propagating to the right), then the other wave ( \(g(m)\)) must be moving down at the exact same rate (due to the fact that it’s propagating to the left). Since both waves are travelling at the same speed, but in opposite directions, the only possible conclusion is that \(f(m)=g(m)\). If you find this hard to believe, write it out with math and prove it that way. But the purpose of this whole post is to try to avoid math where possible, while still getting a correct answer. Since we know the form of the wave at time zero, we know that the shape of the two counter-moving waves must each be half the size of the total wave. So we conclude

$$f(m)=g(m)=\frac{1}{2}\frac{g m^2}{2 M k}.$$

We also can’t forget the time dependence. In particular, when the waves go along their way, the space they leave behind is replaced by a reflected version. An easier way for me to imagine this is that \(f\) and \(g\) are waves extending infinitely in both directions, generated by flip-floping one period of the above expression for \(f\) and \(g\). That is, periodically extend these functions to infinity in both directions, except every other period is mirrored. Now we can imagine these two waves countermoving and superimposing. To achieve this infinite mirror effect, we can use the modulo function, and the following cute trick: shift the parabolic solution over by M, modulo the argument of the function by 2M, and finally shift it back in the opposite direction by M. This will have the effect of looping the function in just the way we wanted. There’s one more remaining step: we need to move back into the non-freefall reference frame by subtracting a \(\frac{1}{2} g t^2\). When the dust settles, the solution works out to be

$$\frac{g \left(\text{Mod}\left[x-\sqrt{k M}t+M,2M\right]-M\right)^2}{4 k M}+\frac{g \left(\text{Mod}\left[x+\sqrt{k M}t+M,2M\right]-M\right)^2}{4 k M}-\frac{1}{2}g t^2.$$

Appreciating (and Visualizing) the Solution

OK, so that’s not very pretty to look at, but the real question is what does the answer look like? Time for a cool animated gif.

On the left is an animation of what our ideal slinky looks like as it falls. The really fascinating thing is that the top of the spring actually passes through itself as it falls. I made one end of it a little narrower than the other so that it would be easier to see this passage.

The second interesting part: we can see very clearly in this animation that the bottom of the spring does in fact remain stationary until the wave reaches it. Further, the position of the ends of the spring are piecewise linear in time, that is, the velocities are constant during any one inversion cycle. Once the next inversion cycle begins, both the velocities change, but are again constant. If you convince yourself that this is true during the first phase, then you must also immediately believe it for any subsequent phase, because once the slinky comes back to its initial configuration, the process is the same except it is falling faster. Thus the speeds will be the same except with a velocity offset.

Perhaps the strangest thing is that this isn’t only true for the endpoints. Look closely at the first phase and you will see that all points on the slinky remain at rest until the wave reaches them from the top. So as strange as it sounds, the slinky never actually changes its shape as it falls, aside from when it undergoes the inversion between phases. The slinky will continue leapfrogging its way downward forever in this way, and aside from some abrupt changes in speed, any point on the slinky remains at a constant velocity as it falls. These abrupt changes in speed ensure that the center of mass of the slinky will continue to fall at the acceleration due to gravity.

Now knowing the answer, is it any easier to convince ourselves of it? I often joke around with a friend of mine about the fact that once you know the answer to a problem, the results seem obvious. In accordance with that principle, there are some ways to understand this phenomenon. If we look at the very bottom of the slinky before it is released, the force of tension in the spring must perfectly cancel the weight of the spring at this point. When the slinky is released, these forces are still in equilibrium, and will remain in equilibrium until the neighboring part of the spring begins to move.

You can also think of it this way: the bottom of the slinky does not begin falling until the wave reaches it. The wave in the slinky carries the information that the top has been released, and since this information travels at a finite speed, there is no way that the bottom of the slinky can know that the top has been released right away. It takes a finite amount of time for this message to propagate through the slinky, and it is not until the wave finally arrives at the bottom that it begins to fall. The bottom of the slinky is like an eager friend of a runner in a race living on the other side of the world in ancient times. Before he can start celebrating, he must wait for the message to get to him via a boat halfway around the world. This principle still holds for points higher up on the slinky, too. Any point on the slinky whatsoever will not begin falling until the wave reaches it. This can also be seen from the slow-motion footage shown in the animated gif!

“Extending” the Solution

As you can see, nowhere in my analysis have I assumed that the slinky is of a certain length. Therefore, this principle should work for any kind of slinky.To see evidence of this, watch the follow-up YouTube video of an even longer slinky. In this video, you can see that the top of the slinky is trying to push through itself! Since it can’t, it begins to tumble after a while. In this video, they also briefly mention the idea of considering the slinky in the absence of gravity, which is equivalent to moving into a freely falling frame as we did in the above derivation. This amounts to removing the \(\frac{-1}{2} g t^2\) term. The solution ends up looking like this:

In the end, if you want to do this successfully with a much longer slinky, and show via experiment that the bottom still doesn’t move, then the radius of your slinky must increase proportionally to the length so that slight imperfections in the way it’s dropped, air currents, etc. do not affect the motion.

Quite some time ago, I came across a riddle, and I’d like to share a variant of it with you:

100 prisoners are incarcerated in a very peculiar jail. The jail has a single room with a light controlled by a switch that is initially off. The prisoners are each told that they will be taken one at a time into this room, and will be allowed to flip the switch if they so desire. They will then be taken back to their room, and another prisoner will be brought to the room, given the same privilege, etc. While there is no specific reasoning whatsoever to the order in which the prisoners are sent to the room, there is only one known fact: the order doesn’t forbid any prisoner from entering the room after some time. That is, if the order were to continue indefinitely, no prisoner would enter the room for his or her last time (this happens almost surely anyway). If any one prisoner at any time correctly claims that every prisoner has been to the room at least once (by shouting “DONE!”), then all the prisoners are set free. If he or she is wrong, they all are killed. Before the prisoners begin to be led into the room, they are allotted a single planning session in order to discuss a strategy. Is there a certain escape strategy for the prisoners?

If you like riddles, think about this one for a while before reading ahead. Keep in mind that the exact time of unanimous entry need not be known, but when a prisoner shouts “DONE!” each prisoner must have visited the room at least once, 100% of the time, always. If you get bored with this one, try to solve the problem assuming you do not know the initial condition of the switch.

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