PDA

View Full Version : Gizmo with Five Swinging Balls


CalMeacham
05-22-2000, 10:38 PM
Interesting thing about "Newton's Cradle" -- physcs teachers say that it proves conservation of energy and momentum, since those laws constrain the motion of the balls completely. If you raise one ball, then after it hits only one ball comes off the other side. Drop two balls, and two balls exit the block of five after the collision, etc. A really neat trick is to raise two one one side and ny one on the other, then dropping them so they hit simultaneously. You get one bouncing of the other side, and two bouncing off the side that had one raised.

The problem is that the explanation isn't correct. As the physics and math types out there all know, you need five equations to fully constrain the motions of five balls tou've limited them to effective one-dimensional travel), and the laws of conservation of momentum and conservation of energy only give you two equations. There are a LOT of possible combinations of motion of the five balls after the collision besides the obvious one (in fact, there is an infinte number of solutions). Clearly something else is going on.

This was addressed about twenty years ago in he American Journal of Physics, but I found their final result vague and unsatisfying.

C K Dexter Haven
05-22-2000, 11:09 PM
Link to the Mailbag Answer is: Who invented the little gizmo with the five swinging balls?
(http://www.straightdope.com/mailbag/mnewtcradle.html)

Chronos
05-23-2000, 12:48 AM
Y'know, it's folks like you who make me feel stupid for not realizing something like that myself... What this means, of course, is that there's three more equations governing that system. Anyone have any idea what they would be? We need something easily generalizable to other numbers of balls, too.
BTW, CK, why the double link?

Bletchley
05-23-2000, 08:36 AM
The Newton's cradle demonstrates "Inertia" an object at rest's tendency to remain at rest. Also an object in motion tends to remain in motion. Swinging one ball overcomes one ball's woth of inertia. Swinging two balls etc...

So if you made a special Newton's Cradle with one of the balls on the end exactly twice as heavy as the others (ie. make the others hollow), you you see two lighter balls flying off the end if you started the contraption in motion by lifting and dropping the single heavy ball.

The friction created between the balls at the point of impact also helps to stabilize the array. This friction in enhanced by using a fixed attachment of the wires to the balls (ie. a weld). This restricts their rotation, which would rather quickly lead to the entire array swinging back and forth together. They end up doing this anyway if you watch them long enough without touching them.

Because of the balanced nature of the device, equations describing the motions of the balls starting with five or more formulas (angular momentum, gravity acceleration, kinetic energy, etc.)will largely cancel out those terms, yieilding a surprisingly simple equation. One that emphasises inertia and friction. So go to it you physics nerds, lets see some equations. Let me get you started.

F=ma

You take it from there.

CalMeacham
05-23-2000, 08:50 AM
Chronos: I'm not that smart. I wish I was, but the fact that you only have two equations to describe the motion of five balls was pointed out in the AJP article. I didn't think of it. Also, it's quite possible (probable, even) that I found the article unsatisfying because I didn't properly understand it.

Bletchley: I don't think you quite understand. Newton's Cradle is often said to demonstrate the law of conservation of momentum (which is what I think you are saying by saying that "inertia" is responsible), but that alone can't explain the observed effect. There are other solutions besides one ball left = one ball right that conserve momentum. Even when you add in the conservation of energy there are other solutions besides one ball left = one ball right. It's true that the ball are constrained to move in arcs by the strings, turning this into essentially a one-dimensional problem (you can demonstrate this effect with pennies sliding on a tabletop, by the way), but "friction" has nothing at all to do with it.

I'll se if I can find the AJP article reference and post it.

Bletchley
05-23-2000, 12:53 PM
Cal: Let me try and be a little more clear. Conservation of energy, mass, momentum, whatever allows you to set up an equation. Then, depending on how many variables we want to take account of, such as friction, we add terms to the equation. You can then do manipulations to the equation(s) that allow you to solve for, lets say, velocity of a given ball with relation to time from start, or really lots of different things to help you understand the physics of what is happening. Thats where calculus and algebra might come in. As I recall, in this type of problem, solving for something you might be asked to do on a test, usually means cancelling out some of the more complex terms. (When that starts happening, it always makes me think, OK, I am gonna get this one right)

As if you hadn't already surmised, it has been a long time since I took physics, but I kinda liked it.

I would like to have a look at that article, if you would be so kind as to link me up to it.

Bletchley
05-23-2000, 12:55 PM
Cal: Let me try and be a little more clear. Conservation of energy, mass, momentum, whatever allows you to set up an equation. Then, depending on how many variables we want to take account of, such as friction, we add terms to the equation. You can then do manipulations to the equation(s) that allow you to solve for, lets say, velocity of a given ball with relation to time from start, or really lots of different things to help you understand the physics of what is happening. Thats where calculus and algebra might come in. As I recall, in this type of problem, solving for something you might be asked to do on a test, usually means cancelling out some of the more complex terms. (When that starts happening, it always makes me think, OK, I am gonna get this one right)

As if you hadn't already surmised, it has been a long time since I took physics, but I kinda liked it.

I would like to have a look at that article, if you would be so kind as to link me up to it.

CalMeacham
05-23-2000, 01:42 PM
OK! I've found the article. It's "Simple Explanation of a Well-Known Collision Experiment" by F. Herrmann and P. Schmalzle (with an umlaut over the "a" -- I don't know how to do that here). I appeared in the American Journal of Physics, vol. 49, #8, pp. 761-764, August 1981. There's a follow-up article, "How does the Ball-Chain Work?" by F. Hermann and M. Seitz, AJP vol. 50, #11, pp. 977-981, November 1982. For extra credit you might want "Energy Transfer in One-imensional Collisions of Many Objects" by John B. Hart and Robert B. Herrmann (evidently a different Herrmann) in AJP Vol. 36 #1 , pp. 46-48.

As Herrmann and Schmalzle explain in the first article, "...it is easy to convince oneself that the conservation laws of energy and momentum are NOT SUFFICIENT (their italics) to explain the observed behavior of the balls." They give a counter-example of a three-ball system in which one ball initially approaches the two stationary balls with a velocity V. Afterwards, the first ball rebounds with velocity _V/3, while the two balls that had been stationary move off with velocity 2V/3. Such an outcome satisfes conservation of momentum and energy, but it's not the observed effect, which is that one of the stationary balls moves off with velocity V, while the other two bals remain fixed.

H and S say that in addition to those conservation laws you have to have a system with no dispersion of the energy. But this isn't satisfying -- it seems to be restating the obvious without telling you why it should happen. My naive expectation is that you can assume perfectly elastic collisions and somehow get the observed result.

In the followup paper H and S look into the physics of the collisions between the balls, arguing that the initial collision is almost perfectly dispersionless. They note that after the collision the mass of balls is actually slightly separated and moving with a slight velocity, so the result that's usually stated for the situation (one ball moves off with the same velocity and the others stay in place!) isn't strictly true. Still seems a bit unsatisfactory

I got these in a university library. I don't think they're available on-line.

CalMeacham
05-23-2000, 01:46 PM
OK! I've found the article. It's "Simple Explanation of a Well-Known Collision Experiment" by F. Herrmann and P. Schmalzle (with an umlaut over the "a" -- I don't know how to do that here). I appeared in the American Journal of Physics, vol. 49, #8, pp. 761-764, August 1981. There's a follow-up article, "How does the Ball-Chain Work?" by F. Hermann and M. Seitz, AJP vol. 50, #11, pp. 977-981, November 1982. For extra credit you might want "Energy Transfer in One-imensional Collisions of Many Objects" by John B. Hart and Robert B. Herrmann (evidently a different Herrmann) in AJP Vol. 36 #1 , pp. 46-48.

As Herrmann and Schmalzle explain in the first article, "...it is easy to convince oneself that the conservation laws of energy and momentum are NOT SUFFICIENT (their italics) to explain the observed behavior of the balls." They give a counter-example of a three-ball system in which one ball initially approaches the two stationary balls with a velocity V. Afterwards, the first ball rebounds with velocity _V/3, while the two balls that had been stationary move off with velocity 2V/3. Such an outcome satisfes conservation of momentum and energy, but it's not the observed effect, which is that one of the stationary balls moves off with velocity V, while the other two bals remain fixed.

H and S say that in addition to those conservation laws you have to have a system with no dispersion of the energy. But this isn't satisfying -- it seems to be restating the obvious without telling you why it should happen. My naive expectation is that you can assume perfectly elastic collisions and somehow get the observed result.

In the followup paper H and S look into the physics of the collisions between the balls, arguing that the initial collision is almost perfectly dispersionless. They note that after the collision the mass of balls is actually slightly separated and moving with a slight velocity, so the result that's usually stated for the situation (one ball moves off with the same velocity and the others stay in place!) isn't strictly true. Still seems a bit unsatisfactory

I got these in a university library. I don't think they're available on-line.

zut
05-23-2000, 03:56 PM
Hmmm, Cal, that's... interesting. I always thought I knew how those things work. I do have a couple questions:

You say, Herrmann and Schmalzle... give a counter-example of a three-ball system in which one ball initially approaches the two stationary balls with a velocity V. Afterwards, the first ball rebounds with velocity _V/3, while the two balls that had been stationary move off with velocity 2V/3.

When they "give a counter-example," what do you mean? Did they actually construct an apparatus that performs as you state? If so, how is it different from the original? Or did they just do a thought experiment, showing that the unexpected outcome is still consistant with laws of conservation?

Later on, you also state,
H and S say that in addition to those conservation laws you have to have a system with no dispersion of the energy.

What does "no dispersion of the energy" mean? I've never run across this term, but it seems to imply conservation of energy, i.e., perfectly elastic collision; however, it appears they are referring to something else, since they say "dispersionless" in addition to conservation of energy.

Finally, you write a couple sentences,
My naive expectation is that you can assume perfectly elastic collisions and somehow get the observed result...
[H and S] note that after the collision the mass of balls is actually slightly separated and moving with a slight velocity, so the result that's usually stated for the situation (one ball moves off with the same velocity and the others stay in place!) isn't strictly true.

Both your assumption and the observation by H and S seem to jibe with how I thought these gizmos worked, namely, there occurs a series of perfectly elastic collisions (one-to-two, two-to-three, etc.) that you can analyze seperately, one at a time, which ultimately transfers momentum from the first ball to the last. The slight separation and movement is because it takes a little time to actually transfer momentum, thus leaving the balls slightly displaced.

Seems to me that you could do the same thing with a line of pool balls on a pool table, a situation where, at least to me, visualization is easier. In this case, it still seem loical that a series of collisions is occuring.

C K Dexter Haven
05-23-2000, 04:17 PM
<< , so the result that's usually stated for the situation (one ball moves off with the same velocity and the others stay in place!) isn't strictly true. >>

Kinda obvious, though, innit? Since if the end ball moved off with the same velocity, you'd have constructed a perpetual motion machine...

RM Mentock
05-23-2000, 04:36 PM
Kinda obvious, though, innit? Since if the end ball moved off with the same velocity, you'd have constructed a perpetual motion machine...


I don't think so. Otherwise, the law of inertia itself would imply some sort of perpetual motion machine--that things just keep going. The name "perpetual motion machine" implies more than that. In order to qualify as perpetual motion machines, the machine has to do real work (power something, or overcome friction) and still keep going.

CalMeacham
05-23-2000, 04:37 PM
Zut, you have the situation as it occurs after the first collision (acording to the second paper). When you start out, though, all the balls are touching, so you can't view it as a two-body collision. H and M do some computer modeling to show what happens, and they conclude that there is little "dispersion" (which they don't define in the papers) during the first interaction. (The numbers are strictly a thought experiment, by the way -- their point is that you DON'T observe the first ball rebounding and the other two moving together with 2/3 the velocity). Afterwards, though, they claim the balls separate a little, so that subsequent interactions are two-ball interactions. Yes, you can do this with billiard balls. As I pointed out above, you can do it with pennies on a tabletop, too. You just have to ignore rolling effects and friction. But the question of why, in the first collision, only one ball comes off the end of a connected series is still a valid and nontrivial one.

CK-- there's no question of perpetual motion. All I say is that some balls rebound with some velocity right after the impact. After all, we really do see some ball coming off with some velocity -- the only question is what velocities they have. The motion isn't perpetual. In the Newton's Cradle model gravity starts pulling the ball down, and eventually air resistance and other dissipative forces drain away the kinetic energy.

JonF
05-23-2000, 06:16 PM
The SHAKE Algorithm Applied to the Lattice Newton's Cradle (http://titania.math.ukans.edu/~bond/996/report/math996.html)

CalMeacham
05-23-2000, 10:36 PM
JonF:

Interesting mathematical model. I have no idea what SHAKE is -- I assume it's their dynamic modeling program for interactions.

The big assumption they make is in the nature of the interaction -- they use a inverse-power repulsion, they say. This is very diffeent from what was used in the papers I cite. suspect it's not a good model for a real metal all, but it gives very pretty results.

Thanks for finding that. And thanks to all those who commented.

--CM

zut
05-24-2000, 08:41 AM
...and I think I understand what the authors are getting at.

In the first paper, they model the Newton's cradle ball arrangement as a system of discrete springs and masses. They then show that to reproduce the symmetric collision behavior the balls exhibit (i.e., n balls collide from the left, and n leave from the right, with the remaining balls stationary), the spring-mass system must be dispersion-free. By dispersion-free, they mean that the wave speed associated with each natural mode is identical (i.e., no dispersion of [spacing between] the wave speeds).

This makes sense if you understand that in the original collision, all modes are excited. In a dispersion-free system, the portion of the disturbance associated with each mode propagates at the same speed, so it kind of "comes back together" at the same spot, pushing the n right-hand balls. If the system is not dispersion free, wave speeds are different, and disturbances associated with each mode bounce back-and-forth without recombining in an orderly fashion.

For those of you who didn't read the paper, an interesting experiment is performed: Four air-track gliders of the same mass are fitted with springs. Three are lined up, and the fourth is slid down the track. It turns out that this arrangement is not dispersion-free, and the system scatters. The authors calculate the parameters for a dispersion-free system (different ratios of springs and mass), run an experiment, and it works like a charm.

OK, now the second paper models the real ball-chain apparatus. In this paper, the authors basically say, "forget the first paper, except for the dispersion concept." They discuss that the "spring" between balls arises from Hertzian contact stresses, which follow the model F=k*x^(3/2), rather than the typical F=kx (like the glider model in the first paper).

Now, here's the important part: They say that a system of balls, or sliders, with slight gaps between each is dispersion-free (i.e., it will perform like a Newton's cradle). Since the slope of the Hertzian contact equation F=k*x^(3/2) is zero at x=0, it's almost like the force is zero for small displacements...kind of like a small gap between the balls. This isn't perfectly dispersion-free, but it's good enough, since nearly all of the momentum is transferred to the last ball. The remaining momentum imparts a small velocity to the balls, so that during the second round of collisions, there really is a small gap between balls, and the system is perfectly dispersion-free.

Now, I still have some questions:
1) I've never heard the term "dispersion-free" before. I wonder if this is some jargon coined by the authors, or if it's been used in other places.
2) The authors never show how "dispersion-free" applies to the system of masses seperated by a small distance; they just state that it is. I can logic out how this works, I think, but it still seems to be a gap in the paper.
3) The authors never really prove that a dispersion-free system is either necessary or sufficient for the symmetric behavior shown by the Newton's cradle. They offer a plausible explanation, which I suppose is good enough for me, but they don't prove it.
4) Their model shows that, for a five-ball Newton's cradle with one ball dropped, the fourth ball has 12% of the velocity post-collision that the first one did pre-collision. This seems kind of high: anyone out there up to some observational experimentation?
5) In their conclusion, the authors say that a 40 micron gap between balls is sufficient for dispersion-free reaction. The chart they refer to clearly shows four microns. Maybe a typo, but in a followup response paper by the authors [AJP, Vol. 52, p.84], they give a figure of 20 microns. Which is it?

Oh, good job on the research, Cal. These papers were a pretty interesting followup to the mailbag item.

zut
05-24-2000, 08:58 AM
I just skimmed the site (cite?) given by JonF, and it looks to me that their method of modeling collision interaction is intended to make computation easier, rather than more realistically portray the physical ball interaction. I suspect they just model individual momentum transfers, one at a time, and let it go at that. (I'll admit I was just skimming, though, and this kind of formulation is right on the edge of my experience).

Chronos
05-24-2000, 02:30 PM
Yes, dispersion is a commonly used term in n-body systems, but it's kind of hard to get a handle on without putting it into context. Just saying that the system is dispersionless doesn't seem to be all that meaningful, but if you specifically say that it's the dispersion in the wave speeds that you're talking about, then it's clear. This also gives us the extra equations I was talking about (three linearly independent relationships between wave speeds), so I, for one, am now satisfied with this problem. I still can't quite follow the explanation completely without more effort than I'm willing to put into this (involving grabbing a bunch of my textbooks, looking stuff up, and doing the math myself), but it's enough for me (for the moment) to know that an explanation exists. :)

Manlob
06-01-2000, 01:06 AM
The spheres are deformable objects, and the motion of Newton's cradle can be explained by accounting for the compression waves that move through the spheres. Evidently, the elastic waves move through a row of contacting spheres in the same way as through a straight rod. So to keep things simple and one-dimensional, think of longitudinal waves (sound waves) moving through a straight rod.

When a moving length of rod impacts the end of a stationary rod, a compression wave is initiated at the point of impact. This region of compressed atoms spreads in both directions away from the point of contact. One wave front moves into the originally stationary segment, and the other wave front travels back into the segment that was originally in motion. Any material passed by the compressive wave front is left in a state of compression (or stored potential energy) until a tensile wave comes through. When a wave front reaches the end of the rod, it is reflected back in the opposite direction and "changes sign". The change of sign makes a compressive wave front into a tensile wave front.

A row of contacting spheres behaves like an elastic rod as long as it is being compressed, but a tensile load cannot be supported by the balls because there is nothing preventing them from being pulled apart. In the case of Newton's cradle, the two compressive waves start at the point of impact travelling in opposite directions, get reflected off the ends, and eventually run into each other. After reflection from an end and passing over compressed material, the resulting tensile wave front uncompresses material releasing potential energy, and leaving behind undeformed material. If the orginally moving segment had length L (which is equal to an integer multiple of sphere diameter), the wave fronts will first meet a distance L from the other end of the rod. When they meet, all the compressive stress will have been undone by the passing of the tensile wave front, so no elastic potential energy remains at that time. At that point in time all the energy is in the form of kinetic energy of a segment of length L opposite from the point of impact- all points in that segment would then be moving at the same velocity, equal to that of the orginal impacting segment. If the rod could support a tensile load the tensile wave fronts would continue into unstressed material where previously they had always entered compressed material. This would cause tensile loads to develop from where the waves crossed paths. However that point is where spheres contact so no tensile load can be supported, and instead of a tensile stress being developed, the segment of length L will separate from the end of the rod and continue on with the same kinetic energy as the original impacting segment.

All of this can be verified by solving the wave equation with appropriate initial conditions. Since waves can be superimposed, and the time it takes for the wave fronts to meet is independent of the the length, L, two different lengths can simulateously impact both ends and the segments with those lengths will exit the other ends.

CalMeacham
06-01-2000, 09:30 PM
I don't deny that you can model the sysem and come up with a solution, and thanks to all f you who provided such a solution. But the point was that the explanation usually given in physics classes -- that you can explain the results solely on the laws of conservation of momentum and conservation of energy -- is wrong. This is somewhat surprising, but undeniably true.

On a naive level, it seems as if you ought to be able to simply eclare all the collisions perfectly elastic and somehow grind out the observed results, but you can't. Faced with that fact, people inevitably descend to the next levl -- modelling the interactions. The equally surprising fact is that you don't invariably get the result you expect. When the authors of one of the papers I cited above tried the experiment on a linear ai track with springs between the masses they did NOT see the reslt that "one mass in" yields "one mass off". Instead, all masses moved off with some velocity. So not only isn't the "correct" result provable using very simple assumptions, it's not even correct.

This is why I love the American Journal of Physics. At least once a year they disprove some "Fact" of physics you were certain was invariably correct. Like "Olaus Roemer measured the speed f light by observing Jupiter's moons", or "The tighst radius f curvature on a conductor produces the greatest electric field" ("The Lightning Rod Fallacy").

Chronos
06-02-2000, 11:15 PM
Thank you, Manlob, that makes a lot of sense, that way. I think that that's the answer that all the rest of us physicists should have posted. :) Stick around, post some more, you're helping the board.

Manlob
06-03-2000, 10:12 PM
Herrmann and Schmalzle... give a counter-example of a three-ball system in which one ball initially approaches the two stationary balls with a velocity V. Afterwards, the first ball rebounds with velocity -V/3, while the two balls that had been stationary move off with velocity 2V/3.

Not only does this example result in conservation of momentum with no loss of kinetic energy (i.e. a perfectly elastic collision), this situation could be obtained in experiments (No, I have not tried it, but did do the math).

Suppose there are 3 white billard balls, and 3 red billard balls. All six balls have the same diameter and mass, and each ball is a solid sphere with no hollow cavities. When a white ball moving with velocity V hits the end of two stationary white ones, a single white one exits the other end with velocity V while the other balls are left stationary. Same thing happens when one red ball strikes two stationary red ones. But after a white ball strikes two stationary red balls, the white one is left moving with velocity -V/3 and the two red balls move together in the opposite direction with velocity 2V/3.

What is the difference (besides color) between the red and the white balls? In the world of freshmen physics, objects are treated as rigid or as point masses, and all the balls would seem to be the same. In the real world, objects are deformable, and as described in a previous post, the waves travelling through the balls determine what happens in the collision. The answer is that the difference in the balls is how they deform-- they have a different modulus of elasticity and therefore waves move at different speeds through the two balls.

To be left with the white ball moving at -V/3 and the two red ones at 2V/3, the speed of sound in the red balls must be precisely twice that of the white ones. With this ratio of wave speeds, the wave fronts initiated from the point of impact later meet again at the impact point, so the balls separate there. See my previous post for an explanation of how the waves interact to cause the balls to separate when all the balls have the same modulus of elasticity. If the wave fronts do not meet exactly at a contact point between balls, the balls will separate, but they will be left "ringing" with residual compression and tension waves bouncing around inside of them. When this happens the collision is not perfectly elastic, because some of the energy is used by those waves.

In practice, it is easy to get the waves fronts to meet between balls when they are all the same, since it does not matter what the speed of sound is. To get an elastic collision in the laboratory of the type described above is more tricky because you need to make balls with a precise ratio of material properties, but it should be possible to get very close by using composite materials, for example.

zut
06-05-2000, 12:22 PM
Manlob writes,
To be left with the white ball moving at -V/3 and the two red ones at 2V/3, the speed of sound in the red balls must be precisely twice that of the white ones.

One nitpick to throw into an otherwise excellent discussion. In the Herrmann and Seitz paper cited above, they discover that the speed of the propagation of the disturbance is based on the time required for the Hertzian contact forces to develop after contact, rather than the speed of sound in the material. In fact, they measure propagation speeds of approx. an order of magnitude less than the speed of sound in the material.

What this means is that, to obtain different collision properties, you could also change the radius of the balls. This idea might work better on the "Newton's cradle" rather than the pool balls, because the Newton's cradle masses could be oblong, with different radii at the points of contact than at other points on the body.

ZenBeam
06-05-2000, 01:39 PM
To get an elastic collision in the laboratory of the type described above is more tricky because you need to make balls with a precise ratio of material properties, but it should be possible to get very close by using composite materials, for example.

Even if you couldn't get a 2:1 ratio, wouldn't any difference show up in some departure from the "only the last ball moves" case?

I wonder if some golfer who reads this could try the experiment using golf balls from different manufacturers as the "red" and "white" balls.