[math-fun] Newton's cradle
I tried to do this by thinking of the balls as springs with a very big constant instead of as rigid bodies. Then the question of space between them doesn't seem so important any more, and I start to think that the behavior of the system will depend more on the structure of the big ball (how does it compress compared with the small balls).
I still don't see how to predict exactly what will happen. In fact, my cluelessness about the way the big spring acts makes me even less clueful about what should happen.
--Joshua Zucker Neil is cheating. He has this amazing physics simulator from http://www.phunland.com/wiki/Home that even models air resistance. The clone feature naturally placed the balls slightly apart, which we deemed a plus, but strangeness ensued! The middle balls soon started moving, then gradually calmed down, then gradually excited again, ...
It turns out to depend on string length! If the strings are short and the gaps are large, the ball collisions are off center. So make the strings really long, and all should be clear. --rwg (Neil wishes to point out that his incorrect answer preceded the software experiment. He may be a poor theoretician but never a poor experimentalist.)
ah! maybe that's why my hybrid gadget with the 1st ball mass 2m is only kicking out one ball. something with string length. or that the shock wave pattern is being disrupted by the ball's bigger diameter. On Fri, Dec 10, 2010 at 8:48 PM, Bill Gosper <billgosper@gmail.com> wrote:
I tried to do this by thinking of the balls as springs with a very big constant instead of as rigid bodies. Then the question of space between them doesn't seem so important any more, and I start to think that the behavior of the system will depend more on the structure of the big ball (how does it compress compared with the small balls).
I still don't see how to predict exactly what will happen. In fact, my cluelessness about the way the big spring acts makes me even less clueful about what should happen.
--Joshua Zucker Neil is cheating. He has this amazing physics simulator from http://www.phunland.com/wiki/Home that even models air resistance. The clone feature naturally placed the balls slightly apart, which we deemed a plus, but strangeness ensued! The middle balls soon started moving, then gradually calmed down, then gradually excited again, ...
It turns out to depend on string length! If the strings are short and the gaps are large, the ball collisions are off center. So make the strings really long, and all should be clear. --rwg (Neil wishes to point out that his incorrect answer preceded the software experiment. He may be a poor theoretician but never a poor experimentalist.) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I've not the slightest doubt that Bill (and several others on the list) know far more about mechanics than I do; but I can't resist adding my two-penn'orth at this point. I've never been convinced that these problems are even well-defined, for the same reason that a simultaneous collision between three bodies in the plane is ill-defined. In that case, it's immediately obvious that you get completely different results according to how you decompose into a sequence of 2-body collisions. The resemblance to the preceding discussion is striking! WFL On 12/11/10, Gary Antonick <gantonick@post.harvard.edu> wrote:
ah! maybe that's why my hybrid gadget with the 1st ball mass 2m is only kicking out one ball. something with string length. or that the shock wave pattern is being disrupted by the ball's bigger diameter.
On Fri, Dec 10, 2010 at 8:48 PM, Bill Gosper <billgosper@gmail.com> wrote:
I tried to do this by thinking of the balls as springs with a very big constant instead of as rigid bodies. Then the question of space between them doesn't seem so important any more, and I start to think that the behavior of the system will depend more on the structure of the big ball (how does it compress compared with the small balls).
I still don't see how to predict exactly what will happen. In fact, my cluelessness about the way the big spring acts makes me even less clueful about what should happen.
--Joshua Zucker Neil is cheating. He has this amazing physics simulator from http://www.phunland.com/wiki/Home that even models air resistance. The clone feature naturally placed the balls slightly apart, which we deemed a plus, but strangeness ensued! The middle balls soon started moving, then gradually calmed down, then gradually excited again, ...
It turns out to depend on string length! If the strings are short and the gaps are large, the ball collisions are off center. So make the strings really long, and all should be clear. --rwg (Neil wishes to point out that his incorrect answer preceded the software experiment. He may be a poor theoretician but never a poor experimentalist.) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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How ill-defined? For example, take the case where all three have equal mass, the central body starts out motionless and the outer two are approaching at the same speed (from opposite directions). The midpoint between the positions of the outer two is X, and isn't changing until the first impact because of the initial conditions I just stated. If the central body starts out at position "X+epsilon", I get two non-simultaneous collisions after which the central body ends up at position "X-epsilon". In the limit as epsilon goes to zero, I get two simultaneous collisions, the central body remains motionless and the outer two simultaneously bounce off it. If I decompose in the other order (starting with a negative "epsilon" so the other impact happens first) the limit still gives me the same answer. Why is the limit ill-defined? Because the velocity of either particle is not defined (and discontinuous) at the moment of impact? Wouldn't that make all individual collision events ill-defined? (Of course this is all assuming the bodies are ideal perfectly rigid, i.e. the speed of sound in their material is infinite and similar idealizations) - Robert On Sat, Dec 11, 2010 at 07:55, Fred lunnon <fred.lunnon@gmail.com> wrote:
[...] I've never been convinced that these problems are even well-defined, for the same reason that a simultaneous collision between three bodies in the plane is ill-defined. In that case, it's immediately obvious that you get completely different results according to how you decompose into a sequence of 2-body collisions.
The resemblance to the preceding discussion is striking! WFL
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com
first of all, we have a serious problem in that one ball appears to be kicking out only one ball. so pretty much everyone might be wrong.. or at least a new group is a bit more right. and now completely perplexed. am epoxying together a larger cradle (1" and 1.25" balls) this moment. second... in defense of Fred (from very much the layperson view on this) it does seem that given three colliding bodies all you know is that the ending velocities (along with the beginning velocities) will lie on a circle at the intersection of a sphere (kinetic energy) and a plane (momentum). I don't see how this endpoint could ever be determinate without another constraint. On Sat, Dec 11, 2010 at 10:45 AM, Robert Munafo <mrob27@gmail.com> wrote:
How ill-defined?
For example, take the case where all three have equal mass, the central body starts out motionless and the outer two are approaching at the same speed (from opposite directions). The midpoint between the positions of the outer two is X, and isn't changing until the first impact because of the initial conditions I just stated.
If the central body starts out at position "X+epsilon", I get two non-simultaneous collisions after which the central body ends up at position "X-epsilon". In the limit as epsilon goes to zero, I get two simultaneous collisions, the central body remains motionless and the outer two simultaneously bounce off it.
If I decompose in the other order (starting with a negative "epsilon" so the other impact happens first) the limit still gives me the same answer.
Why is the limit ill-defined? Because the velocity of either particle is not defined (and discontinuous) at the moment of impact? Wouldn't that make all individual collision events ill-defined?
(Of course this is all assuming the bodies are ideal perfectly rigid, i.e. the speed of sound in their material is infinite and similar idealizations)
- Robert
On Sat, Dec 11, 2010 at 07:55, Fred lunnon <fred.lunnon@gmail.com> wrote:
[...] I've never been convinced that these problems are even well-defined, for the same reason that a simultaneous collision between three bodies in the plane is ill-defined. In that case, it's immediately obvious that you get completely different results according to how you decompose into a sequence of 2-body collisions.
The resemblance to the preceding discussion is striking! WFL
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com
math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Sat, Dec 11, 2010 at 1:45 PM, Robert Munafo <mrob27@gmail.com> wrote:
How ill-defined?
For example, take the case where all three have equal mass, the central body starts out motionless and the outer two are approaching at the same speed (from opposite directions). The midpoint between the positions of the outer two is X, and isn't changing until the first impact because of the initial conditions I just stated.
You happen to have chosen a case where the order doesn't make a difference. But let's try a slightly different example. One ball of mass 2 is stationary very near the origin, but to the left of the origin. Two balls of mass 1, one at -X and one at +X, approach it at speed V. First the ball on the left hits the stationary ball. Now the ball on the left is going left at speed V/3, while the center ball is going right at speed 2V/3. An instant later, the center ball hits the ball on the right. Now the ball on the right is going right with speed 11V/9, while the center ball is moving left with speed 4V/9 Of course, if the large ball started out slightly to the right of the origin, the collision order would be reversed, and we'd end up with the left ball moving left with speed 11V/9, the center ball moving right with speed 4V/9, and the right-hand ball moving right with velocity V/3. So what happens when the center ball is exactly at the origin, and the two collisions happen simultaneously? One might argue from symmetry that the center ball would stay stationary, and the other two would rebound and end moving away from the center at speed V. But experimentally, you can't place the ball *exactly* at the origin, so wouldn't you instead expect one of the two asymmetric outcomes above? And what if the ball coming from the left has mass 1, the center ball has mass 3, and the left hand ball has mass 2, and again, the outer balls are approaching the origin with speed V from equal distances? I know how to calculate what happens if the left-hand collision happens first, and what happens if the right-hand collision happens first. But is there a third "simultaneous collisions", possibility, and if so, in the absence of symmetry, how do you calculate what happens? You have 2 equations (conservation of momentum and energy) in 3 unknowns, so there are infinitely many solutions, and I have no idea how to choose which solution "between" the two non-simultaneous-collision answers is the correct answer! So the answer to the original question of "why don't physicists get the problem of Newton's Cradle with a double-mass ball hitting the stationary balls" may be "It's an ill-posed problem"! If we model it as a series of non-simultaneous collisions, then after the big ball hits the second ball, it's still moving in the same direction, at speed V/3. So it will be involved in more collisions. All the collisions are nearly simultaneous, and the answer will depend on what order they occur in, so I don't see any way to give a unique answer. Andy
If the central body starts out at position "X+epsilon", I get two non-simultaneous collisions after which the central body ends up at position "X-epsilon". In the limit as epsilon goes to zero, I get two simultaneous collisions, the central body remains motionless and the outer two simultaneously bounce off it.
If I decompose in the other order (starting with a negative "epsilon" so the other impact happens first) the limit still gives me the same answer.
Why is the limit ill-defined? Because the velocity of either particle is not defined (and discontinuous) at the moment of impact? Wouldn't that make all individual collision events ill-defined?
(Of course this is all assuming the bodies are ideal perfectly rigid, i.e. the speed of sound in their material is infinite and similar idealizations)
- Robert
On Sat, Dec 11, 2010 at 07:55, Fred lunnon <fred.lunnon@gmail.com> wrote:
[...] I've never been convinced that these problems are even well-defined, for the same reason that a simultaneous collision between three bodies in the plane is ill-defined. In that case, it's immediately obvious that you get completely different results according to how you decompose into a sequence of 2-body collisions.
The resemblance to the preceding discussion is striking! WFL
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com
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-- Andy.Latto@pobox.com
On Sat, Dec 11, 2010 at 12:34 PM, Andy Latto <andy.latto@pobox.com> wrote:
So the answer to the original question of "why don't physicists get the problem of Newton's Cradle with a double-mass ball hitting the stationary balls" may be "It's an ill-posed problem"! If we model it as a series of non-simultaneous collisions, then after the big ball hits the second ball, it's still moving in the same direction, at speed V/3. So it will be involved in more collisions. All the collisions are nearly simultaneous, and the answer will depend on what order they occur in, so I don't see any way to give a unique answer.
How about assuming the balls are springy enough that they compress sufficiently so they're all in contact simultaneously? And if we assume they are close enough to touching that they all start compressing simultaneously? This seems like the natural assumption to make in the absence of more knowledge about the details of the spacing. Maybe there's still some assumption that needs to be made about the spring constant of the bigger ball compared with all the smaller balls. And maybe I should stop babbling and sit down and do some actual math, given the name of this list. Speaking of which, I should send another post on a different topic, with a little math in it. So I will do that. --Joshua
Here's a way to visualize the discontinuity near a 3-body collision of most combinations of masses in 1 dimension: When there are just two bodies, if you use inertial coordinates, there is just one degree of freedom, and using as parameter the distance between the two bodies, the motion is that of a reflection. When there are three bodies, then in an inertial coordinate system there are two degrees of freedom. The parameter space is an angle between two lines. There is a physically natural metric, ds^2 = kinetic energy of system and in this metric, the dynamics follows geodesics that reflect off the walls just like light reflects off a mirror. When the particles all have equal mass, the angle is 60 degrees: two mirrors at a 60 degree angle form a continuous image across the corner, and the dynamics is continuous in a neighborhood of the triple collision. As the mass of the particles varies, the angle between the lines changes. If the mass of the central object is small compared to the other two, it bounces back and forth many times before they all separate: the angle is very small. If the mass in the middle is very large, the angle is almost 90 degrees. Just as for two mirrors, the dynamics is continuous if and only if the angle has the form pi/n. If there are more masses, the geometric figure is wedge formed between n hyperplanes in R^n, linearly equivalent to an orthant (whose coordinates are the distances between masses) but metrically of a variable shape. The angle between two faces that corresponding to collisions of disjoint pairs of particles are at right angles. The dynamics is continuous if and only if it is continuous for the collision of each triple of consecutive particles, i.e. if these angles have the form pi/n. It should be fun to make a Newton's cradle with some of the variants. --- I figure any of the Euclidean Coxeter groups should be doable. Bill Thurston On Dec 11, 2010, at 3:34 PM, Andy Latto wrote:
On Sat, Dec 11, 2010 at 1:45 PM, Robert Munafo <mrob27@gmail.com> wrote:
How ill-defined?
For example, take the case where all three have equal mass, the central body starts out motionless and the outer two are approaching at the same speed (from opposite directions). The midpoint between the positions of the outer two is X, and isn't changing until the first impact because of the initial conditions I just stated.
You happen to have chosen a case where the order doesn't make a difference. But let's try a slightly different example.
One ball of mass 2 is stationary very near the origin, but to the left of the origin. Two balls of mass 1, one at -X and one at +X, approach it at speed V.
First the ball on the left hits the stationary ball. Now the ball on the left is going left at speed V/3, while the center ball is going right at speed 2V/3.
An instant later, the center ball hits the ball on the right. Now the ball on the right is going right with speed 11V/9, while the center ball is moving left with speed 4V/9
Of course, if the large ball started out slightly to the right of the origin, the collision order would be reversed, and we'd end up with the left ball moving left with speed 11V/9, the center ball moving right with speed 4V/9, and the right-hand ball moving right with velocity V/3.
So what happens when the center ball is exactly at the origin, and the two collisions happen simultaneously? One might argue from symmetry that the center ball would stay stationary, and the other two would rebound and end moving away from the center at speed V. But experimentally, you can't place the ball *exactly* at the origin, so wouldn't you instead expect one of the two asymmetric outcomes above?
And what if the ball coming from the left has mass 1, the center ball has mass 3, and the left hand ball has mass 2, and again, the outer balls are approaching the origin with speed V from equal distances? I know how to calculate what happens if the left-hand collision happens first, and what happens if the right-hand collision happens first. But is there a third "simultaneous collisions", possibility, and if so, in the absence of symmetry, how do you calculate what happens? You have 2 equations (conservation of momentum and energy) in 3 unknowns, so there are infinitely many solutions, and I have no idea how to choose which solution "between" the two non-simultaneous-collision answers is the correct answer!
So the answer to the original question of "why don't physicists get the problem of Newton's Cradle with a double-mass ball hitting the stationary balls" may be "It's an ill-posed problem"! If we model it as a series of non-simultaneous collisions, then after the big ball hits the second ball, it's still moving in the same direction, at speed V/3. So it will be involved in more collisions. All the collisions are nearly simultaneous, and the answer will depend on what order they occur in, so I don't see any way to give a unique answer.
Andy
If the central body starts out at position "X+epsilon", I get two non-simultaneous collisions after which the central body ends up at position "X-epsilon". In the limit as epsilon goes to zero, I get two simultaneous collisions, the central body remains motionless and the outer two simultaneously bounce off it.
If I decompose in the other order (starting with a negative "epsilon" so the other impact happens first) the limit still gives me the same answer.
Why is the limit ill-defined? Because the velocity of either particle is not defined (and discontinuous) at the moment of impact? Wouldn't that make all individual collision events ill-defined?
(Of course this is all assuming the bodies are ideal perfectly rigid, i.e. the speed of sound in their material is infinite and similar idealizations)
- Robert
On Sat, Dec 11, 2010 at 07:55, Fred lunnon <fred.lunnon@gmail.com> wrote:
[...] I've never been convinced that these problems are even well-defined, for the same reason that a simultaneous collision between three bodies in the plane is ill-defined. In that case, it's immediately obvious that you get completely different results according to how you decompose into a sequence of 2-body collisions.
The resemblance to the preceding discussion is striking! WFL
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com
math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
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Bill- I have a related question I think. Kinetic energy and momentum for three bodies make a sphere and intersecting plane if plotted in 3D using the three body velocities (am not sure "inertial coordinates" mean but perhaps something similar. would be nice to see some animations of this geometry if you know of any. eg: the first ball has twice the mass of the other two: sphere flattens (which you point out) the third ball always has mass zero (is never struck): sphere turns into cylinder first ball hits 2nd and 3rd ball simultaneously (guess you did this but I need to digest) would be especially interesting to see how this geometry changes with totally inelastic collisions. kinetic energy is lost, so the sphere shrinks with each hit. momentum stays the same, so the plane stays where it is. and eventually all bodies are stuck together traveling at the same velocity at coordinate [v,v,v]. cheers, Gary On Sat, Dec 11, 2010 at 6:27 PM, Bill Thurston <wpthurston@me.com> wrote:
Here's a way to visualize the discontinuity near a 3-body collision of most combinations of masses in 1 dimension:
When there are just two bodies, if you use inertial coordinates, there is just one degree of freedom, and using as parameter the distance between the two bodies, the motion is that of a reflection.
When there are three bodies, then in an inertial coordinate system there are two degrees of freedom. The parameter space is an angle between two lines. There is a physically natural metric, ds^2 = kinetic energy of system and in this metric, the dynamics follows geodesics that reflect off the walls just like light reflects off a mirror.
When the particles all have equal mass, the angle is 60 degrees: two mirrors at a 60 degree angle form a continuous image across the corner, and the dynamics is continuous in a neighborhood of the triple collision.
As the mass of the particles varies, the angle between the lines changes. If the mass of the central object is small compared to the other two, it bounces back and forth many times before they all separate: the angle is very small. If the mass in the middle is very large, the angle is almost 90 degrees.
Just as for two mirrors, the dynamics is continuous if and only if the angle has the form pi/n.
If there are more masses, the geometric figure is wedge formed between n hyperplanes in R^n, linearly equivalent to an orthant (whose coordinates are the distances between masses) but metrically of a variable shape. The angle between two faces that corresponding to collisions of disjoint pairs of particles are at right angles. The dynamics is continuous if and only if it is continuous for the collision of each triple of consecutive particles, i.e. if these angles have the form pi/n. It should be fun to make a Newton's cradle with some of the variants. --- I figure any of the Euclidean Coxeter groups should be doable. Bill Thurston
On Dec 11, 2010, at 3:34 PM, Andy Latto wrote:
On Sat, Dec 11, 2010 at 1:45 PM, Robert Munafo <mrob27@gmail.com> wrote:
How ill-defined?
For example, take the case where all three have equal mass, the central body starts out motionless and the outer two are approaching at the same speed (from opposite directions). The midpoint between the positions of the outer two is X, and isn't changing until the first impact because of the initial conditions I just stated.
You happen to have chosen a case where the order doesn't make a difference. But let's try a slightly different example.
One ball of mass 2 is stationary very near the origin, but to the left of the origin. Two balls of mass 1, one at -X and one at +X, approach it at speed V.
First the ball on the left hits the stationary ball. Now the ball on the left is going left at speed V/3, while the center ball is going right at speed 2V/3.
An instant later, the center ball hits the ball on the right. Now the ball on the right is going right with speed 11V/9, while the center ball is moving left with speed 4V/9
Of course, if the large ball started out slightly to the right of the origin, the collision order would be reversed, and we'd end up with the left ball moving left with speed 11V/9, the center ball moving right with speed 4V/9, and the right-hand ball moving right with velocity V/3.
So what happens when the center ball is exactly at the origin, and the two collisions happen simultaneously? One might argue from symmetry that the center ball would stay stationary, and the other two would rebound and end moving away from the center at speed V. But experimentally, you can't place the ball *exactly* at the origin, so wouldn't you instead expect one of the two asymmetric outcomes above?
And what if the ball coming from the left has mass 1, the center ball has mass 3, and the left hand ball has mass 2, and again, the outer balls are approaching the origin with speed V from equal distances? I know how to calculate what happens if the left-hand collision happens first, and what happens if the right-hand collision happens first. But is there a third "simultaneous collisions", possibility, and if so, in the absence of symmetry, how do you calculate what happens? You have 2 equations (conservation of momentum and energy) in 3 unknowns, so there are infinitely many solutions, and I have no idea how to choose which solution "between" the two non-simultaneous-collision answers is the correct answer!
So the answer to the original question of "why don't physicists get the problem of Newton's Cradle with a double-mass ball hitting the stationary balls" may be "It's an ill-posed problem"! If we model it as a series of non-simultaneous collisions, then after the big ball hits the second ball, it's still moving in the same direction, at speed V/3. So it will be involved in more collisions. All the collisions are nearly simultaneous, and the answer will depend on what order they occur in, so I don't see any way to give a unique answer.
Andy
If the central body starts out at position "X+epsilon", I get two non-simultaneous collisions after which the central body ends up at position "X-epsilon". In the limit as epsilon goes to zero, I get two simultaneous collisions, the central body remains motionless and the outer two simultaneously bounce off it.
If I decompose in the other order (starting with a negative "epsilon" so the other impact happens first) the limit still gives me the same answer.
Why is the limit ill-defined? Because the velocity of either particle is not defined (and discontinuous) at the moment of impact? Wouldn't that make all individual collision events ill-defined?
(Of course this is all assuming the bodies are ideal perfectly rigid, i.e. the speed of sound in their material is infinite and similar idealizations)
- Robert
On Sat, Dec 11, 2010 at 07:55, Fred lunnon <fred.lunnon@gmail.com> wrote:
[...] I've never been convinced that these problems are even well-defined, for the same reason that a simultaneous collision between three bodies in the plane is ill-defined. In that case, it's immediately obvious that you get completely different results according to how you decompose into a sequence of 2-body collisions.
The resemblance to the preceding discussion is striking! WFL
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com
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-- Andy.Latto@pobox.com
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I've been discussing this question with my brother (a biophysicist): we've been gradually developing a clearer picture. A geometric model for n bodies in 1 dimension that takes into account the kinetic energy lost in each collision would be to start with a in R^n-1 (= the subspace center of mass = 0) with an accessible region that is the cone on a simplex (linearly equivalent to an orthant). At each collision, you shorten the velocity in the normal direction by a constant (called the coefficient of restitution, I believe). I.e. the two colliding particles travel more nearly in the same direction. This changes the geometry of reflections, so continuity around a 3-body collision never holds, no matter what the masses. For the idealized elastic case, I've checked out more about special cases that correspond to kaleidoscopic dynamics that are continuous (with help from my brother). If you have two particles (or objects) in a box of masses m1 and m2, say [0,1], then the dynamics is equivalent to a billiard ball in a right triangle with sides in the ratio sqrt(m1)/sqrt(m2). In particular, if the masses are equal, it's a 45-45-90 triangle. If the masses are in the ratio 1:3, then it's a 30-60-90 triangle. In these cases, the dynamics is continuous near the configuration where both particles collide simultaneously with each other and with the end; in all other configurations, the dynamics discontinuous. I think the 1:3 ratio should be particularly nice to make some kind of model (superballs on a track? magnets swinging?). If you start with just one of the masses moving, the dynamics is periodic (for the idealized dynamics). For three particles on a line with masses a, b and c at positions x,y,z, in the 2-dimensional plane center of mass = the origin, there are three lines x=y, y=z and x=z. If you give this plane a metric where vectors of kinetic energy 1 have length 1, then you can figure out the angles by drawing a hexagon with sides parallel to the lines and vertices on these lines. They are the same as the angles of the triangle with side lengths sqrt(a(b+c)), sqrt(b(a+c)) and sqrt(c(a+b)). By varying the masses, you can make any acute-angled triangle. The dynamics confines the arrangement to one particular order <--> one of the six chambers formed by these lines. If you add a 4th particle of mass d, then you get an arrangement of 6 planes in R^3, linearly equivalent to the set of planes through edges of a regular tetrahedron centered at the origin, or equivalently, the planes that go through a pair of opposite edges of a cube centered at the origin. Collisions confine the arrangement to one of the 24 = 4! chambers. The intersection of any two of these planes is either a 3 body collision or a pair of simultaneous collisions of 2 disjoint pairs of particles. In the latter case, the angles are 90 degrees, and in the former case, they are determined by the 3-body case above. Each chamber is the cone on a right-angled spherical angle whose other two angles are acute, and every such triangle can be attained by suitable choices of masses. For instance, to get the pi/2, pi/3, pi/5 Kaleidoscope (the cone on a triangle in the barycentric subdivision of an icosahedron), one solution for the masses is approximately a=55.6, b=13, c=5.8 and d=100. These numbers are imprecise because I got them by a Manipulate command in mathematica, where there are 4 sliders to adjust while you watch the angles change. The angles are 59.9969, 36.0558, and 90.) As the two outer weights become large, the sum of the angles tends toward 90 degrees, and it converges to the configuration of two particles bouncing between two walls. In a configuration with three bodies of masses 2, 1, 1, the angles of the chamber are to the 3-body collisions are 54.7356, 60, 90, ... , which confirms that the Newton's cradle configuration is discontinuous near the set of 3-body collisions where the first mass is double the other two. It's continuous near the particular trajectories where the heavy ball collides with a pair of stationary balls. Bill Thurston On Dec 12, 2010, at 1:11 AM, Gary Antonick wrote:
Bill-
I have a related question I think.
Kinetic energy and momentum for three bodies make a sphere and intersecting plane if plotted in 3D using the three body velocities (am not sure "inertial coordinates" mean but perhaps something similar. would be nice to see some animations of this geometry if you know of any. eg:
the first ball has twice the mass of the other two: sphere flattens (which you point out) the third ball always has mass zero (is never struck): sphere turns into cylinder first ball hits 2nd and 3rd ball simultaneously (guess you did this but I need to digest)
would be especially interesting to see how this geometry changes with totally inelastic collisions. kinetic energy is lost, so the sphere shrinks with each hit. momentum stays the same, so the plane stays where it is. and eventually all bodies are stuck together traveling at the same velocity at coordinate [v,v,v].
cheers,
Gary
On Sat, Dec 11, 2010 at 6:27 PM, Bill Thurston <wpthurston@me.com> wrote:
Here's a way to visualize the discontinuity near a 3-body collision of most combinations of masses in 1 dimension:
When there are just two bodies, if you use inertial coordinates, there is just one degree of freedom, and using as parameter the distance between the two bodies, the motion is that of a reflection.
When there are three bodies, then in an inertial coordinate system there are two degrees of freedom. The parameter space is an angle between two lines. There is a physically natural metric, ds^2 = kinetic energy of system and in this metric, the dynamics follows geodesics that reflect off the walls just like light reflects off a mirror.
When the particles all have equal mass, the angle is 60 degrees: two mirrors at a 60 degree angle form a continuous image across the corner, and the dynamics is continuous in a neighborhood of the triple collision.
As the mass of the particles varies, the angle between the lines changes. If the mass of the central object is small compared to the other two, it bounces back and forth many times before they all separate: the angle is very small. If the mass in the middle is very large, the angle is almost 90 degrees.
Just as for two mirrors, the dynamics is continuous if and only if the angle has the form pi/n.
If there are more masses, the geometric figure is wedge formed between n hyperplanes in R^n, linearly equivalent to an orthant (whose coordinates are the distances between masses) but metrically of a variable shape. The angle between two faces that corresponding to collisions of disjoint pairs of particles are at right angles. The dynamics is continuous if and only if it is continuous for the collision of each triple of consecutive particles, i.e. if these angles have the form pi/n. It should be fun to make a Newton's cradle with some of the variants. --- I figure any of the Euclidean Coxeter groups should be doable. Bill Thurston
On Dec 11, 2010, at 3:34 PM, Andy Latto wrote:
On Sat, Dec 11, 2010 at 1:45 PM, Robert Munafo <mrob27@gmail.com> wrote:
How ill-defined?
For example, take the case where all three have equal mass, the central body starts out motionless and the outer two are approaching at the same speed (from opposite directions). The midpoint between the positions of the outer two is X, and isn't changing until the first impact because of the initial conditions I just stated.
You happen to have chosen a case where the order doesn't make a difference. But let's try a slightly different example.
One ball of mass 2 is stationary very near the origin, but to the left of the origin. Two balls of mass 1, one at -X and one at +X, approach it at speed V.
First the ball on the left hits the stationary ball. Now the ball on the left is going left at speed V/3, while the center ball is going right at speed 2V/3.
An instant later, the center ball hits the ball on the right. Now the ball on the right is going right with speed 11V/9, while the center ball is moving left with speed 4V/9
Of course, if the large ball started out slightly to the right of the origin, the collision order would be reversed, and we'd end up with the left ball moving left with speed 11V/9, the center ball moving right with speed 4V/9, and the right-hand ball moving right with velocity V/3.
So what happens when the center ball is exactly at the origin, and the two collisions happen simultaneously? One might argue from symmetry that the center ball would stay stationary, and the other two would rebound and end moving away from the center at speed V. But experimentally, you can't place the ball *exactly* at the origin, so wouldn't you instead expect one of the two asymmetric outcomes above?
And what if the ball coming from the left has mass 1, the center ball has mass 3, and the left hand ball has mass 2, and again, the outer balls are approaching the origin with speed V from equal distances? I know how to calculate what happens if the left-hand collision happens first, and what happens if the right-hand collision happens first. But is there a third "simultaneous collisions", possibility, and if so, in the absence of symmetry, how do you calculate what happens? You have 2 equations (conservation of momentum and energy) in 3 unknowns, so there are infinitely many solutions, and I have no idea how to choose which solution "between" the two non-simultaneous-collision answers is the correct answer!
So the answer to the original question of "why don't physicists get the problem of Newton's Cradle with a double-mass ball hitting the stationary balls" may be "It's an ill-posed problem"! If we model it as a series of non-simultaneous collisions, then after the big ball hits the second ball, it's still moving in the same direction, at speed V/3. So it will be involved in more collisions. All the collisions are nearly simultaneous, and the answer will depend on what order they occur in, so I don't see any way to give a unique answer.
Andy
If the central body starts out at position "X+epsilon", I get two non-simultaneous collisions after which the central body ends up at position "X-epsilon". In the limit as epsilon goes to zero, I get two simultaneous collisions, the central body remains motionless and the outer two simultaneously bounce off it.
If I decompose in the other order (starting with a negative "epsilon" so the other impact happens first) the limit still gives me the same answer.
Why is the limit ill-defined? Because the velocity of either particle is not defined (and discontinuous) at the moment of impact? Wouldn't that make all individual collision events ill-defined?
(Of course this is all assuming the bodies are ideal perfectly rigid, i.e. the speed of sound in their material is infinite and similar idealizations)
- Robert
On Sat, Dec 11, 2010 at 07:55, Fred lunnon <fred.lunnon@gmail.com> wrote:
[...] I've never been convinced that these problems are even well-defined, for the same reason that a simultaneous collision between three bodies in the plane is ill-defined. In that case, it's immediately obvious that you get completely different results according to how you decompose into a sequence of 2-body collisions.
The resemblance to the preceding discussion is striking! WFL
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Thanks for explaining that. I understand now why that old billiard simulator program didn't work too well (-: And I suppose my answer to the basic question is, when the balls are "almost but not exactly" in alignment, the collisions overlap in time, because the speed of sound inside the balls is finite and the shock waves are all initiated before any of them travels very far. On Sat, Dec 11, 2010 at 15:34, Andy Latto <andy.latto@pobox.com> wrote:
[...] Of course, if the large ball started out slightly to the right of the origin, the collision order would be reversed, and we'd end up with the left ball moving left with speed 11V/9, the center ball moving right with speed 4V/9, and the right-hand ball moving right with velocity V/3.
[...] And what if the ball coming from the left has mass 1, the center ball has mass 3, and the left hand ball has mass 2, and again, the outer balls are approaching the origin with speed V from equal distances? [...] is there a third "simultaneous collisions", possibility, and if so, in the absence of symmetry, how do you calculate what happens? [...]
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While I don't want to discourage the mathematical tangents this thread has generated, since it started as a physics question I think you deserve a physics answer. The role of physics is not just to provide general principles, such as conservation laws. Sometimes it just comes down to numbers, plain and simple. The rules that one should apply to swinging steel balls are very different than what is appropriate to, say a couple of krypton atoms adsorbed on a graphite surface and struck by a third krypton atom. In that case one needs to consider the force of every atom on every other atom as the collision unfolds. By contrast, steel balls are simpler. So here are some numbers. The speed of sound in steel is about 5000 m/s. A ball dropped from 5 cm reaches a speed of about 1 m/s. This tells us that the stresses imparted to the steel balls can be treated as quasi-static to a very good approximation. There are no shock waves. The Young's modulus of steel is E = 200 Gpa. If we uniaxially compress a cylinder with cross sectional area A and length L by x, then the stored elastic energy is E A x^2/(2L). Equating this with the energy of the incident ball, say corresponding to m = 1 gm dropped an equivalent vertical rise of h = 10 cm or mgh = .01 J, I get x = 1 micron when A and L are of order 1 cm^2 and 1 cm. One micron is .01 the diameter of a human hair. Using the Young's modulus again, we find that the pressure at maximum compression, E x/L, is about 10 MPa. By contrast, the pressure supporting the weight of a ball placed on top of another is only about mg/A = 1 kPa -- four orders of magnitude smaller. So even when two balls in Newton's cradle are "in contact" because the strings supporting them make a slight angle with respect to the vertical, they may as well be miles apart with regard to the scale of the forces. To help you digest these numbers, think of a pair of steel balls as two very stiff springs. The springs are in a collision course. The springs have a natural oscillation period but the collision is very slow on this scale. Start playing Smetana's Die Moldau (or Ravel's Bolero, your choice) right when the springs first touch. At about the time the Moldau crescendos into the Elbe the kinetic energy is completely and adiabatically (without the generation of entropy) converted into elastic energy. The springs will have compressed by a fraction of a millimeter. If you picture this in the center-of-mass frame, all motion ceases at an instant that marks the half-way point of the collision; thereafter the rest of the collision unfolds as the time reversal of the first half (and so even anharmonicity is not an issue). I hope these numbers have convinced you that the pairwise collision model is the appropriate approximation to use for Newton's cradle, billiard ball dynamics, etc. Here's a fun application of the pairwise model that you can try at home. Hold a basketball while resting a tennis ball on top of it. Now drop the basketball (the tennis ball should be co-moving and nearly in contact). You'll find that after all the collisions (pairwise!) have finished the tennis ball is shot up to a surprising height. Veit
Thanks for filling in the numbers, and confirming the intuition that thinking of it as a chain of 2-dimensional collisions is a good model. To summarize: the reason 2 balls can't fly out at the other end when a double-mass ball hits is that the intermediate balls can't individually transmit the given momentum with the given amount of energy. An object of half the mass requires twice as much energy to carry a given amount of momentum, hence the initial ball must retain some of the momentum to be transmitted later, creating a more complicated outcome. It's quite different than when two balls are dropped, because in that case instead of one indigestible large dose of momentum, two quick (to us) doses of half the strength are administered. As I said in my earlier message, I think the Newton's cradle system with the initial mass twice the weight is continous at the time of collision near the particular trajectory we've been discussing, where all but the heavy ball are initially stationary. Trajectories for collisions of equal masses in one dimension are continuous near the multiple collision loci. The only way there could be a discontinuity (when the order of collisions changes) would be if the heavy ball and two (or more) others were all converging relative to each other toward a simultaneous collision. This can't happen near the time of the first collision. Later on, when the motion is more chaotic, there could be discontinuities. Question: is there some sequence of masses for Newton's cradle that would exhibit visible discontinuity when the ball at one end is dropped against the others? Perhaps sometimes the initial ball flies out energetically, and sometimes not. Question: what about the break shot in billiards, where initially the balls are racked so they are in contact in a triangular pattern? Is the outcome of the collision discontinuous in this situation, or do the trajectories just vary quickly but continuously depending on the initial impact? Bill Thurston On Dec 13, 2010, at 12:08 PM, Veit Elser wrote:
While I don't want to discourage the mathematical tangents this thread has generated, since it started as a physics question I think you deserve a physics answer.
The role of physics is not just to provide general principles, such as conservation laws. Sometimes it just comes down to numbers, plain and simple. The rules that one should apply to swinging steel balls are very different than what is appropriate to, say a couple of krypton atoms adsorbed on a graphite surface and struck by a third krypton atom. In that case one needs to consider the force of every atom on every other atom as the collision unfolds. By contrast, steel balls are simpler.
So here are some numbers. The speed of sound in steel is about 5000 m/s. A ball dropped from 5 cm reaches a speed of about 1 m/s. This tells us that the stresses imparted to the steel balls can be treated as quasi-static to a very good approximation. There are no shock waves.
The Young's modulus of steel is E = 200 Gpa. If we uniaxially compress a cylinder with cross sectional area A and length L by x, then the stored elastic energy is E A x^2/(2L). Equating this with the energy of the incident ball, say corresponding to m = 1 gm dropped an equivalent vertical rise of h = 10 cm or mgh = .01 J, I get x = 1 micron when A and L are of order 1 cm^2 and 1 cm. One micron is .01 the diameter of a human hair. Using the Young's modulus again, we find that the pressure at maximum compression, E x/L, is about 10 MPa. By contrast, the pressure supporting the weight of a ball placed on top of another is only about mg/A = 1 kPa -- four orders of magnitude smaller. So even when two balls in Newton's cradle are "in contact" because the strings supporting them make a slight angle with respect to the vertical, they may as well be miles apart with regard to the scale of the forces.
To help you digest these numbers, think of a pair of steel balls as two very stiff springs. The springs are in a collision course. The springs have a natural oscillation period but the collision is very slow on this scale. Start playing Smetana's Die Moldau (or Ravel's Bolero, your choice) right when the springs first touch. At about the time the Moldau crescendos into the Elbe the kinetic energy is completely and adiabatically (without the generation of entropy) converted into elastic energy. The springs will have compressed by a fraction of a millimeter. If you picture this in the center-of-mass frame, all motion ceases at an instant that marks the half-way point of the collision; thereafter the rest of the collision unfolds as the time reversal of the first half (and so even anharmonicity is not an issue).
I hope these numbers have convinced you that the pairwise collision model is the appropriate approximation to use for Newton's cradle, billiard ball dynamics, etc.
Here's a fun application of the pairwise model that you can try at home. Hold a basketball while resting a tennis ball on top of it. Now drop the basketball (the tennis ball should be co-moving and nearly in contact). You'll find that after all the collisions (pairwise!) have finished the tennis ball is shot up to a surprising height.
Veit
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am not sure I'm following completely. I've tried this experiment now with several different model cradles (all with 1st ball mass 2m and remaining balls mass m) and get the same result each time: the last ball flies out and the remaining balls barely move. The balls clearly do *not* swing out in a fan pattern predicted by pairwise collisions. then the balls come together and interact again. after this *second* set of collisions the the pairwise model seems accurate (because the balls started off slightly apart)— the balls fan out in the predicted pattern. On Mon, Dec 13, 2010 at 10:20 AM, Bill Thurston <wpt4@cornell.edu> wrote:
Thanks for filling in the numbers, and confirming the intuition that thinking of it as a chain of 2-dimensional collisions is a good model.
To summarize: the reason 2 balls can't fly out at the other end when a double-mass ball hits is that the intermediate balls can't individually transmit the given momentum with the given amount of energy. An object of half the mass requires twice as much energy to carry a given amount of momentum, hence the initial ball must retain some of the momentum to be transmitted later, creating a more complicated outcome. It's quite different than when two balls are dropped, because in that case instead of one indigestible large dose of momentum, two quick (to us) doses of half the strength are administered.
As I said in my earlier message, I think the Newton's cradle system with the initial mass twice the weight is continous at the time of collision near the particular trajectory we've been discussing, where all but the heavy ball are initially stationary. Trajectories for collisions of equal masses in one dimension are continuous near the multiple collision loci. The only way there could be a discontinuity (when the order of collisions changes) would be if the heavy ball and two (or more) others were all converging relative to each other toward a simultaneous collision. This can't happen near the time of the first collision. Later on, when the motion is more chaotic, there could be discontinuities.
Question: is there some sequence of masses for Newton's cradle that would exhibit visible discontinuity when the ball at one end is dropped against the others? Perhaps sometimes the initial ball flies out energetically, and sometimes not.
Question: what about the break shot in billiards, where initially the balls are racked so they are in contact in a triangular pattern? Is the outcome of the collision discontinuous in this situation, or do the trajectories just vary quickly but continuously depending on the initial impact?
Bill Thurston On Dec 13, 2010, at 12:08 PM, Veit Elser wrote:
While I don't want to discourage the mathematical tangents this thread has generated, since it started as a physics question I think you deserve a physics answer.
The role of physics is not just to provide general principles, such as conservation laws. Sometimes it just comes down to numbers, plain and simple. The rules that one should apply to swinging steel balls are very different than what is appropriate to, say a couple of krypton atoms adsorbed on a graphite surface and struck by a third krypton atom. In that case one needs to consider the force of every atom on every other atom as the collision unfolds. By contrast, steel balls are simpler.
So here are some numbers. The speed of sound in steel is about 5000 m/s. A ball dropped from 5 cm reaches a speed of about 1 m/s. This tells us that the stresses imparted to the steel balls can be treated as quasi-static to a very good approximation. There are no shock waves.
The Young's modulus of steel is E = 200 Gpa. If we uniaxially compress a cylinder with cross sectional area A and length L by x, then the stored elastic energy is E A x^2/(2L). Equating this with the energy of the incident ball, say corresponding to m = 1 gm dropped an equivalent vertical rise of h = 10 cm or mgh = .01 J, I get x = 1 micron when A and L are of order 1 cm^2 and 1 cm. One micron is .01 the diameter of a human hair. Using the Young's modulus again, we find that the pressure at maximum compression, E x/L, is about 10 MPa. By contrast, the pressure supporting the weight of a ball placed on top of another is only about mg/A = 1 kPa -- four orders of magnitude smaller. So even when two balls in Newton's cradle are "in contact" because the strings supporting them make a slight angle with respect to the vertical, they may as well be miles apart with regard to the scale of the forces.
To help you digest these numbers, think of a pair of steel balls as two very stiff springs. The springs are in a collision course. The springs have a natural oscillation period but the collision is very slow on this scale. Start playing Smetana's Die Moldau (or Ravel's Bolero, your choice) right when the springs first touch. At about the time the Moldau crescendos into the Elbe the kinetic energy is completely and adiabatically (without the generation of entropy) converted into elastic energy. The springs will have compressed by a fraction of a millimeter. If you picture this in the center-of-mass frame, all motion ceases at an instant that marks the half-way point of the collision; thereafter the rest of the collision unfolds as the time reversal of the first half (and so even anharmonicity is not an issue).
I hope these numbers have convinced you that the pairwise collision model is the appropriate approximation to use for Newton's cradle, billiard ball dynamics, etc.
Here's a fun application of the pairwise model that you can try at home. Hold a basketball while resting a tennis ball on top of it. Now drop the basketball (the tennis ball should be co-moving and nearly in contact). You'll find that after all the collisions (pairwise!) have finished the tennis ball is shot up to a surprising height.
Veit
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On Dec 13, 2010, at 7:33 PM, Gary Antonick wrote:
am not sure I'm following completely. I've tried this experiment now with several different model cradles (all with 1st ball mass 2m and remaining balls mass m) and get the same result each time: the last ball flies out and the remaining balls barely move. The balls clearly do *not* swing out in a fan pattern predicted by pairwise collisions.
then the balls come together and interact again. after this *second* set of collisions the the pairwise model seems accurate (because the balls started off slightly apart)— the balls fan out in the predicted pattern.
Experiments should be taken seriously. By "fan pattern predicted by pairwise collisions" I assume you mean this pattern of final velocities: 2 balls 1/3 4/3 3 balls 1/9 4/9 4/3 4 balls 1/27 4/27 4/9 4/3 etc. Another thing to bear in mind. In the standard cradle the number of collisions for 2,3,4,... balls is 1,2,3,... But for your modification the number of collisions (pairwise model) goes like 1,3,6,10,... so a non-ideal coefficient of restitution degrades the kinetic energy more. Veit
Another thing to bear in mind. In the standard cradle the number of collisions for 2,3,4,... balls is 1,2,3,... But for your modification the number of collisions (pairwise model) goes like 1,3,6,10,... so a non-ideal coefficient of restitution degrades the kinetic energy more.
Veit __________________________________
Is there a good source to find measured values for coefficients of restitution? I found a paper with measurements of the coefficient of restitution for steel balls against a steel plate, at velocities between 2.2 and 3.5. The coefficient of restitution (COR) is velocity-dependent, but the maximum is only .62 at 2.2 meters per second. These seem very low. I suppose the balls are hollow, and this increases the coefficient of restitution(?) If someone with a Newton's cradle could take a video and single-step through, it should be possible to figure it out. To be more explicit about Veit's point, in the Newton's cradle with one ball of double weight, the first ball (4/3 in the perfectly elastic case) emerges as the result of (n-1) collisions, while the second ball (4/9) is the result of those plus (n-2) additional collisions, etc. Even if the coefficient of restitution is say .95, the effect would be severely dampened for say a 5 ball Newton's cradle. I understand that there is a strange behavior of the COR at low velocities, going up and down and up, or something like that. An anomalies of that sort could also ruin the effect. Bill Thurston On Dec 13, 2010, at 8:46 PM, Veit Elser wrote:
...
Experiments should be taken seriously. By "fan pattern predicted by pairwise collisions" I assume you mean this pattern of final velocities:
2 balls 1/3 4/3
3 balls 1/9 4/9 4/3
4 balls 1/27 4/27 4/9 4/3
etc.
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Suppose you have an infinite number of point masses, a_n (for all n in Z), positioned at regular intervals (1 unit) in a straight line, extending infinitely in both directions. Additionally, suppose that (instead of a gravity force), there is an 'antigravity' force (inverse square repulsion force) between any two point masses. Now, the masses will stay at rest, as there is a force of pi²/6 on each side. However, consider what happens when we remove mass a_0. If we consider a_1, for example, there is a force of pi²/6 pushing in one direction, and pi²/6 - 1 pushing in the other direction. Obviously, two masses cannot cross each other (as doing so would require overcoming an infinite repulsive force), so the situation can be modelled by [an infinite number of] second-order ordinary differential equations. Is there an analytic solution to this physical problem, i.e. can you determine the position of a_1 at time t=1, for example? Sincerely, Adam P. Goucher
Veit- I'm still not quite able to reconcile your posts (if I understand them correctly) with what I'm now observing. Velocities are very different depending on whether the balls are touching. I hadn't noticed this before. When the balls are initially separated the final velocities seem to be essentially 1/81, 4/81, 4/27, 4/9 and 4/3. When balls are initially touching the final velocities are clearly *not* 1/81, 4/81, 4/27, 4/9 and 4/3. The velocity of the final ball is somewhat greater than in the previous scenario. The velocities of the first four balls are very close to zero. Here's one approach that seems to predict what I'm observing: "Assumptions: At the moment of collision all the balls are scrunched up and become one entity as far as momentum and kinetic energy are concerned, because of serial compression from left to right. Then when the compression relaxes (at the speed of sound) the last ball is ejected, converting most of the stored potential energy of the compression to kinetic energy. The residue of the energy and momentum is shared between the four remaining balls. The equations for the velocities are as follows: Let the final velocity of balls 1-4 be x and of ball 5 be y. The combined mass of balls 1-4 is 5, and that of ball 5 is 1. Original momentum was 2 (2*1). ∴ Final momentum = 2 ∴ Equation A: 5*x + y =2 Original kinetic energy was 1 (0.5*2*1*1). ∴ Final kinetic energy = 1 ∴Equation B: 0.5*5*x^2 + 0.5*1*y^2 =1 Solving these equations gives one non-physical solution (where ball 5 has a negative velocity) and the other solution is as shown earlier: x = velocity of Balls 1-4 = 0.12251482 y = velocity of Ball 5 = 1.38742589 I think the key reason that the two-ball interactions don't work here is that the heavier ball cannot be completely stopped by any one ball, as opposed to the equal weight Newton's cradle. Hence it continues forward during the collision, inexorably compressing the other balls from left to right." On Mon, Dec 13, 2010 at 5:46 PM, Veit Elser <ve10@cornell.edu> wrote:
On Dec 13, 2010, at 7:33 PM, Gary Antonick wrote:
am not sure I'm following completely. I've tried this experiment now with several different model cradles (all with 1st ball mass 2m and remaining balls mass m) and get the same result each time: the last ball flies out and the remaining balls barely move. The balls clearly do *not* swing out in a fan pattern predicted by pairwise collisions.
then the balls come together and interact again. after this *second* set of collisions the the pairwise model seems accurate (because the balls started off slightly apart)— the balls fan out in the predicted pattern.
Experiments should be taken seriously. By "fan pattern predicted by pairwise collisions" I assume you mean this pattern of final velocities:
2 balls 1/3 4/3
3 balls 1/9 4/9 4/3
4 balls 1/27 4/27 4/9 4/3
etc.
Another thing to bear in mind. In the standard cradle the number of collisions for 2,3,4,... balls is 1,2,3,... But for your modification the number of collisions (pairwise model) goes like 1,3,6,10,... so a non-ideal coefficient of restitution degrades the kinetic energy more.
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In case anybody's still skeptical, I've made an animation of the standard Newton's cradle, Antonick's variation, and an "exponential" type: http://www.youtube.com/watch?v=31WDcngRRUk or on vimeo: http://vimeo.com/18550286 Sorry about the air resistance, but they're 19 tons. Made using Phun at http://www.phunland.com/wiki/Home . --Neil Bickford On 12/14/10, Gary Antonick <gantonick@post.harvard.edu> wrote:
Veit-
I'm still not quite able to reconcile your posts (if I understand them correctly) with what I'm now observing. Velocities are very different depending on whether the balls are touching. I hadn't noticed this before.
When the balls are initially separated the final velocities seem to be essentially 1/81, 4/81, 4/27, 4/9 and 4/3.
When balls are initially touching the final velocities are clearly *not* 1/81, 4/81, 4/27, 4/9 and 4/3. The velocity of the final ball is somewhat greater than in the previous scenario. The velocities of the first four balls are very close to zero.
Here's one approach that seems to predict what I'm observing:
"Assumptions: At the moment of collision all the balls are scrunched up and become one entity as far as momentum and kinetic energy are concerned, because of serial compression from left to right. Then when the compression relaxes (at the speed of sound) the last ball is ejected, converting most of the stored potential energy of the compression to kinetic energy. The residue of the energy and momentum is shared between the four remaining balls.
The equations for the velocities are as follows:
Let the final velocity of balls 1-4 be x and of ball 5 be y. The combined mass of balls 1-4 is 5, and that of ball 5 is 1.
Original momentum was 2 (2*1). ∴ Final momentum = 2 ∴ Equation A: 5*x + y =2
Original kinetic energy was 1 (0.5*2*1*1). ∴ Final kinetic energy = 1
∴Equation B: 0.5*5*x^2 + 0.5*1*y^2 =1
Solving these equations gives one non-physical solution (where ball 5 has a negative velocity) and the other solution is as shown earlier: x = velocity of Balls 1-4 = 0.12251482 y = velocity of Ball 5 = 1.38742589
I think the key reason that the two-ball interactions don't work here is that the heavier ball cannot be completely stopped by any one ball, as opposed to the equal weight Newton's cradle. Hence it continues forward during the collision, inexorably compressing the other balls from left to right."
On Mon, Dec 13, 2010 at 5:46 PM, Veit Elser <ve10@cornell.edu> wrote:
On Dec 13, 2010, at 7:33 PM, Gary Antonick wrote:
am not sure I'm following completely. I've tried this experiment now with several different model cradles (all with 1st ball mass 2m and remaining balls mass m) and get the same result each time: the last ball flies out and the remaining balls barely move. The balls clearly do *not* swing out in a fan pattern predicted by pairwise collisions.
then the balls come together and interact again. after this *second* set of collisions the the pairwise model seems accurate (because the balls started off slightly apart)— the balls fan out in the predicted pattern.
Experiments should be taken seriously. By "fan pattern predicted by pairwise collisions" I assume you mean this pattern of final velocities:
2 balls 1/3 4/3
3 balls 1/9 4/9 4/3
4 balls 1/27 4/27 4/9 4/3
etc.
Another thing to bear in mind. In the standard cradle the number of collisions for 2,3,4,... balls is 1,2,3,... But for your modification the number of collisions (pairwise model) goes like 1,3,6,10,... so a non-ideal coefficient of restitution degrades the kinetic energy more.
Veit _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Neil, This looks great! Nicely done. And.. if you're taking requests.. your simulation mimics a Newton's Cradle with gaps between the balls. Do you think you could model a cradle *without* the gaps? The gaps make a big difference. Cheers, Gary On Fri, Jan 7, 2011 at 4:40 PM, Neil Bickford <techie314@gmail.com> wrote:
In case anybody's still skeptical, I've made an animation of the standard Newton's cradle, Antonick's variation, and an "exponential" type: http://www.youtube.com/watch?v=31WDcngRRUk or on vimeo: http://vimeo.com/18550286
Sorry about the air resistance, but they're 19 tons. Made using Phun at http://www.phunland.com/wiki/Home .
--Neil Bickford
On 12/14/10, Gary Antonick <gantonick@post.harvard.edu> wrote:
Veit-
I'm still not quite able to reconcile your posts (if I understand them correctly) with what I'm now observing. Velocities are very different depending on whether the balls are touching. I hadn't noticed this before.
When the balls are initially separated the final velocities seem to be essentially 1/81, 4/81, 4/27, 4/9 and 4/3.
When balls are initially touching the final velocities are clearly *not* 1/81, 4/81, 4/27, 4/9 and 4/3. The velocity of the final ball is somewhat greater than in the previous scenario. The velocities of the first four balls are very close to zero.
Here's one approach that seems to predict what I'm observing:
"Assumptions: At the moment of collision all the balls are scrunched up and become one entity as far as momentum and kinetic energy are concerned, because of serial compression from left to right. Then when the compression relaxes (at the speed of sound) the last ball is ejected, converting most of the stored potential energy of the compression to kinetic energy. The residue of the energy and momentum is shared between the four remaining balls.
The equations for the velocities are as follows:
Let the final velocity of balls 1-4 be x and of ball 5 be y. The combined mass of balls 1-4 is 5, and that of ball 5 is 1.
Original momentum was 2 (2*1). ∴ Final momentum = 2 ∴ Equation A: 5*x + y =2
Original kinetic energy was 1 (0.5*2*1*1). ∴ Final kinetic energy = 1
∴Equation B: 0.5*5*x^2 + 0.5*1*y^2 =1
Solving these equations gives one non-physical solution (where ball 5 has a negative velocity) and the other solution is as shown earlier: x = velocity of Balls 1-4 = 0.12251482 y = velocity of Ball 5 = 1.38742589
I think the key reason that the two-ball interactions don't work here is that the heavier ball cannot be completely stopped by any one ball, as opposed to the equal weight Newton's cradle. Hence it continues forward during the collision, inexorably compressing the other balls from left to right."
On Mon, Dec 13, 2010 at 5:46 PM, Veit Elser <ve10@cornell.edu> wrote:
On Dec 13, 2010, at 7:33 PM, Gary Antonick wrote:
am not sure I'm following completely. I've tried this experiment now with several different model cradles (all with 1st ball mass 2m and remaining balls mass m) and get the same result each time: the last ball flies out and the remaining balls barely move. The balls clearly do *not* swing out in a fan pattern predicted by pairwise collisions.
then the balls come together and interact again. after this *second* set of collisions the the pairwise model seems accurate (because the balls started off slightly apart)— the balls fan out in the predicted pattern.
Experiments should be taken seriously. By "fan pattern predicted by pairwise collisions" I assume you mean this pattern of final velocities:
2 balls 1/3 4/3
3 balls 1/9 4/9 4/3
4 balls 1/27 4/27 4/9 4/3
etc.
Another thing to bear in mind. In the standard cradle the number of collisions for 2,3,4,... balls is 1,2,3,... But for your modification
the
number of collisions (pairwise model) goes like 1,3,6,10,... so a non-ideal coefficient of restitution degrades the kinetic energy more.
Veit _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Thanks. I made sure to only leave gaps a centimeter at most on the scale of ~100 meters, but the real culprit might be that the heights of the pendulums are not as precise. If I get a chance to, I'll redo the simulation with more precise hinges. --Neil Bickford On Fri, Jan 7, 2011 at 5:50 PM, Gary Antonick <gantonick@post.harvard.edu>wrote:
Neil,
This looks great! Nicely done.
And.. if you're taking requests.. your simulation mimics a Newton's Cradle with gaps between the balls. Do you think you could model a cradle *without* the gaps? The gaps make a big difference.
Cheers,
Gary
On Fri, Jan 7, 2011 at 4:40 PM, Neil Bickford <techie314@gmail.com> wrote:
In case anybody's still skeptical, I've made an animation of the standard Newton's cradle, Antonick's variation, and an "exponential" type: http://www.youtube.com/watch?v=31WDcngRRUk or on vimeo: http://vimeo.com/18550286
Sorry about the air resistance, but they're 19 tons. Made using Phun at http://www.phunland.com/wiki/Home .
--Neil Bickford
On 12/14/10, Gary Antonick <gantonick@post.harvard.edu> wrote:
Veit-
I'm still not quite able to reconcile your posts (if I understand them correctly) with what I'm now observing. Velocities are very different depending on whether the balls are touching. I hadn't noticed this before.
When the balls are initially separated the final velocities seem to be essentially 1/81, 4/81, 4/27, 4/9 and 4/3.
When balls are initially touching the final velocities are clearly *not* 1/81, 4/81, 4/27, 4/9 and 4/3. The velocity of the final ball is somewhat greater than in the previous scenario. The velocities of the first four balls are very close to zero.
Here's one approach that seems to predict what I'm observing:
"Assumptions: At the moment of collision all the balls are scrunched up and become one entity as far as momentum and kinetic energy are concerned, because of serial compression from left to right. Then when the compression relaxes (at the speed of sound) the last ball is ejected, converting most of the stored potential energy of the compression to kinetic energy. The residue of the energy and momentum is shared between the four remaining balls.
The equations for the velocities are as follows:
Let the final velocity of balls 1-4 be x and of ball 5 be y. The combined mass of balls 1-4 is 5, and that of ball 5 is 1.
Original momentum was 2 (2*1). ∴ Final momentum = 2 ∴ Equation A: 5*x + y =2
Original kinetic energy was 1 (0.5*2*1*1). ∴ Final kinetic energy = 1
∴Equation B: 0.5*5*x^2 + 0.5*1*y^2 =1
Solving these equations gives one non-physical solution (where ball 5 has a negative velocity) and the other solution is as shown earlier: x = velocity of Balls 1-4 = 0.12251482 y = velocity of Ball 5 = 1.38742589
I think the key reason that the two-ball interactions don't work here is that the heavier ball cannot be completely stopped by any one ball, as opposed to the equal weight Newton's cradle. Hence it continues forward during the collision, inexorably compressing the other balls from left to right."
On Mon, Dec 13, 2010 at 5:46 PM, Veit Elser <ve10@cornell.edu> wrote:
On Dec 13, 2010, at 7:33 PM, Gary Antonick wrote:
am not sure I'm following completely. I've tried this experiment now with several different model cradles (all with 1st ball mass 2m and remaining balls mass m) and get the same result each time: the last ball flies out and the remaining balls barely move. The balls clearly do *not* swing out in a fan pattern predicted by pairwise collisions.
then the balls come together and interact again. after this *second* set of collisions the the pairwise model seems accurate (because the balls started off slightly apart)— the balls fan out in the predicted pattern.
Experiments should be taken seriously. By "fan pattern predicted by pairwise collisions" I assume you mean this pattern of final velocities:
2 balls 1/3 4/3
3 balls 1/9 4/9 4/3
4 balls 1/27 4/27 4/9 4/3
etc.
Another thing to bear in mind. In the standard cradle the number of collisions for 2,3,4,... balls is 1,2,3,... But for your modification
the
number of collisions (pairwise model) goes like 1,3,6,10,... so a non-ideal coefficient of restitution degrades the kinetic energy more.
Veit _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I think your gaps may be an order of magnitude or so too big. it seems with a normal newton's cradle gaps of as little as 20 micrometers are enough to create the velocity pattern seen in your simulation. I think you're onto something using a very small gap. Am guessing a gap of zero doesn't simulate very well.. that is, the balls will behave erratically and unpredictably. Would be great to see what happens with a reduced gap (and ensure the balls are still when the first ball hits). Could you send a link to the file on the PHUN site? I wasn't sure how to find. On Fri, Jan 7, 2011 at 6:00 PM, Neil Bickford <techie314@gmail.com> wrote:
Thanks. I made sure to only leave gaps a centimeter at most on the scale of ~100 meters, but the real culprit might be that the heights of the pendulums are not as precise. If I get a chance to, I'll redo the simulation with more precise hinges. --Neil Bickford
On Fri, Jan 7, 2011 at 5:50 PM, Gary Antonick <gantonick@post.harvard.edu
wrote:
Neil,
This looks great! Nicely done.
And.. if you're taking requests.. your simulation mimics a Newton's Cradle with gaps between the balls. Do you think you could model a cradle *without* the gaps? The gaps make a big difference.
Cheers,
Gary
On Fri, Jan 7, 2011 at 4:40 PM, Neil Bickford <techie314@gmail.com> wrote:
In case anybody's still skeptical, I've made an animation of the standard Newton's cradle, Antonick's variation, and an "exponential" type: http://www.youtube.com/watch?v=31WDcngRRUk or on vimeo: http://vimeo.com/18550286
Sorry about the air resistance, but they're 19 tons. Made using Phun at http://www.phunland.com/wiki/Home .
--Neil Bickford
On 12/14/10, Gary Antonick <gantonick@post.harvard.edu> wrote:
Veit-
I'm still not quite able to reconcile your posts (if I understand them correctly) with what I'm now observing. Velocities are very different depending on whether the balls are touching. I hadn't noticed this before.
When the balls are initially separated the final velocities seem to be essentially 1/81, 4/81, 4/27, 4/9 and 4/3.
When balls are initially touching the final velocities are clearly *not* 1/81, 4/81, 4/27, 4/9 and 4/3. The velocity of the final ball is somewhat greater than in the previous scenario. The velocities of the first four balls are very close to zero.
Here's one approach that seems to predict what I'm observing:
"Assumptions: At the moment of collision all the balls are scrunched up and become one entity as far as momentum and kinetic energy are concerned, because of serial compression from left to right. Then when the compression relaxes (at the speed of sound) the last ball is ejected, converting most of the stored potential energy of the compression to kinetic energy. The residue of the energy and momentum is shared between the four remaining balls.
The equations for the velocities are as follows:
Let the final velocity of balls 1-4 be x and of ball 5 be y. The combined mass of balls 1-4 is 5, and that of ball 5 is 1.
Original momentum was 2 (2*1). ∴ Final momentum = 2 ∴ Equation A: 5*x + y =2
Original kinetic energy was 1 (0.5*2*1*1). ∴ Final kinetic energy = 1
∴Equation B: 0.5*5*x^2 + 0.5*1*y^2 =1
Solving these equations gives one non-physical solution (where ball 5 has a negative velocity) and the other solution is as shown earlier: x = velocity of Balls 1-4 = 0.12251482 y = velocity of Ball 5 = 1.38742589
I think the key reason that the two-ball interactions don't work here is that the heavier ball cannot be completely stopped by any one ball, as opposed to the equal weight Newton's cradle. Hence it continues forward during the collision, inexorably compressing the other balls from left to right."
On Mon, Dec 13, 2010 at 5:46 PM, Veit Elser <ve10@cornell.edu> wrote:
On Dec 13, 2010, at 7:33 PM, Gary Antonick wrote:
am not sure I'm following completely. I've tried this experiment
now
with several different model cradles (all with 1st ball mass 2m and remaining balls mass m) and get the same result each time: the last ball flies out and the remaining balls barely move. The balls clearly do *not* swing out in a fan pattern predicted by pairwise collisions.
then the balls come together and interact again. after this *second* set of collisions the the pairwise model seems accurate (because the balls started off slightly apart)— the balls fan out in the predicted pattern.
Experiments should be taken seriously. By "fan pattern predicted by pairwise collisions" I assume you mean this pattern of final velocities:
2 balls 1/3 4/3
3 balls 1/9 4/9 4/3
4 balls 1/27 4/27 4/9 4/3
etc.
Another thing to bear in mind. In the standard cradle the number of collisions for 2,3,4,... balls is 1,2,3,... But for your modification the number of collisions (pairwise model) goes like 1,3,6,10,... so a non-ideal coefficient of restitution degrades the kinetic energy more.
Veit _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I keep waiting for Neil to pipe up and say "For heaven's sake, people, stop guessing. Go to http://www.phunland.com/wiki/Download and start experimenting!" However, when I followed my own advice, I needed to call him up to get started. His lesson was roughly: Click on the Box creation tool. Drag diagonally near the top center of the background to create a fairly wide box. Click on Simulate (rightward triangle). The box should crash to the ground. Click Undo (double leftward triangle). The box should relevitate. Click on the Fixator tool (X). Nail the box to the background. Click on Simulate. Nothing should happen (except clouds drifting.) Click off Air resistance. Create a long, moderately thin vertical box for the pendulum "string". To get a Materials menu, you might need to click on the Move tool (crosshairs). Make the string out of helium or something. Nail it to your overhead box with a Hinge tool. Make a large steel ball with the Circle creation tool. With Materials, set its restitution to 1.0, and its mass to something convenient, e.g. 100kg . It might help to set its friction to 0. Nail it to the lower end of the string. With the simulation running, you can swing it out with the Drag tool (hand) and unclick to release it. Select the hinge-string-nail-ball system with the Polygon tool and Group it (with Select) and Clone it three or four times. Move the groups closer together to minimize offcenter collisions.
This simulator will eagerly model imperfections. When I got the equal mass scene working ideally, I saved it and then doubled the mass of the impactor and saved that, intending to avail the two scenes to math-fun to save you the foregoing setup trouble. But as far as I can tell, the save feature doesn't work on the Mac.
Anyway, the double mass impactor sent the last ball on a large excursion and the penultimate ball on a smaller one.
Phun can presumably well-describe a one dimensional sliding track model, which is the infinite-string limit. --rwg PS, an unfinished paper of Neil's informed me of http://www.chris-j.co.uk/rott.php , which you should check out if you've never seen Rott's pendulum.
-----------------
Neil is cheating. He has this amazing physics simulator from http://www.phunland.com/wiki/Home that even models air resistance. The clone feature naturally placed the balls slightly apart, which we deemed a plus, but strangeness ensued! The middle balls soon started moving, then gradually calmed down, then gradually excited again, ...
It turns out to depend on string length! If the strings are short and the gaps are large, the ball collisions are off center. So make the strings really long, and all should be clear. --rwg (Neil wishes to point out that his incorrect answer preceded the software experiment. He may be a poor theoretician but never a poor experimentalist.)
participants (11)
-
Adam P. Goucher -
Andy Latto -
Bill Gosper -
Bill Thurston -
Bill Thurston -
Fred lunnon -
Gary Antonick -
Joshua Zucker -
Neil Bickford -
Robert Munafo -
Veit Elser