Here we have a V-shaped barrier in which a point mass (ball) bounces about, under the influence of a constant vertically downward force (e.g. gravity). There seemingly being nothing new in chaos, this is (perhaps well) known as the gravitational wedge billiard [1].
We consider a right angle wedge as shown below:
As in the previous post, I've assumed that if the ball is below slope there is an outward (towards positive y-axis) restoring force along the normal to the slope. As before, this force is given by Hooke's law. In the previous post a value for q (the normal distance from ball to slope) was worked out, applicable to the right hand side (i.e. x > 0). Extending the thing to both sides of the wedge gives the result:
and, as before m is the mass of the ball and k the spring constant. Hence the above gives the motion of the ball as it hits, descends into, and eventually bounces off the slope. The bounce is elastic. The bigger the value of k, the less spongy the slope is. Here's the patching:
Each of the following results shows the first four minutes' worth of the trajectory (1 second time constant on the integrators), with the ball starting at (0,6). Acceleration due to gravity (acting vertically downwards) is 10 m/s2. The quoted spring constants correspond to a 1 kg mass. Note that, as spring constant k is reduced, the ball sinks further below the x axis, at least for the first few bounces. Although, in theory, one would expect the ball to simply bounce up and down along the y-axis forever, in practice it deviates to one side or the other after two or three bounces, enough to flick it on to one of the sides of the wedge, high up enough to lead to further deviations and ensuing chaos.
An interesting observation: the 'effective' wedge angle reduces as the spring constant reduces: the more spongy the slope, the smaller the wedge angle. What's also interesting / unexpected (perhaps) is that the sides of the triangular wedge trajectory space remains sharply defined.
Finally, if, rather than dropping the ball straight down the wedge's vertical axis, we launch it from part way up the left hand side, towards the right hand side, we achieve a more stable trajectory / orbit. For example, with the initial velocities along x and y as +8 m/s and +6 m/s respectively, and starting at (-4,4) we get:
This looks something like a precessing(?) period two orbit.
References
[1] Classical and quantum chaos of the wedge billiard. I. Classical mechanics, T. Szeredi and D. A. Goodings, Physical review. E, November 1993; 48(5).
We consider a right angle wedge as shown below:
As in the previous post, I've assumed that if the ball is below slope there is an outward (towards positive y-axis) restoring force along the normal to the slope. As before, this force is given by Hooke's law. In the previous post a value for q (the normal distance from ball to slope) was worked out, applicable to the right hand side (i.e. x > 0). Extending the thing to both sides of the wedge gives the result:
and, as before m is the mass of the ball and k the spring constant. Hence the above gives the motion of the ball as it hits, descends into, and eventually bounces off the slope. The bounce is elastic. The bigger the value of k, the less spongy the slope is. Here's the patching:
Each of the following results shows the first four minutes' worth of the trajectory (1 second time constant on the integrators), with the ball starting at (0,6). Acceleration due to gravity (acting vertically downwards) is 10 m/s2. The quoted spring constants correspond to a 1 kg mass. Note that, as spring constant k is reduced, the ball sinks further below the x axis, at least for the first few bounces. Although, in theory, one would expect the ball to simply bounce up and down along the y-axis forever, in practice it deviates to one side or the other after two or three bounces, enough to flick it on to one of the sides of the wedge, high up enough to lead to further deviations and ensuing chaos.
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| k = 200 N/m |
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| k = 20 N/m |
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| k = 10 N/m |
Finally, if, rather than dropping the ball straight down the wedge's vertical axis, we launch it from part way up the left hand side, towards the right hand side, we achieve a more stable trajectory / orbit. For example, with the initial velocities along x and y as +8 m/s and +6 m/s respectively, and starting at (-4,4) we get:
References
[1] Classical and quantum chaos of the wedge billiard. I. Classical mechanics, T. Szeredi and D. A. Goodings, Physical review. E, November 1993; 48(5).
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