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if you take a ball and give it a spin with the axis being roughly perpendicular to the floor, then the direction to which it's spinning will reverse after it bounces. I have ideas but I'm not completely sure why.

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  • $\begingroup$ What are your ideas? The answer will depend somewhat on the nature of the ball and the floor. $\endgroup$ – J. Murray Dec 1 '17 at 20:12
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    $\begingroup$ well my guess is that the ball deforms at the point of contact in a twisting type of way, and so it tries to snap back to how it was previously but because of inertia it goes further and results in a change in direction. $\endgroup$ – Willsbury Dec 1 '17 at 20:15
  • $\begingroup$ Sounds right to me. It's important to note that torque, which is required to change the angular momentum of the ball, would need to be exerted over an area in this case - so the ball would have to be sufficiently "squashed" during the impact time. You would also need a sufficient amount of friction. The reversal would not occur in the limit of low friction or very rigid ball. $\endgroup$ – J. Murray Dec 1 '17 at 20:21
  • $\begingroup$ A proper answer would give better quantitative conditions, but I suspect that an experiment performed with a basketball on slick ice and a ping pong ball on a regular floor would demonstrate the above difference in behavior. $\endgroup$ – J. Murray Dec 1 '17 at 20:22
  • $\begingroup$ ya I know that it doesn't work with all balls and surfaces, should have specified $\endgroup$ – Willsbury Dec 1 '17 at 20:27
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To treat this problem in 2D, let us assume a cylinder with a perfectly rough surface, uniform density, and perfect elasticity. Assume further that it is spinning with an initial velocity $\omega$ (assume the axis to be parallel to the surface of impact) and is dropped with no horizontal velocity.

Upon impact, there will be a time interval $\Delta t$ during which the contact point cannot move (because of perfectly rough surface), but since the ball has an initial angular momentum we need some way to resolve this contradiction.

Because of finite elasticity and density, the information that the cylinder has made contact with the surface is traveling across the cylinder at finite velocity: in other words, there will be parts of the cylinder that are still moving after the initial impact, because the elastic wave has not yet reached all the way across the cylinder. I try to illustrate that here:

enter image description here

After the ball makes contact at point P, an expanding region stops moving: this is indicated with the grey area in the third diagram, and the region marked E corresponds to extension while the region marked C corresponds to compression. In this way, rotational elastic energy is stored in the ball during impact.

Once the grey region reaches the top of the ball, every part of the ball "knows" about the impact and it has reached maximum compression (of course the top of the ball continues to move down until this time as well).

Now that the ball has stopped moving, there are only elastic forces that push the ball away from the surface - and all these elastic forces are still there. Since the ball has completely stopped rotating at this point, the angular momentum is zero; but there continues to be a horizontal force at point P as the ball starts to "unwind". This means that the total angular momentum imparted by the surface will be greater than the initial angular momentum - and the spin will reverse.

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    $\begingroup$ Nice explanation, especially the illustration. But the OP asked for a spin axis perpendicular to the floor. Same basic idea, but a bit harder to illustrate. :) $\endgroup$ – Mike Dec 4 '17 at 15:28
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    $\begingroup$ @Mike ah - I missed that. The same principle applies - it "winds up" until the whole ball is stationary, then "unwinds" again; all the while the floor continues to apply torque to the ball. I may need to completely rewrite this answer... will have to wait until tonight. $\endgroup$ – Floris Dec 4 '17 at 15:31

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