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I Collision of two billiard balls with spin

  1. Nov 16, 2016 #1
    Assume both billiard balls have same mass and same radius.

    So situation 1: ball 1 is moving forward without slipping on a surface and collides elastically with ball 2 which is stationary on that same surface.

    I read that immediately after the impact, ball 1 stops and keeps spinning with its original spin while ball 2 is imparted with the same CM velocity that ball 1 had.

    Does ball 1 not impart any spin to ball 2 because there is no kinetic friction between them?

    Now situation 2.
    Two identical billiard balls of radius R and mass M rolling with velocities ±⃗v collide elastically, head-on. Assume that after the collision they have both reversed motion and are still rolling.

    Below is initial picture before collision.


    Now situation 2 should be similar to situation 1. So why is it that they reverse motion in all aspect? They should only swap linear velocities. But their spin was swapped as well. Ball one is now counter clockwise whereas it wasn't before. But there is no vertical contact force from the collision since there is no friction from the contact of the balls surfaces. That is ball one cannot alter the spin of ball 2 and vice versa. So why does the spin gets traded in this case verse case one where it didn't?
  2. jcsd
  3. Nov 16, 2016 #2


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    Because you assumed it.
  4. Nov 16, 2016 #3
    I didn't assume it. It was a qual question I looked up.

    They wanted me to find the impulse the table delivers. So that implies that the table does the reversion?

    But then why doesn't the table do anything for case one?
  5. Nov 16, 2016 #4
    And in case two, why does the table deliver any impulse?

    During the collision, the two balls slip over eachother. Therefore there shouldn't be any extra pressing on the table compared to the original weight that was always there.
  6. Nov 16, 2016 #5
    Oh I think I get it. in case 2, during the collision, what happens first is that the Vcm is reduced rapidly, hence making the ball slip( since the original spin can no longer keep up with the Vcm). Therefore kinetic friction rapidly increases, applying a torque to counter the original clockwise motion. By the time the Vcm has switched directions, the Kinetic Friction would have done enough work to completely revese the spin to counterclockwise
  7. Nov 16, 2016 #6


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    So is there is friction or not?
  8. Nov 16, 2016 #7
    Because the collision is elastic, there can't be friction between the balls.

    But the is friction between the table and the balls. Otherwise, how to it get to have pure roll in the first place? So during the collision, there is slipping, which means there must be friction to torque the spin
  9. Nov 16, 2016 #8
    I reread the question for situation 1. It wants me to assume that there is no friction during the collision. That is why no spin is traded.

    In situation two, nothing was mentioned about there being friction or not, but because there is a reversal of direction, logic dictates that its friction that does the reversal of angular momentum for a single ball.

    It can't be the normal force since the normal force is in line with the center of the ball and so it can't torque. And the collision is elastic so the balls cannot deform in anyway to make the normal force off the axis pointing to the cm.

    Since normal force cannot be the factor, and there is no friction between the balls themselves, it must be friction from the table
  10. Nov 16, 2016 #9


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    Doesn't follow. Static friction is non-dissipative.

    Pure roll doesn't require friction, and how the balls started rolling is not part of the question.
  11. Nov 16, 2016 #10
    Ok so you are saying that it must be due to the friction between the balls then that caused the reversal of motion? The friction itself cannot cause the reversal since the tangetial velocities are in the same direction

    we can consider the idea the the surfaces of the balls are somewhat jagged and then both balls would boost the angular speed of the other. Hence the tangential speed for any one ball over duration of the collision would double. But angular momentum must be conserved. By Noethers theorem since there is rotational symmetry before the collision. So somehow, the ground would counter with and even greater impulse to completely reverse the angular momentum of each ball so that the total angular momentum stays 0. Because that the balls are faster downwards, the ground must counter. But with what? A normal impulse up?

    So the tangential portion of the collision causes an impluse downwards, making the balls want to dig into the table. As a response the table pushs back with a greater normal force. But this normal force must slightly be off center in order to torque.

    This is all assuming that there is no friction between the table and balls the whole time.
  12. Nov 16, 2016 #11


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    Where did I say that? I was asking you to define the scenario properly.

    What does vertical momentum conservation say about this idea?
  13. Nov 16, 2016 #12
    Oh ok got it. Sorry about that.

    That it has to be the same so it can't be greater.
  14. Nov 16, 2016 #13


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    What can't be greater than what? The balls' centers of mass never have any vertical acceleration, right?
  15. Nov 16, 2016 #14
    But if the response by the normal force isn't greater the increase in vertical momentum, then there is no way to reverse the spin
  16. Nov 16, 2016 #15
    Hmm right. It's because the normal force increased to prevent any acceleration vertically.

    during the collision, a ball will experience a vertical push componant downward due to the other ball rotating down on it. So the normal increases to counter.
  17. Nov 16, 2016 #16
    But the impulse from the ground is just enough to counter any vertically downward tangential impulse from the other ball. Even if the table impulse was off center, the counterclockwise torque it would create would be far less than the clockwise torque from the other ball. Meaning that while the CM would not accelerate downwards, the angular rotation would still be in the same direction but greater.
  18. Nov 16, 2016 #17
    Here is the problem and the solution. I have a hard time understanding it. It seems that friction is involved somehow because the leaver arm to the force from the table is R in the angular impulse expression.


  19. Nov 17, 2016 #18


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    Please justify this claim. And reconcile it with Newton's third law.
  20. Nov 17, 2016 #19
    I thought about it, and it seems like it would not make sense with newtons third law.

    Now i am thinking about an analogy. So if two identical vertical pendulums with the bobs on their tops and drop them in such a way that they crash, they will bounce back, rotating in the opposite direction. kinda like how a head on collision of two same speed and mass particles will just make them bounce back.

    I imagine that the particles on the surface of the ball act the same way. Microsopically, the first particles that hit should hit at an angle slightly above the horizontal, giving a backwards rotatory bounce.
  21. Nov 17, 2016 #20


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