Collision of two billiard balls with spin

  • #1
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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.

billard.png


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?
 

Answers and Replies

  • #2
A.T.
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Assume that after the collision they have both reversed motion and are still rolling.
...
So why is it that they reverse motion in all aspect?
Because you assumed it.
 
  • #3
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Because you assumed it.
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?
 
  • #4
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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.
 
  • #5
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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
 
  • #6
A.T.
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because there is no kinetic friction between them?
the Kinetic Friction would have done enough work
So is there is friction or not?
 
  • #7
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So is there is friction or not?
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
 
  • #8
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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
 
  • #9
A.T.
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Because the collision is elastic, there can't be friction between the balls.
Doesn't follow. Static friction is non-dissipative.

But the is friction between the table and the balls. Otherwise, how to it get to have pure roll in the first place?
Pure roll doesn't require friction, and how the balls started rolling is not part of the question.
 
  • #10
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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.
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.
 
  • #11
A.T.
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Ok so you are saying that it must be due to the friction between the balls then that caused the reversal of motion?
Where did I say that? I was asking you to define the scenario properly.

As a response the table pushs back with a greater normal force.
What does vertical momentum conservation say about this idea?
 
  • #12
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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?
Oh ok got it. Sorry about that.

That it has to be the same so it can't be greater.
 
  • #13
A.T.
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That it has to be the same so it can't be greater.
What can't be greater than what? The balls' centers of mass never have any vertical acceleration, right?
 
  • #14
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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
 
  • #15
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What can't be greater than what? The balls' centers of mass never have any vertical acceleration, right?
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.
 
  • #16
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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.
 
  • #17
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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.

collisionBillard.png


SolutiontoCollision.png
 
  • #18
jbriggs444
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during the collision, a ball will experience a vertical push componant downward due to the other ball rotating down on it.
Please justify this claim. And reconcile it with Newton's third law.
 
  • #19
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Please justify this claim. And reconcile it with Newton's third law.
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.
 
  • #20
jbriggs444
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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.
No.
 
  • #21
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No.
ok so then they collide right at the horizontal, and the y momentum of the two rim particle that collided would be conserved. Well if this collision took place in outer space. In in the pic where red and blue are the rim particle that collides.

RIM Particles .png



Except that on the table, this vertical impulse would cause a impulsive response from the table. And on the table, the table cannot respond x impulse since its directed parallel to the table.
 
  • #22
jbriggs444
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Except that on the table, this vertical impulse would cause a impulsive response from the table. And on the table, the table cannot respond x impulse since its directed parallel to the table.
What vertical impulse? You have two particles with the same vertical velocity colliding. The resulting force on both would be purely horizontal.
 
  • #23
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What vertical impulse? You have two particles with the same vertical velocity colliding. The resulting force on both would be purely horizontal.
Makes sense. The line of contact is horizontal.

In that case, there is no vertical impulse and the normal force from the table remains unchanged.

I was thinking that maybe a slightly off center normal impulse would cause the torque. But the ball is rigid during the elastic collision so that option is impossible.

Therefore kinetic friction from the table must act during the collision to reverse the spins. During the collision the pure roll is destoried because v became smaller than r*w. So this slipping would cause the friction to act.

Though the problem never said that there is kinetic friction from the ground, this must be the case as the horizontal impulse from the collision itself cannot torque.
 
  • #24
jbriggs444
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Therefore kinetic friction from the table must act during the collision to reverse the spins.
If the collision is of short duration than you would require a large torque for a short time. Without an insane friction coefficient, that cannot be. So kinetic friction during the collision cannot explain any such effect.
 
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  • #25
jbriggs444
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I was thinking that maybe a slightly off center normal impulse would cause the torque
You might want to calculate just how far off center that impulse would need to be. You have all the information you need for such a calculation.
 

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