Physics in Billiards
Entry posted by willorn
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Since there's nothing better than rolling spheres and sliding blocks in physics, I figured my first blog post ought to be about one or both of the two. Unfortunately, I can't think of anything with sliding blocks that hasn't already been done, so this post will be about spheres.
The Question I'm trying to answer will be: "If we are supplied with given information about a billiards player's first shot in the game (the "Break") can we determine the resulting force on the two corner billiards balls using momentum and impulse? Assuming that the Triangle Rack is set up perfectly.
Let's just say that the average pool shark can imbue his 160 gram cue ball with 20 Newtons of Force in .05 seconds. Billiard Balls are made of Phenolic Resin Plastic, which has a notoriously low coefficient of friction. Take into account the soft cloth table, and the fact that the slipperiest solid we know of has a coefficient of .02, and we can reasonably assume that the coefficient of friction between the balls and the table could be .15
If the cue ball is to be placed 1 meters from the leading billiard ball as shown below.
[ATTACH=CONFIG]60[/ATTACH]
We can than calculate the velocity the cue ball will have the moment before it contacts the set of billiard balls using common Kinematics equations.
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The Circled Value is the velocity of the cue ball just before it hits the first billiard ball, taking deceleration of friction into account. In this case, the Cue Ball has an instantaneous velocity of 6.005 m/s which we can safely round to 6 m/s. Therefore, because p=mv (where p is rho/momentum), the cue ball strikes the others with .96 N*s.
Therefore, based on our laws of conservation of momentum, we can determine that (since our billiard balls are all perfectly and symmetrically arranged and touching) that the momentum from the cue ball is transfered to the corner balls completely. The fact that our billiard balls are all in contact is important, because any distance between them would be closed by the impulse from the cue ball and a loss of energy (momentum as well) would be lost.
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In the above snapshot, we can see that the momentum of the cue ball just before impact is equal to half of the momentum given to the corner balls. Realistically, the corner balls would receive momentum roughly equivalent to M/2 because for each billiard ball that receives the momentum of the cue ball (imagine the total energy passing through each ball along the outer edges) some of that ball's momentum is transfered to the two balls in front of it, but not equivalently. Another graphic can explain this phenomena.
In the graphic, the ROYGBIV scale denotes--roughly--the vector magnitude, with Red being the greatest. Notice that the black circle indicates a close-up vector addition analysis that betrays how initially, a billiard ball may look as though it doesn't have a trajectory that lines up with the pocket, but that a normal force does indeed cause the ball to be propelled towards the corner pocket (but only of course if you are a billiards master and can hit the cue ball dead-on)!
Regardless, the setup of the game is such that the corner balls receive the most of the original momentum and are propelled towards the corner pockets. For this reason the corner balls are the billiard balls most commonly sunk by the break. In fact, a billiards player would be foolish not to take advantage of this setup on his break, physics says so 
Back to math. The law of conservation of momentum (Mi=Mf) dictates that the momentum of the cue ball must equal the sum of all resulting momentum. Remember, in reality there are other momentum for each ball, but the majority of momentum goes to the corner billiard balls.
Mi=Mf => mvi=(mv)+(mv) => (mvi)=2mv => (mvi)/2m=vi/2=v
Therefore, if you apply a reasonable 20N force to a regulation 160 gram cue ball and you are GOD when it comes to aiming your cue ball for the break, you can expect each corner ball to shoot off with approximate velocity 6/2= 3m/s toward the corner pocket. :einstein)
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