Physics of Pool
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explored at the macroscopic and microscopic levels, giving insight into the way objects interact. Here we will take a look at how physics can be applied to the game of pool. Almost all of the different aspects of pool can be explained through physics, and a better understanding of these fundemental principles can aid in performance playing pool. In the game of pool there are sixteen balls, one cueball and fifteen physically identical numbered balls. While there are many interactions between all the balls on the table, in order to simplify things we will only look at the interactions between the cue ball and the first ball it strikes. These principles can be expanded to evaluate the entire system of balls on the pool table. In order to evaluate the physics of pool, first the measurable values of the system must be collected. Parameter Symbol Value Cue ball mass m1 1.6x101 kg Cue ball radius r1 2.79x102m Ball mass m2m16 1.7x101 kg Ball radii r2r16 2.86x102 m Coefficient of friction µk 0.027 Knowing these values the Motion, Work, and Energy can all be derived. The motion of the balls can be catagorized into two general catagories which are collisions and spin or rotation. Here the focus will be on collisions between the balls. The spin can have a significant effect on the motion of the ball, but due to time constraints and complexity of the science of it it will not be addressed to much detail. "An elastic collision between two objects is one in which the total kinetic energy (as well as total momentum) of the system is the same before and after the collision." (Physics for Scientists and Engineers) In a collision between the cue ball and another ball, since the masses are so similar the resulting angle between the vectors of the balls is approximately ninety degrees. How to Cite this Page
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This is due to the Conservation of
Energy: 1/2 m(cue) * v(cue(i))^2 = 1/2 m(cue) * v(cue(f))^2 + 1/2 m(ball) * v(ball(f))^2 or by dividing out the 1/2 m => v(cue(i))^2 = v(cue(f))^2 + v(ball(f))^2; where m(cue) is the mass of the cue ball, v(cue(i)) is the initial velocity of the cue ball, m(ball) is the mass of the other ball, and v(ball(f)) is the velocity of the ball. Since the velocities we are dealing with occure in a two dimentional plane they can be represented in vector form which allows us to determine the angle between the two balls using the dot product rule. Dot Product: V(cue(f)) • V(ball(f)) = v(cue(f)) * v(ball(f)) cos (Ø + O) v(cue(i))^2 = v(cue(f))^2 + v(ball(f))^2 + 2 * v(cue(f)) * v(ball(f)) cos (Ø + O) Subtracting the first equation from the above yields the following: 0 = 2 * v(cue(f)) * v(ball(f)) cos (Ø + O) dividing by 2 * v(cue(f)) * v(ball(f)) => 0 = cos (Ø + O) Therefore Ø + O = 90 degrees The Work involved in any aspect of physics relates force to distance in the following equation: W = F * d ; where W is work, F is force, and d is distance. Knowing this relationship the force necessary to move the ball a desired distance can be found by the initial force. In pool the initial force is added to the system by the person through the cue stick when striking the cue ball. However, the force is a function of two variables: the mass and acceleration of the cue stick striking the cue ball. F = m * a ; where f(i) is force, m(stick) is mass of the stick, and a(stick) acceleration of the stick To account for the frictional force in the shot, the work of the cue ball on the other ball is the F(i) multiplied by the distance between the cue ball and other ball d(i) minus the work done by the frictional force F(f) over the same distance all multiplied by the ratio of the angle(Ø) between the two balls . F(i) * d(i)  F(f) * d(i) = (90/(90Ø)) * F(ball) * d(ball) Knowing the above equations, the values from the properties, and the distance one wants the ball to travel the needed acceleration of the cue stick can be calculated by the following: a(stick) = [(90 / (90Ø)) * (F(ball) * d(ball)  F(f) * d(i))] / [*m(stick) * d(i)] 
