Billiards experts work their magic on a pool table without any exact knowledge of distances, angles or speeds. Can you?
A regulation billiards table is twice as long as it is wide, has six pockets in the conventional positions, and a total of 18 “rail sights” (the little aiming circles or diamonds along all four sides). Rail sights and pockets form 24 equally spaced divisions around the table. You are at the “head” of the table, and desire to hit the cue ball from against your rail (the head rail), over the “foot spot” (that little dot near the other end), off the far (foot) rail, and back into one of the corner pockets on your end. For non-experts, (including me!), the foot spot is always two rail sight marks from the from the foot rail and centered between the two long sides.
Assume the following:
The ball rolls only – no sliding, no “english”, no leaving the table, therefore all shots go straight since the table is level
Coefficient of friction of the rolling ball on the table is 0.05 (a realistic value)
The ball always makes a perfect collision with the bumper (i.e. no energy loss and no friction while in contact)
Ignore the size of the ball and pockets – treat both as point entities.
Mass of the ball = OOOPS!, some pool shark has switched your cue ball for one of unknown weight (mass).
Q1: Where do you place the cue ball in order to make the bank shot?
Q2: What is the minimum initial speed you must give to the ball on this trajectory in order to make the shot (assume that if the ball stops exactly at the pocket, you succeeded)
(In reply to solution
In response to Kenny's comment, the following is a correction to my original solution. When taking the square root of t^2, I had taken the square root of the coefficient of w, but not of w itself. That is corrected below. Further precision could probably be gained by using the 32.174 value for g, rather than the 32 I used.
The foot spot is 1/4 of the table length from the foot rail, so the path from the foot spot to the foot rail will need to take it 1/5 of the way from the center-line of the table, or 1/10 of the table width, while the trip back to the head rail will take up the remaining 4/10 of the width. Traveling backward from the foot spot to the point where the cue hit the ball will require 3/10 of the width, and thus 2/10 of the width of the table from the opposite pocket to that aimed at.
The total distance to be traveled by the ball is the hypotenuse of a triangle 4 table widths long and .8 table widths wide, or a total length of 4.079215610874228 table widths, to the accuracy shown.
I'm not sure how kilograms of mass are related to kilograms of weight, but in the English system used in the U.S., in which I learned physics, mass, measured in slugs is 1/32 the weight in pounds on the earth's surface (32.174 in the denominator actually).
The frictional force is .05 the weight, so since f = ma, .05w = (w/32) a, and
a = 1.6 ft/s^2
v = 1.6 t ft/s
d = .8 t^2 ft
If w is the table width,
4.079215610874228 w = .8 t^2
t = 2.258100864353226 sqrt(w) s
v = 1.6 * 2.258100864353226 * sqrt(w) = 3.612961382965161 * sqrt(w) ft/s initially.
So it depends on the width of the table, being about 3.613 times the square root of the table width, in feet per second, when the table width is measured in feet.
Posted by Charlie
on 2007-01-06 11:16:34