Anyone for Pool (besides ESPN2)?
Although the following problems have fallen out of favor with the AP C examiners (the last time one was asked was 1980- see problem 2 below), they're fun to do and the students usually enjoy the challenge. In the third newsletter last year it was shown that in an elastic collision between objects of equal mass, with one initially at rest, the angle between their final velocities is 90°, a result that is of more practical use to pool players than the results below.
Besides Newton's second laws for translation (F = m
Dv/Dt) and rotation (t = IDw/Dt), what is needed is that impulse J equals change in linear momentum (J = Dp) and angular impulse (the product of J and the perpendicular distance between J and the center of mass) equals change in angular momentum (in the first figure Jh = DL, where L = Iw). Also, when an object of radius r is rolling without slipping its speed and angular speed are related by v = wr.|
A. At what height h above the center should a ball be struck a horizontal blow so that it begins to roll without slipping (this is also the height the table cushion contacts the ball)? Using the two impulse equations, J = mv and Jh = (2mr2/5)w, yields h = 2r2w/(5v). If it begins to roll without slipping, v = wr, so h = 2r/5. |
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B. If a ball is struck horizontally through its center so that its initial speed is vo, find its speed when it begins to roll without slipping. Since we know that the initial speed is vo and initial angular speed is 0, the second laws can be applied directly. It should be stressed that friction slows the translational speed and increases the rotational speed until they "match" at v = wr. F = ma => -f = m(v - vo)/t, and t = Ia => +fr = (2mr2/5)(w - 0)/t |
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Eliminating f and simplifying, using the fact that v =
wr when the slipping ceases, yieldsv = 5v
o/7. If a coefficient of friction m is assumed, then the distance and time until no slipping occurs can be determined by noting that the acceleration is -mg.
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C. If a ball Is struck horizontally a distance h = 4r/5 above its center, find its speed when it begins to roll without slipping if its initial speed is vo. Since the ball is struck above h = 2r/5 (see A), it will have top spin and friction will act to slow the rotational speed and increase the translation speed until they match at v = wr. This problem requires all four equations: J = mv => J = mvo (1) |
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Jh = I
w => 4rJ/5 = (2mr2/5)wo (2)F = ma
=> +f = m(v - vo)/t (3)t = Ia => -rf = (2mr2/5)(w - wo)/t (4)
Combining Eqs. (1) and (2), (3) and (4), yields v
o = rwo/2 and v - vo = (2r/5)(wo - w). Solving these two along with v = wr leads to v = 9vo/7. If a coefficient of friction m is assumed, then the distance and time until no slipping occurs can be determined by noting that the acceleration is +mg.
D. This has nothing to do with pool, but it might be some help with the Problem of the Month. If a marble is "squeezed" out from under one's thumb with backspin, what must be the relationship between v
o and wo if it comes to a "dead" stop (both translational and rotational speeds reaching zero simultaneously)? Here f acts to reduce both speeds, so the second laws yield -f = m(0 - vo)/t and -fr = (2mr2/5)(0 - wo)/t. Solving simultaneously yields vo = (2/5)rwo.
