A billiard ball that is initially at rest is given a sharp blow by a cue stick.

Question

A billiard ball that is initially at rest is given a sharp blow by a cue stick. The force is horizontal and is applied at a distance h = 2R/3 below the centerline, as shown in the figure below. The speed of the ball just after the blow is v0 and the coefficient of kinetic friction between the ball and the billiard table is ?k. (Use the following as necessary: v0, ?k, m for the mass and R for the radius of the billiard ball.)

a) What is the magnitude of the angular speed of the ball just after the blow? (Note: Write h in terms of R, and assume no friction the instant the ball is struck.)

?0 =

(b) What is the speed of the ball once it begins to roll without slipping?
v =

c) What is the kinetic energy of the ball just after the hit?

Ki =

Explanation / Answer

·         Apply Newton's second law

Tnet = F av ( h- R) = Icm * alpha = I cm = w/ delt

solving

wo = Fav ( h-R) del t/ I cm = F av( h-R) del t/ ( 2/5) mr^2

from the impulse formula

F av del t = m vo

wo = F av( h-R) m vo/ F av/ ( 2/5) mr^2 = 5 vo ( h-r)/ 2r^2

(b)

Apply Newton's second law to the ball

To = fkR = I cm alpha

Fy = Fn - mg = 0

Fx = - fk = ma

solving

alpha = uk mgR/ I cm = uk mgR/ ( 2/5) mR^2 = 5 uk g/ 2R

from the rotational kinematic equation

w = wo + alpha del t = wo + 5 uk g/ 2R) del t

a = - uk g

v = vo+ a del t

= vo- uk gt

R ( wo+ 5 ukg/ 2R * del t) = vo - uk g del t

del t = 16/21 ( vo/ uk g

v = vo - uk g ( 16/21) ( vo/ uk) g = 5/21 v0

(c) initial kineti energy

Ki = Ktran+ K rot = 1/2 mvo^2 + 1/2 I wo^2

= 1/2 mvo^2 + 1/2 ( 2/5* mR^2 ( 5 vo/ 3R)^2

= 19/18 * mvo^2

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