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 µk. (Use the following as necessary: v0, µk, 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

a) Use rotation impulse, Pt = mv0r; r = 2R/3 Pt = Iw0; w0 =(2mv0R/3)/[(2/5)mR2] = 5vo/3R b) Since F is below the center line, the spin is backward, i.e., the ball will slow down. w0 =-5v0/R. w = w0 + (5/2)mkgt/R; v = v0 - mkgt; set wR =v v0 - mkgt = -(5/3)v0 + (5/2)mkgt; t =(16/21)v0/mkg v = (5/21)v0 =0.238vo c) ?The kinetic energy of the ball just after the hit is KE = 0.5*mv02 + 0.5*I?02 KE = 0.5*m*v02 + 0.5*(2/5 mR2 )(25v02/9R2) KE = 0.5*m*v02 + 5/9 * m*v02 KE = 1.056 m*v02

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