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 v_0 and the coefficient of kinetic friction between the ball and the billiard table is µ_k. (Use the following as necessary: v_0, µ_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?
K_i =

Explanation / Answer

The mass of the ball is m

The point of application of force h = 2R/3

The speed of the ball just after the blow v0

The coefficient of kinetic friction is k

Radius of the ball is R

a)

The angular spped of the ball just after the hit is given by

I0 = mv0h

Where I0 is the rotational impulse

0 = mv0h/I

0 = mv0(2R/3)/(2/5mR2)

0 = 5v0/3R

b)

The equation of motion of the rotating ball which is slowing down(application of force is below the centre) is

=0 + 5k gt/2R

R = 0R + 5k gt/2

v = -5v0/3 + 5k gt/2

v0 - k gt = -5v0/3 + 5k gt/2

8v0 /3 = 7k gt/2

k gt = 16v0 /21

Therefore v = v0 - k gt

v = v0 - 16v0 /21

v = 5v0 /21

C)

The kinetic energy of the ball just after the hit is

KE = 0.5*mv02 + 0.5*I02

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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