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.)
(b) What is the speed of the ball once it begins to roll without slipping?
(c) What is the kinetic energy of the ball just after the hit?

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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