A pool cue hits a pool ball of mass m and radius r giving it

A pool cue hits a pool ball of mass m and radius r, giving it a velocity upsilon0. At first the ball slides without rolling. As it slides, it loses velocity due to kinetic friction, muk. What Ls the velocity of the ball when it starts rolling without sliding. How far has it traveled?

Solution

When the ball is slipping, the forces acting on it are:

1) Weight Mg, acting vertically downward through the centre,
2) Normal reaction N, vertically upward through the point of contact
3) Forces of kinetic friction, f = µN, horizontally through the point of contact and opposite to the direction of motion

The only horizontal force f, produces horizontal acceleration, a = force/mass = µN/ M = µg [as N = Mg]. This acceleration is directed opposite to the direction of motion of the ball and hence its velocity V at any instant t is given by: V = V0 – µgt. .................................(1)

Further, the only force that produces a torque about the centre is fk. This torque is of magnitude fkR, acting in anticlockwise direction producing an anticlockwise angular acceleration, , of the ball about its center given as
fR = Icm , or µMgR = (2/5)MR2 , or = 5 µg /2R. This angular acceleration sets the ball rotating with increasing angular velocity in anticlockwise direction whose magnitude , at any instant t, is given by

= t. ....................(2)

When the ball starts pure rolling, V = R ................(3)

Hence using equations (1) , (2) and (3) we get, V0 – µgt = tR = 5 µgt /2, or µgt = 2V0 /7. Putting this value back in equation (1), we get

V = 5V0 /7.

Again, using the equation of kinematics, v2 = u 2 + 2aD, we get (5V0/7)2 = V0 2 - 2µgD or,

D = 12V0² /(49µg).