In a game of billiards, a skillful player hits a cue ball with just enough backspin so that the ball stops after traveling a short distance away from where it was hit. Suppose the initial magnitude of the velocity of the ball was v0 (after being hit). Assume the cue ball is a solid sphere with mass m and radius r.
How much backspin (with an initial magnitude of angular velocity ω0) would be required for this to occur? Here, you may assume that rolling friction may be neglected, but kinetic friction and static friction are relevant. Make a sketch of the system for each part below. Also, choose coordinates and make sure your vector quantities are properly treated as vectors (or components of vectors).
Solve this problem in two ways:
(a) Consider the force of friction and use equations describing the linear kinematics and rotational kinematics.
(b) Find a reference point about which the net torque is zero and use the concept of conservation of angular momentum.

a) w=(5*vo)/(2r)

b) w=(5*vo)/(2*r)

Explanation:

a) according to the attached diagram we have to:

vo is the velocity of the ball after being hit. if we use Newton's second law:

F=m*a (eq.1)

where F is the force of friction, m is the mass of the ball and a is the acceleration. The normal force is equal to:

N=m*g

Also F=u*N (eq. 2), where u is the coefficient of friction. Replacing eq. 1 in eq. 2, we have:

F=m*u*g (eq. 3)

analyzing equation 1 and 3:

m*a=m*u*g

a=u*g

The torque is equal to:

τ=F*r (eq. 4)

τ=m*u*g*r

The relation between torque and angular acceleration is named moment of inertia.

I=(2/5)*m*r^2

if we have:

α=τ/I=(5*u*g)/(2*r)

The time is equal to:

t=(vo-u)/a=vo/(u*g)

the angular velocity is equal to:

w=α*t=((5*u*g)/(2*r))*(vo/(u*g))=(5*vo)/(2r)

b) the angular momentum is equal to:

L=m*v*r

the initial angular momentum is equal to:

Li=-I*w + m*vo*r

The final angular momentum (Lf) is 0

Thus we have:

Li=Lf

Replacing:

-I*w + m*vo*r=0

-((2/5)*m*r^2)*w + (m*vo*r)=0

clearing w:

w=(5*vo)/(2*r)

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