A uniform, spherical bowling ball of mass m and radius R is projected horizontally along the floor at an initial velocity v0 = 6.00 m/s. The ball is not rotating initially, so ω0 = 0. It picks up rotation due to (kinetic) friction as it initially slips along the floor. The coeffcient of

kinetic friction between the ball and the floor is µk . After a time ts , the ball stops slipping and
makes a transition to rolling without slipping at angular speed ωs and translational velocity vs .
Thereafter, it rolls without slipping at constant velocity.

(b) Find an equation for the linear acceleration a of the ball during this time. The acceleration
should be negative, since the ball is slowing down.
(c) Find an equation for the angular acceleration α of the ball while it is slipping. It will be
simpler if you use the sign convention that clockwise rotations are positive, so α > 0.
(d) What constraint on ω and v must take effect at time t = ts , the moment when the ball
stops slipping and begins rolling without slipping?

b) Energy is not conserved do to friction. So you know initial veloicty, and final velocity.

linearacceleartion=(finalveloicyt-initialvelocity)/time
= (vs-vo)/ts

c) alpha= acceleration above/ radius

d) v=w*r at ts

To find the equations for linear acceleration (a) and angular acceleration (α) of the ball during the slipping phase, we can start by considering the forces acting on the ball.

During the slipping phase, the only force acting on the ball is the kinetic friction force (f).

(b) Linear acceleration (a):
The net force acting on the ball can be expressed as the difference between the applied force (zero in this case) and the kinetic friction force:

Net force = 0 - f = -f

Newton's second law tells us that the net force is equal to the mass of the object multiplied by its linear acceleration:

- f = ma

Substituting the expression for the kinetic friction force (f = μk * mg), where μk is the coefficient of kinetic friction and mg is the gravitational force acting on the ball, we get:

-μk * mg = ma

So, the equation for the linear acceleration (a) during the slipping phase is:

a = -μk * g

This acceleration is negative, as the ball is slowing down.

(c) Angular acceleration (α):
During the slipping phase, the ball is rotating due to the slipping motion. The torque causing the angular acceleration is due to the friction force acting at a perpendicular distance equal to the radius of the ball (R) from the center of mass.

The net torque acting on the ball can be expressed as:

Torque = Fr

Where F is the friction force and r is the radius of the ball.

The friction force (F = f) can be expressed as the coefficient of kinetic friction (μk) multiplied by the normal force (N = mg) acting on the ball.

F = μk * N = μk * mg

Substituting the values, we get:

Torque = μk * mg * R

By Newton's second law for rotation, the torque is equal to the moment of inertia (I) of the ball multiplied by the angular acceleration (α):

Torque = I * α

For a uniform spherical ball, the moment of inertia (I) is equal to (2/5) * m * R^2, where m is the mass of the ball.

Thus, we have:

(2/5)mR^2 * α = μk * mg * R

Simplifying, we get:

α = (5/2) * (μk * g) / R

So, the equation for the angular acceleration (α) while the ball is slipping is:

α = (5/2) * (μk * g) / R

(d) Constraint on ω and v at t = ts:
At the moment t = ts, the ball stops slipping and begins rolling without slipping. For rolling without slipping, the linear velocity (v) and angular velocity (ω) are related by the equation:

v = ω * R

Therefore, at t = ts, the linear velocity (vs) and angular velocity (ωs) are related by:

vs = ωs * R

This constraint ensures that the ball rolls without slipping.