A. A Bowler releases his ball at a speed of 3m/sec. If the ball has a diameter of 250 mm, what spin w should be put on the ball so there is no slipping? If he puts on only half of this spin w, keeping the same speed of 3 m/sec., what is the speed after .3 sec.? The ball has a mass of 1.8 kg and a dynamic coefficient of friction with the floor of .1. Neglect wind resistance and rolling resistance (as discussed in Section 7.1).

To determine the spin that should be put on the ball so there is no slipping, we can use the equation for the maximum static friction force (Fs) between two surfaces:

Fs = μs * N

Where μs is the static coefficient of friction and N is the normal force. For a spinning ball, the maximum static friction force is proportional to the spin angular velocity (w). Therefore, we can write:

Fs = k * w

Where k is a constant that depends on the ball and the surface it's rolling on.

Given that the diameter of the ball is 250 mm, we can calculate the radius (r) as:

r = diameter / 2 = 250 mm / 2 = 125 mm = 0.125 m

The normal force (N) acting on the ball can be calculated as the weight (mg) of the ball, where g is the acceleration due to gravity:

N = mg = 1.8 kg * 9.8 m/s^2 = 17.64 N

Now, let's solve for the spin angular velocity (w):

Fs = k * w
μs * N = k * w
0.1 * 17.64 N = k * w
1.764 N = k * w

Since the question doesn't provide a specific value for k, we can't determine the exact spin angular velocity (w). However, we can say that the spin angular velocity should be such that the product of this value and k is equal to 1.764 N, according to the given information.

Now, let's move on to the second part of the question. If the spin angular velocity (w) is decreased to half the initial value while keeping the same initial linear speed (v), we can use the principle of conservation of angular momentum to find the new spin angular velocity.

The angular momentum (L) of the ball can be calculated as:

L = I * w

Where I is the moment of inertia of the ball. For a solid sphere rotating about its diameter, the moment of inertia (I) is given by:

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

Given the mass of the ball (m = 1.8 kg) and the radius (r = 0.125 m), we can calculate the initial angular momentum (L1) as:

L1 = (2/5) * m * r^2 * w

Now, if the spin angular velocity (w) is halved, the new angular momentum (L2) can be calculated as:

L2 = (2/5) * m * r^2 * (w/2) = (1/5) * m * r^2 * w

According to the principle of conservation of angular momentum, L1 = L2. Therefore:

(2/5) * m * r^2 * w = (1/5) * m * r^2 * w

The mass (m) and the radius (r) are the same on both sides of the equation. Therefore, we can cancel them out:

(2/5) * w = (1/5) * w

Simplifying the equation, we find:

w = w/2

This tells us that the new spin angular velocity (w) is half of the initial spin angular velocity.

Finally, to find the speed after 0.3 seconds, we can use the equation for linear motion under constant acceleration:

v_f = v_i + a * t

Where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time.

In this case, the acceleration (a) is caused by the friction force opposing the motion of the ball. The magnitude of this force can be determined using:

F = μ * N

Where μ is the dynamic coefficient of friction.

Considering that the force due to friction (F) is equal to the total force on the ball in the horizontal direction (since there is no other force acting in that direction), we can rewrite the equation as:

μ * N = m * a

Solving for the acceleration (a), we get:

a = μ * N / m

Plugging in the given values:

a = 0.1 * 17.64 N / 1.8 kg = 0.098 m/s^2

Now, we can use the equation for linear motion to find the final velocity:

v_f = 3 m/s + (0.098 m/s^2) * 0.3 s

Simplifying the equation, we find:

v_f = 3 m/s + 0.0294 m/s
v_f = 3.0294 m/s

Therefore, the speed after 0.3 seconds is 3.0294 m/s.