Question

A bowling ball of mass M and radius R = 20.0 cm is released with an initial translational velocity of vi = 14.0 m/s to the right and an initial angular velocity of zero. The bowling ball can be treated as a uniform solid sphere. The coefficient of kinetic friction between the ball and the surface is 0.190. The force of kinetic friction causes a linear acceleration, as well as a torque that causes the ball to spin. The ball rolls while slipping (sliding) along the horizontal surface for some time, and then rolls without slipping at constant velocity after that. Use g = 10 m/s2.(b) What is the magnitude of the acceleration of the ball while it is sliding?(c) What is the magnitude of the angular acceleration of the ball while it is sliding?(d) How far does the ball travel while it is sliding?(e) What is the constant speed of the ball when the ball rolls without slipping?

Answer 1

Answer:

b) a = 1.9m/s^2

c) α = 23.75 rad/s^2

d) d = 25.21m

e) V = 10.01m/s

Explanation:

Since the only force acting on the ball is friction:

Ff = m * a

a = Ff / m = μ * m*g / m = 1.900 m/s^2

With torque, we get:

Ff * R = I * α

μ * m * g * R = 2/5 *m * R^2 * α     Solving for α:

α = 23.75 rad/s^2

The ball will start rolling without sliding when linear velocity matches angular velocity:

Vf = vi - a * t = R * ωf = R *α * t

vi - a * t = R *α * t    Solving for t:

t = 2.1s

With this time we calculate the distance:

d = vi * t - a / 2 * t^2 = 25.21m

As for the speed:

Vf = vi - a * t = 10.01m/s