A bowler throws a bowling ball of radius R=11 cm down a lane. The ball slides on the lane, with initial speed v(of the center of mass)=7 m/s and initial angular speed =0. The coefficient of kinetic friction between the ball and the lane is .25. The kinectic frictional force fk acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. Whenn speed v(of the center of mass) has decreased enough and angular speed has increased wnough, the ball stops sliding and then rolls smoothly.
What then is v in terms of angular speed?
During the sliding what is the balls linear acceleration?
During the sliding what is the balls angular acceleration/
How long does the ball slide?
How far does the ball slide?
What is the speed of the ball when smooth rolling begins?
I will be glad to critique your work
To solve the problem, we can analyze the motion of the ball during the sliding phase and then during the smooth rolling phase.
During the sliding phase:
1. Linear Acceleration: The linear acceleration of the ball can be determined using the equation Fk = m * a, where Fk is the frictional force and m is the mass of the ball. Since the only horizontal force acting on the ball is the frictional force, we have Fk = μ * N, where μ is the coefficient of kinetic friction and N is the normal force. The normal force N is equal to the weight of the ball, N = m * g, where g is the acceleration due to gravity. Therefore, Fk = μ * m * g. Substituting this into the equation Fk = m * a, we get μ * m * g = m * a. Canceling the mass, we find that the linear acceleration a is equal to μ * g. So, the linear acceleration of the ball during the sliding phase is a = μ * g = 0.25 * 9.8 m/s^2.
2. Angular Acceleration: The angular acceleration of the ball can be determined using the equation τ = I * α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Since the ball is sliding, the only torque acting on the ball is due to the frictional force. The torque can be calculated as τ = r * Fk, where r is the radius of the ball. Substituting the value of Fk from the previous step, we get τ = r * μ * m * g. The moment of inertia of a solid sphere rotating about its diameter is I = (2/5) * m * r^2. Substituting these values into the equation τ = I * α, we find r * μ * m * g = (2/5) * m * r^2 * α. Canceling the mass and rearranging the equation, we get α = (5 * μ * g) / (2 * r).
During the smooth rolling phase:
3. Speed v in terms of angular speed: When the ball starts to roll smoothly, the linear speed of the center of mass is equal to the product of the angular speed and the radius of the ball. So, v = ω * r, where v is the linear speed and ω is the angular speed.
4. Duration of sliding: To find the duration of sliding, we need to determine the time it takes for the ball to come to a stop. Since the linear acceleration is constant during the sliding phase, we can use the equation vf = vi + a * t, where vf is the final velocity, vi is the initial velocity, a is the linear acceleration, and t is the time. In this case, vf = 0 m/s (final velocity), vi = 7 m/s (initial velocity), and a = μ * g. Solving for t gives t = (vf - vi) / a = (0 - 7) / (0.25 * 9.8) s.
5. Distance traveled during sliding: The distance traveled during sliding can be calculated using the equation d = vi * t + (1/2) * a * t^2, where d is the distance, vi is the initial velocity, a is the linear acceleration, and t is the time. Substituting the known values, we get d = 7 * t + (1/2) * (0.25 * 9.8) * t^2 m.
6. Speed at the start of smooth rolling: Once the ball starts to roll smoothly, the linear speed of the center of mass is equal to the product of the angular speed and the radius of the ball. So, the speed at the start of smooth rolling is v = ω * r, where v is the linear speed and ω is the angular speed.
Please let me know if anything is unclear or if you need further assistance in solving the problem.