A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides
ID: 2251675 • Letter: A
Question
A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides on the lane with initial speed vcom,0 = 10.5 m/s and initial angular speed ?0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.17. The kinetic 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. When speed vcom has decreased enough and angular speed ? has increased enough, the ball stops sliding and then rolls smoothly.
A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides on the lane with initial speed vcom,0 = 10.5 m/s and initial angular speed ?0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.17. The kinetic 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. When speed vcom has decreased enough and angular speed ? has increased enough, the ball stops sliding and then rolls smoothly. What then is vcm in terms of ?? During the sliding, what is the ball's linear acceleration? m/s2 During the sliding, what is the ball's angular acceleration? rad/s2 How long does the ball slide? s How far does the ball slide? m What is the speed of the ball when smooth rolling begins? m/sExplanation / Answer
the torque = 0.20*m*g*R = [(2/5)mR^2]*(alpha), where alpha is angular acceleration. So alpha is given by alpha = (0.5*g)/R = (0.5*9.8)/0.11 = 44.54 rad/s^2
Let us calculate ? after time t. ? = (0.5*g/R)*t and v (after t) =7 - 0.20*g*t. The sliding stops when ?*R = v or 0.5*g*t = 7 - 0.20*g*t or
t = 7/(0.7*9.8) = 1.02 s. After 10 seconds ball stops sliding
Distance traveled due to translation = 7*1.20 - 0.5*0.20*9.8*(1.02^2) = 6.12 m
Angular distance traveled by ball = 0.5*44.54*(1.02^2) = 23.17 rad; equivalent distance traveled on ground = 23.17*R = 2.55 m. So total distance traveled before ball stops sliding = 6.12 + 2.55 = 8.67 m.
sped of the ball when smooth rolling begins = 0.5*9.8*1.02 = 5.0 m/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.