A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides

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 = 8.0 m/s and initial angular speed ?0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.25. 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) what then is v(center of mass) in terms of w?
m*w

(b) during the sliding, what is the ball's linear acceleration?
m/s^2

(c) during the sliding, what is the ball's angular acceleration?
rad/s^2

(d) how long does the ball slide?
s

(e) how far does the ball slide?
m

(f) what is the speed of the ball when smooth rolling begins?
m/s

Explanation / Answer

forces on it = mass * acceleration f=ma (b) a=kg=2.254m/s2 until the ball starts rolling where f is frictional force=mgk where k is coefficient of friction. now after time t its Vcom be V and initial be U so V=U-gkt for torque, f(R)=Ia (c) a=51.227 rad /sec2 until the ball starts rolling where I=2/5 m(R)^2 after time t let angular speed be ?. so, ?=kmgRt/2/5m(R)^2 =5kgt/2R for rolling V=?R=5kgt/2=U - kgt so t=2U/7kg =1.14 s ..............(d) so final V=6.428 m/s .........(f) distance travelled=U^2-V^2/2a = 8.7m.......(

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