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 = 6.5 m/s and initial angular speed ?0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.27. 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 vcm in terms of ??
_______m·?

(b) During the sliding, what is the ball's linear acceleration?
________m/s2

(c) During the sliding, what is the ball's angular acceleration?
_________rad/s2

(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

I can answer parts b through f, but I can't answer part a because I can't tell what it says. Read part a for yourself: there are two ?? at the end. That happened because Cramster couldn't interpret the characters when you copied and pasted the question (look through your post... you'll see other ? where this happened.)

If I knew what Part a said, I could answer it. But you'll have to send me a private message with more details on that question.

Anyway...

   force of friction = umg

    linear acc = force / m = ug = 0.27 * 9.8 = 2.646

    moment of inertia = 0.4*mr^2 = 0.4 * m * 0.11^2 = 0.00484 m

   torque = force * r = umg*r = 0.27*m*9.8*.11 = 0.29106 m

    ang acc = torque / I = 0.29106 m / 0.00484 m = 60.136

To find the time for the ball's slide, relate the angular speed and linear speed:

   final linear speed = initial speed - acc * time = 6.5 - 2.646 t

also   final linear speed = ang acc * time * r = 60.136 * t * 0.11

set them equal

            60.136 * .11 * t = 6.5 - 2.646 t

   9.261 t = 6.5

           t = 0.7019 seconds

final speed of ball = 6.5 - 2.646*0.7019 = 4.643 m/s

distance = avg speed * time = (1/2)(6.5+4.643)*0.7019 = 3.91 m

so finally

b) -2.646 m/s^2

  c) 60.136 rad/s^2

d) 0.7019 sec

e) 3.91 m

f) 4.643 m/s               

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