A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown d

Question

A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/s. The coefficient of kinetic friction between the sliding ball and the ground is = 0.28. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.

1) What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?

2) What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?

3) How long does it take the bowling ball to begin rolling without slipping?

4) How far does the bowling ball slide before it begins to roll without slipping?

5) What is the magnitude of the final velocity?

Explanation / Answer

1.)1/2 IW^2 is rotational KE

W is angular velocity, I is Moment of Inertia of ball = (2/5)*MR^2

Let u = kinetic friction coefficient = 0.28

1) Fr R = I alpha

alpha = u mg R / (2/5mR^2) = 5 u g / (2R)

Angular acceleration can be calculated to be (5ug)/(2R) = 5*0.28*9.8/(2*0.101) = 67.92 rad/s2

2.) Linear acceleration is given by Fr = m a

a = u m g / m = ug = 0.28*9.8 = 2.744 m/s^2

3) Time Taken is Vo / (alpha R + a ) = (2Vo)/(7ug) = (2*8.7)/(7*0.28*9.8) = 0.905 s

4.) Distance travelled is given by (12Vo^2)/(49ug) = 12*(8.7^2)/(49*0.28*9.8) = 6.75 m

5.) Final Velocity is 5/7*V0 = 6.214 m/s

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