A spherical bowling ball with mass m = 4.3 kg and radius R = 0.102 m is thrown d

Question

A spherical bowling ball with mass m = 4.3 kg and radius R = 0.102 m is thrown down the lane with an initial speed of v = 8.3 m/s. The coefficient of kinetic friction between the sliding ball and the ground is = 0.3. 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?

rad/s2

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

m/s2

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

s

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

m

5)What is the magnitude of the final velocity?

m/s

6)After the bowling ball begins to roll without slipping, compare the rotational and translational kinetic energy of the bowling ball:

KErot < KEtran

KErot = KEtran

KErot > KEtran

could you please help :)

Explanation / Answer

Given mass of the sperical ball m = 4.3 kg , radius R = 0.102 m,
    initial velocity v0= 8.3m/s, coefficient of kinetic friction bwteen ball and ground = 0.3

and ball begins rolling without slipping

here frictional force is equal to the torque on the ball
    k mg R = I*alpha             alpha --- angular acceleration, I-- moment of inertia of ball = 2/5 m R^2
    

1)   alpha = k mg R/(2/5 mR^2)
              = 5/2   k g / R
              = (5/2)(0.3*9.8/ (0.102))
         alpha= 72.05 rad/s^2

2) linear acceleration of ball

   ma = k mg
          a = k*g
            = 0.3*9.8 = 2.94 m/s2
3)

   time taken to the ball to begin rolling without slipping is

   when the linear speed of the ball decreased then angular speed (rolling) strarts to increase

    equating them
                      v= vo - at, w = w0 + alpha t equations of motion
initial angular speed is zero so w = alph*t , and using the condition for ROLLING WITH OUT SLIPPING v = Rw

                      Rw = v0 -a t
                      R*alpha*t = v0 - at
                      t(R*alpha +a) = v0
                      t = v0/(R*alpha +a)
                        = 8.3/(0.102*72.05 +2.94)
                      t = 0.8066 s
4) distance we know time period and initial speed is zero , from eauations of motion

       x= ut + 1/2 at^2
                 = 0 + 1/2 a t^2
                 = 1/2*2.94*(0.8066)^2
                x= 0.95638 m
5) magintude of thefinal velocity
       v = v0 - at
                v = 8.3 - 2.94*0.8066
                v= 5.9285 m/s
6)   Rotational kinetic energy Kr = 1/2 I*w^2= 1/2 (2/5 *m*R^2)w^2
                                 = 1/5*mR^2*w^2
                                 = 1/5*4.3*0.102^2*(72.05*0.8066)^2
                                 = 30.219 J
     linear kinetic energy K.E = 1/2 m*v^2
                               = 1/2 *4.3*5.9285^2
                               = 75.566 J
           

        KErot < KE tran

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