A spherical bowling ball with mass m = 3.2 kg and radius R = 0.106 m is thrown d

Question

A spherical bowling ball with mass m = 3.2 kg and radius R = 0.106 m is thrown down the lane with an initial speed of v = 8.6 m/s. The coefficient of kinetic friction between the sliding ball and the ground is mu = 0.34. Once the ball begins to roll without slipping it moves with a constant velocity down the lane. What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane? rad/s^2 What is magnitude of the linear acceleration of the bowling ball as it slides down the lane? m/s^2 How long does it take the bowling ball to begin rolling without slipping? s How far does the bowling ball slide before it begins to roll without slipping? m What is the magnitude of the final velocity? m/s After the bowling ball begins to roll without slipping. compare the rotational and translational kinetic energy of the bowling ball: KE_rot KE_tran

Explanation / Answer

1.) Ialpha = umgR

alpha = umgR/(2/5)mR^2

= (5ug)/(2R) =2.5*0.34*9.8/0.106= 78.5849 rad/s^2

2.) ug = 3.332 m/s^2

3) 3. When is v(t) = w(t) R (pure rolling)?
v0 - g t = w0 R + 5 g t / 2
(7 g /2) t = v0 + w0 R
t =( 2 v0 + 2 w0 R)/( 7 g )

Assuming w0 = 0 (no spin is given in the throw) , this is

t = 2 v0 / (7 g ) = 2*8.6/(7*3.332) = 0.7374 s

4.) s(t) = v(0) t + 1/2 a t^2
= 2 v0^2 / (7 g ) - 1/2 * ( g) ( 4 v0^2 / (7 g )^2
= 2/7 (v0^2 / ( g) ) - 2/49 (v0^2 / ( g) )
= 12 v0^2/( 49 g)
= 12*8.6^2/(49*3.332)

= 5.436 m

5.) vf = u - at = 8.6 - 0.34*9.8*0.7374 = 6.143 m/s

6.) KErot<KEtran

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