A spherical bowling ball with mass m- 3.6 kg and radius R 0.118 m is thrown down

Question

A spherical bowling ball with mass m- 3.6 kg and radius R 0.118 m is thrown down the lane with an initial speed of v 8.5 m/s. The coefficient of kinetic friction between the sliding ball and the ground is 0.26. 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/S2Submit rad/s Submit 2) What is magnitude of the linear acceleration of the bowling bal as it slides down the lane? m/s Submit 3) How long does it take the bowling ball to begin rolling without slipping? S Submit

Explanation / Answer

1.Moment of inertia

I=(2/5)mR2

angular acceleration

alpha =T/I =ukmgR/(2/5)mR2 = 5ukg/2R

alpha =5*0.26*9.8/2*0.118 =54 rad/s2

2.From newtons second law

a=ukg =0.26*9.8 =2.55 m/s2

3.

From

V=Vo+at

here acceleration is negative ,since it is falling down.

angular velocity

W=alpha*t =(5ukg/2R)t

V=WR =(5ukg/2)t

=>(5ukg/2)t =Vo-ukgt

t=2Vo/7ukg =2*8.5/7*0.26*9.8

t=0.953 s

4.From

S=Vot-(1/2)at2

S =8.5*0.953-(1/2)*2,55*0.9532

S=6.94 m

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