A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s

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A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s

and initial angular speed ω0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.35. The kinetic frictional force f with arrowk 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 ω has increased enough, the ball stops sliding and then rolls smoothly.(a) What then isvinterms ofω? (b) During the sliding, what are the ball’s translational and angularacceleration? (c) How far does the ball slide?

Physics

2 answers:

ankoles [38] · Answer 1 · 2022-07-21T05:56:03+03:00

Answer:

Explanation:

Radius of the ball is $R=11cm=0.11m$

Initial speed of the ball is $v_{com0}=6.0m/s$

Initial angular speed of the ball is $\omega = 0$

Coefficient of kinetic friction between the ball and the lane

is $\mu =0.35$

Due to the presence of frictional force, ball moves with

decreasing velocity.

(a)

velocity $v_{com0}$ in terms of $\omega$ is

$V_{com0} = -R\omega\\\\=-(0.11m)\omega\\\\= (-0.11\omega)m/s$

(b)

Ball's linear acceleration is given by

$a=-\mu g\\\\=-(0.35) (9.8 m/s^2)\\\\= -3.43m/s^2$

(c)

During sliding, ball's angular acceleration is calculated as

$\alpha=-\frac{\tau}{I}\\\\-\frac{\mu mgR}{(\frac{2}{5}mR^2)}\\\\-\frac{2}{5}\frac{\mu g}{R}\\\\-\frac{2}{5}\frac{(0.35)(9.8)}{0.11}\\\\-77.95rad/s^2$

(d)

The time for which the ball slides is calculated from the

equation of motion is

$V_{cm}= V_{cm0} + at\\\\V_{cm} = V_{cm0} + (-\mu g )t\\\\-0.11\omega=6.0m/s -(3.43m/s^2 )t\\\\-(0.11) (\alpha t) =6.0 m/s - (3.43 m/s^2)\\\\- (0.11)(-77.95 rad/s^2)t = 6.0m/s - (3.43 m/s^2 )t\\\\8.5745t + 3.43t= 6.0\\\\12.0045t = 6.0\\\\t= 0.4998s$