Answer:
Explanation:
Radius of the ball is [tex]R=11cm=0.11m[/tex]
Initial speed of the ball is [tex]v_{com0}=6.0m/s[/tex]
Initial angular speed of the ball is [tex]\omega = 0[/tex]
Coefficient of kinetic friction between the ball and the lane
is [tex]\mu =0.35[/tex]
Due to the presence of frictional force, ball moves with
decreasing velocity.
(a)
velocity [tex]v_{com0}[/tex] in terms of [tex]\omega[/tex] is
[tex]V_{com0} = -R\omega\\\\=-(0.11m)\omega\\\\= (-0.11\omega)m/s[/tex]
(b)
Ball's linear acceleration is given by
[tex]a=-\mu g\\\\=-(0.35) (9.8 m/s^2)\\\\= -3.43m/s^2[/tex]
(c)
During sliding, ball's angular acceleration is calculated as
[tex]\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[/tex]
(d)
The time for which the ball slides is calculated from the
equation of motion is
[tex]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[/tex]
(e)
Distance traveled by the ball is
[tex]X= V_{com,0}+ \frac{1}{2}at^2\\\\= (6.0m/s)(0.4998 s)+ 0.5(-3.43m/s^2) (0.4998 s)^2\\\\=2.57m[/tex]
(for)
The speed of the ball when smooth rolling begins is
[tex]V_{cm} = V_{com, 0}+ at\\\\=6.0 m/s +(-3.43m/s^2 )(0.4998 s)\\\\= 4.29m/s[/tex]
