Question

A bowler releases a bowling ball with no spin, sending it sliding straight down the alley toward the pins. The ball continues to slide for some distance before its motion becomes rolling without slipping; what is the magnitude of this distance? Assume the ball maintains an essentially constant speed of 5.90 m/s and that the coefficient of kinetic friction for the polished alley is 0.105.

Answer 1

Answer:

Time, t = 0.197 s

Solution:

As per the question:

Constant speed of the ball, v = 5.90 m/s

Coefficient of Kinetic friction, $\mu_{k} = 0.105$

Now,

the ball will initially rotate with some angular velocity and then slide without rolling.

Friction force, f = $\mu_{k}N$

where

N = mg = normal reaction

Thus

f = $\mu_{k}mg$   (1)

Now, consider the ball to be a solid sphere,

Moment of inertia of the ball, I = $\frac{2}{5}mR^{2}$       (2)

where

R = radius of the ball

Torque is given by:

$\tau = I\alpha$          (3)

where

$\alpha = angular\ acceleration$

Thus from eqn (1), (2) and (3):

$\alpha = \frac{5\mu_{k}g}{2R}$

Now, the time taken is given by kinematic eqn:

$\omega' = \omega + \alpha t$

$\omega' = 0 +\frac{5\mu_{k}g}{2R} t$

$\omega' = \frac{v}{r}$

$\frac{v}{r} = \frac{5\mu_{k}g}{2R} t$

t = $\frac{2\times 5.90}{5\times 1.05\times 9.8} = 0.197\ s$