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

The magnitude of the distance the ball slides before it starts rolling without slipping can be calculated using the formula: distance = (initial velocity²) / (2 × kinetic friction)

Step-by-step explanation:

To determine the magnitude of the distance the ball slides before it starts rolling without slipping, we need to consider the friction force acting on the ball. The friction force can be found using the equation:

friction force = coefficient of kinetic friction × normal force

The normal force is equal to the weight of the ball, which can be calculated using:

normal force = mass × acceleration due to gravity

Then, we can use the formula for kinetic friction to find the magnitude of the distance:

distance = (initial velocity²) / (2 × kinetic friction)

Substituting the given values, we find:

distance = (5.90 m/s)² / (2 × 0.105).

Answer 2

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 =
`(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 = (5\mu_(k)g)/(2R)`

Now, the time taken is given by kinematic eqn:

`\omega' = \omega + \alpha t`

`\omega' = 0 +(5\mu_(k)g)/(2R) t`

`\omega' = (v)/(r)`

`(v)/(r) = (5\mu_(k)g)/(2R) t`

t =
`(2* 5.90)/(5* 1.05* 9.8) = 0.197\ s`