Question

A bowling ball with a circumference of 27 in. weighs 14.8 lb and has a radius of gyration of 3.43 in. If the ball is released with a velocity of 23 ft/sec but with no angular velocity as it touches the alley floor, compute the distance traveled by the ball before it begins to roll without slipping. The coefficient of friction between the ball and the floor is 0.19.

Answer 1

Answer:

Distance covered is 37.63 ft

Solution:

As per the question:

Circumference of the bowling ball, C = 27 in

Weight of the bowling ball, w = 14.8 lb

Radius of gyration, k = 3.43 in

Velocity of release of the ball, v' = 23 ft/s = 276 in/s

Coefficient of friction, $\mu = 0.19$

Now,

We know that the circumference of the circle is given by:

C = $2\pi R$

27 = $2\pi R$

$R = \frac{27}{2\pi } = 4.29\ in$

Also, in case of rolling, we know that:

Angular velocity, $\omega = \frac{v}{R}$

$v = \omega R$

Now, applying the conservation of angular momentum along the floor:

Initial angular momentum = Final angular momentum

$m\omega' = m\omega + I\omega$

Moment of Inertia, I = $\frac{2}{5}mR^{2} = mk^{2}$

$mv'R = mvR + mk^{2}\times \frac{v}{R}$

$v'R = vR + k^{2}\times \frac{v}{R}$

$276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}$

$1184.04 - 756.903 = 4.29v$

v = 99.56 in/s

Now,

Friction force, f = $\mu mg$

Also, acceleration of the ball can be computed as:

$\mu mg = - ma$

$a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}$

Now, the distance, d covered by the ball before rolling without slipping:

$v^{2} = v'^{2} + 2ad$

$99.56^{2} = 276^{2} + 2\times (- 73.357)d$

$99.56^{2} - 276^{2} = 2\times (- 73.357)d$

d = 451.65 in = 37.53 ft