Respuesta :
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, [tex]\mu = 0.19[/tex]
Now,
We know that the circumference of the circle is given by:
C = [tex]2\pi R[/tex]
27 = [tex]2\pi R[/tex]
[tex]R = \frac{27}{2\pi } = 4.29\ in[/tex]
Also, in case of rolling, we know that:
Angular velocity, [tex]\omega = \frac{v}{R}[/tex]
[tex]v = \omega R[/tex]
Now, applying the conservation of angular momentum along the floor:
Initial angular momentum = Final angular momentum
[tex]m\omega' = m\omega + I\omega[/tex]
Moment of Inertia, I = [tex]\frac{2}{5}mR^{2} = mk^{2}[/tex]
[tex]mv'R = mvR + mk^{2}\times \frac{v}{R}[/tex]
[tex]v'R = vR + k^{2}\times \frac{v}{R}[/tex]
[tex]276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}[/tex]
[tex]1184.04 - 756.903 = 4.29v[/tex]
v = 99.56 in/s
Now,
Friction force, f = [tex]\mu mg[/tex]
Also, acceleration of the ball can be computed as:
[tex]\mu mg = - ma[/tex]
[tex]a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}[/tex]
Now, the distance, d covered by the ball before rolling without slipping:
[tex]v^{2} = v'^{2} + 2ad[/tex]
[tex]99.56^{2} = 276^{2} + 2\times (- 73.357)d[/tex]
[tex]99.56^{2} - 276^{2} = 2\times (- 73.357)d[/tex]
d = 451.65 in = 37.53 ft
