A 4500 kg car accelerates from rest to 45.0
1 answer:
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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,
Now,
We know that the circumference of the circle is given by:
C =
27 =
Also, in case of rolling, we know that:
Angular velocity,
Now, applying the conservation of angular momentum along the floor:
Initial angular momentum = Final angular momentum
Moment of Inertia, I =
v = 99.56 in/s
Now,
Friction force, f =
Also, acceleration of the ball can be computed as:
Now, the distance, d covered by the ball before rolling without slipping:
d = 451.65 in = 37.53 ft