An 8-kg, 10-cm radius bowling ball is released with a speed of 5 m/s from a heig
ID: 1775823 • Letter: A
Question
An 8-kg, 10-cm radius bowling ball is released with a speed of 5 m/s from a height of 0.5 m above the floor. When released, the ball is not rotating. Later, the ball is rolling on the floor (h = 0) without slipping at a speed of 4 m/s.
What is the initial kinetic energy of the bowling ball?
What is the initial potential energy of the bowling ball?
What is the rotational inertia of the bowling ball?
What is the magnitude of the final angular velocity of the bowling ball?
What is the final kinetic energy of the bowling ball?
How much work did friction do on the bowling ball?
Explanation / Answer
Ki = m v^2 / 2 = 8 x 5^2 /2 = 100 J ....Ans
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PEi = m g h = 8 x 9.8 x 0.5 = 39.2 J
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I = 2 m r^2 / 5 = 0.032 kg m^2
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wf = v / r = 4 / 0.10 = 40 rad/s
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KEf = (8 x 4^2 / 2) + (0.032 x 40^2 / 2)
=89.6 J
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Work done by friction = (KEf + PEf) - (KEi + PEi)
= (89.6 + 0) - ( 100 + 39.2)
= - 49.6 J ........Ans
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