Questions
A man stands on a rotating platform with his arms stretched holding a 5 kg weight in each hand. The angular speed of the platform is 1.2 rev . The moment of inertia of the man together with the platform may be taken to be constant and equal to 6 kg . If the man brings his
arms close to his chest with the distance of each weight from the axis changing from 100 cm to 20 cm. The new angular speed of the platform is
detailed solution
Correct option is B
Initial moment of inertia, I1 = 6 +[2×5×(1)2] = 16 kg m2Initial angular velocity, ω1 = 1.2 rev s-1Initial angular momentum is L1 = I1ω1Final moment of inertia, I2 = 6+[2×5×(0.2)2 ]= 6.4 kg m2Final angular speed = ω2Final angular momentum is L2 = I2ω2According to law of conservation of angular momentum, L1 = L2 or I1ω1 = I2ω2ω2 = I1ω1I2 = (16 kg m2)(1.2 rev s-1)(6.4 kg m2) = 3 rev s-1Talk to our academic expert!
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