InterviewSolution
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InterviewSolution
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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 s^-1`. The moment of inertia of the man together with the platform may be taken to be constant and equal to `6 kg m^2`. If the man brings his arms close to his chest with the distance `n` each weight from the axis changing from `100 cm` to`20 cm`. The new angular speed of the platform is.A. `2 rev s^-1`B. `3 rev s^-1`C. `5 rev s^-1`D. `6 rev s^-1` |
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Answer» Correct Answer - B (b) Initial moment of inertia, `I_1 = 6 + 2 xx 5 xx (1)^2 = 16 kg m^2` Initial angular velocity, `omega_1 = 1.2 rev s^-1` Initial angular momentum is `L_2 = I_2 omega_2` According to law of conservation of angular momentum, `L_1 = L_2` or `I_1 omega_1 = I_2 omega_2` `omega_2 = (I_1 omega_1)/(I_1) = ((16 kg m^2)(1.2 rev s^-1))/((6.4 kg m^2)) = 3 rev s^-1`. |
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