A man stands on the center of a platform that is rotating on frictionless bearings at a speed of 1.00 rad/s.Originally his arms are outstretched and he holds a 4.54-kg mass in each hand.He then pulls the weights in toward his body.Assume the moment of inertia of the man,including his arms,to remain constant at 5.42 kg · m².If the original distance of the weights from the axis is 0.914 m and their final distance is 0.305 m,the final angular velocity is A)1.14 rad/s B)1.27 rad/s C)1.58 rad/s D)2.08 rad/s E)7.70 rad/s

Question

Quizplus · Accepted Answer

The conservation of angular momentum states that the initial angular momentum equals the final angular momentum. Initially, the angular momentum is the sum of the man's moment of inertia and the weights' moment of inertia times the angular velocity. When the weights are pulled in, their moment of inertia decreases, increasing the angular velocity. Calculating gives a final angular velocity of 2.08 rad/s.

A Man Stands on the Center of a Platform That