A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the weights is to double the rotational inertia of the system. As he is rotating, the man opens his hands and drops the two weights. They fall outside the turntable. What happens to the man's angular velocity?(adsbygoogle = window.adsbygoogle || []).push({});

The answer is that his angular velocity should stay relatively constant. I am confused to why this occurs. This is my reasoning for this:

As the weights are in free fall, they still are roughly same distance from the axis of rotation. Thus I is constant, and hence the nothing in the system changes. Angular velocity is still constant.

However, once the weights fall on the ground, the external friction force exerts a net torque on the weights, stopping them. However, since the system is not closed, the angular momentum need not be conserved, so the man's angular velocity remains unchanged.

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# Angular Momentum Scenario

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