Question

part a
when the rotational speed of a rotating system doubles, its angular momentum
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remains unchanged
doubles
reduces to zero
quadruples
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Explanation:

The formula for angular momentum is $L = I\omega$, where $L$ is angular momentum, $I$ is the moment of inertia (assuming it remains constant in this case) and $\omega$ is the angular - speed. If $\omega$ doubles (new $\omega'=2\omega$) and $I$ is constant, then the new angular momentum $L'=I\omega' = I\times(2\omega)=2L$. So the angular momentum doubles.

Answer:

doubles