Question

learn more
remarks although the angular momentum of the system is conserved, mechanical energy is not conserved because the student does work on the system.
question if the student suddenly releases the weights, how does his own angular momentum, rotational kinetic energy, and angular speed change? (select all that apply.)
his rotational kinetic energy decreases.
his rotational angular speed remains the same.
his angular momentum decreases.
his rotational kinetic energy increases.
his angular momentum increases.
his rotational kinetic energy remains the same.
his rotational angular speed decreases.
his angular momentum remains the same.
his rotational angular speed increases.
consider what angular momentum if any is transferred to the barbells by letting go of them. how would that change the angular momentum of the student himself? based on that, how would it affect the students angular speed and kinetic energy?

Explanation:

When the student releases the weights, no external torque acts on the student - system (since the only forces are internal when considering the student and the weights as a system). By the law of conservation of angular momentum \(L = I\omega\) (where \(I\) is the moment of inertia and \(\omega\) is the angular speed), the angular momentum \(L\) of the student (as part of the system) remains the same because \(L_{total}=L_{student}+L_{weights}\) and when the weights are released, the torque on the student is zero.

The moment of inertia of the student \(I_{student}\) does not change when the weights are released (the student's mass distribution remains the same). Since \(L = I\omega\) and \(L\) is constant for the student (as no external torque), if the moment of inertia of the combined system (student + weights) was previously \(I_{total}=I_{student} + I_{weights}\), after releasing the weights, for the student \(L = I_{student}\omega\). But when the weights are released, the rotational kinetic energy \(K=\frac{1}{2}I\omega^{2}\). The initial rotational kinetic energy of the system (student + weights) was \(K_{i}=\frac{1}{2}(I_{student}+I_{weights})\omega_{i}^{2}\). After releasing the weights, for the student \(K_{f}=\frac{1}{2}I_{student}\omega_{f}^{2}\). Since \(L=(I_{student}+I_{weights})\omega_{i}=I_{student}\omega_{f}\), \(\omega_{f}<\omega_{i}\) (because \(I_{student}I_{student}\), \(\omega_{2}<\omega_{1}\). And \(K=\frac{1}{2}I\omega^{2}\), with \(I\) (student's \(I\)) constant and \(\omega\) decreasing, \(K\) (student's rotational kinetic energy) decreases.

Answer:

His rotational kinetic energy decreases.
His rotational angular speed decreases.
His angular momentum remains the same.