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(B) $1$. The system consists of the triangle and the two beads. The only external forces acting on the system are gravity and the normal force from th

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Total angular momentum and total energy

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(B) $1$. The system consists of the triangle and the two beads. The only external forces acting on the system are gravity and the normal force from the pivot at $A$. Since the pivot force acts at the axis of rotation,its torque is zero. Gravity also exerts no torque about the vertical axis $AO$. Therefore,the net external torque on the system about the axis of rotation is zero,which implies that the total angular momentum $L$ is conserved.$2$. Since there is no friction and the only forces doing work are gravity (a conservative force),the total mechanical energy of the system is conserved.$3$. As the beads slide down,their distance from the axis of rotation increases,which increases the moment of inertia $I$ of the system. Since $L = I\omega$ is constant and $I$ increases,the angular velocity $\omega$ must decrease. Thus,neither $\omega$ nor $I$ are conserved.

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