Question

a merry-go-round spins with negligible friction. a child moves from the center out to the rim of the merry-go-round platform. let the system be the merry-go-round + child. which of these quantities change: angular momentum of the system, angular velocity of the system, rotational kinetic energy of the system? multiple choice both angular momentum, angular velocity, and rotational kinetic energy of the system remain the same. the angular momentum of the system remains the same. both the angular velocity and rotational kinetic energy decrease. none of the quantities remain the same the angular momentum of the system remains the same. both the angular velocity and rotational kinetic energy increase.

Explanation:

1. Angular Momentum: The system (merry - go - round + child) has negligible friction, so there is no external torque. By the law of conservation of angular momentum ($L = I\omega$, and $\tau_{ext}=\frac{dL}{dt} = 0$), angular momentum $L$ of the system remains constant.
2. Moment of Inertia ($I$): When the child moves from the center to the rim, the moment of inertia of the child (and thus the system, since $I_{system}=I_{merry - go - round}+I_{child}$) increases (as $I = mr^{2}$ and $r$ increases for the child).
3. Angular Velocity ($\omega$): From $L = I\omega$, since $L$ is constant and $I$ increases, $\omega=\frac{L}{I}$ must decrease.
4. Rotational Kinetic Energy ($K_{rot}$): The formula for rotational kinetic energy is $K_{rot}=\frac{1}{2}I\omega^{2}=\frac{L^{2}}{2I}$ (since $L = I\omega\Rightarrow\omega=\frac{L}{I}$). Since $L$ is constant and $I$ increases, $K_{rot}=\frac{L^{2}}{2I}$ decreases.

So the angular momentum remains the same, and both angular velocity and rotational kinetic energy decrease.

Answer:

The angular momentum of the system remains the same. Both the angular velocity and rotational kinetic energy decrease.