A horizontal rigid bar of length $2 R$ is pivoted at its center and is free to rotate in a horizontal plane around a vertical axis passing through the pivot. A point mass $M$ is fixed to one end of the bar, while a container is attached to the opposite end. A separate point mass of $M / 3$ approaches perpendicularly to the bar with velocity $v$, collides with the container, and becomes embedded in it. As a result, the entire system begins to rotate about the vertical axis. The masses of the bar and container are considered negligible.

Write down an expression, in terms of $M$, $v$ and $R$, for the angular momentum of the system about the vertical axis just before the collision.

Just after the collision the system begins to rotate about the vertical axis with angular velocity $ω$. Show that the angular momentum of the system is equal to $4/3 MR^2ω$.

Hence, show that $ω = v/(4R)$.

Determine in terms of $M$ and $v$ the energy lost during the collision.

A torque of $0.010 \, Nm$ brings the system to rest after a number of revolutions. For this case $R = 0.50 \, m$, $M = 0.70 \, kg$ and $v = 2.1 \, m \, s^{-1}$. Show that the angular deceleration of the system is $0.043 \, \mathrm{rad \, s}^{-2}$.

Calculate the number of revolutions made by the system before it comes to rest.