Question

A homogeneous (uniform) rod XY of length L and mass M is pivoted at the centre C such that it can rotate freely in a vertical plane. Initially, the rod is horizontal. A blob of wax of the same mass M as that of the rod falls vertically with speed V and sticks to the rod midway between points C and Y. As a result, the rod rotates with angular speed ω. What will be the angular speed in terms of V and L?

Answer 1

The initial angular momentum of the rod is zero.

The initial angular momentum of the falling blob of wax about the point C is (in magnitude)

= mass × speed × perpendicular distance between its direction of motion and point C = MV.\(\frac L4\)

The total initial angular momentum of the rod and blob of wax = \(\frac{MVL}4\)....(1)

After the blob of wax sticks to the rod, and the system rotates with an angular speed ω about the horizontal axis through point C perpendicular to the plane of the figure, the total final angular momentum of the system about this axis

From Eqs (1) and (2), by the principle of conservation of angular momentum.

This gives the required angular speed.