A thin rod of mass $0.9$ kg and length $1$ m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass $0.1$ kg moving in a straight line with velocity $80\ \text{m s}^{-1}$ hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad s$^{-1}$) of the rod immediately after the collision will be ______.
Numerical value type. Enter your answer.
Answer: 20
Angular momentum about the pivot is conserved during the collision (the pivot force has no torque about it).
Before: $L = mvl = 0.1\times80\times1 = 8\ \text{kg m}^2\text{/s}$.
After: $I = \dfrac{Ml^2}{3} + ml^2 = 0.3 + 0.1 = 0.4\ \text{kg m}^2$.
$$\omega = \frac{8}{0.4} = 20\ \text{rad/s}$$
Solution by Sreeraj P, M.Sc Physics