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A thin rod of mass $$0.9\,\text{kg}$$ and length $$1\,\text{m}$$ is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass $$0.1\,\text{kg}$$ 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 $$\text{rad s}^{-1}$$) of the rod immediately after the collision will be...........
Correct Answer: 20
$$L_i = m v L$$
$$I = I_{\text{rod}} + I_{\text{particle}}$$
$$I = \frac{1}{3}ML^2 + mL^2$$
$$I = \left(\frac{0.9}{3} + 0.1\right) \times 1^2 = 0.3 + 0.1 = 0.4\text{ kg m}^2$$
$$L_f = I\omega$$
$$m v L = I\omega \implies 0.1 \times 80 \times 1 = 0.4 \times \omega \implies 8 = 0.4\omega \implies \omega = 20\text{ rad s}^{-1}$$
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