A thin rod of mass 0.9kg and length


A thin rod of mass $0.9 \mathrm{~kg}$ and length $1 \mathrm{~m}$ is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move $0.1 \mathrm{~kg}$ moving in a straight line with velocity $80 \mathrm{~m} / \mathrm{s}$ hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in $\mathrm{rad} / \mathrm{s}$ ) of the rod immediately after the collision will be



$\mathrm{mvL}=\mathrm{I} \omega$

$\mathrm{mvL}=\left(\frac{\mathrm{ML}^{2}}{3}+\mathrm{mL}^{2}\right) \omega$

$0.1 \times 80 \times 1=\left(\frac{0.9 \times 1^{2}}{3}+0.1 \times 1^{2}\right) \omega$

$8=\left(\frac{3}{10}+\frac{1}{10}\right) \omega$

$8=\frac{4}{10} \omega$

$\omega=20 \mathrm{rad} \frac{\mathrm{rad}}{\mathrm{sec}}$


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