A thin rod of mass 0 . 9kg and length 1m is suspended, at rest,...

Given,

Mass of the rod, $M = 0 . 9kg$

Mass of particle, $m = 0 . 1kg$

Length of the rod, $l = 1m$

Velocity of particle, $v = 80m{s}^{- 1}$

No external torque acting on the system of rod and particle, so angular momentum of system about the suspension is conserved, i.e.; ${L}_{i} = {L}_{f}$

$⇒ mvl = I\omega . . . ( 1 )$

Where, $I$ is the momentum of inertia of rod and particle system, $\omega$ is the angular velocity of system just after the collision.

$⇒ mvl = \left(\dfrac{M{l}^{2}}{3} + m{l}^{2}\right)\omega . . . ( 2 )$

Substitute the values in equation (2)

$⇒ 0 . 1 \times 80 \times 1 = \left(\dfrac{0 . 9 \times ( 1{)}^{2}}{3} + 0 . 1 \times ( 1{)}^{2}\right) \times \omega$

$\omega = 20rad{s}^{- 1}$