Rod and a Particle

Moment of Inertia, I for a uniform rod = $\frac{ML^{2}}{12}$ , where M is the mass of the rod and L is its overall length.

Let m be the mass of the particle moving at speed v.

By conservation of linear momentum, mv = MV; where, V is the velocity of the rod after collision w.r.t. the centre of the rod.

$\implies$ V = $\frac{mv}{M}$

Now we can solve this via the conservation of angular momentum, $\alpha$ as net torque is zero.

Before collision:

$\alpha$ = m x v x 2 = 2mv

= 2 x 2 x 6 = 24 $kg m^{2}s^{-1}$ ... (1)

After collision (the particle comes to rest):

$\alpha$ = $I\omega$ + M. R x V ; R is the position vector w.r.t. the centre of the rod. We observe that R || V and hence, the cross product of R and V is zero

= $I\omega$

= $\frac{ML^{2}}{12}$ $\omega$

= $\frac{160}{3}$ $\omega$ ... (2)

Equating (1) and (2), we get:

$\omega$ = $\frac{72}{160}$ = 0.45 $\frac{rad}{s}$