Moment of inertia of uniform rod of mass M and length L about an axis through its centre and perpendicular to its length is given by ML²/12 . Now consider one such rod pivoted at its centre, free to rotate in a vertical plane. The rod is at rest in the vertical position. A bullet of mass M moving horizontally at a speed v strikes and embedded in one end of the rod. The angular velocity of the rod just after the collision will be

The correct option is 3v / 2L.

To find the angular velocity of the rod-bullet system just after the collision, we can use the principle of conservation of angular momentum about the pivot point (the center of the rod). Since the rod is free to rotate in a vertical plane and the pivot is fixed, no external torque acts about the pivot during the very brief collision period.

Let:
- $M$ be the mass of the uniform rod, and $L$ be its length.
- $m=M$ be the mass of the bullet.
- $v$ be the horizontal velocity of the bullet before the collision.
- $\omega$ be the angular velocity of the system just after the collision.

Before the collision, the rod is at rest. The bullet is moving horizontally at speed $v$ and strikes one end of the rod. The perpendicular distance from the pivot (center of the rod) to the line of motion of the bullet is:
$r=\frac{L}{2}$

Thus, the initial angular momentum ($L_i$) of the system about the pivot is solely due to the bullet:
$L_i=M·v·\frac{L}{2}$

After the collision, the bullet becomes embedded in the end of the rod, and the entire system rotates together with an angular velocity $\omega$ about the pivot.
The moment of inertia of the rod about its center is:
$I_{\text{rod}}=\frac{M{L}^2}{12}$

The moment of inertia of the embedded bullet (treated as a point mass $M$ at a distance $\frac{L}{2}$ from the center) is:
$I_{\text{bullet}}=M{\left(\frac{L}{2}\right)}^2=\frac{M{L}^2}{4}$

The total moment of inertia ($I_{\text{total}}$) of the system after the collision is:
$I_{\text{total}}=I_{\text{rod}}+I_{\text{bullet}}=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}$

Finding a common denominator:
$I_{\text{total}}=\frac{M{L}^2+3M{L}^2}{12}=\frac{4M{L}^2}{12}=\frac{M{L}^2}{3}$

The final angular momentum ($L_f$) of the system about the pivot is:
$L_f=I_{\text{total}}·\omega =\frac{M{L}^2}{3}·\omega$

Applying the conservation of angular momentum ($L_i=L_f$):
$Mv\frac{L}{2}=\frac{M{L}^2}{3}\omega$

Dividing both sides by $ML$:
$\frac{v}{2}=\frac{L\omega }{3}$

Solving for $\omega$:
$\omega =\frac{3v}{2L}$