Question

A uniform thin rod of mass $M$ and length $L$ is hinged by a frictionless pivot at its end $O$, as shown in the diagram. A bullet of mass $m$ moving horizontally with a velocity $v$, strikes the free end of the rod and gets embedded in it. The angular velocity of the system about $O$, just after the collision is,

• A. $\frac{mv}{L(M+m)}$
• B. $\frac{2mv}{L(M+2m)}$
• C. $\frac{3mv}{L(M+3m)}$
• D. $\frac{mv}{LM}$

Answer 1

The correct option is C $\frac{3mv}{L(M+3m)}$

Let $\omega$ be the angular velocity acquired by the system (rod + bullet) immediately after the collision.
Since no external torque acts, the angular momentum of the system is conserved, thus

$mvL=I\omega -\left(1\right)$

Where $I$ is the moment of inertia of the system about an axis passing throug $O$ and perpendicular to the rod.

$\therefore I=MOI$ of rod about $O+MOI$ of bullet stuck at its lower end about $O$

$=\frac{1}{3}M{L}^2+m{L}^2=\frac{1}{3}(M+3m)L^2-\left(2\right)$

Using Eq. $\left(1\right)$ in Eq. $\left(2\right)$, we have,

$mvL=\frac{1}{3}(M+3m)L^2\omega$

$\therefore \omega =\frac{3mv}{L(M+3m)}$

Hence, $\left(C\right)$ is the correct answer.