Question

Moment of inertia of a uniform rod of mass M and length L about and axis through its centre and perpendicular to its length is given by $\frac{M{L}^2}{12}$. Now, consider one such rod pivoted at its centre free to rotate in vertical plane. The rod is at rest in vertical position. A bullet of mass M moving horizontally a speed $v$ strikes and is embedded in one end of the rod. The angular velocity of the rod just after collision will be

• A. $\frac{v}{L}$
• B. $\frac{2v}{L}$
• C. $\frac{3v}{2L}$
• D. $\frac{6v}{L}$

Answer 1

The correct option is C $\frac{3v}{2L}$

Angular Momentum Conservation will take place.
Initial Angular Momentum $=Mv\frac{L}{2}$
Final Moment of Inertia, $I=\frac{M{L}^2}{12}+M(\frac{L}{2}{)}^2=\frac{M{L}^2}{3}$
Final Angular Momentum, $L=I\omega =\frac{M{L}^2}{3}\omega$
$\therefore Mv\frac{L}{2}=\frac{M{L}^2}{3}\omega \omega =\frac{3v}{2L}$