Question

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 $\frac{ML^{2}}{12}$. 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

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

Answer 1

Answer: (C)

$\frac{3v}{2L}$

Answer 2

Initial angular momentum of the system = Angular momentum of bullet before collision $= Mv\left( \frac{L}{2} \right)$ .....(i)

let the rod rotates with angular velocity $\omega.$

Final angular momentum of the system

$= \left( \frac{ML^{2}}{12} \right)\omega + M\left( \frac{L}{2} \right)^{2}\omega$ ......(ii)

By equation (i) and (ii) $Mv\frac{L}{2} = \left( \frac{ML^{2}}{12} + \frac{ML^{2}}{4} \right)\omega$