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 $\frac{M{L}^2}{12}$. 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. $\frac{6v}{L}$

Answer 1

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

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 $=\left(\frac{M{L}^2}{12}\right)\omega +M{\left(\frac{L}{2}\right)}^2\omega$ ....(ii)
By equation (i) and (ii) $Mv\frac{L}{2}=(\frac{M{L}^2}{12}+\frac{M{L}^2}{4})\omega or\omega =\frac{3v}{2L}$ ​​​​​​​