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Q. Consider a rod of mass $M$ and length $L$ pivoted at its centre is 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

System of Particles and Rotational Motion

Solution:

Apply conservation of angular momentum about $C$.
$m v \frac{L}{2}=I \omega, \text { where } I=M\left(\frac{L}{2}\right)^{2}+\frac{M L^{2}}{12} $
$ \Rightarrow \omega=\frac{3 v}{2 L}$