Question 705: Centre of Mass and Rotational Motion

Question

Consider a rod of mass $$M$$ and length $$L$$ 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 horizontal at a speed $$v$$ strikes and gets embedded in one end of the rod. The angular velocity of the rod just after the collision will be

Accepted Answer

Apply conservation of angular momentum about $$C$$,
$$\begin{aligned}
m v \frac{L}{2} & =l \omega \
	ext { where, } I & =M\left(\frac{L}{2}ight)^2+\frac{M L^2}{12} \
\Rightarrow \quad \omega & =\frac{3 v}{2 L}
\end{aligned}$$

Answer

$$v / L$$

Answer

$$2 v / 2 L$$

Answer

$$3 v / L$$

Answer

$$6 v / L$$

Centre of Mass and Rotational Motion

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