An uniform rod $\mathrm{AB}$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An…

An uniform rod $\mathrm{AB}$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $P$ applied to the end $B$. The time taken by the rod to turn through a right angle is
  1. $\frac{\pi}{12} \frac{m \ell}{P}$
  2. $2 \pi \frac{m \ell}{P}$
  3. $2 \frac{\pi P}{m \ell}$
  4. $\frac{\pi P}{m \ell}$

Solution

Concept: $($ angular impulse $)=($ change in angular momentum $)$ $I=P \frac{l}{2}=I \omega$ $I=$ moment of inertia of the rod about $\mathrm{O}$. $\begin{aligned} & I=\left(\frac{m l^2}{12}\right) \\ & \therefore P \frac{l}{2}=\frac{m l^2}{12} \omega \\ & \Rightarrow \omega=\frac{6 p}{m l} \end{aligned}$ We know, $\omega=\frac{\Delta Q}{\Delta t} ; \text { For } \Delta \theta=\frac{\pi}{2}$ $\Delta t=\frac{\Delta \theta}{\omega}=\frac{\pi}{2} \cdot \frac{l m}{6 p}=\frac{\pi m l}{12 p}$ ~

Asked in: MHT CET 2022 (08 Aug Shift 2)

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