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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
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Verified Answer
The correct answer is:
$\frac{\pi}{12} \frac{m \ell}{P}$
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,
$\omega=\frac{\Delta Q}{\Delta t} ; \text { For } \Delta \theta=\frac{\pi}{2}$
$\Delta t=\frac{\Delta \theta}{\omega}=\frac{\pi}{2} \cdot \frac{l m}{6 p}=\frac{\pi m l}{12 p}$

$I=P \frac{l}{2}=I \omega$
$I=$ 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,
$\omega=\frac{\Delta Q}{\Delta t} ; \text { For } \Delta \theta=\frac{\pi}{2}$
$\Delta t=\frac{\Delta \theta}{\omega}=\frac{\pi}{2} \cdot \frac{l m}{6 p}=\frac{\pi m l}{12 p}$
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