Physics - Rotation Question with Solution | TestHub

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