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A uniform rod $A B$ of mass $2 \mathrm{~kg}$ and Length $30 \mathrm{~cm}$ at rest on a smooth horizontal surface. An impulse of force $0.2\ \mathrm{Ns}$ is applied to end $B.$ The time taken by the rod to turn through at right angles will be $\frac{\pi}{\mathrm{x}}\ \mathrm{s}$, where X=____
$4$
$5$
$6$
$7$
Solution
Impulse $\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}$
$\mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s}$
Angular impuls ( $\overrightarrow{\mathrm{M}})$
$\overrightarrow{\mathrm{M}}_{\mathrm{c}}=\int \tau \mathrm{dt}$
$=\int \mathrm{F} \frac{\mathrm{L}}{2} \mathrm{dt}$
$=\frac{\mathrm{L}}{2} \int \mathrm{Fdt}=\frac{\mathrm{L}}{2} \times \mathrm{J}v$
$=\frac{0.3}{2} \times 0.2$
$=0.03$
$I_{c m}=\frac{\mathrm{LL}^2}{12}=\frac{2 \times(0.3)^2}{12}=\frac{0.09}{6}$
$\mathrm{M}=\mathrm{I}_{\mathrm{cm}}\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\right)$
$0.03=\frac{0.09}{6}\left(\omega_{\mathrm{f}}\right)$
$\omega_{\mathrm{f}}=2 \mathrm{rad} / \mathrm{s}$
$\theta=\omega \mathrm{t}$
$\mathrm{t}=\frac{\theta}{\omega}=\frac{\pi}{2 \times 2}=\frac{\pi}{4} \mathrm{sec} .$
$\mathrm{X}=4$
