A uniform rod A B of mass 2 mathrm{~kg} and L | vedclass

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Question

Impulse $\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}$

$\mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s}$

Angular impuls ( $\overrightarrow{\mat

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$4$

Vedclass · Answer

Impulse $\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}$

$\mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s}$

Angular impuls ( $\overrightarrow{\mathrm{M}})$

$\overrightarrow{\mathrm{M}}_{\mathrm{c}}=\int 	au \mathrm{dt}$

$=\int \mathrm{F} \frac{\mathrm{L}}{2} \mathrm{dt}$

$=\frac{\mathrm{L}}{2} \int \mathrm{Fdt}=\frac{\mathrm{L}}{2} 	imes \mathrm{J}v$

$=\frac{0.3}{2} 	imes 0.2$

$=0.03$

$I_{c m}=\frac{\mathrm{LL}^2}{12}=\frac{2 	imes(0.3)^2}{12}=\frac{0.09}{6}$

$\mathrm{M}=\mathrm{I}_{\mathrm{cm}}\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}ight)$

$0.03=\frac{0.09}{6}\left(\omega_{\mathrm{f}}ight)$

$\omega_{\mathrm{f}}=2 \mathrm{rad} / \mathrm{s}$

$	heta=\omega \mathrm{t}$

$\mathrm{t}=\frac{	heta}{\omega}=\frac{\pi}{2 	imes 2}=\frac{\pi}{4} \mathrm{sec} .$

$\mathrm{X}=4$

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=____

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