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A thin uniform rod $A B$ of mass ' $\mathrm{m}$ ' and length ' $l$ ' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its end B strikes the ground is ( $g$ = acceleration due to gravity)
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Verified Answer
The correct answer is:
$\sqrt{\frac{3 \mathrm{~g}}{l}}$
We know,
Loss in P.E = Gain in rotational K.E.
When the centre of mass of the rod falls through a distance $\frac{\mathrm{L}}{2}$,
Loss in P.E $=m g \frac{L}{2}$ ...(i)
and,
Gain in Rotational K.E $=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2}\left[\frac{\mathrm{ML}^2}{3}\right] \omega^2$ ...(ii)
Equating (i) and (ii)
$$
\begin{aligned}
& \frac{\mathrm{MgL}}{2}=\frac{1}{2} \frac{\mathrm{ML}^2}{3} \omega^2 \\
& \omega^2=\frac{3 \mathrm{~g}}{\mathrm{~L}} \\
& \omega=\sqrt{\frac{3 \mathrm{~g}}{\mathrm{~L}}}=\sqrt{\frac{3 \mathrm{~g}}{l}} \quad \ldots(\because \text { here } \mathrm{L}=l)
\end{aligned}
$$
$\ldots(\because$ here $\mathrm{L}=l)$
Loss in P.E = Gain in rotational K.E.
When the centre of mass of the rod falls through a distance $\frac{\mathrm{L}}{2}$,
Loss in P.E $=m g \frac{L}{2}$ ...(i)
and,
Gain in Rotational K.E $=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2}\left[\frac{\mathrm{ML}^2}{3}\right] \omega^2$ ...(ii)
Equating (i) and (ii)
$$
\begin{aligned}
& \frac{\mathrm{MgL}}{2}=\frac{1}{2} \frac{\mathrm{ML}^2}{3} \omega^2 \\
& \omega^2=\frac{3 \mathrm{~g}}{\mathrm{~L}} \\
& \omega=\sqrt{\frac{3 \mathrm{~g}}{\mathrm{~L}}}=\sqrt{\frac{3 \mathrm{~g}}{l}} \quad \ldots(\because \text { here } \mathrm{L}=l)
\end{aligned}
$$
$\ldots(\because$ here $\mathrm{L}=l)$
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