A uniform rod of mass m and length l hinged a | vedclass

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Question

Loss in $P.E.$ $=$ gain in $K.E.$

or $\quad \operatorname{mg} \frac{l}{2}=\frac{1}{2} \mathrm{I} \omega^{2}$

or $\quad \mathrm{mg} l=\frac{\math

Vedclass · Accepted Answer

$\frac{{5Mg}}{2}$

Vedclass · Answer

Loss in $P.E.$ $=$ gain in $K.E.$

or $\quad \operatorname{mg} \frac{l}{2}=\frac{1}{2} \mathrm{I} \omega^{2}$

or $\quad \mathrm{mg} l=\frac{\mathrm{m} l^{2}}{3} \omega^{2}$

or $\frac{3 \mathrm{g}}{l}=\omega^{2}$

$\mathrm{F}-\mathrm{mg}=\mathrm{m} \frac{l}{2} \omega^{2}$

or $\quad \mathrm{F}=\mathrm{mg}+\mathrm{m} \frac{l}{2} \omega^{2}$

$\mathrm{F}=\mathrm{mg}+\mathrm{m} \frac{l}{2} \frac{3 \mathrm{g}}{l}=\frac{5 \mathrm{mg}}{2}$

$A$ uniform rod of mass $m$ and length $l$ hinged at its end is released from rest when it is in the horizontal position. The normal reaction at the hinge when the rod becomes vertical is :

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