$W_{\text {gravity }}=\Delta K E$
$m g\left(\frac{l}{2}\right)=\frac{1}{2} I \omega^{2} \Rightarrow \omega=\sqrt{\frac{3 g}{l}}$
With this angular velocity the half rod starts moving upward. From Work Energy Theorem
$W_{\text {gravity }}=\Delta K E$
$-\frac{m}{2} g\left[\frac{l}{4}-\frac{l}{4} \cos \theta\right]=0-\frac{1}{2} I^{\prime} \omega$
$\frac{m g l}{8}(1-\cos \theta)=\frac{1}{2} I^{\prime} \omega^{\prime}$
$\frac{m g l}{8}(1-\cos \theta)=\frac{1}{2}\left[\frac{1}{3} \frac{m}{2}\left(\frac{l}{2}\right)^{2}\right] \times \frac{3 g}{l}$
$1-\cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ}$
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$\left( g =10\,ms ^{-2}\right)$
