A rod hinged at one end is released from the | vedclass

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Question

When rod becomes vertical

$W_{	ext {gravity }}=\Delta K E$

$m g\left(\frac{l}{2}ight)=\frac{1}{2} I \omega^{2} \Rightarrow \omega=\sqrt{\frac

Vedclass · Accepted Answer

$60$

Vedclass · Answer

When rod becomes vertical

$W_{	ext {gravity }}=\Delta K E$

$m g\left(\frac{l}{2}ight)=\frac{1}{2} I \omega^{2} \Rightarrow \omega=\sqrt{\frac{3 g}{l}}$

With this angular velocity the half rod starts moving upward. From Work Energy Theorem

$W_{	ext {gravity }}=\Delta K E$

$-\frac{m}{2} g\left[\frac{l}{4}-\frac{l}{4} \cos 	hetaight]=0-\frac{1}{2} I^{\prime} \omega$

$\frac{m g l}{8}(1-\cos 	heta)=\frac{1}{2} I^{\prime} \omega^{\prime}$

$\frac{m g l}{8}(1-\cos 	heta)=\frac{1}{2}\left[\frac{1}{3} \frac{m}{2}\left(\frac{l}{2}ight)^{2}ight] 	imes \frac{3 g}{l}$

$1-\cos 	heta=\frac{1}{2} \Rightarrow \cos 	heta=\frac{1}{2} \Rightarrow 	heta=60^{\circ}$

$A$ rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical its lower half separates without exerting any reaction at the breaking point. Then the maximum angle $‘\theta ’$ made by the hinged upper half with the vertical is ......... $^o$.

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