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Question

The moment of inertia of the rod about $O$ is 

$\frac{1}{3}m{\ell ^2}$. The maximum angular speed of the rod is when the rod is instantaneousl

Vedclass · Accepted Answer

$\;\frac{1}{6}\frac{{{l^2}{\omega ^2}}}{g}$

Vedclass · Answer

The moment of inertia of the rod about $O$ is 

$\frac{1}{3}m{\ell ^2}$. The maximum angular speed of the rod is when the rod is instantaneously vertical. The energy of the rod in this condition is $\frac{1}{2}I{\omega ^2}\,$ where $I$ is the moment of inertia of the rod about $O.$ When the rod is in extreme portion, its angular velocity is zero momentarily. In this case, the energy of the rod is mgh where h is the maximum height to which the center of mass $(C.M)$ rises

$\begin{array}{l}
	herefore \,mgh = \frac{1}{2}I{\omega ^2} = \frac{1}{2}\left( {\frac{1}{3}m{l^2}} ight){\omega ^2}\
 \Rightarrow h = \frac{{{\ell ^2}{\omega ^2}}}{{6g}}
\end{array}$

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