Question

minimum value of $\omega$ so that centre af rod can complete circular motion.

Answer 1

From conservation of mechanical energy,
$mgl=\frac{1}{2}I{\omega }_{min}^2$
$\therefore {\omega }_{min}=\sqrt{\frac{2mgl}{I}}=\sqrt{\frac{2mgl}{\frac{7}{3}m{l}^2}}=\sqrt{\frac{6}{7}\frac{g}{l}}$