Question

A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor. Initially it stands vertically. It is allowed to fall freely in a vertical plane. If the linear speed of B when it hits the floor is $\sqrt{\frac{xg}{L}}$, what is x?___

Answer 1

When the end B hits the floor, the vertical distance through which C (the mid point) falls is $\frac{L}{2}$. From the law of conservation of energy, Loss of P.E = Loss of K.E.
$Mg\frac{L}{2}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\frac{M{L}^2}{3}{\omega }^2\Rightarrow \sqrt{\frac{3g}{L}}$