Question

A thin uniform rod "AB" of mass $m$ and length $L$ is hinged at one end A to the level floor. Initially it stands vertically and is allowed to fall freely to the floor in the vertical plane. The angular velocity of the rod when its end B strikes the floor is : (g is acceleration to gravity)

• A. $\frac{mg}{L}$
• B. $\sqrt{\frac{mg}{3L}}$
• C. $\frac{g}{L}$
• D. $\sqrt{\frac{3g}{L}}$

Answer 1

Answer: (D)

$\sqrt{\frac{3g}{L}}$

Change in the position of CM is h= L/2

MI about the endpoint is $I=\frac{M{L}^2}{3}$

Using energy conservation $\frac{I{\omega }^2}{2}=Mgh$

So we get $\frac{M{L}^2{\omega }^2}{6}=MgL/2\Rightarrow \omega =\sqrt{\frac{3g}{L}}$