Question

An uniform rod of mass $M$ & length $L$ is hinged about its one end as shown. Initially it is held vertical and then allowed to rotate, the angular velocity of rod when it makes an angle of ${37}^{\circ }$ with the vertical is

• A. $\sqrt{\frac{12g}{5L}}$
• B. $\sqrt{\frac{3g}{5L}}$
• C. $\sqrt{\frac{g}{5L}}$
• D. $\sqrt{\frac{g}{L}}$

Answer 1

The correct option is B $\sqrt{\frac{3g}{5L}}$

Using work energy theorem,
Work done by gravity = change in kinetic energy,
$Mg\frac{L}{2}(1-\cos {37}^{\circ })=\frac{I{\omega }^2}{2}$
$Mg\frac{L}{2}(1-\cos {37}^{\circ })=\frac{M{L}^2}{3}\frac{{\omega }^2}{2}$
$\Rightarrow \frac{MgL}{2}\times \frac{1}{5}=\frac{M{L}^2{\omega }^2}{6}$
$\Rightarrow \omega =\sqrt{\frac{3g}{5L}}$