Question

A rod of length $\frac{2}{3}$ m is fixed at one end. It is raised to some height such that it makes ${37}^{\circ }$ from the horizontal. Find its angular speed in rad/sec, when it passes through the horizontal. [$sin{37}^{\circ }=3/5,g=10m/s^2$]

• A. $3$
• B. $6$
• C. $9\sqrt{3}$
• D. $3\sqrt{3}$

Answer 1

The correct option is D $3\sqrt{3}$

Applying conservation of mechanical energy:
$lossinP.E=\frac{mgl}{2}sin{37}^{\circ }gaininK.E=\frac{1}{2}I{w}^2\frac{\text{/}mgl}{\text{/}2}sin{37}^{\circ }=\frac{1}{\text{/}2}\frac{\text{/}m{l}^2}{3}.w^2\frac{3gl}{5}=\frac{l^2}{3}{w}^2{w}^2=27w=3\sqrt{3}rad/s$