Question

A thin rod of mass $m$ and length $l$ is suspended from one of its ends. It is set into oscillation about a horizontal axis. Its angular speed is $\omega$ while passing through its mean position. How high will its centre of mass rise from its lowest position?

• A. $\frac{\omega^{2} l^{2}}{2 g}$
• B. $\frac{\omega^{2} l^{2}}{3 g}$
• C. $\frac{\omega^{2} l^{2}}{g}$
• D. $\frac{\omega^{2} l^{2}}{6 g}$

Answer 1

Answer: (D)

$\frac{1}{2} \frac{m l^{2}}{3} \cdot \omega^{2}=m g h$ (Energy conservation)
$h=\frac{l^{2} \omega^{2}}{6 g}$