Q. A thin uniform rod of length $l\ $and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to maximum height of
A) $\frac{1}{3}\ \frac{l^{2}\omega^{2}}{g}$
B) $\frac{1}{6}\ \frac{l\omega}{g}$
C) $\frac{1}{2}\ \frac{l^{2}\omega^{2}}{g}$
D) $\frac{1}{6}\ \frac{l^{2}\omega^{2}}{g}$
Explanation
(d) $TE_{i} = TE_{r}$
$\frac{1}{2}I\omega^{2} = mgh$
$\frac{1}{2} \times \frac{1}{3}ml^{2}\omega^{2} = mgh$
$\Rightarrow$ $h = \frac{1}{6}\ \frac{l^{2}\omega^{2}}{g}$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.