Q 11-06-022JEE MainAIEEE 2009Medium
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 a maximum height of
Answer: (D) $\dfrac{1}{6}\dfrac{l^2\omega^2}{g}$
The angular speed is maximum at the lowest position. All of that rotational KE becomes PE of the centre of mass:
$$\frac{1}{2}\left(\frac{ml^2}{3}\right)\omega^2 = mgh \;\Rightarrow\; h = \frac{l^2\omega^2}{6g}$$
Solution by Sreeraj P, M.Sc Physics