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) $\dfrac{1}{6}\dfrac{l^2\omega^2}{g}$
Concept. At the lowest point the rod's KE is greatest; it all turns into PE at the top of the swing: $\tfrac12I\omega^2=mgh_{cm}$, with $I=\tfrac13ml^2$ about the end.
Step 1. Energy. $\tfrac12\cdot\tfrac13ml^2\omega^2=mgh$.
Step 2. Solve. $h=\dfrac{l^2\omega^2}{6g}$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
