A thin uniform rod of length $\ell$ 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

• A

  $\frac{1}{3} \frac{\ell^2 \omega^2}{g}$
• B

  $\frac{1}{6} \frac{\ell \omega}{\mathrm{g}}$
• C

  $\frac{1}{2} \frac{\ell^2 \omega^2}{g}$
• Correct answer
  D

  $\frac{1}{6} \frac{\ell^2 \omega^2}{\mathrm{~g}}$

Solution

$$ \begin{aligned} & \text { } \\ & \text { T.E }=T . E_{\mathrm{f}} \\ & \frac{1}{2} I \omega^2=m g h \\ & \frac{1}{2} \times \frac{1}{3} m \ell^2 \omega^2=m g h \Rightarrow h=\frac{1}{6} \frac{\ell^2 \omega^2}{\mathrm{~g}} \end{aligned} $$

Hence, the correct answer is option: 'D'.