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6.System of Particles and Rotational Motion
hard
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
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
$\frac{1}{6}\frac{{{l^2}{\omega ^2}}}{g}$
Solution
$\mathrm{TE}_{\mathrm{i}}=\mathrm{TE}_{\mathrm{f}}$
$\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}$
$\frac{1}{2} \times \frac{1}{3} \mathrm{m} \ell^{2} \omega^{2}=\mathrm{mgh}$
Or $\mathrm{h}=\frac{1}{6} \frac{\ell^{2} \omega^{2}}{\mathrm{g}}$
Standard 11
Physics
