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6.System of Particles and Rotational Motion
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A thin rod of mass $m$ and length $l$ is oscillating about horizontal axis through its one end. Its maximum angular speed is $\omega$. Its centre of mass will rise upto maximum height :-
A
$\frac{1}{6} \frac{l \omega}{g}$
B
$\frac{1}{2} \frac{l^2 \omega^2}{g}$
C
$\frac{1}{6} \frac{l^2 \omega^2}{g}$
D
$\frac{1}{3} \frac{l^2 \omega^2}{g}$
Solution
$\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}$
$\frac{1}{2} \frac{\mathrm{m} \ell^{2}}{3} \omega^{2}=\mathrm{mgh}$
$\mathrm{h}=\frac{1}{6} \frac{\ell^{2} \omega^{2}}{\mathrm{g}}$
Standard 11
Physics
