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6.System of Particles and Rotational Motion
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A rod of length $50\,cm$ is pivoted at one end. It is raised such that if makes an angle of $30^o$ fro the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\,s^{-1}$ ) will be $(g = 10\,ms^{-2})$
A
$\sqrt \frac {30}{2}$
B
$\sqrt {30}$
C
$\sqrt \frac {20}{2}$
D
$ \frac {\sqrt {30}}{2}$
(JEE MAIN-2019)
Solution
$\begin{array}{l}
mg\frac{\ell }{2}\left( {\frac{1}{2}} \right) = \frac{1}{2}\left( {\frac{{m{\ell ^2}}}{3}} \right){\omega ^2}\\
\Rightarrow \,\,\,\omega = \sqrt {\frac{{3g}}{{2\ell }}} = \sqrt {30}
\end{array}$
Standard 11
Physics
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