JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \( \frac {\sqrt {30}}{2}\)
Answer & Solution
Correct Answer
(B) \(\sqrt {30}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} mg\frac{\ell }{2}\left( {\frac{1}{2}} \right) = \frac{1}{2}\left( {\frac{{m{\ell ^2}}}{3}} \right){\omega ^2}\\ \Rightarrow \,\,\,\omega = \sqrt {\frac{{3g}}{{2\ell }}} = \sqrt {30} \end{array}\)
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