JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A thin uniform rod \(( X )\) of mass M and length L is pivoted at a height \(\left(\frac{ L }{3}\right)\) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is _________ . (g = gravitational acceleration)
- A \(\sqrt{\frac{3}{2} \frac{g}{ L }}\)
- B \(\frac{3}{\sqrt{2}}\sqrt{\frac{g}{L}}\)
- C \(\frac{1}{\sqrt{2}}\sqrt{\frac{g}{L}}\)
- D \(\sqrt{\frac{3g}{L}}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{\frac{3g}{L}}\)
Step-by-step Solution
Detailed explanation
\(mg \frac{\ell}{6}=\frac{1}{2} I \omega^2\) Here \(I=\frac{m \ell^2}{12}+\frac{m \ell^2}{36}=\frac{m \ell^2}{9}\) \(mg \frac{\ell}{6}=\frac{ m \ell^2}{18} \omega^2 \Rightarrow \omega^2=\frac{3 g}{\ell}\) \(\omega=\sqrt{\frac{3 g}{\ell}}\)
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