One end of rod of length L is on horizontal plane. It is inclined at angle α to horizont

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6.System of Particles and Rotational Motion

hard

One end of rod of length $L$ is on horizontal plane. It is inclined at angle $\alpha$ to horizontal plane. When released its angular velocity after coming to horizontal plane is

A

$\sqrt{\frac{3g \sin \alpha}{L}}$

B

$\sqrt{\frac{2L}{3g \sin \alpha}}$

C

$\sqrt{\frac{6g \sin \alpha}{L}}$

D

$\sqrt{\frac{L}{g \sin \alpha}}$

Solution

Energy balance method

$\operatorname{mg} \frac{\mathrm{L}}{2} \sin \alpha=\frac{1}{2} \mathrm{I} \omega^{2}$

$m g \frac{L}{2} \sin \alpha=\frac{1}{2} \frac{m L^{2}}{3} \omega^{2}$

$\omega=\sqrt{\frac{3 g \sin \alpha}{L}}$

Standard 11

Physics

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(KVPY-2013)