Rotational Motion

JEE Main level, single correct, medium

192 of 280

Question 192 of 280

A thin uniform rod (X)(X) of mass MM and length LL is pivoted at a height (L3)\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=\mathrm{g}= gravitational acceleration)

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Answer: (D) 3gL\sqrt{\frac{3 g}{L}}

The rod is pivoted at a distance L3\frac{L}{3} from one end. The centre of mass is at a distance L6\frac{L}{6} from the pivot.
Using energy conservation:
mgℓ6=12Iω2mg\frac{\ell}{6} = \frac{1}{2}I\omega^2
Moment of inertia about the pivot (using parallel axis theorem):
I=mℓ212+m(ℓ6)2=mℓ212+mℓ236=mℓ29I = \frac{m\ell^2}{12} + m\left(\frac{\ell}{6}\right)^2 = \frac{m\ell^2}{12} + \frac{m\ell^2}{36} = \frac{m\ell^2}{9}
mgℓ6=12⋅mℓ29⋅ω2mg\frac{\ell}{6} = \frac{1}{2} \cdot \frac{m\ell^2}{9} \cdot \omega^2
ω2=9g6⋅ℓ2=3gℓ\omega^2 = \frac{9g}{6 \cdot \frac{\ell}{2}} = \frac{3g}{\ell}
ω=3gℓ\omega = \sqrt{\frac{3g}{\ell}}

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