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(B) The rod is hinged at $O$. Let $L$ be the length of the rod. The center of mass is at $L/2$. The potential energy $U$ at an angle $	heta$ with the

Vedclass · Accepted Answer

$\omega_4 = 2 \omega_2$

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(B) The rod is hinged at $O$. Let $L$ be the length of the rod. The center of mass is at $L/2$. The potential energy $U$ at an angle $	heta$ with the vertical is $U = mg(L/2) \cos 	heta$. By conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy: $mg(L/2)(1 - \cos 	heta) = \frac{1}{2} I \omega^2$,where $I = \frac{mL^2}{3}$. Substituting $I$: $mg(L/2)(1 - \cos 	heta) = \frac{1}{2} (\frac{mL^2}{3}) \omega^2 \implies \omega^2 = \frac{3g}{L}(1 - \cos 	heta)$. For position $(2)$,$	heta = 60^{\circ}$,$\cos 60^{\circ} = 0.5$,so $\omega_2^2 = \frac{3g}{L}(1 - 0.5) = \frac{1.5g}{L}$. For position $(4)$,$	heta = 180^{\circ}$,$\cos 180^{\circ} = -1$,so $\omega_4^2 = \frac{3g}{L}(1 - (-1)) = \frac{6g}{L}$. Comparing $\omega_4^2$ and $\omega_2^2$: $\frac{\omega_4^2}{\omega_2^2} = \frac{6g/L}{1.5g/L} = 4$. Therefore,$\omega_4 = 2 \omega_2$.

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