A rod of length 50,cm is pivoted at one end. | vedclass

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(B) Let the length of the rod be $\ell = 0.5\,m$. The center of mass of the rod is at a distance $\frac{\ell}{2}$ from the pivot.When the rod is at an

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$\sqrt{30}$

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(B) Let the length of the rod be $\ell = 0.5\,m$. The center of mass of the rod is at a distance $\frac{\ell}{2}$ from the pivot.When the rod is at an angle of $30^o$ with the horizontal,the vertical height of the center of mass above the horizontal position is $h = \frac{\ell}{2} \sin(30^o) = \frac{\ell}{2} 	imes \frac{1}{2} = \frac{\ell}{4}$.By the law of conservation of mechanical energy,the loss in potential energy equals the gain in rotational kinetic energy:$mg h = \frac{1}{2} I \omega^2$Where $I = \frac{m\ell^2}{3}$ is the moment of inertia of the rod about the pivot.$mg \left( \frac{\ell}{4} ight) = \frac{1}{2} \left( \frac{m\ell^2}{3} ight) \omega^2$$g \frac{\ell}{4} = \frac{\ell^2}{6} \omega^2$$\omega^2 = \frac{6g}{4\ell} = \frac{3g}{2\ell}$Substituting $g = 10\,ms^{-2}$ and $\ell = 0.5\,m$:$\omega^2 = \frac{3 	imes 10}{2 	imes 0.5} = \frac{30}{1} = 30$$\omega = \sqrt{30}\,rad\,s^{-1}$.

$A$ rod of length $50\,cm$ is pivoted at one end. It is raised such that it makes an angle of $30^o$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\,s^{-1}$) will be $(g = 10\,ms^{-2})$.

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