A light rod carries three equal masses A,B,an | vedclass

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(B) By the principle of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy.Let $m$ be the mass of each p

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$\sqrt {\frac{{8g\ell }}{7}} $

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(B) By the principle of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy.Let $m$ be the mass of each particle $A$,$B$,and $C$. The distances of $A$,$B$,and $C$ from the pivot are $\frac{\ell}{3}$,$\frac{2\ell}{3}$,and $\ell$ respectively.Loss in potential energy = $mg(\frac{\ell}{3}) + mg(\frac{2\ell}{3}) + mg(\ell) = mg\ell(1/3 + 2/3 + 1) = 2mg\ell$.Gain in rotational kinetic energy = $\frac{1}{2} I \omega^2$,where $I = m(\frac{\ell}{3})^2 + m(\frac{2\ell}{3})^2 + m(\ell)^2 = m\ell^2(\frac{1}{9} + \frac{4}{9} + 1) = m\ell^2(\frac{14}{9})$.Equating the two: $2mg\ell = \frac{1}{2} (m\ell^2 \frac{14}{9}) \omega^2$.$2g = \frac{7}{9} \ell \omega^2 \Rightarrow \omega^2 = \frac{18g}{7\ell} \Rightarrow \omega = \sqrt{\frac{18g}{7\ell}}$.The velocity of mass $B$ is $v_B = \omega r_B = \omega (\frac{2\ell}{3})$.$v_B = \frac{2\ell}{3} \sqrt{\frac{18g}{7\ell}} = \sqrt{\frac{4\ell^2}{9} \cdot \frac{18g}{7\ell}} = \sqrt{\frac{8g\ell}{7}}$.

$A$ light rod carries three equal masses $A$,$B$,and $C$ as shown in the figure. If the rod is released from the horizontal position,what will be the velocity of mass $B$ when the rod reaches the vertical position?

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