A light rod of length L is pivoted at its upp | vedclass

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(C) Let the velocity imparted to the lower mass be $v_0$. Since the rod is rigid,the angular velocity $\omega$ of the rod is given by $\omega = \frac{

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$\sqrt{\frac{12Lg}{5}}$

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(C) Let the velocity imparted to the lower mass be $v_0$. Since the rod is rigid,the angular velocity $\omega$ of the rod is given by $\omega = \frac{v_0}{L}$.The velocity of the mass at the midpoint is $v_{mid} = \omega \cdot \frac{L}{2} = \frac{v_0}{2}$.By the principle of conservation of mechanical energy,the loss in kinetic energy equals the gain in potential energy as the rod moves from the vertical to the horizontal position.Initial kinetic energy $K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}m\left(\frac{v_0}{2}\right)^2 = \frac{1}{2}mv_0^2 + \frac{1}{8}mv_0^2 = \frac{5}{8}mv_0^2$.Final kinetic energy $K_f = 0$ (at the horizontal position).Gain in potential energy $U_f - U_i = mg\left(\frac{L}{2}\right) + mgL = \frac{3}{2}mgL$.Equating the loss in kinetic energy to the gain in potential energy:$\frac{5}{8}mv_0^2 = \frac{3}{2}mgL$.$v_0^2 = \frac{3}{2} \cdot \frac{8}{5} gL = \frac{12}{5}gL$.$v_0 = \sqrt{\frac{12Lg}{5}}$.

$A$ light rod of length $L$ is pivoted at its upper end. Two masses (each of mass $m$) are attached to the rod,one at its midpoint and the other at its free end. What horizontal velocity must be imparted to the lower end so that the rod reaches the horizontal position?

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