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(B) By the principle of conservation of energy,the potential energy of the rod at the initial vertical position is converted into rotational kinetic e

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$\sqrt{3gL}$

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(B) By the principle of conservation of energy,the potential energy of the rod at the initial vertical position is converted into rotational kinetic energy at the moment it hits the floor.Initial potential energy $U_i = Mg \frac{L}{2}$.Final rotational kinetic energy $K_f = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia of the rod about the end on the floor,$I = \frac{ML^2}{3}$.Equating $U_i = K_f$:$Mg \frac{L}{2} = \frac{1}{2} (\frac{ML^2}{3}) \omega^2$.Simplifying,$gL = \frac{L^2}{3} \omega^2$,which gives $\omega^2 = \frac{3g}{L}$,so $\omega = \sqrt{\frac{3g}{L}}$.The linear velocity $v$ of the other end is given by $v = L \omega$.Substituting $\omega$,$v = L \sqrt{\frac{3g}{L}} = \sqrt{3gL}$.

$A$ thin rod of length $L$ and mass $M$ is held vertically with one end on the floor and is allowed to fall. Find the velocity of the other end when it hits the floor,assuming that the end on the floor does not slip.

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