A thin uniform rod of mass M and length L is | vedclass

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(D) The rod is pivoted at a distance $\frac{L}{3}$ from the lower end. The center of mass of the rod is at a distance $\frac{L}{2}$ from the lower end

Vedclass · Accepted Answer

$\sqrt{\frac{3g}{L}}$

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(D) The rod is pivoted at a distance $\frac{L}{3}$ from the lower end. The center of mass of the rod is at a distance $\frac{L}{2}$ from the lower end.Initially, the center of mass is at a height $h_i = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$ above the pivot point.When the rod hits the table, the center of mass is at the same level as the pivot point, so the final height $h_f = 0$.The change in potential energy is $\Delta PE = Mg(h_f - h_i) = -Mg\frac{L}{6}$.By the principle of conservation of energy, the loss in potential energy equals the gain in rotational kinetic energy: $Mg\frac{L}{6} = \frac{1}{2} I \omega^2$.The moment of inertia $I$ about the pivot point (which is at distance $\frac{L}{3}$ from the end) is calculated using the parallel axis theorem: $I = I_{cm} + M d^2 = \frac{ML^2}{12} + M(\frac{L}{2} - \frac{L}{3})^2 = \frac{ML^2}{12} + M(\frac{L}{6})^2 = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{3ML^2 + ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$.Substituting $I$ into the energy equation: $Mg\frac{L}{6} = \frac{1}{2} (\frac{ML^2}{9}) \omega^2$.$Mg\frac{L}{6} = \frac{ML^2}{18} \omega^2$.$\omega^2 = \frac{MgL}{6} \cdot \frac{18}{ML^2} = \frac{3g}{L}$.$\omega = \sqrt{\frac{3g}{L}}$.

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