One end of a rod of length L is on a horizont | vedclass

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(A) We use the principle of conservation of mechanical energy.Let the mass of the rod be $m$.The initial potential energy of the rod is determined by

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

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(A) We use the principle of conservation of mechanical energy.Let the mass of the rod be $m$.The initial potential energy of the rod is determined by the height of its center of mass from the horizontal plane.The center of mass of the rod is at a distance $L/2$ from the end on the plane.Initial height $h = \frac{L}{2} \sin \alpha$.Initial potential energy $U_i = mgh = mg \frac{L}{2} \sin \alpha$.When the rod hits the horizontal plane,its potential energy is zero $(U_f = 0)$.The entire potential energy is converted into rotational kinetic energy about the end fixed on the plane.Rotational kinetic energy $K_f = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia of the rod about one of its ends,$I = \frac{mL^2}{3}$.By conservation of energy,$U_i = K_f$.$mg \frac{L}{2} \sin \alpha = \frac{1}{2} \left( \frac{mL^2}{3} \right) \omega^2$.$mg \frac{L}{2} \sin \alpha = \frac{mL^2}{6} \omega^2$.Solving for $\omega$:$\omega^2 = \frac{6mgL \sin \alpha}{2mL^2} = \frac{3g \sin \alpha}{L}$.$\omega = \sqrt{\frac{3g \sin \alpha}{L}}$.

One end of a rod of length $L$ is on a horizontal plane. It is inclined at an angle $\alpha$ to the horizontal plane. When released,its angular velocity after coming to the horizontal plane is:

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