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

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(A) By the law of conservation of mechanical energy,the loss in potential energy of the rod is equal to the gain in its rotational kinetic energy.The

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

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(A) By the law of conservation of mechanical energy,the loss in potential energy of the rod is equal to the gain in its rotational kinetic energy.The center of mass of the rod is at a distance $L/2$ from the pivot.When the rod moves from the horizontal position to the vertical position,the center of mass falls by a vertical distance $h = L/2$.Loss in potential energy = $Mgh = Mg(L/2) = MgL/2$.The moment of inertia of a rod about an axis passing through one end is $I = ML^2/3$.Gain in rotational kinetic energy = $\frac{1}{2} I \omega^2 = \frac{1}{2} (ML^2/3) \omega^2 = \frac{ML^2 \omega^2}{6}$.Equating the two: $MgL/2 = \frac{ML^2 \omega^2}{6}$.Solving for $\omega$: $\omega^2 = \frac{6gL}{2L^2} = \frac{3g}{L}$.Therefore,$\omega = \sqrt{\frac{3g}{L}}$.

$A$ uniform rod of mass $M$ and length $L$ is pivoted at one end and is free to rotate in a vertical plane. The rod is released from rest in a horizontal position. What is the angular velocity of the rod when it reaches the vertical position?

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