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

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(D) The rod rotates about an axis passing through its end. The moment of inertia of the rod about this axis is $I = \frac{1}{3}ML^2$.The rotational ki

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$\frac{L^2 \omega^2}{6g}$

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(D) The rod rotates about an axis passing through its end. The moment of inertia of the rod about this axis is $I = \frac{1}{3}ML^2$.The rotational kinetic energy of the rod at its lowest point (maximum angular speed) is $K = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{3}ML^2)\omega^2 = \frac{1}{6}ML^2\omega^2$.At the maximum height $h$,the centre of mass of the rod rises by $h$. The potential energy gained by the rod is $U = Mgh$.By the law of conservation of energy,the rotational kinetic energy at the bottom equals the potential energy at the maximum height:$\frac{1}{6}ML^2\omega^2 = Mgh$.Solving for $h$:$h = \frac{L^2\omega^2}{6g}$.

$A$ thin uniform rod of length $L$ and mass $M$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of (where $g$ is the acceleration due to gravity):

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