A thin rod of length l and mass m oscillates | vedclass

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(D) The moment of inertia of a rod of length $l$ and mass $m$ about an axis passing through one of its ends is $I = \frac{1}{3} ml^2$.The rotational k

Vedclass · Accepted Answer

$\frac{1}{6} \frac{l^2 \omega^2}{g}$

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(D) The moment of inertia of a rod of length $l$ and mass $m$ about an axis passing through one of its ends is $I = \frac{1}{3} ml^2$.The rotational kinetic energy of the rod at its lowest position is $K = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{1}{3} ml^2) \omega^2 = \frac{1}{6} ml^2 \omega^2$.As the rod swings,this kinetic energy is converted into gravitational potential energy of the center of mass.The center of mass of the rod is at a distance $l/2$ from the axis of rotation.Let $h$ be the maximum height reached by the center of mass.By the law of conservation of energy,the change in potential energy equals the initial rotational kinetic energy:$mgh = \frac{1}{6} ml^2 \omega^2$.Solving for $h$,we get $h = \frac{l^2 \omega^2}{6g}$.

$A$ thin rod of length $l$ and mass $m$ oscillates in a vertical plane about a horizontal axis passing through one of its ends. If the maximum angular velocity of the rod is $\omega$,what is the maximum height reached by its center of mass?

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