A thin rod MN,free to rotate in the vertical | vedclass

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Question

(D) Let $l$ be the length of the rod and $m$ be its mass.When the rod is horizontal,its potential energy is taken as zero.When the rod makes an angle

Vedclass · Accepted Answer

$\sqrt{\sin \alpha}$

Vedclass · Answer

(D) Let $l$ be the length of the rod and $m$ be its mass.When the rod is horizontal,its potential energy is taken as zero.When the rod makes an angle $\alpha$ with the horizontal,the center of mass of the rod (at distance $l/2$ from $N$) descends by a vertical distance $h = (l/2) \sin \alpha$.By the law of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy:$mg(l/2) \sin \alpha = \frac{1}{2} I \omega^2$Here,$I$ is the moment of inertia of the rod about the fixed end $N$,which is $I = \frac{ml^2}{3}$.Substituting $I$ into the energy equation:$mg(l/2) \sin \alpha = \frac{1}{2} (\frac{ml^2}{3}) \omega^2$$mg(l/2) \sin \alpha = \frac{ml^2}{6} \omega^2$$\omega^2 = \frac{3g \sin \alpha}{l}$$\omega = \sqrt{\frac{3g \sin \alpha}{l}}$The linear speed $v$ of the end $M$ is given by $v = \omega l$:$v = l \sqrt{\frac{3g \sin \alpha}{l}} = \sqrt{3gl \sin \alpha}$Thus,the speed $v$ is proportional to $\sqrt{\sin \alpha}$.

$A$ thin rod $MN$,free to rotate in the vertical plane about the fixed end $N$,is held horizontal. When the end $M$ is released,the speed of this end,when the rod makes an angle $\alpha$ with the horizontal,will be proportional to (see figure):

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