A uniform rod of length L is free to rotate i | vedclass

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Question

When the rod rotates through angle $	heta,$ the fall $'h'$ of centre of gravity is given by $	heta .$

$\frac{\mathrm{L} / 2-\mathrm{h}}{\

Vedclass · Accepted Answer

$\sqrt {\frac{{6g}}{L}} \,\sin \,\frac{	heta }{2}$

Vedclass · Answer

When the rod rotates through angle $	heta,$ the fall $'h'$ of centre of gravity is given by $	heta .$

$\frac{\mathrm{L} / 2-\mathrm{h}}{\mathrm{L} / 2}=\cos 	heta$

or $\quad \mathrm{h}=\frac{\mathrm{L}}{2}(1-\cos 	heta)$

Decrease in potential energy

$=\mathrm{Mgh}=\mathrm{Mg} \frac{\mathrm{L}}{2}(1-\cos 	heta)$

Now, $KE$ of rotation $=\frac{1}{2} \mathrm{I} \omega^{2}$

$=\frac{1}{2} 	imes \frac{M L^{2}}{3} \omega^{2}$

$\mathrm[{I}=\mathrm{ML}^{2} / 3$ (because rod is rotating about an axis passing through its one end)]

According to law of conservation of energy.

$\mathrm{Mg} \frac{\mathrm{L}}{2}(1-\cos 	heta)=\frac{\mathrm{ML}^{2}}{6} \omega^{2}$

$	herefore$         $\omega=\sqrt{6 g / L} \sin \left(\frac{	heta}{2}ight)$

A uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta $ its angular velocity $\omega $ is given as

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