Physics - Rotation Question with Solution | TestHub

For horizontal acceleration, net force in vertical direction should be zero.



Loss in potential energy = gain in kinetic energy

$mgh=\frac{1}{2}m{v}^2$

$mgl\cos \mathrm{\alpha }=\frac{1}{2}m{v}^2$

$\therefore v=\sqrt{2gl\cos \mathrm{\alpha }}$

$\because$The net force towards O is

$T-mg\cos \mathrm{\alpha }=\frac{m{v}^2}{l}$

$=\frac{m2gl\cos \mathrm{\alpha }\mathrm{ }}{l}=2\mathrm{ }mg\cos \mathrm{\alpha }$

$\therefore T=3mg\cos \mathrm{\alpha }$

$\therefore T\cos \mathrm{\alpha }=mg$

Or$3mg\mathrm{ }\mathrm{c}\mathrm{o}{\mathrm{s}}^2\mathrm{ }\mathrm{\alpha }=mg$

$\therefore \mathrm{\alpha }={\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}\left(\frac{1}{\sqrt{3}}\right)$