Physics - Rotation Question with Solution | TestHub

Moment of inertia about axis$\mathrm{A}\mathrm{A}\text{'}$is



${\mathrm{I}}_{\mathrm{A}\mathrm{A}\text{'}}=\left\{\left(\frac{\mathrm{M}}{2}\right)\frac{{\mathrm{l}}^2}{12}+\left(\frac{\mathrm{M}}{2}\right)\frac{5{\mathrm{l}}^2}{4}+\left(\frac{\mathrm{M}}{2}\right)\frac{{\mathrm{l}}^2}{3}\right\}$

$\therefore \mathrm{ }\mathrm{ }\mathrm{O}\mathrm{D}=\frac{\sqrt{5}\mathrm{l}}{2}$

$\mathrm{O}\mathrm{C}=\frac{\sqrt{10}\mathrm{l}}{4}$( point$\mathrm{C}$is centre of mass of the system )

Torque of$\mathrm{M}\mathrm{g}$will be maximum when$\mathrm{O}\mathrm{C}$is horizontal

$\therefore \mathrm{ }\mathrm{ }\mathrm{\alpha }=\frac{\mathrm{\tau }}{{\mathrm{I}}_{\mathrm{A}\mathrm{A}\text{'}}}=\frac{\frac{\mathrm{M}\mathrm{g}\sqrt{10}\mathrm{l}}{4}}{\frac{5\mathrm{M}{\mathrm{l}}^2}{6}}$

$\mathrm{\alpha }=\frac{3\mathrm{g}}{\sqrt{10}\mathrm{l}}$