Physics - Rotation Question with Solution | TestHub

Explain Question : It is given that ring can rotate in vertical plane about 'O' choose the correct options when ring rotates through an angle $\theta$ while it is released from rest at $\theta =0$. Concept : Rotation of a body about a fixed axis Solution :

$\mathrm{mgR}\cos \theta =2{\mathrm{mR}}^2\alpha \Rightarrow \alpha =\frac{\mathrm{g}\cos \theta }{2\mathrm{R}}$ $\mathrm{mg}\mathrm{R}\sin \theta =\frac{1}{2}\times 2{\mathrm{mR}}^2{\omega }^2\Rightarrow \mathrm{g}\sin \theta =\mathrm{R}{\omega }^2\Rightarrow {\omega }^2=\frac{\mathrm{g}\sin \theta }{\mathrm{R}}$

$\mathrm{mg}\cos \theta +{\mathrm{R}}_{\mathrm{y}}=\mathrm{mR}\left(\frac{\mathrm{g}\cos \theta }{2\mathrm{R}}\right)=\frac{\mathrm{mg}\cos \theta }{2}$ ${\mathrm{R}}_{\mathrm{y}}=\frac{-\mathrm{mg}\cos \theta }{2}$ ${\mathrm{R}}_{\mathrm{x}}-\mathrm{mg}\sin \theta =\mathrm{mR}\times \frac{\mathrm{g}\sin \theta }{\mathrm{R}}$ ${\mathrm{R}}_{\mathrm{x}}=2\mathrm{mg}\sin \theta$ $R_y^2+R_x^2=mg\sqrt{\frac{{\cos }^2\theta }{4}+4{\sin }^2\theta }=\frac{mg}{2}\sqrt{{\cos }^2\theta +16{\sin }^2\theta }=\frac{mg}{2}\sqrt{1+15{\sin }^2\theta }$ Final Answer The correct options are (1),(3)