Physics - Circular Motion Question with Solution | TestHub

$\mathrm{N}=\mathrm{m}\mathrm{g}\cos \mathrm{\theta }+\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{R}}\sin \mathrm{\theta }$

$f_{\max }=\mu \mathrm{N}\Rightarrow$$f_{\mathrm{m}\mathrm{a}\mathrm{x}}={\mu }_s\mathrm{m}\mathrm{g}\cos \mathrm{\theta }+\frac{{\mu }_s\mathrm{m}{\mathrm{v}}^2}{\mathrm{R}}\sin \mathrm{\theta }$

$\mathrm{mg}\sin \theta +f_{\max }=\frac{\mathrm{m}{\mathrm{v}}^2}{R}\cos \theta$

$\mathrm{m}\mathrm{g}\sin \mathrm{\theta }+{\mu }_s\mathrm{m}\mathrm{g}\cos \mathrm{\theta }+\frac{{\mu }_s\mathrm{m}{\mathrm{v}}^2}{\mathrm{R}}\sin \mathrm{\theta }=\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{R}}\cos \mathrm{\theta }$

$\mathrm{g}\sin \mathrm{\theta }+{\mu }_s\mathrm{g}\cos \mathrm{\theta }=\frac{{\mathrm{v}}^2}{\mathrm{R}}\left(\cos \mathrm{\theta }-{\mu }_s\sin \mathrm{\theta }\right)$

$\mathrm{g}\mathrm{R}\left[\frac{\tan \mathrm{\theta }+{\mu }_s}{1-{\mu }_s\tan \mathrm{\theta }}\right]={\mathrm{v}}^2$