Physics - Circular Motion Question with Solution | TestHub

Given $r=30m\mathrm{and}\mu =\mathrm{0.4.}$  $v_{\max }=\sqrt{\mu rg}=\sqrt{0.4\times 30\times 9.8}=10.84m{s}^{-1}$ For a safe turn  $\begin{array}{l}\frac{M{v}^2}{r}\le {\mu }_s\mathrm{Mg}\\ \mathrm{or}{v}^2\le {\mu }_s\mathrm{rg}\end{array}$ where  ${\mu }_s\to$ coefficient of static friction