Physics - Circular Motion Question with Solution | TestHub

In this problem it is assumed that particle although moving in a vertical loop but it speed remain constant

Tension at lowest point$T_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{m{v}^2}{r}+m$g

Tension at highest point$T_{\mathrm{m}\mathrm{i}\mathrm{n}}=\frac{m{v}^2}{r}-m\mathrm{g}$

$\frac{T_{\mathrm{m}\mathrm{a}\mathrm{x}}}{T_{\mathrm{m}\mathrm{i}\mathrm{n}}}=\frac{\frac{m{v}^2}{r}+m\mathrm{g}}{\frac{m{v}^2}{r}-m\mathrm{g}}=\frac{5}{3}$

$\frac{3m{v}^2}{r}+3mg=\frac{5m{v}^2}{r}-5mg$

$\frac{2m{v}^2}{r}=8mg$

$v=\sqrt{4rg}$

$=\sqrt{4\times 2.5\times 9.8}$

$=\sqrt{98} \mathrm{m} {\mathrm{s}}^{-1}$