Physics - Circular Motion Question with Solution | TestHub

$\text{mg cos}\theta -\text{N}=\frac{{\text{mv}}^2}{\text{R}}$



to leave the contact N = 0

$\text{mg cos }\theta =\frac{{\text{mv}}^2}{\text{R}} \Rightarrow \text{g cos }\theta =\frac{{\text{v}}^2}{\text{R}} \dots \left(1\right)$

$\text{cos}\theta =\frac{\left(\text{R}-\text{h}\right)}{\text{R}} \dots \left(2\right)$

from energy conservation$\Rightarrow \frac{1}{2}{\text{mv}}^2=\text{mgh} \dots \left(3\right)$

from 1, 2, 3$\frac{1}{2}\text{gR}\times \frac{\left(\text{R}-\text{h}\right)}{\text{R}}=\text{gh} \Rightarrow \text{R}-\text{h}=2\text{h}\Rightarrow \text{h}=\text{R}/3$