Circular Motion

Given, radius of horizontal circle, $r=\frac{L}{\sqrt{2}}$ Figure illustrating the particle of mass m moving in a horizontal circle, while suspended from a ceiling is shown In equilibrium condition at point A, $T \cos \theta = m g$ ...(i) $T \sin \theta = \frac{m v^{2}}{r}$ ...(ii) Divide Eq. (ii) by Eq. (i), $\tan \theta = \frac{v^{2}}{r g}$ $\Rightarrow v = \sqrt{r g \tan \theta}$ ...(iii) Now, from figure, we can write $\sin \theta = \frac{\frac{L}{\sqrt{2}}}{L} = \frac{1}{\sqrt{2}}$ ⇒ θ = 45° Substituting the value of θ in Eq. (iii), we get $v = \sqrt{r g \tan 45^{\circ}} = \sqrt{r g}$ Thus, the value of speed of particle is $v = \sqrt{r g}$.