Circular Motion

Concept:When tension in a vertical circular motion becomes zero at a point, the radial component of weight provides the centripetal force. Conservation of mechanical energy relates speeds at different positions.Explanation:At point P where tension is zero, centripetal force is given by $mg \sin 30^\circ = \dfrac{mv_P^2}{r}$.Since $\sin 30^\circ = \dfrac12$, we get $v_P^2 = \dfrac{gr}{2}$.Height of P above bottom A: vertical distance from centre is $r \sin 30^\circ = \dfrac{r}{2}$.Total height $h = r + \dfrac{r}{2} = \dfrac{3r}{2}$.Apply energy conservation between A (zero potential) and P:$\dfrac12 m v_A^2 = \dfrac12 m v_P^2 + mgh$.Substitute $v_P^2 = \dfrac{gr}{2}$ and $h = \dfrac{3r}{2}$:$\dfrac{v_A^2}{2} = \dfrac12 \cdot \dfrac{gr}{2} + g \cdot \dfrac{3r}{2} = \dfrac{gr}{4} + \dfrac{3gr}{2}$.Multiply by 2: $v_A^2 = \dfrac{gr}{2} + 3gr = \dfrac{7gr}{2}$.Thus $v_A = \sqrt{\dfrac{7}{2} g r}$.Answer:Option A: $\sqrt{\dfrac{7}{2} g r}$.