In case of vertical circular motion of a particle by a thread of length $r$ if the tension in the thread is zero at an angle $30^\circ$ shown in figure, the velocity at the bottom point $(A)$ of the circular path is (g = gravitational acceleration)
Answer: (A) $\sqrt{\dfrac72gr}$
At the point shown the thread makes $30^\circ$ above the horizontal, so the particle is $r\sin30^\circ = \dfrac r2$ above the centre, i.e. $\dfrac{3r}{2}$ above $A$.
With $T = 0$, only the component of gravity along the thread provides the centripetal force:
$$mg\sin30^\circ = \frac{mv^2}{r}\Rightarrow v^2 = \frac{gr}{2}$$
Energy conservation from $A$:
$$v_A^2 = v^2 + 2g\cdot\frac{3r}{2} = \frac{gr}2 + 3gr = \frac72gr$$
$$v_A = \sqrt{\frac72gr}$$
Solution by Sreeraj P, M.Sc Physics