In case of vertical circular motion of a particle by a thread of length $r$ if the tension in the th

Question

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)

Professor PC Thomas Classes · Answer

Based on standard conventions for this classic problem, if the tension becomes zero when the string makes an angle of $30^\circ$ with the horizontal (which is $60^\circ$ from the topmost vertical position), we apply Newton's second law at that point: $T + mg\cos(60^\circ) = \frac{mv^2}{r}$. Setting $T=0$ yields $v^2 = rg\cos(60^\circ) = \frac{gr}{2}$. The height of this point from the bottom is $h = r + r\sin(30^\circ) = 1.5r$. Applying the conservation of mechanical energy from the bottom (point A) to this point gives $\frac{1}{2}mv_A^2 = \frac{1}{2}mv^2 + mgh \implies v_A^2 = v^2 + 2gh = \frac{gr}{2} + 2g(1.5r) = \frac{gr}{2} + 3gr = \frac{7gr}{2}$. Thus, $v_A = \sqrt{\frac{7}{2}gr}$.