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

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)

Accepted Answer

$${T + mgcos60^{\circ} = \frac{{mV}^{2}}{\mathcal{l}} }{T = 0 }$$$V^{2} = \frac{g\mathcal{l}}{2}$ here V is the speed at point A

M.E.C.

$${\frac{1}{2}{mu}^{2} = mg\left( \mathcal{l} + \mathcal{l}cos60^{\circ} \right) + \frac{1}{2}{mV}^{2} }{u^{2} = 3\text{ }g\mathcal{l} + \frac{g\mathcal{l}}{2} }{u = \sqrt{\frac{7\text{ }g\mathcal{l}}{2}}}$$

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