Circular MotionHard
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)
Options
A.$\sqrt{5gr}$
B.$\sqrt{\frac{7}{2}gr}$
C.$\sqrt{4gr}$
D.$\sqrt{\frac{5}{2}gr}$
Solution
$${T + mgcos60^{\circ} = \frac{{mV}^{2}}{\mathcal{l}} }{T = 0 }$$$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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