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 $30 °$ shown in figure, the velocity at the bottom point (A) of the circular path is

(g = gravitational acceleration) [2026]

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 $30 °$ shown in figure, the velocity at the bottom point (A) of the circular path is

(g = gravitational acceleration) [2026]

Smart Achievers · Accepted Answer

(3)

$T + m g \cos 60 ° = \frac{m V^{2}}{ℓ}$

$T = 0$

$V^{2} = \frac{g ℓ}{2} where V is the speed at point A$

$M.E.C.$

$\frac{1}{2} m u^{2} = m g (ℓ + ℓ \cos 60 °) + \frac{1}{2} m V^{2}$

$u^{2} = 3 g ℓ + \frac{g ℓ}{2}$

$u = \sqrt{\frac{7 g ℓ}{2}}$

Solution

Q.
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 $30 °$ shown in figure, the velocity at the bottom point (A) of the circular path is

(g = gravitational acceleration) [2026]

1 $\sqrt{4 g r}$

2 $\sqrt{5 g r}$

3 $\sqrt{\frac{7}{2} g r}$

4 $\sqrt{\frac{5}{2} g r}$

Ans.
(3)

$T + m g \cos 60 ° = \frac{m V^{2}}{ℓ}$

$T = 0$

$V^{2} = \frac{g ℓ}{2} where V is the speed at point A$

$M.E.C.$

$\frac{1}{2} m u^{2} = m g (ℓ + ℓ \cos 60 °) + \frac{1}{2} m V^{2}$

$u^{2} = 3 g ℓ + \frac{g ℓ}{2}$

$u = \sqrt{\frac{7 g ℓ}{2}}$

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Solution

Q. 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° shown in figure, the velocity at the bottom point (A) of the circular path is (g = gravitational acceleration) [2026]

1 4gr

2 5gr

3 72gr

4 52gr

Ans. (3) T+mgcos60°=mV2ℓ T=0 V2=gℓ2 where V is the speed at point A M.E.C. 12mu2=mg(ℓ+ℓcos60°)+12mV2 u2=3gℓ+gℓ2 u=7gℓ2