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)

Solution

Physics sir · 3 steps

Where the thread goes slack, gravity alone gives the centripetal pull. Then use energy down to A.

1. Slack thread condition. At 30° above the horizontal, the part of gravity along the thread towards the centre is mg sin 30°. With T = 0, this alone provides mv²/r, so v² = gr sin 30° = gr/2.
2. Height above A. The particle is r sin 30° = r/2 above the centre. The centre is r above A, so the particle is r + r/2 = 3r/2 above the bottom point.

   1/2mu² = 1/2mv² + mg(3r/2) ⇒ u² = gr/2 + 3gr = 7/2gr

3. So. The speed at A is u = √(7/2gr), which is option (a).

The idea to keep. Where the string slackens, set T = 0 and keep only the part of gravity along the string. Then link that point to the bottom by energy conservation, measuring height from the bottom.

Answer from the NTA final answer key. Solution by PrepWiser.