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) [figure]
- 72gr
- 4gr
- 52gr
- 5gr
Answer
(1) 72gr
Solution
At the given point the thread makes 30 above the horizontal, so it is at height r 30 = r2 above the centre.
Radial component of gravity (towards centre) = mg 30
With T = 0: mg 30 = mv2r
v2 = gr2
Height of this point above A = r + r2 = 3r2
Energy conservation between A and this point:
12mvA2 = 12mv2 + mg ( 3r2 )
vA2 = gr2 + 3gr = 72gr
vA = 72gr
Answer: (1)