A bead of mass 'm' slides without friction on the wall of a vertical circular hoop of radius 'R' as shown in figure. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is 'R'. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes 'R', would be (spring constant is 'k', g is acceleration due to gravity) [figure]
- 2Rg + kR2m
- 3Rg + kR2m
- 2Rg + 4kR2m
- 2 gR + kR2m
Answer
(2) 3Rg + kR2m
Solution
At the top, spring length = 2R extension x1 = 2R - R = R.
When the spring length is R: the chord from the bottom point B to the bead P has length R.
With centre O: OB = OP = BP = R OBP is equilateral BOP = 60
Height of P above B = R - R 60 = R2
Fall in height h = 2R - R2 = 3R2, and the spring is now at its natural length.
Normal reaction does no work; by conservation of mechanical energy:
12kR2 + mg ( 3R2 ) = 12mv2
v2 = 3gR + kR2m
v = 3Rg + kR2m
Answer: (2)