A bead of '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

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Solution

Q.
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) [2025]

1 $2 \sqrt{g R + \frac{k R^{2}}{m}}$

2 $\sqrt{2 R g + \frac{4 k R^{2}}{m}}$

3 $\sqrt{2 R g + \frac{k R^{2}}{m}}$

4 $\sqrt{3 R g + \frac{k R^{2}}{m}}$

Ans.
(4)

Work done by gravity = mg (2R – R cos 60°) = $\frac{3 m g R}{2}$

Work done by spring $= - \frac{1}{2} k (0^{2} - R^{2}) = \frac{1}{2} k R^{2}$

Net work = Change in kinetic energy

i.e., $\frac{3 m g R}{2} + \frac{k R^{2}}{2} = \frac{1}{2} m v^{2}$

or $v^{2} = 3 g R + \frac{k R^{2}}{m}$

or $v = \sqrt{3 g R + \frac{k R^{2}}{m}}$

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