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 accleration due to gravity)

Solution

Physics sir · 4 steps

Find where the bead is when the spring is back to length R, then do one energy balance from the top.

1. Start at the top. At the top the spring stretches across the full diameter, so its length is 2R. Its natural length is R, so it is stretched by R and stores 1/2kR². The bead starts from rest.
2. Locate the bead. When the spring length is R, the bead sits on the hoop at a chord of length R from the bottom point. A chord equal to the radius makes an equilateral triangle with the centre, so the chord subtends 60° at the centre. The bead is then R - R cos 60° = R/2 above the bottom, so it has dropped 2R - R/2 = 3R/2.

   1/2kR² + mg(3R/2) = 1/2mv²

3. Energy balance. The hoop is frictionless and its normal force does no work, so mechanical energy is conserved. At this point the spring is unstretched and stores nothing. All the spring energy and the lost gravitational energy become kinetic energy.
4. So. v² = 3gR + kR²/m, so v = √(3Rg + kR²/m), which is option (b).

The idea to keep. With a spring tied to a point on a circle, first use geometry to find where the bead is. A chord equal to the radius always subtends 60° at the centre. Then one energy equation finishes the problem.

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