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 the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):
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In problems involving energy conservation, always account for both kinetic and potential energy.
The bead moves under the influence of gravity and the restoring force from the spring. Step 1: Apply energy conservation. At the top of the hoop, the bead has potential energy due to gravity and the spring. At the bottom, the bead will have kinetic energy. Step 2: At the top, the potential energy is: \[ U_{\text{top}} = mgh = mg(2R) \] where \( h = 2R \) is the height of the bead from the bottom. Step 3: The spring potential energy at the top is zero since the spring is at its equilibrium length. Step 4: At the bottom, the kinetic energy is: \[ K_{\text{bottom}} = \frac{1}{2} m v^2 \] The spring at the bottom has a compression of \( R \), so the spring potential energy is: \[ U_{\text{spring}} = \frac{1}{2} k R^2 \] Step 5: Apply conservation of mechanical energy: \[ m g (2R) = \frac{1}{2} m v^2 + \frac{1}{2} k R^2 \] Step 6: Solve for \( v \): \[ v = \sqrt{\frac{2gR + kR^2}{m}} \] Final Conclusion: The velocity of the bead when the spring becomes \( R \) is given by \( \sqrt{\frac{2gR + kR^2}{m}} \), which is Option (3).
