Question

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): 

• A. \( \sqrt{\frac{3Rg + kR^2}{m}} \)
• B. \( \sqrt{\frac{2Rg + kR^2}{m}} \)
• C. \( \sqrt{\frac{2gR + kR^2}{m}} \)
• D. \( \sqrt{\frac{2Rg + 4kR^2}{m}} \)

Hint:

In problems involving energy conservation, always account for both kinetic and potential energy.

Answer 1

The Correct Option is C

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).