Question

The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 

• A. \( \sqrt{\frac{2}{3}\left(\frac{g}{R - r} + \frac{k}{m}\right)} \)
• B. \( \sqrt{\frac{2g}{3(R - r)} + \frac{k}{m}} \)
• C. \( \sqrt{\frac{1}{6}\left(\frac{g}{R - r} + \frac{k}{m}\right)} \)
• D. \( \sqrt{\frac{1}{4}\left(\frac{g}{R - r} + \frac{k}{m}\right)} \) \bigskip

Hint:

For rolling motion inside a ring, consider combined translational and rotational kinetic energy and effective radius for oscillations to find the angular frequency.

Answer 1

The Correct Option is A

Step 1: The disk rolls without slipping inside a ring of radius \( R \). The center of the disk moves on a circle of radius \( R - r \). 
Step 2: The kinetic energy includes translation and rotation: \[ K = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2, \] where \( v = (R - r)\dot{\theta} \). 
Step 3: The potential energy due to gravity and spring for small oscillations is \[ U = \frac{1}{2} m g (R - r) \theta^2 + \frac{1}{2} k (R - r)^2 \theta^2. \] Step 4: The angular frequency \( \omega \) is given by \[ \omega^2 = \frac{\text{restoring torque per unit angle}}{\text{effective moment of inertia}} = \frac{m g (R - r) + k (R - r)^2}{\frac{3}{4} m (R - r)^2} = \frac{4}{3}\left(\frac{g}{R - r} + \frac{k}{m}\right). \] Step 5: Taking square root and matching with options, the correct expression is \[ \boxed{\omega = \sqrt{\frac{2}{3}\left(\frac{g}{R - r} + \frac{k}{m}\right)}}, \] which corresponds to option (A).