Question

A solid disc of mass \( m \), radius \( r \) is resting on a horizontal smooth surface. A spring of stiffness \( k \) is connected to the disc at distance \( e \) directly above the centre of the disc. Another end of the spring is connected to the vertical wall. For small angular displacement of the disc, the natural frequency of the system in radians per second will be:

• A. \( \sqrt{\frac{2k(r+e)}{3mr}} \)
• B. \( \sqrt{\frac{2k(r+e)}{mr}} \)
• C. \( \sqrt{\frac{2k(r+e)^2}{mr^2}} \)
• D. \( \sqrt{\frac{2k(r+e)^2}{3mr^2}} \)

Hint:

When solving problems involving rotational motion, the natural frequency of oscillation depends on both the stiffness of the spring and the moment of inertia of the body. The moment of inertia for a disc about its center is 1/2mr2 , and for angular displacement problems, the effective length from the center must be considered.

Answer 1

The Correct Option is D

The natural frequency of a system with rotational motion can be found using the formula for the torsional pendulum. For small angular displacement, the restoring force is proportional to the angular displacement, and the equivalent moment of inertia for the system is considered. The effective spring constant for rotational motion is adjusted by considering the location of the spring above the centre of mass of the disc.

Using energy principles and angular dynamics, the natural frequency \( \omega \) is given by:

\( \omega = \sqrt{\frac{k(r+e)^2}{I}} \)

where \( I = \frac{1}{2}mr^2 \) is the moment of inertia of the disc. Simplifying the expression:

\( \omega = \sqrt{\frac{2k(r+e)^2}{3mr^2}} \)

Thus, the correct answer is Option (4).