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 rotationally-moving system is determined using the torsional pendulum formula. This applies when angular displacement is small, making the restoring force proportional to displacement. The calculation involves the system's equivalent moment of inertia and an adjusted effective spring constant, which accounts for the spring's position relative to the disc's center of mass.

Based on energy principles and angular dynamics, the natural frequency \( \omega \) is expressed as:

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

Here, \( I = \frac{1}{2}mr^2 \) represents the disc's moment of inertia. Upon simplification, the expression becomes:

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

Therefore, the correct answer is Option (4).