Question

The figure shows a disc of mass \( m \) and radius \( R \) hinged at point 'A' on its periphery and free to oscillate about the axis. Find the time period for small oscillations of the disc: 

• A. \( 2\pi \sqrt{\frac{3R}{2g}} \)
• B. \( 2\pi \sqrt{\frac{R}{g}} \)
• C. \( 2\pi \sqrt{\frac{5R}{4g}} \)
• D. \( 2\pi \sqrt{\frac{R}{4g}} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a case of a physical pendulum. A rigid body oscillating about a fixed horizontal axis has a time period determined by its moment of inertia about that axis and the distance of the center of mass from the hinge.

Step 2: Key Formula or Approach:
Time period of a physical pendulum: \[ T = 2\pi \sqrt{\frac{I}{mgL}} \] where \( I \) is the moment of inertia about the hinge and \( L \) is the distance from the hinge to the center of mass.

Step 3: Detailed Explanation:
1. Distance from hinge (A) to Center of Mass (C): \( L = R \). 2. Moment of inertia about center: \( I_{cm} = \frac{1}{2}mR^2 \). 3. Using Parallel Axis Theorem for hinge A: \[ I_A = I_{cm} + mL^2 = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2 \] 4. Calculate Time Period: \[ T = 2\pi \sqrt{\frac{\frac{3}{2}mR^2}{mgR}} = 2\pi \sqrt{\frac{3R}{2g}} \]

Step 4: Final Answer:
The time period for small oscillations is \( 2\pi \sqrt{\frac{3R}{2g}} \).