Question

Calculate the moment of inertia of a uniform circular disc of mass \(M\) and radius \(R\) about its diameter.

• A. \( \frac{1}{2}MR^2 \)
• B. \( \frac{1}{4}MR^2 \)
• C. \( MR^2 \)
• D. \( \frac{3}{2}MR^2 \)

Hint:

For circular discs, remember: Center perpendicular axis \(=\frac{1}{2}MR^2\). Diameter axis \(=\frac{1}{4}MR^2\) using the perpendicular axis theorem.

Answer 1

The Correct Option is B

Concept: For a uniform circular disc: - Moment of inertia about an axis perpendicular to the plane and passing through the center is \[ I_z = \frac{1}{2}MR^2 \] Using the Perpendicular Axis Theorem \[ I_z = I_x + I_y \] Since the disc is symmetric, \[ I_x = I_y \]

Step 1: Apply the perpendicular axis theorem. \[ \frac{1}{2}MR^2 = I_x + I_y \] \[ \frac{1}{2}MR^2 = 2I_x \]

Step 2: Solve for the moment of inertia about the diameter. \[ I_x = \frac{1}{4}MR^2 \] Final Answer: \[ I = \frac{1}{4}MR^2 \]