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

Step 1: Understanding the Question:
We need to find the Moment of Inertia (M.I.) of a disc for a specific axis (its diameter).
The topic is Rotational Dynamics.
Step 2: Key Formula or Approach:
We use the Perpendicular Axis Theorem: \( I_z = I_x + I_y \).
For a flat disc, \( I_x \) and \( I_y \) represent the M.I. about two perpendicular diameters.
Step 3: Detailed Explanation:
The M.I. of a disc about an axis passing through its center and perpendicular to its plane is known:
\[ I_z = \frac{1}{2}MR^2 \]
Due to circular symmetry, the M.I. about any diameter is the same:
\[ I_x = I_y = I_{\text{diameter}} \]
Substituting these into the theorem:
\[ I_z = I_{\text{diameter}} + I_{\text{diameter}} \]
\[ \frac{1}{2}MR^2 = 2I_{\text{diameter}} \]
Step 4: Final Answer:
\[ I_{\text{diameter}} = \frac{1}{4}MR^2 \]