Question

The moment of inertia of a disc of mass \(m\) and radius \(R\) about an axis, which is tangential to the circumference of the disc and parallel to its diameter is

• A. \(\frac{3}{2} mR^2\)
• B. \(\frac{2}{3} mR^2\)
• C. \(\frac{5}{4} mR^2\)
• D. \(\frac{4}{5} mR^2\)

Hint:

For disc: \(I_{\text{about diameter}} = \frac{1}{4}mR^2\), \(I_{\text{about perpendicular axis through centre}} = \frac{1}{2}mR^2\).

Answer 1

The Correct Option is C

Step 1: Concept:
Use parallel axis theorem: \(I = I_{\text{cm}} + m d^2\). For disc about diameter, \(I_{\text{cm}} = \frac{1}{4}mR^2\).
Step 2: Explanation:
Axis is tangential to circumference and parallel to diameter. Distance from centre to tangent = \(R\).
\(I = I_{\text{cm}} + mR^2 = \frac{1}{4}mR^2 + mR^2 = \frac{1}{4}mR^2 + \frac{4}{4}mR^2 = \frac{5}{4}mR^2\).
Step 3: Conclusion:
Thus, \(I = \frac{5}{4}mR^2\).