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
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
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For disc: \(I_{\text{about diameter}} = \frac{1}{4}mR^2\), \(I_{\text{about perpendicular axis through centre}} = \frac{1}{2}mR^2\).
Updated On: Apr 23, 2026
- \(\frac{3}{2} mR^2\)
- \(\frac{2}{3} mR^2\)
- \(\frac{5}{4} mR^2\)
- \(\frac{4}{5} mR^2\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the 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: Detailed 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: Final Answer:
Thus, \(I = \frac{5}{4}mR^2\).
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: Detailed 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: Final Answer:
Thus, \(I = \frac{5}{4}mR^2\).
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