A semicircular lamina has radius R and diameter D. Determine its moment of inertia about the diametrical axis.
Try it first: pick an option, then read the explanation.
Correct answer: D. πR⁴/8
Asked in SSC JE 10/10/2023 (Morning)
Detailed Explanation
About its diameter, a semicircle has half the moment of inertia of a full circle about that diameter.
\[ I = \frac{\pi D^4}{128} = \frac{\pi (2R)^4}{128} = \frac{16\pi R^4}{128} = \frac{\pi R^4}{8} \]
Why “πR⁴/8” Is Correct
Writing the answer in terms of the radius gives πR⁴/8, option (d).
Why the Other Options Are Wrong
- A. πD⁴/32: πD⁴/32 is the polar moment of a full circle.
- B. πR⁴/16: πR⁴/16 is a quarter circle about one of its straight edges.
- C. πD⁴/64: πD⁴/64 is a full circle about its diameter.
Key Point
Semicircle about diameter: πR⁴/8 = πD⁴/128.
Key Concepts for Students
- Quarter circle about a straight edge: πR⁴/16.
- Check answers by converting D = 2R.
- Halving the area about the same axis halves I when the axis splits it symmetrically.
Practice more: this is question 5 in Moment MCQs – Part 1. Try the whole set and check every answer instantly.