For a semicircular plate of diameter D and radius R, with y as the vertical axis and x as the horizontal axis passing through the diameter, the moment of inertia about the y-axis will be:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 07/06/2024 (Morning)
Detailed Explanation
A full circle has I = πD⁴/64 about any diameter. A semicircle is exactly half of that circle, and both of its main axes through the centre of the diameter — the diameter itself (x) and the vertical axis of symmetry (y) — split the area symmetrically.
\[ I_{x} = I_{y} = \tfrac{1}{2}\times\frac{\pi D^4}{64} = \frac{\pi D^4}{128} = \frac{\pi R^4}{8} \]
Why “πD⁴/128” Is Correct
About the y-axis (and also about the diameter), the semicircle contributes exactly half of the circle's value, giving πD⁴/128. That is option (d).
Why the Other Options Are Wrong
- A. 0.10976 R³: R³ has the wrong dimensions; moment of inertia is length to the fourth power.
- B. πD⁴/16: πD⁴/16 is far larger than even a full circle about its centre (πD⁴/32).
- C. 0.10976 R⁴: 0.1098R⁴ is about the centroidal axis parallel to the diameter, not about the y-axis.
Key Point
Semicircle: I about diameter = I about symmetry axis = πD⁴/128 = πR⁴/8.
Key Concepts for Students
- Centroid of a semicircle: 4R/3π from the diameter.
- Centroidal I (parallel to diameter) = πR⁴/8 − (πR²/2)(4R/3π)² ≈ 0.1098R⁴.
- Units of area moment of inertia: mm⁴ or m⁴.
Practice more: this is question 1 in Moment MCQs – Part 1. Try the whole set and check every answer instantly.