The polar moment of inertia of a semicircular lamina of radius R about its central axis (through the centre of the diameter, perpendicular to the plane) is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 10/10/2023 (Morning)
Detailed Explanation
The perpendicular axis theorem says that for a flat lamina, the moment of inertia about an axis perpendicular to its plane equals the sum of the moments of inertia about two perpendicular axes in its plane through the same point: J = Ix + Iy.
At the centre of the diameter: \(I_x = I_y = \dfrac{\pi R^4}{8}\)
\[ J = I_x + I_y = \frac{\pi R^4}{8} + \frac{\pi R^4}{8} = \frac{\pi R^4}{4} \]
Why “πR⁴/4” Is Correct
The result, πR⁴/4, is exactly half the polar moment of a full circle (πR⁴/2), as expected for half the area.
Why the Other Options Are Wrong
- A. πR⁴/2: πR⁴/2 is the polar moment of a full circle.
- B. πR⁴/8: πR⁴/8 is the moment about one in-plane axis only.
- D. πR⁴/16: πR⁴/16 is a quarter circle about one axis.
Key Point
Polar MI J = Ix + Iy. Semicircle at diameter centre: πR⁴/4.
Key Concepts for Students
- Full circle: J = πD⁴/32 = πR⁴/2.
- Polar moment is used in torsion: τ = Tr/J.
- Perpendicular axis theorem applies only to plane laminae.
Practice more: this is question 4 in Moment MCQs – Part 1. Try the whole set and check every answer instantly.