Centre of gravity of a quadrant (area) of a circle For a plane lamina forming a quadrant of a circle (radius R), the centre of gravity lies along its central radius at approximately what distance from the centre?

Introduction / Context:Centres of gravity of standard plane areas are used constantly in structural analysis, plate design, and balanced machining. A common case is the quadrant of a circle (a 90° sector), where a simple approximate factor is widely memorized for quick checks.

Given Data / Assumptions:

• Plane lamina, uniform thickness and density.
• Area is a quadrant (sector of 90°) of a full circle of radius R.
• We seek distance along the sector’s central radius from the circle center.

Concept / Approach:

The exact centroidal distance for a circular sector of angle 2α (radians) is r̄ = 2R * sin(α) / (3α). For a quadrant, α = π/4, so the distance simplifies to r̄ = 4R / (3π) ≈ 0.424 R. In practice, options often provide rounded values; 0.4 R is the closest listed.

Step-by-Step Solution:

Verification / Alternative check:

Reference tables of centroids list the quadrant centroid at 4R/(3π). Comparing to answers, 0.4 R is the nearest permissible option.

Why Other Options Are Wrong:

(a) 0.2 R and (b) 0.3 R are too close to the center; (d) 0.5 R and (e) 0.6 R are too far from the center for an area concentrated near the corner arc.

Common Pitfalls:

Mixing the centroid of a circular arc with that of a sector; confusing 0.424 R with 0.5 R.

Final Answer:

0.4 R