Centre of gravity of a quadrant of a circle For a plane quadrant of a circle of radius r, the centre of gravity (centroid) lies along the central radius at a distance from the circle's center equal to:
-
A4r / (3π)
-
Br / 2
-
C0.6 r
-
D3r / 4
-
Er / π
Answer
Correct Answer: 4r / (3π)
Explanation
Introduction / Context:Locating centroids of standard plane figures is essential for calculating bending stresses, locating neutral axes, and setting up composite-area problems in mechanics of materials.
Given Data / Assumptions:
- Figure: a quadrant (one-fourth) of a circle, radius r.
- Uniform area density (homogeneous lamina).
- Centroid lies along the central radius (the bisector of the right angle).
Concept / Approach:The centroid of a circular arc, sector, and quadrant are standard results. For a quadrant of a circle (area), the distance of the centroid from the circle center along the bisector is x_c = 4r / (3π) ≈ 0.424 r. This is obtained by integrating the area’s first moments or from tabulated centroid formulae used in engineering handbooks.
Step-by-Step Solution:
Define polar element: dA = r dθ dr over θ ∈ [0, π/2], r ∈ [0, R].Use symmetry to conclude centroid lies on the bisector; only radial distance is unknown.Compute first moment along the bisector and divide by total area (π r^2 / 4).Result simplifies to x_c = 4r / (3π).Verification / Alternative check:Compare with numerical value: 4/(3π) ≈ 0.424 < 0.5, hence centroid is inside the quadrant closer to the corner than to the midpoint of the radius, which matches intuition.
Why Other Options Are Wrong:
- r/2, 0.6 r, 3r/4, r/π: Do not match the standard derived centroid position; several are dimensionally fine but numerically incorrect.
Common Pitfalls:Confusing centroid of a quadrant (area) with that of a circular arc; the arc centroid distance is 2r/π ≈ 0.637 r, not applicable here.
Final Answer:4r / (3π)