Centroid of a semicircular arc (mechanics of solids) For a semicircular arc of radius r, determine the perpendicular distance of the centroid (center of gravity of the arc length) from the diameter along the axis of symmetry. State the result in terms of r, keeping the standard result used in engineering mechanics.

Introduction / Context:
The centroid of line elements such as circular arcs is a staple result in engineering mechanics and is frequently used in structural analysis, machine design, and composite area calculations. For a semicircular arc, the centroid lies on the radius of symmetry midway between the arc ends, at some distance above the diameter. Knowing this exact distance allows quick evaluation of moments and equivalent point loads for distributed cables, rings, or curved bars.

Given Data / Assumptions:

• Semi-circular arc of radius r (thin line element, uniform line density).
• Centroid sought along the symmetry axis perpendicular to the diameter.
• Classical planar statics; thickness of the arc is negligible compared with r.

Concept / Approach:

For line elements, centroid coordinates are obtained by integrating position weighted by infinitesimal length ds, then dividing by total length. For a semicircular arc, symmetry gives x̄ = 0 and the centroid lies on the axis through the circle center. The known closed-form result places the centroid at a distance 2r/π from the circle’s center along the radius of symmetry, equivalently measured from the diameter since the diameter passes through the center.

Step-by-Step Solution:

Verification / Alternative check:

The numeric factor 2/π ≈ 0.6366 lies between 0.5 and 1.0, which is sensible: the centroid of the arc must be closer to the center than the arc itself at distance r, yet above the diameter (0).

Why Other Options Are Wrong:

(b) r/π and (d) 4r/π^2 are too small; (c) πr/4 is too large; (e) r/2 underestimates the known exact value.

Common Pitfalls:

Confusing the centroid of a semicircular area (located at 4r/(3π)) with that of a semicircular arc (2r/π); mixing reference points (from center vs from diameter) leads to the same magnitude here because the diameter passes through the center.

Final Answer:

2r/π