Find the centroid of a uniform circular wire made from a wire with total length of 10x meters in a form of: a. A semi-circle b. Quadrant of a circle

Answer 1

Calculus solution:

1. Place the wire in a convenient coordinate system and position.
2. For a centroid, the mass is always assumed uniformly distributed, so assume we have unit density.
3. Integrate.

a. The radius is r=10x/π. The semicircle can be taken to have 0 ≤ θ ≤ π.

x_G = 0 by symmetry

y_G = (1/π) ∫₀^π y(θ) dθ = (1/π) ∫₀^π r sin θ dθ = -(r/π)cos θ |₀^π = 2r/π = 20x/π²

b. The radius is r=20x/π. The quadrant can be taken to have π/4 ≤ θ ≤ 3π/4.

x_G = 0 by symmetry

y_G = (2/π) ∫³_π^π_/^/₄⁴ y(θ) dθ = (2/π) ∫³_π^π_/^/₄⁴ r sin θ dθ = -(2r/π)cos θ |³_π^π_/^/₄⁴ = (2√2)r/π = (40√2)x/π²

Non-Calculus Solution:

We may use Pappus' Centroid theorem for surface area:

The surface area of the surface of revolution of a wire about an axis equals the length of the wire times circumference swept by the wire's centroid.

We can take the axis to be one of two radii to a wire endpoint. Denote the circumference swept by the centroid by C.

a.

The wire length is 10x. Surface of revolution is a full sphere of radius r=10x/π.

Surface area is 4πr² = 400x²/π = 10xC. So C = 40x/π.

The centroid sweeps a circle of radius C/(2π) = 20x/π²

So the centroid is on the axis of symmetry and a distance 20x/π² inward from the circle center.

b.

The wire length is 10x. Surface of revolution is a hemisphere of radius r=20x/π.

Surface area is 2πr² = 800x²/π = 10xC. So C = 80x/π.

The centroid sweeps a circle of radius C/(2π) = 40x/π²

This is the distance from either radius.

So the centroid is on the axis of symmetry and a distance (40√2)x/π² inward from the circle center.

Notice that both approaches agree.