All subjects · Mathematics · WME02 Mechanics 2

Centres of mass of wire frameworks

Edexcel IAL Mathematics · WME02 Mechanics 2

A framework is made from uniform wire or rods. Each piece acts as a particle at its own centre of mass, with mass proportional to its **length**, not its area. Pearson's examiners report that using areas here is a common error.

Practise this in the app: instant marking and a fix for every mistake. Free, no email needed to start.

Practise this lesson

Mass proportional to length

For uniform wire, mass is proportional to length. A straight piece has its centre of mass at its midpoint. A curved piece uses the arc result r sin α/α (semicircle: 2r/π).

Method

1. List every piece of wire, its length and the position of its midpoint (or arc centre of mass).

2. x̄ = Σ(length × x)/Σ length, and the same for ȳ.

Check you have included every side, including any diameter or radius pieces.

Lamina or framework?

Read the question carefully. "A uniform lamina" means use areas. "A uniform wire" or "framework" means use lengths. A semicircular lamina and a semicircular wire have different centres of mass (4r/(3π) and 2r/π).

Worked example

A uniform wire is bent into a triangle OAB with O(0, 0), A(8, 0), B(0, 6) (units cm). Find its centre of mass.

  1. Lengths: OA = 8, OB = 6, AB = 10. Midpoints: (4, 0), (0, 3), (4, 3).
  2. x̄ = (8×4 + 6×0 + 10×4)/24 = 72/24 = 3.
  3. ȳ = (8×0 + 6×3 + 10×3)/24 = 48/24 = 2.

Answer: (3, 2), which is not the same as the centroid of the triangular lamina, (8/3, 2)

Key idea: Wire: mass ∝ length, each piece at its midpoint (arc: r sin α/α). Lamina: mass ∝ area.

Check yourself

  1. A uniform wire forms three sides of a square of side 2 cm: from (0, 0) to (0, 2), to (2, 2), to (2, 0). Find ȳ, to 3 significant figures.
    Show answer

    1.33 cm. Three pieces of length 2 with midpoints (0, 1), (1, 2), (2, 1). ȳ = (2 + 4 + 2)/6 = 1.33 cm.

  2. A uniform wire is bent into a semicircular arc of radius 5 cm together with its diameter. Find the distance of its centre of mass from the diameter, to 3 significant figures.
    Show answer

    1.94 cm. Arc: length 5π at 10/π from the diameter. Diameter: length 10 at 0. ȳ = (5π × 10/π)/(5π + 10) = 50/(5π + 10) = 1.94 cm.

  3. A uniform wire is bent into the triangle with vertices O(0, 0), A(6, 0), B(6, 8). Find ȳ.
    Show answer

    3. Lengths 6, 8, 10 with midpoints (3, 0), (6, 4), (3, 4). ȳ = (0 + 32 + 40)/24 = 3.

  4. A uniform wire is bent into the boundary of a quarter circle of radius 6 cm (the arc and both radii). Find the distance of its centre of mass from the centre of the circle, measured along the axis of symmetry, to 3 significant figures.
    Show answer

    3.56 cm. Arc: length 3π at 5.40 along the axis. Each radius: length 6, midpoint 3 cm from O at 45° to the axis, so 3 cos 45° along it. Distance = (3π × 5.40 + 12 × 2.12)/(3π + 12) = 3.56 cm.

  5. Find the distance of the centre of mass of a uniform semicircular wire of radius 3 cm from its centre, to 3 significant figures.
    Show answer

    1.91 cm. 2r/π = 6/π = 1.91 cm.

  6. In a framework made of uniform wire, what is the mass of each piece proportional to?
    • Its area
    • The square of its length
    • The angle it makes with the x-axis
    • Its length
    Show answer

    Its length. Uniform wire has the same mass per unit length.

Practise with instant feedback

Sources

Last updated 8 November 2026. Spotted a mistake? Tell us.