All subjects · Mathematics · WME02 Mechanics 2

Centre of mass of a set of particles

Edexcel IAL Mathematics · WME02 Mechanics 2

The centre of mass is the point where the whole mass of a system can be treated as acting. For particles, it is a weighted average of their positions, with the masses as weights.

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One dimension

For particles of mass m_1, m_2, ... on a line at positions x_1, x_2, ...:

x̄ = Σ m x / Σ m.

This comes from moments: the total moment of the separate weights about O equals the moment of the total weight acting at x̄.

Two dimensions

Do the same for each coordinate: x̄ = Σ m x / Σ m and ȳ = Σ m y / Σ m. In vector form, r̄ = Σ m r / Σ m. A table with columns for mass, x and y keeps the working clear.

Unknowns

If the centre of mass is given, set up the same equation with the unknown mass or position and solve. A uniform rod can be treated as a particle at its midpoint with the rod's mass.

Worked example

Particles of mass 2 kg, 3 kg and 5 kg are at (1, 2), (4, 0) and (2, 6). Find their centre of mass.

  1. Σm = 10.
  2. x̄ = (2×1 + 3×4 + 5×2)/10 = 24/10 = 2.4.
  3. ȳ = (2×2 + 3×0 + 5×6)/10 = 34/10 = 3.4.

Answer: (2.4, 3.4)

Key idea: x̄ = Σmx/Σm and ȳ = Σmy/Σm. Treat a uniform rod as a particle at its midpoint.

Check yourself

  1. Particles of mass 3 kg and 5 kg are fixed to a light rod at x = 0 and x = 4 m. Find the distance of the centre of mass from the 3 kg particle.
    Show answer

    2.5 m. x̄ = (3×0 + 5×4)/8 = 2.5 m.

  2. Particles of mass 2 kg, 4 kg and m kg are at x = 0, 3 m and 6 m on a light rod. The centre of mass is at x = 4 m. Find m.
    Show answer

    6 kg. (0 + 12 + 6m)/(6 + m) = 4, so 12 + 6m = 24 + 4m and m = 6.

  3. Particles of mass 1 kg, 2 kg and 3 kg are at (0, 0), (3, 0) and (0, 4). Find the y-coordinate of their centre of mass.
    Show answer

    2. ȳ = (0 + 0 + 12)/6 = 2.

  4. Particles of mass 2 kg, 3 kg and 5 kg have position vectors (i + j), (4i - j) and (2i + 3j) m. The centre of mass is at (2.4i + qj) m. Find q.
    Show answer

    1.4. Σm r = (2 + 12 + 10)i + (2 - 3 + 15)j = 24i + 14j. Dividing by 10: 2.4i + 1.4j, so q = 1.4.

  5. A uniform rod AB of length 2 m and mass 4 kg has a particle of mass 1 kg attached at B. Find the distance of the centre of mass from A.
    Show answer

    1.2 m. Rod as 4 kg at 1 m from A: x̄ = (4×1 + 1×2)/5 = 1.2 m.

  6. A 3 kg particle is at (2, 1) and a 1 kg particle is at (a, 5). Their centre of mass is at (2.5, b). Find a.
    Show answer

    4. (6 + a)/4 = 2.5, so a = 4.

  7. Two equal masses are at the ends of a light rod. Where is their centre of mass?
    • At either end
    • Two thirds of the way along
    • It cannot be found without the length
    • At the midpoint of the rod
    Show answer

    At the midpoint of the rod. Equal weights balance about the midpoint.

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