All subjects · Mathematics · WME02 Mechanics 2

Composite laminas and laminas with holes

Edexcel IAL Mathematics · WME02 Mechanics 2

Most M2 lamina questions use shapes built from rectangles, triangles and circles, or with pieces cut out. You treat each part as a particle at its own centre of mass, with mass proportional to its area.

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Mass proportional to area

For a uniform lamina, mass is proportional to area, so you can use areas in place of masses. Make a table: each part, its area, and the coordinates of its centre of mass. Then x̄ = Σ(area × x)/Σ area.

Removed parts

A hole is treated as a negative area. For a square with a circle cut out: use the whole square (positive) and the circle (negative). The total area is square minus circle.

Choose good axes

Put the origin at a corner, or on an axis of symmetry so one coordinate is known straight away. State which point and which directions you measure from, because later parts (such as hanging the lamina) depend on it.

Worked example

A uniform rectangular lamina OABC has OA = 8 cm along the x-axis and OC = 6 cm along the y-axis. A 2 cm by 2 cm square is removed from the corner at B. Find the centre of mass of the remaining lamina.

  1. Rectangle: area 48, centre (4, 3). Square: area 4, centre (7, 5).
  2. x̄ = (48×4 - 4×7)/44 = 164/44 = 3.73 cm.
  3. ȳ = (48×3 - 4×5)/44 = 124/44 = 2.82 cm.

Answer: (3.73, 2.82)

Key idea: Area acts as mass for a uniform lamina. Holes are negative areas. Tabulate area and centre of each part.

Check yourself

  1. An L-shaped uniform lamina is made of a 6 by 2 rectangle (0 ≤ x ≤ 6, 0 ≤ y ≤ 2) and a 2 by 4 rectangle (0 ≤ x ≤ 2, 2 ≤ y ≤ 6). Find x̄.
    Show answer

    2.2. x̄ = (12×3 + 8×1)/20 = 44/20 = 2.2.

  2. A uniform square lamina of side 10 cm has corners at (0, 0) and (10, 10). A circle of radius 2 cm centred at (7, 5) is cut out. Find x̄, to 3 significant figures.
    Show answer

    4.71 cm. x̄ = (100×5 - 4π×7)/(100 - 4π) = 4.71 cm.

  3. A uniform lamina is made of a 4 cm wide, 3 cm tall rectangle and a semicircle of radius 2 cm whose diameter is the top edge of the rectangle, with the semicircle outside the rectangle. Find the height of its centre of mass above the bottom edge, to 3 significant figures.
    Show answer

    2.31 cm. Rectangle: area 12 at height 1.5. Semicircle: area 2π at height 3 + 8/(3π). ȳ = 2.31 cm.

  4. A uniform square lamina of side 6 cm has corners (0, 0) and (6, 6). The triangle with vertices (0, 0), (6, 0) and (3, 3) is removed. Find ȳ for the remaining lamina, to 3 significant figures.
    Show answer

    3.67 cm. Square: 36 at y = 3. Triangle: 9 at y = (0 + 0 + 3)/3 = 1. ȳ = (108 - 9)/27 = 3.67 cm.

  5. A uniform rectangular lamina of mass 4 kg has corners (0, 0) and (6, 4). A particle of mass 2 kg is attached at (6, 4). Find x̄.
    Show answer

    4. x̄ = (4×3 + 2×6)/6 = 4.

  6. Why can areas be used in place of masses for a uniform lamina?
    • Laminas have no mass
    • Mass is proportional to area when the density is the same throughout
    • The area always equals the mass
    • Gravity does not act on laminas
    Show answer

    Mass is proportional to area when the density is the same throughout. Uniform means equal mass per unit area, so mass = k × area and k cancels.

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