Centres of mass of standard laminas and arcs

Use symmetry first

A uniform lamina's centre of mass lies on every axis of symmetry. So a rectangle's or circle's centre of mass is at its centre, and an isosceles triangle's is on its line of symmetry. Spec 2.2 says integration is not required.

Results from the formulae booklet

• Triangular lamina: 2/3 of the way along a median from the vertex. With coordinates, this is the mean of the three vertices: ((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3).
• Sector of radius r with angle 2α at the centre: 2r sin α / (3α) from the centre, on the axis of symmetry.
• Circular arc (a wire) of radius r with angle 2α at the centre: r sin α / α from the centre.

α is half the angle at the centre and must be in radians.

Useful special cases

• Semicircular lamina (α = π/2): 4r/(3π) from the centre.
• Semicircular arc (α = π/2): 2r/π from the centre.
• Quarter circle lamina (α = π/4): measured along the line of symmetry at 45°.

Worked example

Find the distance of the centre of mass of a uniform sector of radius 6 cm and angle 2π/3 at the centre from the centre of the circle.

1. Angle at the centre 2α = 2π/3, so α = π/3.
2. x̄ = 2r sin α/(3α) = 2(6) sin(π/3)/(3 × π/3) = 12(0.866...)/π.

Answer: 3.31 cm from the centre, on the axis of symmetry

Check yourself

1. A uniform triangular lamina has vertices (0, 0), (6, 0) and (3, 9). Find the y-coordinate of its centre of mass.
   Show answer

   3. ȳ = (0 + 0 + 9)/3 = 3.

2. Find the distance of the centre of mass of a uniform semicircular lamina of radius 9 cm from the centre of its straight edge, to 3 significant figures.
   Show answer

   3.82 cm. 4r/(3π) = 36/(3π) = 3.82 cm.

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

   3.18 cm. 2r/π = 10/π = 3.18 cm.

4. A uniform lamina is a quarter of a circle of radius 6 cm. Find the distance of its centre of mass from the centre of the circle, measured along its axis of symmetry, to 3 significant figures.
   Show answer

   3.60 cm. α = π/4: 2(6) sin(π/4)/(3π/4) = 3.60 cm.

5. A uniform wire is bent into an arc of radius 4 cm subtending an angle of π/3 at the centre. Find the distance of its centre of mass from the centre, to 3 significant figures.
   Show answer

   3.82 cm. α = π/6: r sin α/α = 4(0.5)/(π/6) = 12/π = 3.82 cm.

6. When using 2r sin α/(3α) for a sector, how must α be measured?
   • In degrees, as half the angle at the centre
   • In radians, as the whole angle at the centre
   • In radians, as half the angle at the centre
   • In degrees, as the whole angle
   Show answer

   In radians, as half the angle at the centre. The formula uses α in radians and α is half the sector angle.

7. Where is the centre of mass of a uniform triangular lamina?
   • Two thirds of the way along a median from the vertex
   • Half way along a median
   • One third of the way along a median from the vertex
   • At the midpoint of the longest side
   Show answer

   Two thirds of the way along a median from the vertex. The booklet gives 2/3 along a median from the vertex (so 1/3 from the side).

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M2
• Pearson Edexcel IAL Mechanics M2 (WME02) question paper, June 2025: front-cover instructions on g and accuracy
• Pearson Edexcel IAL Mathematics, Further Mathematics and Pure Mathematics: Mathematical Formulae and Statistical Tables (issue 2), Mechanics M2 section

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