Mini Physics
Part of Mini EducationSwitch subject

UY1: Centre Of Mass Of A Right-Angle Triangle

Find the centre of mass of a uniform right-angle triangular lamina using integration and symmetry-aware setup.

  • University Physics Year 1
On this page

Learning objectives

  • Use vector and differential or integral calculus to express physical change and accumulation.
  • Use complex numbers, linear algebra, and differential equations to solve coupled physical models.
  • Choose an efficient mathematical method and validate the result physically.
Cluster role + pathway links

This page is the archived worked-example derivation in the UY1 mechanics pathway.

1) At a glance

  • Prerequisite: x_CM = 1/M∫ x dm, y_CM = 1/M∫ y dm, similar triangles.
  • Outcomes: compute CM of a uniform right-angle triangular lamina.
  • Key result for the triangle shown:
x_CM = 2a/3, y_CM = b/3
Common traps (geometry consistency)
  • When using strips, express the strip height/length from the same triangle geometry every time (don’t mix up which side is shrinking).
  • Keep your vertex choice consistent: the final (x_CM,y_CM) depends on where you place the axes.
A right-angle triangle with vertices at the origin O, at (a, 0) and at (a, b). The base along the x-axis has length a, the vertical side has height b and the hypotenuse is y = (b/a)x. The centre of mass is marked at (2a/3, b/3).
The hypotenuse y = (b/a)x sets every strip length used below; the centre of mass lies at (2a/3, b/3).

2) Setup (model, assumptions, coordinates/signs)

  • Let the triangle have vertices (0,0), (a,0), (a,b).
  • Uniform area density ρ (constant), so M = ρ((1/2)ab).
  • Hypotenuse equation from similar triangles: y = (b/a)x.
  • Coordinates are measured from the shown axes; positive x rightward, positive y upward.
Fast check: centroid of any uniform triangle

The centroid of a uniform triangular lamina is at the average of its vertices:

(x_CM,y_CM) = ((x₁ + x₂ + x₃)/3,(y₁ + y₂ + y₃)/3)

For (0,0), (a,0), (a,b) this gives (2a/3,b/3), matching the integration result.

3) Core method / derivation

x_CM = 1/M∫ x dm; y_CM = 1/M∫ y dm

For x_CM, use vertical strips of width dx and height y = (b/a)x:

dm = ρ y dx = ρ(b/a)x dx

Hence:

x_CM = 1/M∫₀^a x dm; = (ρ/M)b/a∫₀^a x² dx; = (ρ/M)(b/a)a³/3

Since ρ = 2M/ab:

x_CM = 2a/3
The triangle with a thin vertical strip of width dx at distance x from the origin. The strip's height is y = (b/a)x and its mass is dm.
Every point in a vertical strip has the same x, so each strip contributes x dm to the integral for the x-coordinate of the centre of mass.

For y_CM, use horizontal strips of thickness dy. At height y, strip length is

a-x = a-(a/b)y

so

dm = ρ(a-(a/b)y)dy

Then:

y_CM = 1/M∫₀^b y dm; = ρ/M∫₀^b y(a-(a/b)y)dy; = ((ρ a)/M)[y²/2-y³/3b]₀^b

Substitute ρ = 2M/ab:

y_CM = b/3
The triangle with a thin horizontal strip of thickness dy at height y. The strip runs from the hypotenuse, a distance x from the y-axis, to the vertical side, so its length is a − x. Its mass is dm.
Every point in a horizontal strip has the same y. Its length a − x shrinks to zero at the top vertex, where y = b.

Checks:

  • Units: both x_CM and y_CM are lengths.
  • Sign: both positive because the lamina lies in first quadrant.
  • Limit: if b → 0, then y_CM → 0 (triangle collapses toward x-axis).

4) Worked example

Let a = 6 cm and b = 9 cm.

x_CM = 2a/3 = 4 cm, y_CM = b/3 = 3 cm

Point (4,3) lies inside the triangle and closer to the wider side, which is physically sensible.

5) Practice set (hints + answers)

  1. For a = 12 m, b = 3 m, find (x_CM,y_CM). Hint: use the final formulas directly. Answer: (8 m,1 m).

  2. A similar triangle is scaled by factor k. How do x_CM and y_CM change? Hint: CM coordinates scale linearly with dimensions. Answer: both coordinates multiply by k.

  3. If density doubles everywhere, does CM location change? Hint: density cancels in the ratio. Answer: no.

6) Summary + next steps

  • Choose strips so each strip has a single coordinate value.
  • Build dm from density times strip area.
  • Final CM for this orientation is (2a/3,b/3).

Next steps: