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UY1: What is centre of mass?

Key idea: Understand centre of mass for particles and continuous bodies, and apply the coordinate formulas with clear physical interpretation.

  • University Physics Year 1
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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 core concept primer for the centre-of-mass part of the UY1 mechanics pathway.

Scope and use-case

Use this page when you need a reliable centre-of-mass method for first-pass modelling:

  • discrete masses and coordinate tables,
  • continuous bodies where symmetry reduces the integral setup,
  • quick checks before you move to dynamics in Motion of a System of Particles.

If your current task is torque/rotation rather than CM location, jump to Rotational Kinematics and Moment Of Inertia.

1) At a glance

  • Prerequisite: vectors, weighted averages, basic integration.
  • Outcomes: compute centre of mass (CM) for particles and continuous bodies.
  • Key result:
vector r_CM = 1/M∑ᵢ mᵢ vector rᵢ or vector r_CM = 1/M∫ vector r dm
  • Common trap: confusing geometric center with CM for non-uniform or cut-out objects.

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

  • Choose an origin and coordinate axes first.
  • Define each mass element position vector rᵢ = (xᵢ,yᵢ,zᵢ).
  • Total mass is M = ∑ᵢ mᵢ (or M = ∫ dm), with M > 0.
  • For symmetric uniform bodies, use symmetry before integration.

3) Core method / derivation

For two particles on the x-axis:

x_CM = (m₁x₁ + m₂x₂)/(m₁ + m₂)

General discrete system:

x_CM = (∑ᵢ mᵢxᵢ)/(∑ᵢ mᵢ), y_CM = (∑ᵢ mᵢyᵢ)/(∑ᵢ mᵢ), z_CM = (∑ᵢ mᵢzᵢ)/(∑ᵢ mᵢ)

Vector form:

vector r_CM = (∑ᵢ mᵢ vector rᵢ)/M

Continuous body (take limit of many small mass elements):

vector r_CM = 1/M∫ vector r dm

If density is uniform, dm is proportional to length/area/volume element.

Checks:

  • Units: coordinates of CM must have units of length.
  • Sign: CM coordinates can be negative depending on chosen origin.
  • Limits: if one mass dominates (m₁≫ m₂), CM approaches that mass position.

4) Worked example

Two masses on a line: m₁ = 2 kg at x₁ = 0, m₂ = 3 kg at x₂ = 4 m.

x_CM = (2(0) + 3(4))/(2 + 3) = 12/5 = 2.4 m

CM is closer to the heavier mass at x = 4 m, as expected.

2 kg at 0 m ───── centre of mass at 2.4 m ─── 3 kg at 4 m

The weighted position lies between the masses and closer to the heavier 3 kg mass.

5) Practice set (hints + answers)

  1. Masses 1 kg at x = -2 m and 3 kg at x = 1 m. Find x_CM. Hint: weighted average. Answer: x_CM = (1(-2) + 3(1))/4 = 0.25 m.

  2. For three equal masses at vertices (0,0), (2,0), (0,2), find (x_CM,y_CM). Hint: equal masses give arithmetic mean of coordinates. Answer: (x_CM,y_CM) = (2/3,2/3).

  3. True or false: CM must lie inside the material of an object. Hint: think about rings/frames. Answer: false. For a ring, CM is at the center, where there is no material.

6) Summary + next steps

  • CM is a mass-weighted position, not just geometric center.
  • Discrete and continuous formulas are the same idea in sum vs integral form.
  • Use symmetry first, then integrate only if needed.

If you are revising under time pressure, use this order:

  1. One discrete weighted-average question.
  2. One continuous-body setup question.
  3. One CM interpretation question (inside/outside material, sign convention, symmetry).

Next steps:

Continue with the next resource in this course.

Course and syllabus information
Course
University Physics Year 1
Edition
University Physics Year 1