Centre of Mass — Discrete System

What the centre of mass is

The centre of mass (CM) of a system is the single point that behaves as if the entire mass of the system were concentrated there, as far as external forces are concerned.

When you throw a cricket bat in the air, every point on the bat follows a complicated path — rotating and translating simultaneously. But one special point, the centre of mass, follows a perfect parabola — exactly as a point mass would. That is the physical meaning of the CM.

The formula

For a system of $n$ particles with masses $m_1, m_2, \ldots, m_n$ at positions $\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n$:

$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \cdots + m_n\vec{r}_n}{m_1 + m_2 + \cdots + m_n} = \frac{\sum m_i \vec{r}_i}{M}$

where $M = \sum m_i$ is the total mass.

In component form:

$x_{cm} = \frac{\sum m_i x_i}{M}, \quad y_{cm} = \frac{\sum m_i y_i}{M}, \quad z_{cm} = \frac{\sum m_i z_i}{M}$

Where does this definition come from?

The CM is defined so that the equation $\vec{F}_{ext} = M\vec{a}_{cm}$ holds — so that the whole system responds to external forces as if it were a single particle of mass $M$ at the CM.

Starting from Newton's Second Law for each particle and summing over all particles:

$\sum_i \vec{F}_i = \sum_i m_i \vec{a}_i$

Internal forces cancel in pairs (Newton's Third Law). Only external forces remain:

$\vec{F}_{ext} = \sum_i m_i \vec{a}_i = M \cdot \frac{\sum m_i \vec{a}_i}{M} = M\vec{a}_{cm}$

For this to hold, we must define:

$\vec{a}_{cm} = \frac{\sum m_i \vec{a}_i}{M} \implies \vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{M}$

The CM definition is not arbitrary — it is the unique point that satisfies $\vec{F}_{ext} = M\vec{a}_{cm}$.

Example: Two particles

Masses $m_1 = 2$ kg at $x_1 = 1$ m and $m_2 = 6$ kg at $x_2 = 5$ m:

$x_{cm} = \frac{2 \times 1 + 6 \times 5}{2 + 6} = \frac{2 + 30}{8} = \frac{32}{8} = 4 \text{ m}$

The CM is at $x = 4$ m — closer to the heavier mass. This is always the case: the CM is pulled toward the heavier masses.

Example: Three particles

Masses $m_1 = 1$ kg at $(0, 0)$, $m_2 = 2$ kg at $(3, 0)$, $m_3 = 3$ kg at $(0, 4)$:

$x_{cm} = \frac{1(0) + 2(3) + 3(0)}{6} = \frac{6}{6} = 1 \text{ m}$

$y_{cm} = \frac{1(0) + 2(0) + 3(4)}{6} = \frac{12}{6} = 2 \text{ m}$

CM is at $(1, 2)$ m.

Key properties

CM lies within the body (or system): For a convex body, CM is always inside. For a non-convex or hollow body, CM may be outside the material (e.g., a ring's CM is at its geometric centre, where there is no material).

CM of uniform symmetric bodies: Lies on the axis/plane of symmetry. For a uniform sphere, cube, or cylinder — at the geometric centre.

CM and weight: For a body in a uniform gravitational field, the CM coincides with the centre of gravity. The body balances at the CM.