Centre of Mass

6.2 Centre of Mass

The Core Idea

When we study the motion of a system of particles — whether it's two atoms or a spinning cricket bat — we quickly realise that tracking every single particle is impractical. The centre of mass is the single point that behaves as if the entire mass of the system were concentrated there, and as if all external forces acted at that point. For a rigid body undergoing pure translation, every particle moves exactly as the centre of mass moves. For more complex motion that includes rotation, the centre of mass still follows a simple translational path, while the body rotates about it.

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Two-Particle System Along a Line

Consider two particles of masses $m_1$ and $m_2$ lying on the $x$-axis. Let their positions be $x_1$ and $x_2$ measured from some origin O. The centre of mass of this two-particle system is the point C whose coordinate $X$ is given by

This is a mass-weighted average of the positions. If the two masses are equal ($m_1 = m_2 = m$), then

so the centre of mass lies exactly midway between the two particles.

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Generalisation to $n$ Particles on a Line

For $n$ particles of masses $m_1, m_2, \dots, m_n$ placed along the $x$-axis at positions $x_1, x_2, \dots, x_n$, the centre of mass coordinate $X$ is

The denominator is the total mass $M = \sum m_i$. Using summation notation,

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Centre of Mass in a Plane

For three particles not lying on a straight line, we set up $x$ and $y$ axes in their plane. Let the particles have masses $m_1, m_2, m_3$ and coordinates $(x_1, y_1), (x_2, y_2), (x_3, y_3)$. The centre of mass is at $(X, Y)$ where

If all three masses are equal ($m_1 = m_2 = m_3 = m$), then

These are precisely the coordinates of the centroid of the triangle formed by the three particles. For equal masses, the centre of mass coincides with the centroid.

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Centre of Mass in Space: The Vector Form

For $n$ particles distributed in three-dimensional space, with the $i$th particle of mass $m_i$ at $(x_i, y_i, z_i)$, the centre of mass coordinates are

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

These three scalar equations combine elegantly into a single vector equation. Let $\mathbf{r}_i = x_i \hat{\mathbf{i}} + y_i \hat{\mathbf{j}} + z_i \hat{\mathbf{k}}$ be the position vector of the $i$th particle, and let $\mathbf{R} = X \hat{\mathbf{i}} + Y \hat{\mathbf{j}} + Z \hat{\mathbf{k}}$ be the position vector of the centre of mass. Then

If we choose the centre of mass itself as the origin of our coordinate system, then $\mathbf{R} = \mathbf{0}$, which gives

This condition is often useful in derivations.

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Continuous Mass Distribution

A rigid body is a system of closely packed particles — so many that summing over individual atoms is impossible. Instead, we treat the body as a continuous distribution of mass.

We divide the body into $n$ small elements of mass $\Delta m_1, \Delta m_2, \dots, \Delta m_n$, each located approximately at $(x_i, y_i, z_i)$. The centre of mass coordinates are approximately

As we take $n$ larger and each $\Delta m_i$ smaller, these approximations become exact. The sums become integrals:

and similarly for $y$ and $z$. The exact centre of mass coordinates are therefore

The vector form is

If the centre of mass is taken as the origin, then $\mathbf{R} = \mathbf{0}$, which implies

or equivalently

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Symmetry and the Centre of Mass of Homogeneous Bodies

For homogeneous bodies (uniform mass distribution) of regular shape — rings, discs, spheres, rods — the centre of mass lies at the geometric centre. This follows from reflection symmetry. …