Derivation of the Centre of Mass for a Two-Particle System

Derivation of the Centre of Mass for a Two-Particle System

The centre of mass is the point where the total mass of a system can be treated as concentrated. For two particles, its position follows directly from the definition of mass-weighted average position.

Consider two particles. Particle 1 has mass $m_1$m1​ and position vector $\vec{r}_1$r1​. Particle 2 has mass $m_2$m2​ and position vector $\vec{r}_2$r2​. Both positions are measured from the same origin.

By definition, the position of the centre of mass, $\vec{R}$R, is the weighted average of the individual positions:

$$\vec{R} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$$R=m1​+m2​m1​r1​+m2​r2​​

The denominator is the total mass, $M = m_1 + m_2$M=m1​+m2​. Therefore the same relation can be written as

$$\vec{R} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{M}$$R=Mm1​r1​+m2​r2​​

In Cartesian coordinates this splits into three scalar equations:

$$X = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}, \quad Y = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}, \quad Z = \frac{m_1 z_1 + m_2 z_2}{m_1 + m_2}$$X=m1​+m2​m1​x1​+m2​x2​​,Y=m1​+m2​m1​y1​+m2​y2​​,Z=m1​+m2​m1​z1​+m2​z2​​

For motion along a straight line, only one coordinate is needed:

$$X = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$X=m1​+m2​m1​x1​+m2​x2​​

This formula has a simple meaning. The heavier particle pulls the centre of mass closer to itself. If $m_1 = m_2$m1​=m2​, the centre of mass lies exactly midway between the two particles. If one mass is much larger, the centre of mass lies very near that particle.

The same result can be reached from the idea of balance. Suppose the centre of mass is at a distance $d_1$d1​ from $m_1$m1​ and $d_2$d2​ from $m_2$m2​, with $d_1 + d_2$d1​+d2​ equal to the separation. Taking moments about the centre of mass gives

$$m_1 d_1 = m_2 d_2$$m1​d1​=m2​d2​

Solving for the position then recovers the weighted-average formula above. Thus both the definition and the moment condition lead to the same centre of mass for two particles.