Centre of Mass: Step-by-Step Worked Examples (1)

Question

Two particles of masses $m_1 = 2$ kg and $m_2 = 3$ kg are placed at positions $\vec{r}_1 = (1, 2)$ m and $\vec{r}_2 = (4, -1)$ m. Find the position of the centre of mass.

Solution — Step by Step

Final answer: $\vec{r}_{\text{COM}} = (2.8,\,0.2)$ m.

Why This Works

The centre of mass is the mass-weighted average position. Heavier particles “pull” the COM toward themselves. With $m_2 > m_1$, the COM lies closer to particle 2 — and indeed $2.8$ is closer to $4$ than to $1$ on the $x$-axis.

For more particles, the formula generalises to $\vec{r}_{\text{COM}} = \sum m_i \vec{r}_i / \sum m_i$. For continuous bodies, the sum becomes an integral.

Alternative Method

Vector form: $\vec{r}_{\text{COM}} = (m_1\vec{r}_1 + m_2\vec{r}_2)/(m_1+m_2) = (2(1,2) + 3(4,-1))/5 = ((2+12)/5, (4-3)/5) = (2.8, 0.2)$ m. Same answer in one line.

Common Mistake

Forgetting that mass is a scalar but position is a vector. You cannot add positions directly — you must take a weighted average using the masses as weights. Some students compute “average position” $(\vec{r}_1 + \vec{r}_2)/2$ ignoring the masses, which only works if $m_1 = m_2$.