A rigid body has countless particles, yet it moves as if its entire mass were concentrated at a single point. That point is the centre of mass (CM). When you throw a spanner, every particle follows a complicated path, but one special point traces a neat parabola — exactly the path a single particle of the same mass would follow. Studying that point lets us treat an extended body like a point particle.
For a system of discrete particles of masses $m_1, m_2, \dots, m_n$ at position vectors $\vec{r}_1, \vec{r}_2, \dots, \vec{r}_n$, the centre of mass is the mass-weighted average position:
- $\vec{r}_{cm}=\frac{\sum m_i \vec{r}_i}{\sum m_i}=\frac{m_1\vec{r}_1+m_2\vec{r}_2+\cdots+m_n\vec{r}_n}{m_1+m_2+\cdots+m_n}$.
- In components: $x_{cm}=\frac{\sum m_i x_i}{\sum m_i}$ and $y_{cm}=\frac{\sum m_i y_i}{\sum m_i}$.
- For a continuous body the sums become integrals: $\vec{r}_{cm}=\frac{1}{M}\int \vec{r}\,dm$.
For symmetric, uniform bodies the CM lies at the geometric centre. The CM of a uniform rod is at its midpoint, of a uniform disc or ring at its centre, and of a uniform sphere at its centre. Note that the CM need not lie inside the body — for a ring it lies in the empty central space, and for an L-shaped lamina it can fall outside the material altogether.
Motion of the centre of mass. Differentiating $\vec{r}_{cm}$ gives the velocity and acceleration of the CM:
- $\vec{v}_{cm}=\frac{\sum m_i \vec{v}_i}{M}$, so the total momentum of the system is $\vec{P}=M\vec{v}_{cm}$.
- $M\vec{a}_{cm}=\vec{F}_{ext}$ — the CM accelerates only under the net external force; internal forces cancel in pairs (Newton's third law).
Momentum of a system. Because internal forces cannot shift the CM, if $\vec{F}_{ext}=0$ then $\vec{P}=M\vec{v}_{cm}$ is constant — the law of conservation of linear momentum. This is why a bomb exploding in mid-air has fragments flying apart, yet their CM keeps moving along the original parabola: the explosive forces are purely internal.