Q.A thin uniform rod $AB$ of mass $3\ \text{kg}$ and length $1.2\ \text{m}$ has a small block of mass $1\ \text{kg}$ fixed rigidly at end $B$. Treating the rod's own mass as concentrated at its geometric centre, locate the centre of mass of the rod-plus-block system, measured from end $A$.

The centre of mass (c.m.) is the single hypothetical point at which the entire mass of an extended, finite-sized object (or system of objects) can be treated as concentrated, allowing Newton's laws of motion -- which strictly apply only to point masses -- to be validly applied to real, extended objects. For a system of n particles, its position is the mass-weighted average of the individual particles' positions, $\vec{r}=\frac{\sum m_i\vec{r_i}}{M}$ (or the corresponding integral, $\vec{r}=\frac{\int \vec{r}\,dm}{M}$, for a continuous mass distribution); its velocity and acceleration are similarly the mass-weighted averages of the individual particles' velocities and accelerations.

The centre of mass has many useful properties: it is a location rather than a physical quantity in itself; it can only be shifted by an external unbalanced force, never by internal forces (as in a collision or explosion); it lies on any axis or point of symmetry an object has; it need not coincide with any individual particle, or even lie physically within the body at all (a ring or horseshoe are simple examples); and only the net external force on a system determines the acceleration of its centre of mass, $\vec{F}_{ext}=M\vec{a}_{cm}$, regardless of how complicated the system's internal motions or forces are.