Centre of mass of discrete massesEdexcel International A Level Further Maths: Revision notes

The centre of mass of a system is the point at which the whole mass can be considered to act. For particles of masses $m_1,m_2,\ldots$ at positions $x_1,x_2,\ldots$ on a line, $\bar{x}=\frac{\sum m_ix_i}{\sum m_i}.$ This comes from equating the moment of the total mass about the origin with the sum of the individual moments: $M\bar{x}=m_1x_1+m_2x_2+\cdots$. Example: $2$ kg at $1$ m, $3$ kg at $4$ m and $5$ kg at $7$ m give $\bar{x}=\frac{2+12+35}{10}=4.9$ m.

Choose the origin at a convenient position, such as one end of a rod or a particle's position, so that some terms vanish. Always divide by the total mass: using $\sum mx$ alone, or the unweighted mean of the positions, are the common errors. For a light rod with particles attached, the rod has no mass, so only the particles appear in the formula. A rod supported at the centre of mass of its particles balances. Example: $3$ kg at $A$ and $5$ kg at $B$, with $AB=4$ m, has $\bar{x}=\frac{5\times4}{8}=2.5$ m from $A$, closer to the heavier particle.

For particles at $(x_i,y_i)$, find each coordinate separately with the same masses: $\bar{x}=\frac{\sum m_ix_i}{\sum m_i},\qquad\bar{y}=\frac{\sum m_iy_i}{\sum m_i}.$ In vector form the centre of mass has position vector $\bar{\mathbf{r}}=\dfrac{\sum m_i\mathbf{r}_i}{\sum m_i}$. Example: $2$ kg at $(1,4)$, $3$ kg at $(4,0)$ and $5$ kg at $(3,2)$ give $\bar{x}=\frac{29}{10}=2.9$ and $\bar{y}=\frac{18}{10}=1.8$. The distance from the origin is $\sqrt{2.9^2+1.8^2}=3.41$ m.

When a particle is added or removed, recompute with the new total mass and the new list of moments; the old centre of mass is not simply averaged unless the new particle has the same mass as the old total. If a mass or position is unknown, write the formula with the unknown, set it equal to the given centre of mass, and solve. Example: $3$, $m$ and $5$ kg at $0$, $2$ and $4$ m with $\bar{x}=2.4$: $\frac{2m+20}{8+m}=2.4$ gives $m=2$.

• State the total mass and show each moment, then the division.
• Give coordinates as a pair $(\bar{x},\bar{y})$ and keep fractions exact until the last line.
• For a distance between two points use Pythagoras on the coordinate differences.
• Sense-check: $\bar{x}$ and $\bar{y}$ lie within the range of the given coordinates, and the centre of mass is nearer the heavier particles.