MG1–MG3 - Centres of mass of particles and shapes

You already know

A force's moment is $Fd$, using its perpendicular distance from the moment axis. Signed coordinates or position vectors locate points relative to a chosen origin; in space, keep all three components.

Centre of mass

The centre of mass $G$ is the mass-weighted mean position of a system. In a uniform gravitational field, the system's total weight has the same moment as a single force acting through $G$.

Particles on one coordinate axis

For particles of positive masses $m_i$ at coordinates $x_i$, with total mass $M=\sum_i m_i$,

Here $\bar x$ is the coordinate of $G$. Positions must use one origin and one unit.

Not in the formula booklet: learn it.

Worked example 1: particles on opposite sides of the origin

Question. Particles of masses $2\,\mathrm{kg}$, $3\,\mathrm{kg}$ and $5\,\mathrm{kg}$ have coordinates $-4\,\mathrm m$, $0\,\mathrm m$ and $6\,\mathrm m$. Find their centre of mass.

Plan. Add the signed mass moments, then divide by the total mass.

Why this step? The left-hand particle contributes a negative moment about the origin; using its distance $4$ would put all the mass on the right.

Answer. $G$ has coordinate $11/5\,\mathrm m$.

Check. The moments about $G$ sum to $2(-4-2.2)+3(0-2.2)+5(6-2.2)=0$. Also, $G$ lies between the extreme particles.

Worked example 2: an unknown mass

Question. The masses at $-4\,\mathrm m$ and $0\,\mathrm m$ remain $2\,\mathrm{kg}$ and $3\,\mathrm{kg}$. The mass at $6\,\mathrm m$ is $q\,\mathrm{kg}$. Find $q$ if $\bar x=2\,\mathrm m$.

Plan. Include $q$ in both the total mass and its mass moment.

Why this step? Changing a mass changes the denominator as well as the numerator of the weighted mean.

Answer. $q=9/2$, so the third particle has mass $4.5\,\mathrm{kg}$.

Check. The total mass is $9.5\,\mathrm{kg}$, and its mass moment is $19\,\mathrm{kg\,m}$. Their ratio is $2\,\mathrm m$.

Watch out

Wrong: average the positions of unequal masses by dividing their sum by the number of particles.

Right: divide the sum of mass times coordinate by the total mass. The ordinary arithmetic mean works only when all the masses are equal.

Particle systems in two or three dimensions

For particles at $(x_i,y_i,z_i)$, with $M=\sum_i m_i$,

In two dimensions, omit the $z$-equation. Equivalently, $M\mathbf r_G=\sum_i m_i\mathbf r_i$.

Not in the formula booklet: learn it.

Worked example 3: a system in a plane

Question. Particles of masses $2\,\mathrm{kg}$, $3\,\mathrm{kg}$ and $5\,\mathrm{kg}$ are fixed at $A(0,0)$, $B(6,0)$ and $C(0,4)$. Coordinates are in metres. Find $G$.

Plan. Calculate the weighted means of $x$ and $y$, using total mass $10\,\mathrm{kg}$ for both.

Why this step? The $5\,\mathrm{kg}$ particle has $x=0$, but still contributes to the total mass used to average the $x$-coordinates.

Answer. $G=(9/5,2)$, in metres.

Check. Relative to $G$, the weighted $x$-offsets sum to $18-10(9/5)=0$, and the weighted $y$-offsets sum to $20-10(2)=0$.

Worked example 4: lifting one particle into space

Question. The same masses now occupy $A(0,0,0)$, $B(6,0,0)$ and $C(0,4,4)$. Find their centre of mass.

Plan. Retain the first two equations and add the $z$-equation.

Why this step? The three-dimensional problem needs three coordinate balances, rather than a different rule for the centre of mass.

Answer. $G=(9/5,2,2)$, in metres.

Check. Here every particle has $y=z$, so $G$ must also have $\bar y=\bar z$. Its projection onto the $xy$-plane agrees with example 3.

Watch out

A light frame has negligible mass. Include its attached particles, but do not add the frame's vertices as extra masses. Equally, do not replace unequal vertex masses by the centroid of a uniform triangular lamina.