Centres of mass of rigid bodiesEdexcel A-Level Further Maths: Revision notes

For particles, $\bar x=\frac{\sum mx}{\sum m}$. For a continuous body, divide it into thin slices of mass $\delta m$ and let the sum become an integral: $\bar x=\frac{\int x\,dm}{\int dm}.$ For a non-uniform rod with mass per unit length $\rho(x)$, $dm=\rho\,dx$, so $\bar x=\frac{\int x\rho\,dx}{\int\rho\,dx}$. For a uniform solid of revolution about the $x$-axis, a slice has volume $\pi y^2\,dx$, so $\bar x=\frac{\int xy^2\,dx}{\int y^2\,dx}.$ For a uniform lamina under a curve, $\bar x=\frac{\int xy\,dx}{\int y\,dx}$ and $\bar y=\frac{\frac12\int y^2\,dx}{\int y\,dx}$. Symmetry still helps: if there is an axis of symmetry, the centre of mass lies on it.

Solid of revolution: $y=\sqrt{x}$, $0\le x\le4$, rotated about the $x$-axis. Volume $=\pi\int_0^4x\,dx=8\pi$. Moment $=\pi\int_0^4x\cdot x\,dx=\frac{64\pi}{3}$, so $\bar x=\frac{64\pi/3}{8\pi}=\frac83$. This is $\frac43$ cm from the plane face at $x=4$. Uniform lamina: the region under $y=x^2$ for $0\le x\le2$. Area $=\int_0^2x^2\,dx=\frac83$. $\bar x=\frac{\int x^3\,dx}{8/3}=\frac{4}{8/3}=1.5$. $\bar y=\frac{\frac12\int x^4\,dx}{8/3}=\frac{16/5}{8/3}=1.2$. Non-uniform rod: length 2 m with $\rho=(3+x)$ kg m⁻¹. Mass $=\int_0^2(3+x)\,dx=8$ kg, moment $=\frac{26}{3}$, so $\bar x=\frac{13}{12}$ m. It is more than 1 m because the rod is denser at the far end.

These results are in the formulae book and may be quoted without proof:

• solid cone or pyramid of height $h$: $\frac14h$ from the base (on the axis);
• solid hemisphere of radius $r$: $\frac38r$ from the plane face;
• hemispherical shell: $\frac12r$ from the plane face (centre);
• conical shell: $\frac13h$ from the base;
• uniform cylinder or cuboid: at the geometric centre. Volume formulae you need: cone $\frac13\pi r^2h$, hemisphere $\frac23\pi r^3$, cylinder $\pi r^2h$. Example: a solid hemisphere of radius 6 cm has its centre of mass $\frac38(6)=2.25$ cm from its plane face.

A composite body is made of standard parts. For a uniform material the weight of each part is proportional to its volume, so take moments about a plane (or axis) using volumes: $V\bar x=\sum v_ix_i.$ A part that is removed counts as a negative volume. Choose the reference plane carefully, such as a joining face, and measure distances in a consistent direction. Example (hemisphere and cylinder): the hemisphere has volume $144\pi$ and its centre is 2.25 cm on one side of the joining face. The cylinder (radius 6, height 12) has volume $432\pi$ and its centre is 6 cm on the other side. Then $576\pi\bar x=432\pi(6)-144\pi(2.25)$, so $\bar x=3.94$ cm into the cylinder. Example (frustum): large cone $144\pi$ with centre 3 cm from the base, removed cone $18\pi$ with centre 7.5 cm from the base. Then $126\bar x=144(3)-18(7.5)$, so $\bar x=2.36$ cm.

If the density is not uniform, the weight of each part is its mass, not its volume. For continuous variation, integrate the density: use $\rho\,dx$ for a rod, or $\rho\pi y^2\,dx$ for a solid of revolution. Non-uniform bodies can also be combined with particles. A 2 kg particle at $A$ added to the rod above gives $\bar x=\frac{8(13/12)+0}{10}=\frac{13}{15}$ m. To make a body balance at a chosen point, set up moments about a convenient point and require the centre of mass of the whole system to be at that point. Adding a mass $m$ at $B$ (2 m from $A$) with balance at $x=1$ gives $\frac{26}{3}+2m=10+m$, so $m=\frac43$.