Centres of mass by integration and stability

What you’ll learn

• Use thin strips or discs to calculate a centre of mass.
• Test suspension, sliding and toppling conditions.

Understand the idea

For a continuous body, replace weighted sums by integrals. Choose a thin element whose mass and position are easy to describe: a lamina strip has mass proportional to its area, and a solid-of-revolution disc has mass proportional to its circular cross-section times its thickness. Symmetry supplies the other centre-of-mass coordinates.

A suspended body settles with its centre of mass vertically below the suspension point. A body on a plane begins to topple when the vertical line through its centre of mass reaches the edge of the supporting base. On a rough incline, compare this angle with the sliding condition because sliding may occur first.

Key formulae

$$\bar x=\frac{\int x\,dm}{\int dm}$$

$$dm=\rho\pi y^2\,dx\quad\text{(solid of revolution)}$$

$$\bar x=\frac{\int xy\,dx}{\int y\,dx}\quad\text{(uniform lamina under }y\ge0\text{)}$$

$$\tan\theta_{\mathrm{topple}}=\frac bh\quad\text{(COM height }h\text{, half-base }b\text{)}$$

Worked example

A uniform solid cone has height $h$. Use discs to find the distance of its centre of mass from the vertex.

1. Measure $x$ from the vertex along the axis. By similar triangles, disc radius is proportional to $x$.
2. Hence $dm=Cx^2\,dx$ for a constant $C$.
3. The centre coordinate is $\frac{\int_0^h Cx^3\,dx}{\int_0^h Cx^2\,dx}=\frac{Ch^4/4}{Ch^3/3}$.

Answer: $3h/4$ from the vertex, or $h/4$ from the base.