6.04a–6.04c - Centres of mass of composite bodies

Weight, symmetry and empty space

Can an object's centre of mass be somewhere that contains none of the object? A uniform circular wire has its centre of mass at the centre of the circle, where there is only air. The centre of mass describes a balance of mass, not a specially heavy piece of material.

In a uniform gravitational field, every small piece has a downward weight proportional to its mass. These parallel forces have the same overall force and moment as one downward force through the centre of mass. A body of mass $M$ therefore has weight $Mg$, acting at a point usually labelled $G$.

For linear motion you may model a rigid body as a particle of the same mass at $G$. That preserves the resultant-force equation for its translation. It does not preserve the way forces turn the body: their lines of action still matter in a rigid-body equilibrium problem.

Symmetry is a mass-pairing argument. Reflection in a line pairs equal pieces at equal perpendicular distances on opposite sides, so their contributions cancel in that direction. The centre of mass of a uniform lamina lies on every line of symmetry. Two intersecting symmetry lines therefore fix its position. The corresponding statement uses planes of symmetry in three dimensions. These symmetry principles and the particle model are not in the formula booklet: learn them.

A uniform rectangular lamina with corners $(0,0)$, $(8,0)$, $(8,4)$ and $(0,4)$ has $G=(4,2)$. A uniform rectangular wire frame with those corners has the same centre, despite having no material there. A uniform semicircular lamina has only one symmetry axis: it fixes one coordinate, but not both.

Before calculating, decide what the model's symmetry actually tells you.

Mass-weighted coordinates

Two equal masses balance halfway between them; an unequal pair balances nearer the larger mass. The position is an average in which each mass supplies its own weight in the calculation.

Suppose particles of masses $m_i$ have coordinates $(x_i,y_i)$. Equating the moment of their downward weights to that of the single resultant gives $Mg\bar x=\sum m_i g x_i$. The common $g$ cancels. Turning the coordinate system gives the same equation for the other coordinate.

The equations contain masses because the same gravitational acceleration has cancelled from both sides. That cancellation will not allow you to mix mass and force in a later moments equation.

Change only the third mass, from $1\,\mathrm{kg}$ to $4\,\mathrm{kg}$. The centre must move towards $(0,6)$, so its horizontal coordinate should fall and its vertical coordinate should rise.

A component can itself be a rigid body. Replace it by its mass at its own centre, then use precisely the same sum.

Keep the same three positions and vary that upper mass once more.

Adding and removing laminae

For pieces cut from the same uniform sheet, mass is proportional to area. The common mass per unit area cancels, so you can use signed areas in the mass-weighted equations.

A rectangle is centred at the intersection of its diagonals. A triangular lamina is centred two-thirds of the way along any median from its vertex; equivalently, its coordinates are the averages of its three vertices. The triangular result is in the formula booklet. It does not extend to averaging the vertices of an arbitrary quadrilateral: split that shape into rectangles and triangles instead.

For a semicircular lamina, use the booklet's sector formula with half-angle $\alpha=\pi/2$. Its centre lies $4r/(3\pi)$ from the circle's centre towards the curved edge. This derived semicircle form is not listed separately in the booklet; it follows from the sector result given there.

[DIAGRAM: asset_name: Adding or removing the same semicircle; asset_slug: h245_composite_lamina_semicircle; caption: The same semicircle has a positive area when attached and a negative area when removed; description: Two equal-scale coordinate diagrams show a six by four rectangle. Left: an upper semicircle of radius three is attached above its top edge, centred at (3,4), with its centroid at height four plus four over pi. Right: the same upper semicircle is removed above the bottom edge inside the rectangle, centred at (3,0), with its centroid at height four over pi. The remaining body and component centroids are clearly labelled.]

If a hole was cut from a complete lamina, write “whole body minus missing piece”. This subtracts both its area and its first moment; it is an accounting device, not a claim that the hole has physical negative mass.

First attach a semicircle to a rectangle; the centre should move towards the attached part.

Change the operation from attachment to removal. Put the semicircle inside the rectangle, above its bottom edge; its own centroid is now measured from $y=0$. The subtraction should move the centre away from the missing low material.