M2.2 - Centres of mass

Learning objectives

M2.2.1Centre of mass of discrete distributionsCentre of mass of a discrete mass distribution in one and two dimensions.M2.2.2Centre of mass of uniform planeCentre of mass of uniform plane The use of an axis of symmetry will be acceptable where figures, and simple cases of appropriate.; Use of integration is not required.; Figures may composite plane figures. include the shapes referred to in the formulae book.; Results given in the formulae book may be quoted without proof.M2.2.3Simple cases of equilibriumSimple cases of equilibrium of a The lamina may plane lamina. (i) be suspended from a fixed point; (ii) be free to rotate about a fixed horizontal axis; (iii) be put on an inclined plane.

The centre of mass of discrete particles is the mass-weighted mean of their positions. Choose an origin and signed coordinate axes first: particles farther from an axis contribute a larger moment, but only in proportion to their masses.

$\bar x=\frac{\sum m_i x_i}{\sum m_i},\qquad \bar y=\frac{\sum m_i y_i}{\sum m_i}$

These equations state that the total first moment about an axis is unchanged if the whole distribution is replaced by its total mass at $(\bar x,\bar y)$ . In one dimension use only the relevant coordinate. In two dimensions calculate the two coordinates separately with the same total mass.

Step	Check
set axes	every particle position is measured from the same origin
list data	pair each mass with its signed $x$ and $y$ coordinates
take moments	form $\sum m_ix_i$ and $\sum m_iy_i$ without mixing axes
divide	use the total mass $\sum m_i$ in both coordinates
validate	the centre lies inside the coordinate range when all masses are positive

Particles of masses $2$ , $3$ and $5$ kg are at $(0,0)$ , $(4,0)$ and $(2,6)$ metres. Their total mass is $10$ kg, so $\bar x=\frac{2(0)+3(4)+5(2)}{10}=2.2,qquad \bar y=\frac{2(0)+3(0)+5(6)}{10}=3.$ The centre of mass is therefore $(2.2,3)$ m relative to the chosen origin.

Do not take an unweighted mean unless all masses are equal, use different origins in the same equation, or discard negative coordinates. Multiplying every mass by $g$ gives the same centre because the common factor cancels; it is not necessary to convert masses into weights.

For a uniform plane lamina, mass is proportional to area. Split a composite figure into standard shapes, place each shape's area at its known centre, and take area-weighted moments. An axis of symmetry contains the centre of mass; two independent symmetry axes locate it immediately.

$\bar x=\frac{\sum A_i x_i}{\sum A_i},\qquad \bar y=\frac{\sum A_i y_i}{\sum A_i}$

Feature	Treatment
added uniform piece	positive area at that piece's centre
removed hole or cut-out	negative area at the removed piece's centre
common uniform density and thickness	cancel from numerator and denominator
standard shape	use its formula-book centre without proof where supplied
symmetry axis	use it to fix one coordinate before taking moments for the other

Choose axes that make all component centres easy to measure. Record signed areas and centre coordinates in one table, then take moments about a parallel reference axis. The numerator has dimensions of length cubed and the denominator length squared, so the result must have dimensions of length. No integration is required in this syllabus objective.

A uniform $6\times4$ rectangle has the top-right $2\times2$ square removed. From the lower-left corner, the full rectangle has area $24$ and centre $(3,2)$ ; the removed square has signed area $-4$ and centre $(5,3)$ . Hence $\bar x=\frac{24(3)-4(5)}{20}=2.6,qquad \bar y=\frac{24(2)-4(3)}{20}=1.8.$ The missing top-right mass shifts the centre down and left.

Do not average the component centroids without area weights, treat a hole as positive area, or assume a symmetry line that the completed composite shape does not possess. Formula-book results may be quoted; deriving them by integration is outside the required method here.

A lamina's weight acts vertically downward through its centre of mass. Equilibrium requires zero resultant force and zero total moment. The geometry of the centre of mass therefore determines how a suspended, pivoted or supported lamina can rest.

$\sum\mathbf F=\mathbf0,\qquad \sum M_P=0$

Situation	Equilibrium consequence
freely suspended from a fixed point $P$	the centre of mass lies vertically below $P$ ; the support force has zero moment about $P$
free to rotate about a fixed horizontal axis	take moments about the axis; weight and any applied forces must have balancing turning effects
placed on an inclined plane	include weight and all contact forces, then resolve forces and take moments about a useful contact point
contact is just being lost	the reaction at that contact is zero; the remaining contact acts as the limiting pivot

For free suspension, first locate the centre of mass $G$ in body-fixed coordinates. The lamina rotates until $PG$ is vertical, with $G$ below $P$ . If in a reference orientation $G$ is $3$ cm to the right and $4$ cm below $P$ , the angle through which $PG$ differs from the downward vertical satisfies $\tan\theta=\frac{3}{4},$ so $\theta=36.9^\circ$ . Use the required orientation in the diagram to choose the correct sense.

If a pivoted lamina of weight $W$ has a horizontal moment arm $0.30$ m from its axis, and a horizontal holding force $F$ has perpendicular moment arm $0.50$ m, equilibrium gives $F(0.50)=W(0.30),$ so $F=0.60W$ . The pivot reaction is absent from this moment equation because its line of action passes through the pivot.

A centre of mass need not lie inside a concave lamina, but the weight still acts through that point. Do not infer equilibrium merely because the centre is known, place the centre vertically above a free suspension point, or omit contact-force and moment conditions for a lamina on an incline.

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M2.2 - Centres of mass

Learning objectives

Locate the centre of mass of discrete particles

Build the centre of mass of a composite lamina

Use the centre of mass to balance a lamina

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