3.2 - Centres of mass and symmetry

3.2 - Centres of mass and symmetry

A body's weight is spread throughout its material, yet one force can represent its gravitational effect. In this part of 3.2, you will place that force at the centre of mass, use symmetry to locate it and select the appropriate standard result for a lamina, wire, shell or solid.

A single equivalent weight

Where should you draw the weight arrow for a sloping rod? Drawing it at the lowest point changes its turning effect and can give a wrong support force. The correct point is determined by the distribution of mass.

Centre of mass

The centre of mass, GG, is the point at which the whole mass may be regarded as concentrated when calculating its gravitational effect in a uniform gravitational field. The body's gravity is equivalent to one vertically downward force mgmg acting at GG.

This replacement preserves both the total force and its moment about any point. It does not mean the material is physically concentrated at GG. A ring, for example, has its centre of mass in the empty space at its centre.

For a uniform straight rod, equal lengths have equal masses and GG is the midpoint. If a rod of length LL makes an angle θ\theta above the horizontal, the perpendicular distance from its lower end to the vertical weight line is (L/2)cos⁡θ(L/2)\cos\theta. The distance along the rod is not the moment arm.

Worked example 1: the weight of an inclined rod

Question. A uniform rod ABAB, length 2.4 m2.4\,\mathrm m and weight 80 N80\,\mathrm N, slopes upwards to the right at 60∘60^\circ to the horizontal. Find the moment of its weight about AA.

Plan. Place the weight at the midpoint, then find its horizontal distance from AA.

AG=1.2 md=1.2cos⁡60∘=0.6 m∣MA∣=80(0.6)=48 N m.\begin{aligned} AG&=1.2\,\mathrm m \\ d&=1.2\cos60^\circ=0.6\,\mathrm m \\ |M_A|&=80(0.6)=48\,\mathrm{N\,m}. \end{aligned}

Answer. 48 N m48\,\mathrm{N\,m} clockwise.

Check. Resolving the weight perpendicular to the rod gives 80cos⁡60∘=40 N80\cos60^\circ=40\,\mathrm N. Its moment is 40(1.2)=48 N m40(1.2)=48\,\mathrm{N\,m}.

Reduce only the rod's angle. The mass distribution stays the same, but its weight line moves farther horizontally from AA.

Worked example 2: changing the inclination

Question. The same rod slopes upwards at 30∘30^\circ instead. Find the moment of its weight about AA.

Plan. Keep GG at the midpoint and change the horizontal projection.

d=1.2cos⁡30∘=0.63 m∣MA∣=80(0.63)=483 N m.\begin{aligned} d&=1.2\cos30^\circ=0.6\sqrt3\,\mathrm m \\ |M_A|&=80(0.6\sqrt3)=48\sqrt3\,\mathrm{N\,m}. \end{aligned}

Answer. 483 N m48\sqrt3\,\mathrm{N\,m}, or 83.1 N m83.1\,\mathrm{N\,m}, clockwise.

Check. A horizontal rod would give 80(1.2)=96 N m80(1.2)=96\,\mathrm{N\,m}. The smaller inclination gives a moment closer to that maximum, as expected.

A rotation of the rod changes the moment arm, not the position of GG within the rod.

Watch out

The weight acts vertically downwards at GG, not perpendicular to the rod and not at the end. Draw its line of action before measuring a distance.

Locate the weight first in this check.

Uniform symmetric bodies

A symmetry of the mass distribution must leave the centre of mass unchanged. Otherwise the reflected body would be physically identical but have a different centre of mass. This gives a fast way to locate GG.

Uniform body

A uniform lamina has constant mass per unit area; a uniform wire has constant mass per unit length; a uniform solid has constant mass per unit volume. Geometry alone locates the centre of mass by symmetry only when the mass distribution shares that symmetry.

For a planar body with one line of symmetry, GG lies somewhere on that line. Two intersecting symmetry lines locate it at their intersection. A uniform rectangle therefore has GG at the intersection of its diagonals, and a uniform disc or circular wire has GG at its centre. A uniform sphere has GG at its centre; a uniform right circular cylinder has GG halfway along its axis.

An axis of symmetry alone does not tell you how far along it to go. A cone's centre of mass, for example, lies on its axis but is not halfway up it.

Worked example 3: intersecting symmetry lines

Question. A uniform rectangular lamina occupies 0≤x≤80\le x\le8, 0≤y≤40\le y\le4, with distances in centimetres. Find its centre of mass.

Plan. Identify two symmetry lines and use their intersection.

x=4vertical symmetry liney=2horizontal symmetry line.\begin{aligned} x&=4 &&\text{vertical symmetry line} \\ y&=2 &&\text{horizontal symmetry line}. \end{aligned}

Answer. G=(4,2) cmG=(4,2)\,\mathrm{cm}.

Check. Reflecting (4,2)(4,2) in either symmetry line leaves it fixed, and there is equal mass on opposite sides of each line.

Remove material without breaking those symmetries. The centre of mass can stay fixed even though the total mass changes.

Worked example 4: a symmetric hole

Question. A circular hole of radius 1 cm1\,\mathrm{cm}, centred at (4,2)(4,2), is cut from the lamina in Worked example 3. Find the centre of mass of the remaining uniform lamina.

Plan. Check which symmetries the remaining material retains.

The circle and rectangle both have symmetry lines x=4x=4 and y=2y=2. Removing the circle preserves both lines, so

G=(4,2) cm.G=(4,2)\,\mathrm{cm}.

Answer. The centre of mass is still (4,2) cm(4,2)\,\mathrm{cm}, inside the hole.

Check. Every remaining small piece has an equal-mass partner directly opposite it about (4,2)(4,2), so their moments about the centre cancel.

Moving that hole off centre would usually remove these symmetries. You would then need the mass-weighted method from the next lesson.

Continue this lesson.

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