Centre of Mass

3
3.1 DEFINITIONS AND CONCEPTS
The centre of gravity of a body is the point about which the body is balanced, or the point through
which the weight of the body acts.
The location of the centre of gravity of a body coincides with the centre of mass of the body
when the dimensions of the body are much smaller than those of the earth.
When the density of a body is uniform throughout, the centre of mass and the centroid of the
body are at the same point.

•• If the centre of mass of a body is not positioned above the area over which it is supported, the
body will topple over.
•• The lower the centre of mass of a body, the more stable the body.

3.2 THEORETICAL BACKGROUND
Centre of gravity: The centre of gravity of a body is a point through which the weight of the body
acts or is a point about which the body can be balanced. The location of the centre of gravity of a
body can be determined mathematically using the moment equilibrium condition for a parallel force
system. Physically, a body is composed of an infinite number of particles and if a typical particle has
a weight of dW and is located at (x, y, z), as shown in Figure 3.1, the position of the centre of gravity,
G, at (x, y, z) can be determined using the following equations:

∫ x dW ∫ y dW ∫z dW
x= y= z= (3.1)
∫d W ∫d W ∫d W

Centre of mass: To study problems relating to the motion of a body subject to dynamic loads,
it is necessary to determine the centre of mass of the body. For a particle with weight dW and mass
dm, the relationship between mass and weight is given by dW = gdm, where g is the acceleration due
to gravity. Substituting this relationship into Equation 3.1 and cancelling g from both the numerator
and denominator leads to

∫ x dm ∫ y dm ∫z dm
x= y= z= (3.2)
∫dm ∫dm ∫dm

Equation 3.2 contains no reference to gravitational effects (g) and therefore it defines a unique
point which is a function solely of the distribution of mass. This point is called the centre of mass
of the body. For civil engineering structures, whose dimensions are very small compared with the
dimensions of the earth, g may be considered to be uniform and hence the centre of mass of a body
coincides with the centre of gravity of the body. In practice, no distinction is made between the
centre of mass and the centre of gravity of a body.
The mass, dm, of a particle in a body relates to its volume, dV, and can be expressed as dm = ρ(x, y, z)
dV, where ρ(x, y, z) is the density of the body. Substituting this relationship into Equation 3.2 gives

∫ x ρ( x, y, z)dV ∫ yρ( x, y, z)dV ∫zρ( x, y, z)dV (3.3)

x= y= z=
∫ρ( x, y, z)dV ∫ρ( x, y, z)dV ∫ρ( x, y, z )dV
26  Understanding and Using Structural Concepts

dV G

dW
W
z z
x
x y
y
y x

FIGURE 3.1
Centre of gravity.

Centroid: The centroid C of a body is a point which defines the geometrical centre of the body.
If a body is composed of one material, or the density of the material ρ is constant, it can be removed
from Equation 3.3, and then the centroid of the volume of the body can be defined by

∫ x dV ∫ y dV ∫z dV
x= y= z= (3.4)
∫dV ∫dV ∫dV

Equation 3.4 contains no reference to the mass of the body and therefore defines a point, the
centroid, which is solely a function of geometry.
If the density of the material of a body is uniform, the locations of the centroid and the centre
of mass of the body coincide. However, if the density varies throughout a body, the centroid and the
centre of mass will not be the same in general (Section 3.3.2).
When the centroid of a body, such as a uniform thin flat plate, needs to be determined, the vol-
ume of a small section of the body, dV, can be expressed as dV =tdA, where dA is the surface area
of the volume dV perpendicular to the axis through the thickness, t, of the volume. Substituting the
expression for dV into Equation 3.4 leads to

∫ x dA ∫ y dA ∫ y dA
x= y= z= (3.5)
∫dA ∫dA ∫dA

Terms such as xdA in the numerators of the integrals represent the moments of the area of the
particle, dA, about the centroid. The integrals represent the summations of these moments of areas
for the whole body and are usually called the first moments of area of the body about the axes in
question. The integral in the denominators in Equation 3.5 is the area of the body.
When the centroid of a line, effectively a body with one dimension, needs to be determined,
Equation 3.5 can be simplified to

∫ x dL ∫ y dL ∫z dL
x= y= z= (3.6)
∫dL ∫dL ∫dL

where dL is the length of a short element or part of the line.

Equations 3.4 through 3.6 are based on a coordinate system that may be chosen to suit a par-
ticular situation and in general such a coordinate system should be chosen to simplify, as much as
possible, the integrals. When the geometric shape of an object is simple, the integrals in the equa-
tions can be replaced by simple algebraic calculations.
 3  Centre of Mass  27

EXAMPLE 3.1
y y

F E F E

C2
40 mm y2
O2 C
D C D y1 C
H x H
10 mm C1 O1 x
A I B A I B

(a) 10 mm 60 mm (b)

FIGURE 3.2
(a,b) Centroid of an L-shaped object.

Determine the centroid of the uniform L-shaped object shown in Figure 3.2a.

SOLUTION 1
As the L-shaped object consists of two rectangles, ABCH and HDEF, these areas and
the moments of the areas about the axes can be easily calculated without the need for
integration. The areas and their centroids for the coordinate system chosen are

ABCH: A1 = 70 × 10 = 700 mm2, xC 1 = 35 mm, yC 1 = 5 mm

HDEF: A2 = 40 × 10 = 400 mm2, xC 2 = 5 mm, yC 2 = 30 mm

Using the first two formulae in Equation 3.5 gives

A1 x C 1 + A 2 xC 2 700 × 35 + 400 × 5 26, 500

x= = = = 24.09 mm
A1 + A 2 700 + 400 1,100

A1 y C 1 + A 2 y C 2 700 × 5 + 400 × 30 15, 500

y= = = = 14.09 mm
A1 + A 2 700 + 400 1,100

SOLUTION 2
The centroid of the L-shaped object can also be determined using a geometrical
method. The procedure can be described as follows:

•• Divide the L-shape into two rectangles, ABCH and HDEF, as shown in Figure 3.2b.
Find the centroids of the two rectangles, C1 and C2, and then draw a line linking
C1 and C2. The centroid of the object must lie on the line C1 − C2. The equation of
this line is y1 − 5 = −5(x − 35)/6.
•• Divide the shape into two other rectangles, IBCD and AIEF, as shown in Figure 3.2b.
Find the centres of the two rectangles, O1 and O2, then draw a line joining their
centres. The centroid of the object must also lie on the line O1 − O2, which is
y2 − 5 = −4(x − 40)/7.
28  Understanding and Using Structural Concepts

•• As the centroid of the object lies on both lines C1 − C2 and O1 − O2, it must be
at their intersection, C, as shown in Figure 3.2b. Solving the two simultaneous
equations for x and y leads to x = 24.09 cm and y = 14.09 cm.

It can be seen from Figure 3.2b that the centroid, C, is located outside the bound-
aries of the object. As the density of the L-shaped object is uniform, the centre of
gravity, the centre of mass and the centroid coincide, so all lie outside the boundaries
of the object. From the definitions of centre of mass and centre of gravity given in
Section 3.1, the L-shaped object can be supported by a force applied at the centre of
mass, perpendicular to the plane of the object. In practice, this may be difficult as the
centre of mass lies outside the object. Section 3.3.3 provides a model demonstration
that shows how an L-shaped body can be lifted at its centre of mass.

EXAMPLE 3.2
Determine the centre of mass of a solid, uniform pyramid that has a square base of
a × a and a height of h. The base of the pyramid lies in the x–y plane and the apex of
the pyramid is aligned with the z-axis as shown in Figure 3.3a.

SOLUTION
Due to symmetry, the centroid must lie on the z-axis, for example

x = y =0

The location of the centroid of the pyramid on the z-axis can be obtained by using
the third formula in Equation 3.4 where the integral is taken over the volume of the
pyramid. Consider a slice of thickness dz of the pyramid, which is parallel to the base
with a distance z from the base, as shown in Figure 3.3b. The length of the side of
the square slice is a(1 – z/h), and hence the volume of the slice is dV = a2(1 – z/h)2dz.
Substituting these quantities into the third formula in Equation 3.4 and integrating with
respect to z through the height of the pyramid gives
h 2
 z
z c = 0h
∫ a 2h 2
za2  1 −  dz

= 12
h
=
h
2
 z
2 a h 4
∫
a2  1 −  dz
 h 3
0

Thus, the centre of mass of a uniform square-based pyramid is located on its verti-
cal axis of symmetry at a quarter of its height measured from the base.
z

h
h dz

z
a O x
a
a/2 a/2
(a) (b)

FIGURE 3.3
(a,b) Centroid of a pyramid.
 3  Centre of Mass  29

EXAMPLE 3.3
b

h θ W

A θ

FIGURE 3.4
Centre of mass and stability.

The location of the centre of mass of a uniform block is at a height of h from its
base that has a side length of b. Place the block on an inclined surface as shown in
Figure 3.4 and gradually increase the slope of the surface until the block topples over.
Determine the maximum slope for which the block does not topple (assuming that
there is no sliding between the base of the block and the supporting surface).

SOLUTION
It would normally be assumed that the block would not topple over if the action line of
its weight remains over the area of the base of the block. Thus, the critical position for
the block is when the centre of gravity of the block and the corner of the base lie on
the same vertical line. Therefore, the critical angle of tilt for the block is

b/2 b
θC = tan−1 = tan−1
h 2h (3.7)

If the assumption is correct, Equation 3.7 indicates that if θ < θC the block will not
topple. The assumption can be checked by conducting the simple model tests in
Section 3.3.5. Further details can be seen in [1].

3.3 MODEL DEMONSTRATIONS
3.3.1 Centre of Mass of a Piece of
Cardboard of Arbitrary Shape
This demonstration shows how the centre of mass of a body with an arbitrary shape can be deter-
mined experimentally.
Take a piece of cardboard of any size and shape, such as the one shown in Figure 3.5, and drill
three small holes at arbitrary locations along its edge. Now suspend the cardboard using a drawing
pin through one of the holes and hang from the drawing pin a length of string supporting a weight.
As the diameter of the hole is larger than the diameter of the pin, the cardboard can rotate about
the pin and will be in equilibrium under the action of the self-weight of the cardboard. This means
that the resultant of the gravitational force on the cardboard must pass through the pin. In other
words, the action force (gravitational force) and the reaction force from the pin must be in a vertical
line, which is shown by the string. Now mark a point on the cardboard under the string and draw
a straight line between this point and the hole from which the cardboard is supported. Repeat the
procedure hanging the cardboard from each of the other two holes in turn, and in each case draw a
line between a point on the string and the support hole. It can be seen that the three lines (or their
extensions) are concurrent at a single point, which is the centre of mass of the piece of cardboard
and of course is also the centre of gravity and the centroid of the cardboard.
30  Understanding and Using Structural Concepts

(a) (b) (c)

FIGURE 3.5
(a–c) Locating the centre of mass of a piece of cardboard of arbitrary shape.

3.3.2 Centre of Mass and Centroid of a Body

This model is designed to show that the location of the centre of mass of a body can be different
from that of the centroid of the body.
A 300 mm tall pendulum model is made of wood. The model consists of a supporting base, a
mast fixed at the base and a pendulum pinned to the other end of the mast, as shown in Figure 3.6.
A piece of metal is inserted into the right arm of the pendulum. The centre of mass of the pendulum
must lie on a vertical line passing through the pinned point. The centroid of the pendulum lies on
the intersection of the vertical and horizontal members. Due to the different densities of wood and
metal, the centroid of the pendulum does not coincide with the centre of mass of the body.

FIGURE 3.6
Centre of mass and centroid.
 3  Centre of Mass  31

3.3.3 Centre of Mass of a Body in a Horizontal Plane

This demonstration shows how to locate the centre of mass of a horizontal L-shaped body.
Figure 3.2b shows an L-shaped area whose centre of mass, C, is outside the body. The following
simple experiment, using a fork, a spoon, a toothpick and a wine glass, can be carried out to locate
the centre of mass of an L-shaped body made up from a spoon and a fork.

1. Take a spoon and a fork and insert the spoon into the prongs of the fork, to form an L-shape,
as shown in Figure 3.7a.
2. Then take a toothpick and wedge the toothpick between two prongs of the fork and rest the
head of the toothpick on the spoon, as shown in Figure 3.7b. Make sure the end of the tooth-
pick is firmly in contact with the spoon.
3. The spoon–fork can then be lifted using the toothpick. The toothpick is subjected to an upward
force from the spoon, a downward force from the prong of the fork and forces at the point at
which it is lifted, as shown in Figure 3.7c.

(a) (b)

(c) (d)

(e) (f )

FIGURE 3.7
(a–f) Centre of mass of an L-shaped body in a horizontal plane.
32  Understanding and Using Structural Concepts

4. Place the toothpick on the edge of a wine glass, adjusting its position until the spoon–fork–
toothpick is balanced at the edge of the glass, as shown in Figure 3.7d. According to the defi-
nition of the centre of mass, the contact point between the edge of the glass and the toothpick
is the centre of mass of the spoon–fork–toothpick system. The mass of the toothpick is negli-
gible in comparison with that of the spoon–fork, so the contact point can be considered to be
the centre of mass of the L-shaped spoon–fork system.
5. To reinforce the demonstration, set alight the free end of the toothpick (Figure  3.7e). The
flame goes out when it reaches the edge of the glass as the glass absorbs heat. As shown in
Figure 3.7f, the spoon–fork just balances on the edge of the glass.

If the spoon and fork are made using the same material, the centre of mass and the centroid of
the spoon–fork system will coincide.

3.3.4 Centre of Mass of a Body in a Vertical Plane

This demonstration shows that the lower the centre of mass of a body, the more stable the body.
A cork, two forks, a toothpick, a coin and a wine bottle are used in the demonstration shown in
Figure 3.8a. The procedure for the demonstration is as follows:

1. Push a toothpick into one end of the cork to make a toothpick–cork system as shown in
Figure 3.8b.
2. Try to make the system stand up with the toothpick in contact with the surface of a table. In
practice, this is not possible as the centre of mass of the system is high and easily falls outside
the area of the support point. The weight of the toothpick–cork and the reaction from the
surface of the table that supports the toothpick–cork are not in the same vertical line and this
causes overturning of the system.
3. Push two forks into opposite sides of the cork and place the coin on the top of the bottle.
4. Place the toothpick–cork–fork system on the top of the coin on which the toothpick will stand
in equilibrium as shown in Figure 3.8c.

The addition of the two forks significantly lowers the centre of mass of the toothpick–cork
system in the vertical plane below the contact point of the toothpick on the coin. When the tooth-
pick–cork–fork system is placed on the coin at the top of the bottle, the system rotates until the
action and reaction forces at the contact point of the toothpick are in the same line, producing an
equilibrium configuration.

(a) (b) (c)

FIGURE 3.8
(a–c) Centre of mass of a body in a vertical plane.
 3  Centre of Mass  33

3.3.5 Centre of Mass and Stability

This demonstration shows how the stability of a body relates to the location of its centre of mass
and the size of its base.
Figure 3.9a shows three aluminium blocks with the same height of 150 mm. The square-sectioned
block and the smaller pyramid have the same base area of 29 × 29 mm. The larger pyramid has a
base area of 50 × 50 mm but has the same volume as that of the square-sectioned block. The three
blocks are placed on a board with metal stoppers provided to prevent the blocks from sliding when
the board is inclined. As the board is inclined, its angle of inclination can be measured by the simple
equipment shown in Figure 3.9b. Basic data for the three blocks and the theoretical critical angles
calculated using Equation 3.7 are given in Table 3.1. Theory predicts that the largest critical angle
occurs with the large pyramid and the smallest critical angle occurs with the square-sectioned
block.
The demonstration is as follows:

1. The blocks are placed on the board, as shown in Figure 3.9, in the order of increasing pre-
dicted critical angle.
2. The left-hand end of the board is gradually lifted and the square-sectioned block is the first to
become unstable and topple over (Figure 3.9b). The angle at which the block topples is noted.
3. The board is inclined further and the pyramid with the smaller base is the next to topple
(Figure 3.9c). Although the height of the centre of mass of the two pyramids is the same, the
smaller pyramid has a smaller base and the line of action of its weight lies outside the base
at a lower inclination than is the case for the larger pyramid. The angle at which the smaller
pyramid topples over is noted.
4. When the board is inclined further, the larger pyramid will eventually topple, but its improved
stability over the other two blocks becomes apparent (Figure 3.9d). Once again, the angle at
which the block topples is noted.

(a) (b)

(c) (d)

FIGURE 3.9
(a–d) Centre of mass and stability of three aluminium blocks.
34  Understanding and Using Structural Concepts

TABLE 3.1  Comparison of the Calculated and Measured Critical Angles

Model Cuboid Small Pyramid Large Pyramid

Height of the body (mm) 150 150 150
Height of the centre of mass (mm) 75 37.5 37.5
Width of the base (mm) 29 29 50
Volume (mm3) 1.26 × 105 0.420 × 105 1.25 × 105
Theoretical max. inclination (°) 10.9 21.9 33.7
Measured max. inclination (°) 10 19 31

The angles at which the three blocks toppled are shown in Table 3.1. The results of the demon-
stration, as given in Table 3.1, show that

•• The order in which the blocks topple is as predicted by Equation 3.7 in terms of the measured
inclinations. This confirms that the larger the base, or the lower the centre of mass of a block,
the larger the critical angle that is needed to cause the block to topple.
•• All the measured critical angles are slightly smaller than those predicted by Equation 3.7.

Repeating the experiment several times confirms the measurements and it can be observed
that the bases of the blocks just leave the supporting surface immediately before they topple, which
makes the centres of the masses move outwards. In the theory, the bases of the blocks remain in
contact with the support surface until they topple.
This demonstration shows that the larger the base or the lower the centre of gravity of a body
or both, the larger will be the critical angle needed to cause the body to topple and that this angle
is slightly less than that predicted by theory.

3.3.6 Centre of Mass and Motion

This demonstration shows that a body can appear to move up a slope unaided.
Take two support rails which incline in both the vertical and horizontal planes as shown in
Figure 3.10a. When a doubly conical solid body is placed on the lower ends of the rails, it can be
observed that the body rotates and travels up to the higher end of the rails (Figure 3.10b).
It appears that the body moves against gravity, though, in fact, it moves with gravity. When
the locations of the centre of mass of the body, C, are measured at the lowest and highest ends of

(a) (b)

FIGURE 3.10
(a,b) Centre of mass and motion.
 3  Centre of Mass  35

the rails, it is found that the centre of mass of the body at the lower end is actually higher than that
when the body is at the higher end of the rails. It is gravity that makes the body rotate and appear
to move up the slope.
The reason for this lies in the design parameters of the rail supports and the conical body. The
control condition is that the slope of the conical solid body should be larger than the ratio of the
increased height to a half of the increased width of the rails between the two ends.

3.4 PRACTICAL EXAMPLES
3.4.1 Cranes on Construction Sites
Tower cranes are common sights on construction sites. Such cranes normally have large weights
placed on and around their bases; these weights ensure that the centre of mass of the crane and its
applied loading lie over its base area and that the centre of mass of the crane is low, increasing its
stability. Figure 3.11a shows a typical tower crane on a construction site. Concrete blocks are pur-
posely placed on the base of the crane to lower its centre of mass to keep the crane stable, as shown
in Figure 3.11b.

3.4.2 Eiffel Tower
Building structures are often purposely designed to be larger at their bases and smaller at their
upper parts so that the distribution of the mass of the structure reduces with height. This lowers the
centre of mass of the structure and the greater base dimensions reduce the tendency of the structure
to overturn when subjected to lateral loads, such as wind. A good example of such a structure is the
Eiffel Tower in Paris, as shown in Figure 3.12. The form of the tower is reassuring and appears to
be stable and safe.

3.4.3 Display Unit
Figure 3.13 shows an inclined display unit. The centre of mass of the unit is outside the base of the
unit, as shown in Figure 3.13a. To make the unit stable and prevent it from overturning, an additional
support is fixed to the lower part of the display unit (Figure 3.13b). The added base area effectively
increases the size of the base of the unit to ensure that the centre of mass of the display unit will
be above the area of the new base of the unit. Thus, the inclined display unit becomes stable. Other
safety measures are also applied to ensure that the display unit is stable.

(a) (b)

FIGURE 3.11
(a,b) Cranes on construction sites.
36  Understanding and Using Structural Concepts

FIGURE 3.12
The Eiffel Tower.

(a) (b)

FIGURE 3.13
(a,b) Inclined display unit.
 3  Centre of Mass  37

FIGURE 3.14
One of the two Kio Towers.

3.4.4 Kio Towers
Figure 3.14 shows one of the two 26-storey, 114 m high buildings of the Kio Towers in Madrid,
which are also known as Puerta de Europa (Gateway to Europe). The Kio Towers actually lean
towards each other, each inclined at 15° from the vertical.
The inclinations move the centres of mass of the buildings sideways, tending to cause toppling
effects on the buildings. One of the measures used to reduce these toppling effects was to add mas-
sive concrete counterweights to the basements of the buildings. This measure not only lowered the
centres of the mass of each building but also moved the centre of the mass towards a position above
the centre of the base of the building.

PROBLEMS
1. Reproduce the demonstration shown in Sections 3.3.3 and 3.3.4.
2. Think of one example in everyday life or engineering practice in which the centre
of mass of a body affects the behaviour of the body.
3. The geometry and parameters of the rail support and the conical body shown in
Section 3.4.1 are given in Figure 3.15. Show that the necessary condition that
the conical body moves from the lower end to the higher end of the rail support is

r h tan γ
> or tan θ >
d b tan φ

Identify four other constraint conditions for designing a feasible rail support and
conical body system.
38  Understanding and Using Structural Concepts

t t

φ S2
S1 b

t1 c1 t2

a γ c2 h+a

x1
x2

t2
r
t1
S1 S2
θ r

d e d

FIGURE 3.15
Diagrams of the rail support and conical body system.

REFERENCE
1. Hibbeler, R. C. Mechanics of Materials, 6th edn, Singapore: Prentice-Hall, 2005.
This page intentionally left blank
 CONTENTS
CHAPTER 4 4.1 Definitions and Concepts 41

4.2 Theoretical Background 41

4.3 Model Demonstrations 46

4.4 Practical Examples 47

40