1. CENTRE OF MASS

2. The centre of mass of a body is the point at which its
weight acts.
The centre of mass of a body lies on any line of symmetry.
The principle to find a centre of mass of a composite body
is that, the sum of the moments of the composite parts of
the body about a given axis is equal to the moment of the
whole body about the same axis.

3. Example 1: A uniform wire of length 24cm is bent to form a triangle
ABC where AB = 6cm and BC = 8cm. Find the distance
of the centre of mass from AB and from BC. A B C
Since AB = 6, BC = 8
AC = 10 6m 8m
10m
By Pythagoras’ theorem, the triangle is right angled at B.
( = 102 )
Body
Weight
Distance of G
from AB.
from BC. AB BC AC
Whole
Let the mass per unit length be m,
and the centre of mass be at G. 6m 3 8m 4
The mass of each of the 3 sides
acts at their centres.
10m 4 3 x y
24m
Taking moments about AB:
6m(0) + 8m(4) + 10m(4) = 24m( x ) x = 3
Taking moments about BC:
6m(3) + 8m(0) + 10m(3) = 24m( y ) y = 2

4. Problems involving laminas are treated in the same way, with the
mass of each part acting at its centre of mass:
Example 2: For the lamina ABCDEF shown, find the distance of the
centre of mass from AB and from BC. A C D B E F 4 2 6 5
Let the mass per unit area be m,
and the centre of mass be at G. P
Body
Weight
Distance of G
from AB.
from BC.
ABPF
PCDE
Whole
24m 2 3
10m
6.5 1 y
34m x
Taking moments about AB:
24m(2) + 10m(6.5) = 34m( x ) x =
3.32
(3 sig.figs.)
Taking moments about BC:
24m(3) + 10m(1) = 34m( y ) y =
2.41
(3 sig.figs.)

5. Example 3: For the lamina ABCDEF shown, the section PCDE is made
of a material which is twice as heavy as the material used
for the section ABPF. Find the distance of the centre of mass from AB
and from BC.
Let the mass per unit area be m for ABPF, and
2m for PCDE and the centre of mass be at G. A C D B E F 4 2 6 5 P
Body
Weight
Distance of G
from AB.
from BC.
ABPF
PCDE
Whole
24m 2 3
20m
6.5 1
44m x y
Taking moments about AB:
24m(2) + 20m(6.5) = 44m( x ) x =
4.05
(3 sig.figs.)
Taking moments about BC:
24m(3) + 20m(1) = 44m( y ) y =
2.09
(3 sig.figs.)

6. Example 4: For the lamina ABCDEF shown, the section ABPF is
twice as heavy as the section PCDE. Find the distance of
the centre of mass from AB and from BC. A C D B E F 4 2 6 5 P
Body
Weight
Distance of G
from AB.
from BC.
ABPF
PCDE
Whole 2m 2 3 m
6.5 1 3m x y
Taking moments about AB:
2m(2) + m(6.5) = 3m( x ) x =
3.5
Taking moments about BC:
2m(3) + m(1) = 3m( y ) y =
2.33
(3 sig.figs.)

7. Certain shapes have centres of mass at points which can be quoted.
The centre of mass of a triangular lamina lies one third of the
distance along the line from the centre of a side of the triangle
to the opposite vertex.

8. The centre of mass of a lamina in the form of a sector of a circle is
another standard result. O G r α A B
If OAB is a sector of a circle, radius r, with
angle 2α at the centre then the centre of mass
G lies on the line of symmetry such that:
OG =
2r sin α
3 α
where α is measured in radians. O G r
Hence for a semicircle, α = π 2
2r sin 3 2 π =
4 r 3π
OG =

9. The centre of mass of a uniform wire in the form of an arc of a circle
is one more standard result. G r O α A B
If AB is an arc of a circle, radius r, with
angle 2α at the centre, then the centre of
mass G lies on the line of symmetry such
that:
OG =
r sin α α
where α is measured in radians.

10. Example 5: For the lamina ABCD shown, find the distance of the
centre of mass from AB and from BC. 6 2 3 A B C D
Let the mass per unit area be m,
and the centre of mass be at G. P
Body
Weight
Distance of G
from AB.
from BC.
PBCD
APD
Whole
12m 3 1 9m 2 3
21m x y
Taking moments about AB:
12m(3) + 9m(2) = 21m( x ) x =
2.57
(3 sig.figs.)
Taking moments about BC:
12m(1) + 9m(3) = 21m( y ) y =
1.86
(3 sig.figs.)

11. Example 6: A uniform circular lamina, radius 5a has a circular hole of
radius 2a. The circumference of the hole lies on the centre
of the lamina. Find the distance of the centre of mass of the lamina from
its centre.
Let the mass per unit area be m,
and the centre of mass be at G. x y
Body
Weight
Distance of G
from y-axis. 5a 2a
21a2 π m x
Note: G lies
on the x-axis
by symmetry.
4a2 π m 7a
25a2 π m 5a
Taking moments about the y-axis:
21a2 π m
x +
4a2 π m ( 7a ) =
25a2 π m ( 5a )
x + 28a = 125a 21
The distance of G from its centre is: a 97 21 97 21
a =
5a – 8 21 a
x =

12. Find the angle that DE makes with the vertical.
Example 7: A lamina ABCDE is made by joining a rectangle to an isosceles triangle as shown: a) Find the distance of the centre of mass from CD. The lamina is now suspended from point E.
Find the angle that DE makes with the vertical. C D E B A 3 5 8
Body
Weight
Distance of G
from CD.
Let the mass per
unit area be m,
and the centre of
mass be at G.
12m 6
40m
2.5 y
52m
12m(6) + 40m(2.5) = 52m( y )
Taking moments about CD: y =
3.308
The distance of the centre of mass from CD is 3.31 (3 sig.figs.)

13. Since the only forces acting are equal and opposite, the centre B C
Since the only forces acting are
equal and opposite, the centre
of mass, G must lie vertically
below E. y 5 8 4 θ G
By symmetry, G must also lie
on the line joining A to the
mid-point of CD.
5 – y 4 =
5 – 3.308 4
tan θ =
θ = 22.9°
DE makes an angle 90 – θ with the vertical,
= 67.1°.

14. the plane is sufficiently rough to prevent the lamina sliding down the
Example 7a: The lamina ABCDE in example 7 is now placed on an inclined plane with DC on the plane as shown. Given that
the plane is sufficiently rough to prevent the lamina sliding down the
plane, find the greatest angle of inclination of the plane to the horizontal such that the lamina will not topple. C D B A E
The lamina will be on the point
of toppling when the weight acts
through the point D as shown: G θ 4 y θ
also θ.
If the inclination of the plane is θ,
then the angle between the line of
action of the weight, and the line of
symmetry of the lamina is = 4 y 4
3.308
tan θ =
θ = 50.4°.

15. The centre of mass of a body is the point at which its weight acts.
Summary of key points:
The centre of mass of a body is the point at which its weight acts.
The centre of mass of a body lies on any line of symmetry.
The centre of mass of a triangular lamina lies one third of the
distance along the line from the centre of a side of the triangle
to the opposite vertex. O G r α A B
For a lamina in the form of a sector of a circle: G r O α A B
For an arc of a circle:
OG =
2r sin α
3 α
OG =
r sin α α
For both, α is measured in radians.
This PowerPoint produced by R.Collins ; Updated Sep.2011