Centres of Mass
Chapter objectives
• Find the centre of mass of a system of
particles distributed in one dimension
• Find the centre of mass of a system of
particles distributed in two dimensions
• Use knowledge of standard results to find the
centre of mass of plane figures
• Consider the equilibrium of a lamina which is
suspended from a fixed point or placed on an
inclined plane
Keywords
centre of mass – the point at which the whole mass
of the body can be considered to be concentrated
Lamina – an object which is very thin compared to
it’s other two dimensions is modelled as a lamina
e.g. a sheet of paper
Uniform – mass is evenly distributed Find the
centre of mass of a system of particles distributed
in one dimension
Find the centre of mass of a system of particles
distributed in one dimension
This means your particles are arranged along a straight line
For example
Three particles of mass 6kg, 3kg and 2.5kg are attached
to a light rod PQ of length 3m at the points P, Q and R,
where PR = 0.9m. Find the position of the centre of mass
of the system.
Draw a diagram or 2!
P
Q
R
3g
2.5g
0.9m
6g
Start by adding the centre of mass to the diagram and let the
distance PG be x.
P
Q
G
3g
11.5g
R
0.9m
6g
2.5g
P
Q
G
3g
11.5g
R
0.9m
6g
Taking moments about P gives:
11.5g × x = 3g × 0.9 + 3 × 2.5g
x = 0.89m
2.5g
There is a general rule you can apply to this
𝑛
𝑛
𝑚𝑖𝑦𝑖 = 𝑚
𝑖=1
𝑦𝑖
𝑖=1
And our text book says you should learn it!
Find the centre of mass of a system of particles
distributed in two dimensions
Now our particles are arranged in a plane
For example
Particles of mass 2kg, 4kg, 5kg and 6kg are attached to the corners of
a light rectangular plate PQRS. Given that PQ = 5cm and QR = 12cm
calculate the distance of the centre of mass of the system from
a) PQ
b) PS
Draw a diagram
The question asks us to:
calculate the distance of the centre of mass of the system from
a) PQ
b) PS
calculate the distance of the centre of mass of the system from
a) PQ
b) PS
There are 3 different methods to solve this question
Method 1: taking P as (0, 0) put the information in a table
Separate Masses
Mass
x co-ord
Y co-ord
2
0
0
4
0
5
5
12
5
6
12
0
Total
Mass
17
𝑥
𝑦
calculate the distance of the centre of mass of the system from
a) PQ
b) PS
Total
Mass
17
𝑥
𝑦
Separate Masses
Mass
x co-ord
Y co-ord
2
0
0
4
0
5
To find the distance of the
centre of mass from PQ we
use the formula:
 m x  m X
i
i
i
5
12
5
6
12
0
To find the distance from PS
we use the formula:
 m y  m Y
i
i
i
6 0  7 2  17 X
2 0  2 5  17 Y
X  7 .7 6
Y  2 .6 5
centre of mass is at (7.66, 2.65)
Method 2
Take moments about the Y axis PQ:
5 x 12 + 6 x 12 = 17𝑥
𝑥 = 7.66
n.b. g has been cancelled
Take moments about the X axis PS:
4 x 5 + 5 x 5 = 17 𝑦
𝑦 = 2.65
centre of mass is at (7.66, 2.65)
Method 3
Use position vectors taking P as (0, 0)
2
0
0
0
0
+6
+
12
0
72
0
+
+5
60
25
12
5
+
+4
0
20
0
5
= 17
= 17
7.66
2.65
=
𝑥
𝑦
𝑥
𝑦
𝑥
𝑦
centre of mass is at (7.66, 2.65)
Pick a method and try this example
The diagram below shows a series of particles that make up a
system. The centre of mass of the system is at the point (x,y). Find
the coordinates of the centre of mass of the system
2Kg
Y
3kg
(x,y)
X
2.5kg
4Kg
−2
2
+3
2
−4
4
+
15
6
5
2
+4
+
12
−8
3
−2
+
+ 2.5
2.5
−2.5
25.5
−0.5
2.22
−0.04
1
−1
= 11.5
= 11.5
𝑥
𝑦
= 11.5
𝑥
𝑦
=
centre of mass is at (2.22, -0.04)
𝑥
𝑦
𝑥
𝑦
There is a general rule that goes with all this.
It can be found on page 35.
There is an error in question 9 on page 39
The third side is CD not CA
Use knowledge of standard results to find the centre of
mass of plane figures
There are some standard results to be taken for granted when it comes to
the centre of mass.



Uniform rectangular lamina:- at centre of the shape
Uniform circular disc:- at centre of disc
Uniform triangular lamina:o Equilateral:- at centre
o Isosceles:- at the intersection of the medians
A median is a line that joins a vertex of a triangle to the centre of the side
opposite to the vertex. The centre of mass of a scalene triangle is at a point
one third of the way along the median (from the edge).
The centre of mass of the triangle ABC is at the point G, where EG =
⅓EC.
B
The centre of mass of the triangle ABC
is at the point G, where EG = ⅓EC
D
E
G
A
F
C
If the coordinates of the three vertices of a uniform
triangular lamina are (x1, y1), (x2, y2) and (x3, y3) then the
coordinates of the centre of mass are found by taking the
mean of the coordinates of the 3 vertices.
Calculate the coordinates of the centre of mass of the uniform triangular
lamina ABC if the point A is placed at the origin.
B
12m
18m
C
A
centre of mass G is at the point
0+18+0
3
,
0+0+12
3
= 6, 4

r
2α
Uniform Sector:- centre of mass is on the axis of symmetry at a
distance
2r sin 
3
from the centre, where α is measured in radians
(note the use of α in the formula and 2α in the diagram.)
This formulae is in the booklet
C
G
A
O
B
Find the centre of mass of the lamina with radius 5 cm
The centre of mass lies on the line of symmetry OC
𝜋
Using the result for a sector with r = 5 and α =
2
𝜋
Then OG =
2 𝑥 5 𝑥 𝑠𝑖𝑛 2
3𝜋
2
=
20
3𝜋
The centre of mass lies on the line OC a distance
20
3𝜋
from O.
Application to composite figures
C
B
12cm
9cm
F
A
7.5cm
D
E
You can find the centre of mass for this shape by applying the ideas we have already come
across:
1. Work out the centre of mass for the rectangle
2. Work out the centre of mass for the triangle
3. Find the centre of mass of both ‘particles’
centre of mass for ABCE
Take A as the origin then
(6, 4.5) is the centre of mass
centre of mass for CDE
=
12+12+19.5
3
,
0+9+4.5
3
= (14.5, 4.5)
Because the shapes are uniform the area of each can be used to represent it’s mass
Mass of rectangle = 108 cm²
mass of triangle = 33.75cm²
centre of mass of shape
𝑥
6
14.5
= 108
+ 33.75
= 141.75 𝑦
4.5
4.5
8.02
=
4.5
𝑥
𝑦
centre of mass is 8.02 cm from AB and 4.5 cm from AE
The object below is formed by removing a uniform semi
circular disc of radius 3m from a second uniform semi
circular disc of radius 6m. Calculate the centre of mass
of the object.
6m
3m
By definition the centre of mass will lie on the mirror line.
The formula for the centre of mass for a uniform sector is
2r s in 
3
where the angle at the centre is 2α. Obviously in the above example 2α =
∏.
Using the tabular approach
Separate Masses
Little disc
Mass
Big disc
9
18 
2
y co-ord
2  3  sin
3
2
Total Mass
27 
2

2 
4
3
sin

2
2  6  sin
3

2  8 sin 

2
X
2
the disc we are interested in can be found by subtracting
the small semi circular disc from the larger one.
 m x  m X
i
i
i
8
   9 4
  27

18   sin
 
 sin  
X



2  2

2
2

14 4  18 
27
2
X
X  2 .9 7
So the centre of mass is 2.97cm from the centre along
the axis of symmetry.
The object below is formed by removing two uniform
circular discs from a uniform rectangular lamina.
Calculate the centre of mass of the plate.
16cm
B
C
1.5cm
10cm
3cm
6cm
A
D
12cm
The same principle of adding and subtracting parts can be
applied to this question
(120-4.5∏)
Separate Masses
Total Mass
Rectangular
Circle 1
Circle 2
Plate
Lamina
Mass
x co-ord
Y co-ord
2.25∏
3
6
2.25∏
12
6
120
6
5
120-4.5∏
X
Y
Plate = Rectangular Lamina - Circle 1 - Circle 2
 m x  m X
i
X
X
i
i
= 720 – (7.75∏ + 27∏)
= 5.77 cm
Looking at it vertically:
 m y  m Y
i
i
i
(120-4.5∏) Y = 600 – (13.5∏ + 13.5∏)
Y
= 4.87cm
Therefore the centre of mass is 5.77cm from AB and 4.87cm from AD.
2.5
You can find the centre of mass of a framework by using the centre
of mass of each rod or wire that makes up the framework
The framework ABC is made up of three uniform rods and
a semi circular arc. Find the centre of mass of the
framework assuming that A is at the origin
B
20cm
12cm
16cm
A
C
Remembering that the formula for the centre of mass of a uniform
circular arc is
r sin 

and that  

2
. Also note that the centre of mass
of the arc will have a negative x coordinate.
Separate Masses
Mass
x co-ord
Total Mass
Rod AB
Rod AC
Rod BC
Arc AB
Framework
12m
16m
20m
6∏m
(48 + 6∏)m
0
8
8
 6 sin


2   12

X
2
Y co-ord
6
0
6
6
Y
Looking at the problem horizontally:
 m x  m X
i
i
i
(16  8)  (2 0  8)  (6  
 12

)  ( 4 8  6  )X
X  3 .2 3cm
And now vertically:
 m y  m Y
i
i
i
(12  8)  (2 0  6)  (6   6)  ( 4 8  6  )Y
Y  4 .5 6cm
Therefore the centre of mass is at the point with coordinates
(3.23,4.56).
2.6 Laminas in equilibrium
A suspended lamina will be in equilibrium when its centre
of mass is directly below the point of suspension
It can be suspended from a fixed point or a pivot
string
R
Fixed point
T
O
Point of
suspension
pivot
O
G
W
G
W
Lets look at the example in the book on page 54
Equilibrium of a uniform lamina on an inclined plane.
For an inclined lamina to remain in equilibrium on an inclined plane the
line of action of the weight must fall within the side of the lamina that
is in contact with the plane (as shown in the diagram below).
G
θ
In the second diagram the object will topple over
G
θ
A uniform rectangular lamina is placed on a plane inclined at an angle θ.
Given that the lamina is in limiting equilibrium find the angle θ.
6cm
G
15cm
θ
By simple trigonometry:
T an 
3
7 .5
   2 1.8

Done