We use moments to
find the centre of mass of uniform plane figures and discrete mass
distributions. In M1 we used moments to calculate the centre of mass for non
uniform rods but in M2 we need to consider two dimensions.
M1 Recap
Example 1
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.
Start by adding
the centre of mass to the diagram and let the distance PG be x.
Taking moments
about P gives:
11.5g × x = 3g ×
0.9 + 3 × 2.5g
x = 0.89m
Centre of mass of a system of particles distributed in two dimensions
The principle
applied above is simply applied firstly in the horizontal direction and then
vertically.
Example 2
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
It is very easy to
make simple numerical mistakes with these questions and therefore you are
advised to set the question out in a table. It is assumed that the centre of
mass horizontally is at 
and vertically at
.
|
|
Separate Masses |
Total Mass |
|||
|
Mass |
2 |
4 |
5 |
6 |
17 |
|
x co-ord |
0 |
0 |
12 |
12 |
|
|
Y co-ord |
0 |
5 |
5 |
0 |
|
To find the distance
of the centre of mass from PQ we use the formula:
To find the
distance from PS we use the formula:
Example 3
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.
With an example
such as this it is once again easy to make a simple numerical mistake as
students may miss the minus signs.
Using the same
tabular approach:
|
|
Separate Masses |
Total Mass |
|||
|
Mass |
2 |
2.5 |
4 |
3 |
11.5 |
|
x co-ord |
-2 |
1 |
3 |
5 |
|
|
Y co-ord |
2 |
-1 |
-2 |
2 |
|
Using the formula:
And similarly for
the y direction:
Therefore the centre
of mass of the system is at (2.22,0.04)
Questions 1
1 A light rod PR of length 3.5m has particles
of mass 1.5kg, 3kg and 2.5kg attached to it at P Q and R respectively, where PQ
= 1.5m. Determine the distance of the centre of mass from R.
2 Particles of mass 4kg, 2.5kg, 6kg and 3kg are
attached to the corners of a light rectangular plate PQRS. Given that PQ = 8cm
and QR = 16cm calculate the distance of the centre of mass of the system from
a) PQ
b) PS
3 Four particles of mass 1kg, 2kg, 5kg and
2.5kg lie in the (x,y) plane at the points with coordinates (1,2), (-2,3),
(4,2) and (-3,2) respectively. Calculate the coordinates of the centre of mass
of the system.
4 The system below is made up of three light rods. Three masses of
value 2.5kg, 4kg and 1kg are placed at the vertices A, B and C respectively.
Calculate the distance of the centre of mass from
a) AB
b) AC.
Centre of mass of a uniform plane lamina
Obviously there
are some standard results to be taken for granted.
- Uniform rectangular lamina:- at centre
of the shape
- Uniform circular disc:- at centre of
disc
- Uniform triangular lamina:-
- Equilateral:- at centre
- 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.
- Uniform circular arc:- the centre of mass is along the axis of
symmetry at a distance
from the centre
where α is measured in radians.
- Uniform Sector:-
centre of mass is on the axis of symmetry at a distance
from the centre,
where α is measured in radians.
(note the use of α in the formula and
2α in the diagram.)
Example 4
Calculate the
coordinates of the centre of mass of the uniform triangular lamina ABC if the
point A is placed at the origin.
From the
definitions above the centre of mass is at the point that is one third of the
way along DC.
Note that D has
coordinates (0,6).
Therefore C of M
is at (6,4).
Application to composite figures.
Example 5
The centre of mass
for the shape below can be found by using the same ideas as those set out
above. All you need to do is treat each part separately.
Assume that the
shape ABCDE is a uniform lamina made up of a rectangle and an isosceles
triangle.
One only needs to
consider the distance of the centre of mass from the edge AB as the shape is
symmetrical and therefore the centre of M will lie along the mirror line.
Since the
rectangle ABCE is uniform its centre of mass will be 6 cm from AB. The centre
of mass of the isosceles triangle will be one third of the way along the line
FD which is a distance of 14.5cm from the line AB. By applying the tabular
method we get:-