The centre of mass of an object is the point where the weight acts  it can be replaced by a particle at the centre of mass and the effect of forces and moments would stay the same. If you stick a pin at the centre of mass of an object, it would balance. If you hang the object, the centre of mass will be vertically below the point where it's hung.
Finding the centre of mass of a particle system
A particle system is the fundamental model of finding centres of mass. In such a system, the centre of mass of the system for any direction is the sum of all the particles' positions multiplied by their mass, divided by the total mass of the particles. In more mathematical terms:
For a centre of mass (X, Y):
X = Σm_{i}x_{i} / Σm_{i}
Y = Σm_{i}y_{i} / Σm_{i}
Example
Find the centre of mass of these particles:
 1) Position (1, 1), mass 1
 o 2) Position (2, 3), mass 2
 3) Position (7, 1), mass 5
o O
+
X = (1 * 1) + (2 * 2) + (7 * 5) / (1 + 2 + 5)
= 40 / 8
= 5
Y = (1 * 1) + (3 * 2) + (1 * 5) / (1 + 2 + 5)
= 12 / 8
= 1.5
The heaviest particle at position 3 has tilted the centre of mass towards it. So the centre of mass is located about here:

 o

o ' O
+
If you are dealing with a threedimentional system, the Z coordinate can be found in much the same way.
Finding the centre of mass of uniform plane laminae
A uniform lamina is used to represent anything flat  a piece of paper, a sign, or a CD case, for example.
All axes of reflectional symmetry have been drawn in  for a uniform lamina, the centre of mass must lie on these axes.
For an explanation of the mass formulae given, see common geometric formulas.
Squares and rectangles
A rectangle, as everyone knows, has two lines of symmetry  one horizontal and one vertical. This makes it easy to find the centre of mass:
######################################
######################################
###  ###
###  ###
###  ###
###  ###
###  ###
###+###
###  ###
###  ###
###  ###
###  ###
###  ###
######################################
######################################
The centre of mass is, in this case, in the centre  halfway along the width, and halfway up the height. Squares, being rectangles, follow the same rule.
Circles, segments and sectors
Circles are just as easy. A circular disc has infinitely many diameters, and infinitely many axes of symmetry; the centre of mass is in the centre of the circle.
#############
###################
#####  #####
####  #####
###  ###
###  ###
###  ###
###+###
###  ###
###  ###
###  ###
####  #####
####  ####
###################
###############
For a circular disc, with radius r:
Centre of mass: The centre of the circle
Mass: πr^{2}
#############
##################
#####., ###
#### ., ##
### +, ###
### .,_ ###
###  ###
#####################
###################
The first thing to notice is that there is an axis of symmetry, bisecting the angle at the centre into two angles θ. So the centre of mass must lie on this line.
But where on the line? There is no simple way to work it out; we need to use a formula. That's just how life is sometimes.
For a sector of a circle, with radius r, and angle at centre 2θ:
Centre of Mass: 2r sin θ / 3θ away from the centre
Mass: r^{2}θ / 2
#############
##################
#####., ###
#### +, ####
### .,(########
### #######
### ########
#########
####
I know that my diagram doesn't show it very well, but that's supposed to be a segment of a circle.
The line of symmetry, although it looks the same (sorry), is still there, although not in the same place as it was. Because a segment is a sector with a bit near the centre missing, the centre of mass is going to be pushed away from the centre slightly.
For a segment of a circle, with radius r, and angle at centre 2θ:
Centre of Mass: r sin θ / θ away from the centre
Mass: r^{2} (θ  sin θ) / 2
Triangles
Finding the centre of mass of triangles requires more geometric knowledge than the other shapes: The centre of mass for any triangle is located at ^{2}/_{3} of the distance from any vertex to the midpoint of the opposite side. It looks much nicer on a diagram:
##
####
##,###
## , ###
## ,, ###
## , ###
## ,,.. ###
## +, ###
## .. ,, ###
##. , ###
#####################
#######################
Here, I have drawn two lines from vertex to opposite edge, and the centre of mass will be where the two lines intersect. This method is far easier to do for rightangled triangles than any other, because ^{1}/_{3} of the base and height will give you your answer.
For any triangle, with base b, and height h:
Centre of Mass: ^{2}/_{3} of the distance from a vertex to a midpoint
Mass: bh / 2
Collections of shapes
Consider the following lamina:
######################
#######################
### 3 ###
###  ###
###  ###
###  ###
###  ###
### 4  ###
###  ###
###  ###
###  ###
###  2 ###
###  ###
###################################
####################################
We have a 4x3 rectangle attached to a 4x2 rightangled triangle. It doesn't follow any of the rules for the shapes above. So how can we find the centre of mass of such a rectangletriangle hybrid?
The answer  and this is why particle systems came first in the node  is to turn each shape into a particle at its centre of mass, then find the centre of mass of
that system.
Taking the bottomleft corner as the origin:
 The centre of mass of the rectangle is at half the base and half the height: (^{3}/_{2}, ^{4}/_{2}) = (1.5, 2); its mass is 3*4 = 12.
 Luckily, our triangle is rightangled, so its centre of mass is at a third of the base and the height, plus 3 horizontally (from the rectangle): (3 + ^{2}/_{3}, 1 ^{1}/_{3}); the mass is 2 * 4 / 2 = 4.
From this, we can work out the centre of mass, using the original equations:
X = Σm_{i}x_{i} / Σm_{i}
= (1.5 * 12 + 3^{2}/_{3} * 4) / (12 + 4)
= 32 ^{2}/_{3} / 16
= 2 ^{1}/_{24}
Y = Σm_{i}y_{i} / Σm_{i}
= (2 * 12 + 1^{1}/_{3} * 4) / (12 + 4)
= 29 ^{1}/_{3} / 16
= 1 ^{5}/_{6}
Which leaves us with the centre of mass of the entire lamina of (2 ^{1}/_{24}, 1 ^{5}/_{6}).