The Center of Mass of an object (of mass m) is the single point that moves in the same way as a point mass (of mass m) would move when subjected to the same external forces that act on the object. That is, if the resultant force acting on an object (or system of objects) of mass m is F, the acceleration of the center of mass of the object (or system) is given by acm = F/m

If the object is considered to be composed of tiny masses m1, m2, m3, and so on, at coordinates (x1,y1,z1), (x2,y2,z2), and so on, then the coordinates of the center of mass are given by

where Σ = sigma, the sequence of

The x-coordinate
Xcm = Σ(xi * mi) / Σ(mi)

The y-coordinate
Ycm = Σ(yi * mi) / Σ(mi)

The z-coordinate
Zcm = Σ(zi * mi) / Σ(mi)

where the sums extend over all masses composing the object. In a uniform gravitational field, the center of mass and the center of gravity coincide.

For the integration method of center of mass, check out Center of Mass: Part Two.

Aside from the head area the c.o.m. is also the recommended point of aim for defensive shooting, called the A zone, which is pretty much the center of a human torso.

While instant incapacitation is still your goal in any defensive shooting aiming for the head is not very practical as the high adrenaline levels pumping in your veins combined with a small and highly mobile target does not make for high hit probability.

Because of the human body's amazing stretchability, one shot stops are not all that common as Hollyweird might have you believe. The rule of thumb is keep on firing until the threat is neutralized. As always, remember that you do not shoot to kill, you only shoot to stay alive.

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 = Σmixi / Σmi
Y = Σmiyi / Σmi

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 three-dimentional 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: πr2


            #############
         ##################
      #####.,            ###
    ####     .,           ##
  ###          +,       ###
 ###             .,-_ ###
###               | ###
#####################
 ###################

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: r2θ / 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: r2 (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 right-angled 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 right-angled 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 rectangle-triangle 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 bottom-left 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 right-angled, 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 = Σmixi / Σmi = (1.5 * 12 + 32/3 * 4) / (12 + 4)
= 32 2/3 / 16
= 2 1/24

Y = Σmiyi / Σmi = (2 * 12 + 11/3 * 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).

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