Learn the definition of center of mass and learn how to calculate it.

What is the center of mass?

The center of mass is a position defined relative to an object or system of objects. It is the average position of all the parts of the system, weighted according to their masses.
For simple rigid objects with uniform density, the center of mass is located at the centroid. For example, the center of mass of a uniform disc shape would be at its center. Sometimes the center of mass doesn't fall anywhere on the object. The center of mass of a ring for example is located at its center, where there isn't any material.
Figure 1: Center of mass for some simple geometric shapes (red dots).
Figure 1: Center of mass for some simple geometric shapes (red dots).
For more complicated shapes, we need a more general mathematical definition of the center of mass : it is the unique position at which the weighted position vectors of all the parts of a system sum up to zero.
A weighted position vector is a vector which points from the origin to an object and has magnitude mm, where mm is the mass of the object. I.e. (mr)(m\cdot \mathbf{{r}}) where r\mathbf{{r}} is a position vector pointing from the origin to the object.
For a system of nn objects, the center of mass is the point where
i=1nmiri=0\sum_{i=1}^n m_i \mathbf{{r}}_i = 0
If the origin we have chosen for our vectors happens to already be at the center of mass, then the sum will come to zero. Otherwise, the vector sum S\mathbf{S} will point us to the center of mass. Here MM is the total mass of the system.
S=1Mi=1nmiri\mathbf{S} = \frac{1}{M} \sum_{i=1}^n m_i \mathbf{{r}}_i

What is useful about the center of mass?

The interesting thing about the center of mass of an object or system is that it is the point where any uniform force on the object acts. This is useful because it makes it easy to solve mechanics problems where we have to describe the motion of oddly-shaped objects and complicated systems.
For the purposes of calculation, we can treat an oddly-shaped object as if all its mass is concentrated in a tiny object located at the center of mass. We sometimes call this imaginary object a point mass.
If we push on a rigid object at its center of mass, then the object will always move as if it is a point mass. It will not rotate about any axis, regardless of its actual shape. If the object is subjected to an unbalanced force at some other point, then it will begin rotating about the center of mass.

How can we find the center of mass of any object or system?

In general the center of mass can be found by vector addition of the weighted position vectors which point to the center of mass of each object in a system. One quick technique which lets us avoid the use of vector arithmetic is finding the center of mass separately for components along each axis. I.e:
For object positions along the x axis:
COMx=m1x1+m2x2+m3x3+m1+m2+m3+\mathrm{COM}_x = \frac{m_1 \cdot x_1 + m_2 \cdot x_2 + m_3 \cdot x_3 + \ldots}{m_1 + m_2 + m_3 + \ldots}
And similarly for the y axis:
COMy=m1y1+m2y2+m3y3+m1+m2+m3+\mathrm{COM}_y = \frac{m_1 \cdot y_1 + m_2 \cdot y_2 + m_3 \cdot y_3 + \ldots}{m_1 + m_2 + m_3 + \ldots}
Together, these give the full coordinates (COMx,COMy)(\mathrm{COM}_x, \mathrm{COM}_y) of the center of mass of the system. For example, consider the system of three flat objects of uniform density shown in Figure 2.
Figure 2: A system of three flat objects.
Figure 2: A system of three flat objects.
The location of the center of mass in the xx direction is:
14+16+2121+1+2=8.5\frac{1\cdot 4 + 1\cdot 6 + 2\cdot 12}{1+1+2} = 8.5
and in the yy direction:
15+112+28.51+1+2=8.5\frac{1\cdot 5 + 1\cdot 12 + 2\cdot 8.5}{1+1+2} = 8.5
Complex objects can often be represented as collections of simple shapes, each with uniform mass. We can then represent each component shape as a point mass located at the centroid. Voids within objects can even be accounted for by representing them as shapes with negative mass.
Consider the irregularly-shaped flat, uniform density object shown in Figure 3a.
Figure 3: (a) An irregularly shaped flat object. (b) Object divided into simple shapes.
Figure 3: (a) An irregularly shaped flat object. (b) Object divided into simple shapes.
We can break this object up into four rectangles and one circle as shown in figure 3b. Here we are only interested in the position of the center of mass in the relative units shown in the figure. The material has uniform density so the mass is proportional to the area. For simplicity we can represent the mass of each section in units of 'squares' as shown in the diagram.
In the xx direction, the center of mass is at:
Note that the area of the circular void is π1.527.1\pi \cdot 1.5^2 \simeq 7.1. This is accounted for as a negative mass.
In the yy direction:

What is the center of gravity?

The center of gravity is the point through which the force of gravity acts on an object or system. In most mechanics problems the gravitational field is assumed to be uniform. The center of gravity is then in exactly the same position as the center of mass. The terms center of gravity and center of mass tend to often be used interchangeably since they are often at the same location.

What about determining the center of mass for a real object?

There are a couple of useful experimental tests that can be done to determine the center of mass of rigid physical objects.
The table edge method (Figure 4) can be used to find the center of mass of small rigid objects with at least one flat side. The object is pushed slowly without rotating along the surface of a table towards an edge. At the point where the object is just about to fall, a line is drawn parallel to the table edge. The procedure is repeated with the object rotated 90°. The intersection point of the two lines gives the location of the center of mass in the plane of the table.
Figure 4: Table edge method used for finding the center of mass of an irregular object.
Figure 4: Table edge method used for finding the center of mass of an irregular object.
The plumb line method (Figure 5) is also useful for objects which can be suspended freely about a point of rotation. An irregularly shaped piece of cardboard suspended on a pin-board is a good example of this. The cardboard pivots freely around the pin under gravity and reaches a stable point. A plumb line is hung from the pin and used to mark a line on the object. The pin is moved to another location and the procedure repeated. The center of mass then lies beneath the intersection point of the two lines.
Figure 5: Plumb line method being used to find the center of mass of an irregular object.
Figure 5: Plumb line method being used to find the center of mass of an irregular object.

Center of mass and toppling stability

One useful application of the center of mass is determining the maximum angle that an object can be tilted before it will topple over.
Figure 6a shows a cross section of a truck. The truck has been poorly loaded with many heavy items loaded on the left-hand side. The center of mass is shown as a red dot. A red line extends down from the center of mass, representing the force of gravity. Gravity acts on all the weight of the truck through this line.
If the truck is tipped at an angle θt\theta_\mathrm{t} (as shown in figure 6b) then all the weight of the truck will be supported by the left-most edge of the left wheel. Should the angle be further increased then the point of support will move outside of any point of contact with the road and the truck is guaranteed to topple over. The angle θt\theta_\mathrm{t} is the topple limit.
Figure 6: Topple limit of a poorly loaded truck.
Figure 6: Topple limit of a poorly loaded truck.
Exercise 1: Determine the topple limit for the uniform density object shown in figure 7 as it is tipped to the right.
Figure 7: Exercise 1 toppling object.
Figure 7: Exercise 1 toppling object.
We first find the center of mass of the shape by dividing it into three rectangles:
In the xx direction : 247+246+125.524+24+12=6.3\frac{24\cdot 7 + 24\cdot 6 + 12\cdot 5.5}{24+24+12} = 6.3
In the yy direction : 2411+247+12224+24+12=7.6\frac{24\cdot 11 + 24\cdot 7 + 12\cdot 2}{24+24+12} = 7.6
Figure 8: Calculation of topple angle.
Figure 8: Calculation of topple angle.
Finding the topple limit θt\theta_\mathrm{t} is now a matter of using trigonometry to find the interior angle of the triangle shown in Figure 8.
The rightmost point of contact is located at x=7x=7 and the center of mass is located at x=6.3x=6.3. This leaves a difference of 0.7 which is the opposite side of the triangle.
The angle θt\theta_\mathrm{t} indicated on the diagram must be equal to the interior angle of the triangle because θt+α=90\theta_\mathrm{t} + \alpha = 90^\circ. From the trigonometry of right-angled triangles:
θt=tan1(0.77.6)=5.26\theta_\mathrm{t} = \tan^{-1}\left( \frac{0.7}{7.6} \right) = 5.26^\circ

Center of mass reference frame

When the term reference frame is used in physics it refers to the coordinate system used for calculations. A reference frame has a set of axes and an origin (zero point). In most problems the reference frame is fixed relative to the laboratory and a convenient (but arbitrary) origin point is chosen. This is known as the laboratory reference frame. However, in classical physics it is also possible to use any other reference frame and expect the laws of physics to hold within it. This includes reference frames which are moving relative to the laboratory.
One very useful property of the center of mass is that it can be used to define the origin of a moving reference frame for a system. This reference frame is sometimes called the COM frame. The COM frame is particularly useful in collision problems. It turns out that the momentum of a fully-defined system measured in the COM frame is always zero. This means that calculations can often be much simpler when done in the COM frame compared to the laboratory reference frame. Let us consider a simple example:
Consider two trolleys running along a track in the same direction as shown in Figure 9. The left trolley is traveling faster so there will inevitably be a collision. Let's assume the collision is elastic. What are the velocities after the collision?
Figure 9: Two moving trolleys about to collide : the collision is much easier to analyze in the COM frame.
Figure 9: Two moving trolleys about to collide : the collision is much easier to analyze in the COM frame.
We first need to find the velocity of the center of mass in the laboratory frame. Since velocity is distance/time, we can simply put velocity in the place of position in the center of mass equation and proceed just as for a stationary center of mass. The result will be (position of the center of mass) / (second): i.e., the velocity of the center of mass, vCMv_\mathrm{CM}:
vCM=(215+43) kgm/s(2+4) kg=7 m/s\begin{aligned} v_\mathrm{CM} &= \frac{(2\cdot 15 + 4\cdot 3)~\mathrm{kg\cdot m/s}}{(2 + 4)~\mathrm{kg}} \\&= 7~\mathrm{m/s}\end{aligned}
We can now find the initial (subscript i) velocities and momenta for the trolleys a and b from the point of view of the center of mass reference frame. This is done by subtracting the velocity of the COM frame. Quantities in this frame will be denoted with a ' (prime symbol) to distinguish them from the laboratory frame.
vai=157=8 m/sv'_{ai} = 15 - 7 = 8~\mathrm{m/s}
vbi=37=4 m/sv'_{bi} = 3 - 7 = -4~\mathrm{m/s}
pai=82=16 kgm/sp'_{ai} = 8 \cdot 2 = 16~\mathrm{kg\cdot m/s}
pbi=44=16 kgm/sp'_{bi} = -4 \cdot 4 = -16~\mathrm{kg\cdot m/s}
As can be seen, the momenta of the two colliding objects is equal in magnitude and opposite in direction. This is always the case for such a collision in the COM frame. In this example, we know that the collision is elastic and therefore the momenta simply swap signs after the collision.
paf=16 kgm/sp'_{af} = -16~\mathrm{kg\cdot m/s}
pbf=16 kgm/sp'_{bf} = 16~\mathrm{kg\cdot m/s}
The final velocities are then:
vaf=16/2=8 m/sv'_{af} = -16/2 = -8 ~\mathrm{m/s}
vbf=16/4=4 m/sv'_{bf} = 16/4 = 4~\mathrm{m/s}
Which can be converted back to the laboratory reference frame by adding the velocity of the frame relative to the laboratory:
vaf=8+7=1 m/sv_{af} = -8 + 7 = -1 ~\mathrm{m/s}
vbf=4+7=11 m/sv_{bf} = 4 + 7 = 11 ~\mathrm{m/s}
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