Mathematical expression of COM for a system of particles

Let us now look at a baseball bat closely. It has a peculiar shape. Where should the COM lie? By experience, we can say that it should lie on the heavier side. This perception comes from the realization that heavier part has most of the mass. It gives us the clue about COM that it lies on the heavier side of an irregularly shaped body. Now, let us consider the motion of spherical sphere of uniform density. Here, mass is evenly distributed. Where should the COM lie? Obviously, COM coincides with the center of spherical ball.

It emerges from the discussion that COM is a statement of spatial arrangement of mass i.e. distribution of mass within the system. The position of COM is given a mathematical formulation which involves distribution of mass in a volumetric space, assuming as if all mass is concentrated there :

Figure 3: COM assumes as if mass of all particles is concentrated there.

COM for a system of particles

Total mass x Position of COM = ∑ Mass of individual particle x Position of particle Total mass x Position of COM = ∑ Mass of individual particle x Position of particle

⇒ Position of COM = ∑ Mass of individual particle x Position of particle Total mass ⇒ Position of COM = ∑ Mass of individual particle x Position of particle Total mass

This relation is read as "position of COM is mass weighted average of the positions of particles".

Further, COM is a geometric point in three-dimensional volume and it involves distribution of mass with respect to some reference system. This expression of the position of COM is description of a position in three dimensional volume. Now, substituting with appropriate symbols, we have :

⇒ r COM = ∑ m i r i M ⇒ r COM = ∑ m i r i M

It is important to note here that this formulation is not arbitrary. We shall find subsequently that this mathematical expression of COM, as a matter of fact, leads us to formulation of Newton's second law for a system of particles, while keeping its form intact.

We can express the position of COM in scalar component form as :

⇒ x COM = ∑ m i x i M ⇒ y COM = ∑ m i y i M ⇒ z COM = ∑ m i z i M ⇒ x COM = ∑ m i x i M ⇒ y COM = ∑ m i y i M ⇒ z COM = ∑ m i z i M

We know that there are only two directions involved with each of the axes. As such, we assign positive value for distances in reference direction and negative in opposite direction.

Special cases

(a) Particle system along a straight line :

COM of particle system along a straight line is given by the scalar expression in one dimension:

Figure 4: Scalar COM expression in one dimension

COM for a system of particles along a straight line

⇒ x COM = ∑ m i x i M ⇒ x COM = ∑ m i x i M

Evidently, COM of particles lie on the line containing particles.

(b) Two particles system

The expression of COM reduces for two particles as :

⇒ x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 ⇒ x COM = m 1 x 1 + m 2 x 2 m 1 + m 2

As only one coordinate is involved, it is imperative that COM lies on the line joining two particles. It can also be seen that center of mass lies in between the masses on the line connecting two particles. There are two additional simplified cases for two particles system as :

(i) Origin of coordinate coincides with position of one of the particles :

The expression of COM is simplified when position of one of the particles is the origin. For two particles system, x 1 = 0 , x 2 = x , x 1 = 0 , x 2 = x ,

⇒ x COM = m 2 x m 1 + m 2 ⇒ x COM = m 2 x m 1 + m 2

where "x" is the linear distance between two particles. Since " m 2 m 1 + m 2 m 2 m 1 + m 2 " is a fraction, it follows that center of mass two particles system is indeed a point which lies between the positions of two particles.

(ii) The mass of the particles are equal :

In this case, m 1 = m 2 = m m 1 = m 2 = m (say)

⇒ x COM = x 1 + x 2 2 ⇒ x COM = x 1 + x 2 2

When origin of coordinate coincides with position of one of the particles, x 1 = 0 , x 2 = x , x 1 = 0 , x 2 = x , :

⇒ x COM = x 2 ⇒ x COM = x 2

This result is on expected line. Center of mass (COM) of two particles of equal masses is midway between the particles.

Example 1

Problem : Two particles of 2 kg and 4 kg are placed at (1,4) and (3,-2) in x and y plane. If the coordinates are in meters, then find the position of COM.

Solution : Using expression in scalar component form :

Figure 5: Positions of the particles are shown on "xy" coordinate system.

Two particles system

x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 x COM = m 1 x 1 + m 2 x 2 m 1 + m 2

⇒ x COM = 2 x 1 + 4 x 3 6 = 7 3 m ⇒ x COM = 2 x 1 + 4 x 3 6 = 7 3 m

and

y COM = m 1 y 1 + m 2 y 2 m 1 + m 2 y COM = m 1 y 1 + m 2 y 2 m 1 + m 2

⇒ y COM = 2 x 4 + 4 x ( - 2 ) 6 = 0 ⇒ y COM = 2 x 4 + 4 x ( - 2 ) 6 = 0

Thus, position of COM lies on x - axis at x = 7/3 m.

COM of rigid body

Rigid body is composed of very large numbers of particles. Mass of rigid body is distributed closely. Thus, the distribution of mass can be treated as continuous. The mathematical expression for rigid body, therefore, is modified involving integration. The integral expressions of the components of position of COM in three mutually perpendicular directions are :

⇒ x COM = 1 M ∫ x đ m ⇒ y COM = 1 M ∫ y đ m ⇒ z COM = 1 M ∫ z đ m ⇒ x COM = 1 M ∫ x đ m ⇒ y COM = 1 M ∫ y đ m ⇒ z COM = 1 M ∫ z đ m

Note that the term in the numerator of the expression is nothing but the product of the mass of particle like small volumetric element and its distance from the origin along the axis. Evidently, this terms when integrated is equal to sum of all such products of mass elements constituting the rigid body.

Evaluation of above integrals is simplified, if the density of the rigid body is uniform. In that case,

ρ = M V = đ m đ V ⇒ đ m M = đ V V ρ = M V = đ m đ V ⇒ đ m M = đ V V

Substituting,

⇒ x COM = 1 V ∫ x đ V ⇒ y COM = 1 V ∫ y đ V ⇒ z COM = 1 V ∫ z đ V ⇒ x COM = 1 V ∫ x đ V ⇒ y COM = 1 V ∫ y đ V ⇒ z COM = 1 V ∫ z đ V

We must understand here that once we determine COM of a rigid body, the same can be treated as a particle at COM with all the mass assigned to that particle. This concept helps to find COM of a system of rigid bodies, comprising of many rigid bodies. Similarly, when a portion is removed from a rigid body, the COM of the rigid body can be obtained by treating the "portion removed" and the "remaining body" as particles. We shall see the working of this concept in the example given in the next section.

Symmetry and COM of rigid body

Evaluation of the integrals for determining COM is very difficult for irregularly shaped bodies. On the other hand, symmetry plays important role in determining COM of a regularly shaped rigid body. There are certain simplifying facts about symmetry and COM :

1. If symmetry is about a point, then COM lies on that point. For example, COM of a spherical ball of uniform density is its center.
2. If symmetry is about a line, then COM lies on that line. For example, COM of a cone of uniform density lies on cone axis.
3. If symmetry is about a plane, then COM lies on that plane. For example, COM of a cricket bat lies on the central plane.

The test of symmetry about a straight line or a plane is that the body on one side is replicated on the other side.

Example 2

Problem : A small circular portion of radius r/4 is taken out from a circular disc of uniform thickness having radius "r" and mass "m" as shown in the figure. Determine the COM of the remaining portion of the uniform disc.

Figure 6: A small circular portion of radius r/4 is taken out.

COM of remaining portion of circular disc

Solution : If we start from integral to determine COM of the remaining disc portion, then it would be a really complex proposition. Here, we shall make use of the connection between symmetry and COM. We note that the COM of the given disc is "O" and COM of the smaller disc removed is "O'". How can we use these fact to find center of mass of the remaining portion ?

The main idea here is that we can treat regular bodies with known COM as particles, which are separated by a known distance. Then, we shall employ the expression of COM for two particles to determine the COM of the remaining portion. We must realize that when a portion is removed from the bigger disc on the right side, the COM of the remaining portion shifts towards left side (heavier side).

The test of symmetry about a straight line or a plane is that the body on one side is replicated on the other side. We see here that the remaining portion of the disc is not symmetric about y-axis, but is symmetric about x-axis. It means that the COM of the remaining portion lies on the x-axis on the left side of the center of original disc. It also means that we need to employ the expression of COM for one dimension only.

While employing expression of COM, we use the logic as explained here. The original disc (with known COM) is equivalent to two particles system comprising of (i) remaining portion (with unknown COM) and (ii) smaller disc (with known COM). Now, x-component of the COM of original disc is :

Figure 7: A small circular portion of radius r/4 is taken out.

COM of remaining portion of circular disc

x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 x COM = m 1 x 1 + m 2 x 2 m 1 + m 2

But, x-component of the COM of original disc coincides with origin of the coordinate system. Further, let us denote the remaining disc by subscript "r" and the smaller circular disk removed by subscript "s".

⇒ 0 = m r x r + m s x s m r + m s ⇒ 0 = m r x r + m s x s m r + m s

⇒ m r x r + m s x s = 0 ⇒ x r = - m s x s m r ⇒ m r x r + m s x s = 0 ⇒ x r = - m s x s m r

As evident from figure, x s = r 2 x s = r 2 . We, now, need to find mass of smaller disc, m s m s , and mass of remaining portion, m r m r , using the fact that the density is uniform. The density of the material is :

ρ = m V = m π r 2 t ρ = m V = m π r 2 t

where "t" is thickness of disc.

⇒ m s = V ρ = ( π r 2 16 ) t x m π r 2 t = m 16 ⇒ m r = m - m s = m - m 16 = 15 m 16 ⇒ m s = V ρ = ( π r 2 16 ) t x m π r 2 t = m 16 ⇒ m r = m - m s = m - m 16 = 15 m 16

Putting these values in the expression of the position of COM, we have :

⇒ x r = - m s x s m r = - ( m 16 ) ( r 2 ) 15 m 16 = - r 30 ⇒ x r = - m s x s m r = - ( m 16 ) ( r 2 ) 15 m 16 = - r 30

Acknowledgment

Author wishes to thank Aladesanmi Oladele A for making suggestion to remove error in the module.