Center of mass  (Page 2/3)

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 :

Com for a system of particles

$$\begin{array}{l}\mathrm{Total\; mass}x\mathrm{Position\; of\; COM}=\sum \mathrm{Mass\; of\; individual\; particle}x\mathrm{Position\; of\; particle}\end{array}$$

$$\begin{array}{l}\Rightarrow \mathrm{Position\; of\; COM}=\frac{\sum \mathrm{Mass\; of\; individual\; particle}x\mathrm{Position\; of\; particle}}{\mathrm{Total\; mass}}\end{array}$$

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 :

$$\begin{array}{l}\Rightarrow {\mathbf{r}}_{\mathrm{COM}}=\frac{\sum m_i{\mathbf{r}}_i}{M}\end{array}$$

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 :

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{\sum m_i{x}_i}{M}\\ \Rightarrow y_{\mathrm{COM}}=\frac{\sum m_i{y}_i}{M}\\ \Rightarrow z_{\mathrm{COM}}=\frac{\sum m_i{z}_i}{M}\end{array}$$

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:

Com for a system of particles along a straight line

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{\sum m_i{x}_i}{M}\end{array}$$

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

(b) Two particles system

The expression of COM reduces for two particles as :

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}\end{array}$$

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, $x_1=0,x_2=x,$

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{m_{2}x}{m_1+m_2}\end{array}$$

where "x" is the linear distance between two particles. Since " $\frac{m_2}{m_1+m_2}$ " 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, $m_1=m_2=m$ (say)

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{x_1+x_2}{2}\end{array}$$

When origin of coordinate coincides with position of one of the particles, $x_1=0,x_2=x,$ :

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{x}{2}\end{array}$$

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

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 :

Two particles system

$$\begin{array}{l}{x}_{\mathrm{COM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}\end{array}$$

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{2x1+4x3}{6}=\frac{7}{3}m\end{array}$$

and

$$\begin{array}{l}{y}_{\mathrm{COM}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}\end{array}$$

$$\begin{array}{l}\Rightarrow y_{\mathrm{COM}}=\frac{2x4+4x(-2)}{6}=0\end{array}$$

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

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