Q29P - Mathematical Methods in Physical Sciences

Question

Show that the center of mass of three identical particles situated at the point ${\mathit{z}}_{\mathbf{1}}\mathbf{,}{\mathit{z}}_{\mathbf{2}}\mathbf{,}{\mathit{z}}_{\mathbf{3}}$ is $\frac{\left({\mathit{z}}_{\mathbf{1}}\mathbf{,}{\mathit{z}}_{\mathbf{2}}\mathbf{,}{\mathit{z}}_{\mathbf{3}}\right)}{\mathbf{3}}$ .

Step-by-Step Solution

Answer

It proved that the center of mass of three identical particles situated at the point $z_1,z_2,z_3$ is $\frac{\left(z_1,z_2,z_3\right)}{3}$ .

Step 1: Given Information

To prove that the center of mass of three identical particles situated at the point $z_1,z_2,z_3$ is $\frac{\left(z_1,z_2,z_3\right)}{3}$ .

Step 2: Definition of the complex number

Complex numbers possess real numbers and imaginary numbers; a complex  can be written in the form of: 

z = x + iy 

 

Here x and y are real numbers, and i is the imaginary number which is known as iota, whose value is  $\sqrt{-1}$ .

Step 3: Calculate the center of mass

Assume that the center of   direction is,

 

$x_c=\frac{{\displaystyle \underset{ }{\sum x_n{m}_n}}}{\sum m_n}$                                                                                    ……. (1)

Substitute in equation (1), and we get,

$x_c=\frac{x_1,m_1+x_2{m}_2+x_{3}m3}{m_1+m_2+m_3}$                                                            ........(2)

Use the 4^th assumption as:

$$\begin{gathered} x_c=\frac{m\left(x_1+x_2+x_3\right)}{3m} \\ =\frac{x_1+x_2+x_3}{3} \end{gathered}$$

Same for y :

$y_c=\frac{{\displaystyle \underset{ }{\sum y_n{m}_n}}}{\sum m_n}$                                                                                    ……. (3)

Substitute in equation (1), we get,

          

$y_c=\frac{y_1{m}_1+y_2{m}_2+y_3{m}_3}{m_1+m_2+m_3}$                                                                              ……. (4)

Use the 4^th assumption as:

 

 $$\begin{gathered} y_c=\frac{m\left(y_1+y_2+y_3\right)}{3m} \\ =\frac{y_1+y_2+y_3}{3} \end{gathered}$$

Step 4: Solve it further

The location equation can be written as,

 

 $$\begin{gathered} z_c=x_c+i{y}_c \\ =\frac{x_1+x_2+x_3}{3}+i\left(\frac{y_1+y_2+y_3}{3}\right) \\ =\frac{\left(x_1+i{y}_1\right)+\left(x_2+i{y}_2\right)+\left(x_3+i{y}_3\right)}{3} \\ =\frac{z_1+z_2+z_3}{3} \end{gathered}$$

Hence the center of mass of three identical particles situated at the point $z_1,z_2,z_3$ is 

$\frac{z_1+z_2+z_3}{3}$