Center of Mass Calculator

Note

First Enter the values of the mass separated by commas whose center of mass need to be found Example 2,3,4
Now enter the x-coordinates of these respective masses separated by commas
Now enter the y-coordinates of these respective masses separated by commas
Now enter the z-coordinates of these respective masses separated by commas
If x -coordinate is required only, we can ignore the y and z ones
If x and y coordinate is required only, we can ignore the z ones
Click on the calculate button.

Formula used
$x_{cm} \frac {m_1x_1 + m_2x_2 + m_3x_3 + .... + m_nx_n}{m_1 + m_2 + m_3 + ..... + m_n}$ $y_{cm} \frac {m_1y_1 + m_2y_2 + m_3y_3 + .... + m_ny_n}{m_1 + m_2 + m_3 + ..... + m_n}$ $z_{cm} \frac {m_1z_1 + m_2z_2 + m_3z_3 + .... + m_nz_n}{m_1 + m_2 + m_3 + ..... + m_n}$

Center of Mass Calculator

What is Centre of mass
Centre of mass is the point where whole mass of the system can be supposed to be concentrated and motion of the system can be defined in terms of the centre of mass . It is the mass weighted average of its components
$x_{cm} =\frac {m_1x_1 + m_2x_2 + m_3x_3 + .... + m_nx_n}{m_1 + m_2 + m_3 + ..... + m_n}$
$y_{cm} =\frac {m_1y_1 + m_2y_2 + m_3y_3 + .... + m_ny_n}{m_1 + m_2 + m_3 + ..... + m_n}$
$z_{cm} =\frac {m_1z_1 + m_2z_2 + m_3z_3 + .... + m_nz_n}{m_1 + m_2 + m_3 + ..... + m_n}$

Example of Few questions where you can use this Centre of mass Calculator
Question 1
Four particles of same mass 1 kg lies in x-y plane.The (x,y) coordinates of their positions are (2,2) (3,3),(-1,2) and (-1,-1) respectively. Find the position of the center of mass of the system Solution
Given $m_1=m_2 =m_3=m_4 =1$
$x_1=2$, $x_2=3$,$x_3=-1$, $x_4=-1$
$y_1=2$, $y_2=3$,$y_3=2$, $x_4=-1$
Center of Mass is calculated as
$x_{cm} =\frac {m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}= \frac {1 \times 2 + 1 \times 3 + 1 \times -1 + 1 \times -1}{1 + 1 + 1 + 1}=3/4$
$y_{cm} =\frac {m_1y_1 + m_2y_2 + m_3y_3+ m_4y_4}{m_1 + m_2 + m_3 + m_4}=\frac {1 \times 2 + 1 \times 3 + 1 \times 2 + 1 \times -1}{1 + 1 + 1 + 1}=3/2$

Question 2
Three particles of masses 1 kg,2 kg and 3 kg respectively lies in x-y-z plane.The (x,y,z) coordinates of their positions are (2,2,1) (3,3,1),(-1,2,3) respectively. Find the position of the center of mass of the system
Solution
Given $m_1=1 \ kg,m_2 =2 \ kg ,m_3= 3 \ kg$
$x_1=2$, $x_2=3$,$x_3=-1$
$y_1=2$, $y_2=3$,$y_3=2$
$z_1=1$, $z_2=1$,$z_3=3$
Center of Mass is calculated as
$x_{cm} =\frac {m_1x_1 + m_2x_2 + m_3x_3 }{m_1 + m_2 + m_3 }= \frac {1 \times 2 + 2 \times 3 + 3 \times -1 }{1 + 2 + 3 }=5/6$
$y_{cm} =\frac {m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3 }=\frac {1 \times 2 + 2 \times 3 + 3 \times 2 }{1 + 2 + 3 }=15/6= 5/2$
$z_{cm} =\frac {m_1z_1 + m_2z_2 + m_3z_3}{m_1 + m_2 + m_3 }=\frac {1 \times 1 + 2 \times 1 + 3 \times 3 }{1 + 2 + 3 }=12/6= 2$

How Center Of Mass Calculator Works

if m, x,y,z are given
Center of Mass is calculated as
$x_{cm} =\frac {m_1x_1 + m_2x_2 + m_3x_3 + .... + m_nx_n}{m_1 + m_2 + m_3 + ..... + m_n}$
$y_{cm} =\frac {m_1y_1 + m_2y_2 + m_3y_3 + .... + m_ny_n}{m_1 + m_2 + m_3 + ..... + m_n}$
$z_{cm} =\frac {m_1z_1 + m_2z_2 + m_3z_3 + .... + m_nz_n}{m_1 + m_2 + m_3 + ..... + m_n}$
if m, x,y are given and z coordinate is not given
Center of Mass is calculated as
$x_{cm} =\frac {m_1x_1 + m_2x_2 + m_3x_3 + .... + m_nx_n}{m_1 + m_2 + m_3 + ..... + m_n}$
$y_{cm} =\frac {m_1y_1 + m_2y_2 + m_3y_3 + .... + m_ny_n}{m_1 + m_2 + m_3 + ..... + m_n}$
if m, x and y and z coordinates are not given
Center of Mass is calculated as
$x_{cm} =\frac {m_1x_1 + m_2x_2 + m_3x_3 + .... + m_nx_n}{m_1 + m_2 + m_3 + ..... + m_n}$