Figure shows a three particle system, with masses${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{kg}$,${\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{kg}$,and${\mathbf{m}}_{\mathbf{3}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{kg}$. The scales on the axes are set by ${\mathbf{\text{x}}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{m}$and${\mathbf{y}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{m}$. What is (a) The xcoordinate and (b) The ycoordinate of the system’s center of mass? (c) If is gradually increased, does the center of mass of the system shift toward or away from that particle, or does it remain stationary?

Mass of particle 1, ${\mathrm{m}}_1=3.0\mathrm{kg}$

Mass of particle 2,${\mathrm{m}}_2=4.0\mathrm{kg}$

Mass of particle 3, ${\mathrm{m}}_3=8.0\mathrm{kg}$

The scales on the axis are set by $x_2$and ${\mathrm{y}}_{\mathrm{s}}=2.0\mathrm{m}$

For a system of particles, the whole mass of the system is concentrated at the center of mass of the system. Center of mass is the point where external forces are seems to be applied for the motion of the system as a whole.

The expression for the x coordinates of the center of mass is given as:

$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+...}{m_1+m_2+m_3+...}$ … (i)

Here,${\mathrm{m}}_1,{\mathrm{m}}_2,{\mathrm{m}}_3,...$ are the masses of the particles, and${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,...$are their respective x coordinates.

The expression for the y coordinates of the center of mass is given as:

$y_{com}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+...}{m_1+m_2+m_3+...}$ … (ii)

Here,${\mathrm{y}}_1,{\mathrm{y}}_2,{\mathrm{y}}_{3,...}$are their respective y coordinates

From the figure, the x coordinates for all three particles are:

${\mathrm{x}}_1=0.0\mathrm{m},{\mathrm{x}}_2=2.0\mathrm{m}\mathrm{and}{\mathrm{x}}_3=1.0\mathrm{m}$

Using equation (i), the x coordinate of center of mass is calculated as;

$$\begin{gathered} {\mathrm{x}}_{\mathrm{com}}=\frac{(3.0\mathrm{kg})(0.0\mathrm{m})+(4.0\mathrm{kg})(2.0\mathrm{m})+(8.0\mathrm{kg})(1.0\mathrm{m})}{3.0\mathrm{kg}+4.0\mathrm{kg}+8.0\mathrm{kg}} \\ =\frac{0.0\mathrm{kg}-\mathrm{m}+8.0\mathrm{kg}-\mathrm{m}+8.0\mathrm{kg}-\mathrm{m}}{15.0\mathrm{kg}} \\ \approx 1.10\mathrm{m} \end{gathered}$$

Thus, the x coordinate of the center of mass is $1.10\mathrm{m}$.

From the figure, the y coordinates for all three particles are:

${\mathrm{y}}_1=0.0\mathrm{m},{\mathrm{y}}_2=1.0\mathrm{m}\mathrm{and}{y}_3=2.0\mathrm{m}$

Using equation (ii), the y coordinate of center of mass is calculated as:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{com}}=\frac{(3.0\mathrm{kg})(0.0\mathrm{m})+(4.0\mathrm{kg})(1.0\mathrm{m})+(8.0\mathrm{kg})(2.0\mathrm{m})}{3.0\mathrm{kg}+4.0\mathrm{kg}+8.0\mathrm{kg}} \\ =\frac{0.0\mathrm{kg}-\mathrm{m}+4.0\mathrm{kg}-\mathrm{m}+16.0\mathrm{kg}-\mathrm{m}}{15.0\mathrm{kg}} \\ \approx 1.33\mathrm{m} \end{gathered}$$

Thus, the y coordinate of the center of mass is$1.33\mathrm{m}$.

From the equation of center of mass, it is clear that as the mass of any particle increases, the center of mass would shift towards that particle.