Question: Because gvaries so little over the extent of most structures, any structure’s center of gravity effectively coincides with its center of mass. Here is a fictitious example where gvaries more significantly. Figure 12-25 shows an array of six particles, each with mass m, fixed to the edge of a rigid structure of negligible mass. The distance between adjacent particles along the edge is 2.00 m. The following table gives the value of g (m/s²)at each particle’s location. Using the coordinate system shown, find (a) the xcoordinate x_com and (b) the ycoordinate Y_com of the center of mass of the six-particle system. Then find (c) the xcoordinate x_cog and (d) the ycoordinateY_cogof the center of gravity of the six-particle system.

The mass of each particle is m

The distance between adjacent particles is 2.00.

You can find the components of the center of mass and the center of gravity using the formulae for them which are given below.

$$\begin{gathered} x_{com}=\frac{\sum m_i{x}_i}{M} \\ {y}_{com}=\frac{\sum m_i{y}_i}{M} \\ {x}_{cog}=\frac{\sum m_i{x}_i{g}_i}{\sum m_i{g}_i} \\ {y}_{cog}=\frac{\sum m_i{y}_i{g}_i}{\sum m_i{g}_i} \end{gathered}$$

The x component of the center of mass x_com of the six particle system is,

$$\begin{gathered} x_{com}=\frac{\sum m_i{x}_i}{M} \\ =\frac{0+0+0+m\left(2\right)+m\left(2\right)+m\left(2\right)}{6m} \\ =\frac{6}{6} \\ =1.00\text{m} \end{gathered}$$

The y component of the center of mass y_com of the six particle system is,

$$\begin{gathered} y_{com}=\frac{\sum m_i{y}_i}{M} \\ =\frac{0+0+m\left(2\right)+m\left(2\right)+m\left(4\right)+m\left(4\right)}{6m} \\ =2.00\text{m} \end{gathered}$$

The x component of the center of gravity x_cog of the six particle system is

$$\begin{gathered} x_{cog}=\frac{\sum m_i{x}_i{g}_i}{\sum m_i{g}_i} \\ =\frac{0+0+0+m\left(2\right)\left(7.4\right)+m\left(2\right)\left(7.6\right)+m\left(2\right)\left(7.8\right)}{m\left(8\right)+m\left(7.8\right)+m\left(7.6\right)+m\left(7.4\right)+m\left(7.6\right)+m\left(7.8\right)} \\ =\frac{45.6}{46.2} \\ =0.987\text{m} \end{gathered}$$

The y component of the center of gravity y_cog of the six particle system is,

$$\begin{gathered} y_{cog}=\frac{\sum m_i{y}_i{g}_i}{\sum m_i{g}_i} \\ =\frac{0+0+m\left(2\right)\left(7.8\right)+m\left(4\right)\left(7.6\right)+m\left(4\right)\left(7.4\right)+m\left(2\right)\left(7.6\right)}{m\left(8\right)+m\left(7.8\right)+m\left(7.6\right)+m\left(7.4\right)+m\left(7.6\right)+m\left(7.8\right)} \\ =1.97\text{m} \end{gathered}$$