Figure 9-36 shows a slab with dimensions${\mathbf{d}}_{\mathbf{1}}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{0}\mathbf{cm}$,${\mathbf{d}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{80}\mathbf{cm}$, and${\mathbf{d}}_{\mathbf{3}}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{0}\mathbf{cm}$. Half the slab consists of aluminum$\mathbf{(}\mathbf{density}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{70}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}\mathbf{)}$and half consists of iron$\mathbf{(}\mathbf{density}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{85}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}\mathbf{)}$. What are (a) The xcoordinate,(b) The ycoordinate, and (c) The zcoordinate of the slab’s center of mass?

Dimensions of slab,

$$\begin{gathered} {\mathrm{d}}_1=11.0\mathrm{cm} \\ {\mathrm{d}}_2=2.80\mathrm{cm} \\ {\mathrm{d}}_3=13.0\mathrm{cm} \\ \mathrm{The}\mathrm{density}\mathrm{of}\mathrm{aluminum},{\mathrm{p}}_{\mathrm{A}}=2.70\mathrm{g}/{\mathrm{cm}}^2 \\ \mathrm{The}\mathrm{density}\mathrm{of}\mathrm{iron},{\mathrm{p}}_{\mathrm{i}}=7.85\mathrm{g}/{\mathrm{cm}}^2 \end{gathered}$$

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 density is given as:

… (i)

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

${\overrightarrow{r}}_{com}=\frac{1}{M}\sum _{i=1}^n{m}_i{r}_i$ … (ii)

Here, $M$is the total mass, $m_i$ is the individual mass of ith particle and $r$ is the coordinates of ithparticle.

From the symmetry of the figure, we can say that,

$x_{com}=-\frac{d_3}{2}$

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{x}}_{\mathrm{com}}=-\frac{13\mathrm{cm}}{2} \\ =-6.5\mathrm{cm} \end{gathered}$$

Thus, the x coordinate of the center of mass is$-6.5\mathrm{cm}$.

Using equation (i) and (ii), the y coordinate of center of mass is,

$$\begin{gathered} {\mathrm{y}}_{\mathrm{com}}=\frac{{\mathrm{m}}_{\mathrm{i}}{\mathrm{y}}_{\mathrm{comi}}+{\mathrm{m}}_{\mathrm{a}}{\mathrm{y}}_{\mathrm{coma}}}{{\mathrm{m}}_{\mathrm{i}}+{\mathrm{m}}_{\mathrm{a}}} \\ =\frac{{\mathrm{p}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}{\mathrm{y}}_{\mathrm{comi}}+{\mathrm{p}}_{\mathrm{a}}{\mathrm{v}}_{\mathrm{a}}{\mathrm{y}}_{\mathrm{coma}}}{{\mathrm{p}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}+{\mathrm{p}}_{\mathrm{a}}{\mathrm{v}}_{\mathrm{a}}} \end{gathered}$$

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{y}}_{\mathrm{com}}=\frac{\left\{\right(6.5\left)\right(7.85)+3(6.5\left)\right(2.7\left)\right\}}{7.85+2.7} \\ =8.3\mathrm{cm} \end{gathered}$$

Thus, the y coordinate of the center of mass is $8.3\mathrm{cm}$.

Again by symmetry we get,

$$\begin{gathered} {\mathrm{z}}_{\mathrm{com}}=\frac{{\mathrm{d}}_2}{2} \\ =\frac{2.80\mathrm{cm}}{2} \\ =1.40\mathrm{cm} \end{gathered}$$

Thus, the z coordinate of the center of mass is$1.40\mathrm{cm}$