Question

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?

Short answer

1. x coordinate of center of mass of slab is$-6.5\mathrm{cm}$.
2. y coordinate of center of mass of slab is$8.3\mathrm{cm}$.
3. z coordinate of center of mass of slab is$1.40\mathrm{cm}$.

Step-by-step solution

Given data

Dimensions of slab,

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

Understanding the concept of center of mass

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.

(a) Determination of x coordinate of center of mass of slab

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$-6.5\mathrm{cm}$.

(b) Determination of y coordinate of center of mass of slab

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 $8.3\mathrm{cm}$.

(c) Determination of z coordinates of center of mass of slab

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$1.40\mathrm{cm}$