91Ó°ÊÓ

The value of$\mathrm{L}=5.0\mathrm{cm}$

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 coordinates of the center of mass are given as:

${\stackrel{→}{\mathrm{r}}}_{\mathrm{com}}=\frac{1}{\mathrm{M}}\underset{\mathrm{i}=1}{\overset{\mathrm{n}}{∑}}{\mathrm{m}}_{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}$ … (i)

Here, $\mathrm{M}$is the total mass, ${\mathrm{m}}_{\mathrm{i}}$ is the individual mass of ith particle and ${\mathrm{r}}_{\mathrm{i}}$ is the coordinates of ithparticle.

We can divide the given structure into three symmetrical parts.

First is at extreme left having area,

$$\begin{gathered} {\mathrm{A}}_1=2\mathrm{L}×7\mathrm{L} \\ =14{\mathrm{L}}^2 \end{gathered}$$

Second at the top right having area,

$$\begin{gathered} {\mathrm{A}}_2=4\mathrm{L}×\mathrm{L} \\ =4{\mathrm{L}}^2 \end{gathered}$$

And third at the bottom right having area,

$$\begin{gathered} {\mathrm{A}}_3=2\mathrm{L}×2\mathrm{L} \\ =4{\mathrm{L}}^2 \end{gathered}$$

The percentage of area consumes by first part is,

$$\begin{gathered} {\mathrm{A}}_1\%=\frac{14{\mathrm{L}}^2}{14{\mathrm{L}}^2+4{\mathrm{L}}^2+4{\mathrm{L}}^2}×100 \\ =63.6\% \\ \mathrm{Similarly}, \\ {\mathrm{A}}_2\%=18.2\% \\ {\mathrm{A}}_3\%=18.2\% \end{gathered}$$

Using the symmetry, we can write the center of mass of the first part as, $({\mathrm{x}}_1,{\mathrm{y}}_1)=(-5.0\mathrm{cm},-2.5\mathrm{cm})$

Similarly, using symmetry, we can write the center of mass for the second part as,$({\mathrm{x}}_2,{\mathrm{y}}_2)=(10\mathrm{cm},12.5\mathrm{cm})$

And using symmetry, we can write the center of mass for the third part as,$({\mathrm{x}}_3,{\mathrm{y}}_3)=(5\mathrm{cm},-15\mathrm{cm})$

The x coordinate of the center of mass of the whole plate is calculated as:

$$\begin{gathered} {\mathrm{x}}_{\mathrm{com}}=(0.636){\mathrm{x}}_1+(0.182){\mathrm{x}}_2+(0.182){\mathrm{x}}_3 \\ =(0.636)(-5.0\mathrm{cm})+(0.182)(10.0\mathrm{cm})+(0.182)(5.0\mathrm{cm}) \\ =-0.45\mathrm{cm} \\ \mathrm{Thus},\mathrm{the}x\mathrm{coordinate}\mathrm{of}\mathrm{center}\mathrm{of}\mathrm{mass}\mathrm{is}-0.45\mathrm{cm} \end{gathered}$$

The y coordinate of the center of mass of the whole plate is calculated as:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{com}}=(0.636){\mathrm{y}}_1+(0.182){\mathrm{y}}_2+(0.182){\mathrm{y}}_3 \\ =(0.636)(-2.5\mathrm{cm})+(0.182)(12.5\mathrm{cm})+(0.182)(-15.0\mathrm{cm}) \\ =-2.0\mathrm{cm} \\ \mathrm{Thus},\mathrm{the}\mathrm{y}\mathrm{coordinate}\mathrm{of}\mathrm{center}\mathrm{of}\mathrm{mass}\mathrm{is}-2.0\mathrm{cm} \end{gathered}$$