Use the methods of Challenge Problem 8.104 to calculate the x- and y-coordinates of the center of mass of a semicircular metal plate with uniform density p and thickness t. Let the radius of the plate be a. The mass of the plate is thus$\mathit{M}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{p}\mathit{\pi }{\mathit{a}}^{\mathbf{2}}\mathit{t}$ . Use the coordinate system indicated in Fig. P8.105.

The mass of the plate is,

$\mathrm{M}=\frac{1}{2}{\mathrm{p\pi a}}^2\mathrm{t}$

The coordinates of the center of mass are the weightage average of the point coordinates of the distributed mass.

The coordinates of center of mass is given by,

$$\begin{gathered} {\mathit{x}}_{\mathbf{cm}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{M}}\mathbf{\int }\mathit{x}\mathit{d}\mathit{m} \\ {\mathit{y}}_{\mathbf{cm}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{M}}\mathbf{\int }\mathit{y}\mathit{d}\mathit{m} \end{gathered}$$

Lets divide the plate into large number of thin strips parallel to y-axis. Each strip has dimension y,t,dx and dm .

Since the density is uniform, the mass of one strip is given by,

$$\begin{gathered} dm=pdV \\ =p\left(ytdx\right).......\left(1\right) \end{gathered}$$

The y-coordinates is given by following relationship.

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2 \\ \mathrm{y}=\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2} \end{gathered}$$

Substitute in equation (1).

$$\begin{gathered} dm=pdV \\ =pt\left(\sqrt{a^2-x^2}\right)dx...........\left(2\right) \end{gathered}$$

Find the x-coordinate of the center of mass as follows.

$$\begin{gathered} x_{cm}=\frac{1}{M}{\int }_{-a}^{a}x\left(pt\left(\sqrt{a^2-x^2}\right)dx\right) \\ =\frac{pt}{M}{\int }_{-a}^{a}x\sqrt{a^2-x^2}dx \end{gathered}$$

In order to solve this integral, let $a^2-x^2=u$, so when x =-a, u=0 , and when x=a, u=0 .

If the above expression is integrated with limit$0\to 0$then the integration will equal to zero.

Therefore, the x-coordinate of the center of mass is 0.

For$y_{cm}$ lets divide the plate into large number of thin strips parallel to x-axis. Each strip has dimension 2x, t,dy and dm.

Since the density is uniform, the mass of one strip is given by,

$$\begin{gathered} dm=pdV \\ =p\left(2xtdy\right)...........\left(3\right) \end{gathered}$$

The y-coordinates is given by following relationship.

$$\begin{gathered} x^2+y^2=a^2 \\ x=\sqrt{a^2-y^2} \end{gathered}$$

Substitute$x=\sqrt{a^2-y^2}$ in equation (3).

$$\begin{gathered} dm=pdV \\ =2pt\left(\sqrt{a^2-y^2}\right)dy \end{gathered}$$

Find the -coordinate of the center of mass as follows.

$$\begin{gathered} y_{cm}=\frac{1}{M}{\int }_0^{a}y\left[2pt\left(\sqrt{a^2-y^2}\right)dy\right] \\ =\frac{2pt}{M}{\int }_0^{a}y\sqrt{a^2-y^2}dy.........\left(4\right) \end{gathered}$$

In order to solve this integral, let $a^2-x^2=u$, so when y =0,$u=a^2$ , and when y=a, u=0 .

Also,

$$\begin{gathered} du=-2ydy \\ dy=-\frac{du}{2y} \end{gathered}$$

Substitute all the values in equation (4).

$$\begin{gathered} y_{cm}=\frac{2pt}{M}{\int }_{a^2}^{0}y\sqrt{u}\left(\frac{du}{-2y}\right) \\ =-\frac{pt}{M}{\int }_{a^2}^0\sqrt{u}du \\ =-\frac{pt}{M}{\left(\frac{2u^{{\displaystyle \frac{3}{2}}}}{3}\right)}_{a^2}^0 \\ =\frac{2pt}{M}.\frac{a^3}{3} \end{gathered}$$

Simplify further.

$$\begin{gathered} y_{cm}=\frac{2pt}{{\displaystyle \frac{1}{2}}{\mathrm{\pi pa}}^2\mathrm{t}}.\frac{a^3}{3} \\ =\frac{4a}{3\mathrm{\pi }} \end{gathered}$$

Therefore, the y-coordinate of the center of mass is $\frac{4a}{3\mathrm{\pi }}$.