Chapter 5.4, Problem 5.123P

To determine

To determine the value of $\overline{x}$ for the two volumes.

Answer to Problem 5.123P

The value of $\overline{x}$ for the first component is $\frac{21}{88}h$, and the value of $\overline{x}$ for the second component is $\frac{27}{40}h$.

Explanation of Solution

Refer Fig. P5.21 and Fig. 1.

Let the element of volume of the disk be of radius r and thickness dx.

Find the equation for radius from the equation of the generating curve.

$\begin{array}{c}\frac{x^2}{h^2}+\frac{y^2}{a^2}=1\\ r^2=\frac{a^2}{h^2}\left(h^2-x^2\right)\end{array}$

Here, the y component is equal to the radius, $h$ is the radius of the semi ellipsoid, $x$ is the centroid of the element and $dx$ is the thickness of the element.

Write the expression for the volume of the element.

$\begin{array}{c}dV=\pi r^{2}dx\\ =\pi \frac{a^2}{h^2}\left(h^2-x^2\right)dx\end{array}$ (I)

Find the centroid of the element.

$\overline{x_{EL}}=x$

Here, $\overline{x_{EL}}$ is the centroid of the element.

Write the expression for the $\overline{x}$ for the volume.

$\begin{array}{c}\overline{x}V={\displaystyle \underset{}{\int }\overline{x_{EL}}}dV\\ \overline{x}=\frac{{\displaystyle \underset{}{\int }\overline{x_{EL}}}dV}{V}\end{array}$ (II)

Conclusion:

Refer Fig.2

Calculate the volume of component 1.

$\begin{array}{c}{V}_1={\displaystyle {\int }_0^{h/2}dV}\\ ={\displaystyle {\int }_0^{h/2}\pi \frac{a^2}{h^2}\left(h^2-x^2\right)dx}\\ =\pi \left(\frac{a^2}{h^2}\right){\left[h^{2}x-\frac{x^3}{3}\right]}_0^{h/2}\\ =\frac{11}{24}\pi a^{2}h\end{array}$

Calculate ${\displaystyle \underset{1}{\int }\overline{x_{EL}}}dV$ for component 1.

$\begin{array}{c}{\displaystyle \underset{1}{\int }\overline{x_{EL}}}dV={\displaystyle {\int }_0^{h/2}x\left[\pi \frac{a^2}{h^2}\left(h^2-x^2\right)dx\right]}\\ =\pi \left(\frac{a^2}{h^2}\right){\left[h^2\frac{x^2}{2}-\frac{x^4}{4}\right]}_0^{h/2}\\ =\frac{7}{64}\pi a^2{h}^2\end{array}$

Substitute $\frac{7}{64}\pi a^2{h}^2$ for ${\displaystyle \underset{1}{\int }\overline{x_{EL}}}dV$ and $\frac{11}{24}\pi a^{2}h$ for $V_1$ in equation (II).

$\begin{array}{c}\overline{x_1}=\frac{\frac{7}{64}\pi a^2{h}^2}{\frac{11}{24}\pi a^{2}h}\\ =\frac{21}{88}h\end{array}$

Thus, the value of $\overline{x}$ for the first component is $\frac{21}{88}h$.

Calculate the volume of component 2.

$\begin{array}{c}{V}_2={\displaystyle {\int }_{h/2}^{h}dV}\\ ={\displaystyle {\int }_{h/2}^h\pi \frac{a^2}{h^2}\left(h^2-x^2\right)dx}\\ =\pi \left(\frac{a^2}{h^2}\right){\left[h^{2}x-\frac{x^3}{3}\right]}_{h/2}^h\\ =\pi \left(\frac{a^2}{h^2}\right)\left\{\left[h^2\left(h\right)-\frac{h^3}{3}\right]-\left[h^2\left(\frac{h}{2}\right)-\frac{{\left(\frac{h}{2}\right)}^3}{3}\right]\right\}\\ =\frac{5}{24}\pi a^{2}h\end{array}$

Calculate ${\displaystyle \underset{2}{\int }\overline{x_{EL}}}dV$ for component 2.

$\begin{array}{c}{\displaystyle \underset{2}{\int }\overline{x_{EL}}}dV={\displaystyle {\int }_{a/2}^{a}x\left[\pi \frac{a^2}{h^2}\left(h^2-x^2\right)dx\right]}\\ =\pi \left(\frac{a^2}{h^2}\right){\left[h^2\frac{x^2}{2}-\frac{x^4}{4}\right]}_{h/2}^h\\ =\pi \left(\frac{a^2}{h^2}\right)\left\{\left[h^2\frac{h^2}{2}-\frac{h^4}{4}\right]-\left[h^2\frac{{\left(\frac{h}{2}\right)}^2}{2}-\frac{{\left(\frac{h}{2}\right)}^4}{4}\right]\right\}\\ =\frac{9}{64}\pi a^2{h}^2\end{array}$

Substitute $\frac{9}{64}\pi a^2{h}^2$ for ${\displaystyle \underset{2}{\int }\overline{x_{EL}}}dV$ and $\frac{5}{24}\pi a^{2}h$ for $V_2$ in equation (II).

$\begin{array}{c}\overline{x_2}=\frac{\frac{9}{64}\pi a^2{h}^2}{\frac{5}{24}\pi a^{2}h}\\ =\frac{27}{40}h\end{array}$

Thus, the value of $\overline{x}$ for the second component is $\frac{27}{40}h$.

Therefore, the value of $\overline{x}$ for the first component is $\frac{21}{88}h$, and the value of $\overline{x}$ for the second component is $\frac{27}{40}h$.