Chapter 8, Problem 8.50P

To determine

Centroid of the curved surface

Answer to Problem 8.50P

The centroid $\left(\overline{x},\overline{y},\overline{z}\right)$ of curved surface is $\left[-{\left(\frac{2}{\pi }\right)}^{2}R,\frac{2}{\pi }R,\frac{h}{3}\right]$.

Explanation of Solution

Given information:

The centroid of the curved surface is defined as:

  $\begin{array}{l}\overline{x}=\frac{Q_{yz}}{A}=\frac{{\displaystyle \underset{A}{\int }xdA}}{{\displaystyle \underset{A}{\int }dA}}\\ \overline{y}=\frac{Q_{xz}}{A}=\frac{{\displaystyle \underset{A}{\int }ydA}}{{\displaystyle \underset{A}{\int }dA}}\\ \overline{z}=\frac{Q_{xy}}{A}=\frac{{\displaystyle \underset{A}{\int }zdA}}{{\displaystyle \underset{A}{\int }dA}}\end{array}$

Calculation:

The differential area $dA$ of element shown

  $\begin{array}{l}dA=zRd\theta \\ =\frac{hR}{\pi }\theta d\theta \end{array}$

The area $A$ of the element

  $\begin{array}{l}A={\displaystyle \underset{0}{\overset{\pi }{\int }}dA}\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{hR}{\pi }\theta d\theta }\\ =\frac{\pi }{2}Rh\end{array}$

The centroidal coordinates $\left(\overline{x_{el}},\overline{y_{el}},\overline{z_{el}}\right)$ of element

  $\begin{array}{l}\overline{x_{el}}=R\cos \theta \\ \overline{y_{el}}=R\sin \theta \\ \overline{z_{el}}=\frac{z}{2}=\frac{h}{2\pi }\theta \end{array}$

The first moment $Q_{xy}$ of element

  $\begin{array}{l}{Q}_{xy}={\displaystyle \underset{A}{\int }\overline{z_{el}}dA}\\ =\frac{h^{2}R}{2{\pi }^2}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\theta }^{2}d\theta }\\ =\frac{\pi }{6}{h}^{2}R\end{array}$

The first moment $Q_{yz}$ of element

  $\begin{array}{l}{Q}_{yz}={\displaystyle \underset{A}{\int }\overline{x_{el}}dA}\\ =\frac{h{R}^2}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}\theta \cos \theta d\theta }\\ =\frac{h{R}^2}{\pi }{\left[\cos \theta +\theta sin\theta \right]}_0^{\pi }\\ =-\frac{2h{R}^2}{\pi }\end{array}$

The first moment $Q_{yz}$ of element

  $\begin{array}{l}{Q}_{xz}={\displaystyle \underset{A}{\int }\overline{y_{el}}dA}\\ =\frac{h{R}^2}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}\theta sin\theta d\theta }\\ =\frac{h{R}^2}{\pi }{\left[sin\theta -\theta \cos \theta \right]}_0^{\pi }\\ =h{R}^2\end{array}$

The centroid $\left(\overline{x},\overline{y},\overline{z}\right)$ of curved surface

  $\begin{array}{l}\overline{x}=\frac{Q_{yz}}{A}=\frac{\left(-\frac{2h{R}^2}{\pi }\right)}{\left(\frac{\pi }{2}Rh\right)}=-{\left(\frac{2}{\pi }\right)}^{2}R\\ \overline{y}=\frac{Q_{xz}}{A}=\frac{h{R}^2}{\left(\frac{\pi }{2}Rh\right)}=\frac{2}{\pi }R\end{array}$

  $\overline{z}=\frac{Q_{xy}}{A}=\frac{\frac{\pi }{6}{h}^{2}R}{\left(\frac{\pi }{2}Rh\right)}=\frac{h}{3}$

Conclusion:

The centroid $\left(\overline{x},\overline{y},\overline{z}\right)$ of curved surface is $\left[-{\left(\frac{2}{\pi }\right)}^{2}R,\frac{2}{\pi }R,\frac{h}{3}\right]$.