Chapter 8, Problem 8.51P

To determine

Centroidal points of a curve connecting points A and B

Answer to Problem 8.51P

The centroid $\left(\overline{x},\overline{y},\overline{z}\right)$ of curved surface is $\left[0,\frac{2}{\pi }R,\frac{h}{2}\right]$.

Explanation of Solution

Given information:

The centroidal points of the curve connecting point are 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 length $ds$ of the element

  $ds=\frac{R}{\cos \alpha }d\theta$

The length $L$ of element

  $\begin{array}{l}L={\displaystyle \underset{L}{\int }ds}\\ ={\displaystyle {\int }_0^{\pi }\frac{R}{\cos \alpha }d\theta }\\ L=\frac{\pi R}{\cos \alpha }\end{array}$

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

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

The first moment $Q_{xy}$ of element

  $\begin{array}{l}{Q}_{xy}={\displaystyle \underset{L}{\int }\overline{z_{el}}ds}\\ =\frac{Rh}{\pi \cos \alpha }{\displaystyle \underset{0}{\overset{\pi }{\int }}\theta d\theta }\\ =\frac{\pi Rh}{2\cos \alpha }\end{array}$

The first moment $Q_{yz}$ of element

  $\begin{array}{l}{Q}_{yz}={\displaystyle \underset{L}{\int }\overline{x_{el}}ds}\\ =\frac{R^2}{\cos \alpha }{\displaystyle \underset{0}{\overset{\pi }{\int }}\cos \theta d\theta }\\ =0\end{array}$

The first moment $Q_{xz}$ of element

  $\begin{array}{l}{Q}_{xz}={\displaystyle \underset{L}{\int }\overline{y_{el}}ds}\\ =\frac{R^2}{\cos \alpha }{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta d\theta }\\ =\frac{2R^2}{\cos \alpha }\end{array}$

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

  $\begin{array}{l}\overline{x}=\frac{Q_{yz}}{L}=0\\ \overline{y}=\frac{Q_{xz}}{L}=\frac{\left(\frac{2R^2}{\cos \alpha }\right)}{\left(\frac{\pi R}{\cos \alpha }\right)}=\frac{2}{\pi }R\\ \overline{z}=\frac{Q_{xy}}{L}=\frac{\frac{\pi Rh}{2\cos \alpha }}{\left(\frac{\pi R}{\cos \alpha }\right)}=\frac{h}{2}\end{array}$

Conclusion:

The centroid $\left(\overline{x},\overline{y},\overline{z}\right)$ of curved surface is $\left[0,\frac{2}{\pi }R,\frac{h}{2}\right]$.