Chapter 5.4, Problem 5.133P

Locate the centroid of the section shown, which was cut f a thin circular pipe by two oblique planes.

Fig. P5.133

To determine

The location of the centre of gravity of the bowl.

Answer to Problem 5.133P

The location of the centre of gravity of the bowl is $\overline{x}=0$, $\overline{y}=-121.9\text{mm}$ and $\overline{z}=-\frac{1}{2}R$.

Explanation of Solution

Refer Fig. P5.132 and Fig. 1.

For the coordinate axis given below, using the symmetry of the diagram, determine the x and z coordinates of the centre of gravity.

$\begin{array}{l}\overline{x}=0\\ \overline{z}=0\end{array}$

The bowl can be assumed as a shell, where the centre of gravity coincides with the centroid of the shell.

The element of area is obtained by rotating the arc ds about the y-axis for the walls of the bowl.

Write the expression for the area of the element.

$d{A}_{\text{wall}}=\left(2\pi R\sin \theta \right)\left(Rd\theta \right)$ (I)

Here, $d{A}_{\text{wall}}$ is the area of the element and $R$ is the radius of the shell.

Write the expression for y coordinate of the centroid of the element.

${\left({\overline{y}}_{EL}\right)}_{wall}=-R\cos \theta$ (II)

Here, ${\overline{y}}_{EL}$ is the y coordinate of the centroid of the element.

Write the expression for ${\overline{y}}_{\text{wall}}{A}_{\text{wall}}$ of the element.

${\overline{y}}_{\text{wall}}{A}_{\text{wall}}={\displaystyle \int {\left({\overline{y}}_{EL}\right)}_{\text{wall}}}dA$ (III)

Write the expression for y coordinate of the centre of gravity.

$\overline{y}=\frac{\Sigma \overline{y}A}{\Sigma A}$ (IV)

Here, $\Sigma \overline{y}A$ is the sum of the product of the y coordinate and the area and $\Sigma A$ is the sum of the areas.

Conclusion:

Calculate the area using equation (I).

$\begin{array}{c}{A}_{\text{wall}}={\displaystyle \underset{\pi /6}{\overset{\pi /2}{\int }}dA}\\ ={\displaystyle \underset{\pi /6}{\overset{\pi /2}{\int }}2\pi R^2\sin \theta d\theta }\\ =2\pi R^2{\left[-\cos \theta \right]}_{\pi /6}^{\pi /2}\\ =\pi \sqrt{3}{R}^2\end{array}$ (V)

Substitute (I) and (II) in (III).

$\begin{array}{c}{\overline{y}}_{\text{wall}}{A}_{\text{wall}}={\displaystyle \underset{\pi /6}{\overset{\pi /2}{\int }}\left(-R\cos \theta \right)\left(2\pi R^2\sin \theta d\theta \right)}\\ =\pi R^3{\left[{\cos }^2\theta \right]}_{\pi /6}^{\pi /2}\\ =-\frac{3}{4}\pi R^3\end{array}$ (VI)

From figure 1, find the area of the base and distance of the centroid of the base to the y axis.

$\begin{array}{c}{A}_{\text{base}}=\frac{\pi }{4}{R}^2\\ {\overline{y}}_{\text{base}}=-\frac{\sqrt{3}}{2}R\end{array}$ (VII)

Substitute equations (V), (VI) and (VII) in equation (IV).

$\begin{array}{c}\overline{y}=\frac{\left(-\frac{3}{4}\pi R^3\right)+\left(\frac{\pi }{4}{R}^2\right)\left(-\frac{\sqrt{3}}{2}R\right)}{\pi \sqrt{3}{R}^2+\frac{\pi }{4}{R}^2}\\ =-0.48763R\end{array}$ (VIII)

Substitute $250\text{mm}$ for $R$ in equation (VIII) to find the y coordinate of the centre of gravity.

$\begin{array}{c}\overline{y}=\left(-0.48763\right)\left(250\text{mm}\right)\\ =-121.9\text{mm}\end{array}$

Therefore, the location of the centre of gravity of the bowl is $\overline{x}=0$, $\overline{y}=-121.9\text{mm}$ and $\overline{z}=-\frac{1}{2}R$.

To determine

The location of the centre of gravity of the punch.

Answer to Problem 5.133P

The location of the centre of gravity of the bowl is $\overline{x}=0$, $\overline{y}=-90.2\text{mm}$ and $\overline{z}=-\frac{1}{2}R$.

Explanation of Solution

Refer Fig. P5.132 and Fig. 2.

For the coordinate axis given below, using the symmetry of the diagram, determine the x and z coordinates of the centre of gravity.

$\begin{array}{l}\overline{x}=0\\ \overline{z}=0\end{array}$

The punch can be assumed as homogenous, where the centre of gravity coincides with the centroid of the volume.

The element of volume of the disk has radius x and thickness dy.

Find the expression for $x^2$ of the element from the equation of the curve.

$\begin{array}{c}{x}^2+y^2=R^2\\ x^2=R^2-y^2\end{array}$

Write the expression for the volume of the element.

$\begin{array}{c}dV=\pi x^{2}dy\\ =\pi \left(R^2-y^2\right)dy\end{array}$ (IX)

Here, $R$ is the radius of the punch, and $y$ is distance of the element from the top.

Write the expression for y coordinate of the centre of gravity.

$\overline{y}=\frac{{\displaystyle \int {\overline{y}}_{EL}dV}}{V}$ (X)

Conclusion:

Calculate the volume using equation (IX).

$\begin{array}{c}V={\displaystyle \underset{-\sqrt{3}/2R}{\overset{0}{\int }}dV}\\ ={\displaystyle \underset{-\sqrt{3}/2R}{\overset{0}{\int }}\pi \left(R^2-y^2\right)dy}\\ =\pi {\left[R^{2}y-\frac{1}{3}{y}^3\right]}_{-\sqrt{3}/2R}^0\\ =-\pi \left[R^2\left(-\frac{\sqrt{3}}{2}R\right)-\frac{1}{3}{\left(-\frac{\sqrt{3}}{2}R\right)}^3\right]\\ =\frac{3}{8}\pi \sqrt{3}{R}^3\end{array}$ (XI)

Find ${\displaystyle \int {\overline{y}}_{EL}dV}$.

$\begin{array}{c}{\displaystyle \int {\overline{y}}_{EL}dV}={\displaystyle {\int }_{-\sqrt{3}/2R}^{0}y\left[\pi \left(R^2-y^2\right)dy\right]}\\ =\pi {\left[\frac{1}{2}{R}^2{y}^2-\frac{1}{4}{y}^4\right]}_{-\sqrt{3}/2R}^0\\ =-\pi \left[\frac{1}{2}{R}^2{\left(-\frac{\sqrt{3}}{2}R\right)}^2-\frac{1}{4}{\left(-\frac{\sqrt{3}}{2}R\right)}^4\right]\\ =-\frac{15}{64}\pi R^4\end{array}$ (XII)

Substitute equations (XI), and (XII) in equation (X).

$\begin{array}{c}\overline{y}=\frac{-\frac{15}{64}\pi R^4}{\frac{3}{8}\pi \sqrt{3}{R}^3}\\ =-\frac{5}{8\sqrt{3}}R\end{array}$ (XIII)

Substitute $250\text{mm}$ for $R$ in equation (XIII) to find the y coordinate of the centre of gravity.

$\begin{array}{c}\overline{y}=-\frac{5}{8\sqrt{3}}\left(250\text{mm}\right)\\ =-90.2\text{mm}\end{array}$

Therefore, the location of the centre of gravity of the bowl is $\overline{x}=0$, $\overline{y}=-90.2\text{mm}$ and $\overline{z}=-\frac{1}{2}R$.