Chapter 9, Problem 1FP

Determine the centroid $\left(\bar{x},\bar{y}\right)$ of the shaded area.

Prob. F9-1

To determine

The centroid $\left(\overline{x}\right)$ of the parabolic-shaded area.

Answer to Problem 1FP

The centroid $\left(\overline{x}\right)$ of the parabolic shaded area is $\underset{\_}{0.4\text{m}}$.

Explanation of Solution

Given:

The length of the shaded area is 1 m.

The height of the shaded area is 1 m.

Show the area of the differential element as in Figure 1.

Using Figure 1,

Express the parabolic value.

$\begin{array}{l}y=x^3\\ x=y^{1/3}\end{array}$

Compute the area of the element.

$dA=xdy$

Substitute $y^{1/3}$ for x.

$dA=y^{1/3}dy$

Compute the centroid of the differential element along the x-axis using the formula.

$\tilde{x}=\frac{x}{2}$ (I)

Substitute $y^{1/3}$ for x in Equation (I).

$\tilde{x}=\frac{y^{1/3}}{2}$

Determine the location of the centre of gravity of the homogeneous rod along the x-axis $\left(\overline{x}\right)$ using the relation.

$\overline{x}=\frac{{\displaystyle {\int }_A\tilde{x}dA}}{{\displaystyle {\int }_{A}dA}}$ (II)

Conclusion:

Apply the limits from 0 to 1 m, substitute $\frac{y^{1/3}}{2}$ for $\tilde{x}$ and $y^{1/3}dy$ for dA in Equation (II).

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \underset{0}{\overset{\text{1}\text{m}}{\int }}\left(\frac{y^{1/3}}{2}\right)y^{1/3}dy}}{{\displaystyle \underset{0}{\overset{1\text{m}}{\int }}{y}^{1/3}dy}}\\ =\frac{\frac{1}{2}{\displaystyle \underset{0}{\overset{\text{1}\text{m}}{\int }}{y}^{2/3}dy}}{{\displaystyle \underset{0}{\overset{1\text{m}}{\int }}{y}^{1/3}dy}}\end{array}$ (III)

Calculate the numerator part from Equation (III).

$\begin{array}{c}\frac{1}{2}{\displaystyle \underset{0}{\overset{\text{1}\text{m}}{\int }}{y}^{2/3}dy}=\frac{1}{2}\left[\frac{y^{\frac{2}{3}+1}}{\frac{2}{3}+1}\right]\\ =\frac{1}{2}{\left[\frac{y^{5/3}}{5/3}\right]}_0^{1\text{m}}\\ =\frac{3}{10}{\left[y^{5/3}\right]}_0^{1\text{m}}\\ =\frac{3}{10}{\text{m}}^{5/3}\end{array}$

Calculate the denominator part from Equation (III).

$\begin{array}{c}{\displaystyle \underset{0}{\overset{1\text{m}}{\int }}{y}^{1/3}dy}=\frac{y^{\frac{1}{3}+1}}{\left(\frac{1}{3}+1\right)}\\ =\frac{3}{4}{\left[y^{\frac{4}{3}}\right]}_0^{1\text{m}}\\ =\frac{3}{4}{\text{m}}^{4/3}\end{array}$

Substitute $\frac{3}{4}{\text{m}}^{4/3}$ for ${\displaystyle \underset{0}{\overset{1\text{m}}{\int }}{y}^{1/3}dy}$ and $\frac{3}{10}{\text{m}}^{5/3}$ for $\frac{1}{2}{\displaystyle \underset{0}{\overset{\text{1}\text{m}}{\int }}{y}^{2/3}dy}$ in Equation (III).

$\begin{array}{c}\overline{x}=\frac{\frac{3}{10}{\text{m}}^{5/3}}{\frac{3}{4}{\text{m}}^{4/3}}\\ =0.4\text{m}\end{array}$

Thus, the centroid $\overline{x}$ of the shaded area is $\underset{\_}{0.4\text{m}}$.

To determine

The centroid $\left(\overline{y}\right)$ of the parabolic-shaded area.

Answer to Problem 1FP

The centroid $\left(\overline{y}\right)$ of the parabolic shaded area is $\underset{\_}{0.571\text{m}}$.

Explanation of Solution

Given:

The length of the shaded area is 1 m.

The height of the shaded area is 1 m.

Show the area of the differential element as in Figure 2.

Using Figure 2,

Express the parabolic value.

$\begin{array}{l}y=x^3\\ x=y^{1/3}\end{array}$

Compute the area of the element.

$dA=xdy$

Substitute $y^{1/3}$ for x.

$dA=y^{1/3}dy$

Compute the centroid of the differential element along the y-axis using the formula.

$\tilde{y}=y$

Determine the location of the centre of gravity of the homogeneous rod along the y-axis $\left(\overline{y}\right)$ using the relation.

$\overline{y}=\frac{{\displaystyle {\int }_A\tilde{y}dA}}{{\displaystyle {\int }_{A}dA}}$ (IV)

Conclusion:

Apply the limits from 0 to 1 m, substitute $y$ for $\tilde{y}$ and $y^{1/3}dy$ for dA in Equation (IV).

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \underset{0}{\overset{1\text{m}}{\int }}y\cdot y^{1/3}dy}}{{\displaystyle {\int }_0^{1\text{m}}{y}^{1/3}dy}}\\ =\frac{{\displaystyle \underset{0}{\overset{1\text{m}}{\int }}{y}^{4/3}dy}}{{\displaystyle {\int }_0^{1\text{m}}{y}^{1/3}dy}}\end{array}$ (V)

Calculate the numerator part from Equation (V).

$\begin{array}{c}{\displaystyle \underset{0}{\overset{1\text{m}}{\int }}{y}^{4/3}dy}={\left[\frac{y^{\frac{4}{3}+1}}{\frac{4}{3}+1}\right]}_0^{1\text{m}}\\ =\frac{3}{7}{\text{m}}^{7/3}\end{array}$

Calculate the denominator part from Equation (V).

$\begin{array}{c}{\displaystyle \underset{0}{\overset{1\text{m}}{\int }}{y}^{1/3}dy}={\left[\frac{y^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]}_0^{1\text{m}}\\ =\frac{3}{4}{\text{m}}^{4/3}\end{array}$

Substitute $\frac{3}{4}{\text{m}}^{4/3}$ for ${\displaystyle \underset{0}{\overset{1\text{m}}{\int }}{y}^{1/3}dy}$ and $\frac{3}{7}{\text{m}}^{7/3}$ for ${\displaystyle \underset{0}{\overset{1\text{m}}{\int }}{y}^{4/3}dy}$ in Equation (V).

$\begin{array}{c}\overline{y}=\frac{\frac{3}{7}{\text{m}}^{7/3}}{\frac{3}{4}{\text{m}}^{4/3}}\\ =0.571\text{m}\end{array}$

Thus, the centroid $\overline{y}$ of the shaded area is $\underset{\_}{0.571\text{m}}$.