Chapter 5.1, Problem 5.17P

Show that as r₁ approaches r₂, the location of the centroid approaches that for an arc of circle of radius (r₁ + r₂)/2.

To determine

To show that as $r_1$ approaches $r_2$ the location of the centroid approaches that for an arc of circle of radius $\frac{r_1+r_2}{2}$.

Answer to Problem 5.17P

The location of the centroid for an arc of circle of radius $\frac{r_1+r_2}{2}$ is $\underset{\_}{\left(\frac{1}{2}\right)\left(r_1+r_2\right)\left(\frac{\cos \alpha }{90°-\alpha }\right)}$.

Explanation of Solution

Write the expression for the y-coordinate of the centroid of the sector with radius $r_2$.

$\begin{array}{c}\overline{y_2}=\left(\frac{2}{3}\right)r_2\left(\frac{\sin \left(90°-\alpha \right)}{90°-\alpha }\right)\\ =\left(\frac{2}{3}\right)r_2\left(\frac{\cos \alpha }{90°-\alpha }\right)\end{array}$ (I)

Here $\overline{y_2}$ is the y-coordinate of the centroid of the sector with radius $r_2$.

Write the expression for the area of the sector with radius $r_2$.

$A=\left(90°-\alpha \right)r_2{}^2$ (II)

$A$ is the radius of the sector with radius $r_2$.

Write the expression for the y-coordinate of the centroid of the sector with radius $r_1$.

$\begin{array}{c}\overline{y_1}=\left(\frac{2}{3}\right)r_1\left(\frac{\sin \left(90°-\alpha \right)}{90°-\alpha }\right)\\ =\left(\frac{2}{3}\right)r_1\left(\frac{\cos \alpha }{90°-\alpha }\right)\end{array}$ (III)

Here $\overline{y_1}$ is the y-coordinate of the centroid of the sector with radius $r_1$.

Write the expression for the area of the sector with radius $r_1$.

$A^{\prime }=\left(90°-\alpha \right)r_1{}^2$ (IV)

$A^{\prime }$ is the radius of the sector with radius $r_1$.

Write the expression for ${\displaystyle \sum \overline{y}}A$.

${\displaystyle \sum \overline{y}}A=\left(\overline{y_2}\right)A-\left(\overline{y_1}\right)A^{\prime }$

Substitute (I), (II), (III) and (IV) in the above equation to calculate ${\displaystyle \sum \overline{y}}A$.

$\begin{array}{c}{\displaystyle \sum \overline{y}}A=\left(\left(\frac{2}{3}\right)r_2\left(\frac{\cos \alpha }{90°-\alpha }\right)\right)\left(90°-\alpha \right)r_2{}^2-\left(\left(\frac{2}{3}\right)r_1\left(\frac{\cos \alpha }{90°-\alpha }\right)\right)\left(90°-\alpha \right)r_1{}^2\\ =\left(\frac{2}{3}\right)\left(r_2{}^3-r_1{}^3\right)\cos \alpha \end{array}$ (V)

Write the expression for ${\displaystyle \sum A}$.

${\displaystyle \sum A}=A-A^{\prime }$

Substitute (II) and (IV) in the above equation to calculate ${\displaystyle \sum A}$.

$\begin{array}{c}{\displaystyle \sum A}=\left(90°-\alpha \right)r_2{}^2-\left(90°-\alpha \right)r_1{}^2\\ =\left(90°-\alpha \right)\left(r_2{}^2-r_1{}^2\right)\end{array}$ (VI)

Write the expression to calculate the y-coordinate of the centroid of the given area.

$\overline{Y}{\displaystyle \sum A}={\displaystyle \sum \overline{y}}A$

$\overline{Y}$ is the y-coordinate of the centroid of the given area.

Substitute (V) and (VI) in the above equation to calculate $\overline{Y}$.

$\begin{array}{c}\overline{Y}\left(\left(90°-\alpha \right)\left(r_2{}^2-r_1{}^2\right)\right)=\left(\frac{2}{3}\right)\left(r_2{}^3-r_1{}^3\right)\cos \alpha \\ \overline{Y}=\left(\frac{2}{3}\right)\frac{\left(r_2{}^3-r_1{}^3\right)}{\left(r_2{}^2-r_1{}^2\right)}\left(\frac{\cos \alpha }{90°-\alpha }\right)\end{array}$ (VII)

Write the expression for the y-coordinate of the centroid with an arc of radius $\frac{r_1+r_2}{2}$.

$\begin{array}{c}\overline{y}=\left(\frac{r_1+r_2}{2}\right)\left(\frac{\sin \left(90°-\alpha \right)}{90°-\alpha }\right)\\ =\left(\frac{r_1+r_2}{2}\right)\left(\frac{\cos \alpha }{90°-\alpha }\right)\end{array}$ (VIII)

Rewrite the expression $\frac{\left(r_2{}^3-r_1{}^3\right)}{\left(r_2{}^2-r_1{}^2\right)}$.

$\begin{array}{c}\frac{\left(r_2{}^3-r_1{}^3\right)}{\left(r_2{}^2-r_1{}^2\right)}=\frac{\left(r_2-r_1\right)\left(r_2{}^2+r_1{r}_2+r_1{}^2\right)}{\left(r_2-r_1\right)\left(r_2+r_1\right)}\\ =\frac{\left(r_2{}^2+r_1{r}_2+r_1{}^2\right)}{\left(r_2+r_1\right)}\end{array}$ (IX)

Let $r_2=r+\Delta$ and $r_1=r+\Delta$, where $\Delta$ is a small variation in r.

Substitute the above two expressions in (VIII) to rewrite.

$\begin{array}{c}\frac{\left(r_2{}^3-r_1{}^3\right)}{\left(r_2{}^2-r_1{}^2\right)}=\frac{\left({\left(r+\Delta \right)}^2+\left(r+\Delta \right)\left(r-\Delta \right)+{\left(r-\Delta \right)}^2\right)}{\left(r+\Delta +r-\Delta \right)}\\ =\frac{r^2+2r\Delta +{\Delta }^2+r^2+r\Delta -\Delta r-{\Delta }^2+r^2-2r\Delta +{\Delta }^2}{2r}\\ =\frac{3r^2+{\Delta }^2}{2r}\end{array}$ (X)

Apply the limits $\Delta \to 0$ $\left(r_1=r_2\right)$ in the above expression to rewrite.

$\begin{array}{c}\frac{\left(r_2{}^3-r_1{}^3\right)}{\left(r_2{}^2-r_1{}^2\right)}=\frac{3r^2+0^2}{2r}\\ =\frac{3}{2}r\end{array}$ (XI)

Rewrite r using $r_1$ and $r_2$ in the above equation.

$\begin{array}{c}\frac{\left(r_2{}^3-r_1{}^3\right)}{\left(r_2{}^2-r_1{}^2\right)}=\frac{3}{2}\left(\frac{1}{2}\right)\left(r_1+r_2\right)\\ =\frac{3}{4}\left(r_1+r_2\right)\end{array}$ (XII)

Conclusion:

Substitute (XII) in (VII) to calculate $\overline{Y}$.

$\begin{array}{c}\overline{Y}=\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\left(r_1+r_2\right)\left(\frac{\cos \alpha }{90°-\alpha }\right)\\ =\left(\frac{1}{2}\right)\left(r_1+r_2\right)\left(\frac{\cos \alpha }{90°-\alpha }\right)\end{array}$

Therefore, the above expression is same as that of the expression given in (VIII).

Thus, the location of the centroid for an arc of circle of radius $\frac{r_1+r_2}{2}$ is $\underset{\_}{\left(\frac{1}{2}\right)\left(r_1+r_2\right)\left(\frac{\cos \alpha }{90°-\alpha }\right)}$.