Chapter 5.1, Problem 5.16P

PROBLEM 5.16

Determine the y coordinate of the centroid of the shaded area in terms of r₁, r₂, and α.

To determine

The y coordinate of the centroid of the shaded area in terms of $r_1,r_2$ and $\alpha$.

Answer to Problem 5.16P

The y-coordinate of the centroid of the plane area is $\underset{\_}{\left(\frac{2}{3}\right)\frac{\left(r_2{}^3-r_1{}^3\right)}{\left(r_2{}^2-r_1{}^2\right)}\left(\frac{\cos \alpha }{90°-\alpha }\right)}$.

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.

Conclusion:

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

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

Thus, the y-coordinate of the centroid of the plane area is $\underset{\_}{\left(\frac{2}{3}\right)\frac{\left(r_2{}^3-r_1{}^3\right)}{\left(r_2{}^2-r_1{}^2\right)}\left(\frac{\cos \alpha }{90°-\alpha }\right)}$.