Chapter 9.1, Problem 9.23P

9.23 and 9.24 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to point P.

Fig. P9.23

To determine

Find the polar moment of inertia and polar radius of gyration of the shaded area shown with respect to point P.

Answer to Problem 9.23P

The polar moment of inertia of the shaded area shown with respect to point P is $\underset{\_}{\frac{64}{15}{a}^4}$.

The polar radius of gyration of the shaded area shown with respect to point P is $\underset{\_}{1.265a}$.

Explanation of Solution

Given information:

The equation of the curve is $y_2=c+k_2{x}^2$

The equation of the curve is $y_1=k_1{x}^2$

Calculation:

Sketch the vertical strip shaded portion as shown in Figure 1.

Write the curve $\left(y_1\right)$ Equation as follows:

$y_1=k_1{x}^2$ (1)

Write the curve $\left(y_1\right)$ Equation as follows:

$y_2=c+k_2{x}^2$ (2)

Refer to Figure 1.

At $y_1$

For $x=2a$, $y_1=2a$.

Find the constant $\left(k_1\right)$ using the relation:

Substitute $2a$ for $y_1$ and $2a$ for $x$ in Equation (1).

$\begin{array}{l}2a=k_1{\left(2a\right)}^2\\ k_1=\frac{2a}{{\left(2a\right)}^2}\\ k_1=\frac{1}{2a}\end{array}$

Substitute 0 for x and a for y in Equation (2).

$\begin{array}{l}a=c+k_2{\left(0\right)}^2\\ a=c+0\\ a=c\end{array}$

Refer to Figure 1.

At $y_2$

For $x=2a$, $y=2a$.

Find the constant $\left(k_2\right)$ using the relation:

Substitute $2a$ for $y_2$, a for c, and $2a$ for $x$ in Equation (2).

$\begin{array}{l}2a=a+k_2{\left(2a\right)}^2\\ k_2=\frac{2a}{4a^2}\\ k_2=\frac{1}{4a}\end{array}$

Substitute $\frac{1}{2a}$ for $k_1$ in Equation (1).

$y_1=\left(\frac{1}{2a}\right)x^2$

Substitute $\frac{1}{4a}$ for $k_2$ and a for c in Equation (2).

$\begin{array}{c}{y}_2=a+\left(\frac{1}{4a}\right)x^2\\ =a+\frac{1}{4a}{x}^2\end{array}$

Determine the area of the strip element $dA$ as shown in below:

$dA=\left(y_2-y_1\right)dx$ (3)

Substitute $a+\frac{1}{4a}{x}^2$ for $y_2$ and $\left(\frac{1}{2a}\right)x^2$ for $y_1$ in Equation (3).

$\begin{array}{c}dA=\left(a+\frac{1}{4a}{x}^2-\left(\frac{1}{2a}\right)x^2\right)dx\\ =\left(\frac{1}{4a}\left(4a^2+x^2\right)-\left(\frac{1}{2a}\right)x^2\right)dx\\ =\frac{1}{4a}\left(4a^2-x^2\right)dx\end{array}$

Find the shaded area (A) using the relation:

$A={\displaystyle \int dA}$ (4)

Substitute $\frac{1}{4a}\left(4a^2-x^2\right)dx$ for $dA$ in Equation (4).

$\begin{array}{c}A={\displaystyle \int \frac{1}{4a}\left(4a^2-x^2\right)dx}\\ =2{\displaystyle \underset{0}{\overset{2a}{\int }}\frac{1}{4a}\left(4a^2-x^2\right)dx}\\ =\frac{1}{2a}{\left[4a^{2}x-\frac{x^3}{3}\right]}_0^{2a}\\ =\frac{1}{2a}\left[4a^2\left(2a\right)-\frac{{\left(2a\right)}^3}{3}\right]\end{array}$

$\begin{array}{c}A=\frac{1}{2a}\left[8a^3-\frac{8a^3}{3}\right]\\ =\frac{1}{2a}\left[\frac{16a^3}{3}\right]\\ =\frac{8}{3}{a}^2\end{array}$

Determine the moment of inertia $\left(d{I}_x\right)$ with respect to x axis:

Refer Equation 9.2 in section 9.1B determining the moment of inertia of an area by integration:

$\begin{array}{c}d{I}_x=\frac{1}{3}{y}^{3}dx\\ =\left(\frac{1}{3}{y}_2^3-\frac{1}{3}{y}_1^3\right)dx\end{array}$ (5)

Substitute $a+\frac{1}{4a}{x}^2$ for $y_2$ and $\left(\frac{1}{2a}\right)x^2$ for $y_1$ in Equation (5).

$\begin{array}{c}d{I}_x=\left(\frac{1}{3}{\left(a+\frac{1}{4a}{x}^2\right)}^3-\frac{1}{3}{\left(\left(\frac{1}{2a}\right)x^2\right)}^3\right)dx\\ =\frac{1}{3}\left({\left[\frac{1}{4a}{\left(4a^2+x^3\right)}^3\right]}^3-{\left[\frac{1}{2a}{x}^2\right]}^2\right)dx\\ =\frac{1}{3}\left(\frac{1}{64a^3}\left(64a^6+48a^4{x}^2+12a^2{x}^4+x^6\right)-\frac{1}{8a^3}{x}^6\right)dx\\ =\frac{1}{192a^3}\left(64a^6+48a^4{x}^2+12a^2{x}^4-7x^6\right)dx\end{array}$ (6)

Integrate Equation (6) with respect to x.

$\begin{array}{c}{I}_x={\displaystyle \int d{I}_x}\\ I_x={\displaystyle \int \frac{1}{192a^3}\left(64a^6+48a^4{x}^2+12a^2{x}^4-7x^6\right)dx}\\ =2{\displaystyle \underset{0}{\overset{2a}{\int }}\frac{1}{192a^3}\left(64a^6+48a^4{x}^2+12a^2{x}^4-7x^6\right)dx}\\ =\frac{1}{96a^3}{\left[64a^{6}x+48a^4\frac{x^3}{3}+12a^2\frac{x^5}{5}-7\frac{x^7}{7}\right]}_0^{2a}\end{array}$

$\begin{array}{c}{I}_x=\frac{1}{96a^3}{\left[64a^{6}x+16a^4{x}^3+\frac{12}{5}{a}^2{x}^5-x^7\right]}_0^{2a}\\ =\frac{1}{96a^3}\left[64a^6\left(2a\right)+16a^4{\left(2a\right)}^3+\frac{12}{5}{a}^2{\left(2a\right)}^5-{\left(2a\right)}^7\right]\\ =\frac{1}{96}{a}^4\left[128+128+\frac{12}{5}\times 32-128\right]\\ =\frac{32}{15}{a}^4\end{array}$

Determine the moment of inertia $\left(d{I}_y\right)$ with respect to y axis:

$\begin{array}{c}\left(d{I}_y\right)=x^{2}dA\\ =x^2\left[\frac{1}{4a}\left(4a^2-x^2\right)dx\right]\end{array}$ (7)

Integrate Equation (7) with respect to x.

$\begin{array}{c}{I}_y={\displaystyle \int d{I}_y}\\ =2{\displaystyle \underset{0}{\overset{2}{\int }}\frac{1}{4a}{x}^2\left(4a^2-x^2\right)dx}\\ =\frac{1}{2a}{\left[\frac{4}{3}{a}^2{x}^3-\frac{1}{5}{x}^5\right]}_0^{2a}\\ =\frac{1}{2a}\left[\frac{4}{3}{a}^2{\left(2a\right)}^3-\frac{1}{5}{\left(2a\right)}^5\right]\end{array}$

$\begin{array}{c}=\frac{1}{2a}\left[\frac{4}{3}{a}^2\left(8a^3\right)-\frac{1}{5}\left(32a^5\right)\right]\\ =\frac{1}{2a}\left[\frac{32a^5}{3}-\frac{32a^5}{5}\right]\\ =\frac{1}{2a}\left[a^5\left(\frac{32}{3}-\frac{32}{5}\right)\right]\\ =\frac{1}{2a}\left[a^5\left(\frac{160-96}{15}\right)\right]\end{array}$

$\begin{array}{c}=\frac{1}{2a}\left[a^5\left(\frac{64}{15}\right)\right]\\ I_y=\frac{32a^4}{64}\end{array}$

Find the polar moment of inertia $\left(J_P\right)$ of the shaded area shown with respect to point P.

$\left(J_P\right)=I_x+I_y$ (8)

Here, $I_x$ is moment of inertia about x axis and $I_y$ is moment of inertia about y axis.

Substitute $\frac{32a^4}{64}$ for $I_y$ and $\frac{32}{15}{a}^4$ for $I_x$ in Equation (8).

$\begin{array}{c}\left(J_P\right)=\frac{32}{15}{a}^4+\frac{32a^4}{64}\\ =\frac{64}{15}{a}^4\end{array}$

Thus, the polar moment of inertia of the shaded area shown with respect to point P is $\underset{\_}{\frac{64}{15}{a}^4}$.

Find the polar radius of gyration of the shaded area shown with respect to point P.

$\left(k_P^2\right)=\frac{J_P}{A}$ (9)

Here, $J_P$ is polar moment of inertia.

Substitute $\frac{64}{15}{a}^4$ for $J_P$ and $\frac{8}{3}{a}^2$ for A in Equation (9).

$\begin{array}{c}\left(k_P^2\right)=\frac{\frac{64}{15}{a}^4}{\frac{8}{3}{a}^2}\\ =\frac{64}{15}{a}^4\times \frac{3}{8}{a}^2\\ =\frac{8}{5}{a}^2\\ k_p=1.265a\end{array}$

Thus, the polar radius of gyration of the shaded area shown with respect to point P is $\underset{\_}{1.265a}$.