Chapter 9.1, Problem 9.24P

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. P9.24

To determine

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

Answer to Problem 9.24P

The polar moment of inertia of the shaded area with respect to point P is $\underset{\_}{1.155r^4}$

The polar radius of gyration of the shaded area with respect to point P is $\underset{\_}{0.676r}$

Explanation of Solution

Calculation:

Sketch the horizontal strip along circular portion as shown in Figure 1.

Write the curve equation of circle as follows:

$x^2+y^2=r^2$ (1)

Modify Equation (1).

$\begin{array}{l}{x}^2+y^2=r^2\\ x^2=r^2-y^2\\ x=\sqrt{r^2-y^2}\end{array}$

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

$dA=xdy$ (2)

Substitute $\sqrt{r^2-y^2}$ for x in Equation (2).

$dA=\left(\sqrt{r^2-y^2}\right)dy$

Find the shaded area (A) using the relation:

$A={\displaystyle \int dA}$ (3)

Substitute $\left(\sqrt{r^2-y^2}\right)dy$ for $dA$ and apply the limits in Equation (3).

$\begin{array}{c}A={\displaystyle \int \left(\sqrt{r^2-y^2}\right)dy}\\ =2{\displaystyle \underset{-\frac{r}{2}}{\overset{r}{\int }}\sqrt{r^2-y^2}dy}\end{array}$ (4)

Consider $y=r\sin \theta$

Differentiate both sides of the Equation.

$dy=r\cos \theta d\theta$

Substitute $r\sin \theta$ for y and $r\cos \theta d\theta$ for $dy$ and apply the limits in Equation (4).

$\begin{array}{c}A=2{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}\sqrt{r^2-{\left(r\sin \theta \right)}^2}r\cos \theta d\theta }\\ =2{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}{r}^2{\cos }^2\theta d\theta }\\ =2r^2{\left[\frac{\theta }{2}+\frac{\sin 2\theta }{4}\right]}_{-\frac{\pi }{6}}^{\frac{\pi }{2}}\\ =2r^2\left[\frac{\frac{\pi }{2}}{2}+\frac{\sin 2\left(\frac{\pi }{2}\right)}{4}-\left(-\frac{\frac{\pi }{6}}{2}\right)+\frac{\sin 2\left(\frac{\pi }{6}\right)}{4}\right]\end{array}$

$\begin{array}{c}A=2r^2\left[\frac{\frac{\pi }{2}}{2}-\left(-\frac{\frac{\pi }{6}}{2}\right)+\frac{\sin 2\left(\frac{\pi }{6}\right)}{4}\right]\\ =2r^2\left(\frac{\pi }{3}+\frac{\sqrt{3}}{8}\right)\\ =2.5274r^2\end{array}$

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

$d{I}_x=y^{2}dA$

Substitute $\left(\sqrt{r^2-y^2}\right)dy$ for $dA$.

$d{I}_x=y^2\left(\left(\sqrt{r^2-y^2}\right)dy\right)$ (5)

Integrate Equation (5) with respect to y.

$\begin{array}{c}{I}_x={\displaystyle \int d{I}_x}\\ =2{\displaystyle \underset{-\frac{r}{2}}{\overset{r}{\int }}{y}^2\left(\left(\sqrt{r^2-y^2}\right)dy\right)}\end{array}$ (6)

Consider $y=r\sin \theta$

Differentiate both sides of the Equation.

$dy=r\cos \theta d\theta$

Substitute $r\sin \theta$ for y and $r\cos \theta d\theta$ for $dy$ and apply the limits in Equation (6).

$\begin{array}{c}{I}_x=2{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}r\sin {\theta }^2\left(\left(\sqrt{r^2-{\left(r\sin \theta \right)}^2}\right)\left(r\cos \theta d\theta \right)\right)}\\ =2{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}{r}^2{\sin }^2\theta \left(r\cos \theta \right)r\cos \theta d\theta }\\ =2{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}{r}^4{\sin }^2\theta {\cos }^2\theta d\theta }\\ =2{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}{r}^4\left(\frac{1}{4}{\sin }^{2}2\theta \right)d\theta }\end{array}$

$\begin{array}{c}{I}_x=\frac{r^4}{2}{\left[\frac{\theta }{2}-\frac{\sin 4\theta }{8}\right]}_{-\frac{\pi }{6}}^{\frac{\pi }{2}}\\ =\frac{r^4}{2}\left[\frac{\frac{\pi }{2}}{2}-\frac{\sin 4\left(\frac{\pi }{2}\right)}{8}-\left(\frac{-\frac{\pi }{6}}{2}-\frac{\sin 4\left(-\frac{\pi }{6}\right)}{8}\right)\right]\\ =\frac{r^4}{2}\left[\frac{\frac{\pi }{2}}{2}-\left(\frac{\frac{\pi }{6}}{2}-\frac{\sin \left(-\frac{2\pi }{3}\right)}{8}\right)\right]\\ =\frac{r^4}{2}\left(\frac{\pi }{3}-\frac{\sqrt{3}}{16}\right)\end{array}$

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

$d{I}_y=\frac{1}{3}{x}^{3}dy$ (7)

Substitute $\sqrt{r^2-y^2}$ for x in Equation (7).

$d{I}_y=\frac{1}{3}{\left(\sqrt{r^2-y^2}\right)}^{3}dy$ (8)

Integrate Equation (8) with respect to y.

$\begin{array}{c}{I}_y={\displaystyle \int d{I}_y}\\ ={\displaystyle \underset{-\frac{r}{2}}{\overset{r}{\int }}\frac{1}{3}{\left(\sqrt{r^2-y^2}\right)}^{3}dy}\end{array}$ (9)

Consider $y=r\sin \theta$

Differentiate both sides of the Equation.

$dy=r\cos \theta d\theta$

Substitute $r\sin \theta$ for y and $r\cos \theta d\theta$ for $dy$ and apply the limits in Equation (9).

$\begin{array}{c}{I}_y={\displaystyle \underset{-\frac{r}{2}}{\overset{r}{\int }}\frac{1}{3}{\left(\sqrt{r^2-{\left(r\sin \theta \right)}^2}\right)}^{3}r\cos \theta d\theta }\\ =\frac{2}{3}{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}{\left[r^2-{\left(r\sin \theta \right)}^2\right]}^{\frac{3}{2}}r\cos \theta d\theta }\\ =\frac{2}{3}{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}\left[r^3{\cos }^3\theta \right]r\cos \theta d\theta }\\ =\frac{2}{3}{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}{r}^4{\cos }^4\theta d\theta }\end{array}$

$\begin{array}{c}=\frac{2}{3}{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}{r}^4{\cos }^2\theta \left(1-{\sin }^2\theta \right)}d\theta \\ =\frac{2}{3}{\displaystyle \underset{-\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}{r}^4\left({\cos }^2\theta -\frac{1}{4}{\sin }^{2}2\theta \right)}d\theta \\ =\frac{2}{3}{r}^4{\left[\left(\frac{\theta }{2}+\frac{\sin 2\theta }{4}\right)-\frac{1}{4}\left(\frac{\theta }{2}-\frac{\sin 4\theta }{8}\right)\right]}_{-\frac{\pi }{6}}^{\frac{\pi }{2}}\\ =\frac{2}{3}{r}^4\left[\left(\frac{\frac{\pi }{2}}{2}+\frac{\sin 2\left(\frac{\pi }{2}\right)}{4}\right)-\frac{1}{4}\left(\frac{\left(-\frac{\pi }{6}\right)}{2}-\frac{\sin 4\left(-\frac{\pi }{6}\right)}{8}\right)\right]\end{array}$

$\begin{array}{c}{I}_y=\frac{2}{3}{r}^4\left[\frac{\frac{\pi }{2}}{2}-\frac{1}{4}\left(\frac{\frac{\pi }{2}}{2}\right)-\left(\frac{\left(-\frac{\pi }{6}\right)}{2}+\frac{\sin \left(-\frac{\pi }{3}\right)}{4}-\frac{1}{4}\left(\frac{-\frac{\pi }{6}}{2}-\frac{\sin \left(-\frac{2\pi }{3}\right)}{8}\right)\right)\right]\\ =\frac{2}{3}{r}^4\left[\frac{\pi }{4}-\frac{\pi }{16}+\frac{\pi }{12}+\frac{1}{4}\left(\frac{\sqrt{3}}{2}\right)-\frac{\pi }{48}+\frac{1}{32}\left(\frac{\sqrt{3}}{2}\right)\right]\\ =\frac{2}{3}{r}^4\left(\frac{\pi }{4}+\frac{9\sqrt{3}}{64}\right)\end{array}$

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

$\left(J_P\right)=I_x+I_y$ (10)

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

Substitute $\frac{2}{3}{r}^4\left(\frac{\pi }{4}+\frac{9\sqrt{3}}{64}\right)$ for $I_y$ and $\frac{r^4}{2}\left(\frac{\pi }{3}-\frac{\sqrt{3}}{16}\right)$ for $I_x$ in Equation (10).

$\begin{array}{c}\left(J_P\right)=\frac{r^4}{2}\left(\frac{\pi }{3}-\frac{\sqrt{3}}{16}\right)+\frac{2}{3}{r}^4\left(\frac{\pi }{4}+\frac{9\sqrt{3}}{64}\right)\\ =r^4\left(\frac{\pi }{3}+\frac{\sqrt{3}}{16}\right)\\ =r^4\left(1.047+0.10825\right)\\ =1.1545r^4\end{array}$

Thus, the polar moment of inertia of the shaded area with respect to point P is $\underset{\_}{1.155r^4}$

Find the polar radius of gyration $\left(k_P\right)$ of the shaded area with respect to point P as shown below:

$\left(k_P^2\right)=\frac{J_P}{A}$ (11)

Here, $J_P$ is polar moment of inertia.

Substitute $1.1545r^4$ for $J_P$ and $2.5274r^2$ for A in Equation (11).

$\begin{array}{c}\left(k_P^2\right)=\frac{1.1545r^4}{2.5274r^2}\\ =0.457r^2\\ k_P=0.676r\end{array}$

Thus, the polar radius of gyration of the shaded area with respect to point P is $\underset{\_}{0.676r}$.