Polar Moment of Inertia – Fusion of Engineering, Control, Coding, Machine Learning, and Science

The polar moment of inertia is used to compute the maximal shear stress when an element is subject to torsional load. Consequently, it is important to know how to calculate the polar moment of inertia for simple geometries.

Consider a circle shown in the figure below.

Figure 1: Calculation of the polar moment of intertia.

The polar moment of inertia is defined as

(1) $\begin{align*}J=\iint_{A} r^{2} dA\end{align*}$

where $r$ is a polar coordinate shown in the figure, and $dA$ is loosely speaking, an area of an infinitely small segment shown in the figure. Our first goal is to express dA using polar coordinates $r,\varphi$ and associated differentials $dr,d\varphi$ . Consider the next figure

Figure 2: Sketch of a small surface and associated arc lengths and side dimensions.

The length of the arc segment $\overline{PQ}$ is

(2) $\begin{align*} \overline{PQ} = r\cdot d\varphi\end{align*}$

where $d\varphi$ is measured in radians. Similarly, the length of the arc segment $\overline{ML}$ is

(3) $\begin{align*} \overline{ML} = (r+dr)\cdot d\varphi\end{align*}$

On the other hand, since $dr$ is very small, we have that $r\cdot d\varphi \approx (r+dr)\cdot d\varphi$ .

Furthermore, since the angle $d\varphi$ and the length $dr$ are small, the segment PQLM can be approximated by a rectangle shown in Fig. 3 below.

From Fig. 2, we can see that $dA$ can be approximated by an area of the rectangle, that is,

(4) $\begin{align*}dA\approx r\cdot d\varphi \cdot dr\end{align*}$

By substituting this approximation in (1), we have

(5) $\begin{align*} J=\int_{r=0}^{r=R} \int_{\varphi = 0}^{ \varphi = 2\pi } r^{2} \cdot r\cdot d\varphi \cdot dr = \int_{r=0}^{r=R} \int_{\varphi = 0}^{ \varphi = 2\pi } r^{3} \cdot d\varphi \cdot dr \end{align*}$

In (5), we have substituted the surface bounds. The variable $r$ goes from $r=0$ to $r=R$ . On the other hand, the variable $\varphi$ goes from $\varphi =0$ to $\varphi =2\pi$ . Next, we solve the integral

(6) $\begin{align*}& J= \int_{r=0}^{r=R} \int_{\varphi = 0}^{ \varphi = 2\pi } r^{3} \cdot d\varphi \cdot dr \\& = \int_{r=0}^{r=R} \Big( \int_{\varphi = 0}^{ \varphi = 2\pi } 1\cdot d\varphi \Big) r^{3} \cdot dr \\ & = \int_{r=0}^{r=R} \varphi \bigg|_{0}^{2\pi} r^{3} \cdot dr \\& = \int_{r=0}^{r=R} \Big( 2\pi - 0 \Big) r^{3} \cdot dr \\& = 2\pi \int_{r=0}^{r=R} r^{3} \cdot dr \\& = 2\pi \frac{r^{4}}{4} \bigg|_{0}^{R} \\& = 2\pi \Big(\frac{R^{4}}{4} - 0 \Big)=\pi \frac{R^{4}}{2} \end{align*}$