# 3.4.3: Examples of Moment of Inertia

Example 3.2

Calculate the moment of inertia for the mass of the cylinder about center axis which height of $$h$$ and radius, $$r_0$$, as shown in Figure 3.6. The material is with an uniform density and homogeneous.

Solution 3.2

The element can be calculated using cylindrical coordinate. Here the convenient element is a shell of thickness $$dr$$ which shown in Figure 3.6 as

$I_{rr} = \rho \int_V r^2 dm = \rho \int_0^{r_0}r^2\; \overbrace{h\,2\,\pi\,r\, dr}^{dV} = \rho \, h \, 2\,\pi \dfrac ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/03:_Review_of_Mechanics/3.4:_Moment_of_Inertia/3.4.3:_Examples_of_Moment_of_Inertia), /content/body/div[2]/div/p[3]/span, line 1, column 4  = \dfrac{r_0} {\sqrt{2}} \tag{30}$

Fig. 3.7 Description of rectangular in x–y plane for calculation of moment of inertia.

Example 3.3

Calculate the moment of inertia of the rectangular shape shown in Figure 3.7 around $$\mathbf{x}$$ coordinate.

Solution 3.3

The moment of inertia is calculated utilizing equation (20) as following
$I_{xx} = \int_A \left( \overbrace{y^2}^{0} + z^2 \right) dA = \int_0^a z^2 \overbrace{bdz}^{dA} = \dfrac{a^3\,b}{3} \tag{31}$
This value will be used in later examples.

Example 3.4

To study the assumption of zero thickness, consider a simple shape to see the effects of this assumption. Calculate the moment of inertia about the center of mass of a square shape with a thickness, $$t$$ compare the results to a square shape with zero thickness.

Solution 3.4

The moment of inertia of transverse slice about $$y^{′}$$ (see Figure 3.8) is
${dI_{xx}}_m = \rho \overbrace{dy}^{t} \overbrace{\dfrac{ b\, a^3}{12}}^{I_{xx}} \label{mech:eq:transverseElement} \tag{32}$

Fig. 3.8 A square element for the calculations of inertia of two-dimensional to three–dimensional deviations.

The transformation into from local axis $$x$$ to center axis, $$x^{′}$$ can be done as following
${dI_{x^{'}x^{'}}}_m = \rho dy \left(\overbrace{\dfrac{ b\, a^3}{12}}^{I_{xx}} + \overbrace{ \underbrace{z^2}_{r^2} \; \underbrace{b\,a }_{A} }^{r^2\,A} \right) \label{mech:eq:transverseElementM} \tag{33}$
The total moment of inertia can be obtained by integration of equation (33) to write as
${I_{xx}}_m = \rho \int_{-t/2}^{t/2} \left( \dfrac{ b\, a^3}{12} + z^2 \,b\,a \right) dz = \rho\, t\, \dfrac{a\,b\,{t}^{2}+{a}^{3}\,b}{12} \label{mech:eq:transverseElementC} \tag{34}$

Fig. 3.9 The ratio of the moment of inertia of two-dimensional to three–dimensional.

Comparison with the thin body results in

$\dfrac{I_{xx}\,\rho \, t}{{I_{xx}}_m} = \dfrac{b\,a^3}{t^2\,b\,a + b\,a^3} = \dfrac{1}{1 + \dfrac{t^2}{a^2} } \label{mech:eq:IxxRecR} \tag{35}$

It can be noticed right away that equation (35) indicates that ratio approaches one when thickness ratio is approaches zero, $${I_{xx}}_m (t\rightarrow 0) \rightarrow 1$$. Additionally it can be noticed that the ratio $$a^{2}/t^{2}$$ is the only contributor to the error. The results are present in Figure 3.9. I can be noticed that the error is significant very fast even for small values of $$t/a$$ while the with of the box, $$b$$ has no effect on the error.

Example 3.5

Fig. 3.10 Rectangular Moment of inertia.

Calculate the rectangular moment of Inertia for the rotation trough center in $$zz$$ axis (axis of rotation is out of the page). Hint, construct a small element and build longer build out of the small one. Using this method calculate the entire rectangular.

Solution 3.5

The moment of inertia for a long element with a distance $$y$$ shown in Figure 3.10 is
$\label{Izz:element} \left. d\, I_{zz} \right|_{dy} = \int_{-a}^{a} \overbrace{\left(y^2+x^2\right)}^{r^2} \,dy\,dx = \dfrac{2\,\left( 3\,a\,{y}^{2}+{a}^{3}\right) }{3} \,dy \tag{36}$
The second integration (no need to use (26), why?) is
$\label{Izz:elementY} I_{zz} = \int_{-b}^{b} \dfrac{2\,\left( 3\,a\,{y}^{2}+{a}^{3}\right) }{3}\,dy \tag{37}$
Results in
$\label{Izz:totalC} I_{zz} = \dfrac{a\,\left( 2\,a\,{b}^{3}+2\,{a}^{3}\,b\right) }{3} = \overbrace{A}^{4\,a\,b} \left( \dfrac{(2a)^2+(2b)^2}{12} \right) \tag{38}$
Or

Example 3.6

Fig. 3.11 Parabola for calculations of moment of inertia.

Calculate the center of area and moment of inertia for the parabola, $$y = \alpha x^2$$, depicted in Figure 3.11. Hint, calculate the area first. Use this area to calculate moment of inertia. There are several ways to approach the calculation (different infinitesimal area).

Solution 3.6

For $$y=b$$ the value of $$x = \sqrt ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/03:_Review_of_Mechanics/3.4:_Moment_of_Inertia/3.4.3:_Examples_of_Moment_of_Inertia), /content/body/div[10]/div/p[2]/span[1], line 1, column 2 $$. First the area inside the parabola calculated as
$A = 2\, \int_0^{\sqrt ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/03:_Review_of_Mechanics/3.4:_Moment_of_Inertia/3.4.3:_Examples_of_Moment_of_Inertia), /content/body/div[10]/div/p[2]/span[2], line 1, column 2  } \overbrace{(b - \alpha \xi^2) d\xi}^{dA/2} = \dfrac{2(3\,\alpha-1)}{3} \; \left( \dfrac{b}{\alpha}\right) ^{\dfrac{3}{2}} \tag{39}$
The center of area can be calculated utilizing equation (12). The center of every element is at, $$\left(\alpha\,\xi^2 +\dfrac{b-\alpha\xi^2}{2}\right)$$ the element area is used before and therefore
$x_c = \dfrac{1}{A} \int_0^{\sqrt ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/03:_Review_of_Mechanics/3.4:_Moment_of_Inertia/3.4.3:_Examples_of_Moment_of_Inertia), /content/body/div[10]/div/p[2]/span[3], line 1, column 2  } \overbrace{\left(\alpha\xi^2 +\dfrac{(b-\alpha\xi^2)}{2}\right)} ^{x_c} \overbrace{(b - \alpha \xi^2) d\xi}^{dA} = \dfrac{3\,\alpha\,b}{15\,\alpha-5} \label{mech:eq:xcParabola} \tag{40}$
The moment of inertia of the area about the center can be found using in equation (40) can be done in two steps first calculate the moment of inertia in this coordinate system and then move the coordinate system to center. Utilizing equation (20) and doing the integration from 0 to maximum y provides
$I_{x^{'}x^{'}} = 4\, \int_0^{b} \xi^2 \overbrace{\sqrt{\dfrac{\xi}{\alpha}}\;d\xi}^{dA} = \dfrac{2\,{b}^{7/2}}{7\,\sqrt{\alpha}} \tag{41}$
Utilizing equation (26)
$I_{xx} = I_{x^{'}x^{'}} - A\; {\Delta x}^2 = \overbrace{\dfrac{4\,{b}^{7/2}}{7\,\sqrt{\alpha}}} ^{I_{x^{'}x^{'}}} - \overbrace{ \dfrac{3\,\alpha-1}{3} \; \left( \dfrac{b}{\alpha}\right) ^{\dfrac{3}{2}}}^{A} \overbrace{\left(\dfrac{3\,\alpha\,b}{15\,\alpha-5}\right)^2} ^{{\left( \Delta x = x_c\right) }^2} \tag{42}$

or after working the details results in

$I_{xx} = \dfrac{\sqrt{b}\,\left( 20\,{b}^{3}-14\,{b}^{2}\right) } {35\,\sqrt{\alpha}} \tag{43}$

Example 3.7

Fig. 3.12 Triangle for example.

Calculate the moment of inertia of strait angle triangle about its $$y$$ axis as shown in the Figure on the right. Assume that base is $$mathbf{a}$$ and the height is $$mathbf{h}$$. What is the moment when a symmetrical triangle is attached on left? What is the moment when a symmetrical triangle is attached on bottom? What is the moment inertia when $$a\longrightarrow 0$$? What is the moment inertia when $$h\longrightarrow 0$$?

Solution 3.7

$\dfrac{y}{h} = \left( 1 - \dfrac{x}{a} \right) \tag{44}$
or
$\dfrac{x}{a} = \left( 1 - \dfrac{y}{h} \right) \tag{45}$
Now using the moment of inertia of rectangle on the side ($$\mathbf{y}$$)
coordinate (see example 3.3)
$\int_0^h \dfrac{ a\,\left( 1 - \dfrac{y}{h} \right)^3 dy } {3} = \dfrac {a^3\,h}{4} \tag{46}$
For two triangles attached to each other the moment of inertia will be sum
as $$\dfrac {a^3\,h}{2}$$ The rest is under construction.

## Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.