Centre of gravity, static moments and moments of inertia

Multiple integrals: Applications of multiple integrals

Centre of gravity, static moments and moments of inertia

If the continuous density $\rho(x,y,z)$ is given in every point of a rigid body $R$ , then for a volume element with very small dimensions $\dd (x,y,z)$ it is true that its mass is approximately equal to $\rho(x,y,z)\,\dd(x, y,z)$ . The total mass $M$ of $R$ is therefore given by $M=\iiint_R \rho(x,y,z)\,\dd(x, y,z)$

The 3D static moments $M_{yz}$ , $M_{xz}$ and $M_{xy}$ of $R$ relative to the $yz$ -plane, the $xz$ -plane and the $xy$ -plane, respectively, also called first (mass)moments, are defined as follows:
$\begin{aligned} M_{yz} &=\iiint_R x\,\rho(x,y,z)\,\dd(x,y,z)\\[0.25cm] M_{xz} &=\iiint_R y\,\rho(x,y,z)\,\dd(x,y,z)\\[0.25cm] M_{xy} &=\iiint_R z\,\rho(x,y,z)\,\dd(x,y,z)\end{aligned}$

The 3D centre of gravity, usually called the centroid, $(\bar{x},\bar{y},\bar{z})$ of $R$ is given by: $\bar{x}=\frac{M_{yz}}{M},\quad \bar{y}=\frac{M_{xz}}{M},\quad \bar{z}=\frac{M_{xy}}{M}$

Determine the centre of gravity of the rigid body $R$ sketched below and enclosed by the plane $z=0$ and the paraboloid $z=16-x^2-y^2$ under the assumption that the density is a constant $\rho$ .

GeoGebra

$\cv{\bar{x} \\ \bar{y} \\ \bar{z}} ={}$ $\cv{0\\ 0\\ {{16}\over{3}}}$

By symmetry, we know that $\bar{x}=0$ and $\bar{y}=0$ . We can therefore limit ourselves to the calculation of $\bar{z}$ .

First we calculate the total mass $M$ : $\begin{aligned} M &= \iiint_R \rho\,\dd(x,y,z)\\[0.25cm] &= \rho\, \int_{z=0}^{z=16}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{r=0}^{r=\sqrt{16-z}}r\,\dd r\right)\dd\varphi\right)\dd z\\[0.25cm]&\phantom{abcuva} \blue{\text{in cylindrical coordinates}}\\[0.25cm] &= \rho\, \int_{z=0}^{z=16}\left(\int_{\varphi=0}^{\varphi=2\pi}\biggl[\frac{1}{2}r^2\biggl]_{r=0}^{r=\sqrt{16-z}}\;\dd\varphi\right)\dd z\\[0.25cm] &=\frac{1}{2}\rho\int_{z=0}^{z=16}\left(\int_{\varphi=0}^{\varphi=2\pi}(16-z)\,\dd \varphi\right)\dd z\\[0.25cm] &=\frac{1}{2}\rho\int_{z=0}^{z=16}\biggl[(16-z)\,\varphi\biggr]_{\varphi=0}^{\varphi=2\pi}\;\dd z \\[0.25cm] &= \pi\,\rho\int_{z=0}^{z=16}(16-z)\,\dd z\\[0.25cm] &= \pi\,\rho\,\biggl[-\frac{1}{2}(16-z)^2\biggr]_{z=0}^{z=16}\\[0.25cm] &=\pi\,\rho\, \left(0-\Bigl(-\frac{1}{2}\,16^2\Bigr)\right)\\[0.25cm] &= 128\,\pi\,\rho\end{aligned}$ Then we determine the static moment in a similar way $M_{xy}$ : $\begin{aligned} M_{xy} &= \iiint_R z\,\rho\,\dd(x,y,z)\\[0.25cm] &= \rho\, \int_{z=0}^{z=16}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{r=0}^{r=\sqrt{16-z}}z\,r\,\dd r\right)\dd\varphi\right)\dd z\\[0.25cm]&\phantom{abcuva} \blue{\text{in cylindrical coordinates}}\\[0.25cm] &= \rho\, \int_{z=0}^{z=16}\left(\int_{\varphi=0}^{\varphi=2\pi}\biggl[\frac{1}{2}r^2\,z\biggl]_{r=0}^{r=\sqrt{16-z}}\;\dd\varphi\right)\dd z\\[0.25cm] &=\frac{1}{2}\rho\int_{z=0}^{z=16}\left(\int_{\varphi=0}^{\varphi=2\pi}(16-z)\,z\,\dd \varphi\right)\dd z\\[0.25cm] &=\frac{1}{2}\rho\int_{z=0}^{z=16}\biggl[(16-z)\,z\,\varphi\biggr]_{\varphi=0}^{\varphi=2\pi}\;\dd z \\[0.25cm] &= \pi\,\rho\int_{z=0}^{z=16}(16-z)\,z\,\dd z\\[0.25cm] &= \pi\,\rho\,\biggl[8z^2-\frac{1}{3}z^3\biggr]_{z=0}^{z=16}\\[0.25cm] &=\pi\,\rho\, \left(\frac{1}{2}\,16^2-\frac{1}{3}\,16^3-0\right)\\[0.25cm] &= {{2048}\over{3}}\,\pi\,\rho\end{aligned}$ The $z$ -coordinate of the centre of gravity can now be calculated as a quotient: $\begin{aligned} \bar{z}&=\frac{M_{xy}}{M}\\[0.25cm] &= \frac{{{2048}\over{3}}\,\pi\,\rho}{128\,\pi\,\rho}\\[0.25cm] &= {{16}\over{3}}\end{aligned}$

New example

The 3D moment of inertia $M_{\ell}$ of $R$ relative to the line $\ell$ is defined as follows: $I_{\ell} = \iiint \delta(x,y,z)^2\,\rho(x,y,z)\,\dd(x,y,z)$ where $\delta(x,y,z)^2=\text{distance from the point }(x,y,z)\text{ to the line }\ell\text.$ Three special cases are the moments of inertia relative to the coordinate axes: $\begin{aligned} I_x &=\iiint_R (y^2+z^2)\,\rho(x,y,z)\,\dd(x,y,z)\\[0.25cm] I_{y} &=\iiint_R (x^2+z^2) \,\rho(x,y,z)\,\dd(x,y,z)\\[0.25cm] I_{z} &=\iiint_R (x^2+y^2)\,\rho(x,y,z)\,\dd(x,y,z)\end{aligned}$ The radius of inertia $k_{\ell}$ relative to a line $\ell$ is defined in terms of the moment of inertia $I_\ell$ and the total mass of $M$ of $R$ as follows: $k_{\ell}=\sqrt{\frac{I_{\ell}}{M}}$

In practice, the following theorem is used to calculate the moment of inertia $I_{\ell}$ with respect to a line $\ell$ .

The moment of inertia $I_{\ell}$ with respect to a line $\ell$ for a solid can be calculated as $I_{\ell} = I_0+ Md^2$ where $M$ is the mass of the solid, $I_0$ is the moment of inertia with respect to the line $\ell_0$ through the centroid and parallel to $\ell$ , and $d$ is the distance between the two lines $\ell$ and $\ell_0$ .

For a cylinder with radius $a$ , height $h$ , constant density and total mass $M$ , the moment of inertia $I_0$ with respect to a line $\ell_0$ through the centroid and perpendicular the cylindrical axis is equal $I_0 =\frac{1}{12}M(3a^2+h^2)$ Check this via the definition of the moment of inertia. If you want to calculate the moment of inertia $I_\ell$ at the end of the cylinder with respect to a line $\ell_0$ parallel to $\ell$ , then Steiner's theorem states: $\begin{aligned} I_\ell &= I_0 +M\left(\frac{h}{2}\right)^2\\[0.25cm] &= \frac{1}{4}M(a^2+h^2)\end{aligned}$

Similar formulas hold for a two-dimensional bounded region by simply omitting $z$ . In Steiner's theorem, mass must be replaced by area.

If for a two-dimensional region $R$ the continuous density $\rho(x,y)$ is given in every point of $R$ , then for an area element with very small dimensions $\dd (x,y)$ its mass is approximately equal to $\rho(x,y)\,\dd(x, y)$ . $M=\iint_R \rho(x,y)\,\dd(x, y)$ therefore is true for the total mass $M$ of $R$

The 2D static moments $M_{y}$ and $M_{x}$ of $R$ , respectively, relative to the $x$ and $y$ axes, also called first moments of area, are defined as follows:
$\begin{aligned} M_{y} &=\iint_R x\,\rho(x,y)\,\dd(x,y)\\[0.25cm] M_{x} &=\iiint_R y\,\rho(x,y)\,\dd(x,y)\end{aligned}$

The 2D centre of mass, more commonly known as centre of gravity, $(\bar{x},\bar{y})$ of $R$ is given by: $\bar{x}=\frac{M_{y}}{M},\quad \bar{y}=\frac{M_{x}}{M}$

Determine the centroid of the region $R$ sketched below in the plane and emclosed by the curve $y=x^2$ and the line $y=3x$ , assuming that the density is a constant $\rho$ . The centre of gravity is indicated in the figure below with a red dot.

GeoGebra

$\cv{\bar{x} \\ \bar{y}} ={}$ $\cv{{{3}\over{2}}\\ {{18}\over{5}}}$

First we calculate the total mass $M$ : $\begin{aligned} M &= \iint_R \rho\,\dd(x,y)\\[0.25cm] &= \rho\, \int_{x=0}^{x=3}\left(\int_{y=x^2}^{y=3 x}\dd y\right)\dd x\\[0.25cm] &= \rho \int_{x=0}^{x=3}\biggl[y\biggr]_{y=x^2}^{y=3 x}\\[0.25cm] &=\rho \int_{x=0}^{x=3} \left(3 x-x^2\right)\,\dd x\\[0.25cm] &= \rho\cdot \biggl[{{3}\over{2}}x^2-\frac{1}{3}x^3\biggr]_{x=0}^{x=3}\\[0.25cm] &= {{9}\over{2}}\rho \end{aligned}$ Next we determine in a similar way the static moments $M_{y}$ and $M_{x}$ : $\begin{aligned} M_{y} &= \iint_R x\,\rho\,\dd(x,y)\\[0.25cm] &= \rho\int_{x=0}^{x=3}\left(\int_{y=x^2}^{y=3 x}x\, \dd y\right)\dd x\\[0.25cm] &= \rho \int_{x=0}^{x=3}\biggl[x\,y\biggr]_{y=x^2}^{y=3 x}\;\dd x\\[0.25cm] &=\rho \int_{x=0}^{x=3} \left(3 x^2-x^3\right)\,\dd x\\[0.25cm] &= \rho\cdot \biggl[x^3-\frac{1}{4}x^4\biggr]_{x=0}^{x=3}\\[0.25cm] &= {{27}\over{4}}\,\rho\\[1.0cm] M_{x} &= \iint_R y\,\rho\,\dd(x,y)\\[0.25cm] &= \rho\int_{x=0}^{x=3}\left(\int_{y=x^2}^{y=3 x}y\, \dd y\right)\dd x\\[0.25cm] &= \rho \int_{x=0}^{x=3}\biggl[\frac{1}{2}y^2\biggr]_{y=x^2}^{y=3 x}\\[0.25cm] &=\rho \int_{x=0}^{x=3} \left({{9}\over{2}} x^2-\frac{1}{2}x^4\right)\,\dd x\\[0.25cm] &= \rho\cdot \biggl[{{3}\over{2}}x^3-\frac{1}{10}x^5\biggr]_{x=0}^{x=3}\\[0.25cm] &= {{81}\over{5}}\,\rho \end{aligned}$ The coordinates of the centre of gravity can now be calculated as quotients: $\begin{aligned} \bar{x}&=\frac{M_{y}}{M}\\[0.25cm] &= \frac{{{27}\over{4}}\,\rho}{{{9}\over{2}}\,\rho}\\[0.25cm] &= {{3}\over{2}}\\[0.5cm] \bar{y}&=\frac{M_{x}}{M}\\[0.25cm] &= \frac{{{81}\over{5}}\,\rho}{{{9}\over{2}}\,\rho}\\[0.25cm] &= {{18}\over{5}}\end{aligned}$

New example

The 2D moment of inertia $M_{\ell}$ of $R$ relative to. the line $\ell$ , also called area moment of inertia or second moment of area, is defined as follows: $I_{\ell} = \iiint \delta(x,y)^2\,\rho(x,y)\,\dd(x,y,z)$ where $\delta(x,y)^2=\text{distance of the point }(x,y)\text{ to the line }\ell\text.$ Two special cases are the moments of inertia relative to the coordinate axes: $\begin{aligned} I_x &=\iint_R y^2\,\rho(x,y)\,\dd(x,y)\\[0.25cm] I_{y} &=\iint_R x^2\,\rho(x,y,z)\,\dd(x,y)\end{aligned}$ The moment of inertia $I_O$ about the origin, also called the polar moment of inertia, is defined as $\begin{aligned}I_O&=I_x+I_y\\[0.25cm] &=\iint \left((x^2+y^2\right)\,\rho(x,y)\,\dd(x,y)\end{aligned}$