5.6 Calculating centers of mass and moments of inertia  (Page 3/8)

Finding a centroid

Find the centroid of the region under the curve $y=e^x$ over the interval $1\le x\le 3$ (see the following figure).

To compute the centroid, we assume that the density function is constant and hence it cancels out:

$\begin{array}{}\\ \\ \\ \\ x_c=\frac{M_y}{m}=\frac{{\displaystyle \underset{R}{\iint }xdA}}{{\displaystyle \underset{R}{\iint }dA}}\text{and}{y}_c=\frac{M_x}{m}=\frac{{\displaystyle \underset{R}{\iint }ydA}}{{\displaystyle \underset{R}{\iint }dA}},\hfill \\ x_c=\frac{M_y}{m}=\frac{{\displaystyle \underset{R}{\iint }xdA}}{{\displaystyle \underset{R}{\iint }dA}}=\frac{{\displaystyle \underset{x=1}{\overset{x=3}{\int }}{\displaystyle \underset{y=0}{\overset{y=e^x}{\int }}xdydx}}}{{\displaystyle \underset{x=1}{\overset{x=3}{\int }}{\displaystyle \underset{y=0}{\overset{y=e^x}{\int }}dydx}}}=\frac{{\displaystyle \underset{x=1}{\overset{x=3}{\int }}x{e}^{x}dx}}{{\displaystyle \underset{x=1}{\overset{x=3}{\int }}{e}^{x}dx}}=\frac{2e^3}{e^3-e}=\frac{2e^2}{e^2-1},\hfill \\ y_c=\frac{M_x}{m}=\frac{{\displaystyle \underset{R}{\iint }ydA}}{{\displaystyle \underset{R}{\iint }dA}}=\frac{{\displaystyle \underset{x=1}{\overset{x=3}{\int }}{\displaystyle \underset{y=0}{\overset{y=e^x}{\int }}ydydx}}}{{\displaystyle \underset{x=1}{\overset{x=3}{\int }}{\displaystyle \underset{y=0}{\overset{y=e^x}{\int }}dydx}}}=\frac{{\displaystyle \underset{x=1}{\overset{x=3}{\int }}\frac{e^{2x}}{2}dx}}{{\displaystyle \underset{x=1}{\overset{x=3}{\int }}{e}^{x}dx}}=\frac{\frac{1}{4}{e}^2\left(e^4-1\right)}{e\left(e^2-1\right)}=\frac{1}{4}e\left(e^2+1\right).\hfill \end{array}$

Thus the centroid of the region is

$\left(x_c,y_c\right)=\left(\frac{2e^2}{e^2-1},\frac{1}{4}e\left(e^2+1\right)\right).$

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Moments of inertia

For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section $6.6.$ The moment of inertia of a particle of mass $m$ about an axis is $m{r}^2,$ where $r$ is the distance of the particle from the axis. We can see from [link] that the moment of inertia of the subrectangle $R_{ij}$ about the $x\text{-axis}$ is ${(y_{ij}^{*})}^2\rho (x_{ij}^{*},y_{ij}^{*})\text{Δ}A.$ Similarly, the moment of inertia of the subrectangle $R_{ij}$ about the $y\text{-axis}$ is ${(x_{ij}^{*})}^2\rho (x_{ij}^{*},y_{ij}^{*})\text{Δ}A.$ The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis.

The moment of inertia $I_x$ about the $x\text{-axis}$ for the region $R$ is the limit of the sum of moments of inertia of the regions $R_{ij}$ about the $x\text{-axis}\text{.}$ Hence

Similarly, the moment of inertia $I_y$ about the $y\text{-axis}$ for $R$ is the limit of the sum of moments of inertia of the regions $R_{ij}$ about the $y\text{-axis}\text{.}$ Hence

Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. We denote this by $I_0$ and obtain it by adding the moments of inertia $I_x$ and $I_y.$ Hence

All these expressions can be written in polar coordinates by substituting $x=r\text{cos}\theta ,$ $y=r\text{sin}\theta ,$ and $dA=rdrd\theta .$ For example, $I_0={\displaystyle \underset{R}{\iint }{r}^2\rho \left(r\text{cos}\theta ,r\text{sin}\theta \right)}dA.$

As mentioned earlier, the moment of inertia of a particle of mass $m$ about an axis is $m{r}^2$ where $r$ is the distance of the particle from the axis, also known as the radius of gyration    .

Hence the radii of gyration with respect to the $x\text{-axis,}$ the $y\text{-axis,}$ and the origin are