Notes: `y=x^(2/3), y=4` Find the x and y moments of inertia and center of mass for the laminas of uniform density `p` bounded by the graphs of the...

Monday, 1 September 2014

$\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}},{y}={4}$ Find the x and y moments of inertia and center of mass for the laminas of uniform density $\displaystyle{p}$ bounded by the graphs of the...

Consider an irregularly shaped planar lamina of uniform density $\displaystyle\rho$ , bounded by graphs $\displaystyle{y}={f{{\left({x}\right)}}},{y}={g{{\left({x}\right)}}}$ , and $\displaystyle{a}<{x}<{b}$ .The mass $\displaystyle{m}$ of this region is given by:

$\displaystyle{m}=\rho{\int_{{a}}^{{b}}}{\left[{f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right]}{\left.{d}{x}\right.}$

$\displaystyle{m}=\rho{A}$ where A is the area of the region.

The moments about the x- and y-axes are given by:

$\displaystyle{M}_{{x}}=\rho{\int_{{a}}^{{b}}}\frac{{1}}{{2}}{\left({{\left[{f{{\left({x}\right)}}}\right]}}^{{2}}-{{\left[{g{{\left({x}\right)}}}\right]}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{\int_{{a}}^{{b}}}{x}{\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$

The center of mass $\displaystyle{\left({\overline{{x}}},{\overline{{y}}}\right)}$ is given by ;

$\displaystyle{\overline{{x}}}=\frac{{M}_{{y}}}{{m}}$

$\displaystyle{\overline{{y}}}=\frac{{M}_{{x}}}{{m}}$

Given: $\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}},{y}={4}$

Refer to attached image. The plot of $\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}}$ is red in color.

The graphs...

$\displaystyle{m}=\rho{\int_{{a}}^{{b}}}{\left[{f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right]}{\left.{d}{x}\right.}$

$\displaystyle{m}=\rho{A}$ where A is the area of the region.

The moments about the x- and y-axes are given by:

$\displaystyle{M}_{{x}}=\rho{\int_{{a}}^{{b}}}\frac{{1}}{{2}}{\left({{\left[{f{{\left({x}\right)}}}\right]}}^{{2}}-{{\left[{g{{\left({x}\right)}}}\right]}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{\int_{{a}}^{{b}}}{x}{\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$

The center of mass $\displaystyle{\left({\overline{{x}}},{\overline{{y}}}\right)}$ is given by ;

$\displaystyle{\overline{{x}}}=\frac{{M}_{{y}}}{{m}}$

$\displaystyle{\overline{{y}}}=\frac{{M}_{{x}}}{{m}}$

Given: $\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}},{y}={4}$

Refer to attached image. The plot of $\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}}$ is red in color.

The graphs intersect at the coordinates $\displaystyle{\left(-{8},{4}\right)},{\left({8},{4}\right)}$

Let's evaluate the area of the bounded region,

$\displaystyle{A}={\int_{{-{8}}}^{{8}}}{\left({4}-{{x}}^{{\frac{{2}}{{3}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={2}{\int_{{0}}^{{8}}}{\left({4}-{{x}}^{{\frac{{2}}{{3}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={2}{{\left[{4}{x}-\frac{{{x}}^{{\frac{{2}}{{3}}+{1}}}}{{\frac{{2}}{{3}}+{1}}}\right]}_{{0}}^{{8}}}$

$\displaystyle{A}={2}{{\left[{4}{x}-\frac{{3}}{{5}}{{x}}^{{\frac{{5}}{{3}}}}\right]}_{{0}}^{{8}}}$

$\displaystyle{A}={2}{\left[{4}\cdot{8}-\frac{{3}}{{5}}{{\left({8}\right)}}^{{\frac{{5}}{{3}}}}\right]}$

$\displaystyle{A}={2}{\left[{32}-\frac{{3}}{{5}}{{\left({{2}}^{{3}}\right)}}^{{\frac{{5}}{{3}}}}\right]}$

$\displaystyle{A}={2}{\left[{32}-\frac{{3}}{{5}}{{\left({2}\right)}}^{{5}}\right]}$

$\displaystyle{A}={2}{\left[{32}-\frac{{96}}{{5}}\right]}$

$\displaystyle{A}={2}{\left[\frac{{{160}-{96}}}{{5}}\right]}$

$\displaystyle{A}=\frac{{{2}{\left({64}\right)}}}{{5}}$

$\displaystyle{A}=\frac{{128}}{{5}}$

Now let's evaluate the moments about the x- and y-axes,

Since the graph is symmetrical about the y-axis,

So, $\displaystyle{M}_{{y}}={0}$ and $\displaystyle{\overline{{x}}}={0}$ ,

$\displaystyle{M}_{{x}}=\rho{\int_{{-{8}}}^{{8}}}\frac{{1}}{{2}}{\left({\left[{{4}}^{{2}}\right]}-{{\left[{{x}}^{{\frac{{2}}{{3}}}}\right]}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\int_{{-{8}}}^{{8}}}{\left({16}-{{x}}^{{\frac{{4}}{{3}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left({2}\right)}{\int_{{0}}^{{8}}}{\left({16}-{{x}}^{{\frac{{4}}{{3}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\rho{{\left[{16}{x}-{\left(\frac{{{x}}^{{\frac{{4}}{{3}}+{1}}}}{{\frac{{4}}{{3}}+{1}}}\right)}\right]}_{{0}}^{{8}}}$

$\displaystyle{M}_{{x}}=\rho{{\left[{16}{x}-\frac{{3}}{{7}}{{x}}^{{\frac{{7}}{{3}}}}\right]}_{{0}}^{{8}}}$

$\displaystyle{M}_{{x}}=\rho{\left[{16}\cdot{8}-\frac{{3}}{{7}}{{\left({8}\right)}}^{{\frac{{7}}{{3}}}}\right]}$

$\displaystyle{M}_{{x}}=\rho{\left[{128}-\frac{{3}}{{7}}{\left({{2}}^{{7}}\right)}\right]}$

$\displaystyle{M}_{{x}}=\rho{\left[{128}-\frac{{3}}{{7}}{\left({128}\right)}\right]}$

$\displaystyle{M}_{{x}}=\frac{{512}}{{7}}\rho$

$\displaystyle{\overline{{y}}}=\frac{{M}_{{x}}}{{m}}=\frac{{M}_{{x}}}{{\rho{A}}}$

$\displaystyle{\overline{{y}}}=\frac{{\frac{{512}}{{7}}\rho}}{{\rho\frac{{128}}{{5}}}}$

$\displaystyle{\overline{{y}}}={\left(\frac{{512}}{{7}}\right)}{\left(\frac{{5}}{{128}}\right)}$

$\displaystyle{\overline{{y}}}=\frac{{20}}{{7}}$

The coordinates of the center of mass are $\displaystyle{\left({0},\frac{{20}}{{7}}\right)}$

Notes

Monday, 1 September 2014

$\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}},{y}={4}$ Find the x and y moments of inertia and center of mass for the laminas of uniform density $\displaystyle{p}$ bounded by the graphs of the...

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Monday, 1 September 2014

Find the x and y moments of inertia and center of mass for the laminas of uniform density bounded by the graphs of the...

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}},{y}={4}$ Find the x and y moments of inertia and center of mass for the laminas of uniform density $\displaystyle{p}$ bounded by the graphs of the...

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?