Notes: `y=sqrt(x) , y =1/2x` Find the x and y moments of inertia and center of mass for the laminas of uniform density `p` bounded by the graphs of...

For an irregularly shaped planar lamina of uniform density $\displaystyle{\left(\rho\right)}$ , bounded by graphs $\displaystyle{y}={f{{\left({x}\right)}}},{y}={g{{\left({x}\right)}}}$ and $\displaystyle{a}\le{x}\le{b}$ , the mass $\displaystyle{\left({m}\right)}$ 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:

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

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

We are given: $\displaystyle{y}=\sqrt{{{x}}},{y}=\frac{{1}}{{2}}{x}$

Please refer to the attached image. Plot of $\displaystyle{y}=\sqrt{{{x}}}$ is red in...

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

$\displaystyle{m}=\rho{A}$ , 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:

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

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

We are given: $\displaystyle{y}=\sqrt{{{x}}},{y}=\frac{{1}}{{2}}{x}$

Please refer to the attached image. Plot of $\displaystyle{y}=\sqrt{{{x}}}$ is red in color and plot of $\displaystyle{y}=\frac{{1}}{{2}}{x}$ is blue in color. The curves intersect at $\displaystyle{\left({0},{0}\right)}$ and $\displaystyle{\left({4},{2}\right)}$ .

First let's find the area of the region,

$\displaystyle{A}={\int_{{0}}^{{4}}}{\left(\sqrt{{{x}}}-\frac{{1}}{{2}}{x}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={\int_{{0}}^{{4}}}{\left({{x}}^{{\frac{{1}}{{2}}}}-\frac{{1}}{{2}}{x}\right)}{\left.{d}{x}\right.}$

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

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

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

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

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

$\displaystyle{A}={\left[\frac{{16}}{{3}}-{4}\right]}$

$\displaystyle{A}=\frac{{4}}{{3}}$

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

$\displaystyle{M}_{{x}}=\rho{\int_{{0}}^{{4}}}\frac{{1}}{{2}}{\left({{\left(\sqrt{{{x}}}\right)}}^{{2}}-{{\left(\frac{{1}}{{2}}{x}\right)}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\rho{\int_{{0}}^{{4}}}\frac{{1}}{{2}}{\left({x}-\frac{{{x}}^{{2}}}{{4}}\right)}{\left.{d}{x}\right.}$

Take the constant out,

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

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{{\left[\frac{{{x}}^{{2}}}{{2}}-\frac{{1}}{{4}}{\left(\frac{{{x}}^{{3}}}{{3}}\right)}\right]}_{{0}}^{{4}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{{\left[\frac{{{x}}^{{2}}}{{2}}-\frac{{{x}}^{{3}}}{{12}}\right]}_{{0}}^{{4}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[\frac{{{4}}^{{2}}}{{2}}-\frac{{{4}}^{{3}}}{{12}}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[\frac{{16}}{{2}}-\frac{{64}}{{12}}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[{8}-\frac{{16}}{{3}}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left(\frac{{8}}{{3}}\right)}$

$\displaystyle{M}_{{x}}=\frac{{4}}{{3}}\rho$

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{4}}}{x}{\left(\sqrt{{{x}}}-\frac{{1}}{{2}}{x}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{4}}}{\left({x}\sqrt{{{x}}}-\frac{{1}}{{2}}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{4}}}{\left({{x}}^{{\frac{{3}}{{2}}}}-\frac{{1}}{{2}}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{{\left[\frac{{{x}}^{{\frac{{3}}{{2}}+{1}}}}{{\frac{{3}}{{2}}+{1}}}-\frac{{1}}{{2}}{\left(\frac{{{x}}^{{3}}}{{3}}\right)}\right]}_{{0}}^{{4}}}$

$\displaystyle{M}_{{y}}=\rho{{\left[\frac{{2}}{{5}}{{x}}^{{\frac{{5}}{{2}}}}-\frac{{{x}}^{{3}}}{{6}}\right]}_{{0}}^{{4}}}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{2}}{{5}}{{\left({4}\right)}}^{{\frac{{5}}{{2}}}}-\frac{{{4}}^{{3}}}{{6}}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{2}}{{5}}{{\left({{2}}^{{2}}\right)}}^{{\frac{{5}}{{2}}}}-\frac{{64}}{{6}}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{2}}{{5}}{{\left({2}\right)}}^{{5}}-\frac{{32}}{{3}}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{64}}{{5}}-\frac{{32}}{{3}}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{{192}-{160}}}{{15}}\right]}$

$\displaystyle{M}_{{y}}=\frac{{32}}{{15}}\rho$

The center of mass can be evaluated by plugging in the moments and area as below:

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

$\displaystyle{\overline{{x}}}=\frac{{\frac{{32}}{{15}}\rho}}{{\rho\frac{{4}}{{3}}}}$

$\displaystyle{\overline{{x}}}={\left(\frac{{32}}{{15}}\right)}{\left(\frac{{3}}{{4}}\right)}$

$\displaystyle{\overline{{x}}}=\frac{{8}}{{5}}$

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

$\displaystyle{\overline{{y}}}=\frac{{\frac{{4}}{{3}}\rho}}{{\rho\frac{{4}}{{3}}}}$

$\displaystyle{\overline{{y}}}={1}$

The coordinates of the center of mass are $\displaystyle{\left(\frac{{8}}{{5}},{1}\right)}$

Notes

Thursday, 4 December 2014

$\displaystyle{y}=\sqrt{{{x}}},{y}=\frac{{1}}{{2}}{x}$ 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...

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

Thursday, 4 December 2014

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

No comments:

Post a Comment

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

$\displaystyle{y}=\sqrt{{{x}}},{y}=\frac{{1}}{{2}}{x}$ 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...

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