Notes: `y=sqrt(x) +1 , y=1/3x + 1` Find the x and y moments of inertia and center of mass for the laminas of uniform density `p` bounded by the...

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: $\displaystyle{y}=\sqrt{{{x}}}+{1},{y}=\frac{{1}}{{3}}{x}+{1}$

Refer to the attached image, plot of $\displaystyle{y}=\sqrt{{{x}}}+{1}$ is red in color and...

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

$\displaystyle{m}=\rho{A}$ , 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: $\displaystyle{y}=\sqrt{{{x}}}+{1},{y}=\frac{{1}}{{3}}{x}+{1}$

Refer to the attached image, plot of $\displaystyle{y}=\sqrt{{{x}}}+{1}$ is red in color and plot of $\displaystyle{y}=\frac{{1}}{{3}}{x}+{1}$ is blue in color.The curves intersect at $\displaystyle{\left({0},{1}\right)}$ and $\displaystyle{\left({9},{4}\right)}$ .

Now let's evaluate the area of the region bounded by the graphs of the given equations,

$\displaystyle{A}={\int_{{0}}^{{9}}}{\left({\left(\sqrt{{{x}}}+{1}\right)}-{\left(\frac{{1}}{{3}}{x}+{1}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={\int_{{0}}^{{9}}}{\left(\sqrt{{{x}}}+{1}-\frac{{1}}{{3}}{x}-{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={\int_{{0}}^{{9}}}{\left(\sqrt{{{x}}}-\frac{{1}}{{3}}{x}\right)}{\left.{d}{x}\right.}$

Evaluate using power rule,

$\displaystyle{A}={{\left[\frac{{{x}}^{{\frac{{1}}{{2}}+{1}}}}{{\frac{{1}}{{2}}+{1}}}-\frac{{1}}{{3}}{\left(\frac{{{x}}^{{2}}}{{2}}\right)}\right]}_{{0}}^{{9}}}$

$\displaystyle{A}={{\left[\frac{{2}}{{3}}{{x}}^{{\frac{{3}}{{2}}}}-\frac{{1}}{{6}}{{x}}^{{2}}\right]}_{{0}}^{{9}}}$

$\displaystyle{A}={\left[\frac{{2}}{{3}}{{\left({9}\right)}}^{{\frac{{3}}{{2}}}}-\frac{{1}}{{6}}{{\left({9}\right)}}^{{2}}\right]}$

$\displaystyle{A}={\left[\frac{{2}}{{3}}{{\left({{3}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}-\frac{{1}}{{6}}{\left({81}\right)}\right]}$

$\displaystyle{A}={\left[\frac{{2}}{{3}}{{\left({3}\right)}}^{{3}}-\frac{{27}}{{2}}\right]}$

$\displaystyle{A}={\left[{18}-\frac{{27}}{{2}}\right]}$

$\displaystyle{A}=\frac{{9}}{{2}}$

Now let's find the moments about the x- and y-axes using the formulas stated above,

$\displaystyle{M}_{{x}}=\rho{\int_{{0}}^{{9}}}\frac{{1}}{{2}}{\left({{\left[\sqrt{{{x}}}+{1}\right]}}^{{2}}-{{\left[\frac{{1}}{{3}}{x}+{1}\right]}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\rho{\int_{{0}}^{{9}}}\frac{{1}}{{2}}{\left({\left[{x}+{2}\sqrt{{{x}}}+{1}\right]}-{\left[{{\left(\frac{{1}}{{3}}{x}\right)}}^{{2}}+{2}{\left(\frac{{1}}{{3}}{x}\right)}{\left({1}\right)}+{{1}}^{{2}}\right]}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\rho{\int_{{0}}^{{9}}}\frac{{1}}{{2}}{\left({x}+{2}\sqrt{{{x}}}+{1}-\frac{{{x}}^{{2}}}{{9}}-\frac{{2}}{{3}}{x}-{1}\right)}{\left.{d}{x}\right.}$

Take the constant out,

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\int_{{0}}^{{9}}}{\left({x}-\frac{{2}}{{3}}{x}+{2}\sqrt{{{x}}}-\frac{{{x}}^{{2}}}{{9}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\int_{{0}}^{{9}}}{\left(\frac{{x}}{{3}}+{2}{{\left({x}\right)}}^{{\frac{{1}}{{2}}}}-\frac{{{x}}^{{2}}}{{9}}\right)}{\left.{d}{x}\right.}$

Apply the basic integration rules i.e sum and power rules,

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{{\left[\frac{{1}}{{3}}{\left(\frac{{{x}}^{{2}}}{{2}}\right)}+{2}\frac{{{x}}^{{\frac{{1}}{{2}}+{1}}}}{{\frac{{1}}{{2}}+{1}}}-\frac{{1}}{{9}}{\left(\frac{{{x}}^{{3}}}{{3}}\right)}\right]}_{{0}}^{{9}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{{\left[\frac{{{x}}^{{2}}}{{6}}+{2}{\left(\frac{{2}}{{3}}\right)}{{x}}^{{\frac{{3}}{{2}}}}-\frac{{{x}}^{{3}}}{{27}}\right]}_{{0}}^{{9}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{{\left[\frac{{{x}}^{{2}}}{{6}}+\frac{{4}}{{3}}{{x}}^{{\frac{{3}}{{2}}}}-\frac{{{x}}^{{3}}}{{27}}\right]}_{{0}}^{{9}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[\frac{{{9}}^{{2}}}{{6}}+\frac{{4}}{{3}}{{\left({9}\right)}}^{{\frac{{3}}{{2}}}}-\frac{{{9}}^{{3}}}{{27}}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[\frac{{81}}{{6}}+\frac{{4}}{{3}}{{\left({{3}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}-\frac{{{9}}^{{3}}}{{{9}\cdot{3}}}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[\frac{{27}}{{2}}+\frac{{4}}{{3}}{\left({{3}}^{{3}}\right)}-{27}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[\frac{{27}}{{2}}+{36}-{27}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[\frac{{27}}{{2}}+{9}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left(\frac{{{27}+{18}}}{{2}}\right)}$

$\displaystyle{M}_{{x}}=\frac{{45}}{{4}}\rho$

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{9}}}{x}{\left({\left(\sqrt{{{x}}}+{1}\right)}-{\left(\frac{{1}}{{3}}{x}+{1}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{9}}}{x}{\left(\sqrt{{{x}}}+{1}-\frac{{1}}{{3}}{x}-{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{9}}}{x}{\left({{x}}^{{\frac{{1}}{{2}}}}-\frac{{1}}{{3}}{x}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{9}}}{\left({{x}}^{{\frac{{3}}{{2}}}}-\frac{{1}}{{3}}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{{\left[\frac{{{x}}^{{\frac{{3}}{{2}}+{1}}}}{{\frac{{3}}{{2}}+{1}}}-\frac{{1}}{{3}}{\left(\frac{{{x}}^{{3}}}{{3}}\right)}\right]}_{{0}}^{{9}}}$

$\displaystyle{M}_{{y}}=\rho{{\left[\frac{{2}}{{5}}{{x}}^{{\frac{{5}}{{2}}}}-\frac{{1}}{{9}}{{x}}^{{3}}\right]}_{{0}}^{{9}}}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{2}}{{5}}{{\left({9}\right)}}^{{\frac{{5}}{{2}}}}-\frac{{1}}{{9}}{{\left({9}\right)}}^{{3}}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{2}}{{5}}{{\left({{3}}^{{2}}\right)}}^{{\frac{{5}}{{2}}}}-{81}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{2}}{{5}}{{\left({3}\right)}}^{{5}}-{81}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{2}}{{5}}{\left({243}\right)}-{81}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{486}}{{5}}-{81}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{{486}-{405}}}{{5}}\right]}$

$\displaystyle{M}_{{y}}=\frac{{81}}{{5}}\rho$

Now let's evaluate the coordinates of the center of mass by plugging in the values of the moments and area,

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

$\displaystyle{\overline{{x}}}=\frac{{\frac{{81}}{{5}}\rho}}{{\rho\frac{{9}}{{2}}}}$

$\displaystyle{\overline{{x}}}={\left(\frac{{81}}{{5}}\right)}{\left(\frac{{2}}{{9}}\right)}$

$\displaystyle{\overline{{x}}}=\frac{{18}}{{5}}$

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

$\displaystyle{\overline{{y}}}=\frac{{\frac{{45}}{{4}}\rho}}{{\rho\frac{{9}}{{2}}}}$

$\displaystyle{\overline{{y}}}={\left(\frac{{45}}{{4}}\right)}{\left(\frac{{2}}{{9}}\right)}$

$\displaystyle{\overline{{y}}}=\frac{{5}}{{2}}$

The center of mass is $\displaystyle{\left(\frac{{18}}{{5}},\frac{{5}}{{2}}\right)}$

Notes

Sunday, 22 November 2015

$\displaystyle{y}=\sqrt{{{x}}}+{1},{y}=\frac{{1}}{{3}}{x}+{1}$ Find the x and y moments of inertia and center of mass for the laminas of uniform density $\displaystyle{p}$ bounded by the...

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

Sunday, 22 November 2015

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

No comments:

Post a Comment

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

$\displaystyle{y}=\sqrt{{{x}}}+{1},{y}=\frac{{1}}{{3}}{x}+{1}$ Find the x and y moments of inertia and center of mass for the laminas of uniform density $\displaystyle{p}$ bounded by the...

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