Notes: `y=x^(2/3), y=0, x=8` 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 $\displaystyle\rho$ , bounded by graphs $\displaystyle{y}={f{{\left({x}\right)}}}$ , $\displaystyle{y}={g{{\left({x}\right)}}}$ and $\displaystyle{a}\le{x}\le{b}$ , the mass (m) of this region is given by:

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

where A is the area of the region.

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

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

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

The center of mass $\displaystyle{\left({\overline{{x}}},{\overline{{y}}}\right)}$ is given by $\displaystyle{\overline{{x}}}=\frac{{{M}_{{y}}}}{{m}}$ and $\displaystyle{\overline{{y}}}=\frac{{M}_{{x}}}{{m}}$ ,

$\displaystyle{\overline{{x}}}=\frac{{1}}{{A}}{\int_{{a}}^{{b}}}{x}{\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$

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

Now we given $\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}},{y}={0},{x}={8}$

The plot of the functions is attached...

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

where A is the area of the region.

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

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

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

$\displaystyle{\overline{{x}}}=\frac{{1}}{{A}}{\int_{{a}}^{{b}}}{x}{\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$

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

Now we given $\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}},{y}={0},{x}={8}$

The plot of the functions is attached as image and the bounds of the limits can be found from the same.

Area of the region A = $\displaystyle{\int_{{0}}^{{8}}}{{x}}^{{\frac{{2}}{{3}}}}{\left.{d}{x}\right.}$

Use the power rule,

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

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

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

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

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

$\displaystyle{A}={\left(\frac{{3}}{{5}}{\left({32}\right)}\right)}$

$\displaystyle{A}=\frac{{96}}{{5}}$

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

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

$\displaystyle=\rho{\int_{{0}}^{{8}}}{\left[\frac{{1}}{{2}}{{\left({{x}}^{{\frac{{2}}{{3}}}}\right)}}^{{2}}\right]}{\left.{d}{x}\right.}$

$\displaystyle=\rho{\int_{{0}}^{{8}}}\frac{{1}}{{2}}{{x}}^{{\frac{{4}}{{3}}}}{\left.{d}{x}\right.}$

Take the constant out and apply the power rule,

$\displaystyle=\frac{\rho}{{2}}{\int_{{0}}^{{8}}}{{x}}^{{\frac{{4}}{{3}}}}{\left.{d}{x}\right.}$

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

$\displaystyle=\frac{\rho}{{2}}{{\left[\frac{{3}}{{7}}{{x}}^{{\frac{{7}}{{3}}}}\right]}_{{0}}^{{8}}}$

$\displaystyle=\frac{\rho}{{2}}{\left[\frac{{3}}{{7}}{{\left({8}\right)}}^{{\frac{{7}}{{3}}}}\right]}$

$\displaystyle=\frac{\rho}{{2}}{\left[\frac{{3}}{{7}}{{\left({{2}}^{{3}}\right)}}^{{\frac{{7}}{{3}}}}\right]}$

$\displaystyle=\frac{\rho}{{2}}{\left[\frac{{3}}{{7}}{{\left({2}\right)}}^{{7}}\right]}$

$\displaystyle=\rho{\left(\frac{{3}}{{7}}\right)}{{\left({2}\right)}}^{{6}}$

$\displaystyle=\rho{\left(\frac{{3}}{{7}}\right)}{\left({64}\right)}$

$\displaystyle=\frac{{192}}{{7}}\rho$

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

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

$\displaystyle=\rho{\int_{{0}}^{{8}}}{{x}}^{{\frac{{5}}{{3}}}}{\left.{d}{x}\right.}$

$\displaystyle=\rho{{\left[\frac{{{x}}^{{\frac{{5}}{{3}}+{1}}}}{{\frac{{5}}{{3}}+{1}}}\right]}_{{0}}^{{8}}}$

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

$\displaystyle=\rho{\left[\frac{{3}}{{8}}{{\left({8}\right)}}^{{\frac{{8}}{{3}}}}\right]}$

$\displaystyle=\rho{\left[\frac{{3}}{{8}}{{\left({{2}}^{{3}}\right)}}^{{\frac{{8}}{{3}}}}\right]}$

$\displaystyle=\rho{\left[\frac{{3}}{{8}}{\left({{2}}^{{8}}\right)}\right]}$

$\displaystyle=\rho{\left(\frac{{3}}{{8}}\right)}{\left({256}\right)}$

$\displaystyle={96}\rho$

Now let's find the center of mass,

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

Plug in the value of $\displaystyle{M}_{{y}}$ and $\displaystyle{A}$ ,

$\displaystyle{\overline{{x}}}=\frac{{{96}\rho}}{{\rho{\left(\frac{{96}}{{5}}\right)}}}$

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

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

Plug in the values of $\displaystyle{M}_{{x}}$ and $\displaystyle{A}$ ,

$\displaystyle{\overline{{y}}}=\frac{{\frac{{192}}{{7}}\rho}}{{\rho{\left(\frac{{96}}{{5}}\right)}}}$

$\displaystyle{\overline{{y}}}={\left(\frac{{192}}{{7}}\right)}{\left(\frac{{5}}{{96}}\right)}$

$\displaystyle{\overline{{y}}}=\frac{{10}}{{7}}$

The coordinates of the center of mass are, $\displaystyle{\left({5},\frac{{10}}{{7}}\right)}$

Notes

Wednesday, 24 August 2016

$\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}},{y}={0},{x}={8}$ 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?

Wednesday, 24 August 2016

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?

$\displaystyle{y}={{x}}^{{\frac{{2}}{{3}}}},{y}={0},{x}={8}$ 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?