Notes: `y=-x^2+4x+2 , y=x+2` 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, $\displaystyle{y}=-{{x}}^{{2}}+{4}{x}+{2},{y}={x}+{2}$

Refer to the attached image. The plot of $\displaystyle{y}=-{{x}}^{{2}}+{4}{x}+{2}$ is in red color...

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

$\displaystyle{m}=\rho{A}$ , 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, $\displaystyle{y}=-{{x}}^{{2}}+{4}{x}+{2},{y}={x}+{2}$

Refer to the attached image. The plot of $\displaystyle{y}=-{{x}}^{{2}}+{4}{x}+{2}$ is in red color and the plot of $\displaystyle{y}={x}+{2}$ is in blue color. The curves intersect at $\displaystyle{\left({0},{2}\right)}$ and $\displaystyle{\left({3},{5}\right)}$ .

Now let's evaluate the area (A) of the region,

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

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

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

Using basic integration properties:

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

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

$\displaystyle{A}={\left[-{9}+\frac{{27}}{{2}}\right]}$

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

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

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

$M_x=1/2rhoint_0^3{[(-x^2+4x+2)+(x+2)][(-x^2+4x+2)-(x+2)]}dx$

$M_x=1/2rhoint_0^3{[-x^2+5x+4][-x^2+3x]}dx$

$\displaystyle{M}_{{x}}=\frac{{1}}{{2}}\rho{\int_{{0}}^{{3}}}{\left({{x}}^{{4}}-{3}{{x}}^{{3}}-{5}{{x}}^{{3}}+{15}{{x}}^{{2}}-{4}{{x}}^{{2}}+{12}{x}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\frac{{1}}{{2}}\rho{\int_{{0}}^{{3}}}{\left({{x}}^{{4}}-{8}{{x}}^{{3}}+{11}{{x}}^{{2}}+{12}{x}\right)}{\left.{d}{x}\right.}$

Evaluate using the basic integration rules:

$\displaystyle{M}_{{x}}=\frac{{1}}{{2}}\rho{{\left[\frac{{{x}}^{{5}}}{{5}}-{8}{\left(\frac{{{x}}^{{4}}}{{4}}\right)}+{11}{\left(\frac{{{x}}^{{3}}}{{3}}\right)}+{12}{\left(\frac{{{x}}^{{2}}}{{2}}\right)}\right]}_{{0}}^{{3}}}$

$\displaystyle{M}_{{x}}=\frac{{1}}{{2}}\rho{{\left[\frac{{{x}}^{{5}}}{{5}}-{2}{{x}}^{{4}}+\frac{{11}}{{3}}{{x}}^{{3}}+{6}{{x}}^{{2}}\right]}_{{0}}^{{3}}}$

$\displaystyle{M}_{{x}}=\frac{{1}}{{2}}\rho{\left[\frac{{{\left({3}\right)}}^{{5}}}{{5}}-{2}{{\left({3}\right)}}^{{4}}+\frac{{11}}{{3}}{{\left({3}\right)}}^{{3}}+{6}{{\left({3}\right)}}^{{2}}\right]}$

$\displaystyle{M}_{{x}}=\frac{{1}}{{2}}\rho{\left[\frac{{243}}{{5}}-{162}+{99}+{54}\right]}$

$\displaystyle{M}_{{x}}=\frac{{1}}{{2}}\rho{\left[\frac{{243}}{{5}}-{9}\right]}$

$\displaystyle{M}_{{x}}=\frac{{1}}{{2}}\rho{\left[\frac{{{243}-{45}}}{{5}}\right]}$

$\displaystyle{M}_{{x}}=\frac{{1}}{{2}}\rho{\left(\frac{{198}}{{5}}\right)}$

$\displaystyle{M}_{{x}}=\frac{{99}}{{5}}\rho$

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

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

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

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

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

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

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

$\displaystyle{M}_{{y}}=\rho{\left[-\frac{{1}}{{4}}{\left({81}\right)}+{27}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{-{81}+{108}}}{{4}}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left(\frac{{27}}{{4}}\right)}$

$\displaystyle{M}_{{y}}=\frac{{27}}{{4}}\rho$

Now evaluate the center of mass by plugging in the values of moments and area as below:

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

$\displaystyle{\overline{{x}}}=\frac{{\frac{{27}}{{4}}\rho}}{{\rho\frac{{9}}{{2}}}}$

$\displaystyle{\overline{{x}}}={\left(\frac{{27}}{{4}}\right)}{\left(\frac{{2}}{{9}}\right)}$

$\displaystyle{\overline{{x}}}=\frac{{3}}{{2}}$

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

$\displaystyle{\overline{{y}}}=\frac{{\frac{{99}}{{5}}\rho}}{{\rho\frac{{9}}{{2}}}}$

$\displaystyle{\overline{{y}}}={\left(\frac{{99}}{{5}}\right)}{\left(\frac{{2}}{{9}}\right)}$

$\displaystyle{\overline{{y}}}=\frac{{22}}{{5}}$

The center of mass $\displaystyle{\left({\overline{{x}}},{\overline{{y}}}\right)}$ are $\displaystyle{\left(\frac{{3}}{{2}},\frac{{22}}{{5}}\right)}$

Notes

Tuesday, 11 October 2016

$\displaystyle{y}=-{{x}}^{{2}}+{4}{x}+{2},{y}={x}+{2}$ 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?

Tuesday, 11 October 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?

$\displaystyle{y}=-{{x}}^{{2}}+{4}{x}+{2},{y}={x}+{2}$ 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?