Notes: `x=-y , x=2y-y^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 the...

For an irregularly shaped planar lamina of uniform density $\displaystyle{\left(\rho\right)}$ bounded by graphs $\displaystyle{x}={f{{\left({y}\right)}}},{x}={g{{\left({y}\right)}}}$ and $\displaystyle{c}\le{y}\le{d}$ , the mass $\displaystyle{\left({m}\right)}$ of this region is given by,

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

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

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

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

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

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: $\displaystyle{x}=-{y},{x}={2}{y}-{{y}}^{{2}}$

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

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

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

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

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

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

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: $\displaystyle{x}=-{y},{x}={2}{y}-{{y}}^{{2}}$

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

Let's first evaluate the area of the bounded region,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle{M}_{{x}}=\rho{\left[{27}-\frac{{81}}{{4}}\right]}$

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

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

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

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

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

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

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

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

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

$\displaystyle{M}_{{y}}=\frac{\rho}{{2}}{\left[\frac{{243}}{{5}}-{81}+{27}\right]}$

$\displaystyle{M}_{{y}}=\frac{\rho}{{2}}{\left[\frac{{243}}{{5}}-{54}\right]}$

$\displaystyle{M}_{{y}}=\frac{\rho}{{2}}{\left[\frac{{{243}-{270}}}{{5}}\right]}$

$\displaystyle{M}_{{y}}=-\frac{{27}}{{10}}\rho$

Now let's find the center of the mass by plugging the moments and and the area evaluated above,

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

$\displaystyle{\overline{{x}}}=\frac{{-\frac{{27}}{{10}}\rho}}{{\rho\frac{{9}}{{2}}}}$

$\displaystyle{\overline{{x}}}={\left(-\frac{{27}}{{10}}\right)}{\left(\frac{{2}}{{9}}\right)}$

$\displaystyle{\overline{{x}}}=\frac{{-{3}}}{{5}}$

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

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

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

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

The coordinates of the center of mass are $\displaystyle{\left(\frac{{-{3}}}{{5}},\frac{{3}}{{2}}\right)}$

Notes

Saturday, 23 April 2016

$\displaystyle{x}=-{y},{x}={2}{y}-{{y}}^{{2}}$ 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 the...

No comments:

Post a Comment

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

Saturday, 23 April 2016

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?

$\displaystyle{x}=-{y},{x}={2}{y}-{{y}}^{{2}}$ 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 the...

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