Notes: `x=4-y^2 , x=0` 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...

Wednesday, 30 April 2014

$\displaystyle{x}={4}-{{y}}^{{2}},{x}={0}$ 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...

Consider an irregularly shaped planar lamina of uniform density $\displaystyle\rho$ , bounded by graphs $\displaystyle{x}={f{{\left({y}\right)}}}$ , $\displaystyle{x}={g{{\left({y}\right)}}}$ and $\displaystyle{c}\le{y}\le{d}$ . The mass $\displaystyle{m}$ 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 $\displaystyle{x}={4}-{{y}}^{{2}}$ , $\displaystyle{x}={0}$

Refer to the attached image. The...

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

$\displaystyle{m}=\rho{A}$ , 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 $\displaystyle{x}={4}-{{y}}^{{2}}$ , $\displaystyle{x}={0}$

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

Let's evaluate the area of the region,

$\displaystyle{A}={\int_{{-{2}}}^{{2}}}{\left({4}-{{y}}^{{2}}\right)}{\left.{d}{y}\right.}$

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

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

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

$\displaystyle{A}={2}{\left[{8}-\frac{{8}}{{3}}\right]}$

$\displaystyle{A}=\frac{{32}}{{3}}$

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

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

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{2}}}{\left({16}-{8}{{y}}^{{2}}+{{y}}^{{4}}\right)}{\left.{d}{y}\right.}$

$\displaystyle{M}_{{y}}=\rho{{\left[{16}{y}-{8}{\left(\frac{{{y}}^{{3}}}{{3}}\right)}+\frac{{{y}}^{{5}}}{{5}}\right]}_{{0}}^{{2}}}$

$\displaystyle{M}_{{y}}=\rho{\left[{16}\cdot{2}-\frac{{8}}{{3}}{\left({{2}}^{{3}}\right)}+\frac{{1}}{{5}}{\left({{2}}^{{5}}\right)}\right]}$

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

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{{480}-{320}+{96}}}{{15}}\right]}$

$\displaystyle{M}_{{y}}=\frac{{256}}{{15}}\rho$

By symmetry, $\displaystyle{M}_{{x}}={0},{\overline{{y}}}={0}$

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

$\displaystyle{\overline{{x}}}=\frac{{\frac{{256}}{{15}}\rho}}{{\rho\frac{{32}}{{3}}}}$

$\displaystyle{\overline{{x}}}={\left(\frac{{256}}{{15}}\right)}{\left(\frac{{3}}{{32}}\right)}$

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

The coordinates of the center of mass are $\displaystyle{\left({\overline{{x}}},{\overline{{y}}}\right)}$ are $\displaystyle{\left(\frac{{8}}{{5}},{0}\right)}$

Notes

Wednesday, 30 April 2014

$\displaystyle{x}={4}-{{y}}^{{2}},{x}={0}$ 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...

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

Wednesday, 30 April 2014

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?

$\displaystyle{x}={4}-{{y}}^{{2}},{x}={0}$ 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?