Notes: `y=sqrt(x), y=0 , x=4` Find the x and y moments of inertia and center of mass for the laminas of uniform density `p` bounded by the graphs of...

Friday, 12 August 2016

$\displaystyle{y}=\sqrt{{{x}}},{y}={0},{x}={4}$ 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...

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,

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

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

The center of mass $\displaystyle{\left({\overline{{x}}},{\overline{{y}}}\right)}$ is given by:

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

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

Now we are given $\displaystyle{y}=\sqrt{{{x}}},{y}={0},{x}={4}$

The attached image shows the region bounded by the functions,

Now let's find...

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

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

The moments about the x- and y-axes are,

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

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

The center of mass $\displaystyle{\left({\overline{{x}}},{\overline{{y}}}\right)}$ is given by:

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

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

Now we are given $\displaystyle{y}=\sqrt{{{x}}},{y}={0},{x}={4}$

The attached image shows the region bounded by the functions,

Now let's find the area of the region,

$\displaystyle{A}={\int_{{0}}^{{4}}}\sqrt{{{x}}}{\left.{d}{x}\right.}$

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

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

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

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

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

$\displaystyle{A}=\frac{{16}}{{3}}$

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

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

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

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{{\left[\frac{{{x}}^{{2}}}{{2}}\right]}_{{0}}^{{4}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[\frac{{{4}}^{{2}}}{{2}}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left(\frac{{16}}{{2}}\right)}$

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

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{4}}}{x}\sqrt{{{x}}}{\left.{d}{x}\right.}$

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

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

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

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

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

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

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{2}}{{5}}{\left({32}\right)}\right]}$

$\displaystyle{M}_{{y}}=\frac{{64}}{{5}}\rho$

The coordinates of the center of the mass are given by,

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

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

$\displaystyle{\overline{{x}}}=\frac{{\frac{{64}}{{5}}\rho}}{{\rho\frac{{16}}{{3}}}}$

$\displaystyle{\overline{{x}}}=\frac{{64}}{{5}}{\left(\frac{{3}}{{16}}\right)}$

$\displaystyle{\overline{{x}}}=\frac{{12}}{{5}}$

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

$\displaystyle{\overline{{y}}}=\frac{{{4}\rho}}{{\rho\frac{{16}}{{3}}}}$

$\displaystyle{\overline{{y}}}={4}{\left(\frac{{3}}{{16}}\right)}$

$\displaystyle{\overline{{y}}}=\frac{{3}}{{4}}$

The coordinates of the center of mass are $\displaystyle{\left(\frac{{12}}{{5}},\frac{{3}}{{4}}\right)}$

Notes

Friday, 12 August 2016

$\displaystyle{y}=\sqrt{{{x}}},{y}={0},{x}={4}$ 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?

Friday, 12 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?

$\displaystyle{y}=\sqrt{{{x}}},{y}={0},{x}={4}$ 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?