Notes: `y=x^2, y=x^3` 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, 17 September 2014

$\displaystyle{y}={{x}}^{{2}},{y}={{x}}^{{3}}$ 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...

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 $\displaystyle{y}={{x}}^{{2}},{y}={{x}}^{{3}}$

Refer the attached image. Plot in red color is of $\displaystyle{y}={{x}}^{{2}}$ and blue...

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

$\displaystyle{m}=\rho{A}$ , 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 $\displaystyle{y}={{x}}^{{2}},{y}={{x}}^{{3}}$

Refer the attached image. Plot in red color is of $\displaystyle{y}={{x}}^{{2}}$ and blue color is of $\displaystyle{y}={{x}}^{{3}}$

Curves intersect at (1,1)

Now let's evaluate the area of the region,

$\displaystyle{A}={\int_{{0}}^{{1}}}{\left({{x}}^{{2}}-{{x}}^{{3}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={{\left[\frac{{{x}}^{{3}}}{{3}}-\frac{{{x}}^{{4}}}{{4}}\right]}_{{0}}^{{1}}}$

$\displaystyle{A}={\left[\frac{{{1}}^{{3}}}{{3}}-\frac{{{1}}^{{4}}}{{4}}\right]}$

$\displaystyle{A}={\left(\frac{{1}}{{3}}-\frac{{1}}{{4}}\right)}=\frac{{{4}-{3}}}{{12}}$

$\displaystyle{A}=\frac{{1}}{{12}}$

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

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

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\int_{{0}}^{{1}}}{\left({{x}}^{{4}}-{{x}}^{{6}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{{\left[\frac{{{x}}^{{5}}}{{5}}-\frac{{{x}}^{{7}}}{{7}}\right]}_{{0}}^{{1}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[\frac{{{1}}^{{5}}}{{5}}-\frac{{{1}}^{{7}}}{{7}}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left(\frac{{1}}{{5}}-\frac{{1}}{{7}}\right)}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}\frac{{{7}-{5}}}{{{35}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{35}}$

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

$\displaystyle{M}_{{y}}=\rho{\int_{{0}}^{{1}}}{\left({{x}}^{{3}}-{{x}}^{{4}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\rho{{\left[\frac{{{x}}^{{4}}}{{4}}-\frac{{{x}}^{{5}}}{{5}}\right]}_{{0}}^{{1}}}$

$\displaystyle{M}_{{y}}=\rho{\left[\frac{{{1}}^{{4}}}{{4}}-\frac{{{1}}^{{5}}}{{5}}\right]}$

$\displaystyle{M}_{{y}}=\rho\frac{{{5}-{4}}}{{{20}}}$

$\displaystyle{M}_{{y}}=\frac{\rho}{{20}}$

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

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

$\displaystyle{\overline{{x}}}=\frac{{\frac{\rho}{{20}}}}{{\rho\frac{{1}}{{12}}}}$

$\displaystyle{\overline{{x}}}=\frac{{12}}{{20}}$

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

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

$\displaystyle{\overline{{y}}}=\frac{{\frac{\rho}{{35}}}}{{\rho\frac{{1}}{{12}}}}$

$\displaystyle{\overline{{y}}}=\frac{{12}}{{35}}$

The coordinates of the center of mass are $\displaystyle{\left(\frac{{3}}{{5}},\frac{{12}}{{35}}\right)}$

Notes

Wednesday, 17 September 2014

$\displaystyle{y}={{x}}^{{2}},{y}={{x}}^{{3}}$ 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, 17 September 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?

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