Notes: `y = 6-x , y=0, 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...

Tuesday, 24 May 2016

$\displaystyle{y}={6}-{x},{y}={0},{x}={0}$ 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...

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: $\displaystyle{y}={6}-{x},{y}={0},{x}={0}$

Refer to the attached image for the bounded region.

Let's first evaluate...

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

$\displaystyle{m}=\rho{A}$ , 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: $\displaystyle{y}={6}-{x},{y}={0},{x}={0}$

Refer to the attached image for the bounded region.

Let's first evaluate the area of the region,

$\displaystyle{A}={\int_{{0}}^{{6}}}{\left({6}-{x}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={{\left[{6}{x}-\frac{{{x}}^{{2}}}{{2}}\right]}_{{0}}^{{6}}}$

$\displaystyle{A}={\left[{6}{\left({6}\right)}-\frac{{{6}}^{{2}}}{{2}}\right]}$

$\displaystyle{A}={\left[{36}-\frac{{36}}{{2}}\right]}$

$\displaystyle{A}={18}$

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

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

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

Take the constant out and simplify,

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\int_{{0}}^{{6}}}{\left({36}-{12}{x}+{{x}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{{\left[{36}{x}-{12}\frac{{{x}}^{{2}}}{{2}}+\frac{{{x}}^{{3}}}{{3}}\right]}_{{0}}^{{6}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{{\left[{36}{x}-{6}{{x}}^{{2}}+\frac{{{x}}^{{3}}}{{3}}\right]}_{{0}}^{{6}}}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[{36}{\left({6}\right)}-{6}{{\left({6}\right)}}^{{2}}+\frac{{{6}}^{{3}}}{{3}}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left[{216}-{216}+\frac{{216}}{{3}}\right]}$

$\displaystyle{M}_{{x}}=\frac{\rho}{{2}}{\left({72}\right)}$

$\displaystyle{M}_{{x}}={36}\rho$

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

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

$\displaystyle{M}_{{y}}=\rho{{\left[{6}\frac{{{x}}^{{2}}}{{2}}-\frac{{{x}}^{{3}}}{{3}}\right]}_{{0}}^{{6}}}$

$\displaystyle{M}_{{y}}=\rho{{\left[{3}{{x}}^{{2}}-\frac{{{x}}^{{3}}}{{3}}\right]}_{{0}}^{{6}}}$

$\displaystyle{M}_{{y}}=\rho{\left[{3}{{\left({6}\right)}}^{{2}}-\frac{{{6}}^{{3}}}{{3}}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left[{108}-\frac{{216}}{{3}}\right]}$

$\displaystyle{M}_{{y}}=\rho{\left({108}-{72}\right)}$

$\displaystyle{M}_{{y}}={36}\rho$

Now let's find the coordinates of the center of mass,

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

$\displaystyle{\overline{{x}}}=\frac{{{36}\rho}}{{\rho{18}}}$

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

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

$\displaystyle{\overline{{y}}}=\frac{{{36}\rho}}{{\rho{18}}}$

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

The center of the mass is $\displaystyle{\left({2},{2}\right)}$

Notes

Tuesday, 24 May 2016

$\displaystyle{y}={6}-{x},{y}={0},{x}={0}$ 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...

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

Tuesday, 24 May 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?

$\displaystyle{y}={6}-{x},{y}={0},{x}={0}$ 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...

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