Notes: `x = y+2, x=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...

Tuesday, 18 October 2016

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

First lets find the bounds of integration. When looking at the graph the furthest that the lamina is bounded on the y-axis is where the curves interest. Lets find those points.

$\displaystyle{y}+{2}={{y}}^{{2}}$

$\displaystyle{0}={\left({y}+{1}\right)}{\left({y}-{2}\right)}$

Therefore the y bounds are y=-1 and y=2. Then we will integrate between the furthest right curve (x=y) and the furthest left curve.

The center of Mass is:

$\displaystyle{\left({x}_{{{c}{m}}},{y}_{{{c}{m}}}\right)}={\left(\frac{{M}_{{y}}}{{M}},\frac{{M}_{{x}}}{{M}}\right)}$

Where the moments of mass are defined as:

$\displaystyle{M}_{{x}}=\int\int_{{A}}\rho{\left({x},{y}\right)}\cdot{y}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$

$\displaystyle{M}_{{y}}=\int\int_{{A}}\rho{\left({x},{y}\right)}\cdot{x}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$

The total mass is defined as:

$\displaystyle{M}=\int\int_{{A}}\rho{\left({x},{y}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$

First, lets find the total mass.

$\displaystyle{M}={\int}^{{2}}_{-{{1}}}{\left[{\int}^{{{x}={y}+{2}}}_{\left({x}={{y}}^{{2}}\right)}\rho{\left.{d}{x}\right.}\right]}{\left.{d}{y}\right.}$

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

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

$\displaystyle{M}=\frac{{9}}{{2}}\rho$

Now lets find the x moment of mass.

$\displaystyle{M}={\int}^{{2}}_{-{{1}}}{y}\cdot{\left[{\int}^{{{x}={y}+{2}}}_{\left({x}={{y}}^{{2}}\right)}\rho{\left.{d}{x}\right.}\right]}{\left.{d}{y}\right.}$

$\displaystyle{M}=\rho{\int}^{{2}}_{-{{1}}}{y}\cdot{\left[{\left({y}+{2}\right)}-{\left({{y}}^{{2}}\right)}\right]}{\left.{d}{y}\right.}$

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

$\displaystyle{M}=\rho{\left({\left(\frac{{1}}{{3}}\right)}{{y}}^{{3}}+{{y}}^{{2}}-{\left(\frac{{1}}{{4}}\right)}{{y}}^{{4}}\right)}{{\mid}}^{{2}}_{-{{1}}}$

$\displaystyle{M}_{{x}}=\frac{{9}}{{4}}\rho$

Now the y moment of mass.

$\displaystyle{M}={\int}^{{2}}_{-{{1}}}{\left[{\int}^{{{x}={y}+{2}}}_{\left({x}={{y}}^{{2}}\right)}\rho{x}{\left.{d}{x}\right.}\right]}{\left.{d}{y}\right.}$

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

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

$\displaystyle{M}=\frac{\rho}{{2}}{\left({\left(\frac{{1}}{{3}}\right)}{{y}}^{{3}}+{2}{{y}}^{{2}}+{4}{y}-{\left(\frac{{1}}{{5}}\right)}{{y}}^{{5}}\right)}{{\mid}}^{{2}}_{-{{1}}}$

$\displaystyle{M}_{{y}}=\frac{{36}}{{5}}\rho$

Therefore the center of mass is:

$\displaystyle{\left({x}_{{{c}{m}}},{y}_{{{c}{m}}}\right)}={\left(\frac{{M}_{{y}}}{{M}},\frac{{M}_{{x}}}{{M}}\right)}={\left(\frac{{\frac{{36}}{{5}}\rho}}{{\frac{{9}}{{2}}\rho}},\frac{{\frac{{9}}{{4}}\rho}}{{\frac{{9}}{{2}}\rho}}\right)}={\left(\frac{{8}}{{5}},\frac{{1}}{{2}}\right)}$

The moments of inerita or the second moments of the lamina are:

$\displaystyle{I}_{{x}}=\int\int_{{A}}\rho{\left({x},{y}\right)}\cdot{{y}}^{{2}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$

$\displaystyle{I}_{{y}}=\int\int_{{A}}\rho{\left({x},{y}\right)}\cdot{{x}}^{{2}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$

I won't solve these integrals step by step since they are very similar to the others, but you will find that:

$\displaystyle{I}_{{x}}=\frac{{63}}{{20}}\rho$

$\displaystyle{I}_{{y}}=\frac{{423}}{{28}}\rho$

Notes

Tuesday, 18 October 2016

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

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

Tuesday, 18 October 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?

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