Rotation system

In mathematics, a planar lamina is a closed set in a plane of mass $m$ and surface density $\rho \ (x,y)$ such that:

$m=\int \int _{}{}\rho \ (x,y)\,dx\,dy$, over the closed set.

The center of mass of the lamina is at the point

$\left({\frac {M_{y}}{m}},{\frac {M_{x}}{m}}\right)$

where $M_{y}$ moment of the entire lamin about the x-axis and $M_{x}$ moment of the entire lamin about the y-axis.

$M_{y}=\lim _{m,n\to \infty }\,\sum _{i=1}^{m}\,\sum _{j=1}^{n}\,x{_{ij}}^{*}\,\rho \ (x{_{ij}}^{*},y{_{ij}}^{*})\,\Delta \mathrm {A} =\iint _{}{}x\,\rho \ (x,y)\,dx\,dy$, over the closed surface.

$M_{x}=\lim _{m,n\to \infty }\,\sum _{i=1}^{m}\,\sum _{j=1}^{n}\,y{_{ij}}^{*}\,\rho \ (x{_{ij}}^{*},y{_{ij}}^{*})\,\Delta \mathrm {A} =\iint _{}{}y\,\rho \ (x,y)\,dx\,dy$, over the closed surface.

Example 1.

Find the center of mass of a lamina with edges given by the lines $x=0$, $x=y$ and $y=4-x$, where the density is given as $\rho \ (x,y)\,=2x+3y+2$.

$m=\int _{0}^{2}{\int _{x}^{4-x}}_{}{}\,2x+3y+2\,dy\,dx$

$m=\int _{0}^{2}(2xy+{\frac {3y^{2}}{2}}+2y)|_{x}^{4-x}\,dx$

$m=\int _{0}^{2}-4x^{2}-8x+32\,dx$

$m=({\frac {-4x^{3}}{3}}-4x^{2}+32x)|_{0}^{2}$

$m={\frac {112}{3}}$

$M_{y}=\int _{0}^{2}{\int _{x}^{4-x}}{}{}x\,(2x+3y+2)\,dy\,dx$

$M_{y}=\int _{0}^{2}(2x^{2}y+{\frac {3xy^{2}}{2}}+2xy)|_{x}^{4-x}\,dx$

$M_{y}=\int _{0}^{2}-4x^{3}-8x^{2}+32x\,dx$

$M_{y}=(-x^{4}-{\frac {8x^{3}}{3}}+16x^{2})|_{0}^{2}$

$M_{y}={\frac {80}{3}}$

$M_{x}=\int _{0}^{2}{\int _{x}^{4-x}}{}{}y\,(2x+3y+2)\,dy\,dx$

$M_{x}=\int _{0}^{2}(xy^{2}+y^{3}+y^{2})|_{x}^{4-x}\,dx$

$M_{x}=\int _{0}^{2}(-2x^{3}+4x^{2}-40x+80\,dx$

$M_{x}=\left.\left({\frac {-x^{4}}{2}}+{\frac {4x^{3}}{3}}-20x^{2}+80x\right)\right|_{0}^{2}$

$M_{x}={\frac {248}{3}}$

center of mass is at the point

$\left({\frac {\frac {80}{3}}{\frac {112}{3}}},{\frac {\frac {248}{3}}{\frac {112}{3}}}\right)=\left({\frac {5}{7}},{\frac {31}{14}}\right)$

Planar laminas can be used to determine moments of inertia, or center of mass.