Planar lamina

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 lamina about the x-axis and $M_x$ moment of the entire lamina about the y-axis.

$M_y = \lim_{m,n \to \infty}\,\sum_{i=1}^{m}\,\sum_{j=1}^{n}\,x{_{ij}}^{*}\,\rho\ (x{_{ij}}^{*},y{_{ij}}^{*})\,\Delta\Alpha = \iint_{}{} x\, \rho\ (x,y)\,dx\,dy$, over the closed surface.

$M_x = \lim_{m,n \to \infty}\,\sum_{i=1}^{m}\,\sum_{j=1}^{n}\,y{_{ij}}^{*}\,\rho\ (x{_{ij}}^{*},y{_{ij}}^{*})\,\Delta\Alpha = \iint_{}{} y\, \rho\ (x,y)\,dx\,dy$, over the closed surface.

Example 1.

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

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

Integrate 2x + 3y + 2 with respect to y and substitute the limits 4-x and x

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

${\displaystyle m=\int _{0}^{2}\left({\Big [}2x(4-x)+{\frac {3(4-x)^{2}}{2}}+2(4-x){\Big ]}-{\Big [}2x(x)+{\frac {3(x)^{2}}{2}}+2(x){\Big ]}\right)\,dx}$

${\displaystyle m=\int _{0}^{2}\left(8x-2x^{2}+{\frac {3x^{2}-24x+48}{2}}+8-2x-2x^{2}-{\frac {3x^{2}}{2}}-2x\right)\,dx}$

${\displaystyle m=\int _{0}^{2}\left(8x-2x^{2}+{\frac {3}{2}}x^{2}-12x+24+8-2x-2x^{2}-{\frac {3}{2}}x^{2}-2x\right)\,dx}$

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

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

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

$M_y = \int_0^2{\int_x^{4-x}}{}{}x\,(2x+3y+2)\,dy\,dx$

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

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

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

$M_y= \frac{80}{3}$

$M_x = \int_0^2{\int_x^{4-x}}{}{}y\,(2x+3y+2)\,dy\,dx$

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

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

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

$M_x= \frac{248}{3}$

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.