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:

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

The center of mass of the lamina is at the point

$(\frac{M_y}{m},\frac{M_x}{m})$

where $M_y$ moment of the entire lamin about the x-axis and $M_x$ moment of the entire lamin about the y-axis.

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

$M_x=\underset{m,n\to \mathrm{\infty }}{lim}\sum _{i=1}^m\sum _{j=1}^{n}y{{}_{ij}}^{\ast }\rho (x{{}_{ij}}^{\ast },y{{}_{ij}}^{\ast })\Delta \mathrm{A}={\iint }_{}y\rho (x,y)dxdy$, 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 $\rho (x,y)=2x+3y+2$.

$m={\int }_0^2{{\int }_x^{4-x}}_{}2x+3y+2dydx$

$m={\int }_0^2(2xy+\frac{3y^2}{2}+2y){|}_x^{4-x}dx$

$m={\int }_0^2-4x^2-8x+32dx$

$m=(\frac{-4x^3}{3}-4x^2+32x){|}_0^2$

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

$M_y={\int }_0^2{\int }_x^{4-x}x(2x+3y+2)dydx$

$M_y={\int }_0^2(2x^{2}y+\frac{3x{y}^2}{2}+2xy){|}_x^{4-x}dx$

$M_y={\int }_0^2-4x^3-8x^2+32xdx$

$M_y=(-x^4-\frac{8x^3}{3}+16x^2){|}_0^2$

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

$M_x={\int }_0^2{\int }_x^{4-x}y(2x+3y+2)dydx$

$M_x={\int }_0^2(x{y}^2+y^3+y^2){|}_x^{4-x}dx$

$M_x={\int }_0^2(-2x^3+4x^2-40x+80dx$

$M_x={(\frac{-x^4}{2}+\frac{4x^3}{3}-20x^2+80x)|}_0^2$

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

center of mass is at the point

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

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