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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.

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

Properties

The center of mass of the lamina is at the point

\left(\frac{M_y}{m},\frac{M_x}{m}\right)

where $M_y$ is the moment of the entire lamina about the y-axis and $M_x$ is the moment of the entire lamina about the x-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

Bound region.jpg

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 $\rho\ (x,y)\,=2x+3y+2$ .

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

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

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

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

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

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

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

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

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

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

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

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

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)

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