Question #236118

A triangular lamina in the xy -plane such that its vertices are (0,0), (0,1) and (1,0). Suppose that the density function of the lamina is defined by p(x,y)=30xy gram per cubic centimetre. What is the total mass of the lamina and the center of gravity. The moment about the y-axis of the lamina.the moment about the x axis of the lamina


1
Expert's answer
2021-09-12T23:52:32-0400

Let A=(0,1),B=(1,0),O=(0,0).A=(0,1), B=(1, 0), O=(0,0).

Line OA: x=0,0y1x=0, 0\leq y\leq1

LineAB: y=x+1,0x1y=-x+1, 0\leq x\leq1

Line OB: y=0,0x1y=0, 0\leq x\leq 1


Given ρ(x,y)=30xy\rho(x,y)=30xy

Find the mass of the lamina


m=Dρ(x,y)dA=0101x30xydydx=m=\iint_D\rho(x,y)dA=\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}30xydydx==3001x[y22]1x0dx=1501(x2x2+x3)dx==30\displaystyle\int_{0}^1x\big[{y^2\over 2}\big]\begin{matrix} 1-x \\ 0 \end{matrix}dx=15\displaystyle\int_{0}^1(x-2x^2+x^3 )dx==15[x222x33+x44]10=54(units of mass)=15\big[{x^2 \over 2}-{2x^3 \over 3}+{x^4 \over 4}\big]\begin{matrix} 1 \\ 0 \end{matrix}={5 \over 4} (units\ of\ mass)

Mass of the lamina is 54\dfrac{5}{4} units of mass.


The moment of the lamina about the x-axis


Mx=Dyρ(x,y)dA=300101xxy2dydx=M_x=\iint_Dy\rho(x,y)dA=30\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}xy^2dydx=




=3001x[y33]1x0dx=30\displaystyle\int_{0}^1x\big[{y^3\over 3}\big]\begin{matrix} 1-x \\ 0 \end{matrix}dx




=1001(x3x2+3x3x4)dx=10\displaystyle\int_{0}^1(x-3x^2+3x^3-x^4 )dx




=10[x223x33+3x44x55]10=10\big[{x^2 \over 2}-{3x^3 \over 3}+{3x^4 \over 4}-{x^5 \over 5}\big]\begin{matrix} 1 \\ 0 \end{matrix}

=510+1522=12=5-10+\dfrac{15}{2}-2={1 \over2}

The moment of the lamina about the y-axis


My=Dxρ(x,y)dA=0101x30x2ydydxM_y=\iint_Dx\rho(x,y)dA=\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}30x^2ydydx




=3001x2[y22]1x0dx=30\displaystyle\int_{0}^1x^2\big[{y^2\over 2}\big]\begin{matrix} 1-x \\ 0 \end{matrix}dx




=1501(x22x3+x4)dx=15\displaystyle\int_{0}^1(x^2-2x^3+x^4 )dx




=15[x332x44+x55]10=15\big[{x^3 \over 3}-{2x^4 \over 4}+{x^5 \over 5}\big]\begin{matrix} 1 \\ 0 \end{matrix}


=5152+3=12=5-\dfrac{15}{2}+3=\dfrac{1}{2}

Mx=12,My=12M_x=\dfrac{1}{2}, M_y=\dfrac{1}{2}

Find the coordinates of the center of mass


xˉ=Mym=1254=25\bar{x}={M_y\over m}=\dfrac{\dfrac{1}{2}}{\dfrac{5}{4}}=\dfrac{2}{5}



yˉ=Mxm=1254=25\bar{y}={M_x\over m}=\dfrac{\dfrac{1}{2}}{\dfrac{5}{4}}=\dfrac{2}{5}



Center of gravity is(25,25).\bigg(\dfrac{2}{5},\dfrac{2}{5}\bigg).



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