Answer :
Answer:
m = 2/9, M(x) = 0, M(y) = 1/8, COM = (9/16, 0)
Step-by-step explanation:
Given that the density of the uniform plate is 1 and it is bound by the curves y = x²lnx and y = -x²lnx and the lines x = 0 and x = 1
Mass of the given lamina is
m = ∫∫ ρ(x,y) dA
m = ∫(0,1)∫(x²lnx,-x²lnx) 1 dy.dx
m = ∫(0,1) [y]x²lnx, -x²lnx dx
m = ∫(0,1) (-x²lnx – x²lnx) dx
m = ∫(0,1) (-2x²lnx) dx
m = 2∫(0,1) (-x²lnx) dx
m = 2[-((x³lnx)/³) + x³/9 ](0,1)
m = 2[(1³/9)ln1 + 1/9 - 0]
m = 2/9
Now, M(x) =∫∫ y.ρ(x,y) dA and M(y) =∫∫ x.ρ(x,y) dA
M(x) = ∫∫(D) y.ρ(x,y) dA
M(x) = ∫(0,1)∫(x²lnx,-x²lnx) y dy.dx
M(x) = ∫(0,1) [y²/2] x²lnx,-x²lnx dx
M(x) = ½ ∫(0,1) {(-x²lnx)² – (x²lnx)²} dx
M(x) = ½ ∫(0,1) (x⁴ (lnx)² – x⁴(lnx)²) dx
M(x) = 0
M(y) = ∫(0,1)∫(x²lnx,-x²lnx) x dy.dx
M(y) = ∫(0,1) x[y] x²lnx,-x²lnx dx
M(y) = ∫(0,1) x(-x²lnx -x²lnx) dx
M(y) = 2∫(0,1) (-x³lnx) dx
M(y) = 2[-x⁴lnx/4 + x⁴/16]0,1
M(y) = 2[-ln1/4 + 1/16 - 0]
M(y) = 2/16 = 1/8
x’ = M(y)/m = (1/8) / (2/9) = 9/16
y’ = M(x)/m = (0/2) / (2/9) = 0
Hence the center of mass is at the point (x’,y’) = (9/16 , 0)
The mass of the uniform flat plate is [tex]\frac{2}{9}[/tex].
The mass moment respect to the y-axis is [tex]\frac{9}{16}[/tex].
The mass moment respect to the x-axis is [tex]0[/tex].
The center of mass of the uniform flat plate is [tex](x,y) = \left(\frac{9}{16}, 0 \right)[/tex].
How to find the center of mass of a plate
In this question we shall use the integral formulas of mass ([tex]m[/tex]) and mass moments ([tex]M_{x}[/tex], [tex]M_{y}[/tex]) to determine the location of the center of mass with respect to origin ([tex]\bar{x}[/tex], [tex]\bar y[/tex]). The mass of the uniform flat plate is obtained after the following formula:
[tex]m = \int\limits^{A}_{0} {\rho(x, y)} \, dA[/tex] (1)
Where:
- [tex]\rho(x,y)[/tex] - Density function.
- [tex]A[/tex] - Area
By definition of rectangular coordinates, we expanded the formula presented above:
[tex]m = \int\limits^1_0 {\int\limits^{-x^{2}\cdot \ln x}_{x^{2}\cdot \ln x}\, dy } \, dx[/tex] (1b)
Whose solution is:
[tex]m = -2\cdot \left[\frac{x^{3}}{3}\cdot \left(\ln x - \frac{1}{3} \right) \right] \left |_{0}^{1}[/tex]
[tex]m = \frac{2}{9}[/tex]
The mass of the uniform flat plate is [tex]\frac{2}{9}[/tex]. [tex]\blacksquare[/tex]
And the mass moments are described by the following formulae:
[tex]M_{y} = \int\limits^1_0 {x\int\limits^{-x^{2}\cdot \ln x}_{x^{2}\cdot \ln x}\, dy } \, dx[/tex] (2)
Whose solution is:
[tex]M_{y} = -2\cdot \left[\frac{x^{4}}{4}\cdot \left(\ln x - \frac{1}{4} \right) \right] \left|_{0}^{1}[/tex]
[tex]M_{y} = \frac{1}{8}[/tex]
The mass moment respect to the y-axis is [tex]\frac{9}{16}[/tex]. [tex]\blacksquare[/tex]
[tex]M_{x} = \int\limits^{1}_{0} {\int\limits^{-x^{2}\cdot \ln x}_{x^{2}\cdot \ln x} {y} \, dy } \, dx[/tex] (3)
Whose solution is:
[tex]M_{x} = 0[/tex]
The mass moment respect to the x-axis is [tex]0[/tex]. [tex]\blacksquare[/tex]
Lastly, the coordinates of the center of mass are found by using the following expressions:
[tex]\bar x = \frac{M_{y}}{m}[/tex] (4)
[tex]\bar x = \frac{\frac{1}{8} }{\frac{2}{9} }[/tex]
[tex]\bar x = \frac{9}{16}[/tex]
[tex]\bar y = \frac{M_{x}}{m}[/tex] (5)
[tex]\bar y = \frac{0}{\frac{2}{9} }[/tex]
[tex]\bar y = 0[/tex]
The center of mass of the uniform flat plate is [tex](x,y) = \left(\frac{9}{16}, 0 \right)[/tex]. [tex]\blacksquare[/tex]
To learn more on centers of mass, we kindly invite to check this verified question: https://brainly.com/question/8662931
