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Answer:

[tex]M_x=\dfrac{1}{2}[/tex]

[tex]M_y=\dfrac{3}{2}[/tex]

Centre of mass: [tex]\left(\dfrac{2}{\pi+2},\dfrac{2}{3(\pi+2)}\right)[/tex]

Step-by-step explanation:

The density of shape is 3.

First we find the mass of lamina.

Mass = Density x Area

Lamina is form by quarter circle and triangle.

Area of quarter circle [tex]=\dfrac{1}{4}\times \pi\times 1^2 = \dfrac{\pi}{4}[/tex]

Area of triangular region [tex]=\dfrac{1}{2}\times 1\times 1 = \dfrac{1}{2}[/tex]

Total area of lamina [tex]=\dfrac{\pi+2}{4}[/tex]

Mass of lamina [tex]=\dfrac{3}{4}(\pi+2)[/tex]

Now we will find moments about x-axis, Mx

[tex]M_x=\int_0^1\int_{x-1}^{\sqrt{1-x^2}}3ydydx[/tex]

[tex]M_x=\dfrac{3}{2}\int_0^1y^2|_{x-1}^{\sqrt{1-x^2}}dydx[/tex]

[tex]M_x=\int _0^13x\left(-x+1\right)dx[/tex]

[tex]M_x=3\left(-\frac{x^3}{3}+\frac{x^2}{2}\right)| _0^1[/tex]

[tex]M_x=\dfrac{1}{2}[/tex]

Now we will find moments about y-axis, My

[tex]M_y=\int_0^1\int_{x-1}^{\sqrt{1-x^2}}3xdydx[/tex]

[tex]M_y=3\int_0^1xy|_{x-1}^{\sqrt{1-x^2}}dydx[/tex]

[tex]M_y=\int _0^13x\left(\sqrt{1-x^2}-(x-1)\right)dx[/tex]

[tex]M_y=-\left(1-x^2\right)^{\frac{3}{2}}-x^3+\frac{3x^2}{2}| _0^1[/tex]

[tex]M_y=\dfrac{3}{2}[/tex]

Centre of mass: [tex](\bar x,\bar y)[/tex]

[tex]\bar x=\dfrac{M_y}{M}=\dfrac{3/2}{3(\pi+2)/4)}=\dfrac{2}{\pi+2}[/tex]

[tex]\bar y=\dfrac{M_x}{M}=\dfrac{1/2}{3(\pi+2)/4)}=\dfrac{2}{3(\pi+2)}[/tex]

Hence, [tex]M_x=\dfrac{1}{2}[/tex],   [tex]M_y=\dfrac{3}{2}[/tex]  and Centre of mass [tex]\left(\dfrac{2}{\pi+2},\dfrac{2}{3(\pi+2)}\right)[/tex]

tifanihayyu

The moments Mx and My and the center of mass of a lamina with the given density and shape.  ρ = 3 are Mx = 1/2 and My = 3/2 and the center of mass = [tex](\frac{2}{\pi +2} , \frac{2}{3\pi+6} )[/tex]

Explanation:

 The equation of the first function is

[tex]f(x) = \sqrt{1-x^2}[/tex]

Whereas the second function equation is

[tex]g(x)=x-1[/tex]

The moments of My is

[tex]M_y=  \rho  \int\limits^a_b {x} \, (f(x) - g(x)) dx[/tex]

[tex]M_y=  3  \int\limits^1_0 {x} \, (\sqrt{1-x^2} - (x-1) ) dx[/tex]

By inserting the function f(x) and g(x) with the corner from 0 to 1 as shown in the picture, where   ρ = 3

[tex]M_y=  3  [\int\limits^1_0 {x} \, \sqrt{1-x^2} - dx \int\limits^1_0 {x^2} \, dx + \int\limits^1_0 {x} \, dx  ][/tex]

Then we do simplification by using u substitution [tex]u = 1-x^2[/tex] and [tex]du = -2xdx[/tex]

Then the integral is

[tex]M_y = 3[-\frac{1}{3}(1-x^2)^{\frac{3}{2}} - \frac{1}{3} x^3 + \frac{1}{2}x^2 ]_0^1[/tex]

Whereas the moments of Mx is

[tex]M_x = \rho \int\limits^a_b {(f(x)^2 - g(x)^2)} \, dx[/tex]

We insert function f(x) and g(x)  with the corner from 0 to 1 as shown in the picture, where   ρ = 3

[tex]M_x=  3  \int\limits^1_0 {\frac{1}2} } \, ((\sqrt{1-x^2})^2 - (x-1)^2 ) dx[/tex]

Then we simplify the integration

[tex]M_x = \frac{3}{2} \int\limits^1_0 {(-2x^2 + 2x)} \, dx[/tex]

[tex]M_x = \frac{3}{2} [-\frac{2}{3}x^3 + x^2 ]_0^1[/tex]

[tex]M_x = \frac{1}{2}[/tex]

For the center of mass of a lamina

The centre of mass is equal to the shape centroid.

[tex](\frac{M_y}{A\rho} , \frac{M_x}{A\rho} )[/tex]

Where

[tex]A = \int\limits^1_0 {\sqrt{1-x^2} - (x-1) } \, dx[/tex]

Then integrate above where the term [tex]\sqrt{1-x^2}[/tex] is in the form [tex]\sqrt{a-u^2}[/tex]

[tex]A = [\frac{x}{2} \sqrt{1-x^2}+ \frac{1}{2} sin^{-1}x - \frac{1}{2} x^2 + x ]^1_0[/tex]

[tex]A =\frac{\pi+2}{4}[/tex]

Then the center of mass is

[tex](\frac{\frac{3}{2} }{(3)\frac{\pi+2}{4} } , \frac{\frac{1}{2} }{(3)\frac{\pi+2}{4} } )\\(\frac{2}{\pi+2}, \frac{2}{3\pi+6} )[/tex]

Learn more about the center of mass of a lamina https://brainly.com/question/13177184

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