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isyllus

Answer:

\(M_x=\dfrac{1}{2}\)

\(M_y=\dfrac{3}{2}\)

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

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 \(=\dfrac{1}{4}\times \pi\times 1^2 = \dfrac{\pi}{4}\)

Area of triangular region \(=\dfrac{1}{2}\times 1\times 1 = \dfrac{1}{2}\)

Total area of lamina \(=\dfrac{\pi+2}{4}\)

Mass of lamina \(=\dfrac{3}{4}(\pi+2)\)

Now we will find moments about x-axis, Mx

\(M_x=\int_0^1\int_{x-1}^{\sqrt{1-x^2}}3ydydx\)

\(M_x=\dfrac{3}{2}\int_0^1y^2|_{x-1}^{\sqrt{1-x^2}}dydx\)

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

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

\(M_x=\dfrac{1}{2}\)

Now we will find moments about y-axis, My

\(M_y=\int_0^1\int_{x-1}^{\sqrt{1-x^2}}3xdydx\)

\(M_y=3\int_0^1xy|_{x-1}^{\sqrt{1-x^2}}dydx\)

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

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

\(M_y=\dfrac{3}{2}\)

Centre of mass: \((\bar x,\bar y)\)

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

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

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

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 = \((\frac{2}{\pi +2} , \frac{2}{3\pi+6} )\)

Explanation:

 The equation of the first function is

\(f(x) = \sqrt{1-x^2}\)

Whereas the second function equation is

\(g(x)=x-1\)

The moments of My is

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

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

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

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

Then we do simplification by using u substitution \(u = 1-x^2\) and \(du = -2xdx\)

Then the integral is

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

Whereas the moments of Mx is

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

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

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

Then we simplify the integration

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

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

\(M_x = \frac{1}{2}\)

For the center of mass of a lamina

The centre of mass is equal to the shape centroid.

\((\frac{M_y}{A\rho} , \frac{M_x}{A\rho} )\)

Where

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

Then integrate above where the term \(\sqrt{1-x^2}\) is in the form \(\sqrt{a-u^2}\)

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

\(A =\frac{\pi+2}{4}\)

Then the center of mass is

\((\frac{\frac{3}{2} }{(3)\frac{\pi+2}{4} } , \frac{\frac{1}{2} }{(3)\frac{\pi+2}{4} } )\\(\frac{2}{\pi+2}, \frac{2}{3\pi+6} )\)

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

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