Question #193303

Find the centre of gravity of a thin sheet with density δ(x, y) = y, bounded by the 

curves

y = 4x² and x = 4


1
Expert's answer
2021-05-18T08:40:02-0400

Given, densityδ(x,y)=y and bounded by y=4x2andx=4.mass,m=δdA=ydxdy=x=04y=0x2ydxdy=12x=04x4dxdy=102410=102.4Now, calculate momentsMx=yδdA=y2dxdy=x=04y=0x2y2dxdy=13x=04x6dxdy=1638421=780.19My=xδdA=xydxdy=x=04y=0x2xydxdy=12x=04x5dxdy=409612=341.33Let (X, Y) be the center of mass.X=Mym=341.33102.4=3.33Y=Mxm=780.19102.4=7.62Center of mass is(3.33,7.62).Therefore, center of gravity is (3.33,7.62).\text{Given, density} \delta(x,y)=y\text{ and bounded by }y=4x^2and x=4.\newline \text{mass},m=\int\int\delta dA=\int\int y dxdy=\int_{x=0}^4\int_{y=0}^{x^2}y dxdy=\frac{1}{2}\int_{x=0}^4x^4dxdy=\frac{1024}{10}=102.4\newline \text{Now, calculate moments}\newline M_x=\int\int y\delta dA=\int\int y^2dxdy=\int_{x=0}^4\int_{y=0}^{x^2}y^2dxdy=\frac{1}{3}\int_{x=0}^4x^6dxdy=\frac{16384}{21}=780.19\newline M_y=\int\int x\delta dA=\int\int xydxdy=\int_{x=0}^4\int_{y=0}^{x^2}xydxdy=\frac{1}{2}\int_{x=0}^4 x^5dxdy=\frac{4096}{12}=341.33\newline \text{Let (X, Y) be the center of mass.}\newline X=\frac{M_y}{m}=\frac{341.33}{102.4}=3.33\newline Y=\frac{M_x}{m}=\frac{780.19}{102.4}=7.62\newline \text{Center of mass is} (3.33, 7.62).\newline \text{Therefore, center of gravity is }(3.33, 7.62).


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