Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 110739 by ajfour last updated on 30/Aug/20

Commented by ajfour last updated on 30/Aug/20

$$Findheightofcenterofmassof$$$$hemisphereifitsdensity$$$$\mathit{\rho }={\mathit{\rho }}_0(1+\frac{y}{R}).$$

Answered by mr W last updated on 30/Aug/20

$$r^2=R^2-y^2$$$$y_S=\frac{\int \rho ydA}{\int \rho dA}=\frac{{\int }_0^R{\rho }_0(1+\frac{y}{R})y\pi (R^2-y^2)dy}{{\int }_0^R{\rho }_0(1+\frac{y}{R})\pi (R^2-y^2)dy}$$$$=\frac{{\int }_0^{R}y(R+y)(R^2-y^2)dy}{{\int }_0^R(R+y)(R^2-y^2)dy}=\frac{A}{B}$$$$B={\int }_0^R(R+y)(R^2-y^2)dy$$$$={\int }_0^R(R^3+R^{2}y-{Ry}^2-y^3)dy$$$$=R^{3}R+R^2\frac{R^2}{2}-R\frac{R^3}{3}-\frac{R^4}{4}=\frac{11R^4}{12}$$$$A={\int }_0^{R}y(R+y)(R^2-y^2)dy$$$$={\int }_0^R(R^{3}y+R^2{y}^2-{Ry}^3-y^4)dy$$$$=R^3\frac{R^2}{2}+R^2\frac{R^3}{3}-R\frac{R^4}{4}-\frac{R^5}{5}=\frac{23R^5}{60}$$$$\Rightarrow y_S=\frac{23R^5}{60}\times \frac{12}{11R^4}=\frac{23}{55}R$$

Commented by ajfour last updated on 30/Aug/20

$$Whatapresentation,Sir!$$$$Thanksalot.Incredible!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com