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

Find height of center of mass of hemisphere if its density 𝛒= 𝛒_0 (1+(y/R)).

$${Find}\:{height}\:{of}\:{center}\:{of}\:{mass}\:{of} \\ $$$${hemisphere}\:{if}\:{its}\:{density}\: \\ $$$$\boldsymbol{\rho}=\:\boldsymbol{\rho}_{\mathrm{0}} \left(\mathrm{1}+\frac{{y}}{{R}}\right). \\ $$

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

r^2 =R^2 −y^2 y_S =((∫ρydA)/(∫ρdA))=((∫_0 ^R ρ_0 (1+(y/R))yπ(R^2 −y^2 )dy)/(∫_0 ^R ρ_0 (1+(y/R))π(R^2 −y^2 )dy)) =((∫_0 ^R y(R+y)(R^2 −y^2 )dy)/(∫_0 ^R (R+y)(R^2 −y^2 )dy))=(A/B) B=∫_0 ^R (R+y)(R^2 −y^2 )dy =∫_0 ^R (R^3 +R^2 y−Ry^2 −y^3 )dy =R^3 R+R^2 (R^2 /2)−R(R^3 /3)−(R^4 /4)=((11R^4 )/(12)) A=∫_0 ^R y(R+y)(R^2 −y^2 )dy =∫_0 ^R (R^3 y+R^2 y^2 −Ry^3 −y^4 )dy =R^3 (R^2 /2)+R^2 (R^3 /3)−R(R^4 /4)−(R^5 /5)=((23R^5 )/(60)) ⇒y_S =((23R^5 )/(60))×((12)/(11R^4 ))=((23)/(55))R

$${r}^{\mathrm{2}} ={R}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$${y}_{{S}} =\frac{\int\rho{ydA}}{\int\rho{dA}}=\frac{\int_{\mathrm{0}} ^{{R}} \rho_{\mathrm{0}} \left(\mathrm{1}+\frac{{y}}{{R}}\right){y}\pi\left({R}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dy}}{\int_{\mathrm{0}} ^{{R}} \rho_{\mathrm{0}} \left(\mathrm{1}+\frac{{y}}{{R}}\right)\pi\left({R}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dy}} \\ $$$$=\frac{\int_{\mathrm{0}} ^{{R}} {y}\left({R}+{y}\right)\left({R}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dy}}{\int_{\mathrm{0}} ^{{R}} \left({R}+{y}\right)\left({R}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dy}}=\frac{{A}}{{B}} \\ $$$${B}=\int_{\mathrm{0}} ^{{R}} \left({R}+{y}\right)\left({R}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dy} \\ $$$$=\int_{\mathrm{0}} ^{{R}} \left({R}^{\mathrm{3}} +{R}^{\mathrm{2}} {y}−{Ry}^{\mathrm{2}} −{y}^{\mathrm{3}} \right){dy} \\ $$$$={R}^{\mathrm{3}} {R}+{R}^{\mathrm{2}} \frac{{R}^{\mathrm{2}} }{\mathrm{2}}−{R}\frac{{R}^{\mathrm{3}} }{\mathrm{3}}−\frac{{R}^{\mathrm{4}} }{\mathrm{4}}=\frac{\mathrm{11}{R}^{\mathrm{4}} }{\mathrm{12}} \\ $$$${A}=\int_{\mathrm{0}} ^{{R}} {y}\left({R}+{y}\right)\left({R}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dy} \\ $$$$=\int_{\mathrm{0}} ^{{R}} \left({R}^{\mathrm{3}} {y}+{R}^{\mathrm{2}} {y}^{\mathrm{2}} −{Ry}^{\mathrm{3}} −{y}^{\mathrm{4}} \right){dy} \\ $$$$={R}^{\mathrm{3}} \frac{{R}^{\mathrm{2}} }{\mathrm{2}}+{R}^{\mathrm{2}} \frac{{R}^{\mathrm{3}} }{\mathrm{3}}−{R}\frac{{R}^{\mathrm{4}} }{\mathrm{4}}−\frac{{R}^{\mathrm{5}} }{\mathrm{5}}=\frac{\mathrm{23}{R}^{\mathrm{5}} }{\mathrm{60}} \\ $$$$\Rightarrow{y}_{{S}} =\frac{\mathrm{23}{R}^{\mathrm{5}} }{\mathrm{60}}×\frac{\mathrm{12}}{\mathrm{11}{R}^{\mathrm{4}} }=\frac{\mathrm{23}}{\mathrm{55}}{R} \\ $$

Commented by ajfour last updated on 30/Aug/20

What a presentation, Sir! Thanks a lot. Incredible!

$${What}\:{a}\:{presentation},\:{Sir}! \\ $$$${Thanks}\:{a}\:{lot}.\:{Incredible}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com