Question-110739

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

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

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