Question Number 219789 by Firhad last updated on 01/May/25
Commented by Firhad last updated on 02/May/25
$${can}\:{you}\:{help}\:{me} \\ $$
Commented by mr W last updated on 02/May/25
$${do}\:{you}\:{mean}\:{the}\:{following}\:{solid}? \\ $$
Commented by mr W last updated on 02/May/25
Commented by Firhad last updated on 02/May/25
$${yes}\:{sir} \\ $$
Answered by mr W last updated on 02/May/25
Commented by mr W last updated on 02/May/25
![due to symmetry, the center of mass is at C(0,y_S ,z_S ) R=2 b=(√(R^2 −y^2 )) h=y dM=ρdV=ρ×2bhdy=2ρy(√(R^2 −y^2 ))dy M=2ρ∫_0 ^R y(√(R^2 −y^2 ))dy =2ρR^3 ∫_0 ^(π/2) sin θ cos^2 θ dθ =(2/3)ρR^3 [cos^3 θ]_(π/2) ^0 =(2/3)ρR^3 My_S =∫ydM =2ρ∫_0 ^R y^2 (√(R^2 −y^2 ))dy =2ρR^4 ∫_0 ^(π/2) sin^2 θ cos^2 θ dθ =(1/4)ρR^4 ∫_0 ^(π/2) (1−cos 4θ)dθ =(π/8)ρR^4 ⇒y_S =((πρR^4 ×3)/(8×2ρR^3 ))=((3πR)/(16)) ✓ Mz_S =∫(h/2)dM =ρ∫_0 ^R y^2 (√(R^2 −y^2 ))dy =((πρR^4 )/(16)) ⇒z_S =((3πR)/(32)) ✓](https://www.tinkutara.com/question/Q219842.png)
$${due}\:{to}\:{symmetry},\:{the}\:{center}\:{of}\:{mass} \\ $$$${is}\:{at}\:{C}\left(\mathrm{0},{y}_{{S}} ,{z}_{{S}} \right) \\ $$$${R}=\mathrm{2} \\ $$$${b}=\sqrt{{R}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$${h}={y} \\ $$$${dM}=\rho{dV}=\rho×\mathrm{2}{bhdy}=\mathrm{2}\rho{y}\sqrt{{R}^{\mathrm{2}} −{y}^{\mathrm{2}} }{dy} \\ $$$${M}=\mathrm{2}\rho\int_{\mathrm{0}} ^{{R}} {y}\sqrt{{R}^{\mathrm{2}} −{y}^{\mathrm{2}} }{dy} \\ $$$$\:\:\:\:=\mathrm{2}\rho{R}^{\mathrm{3}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta\:{d}\theta \\ $$$$\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\rho{R}^{\mathrm{3}} \left[\mathrm{cos}^{\mathrm{3}} \:\theta\right]_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \\ $$$$\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\rho{R}^{\mathrm{3}} \\ $$$${My}_{{S}} =\int{ydM} \\ $$$$\:\:\:\:\:=\mathrm{2}\rho\int_{\mathrm{0}} ^{{R}} {y}^{\mathrm{2}} \sqrt{{R}^{\mathrm{2}} −{y}^{\mathrm{2}} }{dy} \\ $$$$\:\:\:\:\:=\mathrm{2}\rho{R}^{\mathrm{4}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta\:{d}\theta \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\rho{R}^{\mathrm{4}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}\theta\right){d}\theta \\ $$$$\:\:\:\:\:=\frac{\pi}{\mathrm{8}}\rho{R}^{\mathrm{4}} \\ $$$$\Rightarrow{y}_{{S}} =\frac{\pi\rho{R}^{\mathrm{4}} ×\mathrm{3}}{\mathrm{8}×\mathrm{2}\rho{R}^{\mathrm{3}} }=\frac{\mathrm{3}\pi{R}}{\mathrm{16}}\:\checkmark \\ $$$${Mz}_{{S}} =\int\frac{{h}}{\mathrm{2}}{dM} \\ $$$$\:\:\:\:\:=\rho\int_{\mathrm{0}} ^{{R}} {y}^{\mathrm{2}} \sqrt{{R}^{\mathrm{2}} −{y}^{\mathrm{2}} }{dy} \\ $$$$\:\:\:\:\:=\frac{\pi\rho{R}^{\mathrm{4}} }{\mathrm{16}} \\ $$$$\Rightarrow{z}_{{S}} =\frac{\mathrm{3}\pi{R}}{\mathrm{32}}\:\checkmark \\ $$