Question Number 71538 by mr W last updated on 16/Oct/19
Commented by mr W last updated on 16/Oct/19
$${find}\:{the}\:{position}\:{of}\:{center}\:{of}\:{mass} \\ $$$${of}\:{the}\:{parabolic}\:{arc} \\ $$
Answered by ajfour last updated on 17/Oct/19
$${y}={Ax}^{\mathrm{2}} \\ $$$${dl}=\left({dx}\right)\sqrt{\mathrm{1}+\mathrm{4}{A}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$${dm}=\rho{dl} \\ $$$${y}_{{cm}} =\frac{\int_{−{a}} ^{\:\:{a}} \left({Ax}^{\mathrm{2}} \right)\rho\sqrt{\mathrm{1}+\mathrm{4}{A}^{\mathrm{2}} {x}^{\mathrm{2}} }{dx}}{\int_{−{a}} ^{\:\:{a}} \rho\sqrt{\mathrm{1}+\mathrm{4}{A}^{\mathrm{2}} {x}^{\mathrm{2}} }\:{dx}} \\ $$
Answered by mr W last updated on 17/Oct/19
![y=((4Hx^2 )/L^2 )=Ax^2 A=((4H)/L^2 ) let λ=AL=((4H)/L) h_S =((∫_0 ^(L/2) Ax^2 (√(1+(2Ax)^2 ))dx)/(∫_0 ^(L/2) (√(1+(2Ax)^2 ))dx)) h_S =(1/(4A))×((∫_0 ^(L/2) (2Ax)^2 (√(1+(2Ax)^2 ))d(2Ax))/(∫_0 ^(L/2) (√(1+(2Ax)^2 ))d(2Ax))) h_S =(1/(4A))×((∫_0 ^(AL) t^2 (√(1+t^2 ))dt)/(∫_0 ^(AL) (√(1+t^2 ))dt)) ∫_0 ^λ (√(1+t^2 ))dt=[((ln ((√(1+t^2 ))+t)+t(√(1+t^2 )))/2)]_0 ^λ =((ln ((√(1+λ^2 ))+λ)+λ(√(1+λ^2 )))/2) ∫_0 ^λ t^2 (√(1+t^2 ))dt=[((t(2t^2 +1)(√(1+t^2 ))−ln ((√(1+t^2 ))+t))/8)]_0 ^λ =((λ(2λ^2 +1)(√(1+λ^2 ))−ln ((√(1+λ^2 ))+λ))/8) h_S =(L/(4λ))×((∫_0 ^λ t^2 (√(1+t^2 ))dt)/(∫_0 ^λ (√(1+t^2 ))dt)) =(L/(4λ))×((λ(2λ^2 +1)(√(1+λ^2 ))−ln ((√(1+λ^2 ))+λ))/8)×(2/(ln ((√(1+λ^2 ))+λ)+λ(√(1+λ^2 )))) =(L/(16))×(((2λ^2 +1)(√(1+λ^2 ))−((ln ((√(1+λ^2 ))+λ))/λ))/((√(1+λ^2 ))+((ln ((√(1+λ^2 ))+λ))/λ))) ⇒(h_S /H)=(1/(4λ))[((2(1+λ^2 )^(3/2) )/((√(1+λ^2 ))+((ln ((√(1+λ^2 ))+λ))/λ)))−1]](Q71594.png)
$${y}=\frac{\mathrm{4}{Hx}^{\mathrm{2}} }{{L}^{\mathrm{2}} }={Ax}^{\mathrm{2}} \\ $$$${A}=\frac{\mathrm{4}{H}}{{L}^{\mathrm{2}} } \\ $$$${let}\:\lambda={AL}=\frac{\mathrm{4}{H}}{{L}} \\ $$$${h}_{{S}} =\frac{\int_{\mathrm{0}} ^{\frac{{L}}{\mathrm{2}}} {Ax}^{\mathrm{2}} \sqrt{\mathrm{1}+\left(\mathrm{2}{Ax}\right)^{\mathrm{2}} }{dx}}{\int_{\mathrm{0}} ^{\frac{{L}}{\mathrm{2}}} \sqrt{\mathrm{1}+\left(\mathrm{2}{Ax}\right)^{\mathrm{2}} }{dx}} \\ $$$${h}_{{S}} =\frac{\mathrm{1}}{\mathrm{4}{A}}×\frac{\int_{\mathrm{0}} ^{\frac{{L}}{\mathrm{2}}} \left(\mathrm{2}{Ax}\right)^{\mathrm{2}} \sqrt{\mathrm{1}+\left(\mathrm{2}{Ax}\right)^{\mathrm{2}} }{d}\left(\mathrm{2}{Ax}\right)}{\int_{\mathrm{0}} ^{\frac{{L}}{\mathrm{2}}} \sqrt{\mathrm{1}+\left(\mathrm{2}{Ax}\right)^{\mathrm{2}} }{d}\left(\mathrm{2}{Ax}\right)} \\ $$$${h}_{{S}} =\frac{\mathrm{1}}{\mathrm{4}{A}}×\frac{\int_{\mathrm{0}} ^{{AL}} {t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}}{\int_{\mathrm{0}} ^{{AL}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}} \\ $$$$\int_{\mathrm{0}} ^{\lambda} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\left[\frac{\mathrm{ln}\:\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+{t}\right)+{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}}\right]_{\mathrm{0}} ^{\lambda} \\ $$$$=\frac{\mathrm{ln}\:\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }+\lambda\right)+\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\lambda} {t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\left[\frac{{t}\left(\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{ln}\:\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+{t}\right)}{\mathrm{8}}\right]_{\mathrm{0}} ^{\lambda} \\ $$$$=\frac{\lambda\left(\mathrm{2}\lambda^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\mathrm{ln}\:\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }+\lambda\right)}{\mathrm{8}} \\ $$$${h}_{{S}} =\frac{{L}}{\mathrm{4}\lambda}×\frac{\int_{\mathrm{0}} ^{\lambda} {t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}}{\int_{\mathrm{0}} ^{\lambda} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}} \\ $$$$=\frac{{L}}{\mathrm{4}\lambda}×\frac{\lambda\left(\mathrm{2}\lambda^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\mathrm{ln}\:\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }+\lambda\right)}{\mathrm{8}}×\frac{\mathrm{2}}{\mathrm{ln}\:\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }+\lambda\right)+\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }} \\ $$$$=\frac{{L}}{\mathrm{16}}×\frac{\left(\mathrm{2}\lambda^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\frac{\mathrm{ln}\:\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }+\lambda\right)}{\lambda}}{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }+\frac{\mathrm{ln}\:\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }+\lambda\right)}{\lambda}} \\ $$$$\Rightarrow\frac{{h}_{{S}} }{{H}}=\frac{\mathrm{1}}{\mathrm{4}\lambda}\left[\frac{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }+\frac{\mathrm{ln}\:\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }+\lambda\right)}{\lambda}}−\mathrm{1}\right] \\ $$