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Question Number 71538 by mr W last updated on 16/Oct/19

Commented by mr W last updated on 16/Oct/19

$$findthepositionofcenterofmass$$$$oftheparabolicarc$$

Answered by ajfour last updated on 17/Oct/19

$$y={Ax}^2$$$$dl=\left(dx\right)\sqrt{1+4A^2{x}^2}$$$$dm=\rho dl$$$$y_{cm}=\frac{{\int }_{-a}^a\left({Ax}^2\right)\rho \sqrt{1+4A^2{x}^2}dx}{{\int }_{-a}^a\rho \sqrt{1+4A^2{x}^2}dx}$$

Answered by mr W last updated on 17/Oct/19

$$y=\frac{4{Hx}^2}{L^2}={Ax}^2$$$$A=\frac{4H}{L^2}$$$$let\lambda =AL=\frac{4H}{L}$$$$h_S=\frac{{\int }_0^{\frac{L}{2}}{Ax}^2\sqrt{1+{\left(2Ax\right)}^2}dx}{{\int }_0^{\frac{L}{2}}\sqrt{1+{\left(2Ax\right)}^2}dx}$$$$h_S=\frac{1}{4A}\times \frac{{\int }_0^{\frac{L}{2}}{\left(2Ax\right)}^2\sqrt{1+{\left(2Ax\right)}^2}d\left(2Ax\right)}{{\int }_0^{\frac{L}{2}}\sqrt{1+{\left(2Ax\right)}^2}d\left(2Ax\right)}$$$$h_S=\frac{1}{4A}\times \frac{{\int }_0^{AL}{t}^2\sqrt{1+t^2}dt}{{\int }_0^{AL}\sqrt{1+t^2}dt}$$$${\int }_0^{\lambda }\sqrt{1+t^2}dt={\left[\frac{\ln (\sqrt{1+t^2}+t)+t\sqrt{1+t^2}}{2}\right]}_0^{\lambda }$$$$=\frac{\ln (\sqrt{1+{\lambda }^2}+\lambda )+\lambda \sqrt{1+{\lambda }^2}}{2}$$$${\int }_0^{\lambda }{t}^2\sqrt{1+t^2}dt={\left[\frac{t(2t^2+1)\sqrt{1+t^2}-\ln (\sqrt{1+t^2}+t)}{8}\right]}_0^{\lambda }$$$$=\frac{\lambda (2{\lambda }^2+1)\sqrt{1+{\lambda }^2}-\ln (\sqrt{1+{\lambda }^2}+\lambda )}{8}$$$$h_S=\frac{L}{4\lambda }\times \frac{{\int }_0^{\lambda }{t}^2\sqrt{1+t^2}dt}{{\int }_0^{\lambda }\sqrt{1+t^2}dt}$$$$=\frac{L}{4\lambda }\times \frac{\lambda (2{\lambda }^2+1)\sqrt{1+{\lambda }^2}-\ln (\sqrt{1+{\lambda }^2}+\lambda )}{8}\times \frac{2}{\ln (\sqrt{1+{\lambda }^2}+\lambda )+\lambda \sqrt{1+{\lambda }^2}}$$$$=\frac{L}{16}\times \frac{(2{\lambda }^2+1)\sqrt{1+{\lambda }^2}-\frac{\ln (\sqrt{1+{\lambda }^2}+\lambda )}{\lambda }}{\sqrt{1+{\lambda }^2}+\frac{\ln (\sqrt{1+{\lambda }^2}+\lambda )}{\lambda }}$$$$\Rightarrow \frac{h_S}{H}=\frac{1}{4\lambda }[\frac{2{(1+{\lambda }^2)}^{\frac{3}{2}}}{\sqrt{1+{\lambda }^2}+\frac{\ln (\sqrt{1+{\lambda }^2}+\lambda )}{\lambda }}-1]$$

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