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Question Number 105073 by Dwaipayan Shikari last updated on 25/Jul/20

$$Findthecentreofthemassofaconeofheight{}^{\prime }{H}^{\prime }andRadius$$$${}^{\prime }{R}^{\prime }$$

Commented by ajfour last updated on 25/Jul/20

$$solidconeorhollow?$$

Commented by Dwaipayan Shikari last updated on 25/Jul/20

$$Solidsir$$

Commented by mr W last updated on 25/Jul/20

Commented by mr W last updated on 25/Jul/20

$$forasolidconethedistributionof$$$$massisaparabolaofsecondorder,$$$$therefore{y}_S=\frac{H}{4}.$$

Answered by ajfour last updated on 25/Jul/20

Commented by ajfour last updated on 25/Jul/20

$$y_{cone}=\frac{{\int }_0^h\rho (h-y)\pi r^{2}dy}{\rho \left(\frac{1}{3}\right)\pi R^{2}h}$$$$\frac{r}{y}=\frac{R}{h}\Rightarrow r^2={\left(\frac{R}{h}\right)}^2{y}^2$$$$y_{cone}=\frac{{\int }_0^h\rho (h-y)\pi {\left(\frac{R}{h}\right)}^2{y}^{2}dy}{\rho \left(\frac{1}{3}\right)\pi R^{2}h}$$$$=\frac{3}{h^3}{\int }_0^h({hy}^2-y^3)dy$$$$=\frac{3}{h^3}(\frac{{hy}^3}{3}-\frac{y^4}{4}){\mid }_0^h$$$$y_{cone}=\frac{3}{h^3}(\frac{h^4}{3}-\frac{h^4}{4})=\frac{h}{4}.$$$$$$

Commented by Dwaipayan Shikari last updated on 25/Jul/20

$$Kindlycanyoutellmehowtodrawsuchdiagrams?⊛$$

Commented by ajfour last updated on 25/Jul/20

$$iuselekhdiagramapp.$$

Commented by Dwaipayan Shikari last updated on 25/Jul/20

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