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Question Number 28027 by ajfour last updated on 18/Jan/18

Answered by mrW2 last updated on 19/Jan/18

$$radiusofbasecircle=R$$$$radiusofdiscindistancexfromthe$$$$baseisr=\frac{h-x}{h}\times R=(1-\frac{x}{h})R$$$$massofdiscisdm=\rho \pi r^{2}dx$$$$momentofinertiaofdiscaboutitsowndiameter$$$$is{I}_0=\frac{r^{2}dm}{4},andaboutthediameterof$$$$baseisthendI=x^{2}dm+\frac{r^{2}dm}{4}$$$$I={\int }^{}(x^{2}dm+\frac{r^{2}dm}{4})=\int (x^2+\frac{r^2}{4})dm$$$$={\int }_0^h(x^2+\frac{r^2}{4})\rho \pi r^{2}dx$$$$=\frac{\rho \pi }{4}{\int }_0^h(4x^2+r^2)r^{2}dx$$$$=\frac{\rho \pi }{4}{\int }_0^h[4x^2+{(1-\frac{x}{h})}^2{R}^2]{(1-\frac{x}{h})}^2{R}^{2}dx$$$$=\frac{\rho \pi {hR}^4}{4}{\int }_0^h[\frac{4h^2}{R^2}{\left(\frac{x}{h}\right)}^2+{(1-\frac{x}{h})}^2]{(1-\frac{x}{h})}^{2}d\left(\frac{x}{h}\right)$$$$=\frac{\rho \pi {hR}^4}{4}{\int }_0^1[\frac{4h^2}{R^2}{(1-s)}^2+s^2]s^{2}ds$$$$=\frac{\rho \pi {hR}^4}{4}{\int }_0^1[\frac{4h^2}{R^2}(s^2-2s^3+s^4)+s^4]ds$$$$=\frac{\rho \pi {hR}^4}{4}{[\frac{4h^2}{R^2}(\frac{s^3}{3}-\frac{s^4}{2}+\frac{s^5}{5})+\frac{s^5}{5}]}_0^1$$$$=\frac{\rho \pi {hR}^4}{4}[\frac{4h^2}{R^2}(\frac{1}{3}-\frac{1}{2}+\frac{1}{5})+\frac{1}{5}]$$$$=\frac{\rho \pi {hR}^4}{4}(\frac{2h^2}{15R^2}+\frac{1}{5})$$$$=\frac{\rho \pi {hR}^2(2h^2+3R^2)}{60}$$$$=\frac{\rho \pi {hR}^2}{3}\times \frac{(2h^2+3R^2)}{20}$$$$\Rightarrow I=\frac{M(2h^2+3R^2)}{20}$$

Commented by ajfour last updated on 19/Jan/18

$$GreatsoquickSir!youare$$$$wonderful.Answerisrightsir.$$$$Ihavefolloweditcarefully,Sir.$$$$theintegralis{e}^{x}cellentlyhandled!$$

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