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Question Number 144554 by imjagoll last updated on 26/Jun/21

Answered by Olaf_Thorendsen last updated on 26/Jun/21

$$a)\mathrm{\Delta }=\left[\mathrm{O}x\right)$$$${\mathrm{I}}_{\mathrm{\Delta }}=\int r^{2}dm=\int r^2\delta dS$$$${\mathrm{I}}_{\mathrm{\Delta }}=\delta {\int }_0^2{y}^2(1-\frac{1}{2}y)dy$$$${\mathrm{I}}_{\mathrm{\Delta }}=\delta {[\frac{y^3}{3}-\frac{1}{4}{y}^2]}_0^1$$$${\mathrm{I}}_{\mathrm{\Delta }}=\delta (\frac{1}{3}-\frac{1}{4})=\frac{\delta }{12}=0,25gm.{cm}^2$$$$b)\mathrm{M}=\delta \times \mathrm{S}=3\times \left(\frac{1}{2}\times 1\times 2\right)=3gm$$$$c)y_{\mathrm{G}}=\frac{1}{\mathrm{M}}{\int }_{S}ydm$$$$y_{\mathrm{G}}=\frac{1}{\mathrm{M}}{\int }_{S}y\left(\delta ydx\right)$$$$y_{\mathrm{G}}={\int }_0^{1}2x\left(2xdx\right)$$$$y_{\mathrm{G}}={\int }_0^{1}4x^{2}dx$$$$y_{\mathrm{G}}={\left[4\frac{x^3}{3}\right]}_0^1=\frac{4}{3}$$

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