Question Number 144554 by imjagoll last updated on 26/Jun/21
Answered by Olaf_Thorendsen last updated on 26/Jun/21
![a) Δ = [Ox) I_Δ = ∫r^2 dm = ∫r^2 δdS I_Δ = δ∫_0 ^2 y^2 (1−(1/2)y)dy I_Δ = δ[(y^3 /3)−(1/4)y^2 ]_0 ^1 I_Δ = δ((1/3)−(1/4)) = (δ/(12)) = 0,25 gm.cm^2 b) M = δ×S = 3×((1/2)×1×2) = 3 gm c) y_G = (1/M)∫_S y dm y_G = (1/M)∫_S y (δydx) y_G = ∫_0 ^1 2x (2xdx) y_G = ∫_0 ^1 4x^2 dx y_G = [4(x^3 /3)]_0 ^1 = (4/3)](https://www.tinkutara.com/question/Q144590.png)
$$\left.{a}\right)\:\Delta\:=\:\left[\mathrm{O}{x}\right) \\ $$$$\mathrm{I}_{\Delta} \:=\:\int{r}^{\mathrm{2}} {dm}\:=\:\int{r}^{\mathrm{2}} \delta{dS} \\ $$$$\mathrm{I}_{\Delta} \:=\:\delta\int_{\mathrm{0}} ^{\mathrm{2}} {y}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{y}\right){dy} \\ $$$$\mathrm{I}_{\Delta} \:=\:\delta\left[\frac{{y}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}{y}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\mathrm{I}_{\Delta} \:=\:\delta\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\:\frac{\delta}{\mathrm{12}}\:=\:\mathrm{0},\mathrm{25}\:{gm}.{cm}^{\mathrm{2}} \\ $$$$\left.{b}\right)\:\mathrm{M}\:=\:\delta×\mathrm{S}\:=\:\mathrm{3}×\left(\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\mathrm{2}\right)\:=\:\mathrm{3}\:{gm} \\ $$$$\left.{c}\right)\:{y}_{\mathrm{G}} \:=\:\frac{\mathrm{1}}{\mathrm{M}}\int_{{S}} {y}\:{dm} \\ $$$${y}_{\mathrm{G}} \:=\:\frac{\mathrm{1}}{\mathrm{M}}\int_{{S}} {y}\:\left(\delta{ydx}\right) \\ $$$${y}_{\mathrm{G}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{x}\:\left(\mathrm{2}{xdx}\right) \\ $$$${y}_{\mathrm{G}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} {dx} \\ $$$${y}_{\mathrm{G}} \:=\:\left[\mathrm{4}\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$