Question Number 26799 by 803jaideep@gmail.com last updated on 29/Dec/17
Answered by mrW1 last updated on 29/Dec/17
![I=∫d^2 dm dm=ρdV=ρrdrdθdz d^2 =z^2 +(rcos θ)^2 =z^2 +r^2 cos^2 θ I=∫_(−(L/2)) ^( (L/2)) ∫_0 ^( 2π) ∫_0 ^( R) (z^2 +r^2 cos^2 θ)ρrdrdθdz =2ρ∫_0 ^( (L/2)) ∫_0 ^( 2π) ∫_0 ^( R) (z^2 +r^2 cos^2 θ)rdrdθdz =2ρ∫_0 ^( (L/2)) ∫_0 ^( 2π) (z^2 ×(R^2 /2)+(R^4 /4)cos^2 θ)dθdz =((ρR^2 )/2)∫_0 ^( (L/2)) ∫_0 ^( 2π) (2z^2 +R^2 cos^2 θ)dθdz =((ρR^2 )/2)∫_0 ^( (L/2)) ∫_0 ^( 2π) (2z^2 +(R^2 /2)+(R^2 /2)cos 2θ)dθdz =((ρR^2 )/2)∫_0 ^( (L/2)) (2z^2 +(R^2 /2))2πdz =πρR^2 ∫_0 ^( (L/2)) (2z^2 +(R^2 /2))dz =πρR^2 [(2/3)((L/2))^3 +(R^2 /2)×(L/2)] =πρR^2 L×(1/(12))(L^2 +3R^2 ) =(M/(12))(L^2 +3R^2 )](Q26802.png)
$${I}=\int{d}^{\mathrm{2}} {dm} \\ $$$${dm}=\rho{dV}=\rho{rdrd}\theta{dz} \\ $$$${d}^{\mathrm{2}} ={z}^{\mathrm{2}} +\left({r}\mathrm{cos}\:\theta\right)^{\mathrm{2}} ={z}^{\mathrm{2}} +{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta \\ $$$${I}=\int_{−\frac{{L}}{\mathrm{2}}} ^{\:\frac{{L}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \int_{\mathrm{0}} ^{\:{R}} \left({z}^{\mathrm{2}} +{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta\right)\rho{rdrd}\theta{dz} \\ $$$$=\mathrm{2}\rho\int_{\mathrm{0}} ^{\:\frac{{L}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \int_{\mathrm{0}} ^{\:{R}} \left({z}^{\mathrm{2}} +{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta\right){rdrd}\theta{dz} \\ $$$$=\mathrm{2}\rho\int_{\mathrm{0}} ^{\:\frac{{L}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \left({z}^{\mathrm{2}} ×\frac{{R}^{\mathrm{2}} }{\mathrm{2}}+\frac{{R}^{\mathrm{4}} }{\mathrm{4}}\mathrm{cos}^{\mathrm{2}} \:\theta\right){d}\theta{dz} \\ $$$$=\frac{\rho{R}^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\:\frac{{L}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \left(\mathrm{2}{z}^{\mathrm{2}} +{R}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta\right){d}\theta{dz} \\ $$$$=\frac{\rho{R}^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\:\frac{{L}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \left(\mathrm{2}{z}^{\mathrm{2}} +\frac{{R}^{\mathrm{2}} }{\mathrm{2}}+\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta{dz} \\ $$$$=\frac{\rho{R}^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\:\frac{{L}}{\mathrm{2}}} \left(\mathrm{2}{z}^{\mathrm{2}} +\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\right)\mathrm{2}\pi{dz} \\ $$$$=\pi\rho{R}^{\mathrm{2}} \int_{\mathrm{0}} ^{\:\frac{{L}}{\mathrm{2}}} \left(\mathrm{2}{z}^{\mathrm{2}} +\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\right){dz} \\ $$$$=\pi\rho{R}^{\mathrm{2}} \left[\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{{L}}{\mathrm{2}}\right)^{\mathrm{3}} +\frac{{R}^{\mathrm{2}} }{\mathrm{2}}×\frac{{L}}{\mathrm{2}}\right] \\ $$$$=\pi\rho{R}^{\mathrm{2}} {L}×\frac{\mathrm{1}}{\mathrm{12}}\left({L}^{\mathrm{2}} +\mathrm{3}{R}^{\mathrm{2}} \right) \\ $$$$=\frac{{M}}{\mathrm{12}}\left({L}^{\mathrm{2}} +\mathrm{3}{R}^{\mathrm{2}} \right) \\ $$
Commented by mrW1 last updated on 29/Dec/17
Commented by 803jaideep@gmail.com last updated on 30/Dec/17
$$\mathrm{wow}\:\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot} \\ $$