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Question Number 26799 by 803jaideep@gmail.com last updated on 29/Dec/17

Answered by mrW1 last updated on 29/Dec/17

$$I=\int d^{2}dm$$$$dm=\rho dV=\rho rdrd\theta dz$$$$d^2=z^2+{\left(r\cos \theta \right)}^2=z^2+r^2{\cos }^2\theta$$$$I={\int }_{-\frac{L}{2}}^{\frac{L}{2}}{\int }_0^{2\pi }{\int }_0^R(z^2+r^2{\cos }^2\theta )\rho rdrd\theta dz$$$$=2\rho {\int }_0^{\frac{L}{2}}{\int }_0^{2\pi }{\int }_0^R(z^2+r^2{\cos }^2\theta )rdrd\theta dz$$$$=2\rho {\int }_0^{\frac{L}{2}}{\int }_0^{2\pi }(z^2\times \frac{R^2}{2}+\frac{R^4}{4}{\cos }^2\theta )d\theta dz$$$$=\frac{\rho R^2}{2}{\int }_0^{\frac{L}{2}}{\int }_0^{2\pi }(2z^2+R^2{\cos }^2\theta )d\theta dz$$$$=\frac{\rho R^2}{2}{\int }_0^{\frac{L}{2}}{\int }_0^{2\pi }(2z^2+\frac{R^2}{2}+\frac{R^2}{2}\cos 2\theta )d\theta dz$$$$=\frac{\rho R^2}{2}{\int }_0^{\frac{L}{2}}(2z^2+\frac{R^2}{2})2\pi dz$$$$=\pi \rho R^2{\int }_0^{\frac{L}{2}}(2z^2+\frac{R^2}{2})dz$$$$=\pi \rho R^2[\frac{2}{3}{\left(\frac{L}{2}\right)}^3+\frac{R^2}{2}\times \frac{L}{2}]$$$$=\pi \rho R^{2}L\times \frac{1}{12}(L^2+3R^2)$$$$=\frac{M}{12}(L^2+3R^2)$$

Commented by mrW1 last updated on 29/Dec/17

Commented by 803jaideep@gmail.com last updated on 30/Dec/17

$$\mathrm{wow}\mathrm{thanks}\mathrm{a}\mathrm{lot}$$

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