The variation of density of a cylindrical thi | vedclass

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(D) The center of mass $x_{cm}$ for a continuous body is given by $x_{cm} = \frac{\int x dm}{\int dm}$.Given the density $\rho(x) = \rho_0 \frac{x^2}{

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$3L/4$

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(D) The center of mass $x_{cm}$ for a continuous body is given by $x_{cm} = \frac{\int x dm}{\int dm}$.Given the density $\rho(x) = \rho_0 \frac{x^2}{L^2}$,the mass element $dm$ for a cross-sectional area $A$ is $dm = \rho(x) A dx = \rho_0 \frac{x^2}{L^2} A dx$.The total mass $M = \int_0^L dm = \int_0^L \rho_0 \frac{x^2}{L^2} A dx = \frac{\rho_0 A}{L^2} \left[ \frac{x^3}{3} \right]_0^L = \frac{\rho_0 A L}{3}$.The moment of mass about the origin is $\int_0^L x dm = \int_0^L x \left( \rho_0 \frac{x^2}{L^2} A \right) dx = \frac{\rho_0 A}{L^2} \int_0^L x^3 dx = \frac{\rho_0 A}{L^2} \left[ \frac{x^4}{4} \right]_0^L = \frac{\rho_0 A L^2}{4}$.Therefore,$x_{cm} = \frac{\frac{\rho_0 A L^2}{4}}{\frac{\rho_0 A L}{3}} = \frac{3L}{4}$.

The variation of density of a cylindrical thick and long rod is $\rho = \rho_0 \frac{x^2}{L^2}$. The position of its centre of mass from the $x = 0$ end is:

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