The variation of density of a solid cylindric | vedclass

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(D) The density of the rod varies as $\rho = \rho_0 \frac{x^2}{L^2}$.Consider a small elemental disc of thickness $dx$ at a distance $x$ from the end

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$3L/4$

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(D) The density of the rod varies as $\rho = \rho_0 \frac{x^2}{L^2}$.Consider a small elemental disc of thickness $dx$ at a distance $x$ from the end $x=0$.The mass of this element is $dm = \rho \cdot dV = \rho \cdot (\alpha dx) = \left( \rho_0 \frac{x^2}{L^2} \right) \alpha dx$.The position of the centre of mass $X_{cm}$ is given by the formula:$X_{cm} = \frac{\int x dm}{\int dm}$Substituting the values:$X_{cm} = \frac{\int_0^L x \left( \rho_0 \frac{x^2}{L^2} \alpha dx \right)}{\int_0^L \left( \rho_0 \frac{x^2}{L^2} \alpha dx \right)}$$X_{cm} = \frac{\frac{\rho_0 \alpha}{L^2} \int_0^L x^3 dx}{\frac{\rho_0 \alpha}{L^2} \int_0^L x^2 dx}$$X_{cm} = \frac{[x^4/4]_0^L}{[x^3/3]_0^L} = \frac{L^4/4}{L^3/3} = \frac{3}{4} L$.

The variation of density of a solid cylindrical rod of cross-sectional area $\alpha$ and length $L$ is given by $\rho = \rho_0 \frac{x^2}{L^2}$, where $x$ is the distance from one end of the rod. The position of its centre of mass from that end $(x=0)$ is:

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