The distance of centre of mass from end $A$ of a one dimensional rod $\left(AB\right)$ having mass density $\rho ={\rho }_0\left(1-\frac{x^2}{L^2}\right) \mathrm{kg} {\mathrm{m}}^{-1}$ and length $L$ (in meter) is $\frac{3L}{\alpha } \mathrm{m}$. The value of $\alpha$ is _____ (where $x$ is the distance form end $A$)

Given here:

Mass density, $\rho ={\rho }_0\left(1-\frac{x^2}{L^2}\right) \mathrm{kg} {\mathrm{m}}^{-1}$

Let us suppose there is an element of length $dx$ at a distance $x$ from left must end.

Mass of this element is $dm=\rho dx={\rho }_0\left(1-\frac{x^2}{L^2}\right)$

The centre of mass of this element is ${\mathrm{X}}_{\mathrm{cm}}=\frac{\int \mathrm{xdm}}{\int \mathrm{dm}}$

$=\frac{{\rho }_0{\int }_0^l\left(x-\frac{x^2}{L^2}\right)xdx}{{\int }_0^l{\rho }_0\left(1-\frac{x^2}{L^2}\right)dx}$

$=\frac{\frac{L^2}{2}-\frac{L^4}{4L^2}}{L-\frac{L^3}{3L^2}}=\frac{{\displaystyle \frac{L^2}{4}}}{{\displaystyle \frac{2L}{3}}}=\frac{3L}{8}$

Hence, the value of $\alpha =8$.