Question

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

Answer 1

Answer: 8

$dm=\lambda \cdot dx={\lambda }_0\left(1-\frac{x^2}{{\ell }^2}\right)$
$X_{cm}=\frac{\int xdm}{\int d{m}_{\ell }}$
$=\frac{{\lambda }_0\underset{0}{\overset{\ell }{\int }}\left(x-\frac{x^3}{{\ell }^2}\right)dx}{\ell }=\frac{\frac{{\ell }^2}{2}-\frac{{\ell }^4}{4{\ell }^2}}{\underset{0}{\overset{\ell }{\int }}{\lambda }_0\left(1-\frac{x^2}{{\ell }^2}\right)dx}=\frac{3\ell }{8}$