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 $\frac{3 L }{\alpha} m$. The value of $\alpha$ is .....(where $x$ is the distance form end $A$)

Answer 1

$dm =\lambda \cdot dx =\lambda_0\left(1-\frac{ x ^2}{\ell^2}\right)$
$X _{ cm }=\frac{\int xdm }{\int dm _{\ell}}$
$=\frac{\lambda_0 \int\limits_0^{\ell}\left( x -\frac{ x ^3}{\ell^2}\right) dx }{\ell}=\frac{\frac{\ell^2}{2}-\frac{\ell^4}{4 \ell^2}}{\int\limits_0^{\ell} \lambda_0\left(1-\frac{ x ^2}{\ell^2}\right) dx }=\frac{3 \ell}{8}$