Question

The distance of centre of mass from end $A$ of a one dimensional rod ($AB$) having mass density $ρ=ρ_0(1-\frac{x^2}{L^2})kg/m$ and length L. (in meter) is $\frac{3L}{α} m$. The value of $α$ is......... (where $x$ is the distance form end $A$)

Answer 1

$dm=λ.dx=λ_01-\frac{x^2}{L^2}$

$X_{cm}=\frac{∫xdm}{∫dm}$

=\frac{λ_0\int\limits_0^Lx-\frac{x^3}{L^2}dx}{\int\limits_0^Lλ_01-\frac{x^2}{L^2}dx}=\frac{\frac{L^2}{2}-\frac{L^4}{4L^2}}{L-\frac{L^3}{3L^2}}$$=\frac{3L}{8}

⇒ $α=8$