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

\(dm=λ.dx=λ_01-\frac{x^2}{L^2}\)

\(X_{cm}=\frac{∫xdm}{∫dm}\)

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

⇒ \(α=8\)