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

Given : Linear mass density \( \rho = \rho_0\left(1-\frac{x^2}{L^2}\right) \)

Centre of mass

\(\begin{array}{l} x_{cm} = \frac{\int x\,dm}{\int dm} \end{array}\)

\(\begin{array}{l} dm = \rho\,dx = \rho_0\left(1-\frac{x^2}{L^2}\right)dx \end{array}\)

Denominator :

\(\begin{array}{l} \int_0^L dm = \rho_0 \int_0^L \left(1-\frac{x^2}{L^2}\right)dx \end{array}\)

\(\begin{array}{l} = \rho_0 \left[x - \frac{x^3}{3L^2}\right]_0^L = \rho_0 \left(L - \frac{L}{3}\right) = \frac{2\rho_0 L}{3} \end{array}\)

Numerator :

\(\begin{array}{l} \int_0^L x\,dm = \rho_0 \int_0^L x\left(1-\frac{x^2}{L^2}\right)dx \end{array}\)

\(\begin{array}{l} = \rho_0 \left[\frac{x^2}{2} - \frac{x^4}{4L^2}\right]_0^L = \rho_0 \left(\frac{L^2}{2} - \frac{L^2}{4}\right) = \frac{\rho_0 L^2}{4} \end{array}\)

Thus,

\(\begin{array}{l} x_{cm} = \frac{\frac{\rho_0 L^2}{4}}{\frac{2\rho_0 L}{3}} = \frac{3L}{8} \end{array}\)

Given \(x_{cm} = \frac{3L}{\alpha}\)

\(\begin{array}{l} \frac{3L}{\alpha} = \frac{3L}{8} \Rightarrow \alpha = 8 \end{array}\)

Hence, \(\alpha = \mathbf{8}\).