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$$^{-1}$$ and length $$L$$ (in meter) is $$\frac{3L}{\alpha}$$ m. The value of $$\alpha$$ is _____ (where $$x$$ is the distance from end A)

We have a one-dimensional rod AB of length $$L$$ with mass density $$\rho = \rho_0\left(1 - \frac{x^2}{L^2}\right)$$ kg/m, where $$x$$ is measured from end A. We need to find the distance of the centre of mass from end A.

The centre of mass is given by $$x_{\text{cm}} = \frac{\int_0^L x \, \rho \, dx}{\int_0^L \rho \, dx}$$.

First, let us compute the denominator (total mass). We have $$\int_0^L \rho_0\left(1 - \frac{x^2}{L^2}\right) dx = \rho_0\left[x - \frac{x^3}{3L^2}\right]_0^L = \rho_0\left(L - \frac{L}{3}\right) = \frac{2\rho_0 L}{3}$$.

Now for the numerator, $$\int_0^L x \cdot \rho_0\left(1 - \frac{x^2}{L^2}\right) dx = \rho_0\int_0^L \left(x - \frac{x^3}{L^2}\right) dx = \rho_0\left[\frac{x^2}{2} - \frac{x^4}{4L^2}\right]_0^L = \rho_0\left(\frac{L^2}{2} - \frac{L^2}{4}\right) = \frac{\rho_0 L^2}{4}$$.

So the centre of mass position is $$x_{\text{cm}} = \frac{\rho_0 L^2 / 4}{2\rho_0 L / 3} = \frac{L^2}{4} \times \frac{3}{2L} = \frac{3L}{8}$$.

Comparing with the given expression $$\frac{3L}{\alpha}$$, we get $$\alpha = 8$$.

Hence, the value of $$\alpha$$ is $$\textbf{8}$$.