For which of the following situations is the centre of mass of a...

Let the rod be placed along the x-axis from $x = -L/2$ to $x = L/2$. The geometric centre is at $x = 0$.

The position of the centre of mass is given by $x_{cm} = \dfrac{\int_{-L/2}^{L/2} x \lambda(x) dx}{\int_{-L/2}^{L/2} \lambda(x) dx}$.

If the density $\lambda(x)$ increases continuously from left to right, then $\lambda(x) > \lambda(-x)$ for all $x > 0$.

The numerator of $x_{cm}$ can be written as $\int_{0}^{L/2} x (\lambda(x) - \lambda(-x)) dx$. Since $\lambda(x) > \lambda(-x)$ for $x > 0$, this integral is strictly positive, meaning $x_{cm} > 0$. Thus, the centre of mass is definitively to the right of the geometric centre.

Similarly, if the density decreases continuously from left to right, $\lambda(x) < \lambda(-x)$ for all $x > 0$. The integral becomes strictly negative, meaning $x_{cm} < 0$. Thus, the centre of mass is definitively to the left of the geometric centre.

For the cases where density increases or decreases up to the centre and then reverses its trend, the density distribution can be symmetric about the centre (i.e., $\lambda(x) = \lambda(-x)$). In such symmetric cases, the integral is zero, and the centre of mass is exactly at the geometric centre. Thus, it is not definitively away from the geometric centre.