A thin uniform rod of length $$2$$ m, cross sectional area $$A$$ and density $$d$$ is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $$\omega$$. If value of $$\omega$$ in terms of its rotational kinetic energy $$E$$ is $$\sqrt{\frac{\alpha E}{Ad}}$$, then the value of $$\alpha$$ is ______.

A thin uniform rod of length $$L = 2$$ m, cross-sectional area $$A$$, and density $$d$$ rotates about its center. To determine the constant $$\alpha$$ in the relation $$\omega = \sqrt{\frac{\alpha E}{Ad}}$$, we first compute the mass of the rod and its moment of inertia about the center.

The mass of the rod is given by $$m = A \times d \times L = 2Ad$$. Substituting this into the standard formula for the moment of inertia of a rod about its center, we have $$I = \frac{mL^2}{12} = \frac{2Ad \times 4}{12} = \frac{2Ad}{3}$$.

Next, we express the rotational kinetic energy in terms of the angular speed $$\omega$$: $$E = \frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2Ad}{3} \times \omega^2 = \frac{Ad\omega^2}{3}$$. Solving for $$\omega^2$$ gives $$\omega^2 = \frac{3E}{Ad}$$, so that $$\omega = \sqrt{\frac{3E}{Ad}}$$.

By comparing this result with the given form $$\omega = \sqrt{\frac{\alpha E}{Ad}}$$, it follows that $$\alpha = \boxed{3}$$.