A uniform rod having a mass of 300 g and a length of 50 cm rotates...

Mass of the rod, $m = 300 \text{ g} = 0.3 \text{ kg}$

Length of the rod, $l = 50 \text{ cm} = 0.5 \text{ m}$

Angular speed, $\omega = 2 \text{ rad/s}$

The moment of inertia of a uniform rod about an axis passing through one end and perpendicular to its length is:

$I = \dfrac{m l^2}{3}$

Substituting the given values:

$I = \dfrac{0.3 \times (0.5)^2}{3} = 0.1 \times 0.25 = 0.025 \text{ kg m}^2$

The rotational kinetic energy of the rod is given by:

$K = \dfrac{1}{2} I \omega^2$

Substituting the values of $I$ and $\omega$:

$K = \dfrac{1}{2} \times 0.025 \times (2)^2$

$K = 0.025 \times 2 = 0.05 \text{ J}$