$$E_{rotational} = \frac1 2 I*\omega^2$$

Consider a particle with mass $m$ revolving about a pivot $O$ with a constant distance $r$ between them. It is revolving with constant angular velocity $\omega$ Its moment of inertia is: $$I = mr^2$$ Thus $$E_{rotational} = \frac1 2 mr^2*\omega^2$$

Since $\omega = \frac v r$

$$E{rotational} = \frac1 2 mr^2\frac{v^2}{r^2}$$ $$E{rotational} = \frac1 2 m*v^2$$

We could see this looks exactly as the Translational Kinetic Energy, which is:

Of an any given moment, the instantaneous velocity of this particle is $v$, so we can calculate its Translational Kinetic Energy with the formula

$$E_{translational} = \frac1 2 m*v^2$$