Energy of Rod about pivot

Consider a rod, with mass $m$ and length $l$, that is fixed about a pivot point at one of its ends.

Now imagine that we flick the rod so that the end of it attains a velocity of $v_{0}$.

Now, if we look at the instant of impact, the rod appears to be moving towards the right with a velocity of $v_{0}$ or with a kinetic energy of

$KE=frac{1}{2}mv_{0}^2$

(I draw upon this idea as I often use this technique to find the velocity of a rod after a collision using conservation of momentum)

However, if we calculate the kinetic energy of the rod with respect to its moment of inertia and angular velocity, we attain a different answer:

$KE = frac{1}{2}Iomega^2$

$KE = frac{1}{2}frac{1}{3}ml^2omega^2$ (moment of inertia of a rod about a fixed end is $frac{1}{3}ml^2$)

$KE = frac{1}{2}frac{1}{3}ml^2(frac{v}{l})^2$

$KE = frac{1}{6}mv^2$

However, these two values do not equal each other. What went wrong? Is it incorrect to imagine the rod as not being attached to the pivot for an instant? If so, why does conservation of linear momentum still hold when using this trick?