Kinetic and rotational kinetic energy When a body is rotating on its own axis, and at the same time it is moving, does it possess both KE and RKE?
So consider the case of the moon. The moon rotates on its own axis, and at the same time rotates around the Earth. So the moon posses 2 different types of RKE? Or RKE and KE? Or RKE And RKE and KE? I'm getting confused here.
 A: RKE is just a convenient way to describe KE in rotational systems.  We have $1/2I\omega^2$ and $1/2mv^2$.  These are equivalent, it just depends on what is convenient we have.  Let's model the Moon orbiting the Earth as a point in orbit.
$I = mr^2$ for this orbit
We also note $v = \omega r$ therefore $\omega = \frac{v}{r}$
Taking $1/2I\omega^2$ We put in our expression for $I$
$1/2mr^2\omega^2$
and then put in our expression for omega
$1/2mv^2$
Which is the expression for KE.  So it is apparent that KE and RKE are equivalent.
The same is true for the moon rotating about its own axis, however, it is much harder to show because the velocity is different for a piece of mass on the surface of the moon and a piece of mass closer to the center.  You could integrate this and calculate the energy using $1/2mv^2$, however, it is much easier to look up $I$ and $\omega$. and calculate using $1/2I\omega^2$.
The moon therefore has two types of RKE or two types of KE or one KE and one RKE depending on how you want to look at it.  All are equivalent, but some are easier to calculate.
