# Rotational kinetic energy during vertical circular motion of a particle

Why is it not necessary to take into account rotational kinetic energy when using the Law of Conservation of Mechanical Energy to solve vertical circular motion problems? After all, the particle is rotating about the centre of the circle and does have rotational KE, doesn't it?

All the examples I have seen just use KE= 1/2 $mv^2$, e.g. here: http://www.physicsforums.com/showpost.php?p=2312566&postcount=4

-
I think I figured it out. 1/2 $mv^2$ in the circular motion examples clearly does refer to the rotational kinetic energy of the particle, not its translational kinetic energy, since $v$ here is the particle's tangential speed (tangential to the circle). The confusion arose due to 1/2 $mv^2$ being expressed in linear form instead of angular form 1/2 $mr^2ω^2$. In short, while any quantity (e.g. velocity,momentum,KE) expressed in angular form necessarily describes rotational motion, when expressed in linear form, the quantity may be describing either translational or rotational motion. –  Ryan Apr 27 '12 at 5:09

For a point particle, Translational KE is rotational KE:

$$\frac12I\omega^2=\frac12mr^2\omega=\frac12mv^2$$

The formula for rotational KE ($\frac12I\omega^2$) is derived by adding up the KE of each particle in a rigid body in pure rotation. When body has both rotation and translation, we can derive that:

$$KE=KE_R+KE_T=\frac12I_{com}\omega_{com}^2+\frac12mv_{com}^2$$

In this case, the point particle has no $\omega$ about its center of mass, so no problem. Though we can still apply the pure rotation formula.

Just that we can't say it has both rotational and translational mortion.

-
Thanks, Manishearth. –  Ryan Apr 27 '12 at 5:11

When dealing with point particles, rather than bulk (extended) matter, there is no need for the concept of angular kinetic energy, (regular) $\frac{1}{2}mv^2$ kinetic energy (in addition to potential energy) is the relevant conserved quantity. More accurately, when we deal with bulk matter in classical kinematics problems, we ignore the internal forces between the particles that constitute the matter, and consider their orientation fixed. This lets us separate the energy deriving from the motion of the center of mass from the rotation of that mass about it's center. In this case you have two terms for the energy because you have two parameters for it's "speed", velocity of the center of mass, and rate of angular rotation.

-
Thanks your your answer! BTW, after having answered my own question (see my comment in the main question), I now dislike the term "angular kinetic energy" because of its analogue "linear kinetic energy", which refers to non-rotational motion. But if one were careless like me, one would be quick to conclude that 1/2 $mv^2$ (where $v$ = linear velocity) also necessarily refers to non-rotational motion, which would be a mistake and a source of momentary confusion. –  Ryan Apr 27 '12 at 5:29