So again I have two equations:

$$K= \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$


$$K_w = \frac{1}{2} I \omega^2 $$

What's the difference between these two? Thanks.

  • 1
    $\begingroup$ Easy, $\frac{1}{2} m v^2$. Seriously though, the terms are for kinetic energy and rotational energy. The latter is still kinetic energy, just the kinetic energy due to the infinitesimal part of the object moving around the axis of rotation. $\endgroup$ – Alan Rominger Apr 23 '14 at 3:00
  • $\begingroup$ @AlanSE I'd suggest posting that as an answer $\endgroup$ – David Z Apr 23 '14 at 3:05

When you have a sphere, let's say, rolling down a ramp, the gravitational potential energy will be converted into two energies: Rotational kinetic energy and translational kinetic energy. You use energy to keep it spinning (AKA moving angularly) in addition to keeping it moving translationally. Therefore, the sphere's total energy at the bottom of the ramp will be the first equation you labeled: $K=\frac12 mv^2 + \frac12 I\omega^2$, the addition of translational and rotational kinetic energy. Of course the second equation is the rotational kinetic energy, $K_w=\frac12 I\omega^2$. This alone is the energy that moves it angularly. If you had just that, the ball would be spinning. If you had only $\frac12 mv^2$, on the other hand, the ball would be just moving without spinning.

This is basically it. At least as much as you need to know for your level.


If you are looking for an intuitive understanding think of it in terms of the two types of motion: translation (depends on $v$) and rotation (depends on $\omega$). An object can have either or both types of motions at a given time. Your first equation is more general; it has both translation and rotation.


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