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Many students have asked about when to include center of mass motion when calculating angular momentum and energy.  In this page, we give a discussion of some of the main issues.

h4. Issue 1:  Pure Rotation and the Parallel Axis Theorem

h5. Recognizing Pure Rotation

One key idea in determining how to describe rotational problems is recognizing whether the problem can be treated as pure rotation.  Whenever an object is spinning around a fixed axle, we can *choose* to describe that motion as pure rotation about the axle, *even if the center of mass is not located at the axle*.  

h5. Two Descriptions of Pure Rotation:  The Parallel Axis Theorem vs. Rotation and Translation

An object moving in pure rotation about an axle can be described in *two different but equivalent ways*.  

# The object is *only rotating*, but the moment of inertia about the fixed axle must be found using the parallel axis theorem. 
# The object is *rotating about the center of mass and its center of mass is translating*.  

To see why these descriptions are equivalent, consider how we would find the angular momentum and kinetic energy of a bar pivoted at one end.

h6. Method 1:  The bar is purely rotating.  We do not include any contribution from the CM motion.

{latex}\begin{large}\[ L = I\omega = \frac{1}{3}mL^{2}\omega \]\end{large}{latex}\\{latex}\begin{large}\[ K = \frac{1}{2}I\omega^{2} = \frac{1}{6}mL^{2}\omega^{2}\]\end{large}{latex}

h6. Method 2:  The bar is rotating about its CM and the CM is translating. 

{latex}\begin{large}\[ L = \vec{R}\times m\vec{v}_{CM} + I_{CM}\omega = \frac{L}{2}mv_{CM} + \frac{1}{12}mL^{2}\omega \]\end{large}{latex}\\{latex}\begin{large}\[ K = \frac{1}{2}mv_{CM}^{2} + \frac{1}{2}I_{CM}\omega^{2} \]\end{large}{latex}

To see that Method 2 is equivalent to Method 1, we must realize that the center of mass speed is related to the rotational velocity of the bar by the distance of the CM from the axis of rotation:

{latex}\begin{large}\[ \omega \frac{L}{2} = v_{CM} \]\end{large}{latex}

You can now substitute this into the expressions for L and K:

{latex}\begin{large}\[ L = \left(\frac{L}{2}\right)^{2} m\omega + \frac{1}{12}mL^{2} \omega = \frac{1}{3}mL^{2}\omega \]\end{large}{latex}\\{latex}\begin{large}\[ K = \frac{1}{2}m\left(\frac{L}{2}\right)^{2} \omega^{2} + \frac{1}{2} \frac{1}{12}mL^{2}\omega^{2} = \frac{1}{6}mL^{2}\omega^{2}\]\end{large}{latex}

which are equivalent to the expressions found in Method 1.

h5. Parallel Axis Theorem is Preferred

Generally it is advantageous use the parallel axis theorem and consider the motion to be pure rotation.  The problem of the rod pivoted about one end is a good example of why.  If you are asked to find the angular acceleration of the rod or to find the frequency of small oscillations, you will need to compute the torque around the pivot and set that torque equal to Iα.  The appropriate moment of inertia to use is the moment of inertia about the end of the rod.

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h4. Issue 2:  Rotation andplus Translation and External vs.or CM AxesAxis

h5.