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{excerpt}A relationship between the moment of inertia of a rigid body about an axis passing through the body's center of mass and the moment of inertia about any parallel axis.{excerpt}

h4. Statement of the Theorem

The parallel axis theorem states that if the moment of inertia of a rigid body about an axis passing through the body's center of mass is _I_~cm~ then the moment of inertia of the body about any parallel axis can be found by evaluating the sum:

{latex}\begin{large}\[ I_{||} = I_{cm} + Md^{2} \] \end{large}{latex}

where _d_ is the (perpendicular) distance between the original center of mass axis and the new parallel axis.

h4. Motivation for the Theorem

We know that the [angular momentum about a single axis] of a [rigid body] that is translating and rotating with respect to a (non-accelerating) axis can be written:

{latex}\begin{large}\[ L = m\vec{r}_{\rm cm,axis}\times\vec{v}_{cm} + I_{cm}\omega_{cm} \]\end{large}{latex}

Now suppose that the rigid body is executing pure rotation about the axis.  In that case, the velocity of the center of mass will be perpendicular to the displacement vector from the axis of rotation to the center of mass.  Calling the magnitude of that displacement _d_, to make contact with the form of the theorem, we then have:

{latex}\begin{large}\[ L \mbox{ (pure rotation)} = mv_{cm}d + I_{cm}\omega_{cm} \]\end{large}{latex}

If the body is purely rotating, we can also define an angular speed for rotation about the new parallel axis.  The angular speed must satisfy (consider that the center of mass is describing a circle of radius _d_ about the axis):

{latex}\begin{large} \[ \omega_{\rm axis} = \frac{v_{cm}}{d} \] \end{large}{latex}

Futher, the rotation rate of the object about its center of mass must equal the rotation rate about the parallel axis, since when the object has completed a revolution about the parallel, its oritentation must be the same if it is executing pure rotation.  Thus, we can write:

{latex}\begin{large}\[ L \mbox{ (pure rotation)} = m\omega_{\rm axis} d^{2} + I_{cm}\omega_{\rm axis} \]\end{large}{latex}

which implies the parallel axis theorem holds.

h4. Derivation of the Theorem

From the definition of the moment of inertia:

{latex}\begin{large}\[ I = \int r^{2} dm \] \end{large}{latex}

The center of mass is at a position _r_~cm~ with respect to the desired axis of rotation.  We define new coordinates:

{latex}\begin{large}\[ \vec{r} = \vec{r}\:' + \vec{r}_{cm}\]\end{large}{latex}

where _r'_ measures the positions relative to the object's center of mass.  Substituting into the moment of inertia formula: 

{latex}\begin{large}\[ I = \int\: (r\:'^{2} + 2\vec{r}\:'\cdot\vec{r}_{cm} + r_{cm}^{2})\: dm \]\end{large}{latex}

The _r_~cm~ is a constant within the integral over the body's mass elements.  Thus, the middle term can be written:

{latex}\begin{large}\[ \int\:\vec{r}\:'\cdot \vec{r}_{cm}\:dm = r_{cm}\cdot \int\:\vec{r}\:'\:dm \]\end{large}{latex}

Since the _r_' measure deviations from the center of mass position, the integral _r_'_dm_ must give zero (the position of the center of mass in the _r_' system).  Thus, we are left with:

{latex}\begin{large}\[ I = \int\: r\:'^{2}\:dm + r_{cm}^{2} \int\:dm = I_{cm} + Mr_{cm}^{2}\]\end{large}{latex}

Which is the parallel axis theorem.