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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. |
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 Icm then the moment of inertia of the body about any parallel axis can be found by evaluating the sum:
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| Wiki Markup |
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 |
where d is the (perpendicular)
...
distance
...
between
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the
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original
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center
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of
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mass
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axis
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and
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the
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new
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parallel
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axis.
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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)
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axis
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can
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be
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written:
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}\begin{large}\[ L = m\vec{r}_{\rm cm,axis}\times\vec{v}_{cm} + I_{cm}\omega_{cm} \]\end{large}{latex} |
Now
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suppose
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that
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the
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rigid
...
body
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is
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executing
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pure
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rotation
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about
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the
...
axis.
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In
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that
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case,
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the
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velocity
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of
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the
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center
...
of
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mass
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will
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be
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perpendicular
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to
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the
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displacement
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vector
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from
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the
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axis
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of
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rotation
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to
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the
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center
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of
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mass.
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Calling
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the
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magnitude
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of
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that
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displacement
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d
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,
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to
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make
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contact
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with
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the
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form
...
of
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the
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theorem,
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we
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then
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have:
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}\begin{large}\[ L \mbox{ (pure rotation)} = mv_{cm}d + I_{cm}\omega_{cm} \]\end{large}{latex} |
If
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the
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body
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is
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purely
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rotating,
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we
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can
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also
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define
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an
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angular
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speed
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for
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rotation
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about
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the
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new
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parallel
...
axis.
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The
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angular
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speed
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must
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satisfy
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(consider
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that
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the
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center
...
of
...
mass
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is
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describing
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a
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circle
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of
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radius
...
d
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about
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the
...
axis):
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}\begin{large} \[ \omega_{\rm axis} = \frac{v_{cm}}{d} \] \end{large}{latex} |
Futher,
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the
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rotation
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rate
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of
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the
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object
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about
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its
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center
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of
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mass
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must
...
equal
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the
...
rotation
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rate
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about
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the
...
parallel
...
axis,
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since
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when
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the
...
object
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has
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completed
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a
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revolution
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about
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the
...
parallel,
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its
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oritentation
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must
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be
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the
...
same
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if
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it
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is
...
executing
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pure
...
rotation.
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Thus,
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we
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can
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write:
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}\begin{large}\[ L \mbox{ (pure rotation)} = m\omega_{\rm axis} d^{2} + I_{cm}\omega_{\rm axis} \]\end{large} |
which implies the parallel axis theorem holds.
Derivation of the Theorem
From the definition of the moment of inertia:
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\begin{large}\[ I = \int r^{2} dm \] \end{large} |
The center of mass is at a position rcm with respect to the desired axis of rotation. We define new coordinates:
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\begin{large}\[ \vec{r} = \vec{r}\:' + \vec{r}_{cm}\]\end{large} |
where r' measures the positions relative to the object's center of mass. Substituting into the moment of inertia formula:
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\begin{large}\[ I = \int\: (r\:'^{2} + 2\vec{r}\:'\cdot\vec{r}_{cm} + r_{cm}^{2})\: dm \]\end{large} |
The rcm is a constant within the integral over the body's mass elements. Thus, the middle term can be written:
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\begin{large}\[ \int\:\vec{r}\:'\cdot \vec{r}_{cm}\:dm = r_{cm}\cdot \int\:\vec{r}\:'\:dm \]\end{large} |
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:
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\begin{large}\[ I = \int\: r\:'^{2}\:dm + r_{cm}^{2} \int\:dm = I_{cm} + Mr_{cm}^{2}\]\end{large} |
Which is the parallel axis theorem.