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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 |
{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} ||Page Contents|| |{toc:indent=10px|style=none}| ---- h2. 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 |
where d is the (perpendicular)
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distance
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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
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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
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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
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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
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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
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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
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of
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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
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d
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about
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the
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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
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equal
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the
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rotation
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rate
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about
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the
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parallel
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axis,
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since
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when
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the
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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
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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
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same
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if
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it
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is
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executing
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pure
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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}{latex} |
which
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implies
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the
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parallel
...
axis
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theorem
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holds.
Derivation of the Theorem
From the definition of the moment of inertia:
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---- h2. Derivation of the Theorem From the definition of the moment of inertia: {latex}\begin{large}\[ I = \int r^{2} dm \] \end{large}{latex} |
The
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center
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of
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mass
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is
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at
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a
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position
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r
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cm with
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respect
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to
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the
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desired
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axis
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of
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rotation.
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We
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define
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new
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coordinates:
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}\begin{large}\[ \vec{r} = \vec{r}\:' + \vec{r}_{cm}\]\end{large}{latex} where _ |
where r'
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measures
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the
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positions
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relative
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to
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the
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object's
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center
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of
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mass.
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Substituting
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into
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the
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moment
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of
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inertia
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formula:
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{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 |
The rcm is a constant within the integral over the body's
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mass
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elements.
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Thus,
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the
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middle
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term
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can
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be
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written:
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}\begin{large}\[ \int\:\vec{r}\:'\cdot \vec{r}_{cm}\:dm = r_{cm}\cdot \int\:\vec{r}\:'\:dm \]\end{large}{latex} |
Since
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the
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r
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'
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measure
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deviations
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from
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the
...
center
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of
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mass
...
position,
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the
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integral
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r
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'
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dm
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must
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give
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zero
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(the
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position
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of
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the
...
center
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of
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mass
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in
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the
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r
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'
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system).
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Thus,
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we
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are
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left
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with:
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}\begin{large}\[ I = \int\: r\:'^{2}\:dm + r_{cm}^{2} \int\:dm = I_{cm} + Mr_{cm}^{2}\]\end{large}{latex} |
Which
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is
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the
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parallel
...
axis
...
theorem.
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