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Integrate to find the moment of inertia of a solid sphere. |
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| width | 70% |
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Calculate
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the
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...
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of
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a
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solid
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sphere
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of
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radius
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R
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and
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uniform
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density
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ρ rotated
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about
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an
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axis
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passing
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through
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its
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center.
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Solution
System,
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Interactions
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and
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Model:
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Not
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applicable.
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We
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solve
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this
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problem
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using
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the
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definition
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of
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the
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...
...
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.
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Approach:
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We
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will
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have
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to
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perform
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an
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integral
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over
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the
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volume
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of
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the
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sphere
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to
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calculate
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the
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moment
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of
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inertia.
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The
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integral
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is
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best
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performed
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in
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cylindrical
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coordinates
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with
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the
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z-axis
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chosen
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to
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lie
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along
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the
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axis
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of
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rotation,
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since
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it
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takes
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the
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form:
| Latex |
|---|
}\begin{large}\[ I =\int\int\int r^{2}\:\rho\:r\:dr\:d\theta\:dz \] \end{large}{latex}
{info}In spherical |
| Info |
|---|
In spherical coordinates, the integrand is complicated by the fact that the radial coordinate measures deviation from the center point instead of the axis of rotation. {info} |
The
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best
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way
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to
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configure
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this
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integral
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is
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to
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integrate
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over
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angle
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first,
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then
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radius,
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and
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finally
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the
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z-coordinate.
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The
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sphere
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is
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continuous
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so
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the
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limits
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on
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theta
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are
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always
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zero
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to
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2
| Latex |
|---|
}$\pi${latex} |
,
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and
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the
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sphere
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is
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axially
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symmetric
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so
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the
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integrand
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has
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no
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dependence
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on
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θ. Thus,
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we
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can
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simply
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perform
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this
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integral
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to
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obtain:
| Latex |
|---|
}\begin{large} \[ I = 2\pi \int \int r^{3}\:\rho\:dr\:dz \]\end{large}{latex}
|
It
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is
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simplest
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to
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next
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perform
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the
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integral
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over
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r
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.
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Here
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we
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have
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some
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trouble,
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however,
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since
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the
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limits
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on
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r
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will
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be
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a
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function
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of
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z
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.
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Basically,
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we
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are
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cutting
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the
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sphere
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up
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into
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a
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stack
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of
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circles
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perpendicular
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to
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the
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z-axis.
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If
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we
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move
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through
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the
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stack
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from
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bottom
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to
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top,
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the
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circles
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first
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grow
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larger
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(as
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we
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near
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the
...
center)
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and
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then
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reduce
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in
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size
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again
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until
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their
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radius
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disappears
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at
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the
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"north
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pole"
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(the
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top
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of
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the
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sphere).
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Based
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upon
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the
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well-known
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definition
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of
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a
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spherical
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surface
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in
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rectangular
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coordinates:
| Latex |
|---|
}\begin{large}\[ x^2 + y^2 + z^2 = R^2 \]\end{large}{latex}
|
and
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using
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the
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relationship
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between
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cylindrical
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coordinates
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and
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rectangular
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coordinates:
| Latex |
|---|
{latex}\begin{large}\[ r^{2} = x^2 + y^2\]\end{large}{latex}
|
we
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can
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see
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that
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the
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radius
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of
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the
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circle
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perpendicular
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to
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the
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z-axis
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at
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a
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given
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z-value
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will
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be:
| Latex |
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}\begin{large} \[ r_{max} = \sqrt{R^{2} - z^{2}} \] \end{large}{latex}
|
Thus,
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we
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can
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set
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up
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the
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limits
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of
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our
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integral
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as:
| Latex |
|---|
}\begin{large}\[ I = 2\pi \int_{-R}^{R}\int_{0}^{\sqrt{R^{2}-z^{2}}} \:r^{3}\:\rho\:dr\:dz \]\end{large}{latex}
|
Performing
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the
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r
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integral
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gives:
| Latex |
|---|
}\begin{large} \[ I = 2\pi \int_{-R}^{R} \:\frac{1}{4}(R^{2} - z^{2})^{2} \:\rho\:dz = 2\pi \int_{-R}^{R} \:\frac{1}{4}(R^{4} - 2R^{2}z^{2} + z^{4}) \:\rho\:dz \] \end{large}{latex}
|
and
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finishing
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with
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the
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z
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integral:
| Latex |
|---|
}\begin{large}\[ I = \pi \rho (R^{5} - \frac{2}{3} R^{5} + \frac{1}{5} R^{5}) = \frac{8}{15}\rho R^{5} \]\end{large}{latex}
|
The
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answer
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can
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be
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put
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in
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terms
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of
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the
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mass
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of
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the
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sphere
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by
...
noting
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that
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for
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a
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uniform
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sphere:
| Latex |
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}\begin{large}\[ M = \frac{4\pi}{3} \rho R^{3}\]\end{large}{latex}
|
so:
| Latex | ||
|---|---|---|
}\begin{large} \[ I = \frac{2}{5} MR^{2}\]\end{large}{latex}
Column | | width | 35%
| unmigrated-inline-wiki-markup |
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