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