Rotational Inertia of a Solid CylinderRotationalInertiaOfASolidCylinder2006-03-28 11:22:142006-03-29 13:13:26Definition% this is the default PlanetMath preamble. as your knowledge
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% define commands hereThe rotational inertia or moment of inertia of a solid cylinder rotating about the central axis or the z axis as shown in the figure is
\begin{equation}
I = \frac{1}{2} M R^2
\end{equation}
for other axes, such as rotation about x or y, the moment of inertia is given as
\begin{equation}
I = \frac{1}{4} M R^2 + \frac{1}{12} M L^2
\end{equation}
\begin{figure}
\includegraphics[scale=.6]{SolidCylinder.eps}
\caption{Rotational inertia of a solid cylinder}
\end{figure}
For the moment of inertia about the z axis, the integration in cylindrical coordinates is straight forward, since r in cylindrical coordinates is the same as in the inertia calculation so we have
$$ I = \int r^2 dm $$
Assuming constant density throughout the cylinder leads to
$$ dm = \rho dV $$
and in cylindrical coordinates the infinitesmal volume dV is given by
$$ dV = r \, dr d\phi dz $$
giving the equation to integrate as
$$ I = \rho \int_{-L/2}^{L/2} \int_0^{2\pi} \int_0^R r^3 \, dr d\phi dz $$
Integrating the r term yields
$$ I = \frac{R^4 \rho}{4} \int_{-L/2}^{L/2} \int_0^{2\pi} \, d\phi dz $$
and ingtegrating the $\phi$ term gives
$$ I = \frac{2 \pi R^4 \rho}{4} \int_{-L/2}^{L/2} \, dz $$
Next, integrating the z term and putting in the limits simplifies to
$$ I = \frac{\pi R^4 \rho L}{2} $$
Finally, plugging in the equation for density and volume of a cylinder
$$ \rho = \frac{M}{V} $$
$$ V = \pi R^2 L $$
leaves us with equation (1)
$$ I = \frac{1}{2} M R^2 $$
In order to derive the rotational inertia about the x and y axes, one needs to reference the inertia tensor to make things easy on us. Essentially, we are trying to calculate $I_{11}$ and $I_{22}$ which correspond to the moments of inertia about the x and y axes in this case. Turning the sums into integrals for our continuous example to work with these equations
$$ I_{11} = \int (r^2 - x^2) dm $$
$$ I_{22} = \int (r^2 - y^2) dm $$
before we can dive into the integration, we need to convert to cylindrical coordinates. First we note that
$$ r^2 = x^2 + y^2 + z^2 $$
which gives us
$$ I_{11} = \int (y^2 + z^2) dm $$
$$ I_{22} = \int (x^2 + z^2) dm $$
Next, we see that in cylindrical coordinates that
$$ x = r \cos \phi $$
$$ y = r \sin \phi $$
$$ z = z $$
the z coordinate is obvious, but to see the x and y coordinates see the below figure which shows a slice out of the cylinder
\begin{figure}
\includegraphics[scale=.4]{CylinderSlice.eps}
\caption{Cylinder Slice}
\end{figure}
It might not be obvious now but the integrals for x and y will come out to the same answer and we shall show this shortly. So the switch to cylindrical coordinates is complete once we change $dm$ to $\rho dV$ giving
$$ I_{11} = \int (r^2 \sin^2 \phi + z^2) \rho dV $$
$$ I_{22} = \int (r^2 \cos^2 \phi + z^2) \rho dV $$
Once again in cylindrical coordinates the infinitesmal volume dV is given by
$$ dV = r \, dr d\phi dz $$
so we must integrate
$$ I_{11} = \rho \int_{-L/2}^{L/2} \int_0^{2\pi} \int_0^R (r^3 \sin^2 \phi + r z^2) \, dr d\phi dz $$
$$ I_{22} = \rho \int \int \int_0^R (r^3 \cos^2 \phi + r z^2) \, dr d\phi dz $$
Let us break up the integral and start with the $r z^2$ term so first integrate $dr$ to get
$$ \int_{-L/2}^{L/2} \int_0^{2\pi} \frac{1}{2}R^2 z^2 d\phi dz $$
the $\phi$ term leaves us with
$$ \frac{2\pi}{2}R^2\int_{-L/2}^{L/2} z^2 dz $$
Finally, integrating the $z$ term gives us
$$ \frac{ \pi R^2 L^3}{12}$$
\subsection{References}
[1] Halliday, D., Resnick, R., Walker, J.: "Fundamentals of Physics".\, 5th Edition, John Wiley \& Sons, New York, 1997.