# PlanetPhysics/Rotational Inertia of a Solid Cylinder

The 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
   $I={\frac {1}{2}}MR^{2}$
for other axes, such as rotation about x or y, the moment of inertia is given as
   $I={\frac {1}{4}}MR^{2}+{\frac {1}{12}}ML^{2}$
\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
$dV=r\,drd\phi dz$
giving the equation to integrate as
$I=\rho \int _{-L/2}^{L/2}\int _{0}^{2\pi }\int _{0}^{R}r^{3}\,drd\phi dz$
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)
$I={\frac {1}{2}}MR^{2}$
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
$dV=r\,drd\phi dz$
so we must integrate
$I_{11}=\rho \int _{-L/2}^{L/2}\int _{0}^{2\pi }\int _{0}^{R}(r^{3}\sin ^{2}\phi +rz^{2})\,drd\phi dz$ $I_{22}=\rho \int _{-L/2}^{L/2}\int _{0}^{2\pi }\int _{0}^{R}(r^{3}\cos ^{2}\phi +rz^{2})\,drd\phi dz$
Let us break up the integral and start with the $rz^{2}$ 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
   ${\frac {\pi R^{2}L^{3}}{12}}$
Next up is the $r^{3}\sin ^{2}$ term, so first integrate $dr$ to get
$\int _{-L/2}^{L/2}\int _{0}^{2\pi }{\frac {1}{4}}R^{4}\sin ^{2}\phi d\phi dz$
to integrate the $\phi$ term use the trigonometric identity that
$\sin ^{2}\phi =1-\cos ^{2}\phi$
and then use another trigonometric identity
$\cos ^{2}\phi ={\frac {1}{2}}(1+\cos(2\phi )$
so the integration becomes
$\int _{-L/2}^{L/2}\int _{0}^{2\pi }{\frac {1}{4}}R^{4}(1-1/2+1/2\cos(2\phi ))d\phi dz$
Use u substitution to solve this so
$u=2\phi$ $du=2d\phi$ $d\phi ={\frac {du}{2}}$
and we carry out the integration of
$\int _{0}^{2\pi }\cos u\,du$
and this integrates to zero and we are left with
$\int _{-L/2}^{L/2}\int _{0}^{2\pi }{\frac {1}{8}}R^{4}d\phi dz$
This integration is simple now and we get
$\int _{-L/2}^{L/2}{\frac {\pi }{4}}R^{4}dz$
Finally, the $z$ term gives us
   ${\frac {\pi }{4}}R^{4}L$
Plugging equations (5) and (6) into (3) gives us
   $I_{11}=\rho ({\frac {\pi }{4}}R^{4}L+{\frac {\pi R^{2}L^{3}}{12}})$
Using the volume of a cylinder
$V_{cyl}=\pi R^{2}L$
we get the expression for the density
$\rho ={\frac {M}{\pi R^{2}L}}$
and plugging this into seven and simplifying gives us the moment of inertia about the x axis, which was stated in (1)
   $I_{11}=\left({\frac {1}{4}}MR^{2}+{\frac {1}{12}}ML^{2}\right)$