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
![{\displaystyle I={\frac {1}{2}}MR^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42051f51fac8a058905043317190d33798263863)
for other axes, such as rotation about x or y, the moment of inertia is given as
![{\displaystyle I={\frac {1}{4}}MR^{2}+{\frac {1}{12}}ML^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3650eeb516d05160c68bee517986d4a8d3acc0d)
\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
Assuming constant density throughout the cylinder leads to
and in cylindrical coordinates the infinitesmal volume dV is given by
giving the equation to integrate as
Integrating the r term yields
and ingtegrating the
term gives
Next, integrating the z term and putting in the limits simplifies to
Finally, plugging in the equation for density and volume of a cylinder
leaves us with equation (1)
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
and
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
before we can dive into the integration, we need to convert to cylindrical coordinates. First we note that
which gives us
Next, we see that in cylindrical coordinates that
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
to
giving
![{\displaystyle I_{11}=\int (r^{2}\sin ^{2}\phi +z^{2})\rho dV}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a88cf79b3fcf5e0734b5c2811bc506fe30d2197)
![{\displaystyle I_{22}=\int (r^{2}\cos ^{2}\phi +z^{2})\rho dV}](https://wikimedia.org/api/rest_v1/media/math/render/svg/956d6feaafb816f0f96f9b865c9edb8a9fb47cf4)
Once again in cylindrical coordinates the infinitesmal volume dV is given by
so we must integrate
Let us break up the integral and start with the
term so first integrate
to get
the
term leaves us with
Finally, integrating the
term gives us
![{\displaystyle {\frac {\pi R^{2}L^{3}}{12}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81e69a07ad567ab046128ff1c07aa3d375127877)
Next up is the
term, so first integrate
to get
to integrate the
term use the trigonometric identity that
and then use another trigonometric identity
so the integration becomes
Use u substitution to solve this so
and we carry out the integration of
and this integrates to zero and we are left with
This integration is simple now and we get
Finally, the
term gives us
![{\displaystyle {\frac {\pi }{4}}R^{4}L}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4c81bf3892718214a048c627323c49c49d91808)
Plugging equations (5) and (6) into (3) gives us
![{\displaystyle I_{11}=\rho ({\frac {\pi }{4}}R^{4}L+{\frac {\pi R^{2}L^{3}}{12}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d3f721ff62914e6d8c30a8196ff386e2fbded9a)
Using the volume of a cylinder
we get the expression for the density
and plugging this into seven and simplifying gives us the moment of inertia about the x axis, which was stated in (1)
![{\displaystyle I_{11}=\left({\frac {1}{4}}MR^{2}+{\frac {1}{12}}ML^{2}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b60d805ee811548d3b5bec23f6c3f8bdae3dc680)