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)

$I_{11}=\left({\frac {1}{4}}MR^{2}+{\frac {1}{12}}ML^{2}\right)$

References

[1] Halliday, D., Resnick, R., Walker, J.: "fundamentals of physics".\, 5th Edition, John Wiley \& Sons, New York, 1997.