PlanetPhysics/Rotational Inertia of a Solid Cylinder

The [[../MomentOfInertia/|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

$I=\frac{1}{2}M{R}^2$

for other axes, such as rotation about x or y, the moment of inertia is given as

$I=\frac{1}{4}M{R}^2+\frac{1}{12}M{L}^2$

\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/|volume]] dV is given by

$dV=rdrd\varphi dz$

giving the equation to integrate as

$I=\rho \underset{-L/2}{\overset{L/2}{\int }}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{R}{\int }}{r}^{3}drd\varphi dz$

Integrating the r term yields

$I=\frac{R^4\rho }{4}\underset{-L/2}{\overset{L/2}{\int }}\underset{0}{\overset{2\pi }{\int }}d\varphi dz$

and ingtegrating the $\varphi$ term gives

$I=\frac{2\pi R^4\rho }{4}\underset{-L/2}{\overset{L/2}{\int }}dz$

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 [[../InertiaTensor/|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/|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

$x=r\cos \varphi$, $y=r\sin \varphi$, $z=z$

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

$I_{11}=\int (r^2{\sin }^2\varphi +z^2)\rho dV$

$I_{22}=\int (r^2{\cos }^2\varphi +z^2)\rho dV$

Once again in cylindrical coordinates the infinitesmal volume dV is given by

$dV=rdrd\varphi dz$

so we must integrate

$I_{11}=\rho \underset{-L/2}{\overset{L/2}{\int }}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{R}{\int }}(r^3{\sin }^2\varphi +r{z}^2)drd\varphi dz$ $I_{22}=\rho \underset{-L/2}{\overset{L/2}{\int }}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{R}{\int }}(r^3{\cos }^2\varphi +r{z}^2)drd\varphi dz$

Let us break up the integral and start with the $r{z}^2$ term so first integrate $dr$ to get

$\underset{-L/2}{\overset{L/2}{\int }}\underset{0}{\overset{2\pi }{\int }}\frac{1}{2}{R}^2{z}^{2}d\varphi dz$

the $\varphi$ term leaves us with

$\frac{2\pi }{2}{R}^2\underset{-L/2}{\overset{L/2}{\int }}{z}^{2}dz$

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

$\underset{-L/2}{\overset{L/2}{\int }}\underset{0}{\overset{2\pi }{\int }}\frac{1}{4}{R}^4{\sin }^2\varphi d\varphi dz$

to integrate the $\varphi$ term use the trigonometric [[../Cod/|identity]] that

${\sin }^2\varphi =1-{\cos }^2\varphi$

and then use another trigonometric identity

${\cos }^2\varphi =\frac{1}{2}(1+\cos (2\varphi )$

so the integration becomes

$\underset{-L/2}{\overset{L/2}{\int }}\underset{0}{\overset{2\pi }{\int }}\frac{1}{4}{R}^4(1-1/2+1/2\cos (2\varphi ))d\varphi dz$

Use u substitution to solve this so

$u=2\varphi$ $du=2d\varphi$ $d\varphi =\frac{du}{2}$

and we carry out the integration of

$\underset{0}{\overset{2\pi }{\int }}\cos udu$

and this integrates to zero and we are left with

$\underset{-L/2}{\overset{L/2}{\int }}\underset{0}{\overset{2\pi }{\int }}\frac{1}{8}{R}^{4}d\varphi dz$

This integration is simple now and we get

$\underset{-L/2}{\overset{L/2}{\int }}\frac{\pi }{4}{R}^{4}dz$

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)

$I_{11}=(\frac{1}{4}M{R}^2+\frac{1}{12}M{L}^2)$

References

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