Talk:PlanetPhysics/Rotational Inertia of a Solid Cylinder

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The \htmladdnormallink{Rotational Inertia}{http://planetphysics.us/encyclopedia/MomentOfInertia.html} 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 \htmladdnormallink{volume}{http://planetphysics.us/encyclopedia/Volume.html} 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 \htmladdnormallink{inertia tensor}{http://planetphysics.us/encyclopedia/InertiaTensor.html} 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 \htmladdnormallink{work}{http://planetphysics.us/encyclopedia/Work.html} 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 \begin{equation} I_{11} = \int (r^2 \sin^2 \phi + z^2) \rho dV \end{equation} \begin{equation} I_{22} = \int (r^2 \cos^2 \phi + z^2) \rho dV \end{equation}

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

$$ dV = r \, dr d\phi dz $$

so we must integrate

$$ I_{11} = \rho \int_{-L/2}^{L/2} \int_0^{2\pi} \int_0^R (r^3 \sin^2 \phi + r z^2) \, dr d\phi dz $$ $$ I_{22} = \rho \int_{-L/2}^{L/2} \int_0^{2\pi} \int_0^R (r^3 \cos^2 \phi + r z^2) \, dr d\phi dz $$

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

\begin{equation} \frac{ \pi R^2 L^3}{12} \end{equation}

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 \htmladdnormallink{identity}{http://planetphysics.us/encyclopedia/Cod.html} 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

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

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

\begin{equation} \frac{\pi}{4}R^4 L \end{equation}

Plugging equations (5) and (6) into (3) gives us

\begin{equation} I_{11} = \rho ( \frac{\pi}{4}R^4 L + \frac{ \pi R^2 L^3}{12} ) \end{equation}

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)

\begin{equation} I_{11} = \left( \frac{1}{4}M R^2 + \frac{1}{12}M L^2 \right) \end{equation}

\subsection{References}

[1] Halliday, D., Resnick, R., Walker, J.: "\htmladdnormallink{fundamentals of physics".\,}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} 5th Edition, John Wiley \& Sons, New York, 1997.

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