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

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

\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

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

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

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

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)

References

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