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[parent] Rotational Inertia of a Solid Cylinder (Definition)

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

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

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

$\displaystyle I = \frac{1}{4} M R^2 + \frac{1}{12} M L^2$ (2)
Figure: Rotational inertia of a solid cylinder
\includegraphics[scale=.6]{SolidCylinder.eps}

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

$\displaystyle I = \int r^2 dm $

Assuming constant density throughout the cylinder leads to

$\displaystyle dm = \rho dV $

and in cylindrical coordinates the infinitesmal volume dV is given by

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

giving the equation to integrate as

$\displaystyle 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

$\displaystyle I = \frac{R^4 \rho}{4} \int_{-L/2}^{L/2} \int_0^{2\pi} \, d\phi dz $

and ingtegrating the $\phi$ term gives

$\displaystyle 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

$\displaystyle I = \frac{\pi R^4 \rho L}{2} $

Finally, plugging in the equation for density and volume of a cylinder

$\displaystyle \rho = \frac{M}{V} $

$\displaystyle V = \pi R^2 L $

leaves us with equation (1)

$\displaystyle 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 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

$\displaystyle I_{11} = \int (r^2 - x^2) dm $

$\displaystyle 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

$\displaystyle r^2 = x^2 + y^2 + z^2 $

which gives us

$\displaystyle I_{11} = \int (y^2 + z^2) dm $

$\displaystyle I_{22} = \int (x^2 + z^2) dm $

Next, we see that in cylindrical coordinates that

$\displaystyle x = r \cos \phi $

$\displaystyle y = r \sin \phi $

$\displaystyle 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

Figure: Cylinder Slice
\includegraphics[scale=.4]{CylinderSlice.eps}

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

$\displaystyle I_{11} = \int (r^2 \sin^2 \phi + z^2) \rho dV$ (3)
$\displaystyle I_{22} = \int (r^2 \cos^2 \phi + z^2) \rho dV$ (4)

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

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

so we must integrate

$\displaystyle 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 $

$\displaystyle 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

$\displaystyle \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

$\displaystyle \frac{2\pi}{2}R^2\int_{-L/2}^{L/2} z^2 dz $

Finally, integrating the $z$ term gives us

$\displaystyle \frac{ \pi R^2 L^3}{12}$ (5)

Next up is the $r^3 \sin^2$ term, so first integrate $dr$ to get

$\displaystyle \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

$\displaystyle \sin^2 \phi = 1 - \cos^2 \phi $

and then use another trigonometric identity

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

so the integration becomes

$\displaystyle \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

$\displaystyle u = 2 \phi $

$\displaystyle du = 2 d\phi $

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

and we carry out the integration of

$\displaystyle \int_0^{2\pi} \cos u \, du $

and this integrates to zero and we are left with

$\displaystyle \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

$\displaystyle \int_{-L/2}^{L/2} \frac{\pi}{4}R^4 dz $

Finally, the $z$ term gives us

$\displaystyle \frac{\pi}{4}R^4 L$ (6)

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

$\displaystyle I_{11} = \rho ( \frac{\pi}{4}R^4 L + \frac{ \pi R^2 L^3}{12} )$ (7)

Using the volume of a cylinder

$\displaystyle V_{cyl} = \pi R^2 L $

we get the expression for the density

$\displaystyle \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)

$\displaystyle I_{11} = \left( \frac{1}{4}M R^2 + \frac{1}{12}M L^2 \right)$ (8)

References

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



"Rotational Inertia of a Solid Cylinder" is owned by bloftin.

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See Also: rotational inertia of a solid sphere

Other names:  moment of inertia of a solid cylinder

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Cross-references: fundamentals of physics, identity, work, inertia tensor, volume, Rotational Inertia
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This is version 8 of Rotational Inertia of a Solid Cylinder, born on 2006-03-28, modified 2006-07-10.
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Physics Classification: 45.40.-f (Dynamics and kinematics of rigid bodies)
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