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[parent] moment of inertia of a circular disk (Example)

Here we look at two cases for the moment of inertia of a homogeneous circular disk

(a) about its geometrical axis,

(b) about one of the elements of its lateral surface.

Let m be the mass, a the radius, l the thickness, and $\tau$ the density of the disk. Then choosing a circular ring for the element of mass we have

$\displaystyle dm = \tau \cdot l \cdot 2\pi r \cdot dr$

where $r$ is the radius of the ring and $dr$ its thickness.

\includegraphics[scale=.6]{Fig87.eps}

Therfore the moment of inertia about the axis of the disk is

$\displaystyle I = 2\pi l \tau \int_0^a r^3 dr$

$\displaystyle I = \frac{\tau l \pi a^4}{2}$

$\displaystyle I = \frac{ma^2}{2}$

The moment of inertia about the element is obtained easily by the help of theorem II. Thus

$\displaystyle I' = I + ma^2$

$\displaystyle I' = \frac{3}{2}ma^2$

It will be noticed that the thickness of the disk does not enter into the expressions for $I$ and $I'$ except through the mass of the disk. Therefore these expressions hold good whether the disk is thick enough to be called a cylinder or thin enough to be called a circular lamina.

References

This article is a derivative of the public domain book, "Analytical mechanics" by Haroutune M. Dadourian, 1913. Made available by the internet archive



"moment of inertia of a circular disk" is owned by bloftin.
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Other names:  rotational inertia of a circular disk

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Cross-references: mechanics, domain, theorem, mass, moment of inertia

This is version 3 of moment of inertia of a circular disk, born on 2006-07-11, modified 2006-07-11.
Object id is 197, canonical name is MomentOfInertiaOfACircularDisk.
Accessed 7464 times total.

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Physics Classification: 45.40.-f (Dynamics and kinematics of rigid bodies)
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