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## Moment of inertia

Consider an extended object which is made up of elements. Let the th element possess mass , position vector , and velocity . The total kinetic energy of the object is written
 (334)

Suppose that the motion of the object consists merely of rigid rotation at angular velocity . It follows, from Sect. 8.4, that
 (335)

Let us write
 (336)

where is a unit vector aligned along the axis of rotation (which is assumed to pass through the origin of our coordinate system). It follows from the above equations that the kinetic energy of rotation of the object takes the form
 (337)

or
 (338)

Here, the quantity is termed the moment of inertia of the object, and is written
 (339)

where is the perpendicular distance from the th element to the axis of rotation. Note that for translational motion we usually write
 (340)

where represents mass and represents speed. A comparison of Eqs. (338) and (340) suggests that moment of inertia plays the same role in rotational motion that mass plays in translational motion.

For a continuous object, analogous arguments to those employed in Sect. 8.5 yield

 (341)

where is the mass density of the object, is the perpendicular distance from the axis of rotation, and is a volume element. Finally, for an object of constant density, the above expression reduces to
 (342)

Here, is the total mass of the object. Note that the integrals are taken over the whole volume of the object.

The moment of inertia of a uniform object depends not only on the size and shape of that object but on the location of the axis about which the object is rotating. In particular, the same object can have different moments of inertia when rotating about different axes.

Unfortunately, the evaluation of the moment of inertia of a given body about a given axis invariably involves the performance of a nasty volume integral. In fact, there is only one trivial moment of inertia calculation--namely, the moment of inertia of a thin circular ring about a symmetric axis which runs perpendicular to the plane of the ring. See Fig. 75. Suppose that is the mass of the ring, and is its radius. Each element of the ring shares a common perpendicular distance from the axis of rotation--i.e., . Hence, Eq. (342) reduces to

 (343)

In general, moments of inertia are rather tedious to calculate. Fortunately, there exist two powerful theorems which enable us to simply relate the moment of inertia of a given body about a given axis to the moment of inertia of the same body about another axis. The first of these theorems is called the perpendicular axis theorem, and only applies to uniform laminar objects. Consider a laminar object (i.e., a thin, planar object) of uniform density. Suppose, for the sake of simplicity, that the object lies in the - plane. The moment of inertia of the object about the -axis is given by

 (344)

where we have suppressed the trivial -integration, and the integral is taken over the extent of the object in the - plane. Incidentally, the above expression follows from the observation that when the axis of rotation is coincident with the -axis. Likewise, the moments of inertia of the object about the - and - axes take the form
 (345) (346)

respectively. Here, we have made use of the fact that inside the object. It follows by inspection of the previous three equations that
 (347)

See Fig. 76.

Let us use the perpendicular axis theorem to find the moment of inertia of a thin ring about a symmetric axis which lies in the plane of the ring. Adopting the coordinate system shown in Fig. 77, it is clear, from symmetry, that . Now, we already know that , where is the mass of the ring, and is its radius. Hence, the perpendicular axis theorem tells us that

 (348)

or
 (349)

Of course, , because when the ring spins about the -axis its elements are, on average, farther from the axis of rotation than when it spins about the -axis.

The second useful theorem regarding moments of inertia is called the parallel axis theorem. The parallel axis theorem--which is quite general--states that if is the moment of inertia of a given body about an axis passing through the centre of mass of that body, then the moment of inertia of the same body about a second axis which is parallel to the first is

 (350)

where is the mass of the body, and is the perpendicular distance between the two axes.

In order to prove the parallel axis theorem, let us choose the origin of our coordinate system to coincide with the centre of mass of the body in question. Furthermore, let us orientate the axes of our coordinate system such that the -axis coincides with the first axis of rotation, whereas the second axis pieces the - plane at . From Eq. (328), the fact that the centre of mass is located at the origin implies that

 (351)

where the integrals are taken over the volume of the body. From Eq. (342), the expression for the first moment of inertia is
 (352)

since is the perpendicular distance of a general point from the -axis. Likewise, the expression for the second moment of inertia takes the form
 (353)

The above equation can be expanded to give
 (354)

It follows from Eqs. (351) and (352) that
 (355)

which proves the theorem.

Let us use the parallel axis theorem to calculate the moment of inertia, , of a thin ring about an axis which runs perpendicular to the plane of the ring, and passes through the circumference of the ring. We know that the moment of inertia of a ring of mass and radius about an axis which runs perpendicular to the plane of the ring, and passes through the centre of the ring--which coincides with the centre of mass of the ring--is . Our new axis is parallel to this original axis, but shifted sideways by the perpendicular distance . Hence, the parallel axis theorem tells us that

 (356)

See Fig. 78.

As an illustration of the direct application of formula (342), let us calculate the moment of inertia of a thin circular disk, of mass and radius , about an axis which passes through the centre of the disk, and runs perpendicular to the plane of the disk. Let us choose our coordinate system such that the disk lies in the - plane with its centre at the origin. The axis of rotation is, therefore, coincident with the -axis. Hence, formula (342) reduces to

 (357)

where the integrals are taken over the area of the disk, and the redundant -integration has been suppressed. Let us divide the disk up into thin annuli. Consider an annulus of radius and radial thickness . The area of this annulus is simply . Hence, we can replace in the above integrals by , so as to give
 (358)

The above expression yields
 (359)

Similar calculations to the above yield the following standard results:

• The moment of inertia of a thin rod of mass and length about an axis passing through the centre of the rod and perpendicular to its length is

• The moment of inertia of a thin rectangular sheet of mass and dimensions and about a perpendicular axis passing through the centre of the sheet is

• The moment of inertia of a solid cylinder of mass and radius about the cylindrical axis is

• The moment of inertia of a thin spherical shell of mass and radius about a diameter is

• The moment of inertia of a solid sphere of mass and radius about a diameter is

Next: Torque Up: Rotational motion Previous: Centre of mass
Richard Fitzpatrick 2006-02-02