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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
Moment of inertia
Suppose that the motion of the object consists merely of rigid rotation at angular
. It follows, from Sect. 8.4, that
Let us write
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
Here, the quantity is termed the moment of inertia of the object, and
is the perpendicular distance from the th element to the axis of
rotation. Note that for translational motion we usually write
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
where is the mass density of the object,
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
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
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)
The moment of inertia of a ring about a perpendicular symmetric axis.
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
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
Here, we have made use of the fact that inside the object. It follows by inspection
of the previous three equations that
See Fig. 76.
The perpendicular axis theorem.
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
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 moment of inertia of a ring about a coplanar symmetric 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
where is the mass of the body, and is the perpendicular distance between the
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
where the integrals are taken over the volume of the body. From Eq. (342),
the expression for the first moment of inertia is
since is the perpendicular distance of a general point from the -axis.
Likewise, the expression for the second moment of inertia takes the
The above equation can be expanded to give
It follows from Eqs. (351) and (352) that
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
See Fig. 78.
An application of the parallel axis theorem.
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
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
and radial thickness . The area of this annulus is simply
. Hence, we can replace in the above integrals by
, so as to give
The above expression yields
Similar calculations to the above yield the following standard results:
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