Moment of inertia

(334) |

(335) |

(336) |

(337) |

Here, the quantity is termed the

(339) |

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) |

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) |

(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) |

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) |

(349) |

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) |

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

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) |

(354) |

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

(355) |

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) |

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) |

(358) |

(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