Moment of inertia

Consider an extended object which is made up of $N$ elements. Let the $i$th element possess mass $m_i$, position vector ${\bf r}_i$, and velocity ${\bf v}_i$. The total kinetic energy of the object is written

$$K = \sum_{i=1,N} \frac{1}{2} m_i v_i^{ 2}.$$ (334)

Suppose that the motion of the object consists merely of rigid rotation at angular velocity $\mbox{\boldmath$\omega$}$. It follows, from Sect. 8.4, that

$${\bf v}_i = \mbox{\boldmath$\omega$}\times {\bf r}_i.$$ (335)

Let us write

$$\mbox{\boldmath$\omega$}= \omega {\bf k},$$ (336)

where ${\bf k}$ 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

$$K = \sum_{i=1,N} \frac{1}{2} m_i \vert{\bf k}\times {\bf r}_i\vert^2 \omega^2,$$ (337)

or

$$K = \frac{1}{2} I \omega^2.$$ (338)

Here, the quantity $I$ is termed the moment of inertia of the object, and is written

$$I = \sum_{i=1,N} m_i \vert{\bf k}\times {\bf r}_i\vert^2= \sum_{i=1,N} m_i \sigma_i^{ 2},$$ (339)

where $\sigma_i=\vert{\bf k}\times {\bf r}_i\vert$ is the perpendicular distance from the $i$th element to the axis of rotation. Note that for translational motion we usually write

$$K = \frac{1}{2} M v^2,$$ (340)

where $M$ represents mass and $v$ 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

$$I = \int\!\int\!\int \rho \sigma^2 dV,$$ (341)

where $\rho({\bf r})$ is the mass density of the object, $\sigma = \vert{\bf k}\times {\bf r}\vert$ is the perpendicular distance from the axis of rotation, and $dV$ is a volume element. Finally, for an object of constant density, the above expression reduces to

$$I = M \frac{\int\!\int\!\int \sigma^2 dV} {\int\!\int\!\int dV}.$$ (342)

Here, $M$ 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 $M$ is the mass of the ring, and $b$ is its radius. Each element of the ring shares a common perpendicular distance from the axis of rotation--i.e., $\sigma=b$. Hence, Eq. (342) reduces to

$$I = M b^2.$$ (343)

Figure 75: 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 $x$-$y$ plane. The moment of inertia of the object about the $z$-axis is given by

$$I_z = M \frac{\int \int (x^2+y^2) dx dy}{\int \int dx dy},$$ (344)

where we have suppressed the trivial $z$-integration, and the integral is taken over the extent of the object in the $x$-$y$ plane. Incidentally, the above expression follows from the observation that $\sigma^2 = x^2+y^2$ when the axis of rotation is coincident with the $z$-axis. Likewise, the moments of inertia of the object about the $x$- and $y$- axes take the form

$$I_x = M \frac{\int \int y^2 dx dy}{\int \int dx dy},$$ (345)

$$I_y = M \frac{\int \int x^2 dx dy}{\int \int dx dy},$$ (346)

respectively. Here, we have made use of the fact that $z=0$ inside the object. It follows by inspection of the previous three equations that

$$I_z = I_x + I_y.$$ (347)

See Fig. 76.

Figure 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 $I_x=I_y$. Now, we already know that $I_z=M b^2$, where $M$ is the mass of the ring, and $b$ is its radius. Hence, the perpendicular axis theorem tells us that

$$2 I_x = I_z,$$ (348)

or

$$I_x = \frac{I_z}{2} = \frac{1}{2} M b^2.$$ (349)

Of course, $I_z>I_x$, because when the ring spins about the $z$-axis its elements are, on average, farther from the axis of rotation than when it spins about the $x$-axis.

Figure 77: 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 $I$ 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 $I'$ of the same body about a second axis which is parallel to the first is

$$I' = I + M d^2,$$ (350)

where $M$ is the mass of the body, and $d$ 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 $z$-axis coincides with the first axis of rotation, whereas the second axis pieces the $x$-$y$ plane at $x=d, y=0$. From Eq. (328), the fact that the centre of mass is located at the origin implies that

$$\int\!\int\!\int x dx dy dz = \int\!\int\!\int y dx dy dz=\int\!\int\!\int z dx dy dz=0,$$ (351)

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

$$I = M \frac{\int\!\int\!\int (x^2+y^2) dx dy dz} {\int\!\int\!\int dx dy dz},$$ (352)

since $x^2+y^2$ is the perpendicular distance of a general point $(x,y,z)$ from the $z$-axis. Likewise, the expression for the second moment of inertia takes the form

$$I' = M \frac{\int\!\int\!\int [(x-d)^2+y^2] dx dy dz} {\int\!\int\!\int dx dy dz}.$$ (353)

The above equation can be expanded to give

$$I' = M \frac{\int\!\int\!\int [(x^2+y^2) - 2 d x + d^2] dx dy dz} {\int\!\int\!\int dx dy dz}$$

$$= M\frac{\int\!\int\!\int (x^2+y^2) dx dy dz} {\int\!\int\!\int ... ...z}-2 d M\frac{\int\!\int\!\int x dx dy dz} {\int\!\int\!\int dx dy dz}$$

$$+ d^2 M \frac{\int\!\int\!\int dx dy dz} {\int\!\int\!\int dx dy dz}.$$ (354)

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

$$I' = I + M d^2,$$ (355)

which proves the theorem.

Let us use the parallel axis theorem to calculate the moment of inertia, $I'$, 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 $M$ and radius $b$ 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 $I=M b^2$. Our new axis is parallel to this original axis, but shifted sideways by the perpendicular distance $b$. Hence, the parallel axis theorem tells us that

$$I' = I + M b^2 = 2 M b^2.$$ (356)

See Fig. 78.

Figure 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 $M$ and radius $b$, 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 $x$-$y$ plane with its centre at the origin. The axis of rotation is, therefore, coincident with the $z$-axis. Hence, formula (342) reduces to

$$I = M \frac{\int\!\int (x^2+y^2) dx dy}{\int\!\int dx dy},$$ (357)

where the integrals are taken over the area of the disk, and the redundant $z$-integration has been suppressed. Let us divide the disk up into thin annuli. Consider an annulus of radius $\sigma=\sqrt{x^2+y^2}$ and radial thickness $d\sigma$. The area of this annulus is simply $2\pi \sigma d\sigma$. Hence, we can replace $dx dy$ in the above integrals by $2\pi \sigma d\sigma$, so as to give

$$I = M \frac{\int_0^b 2\pi \sigma^3 d\sigma}{\int_0^b 2\pi \sigma d\sigma}.$$ (358)

The above expression yields

$$I = M \frac{\left[2 \pi \sigma^4/4\right]_0^b}{\left[2 \pi \sigma^2/2\right]_0^b}= \frac{1}{2} M b^2.$$ (359)

Similar calculations to the above yield the following standard results:

• The moment of inertia of a thin rod of mass $M$ and length $l$ about an axis passing through the centre of the rod and perpendicular to its length is
  
  $$I = \frac{1}{12} M l^2.$$
  
  

• The moment of inertia of a thin rectangular sheet of mass $M$ and dimensions $a$ and $b$ about a perpendicular axis passing through the centre of the sheet is
  
  $$I = \frac{1}{12} M (a^2+b^2).$$
  
  

• The moment of inertia of a solid cylinder of mass $M$ and radius $b$ about the cylindrical axis is
  
  $$I = \frac{1}{2} M b^2.$$
  
  

• The moment of inertia of a thin spherical shell of mass $M$ and radius $b$ about a diameter is
  
  $$I = \frac{2}{3} M b^2.$$
  
  

• The moment of inertia of a solid sphere of mass $M$ and radius $b$ about a diameter is
  
  $$I = \frac{2}{5} M b^2.$$