Kinetic Energy of Rotation; Moment of Inertia 303
As for any system of particles, the total kinetic energy K of a rotating rigid body is simply the sum of the individual kinetic energies of all the particles. If the particles have masses mi and velocities vi (where i = 1, 2, . . . , n), then
In a rigid body rotating about a given axis, all the particles move with the same angular velocity w along circular paths. The speeds of the particles along their paths are proportional to their radial distances:
and hence the total kinetic energy is
We will write this as
Note that Eq. (21) has a mathematical form reminiscent of the familiar expression ½mv2 for the kinetic energy of a single particle -- the moment of inertia replaces the mass and the angular velocity replaces the translational velocity. As we will see in the next chapter, the moment of inertia is a measure of the resistance that a body offers to changes in its rotational motion, just as mass is a measure of the resistance that a body offers to changes in its translational motion. These analogies between rotational and translational quantities make it easy to remember the rotational equations.
Equation (22) shows that the moment of inertia -- and consequently the kinetic energy for a given value of w -- is large if most of the mass of the body is at a large distance from the axis of rotation. This is very reasonable: for a given value of w, particles at a large distance from the axis move with high speeds and therefore have a large kinetic energies.
EXAMPLE 5. Suppose that the molecule of potassium bromide (KBr) described in Example 10.5 rotates rigidly about its center of mass (Figure 12.11). What is the moment of inertia of the molecule about this axis? Suppose that the molecule rotates with an angular velocity of 1.0 X 1012 radian/s. What is the rotational kinetic energy?
SOLUTION: We can regard the molecule as a rigid body, consisting of two particles joined by a massless rod. It is best to place the origin of coordinates at the center of mass (see Figure 12.11). According to the results obtained in Example 10.5, the distance of the Br atom is then R1 = 0.93 Å, and that of the K atom is R2 =1.89 Å. The corresponding masses are m1 = 79.9 u and m2 = 39.1 u. The moment of inertia of the molecule is
COMMENTS AND SUGGESTIONS: This kinetic energy could equally well have been obtained by first calculating the individual velocities of the atoms (v1 = R1w, v2 = R2w) and then adding the corresponding individual kinetic energies.
If we regard the mass of a solid body as continuously distributed over its volume, then we can calculate the moment of inertia by converting Eq. (22) into an integral. As in the calculation of the position of the center of mass [see Eqs. (10.21)-(10.26)], we imagine that the volume is divided into small volume elements with masses
In some simple cases, it is possible to find the moment of inertia without having recourse to an explicit calculation of the integral in Eq. (25). For example, if the rigid body is a thin hoop or a thin cylindrical shell of radius R0 (Figures 12.12 and 12.13) rotating about its axis of symmetry, then all of the mass of the body is at the same distance from the axis of rotation -- the moment of inertia is then simply the total mass M of the hoop or shell multiplied by its radius R0 squared,
The following are some examples of explicit evaluations of the integral in Eq. (25).
EXAMPLE 6. Find the moment of inertia of a disk of uniform density of mass M and radius R0 rotating about its axis.
SOLUTION: Figure 12.14 shows the disk with the z axis along the axis of rotation. Suppose the disk has a thickness l. The disk can be regarded as made up of a large number of thin concentric hoops fitting one around another. Figure 12.14 shows one such hoop of radius R, width dR, and thickness l. The volume of the hoop is dV = 2p Rl dR; with this, Eq. (25) gives
for the moment of inertia of the disk.
COMMENTS AND SUGGESTIONS: This formula also gives the moment of inertia of a cylinder, since such a body is simply a very thick disk. The above calculation is equally valid for thin and for thick disks, and the final formula is independent of the thickness l of the disk.
EXAMPLE 7. Find the moment of inertia of a uniform thin rod of length l and mass M rotating about an axis perpendicular to the rod and through its center.
SOLUTION: Figure 12.15 shows the rod lying along the x axis; the axis of rotation is the z axis. Consider a small slice dx of the rod. If the cross-sectional area of the rod is A, then dV = A dx and Eq. (25) becomes
Since Al is the volume of the rod, rAl is its total mass and therefore the moment of inertia can be written
EXAMPLE 8. Repeat the calculation of the preceding example for an axis through one end of the rod.
SOLUTION: Figure 12.16 shows the rod and the axis of rotation. The rod extends from x = 0 to x = l; hence instead of Eq. (29) we obtain
and instead of Eq. (30), we obtain
Comparison of Eqs. (30) and (32) makes it very clear that the value of the moment of inertia depends on the location of the axis of rotation. The moment of inertia is small if the axis passes through the center of mass, and large if it passes through the end of the rod.
We can prove a general theorem, the parallel-axis theorem, that relates the moment of inertia ICM about an axis through the center of mass to the moment of inertia I about a parallel axis through some other point. The theorem asserts that
where M is the total mass of the body and d the distance between the two axes. It is a corollary of Eq. (33) that the moment of inertia about an axis through the center of mass is always less than that about any other parallel axis.
A simple and instructive proof of the theorem is as follows: Figure 12.17 shows the body rotating about a fixed axis and also shows an alternative, parallel axis through the center of mass. The kinetic energy of the body rotating about the fixed axis is
However, we can express the kinetic energy in an alternative form. According to Eq. (10.44) the kinetic energy of an arbitrary system of particles is the sum of the translational energy of motion of the center of mass and the "internal" energy of motion relative to the center of mass,
In Figure 12.17 the center of mass moves along a circle of radius d around the axis. Hence
Let us next look at the second term on the right side of Eq. (35). Although the rotational motion of the body in Figure 12.17 is most simply described as a rotation about the fixed axis, it can also be described as a rotation about the moving axis through the center of mass. Each time the body completes one revolution about the fixed axis, it also completes one revolution about this center-of-mass axis; consequently, the angular velocity of the rotation around the center-of-mass axis is also w. The kinetic energy of the rotational motion relative to the center of mass is then
The values given by Eqs. (34) and (39) must be equal; this implies that
which is the result we wanted to obtain.
We can also prove another theorem, the perpendicular-axis theorem, which relates the moments of inertia of a thin flat plate (for example, a sheet of metal) about three mutually perpendicular axes. Figure 12.18 shows a flat plate of arbitrary shape in the x-y plane. The plate may rotate either about an axis perpendicular to the plate (the z axis), or about an axis in the plane of the plate (the x axis or the y axis). We will call the moments of inertia for rotation about these alternative axes Iz, Ix, and Iy, respectively. Then the perpendicular-axis theorem asserts that
The proof of the theorem is very simple: For rotation about the z axis, the distance from the axis of rotation to a particle in the plate is
Comparison of these three formulas makes it obvious that the sum of Ix and Iy, equals Iz.
Table 12.1 lists the moments of inertia of a variety of rigid bodies about axes through their centers of mass; all the bodies are assumed to have uniform density. The parallel-axis theorem can be used to find the moment of inertia about any axis parallel to the center-of-mass axis. Furthermore, by means of combinations of the formulas of Table 12.1, it is possible to find the moments of inertia of some other bodies.
EXAMPLE 9. Find the moment of inertia of a spherical shell, of inner radius R1 and outer radius R2 about an axis through the center (Figure 12.19).
SOLUTION: According to Table 12.1, the moment of inertia of a solid sphere of radius R2 is
The shell of Figure 12.19 is a solid sphere with a concentric spherical hole. Correspondingly, the inertia is that of a solid sphere of radius R2 minus that of a concentric sphere of radius R1:
The volume of the shell is
and hence the density is
where M is now the mass of the shell. This leads to
for the inertia of the spherical shell.