Moment of inertia

Figure 1.20 shows a simple case of rotation about a fixed axis. The rotating rigid body consists of a single particle of mass m fastened to the end of a massless rod of length r. A force acts as shown, causing the particle to move in a circle about the axis. The particle has a tangential component of acceleration , governed by Newton's second law:

The torque acting on the particle is

Taking into account that we can write

The quantity in parentheses on the right side of this equation tells us how the mass of the rotating body is distributed about its axis of rotation. We call that quantity the rotational inertia (or moment of inertia) I of the body with respect to the axis of rotation. It is constant for a particular rigid body and for a particular rotation axis. Moment of inertia is measured by kg×m².

We may now write

and substitute into Equation for torque, obtaining

as the expression we seek.

Although we derived this equation for the special case of a single particle rotating about a fixed axis, it suits for any rigid body rotating about a fixed axis, because any such body can be analyzed as an assembly of single particles.

For the situation in which more than one force is applied to the particle, we can extend obtained equation to

where is the net torque (the sum of all external torques) acting on the particle. This equation is the angular (or rotational) form of Newton's second law.

If a rigid body is made up of discrete particles we can calculate its rotational inertia I. If the body is continuous, we can replace the sum of inertia moment with an integral, and the definition of rotational inertia becomes

In the sample problems that follow this section, we calculate I for bodies of both kinds. In general, the rotational inertia of any rigid body with respect to a rotation axis depends on (1) the shape of the body, (2) the perpendicular distance from the axis to the body's center of mass, and (3) the orientation of the body with respect to the axis.

Table 2 gives the rotational inertias of several common bodies about various axes. Note how the distribution of mass relative to the rotational axis affects the value of the rotational inertia I.

Table 1.5

— moment of inertia of a mass point:
— moment of inertia of a uniform rod of length about axis through its centre of mass:
— moment of inertia of a ring with radius about the axis through its centre, which is at right angles to the plane of the ring:
— moment of inertia of a uniform disc with radius about the axis through its centre, which is at right angles to the plane of the disc:
— moment of inertia of a sphere with radius about an axis through its centre:

Fig 1.21

The parallel-axis theorem (theorem of Schteiner)

If you know the rotational inertia of a body about any axis that passes through its center of mass, you can find its rotational inertia about any other axis parallel to that axis according to the parallel-axis theorem (fig.1.21):

here m is the mass of the body and d is the perpendicular distance between the two (parallel) axes. In other words, this theorem can be stated as follows:

The rotational inertia of a body about any axis is equal to the rotational inertia
it would have about that axis if all its mass was concentrated at its center of mass
plus its rotational inertia about a parallel axis through its center of mass.

Angular momentum

Like all other linear quantities, linear momentum has its angular counterpart. Figure 1.22 shows a particle with linear momentum located at point P in the XY plane. The angular momentum of this particle with respect to the origin O is a vector quantity defined as

where is the position vector of the particle with respect to O. As the particle moves relative to O, in the direction of its momentum , position vector rotates around O. Note carefully that to have angular momentum about O, the particle does not have to rotate itself around O. The SI unit of angular momentum is the kilogram-meter-squared per second (kg∙m²/s), equivalent to the joule-second (J∙s).

To find the direction of the angular momentum vector in Fig. 1.22 we slide the vector until its tail is at origin O. Then we use the right-hand rule for vector products sweeping the fingers from into . The outstretched thumb then shows that vector points in the positive direction of the Z axis in Fig. 1.22. This positive direction is consistent with the counterclockwise rotation of the particle’s position vector about the Z axis, as the particle continues to move. (A negative direction of would be consistent with a clockwise rotation of about the Z axis).

To find the magnitude of , we will use the following equation:

where is the angle between and when these two vectors are tail to tail. From Fig.1.22a, we see that equation for angular momentum can be rewritten as

where is the component of perpendicular to and From Fig. 1.22 b, we see that equation for angular momentum can also be rewritten as

where is the perpendicular distance between O and an extension of .

As it is true for torque, angular momentum has value only with respect to a specified origin. Moreover, if the particle in Fig. 1.22 does not lie in the XY plane, or if the linear momentum of the particle did not also lie in that plane, the angular momentum will not be parallel to the Z axis. The direction of the angular momentum vector is always perpendicular to the plane formed by the vectors and .

As it was mentioned above Newton’s second law for translational moving particle has the form:

which expresses the close relation between force and linear momentum for a single particle. We have seen enough of the parallelism between linear and angular quantities to be pretty sure that there is also a close relation between torque and angular momentum. Guided by Newton’s second law for translational moving particle, we can even guess that it must be

This equation is indeed an angular form of Newton’s second law for a single particle (fundamental law of dynamics of rotational motion for a single particle):

The (vector) sum of all torques acting on a particle
is equal to the time rate of change of the angular momentum of that particle.

Angular form of Newton’s second law for a single particle has no meaning unless the torques and the angular momentum are defined with respect to the same origin.

Now let us prove this equation. We start with the definition of the angular momentum of a particle:

where is the position vector of the particle and is the velocity of the particle. Differentiating each side with respect to time t we get:

As and

then

Vector product of two velocities equals zero and

Thus

It is the relation that we set out to prove.

Now we pay our attention to the motion of a system of particles. Note that "a system of particles" includes a rigid body as a special case. The total angular momentum of a system of particles is the (vector) sum of the angular momenta of the particles

in which j = 1, 2, 3,…… labels the particles.

Along time, the angular momenta of individual particles may change, either due to the interactions within the system (between the individual particles) or due to influences that may act on the system from the outside. We can find the change in as

where is only , the (vector) sum of the torques that act on the i -th particle.

Some torques are internal associated with forces acting among the particles within the system, other torques are external associated with forces acting from outside the system. The internal forces, due to Newton's law of action and reaction, cancel in pairs. So, to add the torques, we have to consider only those associated with external forces. Thus

where

in which i = 1, 2, 3,…… labels the particles.

This equation represents Newton's second law for rotation in an angular form (fundamental law of dynamics for a rotational motion), expressed for a system of particles, it is analogous to . Last equation tells us that

the (vector) sum of the external torques acting on a system
of particles is equal to the tune rate of change
of the angular momentum of that system.

It has the meaning only if the torque and angular momentum vectors are referred to the same origin. In an inertial reference frame, it can be applied to any point. In an accelerating frame, fundamental law of dynamics of rotational motion can be applied only to the center of mass of the system.

Next we evaluate the angular momentum of a system of particles that form a rigid body which rotates around a fixed axis. Figure 1.23a shows such a body. The fixed axis of rotation is the Z axis, and the body rotates around it at constant angular speed . We want to find the angular momentum of the body around the axis of rotation.

We can find the angular momentum by summing the Z components of the angular momenta of the mass elements in the body. In Fig. 1.23a, a typical mass element , of the body moves around the Z axis in a circular path. The position of the mass element is located relative to the origin 0 by position vector . The radius of the mass element's circular path is the perpendicular distance between the element and the Z axis.

The magnitude of the angular momentum of this mass element, with respect to O, is given by the equation:

where and , are the linear momentum and linear speed of the mass element, and 90° is the angle between , and . The angular momentum vector , for the mass element in Fig 1.23a is shown in Fig 1.23b; its direction must be perpendicular to those of , and .

We are interested in the component of which is parallel to the rotation axis, here the Z axis. That Z component is

The Z component of the angular momentum for the rotating rigid body as a whole is found by adding up the contributions of all the mass elements which make up the body. Thus, because , we may write

We can remove from the here because it is constant: it has the same value for all points of the rotating rigid body.

The quantity is the rotational inertia I of the body about the fixed axis. Thus

We have dropped the subscript Z, but you must remember that the angular momentum is the angular momentum about the rotation axis. Also, I in that equation is the rotational inertia about that the same axis.

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