Introduction
Up to now, we have only analyzed the dynamics of point masses (i.e., objects whose spatial extent is either negligible or plays no role in their motion). Let us now broaden our approach in order to take extended objects into account. Now, the only type of motion which a point mass object can exhibit is translational motion: i.e., motion by which the object moves from one point in space to another. However, an extended object can exhibit another, quite distinct, type of motion by which it remains located (more or less) at the same spatial position, but constantly changes its orientation with respect to other fixed points in space. This new type of motion is called rotation. Let us investigate rotational motion.
Rigid body rotation
Consider a rigid body executing pure rotational motion (i.e., rotational motion which has no translational component). It is possible to define an axis of rotation (which, for the sake of simplicity, is assumed to pass through the body)--this axis corresponds to the straight-line which is the locus of all points inside the body which remain stationary as the body rotates. A general point located inside the body executes circular motion which is centred on the rotation axis, and orientated in the plane perpendicular to this axis. In the following, we tacitly assume that the axis of rotation remains fixed.
Figure 67: Rigid body rotation. Figure 67 shows a typical rigidly rotating body. The axis of rotation is the line . A general point lying within the body executes a circular orbit, centred on , in the plane perpendicular to . Let the line be a radius of this orbit which links the axis of rotation to the instantaneous position of at time . Obviously, this implies that is normal to .Suppose that at time point has moved to , and the radius has rotated through an angle . The instantaneous angular velocity of the body is defined
(309)
Note that if the body is indeed rotating rigidly, then the calculated value of should be the same for all possible points lying within the body (except for those points lying exactly on the axis of rotation, for which is ill-defined). The rotation speed of point is related to the angular velocity of the body via
(310)
where is the perpendicular distance from the axis of rotation to point . Thus, in a rigidly rotating body, the rotation speed increases linearly with (perpendicular) distance from the axis of rotation.
It is helpful to introduce the angular acceleration of a rigidly rotating body: this quantity is defined as the time derivative of the angular velocity. Thus,
(311)
where is the angular coordinate of some arbitrarily chosen point reference within the body, measured with respect to the rotation axis. Note that angular velocities are conventionally measured in radians per second, whereas angular accelerations are measured in radians per second squared.
For a body rotating with constant angular velocity, , the angular acceleration is zero, and the rotation angle increases linearly with time:
(312)
where . Likewise, for a body rotating with constant angular acceleration, , the angular velocity increases linearly with time, so that
(313)
and the rotation angle satisfies
(314)
Here, . Note that there is a clear analogy between the above equations, and the equations of rectilinear motion at constant acceleration introduced in Sect. 2.6--rotation angle plays the role of displacement, angular velocity plays the role of (regular) velocity, and angular acceleration plays the role of (regular) acceleration.
Multi-component systems
Figure 48: A 3-dimensional dynamical system consisting of many point mass objects. Consider a system of mutually interacting point mass objects which move in 3-dimensions. See Fig. 48. Let the th object, whose mass is , be located at vector displacement . Suppose that this object exerts a force on the th object. By Newton's third law of motion, the force exerted by the th object on the th is given by
(195)
Finally, suppose that the th object is subject to an external force .
Newton's second law of motion applied to the th object yields
(196)
Note that the summation on the right-hand side of the above equation excludes the case , since the th object cannot exert a force on itself. Let us now take the above equation and sum it over all objects. We obtain
(197)
Consider the sum over all internal forces: i.e., the first term on the right-hand side. Each element of this sum--, say--can be paired with another element--, in this case--which is equal and opposite. In other words, the elements of the sum all cancel out in pairs. Thus, the net value of the sum is zero. It follows that the above equation can be written
(198)
where is the total mass, and is the net external force. The quantity is the vector displacement of the centre of mass. As before, the centre of mass is an imaginary point whose coordinates are the mass weighted averages of the coordinates of the objects which constitute the system. Thus,
(199)
According to Eq. (198), the motion of the centre of mass is equivalent to that which would be obtained if all the mass contained in the system were collected at the centre of mass, and this conglomerate mass were then acted upon by the net external force. As before, the motion of the centre of mass is likely to be far simpler than the motions of the component masses.
Suppose that there is zero net external force acting on the system, so that . In this case, Eq. (198) implies that the centre of mass moves with uniform velocity in a straight-line. In other words, the velocity of the centre of mass,
(200)
is a constant of the motion. Now, the momentum of the th object takes the form . Hence, the total momentum of the system is written
(201)
A comparison of Eqs. (200) and (201) suggests that is also a constant of the motion when zero net external force acts on the system. Finally, Eq. (198) can be rewritten
(202)
In other words, the time derivative of the total momentum is equal to the net external force acting on the system.
It is clear, from the above discussion, that most of the important results obtained in the previous section, for the case of a two-component system moving in 1-dimension, also apply to a multi-component system moving in 3-dimensions.
Figure 49: The unfortunate history of the planet Krypton. As an illustration of the points raised in the above discussion, let us consider the unfortunate history of the planet Krypton. As you probably all know, Krypton--Superman's home planet--eventually exploded. Note, however, that before, during, and after this explosion the net external force acting on Krypton, or the fragments of Krypton--namely, the gravitational attraction to Krypton's sun--remained the same. In other words, the forces responsible for the explosion can be thought of as large, transitory, internal forces. We conclude that the motion of the centre of mass of Krypton, or the fragments of Krypton, was unaffected by the explosion. This follows, from Eq. (198), since the motion of the centre of mass is independent of internal forces. Before the explosion, the planet Krypton presumably executed a standard elliptical orbit around Krypton's sun. We conclude that, after the explosion, the fragments of Krypton (or, to be more exact, the centre of mass of these fragments) continued to execute exactly the same orbit. See Fig. 49.
Translational motion versus rotational motion
It should be clear, by now, that there is a strong analogy between rotational motion and standard translational motion. Indeed, each physical concept used to analyze rotational motion has its translational concomitant. Likewise, every law of physics governing rotational motion has a translational equivalent. The analogies between rotational and translational motion are summarized in Table 3.
Table 3: The analogies between translational and rotational motion. Translational motion
Rotational motion
Displacement
Angular displacement
Velocity
Angular velocity
Acceleration
Angular acceleration
Mass
Moment of inertia
Force
Torque
Work
Work
Power
Power
Kinetic energy
Kinetic energy
Worked example 8.7: Rotating cylinder
Question: A uniform cylinder of radius is given an angular speed of about an axis, parallel to its length, which passes through its centre. The cylinder is gently lowered onto a horizontal frictional surface, and released. The coefficient of friction of the surface is . How long does it take before the cylinder starts to roll without slipping? What distance does the cylinder travel between its release point and the point at which it commences to roll without slipping?
Answer: Let be the velocity of the cylinder's centre of mass, the cylinder's angular velocity, the frictional force exerted by the surface on the cylinder, the cylinder's mass, and the cylinder's moment of inertia. The cylinder's translational equation of motion is written
Note that the friction force acts to accelerate the cylinder's translational motion. Likewise, the cylinder's rotational equation of motion takes the form
since the perpendicular distance between the line of action of and the axis of rotation is the radius, , of the cylinder. Note that the friction force acts to decelerate the cylinder's rotational motion. If the cylinder is slipping with respect to the surface, then the friction force, , is equal to the coefficient of friction, , times the normal reaction, , at the surface:
Finally, the moment of inertia of the cylinder is
The above equations can be solved to give
Given that (i.e., the cylinder is initially at rest) and at time , the above expressions can be integrated to give
which yields
Now, the cylinder stops slipping as soon as the ``no slip'' condition,
is satisfied. This occurs when
Whilst it is slipping, the cylinder travels a distance