This section of the course deals with a different way of representing the laws of motion, based on the motion about some fixed line. In the next section, called rotational dynamics, we will build a model based on rotation, allowing you to predict the future of rotating objects. In this section we will introduce the parameters that describe rotation and cover enough of the basics to help you develop some intuition about general motion of many particle systems.

In the translational motion which we have been studying, every particle undergoes the same motion. Now we will consider a kind of motion which is fundamentally different. We will give the name "rotation" to the motion of a solid object in which every particle follows a circular path such that the center of the circle that each particle traces lies on a straight line. That line which contains the centers of each particle's circular trajectory we will call the "axis of rotation". The axis of rotation may pass through the object or not.

The obvious difference between rotation and translation is that in translation the object as a whole goes somewhere, with no guarantee that it will ever come back. In pure rotation it remains in the vicinity of the axis of rotation and periodically returns to its original position. I use the term "pure" rotation because it is possible that an object undergo both translation and rotation simultaneously. This is illustrated in the Combined Motion display. In this display we have a solid object made up of ten visible particles. Remember that being a solid means that the particles maintain their original separations. Because this is a computer model rather than a real object, we can also show the center of mass of the object. It is the larger white spot.

What we will do next is develop the tools necessary to predict the future of an object undergoing this sort of motion. A different set of variables is used to describe rotation. The displacement in translation was measured along the coordinate axes, or as a distance and direction. In rotation we will measure the angle from a reference line to the position of interest and call it the angular displacement theta(the Greek letter q). Angular displacement is evidently a scalar quantity since one number completely defines it. If the reference line is the x axis, the angular displacement is the angular position.

The units of angular measure will be the "radian". A radian is the ratio or the length of the arc cut by an angle, to the length of the radius of that arc. See the Angular Displacement display for an illustration. Since it is the ratio of two lengths, the radian is unitless.

We also have measures in rotation analogous to velocity and acceleration. Velocity in angular measure is the rate of change of the angular position with respect to time, Dq/Dt. Since the angular position as we have defined it is a scalar quantity, so is average the angular velocity, being a scalar divided by another scalar, the difference in time. The instantaneous angular velocity is the limit as Dt approaches zero. We will use the Greek letter omega, w, to symbolize the instantaneous angular velocity.

In thinking about angular velocity, we are faced with one of the things that makes introductory physics courses skip over much of the study of rotation. I argued convincingly a while back that average angular velocity, Dq/Dt, was a scalar quantity. When we let Dt approach zero to get the instantaneous angular velocity, the Dq becomes a vector. In fact it becomes a vector pointing off in a direction not even in the plane in which the rotation is taking place. It boggles the mind.

Rather than interrupt our life at this point to ponder this perversion, consider the following experiment to help you believe that finite rotations behave differently than infinitesimal ones. Take a beer stein and hold it with the handle to your right and the opening facing upward. We are going to consider two kinds of rotation of the stein. An "A" rotation will move the opening directly away from you. A "B" rotation will rotate the stein clockwise as you look down on it.

Now perform a 90-degree A rotation, followed by a 90 degree B rotation. You should end up with the stein's handle toward you and the opening to your right. Return the stein to its original position and then give it a 90 degree B rotation followed by a 90 degree A rotation. Now you see the handle on top and the opening away from you. Evidently for large rotations the order is important. For the mathematicians among you, the finite rotation operators A and B do not commute.

Now repeat the experiment with 10-degree rotations. The difference in the final positions in the two cases when the rotation is small is well nigh undetectable. In the limit of infinitely small rotations, the operators do in fact commute. This does not prove that instantaneous angular velocity and acceleration are vectors, but lends credibility that the instantaneous values might be qualitatively different than the average ones.

The average angular acceleration is the rate of change of Dq/Dt over some time difference so it too is a scalar. The instantaneous angular acceleration is a vector symbolized by the Greek letter alpha, a.

The Angular Velocity and Acceleration display shows the successive angular displacement of an object undergoing constant angular acceleration where a = 2 radians per second squared, r/s². The display shows 125 steps of .02 seconds each, about enough to complete one revolution starting from rest. Notice the angular displacement is larger in each step as w increases.

Because of the way we have defined these angular terms, the relationships among q, w and a are the same as the relationships among x, v and a in translational motion. The time slice techniques we developed in going from acceleration to velocity to displacement also apply in going from a to w to q. The Angular Acceleration Time Slice display illustrates this.

So... Let us consider the instantaneous angular velocity, w, to be a vector whose magnitude is the limit as Dt approaches zero of Dq/Dt, and whose direction is given by the direction of your right thumb when your right fingers curl in the direction of the rotation of the object. I am not kidding. That is really the way it is defined. Actually that choice of direction does make a certain amount of sense since it coincides with the axis of rotation. The axis of rotation is the only line uniquely defining the orientation plane in which the rotation of an object takes place.

The instantaneous angular acceleration, a, which is the change in instantaneous angular velocity divided by the time for that change, is of course a vector in the same direction whose magnitude is the limit of Dw/Dt as the change in time approaches zero.

Now let's look at a particle rotating in the (x,y) plane about the origin at a radius r. The relationship of its angular position to its displacement s along the arc of its circular path is s = r * q. The definition of a radian is s/r so with q measured in radians, this relationship comes directly from the definition. A tiny change in s, Ds, then would equal r times the corresponding tiny change in q, Dq. Dividing both sides of that equation by the increment of time Dt over which the changes take place we get Ds/Dt = r * Dq/Dt. But Ds/Dt is just the velocity tangent to the circle and Dq/Dt is just w, so the linear speed, v, of the particle is equal to the radius times the angular speed so v_t = r * w. The same argument we just made for velocity holds for the acceleration as well so the tangential component of a particle's acceleration equals the radius times the angular acceleration so a_t = r * a. In the section on circular motion of a particle we found the radial component of the acceleration of a particle in circular motion of radius r to be v_t² / r which in terms of angular quantities is a_r = (r * w)² / r . So now we have equations relating each of the linear quantities to the angular ones.

Now if you have been paying close attention, you may have noticed that I seem to have talked myself out of the necessity for introducing the angular quantities at all. It seems that we can express even circular motion of a particle by use of the linear quantities. The real reason for the angular quantities is that different points in a rotating body may have different linear displacement, velocity and acceleration. In a body rotating around a fixed axis, each point, and therefore the body as a whole will have the same angular displacement, velocity and acceleration. By use of the angular quantities the motion of the whole body may be described in a simple way.

So far we have developed the necessary tools to describe rotation. Now we will work on what causes rotation. In the case of translation we associated a force with the linear acceleration of an object. In rotational motion, what quantity should we associate with angular acceleration? It must be something other than simple force because the same force, depending on where it is applied will produce different angular accelerations. Think of trying to push open a heavy door by exerting a force near the hinge. The same force applied at the doorknob might work just fine.

To help get a grip on the effect of the position vector r and the force vector f on the resulting torque on a particle, let's set up a reference frame with the (x,y) plane being the plane defined by the two vectors, r and f. Then torque will be directed along the positive or negative z direction. This situation is illustrated in the Torque on a Particle display.

Torque used to be a confusing concept for me. Here are some of the things that bothered me about it. Notice that the units on torque are force times distance. These are the same units that we attached to work in the case of translational motion. Torque and work though are two different physical quantities. One obvious difference is that torque is a vector and work is a scalar.

The vector nature of torque is another issue. Why a vector? And for that matter, where a vector? In our display the torque either points straight out of the screen if positive and straight in negative. In other words, straight along the z Axis. So its magnitude and direction are fixed but what about its location. Customarily we see the torque vector drawn at the origin of the r vector so it is visually more associated with the reference frame than with the particle, still we say it is the torque on the particle. Of course the rest of the story is, that it is the torque on the particle... about the chosen origin. Both the location of the particle and the choice of the axis of rotation affect the torque.

And another thing. The torque vector does not point in a direction in which anything is happening The force, the radius and the motion are all in the (x,y) plane. Why does the torque refuse to cooperate and insist on running at right angles to everything else?

It is best for now to let go of these troubling issues and press on with the study of rotation. Let us for the moment accept that it is torque which produces rotation of a particle about an axis and not stew about the perverse vector nature of the thing.

You might observe that rotation about an axis not passing through the object begins to look like the translation - rotation motion combined, except that the translation is in a circular path. The distinction between translation and rotation is really about whether the object travels in a circle comes back during the time of our observation. In the rotation about the z axis case, it may, so we call that rotation.

Since each particle moves around its orbit with constant speed, the angular acceleration is zero so evidently the torque applied to the object must also be zero since any torque results in angular acceleration. Does this imply that the particles of the system or the system as a whole experiences no external force? Clearly not the case. The center of mass of this system follows a circular path so the object as a whole must be subject to a force. Otherwise it would fly off in a straight line. What the zero torque condition tells us is that the direction of the force on each particle in the system is exactly towards the z axis.

What if we tried to describe the motion of this object where the torque and angular measures were taken about some arbitrary point in the object, say one of the particles for example? Well in theory we could write expressions relating torque to angular acceleration in such a system. The problem with an arbitrary point as a reference for angular quantities is that the motion of the point itself can be quite complicated. It may be rotating about any combination of three axes and also translating, not to mention vibrating. This is going to make a real mess out of anything measured relative to such a point.

There is a point in a collection of particles which is not as badly behaved as an arbitrary point. That is the center of mass. Lets place a little reference frame at the center of mass and picture the movement of our collection of particles in that reference frame. Since this frame travels in a circular path, it is not an inertial reference frame but for purposes of this demonstration that is not an issue. We will deal with non-inertial reference frames in the run-time book, Physics_T. The Center of Mass Frame display adds the center of mass frame to the previous display, with the orientation of that frame fixed in the larger reference frame. In discussing the previous display we said that the net torque was zero. Does that mean that the object is not rotating around the center of mass as it rotates about the z-axis? Have that question in mind as you play with this next display.

As you can see, motion of the object includes its travel around the z axis and a secondary rotation relative to the center of mass frame of reference. In this instance these two rotations are synchronized so as to keep the same side of the object always facing the origin of the reference frame. Like the Moon in its orbit around the Earth. If the object was not rotating on its own axis, as we view it from the reference frame origin it would appear to be rotating once per the period of its orbit around the origin.

Suppose we were observing the object in the reference frame with its origin at the center of mass. Then we would only see the rotation about the object's axis as shown in the Rotation Axis Through Center of Mass display.

Any motion that a rigid object undergoes may be resolved into motion of the center of mass and rotation about the center of mass. Forces that act through the center of mass contribute nothing to the rotation about the center of mass since the torque produced by such forces is zero. Other forces may contribute to both rotation and translation. In the next section on rotational dynamics, we will work out some of the details.

Are there any questions?