Conservation of Angular Momentum

Having defined the Universe as the set of all things that exist and having identified space as an element of that set, we must proclaim that nothing exists outside the Universe. Indeed, the concept "outside the Universe" has no meaning because it depends upon something like space existing beyond whatever boundary we take as enclosing the Universe. That proposition, in turn, has led us to state that the Universe cannot move and to deduce from that statement the laws of motion originally devised by Isaac Newton. Now we draw from that proposition the statement that the Universe has no rotary motion and will derive from that statement the laws pertaining to rotary motion that are analogous to Newton's laws of linear motion.

I point out the fact that the nonexistence of space outside the Universe necessarily entails the nonexistence of direction outside the Universe. If we cannot orient the Universe in some direction, then we certainly cannot change the Universe's orientation, which means that the Universe cannot have any rotary motion. But within the Universe, within space, bodies can rotate and revolve about other bodies and thereby display rotary motion. As we did with linear motion, we make the statement that the Universe having no rotary motion remains true to Reality if and only if all of the rotary motions within the Universe always add up to a net zero.

That latter proposition necessitates two rules pertaining to rotary motion. First, the rotary motion of a body or system of bodies must remain unchanged unless the body or system of bodies interacts with some other body or system of bodies capable of having rotary motion. And second, a change in the rotary motion of one body or system of bodies necessitates an equal and oppositely directed change in the rotary motion of some other body or system of bodies.

We know that something called rotary motion has something to do with things going round and round, but does what I call rotary motion congrue with what physicists call angular momentum? In order to answer that question I will lead you through a consideration of the simplest rotating system - two bodies connected to each other by a thread and moving in opposite directions. Actually, I want to connect the bodies, of mass m1 and m2 and moving at speeds v1 and v2, to each other by two threads, a short one of length l and a long one of length L. And because I want to consider the dynamic properties of only the bodies, I feign to have made the threads of an unstretchable fantasy material so light that we may regard the threads as effectively having no mass.

Because the threads cannot stretch, they constrain the bodies to move in circles. To make the situation as simple as possible, I assume that the two bodies possess zero net linear momentum. That assumption necessitates that the bodies possess at any given instant equal and oppositely directed linear momenta, that the system as a whole does not drift through space, and that the bodies revolve about one fixed point in space. Contrary to what Aristotle taught, bodies do not move in circles out of their own propensities: something must force bodies to move in circles. In this case tension in the taut thread produces the force that pulls each body onto a circular path. Now conservation of linear momentum necessitates that the taut thread exert equal and oppositely directed forces upon the two bodies and that necessity, in turn, determines how the bodies move.

If a body moves on a circle with unchanging speed, an arrow representing the body's velocity must rotate in the same manner in which an arrow representing the body's distance from the center of the circle rotates. But in a vector representation the velocity arrow expresses the ratio between a change in the radius vector and the elapse of time (v = dr/dt), so we know that we must find a right angle between r and dr at all times. Because we consider only instantaneous values of the relevant changes, we must make dr so short that it adds nothing to the length of r as it indicates how r rotates. Now dr itself must also rotate, so we have a change in dr, represented by the arrow d2r, which has the same proportion to dr that dr has to r, so we have the equality of the ratios

d2r/dr = dr/r.

(Eq'n 1)

If I multiply that equation by dr and divide it by the square of the time interval dt, then I can calculate the rate at which the body must accelerate toward the center of the circle in order to continue its motion on the circle. I obtain

d2r/dt2 = a = v2/r.

(Eq'n 2)

That equation gives us the basic kinematics of simple circular motion.

If I multiply that equation by the mass of the body, I calculate the force that something must exert upon the body to keep it moving on the circle. I now turn my attention back to my two corotating bodies and note that whatever keeps them moving on their respective circles must exert forces of equal magnitude upon them, so I have

m1v12/r1 = m2v22/r2,

(Eq'n 3)

in which equation r1 and r2 represent the distances of Body-1 and Body-2, respectively, from the common center of their circles, the unique point on the string connecting the two bodies that does not move. We also have r1+r2=l, the length of the thread. Because both bodies must go around their circles in the same time, they must have the same angular velocity, so we also have

ω = v1/r1 = v2/r2.

(Eq'n 4)

If I substitute from that equation into Equation 3 and divide out the common factors, I obtain the statement that m1v1=m2v2, which tells us that at any given instant the bodies possess equal (and oppositely directed) linear momenta. We knew that fact already, but this gives us a confirmation that our reasoning has gone in the right direction so far.

We can interpret the equality of the bodies' linear momenta via Equation 4 (in the form v1=r1 and v2=r2) to get

m1r1 = m2r2.

(Eq'n 5)

Using r2 = l-r1 to eliminate r2, I can rewrite that equation as

r1 = m2l/(m1+m2).

(Eq'n 6)

This equation and/or the similar one for r2 tells us how to locate on the thread the unique point that does not move. We call that point the center of mass of the system.

Now I want to consider what happens when the short thread breaks. What property of the system remains unchanged when that happens?

After the short thread breaks both bodies travel on straight lines until the long thread goes taut, when the bodies resume moving on circles about their common center of mass. In going taut the long thread absorbs from both bodies that part of their linear momenta parallel to the orientation of the thread at that instant. That act leaves only the parts of the bodies' linear momenta perpendicular to the thread at that instant. Now I want to determine the relationship between that remnant momentum of one of the bodies and the body's original momentum.

For each body we can draw two similar right triangles, a distance triangle and a momentum triangle. Comparing the ratios of the sides of those triangles will give us our desired relationship.

In drawing the distance triangle I take the line defined by the short thread, from the system's center of mass to the body, at the instant the short thread breaks as the base. I take the line defined by the long thread, from the system's center of mass to the body, as the hypotenuse. And I take the straight line traced by the body between the instant the short thread breaks and the long thread goes taut as the altitude.

I draw the momentum triangle with straight lines representing the body's linear momentum at the instant the short thread breaks and representing two of its component vectors. In particular I want to use the parts of the momentum vector that extend parallel to and perpendicular to the long thread at the instant that thread goes taut. Because the lines representing those components necessarily meet at a right angle, we must take the line representing the initial momentum as the hypotenuse of the momentum triangle. We can see in our imaginations that the longer I make the long thread relative to the short thread, the longer I must draw the altitude of the distance triangle and the more of the initial momentum the long thread will subtract from the body when it goes taut. Thus, in drawing the momentum triangle I must take the line representing the momentum absorbed by the long thread as the altitude and take the line representing the momentum left over, the component of momentum perpendicular to the long thread at the instant it goes taut, as the base of the triangle.

Because we have two similar triangles we know that we can equate the ratios of corresponding sides to each other. In particular we can equate the ratio of the hypotenuses to the ratio of the bases. We write that statement algebraically as

rl/mvi = rs/mvr,

(Eq'n 7)

in which equation rl represents the length of the long thread (measured from the system's center of mass to the body), rs represents the length of the short thread (measured from the system's center of mass to the body), vi represents the body's velocity when the short thread breaks, and vr represents the body's remnant velocity, the body's velocity immediately after the long thread goes taut. I can rewrite that equation, via the rules of algebra, as

mvrrl = mvirs.

(Eq'n 8)

That equation tells us that what has remained unchanged in this scenario coincides with the product of the body's mass, the body's velocity, and the body's distance from the system's center of mass. Of course, we can make the same statement of the other body.

But that statement, as it stands, is not strictly true to Reality. In the interval between the breaking of the short thread and the tautening of the long thread the product of the body's linear momentum and the body's distance from the system's center of mass increases. In that interval we cannot use Equation 8 to describe the body's rotary motion if we insist that the rotary motion remains unchanged. However we can repair that situation if we re-examine our triangles in light of trigonometry and discern that at the instant that the long thread goes taut we have

vr = viSinq,

(Eq'n 9)

in which equation q represents the angle between the arrow representing vi and the arrow representing the body's distance from the system's center of mass (the vector corresponding to rl at that instant). That relationship holds true to Reality as we trace the body's movement backward to the point where the short thread broke, so now we know that the amount of rotary motion in the body conforms to the product

L = mvrSinq.

(Eq'n 10)

That product has the same magnitude as does the vector cross product

L = mv×r.

(Eq'n 11)

What more do we need in order to identify L as the quantity of rotary motion possessed by our body? We should notice here that the arrow representing L points in a direction perpendicular to both of the arrows representing v and r. As our body goes round and round the vectors v and r rotate in the same way, but while they rotate the vector L remains pointing steadfastly in one direction. We require just such steadfastness of the mathematical description of rotary motion, something that provides a unique direction to the expression of Equation 10 while remaining unchanged until forced to change.

With those facts in mind we can say the Equation 11 describing the quantity of a body's rotary motion does, indeed, describe the angular momentum of that body as it revolves about an axis. Thus Rule #2 corresponds precisely to the law of conservation of angular momentum. But just to make certain of the truth of that statement I want to conduct some more imaginary experiments.

Equation 11 describes the angular momentum of a single body. We can also apply it to the description of a multiple body system by using it to calculate the angular momentum of each component body of the system and then adding up all of the contributions. For our two-body bolo we have

L = m1v1r1 + m2v2r2.

(Eq'n 12)

But in this example m1v1 = m2v2 = p, so we also have

L = pr1 + pr2 = pr12;

(Eq'n 13)

that is, the total angular momentum of the bolo equals the angular momentum one of the bodies would have if it revolved alone at one end of the thread connecting the two bodies with the other end of the thread connected to a freely rotating, otherwise immovable axle.

That fact suggests a new question. Define a point A situated a distance d from the center of mass of our bolo in the same plane in which the two bodies move. Draw through that point A a straight line parallel to the arrow representing the bolo's angular momentum (and perpendicular to the plane). Does the bolo have an angular momentum relative to that line; that is, can we rightly regard that line as an axis about which the bodies comprising the bolo have a rotary motion?

At any given instant one body's linear momentum vector makes an angle with the line d. If we draw a line extending that linear momentum vector, that extended line passes closest to the point A at a distance equal to dSinq+r1 or dSinq-r2, depending upon which body we consider. We identify those minimum distances as the moment arms on which the linear momenta of the bodies manifest an angular momentum. Thus, at that instant the bodies possess, relative to the point A, angular momenta L1 = p1(dSinq+r1) and L2 = p2(dSinq-r2). But we have p1 = -p2 true to Reality, so our bolo must have a net angular momentum relative to the point A of L = p(r1+r2), identical to the result in Equation 13.

We have found, then, that the angular momentum of a rotating system does not depend upon the location of the axis to which we refer it. This fact conforms to the requirement that the conservation law have absolute control over all rotary motions in the Universe. We will not contrive any manipulation of rotating systems that will violate the conservation law, certainly not by using systems with shifting axes of rotation.

From here the discussion of rotary motion wanders farther from the conservation law and follows the standard presentation of the lower division course in classical mechanics. We gain the concepts and mathematical expressions for moment of inertia and torque. I won't waste your time by covering that material here.

Obligatory Forces

Imagine that we have a compact bead free to slide on a circular loop of wire whose radius we may expand or contract at will. We know that the wire exerts a centripetally directed force on the bead to keep it moving in a circle and we know, from Newton's third law of motion, that the bead exerts an equal and centrifugally directed force upon the wire. Further, we know that the product of the radius of the circle and the speed of the bead's motion on the wire must remain constant, so if the radius of the wire changes, the bead's speed must also change: the bead moves as if a force acted to accelerate or decelerate its motion on the wire. Physicists and engineers conventionally call those forces, the centrifugal force and the Coriolis force, fictitious forces, largely because those forces do not originate in some property of matter as do, for example, the electric force or gravity. But I see nothing fictitious about those forces: they originate not in properties of matter but rather in the obligations that the structure of Reality impose upon matter in motion. For that reason I believe that we would name those forces more accurately if we call them obligatory forces.

You may object that "a rose by any other name would smell as sweet", but I will point out that, while he made a good point, Shakespeare knew nothing of the Sapir-Whorf Hypothesis. As I will show in another essay, words do indeed condition our thoughts and, if poorly chosen, degrade our understanding of Reality.

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