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Unformatted text preview: 18 Rotation in Two Dimensions ___________________—————————— 18—1 The center of mass In the previous chapters we have been studying the mechanics of points, or
small particles whose internal structurc’does not concern us. For the next few
chapters we shall study the application of Newton’s laws to more complicated
things. When the world becomes more complicated, it also becomes more inter
esting, and we shall ﬁnd that the phenomena associated with the mechanics of a
more complex object than just a point are really quite striking. Of course these
phenomena involve nothing but combinations of Newton’s laws, but it is some
times hard to believe that only F = ma is at work. The more complicated objects we deal with can be of several kinds: water
ﬂowing, galaxies whirling, and so on. The simplest “complicated” object to analyze,
at the start, is what we call a rigid body, a solid object that is turning as it moves
about. However, even such a simple object may have a most complex motion, and
we shall therefore ﬁrst consider the simplest aspects of such motion, in which an
extended body rotates about a ﬁxed axis. A given point on such a body then moves
in a plane perpendicular to this axis. Such rotation of a body about a ﬁxed axis is
called plane rotation or rotation in two dimensions. We shall later generalize the
results to three dimensions, but in doing so we shall ﬁnd that, unlike the case of
ordinary particle mechanics, rotations are subtle and hard to understand unless
we ﬁrst get a solid grounding in two dimensions. The ﬁrst interesting theorem concerning the motion of complicated objects
can be observed at work if we throw an object made of a lot of blocks and spokes,
held together by strings, into the air. Of course we know it goes in a parabola,
because we studied that for a particle. But now our object is not a particle; it
wobbles and it jiggles, and so on. It does go in a parabola though; one can see
that. What goes in a parabola? Certainly not the point on the corner of the block,
because that is jiggling about; neither is it the end of the wooden stick, or the middle
of the wooden stick, or the middle of the block. But something goes in a parabola,
there is an eiTective “center” which moves in a parabola. So our ﬁrst theorem
about complicated objects is to demonstrate that there is a mean position which is
mathematically deﬁnable, but not necessarily a point of the material itself, which
goes in a parabola. That is called the theorem of the center of the mass, and the
proof of it is as follows. We may consider any object as being made of lots of little particles, the atoms,
with various forces among them. Let i represent an index which deﬁnes one of the
particles. (There are millions of them, so i goes to 1023, or something.) Then the
force on the ith particle is, of course, the mass times the acceleration of that
particle: F, = mi(d2r,/dt2). (18.1) In the next few chapters our moving objects will be ones in which all the
parts are moving at speeds very much slower than the speed of light, and we shall
use the nonrelativistic approximation for all quantities. In these circumstances
the mass is constant, so that F1; = d2(m.r,)/dt2 (18.2) If we now add the force on all the particles, that is, if we take the sum of all the
Fi’s for all the different indexes, we get the total force, F. On the other side of the 18—1 18—1 The center of mass
18—2 Rotation of a rigid body
18—3 Angular momentum 184 Conservation of angular
momentum equation, we get the same thing as though we added before the d1ﬁerent1atlon: 2 F, = F = d———2(Zdi’2"m) (18.3)
i 1 Therefore the total force is the second derivative of the masses times their positions,
added together. Now the total force on all the particles is the same as the external force. Why?
Although there are all kinds of forces on the particles because of the strings, the
wigglings, the pullings and pushings, and the atomic forces, and who knows what,
and we have to add all these together, we are rescued by Newton’s Third Law.
Between any two particles the action and reaction are equal, so that when we add
all the equations together, if any two particles have forces between them it cancels
out in the sum; therefore the net result is only those forces which arise from other
particles which are not included in whatever object we decide to sum over. So if
Eq. (18.3) is the sum over a certain number of the particles, which together are
called “the object,” then the external force on the total object is equal to the sum
of all the forces on all its constituent particles. Now it would be nice if we could write Eq. (18.3) as the total mass times some
acceleration. We can. Let us say M is the sum of all the masses, i.e., the total mass.
Then if we deﬁne a certain vector R to be R = Z min/M, (18.4)
Eq. (18.3) will be simply "'
F = d2(MR)/dt2 = M(d2R/dt2), (18.5) since M is a constant. Thus we ﬁnd that the external force is the total mass times
the acceleration of an imaginary point whose location is R. This point is called
the center of mass of the body. It is a point somewhere in the “middle” of the
object, a kind of average r in which the different ri’s have weights or importances
proportional to the masses. We shall discuss this important theorem in more detail in a later chapter, and
we shall therefore limit our remarks to two points: First, if the external forces are
zero, if the object were ﬂoating in empty space, it might whirl, and jiggle, and twist,
and do all kinds of things. But the center of mass, this artiﬁcially invented, cal
culated position, somewhere in the middle, will move with a constant velocity.
In particular, if it is initially at rest, it will stay at rest. So if we have some kind
of a box, perhaps a space ship, with people in it, and we calculate the location of
the center of mass and ﬁnd it is standing still, then the center of mass will continue
to stand still if no external forces are acting on the box. Of course, the space ship
may move a little in space, but that is because the people are walking back and
forth inside; when one walks toward the front, the ship goes toward the back so as
to keep the average position of all the masses in exactly the same place. Is rocket propulsion therefore absolutely impossible because one cannot move
the center of mass? No; but of course we ﬁnd that to propel an interesting part
of the rocket, an uninteresting part must be thrown away. In other words, if we
start with a rocket at zero velocity and we spit some gas out the back end, then this
little blob of gas goes one way as the rocket ship goes the other, but the center of
mass is still exactly where it was before. So we simply move the part that we are
interested in against the part we are not interested in. The second point concerning the center of mass, which is the reason we
introduced it into our discussion at this time, is that it may be treated separately
from the “internal” motions of an object, and may therefore be ignored in our
discussion of rotation. 18—2 Rotation of a rigid body Now let us discuss rotations. Of course an ordinary object does not simply
rotate, it wobbles, shakes, and bends, so to simplify matters we shall discuss the
motion of a nonexistent ideal object which we call a rigid body. This means an 18—2 object in which the forces between the atoms are so strong, and of such character,
that the little forces that are needed to move it do not bend it. Its shape stays
essentially the same as it moves about. If we wish to study the motion of such a
body, and agree to ignore the motion of its center of mass, there is only one thing
left for it to do, and that is to turn. We have to describe that. How? Suppose
there is some line in the body which stays put (perhaps it includes the center of
mass and perhaps not), and the body is rotating about this particular line as an
axis. How do we deﬁne the rotation? That is easy enough, for if we mark a point
somewhere on the object, anywhere except on the axis, we can always tell exactly
where the object is, if we only know where this point has gone to. The only thing
needed to describe the position of that point is an angle. So rotation consists of a
study of the variations of the angle with time. In order to study rotation, we observe the angle through which a body has
turned. Of course, we are not referring to any particular angle inside the object
itself; it is not that we draw some angle on the object. We are talking about the
angular change of the position of the whole thing, from one time to another. First, let us study the kinematics of rotations. The angle will change with time,
and just as we talked about position and velocity in one dimension, we may talk
about angular position and angular velocity in plane rotation. In fact, there is a
very interesting relationship between rotation in two dimensions and onedimen
sional displacement, in which almost every quantity has its analog. First, we have
the angle 6 which deﬁnes how far the body has gone around; this replaces the
distance y, which deﬁnes how far it has gone along. In the same manner, we have a
velocity of turning, to = do/dt, which tells us how much the angle changes in a
second, just as v = ds/dt describes how fast a thing moves, or how far it moves
in a second. If the angle is measured in radians, then the angular velocity to will
be so and so many radians per second. The greater the angular velocity, the faster
the object is turning, the faster the angle changes. We can go on: we can differ
entiate the angular velocity with respect to time, and we can call a = dw/dt =
alzti/dt2 the angular acceleration. That would be the analog of the ordinary accel
eration. Now of course we shall have to relate the dynamics of rotation to the laws of
dynamics of the particles of which the object is made, so we must ﬁnd out how a
particular particle moves when the angular velocity is such and such. To do this,
let us take a certain particle which is located at a distance r from the axis and say
it is in a certain location P(x, y) at a given instant, in the usual manner (Fig. 18—1).
If at a moment At later the angle of the whole object has turned through A0, then
this particle is carried with it. It is at the same radius away from 0 as it was before,
but is carried to Q. The ﬁrst thing we would like to know is how much the distance
x changes and how much the distance y changes. If OP is called I, then the length
PQ is r A0, because of the way angles are deﬁned. The change in x, then, is simply
the projection of r A0 in the x—direction: Ax = —PQ sin 0 = —rA0 (y/r) = —yA0. (18.6)
Similarly,
Ay = +xA0. (18.7) If the object is turning with a given angular velocity to, we ﬁnd, by dividing both
sides of (18.6) and (18.7) by At, that the velocity of the particle is 22,, = —wy and 1),, = +wx. (18.8) Of course if we want to ﬁnd the magnitude of the velocity, we just write
2) = V05 + v; = Vw2y2 + w2x2 = m/x2 + y2 = wr (18.9) It should not be mysterious that the value of the magnitude of this velocity is
wr; in fact, it should be selfevident, because the distance that it moves is r A0 and
the distance it moves per second is r AO/At, or rw. 18—3 Fig. 18—1.
sional rotation. Kinematics of twodimen Let us now move on to consider the dynamicsof rotation. Here a new
concept, force, must be introduced. Let us inquire whether we can invent something
which we shall call the torque (L. torquere, to twist) which bears the same relation
ship to rotation as force does to linear movement. A force is the thing that is
needed to make linear motion, and the thing that makes something rotate is a
“rotary force” or a “twisting force,” Le, a torque. Qualitatively, a torque is a
“twist”; what is a torque quantitatively? We shall get to the theory of torques
quantitatively by studying the work done in turning an object, for one very nice
way of deﬁning a force is to say how much work it does when it acts through a
given displacement. We are going to try to maintain the analogy between linear
and angular quantities by equating the work that We do when we turn something a
little bit when there are forces acting on it, to the torque times the angle it turns
through. In other words, the deﬁnition of the torque is going to be so arranged
that the theorem of work has an absolute analog: force times distance is work, and
torque times angle is going to be work. That tells us what torque is. Consider,
for instance, a rigid body of some kind with various forces acting on it, and an
axis about which the body rotates. Let us at ﬁrst concentrate on one force and
suppose that this force is applied at a certain point (x, y). How much work would
be done if we were to turn the object through a very small angle? That is easy.
The work done is AW = F,c Ax + F1, Ay. (18.10) We need only to substitute Eqs. (18.6) and (18.7) for Ax and Ay to obtain
AW = (xFy — ny) A0. (18.11) That is, the amount of work that we have done is, in fact, equal to the angle through
which we have turned the object, multiplied by a strangelooking combination of
the force and the distance. This “strange combination” is what we call the torque.
So, deﬁning the change in work as the torque times the angle, we now have the
formula for torque in terms of the forces. (Obviously, torque is not a completely
new idea independent of Newtonian mechanics—torque must have a deﬁnite
deﬁnition in terms of the force.) When there are several forces acting, the work that is done is, of course, the
sum of the works done by all the forces, so that AW will be a whole lot of terms,
all added together, for all the forces, each of which is proportional, however, to A0.
We can take the A0 outside and therefore can say that the change in the work is
equal to the sum of all the torques due to all the different forces that are acting,
times A9. This sum we might call the total torque, 1. Thus torques add by the
ordinary laws of algebra, but we shall later see that this is only because we are
working in a plane. It is like onedimensional kinematics, where the forces simply
add algebraically, but only because they are all in the same direction. It is more
complicated in three dimensions. Thus, for twodimensional rotation, 7', = xiFyi  yini (18.12)
and T = 27,. (18.13) It must be emphasized that the torque is about a given axis. If a diﬂ‘erent.axis is chosen, so that all the x, and y, are changed, the value of the torque is (usually)
changed too. Now we pause brieﬂy to note that our foregoing introduction of torque,
through the idea of work, gives us a most important result for an object in equilib
rium: if all the forces on an object are in balance both for translation and rotation,
not only is the net force zero, but the total of all the torques is also zero, because
if an object is in equilibrium, no work is done by the forces for a small displacement.
Therefore, since AW = 1A0 = 0, the sum of all the torques must be zero. So
there are two conditions for equilibrium: that the sum of the forces is zero, and
that the sum of the torques is zero. Prove that it sufﬁces to be sure that the sum
of torques about any one axis (in two dimensions) is zero. 18—4 Now let us consider a single force, and try to ﬁgure out, geometrically, what
this strange thing xFy — ny amounts to. In Fig. 18—2 we see a force F acting at
a point r. When the object has rotated through a small angle A0, the work done,
of course, is the component of force in the direction of the displacement times the
displacement. In other words, it is only the tangential component of the force
that counts, and this must be multiplied by the distance rA0. Therefore we see
thatlthe torque is also equal to the tangential component of force (perpendicular
to the radius) times the radius. That makes sense in terms of our ordinary idea
of the torque, because if the force were completely radial, it would not put any
“twist” on the body; it is evident that the twisting effect should involve only the
part of the force which is not pulling out from the center, and that means the
tangential component. Furthermore, it is clear that a given force is more effective
on a long arm than near the axis. In fact, if we take the case where we push right
on the axis, we are not twisting at all! So it makes sense that the amount of twist,
or torque, is proportional both to the radial distance and to the tangential com
ponent of the force. There is still a third formula for the torque which is very interesting. We have
just seen that the torque is the force times the radius times the sine of the angle a,
in Fig. 18—2. But if we extend the line of action of the force and draw the line OS,
the perpendicular distance to the line of action of the force (the lever arm of the
force) we notice that this lever arm is shorter than r in just the same proportion
as the tangential part of the force is less than the total force. Therefore the formula
for the torque can also be written as the magnitude of the force times the length
of the lever arm. The torque is also often called the moment of the force. The origin of this
term is obscure, but it may be related to the fact that “moment” is derived from
the Latin movimentum, and that the capability of a force to move an object (using
the force on a lever or crowbar) increases with the length of the lever arm. In
mathematics ‘moment” means weighted by how far away it is from an axis. 18—3 Angular momentum Although we have so far considered only the special case of a rigid body, the
properties of torques and their mathematical relationships are interesting also
even when an object is not rigid. In fact, we can prove a very remarkable theorem:
just as external force is the rate of change of a quantity p, which we call the total
momentum of a collection of particles, so the external torque is the rate of change
of a quantity L which we call the angular momentum of the group of particles. To prove this, we shall suppose that there is a system of particles on which
there are some forces acting and ﬁnd out what happens to the system as a result of
the torques due to these forces. First, of course, we should consider just one
particle. In Fig. 18—3 is one particle of mass m, and an axis 0; the particle is not
necessarily rotating in a circle about 0, it may be moving in an ellipse, like a planet
going around the sun, or in some other curve. It is moving somehow, and there
are forces on it, and it accelerates according to the usual formula that the xcom
ponent of force is the mass times the xcomponent of acceleration, etc. But let us
see what the torque does. The torque equals xFy — ny, and the force in the
x or yd1rection is the mass times the acceleration in the x or ydirection: T = xFy — ny
= xm(d2y/dt2) — ym(d2x/dt2). (18.14) Now, although this does not appear to be the derivative of any simple quantity, it
is in fact the derivative of the quantity xm(dy/dt) — ym(dx/dt): d dy dx _ dzy dx dy
a l” (a) ‘ y’” (27)] ‘ x’" (W) + (73) m (a) (1815)
d2x) dy) dx) _ _ any) ﬁx)
— ym dt2 _ dt m dt xm “’12 — ym W I 1 8—5 Fig. 18—2. The torque produced by
a force. Fig. 18—3. A particle moves about
an axis 0. So it is true that the torque is the rate of change of something with time! So we
pay attention to the “something,” we give it a name: we call it L, the angular
momentum: L = xm(dy/dt) — ym(dx/dt)
= xpy  mp. (1816) Although our present discussion is nonrelativistic, the second form for L
given above is relativistically correct. So we have found that there is also a rota
tional analog for the momentum, and that this analog, the angular momentum, is
given by an expression in terms of the components of linear momentum that is
just like the formula for torque in terms of the force components! Thus, if we want
to know the angular momentum of a particle about an axis, we take only the
component of the momentum that is tangential, and multiply it by the radius. In
other words, what counts for angular momentum is not how fast it is going away
from the origin, but how much it is going around the origin. Only the tangential
part of the momentum counts for angular momentum. Furthermore, the farther
out the line of the momentum extends, the greater the angular momentum. And
also, because the geometrical facts are the same whether the quantity is labeled
p or F, it is true that there is a lever arm (not the same as the lever arm of the force
on the particle!) which is obtained by extending the line of the momentum and
ﬁnding the perpendicular distance to the axis. Thus the angular momentum is the
magnitude of the momentum times the momentum lever arm. So we have three
formulas for angular momentum, just as we have three formulas for the torque: L = xpy — ypz
= rptang = p  lever arm. (18.17) Like torque, angular momentum depends upon the position of the axis about
which it is to be calculated. Before proceeding to a treatment of more than one particle, let us apply the
above results to a planet going around the sun. In which direction is the force?
The force is toward the sun. What, then, is the torque on the object? Of course,
this depends upon where we take the axis, but we get a very simple result if we
take it at the sun itself, for the torque is the force times the lever arm, or the com
ponent of force perpendicular to r, times r. But there is no tangential force, so
there is no torque about an axis at the sun! Therefore, the angular momentum of
the planet going around the sun must remain constant. Let us see what that means.
The tangential component of velocity, times the mass, times the radius, will be
constant, because that is the angular momentum, and the rate of change of the
angular momentum is the torque, and, in this problem, the torque is zero. Of
course since the mass is also a constant, this means that the tangential velocity
times the radius is a constant. But this is something we already knew for the motion
of a planet. Suppose we consider a small amount of time At. How far will the
planet move when it moves from P to Q (Fig. 18—3)? How much area will it sweep
through? Disregarding the very tiny area QQ’P compared with the much larger
area OPQ, it is simply half the base PQ times the height, 0R. In other words, the
area that is swept through in unit time will be equal to the velocity times the lever
arm of the velocity (times onehalf). Thus the rate of change of area is proportional
to the angular momentum, which is constant. So Kepler’s law about equal areas
in equal times is a word description of the statement of the law of conservation of
angular momentum, when there is no torque produced by the force. 18—4 Conservation of angular momentum Now we shall go on to consider what happens when there is a large number
of particles, when an object is made of many pieces with many forces acting between
them and on them from the outside. Of course, we already know that, about any
given ﬁxed axis, the torque on the ith particle (which is the force on the ith particle 18—6 times the lever arm of that force) is equal to the rate of change of the angular
momentum of that particle, and that the angular momentum of the ith particle
is its momentum times its momentum lever arm. Now suppose we add the torques
T, for all the particles and call it the total torque 1'. Then this will be the rate of
change of the sum of the angular momenta of all the particles L,, and that deﬁnes
a new quantity which we call the total angular momentum L. Just as the total
momentum of an object is the sum of the momenta of all the parts, so the angular
momentum is the sum of the angular momenta of all the parts. Then the rate of
change of the total L is the total torque: dL. * dL
7' _ Zn _ dt — E‘ (18.18) Now it might seem that the total torque is a complicated thing. There are all
those internal forces and all the outside forces to be considered. But, if we take
Newton’s law of action and reaction to say, not simply that the action and reaction
are equal, but also that they are directed exactly oppositely along the same line
(Newton may or may not actually have said this, but he tacitly assumed it), then
the two torques on the reacting objects, due to their mutual interaction, will be
equal and opposite because the lever arms for any axis are equal. Therefore the
internal torques balance out pair by pair, and so we have the remarkable theorem
that the rate of change of the total angular momentum about any axis is equal to
the external torque about that axis! ‘7' = Z“ = r,,, = dL/dt. (18.19) Thus we have a very powerful theorem concerning the motion of large collections
of particles, which permits us to study the overall motion without having to look
at the detailed machinery inside. This theorem is true for any collection of objects,
whether they form a rigid body or not. One extremely important case of the above theorem is the law of conservation
of angular momentum: if no external torques act upon a system of particles, the
angular momentum remains constant. A special case of great importance is that of a rigid body, that is, an object of a
deﬁnite shape that is just turning around. Consider an object that is ﬁxed in its
geometrical dimensions, and which is rotating about a ﬁxed axis. Various parts of
the object bear the same relationship to one another at all times. Now let us try
to ﬁnd the total angular momentum of this object. If the mass of one of its particles
is m), and its position or location is at (x,, y,~), then the problem is to ﬁnd the
angular momentum of that particle, because the total angular momentum is the
sum of the angular momenta of all such particles in the body. For an object going
around in a circle, the angular momentum, of course, is the mass times the velocity times the distance from the axis, and the velocity is equal to the angular velocity
times the distance from the axis: Li = miviri = mtﬁw, (18.20) or, summing over all the particles i, we get L = In), (18.21)
where 1 = Z mug. (18.22) This is the analog of the law that the momentum is mass times velocity. Velocity is replaced by angular velocity, and we see that the mass is replaced by
a new thing which we call the moment of inertia I, which is analogous to the mass.
Equations (18.21) and (18.22) say that a body has inertia for turning which depends,
not just on the masses, but on how far away they are from the axis. So, if we have
two objects of the same mass, when we put the masses farther away from the axis,
the inertia for turning will be higher. This is easily demonstrated by the apparatus 18—7 Fig. 18—4. The “inertia for turning"
depends upon the lever arm of the masses. shown in Fig. 18—4, where a weight M is kept from falling very fast because it has
to turn the large weighted rod. At ﬁrst, the masses m are close to the axis, and
M speeds up at a certain rate. But when we change the moment of inertia by
putting the two masses m much farther away from the axis, then we see that M
accelerates much less rapidly than it did before, because the body has much more
inertia against turning. The moment of inertia is the inertia against turning, and
is the sum of the contributions of all the masses, times their distances squared,
from the axis. There is one important difference between mass and moment of inertia which
is very dramatic. The mass of an object never changes, but its moment of inertia
can be changed. If we stand on a frictionless rotatable stand with our arms out
stretched, and hold some weights in our hands as we rotate slowly, we may change
our moment of inertia by drawing our arms in, but our mass does not change.
When we do this, all kinds of wonderful things happen, because of the law of the
conservation of angular momentum: If the external torque is zero, then the angular
momentum, the moment of inertia t1mes omega, remains constant. Initially, we
were rotating with a large moment of inertia 11 at a low angular velocity (.01, and
the angular momentum was I 10.: 1. Then we changed our moment of inertia by
pulling our arms in, say to a smaller value 12. Then the product Iw, which has to
stay the same because the total angular momentum has to stay the same, was I2w2. So 110:1 = Ing. That is, if we reduce the moment of inertia, we have to
increase the angular velocity. 18—8 ...
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 Spring '09
 LeeKinohara
 Physics

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