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Unformatted text preview: 19 Center of Mass; Moment of Inertia 19—1 Properties of the center of mass In the previous chapter we found that if a great many forces are acting on a
complicated mass of particles, whether the particles comprise a rigid or a nonrigid
body, or a cloud of stars, or anything else, and we ﬁnd the sum of all the forces
(that is, of course, the external forces, because the internal forces balance out), then
if we consider the body as a whole, and say it has a total mass M, there is a certain
point “inside” the body, called the center of mass, such that the net resulting
external force produces an acceleration of this point, just as though the whole
mass were concentrated there. Let us now discuss the center of mass in a little
more detail. The location of the center of mass (abbreviated CM) is given by the equation (19.1) This is, of course, a vector equation which is really three equations, one for each of
the three directions. We shall consider only the x—direction, because if we can
understand that one, we can understand the other two. What does XCM =
Emmi/2m, mean? Suppose for a moment that the object is divided into little
pieces, all of which have the same mass m; then the total mass is simply the number
N of pieces times the mass of one piece, say one gram, or any unit. Then this
equation simply says that we add all the x’s, and then divide by the number of
things that we have added: XCM = min/mN = in/N. In other words,
XCM is the average of all the x’s, if the masses are equal. But suppose one of them
were twice as heavy as the others. Then in the sum, that x would come in twice.
This is easy to understand, for we can think of this double mass as being split
into two equal ones, just like the others; then in taking the average, of course, we
have to count that x twice because there are two masses there. Thus X is the
average position, in the xdirection, of all the masses, every mass being counted a
number of times proportional to the mass, as though it were divided into “little
grams.” From this it is easy to prove that X must be somewhere between the
largest and the smallest x, and, therefore lies inside the envelope including the
entire body. It does not have to be in the material of the body, for the body could
be a circle, like a hoop, and the center of mass is in the center of the hoop, not in
the hoop itself. Of course, if an object is symmetrical in some way, for instance, a rectangle,
so that it has a plane of symmetry, the center of mass lies somewhere on the plane
of symmetry. In the case of a rectangle there are two planes, and that locates it
uniquely. But if it is just any symmetrical object, then the center of gravity lies
somewhere on the axis of symmetry, because in those circumstances there are as
many positive as negative x’s. Another interesting proposition is the following very curious one. Suppose
that we imagine an object to be made of two pieces, A and B (Fig. 19—1). Then
the center of mass of the whole object can be calculated as follows. First, ﬁnd the
center of mass of piece A, and then of piece B. Also, ﬁnd the total mass of each
piece, M A and M 3. Then consider a new problem, in which a point mass M A is
at the center of mass of object A, and another point mass M B is at the center of
mass of object B. The center of mass of these two point masses is then the center
of mass of the whole object. In other words, if the centers of mass of various parts 19—1 19—1 Properties of the center of mass
19—2 Locating the center of mass
19—3 Finding the moment of inertia 19—4 Rotational kinetic energy Fig. 19—1. The CM of a compound
body lies on the line joining the CM’s of
the two composite parts. of an object have been worked out, we do not have to start all over again to ﬁnd
the center of mass of the whole object; we just have to put the pieces together,
treating each one as a point mass situated at the center of mass of that piece.
Let us see why that is. Suppose that we wanted to calculate the center of mass of
a complete object, some of whose particles are considered to be members of
object A and some members of object B. The total sum Zmixi can then be split
into two pieces—the sum ZAmix, for the A object only, and the sum 23mm
for object B only. Now if we were computing the center of mass of object A alone,
we would have exactly the ﬁrst of these sums, and we know that this by itself is
M A X A, the total mass of all the particles in A times the position of the center of
mass of A, because that is the theorem of the center of mass, applied to object A.
In the same manner, just by looking at object B, we get M BX B, and of course, adding the two yields MX:
z mixi + Z mixi
A B MXCM
= MAXA + MBXB. (19.2) Now since M is evidently the sum of M A and M B, we see that Eq. (19.2) can be
interpreted as a special example of the center of mass formula for two point objects,
one of mass M A located at X A and the other of mass M 3 located at X B. The theorem concerning the motion of the center of mass is very interesting,
and has played an important part in the development of our understanding of
physics. Suppose we assume that Newton’s law is right for the small component
parts of a much larger object. Then this theorem shows that Newton’s law is also
correct for the larger object, even if we do not study the details of the object, but
only the total force acting on it and its mass. In other words, Newton’s law has
thepeculiar property that if it is right on a certain small scale, then it will be right
on a larger scale. If we do not consider a baseball as a tremendously complex
thing, made of myriads of interacting particles, but study only the motion of the
center of mass and the external forces on the ball, we ﬁnd F= ma, where F is the
external force on the baseball, m is its mass, and a is the acceleration of its center
of mass. So F = ma is a law which reproduces itself on a larger scale. (There
ought to be a good word, out of the Greek, perhaps, to describe a law which
reproduces the same law on a larger scale.) Of course, one might suspect that the ﬁrst laws that would be discovered by
human beings would be those that would reproduce themselves on a larger scale.
Why? Because the actual scale of the fundamental gears and wheels of the universe
are of atomic dimensions, which are so much ﬁner than our observations that we
are nowhere near that scale in our ordinary observations. So the ﬁrst things that we
would discover must be true for objects of no special size relative to an atomic
scale. If the laws for small particles did not reproduce themselves on a larger scale,
we would not discover those laws very easily. What about the reverse problem?
Must the laws on a small scale be the same as those on a larger scale? Of course
it is not necessarily so in nature, that at an atomic level the laws have to be the same
as on a large scale. Suppose that the true laws of motion of atoms were given by
some strange equation which does not have the property that when we go to a
larger scale we reproduce the same law, but instead has the property that if we
go to a larger scale, we can approximate it by a certain expression such that, if we
extend that expression up and up, it keeps reproducing itself on a larger and larger
scale. That is possible, and in fact that is the way it works. Newton’s laws are the
“tail end” of the atomic laws, extrapolated to a very large size. The actual laws of
motion of particles on a ﬁne scale are very peculiar, but if we take large numbers of
them and compound them, they approximate, but only approximate, Newton’s
laws. Newton’s laws then permit us to go on to a higher and higher scale, and
it still seems to be the same law. In fact, it becomes more and more accurate as
the scale gets larger and larger. This selfreproducing factor of Newton’s laws is
thus really not a fundamental feature of nature, but is an important historical
feature. We would never discover the fundamental laws of the atomic particles at
ﬁrst observation because the ﬁrst observations are much too crude. In fact, it turns 1 9—2 out that the fundamental atomic laws, which we call quantum mechanics, are quite
different from Newton’s laws, and are difﬁcult to understand because all our direct
experiences are with largescale objects and the smallscale atoms behave like
nothing we see on a large scale. So we cannot say, “An atom is just like a planet
going around the sun,” or anything like that. It is like nothing we are familiar with
because there is nothing like it. As we apply quantum mechanics to larger and larger
things, the laws about the behavior of many atoms together do not reproduce
themselves, but produce new laws, which are Newton’s laws, which then continue
to reproduce themselves from, say, micromicrogram size, which still is billions
and billions of atoms, on up to the size of the earth, and above. Let us now return to the center of mass. The center of mass is sometimes
called the center of gravity, for the reason that, in many cases, gravity may be
considered uniform. Let us suppose that we have small enough dimensions that
the gravitational force is not only proportional to the mass, but is everywhere
parallel to some ﬁxed line. Then consider an object in which there are gravitational
forces on each of its constituent masses. Let m, be the mass of one part. Then the
gravitational force on that part is m, times g. Now the question is, where can we
apply a single force to balance the gravitational force on the whole thing, so that the
entire object, if it is a rigid body, will not turn? The answer is that this force must
go through the center of mass, and we show this in the following way. In order that
the body will not turn, the torque produced by all the forces must add up to zero,
because if there is a torque, there is a change of angular momentum, and thus a
rotation. So we must calculate the total of all the torques on all the particles, and
see how much torque there is about any given axis; it should be zero if this axis is
at the center of mass. Now, measuring x horizontally and y vertically, we know
that the torques are the forces in the ydirection, times the lever arm x (that is to
say, the force times the lever arm around which we want to measure the torque).
Now the total torque is the sum 7' = 2 migxi = 82 mixis (193) so if the total torque is to be zero, the sum 2mm must be zero. But 2mm, = M X,
the total mass times the distance of the center of mass from the axis. Thus the
x—distance of the center of mass from the axis is zero. Of course, we have checked the result only for the xdistance, but if we use
the true center of mass the object will balance in any position, because if we turned
it 90 degrees, we would have y’s instead of x’s. In other words, when an object
is supported at its center of mass, there is no torque on it because of a parallel
gravitational ﬁeld. In case the object is so large that the nonparallelism of the
gravitational forces is signiﬁcant, then the center where one must apply the balanc
ing force is not simple to describe, and it departs slightly from the center of mass.
That is why one must distinguish between the center of mass and the center of
gravity. The fact that an object supported exactly at the center of mass will balance
in all positions has another interesting consequence. If, instead of gravitation,
we have a pseudoforce due to acceleration, we may use exactly the same mathe
matical procedure to ﬁnd the position to support it so that there are no torques
produced by the inertial force of acceleration. Suppose that the object is held in
some manner inside a box, and that the box, and everything contained in it, is
accelerating. We know that, from the point of View of someone at rest relative to
this accelerating box, there will be an effective force due to inertia. That is, to
make the object go along with the box, we have to push on it to accelerate it, and
this force is “balanced” by the “force of inertia,” which is a pseudoforce equal to
the mass times the acceleration of the box. To the man in the box, this is the same
situation as if the object were in a uniform gravitational ﬁeld whose “g” value is
equal to the acceleration a. Thus the inertial force due to accelerating an object has
no torque about the center of mass. This fact has a very interesting consequence. In an inertial frame that is not
accelerating, the torque is always equal to the rate of change of the angular mo
mentum. However, about an axis through the center of mass of an object which 19—3 Fig. l9—2. A right triangle and a
right circular cone generated by ro
tating the triangle. is accelerating, it is still true that the torque is equal to the rate of change of the
angular momentum. Even if the center of mass is accelerating, we may still choose
one special axis, namely, one passing through the center of mass, such that it will
still be true that the torque is equal to the rate of change of angular momentum
around that axis. Thus the theorem that torque equals the rate of change of angular
momentum is true in two general cases: (1) a ﬁxed axis in inertial space, (2) an axis
through the center of mass, even though the object may be accelerating. 19—2 Locating the center of mass The mathematical techniques for the calculation of centers of mass are in the
province of a mathematics course, and such problems provide good exercise in
integral calculus. After one has learned calculus, however, and wants to know
how to locate centers of mass, it is nice to know certain tricks which can be used
to do so. One such trick makes use of what is called the theorem of Pappus. It
works like this: if we take any closed area in a plane and generate a solid by moving
it through space such that each point is always moved perpendicular to the plane
of the area, the resulting solid has a total volume equal to the area of the cross
section times the distance that the center of mass moved! Certainly this is true if
we move the area in a straight line perpendicular to itself, but if we move it in a
circle or in some other curve, then it generates a rather peculiar volume. For a
curved path, the outside goes around farther, and the inside goes around less, and
these effects balance out. So if we want to locate the center of mass of a plane
sheet of uniform density, we can remember that the volume generated by spinning
it about an axis is the distance that the center of mass goes around, times the area
of the sheet. For example, if we wish to ﬁnd the center of mass of a right triangle of base
D and height H (Fig. 19—2), we might solve the problem in the following way.
Imagine an axis along H, and rotate the triangle about that axis through a full
360 degrees. This generates a cone. The distance that the xcoordinate of the
center of mass has moved is 21rx. The area which is being moved is the area of
the triangle, %HD. So the xdistance of the center of mass times the area of the
triangle is the volume swept out, which is of course 7rD2H / 3. Thus (21rx)(%HD) =
l/37rD2H, or x = D/ 3. In a similar manner, by rotating about the other axis, or by
symmetry, we ﬁnd y = H /3. In fact, the center of mass of any uniform triangular
area is where the three medians, the lines from the vertices through the centers of
the opposite sides, all meet. That point is 1/3 of the way along each median. Clue:
Slice the triangle up into a lot of little pieces, each parallel to a base. Note that the
median line bisects every piece, and therefore the center of mass must lie on this line. Now let us try a more complicated ﬁgure. Suppose that it is desired to ﬁnd
the position of the center of mass of a uniform semicircular disc—a disc sliced in
half. Where is the center of mass? For a full disc, it is at the center, of course, but
a halfdisc is more difﬁcult. Let r be the radius and x be the distance of the center
of mass from the straight edge of the. disc. Spin it around this edge as axis to
generate a sphere. Then the center of mass has gone around 27rx, the area is
1rr2/2 (because it is only half a circle). The volume generated is, of course, 47rr3/3,
from which we ﬁnd that (27rx)(%1rr2) = 47rr3/3,
or
x = 4r/37r. There is another theorem of Pappus which is a special case of the above one,
and therefore equally true. Suppose that, instead of the solid semicircular disc,
we have a semicircular piece of wire with uniform mass density along the wire,
and we want to ﬁnd its center of mass. In this case there is no mass in the interior,
only on the wire. Then it turns out that the area which is swept by a plane curved
line, when it moves as before, is the distance that the center of mass moves times
the length of the line. (The line can be thought of as a very narrow area, and the
previous theorem can be applied to it.) 19—4 19—3 Finding the moment of inertia Now let us discuss the problem of ﬁnding the moments of inertia of various
objects. The formula for the moment of inertia about the zaxis of an object is I = 2 mix? + y?) 01' I = [(x2 + yz) dm = [(x2 + y2)p dv. (19.4)
That is, we must sum the masses, each one multiplied by the square of its distance
(x,2 + y?) from the axis. Note that it is not the threedimensional distance, only
the twodimensional distance squared, even for a threedimensional object. For
the most part, we shall restrict ourselves to twodimensional objects, but the
formula for rotation about the zaxis is just the same in three dimensions. As a simple example, consider a rod rotating about a perpendicular axis
through one end (Fig. 19—3). Now we must sum all the masses times the xdistances
squared (the y’s being all zero in this case). What we mean by “the sum,” of
course, is the integral of x2 times the little elements of mass. If we divide the rod
into small elements of length dx, the corresponding elements of mass are propor
tional to dx, and if dx were the length of the whole rod the mass would be M.
Therefore dm = de/L
and so
L Md M L ML2
_ 2 x — _ 2 = 
I—Ax L —L/0xdx 3 (19.5) The dimensions of moment of inertia are always mass times length squared, so
all we really had to work out was the factor 1/3. Now what is I if the rotation axis is at the center of the rod? We could just
do the integral over again, letting x range from —%L to +%L. But let us notice a
few things about the moment of inertia. We can imagine the rod as two rods,
each of mass M/2 and length L/2; the moments of inertia of the two small rods
are equal, and are both given by the formula (19.5). Therefore the moment of
inertia is
_ 2(M/2)(L/2)2 _ ML2 I 3 _12 (19.6) Thus it is much easier to turn a rod about its center, than to swing it around an end. Of course, we could go on to compute the moments of inertia of various other
bodies of interest. However, while such computations provide a certain amount
of important exercise in the calculus, they are not basically of interest to us as
such. There is, however, an interesting theorem which is very useful. Suppose
we have an object, and we want to ﬁnd its moment of inertia around some axis.
That means we want the inertia needed to carry it by rotation about that axis.
Now if we support the object on pivots at the center of mass, so that the object
does not turn as it rotates about the axis (because there is no torque on it from
inertial effects, and therefore it will not turn when we start moving it), then the
forces needed to swing it around are the same as though all the mass were concen
trated at the center of mass, and the moment of inertia would be simply 11 =
MREM, where RCM is the distance from the axis to the center of mass. But of
course that is not the right formula for the moment of inertia of an object which
is really being rotated as it revolves, because not only is the center of it moving in
a circle, which would contribute an amount 11 to the moment of inertia, but also
we must turn it about its center of mass. So it is not unreasonable that we must
add to 11 the moment of inertia I, about the center of mass. So it is a good guess that the total moment of inertia about any axis will be
I = 16 + MR?;M. (19.7) 19—5 Fig. 19—3. A straight rod of length
L rotating about an axis through one end. This theorem is called the parallelaxis theorem, and may be easily proved.
The moment of inertia about any axis is the mass times the sum of the xi’s and
the yi’s, each squared: I = Z (x,2 + yf)m,. We shall concentrate on the x’s, but
of course the y’s work the same way. Now x is the distance of a particular point
mass from the origin, but let us consider how it would look if we measured x’ from
the CM, instead of x from the origin. To get ready for this analysis, we write xi = X'r + XCM
Then we just square this to ﬁnd
7612' = x22 + ZXCMx'i + X(2)M So, when this is multiplied by m, and summed over all 1', what happens? Taking
the constants outside the summation sign, we get [2: = Z mzx’z2 + 2XCM Z mzx’r + X32114 2 mi The third sum is easy; it is just M XgM. In the second sum there are two pieces, one
of them is 2 mix; which is the total mass times the x’—coordinate of the center of
mass. But this contributes nothing, because x’ is measured from the center of mass,
and in these axes the average position of all the particles, weighted by the masses,
is zero. The ﬁrst sum, of course, is the x part of 16. Thus we arrive at Eq. (19.7),
just as we guessed. Let us check (19.7) for one example. Let us just see whether it works for the
rod. For an axis through one end, the moment of inertia should be mL2/3, for
we calculated that. The center of mass of a rod, of course, is in the center of the
rod, at a distance L/2. Therefore we should ﬁnd that ML2/3 = MLZ/ 12 +
M(L/2)2. Since onequarter plus onetwelfth is onethird, we have made no
fundamental error. Incidentally, we did not really need to use an integral to ﬁnd the moment of
inertia (19.5). If we simply assume that it is ML2 times 7, an unknown coefﬁcient,
and then use the argument about the two halves to get iv for (19.6), then from our
argument about transferring the axes we could prove that ‘Y = iv + i, so 7
must be 1/3. There is always another way to do it! In applying the parallelaxis theorem, it is of course important to remember
that the axis for I, must be parallel to the axis about which the moment of inertia
is wanted. One further property of the moment of inertia is worth mentioning because
it is often helpful in ﬁnding the moment of inertia of certain kinds of objects.
This property is that if one has a plane ﬁgure and a set of coordinate axes with
origin in the plane and z—axis perpendicular to the plane, then the moment of
inertia of this ﬁgure about the zaxis is equal to the sum of the moments of inertia
about the x and yaxes. This is easily proved by noting that 11: = 2 mm + 222) = 2 my?
(since 2, = 0). Similarly, 1a = Z mz(xf + 221) = 2 mix? I, = Z m,(x§ + yzi) = 2 mix? ‘1‘ 2 mil}?
=n+n As an example, the moment of inertia of a uniform rectangular plate of mass
M, width w, and length L, about an axis perpendicular to the plate and through its
center is simply but I = M(w2 + L2)/12, because its moment of inertia about an axis in its plane and parallel to its length
is Mw2/12, i.e., just as for a rod of length w, and the moment of inertia about the
other axis in its plane is MLZ/ 12, just as for a rod of length L. 19—6 To summarize, the moment of inertia of an object about a given axis, which
we shall call the zaxis, has the following properties: (1) The moment of inertia is 12 = Z mi(x% + y%) = [(x2 + y?) dm.
(2) If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia
of the pieces. (3) The moment of inertia about any given axis is equal to the moment of
inertia about a parallel axis through the CM plus the total mass times the
square of the distance from the axis to the CM. (4) If the object is a plane ﬁgure, the moment of inertia about an axis perpendicu
lar to the plane is equal to the sum of the moments of inertia about any two
mutually perpendicular axes lying in the plane and intersecting at the
perpendicular axis. The moments of inertia of a number of elementary shapes having uniform mass
densities are given in Table 19—1, and the moments of inertia of some other objects, which may be deduced from Table 19—1, using the above properties, are given in
Table 19—2. Table 191
Objéct zaxis I,
Thin rod, length L _L rod at center ML2/ 12
Thin concentric
circular ring, radii J. ring at center M(r“{ + r§)/2
r1 and r2
Sphere, radius r through center 2Mr2/5
Table 19—2
Object zaxis I,
Rect. sheet, sides a, b [I b at center Maz/ 12
Rect. sheet, sides a, b J. sheet at M(a2 + b2)/ 12
center
Thin annular ring, any diameter M0? + r§)/4
radii r 1, r2
Rect. parallelepiped, H c, through M(a2 + b2)/ 12
sides a, b, c center
Rt. circ. cyl., radius [I L, through Mr2/2
r, length L center
Rt. circ. cyl., radius .1. L, through M(r2/4 + L2/ 12)
r, length L center 19—4 Rotational kinetic energy Now let us go on to discuss dynamics further. In the analogy between linear
motion and angular motion that we discussed in Chapter 18, we used the work
theorem, but we did not talk about kinetic energy. What is the kinetic energy of a
rigid body, rotating about a certain axis with an angular velocity as? We can im
mediately guess the correct answer by using our analogies. The moment of inertia
corresponds to the mass, angular velocity corresponds to velocity, and so the
kinetic energy ought to be %Iw2, and indeed it is, as will now be demonstrated.
Suppose the object is rotating about some axis so that each point has a velocity
whose magnitude is can, where r, is the radius from the particular point to the axis. 1 9*7 Then if m, is the mass of that point, the total kinetic energy of the whole thing is
just the sum of the kinetic energies of all of the little pieces: T = % 2 mi”? = % Z mi(riw)2
Now 0:2 is a constant, the same for all points. Thus
T = ME mm? = gm. (19.8) At the end of Chapter 18 we pointed out that there are some interesting
phenomena associated with an object which is not rigid, but which changes from
one rigid condition with a deﬁnite moment of inertia, to another rigid condition.
Namely, in our example of the turntable, we had a certain moment of inertia 11
with our arms stretched out, and a certain angular velocity wl. When we pulled
our arms in, we had a different moment of inertia, 12, and a different angular veloc
ity, (.02, but again we were “rigid.” The angular momentum remained constant,
since there was no torque about the vertical axis of the turntable. This means that
Ilwl = 12w2. Now what about the energy? That is an interesting question.
With our arms pulled in, we turn faster, but our moment of inertia is less, and it
looks as though the energies might be equal. But they are not, because what does
balance is 1w, not Iw2. So if we compare the kinetic energy before and after, the
kinetic energy before is %1 1w? = éLwl, where L = 11m = 12092 is the angular
momentum. Afterward, by the same argument, we have T = %Lw2, and since
(.02 > (.01 the kinetic energy of rotation is greater than it was before. So we had a
certain energy when our arms were out, and when we pulled them in, we were turn
ing faster and had more kinetic energy. What happened to the theorem of the
conservation of energy? Somebody must have done some work. We did work!
When did we do any work? When we move a weight horizontally, we do not do
any work. If we hold a thing out and pull it in, we do not do any work. But that
is when we are not rotating! When we are rotating, there is centrifugal force on
the weights. They are trying to ﬂy out, so when we are going around we have to
pull the weights in against the centrifugal force. So, the work we do against the
centrifugal force ought to agree with the difference in rotational energy, and of
course it does. That is where the extra kinetic energy comes from. There is still another interesting feature which we can treat only descriptively,
as a matter of general interest. This feature is a little more advanced, but is worth
pointing out because it is quite curious and produces many interesting effects. Consider that turntable experiment again. Consider the body and the arms
separately, from the point of view of the man who is rotating. After the weights
are pulled in, the whole object is spinning faster, but observe, the central part of
the body is not changed, yet 1t is turning faster after the event than before. So, if
we were to draw a circle around the inner body, and consider only objects inside the
circle, their angular momentum would change; they are going faster. Therefore
there must be a torque exerted on the body while we pull in our arms. No torque
can be exerted by the centrifugal force, because that is radial. So that means
that among the forces that are developed in a rotating system, centrifugal force is
not the entire story, there is another force. This other force is called Coriolis force,
and it has the very strange property that when we move something in a rotating
system, it seems to be pushed sidewise. Like the centrifugal force, it is an apparent
force. But if we live in a system that is rotating, and move something radially, we
ﬁnd that we must also push it sidewise to move it radially. This sidewise push which
we have to exert is what turned our body around. Now let us develop a formula to show how this Coriolis force really works.
Suppose Moe is sitting on a carousel that appears to him to be stationary. But from
the point of view of Joe, who is standing on the ground and who knows the right
laws of mechanics, the carousel is going around. Suppose that we have drawn a
radial line on the carousel, and that Moe is moving some mass radially along this
line. We would like to demonstrate that a sidewise force is required to do that.
We can do this by paying attention to the angular momentum of the mass. It is 1 9—8 always going around with the same angular velocity w, so that the angular mo
mentum is L = mvtangr = mwr'r = mwr2.
So when the mass is close to the center, it has relatively little angular momentum,
but if we move it to a new position farther out, if we increase r, m has more angular
momentum, so a torque must be exerted in order to move it along the radius.
(To walk along the radius in a carousel, one has to lean over and push sidewise.
Try it sometime.) The torque that is required is the rate of change of L with time
as m moves along the radius. If m moves only along the radius, omega stays con stant, so that the torque is 2 T = Fcr = ail—l; = ——d(r:l;:r ) = mer%,
where F c is the Coriolis force. What we really want to know is what sidewiseforce
has to be exerted by Moe in order to move m out at speed 0, = dr/dt. This is
F, = 'r/r = 2mwvr. Now that we have a formula for the Coriolis force, let us look at the situation
a little more carefully, to see whether we can understand the origin of this force
from a more elementary point of view. We note that the Coriolis force is the same
at every radius, and is evidently present even at the origin! But it is especially
easy to understand it at the origin, just by looking at what happens from the in
ertial system of Joe, who is standing on the ground. Figure 19—4 shows three
successive views of m just as it passes the origin at t = 0. Because of the rotation
of the carousel, we see that m does not move in a straight line, but in a curved path
tangent to a diameter of the carousel where r = 0. In order for m to go in a curve,
there must be a force to accelerate it in absolute space. This is the Coriolis force. This is not the only case in which the Coriolis force occurs. We can also
show that if an object is moving with constant speed around the circumference of
a circle, there is also a Coriolis force. Why? Moe sees a velocity 22M around the
circle. On the other hand, Joe sees m going around the circle with the velocity
UJ = 12M + wr, because m is also carried by the carousel. Therefore we know what
the force really is, namely, the total centripetal force due to the velocity v], or
mpg/r; that is the actual force. Now from Moe’s point of view, this centripetal
force has three pieces. We may write it all out as follows: 2 my my
Fr=__i=_ M
r r — 2vaw — mwzr. Now, F r is the force that Moe would see. Let us try to understand it. Would Moe
appreciate the ﬁrst term? “Yes,” he would say, “even if I were not turning, there
would be a centripetal force if I were to run around a circle with velocity 21M.”
This is simply the centripetal force that Moe would expect, having/nothing to do
with rotation. In addition, Moe is quite aware that there is another centripetal
force that would act even on objects which are standing still on his carousel. This
is the third term. But there is another term in addition to these, namely the second
term, which is again 2mm). The Coriolis force F c was tangential when the velocity
was radial, and now it is radial when the velocity is tangential. In fact, one ex
pression has a minus sign relative to the other. The force is always in the same
direction, relative to the velocity, no matter in which direction the velocity is.
The force is at right angles to the velocity, and of magnitude mev. 1 9—9 Fig. 19—4. Three successive views of
a point moving radially on a rotating
turntable. ...
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This note was uploaded on 06/18/2009 for the course PHYSICS none taught by Professor Leekinohara during the Spring '09 term at Uni. Nottingham  Malaysia.
 Spring '09
 LeeKinohara
 Physics, Center Of Mass, Inertia, Mass

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