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Suppose the objJect is rotating about some axis so that each point has a velocity
whose magnitude is œ¿r¿, where r¿ is the radius from the particular poin$ to the
axis. Thhen If rm¿ 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 littÌe pieces:
T=j » TUỆ = 5 » m¿(r¿o)Ÿ.
Now ¿2 is a constant, the same for all points. Thus
T= 3® mịn? = 310. (19.8)
At the end of Chapter 1S 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 defnite moment of inertia, to another rigid condition.
Namely, in our example of the turntable, we had a certain moment of inertia Ï
with our arms stretched out, and a certain angular velocity œị. When we pulled
our arms in, we had a diferent moment of inertia, f¿, and a diferent angular
velocity, œ2, bu again we were “rigid.” The angular momentum remained constant,
since there was no torque about the vertical axis of the turntable. 'Phis means
that Tới = Taøa¿. 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 balanee is Tự, not Tư?. So if we compare the kinetie energy before and
after, the kinetic energy before is shư? = s0, where Ù = lịư1 = Ï2ús is the
angular momentum. Afterward, by the same argument, we have 7 = 3a and
since œa > œ0 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 turning faster and had more kinetic energy. What happened to the theorem
of the conservation of energy? Somebody must have done some work. We did
workl 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 rotatingl When we are rotating, there is centrifugal
force on the weights. 'Phey are trying to y 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 i% does. That is where the extra kinetic energy comes Írom.
--- Trang 359 ---
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
cfects.
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, #he centrol part
0ƒ the bod is not changed, yet 1 1s turning faster after the event than before.
So, 1Ý we were to draw a circle around the inner body, and consider only obJects
inside the circle, /he#r angular momentum would chønge; 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. 5o that
means that among the forces that are developed in a rotating system, centrifugal
force is not the entire story, £here is œnother force. This other force is called
Coriolis [orce, and i9 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 IÝ we live in a system that is rotating, and move
something radially, we fnd that we must also push ït sidewise to move it radially.
'This sidewise push which we have to exert 1s 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
trom the point of view of jJoe, 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. Ït is always going around with the same angular velocity œ, so that the
angular momentum is
= ThU‡angT — THUT ©† = m2.
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 z, rm has more
angular momentum, so a #orque rnust be ezerted 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 Ù
with tỉme as ?w moves along the radius. If m moves only along the radius, omega
--- Trang 360 ---
stays constant, so that the torque is
T= F(r= n = —— ) = 2m¿ur m
where #4 is the Coriolis force. What we really want to know is what sidewise
ƒorce has to be exerted by Moe in order to move ?n out at speed „ = dr/df. Thịs
1s Fạ = TÍr = 2m0.
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 evidentÌy present even at the originl But it is especially
easy to understand it at the origin, just by looking at what happens from the
Inertial system of Joe, who is standing on the ground. Figure 19-4 shows three
Successive views of mm Just as it passes the origin at ý = 0. Because of the rotation
of the carousel, we see that rm does not move in a straight line, but in a curued
pa‡h tangent to a diameter of the carousel where z = 0. In order for ?nw to gO
in a curve, there must be a force to accelerate i% in absolute space. This is the
Coriolis force.
1 ¡ 3 3
Fig. 19-4. Three successive views of a point moving radially on a
rotating turntable.
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 cireumference
of a circle, there is also a Coriolis force. Why? Moe sees a velocity 0a; around
the circle. On the other hand, .Joe sees rm going around the circle with the
velocitY 0 — 0; + œr, because m is also carried by the carousel. “Therefore
we know what the force really is, namely, the total centripetal force due to the
velocitV 0, Or mu} /r; that is the actual force. Now from Moe”s poinb oŸ view,
this centripetal force has three pieces. We may write it all out as follows:
hạ — "....... 21mUjd0 — ThuỶr.
--- Trang 361 ---
Now, #¿ 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 Ï were not turning, there