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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 |
--- Trang 349 --- |
problem? Must the laws on a small scale be the same as those on a larger scale? |
OŸÝ 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 Ͽ# have the property that when |
we øo to a larger scale we reproduce the same law, but instead has the property |
that if we go to a larger scale, we can øpprozimeite ?t bụ a certain ezpression such |
that, if we extend that expression up and up, ? keeps reproducing ïtself 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 |
1ƒ we take large numbers of them and compound them, they approximate, but |
on 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 self-reproducing |
factor of Newton's laws is thus really not a fundamental feature of nature, but |
1s an Iimportant 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, i% turns out that the fundamental atomic laws, |
which we call quantum mechanics, are quite diferent from Newton”s laws, and |
are difficult to understand because all our direct experiences are with large-scale |
objects and the small-scale atoms behave like nothing we see on a large scale. 5o |
we cannot say, “An atom ¡is just like a planet going around the sun,” or anything |
like that. It is like nof#h#ng we are familiar with because there is noứh#ng like |
z‡. As we apply quantum mechanics to larger and larger things, the laws about |
the behavior of many atoms together do øoø reproduce themselves, but produce |
neu laus, which are Ñewton'”s laws, which then continue to reproduce themselves |
from, say, micro-microgram 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. 'Phe 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 foree is not only proportional to the mass, but is everywhere parallel |
to some fñxed line. Then consider an object in which there are gravitational Íorces |
on each of its constituent masses. Let ?m¿ be the mass of one part. Then the |
gravitational force on that part 1s ?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 |
--- Trang 350 --- |
mmust 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; ¡§ should be |
zero 1ƒ this axis is at the center of mass. Now, measuring z horizontally and |
vortically, we know that the torques are the forces in the -direction, times the |
lever arm ø# (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 |
T= Àmga; = gà” T¿, (19.3) |
so if the total torque is to be zero, the sum À `7n¿#¿ must be zero. But È) m¿#¿ = |
1M Xe, the total mass times the distance of the center of mass from the axis. |
'Thus the z-distance of the center of mass from the axis is zero. |
Of course, we have checked the result only for the z-distance, but IÝ we use |
the true center of mass the object will balance in any position, because IÝ we |
turned ï© 90 degrees, we would have zs instead of zøs. In other words, when an |
object is supported at its center of mass, there is no torque on i§ because of a |
parallel gravitational fñield. In case the object is so large that the nonparallelism |
of the gravitational forces is significant, then the center where one must apply |
the balancing force is not simple to describe, and ¡it departs slightly from the |
center of mass. hat is why one must distinguish between the center oŸ 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. ÏÝ, instead |
of gravitation, we have a pseudo force due to acceleration, we may use exactÌy |
the same mathematical procedure to fnd the position to support it so that there |
are no torques produeced by the inertial force of acceleration. Suppose that the |
object is held in some mamner inside a box, and that the box, and everything |
contained ïn 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 |
pseudo force equal to the mass times the acceleration of the box. To the man in |
the box, this is the same situation as ïif the object were in a uniform gravitational |
fñeld whose “ø” value is equal to the acceleration ø. 'Phus the inertial force due |
to accelerating an obJect has no torque about the center of mass. |
--- Trang 351 --- |
This fact has a very interesting consequence. Ín an inertial frame that is |
not accelerating, the torque is always equal to the rate of change of the angular |
momentum. However, about an axis through the center of mass of an object |
which ¿s accelerating, 1t is sfil true that the torque is equal to the rate of change |
of the angular momentum. ven ïf 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 |
qmomentum around that axis. Thhus the theorem that torque equals the rate of |
change of angular momentum is true in two general cases: (1) a ñxed axis in |
inertial space, (2) an axis through the center oŸ mass, even though the object |
may be accelerating. |
19-2 Locating the center of mass |
'The mathematical techniques for the calculation of centers oŸ 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. Ône such trick makes use of what is called the theorem of Pappus. lt |
works like this: if we take any closed area in a plane and generate a solid by |
moving it through space such that each poïnt 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 movedl Certainly |
this is true If we move the area in a straight line perpendicular to itself, but 1f |
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