Now we pause briefly to note that ourforegoing introduction of torque, through the idea of work, gives us a mostimportant result for an object in equilibrium: if all the forces on an objectare in balance both for translation and rotation, not only is the net forcezero, but the total of all the torques is also zero, because if anobject is in equilibrium, no work is done by the forces for a small displacement.Therefore, since ΔW=τΔθ=0 , the sum of all the torques must be zero. So there are two conditionsfor equilibrium: that the sum of the forces is zero, and that the sum of thetorques is zero. Prove that it suffices to be sure that the sum of torques aboutany one axis (in two dimensions) is zero.
现在，我们暂停，注意一下，前面，我们通过功这个想法，对力矩进行了介绍，对于一个处在平衡中的对象，此介绍，给我们提供了最重要的结果：对于一个对象，如果所有作用于其上的力，无论对平移还是旋转来说，都处于平衡，那么，不仅净的力是零，且总的力矩也是零，因为，如果一个对象处于平衡，那么，对于一个小的位移来说，诸力并未做功。因此，由于ΔW=τΔθ=0，所以，所有力矩之和，应为零。所以，对于平衡，有两个条件：诸力之和为零，诸力矩之和为零。这足以证明，对于任何一个轴（在二维中），诸力矩之和为零{？}。

Fig. 18–2.The torque produced by a force. 图18-2 一个力所产生的力矩。
Now let us consider a single force, and tryto figure out, geometrically, what this strange thing xFy−yFxamounts to. In Fig. 18–2 wesee a force F acting at a point r{？P} . When the object has rotated through a small angle Δθ , the work done, of course, is the component of force in the directionof the displacement times the displacement. In other words, it is only thetangential component of the force that counts, and this must be multiplied bythe distance rΔθ . Therefore we see that the torque is also equal to the tangential componentof force (perpendicular to the radius) times the radius. That makes sense interms of our ordinary idea of the torque, because if the force were completelyradial, it would not put any “twist” on the body; it is evident that thetwisting effect should involve only the part of the force which is not pullingout from the center, and that means the tangential component. Furthermore, itis 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 nottwisting at all! So it makes sense that the amount of twist, or torque, isproportional both to the radial distance and to the tangential component of theforce.
现在，让我们考虑一个单独的力，尝试搞清楚，这个奇怪的事物，其几何意义为何。在图18-2中，我们看到，力 F作用于点P。当对象旋转过一个小的角度Δθ是，所做功，当然就是力在位移的方向的投影，乘以位移。换句话说，只有力的切线分量，才算数，这应该乘以距离rΔθ。因此，我们就看到，力矩也等于，力的切线分量（垂直于半径的），乘以半径。用我们普通的想法，来描述力矩，这也是合理的，因为，如果力是完全径向的，那么，它不会给物体，带来任何“扭动”；很明显，扭动效果，不应牵扯到从中心向外的那部分力，意思就是，只牵扯到切线分量。另外，很清楚，对于一个给定的力，在力臂大的地方，要比靠近轴的地方，更有效。事实上，如果我们直接推轴，则根本不会带来任何扭动！所以，扭动的量、或力矩，正比于半径的距离，及力的切线分量，这是合理的。

There is still a third formula for thetorque which is very interesting. We have just seen that the torque is theforce times the radius times the sine of the angle α , in Fig. 18–2. But ifwe 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 leverarm 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 isless than the total force. Therefore the formula for the torque can also bewritten as the magnitude of the force times the length of the lever arm.
对于力矩，还有第三个公式，它很有趣。刚刚，我们已经看到，力矩，就是力、乘以半径、乘以角度α的正弦，见图18-2。但是，如果我们扩展力的作用线，然后，画线OS，为O到力的作用线的垂直距离（力的杠杆臂），我们注意到，这个杠杆臂，比r要短，它与r的比，与力的切线部分与总的力的比，是一样的。从而，力矩的公式，也可以写为，力的大小，乘以杠杆臂的长度。

The torque is also often called the momentof the force. The origin of this term is obscure, but it may be related to thefact that “moment” is derived from the Latin movimentum, and that thecapability 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” meansweighted by how far away it is from an axis.
扭力通常有被称为距。此词起源，比较模糊，但是，它可与如下事实有关，即“距”是从拉丁文movimentum来的，而一个力移动对象的能力（用力作用于一个杠杆或撬棍上），随着杠杆臂的长度而增加。在数学中，“距”意味着，通过它离轴的距离有多远，来衡量。

18–3Angular momentum 18-3 角动量
Although we have so far considered only thespecial case of a rigid body, the properties of torques and their mathematicalrelationships 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 ofchange of a quantity p , which we call the total momentum of a collection of particles, sothe external torque is the rate of change of a quantity L which we call the angular momentum of the group of particles.
虽然到目前为止，我们考虑的，只是刚体这种特殊情况，但即便一个对象不是刚体，扭力的特性及其数学关系，也依然有趣。事实上，我们可以证明一个非常著名的定理：一批粒子的总动量，我们称为 p，一组离子的角动量，我们称为L，正如外力是量p的变化率，外扭力则是量L的变化率。

Fig. 18–3.A particle moves about anaxis O . 图18-3 一个粒子，绕轴O运动。
To prove this, we shall suppose that thereis a system of particles on which there are some forces acting and find outwhat 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 isone particle of mass m , and an axis O ; the particle is not necessarily rotating in a circle about O, 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, andit accelerates according to the usual formula that the x -component of force is the mass times the x -component of acceleration, etc. But let us see what the torquedoes. The torque equals xFy−yFx , and the force in the x - or y -direction is the mass times the acceleration in the x - or y -direction:
Τ = xFy−yFx =xm(d2y/dt2)−ym(d2x/dt2). (18.14)
Now, although this does not appear to be the derivative of any simplequantity, it is in fact the derivative of the quantity xm(dy/dt)−ym(dx/dt):
(18.15)
So it is true that the torque is the rate of change of something withtime! So we pay attention to the “something,” we give it a name: we callit L , the angular momentum:
L=xm(dy/dt)−ym(dx/dt)=xpy−ypx. (18.16)
要证明这点，我们将建议，有一个粒子的系统，有一些力，作用于其上，力矩可归于这些力，我们将找出，此系统在扭力的作用下，会发生什么。当然，首先我们只考虑一个粒子。在图18-3中，是一个质量为m的粒子，O为轴；粒子绕轴旋转，但运动的轨迹，并不必然是一个圆，也可能是一个抛物线，如行星绕太阳那样，或是其他曲线。它以某种方式在运动，有力作用于其上，它按通常的公式，被加速，即力的x方向分量，就是质量，乘以加速度的x方向分量，等等。但是，让我们看看，扭力做了些什么。扭力等于xFy−yFx，力的x或y方向的分量，就是质量，乘以加速度的x或y方向的分量：
Τ = xFy−yFx =xm(d2y/dt2)−ym(d2x/dt2). (18.14)
现在，虽然看上去，它并不像任何简单的量的导数，但事实上，它是量xm(dy/dt)−ym(dx/dt)的导数
(18.15)
于是，下点为真：扭力就是某物随时间的变化！所以，我们要注意这个“某物”，给它起个名字，称之为角动量L：
L=xm(dy/dt)−ym(dx/dt)=xpy−ypx. (18.16)

Before proceeding to a treatment of morethan one particle, let us apply the above results to a planet going around thesun. 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 theaxis, but we get a very simple result if we take it at the sun itself, for thetorque is the force times the lever arm, or the component of forceperpendicular to r , times r . But there is no tangential force, so there is no torque about anaxis at the sun! Therefore, the angular momentum of the planet going around thesun must remain constant. Let us see what that means. The tangential componentof velocity, times the mass, times the radius, will be constant, because thatis the angular momentum, and the rate of change of the angular momentum is thetorque, and, in this problem, the torque is zero. Of course since the mass isalso a constant, this means that the tangential velocity times the radius is aconstant. But this is something we already knew for the motion of a planet. Supposewe consider a small amount of time Δt . How far will the planet move when it moves from P to Q (Fig. 18–3)? Howmuch area will it sweep through? Disregarding the very tiny area QQ′Pcompared with the much larger area OPQ , it is simply half the base PQ times the height, OR . In other words, the area that is swept through in unit time will beequal to the velocity times the lever arm of the velocity (times one-half). Thusthe rate of change of area is proportional to the angular momentum, which isconstant. So Kepler’s law about equal areas in equal times is a word descriptionof the statement of the law of conservation of angular momentum, when there isno torque produced by the force.
在我们要处理多个粒子之前，让我们把上面的结果，应用于一个绕太阳旋转的行星。力在哪个方向呢？力是向着太阳的。因此，对象上的扭力是什么？当然，这依赖于我们把轴，取在何处，如果我们把轴取在太阳本身，那么，我们将得到一个简单的结果，因为扭力，是力乘以杠杆臂，或是，力的垂直于r的分量，乘以r。但是，没有切线方向的力，于是，对于处于太阳的轴，没有扭力！因此，绕太阳运行的行星的角动量，应为常数。让我们看看，这意味着什么。矢速的切线分量，乘以质量，再乘以半径，将是常数，因为，这就是角动量，而角动量的变化率就是扭力，在这个问题中，扭力为零。当然，由于质量也是一个常数，这就意味着，切线矢速乘以半径，是一个常数。但是，对于行星的运动来说，这件事，我们已经知道。假设我们考虑一个时间小量 Δt。当行星从P运动到 Q，它走了多远呢？非常小的面积QQ′P ，与大面积OPQ相比，可以忽略，面积OPQ等于PQ乘以高度OR，除以2。换句话说，单位时间内所扫过的面积，将等于矢速，乘以矢速的杠杆臂（乘以一半）。这样，面积的变化率，就正比于角动量，它是一个常数。开普勒的一条规律说，在相等的时间，扫过的面积相等；所以，当由力所产生的扭力，不存在时，此开普勒规律，就是角动量守恒规律的文字描述。

18–4Conservation of angular momentum 18-4角动量的守恒
Now we shall go on to consider what happenswhen there is a large number of particles, when an object is made of manypieces with many forces acting between them and on them from the outside. Ofcourse, we already know that, about any given fixed axis, the torque on the ith particle (which is the force on the i th particle times the lever arm of that force) is equal to therate of change of the angular momentum of that particle, and that the angularmomentum of the i th particle is its momentum times its momentum lever arm. Nowsuppose we add the torques τi for all the particles and call it the total torque τ . Then this will be the rate of change of the sum of the angularmomenta of all the particles Li , and that defines a new quantity which we call the total angularmomentum L . Just as the total momentum of an object is the sum of themomenta of all the parts, so the angular momentum is the sum of the angularmomenta of all the parts. Then the rate of change of the total L is the total torque:
我们将考虑，当有大量的粒子、当一个对象是由很多块组成，且块之间，有很多力，外部也有很多力，作用于这些粒子之上时，会发生什么。当然，我们已经知道，对于任何被给予的固定轴，作用于第i个粒子上的扭力（就是作用于第i个粒子上的力，乘以该力的杠杆臂），就等于该粒子的角动量的变化率，而第i个粒子的角动量，就是它的动量，乘以动量的杠杆臂。现在，假定我们把所有的扭力τi都加在一起，称其为总扭力τ。因此，这将是所有粒子的角动量总和的变化率，这就定义了一个新的量，我们称之为总的角动量L。正如一个对象的总动量，就是其所有部分的动量之和一样，所以，总角动量就是所有部分的角动量之和。因此，总角动量L的变化率，就是总扭力：
(18.18)

18–4Conservation of angular momentum 18-4角动量的守恒
Now we shall go on to consider what happenswhen there is a large number of particles, when an object is made of manypieces with many forces acting between them and on them from the outside. Ofcourse, we already know that, about any given fixed axis, the torque on the ith particle (which is the force on the i th particle times the lever arm of that force) is equal to therate of change of the angular momentum of that particle, and that the angularmomentum of the i th particle is its momentum times its momentum lever arm. Nowsuppose we add the torques τi for all the particles and call it the total torque τ . Then this will be the rate of change of the sum of the angularmomenta of all the particles Li , and that defines a new quantity which we call the total angularmomentum L . Just as the total momentum of an object is the sum of themomenta of all the parts, so the angular momentum is the sum of the angularmomenta of all the parts. Then the rate of change of the total L is the total torque:
(18.18)
Now it might seem that the total torque is a complicated thing. Thereare 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 actionand reaction are equal, but also that they are directed exactly oppositelyalong the same line (Newton may or may not actually have said this, but hetacitly assumed it), then the two torques on the reacting objects, dueto their mutual interaction, will be equal and opposite because the lever armsfor any axis are equal. Therefore the internal torques balance out pair bypair, and so we have the remarkable theorem that the rate of change of thetotal angular momentum about any axis is equal to the external torque aboutthat axis!
(18.19)
Thus we have a very powerful theorem concerning the motion of largecollections of particles, which permits us to study the overall motion withouthaving to look at the detailed machinery inside. This theorem is true for anycollection of objects, whether they form a rigid body or not.
我们将考虑，当有大量的粒子、当一个对象是由很多块组成，且块之间，有很多力，外部也有很多力，作用于这些粒子之上时，会发生什么。当然，我们已经知道，对于任何被给予的固定轴，作用于第i个粒子上的扭力（就是作用于第i个粒子上的力，乘以该力的杠杆臂），就等于该粒子的角动量的变化率，而第i个粒子的角动量，就是它的动量，乘以动量的杠杆臂。现在，假定我们把所有的扭力τi都加在一起，称其为总扭力τ。因此，这将是所有粒子的角动量总和的变化率，这就定义了一个新的量，我们称之为总的角动量L。正如一个对象的总动量，就是其所有部分的动量之和一样，所以，总角动量就是所有部分的角动量之和。因此，总角动量L的变化率，就是总扭力：
(18.18)
现在，似乎总的扭力，就是一个复杂的事情。所有的内力，和所有的外力，都要考虑。但是，如果我们拿牛顿的作用与反作用规律来说，那么，不仅作用与反作用，是相等的，且它们是在一条直线上，方向相反（牛顿可能没有事实上说过，但他默认了这点），因此，在两个相互作用的对象上的扭力，由于它们的相互作用，将是相等的、及相反的，因为，对于任何轴而言，它们的杠杆臂，都相等。因此，内部的扭力，一对一对地，相互平衡了，所以，我们就有了著名的定理，即关于任何轴的总的角动量的变化率，就等于，关于该轴的外扭力！
(18.19)
这样，我们就有了一个非常著名的定理，它与大群粒子集合的运动有关，它允许我们，不用去查看内部的力学细节，就可研究整体运动。这个定理，对于任何对象的集合来说，无论其是否为刚体，都为真。

One extremely important case of the abovetheorem is the law of conservation of angular momentum: if no externaltorques act upon a system of particles, the angular momentum remains constant.
关于上面定理，有一个极为重要的情况，就是角动量守恒的规律：如果没有外部的扭力，作用于粒子系统，那么，角动量将保持为常数。

A special case of great importance is thatof a rigid body, that is, an object of a definite shape that is just turningaround. Consider an object that is fixed in its geometrical dimensions, andwhich is rotating about a fixed axis. Various parts of the object bear the samerelationship to one another at all times. Now let us try to find the totalangular momentum of this object. If the mass of one of its particles is mi, and its position or location is at (xi,yi), 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 suchparticles in the body. For an object going around in a circle, the angularmomentum, of course, is the mass times the velocity times the distance from theaxis, and the velocity is equal to the angular velocity times the distance fromthe axis:
Li=miviri=mir2iω, (18.20)
or, summing over all the particles i , we get
L=Iω, (18.21)
where
(18.22)
关于刚体，有一个非常重要的情况，即一个确定形状的对象，正在旋转运动。考虑一个对象，其几何大小，是固定的，且绕着一个固定轴旋转。对象的不同部分，相互之间的关系，在所有时间内，都是同样的。现在，让我们找出，这个对象的总角动量。如果它的一个粒子的质量是mi，其位置或坐标是 (xi,yi)，那么，问题就是找出这个粒子的角动量，因为总角动量，就是物体中的所有粒子的角动量之和。对于一个做着圆运动的对象，其角动量，当然就是，质量、乘以速度、乘以到轴的距离，而矢速，就等于，角动量、乘以到轴的距离：
Li=miviri=mir2iω, (18.20)
或者，对所有的粒子求和，我们得到：
L=Iω, (18.21)
这里
(18.22)

This is the analog of the law that the momentum is mass timesvelocity. Velocity is replaced by angular velocity, and we see that the mass isreplaced 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 objectsof the same mass, when we put the masses farther away from the axis, theinertia for turning will be higher. This is easily demonstrated by theapparatus shown in Fig. 18–4,where a weight M is kept from falling very fast because it has to turn the largeweighted 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 inertiaby 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 hasmuch more inertia against turning. The moment of inertia is the inertia againstturning, and is the sum of the contributions of all the masses, times theirdistances squared, from the axis.
动量，是质量乘以矢速，这是一条规律，而上面，就是对此规律的类比。矢速被角速度代替了，我们看到，质量被一个新事物代替了，我们称其为惯性力矩 I，它可类比于质量。方程（18.21）和（18.22）说，一个具有转动惯性的物体，其惯性，不仅依赖于质量，而且依赖于它们距轴有多远。所以，如果我们有两个对象，质量相同，如其位置，距轴更远，则其转动惯性，将会更高。这一点，用图（18-4）所示仪器，很易演证，在其中，让重量M持续地快速下落，因为，它带动了那条大的重的杆子。最初，质量m，距轴较近，而让M加速到一定的速率。但是，当我们把两个质量m，放的距轴很远的地方时，惯性力矩被改变了，然后，我们看到，M加速时，明显比以前变慢，因为，物体惯性增多，妨碍转动。惯性力矩，就是妨碍转动的惯性，它是所有质量的贡献之和，即所有质量乘以它们到轴距离的平方之和。

Fig. 18–4.The “inertia for turning” dependsupon the lever arm of the masses. 图18-4 “旋转的惯性”，依赖于质量的杠杆臂。

There is one important difference betweenmass and moment of inertia which is very dramatic. The mass of an object neverchanges, but its moment of inertia can be changed. If we stand on a frictionlessrotatable stand with our arms outstretched, and hold some weights in our handsas we rotate slowly, we may change our moment of inertia by drawing our armsin, but our mass does not change. When we do this, all kinds of wonderfulthings happen, because of the law of the conservation of angular momentum: Ifthe external torque is zero, then the angular momentum, the moment of inertiatimes omega, remains constant. Initially, we were rotating with a large momentof inertia I1 at a low angular velocity ω1 , and the angular momentum was I1ω1. Then we changed our moment of inertia by pulling our arms in, say toa smaller value I2 . Then the product Iω , which has to stay the same because the total angular momentum has tostay the same, was I2ω2 . So I1ω1=I2ω2. That is, if we reduce the moment of inertia, we have to increasethe angular velocity.
在质量和惯性力矩之间，有一个非常重要的区别，很有戏剧性。对象的质量，永远不变，但其力矩，则可被变。如果我们站在一个无摩擦的旋转平台上，手臂外展，手中抓着某种重量，在我们慢慢旋转时，我们可以通过划动我们的手臂，来改变惯性力矩，但我们的质量，并未改变。当我们做此事时，由于角动量守恒规律，所有奇妙的事情，都会发生：如果外部扭力是零，那么，角动量，即惯性力矩乘以欧米伽Iω，保持为常数。初始，我们旋转时，有一个大的惯性力矩I1，和一个小的角速度ω1，角动量是 I1ω1。然后，通过外展我们的手臂，我们改变了我们的惯性力矩，比如说，达到一个更小的值I2。然后，因为总的角动量，应该保持一样，则乘积Iω就是I2ω2。于是I1ω1=I2ω2 。也就是说，如果我们减少了惯性力矩，我们就必须增加角速度。

Chapter19.Centre of Mass;Moment of Inertia第19章 质量中心；惯性的力矩
19–1Properties of the center of mass 19-1质量中心的特性
In the previous chapter we found that if agreat many forces are acting on a complicated mass of particles, whether theparticles comprise a rigid or a nonrigid body, or a cloud of stars, or anythingelse, and we find the sum of all the forces (that is, of course, the externalforces, because the internal forces balance out), then if we consider the bodyas a whole, and say it has a total mass M , there is a certain point “inside” the body, called the center ofmass, such that the net resulting external force produces an accelerationof this point, just as though the whole mass were concentrated there. Let usnow discuss the center of mass in a little more detail.
在上一章，我们发现，如果大量的力，作用于一个由复杂的粒子所构成的质量，无论粒子所组成的，是一个刚体、还是非刚体、或者是一个星云、或任何其他事情，我们寻找所有力之和（当然，是指外力，因为内力相互平衡了），因此，如果我们把物体，当作一个整体来考虑，认为其总质量为M，则在物体内，有一个确定的点，称为质量中心，这样，净的外力，会对这个点，产生一个加速度，就好像全部质量，都集中在此一样。现在，让我们更仔细地讨论质量中心。

The location of the center of mass(abbreviated CM) is given by the equation
RCM=∑miri / ∑mi. (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 understandthe other two. What does XCM=∑mixi /∑mi mean? Suppose for a moment that the object is divided into littlepieces, 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. Thenthis equation simply says that we add all the x ’s, and then divide by the number of things that we have added: XCM=m∑xi/mN= ∑xi / 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 asheavy as the others. Then in the sum, that x would come in twice. This is easy to understand, for we can think ofthis double mass as being split into two equal ones, just like the others; thenin 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 x -direction, of all the masses, every mass being counted a number oftimes 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 couldbe a circle, like a hoop, and the center of mass is in the center of the hoop,not in the hoop itself.
质量中心(缩写为CM)的位置，由下式给出:
RCM=∑miri / ∑mi. (19.1)
这当然是一个矢量方程，真正说来，包含三个方程，每个方向一个。我们将只考虑x方向的方程，因为，如果我们可以理解这个，那么，其它两个，也可理解。XCM=∑mixi/ ∑mi的意思是什么呢？假设，在某段时间，对象被分成N小块，每块质量，都是m，那么，总的质量，就是N乘以一块的质量，比如说一克，或其他任何单位。因此，这个方程说的就是，我们把所有x方向的量，搜加在一起，然后，除以所加的量的个数：XCM= m∑xi/mN= ∑xi / N。换句话说，如果质量相等，则XCM就是所有x的平均值。但是，假设其中一块，是其它块的两倍重。那么，在总和中，这个x就应被计数两次。这一点很易理解，因为，我们可以把这个双倍的质量，想象分裂成两个相等的质量，就像其它的质量一样；因此，在取平均值时，我们当然就应该把这个x，计数两次，因为，那里有两个质量。这样，X就是所有质量在x方向的平均位置，每个质量，所计次数，正比于该质量，就好像它被分成了“小的克数”一样。从此出发，很容易证明，X是在最大x和最小x之间，因而，就处于包含着全部物体的信封之中{范围内}。它不一定非得在物体的材料中，因为物体可能是一个圆，比如一个铁环，其质量中心，是在铁环的中心，而不是在铁环本身中。