As an example, the moment of inertia of a uniform rectangular plate ofmass M , width w , and length L , about an axis perpendicular to the plate and through its center issimply
I=M(w2+L2)/12,
because its moment of inertia about an axis in its plane and parallelto 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 ML2/12, just as for a rod of length L .
例如，一个矩形板子，均匀，质量为M，宽度w ,长度 L，一轴垂直于它，过其中心，那么，关于此轴的惯性力距，简单就是
I=M(w2+L2)/12,
因为，设有一轴，在其平面内，平行于其长度，则关于此轴的惯性力矩，就是Mw2/12，亦即，正如一个长度为w的杆一样；同理，设有一轴，在其平面内，平行于其宽度，则关于此轴的惯性力矩，就是ML2/12，亦即，正如一个长度为L的杆一样。

To summarize, the moment of inertia of anobject about a given axis, which we shall call the z -axis, has the following properties:
1. The moment of inertia is
2. If the object is made of a number of parts, each of whose moment ofinertia is known, the total moment of inertia is the sum of the moments ofinertia of the pieces.
3. The moment of inertia about any given axis is equal to the moment ofinertia about a parallel axis through the CM plus the total mass times thesquare of the distance from the axis to the CM.
4. If the object is a plane figure, the moment of inertia about an axisperpendicular to the plane is equal to the sum of the moments of inertia aboutany two mutually perpendicular axes lying in the plane and intersecting at theperpendicular axis.
设有一轴，我们称之为z轴，一个对象关于此轴，有惯性力矩，其特性，总结如下：
1、惯性力矩是：

2、如果对象，由若干部分组成，每部分的惯性力矩，都是已知，那么，总的惯性力距，就是各部分的惯性力矩之和。
3、给定一轴，关于此轴的惯性力矩，等于两部分之和：一、通过质心，另有一轴，关于此轴的惯性力矩；二、所有质量，乘以给定轴到质心距离的平方。
4，如果对象，是一个平面图形，那么，设有一轴，垂直于此平面，另设平面内，另有两轴，相互垂直，都与前轴，垂直相交，那么，关于前轴的惯性力矩，就等于，关于后两轴的惯性力矩之和。

The moments of inertia of a number ofelementary shapes having uniform mass densities are given in Table 19–1, andthe moments of inertia of some other objects, which may be deduced fromTable 19–1,using the above properties, are given in Table 19–2.
表19-1，是一些对象的惯性力距，这些对象，具有基本形状，质量密度均匀；有些对象的惯性力矩，可以使用上面的特性，从表19-1中推出来，见表19-2。
Table 19–1 表19-1

Table 19–2 表19-2

19–4Rotational kinetic energy 19-4 旋转的动能
Now let us go on to discuss dynamicsfurther. In the analogy between linear motion and angular motion that wediscussed in Chapter 18,we used the work theorem, but we did not talk about kinetic energy. What is thekinetic energy of a rigid body, rotating about a certain axis with an angularvelocity ω ? We can immediately guess the correct answer by using our analogies.The moment of inertia corresponds to the mass, angular velocity corresponds tovelocity, and so the kinetic energy ought to be Iω2/2, and indeed it is, as will now be demonstrated. Suppose the object isrotating about some axis so that each point has a velocity whose magnitudeis ωri , where ri is the radius from the particular point to the axis. Then if miis the mass of that point, the total kinetic energy of the whole thingis just the sum of the kinetic energies of all of the little pieces:
Now ω2 is a constant, the same for all points. Thus
(19.8)
现在，让我们进一步讨论动力学。在18章，我们讨论了，线性运动与角运动的类比，那里，我们用的是做功定理，但并没有谈到动能。一个刚体，绕某轴旋转，角速度为 ω，其动能为何？通过使用我们的类比，我们立刻就能猜出答案。惯性力矩，相应于质量，角速度，相应于速度，于是，动能就应该是 Iω2/2，也确实如此，我们将演证。假设对象，绕某轴旋转，于是，每一个点，都有矢速，大小为ωri ，这里，ri是某具体的点，到轴的距离。因此，如果mi是该点的质量，则整个事物的总动能，就是所有小部分的动能之和：

现在，ω2是一个常数，对所有的点都一样。这样：
(19.8)

There is still another interesting featurewhich we can treat only descriptively, as a matter of general interest. Thisfeature is a little more advanced, but is worth pointing out because it isquite curious and produces many interesting effects.
还有另外一种有趣的特性，我们只能叙述性地来对待，作为一种一般兴趣。{？}这个特性，有一点儿先进，但值得指出，因为它很奇怪，并产生了很多有趣的影响。

Consider that turntable experiment again.Consider the body and the arms separately, from the point of view of the manwho is rotating. After the weights are pulled in, the whole object is spinningfaster, but observe, the central part of the body is not changed, yet itis turning faster after the event than before. So, if we were to draw a circlearound the inner body, and consider only objects inside the circle, theirangular momentum would change; they are going faster. Therefore theremust be a torque exerted on the body while we pull in our arms. No torque canbe exerted by the centrifugal force, because that is radial. So that means thatamong the forces that are developed in a rotating system, centrifugal force isnot the entire story, there is another force. This other force is calledCoriolis force, and it has the very strange property that when we movesomething in a rotating system, it seems to be pushed sidewise. Like thecentrifugal force, it is an apparent force. But if we live in a system that isrotating, and move something radially, we find that we must also push itsidewise to move it radially. This sidewise push which we have to exert is whatturned our body around.
再次考虑，转盘实验。从那个旋转着的人的观点出发，把身体与手臂，分开考虑。在重量被收回之后，整个对象，旋转更快，但要观察，身体的中心部分，并未改变，尽管手臂收回这个事件发生之后，比以前转得更快了。于是，如果我们想给内部的身体，画一个圆，且只考虑此圆内的对象，那么，其角动量，将会改变；它走得更快了。因此，当我们收回手臂时，应该有一个扭力，作用于身体上。通过离心力，不可能施加扭力，因它是径向的。这就意味着，一个旋转系统中，会产生各种力，其中离心力，并非唯一，还有另外的力。这个另外的力，被称为科里奥利力，它有很奇怪的属性，即当我们在一个旋转系统中，移动某物时，它似乎是在往旁边推。就像离心力一样，它是一个表观力。但是，如果我们生活在一个旋转系统中，要径向地移动某物，那么，我们就会发现，我们同时必须也把它往旁边推，才能让它沿着径向走。这个往旁边的推力，是我们必须施加的；它就是转动我们身体的力。

Now let us develop a formula to show howthis Coriolis force really works. Suppose Moe is sitting on a carousel thatappears to him to be stationary. But from the point of view of Joe, who isstanding on the ground and who knows the right laws of mechanics, the carouselis going around. Suppose that we have drawn a radial line on the carousel, andthat Moe is moving some mass radially along this line. We would like to demonstratethat a sidewise force is required to do that. We can do this by payingattention to the angular momentum of the mass. It is always going around withthe same angular velocity ω , so that the angular momentum is
L=mvtangr=mωr⋅r=mωr2.
So when the mass is close to the center, it has relatively littleangular momentum, but if we move it to a new position farther out, if weincrease r , m has more angular momentum, so a torque must be exerted in orderto move it along the radius. (To walk along the radius in a carousel, one hasto lean over and push sidewise. Try it sometime.) The torque that is requiredis the rate of change of L with time as m moves along the radius. If m moves only along the radius, omega stays constant, so that the torqueis
τ=Fcr=dL/dt=d(mωr2)dt=2mωrdr/dt,
where Fc is the Coriolis force. What we really want to know is what sidewise forcehas to be exerted by Moe in order to move m out at speed vr=dr/dt . This is Fc= τ/r= 2mωvr .
现在，让我们研发一个公式，来指出，这个科里奥利力，是如何起作用的。假设Moe，坐在旋转木马上，且木马对他来说，是静止的。而Joe站在地面上，他知道正确的力学规律，从他的观点看，木马正在转动。设我们已经在木马上，画了一条径向的线，而Moe，正沿着这条线，移动某些质量。我们希望验证，要做此事，需要一个往旁边的力。通过注意此质量的角动量，我们可以做此事。它总是以同样的角速度ω在转动，所以，角动量就是：
L=mvtangr=mωr⋅r=mωr2.
于是，当质量离中心较近时，角动量相对较小，但是，如果我们把它向外移到一个新的、较远的位置，如果我们增加r，m将有更多的角动量，所以，必须施加一个扭力，以让它沿着半径移动（在一个旋转木马上，要沿着半径走，人必须斜着，往旁边推；有时间试一下）。所需要的扭力，就是当m沿着半径移动时，L随时间的变化率。如果m只是沿着半径移动，则ω保持为常数，于是，扭力就是：
τ=Fcr=dL/dt=d(mωr2)dt=2mωrdr/dt,
这里，Fc就是科里奥利力。我们真正想知道的是：为了以速度vr=dr/dt，把m往外移动，由Moe所施加的往旁边的力，是多少。这就是Fc= τ/r= 2mωvr。

Now that we have a formula for the Coriolisforce, let us look at the situation a little more carefully, to see whether wecan 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 evidentlypresent even at the origin! But it is especially easy to understand it at theorigin, just by looking at what happens from the inertial system of Joe, who isstanding on the ground. Figure 19–4 showsthree 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 tangentto 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 absolutespace. This is the Coriolis force.
现在，对于科里奥利力，我们就有了一个公式，让我们更仔细地查看一下这个情况，即对于这个力的起源，我们是否可以从一个更基础的观点，来理解之。我们注意到，科里奥利力，在每个半径上，都是一样，甚至在原点，表现都很明显。但是，在原点，要理解它，特别容易，只要从Joe的最惯性系统中，看一下会发生什么就行，他站在地面上。图19-4指出， 在t=0，当m移过原点时，关于m的三个成功的视图。因为木马的旋转，我们看到，m并不是在一条直线上移动，而是一条曲线，曲线与木马的直径，在r=0处相切。为了让m走一条曲线，就应该有一个力，在绝对的空间对它加速。这就是科里奥利力。

Fig. 19–4.Three successive views of a pointmoving radially on a rotating turntable. 图19-4 一个点，在一个旋转着的转盘上，沿径向移动，关于它的三个成功的视图。

This is not the only case in which theCoriolis force occurs. We can also show that if an object is moving with constantspeed around the circumference of a circle, there is also a Coriolis force.Why? Moe sees a velocity vM around the circle. On the other hand, Joe sees m going around the circle with the velocity vJ=vM+ωr, because m is also carried by the carousel. Therefore we know what the forcereally is, namely, the total centripetal force due to the velocity vJ, or mv2J /r ; that is the actual force. Now from Moe’s point of view, thiscentripetal force has three pieces. We may write it all out as follows:
Fr = −mv2J/ r = −mv2M / r −2mvMω−mω2r.
Now, Fr is the force that Moe would see. Let us try to understand it. WouldMoe appreciate the first term? “Yes,” he would say, “even if I were notturning, there would be a centripetal force if I were to run around a circlewith velocity vM .” This is simply the centripetal force that Moe would expect, havingnothing to do with rotation. In addition, Moe is quite aware that there isanother centripetal force that would act even on objects which are standingstill on his carousel. This is the third term. But there is another term inaddition to these, namely the second term, which is again 2mωv . The Coriolis force Fc was tangential when the velocity was radial, and now it is radial whenthe velocity is tangential. In fact, one expression has a minus sign relativeto 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 tothe velocity, and of magnitude 2mωv .
科里奥利力，并不只出现在这种情况中。我们还可以指出，如果一个对象，以恒速绕着一个圆周走，也会有科里奥利力。为什么？Moe看到，绕圆的矢速为vM。另一方面，Joe看到，m绕圆的矢速为vJ=vM+ωr 。因为，m同时也被旋转木马带着走。因此，我们就知道了，这个力真正是什么，即可归于矢速vJ、或 mv2J /r 的总向心力，这是一个实际的力。现在，从Moe的观点，这个向心力，分三部分。我们把它们写作：
Fr = −mv2J/ r = −mv2M / r −2mvMω−mω2r.
现在，Fr就是Moe会看到的力。让我们尝试去理解它。Moe会认可第一项吗？他会说：“是的，即使我没有在旋转，如果我绕着一个圆，以矢速vM奔跑，也会有一个向心力。”这个向心力，只是Moe所期待的，与旋转无关。另外，Moe完全可以意识到，还有另外一个向心力，会作用于那些静止在他的旋转木马上的对象。这就是第三项。但是，除此之外，还有另外一项，即第二项，它又是2mωv。当矢速是径向的时候，科里奥利力就是切向的，现在矢速是切向的，它就是径向的。事实上，这个表达式，相对于另一个，有个负号{？}。相对于矢速来说，力总是在同一个方向，且无论矢速，在哪个方向。力与矢速，成直角，大小为2mωv。

1 Chapter20.Rotation in space第20章 空间中的旋转
20–1Torques in three dimensions 20-1 三维中的扭力
In this chapter we shall discuss one of themost remarkable and amusing consequences of mechanics, the behavior of arotating wheel. In order to do this we must first extend the mathematicalformulation of rotational motion, the principles of angular momentum, torque,and so on, to three-dimensional space. We shall not use these equationsin all their generality and study all their consequences, because this wouldtake many years, and we must soon turn to other subjects. In an introductorycourse we can present only the fundamental laws and apply them to a very fewsituations of special interest.
力学之中，有很多著名且好玩的结果，其中之一，就是旋转中的轮子，在本章，我们将讨论它。为做此事，我们必须首先把旋转运动的数学公式、角动量、扭力等的原理，扩展到三维空间。我们并不是在它们所有的普遍性中，使用这些方程，及研究所有的后果，因为，这将会花很多年，我们应该很快转向其他主题。本课程，是介绍性的，我们将只提供基本规律，并把它们应用于：几个特别感兴趣的情况中。

That these three equations can be deducedfor the motion of a single particle is quite clear. Furthermore, if we addedsuch things as xpy−ypx together for many particles and called it the total angular momentum,we would have three kinds for the three planes xy , yz , and zx , and if we did the same with the forces, we would talk about thetorque in the planes xy , yz , and zx also. Thus we would have laws that the external torque associated withany plane is equal to the rate of change of the angular momentum associatedwith that plane. This is just a generalization of what we wrote in twodimensions.
对于一个粒子，可以为其运动，推出这三个方程，这一点，很清楚。进一步，如果对于很多粒子，我们把xpy−ypx这种东西加起来，并称其为总角动量，那么，对于xy , yz , 和 zx这三个平面，我们将有三个这种方程；如果对力，我们做同样的事情，那么，我们就可以谈论这三个平面中的扭力了。这样，我们就有了规律：对于任一平面，与其相联的外部扭力，就等于，与其相联的角动量的变化率。我们在二维中，写过相关规律，这正是它们的普遍化。

But now one may say, “Ah, but there aremore planes; after all, can we not take some other plane at some angle, andcalculate the torque on that plane from the forces? Since we would have towrite another set of equations for every such plane, we would have a lot ofequations!” Interestingly enough, it turns out that if we were to work out thecombination x′Fy′−y′Fx′for another plane, measuring the x′ , F y′ , etc., in that plane, the result can be written as some combinationof the three expressions for the xy -, yz - and zx -planes. There is nothing new. In other words, if we know what thethree torques in the xy -, yz -, and zx -planes are, then the torque in any other plane, and correspondinglythe angular momentum also, can be written as some combination of these: sixpercent of one and ninety-two percent of another, and so on. This property weshall now analyze.
但是现在，有人会说：“啊，毕竟还有更多平面；我们能以某角度，取个平面，然后去计算，力在此平面上所产生的扭力吗？由于，对于每一个这种平面，我们都必须写另外一套方程，那么，我们就要有很多方程了”。有趣的结果则是，对于另外的平面，如果我们要为其得出组合 x′Fy′−y′Fx′，即去测量那个平面上的x′ , F y′等，那么，最终结果，则可以写为xy -, yz - 和 zx这三个平面上的表达式的组合。没有任何新的东西。换句话说，如果我们知道了xy -, yz - 和 zx这三个平面上的扭力，那么，任何其他平面的扭力、及相应的角动量，都可被写作其组合：一个的6%，和另一个的92%，等等。现在，我们将分析这个特性。

Suppose that in the xyz -axes, Joe has worked out all his torques and his angular momenta inhis planes. But Moe has axes x′,y′,z′ in some other direction. To make it a little easier, we shall supposethat only the x - and y -axes have been turned. Moe’s x′ and y′ are new, but his z′ happens to be the same. That is, he has new planes, let us say, for yzand zx . He therefore has new torques and angular momenta which he would workout. For example, his torque in the x′y′ -plane would be equal to x′Fy′−y′Fx′and so forth. What we must now do is to fid the relationship between thenew torques and the old torques, so we will be able to make a connection fromone set of axes to the other. Someone may say, “That looks just like what wedid with vectors.” And indeed, that is exactly what we are intending to do.Then he may say, “Well, isn’t torque just a vector?” It does turn out tobe a vector, but we do not know that right away without making an analysis. Soin the following steps we shall make the analysis. We shall not discuss everystep in detail, since we only want to illustrate how it works. The torquescalculated by Joe are
(20.1)
假设在xyz坐标轴中，Joe已经得到了他的平面中的所有扭力，及角动量。但是，Moe的坐标轴是x′,y′,z′，在其他的方向。为了让事情更容易些，我们将假设，只有x和 y轴，被转动了。Moe的 x′ 和 y′ 是新的，但他的z′ 是一样的。也就是说，对于yz和 zx，他有新的平面。因此，他将得出，新的扭力和角动量。例如，他在x′y′平面中的扭力，将等于x′Fy′−y′Fx′，等等。我们现在要做的，就是找出，新的扭力，与旧的扭力之间的关系，这样，我们就可以在两组坐标轴之间，建立联系。有人会说：“这看上去，像是我们对矢量所做的事情”。确实如此，这正是我们要做的。然后，他可能会说：“好吧，那么，扭力是不是一个矢量呢？”结果将显示，它是一个矢量，但是，没有做任何分析的话，我们并不能立刻知道这一点。所以，我们将按下面的步骤，做分析。我们不会仔细地讨论每一步，由于，我们只是想展示，事情是如何进行的。Joe所计算的扭力是：
(20.1)

We digress at this point to note that insuch cases as this one may get the wrong sign for some quantity if thecoordinates are not handled in the right way. Why not write τyz=zFy−yFz? The problem arises from the fact that a coordinate system may beeither “right-handed” or “left-handed.” Having chosen (arbitrarily) a sign for,say τxy , then the correct expressions for the other two quantities may alwaysbe found by interchanging the letters xyz in either order
在这里，我们先岔开话题，去注意，如果坐标系没有掌握好，那么，在类似这种的情况下，对于某些量，可能会得到错误的符号。为什么不写成τyz=zFy−yFz呢？这个问题，产生于以下事实，即一个坐标系，可以是“右手的”，也可是“左手的”。比如，在为τxy选（任意地）了一个符号之后，那么，关于另外两个量的正确的表达，总是可以通过以下面两种方式之一，交换字母xyz的顺序，来得到：

Moe now calculates the torques in hissystem:
现在Moe，在他的坐标系中，计算扭力：
(20.2)