14–2Constrained motion 14-2 受限的运动
Another interesting feature of forces andwork is this: suppose that we have a sloping or a curved track, and a particlethat must move along the track, but without friction. Or we may have a pendulumwith a string and a weight; the string constrains the weight to move in acircle about the pivot point. The pivot point may be changed by having thestring hit a peg, so that the path of the weight is along two circles of differentradii. These are examples of what we call fixed, frictionless constraints.
力和功的另外一个有趣特征就是：假设我们有一个斜的或弯曲的轨道，一个粒子必须沿着这个轨道运动，但没有摩擦。或者，我们可以有一个单摆，由一根绳子和一个重量组成，这个绳子限制着重量，让重量绕着一个轴心点，做圆运动。通过让轴心点撞一个钉子，可以让轴心点发生改变，于是，重量的路径，就是沿着两个不同半径的圆。这些就是我们称为固定的、无摩擦限制的例子。

In motion with a fixed frictionlessconstraint, no work is done by the constraint because the forces of constraintare always at right angles to the motion. By the “forces of constraint” we meanthose forces which are applied to the object directly by the constraintitself—the contact force with the track, or the tension in the string.
在些运动，是有限制{条件}的，这种限制，是固定的和无摩擦的，限制并未做功，因为限制的力，总垂直于运动。所谓“限制的力”，我们的意思是指：那些通过限制本身，而被直接应用到对象上的力—即与轨道接触的力，或绳子的张力。

The forces involved in the motion of aparticle on a slope moving under the influence of gravity are quite complicated,since there is a constraint force, a gravitational force, and so on. However,if we base our calculation of the motion on conservation of energy and thegravitational force alone, we get the right result. This seems ratherstrange, because it is not strictly the right way to do it—we should use the resultantforce. Nevertheless, the work done by the gravitational force alone will turnout to be the change in the kinetic energy, because the work done by the constraintpart of the force is zero (Fig. 14–1).
一个粒子，在重力的影响下，在一个斜坡上运动，所牵扯到的力，相当复杂，由于有一个限制的力、一个重力等。然而，如果让我们对运动的计算，只基于能量守恒和万有引力，那么，我们还是会得到正确的结果。这似乎相当奇怪，因为，严格讲，这并不是做此事的正确方法--我们应该用合力。尽管如此，将会证明，仅万有引力做的功，将会是动能变化的原因，因为限制部分的力，所做功为零（图14-1）。

Fig. 14–1.Forces acting on a sliding body(no friction). 图14-1 一个滑动物体（无摩擦力）上的力。

The important feature here is that if aforce can be analyzed as the sum of two or more “pieces” then the work done bythe resultant force in going along a certain curve is the sum of the works doneby the various “component” forces into which the force is analyzed. Thus if we analyzethe force as being the vector sum of several effects, gravitational plusconstraint forces, etc., or the x -component of all forces and the y -component of all forces, or any other way that we wish to split itup, then the work done by the net force is equal to the sum of the works doneby all the parts into which we have divided the force in making the analysis.
这里的重要特征就是，如果一个力，可以被分析为多个部分的总和，那么，合力沿着曲线所做的功，就是‘力所分解成的各分力’所做功的总和。这样，如果我们把力，分析为几个效力的矢量总和，如：重力加上限制力等等，或者是所有力在x方向的分量、在y方向的分量，或任何其他我们可以把它分开的方法，那么，净力所做的功，就等于所有分力所做的功；分力就是在分析时，我们把力所分成的部分。

14–3 Conservative forces 14-3 保守的力
In nature there are certain forces, that ofgravity, for example, which have a very remarkable property which we call“conservative” (no political ideas involved, it is again one of those “crazywords”). If we calculate how much work is done by a force in moving an objectfrom one point to another along some curved path, in general the work depends uponthe curve, but in special cases it does not. If it does not depend upon thecurve, we say that the force is a conservative force. In other words, if theintegral of the force times the distance in going from position 1 to position 2 in Fig. 14–2 iscalculated along curve A and then along B , we get the same number of joules, and if this is true for this pairof points on every curve, and if the same proposition works no matterwhich pair of points we use, then we say the force is conservative. In suchcircumstances, the work integral going from 1 to 2 can be evaluated in a simple manner, and we can give a formula for theresult. Ordinarily it is not this easy, because we also have to specify the curve,but when we have a case where the work does not depend on the curve, then, ofcourse, the work depends only upon the positions of 1 and 2 .
在自然中，有某些列，例如重力，它们又一个非常著名的性质，我们称之为“保守的”（没有牵扯到政治意义，这又是那种“疯狂词汇”。）一个对象，被一个力作用，沿着某个曲线路径，从一个点运动到另一个点，如果我们计算，此力做了多少功，那么，通常此功依赖于曲线，但在特殊案例中，则不然。如果它不依赖于曲线，那么，我们就说，此力是一个保守的力。换句话说，如图14-2，从位置1到位置2，对时间乘以距离进行积分，如果此积分，无论是沿着曲线A，还是沿着曲线B，我们都得到同样的焦耳数，如果对于曲线上的每对点来说，这个积分都为真，且如果无论我们使用的这一对点是什么，这个命题都成立，那么，我们就说，力是保守的。在这种情形下，从1到2，积分出来的功，可以用一种简单的方式来估算，且我们可以为结果，给出一个公式。通常，这并不容易，因为，我们还必须具体指定曲线，但是，当我们的案例不依赖于曲线时，那么，当然，功就只依赖于位置1和2.

Fig. 14–2.Possible paths between two pointsin a field of force. 图14-2 一个力场中的两个点之间的可能路径

To demonstrate this idea, consider the following.We take a “standard” point P , at an arbitrary location (Fig. 14–2).Then, the work line-integral from 1 to 2 , which we want to calculate, can be evaluated as the work done ingoing from 1 to P plus the work done in going from P to 2 , because the forces are conservative and the work does not dependupon the curve. Now, the work done in going from position P to a particular position in space is a function of that position inspace. Of course it really depends on P also, but we hold the arbitrary point P fixed permanently for the analysis. If that is done, then the workdone in going from point P to point 2 is some function of the final position of 2 . It depends upon where 2 is; if we go to some other point we get a different answer.
为了验证这个想法，考虑下面。我们在任意位置，取一个“标准的”点P（图14-2）。我们想计算的，是从1到2的线性积分的功，就可以被估算为从1到P的所做的功，加上从P到2所做的功，因为力是保守的，且功不依赖于曲线。现在，从位置P到空间中的一个具体位置所做的功，就是那个空间中的位置的函数。当然，它也确实依赖于P，但是，为了分析，我们设任意点P为固定的。如果这定了，那么，从点P到点2所做的功，就是2的最终位置的函数。它依赖于2在何处；如果我们去往另外一个点，我们就得到一个不同的答案。

We shall call this function of position −U(x,y,z), and when we wish to refer to some particular point 2 whose coordinates are (x2,y2,z2), we shall write U(2) , as an abbreviation for U(x2,y2,z2). The work done in going from point 1 to point P can be written also by going the other way along the integral,reversing all the ds ’s. That is, the work done in going from 1 to P is minus the work done in going from the point P to 1 :
Thus the work done in going from P to 1 is −U(1) , and from P to 2 the work is −U(2) . Therefore the integral from 1 to 2 is equal to −U(2) plus [−U(1) backwards], or +U(1)−U(2) :
(14.1)
The quantity U(1)−U(2) is called the change in the potential energy, and we call Uthe potential energy. We shall say that when the object is located atposition 2 , it has potential energy U(2) and at position 1 it has potential energy U(1) . If it is located at position P , it has zero potential energy. If we had used any other point,say Q , instead of P , it would turn out (and we shall leave it to you to demonstrate) thatthe potential energy is changed only by the addition of a constant.Since the conservation of energy depends only upon changes, it does notmatter if we add a constant to the potential energy. Thus the point Pis arbitrary.
我们把这个位置的函数，称为−U(x,y,z)，设某个具体的点2的坐标为(x2,y2,z2)，当我们希望引用它时，我们将写成 U(2)，作为 U(x2,y2,z2)的缩写。从点1到点P所做的功，也按积分线的反方向来写，所有的ds都取反了。也就是说，从点到P所做的功，就是负的从点P到点1所做的功：

这样，从P到1所做的功，就是−U(1)，从P到2所做的功，就是 −U(2)。因此，从1到2的积分，就等于−U(2) 加上[−U(1) 反向], or +U(1)−U(2):

(14.1)
量U(1)−U(2)，被称为势能的变化，我们称U为势能。当对象处于位置2时，我们说，其势能为U(2)，处于位置1时，我们说，其势能为U(1)。如果在位置P，其势能为零，如果我们用另一个点，比如说Q，来代替P，那么,结果就是，势能的变化，只是加了一个常数。（我们将把这个留给读者去演证）。由于势能的守恒，只依赖于变化，与加不加一个常数无关。这样，点P就是任意的。

Now, we have the following twopropositions: (1) that the work done by a force is equal to the change inkinetic energy of the particle, but (2) mathematically, for a conservativeforce, the work done is minus the change in a function U which we call the potential energy. As a consequence of these two, wearrive at the proposition that if only conservative forces act, the kineticenergy T plus the potential energy U remains constant:
T+U=constant. (14.2)
现在，我们就有下面两个命题：（1）一个力所做的功，等于这个粒子的动能的变化，但是，（2）从数学上看，对于一个保守的力，它所做的功，就是负的函数U的变化，U我们称为势能。作为这两个命题的结果，我们得到一个命题：如果只有保守力在做功，动能T加上势能U，恒为常数：
T+U=constant. (14.2)

Let us now discuss the formulas for the potential energy for a numberof cases. If we have a gravitational field that is uniform, if we are not goingto heights comparable with the radius of the earth, then the force is a constantvertical force and the work done is simply the force times the verticaldistance. Thus
U(z)=mgz, (14.3)
and the point P which corresponds to zero potential energy happens to be any point inthe plane z=0 . We could also have said that the potential energy is mg(z−6)if we had wanted to—all the results would, of course, be the same inour analysis except that the value of the potential energy at z=0 would be −6mg . It makes no difference, because only differences in potentialenergy count.
现在，我们来看几个案例，讨论它们势能的公式。如果我们有一个均匀的万有引力场，如果我们所达到的高度，不足以与这个地球的半径相比，那么，力就恒为垂直的，而功就只是力乘以垂直距离。这样：
U(z)=mgz, (14.3)
且相应于零势能的点P，刚好就在平面z=0处。如果我们想的话，我们也可以说，势能是mg(z−6)，--当然，在我们的分析中，所有的结果，将是同样的，除了在z=0处的势能的值，将是−6mg。但这并没有任何区别，因为，只有势能的差，才被用到。

The energy needed to compress a linearspring a distance x from an equilibrium point is
U(x)= kx2/2 (14.4)
and the zero of potential energy is at the point x=0 , the equilibrium position of the spring. Again we could add anyconstant we wish.
把一个线性的弹簧，从平衡点压缩距离x，所需的能量就是：
U(x)= kx2/2 (14.4)
且势能为零的位置，就是点x=0，即弹簧的平衡位置。我们还是加上任何我们希望的常数。

The potential energy of gravitation forpoint masses M and m , a distance r apart, is
U(r)=−GMm/r. (14.5)
The constant has been chosen here so that the potential is zero atinfinity. Of course the same formula applies to electrical charges, because itis the same law:
U(r)=q1q2/4πϵ0r. (14.6)
在万有引力场中，有质量点M和 m，其距离为r，那么，势能就是：
U(r)=−GMm/r。 (14.5)
这里，常数已经被选择了，这样在无穷远处，势能为零。当然，同样的公式，可以用于电荷，因为规律一样：
U(r)=q1q2/4πϵ0r. (14.6)

Now let us actually use one of these formulas, to see whether weunderstand what it means. Question: How fast do we have to shoot arocket away from the earth in order for it to leave? Solution: Thekinetic plus potential energy must be a constant; when it “leaves,” it will bemillions of miles away, and if it is just barely able to leave, we may supposethat it is moving with zero speed out there, just barely going. Let a be the radius of the earth, and M its mass. The kinetic plus potential energy is then initially givenby mv2/2 −GmM/a . At the end of the motion the two energies must be equal. The kineticenergy is taken to be zero at the end of the motion, because it is supposed tobe just barely drifting away at essentially zero speed, and the potentialenergy is GmM divided by infinity, which is zero. So everything is zero on one sideand that tells us that the square of the velocity must be 2GM/a. But GM/a2 is what we call the acceleration of gravity, g . Thus
v2=2ga.
现在，让我们实际用一个公式，来看看，我们能否理解它的意思。问题：为了让一个火箭发射后，能离开地球，发射时的速度应该是多少？解答：动能加上势能，应该是一个常数。当它“离开”时，应该是百万英里之外了，如果它只是刚刚能够离开我们，我们可以假设，它在那里的速度为零，只是勉强能走。设a是地球的半径，M是地球的质量。动能加势能，最初就是通过mv2/2−GmM/a，而被给予。在运动结束时，这两个能量，应该相等。在运动的结束时，动能被当做零，因为，它被假设，只是勉强漂走，本质上速度为零，而势能则是GmM除以无穷大，也是零。所以，两边都是零，这就告诉我们，速度的平方应该是2GM/a。但是，我们称GM/a2为重力加速度。这样：
v2=2ga。

At what speed must a satellite travel in order to keep going aroundthe earth? We worked this out long ago and found that v2=GM/a. Therefore to go away from the earth, we need 2–√ times the velocity we need to just go around the earthnear its surface. We need, in other words, twice as much energy (becauseenergy goes as the square of the velocity) to leave the earth as we do to goaround it. Therefore the first thing that was done historically with satelliteswas to get one to go around the earth, which requires a speed of five miles persecond. The next thing was to send a satellite away from the earth permanently;this required twice the energy, or about seven miles per second.
要让卫星绕着地球转，它的速度应该是多少？我们很久以前，就得出为v2=GM/a。因此，要离开地球，我们就需要‘根号2’乘以一个速度，该速度，是在地球表面，绕着地球转的速度。换句话说，我们需要两倍的能量，因为我们离开地球所需的能量，是绕着地球走所需能量的两倍（因为能量与矢速的平方成正比）。因此，历史上对卫星做的第一件事情，就是让一个卫星，绕着地球转，这要求的速度，是五英里每秒。第二件事情，就是发射一颗卫星，永远地离开地球，这要求两倍的能量，或者，七英里每秒。

Now, continuing our discussion of thecharacteristics of potential energy, let us consider the interaction of twomolecules, or two atoms, two oxygen atoms for instance. When they are very farapart, the force is one of attraction, which varies as the inverse seventhpower of the distance, and when they are very close the force is a very largerepulsion. If we integrate the inverse seventh power to find the work done, wefind that the potential energy U , which is a function of the radial distance between the two oxygenatoms, varies as the inverse sixth power of the distance for large distances.
现在，继续我们关于势能特点的讨论，让我们考虑两个分子之间的交互影响，也可以是两个原子，比如说两个氧原子。当它们离得很远时，力是吸引力的一种，随着距离的7次方的倒数而变化，当它们很近时，力就是很大的排斥力。如果我们积分这个七次方的倒数，以找出所做的功，我们会发现，势能 U，是两个氧原子之间径向距离的函数，对于大的距离来说，随着距离的6次方的倒数而变化。

Fig. 14–3.The potential energy between twoatoms as a function of the distance between them. 图14-3 两个原子之间的势能，作为它们之间距离的函数。
If we sketch the curve of the potentialenergy U(r) as in Fig. 14–3, wethus start out at large r with an inverse sixth power, but if we come in sufficiently near wereach a point d where there is a minimum of potential energy. The minimum of potentialenergy at r=d means this: if we start at d and move a small distance, a very small distance, the work done, whichis the change in potential energy when we move this distance, is nearly zero,because there is very little change in potential energy at the bottom of thecurve. Thus there is no force at this point, and so it is the equilibriumpoint. Another way to see that it is the equilibrium point is that it takeswork to move away from d in either direction. When the two oxygen atoms have settled down, sothat no more energy can be liberated from the force between them, they are in thelowest energy state, and they will be at this separation d .This is the way an oxygen molecule looks when it is cold. When weheat it up, the atoms shake and move farther apart, and we can in fact breakthem apart, but to do so takes a certain amount of work or energy, which is thepotential energy difference between r=d and r=∞ . When we try to push the atoms very close together the energy goes upvery rapidly, because they repel each other.
如果我们画出势能U(r)的曲线的草图，如图14-3，这样，我们就从大的r出发，按6次方的倒数走，但是，当我们来到足够近时，我们会达到一个点d，这里势能最小。在r=d 处，势能最小，其意思为：如果我们从d开始，移动一个小的距离，一个非常小的距离，那么，所做的功，即当我们移过这个距离时的势能变化，几乎是零，因为，在曲线底部，势能变化非常小。这样，在这个点，没有力，所以，它就是平衡点。当两个氧原子定居下来，这样，没有其他能量，可以把它们从它们之间的力中解放出来，它们就是处于最低状态，就是处于这个分开的距离d。当氧原子是看上去是冷的时候，它就处于这种方式。当我们加热它时，原子摇动，相互距离加大，事实上我们也能让它们分开，但要这样做，要一定量的功和能量，这就是在r=d 和 r=∞之间的势能差。当我们尝试把原子推得非常近时，能量上升非常快，因为它们互斥。