Of course, if an object is symmetrical insome way, for instance, a rectangle, so that it has a plane of symmetry, thecenter of mass lies somewhere on the plane of symmetry. In the case of arectangle there are two planes, and that locates it uniquely. But if it is justany symmetrical object, then the center of gravity lies somewhere on the axisof symmetry, because in those circumstances there are as many positive asnegative x ’s.
当然，如果一个对象，在某种意义上，是对称的，例如一个矩形，那么，它就有一个对称的平面，则质量中心，就处于对称平面的某处，在矩形这种情形下，有两个平面，这就可以唯一地定位质量中心。但是，如果对象，是任何对称的对象，那么，重量的中心，就处于对称轴的某处，因为，在这些情形中，正的x的数目，与负的x的数目，一样多。{？}

Fig. 19–1.The CM of a compound body lies onthe line joining the CM’s of the two composite parts. 图19-1 一个复合物体，由两部分构成，其质量中心，处于这两部分的质量中心的连线上。
Another interesting proposition is thefollowing very curious one. Suppose that we imagine an object to be made of twopieces, A and B (Fig. 19–1).Then the center of mass of the whole object can be calculated as follows.First, find the center of mass of piece A , and then of piece B . Also, find the total mass of each piece, MA and MB . Then consider a new problem, in which a point mass MAis at the center of mass of object A , and another point mass MB is at the center of mass of object B . The center of mass of these two point masses is then the center ofmass of the whole object. In other words, if the centers of mass of variousparts of an object have been worked out, we do not have to start all over againto find the center of mass of the whole object; we just have to put the piecestogether, treating each one as a point mass situated at the center of mass ofthat piece. Let us see why that is. Suppose that we wanted to calculate the centerof mass of a complete object, some of whose particles are considered to bemembers of object A and some members of object B . The total sum ∑mixi can then be split into two pieces—the sum ∑A mixifor the A object only, and the sum ∑B mixifor object B only. Now if we were computing the center of mass of object Aalone, we would have exactly the first of these sums, and we know thatthis by itself is MAXA , the total mass of all the particles in A times the position of the center of mass of A , because that is the theorem of the center of mass, applied toobject A . In the same manner, just by looking at object B , we get MBXB , and of course, adding the two yields MXCM :
(19.2)
Now since M is evidently the sum of MA and MB , we see that Eq. (19.2)can be interpreted as a special example of the center of mass formula for twopoint objects, one of mass MA located at XA and the other of mass MB located at XB .
另外一个有趣的命题，就是下面这个，非常奇特。想象一个对象，由两部分A和B组成（图19-1）。因此，整个对象的质量中心，就可如下计算。首先，找出A部分的质量中心，然后是B的。另外，找出每一部分的总质量MA 和MB。然后，考虑一个新的问题，在其中，点质量MA，位于A的质量中心，点质量MB，位于B的质量中心。因此，这两个点质量的质量中心，就是整个对象的质量中心。换句话说，如果一个对象的各部分的质量中心，已经得到，那么，要找整个对象的质量中心，我们没有必要，从头开始，我们只需把各部分，放在一起，把每一部分，都当做一个点质量，且位于该部分的质量中心处。让我们看看，为什么是这样。对于一个完整的对象，假设我们想计算其质量中心，其粒子的一部分，被认为是对象A的成员，另一部分，被认为是对象B的成员。因此，总和∑mixi，就可被分成两部分—只关于对象A的和∑A mixi，与只关于对象B的和 ∑B mixi。现在，如果我们只计算对象A的质量中心，我们刚好有这些和的第一部分{？}，我们知道这本身就是MAXA，A中的所有粒子的总质量，乘以A的质量中心的位置，因为，这就是质量中心定理，被应用于对象A。同理可得MBXB，当然，两者相加，可得MXCM：
(19.2)
现在，由于M，显然就是MA与MB之和，我们看到，对于两个点对象A与B，质量MA位于XA ，质量MB 位于XB，要求这两个点对象的质量中心，需要公式，方程（19.2），就可被解释为此类公式的特殊例子。

Let us now return to the center of mass.The center of mass is sometimes called the center of gravity, for the reasonthat, in many cases, gravity may be considered uniform. Let us suppose that wehave small enough dimensions that the gravitational force is not only proportionalto the mass, but is everywhere parallel to some fixed line. Then consider an objectin which there are gravitational forces on each of its constituent masses. Let mibe the mass of one part. Then the gravitational force on that part is mitimes g . Now the question is, where can we apply a single force to balancethe gravitational force on the whole thing, so that the entire object, if it isa rigid body, will not turn? The answer is that this force must go through thecenter of mass, and we show this in the following way. In order that the bodywill 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 arotation. So we must calculate the total of all the torques on all theparticles, and see how much torque there is about any given axis; it should be zeroif this axis is at the center of mass. Now, measuring x horizontally and y vertically, we know that the torques are the forces in the y -direction, times the lever arm x (that is to say, the force times the lever arm around which we want tomeasure the torque). Now the total torque is the sum
(19.3)
so if the total torque is to be zero, the sum ∑miximust be zero. But ∑mixi=MXCM, the total mass times the distance of the center of mass from theaxis. Thus the x -distance of the center of mass from the axis is zero.
现在，让我们返回质心。质心有时又被称为重心，因为在很多情况下，重心可被考虑为统一的。我们假设，我们有足够小的尺寸，这样，重力不仅正比于质量，而且，在任何地方，都平行于某些固定的线。因此，考虑一个对象，在其中，有重力作用于组成它的每个质量。设mi是某一部分的质量。那么，作用于这部分的重力，就是mi乘以g。现在的问题就是，整个事物，受着一个重力，对于此重力，我们要在什么地方，应用一个单一的力，才能与之平衡，这样，整个对象，如果它是一个刚体的话，就不会翻转了。答案就是，这个力，应该通过质心。我们将以如下方式，指明这点。为了让物体不翻转，由所有力所产生的扭力，加起来，就应该为零，因为，如果有一个扭力，那么，就有角动量的改变，从而就会有旋转。于是，我们应该计算，所有粒子上的所有扭力的总和，然后看看，对于任何被给予的轴，其扭力为多少；如果此轴在质心，那么，扭力就应为零。现在，水平地测量x，垂直此测量y，我们知道，扭力就是y方向的力，乘以x方向的杠杆臂(也就是说,力乘以杠杆臂，我们想测量的扭力，就是对此臂而言的)。现在，总得扭力，就是下面的和：
(19.3)
于是，如果总的扭力为零，那么，和∑mixi就应为零。但是，∑mixi=MXCM，即总的质量，乘以轴到质心的距离。这样，在x方向，轴都质心的距离，就是零。

This fact has a very interesting consequence.In an inertial frame that is not accelerating, the torque is always equal tothe rate of change of the angular momentum. However, about an axis through thecenter of mass of an object which is accelerating, it is still truethat the torque is equal to the rate of change of the angular momentum. Even ifthe 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 truethat the torque is equal to the rate of change of angular momentum around thataxis. Thus the theorem that torque equals the rate of change of angular momentumis true in two general cases: (1) a fixed axis in inertial space,(2) an axis through the center of mass, even though the object may beaccelerating.
这个事实，有个后果，非常有趣。在一个没有加速的惯性框架中，扭力总是等于角动量的变化率。然而，对于一个正在加速的对象，设有一轴，过其质心，那么，关于此轴：扭力等于角动量的变化率，仍然正确。即便质心，仍在加速，我们仍可选一特殊之轴，比如说，一个过质心的轴，这样，扭力等于绕此轴的角动量的变化率，仍然正确。这样，定理：扭力等于角动量的变化率，在两种情况下，为真（1）在惯性空间中的一个固定的轴，（2）一个过质心的轴，虽然对象可以在加速。

19–2Locating the center of mass 19-2 定位质量中心
The mathematical techniques for thecalculation of centers of mass are in the province of a mathematics course, andsuch problems provide good exercise in integral calculus. After one has learnedcalculus, however, and wants to know how to locate centers of mass, it is niceto know certain tricks which can be used to do so. One such trick makes use ofwhat is called the theorem of Pappus. It works like this: if we take any closedarea in a plane and generate a solid by moving it through space such that eachpoint is always moved perpendicular to the plane of the area, the resultingsolid has a total volume equal to the area of the cross section times the distancethat the center of mass moved! Certainly this is true if we move the area in astraight line perpendicular to itself, but if we move it in a circle or in someother curve, then it generates a rather peculiar volume. For a curved path, theoutside goes around farther, and the inside goes around less, and these effectsbalance out. So if we want to locate the center of mass of a plane sheet ofuniform density, we can remember that the volume generated by spinning it aboutan axis is the distance that the center of mass goes around, times the area ofthe sheet.
计算质心所需的数学技术，在数学课程的范围内，这类问题，给微积分，提供了好的练习。然而，在一个人学习了微积分之后，且想知道，如何定位质心，那么，了解一些用来做此事的技巧，将很有帮助。一个这种技巧，使得帕普斯定理（Pappus），变得有用。它是这样起作用的：在一个平面上，取一个封闭面积，通过在此空间，移动它，产生一个立体图形，这样，每个点在移动时，都垂直于此面积的平面，最终，产生一个立体图形，其体积，等于横截面的面积，乘以质心所移过的距离！当然，如果我们是在一条垂直于此面积的直线上，移动它，这当然是真的，但是，如果是在一个圆、或某种曲线上移动，那么，产生的体积，将相当独特。对于一个曲线路径，外部走的多些，内部走的少些，这些效果，相互抵消。于是，对于一个均匀密度的平面薄板，如果我们想定位其质心，我们可以记住，它绕轴旋转，所产生的体积，就是质心所走的距离，乘以此板的面积。

Fig. 19–2.A right triangle and a rightcircular cone generated by rotating the triangle. 图19-2 一个直角，与旋转它所产生的一个直立圆锥。
For example, if we wish to find the centerof mass of a right triangle of base D and height H (Fig. 19–2), wemight solve the problem in the following way. Imagine an axis along H, and rotate the triangle about that axis through a full 360 degrees. This generates a cone. The distance that the x -coordinate of the center of mass has moved is 2πx . The area which is being moved is the area of the triangle, HD/2. So the x -distance of the center of mass times the area of the triangle is thevolume swept out, which is of course πD2H/3 . Thus (2πx)(HD/2)=πD2H/3 , or x=D/3 . In a similar manner, by rotating about the other axis, or bysymmetry, we find y=H/3 . In fact, the center of mass of any uniform triangular area is wherethe three medians, the lines from the vertices through the centers of the oppositesides, all meet. That point is 1/3 of the way along each median. Clue: Slice the triangle up intoa lot of little pieces, each parallel to a base. Note that the median linebisects every piece, and therefore the center of mass must lie on this line.
例如，对于一个直角三角形，底为D，高为H（图19-2），如果我们希望，找到其质心，我们可用下面的方式，解决此问题。想象沿着H，有一个轴，让此三角形，绕这个轴，旋转360度。这会产生一个圆锥。质心的x坐标，移动的距离，是 2πx。被移动的面积，就是三角形的面积HD/2。于是，质心的x方向的距离，乘以三角形的面积，就是所扫过的体积，它当然就是πD2H/3。这样(2πx)(HD/2)=πD2H/3，或 x=D/3。类似地，通过绕其他的轴旋转、或通过对称，我们发现y=H/3。事实上，对于任何均匀的三角形，其面积的质心，就是其三条中线的交点，中线，就是顶点与对边中点的连线。这个点，在每条中线的1/3处。提示：把此三角形，从下到上，切成细条，每一条都平行于底边。注意，中线等分每个小条，因此，质心必在这条线上。

Now let us try a more complicated figure.Suppose that it is desired to find the position of the center of mass of auniform semicircular disc—a disc sliced in half. Where is the center of mass?For a full disc, it is at the center, of course, but a half-disc is moredifficult. Let r be the radius and x be the distance of the center of mass from the straight edge of thedisc. Spin it around this edge as axis to generate a sphere. Then the center ofmass has gone around 2πx , the area is πr2/2 (because it is only half a circle). The volume generated is, ofcourse, 4πr3/3 , from which we find that
(2πx)(πr2/2)=4πr3/3,
or
x=4r/3π.
现在，让我们看一个更复杂的图形。假设要找出质心的位置，一个均匀的半圆盘—一个圆盘，分成两半。质心何在？对于一个完整的圆盘，它当然在中心，但是，半圆盘，要更复杂些。设r为半径，x是质心到圆盘直边的距离。以这个直边为轴，转它，就会产生一个球体。因此，质心走过的距离就是2πx，面积是πr2/2（因为它是一个半圆）。所产生的体积，当然就是4πr3/3，由此我们可以得出：
(2πx)(πr2/2)=4πr3/3,
或
x=4r/3π.

There is another theorem of Pappus which is a special case of theabove one, and therefore equally true. Suppose that, instead of the solidsemicircular disc, we have a semicircular piece of wire with uniform massdensity along the wire, and we want to find its center of mass. In this casethere is no mass in the interior, only on the wire. Then it turns out that the areawhich is swept by a plane curved line, when it moves as before, is the distancethat the center of mass moves times the length of the line. (The linecan be thought of as a very narrow area, and the previous theorem can be appliedto it.)
还有另外一条帕普斯定理，它是上面情况的特例，因此，也同样为真。假设，我们有的，不是一个固体的半圆盘，而是一条匀质的金属线，它构成一个半圆，我们想找出其质心。在此特例下，线内没有质量，只有线有质量。因此，当线的平面，与前面一样移动时，所扫过的面积，就是质心移过的距离，乘以线的长度（线可被认为，是一个非常窄的面积，前面的定理，也可适用于它。）

19–3Finding the moment of inertia 19-3 寻找惯性力矩
Now let us discuss the problem of findingthe moments of inertia of various objects. The formula for the moment ofinertia about the z -axis of an object is
I=∑mi(x2i+y2i)
or
I=∫(x2+y2)dm=∫(x2+y2)ρdV. (19.4)
That is, we must sum the masses, each one multiplied by the square ofits distance (x2i+y2i)from the axis. Note that it is not the three-dimensional distance,only the two-dimensional distance squared, even for a three-dimensional object.For the most part, we shall restrict ourselves to two-dimensional objects, butthe formula for rotation about the z -axis is just the same in three dimensions.
现在，让我们讨论问题：如何找出不同对象的惯性力矩。一个对象关于z轴的惯性力矩的公式就是：
I=∑mi(x2i+y2i)
或
I=∫(x2+y2)dm=∫(x2+y2)ρdV. (19.4)
也就是说，我们必须把质量加起来，且每个质量，都要乘以它到轴的距离的平方。注意，虽然对象是三维的，但是，这不是三维的距离，而是二维的距离的平方。大部分情况下，我们将把我们自己，限制在二维的对象上，但是，关于z轴的旋转公式，在三维中，是一样的。

Fig. 19–3.A straight rod of length Lrotating about an axis through one end. 图19-3 一个直杆，长度为L，通过一端，绕轴旋转。
As a simple example, consider a rodrotating about a perpendicular axis through one end (Fig. 19–3). Nowwe must sum all the masses times the x -distances squared (the y ’s being all zero in this case). What we mean by “the sum,” of course,is the integral of x2 times the little elements of mass. If we divide the rod into smallelements of length dx , the corresponding elements of mass are proportional to dx, and if dx were the length of the whole rod the mass would be M . Therefore
dm=M dx/L
and so
(19.5)
The dimensions of moment of inertia are always mass times lengthsquared, so all we really had to work out was the factor 1/3 .
举一个简单的例子，考虑一个杆，以一端，绕一个垂直轴旋转（图19-3）。现在，我们必须把所有‘质量乘以x方向距离的平方’，加起来，（在此情况下，y方向的距离都是零）。我们通过“加”，意味着什么呢？当然，就是‘x2乘以小的质量元素’的积分。如果我们把杆，分成长度为dx的小元素，那么，相应的质量元素，就正比于dx，如果dx是整个杆的长度，则质量就是 M。
因此：
dm=M dx/L
于是:
(19.5)
惯性力矩的大小，总是：质量乘以长度的平方；所以，我们真正要得出的，就是因子1/3。

Now what is I if the rotation axis is at the center of the rod? We could just do theintegral over again, letting x range from – L /2 to + L /2 . But let us notice a few things about the moment of inertia. We canimagine the rod as two rods, each of mass M/2 and length L/2 ; the moments of inertia of the two small rods are equal, and are bothgiven by the formula (19.5).Therefore the moment of inertia is
I=2(M/2)(L/2)2 /3=ML2/12. (19.6)
Thus it is much easier to turn a rod about its center, than to swingit around an end.
现在，如果旋转轴，在杆的中心，那么，I是多少？我们可以把这个积分，重做一遍，让x的变化范围，是从– L /2 ，到 + L /2。但是，让我们注意惯性力矩的一些事情。我们可以想象，把杆分成两部分，每部分的质量都是M/2，长度都是L/2；两个小杆的惯性力矩相等，都可通过公式（19.5）给出。因此，惯性力矩就是：
I=2(M/2)(L/2)2 /3=ML2/12. (19.6)
这样，让一个杆，绕着中心旋转，比绕着一端旋转，要容易地多。

This theorem is called the parallel-axis theorem, and may beeasily proved. The moment of inertia about any axis is the mass times the sumof the x i ’s and the y i ’s, each squared: I=∑(x2i+y2i)m i . We shall concentrate on the x ’s, but of course the y ’s work the same way. Now x is the distance of a particular point mass from the origin, but let usconsider how it would look if we measured x′ from the CM, instead of x from the origin. To get ready for this analysis, we write
xi=x′I +XCM.
Then we just square this to find
So, when this is multiplied by mi and summed over all i , what happens? Taking the constants outside the summation sign, weget
The third sum is easy; it is just MX2CM . In the second sum there are two pieces, one of them is ∑mix′i, which is the total mass times the x′ -coordinate of the center of mass. But this contributes nothing,because x′ is measured from the center of mass, and in these axes theaverage position of all the particles, weighted by the masses, is zero. Thefirst sum, of course, is the x part of Ic . Thus we arrive at Eq. (19.7),just as we guessed.
这个定理，被称为平行轴定理，很易证明。关于任何轴的惯性力矩，就是质量，乘以x i的平方加上y i的平方，并对i求和，即I=∑(x2i+y2i)m I 。我们将集中研究x，当然，y同理。现在x是某一个具体的点质量，到原点的距离，但是，让我们考虑，如果我们从CM开始，测量x′，用它来表示x，我们看，结果如何。为了准备这个分析，我们写
xi=x′I +XCM.
对此事两边取平方，得到：

于是，两边乘以mi，并对所有的 i求和，会发什么呢？把常数从求和符号中提出，我们就得到：

第三项很简单，它就是MX2CM。第二项中，有个，一个是∑mix′i，它是总质量，乘以质心的x′ 坐标。但是这项，没有贡献，因为x′ 是从质心开始测量的，在这些轴中，所有粒子的通过质量加权{？}的平均位置，是零。第一项，当然就是 Ic 的x部分。这样，我们就得到了方程（19.7），如所预测。

Let us check (19.7)for one example. Let us just see whether it works for the rod. For an axisthrough one end, the moment of inertia should be ML2/3 , for we calculated that. The center of mass of a rod, of course, isin the center of the rod, at a distance L/2 . Therefore we should find that ML2/3=ML2/12+M(L/2)2. Since one-quarter plus one-twelfth is one-third, we have made no fundamentalerror.
让我们用一个例子，来检查（19.7）。让们看看，对于杆子，它是否起作用。如果轴，通过杆子的一端，那么，惯性力矩就应该是ML2/3，因为我们已经计算过了。杆子的质量中心，当然就是在杆子的中间，在距离 L/2处。因此，我们应该得出，ML2/3=ML2/12+M(L/2)2。由于四分之一，加十二分分之一，就是三分之一，可见，我们并没有犯任何根本性的错误。

Incidentally, we did not really need to usean integral to find the moment of inertia (19.5).If we simply assume that it is ML2 times γ , an unknown coefficient, and then use the argument about the twohalves to get γ/4 for (19.6),then from our argument about transferring the axes we could prove that γ=γ/4+1/4, so γ must be 1/3 . There is always another way to do it!
顺便说一下，我们确实并不需要使用积分，以得到惯性力矩（19.5）。如果我们只需简单地认为，它就是ML2乘以 γ，作为未知的系数，然后，对于两个一半，使用参数，为（19.6）得到γ/4，那么，从我们转换轴后所得到的参数，我们就可以证明，γ=γ/4+1/4 , 故γ必为1/3。做此事，总有其他办法。{？}

In applying the parallel-axis theorem, itis of course important to remember that the axis for Ic must be parallel to the axis about which the moment of inertiais wanted.
在应用平行轴定理时，有一点，当然非常重要，即我们所要求的惯性力距，有一个轴，而Ic的轴，应与此轴平行。