Center of mass

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A Center of mass is a weighted mean of the position vectors of a discrete distribution of N-particles with masses.

[math]\displaystyle{ R_{CM}=\frac{\sum_i^N m_ir_i}{\sum_i^N m_i} \quad\textrm{for}\quad i=1,2,...N }[/math]
where [math]\displaystyle{ N }[/math] is the total number of particles, [math]\displaystyle{ r_i }[/math] and [math]\displaystyle{ m_i }[/math] are the individual position vectors and masses, respectively.
For a continuous distribution of mass, i.e. a rigid object of mass M, this expression becomes:

[math]\displaystyle{ R_{CM}= lim_{\Delta m \rightarrow 0}\frac{\sum_i^N \Delta m_ir_i}{M}=\frac{1}{M}\int_0^Mr\;dm }[/math]

[math]\displaystyle{ R_{CM}=\frac{m \sum_0^Mr_i}{Nm}=\frac{\sum_0^Mr_i}{N} }[/math]
[math]\displaystyle{ R_{CM}=\frac{1}{M}\int_0^Mxdm=\frac{1}{M}\int_0^Lx \frac{M}{L}x\;dx=\frac{L}{2} }[/math]


References

2016

2005

[math]\displaystyle{ x_{cm}=\frac{m_1x_1+m_2x_2}{m1+m2} }[/math]

If you are making measurements from the center of mass point for a two-mass system then the center of mass condition can be expressed as
[math]\displaystyle{ m_1=m_2\frac{r_2}{r_1} }[/math]
where [math]\displaystyle{ r_1 }[/math] and [math]\displaystyle{ r_2 }[/math] locate the masses. The center of mass lies on the line connecting the two masses.

1996

1963

  • (Feynman et al., 1963) ⇒ Richard P. Feynman, Robert B. Leighton and Matthew Sands (1963, 1977, 2006, 2010, 2013) "The Feynman Lectures on Physics": New Millennium Edition is now available online by the California Institute of Technology, Michael A. Gottlieb, and Rudolf Pfeiffer ⇒ http://www.feynmanlectures.caltech.edu/
    • Chapter 19: Suppose for a moment that the object is divided into little pieces, 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. Then this equation simply says that we add all the x’s, and then divide by the number of things that we have added [math]\displaystyle{ X_CM=m\sum x_i/mN = \sum x_i/N }[/math]. In other words, [math]\displaystyle{ X_{CM} }[/math] is the average of all the x’s, if the masses are equal. But suppose one of them were twice as heavy as the others. Then in the sum, that x would come in twice. This is easy to understand, for we can think of this double mass as being split into two equal ones, just like the others; then in 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 of times 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 could be a circle, like a hoop, and the center of mass is in the center of the hoop, not in the hoop itself.

      Of course, if an object is symmetrical in some way, for instance, a rectangle, so that it has a plane of symmetry, the center of mass lies somewhere on the plane of symmetry. In the case of a rectangle there are two planes, and that locates it uniquely. But if it is just any symmetrical object, then the center of gravity lies somewhere on the axis of symmetry, because in those circumstances there are as many positive as negative x’s(...)

      Suppose that we imagine an object to be made of two pieces, 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, [math]\displaystyle{ M_A }[/math] and [math]\displaystyle{ M_B }[/math]. Then consider a new problem, in which a point mass [math]\displaystyle{ M_A }[/math] is at the center of mass of object A, and another point mass [math]\displaystyle{ M_B }[/math] is at the center of mass of object B. The center of mass of these two point masses is then the center of mass of the whole object(...)

      The 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 force is not only proportional to the mass, but is everywhere parallel to some fixed line. Then consider an object in which there are gravitational forces on each of its constituent masses. Let [math]\displaystyle{ m_i }[/math] be the mass of one part. Then the gravitational force on that part is [math]\displaystyle{ m_i }[/math] 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 must 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 (...)

      In 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 is accelerating, it is still true that the torque is equal to the rate of change of the angular momentum. Even if 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 momentum around that axis. Thus the theorem that torque equals the rate of change of angular momentum is 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 be accelerating.