Eljuicio contra José Antonio Primo de Rivera'

We are going to manipulate Eqns. 8.3 and 8.4 to get some interesting results. Let us add their left-hand sides and equate them to the sum of the right-hand sides:

m1¨x1+ m2¨x2= F12+ F1e+ F21+ F2e. (8.5) I now invoke Newton’s third law,

F12= −F21, (8.6)

whatever the underlying force: gravity, spring, electrostatic, and so forth.

This whole chapter is about milking this one simple result, this cancella-tion of F12and F21. Next, I lump all the external forces F1eand F2einto one external force Fe:

Fe= F1e+ F2e. (8.7)

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I then have this equation:

m1¨x1+ m2¨x2= Fe. (8.8)

I multiply and divide the left-hand side by the total mass

M= m1+ m2 (8.9) is called the center-of-mass coordinate or the CM.

What I have done is correct, but why did I do that? I have introduced a fictitious entity, the center of mass. The center of mass has a location X, which is a weighted average of x1and x2: vice versa. The weighted sum gives a certain coordinate, but there is noth-ing there. All the stuff is either at x1or x2. The center of mass is the location of a mathematical entity. It’s not a physical entity. But we care, because it behaves like a body. After all, if you were shown only Eqn. 8.11, you would say, “Well, this guy’s talking about a body of mass M undergoing some acceleration due to the force Fe.” Thus, the center of mass is a fictitious body, whose mass is the total mass of these two particles, and whose accel-eration is controlled by only the total external force. This is the key. All the internal forces have canceled out, and what remains is the external force.

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If you have three bodies, you can do a similar manipulation with extra forces like F23= −F32and so on, and you will end up with Eqn. 8.11

We write such sums in a compact notation:

X=

which can be generalized to N particles by replacing the 3 by N.

Once more with feeling: the CM responds only to the total external force; it doesn’t care about internal forces. I’ll give an example. A cou-ple of samurai are having a fight in an airplane, punching each other and so on. It is a flight and fight situation. The rest of the passengers get fed up and throw them out. So, they’re falling down, affecting each other’s dynamics. This samurai will feel a force due to that samurai, that samurai will feel a force due to this samurai, but the center of mass is still going to drop like a rock. It’s going to feel a force (m1+ m2)g, and it will have an acceleration g.

Suppose at some point one falling samurai cuts the other into two pieces. So now we have three bodies: the first protagonist and the other two, who used to be one. You can take these three bodies, find their center of mass, and it will be the same story—the center of mass will just keep accelerating at the same g, as if nothing happened.

Indeed, if you were following the CM alone, you would see no sign of all this violence and the involuntary partitioning of the second samurai.

The system is becoming more and more complicated, but nothing changes the dynamics of the center of mass, undergoing free fall under gravity.

In summary, the CM can accelerate only due to external forces, like gravity in this example. If there were no external forces, then the center of mass would behave like a free particle. If it was not moving to begin with, it won’t move later. If it was moving to begin with, it will maintain the initial velocity.

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Figure 8.2 The CM vector R lies on r₂− r1, the line joining the tips of r₁and r₂.

If you’re living in two dimensions, you define a CM vector R for two particles as follows and as indicated in Figure 8.2:

R= iX + jY =m1r1+ m2r2

m1+ m2

, (8.17)

which is equal to two relations X=m1x1+ m2x2

m1+ m2

(8.18) Y=m1y1+ m2y2

m1+ m2

. (8.19)

I have shown the CM situated on the line connecting the masses.

Why? One way to understand this is to choose a new x^-axis that passes through the masses and a new y^-axis perpendicular to it. Since the masses only have an x^coordinate, the weighted average must also have only an x^ coordinate.

Here’s another example. You take a complicated collection of masses and springs, connected by ropes and chains and whatnot. You throw the whole mess into the air. All the different parts of it are jiggling and doing complicated movements, but if you follow the center of mass, in other words, at every instant you compute

R=

N i=1miri

N i=1mi

, (8.20)

it will simply follow the parabolic path of a body curving under gravity.

If at some point this complicated object fragments into two disconnected chunks that fly off on their own paths, the CM will continue as before.

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Here is a very useful alternative for finding the CM of a collection of masses.

1. Take a subset of them and replace them with all their mass sitting at their center of mass. Replace the rest by their total mass sitting at their center of mass.

2. Find the center of mass of these two centers of mass, properly weighted.

Let us verify the equivalence of this to the standard definition, for the case of three bodies. The original recipe to compute X was

X=m1x1+ m2x2+ m3x3

m1+ m2+ m3

. (8.21)

Next we follow the new recipe and divide the system into two parts, one made of masses numbered 1 and 2, of total mass M12= m1+ m2, and the third one by itself. We first compute the CM of 1+ 2 following the standard definition:

X12=m1x1+ m2x2

M12

(8.22) and combine it with object 3 and see what happens if we follow the recipe

XRecipe =m3x3+ M12X12

Clearly, in a general case, we can subdivide the masses into more than two subsets and follow the same procedure.

As long as you are dealing with a countable number of masses, find-ing the CM is just plug and chug, in one or higher dimensions. Thfind-ings become more interesting if I give you an extended body, like a rod of mass M and length L, shown in the upper part of Figure 8.3. Where is the center of mass? We have to adapt the previous definition for discrete masses. The

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Figure 8.3 (Top) A rod of length L and mass M. (Bottom) An L-shaped object made of two rectangles. The points 1 and 2 are the CMs where all their mass may be assumed to be concentrated. The point marked CM is the weighted sum of 1 and 2.

trick is to partition the rod into tiny pieces of length dx located a distance x from the origin, chosen to be at the left end of the rod. The mass of this segment is ^M_Ldx, the product of the mass per unit length and the length of the sliver. Its location is x.

Now, you might object that the segment extends from x to x+ dx and doesn’t have a definite coordinate. This objection is valid for any finite dx, but in the end, we will let dx→ 0 and the objection will disappear in that limit. The CM is now given by the ratio of integrals, rather than sums as in Eqn. 8.20:

So the center of mass of this rod, to nobody’s surprise, is right at the midpoint. We knew this before doing the integral. What was behind that intuition? If you take the origin at the midpoint of the rod, you can argue that for every sliver with coordinate x, you have another equally mas-sive one at−x, and that the weighted average of these two points is zero.

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The CM, which is the weighted average of all these zeros, also lies at the geometric center.

This symmetry argument applies even for a two-dimensional body, like a rectangle. We can argue that its CM is at its geometric center. This is because for each tiny square segment of size dx· dy at (x, y) (measured from the geometric center), we can find an equal one at (−x, −y), and the weighted average of the two will be at the origin (0, 0). The weighted average of all these zeros will also be (0, 0).

Now consider a rod whose linear mass density, or mass per unit length, varies as some functionρ(x). Then by the usual argument

X=

L

0 ρ(x)xdx

L

0 ρ(x)dx . (8.29)

For example, ifρ(x) = Ax, where A is some constant

X=

which is biased to the right as you would expect.

Consider next an L-shaped object shown in the lower part of Figure 8.3. Where is its CM? We resort to the trick of first reducing each rectangle to its CM and then doing the weighted sum of the two CMs. By the symmetry argument, the CM of each rectangle is at its geometric cen-ter, the points numbered 1 and 2. We may imagine their total masses M1

and M2to be concentrated at these points. The CM is the weighted sum of these two point masses, and it lies along the line joining them. Its precise location is easily found if the masses and dimensions of the rectangles are known.

Now for one final CM calculation, the most difficult one you are sup-posed to know. The object is a triangle of mass M, base 2w, and height h, as shown in Figure 8.4. It has an area

A=1

22wh= wh (8.32)

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Figure 8.4 The CM calculation of a triangle of base 2w and height h. It is viewed as a weighted sum over rods of width dx and height 2y(x).

and a mass per unit area or areal density ρ =M

A = M

wh. (8.33)

Where is the center of mass of this object? Again, by symmetry, you can tell that Y, the y coordinate of the center of mass, must be zero. For every tiny square dxdy with some coordinate (x, y), there is a matching one with coordinate (x,−y). For X, you have to do some honest work. We will divide and conquer.

Let us imagine the triangle as composed of thin rectangles of width dx and height 2y(x), as indicated. (Each strip is not quite a rectangle, because the edges are slightly tapered, but when dx→ 0, they will reduce to rectangles.) The mass dm of the rectangle at a given x is

dm=M

A2y(x)dx= M

wh2y(x)dx, (8.34)

which is just the product of the mass per unit area ^M_A and the area of the strip 2y(x)dx. We find y(x) using similar triangles:

y(x) w =x

h which means y(x)=wx

h . (8.35)

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The weighted average of x is then

X= 1

We could have anticipated that X would be skewed to the right, and this formula quantifies that intuition. Note in Eqn. 8.37 that this two-dimensional problem maps onto a one-two-dimensional one, with a linear density proportional to x, that is, ρ(x) ∝ x. This is because each vertical strip may be replaced by a point mass on the x-axis proportional to y(x), which in turn grows linearly with x.

To summarize, when we work with extended bodies or more than one body, we can replace the entire body by a single point for certain purposes. The single point is called a center of mass or CM. The CM is fictitious. It has a mass equal to the total mass. It has a location R that moves in response to the total external force:

Md²R

dt² = Fe. (8.40)

The center of mass is not aware of internal forces, and that’s what we want to exploit.

One class of problems has a net external force Fe, and there we know that the CM responds as a point to Fe, regardless of its constituents.

For example, a jumbled mass of constituents tossed in the air follows the parabolic trajectory of a point mass, in response to gravity. This is just a one-body problem, which we have studied extensively. So we move on.

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8.3 Law of conservation of momentum