Find the coordinates of the center of mass of a homogeneous plate. How to calculate the center of gravity of a flat bounded figure using a double integral? How to perform a typical calculation

– calculating the center of gravity of a flat bounded figure. Many readers intuitively understand what the center of gravity is, but, nevertheless, I recommend repeating the material from one of the lessons analytical geometry, where I dismantled the problem of the center of gravity of a triangle and in an accessible form deciphered the physical meaning of this term.

In independent and control tasks, as a rule, it is proposed to solve simplest case– flat limited homogeneous a figure, that is, a figure of constant physical density - glass, wooden, tin cast-iron toys, difficult childhood, etc. Further, by default, we will only talk about such figures =)

First rule and the simplest example : if a flat figure has center of symmetry, then it is the center of gravity of this figure. For example, the center of a round homogeneous plate. It is logical and worldly clear - the mass of such a figure is "fairly distributed in all directions" relative to the center. Believe - I don't want to.

However, in harsh realities, you are unlikely to be thrown a sweet elliptical chocolate bar, so you have to arm yourself with a serious kitchen tool:

The coordinates of the center of gravity of a flat homogeneous limited figure are calculated by the following formulas:

, or:

, where is the area of ​​the region (figure); or very short:

, where

We will conditionally call the integral the “X” integral, and the integral the “Y” integral.

Note-help : for flat limited heterogeneous figure, the density of which is given by the function, the formulas are more complex:
, where - the mass of the figure;in the case of uniform density, they are simplified to the above formulas.

On the formulas, in fact, all the novelty ends, the rest is your ability solve double integrals By the way, now is a great opportunity to practice and improve your technique. And perfection, as you know, there is no limit =)

Let's throw in an invigorating portion of parabolas:

Example 1

Find the coordinates of the center of gravity of a homogeneous flat figure bounded by lines.

Solution: the lines here are elementary: it sets the abscissa axis, and the equation - a parabola, which is easily and quickly built using geometric transformations of graphs:

– parabola, shifted 2 units to the left and 1 unit down.

I will complete the entire drawing at once with the finished point of the center of gravity of the figure:

Rule Two: if the figure has axis of symmetry, then the center of gravity of this figure necessarily lies on this axis.

In our case, the figure is symmetrical about straight, that is, in fact, we already know the "x" coordinate of the point "em".

Also note that vertically the center of gravity is shifted closer to the x-axis, since the figure is more massive there.

Yes, perhaps not everyone has yet fully understood what the center of gravity is: please raise your index finger and mentally place the shaded “sole” on it with a dot. Theoretically, the figure should not fall.

We find the coordinates of the center of gravity of the figure by the formulas , where .

The order of traversing the area (shape) is obvious here:

Attention! Determining the most profitable traversal order once- and use it for all integrals!

1) First, calculate the area of ​​the figure. In view of the relative simplicity of the integral, the solution can be formulated compactly, the main thing is not to get confused in the calculations:

We look at the drawing and estimate the area by cells. It turned out about the case.

2) The x-coordinate of the center of gravity has already been found by the "graphical method", so you can refer to the symmetry and go to the next point. However, I still don’t advise doing this - it is likely that the solution will be rejected with the wording “use the formula”.

Note that here you can get by with exclusively oral calculations - sometimes it is not at all necessary to bring fractions to a common denominator or torment the calculator.

In this way:
which is what was required.

3) Find the ordinate of the center of gravity. Let's calculate the "game" integral:

And here it would be hard without a calculator. Just in case, I will comment that as a result of multiplication of polynomials, 9 terms are obtained, and some of them are similar. I gave similar terms orally (as is usually done in similar cases) and immediately wrote down the final amount.

As a result:
which is very, very close to the truth.

At the final stage, we mark a point on the drawing. According to the condition, it was not required to draw anything, but in most problems we willy-nilly are forced to draw a figure. But there is an absolute plus - a visual and quite effective check of the result.

Answer:

The following two examples are for independent solution.

Example 2

Find the coordinates of the center of gravity of a homogeneous plane figure bounded by lines

By the way, if you imagine how the parabola is located and saw the points at which it intersects the axis, then here you can actually do without a drawing.

And more difficult:

Example 3

Find the center of gravity of a homogeneous plane figure bounded by lines

If you have difficulty plotting, study (review) lesson on parabolas and / or Example No. 11 of the article Double integrals for dummies.

Sample solutions at the end of the lesson.

In addition, a dozen or two similar examples can be found in the corresponding archive on the page Ready-made solutions for higher mathematics.

Well, I can't help but please the lovers higher mathematics who often ask me to sort out difficult problems:

Example 4

Find the center of gravity of a homogeneous flat figure bounded by lines. Draw the figure and its center of gravity on the drawing.

Solution: the condition of this task already categorically requires the execution of a drawing. But the requirement is not so formal! - even a person with an average level of training can imagine this figure in his mind:

A straight line cuts the circle into 2 parts, and an additional clause (cm. linear inequalities) indicates that we are talking about a small shaded piece.

The figure is symmetrical about a straight line (depicted by a dotted line), so the center of gravity must lie on this line. And obviously its coordinates are modulo. An excellent guideline that practically excludes an erroneous answer!

Now the bad news =) An unpleasant integral from the root looms on the horizon, which we analyzed in detail in Example No. 4 of the lesson Efficient methods for solving integrals. And who knows what else will be drawn there. It would seem that due to the presence circles profitable, but not everything is so simple. The straight line equation is converted to the form and integrals will also turn out not sugar (although fans trigonometric integrals appreciate). In this regard, it is more prudent to dwell on Cartesian coordinates.

Shape traversal order:

1) Calculate the area of ​​the figure:

It is more rational to take the first integral subsuming under the sign of the differential:

And in the second integral we will carry out the standard replacement:

Let us calculate the new limits of integration:

2) Let's find .

Here in the 2nd integral was again used method of bringing a function under a differential sign. Practice and adopt these optimal (in my opinion) methods for solving typical integrals.

After difficult and lengthy calculations, we again turn our attention to the drawing (remember that the points we don't know yet! ) and we get deep moral satisfaction from the found value.

3) Based on the analysis carried out earlier, it remains to make sure that .

Excellent:

Let's draw a point on the drawing. In accordance with the formulation of the condition, we write it as the final answer:

A similar task for an independent solution:

Example 5

Find the center of gravity of a homogeneous flat figure bounded by lines. Execute the drawing.

This task is interesting because it contains a figure of sufficiently small sizes, and if you make a mistake somewhere, then there is a high probability of not getting into the area at all. Which, of course, is good in terms of decision control.

Sample design at the end of the lesson.

Sometimes useful transition to polar coordinates in double integrals. It depends on the figure. I searched and searched for a good example, but I didn’t find it, so I’ll demonstrate the solution on the 1st demo task of the lesson above:


Recall that in that example, we switched to polar coordinates, found out the procedure for bypassing the area and calculate its area

Let's find the center of gravity of this figure. The schema is the same: . The value is visible directly from the drawing, and the “x” coordinate should be shifted a little closer to the y-axis, since the more massive part of the semicircle is located there.

In integrals, we use standard transition formulas:

It's likely that they weren't wrong.

3 Applications of double integrals

3.1 Theoretical introduction

Consider Applications double integral to the solution of a number of geometric problems and problems of mechanics.

3.1.1 Calculating the area and mass of a flat plate

Consider a thin material plate D located in the plane Ohu. Square S this plate can be found using the double integral formula:

3.1.2 Static moments. Center of mass of a flat plate

static moment M x about the axis Ox material point P(x;y) lying in the plane Oxy and having a mass m, is called the product of the mass of a point and its ordinate, i.e. M x = my. The static moment is defined similarly M y about the axis Oy: ­ ­ ­ M y = mx. Static moments flat plate with surface density γ = γ (x, y) are calculated by the formulas:

As is known from mechanics, the coordinates x c ,y c the center of mass of a flat material system are determined by the equalities:

where m is the mass of the system, and M x and M y are the static moments of the system. Flat plate weight m is determined by formula (1), the static moments of a flat plate can be calculated by formulas (3) and (4). Then, according to formulas (5), we obtain an expression for the coordinates of the center of mass of a flat plate:

A typical calculation contains two tasks. In each problem, a flat plate is given D, bounded by the lines specified in the condition of the problem. G(x,y) is the surface density of the plate D. For this plate find: 1. S- square; 2. m- mass; 3. M y , M x– static moments about the axes Oy and Oh respectively; 4. , are the coordinates of the center of mass.

3.3 How to perform a typical calculation

When solving each problem, you must: 1. Make a drawing of a given area. Select the coordinate system in which the double integrals will be calculated. 2. Record the area as a system of inequalities in the selected coordinate system. 3. Calculate area S and mass m plates according to formulas (1) and (2). 4. Calculate static moments M y , M x according to formulas (3) and (4). 5. Calculate the coordinates of the center of mass , according to formulas (6). Put the center of mass on the drawing. In this case, there is a visual (qualitative) control of the results obtained. Numeric responses should be received with three significant figures.

3.4 Sample calculation examples

Task 1. plate D limited by lines: y = 4 – x 2 ; X = 0; y = 0 (x ≥ 0; y≥ 0) Areal density γ 0 = 3. Solution. The area specified in the problem is limited by a parabola y = 4 – x 2 , coordinate axes and lies in the first quarter (Fig. 1). The problem will be solved in the Cartesian coordinate system. This area can be described by a system of inequalities:

Rice. one

Square S plate is equal to (1): Since the plate is homogeneous, its mass m = γ 0 S= 3 = 16. Using formulas (3), (4), we find the static moments of the plate: The coordinates of the center of mass are found by the formula (6): Answer: S ≈ 5,33; m = 16; M x = 25,6; M y = 12; = 0,75; = 1,6.

Task 2. plate D limited by lines: X 2 + at 2 = 4; X = 0, at = X (X ≥ 0, at≥ 0). Surface density γ (x,y) = at. Solution. The plate is bounded by a circle and straight lines passing through the origin (Fig. 2). Therefore, to solve the problem, it is convenient to use the polar coordinate system. polar angle φ varies from π/4 to π/2. A beam drawn from the pole through the plate "enters" it at ρ = 0 and "leaves" the circle, the equation of which is: X 2 + at 2 = 4 <=>p = 2.

Rice. 2

Therefore, the given area can be written as a system of inequalities: The area of ​​the plate is found by formula (1): We find the mass of the plate by formula (2), substituting γ (x,y) = y = ρ sin φ :
To calculate the static moments of the plate, we use formulas (3) and (4):
We obtain the coordinates of the center of mass using formulas (6): Answer: S ≈ 1,57; m ≈ 1,886; M x = 2,57; M y = 1; = 0,53; = 1,36.

3.5 Reporting

The report should contain all the calculations performed, neatly executed drawings. Numeric responses should be received with three significant figures.

Let us give an example of determining the center of mass of a body by dividing it into separate bodies, the centers of mass of which are known.

Example 1. Determine the coordinates of the center of mass of a homogeneous plate (Fig. 9). Dimensions are given in millimeters in Figure 9.

Solution: Show the coordinate axes and . We divide the plate into parts, which are formed by three rectangles. For each rectangle, we draw diagonals, the intersection points of which determine the positions of the centers of mass of each rectangle. In the accepted coordinate system, it is easy to find the values ​​of the coordinates of these points. Namely:

(-1; 1), (1; 5), (5; 9). The areas of each body are respectively equal to:

; ; .

The area of ​​the entire plate is:

To determine the coordinates of the center of mass of a given plate, we use expressions (21). Substitute the values ​​of all known quantities in this equation, we get

According to the obtained values ​​of the coordinates of the center of mass of the plate, we indicate the point C in the figure. As you can see, the center of mass (geometric point) of the plate is outside it.

Addition method. This method is a partial case of the separation method. It can be applied to bodies that have notches (voids). Moreover, without the cut out part, the position of the center of mass of the body is known. Consider, for example, the application of such a method.

Example 2 Determine the position of the center of mass of the weight of a round plate with radius R, in which there is a cutout with radius r (Fig. 10). Distance .

Solution: As you can see, from Fig. 10, the center of mass of the plate lies on the axis of symmetry of the plate, that is, on the straight line, since this line is the axis of symmetry. Thus, to determine the position of the center of mass of this plate, it is necessary to determine only one coordinate, since the second coordinate will be located on the axis of symmetry and balance the zero ones. Let's show the coordinate axes , . Let us assume that the plate is made up of two bodies - from a full circle (as if without a cutout) and a body that seems to be made with a cutout. In the adopted coordinate system, the coordinates for the indicated bodies will be: .The areas of the bodies are: ; . The total area of ​​the whole body will be equal to the difference between the areas of the first and second bodies, namely