Methods for determining the coordinates of the center of gravity. How to calculate the center of gravity of a flat bounded figure using a double integral? Determination of the center of gravity of bodies of complex shape

Author: Let's take an arbitrary shape body. Is it possible to hang it on a thread so that after hanging it retains its position (i.e. does not begin to turn) when any initial orientation (fig. 27.1)?

In other words, is there such a point, relative to which the sum of the moments of the forces of gravity acting on different parts of the body, would be equal to zero at any orientation of the body in space?

Reader: Yes, I think so. Such a point is called the center of gravity of the body.

Proof. For simplicity, consider a body in the form of a flat plate of arbitrary shape arbitrarily oriented in space (Fig. 27.2). Take the coordinate system X 0at with the origin at the center of mass - a point With, then x C = 0, at C = 0.

We represent this body as a collection of a large number of point masses m i, the position of each of which is given by the radius vector .

By definition of the center of mass , and the coordinate x C = .

Since in our coordinate system x C= 0, then . Let's multiply this equation by g and get

As can be seen from fig. 27.2, | x i| is the shoulder of strength. And if x i> 0, then the moment of force M i> 0, and if x j < 0, то Mj < 0, поэтому с учетом знака можно утверждать, что для любого x i moment of force will be M i = m i gx i . Then equality (1) is equivalent to , where M i is the moment of gravity. And this means that with an arbitrary orientation of the body, the sum of the moments of the forces of gravity acting on the body will be equal to zero relative to its center of mass.

In order for the body we are considering to be in equilibrium, it is necessary to apply to it at a point With strength T = mg pointing vertically upward. The moment of this force about the point With equals zero.

Since our reasoning did not depend in any way on how exactly the body is oriented in space, we proved that the center of gravity coincides with the center of mass, which was what was required to be proved.

Problem 27.1. Find the center of gravity of a weightless rod of length l, at the ends of which two point masses are fixed t 1 and t 2 .

t 1 t 2 l	Decision. We will look not for the center of gravity, but for the center of mass (since they are one and the same). Let's introduce the axis X(Fig. 27.3). Rice. 27.3
x C =?

Answer: away from mass t 1 .

STOP! Decide for yourself: B1-B3.

Statement 1 . If a homogeneous flat body has an axis of symmetry, the center of gravity is on this axis.

Indeed, for any point mass m i, located to the right of the axis of symmetry, there is the same point mass located symmetrically with respect to the first (Fig. 27.4). In this case, the sum of the moments of forces .

Since the whole body can be represented as divided into similar pairs of points, the total moment of gravity relative to any point lying on the axis of symmetry is zero, which means that the center of gravity of the body is also located on this axis. This leads to an important conclusion: if the body has several axes of symmetry, then the center of gravity lies at the intersection of these axes(Fig. 27.5).

Rice. 27.5

Statement 2. If two bodies with masses t 1 and t 2 are connected into one, then the center of gravity of such a body will lie on a straight line connecting the centers of gravity of the first and second bodies (Fig. 27.6).

Rice. 27.6 Rice. 27.7

Proof. Let us arrange the composite body so that the segment connecting the centers of gravity of the bodies is vertical. Then the sum of the moments of gravity of the first body with respect to the point With 1 is equal to zero, and the sum of the moments of gravity of the second body about the point With 2 is zero (Fig. 27.7).

notice, that shoulder gravity of any point mass t i the same with respect to any point on the segment With 1 With 2 , and hence the moment of gravity relative to any point lying on the segment With 1 With 2 are the same. Therefore, the gravity of the whole body is zero with respect to any point on the segment With 1 With 2. Thus, the center of gravity of the composite body lies on the segment With 1 With 2 .

Statement 2 implies an important practical conclusion, which is clearly formulated in the form of an instruction.

instruction,

how to find the center of gravity of a rigid body if it can be broken

into parts, the positions of the centers of gravity of each of which are known

1. Replace each part with a mass located at the center of gravity of that part.

2. Find center of gravity(and this is the same as the center of gravity) of the resulting system of point masses, choosing a convenient coordinate system X 0at, according to the formulas:

Indeed, let us position the compound body in such a way that the segment With 1 With 2 was horizontal, and we will hang it on threads at points With 1 and With 2 (Fig. 27.8, a). It is clear that the body will be in equilibrium. And this balance will not be disturbed if we replace each body with point masses t 1 and t 2 (Fig. 27.8, b).

Rice. 27.8

STOP! Decide for yourself: C3.

Problem 27.2. Balls of mass are placed at two vertices of an equilateral triangle t everyone. The third vertex contains a ball of mass 2 t(Fig. 27.9, a). Triangle side a. Determine the center of gravity of this system.

t 2t a	Rice. 27.9
x C = ? at C = ?

Decision. We introduce the coordinate system X 0at(Fig. 27.9, b). Then

,

.

Answer: x C = a/2; ; the center of gravity lies at half the height AD.

Based on the general formulas obtained above, it is possible to indicate specific methods for determining the coordinates of the centers of gravity of bodies.

1. If a homogeneous body has a plane, axis or center of symmetry, then its center of gravity lies respectively either in the plane of symmetry, or on the axis of symmetry, or in the center of symmetry.

Suppose, for example, that a homogeneous body has a plane of symmetry. Then, by this plane, it is divided into two such parts, the weights of which and are equal to each other, and the centers of gravity are at equal distances from the plane of symmetry. Consequently, the center of gravity of the body as a point through which the resultant of two equal and parallel forces passes will indeed lie in the plane of symmetry. A similar result is obtained in cases where the body has an axis or center of symmetry.

It follows from the properties of symmetry that the center of gravity of a homogeneous round ring, a round or rectangular plate, a rectangular parallelepiped, a ball and other homogeneous bodies with a center of symmetry lies in the geometric center (center of symmetry) of these bodies.

2. Partitioning. If the body can be divided into a finite number of such parts, for each of which the position of the center of gravity is known, then the coordinates of the center of gravity of the whole body can be directly calculated using formulas (59) - (62). In this case, the number of terms in each of the sums will be equal to the number of parts into which the body is divided.

Problem 45. Determine the coordinates of the center of gravity of the homogeneous plate shown in fig. 106. All measurements are in centimeters.

Decision. We draw the x, y axes and divide the plate into three rectangles (cut lines are shown in Fig. 106). We calculate the coordinates of the centers of gravity of each of the rectangles and their area (see table).

Whole plate area

Substituting the calculated quantities into formulas (61), we obtain:

The found position of the center of gravity C is shown in the drawing; point C is outside the plate.

3. Addition. This method is a special case of the partitioning method. It applies to bodies with cutouts if the centers of gravity of the body without the cutout and the cutout are known.

Problem 46. Determine the position of the center of gravity of a round plate of radius R with a radius cut (Fig. 107). Distance

Decision. The center of gravity of the plate lies on the line, since this line is the axis of symmetry. Draw coordinate axes. To find the coordinate, we supplement the area of ​​the plate to a full circle (part 1), and then subtract the area of ​​the cut circle from the resulting area (part 2). In this case, the area of ​​\u200b\u200bpart 2, as subtracted, should be taken with a minus sign. Then

Substituting the found values ​​into formulas (61), we obtain:

The found center of gravity C, as you can see, lies to the left of the point

4. Integration. If the body cannot be divided into several finite parts, the positions of the centers of gravity of which are known, then the body is first divided into arbitrary small volumes for which formulas (60) take the form

where are the coordinates of some point lying inside the volume. Then, in equalities (63), they pass to the limit, tending everything to zero, i.e., contracting these volumes into points. Then the sums in the equalities turn into integrals extended over the entire volume of the body, and formulas (63) give in the limit:

Similarly, for the coordinates of the centers of gravity of areas and lines, we obtain in the limit from formulas (61) and (62):

An example of applying these formulas to determining the coordinates of the center of gravity is considered in the next paragraph.

5. Experimental method. The centers of gravity of inhomogeneous bodies of complex configuration (aircraft, steam locomotive, etc.) can be determined experimentally. One of the possible experimental methods (suspension method) is that the body is suspended on a thread or cable at its various points. The direction of the thread on which the body is suspended will each time give the direction of gravity. The point of intersection of these directions determines the center of gravity of the body. Another possible way to experimentally determine the center of gravity is the weighing method. The idea behind this method is clear from the example below.

center of gravity A rigid body is a geometric point that is rigidly connected with this body and is the center of parallel gravity forces applied to individual elementary particles of the body (Figure 1.6).

Radius vector of this point

Figure 1.6

For a homogeneous body, the position of the center of gravity of the body does not depend on the material, but is determined by the geometric shape of the body.

If the specific gravity of a homogeneous body γ , the weight of the elementary particle of the body

P k = γΔV k (P = γV ) substitute into the formula to determine r C , we have

From where, projecting onto the axes and passing to the limit, we obtain the coordinates of the center of gravity of a homogeneous volume

Similarly, for the coordinates of the center of gravity of a homogeneous surface with an area S (Figure 1.7, a)

Figure 1.7

For the coordinates of the center of gravity of a homogeneous line of length L (Figure 1.7, b)

Methods for determining the coordinates of the center of gravity

Based on the general formulas obtained earlier, it is possible to indicate methods for determining the coordinates of the centers of gravity of solid bodies:

1 Analytical(by integration).

2 Symmetry method. If the body has a plane, axis or center of symmetry, then its center of gravity lies respectively in the plane of symmetry, axis of symmetry or in the center of symmetry.

3 Experimental(body suspension method).

4 splitting. The body is divided into a finite number of parts, for each of which the position of the center of gravity C and area S known. For example, the projection of a body onto a plane xOy (Figure 1.8) can be represented as two flat figures with areas S 1 and S 2 (S=S 1 + S 2 ). The centers of gravity of these figures are at the points C 1 (x 1 ,y 1 ) and C 2 (x 2 ,y 2 ) . Then the coordinates of the center of gravity of the body are

Figure 1.8

5Addition(method of negative areas or volumes). A special case of the partitioning method. It applies to bodies with cutouts if the centers of gravity of the body without the cutout and the cutout are known. For example, you need to find the coordinates of the center of gravity of a flat figure (Figure 1.9):

Figure 1.9

Centers of gravity of the simplest figures

Figure 1.10

1 triangle

The center of gravity of the triangle area coincides with the point of intersection of its medians (Figure 1.10, a).

DM=MB , CM= (1/3)AM .

2 Arc of a circle

The arc has an axis of symmetry (Figure 1.10, b). The center of gravity lies on this axis, i.e. y C = 0 .

dl – arc element, dl = Rdφ , R is the radius of the circle, x = Rcosφ , L= 2aR ,

Hence:

x C = R(sinα/α) .

3 Circular sector

Radius sector R with central angle 2 α has an axis of symmetry Ox , on which the center of gravity is located (Figure 1.10, c).

We divide the sector into elementary sectors, which can be considered triangles. The centers of gravity of elementary sectors are located on the arc of a circle of radius (2/3) R .

The center of gravity of the sector coincides with the center of gravity of the arc AB :

14. Methods for specifying the movement of a point.

With the vector method of specifying motion, the position of a point is determined by the radius vector drawn from a fixed point in the selected reference system.

With the coordinate method of specifying motion, the coordinates of a point are specified as a function of time:

These are the parametric equations of the trajectory of a moving point, in which time plays the role of a parameter t . To write down its equation in an explicit form, it is necessary to exclude from them t .

With the natural method of specifying the movement, the trajectory of the point, the origin on the trajectory with the indication of the positive direction of reference, the law of change of the arc coordinate are set: s=s(t) . This method is convenient to use if the trajectory of the point is known in advance.

15. 1.2 Point speed

Consider the movement of a point over a small period of time Δt :

average speed of a point over a period of time Dt . The speed of a point at a given time

Point speed is a kinematic measure of its motion, equal to the time derivative of the radius vector of this point in the reference frame under consideration. The velocity vector is directed tangentially to the trajectory of the point in the direction of motion.

The center of gravity is the point through which the line of action of the resultant elementary forces of gravity passes. It has the property of the center of parallel forces (E. M. Nikitin, § 42). So formulas for determining the position of the center of gravity of various bodies look like:
x c = (∑ G i x i) / ∑ G i ;
(1) y c = (∑ G i y i) / ∑ G i ;
z c = (∑ G i z i) / ∑ G i .

If the body whose center of gravity needs to be determined can be identified with a figure made up of lines (for example, a closed or open contour made of wire, as in Fig. 173), then the weight G i of each segment l i can be represented as a product
G i \u003d l i d,
where d is the weight of a unit length of the material that is constant for the entire figure.

After substituting into formulas (1) instead of G i their values ​​l i d, the constant factor d in each term of the numerator and denominator can be put out of brackets (outside the sign of the sum) and reduced. Thus, formulas for determining the coordinates of the center of gravity of a figure composed of line segments, will take the form:
x c = (∑ l i x i) / ∑ l i ;
(2) y c = (∑ l i y i) / ∑ l i ;
z c = (∑ l i z i) / ∑ l i .

If the body has the form of a figure composed of planes or curved surfaces located in various ways (Fig. 174), then the weight of each plane (surface) can be represented as follows:
G i = F i p,
where F i are the areas of each surface, and p is the weight per unit area of ​​the figure.

After substituting this value of G i into formulas (1), we obtain formulas for the coordinates of the center of gravity of a figure composed of areas:
x c = (∑ F i x i) / ∑ F i ;
(3) y c = (∑ F i y i) / ∑ F i ;
z c = (∑ F i z i) / ∑ F i .

If a homogeneous body can be divided into simple parts of a certain geometric shape (Fig. 175), then the weight of each part
G i = V i γ,
where V i is the volume of each part, and γ is the weight per unit volume of the body.

After substituting the values ​​of G i into formulas (1), we obtain formulas for determining the coordinates of the center of gravity of a body composed of homogeneous volumes:
x c = (∑ V i x i) / ∑ V i ;
(4) y c = (∑ V i y i) / ∑ V i ;
z c = (∑ V i z i) / ∑ V i .

When solving some problems to determine the position of the center of gravity of bodies, it is sometimes necessary to know where the center of gravity of an arc of a circle, a circular sector or a triangle is located.

If the radius of the arc r and the central angle 2α, contracted by the arc and expressed in radians, are known, then the position of the center of gravity C (Fig. 176, a) relative to the center of the arc O is determined by the formula:
(5) x c = (r sin α)/α.

If the chord AB=b of the arc is given, then in formula (5) it is possible to make the replacement
sinα = b/(2r)
and then
(5a) x c = b/(2α).

In a special case for a semicircle, both formulas will take the form (Fig. 176, b):
(5b) x c = OC = 2r/π = d/π.

The position of the center of gravity of the circular sector, if its radius r is given (Fig. 176, c), is determined using the formula:
(6) x c = (2r sin α)/(3α).

If the chord of the sector is given, then:
(6a) x c = b/(3α).

In a special case for a semicircle, both last formulas will take the form (Fig. 176, d)
(6b) x c = OC = 4r/(3π) = 2d/(3π).

The center of gravity of the area of ​​any triangle is located from any side at a distance equal to one third of the corresponding height.

In a right triangle, the center of gravity is at the intersection of perpendiculars raised to the legs from points located at a distance of one third of the length of the legs, counting from the top of the right angle (Fig. 177).

When solving problems to determine the position of the center of gravity of any homogeneous body, composed either of thin rods (lines), or of plates (areas), or of volumes, it is advisable to adhere to the following order:

1) draw a body, the position of the center of gravity of which needs to be determined. Since all body dimensions are usually known, scale must be observed;

2) break the body into component parts (line segments or areas, or volumes), the position of the centers of gravity of which is determined based on the size of the body;

3) determine either the lengths, or the areas, or the volumes of the constituent parts;

4) choose the location of the coordinate axes;

5) determine the coordinates of the centers of gravity of the constituent parts;

6) the found values ​​of the lengths or areas, or volumes of individual parts, as well as the coordinates of their centers of gravity, substitute into the appropriate formulas and calculate the coordinates of the center of gravity of the whole body;

7) according to the found coordinates, indicate in the figure the position of the center of gravity of the body.

§ 23. Determining the position of the center of gravity of a body composed of thin homogeneous rods

§ 24. Determination of the position of the center of gravity of figures composed of plates

In the last problem, as well as in the problems given in the previous paragraph, the division of figures into component parts does not cause much difficulty. But sometimes the figure has such a form that allows you to divide it into its component parts in several ways, for example, a thin rectangular plate with a triangular cut (Fig. 183). When determining the position of the center of gravity of such a plate, its area can be divided into four rectangles (1, 2, 3 and 4) and one right triangle 5 in several ways. Two options are shown in Fig. 183, a and b.

The most rational is the way of dividing the figure into its component parts, in which the smallest number of them is formed. If the figure has cutouts, then they can also be included in the number of component parts of the figure, but the area of ​​the cut out part is considered negative. Therefore, this division is called the method of negative areas.

The plate in fig. 183, c is divided using this method into only two parts: rectangle 1 with the area of ​​​​the entire plate, as if it were whole, and triangle 2 with an area that we consider negative.

§ 26. Determination of the position of the center of gravity of a body composed of parts having a simple geometric shape

To solve problems of determining the position of the center of gravity of a body made up of parts that have a simple geometric shape, it is necessary to have the skills to determine the coordinates of the center of gravity of figures made up of lines or areas.

Determining the center of gravity of an arbitrary body by successively adding up the forces acting on its individual parts is a difficult task; it is facilitated only for bodies of comparatively simple form.

Let the body consist of only two weights of mass and connected by a rod (Fig. 125). If the mass of the rod is small compared to the masses and , then it can be neglected. Each of the masses is affected by gravity equal to and respectively; both of them are directed vertically down, that is, parallel to each other. As we know, the resultant of two parallel forces is applied at the point , which is determined from the condition

Rice. 125. Determination of the center of gravity of a body consisting of two loads

Therefore, the center of gravity divides the distance between two loads in a ratio inverse to the ratio of their masses. If this body is suspended at a point , it will remain in equilibrium.

Since two equal masses have a common center of gravity at a point dividing the distance between these masses in half, it is immediately clear that, for example, the center of gravity of a homogeneous rod lies in the middle of the rod (Fig. 126).

Since any diameter of a homogeneous round disk divides it into two completely identical symmetrical parts (Fig. 127), the center of gravity must lie on each disk diameter, that is, at the point of intersection of the diameters - in the geometric center of the disk. Arguing in a similar way, we can find that the center of gravity of a homogeneous ball lies in its geometric center, the center of gravity of a homogeneous rectangular parallelepiped lies at the intersection of its diagonals, etc. The center of gravity of a hoop or ring lies in its center. The last example shows that the center of gravity of a body can lie outside the body.

Rice. 126. The center of gravity of a homogeneous rod lies in its middle

Rice. 127. The center of a homogeneous disk lies at its geometric center

If the body has an irregular shape or if it is inhomogeneous (for example, it has voids), then the calculation of the position of the center of gravity is often difficult and this position is more convenient to find through experience. Let, for example, it is required to find the center of gravity of a piece of plywood. Let's hang it on a thread (Fig. 128). Obviously, in the equilibrium position, the center of gravity of the body must lie on the continuation of the thread, otherwise the force of gravity will have a moment relative to the point of suspension, which would begin to rotate the body. Therefore, drawing a straight line on our piece of plywood, representing the continuation of the thread, we can assert that the center of gravity lies on this straight line.

Indeed, by suspending the body at different points and drawing vertical lines, we will make sure that they all intersect at one point. This point is the center of gravity of the body (since it must lie simultaneously on all such lines). In a similar way, one can determine the position of the center of gravity not only of a flat figure, but also of a more complex body. The position of the center of gravity of the aircraft is determined by rolling it with wheels onto the scale platform. The resultant of the weight forces on each wheel will be directed vertically, and you can find the line along which it acts by the law of addition of parallel forces.

Rice. 128. The point of intersection of vertical lines drawn through the points of suspension is the center of gravity of the body

When the masses of individual parts of the body change or when the shape of the body changes, the position of the center of gravity changes. So, the center of gravity of an aircraft moves when fuel is consumed from the tanks, when luggage is loaded, etc. For a visual experiment illustrating the movement of the center of gravity when the shape of the body changes, it is convenient to take two identical bars connected by a hinge (Fig. 129). In the case when the bars form a continuation of one another, the center of gravity lies on the axis of the bars. If the bars are bent at the hinge, then the center of gravity is outside the bars, on the bisector of the angle they form. If an additional load is put on one of the bars, then the center of gravity will move towards this load.