Center of inertia of a system of material points. Center of mass of a system of material points. Body center of mass

Draw a diagram of the system and mark the center of gravity on it. If the found center of gravity is outside the object system, you received an incorrect answer. You may have measured distances from different reference points. Repeat the measurements.

• For example, if children are sitting on a swing, the center of gravity will be somewhere between the children, and not to the right or left of the swing. Also, the center of gravity will never coincide with the point where the child is sitting.
• These arguments are valid in two-dimensional space. Draw a square that will contain all the objects of the system. The center of gravity should be inside this square.

Check mathematical calculations, if you get a small result. If the reference point is at one end of the system, a small result places the center of gravity near the end of the system. This may be the correct answer, but in the vast majority of cases this result indicates an error. When you calculated the moments, did you multiply the corresponding weights and distances? If instead of multiplying you added the weights and distances, you would get a much smaller result.

Correct the error if you found multiple centers of gravity. Each system has only one center of gravity. If you found multiple centers of gravity, you most likely did not add up all the moments. The center of gravity is equal to the ratio of the “total” moment to the “total” weight. There is no need to divide “every” moment by “every” weight: this way you will find the position of each object.

Differential equations of system motion

Let us consider a system consisting of $n$ material points. Let us select some point of the system with mass $m_(k).$ We denote the resultant of all external forces applied to the point (both active and constraint reactions) by $\overline(F)_(k)^(e) $, and the resultant all internal forces - through $\overline(F)_(k)^(l) $. If the point has an acceleration $\overline(a_(k) )$, then according to the basic law of dynamics:

We get a similar result for any point. Therefore, for the entire system it will be:

Equations (1) are differential equations movement of the system in vector form.

Projecting equalities (1) onto the coordinate axes, we obtain the equations of motion of the system in differential form in projections onto these axes.

However, when solving many specific tasks the need to find the law of motion of each of the points of the system does not arise, but sometimes it is enough to find the characteristics that determine the movement of the entire system as a whole.

Theorem on the motion of the center of mass of the system

To determine the nature of the motion of a system, it is necessary to know the law of motion of its center of mass. The center of mass or center of inertia of a system is such an imaginary point, the radius vector $R$ of which is expressed through the radius vectors $r_(1) ,r_(2) ,...$of material points according to the formula:

$R=\frac(m_(1) r_(1) +m_(2) r_(2) +...+m_(n) r_(n) )(m) $, (2)

where $m=m_(1) +m_(2) +...+m_(n) $ is the total mass of the entire system.

To find this law, let us turn to the equations of motion of system (1) and add their left and right sides term by term. Then we get:

$\sum m_(k) \overline(a)_(k) =\sum \overline(F)_(k)^(e) +\sum \overline(F)_(k)^(l) $. (3)

From formula (2) we have:

Taking the second derivative with respect to time, we get:

$\sum m_(k) \overline(a)_(k) =M\overline(a)_(c) $, (4)

where $\overline(a)_(c) $ is the acceleration of the center of mass of the system.

Since, by the property of internal forces in the system, $\sum \overline(F)_(k)^(l) =0$, we finally obtain from equality (3), taking into account (4):

$M\overline(a)_(c) =\sum \overline(F)_(k)^(e) $. (5)

Equation (5) expresses the theorem on the motion of the center of mass of the system: the product of the mass of the system and the acceleration of its center of mass is equal to the geometric sum of all external forces acting on the system, or the center of mass of the system moves like a material point, the mass of which is equal to the mass of the entire system and to which all external forces are applied forces acting on the system.

Projecting both sides of equality (5) onto the coordinate axes, we obtain:

$M\ddot(x)_(c) =\sum \overline(F)_(kx)^(e) $, $M\ddot(y)_(c) =\sum \overline(F)_( ky)^(e) $, $M\ddot(z)_(c) =\sum \overline(F)_(kz)^(e) $. (6)

These equations are differential equations of motion of the center of mass in projections on the axes of the Cartesian coordinate system.

The meaning of the theorem is as follows:

Theorem

• A forward moving body can always be considered as a material point with a mass equal to the mass of the body. In other cases, the body can be considered as a material point only when, in practice, to determine the position of the body it is enough to know the position of its center of mass and it is permissible, according to the conditions of the problem, not to take into account the rotational part of the body’s movement;
• The theorem allows us to exclude from consideration all previously unknown internal forces. This is its practical value.

Example

A metal ring suspended on a thread to the axis of a centrifugal machine rotates uniformly with an angular velocity $\omega $. The thread makes an angle $\alpha $ with the axis. Find the distance from the center of the ring to the axis of rotation.

\[\omega \] \[\alpha \]

Our system is affected by the force of gravity $\overline(N)$ $\overline(N)$ $\alpha \alpha$, the tension force of the thread and centripetal acceleration.

Let's write down Newton's second law for our system:

Let's project both parts onto the x and y axes:

\[\left\( \begin(array)(c) N\sin \alpha =ma; \\ N\cos \alpha =mg; \end(array) \right.(4)\]

Dividing one equation by the other, we get:

Since $a=\frac(v^(2) )(R) ;$$v=\omega R$, we find the required distance:

Answer: $R=\frac(gtg\alpha )(\omega ^(2) ) $

The motion of the system, in addition to the acting forces, also depends on its total mass and mass distribution. The mass of the system (denoted by M or ) is equal to the arithmetic sum of the masses of all points or bodies forming the system.

the distribution of masses in the system is determined by the values ​​of the masses of its points and their relative positions, i.e., their coordinates. However, it turns out that when solving those problems of dynamics that we will consider, in particular the dynamics of a rigid body, to take into account the distribution of masses it is not enough to know all the quantities , and some, the summary characteristics expressed through them. They are: the coordinates of the center of mass (expressed through the sum of the products of the masses of the points of the system by their coordinates), axial moments of inertia (expressed through the sum of the products of the masses of the points of the system by the squares of their coordinates) and centrifugal moments of inertia (expressed through the sum of the products of the masses of the points of the system and two of them). their coordinates). We will consider these characteristics in this chapter.

Center of mass In a uniform gravitational field, for which g=const, the weight of any particle of the body is proportional to its mass. Therefore, the distribution of masses in a body can be judged by the position of its center of gravity. Let us transform formulas (59) from § 32, which determine the coordinates of the center of gravity of the body, to a form that explicitly contains mass. To do this, let us put in the above formulas, after which, reducing by g, we find:

The resulting equalities now include the masses of the material points (particles) forming the body and the coordinates of these points. Consequently, the position of a point really characterizes the distribution of masses in a body or in any mechanical system, if we mean, respectively, the masses and coordinates of the points of the system.

The geometric point C, the coordinates of which are determined by formulas (1), is called the center of mass or center of inertia of a mechanical system.

If the position of the center of mass is determined by its radius vector, then from equalities (1) for we obtain the formula

where are the radius vectors of the points forming the system.

From the results obtained it follows that for a rigid body located in a uniform gravitational field, the positions of the center of mass and the center of gravity coincide. But unlike the center of gravity, the concept of the center of mass retains its meaning for a body located in any force field (for example, in the central gravitational field), and, in addition, as a characteristic of the distribution of masses, it makes sense not only for a solid body, but also for any mechanical system.

The term “center of mass” is used not only in mechanics and in calculations of motion, but also in everyday life. It’s just that people don’t always think about what laws of nature are manifested in a given situation. For example, figure skaters in pair skating actively use the center of mass of the system when they spin holding hands.

The concept of center of mass is also used in ship design. It is necessary to take into account not just two bodies, but a huge number of them and bring everything to a single denominator. Errors in calculations mean a lack of stability of the ship: in one case, it will be excessively submerged in water, risking sinking with the slightest waves; and in another they are too elevated above sea level, creating the danger of turning over on their side. By the way, this is why every thing on board must be in its place, as specified by the calculations: the most massive ones are at the very bottom.

The center of mass is used not only in relation to celestial bodies and the design of mechanisms, but also in the study of the “behavior” of particles of the microworld. For example, many of them are born in pairs (electron-positron). Possessing initial rotation and obeying the laws of attraction/repulsion, they can be considered as a system with a common center of mass.

Lesson "Center of Mass"

Schedule: 2 lessons

Target: Introduce students to the concept of “center of mass” and its properties.

Equipment: figures made of cardboard or plywood, tumbler, penknife, pencils.

Lesson Plan

Lesson stages time methods and techniques

I Introduction to students 10 frontal survey, students’ work at the blackboard.

to the lesson problem

II. Learning something new 15-20 Teacher’s story, problem solving,

material: 10 experimental task

III Practicing new 10 student messages

material: 10-15 problem solving,

15 frontal poll

IV. Conclusions. Homework 5-10 Oral summary of the material by the teacher.

task Writing on the board

Progress of the lesson.

I Repetition 1. Frontal survey: leverage of force, moment of force, equilibrium condition, types of equilibrium

Epigraph: The center of gravity of each body is a certain point located inside it - such that if you mentally hang the body from it, then it remains at rest and maintains its original position.

II. Explanationnew material

Let a body or system of bodies be given. Let us mentally divide the body into arbitrarily small parts with masses m1, m2, m3... Each of these parts can be considered as a material point. The position in space of the i-th material point with mass mi is determined by the radius vector ri(Fig. 1.1). The mass of a body is the sum of the masses of its individual parts: m = ∑ mi.

The center of mass of a body (system of bodies) is such a point C, the radius vector of which is determined by the formula

r= 1/m∙∑ mi ri

It can be shown that the position of the center of mass relative to the body does not depend on the choice of the origin O, i.e. The definition of the center of mass given above is unambiguous and correct.

The center of mass of homogeneous symmetrical bodies is located in their geometric center or on the axis of symmetry; the center of mass of a flat body in the form of an arbitrary triangle is located at the intersection of its medians.

Problem solution

PROBLEM 1. Homogeneous balls with masses m1 = 3 kg, m2 = 2 kg, m3 = 6 kg, and m4 = 3 kg are attached to a light rod (Fig. 1.2). Distance between the centers of any nearby balls

a = 10 cm. Find the position of the center of gravity and the center of mass of the structure.

SOLUTION. The position of the center of gravity of the structure relative to the balls does not depend on the orientation of the rod in space. To solve the problem, it is convenient to place the rod horizontally, as shown in Figure 2. Let the center of gravity be on the rod at a distance L from the center of the left ball, i.e. from t. A. At the center of gravity, the resultant of all gravity forces is applied and its moment relative to axis A is equal to the sum of the moments of gravity of the balls. We have r = (m1 + m2 + m3 + m4) g ,

R L = m2gα + m 3 g 2 a + m 4 g 3 a.

Hence L=α (m1 +2m3 + 3m4)/ (m1 + m2 + m3 + m4) ≈ 16.4 cm

ANSWER. The center of gravity coincides with the center of mass and is located at point C at a distance L = 16.4 cm from the center of the left ball.

It turns out that the center of mass of a body (or system of bodies) has a number of remarkable properties. In dynamics it is shown that the momentum of an arbitrarily moving body is equal to the product of the mass of the body and the speed of its center of mass and that the center of mass moves as if all external forces acting on the body were applied at the center of mass, and the mass of the entire body was concentrated in it.

The center of gravity of a body located in the gravitational field of the Earth is called the point of application of the resultant of all gravity forces acting on all parts of the body. This resultant is called the force of gravity acting on the body. The force of gravity applied at the center of gravity of the body has the same effect on the body as the forces of gravity acting on individual parts of the body.

An interesting case is when the size of the body is much smaller than the size of the Earth. Then we can assume that parallel gravity forces act on all parts of the body, i.e. the body is in a uniform gravitational field. Parallel and identically directed forces always have a resultant force, which can be proven. But at a certain position of the body in space, it is possible to indicate only the line of action of the resultant of all parallel forces of gravity; the point of its application will remain undetermined for now, because for a solid body, any force can be transferred along the line of its action. What about the application point?

It can be shown that for any position of the body in a uniform field of gravity, the line of action of the resultant of all gravitational forces acting on individual parts of the body passes through the same point, motionless relative to the body. At this point the equal force is applied, and the point itself will be the center of gravity of the body.

The position of the center of gravity relative to the body depends only on the shape of the body and the distribution of mass in the body and does not depend on the position of the body in a uniform field of gravity. The center of gravity is not necessarily located in the body itself. For example, a hoop in a uniform field of gravity has its center of gravity at its geometric center.

In a uniform field of gravity, the center of gravity of a body coincides with its center of mass.

In the overwhelming majority of cases, one term can be painlessly replaced by another.

But: the center of mass of a body exists regardless of the presence of a gravitational field, and we can talk about the center of gravity only in the presence of gravity.

It is convenient to find the location of the center of gravity of the body, and therefore the center of mass, taking into account the symmetry of the body and using the concept of moment of force.

If the arm of the force is zero, then the moment of the force is zero and such a force does not cause rotational motion of the body.

Consequently, if the line of action of the force passes through the center of mass, then it moves translationally.

Thus, the center of mass of any flat figure can be determined. To do this, you need to fix it at one point, giving it the opportunity to rotate freely. It will be installed so that the force of gravity, turning it, passes through the center of mass. At the point where the figure is secured, hang a thread with a load (nut), draw a line along the suspension (i.e., the line of gravity). Let's repeat the steps, securing the figure at another point. The intersection of the lines of action of gravity forces is the center of mass of the body

Experimental task: determine the center of gravity of a flat figure (based on the figures prepared earlier by students from cardboard or plywood).

Instructions: fix the figure on a tripod. We hang a plumb line from one of the corners of the figure. We draw the line of action of gravity. Rotate the figure and repeat the action. The center of mass lies at the point of intersection of the lines of action of gravity.

Students who quickly complete the task can be given an additional task: attach a weight (metal bolt) to the figure and determine the new position of the center of mass. Draw a conclusion.

The study of the remarkable properties of “centers”, which are more than two thousand years old, turned out to be useful not only for mechanics - for example, in the design vehicles And military equipment, calculating the stability of structures or for deriving equations of motion for jet vehicles. It is unlikely that Archimedes could even imagine that the concept of the center of mass would be very convenient for research in nuclear physics or in the physics of elementary particles.

Student messages:

In his work “On the Equilibrium of Flat Bodies,” Archimedes used the concept of the center of gravity without actually defining it. Apparently, it was first introduced by an unknown predecessor of Archimedes or by himself, but in an earlier work that has not reached us.

Seventeen long centuries had to pass before science added new results to Archimedes’ research on centers of gravity. This happened when Leonardo da Vinci managed to find the center of gravity of the tetrahedron. He, reflecting on the stability of Italian inclined towers, including the Pisa tower, came to the “theorem about the support polygon.”

The conditions of equilibrium of floating bodies, discovered by Archimedes, subsequently had to be rediscovered. This was done at the end of the 16th century by the Dutch scientist Simon Stevin, who used, along with the concept of the center of gravity, the concept of “center of pressure” - the point of application of the pressure force of the water surrounding the body.

Torricelli's principle (and the formulas for calculating the center of mass are also named after him), it turns out, was anticipated by his teacher Galileo. In turn, this principle formed the basis of Huygens’s classic work on pendulum clocks, and was also used in Pascal’s famous hydrostatic studies.

The method that allowed Euler to study the motion of a rigid body under the action of any forces was to decompose this motion into the displacement of the center of mass of the body and rotation around the axes passing through it.

To keep objects in a constant position when their support moves, the so-called cardan suspension has been used for several centuries - a device in which the center of gravity of a body is located below the axes around which it can rotate. An example is a ship's kerosene lamp.

Although the gravity on the Moon is six times less than on Earth, it would be possible to increase the high jump record there “only” by four times. Calculations based on changes in the height of the center of gravity of the athlete’s body lead to this conclusion.

In addition to the daily rotation around its axis and the annual revolution around the Sun, the Earth takes part in another circular motion. Together with the Moon, it “spins” around a common center of mass, located approximately 4,700 kilometers from the center of the Earth.

Some artificial Earth satellites are equipped with a folding rod several or even tens of meters long, weighted at the end (the so-called gravitational stabilizer). The fact is that an elongated satellite, when moving in orbit, tends to rotate around its center of mass so that its longitudinal axis is vertical. Then it, like the Moon, will always be facing the Earth with one side.

Observations of the movement of some visible stars indicate that they are part of binary systems in which “celestial partners” rotate around a common center of mass. One of the invisible companions in such a system could be a neutron star or, possibly, a black hole.

Teacher's explanation

Center of mass theorem: the center of mass of a body can change its position only under the influence of external forces.

Corollary of the theorem on the center of mass: the center of mass of a closed system of bodies remains motionless during any interactions of the bodies of the system.

Solving the problem (at the board)

PROBLEM 2. The boat is standing motionless in still water. The person in the boat moves from the bow to the stern. At what distance h will the boat move if the mass of a person is m = 60 kg, the mass of the boat is M = 120 kg, and the length of the boat is L = 3 m? Neglect water resistance.

SOLUTION. Let us use the condition of the problem that the initial velocity of the center of mass is zero (the boat and the man were initially at rest) and there is no water resistance (no external forces in the horizontal direction act on the “man-boat” system). Consequently, the coordinate of the center of mass of the system in the horizontal direction has not changed. Figure 3 shows the initial and final positions of the boat and the person. Initial coordinate x0 of the center of mass x0 = (mL+ML/2)/(m+M)

Final coordinate x of the center of mass x = (mh+M(h+L/2))/(m+M)

Equating x0 = x, we find h= mL/(m+M) =1m

Additionally: collection of problems by Stepanova G.N. No. 393

Teacher's explanation

Recalling the equilibrium conditions, we found that

For bodies with a support area, stable equilibrium is observed when the line of action of gravity passes through the base.

Consequence: than larger area supports and the lower the center of gravity, the more stable the equilibrium position.

Demonstration

Place the children's toy tumbler (Vanka - Vstanka) on a rough board and lift the right edge of the board. In what direction will the “head” of the toy deviate while maintaining its balance?

Explanation: The center of gravity C of the tumbler is below the geometric center O of the spherical surface of the “torso”. In the equilibrium position, point C and point of contact A of a toy with an inclined plane should be on the same vertical; therefore, the tumbler’s “head” will deviate to the left

How to explain the preservation of equilibrium in the case shown in the figure?

Explanation: The center of gravity of the pencil-knife system lies below the fulcrum

IIIConsolidation. Frontal survey

Questions and tasks

1. When a body moves from the equator to the pole, the force of gravity acting on it changes. Does this affect the position of the body's center of gravity?

Answer: no, because the relative changes in the force of gravity of all elements of the body are the same.

2. Is it possible to find the center of gravity of a “dumbbell” consisting of two massive balls connected by a weightless rod, provided that the length of the “dumbbell” is comparable to the diameter of the Earth?

Answer: no. The condition for the existence of a center of gravity is the uniformity of the gravitational field. In a non-uniform gravitational field, rotations of the “dumbbell” around its center of mass lead to the fact that the lines of action L1 and L2, the resultant forces of gravity applied to the balls, do not have a common point

3. Why does the front part of a car drop when you brake sharply?

Answer: when braking, a frictional force acts on the wheels on the road side, creating a torque around the center of mass of the car.

4. Where is the center of gravity of the donut?

Answer: in the hole!

5. Water is poured into a cylindrical glass. How will the position of the center of gravity of the glass - water system change?

Answer: The center of gravity of the system will first decrease and then increase.

6. What length of end must be cut from a homogeneous rod so that its center of gravity shifts by ∆ℓ?

Answer: length 2∆ℓ.

7. A homogeneous rod was bent in the middle at a right angle. Where was his center of gravity now?

Answer: at point O - the middle of the segment O1O2 connecting the midpoints of sections AB and BC of the rod

9. The stationary space station is a cylinder. The astronaut begins a circular walk around the station along its surface. What will happen to the station?

Answer: With the station will begin to rotate in the opposite direction, and its center will describe a circle around the same center of mass as the astronaut.

11. Why is it difficult to walk on stilts?

Answer: the center of gravity of a person on stilts increases significantly, and the area of ​​his support on the ground decreases.

12. When is it easier for a tightrope walker to maintain balance - during normal movement along a rope or when carrying a strongly curved beam loaded with buckets of water?

Answer: In the second case, since the center of mass of the rope walker with buckets lies lower, i.e. closer to the support - the rope.

IVHomework:(performed by those who wish - the tasks are difficult, those who solve them receive a “5”).

*1. Find the center of gravity of the system of balls located at the vertices of the equilateral weightless triangle shown in the figure

Answer: the center of gravity lies in the middle of the bisector of the angle at the vertex of which there is a ball of mass 2m

*2. The depth of the hole in the board into which the ball is inserted is half the radius of the ball. At what angle of the board's inclination to the horizon will the ball jump out of the hole?