Basic theorems of dynamics. General theorems of dynamics

Lecture 3 General theorems of dynamics

Dynamics of the system of material points is an important branch of theoretical mechanics. Here, we mainly consider the problems of the motion of mechanical systems (systems of material points) with finite number degrees of freedom - the maximum number of independent parameters that determine the position of the system. the main task system dynamics - the study of the laws of motion of a rigid body and mechanical systems.

The simplest approach to studying the motion of a system, consisting of N material points, is reduced to the consideration of the movements of each individual point of the system. In this case, all forces acting on each point of the system, including the forces of interaction between points, must be determined.

Determining the acceleration of each point in accordance with Newton's second law (1.2), we obtain for each point three scalar differential laws of motion of the second order, i.e. 3 N differential law of motion for the entire system.

To find the equations of motion mechanical system given forces and initial conditions for each point of the system, obtained differential laws needs to be integrated. This problem is difficult even in the case of two material points that move only under the action of forces of interaction according to the law of universal attraction (the problem of two bodies), and is extremely difficult in the case of three interacting points (the problem of three bodies).

Therefore, it is necessary to find such methods for solving problems that would lead to solvable equations and give an idea of ​​the motion of a mechanical system. General theorems of dynamics, being a consequence of the differential laws of motion, make it possible to avoid the complexity that arises during integration and obtain the necessary results.

3.1. General remarks

The points of the mechanical system will be numbered by indices i, j, k etc. that run through all values 1, 2, 3… N, where N is the number of system points. Physical quantities related to k th point are denoted by the same index as the point. For example, they express respectively the radius vector and the speed k-th point.

Forces of two origin act on each of the points of the system: firstly, forces whose sources lie outside the system, called external forces and denoted by ; secondly, forces from other points of this system, called internal forces and denoted by . Internal forces satisfy Newton's third law. Consider the simplest properties of internal forces acting on the entire mechanical system in any of its states.

First property. The geometric sum of all internal forces of the system (the main vector of internal forces) is equal to zero.

Indeed, if we consider any two arbitrary points of the system, for example, and (Fig. 3.1), then for them , because the forces of action and reaction are always equal in absolute value, they act along one line of action in the opposite direction, which connects the interacting points. The main vector of internal forces consists of pairs of forces of interacting points, therefore

(3.1)

Second property. The geometric sum of the moments of all internal forces relative to an arbitrary point in space is zero.

Consider the system of moments of forces and with respect to the point O(Fig. 3.1). From (Fig. 3.1). it's clear that

,

because both forces have the same arms and opposite directions of vector moments. The main moment of internal forces about the point O consists of the vector sum of such expressions and is equal to zero. Hence,

Let external and internal forces acting on a mechanical system consisting of N points (Fig. 3.2). If the resultant of external forces and the resultant of all internal forces are applied to each point of the system, then for any k th point of the system, one can compose differential equations of motion. In total such equations will be N:

and in projections onto fixed coordinate axes 3 N:

(3.4)

Vector equations (3.3) or equivalent scalar equations (3.4) represent the differential laws of motion of material points of the entire system. If all points move parallel to one plane or one straight line, then the number of equations (3.4) in the first case will be 2 N, in the second N.

Example 1 Two loads of mass and are interconnected by an inextensible cable thrown over a block (Fig. 3.3). Neglecting the forces of friction, as well as the mass of the block and the cable, determine the law of movement of goods and the tension of the cable.

Decision. The system consists of two material bodies (connected by an inextensible cable) moving parallel to one axis X. Let us write down the differential laws of motion in projections onto the axis X for every body.

Let the right weight descend with acceleration , then the left weight will rise with acceleration . We mentally free ourselves from the connection (cable) and replace it with reactions and (Fig. 3.3). Assuming that the bodies are free, we will compose the differential laws of motion in the projection onto the axis X(meaning that the thread tensions are internal forces, and the weight of the loads are external):

Since and (the bodies are connected by an inextensible cable), we obtain

Solving these equations for the acceleration and tension of the rope T, we get

.

Note that the cable tension at is not equal to the gravity of the corresponding load.

3. 2. The theorem on the motion of the center of mass

It is known that a rigid body and a mechanical system in a plane can move quite difficult. The first theorem on the motion of a body and a mechanical system can be arrived at in the following way: drop the c.-l. an object consisting of many solid bodies fastened together. It is clear that he will fly in a parabola. This was revealed when studying the motion of a point. However, now the object is not a point. It turns, sways in the process of flying around some effective center, which moves along a parabola. First motion theorem difficult subjects says that a certain effective center is the center of mass of a moving object. The center of mass is not necessarily located in the body itself, it can lie somewhere outside it.

Theorem. The center of mass of a mechanical system moves as a material point with a mass equal to the mass the entire system to which all external forces acting on the system are applied.

To prove the theorem, we rewrite the differential laws of motion (3.3) in following form:

(3.5)

where N is the number of system points.

Let's add the equations together term by term:

(a)

The position of the center of mass of the mechanical system relative to the selected coordinate system is determined by formula (2.1): where M is the mass of the system. Then the left side of equality (a) is written

The first sum, standing on the right side of equality (a), is equal to the main vector of external forces, and the last one, by the property of internal forces, is equal to zero. Then equality (a), taking into account (b), will be rewritten

, (3.6)

those. the product of the mass of the system and the acceleration of its center of mass is equal to geometric sum all external forces acting on the system.

It follows from equation (3.6) that internal forces do not directly affect the motion of the center of mass. However, in some cases they are the cause of the appearance of external forces applied to the system. Thus, the internal forces that rotate the driving wheels of the car cause the action on it of an external adhesion force applied to the wheel rim.

Example 2 The mechanism, located in a vertical plane, is installed on a horizontal smooth plane and attached to it with bars rigidly fixed to the surface. To and L (Fig. 3.4).

Disc 1 radius R motionless. Disc 2 mass m and radius r fastened with a crank, length R+ r at the point From 2. The crank rotates at a constant

angular speed. AT initial moment the crank occupied the right horizontal position. Neglecting the mass of the crank, determine the maximum horizontal and vertical forces acting on the bars, if total weight bed and wheel 1 is equal to M. Also consider the behavior of the mechanism in the absence of bars.

Decision. The system consists of two masses ( N=2 ): a fixed disk 1 with a frame and a movable disk 2. Let's direct the axis at through the center of gravity of the fixed disk vertically upwards, the axis X- along horizontal plane.

We write the theorem on the motion of the center of mass (3.6) in the coordinate form

The external forces of this system are: the weight of the frame and the fixed disk - mg, movable disc weight mg, - the total horizontal reaction of the bolts, - the normal total reaction of the plane. Hence,

Then the laws of motion (b) are rewritten

Let us calculate the coordinates of the center of mass of the mechanical system:

; (G)

as seen from (Fig. 3.4), , , (angle of rotation of the crank), . Substituting these expressions in (r) and calculating the second derivatives with respect to time t from , , we get that

(e)

Substituting (c) and (e) into (b), we find

The horizontal pressure acting on the bars has the greatest and smallest value, when cos = 1 respectively, i.e.

The pressure of the mechanism on the horizontal plane has the highest and lowest values ​​when sin respectively, i.e.

In fact, the first problem of dynamics has been solved: according to the known equations of motion of the center of mass of the system (e), the forces involved in the motion are restored.

In the absence of bars K and L (Fig. 3.4), the mechanism may begin to bounce above the horizontal plane. This will take place when , i.e. when , it follows that the angular velocity of rotation of the crank, at which the mechanism bounces, must satisfy the equality

.

3. 3. Law of conservation of motion of the center of mass

If the main vector of external forces acting on the system is equal to zero, i.e. , then from(3.6)it follows that the acceleration of the center of mass is zero, therefore, the velocity of the center of mass is constant in magnitude and direction. If, in particular, at the initial moment the center of mass is at rest, then it is at rest during the entire time until the main vector of external forces is equal to zero.

Several corollaries follow from this theorem.

· Internal forces alone cannot change the nature of the movement of the center of mass of the system.

· If the main vector of external forces acting on the system is equal to zero, then the center of mass is at rest or moves uniformly and rectilinearly.

· If the projection of the main vector of the external forces of the system on some fixed axis is equal to zero, then the projection of the velocity of the center of mass of the system on this axis does not change.

· A couple of forces applied to a rigid body cannot change the motion of its center of mass (it can only cause the body to rotate around the center of mass).

Let's consider an example illustrating the law of conservation of motion of the center of mass.

Example 3 Two weights with masses and are connected by an inextensible thread thrown over a block (Fig. 3.5), fixed on a wedge with mass M. The wedge rests on a smooth horizontal plane. Initially, the system was at rest. Find the displacement of the wedge along the plane when the first load is lowered to a height N. Ignore the mass of the block and the thread.

Decision. The external forces acting on the wedge together with the weights are the forces of gravity , and mg, as well as normal reaction smooth horizontal surface N. Therefore,

Since the system was at rest at the initial moment, we have .

Let us calculate the coordinate of the center of mass of the system at and at the moment t 1 when the weight of the load g descend to a height H.

For a moment:

,

where , , X- respectively, the coordinates of the center of mass of loads weighing g, g and wedge weighing Mg.

Let us assume that the wedge at the moment of time moves in the positive direction of the axis Ox by the amount L if the weight of the load falls to a height N. Then, for a moment

because loads together with the wedge will move to L to the right, a the weight will move a distance up the wedge. Since , after calculations we get

.

3.4. Quantity of movement system

3.4.1. Computing the momentum of a system

The momentum of a material point is a vector quantity, equal to the product mass of a point on its velocity vector

Unit of measurement of the amount of movement -

The momentum of a mechanical system is called the vector sum of the momentum of the individual points of the system, i.e.

where N is the number of system points.

The momentum of a mechanical system can be expressed in terms of the mass of the system M and the speed of the center of mass. Really,

those. the momentum of the system is equal to the product of the mass of the entire system and the velocity of its center of mass. The direction is the same as the direction (Fig. 3.6)

In projections onto rectangular axes, we have

where , , - projections of the velocity of the center of mass of the system.

Here M is the mass of the mechanical system; does not change as the system moves.

It is especially convenient to use these results when calculating the momenta of rigid bodies.

It can be seen from formula (3.7) that if a mechanical system moves in such a way that its center of mass remains stationary, then the momentum of the system remains equal to zero.

3.4.2. Elemental and full force impulse

The action of a force on a material point over time dt can be characterized by an elementary impulse. Total impulse of force in time t, or force impulse , is determined by the formula

or in projections onto the coordinates of the axis

(3.8a)

The unit of force impulse is .

3.4.3. Theorem on the change in the momentum of the system

Let external and internal forces be applied to the points of the system. Then, for each point of the system, we can apply the differential laws of motion (3.3), bearing in mind that :

.

Summing over all points of the system, we obtain

By the property of internal forces and by definition we have

(3.9)

Multiplying both sides of this equation by dt, we obtain a theorem on the change in the momentum in differential form:

, (3.10)

those. the differential of the momentum of a mechanical system is equal to the vector sum of the elementary impulses of all external forces acting on the points of the mechanical system.

Calculating the integral of both parts of (3.10) over time from 0 to t, we obtain the theorem in finite or integral form

(3.11)

In projections onto the coordinate axes, we will have

Change in momentum of a mechanical system over timet, is equal to the vector sum of all impulses of external forces acting on the points of the mechanical system in the same time.

Example 4 Load of mass m going down the inclined plane from rest under the action of a force F, proportional to time: , where (Fig. 3.7). What is the speed of the body after t seconds after the start of movement, if the coefficient of sliding friction of the load on the inclined plane is equal to f.

Decision. Let's depict the forces applied to the load: mg - gravity of the load, N is the normal reaction of the plane, is the sliding friction force of the load on the plane, and . The direction of all forces is shown in (Fig. 3.7).

Let's direct the axis X down an inclined plane. Let us write the theorem on the change in momentum (3.11) in the projection onto the axis X:

(a)

By condition, because at the initial moment of time, the load was at rest. The sum of the projections of the impulses of all forces on the x-axis is

Hence,

,

.

3.4.4. Laws of conservation of momentum

Conservation laws are obtained as special cases of the momentum change theorem. Two special cases are possible.

· If the vector sum of all external forces applied to the system is equal to zero, i.e. , then it follows from the theorem (3.9) , what ,

those. if the main vector of external forces of the system is equal to zero, then the momentum of the system is constant in magnitude and direction.

· If the projection of the main vector of external forces on any coordinate axis is equal to zero, for example Oh, i.e. , then the projection of the amount of motion on this axis is constant.

Consider an example of applying the law of conservation of momentum.

Example 5 A ballistic pendulum is a body of mass , suspended on a long string (Fig. 3.8).

A bullet of mass moving at a speed V and falling into a motionless body, gets stuck in it, and the body is deflected. What was the speed of the bullet if the body rose to a height h ?

Decision. Let the body with the stuck bullet acquire speed . Then, using the law of conservation of momentum in the interaction of two bodies, we can write .

Velocity can be calculated using the conservation law mechanical energy . Then . As a result, we find

.

Example 6. Water enters a fixed channel (Fig. 3.9) variable section with a speed at an angle to the horizon; square cross section channel at the entrance; the speed of the water at the outlet of the channel and makes an angle with the horizon.

Determine the horizontal component of the reaction that water exerts on the walls of the channel. Density of water .

Decision. We will determine the horizontal component of the reaction exerted by the channel walls on water. This force is equal in absolute value and opposite in sign to the desired force. We have, according to (3.11a),

. (a)

We calculate the mass of the volume of liquid entering the channel during the time t:

The value of rAV 0 is called second mass - the mass of liquid flowing through any section of the pipe per unit time.

The same amount of water leaves the canal in the same time. The initial and final speeds are given in the condition.

Compute right side equality (a) which determines the sum of projections onto the horizontal axis of external forces applied to the system (water). The only horizontal force is the horizontal component of the resultant reaction of the walls R x. This force is constant during the steady motion of water. So

. (in)

Substituting (b) and (c) into (a), we get

3.5. Kinetic moment of the system

3.5.1. Principal moment of momentum of the system

Let be the radius vector of a point with the mass of the system relative to some point A, called the center (Fig. 3.10).

Moment of momentum (kinetic moment) of a point relative to the center A called vector , determined by the formula

. (3.12)

In this case, the vector directed perpendicular to the plane passing through the center BUT and vector .

Moment of momentum (kinetic moment) of a point about an axis is called the projection onto this axis of the angular momentum of the point relative to any center chosen on this axis.

The main moment of momentum (kinetic moment) of the system relative to the center A is called the quantity

(3.13)

The main moment of momentum (kinetic moment) of the system about the axis is called the projection onto this axis of the main moment of the momentum of the system relative to any chosen on the given center axis.

3.5.2. Momentum of a rotating rigid body about the axis of rotation

Compatible fixed point O body lying on the axis of rotation Oz, with the origin of the coordinate system Ohuz, whose axes will rotate with the body (Fig. 3.11). Let be the radius-vector of the point of the body relative to the origin of coordinates, its projections on the axes will be denoted by , , . Vector projections angular velocity bodies on the same axes will be denoted by 0, 0, ().

Quite often it is possible to distinguish important features motion of a mechanical system without resorting to integration of the system of differential equations of motion. This is achieved by applying general theorems of dynamics.

5.1. Basic concepts and definitions

External and internal forces. Any force acting on a point in a mechanical system is necessarily either active force, or a bond reaction. The entire set of forces acting on the points of the system can be divided into two classes differently: into external forces and internal forces (subscripts e and i - from Latin words externus - external and internus - internal). External forces are called forces acting on points of the system from points and bodies that are not part of the system under consideration. Forces of interaction between points and bodies of the considered system are called internal.

This division depends on what material points and bodies are included by the researcher in the considered mechanical system. If the composition of the system is expanded to include additional points and bodies, then some forces that were external for the previous system may become internal for the expanded system.

Properties of internal forces. Since these forces are forces of interaction between parts of the system, they are included in the complete system of internal forces in "twos" organized in accordance with the action-reaction axiom. Each such "two" of forces

main vector and main point relative to an arbitrary center are equal to zero. Since the complete system of internal forces consists only of "twos", then

1) the main vector of the system of internal forces is equal to zero,

2) the main moment of the system of internal forces relative to an arbitrary point is equal to zero.

The mass of the system is arithmetic sum mass mk of all points and bodies forming the system:

center of gravity(center of inertia) of a mechanical system is called geometric point C, the radius vector and coordinates of which are determined by the formulas

where are the radius vectors and the coordinates of the points that form the system.

For a rigid body in a uniform gravitational field, the positions of the center of mass and the center of gravity coincide; in other cases, these are different geometric points.

Together with the inertial frame of reference, one often considers simultaneously a non-inertial frame of reference moving forward. Its coordinate axes (Koenig axes) are chosen so that the reference point C always coincides with the center of mass of the mechanical system. In accordance with the definition, the center of mass is fixed in the Koenig axes and is located at the origin of coordinates.

The moment of inertia of the system about the axis is called scalar equal to the sum the products of the masses mk of all points of the system by the squares of their distances to the axis:

If the mechanical system is a rigid body, to find 12, you can use the formula

where is the density, the volume occupied by the body.

Consider the motion of a certain system of material volumes relative to a fixed coordinate system. When the system is not free, then it can be considered as free, if we discard the constraints imposed on the system and replace their action with the corresponding reactions.

Let us divide all the forces applied to the system into external and internal ones; both may include reactions of discarded

connections. Denote by and the main vector and the main moment of external forces relative to point A.

1. Theorem on the change in momentum. If is the momentum of the system, then (see )

i.e., the theorem is valid: the time derivative of the momentum of the system is equal to the main vector of all external forces.

Replacing the vector through its expression where is the mass of the system, is the velocity of the center of mass, equation (4.1) can be given a different form:

This equality means that the center of mass of the system moves as a material point whose mass is equal to the mass of the system and to which a force is applied that is geometrically equal to the main vector of all external forces of the system. The last statement is called the theorem on the motion of the center of mass (center of inertia) of the system.

If then from (4.1) it follows that the momentum vector is constant in magnitude and direction. Projecting it on the coordinate axis, we obtain three scalar first integrals of the differential equations of the system's double chain:

These integrals are called momentum integrals. When the speed of the center of mass is constant, i.e., it moves uniformly and rectilinearly.

If the projection of the main vector of external forces on any one axis, for example, on the axis, is equal to zero, then we have one first integral, or if two projections of the main vector are equal to zero, then there are two integrals of the momentum.

2. Theorem on the change of the kinetic moment. Let A be some arbitrary point in space (moving or stationary), which does not necessarily coincide with any particular material point of the system during the entire time of movement. Its velocity in a fixed system of coordinates will be denoted by Theorem on the change in angular momentum material system with respect to point A has the form

If point A is fixed, then equality (4.3) takes a simpler form:

This equality expresses the theorem on the change in the angular momentum of the system with respect to fixed point: the time derivative of the angular momentum of the system, calculated with respect to some fixed point, is equal to the principal moment of all external forces with respect to this point.

If then, according to (4.4), the angular momentum vector is constant in magnitude and direction. Projecting it on the coordinate axis, we obtain the scalar first integrals of the differential equations of the motion of the system:

These integrals are called the integrals of the angular momentum or the integrals of the areas.

If point A coincides with the center of mass of the system, Then the first term on the right side of equality (4.3) vanishes and the theorem on the change in angular momentum has the same form (4.4) as in the case of a fixed point A. Note (see 4 § 3) that in the case under consideration the absolute angular momentum of the system on the left side of equality (4.4) can be replaced by the equal angular momentum of the system in its motion relative to the center of mass.

Let be some constant axis or an axis of constant direction passing through the center of mass of the system, and let be the angular momentum of the system relative to this axis. From (4.4) it follows that

where is the moment of external forces about the axis. If during the whole time of motion then we have the first integral

In the works of S. A. Chaplygin, several generalizations of the theorem on the change in angular momentum were obtained, which were then applied in solving a number of problems on the rolling of balls. Further generalizations of the theorem on the change of the kpnetological moment and their applications in problems of the dynamics of a rigid body are contained in the works. The main results of these works are related to the theorem on the change in the kinetic moment relative to the moving one, constantly passing through some moving point A. Let - unit vector directed along this axis. Multiplying scalarly by both sides of equality (4.3) and adding the term to both its parts, we obtain

When the kinematic condition is met

equation (4.5) follows from (4.7). And if condition (4.8) is satisfied during the whole time of motion, then the first integral (4.6) exists.

If the connections of the system are ideal and allow rotation of the system as a rigid body around the axis and in the number of virtual displacements, then the main moment of reactions about the axis and is equal to zero, and then the value on the right side of equation (4.5) is the main moment of all external active forces about the axis and . The equality to zero of this moment and the satisfiability of relation (4.8) will be in the case under consideration sufficient conditions for the existence of the integral (4.6).

If the direction of the axis and is unchanged, then condition (4.8) can be written as

This equality means that the projections of the velocity of the center of mass and the velocity of point A on the axis and on the plane perpendicular to this are parallel. In the work of S. A. Chaplygin, instead of (4.9), it is required that less than general condition where X is an arbitrary constant.

Note that condition (4.8) does not depend on the choice of a point on . Indeed, let P be an arbitrary point on the axis. Then

and hence

In conclusion, we note Resal's geometric interpretation of equations (4.1) and (4.4): the vectors absolute speeds the ends of the vectors and are equal respectively to the main vector and the main moment of all external forces relative to point A.

MINISTRY OF AGRICULTURE AND FOOD OF THE REPUBLIC OF BELARUS

Educational Institution "BELARUSIAN STATE AGRARIAN

TECHNICAL UNIVERSITY"

Department of Theoretical Mechanics and Theory of Mechanisms and Machines

THEORETICAL MECHANICS

methodological complex for students of the group of specialties

74 06 Agricultural engineering

In 2 parts Part 1

UDC 531.3(07) LBC 22.213ya7 T 33

Compiled by:

Candidate of Physical and Mathematical Sciences, Associate Professor Yu. S. Biza, candidate technical sciences, Associate Professor N. L. Rakova, Senior LecturerI. A. Tarasevich

Reviewers:

Department of Theoretical Mechanics of the Educational Establishment "Belarusian National Technical University» (head

Department of Theoretical Mechanics BNTU Doctor of Physical and Mathematical Sciences, Professor A. V. Chigarev);

Leading Researcher of the Laboratory "Vibroprotection of Mechanical Systems" State Scientific Institution "Joint Institute of Mechanical Engineering

National Academy of Sciences of Belarus”, Candidate of Technical Sciences, Associate Professor A. M. Goman

Theoretical mechanics. Section "Dynamics": educational

T33 method. complex. In 2 parts. Part 1 / comp.: Yu. S. Biza, N. L. Rakova, I. A. Tarasevich. - Minsk: BGATU, 2013. - 120 p.

ISBN 978-985-519-616-8.

AT educational and methodical complex presents materials on the study of the section "Dynamics", part 1, which is part of the discipline "Theoretical Mechanics". Includes a course of lectures, basic materials for the implementation practical exercises, tasks and samples of tasks for independent work and control learning activities full-time and correspondence forms learning.

UDC 531.3(07) LBC 22.213ya7

INTRODUCTION .................................................. .........................................
1. SCIENTIFIC AND THEORETICAL CONTENT OF EDUCATIONAL
OF THE METHODOLOGICAL COMPLEX .............................................. ..
1.1. Glossary................................................. ................................
1.2. Topics of lectures and their content .............................................. ..
Chapter 1. Introduction to dynamics. Basic concepts
classical mechanics .................................................................. ....................
Topic 1. Dynamics of a material point..............................................
1.1. Laws of material point dynamics
(laws of Galileo - Newton) .............................................. ..........
1.2. Differential Equations movements
1.3. Two main tasks of dynamics ..........................................................
Topic 2. Dynamics of relative motion
material point ................................................................ .........................
Review questions .................................................................. .............
Topic 3. Dynamics of a mechanical system ..............................................
3.1. Mass geometry. Center of mass of a mechanical system......
3.2. Internal Forces .................................................................. .................
Review questions .................................................................. .............
Topic 4. Moments of inertia of a rigid body .......................................
4.1. Moments of inertia of a rigid body
relative to the axis and pole .................................................................. .....
4.2. Theorem on the moments of inertia of a rigid body
about parallel axes
(Huygens-Steiner theorem) .............................................. ....
4.3. Centrifugal moments of inertia ...............................................
Review questions .................................................................. ............
Chapter 2

Topic 5. The theorem on the motion of the center of mass of the system ...............................
Review questions .................................................................. .............
Tasks for self-study ....................................................
Topic 6. The amount of movement of a material point
and mechanical system ............................................................... ...................
6.1. Quantity of movement of a material point 43
6.2. Impulse of force .................................................. .......................
6.3. Theorem on the change in momentum
material point ................................................................ ....................
6.4. Principal vector change theorem
momentum of a mechanical system ..........................................
Review questions .................................................................. .............
Tasks for self-study ....................................................
Topic 7. Moment of momentum of a material point
and mechanical system relative to the center and axis ..................................
7.1. Moment of momentum of a material point
relative to the center and axis .............................................................. ...........
7.2. Theorem on the change in angular momentum
material point relative to the center and axis .......................
7.3. Theorem on the change of the kinetic moment
mechanical system relative to the center and axis ..................................
Review questions .................................................................. .............
Tasks for self-study ....................................................
Topic 8. Work and power of forces ....................................... ............
Review questions .................................................................. .............
Tasks for self-study ....................................................
Topic 9. Kinetic energy of a material point
and mechanical system ............................................................... ...................
9.1. Kinetic energy of a material point
and mechanical system. Koenig's theorem...............................
9.2. Kinetic energy of a rigid body
with different movements .................................................................. .............
9.3. Change theorem kinetic energy
material point ................................................................ ....................

9.4. Kinetic energy change theorem
mechanical system .................................................................. ................
Review questions .................................................................. .............
Tasks for self-study ....................................................
Topic 10. Potential force field
and potential energy ............................................................... .................
Review questions .................................................................. .............
Topic 11. Dynamics of a rigid body.................................................... .......
Review questions .................................................................. .............
2. MATERIALS FOR CONTROL
BY MODUL................................................... ...................................
INDEPENDENT WORK OF STUDENTS ..............................
4. REQUIREMENTS FOR THE DESIGN OF CONTROL
WORKS FOR FULL-TIME AND CORRESPONDENCE STUDENTS
FORMS OF TRAINING ................................................................ .........................
5. LIST OF PREPARATION QUESTIONS
TO THE EXAM (STUDY) OF STUDENTS
FULL-TIME AND CORRESPONDENCE EDUCATION..................................................
6. LIST OF REFERENCES ............................................... ............

INTRODUCTION

Theoretical mechanics - the science of general laws mechanical movement, equilibrium and interaction of material bodies.

This is one of the fundamental general scientific physical and mathematical disciplines. It is the theoretical basis of modern technology.

The study of theoretical mechanics, along with other physical and mathematical disciplines, contributes to the expansion of scientific horizons, forms the ability to concrete and abstract thinking and contributes to the improvement of the general technical culture of the future specialist.

Theoretical mechanics, being the scientific basis of all technical disciplines, contributes to the development of skills rational decisions engineering tasks associated with the operation, repair and design of agricultural and reclamation machinery and equipment.

According to the nature of the tasks under consideration, mechanics is divided into statics, kinematics and dynamics. Dynamics is a section of theoretical mechanics that studies the motion of material bodies under the action of applied forces.

AT educational and methodical complex (UMK) presents materials on the study of the section "Dynamics", which includes a course of lectures, basic materials for conducting practical work, tasks and samples of execution for independent work and control of educational activities of full-time part-time students.

AT as a result of studying the "Dynamics" section, the student must learn theoretical basis dynamics and master the basic methods for solving problems of dynamics:

Know methods for solving problems of dynamics, general theorems of dynamics, principles of mechanics;

To be able to determine the laws of motion of a body depending on the forces acting on it; apply the laws and theorems of mechanics to solve problems; determine the static and dynamic reactions of the bonds that limit the movement of bodies.

The curriculum of the discipline "Theoretical Mechanics" provides for a total number of classroom hours - 136, including 36 hours for studying the section "Dynamics".

1. SCIENTIFIC AND THEORETICAL CONTENT OF THE EDUCATIONAL AND METHODOLOGICAL COMPLEX

1.1. Glossary

Statics is a section of mechanics that outlines the general doctrine of forces, the reduction is studied complex systems forces to the simplest form and equilibrium conditions are established various systems forces.

Kinematics is a section of theoretical mechanics in which the movement of material objects is studied, regardless of the causes that cause this movement, i.e., regardless of the forces acting on these objects.

Dynamics is a section of theoretical mechanics that studies the motion of material bodies (points) under the action of applied forces.

Material point- a material body, the difference in the movement of points of which is insignificant.

The mass of a body is a scalar positive value that depends on the amount of matter contained in a given body and determines its measure of inertia at forward movement.

Reference system - a coordinate system associated with the body, in relation to which the motion of another body is being studied.

inertial system- a system in which the first and second laws of dynamics are fulfilled.

The momentum of a force is a vector measure of the action of a force over some time.

Quantity of movement of a material point is the vector measure of its motion, which is equal to the product of the mass of the point and the vector of its velocity.

Kinetic energy is a scalar measure of mechanical motion.

Elementary work of force is an infinitesimal scalar value equal to dot product vector of force to the vector of infinitesimal displacement of the point of application of the force.

Kinetic energy is a scalar measure of mechanical motion.

The kinetic energy of a material point is a scalar

a positive value equal to half the product of the mass of a point and the square of its speed.

The kinetic energy of a mechanical system is an arithme-

the kinetic sum of the kinetic energies of all material points of this system.

Force is a measure of the mechanical interaction of bodies, characterizing its intensity and direction.

1.2. Lecture topics and their content

Section 1. Introduction to dynamics. Basic concepts

classical mechanics

Topic 1. Dynamics of a material point

The laws of dynamics of a material point (the laws of Galileo - Newton). Differential equations of motion of a material point. Two main tasks of dynamics for a material point. Solution of the second problem of dynamics; integration constants and their determination from initial conditions.

References:, pp. 180-196, , pp. 12-26.

Topic 2. Dynamics of the relative motion of the material

Relative motion of a material point. Differential equations of relative motion of a point; portable and Coriolis forces of inertia. The principle of relativity in classical mechanics. A case of relative rest.

References: , pp. 180-196, , pp. 127-155.

Topic 3. Geometry of masses. Center of mass of a mechanical system

Mass of the system. The center of mass of the system and its coordinates.

Literature:, pp. 86-93, pp. 264-265

Topic 4. Moments of inertia of a rigid body

Moments of inertia of a rigid body about the axis and pole. Radius of inertia. Theorem about moments of inertia about parallel axes. Axial moments of inertia of some bodies.

Centrifugal moments of inertia as a characteristic of body asymmetry.

References: , pp. 265-271, , pp. 155-173.

Section 2. General theorems of the dynamics of a material point

and mechanical system

Topic 5. The theorem on the motion of the center of mass of the system

The theorem on the motion of the center of mass of the system. Consequences from the theorem on the motion of the center of mass of the system.

References: , pp. 274-277, , pp. 175-192.

Topic 6. The amount of movement of a material point

and mechanical system

Quantity of motion of a material point and a mechanical system. Elementary impulse and impulse of force for a finite period of time. Theorem on the change in the momentum of a point and a system in differential and integral forms. Law of conservation of momentum.

Literature: , pp. 280-284, , pp. 192-207.

Topic 7. Moment of momentum of a material point

and mechanical system relative to the center and axis

The moment of momentum of a point about the center and axis. The theorem on the change in the angular momentum of a point. Kinetic moment of a mechanical system about the center and axis.

The angular momentum of a rotating rigid body about the axis of rotation. Theorem on the change in the kinetic moment of the system. Law of conservation of momentum.

References: , pp. 292-298, , pp. 207-258.

Topic 8. Work and power of forces

Elementary work of force, its analytical expression. The work of the force on final path. The work of gravity, elastic force. Equality to zero of the sum of the work of internal forces acting in a solid. The work of forces applied to a rigid body rotating around a fixed axis. Power. Efficiency.

References: , pp. 208-213, , pp. 280-290.

Topic 9. Kinetic energy of a material point

and mechanical system

Kinetic energy of a material point and a mechanical system. Calculation of the kinetic energy of a rigid body in various cases of its motion. Koenig's theorem. Theorem on the change in the kinetic energy of a point in differential and integral forms. Theorem on the change in the kinetic energy of a mechanical system in differential and integral forms.

References: , pp. 301-310, , pp. 290-344.

Topic 10. Potential force field and potential

The concept of a force field. Potential force field and force function. The work of a force on the final displacement of a point in a potential force field. Potential energy.

References: , pp. 317-320, , pp. 344-347.

Topic 11. Rigid body dynamics

Differential equations of translational motion of a rigid body. Differential equation rotary motion rigid body around a fixed axis. physical pendulum. Differential equations of plane motion of a rigid body.

References: , pp. 323-334, , pp. 157-173.

Section 1. Introduction to dynamics. Basic concepts

classical mechanics

Dynamics is a section of theoretical mechanics that studies the motion of material bodies (points) under the action of applied forces.

material body- a body that has mass.

Material point- a material body, the difference in the movement of points of which is insignificant. This can be either a body, the dimensions of which can be neglected during its movement, or a body of finite dimensions, if it moves forward.

Particles are also called material points, into which a solid body is mentally divided when determining some of its dynamic characteristics. Examples of material points (Fig. 1): a - the movement of the Earth around the Sun. The earth is a material point; b is the translational motion of a rigid body. Solid- mother-

al point, since V B \u003d V A; a B = a A ; c - rotation of the body around the axis.

A body particle is a material point.

Inertia is the property of material bodies to change the speed of their movement faster or slower under the action of applied forces.

The mass of a body is a scalar positive value that depends on the amount of matter contained in a given body and determines its measure of inertia during translational motion. In classical mechanics, mass is a constant.

Force - quantitative measure mechanical interaction between bodies or between a body (point) and a field (electric, magnetic, etc.).

Force is a vector quantity characterized by magnitude, point of application and direction (line of action) (Fig. 2: A - point of application; AB - line of action of the force).

Rice. 2

In dynamics, along with constant forces, there are also variable forces that can depend on time t, speed ϑ, distance r, or on a combination of these quantities, i.e.

F = const;

F = F(t);

F = F(ϑ ) ;

F = F(r) ;

F = F(t, r, ϑ ) .

	Examples of such forces are shown in Figs. 3: a										- body weight;
			(ϑ) – air resistance force;b ​​−		T =						- traction force

electric locomotive; c − F = F (r) is the force of repulsion from the center O or attraction to it.

Reference system - a coordinate system associated with the body, in relation to which the motion of another body is being studied.

An inertial system is a system in which the first and second laws of dynamics are fulfilled. This is a fixed coordinate system or a system moving uniformly and rectilinearly.

Movement in mechanics is a change in the position of a body in space and time in relation to other bodies.

The space in classical mechanics is three-dimensional, obeying Euclidean geometry.

Time is a scalar quantity that flows in the same way in any reference systems.

A system of units is a set of units of measurement physical quantities. To measure all mechanical quantities, three basic units are sufficient: units of length, time, mass or force.

Mechanical	Dimension		Notation	Dimension	Notation
magnitude
				centimeter
	kilogram-

All other units of measurement of mechanical quantities are derivatives of these. Two types of systems of units are used: international system SI units (or smaller - CGS) and the technical system of units - MKGSS.

Topic1. Material point dynamics

1.1. The laws of dynamics of a material point (the laws of Galileo - Newton)

The first law (of inertia).

isolated from external influences a material point maintains its state of rest or moves uniformly and rectilinearly until the applied forces force it to change this state.

The movement made by a point in the absence of forces or under the action of a balanced system of forces is called inertia motion.

For example, the movement of a body along a smooth (friction force is zero) go-

horizontal surface (Fig. 4: G - body weight; N - normal reaction of the plane).

Since G = − N , then G + N = 0.

When ϑ 0 ≠ 0 the body moves at the same speed; at ϑ 0 = 0 the body is at rest (ϑ 0 is the initial velocity).

The second law (basic law of dynamics).

The product of the mass of a point and the acceleration that it receives under the action of a given force is equal in absolute value to this force, and its direction coincides with the direction of acceleration.

a b

Mathematically, this law is expressed by the vector equality

For F = const,

a = const - the motion of the point is uniform. EU-

whether a ≠ const, α

- slow motion (Fig. 5, but);

a ≠ const,

a -

– accelerated motion (Fig. 5, b); m – point mass;

acceleration vector;

– vector force; ϑ 0 is the velocity vector).

At F = 0,a 0 = 0 = ϑ 0 = const - the point moves uniformly and rectilinearly, or at ϑ 0 = 0 - it is at rest (the law of inertia). Second

the law allows you to establish a relationship between the mass m of a body located near earth's surface, and its weight G .G = mg , where g is

acceleration of gravity.

The third law (the law of equality of action and reaction). Two material points act on each other with forces equal in magnitude and directed along the straight line connecting

these points, in opposite directions.

Since the forces F 1 = − F 2 are applied to different points, then the system of forces (F 1 , F 2 ) is not balanced, i.e. (F 1 , F 2 )≈ 0 (Fig. 6).

In its turn

m a = m a

- attitude

the masses of the interacting points are inversely proportional to their accelerations.

The fourth law (the law of the independence of the action of forces). The acceleration received by a point under the action of a simultaneous

but of several forces, is equal to the geometric sum of those accelerations that a point would receive under the action of each force separately on it.

Explanation (Fig. 7).

t a n

a 1 a kF n

The resultant R forces (F 1 ,...F k ,...F n ) .

Since ma = R ,F 1 = ma 1 , ...,F k = ma k , ...,F n = ma n , then

a = a 1 + ...+ a k + ...+ a n = ∑ a k , i.e. the fourth law is equivalent to

k = 1

the rule of addition of forces.

1.2. Differential equations of motion of a material point

Let several forces act simultaneously on a material point, among which there are both constants and variables.

We write the second law of dynamics in the form

= ∑

(t ,

k = 1

, ϑ=

r is the radius vector of the moving

points, then (1.2) contains derivatives of r and is a differential equation of motion of a material point in vector form or the basic equation of the dynamics of a material point.

Projections of vector equality (1.2): - on the axis of Cartesian coordinates (Fig. 8, but)

max=md
					= ∑Fkx;
					k = 1
may=md
					= ∑Fky;	(1.3)
					k = 1
maz=m					= ∑Fkz;
					k = 1

On the natural axis (Fig. 8, b)

mat				= ∑ Fk τ ,
				k = 1
				= ∑ F k n ;
				k = 1

mab = m0 = ∑ Fk b

k = 1

M t oM oa

b on o

Equations (1.3) and (1.4) are differential equations of motion of a material point in the Cartesian coordinate axes and natural axes, respectively, i.e., natural differential equations that are usually used for curvilinear motion of a point if the trajectory of the point and its radius of curvature are known.

1.3. Two main problems of dynamics for a material point and their solution

The first (direct) task.

Knowing the law of motion and the mass of the point, determine the force acting on the point.

To solve this problem, you need to know the acceleration of the point. In problems of this type, it can be given directly, or the law of motion of a point is given, in accordance with which it can be determined.

1. So, if the movement of a point is given in Cartesian coordinates

x \u003d f 1 (t) , y \u003d f 2 (t) and z \u003d f 3 (t) then the projections of the acceleration are determined

on the coordinate axis x =		d2x			d2y			d2z		And then - project-
F x ,F y and F z forces on these axes:
				,k ) = F F z . (1.6) 2. If the point commits curvilinear motion and the law of motion is known s = f (t), the trajectory of the point and its radius of curvature ρ, then it is convenient to use natural axes, and the acceleration projections on these axes are determined by the well-known formulas: Tangential axis a τ = d ϑ = d 2 2 s – tangential acceleration;dt dt HomeNormal ds 2 a n = ϑ 2 = dt is normal acceleration. The projection of the acceleration onto the binormal is zero. Then the projections of the force on the natural axes F=m F=m The modulus and direction of the force are determined by the formulas: F \u003d F τ 2 + F n 2; cos ( ; cos( The second (inverse) task. Knowing the forces acting on the point, its mass and initial conditions movement, determine the law of motion of a point or any other of its kinematic characteristics. The initial conditions for the movement of a point in the Cartesian axes are the coordinates of the point x 0, y 0, z 0 and the projection of the initial velocity ϑ 0 onto these axes ϑ 0 x \u003d x 0, ϑ 0 y \u003d y 0 and ϑ 0 z \u003d z 0 at the time corresponding to giving the beginning of the point motion and taken equal to zero. Solving problems of this type is reduced to compiling a differential differential equations (or one equation) of motion of a material point and their subsequent solution by direct integration or using the theory of differential equations. Review questions 1. What does dynamics study? 2. What kind of motion is called inertial motion? 3. Under what condition will a material point be at rest or move uniformly and rectilinearly? 4. What is the essence of the first main problem of the dynamics of a material point? Second task? 5. Write down the natural differential equations of motion of a material point. Tasks for self-study 1. A point of mass m = 4 kg moves along a horizontal straight line with an acceleration a = 0.3 t. Determine the module of the force acting on the point in the direction of its movement at the time t = 3 s. 2. A part of mass m = 0.5 kg slides down the tray. At what angle to the horizontal plane should the tray be located so that the part moves with an acceleration a = 2 m / s 2? Angle express in degrees. 3. A point with a mass m = 14 kg moves along the Ox axis with an acceleration a x = 2 t . Determine the modulus of the force acting on the point in the direction of motion at time t = 5 s.

(MECHANICAL SYSTEMS) - IV option

1. The basic equation of the dynamics of a material point, as is known, is expressed by the equation . Differential equations of motion arbitrary points of a non-free mechanical system according to two ways of dividing forces can be written in two forms:

(1) , where k=1, 2, 3, … , n is the number of points of the material system.

(2)

where is the mass of the k-th point; - radius vector of the k-th point, - given (active) force acting on the k-th point or the resultant of all active forces acting on the k-th point. - the resultant of the reaction forces of the bonds, acting on the k-th point; - resultant of internal forces acting on the k-th point; - the resultant of external forces acting on the k-th point.

Equations (1) and (2) can be used to solve both the first and second problems of dynamics. However, the solution of the second problem of dynamics for the system becomes very complicated not only with mathematical point vision, but also because we face fundamental difficulties. They lie in the fact that both for system (1) and for system (2) the number of equations is significantly less than number unknown.

So, if we use (1), then the known for the second (inverse) problem of dynamics will be and , and the unknowns will be and . Vector equations will be " n", and unknown - "2n".

If we proceed from the system of equations (2), then the known and part of the external forces . Why a part? The fact is that the number of external forces includes external reactions connections that are unknown. In addition, there will also be unknowns.

Thus, both system (1) and system (2) are OPEN. We need to add equations, taking into account the equations of relations, and perhaps we still need to impose some restrictions on the relations themselves. What to do?

If we proceed from (1), then we can follow the path of compiling the Lagrange equations of the first kind. But this way is not rational because easier task(less degrees of freedom), the more difficult it is to solve it from the point of view of mathematics.

Then let's pay attention to the system (2), where - are always unknown. The first step in solving the system is to eliminate these unknowns. It should be borne in mind that, as a rule, we are not interested in internal forces during the movement of the system, that is, when the system moves, it is not necessary to know how each point of the system moves, but it is enough to know how the system as a whole moves.

Thus, if different ways exclude from the system (2) unknown forces, then we obtain some relations, i.e., some General characteristics for the system, the knowledge of which makes it possible to judge how the system moves in general. These characteristics are introduced using the so-called general theorems of dynamics. There are four such theorems:

1. Theorem about movement of the center of mass of the mechanical system;

2. Theorem about change in the momentum of a mechanical system;

3. Theorem about change in the angular momentum of a mechanical system;

4. Theorem about change in the kinetic energy of a mechanical system.