Let us consider the movement of a certain system of material objects relative to a fixed coordinate system. When the system is not free, then it can be considered as free if we discard the connections imposed on the system and replace their action with corresponding reactions.
Let us divide all the forces applied to the system into external and internal; both may include reactions of discarded
connections. Let and denote the main vector and main point external forces relative to point A.
1. Theorem on the change in momentum. If is the amount of motion of the system, then (see)
that is, the theorem is valid: the time derivative of the momentum of the system is equal to the main vector of all external forces.
By 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 like 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 axes, we obtain three scalar first integrals, differential equations system cap:
These integrals are called momentum integrals. When the speed of the center of mass is constant, that is, it moves uniformly and rectilinearly.
If the projection of the main vector of external forces on any one axis, for example on an 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 momentum.
2. Theorem on the change in angular momentum. Let A be some arbitrary point in space (moving or stationary), which does not necessarily coincide with any specific material point of the system during the entire time of movement. We denote its speed in a fixed coordinate system by The theorem on the change in the kinetic moment of a material system relative to point A has the form
If point A is fixed, then equality (4.3) takes on a simpler form:
This equality expresses the theorem about the variation of the angular momentum of a system relative to a fixed point: the time derivative of the angular momentum of the system, calculated relative to some fixed point, is equal to the principal moment of all external forces relative to this point.
If then according to (4.4) the angular momentum vector is constant in magnitude and direction. Projecting it on the coordinate axes, we obtain the scalar first integrals of the differential equations of the double system:
These integrals are called momentum integrals or area integrals.
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 of writing (4.4) as in the case of a fixed point A. Note (see. p. 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 axis of constant direction passing through the center of mass of the system, and let be the kinetic moment of the system relative to this axis. From (4.4) it follows that
where is the moment of external forces relative to the axis. If during the entire movement we have the first integral
In the works of S.A. Chaplygin, several generalizations of the theorem on the change in kinetic momentum were obtained, which were then applied to solve a number of problems on rolling balls. Further generalizations of the theorem on the change in the mechanical moment and their applications in problems of dynamics solid contained in the works. The main results of these works are related to the theorem on the change in kinetic momentum relative to a moving one, constantly passing through some moving point A. Let be a unit vector directed along this axis. Multiplying scalarly by both sides of equality (4.3) and adding the term to its two parts we get
When the kinematic condition is met
Equation (4.5) follows from (4.7). And if condition (4.8) is satisfied during the entire movement, then the first integral (4.6) exists.
If the connections of the system are ideal and allow, among virtual displacements, rotation of the system as a rigid body around the axis and, then the main moment of the reactions relative to the axis and is equal to zero, and then the value on the right side of equation (4.5) represents the main moment of all external active forces relative to the axis and . The equality to zero of this moment and the validity 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 constant, then condition (4.8) will be written in the form
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 a plane perpendicular to this are parallel. In the work of S.A. Chaplygin, instead of (4.9), the fulfillment of a less general condition is required where X is an arbitrary constant value.
Note that condition (4.8) does not depend on the choice of point on . Indeed, let P be an arbitrary point on the axis. Then
and therefore
In conclusion, we note Rézal’s geometric interpretation of equations (4.1) and (4.4): the absolute velocity vectors of 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.
Formulate a theorem on the motion of the center of mass of the system.
The center of mass of a mechanical system moves like a material point with mass, equal mass the entire system to which all the forces acting on the system are applied.
What motion of a rigid body can be considered as the motion of a material point having the mass of a given body, and why?
The translational motion of a rigid body is completely determined by the movement of one of its points. Consequently, having solved the problem of the movement of the center of mass of a body as a material point with body mass, we can determine forward movement of the whole body.
Under what conditions is the center of mass of the system at rest and under what conditions does it move uniformly and in a straight line?
If the main vector of external forces remains equal to zero all the time and the initial velocity of the center of mass is zero, then the center of mass is at rest.
If the main vector of external forces remains equal to zero all the time and the initial speed 
, then the center of mass moves uniformly and rectilinearly.
Under what conditions does the center of mass of the system not move along a certain axis?
If the projection of the main vector of external forces onto any axis remains equal to zero all the time and the projection of velocity onto this axis is equal to zero, then the coordinate of the center of mass along this axis remains constant.
What effect does a pair of forces applied to it have on a free solid body?
If you apply a pair of forces to a free rigid body that is at rest, then under the action of this pair of forces the body will begin to rotate around its center of mass.
Theorem on the change of momentum.
How is an impulse of variable force determined over a finite period of time? What characterizes a force impulse?
Variable impulse 
for a finite period of time 
equals
.
The impulse of force characterizes the transfer of mechanical motion to a body from the bodies acting on it over a given period of time.
What are the projections of constant and variable force impulses on the coordinate axes?
The projections of the variable force impulse on the coordinate axes are equal to
,
,
.
Projections of a constant force impulse on the coordinate axes over a period of time 
equal
,
,
.
What is the impulse of the resultant?
The impulse of the resultant of several forces over a certain period of time is equal to the geometric sum of the impulses of the component forces over the same period of time
.
How does the momentum of a point moving uniformly around a circle change?
When a point moves uniformly around a circle, the direction of the momentum changes 
, but its module is preserved 
.
What is momentum called? mechanical system?
The amount of motion of a mechanical system is a vector equal to the geometric sum (principal vector) of the amounts of motion of all points of the system
.
What is the momentum of a flywheel rotating around a fixed axis passing through its center of gravity?
The amount of motion of a flywheel rotating around a fixed axis passing through its center of gravity is zero, because 
.
Formulate theorems on the change in momentum of a material point and a mechanical system in differential and finite forms. Express each of these theorems with a vector equation and three equations in projections on the coordinate axes.
The differential momentum of a material point is equal to the elementary impulse of the forces acting on the point
.
The change in the number of movements of a point over a certain period of time is equal to the geometric sum of the impulses of forces applied to the point over the same period of time
.
In projections, these theorems have the form
,
,
,
,
.
The time derivative of the momentum of a mechanical system is geometrically equal to the main vector of external forces acting on the system
.
The time derivative of the projection of the momentum of a mechanical system onto any axis is equal to the projection of the main vector of external forces onto the same axis
,
,
.
The change in the momentum of the system over a certain period of time is equal to the geometric sum of the impulses of external forces applied to the system over the same period
.
The change in the projection of the momentum of the system onto any axis is equal to the sum of the projections of the impulses of all external forces acting on the system onto the same axis
,
,
.
Under what conditions does the momentum of a mechanical system not change? Under what conditions does its projection onto a certain axis not change?
If the main vector of external forces for the considered period of time is equal to zero, then the amount of motion of the system is constant.
If the projection of the main vector of external forces onto any axis is zero, then the projection of the momentum onto this axis is constant.
Why does the gun roll back when fired?
The rollback of a gun when fired in a horizontal direction is due to the fact that the projection of the momentum onto the horizontal axis 
does not change in the absence of horizontal forces
,
.
Can internal forces change the momentum of a system or the momentum of a part of it?
Since the main vector of internal forces is zero, they cannot change the amount of motion of the system.
Dynamics:
Dynamics of a material system
§ 35. Theorem on the motion of the center of mass of a material system
Problems with solutions
35.1 Determine the main vector of external forces acting on the flywheel M, rotating around the axis AB. The AB axis, mounted in a circular frame, in turn rotates around the DE axis. The center of mass C of the flywheel is at the intersection point of the AB and DE axes.
SOLUTION
35.2 Determine the main vector of external forces applied to the ruler AB of the ellipsograph shown in the figure. The crank OC rotates at a constant angular velocity ω; the mass of the ruler AB is equal to M; OC=AC=BC=l.
SOLUTION
35.3 Determine the main vector of external forces acting on a wheel of mass M rolling down from an inclined plane if its center of mass C moves according to the law xC=at2/2.
SOLUTION
35.4 A wheel slides along a horizontal line under the action of a force F shown in the figure. Find the law of motion of the center of mass C of the wheel if the coefficient of sliding friction is f, a F=5fP, where P is the weight of the wheel. At the initial moment the wheel was at rest.
SOLUTION
35.5 A wheel slides along a horizontal line under the influence of a torque applied to it. Find the law of motion of the center of mass C of the wheel if the sliding friction coefficient is equal to f. At the initial moment the wheel was at rest.
SOLUTION
35.6 A tram car performs vertical harmonic oscillations on the springs with an amplitude of 2.5 cm and a period of T=0.5 s. The mass of the body with a load is 10 tons, the mass of the bogie and wheels is 1 ton. Determine the force of pressure of the car on the rails.
SOLUTION
35.7 Determine the force of pressure on the ground of a pump for pumping out water when it is running idle, if the mass of the stationary parts of the body D and the foundation E is equal to M1, the mass of the crank OA=a is equal to M2, the mass of the linkage B and piston C is equal to M3. Crank OA, rotating uniformly with angular velocity ω, is considered to be a homogeneous rod.
SOLUTION
35.8 Using the data from the previous problem, assume that the pump is installed on an elastic base, the elasticity coefficient of which is equal to c. Find the law of motion of the O axis of the crank OA vertically, if at the initial moment the O axis was in a position of static equilibrium and it was imparted a vertical downward speed v0. Take the origin of the x-axis, directed vertically downwards, in the position of static equilibrium of the O-axis. Neglect the resistance forces.
SOLUTION
35.9 Shears for cutting metal consist of a crank-slider mechanism OAB, to the slider B of which a movable knife is attached. The fixed knife is fixed on the foundation C. Determine the pressure of the foundation on the ground if the length of the crank r, the mass of the crank M1, the length of the connecting rod l, the mass of the slider B with the movable knife M2, the mass of the foundation C and the body D is equal to M3. Neglect the mass of the connecting rod. Crank OA, uniformly rotating with angular velocity ω, is considered to be a homogeneous rod.
SOLUTION
35.10 Electric mass motor M1 is installed without fastenings on a smooth horizontal foundation; a homogeneous rod of length 2l and mass M2 is fixed at a right angle to the motor shaft at one end; a point load of mass M3 is mounted on the other end of the rod; the angular velocity of the shaft is ω. Determine: 1) horizontal movement of the motor; 2) the greatest horizontal force R acting on the bolts if they secure the electric motor casing to the foundation.
SOLUTION
35.11 Based on the conditions of the previous problem, calculate the angular velocity ω of the electric motor shaft at which the electric motor will bounce above the foundation without being bolted to it.
SOLUTION
35.12 When assembling the electric motor, its rotor B was eccentrically mounted on the axis of rotation C1 at a distance C1C2=a, where C1 is the center of mass of stator A, and C2 is the center of mass of rotor B. The rotor rotates uniformly with angular velocity ω. The electric motor is installed in the middle of an elastic beam, the static deflection of which is equal to Δ; M1 is the stator mass, M2 is the rotor mass. Find the equation of motion of point C1 vertically if at the initial moment it was at rest in a position of static equilibrium. Neglect resistance forces. The x-axis origin is taken at the static equilibrium position of point C1.
SOLUTION
35.13 An electric motor of mass M1 is mounted on a beam whose stiffness is equal to c. A mass of mass M2 is mounted on the motor shaft at a distance l from the shaft axis. Angular velocity motor ω=const. Determine amplitude forced oscillations motor and the critical number of its revolutions per minute, neglecting the mass of the beam and the resistance to movement.
SOLUTION
35.14 The figure shows a crane trolley A of mass M1, which is braked in the middle of a beam BD. At the center of mass C1 of the cart, a cable of length l is suspended with a load C2 of mass M2 attached to it. A cable with a load performs harmonic oscillations in the vertical plane. Determine: 1) the total vertical reaction of the beam BD, considering it rigid; 2) the law of motion of point C1 in the vertical direction, considering the beam elastic with an elasticity coefficient equal to c. At the initial moment, the beam, being undeformed, was at rest in a horizontal position. Considering the vibrations of the cable to be small, accept: sin φ≈φ, cos φ≈1. The y-axis origin is taken at the static equilibrium position of point C1. Neglect the mass of the cable and the dimensions of the trolley compared to the length of the beam.
SOLUTION
35.15 Keeping the data from the previous problem and considering beam BD rigid, determine: 1) the total horizontal reaction of the rails; 2) assuming that the cart is not braked, the law of motion of the center of mass C1 of cart A along the x axis. At the initial moment, point C1 was at rest at the origin of the x-axis. The cable oscillates according to the law φ=φ0 cos ωt.
SOLUTION
35.16 On the middle bench of the boat, which was at rest, two people were sitting. One of them, mass M1=50 kg, moved to the right to the bow of the boat. In what direction and what distance must the second person of mass M2=70 kg move in order for the boat to remain at rest? The length of the boat is 4 m. Neglect the resistance of the water to the movement of the boat.
SOLUTION
35.17 A homogeneous prism B is placed on a homogeneous prism A lying on a horizontal plane; cross sections of prisms right triangles, the mass of prism A is three times greater than the mass of prism B. Assuming that the prisms and the horizontal plane are ideally smooth, determine the length l by which prism A will move when prism B, descending along A, reaches the horizontal plane.
SOLUTION
35.18 On a horizontal goods platform with a length of 6 m and a mass of 2700 kg, which was initially at rest, two workers roll a heavy casting from the left end of the platform to the right. In which direction and how much will the platform move if the total mass of the cargo and workers is 1800 kg? Neglect the forces of resistance to the movement of the platform.
SOLUTION
35.19 Two loads M1 and M2, respectively masses M1 and M2, connected by an inextensible thread thrown over block A, slide along the smooth sides of a rectangular wedge resting with its base BC on a smooth horizontal plane. Find the displacement of the wedge along the horizontal plane when lowering the load M1 to a height of h=10 cm. Mass of the wedge M=4M1=16M2; Neglect the mass of the thread and block.
SOLUTION
35.20 Three masses of mass M1=20 kg, M2=15 kg and M3=10 kg are connected by an inextensible thread thrown through fixed blocks L and N. When mass M1 is lowered down, mass M2 moves along the upper base of a quadrangular truncated pyramid ABCD of mass M=100 kg to the right, and the load M3 rises along the side edge AB upward. Neglecting the friction between the truncated pyramid ABCD and the floor, determine the displacement of the truncated pyramid ABCD relative to the floor if the load M1 moves down 1 m. Neglect the mass of the thread.
SOLUTION
35.21 A movable rotary crane for repairing a street electrical network is installed on a vehicle weighing 1 ton. The cradle K of the crane, mounted on a rod L, can be rotated around a horizontal axis O, perpendicular to the plane of the drawing. At the initial moment, the crane, which occupied a horizontal position, and the car were at rest. Determine the displacement of the unbraked vehicle if the crane is rotated 60°. The mass of a homogeneous rod L of length 3 m is 100 kg, and the cradle K is 200 kg. The center of mass C of the cradle K is located at a distance OC=3.5 m from the axis O. Neglect the resistance to movement.
MOMENTUM THEOREM (in differential form).
1. For a point: the derivative of the momentum of the point with respect to time is equal to the resultant of the forces applied to the point:
or in coordinate form:
2. For a system: the derivative of the momentum of the system with respect to time is equal to the main vector of external forces of the system (vector sum of external forces applied to the system):
or in coordinate form:
MOMENTUM THEOREM (momentum theorem in final form).
1. For a point: the change in the momentum of the point over a finite period of time is equal to the sum of the impulses applied to the force point (or the resultant impulse of the forces applied to the point)
or in coordinate form:
2. For a system: the change in the momentum of the system over a finite period of time is equal to the sum of the impulses of external forces:
or in coordinate form:
Consequences: in the absence of external forces, the amount of motion of the system is a constant value; if the external forces of the system are perpendicular to a certain axis, then the projection of the momentum onto this axis is a constant value.
MOMENTUM THEOREM
1. For a point: The time derivative of the moment of momentum of the point relative to some center (axis) is equal to the sum of the moments of forces applied to the point relative to the same center (axis):
2. For the system:
The time derivative of the moment of momentum of the system relative to some center (axis) is equal to the sum of the moments of the external forces of the system relative to the same center (axis):
Consequences: if the external forces of the system do not provide a moment relative to a given center (axis), then the angular momentum of the system relative to this center (axis) is a constant value.
If the forces applied to a point do not produce a moment relative to a given center, then the angular momentum of the point relative to this center is a constant value and the point describes a flat trajectory.
KINETIC ENERGY THEOREM
1. For point: change kinetic energy point on its final displacement is equal to the work of the active forces applied to it (the tangential components of the reactions of non-ideal connections are included in the number of active forces):
For the case of relative motion: change in the kinetic energy of a point at relative motion equal to the work of the active forces applied to it and the transferring force of inertia (see “Special cases of integration”):
2. For a system: the change in the kinetic energy of the system at a certain displacement of its points is equal to the work of the external active forces applied to it and the internal forces applied to the points of the system, the distance between which changes:
If the system is immutable (solid body), then ΣA i =0 and the change in kinetic energy is equal to the work of only external active forces.
THEOREM ABOUT THE MOTION OF THE CENTER OF MASS OF A MECHANICAL SYSTEM. The center of mass of a mechanical system moves as a point whose mass is equal to the mass of the entire system M=Σm i , to which all external forces of the system are applied:
or in coordinate form:
where is the acceleration of the center of mass and its projection on the axis Cartesian coordinates; external force and its projections on the Cartesian coordinate axes.
MOMENTUM THEOREM FOR THE SYSTEM, EXPRESSED IN THROUGH THE MOTION OF THE CENTER OF MASS.
The change in the speed of the center of mass of the system over a finite period of time is equal to the impulse of the external forces of the system over the same period of time, divided by the mass of the entire system.
Ministry of Education and Science of the Russian Federation
Federal State Budgetary Educational Institution of Higher Professional Education
"Kuban State Technological University"
Theoretical mechanics
Part 2 dynamics
Approved by the Editorial and Publishing Committee
university council as
Krasnodar
UDC 531.1/3 (075)
Theoretical mechanics. Part 2. Dynamics: Textbook / L.I. Draiko; Kuban. state technol.un-t. Krasnodar, 2011. 123 p.
ISBN 5-230-06865-5
The theoretical material is presented in a brief form, examples of problem solving are given, most of which reflect real technical issues, and attention is paid to the choice of a rational method of solution.
Designed for bachelors of correspondence and distance learning in construction, transport and mechanical engineering.
Table 1 Ill. 68 Bibliography 20 titles
Scientific editor Candidate of Technical Sciences, Associate Professor. V.F.Melnikov
Reviewers: Head of the Department of Theoretical Mechanics and Theory of Mechanisms and Machines, Kuban Agrarian University prof. F.M. Kanarev; Associate Professor, Department of Theoretical Mechanics, Kuban State Technological University M.E. Multykh
Published by decision of the Editorial and Publishing Council of the Kuban State Technological University.
Reissue
ISBN 5-230-06865-5 KubSTU 1998
Preface
This textbook is intended for part-time students of construction, transport and mechanical engineering specialties, but can be used when studying the “Dynamics” section of the theoretical mechanics course by part-time students of other specialties, as well as full-time students working independently.
The manual is compiled in accordance with the current syllabus of the theoretical mechanics course and covers all the issues of the main part of the course. Each section contains brief theoretical material, accompanied by illustrations and methodological recommendations for its use in solving problems. The manual contains solutions to 30 problems that reflect real technical issues and correspond to test tasks for independent decision. For each problem, a calculation diagram is presented that clearly illustrates the solution. The formatting of the solution meets the requirements for the formatting of test papers for part-time students.
The author expresses deep gratitude to the teachers of the Department of Theoretical Mechanics and Theory of Mechanisms and Machines of the Kuban Agrarian University for their great work in reviewing the textbook, as well as the teachers of the Department of Theoretical Mechanics of the Kuban State Technological University for valuable comments and advice on preparing the textbook for publication.
All critical comments and suggestions will be accepted with gratitude by the author in the future.
Introduction
Dynamics is the most important section of theoretical mechanics. Most of the specific problems encountered in engineering practice relate to dynamics. Using the conclusions of statics and kinematics, dynamics establishes the general laws of motion of material bodies under the action of applied forces.
The simplest material object is a material point. A material body of any shape can be taken as a material point, the dimensions of which can be neglected in the problem under consideration. Behind material point a body of finite dimensions can be accepted if the difference in the movement of its points is not significant for a given problem. This happens when the dimensions of the body are small compared to the distances covered by the points of the body. Each particle of a solid body can be considered a material point.
Forces applied to a point or a material body are dynamically assessed by their dynamic impact, i.e., by how they change the characteristics of the movement of material objects.
The movement of material objects over time occurs in space relative to a certain frame of reference. In classical mechanics, based on Newton's axioms, space is considered three-dimensional, its properties do not depend on the material objects moving in it. The position of a point in such a space is determined by three coordinates. Time is not related to space and the movement of material objects. It is considered the same for all reference systems.
The laws of dynamics describe the movement of material objects in relation to absolute coordinate axes, conventionally accepted as stationary. The origin of the absolute coordinate system is taken to be at the center of the Sun, and the axes are directed to distant, conditionally stationary stars. When solving many technical problems, the coordinate axes connected to the Earth can be considered conditionally immovable.
Options mechanical movement material objects in dynamics are established by mathematical derivations from the basic laws of classical mechanics.
First law (law of inertia):
A material point maintains a state of rest or uniform and linear motion until the action of some forces takes it out of this state.
The uniform and linear motion of a point is called motion by inertia. Rest is a special case of motion by inertia, when the speed of a point is zero.
Every material point has inertia, that is, it strives to maintain a state of rest or uniform linear motion. The reference system in relation to which the law of inertia holds is called inertial, and the motion observed in relation to this system is called absolute. Any reference system that performs translational rectilinear and uniform motion relative to an inertial system will also be an inertial system.
Second law (basic law of dynamics):
The acceleration of a material point relative to the inertial frame of reference is proportional to the force applied to the point and coincides with the force in the direction: 
.
From the basic law of dynamics it follows that with force 
acceleration 
. The mass of a point characterizes the degree of resistance of a point to changes in its speed, that is, it is a measure of the inertia of a material point.
Third Law (Law of Action and Reaction):
The forces with which two bodies act on each other are equal in magnitude and directed along one straight line in opposite directions.
The forces called action and reaction are applied to different bodies and therefore do not form a balanced system.
Fourth law (law of independence of forces):
With the simultaneous action of several forces, the acceleration of a material point is equal to the geometric sum of the accelerations that the point would have under the action of each force separately:
, Where 
,
,…,
.
