The amount of motion is a measure of mechanical motion if the mechanical motion becomes mechanical. For example, the mechanical movement of a billiard ball (Fig. 22) before the impact passes into the mechanical movement of the balls after the impact. For a point, the momentum is equal to the product.

The measure of the action of the force in this case is the momentum of the force

. (9.1)

Momentum determines the action of force for a period of time . For a material point, the momentum change theorem can be used in differential form
(9.2) or integral (finite) form
. (9.3)

The change in the momentum of a material point over a certain period of time is equal to the momentum of all forces applied to the point in the same time.

Figure 22

When solving problems, theorem (9.3) is more often used in projections onto the coordinate axes
;

; (9.4)

.

Using the theorem on the change in the momentum of a point, it is possible to solve problems in which a point or a body moving translationally is subject to constant or variable forces that depend on time, and the number of given and sought values ​​includes the time of movement and speed at the beginning and end of the movement. Problems using the theorem are solved in the following sequence:

1. choose a coordinate system;

2. depict all given (active) forces and reactions acting on a point;

3. write down the theorem on the change in the momentum of a point in projections onto the selected coordinate axes;

4. determine the desired values.

EXAMPLE 12.

A hammer weighing G=2t falls from a height h=1m onto a workpiece in a time t=0.01s and stamps the part (Fig. 23). Determine the average force of the hammer on the workpiece.

SOLUTION.

1. Hammer gravity acts on the workpiece and support reaction . The value of the support reaction changes with time, so consider its average value
.

2. direct the coordinate axis y vertically down and apply the theorem on the change in the momentum of a point in projection onto this axis:
, (1) where - speed of the hammer at the end of the blow;

- the initial speed of the hammer at the moment of contact with the workpiece.

3. To determine the speed we compose the differential equation of motion of the hammer in projection onto the y-axis:

. (2)

Separate the variables, integrate equation (2) twice:
;

;

. The integration constants С 1 , С 2 can be found from the initial conditions. At t=0 V y =0, then C 1 =0; y \u003d 0, then C 2 \u003d 0. Therefore, the hammer moves according to the law
, (3) and the speed of the hammer changes according to the law
. (4) We will express the time of movement of the hammer from (3) and substitute into (4)
;
. (5)

4. We find the projection of the momentum of external forces on the y-axis by the formula:
. (6) Substitute (5) and (6) into (1):
, from where we find the reaction of the support, and, consequently, the desired pressure of the hammer on the workpiece
T.

Figure 24

TO

where M is the mass of the system, V c is the speed of the center of mass. The theorem on the change in the momentum of a mechanical system can be written in differential and finite (integral) form:
;

. (9.7)

The amount of movement of a mechanical system can be defined as the sum of the amounts of movement of the points of the system
. (9.5) The amount of motion of a system or a rigid body can be determined knowing the mass of the system and the velocity of the center of mass
, (9.6)

The change in the amount of motion of a mechanical system over a certain period of time is equal to the sum of the impulses of external forces acting for the same time. Sometimes it is more convenient to use the theorem on the change in momentum in the projection onto the coordinate axes
; (9.8)
. (9.9)

The law of conservation of momentum establishes that in the absence of external forces, the momentum of a mechanical system remains constant. The action of internal forces cannot change the momentum of the system. Equation (9.6) shows that for
,
.

If
, then
or
.

D

propeller or propeller, jet propulsion. Squids move in jerks, throwing water out of the muscular sac according to the principle of a water cannon (Fig. 25). The repelled water has a known amount of backward motion. The squid gains the corresponding speed forward movement due to reactive thrust , because before the squid jumps out, the force balanced by gravity .

the operation of the law of conservation of the momentum of a mechanical system can be illustrated by the example of the phenomenon of recoil or rollback when shooting, work

Application of the momentum change theorem makes it possible to exclude from consideration all internal forces.

EXAMPLE 13.

On a railway platform, free-standing on the rails, a winch A with a drum of radius r is installed (Fig. 26). The winch is designed to move on the platform of cargo B with mass m 1 . Weight of platform with winch m 2 . The winch drum rotates according to the law
. At the initial moment of time, the system was mobile. Neglecting friction, find the law of change in the speed of the platform after turning on the winch.

R DECISION.

1. Consider the platform, the winch and the load as a single mechanical system, which is affected by external forces: the force of gravity of the load and platforms and reactions And
.

2. Since all external forces are perpendicular to the x axis, i.e.
, we apply the law of conservation of the momentum of a mechanical system in projection onto the x-axis:
. At the initial moment of time, the system was stationary, therefore,

Let us express the amount of motion of the system at an arbitrary point in time. The platform moves forward at a speed , the load performs a complex movement, consisting of a relative movement along the platform with a speed and portable movement together with the platform at a speed ., where
. The platform will move in the direction opposite to the relative movement of the load.

EXAMPLE 14.

M

SOLUTION.

1. Apply the theorem on the change in the momentum of a mechanical system in projection onto the x-axis. Since all external forces acting on the system are vertical, then
, then
, where
. (1)

2. We express the projection of the amount of motion on the x-axis for the considered mechanical system
,

The mechanical system consists of a rectangular vertical plate 1 with a mass m 1 =18kg, moving along horizontal guides and a load D with a mass m 2 =6kg. At the time t 0 =0, when the plate was moving at a speed u 0 =2m/s, the load began to move along the chute in accordance with the equation S=AD=0.4sin( t 2) (S-in meters, t-in seconds), (Fig. 26). Determine the speed of the plate at time t 1 =1s, using the theorem on the change in the momentum of the mechanical system.

where ,
-- the amount of movement of the plate and the load, respectively.

;
, where --absolute speed of the loadD. From equality (1) it follows that K 1x + K 2x \u003d C 1 or m 1 u x + m 2 V Dx \u003d C 1. (2) To determine V Dx, we consider the movement of the load D as complex, considering its movement relative to the plate to be relative, and the movement of the plate itself to be portable, then
, (3)
; or in the projection on the x-axis: . (4) Substitute (4) into (2):
. (5) The integration constant C 1 is determined from the initial conditions: at t=0 u=u 0 ; (m 1 +m 2)u 0 \u003d C 1. (6) Substituting the value of the constant C 1 into equation (5), we obtain

m/s.

Consisting of n material points. Let us single out some point from this system Mj with mass mj. It is known that external and internal forces act on this point.

Apply to a point Mj resultant of all internal forces F j i and the resultant of all external forces F j e(Figure 2.2). For selected material point Mj(as for a free point) we write the theorem on the change in momentum in differential form (2.3):

We write similar equations for all points of the mechanical system (j=1,2,3,…,n).

Figure 2.2

Let's put everything together n equations:

∑d(m j ×V j)/dt = ∑F j e + ∑F j i, (2.9)

d∑(m j ×V j)/dt = ∑ F j e + ∑ F j i. (2.10)

Here ∑mj ×Vj =Q is the momentum of the mechanical system;
∑ F j e = R e is the main vector of all external forces acting on the mechanical system;
∑ F j i = R i =0- the main vector of the internal forces of the system (according to the property of internal forces, it is equal to zero).

Finally, for the mechanical system, we obtain

dQ/dt = Re. (2.11)

Expression (2.11) is a theorem on the change in the momentum of a mechanical system in differential form (in vector expression): the time derivative of the momentum vector of a mechanical system is equal to the main vector of all external forces acting on the system.

Projecting the vector equality (2.11) onto the Cartesian coordinate axes, we obtain expressions for the theorem on the change in the momentum of a mechanical system in a coordinate (scalar) expression:

dQ x /dt = R x e;

dQ y /dt = R y e;

dQ z /dt = R z e, (2.12)

those. the time derivative of the projection of the momentum of a mechanical system onto any axis is equal to the projection onto this axis of the main vector of all external forces acting on this mechanical system.

Multiplying both sides of equality (2.12) by dt, we obtain the theorem in another differential form:

dQ = R e ×dt = δS e, (2.13)

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

Integrating equality (2.13) within the time range from 0 to t, we obtain a theorem on the change in the momentum of a mechanical system in a finite (integral) form (in vector expression):

Q - Q 0 \u003d S e,

those. the change in the amount of motion of a mechanical system over a finite period of time is equal to the total impulse of the main vector (the sum of the total impulses) of all external forces acting on the system over the same period of time.

Projecting the vector equality (2.14) onto the Cartesian coordinate axes, we obtain expressions for the theorem in projections (in a scalar expression):

those. the change in the projection of the momentum of the mechanical system on any axis over a finite period of time is equal to the projection on the same axis of the total impulse of the main vector (the sum of the total impulses) of all external forces acting on the mechanical system for the same period of time.

From the considered theorem (2.11) - (2.15) follow the following corollaries:

1. If R e = ∑ F j e = 0, then Q = const– we have the law of conservation of the momentum vector of the mechanical system: if the main vector Re of all external forces acting on a mechanical system is equal to zero, then the momentum vector of this system remains constant in magnitude and direction and equal to its initial value Q0, i.e. Q = Q0.
2. If R x e = ∑X j e =0 (R e ≠ 0), then Q x = const- we have the law of conservation of the projection onto the axis of the momentum of the mechanical system: if the projection of the main vector of all forces acting on the mechanical system on any axis is zero, then the projection onto the same axis of the momentum vector of this system will be a constant value and equal to the projection onto this axis initial momentum vector, i.e. Qx = Q0x.

The differential form of the theorem on the change in momentum of a material system has important and interesting applications in continuum mechanics. From (2.11) one can obtain Euler's theorem.

The amount of movement of a material point is called a vector quantity mv, equal to the product of the mass of the point and the vector of its velocity. Vector mV attached to a moving point.

Quantity of system movement is called a vector quantity Q, equal to the geometric sum (principal vector) of the momentum of all points of the system:

Vector Q is a free vector. In the SI system of units, the momentum modulus is measured in kg m/s or N s.

As a rule, the velocities of all points of the system are different (see, for example, the distribution of velocities of the points of a rolling wheel shown in Fig. 6.21), and therefore the direct summation of the vectors on the right side of equality (17.2) is difficult. Let us find a formula with the help of which the quantity Q much easier to calculate. It follows from equality (16.4) that

Taking the time derivative of both parts, we get Hence, taking into account equality (17.2), we find that

i.e., the amount of motion of the system is equal to the product of the mass of the entire system and the speed of its center of mass.

Note that the vector Q, like the main vector of forces in statics, is some generalized vector characteristic of the motion of the entire mechanical system. In the general case of motion of a system, its momentum is Q can be considered as a characteristic of the translational part of the motion of the system together with its center of mass. If during the movement of the system (body) the center of mass is stationary, then the momentum of the system will be equal to zero. Such, for example, is the momentum of a body rotating around a fixed axis passing through its center of mass.

Example. Determine the amount of motion of the mechanical system (Fig. 17.1, but), consisting of cargo BUT weight t A - 2 kg, homogeneous block IN weighing 1 kg and wheels D weight mD-4 kg. Cargo BUT moving at a speed V A - 2 m/s, wheel D rolls without slipping, the thread is inextensible and weightless. Solution. The amount of movement of the body system

Body BUT moving forward and Q A \u003d m A V A(numerically Q A= 4 kg m/s, vector direction Q A coincides with the direction VA). Block IN performs rotational motion around a fixed axis passing through its center of mass; Consequently, QB- 0. Wheel D makes a plane-parallel

motion; its instantaneous center of velocities is at the point TO, so the speed of its center of mass (points E) is equal to V E = V A /2= 1 m/s. Number of wheel movement Q D - m D V E - 4 kg m/s; vector Q D directed horizontally to the left.

Depicting vectors Q A And Q D in fig. 17.1, b, find the momentum Q systems according to formula (a). Taking into account the directions and numerical values ​​of the quantities, we obtain Q ~^Q A +Q E=4l/2~kg m/s, vector direction Q shown in fig. 17.1, b.

Given that a-dV/dt, equation (13.4) of the basic law of dynamics can be represented as

Equation (17.4) expresses the theorem on the change in the momentum of a point in differential form: at each moment of time, the time derivative of the momentum of a point is equal to the force acting on the point. (In essence, this is another formulation of the basic law of dynamics, close to the one given by Newton.) If several forces act on a point, then on the right side of equality (17.4) there will be a resultant of the forces applied to the material point.

If both sides of the equation are multiplied by dt, then we get

The vector value on the right side of this equality characterizes the action exerted on the body by force in an elementary period of time dt this value is denoted dS and call elementary impulse of force, i.e.

Pulse S strength F over a finite time interval /, - / 0 is defined as the limit of the integral sum of the corresponding elementary impulses, i.e.

In a particular case, if the force F constant in modulus and in direction, then S = F(t| -/0) and S- F(t l -/ 0). In the general case, the modulus of the force impulse can be calculated from its projections onto the coordinate axes:

Now, integrating both sides of equality (17.5) with T= const, we get

Equation (17.9) expresses the theorem on changing the momentum of a point in finite (integral) form: the change in the momentum of a point over a certain period of time is equal to the momentum of the force acting on the point (or the momentum of the resultant of all forces applied to it) for the same period of time.

When solving problems, the equations of this theorem are used in projections onto the coordinate axes

Now consider a mechanical system consisting of P material points. Then, for each point, we can apply the momentum change theorem in the form (17.4), taking into account the external and internal forces applied to the points:

Summing up these equalities and taking into account that the sum of derivatives is equal to the derivative of the sum, we obtain

Since by the property of internal forces H.F.k=0 and by definition of momentum ^fn k V/ c = Q, then we finally find

Equation (17.11) expresses the theorem on the change in the momentum of the system in differential form: at each moment of time, the time derivative of the momentum of the system is equal to the geometric sum of all external forces acting on the system.

Projecting equality (17.11) onto the coordinate axes, we obtain

Multiplying both sides of (17.11) by dt and integrating, we get

where 0, Q0 - the amount of motion of the system at times, respectively, and / 0 .

Equation (17.13) expresses the theorem on the change in the momentum of the system in integral form: the change in the momentum of the system over any time is equal to the sum of the impulses of all external forces acting on the system over the same time.

In projections onto the coordinate axes, we get

From the theorem on the change in the momentum of the system, the following important consequences can be obtained, which express law of conservation of momentum of the system.

• 1. If the geometric sum of all external forces acting on the system is equal to zero (LF k=0), then from equation (17.11) it follows that in this case Q= const, i.e. the momentum vector of the system will be constant in magnitude and direction.
• 2. If the external forces acting on the system are such that the sum of their projections on any axis is zero (for example, I e kx = 0), then from equations (17.12) it follows that in this case Q x = const, i.e. the projection of the momentum of the system on this axis remains unchanged.

Note that the internal forces of the system do not participate in the equation of the theorem on the change in the momentum of the system. These forces, although they affect the momentum of individual points of the system, cannot change the momentum of the system as a whole. Given this circumstance, when solving problems, it is expedient to choose the system under consideration so that the unknown forces (all or part of them) are internal.

The law of conservation of momentum is convenient to apply in cases where the change in the speed of one part of the system is necessary to determine the speed of another part of it.

Problem 17.1. TO trolley weighing t x- 12 kg moving on a smooth horizontal plane, at a point BUT a weightless rod is attached with the help of a cylindrical hinge AD length /= 0.6 m with load D weight t 2 - 6 kg at the end (Fig. 17.2). At time / 0 = 0, when the speed of the trolley And () - 0.5 m/s, rod AD starts to rotate around the axis BUT, perpendicular to the plane of the drawing, according to the law φ \u003d (tg / 6) (3 ^ 2 - 1) rad (/- in seconds). Define: u=f.

§ 17.3. Theorem on the motion of the center of mass

The theorem on the change in the momentum of a mechanical system can be expressed in another form, which is called the theorem on the motion of the center of mass.

Substituting into equation (17.11) the equality Q=MV C , we get

If mass M system is constant, we get

where and with - acceleration of the center of mass of the system.

Equation (17.15) 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.

Projecting equality (17.15) onto the coordinate axes, we obtain

where x c , y c , z c - coordinates of the center of mass of the system.

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

Let's discuss the results. Let us preliminarily recall that the center of mass of the system is a geometric point, sometimes located outside the geometric boundaries of the body. The forces acting on the mechanical system (external and internal) are applied to all material points of the system. Equations (17.15) make it possible to determine the motion of the center of mass of the system without determining the motion of its individual points. Comparing the equations (17.15) of the theorem on the motion of the center of mass and the equation (13.5) of Newton's second law for a material point, we come to the conclusion: the center of mass of a mechanical system moves as a material point, the mass of which is equal to the mass of the entire system, and as if all external forces acting on the system are applied to this point. Thus, the solutions that we obtain by considering a given body as a material point determine the law of motion of the center of mass of this body.

In particular, if the body moves forward, then the kinematic characteristics of all points of the body and its center of mass are the same. That's why a progressively moving body can always be considered as a material point with a mass equal to the mass of the entire body.

As can be seen from (17.15), the internal forces acting on the points of the system do not affect the motion of the center of mass of the system. Internal forces can influence the movement of the center of mass in those cases when external forces change under their influence. Examples of this will be given below.

From the theorem on the motion of the center of mass, the following important consequences can be obtained, which express the law of conservation of the motion of the center of mass of the system.

1. If the geometric sum of all external forces acting on the system is zero (LF k=0), then it follows from equation (17.15),

what about a c = 0 or V c = const, i.e. the center of mass of this system

moves with a constant speed in magnitude and direction (otherwise, uniformly and rectilinearly). In a special case, if at the beginning the center of mass was at rest ( Vc=0), then it will remain at rest; where

track predicts that its position in space will not change, i.e. rc = const.

2. If the external forces acting on the system are such that the sum of their projections on some axis (for example, the axis X) zero (?F e kx= 0), then from equation (17.16) it follows that in this case x s=0 or V Cx \u003d x c \u003d const, i.e., the projection of the velocity of the center of mass of the system onto this axis is a constant value. In a special case, if at the initial moment Vex= 0, then at any subsequent time this value will be preserved, and hence it follows that the coordinate x s the center of mass of the system will not change, i.e. x s - const.

Consider examples illustrating the law of motion of the center of mass.

Examples. 1. As noted, the movement of the center of mass depends only on external forces; internal forces cannot change the position of the center of mass. But the internal forces of the system can cause external influences. So, the movement of a person on a horizontal surface occurs under the action of friction forces between the soles of his shoes and the road surface. With the strength of his muscles (internal forces), a person pushes off the road surface with his feet, which causes a friction force (external for a person) at the points of contact with the road, directed in the direction of his movement.

• 2. The car moves in the same way. The internal pressure forces in its engine make the wheels rotate, but since the latter have traction, the friction forces that arise “push” the car forward (as a result, the wheels do not rotate, but move in a plane-parallel way). If the road is absolutely smooth, then the center of mass of the car will be stationary (at zero initial speed) and the wheels, in the absence of friction, will slip, i.e., rotate.
• 3. Movement with the help of a propeller, propeller, oars occurs due to the rejection of a certain mass of air (or water). If we consider the discarded mass and the moving body as one system, then the forces of interaction between them, as internal, cannot change the total momentum of this system. However, each of the parts of this system will move, for example, the boat forward, and the water that the oars throw back.
• 4. In airless space, when the rocket is moving, the “discarded mass” should be “taken with you”: the jet engine informs the rocket about the movement by throwing back the combustion products of the fuel that the rocket is filled with.
• 5. When descending on a parachute, you can control the movement of the center of mass of the man-parachute system. If by muscular effort a person pulls the parachute lines in such a way that the shape of its canopy or the angle of attack of the air flow changes, then this will cause a change in the external influence of the air flow, and thereby affect the movement of the entire system.

Problem 17.2. IN task 17.1 (see Figure 17.2) determine: 1) law of motion of the trolley X (= /)(/), if it is known that at the initial moment of time t 0 = About the system was at rest and the coordinate x 10 = 0; 2) the law of change with time of the total value of the normal reaction N(N = N" + N") horizontal plane, i.e. N=f 2 (t).

Solution. Here, as in problem 17.1, we consider a system consisting of a trolley and a load D, in an arbitrary position under the action of external forces applied to it (see Fig. 17.2). Coordinate axes Ohu draw so that the x-axis is horizontal and the x-axis at passed through the point A 0 , i.e. the location of the point BUT at the time t-t 0 - 0.

1. Determination of the law of motion of the cart. To determine x, = /, (0, we use the theorem on the motion of the center of mass of the system. Let us compose a differential equation of its motion in projection onto the x axis:

Since all external forces are vertical, then T, F e kx = 0, and therefore

Integrating this equation, we find that Mx c \u003d B, i.e., the projection of the velocity of the center of mass of the system on the x-axis is a constant value. Since at the initial moment of time

Integrating the equation Mx s= 0, we get

i.e. coordinate x s the center of mass of the system is constant.

Let's write the expression Mx s for an arbitrary position of the system (see Fig. 17.2), taking into account that x A - x { , x D - x 2 And x 2 - x ( - I sin f. In accordance with formula (16.5), which determines the coordinate of the center of mass of the system, in this case Mx s - t(x( + t 2 x 2".

for an arbitrary point in time

for time point / () = 0, X (= 0 and

In accordance with equality (b), the coordinate x s the center of mass of the entire system remains unchanged, i.e. x c (t). Therefore, by equating expressions (c) and (d), we obtain the dependence of the x coordinate on time.

Answer: X - 0.2 m, where t- in seconds.

2. Reaction definition N. For determining N=f 2 (t) we compose the differential equation of motion of the center of mass of the system in projection onto the vertical axis at(see fig. 17.2):

Hence, denoting N=N+N", we get

According to the formula that determines the ordinate u s the center of mass of the system, Mu s = t (y x + t 2 y 2, where y, = at C1,at 2= yD = Atbut ~ 1 cos Ф» we get

Differentiating this equality twice with respect to time (taking into account that at C1 And at A the quantities are constant and, consequently, their derivatives are equal to zero), we find

Substituting this expression into equation (e), we determine the required dependence N from t.

Answer: N- 176,4 + 1,13,

where φ \u003d (i / 6) (3 / -1), t- in seconds N- in newtons.

Problem 17.3. Electric motor mass t x attached to the horizontal surface of the foundation with bolts (Fig. 17.3). On the motor shaft at right angles to the axis of rotation, a weightless rod of length l is fixed at one end, and a point weight is mounted on the other end of the rod BUT weight t 2 . The shaft rotates uniformly at an angular velocity o. Find the horizontal pressure of the motor on the bolts. Solution. Consider a mechanical system consisting of a motor and a point weight BUT, in an arbitrary position. Let us depict the external forces acting on the system: gravity R x, R 2, foundation reaction in the form of a vertical force N and horizontal force R. Draw the x-axis horizontally.

To determine the horizontal pressure of the motor on the bolts (and it will be numerically equal to the reaction R and directed opposite to the vector R ), we compose the equation of the theorem on the change in the momentum of the system in projection onto the horizontal axis x:

For the system under consideration in its arbitrary position, given that the amount of motion of the motor housing is zero, we obtain Q x = - t 2 U A col. Taking into account that V A = a s/, φ = ω/ (uniform rotation of the motor), we get Q x - - m 2 co/cos co/. differentiating Q x in time and substituting into equality (a), we find R- m 2 co 2 /sin co/.

Note that it is precisely such forces that are forcing (see § 14.3), when they act, forced vibrations of structures occur.

Exercises for independent work

• 1. What is called the momentum of a point and a mechanical system?
• 2. How does the momentum of a point moving uniformly around a circle change?
• 3. What characterizes the impulse of force?
• 4. Do the internal forces of the system affect its momentum? On the movement of its center of mass?
• 5. How do couples of forces applied to it affect the motion of the center of mass of the system?
• 6. Under what conditions is the center of mass of the system at rest? moving uniformly and in a straight line?

7. In a stationary boat, in the absence of water flow, an adult sits at the stern, and a child sits at the bow of the boat. In which direction will the boat move if they switch places?

In which case will the displacement module of the boat be large: 1) if the child goes to the adult in the stern; 2) if an adult goes to the child on the bow of the boat? What will be the displacements of the center of mass of the “boat and two people” system during these movements?

Differential equation of motion of a material point under the action of a force F can be represented in the following vector form:

Since the mass of a point m is assumed to be constant, then it can be introduced under the sign of the derivative. Then

Formula (1) expresses the theorem on the change in the momentum of a point in differential form: the first time derivative of the momentum of a point is equal to the force acting on the point.

In projections onto the coordinate axes (1) can be represented as

If both sides of (1) are multiplied by dt, then we obtain another form of the same theorem - the momentum theorem in differential form:

those. the differential of the momentum of a point is equal to the elementary impulse of the force acting on the point.

Projecting both parts of (2) onto the coordinate axes, we obtain

Integrating both parts of (2) from zero to t (Fig. 1), we have

where is the speed of the point at the moment t; - speed at t = 0;

S- momentum of force over time t.

The expression in the form (3) is often called the momentum theorem in finite (or integral) form: the change in the momentum of a point over any period of time is equal to the momentum of the force over the same period of time.

In projections onto the coordinate axes, this theorem can be represented in the following form:

For a material point, the theorem on the change in the amount of motion in any of the forms, in essence, does not differ from the differential equations of motion of a point.

Theorem on the change in the momentum of the system

The amount of motion of the system will be called the vector quantity Q, equal to the geometric sum (principal vector) of the momentum of all points of the system.

Consider a system consisting of n material points. Let us compose differential equations of motion for this system and add them term by term. Then we get:

The last sum by the property of internal forces is equal to zero. Besides,

We finally find:

Equation (4) expresses the theorem on the change in the momentum of the system in differential form: the time derivative of the momentum of the system is equal to the geometric sum of all external forces acting on the system.

Let's find another expression of the theorem. Let at the moment t= 0 the momentum of the system is Q0, and at the moment of time t1 becomes equal Q1. Then, multiplying both sides of equality (4) by dt and integrating, we get:

Or where:

(S- force impulse)

since the integrals on the right give the impulses of external forces,

equation (5) expresses the theorem on the change in the momentum of the system in integral form: the change in the amount of motion of the system over a certain period of time is equal to the sum of the impulses of external forces acting on the system over the same period of time.

In projections on the coordinate axes, we will have:

Law of conservation of momentum

From the theorem on the change in the momentum of the system, the following important consequences can be obtained:

1. Let the sum of all external forces acting on the system be equal to zero:

Then it follows from Eq. (4) that, in this case, Q=const.

In this way, if the sum of all external forces acting on the system is equal to zero, then the momentum vector of the system will be constant in modulus and direction.

2. Let the external forces acting on the system be such that the sum of their projections on some axis (for example, Ox) is equal to zero:

Then it follows from equations (4`) that in this case Q = const.

In this way, if the sum of the projections of all acting external forces on some axis is equal to zero, then the projection of the momentum of the system on this axis is a constant value.

These results express law of conservation of momentum of the system. It follows from them that internal forces cannot change the total momentum of the system.

Let's look at some examples:

· P h e n i e o f recoil or recoil. If we consider a rifle and a bullet as one system, then the pressure of the powder gases when fired will be an internal force. This force cannot change the total momentum of the system. But since the propellant gases, acting on the bullet, give it a certain amount of movement directed forward, they must simultaneously tell the rifle the same amount of movement in the opposite direction. This will cause the rifle to move backward, i.e. so-called return. A similar phenomenon occurs when firing from a gun (rollback).

· Operation of the propeller (propeller). The propeller tells some mass of air (or water) to move along the axis of the propeller, throwing this mass back. If we consider the ejected mass and the aircraft (or ship) as one system, then the interaction forces of the propeller and the medium as internal cannot change the total momentum of this system. Therefore, when a mass of air (water) is thrown back, the aircraft (or vessel) obtains the corresponding forward speed, such that the total momentum of the system under consideration remains equal to zero, since it was zero before the start of movement.

A similar effect is achieved by the action of oars or paddle wheels.

· Rocket propulsion. In a rocket projectile (rocket), gaseous products of fuel combustion are ejected at high speed from a hole in the tail of the rocket (from the nozzle of a jet engine). The pressure forces acting in this case will be internal forces and they cannot change the total momentum of the rocket-powder gases system. But since the escaping gases have a certain amount of motion directed backwards, the rocket receives in this case the corresponding forward speed.

Theorem of moments about the axis.

Consider a material point of mass m moving under the influence of a force F. Let us find for it the dependence between the moment of the vectors mV And F about some fixed Z-axis.

m z (F) = xF - yF (7)

Similarly for the quantity m (mV), if taken out m bracket will be

m z (mV) \u003d m (xV - yV)(7`)

Taking time derivatives of both sides of this equality, we find

On the right side of the resulting expression, the first parenthesis is 0, since dx/dt=V and dу/dt=V, while the second bracket according to formula (7) is equal to

m z (F), since according to the basic law of dynamics:

Finally we will have (8)

The resulting equation expresses the theorem of moments about the axis: the time derivative of the angular momentum of a point about some axis is equal to the moment of the acting force about the same axis. A similar theorem also holds for moments about any center O.

(Fragments of a mathematical symphony)

The connection of the force impulse with the basic equation of Newtonian dynamics is expressed by the theorem on the change in the momentum of a material point.

Theorem. The change in the momentum of a material point for a certain period of time is equal to the impulse of the force () acting on the material point for the same period of time. The mathematical proof of this theorem can be called a fragment of a mathematical symphony. Here he is.

The differential momentum of a material point is equal to the elementary impulse of the force acting on the material point. Integrating the expression (128) for the momentum differential of a material point, we have

(129)

The theorem is proved and mathematicians consider their mission completed, and engineers, whose fate is to sacredly believe mathematicians, have questions when using the proven equation (129). But they are firmly blocked by the sequence and beauty of mathematical actions (128 and 129), which fascinate and encourage us to call them a fragment of a mathematical symphony. How many generations of engineers agreed with mathematicians and trembled at the mystery of their mathematical symbols! But then there was an engineer who disagreed with the mathematicians and asked them questions.

Dear mathematicians! Why is it that none of your textbooks on theoretical mechanics discusses the process of applying your symphonic result (129) in practice, for example, when describing the process of accelerating a car? The left side of equation (129) is extremely clear. The car starts acceleration from a speed and finishes it, for example, at a speed of . It is quite natural that equation (129) becomes

And the first question immediately arises: how can we determine the force from equation (130), under the influence of which the car is accelerated to a speed of 10 m/s? There is no answer to this question in any of the innumerable textbooks on theoretical mechanics. Let's go further. After acceleration, the car begins to move uniformly with the achieved speed of 10m/s. What is the force that drives the car? I have no choice but to blush along with the mathematicians. The first law of Newtonian dynamics states that when a car moves uniformly, no forces act on it, and the car, figuratively speaking, sneezes on this law, consumes gasoline and does work, moving, for example, a distance of 100 km. And where is the force that has done the work to move the car 100 km? The symphonic mathematical equation (130) is silent, but life goes on and requires an answer. We start looking for it.

Since the car is moving in a straight line and uniformly, the force that moves it is constant in magnitude and direction, and equation (130) becomes

(131)

So, equation (131) in this case describes the accelerated motion of the body. What is the force equal to? How to express its change over time? Mathematicians prefer to sidestep this question and leave it to the engineers, believing that they should look for the answer to this question. Engineers have one possibility left - to take into account that if, after the completion of the accelerated motion of the body, a phase of uniform motion begins, which is accompanied by a constant force, represent equation (131) for the moment of transition from accelerated to uniform motion in this form

(132)

The arrow in this equation does not mean the result of integrating this equation, but the process of transition from its integral form to a simplified form. The force in this equation is equivalent to the average force that changed the momentum of the body from zero to the final value . So, dear mathematicians and theoretical physicists, the absence of your method for determining the magnitude of your momentum forces us to simplify the procedure for determining the force, and the lack of a method for determining the duration of this force generally puts us in a hopeless situation and we are forced to use the expression to analyze the process of changing the momentum of the body . As a result, the longer the force acts, the greater its momentum. This clearly contradicts the long-established ideas that the impulse of force is greater, the shorter the time of its action.

Let us pay attention to the fact that the change in the momentum of a material point (force impulse) during its accelerated movement occurs under the action of the Newtonian force and the forces of resistance to movement, in the form of forces formed by mechanical resistances and the force of inertia. But Newtonian dynamics in the vast majority of problems ignores the force of inertia, and Mechanodynamics asserts that the change in the momentum of a body during its accelerated movement occurs due to the excess of the Newtonian force over the forces of resistance to movement, including the force of inertia.

When a body moves in slow motion, for example, a car with a gear off, there is no Newtonian force, and the change in the momentum of the car occurs due to the excess of the forces of resistance to movement over the force of inertia that moves the car during its slow motion.

How now to return the results of the noted "symphonic" mathematical operations (128) to the channel of cause-and-effect relationships? There is only one way out - to find a new definition for the concepts of "impulse of force" and "impact force". To do this, we divide both sides of equation (132) by the time t. As a result, we will have

. (133)

Let's pay attention to the fact that the expression mV / t is the rate of change in the momentum (mV / t) of a material point or body. If we take into account that V / t is acceleration, then mV / t is a force that changes the momentum of the body. The same dimension on the left and right of the equal sign gives us the right to call the force F the impact force and designate it with the symbol , and the impulse S - the impact impulse and designate it with the symbol . From this follows a new definition of impact force. Impact force, acting on a material point or body, is equal to the ratio of the change in the momentum of the material point or body to the time of this change.

Let us pay special attention to the fact that only the Newtonian force is involved in the formation of the shock impulse (134), which changed the speed of the car from zero to the maximum value - , therefore, equation (134) belongs entirely to Newtonian dynamics. Since it is much easier to fix the speed value experimentally than accelerations, formula (134) is very convenient for calculations.

Equation (134) implies such an unusual result.

Let us pay attention to the fact that, according to the new laws of mechanodynamics, the generator of the force impulse during the accelerated movement of a material point or body is the Newtonian force. It generates an acceleration of the movement of a point or body, at which an inertia force automatically arises, directed opposite to the Newtonian force, and the impact Newtonian force must overcome the action of the inertia force, therefore the inertia force must be represented in the balance of forces on the left side of equation (134). Since the force of inertia is equal to the mass of a point or body, multiplied by the deceleration , which it forms, then equation (134) becomes

(136)

Dear mathematicians! You can see what form the mathematical model has taken, describing the shock impulse, which accelerates the movement of the hit body from zero speed to maximum V (11). Now let's check its work in determining the impact impulse , which is equal to the impact force that fired the 2nd UGS power unit (Fig. 120), and we will leave your useless equation (132) to you. In order not to complicate the presentation, we will leave formula (134) alone for the time being and use the formulas that give the averaged values ​​of the forces. You see in what position you put an engineer seeking to solve a specific problem.

Let's start with Newtonian dynamics. The experts found that the 2nd power unit rose to a height of 14m. Since it was rising in the field of gravity, then at a height h = 14m its potential energy turned out to be equal to

and the average kinetic energy was

Rice. 120. Photo of the engine room before the disaster

From the equality of the kinetic (138) and potential (137) energies, the average lifting speed of the power unit follows (Fig. 121, 122)

Rice. 121. Photon of the machine room after the disaster

According to the new laws of mechanodynamics, the rise of the power unit consisted of two phases (Fig. 123): the first phase OA - an accelerated rise and the second phase AB - a slow rise , , .

The time and distance of their action are approximately equal to (). Then the kinematic equation of the accelerated lifting phase of the power unit will be written as

. (140)

Rice. 122. View of the well of the power unit and the power unit itself after the disaster

The law of change in the lifting speed of the power unit in the first phase has the form

. (141)

Rice. 123. The pattern of change in the speed V of the flight of the power unit

Substituting the time from equation (140) into equation (141), we have

. (142)

The block lifting time in the first phase is determined from the formula (140)

. (143)

Then the total time of lifting the power unit to a height of 14m will be equal to . The mass of the power unit and the cover is 2580 tons. According to Newton's dynamics, the force that lifted the power unit is equal to

Dear mathematicians! We follow your symphonic mathematical results and write down your formula (129), which follows from Newton's dynamics, to determine the shock impulse that fired the 2nd power unit

and ask an elementary question: how to determine the duration of the shock pulse that fired the 2nd power unit????????????

Dear!!! Remember how much chalk the generations of your colleagues wrote on the educational boards, abstrusely teaching students how to determine the impact impulse and no one explained how to determine the duration of the impact impulse in each specific case. You will say the duration of the shock impulse is equal to the time interval for changing the speed of the power unit from zero to, we assume, the maximum value of 16.75 m/s (139). It is in formula (143) and is equal to 0.84 s. We agree with you for now and determine the average value of the shock impulse

The question immediately arises: why is the magnitude of the shock impulse (146) less than the Newtonian force of 50600 tons? The answer, you, dear mathematicians, no. Let's go further.

According to Newton's dynamics, the main force that resisted the lifting of the power unit is gravity. Since this force is directed against the movement of the power unit, it generates a deceleration, which is equal to the free fall acceleration. Then the gravitational force acting on the power unit flying upwards is equal to

The dynamics of Newton does not take into account other forces that prevented the action of the Newtonian force of 50600 tons (144), and mechanodynamics claims that the inertia force equal to

The question immediately arises: how to find the magnitude of the deceleration of the movement of the power unit? Newton's dynamics is silent, and mechanodynamics answers: at the moment of action of the Newtonian force that lifted the power unit, it was resisted: gravity and inertia, so the equation of forces acting on the power unit at that moment is written as follows.