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

The amount of movement of the system called a vector quantity Q, equal to geometric sum(main vector) quantities of motion of all points of the system:

Vector Q is a free vector. In the SI system of units, the modulus of momentum 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 points of a rolling wheel, shown in Fig. 6.21), and therefore direct summation of vectors on the right side of equality (17.2) is difficult. Let us find a formula using which the quantity Q much easier to calculate. From equality (16.4) it follows that

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

that is, the momentum 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, it is some generalized vector characteristic of the movement 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 system’s motion together with its center of mass. If, when the system (body) moves, the center of mass is stationary, then the amount of motion of the system will be equal to zero. This is, for example, the momentum of a body rotating around fixed axis passing through its center of mass.

Example. Determine the amount of motion of the mechanical system (Fig. 17.1, A), consisting of cargo A mass t A - 2 kg, homogeneous block IN weighing 1 kg and wheels D mass m D - 4 kg. Cargo A moves at speed V A - 2 m/s, wheel D rolls without slipping, the thread is inextensible and weightless. Solution. Quantity of motion of a system of bodies

Body A moves forward and Q A =m A V A(numerically Q A= 4 kg m/s, vector direction Q A coincides with the direction V A). Block IN commits rotational movement around a fixed axis passing through its center of mass; hence, QB- 0. Wheel D makes a plane-parallel

movement; its instantaneous velocity center is at the point TO, therefore the speed of its center of mass (point E) equal to V E = V A /2= 1 m/s. Wheel movement amount Q D - m D V E - 4 kg m/s; vector QD directed horizontally to the left.

By depicting the vectors Q A And QD in Fig. 17.1, b, find the amount of motion 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.

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

Equation (17.4) expresses the theorem about the change in the momentum of a point in differential form: at each instant of time, the time derivative of the momentum of a point is equal to the force acting on the point. (Essentially, this is another formulation of the fundamental 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 equality are multiplied by dt, then we get

The vector quantity on the right side of this equality characterizes the action exerted on the body by a 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 for a finite period of time /, - / 0 is defined as the limit of the integral sum of the corresponding elementary impulses, i.e.

In the special case, if the force F is constant in magnitude and direction, then S = F(t| -/ 0) and S- F(t l -/ 0). In the general case, the magnitude 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 about the change in 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 impulse of the force acting on the point (or the impulse of the resultant of all forces applied to it) over the same period of time.

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

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

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

Since by the nature of internal forces HF k=0 and by definition of momentum ^fn k V/ c = Q, then we finally find

Equation (17.11) expresses the theorem about the change in 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 (17.11) by dt and integrating, we get

where 0, Q 0 - the amount of motion of the system at moments of time respectively and / 0.

Equation (17.13) expresses the theorem about the change in 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 during the same time.

In projections onto the coordinate axes we get

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

• 1. If the geometric sum of all external forces acting on the system is 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 onto 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 system's momentum onto 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 influence the momentum of individual points of the system, cannot change the momentum of the system as a whole. Taking this circumstance into account, when solving problems, it is advisable to choose the system in question so that unknown forces(all or part of them) to be made internal.

The law of conservation of momentum is convenient to apply in cases where, by changing the speed of one part of the system, it is necessary to determine the speed of its other part.

Problem 17.1. TO cart weighing t x- 12 kg moving on smooth horizontal plane, at point A a weightless rod is attached using a cylindrical hinge AD length /= 0.6 m with load D mass t 2 - 6 kg at the end (Fig. 17.2). At time / 0 = 0, when the speed of the cart And () - 0.5 m/s, rod AD begins to rotate around an axis A, perpendicular to the plane drawing, according to the law f = (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, called the theorem on the motion of the center of mass.

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

If the mass M system is constant, we get

Where and with - acceleration of the system's center of mass.

Equation (17.15) expresses the theorem on the movement of the center of mass of the system: the product of the mass of a 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 system's center of mass.

These equations are differential equations of motion of the center of mass in projections on the axis Cartesian system coordinates

Let's discuss the results obtained. Let us first recall that the center of mass of the system is 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 movement of the center of mass of the system without determining the movement of its individual points. Comparing equations (17.15) of the theorem on the motion of the center of mass and equations (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 like 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 were 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 a body moves translationally, then the kinematic characteristics of all points of the body and its center of mass are the same. That's why a translationally 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 movement of the center of mass of the system. Internal forces can influence the movement of the center of mass in cases where 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 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 from equation (17.15) it follows,

what about this 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 (in other words, uniformly and rectilinearly). In the special case, if at first the center of mass was at rest ( V c=0), then it will remain at rest; where

track You know that its position in space will not change, i.e. r c = const.

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

Let us 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. Thus, a person’s movement on a horizontal surface occurs under the influence of friction forces between the soles of his shoes and the surface of the road. With the strength of his muscles (internal forces), a person pushes off the road surface with his feet, which is why a friction force (external to the person) arises at the points of contact with the road, directed in the direction of his movement.

• 2. The car moves in a similar way. The internal pressure forces in its engine force the wheels to rotate, but since the latter have traction with the road, the resulting frictional forces “push” the car forward (as a result, the wheels do not rotate, but move plane-parallel). 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., perform a rotational movement.
• 3. Movement with the help of a propeller, propeller, or oars occurs due to the rejection of a certain mass of air (or water). If we consider the thrown mass and the moving body as one system, then the interaction forces between them, as internal ones, cannot change the total amount of motion of this system. However, each part of this system will move, for example, the boat forward, and the water that the oars throw back.
• 4. In airless space, when a rocket moves, the “thrown mass” should be “taken with you”: the jet engine imparts movement to the rocket by throwing back the combustion products of the fuel with which the rocket is filled.
• 5. When descending by parachute, you can control the movement of the center of mass of the man-parachute system. If, through muscular efforts, a person tightens the parachute lines so 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 influence the movement of the entire system.

Problem 17.2. IN Problem 17.1 (see Fig. 17.2) determine: 1) law of trolley motion X (= /)(/), if it is known that at the initial moment of time t 0 = O the system was at rest and the coordinate x 10 = 0; 2) the law of change over 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 cart and a load D, in an arbitrary position under the action of external forces applied to it (see Fig. 17.2). Coordinate axes Ohoo draw it so that the x axis is horizontal and the axis at passed through the point A 0, i.e. the location of the point A at a point in time t-t 0 - 0.

1. Determination of the law of motion of the trolley. To determine x, = /,(0, we use the theorem on the motion of the center of mass of the system. Let’s create 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 = B, that is, the projection of the velocity of the center of mass of the system onto the x-axis is a constant value. Since at the initial moment of time

Integrating Eq. Mx s= 0, we get

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

Let's write down 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 the moment of time / () = 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. xD^,) = xc(t). Consequently, 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. Definition of reaction N. To determine N=f 2 (t) let's create a 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 y s center of mass of the system, Mu s = t ( y x + t 2 y 2, where y, = at C1,at 2= y D = UA ~ 1 cos Ф" we get

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

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

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

where f = (i/6)(3/ -1), t- in seconds, N- in newtons.

Problem 17.3. Electric motor weight t x attached to the horizontal surface of the foundation with bolts (Fig. 17.3). A weightless rod of length l is fixed at one end to the motor shaft at right angles to the axis of rotation, and a point weight is mounted on the other end of the rod. A mass t 2. The shaft rotates uniformly at an angular velocity c. Find the horizontal pressure of the motor on the bolts. Solution. Consider a mechanical system consisting of a motor and a point weight A, in any position. Let us depict the external forces acting on the system: gravity R x, R 2, reaction of the foundation in the form of vertical force N and horizontal force R. Let's 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 will compose the equation of the theorem on the change in the momentum of the system in projection onto the horizontal x axis:

For the system under consideration in its arbitrary position, taking into account that the amount of motion of the motor body is zero, we obtain Q x = - t 2 U A soc. Taking into account that V A = a z/, f = co/ (motor rotation is uniform), 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 arise.

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 a force impulse?
• 4. Do the internal forces of a system affect its momentum? On the movement of its center of mass?
• 5. How do couples of forces applied to it affect the movement of the center of mass of the system?
• 6. Under what conditions is the center of mass of the system at rest? moves uniformly and in a straight line?

7. In a stationary boat with no water flow, an adult sits at the stern, and a child sits at the bow of the boat. In what direction will the boat move if they switch places?

In what case will the module of movement of the boat be large: 1) if the child moves to the adult’s stern; 2) if an adult goes to the child at the bow of the boat? What will be the displacement of the center of mass of the “boat and two people” system during these movements?

The system discussed in the theorem can be any mechanical system consisting of any bodies.

Statement of the theorem

The amount of motion (impulse) of a mechanical system is a quantity equal to the sum of the amounts of motion (impulses) of all bodies included in the system. The impulse of external forces acting on the bodies of the system is the sum of the impulses of all external forces acting on the bodies of the system.

( kg m/s)

The theorem on the change in momentum of a system states

The change in the momentum of the system over a certain period of time is equal to the impulse of external forces acting on the system over the same period of time.

Law of conservation of momentum of a system

If the sum of all external forces acting on the system is zero, then the amount of motion (momentum) of the system is a constant quantity.

, we obtain the expression of the theorem on the change in the momentum of the system in differential form:

Having integrated both sides of the resulting equality over an arbitrarily taken period of time between some and , we obtain the expression of the theorem on the change in the momentum of the system in integral form:

Law of conservation of momentum (Law of conservation of momentum) states that vector sum the impulses of all bodies of the system are a constant value if the vector sum of external forces acting on the system is equal to zero.

(moment of momentum m 2 kg s −1)

Theorem on the change in angular momentum relative to the center

derivative with respect to time of momentum ( kinetic moment) of a material point relative to any fixed center is equal to the moment of force acting on the point relative to the same center.

dk 0 /dt = M 0 (F ) .

Theorem on the change in angular momentum relative to an axis

the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed axis is equal to the moment of the force acting on this point relative to the same axis.

dk x /dt = M x (F ); dk y /dt = M y (F ); dk z /dt = M z (F ) .

Consider a material point M mass m , moving under the influence of force F (Figure 3.1). Let's write down and construct the vector of angular momentum (kinetic momentum) M 0 material point relative to the center O :

Let us differentiate the expression for the angular momentum (kinetic moment k 0) by time:

Because dr /dt = V , That vector product V ⊗ m ⋅ V (collinear vectors V And m ⋅ V ) is equal to zero. At the same time d(m ⋅ V) /dt = F according to the theorem on the momentum of a material point. Therefore we get that

dk 0 /dt = r ⊗F , (3.3)

Where r ⊗F = M 0 (F ) – vector-moment of force F relative to a fixed center O . Vector k 0 ⊥ plane ( r , m ⊗V ), and the vector M 0 (F ) ⊥ plane ( r ,F ), we finally have

dk 0 /dt = M 0 (F ) . (3.4)

Equation (3.4) expresses the theorem about the change in angular momentum (angular momentum) of a material point relative to the center: the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed center is equal to the moment of force acting on the point relative to the same center.

Projecting equality (3.4) onto the axes of Cartesian coordinates, we obtain

dk x /dt = M x (F ); dk y /dt = M y (F ); dk z /dt = M z (F ) . (3.5)

Equalities (3.5) express the theorem about the change in angular momentum (kinetic momentum) of a material point relative to the axis: the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed axis is equal to the moment of the force acting on this point relative to the same axis.

Let us consider the consequences following from Theorems (3.4) and (3.5).

Corollary 1. Consider the case when the force F during the entire movement of the point passes through the stationary center O (case of central force), i.e. When M 0 (F ) = 0. Then from Theorem (3.4) it follows that k 0 = const ,

those. in the case of a central force, the angular momentum (kinetic moment) of a material point relative to the center of this force remains constant in magnitude and direction (Figure 3.2).

Figure 3.2

From the condition k 0 = const it follows that the trajectory of a moving point is a flat curve, the plane of which passes through the center of this force.

Corollary 2. Let M z (F ) = 0, i.e. force crosses the axis z or parallel to it. In this case, as can be seen from the third of equations (3.5), k z = const ,

those. if the moment of force acting on a point relative to any fixed axis is always zero, then the angular momentum (kinetic moment) of the point relative to this axis remains constant.

Proof of the theorem on the change in momentum

Let the system consist of material points with masses and accelerations. We divide all forces acting on the bodies of the system into two types:

External forces are forces acting from bodies not included in the system under consideration. The resultant of external forces acting on a material point with number i let's denote

Internal forces are the forces with which the bodies of the system itself interact with each other. The force with which on the point with the number i the point with the number is valid k, we will denote , and the force of influence i th point on k th point - . Obviously, when , then

Using the introduced notation, we write Newton’s second law for each of the material points under consideration in the form

Considering that and summing up all the equations of Newton’s second law, we get:

The expression represents the sum of all internal forces acting in the system. According to Newton’s third law, in this sum, each force corresponds to a force such that, therefore, it holds Since the entire sum consists of such pairs, the sum itself is zero. Thus, we can write

Using the notation for the momentum of the system, we obtain

By introducing into consideration the change in the momentum of external forces , we obtain the expression of the theorem on the change in the momentum of the system in differential form:

Thus, each of the last equations obtained allows us to state: a change in the momentum of the system occurs only as a result of the action of external forces, and internal forces cannot have any influence on this value.

Having integrated both sides of the resulting equality over an arbitrarily taken time interval between some and , we obtain the expression of the theorem on the change in the momentum of the system in integral form:

where and are the values ​​of the amount of motion of the system at moments of time and, respectively, and is the impulse of external forces over a period of time. In accordance with what was said earlier and the introduced notations,

(Fragments of a mathematical symphony)

The connection between the impulse of force and 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 over a certain period of time is equal to the impulse of the force () acting on the material point over 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 expression (128) for the differential momentum of a material point, we have

(129)

The theorem has been proven and mathematicians consider their mission completed, but engineers, whose destiny is to sacredly believe in mathematicians, have questions when using the proven equation (129). But they are firmly blocked by consistency and beauty mathematical operations(128 and 129), which fascinate and encourage you to call them a fragment of a mathematical symphony. How many generations of engineers agreed with mathematicians and were in awe of the mystery of their mathematical symbols! But then there was an engineer who disagreed with the mathematicians and asked them questions.

Dear mathematicians! Why in none of your textbooks on theoretical mechanics is the process of applying your symphonic result (129) in practice, for example, when describing the process of accelerating a car, not considered? The left side of equation (129) is very clear. The car starts acceleration from speed and ends it, for example, at speed. It is quite natural that equation (129) becomes

And the first question immediately arises: how can we determine from equation (130) the force under the influence of which the car is accelerated to a speed of 10 m/s? The answer to this question is not found in any of the countless textbooks on theoretical mechanics. Let's move on. After acceleration, the car begins to move uniformly at a speed of 10 m/s. What force moves 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 at this law, consumes gasoline and does work, moving, for example, a distance of 100 km. Where is the force that did the work to move the car 100 km? The symphonic mathematical equation (130) is silent, but life goes on and demands an answer. We start looking for him.

Since the car moves rectilinearly and uniformly, the force moving 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 bypass this question and leave it to engineers, believing that they must seek the answer to this question. Engineers have only one option left - to take into account that if, after completing the accelerated motion of a body, a phase of uniform motion begins, which is accompanied by the action of a constant force, present equation (131) for the moment of transition from accelerated to uniform movement 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 a final value. So, dear mathematicians and theoretical physicists, the absence of your method for determining the magnitude of your impulse forces us to simplify the procedure for determining force, and the absence of a method for determining the time of action of this force generally puts us in a hopeless position and we are forced to use an expression to analyze the process of changing the momentum of a body . The result is that the longer the force acts, the greater its impulse. This clearly contradicts the long-established idea that the shorter the duration of its action, the greater the force impulse.

Let us draw attention to the fact that the change in the momentum of a material point (impulse of force) during its accelerated motion occurs under the action of Newtonian force and forces of resistance to motion, in the form of forces generated by mechanical resistances and the force of inertia. But Newtonian dynamics in the vast majority of problems ignores the force of inertia, and Mechanodynamics states that a change in the momentum of a body during its accelerated motion 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 the gear turned off, there is no Newtonian force, and the change in the momentum of the car occurs due to the excess of the motion resistance forces over the inertia force that moves the car when it moves slowly.

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

. (133)

Let us pay attention to the fact that the expression mV/t is the rate of change of momentum (mV/t) of a material point or body. If we take into account that V/t is acceleration, then mV/t is the force that changes the amount of motion of the body. The same dimension on the left and right of the equal sign gives us the right to call the force F a shock force and denote it by the symbol, and the impulse S - a shock impulse and denote it by the symbol. This leads to a new definition of impact force. The 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 participates in the formation of the shock impulse (134), which changed the speed of the car from zero to the maximum - , therefore equation (134) belongs entirely to Newtonian dynamics. Since it is much easier to determine the magnitude of velocity experimentally than it is to determine acceleration, formula (134) is very convenient for calculations.

This unusual result follows from equation (134).

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 Newtonian force. It forms the acceleration of the movement of a point or body, at which an inertial force automatically arises, directed opposite to the Newtonian force and the impact Newtonian force must overcome the action of the inertial force, therefore the inertial force must be represented in the balance of forces on the left side of equation (134). Since the inertial force is equal to the mass of the point or body multiplied by the deceleration that it forms, then equation (134) becomes

(136)

Dear mathematicians! See what form it took mathematical model, describing the shock impulse, which accelerates the movement of the impacted 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 power unit of the SShG (Fig. 120), and we will leave you with your useless equation (132). In order not to complicate the presentation, we will leave formula (134) alone for now and use formulas that give average values ​​of forces. You see in what position you put an engineer trying to solve a specific problem.

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

and the average kinetic energy was equal to

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

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

Rice. 121. Photon of the turbine 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 - accelerated rise and the second phase AB - slow rise , , .

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

. (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 rate of rise of the power unit in the first phase has the form

. (141)

Rice. 123. Regularity of changes in flight speed V of a power unit

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

. (142)

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

. (143)

Then the total time for raising the power unit to a height of 14 m will be equal to . The mass of the power unit and cover is 2580 tons. According to Newtonian 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), following from Newtonian dynamics, to determine the shock pulse that fired the 2nd power unit

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

Dear!!! Remember how much chalk was written on the blackboards by generations of your colleagues, abstrusely teaching students how to determine the shock impulse, and no one explained how to determine the duration of the shock impulse in each specific case. You will say that the duration of the shock pulse is equal to the time interval of the change in the speed of the power unit from zero to, we will 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? You, dear mathematicians, have no answer. Let's move on.

According to Newtonian dynamics, the main force that resisted the rise of the power unit was gravity. Since this force is directed against the movement of the power unit, it generates deceleration, which is equal to acceleration free fall. Then the gravitational force acting on the power unit flying upward is equal to

Newton's dynamics does not take into account other forces that prevented the action of the Newtonian force of 50,600 tons (144), and mechanodynamics states that the rise of the power unit was also resisted by an inertial force equal to

The question immediately arises: how to find the amount of deceleration in the movement of the power unit? Newtonian dynamics is silent, but mechanodynamics answers: at the moment of the action of the Newtonian force, which lifted the power unit, it was resisted by: the force of gravity and the force of inertia, therefore the equation of the forces acting on the power unit at that moment is written as follows.

Theorem on the change in momentum of a point

Since the mass of a point is constant, and its acceleration, the equation expressing the basic law of dynamics can be represented in the form

The equation simultaneously expresses the theorem about the change in the momentum of a point in differential form: time derivative of the momentum of a point is equal to the geometric sum of the forces acting on the point.

Let's integrate this equation. Let the mass point m, moving under the influence of force (Fig. 15), has at the moment t=0 speed, and at the moment t 1-speed.

Fig.15

Let us then multiply both sides of the equality by and take from them definite integrals. In this case, on the right, where integration occurs over time, the limits of the integrals will be 0 and t 1, and on the left, where the speed is integrated, the limits of the integral will be the corresponding values ​​of speed and . Since the integral of equals , then as a result we get:

.

The integrals on the right represent the impulses of the acting forces. Therefore, we will finally have:

.

The equation expresses the theorem about the change in the momentum of a point in final form: the change in the momentum of a point over a certain period of time is equal to the geometric sum of the impulses of all forces acting on the point over the same period of time ( rice. 15).

When solving problems, equations in projections are often used instead of vector equations.

In the case of rectilinear motion occurring along the axis Oh the theorem is expressed by the first of these equations.

Example 9. Find the law of motion of a material point of mass m, moving along the axis X under the influence of a force constant in modulus F(Fig. 16) at initial conditions: , at .

Fig.16

Solution. Let's create a differential equation for the motion of a point in projection onto the axis X: . Integrating this equation, we find: . The constant is determined from the initial condition for the speed and is equal to . Finally

.

Further, taking into account that v = dx/dt, we arrive at the differential equation: , integrating which we obtain

The constant is determined from the initial condition for the coordinate of the point. It is equal. Consequently, the law of motion of a point has the form

Example 10. Weight load R(Fig. 17) begins to move from rest along a smooth horizontal plane under the influence of force F = kt. Find the law of motion of the load.

Fig.17

Solution. Let us choose the origin of the coordinate system ABOUT in the initial position of the load and direct the axis X in the direction of movement (Fig. 17). Then the initial conditions have the form: x(t = 0) = 0,v( t = 0) = 0. Forces act on the load F,P and plane reaction force N. Projections of these forces onto the axis X have meanings Fx = F = kt, Rx = 0, Nx= 0, so the corresponding equation of motion can be written as follows: . Separating the variables in this differential equation and then integrating, we get: v = gkt 2 /2P + C 1. Substituting the initial data ( v(0) = 0), we find that C 1 = 0, and we get the law of speed change .

The last expression, in turn, is a differential equation, integrating which we find the law of motion of a material point: . The constant included here is determined from the second initial condition X(0) = 0. It is easy to verify that . Finally

Example 11. On a load at rest on a horizontal smooth plane (see Fig. 17) at a distance a from the origin, begins to act at positive direction axes x strength F = k 2 (P/g)x, Where R – cargo weight. Find the law of motion of the load.

Solution. Equation of motion of the load under consideration (material point) in projection onto the axis X

The initial conditions of equation (1) have the form: x(t = 0) = a, v( t = 0) = 0.

Let us represent the time derivative of speed included in equation (1) as follows:

.

Substituting this expression into equation (1) and reducing by ( P/g), we get

Separating the variables in the last equation, we find that . Integrating the latter, we have: . Using initial conditions , we get , and, therefore,

, . (2)

Since the force acts on the load in the positive direction of the axis X, then it is clear that he should move in the same direction. Therefore, in solution (2) the plus sign should be chosen. Replacing further in the second expression (2) with , we obtain a differential equation for determining the law of motion of the load. Whence, separating the variables, we have

.

Integrating the latter, we find: . After finding the constant we finally get

Example 12. Ball M masses m(Fig. 18) falls without initial speed under the influence of gravity. As the ball falls, it experiences resistance, where – constant resistance coefficient. Find the law of motion of the ball.

Fig.18

Solution. Let us introduce a coordinate system with the origin at the ball’s location point at t = 0, directing the axis at vertically down (Fig. 18). Differential equation movement of the ball in projection onto the axis at then has the form

The initial conditions for the ball are written as follows: y(t = 0) = 0, v( t = 0) = 0.

Separating the variables in equation (1)

and integrating, we find: , where . Or after finding a constant

or . (2)

It follows that the maximum speed, i.e. speed at , is equal to .

To find the law of motion, replace v in equation (2) by dy/dt. Then, integrating the resulting equation taking into account the initial condition, we finally find

.

Example 13. Research submarine of spherical shape and mass m= = 1.5×10 5 kg begins to dive with the engines turned off, having a horizontal speed v X 0 = 30 m/s and negative buoyancy R 1 = 0.01mg, Where – vector sum of the Archimedean buoyancy force Q and gravity mg, acting on the boat (Fig. 20). Water resistance force , kg/s. Determine the equations of motion of the boat and its trajectory.

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Quantity of movement

Momentum of a material point - a vector quantity equal to the product of the mass of a point and its velocity vector.

The unit of measurement for momentum is (kg m/s).

Mechanical system momentum - a vector quantity equal to the geometric sum (principal vector) of the momentum of a mechanical system is equal to the product of the mass of the entire system and the speed of its center of mass.

When a body (or system) moves so that its center of mass is stationary, then the amount of motion of the body is equal to zero (for example, rotation of the body around a fixed axis passing through the center of mass of the body).

In the case of complex motion, the amount of motion of the system will not characterize the rotational part of the motion when rotating around the center of mass. That is, the amount of motion characterizes only the translational motion of the system (together with the center of mass).

Impulse force

The impulse of a force characterizes the action of a force over a certain period of time.

Force impulse over a finite period of time is defined as the integral sum of the corresponding elementary impulses.

Theorem on the change in momentum of a material point

(in differential forms e ):

The time derivative of the momentum of a material point is equal to the geometric sum of the forces acting on the points.

(V integral form ):

The change in the momentum of a material point over a certain period of time is equal to the geometric sum of the impulses of forces applied to the point during this period of time.

Theorem on the change in momentum of a mechanical 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.

(in integral form ):

The change in the momentum of a system over a certain period of time is equal to the geometric sum of the impulses of external forces acting on the system during this period of time.

The theorem allows one to exclude obviously unknown internal forces from consideration.

The theorem on the change in momentum of a mechanical system and the theorem on the motion of the center of mass are two in different forms one theorem.

Law of conservation of momentum of a system

1. If the sum of all external forces acting on the system is equal to zero, then the vector of the system's momentum will be constant in direction and magnitude.
2. If the sum of the projections of all acting external forces onto any arbitrary axis is equal to zero, then the projection of the momentum onto this axis is a constant value.

Conclusions:

1. Conservation laws indicate that internal forces cannot change the total amount of motion of the system.
2. The theorem on the change in momentum of a mechanical system does not characterize the rotational motion of a mechanical system, but only the translational one.

An example is given: Determine the momentum of a disk of a certain mass if its angular velocity and size are known.

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