Theorem on the change in momentum of a mechanical system. Quantity of movement

§1. System momentum (system impulse)

Quantity of motion (body impulse) – vector physical quantity equal to the product of a body’s mass and its speed:

Impulse (amount of movement) is one of the most fundamental characteristics of the movement of a body or system of bodies.

Let's write II Newton's law in a different form, given that acceleration Then therefore

The product of a force and the time of its action is equal to the increment in the momentum of the body:

Where- an impulse of force, which shows that the result of the force depends not only on its value, but also on the duration of its action.

The amount of motion of the system (impulse) will be called the vector quantity , equal to the geometric sum (principal vector) of the amounts of motion (impulses) of all points of the system (Fig.2):

It is clear from the drawing that, regardless of the values of the velocities of the points of the system (unless these velocities are parallel), the vectorcan take any values and even be equal to zero when a polygon constructed from vectors, will close. Therefore, in sizeit is impossible to fully judge the nature of the system’s movement.

Fig.2. System movement quantity

§2. Theorem about the change in momentum (momentum)

Let a force act on a body of mass m for a certain short period of time Δt. Under the influence of this force, the speed of the body changes by Consequently, during the time Δt the body moved with acceleration:

From the basic law of dynamics(Newton's second law) follows:

§3. Law of conservation of momentum (law of conservation of momentum)

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

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

Then from Eq. it follows that Q = = const. Thus, if the sum of all external forces acting on a closed system is equal to zero, then the vector of momentum (momentum) of the system will be constant in magnitude and direction.

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

Then from Eq.it follows that in this caseQx= const. Thus, if the sum of the projections of all acting external forces onto any axis is equal to zero, then the projection of the amount of motion (momentum) of the system onto this axis is a constant value.

These results express law of conservation of momentum of the system: for any nature of interaction between bodies forming a closed system, the vector of the total momentum of this system remains constant all the time.

It follows from them that internal forces cannot change the total amount of motion of the system.

The law of conservation of total momentum of an isolated system is a universal law of nature. In the more general case, when the system is not closed, fromit follows that the total momentum of an open-loop system does not remain constant. Its change per unit time is equal to the geometric sum of all external forces.

Let's look at some examples:

a) The phenomenon of recoil or recoil. If we consider the rifle and the bullet as one system, then the pressure of the powder gases during a shot will be an internal force. This force cannot change the total momentum of the system. But since the powder gases, acting on the bullet, impart to it a certain amount of motion directed forward, they must simultaneously impart to the rifle the same amount of motion in the opposite direction. This will cause the rifle to move backwards, i.e. the so-called return. A similar phenomenon occurs when firing a gun (rollback).

b) Operation of the propeller (propeller). The propeller imparts movement to a certain mass of air (or water) along the axis of the propeller, throwing this mass back. If we consider the thrown mass and the aircraft (or ship) as one system, then the forces of interaction between the propeller and the environment, as internal ones, cannot change the total amount of motion of this system. Therefore, when a mass of air (water) is thrown back, the aircraft (or ship) receives a corresponding forward speed, such that the total amount of motion of the system under consideration will remain equal to zero, since it was zero before the movement began.

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

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

Self-test questions:

How is the theorem about the change in momentum of a system formulated?

Write down the mathematical expression of the theorem on the change in momentum of a mechanical system in differential and integral form.

In which case does the momentum of a mechanical system not change?

How is an impulse of variable force determined over a finite period of time? What characterizes a force impulse?

What are the projections of constant and variable force impulses on the coordinate axes?

What is the impulse of the resultant?

How does the momentum of a point moving uniformly around a circle change?

What is the momentum of a mechanical system?

What is the momentum of a flywheel rotating around a fixed axis passing through its center of gravity?

Under what conditions does the momentum of a mechanical system not change? Under what conditions does its projection onto a certain axis not change?

Why does the gun roll back when fired?

Can internal forces change the momentum of a system or the momentum of a part of it?

What factors determine the speed of free motion of a rocket?

Does the final speed of a rocket depend on the fuel combustion time?

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Mechanical system of material points or bodies is such a collection of them in which the position and movement of each point (or body) depends on the position and movement of the others.
A material body is considered as a system of material points (particles) that form this body.
By external forces are those forces that act on points or bodies of a mechanical system from points or bodies that do not belong to this system.
By internal forces, are the forces that act on points or bodies of a mechanical system from points or bodies of the same system, i.e. with which the points or bodies of a given system interact with each other.
External and internal forces of the system, in turn, can be active and reactive
System weight equals the algebraic sum of the masses of all points or bodies of the system in a uniform gravitational field, for which the weight of any particle of the body is proportional to its mass. Therefore, the distribution of masses in a body can be determined by the position of its center of gravity - the geometric point WITH, the coordinates of which are called the center of mass or center of inertia of a mechanical system
Theorem on the motion of the center of mass of a mechanical system: the center of mass of a mechanical system moves as a material point whose mass is equal to the mass of the system, and to which all external forces acting on the system are applied
Conclusions:

A mechanical system or a rigid body can be considered as a material point depending on the nature of its motion, and not on its size.
Internal forces are not taken into account by the theorem on the motion of the center of mass.
The theorem on the motion of the center of mass does not characterize the rotational motion of a mechanical system, but only the translational one

Law on the conservation of motion of the center of mass of the system:
1. If the sum of external forces (the main vector) is constantly equal to zero, then the center of mass of the mechanical system is at rest or moves uniformly and rectilinearly.
2. If the sum of the projections of all external forces onto any axis is equal to zero, then the projection of the velocity of the center of mass of the system onto the same axis is a constant value.

Theorem on the change of momentum.

The amount of motion of a material point and is a vector quantity that is 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 all points of the system. Or 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
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 that passes through the center of mass of the body).
If the motion of the body is complex, then it 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 characterizes the action of a force over a certain period of time.
The force impulse for 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 form): The derivative over time of the momentum of a material point is equal to the geometric sum of the forces acting on the points
(in integral form): The change in momentum over a certain period of time is equal to the geometric sum of the impulses of forces applied to a point over the same 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 the system over a certain period of time is equal to the geometric sum of the impulses acting on the system of external forces over the same 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 different forms of the same theorem.
Law of conservation of momentum of a system.

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.
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.

Conservation laws indicate that internal forces cannot change the total amount of motion of the system.

Classification of forces acting on a mechanical system
Properties of internal forces
System mass. Center of mass
Differential equations of motion of a mechanical system
Theorem on the motion of the center of mass of a mechanical system
Law on the conservation of motion of the center of mass of a system
Momentum change theorem
Law of conservation of momentum of a system

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The amount of movement of the system call the geometric sum of the quantities of motion of all material points of the system

To clarify the physical meaning of (70), let us calculate the derivative of (64)

. (71)

Solving (70) and (71) together, we obtain

. (72)

Thus, the vector of momentum of a mechanical system is determined by the product of the mass of the system and the speed of its center of mass.

Let's calculate the derivative of (72)

. (73)

Solving (73) and (67) together, we obtain

. (74)

Equation (74) expresses the following theorem.

Theorem: The time derivative of the system's momentum vector is equal to the geometric sum of all external forces of the system.

When solving problems, equation (74) must be projected onto the coordinate axes:

. (75)

From the analysis of (74) and (75) the following follows: law of conservation of momentum of a system: If the sum of all forces of the system is zero, then its momentum vector retains its magnitude and direction.

If
, That
,Q = const . (76)

In a particular case, this law can be fulfilled along one of the coordinate axes.

If
, That, Q z = const. (77)

It is advisable to use the theorem on the change in momentum in cases where the system includes liquid and gaseous bodies.

Theorem on the change in angular momentum of a mechanical system

The amount of motion characterizes only the translational component of motion. To characterize the rotational motion of a body, the concept of the main angular momentum of the system relative to a given center (kinetic moment) has been introduced.

Kinetic moment of the system relative to a given center is the geometric sum of the moments of the quantities of motion of all its points relative to the same center

. (78)

By projecting (22) on the coordinate axes, we can obtain an expression for the kinetic moment relative to the coordinate axes

. (79)

Kinetic moment of the body relative to the axes equal to the product of the moment of inertia of the body relative to this axis and the angular velocity of the body

. (80)

From (80) it follows that the kinetic moment characterizes only the rotational component of motion.

A characteristic of the rotational action of a force is its moment relative to the axis of rotation.

The theorem on the change in angular momentum establishes the relationship between the characteristic of rotational motion and the force causing this motion.

Theorem: The time derivative of the vector of the angular momentum of the system relative to some center is equal to the geometric sum of the moments of all external forces of the system relative tothe same center

. (81)

When solving engineering problems (81), it is necessary to design on the coordinate axes

Their analysis of (81) and (82) implies law of conservation of angular momentum: If the sum of the moments of all external forces relative to the center (or axis) is equal to zero, then the kinetic moment of the system relative to this center (or axis) retains its magnitude and direction.

The kinetic moment cannot be changed by the action of the internal forces of the system, but due to these forces it is possible to change the moment of inertia, and therefore the angular velocity.

In the same way as for one material point, we will derive a theorem on the change in momentum for the system in various forms.

Let's transform the equation (theorem on the movement of the center of mass of a mechanical system)

in the following way:

;

The resulting equation expresses the theorem about the change in the momentum of a mechanical system in differential form: the derivative of the momentum of a mechanical system with respect to time is equal to the main vector of external forces acting on the system .

In projections onto Cartesian coordinate axes:

; ; .

Taking the integrals of both sides of the last equations over time, we obtain a theorem about the change in the momentum of a mechanical system in integral form: the change in the momentum of a mechanical system is equal to the momentum of the main vector of external forces acting on the system .

Or in projections onto Cartesian coordinate axes:

; ; .

Corollaries from the theorem (laws of conservation of momentum)

The law of conservation of momentum is obtained as special cases of the theorem on the change in momentum for a system depending on the characteristics of the system of external forces. Internal forces can be any, since they do not affect changes in momentum.

There are two possible cases:

1. If the vector sum of all external forces applied to the system is equal to zero, then the amount of motion of the system is constant in magnitude and direction

2. If the projection of the main vector of external forces onto any coordinate axis and/or and/or is equal to zero, then the projection of the momentum on these same axes is a constant value, i.e. and/or and/or respectively.

Similar entries can be made for a material point and for a material point.

The task. From a gun whose mass M, a projectile of mass flies out in a horizontal direction m with speed v. Find speed V guns after firing.

Solution. All external forces acting on the mechanical weapon-projectile system are vertical. This means, based on the corollary to the theorem on the change in the momentum of the system, we have: .

The amount of movement of the mechanical system before firing:

The amount of movement of the mechanical system after the shot:

Equating the right-hand sides of the expressions, we obtain that

The “-” sign in the resulting formula indicates that after firing the gun will roll back in the direction opposite to the axis Ox.

EXAMPLE 2. A stream of liquid with density flows at a speed V from a pipe with cross-sectional area F and hits a vertical wall at an angle. Determine the fluid pressure on the wall.

SOLUTION. Let us apply the theorem on the change in momentum in integral form to a volume of liquid with a mass m hitting a wall over a period of time t.

MESHCHERSKY EQUATION

(basic equation of the dynamics of a body of variable mass)

In modern technology, cases arise when the mass of a point and a system does not remain constant during movement, but changes. So, for example, during the flight of space rockets, due to the ejection of combustion products and individual unnecessary parts of the rockets, the change in mass reaches 90-95% of the total initial value. But not only space technology can be an example of the dynamics of variable mass motion. In the textile industry, there are significant changes in the mass of various spindles, bobbins, rolls at modern operating speeds of machines and machines.

Let us consider the main features associated with changes in mass, using the example of the translational motion of a body of variable mass. The basic law of dynamics cannot be directly applied to a body of variable mass. Therefore, we obtain differential equations of motion of a point of variable mass, applying the theorem on the change in the momentum of the system.

Let the point have mass m+dm moves at speed. Then a certain particle with a mass is separated from the point dm moving at speed.

The amount of motion of the body before the particle comes off:

The amount of motion of a system consisting of a body and a detached particle after its separation:

Then the change in momentum:

Based on the theorem about the change in momentum of the system:

Let us denote the quantity - the relative velocity of the particle:

Let's denote

Size R called reactive force. Reactive force is the engine thrust caused by the ejection of gas from the nozzle.

Finally we get

This formula expresses the basic equation of the dynamics of a body of variable mass (Meshchersky formula). From the last formula it follows that the differential equations of motion of a point of variable mass have the same form as for a point of constant mass, except for the additional reactive force applied to the point due to the change in mass.

The basic equation for the dynamics of a body of variable mass indicates that the acceleration of this body is formed not only due to external forces, but also due to the reactive force.

Reactive force is a force similar to that felt by the person shooting - when shooting from a pistol, it is felt by the hand; When shooting from a rifle, it is perceived by the shoulder.

Tsiolkovsky's first formula (for a single-stage rocket)

Let a point of variable mass or a rocket move in a straight line under the influence of only one reactive force. Since for many modern jet engines , where is the maximum reactive force allowed by the engine design (engine thrust); - the force of gravity acting on the engine located on the earth's surface. Those. the above allows us to neglect the component in the Meshchersky equation and accept this equation in the form for further analysis: ,

Let's denote:

Fuel reserve (for liquid jet engines - the dry mass of the rocket (its remaining mass after burning out all the fuel);

The mass of particles separated from the rocket; is considered as a variable value, varying from to .

Let us write the equation of rectilinear motion of a point of variable mass in the following form:

Since the formula for determining the variable mass of a rocket is

Therefore, the equations of motion of a point Taking the integrals of both sides we get

Where - characteristic speed- this is the speed that a rocket acquires under the influence of thrust after all particles have been erupted from the rocket (for liquid jet engines - after all the fuel has burned out).

Placed outside the integral sign (which can be done on the basis of the mean value theorem known from higher mathematics) is the average speed of particles ejected from the rocket.

and mechanical system

The momentum of a material point is a vector measure of mechanical motion, equal to the product of the mass of the point and its speed, . The unit of measurement of momentum in the SI system is
. The amount of motion of a mechanical system is equal to the sum of the amounts of motion of all material points forming the system:

. (5.2)

Let's transform the resulting formula

According to formula (4.2)
, That's why

Thus, the momentum of a mechanical system is equal to the product of its mass and the speed of the center of mass:

. (5.3)

Since the amount of motion of a system is determined by the motion of only one of its points (the center of mass), it cannot be a complete characteristic of the motion of the system. Indeed, for any movement of the system, when its center of mass remains stationary, the momentum of the system is zero. For example, this occurs when a rigid body rotates around a fixed axis passing through its center of mass.

Let's introduce a reference system Cxyz, having its origin at the center of mass of the mechanical system WITH and moving translationally relative to the inertial system
(Fig. 5.1). Then the movement of each point
can be considered as complex: portable movement together with axes Cxyz and movement relative to these axes. Due to the progressive movement of the axes Cxyz the portable speed of each point is equal to the speed of the center of mass of the system, and the amount of motion of the system, determined by formula (5.3), characterizes only its translational portable motion.

5.3. Impulse force

To characterize the action of a force over a certain period of time, a quantity called impulse of force . An elementary impulse of a force is a vector measure of the action of a force, equal to the product of the force by the elementary time interval of its action:

. (5.4)

The SI unit of force impulse is
, i.e. The dimensions of force impulse and momentum are the same.

Force impulse over a finite period of time
is equal to a certain integral of the elementary momentum:

. (5.5)

The impulse of a constant force is equal to the product of the force and the time of its action:

. (5.6)

In general, the force impulse can be determined by its projections onto the coordinate axes:

. (5.7)

5.4. Momentum change theorem

material point

In the basic equation of dynamics (1.2), the mass of a material point is a constant quantity, its acceleration
, which makes it possible to write this equation in the form:

. (5.8)

The resulting relationship allows us to formulate theorem on the change in momentum of a material point in differential form: The time derivative of the momentum of a material point is equal to the geometric sum (principal vector) of the forces acting on the point.

Now we obtain the integral form of this theorem. From relation (5.8) it follows that

Let's integrate both sides of the equality within the limits corresponding to the moments of time And ,

. (5.9)

The integrals on the right side represent the impulses of the forces acting on the point, so after integrating the left side we get

. (5.10)

Thus it is proven theorem on the change in momentum of a material point in 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 the forces acting on the point over the same period of time.

The vector equation (5.10) corresponds to a system of three equations in projections onto the coordinate axes:

;

; (5.11)

Example 1. The body moves translationally along an inclined plane forming an angle α with the horizon. At the initial moment of time it had a speed , directed upward along an inclined plane (Fig. 5.2).

After what time does the speed of the body become equal to zero if the coefficient of friction is equal to f ?

Let us take a translationally moving body as a material point and consider the forces acting on it. It's gravity
, normal plane reaction and friction force . Let's direct the axis x along the inclined plane upward and write the 1st equation of the system (5.11)

where are the projections of quantities of motion, and are the projections of impulses of constant forces
,And are equal to the products of the projections of forces and the time of movement:

Since the acceleration of the body is directed along the inclined plane, the sum of the projections onto the axis y of all forces acting on the body is equal to zero:
, from which it follows that
. Let's find the friction force

and from equation (5.12) we obtain

from where we determine the time of movement of the body