Formulate the law of conservation of angular momentum. §2

The laws of conservation of kinetic energy and momentum competed with each other for a long time, claiming a leading role, since neither one nor the other law has a strict justification. However, scientists have long suspected the existence of a connection between them, as H. Huygens (1629-1695) spoke about. According to Huygens, this connection means that the conservation of mechanical energy in any uniformly moving system entails the conservation of momentum. Therefore, after lengthy debate, scientists have come to the conclusion that these laws are equivalent. So, for example, d’Alembert made the following statement on this matter: “Everyone must be given the freedom to resolve this issue at his own discretion. Moreover, the question raised is nothing more than a completely fruitless metaphysical dispute about words, unworthy of the attention of philosophers.”
The connection between the laws of conservation of kinetic energy and momentum was established by W. Pauli (1900-1958). To prove this connection he uses Huygens' idea. We quote from: “In a system consisting of colliding particles with masses, the velocities of the particles change after impacts to velocities. Conservation of energy is expressed by the equation:

Let the system gain additional speed V. The particle velocities before the impact will now be equal to , and after the impact, and the conservation of energy is now expressed by the relation:
,

Hence:

Speed V- is arbitrary, therefore the written equality will be valid only if:

In other words, the momentum of the system before the collision of particles, equal to the expression on the left, is conserved after the collision.”
We will also consider this issue in view of its special importance using the example of the collision of balls, but in a slightly different interpretation (Fig. 1).
Let the balls move in an arbitrary inertial frame of reference x-y in the same direction (Fig. 1, a) with speeds and . After the impact, the velocities of the balls will take on the values ​​and . In accordance with the law of conservation of energy, the following expression will be valid:
, (1)

Now consider the relative motion, taking one of the balls as a frame of reference. To do this, we use the principle of motion reversal, that is, we give both balls the same speed, for example, which will lead to the first ball stopping, since its total speed will be zero. The speed of the second ball will be equal to the relative speed:
(2)
The law of conservation of kinetic energy in this case will take the form:
(3)

(4)
Solving equations (1) and (4) together, we obtain the expression:
, (5)

(7)
Thus, an interesting result is obtained: the law of conservation of momentum follows from the law of conservation of energy. It should also be noted that the result obtained does not depend on the choice of reference system.
If we consider the counter-movement of the balls (Fig. 1, b), then to obtain the correct result, the speed should be subtracted from the speed, that is, the relative speed should be found in accordance with expression (2), although, as can be seen from the figure, these speeds should be added . This circumstance is due to the fact that the speeds of movement of all bodies are vectors, which means that even when subtracting their values ​​they can be summed up.
Thus, expressions (2), (5) and (7) should be considered as vector ones.
Solving expressions (1) and (5) together, as well as (3) and (7), we find the velocities of the balls after impact, considering them as vectors:
; (8)
; (9)
; (10)
(11)
Using these expressions, we find the relative velocities of the balls after impact:
; (12)
(13)
Thus, during an elastic impact, the relative velocities of the balls will only change their direction.
Expression (1), characterizing the law of conservation of energy, can be presented in another form:
(14)

; (15)
, (16)

; (17)
, (18)

• whence it follows that the energy acquired by the first ball is equal to the energy given by the second ball.

Substituting the values ​​of speeds and into expressions (7) and (8), we obtain:
; (19)
(20)
Let us now see how the connection between the laws of conservation of energy and momentum will be fulfilled for a more complex case of impact - an oblique impact, when the velocities of moving balls are directed at an angle to each other (Fig. 2). In the figure, the balls are separated to better show their velocity patterns. We assume that the speed coincides with the direction of the axis x.
To solve the problem, we use the method of motion reversal, giving both balls a speed , that is, as a frame of reference in relative motion, we select the first ball, the total speed of which will be equal to zero. Let us also assume, to simplify the problem, that the resulting speed will be directed along the line connecting the centers of the balls. Then, using the known values ​​of the velocities for the second ball, a parallelogram is constructed, with the help of which a connection is established between these velocities and the speed in relative motion, and the angle can also be found, since the angle is given.
Using a parallelogram, using the cosine theorem we obtain the expression:
(21)

• which we transform to the form:

(22)
From this equation we find the speed in relative motion before the start of the impact -:
(23)
The angle characterizing the direction of the vector is found from the expression obtained using the cosine theorem:
, (24)

• from where we get:

(25)
Thus, as a result of the operations performed, we obtain the usual collision of a moving and stationary ball in the direction of the line of their centers with an initial relative speed .
Before determining the velocities of the balls after their collision, let us establish a connection between the kinetic energies of the balls in absolute and relative motion:
; (26)
(27)
Because
(28)

• Accordingly, other speeds in relative motion will be determined:

; (29)
(30)
Substituting these values ​​of relative velocities into expression (27), we obtain:
(31)
Reducing by two and squaring the speed difference, we transform expression (31) to the form:
, (32)

By adding to the first term on the right side of the expression, you can eliminate the terms corresponding to expression (26), as a result of which expression (32) will take the form:
(33)
Reducing this expression by and grouping the terms, we get:
(34)
Having determined the speeds , and in accordance with expressions (28) – (32):
(35)

• and substituting them into expression (34), we transform it to the form:

(36)
Thus, we have established a connection between the laws of conservation of energy and momentum in the absolute and relative motion of balls during an oblique impact.
Solving equations (27) and (36) together, we find the velocities of the balls in their relative motion:
; (37)
, (38)

When solving equations to obtain a solution in vector form, the squares of the velocities should be represented as the scalar product of two identical vectors.
The velocities of the balls in absolute motion can be found using the cosine theorem from parallelograms presented in Fig. 2.
For the first ball, the velocity module is determined by the expression:
, (39)

• from where we get:

(40)
For the second ball, the velocity module will be equal to:
, (41)

• where can we find it:

(42)
The angles and , characterizing the directions of the vectors and with respect to the vectors and , are also found using the cosine theorem:
; (43)
(44)
Substituting the values ​​of velocities and from formulas (39) and (41) into these expressions, we obtain:
; (45)
(46)
To check the solutions obtained, you can find the values ​​of the kinetic energy of the balls after the impact, since before the impact their energy was equal to:
, (47)

• and after the hit it will be:

(48)
Substituting the values ​​of the squared velocities into expression (48) and from expressions (39) and (41), we obtain:
(49)
Now we use the values ​​of the velocity modules and from expressions (37) and (38):
(50)
Substituting the value of the velocity modulus into this expression in accordance with formula (23) and making transformations, we ultimately obtain that , that is, the law of conservation of energy will be fulfilled.
Let us now consider the inelastic collision of two balls. In this case, part of the energy will be spent on structural changes (inelastic deformations in the balls) and on their heating, that is, a change in internal energy. Therefore, the expressions of the laws of conservation of energy in two reference systems will take the form:
; (51)
(52)

By solving this system of equations together, we obtain the law of conservation of momentum in its usual form:
, (53)

• that is, energy losses during the interaction of bodies do not affect the form of this law.

Using equations (51) and (53), we find the velocities of the balls after their inelastic collision:
; (54)
(55)
Obviously, expressions (54) and (55) will have a physical meaning only if the radical expression has a positive value. From this condition, you can find the value at which the law of conservation of momentum will still be satisfied by equating the radical expression to zero:
(56)

, (57)

(58)
Expressions (54) and (56), taking into account formula (57), can be represented as:
; (59)
, (60)

(61)
In relative motion, the expressions for velocities will take the form:
; (62)
(63)
From the above expressions it follows that the speeds of the balls will be equal and they will move together as one.
If the coefficient is greater than one, then the radical expression will be negative and the expressions for velocities will lose their physical meaning. Since at , the balls will move as one unit, one equation is sufficient to determine the speed of their movement. When you can still use the law of conservation of momentum, when you should use only the law of conservation of energy, although in mathematical terms the law of conservation of momentum will be satisfied in this case. Thus, the law of conservation of momentum has limits to its use. This once again confirms the priority role of the law of conservation of energy in relation to the law of conservation of momentum. However, in principle, it is possible that the values ​​of the coefficient cannot be greater than one, then both laws will always be valid, but this statement requires experimental verification.
Since the balls will move as a single whole with the same speed, the law of conservation of energy will take the form:
, (64)

• where, in accordance with expression (61),

(65)
Solving equation (64), we obtain:
(66)

• or in relative motion:

(67)
If all the impact energy is spent on losses, that is, when the relation is satisfied:
, (68)

(69)
True, doubts remain as to whether such a case is actually possible.
In §5 of the first chapter, it was shown that the amount of motion characterizes the inertia of a body and is determined by the ratio, that is, the ratio of the change in the kinetic energy of the body and the change in its speed. In connection with this definition of the inertia of a body, another conclusion can be given to the law of conservation of momentum. To do this, we use expressions (15), (17) and (18), dividing them by the change in the speed of the first body: :
(70)
Let us transform the resulting expression to the form:
(71)
Using the speed ratio (12) in the form:
, (72)

• Let us transform expression (71) to the form:

(73)

• whence follows the law of conservation of momentum:

The laws of conservation of energy and momentum are widely used in solving various problems of mechanics. However, in view of the fact that these laws are integral, since they take into account the states of bodies only before and after their interaction, but not at the moment of the interaction itself, there is a danger of losing the physical meaning of the interaction itself, avoiding the explanation of this physical meaning due to the lack of it understanding, although the end result will be correct.
Let us prove this statement using the example of the movement of a boat when a person in it throws a stone into the water (Fig. 3). There is no doubt that the boat will move in the direction opposite to the throw. To solve the problem, the law of conservation of momentum is used, which, taking into account the direction of velocities, will have the form:
, (74)

, (75)

• that is, the greater the mass of the stone and its speed, the greater the speed of the boat.

If you ask mechanics teachers what reason makes a boat move, most of them will answer that the boat will move because the law of conservation of momentum must be satisfied. They give such an answer because they cannot explain the actual cause of movement, although they know very well that movement can only occur under the influence of force. So what force will make the boat move?
Obviously, here we need to understand the interaction between the human hands and the stone at the moment of throwing. The only reason for the appearance of force acting on a person, and through him on the boat, is the impact from the stone. This force will appear if the stone moves accelerated at the moment of the throw. Then it will deform and elastic forces will arise in it, which will act on the person’s hands. These forces, as we already know, are forces of inertia and their magnitude will be equal to the product of the mass of the stone and its acceleration. You can also say that a person is pushing away from a stone. However, solving this problem using Newton's second law is almost impossible, since we will not be able to find the acceleration of the stone at the moment of the throw. The speed of its movement in the first moments of movement is much easier to find. So the use of integral laws of motion significantly simplifies the solution of many problems in mechanics. True, one should not forget about the physical essence of the phenomena under consideration. In this case, the mathematical power of the integral conservation laws will be revealed even more clearly.
Now let's consider a more complex problem about the movement of a cart on which two loads are located, rotating in different directions with the same angular velocity (Fig. 4). This problem is also solved using the law of conservation of momentum:
, (76)

From expression (76) it follows:
, (77)

• that is, the cart will perform harmonic oscillations. But what is the reason for these fluctuations? It cannot be said that the cart obeys the law of conservation of momentum. A force must make the cart oscillate, but what kind of force? The only candidate for this role can only be the centrifugal force of inertia acting on rotating loads:

(78)
Under the influence of two inertia forces, the cart will move along the axis y. The nature of the cart's motion can be found using Newton's second law:
(79)
The speed of the cart is determined by integrating this expression:
, (80)

• Where WITH– integration constant.

To determine the speed of the cart, it is necessary to use initial conditions. However, a problem arises here: what will the speed of the cart be equal to? Let us assume that at the initial moment of time the unsecured cart and the loads were stationary, and then the loads were immediately set into rotation at a constant angular velocity, that is, there will be no transitional mode of motion. Thus, the magnitude of the inertia forces will immediately take on the final value determined by expression (78). Under the influence of inertial forces, the cart would have to move immediately in a positive direction. However, it must be borne in mind that with the instantaneous appearance of the speed of movement of the loads, a theoretically infinite, but practically very large acceleration in the direction of the axis will appear y, if the loads were located along the axis x, and the corresponding inertial force in the opposite direction, which will make the cart move in the direction of its action in the negative direction of the axis y, that is, there will actually be an impact on the cart.
Let us assume that the initial speed of the cart will be equal to , then from equation (80) we obtain:
,

• where do we find the constant of integration WITH:

(81)
In accordance with this, the speed of the cart will be:
(82)
By integrating this expression, we find the displacement of the cart along the axis y:
(83)
Under the given conditions, the motion of the cart will be harmonic, so the expression in parentheses must equal zero. Then the law of motion of the cart will take the form:
, (84)

(85)
Then the speed of the trolley as a function of the angle of rotation will be determined from expression (80):
,

• which corresponds to expression (77).

However, a second solution to this problem is also possible, if we assume that at first the cart is fixed and the loads rotate at a constant speed. Then, when the loads take a position along the axis x, the trolley is released. Under such conditions, the inertial forces in the direction of the axis y will be absent, since the value of the speed of rotation of the loads will not change, therefore there will be no impact on the cart in the negative direction of the axis y and its initial speed will be zero. Then from equation (80) it follows that the integration constant WITH will be equal to:
, (86)

• therefore, the speed of the cart as a function of time will have the form:

(87)
Integrating this expression over time, we find the movement of the cart along the y-axis:
(88)

, (89)

; (90)
(91)
Thus, the periodically changing projection of the inertia forces of the loads onto the axis y makes the cart perform harmonic oscillations and even move along the axis y depending on the initial driving conditions. An unsecured cart will perform only harmonic oscillations, while a cart that is fixed and then released will perform a rectilinear motion, on which harmonic oscillations will be superimposed.
The analysis we carried out would have been impossible without taking into account the forces acting on the cart, which in this case are the inertial forces. If the movement of the cart is explained by the need to fulfill the law of conservation of momentum, then this means saying nothing on the merits of the matter. Therefore, it is advisable to combine the use of conservation laws with a detailed force analysis of the problem under consideration.

From the theorem on the change in momentum of a 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 from equation (20) it follows that in this case Thus, 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 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, ) is equal to zero:

Then from equations (20) it follows that in this case Thus, if the sum of the projections of all acting external forces onto any axis is equal to zero, then the projection of the momentum of the system onto this axis is a constant value.

These results express the law of conservation of momentum of the system. It follows from them that internal forces cannot change the amount of motion of the system. Let's look at some examples.

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 amount of motion of the system, equal to the shot of the slug. 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, known as recoil. A similar phenomenon occurs when firing a gun (rollback).

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 remains equal to zero, since it was zero before the movement began.

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

Jet propulsion. In a rocket (rocket), gaseous combustion products of fuel are ejected at high speed from an opening in the tail of the rocket (from the rocket engine nozzle). The pressure forces acting in this case will be internal forces and cannot change the momentum of the rocket system - fuel combustion products. But since the escaping gases have a certain amount of motion directed backward, the rocket receives a corresponding speed directed forward. The magnitude of this speed will be determined in § 114.

Please note that a propeller engine (previous example) imparts movement to an object, such as an airplane, by throwing back particles of the medium in which it is moving. In airless space such movement is impossible. A jet engine imparts motion by throwing back the masses generated in the engine itself (combustion products). This movement is equally possible both in the air and in airless space.

When solving problems, the application of the theorem allows us to exclude all internal forces from consideration. Therefore, we must try to choose the system under consideration in such a way that all (or part of) the previously unknown forces are made internal.

The law of conservation of momentum is convenient to apply in cases where, by changing the translational speed of one part of the system, it is necessary to determine the speed of another part. In particular, this law is widely used in impact theory.

Problem 126. A bullet of mass , flying horizontally with a speed and, hits a box of sand mounted on a trolley (Fig. 289). At what speed will the cart begin to move after the impact, if the mass of the cart together with the box is equal to

Solution. We will consider the bullet and the cart as one system. This will allow us to eliminate the forces that arise when the bullet hits the box when solving the problem. The sum of the projections of external forces applied to the system onto the horizontal axis Ox equals zero. Therefore, or where is the amount of motion of the system before impact; - after the blow.

Since the cart is motionless before the impact, then .

After the impact, the cart and the bullet move with a common speed, which we denote by v. Then .

Equating the right-hand sides of the expressions, we find

Problem 127. Determine the free recoil speed of the gun if the weight of the recoil parts is equal to P, the weight of the projectile is , and the speed of the projectile relative to the barrel is equal to at the moment of departure.

Solution. To eliminate unknown pressure forces of powder gases, consider the projectile and the recoil parts as one system.

Let us consider the action on each other of two isolated bodies that do not interact with other bodies. We will assume that the forces are constant throughout the interaction. In accordance with the second law of dynamics, the change in momentum of the first body is:

where is the interaction time interval.

Change in momentum of the second body:

where is the force acting from the first body on the second.

According to Newton's third law

and besides, obviously

Hence,

Regardless of the nature of the interaction forces and the duration of their action, the total momentum of two isolated bodies remains constant.

The result obtained can be extended to any number of interacting bodies and to forces that change over time. To do this, we divide the time interval during which the interaction of bodies occurs into such small intervals during each of which the force can be considered constant with a given degree of accuracy. During each period of time, relation (1.8) will be satisfied. Therefore, it will be valid for the entire time interval

To generalize the conclusion to interacting bodies, we introduce the concept of a closed system.

Closed is a system of bodies for which the resultant external forces are equal to zero.

Let the masses of material points form a closed system. The change in the momentum of each of these points as a result of its interaction with all other points of the system, respectively:

Let us denote the internal forces acting on a point by mass from other points, by the point by mass, etc. (The first index indicates the point on which the force acts; the second index indicates the point on the axis of which the force acts.)

Let us write in the accepted notation the second law of dynamics for each point separately:

The number of equations is equal to the number of bodies in the system. To find the total change in the momentum of the system, you need to calculate the geometric sum of the changes in the momentum of all points of the system. Having summed up equalities (1.9), we obtain on the left side the complete vector of changes in the momentum of the system over time, and on the right side - the elementary impulse of the resultant of all forces acting in the system. But since the system is closed, the resulting forces are zero. In fact, according to the third law of dynamics, each force in equalities (1.9) corresponds to a force and

i.e. etc.,

and the resultant of these forces is zero. Consequently, in the entire closed system the change in momentum is zero:

the total momentum of a closed system is a constant quantity throughout the entire movement (the law of conservation of momentum).

The law of conservation of momentum is one of the fundamental laws of physics, valid both for systems of macroscopic bodies and for systems formed by microscopic bodies: molecules, atoms, etc.

If external forces act on the points of the system, then the amount of motion possessed by the system changes.

Let us write equations (1.9), including in them the resultant external forces acting respectively on the first, second, etc. Up to the th point:

Adding the left and right sides of the equations, we get: on the left - the complete vector of changes in the momentum of the system; on the right - the impulse of the resulting external forces:

or, denoting the resultant external forces:

the change in the total momentum of a system of bodies is equal to the impulse of the resulting external forces.

Equality (1.13) can be written in another form:

the time derivative of the total amount of motion of a system of points is equal to the resultant external forces acting on the points of the system.

Projecting the vectors of momentum of the system and external forces onto three mutually perpendicular axes, instead of vector equality (6.14), we obtain three scalar equations of the form:

If along any axis, say, the component of the resultant external forces is equal to zero, then the amount of motion along this axis does not change, i.e., being generally open, in the direction the system can be considered as closed.

We examined the transfer of mechanical motion from one body to another without its transition to other forms of motion of matter.

The quantity “mv turns out to be a measure of simply transferred, i.e., ongoing, movement...”.

Application of the law of change in momentum to the problem of the motion of a system of bodies allows us to exclude all internal forces from consideration, which simplifies theoretical research and solving practical problems.

1. Let a person stand motionless on a stationary cart (Fig. 2.a). The momentum of the man-cart system is zero. Is this system closed? It is acted upon by external forces - gravity and friction between the wheels of the cart and the floor. Generally speaking, the system is not closed. However, by placing the cart on the rails and treating the surface of the rails and wheels accordingly, i.e., significantly reducing the friction between them, the friction force can be neglected.

The force of gravity, directed vertically downward, is balanced by the reaction of the deformed rails, and the resultant of these forces cannot impart horizontal acceleration to the system, i.e., cannot change the speed, and therefore the momentum of the system. Thus, we can, with a certain degree of approximation, consider this system to be closed.

Let us now assume that a person leaves the cart to the left (Fig. 2.b), having speed. To acquire this speed, a person must, by contracting his muscles, act with his feet on the platform of the cart and deform it. The force acting from the side of the deformed platform on the person’s feet imparts acceleration to the human body to the left, and the force acting from the side of the deformed feet of the person (in accordance with the third law of dynamics) imparts acceleration to the cart to the right. As a result, when the interaction stops (the person gets off the cart), the cart gains some speed.

To find velocities using the basic laws of dynamics, it would be necessary to know how the forces of interaction between a person and a cart change over time and where these forces are applied. The law of conservation of momentum allows you to immediately find the ratio of the speeds of a person and a cart, as well as indicate their mutual direction, if the values ​​of the masses of a person and a cart are known.

While the person stands motionless on the cart, the total amount of motion of the system remains equal to zero:

The speeds acquired by a person and a cart are inversely proportional to their masses. The minus sign indicates their opposite direction.

2. If a person, moving at speed, runs onto a stationary cart and stops on it, then the cart begins to move, so that the total amount of motion of it and the person turns out to be equal to the amount of motion that the person alone had before:

3. A person moving at speed runs onto a cart moving towards him at speed and stops on it. Next, the man-cart system moves with a common speed. The total amount of motion of the person and the cart is equal to the sum of the amounts of motion that they each possessed separately:

4. Using the fact that the cart can only move along the rails, we can demonstrate the vector nature of the change in momentum. If a person enters and stops on a previously stationary cart once along the direction of its possible movement, the second time - at an angle of 45°, and the third time - at an angle of 90° to this direction, then in the second case the speed acquired by the cart is approximately one and a half times less, than in the first, and in the third case the cart is motionless.

Let's consider the most general laws of conservation, which govern the entire material world and which introduce a number of fundamental concepts into physics: energy, momentum (momentum), angular momentum, charge.

Law of conservation of momentum

As is known, the quantity of motion, or impulse, is the product of speed and the mass of a moving body: p = mv This physical quantity allows you to find the change in the motion of a body over a certain period of time. To solve this problem, one would have to apply Newton's second law countless times, at all intermediate moments of time. The law of conservation of momentum (momentum) can be obtained using Newton's second and third laws. If we consider two (or more) material points (bodies) interacting with each other and forming a system isolated from the action of external forces, then during the movement the impulses of each point (body) can change, but the total impulse of the system must remain unchanged:

m 1 v+m 1 v 2 = const.

Interacting bodies exchange impulses while maintaining the total impulse.

In the general case we get:

where P Σ is the total, total impulse of the system, m i v i– impulses of individual interacting parts of the system. Let us formulate the law of conservation of momentum:

If the sum of external forces is zero, the momentum of the system of bodies remains constant during any processes occurring in it.

An example of the operation of the law of conservation of momentum can be considered in the process of interaction of a boat with a person, which has buried its nose on the shore, and the person in the boat quickly walks from stern to bow at a speed v 1 . In this case, the boat will move away from the shore at a speed v 2 :

A similar example can be given with a projectile that exploded in the air into several parts. The vector sum of the impulses of all fragments is equal to the impulse of the projectile before the explosion.

Law of conservation of angular momentum

It is convenient to characterize the rotation of rigid bodies by a physical quantity called angular momentum.

When a rigid body rotates around a fixed axis, each individual particle of the body moves in a circle with a radius r i at some linear speed v i. Speed v i and momentum p = m i v i perpendicular to the radius r i. Product of impulse p = m i v i per radius r i is called the angular momentum of the particle:

L i= m i v i r i= P i r i·

Whole body angular momentum:

If we replace the linear velocity with the angular velocity (v i = ωr i), then

where J = mr 2 – moment of inertia.

The angular momentum of a closed system does not change over time, that is L= const and Jω = const.

In this case, the angular momentum of individual particles of a rotating body can change as desired, but the total angular momentum (the sum of the angular momentum of individual parts of the body) remains constant. The law of conservation of angular momentum can be demonstrated by observing a skater spinning on skates with his arms extended to the sides and with his arms raised above his head. Since Jω = const, then in the second case the moment of inertia J decreases, which means that the angular velocity u must increase, since Jω = const.

Law of energy conservation

Energy is a universal measure of various forms of movement and interaction. The energy given by one body to another is always equal to the energy received by the other body. To quantify the process of energy exchange between interacting bodies, mechanics introduces the concept of the work of a force that causes movement.

The kinetic energy of a mechanical system is the energy of mechanical motion of this system. The force causing the movement of a body does work, and the energy of a moving body increases by the amount of work expended. As is known, a body of mass m, moving at speed v, has kinetic energy E=mv 2 /2.

Potential energy is the mechanical energy of a system of bodies that interact through force fields, for example through gravitational forces. The work done by these forces when moving a body from one position to another does not depend on the trajectory of movement, but depends only on the initial and final position of the body in the force field.

Such force fields are called potential, and the forces acting in them are called conservative. Gravitational forces are conservative forces, and the potential energy of a body of mass m, raised to a height h above the Earth's surface is equal to

E sweat = mgh,

Where g- acceleration of gravity.

Total mechanical energy is equal to the sum of kinetic and potential energy:

E= E kin + E sweat

Law of conservation of mechanical energy(1686, Leibniz) states that in a system of bodies between which only conservative forces act, the total mechanical energy remains unchanged in time. In this case, transformations of kinetic energy into potential energy and vice versa can occur in equivalent quantities.

There is another type of system in which mechanical energy can be reduced by conversion into other forms of energy. For example, when a system moves with friction, part of the mechanical energy is reduced due to friction. Such systems are called dissipative, that is, systems that dissipate mechanical energy. In such systems, the law of conservation of total mechanical energy is not valid. However, when mechanical energy decreases, an amount of energy of a different type always appears equivalent to this decrease. Thus, energy never disappears or reappears, it only changes from one type to another. Here the property of indestructibility of matter and its movement is manifested.

In classical mechanics, there are two conservation laws: the law of conservation of momentum and the law of conservation of energy.

Body impulse

The concept of momentum was first introduced by a French mathematician, physicist, and mechanic. and the philosopher Descartes, who called impulse amount of movement .

From Latin, “impulse” is translated as “push, move.”

Any body that moves has momentum.

Let's imagine a cart standing still. Its momentum is zero. But as soon as the cart starts moving, its momentum will no longer be zero. It will begin to change as the speed changes.

Momentum of a material point, or amount of movement – a vector quantity equal to the product of the mass of a point and its speed. The direction of the point's momentum vector coincides with the direction of the velocity vector.

If we are talking about a solid physical body, then the momentum of such a body is called the product of the mass of this body and the speed of the center of mass.

How to calculate the momentum of a body? One can imagine that a body consists of many material points, or a system of material points.

If - the impulse of one material point, then the impulse of a system of material points

That is, momentum of a system of material points is the vector sum of the momenta of all material points included in the system. It is equal to the product of the masses of these points and their speed.

The unit of impulse in the international system of units SI is kilogram-meter per second (kg m/sec).

Impulse force

In mechanics, there is a close connection between the momentum of a body and force. These two quantities are connected by a quantity called impulse of force .

If a constant force acts on a bodyF over a period of time t , then according to Newton's second law

This formula shows the relationship between the force that acts on a body, the time of action of this force and the change in the speed of the body.

The quantity equal to the product of the force acting on a body and the time during which it acts is called impulse of force .

As we see from the equation, the impulse of force is equal to the difference between the impulses of the body at the initial and final moments of time, or the change in impulse over some time.

Newton's second law in momentum form is formulated as follows: the change in the momentum of a body is equal to the momentum of the force acting on it. It must be said that Newton himself originally formulated his law in exactly this way.

Force impulse is also a vector quantity.

The law of conservation of momentum follows from Newton's third law.

It must be remembered that this law operates only in a closed, or isolated, physical system. A closed system is a system in which bodies interact only with each other and do not interact with external bodies.

Let us imagine a closed system of two physical bodies. The forces of interaction of bodies with each other are called internal forces.

The force impulse for the first body is equal to

According to Newton's third law, the forces that act on bodies during their interaction are equal in magnitude and opposite in direction.

Therefore, for the second body the momentum of the force is equal to

By simple calculations we obtain a mathematical expression for the law of conservation of momentum:

Where m 1 And m 2 – body masses,

v 1 And v 2 – velocities of the first and second bodies before interaction,

v 1" And v 2" – velocities of the first and second bodies after interaction .

p 1 = m 1 · v 1 - momentum of the first body before interaction;

p 2 = m 2 · v 2 - momentum of the second body before interaction;

p 1 "= m 1 · v 1" - momentum of the first body after interaction;

p 2 "= m 2 · v 2" - momentum of the second body after interaction;

That is

p 1 + p 2 = p 1" + p 2"

In a closed system, bodies only exchange impulses. And the vector sum of the momenta of these bodies before their interaction is equal to the vector sum of their momenta after the interaction.

So, as a result of firing a gun, the momentum of the gun itself and the momentum of the bullet will change. But the sum of the impulses of the gun and the bullet in it before the shot will remain equal to the sum of the impulses of the gun and the flying bullet after the shot.

When firing a cannon, there is recoil. The projectile flies forward, and the gun itself rolls back. The projectile and the gun are a closed system in which the law of conservation of momentum operates.

The momentum of each body in a closed system can change as a result of their interaction with each other. But the vector sum of the impulses of bodies included in a closed system does not change when these bodies interact over time, that is, it remains constant. That's what it is law of conservation of momentum.

More precisely, the law of conservation of momentum is formulated as follows: the vector sum of the impulses of all bodies of a closed system is a constant value if there are no external forces acting on it, or their vector sum is equal to zero.

The momentum of a system of bodies can change only as a result of the action of external forces on the system. And then the law of conservation of momentum will not apply.

It must be said that closed systems do not exist in nature. But, if the time of action of external forces is very short, for example, during an explosion, shot, etc., then in this case the influence of external forces on the system is neglected, and the system itself is considered as closed.

In addition, if external forces act on the system, but the sum of their projections onto one of the coordinate axes is zero (that is, the forces are balanced in the direction of this axis), then the law of conservation of momentum is satisfied in this direction.

The law of conservation of momentum is also called law of conservation of momentum .

The most striking example of the application of the law of conservation of momentum is jet motion.

Jet propulsion

Reactive motion is the movement of a body that occurs when some part of it is separated from it at a certain speed. The body itself receives an oppositely directed impulse.

The simplest example of jet propulsion is the flight of a balloon from which air escapes. If we inflate a balloon and release it, it will begin to fly in the direction opposite to the movement of the air coming out of it.

An example of jet propulsion in nature is the release of liquid from the fruit of a crazy cucumber when it bursts. At the same time, the cucumber itself flies in the opposite direction.

Jellyfish, cuttlefish and other inhabitants of the deep sea move by taking in water and then throwing it out.

Jet thrust is based on the law of conservation of momentum. We know that when a rocket with a jet engine moves, as a result of fuel combustion, a jet of liquid or gas is ejected from the nozzle ( jet stream ). As a result of the interaction of the engine with the escaping substance, Reactive force . Since the rocket, together with the emitted substance, is a closed system, the momentum of such a system does not change with time.

Reactive force arises from the interaction of only parts of the system. External forces have no influence on its appearance.

Before the rocket began to move, the sum of the impulses of the rocket and the fuel was zero. Consequently, according to the law of conservation of momentum, after turning on the engines, the sum of these impulses is also zero.

where is the mass of the rocket

Gas flow rate

Changing rocket speed

∆mf - fuel consumption

Suppose the rocket operated for a period of time t .

Dividing both sides of the equation by ∆ t, we get the expression

According to Newton's second law, the reactive force is equal to

Reaction force, or jet thrust, ensures the movement of the jet engine and the object associated with it in the direction opposite to the direction of the jet stream.

Jet engines are used in modern aircraft and various missiles, military, space, etc.