Body impulse. Law of conservation of momentum. §2. Law of conservation of momentum Theorem on the change in angular momentum of a point

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 the system be equal to zero:

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 Oh) is equal to zero:

Then it follows from the equation 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 amount of motion of the system onto this axis is a constant value.

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

D'Alembert's principle.

All methods for solving problems of dynamics that we have considered so far are based on equations that follow either directly from Newton’s laws, or from general theorems that are consequences of these laws. However, this path is not the only one. It turns out that the equations of motion or the equilibrium conditions of a mechanical system can be obtained by basing it on other general principles, called the principles of mechanics, instead of Newton’s laws. In a number of cases, the application of these principles allows, as we will see, to find more effective methods for solving the corresponding problems. This chapter will examine one of the general principles of mechanics, called d'Alembert's principle.

Let us have a system consisting of n material points. Let us select one of the points of the system with mass . Under the influence of external and internal forces applied to it (which include both active forces and coupling reactions), the point receives some acceleration relative to the inertial reference frame.

Let us introduce into consideration the quantity

having the dimension of force. A vector quantity equal in magnitude to the product of the mass of a point and its acceleration and directed opposite to this acceleration is called the inertial force of the point (sometimes the d’Alembert inertial force).

Then it turns out that the motion of a point has the following general property: if at each moment of time we add the force of inertia to the forces actually acting on the point, then the resulting system of forces will be balanced, i.e. will

.

This expression expresses d'Alembert's principle for one material point. It is easy to see that it is equivalent to Newton's second law and vice versa. In fact, Newton's second law for the point in question gives . Moving the term here to the right side of the equality, we arrive at the last relation.

Repeating the above reasoning in relation to each of the points of the system, we arrive at the following result, expressing D'Alembert's principle for the system: if at any moment of time the corresponding inertial forces are applied to each of the points of the system, in addition to the external and internal forces actually acting on it, then the resulting system of forces will be in equilibrium and all static equations can be applied to it.

The significance of d'Alembert's principle lies in the fact that when directly applied to problems of dynamics, the equations of motion of the system are compiled in the form of well-known equilibrium equations; which makes a uniform approach to solving problems and usually greatly simplifies the corresponding calculations. In addition, in combination with the principle of possible displacements, which will be discussed in the next chapter, d'Alembert's principle allows us to obtain a new general method for solving problems of dynamics.

When applying d'Alembert's principle, it should be borne in mind that the point of a mechanical system, the movement of which is being studied, is acted upon only by external and internal forces and , arising as a result of the interaction of points of the system with each other and with bodies not included in the system; under the influence of these forces, the points of the system move with corresponding accelerations. The forces of inertia, which are discussed in D'Alembert's principle, do not act on moving points (otherwise, these points would be at rest or moving without acceleration, and then there would be no inertial forces themselves). The introduction of inertial forces is just a technique that allows one to compose dynamic equations using simpler statics methods.

If the sum of all external forces acting on the system is zero:

Then from equation (8.14) it follows that:

, i.e.:
,

which means that
, i.e.
.

Thus, if the sum of all external forces acting on the system is zero, then the vector of the system’s momentum will be constant in magnitude and direction.

If the external forces acting on the system are such that the sum of their projections onto some axis (for example, OX) is equal to zero:

.

Then the projection of the momentum of the system onto this axis is a constant quantity:

.

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

When solving problems using the law of conservation of the main vector of momentum, you should adhere to the following sequence:

Problem 8.2 (36.3)

Determine the principal vector of the quantities of motion of a pendulum consisting of a homogeneous rod OA weight R 1 length 4 r and homogeneous disk IN weight R 2 radii r, if the angular velocity of the pendulum at the moment is equal to ω .

In this problem, the system consists of two bodies: a rod of length 4r and a uniform disk of radius r. The center of mass of the rod is located at the geometric center (point C), and OS = CA, the center of mass of the disk is located at its geometric center (point B), since the bodies are homogeneous. Then the vector of momentum for the rod can be calculated:

Because
, then the modulus of the vector of the quantities of motion of the rod will be:

.

Vector directed perpendicular to the rod OA. For a disk, the vector of quantities of motion is equal to:

.

Speed ​​at point IN can be determined:

.

Then the module will be equal to:

.

The module of the system's momentum vector is determined as follows:

, Then

Answer:
, the vector of quantities of motion is directed perpendicular to the rod OA.

Questions for self-control:

What is the momentum of a material point and a mechanical system?

Momentum change theorem in differential form?

Momentum change theorem in integral form?

Literature: – .

Lecture 9

1. Theorem on the change in angular momentum of a point

Vector moment
relative to a given center O or axis Z is designated accordingly
And
called angular momentum or angular momentum of a point relative to the center or axis.

The moment of the vector is calculated
as well as the moment of force.

– for the moment of the vector
relative to the center:

.

– for the moment of the vector
relative to the axis:

,

Where – the shortest distance between the point of application of the vector
and axis or center;

Let us turn to the basic equation of the dynamics of rotational motion

and consider a special case when the body is either not acted upon by external forces at all, or they are such that their resultant does not produce a moment relative to the axis of rotation. Then

But if the change in the quantity is zero, then, consequently, the quantity itself remains constant:

Rice. 66. Somersault.

So, if no external forces act on the body (or their resulting moment relative to the axis of rotation is zero), then the angular momentum of the body relative to the axis of rotation remains unchanged. This law is called the law of conservation of angular momentum relative to the axis of rotation

Let us give several examples illustrating the law of conservation of angular momentum.

During an overhead jump (Fig. 66), the gymnast presses his arms and legs to his body. This reduces its moment of inertia,

and since the product must remain unchanged, the angular velocity of rotation increases, and in a short period of time while the gymnast is in the air, he manages to make a full rotation.

The ball is tied to a thread wound around a stick; as the length of the thread decreases, the moment of inertia of the ball decreases and, therefore, the angular velocity increases.

Rice. 67 Rotation of a man standing on the Zhukovsky bench. will speed up if he lowers his arms and slow down if he raises them.

Rice. 68. If we lift a bicycle wheel above our heads and set it into rotation, then we ourselves, together with the platform, will begin to rotate in the opposite direction.

A number of interesting experiments can be performed by standing on a platform rotating on a ball bearing (Zhukovsky bench). In Fig. Figures 67 and 68 depict some of these experiments.

Comparing the equations derived in the last paragraphs with the laws of rectilinear translational motion, it is easy to notice that the formulas that determine rotational motion about a fixed axis are similar to the formulas for rectilinear translational motion.

The following table compares the basic quantities and equations that determine these movements:

(see scan)

Gyroscopes. Reactive gyroscopic effect. A rigid body rotating at a high angular velocity around an axis of complete symmetry (free axis) is called a gyroscope. According to the law of conservation of the angular momentum vector, the gyroscope strives to maintain the direction of its rotation axis unchanged in space and exhibits the greater stability (i.e., the greater resistance to rotation of the rotation axis), the greater its moment of inertia and the greater the angular velocity of rotation.

When we, holding some massive stationary body in our outstretched arms, impart movement to it, for example, from left to right, then the inertial force developed by the body moves us in the opposite direction. The manifestation of the inertial forces of a rotating gyroscope when we turn its axis of rotation turns out to be more complex and, at first glance, unexpected. So, if we, holding the horizontally directed axis of rotation of the gyroscope in our hands, begin to raise one end of the axis and lower the other, i.e., rotate the axis in the vertical plane, then we will feel that the axis puts pressure on the hands not in the vertical, but in the horizontal plane, pressing one of our hands and pulling the other. If, when viewed from the right, the rotation of the gyroscope is seen to occur in a clockwise direction (i.e., the angular momentum of the gyroscope is directed horizontally to the left), then an attempt to raise the left end of the axis, lowering the right one down, causes the left end of the axis to move in a horizontal plane away from us, and the right - on us.

This reaction of the gyroscope (the so-called gyroscopic effect) is explained by the desire of the gyroscope to keep its angular momentum unchanged and, moreover, to keep it unchanged not only in magnitude, but also in direction. Indeed, in order for the angular momentum to remain geometrically unchanged when the axis of rotation of the gyroscope in the vertical plane is rotated in the vertical plane by an angle a (Fig. 69) described above, the gyroscope must acquire additional rotation around the vertical axis with an angular momentum such that geometrically

For this reason, a rotating gyroscope, balanced on a movable axis with a weight (Fig. 70), acquires additional

rotation around a vertical axis if the weight that balanced the gyroscope is moved slightly away from the fulcrum of the axis (by rebalancing, the weight imparts a certain tilt to the axis, which causes the gyroscope axis to rotate around the fulcrum in the direction that corresponds to the direction of the vector in Fig. 69).

For the same reason, the axis of the top acquires, due to the overturning action of gravity, a circular motion, which is called precession (Fig. 71).

So, if a couple of forces are applied to a rotating gyroscope, tending to rotate it about an axis perpendicular to the axis of rotation, then the gyroscope will indeed turn, but only around the third axis, perpendicular to the first two. To rotate a rotating gyroscope (for example, in the direction as shown in Fig. 72), you need to apply a torque to the axis of the gyroscope in a plane perpendicular to the direction of rotation.

Rice. 71. Scheme of the movement of the top.

A more detailed analysis of phenomena similar to those described above shows that the gyroscope tends to position its axis of rotation in such a way that it forms the smallest possible angle with the axis of the forced rotation and that both rotations occur in the same direction.

This property of the gyroscope is used in the gyroscopic compass, which has become widespread especially in the navy. The gyrocompass is a rapidly rotating top (a three-phase current motor operating at up to 25,000 rpm), which floats on a special float in a vessel with mercury and whose axis is set in the plane of the meridian. In this case, the source of external torque is the daily rotation of the Earth around its axis. Under its action, the axis of rotation of the gyroscope tends to coincide in direction with the axis of rotation of the Earth, and since the rotation of the Earth acts on the gyroscope continuously, the axis of the gyroscope finally takes this position, i.e. it is established along the meridian, and continues to remain completely in it just like a regular magnetic needle.

Gyroscopes are often used as stabilizers. They are installed to reduce pitching on ocean-going ships.

Stabilizers for single-rail railways were also designed; A massive, rapidly rotating gyroscope placed inside a single-rail car prevents the car from tipping over. Rotors for gyroscopic stabilizers are manufactured weighing from 1 to 100 tons or more.

In torpedoes, gyroscopic devices, automatically acting on the steering, ensure the straightness of the torpedo's movement in the direction of the shot.

Rice. 73. Precession of the earth's axis.

The daily rotation of the Earth makes it similar to a gyroscope. Since the Earth is not a sphere, but a figure close to an ellipsoid, the attraction of the Sun creates a resultant force that does not pass through the Earth’s center of mass (as would be the case in the case of a sphere). As a result, a torque arises that tends to rotate the Earth’s axis of rotation perpendicular to the plane of its orbit (Fig. 73). In this regard, the earth's axis experiences precessional movement (with a full rotation in approximately 25,800 years).

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 the 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, one should 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.

1. If the main vector of all external forces of 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 all external forces of the system onto any axis is equal to zero (
), then the projection of the momentum of the system onto this axis is a constant value.

Theorem on the motion of the center of mass.

Theorem The center of mass of the system moves in the same way as a material point, the mass of which is equal to the mass of the entire system, if all external forces applied to the mechanical system in question act on the point.

, hence

Momentum of the system.

Momentum systems of material points relative to some center is the vector sum of the angular momentum of individual points of this system relative to the same center

Momentum systems of material points
relative to any axis
passing through the center , is called the projection of the momentum vector
to this axis
.

The moment of momentum of a rigid body relative to the axis of rotation during the rotational motion of a rigid body.

Let us calculate the angular momentum of a rigid body relative to the axis of rotation.

The moment of momentum of a rigid body relative to the axis of rotation during rotational motion is equal to the product of the angular velocity of the body by its moment of inertia relative to the axis of rotation.

Theorem on the change in angular momentum of a system.

Theorem. The time derivative of the moment of momentum of the system, taken relative to some center, is equal to the vector sum of the moments of external forces acting on the system relative to the same center.

(6.3)

Proof: Theorem on the change in angular momentum for
points looks like:

,

Let's add it all up equations and we get:

or
,

Q.E.D.

Theorem. The time derivative of the moment of momentum of the system, taken relative to any axis, is equal to the vector sum of the moments of external forces acting on the system relative to the same axis.

To prove it, it is enough to project the vector equation (6.3) onto this axis. For axis
it will look like this:.

(6.4)

Theorem on the change in the angular momentum of a system relative to the center of mass. (no proof)

For axes moving translationally together with the center of mass of the system, the theorem on the change in the angular momentum of the system relative to the center of mass retains the same form as with respect to a stationary center.

Module 2. Strength of materials.

Topic 1: tension-compression, torsion, bending.

Deformations of the body (structural elements) under consideration arise from the application of an external force. In this case, the distances between the particles of the body change, which in turn leads to a change in the forces of mutual attraction between them. Hence, as a consequence, internal efforts arise. In this case, internal forces are determined by the universal method of sections (or cutting method).

It is known that there are external forces and internal forces. External forces (loads) are a quantitative measure of the interaction of two different bodies. These also include reactions in connections. Internal forces are a quantitative measure of the interaction of two parts of one body located on opposite sides of the section and caused by the action of external forces. Internal forces arise directly in the deformable body.

Figure 1 shows the design diagram of a beam with an arbitrary combination of external load forming an equilibrium system of forces:

From top to bottom: elastic body, left cut-off part, right cut-off part Fig.1. Section method.

In this case, the bond reactions are determined from the known equilibrium equations of the statics of a solid body:

where x 0, y 0, z 0 is the base coordinate system of the axes.

Mentally cutting a beam into two parts with an arbitrary section A (Fig. 1 a) leads to equilibrium conditions for each of the two cut parts (Fig. 1 b, c). Here ( S') And ( S"} - internal forces arising respectively in the left and right cut off parts due to the action of external forces.

When composing mentally cut-off parts, the condition of body equilibrium is ensured by the relation:

Since the initial system of external forces (1) is equivalent to zero, we obtain:

{S ’ } = – {S ” } (3)

This condition corresponds to the fourth axiom of statics about the equality of action and reaction forces.

Using the general methodology of the theorem Poinsot on bringing an arbitrary system of forces to a given center and choosing the center of mass as the reduction pole, sections A " , point WITH " , system of internal forces for the left side ( S ’ ) we reduce to the main vector and the main moment of internal efforts. The same is done for the right cut off part, where the position of the center of mass of the section A"; is determined, respectively, by the point WITH" (Fig. 1 b,c).

Thus, the main vector and the main moment of the system of internal forces arising in the left, conditionally cut off part of the beam are equal in magnitude and opposite in direction to the main vector and the main moment of the system of internal forces arising in the right conditionally cut off part.

The graph (diagram) of the distribution of the numerical values ​​of the main vector and the main moment along the longitudinal axis of the beam determines, first of all, specific issues of strength, rigidity and reliability of structures.

Let us determine the mechanism for the formation of components of internal forces that characterize simple types of resistance: tension-compression, shear, torsion and bending.

At the centers of mass of the sections under study WITH" or WITH" let's ask the left one accordingly (c", x", y", z") or right (c", x", y", z”) systems of coordinate axes (Fig. 1 b, c), which, unlike the base coordinate system x, y, z We will call them “followers”. The term is due to their functional purpose. Namely: tracking changes in the position of section A (Fig. 1 a) when it is conditionally displaced along the longitudinal axis of the beam, for example when: 0 x’ 1 a, ax’ 2 b etc., where A And b- linear dimensions of the boundaries of the studied sections of the timber.

Let us set the positive directions of the projections of the main vector or the main moment or the coordinate axes of the tracking system (Fig. 1 b, c):

In this case, the positive directions of projections of the main vector and the main moment of internal forces on the axis of the servo coordinate system correspond to the rules of statics in theoretical mechanics: for force - along the positive direction of the axis, for moment - counterclockwise rotation when observed from the end of the axis. They are classified as follows:

N x- normal strength, a sign of central tension or compression;

M x - internal torque, occurs during torsion;

Q z ,Q at- transverse or shearing forces – a sign of shear deformations,

M at , M z- internal bending moments, corresponding to bending.

The connection of the left and right mentally cut off parts of the beam leads to the well-known (3) principle of equality in magnitude and opposite direction of all components of the same name of internal forces, and the condition for the equilibrium of the beam is defined as:

As a natural consequence of relations 3,4,5, the resulting condition is necessary for the same components of internal forces to form subsystems of forces equivalent to zero in pairs:

The total number of internal forces (six) in statically definable problems coincides with the number of equilibrium equations for a spatial system of forces and is associated with the number of possible mutual movements of one conditionally cut-off part of the body in relation to another.

The required forces are determined from the corresponding equations for any of the cut-off parts in the tracking system of coordinate axes. Thus, for any cut-off part, the corresponding equilibrium equations take the form;

Here, for simplicity of coordinate system notation c" x" y" z" And c"x"y"t" replaced by a single oxyz.